diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/060_4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/060_4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM.md deleted file mode 100644 index 4536548f0f245f1f6efee455862143b0bfdcf8b9..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/060_4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM.md +++ /dev/null @@ -1,266 +0,0 @@ -## **4.8 FREQUENCY [RESPONSE OF AN](#page-10-0) LTIC SYSTEM** - -Filtering is an important area of signal processing. Filtering characteristics of a system are indicated by its response to sinusoids of various frequencies varying from 0 to ∞. Such characteristics are called the frequency response of the system. In this section, we shall find the frequency response of LTIC systems. - -In Sec. 2.4-4 we showed that an LTIC system response to an everlasting exponential input *x*(*t*) = *est* is also an everlasting exponential *H*(*s*)*est*. As before, we use an arrow directed from the input to the output to represent an input–output pair: - -$$ -e^{st} \Longrightarrow H(s)e^{st} \tag{4.40} -$$ - -Setting *s* = *j*ω in this relationship yields - -$$ -e^{j\omega t} \Longrightarrow H(j\omega)e^{j\omega t} \tag{4.41} -$$ - -Noting that cosω*t* is the real part of *ej*ω*t* , use of Eq. (2.31) yields - -$$ -\cos \omega t \Longrightarrow \text{Re}[H(j\omega)e^{j\omega t}] \tag{4.42} -$$ - -We can express *H*(*j*ω) in the polar form as - -$$ -H(j\omega) = |H(j\omega)|e^{j\angle H(j\omega)} -$$ - -With this result, Eq. (4.42) becomes - -$$ -\cos \omega t \Longrightarrow |H(j\omega)| \cos [\omega t + \angle H(j\omega)] -$$ - -In other words, the system response *y*(*t*) to a sinusoidal input cosω*t* is given by - -$$ -y(t) = |H(j\omega)| \cos[\omega t + \angle H(j\omega)] -$$ - -Using a similar argument, we can show that the system response to a sinusoid cos(ω*t* +θ ) is - -$$ -y(t) = |H(j\omega)|\cos[\omega t + \theta + \angle H(j\omega)]\tag{4.43} -$$ - -This result is valid only for BIBO-stable systems. The frequency response is meaningless for BIBO-unstable systems. This follows from the fact that the frequency response in Eq. (4.41) is obtained by setting *s* = *j*ω in Eq. (4.40). But, as shown in Sec. 2.4-4 [Eqs. (2.38) and (2.39)], Eq. (4.40) applies only for the values of *s* for which *H*(*s*) exists. For BIBO-unstable systems, the ROC for *H*(*s*) does not include the ω axis where *s* = *j*ω [see Eq. (4.10)]. This means that *H*(*s*) when *s* = *j*ω is meaningless for BIBO-unstable systems.† - -Equation (4.43) shows that for a sinusoidal input of radian frequency ω, the system response is also a sinusoid of the same frequency ω. *The amplitude of the output sinusoid is* |*H*(*j*ω)| *times the input amplitude, and the phase of the output sinusoid is shifted by H*(*j*ω) *with respect to the input phase* (see later Fig. 4.38 in Ex. 4.27). For instance, a certain system with |*H*(*j*10)| = 3 and *H*(*j*10) = −30◦ amplifies a sinusoid of frequency ω = 10 by a factor of 3 and delays its phase by 30◦. The system response to an input 5cos(10*t* + 50◦) is 3 × 5 cos(10*t* + 50◦ − 30◦) = 15 cos(10*t* +20◦). - -Clearly |*H*(*j*ω)| is the amplitude *gain* of the system, and a plot of |*H*(*j*ω)| versus ω shows the amplitude gain as a function of frequency ω. We shall call |*H*(*j*ω)| the *amplitude response*. It also goes under the name *magnitude response*. ‡ Similarly, *H*(*j*ω) is the *phase response*, and a plot of *H*(*j*ω) versus ω shows how the system modifies or changes the phase of the input sinusoid. Plots of the magnitude response |*H*(*j*ω)| and phase response *H*(*j*ω) show at a glance how a system responds to sinusoids of various frequencies. Observe that *H*(*j*ω) has the information of |*H*(*j*ω)| and *H*(*j*ω) and is therefore termed the *frequency response* of the system. Clearly, the frequency response of a system represents its filtering characteristics. - -### **EXAMPLE 4.27 Frequency Response** - -Find the frequency response (amplitude and phase responses) of a system whose transfer function is - -$$ -H(s) = \frac{s+0.1}{s+5} -$$ - -Also, find the system response *y*(*t*) if the input *x*(*t*) is - -**(a)** cos 2*t* - -**(b)** cos(10*t* −50◦) - -In this case, - -$$ -H(j\omega) = \frac{j\omega + 0.1}{j\omega + 5} -$$ - -† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cosω0*t* or *eat* cosω0*t* (*a* > 0). Hence, the response of such a system to a sinusoid cosω*t* will contain not just the sinusoid of frequency ω, but also nondecaying natural modes, rendering the concept of frequency response meaningless. - -‡ Strictly speaking, |*H*(ω)| is magnitude response. There is a fine distinction between amplitude and magnitude. Amplitude *A* can be positive and negative. In contrast, the magnitude |*A*| is always nonnegative. We refrain from relying on this useful distinction between amplitude and magnitude in the interest of avoiding proliferation of essentially similar entities. This is also why we shall use the "amplitude" (instead of "magnitude") spectrum for |*H*(ω)|. - -### 414 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Therefore, - -$$ -|H(j\omega)| = \frac{\sqrt{\omega^2 + 0.01}}{\sqrt{\omega^2 + 25}} \quad \text{and} \quad \angle H(j\omega) = \tan^{-1}\left(\frac{\omega}{0.1}\right) - \tan^{-1}\left(\frac{\omega}{5}\right) -$$ - -Both the amplitude and the phase response are depicted in Fig. 4.38a as functions of ω. These plots furnish the complete information about the frequency response of the system to sinusoidal inputs. - -**(a)** For the input *x*(*t*) = cos 2*t*, ω = 2, and - -$$ -|H(j2)| = \frac{\sqrt{(2)^2 + 0.01}}{\sqrt{(2)^2 + 25}} = 0.372 -$$ - -\n -$$ -\angle H(j2) = \tan^{-1}\left(\frac{2}{0.1}\right) - \tan^{-1}\left(\frac{2}{5}\right) = 87.1^{\circ} - 21.8^{\circ} = 65.3^{\circ} -$$ - -**Figure 4.38** Responses for the system of Ex. 4.27. - -We also could have read these values directly from the frequency response plots in Fig. 4.38a corresponding to ω = 2. This result means that for a sinusoidal input with frequency ω = 2, the amplitude gain of the system is 0.372, and the phase shift is 65.3◦. In other words, the output amplitude is 0.372 times the input amplitude, and the phase of the output is shifted with respect to that of the input by 65.3◦. Therefore, the system response to the input cos 2*t* is - -$$ -y(t) = 0.372 \cos(2t + 65.3^{\circ}) -$$ - -The input cos 2*t* and the corresponding system response 0.372cos(2*t* + 65.3◦) are illustrated in Fig. 4.38b. - -**(b)** For the input cos(10*t* − 50◦), instead of computing the values |*H*(*j*ω)| and *H*(*j*ω) as in part (a), we shall read them directly from the frequency response plots in Fig. 4.38a corresponding to ω = 10. These are - -$$ -|H(j10)| = 0.894 -$$ - and $\angle H(j10) = 26^{\circ}$ - -Therefore, for a sinusoidal input of frequency ω = 10, the output sinusoid amplitude is 0.894 times the input amplitude, and the output sinusoid is shifted with respect to the input sinusoid by 26◦. Therefore, the system response *y*(*t*) to an input cos(10*t* −50◦) is - -$$ -y(t) = 0.894 \cos (10t - 50^\circ + 26^\circ) = 0.894 \cos (10t - 24^\circ) -$$ - -If the input were sin(10*t* − 50◦), the response would be 0.894 sin(10*t* − 50◦ + 26◦) = 0.894 sin(10*t* −24◦). - -The frequency response plots in Fig. 4.38a show that the system has highpass filtering characteristics; it responds well to sinusoids of higher frequencies (ω well above 5), and suppresses sinusoids of lower frequencies (ω well below 5). - -### PLOTTING FREQUENCY RESPONSE WITH MATLAB - -It is simple to use MATLAB to create magnitude and phase response plots. Here, we consider two methods. In the first method, we use an anonymous function to define the transfer function *H*(*s*) and then obtain the frequency response plots by substituting *j*ω for *s*. - -``` ->> H = @(s) (s+0.1)./(s+5); omega = 0:.01:20; ->> subplot(1,2,1); plot(omega,abs(H(1j*omega)),'k-'); ->> subplot(1,2,2); plot(omega,angle(H(1j*omega))*180/pi,'k-'); -``` - -In the second method, we define vectors that contain the numerator and denominator coefficients of *H*(*s*) and then use the freqs command to compute frequency response. - -``` ->> B = [1 0.1]; A = [1 5]; H = freqs(B,A,omega); omega = 0:.01:20; ->> subplot(1,2,1); plot(omega,abs(H),'k-'); ->> subplot(1,2,2); plot(omega,angle(H)*180/pi,'k-'); -``` - -Both approaches generate plots that match Fig. 4.38a. - -### **EXAMPLE 4.28 Frequency Responses of Delay, Differentiator, and Integrator Systems** - -Find and sketch the frequency responses (magnitude and phase) for **(a)** an ideal delay of *T* seconds, **(b)** an ideal differentiator, and **(c)** an ideal integrator. - -**(a) Ideal delay of** *T* **seconds.** The transfer function of an ideal delay is [see Eq. (4.30)] - -$$ -H(s) = e^{-sT} -$$ - -Therefore, - -$$ -H(j\omega) = e^{-j\omega T} -$$ - -Consequently, - -$$ -|H(j\omega)| = 1 \quad \text{and} \quad \angle H(j\omega) = -\omega T -$$ - -These amplitude and phase responses are shown in Fig. 4.39a. The amplitude response is constant (unity) for all frequencies. The phase shift increases linearly with frequency with a slope of −*T*. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal delay of *T* seconds, the output is cosω(*t* − *T*). The output sinusoid amplitude is the same as that of the input for all values of ω. Therefore, the amplitude response (gain) is unity for all frequencies. Moreover, the output cosω(*t* − *T*) = cos(ω*t* − ω*T*) has a phase shift −ω*T* with respect to the input cosω*t*. Therefore, the phase response is linearly proportional to the frequency ω with a slope −*T*. - -**(b) An ideal differentiator.** The transfer function of an ideal differentiator is [see Eq. (4.31)] - -*H*(*s*) = *s* - -Therefore, - -*H*(*j*ω) = *j*ω = ω*ej*π/2 - -Consequently, - -$$ -|H(j\omega)| = \omega -$$ - and $\angle H(j\omega) = \frac{\pi}{2}$ - -These amplitude and phase responses are depicted in Fig. 4.39b. The amplitude response increases linearly with frequency, and phase response is constant (π/2) for all frequencies. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal differentiator, the output is −ωsin ω*t* = ωcos[ω*t* + (π/2)]. Therefore, the output sinusoid amplitude is ω times the input amplitude; that is, the amplitude response (gain) increases linearly with frequency ω. Moreover, the output sinusoid undergoes a phase shift π/2 with respect to the input cosω*t*. Therefore, the phase response is constant (π/2) with frequency. - -**Figure 4.39** Frequency response of an ideal **(a)** delay, **(b)** differentiator, and **(c)** integrator. - -In an ideal differentiator, the amplitude response (gain) is proportional to frequency [|*H*(*j*ω)| = ω] so that the higher-frequency components are enhanced (see Fig. 4.39b). All practical signals are contaminated with noise, which, by its nature, is a broadband (rapidly varying) signal containing components of very high frequencies. A differentiator can increase the noise disproportionately to the point of drowning out the desired signal. This is why ideal differentiators are avoided in practice. - -**(c) An ideal integrator.** The transfer function of an ideal integrator is [see Eq. (4.32)] - -$$ -H(s) = \frac{1}{s} -$$ - -Therefore, - -$$ -H(j\omega) = \frac{1}{j\omega} = \frac{-j}{\omega} = \frac{1}{\omega}e^{-j\pi/2} -$$ - -Consequently, - -$$ -|H(j\omega)| = \frac{1}{\omega} \quad \text{and} \quad \angle H(j\omega) = -\frac{\pi}{2} -$$ - -These amplitude and phase responses are illustrated in Fig. 4.39c. The amplitude response is inversely proportional to frequency, and the phase shift is constant (−π/2) with frequency. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal integrator, the output is (1/ω)sin ω*t* = (1/ω) cos[ω*t* −(π/2)]. Therefore, the amplitude response is inversely proportional to ω, and the phase response is constant (−π/2) - -### 418 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -with frequency.† Because its gain is 1/ω, the ideal integrator suppresses higher-frequency components but enhances lower-frequency components with ω < 1. Consequently, noise signals (if they do not contain an appreciable amount of very-low-frequency components) are suppressed (smoothed out) by an integrator. - -### **DR ILL 4.15 Sinusoidal Response of an LTIC System** - -Find the response of an LTIC system specified by - -$$ -\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt} + 5x(t) -$$ - -if the input is a sinusoid 20 sin(3*t* +35◦). - -### **ANSWER** - -10.23 sin(3*t* −61.91◦) - -### **[4.8-1 Steady-State Response to Causal Sinusoidal Inputs](#page-10-0)** - -So far we have discussed the LTIC system response to everlasting sinusoidal inputs (starting at *t* = −∞). In practice, we are more interested in causal sinusoidal inputs (sinusoids starting at *t* = 0). Consider the input *ej*ω*t u*(*t*), which starts at *t* = 0 rather than at *t* = −∞. In this case *X*(*s*) = 1/(*s* + *j*ω). Moreover, according to Eq. (4.27), *H*(*s*) = *P*(*s*)/*Q*(*s*), where *Q*(*s*) is the - -† A puzzling aspect of this result is that in deriving the transfer function of the integrator in Eq. (4.32), we have assumed that the input starts at *t* = 0. In contrast, in deriving its frequency response, we assume that the everlasting exponential input *ej*ω*t* starts at *t* = −∞. There appears to be a fundamental contradiction between the everlasting input, which starts at *t* = −∞, and the integrator, which opens its gates only at *t* = 0. Of what use is everlasting input, since the integrator starts integrating at *t* = 0? The answer is that the integrator gates are always open, and integration begins whenever the input starts. We restricted the input to start at *t* = 0 in deriving Eq. (4.32) because we were finding the transfer function using the unilateral transform, where the inputs begin at *t* = 0. So the integrator starting to integrate at *t* = 0 is restricted because of the limitations of the unilateral transform method, not because of the limitations of the integrator itself. If we were to find the integrator transfer function using Eq. (2.40), where there is no such restriction on the input, we would still find the transfer function of an integrator as 1/*s*. Similarly, even if we were to use the bilateral Laplace transform, where *t* starts at −∞, we would find the transfer function of an integrator to be 1/*s*. The transfer function of a system is the property of the system and does not depend on the method used to find it. - -characteristic polynomial given by *Q*(*s*) = (*s*−λ1)(*s*−λ2)··· (*s*−λ*N*). † Hence, - -$$ -Y(s) = X(s)H(s) = \frac{P(s)}{(s - \lambda_1)(s - \lambda_2) \cdots (s - \lambda_N)(s - j\omega)} -$$ - -In the partial fraction expansion of the right-hand side, let the coefficients corresponding to the *N* terms (*s* − λ1), (*s* − λ2), ... , (*s* − λ*N*) be *k*1, *k*2, ... , *kN*. The coefficient corresponding to the last term (*s*−*j*ω) is *P*(*s*)/*Q*(*s*)|*s*=*j*ω = *H*(*j*ω). Hence, - -$$ -Y(s) = \sum_{i=1}^{n} \frac{k_i}{s - \lambda_i} + \frac{H(j\omega)}{s - j\omega} -$$ - -and - -$$ -y(t) = \underbrace{\sum_{i=1}^{n} k_i e^{\lambda_i t} u(t)}_{\text{transient component } y_{\text{tr}}(t)} + \underbrace{H(j\omega)e^{j\omega t} u(t)}_{\text{steady-state component } y_{\text{ss}}(t)} -$$ - -For an asymptotically stable system, the characteristic mode terms *e*λ*it* decay with time, and, therefore, constitute the so-called *transient* component of the response. The last term *H*(*j*ω)*ej*ω*t* persists forever, and is the *steady-state* component of the response given by - -$$ -y_{ss}(t) = H(j\omega)e^{j\omega t}u(t) -$$ - -This result also explains why an everlasting exponential input *ej*ω*t* results in the total response *H*(*j*ω)*ej*ω*t* for BIBO systems. Because the input started at *t* = −∞, at any finite time the decaying transient component has long vanished, leaving only the steady-state component. Hence, the total response appears to be *H*(*j*ω)*ej*ω*t* . - -From the argument that led to Eq. (4.43), it follows that for a causal sinusoidal input cosω*t*, the steady-state response *yss*(*t*) is given by - -$$ -y_{ss}(t) = |H(j\omega)| \cos[\omega t + \angle H(j\omega)]u(t) -$$ - -In summary, |*H*(*j*ω)| cos[ω*t* + *H*(*j*ω)] is the total response to everlasting sinusoid cosω*t*. In contrast, it is the steady-state response to the same input applied at *t* = 0. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/061_4.9 BODE PLOTS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/061_4.9 BODE PLOTS.md deleted file mode 100644 index 35f59b0ae88b3db4aa54149b0ca13a7dc930afe1..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/061_4.9 BODE PLOTS.md +++ /dev/null @@ -1,417 +0,0 @@ -## **[4.9 BODE](#page-10-0) PLOTS** - -Sketching frequency response plots (|*H*(*j*ω)| and *H*(*j*ω) versus ω) is considerably facilitated by the use of logarithmic scales. The amplitude and phase response plots as a function of ω on a logarithmic scale are known as *Bode plots*. By using the asymptotic behavior of the amplitude and the phase responses, we can sketch these plots with remarkable ease, even for higher-order transfer functions. - -† For simplicity, we have assumed nonrepeating characteristic roots. The procedure is readily modified for repeated roots, and the same conclusion results. - -#### 420 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Let us consider a system with the transfer function - -$$ -H(s) = \frac{K(s+a_1)(s+a_2)}{s(s+b_1)(s^2+b_2s+b_3)} -$$ -\n(4.44) - -where the second-order factor (*s*2 + *b*2*s* + *b*3) is assumed to have complex conjugate roots.† We shall rearrange Eq. (4.44) in the form - -$$ -H(s) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(\frac{s}{a_1} + 1\right)\left(\frac{s}{a_2} + 1\right)}{s\left(\frac{s}{b_1} + 1\right)\left(\frac{s^2}{b_3} + \frac{b_2}{b_3} s + 1\right)} -$$ - -and - -$$ -H(j\omega) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(1 + \frac{j\omega}{a_1}\right)\left(1 + \frac{j\omega}{a_2}\right)}{j\omega\left(1 + \frac{j\omega}{b_1}\right)\left[1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right]} -$$ - -This equation shows that *H*(*j*ω) is a complex function of ω. The amplitude response |*H*(*j*ω)| and the phase response *H*(*j*ω) are given by - -$$ -|H(j\omega)| = \left| \frac{Ka_1a_2}{b_1b_3} \right| \frac{\left| 1 + \frac{j\omega}{a_1} \right| \left| 1 + \frac{j\omega}{a_2} \right|}{|j\omega| \left| 1 + \frac{j\omega}{b_1} \right| \left| 1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3} \right|} -$$ -(4.45) - -and - -$$ -\angle H(j\omega) = \angle \left(\frac{Ka_1a_2}{b_1b_3}\right) + \angle \left(1 + \frac{j\omega}{a_1}\right) + \angle \left(1 + \frac{j\omega}{a_2}\right) -$$ -$$ -- \angle j\omega - \angle \left(1 + \frac{j\omega}{b_1}\right) - \angle \left[1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right] -$$ -(4.46) - -From Eq. (4.46) we see that the phase function consists of the addition of terms of four kinds: (i) the phase of a constant, (ii) the phase of *j*ω, which is 90◦ for all values of ω, (iii) the phase for the first-order term of the form 1+*j*ω/*a*, and (iv) the phase of the second-order term - -$$ -\[1+\frac{jb_2\omega}{b_3}+\frac{(j\omega)^2}{b_3}\] -$$ - -We can plot these basic phase functions for ω in the range 0 to ∞ and then, using these plots, we can construct the phase function of any transfer function by properly adding these basic responses. Note that if a particular term is in the numerator, its phase is added, but if the term is in the - -† Coefficients *a*1, *a*2 and *b*1, *b*2, *b*3 used in this section are not to be confused with those used in the representation of *N*th-order LTIC system equations given earlier [Eqs. (2.1) or (4.26)]. - -denominator, its phase is subtracted. This makes it easy to plot the phase function *H*(*j*ω) as a function of ω. Computation of |*H*(*j*ω)|, unlike that of the phase function, however, involves the multiplication and division of various terms. This is a formidable task, especially when we have to plot this function for the entire range of ω (0 to ∞). - -We know that a log operation converts multiplication and division to addition and subtraction. So, instead of plotting |*H*(*j*ω)|, why not plot log |*H*(*j*ω)| to simplify our task? We can take advantage of the fact that logarithmic units are desirable in several applications, where the variables considered have a very large range of variation. This is particularly true in frequency response plots, where we may have to plot frequency response over a range from a very low frequency, near 0, to a very high frequency, in the range of 1010 or higher. A plot on a linear scale of frequencies for such a large range will bury much of the useful information at lower frequencies. Also, the amplitude response may have a very large dynamic range from a low of 10−6 to a high of 106 . A linear plot would be unsuitable for such a situation. Therefore, logarithmic plots not only simplify our task of plotting, but, fortunately, they are also desirable in this situation. - -There is another important reason for using logarithmic scale. The Weber–Fechner law (first observed by Weber in 1834) states that human senses (sight, touch, hearing, etc.) generally respond in a logarithmic way. For instance, when we hear sound at two different power levels, we judge one sound twice as loud when the ratio of the two sound powers is 10. Human senses respond to equal ratios of power, not equal increments in power [10]. This is clearly a logarithmic response.† - -The logarithmic unit is the *decibel* and is equal to 20 times the logarithm of the quantity (log to the base 10). Therefore, 20log10 |*H*(*j*ω)| is simply the log amplitude in decibels (dB).‡ Thus, instead of plotting |*H*(*j*ω)|, we shall plot 20log10 |*H*(*j*ω)| as a function of ω. These plots (log amplitude and phase) are called *Bode plots*. For the transfer function in Eq. (4.45), the *log amplitude* is - -$$ -20\log|H(j\omega)| = 20\log\left|\frac{Ka_1a_2}{b_1b_3}\right| + 20\log\left|1 + \frac{j\omega}{a_1}\right| + 20\log\left|1 + \frac{j\omega}{a_2}\right| - 20\log|j\omega| -$$ -$$ --20\log\left|1 + \frac{j\omega}{b_1}\right| - 20\log\left|1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right| \tag{4.47} -$$ - -The term 20log(*Ka*1*a*2/*b*1*b*3) is a constant. We observe that the log amplitude is a sum of four basic terms corresponding to a constant, a pole or zero at the origin (20log|*j*ω|), a first-order pole or zero (20log|1+*j*ω/*a*|), and complex-conjugate poles or zeros (20log|1+*j*ω*b*2/*b*3 +(*j*ω)2/*b*3|). - -† Observe that the frequencies of musical notes are spaced logarithmically (not linearly). The octave is a ratio of 2. The frequencies of the same note in the successive octaves have a ratio of 2. On the Western musical scale, there are 12 distinct notes in each octave. The frequency of each note is about 6% higher than the frequency of the preceding note. Thus, the successive notes are separated not by some constant frequency, but by constant ratio of 1.06. - -‡ Originally, the unit *bel* (after the inventor of telephone, Alexander Graham Bell) was introduced to represent power ratio as log10 *P*2/*P*1 bels. A tenth of this unit is a decibel, as in 10 log10 *P*2/*P*1 decibels. Since the power ratio of two signals is proportional to the amplitude ratio squared, or |*H*(*j*ω)| 2, we have 10 log10 *P*2/*P*1 = 10 log10 |*H*(*j*ω)| 2 = 20 log10 |*H*(*j*ω)| dB. - -### 422 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -We can sketch these four basic terms as functions of ω and use them to construct the log-amplitude plot of any desired transfer function. Let us discuss each of the terms. - -## **[4.9-1 Constant](#page-10-0)** *Ka***1***a***2***/b***1***b***3** - -The log amplitude of the constant *Ka*1*a*2/*b*1*b*2 term is also a constant, 20log|*Ka*1*a*2/*b*1*b*3|. The phase contribution from this term is zero for positive value and π for negative value of the constant (complex constants can have different phases). - -### **[4.9-2 Pole \(or Zero\) at the Origin](#page-10-0)** - -LOG MAGNITUDE - -A pole at the origin gives rise to the term −20log|*j*ω|, which can be expressed as - -$$ --20\log|j\omega| = -20\log\omega -$$ - -This function can be plotted as a function of ω. However, we can effect further simplification by using the logarithmic scale for the variable ω itself. Let us define a new variable *u* such that - -*u* = logω - -Hence, - -$$ --20\log\omega = -20u -$$ - -The log-amplitude function −20*u* is plotted as a function of *u* in Fig. 4.40a. This is a straight line with a slope of −20. It crosses the *u* axis at *u* = 0. The ω-scale (*u* = logω) also appears in Fig. 4.40a. Semilog graphs can be conveniently used for plotting, and we can directly plot ω on semilog paper. A ratio of 10 is a *decade*, and a ratio of 2 is known as an *octave*. Furthermore, a decade along the ω scale is equivalent to 1 unit along the *u* scale. We can also show that a ratio of 2 (an octave) along the ω scale equals to 0.3010 (which is log10 2) along the *u* scale.† - -$$ -u_2 - u_1 = \log_{10} \omega_2 - \log_{10} \omega_1 = \log_{10} (\omega_2/\omega_1) -$$ - -Thus, if - -(ω2/ω1) = 10 (which is a decade) - -then - -*u*2 −*u*1 = log10 10 = 1 - -and if - -(ω2/ω1) = 2 (which is an octave) - -then - -$$ -u_2 - u_1 = \log_{10} 2 = 0.3010 -$$ - -† This point can be shown as follows. Let ω1 and ω2 along the ω scale correspond to *u*1 and *u*2 along the *u* scale so that logω1 = *u*1 and logω2 = *u*2. Then - -**Figure 4.40 (a)** Amplitude and **(b)** phase responses of a pole or a zero at the origin. - -Note that equal increments in *u* are equivalent to equal ratios on the ω scale. Thus, 1 unit along the *u* scale is the same as one decade along the ω scale. This means that the amplitude plot has a slope of −20 dB/decade or −20(0.3010) = −6.02 dB/octave (commonly stated as −6 dB/octave). Moreover, the amplitude plot crosses the ω axis at ω = 1, since *u* = log10ω = 0 when ω = 1. - -For the case of a zero at the origin, the log-amplitude term is 20 log ω. This is a straight line passing through ω = 1 and having a slope of 20 dB/decade (or 6 dB/octave). This plot is a mirror image about the ω axis of the plot for a pole at the origin and is shown dashed in Fig. 4.40a. - -### PHASE - -The phase function corresponding to the pole at the origin is − *j*ω [see Eq. (4.46)]. Thus, - -$$ -\angle H(j\omega) = -\angle j\omega = -90^{\circ} -$$ - -The phase is constant (−90◦) for all values of ω, as depicted in Fig. 4.40b. For a zero at the origin, the phase is *j*ω = 90◦. This is a mirror image of the phase plot for a pole at the origin and is shown dashed in Fig. 4.40b. - -### **[4.9-3 First-Order Pole \(or Zero\)](#page-10-0)** - -### THE LOG MAGNITUDE - -The log amplitude of a first-order pole at −*a* is −20log|1+*j*ω/*a*|. Let us investigate the asymptotic behavior of this function for extreme values of ω (ω *a* and ω *a*). - -**(a)** For ω *a*, - -$$ --20\log\left|1+\frac{j\omega}{a}\right|\approx-20\log 1=0 -$$ - -Hence, the log-amplitude function → 0 asymptotically for ω *a* (Fig. 4.41a). - -**(a)** For the other extreme case, where ω *a*, - -$$ --20\log\left|1+\frac{j\omega}{a}\right| \approx -20\log\left(\frac{\omega}{a}\right) = -20\log\omega + 20\log a = -20u + 20\log a -$$ - -This represents a straight line (when plotted as a function of *u*, the log of ω) with a slope of −20 dB/decade (or −6 dB/octave). When ω = *a*, the log amplitude is zero. Hence, this line crosses the ω axis at ω = *a*, as illustrated in Fig. 4.41a. Note that the asymptotes in (a) and (b) meet at ω = *a*. - -The exact log amplitude for this pole is - -$$ --20\log\left|1+\frac{j\omega}{a}\right| = -20\log\left(1+\frac{\omega^2}{a^2}\right)^{1/2} = -10\log\left(1+\frac{\omega^2}{a^2}\right) -$$ - -This exact log magnitude function also appears in Fig. 4.41a. Observe that the actual and the asymptotic plots are very close. A maximum error of 3 dB occurs at ω = *a*. This frequency is known as the *corner frequency* or *break frequency*. The error everywhere else is less than 3 dB. A plot of the error as a function of ω is shown in Fig. 4.42a. This figure shows that the error at 1 octave above or below the corner frequency is 1 dB and the error at 2 octaves above or below the corner frequency is 0.3 dB. The actual plot can be obtained by adding the error to the asymptotic plot. - -The amplitude response for a zero at −*a* (shown dotted in Fig. 4.41a) is identical to that of the pole at −*a* with a sign change and therefore is the mirror image (about the 0 dB line) of the amplitude plot for a pole at −*a*. - -### PHASE - -The phase for the first-order pole at −*a* is - -$$ -\angle H(j\omega) = -\angle \left(1 + \frac{j\omega}{a}\right) = -\tan^{-1}\left(\frac{\omega}{a}\right) -$$ - -Let us investigate the asymptotic behavior of this function. For ω *a*, - -$$ --\tan^{-1}\left(\frac{\omega}{a}\right) \approx 0 -$$ - -and, for ω *a*, - -$$ --\tan^{-1}\left(\frac{\omega}{a}\right) \approx -90^{\circ} -$$ - -**Figure 4.41 (a)** Amplitude and **(b)** phase responses of a first-order pole or zero at *s* = −*a*. - -The actual plot along with the asymptotes is depicted in Fig. 4.41b. In this case, we use a three-line segment asymptotic plot for greater accuracy. The asymptotes are a phase angle of 0◦ for ω ≤ *a*/10, a phase angle of −90◦ for ω ≥ 10*a*, and a straight line with a slope −45◦/decade connecting these two asymptotes (from ω = *a*/10 to 10*a*) crossing the ω axis at ω = *a*/10. It can be seen from Fig. 4.41b that the asymptotes are very close to the curve and the maximum error is 5.7◦. Figure 4.42b plots the error as a function of ω; the actual plot can be obtained by adding the error to the asymptotic plot. - -**Figure 4.42** Errors in asymptotic approximation of a first-order pole at *s* = −*a*. - -The phase for a zero at −*a* (shown dotted in Fig. 4.41b) is identical to that of the pole at −*a* with a sign change, and therefore is the mirror image (about the 0◦ line) of the phase plot for a pole at −*a*. - -### **[4.9-4 Second-Order Pole \(or Zero\)](#page-10-0)** - -Let us consider the second-order pole in Eq. (4.44). The denominator term is *s*2 + *b*2*s* + *b*3. We shall introduce the often-used standard form *s*2 + 2ζω*ns* + ω2 *n* instead of *s*2 + *b*2*s* + *b*3. With this form, the log amplitude function for the second-order term in Eq. (4.47) becomes - -$$ --20\log\left|1+2j\zeta\frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right| -$$ - -and the phase function is - -$$ --\angle \left[1+2j\zeta \frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right] -$$ -\n(4.48) - -### THE LOG MAGNITUDE - -The log amplitude is given by - -log amplitude = -$$ --20 \log \left| 1 + 2j\zeta \left( \frac{\omega}{\omega_n} \right) + \left( \frac{j\omega}{\omega_n} \right)^2 \right| -$$ - (4.49) - -For ω ω*n*, the log amplitude becomes - -$$ -log amplitude \approx -20 log 1 = 0 -$$ - -For ω ω*n*, the log amplitude is - -$$ -\log amplitude \approx -20 \log \left| \left( -\frac{\omega}{\omega_n} \right)^2 \right| = -40 \log \left( \frac{\omega}{\omega_n} \right) -$$ - -= -40 log $\omega$ - 40 log $\omega_n$ = -40u - 40 log $\omega_n$ (4.50) - -The two asymptotes are zero for ω<ω*n* and −40*u*−40logω*n* for ω>ω*n*. The second asymptote is a straight line with a slope of −40 dB/decade (or −12 dB/octave) when plotted against the log ω scale. It begins at ω = ω*n* [see Eq. (4.50)]. The asymptotes are depicted in Fig. 4.43a. The exact log amplitude is given by [see Eq. (4.49)] - -log amplitude = -$$ --20 \log \left\{ \left[ 1 - \left( \frac{\omega}{\omega_n} \right)^2 \right]^2 + 4 \zeta^2 \left( \frac{\omega}{\omega_n} \right)^2 \right\}^{1/2} -$$ - (4.51) - -The log amplitude in this case involves a parameter ζ , resulting in a different plot for each value of ζ . For complex-conjugate poles,† ζ < 1. Hence, we must sketch a family of curves for a number of values of ζ in the range 0 to 1. This is illustrated in Fig. 4.43a. The error between the actual plot and the asymptotes is shown in Fig. 4.44. The actual plot can be obtained by adding the error to the asymptotic plot. - -For second-order zeros (complex-conjugate zeros), the plots are mirror images (about the 0 dB line) of the plots depicted in Fig. 4.43a. Note the resonance phenomenon of the complex-conjugate poles. This phenomenon is barely noticeable for ζ > 0.707 but becomes pronounced as ζ → 0. - -### PHASE - -The phase function for second-order poles, as apparent in Eq. (4.48), is - -$$ -\angle H(j\omega) = -\tan^{-1}\left[\frac{2\zeta\left(\frac{\omega}{\omega_n}\right)}{1 - \left(\frac{\omega}{\omega_n}\right)^2}\right] -$$ -(4.52) - -For ω ω*n*, - -$$ -\angle H(j\omega) \approx 0 -$$ - -† For ζ ≥ 1, the two poles in the second-order factor are no longer complex but real, and each of these two real poles can be dealt with as a separate first-order factor. - -**Figure 4.43** Amplitude and phase response of a second-order pole. - -For ω ω*n*, - -$$ -\angle H(j\omega) \simeq -180^\circ -$$ - -Hence, the phase → −180◦ as ω → ∞. As in the case of amplitude, we also have a family of phase plots for various values of ζ , as illustrated in Fig. 4.43b. A convenient asymptote for the phase of complex-conjugate poles is a step function that is 0◦ for ω<ω*n* and −180◦ for ω>ω*n*. - -**Figure 4.44** Errors in the asymptotic approximation of a second-order pole. - -Error plots for such an asymptote are shown in Fig. 4.44 for various values of ζ . The exact phase is the asymptotic value plus the error. - -For complex-conjugate zeros, the amplitude and phase plots are mirror images of those for complex conjugate-poles. - -We shall demonstrate the application of these techniques with two examples. - -### **EXAMPLE 4.29 Bode Plots for Second-Order Transfer Function with Real Roots** - -Sketch Bode plots for the transfer function - -$$ -H(s) = \frac{20s(s+100)}{(s+2)(s+10)} -$$ - -### MAGNITUDE PLOT - -First, we write the transfer function in normalized form - -$$ -H(s) = \frac{20 \times 100}{2 \times 10} \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)} = 100 \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)} -$$ - -Here, the constant term is 100; that is, 40 dB (20 log 100 = 40). This term can be added to the plot by simply relabeling the horizontal axis (from which the asymptotes begin) as the 40 dB line (see Fig. 4.45a). Such a step implies shifting the horizontal axis upward by 40 dB. This is precisely what is desired. - -In addition, we have two first-order poles at −2 and −10, one zero at the origin, and one zero at −100. - -**Step 1.** For each of these terms, we draw an asymptotic plot as follows (shown in Fig. 4.45a by dashed lines): - -- **(a)** For the zero at the origin, draw a straight line with a slope of 20 dB/decade passing through ω = 1. -- **(b)** For the pole at −2, draw a straight line with a slope of −20 dB/decade (for ω > 2) beginning at the corner frequency ω = 2. -- **(c)** For the pole at −10, draw a straight line with a slope of −20 dB/decade beginning at the corner frequency ω = 10. -- **(d)** For the zero at −100, draw a straight line with a slope of 20 dB/decade beginning at the corner frequency ω = 100. -- **Step 2.** Add all the asymptotes, as depicted in Fig. 4.45a by solid line segments. -- **Step 3.** Apply the following corrections (see Fig. 4.42a): - - **(a)** The correction at ω = 1 because of the corner frequency at ω = 2 is −1 dB. The correction at ω = 1 because of the corner frequencies at ω = 10 and ω = 100 is quite small (see Fig. 4.42a) and may be ignored. Hence, the net correction at ω = 1 is −1 dB. - -**Figure 4.45 (a)** Amplitude and **(b)** phase responses of the second-order system. - -- **(b)** The correction at ω = 2 because of the corner frequency at ω = 2 is −3 dB, and the correction because of the corner frequency at ω = 10 is −0.17 dB. The correction because of the corner frequency ω = 100 can be safely ignored. Hence the net correction at ω = 2 is −3.17 dB. -- **(c)** The correction at ω = 10 because of the corner frequency at ω = 10 is −3 dB, and the correction because of the corner frequency at ω = 2 is −0.17 dB. The correction because of ω = 100 can be ignored. Hence the net correction at ω = 10 is −3.17 dB. - -- **(d)** The correction at ω = 100 because of the corner frequency at ω = 100 is 3 dB, and the corrections because of the other corner frequencies may be ignored. -- **(e)** In addition to the corrections at corner frequencies, we may consider corrections at intermediate points for more accurate plots. For instance, the corrections at ω = 4 because of corner frequencies at ω = 2 and 10 are −1 and about −0.65, totaling −1.65 dB. In the same way, the corrections at ω = 5 because of corner frequencies at ω = 2 and 10 are −0.65 and −1, totaling −1.65 dB. - -With these corrections, the resulting amplitude plot is illustrated in Fig. 4.45a. - -### PHASE PLOT - -We draw the asymptotes corresponding to each of the four factors: - -- **(a)** The zero at the origin causes a 90◦ phase shift. -- **(b)** The pole at *s* = −2 has an asymptote with a zero value for −∞ <ω< 0.2 and a slope of −45◦/decade beginning at ω = 0.2 and going up to ω = 20. The asymptotic value for ω > 20 is −90◦. -- **(c)** The pole at *s* = −10 has an asymptote with a zero value for −∞ <ω< 1 and a slope of −45◦/decade beginning at ω = 1 and going up to ω = 100. The asymptotic value for ω > 100 is −90◦. -- **(d)** The zero at *s* = −100 has an asymptote with a zero value for −∞ <ω< 10 and a slope of 45◦/decade beginning at ω = 10 and going up to ω = 1000. The asymptotic value for ω > 1000 is 90◦. All the asymptotes are added, as shown in Fig. 4.45b. The appropriate corrections are applied from Fig. 4.42b, and the exact phase plot is depicted in Fig. 4.45b. - -### **EXAMPLE 4.30 Bode Plots for Second-Order Transfer Function with Complex Poles** - -Sketch the amplitude and phase response (Bode plots) for the transfer function - -$$ -H(s) = \frac{10(s+100)}{s^2 + 2s + 100} = 10 \frac{1 + \frac{s}{100}}{1 + \frac{s}{50} + \frac{s^2}{100}} -$$ - -### MAGNITUDE PLOT - -Here, the constant term is 10: that is, 20 dB(20 log 10 = 20). To add this term, we simply label the horizontal axis (from which the asymptotes begin) as the 20 dB line, as before (see Fig. 4.46a). - -**Figure 4.46 (a)** Amplitude and **(b)** phase responses of the second-order system. - -In addition, we have a real zero at *s* = −100 and a pair of complex conjugate poles. When we express the second-order factor in standard form, - -$$ -s^2 + 2s + 100 = s^2 + 2\zeta \omega_n s + \omega_n^2 -$$ - -### 434 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -we have - -$$ -\omega_n = 10 \quad \text{and} \quad \zeta = 0.1 -$$ - -**Step 1.** Draw an asymptote of −40 dB/decade (−12 dB/octave) starting at ω = 10 for the complex conjugate poles, and draw another asymptote of 20 dB/decade starting at ω = 100 for the (real) zero. - -**Step 2.** Add both asymptotes. - -**Step 3.** Apply the correction at ω = 100, where the correction because of the corner frequency ω = 100 is 3 dB. The correction because of the corner frequency ω = 10, as seen from Fig. 4.44a for ζ = 0.1, can be safely ignored. Next, the correction at ω = 10 because of the corner frequency ω = 10 is 13.90 dB (see Fig. 4.44a for ζ = 0.1). The correction because of the real zero at −100 can be safely ignored at ω = 10. We may find corrections at a few more points. The resulting plot is illustrated in Fig. 4.46a. - -### PHASE PLOT - -The asymptote for the complex conjugate poles is a step function with a jump of −180◦ at ω = 10. The asymptote for the zero at *s* = −100 is zero for ω ≤ 10 and is a straight line with a slope of 45◦/decade, starting at ω = 10 and going to ω = 1000. For ω ≥ 1000, the asymptote is 90◦. The two asymptotes add to give the sawtooth shown in Fig. 4.46b. We now apply the corrections from Figs. 4.42b and 4.44b to obtain the exact plot. - -**Figure 4.47** MATLAB-generated Bode plots for Ex. 4.30. - -### BODE PLOTS WITH MATLAB - -Bode plots make it relatively simple to hand-draw straight-line approximations to a system's magnitude and frequency responses. To produce exact Bode plots, we turn to MATLAB and its bode command. - ->> bode(tf([10 1000],[1 2 100]),'k-'); - -The resulting MATLAB plots, shown in Fig. 4.47, match the plots shown in Fig. 4.46. - -**Comment.** These two examples demonstrate that actual frequency response plots are very close to asymptotic plots, which are so easy to construct. Thus, by mere inspection of *H*(*s*) and its poles and zeros, one can rapidly construct a mental image of the frequency response of a system. This is the principal virtue of Bode plots. - -### POLES AND ZEROS IN THE RIGHT HALF-PLANE - -In our discussion so far, we have assumed the poles and zeros of the transfer function to be in the left half-plane. What if some of the poles and/or zeros of *H*(*s*) lie in the RHP? If there is a pole in the RHP, the system is unstable. Such systems are useless for any signal-processing application. For this reason, we shall consider only the case of the RHP zero. The term corresponding to RHP zero at *s* = *a* is (*s*/*a*) −1, and the corresponding frequency response is (*j*ω/*a*) −1. The amplitude response is - -$$ -\left|\frac{j\omega}{a} - 1\right| = \left(\frac{\omega^2}{a^2} + 1\right)^{1/2} -$$ - -This shows that the amplitude response of an RHP zero at *s* = *a* is identical to that of an LHP zero or *s* = −*a*. Therefore, the log amplitude plots remain unchanged whether the zeros are in the LHP or the RHP. However, the phase corresponding to the RHP zero at *s* = *a* is - -$$ -\angle \left(\frac{j\omega}{a} - 1\right) = \angle -\left(1 - \frac{j\omega}{a}\right) = \pi + \tan^{-1}\left(\frac{-\omega}{a}\right) = \pi - \tan^{-1}\left(\frac{\omega}{a}\right) -$$ - -whereas the phase corresponding to the LHP zero at *s* = −*a* is tan−1(ω/*a*). - -The complex-conjugate zeros in the RHP give rise to a term *s*2−2ζω*ns*+ω2 *n*, which is identical to the term *s*2 +2ζω*ns*+ω2 *n* with a sign change in ζ . Hence, from Eqs. (4.51) and (4.52), it follows that the amplitudes are identical, but the phases are of opposite signs for the two terms. - -Systems whose poles and zeros are restricted to the LHP are classified as *minimum phase* systems. Minimum phase systems are particularly desirable because the system *and its inverse* are both stable. - -### **[4.9-5 The Transfer Function from the Frequency Response](#page-10-0)** - -In the preceding section we were given the transfer function of a system. From a knowledge of the transfer function, we developed techniques for determining the system response to sinusoidal inputs. We can also reverse the procedure to determine the transfer function of a minimum phase - -### 436 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -system from the system's response to sinusoids. This application has significant practical utility. If we are given a system in a black box with only the input and output terminals available, the transfer function has to be determined by experimental measurements at the input and output terminals. The frequency response to sinusoidal inputs is one of the possibilities that is very attractive because the measurements involved are so simple. One needs only to apply a sinusoidal signal at the input and observe the output. We find the amplitude gain |*H*(*j*ω)| and the output phase shift *H*(*j*ω) (with respect to the input sinusoid) for various values of ω over the entire range from 0 to ∞. This information yields the frequency response plots (Bode plots) when plotted against log ω. From these plots we determine the appropriate asymptotes by taking advantage of the fact that the slopes of all asymptotes must be multiples of ±20 dB/decade if the transfer function is a rational function (function that is a ratio of two polynomials in *s*). From the asymptotes, the corner frequencies are obtained. Corner frequencies determine the poles and zeros of the transfer function. Because of the ambiguity about the location of zeros since LHP and RHP zeros (zeros at *s* = ±*a*) have identical magnitudes, this procedure works only for minimum phase systems. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/062_4.10 FILTER DESIGN BY PLACEMENT OF POLES AND ZEROS OF H(s).md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/062_4.10 FILTER DESIGN BY PLACEMENT OF POLES AND ZEROS OF H(s).md deleted file mode 100644 index 5d374402ad3267be93cd930cbff202c1da63810d..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/062_4.10 FILTER DESIGN BY PLACEMENT OF POLES AND ZEROS OF H(s).md +++ /dev/null @@ -1,164 +0,0 @@ -## **4.10 FILTER DESIGN BY [PLACEMENT OF](#page-10-0) POLES AND ZEROS OF** *H(s)* - -In this section we explore the strong dependence of frequency response on the location of poles and zeros of *H*(*s*). This dependence points to a simple intuitive procedure to filter design. - -### **[4.10-1 Dependence of Frequency Response on Poles](#page-10-0) and Zeros of** *H(s)* - -Frequency response of a system is basically the information about the filtering capability of the system. A system transfer function can be expressed as - -$$ -H(s) = \frac{P(s)}{Q(s)} = b_0 \frac{(s - z_1)(s - z_2) \cdots (s - z_N)}{(s - \lambda_1)(s - \lambda_2) \cdots (s - \lambda_N)} -$$ - -where *z*1, *z*2, ... , *zN* are λ1, λ2, ... , λ*N* are the poles of *H*(*s*). Now the value of the transfer function *H*(*s*) at some frequency *s* = *p* is - -$$ -H(s)|_{s=p} = b_0 \frac{(p-z_1)(p-z_2)\cdots(p-z_N)}{(p-\lambda_1)(p-\lambda_2)\cdots(p-\lambda_N)} -$$ -(4.53) - -This equation consists of factors of the form *p*−*zi* and *p*−λ*i*. The factor *p*−*zi* is a complex number represented by a vector drawn from point *z* to the point *p* in the complex plane, as illustrated in Fig. 4.48a. The length of this line segment is |*p* − *zi*|, the magnitude of *p* − *zi*. The angle of this directed line segment (with the horizontal axis) is (*p* − *zi*). To compute *H*(*s*) at *s* = *p*, we draw line segments from all poles and zeros of *H*(*s*) to the point *p*, as shown in Fig. 4.48b. The vector connecting a zero *zi* to the point *p* is *p* − *zi*. Let the length of this vector be *ri*, and let its angle with the horizontal axis be φ*i*. Then *p*−*zi* = *riej*φ*i* . Similarly, the vector connecting a pole λ*i* to the point *p* is *p* − λ*i* = *diej*θ*i* , where *di* and θ*i* are the length and the angle (with the horizontal axis), - -**Figure 4.48** Vector representations of **(a)** complex numbers and **(b)** factors of *H*(*s*). - -respectively, of the vector *p*−λ*i*. Now from Eq. (4.53) it follows that - -$$ -H(s)|_{s=p} = b_0 \frac{(r_1 e^{j\phi_1})(r_2 e^{j\phi_2}) \cdots (r_N e^{j\phi_N})}{(d_1 e^{j\theta_1})(d_2 e^{j\theta_2}) \cdots (d_N e^{j\theta_N})} -$$ - -= $b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} e^{j[(\phi_1 + \phi_2 + \cdots + \phi_N) - (\theta_1 + \theta_2 + \cdots + \theta_N)]}$ - -Therefore - -$$ -|H(s)|_{s=p} = b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} = b_0 \frac{\text{product of distances of zeros to } p}{\text{product of distances of poles to } p} -$$ -(4.54) - -and - -$$ -\angle H(s)|_{s=p} = (\phi_1 + \phi_2 + \dots + \phi_N) - (\theta_1 + \theta_2 + \dots + \theta_N) -$$ - -= sum of angles of zeros to $p$ – sum of angles of poles to $p$ (4.55) - -Here, we have assumed positive *b*0. If *b*0 is negative, there is an additional phase π. Using this procedure, we can determine *H*(*s*) for any value of *s*. To compute the frequency response *H*(*j*ω), we use *s* = *j*ω (a point on the imaginary axis), connect all poles and zeros to the point *j*ω, and determine |*H*(*j*ω)| and *H*(*j*ω) from Eqs. (4.54) and (4.55). We repeat this procedure for all values of ω from 0 to ∞ to obtain the frequency response. - -### GAIN ENHANCEMENT BY A POLE - -To understand the effect of poles and zeros on the frequency response, consider a hypothetical case of a single pole −α + *j*ω0, as depicted in Fig. 4.49a. To find the amplitude response |*H*(*j*ω)| for a certain value of ω, we connect the pole to the point *j*ω (Fig. 4.49a). If the length of this line is *d*, then |*H*(*j*ω)| is proportional to 1/*d*, - -$$ -|H(j\omega)| = \frac{K}{d} \tag{4.56} -$$ - -where the exact value of constant *K* is not important at this point. As ω increases from zero, *d* decreases progressively until ω reaches the value ω0. As ω increases beyond ω0, *d* increases - -**Figure 4.49** The role of poles and zeros in determining the frequency response of an LTIC system. - -progressively. Therefore, according to Eq. (4.56), the amplitude response |*H*(*j*ω)| increases from ω = 0 until ω = ω0, and it decreases continuously as ω increases beyond ω0, as illustrated in Fig. 4.49b. Therefore, a pole at −α + *j*ω0 results in a frequency-selective behavior that enhances the gain at the frequency ω0 (resonance). Moreover, as the pole moves closer to the imaginary axis (as α is reduced), this enhancement (resonance) becomes more pronounced. This is because α, the distance between the pole and *j*ω0 (*d* corresponding to *j*ω0), becomes smaller, which increases the gain *K*/*d*. In the extreme case, when α = 0 (pole on the imaginary axis), the gain at ω0 goes to infinity. Repeated poles further enhance the frequency-selective effect. To summarize, we can enhance a gain at a frequency ω0 by placing a pole opposite the point *j*ω0. The closer the pole is to *j*ω0, the higher is the gain at ω0, and the gain variation is more rapid (more frequency selective) in the vicinity of frequency ω0. Note that a pole must be placed in the LHP for stability. - -Here we have considered the effect of a single complex pole on the system gain. For a real system, a complex pole −α + *j*ω0 must accompany its conjugate −α − *j*ω0. We can readily show that the presence of the conjugate pole does not appreciably change the frequency-selective behavior in the vicinity of ω0. This is because the gain in this case is *K*/*dd* , where *d* is the distance of a point *j*ω from the conjugate pole −α − *j*ω0. Because the conjugate pole is far from *j*ω0, there is no dramatic change in the length *d* as ω varies in the vicinity of ω0. There is a gradual increase in the value of *d* as ω increases, which leaves the frequency-selective behavior as it was originally, with only minor changes. - -### GAIN SUPPRESSION BY A ZERO - -Using the same argument, we observe that zeros at −α ± *j*ω0 (Fig. 4.49d) will have exactly the opposite effect of suppressing the gain in the vicinity of ω0, as shown in Fig. 4.49e). A zero on the imaginary axis at *j*ω0 will totally suppress the gain (zero gain) at frequency ω0. Repeated zeros will further enhance the effect. Also, a closely placed pair of a pole and a zero (dipole) tend to cancel out each other's influence on the frequency response. Clearly, a proper placement of poles and zeros can yield a variety of frequency-selective behavior. We can use these observations to design lowpass, highpass, bandpass, and bandstop (or notch) filters. - -Phase response can also be computed graphically. In Fig. 4.49a, angles formed by the complex conjugate poles −α±*j*ω0 at ω =0 (the origin) are equal and opposite. As ω increases from 0 up, the angle θ1 (due to the pole −α +*j*ω0), which has a negative value at ω = 0, is reduced in magnitude; the angle θ2 because of the pole −α − *j*ω0, which has a positive value at ω = 0, increases in magnitude. As a result, θ1 + θ2, the sum of the two angles, increases continuously, approaching a value π as ω → ∞. The resulting phase response *H*(*j*ω) = −(θ1 +θ2) is illustrated in Fig. 4.49c. Similar arguments apply to zeros at −α ± *j*ω0. The resulting phase response *H*(*j*ω) = (φ1 + φ2) is depicted in Fig. 4.49f. - -We now focus on simple filters, using the intuitive insights gained in this discussion. The discussion is essentially qualitative. - -### **[4.10-2 Lowpass Filters](#page-10-0)** - -A typical lowpass filter has a maximum gain at ω = 0. Because a pole enhances the gain at frequencies in its vicinity, we need to place a pole (or poles) on the real axis opposite the origin (*j*ω = 0), as shown in Fig. 4.50a. The transfer function of this system is - -$$ -H(s) = \frac{\omega_c}{s + \omega_c} -$$ - -We have chosen the numerator of *H*(*s*) to be ω*c* to normalize the dc gain *H*(0) to unity. If *d* is the distance from the pole −ω*c* to a point *j*ω (Fig. 4.50a), then - -$$ -|H(j\omega)| = \frac{\omega_c}{d} -$$ - -with *H*(0) = 1. As ω increases, *d* increases and |*H*(*j*ω)| decreases monotonically with ω, as illustrated in Fig. 4.50d with label *N* = 1. This is clearly a lowpass filter with gain enhanced in the vicinity of ω = 0. - -### WALL OF POLES - -An ideal lowpass filter characteristic (shaded in Fig. 4.50d) has a constant gain of unity up to frequency ω*c*. Then the gain drops suddenly to 0 for ω>ω*c*. To achieve the ideal lowpass - -**Figure 4.50** Pole-zero configuration and the amplitude response of a lowpass (Butterworth) filter. - -characteristic, we need enhanced gain over the entire frequency band from 0 to ω*c*. We know that to enhance a gain at any frequency ω, we need to place a pole opposite ω. To achieve an enhanced gain for all frequencies over the band (0 to ω*c*), we need to place a pole opposite every frequency in this band. In other words, we need a *continuous wall of poles* facing the imaginary axis opposite the frequency band 0 to ω*c* (and from 0 to −ω*c* for conjugate poles), as depicted in Fig. 4.50b. At this point, the optimum shape of this wall is not obvious because our arguments are qualitative and intuitive. Yet, it is certain that to have enhanced gain (constant gain) at every frequency over this range, we need an infinite number of poles on this wall. We can show that for a maximally flat† response over the frequency range (0 to ω*c*), the wall is a semicircle with an infinite number of poles uniformly distributed along the wall [11]. In practice, we compromise by using a finite number (*N*) of poles with less-than-ideal characteristics. Figure 4.50c shows the pole configuration for a fifth-order (*N* = 5) filter. The amplitude response for various values of *N* is illustrated in Fig. 4.50d. As *N* → ∞, the filter response approaches the ideal. This family of filters is known as the *Butterworth* filters. There are also other families. In *Chebyshev* filters, the wall shape is a semiellipse rather than a semicircle. The characteristics of a Chebyshev filter are inferior to those of Butterworth over the passband (0,ω*c*), where the characteristics show a rippling effect - -† Maximally flat amplitude response means the first 2*N* − 1 derivatives of |*H*(*j*ω)| with respect to ω are zero at ω = 0. - -instead of the maximally flat response of Butterworth. But in the stopband (ω>ω*c*), Chebyshev behavior is superior in the sense that Chebyshev filter gain drops faster than that of the Butterworth. - -### **[4.10-3 Bandpass Filters](#page-10-0)** - -The shaded characteristic in Fig. 4.51b shows the ideal bandpass filter gain. In the bandpass filter, the gain is enhanced over the entire passband. Our earlier discussion indicates that this can be realized by a wall of poles opposite the imaginary axis in front of the passband centered at ω0. (There is also a wall of conjugate poles opposite −ω0.) Ideally, an infinite number of poles is required. In practice, we compromise by using a finite number of poles and accepting less-than-ideal characteristics (Fig. 4.51). - -### **[4.10-4 Notch \(Bandstop\) Filters](#page-10-0)** - -An ideal notch filter amplitude response (shaded in Fig. 4.52b) is a complement of the amplitude response of an ideal bandpass filter. Its gain is zero over a small band centered at some frequency ω0 and is unity over the remaining frequencies. Realization of such a characteristic requires an infinite number of poles and zeros. Let us consider a practical second-order notch filter to obtain zero gain at a frequency ω = ω0. For this purpose, we must have zeros at ±*j*ω0. The requirement of unity gain at ω = ∞ requires the number of poles to be equal to the number of zeros (*M* = *N*). This ensures that for very large values of ω, the product of the distances of poles from ω will be equal to the product of the distances of zeros from ω. Moreover, unity gain at ω = 0 requires a pole and the corresponding zero to be equidistant from the origin. For example, if we use two (complex-conjugate) zeros, we must have two poles; the distance from the origin of the poles and of the zeros should be the same. This requirement can be met by placing the two conjugate poles on the semicircle of radius ω0, as depicted in Fig. 4.52a. The poles can be anywhere on the semicircle to satisfy the equidistance condition. Let the two conjugate poles be at angles ±θ with respect to the negative real axis. Recall that a pole and a zero in the same vicinity tend to cancel out - -**Figure 4.51 (a)** Pole-zero configuration and **(b)** the amplitude response of a bandpass filter. - -**Figure 4.52 (a)** Pole-zero configuration and **(b)** the amplitude response of a bandstop (notch) filter. - -each other's influences. Therefore, placing poles closer to zeros (selecting θ closer to π/2) results in a rapid recovery of the gain from value 0 to 1 as we move away from ω0 in either direction. Figure 4.52b shows the gain |*H*(*j*ω)| for three different values of θ. - -### **EXAMPLE 4.31 Notch Filter Design** - -Design a second-order notch filter to suppress 60 Hz hum in a radio receiver. - -We use the poles and zeros in Fig. 4.52a with ω0 = 120π. The zeros are at *s* = ±*j*ω0. The two poles are at −ω0 cos θ ±*j*ω0 sin θ. The filter transfer function is (with ω0 = 120π) - -$$ -H(s) = \frac{(s - j\omega_0)(s + j\omega_0)}{(s + \omega_0 \cos \theta + j\omega_0 \sin \theta)(s + \omega_0 \cos \theta - j\omega_0 \sin \theta)} -$$ - -= -$$ -\frac{s^2 + \omega_0^2}{s^2 + (2\omega_0 \cos \theta)s + \omega_0^2} = \frac{s^2 + 142122.3}{s^2 + (753.98 \cos \theta)s + 142122.3} -$$ - -and - -$$ -|H(j\omega)| = \frac{-\omega^2 + 142122.3}{\sqrt{(-\omega^2 + 142122.3)^2 + (753.98\omega\cos\theta)^2}} -$$ - -The closer the poles are to the zeros (the closer θ is to π/2), the faster the gain recovery from 0 to 1 on either side of ω0 = 120π. Figure 4.52b shows the amplitude response for three different values of θ. This example is a case of very simple design. To achieve zero gain over a band, we need an infinite number of poles as well as an infinite number of zeros. - -MATLAB easily computes and plots the magnitude response curves of Fig. 4.52b. To illustrate, let us plot the magnitude response using θ = 60◦ over a frequency range of 0 ≤ *f* ≤ 150 Hz. The result, shown in Fig. 4.53, matches the θ = 60◦ case of Fig. 4.52b. - ->> f = (0:.01:150); omega0 = 2\*pi\*60; theta = 60\*pi/180; - -- >> H = @(s) (s.^2+omega0^2)./(s.^2+2\*omega0\*cos(theta)\*s+omega0^2); -- >> plot(f,abs(H(1j\*2\*pi\*f)),'k-'); -- >> xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); - -## **DR ILL 4.16 Magnitude Response from Pole-Zero Plots** - -Use the qualitative method of sketching the frequency response to show that the system with the pole-zero configuration in Fig. 4.54a is a highpass filter and the configuration in Fig. 4.54b is a bandpass filter. - -### **[4.10-5 Practical Filters and Their Specifications](#page-10-0)** - -For ideal filters, everything is black and white; the gains are either zero or unity over certain bands. As we saw earlier, real life does not permit such a worldview. Things have to be gray or shades of gray. In practice, we can realize a variety of filter characteristics that can only approach ideal characteristics. - -An ideal filter has a passband (unity gain) and a stopband (zero gain) with a sudden transition from the passband to the stopband. There is no transition band. For practical (or realizable) filters, on the other hand, the transition from the passband to the stopband (or vice versa) is gradual and takes place over a finite band of frequencies. Moreover, for realizable filters, the gain cannot be zero over a finite band (Paley–Wiener condition). As a result, there can be no true stopband for practical filters. We therefore define a *stopband* to be a band over which the gain is below some small number *Gs*, as illustrated in Fig. 4.55. Similarly, we define a *passband* to be a band over which the gain is between 1 and some number *Gp* (*Gp* < 1), as shown in Fig. 4.55. We have selected the passband gain of unity for convenience. It could be any constant. Usually the gains are specified in terms of decibels. This is simply 20 times the log (to base 10) of the gain. Thus, - -$$ -G(\text{dB}) = 20\log_{10} G -$$ - -A gain of unity is 0 dB and a gain of √ 2 is 3.01 dB, usually approximated by 3 dB. Sometimes the specification may be in terms of attenuation, which is the negative of the gain in dB. Thus, a gain of 1/ √2, that is, 0.707, is −3 dB, but is an attenuation of 3 dB. - -**Figure 4.55** Passband, stopband, and transition band in filters of various types. - -In a typical design procedure, *Gp* (*minimum passband gain*) and *Gs* (*maximum stopband gain*) are specified. Figure 4.55 shows the passband, the stopband, and the transition band for typical lowpass, bandpass, highpass, and bandstop filters. Fortunately, the highpass, bandpass, and bandstop filters can be obtained from a basic lowpass filter by simple frequency transformations. For example, replacing *s* with ω*c*/*s* in the lowpass filter transfer function results in a highpass filter. Similarly, other frequency transformations yield the bandpass and bandstop filters. Hence, it is necessary to develop a design procedure only for a basic lowpass filter. Then, by using appropriate transformations, we can design filters of other types. The design procedures are beyond our scope here and will not be discussed. The interested reader is referred to [1]. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/063_4.11 THE BILATERAL LAPLACE TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/063_4.11 THE BILATERAL LAPLACE TRANSFORM.md deleted file mode 100644 index 07ad310c54fd3a3903d836236ce7586792e1e9bf..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/063_4.11 THE BILATERAL LAPLACE TRANSFORM.md +++ /dev/null @@ -1,397 +0,0 @@ -## **4.11 THE BILATERAL LAPLACE [TRANSFORM](#page-10-0)** - -Situations involving noncausal signals and/or systems cannot be handled by the (unilateral) Laplace transform discussed so far. These cases can be analyzed by the *bilateral* (or *two-sided*) Laplace transform defined by - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ - -and *x*(*t*) can be obtained from *X*(*s*) by the inverse transformation - -$$ -x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds -$$ - -Observe that the unilateral Laplace transform discussed so far is a special case of the bilateral Laplace transform, where the signals are restricted to the causal type. Basically, the two transforms are the same. For this reason we use the same notation for the bilateral Laplace transform. - -Earlier we showed that the Laplace transforms of *e*−*atu*(*t*) and of −*e*−*atu*(−*t*) are identical. The only difference is in their regions of convergence (ROC). The ROC for the former is Re*s* > −*a*; that for the latter is Re *s* < −*a*, as illustrated in Fig. 4.1. Clearly, the inverse Laplace transform of *X*(*s*) is not unique unless the ROC is specified. If we restrict all our signals to the causal type, however, this ambiguity does not arise. The inverse transform of 1/(*s*+*a*) is *e*−*atu*(*t*). Thus, in the unilateral Laplace transform, we can ignore the ROC in determining the inverse transform of *X*(*s*). - -We now show that any bilateral transform can be expressed in terms of two unilateral transforms. It is, therefore, possible to evaluate bilateral transforms from a table of unilateral transforms. - -Consider the function *x*(*t*) appearing in Fig. 4.56a. We separate *x*(*t*) into two components, *x*1(*t*) and *x*2(*t*), representing the positive time (*causal*) component and the negative time (*anticausal*) component of *x*(*t*), respectively (Figs. 4.56b and 4.56c): - -$$ -x_1(t) = x(t)u(t) -$$ - and $x_2(t) = x(t)u(-t)$ - -**Figure 4.56** Expressing a signal as a sum of causal and anticausal components. - -The bilateral Laplace transform of *x*(*t*) is given by - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ - -= -$$ -\int_{-\infty}^{0^-} x_2(t)e^{-st} dt + \int_{0^-}^{\infty} x_1(t)e^{-st} dt -$$ - -= -$$ -X_2(s) + X_1(s) -$$ - (4.57) - -where *X*1(*s*) is the Laplace transform of the causal component *x*1(*t*), and *X*2(*s*) is the Laplace transform of the anticausal component *x*2(*t*). Consider *X*2(*s*), given by - -$$ -X_2(s) = \int_{-\infty}^{0^-} x_2(t)e^{-st} dt = \int_{0^+}^{\infty} x_2(-t)e^{st} dt -$$ - -Therefore, - -$$ -X_2(-s) = \int_{0^+}^{\infty} x_2(-t)e^{-st} dt -$$ - -If *x*(*t*) has any impulse or its derivative(s) at the origin, they are included in *x*1(*t*). Consequently, *x*2(*t*) = 0 at the origin; that is, *x*2(0) = 0. Hence, the lower limit on the integration in the preceding equation can be taken as 0− instead of 0+. Therefore, - -$$ -X_2(-s) = \int_{0^-}^{\infty} x_2(-t)e^{-st} dt -$$ - -Because *x*2(−*t*) is causal (Fig. 4.56d), *X*2(−*s*) can be found from the unilateral transform table. Changing the sign of *s* in *X*2(−*s*) yields *X*2(*s*). - -To summarize, the bilateral transform *X*(*s*) in Eq. (4.57) can be computed from the unilateral transforms in two steps: - -- 1. Split *x*(*t*) into its causal and anticausal components, *x*1(*t*) and *x*2(*t*), respectively. -- 2. Since the signals *x*1(*t*) and *x*2(−*t*) are both causal, take the (unilateral) Laplace transform of *x*1(*t*) and add to it the (unilateral) Laplace transform of *x*2(−*t*), with *s* replaced by −*s*. This procedure gives the (bilateral) Laplace transform of *x*(*t*). - -Since *x*1(*t*) and *x*2(−*t*) are both causal, *X*1(*s*) and *X*2(−*s*) are both unilateral Laplace transforms. Let σ*c*1 and σ*c*2 be the abscissas of convergence of *X*1(*s*) and *X*2(−*s*), respectively. This statement implies that *X*1(*s*) exists for all *s* with Re *s* > σ*c*1, and *X*2(−*s*) exists for all *s* with Re*s* > σ*c*2. Therefore, *X*2(*s*) exists for all *s* with Re *s* < −σ*c*2. † Therefore, *X*(*s*) = *X*1(*s*) + *X*2(*s*) exists for all *s* such that - -$$ -\sigma_{c1} < \text{Re}\,s < -\sigma_{c2} -$$ - -The regions of convergence of *X*1(*s*), *X*2(*s*), and *X*(*s*) are shown in Fig. 4.57. Because *X*(*s*) is finite for all values of *s* lying in the strip of convergence (σ*c*1 < Re *s* < −σ*c*2), poles of *X*(*s*) must lie outside this strip. The poles of *X*(*s*) arising from the causal component *x*1(*t*) lie to the left of the *strip* (region) *of convergence*, and those arising from its anticausal component *x*2(*t*) lie to its right (see Fig. 4.57). This fact is of crucial importance in finding the inverse bilateral transform. - -This result can be generalized to left-sided and right-sided signals. We define a signal *x*(*t*) as a *right-sided* signal if *x*(*t*) = 0 for *t* < *T*1 for some finite positive or negative number *T*1. A causal signal is always a right-sided signal, but the converse is not necessarily true. A signal is said to *left-sided* if it is zero for *t* > *T*2 for some finite, positive, or negative number *T*2. An anticausal signal is always a left-sided signal, but the converse is not necessarily true. A *two-sided* signal is of infinite duration on both positive and negative sides of *t* and is neither right-sided nor left-sided. - -We can show that the conclusions for ROC for causal signals also hold for right-sided signals, and those for anticausal signals hold for left-sided signals. In other words, if *x*(*t*) is causal or - -† For instance, if *x*(*t*) exists for all *t* &gt; 10, then *x*(−*t*), its time-inverted form, exists for *t* &lt; −10. - -**Figure 4.57** Regions of convergence for causal, anticausal, and combined signals. - -right-sided, the poles of *X*(*s*) lie to the left of the ROC, and if *x*(*t*) is anticausal or left-sided, the poles of *X*(*s*) lie to the right of the ROC. - -To prove this generalization, we observe that a right-sided signal can be expressed as *x*(*t*) + *xf*(*t*), where *x*(*t*) is a causal signal and *xf*(*t*) is some finite-duration signal. The ROC of any finite-duration signal is the entire *s*-plane (no finite poles). Hence, the ROC of the right-sided signal *x*(*t*) + *xf*(*t*) is the region common to the ROCs of *x*(*t*) and *xf*(*t*), which is same as the ROC for *x*(*t*). This proves the generalization for right-sided signals. We can use a similar argument to generalize the result for left-sided signals. Let us find the bilateral Laplace transform of - -$$ -x(t) = e^{bt}u(-t) + e^{at}u(t) -$$ -\n(4.58) - -We already know the Laplace transform of the causal component - -$$ -e^{at}u(t) \Longleftrightarrow \frac{1}{s-a} \qquad \text{Re}\,s > a \tag{4.59} -$$ - -For the anticausal component, *x*2(*t*) = *ebtu*(−*t*), we have - -$$ -x_2(-t) = e^{-bt}u(t) \Longleftrightarrow \frac{1}{s+b} \qquad \text{Re}\, s > -b -$$ - -so that - -$$ -X_2(s) = \frac{1}{-s+b} = \frac{-1}{s-b} \qquad \text{Re}\, s < b -$$ - -Therefore, - -$$ -e^{bt}u(-t) \Longleftrightarrow \frac{-1}{s-b} \qquad \text{Re}\,s < b \tag{4.60} -$$ - -and the Laplace transform of *x*(*t*) in Eq. (4.58) is - -$$ -X(s) = -\frac{1}{s-b} + \frac{1}{s-a} -$$ - Res > a and Res < b -= -$$ -\frac{a-b}{(s-b)(s-a)} -$$ - Res < b -(4.61) - -Figure 4.58 shows *x*(*t*) and the ROC of *X*(*s*) for various values of *a* and *b*. Equation (4.61) indicates that the ROC of *X*(*s*) does not exist if *a* > *b*, which is precisely the case in Fig. 4.58f. Observe that the poles of *X*(*s*) are outside (on the edges) of the ROC. The poles of *X*(*s*) because of the anticausal component of *x*(*t*) lie to the right of the ROC, and those due to the causal component of *x*(*t*) lie to its left. - -When *X*(*s*) is expressed as a sum of several terms, the ROC for *X*(*s*) is the intersection of (region common to) the ROCs of all the terms. In general, if *x*(*t*) = %*k i*=1 *xi*(*t*), then the ROC for *X*(*s*) is the intersection of the ROCs (region common to all ROCs) for the transforms *X*1(*s*), *X*2(*s*), ... , *Xk*(*s*). - -**Figure 4.58** Various two exponential signals and their regions of convergence. - -### **EXAMPLE 4.32 Inverse Bilateral Laplace Transform** - -Find the inverse bilateral Laplace transform of - -$$ -X(s) = \frac{-3}{(s+2)(s-1)} -$$ - -if the ROC is **(a)** −2 < Re *s* < 1, **(b)** Re *s* > 1, and **(c)** Re *s* < −2. - -**(a)** - -$$ -X(s) = \frac{1}{s+2} - \frac{1}{s-1} -$$ - -Now, *X*(*s*) has poles at −2 and 1. The strip of convergence is −2 < Re *s* < 1. The pole at −2, being to the left of the strip of convergence, corresponds to a causal signal. The pole at 1, being to the right of the strip of convergence, corresponds to an anticausal signal. Equations (4.59) and (4.60) yield - -$$ -x(t) = e^{-2t}u(t) + e^t u(-t) -$$ - -**(b)** Both poles lie to the left of the ROC, so both poles correspond to causal signals. Therefore, - -)*u*(*t*) - -*x*(*t*) = (*e*−2*t* −*et* - -**Figure 4.59** Three possible inverse transforms of −3/((*s* +2)(*s*−1)). - -**(c)** Both poles lie to the right of the region of convergence, so both poles correspond to anticausal signals, and - -$$ -x(t) = (-e^{-2t} + e^t)u(-t) -$$ - -Figure 4.59 shows the three inverse transforms corresponding to the same *X*(*s*) but with different regions of convergence. - -### **[4.11-1 Properties of the Bilateral Laplace Transform](#page-11-0)** - -Properties of the bilateral Laplace transform are similar to those of the unilateral transform. We shall merely state the properties here without proofs. Let the ROC of *X*(*s*) be *a* < Re *s* < *b*. Similarly, let the ROC of *Xi*(*s*) be *ai* < Re *s* < *bi* for (*i* = 1, 2). - -LINEARITY - -$$ -a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(s) + a_2X_2(s) -$$ - -The ROC for *a*1*X*1(*s*) + *a*2*X*2(*s*) is the region common to (intersection of) the ROCs for *X*1(*s*) and *X*2(*s*). - -TIME SHIFT - -$$ -x(t-T) \Longleftrightarrow X(s)e^{-sT} -$$ - -The ROC for *X*(*s*)*e*−*sT* is identical to the ROC for *X*(*s*). - -FREQUENCY SHIFT - -$$ -x(t)e^{s_0t} \Longleftrightarrow X(s-s_0) -$$ - -The ROC for *X*(*s*−*s*0) is *a*+*c* < Re *s* < *b*+*c*, where *c* = Re *s*0. - -TIME DIFFERENTIATION - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow sX(s) -$$ - -The ROC for *sX*(*s*) contains the ROC for *X*(*s*) and may be larger than that of *X*(*s*) under certain conditions [e.g., if *X*(*s*) has a first-order pole at *s* = 0, it is canceled by the factor *s* in *sX*(*s*)]. - -TIME INTEGRATION - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(s)/s -$$ - -The ROC for *sX*(*s*) is max (*a*, 0) < Re *s* < *b*. - -TIME SCALING - -$$ -x(\beta t) \Longleftrightarrow \frac{1}{|\beta|}X\left(\frac{s}{\beta}\right) -$$ - -The ROC for *X*(*s*/β) is β*a* < Re *s* < β*b*. For β > 1, *x*(β*t*) represents time compression and the corresponding ROC expands by factor β. For 0 >β> 1, *x*(β*t*) represents time expansion and the corresponding ROC is compressed by factor β. - -TIME CONVOLUTION - -$$ -x_1(t) * x_2(t) \Longleftrightarrow X_1(s)X_2(s) -$$ - -The ROC for *X*1(*s*)*X*2(*s*) is the region common to (intersection of ) the ROCs for *X*1(*s*) and *X*2(*s*). - -FREQUENCY CONVOLUTION - -$$ -x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty} X_1(w)X_2(s-w) dw -$$ - -The ROC for *X*1(*s*) ∗ *X*2(*s*) is *a*1 +*a*2 < Re *s* < *b*1 +*b*2. - -TIME REVERSAL - -$$ -x(-t) \Longleftrightarrow X(-s) -$$ - -The ROC for *X*(−*s*) is −*b* < Re *s* < −*a*. - -### **[4.11-2 Using the Bilateral Transform for Linear System Analysis](#page-11-0)** - -Since the bilateral Laplace transform can handle noncausal signals, we can analyze noncausal LTIC systems using the bilateral Laplace transform. We have shown that the (zero-state) output *y*(*t*) is given by - -$$ -y(t) = \mathcal{L}^{-1}[X(s)H(s)] -$$ - -This expression is valid only if *X*(*s*)*H*(*s*) exists. The ROC of *X*(*s*)*H*(*s*) is the region in which both *X*(*s*) and *H*(*s*) exist. In other words, the ROC of *X*(*s*)*H*(*s*) is the region common to the regions of convergence of both *X*(*s*) and *H*(*s*). These ideas are clarified in the following examples. - -Find the current *y*(*t*) for the *RC* circuit in Fig. 4.60a if the voltage *x*(*t*) is - -$$ -x(t) = e^t u(t) + e^{2t} u(-t) -$$ - -**Figure 4.60** Response of a circuit to a noncausal input. - -The transfer function *H*(*s*) of the circuit is given by - -$$ -H(s) = \frac{s}{s+1} \qquad \text{Re}\, s > -1 -$$ - -Because *h*(*t*) is a causal function, the ROC of *H*(*s*) is Re *s* > −1. Next, the bilateral Laplace transform of *x*(*t*) is given by - -$$ -X(s) = \frac{1}{s-1} - \frac{1}{s-2} = \frac{-1}{(s-1)(s-2)} \qquad 1 < \text{Re } s < 2 -$$ - -The response *y*(*t*) is the inverse transform of *X*(*s*)*H*(*s*): - -$$ -y(t) = \mathcal{L}^{-1} \left[ \frac{-s}{(s+1)(s-1)(s-2)} \right] = \mathcal{L}^{-1} \left[ \frac{1}{6} \frac{1}{s+1} + \frac{1}{2} \frac{1}{s-1} - \frac{2}{3} \frac{1}{s-2} \right] -$$ - -The ROC of *X*(*s*)*H*(*s*) is that ROC common to both *X*(*s*) and *H*(*s*). This is 1 < Re *s* < 2. The poles *s* = ±1 lie to the left of the ROC and, therefore, correspond to causal signals; the pole - -### 454 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -*s* = 2 lies to the right of the ROC and thus represents an anticausal signal. Hence, - -$$ -y(t) = \frac{1}{6}e^{-t}u(t) + \frac{1}{2}e^{t}u(t) + \frac{2}{3}e^{2t}u(-t) -$$ - -Figure 4.60c shows *y*(*t*). Note that in this example, if - -$$ -x(t) = e^{-4t}u(t) + e^{-2t}u(-t) -$$ - -then the ROC of *X*(*s*) is −4 < Re *s* < −2. Here no region of convergence exists for *X*(*s*)*H*(*s*). Hence, the response *y*(*t*) goes to infinity. - -### **EXAMPLE 4.34 Response of a Noncausal System** - -Find the response *y*(*t*) of a noncausal system with the transfer function - -$$ -H(s) = \frac{-1}{s-1} \qquad \text{Re}\, s < 1 -$$ - -to the input *x*(*t*) = *e*−2*t u*(*t*). - -We have - -$$ -X(s) = \frac{1}{s+2} \qquad \text{Re}\, s > -2 -$$ - -and - -$$ -Y(s) = X(s)H(s) = \frac{-1}{(s-1)(s+2)} -$$ - -The ROC of *X*(*s*)*H*(*s*) is the region −2 < Re *s* < 1. By partial fraction expansion, - -$$ -Y(s) = \frac{-1/3}{s-1} + \frac{1/3}{s+2} \qquad -2 < \text{Re}\, s < 1 -$$ - -and - -$$ -y(t) = \frac{1}{3} [e^t u(-t) + e^{-2t} u(t)] -$$ - -Note that the pole of *H*(*s*) lies in the RHP at 1. Yet the system is not unstable. The pole(s) in the RHP may indicate instability or noncausality, depending on its location with respect to the region of convergence of *H*(*s*). For example, if *H*(*s*) = −1/(*s*−1) with Re *s* > 1, the system is causal and unstable, with *h*(*t*) = −*et u*(*t*). In contrast, if *H*(*s*) = −1/(*s* − 1) with Re *s* < 1, the system is noncausal and stable, with *h*(*t*) = *et u*(−*t*). - -### **EXAMPLE 4.35 System Response to a Noncausal Input** - -Find the response *y*(*t*) of a system with the transfer function - -$$ -H(s) = \frac{1}{s+5} \qquad \text{Re}\, s > -5 -$$ - -and the input - -$$ -x(t) = e^{-t}u(t) + e^{-2t}u(-t) -$$ - -The input *x*(*t*) is of the type depicted in Fig. 4.58f, and the region of convergence for *X*(*s*) does not exist. In this case, we must determine separately the system response to each of the two input components, *x*1(*t*) = *e*−*t u*(*t*) and *x*2(*t*) = *e*−2*t u*(−*t*). - -$$ -X_1(s) = \frac{1}{s+1} \qquad \text{Re}\, s > -1 -$$ -\n -$$ -X_2(s) = \frac{-1}{s+2} \qquad \text{Re}\, s < -2 -$$ - -If *y*1(*t*) and *y*2(*t*) are the system responses to *x*1(*t*) and *x*2(*t*), respectively, then - -$$ -Y_1(s) = \frac{1}{(s+1)(s+5)} = \frac{1/4}{s+1} - \frac{1/4}{s+5} -$$ - Re $s > -1$ - -so that - -$$ -y_1(t) = \frac{1}{4}(e^{-t} - e^{-5t})u(t) -$$ - -and - -$$ -Y_2(s) = \frac{-1}{(s+2)(s+5)} = \frac{-1/3}{s+2} + \frac{1/3}{s+5} \qquad -5 < \text{Re}\,s < -2 -$$ - -so that - -$$ -y_2(t) = \frac{1}{3} [e^{-2t}u(-t) + e^{-5t}u(t)] -$$ - -Therefore, - -$$ -y(t) = y_1(t) + y_2(t) = \frac{1}{3}e^{-2t}u(-t) + \left(\frac{1}{4}e^{-t} + \frac{1}{12}e^{-5t}\right)u(t) -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/064_4.12 MATLAB - CONTINUOUS-TIME FILTERS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/064_4.12 MATLAB - CONTINUOUS-TIME FILTERS.md deleted file mode 100644 index c2d12983097295a8c77178b236fd3303053bdba2..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/064_4.12 MATLAB - CONTINUOUS-TIME FILTERS.md +++ /dev/null @@ -1,360 +0,0 @@ -## **[4.12 MATLAB: CONTINUOUS-TIME](#page-11-0) FILTERS** - -Continuous-time filters are essential to many if not most engineering systems, and MATLAB is an excellent assistant for filter design and analysis. Although a comprehensive treatment of continuous-time filter techniques is outside the scope of this book, quality filters can be designed and realized with minimal additional theory. - -A simple yet practical example demonstrates basic filtering concepts. Telephone voice signals are often lowpass-filtered to eliminate frequencies above a cutoff of 3 kHz, or ω*c* = 3000(2π ) ≈ 18,850 rad/s. Filtering maintains satisfactory speech quality and reduces signal bandwidth, thereby increasing the phone company's call capacity. How, then, do we design and realize an acceptable 3 kHz lowpass filter? - -### **[4.12-1 Frequency Response and Polynomial Evaluation](#page-11-0)** - -Magnitude response plots help assess a filter's performance and quality. The magnitude response of an ideal filter is a brick-wall function with unity passband gain and perfect stopband attenuation. For a lowpass filter with cutoff frequency ω*c*, the ideal magnitude response is - -$$ -|H_{\text{ideal}}(j\omega)| = \begin{cases} 1 & |\omega| \le \omega_c \\ 0 & |\omega| > \omega_c \end{cases} -$$ - -Unfortunately, ideal filters cannot be implemented in practice. Realizable filters require compromises, although good designs will closely approximate the desired brick-wall response. - -A realizable LTIC system often has a rational transfer function that is represented in the *s*-domain as - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{B(s)}{A(s)} = \frac{\sum_{k=0}^{M} b_{k+N-M} s^{M-k}}{\sum_{k=0}^{N} a_k s^{N-k}} -$$ - -Frequency response *H*(*j*ω) is obtained by letting *s* = *j*ω, where frequency ω is in radians per second. - -MATLAB is ideally suited to evaluate frequency response functions. Defining a length-(*N* + 1) coefficient vector **A** = [*a*0,*a*1,...,*aN*] and a length-(*M* + 1) coefficient vector **B** = [*bN*−*M*,*bN*−*M*+1,..., *bN*], program CH4MP1 computes *H*(*j*ω) for each frequency in the input vector *ω*. - -``` -function [H] = CH4MP1(B,A,omega); -% CH4MP1.m : Chapter 4, MATLAB Program 1 -% Function M-file computes frequency response for LTIC system -% INPUTS: B = vector of feedforward coefficients -% A = vector of feedback coefficients -% omega = vector of frequencies [rad/s]. -% OUTPUTS: H = frequency response -``` - -``` -H = polyval(B,j*omega)./polyval(A,j*omega); -``` - -The function polyval efficiently evaluates simple polynomials and makes the program nearly trivial. For example, when A is the vector of coefficients [*a*0,*a*1,...,*aN*], polyval (A,j\*omega) computes - -$$ -\sum_{k=0}^N a_k (j\omega)^{N-k} -$$ - -for each value of the frequency vector omega. It is also possible to compute frequency responses by using the signal-processing toolbox function freqs. - -### DESIGN AND EVALUATION OF A SIMPLE *RC* FILTER - -One of the simplest lowpass filters is realized by using an *RC* circuit, as shown in Fig. 4.61. This one-pole system has transfer function *HRC*(*s*) = (*RCs* + 1)−1 and magnitude response |*HRC*(*j*ω)|=|(*j*ω*RC* + 1)−1| = 1/ 1+(*RC*ω)2. Independent of component values *R* and *C*, this circuit has many desirable characteristics, such as unity gain at ω = 0 and magnitude response that monotonically decreases to zero as ω → ∞. - -Components *R* and *C* are chosen to set the desired 3 kHz cutoff frequency. For many filter types, the cutoff frequency corresponds to the half-power point, or |*HRC*(*j*ω*c*)| = 1/ √2. Assign *C* a realistic capacitance of 1 nF, then the required resistance is computed by *R* = 1/ *C*2ω2 *c* = 1/ (10−9)2(2π3000)2. - -``` ->> omega_c = 2*pi*3000; C = 1e-9; R = 1/sqrt(C^2*omega_c^2) - R = 5.3052e+004 -``` - -The root of this first-order *RC* filter is directly related to the cutoff frequency, λ = −1/*RC* = −18,850 = −ω*c*. - -To evaluate the *RC* filter performance, the magnitude response is plotted over the mostly audible frequency range (0 ≤ *f* ≤ 20 kHz). - -``` ->> f = linspace(0,20000,200); Hmag_RC = abs(CH4MP1([1],[R*C 1],f*2*pi)); -``` - -``` ->> plot(f,abs(f*2*pi)<=omega_c,'k-',f,Hmag_RC,'k--'); -``` - -``` ->> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); -``` - -``` ->> legend('Ideal','First-order RC','location','best'); -``` - -The linspace(X1,X2,N) command generates an N-length vector of linearly spaced points between X1 and X2. - -As shown in Fig. 4.62, the first-order *RC* response is indeed lowpass with a half-power cutoff frequency equal to 3 kHz. It rather poorly approximates the desired brick-wall response: the passband is not very flat, and stopband attenuation increases very slowly to less than 20 dB at 20 kHz. - -**Figure 4.62** Magnitude response |*HRC*(*j*2π*f*)| of a first-order *RC* filter. - -**Figure 4.63** A cascaded *RC* filter. - -### A CASCADED *RC* FILTER AND POLYNOMIAL EXPANSION - -A first-order *RC* filter is destined for poor performance; one pole is simply insufficient to obtain good results. A cascade of *RC* circuits increases the number of poles and improves the filter response. To simplify the analysis and prevent loading between stages, we employ op-amp followers to buffer the output of each stage, as shown in Fig. 4.63. A cascade of *N* stages results in an *N*th-order filter with transfer function given by - -$$ -H_{\text{cascade}}(s) = [H_{RC}(s)]^N = (RCs + 1)^{-N} -$$ - -Upon choosing a cascade of 10 stages and *C* = 1 nF, a 3 kHz cutoff frequency is obtained by setting *R* = √ 21/10 −1/(*C*ω*c*) = √ 21/10 −1/(6π(10)−6). - ->> R = sqrt(2^(1/10)-1)/(C\*omega\_c) R = 1.4213e+004 - -This cascaded filter has a 10th-order pole at λ = −1/*RC* and no finite zeros. To compute the magnitude response, polynomial coefficient vectors **A** and **B** are needed. Setting **B** = [1] ensures there are no finite zeros or, equivalently, that all zeros are at infinity. The poly command, which expands a vector of roots into a corresponding vector of polynomial coefficients, is used to obtain **A**. - -- >> B = 1; A = poly(-1/(R\*C)\*ones(10,1));A = A/A(end); -- >> Hmag\_cascade = abs(CH4MP1(B,A,f\*2\*pi)); -- >> plot(f,abs(f\*2\*pi)<=omega\_c,'k-',f,Hmag\_cascade,'k--'); -- >> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); -- >> legend('Ideal','Tenth-order RC cascade','location','best'); - -Notice that scaling a polynomial by a constant does not change its roots. Conversely, the roots of a polynomial specify a polynomial within a scale factor. The command A = A/A(end) properly scales the denominator polynomial to ensure unity gain at ω = 0. - -The magnitude response plot of the tenth-order *RC* cascade is shown in Fig. 4.64. Compared with the simple *RC* response of Fig. 4.62, the passband remains relatively unchanged, but stopband attenuation is greatly improved to over 60 dB at 20 kHz. - -**Figure 4.64** Magnitude response |*H*cascade(*j*2π*f*)| of a tenth-order *RC* cascade. - -## **[4.12-2 Butterworth Filters and the](#page-11-0)** Find **Command** - -The pole location of a first-order lowpass filter is necessarily fixed by the cutoff frequency. There is little reason, however, to place all the poles of a 10th-order filter at one location. Better pole placement will improve our filter's magnitude response. One strategy, discussed in Sec. 4.10, is to place a wall of poles opposite the passband frequencies. A semicircular wall of poles leads to the Butterworth family of filters, and a semi-elliptical shape leads to the Chebyshev family of filters. Butterworth filters are considered first. - -To begin, notice that a transfer function *H*(*s*) with real coefficients has a squared magnitude response given by |*H*(*j*ω)| 2 = *H*(*j*ω)*H*∗(*j*ω) = *H*(*j*ω)*H*(−*j*ω) = *H*(*s*)*H*(−*s*)|*s*=*j*ω. Thus, half the poles of |*H*(*j*ω)| 2 correspond to the filter *H*(*s*) and the other half correspond to *H*(−*s*). Filters that are both stable and causal require *H*(*s*) to include only left-half-plane poles. - -The squared magnitude response of a Butterworth filter is - -$$ -|H_{\text{BW}}(j\omega)|^2 = \frac{1}{1 + (j\omega/j\omega_c)^{2N}} -$$ - -This function has the same appealing characteristics as the first-order *RC* filter: a gain that is unity at ω = 0 and monotonically decreases to zero as ω → ∞. By construction, the half-power gain - -occurs at ω*c*. Perhaps most importantly, however, the first 2*N* − 1 derivatives of |*H*BW(*j*ω)| with respect to ω are zero at ω = 0. Put another way, the passband is constrained to be very flat for low frequencies. For this reason, Butterworth filters are sometimes called maximally flat filters. - -As discussed in Sec. B.7, the roots of minus 1 must lie equally spaced on a circle centered at the origin. Thus, the 2*N* poles of |*H*BW(*j*ω)| 2 naturally lie equally spaced on a circle of radius ω*c* centered at the origin. Figure 4.65 displays the 20 poles corresponding to the case *N* = 10 and ω*c* = 3000(2π ) rad/s. An *N*th-order Butterworth filter that is both causal and stable uses the *N* left-half-plane poles of |*H*BW(*j*ω)| 2. - -To design a 10th-order Butterworth filter, we first compute the 20 poles of |*H*BW(*j*ω)| 2: - ->> N=10; poles = roots([(1j\*omega\_c)^(-2\*N),zeros(1,2\*N-1),1]); - -The find command is a powerful and useful function that returns the indices of a vector's nonzero elements. Combined with relational operators, the find command allows us to extract the 10 left-half-plane roots that correspond to the poles of our Butterworth filter. - ->> BW\_poles = poles(find(real(poles)<0)); - -To compute the magnitude response, these roots are converted to coefficient vector **A**. - -``` ->> A = poly(BW_poles); A = A/A(end); Hmag_BW = abs(CH4MP1(B,A,f*2*pi)); -``` - -``` ->> plot(f,abs(f*2*pi)<=omega_c,'k-',f,Hmag_BW,'k--'); -``` - -``` ->> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); -``` - -``` ->> legend('Ideal','Tenth-order Butterworth','location','best'); -``` - -The magnitude response plot of the Butterworth filter is shown in Fig. 4.66. The Butterworth response closely approximates the brick-wall function and provides excellent filter characteristics: flat passband, rapid transition to the stopband, and excellent stopband attenuation (>40 dB at 5 kHz). - -**Figure 4.66** Magnitude response |*H*BW(*j*2π*f*)| of a tenth-order Butterworth filter. - -### **[4.12-3 Using Cascaded Second-Order Sections](#page-11-0) for Butterworth Filter Realization** - -For our *RC* filters, realization preceded design. For our Butterworth filter, however, design has preceded realization. For our Butterworth filter to be useful, we must be able to implement it. - -Since the transfer function *H*BW(*s*) is known, the differential equation is also known. Therefore, it is possible to try to implement the design by using op-amp integrators, summers, and scalar multipliers. Unfortunately, this approach will not work well. To understand why, consider the denominator coefficients *a*0 = 1.766×10−43 and *a*10 = 1. The smallest coefficient is 43 orders of magnitude smaller than the largest coefficient! It is practically impossible to accurately realize such a broad range in scale values. To understand this, skeptics should try to find realistic resistors such that *Rf* /*R* = 1.766×10−43. Additionally, small component variations will cause large changes in actual pole location. - -A better approach is to cascade five second-order sections, where each section implements one complex conjugate pair of poles. By pairing poles in complex conjugate pairs, each of the resulting second-order sections has real coefficients. With this approach, the smallest coefficients are only about nine orders of magnitude smaller than the largest coefficients. Furthermore, pole placement is typically less sensitive to component variations for cascaded structures. - -The Sallen–Key circuit shown in Fig. 4.67 provides a good way to realize a pair of complex-conjugate poles.† The transfer function of this circuit is - -$$ -H_{SK}(s) = \frac{\frac{1}{R_1R_2C_1C_2}}{s^2 + \left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1}\right)s + \frac{1}{R_1R_2C_1C_2}} = \frac{\omega_0^2}{s^2 + \left(\frac{\omega_0}{Q}\right)s + \omega_0^2} -$$ - -Geometrically, ω0 is the distance from the origin to the poles and *Q* = 1/2 cosψ, where ψ is the angle between the negative real axis and the pole. Termed the "quality factor" of a circuit, *Q* - -† A more general version of the Sallen–Key circuit has a resistor *Ra* from the negative terminal to ground and a resistor *Rb* between the negative terminal and the output. In Fig. 4.67, *Ra* = ∞ and *Rb* = 0. - -provides a measure of the peakedness of the response. High-*Q* filters have poles close to the ω axis, which boost the magnitude response near those frequencies. - -Although many ways exist to determine suitable component values, a simple method is to assign *R*1 a realistic value and then let *R*2 = *R*1, *C*1 = 2*Q*/ω0*R*1, and *C*2 = 1/2*Q*ω0*R*2. Butterworth poles are a distance ω*c* from the origin, so ω0 = ω*c*. For our 10th-order Butterworth filter, the angles ψ are regularly spaced at 9, 27, 45, 63, and 81 degrees. MATLAB program CH4MP2 automates the task of computing component values and magnitude responses for each stage. - -``` -% CH4MP2.m : Chapter 4, MATLAB Program 2 -% Script M-file computes Sallen-Key component values and magnitude -% responses for each of the five cascaded second-order filter sections. -omega_0 = 3000*2*pi; % Filter cut-off frequency -psi = [9 27 45 63 81]*pi/180; % Butterworth pole angles -f = linspace(0,6000,200); % Frequency range for magnitude response calculations -Hmag_SK = zeros(5,200); % Pre-allocate array for magnitude responses -for stage = 1:5, - Q = 1/(2*cos(psi(stage))); % Compute Q for current stage - % Compute and display filter components to the screen: - disp(['Stage ',num2str(stage),... - ' (Q = ',num2str(Q),... - '): R1 = R2 = ',num2str(56000),... - ', C1 = ',num2str(2*Q/(omega_0*56000)),... - ', C2 = ',num2str(1/(2*Q*omega_0*56000))]); - B = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute filter coefficients - Hmag_SK(stage,:) = abs(CH4MP1(B,A,2*pi*f)); % Compute magnitude response -end -plot(f,Hmag_SK,'k',f,prod(Hmag_SK),'k:') -xlabel('f [Hz]'); ylabel('Magnitude Response') -``` - -The disp command displays a character string to the screen. Character strings must be enclosed in single quotation marks. The num2str command converts numbers to character strings and facilitates the formatted display of information. The prod command multiplies along the columns of a matrix; it computes the total magnitude response as the product of the magnitude responses of the five stages. - -Executing the program produces the following output: - -``` ->> CH4MP2 - Stage 1 (Q = 0.50623): R1 = R2 = 56000, C1 = 9.5916e-10, C2 = 9.3569e-10 - Stage 2 (Q = 0.56116): R1 = R2 = 56000, C1 = 1.0632e-09, C2 = 8.441e-10 - Stage 3 (Q = 0.70711): R1 = R2 = 56000, C1 = 1.3398e-09, C2 = 6.6988e-10 - Stage 4 (Q = 1.1013): R1 = R2 = 56000, C1 = 2.0867e-09, C2 = 4.3009e-10 - Stage 5 (Q = 3.1962): R1 = R2 = 56000, C1 = 6.0559e-09, C2 = 1.482e-10 -``` - -**Figure 4.68** Magnitude responses for Sallen–Key filter stages. - -Since all the component values are practical, this filter is possible to implement. Figure 4.68 displays the magnitude responses for all five stages (solid lines). The total response (dotted line) confirms a 10th-order Butterworth response. Stage 5, which has the largest *Q* and implements the pair of conjugate poles nearest the ω axis, is the most peaked response. Stage 1, which has the smallest *Q* and implements the pair of conjugate poles furthest from the ω axis, is the least peaked response. In practice, it is best to order high-*Q* stages last; this reduces the risk that the high gains will saturate the filter hardware. - -### **[4.12-4 Chebyshev Filters](#page-11-0)** - -Like an order-*N* Butterworth lowpass filter (LPF), an order-*N* Chebyshev LPF is an all-pole filter that possesses many desirable characteristics. Compared with an equal-order Butterworth filter, the Chebyshev filter achieves better stopband attenuation and reduced transition bandwidth by allowing an adjustable amount of ripple within the passband. - -The squared magnitude response of a Chebyshev filter is - -$$ -|H_{\rm C}(j\omega)|^2 = \frac{1}{1 + \epsilon^2 C_N^2(\omega/\omega_c)} -$$ - -where controls the passband ripple, *CN*(ω/ω*c*) is a degree-*N* Chebyshev polynomial, and ω*c* is the radian cutoff frequency. Several characteristics of Chebyshev LPFs are noteworthy: - -• An order-*N* Chebyshev LPF is equi-ripple in the passband (|ω| ≤ ω*c*), has a total of *N* maxima and minima over (0 ≤ ω ≤ ω*c*), and is monotonic decreasing in the stopband (|ω| > ω*c*). - -### 464 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -- In the passband, the maximum gain is 1 and the minimum gain is 1/ √1+ 2. For odd-valued *N*, |*H*(*j*0)| = 1. For even-valued *N*, |*H*C(*j*0)| = 1/ √1+ 2. -- Ripple is controlled by setting = √ 10*R*/10 −1, where *R* is the allowable passband ripple expressed in decibels. Reducing adversely affects filter performance (see Prob. 4.12-10). -- Unlike Butterworth filters, the cutoff frequency ω*c* rarely specifies the 3 dB point. For = 1, |*H*C(*j*ω*c*)| 2 = 1/(1+ 2) = 0.5. The cutoff frequency ω*c* simply indicates the frequency after which |*H*C(*j*ω)| < 1/ √ 1+ 2. - -The Chebyshev polynomial *CN*(*x*) is defined as - -$$ -C_N(x) = \cos[N\cos^{-1}(x)] = \cosh[N\cosh^{-1}(x)] -$$ - -In this form, it is difficult to verify that *CN*(*x*) is a degree-*N* polynomial in *x*. A recursive form of *CN*(*x*) makes this fact more clear (see Prob. 4.12-13). - -$$ -C_N(x) = 2xC_{N-1}(x) - C_{N-2}(x) -$$ - -With *C*0(*x*) = 1 and *C*1(*x*) = *x*, the recursive form shows that any *CN* is a linear combination of degree-*N* polynomials and is therefore a degree-*N* polynomial itself. For *N* ≥ 2, MATLAB program CH4MP3 generates the (*N* +1) coefficients of Chebyshev polynomial *CN*(*x*). - -``` -function [C_N] = CH4MP3(N); -% CH4MP3.m : Chapter 4, MATLAB Program 3 -% Function M-file computes Chebyshev polynomial coefficients -% using the recursion relation C_N(x) = 2xC_{N-1}(x) - C_{N-2}(x) -% INPUTS: N = degree of Chebyshev polynomial -% OUTPUTS: C_N = vector of Chebyshev polynomial coefficients -C_Nm2 = 1; C_Nm1 = [1 0]; % Initial polynomial coefficients: -for t = 2:N; - C_N = 2*conv([1 0],C_Nm1)-[zeros(1,length(C_Nm1)-length(C_Nm2)+1),C_Nm2]; - C_Nm2 = C_Nm1; C_Nm1 = C_N; -``` - -``` -end -``` - -``` -As examples, consider C2(x) = 2xC1(x) − C0(x) = 2x(x) − 1 = 2x2 − 1 and C3(x) = 2xC2(x) − -C1(x) = 2x(2x2 −1)−x = 4x3 −3x. CH4MP3 easily confirms these cases. -``` - -``` ->> CH4MP3(2) - ans = 2 0 -1 ->> CH4MP3(3) - ans = 4 0 -3 0 -``` - -Since *CN*(ω/ω*c*) is a degree-*N* polynomial, |*H*C(*j*ω)| 2 is an all-pole rational function with 2*N* finite poles. Similar to the Butterworth case, the *N* poles specifying a causal and stable Chebyshev filter can be found by selecting the *N* left-half-plane roots of 1+ 2*C*2 *N*[*s*/(*j*ω*c*)]. - -Root locations and dc gain are sufficient to specify a Chebyshev filter for a given *N* and . To demonstrate, consider the design of an order-8 Chebyshev filter with cutoff frequency *fc* = 1 kHz and allowable passband ripple *R* = 1 dB. First, filter parameters are specified. - -``` ->> omega_c = 2*pi*1000; R = 1; N = 8; ->> epsilon = sqrt(10^(R/10)-1); -``` - -The coefficients of *CN*[*s*/(*j*ω*c*)] are obtained with the help of CH4MP3, and then the coefficients of [1+ 2*C*2 *N*(*s*/(*j*ω*c*))] are computed by using convolution to perform polynomial multiplication. - -``` ->> CN = CH4MP3(N).*((1/(1j*omega_c)).^[N:-1:0]); ->> CP = epsilon^2*conv(CN,CN); CP(end) = CP(end)+1; -``` - -Next, the polynomial roots are found, and the left-half-plane poles are retained and plotted. - -``` ->> poles = roots(CP); i = find(real(poles)<0); C_poles = poles(i); ->> plot(real(C_poles),imag(C_poles),'kx'); axis equal; ->> axis(omega_c*[-1.1 1.1 -1.1 1.1]); ->> xlabel('Real'); ylabel('Imaginary'); -``` - -As shown in Fig. 4.69, the roots of a Chebyshev filter lie on an ellipse† (see Prob. 4.12-14). - -**Figure 4.69** Pole-zero plot for an order-8 Chebyshev LPF with *fc* = 1 kHz and *R* = 1 dB. - -To compute the filter's magnitude response, the poles are expanded into a polynomial, the dc gain is set based on the even value of *N*, and CH4MP1 is used. - -``` ->> A = poly(C_poles); B = A(end)/sqrt(1+epsilon^2); -``` - ->> omega = linspace(0,2\*pi\*2000,2001); H\_C = CH4MP1(B,A,omega); - -``` ->> plot(omega/2/pi,abs(H_C),'k'); axis([0 2000 0 1.1]); -``` - -``` ->> xlabel('f [Hz]'); ylabel('|H_C(j2\pi f)|'); -``` - -† E. A. Guillemin demonstrates a wonderful relationship between the Chebyshev ellipse and the Butterworth circle in his book *Synthesis of Passive Networks* (Wiley, New York, 1957). - -**Figure 4.70** Magnitude responses for an order-8 Chebyshev LPF with *fc* = 1 kHz and *R* = 1 dB. - -As seen in Fig. 4.70, the magnitude response exhibits correct Chebyshev filter characteristics: passband ripples are equal in height and never exceed *R* = 1 dB; there are a total of *N* = 8 maxima and minima in the passband; and the gain rapidly and monotonically decreases after the cutoff frequency of *fc* = 1 kHz. - -For higher-order filters, polynomial rooting may not provide reliable results. Fortunately, Chebyshev roots can also be determined analytically. For - -$$ -\phi_k = \frac{2k+1}{2N}\pi \quad \text{and} \quad \xi = \frac{1}{N}\sinh^{-1}\left(\frac{1}{\epsilon}\right) -$$ - -the Chebyshev poles are - -$$ -p_k = \omega_c \sinh(\xi) \sin(\phi_k) + j\omega_c \cosh(\xi) \cos(\phi_k) -$$ - -Continuing the same example, the poles are recomputed and again plotted. The result is identical to Fig. 4.69. - -``` ->> k = [1:N]; xi = 1/N*asinh(1/epsilon); phi = (k*2-1)/(2*N)*pi; ->> C_poles = omega_c*(-sinh(xi)*sin(phi)+1j*cosh(xi)*cos(phi)); ->> plot(real(C_poles),imag(C_poles),'kx'); axis equal; ->> axis(omega_c*[-1.1 1.1 -1.1 1.1]); ->> xlabel('Real'); ylabel('Imaginary'); -``` - -As in the case of high-order Butterworth filters, a cascade of second-order filter sections facilitates practical implementation of Chebyshev filters. Problems 4.12-5 and 4.12-8 use second-order Sallen–Key circuit stages to investigate such implementations. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/065_4.13 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/065_4.13 SUMMARY.md deleted file mode 100644 index 2629a2955865603979a934fafaeac304cedf12ec..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/065_4.13 SUMMARY.md +++ /dev/null @@ -1,65 +0,0 @@ -## **[4.13 SUMMARY](#page-11-0)** - -This chapter discusses analysis of LTIC (linear, time-invariant, continuous-time) systems by the Laplace transform, which transforms integro-differential equations of such systems into algebraic equations. Therefore solving these integro-differential equations reduces to solving algebraic equations. The Laplace transform method cannot be used for time-varying-parameter systems or for nonlinear systems in general. - -The transfer function *H*(*s*) of an LTIC system is the Laplace transform of its impulse response. It may also be defined as a ratio of the Laplace transform of the output to the Laplace transform of the input when all initial conditions are zero (system in zero state). If *X*(*s*) is the Laplace transform of the input *x*(*t*) and *Y*(*s*) is the Laplace transform of the corresponding output *y*(*t*) (when all initial conditions are zero), then *Y*(*s*) = *X*(*s*)*H*(*s*). For an LTIC system described by an *N*th-order differential equation *Q*(*D*)*y*(*t*) = *P*(*D*)*x*(*t*), the transfer function *H*(*s*) = *P*(*s*)/*Q*(*s*). Like the impulse response *h*(*t*), the transfer function *H*(*s*) is also an external description of the system. - -Electrical circuit analysis can also be carried out by using a transformed circuit method, in which all signals (voltages and currents) are represented by their Laplace transforms, all elements by their impedances (or admittances), and initial conditions by their equivalent sources (initial condition generators). In this method, a network can be analyzed as if it were a resistive circuit. - -Large systems can be depicted by suitably interconnected subsystems represented by blocks. Each subsystem, being a smaller system, can be readily analyzed and represented by its input–output relationship, such as its transfer function. Analysis of large systems can be carried out with the knowledge of input–output relationships of its subsystems and the nature of interconnection of various subsystems. - -LTIC systems can be realized by scalar multipliers, adders, and integrators. A given transfer function can be synthesized in many different ways, such as canonic, cascade, and parallel. Moreover, every realization has a transpose, which also has the same transfer function. In practice, all the building blocks (scalar multipliers, adders, and integrators) can be obtained from operational amplifiers. - -The system response to an everlasting exponential *est* is also an everlasting exponential *H*(*s*)*est*. Consequently, the system response to an everlasting exponential *ej*ω*t* is *H*(*j*ω) *ej*ω*t* . Hence, *H*(*j*ω) is the frequency response of the system. For a sinusoidal input of unit amplitude and having frequency ω, the system response is also a sinusoid of the same frequency (ω) with amplitude |*H*(*j*ω)|, and its phase is shifted by *H*(*j*ω) with respect to the input sinusoid. For this reason |*H*(*j*ω)| is called the amplitude response (gain) and *H*(*j*ω) is called the phase response of the system. Amplitude and phase response of a system indicate the filtering characteristics of the system. The general nature of the filtering characteristics of a system can be quickly determined from a knowledge of the location of poles and zeros of the system transfer function. - -Most of the input signals and practical systems are causal. Consequently we are required most of the time to deal with causal signals. When all signals must be causal, the Laplace transform analysis is greatly simplified; the region of convergence of a signal becomes irrelevant to the analysis process. This special case of the Laplace transform (which is restricted to causal signals) is called the unilateral Laplace transform. Much of the chapter deals with this variety of Laplace transform. Section 4.11 discusses the general Laplace transform (the bilateral Laplace transform), which can handle causal and noncausal signals and systems. In the bilateral transform, the inverse transform of *X*(*s*) is not unique but depends on the region of convergence of *X*(*s*). Thus, the region of convergence plays a very crucial role in the bilateral Laplace transform. - -### 468 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -### **[REFERENCES](#page-11-0)** - -- 1. Lathi, B. P. *Signal Processing and Linear Systems*, 1st ed. Oxford University Press, New York, 1998. -- 2. Doetsch, G. *Introduction to the Theory and Applications of the Laplace Transformation with a Table of Laplace Transformations.* Springer-Verlag, New York, 1974. -- 3. LePage, W. R. *Complex Variables and the Laplace Transforms for Engineers*. McGraw-Hill, New York, 1961. -- 4. Durant, Will, and Ariel Durant. *The Age of Napoleon*, Part XI in *The Story of Civilization Series*. Simon & Schuster, New York, 1975. -- 5. Bell, E. T. *Men of Mathematics*. Simon & Schuster, New York, 1937. -- 6. Nahin, P. J. "Oliver Heaviside: Genius and Curmudgeon." *IEEE Spectrum*, vol. 20, pp. 63–69, July 1983. -- 7. Berkey, D. *Calculus*, 2nd ed. Saunders, Philadelphia, 1988. -- 8. Encyclopaedia Britannica. *Micropaedia IV*, 15th ed., p. 981, Chicago, 1982. -- 9. Churchill, R. V. *Operational Mathematics*, 2nd ed. McGraw-Hill, New York, 1958. -- 10. Truxal, J. G. *The Age of Electronic Messages*. McGraw-Hill, New York, 1990. -- 11. Van Valkenberg, M. *Analog Filter Design*. Oxford University Press, New York, 1982. - -## **[PROBLEMS](#page-11-0)** - -- **4.1-1** By direct integration [Eq. (4.1)] find the Laplace transforms and the region of convergence of the following functions: - - (a) *u*(*t*)−*u*(*t* −1) - - (b) *te*−*t u*(*t*) - - (c) *t* cos ω0*t u*(*t*) - - (d) (*e*2*t* −2*e*−*t* )*u*(*t*) - - (e) cos ω1*t* cos ω2*t u*(*t*) - - (f) cosh(*at*)*u*(*t*) - - (g) sinh(*at*)*u*(*t*) - - (h) *e*−2*t* cos(5*t* +θ )*u*(*t*) -- **4.1-2** By direct integration [Eq. (4.1)] find the Laplace transforms and the region of convergence of the following functions: - -(a) -$$ -e^{-2t}u(t-5) + \delta(t-1) -$$ - -(b) -$$ -\pi e^{3t} u(t+5) - \delta(2t) -$$ - -(c) -$$ -\sum_{k=0}^{\infty} \delta(t - kT), T > 0 -$$ - -- **4.1-3** By direct integration find the Laplace transforms of the signals shown in Fig. P4.1-3. -- **4.1-4** Find the inverse (unilateral) Laplace transforms of the following functions: - -(a) -$$ \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/066_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/066_REFERENCES.md deleted file mode 100644 index 34493acda776d35b06ae6351ea1318ac9869fb23..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/066_REFERENCES.md +++ /dev/null @@ -1,1159 +0,0 @@ -\frac{2s+5}{s^2+5s+6} -$$ - -\n(b) -$$ -\frac{3s+5}{s^2+4s+13} -$$ - -\n(c) -$$ -\frac{(s+1)^2}{s^2-s-6} -$$ - -\n(d) -$$ -\frac{5}{s^2(s+2)} -$$ - -**Figure P4.1-3** - -(e) -$$ -\frac{2s+1}{(s+1)(s^2+2s+2)} -$$ - -\n(f) -$$ -\frac{s+2}{s(s+1)^2} -$$ - -\n(g) -$$ -\frac{1}{(s+1)(s+2)^4} -$$ - -\n(h) -$$ -\frac{s+1}{s(s+2)^2(s^2+4s+5)} -$$ - -(i) -$$ -\frac{s(s+2)^2(s^2+4s+5)}{(s+1)^2(s^2+2s+5)} -$$ - -- **4.2-1** Suppose a CT signal *x*(*t*)=2[*u*(*t* −2)−*u*(*t* +1)] has a transform *X*(*s*). - - (a) If *Y*a(*s*) = *e*−5*s sX s* + 1 2 , determine and sketch the corresponding signal *y*a(*t*). - - (b) If *Y*b(*s*) = 2−*s sX*(*s* −2), determine and sketch the corresponding signal *y*b(*t*). -- **4.2-2** Find the Laplace transforms of the following functions using only Table 4.1 and the time-shifting property (if needed) of the unilateral Laplace transform: - - (a) *u*(*t*) −*u*(*t* −1) - - (b) *e*−(*t*−τ )*u*(*t* −τ ) - - (c) *e*−(*t*−τ )*u*(*t*) - - (d) *e*−*t u*(*t* −τ ) - - (e) *te*−*t u*(*t* −τ ) - - (f) sin[ω0(*t* −τ )]*u*(*t* −τ ) - - (g) sin[ω0(*t* −τ )]*u*(*t*) - - (h) sin ω0*t u*(*t* −τ ) - - (i) *t*sin(*t*)*u*(*t*) - - (j) (1−*t*) cos(*t* −1)*u*(*t* −1) -- **4.2-3** Using only Table 4.1 and the time-shifting property, determine the Laplace transform of the signals in Fig. P4.1-3. [*Hint:* See Sec. 1.4 for discussion of expressing such signals analytically.] -- **4.2-4** Prove the frequency-differentiation property, −*tx*(*t*) ⇐⇒ *d dsX*(*s*). This property holds for both the unilateral and bilateral Laplace transforms. -- **4.2-5** Consider the signal *x*(*t*) = *te*−2(*t*−3) *u*(*t*−2). - - (a) Determine the *unilateral* Laplace transform *X*u(*s*) = *L*u {*x*(*t*)}. - -- (b) Determine the *bilateral* Laplace transform *X*(*s*) = *L*{*x*(*t*)}. -- **4.2-6** Consider the signals *x*(*t*) and *y*(*t*), as shown in Fig. P4.2-6. - - (a) Using the definition, compute *X*(*s*), the bilateral Laplace transform of *x*(*t*). - - (b) Using Laplace transform properties, express *Y*(*s*), the bilateral Laplace transform of *y*(*t*), as a function of *X*(*s*), the bilateral Laplace transform of *x*(*t*). Simplify as much as possible without substituting your answer from part (a). -- **4.2-7** Find the inverse Laplace transforms of the following functions: - -(a) -$$ -\frac{(2s+5)e^{-2s}}{s^2+5s+6} -$$ - -(b) -$$ -\frac{se^{-3s}+2}{s^2+2s+2} -$$ - -(c) -$$ -\frac{e^{-(s-1)}+3}{2(2s+5)} -$$ - -$$ -\begin{array}{c}\n\text{(c)} \quad s^2 - 2s + 5 \\ -\text{(d)} \quad \frac{e^{-s} + e^{-2s} + 1}{s^2 + 3s + 2}\n\end{array} -$$ - -- **4.2-8** Using ROC σ > 0, determine the inverse Laplace transform of *X*(*s*) = *s*−1 *d ds e*−2*s s* . -- **4.2-9** The Laplace transform of a causal periodic signal can be determined from the knowledge of the Laplace transform of its first cycle (period). - - (a) If the Laplace transform of *x*(*t*) in Fig. P4.2-9a is *X*(*s*), then show that *G*(*s*), the Laplace transform of *g*(*t*) (Fig. P4.2-9b), is - -$$ -G(s) = \frac{X(s)}{1 - e^{-sT_0}} \qquad \text{Re}\, s > 0 -$$ - -- (b) Use this result to find the Laplace transform of the signal *p*(*t*) illustrated in Fig. P4.2-9c. -- **4.2-10** Starting only with the fact that δ(*t*) ⇐⇒ 1, build pairs 2 through 10b in Table 4.1, using various properties of the Laplace transform. - -#### **Figure P4.2-9** - -- **4.2-11** (a) Find the Laplace transform of the pulses in Fig. 4.2 by using only the time-different iation property, the time-shifting property, and the fact that δ(*t*) ⇐⇒ 1. - - (b) In Ex. 4.9, the Laplace transform of *x*(*t*) is found by finding the Laplace transform of *d*2*x*/*dt*2. Find the Laplace transform of *x*(*t*) in that example by finding the Laplace transform of *dx*/*dt* and using Table 4.1, if necessary. -- **4.2-12** Determine the inverse unilateral Laplace transform of - -$$ -X(s) = \frac{1}{e^{s+3}} \frac{s^2}{(s+1)(s+2)} -$$ - -- **4.2-13** Since 13 is such a lucky number, determine the inverse Laplace transform of *X*(*s*) = 1/(*s* +1)13 given region of convergence σ > −1. [*Hint:* What is the *n*th derivative of 1/(*s* +*a*)?] -- **4.2-14** It is difficult to compute the Laplace transform *X*(*s*) of signal - -$$ -x(t) = \frac{1}{t}u(t) -$$ - -by using direct integration. Instead, properties provide a simpler method. - -- (a) Use Laplace transform properties to express the Laplace transform of *tx*(*t*) in terms of the unknown quantity *X*(*s*). -- (b) Use the definition to determine the Laplace transform of *y*(*t*) = *tx*(*t*). - -- (c) Solve for *X*(*s*) by using the two pieces from **()**(a) and **()**(b). Simplify your answer. -- **4.3-1** Use the Laplace transform to solve the following differential equations: - - (a) (*D*2 + 3*D* + 2)*y*(*t*) = *Dx*(*t*) if *y*(0−) = *y*˙(0−) = 0 and *x*(*t*) = *u*(*t*) - - (b) (*D*2 + 4*D* + 4)*y*(*t*) = (*D* + 1)*x*(*t*) if *y*(0−) = 2, *y*˙(0−) = 1 and *x*(*t*) = *e*−*t u*(*t*) - - (c) (*D*2 +6*D*+25)*y*(*t*) = (*D*+2)*x*(*t*) if *y*(0−) = *y*˙(0−) = 1 and *x*(*t*) = 25*u*(*t*) -- **4.3-2** Solve the differential equations in Prob. 4.3-1 using the Laplace transform. In each case determine the zero-input and zero-state components of the solution. -- **4.3-3** Consider a causal LTIC system described by the differential equation - -$$ -2\dot{y}(t) + 6y(t) = \dot{x}(t) - 4x(t) -$$ - -- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*(0−) = −3. -- (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *e*δ(*t* −π ). -- **4.3-4** Consider a causal LTIC system described by the differential equation - -$$ -\ddot{y}(t) + 3\dot{y}(t) + 2y(t) = 2\dot{x}(t) - x(t) -$$ - -- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*˙(0−) = 2 and *y*(0−) = −3. -- (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *u*(*t*). -- **4.3-5** Solve the following simultaneous differential equations using the Laplace transform, assuming all initial conditions to be zero and the input *x*(*t*) = *u*(*t*): - - (a) (*D*+3)*y*1(*t*)−2*y*2(*t*) = *x*(*t*) −2*y*1(*t*) +(2*D*+4)*y*2(*t*) = 0 - - (b) (*D*+2)*y*1(*t*)−(*D*+1)*y*2(*t*) = 0 −(*D*+1)*y*1(*t*) +(2*D*+1)*y*2(*t*) = *x*(*t*) - -Determine the transfer functions relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*). - -- **4.3-6** Consider a causal LTIC system described by *y*˙(*t*) +2*y*(*t*) = ˙*x*(*t*). - - (a) Determine the transfer function *H*(*s*) for this system. - - (b) Using your result from part (a), determine the impulse response *h*(*t*) for this system. - - (c) Using Laplace transform techniques, determine the output *y*(*t*) if the input is *x*(*t*) = *e*−*t u*(*t*) and *y*(0−) = √ 2. -- **4.3-7** Repeat Prob. 4.3-6 for a causal LTIC system described by 3*y*(*t*) + ˙*y*(*t*)+ ˙*x*(*t*) = 0. -- **4.3-8** For the circuit in Fig. P4.3-8, the switch is in the open position for a long time before *t* = 0, when it is closed instantaneously. - - (a) Write loop equations (in time domain) for *t* ≥ 0. - - (b) Solve for *y*1(*t*) and *y*2(*t*) by taking the Laplace transform of loop equations found in part (a). - -**4.3-9** For each of the systems described by the following differential equations, find the system transfer function: - -(a) -$$ -\frac{d^2y(t)}{dt^2} + 11\frac{dy(t)}{dt} + 24y(t) = 5\frac{dx(t)}{dt} + 3x(t) -$$ - -\n(b) -$$ -\frac{d^3y(t)}{dt^3} + 6\frac{d^2y(t)}{dt^2} - 11\frac{dy(t)}{dt} + 6y(t) -$$ -$$ -= 3\frac{d^2x(t)}{dt^2} + 7\frac{dx(t)}{dt} + 5x(t) -$$ - -\n(c) -$$ -\frac{d^4y(t)}{dt^4} + 4\frac{dy(t)}{dt} = 3\frac{dx(t)}{dt} + 2x(t) -$$ - -\n(d) -$$ -\frac{d^2y(t)}{dt^2} - y(t) = \frac{dx(t)}{dt} - x(t) -$$ - -**4.3-10** For each of the systems specified by the following transfer functions, find the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable: - -(a) -$$ -H(s) = \frac{s+5}{s^2 + 3s + 8} -$$ - -\n(b) $H(s) = \frac{s^2 + 3s + 5}{s^3 + 8s^2 + 5s + 7}$ -\n(c) $H(s) = \frac{5s^2 + 7s + 2}{s^2 - 2s + 5}$ - -**4.3-11** For a system with transfer function - -$$ -H(s) = \frac{2s+3}{s^2+2s+5} -$$ - -- (a) Find the (zero-state) response for inputs *x*1(*t*) = 10*u*(*t*) and *x*2(*t*) = *u*(*t* −5). -- (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable. -- **4.3-12** For a system with transfer function - -$$ -H(s) = \frac{s}{s^2 + 9} -$$ - -- (a) Find the (zero-state) response if the input *x*(*t*) = (1−*e*−*t* )*u*(*t*) -- (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable. -- **4.3-13** Consider a system with transfer function - -$$ -H(s) = \frac{s+5}{s^2 + 5s + 6} -$$ - -Find the (zero-state) response for the following inputs: - -- (a) *x*a(*t*) = *e*−3*t u*(*t*) -- (b) *x*b(*t*) = *e*−4*t u*(*t*) -- (c) *x*c(*t*) = *e*−4(*t*−5) *u*(*t* −5) -- (d) *x*d(*t*) = *e*−4(*t*−5) *u*(*t*) -- (e) *x*e(*t*) = *e*−4*t u*(*t* −5) - -Assuming that the system *H*(*s*) is controllable and observable, - -- (f) write the differential equation relating the output *y*(*t*) to the input *x*(*t*). -- **4.3-14** An LTI system has a step response given by *s*(*t*) = *e*−*t u*(*t*) − *e*−2*t u*(*t*). Determine the output of this system *y*(*t*) given an input *x*(*t*) = δ(*t* − π )−cos( √3)*u*(*t*). -- **4.3-15** For an LTIC system with zero initial conditions (system initially in zero state), if an input *x*(*t*) produces an output *y*(*t*), then using the Laplace transform, show the following: - - (a) The input *dx*/*dt* produces an output *dy*/*dt*. - - (b) The input \$ *t* 0 *x*(τ )*d*τ produces an output \$ *t* 0 *y*(τ )*d*τ . Hence, show that the unit step response of a system is an integral of the impulse response; that is, \$ *t* 0 *h*(τ )*d*τ . -- **4.3-16** Discuss asymptotic and BIBO stabilities for the systems described by the following transfer functions, assuming that the systems are controllable and observable: - -(a) -$$ -\frac{(s+5)}{s^2+3s+2} -$$ - -(b) -$$ -\frac{s+5}{s^2(s+2)} -$$ - -$$ -(c) \frac{s(s+2)}{s+5} -$$ - -(d) -$$ -\frac{s+5}{s(s+2)} -$$ - -(e) -$$ -\frac{s+5}{s^2-2s+3} -$$ - -- **4.3-17** Repeat Prob. 4.3-16 for systems described by the following differential equations. Systems may be uncontrollable and/or unobservable. - - (a) (*D*2 +3*D*+2)*y*(*t*) = (*D*+3)*x*(*t*) - - (b) (*D*2 +3*D*+2)*y*(*t*) = (*D*+1)*x*(*t*) - - (c) (*D*2 +*D*−2)*y*(*t*) = (*D*−1)*x*(*t*) - - (d) (*D*2 −3*D*+2)*y*(*t*) = (*D*−1)*x*(*t*) -- **4.4-1** The circuit shown in Fig. P4.4-1 has system function given by *H*(*s*) = 1 1+*RCs*. Let *R* = 2 and *C* = 3 and use Laplace transform techniques to solve the following. - - (a) Find the output *y*(*t*) given an initial capacitor voltage of *y*(0−) = 3 and an input *x*(*t*) = *u*(*t*). - - (b) Given an input *x*(*t*) = *u*(*t* − 3), determine the initial capacitor voltage *y*(0−) so that the output *y*(*t*) is 1 volt at *t* = 6 seconds. - -- **4.4-2** Consider the circuit shown in Fig. P4.4-2. Use Laplace transform techniques to solve the following. - - (a) Determine the standard-form, constantcoefficient differential equation description of this circuit. - - (b) Letting *R* = *C* = 1, determine the total response *y*(*t*) to input *x*(*t*) = 3*e*−*t u*(*t*) and initial capacitor voltage of *vC*(0−) = 5. - -**4.4-3** Find the zero-state response *y*(*t*) of the network in Fig. P4.4-3 if the input voltage *x*(*t*) = *te*−*t u*(*t*). Find the transfer function relating the output *Y*(*s*) to the input *X*(*s*). From the transfer function, write the differential equation relating *y*(*t*) to *x*(*t*). - -### **Figure P4.4-3** - -**4.4-4** The switch in the circuit of Fig. P4.4-4 is closed for a long time and then opened instantaneously at *t* = 0. Find and sketch the current *y*(*t*). - -#### **Figure P4.4-4** - -**4.4-5** Find the current *y*(*t*) for the parallel resonant circuit in Fig. P4.4-5 if the input is: (a) *x*(*t*) = *A*cos ω0*t u*(*t*) (b) *x*(*t*) = *A*sin ω0*t u*(*t*) Assume all initial conditions to be zero and, in both cases, ω2 0 = 1/*LC*. - -**Figure P4.4-5** - -**Figure P4.4-6** - -- **4.4-6** Find the loop currents *y*1(*t*) and *y*2(*t*) for *t* ≥ 0 in the circuit of Fig. P4.4-6a for the input *x*(*t*) in Fig. P4.4-6b. -- **4.4-7** For the network in Fig. P4.4-7, the switch is in a closed position for a long time before *t* = 0, when it is opened instantaneously. Find *y*1(*t*) and *vs*(*t*) for *t* ≥ 0. - -#### **Figure P4.4-7** - -- **4.4-8** Find the output voltage *v*0(*t*) for *t* ≥ 0 for the circuit in Fig. P4.4-8, if the input *x*(*t*) = 100*u*(*t*). The system is in the zero state initially. -- **4.4-9** Find the output voltage *y*(*t*) for the network in Fig. P4.4-9 for the initial conditions *iL*(0) = 1 A and *vC*(0) = 3 V. -- **4.4-10** For the network in Fig. P4.4-10, the switch is in position *a* for a long time and then is moved to position *b* instantaneously at *t* = 0. Determine the current *y*(*t*) for *t* > 0. - -**4.4-11** Consider the circuit of Fig. P4.4-11. - -- (a) Using transform-domain techniques, determine the system's standard-form transfer function *H*(*s*). -- (b) Using transform-domain techniques and letting *R* = *L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*t u*(*t* −1). - -**Figure P4.4-10** - -(c) Using transform-domain techniques and letting *R* = 2*L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*t u*(*t* −1). - -**4.4-12** Show that the transfer function that relates the output voltage *y*(*t*) to the input voltage *x*(*t*) for the op-amp circuit in Fig. P4.4-12a is given by - -$$ -H(s) = \frac{Ka}{s+a} \quad \text{where} -$$ - -$$ -K = 1 + \frac{R_b}{R_a} \quad \text{and} \quad a = \frac{1}{RC} -$$ - -and that the transfer function for the circuit in Fig. P4.4-12b is given by - -$$ -H(s) = \frac{Ks}{s+a} -$$ - -**4.4-13** For the second-order op-amp circuit in Fig. P4.4-13, show that the transfer function *H*(*s*) relating the output voltage *y*(*t*) to the input - -### **Figure P4.4-14** - -voltage *x*(*t*) is given by - -$$ -H(s) = \frac{-s}{s^2 + 8s + 12} -$$ - -- **4.4-14** Consider the op-amp circuit of Fig. P4.4-14. - - (a) Determine the standard-form transfer function *H*(*s*) of this system. - - (b) Determine the standard-form constant coefficient linear differential equation description of this circuit. - -- (c) Using transform-domain techniques, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*2*t u*(*t* +1). -- (d) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0− capacitor voltage (first op-amp output voltage) is 3 volts. -- **4.4-15** We desire the op-amp circuit of Fig. P4.4-15 to behave as *y*˙(*t*)−1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*). - - (a) Determine resistors *R*1, *R*2, and *R*3 so that the circuit's input–output behavior follows - -the desired differential equation of *y*˙(*t*) − 1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*). - -- (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltage (first op-amp output voltage) is 2 volts. -- (c) Using transform-domain techniques, determine the impulse response *h*(*t*) of this circuit. -- (d) Determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *u*(*t* −2). -- **4.4-16** We desire the op-amp circuit of Fig. P4.4-16 to behave as \$ \$ *y*(*t*)+ 2 5 \$ *y*(*t*)+ 1 5 *y*(*t*) = \$ \$ *x*(*t*)− \$ *x*(*t*) - - (a) Determine the resistors *R*1, *R*2, and *R*3 to produce the desired behavior. - - (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltages (first two op-amp outputs) are each 1 volt. - -**4.4-17** (a) Using the initial and final value theorems, find the initial and final value of the zero-state response of a system with the transfer function - -$$ -H(s) = \frac{6s^2 + 3s + 10}{2s^2 + 6s + 5} -$$ - -and input *x*(*t*) = *u*(*t*). - -(b) Repeat part (a) for the input *x*(*t*) = *e*−*t u*(*t*). - -(c) Find y(0+) and y( -$$ -\infty -$$ -) if $Y(s) = \frac{s^2 + 5s + 6}{s^2 + 3s + 2}$ - -. - -- (d) Find *y*(0+) and *y*(∞) if *Y*(*s*) = *s*3 +4*s*2 +10*s*+7 *s*2 +2*s* +3 . -- **4.5-1** Consider two LTIC systems. The first has transfer function *H*1(*s*) = 2*s s*+1 , and the second has transfer function *H*2(*s*) = 1 *se*3(*s*−1) . - -**Figure P4.5-2** - -- (a) Determine the overall impulse response *h*s(*t*) if the two systems are connected in series. -- (b) Determine the overall impulse response *h*p(*t*) if the two systems are connected in parallel. -- **4.5-2** Figure P4.5-2a shows two resistive ladder segments. The transfer function of each segment (ratio of output to input voltage) is 1/2. Figure P4.5-2b shows these two segments connected in cascade. - - (a) Is the transfer function (ratio of output to input voltage) of this cascaded network (1/2)(1/2) = 1/4? - - (b) If your answer is affirmative, verify the answer by direct computation of the transfer function. Does this computation confirm the earlier value 1/4? If not, why? - - (c) Repeat the problem with *R*3 = *R*4 = 20 k. Does this result suggest the answer to the problem in part (b)? -- **4.5-3** In communication channels, transmitted signal is propagated simultaneously by several paths of varying lengths. This causes the signal to reach the destination with varying time delays and varying gains. Such a system generally distorts the received signal. For error-free communication, it is necessary to undo this distortion as - -For simplicity, let us assume that a signal is propagated by two paths whose time delays differ by τ seconds. The channel over the intended path has a delay of *T* seconds and unity gain. The signal over the unintended path has a delay of *T* + τ seconds and gain *a*. Such a channel can be modeled, as shown in Fig. P4.5-3. Find the inverse system transfer function to correct the delay distortion and show that the inverse system can be realized by a feedback system. The inverse system should be causal to be realizable. [*Hint:* We want to correct only the distortion caused by the relative delay τ seconds. For distortionless transmission, the signal may be delayed. What is important is to maintain the shape of *x*(*t*). Thus, a received signal of the form *c x*(*t* −*T*) is considered to be distortionless.] - -**4.5-4** Discuss BIBO stability of the feedback systems depicted in Fig. P4.5-4. For the system in - -**Figure P4.5-4** - -Fig. P4.5-4b, consider three cases: **(a)** *K* = 10, **(b)** *K* = 50, and **(c)** *K* = 48. - -**4.6-1** Realize - -$$ -H(s) = \frac{s(s+2)}{(s+1)(s+3)(s+4)} -$$ - -by canonic direct, series, and parallel forms. - -- **4.6-2** Realize the transfer function in Prob. 4.6-1 by using the transposed form of the realizations found in Prob. 4.6-1. -- **4.6-3** Repeat Prob. 4.6-1 for (a) *H*(*s*) = 3*s*(*s* +2) (*s* +1)(*s*2 +2*s*+2) (b) *H*(*s*) = 2*s* −4 (*s* +2)(*s*2 +4) -- **4.6-4** Realize the transfer functions in Prob. 4.6-3 by using the transposed form of the realizations found in Prob. 4.6-3. -- **4.6-5** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{2s+3}{5s(s+2)^2(s+3)} -$$ - -- **4.6-6** Realize the transfer function in Prob. 4.6-5 by using the transposed form of the realizations found in Prob. 4.6-5. -- **4.6-7** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{s(s+1)(s+2)}{(s+5)(s+6)(s+8)} -$$ - -- **4.6-8** Realize the transfer function in Prob. 4.6-7 by using the transposed form of the realizations found in Prob. 4.6-7. -- **4.6-9** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{s^3}{(s+1)^2(s+2)(s+3)} -$$ - -- **4.6-10** Realize the transfer function in Prob. 4.6-9 by using the transposed form of the realizations found in Prob. 4.6-9. -- **4.6-11** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{s^3}{(s+1)(s^2+4s+13)} -$$ - -**4.6-12** Realize the transfer function in Prob. 4.6-11 by using the transposed form of the realizations found in Prob. 4.6-11. - -- **4.6-13** Draw a TDFII block realization of a causal LTIC system with transfer function *H*(*s*) = (*s*−2*j*)(*s*+2*j*) (*s*−*j*)(*s*+*j*)(*s*+2) . Give two reasons why TDFII tends to be a good structure. -- **4.6-14** Consider a causal LTIC system with transfer function *H*(*s*) = (*s*−2*j*)(*s*+2*j*)(*s*−3*j*)(*s*+3*j*) 9(*s*+1)(*s*+2)(*s*+1−*j*)(*s*+1+*j*) . - - (a) Realize *H*(*s*) using a single fourth-order real TDFII structure. Is this block realization unique? Explain. - - (b) Realize *H*(*s*) using a cascade of secondorder real DFII structures. Is this block realization unique? Explain. - - (c) Realize *H*(*s*) using a parallel connection of second-order real DFI structures. Is this block realization unique? Explain. -- **4.6-15** In this problem we show how a pair of complex conjugate poles may be realized by using a cascade of two first-order transfer functions and feedback. Show that the transfer functions of the block diagrams in Figs. P4.6-15a and P4.6-15b are: (a) - -$$ -H_a(s) = \frac{1}{(s+a)^2 + b^2} -$$ - -= -$$ -\frac{1}{s^2 + 2as + (a^2 + b^2)} -$$ - -(b) - -$$ -Hb(s) = \frac{s+a}{(s+a)^2 + b^2} -$$ - -= -$$ -\frac{s+a}{s^2 + 2as + (a^2 + b^2)} -$$ - -Hence, show that the transfer function of the block diagram in Fig. P4.6-15c is (c) - -$$ -H_c(s) = \frac{As + B}{(s+a)^2 + b^2} -$$ - -= -$$ -\frac{As + B}{s^2 + 2as + (a^2 + b^2)} -$$ - -**4.6-16** Show op-amp realizations of the following transfer functions: - -(a) -$$ -\frac{-10}{s+5} -$$ - -\n(b) $\frac{10}{s+5}$ -\n(c) $\frac{s+2}{s+5}$ - -(c) - -*b*2 - -**4.6-17** Show two different op-amp circuit realizations of the transfer function - -$$ -H(s) = \frac{s+2}{s+5} = 1 - \frac{3}{s+5} -$$ - -**4.6-18** Show an op-amp canonic direct realization of the transfer function - -$$ -H(s) = \frac{3s + 7}{s^2 + 4s + 10} -$$ - -**4.6-19** Show an op-amp canonic direct realization of the transfer function - -$$ -H(s) = \frac{s^2 + 5s + 2}{s^2 + 4s + 13} -$$ - -**4.6-20** Consider a system described by a constantcoefficient linear differential equation as *d dt y*(*t*) - -+ 2*y*(*t*) = *x*(*t*) − 3 *d dt x*(*t*). Draw an op-amp realization of this system if resistors and inductors are available but not capacitors. Would using inductors rather than capacitors in this circuit pose any problem? Explain. - -- **4.7-1** Feedback can be used to increase (or decrease) the system bandwidth. Consider the system in Fig. P4.7-1a with transfer function *G*(*s*) = ω*c*/(*s* +ω*c*). - - (a) Show that the 3 dB bandwidth of this system is ω*c* and the dc gain is unity; that is, |*H*(*j*0)| = 1. - - (b) To increase the bandwidth of this system, we use negative feedback with *H*(*s*) = 9, as depicted in Fig. P4.7-1b. Show that the 3 dB bandwidth of this system is 10ω*c*. What is the dc gain? - -- (c) To decrease the bandwidth of this system, we use positive feedback with *H*(*s*) = −0.9, as illustrated in Fig. P4.7-1c. Show that the 3 dB bandwidth of this system is ω*c*/10. What is the dc gain? -- (d) The system gain at dc times its 3 dB bandwidth is the *gain-bandwidth product* of a system. Show that this product is the same for all the three systems in Fig. P4.7-1. This result shows that if we increase the bandwidth, the gain decreases and vice versa. -- **4.8-1** Suppose an engineer builds a controllable, observable LTIC system with transfer function *H*(*s*) = *s*2+4 2*s*2+4*s*+4 . - - (a) By direct calculation, compute the magnitude response at frequencies ω = 0, 1, 2, 3, 5, 10, and ∞. Use these calculations to roughly sketch the magnitude response over 0 ≤ ω ≤ 10. - - (b) To test the system, the engineer connects a signal generator to the system in hopes to measure the magnitude response using a standard oscilloscope. What type of signal should the engineer input into the system to make the measurements? How should the engineer make the measurements? Provide sufficient detail to fully justify your answers. - - (c) Suppose the engineer accidentally constructs the system *H*−1(*s*)= 1 *H*(*s*) = 2*s*2+4*s*+4 *s*2+4 . What impact will this mistake have on his tests? -- **4.8-2** For an LTIC system described by the transfer function - -$$ -H(s) = \frac{s+2}{s^2 + 5s + 4} -$$ - -find the response to the following everlasting sinusoidal inputs: - -- (a) 5 cos(2*t* +30◦) -- (b) 10 sin(2*t* +45◦) -- (c) 10 cos(3*t* +40◦) - -Observe that these are everlasting sinusoids. - -**4.8-3** For an LTIC system described by the transfer function - -$$ -H(s) = \frac{s+3}{(s+2)^2} -$$ - -find the steady-state system response to the following inputs: - -- (a) 10*u*(*t*) -- (b) cos(2*t* +60◦)*u*(*t*) -- (c) sin(3*t* −45◦)*u*(*t*) -- (d) *ej*3*t u*(*t*) -- **4.8-4** For an allpass filter specified by the transfer function - -$$ -H(s) = \frac{-(s-10)}{s+10} -$$ - -find the system response to the following (everlasting) inputs: - -- (a) *ej*ω*t* -- (b) cos(ω*t* +θ ) -- (c) cos *t* -- (d) sin 2*t* -- (e) cos 10*t* -- (f) cos 100*t* - -Comment on the filter response. - -- **4.8-5** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.8-5. The dc response of this system is minus 1, *H*(*j*0) = −1. - - (a) Letting *H*(*s*) = *k*(*s*2 +*b*1*s*+*b*2)/(*s*2 +*a*1*s*+ *a*2), determine the constants *k*, *b*1, *b*2, *a*1, and *a*2. - - (b) What is the output *y*(*t*) of this system in response to the input *x*(*t*) = 4 + cos(*t*/2 + π/3)? - -### **Figure P4.8-5** - -- **4.8-6** Consider a CT system described by (*D*+1)(*D*+ 2){*y*(*t*)} = *x*(*t* − 1). Notice that this differential equation is in terms of *x*(*t* −1), not *x*(*t*)! - - (a) Determine the output *y*(*t*) given input *x*(*t*) = 1. - -- (b) Determine the output *y*(*t*) given input *x*(*t*) = cos(*t*). -- **4.8-7** An LTIC system has transfer function *H*(*s*) = 4*s s*2+2*s*+37 = 4*s* (*s*+1+6*j*)(*s*+1−6*j*). Determine the steady-state output in response to input *x*(*t*) = 1 3 *ej*(6*t*+π/3) *u*(6*t* +π/3). -- **4.9-1** Suppose a real first-order lowpass system *H*(*s*) has unity gain in the passband, one finite pole at *s* = −2, and one finite zero at an unspecified location. - - (a) Determine the location of the system zero so that the filter achieves 40 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response. - - (b) Determine the location of the system zero so that the filter achieves 30 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response. -- **4.9-2** Repeat Prob. 4.9-1 for a highpass rather than a lowpass system. -- **4.9-3** Repeat Prob. 4.9-1 for a second-order system that has a pair of repeated poles and a pair of repeated zeros. -- **4.9-4** Sketch Bode plots for the following transfer functions: - -(a) -$$ -\frac{s(s+100)}{(s+2)(s+20)} -$$ - -(b) -$$ -\frac{(s+10)(s+20)}{s^2(s+100)} -$$ - -(c) -$$ -\frac{(s+10)(s+200)}{(s+20)^2(s+1000)} -$$ - -$$ -4.9-5 \quad \text{Repeat Prob. } 4.9-4 \text{ for} -$$ - -(a) -$$ -\frac{s^2}{(s+1)(s^2+4s+16)} -$$ - -\n(b) -$$ -\frac{s}{(s+1)(s^2+14.14s+100)} -$$ - -\n(c) -$$ -\frac{(s+10)}{s(s^2+14.14s+100)} -$$ - -- **4.9-6** Using the lowest order possible, determine a system function *H*(*s*) with real-valued roots that matches the frequency response in Fig. P4.9-6. Verify your answer with MATLAB. -- **4.9-7** A graduate student recently implemented an analog phase lock loop (PLL) as part of his thesis. His PLL consists of four basic components: a phase/frequency detector, a charge pump, a loop filter, and a voltage-controlled oscillator. This problem considers only the loop filter, which is shown in Fig. P4.9-7a. The loop filter input is the current *x*(*t*), and the output is the voltage *y*(*t*). - - (a) Derive the loop filter's transfer function *H*(*s*). Express *H*(*s*) in standard form. - - (b) Figure P4.9-7b provides four possible frequency response plots, labeled A through D. Each log-log plot is drawn to the same scale, and line slopes are either 20 dB/decade, 0 dB/decade, or −20 dB/decade. Clearly identify which plot(s), if any, could represent the loop filter. - - (c) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for low-frequency inputs? - -**Figure P4.9-6** - -(d) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for high-frequency inputs? - -**Figure P4.9-7** - -- **4.10-1** A causal LTIC system *H*(*s*) = 2(*s*−4*j*)(*s*+4*j*) (*s*+1+2*j*)(*s*+1−2*j*) has input *x*(*t*) = −1 + 2 cos(2*t*) − 3 sin(4*t* + π/3) + 4 cos(10*t*). Below, perform accurate calculations at ω = 0, ±2, ±4, and ±10. - - (a) Using the graphical method of Sec. 4.10-1, accurately sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - - (b) Using the graphical method of Sec. 4.10-1, accurately sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10. - - (c) Approximate the system output *y*(*t*) in response to the input *x*(*t*). -- **4.10-2** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.10-2. The dc response of this system is minus 2, *H*(*j*0) = −2. - - (a) Letting *H*(*s*) = *k s*2+*b*1*s*+*b*2 *s*2+*a*1*s*+*a*2 , determine the constants *k*, *b*1, *b*2, *a*1, and *a*2. - -- (b) Using the graphical method of Sec. 4.10-1, hand-sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB. -- (c) Using the graphical method of Sec. 4.10-1, hand-sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB. -- (d) What is the output *y*(*t*) in response to input *x*(*t*) = −3+cos(3*t*+π/3)−sin(4*t*−π/8)? - -**Figure P4.10-2** - -**4.10-3** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of an LTIC system described by the transfer function - -$$ -H(s) = \frac{s^2 - 2s + 50}{s^2 + 2s + 50} -$$ - -= -$$ -\frac{(s - 1 - j7)(s - 1 + j7)}{(s + 1 - j7)(s + 1 + j7)} -$$ - -What kind of filter is this? - -**4.10-4** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of LTIC systems whose pole-zero plots are shown in Fig. P4.10-4. - -**Figure P4.10-4** - -- **4.10-5** A causal LTIC system *H*(*s*) = (*s*−3*j*)(*s*+3*j*) 3(*s*+2+*j*)(*s*+2−*j*) has input *x*(*t*) = cos(*t*) + sin(3*t* + π/3) + cos(100*t*). - - (a) Using the graphical method of Sec. 4.10-1, sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - - (b) Determine the system output *y*(*t*) in response to the input *x*(*t*). - - (c) Suppose we create a second causal system with transfer function *H*2(*s*) = *H*(−*s*). Sketch this system's pole/zero plot. What is the response *y*2(*t*) of system *H*2(*s*) to input *x*(*t*)? -- **4.10-6** Design a second-order bandpass filter with center frequency ω = 10. The gain should be zero at ω = 0 and at ω = ∞. Select poles at −*a* ± *j*10. Leave your answer in terms of *a*. Explain the influence of *a* on the frequency response. -- **4.10-7** The LTIC system described by *H*(*s*) = (*s*−1)/(*s* +1) has unity magnitude response |*H*(*j*ω)| = 1. Positive Pat claims that the output *y*(*t*) of this system is equal the input *x*(*t*), since the system is allpass. Cynical Cynthia doesn't think so. "This is *signals and systems* class," she complains. "It *has* to be more complicated!" Who is correct, Pat or Cynthia? Justify your answer. -- **4.10-8** Two students, Amy and Jeff, disagree about an analog system function given by *H*1(*s*) = *s*. Sensible Jeff claims the system has a zero at *s* = 0. Rebellious Amy, however, notes that the system function can be rewritten as *H*1(*s*) = 1/*s*−1 and claims that this implies a system pole - -at *s* = ∞. Who is correct? Why? What are the poles and zeros of the system *H*2(*s*) = 1/*s*? - -- **4.10-9** A rational transfer function *H*(*s*) is often used to represent an analog filter. Why must *H*(*s*) be strictly proper for lowpass and bandpass filters? Why must *H*(*s*) be proper for highpass and bandstop filters? -- **4.10-10** For a given filter order *N*, why is the stopband attenuation rate of an all-pole lowpass filter better than filters with finite zeros? -- **4.10-11** Is it possible, with real coefficients ([*k*,*b*1,*b*2,*a*1,*a*2] ∈ *R*), for a system - -$$ -H(s) = k \frac{s^2 + b_1 s + b_2}{s^2 + a_1 s + a_2} -$$ - -to function as a lowpass filter? Explain your answer. - -- **4.10-12** Nick recently built a simple second-order Butterworth lowpass filter for his home stereo. Although the system performs fairly well, Nick is an overachiever and hopes to improve the system performance. Unfortunately, Nick is lazy and doesn't want to design another filter. Thinking "Twice the filtering gives twice the performance," he suggests filtering the audio signal not once but twice with a cascade of two identical filters. His overworked, underpaid signals professor is skeptical and states, "If you are using *identical* filters, it makes no difference whether you filter once or twice!" Who is correct? Why? -- **4.10-13** An LTIC system impulse response is given by *h*(*t*) = *u*(*t*)−*u*(*t* −1). - - (a) Determine the transfer function *H*(*s*). Using *H*(*s*), determine and plot the magnitude response |*H*(*j*ω)|. Which type of filter most accurately describes the behavior of this system: lowpass, highpass, bandpass, or bandstop? - - (b) What are the poles and zeros of *H*(*s*)? Explain your answer. - - (c) Can you determine the impulse response of the inverse system? If so, provide it. If not, suggest a method that could be used to approximate the impulse response of the inverse system. -- **4.10-14** An ideal lowpass filter *H*LP(*s*) has magnitude response that is unity for low frequencies - -and zero for high frequencies. An ideal highpass filter *H*HP(*s*) has an opposite magnitude response: zero for low frequencies and unity for high frequencies. A student suggests a possible lowpass-to-highpass filter transformation: *H*HP(*s*) = 1 − *H*LP(*s*). In general, will this transformation work? Explain your answer. - -- **4.10-15** An LTIC system has a rational transfer function *H*(*s*). When appropriate, assume that all initial conditions are zero. - - (a) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = cos(100π*t*)*u*(*t*)? Explain. - - (b) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = sin(50π*t*)*u*(*t*)? Explain. - - (c) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = cos(100π*t*)? Explain. - - (d) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = sin(50π*t*)? Explain. -- **4.11-1** Find the ROC, if it exists, of the (bilateral) Laplace transform of the following signals: - - (a) *etu*(*t*) (b) *e*−*tu*(*t*) - -(c) -$$ -\frac{1}{1+t^2} -$$ - -(d) -$$ -\frac{1}{1+e^t} -$$ - -(e) $e^{-kt^2}$ - -- **4.11-2** Using the definition and direct integration, find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals. If the Laplace transform does not exist, carefully explain why. - - (a) *x*a(*t*) = *e*(−1−*j*)*t u*(1−*t*) - - (b) *x*b(*t*) = *j* (*t*+1) *u*(−*t* −1) - - (c) *x*c(*t*) = *ej*π/3*u*(2−*t*) +*j*δ(*t* −5) - - (d) *x*d(*t*) = 1+1 = 2 - - (e) *x*e(*t*) = 3*u*(−*t*) +*e*−2*t* [*u*(*t*)−*u*(*t* −10)] - - (f) *x*f(*t*) = *et*−2*u*(1−*t*) +*e*−2*t u*(*t* +1) -- **4.11-3** Determine the bilateral Laplace transform *X*(*s*) of the signal - -$$ -x(t) = \left[e^t u(-t)\right] * \left[t\cos(2t)u(t)\right] -$$ - -**4.11-4** A signal has bilateral Laplace transform *X*(*s*) = (*s*+1) (*s*−2)(*s*−3) but unknown region of convergence. What ROC results in the smallest maximum amplitude of *x*(*t*)? Justify your answer. - -- **4.11-5** Find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals: - - (a) *e*−|*t*| - - (b) *e*−|*t*| cos *t* - -(c) -$$ -e^t u(t) + e^{2t} u(-t) -$$ - -(d) $e^{-tu(t)}$ - -$$ -\begin{array}{cc}\n(e) & e^{-tu(-t)} \\ -(e) & e^{-tu(-t)}\n\end{array} -$$ - -- (f) cosω0*t u*(*t*) +*et u*(−*t*) -- **4.11-6** Find the inverse (bilateral) Laplace transforms of the following functions: - -(a) -$$ -\frac{2s+5}{(s+2)(s+3)} -$$ - -\n -$$ --3 < \sigma < -2 -$$ - -\n(b) -$$ -\frac{2s-5}{(s-2)(s-3)} -$$ - -\n -$$ -2 < \sigma < 3 -$$ - -\n(c) -$$ -\frac{2s+3}{(s+1)(s+2)} -$$ - -\n -$$ -\sigma > -1 -$$ - -\n(d) -$$ -\frac{2s+3}{(s+1)(s+2)} -$$ - -\n -$$ -\sigma < -2 -$$ - -(d) -$$ -\frac{2s+6}{(s+1)(s+2)} \quad \sigma < -2 -$$ - -(e) -$$ -\frac{3s^2 - 2s - 17}{(s+1)(s+3)(s-5)} \quad -1 < \sigma < 5 -$$ - -**4.11-7** Find - -$$ -\mathcal{L}^{-1}\left[\frac{2s^2 - 2s - 6}{(s+1)(s-1)(s+2)}\right] -$$ - -if the ROC is - -- (a) Re *s* > 1 -- (b) Re *s* < −2 -- (c) −1 < Re*s* < 1 -- (d) −2 < Re*s* < −1 -- **4.11-8** For a causal LTIC system having a transfer function *H*(*s*) = 1/(*s*+1), find the output *y*(*t*) if the input *x*(*t*) is given by (a) *e*−|*t*|/2 (b) *et u*(*t*)+*e*2*t u*(−*t*) (c) *e*−*t*/2*u*(*t*) +*e*−*t*/4*u*(−*t*) (d) *e*2*t u*(*t*)+*et u*(−*t*) (e) *e*−*t*/4*u*(*t*) +*e*−*t*/2*u*(−*t*) - - (f) *e*−3*t u*(*t*)+*e*−2*t u*(−*t*) -- **4.11-9** The autocorrelation function *rxx*(*t*) of a signal *x*(*t*) is given by - -$$ -r_{xx}(t) = \int_{-\infty}^{\infty} x(\tau) x(\tau + t) d\tau -$$ - -Derive an expression for *Rxx*(*s*) = *L*(*rxx*(*t*)) in terms of *X*(*s*), where *X*(*s*) = *L*(*x*(*t*)). - -**4.11-10** Determine the inverse Laplace transform of - -$$ -X(s) = \frac{2}{s} + \frac{s}{2} -$$ - -given that the region of convergence is σ < 0. - -**4.11-11** An absolutely integrable signal *x*(*t*) has a pole at *s* = π. It is possible that other poles may be present. Recall that an absolutely integrable signal satisfies - -$$ -\int_{-\infty}^{\infty} |x(t)| dt < \infty -$$ - -- (a) Can *x*(*t*) be left-sided? Explain. -- (b) Can *x*(*t*) be right-sided? Explain. -- (c) Can *x*(*t*) be two-sided? Explain. -- (d) Can *x*(*t*) be of finite duration? Explain. -- **4.11-12** Using ROC σ < 0, determine the inverse Laplace transform of *X*(*s*) = *s d ds e*−2*s s* . [*Hint:* Use Laplace transform properties to avoid tedious calculus.] -- **4.11-13** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal - -$$ -X(s) = \frac{2}{e^s} + \frac{1}{s} \left[ e^s \frac{4}{\frac{s}{3} + 2} \right] -$$ - -where the ROC is −6 < Re{*s*} < 0. - -**4.11-14** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal - -$$ -X(s) = \frac{d^7}{ds^7} \left[ \frac{e^{-4s}}{(s+2)(s+3)} \right] -$$ - -where the ROC is −3 < Re{*s*} < −2. - -**4.11-15** Using the definition, compute the bilateral Laplace transform, including the region of convergence (ROC), of the following complexvalued functions: - -(a) -$$ -x_1(t) = (j + e^{jt})u(t) -$$ - -- (b) *x*2(*t*) = *j* cosh(*t*)*u*(−*t*) -- (c) *x*3(*t*) = *ej*( π 4 ) *u*(−*t* +1)+*j*δ(*t* −5) -- (d) *x*4(*t*) = *j t u*(−*t*) +δ(*t* −π ) - -**4.11-16** A bounded-amplitude signal *x*(*t*) has bilateral Laplace transform *X*(*s*) given by - -$$ -X(s) = \frac{s2^s}{(s-1)(s+1)} -$$ - -- (a) Determine the corresponding region of convergence. -- (b) Determine the time-domain signal *x*(*t*). -- **4.12-1** Express the polynomial *C*20(*x*) in standard form. That is, determine the coefficients % *ak* of *C*20(*x*)= 20 *k*=0 *akx*20−*k*. -- **4.12-2** Consider an LTIC system with - -$$ -H(s) = \frac{1}{s^3 + 4s^2 + 8s + 8} = \frac{1}{(s^2 + 2s + 4)(s + 2)} -$$ - -= -$$ -\frac{1}{(s - 2e^{j2\pi/3})(s - 2e^{-j2\pi/3})(s + 2)}. -$$ - -- (a) Write MATLAB code that accurately plots the system magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. -- (b) Write MATLAB code that accurately plots the system phase response over −10 ≤ ω ≤ 10. -- (c) Determine the max value *y*max of output *y*(*t*) in response to input *x*(*t*) = 2−sin(2*t*+π/3). -- (d) Draw a parallel representation of this system using real DFI structures of order 2 or less. [*Hint:* Use MATLAB to perform a partial fraction expansion of *H*(*s*).] -- **4.12-3** Consider the op-amp circuit of Fig. P4.12-3. Further, let *RC* = 1. - - (a) From Fig. P4.12-3, determine the (simplified, standard form, rational) transfer function *H*(*s*). - - (b) Use MATLAB to accurately plot |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - - (c) Determine the output of this system in response to *x*(*t*) = cos(10*t*)−1. - - (d) The circuit of Fig. P4.12-3 contains two capacitors. Suppose one capacitor must be a 25% tolerance part, while the other must be a 10% tolerance part. If the goal is to preserve the original magnitude response, should you use the 25% tolerance capacitor with the first op-amp or the second op-amp? Justify your answer with appropriate MAT-LAB simulations. - -**Figure P4.12-3** - -- **4.12-4** Design an order-12 Butterworth lowpass filter with a cutoff frequency of ω*c* = 2π5000 by completing the following. - - (a) Locate and plot the filter's poles and zeros in the complex plane. Plot the corresponding magnitude response |*H*LP(*j*ω)| to verify proper design. - - (b) Setting all resistor values to 100,000, determine the capacitor values to implement the filter using a cascade of six second-order Sallen–Key circuit sections. The form of a Sallen–Key stage is shown in Fig. P4.12-4. On a single plot, plot the magnitude response of each section as well as the overall magnitude response. Identify the poles that correspond to each section's magnitude response curve. Are the capacitor values realistic? -- **4.12-5** Rather than a Butterworth filter, repeat Prob. 4.12-4 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each Sallen–Key stage is constrained to have unity gain at dc, an overall gain error of 1/ √ 1+ 2 is acceptable. -- **4.12-6** An analog lowpass filter with cutoff frequency ω*c* can be transformed into a highpass filter - -with cutoff frequency ω*c* by using an *RC*–*CR* transformation rule: each resistor *Ri* is replaced by a capacitor *C i* = 1/*Ri*ω*c* and each capacitor *Ci* is replaced by a resistor *R i* = 1/*Ci*ω*c*. - -Use this rule to design an order-8 Butterworth highpass filter with ω*c* = 2π4000 by completing the following. - -- (a) Design an order-8 Butterworth lowpass filter with ω*c* = 2π4000 by using four second-order Sallen–Key circuit stages, the form of which is shown in Fig. P4.12-4. Give resistor and capacitor values for each stage. Choose the resistors so that the *RC*–*CR* transformation will result in 1 nF capacitors. At this point, are the component values realistic? -- (b) Draw an *RC*–*CR* transformed Sallen–Key circuit stage. Determine the transfer function *H*(*s*) of the transformed stage in terms of the variables *R* 1, *R* 2, *C* 1, and *C* 2. -- (c) Transform the LPF designed in part (a) by using an *RC*–*CR* transformation. Give the resistor and capacitor values for each stage. Are the component values realistic? Using *H*(*s*) derived in part (b), plot the - -magnitude response of each section as well - -as the overall magnitude response. Does the overall response look like a highpass Butterworth filter? - -Plot the HPF system poles and zeros in the complex *s* plane. How do these locations compare with those of the Butterworth LPF? - -- **4.12-7** Repeat Prob. 4.12-6, using ω*c* = 2π1500 and an order-16 filter. That is, eight second-order stages need to be designed. -- **4.12-8** Rather than a Butterworth filter, repeat Prob. 4.12-6 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each transformed Sallen–Key stage is constrained to have unity gain at ω = ∞, an overall gain error of 1/ √ 1+ 2 is acceptable. -- **4.12-9** The MATLAB signal-processing toolbox function butter helps design analog Butterworth filters. Use MATLAB help to learn how butter works. For each of the following cases, design the filter, plot the filter's poles and zeros in the complex *s* plane, and plot the decibel magnitude response 20log10 |*H*(*j*ω)|: - - (a) Design a sixth-order analog lowpass filter with ω*c* = 2π3500. - - (b) Design a sixth-order analog highpass filter with ω*c* = 2π3500. - - (c) Design a sixth-order analog bandpass filter with a passband between 2 and 4 kHz. - - (d) Design a sixth-order analog bandstop filter with a stopband between 2 and 4 kHz. -- **4.12-10** The MATLAB signal-processing toolbox function cheby1 helps design analog Chebyshev - -type I filters. A Chebyshev type I filter has a passband ripple and a smooth stopband. Setting the passband ripple to *Rp* = 3 dB, repeat Prob. 4.12-9 using the cheby1 command. With all other parameters held constant, what is the general effect of reducing *Rp*, the allowable passband ripple? - -- **4.12-11** The MATLAB signal-processing toolbox function cheby2 helps design analog Chebyshev type II filters. A Chebyshev type II filter has a smooth passband and ripple in the stopband. Setting the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the cheby2 command. With all other parameters held constant, what is the general effect of increasing *Rs*, the minimum stopband attenuation? -- **4.12-12** The MATLAB signal-processing toolbox function ellip helps design analog elliptic filters. An elliptic filter has ripple in both the passband and the stopband. Setting the passband ripple to *Rp* = 3 dB and the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the ellip command. -- **4.12-13** Using the definition *CN*(*x*)=cosh(*N* cosh−1(*x*)), prove the recursive relation *CN*(*x*) = 2*xCN*−1(*x*) −*CN*−2(*x*). -- **4.12-14** Prove that the poles of a Chebyshev filter, which are located at *pk* = ω*c* sinh(ξ )sin(φ*k*) + *j*ω*c* cosh(ξ ) cos(φ*k*), lie on an ellipse. [*Hint:* The equation of an ellipse in the *x*–*y* plane is (*x*/*a*) 2+ (*y*/*b*) 2 = 1, where constants *a* and *b* define the major and minor axes of the ellipse.] - -# **[DISCRETE-TIME](#page-11-0) SYSTEM ANALYSIS USING THE** *z***-TRANSFORM** - -The counterpart of the Laplace transform for discrete-time systems is the *z*-transform. The Laplace transform converts integro-differential equations into algebraic equations. In the same way, the *z*-transforms changes difference equations into algebraic equations, thereby simplifying the analysis of discrete-time systems. The *z*-transform method of analysis of discrete-time systems parallels the Laplace transform method of analysis of continuous-time systems, with some minor differences. In fact, we shall see that *the z-transform is the Laplace transform in disguise*. - -The behavior of discrete-time systems is similar to that of continuous-time systems (with some differences). The frequency-domain analysis of discrete-time systems is based on the fact (proved in Sec. 3.8-2) that the response of a linear, time-invariant, discrete-time (LTID) system to an everlasting exponential *zn* is the same exponential (within a multiplicative constant) given by *H*[*z*]*zn*. We then express an input *x*[*n*] as a sum of (everlasting) exponentials of the form *zn*. The system response to *x*[*n*] is then found as a sum of the system's responses to all these exponential components. The tool that allows us to represent an arbitrary input *x*[*n*] as a sum of (everlasting) exponentials of the form *zn* is the *z*-transform. - -## **5.1 THE** *z***[-TRANSFORM](#page-11-0)** - -We define *X*[*z*], the direct *z*-transform of *x*[*n*], as - -$$ -X[z] = \sum_{n = -\infty}^{\infty} x[n]z^{-n} -$$ - (5.1) - -where *z* is a complex variable. The signal *x*[*n*], which is the inverse *z*-transform of *X*[*z*], can be obtained from *X*[*z*] by using the following inverse *z*-transformation: - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ -\n(5.2) - -The symbol 6 indicates an integration in counterclockwise direction around a closed path in the complex plane (see Fig. 5.1). We derive this *z*-transform pair later, in Ch. 9, as an extension of the discrete-time Fourier transform pair. - -CHAPTER - -**5** - -As in the case of the Laplace transform, we need not worry about this integral at this point because inverse *z*-transforms of many signals of engineering interest can be found in a *z*-transform table. The direct and inverse *z*-transforms can be expressed symbolically as - -$$ -X[z] = \mathcal{Z}\{x[n]\} \qquad \text{and} \qquad x[n] = \mathcal{Z}^{-1}\{X[z]\} -$$ - -or simply as - -*x*[*n*] ⇐⇒ *X*[*z*] - -Note that - -*Z*−1 [*Z*{*x*[*n*]}] = *x*[*n*] and *Z*[*Z*−1 {*X*[*z*]}] = *X*[*z*] - -### LINEARITY OF THE *z*-TRANSFORM - -Like the Laplace transform, the *z*-transform is a linear operator. If - -*x*1[*n*] ⇐⇒ *X*1[*z*] and *x*2[*n*] ⇐⇒ *X*2[*z*] - -then - -*a*1*x*1[*n*] +*a*2*x*2[*n*] ⇐⇒ *a*1*X*1[*z*] +*a*2*X*2[*z*] - -The proof is trivial and follows from the definition of the *z*-transform. This result can be extended to finite sums. - -### THE UNILATERAL *z*-TRANSFORM - -For the same reasons discussed in Ch. 4, we find it convenient to consider the unilateral *z*-transform. As seen for the Laplace case, the bilateral transform has some complications because of the non-uniqueness of the inverse transform. In contrast, the unilateral transform has a unique inverse. This fact simplifies the analysis problem considerably, but at a price: the unilateral version can handle only causal signals and systems. Fortunately, most of the practical cases are causal. The more general *bilateral z-transform* is discussed later, in Sec. 5.8. In practice, the term *z-transform* generally means *the unilateral z-transform*. - -In a basic sense, there is no difference between the unilateral and the bilateral *z*-transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at *n* = 0 (causal signals). Hence, the definition of the unilateral transform is the same as that of the bilateral [Eq. (5.1)], except that the limits of the sum are from 0 to ∞: - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - (5.3) - -The expression for the inverse *z*-transform in Eq. (5.2) remains valid for the unilateral case also. - -## THE REGION OF CONVERGENCE (ROC) OF *X*[*z*] - -The sum in Eq. (5.1) [or Eq. (5.3)] defining the direct *z*-transform *X*[*z*] may not converge (exist) for all values of *z*. The values of *z* (the region in the complex plane) for which the sum in Eq. (5.1) converges (or exists) are called the *region of existence,* or more commonly the *region of convergence* (ROC), for *X*[*z*]. This concept will become clear in the following example. - -### **EXAMPLE 5.1 Bilateral** *z***-Transform of a Causal Exponential** - -Find the *z*-transform and the corresponding ROC for the signal γ *nu*[*n*]. - -By definition, - -$$ -X[z] = \sum_{n=0}^{\infty} \gamma^n u[n] z^{-n} -$$ - -Since *u*[*n*] = 1 for all *n* ≥ 0, - -$$ -X[z] = \sum_{n=0}^{\infty} \left(\frac{\gamma}{z}\right)^n = 1 + \left(\frac{\gamma}{z}\right) + \left(\frac{\gamma}{z}\right)^2 + \left(\frac{\gamma}{z}\right)^3 + \dots + \dots -$$ - (5.4) - -It is helpful to remember the geometric progression and its sum [see Sec. B.8-3]: - -$$ -1 + x + x2 + x3 + \dots = \frac{1}{1 - x} \quad \text{if} \quad |x| < 1 -$$ - -Applying this relationship to Eq. (5.4) yields - -$$ -X[z] = \frac{1}{1 - \frac{\gamma}{z}} \qquad \left| \frac{\gamma}{z} \right| < 1 -$$ -\n -$$ -= \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.5} -$$ - -Observe that *X*[*z*] exists only for |*z*| > |γ |. For |*z*| < |γ |, the sum in Eq. (5.4) does not converge; it goes to infinity. Therefore, the ROC of *X*[*z*] is the shaded region outside the circle of radius |γ |, centered at the origin, in the *z*-plane, as depicted in Fig. 5.1b. - -**Figure 5.1** γ *nu*[*n*] and the region of convergence of its *z*-transform. - -Later in Eq. (5.52), we show that the *z*-transform of another signal, −γ *nu*[−(*n* + 1)], is also *z*/(*z* − γ ). However, the ROC in this case is |*z*| < |γ |. Clearly, the inverse *z*-transform of *z*/(*z* − γ ) is not unique. However, if we restrict the inverse transform to be causal, then the inverse transform is unique, namely, γ *nu*[*n*]. - -The ROC is required for evaluating *x*[*n*] from *X*[*z*], according to Eq. (5.2). The integral in Eq. (5.2) is a contour integral, implying integration in a counterclockwise direction along a closed path centered at the origin and satisfying the condition |*z*| > |γ |. Thus, any circular path centered at the origin and with a radius greater than |γ | (Fig. 5.1b) will suffice. We can show that the integral in Eq. (5.2) along any such path (with a radius greater than |γ |) yields the same result, namely, *x*[*n*]. † Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of *z*-transforms (Table 5.1), where *z*-transform pairs are tabulated for a variety of signals. To find the inverse *z*-transform of say, *z*/(*z* − γ ), instead of using the complex integration in Eq. (5.2), we consult the table and find the inverse *z*-transform of *z*/(*z*−γ ) as γ *nu*[*n*]. Because of the uniqueness property of the unilateral *z*-transform, there is only one inverse for each *X*[*z*]. Although the table given here is rather short, it comprises the functions of most practical interest. - -The situation of the *z*-transform regarding the uniqueness of the inverse transform is parallel to that of the Laplace transform. For the bilateral case, the inverse *z*-transform is not unique unless the ROC is specified. For the unilateral case, the inverse transform is unique; the region of convergence need not be specified to determine the inverse *z*-transform. For this reason, we shall ignore the ROC in the unilateral *z*-transform Table 5.1. - -## EXISTENCE OF THE *z*-TRANSFORM - -By definition, - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} = \sum_{n=0}^{\infty} \frac{x[n]}{z^n} -$$ - -The existence of the *z*-transform is guaranteed if - -$$ -|X[z]| \leq \sum_{n=0}^{\infty} \frac{|x[n]|}{|z|^n} < \infty -$$ - -for some |*z*|. Any signal *x*[*n*] that grows no faster than an exponential signal *rn* 0, for some *r*0, satisfies this condition. Thus, if - -$$ -|x[n]| \le r_0^n \qquad \text{for some } r_0 \tag{5.6} -$$ - -† Indeed, the path need not even be circular. It can have any odd shape, as long as it encloses the pole(s) of *X*[*z*] and the path of integration is counterclockwise. - -| No. | x[n] | X[z] | -|-----|----------------------------------------------------|-----------------------------------------------------------------| -| 1 | δ[n−k] | z−k | -| 2 | u[n] | z
z−1 | -| 3 | nu[n] | z
(z−1)2 | -| 4 | n2u[n] | z(z+1)
(z−1)3 | -| 5 | n3u[n] | z(z2 +4z+1)
(z−1)4 | -| 6 | γ nu[n] | z
z−γ | -| 7 | γ n−1u[n−1] | 1
z−γ | -| 8 | nγ nu[n] | γ z
(z−γ )2 | -| 9 | n2γ
nu[n] | γ z(z+γ )
(z−γ )3 | -| 10 | n(n−1)(n−2)···(n−m+1)
γ nu[n]
γ mm! | z
(z−γ )m+1 | -| 11a | n cos
γ
βn u[n] | z(z− γ cos β)
z2 −(2 γ
2
cos β)z+ γ | -| 11b | n sin
γ
βn u[n] | z γ sin β
z2 −(2 γ
2
cos β)z+ γ | -| 12a | n cos(βn+θ
r γ
)u[n] | rz[z cos θ − γ cos(β −θ )]
z2 −(2 γ
2
cos β)z+ γ | -| 12b | n cos(βn+θ
γ = γ ejβ
r γ
)u[n] | (0.5rejθ )z
(0.5re−jθ )z
+
z−γ
z−γ ∗ | -| 12c | n cos(βn+θ
r γ
)u[n] | z(Az+B)
z2 +2az+ γ
2
| -| |
A2 γ
2 +B2 −2AaB

r =
2 −a2
γ | | -| | β = cos−1 −a
γ | | -| | Aa−B
θ = tan−1

2 −a2
A
γ | | - -**TABLE 5.1** Select (Unilateral) *z*-Transform Pairs - -$$ -|X[z]| \le \sum_{n=0}^{\infty} \left(\frac{r_0}{|z|}\right)^n = \frac{1}{1 - \frac{r_0}{|z|}} \qquad |z| > r_0 -$$ - -Therefore, *X*[*z*] exists for |*z*| > *r*0. Almost all practical signals satisfy Eq. (5.6) and are therefore *z*-transformable. Some signal models (e.g., γ *n*2 ) grow faster than the exponential signal *rn* 0 (for any *r*0) and do not satisfy Eq. (5.6) and therefore are not *z*-transformable. Fortunately, such signals are of little practical or theoretical interest. Even such signals over a finite interval are *z*-transformable. - -then - -This geometric sum simplifies [see Sec. B.8-3] to - -$$ -X[z] = \frac{1}{1 - \frac{1}{z}} \qquad \left| \frac{1}{z} \right| < 1 -$$ -\n -$$ -= \frac{z}{z - 1} \qquad |z| > 1 -$$ - -Therefore, - -$$ -u[n] \Longleftrightarrow \frac{z}{z-1} \qquad |z| > 1 -$$ - -**(c)** Recall that cos β*n* = (*ej*β*n* +*e*−*j*β*n*)/2. Moreover, according to Eq. (5.5), - -$$ -e^{\pm j\beta n}u[n] \Longleftrightarrow \frac{z}{z - e^{\pm j\beta}} \qquad |z| > |e^{\pm j\beta}| = 1 -$$ - -Therefore, - -$$ -X[z] = \frac{1}{2} \left[ \frac{z}{z - e^{j\beta}} + \frac{z}{z - e^{-j\beta}} \right] = \frac{z(z - \cos \beta)}{z^2 - 2z \cos \beta + 1} \qquad |z| > 1 -$$ - -**(d)** Here *x*[0] = *x*[1] = *x*[2] = *x*[3] = *x*[4] = 1 and *x*[5] = *x*[6]=···= 0. Therefore, according to Eq. (5.7), - -$$ -X[z] = 1 + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + \frac{1}{z^4} = \frac{z^4 + z^3 + z^2 + z + 1}{z^4} -$$ - for all $z \neq 0$ - -We can also express this result in a more compact form by summing the geometric progression on the right-hand side of the foregoing equation. From the result in Sec. B.8-3 with *r* =1/*z*,*m*= 0, and *n* = 4, we obtain - -$$ -X[z] = \frac{\left(\frac{1}{z}\right)^5 - \left(\frac{1}{z}\right)^0}{\frac{1}{z} - 1} = \frac{z}{z - 1} (1 - z^{-5}) -$$ - -### **DR ILL 5.1 Bilateral** *z***-Transform** - -- **(a)** Find the *z*-transform of a signal shown in Fig. 5.3. -- **(b)** Use pair 12a (Table 5.1) to find the *z*-transform of *x*[*n*] = 20.65( √2)*n* cos[(π/4)*n* − 1.415]*u*[*n*]. - - - -### **[5.1-1 Inverse Transform by Partial Fraction Expansion and Tables](#page-11-0)** - -As in the Laplace transform, we shall avoid the integration in the complex plane required to find the inverse *z*-transform [Eq. (5.2)] by using the (unilateral) transform table (Table 5.1). Many of the transforms *X*[*z*] of practical interest are rational functions (ratio of polynomials in *z*), which can be expressed as a sum of partial fractions, whose inverse transforms can be readily found in a table of transform. The partial fraction method works because for every transformable *x*[*n*] defined for *n* ≥ 0, there is a corresponding unique *X*[*z*] defined for |*z*| > *r*0 (where *r*0 is some constant), and vice versa. - -### **EXAMPLE 5.3 Inverse** *z***-Transform by Partial Fraction Expansion** - -Find the inverse *z*-transforms of - -(a) -$$ -\frac{8z-19}{(z-2)(z-3)} -$$ - -\n(b) -$$ -\frac{z(2z^2-11z+12)}{(z-1)(z-2)^3} -$$ - -\n(c) -$$ -\frac{2z(3z+17)}{(z-1)(z^2-6z+25)} -$$ - -**(a)** Expanding *X*[*z*] into partial fractions yields - -$$ -X[z] = \frac{8z - 19}{(z - 2)(z - 3)} = \frac{3}{z - 2} + \frac{5}{z - 3} -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/067_5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/067_5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM.md deleted file mode 100644 index ddef4342eff33d33e33a833291a6c42ed045f66d..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/067_5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM.md +++ /dev/null @@ -1,304 +0,0 @@ -### 496 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -From Table 5.1, pair 7, we obtain - -$$ -x[n] = [3(2)^{n-1} + 5(3)^{n-1}]u[n-1] -$$ -\n(5.8) - -If we expand rational *X*[*z*] into partial fractions directly, we shall always obtain an answer that is multiplied by *u*[*n* − 1] because of the nature of pair 7 in Table 5.1. This form is rather awkward as well as inconvenient. We prefer the form that contains *u*[*n*] rather than *u*[*n*−1]. A glance at Table 5.1 shows that the *z*-transform of every signal that is multiplied by *u*[*n*] has a factor *z* in the numerator. This observation suggests that we expand *X*[*z*] into *modified partial fractions*, where each term has a factor *z* in the numerator. This goal can be accomplished by expanding *X*[*z*]/*z* into partial fractions and then multiplying both sides by *z*. We shall demonstrate this procedure by reworking part (a). For this case, - -$$ -\frac{X[z]}{z} = \frac{8z - 19}{z(z - 2)(z - 3)} = \frac{(-19/6)}{z} + \frac{(3/2)}{z - 2} + \frac{(5/3)}{z - 3} -$$ - -Multiplying both sides by *z* yields - -$$ -X[z] = -\frac{19}{6} + \frac{3}{2} \left( \frac{z}{z-2} \right) + \frac{5}{3} \left( \frac{z}{z-3} \right) -$$ - -From pairs 1 and 6 in Table 5.1, it follows that - -$$ -x[n] = -\frac{19}{6}\delta[n] + \left[\frac{3}{2}(2)^n + \frac{5}{3}(3)^n\right]u[n] \tag{5.9} -$$ - -The reader can verify that this answer is equivalent to that in Eq. (5.8) by computing *x*[*n*] in both cases for *n* = 0, 1, 2, 3,..., and comparing the results. The form in Eq. (5.9) is more convenient than that in Eq. (5.8). For this reason, we shall always expand *X*[*z*]/*z* rather than *X*[*z*] into partial fractions and then multiply both sides by *z* to obtain modified partial fractions of *X*[*z*], which have a factor *z* in the numerator. - -**(b)** - -$$ -X[z] = \frac{z(2z^2 - 11z + 12)}{(z - 1)(z - 2)^3} -$$ - -and - -$$ -\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{k}{z - 1} + \frac{a_0}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)} -$$ - -where - -$$ -k = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=1} = -3 -$$ - -$$ -a_0 = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=2} = -2 -$$ - -Therefore, - -$$ -\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{-3}{z - 1} - \frac{2}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)} -$$ -(5.10) - -We can determine *a*1 and *a*2 by clearing fractions. Or we may use a shortcut. For example, to determine *a*2, we multiply both sides of Eq. (5.10) by *z* and let *z* → ∞. This yields - -$$ -0 = -3 - 0 + 0 + a_2 \implies a_2 = 3 -$$ - -This result leaves only one unknown, *a*1, which is readily determined by letting *z* take any convenient value, say, *z* = 0, on both sides of Eq. (5.10). This produces - -$$ -\frac{12}{8} = 3 + \frac{1}{4} + \frac{a_1}{4} - \frac{3}{2} -$$ - -which yields *a*1 = −1. Therefore, - -$$ -\frac{X[z]}{z} = \frac{-3}{z-1} - \frac{2}{(z-2)^3} - \frac{1}{(z-2)^2} + \frac{3}{z-2} -$$ - -and - -$$ -X[z] = -3\frac{z}{z-1} - 2\frac{z}{(z-2)^3} - \frac{z}{(z-2)^2} + 3\frac{z}{z-2} -$$ - -Now the use of Table 5.1, pairs 6 and 10, yields - -$$ -x[n] = \left[ -3 - 2\frac{n(n-1)}{8} (2)^n - \frac{n}{2} (2)^n + 3(2)^n \right] u[n] -$$ - -= -$$ -- \left[ 3 + \frac{1}{4} (n^2 + n - 12) 2^n \right] u[n] -$$ - -**(c) Complex Poles.** - -$$ -X[z] = \frac{2z(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2z(3z+17)}{(z-1)(z-3-j4)(z-3+j4)} -$$ - -The poles of *X*[*z*] are 1, 3 + *j*4, and 3 − *j*4. Whenever there are complex-conjugate poles, the problem can be worked out in two ways. In the first method we expand *X*[*z*] into (modified) first-order partial fractions. In the second method, rather than obtain one factor corresponding to each complex-conjugate pole, we obtain quadratic factors corresponding to each pair of complex-conjugate poles. This procedure is explained next. - -MENT-ORDER FACTORS - -\n -$$ -\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2(3z+17)}{(z-1)(z-3-j4)(z-3+j4)} -$$ - -We find the partial fraction of *X*[*z*]/*z* using the Heaviside "cover-up" method: - -$$ -\frac{X[z]}{z} = \frac{2}{z-1} + \frac{1.6e^{-j2.246}}{z-3-j4} + \frac{1.6e^{j2.246}}{z-3+j4} -$$ - -and - -$$ -X[z] = 2\frac{z}{z-1} + (1.6e^{-j2.246})\frac{z}{z-3-j4} + (1.6e^{j2.246})\frac{z}{z-3+j4} -$$ - -The inverse transform of the first term on the right-hand side is 2*u*[*n*]. The inverse transform of the remaining two terms (complex conjugate poles) can be obtained from pair 12b (Table 5.1) by identifying *r*/2 = 1.6, θ = −2.246 rad, γ = 3 + *j*4 = 5*ej*0.927, so that |γ | = 5, β = 0.927. Therefore, - -$$ -x[n] = [2 + 3.2(5)n \cos(0.927n - 2.246)]u[n] -$$ - -### METHOD OF QUADRATIC FACTORS - -$$ -\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{Az+B}{z^2-6z+25} -$$ - -Multiplying both sides by *z* and letting *z* → ∞, we find - -$$ -0 = 2 + A \Longrightarrow A = -2 -$$ - -and - -$$ -\frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{-2z+B}{z^2-6z+25} -$$ - -To find *B*, we let *z* take any convenient value, say, *z* = 0. This step yields - -$$ -\frac{-34}{25} = -2 + \frac{B}{25} \Longrightarrow B = 16 -$$ - -Therefore, - -$$ -\frac{X[z]}{z} = \frac{2}{z-1} + \frac{-2z+16}{z^2-6z+25} -$$ - -and - -$$ -X[z] = \frac{2z}{z-1} + \frac{z(-2z+16)}{z^2 - 6z + 25} -$$ - -We now use pair 12c, where we identify *A* = −2, *B* = 16, |γ | = 5, and *a* = −3. Therefore, - -$$ -r = \sqrt{\frac{100 + 256 - 192}{25 - 9}} = 3.2 -$$ -, $\beta = \cos^{-1}\left(\frac{3}{5}\right) = 0.927$ rad - -and - -$$ -\theta = \tan^{-1}\left(\frac{-10}{-8}\right) = -2.246 \,\text{rad} -$$ - -so that - -$$ -x[n] = [2 + 3.2(5)n \cos (0.927n - 2.246)]u[n] -$$ - -### **DR ILL 5.2 Inverse** *z***-Transform by Partial Fraction Expansion** - -Find the inverse *z*-transform of the following functions: - -(a) -$$ -\frac{z(2z-1)}{(z-1)(z+0.5)} -$$ - -\n(b) -$$ -\frac{1}{(z-1)(z+0.5)} -$$ - -\n(c) -$$ -\frac{9}{(z+2)(z-0.5)^2} -$$ - -\n(d) -$$ -\frac{5z(z-1)}{z^2-1.6z+0.8} -$$ - -\n[*Hint:* $\sqrt{0.8} = 2/\sqrt{5}$ .] -\n**ANSWERS** -\n(a) $\left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]$ -\n(b) $-2\delta[n] + \left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]$ - -- **(c)** 18δ[*n*]−[0.72(−2)*n* +17.28(0.5)*n* −14.4*n*(0.5)*n*]*u*[*n*] -- **(d)** 5 √5 2 √ 2 5 *n* cos(0.464*n*+0.464)*u*[*n*] - -## **5.1-2 Inverse** *z***[-Transform by Power Series Expansion](#page-11-0)** - -By definition, - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - -= $x[0] + \frac{x[1]}{z} + \frac{x[2]}{z^2} + \frac{x[3]}{z^3} + \cdots$ -= $x[0]z^0 + x[1]z^{-1} + x[2]z^{-2} + x[3]z^{-3} + \cdots$ - -This result is a power series in *z*−1. Therefore, if we can expand *X*[*z*] into the power series in *z*−1, the coefficients of this power series can be identified as *x*[0], *x*[1], *x*[2], *x*[3], .... A rational *X*[*z*] can be expanded into a power series of *z*−1 by dividing its numerator by the denominator. Consider, for example, - -$$ -X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = \frac{7z^3 - 2z^2}{z^3 - 1.7z^2 + 0.8z - 0.1} -$$ - -### 500 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -To obtain a series expansion in powers of *z*−1, we divide the numerator by the denominator as follows: - -$$ -z^{3}-1.7z^{2}+0.8z-0.1\overline{)7z^{3}-2z^{2}} -$$ - -$$ -\underline{7z^{3}-11.9z^{2}+5.60z-0.7} -$$ - -$$ -\underline{7z^{3}-11.9z^{2}+5.60z-0.7} -$$ - -$$ -\underline{9.9z^{2}-5.60z+0.7} -$$ - -$$ -\underline{9.9z^{2}-16.83z+7.92-0.99z^{-1}} -$$ - -$$ -\underline{11.23z-7.22+0.99z^{-1}} -$$ - -$$ -\underline{11.23z-19.09+8.98z^{-1}} -$$ - -$$ -\underline{11.87-7.99z^{-1}} -$$ - -Thus, - -$$ -X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = 7 + 9.9z^{-1} + 11.23z^{-2} + 11.87z^{-3} + \cdots -$$ - -Therefore, - -$$ -x[0] = 7, x[1] = 9.9, x[2] = 11.23, x[3] = 11.87, \dots -$$ - -Although this procedure yields *x*[*n*] directly, it does not provide a closed-form solution. For this reason, it is not very useful unless we want to know only the first few terms of the sequence *x*[*n*]. - -### **DR ILL 5.3 Inverse** *z***-Transform by Long Division** - -Using long division to find the power series in *z*−1, show that the inverse *z*-transform of *z*/(*z*−0.5) is (0.5)*nu*[*n*] or (2)−*nu*[*n*]. - -## RELATIONSHIP BETWEEN *h*[*n*] AND *H*[*z*] - -For an LTID system, if *h*[*n*] is its unit impulse response, then from Eq. (3.39), where we defined *H*[*z*], the system transfer function, we write - -$$ -H[z] = \sum_{n=-\infty}^{\infty} h[n]z^{-n} -$$ - (5.11) - -For causal systems, the limits on the sum are from *n* = 0 to ∞. This equation shows that the transfer function *H*[*z*] is the *z*-transform of the impulse response *h*[*n*] of an LTID system; that is, - -$$ -h[n] \Longleftrightarrow H[z] -$$ - -This important result relates the time-domain specification *h*[*n*] of a system to *H*[*z*], the frequency-domain specification of a system. The result is parallel to that for LTIC systems. - -### **DR ILL 5.4 Impulse Response by Inverse** *z***-Transform** - -Redo Drill 3.14 by taking the inverse *z*-transform of *H*[*z*], as given by Eq. (3.41). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/068_5.2 SOME PROPERTIES OF THE z-TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/068_5.2 SOME PROPERTIES OF THE z-TRANSFORM.md deleted file mode 100644 index b7675ff1052525875a6a1fbae2bb962439cbfa94..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/068_5.2 SOME PROPERTIES OF THE z-TRANSFORM.md +++ /dev/null @@ -1,924 +0,0 @@ -## **5.2 SOME [PROPERTIES OF THE](#page-11-0)** *z***-TRANSFORM** - -The *z*-transform properties are useful in the derivation of *z*-transforms of many functions and also in the solution of linear difference equations with constant coefficients. Here we consider a few important properties of the *z*-transform. - -In our discussion, the variable *n* appearing in signals, such as *x*[*n*] and *y*[*n*], may or may not stand for time. However, in most applications of our interest, *n* is proportional to time. For this reason, we shall loosely refer to the variable *n* as time. - -### **[5.2-1 Time-Shifting Properties](#page-11-0)** - -In the following discussion of the shift property, we deal with shifted signals *x*[*n*]*u*[*n*], *x*[*n*−*k*]*u*[*n*− *k*], *x*[*n*−*k*]*u*[*n*], and *x*[*n*+*k*]*u*[*n*]. Unless we physically understand the meaning of such shifts, our understanding of the shift property remains mechanical rather than intuitive or heuristic. For this reason, using a hypothetical signal *x*[*n*], we have illustrated various shifted signals for *k* = 1 in Fig. 5.4. - -## RIGHT SHIFT (DELAY) If - -*x*[*n*]*u*[*n*] ⇐⇒ *X*[*z*] - -then - -$$ -x[n-1]u[n-1] \Longleftrightarrow \frac{1}{z}X[z] \tag{5.12} -$$ - -In general, - -$$ -x[n-m]u[n-m] \Longleftrightarrow \frac{1}{z^m}X[z] -$$ -\n(5.13) - -Moreover, - -$$ -x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1] -$$ -\n(5.14) - -Repeated application of this property yields - -$$ -x[n-2]u[n] \Longleftrightarrow \frac{1}{z} \left[ \frac{1}{z} X[z] + x[-1] \right] + x[-2] = \frac{1}{z^2} X[z] + \frac{1}{z} x[-1] + x[-2] -$$ - -In general, for integer value of *m*, - -$$ -x[n-m]u[n] \Longleftrightarrow z^{-m}X[z] + z^{-m}\sum_{n=1}^{m}x[-n]z^{n} -$$ -\n(5.15) - -A look at Eqs. (5.12) and (5.14) shows that they are identical except for the extra term *x*[−1] in Eq. (5.14). We see from Figs. 5.4c and 5.4d that *x*[*n* − 1]*u*[*n*] is the same as *x*[*n* − 1]*u*[*n* − 1] plus *x*[−1]δ[*n*]. Hence, the difference between their transforms is *x*[−1]. - -### 5.2 Some Properties of the *z*-Transform 503 - -**Proof.** For the integer value of *m*, - -$$ -\mathcal{Z}{x[n-m]u[n-m]} = \sum_{n=0}^{\infty} x[n-m]u[n-m]z^{-n} -$$ - -Recall that *x*[*n* − *m*]*u*[*n* − *m*] = 0 for *n* < *m* so that the limits on the summation on the right-hand side can be taken from *n* = *m* to ∞. Therefore, - -$$ -\mathcal{Z}\{x[n-m]u[n-m]\} = \sum_{n=m}^{\infty} x[n-m]z^{-n} -$$ -$$ -= \sum_{r=0}^{\infty} x[r]z^{-(r+m)} -$$ -$$ -= \frac{1}{z^m} \sum_{r=0}^{\infty} x[r]z^{-r} = \frac{1}{z^m} X[z] -$$ - -To prove Eq. (5.15), we have - -$$ -\mathcal{Z}\{x[n-m]u[n]\} = \sum_{n=0}^{\infty} x[n-m]z^{-n} = \sum_{r=-m}^{\infty} x[r]z^{-(r+m)} -$$ -$$ -= z^{-m} \left[ \sum_{r=-m}^{-1} x[r]z^{-r} + \sum_{r=0}^{\infty} x[r]z^{-r} \right] -$$ -$$ -= z^{-m} \sum_{n=1}^{m} x[-n]z^{n} + z^{-m}X[z] -$$ - -LEFT SHIFT (ADVANCE) If - -$$ -x[n]u[n] \Longleftrightarrow X[z] -$$ - -then - -$$ -x[n+1]u[n] \Longleftrightarrow zX[z] - zx[0] -$$ - -Repeated application of this property yields - -$$ -x[n+2]u[n] \Longleftrightarrow z\{z(X[z] - zx[0]) - x[1]\} = z^2 X[z] - z^2 x[0] - zx[1] -$$ - -and for the integer value of *m*, - -$$ -x[n+m]u[n] \Longleftrightarrow z^m X[z] - z^m \sum_{n=0}^{m-1} x[n]z^{-n} -$$ -\n(5.16) - -### 504 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -**Proof.** By definition, - -$$ -\mathcal{Z}\{x[n+m]u[n]\} = \sum_{n=0}^{\infty} x[n+m]z^{-n} -$$ -$$ -= \sum_{r=m}^{\infty} x[r]z^{-(r-m)} -$$ -$$ -= z^m \sum_{r=m}^{\infty} x[r]z^{-r} -$$ -$$ -= z^m \left[ \sum_{r=0}^{\infty} x[r]z^{-r} - \sum_{r=0}^{m-1} x[r]z^{-r} \right] -$$ -$$ -= z^m X[z] - z^m \sum_{r=0}^{m-1} x[r]z^{-r} -$$ - -### **EXAMPLE 5.4** *z***-Transform Using the Right-Shift Property** - -The signal *x*[*n*] can be expressed as a product of *n* and a gate pulse *u*[*n*] − *u*[*n* − 6]. Therefore, - -$$ -x[n] = n\{u[n] - u[n-6]\} = nu[n] - nu[n-6] -$$ - -We cannot find the *z*-transform of *nu*[*n* − 6] directly by using the right-shift property [Eq. (5.13)]. So we rearrange it in terms of (*n*−6)*u*[*n*−6] as follows: - -$$ -x[n] = nu[n] - (n - 6 + 6)u[n - 6] -$$ - -= $nu[n] - (n - 6)u[n - 6] - 6u[n - 6]$ - -We can now find the *z*-transform of the bracketed term by using the right-shift property [Eq. (5.13)]. Because *u*[*n*] ⇐⇒ *z*/(*z*−1), - -$$ -u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{z-1} = \frac{1}{z^5(z-1)} -$$ - -Also, because *nu*[*n*] ⇐⇒ *z*/(*z*−1)2, - -$$ -(n-6)u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{(z-1)^2} = \frac{1}{z^5(z-1)^2} -$$ - -Therefore, - -$$ -X[z] = \frac{z}{(z-1)^2} - \frac{1}{z^5(z-1)^2} - \frac{6}{z^5(z-1)} = \frac{z^6 - 6z + 5}{z^5(z-1)^2} -$$ - -### **DR ILL 5.5** *z***-Transform Using the Right-Shift Property** - -Using only the fact that *u*[*n*] ⇐⇒ *z*/(*z*−1) and the right-shift property [Eq. (5.13)], find the *z*-transforms of the signals in Figs. 5.2 and 5.3. - -### **ANSWERS** - -See Ex. 5.2d and Drill 5.1a. - -## **5.2-2** *z***[-Domain Scaling Property \(Multiplication by](#page-11-0)** *γ n***)** - -Scaling in the *z*-domain is equivalent to multiplying a time-domain signal by an exponential. That is, if - -$$ -x[n]u[n] \Longleftrightarrow X[z] -$$ - -then - -$$ -\gamma^n x[n]u[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right] \tag{5.17} -$$ - -**Proof.** - -$$ -\mathcal{Z}\{\gamma^{n}x[n]u[n]\} = \sum_{n=0}^{\infty} \gamma^{n}x[n]z^{-n} = \sum_{n=0}^{\infty} x[n]\left(\frac{z}{\gamma}\right)^{-n} = X\left[\frac{z}{\gamma}\right] -$$ - -### **DR ILL 5.6 Using the** *z***-Domain Scaling Property** - -Use Eq. (5.17) to derive pairs 6 and 8 in Table 5.1 from pairs 2 and 3, respectively. - -## **5.2-3** *z***[-Domain Differentiation Property \(Multiplication by](#page-11-0)** *n***)** - -Multiplying a signal by *n* in the time domain produces differentiation in the *z*-domain. That is, if - -$$ -x[n]u[n] \Longleftrightarrow X[z] -$$ - -then - -$$ -nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z] \tag{5.18} -$$ - -**Proof.** - -$$ --z\frac{d}{dz}X[z] = -z\frac{d}{dz}\sum_{n=0}^{\infty}x[n]z^{-n} -$$ -$$ -= -z\sum_{n=0}^{\infty} -nx[n]z^{-n-1} -$$ -$$ -= \sum_{n=0}^{\infty}nx[n]z^{-n} = \mathcal{Z}\{nx[n]u[n]\} -$$ - -### **DR ILL 5.7 Using the** *z***-Domain Differentiation Property** - -Use Eq. (5.18) to derive pairs 3 and 4 in Table 5.1 from pair 2. Similarly, derive pairs 8 and 9 from pair 6. - -### **[5.2-4 Time-Reversal Property](#page-11-0)** - -If - -$$ -x[n] \Longleftrightarrow X[z] -$$ - -then† - -$$ -x[-n] \Longleftrightarrow X[1/z] -$$ - -**Proof.** - -$$ -\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[-n]z^{-n} -$$ - -$$ -x^*[-n] \Longleftrightarrow X^*[1/z^*] -$$ - -† For complex signal *x*[*n*], the time-reversal property is modified as follows: - -Changing the sign of the dummy variable *n* yields - -$$ -\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[n]z^n -$$ -$$ -= \sum_{n=-\infty}^{\infty} x[n](1/z)^{-n} -$$ -$$ -= X[1/z] -$$ - -The region of convergence is also inverted; that is, if the ROC of *x*[*n*] is |*z*| > |γ |, then the ROC of *x*[−*n*] is |*z*| < 1/|γ |. - -### **DR ILL 5.8 Using the Time-Reversal Property** - -Use the time-reversal property and pair 2 in Table 5.1 to show that *u*[−*n*] ⇐⇒ −1/(*z*−1) with the ROC |*z*| < 1. - -### **[5.2-5 Convolution Property](#page-11-0)** - -The time-convolution property states that if‡ - -$$ -x_1[n] \Longleftrightarrow X_1[z] -$$ - and $x_2[n] \Longleftrightarrow X_2[z]$ , - -then (*time convolution*) - -$$ -x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z] \tag{5.19} -$$ - -**Proof.** This property applies to causal as well as noncausal sequences. We shall prove it for the more general case of noncausal sequences, where the convolution sum ranges from −∞ to ∞. - -We have - -$$ -\mathcal{Z}\{x_1[n] * x_2[n]\} = \mathcal{Z}\left[\sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m]\right] -$$ -$$ -= \sum_{n=-\infty}^{\infty} z^{-n} \sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m] -$$ - -‡ There is also the frequency-convolution property, which states that - -$$ -x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi j} \oint X_1[u]X_2\left[\frac{z}{u}\right]u^{-1} du -$$ - -### 508 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -Interchanging the order of summation, we have - -$$ -\mathcal{Z}[x_1[n]*x_2[n]] = \sum_{m=-\infty}^{\infty} x_1[m] \sum_{n=-\infty}^{\infty} x_2[n-m]z^{-n} -$$ -$$ -= \sum_{m=-\infty}^{\infty} x_1[m] \sum_{r=-\infty}^{\infty} x_2[r]z^{-(r+m)} -$$ -$$ -= \sum_{m=-\infty}^{\infty} x_1[m]z^{-m} \sum_{r=-\infty}^{\infty} x_2[r]z^{-r} -$$ -$$ -= X_1[z]X_2[z] -$$ - -### LTID SYSTEM RESPONSE - -It is interesting to apply the time-convolution property to the LTID input–output equation *y*[*n*] = *x*[*n*] ∗ *h*[*n*]. Since *h*[*n*] ⇐⇒ *H*[*z*], it follows from Eq. (5.19) that - -$$ -Y[z] = X[z]H[z] \tag{5.20} -$$ - -### **DR ILL 5.9 Using the Convolution Property** - -Use the time-convolution property and appropriate pairs in Table 5.1 to show that *u*[*n*] ∗ *u*[*n*−1] = *nu*[*n*]. - -### INITIAL AND FINAL VALUES - -For a causal *x*[*n*], the initial value theorem states that - -$$ -x[0] = \lim_{z \to \infty} X[z] -$$ - -This result follows immediately from Eq. (5.7). - -If (*z*−1)*X*[*z*] has no poles outside the unit circle, then the final value theorem states that - -$$ -\lim_{N \to \infty} x[N] = \lim_{z \to 1} (z - 1)X[z] -$$ - -This can be shown from the fact that - -$$ -x[n] - x[n-1] \Longleftrightarrow \left\{1 - \frac{1}{z}\right\} X[z] = \frac{(z-1)X[z]}{z} -$$ - -and - -$$ -\frac{(z-1)X[z]}{z} = \sum_{n=-\infty}^{\infty} \{x[n] - x[n-1]\}z^{-n} -$$ - -$$ -\lim_{z \to 1} \frac{(z-1)X[z]}{z} = \lim_{z \to 1} (z-1)X[z] = \lim_{z \to 1} \lim_{N \to \infty} \sum_{n=-\infty}^{N} \{x[n] - x[n-1]\}z^{-n} = \lim_{N \to \infty} x[N] -$$ - -All these properties of the *z*-transform are listed in Table 5.2. - -| Operation | x[n] | X[z] | -|-----------------------|--------------------|-------------------------------------------------------------| -| Addition | x1[n] +x2[n] | X1[z] +X2[z] | -| Scalar multiplication | ax[n] | aX[z] | -| Right shifting | x[n−m]u[n−m] | 1
zm X[z] | -| | x[n−m]u[n] | "m
1
1
n
zm X[z] +
x[−n]z
zm
n=1 | -| | x[n−1]u[n] | 1
X[z] +x[−1]
z | -| | x[n−2]u[n] | 1
1
z2 X[z] +
x[−1] +x[−2]
z | -| | x[n−3]u[n] | 1
1
1
z3 X[z] +
z2 x[−1] +
x[−2] +x[−3]
z | -| Left shifting | x[n+m]u[n] | m
"−1
−n
zmX[z]
−zm
x[n]z | -| | x[n+1]u[n] | n=0
zX[z] −zx[0] | -| | x[n+2]u[n] | z2X[z]
−z2x[0]
−zx[1] | -| | x[n+3]u[n] | z3X[z]
−z3x[0]
−z2x[1]
−zx[2] | -| Multiplication by γ n | γ nx[n]u[n] | z
!
X
γ | -| Multiplication by n | nx[n]u[n] | d
−z
X[z]
dz | -| Time reversal | x[−n] | X[1/z] | -| Time convolution | x1[n] ∗ x2[n] | X1[z]X2[z] | -| Initial value | x[0] | lim
X[z]
z→∞ | -| Final value | lim
x[N]
N→∞ | lim
(z−1)X[z]
Poles of (z −1)X[z]
z→1 | -| | | inside the unit circle | - -**TABLE 5.2** *z*-Transform Properties - -and - -## **5.3** *z***-TRANSFORM [SOLUTION OF](#page-11-0) LINEAR DIFFERENCE EQUATIONS** - -The time-shifting (left-shift or right-shift) property has set the stage for solving linear difference equations with constant coefficients. As in the case of the Laplace transform with differential equations, the *z*-transform converts difference equations into algebraic equations that are readily solved to find the solution in the *z* domain. Taking the inverse *z*-transform of the *z*-domain solution yields the desired time-domain solution. The following examples demonstrate the procedure. - -### **EXAMPLE 5.5** *z***-Transform Solution of a Linear Difference Equation** - -Solve - -$$ -y[n+2] - 5y[n+1] + 6y[n] = 3x[n+1] + 5x[n] -$$ - -if the initial conditions are *y*[−1] = 11/6, *y*[−2] = 37/36, and the input *x*[*n*] = (2)−*nu*[*n*]. - -As we shall see, difference equations can be solved by using the right-shift or the leftshift property. Because the difference equation here is in advance form, the use of the left-shift property in Eq. (5.16) may seem appropriate for its solution. Unfortunately, this left-shift property requires a knowledge of auxiliary conditions *y*[0], *y*[1], ... , *y*[*N* − 1] rather than of the initial conditions *y*[−1], *y*[−2], ... , *y*[−*n*], which are generally given. This difficulty can be overcome by expressing the difference equation in delay form (obtained by replacing *n* with *n*−2) and then using the right-shift property.† The resulting delay-form difference equation is - -$$ -y[n] - 5y[n-1] + 6y[n-2] = 3x[n-1] + 5x[n-2] -$$ -\n(5.21) - -We now use the right-shift property to take the *z*-transform of this equation. But before proceeding, we must be clear about the meaning of a term like *y*[*n* − 1] here. Does it mean *y*[*n* − 1]*u*[*n* − 1] or *y*[*n* − 1]*u*[*n*]? In any equation, we must have some time reference *n* = 0, and every term is referenced from this instant. Hence, *y*[*n*−*k*] means *y*[*n*−*k*]*u*[*n*]. Remember also that although we are considering the situation for *n* ≥ 0, *y*[*n*] is present even before *n* = 0 (in the form of initial conditions). Now - -$$ -y[n]u[n] \Longleftrightarrow Y[z] -$$ - -\n -$$ -y[n-1]u[n] \Longleftrightarrow \frac{1}{z}Y[z] + y[-1] = \frac{1}{z}Y[z] + \frac{11}{6} -$$ - -\n -$$ -y[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}Y[z] + \frac{1}{z}y[-1] + y[-2] = \frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36} -$$ - -Noting that for causal input *x*[*n*], - -$$ -x[-1] = x[-2] = \cdot \cdot \cdot = x[-n] = 0 -$$ - -† Another approach is to find *y*[0], *y*[1], *y*[2], ... , *y*[*N* − 1] from *y*[−1], *y*[−2], ... , *y*[−*n*] iteratively, as in Sec. 3.5-1, and then apply the left-shift property to the advance-form difference equation. - -We obtain - -$$ -x[n] = (2)^{-n}u[n] = (2^{-1})^{n}u[n] = (0.5)^{n}u[n] \Longleftrightarrow \frac{z}{z - 0.5} -$$ - -$$ -x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1] = \frac{1}{z}\frac{z}{z-0.5} + 0 = \frac{1}{z-0.5} -$$ -$$ -x[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}X[z] + \frac{1}{z}x[-1] + x[-2] = \frac{1}{z^2}X[z] + 0 + 0 = \frac{1}{z(z-0.5)} -$$ - -In general, - -$$ -x[n - r]u[n] \Longleftrightarrow \frac{1}{z^r}X[z] -$$ - -Taking the *z*-transform of Eq. (5.21) and substituting the foregoing results, we obtain - -$$ -Y[z] - 5\left[\frac{1}{z}Y[z] + \frac{11}{6}\right] + 6\left[\frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36}\right] = \frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)} -$$ - -or - -1− 5 *z* + 6 *z*2 *Y*[*z*] − 3− 11 *z* = 3 *z*−0.5 + 5 *z*(*z*−0.5) (5.22) - -from which we obtain - -$$ -(z2 - 5z + 6)Y[z] = \frac{z(3z2 - 9.5z + 10.5)}{(z - 0.5)} -$$ - -so that - -$$ -Y[z] = \frac{z(3z^2 - 9.5z + 10.5)}{(z - 0.5)(z^2 - 5z + 6)} -$$ - -and - -$$ -\frac{Y[z]}{z} = \frac{3z^2 - 9.5z + 10.5}{(z - 0.5)(z - 2)(z - 3)} = \frac{(26/15)}{z - 0.5} - \frac{(7/3)}{z - 2} + \frac{(18/5)}{z - 3} -$$ - -Therefore, - -$$ -Y[z] = \frac{26}{15} \left(\frac{z}{z - 0.5}\right) - \frac{7}{3} \left(\frac{z}{z - 2}\right) + \frac{18}{5} \left(\frac{z}{z - 3}\right) -$$ -$$ -y[n] = \left[\frac{26}{15} (0.5)^n - \frac{7}{3} (2)^n + \frac{18}{5} (3)^n\right] u[n] \tag{5.23} -$$ - -and - -This example demonstrates the ease with which linear difference equations with constant coefficients can be solved by the *z*-transform. This method is general: it can be used to solve a single difference equation or a set of simultaneous difference equations of any order as long as the equations are linear with constant coefficients. - -### **Comment.** - -Sometimes, instead of initial conditions *y*[−1], *y*[−2], ... , *y*[−*n*], auxiliary conditions *y*[0], *y*[1], ... , *y*[*N* − 1] are given to solve a difference equation. In this case, the equation can be solved by expressing it in the advance form and then using the left-shift property (see Drill 5.11). - -### **DR ILL 5.10** *z***-Transform Solution of a Linear Difference Equation** - -Solve the following equation if the initial conditions *y*[−1] = 2, *y*[−2] = 0, and the input *x*[*n*] = *u*[*n*]: - -$$ -y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n] -$$ - -### **ANSWER** - -*y*[*n*] = 12−15 1 2 *n* + 14 3 1 3 *n u*[*n*] - -## **DR ILL 5.11 Difference Equation Solution Using** *y*[**0**]**,** *y*[**1**]**,** ... **,** *y*[*N* −**1**] - -Solve the following equation if the auxiliary conditions are *y*[0] = 1, *y*[1] = 2, and the input *x*[*n*] = *u*[*n*]: - -*y*[*n*] +3*y*[*n*−1] +2*y*[*n*−2] = *x*[*n*−1] +3*x*[*n*−2] - -### **ANSWER** - -*y*[*n*] = 2 3 +2(−1)*n* − 5 3 (−2)*n u*[*n*] - -### ZERO-INPUT AND ZERO-STATE COMPONENTS - -In Ex. 5.5 we found the total solution of the difference equation. It is relatively easy to separate the solution into zero-input and zero-state components. All we have to do is to separate the response into terms arising from the input and terms arising from initial conditions (IC). We can separate the response in Eq. (5.22) as follows: - -$$ -\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] - \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} = \underbrace{\frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)}}_{\text{input terms}} -$$ - -Therefore, - -$$ -\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] = \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} + \underbrace{\frac{(3z + 5)}{z(z - 0.5)}}_{\text{input terms}} -$$ - -Multiplying both sides by *z*2 yields - -$$ -(z2 - 5z + 6)Y[z] = \underbrace{z(3z - 11)}_{\text{IC terms}} + \underbrace{\frac{z(3z + 5)}{z - 0.5}}_{\text{input terms}} -$$ - -and - -$$ -Y[z] = \underbrace{\frac{z(3z-11)}{z^2 - 5z + 6}}_{\text{zero-input response}} + \underbrace{\frac{z(3z+5)}{(z-0.5)(z^2 - 5z + 6)}}_{\text{zero-state response}} -$$ - -We expand both terms on the right-hand side into modified partial fractions to yield - -$$ -Y[z] = \underbrace{\left[5\left(\frac{z}{z-2}\right) - 2\left(\frac{z}{z-3}\right)\right]}_{\text{zero-input response}} + \underbrace{\left[\frac{26}{15}\left(\frac{z}{z-0.5}\right) - \frac{22}{3}\left(\frac{z}{z-2}\right) + \frac{28}{5}\left(\frac{z}{z-3}\right)\right]}_{\text{zero-state response}} -$$ - -and - -$$ -y[n] = \underbrace{(5(2)^n - 2(3)^n) u[n]}_{\text{zero-input response}} + \underbrace{\left(\frac{26}{15}(0.5)^n - \frac{22}{3}(2)^n + \frac{28}{5}(3)^n\right) u[n]}_{\text{zero-state response}} -$$ - -= -$$ -\left[-\frac{7}{3}(2)^n + \frac{18}{5}(3)^n + \frac{26}{15}(0.5)^n\right] u[n] -$$ - -which agrees with the result in Eq. (5.23). - -### **DR ILL 5.12 Separating Zero-Input and Zero-State Responses** - -Solve - -$$ -y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n] -$$ - -if the initial conditions are *y*[−1] = 2, *y*[−2] = 0, and the input *x*[*n*] = *u*[*n*]. Separate the response into zero-input and zero-state responses. - -### **ANSWER** - -$$ -y[n] = \underbrace{\left(3\left(\frac{1}{2}\right)^n - \frac{4}{3}\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-input response}} + \underbrace{\left(12 - 18\left(\frac{1}{2}\right)^n + 6\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-state response}} -$$ -\n -$$ -= \left[12 - 15\left(\frac{1}{2}\right)^n + \frac{14}{3}\left(\frac{1}{3}\right)^n\right]u[n] -$$ - -### **[5.3-1 Zero-State Response of LTID Systems: The Transfer Function](#page-11-0)** - -Consider an *N*th-order LTID system specified by the difference equation - -$$ -Q[E]y[n] = P[E]x[n] -$$ - -or - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] -$$ - -= (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] - -or - -$$ -y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n] -$$ - -= $b_0x[n+N] + \cdots + b_{N-1}x[n+1] + b_Nx[n]$ (5.24) - -We now derive the general expression for the zero-state response: that is, the system response to input *x*[*n*] when all the initial conditions *y*[−1] = *y*[−2]=···= *y*[−*N*] = 0 (zero state). The input *x*[*n*] is assumed to be causal so that *x*[−1] = *x*[−2]=···= *x*[−*N*] = 0. - -Equation (5.24) can be expressed in delay form as - -$$ -y[n] + a_1y[n-1] + \dots + a_Ny[n-N] = b_0x[n] + b_1x[n-1] + \dots + b_Nx[n-N] -$$ - (5.25) - -Because *y*[−*r*] = *x*[−*r*] = 0 for *r* = 1, 2,...,*N*, - -$$ -y[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}Y[z] -$$ - -$$ -x[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}X[z] \qquad m=1,2,\ldots,N -$$ - -Now the *z*-transform of Eq. (5.25) is given by - -$$ -\left(1+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots+\frac{a_N}{z^N}\right)Y[z] = \left(b_0+\frac{b_1}{z}+\frac{b_2}{z^2}+\cdots+\frac{b_N}{z^N}\right)X[z] -$$ - -Multiplication of both sides by *zN* yields - -$$ -(zN + a1zN-1 + \dots + aN-1z + aN)Y[z] -$$ - -= (b0zN + b1zN-1 + \dots + bN-1z + bN)X[z] - -Therefore, - -$$ -Y[z] = \left(\frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}\right) X[z] -$$ - -= $\frac{P[z]}{Q[z]} X[z]$ - -We have shown in Eq. (5.20) that *Y*[*z*] = *X*[*z*]*H*[*z*]. Hence, it follows that - -$$ -H[z] = \frac{P[z]}{Q[z]} = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N} -$$ -(5.26) - -As in the case of LTIC systems, this result leads to an alternative definition of the LTID system transfer function as the ratio of *Y*[*z*] to *X*[*z*] (assuming all initial conditions zero). - -$$ -H[z] \equiv \frac{Y[z]}{X[z]} = \frac{\mathcal{Z}[zero-state response]}{\mathcal{Z}[input]} -$$ - -### ALTERNATE INTERPRETATION OF THE *z*-TRANSFORM - -So far we have treated the *z*-transform as a machine that converts linear difference equations into algebraic equations. There is no physical understanding of how this is accomplished or what it means. We now discuss more intuitive interpretation and meaning of the *z*-transform. - -In Ch. 3, Eq. (3.38), we showed that the LTID system response to an everlasting exponential *zn* is *H*[*z*]*zn*. If we could express every discrete-time signal as a linear combination of everlasting exponentials of the form *zn*, we could readily obtain the system response to any input. For example, if - -$$ -x[n] = \sum_{k=1}^{K} X[z_k] z_k^n -$$ -\n(5.27) - -the response of an LTID system to this input is given by - -$$ -y[n] = \sum_{k=1}^{K} X[z_k]H[z_k]z_k^n -$$ - -Unfortunately, a very small class of signals can be expressed in the form of Eq. (5.27). However, we can express almost all signals of practical utility as a sum of everlasting exponentials over a continuum of values of *z*. This is precisely what the *z*-transform in Eq. (5.2) does. - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ -\n(5.28) - -Invoking the linearity property of the *z*-transform, we can find the system response *y*[*n*] to input *x*[*n*] in Eq. (5.28) as† - -$$ -y[n] = \frac{1}{2\pi j} \oint X[z]H[z]z^{n-1} dz = \mathcal{Z}^{-1}{X[z]H[z]} -$$ - -Clearly, - -*Y*[*z*] = *X*[*z*]*H*[*z*] - -This viewpoint of finding the response of LTID system is illustrated in Fig. 5.6a. Just as in continuous-time systems, we can model discrete-time systems in the transformed manner by - -† In computing *y*[*n*], the contour along which the integration is performed is modified to consider the ROC of *X*[*z*] as well as *H*[*z*]. We ignore this consideration in this intuitive discussion. - -**Figure 5.6** The transformed representation of an LTID system. - -representing all signals by their *z*-transforms and all system components (or elements) by their transfer functions, as shown in Fig. 5.6b. - -The result *Y*[*z*] = *H*[*z*]*X*[*z*] greatly facilitates derivation of the system response to a given input. We shall demonstrate this assertion by an example. - -### **EXAMPLE 5.6 Transfer Function to Find the Zero-State Response** - -Find the response *y*[*n*] of an LTID system described by the difference equation - -$$ -y[n+2] + y[n+1] + 0.16y[n] = x[n+1] + 0.32x[n] -$$ - -or - -$$ -(E2 + E + 0.16)y[n] = (E + 0.32)x[n] -$$ - -for the input *x*[*n*] = (−2)−*nu*[*n*] and with all the initial conditions zero (system in the zero state). - -From the difference equation, we find - -$$ -H[z] = \frac{P[z]}{Q[z]} = \frac{z + 0.32}{z^2 + z + 0.16} -$$ - -For the input *x*[*n*] = (−2)−*nu*[*n*]=[(−2)−1] *nu*(*n*) = (−0.5)*nu*[*n*], - -$$ -X[z] = \frac{z}{z+0.5} -$$ - -and - -$$ -Y[z] = X[z]H[z] = \frac{z(z+0.32)}{(z^2 + z + 0.16)(z+0.5)} -$$ - -Therefore, - -$$ -\frac{Y[z]}{z} = \frac{(z+0.32)}{(z^2+z+0.16)(z+0.5)} = \frac{(z+0.32)}{(z+0.2)(z+0.8)(z+0.5)} -$$ -$$ -= \frac{2/3}{z+0.2} - \frac{8/3}{z+0.8} + \frac{2}{z+0.5} -$$ - -so that - -$$ -Y[z] = \frac{2}{3} \left( \frac{z}{z+0.2} \right) - \frac{8}{3} \left( \frac{z}{z+0.8} \right) + 2 \left( \frac{z}{z+0.5} \right) -$$ - -and - -$$ -y[n] = \left[\frac{2}{3}(-0.2)^n - \frac{8}{3}(-0.8)^n + 2(-0.5)^n\right]u[n] -$$ - -### **EXAMPLE 5.7 Transfer Function of a Unit Delay** - -Show that the transfer function of a unit delay is 1/*z*. - -If the input to the unit delay is *x*[*n*]*u*[*n*], then its output (Fig. 5.7) is given by - -*y*[*n*] = *x*[*n*−1]*u*[*n*−1] - -The *z*-transform of this equation yields [see Eq. (5.12)] - -$$ -Y[z] = \frac{1}{z}X[z] = H[z]X[z] -$$ - -It follows that the transfer function of the unit delay is - -$$ -H[z] = \frac{1}{z} -$$ - -*x*[*n*]*u*[*n*] *X*[*z*] *x*[*n* - 1]*u*[*n* - 1] *Y*[*z*] *X*[*z*] 1 *z* 1 *z* **Figure 5.7** Ideal unit delay and its transfer function. - -### **DR ILL 5.13 Transfer Function to Find Zero-State Response and Difference Equation** - -A discrete-time system is described by the following transfer function: - -$$ -H[z] = \frac{z - 0.5}{(z + 0.5)(z - 1)} -$$ - -- **(a)** Find the system response to input *x*[*n*] = 3−(*n*+1) *u*[*n*] if all initial conditions are zero. -- **(b)** Write the difference equation relating the output *y*[*n*] to input *x*[*n*] for this system. - -### **ANSWERS** - -- **(a)** *y*[*n*] = 1 3 1 2 −0.8(−0.5)*n* +0.3 1 3 *n u*[*n*] -- **(b)** *y*[*n*+2] −0.5*y*[*n*+1] −0.5*y*[*n*] = *x*[*n*+1] −0.5*x*[*n*] - -### **[5.3-2 Stability](#page-11-0)** - -Equation (5.26) shows that the denominator of *H*[*z*] is *Q*[*z*], which is apparently identical to the characteristic polynomial *Q*[γ ] defined in Ch. 3. Does this mean that the denominator of *H*[*z*] is the characteristic polynomial of the system? This may or may not be the case: if *P*[*z*] and *Q*[*z*] in Eq. (5.26) have any common factors, they cancel out, and the effective denominator of *H*[*z*] is not necessarily equal to *Q*[*z*]. Recall also that the system transfer function *H*[*z*], like *h*[*n*], is defined in terms of measurements at the external terminals. Consequently, *H*[*z*] and *h*[*n*] are both external descriptions of the system. In contrast, the characteristic polynomial *Q*[*z*] is an internal description. Clearly, we can determine only external stability, that is, BIBO stability, from *H*[*z*]. If all the poles of *H*[*z*] are within the unit circle, all the terms in *h*[*n*] are decaying exponentials, and as shown in Sec. 3.9, *h*[*n*] is absolutely summable. Consequently, the system is BIBO-stable. Otherwise the system is BIBO-unstable. - -If *P*[*z*] and *Q*[*z*] do not have common factors, then the denominator of *H*[*z*] is identical to *Q*[*z*]. † The poles of *H*[*z*] are the characteristic roots of the system. We can now determine internal stability. The internal stability criterion in Sec. 3.9-2 can be restated in terms of the poles of *H*[*z*], as follows. - -- 1. An LTID system is asymptotically stable if and only if all the poles of its transfer function *H*[*z*] are within the unit circle. The poles may be repeated or simple. -- 2. An LTID system is unstable if and only if either one or both of the following conditions exist: (i) at least one pole of *H*[*z*] is outside the unit circle; (ii) there are repeated poles of *H*[*z*] on the unit circle. - -† There is no way of determining whether any common factors in *P*[*z*] and *Q*[*z*] were canceled out. This is because in our derivation of *H*[*z*], we generally get the final result after the cancellations have been effected. When we use internal description of the system to derive *Q*[*z*], however, we find pure *Q*[*z*] unaffected by any common factor in *P*[*z*]. - -3. An LTID system is marginally stable if and only if there are no poles of *H*[*z*] outside the unit circle, and there are some simple poles on the unit circle. - -### **DR ILL 5.14 Transfer Function to Determine Stability** - -Show that an *accumulator* whose impulse response is *h*[*n*] = *u*[*n*] is marginally stable but BIBO-unstable. - -### **[5.3-3 Inverse Systems](#page-11-0)** - -If *H*[*z*] is the transfer function of a system *S*, then *Si*, its inverse system, has a transfer function *Hi*[*z*] given by - -$$ -H_i[z] = \frac{1}{H[z]} -$$ - -This follows from the fact the inverse system *Si* undoes the operation of *S*. Hence, if *H*[*z*] is placed in cascade with *Hi*[*z*], the transfer function of the composite system (identity system) is unity. For example, an *accumulator* whose transfer function is *H*[*z*] = *z*/(*z* − 1) and a *backward difference system* whose transfer function is *Hi*[*z*] = (*z*−1)/*z* are inverse of each other. Similarly if - -$$ -H[z] = \frac{z - 0.4}{z - 0.7} -$$ - -its inverse system transfer function is - -$$ -H_i[z] = \frac{z - 0.7}{z - 0.4} -$$ - -as required by the property *H*[*z*]*Hi*[*z*] = 1. Hence, it follows that - -$$ -h[n] * h_i[n] = \delta[n] -$$ - -### **DR ILL 5.15 Inverse Systems** - -Find the impulse responses of an accumulator and a first-order backward difference system. Show that the convolution of the two impulse responses yields δ[*n*]. - -## **5.4 SYSTEM [REALIZATION](#page-11-0)** - -Because of the similarity between LTIC and LTID systems, conventions for block diagrams and rules of interconnection for LTID are identical to those for continuous-time (LTIC) systems. It is not necessary to rederive these relationships. We shall merely restate them to refresh the reader's memory. - -The block diagram representations of the basic operations, such as an adder, a scalar multiplier, unit delay, and pickoff points, is shown in Fig. 3.13. In our development, the unit delay, which is represented by a box marked D in Fig. 3.13, will be represented by its transfer function 1/*z*. All the signals will also be represented in terms of their *z*-transforms. Thus, the input and the output will be labeled *X*[*z*] and *Y*[*z*], respectively. - -When two systems with transfer functions *H*1[*z*] and *H*2[*z*] are connected in cascade (as in Fig. 4.18b), the transfer function of the composite system is *H*1[*z*]*H*2[*z*]. If the same two systems are connected in parallel (as in Fig. 4.18c), the transfer function of the composite - -**Figure 5.8** Realization of an *N*th-order causal LTID system transfer function by using **(a)** DFI, **(b)** canonic direct (DFII), and **(c)** the transpose form of DFII. - -system is *H*1[*z*] + *H*2[*z*]. For a feedback system (as in Fig. 4.18d), the transfer function is *G*[*z*]/(1+*G*[*z*]*H*[*z*]). - -We now consider a systematic method for realization (or simulation) of an arbitrary *N*th-order LTID transfer function. Since realization is basically a synthesis problem, there is no unique way of realizing a system. A given transfer function can be realized in many different ways. We present here the two forms of *direct realization*. Each of these forms can be executed in several other ways, such as cascade and parallel. Furthermore, a system can be realized by the transposed version of any known realization of that system. This artifice doubles the number of system realizations. A transfer function *H*[*z*] can be realized by using time delays along with adders and multipliers. - -We shall consider a realization of a general *N*th-order causal LTID system, whose transfer function is given by - -$$ -H[z] = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N} -$$ -(5.29) - -This equation is identical to the transfer function of a general *N*th-order proper LTIC system given in Eq. (4.36). The only difference is that the variable *z* in the former is replaced by the variable *s* in the latter. Hence, the procedure for realizing an LTID transfer function is identical to that for the LTIC transfer function with the basic element 1/*s* (integrator) replaced by the element 1/*z* (unit delay). The reader is encouraged to follow the steps in Sec. 4.6 and rederive the results for the LTID transfer function in Eq. (5.29). Here we shall merely reproduce the realizations from Sec. 4.6 with integrators (1/*s*) replaced by unit delays (1/*z*). - -The direct form I (DFI) is shown in Fig. 5.8a, the canonic direct form (DFII) is shown in Fig. 5.8b and the transpose of canonic direct is shown in Fig. 5.8c. The DFII and its transpose are canonic because they require *N* delays, which is the minimum number needed to implement the *N*th-order LTID transfer function in Eq. (5.29). In contrast, the form DFI is a noncanonic because it generally requires 2*N* delays. The DFII realization in Fig. 5.8b is also called a *canonic direct* form. - -### **EXAMPLE 5.8 Canonical Realizations of Transfer Functions** - -Find the canonic direct and the transposed canonic direct realizations of the following transfer functions: **(a)** 2 *z*+5 , **(b)** 4*z*+28 *z*+1 , **(c)** *z z*+7 , and **(d)** 4*z*+28 *z*2 +6*z*+5 . - -All four of these transfer functions are special cases of *H*[*z*] in Eq. (5.29). **(a)** - -$$ -H[z] = \frac{2}{z+5} -$$ - -For this case, the transfer function is of the first order (*N* = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are - -$$ -a_1 = 5 -$$ - and $b_0 = 0$ , $b_1 = 2$ - -### 522 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -We use Fig. 5.8 as our model and reduce it to the case of *N* = 1. Figure 5.9a shows the canonic direct (DFII) form, and Fig. 5.9b its transpose. The two realizations are almost the same. The minor difference is that in the DFII form, the gain 2 is provided at the output, and in the transpose, the same gain is provided at the input. - -**Figure 5.9** Realization of transfer function 2/(*z*+5): **(a)** canonic direct form and **(b)** its transpose. - -In a similar way, we realize the remaining transfer functions. **(b)** - -$$ -H[z] = \frac{4z + 28}{z + 1} -$$ - -In this case also, the transfer function is of the first order (*N* = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are - -$$ -a_1 = 1 -$$ - and $b_0 = 4$ , $b_1 = 28$ - -Figure 5.10 illustrates the canonic direct and its transpose for this case.† - -**Figure 5.10** Realization of (4*z* +28)/(*z* +1): **(a)** canonic direct form and **(b)** its transpose. - -$$ -H[z] = \frac{4z + 28}{z+1} = 4 + \frac{24}{z+1} -$$ - -Hence, this transfer function can also be realized as two transfer functions in parallel. - -† Transfer functions with *N* = *M* may also be expressed as a sum of a constant and a strictly proper transfer function. For example, - -$$ -H[z] = \frac{z}{z+7} -$$ - -Here *N* = 1 and *b*0 = 1,*b*1 = 0 and *a*1 = 7. Figure 5.11 shows the direct and the transposed realizations. Observe that the realizations are almost alike. - -**Figure 5.11** Realization of *z*/(*z* +7): **(a)** canonic direct form and **(b)** its transpose. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/069_5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/069_5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS.md deleted file mode 100644 index 64ad22e1b9307b2c1415ca5c59534dd4bf81d861..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/069_5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS.md +++ /dev/null @@ -1,55 +0,0 @@ -**(d)** - -$$ -H[z] = \frac{4z + 28}{z^2 + 6z + 5} -$$ - -This is a second-order system (*N* = 2) with *b*0 = 0, *b*1 = 4, *b*2 = 28, *a*1 = 6, *a*2 = 5. Figure 5.12 shows the canonic direct and transposed canonic direct realizations. - -**Figure 5.12** Realization of (4*z* +28)/(*z*2 +6*z*+5): **(a)** canonic direct form and **(b)** its transpose. - -**(c)** - -### **DR ILL 5.16 Realization of a Second-Order Transfer Function** - -Realize the transfer function - -$$ -H[z] = \frac{2z}{z^2 + 6z + 25} -$$ - -### REALIZATION OF FINITE IMPULSE RESPONSE (FIR) FILTERS - -So far we have been quite general in our development of realization techniques. They can be applied to infinite impulse response (IIR) or FIR filters. For FIR filters, the coefficients *ai* = 0 for all *i* = 0.† Hence, FIR filters can be readily implemented by means of the schemes developed so far by eliminating all branches with *ai* coefficients. The condition *ai* = 0 implies that all the poles of a FIR filter are at *z* = 0. - -### **EXAMPLE 5.9 Realization of an FIR Filter** - -Realize *H*[*z*] = (*z*3 +4*z*2 +5*z*+2)/*z*3 using canonic direct and transposed forms. - -† This statement is true for all *i* = 0 because *a*0 is assumed to be unity. - -For *H*[*z*], *b*0 = 1, *b*1 = 4, *b*2 = 5, and *b*3 = 2. Hence, we obtain the canonic direct realization, shown in Fig. 5.13a. We have shown the horizontal orientation because it is easier to see that this filter is basically a tapped delay line. That is why this structure is also known as a *tapped delay line* or *transversal filter*. Figure 5.13b shows the corresponding transposed implementation. - -## CASCADE AND PARALLEL REALIZATIONS, COMPLEX AND REPEATED POLES - -The considerations and observations for cascade and parallel realizations as well as complex and multiple poles are identical to those discussed for LTIC systems in Sec. 4.6-3. - -## **DR ILL 5.17 Cascade and Parallel Realizations of a Transfer Function** - -Find canonic direct realizations of the following transfer function by using the cascade and parallel forms. The specific cascade decomposition is as follows: - -$$ -H[z] = \frac{z+3}{z^2 + 7z + 10} = \left(\frac{z+3}{z+2}\right)\left(\frac{1}{z+5}\right) -$$ - -### DO ALL REALIZATIONS LEAD TO THE SAME PERFORMANCE? - -For a given transfer function, we have presented here several possible different realizations (DFI, canonic form DFII, and its transpose). There are also cascade and parallel versions, and there are many possible grouping of the factors in the numerator and the denominator of *H*[*z*], leading to different realizations. We can also use various combinations of these forms in implementing different subsections of a system. Moreover, the transpose of each version doubles the number. However, this discussion by no means exhausts all the possibilities. Transforming variables affords limitless potential realizations of the same transfer function. - -Theoretically, all these realizations are equivalent; that is, they lead to the same transfer function. This, however, is true only when we implement them with infinite precision. In practice, finite wordlength restriction causes each realization to behave differently in terms of sensitivity to parameter variation, stability, frequency response distortion error, and so on. These effects are serious for higher-order transfer functions, which require correspondingly higher numbers of delay elements. The finite wordlength errors that plague these implementations are coefficient quantization, overflow errors, and round-off errors. From a practical viewpoint, parallel and cascade forms using low-order filters minimize the effects of finite wordlength. Parallel and certain cascade forms are numerically less sensitive than the canonic direct form to small parameter variations in the system. In the canonic direct form structure with large *N*, a small change in a filter coefficient due to parameter quantization results in a large change in the location of the poles and the zeros of the system. Qualitatively, this difference can be explained by the fact that - -### 526 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -in a direct form (or its transpose), all the coefficients interact with each other, and a change in any coefficient will be magnified through its repeated influence from feedback and feedforward connections. In a parallel realization, in contrast, a change in a coefficient will affect only a localized segment; the case of a cascade realization is similar. For this reason, the most popular technique for minimizing finite wordlength effects is to design filters by using cascade or parallel forms employing low-order filters. In practice, high-order filters are realized by using multiple second-order sections in cascade, because second-order filters not only are easier to design but are less susceptible to coefficient quantization and round-off errors, and their implementations allow easier data word scaling to reduce the potential overflow effects of data word-size growth. A cascaded system using second-order building blocks usually requires fewer multiplications for a given filter frequency response [1]. - -There are several ways to pair the poles and zeros of an *N*th-order *H*[*z*] into a cascade of second-order sections, and several ways to order the resulting sections. Quantizing error will be different for each combination. Although several papers published provide guidelines in predicting and minimizing finite wordlength errors, it is advisable to resort to computer simulation of the filter design. This way, one can vary filter hardware characteristic, such as coefficient wordlengths, accumulator register sizes, sequencing of cascaded sections, and input signal sets. Such an approach is both reliable and economical [1]. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/070_5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/070_5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS.md deleted file mode 100644 index e6b783d7f59961497e20b2a13c2a736640c62087..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/070_5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS.md +++ /dev/null @@ -1,406 +0,0 @@ -## **5.5 FREQUENCY RESPONSE OF [DISCRETE-TIME](#page-11-0) SYSTEMS** - -For (asymptotically or BIBO-stable) continuous-time systems, we showed that the system response to an input *ej*ω*t* is *H*(*j*ω)*ej*ω*t* and that the response to an input cos ω*t* is |*H*(*j*ω)| cos[ω*t* + *H*(*j*ω)]. Similar results hold for discrete-time systems. We now show that for an (asymptotically or BIBO-stable) LTID system, the system response to an input *ejn* is *H*[*ej*]*ejn* and the response to an input cos *n* is |*H*[*ej*]| cos(*n*+ *H*[*ej*]). - -The proof is similar to the one used for continuous-time systems. In Sec. 3.8-2, we showed that an LTID system response to an (everlasting) exponential *zn* is also an (everlasting) exponential *H*[*z*]*zn*. This result is valid only for values of *z* for which *H*[*z*], as defined in Eq. (5.11), exists (converges). As usual, we represent this input–output relationship by a directed arrow notation as - -$$ -z^n \Longrightarrow H[z]z^n \tag{5.30} -$$ - -Setting *z* = *ej* in this relationship yields - -$$ -e^{i\Omega n} \Longrightarrow H[e^{i\Omega}]e^{i\Omega n} \tag{5.31} -$$ - -Noting that cos *n* is the real part of *ejn*, use of Eq. (3.34) yields - -$$ -\cos \Omega n \Longrightarrow \text{Re}\{H[e^{i\Omega}]e^{i\Omega n}\}\tag{5.32} -$$ - -Expressing *H*[*ej*] in the polar form - -$$ -H[e^{i\Omega}] = |H[e^{i\Omega}]|e^{i\angle H[e^{i\Omega}]} -$$ - -Eq. (5.32) can be expressed as - -$$ -\cos \Omega n \Longrightarrow |H[e^{i\Omega}]|\cos(\Omega n + \angle H[e^{i\Omega}]) -$$ - -In other words, the system response *y*[*n*] to a sinusoidal input cos *n* is given by - -$$ -y[n] = |H[e^{j\Omega}]|\cos(\Omega n + \angle H[e^{j\Omega}]) -$$ - -Following the same argument, the system response to a sinusoid cos(*n*+θ ) is - -$$ -y[n] = |H[e^{j\Omega}]\cos(\Omega n + \theta + \angle H[e^{j\Omega}])\tag{5.33} -$$ - -This result is valid only for BIBO-stable or asymptotically stable systems. The frequency response is meaningless for BIBO-unstable systems (which include marginally stable and asymptotically unstable systems). This follows from the fact that the frequency response in Eq. (5.31) is obtained by setting *z*=*ej* in Eq. (5.30). But, as shown in Sec. 3.8-2 [Eqs. (3.38) and (3.39)], the relationship of Eq. (5.30) applies only for values of *z* for which *H*[*z*] exists. For BIBO-unstable systems, the ROC for *H*[*z*] does not include the unit circle where *z* = *ej*. This means, for BIBO-unstable systems, that *H*[*z*] is meaningless when *z* = *ej*. † - -This important result shows that the response of an asymptotically or BIBO-stable LTID system to a discrete-time sinusoidal input of frequency is also a discrete-time sinusoid of the same frequency. *The amplitude of the output sinusoid is* |*H*[*ej*]| *times the input amplitude, and the phase of the output sinusoid is shifted by H*[*ej*] *with respect to the input phase*. Clearly, |*H*[*ej*]| is the amplitude gain, and a plot of |*H*[*ej*]| versus is the amplitude response of the discrete-time system. Similarly, *H*[*ej*] is the phase response of the system, and a plot of *H*[*ej*] versus shows how the system modifies or shifts the phase of the input sinusoid. Note that *H*[*ej*] incorporates the information of both amplitude and phase responses and therefore is called the *frequency responses* of the system. - -### STEADY-STATE RESPONSE TO CAUSAL SINUSOIDAL INPUT - -As in the case of continuous-time systems, we can show that the response of an LTID system to a causal sinusoidal input cos *n u*[*n*] is *y*[*n*] in Eq. (5.33), plus a natural component consisting of the characteristic modes (see Prob. 5.5-9). For a stable system, all the modes decay exponentially, and only the sinusoidal component in Eq. (5.33) persists. For this reason, this component is called the sinusoidal *steady-state* response of the system. Thus, *yss*[*n*], the steady-state response of a system to a causal sinusoidal input cos *n u*[*n*], is - -$$ -y_{ss}[n] = |H[e^{j\Omega}]\cos{(\Omega n + \angle H[e^{j\Omega}])}u[n] -$$ - -### SYSTEM RESPONSE TO SAMPLED CONTINUOUS-TIME SINUSOIDS - -So far we have considered the response of a discrete-time system to a discrete-time sinusoid cos *n* (or exponential *ejn*). In practice, the input may be a sampled continuous-time sinusoid cos ω*t* (or an exponential *ej*ω*t* ). When a sinusoid cos ω*t* is sampled with sampling interval *T*, the resulting - -† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cos0*n* or γ *n* cos0*n* (γ > 1). Hence, the response of such a system to a sinusoid cos*n* will contain not just the sinusoid of frequency but also nondecaying natural modes, rendering the concept of frequency response meaningless. Alternately, we can argue that when *z*=*ej*, a BIBO-unstable system violates the dominance condition |γ*i*| &lt; |*ej*| for all *i*, where γ*i* represents *i*th characteristic root of the system. - -### 528 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -signal is a discrete-time sinusoid cos ω*nT*, obtained by setting *t* = *nT* in cosω*t*. Therefore, all the results developed in this section apply if we substitute ω*T* for : - -$$ -\Omega = \omega T \tag{5.34} -$$ - -### **EXAMPLE 5.10 Sinusoidal Response of a Difference Equation System** - -For a system specified by the equation - -$$ -y[n+1] - 0.8y[n] = x[n+1] -$$ - -find the system response to the inputs - -(a) -$$ -1^n = 1 -$$ - -(b) $\cos\left[\frac{\pi}{6}n - 0.2\right]$ - -**(c)** a sampled sinusoid cos 1500*t* with sampling interval *T* = 0.001 - -The system equation can be expressed as - -$$ -(E - 0.8)y[n] = Ex[n] -$$ - -Therefore, the transfer function of the system is - -$$ -H[z] = \frac{z}{z - 0.8} = \frac{1}{1 - 0.8z^{-1}} -$$ - -The frequency response is - -$$ -H[e^{i\Omega}] = \frac{1}{1 - 0.8e^{-i\Omega}} = \frac{1}{(1 - 0.8\cos\Omega) + i0.8\sin\Omega} -$$ - -Therefore, - -$$ -|H[e^{i\Omega}]| = \frac{1}{\sqrt{(1 - 0.8 \cos \Omega)^2 + (0.8 \sin \Omega)^2}} = \frac{1}{\sqrt{1.64 - 1.6 \cos \Omega}} -$$ -(5.35) - -and - -$$ -\angle H[e^{i\Omega}] = -\tan^{-1}\left[\frac{0.8\sin\Omega}{1 - 0.8\cos\Omega}\right] -$$ -\n(5.36) - -The amplitude response |*H*[*ej*]| can also be obtained by observing that |*H*| 2 = *HH*∗. Since our system is real, we therefore see that - -$$ -|H[e^{i\Omega}]|^{2} = H[e^{i\Omega}]H^{*}[e^{i\Omega}] = H[e^{i\Omega}]H[e^{-i\Omega}] -$$ -\n(5.37) - -Substituting for *H*[*ej*], it follows that - -$$ -|H[e^{i\Omega}]|^2 = \left(\frac{1}{1 - 0.8e^{-i\Omega}}\right)\left(\frac{1}{1 - 0.8e^{i\Omega}}\right) = \frac{1}{1.64 - 1.6\cos\Omega} -$$ - -which matches the result found earlier. - -Figure 5.14 shows plots of amplitude and phase response as functions of . We now compute the amplitude and the phase response for the various inputs. - -**Figure 5.14** Frequency response of the LTID system. - -**(a)** Since 1*n* = (*ej*)*n* with = 0, the amplitude response is *H*[*ej*0]. From Eq. (5.35) we obtain - -$$ -H[e^{i0}] = \frac{1}{\sqrt{1.64 - 1.6 \cos(0)}} = \frac{1}{\sqrt{0.04}} = 5 = 5 \angle 0 -$$ - -Therefore, - -$$ -|H[e^{j0}]| = 5 \quad \text{and} \quad \angle H[e^{j0}] = 0 -$$ - -These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = 0. Therefore, the system response to input 1 is - -$$ -y[n] = 5(1^n) = 5 \qquad \text{for all } n -$$ - -**(b)** For *x*[*n*] = cos[(π/6)*n*−0.2], = π/6. According to Eqs. (5.35) and (5.36), - -$$ -|H[e^{j\pi/6}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos \frac{\pi}{6}}} = 1.983 -$$ - -$$ -\angle H[e^{j\pi/6}] = -\tan^{-1} \left[ \frac{0.8 \sin \frac{\pi}{6}}{1 - 0.8 \cos \frac{\pi}{6}} \right] = -0.916 \text{ rad} -$$ - -These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = π/6. Therefore, - -$$ -y[n] = 1.983 \cos\left(\frac{\pi}{6}n - 0.2 - 0.916\right) = 1.983 \cos\left(\frac{\pi}{6}n - 1.116\right) -$$ - -Figure 5.15 shows the input *x*[*n*] and the corresponding system response. - -**Figure 5.15** Sinusoidal input and the corresponding output of the LTID system. - -**(c)** A sinusoid cos 1500*t* sampled every *T* seconds (*t* = *nT*) results in a discrete-time sinusoid - -$$ -x[n] = \cos 1500nT -$$ - -For *T* = 0.001, the input is - -*x*[*n*] = cos(1.5*n*) - -In this case, = 1.5. According to Eqs. (5.35) and (5.36), - -$$ -|H[e^{j1.5}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos(1.5)}} = 0.809 -$$ - -$$ -\angle H[e^{j1.5}] = -\tan^{-1} \left[ \frac{0.8 \sin(1.5)}{1 - 0.8 \cos(1.5)} \right] = -0.702 \text{ rad} -$$ - -These values also could be read directly from Fig. 5.14 corresponding to = 1.5. Therefore, - -$$ -y[n] = 0.809 \cos(1.5n - 0.702) -$$ - -### FREQUENCY RESPONSE PLOTS USING MATLAB - -MATLAB makes it easy to compute and plot magnitude and phase responses directly using a system's transfer function. As the following code demonstrates, there is no need to derive separate expressions for the magnitude and phase responses. - ->> Omega = linspace(-pi,pi,400); H = @(z) z./(z-0.8); - ->> subplot(1,2,1); plot(Omega,abs(H(exp(1j\*Omega))),'k'); axis tight; - ->> xlabel('\Omega'); ylabel('|H[e^{j \Omega}]|'); - ->> subplot(1,2,2); plot(Omega,angle(H(exp(1j\*Omega))\*180/pi),'k'); axis tight; - -``` ->> xlabel('\Omega'); ylabel('\angle H[e^{j \Omega}] [deg]'); -``` - -The resulting plots, shown in Fig. 5.16, confirm the earlier results of Fig. 5.14. - -### 532 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -**Comment:** Figures 5.14 and 5.16 show amplitude and phase response plots as functions of . These plots as well as Eqs. (5.35) and (5.36) indicate that the frequency response of a discrete-time system is a continuous (rather than discrete) function of frequency . There is no contradiction here. This behavior is merely an indication of the fact that the frequency variable is continuous (takes on all possible values) and therefore the system response exists at every value of . - -### **DR ILL 5.18 Frequency Response of Difference Equation** - -For a system specified by the equation - -$$ -y[n+1] - 0.5y[n] = x[n] -$$ - -find the amplitude and the phase response. Find the system response to sinusoidal input cos[1000*t* −(π/3)] sampled every *T* = 0.5 ms. - -### **ANSWER** - -$$ -|H[e^{i\Omega}]| = \frac{1}{\sqrt{1.25 - \cos \Omega}} -$$ - -\n -$$ -\angle H[e^{i\Omega}] = -\tan^{-1} \left[ \frac{\sin \Omega}{\cos \Omega - 0.5} \right] -$$ - -\n -$$ -y[n] = 1.639 \cos \left( 0.5n - \frac{\pi}{3} - 0.904 \right) = 1.639 \cos (0.5n - 1.951) -$$ - -### **DR ILL 5.19 Frequency Response of an Ideal Delay System** - -Show that for an ideal delay (*H*[*z*] = 1/*z*), the amplitude response |*H*[*ej*]| = 1, and the phase response *H*[*ej*]=−. Thus, a pure time delay does not affect the amplitude gain of sinusoidal input, but it causes a phase shift (delay) of radians in a discrete sinusoid of frequency . Thus, for an ideal delay, the phase shift of the output sinusoid is proportional to the frequency of the input sinusoid (linear phase shift). - -### **[5.5-1 The Periodic Nature of Frequency Response](#page-11-0)** - -In Ex. 5.10 and Fig. 5.14, we saw that the frequency response *H*[*ej*] is a periodic function of . This is not a coincidence. Unlike continuous-time systems, all LTID systems have periodic frequency response. This is seen clearly from the nature of the expression of the frequency response of an LTID system. Because *e*±*j*2π*m* = 1 for all integer values of *m* [see Eq. (B.10)], - -$$ -H[e^{j\Omega}] = H[e^{j(\Omega + 2\pi m)}] \qquad m \text{ integer} -$$ - -Therefore, the frequency response *H*[*ej*] is a periodic function of with a period 2π. This is the mathematical explanation of the periodic behavior. The physical explanation that follows provides a much better insight into the periodic behavior. - -### NON-UNIQUENESS OF DISCRETE-TIME SINUSOID WAVEFORMS - -A continuous-time sinusoid cos ω*t* has a unique waveform for every real value of ω in the range 0 to ∞. Increasing ω results in a sinusoid of ever-increasing frequency. Such is not the case for the discrete-time sinusoid cos *n* because - -$$ -\cos[(\Omega \pm 2\pi m)n] = \cos \Omega n \qquad m \text{ integer} -$$ - -and - -$$ -e^{j(\Omega \pm 2\pi m)n} = e^{j\Omega n} \qquad m \text{ integer} -$$ - -This shows that the discrete-time sinusoids cos *n* (and exponentials *ejn*) separated by values of in integral multiples of 2π are identical. The reason for the periodic nature of the frequency response of an LTID system is now clear. Since the sinusoids (or exponentials) with frequencies separated by interval 2π are identical, the system response to such sinusoids is also identical and, hence, is periodic with period 2π. - -This discussion shows that the discrete-time sinusoid cos *n* has a unique waveform only for the values of in the range −π to π. This band is called the *fundamental band*. Every frequency , no matter how large, is identical to some frequency, *a*, in the fundamental band (−π ≤ *a* < π), where - -$$ -\Omega_a = \Omega - 2\pi m \qquad -\pi \le \Omega_a < \pi \quad \text{and} \quad m \text{ integer} \tag{5.38} -$$ - -The integer *m* can be positive or negative. We use Eq. (5.38) to plot the fundamental band frequency *a* versus the frequency of a sinusoid (Fig. 5.17a). The frequency *a* is modulo 2π value of . - -All these conclusions are also valid for exponential *ejn*. - -### ALL DISCRETE-TIME SIGNALS ARE INHERENTLY BANDLIMITED - -This discussion leads to the surprising conclusion that all discrete-time signals are inherently bandlimited, with frequencies lying in the range −π to π radians per sample. In terms of frequency *F* = /2π, where *F* is in cycles per sample, all frequencies *F* separated by an integer number are identical. For instance, all discrete-time sinusoids of frequencies 0.3, 1.3, 2.3, ... cycles per sample are identical. The fundamental range of frequencies is −0.5 to 0.5 cycles per sample. - -Any discrete-time sinusoid of frequency beyond the fundamental band, when plotted, appears and behaves, in every way, like a sinusoid having its frequency in the fundamental band. It is impossible to distinguish between the two signals. Thus, in a basic sense, discrete-time frequencies beyond || = π or |*F*| = 1/2 do not exist. Yet, in a "mathematical" sense, we must admit the existence of sinusoids of frequencies beyond = π. What does this mean? - -**Figure 5.17 (a)** Actual frequency versus **(b)** apparent frequency. - -### A MAN NAMED ROBERT - -To give an analogy, consider a fictitious person Mr. Robert Thompson. His mother calls him Robby; his acquaintances call him Bob, his close friends call him by his nickname, Shorty. Yet, Robert, Robby, Bob, and Shorty are one and the same person. However, we cannot say that only Mr. Robert Thompson exists, or only Robby exists, or only Shorty exists, or only Bob exists. All these four persons exist, although they are one and the same individual. In a same way, we cannot say that the frequency π/2 exists and frequency 5π/2 does not exist; they are both the same entity, called by different names. - -It is in this sense that we have to admit the existence of frequencies beyond the fundamental band. Indeed, mathematical expressions in the frequency domain automatically cater to this need by their built-in periodicity. As seen earlier, the very structure of the frequency response is 2π-periodic. We shall also see later, in Ch. 9, that discrete-time signal spectra are also 2π-periodic. - -Admitting the existence of frequencies beyond π also serves mathematical and computational convenience in digital signal-processing applications. Values of frequencies beyond π may also originate naturally in the process of sampling continuous-time sinusoids. Because there is no upper limit on the value of ω, there is no upper limit on the value of the resulting discrete-time frequency = ω*T* either.† - -The highest possible frequency is π and the lowest frequency is 0 (dc or constant). Clearly, the high frequencies are those in the vicinity of = (2*m* + 1)π and the low frequencies are those in the vicinity of = 2π*m* for all positive or negative integer values of *m*. - -† However, if goes beyond π, the resulting aliasing reduces the apparent frequency to *a* < π. - -### FURTHER REDUCTION IN THE FREQUENCY RANGE - -Because cos(−*n* + θ ) = cos(*n* − θ ), a frequency in the range −π to 0 is identical to the frequency (of the same magnitude) in the range 0 to π (but with a change in phase sign). Consequently the *apparent frequency* for a discrete-time sinusoid of any frequency is equal to some value in the range 0 to π. Thus, cos(8.7π*n* + θ ) = cos(0.7π*n* + θ ), and the apparent frequency is 0.7π. Similarly, - -$$ -\cos(9.6\pi n + \theta) = \cos(-0.4\pi n + \theta) = \cos(0.4\pi n - \theta) -$$ - -Hence, the frequency 9.6π is identical (in every respect) to frequency −0.4π, which, in turn, is equal (within the sign of its phase) to frequency 0.4π. In this case, the apparent frequency reduces to |*a*| = 0.4π. We can generalize the result to say that the apparent frequency of a discrete-time sinusoid is |*a*|, as found from Eq. (5.38), and if *a* <0, there is a phase reversal. Figure 5.17b plots versus the apparent frequency |*a*|. The shaded bands represent the ranges of for which there is a phase reversal, when represented in terms of |*a*|. For example, the apparent frequency for both the sinusoids cos(2.4π + θ ) and cos(3.6π + θ ) is |*a*| = 0.4π, as seen from Fig. 5.17b. But 2.4π is in a clear band and 3.6π is in a shaded band. Hence, these sinusoids appear as cos(0.4π +θ ) and cos(0.4π −θ ), respectively. - -Although every discrete-time sinusoid can be expressed as having frequency in the range from 0 to π, we generally use the frequency range from −π to π instead of 0 to π for two reasons. First, exponential representation of sinusoids with frequencies in the range 0 to π requires a frequency range −π to π. Second, even when we are using a trigonometric representation, we generally need the frequency range −π to π to have exact identity (without phase reversal) of a higher-frequency sinusoid. - -For certain practical advantages, in place of the range −π to π, we often use other contiguous ranges of width 2π. The range 0 to 2π, for instance, is used in many applications. It is left as an exercise for the reader to show that the frequencies in the range from π to 2π are identical to those in the range from −π to 0. - -### **EXAMPLE 5.11 Apparent Frequency** - -Express the following signals in terms of their apparent frequencies: **(a)** cos(0.5π*n* + θ ), **(b)** cos(1.6π*n*+θ ), **(c)** sin(1.6π*n*+θ ), **(d)** cos(2.3π*n*+θ ), and **(e)** cos(34.699*n*+θ ). - -**(a)** =0.5π is in the reduced range already. This is also apparent from Fig. 5.17a or 5.17b. Because *a* = 0.5π, there is no phase reversal, and the apparent sinusoid is cos(0.5π*n*+θ ). - -**(b)** We express 1.6π = −0.4π + 2π so that *a* = −0.4π and |*a*| = 0.4. Also, *a* is negative, implying sign change for the phase. Hence, the apparent sinusoid is cos(0.4π*n*−θ ). This fact is also apparent from Fig. 5.17b. - -**(c)** We first convert the sine form to cosine form as sin(1.6π*n*+θ ) = cos(1.6π*n* − (π/2) + θ ). In part **(b)**, we found *a* = −0.4π. Hence, the apparent sinusoid is cos(0.4π*n* + (π/2)−θ ) = −sin(0.4π*n*−θ ). In this case, both the phase and the amplitude change signs. - -**(d)** 2.3π = 0.3π +2π so that *a* = 0.3π. Hence, the apparent sinusoid is cos(0.3π*n*+θ ). - -**(e)** We have 34.699 = −3+6(2π ). Hence, *a* = −3, and the apparent frequency |*a*| = 3 rad/sample. Because *a* is negative, there is a sign change of the phase. Hence, the apparent sinusoid is cos(3*n*−θ ). - -### **DR ILL 5.20 Apparent Frequency** - -Show that the sinusoids having frequencies of **(a)** 2π, **(b)** 3π, **(c)** 5π, **(d)** 3.2π, **(e)** 22.1327, and **(f)** π + 2 can be expressed, respectively, as sinusoids of frequencies **(a)** 0, **(b)** π, **(c)** π, **(d)** 0.8π, **(e)** 3, and **(f)** π −2. Show that in cases (d), (e), and (f), phase changes sign. - -### **[5.5-2 Aliasing and Sampling Rate](#page-11-0)** - -The non-uniqueness of discrete-time sinusoids and the periodic repetition of the same waveforms at intervals of 2π may seem innocuous, but in reality it leads to a serious problem for processing continuous-time signals by digital filters. A continuous-time sinusoid cosω*t* sampled every *T* seconds (*t* = *nT*) results in a discrete-time sinusoid cosω*nT*, which is cos*n* with = ω*T*. The discrete-time sinusoids cos*n* have unique waveforms only for the values of frequencies in the range <π or ω*T* < π. Therefore, samples of continuous-time sinusoids of two (or more) different frequencies can generate the same discrete-time signal, as shown in Fig. 5.18. *This phenomenon is known as aliasing because through sampling, two entirely different analog sinusoids take on the same "discrete-time" identity*. † - -Aliasing causes ambiguity in digital signal processing, which makes it impossible to determine the true frequency of the sampled signal. Consider, for instance, digitally processing - -**Figure 5.18** Demonstration of the aliasing effect. - -† Figure 5.18 shows samples of two sinusoids cos 12π*t* and cos 2π*t* taken every 0.2 second. The corresponding discrete-time frequencies ( = ω*T* = 0.2ω) are cos 2.4π and cos 0.4π. The apparent frequency of 2.4π is 0.4π, identical to the discrete-time frequency corresponding to the lower sinusoid. This shows that the samples of both these continuous-time sinusoids at 0.2-second intervals are identical, as verified from Fig. 5.18. - -a continuous-time signal that contains two distinct components of frequencies ω1 and ω2. The samples of these components appear as discrete-time sinusoids of frequencies 1 = ω1*T* and 2 = ω2*T*. If 1 and 2 happen to differ by an integer multiple of 2π (if ω2 − ω1 = 2*k*π/*T*), the two frequencies will be read as the same (lower of the two) frequency by the digital processor.‡ As a result, the higher-frequency component ω2 not only is lost for good (by losing its identity to ω1), but also it reincarnates as a component of frequency ω1, thus distorting the true amplitude of the original component of frequency ω1. Hence, the resulting processed signal will be distorted. Clearly, aliasing is highly undesirable and should be avoided. To avoid aliasing, the frequencies of the continuous-time sinusoids to be processed should be kept within the fundamental band ω*T* ≤ π or ω ≤ π/*T*. Under this condition the question of ambiguity or aliasing does not arise because any continuous-time sinusoid of frequency in this range has a unique waveform when it is sampled. Therefore, if ω*h* is the highest frequency to be processed, then, to avoid aliasing, - -$$ -\omega_h < \frac{\pi}{T} -$$ - -If *fh* is the highest frequency in hertz, *fh* = ω*h*/2π, and we avoid aliasing if - -$$ -f_h < \frac{1}{2T} \qquad \text{or} \qquad T < \frac{1}{2f_h} \tag{5.39} -$$ - -This shows that discrete-time signal processing places the limit on the highest frequency *fh* that can be processed for a given value of the sampling interval *T*. Fortunately, we can process a signal of any frequency (without aliasing) by choosing a suitably small value of *T*. Since the sampling frequency *fs* is the reciprocal of the sampling interval *T*, we can also express Eq. (5.39) as - -$$ -f_s = \frac{1}{T} > 2f_h -$$ - or $f_h < \frac{f_s}{2}$ (5.40) - -This result is a special case of the well-known *sampling theorem* (to be proved in Ch. 8). It states that for a discrete-time system to process a continuous-time sinusoid, the sampling rate must be greater than twice the frequency (in hertz) of the sinusoid. In short, *a sampled sinusoid must have a minimum of two samples per cycle*. † For sampling rates below this minimum value, the output signal will be aliased, which means it will be mistaken for a sinusoid of lower frequency. - -### ANTI-ALIASING FILTER - -If the sampling rate fails to satisfy Eq. (5.40), aliasing occurs, causing the frequencies beyond *fs*/2 Hz to masquerade as lower frequencies to corrupt the spectrum at frequencies below *fs*/2. To avoid such a corruption, a signal to be sampled is passed through an *anti-aliasing* filter of bandwidth *fs*/2 prior to sampling. This operation ensures the condition of Eq. (5.40). The drawback of such a filter is that we lose the spectral components of the signal beyond frequency *fs*/2, which is preferable to the aliasing corruption of the signal at frequencies below *fs*/2. Chapter 8 presents a detailed analysis of the aliasing problem. - -‡ In the case shown in Fig. 5.18, ω1 = 12π, ω2 = 2π, and *T* = 0.2. Hence, ω2 − ω1 = 10π*T* = 2π, and the two frequencies are read as the same frequency = 0.4π by the digital processor. - -† Strictly speaking, we must have more than two samples per cycle. - -### **EXAMPLE 5.12 Maximum Sampling Interval** - -Determine the maximum sampling interval *T* that can be used in a discrete-time oscillator that generates a sinusoid of 50 kHz. - -Here the highest significant frequency *fh* = 50 kHz. Therefore from Eq. (5.39), - -$$ -T < \frac{1}{2f_h} = 10\,\mu\,\mathrm{s} -$$ - -The sampling interval must be less than 10µs. The sampling frequency is *fs* = 1/*T* > 100 kHz. - -### **EXAMPLE 5.13 Maximum Frequency Without Aliasing** - -A discrete-time amplifier uses a sampling interval *T* = 25µs. What is the highest frequency of a signal that can be processed with this amplifier without aliasing? - -From Eq. (5.39) - -$$ -f_h < \frac{1}{2T} = 20 \, \text{kHz} -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/071_5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/071_5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS.md deleted file mode 100644 index 7829eb988207bd40b5a5c043ecf271ea69665ff7..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/071_5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS.md +++ /dev/null @@ -1,170 +0,0 @@ -## **5.6 FREQUENCY [RESPONSE FROM](#page-11-0) POLE-ZERO LOCATIONS** - -The frequency responses (amplitude and phase responses) of a system are determined by pole-zero locations of the transfer function *H*[*z*]. Just as in continuous-time systems, it is possible to determine quickly the amplitude and the phase response and to obtain physical insight into the filter characteristics of a discrete-time system by using a graphical technique. The general *N*th-order transfer function *H*[*z*] in Eq. (5.26) can be expressed in factored form as - -$$ -H[z] = b_0 \frac{(z - z_1)(z - z_2) \cdots (z - z_N)}{(z - \gamma_1)(z - \gamma_2) \cdots (z - \gamma_N)} -$$ - -We can compute *H*[*z*] graphically by using the concepts discussed in Sec. 4.10. The directed line segment from *zi* to *z* in the complex plane (Fig. 5.19a) represents the complex number *z* − *zi*. The length of this segment is |*z*−*zi*| and its angle with the horizontal axis is (*z*−*zi*). - -To compute the frequency response *H*[*ej*] we evaluate *H*[*z*] at *z* = *ej*. But for *z* = *ej*, |*z*| = 1 and *z* = so that *z* = *ej* represents a point on the unit circle at an angle with the horizontal. We now connect all zeros (*z*1, *z*2,...,*zN*) and all poles (γ1, γ2, ... , γ*N*) to the point *ej*, as indicated in - -**Figure 5.19** Vector representations of **(a)** complex numbers and **(b)** factors of *H*[*z*]. - -Fig. 5.19b. Let *r*1, *r*2, ... , *rN* be the lengths and φ1, φ2, ... , φ*N* be the angles, respectively, of the straight lines connecting *z*1, *z*2, ... , *zN* to the point *ej*. Similarly, let *d*1, *d*2, ... , *dN* be the lengths and θ1, θ2, ... , θ*N* be the angles, respectively, of the lines connecting γ1, γ2, ... , γ*N* to *ej*. Then - -$$ -H[e^{j\Omega}] = H[z]|_{z=e^{j\Omega}} = b_0 \frac{(r_1 e^{j\phi_1})(r_2 e^{j\phi_2}) \cdots (r_N e^{j\phi_N})}{(d_1 e^{j\theta_1})(d_2 e^{j\theta_2}) \cdots (d_N e^{j\theta_N})} -$$ - -= $b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} e^{j[(\phi_1 + \phi_2 + \cdots + \phi_N) - (\theta_1 + \theta_2 + \cdots + \theta_N)]}$ - -Therefore (assuming *b*0 > 0), - -$$ -|H[e^{j\Omega}]| = b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} = b_0 \frac{\text{product of the distances of zeros to } e^{j\Omega}}{\text{product of distances of poles to } e^{j\Omega}} -$$ -(5.41) - -and - -$$ -\angle H[e^{i\Omega}] = (\phi_1 + \phi_2 + \dots + \phi_N) - (\theta_1 + \theta_2 + \dots + \theta_N) -$$ - -= sum of zero angles to $e^{i\Omega}$ - sum of pole angles to $e^{i\Omega}$ - -In this manner, we can compute the frequency response *H*[*ej*] for any value of by selecting the point on the unit circle at an angle . This point is *ej*. To compute the frequency response *H*[*ej*], we connect all poles and zeros to this point and use the foregoing equations to determine |*H*[*ej*]| and *H*[*ej*]. We repeat this procedure for all values of from 0 to π to obtain the frequency response. - -### CONTROLLING GAIN BY PLACEMENT OF POLES AND ZEROS - -The nature of the influence of pole and zero locations on the frequency response is similar to that observed in continuous-time systems, with minor differences. In place of the imaginary axis of the continuous-time systems, we have the unit circle in the discrete-time case. The nearer the pole (or zero) is to a point *ej* (on the unit circle) representing some frequency , the more influence that pole (or zero) wields on the amplitude response at that frequency because the length of the vector joining that pole (or zero) to the point *ej* is small. The proximity of a pole (or a zero) has a similar effect on the phase response. From Eq. (5.41), it is clear that to enhance the amplitude response at a frequency , we should place a pole as close as possible to the point *ej* (which is on the unit circle).† Similarly, to suppress the amplitude response at a frequency , we should place a zero as close as possible to the point *ej* on the unit circle. Placing repeated poles or zeros will further enhance their influence. - -Total suppression of signal transmission at any frequency can be achieved by placing a zero on the unit circle at a point corresponding to that frequency. This observation is used in the notch (bandstop) filter design. - -Placing a pole or a zero at the origin does not influence the amplitude response because the length of the vector connecting the origin to any point on the unit circle is unity. However, a pole (or a zero) at the origin adds angle − (or ) to *H*[*ej*]. Hence, the phase spectrum − (or ) is a linear function of frequency and therefore represents a pure time delay (or time advance) of *T* seconds (see Drill 5.19). Therefore, a pole (a zero) at the origin causes a time delay (or a time advance) of *T* seconds in the response. There is no change in the amplitude response. - -For a stable system, all the poles must be located inside the unit circle. The zeros may lie anywhere. Also, for a physically realizable system, *H*[*z*] must be a proper fraction, that is, *N* ≥ *M*. If, to achieve a certain amplitude response, we require *M* > *N*, we can still make the system realizable by placing a sufficient number of poles at the origin to make *N* = *M*. This will not change the amplitude response, but it will increase the time delay of the response. - -In general, a pole at a point has the opposite effect of a zero at that point. Placing a zero closer to a pole tends to cancel the effect of that pole on the frequency response. - -### LOWPASS FILTERS - -A lowpass filter generally has a maximum gain at or near = 0, which corresponds to point *ej*0 = 1 on the unit circle. Clearly, placing a pole inside the unit circle near the point *z* = 1 (Fig. 5.20a) would result in a lowpass response.‡ The corresponding amplitude and phase response appear in Fig. 5.20a. For smaller values of , the point *ej* (a point on the unit circle at an angle ) is closer to the pole, and consequently the gain is higher. As increases, the distance of the point *ej* from the pole increases. Consequently the gain decreases, resulting in a lowpass characteristic. Placing a zero at the origin does not change the amplitude response but it does modify the phase response, as illustrated in Fig. 5.20b. Placing a zero at *z* = −1, however, changes both the amplitude and the phase response (Fig. 5.20c). The point *z* = −1 corresponds to frequency - -† The closest we can place a pole is on the unit circle at the point representing . This choice would lead to infinite gain, but should be avoided because it will render the system marginally stable (BIBO-unstable). The closer the point to the unit circle, the more sensitive the system gain to parameter variations. - -‡ Placing the pole at *z* = 1 results in maximum (infinite) gain but renders the system BIBO-unstable, hence should be avoided. - -**Figure 5.20** Various pole-zero configurations and the corresponding frequency responses. - -### 542 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - - = π (*z* = *ej* = *ej*π = −1). Consequently, the amplitude response now becomes more attenuated at higher frequencies, with a zero gain at = π. We can approach ideal lowpass characteristics by using more poles staggered near *z* = 1 (but within the unit circle). Figure 5.20d shows a third-order lowpass filter with three poles near *z* = 1 and a third-order zero at *z* = −1, with corresponding amplitude and phase response. For an ideal lowpass filter, we need an enhanced gain at every frequency in the band (0, *c*). This can be achieved by placing a continuous wall of poles (requiring an infinite number of poles) opposite this band. - -### HIGHPASS FILTERS - -A highpass filter has a small gain at lower frequencies and a high gain at higher frequencies. Such a characteristic can be realized by placing a pole or poles near *z* = −1 because we want the gain at = π to be the highest. Placing a zero at *z* = 1 further enhances suppression of gain at lower frequencies. Figure 5.20e shows a possible pole-zero configuration of the third-order highpass filter with corresponding amplitude and phase responses. - -In the following two examples, we shall realize analog filters by using digital processors and suitable interface devices (C/D and D/C), as shown in Fig. 3.2. At this point, we shall examine the design of a digital processor with the transfer function *H*[*z*] for the purpose of realizing bandpass and bandstop filters in the following examples. - -As Fig. 3.2 shows, the C/D device samples the continuous-time input *x*(*t*) to yield a discrete-time signal *x*[*n*], which serves as the input to *H*[*z*]. The output *y*[*n*] of *H*[*z*] is converted to a continuous-time signal *y*(*t*) by a D/C device. We also saw in Eq. (5.34) that a continuous-time sinusoid of frequency ω, when sampled, results in a discrete-time sinusoid = ω*T*. - -### **EXAMPLE 5.14 Bandpass Filter by Pole-Zero Placement** - -By trial and error, design a tuned (bandpass) analog filter with zero transmission at 0 Hz and also at the highest frequency *fh* = 500 Hz. The resonant frequency is required to be 125 Hz. - -Because *fh* = 500, we require *T* < 1/1000 [see Eq. (5.39)]. Let us select *T* = 10−3. † Recall that the analog frequencies ω correspond to digital frequencies = ω*T*. Hence, analog frequencies ω = 0 and 1000π correspond to = 0 and π, respectively. The gain is required to be zero at these frequencies. Hence, we need to place zeros at *ej* corresponding to = 0 and = π. For = 0, *z* = *ej* = 1; for = π, *ej* = −1. Hence, there must be zeros at *z* = ±1. Moreover, we need enhanced response at the resonant frequency ω = 250π, which corresponds to = π/4, which, in turn, corresponds to *z* = *ej* = *ej*π/4. Therefore, to enhance the frequency response at ω = 250π, we place a pole in the vicinity of *ej*π/4. Because this is a complex pole, we also need its conjugate near *e*−*j*π/4, as indicated in Fig. 5.21a. Let us choose these poles γ1 and γ2 as - -γ1 = |γ |*ej*π/4 and γ2 = |γ |*e*−*j*π/4 - -† Strictly speaking, we need *T* < 0.001. However, we shall show in Ch. 8 that if the input does not contain a finite amplitude component of 500 Hz, *T* = 0.001 is adequate. Generally, practical signals satisfy this condition. - -where |γ | < 1 for stability. The closer γ is to the unit circle, the more sharply peaked is the response around ω = 250π. We also have zeros at ±1. Hence, - -$$ -H[z] = K \frac{(z-1)(z+1)}{(z-|\gamma|e^{j\pi/4})(z-|\gamma|e^{-j\pi/4})} = K \frac{z^2 - 1}{z^2 - \sqrt{2}|\gamma|z+|\gamma|^2} -$$ - -For convenience, we shall choose *K* = 1. The amplitude response is given by - -$$ -|H[e^{i\Omega}]| = \frac{|e^{i2\Omega} - 1|}{|e^{i\Omega} - |\gamma|e^{i\pi/4}||e^{i\Omega} - |\gamma|e^{-i\pi/4}|} -$$ - -Now, by using Eq. (5.37), we obtain - -$$ -|H[e^{i\Omega}]|^2 = \frac{2(1 - \cos 2\Omega)}{\left[1 + |\gamma|^2 - 2|\gamma|\cos\left(\Omega - \frac{\pi}{4}\right)\right] \left[1 + |\gamma|^2 - 2|\gamma|\cos\left(\Omega + \frac{\pi}{4}\right)\right]} -$$ - -**Figure 5.21** Designing a bandpass filter. - -### 544 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -Figure 5.21b shows the amplitude response as a function of ω, as well as = ω*T* = 10−3ω for values of |γ | = 0.83, 0.96, and 1. As expected, the gain is zero at ω = 0 and at 500 Hz (ω = 1000π). The gain peaks at about 125 Hz (ω = 250π). The resonance (peaking) becomes pronounced as |γ | approaches 1. Figure 5.21c shows a canonical realization of this filter, which follows from the transfer function *H*[*z*]. - -## MULTIPLE MAGNITUDE RESPONSE CURVES USING MATLAB - -By defining an anonymous function of the two variables *z* and γ in MATLAB, it is straightforward to duplicate the three magnitude response curves in Fig. 5.21b, corresponding to the cases γ = 0.83, 0.96, and 1. - -``` ->> Omega = linspace(0,pi,400); ->> H = @(z,gamma_m) (z.^2-1)./(z.^2-sqrt(2)*gamma_m*z+gamma_m^2); ->> plot(Omega,abs(H(exp(1j*Omega),0.83)),... ->> Omega,abs(H(exp(1j*Omega),0.96)),... ->> Omega,abs(H(exp(1j*Omega),0.99))); ->> text(.27*pi,35,'|\gamma|=1'); ->> text(.28*pi,25.5,'|\gamma|=0.96'); ->> text(.35*pi,6.41,'|\gamma|=0.83'); ->> set(gca,'xtick',0:pi/4:pi,'ytick',[0 6.41,25.5]); ->> axis([0 pi 0 40]); xlabel('\Omega'); ylabel('|H[e^{j \Omega}]|'); -``` - -The result, shown in Fig. 5.22, confirms the earlier result of Fig. 5.21b. Phase response curves can be generated with minor modification to the MATLAB code. - -**Figure 5.22** MATLAB-generated magnitude response curves for Ex. 5.14. - -### **EXAMPLE 5.15 Bandstop Filter by Pole-Zero Placement** - -Design a second-order notch filter to have zero transmission at 250 Hz and a sharp recovery of gain to unity on both sides of 250 Hz. The highest significant frequency to be processed is *fh* = 400 Hz. - -In this case, *T* < 1/2*fh* = 1.25 × 10−3. Let us choose *T* = 10−3. For the frequency 250 Hz, = 2π(250)*T* = π/2. Thus, the frequency 250 Hz is represented by a point *ej* = *ej*π/2 = *j* on the unit circle, as depicted in Fig. 5.23a. Since we need zero transmission at this frequency, we must place a zero at *z* = *ej*π/2 = *j* and its conjugate at *z* = *e*−*j*π/2 = −*j*. We also require a sharp recovery of gain on both sides of frequency 250 Hz. To accomplish this goal, we place two poles close to the two zeros, to cancel out the effect of the two zeros as we move away from the point *j* (corresponding to frequency 250 Hz). For this reason, let us use poles at ±*ja* with *a* < 1 for stability. The closer the poles are to zeros (the closer the *a* to 1), the faster is the gain recovery on either side of 250 Hz. The resulting transfer function is - -$$ -H[z] = K \frac{(z-j)(z+j)}{(z-ja)(z+ja)} = K \frac{z^2+1}{z^2+a^2} -$$ - -The dc gain (gain at = 0, or *z* = 1 ) of this filter is - -$$ -H[1] = K \frac{2}{1 + a^2} -$$ - -Because we require a dc gain of unity, we must select *K* = (1+*a*2)/2. The transfer function is therefore - -$$ -H[z] = \frac{(1+a^2)(z^2+1)}{2(z^2+a^2)} -$$ - -and according to Eq. (5.37), - -$$ -|H[e^{j\Omega}]|^{2} = \frac{(1+a^{2})^{2}}{4} \frac{(e^{j2\Omega}+1)(e^{-j2\Omega}+1)}{(e^{j2\Omega}+a^{2})(e^{-j2\Omega}+a^{2})} -$$ -$$ -= \frac{(1+a^{2})^{2}(1+\cos 2\Omega)}{2(1+a^{4}+2a^{2}\cos 2\Omega)} -$$ - -Figure 5.23b shows |*H*[*ej*]| for values of *a* = 0.3, 0.6, and 0.95. Figure 5.23c shows a realization of this filter. - -### **DR ILL 5.21 Highpass Filter by Pole-Zero Placement** - -Use the graphical argument to show that a filter with transfer function - -$$ -H[z] = \frac{z - 0.9}{z} -$$ - -acts like a highpass filter. Make a rough sketch of the amplitude response. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/072_5.7 DIGITAL PROCESSING OF ANALOG SIGNALS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/072_5.7 DIGITAL PROCESSING OF ANALOG SIGNALS.md deleted file mode 100644 index 735251f6c401f88e8ba15afb0128d02fe3df85ba..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/072_5.7 DIGITAL PROCESSING OF ANALOG SIGNALS.md +++ /dev/null @@ -1,233 +0,0 @@ -## **5.7 DIGITAL [PROCESSING OF](#page-11-0) ANALOG SIGNALS** - -An analog (meaning continuous-time) signal can be processed digitally by sampling the analog signal and processing the samples by a digital (meaning discrete-time) processor. The output of the processor is then converted back to analog signal, as shown in Fig. 5.24a. We saw some simple cases of such processing in Exs. 3.8, 3.9, 5.14, and 5.15. In this section, we shall derive a criterion for designing such a digital processor for a general LTIC system. - -Suppose that we wish to realize an equivalent of an analog system with transfer function *Ha*(*s*), shown in Fig. 5.24b. Let the digital processor transfer function in Fig. 5.24a that realizes this desired *Ha*(*s*) be *H*[*z*]. In other words, we wish to make the two systems in Fig. 5.24 equivalent (at least approximately). - -By "equivalence" we mean that for a given input *x*(*t*), the systems in Fig. 5.24 yield the same output *y*(*t*). Therefore, *y*(*nT*), the samples of the output in Fig. 5.24b, are identical to *y*[*n*], the output of *H*[*z*] in Fig. 5.24a. - -**Figure 5.24** Analog filter realization with a digital filter. - -### 548 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -For the sake of generality, we are assuming a noncausal system. The argument and the results are also valid for causal systems. The output *y*(*t*) of the system in Fig. 5.24b is - -$$ -y(t) = \int_{-\infty}^{\infty} x(\tau)h_a(t-\tau) d\tau = \lim_{\Delta \tau \to 0} \sum_{m=-\infty}^{\infty} x(m\Delta \tau)h_a(t-m\Delta \tau) \Delta \tau -$$ - -For our purpose, it is convenient to use the notation *T* for τ . Assuming *T* (the sampling interval) to be small enough, such a change of notation yields - -$$ -y(t) = T \sum_{m=-\infty}^{\infty} x(mT)h_a(t - mT) -$$ - -The response at the *n*th sampling instant is *y*(*nT*) obtained by setting *t* = *nT* in the equation is - -$$ -y(nT) = T \sum_{m=-\infty}^{\infty} x(mT)h_a[(n-m)T] -$$ -\n(5.42) - -In Fig. 5.24a, the input to *H*[*z*] is *x*(*nT*) = *x*[*n*]. If *h*[*n*] is the unit impulse response of *H*[*z*], then *y*[*n*], the output of *H*[*z*], is given by - -$$ -y[n] = \sum_{m = -\infty}^{\infty} x[m]h[n-m] -$$ -\n(5.43) - -If the two systems are to be equivalent, *y*(*nT*) in Eq. (5.42) must be equal to *y*[*n*] in Eq. (5.43). Therefore, - -$$ -h[n] = Th_a(nT) \tag{5.44} -$$ - -This is the time-domain criterion for equivalence of the two systems.† According to this criterion, *h*[*n*], the unit impulse response of *H*[*z*] in Fig. 5.24a, should be *T* times the samples of *ha*(*t*), the unit impulse response of the system in Fig. 5.24b. This is known as the *impulse invariance criterion* of filter design. - -Strictly speaking, this realization guarantees the output equivalence only at the sampling instants, that is, *y*(*nT*) = *y*[*n*], and that also requires the assumption that *T* → 0. Clearly, this criterion leads to an approximate realization of *Ha*(*s*). However, it can be shown that when the frequency response of |*Ha*(*j*ω)| is bandlimited, the realization is exact [2], provided the sampling rate is high enough to avoid any aliasing (*T* < 1/2*fh*). - -### REALIZATION OF RATIONAL *H*(*s*) - -If we wish to realize an analog filter with transfer function - -$$ -H_a(s) = \frac{c}{s - \lambda} -$$ - -† Because *T* is a constant, some authors ignore the factor *T*, which yields a simplified criterion *h*[*n*] = *ha*(*nT*). Ignoring *T* merely scales the amplitude response of the resulting filter. - -The impulse response *h*(*t*), given by the inverse Laplace transform of *Ha*(*s*), is - -$$ -h_a(t) = ce^{\lambda t}u(t) -$$ - -The corresponding digital filter unit impulse response *h*[*n*], per Eq. (5.44), is - -$$ -h[n] = Th_a(nT) = Tce^{n\lambda T} -$$ - -Figure 5.25 shows *ha*(*t*) and *h*[*n*]. The corresponding *H*[*z*], the *z*-transform of *h*[*n*], as found from Table 5.1, is - -$$ -H[z] = \frac{Tcz}{z - e^{\lambda T}} -$$ -\n(5.45) - -**Figure 5.25** Impulse response for analog and digital systems in the impulse invariance method of filter design. - -| No. | Ha(s) | ha(t) | h[n] | H[z] | -|-----|------------------|-------------------------------|-----------------------------|-----------------------------------------------------------------| -| 1 | K | Kδ(t) | TKδ[n] | TK | -| 2 | 1
s | u(t) | Tu[n] | Tz
z−1 | -| 3 | 1
s2 | t | nT2 | T2z
(z−1)2 | -| 4 | 1
s3 | 2
t
2 | k2T3
2 | T3z(z
+1)
2(z−1)3 | -| 5 | 1
s −λ | eλt | TeλnT | Tz
z−eλT | -| 6 | 1
(s −λ)2 | teλt | nT2eλnT | T2zeλT
(z−eλT )2 | -| 7 | As+B
s2+2as+c | Tre−at cos(bt+θ
) | Tre−anT cos(bnT+θ
) | Trz[z cos θ −e−aT cos(bT−θ
)]
z2−(2e−aT cos
bT)z+e−2aT | -| | r = |
A2c+B2 −2ABa
,
c−a2 | b = √
θ = tan−1
c−a2, | Aa−B
√
c−a2
A | - -**TABLE 5.3** Select Impulse-Invariance Pairs - -The procedure of finding *H*[*z*] can be systematized for any *N*th-order system. First we express an *N*th-order analog transfer function *Ha*(*s*) as a sum of partial fractions as‡ - -$$ -H_a(s) = \sum_{i=1}^n \frac{c_i}{s - \lambda_i} -$$ - -Then the corresponding *H*[*z*] is given by - -$$ -H[z] = T \sum_{i=1}^{n} \frac{c_i z}{z - e^{\lambda_i T}} -$$ - -This transfer function can be readily realized, as explained in Sec. 5.4. Table 5.3 lists several pairs of *Ha*(*s*) and their corresponding *H*[*z*]. For instance, to realize a digital integrator, we examine its *Ha*(*s*) = 1/*s*. From Table 5.3, corresponding to *Ha*(*s*) = 1/*s* (pair 2), we find *H*[*z*] = *Tz*/(*z* − 1). This is exactly the result we obtained in Ex. 3.9 using another approach. - -Note that the frequency response *Ha*(*j*ω) of a practical analog filter cannot be bandlimited. Consequently, all these realizations are approximate. - -### CHOOSING THE SAMPLING INTERVAL *T* - -The impulse-invariance criterion (5.44) was derived under the assumption that *T* → 0. Such an assumption is neither practical nor necessary for satisfactory design. Avoiding of aliasing is the most important consideration for the choice of *T*. In Eq. (5.39), we showed that for a sampling interval *T* seconds, the highest frequency that can be sampled without aliasing is 1/2*T* Hz or π/*T* radians per second. This implies that *Ha*(*j*ω), the frequency response of the analog filter in Fig. 5.24b should not have spectral components beyond frequency π/*T* radians per second. In other words, to avoid aliasing, the frequency response of the system *Ha*(*s*) must be bandlimited to π/*T* radians per second. We shall see later in Ch. 7 that frequency response of a realizable LTIC system cannot be bandlimited; that is, the response generally exists for all frequencies up to ∞. Therefore, it is impossible to digitally realize an LTIC system exactly without aliasing. The saving grace is that the frequency response of every realizable LTIC system decays with frequency. This allows for a compromise in digitally realizing an LTIC system with an acceptable level of aliasing. The smaller the value of *T*, the smaller the aliasing, and the better the approximation. Since it is impossible to make |*Ha*(*j*ω)| zero, we are satisfied with making it negligible beyond the frequency π/*T*. As a rule of thumb [3], we choose *T* such that |*Ha*(*j*ω)| at the frequency ω = π/*T* is less than a certain fraction (often taken as 1%) of the peak value of |*Ha*(*j*ω)|. This ensures that aliasing is negligible. The peak |*Ha*(*j*ω)| usually occurs at ω = 0 for lowpass filters and at the band center frequency ω*c* for bandpass filters. - -‡ Assuming *Ha*(*s*) has simple poles. For repeated poles, the form changes accordingly. Entry 6 in Table 5.3 is suitable for repeated poles. - -### **EXAMPLE 5.16 Butterworth Filter Design by the Impulse-Invariance Method** - -Design a digital filter to realize a first-order lowpass Butterworth filter with the transfer function - -$$ -H_a(s) = \frac{\omega_c}{s + \omega_c} \qquad \omega_c = 10^5 \tag{5.46} -$$ - -For this filter, we find the corresponding *H*[*z*] according to Eq. (5.45) (or pair 5 in Table 5.3) as - -$$ -H[z] = \frac{\omega_c T z}{z - e^{-\omega_c T}} -$$ -\n(5.47) - -Next, we select the value of *T* by means of the criterion according to which the gain at ω = π/*T* drops to 1% of the maximum filter gain. However, this choice results in such a good design that aliasing is imperceptible. The resulting amplitude response is so close to the desired response that we can hardly notice the aliasing effect in our plot. For the sake of demonstrating the aliasing effect, we shall deliberately select a 10% criterion (instead of 1%). We have - -$$ -|H_a(j\omega)| = \left|\frac{\omega_c}{\sqrt{\omega^2 + \omega_c^2}}\right| -$$ - -In this case |*Ha*(*j*ω)|max = 1, which occurs at ω = 0. Use of 10% criterion leads to |*Ha*(π/*T*)| = 0.1. Observe that - -$$ -|H_a(j\omega)| \approx \frac{\omega_c}{\omega} \qquad \omega \gg \omega_c -$$ - -Hence, - -$$ -|H_a(\pi/T)| \approx \frac{\omega_c}{\pi/T} = 0.1 \quad \Longrightarrow \quad \pi/T = 10\omega_c = 10^6 -$$ - -Thus, the 10% criterion yields *T* = 10−6π. The 1% criterion would have given *T* = 10−7π. Substitution of *T* = 10−6π in Eq. (5.47) yields - -$$ -H[z] = \frac{0.3142z}{z - 0.7304} -$$ -\n(5.48) - -A canonical realization of this filter is shown in Fig. 5.26a. - -To find the frequency response of this digital filter, we rewrite *H*[*z*] as - -$$ -H[z] = \frac{0.3142}{1 - 0.7304z^{-1}} -$$ - -Therefore, - -$$ -H[e^{j\omega T}] = \frac{0.3142}{1 - 0.7304e^{-j\omega T}} = \frac{0.3142}{(1 - 0.7304 \cos \omega T) + j0.7304 \sin \omega T} -$$ - -**Figure 5.26** An example of filter design by the impulse-invariance method: **(a)** filter realization, **(b)** amplitude response, and **(c)** phase response. - -The corresponding magnitude response is - -$$ -|H[e^{j\omega T}]| = \frac{0.3142}{\sqrt{(1 - 0.7304 \cos \omega T)^2 + (0.7304 \sin \omega T)^2}} -$$ -$$ -= \frac{0.3142}{\sqrt{1.533 - 1.4608 \cos \omega T}} -$$ -(5.49) - -### 5.7 Digital Processing of Analog Signals 553 - -and the phase response is - -$$ -\angle H[e^{j\omega T}] = -\tan^{-1}\left(\frac{0.7304 \sin \omega T}{1 - 0.7304 \cos \omega T}\right) -$$ - (5.50) - -This frequency response differs from the desired response *Ha*(*j*ω) because aliasing causes frequencies above π/*T* to appear as frequencies below π/*T*. This generally results in increased gain for frequencies below π/*T*. For instance, the realized filter gain at ω = 0 is *H*[*ej*0] = *H*[1]. This value, as obtained from Eq. (5.48), is 1.1654 instead of the desired value 1. We can partly compensate for this distortion by multiplying *H*[*z*] or *H*[*ej*ω*T* ] by a normalizing constant *K* = *Ha*(0)/*H*[1] = 1/1.1654 = 0.858. This forces the resulting gain of *H*[*ej*ω*T* ] to be equal to 1 at ω = 0. The normalized *Hn*[*z*] = 0.858*H*[*z*] = 0.858(0.1π*z*/(*z*−0.7304)). The amplitude response in Eq. (5.49) is multiplied by *K* = 0.858 and plotted in Fig. 5.26b over the frequency range 0 ≤ ω ≤ π/*T* = 106. The multiplying constant *K* has no effect on the phase response in Eq. (5.50), which is shown in Fig. 5.26c. - -Also, the desired frequency response, according to Eq. (5.46) with ω*c* = 105, is - -$$ -H_a(j\omega) = \frac{\omega_c}{j\omega + \omega_c} = \frac{10^5}{j\omega + 10^5} -$$ - -Therefore, - -$$ -|H_a(j\omega)| = \frac{10^5}{\sqrt{\omega^2 + 10^{10}}} \quad \text{and} \quad \angle H_a(j\omega) = -\tan^{-1}\frac{\omega}{10^5} -$$ - -This desired amplitude and phase response are plotted (dotted) in Figs. 5.26b and 5.26c for comparison with realized digital filter response. Observe that the amplitude response behavior of the analog and the digital filter is very close over the range ω ≤ ω*c* = 105. However, for higher frequencies, there is considerable aliasing, especially in the phase spectrum. Had we used the 1% rule, the realized frequency response would have been closer over another decade of the frequency range. - -### IMPULSE INVARIANCE BY MATLAB - -We can readily use the MATLAB impinvar command to confirm our digital filter designed by the impulse-invariance method. - -``` ->> omegac = 10^5; Ba = [omegac]; Aa = [1 omegac]; Fs = 10^6/pi; ->> [B,A] = impinvar(Ba,Aa,Fs) - B = 0.3142 - A = 1.0000 -0.7304 -``` - -This confirms our earlier result of Eq. (5.48) that the digital filter transfer function is - -$$ -H[z] = \frac{0.3142z}{z - 0.7304} -$$ - -### **DR ILL 5.22 Filter Design by the Impulse-Invariance Method** - -Design a digital filter to realize an analog transfer function - -$$ -H_a(s) = \frac{20}{s+20} -$$ - -**ANSWER** - -*H*[*z*] = 20*Tz z*−*e*−20*T* with *T* = π 2000 diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/073_5.8 THE BILATERAL z-TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/073_5.8 THE BILATERAL z-TRANSFORM.md deleted file mode 100644 index a16d9cfce2d25aea50b8bf32a09ae2b0fd333285..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/073_5.8 THE BILATERAL z-TRANSFORM.md +++ /dev/null @@ -1,361 +0,0 @@ -## **5.8 THE BILATERAL** *z***[-TRANSFORM](#page-11-0)** - -Situations involving noncausal signals or systems cannot be handled by the (unilateral) *z*-transform discussed so far. Such cases can be analyzed by the *bilateral* (or two-sided) *z*-transform defined in Eq. (5.1) as - -$$ -X[z] = \sum_{n=-\infty}^{\infty} x[n]z^{-n} -$$ - -As in Eq. (5.2), the inverse *z*-transform is given by - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ - -These equations define the bilateral *z*-transform. Earlier, we showed that - -$$ -\gamma^n u[n] \Longleftrightarrow \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.51} -$$ - -In contrast, the *z*-transform of the signal −γ *nu*[−(*n*+1)], illustrated in Fig. 5.27a, is - -$$ -\mathcal{Z}\{-\gamma^n u[-(n+1)]\} = \sum_{-\infty}^{-1} -\gamma^n z^{-n} = \sum_{-\infty}^{-1} -\left(\frac{\gamma}{z}\right)^n -$$ -$$ -= -\left[\frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right] -$$ -$$ -= 1 - \left[1 + \frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right] -$$ -$$ -= 1 - \frac{1}{1 - \frac{z}{\gamma}} \qquad \left|\frac{z}{\gamma}\right| < 1 -$$ -$$ -= \frac{z}{z - \gamma} \qquad |z| < |\gamma| -$$ - -**Figure 5.27 (a)** −γ *nu*[−(*n*+1)] and **(b)** the region of convergence (ROC) of its *z*-transform. - -Therefore, - -$$ -\mathcal{Z}\{-\gamma^{n}u[-(n+1)]\} = \frac{z}{z-\gamma} \qquad |z| < |\gamma| \tag{5.52} -$$ - -A comparison of Eqs. (5.51) and (5.52) shows that the *z*-transform of γ *nu*[*n*] is identical to that of −γ *nu*[−(*n* + 1)]. The regions of convergence, however, are different. In the former case, *X*[*z*] converges for |*z*| > |γ |; in the latter, *X*[*z*] converges for |*z*| < |γ | (see Fig. 5.27b). Clearly, the inverse transform of *X*[*z*] is not unique unless the region of convergence is specified. If we add the restriction that all our signals be causal, however, this ambiguity does not arise. The inverse transform of *z*/(*z* − γ ) is γ *nu*[*n*] even without specifying the ROC. Thus, in the unilateral transform, we can ignore the ROC in determining the inverse *z*-transform of *X*[*z*]. - -As in the case of the bilateral Laplace transform, if *x*[*n*] = %*k i*=1 *xi*[*n*], then the ROC for *X*[*z*] is the intersection of the ROCs (region common to all ROCs) for the transforms *X*1[*z*],*X*2[*z*],...,*Xk*[*z*]. - -The preceding results lead to the conclusion (similar to that for the Laplace transform) that if *z* = β is the largest magnitude pole for a causal sequence, its ROC is |*z*| > |β|. If *z* = α is the smallest magnitude nonzero pole for an anticausal sequence, its ROC is |*z*| < |α|. - -### REGION OF CONVERGENCE FOR LEFT-SIDED AND RIGHT-SIDED SEQUENCES - -Let us first consider a finite duration sequence *xf*[*n*], defined as a sequence that is nonzero for *N*1 ≤ *n* ≤ *N*2, where both *N*1 and *N*2 are finite numbers and *N*2 > *N*1. Also, - -$$ -X_f[z] = \sum_{n=N_1}^{N_2} x_f[n]z^{-n} -$$ - -For example, if *N*1 = −2 and *N*2 = 1, then - -$$ -X_f[z] = x_f[-2]z^2 + x_f[-1]z + x_f[0] + \frac{x_f[1]}{z} -$$ - -Assuming all the elements in *xf*[*n*] are finite, we observe that *Xf*[*z*] has two poles at *z* = ∞ because of terms *xf*[−2]*z*2 +*xf*[−1]*z* and one pole at *z* = 0 because of term *xf*[1]/*z*. Thus, a finite-duration sequence could have poles at *z* = 0 and *z* = ∞. Observe that *Xf*[*z*] converges for all values of *z* except possibly *z* = 0 and *z* = ∞. - -This means that the ROC of a general signal *x*[*n*] + *xf*[*n*] is the same as the ROC of *x*[*n*] with the possible exception of *z* = 0 and *z* = ∞. - -A *right-sided* sequence is zero for *n* < *N*2 < ∞ and a left-sided sequence is zero for *n* > *N*1 > −∞. A causal sequence is always a right-sided sequence, but the converse is not necessarily true. An anticausal sequence is always a left-sided sequence, but the converse is not necessarily true. A *two-sided* sequence is of infinite duration and is neither right-sided nor left-sided. - -A right-sided sequence *xr*[*n*] can be expressed as *xr*[*n*] = *xc*[*n*] +*xf*[*n*], where *xc*[*n*] is a causal signal and *xf*[*n*] is a finite-duration signal. Therefore, the ROC for *xr*[*n*] is the same as the ROC for *xc*[*n*] except possibly *z* = ∞. If *z* = β is the largest magnitude pole for a right-sided sequence *xr*[*n*], its ROC is |β| < |*z*|≤∞. Similarly, a left-sided sequence can be expressed as *xl*[*n*] = *xa*[*n*]+*xf*[*n*], where *xa*[*n*] is an anticausal sequence and *xf*[*n*] is a finite-duration signal. Therefore, the ROC for *xl*[*n*] is the same as the ROC for *xa*[*n*] except possibly *z*=0. Thus, if*z*=α is the smallest magnitude nonzero pole for a left-sided sequence, its ROC is 0 ≤ |*z*| < |α|. - -### **EXAMPLE 5.17 Bilateral** *z***-Transform** - -Determine the bilateral *z*-transform of - -$$ -x[n] = \underbrace{(0.9)^n u[n]}_{x_1[n]} + \underbrace{(1.2)^n u[-(n+1)]}_{x_2[n]} -$$ - -From the results in Eqs. (5.51) and (5.52), we have - -$$ -X_1[z] = \frac{z}{z - 0.9} \qquad |z| > 0.9 -$$ - -$$ -X_2[z] = \frac{-z}{z - 1.2} \qquad |z| < 1.2 -$$ - -The common region where both *X*1[*z*] and *X*2[*z*] converge is 0.9 < |*z*| < 1.2 (Fig. 5.28b). Hence, - -$$ -X[z] = X_1[z] + X_2[z] -$$ - -= $\frac{z}{z - 0.9} - \frac{z}{z - 1.2}$ -= $\frac{-0.3z}{(z - 0.9)(z - 1.2)}$ 0.9 < |z| < 1.2 - -The sequence *x*[*n*] and the ROC of *X*[*z*] are depicted in Fig. 5.28. - -### **EXAMPLE 5.18 Inverse Bilateral** *z***-Transform** - -Find the inverse bilateral *z*-transform of - -$$ -X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)} -$$ - -if the ROC is **(a)** |*z*| > 2, **(b)** |*z*| < 0.8, and **(c)** 0.8 < |*z*| < 2. - -**(a)** - -$$ -\frac{X[z]}{z} = \frac{-(z+0.4)}{(z-0.8)(z-2)} = \frac{1}{z-0.8} - \frac{2}{z-2} -$$ - -and - -$$ -X[z] = \frac{z}{z - 0.8} - 2\frac{z}{z - 2} -$$ - -Since the ROC is |*z*| > 2, both terms correspond to causal sequences and - -$$ -x[n] = [(0.8)^n - 2(2)^n]u[n] -$$ - -This sequence appears in Fig. 5.29a. - -**(b)** In this case, |*z*| < 0.8, which is less than the magnitudes of both poles. Hence, both terms correspond to anticausal sequences, and - -$$ -x[n] = [-(0.8)^n + 2(2)^n]u[-(n+1)] -$$ - -This sequence appears in Fig. 5.29b. - -**(c)** In this case, 0.8 < |*z*| < 2; the part of *X*[*z*] corresponding to the pole at 0.8 is a causal sequence, and the part corresponding to the pole at 2 is an anticausal sequence: - -$$ -x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)] -$$ - -This sequence appears in Fig. 5.29c. - -### **DR ILL 5.23 Inverse Bilateral** *z***-Transform** - -Find the inverse bilateral *z*-transform of - -$$ -X[z] = \frac{z}{z^2 + \frac{5}{6}z + \frac{1}{6}} \qquad \frac{1}{2} > |z| > \frac{1}{3} -$$ - -### **ANSWER** - - − 1 3 *n u*[*n*] +6 − 1 2 *n u*[−(*n*+1)] - -INVERSE TRANSFORM BY EXPANSION OF *X*[*z*] IN POWER SERIES OF *z* We have - -$$ -X[z] = \sum_{n} x[n]z^{-n} -$$ - -For an anticausal sequence, which exists only for *n* ≤ −1, this equation becomes - -$$ -X[z] = x[-1]z + x[-2]z^{2} + x[-3]z^{3} + \cdots -$$ - -We can find the inverse *z*-transform of *X*[*z*] by dividing the numerator polynomial by the denominator polynomial, both in ascending powers of *z*, to obtain a polynomial in ascending powers of *z*. Thus, to find the inverse transform of *z*/(*z* − 0.5) (when the ROC is |*z*| < 0.5), we divide *z* by −0.5+*z* to obtain −2*z*−4*z*2−8*z*3−· · ·. Hence, *x*[−1]=−2, *x*[−2]=−4, *x*[−3]=−8, and so on. - -## **[5.8-1 Properties of the Bilateral](#page-11-0)** *z***-Transform** - -Properties of the bilateral *z*-transform are similar to those of the unilateral transform. We shall merely state the properties here, without proofs, for *xi*[*n*] ⇐⇒ *Xi*[*z*]. - -LINEARITY - -$$ -a_1x_1[n]+a_2x_2[n] \Longleftrightarrow a_1X_1[z]+a_2X_2[z] -$$ - -The ROC for *a*1*X*1[*z*] + *a*2*X*2[*z*] is the region common to (intersection of) the ROCs for *X*1[*z*] and *X*2[*z*]. - -SHIFT - -*x*[*n*−*m*] ⇐⇒ 1 *zm X*[*z*] *m* is positive or negative integer - -The ROC for *X*[*z*]/*zm* is the ROC for *X*[*z*] except for the addition or deletion of *z* = 0 or *z* = ∞ caused by the factor 1/*zm*. - -CONVOLUTION - -$$ -x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z] -$$ - -The ROC for *X*1[*z*]*X*2[*z*] is the region common to (intersection of) the ROCs for *X*1[*z*] and *X*2[*z*]. - -MULTIPLICATION BY γ *n* - -$$ -\gamma^n x[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right] -$$ - -If the ROC for *X*[*z*] is |γ1| < |*z*| < |γ2|, then the ROC for *X*[*z*/γ ] is |γ γ1| < |*z*| < |γ γ2|, indicating that the ROC is scaled by the factor |γ |. - -MULTIPLICATION BY *n* - -$$ -nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z] -$$ - -The ROC for −*z*(*dX*/*dz*) is the same as the ROC for *X*[*z*]. - -TIME REVERSAL - -$$ -x[-n] \Longleftrightarrow X[1/z] -$$ - -If the ROC for *X*[*z*] is |γ1| < |*z*| < |γ2|, then the ROC for *X*[1/*z*] is 1/|γ1| > |*z*| > |1/γ2|. - -COMPLEX CONJUGATION - -*x*∗[*n*] ⇐⇒ *X*∗[*z* ∗] - -The ROC for *X*∗[*z*∗] is the same as the ROC for *X*[*z*]. - -## **5.8-2 Using the Bilateral** *z***[-Transform for Analysis of LTID Systems](#page-11-0)** - -Because the bilateral *z*-transform can handle noncausal signals, we can use this transform to analyze noncausal linear systems. The zero-state response *y*[*n*] is given by - -$$ -y[n] = \mathcal{Z}^{-1}\{X[z]H[z]\} -$$ - -provided *X*[*z*]*H*[*z*] exists. The ROC of *X*[*z*]*H*[*z*] is the region in which both *X*[*z*] and *H*[*z*] exist, which means that the region is the common part of the ROC of both *X*[*z*] and *H*[*z*]. - -### **EXAMPLE 5.19 Zero-State Response by Bilateral** *z***-Transform** - -For a causal system specified by the transfer function - -$$ -H[z] = \frac{z}{z - 0.5} -$$ - -find the zero-state response to input - -$$ -x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)] -$$ - -$$ -X[z] = \frac{z}{z - 0.8} - \frac{2z}{z - 2} = \frac{-z(z + 0.4)}{(z - 0.8)(z - 2)} -$$ - -The ROC corresponding to the causal term is |*z*| > 0.8, and that corresponding to the anticausal term is |*z*| < 2. Hence, the ROC for *X*[*z*] is the common region, given by 0.8 < |*z*| < 2. Hence, - -$$ -X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)} \qquad 0.8 < |z| < 2 -$$ - -Therefore, - -$$ -Y[z] = X[z]H[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)} -$$ - -Since the system is causal, the ROC of *H*[*z*] is |*z*| > 0.5. The ROC of *X*[*z*] is 0.8 < |*z*| < 2. The common region of convergence for *X*[*z*] and *H*[*z*] is 0.8 < |*z*| < 2. Therefore, - -$$ -Y[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)} \qquad 0.8 < |z| < 2 -$$ - -Expanding *Y*[*z*] into modified partial fractions yields - -$$ -Y[z] = -\frac{z}{z - 0.5} + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) - \frac{8}{3} \left( \frac{z}{z - 2} \right) \qquad 0.8 < |z| < 2 -$$ - -Since the ROC extends outward from the pole at 0.8, both poles at 0.5 and 0.8 correspond to causal sequence. The ROC extends inward from the pole at 2. Hence, the pole at 2 corresponds to anticausal sequence. Therefore, - -$$ -y[n] = \left[ -(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] + \frac{8}{3}(2)^n u[-(n+1)] -$$ - -### **EXAMPLE 5.20 Zero-State Response for an Input with No** *z***-Transform** - -For the system in Ex. 5.19, find the zero-state response to input - -$$ -x[n] = \underbrace{(0.8)^n u[n]}_{x_1[n]} + \underbrace{(0.6)^n u[-(n+1)]}_{x_2[n]} -$$ - -The *z*-transforms of the causal and anticausal components *x*1[*n*] and *x*2[*n*] of the output are - -$$ -X_1[z] = \frac{z}{z - 0.8} \qquad |z| > 0.8 -$$ - -$$ -X_2[z] = \frac{-z}{z - 0.6} \qquad |z| < 0.6 -$$ - -Observe that a common ROC for *X*1[*z*] and *X*2[*z*] does not exist. Therefore, *X*[*z*] does not exist. In such a case we take advantage of the superposition principle and find *y*1[*n*] and *y*2[*n*], the system responses to *x*1[*n*] and *x*2[*n*], separately. The desired response *y*[*n*] is the sum of *y*1[*n*] and *y*2[*n*]. Now - -$$ -H[z] = \frac{z}{z - 0.5} -$$ - $|z| > 0.5$ -\n -$$ -Y_1[z] = X_1[z]H[z] = \frac{z^2}{(z - 0.5)(z - 0.8)} -$$ - $|z| > 0.8$ -\n -$$ -Y_2[z] = X_2[z]H[z] = \frac{-z^2}{(z - 0.5)(z - 0.6)} -$$ - $0.5 < |z| < 0.6$ - -Expanding *Y*1[*z*] and *Y*2[*z*] into modified partial fractions yields - -$$ -Y_1[z] = -\frac{5}{3} \left( \frac{z}{z - 0.5} \right) + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) \qquad |z| > 0.8 -$$ - -$$ -Y_2[z] = 5 \left( \frac{z}{z - 0.5} \right) - 6 \left( \frac{z}{z - 0.6} \right) \qquad 0.5 < |z| < 0.6 -$$ - -Therefore, - -$$ -y_1[n] = \left[ -\frac{5}{3}(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] -$$ - -$$ -y_2[n] = 5(0.5)^n u[n] + 6(0.6)^n u[-(n+1)] -$$ - -and - -$$ -y[n] = y_1[n] + y_2[n] = \left[\frac{10}{3}(0.5)^n + \frac{8}{3}(0.8)^n\right]u[n] + 6(0.6)^nu[-(n+1)] -$$ - -### **DR ILL 5.24 Zero-State Response by Bilateral** *z***-Transform** - -For the causal system in Ex. 5.19, find the zero-state response to input - -$$ -x[n] = \left(\frac{1}{4}\right)^n u[n] + 5(3)^n u[-(n+1)] -$$ - -**ANSWER** - -− 1 4 *n* +3 1 2 *n u*[*n*] +6(3)*nu*[−(*n*+1)] diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/074_5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/074_5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS.md deleted file mode 100644 index f6eb127f30965b232e7137db3ade9610a7772c65..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/074_5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS.md +++ /dev/null @@ -1,75 +0,0 @@ -## **[5.9 CONNECTING THE](#page-11-0) LAPLACE AND** *z***-TRANSFORMS** - -We now show that discrete-time systems also can be analyzed by means of the Laplace transform. In fact, we shall see that *the z-transform is the Laplace transform in disguise* and that discrete-time systems can be analyzed as if they were continuous-time systems. - -So far we have considered the discrete-time signal as a sequence of numbers and not as an electrical signal (voltage or current). Similarly, we considered a discrete-time system as a mechanism that processes a sequence of numbers (input) to yield another sequence of numbers (output). The system was built by using delays (along with adders and multipliers) that delay sequences of numbers. A digital computer is a perfect example: every signal is a sequence of numbers, and the processing involves delaying sequences of numbers (along with addition and multiplication). - -Now suppose we have a discrete-time system with transfer function *H*[*z*] and input *x*[*n*]. Consider a continuous-time signal *x*(*t*) such that its *n*th sample value is *x*[*n*], as shown in Fig. 5.30.† Let the sampled signal be *x*(*t*), consisting of impulses spaced *T* seconds apart with the *n*th impulse of strength *x*[*n*]. Thus, - -$$ -\bar{x}(t) = \sum_{n=0}^{\infty} x[n]\delta(t - nT) -$$ - -Figure 5.30 shows *x*[*n*] and the corresponding *x*(*t*). The signal *x*[*n*] is applied to the input of a discrete-time system with transfer function *H*[*z*], which is generally made up of delays, adders, and scalar multipliers. Hence, processing *x*[*n*] through *H*[*z*] amounts to operating on the sequence *x*[*n*] by means of delays, adders, and scalar multipliers. Suppose for *x*(*t*) samples, we perform operations identical to those performed on the samples of *x*[*n*] by *H*[*z*]. For this purpose, we need a continuous-time system with transfer function *H*(*s*) that is identical in structure to the discrete-time system *H*[*z*] except that the delays in *H*[*z*] are replaced by elements that delay continuous-time signals (such as voltages or currents). There is no other difference between realizations of *H*[*z*] and *H*(*s*). If a continuous-time impulse δ(*t*) is applied to such a delay of *T* seconds, the output will be δ(*t* −*T*). The continuous-time transfer function of such a delay is *e*−*sT* [see Eq. (4.30)]. Hence, the delay elements with transfer function 1/*z* in the realization of *H*[*z*] will be replaced by the delay elements with transfer function *e*−*sT* in the realization of the corresponding *H*(*s*). This is the same - -† We can construct such *x*(*t*) from the sample values, as will be explained in Ch. 8. - -**Figure 5.30** Connection between the Laplace transform and the *z*-transform. - -as *z* being replaced by *esT* . Therefore, *H*(*s*) = *H*[*esT* ]. Let us now apply *x*[*n*] to the input of *H*[*z*] and apply *x*(*t*) at the input of *H*[*esT* ]. Whatever operations are performed by the discrete-time system *H*[*z*] on *x*[*n*] (Fig. 5.30a) are also performed by the corresponding continuous-time system *H*[*esT* ] on the impulse sequence *x*(*t*) (Fig. 5.30b). The delaying of a sequence in *H*[*z*] would amount to delaying of an impulse train in *H*[*esT* ]. Adding and multiplying operations are the same in both cases. In other words, one-to-one correspondence of the two systems is preserved in every aspect. Therefore if *y*[*n*] is the output of the discrete-time system in Fig. 5.30a, then *y*(*t*), the output of the continuous-time system in Fig. 5.30b, would be a sequence of impulse whose *n*th impulse strength is *y*[*n*]. Thus, - -$$ -\bar{y}(t) = \sum_{n=0}^{\infty} y[n]\delta(t - nT) -$$ - -The system in Fig. 5.30b, being a continuous-time system, can be analyzed via the Laplace transform. If - -$$ -\overline{x}(t) \Longleftrightarrow \overline{X}(s) \quad \text{and} \quad \overline{y}(t) \Longleftrightarrow \overline{Y}(s) -$$ - -then - -$$ -\overline{Y}(s) = H[e^{sT}]\overline{X}(s) -$$ -\n(5.53) - -Also, - -$$ -\overline{X}(s) = \mathcal{L}\left[\sum_{n=0}^{\infty} x[n]\delta(t - nT)\right] -$$ - -Now because the Laplace transform of δ(*t* −*nT*) is *e*−*snT* , - -$$ -\overline{X}(s) = \sum_{n=0}^{\infty} x[n]e^{-snT} \quad \text{and} \quad \overline{Y}(s) = \sum_{n=0}^{\infty} y[n]e^{-snT} -$$ - -Substitution of these expressions into Eq. (5.53) yields - -$$ -\sum_{n=0}^{\infty} y[n]e^{-snT} = H[e^{sT}]\left[\sum_{n=0}^{\infty} x[n]e^{-snT}\right] -$$ - -By introducing a new variable *z* = *esT* , this equation can be expressed as - -$$ -\sum_{n=0}^{\infty} y[n]z^{-n} = H[z] \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - -or - -$$ -Y[z] = H[z]X[z] -$$ - -where - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - and $Y[z] = \sum_{n=0}^{\infty} y[n]z^{-n}$ - -It is clear from this discussion that the *z*-transform can be considered to be the Laplace transform with a change of variable *z* = *esT* or *s* = (1/*T*)ln*z*. Note that the transformation *z* = *esT* transforms the imaginary axis in the *s* plane (*s* = *j*ω) into a unit circle in the *z* plane (*z* = *esT* = *ej*ω*T* , or |*z*| = 1). The LHP and RHP in the *s*-plane map into the inside and the outside, respectively, of the unit circle in the *z* plane. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/075_5.10 MATLAB - DISCRETE-TIME IIR FILTERS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/075_5.10 MATLAB - DISCRETE-TIME IIR FILTERS.md deleted file mode 100644 index 3c407e10009d9834ab05247894ee4953c5798df0..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/075_5.10 MATLAB - DISCRETE-TIME IIR FILTERS.md +++ /dev/null @@ -1,321 +0,0 @@ -## **[5.10 MATLAB: DISCRETE-TIME](#page-11-0) IIR FILTERS** - -Recent technological advancements have dramatically increased the popularity of discrete-time filters. Unlike their continuous-time counterparts, the performance of discrete-time filters is not affected by component variations, temperature, humidity, or age. Furthermore, digital hardware is easily reprogrammed, which allows convenient change of device function. For example, certain digital hearing aids are individually programmed to match the required response of a user. - -Typically, discrete-time filters are categorized as infinite-impulse response (IIR) or finite-impulse response (FIR). A popular method to obtain a discrete-time IIR filter is by transformation of a corresponding continuous-time filter design. MATLAB greatly assists this process. Although discrete-time IIR filter design is the emphasis of this section, methods for discrete-time FIR filter design are considered in Sec. 9.7. - -### **[5.10-1 Frequency Response and Pole-Zero Plots](#page-11-0)** - -Frequency response and pole-zero plots help characterize filter behavior. Similar to continuous-time systems, rational transfer functions for realizable LTID systems are represented in the *z*-domain as - -$$ -H[z] = \frac{Y[z]}{X[z]} = \frac{B[z]}{A[z]} = \frac{\sum_{k=0}^{N} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} = \frac{\sum_{k=0}^{N} b_k z^{N-k}}{\sum_{k=0}^{N} a_k z^{N-k}} -$$ -(5.54) - -When only the first (*N*1 + 1) numerator coefficients are nonzero and only the first (*N*2 + 1) denominator coefficients are nonzero, Eq. (5.54) simplifies to - -$$ -H[z] = \frac{Y[z]}{X[z]} = \frac{B[z]}{A[z]} = \frac{\sum_{k=0}^{N_1} b_k z^{-k}}{\sum_{k=0}^{N_2} a_k z^{-k}} = \frac{\sum_{k=0}^{N_1} b_k z^{N_1 - k}}{\sum_{k=0}^{N_2} a_k z^{N_2 - k}} z^{N_2 - N_1} -$$ -(5.55) - -The form of Eq. (5.55) has many advantages. It can be more efficient than Eq. (5.54); it still works when *N*1 = *N*2 = *N*; and it more closely conforms to the notation of built-in MATLAB discrete-time signal-processing functions. - -The right-hand side of Eq. (5.55) is a form that is convenient for MATLAB computations. The frequency response *H*[*ej*] is obtained by letting *z* = *ej*, where has units of radians. Often, = ω*T*, where ω is the continuous-time frequency in radians per second and *T* is the sampling period in seconds. Defining length-(*N*2 + 1) coefficient vector **A** = [*a*0,*a*1,...,*aN*2 ] and length-(*N*1+1) coefficient vector **B** = [*b*0,*b*1,...,*bN*1 ], program CH5MP1 computes *H*[*ej*] by using Eq. (5.55) for each frequency in the input vector . - -``` -function [H] = CH5MP1(B,A,Omega); -% CH5MP1.m : Chapter 5, MATLAB Program 1 -% Function M-file computes frequency response for LTID systems -% INPUTS: B = vector of feedforward coefficients -% A = vector of feedback coefficients -% Omega = vector of frequencies [rad], typically -pi<=Omega<=pi -% OUTPUTS: H = frequency response -N_1 = length(B)-1; N_2 = length(A)-1; -H = polyval(B,exp(1j*Omega))./polyval(A,exp(1j*Omega)).*... - exp(1j*Omega*(N_2-N_1)); -``` - -Note that owing to MATLAB's indexing scheme, A(k) corresponds to coefficient *ak*−1 and B(k) corresponds to coefficient *bk*−1. It is also possible to use the signal-processing toolbox function freqz to evaluate the frequency response of a system described by Eq. (5.55). Under special circumstances, the control system toolbox function bode can also be used. - -Program CH5MP2 computes and plots the poles and zeros of an LTID system described by Eq. (5.55), again using vectors **B** and **A**. - -``` -function [p,z] = CH5MP2(B,A); -% CH5MP2.m : Chapter 5, MATLAB Program 2 -% Function M-file computes and plots poles and zeros for LTID systems -% INPUTS: B = vector of feedforward coefficients -% A = vector of feedback coefficients -N_1 = length(B)-1; N_2 = length(A)-1; -p = roots([A,zeros(1,N_1-N_2)]); z = roots([B,zeros(1,N_2-N_1)]); -ucirc = exp(1j*linspace(0,2*pi,200)); % Compute unit circle for plot -plot(real(p),imag(p),'xk',real(z),imag(z),'ok',real(ucirc),imag(ucirc),'k:'); -xlabel('Real'); ylabel('Imag'); -ax = axis; dx = 0.05*(ax(2)-ax(1)); dy = 0.05*(ax(4)-ax(3)); -axis(ax+[-dx,dx,-dy,dy]); axis equal; -``` - -The right-hand side of Eq. (5.55) helps explain how the roots are computed. When *N*1 = *N*2, the term *zN*2−*N*1 implies additional roots at the origin. If *N*1 > *N*2, the roots are poles, which are added by concatenating A with zeros(N\_1-N\_2,1); since *N*2 − *N*1 ≤ 0, zeros(N\_2-N\_1,1) produces the empty set and B is unchanged. If *N*2 > *N*1, the roots are zeros, which are added by concatenating B with zeros(N\_2-N\_1,1); since *N*1 − *N*2 ≤ 0, zeros(N\_1-N\_2,1) produces the empty set and A is unchanged. Poles and zeros are indicated with black x's and o's, respectively. For visual reference, the unit circle is also plotted. The last two lines in CH5MP2 expand the plot axis box so that root locations are not obscured and also ensure that the real and imaginary axes are drawn to the same scale. - -### **[5.10-2 Transformation Basics](#page-11-0)** - -Transformation of a continuous-time filter to a discrete-time filter begins with the desired continuous-time transfer function - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{B(s)}{A(s)} = \frac{\sum_{k=0}^{M} b_{k+N-M} s^{M-k}}{\sum_{k=0}^{N} a_k s^{N-k}} -$$ - -As a matter of convenience, *H*(*s*) is represented in factored form as - -$$ -H(s) = \frac{b_{N-M}}{a_0} \frac{\prod_{k=1}^{M} (s - z_k)}{\prod_{k=1}^{N} (s - p_k)} -$$ -(5.56) - -where *zk* and *pk* are the system poles and zeros, respectively. - -A mapping rule converts the rational function *H*(*s*) to a rational function *H*[*z*]. Requiring that the result be rational ensures that the system realization can proceed with only delay, sum, and multiplier blocks. There are many possible mapping rules. For obvious reasons, good transformations tend to map the ω axis to the unit circle, ω = 0 to *z* = 1, ω = ∞ to *z* = −1, and the left half-plane to the interior of the unit circle. Put another way, sinusoids map to sinusoids, zero frequency maps to zero frequency, high frequency maps to high frequency, and stable systems map to stable systems. - -### 568 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -Section 5.9 suggests that the *z*-transform can be considered to be a Laplace transform with a change of variable *z* = *esT* or *s* = (1/*T*)ln*z*, where *T* is the sampling interval. It is tempting, therefore, to convert a continuous-time filter to a discrete-time filter by substituting *s* = (1/*T*)ln*z* into *H*(*s*), or *H*[*z*] = *H*(*s*)|*s*=(1/*T*)ln*z*. This approach is impractical, however, since the resulting *H*[*z*] is not rational and therefore cannot be implemented by using standard blocks. Although not considered here, the so-called matched-*z* transformation relies on the relationship *z* = *esT* to transform system poles and zeros, so the connection is not completely without merit. - -### **[5.10-3 Transformation by First-Order Backward Difference](#page-11-0)** - -Consider the transfer function *H*(*s*) = *Y*(*s*)/*X*(*s*) = *s*, which corresponds to the first-order continuous-time differentiator - -$$ -y(t) = \frac{d}{dt}x(t) -$$ - -An approximation that resembles the fundamental theorem of calculus is the first-order backward difference - -$$ -y(t) = \frac{x(t) - x(t - T)}{T} -$$ - -For sampling interval *T* and *t* = *nT*, the corresponding discrete-time approximation is - -$$ -y[n] = \frac{x[n] - x[n-1]}{T} -$$ - -which has transfer function - -$$ -H[z] = Y[z]/X[z] = \frac{1 - z^{-1}}{T} -$$ - -This implies a transformation rule that uses the change of variable *s* = (1−*z*−1)/*T* or *z* = 1/(1−*sT*). This transformation rule is appealing since the resulting *H*[*z*] is rational and has the same number of poles and zeros as *H*(*s*). Section 3.4 discusses this transformation strategy in a different way in describing the kinship of difference equations to differential equations. - -After some algebra, substituting *s* = (1−*z*−1)/*T* into Eq. (5.56) yields - -$$ -H[z] = \left(\frac{b_{N-M} \prod_{k=1}^{M} (1/T - z_k)}{a_0 \prod_{k=1}^{N} (1/T - p_k)}\right) \frac{\prod_{k=1}^{M} \left(1 - \frac{1}{1 - Tz_k} z^{-1}\right)}{\prod_{k=1}^{N} \left(1 - \frac{1}{1 - Tp_k} z^{-1}\right)} -$$ -(5.57) - -The discrete-time system has *M* zeros at 1/(1−*Tzk*) and *N* poles at 1/(1−*Tpk*). This transformation rule preserves system stability but does not map the ω axis to the unit circle (see Prob. 5.7-10). - -MATLAB program CH5MP3 uses the first-order backward difference method of Eq. (5.57) to convert a continuous-time filter described by coefficient vectors **A** = [*a*0,*a*1,...,*aN*] and **B** = [*bN*−*M*,*bN*−*M*+1,...,*bN*] into a discrete-time filter. The form of the discrete-time filter follows Eq. (5.55). - -function [Bd,Ad] = CH5MP3(B,A,T); % CH5MP3.m : Chapter 5, MATLAB Program 3 % Function M-file first-order backward difference transformation - -``` -% of a continuous-time filter described by B and A into a discrete-time filter. -% INPUTS: B = vector of continuous-time filter feedforward coefficients -% A = vector of continuous-time filter feedback coefficients -% T = sampling interval -% OUTPUTS: Bd = vector of discrete-time filter feedforward coefficients -% Ad = vector of discrete-time filter feedback coefficients -z = roots(B); p = roots(A); % s-domain roots -gain = B(1)/A(1)*prod(1/T-z)/prod(1/T-p); -zd = 1./(1-T*z); pd = 1./(1-T*p); % z-domain roots -Bd = gain*poly(zd); Ad = poly(pd); -``` - -### **[5.10-4 Bilinear Transformation](#page-11-0)** - -The bilinear transformation is based on a better approximation than first-order backward differences. Again, consider the continuous-time integrator - -$$ -y(t) = \frac{d}{dt}x(t) -$$ - -Represent signal *x*(*t*) as - -$$ -x(t) = \int_{t-T}^{t} \frac{d}{d\tau} x(\tau) d\tau + x(t-T) -$$ - -Letting *t* = *nT* and replacing the integral with a trapezoidal approximation yield - -$$ -x(nT) = \frac{T}{2} \left[ \frac{d}{dt} x(nT) + \frac{d}{dt} x(nT - T) \right] + x(nT - T) -$$ - -Substituting *y*(*t*) for (*d*/*dt*)*x*(*t*), the equivalent discrete-time system is - -$$ -x[n] = \frac{T}{2}(y[n] + y[n-1]) + x[n-1] -$$ - -From *z*-transforms, the transfer function is - -$$ -H[z] = \frac{Y[z]}{X[z]} = \frac{2(1 - z^{-1})}{T(1 + z^{-1})} -$$ - -The implied change of variable *s* = 2(1−*z*−1)/*T*(1+*z*−1) or *z* = (1+*sT*/2)/(1−*sT*/2) is called the bilinear transformation. Not only does the bilinear transformation result in a rational function *H*[*z*], the ω axis is correctly mapped to the unit circle (see Prob. 5.6-18a). - -After some algebra, substituting *s* = 2(1−*z*−1)/*T*(1+*z*−1) into Eq. (5.56) yields - -$$ -H[z] = \left(\frac{b_{N-M} \prod_{k=1}^{M} (2/T - z_k)}{a_0 \prod_{k=1}^{N} (2/T - p_k)}\right) \frac{\prod_{k=1}^{M} \left(1 - \frac{1 + z_k T/2}{1 - z_k T/2} z^{-1}\right)}{\prod_{k=1}^{N} \left(1 - \frac{1 + p_k T/2}{1 - p_k T/2} z^{-1}\right)} (1 + z^{-1})^{N-M} \tag{5.58} -$$ - -In addition to the *M* zeros at (1+*zkT*/2)/(1−*zkT*/2) and *N* poles at (1+*pkT*/2)/(1−*pkT*/2), there are *N*−*M* zeros at minus 1. Since practical continuous-time filters require *M* ≤ *N* for stability, the number of added zeros is thankfully always nonnegative. - -MATLAB program CH5MP4 converts a continuous-time filter described by coefficient vectors **A** = [*a*0,*a*1,...,*aN*] and **B** = [*bN*−*M*,*bN*−*M*+1,...,*bN*] into a discrete-time filter by using the bilinear transformation of Eq. (5.58). The form of the discrete-time filter follows Eq. (5.55). If available, it is also possible to use the signal-processing toolbox function bilinear to perform the bilinear transformation. - -``` -function [Bd,Ad] = CH5MP4(B,A,T); -% CH5MP4.m : Chapter 5, MATLAB Program 4 -% Function M-file bilinear transformation of a continuous-time filter -% described by vectors B and A into a discrete-time filter. -% Length of B must not exceed A. -% INPUTS: B = vector of continuous-time filter feedforward coefficients -% A = vector of continuous-time filter feedback coefficients -% T = sampling interval -% OUTPUTS: Bd = vector of discrete-time filter feedforward coefficients -% Ad = vector of discrete-time filter feedback coefficients -if (length(B)>length(A)), - disp('Numerator order must not exceed denominator order.'); - return -end -z = roots(B); p = roots(A); % s-domain roots -gain = real(B(1)/A(1)*prod(2/T-z)/prod(2/T-p)); -zd = (1+z*T/2)./(1-z*T/2); pd = (1+p*T/2)./(1-p*T/2); % z-domain roots -Bd = gain*poly([zd;-ones(length(A)-length(B),1)]); Ad = poly(pd); -As with most high-level languages, MATLAB supports general if-structures: -``` - -``` -if expression, - statements; -elseif expression, - statements; -else, - statements; -end -``` - -In the program CH5MP4, the if statement tests *M* > *N*. When true, an error message is displayed and the return command terminates program execution to prevent errors. - -### **[5.10-5 Bilinear Transformation with Prewarping](#page-11-0)** - -The bilinear transformation maps the entire infinite-length ω axis onto the finite-length unit circle (*z* = *ej*) according to ω = (2/*T*)tan(/2) (see Prob. 5.6-18b). Equivalently, = 2arctan(ω*T*/2). The nonlinearity of the tangent function causes a frequency compression, commonly called frequency warping, that distorts the transformation. - -To illustrate the warping effect, consider the bilinear transformation of a continuous-time lowpass filter with cutoff frequency ω*c* = 2π3000 rad/s. If the target digital system uses a sampling rate of 10 kHz, then *T* = 1/(10,000) and ω*c* maps to *c* = 2arctan(ω*cT*/2) = 1.5116. Thus, the transformed cutoff frequency is short of the desired *c* = ω*cT* = 0.6π = 1.8850. - -Cutoff frequencies are important and need to be as accurate as possible. By adjusting the parameter *T* used in the bilinear transform, one continuous-time frequency can be exactly mapped to one discrete-time frequency; the process is called prewarping. Continuing the last example, adjusting *T* = (2/ω*c*)tan(*c*/2) ≈ 1/6848 achieves the appropriate prewarping to ensure ω*c* = 2π3000 maps to *c* = 0.6π. - -### **[5.10-6 Example: Butterworth Filter Transformation](#page-11-0)** - -To illustrate the transformation techniques, consider a continuous-time 10th-order Butterworth lowpass filter with cutoff frequency ω*c* = 2π3000, as designed in Sec. 4.12. First, we determine continuous-time coefficient vectors **A** and **B**. - -``` ->> omega_c = 2*pi*3000; N=10; ->> poles = roots([(1j*omega_c)^(-2*N),zeros(1,2*N-1),1]); ->> poles = poles(find(poles<0)); ->> B = 1; A = poly(poles); A = A/A(end); -``` - -Programs CH5MP3 and CH5MP4 are used to perform first-order forward difference and bilinear transformations, respectively. - -``` ->> Omega = linspace(0,pi,200); T = 1/10000; Omega_c = omega_c*T; ->> [B1,A1] = CH5MP3(B,A,T); % First-order backward difference transformation ->> [B2,A2] = CH5MP4(B,A,T); % Bilinear transformation ->> [B3,A3] = CH5MP4(B,A,2/omega_c*tan(Omega_c/2)); % Bilinear with prewarping -``` - -Magnitude responses are computed using CH5MP1 and then plotted. - -``` ->> H1mag = abs(CH5MP1(B1,A1,Omega)); ->> H2mag = abs(CH5MP1(B2,A2,Omega)); ->> H3mag = abs(CH5MP1(B3,A3,Omega)); ->> plot(Omega,(Omega<=Omega_c),'k',Omega,H1mag,'k-.',... ->> Omega,H2mag,'k--',Omega,H3mag,'k:'); ->> axis([0 pi -.05 1.5]); ->> xlabel('\Omega [rad]'); ylabel('Magnitude Response'); ->> legend('Ideal','FOBD','BLT','Prewarp BLT','location','best'); -``` - -The result of each transformation method is shown in Fig. 5.31, where FOBD and BLT stand for first-order backward difference and bilinear transformation, respectively. - -Although the first-order backward difference results in a lowpass filter, the method causes significant distortion that makes the resulting filter unacceptable with regard to cutoff frequency. The bilinear transformation is better, but, as predicted, the cutoff frequency falls short of the desired value. Bilinear transformation with prewarping properly locates the cutoff frequency and produces a very acceptable filter response. - -**Figure 5.31** Comparison of various transformation techniques. - -### **[5.10-7 Problems Finding Polynomial Roots](#page-12-0)** - -Numerically, it is difficult to accurately determine the roots of a polynomial. Consider, for example, a simple polynomial that has a root at minus 1 repeated four times, (*s* + 1)4 = *s*4 + 4*s*3 +6*s*2 +4*s*+1. The MATLAB roots command returns a surprising result: - -``` ->> roots([1464 1])' - ans = -1.0002 -1.0000-0.0002i -1.0000+0.0002i -0.9998 -``` - -Even for this low-degree polynomial, MATLAB does not return the true roots. - -The problem worsens as polynomial degree increases. The bilinear transformation of the 10th-order Butterworth filter, for example, should have 10 zeros at minus 1. Figure 5.32 shows that the zeros, computed by CH5MP2 with the roots command, are not correctly located. - -When possible, programs should avoid root computations that may limit accuracy. For example, results from the transformation programs CH5MP3 and CH5MP4 are more accurate if the true transfer function poles and zeros are passed directly as inputs rather than the polynomial coefficient vectors. When roots must be computed, result accuracy should always be verified. - -### **[5.10-8 Using Cascaded Second-Order Sections to Improve Design](#page-12-0)** - -The dynamic range of high-degree polynomial coefficients is often large. Adding the difficulties associated with factoring a high-degree polynomial, it is little surprise that high-order designs are difficult. - -As with continuous-time filters, performance is improved by using a cascade of second-order sections to design and realize a discrete-time filter. Cascades of second-order sections are also more robust to the coefficient quantization that occurs when discrete-time filters are implemented on fixed-point digital hardware. - -To illustrate the performance possible with a cascade of second-order sections, consider a 180th-order transformed Butterworth discrete-time filter with cutoff frequency *c* = 0.6π ≈ 1.8850. Program CH5MP5 completes this design, taking care to initially locate poles and zeros without root computations. - -**Figure 5.32** Pole-zero plot computed by using roots. - -``` -% CH5MP5.m : Chapter 5, MATLAB Program 5 -% Script M-file designs a 180th-order Butterworth lowpass discrete-time filter -% with cutoff Omega_c = 0.6*pi using 90 cascaded second-order filter sections. -omega_0 = 1; % Use normalized cutoff frequency for analog prototype -psi = [0.5:1:90]*pi/180; % Butterworth pole angles -Omega_c = 0.6*pi; % Discrete-time cutoff frequency -Omega = linspace(0,pi,1000); % Frequency range for magnitude response -Hmag = zeros(90,1000); p = zeros(1,180); z = zeros(1,180); % Pre-allocation -for stage = 1:90, - Q = 1/(2*cos(psi(stage))); % Compute Q for stage - B = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute stage coefficients - [B1,A1] = CH5MP4(B,A,2/omega_0*tan(0.6*pi/2)); % Transform stage to DT - p(stage*2-1:stage*2) = roots(A1); % Compute z-domain poles for stage - z(stage*2-1:stage*2) = roots(B1); % Compute z-domain zeros for stage - Hmag(stage,:) = abs(CH5MP1(B1,A1,Omega)); % Compute stage mag response -end -ucirc = exp(j*linspace(0,2*pi,200)); % Compute unit circle for pole-zero plot -figure; -plot(real(p),imag(p),'kx',real(z),imag(z),'ok',real(ucirc),imag(ucirc),'k:'); -axis equal; xlabel('Real'); ylabel('Imag'); -figure; plot(Omega,prod(Hmag),'k'); axis([0 pi -0.05 1.05]); -xlabel('\Omega [rad]'); ylabel('Magnitude Response'); -``` - -The figure command preceding each plot command opens a separate window for each plot. - -The filter's pole-zero plot is shown in Fig. 5.33, along with the unit circle, for reference. All 180 zeros of the cascaded design are properly located at minus 1. The wall of poles provides an amazing approximation to the desired brick-wall response, as shown by the magnitude response in Fig. 5.34. It is virtually impossible to realize such high-order designs with continuous-time filters, which adds another reason for the popularity of discrete-time filters. Still, the design is not - -**Figure 5.33** Pole-zero plot for 180th-order discrete-time Butterworth filter. - -**Figure 5.34** Magnitude response for a 180th-order discrete-time Butterworth filter. - -trivial; even functions from the MATLAB signal-processing toolbox fail to properly design such a high-order discrete-time Butterworth filter. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/076_5.11 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/076_5.11 SUMMARY.md deleted file mode 100644 index d7244271fe65f7f5d317607bb5d0800103f95176..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/076_5.11 SUMMARY.md +++ /dev/null @@ -1,59 +0,0 @@ -## **[5.11 SUMMARY](#page-12-0)** - -In this chapter we discussed the analysis of linear, time-invariant, discrete-time (LTID) systems by means of the *z*-transform. The *z*-transform changes the difference equations of LTID systems into algebraic equations. Therefore, solving these difference equations reduces to solving algebraic equations. - -The transfer function *H*[*z*] of an LTID system is equal to the ratio of the *z*-transform of the output to the *z*-transform of the input when all initial conditions are zero. Therefore, if *X*[*z*] is the *z*-transform of the input *x*[*n*] and *Y*[*z*] is the *z*-transform of the corresponding output *y*[*n*] (when all initial conditions are zero), then *Y*[*z*] = *H*[*z*]*X*[*z*]. For an LTID system specified by the difference equation *Q*[*E*]*y*[*n*] = *P*[*E*]*x*[*n*], the transfer function *H*[*z*] = *P*[*z*]/*Q*[*z*]. Moreover, *H*[*z*] is the *z*-transform of the system impulse response *h*[*n*]. We showed in Ch. 3 that the system response to an everlasting exponential *zn* is *H*[*z*]*zn*. - -We may also view the *z*-transform as a tool that expresses a signal *x*[*n*] as a sum of exponentials of the form *zn* over a continuum of the values of *z*. Using the fact that an LTID system response to *zn* is *H*[*z*]*zn*, we find the system response to *x*[*n*] as a sum of the system's responses to all the components of the form *zn* over the continuum of values of *z*. - -LTID systems can be realized by scalar multipliers, adders, and time delays. A given transfer function can be synthesized in many different ways. We discussed canonical, transposed canonical, cascade, and parallel forms of realization. The realization procedure is identical to that for continuous-time systems with 1/*s* (integrator) replaced by 1/*z* (unit delay). - -The majority of the input signals and practical systems are causal. Consequently, we are required to deal with causal signals most of the time. Restricting all signals to the causal type greatly simplifies *z*-transform analysis; the ROC of a signal becomes irrelevant to the analysis process. This special case of *z*-transform (which is restricted to causal signals) is called the unilateral *z*-transform. Much of the chapter deals with this transform. Section 5.8 discusses the general variety of the *z*-transform (bilateral *z*-transform), which can handle causal and noncausal signals and systems. In the bilateral transform, the inverse transform of *X*[*z*] is not unique, but depends on the ROC of *X*[*z*]. Thus, the ROC plays a crucial role in the bilateral *z*-transform. - -In Sec. 5.9, we showed that discrete-time systems can be analyzed by the Laplace transform as if they were continuous-time systems. In fact, we showed that the *z*-transform is the Laplace transform with a change in variable. - -### **[REFERENCES](#page-12-0)** - -- 1. Lyons, R. G. *Understanding Digital Signal Processing*. Addison-Wesley, Reading, MA, 1997. -- 2. Oppenheim, A. V., and R. W. Schafer. *Discrete-Time Signal Processing*, 2nd ed. Prentice-Hall, Upper Saddle River, NJ, 1999. -- 3. Mitra, S. K. *Digital Signal Processing*, 2nd ed. McGraw-Hill, New York, 2001. - -## **[PROBLEMS](#page-12-0)** - -- **5.1-1** Using the definition, compute the *z*-transform of *x*[*n*] = (−1)*n*(*u*[*n*] − *u*[*n* − 8]). Sketch the poles and zeros of *X*[*z*] in the *z* plane. No calculator is needed to do this problem! -- **5.1-2** Determine the unilateral *z*-transform *X*[*z*] of the signal *x*[*n*] shown in Fig. P5.1-2. As the picture suggests, *x*[*n*]=−3 for all *n* ≥ 9 and *x*[*n*] = 0 for all *n* < 3. -- **5.1-3** (a) A causal signal has *z*-transform given by *X*[*z*] = *z*2 *z*3−1 . Determine the time-domain signal *x*[*n*] and sketch *x*[*n*] over −4 ≤ *n* ≤ 11. [*Hint:* No complex arithmetic is needed to solve this problem!] - -**Figure P5.1-2** - -### 576 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -- (b) Consider the causal semiperiodic signal *y*[*n*] shown in Fig. P5.1-3. Notice, *y*[*n*] continually repeats the sequence [1, 2, 3] for *n* ≥ 0. Determine the unilateral *z*-transform *Y*[*z*] of this signal. If possible, express your result as a rational function in standard form. -- **5.1-4** Using the definition of the *z*-transform, find the *z*-transform and the ROC for each of the following signals. - - (a) *u*[*n*− *m*] - - (b) γ *n* sinπ*n u*[*n*] - - (c) γ *n* cosπ*n u*[*n*] - -**Figure P5.1-3** - -(d) -$$ -\gamma^n \sin \frac{\pi n}{2} u[n] -$$ - -\n(e) $\gamma^n \cos \frac{\pi n}{2} u[n]$ -\n(f) $\sum_{k=0}^{\infty} 2^{2k} \delta[n-2k]$ -\n(g) $\gamma^{n-1} u[n-1]$ - -(h) *n*γ *n u*[*n*] - -$$ -(i) \; n \, u[n] -$$ - -(j) -$$ -\frac{\gamma^n}{n!}u[n] -$$ - -(k) \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/077_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/077_REFERENCES.md deleted file mode 100644 index 77f9e80c4653d680a6631213bcf4c99024ada98b..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/077_REFERENCES.md +++ /dev/null @@ -1,951 +0,0 @@ -$$ -[2^{n-1} - (-2)^{n-1}]u[n] -$$ - -(l) $\frac{(\ln \alpha)^n}{n!}u[n]$ - -- **5.1-5** Showing all work, evaluate \$∞ *n*=0 *n*(−3/2)−*n*. -- **5.1-6** Using only the *z*-transforms of Table 5.1, determine the *z*-transform of each of the following signals. - -(a) *u*[*n*] −*u*[*n*−2] (b) γ *n*−2*u*[*n*− 2] (c) 2*n*+1*u*[*n*− 1] +*en*−1*u*[*n*] (d) 2−*n* cosπ 3 *n u*[*n*−1] (e) *n*γ *nu*[*n*− 1] (f) *n*(*n*−1)(*n*−2)2*n*−3*u*[*n*−*m*] for *m* = 0, 1, 2, 3 (g) (−1)*nnu*[*n*] - -(h) -$$ -\sum_{k=0}^{\infty} k\delta(n-2k+1) -$$ - -**5.1-7** Find the inverse unilateral *z*-transform of each of the following: (a) *z*(*z* −4) - -(a) -$$ -z^2 - 5z + 6 -$$ - -\n(b) $\frac{z-4}{z^2 - 5z + 6}$ -\n(c) $\frac{(e^{-2} - 2)z}{(z - e^{-2})(z - 2)}$ -\n(d) $\frac{(z - 1)^2}{z^3}$ -\n(e) $\frac{z(2z + 3)}{(z - 1)(z^2 - 5z + 6)}$ -\n(f) $\frac{z(-5z + 22)}{(z + 1)(z - 2)^2}$ -\n(g) $\frac{z(1.4z + 0.08)}{(z - 0.2)(z - 0.8)^2}$ - -(h) -$$ -\frac{z(z-2)}{z^2 - z + 1} -$$ - -\n(i) -$$ -\frac{2z^2 - 0.3z + 0.25}{z^2 + 0.6z + 0.25} -$$ - -\n(j) -$$ -\frac{2z(3z-23)}{(z-1)(z^2 - 6z + 25)} -$$ - -\n(k) -$$ -\frac{z(3.83z + 11.34)}{(z-2)(z^2 - 5z + 25)} -$$ - -\n(k) -$$ -\frac{z^2(-2z^2 + 8z - 7)}{(z^2 - 2z^2 + 8z - 7)} -$$ - -$$ -(1) \frac{z^2(-2z^2+8z-7)}{(z-1)(z-2)^3} -$$ - -**5.1-8** (a) Expanding *X*[*z*] as a power series in *z*−1, find the first three terms of *x*[*n*] if - -$$ -X[z] = \frac{2z^3 + 13z^2 + z}{z^3 + 7z^2 + 2z + 1} -$$ - -(b) Extend the procedure used in part (a) to find the first four terms of *x*[*n*] if - -$$ -X[z] = \frac{2z^4 + 16z^3 + 17z^2 + 3z}{z^3 + 7z^2 + 2z + 1} -$$ - -- **5.1-9** A right-sided signal *x*[*n*] has *z*-transform given by *X*[*z*] = *z*6+2*z*5+3*z*4+4*z*3 *z*4−1 . Using a power series expansion of *X*[*z*], determine *x*[*n*] over −5 ≤ *n* ≤ 5. -- **5.1-10** Find *x*[*n*] by expanding - -$$ -X[z] = \frac{\gamma z}{(z - \gamma)^2} -$$ - -as a power series in *z*−1. - -- **5.1-11** (a) In Table 5.1, if the numerator and the denominator powers of *X*[*z*] are *M* and *N*, respectively, explain why in some cases *N* − *M* = 0, while in others *N* − *M* = 1 or *N* − *M* = *m* (*m* any positive integer). - - (b) Without actually finding the *z*-transform, state what is *N* − *M* for *X*[*z*] corresponding to *x*[*n*] = γ *nu*[*n*−4]. -- **5.2-1** For a discrete-time signal shown in Fig. P5.2-1, show that - -$$ -X[z] = \frac{1 - z^{-m}}{1 - z^{-1}} -$$ - -Find your answer by using the definition in Eq. (5.1) and by using Table 5.1 and an appropriate property of the *z*-transform. - -**Figure P5.2-1** - -- **5.2-2** Determine the unilateral *z*-transform of signal *x*[*n*] = (1−*n*) cos π 2 (*n*−1) *u*[*n*−1]. -- **5.2-3** Suppose a DT signal *x*[*n*] = 2(*u*[*n*−10] −*u*[*n*− 6]) has a transform *X*(*z*). Define *Y*(*z*) = 1 2*z*−3 *d dzX*(2*z*). Using graphic plot or vector notation, determine the corresponding signal *y*[*n*]. -- **5.2-4** Suppose a DT signal *x*[*n*] = 3(*u*[*n*] −*u*[*n*−5]) has a transform *X*(*z*). Define *Y*(*z*) = 2*z*−4 *d dzX z* 2 . Using graphic plot or vector notation, determine the corresponding signal *y*[*n*]. -- **5.2-5** Find the *z*-transform of the signal illustrated in Fig. P5.2-5. Solve this problem in two ways, as in Exs. 5.2d and 5.4. Verify that the two answers are equivalent. -- **5.2-6** Using *z*-transform techniques and properties (no time-domain convolution sum!), determine the convolution *y*[*n*] =( 1 2 )*nu*[*n*−3]∗( 1 3 )*n*−6*u*[*n*−4]. Express your answer in the form *y*[*n*] = *c*1γ *n*−*N*1 1 *u*[*n* − *N*1] + *c*2γ *n*−*N*2 2 *u*[*n* − *N*2], making sure to clearly identify the constants *c*1, *c*2, γ1, γ2, *N*1, and *N*2. - -**Figure P5.2-5** - -**5.2-7** Determine the inverse unilateral *z*-transform *x*[*n*] of the signal - -$$ -X[z] = \frac{d^7}{dz^7} \left[ \frac{z^{-4}}{(z - \frac{1}{2})(z + 3)} \right] -$$ - -- **5.2-8** Using only the fact that γ *nu*[*n*]⇐⇒*z*/(*z*−γ ) and properties of the *z*-transform, find the *z*-transform of each of the following: - - (a) *n*2*u*[*n*] - - (b) *n*2γ *nu*[*n*] - - (c) *n*3*u*[*n*] - - (d) *an*[*u*[*n*] −*u*[*n*−*m*]] - - (e) *ne*−2*nu*[*n*−*m*] - - (f) (*n*−2)(0.5)*n*−3 *u*[*n*−4] -- **5.2-9** Using only pair 1 in Table 5.1 and appropriate properties of the *z*-transform, derive iteratively pairs 2 through 9. In other words, first derive pair 2. Then use pair 2 (and pair 1, if needed) to derive pair 3, and so on. -- **5.2-10** Find the *z*-transform of cos(π*n*/4)*u*[*n*] using only pairs 1 and 11b in Table 5.1 and a suitable property of the *z*-transform. -- **5.2-11** Apply the time-reversal property to pair 6 of Table 5.1 to show that γ *nu*[−(*n* + 1)] ⇐⇒ −*z*/(*z*−γ ) and the ROC is given by |*z*| < |γ |. -- **5.2-12** (a) If *x*[*n*] ⇐⇒ *X*[*z*], then show that (−1)*nx*[*n*] ⇐⇒ *X*[−*z*]. - - (b) Use this result to show that (−γ )*nu*[*n*] ⇐⇒ *z*/(*z* +γ ). - - (c) Use these results to find the *z*-transforms of *x*i[*n*]=[2*n*−1 −(−2)*n*−1]*u*[*n*] and *x*−ii[*n*] = γ *n* cosπ*n u*[*n*] -- **5.2-13** (a) If *x*[*n*] ⇐⇒ *X*[*z*], then show that - -$$ -\sum_{k=0}^{n} x[k] \Longleftrightarrow \frac{zX[z]}{z-1} -$$ - -### 578 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -- (b) Use this result to derive pair 2 from pair 1 in Table 5.1. -- **5.2-14** A number of causal time-domain functions are shown in Fig. P5.2-14. List the function of time that corresponds to each of the following functions of *z*. Few or no calculations are necessary! Be careful, the graphs may be scaled differently. - -(a) -$$ -\frac{z^2}{(z-0.75)^2} -$$ - -\n(b) $\frac{z^2 - 0.9z/\sqrt{2}}{z^2 - 0.9\sqrt{2}z + 0.81}$ -\n(c) $\sum_{k=1}^{4} z^{-2k}$ - -$$ -k=0 -$$ -\n(d) - -\n -$$ -\frac{z^{-5}}{1-z^{-1}} -$$ - -(e) -$$ -\frac{z^2}{z^4 - 1} -$$ - -(c) 0.75z - -(f) -$$ -\frac{1}{(z-0.75)^2} -$$ - -\n(g) $\frac{z^2 - z/\sqrt{2}}{z-\sqrt{2}}$ - -(g) -$$ -\frac{z^2 - \sqrt{2}z + 1}{z^2 - 5z^{-5} + 4z^{-6}} -$$ - -\n(h) -$$ -\frac{z^{-1} - 5z^{-5} + 4z^{-6}}{5(1 - z^{-1})^2} -$$ - -\n(i) -$$ -\frac{z}{z - 1.1} -$$ - -(j) -$$ -\frac{0.25z^{-1}}{(1-z^{-1})(1-0.75z^{-1})} -$$ - -- **5.2-15** Suppose we upsample a causal signal *x*[*n*] by factor *N* to produce signal *y*[*n*]. Express *Y*(*z*) in terms of *X*(*z*), taking care to mathematically justify your result. -- **5.3-1** Using *z*-transform techniques, find the output *y*[*n*] of an LTID system specified by the equation *y*[*n*]− 1 3 *y*[*n*−1] = *x*[*n*−1] if the initial condition is *y*[−1] = 2 and the input is *x*[*n*]=−*u*[*n*]. -- **5.3-2** Consider an LTID system *y*[*n*] − *y*[*n* − 2] = *x*[*n*] with *y*[−1] = 0, *y*[−2] = 1, and *x*[*n*] = *u*[*n*]. - - (a) Determine *Y*[*z*], expressed as a rational function in standard factored form. - - (b) Use *Y*[*z*] from part (a) to solve for the system output *y*[*n*]. -- **5.3-3** Solve Prob. 3.8-23 by the *z*-transform method. -- **5.3-4** Consider a DT system with transfer function *H*[*z*] = 2*z*−2 *z*−0.5 . Assuming the system is both controllable and observable, determine the ZIR *y*zir[*n*] given *y*[−1] = 1. -- **5.3-5** (a) Solve - -$$ -y[n+1] + 2y[n] = x[n+1] -$$ - -when -$$ -y[0] = 1 -$$ - and $x[n] = e^{-(n-1)}u[n]$ - -- (b) Find the zero-input and the zero-state components of the response. -- **5.3-6** Consider a LTID system that is described by the difference equation *y*[*n*] − 1 4 *y*[*n* − 2] = *x*[*n*−1]. - -- (a) Use transform-domain techniques to determine the zero-state response *y*zsr[*n*] to input *x*[*n*] = 3*u*[*n*−5]. -- (b) Use transform-domain techniques to determine the zero-input response *y*zir[*n*] given *y*zir[−2] = *y*zir[−1] = 1. -- **5.3-7** (a) Find the output *y*[*n*] of an LTID system specified by the equation - -$$ -2y[n+2] - 3y[n+1] + y[n] -$$ - -= $4x[n+2] - 3x[n+1]$ - -for input *x*[*n*] = (4)−*nu*[*n*] and initial conditions *y*[−1] = 0 and *y*[−2] = 1. - -- (b) Find the zero-input and the zero-state components of the response. -- (c) Find the transient and the steady-state components of the response. -- **5.3-8** Solve Prob. 5.3-7 if initial conditions *y*[−1] and *y*[−2] are instead replaced with auxiliary conditions *y*[0] = 3/2 and *y*[1] = 35/4. -- **5.3-9** (a) Solve - -$$ -4y[n+2] + 4y[n+1] + y[n] = x[n+1] -$$ - -with *y*[−1] = 0, *y*[−2] = 1, and *x*[*n*] = *u*[*n*]. - -- (b) Find the zero-input and the zero-state components of the response. -- (c) Find the transient and the steady-state components of the response. -- **5.3-10** Solve - -$$ -y[n+2] - 3y[n+1] + 2y[n] = x[n+1] -$$ - -if *y*[−1] = 2, *y*[−2] = 3, and *x*[*n*] = (3)*nu*[*n*]. - -**5.3-11** Solve - -$$ -y[n+2] - 2y[n+1] + 2y[n] = x[n] -$$ - -with *y*[−1] = 1, *y*[−2] = 0, and *x*[*n*] = *u*[*n*]. - -- **5.3-12** Consider a causal LTID system described as *H*(*z*) = 21(*z*2+1) 16(*z*2+ 1 4 *z*− 3 8 ) . - - (a) Determine the standard delay-form difference equation description of this system. - - (b) Using transform-domain techniques, determine the system impulse response *h*[*n*]. - - (c) Using transform-domain techniques, determine *y*zir[*n*] given *y*[−1] = 16 and *y*[−2] = 8. - -- **5.3-13** Consider a causal LTID system described as *y*[*n*] − 5 6 *y*[*n* − 1] + 1 6 *y*[*n* − 2] = 3 2 *x*[*n* − 1] + 3 2 *x*[*n*−2]. - - (a) Determine the (standard-form) system transfer function *H*(*z*) and sketch the system pole-zero plot. - - (b) Using transform-domain techniques, determine *y*zir[*n*] given *y*[−1] = 2 and *y*[−2]=−2. -- **5.3-14** Solve - -$$ -y[n] + 2y[n-1] + 2y[n-2] -$$ - -= $x[n-1] + 2x[n-2]$ - -with *y*[0] = 0, *y*[1] = 1, and *x*[*n*] = *enu*[*n*]. - -- **5.3-15** A system with impulse response *h*[*n*] = 2(1/3)*nu*[*n* − 1] produces an output *y*[*n*] = (−2)*nu*[*n* − 1]. Determine the corresponding input *x*[*n*]. -- **5.3-16** A professor recently received an unexpected \$10 (a futile bribe attached to a test). Being the savvy investor that she is, the professor decides to invest the \$10 into a savings account that earns 0.5% interest compounded monthly (6.17% APY). Furthermore, she decides to supplement this initial investment with an additional \$5 deposit made every month, beginning the month immediately following her initial investment. - - (a) Model the professor's savings account as a constant coefficient linear difference equation. Designate *y*[*n*] as the account balance at month *n*, where *n* = 0 corresponds to the first month that interest is awarded (and that her \$5 deposits begin). - - (b) Determine a closed-form solution for *y*[*n*]. That is, you should express *y*[*n*] as a function only of *n*. - - (c) If we consider the professor's bank account as a system, what is the system impulse response *h*[*n*]? What is the system transfer function *H*[*z*]? - - (d) Explain this fact: if the input to the professor's bank account is the everlasting exponential *x*[*n*] = 1*n* = 1, then the output is **not** *y*[*n*] = 1*nH*[1] = *H*[1]. -- **5.3-17** Sally deposits \$100 into her savings account on the first day of every month except for each December, when she uses her money to buy - -holiday gifts. Define *b*[*m*] as the balance in Sally's account on the first day of month *m*. Assume Sally opens her account in January (*m* = 0), continues making monthly payments forever (except each December!), and that her monthly interest rate is 1%. Sally's account balance satisfies a simple difference equation *b*[*m*] = (1.01)*b*[*m* − 1] + *p*[*m*], where *p*[*m*] designates Sally's monthly deposits. Determine a closed-form expression for *b*[*m*] that is only a function of the month *m*. - -- **5.3-18** For each impulse response, determine the number of system poles, whether the poles are real or complex, and whether the system is BIBO-stable. - - (a) *h*1[*n*] = (−1+(0.5)*n*)*u*[*n*] - - (b) *h*2[*n*] = (*j*)*n*(*u*[*n*] −*u*[*n*−10]) -- **5.3-19** Find the following sums: - -(a) -$$ -\sum_{k=0}^{n} k -$$ - -(b) -$$ -\sum_{k=0}^{n} k^2 -$$ - -*k*=0 [*Hint:* Consider a system whose output *y*[*n*] is the desired sum. Examine the relationship between *y*[*n*] and *y*[*n* − 1]. Note also that *y*[0] = 0.] - -### **5.3-20** Find the following sum: - -$$ -\sum_{k=0}^{n} k^3 -$$ - -[*Hint:* See the hint for Prob. 5.3-19.] - -**5.3-21** Find the following sum: - -$$ -\sum_{k=0}^{n} ka^k \qquad a \neq 1 -$$ - -[*Hint:* See the hint for Prob. 5.3-19.] - -- **5.3-22** Redo Prob. 5.3-19 using the result in Prob. 5.2-13a. -- **5.3-23** Redo Prob. 5.3-20 using the result in Prob. 5.2-13a. -- **5.3-24** Redo Prob. 5.3-21 using the result in Prob. 5.2-13a. - -**5.3-25** (a) Find the zero-state response of an LTID system with transfer function - -$$ -H[z] = \frac{z}{(z+0.2)(z-0.8)} -$$ - -and the input *x*[*n*] = *e*(*n*+1) *u*[*n*]. - -(b) Write the difference equation relating the output *y*[*n*] to input *x*[*n*]. - -**5.3-26** Repeat Prob. 5.3-25 for *x*[*n*] = *u*[*n*] and - -$$ -H[z] = \frac{2z+3}{(z-2)(z-3)} -$$ - -**5.3-27** Repeat Prob. 5.3-25 for - -$$ -H[z] = \frac{6(5z - 1)}{6z^2 - 5z + 1} -$$ - -- and the input *x*[*n*] is (a) (4)−*nu*[*n*] (b) (4)−(*n*−2) *u*[*n*−2] (c) (4)−(*n*−2) *u*[*n*] (d) (4)−*nu*[*n*−2] -- **5.3-28** Repeat Prob. 5.3-25 for *x*[*n*] = *u*[*n*] and - -$$ -H[z] = \frac{2z - 1}{z^2 - 1.6z + 0.8} -$$ - -- **5.3-29** Find the transfer functions corresponding to each of the systems specified by difference equations in Probs. 5.3-5, 5.3-7, 5.3-9, and 5.3-14. -- **5.3-30** Find *h*[*n*], the unit impulse response of the systems described by the following equations: - - (a) *y*[*n*] +3*y*[*n*−1] +2*y*[*n*−2] = *x*[*n*] +3*x*[*n*− 1] +3*x*[*n*−2] - - (b) *y*[*n* + 2] + 2*y*[*n* + 1] + *y*[*n*] = 2*x*[*n* + 2] − *x*[*n*+1] - - (c) *y*[*n*]−*y*[*n*−1]+0.5*y*[*n*−2] = *x*[*n*]+2*x*[*n*− 1] -- **5.3-31** Find *h*[*n*], the unit impulse response of the systems in Probs. 5.3-25, 5.3-26, and 5.3-28. -- **5.3-32** A system has impulse response *h*[*n*] = *u*[*n* − 3]. - - (a) Determine the impulse response of the inverse system *h*−1[*n*]. - - (b) Is the inverse stable? Is the inverse causal? - - (c) Your boss asks you to implement *h*−1[*n*] to the best of your ability. Describe your realizable design, taking care to identify any deficiencies. - -**5.4-1** A system has impulse response given by - -$$ -h[n] = \left[ \left( \frac{1+j}{\sqrt{8}} \right)^n + \left( \frac{1-j}{\sqrt{8}} \right)^n \right] u[n] -$$ - -This system can be implemented according to Fig. P5.4-1. - -**Figure P5.4-1** - -- (a) Determine the coefficients *A*1 and *A*2 to implement *h*[*n*] using the structure shown in Fig. P5.4-1. -- (b) What is the zero-state response *y*0[*n*] of this system, given a shifted unit step input *x*[*n*] = *u*[*n*+3]? -- **5.4-2** (a) Show the canonic direct form, a cascade, and a parallel realization of - -$$ -H[z] = \frac{z(3z - 1.8)}{z^2 - z + 0.16} -$$ - -- (b) Find the transpose of the realizations obtained in part (a). -- **5.4-3** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{5z + 2.2}{z^2 + z + 0.16} -$$ - -**5.4-4** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{3.8z - 1.1}{(z - 0.2)(z^2 - 0.6z + 0.25)} -$$ - -**5.4-5** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{z(1.6z - 1.8)}{(z - 0.2)(z^2 + z + 0.5)} -$$ - -**5.4-6** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{z(2z^2 + 1.3z + 0.96)}{(z + 0.5)(z - 0.4)^2} -$$ - -**5.4-7** Consider the LTID system shown in Fig. P5.4-7. - -$$ -x[n] -$$ - $\rightarrow$ $z^{-1}$ $\rightarrow$ $z^{-1}$ $\rightarrow$ $z^{-1}$ $\rightarrow$ $y[n]$ - -- **Figure P5.4-7** -- (a) Determine the standard delay-form difference equation description of this system. -- (b) Determine the impulse response *h*[*n*] of this system. -- (c) Is this realization canonical? Explain. -- (d) Is this system stable? Explain. -- (e) Is this system causal? Explain. -- **5.4-8** Realize a system whose transfer function is - -$$ -H[z] = \frac{2z^4 + z^3 + 0.8z^2 + 2z + 8}{z^4} -$$ - -**5.4-9** Realize a system whose transfer function is given by - -$$ -H[z] = \sum_{n=0}^{6} nz^{-n} -$$ - -**5.4-10** Consider the LTID system shown in Fig. P5.4-10, where parameter *c* is an arbitrary, real constant. - -**Figure P5.4-10** - -- (a) Determine the system transfer function *H*[*z*], expressed in standard rational form. -- (b) Determine all system poles and all system zeros. -- (c) Is the system of Fig. P5.4-10 canonical? Explain. -- (d) What constraints, if any, exist on parameter *c* to ensure that the system is stable? -- **5.4-11** This problem demonstrates the enormous number of ways of implementing even a relatively low-order transfer function. A second-order transfer function has two real zeros and two real poles. Discuss various ways of realizing - -such a transfer function. Consider canonic direct, cascade, parallel, and the corresponding transposed forms. Note also that interchange of cascaded sections yields a different realization. - -**5.4-12** Consider a digital audio system: an input analog-to-digital converter (ADC) is used to collect input samples at a CD-quality rate *Fs* = 44 kHz. Input samples are processed with a digital filter to generate output samples, which are sent to a digital-to-analog converter (DAC) at the same rate *Fs*. Every sample interval *T* = 1/*Fs*, the digital processor executes the following MATLAB-compatible code: - -``` -% Read input sample from the ADC -x = read_ADC; -% Process input and... -mem(1) = x - mem(3)*9/16; -% ...compute output sample -y = mem(1)*7/16 - mem(3)*7/16; -% Send output sample to the DAC -write_DAC = y; -% Update memory for next iteration -mem(3) = mem(2); -mem(2) = mem(1); -``` - -- (a) Does the code implement DFI, DFII, TDFI, or TDFII? Support your answer by drawing the appropriate block diagram labeled in a manner that is consistent with the code. -- (b) Determine the transfer function *H*[*z*] of this system. -- (c) What is the basic filtering function of this system: LP, HP, BP, or BS? Justify your answer. -- (d) Determine the transfer function *H*−1[*z*] of the inverse system to *H*[*z*] and draw its DFI block implementation. How well will the inverse system operate? -- **5.4-13** Repeat Prob. 5.4-12 but instead use the code: - -``` -% Read input sample from the ADC -x = read_ADC; -% Compute output sample -y = x*7/32+mem(1); -% Send output sample to the DAC -write_DAC = y; -% Update memory for next iteration -mem(1) = mem(2); -mem(2) = x*7/32 + y*9/16; -``` - -- **5.5-1** A CT sinusoid *x*(*t*) = cos(ω*t*) is sampled at a greater-than-Nyquist rate *F*s = 1000 Hz to produce a DT sinusoid *x*(*t*) = cos(*n*). Determine the analog frequency ω if (a) = π 4 - - (b) = 2π 3 - - (c) = 7 8 -- **5.5-2** Find the amplitude and phase response of the digital filters depicted in Fig. P5.5-2. -- **5.5-3** A causal LTID system *H*(*z*) = 21(*z*−*j*)(*z*+*j*) 16(*z*− 1 2 )(*z*+ 3 4 ) has a periodic input *x*[*n*] that toggles between the values 1 and 2. That is, *x*[*n*]=[..., 1, 2, 1, ↓ 2 , 1, 2, 1, ...], where *x*[0] = 2. - - (a) Plot the magnitude response |*H*(*ej*)| over −2π ≤ ≤ 2π. - - (b) Plot the phase response *H*(*ej*) over −2π ≤ ≤ 2π. - - (c) Determine the system output *y*[*n*] in response to the periodic input *x*[*n*]. - -**5.5-4** A causal LTID system *H*(*z*) = −7(*z*+1) 32(*z*−*j* 3 4 )(*z*+*j* 3 4 ) has a periodic input *x*[*n*] that cycles through the 4 values 3, 2, 1, and 2. That is, *x*[*n*] = [..., 3, 2, 1, 2, ↓ 3, 2, 1, 2, ...], where *x*[0] = 3. - -- (a) Plot the magnitude response |*H*(*ej*)| over −2π ≤ ≤ 2π. -- (b) Plot the phase response *H*(*ej*) over −2π ≤ ≤ 2π. -- (c) Determine the system output *y*[*n*] in response to the periodic input *x*[*n*]. -- **5.5-5** Find the amplitude and the phase response of the filters shown in Fig. P5.5-5. [*Hint:* Express *H*[*ej*] as *e*−*j*2.5*Ha*[*ej*].] -- **5.5-6** Find the frequency response for the moving-average system in Prob. 3.4-3. The input–output equation of this system is given by - -$$ -y[n] = \frac{1}{5} \sum_{k=0}^{4} x[n-k] -$$ - -**5.5-7** (a) Input–output relationships of two filters are described by - -(i) *y*[*n*]=−0.9*y*[*n*−1] +*x*[*n*] - -(ii) *y*[*n*] = 0.9*y*[*n*−1] +*x*[*n*] - -For each case, find the transfer function, the amplitude response, and the phase response. Sketch the amplitude response, and state the type (highpass, lowpass, etc.) of each filter. - -- (b) Find the response of each of these filters to a sinusoid *x*[*n*] = cos*n* for = 0.01π and 0.99π. In general, show that the gain (amplitude response) of filter (i) at frequency 0 is the same as the gain of filter (ii) at frequency π −0. -- **5.5-8** For an LTID system specified by the equation - -$$ -y[n+1] - 0.5y[n] = x[n+1] + 0.8x[n] -$$ - -(a) Find the amplitude and the phase response. - -- (b) Find the system response *y*[*n*] for the input *x*[*n*] = cos(0.5*k* −(π/3)). -- **5.5-9** For an asymptotically stable LTID system, show that the steady-state response to input *ejnu*[*n*] is *H*[*ej*]*ejnu*[*n*]. The steady-state response is that part of the response which does not decay with time and persists forever. -- **5.5-10** Express the following signals in terms of apparent frequencies: - - (a) cos(0.8π*n*+θ ) - - (b) sin(1.2π*n*+θ ) - - (c) cos(6.9*n*+θ ) - -**Figure P5.5-2** - -(b) - -**Figure P5.5-5** - -- (d) cos(2.8π*n*+θ ) +2 sin(3.7π*n*+θ ) -- (e) sinc(π*n*/2) -- (f) sinc(3π*n*/2) -- (g) sinc(2π*n*) -- **5.5-11** Show that cos √ (0.6π*n* + (π/6)) + 3 cos(1.4π*n* + (π/3)) = 2 cos(0.6π*n* − (π/6)). -- **5.5-12** (a) A digital filter has the sampling interval *T* = 50µs. Determine the highest frequency that can be processed by this filter without aliasing. - - (b) If the highest frequency to be processed is 50 kHz, determine the minimum value of the sampling frequency *Fs* and the maximum value of the sampling interval *T* that can be used. -- **5.5-13** Consider the discrete-time system represented by - -$$ -y[n] = \sum_{k=0}^{\infty} (0.5)^k x[n-k] -$$ - -- (a) Determine and plot the magnitude response |*H*[*ej*]| of the system. -- (b) Determine and plot the phase response *H*[*ej*] of the system. -- (c) Find an efficient block representation that implements this system. -- **5.6-1** Pole-zero configurations of certain filters are shown in Fig. P5.6-1. Sketch roughly the amplitude response of these filters. -- **5.6-2** Figure P5.6-2 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[−1]=−1. - -**Figure P5.6-2** - -- (a) Determine the five constants *b*0, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *b*0*z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−2π ≤ ≤ 0). -- (c) Determine the output *y*[*n*] of this system if the input is *x*[*n*] = sin π*n* 2 . -- **5.6-3** Repeat Prob. 5.6-2 if the zero at *z* = 1 is moved to *z* = −1 and *H*[1]=−1 is specified rather than *H*[−1]=−1. -- **5.6-4** Figure P5.6-4 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[−1] = 1. - - (a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *k z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . - -**Figure P5.6-4** - -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−π ≤ ≤ π ). -- (c) A signal *x*(*t*) = cos(100π*t*) + cos(500π*t*) is sampled at a greater than Nyquist rate *F*s Hz and then input into the above LTID system to produce DT output *y*[*n*] = β cos(0*n* + θ ). Determine *F*s and 0. You do not need to find constants β and θ. -- (d) Is the inpulse response *h*[*n*] of this system absolutely summable? Justify your answer. -- **5.6-5** Figure P5.6-5 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[1]=−1. - -(a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *k z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . - -**Figure P5.6-5** - -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−π ≤ ≤ π ). -- (c) A signal *x*(*t*) = cos(2π*ft*) is sampled at a rate *F*s = 1 kHz and then input into the above LTID system to produce DT output *y*[*n*]. Determine, if possible, the frequency or frequencies *f* that will produce zero output, *y*[*n*] = 0. -- **5.6-6** The system *y*[*n*] − *y*[*n* − 1] = *x*[*n*] − *x*[*n* − 1] is an all-pass system that has zero phase response. Is there any difference between this system and the system *y*[*n*] = *x*[*n*]? Justify your answer. -- **5.6-7** Figure P5.6-7 displays the pole-zero plot of a second-order real, causal LTID system that has a repeated zero and *H*[1] = 4. The solid circle is the unit circle. - -### **Figure P5.6-7** - -- (a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *k z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−π ≤ ≤ π ). -- (c) Determine the steady-state output *y*ss[*n*] of this system if the input is *x*[*n*] = cos( 3π*n* 4 )*u*[*n*]. -- (d) State whether this system is LP, HP, BP, BS, or other. If the digital system operates at *F*s = 8 kHz, what is the approximate hertzian cutoff frequency (or frequencies) of this system? -- **5.6-8** The magnitude and phase responses of a real, stable, LTI system are shown in Fig. P5.6-8. - - (a) What type of system is this: lowpass, highpass, bandpass, or bandstop? - - (b) What is the output of this system in response to - -$$ -x_1[n] = 2\sin\left(\frac{\pi}{2}n + \frac{\pi}{4}\right) -$$ - -(c) What is the output of this system in response to - -$$ -x_2[n] = \cos\left(\frac{7\pi}{4}n\right) -$$ - -- **5.6-9** Consider an LTID system with system function *H*[*z*] = *b*0 *z*2+1 *z*2−9/16 . - - (a) Determine the constant *b*0 so that the system frequency response at = −π is −1. - - (b) Accurately sketch the system poles and zeros. - - (c) Using the locations of the system poles and zeros, sketch |*H*[*ej*]| over 0 ≤ ≤ 2π. - -- (d) Determine the response *y*[*n*] to the input *x*[*n*] = (−1+*j*)+*j n* +(1−*j*)sin(π*n*+1). -- (e) Draw an appropriate block diagram representation of this system. -- **5.6-10** Do Prob. 5.10-3 by graphical procedure. Do the sketches approximately, without using MATLAB. -- **5.6-11** Do Prob. 5.10-8 by graphical procedure. Do the sketches approximately, without using MATLAB. -- **5.6-12** (a) Realize a digital filter whose transfer function is given by - -$$ -H[z] = K \frac{z+1}{z-a} -$$ - -- (b) Sketch the amplitude response of this filter, assuming |*a*| < 1. -- (c) The amplitude response of this lowpass filter is maximum at = 0. The 3 dB bandwidth is the frequency at which the amplitude response drops to 0.707 (or 1/ √2) times its maximum value. Determine the 3 dB bandwidth of this filter when *a* = 0.2. -- **5.6-13** Design a digital notch filter to reject frequency 5000 Hz completely and to have a sharp recovery on either side of 5000 Hz to a gain of unity. The highest frequency to be processed is 20 kHz (*Fh* = 20,000). [*Hint:* See Ex. 5.15. The zeros should be at *e*±*j*ω*T* for ω corresponding to 5000 Hz, and the poles are at *ae*±*j*ω*T* with *a* < 1. Leave your answer in terms of *a*. Realize this filter using the canonical form. Find the amplitude response of the filter.] - -**5.6-14** Consider the desired DT system magnitude response |*H*[*ej*]| in Fig. P5.6-14. - -**Figure P5.6-14** - -- (a) Is the filter LP, HP, BP, BS, or other? Explain. -- (b) Sketch the pole-zero plot of a 2nd-order system that behaves as a reasonable approximation of Fig. P5.6-14. What is the coefficient *b*0? -- **5.6-15** Show that a first-order LTID system with a pole at *z* = *r* and a zero at *z* = 1/*r* (*r* ≤ 1) is an allpass filter. In other words, show that the amplitude response |*H*[*ej*]| of a system with the transfer function - -$$ -H[z] = \frac{z - \frac{1}{r}}{z - r} \qquad r \le 1 -$$ - -is constant with frequency. This is a first-order allpass filter. [*Hint:* Show that the ratio of the distances of any point on the unit circle from the zero (at *z* = 1/*r*) and the pole (at *z* = *r*) is a constant 1/*r*.] - -Generalize this result to show that an LTID system with two poles at *z* = *re*±*j*θ and two zeros at *z* = (1/*r*)*e*±*j*θ (*r* ≤ 1) is an allpass filter. In other words, show that the amplitude response of a system with the transfer function - -$$ -H[z] = \frac{\left(z - \frac{1}{r}e^{j\theta}\right)\left(z - \frac{1}{r}e^{-j\theta}\right)}{(z - re^{j\theta})(z - re^{-j\theta})} -$$ -$$ -= \frac{z^2 - \left(\frac{2}{r}\cos\theta\right)z + \frac{1}{r^2}}{z^2 - (2r\cos\theta)z + r^2} \qquad r \le 1 -$$ - -is constant with frequency. - -**5.6-16** (a) If *h*1[*n*] and *h*2[*n*], the impulse responses of two LTID systems are related by *h*2[*n*] = (−1)*nh*1[*n*], then show that - -$$ -H_2[e^{j\Omega}] = H_1[e^{j(\Omega \pm \pi)}] -$$ - -How is the frequency response spectrum *H*2[*ej*] related to the *H*1[*ej*]? - -- (b) If *H*1[*z*] represents an ideal lowpass filter with cutoff frequency *c*, sketch *H*2[*ej*]. What type of filter is *H*2[*ej*]? -- **5.6-17** Mappings such as the bilinear transformation are useful in the conversion of continuous-time filters to discrete-time filters. Another useful type of transformation is one that converts a discrete-time filter into a different type of discrete-time filter. Consider a transformation that replaces *z* with −*z*. - - (a) Show that this transformation converts lowpass filters into highpass filters and highpass filters into lowpass filters. - - (b) If the original filter is an FIR filter with impulse response *h*[*n*], what is the impulse response of the transformed filter? -- **5.6-18** The bilinear transformation is defined by the rule *s* = 2(1−*z*−1)/*T*(1+*z*−1). - - (a) Show that this transformation maps the ω axis in the *s* plane to the unit circle *z* = *ej* in the *z* plane. - - (b) Show that this transformation maps to 2 arctan(ω*T*/2). -- **5.7-1** In Ch. 3, we used another approximation to find a digital system to realize an analog system. We showed that an analog system specified by Eq. (3.12) can be realized by using the digital system specified by Eq. (3.13). Compare that solution with the one resulting from the - -impulse-invariance method. Show that one result is a close approximation of the other and that the approximation improves as *T* → 0. - -- **5.7-2** A CT system has impulse response *h*ct(*t*) = *e*−*t u*(*t*). Draw the DFI realization of the corresponding DT system designed by the impulse-invariance method with *T* = 0.1. -- **5.7-3** (a) Using the impulse-invariance criterion, design a digital filter to realize an analog filter with transfer function - -$$ -H_a(s) = \frac{7s + 20}{2(s^2 + 7s + 10)} -$$ - -- (b) Show a canonical and a parallel realization of the filter. Use a 1% criterion for the choice of *T*. -- **5.7-4** Use the impulse-invariance criterion to design a digital filter to realize the second-order analog Butterworth filter with transfer function - -$$ -H_a(s) = \frac{1}{s^2 + \sqrt{2}s + 1} -$$ - -Use a 1% criterion for the choice of *T*. - -- **5.7-5** Design a digital integrator using the impulse-invariance method. Find and give a rough sketch of the amplitude response, and compare it with that of the ideal integrator. If this integrator is used primarily for integrating audio signals (whose bandwidth is 20 kHz), determine a suitable value for *T*. -- **5.7-6** An oscillator by definition is a source (no input) that generates a sinusoid of a certain frequency ω0. Therefore, an oscillator is a system whose zero-input response is a sinusoid of the desired frequency. Find the transfer function of a digital oscillator to oscillate at 10 kHz by the methods described in parts (a) and (b). In both methods, select *T* so that there are 10 samples in each cycle of the sinusoid. - - (a) Choose *H*[*z*] directly so that its zero-input response is a discrete-time sinusoid of frequency =ω*T* corresponding to 10 kHz. - - (b) Choose *Ha*(*s*) whose zero-input response is an analog sinusoid of 10 kHz. Now use the impulse invariance method to determine *H*[*z*]. - - (c) Show a canonical realization of the oscillator. - -- **5.7-7** A variant of the impulse invariance method is the *step-invariance* method of digital filter synthesis. In this method, for a given *Ha*(*s*), we design *H*[*z*] in Fig. 5.24a such that *y*(*nT*) in Fig. 5.24b is identical to *y*[*n*] in Fig. 5.24a when *x*(*t*) = *u*(*t*). - - (a) Show that, in general, - -$$ -H[z] = \frac{z-1}{z} \mathcal{Z} \left[ \left( \mathcal{L}^{-1} \frac{H_a(s)}{s} \right)_{t=kT} \right] -$$ - -(b) Use this method to design *H*[*z*] for - -$$ -H_a(s) = \frac{\omega_c}{s + \omega_c} -$$ - -- (c) Use the step-invariance method to synthesize a discrete-time integrator and compare its amplitude response with that of the ideal integrator. -- **5.7-8** Use the *ramp-invariance* method to synthesize a discrete-time differentiator and integrator. In this method, for a given *Ha*(*s*), we design *H*[*z*] such that *y*(*nT*) in Fig. 5.24b is identical to *y*[*n*] in Fig. 5.24a when *x*(*t*) = *tu*(*t*). -- **5.7-9** In an impulse-invariance design, show that if *Ha*(*s*) is a transfer function of a stable system, the corresponding *H*[*z*] is also a transfer function of a stable system. -- **5.7-10** First-order backward differences provide the transformation rule *s* = (1−*z*−1)/*T*. - - (a) Show that this transformation maps the ω axis in the *s* plane to a circle of radius 1/2 centered at (1/2, 0) in the *z* plane. - - (b) Show that this transformation maps the left-half *s* plane to the interior of the unit circle in the *z* plane, which ensures that stability is preserved. -- **5.8-1** Find the *z*-transform (if it exists) and the corresponding ROC for each of the following signals: - - (a) (0.8)*nu*[*n*] +2*nu*[−(*n*+1)] - -(b) -$$ -2^n u[n] - 3^n u[-(n+1)] -$$ - -(c) -$$ -(-2)^{n+3}u[-n] + \sum_{k=0}^{\infty} (0.5)^{k-1} \delta(n-2k) -$$ - -(d) -$$ -(0.8)^n u[n] + (0.9)^n u[-(n+1)] -$$ - -(e) -$$ -[(0.8)^n + 3(0.4)^n]u[-(n+1)] -$$ - -(f) [(0.8)*n* +3(0.4)*n*]*u*[*n*] - -(g) -$$ -(0.8)^n u[n] + 3(0.4)^n u[-(n+1)] -$$ - -$$ -(0.6) \quad u[n] + 5(0.4) \quad u[ -$$ - -(h) -$$ -(0.5)^{|n|} -$$ - -$$ -(i) \t n u[-(n+1)] -$$ - -- **5.8-2** Using the definition, compute the bilateral *z*-transform *X*(*z*) of (a) *x*[*n*] = 3*nu*[−*n*] (b) *x*[*n*] = ( 1 3 )*nu*[*n*] Express your answers in standard rational form. -- **5.8-3** Determine the inverse *z*-transform *x*[*n*] of *X*[*z*] = *z*2− 1 3 *z* (*z*−1)(*z*+2) with ROC 1 &lt; |*z*| &lt; 2. - -**5.8-4** Find the inverse -$$ -z -$$ --transform of - -$$ -X[z] = \frac{(e^{-2} - 2)z}{(z - e^{-2})(z - 2)} -$$ - -when the ROC is, -\n(a) -$$ -|z| > 2 -$$ - -\n(b) $e^{-2} < |z| < 2$ -\n(c) $|z| < e^{-2}$ - -**5.8-5** Use partial fraction expansions, *z*-transform tables, and a region of convergence (|*z*| < 1/2) to determine the inverse *z*-transform of - -$$ -X(z) = \frac{1}{(2z+1)(z+1)(z+\frac{1}{2})} -$$ - -- **5.8-6** Using *z*-transform techniques and properties (no time-domain convolution sum!), determine the convolution *y*[*n*] = ( 1 3 )*n*−3*u*[*n* − 2] ∗ (2)*nu*[−*n*]. Express your answer in the form *y*[*n*] = *c*1γ *n* 1 *u*[*n* + *N*1] + *c*2γ *n* 2 *u*[−*n* + *N*2], making sure to clearly identify the constants *c*1, *c*2, γ1, γ2, *N*1, and *N*2. -- **5.8-7** Using partial fraction expansions, *z*-transform tables, and the fact that *h*[*n*] is stable, determine the inverse *z*-transform of - -$$ -H[z] = \frac{z^4 + z^3}{(z - 2)(z + \frac{1}{2})}. -$$ - -**5.8-8** Consider the system - -$$ -H[z] = \frac{z(z - \frac{1}{2})}{(z^3 - \frac{27}{8})} -$$ - -- (a) Draw the pole-zero diagram for *H*[*z*] and identify all possible regions of convergence. -- (b) Draw the pole-zero diagram for *H*−1[*z*] and identify all possible regions of convergence. -- **5.8-9** A discrete-time signal *x*[*n*] has a rational *z*-transform that contains a pole at *z* = 0.5. Given *x*1[*n*] = (1/3)*nx*[*n*] is absolutely summable and - -Problems 589 - -*x*2[*n*] = (1/4)*nx*[*n*] is **not** absolutely summable, determine whether *x*[*n*] is left-sided, right-sided, or two-sided. Justify your answer! - -- **5.8-10** Let *x*[*n*] be an absolutely summable signal with rational *z*-transform *X*[*z*]. *X*[*z*] is known to have a pole at *z* = (0.75+0.75*j*), and other poles may be present. Recall that an absolutely summable signal satisfies %∞ −∞ |*x*[*n*]| < ∞. - - (a) Can *x*[*n*] be left-sided? Explain. - - (b) Can *x*[*n*] be right-sided? Explain. - - (c) Can *x*[*n*] be two-sided? Explain. - - (d) Can *x*[*n*] be of finite duration? Explain. -- **5.8-11** Consider a causal system that has transfer function - -$$ -H[z] = \frac{z - 0.5}{z + 0.5} -$$ - -When appropriate, assume initial conditions of zero. - -- (a) Determine the output *y*1[*n*] of this system in response to *x*1[*n*] = (3/4)*nu*[*n*]. -- (b) Determine the output *y*2[*n*] of this system in response to *x*2[*n*] = (3/4)*n*. -- (c) Determine the output *y*3[*n*] of this system in response to *x*3[*n*] = (3/4)*nu*[−*n*−1]. -- **5.8-12** Let *x*[*n*] =(−1)*nu*[*n*−*n*0]+α*nu*[−*n*]. Determine the constraints on the complex number α and the integer *n*0 so that the *z*-transform *X*[*z*] exists with region of convergence 1 < |*z*| < 2. -- **5.8-13** Using the definition, compute the bilateral *z*-transform, including the region of convergence (ROC), of the following complex-valued functions: - - (a) *x*1[*n*] = (−*j*)−*nu*[−*n*] +δ[−*n*] - - (b) *x*2[*n*] = (*j*)*n* cos(*n*+1)*u*[*n*] - - (c) *x*3[*n*] = *j*sinh(*n*)*u*[−*n*+1] - - (d) *x*4[*n*] = %0 *k*=−∞(2*j*)*n*δ[*n*−2*k*] -- **5.8-14** Use partial fraction expansions, *z*-transform tables, and a region of convergence (0.5 < |*z*| < 2) to determine the inverse *z*-transform of - - (a) *X*1[*z*] = 1 1+ 13 6 *z*−1 + 1 6 *z*−2 − 1 3 *z*−3 1 - -(b) -$$ -X_2[z] = \frac{1}{z^{-3}(2 - z^{-1})(1 + 2z^{-1})} -$$ - -**5.8-15** Use partial fraction expansions, *z*-transform tables, and the fact that the systems are stable to determine the inverse *z*-transform of - -(a) -$$ -H_1[z] = \frac{z^{-1}}{(z - \frac{1}{2})(1 + \frac{1}{2}z^{-1})} -$$ - -(b) -$$ -H_2[z] = \frac{z+1}{z^3(z-2)(z+\frac{1}{2})} -$$ - -**5.8-16** By inserting *N* − 1 zeros between every sample of a unit step, we obtain a signal - -$$ -h[n] = \sum_{k=0}^{\infty} \delta[n - Nk] -$$ - -Determine *H*[*z*], the bilateral *z*-transform of *h*[*n*]. Identify the number and location(s) of the poles of *H*[*z*]. - -- **5.8-17** Using transform-domain techniques, determine the zero-state response *y*zsr[*n*] of LTID system *y*[*n*] − 1 4 *y*[*n* − 2] = *x*[*n*] to the noncausal input *x*[*n*] = 2*nu*[2−*n*]. -- **5.8-18** Determine the zero-state response of a system having a transfer function - -$$ -H[z] = \frac{z}{(z+0.2)(z-0.8)} \qquad |z| > 0.8 -$$ - -and an input *x*[*n*] given by - -- (a) *x*[*n*] = *enu*[*n*] -- (b) *x*[*n*] = 2*nu*[−(*n*+1)] -- (c) *x*[*n*] = *enu*[*n*] +2*nu*[−(*n*+1)] -- (d) *x*[*n*] = 2*nu*[*n*] +*u*[−(*n*+1)] -- (e) *x*[*n*] = *e*−2*nu*[−(*n*+1)] -- **5.8-19** The discrete cross-correlation between real signal *x*[*n*] and real signal *y*[*n*] is - -$$ -c_{xy}[n] = \sum_{k=-\infty}^{\infty} x[k]y[k-n] -$$ - -Let signal *x*[*n*] have *z*-transform *X*[*z*] with ROC *Rx*, and let signal *y*[*n*] have *z*-transform *Y*[*z*] with ROC *Ry*. Determine *CXY* [*z*] (the bilateral *z*-transform of *cxy*[*n*]) in terms of the *z*-transforms of *x*[*n*] and *y*[*n*]. - -**5.8-20** Transform properties can be very useful. The accumulation property states - -$$ -\sum_{k=-\infty}^{n} x[k] \Longleftrightarrow \frac{z}{z-1} X[z] -$$ - -This property is the DT version of the Laplace transform "integration in time" property. Given *x*[*n*] ⇐⇒ *X*[*z*], prove the accumulation property. [*Hint:* Polynomial long division can be helpful.] **5.10-1** Use MATLAB to generate pole-zero plots for the causal systems with the following transfer functions: - -(a) -$$ -H_a[z] = \frac{z^4 - \sqrt{2}z^2 + 1}{z^4 + 0.4096} -$$ - -(b) $H_b[z] = \frac{-3z^{-1} + \frac{3}{4}z^{-3}}{3 + \frac{3}{8}z^{-2} + 2z^{-4}}$ - -3+ 3 2 *z*−2+2*z*−4 In each case, determine whether the system is - -- stable. -- **5.10-2** For each of the following stable LTID systems, use MATLAB to generate magnitude and phase response plots over −π ≤ <π. (a) *H*a[*z*] = cos(*z*) *z*−0.5 - - (b) *H*b[*z*] = *z*3 sin(*z*−1) -- **5.10-3** Consider an LTID system described by the difference equation 4*y*[*n* + 2] − *y*[*n*] = *x*[*n* + 2] +*x*[*n*]. - - (a) Plot the pole-zero diagram for this system. - - (b) Plot the system's magnitude response |*H*[*ej*]| over −π ≤ ≤ π. - - (c) What type of system is this: lowpass, highpass, bandpass, or bandstop? - - (d) Is this system stable? Justify your answer. - - (e) Is this system real? Justify your answer. - - (f) If the system input is of the form *x*[*n*] = cos(*n*), what is the greatest possible amplitude of the output? Justify your answer. - - (g) Draw an efficient, causal implementation of this system using only add, scale, and delay blocks. -- **5.10-4** Consider an LTID system with system function *H*(*z*) = *b*0 *z*2+1 *z*2−4/9 . - - (a) Determine the constant *b*0 so that the system frequency response at = −π is −1. - - (b) Plot the pole-zero diagram for this system. - - (c) Plot the system's magnitude response |*H*[*ej*]| over −π ≤ ≤ π. - - (d) Plot the system's phase response *H*[*ej*] over −π ≤ ≤ π. - - (e) What type of system is this: lowpass, highpass, bandpass, or bandstop? - - (f) Determine the response *y*[*n*] to the input *x*[*n*] = (−1 − *j*) + (−*j*)*n* + (1 − *j*) cos(π*n* + 1 3 ). - - (g) Draw a TDFII block diagram representation of this system. - -**5.10-5** Consider the LTID system shown in Fig. P5.10-5, where parameters *c*1 and *c*2 are arbitrary constants. - -### **Figure P5.10-5** - -- (a) Determine the system function *H*[*z*], expressed in standard rational form. -- (b) What is the order *N* of this system? -- (c) Determine the *N* poles and *N* zeros of this system. Use MATLAB to create the corresponding pole-zero plot. -- (d) What constraints, if any, exist on parameters *c*1 and *c*2 to ensure that the system is stable? -- (e) Determine *c*1 and *c*2 so that this system functions as an LPF with narrow passband. Use MATLAB to generate the corresponding magnitude response |*H*[*ej*]| over −π ≤ ≤ π. -- (f) Ms. Zeroine, the heroine of DT systems, believes that if *x*[*n*] = 0, then *y*[*n*] = 0 also. Is Ms. Zeroine correct? Fully justify your answer. -- (g) Dr. Strange suggests that by setting *c*1 = −1 and *c*2 = −2, the system will act as a highpass filter. Is Dr. Strange right? Fully justify your answer. -- **5.10-6** One interesting and useful application of discrete systems is the implementation of complex (rather than real) systems. A complex system is one in which a real-valued input can produce a complex-valued output. Complex systems that are described by constant coefficient difference equations require at least one complex-valued coefficient, and they are capable of operating on complex-valued inputs. Consider the complex discrete-time system - -$$ -H[z] = \frac{z^2 - j}{z - 0.9e^{j3\pi/4}} -$$ - -(a) Determine and plot the system zeros and poles. - -- (b) Sketch the magnitude response |*H*[*ej*ω]| of this system over −2π ≤ ω ≤ 2π. Comment on the system's behavior. -- **5.10-7** Consider the complex system - -$$ -H[z] = \frac{z^4 - 1}{2(z^2 + 0.81j)} -$$ - -Refer to Prob. 5.10-6 for an introduction to complex systems. - -- (a) Plot the pole-zero diagram for *H*[*z*]. -- (b) Plot the system's magnitude response |*H*[*ej*]| over −π ≤ ≤ π. -- (c) Explain why *H*[*z*] is a noncausal system. Do not give a general definition of causality; specifically identify what makes this system noncausal. -- (d) One way to make this system causal is to add two poles to *H*[*z*]. That is, - -$$ -H_{\text{causal}}[z] = H[z] \frac{1}{(z-a)(z-b)} -$$ - -Find poles *a* and *b* such that |*H*causal[*ej*]| = |*H*[*ej*]|. - -- (e) Draw an efficient block implementation of *H*causal[*z*]. -- **5.10-8** A discrete-time LTI system is shown in Fig. P5.10-8. - - (a) Determine the difference equation that describes this system. - - (b) Determine the magnitude response |*H*[*ej*]| for this system and simplify your answer. Plot the magnitude response over −π ≤ ≤ π. What type of standard filter (lowpass, highpass, bandpass, or bandstop) best describes this system? - - (c) Determine the impulse response *h*[*n*] of this system. - -**Figure P5.10-8** - -- **5.10-9** Determine the impulse response *h*[*n*] for the system shown in Fig. P5.10-9. Is the system stable? Is the system causal? -- **5.10-10** An LTID filter has an impulse response function given by *h*[*n*] = δ[*n* − 1] + δ[*n* + 1]. Determine and carefully sketch the magnitude response |*H*[*ej*]| over the range −π ≤ ≤ π. For this range of frequencies, is this filter lowpass, highpass, bandpass, or bandstop? -- **5.10-11** A causal, stable discrete system has the rather strange transfer function *H*[*z*] = cos(*z*−1). - - (a) Write MATLAB code that will compute and plot the magnitude response of this system over an appropriate range of digital frequencies . Comment on the system. - - (b) Determine the impulse response *h*[*n*]. Plot *h*[*n*] over (0 ≤ *n* ≤ 10). - - (c) Determine a difference equation description for an FIR filter that closely approximates the system *H*[*z*] = cos(*z*−1). To verify proper behavior, plot the FIR filter's magnitude response and compare it with the magnitude response computed in part (a). -- **5.10-12** The MATLAB signal-processing toolbox function butter helps design digital Butterworth filters. Use MATLAB help to learn how butter works. For each of the following cases, design the filter, plot the filter's poles and zeros in the complex *z* plane, and plot the decibel magnitude response 20log10 |*H*[*ej*]|. - - (a) Design an eighth-order digital lowpass filter with *c* = π/3. - - (b) Design an eighth-order digital highpass filter with *c* = π/3. - - (c) Design an eighth-order digital bandpass filter with passband between 5π/24 and 11π/24. - - (d) Design an eighth-order digital bandstop filter with stopband between 5π/24 and 11π/24. -- **5.10-13** The MATLAB signal-processing toolbox function cheby1 helps design digital Chebyshev - -type I filters. A Chebyshev type I filter has passband ripple and smooth stopband. Setting the passband ripple to *Rp* = 3 dB, repeat Prob. 5.10-12 using the cheby1 command. With all other parameters held constant, what is the general effect of reducing *Rp*, the allowable passband ripple? - -**5.10-14** The MATLAB signal-processing toolbox function cheby2 helps design digital Chebyshev type II filters. A Chebyshev type II filter has smooth passband and ripple in the stopband. Setting the stopband ripple *Rs* = 20 dB down, repeat Prob. 5.10-12 using the cheby2 command. With all other parameters held constant, what is the general effect of increasing *Rs*, the minimum stopband attenuation? - -**5.10-15** The MATLAB signal-processing toolbox function ellip helps design digital elliptic filters. An elliptic filter has ripple in both the passband and the stopband. Setting the passband ripple to *Rp* = 3 dB and the stopband ripple *Rs* = 20 dB down, repeat Prob. 5.10-12 using the ellip command. - - diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/078_6 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER SERIES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/078_6 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER SERIES.md deleted file mode 100644 index 8e89d6dd193f90d700816d1518f7f7a3ed6e3c79..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/078_6 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER SERIES.md +++ /dev/null @@ -1,5 +0,0 @@ -# **[CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER SERIES** - -Electrical engineers instinctively think of signals in terms of their frequency spectra and think of systems in terms of their frequency response. Most teenagers know about the audible portion of audio signals having a bandwidth of about 20 kHz and the need for good-quality speakers to respond up to 20 kHz. This is basically thinking in the frequency domain. In Chs. 4 and 5 we discussed extensively the frequency-domain representation of systems and their spectral response (system response to signals of various frequencies). In Chs. 6 through 9, we discuss spectral representation of signals, where signals are expressed as a sum of sinusoids or exponentials. Actually, we touched on this topic in Chs. 4 and 5. Recall that the Laplace transform of a continuous-time signal is its spectral representation in terms of exponentials (or sinusoids) of complex frequencies. Similarly the *z*-transform of a discrete-time signal is its spectral representation in terms of discrete-time exponentials. However, in the earlier chapters we were concerned mainly with system representation; the spectral representation of signals was incidental to the system analysis. Spectral analysis of signals is an important topic in its own right, and now we turn to this subject. - -In this chapter we show that a periodic signal can be represented as a sum of sinusoids (or exponentials) of various frequencies. These results are extended to aperiodic signals in Ch. 7 and to discrete-time signals in Ch. 9. The fascinating subject of sampling of continuous-time signals is discussed in Ch. 8, leading to A/D (analog-to-digital) and D/A conversion. Chapter 8 forms the bridge between the continuous-time and the discrete-time worlds. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/079_6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/079_6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES.md deleted file mode 100644 index 3fce33c65ff104ac51346cd6c516c45b0902687a..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/079_6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES.md +++ /dev/null @@ -1,599 +0,0 @@ -## **6.1 PERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY TRIGONOMETRIC FOURIER SERIES** - -As seen in Sec. 1.3-3 [Eq. (1.7)], a periodic signal *x*(*t*) with period *T*0 (Fig. 6.1) has the property - -$$ -x(t) = x(t + T_0) \qquad \text{for all } t -$$ - -The *smallest* value of *T*0 that satisfies this periodicity condition is the *fundamental period* of *x*(*t*). As argued in Sec. 1.3-3, this equation implies that *x*(*t*) starts at −∞ and continues to ∞. Moreover, the area under a periodic signal *x*(*t*) over any interval of duration *T*0 is the same; that is, for any - -**Figure 6.1** A periodic signal of period *T*0. - -real numbers *a* and *b* - -$$ -\int_{a}^{a+T_0} x(t) dt = \int_{b}^{b+T_0} x(t) dt -$$ - -This result follows from the fact that a periodic signal takes the same values at intervals of *T*0. Hence, the values over any segment of duration *T*0 are repeated in any other interval of the same duration. For convenience, the area under *x*(*t*) over any interval of duration *T*0 will be denoted by - -$$ -\int_{T_0} x(t) \, dt -$$ - -The frequency of a sinusoid cos 2π*f*0*t* or sin 2π*f*0*t* is *f*0, and the period is *T*0 = 1/*f*0. These sinusoids can also be expressed as cosω0*t* or sinω0*t*, where ω0 = 2π*f*0 is the *radian frequency*, although for brevity, it is often referred to as frequency (see Sec. B.2). A sinusoid of frequency *nf*0 is said to be the *nth harmonic* of the sinusoid of frequency *f*0. - -Let us consider a signal *x*(*t*) made up of a sines and cosines of frequency ω0 and all of its harmonics (including the zeroth harmonic; i.e., dc) with arbitrary amplitudes† : - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.1} -$$ - -The frequency ω0 is called the *fundamental frequency.* - -We now prove an extremely important property: *x*(*t*) in Eq. (6.1) is a periodic signal with the same period as that of the fundamental, regardless of the values of the amplitudes *an* and *bn*. Note that the period *T*0 of the fundamental satisfies - -$$ -T_0 = \frac{1}{f_0} = \frac{2\pi}{\omega_0} \quad \text{and} \quad \omega_0 T_0 = 2\pi \tag{6.2} -$$ - -† In Eq. (6.1), the constant term *a*0 corresponds to the cosine term for *n* = 0 because cos(0 × ω0)*t* = 1. However, sin(0×ω0)*t* = 0. Hence, the sine term for *n* = 0 is nonexistent. - -To prove the periodicity of *x*(*t*), all we need is to show that *x*(*t*) = *x*(*t* +*T*0). From Eq. (6.1), - -$$ -x(t+T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 (t+T_0) + b_n \sin n\omega_0 (t+T_0) -$$ - -= $a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + n\omega_0 T_0) + b_n \sin(n\omega_0 t + n\omega_0 T_0)$ - -From Eq. (6.2), we have *n*ω0*T*0 = 2π*n*, and - -$$ -x(t + T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + 2\pi n) + b_n \sin(n\omega_0 t + 2\pi n) -$$ - -= $a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t = x(t)$ - -We could also infer this result intuitively. In one fundamental period *T*0, the *n*th harmonic executes *n* complete cycles. Hence, every sinusoid on the right-hand side of Eq. (6.1) executes a complete number of cycles in one fundamental period *T*0. Therefore, at *t* = *T*0, every sinusoid starts as if it were the origin and repeats the same drama over the next *T*0 seconds, and so on, ad infinitum. Hence, the sum of such harmonics results in a periodic signal of period *T*0. - -This result shows that any combination of sinusoids of frequencies 0, *f*0, 2*f*0, ..., *kf*0 is a periodic signal of period *T*0 = 1/*f*0 regardless of the values of amplitudes *ak* and *bk* of these sinusoids. By changing the values of *ak* and *bk* in Eq. (6.1), we can construct a variety of periodic signals, all of the same period *T*0 (*T*0 = 1/*f*0 = 2π/ω0). - -The converse of this result is also true. We shall show in Sec. 6.5-4 that *a periodic signal x*(*t*) *with a period T*0 *can be expressed as a sum of a sinusoid of frequency f*0 *(f*0 = 1/*T*0*) and all its harmonics, as shown in Eq. (6.1)*. † The infinite series on the right-hand side of Eq. (6.1) is known as the *trigonometric Fourier series* of a periodic signal *x*(*t*). - -### COMPUTING THE COEFFICIENTS OF A FOURIER SERIES - -To determine the coefficients of a Fourier series, consider an integral *I* defined by - -$$ -I = \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt -$$ - -where \$ *T*0 stands for integration over any contiguous interval of *T*0 seconds. By using a trigonometric identity (see Sec. B.8-6), this integral can be expressed as - -$$ -I = \frac{1}{2} \left[ \int_{T_0} \cos(n+m)\omega_0 t \, dt + \int_{T_0} \cos(n-m)\omega_0 t \, dt \right] \tag{6.3} -$$ - -† Strictly speaking, this statement applies only if a periodic signal *x*(*t*) is a continuous function of *t*. However, Sec. 6.5-4 shows that it can be applied even for discontinuous signals, if we interpret the equality in Eq. (6.1) in the mean-square sense instead of in the ordinary sense. This means that the power of the difference between the periodic signal *x*(*t*) and its Fourier series on the right-hand side of Eq. (6.1) approaches zero as the number of terms in the series approaches infinity. - -#### 596 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -Because cos ω0*t* executes one complete cycle during any interval of duration *T*0, cos(*n* + *m*)ω0*t* executes (*n* + *m*) complete cycles during any interval of duration *T*0. Therefore, the first integral in Eq. (6.3), which represents the area under *n*+*m* complete cycles of a sinusoid, equals zero. The same argument shows that the second integral in Eq. (6.3) is also zero, except when *n* = *m*. Hence, *I* in Eq. (6.3) is zero for all *n* = *m*. When *n* = *m*, the first integral in Eq. (6.3) is still zero, but the second integral yields - -$$ -I = \frac{1}{2} \int_{T_0} dt = \frac{T_0}{2} -$$ - -Thus, - -$$ -\int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & m = n \neq 0 \end{cases} \tag{6.4} -$$ - -Using similar arguments, we can show that - -$$ -\int_{T_0} \sin n\omega_0 t \sin m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & n = m \neq 0 \end{cases} \tag{6.5} -$$ - -and - -$$ -\int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt = 0 \qquad \text{for all } n \text{ and } m \tag{6.6} -$$ - -To determine *a*0 in Eq. (6.1), we integrate both sides of Eq. (6.1) over one period *T*0 to yield - -$$ -\int_{T_0} x(t) dt = a_0 \int_{T_0} dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t dt + b_n \int_{T_0} \sin n\omega_0 t dt \right] -$$ - -Recall that *T*0 is the period of a sinusoid of frequency ω0. Therefore, functions cos *n*ω0*t* and sin *n*ω0*t* execute *n* complete cycles over any interval of *T*0 seconds so that the area under these functions over an interval *T*0 is zero, and the last two integrals on the right-hand side of the foregoing equation are zero. This yields - -$$ -\int_{T_0} x(t) dt = a_0 \int_{T_0} dt = a_0 T_0 \quad \text{and} \quad a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt -$$ - -Next we multiply both sides of Eq. (6.1) by cos*m*ω0*t* and integrate the resulting equation over an interval *T*0: - -$$ -\int_{T_0} x(t) \cos m\omega_0 t \, dt = a_0 \int_{T_0} \cos m\omega_0 t \, dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt + b_n \int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt \right] -$$ - -The first integral on the right-hand side is zero because it is an area under *m* integral number of cycles of a sinusoid. Also, the last integral on the right-hand side vanishes because of Eq. (6.6). This leaves only the middle integral, which is also zero for all *n* = *m* because of Eq. (6.4). But *n* takes on all values from 1 to ∞, including *m*. When *n* = *m*, this integral is *T*0/2, according to Eq. (6.4). Therefore, from the infinite number of terms on the right-hand side, only one term survives to yield *anT*0/2 = *amT*0/2 (recall that *n* = *m*). Therefore, - -$$ -\int_{T_0} x(t) \cos m\omega_0 t \, dt = \frac{a_m T_0}{2} \qquad \text{and} \qquad a_m = \frac{2}{T_0} \int_{T_0} x(t) \cos m\omega_0 t \, dt -$$ - -Similarly, by multiplying both sides of Eq. (6.1) by sin *n*ω0*t* and then integrating over an interval *T*0, we obtain - -$$ -b_m = \frac{2}{T_0} \int_{T_0} x(t) \sin m\omega_0 t \, dt -$$ - -To sum up our discussion, which applies to real or complex *x*(*t*), we have shown that a periodic signal *x*(*t*) with period *T*0 can be expressed as a sum of a sinusoid of period *T*0 and its harmonics: - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.7} -$$ - -where ω0 = 2π*f*0 = 2π *T*0 and - -$$ -a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{T_0} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{T_0} x(t) \sin n\omega_0 t dt \tag{6.8} -$$ - -### COMPACT FORM OF FOURIER SERIES - -The results derived so far are general and apply whether *x*(*t*) is a real or a complex function of *t*. However, when *x*(*t*) is real, coefficients *an* and *bn* are real for all *n*, and the trigonometric Fourier series can be expressed in a *compact form,* using the results in Eq. (B.16): - -$$ -x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \theta_n) -$$ -\n(6.9) - -where *Cn* and θ*n* are related to *an* and *bn*, as [see Eq. (B.17)] - -$$ -C_0 = a_0 -$$ -, $C_n = \sqrt{a_n^2 + b_n^2}$ , and $\theta_n = \tan^{-1} \left( \frac{-b_n}{a_n} \right)$ (6.10) - -These results are summarized in Table 6.1. - -The compact form in Eq. (6.9) uses the cosine form. We could just as well have used the sine form, with terms sin(*n*ω0*t* + θ*n*) instead of cos(*n*ω0*t* + θ*n*). The literature overwhelmingly favors the cosine form, for no apparent reason except possibly that the cosine phasor is represented by the horizontal axis, which happens to be the reference axis in phasor representation. - -Equation (6.8) shows that *a*0 (or *C*0) is the average value of *x*(*t*) (averaged over one period). This value can often be determined by inspection of *x*(*t*). - -Because *an* and *bn* are real, *Cn* and θ*n* are also real. In the following discussion of trigonometric Fourier series, we shall assume real *x*(*t*), unless mentioned otherwise. - -| Series Form | Coefficient Computation | Conversion Formulas | -|--------------------------------------------------------|----------------------------------------------------|-------------------------------| -| Trigonometric | #
1
a0
f(t)dt
=
T0
T0 | a0
= C0
= D0 | -| "∞
f(t) = a0+
an
cosnω0t+bn
sinnω0t
n=1 | #
2
an
f(t) cosnω0t dt
=
T0
T0 | = Cnejθn =
an−jbn
2Dn | -| | #
2
bn
f(t)sinnω0t dt
=
T0 | = Cne−jθn =
an+jbn
2D−n | -| Compact trigonometric | T0
C0
= a0 | C0
= D0 | -| "∞
f(t) = C0
Cn
cos(nω0t +θn)
+ | =
2
2 +bn
Cn
an | Cn
= 2 Dn
n ≥ 1 | -| n=1 | −bn
= tan−1
θn
an | θn
= Dn | -| Exponential | | | -| f(t) = "∞
Dnejnω0t
n=−∞ | #
1
f(t)e−jnω0t
Dn
dt
=
T0
T0 | | - -**TABLE 6.1** Fourier Series Representation of a Periodic Signal of Period *T*0 (ω0 = 2π/*T*0) - -### **[6.1-1 The Fourier Spectrum](#page-12-0)** - -The compact trigonometric Fourier series in Eq. (6.9) indicates that a periodic signal *x*(*t*) can be expressed as a sum of sinusoids of frequencies 0 (dc), ω0, 2ω0, ..., *n*ω0, ..., whose amplitudes are *C*0, *C*1, *C*2, ..., *Cn*, ..., and whose phases are 0, θ1, θ2, ..., θ*n*, ..., respectively. We can readily plot amplitude *Cn* versus *n* (*the amplitude spectrum*) and θ*n* versus *n* (the *phase spectrum*).† Because *n* is proportional to the frequency *n*ω0, these plots are scaled plots of *Cn* versus ω and θ*n* versus ω. The two plots together are the *frequency spectra* of *x*(*t*). These spectra show at a glance the frequency contents of the signal *x*(*t*) with their amplitudes and phases. Knowing these spectra, we can reconstruct or synthesize the signal *x*(*t*) according to Eq. (6.9). Therefore, frequency spectra, which are an alternative way of describing a periodic signal *x*(*t*), are in every way equivalent to the plot of *x*(*t*) as a function of *t*. The frequency spectra of a signal constitute the *frequency-domain description* of *x*(*t*), in contrast to the *time-domain description,* where *x*(*t*) is specified as a function of time. - -In computing θ*n*, the phase of the *n*th harmonic from Eq. (6.10), the quadrant in which θ*n* lies should be determined from the signs of *an* and *bn*. For example, if *an* = −1 and *bn* = 1, θ*n* lies in the third quadrant, and - -$$ -\theta_n = \tan^{-1}\left(\frac{-1}{-1}\right) = -135^\circ -$$ - -Observe that - -$$ -\tan^{-1}\left(\frac{-1}{-1}\right) \neq \tan^{-1}(1) = 45^{\circ} -$$ - -† The amplitude *Cn*, by definition here, is nonnegative. Some authors define amplitude *An* that can take positive or negative values and magnitude *Cn* = |*An*| that can only be nonnegative. Thus, what we call amplitude spectrum becomes magnitude spectrum. The distinction between amplitude and magnitude, although useful, is avoided in this book in the interest of keeping definitions of essentially similar entities to a minimum. - -Although *Cn*, the amplitude of the *n*th harmonic as defined in Eq. (6.10), is positive, we shall find it convenient to allow *Cn* to take on negative values when *bn* = 0. This will become clear in later examples. - -## **EXAMPLE 6.1 Compact Trigonometric Fourier Series of Periodic Exponential Wave** - -Find the compact trigonometric Fourier series for the periodic signal *x*(*t*) shown in Fig. 6.2a. Sketch the amplitude and phase spectra for *x*(*t*). - -**Figure 6.2 (a)** A periodic signal and **(b, c)** its Fourier spectra. - -In this case the period *T*0 = π and the fundamental frequency *f*0 = 1/*T*0 = 1/π Hz, and - -$$ -\omega_0 = \frac{2\pi}{T_0} = 2 \,\text{rad/s} -$$ - -Therefore, - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos 2nt + b_n \sin 2nt -$$ - -where - -$$ -a_0 = \frac{1}{\pi} \int_{T_0} x(t) dt -$$ - -In this example the obvious choice for the interval of integration is from 0 to π. Hence, - -$$ -a_0 = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} dt = 0.504 -$$ - -$$ -a_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \cos 2nt dt = 0.504 \left(\frac{2}{1 + 16n^2}\right) -$$ - -and - -$$ -b_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \sin 2nt \, dt = 0.504 \left( \frac{8n}{1 + 16n^2} \right) -$$ - -Therefore, - -$$ -x(t) = 0.504 \left[ 1 + \sum_{n=1}^{\infty} \frac{2}{1 + 16n^2} (\cos 2nt + 4n \sin 2nt) \right] -$$ - -Also from Eq. (6.10), - -$$ -C_0 = a_0 = 0.504 -$$ - -\n -$$ -C_n = \sqrt{a_n^2 + b_n^2} = 0.504 \sqrt{\frac{4}{(1 + 16n^2)^2} + \frac{64n^2}{(1 + 16n^2)^2}} = 0.504 \left(\frac{2}{\sqrt{1 + 16n^2}}\right) -$$ - -\n -$$ -\theta_n = \tan^{-1}\left(\frac{-b_n}{a_n}\right) = \tan^{-1}(-4n) = -\tan^{-1} 4n -$$ - -Amplitude and phases of the dc and the first seven harmonics are computed from the above equations as - -| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | -|----|-------|---------|---------|---------|---------|---------|---------|---------| -| Cn | 0.504 | 0.244 | 0.125 | 0.084 | 0.063 | 0.0504 | 0.042 | 0.036 | -| θn | 0◦ | −75.96◦ | −82.87◦ | −85.24◦ | −86.42◦ | −87.14◦ | −87.61◦ | −87.95◦ | - -We can use these numerical values to express *x*(*t*) as - -$$ -x(t) = 0.504 + 0.504 \sum_{n=1}^{\infty} \frac{2}{\sqrt{1 + 16n^2}} \cos(2nt - \tan^{-1} 4n) -$$ - -= 0.504 + 0.244 cos (2t - 75.96°) + 0.125 cos (4t - 82.87°) -+ 0.084 cos (6t - 85.24°) + 0.063 cos (8t - 86.42°) + ... (6.11) - -### PLOTTING FOURIER SERIES SPECTRA USING MATLAB - -MATLAB is well suited to compute and plot Fourier series spectra. The results in Fig. 6.3, which plot *Cn* and θ*n* as functions of *n*, match Figs. 6.2b and 6.2c, which plot *Cn* and θ*n* as functions of ω = *n*ω0 = 2*n*. Plots of *an* and *bn* are similarly simple to generate. - -The amplitude and phase spectra for *x*(*t*), in Figs. 6.2b and 6.2c, tell us at a glance the frequency composition of *x*(*t*), that is, the amplitudes and phases of various sinusoidal components of *x*(*t*). Knowing the frequency spectra, we can reconstruct *x*(*t*), as shown on the right-hand side of Eq. (6.11). Therefore the frequency spectra (Figs. 6.2b, 6.2c) provide an alternative description—the frequency-domain description of *x*(*t*). The time-domain description of *x*(*t*) is shown in Fig. 6.2a. *A signal, therefore, has a dual identity: the time-domain identity x*(*t*) *and the frequency-domain identity (Fourier spectra). The two identities complement each other; taken together, they provide a better understanding of a signal.* - -An interesting aspect of Fourier series is that whenever there is a jump discontinuity in *x*(*t*), the series at the point of discontinuity converges to an average of the left-hand and right-hand limits of *x*(*t*) at the instant of discontinuity.† In the present example, for instance, *x*(*t*) is discontinuous at *t* = 0 with *x*(0+) = 1 and *x*(0−) = *x*(π ) = *e*−π/2 = 0.208. The corresponding Fourier series converges to a value (1 + 0.208)/2 = 0.604 at *t* = 0. This is easily verified from Eq. (6.11) by setting *t* = 0. - -† This behavior of the Fourier series is dictated by its convergence in the mean, discussed later in Secs. 6.2 and 6.5. - -### 602 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -### **EXAMPLE 6.2 Compact Trigonometric Fourier Series of a Periodic Triangle Wave** - -Find the compact trigonometric Fourier series for the triangular periodic signal *x*(*t*) shown in Fig. 6.4a, and sketch the amplitude and phase spectra for *x*(*t*). - -**Figure 6.4 (a)** A triangular periodic signal and **(b, c)** its Fourier spectra. - -In this case the period *T*0 = 2. Hence, - -$$ -\omega_0 = \frac{2\pi}{2} = \pi -$$ - -and - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\pi t + b_n \sin n\pi t -$$ - -where - -$$ -x(t) = \begin{cases} 2At & |t| < \frac{1}{2} \\ 2A(1-t) & \frac{1}{2} < t < \frac{3}{2} \end{cases} -$$ - -Here it will be advantageous to choose the interval of integration from −1/2 to 3/2 rather than 0 to 2. - -A glance at Fig. 6.4a shows that the average value (dc) of *x*(*t*) is zero so that *a*0 = 0. Also, - -$$ -a_n = \frac{2}{2} \int_{-1/2}^{3/2} x(t) \cos n\pi t dt -$$ - -= -$$ -\int_{-1/2}^{1/2} 2A t \cos n\pi t dt + \int_{1/2}^{3/2} 2A(1-t) \cos n\pi t dt -$$ - -Detailed evaluation of these integrals shows that both have a value of zero. Therefore *an* = 0. Next, - -$$ -b_n = \int_{-1/2}^{1/2} 2At \sin n\pi t \, dt + \int_{1/2}^{3/2} 2A(1-t) \sin n\pi t \, dt -$$ - -Detailed evaluation of these integrals yields, in turn, - -$$ -b_n = \frac{8A}{n^2 \pi^2} \sin\left(\frac{n\pi}{2}\right) = \begin{cases} 0 & n \text{ even} \\ \frac{8A}{n^2 \pi^2} & n = 1, 5, 9, 13, \dots \\ -\frac{8A}{n^2 \pi^2} & n = 3, 7, 11, 15, \dots \end{cases} -$$ - -Therefore, - -$$ -x(t) = \frac{8A}{\pi^2} \left[ \sin \pi t - \frac{1}{9} \sin 3\pi t + \frac{1}{25} \sin 5\pi t - \frac{1}{49} \sin 7\pi t + \dots \right] -$$ - (6.12) - -To plot Fourier spectra, the series must be converted into compact trigonometric form as in Eq. (6.9). In this case this is readily done by converting sine terms into cosine terms with a suitable phase shift. For example, - -$$ -\sin kt = \cos (kt - 90^\circ) \qquad \text{and} \qquad -\sin kt = \cos (kt + 90^\circ) -$$ - -By using these identities, Eq. (6.12) can be expressed as - -$$ -x(t) = \frac{8A}{\pi^2} \bigg[ \cos(\pi t - 90^\circ) + \frac{1}{9} \cos(3\pi t + 90^\circ) + \frac{1}{25} \cos(5\pi t - 90^\circ) + \frac{1}{49} \cos(7\pi t + 90^\circ) + \cdots \bigg] -$$ - -In this series all the even harmonics are missing. The phases of the odd harmonics alternate from −90◦ to 90◦. Figure 6.4 shows amplitude and phase spectra for *x*(*t*). - -## **EXAMPLE 6.3 Converting a Trigonometric FS to a Compact Trigonometric FS** - -A periodic signal *x*(*t*) is represented by a trigonometric Fourier series - -$$ -x(t) = 2 + 3\cos 2t + 4\sin 2t + 2\sin (3t + 30^\circ) - \cos (7t + 150^\circ) -$$ - -Express this series as a compact trigonometric Fourier series, and sketch amplitude and phase spectra for *x*(*t*). - -In compact trigonometric Fourier series, the sine and cosine terms of the same frequency are combined into a single term and all terms are expressed as cosine terms with positive amplitudes. Using Eqs. (6.9) and (6.10), we have - -$$ -3\cos 2t + 4\sin 2t = 5\cos (2t - 53.13^{\circ}) -$$ - -Also, - -$$ -\sin(3t + 30^{\circ}) = \cos(3t + 30^{\circ} - 90^{\circ}) = \cos(3t - 60^{\circ}) -$$ - -and - -$$ --\cos(7t + 150^{\circ}) = \cos(7t + 150^{\circ} - 180^{\circ}) = \cos(7t - 30^{\circ}) -$$ - -Therefore, - -$$ -x(t) = 2 + 5\cos(2t - 53.13^{\circ}) + 2\cos(3t - 60^{\circ}) + \cos(7t - 30^{\circ}) -$$ - -**Figure 6.5** Fourier spectra of the signal. - -In this case only four components (including dc) are present. The amplitude of dc is 2. The remaining three components are of frequencies ω = 2, 3, and 7 with amplitudes 5, 2, and 1 and phases −53.13◦, −60◦, and −30◦, respectively. The amplitude and phase spectra for this signal are shown in Figs. 6.5a and 6.5b, respectively. - -### **EXAMPLE 6.4 Compact Trigonometric Fourier Series of a Periodic Square Wave** - -Find the compact trigonometric Fourier series for the square-pulse periodic signal shown in Fig. 6.6a and sketch its Fourier spectrum. - -**Figure 6.6 (a)** A square pulse periodic signal and **(b)** its Fourier spectrum. - -Here the period is *T*0 = 2π and ω0 = 2π/*T*0 = 1. Therefore, - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos nt + b_n \sin nt -$$ - -where - -$$ -a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt -$$ - -From Fig. 6.6a, it is clear that a proper choice of region of integration is from −π to π. But since *x*(*t*) = 1 only over (−π/2, π/2), and *x*(*t*) = 0 over the remaining segment, - -$$ -a_0 = \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} dt = \frac{1}{2} -$$ - -We could have found *a*0, the average value of *x*(*t*), to be 1/2 merely by inspection of *x*(*t*) in Fig. 6.6a. Also, - -$$ -a_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos nt \, dt = \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right) -$$ - -= -$$ -\begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n = 1, 5, 9, 13, \dots \\ -\frac{2}{\pi n} & n = 3, 7, 11, 15, \dots \end{cases} -$$ - -$$ -b_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \sin nt \, dt = 0 -$$ - -Therefore - -$$ -x(t) = \frac{1}{2} + \frac{2}{\pi} \left( \cos t - \frac{1}{3} \cos 3t + \frac{1}{5} \cos 5t - \frac{1}{7} \cos 7t + \cdots \right) -$$ -(6.13) - -Observe that *bn* = 0 and all the sine terms are zero. Only the cosine terms appear in the trigonometric series. The series is therefore already in the compact form except that the amplitudes of alternating harmonics are negative. Now by definition, amplitudes *Cn* are positive [see Eq. (6.10)]. The negative sign can be accommodated by associating a proper phase, as seen from the trigonometric identity† - -−cos *x* = cos(*x* −π ) - -Using this fact, we can express the series in Eq. (6.13) as - -$$ -x(t) = \frac{1}{2} + \frac{2}{\pi} \left[ \cos \omega_0 t + \frac{1}{3} \cos (3\omega_0 t - \pi) + \frac{1}{5} \cos 5\omega_0 t + \frac{1}{7} \cos (7\omega_0 t - \pi) + \frac{1}{9} \cos 9\omega_0 t + \cdots \right] -$$ - -† Because cos(*x*±π ) = −cos *x*, we could have chosen the phase π or −π. In fact, cos(*x*±*N*π ) = −cos *x* for any odd integral value of *N*. Therefore the phase can be chosen as ±*N*π, where *N* is any convenient odd integer. - -This is the desired form of the compact trigonometric Fourier series. The amplitudes are - -$$ -C_0 = \frac{1}{2} -$$ - and $C_n = \begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n \text{ odd} \end{cases}$ - -The phases are - -$$ -\theta_n = \begin{cases} 0 & \text{for all } n \neq 3, 7, 11, 15, \dots \\ -\pi & n = 3, 7, 11, 15, \dots \end{cases} -$$ - -We might use these values to plot amplitude and phase spectra. However, we can simplify our task in this special case if we allow amplitude *Cn* to take on negative values. If this is allowed, we do not need a phase of −π to account for the sign as seen from Eq. (6.13). This means that phases of all components are zero, and we can discard the phase spectrum and manage with only the amplitude spectrum, as shown in Fig. 6.6b. Observe that there is no loss of information in doing so and that the amplitude spectrum in Fig. 6.6b has the complete information about the Fourier series in Eq. (6.13). *Therefore, whenever all sine terms vanish (bn* = 0*), it is convenient to allow Cn to take on negative values*. This permits the spectral information to be conveyed by a single spectrum.† - -Let us investigate the behavior of the series at the points of discontinuities. For the discontinuity at *t* =π/2, the values of *x*(*t*) on either sides of the discontinuity are *x*((π/2)−)=1 and *x*((π/2)+) = 0. We can verify by setting *t* = π/2 in Eq. (6.13) that *x*(π/2) = 0.5, which is a value midway between the values of *x*(*t*) on either side of the discontinuity at *t* = π/2. - -### **[6.1-2 The Effect of Symmetry](#page-12-0)** - -The Fourier series for the signal *x*(*t*) in Fig. 6.2a (Ex. 6.1) consists of sine and cosine terms, but the series for the signal *x*(*t*) in Fig. 6.4a (Ex. 6.2) consists of sine terms only, and the series for the signal *x*(*t*) in Fig. 6.6a (Ex. 6.4) consists of cosine terms only. This is no accident. We can show that the Fourier series of any even periodic function *x*(*t*) consists of cosine terms only and the series for any odd periodic function *x*(*t*) consists of sine terms only. Moreover, because of symmetry (even or odd), the information of one period of *x*(*t*) is implicit in only half the period, as seen in Figs. 6.4a and 6.6a. In these cases, knowing the signal over a half-period and knowing the kind of symmetry (even or odd), we can determine the signal waveform over a complete period. For this reason, the Fourier coefficients in these cases can be computed by integrating over only half the period rather than a complete period. To prove this result, recall that - -$$ -a_0 = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \sin n\omega_0 t dt -$$ - -Recall also that cos*n*ω0*t* is an even function and sin*n*ω0*t* is an odd function of *t*. If *x*(*t*) is an even function of *t*, then *x*(*t*) cos*n*ω0*t* is also an even function and *x*(*t*)sin *n*ω0*t* is an odd function of *t* - -† Here, the distinction between amplitude *An* and magnitude *Cn* = |*An*| would have been useful. But, for the reasons mentioned in the footnote on page 598, we refrain from this distinction formally. - -#### 608 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -(see Sec. 1.5-1). Therefore, following from Eq. (1.16), - -$$ -a_0 = \frac{2}{T_0} \int_0^{T_0/2} x(t) dt, \quad a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = 0 \tag{6.14} -$$ - -Similarly, if *x*(*t*) is an odd function of *t*, then *x*(*t*) cos *n*ω0*t* is an odd function of *t* and *x*(*t*)sin *n*ω0*t* is an even function of *t*. Therefore, - -$$ -a_n = 0 -$$ - and $b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t dt$ (6.15) - -Observe that because of symmetry, the integration required to compute the coefficients need be performed over only half the period. - -If a periodic signal *x*(*t*) shifted by half the period remains unchanged except for a sign—that is, if - -$$ -x\left(t - \frac{T_0}{2}\right) = -x(t) -$$ - -then the signal is said to have a *half-wave* symmetry. It can be shown that for a signal with a half-wave symmetry, all the even-numbered harmonics vanish (see Prob. 6.1-6). The signal in Fig. 6.4a is an example of such a symmetry. The signal in Fig. 6.6a also has this symmetry, although it is not obvious owing to a dc component. If we subtract the dc component of 0.5 from this signal, the remaining signal has half-wave symmetry. For this reason, this signal has only odd harmonics and a dc component of 0.5. - -### **DR ILL 6.1 Compact Trigonometric Fourier Series** - -Find the compact trigonometric Fourier series for periodic signals shown in Fig. 6.7. Sketch their amplitude and phase spectra. Allow *Cn* to take on negative values if *bn* = 0 so that the phase spectrum can be eliminated. [*Hint:* Use Eqs. (6.14) and (6.15) for appropriate symmetry conditions.] - -### **ANSWERS** - -(a) -$$ -x(t) = \frac{1}{3} - \frac{4}{\pi^2} \left( \cos \pi t - \frac{1}{4} \cos 2\pi t + \frac{1}{9} \cos 3\pi t - \frac{1}{16} \cos 4\pi t + \cdots \right) -$$ - -\n -$$ -= \frac{1}{3} + \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos n\pi t -$$ -\n(b) $x(t) = \frac{2A}{\pi} \left[ \sin \pi t - \frac{1}{2} \sin 2\pi t + \frac{1}{3} \sin 3\pi t - \frac{1}{4} \sin 4\pi t + \cdots \right]$ -\n -$$ -= \frac{2A}{\pi} \left[ \cos (\pi t - 90^\circ) + \frac{1}{2} \cos (2\pi t + 90^\circ) + \frac{1}{3} \cos (3\pi t - 90^\circ) + \frac{1}{4} \cos (4\pi t + 90^\circ) + \cdots \right] -$$ - - - -### **[6.1-3 Determining the Fundamental Frequency and Period](#page-12-0)** - -We have seen that every periodic signal can be expressed as a sum of sinusoids of a fundamental frequency ω0 and its harmonics. One may ask whether a sum of sinusoids of *any* frequencies represents a periodic signal. If so, how does one determine the period? Consider the following three functions: - -$$ -x_1(t) = 2 + 7\cos(\frac{1}{2}t + \theta_1) + 3\cos(\frac{2}{3}t + \theta_2) + 5\cos(\frac{7}{6}t + \theta_3) -$$ - -\n -$$ -x_2(t) = 2\cos(2t + \theta_1) + 5\sin(\pi t + \theta_2) -$$ - -\n -$$ -x_3(t) = 3\sin(3\sqrt{2}t + \theta) + 7\cos(6\sqrt{2}t + \phi) -$$ - -Recall that every frequency in a periodic signal is an integer multiple of the fundamental frequency ω0. Therefore the ratio of any two frequencies is of the form *m*/*n*, where *m* and *n* are integers. This means that the ratio of any two frequencies is a rational number. When the ratio of two frequencies is a rational number, the frequencies are said to be *harmonically* related. - -The largest number of which all the frequencies are integer multiples is the fundamental frequency. In other words, the fundamental frequency is the *greatest common factor* (GCF) of all the frequencies in the series. The frequencies in the spectrum of *x*1(*t*) are 1/2, 2/3, and 7/6 (we do not consider dc). The ratios of the successive frequencies are 3:4 and 4:7, respectively. Because both these numbers are rational, all the three frequencies in the spectrum are harmonically related, and the signal *x*1(*t*) is periodic. The GCF, that is, the greatest number of which 1/2, 2/3, and 7/6 are integer multiples, is 1/6.† Moreover, 3(1/6) = 1/2, 4(1/6) = 2/3, and 7(1/6) = 7/6. Therefore the fundamental frequency is 1/6, and the three frequencies in the spectrum are the third, fourth, and seventh harmonics. Observe that the fundamental frequency component is absent in this Fourier series. - -† The greatest common factor of *a*1/*b*1, *a*2/*b*2, ..., *am*/*bm* is the ratio of the GCF of the numerators set (*a*1,*a*2,...,*am*) to the LCM (least common multiple) of the denominator set (*b*1,*b*2,...,*bm*). For instance, for the set (2/3, 6/7, 2), the GCF of the numerator set (2, 6, 2) is 2; the LCM of the denominator set (3, 7, 1) is 21. Therefore, 2/21 is the largest number of which 2/3, 6/7, and 2 are integer multiples. - -#### 610 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -The signal *x*2(*t*) is not periodic because the ratio of two frequencies in the spectrum is 2/π, which is not a rational number. The signal *x*3(*t*) is periodic because the ratio of frequencies 3 √ 2 and 6√ 2 is 1/2, a rational number. The greatest common factor of 3√2 and 6√2 is 3√2. Therefore, the fundamental frequency ω0 = 3 √2, and the period - -$$ -T_0 = \frac{2\pi}{(3\sqrt{2})} = \frac{\sqrt{2}}{3}\pi -$$ - -### **DR ILL 6.2 Determining Periodicity, Fundamental Frequency, and Harmonic Content** - -Determine whether the signal - -$$ -x(t) = \cos\left(\frac{2}{3}t + 30^{\circ}\right) + \sin\left(\frac{4}{5}t + 45^{\circ}\right) -$$ - -is periodic. If it is periodic, find the fundamental frequency and the period. What harmonics are present in *x*(*t*)? - -### **ANSWERS** - -Periodic with ω0 =2/15 and period *T*0 =15π. Signal *x*(*t*) contains the fifth and sixth harmonics. - -## A HISTORICAL NOTE: BARON JEAN-BAPTISTE-JOSEPH FOURIER (1768–1830) - -The Fourier series and integral comprise a most beautiful and fruitful development, which serves as an indispensable instrument in the treatment of many problems in mathematics, science, and engineering. Maxwell was so taken by the beauty of the Fourier series that he called it a great mathematical poem. In electrical engineering, it is central to the areas of communication, signal processing, and several other fields, including antennas, but its initial reception by the scientific world was not enthusiastic. In fact, Fourier could not get his results published as a paper. - -Fourier, a tailor's son, was orphaned at age 8 and educated at a local military college (run by Benedictine monks), where he excelled in mathematics. The Benedictines prevailed upon the young genius to choose the priesthood as his vocation, but revolution broke out before he could take his vows. Fourier joined the people's party. But in its early days, the French Revolution, like most such upheavals, liquidated a large segment of the intelligentsia, including prominent scientists such as Lavoisier. Observing this trend, many intellectuals decided to leave France to save themselves from a rapidly rising tide of barbarism. Fourier, despite his early enthusiasm for the Revolution, narrowly escaped the guillotine twice. It was to the everlasting credit of Napoleon that he stopped the persecution of the intelligentsia and founded new schools to replenish their ranks. The 26-year-old Fourier was appointed chair of mathematics at the newly created École Normale in 1794 [1]. - -Jean-Baptiste-Joseph Fourier and Napoleon - -Napoleon was the first modern ruler with a scientific education, and he was one of the rare persons who are equally comfortable with soldiers and scientists. The age of Napoleon was one of the most fruitful in the history of science. Napoleon liked to sign himself as "member of *Institut de France*" (a fraternity of scientists), and he once expressed to Laplace his regret that "force of circumstances has led me so far from the career of a scientist" [2]. Many great figures in science and mathematics, including Fourier and Laplace, were honored and promoted by Napoleon. In 1798 he took a group of scientists, artists, and scholars—Fourier among them—on his Egyptian expedition, with the promise of an exciting and historic union of adventure and research. Fourier proved to be a capable administrator of the newly formed Institut d'Égypte, which, incidentally, was responsible for the discovery of the Rosetta Stone. The inscription on this stone in two languages and three scripts (hieroglyphic, demotic, and Greek) enabled Thomas Young and Jean-François Champollion, a protégé of Fourier, to invent a method of translating hieroglyphic writings of ancient Egypt—the only significant result of Napoleon's Egyptian expedition. - -Back in France in 1801, Fourier briefly served in his former position as professor of mathematics at the École Polytechnique in Paris. In 1802 Napoleon appointed him the prefect of Isère (with its headquarters in Grenoble), a position in which Fourier served with distinction. Fourier was named Baron of the Empire by Napoleon in 1809. Later, when Napoleon was exiled to Elba, his route was to take him through Grenoble. Fourier had the route changed to avoid meeting Napoleon, which would have displeased Fourier's new master, King Louis XVIII. Within a year, Napoleon escaped from Elba and returned to France. At Grenoble, Fourier was brought before him in chains. Napoleon scolded Fourier for his ungrateful behavior but reappointed him the prefect of Rhône at Lyons. Within four months Napoleon was defeated at Waterloo and was exiled to St. Helena, where he died in 1821. Fourier once again was in disgrace as a Bonapartist and had to pawn his possessions to keep himself alive. But through the intercession of a former student, who was now a prefect of Paris, he was appointed director of the statistical bureau of the Seine, a position that allowed him ample time for scholarly pursuits. Later, in 1827, he was elected to the powerful position of perpetual secretary of the Paris Academy of Science, a section of the Institut de France [3]. - -While serving as the prefect of Grenoble, Fourier carried on his elaborate investigation of the propagation of heat in solid bodies, which led him to the Fourier series and the Fourier integral. On December 21, 1807, he announced these results in a prize paper on the theory of heat. Fourier claimed that an arbitrary function (continuous or with discontinuities) defined in a finite interval by an arbitrarily capricious graph can always be expressed as a sum of sinusoids (Fourier series). The judges, who included the great French mathematicians Laplace, Lagrange, Legendre, Monge, and LaCroix, admitted the novelty and importance of Fourier's work but criticized it for lack of mathematical rigor and generality. Lagrange thought it incredible that a sum of sines and cosines could add up to anything but an infinitely differentiable function. Moreover, one of the properties of an infinitely differentiable function is that if we know its behavior over an arbitrarily small interval, we can determine its behavior over the entire range (the Taylor–Maclaurin series). Such a function is far from an arbitrary or a capriciously drawn graph [4]. Laplace had additional reason to criticize Fourier's work. Laplace and his students had already approached the problem of heat conduction from a different angle, and Laplace was reluctant to accept the superiority of Fourier's method [5]. Fourier thought the criticism unjustified but was unable to prove his claim because the tools required for operations with infinite series were not available at the time. However, posterity has proved Fourier to be closer to the truth than his critics. This is the classic conflict between pure mathematicians and physicists or engineers, as we saw earlier (Ch. 4) in the life of Oliver Heaviside. In 1829 Dirichlet proved Fourier's claim concerning capriciously drawn functions with a few restrictions (Dirichlet conditions). - -Although three of the four judges were in favor of publication, Fourier's paper was rejected because of vehement opposition by Lagrange. Fifteen years later, after several attempts and disappointments, Fourier published the results in expanded form as a text, *Théorie analytique de la chaleur,* which is now a classic. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/080_6.2 EXISTENCE AND CONVERGENCE OF THE FOURIER SERIES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/080_6.2 EXISTENCE AND CONVERGENCE OF THE FOURIER SERIES.md deleted file mode 100644 index 301e0e0e2d3855f8820d2d8f730cfbd6aa6943a4..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/080_6.2 EXISTENCE AND CONVERGENCE OF THE FOURIER SERIES.md +++ /dev/null @@ -1,261 +0,0 @@ -## **[6.2 EXISTENCE AND](#page-12-0) CONVERGENCE OF THE FOURIER SERIES** - -For the existence of the Fourier series, coefficients *a*0,*an*, and *bn* in Eq. (6.8) must be finite. It follows from Eq. (6.8) that the existence of these coefficients is guaranteed if *x*(*t*) is absolutely integrable over one period; that is, - -$$ -\int_{T_0} |x(t)| \, dt < \infty \tag{6.16} -$$ - -However, existence, by itself, does not inform us about the nature and the manner in which the series converges. We shall first discuss the notion of convergence. - -### **[6.2-1 Convergence of a Series](#page-12-0)** - -The key to many puzzles lies in the nature of the convergence of the Fourier series. Convergence of infinite series is a complex problem. It took mathematicians several decades to understand the convergence aspect of the Fourier series. We shall barely scratch the surface here. - -Nothing annoys a student more than the discussion of convergence. "Have we not proved," they ask, "that a periodic signal *x*(*t*) can be expressed as a Fourier series"? Then why spoil the fun by this annoying discussion? All we have shown so far is that a signal represented by a Fourier series in Eq. (6.1) is periodic. We have not proved the converse, that every periodic signal can be expressed as a Fourier series. This issue will be tackled later, in Sec. 6.5-4, where it will be shown that a periodic signal can be represented by a Fourier series, as in Eq. (6.1), where the equality of the two sides of the equation is not in the ordinary sense, but in the mean-square sense (explained later in this discussion). But the astute reader should have been skeptical of the claims of the Fourier series to represent discontinuous functions in Figs. 6.2a and 6.6a. If *x*(*t*) has a jump discontinuity, say, at *t* = 0, then *x*(0+), *x*(0), and *x*(0−) are generally different. How could a series consisting of the sum of continuous functions of the smoothest type (sinusoids) add to one value at *t* = 0− and a different value at *t* = 0 and yet another value at *t* = 0+? The demand is impossible to satisfy unless the math involved executes some spectacular acrobatics. How does a Fourier series act under such conditions? Precisely for this reason, the great mathematicians Lagrange and Laplace, two of the judges examining Fourier's paper, were skeptical of Fourier's claims and voted against publication of the paper that later became a classic. - -There are also other issues. In any practical application, we can use only a finite number of terms in a series. If, with a fixed number of terms, the series guarantees convergence within an arbitrarily small error at every value of *t*, such a series is highly desirable and is called a *uniformly convergent* series. If a series converges at every value of *t*, but to guarantee convergence within a given error requires a different number of terms at different *t*, then the series is still convergent, but less desirable. It goes under the name *pointwise convergent* series. - -Finally, we have the case of a series that refuses to converge at some *t*, no matter how many terms are added. But the series may *converge in the mean*; that is, the energy of the difference between *x*(*t*) and the corresponding finite term series approaches zero as the number of terms approaches infinity.† To explain this concept, let us consider representation of a function *x*(*t*) by an infinite series - -$$ -x(t) = \sum_{n=1}^{\infty} z_n(t) -$$ - -Let the partial sum of the first *N* terms of the series on the right-hand side be denoted by *xN*(*t*), that is, - -$$ -x_N(t) = \sum_{n=1}^N z_n(t) -$$ - -† The behavior is called "convergence in the mean" because minimizing the error energy over a certain interval is equivalent to minimizing the mean-square value of the error over the same interval. - -#### 614 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -If we approximate *x*(*t*) by *xN*(*t*) (the partial sum of the first *N* terms of the series), the *error* in the approximation is the difference *x*(*t*) − *xN*(*t*). The series converges *in the mean* to *x*(*t*) in the interval (0, *T*0) if - -$$ -\int_0^{T_0} |x(t) - x_N(t)|^2 dt \to 0 \quad \text{as} \quad N \to \infty -$$ - -Hence, the energy of the error *x*(*t*)−*xN*(*t*) approaches zero as *N* → ∞. This form of convergence does not require the series to be equal to *x*(*t*) for all *t*. It just requires the energy of the difference (area under |*x*(*t*)−*xN*(*t*)| 2) to vanish as *N* → ∞. Superficially it may appear that if the energy of a signal over an interval is zero, the signal (the error) must be zero everywhere. This is not true. The signal energy can be zero even if there are nonzero values at a finite number of isolated points. This is because although the signal is nonzero at a point (and zero everywhere else), the area under its square is still zero. Thus, a series that converges in the mean to *x*(*t*) need not converge to *x*(*t*) at a finite number of points. This is precisely what happens to the Fourier series when *x*(*t*) has jump discontinuities. This is also what makes Fourier series convergence compatible with the Gibbs phenomenon, to be discussed later in this section. - -There is a simple criterion for ensuring that a periodic signal *x*(*t*) has a Fourier series that converges in the mean. The Fourier series for *x*(*t*) converges to *x*(*t*) in the mean if *x*(*t*) has a finite energy over one period, that is, - -$$ -\int_{T_0} |x(t)|^2 dt < \infty \tag{6.17} -$$ - -Thus, the periodic signal *x*(*t*), having a finite energy over one period, guarantees the convergence in the mean of its Fourier series. In all the examples discussed so far, Eq. (6.17) is satisfied; hence the corresponding Fourier series converges in the mean. Equation (6.17), like Eq. (6.16), guarantees that the Fourier coefficients are finite. - -We shall now discuss an alternate set of criteria, due to Dirichlet, for convergence of the Fourier series. - -### DIRICHLET CONDITIONS - -Dirichlet showed that if *x*(*t*) satisfies certain conditions (*Dirichlet conditions*), its Fourier series is guaranteed to converge pointwise at all points where *x*(*t*) is continuous. Moreover, at the points of discontinuities, *x*(*t*) converges to the value midway between the two values of *x*(*t*) on either side of the discontinuity. These conditions are: - -- 1. The function *x*(*t*) must be absolutely integrable; that is, it must satisfy Eq. (6.16). -- 2. The function *x*(*t*) must have only a finite number of finite discontinuities in one period. -- 3. The function *x*(*t*) must contain only a finite number of maxima and minima in one period. - -All practical signals, including those in Exs. 6.1, 6.2, 6.3, and 6.4, satisfy these conditions. - -### **[6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping](#page-12-0)** - -The trigonometric Fourier series of a signal *x*(*t*) shows explicitly the sinusoidal components of *x*(*t*). We can synthesize *x*(*t*) by adding the sinusoids in the spectrum of *x*(*t*). Let us synthesize the square-pulse periodic signal *x*(*t*) of Fig. 6.6a by adding successive harmonics in its spectrum step by step and observing the similarity of the resulting signal to *x*(*t*). The Fourier series for this function as found in Eq. (6.13) is - -$$ -x(t) = \frac{1}{2} + \frac{2}{\pi} \left( \cos t - \frac{1}{3} \cos 3t + \frac{1}{5} \cos 5t - \frac{1}{7} \cos 7t + \cdots \right) -$$ - -We start the synthesis with only the first term in the series (*n* = 0), a constant 1/2 (dc); this is a gross approximation of the square wave, as shown in Fig. 6.8a. In the next step we add the dc (*n* = 0) and the first harmonic (fundamental), which results in a signal shown in Fig. 6.8b. Observe that the synthesized signal somewhat resembles *x*(*t*). It is a smoothed-out version of *x*(*t*). The sharp corners in *x*(*t*) are not reproduced in this signal because sharp corners mean rapid changes, and their reproduction requires rapidly varying (i.e., higher-frequency) components, which are excluded. Figure 6.8c shows the sum of dc, first, and third harmonics (even harmonics are absent). As we increase the number of harmonics progressively, as shown in Figs. 6.8d (sum up to the fifth harmonic) and 6.8e (sum up to the nineteenth harmonic), the edges of the pulses become sharper and the signal resembles *x*(*t*) more closely. - -### ASYMPTOTIC RATE OF AMPLITUDE SPECTRUM DECAY - -Figure 6.8 brings out one interesting aspect of the Fourier series. Lower frequencies in the Fourier series affect the large-scale behavior of *x*(*t*), whereas the higher frequencies determine the fine structure such as rapid wiggling. Hence, sharp changes in *x*(*t*), being a part of fine structure, necessitate higher frequencies in the Fourier series. The sharper the change [the higher the time derivative *x*˙(*t*)], the higher are the frequencies needed in the series. - -The amplitude spectrum indicates the amounts (amplitudes) of various frequency components of *x*(*t*). If *x*(*t*) is a smooth function, its variations are less rapid. Synthesis of such a function requires predominantly lower-frequency sinusoids and relatively small amounts of rapidly varying (higher-frequency) sinusoids. The amplitude spectrum of such a function would decay swiftly with frequency. To synthesize such a function, we require fewer terms in the Fourier series for a good approximation. On the other hand, a signal with sharp changes, such as jump discontinuities, contains rapid variations, and its synthesis requires a relatively large amount of high-frequency components. The amplitude spectrum of such a signal would decay slowly with frequency, and to synthesize such a function, we require many terms in its Fourier series for a good approximation. The square wave *x*(*t*) is a discontinuous function with jump discontinuities, and therefore its amplitude spectrum decays rather slowly, as 1/*n* [see Eq. (6.13)]. On the other hand, the triangular-pulse periodic signal in Fig. 6.4a is smoother because it is a continuous function (no jump discontinuities). Its spectrum decays rapidly with frequency as 1/*n*2 [see Eq. (6.12)]. - -We can show that if the first *k* − 1 derivatives of a periodic signal *x*(*t*) are continuous and the *k*th derivative is discontinuous, then its amplitude spectrum *Cn* decays with frequency at least as rapidly as 1/*nk*+1 [6]. This result provides a simple and useful means for predicting the asymptotic - -**Figure 6.8** Synthesis of a square-pulse periodic signal by successive addition of its harmonics. - -rate of convergence of the Fourier series. In the case of the square-wave signal (Fig. 6.6a), the zeroth derivative of the signal (the signal itself) is discontinuous so that *k* = 0. For the triangular periodic signal in Fig. 6.4a, the first derivative is discontinuous; that is, *k* = 1. For this reason, the spectra of these signals decay as 1/*n* and 1/*n*2, respectively. - -## **EXAMPLE 6.5 Square-Wave Synthesis by Truncated Fourier Series Using MATLAB** - -Use MATLAB to synthesize and plot the square wave of Fig. 6.8a using a Fourier series that is truncated to the 19th harmonic. The result should match Fig. 6.8e. - -To synthesize the waveform, we use the Fourier series of Eq. (6.13). - -``` ->> x = @(t) 1.0*(mod(t+pi/2,2*pi)<=pi); ->> t = linspace(-2*pi,2*pi,10001); ->> x19 = 0.5*ones(size(t)); ->> for n=1:19, x19 = x19+2/(pi*n)*sin(pi*n/2)*cos(n*t); end ->> plot(t,x19,'k-'); axis([-2*pi 2*pi -0.2 1.2]); ->> xlabel('t'); ylabel('x_{19}(t)'); -``` - -As expected, the result of Fig. 6.9 matches Fig. 6.8e. - -### PHASE SPECTRUM: THE WOMAN BEHIND A SUCCESSFUL MAN - -The role of the amplitude spectrum in shaping the waveform *x*(*t*) is quite clear. However, the role of the phase spectrum in shaping this waveform is less obvious. Yet, the phase spectrum, like the woman behind a successful man,† plays an equally important role in waveshaping. We can explain this role by considering a signal *x*(*t*) that has rapid changes, such as jump discontinuities. To synthesize an instantaneous change at a jump discontinuity, the phases of the various sinusoidal components in its spectrum must be such that all (or most) of the harmonic components will have - -† Or, to keep up with the times, the man behind a successful woman. - -#### 618 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -one sign before the discontinuity and the opposite sign after the discontinuity. This will result in a sharp change in *x*(*t*) at the point of discontinuity. We can verify this fact in any waveform with jump discontinuity. Consider, for example, the sawtooth waveform in Fig. 6.7b. This waveform has a discontinuity at *t* = 1. The Fourier series for this waveform, as given in Drill 6.1b, is - -$$ -x(t) = \frac{2A}{\pi} \left[ \cos(\pi t - 90^\circ) + \frac{1}{2} \cos(2\pi t + 90^\circ) + \frac{1}{3} \cos(3\pi t - 90^\circ) + \frac{1}{4} \cos(4\pi t + 90^\circ) + \cdots \right] -$$ - -Figure 6.10 shows the first three components of this series. The phases of all the (infinite) components are such that all the components are positive just before *t* = 1 and turn negative just after *t* = 1, the point of discontinuity. The same behavior is also observed at *t* = −1, where a similar discontinuity occurs. This sign change in all the harmonics adds up to produce very nearly a jump discontinuity. The role of the phase spectrum is crucial in achieving a sharp change in the waveform. If we ignore the phase spectrum when trying to reconstruct this signal, the result will be a smeared and spread-out waveform. In general, the phase spectrum is just as crucial as the amplitude spectrum in determining the waveform. *The synthesis of any signal x*(*t*) *is achieved by using a proper combination of amplitudes and phases of various sinusoids. This unique combination is the Fourier spectrum of x*(*t*)*.* - -**Figure 6.10** Role of the phase spectrum in shaping a periodic signal. - -### FOURIER SYNTHESIS OF DISCONTINUOUS FUNCTIONS: THE GIBBS PHENOMENON - -Figure 6.8 showed the square function *x*(*t*) and its approximation by a truncated trigonometric Fourier series that includes only the first *N* harmonics for *N* = 1, 3, 5, and 19. The plot of the truncated series approximates closely the function *x*(*t*) as *N* increases, and we expect that the series will converge exactly to *x*(*t*) as *N* → ∞. Yet the curious fact, as seen from Fig. 6.8, is that even for large *N*, the truncated series exhibits an oscillatory behavior and an overshoot approaching a value of about 9% in the vicinity of the discontinuity at the nearest peak of oscillation.† Regardless of the value of *N*, the overshoot remains at about 9%. Such strange behavior certainly would undermine anyone's faith in the Fourier series. In fact, this behavior puzzled many scholars at the turn of the century. Josiah Willard Gibbs, an eminent mathematical physicist who was the inventor of vector analysis, gave a mathematical explanation of this behavior (now called the *Gibbs phenomenon*). - -We can reconcile the apparent aberration in the behavior of the Fourier series by observing from Fig. 6.8 that the frequency of oscillation of the synthesized signal is *Nf*0, so the width of the spike with 9% overshoot is approximately 1/2*Nf*0. As we increase *N*, the frequency of oscillation increases and the spike width 1/2*Nf*0 diminishes. As *N* → ∞, the error power → 0 because the error consists mostly of the spikes, whose widths → 0. Therefore, as *N* → ∞, the corresponding Fourier series differs from *x*(*t*) by about 9% at the immediate left and right of the points of discontinuity, and yet the error power →0. The reason for all this confusion is that in this case, the Fourier series converges in the mean. When this happens, all we promise is that the error energy (over one period) → 0 as *N* → ∞. Thus, the series may differ from *x*(*t*) at some points and yet have the error signal power zero, as verified earlier. Note that the series, in this case, also converges pointwise at all points except the points of discontinuity. It is precisely at the discontinuities that the series differs from *x*(*t*) by 9%.‡ - -When we use only the first *N* terms in the Fourier series to synthesize a signal, we are abruptly terminating the series, giving a unit weight to the first *N* harmonics and zero weight to all the remaining harmonics beyond *N*. This abrupt termination of the series causes the Gibbs phenomenon in synthesis of discontinuous functions. Section 7.8 offers more discussion on the Gibbs phenomenon, its ramifications, and cure. - -The Gibbs phenomenon is present only when there is a jump discontinuity in *x*(*t*). When a continuous function *x*(*t*) is synthesized by using the first *N* terms of the Fourier series, the synthesized function approaches *x*(*t*) for all *t* as *N* → ∞. No Gibbs phenomenon appears. This can be seen in Fig. 6.11, which shows one cycle of a continuous periodic signal being synthesized from the first 19 harmonics. Compare the similar situation for a discontinuous signal in Fig. 6.8. - -### **DR ILL 6.3 Rate of Spectral Decay** - -By inspection of signals in Figs. 6.2a, 6.7a, and 6.7b, determine the asymptotic rate of decay of their amplitude spectra. - -† There is also an undershoot of 9% at the other side [at *t* = (π/2)+] of the discontinuity. - -‡ Actually, at discontinuities, the series converges to a value midway between the values on either side of the discontinuity. The 9% overshoot occurs at *t* = (π/2)− and 9% undershoot occurs at *t* = (π/2)+. - -**Figure 6.11** Fourier synthesis of a continuous signal using first 19 harmonics. - -### **ANSWERS** - -1/*n*, 1/*n*2, and 1/*n*, respectively. - -### A HISTORICAL NOTE ON THE GIBBS PHENOMENON - -Normally speaking, troublesome functions with strange behavior are invented by mathematicians; we rarely see such oddities in practice. In the case of the Gibbs phenomenon, however, the tables were turned. A rather puzzling behavior was observed in a mundane object, a mechanical wave synthesizer, and then well-known mathematicians of the day were dispatched on the scent of it to discover its hideout. - -Albert Michelson (of Michelson–Morley fame) was an intense, practical man who developed ingenious physical instruments of extraordinary precision, mostly in the field of optics. His harmonic analyzer, developed in 1898, could compute the first 80 coefficients of the Fourier series of a signal *x*(*t*) specified by any graphical description. The instrument could also be used as a harmonic synthesizer, which could plot a function *x*(*t*) generated by summing the first 80 harmonics (Fourier components) of arbitrary amplitudes and phases. This analyzer, therefore, had the ability of self-checking its operation by analyzing a signal *x*(*t*) and then adding the resulting 80 components to see whether the sum yielded a close approximation of *x*(*t*). - -Michelson found that the instrument checked very well with most of signals analyzed. However, when he tried a discontinuous function, such as a square wave,† a curious behavior was observed. The sum of 80 components showed oscillatory behavior (ringing), with an overshoot of 9% in the vicinity of the points of discontinuity. Moreover, this behavior was a constant feature regardless of the number of terms added. A larger number of terms made the oscillations proportionately faster, but regardless of the number of terms added, the overshoot remained 9%. This puzzling behavior caused Michelson to suspect some mechanical defect in his synthesizer. He wrote about his observation in a letter to *Nature* (December 1898). Josiah Willard Gibbs, who was a professor at Yale, investigated and clarified this behavior for a sawtooth periodic signal in a letter to *Nature* [7]. Later, in 1906, Bôcher generalized the result for any function with discontinuity [8]. - -† Actually, it was a periodic sawtooth signal. - -Albert Michelson and Josiah Willard Gibbs - -It was Bôcher who gave the name *Gibbs phenomenon* to this behavior. Gibbs showed that the peculiar behavior in the synthesis of a square wave was inherent in the behavior of the Fourier series because of nonuniform convergence at the points of discontinuity. - -This, however, is not the end of the story. Both Bôcher and Gibbs were under the impression that this property had remained undiscovered until Gibbs's work published in 1899. It is now known that what is called the Gibbs phenomenon had been observed in 1848 by Wilbraham of Trinity College, Cambridge, who clearly saw the behavior of the sum of the Fourier series components in the periodic sawtooth signal later investigated by Gibbs [9]. Apparently, this work was not known to most people, including Gibbs and Bôcher. - -## **[6.3 EXPONENTIAL](#page-12-0) FOURIER SERIES** - -By using Euler's equality, we can express cos *n*ω0*t* and sin *n*ω0*t* in terms of exponentials *ejn*ω0*t* and *e*−*jn*ω0*t* . Clearly, we should be able to express the trigonometric Fourier series in Eq. (6.7) in terms of exponentials of the form *ejn*ω0*t* with the index *n* taking on all integer values from −∞ to ∞, including zero. Derivation of the exponential Fourier series from the results already derived for the trigonometric Fourier series is straightforward, involving conversion of sinusoids to exponentials. We shall, however, derive them here independently, without using the prior results of the trigonometric series. - -This discussion shows that the *exponential Fourier series* for a periodic signal *x*(*t*) can be expressed as - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} -$$ - -#### 622 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -To derive the coefficients *Dn*, we multiply both sides of this equation by *e*−*jm*ω0*t* (*m* integer) and integrate over one period. This yields - -$$ -\int_{T_0} x(t) e^{-jm\omega_0 t} dt = \sum_{n=-\infty}^{\infty} D_n \int_{T_0} e^{j(n-m)\omega_0 t} dt -$$ - -To simplify this expression, we use the *orthogonality* property of exponentials, which states that† - -$$ -\int_{T_0} e^{jn\omega_0 t} e^{-jm\omega_0 t} dt = \begin{cases} 0 & m \neq n \\ T_0 & m = n \end{cases} -$$ -\n(6.18) - -Thus, - -$$ -\int_{T_0} x(t) e^{-jm\omega_0 t} dt = D_m T_0 -$$ - -from which we obtain - -$$ -D_m = \frac{1}{T_0} \int_{T_0} x(t) e^{-jm\omega_0 t} dt. -$$ - -To summarize, the exponential Fourier series can be expressed as - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \text{where} \qquad D_n = \frac{1}{T_0} \int_{T_0} x(t) e^{-jn\omega_0 t} dt \tag{6.19} -$$ - -Observe the compactness of Eq. (6.19) and compare it with the trigonometric Fourier series expression. Such a comparison demonstrates very clearly the principal virtue of the exponential Fourier series. First, the form of the series is most compact. Second, the mathematical expression for deriving the coefficients of the series is also compact. It is much more convenient to handle the exponential series than the trigonometric one. For these reasons we shall use the exponential (rather than trigonometric) representation of signals in the rest of the book. - -We can now relate *Dn* to trigonometric series coefficients *an* and *bn*. Setting *n*=0 in Eq. (6.19), we obtain - -$$ -D_0=a_0 -$$ - -Moreover, for *n* = 0, - -$$ -D_n = \frac{1}{T_0} \int_{T_0} x(t) \cos n\omega_0 t \, dt - \frac{j}{T_0} \int_{T_0} x(t) \sin n\omega_0 t \, dt = \frac{1}{2} (a_n - jb_n) \tag{6.20} -$$ - -$$ -\int_{T_0} e^{j(n-m)\omega_0 t} dt = \int_{T_0} \cos{(n-m)\omega_0 t} dt + j \int_{T_0} \sin{(n-m)\omega_0 t} dt -$$ - -Both the integrals on the right-hand side represent area under *n* − *m* number of cycles. Because *n* − *m* is an integer, both the areas are zero. Hence, Eq. (6.18) follows. - -† We can readily prove this property as follows. For the case of *m* = *n*, the integrand in Eq. (6.18) is unity and the integral is *T*0. When *m* = *n*, the integral on the left-hand side of Eq. (6.18) can be expressed as - -and - -$$ -D_{-n} = \frac{1}{T_0} \int_{T_0} x(t) \cos n\omega_0 t \, dt + \frac{j}{T_0} \int_{T_0} x(t) \sin n\omega_0 t \, dt = \frac{1}{2} (a_n + jb_n) \tag{6.21} -$$ - -These results are valid for general *x*(*t*), real or complex. When *x*(*t*) is real, *an* and *bn* are real, and Eqs. (6.20) and (6.21) show that *Dn* and *D*−*n* are conjugates. - -$$ -D_{-n}=D_n^* -$$ - -Moreover, from Eq. (6.10), we observe that - -$$ -a_n - jb_n = \sqrt{a_n^2 + b_n^2} \, e^{j \tan^{-1} \left( \frac{-b_n}{a_n} \right)} = C_n e^{j \theta_n} -$$ - -Hence, - -$$ -D_0=a_0=C_0 -$$ - -and - -$$ -D_n = \frac{1}{2} C_n e^{j\theta_n} \qquad D_{-n} = \frac{1}{2} C_n e^{-j\theta_n} -$$ - -Therefore, for *n* = 0, - -$$ -|D_n| = |D_{-n}| = \frac{1}{2}C_n, \quad \angle D_n = \theta_n, \quad \text{and} \angle D_{-n} = -\theta_n \tag{6.22} -$$ - -Note that |*Dn*| are the amplitudes and *Dn* are the angles of various exponential components. From Eq. (6.22) it follows that when *x*(*t*) is real, the amplitude spectrum (|*Dn*| versus ω) is an even function of ω and the angle spectrum ( *Dn* versus ω) is an odd function of ω. For complex *x*(*t*), *Dn* and *D*−*n* are generally not conjugates. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/081_6.3 EXPONENTIAL FOURIER SERIES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/081_6.3 EXPONENTIAL FOURIER SERIES.md deleted file mode 100644 index b981fb1f74d5be26fedb6e8ebc1992cd4dd467f4..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/081_6.3 EXPONENTIAL FOURIER SERIES.md +++ /dev/null @@ -1,417 +0,0 @@ -### **EXAMPLE 6.6 Exponential Fourier Series of Periodic Exponential Wave** - -Find the exponential Fourier series for the signal of Fig. 6.2a from Ex. 6.1. - -In this case *T*0 = π, ω0 = 2π/*T*0 = 2, and - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{j2nt} -$$ - -where - -$$ -D_n = \frac{1}{T_0} \int_{T_0} x(t) e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-(1/2 + j2n)t} dt -$$ - -= $\frac{-1}{\pi (\frac{1}{2} + j2n)} e^{-(1/2 + j2n)t} \Big|_0^{\pi} = \frac{0.504}{1 + j4n}$ - -$$ -x(t) = 0.504 \sum_{n=-\infty}^{\infty} \frac{1}{1+j4n} e^{j2nt} -$$ - -= 0.504 $\left[ 1 + \frac{1}{1+j4} e^{j2t} + \frac{1}{1+j8} e^{j4t} + \frac{1}{1+j12} e^{j6t} + \cdots + \frac{1}{1-j4} e^{-j2t} + \frac{1}{1-j8} e^{-j4t} + \frac{1}{1-j12} e^{-j6t} + \cdots \right]$ - -Observe that the coefficients *Dn* are complex. Moreover, *Dn* and *D*−*n* are conjugates, as expected. - -### **[6.3-1 Exponential Fourier Spectra](#page-12-0)** - -In exponential spectra, we plot coefficients *Dn* as a function of ω. But since *Dn* is complex in general, we need both parts of one of two sets of plots: the real and the imaginary parts of *Dn*, or the magnitude and the angle of *Dn*. We prefer the latter because of its close connection to the amplitudes and phases of corresponding components of the trigonometric Fourier series. We therefore plot |*Dn*| versus ω and *Dn* versus ω. This requires that the coefficients *Dn* be expressed in polar form as |*Dn*|*ej Dn* , where |*Dn*| are the amplitudes and *Dn* are the angles of various exponential components. Equation (6.22) shows that for real *x*(*t*), the amplitude spectrum (|*Dn*| versus ω) is an even function of ω and the angle spectrum ( *Dn* versus ω) is an odd function of ω. - -For the series in Ex. 6.6, for instance, - -$$ -D_0 = 0.504 -$$ - -\n -$$ -D_1 = \frac{0.504}{1+j4} = 0.122e^{-j75.96^\circ} \implies |D_1| = 0.122, \angle D_1 = -75.96^\circ -$$ - -\n -$$ -D_{-1} = \frac{0.504}{1-j4} = 0.122e^{j75.96^\circ} \implies |D_{-1}| = 0.122, \angle D_{-1} = 75.96^\circ -$$ - -and - -$$ -D_2 = \frac{0.504}{1+j8} = 0.0625e^{-j82.87^\circ} \implies |D_2| = 0.0625, \ \angle D_2 = -82.87^\circ -$$ - -$$ -D_{-2} = \frac{0.504}{1-j8} = 0.0625e^{j82.87^\circ} \implies |D_{-2}| = 0.0625, \ \angle D_{-2} = 82.87^\circ -$$ - -and so on. Note that *Dn* and *D*−*n* are conjugates, as expected [see Eq. (6.22)]. - -Figure 6.12 shows the frequency spectra (amplitude and angle) of the exponential Fourier series for the periodic signal *x*(*t*) in Fig. 6.2a. - -We notice some interesting features of these spectra. First, the spectra exist for positive as well as negative values of ω (the frequency). Second, the amplitude spectrum is an even function of ω and the angle spectrum is an odd function of ω. - -and - -**Figure 6.12** Exponential Fourier spectra for the signal in Fig. 6.2a. - -At times it may appear that the phase spectrum of a real periodic signal fails to satisfy the odd symmetry: for example, when *Dk* = *D*−*k* = −10. In this case, *Dk* = 10*ej*π , and therefore, *D*−*k* = 10*e*−*j*π . Recall that *e*±*j*π = −1. Here, although *Dk* = *D*−*k*, their phases should be taken as π and −π. - -### **EXAMPLE 6.7 Plotting Fourier Series Spectra with MATLAB** - -Using MATLAB and the results of Ex. 6.6, compute and plot the exponential Fourier spectra for the periodic signal *x*(*t*) shown in Fig. 6.2a. The result should match Fig. 6.12. - -The expression for *Dn* is derived in Ex. 6.6. - -``` ->> clf; n = (-5:5); D_n = 0.504./(1+4j*n); -``` - -- >> subplot(1,2,1); stem(n,abs(D\_n),'.k'); -- >> xlabel('n'); ylabel('|D\_n|'); -- >> subplot(1,2,2); stem(n,angle(D\_n),'.k'); -- >> xlabel('n'); ylabel('\angle D\_n [rad]'); - -Except that it is plotted as a function of *n* rather than ω, the result in Fig. 6.13 matches Fig. 6.12. - -### WHAT IS A NEGATIVE FREQUENCY? - -The existence of the spectrum at negative frequencies is somewhat disturbing because, by definition, the frequency (number of repetitions per second) is a positive quantity. How do we interpret a negative frequency? We can use a trigonometric identity to express a sinusoid of a negative frequency −ω0 as - -$$ -\cos(-\omega_0 t + \theta) = \cos(\omega_0 t - \theta) -$$ - -This equation clearly shows that the frequency of a sinusoid cos(ω0*t* + θ ) is |ω0|, which is a positive quantity. The same conclusion is reached by observing that - -$$ -e^{\pm j\omega_0 t} = \cos \omega_0 t \pm j \sin \omega_0 t -$$ - -Thus, the frequency of exponentials *e*±*j*ω0*t* is indeed |ω0|. How do we then interpret the spectral plots for negative values of ω? A more satisfying way of looking at the situation is to say that *exponential spectra are a graphical representation of coefficients Dn as a function of* ω*. Existence of the spectrum at* ω = −*n*ω0 *is merely an indication that an exponential component e*−*jn*ω0*t exists in the series*. We know that a sinusoid of frequency *n*ω0 can be expressed in terms of a pair of exponentials *ejn*ω0*t* and *e*−*jn*ω0*t* . - -We see a close connection between the exponential spectra in Fig. 6.12 and the spectra of the corresponding trigonometric Fourier series for *x*(*t*) (Figs. 6.2b, 6.2c). Equation (6.22) explains the reason for the close connection, for real *x*(*t*), between the trigonometric spectra (*Cn* and θ*n*) with exponential spectra (|*Dn*| and *Dn*). The dc components *D*0 and *C*0 are identical in both spectra. Moreover, the exponential amplitude spectrum |*Dn*| is half the trigonometric amplitude spectrum *Cn* for *n* ≥ 1. The exponential angle spectrum *Dn* is identical to the trigonometric phase spectrum θ*n* for *n* ≥ 0. We can therefore produce the exponential spectra merely by inspection of trigonometric spectra, and vice versa. The following example demonstrates this feature. - -### **EXAMPLE 6.8 Relating Exponential to Trigonometric Fourier Series Spectra** - -The trigonometric Fourier spectra of a certain periodic signal *x*(*t*) are shown in Fig. 6.14a. After inspecting these spectra, sketch the corresponding exponential Fourier spectra and verify your results analytically. - -**Figure 6.14** Fourier series spectra for Ex. 6.8. - -The trigonometric spectral components exist at frequencies 0, 3, 6, and 9. The exponential spectral components exist at 0, 3, 6, 9, and −3, −6, −9. Consider first the amplitude spectrum. The dc component remains unchanged: that is, *D*0 = *C*0 = 16. Now |*Dn*| is an even function of ω and |*Dn*|=|*D*−*n*| = *Cn*/2. Thus, all the remaining spectrum |*Dn*| for positive *n* is half the trigonometric amplitude spectrum *Cn*, and the spectrum |*Dn*| for negative *n* is a reflection about the vertical axis of the spectrum for positive *n*, as shown in Fig. 6.14b. - -The angle spectrum is *Dn* = θ*n* for positive *n* and is −θ*n* for negative *n*, as depicted in Fig. 6.14b. We shall now verify that both sets of spectra represent the same signal. - -Signal *x*(*t*), whose trigonometric spectra are shown in Fig. 6.14a, has four spectral components of frequencies 0, 3, 6, and 9. The dc component is 16. The amplitude and the phase of the component of frequency 3 are 12 and −π/4, respectively. Therefore, this component can be expressed as 12cos(3*t* − π/4). Proceeding in this manner, we can write the Fourier series for *x*(*t*) as - -$$ -x(t) = 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right) -$$ - -Consider now the exponential spectra in Fig. 6.14b. They contain components of frequencies 0 (dc), ±3, ±6, and ±9. The dc component is *D*0 = 16. The component *ej*3*t* (frequency 3) has magnitude 6 and angle −π/4. Therefore, this component strength is 6*e*−*j*π/4, and it can be expressed as (6*e*−*j*π/4)*ej*3*t* . Similarly, the component of frequency −3 is (6*ej*π/4)*e*−*j*3*t* . Proceeding in this manner, *x*ˆ(*t*), the signal corresponding to the spectra in Fig. 6.14b, is - -$$ -\hat{x}(t) = 16 + \left[6e^{-j\pi/4}e^{j3t} + 6e^{j\pi/4}e^{-j3t}\right] + \left[4e^{-j\pi/2}e^{j6t} + 4e^{j\pi/2}e^{-j6t}\right] -$$ -\n -$$ -+ \left[2e^{-j\pi/4}e^{j9t} + 2e^{j\pi/4}e^{-j9t}\right] -$$ -\n -$$ -= 16 + 6\left[e^{j(3t - \pi/4)} + e^{-j(3t - \pi/4)}\right] + 4\left[e^{j(6t - \pi/2)} + e^{-j(6t - \pi/2)}\right] -$$ -\n -$$ -+ 2\left[e^{j(9t - \pi/4)} + e^{-j(9t - \pi/4)}\right] -$$ -\n -$$ -= 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right) -$$ - -Clearly both sets of spectra represent the same periodic signal. - -### BANDWIDTH OF A SIGNAL - -The difference between the highest and the lowest frequencies of the spectral components of a signal is the *bandwidth* of the signal. The bandwidth of the signal whose exponential spectra are shown in Fig. 6.14b is 9 (in radians). The highest and lowest frequencies are 9 and 0, respectively. Note that the component of frequency 12 has zero amplitude and is nonexistent. Moreover, the lowest frequency is 0, not −9. Recall that the frequencies (in the conventional sense) of the spectral components at ω = −3, −6, and −9 in reality are 3, 6, and 9.† The bandwidth can be more readily seen from the trigonometric spectra in Fig. 6.14a. - -### **EXAMPLE 6.9 Fourier Series Spectra of an Impulse Train** - -Find the exponential Fourier series and sketch the corresponding spectra for the impulse train δ*T*0 (*t*) depicted in Fig. 6.15a. From this result, sketch the trigonometric spectrum and write the trigonometric Fourier series for δ*T*0 (*t*). - -† Some authors *do* define bandwidth as the difference between the highest and the lowest (negative) frequency in the exponential spectrum. The bandwidth according to this definition is twice that defined here. In reality, this phrasing defines not the signal bandwidth but the *spectral width* (width of the exponential spectrum of the signal). - -The unit impulse train shown in Fig. 6.15a can be expressed as - -$$ -\sum_{n=-\infty}^{\infty} \delta(t - nT_0) -$$ - -Following Papoulis, we shall denote this function as δ*T*0 (*t*) for the sake of notational brevity. The exponential Fourier series is given by - -$$ -\delta_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ -\n(6.23) - -where - -$$ -D_n = \frac{1}{T_0} \int_{T_0} \delta_{T_0}(t) e^{-jn\omega_0 t} dt -$$ - -Choosing the interval of integration (−*T*0/2,*T*0/2) and recognizing that over this interval δ*T*0 (*t*) = δ(*t*), we get - -$$ -D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} \delta(t) e^{-jn\omega_0 t} dt -$$ - -### 630 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -In this integral, the impulse is located at *t* = 0. From the sampling property of Eq. (1.11), the integral on the right-hand side is the value of *e*−*jn*ω0*t* at *t* = 0 (where the impulse is located). Therefore, - -$$ -D_n = \frac{1}{T_0} \tag{6.24} -$$ - -From this result, we see that the exponential spectrum is constant for all frequencies, as shown in Fig. 6.15b. The spectrum, being real, requires only the amplitude plot. All phases are zero. - -Substituting *Dn* = 1 *T*0 into Eq. (6.23) yields the desired exponential Fourier series - -$$ -\delta_{T_0}(t) = \frac{1}{T_0} \sum_{n = -\infty}^{\infty} e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -To sketch the trigonometric spectrum, we use Eq. (6.22) to obtain - -$$ -C_0 = D_0 = \frac{1}{T_0} -$$ - -\n -$$ -C_n = 2|D_n| = \frac{2}{T_0} \qquad n = 1, 2, 3, ... -$$ - -\n -$$ -\theta_n = 0 -$$ - -Figure 6.15c shows the trigonometric Fourier spectrum. From this spectrum we can express δ*T*0 (*t*) as - -$$ -\delta_{T_0}(t) = \frac{1}{T_0} \left[ 1 + 2(\cos \omega_0 t + \cos 2\omega_0 t + \cos 3\omega_0 t + \cdots) \right] \qquad \omega_0 = \frac{2\pi}{T_0} \tag{6.25} -$$ - -### EFFECT OF SYMMETRY IN EXPONENTIAL FOURIER SERIES - -When *x*(*t*) has an even symmetry, *bn* = 0, and from Eq. (6.20), *Dn* = *an*/2, which is real (positive or negative). Hence, *Dn* can only be 0 or ±π. Moreover, we may compute *Dn* = *an*/2 by using Eq. (6.14), which requires integration over a half-period only. Similarly, when *x*(*t*) has an odd symmetry, *an* = 0, and *Dn* = −*jbn*/2 is imaginary (positive or negative). Hence, *Dn* can only be 0 or ±π/2. Moreover, we may compute *Dn* = −*jbn*/2 by using Eq. (6.15), which requires integration over a half-period only. Note, however, that in the exponential case, we are using the symmetry property indirectly by finding the trigonometric coefficients. We cannot apply it directly in finding *Dn* from Eq. (6.19) since the function *ejn*ω0*t* is neither even nor odd. - -## **DR ILL 6.4 Relating Trigonometric to Exponential Fourier Series Spectra** - -The exponential Fourier spectra of a certain periodic signal *x*(*t*) are shown in Fig. 6.16. Determine and sketch the trigonometric Fourier spectra of *x*(*t*) by inspection of Fig. 6.16. Now write the (compact) trigonometric Fourier series for *x*(*t*). - -## **DR ILL 6.5 Fourier Series Spectrum of a Full-Wave Rectified Sine Wave** - -Find the exponential Fourier series and sketch the corresponding Fourier spectrum *Dn* versus ω for the full-wave rectified sine wave depicted in Fig. 6.17. - -### **ANSWER** - -### **DR ILL 6.6 Exponential Fourier Series and Spectra** - -Find the exponential Fourier series and sketch the corresponding Fourier spectra for the periodic signals shown in Fig. 6.7. - -#### **ANSWERS (a)** *x*(*t*) = 1 3 + 2 π2 "∞ *n*=−∞(*n*=0) (−1)*n n*2 *ejn*π*t* **(b)** *x*(*t*) = *jA* π "∞ *n*=−∞(*n*=0) (−1)*n n ejn*π*t* - -### **[6.3-2 Parseval's Theorem](#page-12-0)** - -The trigonometric Fourier series of a periodic signal *x*(*t*) is given by - -$$ -x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \theta_n) -$$ - -Every term on the right-hand side of this equation is a power signal. As shown in Ex. 1.2, Eq. (1.3), the power of *x*(*t*) is equal to the sum of the powers of all the sinusoidal components on the right-hand side. - -$$ -P_x = C_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} C_n^2 -$$ -\n(6.26) - -This result is one form of *Parseval's theorem,* as applied to power signals. It states that the power of a periodic signal is equal to the sum of the powers of its Fourier components. - -We can apply the same argument to the exponential Fourier series (see Prob. 1.1-11). The power of a periodic signal *x*(*t*) can be expressed as a sum of the powers of its exponential components. In Eq. (1.4), we showed that the power of an exponential *Dej*ω0*t* is |*D*2|. We can use this result to express the power of a periodic signal *x*(*t*) in terms of its exponential Fourier series coefficients as - -$$ -P_x = \sum_{n=-\infty}^{\infty} |D_n|^2 \tag{6.27} -$$ - -For a real *x*(*t*), |*D*−*n*|=|*Dn*|. Therefore, - -$$ -P_x = D_0^2 + 2\sum_{n=1}^{\infty} |D_n|^2 -$$ -\n(6.28) - -### **EXAMPLE 6.10 Harmonic Distortion of Clipped Sinusoid** - -The input signal to an audio amplifier of gain 100 is given by *x*(*t*) = 0.1 cosω0*t*. Hence, the output is a sinusoid 10 cos ω0*t*. However, the amplifier, being nonlinear at higher amplitude levels, clips all amplitudes beyond ±8 volts, as shown in Fig. 6.18a. We shall determine the harmonic distortion incurred in this operation. - -**Figure 6.18 (a)** A clipped sinusoid cos ω0*t*. **(b)** The distortion component *xd*(*t*) of the signal in (a). - -The output *y*(*t*) is the clipped signal in Fig. 6.18a. The distortion signal *yd*(*t*), shown in Fig. 6.18b, is the difference between the undistorted sinusoid 10cosω0*t* and the output signal *y*(*t*). The signal *yd*(*t*), whose period is *T*0 [the same as that of *y*(*t*)], can be described over the first cycle as - -$$ -y_d(t) = \begin{cases} 10 \cos \omega_0 t - 8 & |t| \le 0.1024T_0 \\ 10 \cos \omega_0 t + 8 & \frac{T_0}{2} - 0.1024T_0 \le |t| \le \frac{T_0}{2} + 0.1024T_0 \\ 0 & \text{everywhere else} \end{cases} -$$ - -Observe that *yd*(*t*) is an even function of *t* and its mean value is zero. Hence, *a*0 = *C*0 = 0, and *bn* = 0. Thus, *Cn* = *an* and the Fourier series for *yd*(*t*) can be expressed as - -$$ -y_d(t) = \sum_{n=1}^{\infty} C_n \cos n\omega_0 t -$$ - -As usual, we can compute the coefficients *Cn* (which is equal to *an*) by integrating *yd*(*t*) cos *n*ω0*t* over one cycle (and then dividing by 2/*T*0). Because *yd*(*t*) has even symmetry, we can find *an* by integrating the expression over a half-cycle only using Eq. (6.14). The - -#### 634 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -straightforward evaluation of the appropriate integral yields† - -$$ -C_n = \begin{cases} \frac{20}{\pi} \left[ \frac{\sin\left[0.6435(n+1)\right]}{n+1} + \frac{\sin\left[0.6435(n-1)\right]}{n-1} \right] - \frac{32}{\pi} \left[ \frac{\sin\left(0.6435n\right)}{n} \right] & n \text{ odd} \\ 0 & n \text{ even} \end{cases} -$$ - -Computing the coefficients *C*1, *C*2, *C*3, ... from this expression, we can write - -*yd*(*t*) = 1.04 cos ω0*t* +0.733 cos 3ω0*t* +0.311 cos 5ω0*t* +··· - -### COMPUTING HARMONIC DISTORTION - -We can compute the amount of harmonic distortion in the output signal by computing the power of the distortion component *yd*(*t*). Because *yd*(*t*) is an even function of *t* and because the energy in the first half-cycle is identical to the energy in the second half-cycle, we can compute the power by averaging the energy over a quarter-cycle. Thus, - -$$ -P_{y_d} = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} y_d^2(t) dt = \frac{1}{T_0/4} \int_0^{T_0/4} y_d^2(t) dt -$$ - -= $\frac{4}{T_0} \int_0^{0.1024T_0} (10 \cos \omega_0 t - 8)^2 dt = 0.865$ - -The power of the desired signal 10cos ω0*t* is (10)2/2 = 50. Hence, the total harmonic distortion is‡ - -$$ -D_{\text{tot}} = \frac{0.865}{50} \times 100 = 1.73\% -$$ - -The power of the third harmonic components of *yd*(*t*) is (0.733)2/2 = 0.2686. The third harmonic distortion is - -$$ -D_3 = \frac{0.2686}{50} \times 100 = 0.5372\% -$$ - -$$ -C_n = a_n = \frac{8}{T_0} \int_0^{0.1024T_0} [10 \cos \omega_0 t - 8] \cos n\omega_0 t dt -$$ - -† In addition, *yd*(*t*) exhibits half-wave symmetry (see Prob. 6.1-6), where the second half-cycle is the negative of the first. Because of this property, all the even harmonics vanish, and the odd harmonics can be computed by integrating the appropriate expressions over the first half-cycle only (from −*T*0/4 to *T*0/4) and doubling the resulting values. Moreover, because of even symmetry, we can integrate the appropriate expressions over 0 to *T*0/4 (instead of from −*T*0/4 to *T*0/4) and double the resulting values. In essence, this allows us to compute *Cn* by integrating the expression over the quarter-cycle only and then quadrupling the resulting values. Thus, - -‡ In the literature, the harmonic distortion often refers to the rms distortion rather than the power distortion. The rms values are the square-root values of the corresponding powers. Thus, the third harmonic distortion in this sense is √(0.2686/50) × 100 = 7.33%. Alternately, we may also compute this value directly from the amplitudes of the third harmonic 0.733 and that of the fundamental as 10. The ratio of the rms values is (0.733/ √2) : (10/ √ 2) = 0.0733 and the percentage distortion is 7.33%. - -### **[6.3-3 Properties of the Fourier Series](#page-12-0)** - -As with the Laplace and *z*-transforms, the Fourier series has a variety of properties that can simplify work and help provide a more intuitive understanding of signals. Table 6.2 provides the most important properties of the Fourier series for a periodic signal *x*(*t*) and its spectrum *Dn*. Properties that involve two signals require that the two signals have a common fundamental frequency ω0. While not given here, the proofs of these properties are straightforward and parallel the proofs of the Fourier transform properties given in Ch. 7. - -To demonstrate the utility of Fourier series properties, let us consider an example where we use a selection of properties to simplify the work of finding a piecewise polynomial signal's spectrum. - -| Operation | x(t) | Dn | -|-----------------------|------------------------------|----------------| -| Scalar multiplication | kx(t) | kDn | -| Addition | x1(t)+x2(t) | D1,n
+D2,n | -| | x1(t), x2(t) require same ω0 | | -| Conjugation | x∗(t) | D∗
−n | -| Reversal | x(−t) | D−n | -| Time shifting | x(t −t0) | Dne−jnω0t0 | -| Frequency shifting | x(t)ejn0ω0t | Dn−n0 | -| Frequency convolution | x1(t)x2(t) | D1,n
∗ D2,n | -| | x1(t), x2(t) require same ω0 | | -| Time differentiation | dkx(t)
dtk | (jnω0)kDn | - -**TABLE 6.2** Selected Fourier Series Properties - -### **EXAMPLE 6.11 Using Fourier Series Properties** - -Use properties rather than integration to compute the exponential Fourier series coefficients *Dn* of the triangular signal *x*(*t*) shown in Fig. 6.4. Verify the correctness of *Dn* for *A* = 1 by synthesizing *x*(*t*) with a suitable truncation of Eq. (6.19). - -From Fig. 6.4, we see that *x*(*t*) is a piecewise linear function that is *T*0 = 2 periodic. To compute *Dn* directly using Eq. (6.19) would therefore require tedious integration by parts. Fortunately, we can compute *Dn* without integration by instead using Fourier series properties. First, however, we must compute the dc component *D*0 separately from other *Dn*. By simple inspection of Fig. 6.4, we see that *x*(*t*) has no dc component, so *D*0 = 0. - -To determine the remaining *Dn*, we begin by noting that *x*(*t*) has a constant slope of either 2*A* or −2*A*. Thus, differentiating *x*(*t*) once yields a square wave with amplitudes ±2*A*. Here, differentiation reduces *x*(*t*) from a piecewise linear to a piecewise constant function, - -thereby eliminating the need for integration by parts.† Differentiating *x*(*t*) twice yields a pair of shifted impulse trains, weighted by ±4*A*. This second differentiation eliminates the need for any integration whatsoever, since the Fourier series coefficients of an impulse train are known to be 1 *T*0 (see Ex. 6.9). - -Stated mathematically, we see that - -$$ -\frac{d^2}{dt^2}x(t) = 4A\delta_2(t + \frac{1}{2}) - 4A\delta_2(t - \frac{1}{2}) -$$ - -Transforming this expression and using the Fourier series properties of scalar multiplication, addition, frequency shifting, and time differentiation yield (for *n* = 0) - -$$ -(in\pi)^2 D_n = 4A \frac{e^{jn\omega_0/2}}{2} - 4A \frac{e^{-jn\omega_0/2}}{2} -$$ - -Substituting ω0 = 2π/*T*0 = π and solving for *Dn* yield - -$$ -D_n = 4A \frac{e^{in\pi/2} - e^{-in\pi/2}}{-2n^2\pi^2} -$$ - -Combining with the dc component and simplifying expressions, the final result is - -$$ -D_n = \begin{cases} 0 & n = 0\\ \frac{-A4j\sin(n\pi/2)}{n^2\pi^2} & n \neq 0 \end{cases} -$$ - -Overall, this is a neat way to determine the signal's spectrum. Through the selective use of properties, we have determined *Dn* without any integration. With a little care, this basic approach can yield the spectrum of *any piecewise polynomial periodic function*. Since all periodic functions of practical interest to engineers can, to an arbitrary level of accuracy, be represented as piecewise polynomial functions, we can always find their spectra without integration except for the dc term *D*0. - -### USING A TRUNCATED FOURIER SERIES TO VERIFY SPECTRUM CORRECTNESS - -While a signal's spectrum *Dn* provides useful insight into signal character, it can be difficult to look at *Dn* and know that it is correct for a particular signal *x*(*t*). For example, is it at all obvious that *Dn* = 4*Aj*sin(*n*π/2) −*n*2π2 is the spectrum for the triangle wave of Fig. 6.4? Probably not. - -There is an easy way, however, to verify a signal's spectrum: synthesize *x*(*t*) using a suitable truncation of Eq. (6.19). If the synthesized signal closely matches the original, we can be relatively certain that the spectrum *Dn* is correct. What is a suitable truncation? Well, it depends. All significant *Dn* terms need to be included, but not so many as to make the reconstruction impractical to compute. Since in the present case *Dn* decays at a rate 1/*n*2, a good approximation is possible with a relatively few number of terms; a 10-harmonic truncation should be just fine. Let us use MATLAB to synthesize *x*(*t*) using a 10-harmonic truncated Fourier series. - -† Differentiation also destroys the dc component of the signal, providing further justification as to why the dc component needs to be separately computed. - -To begin, we define *A*, *Dn*, *T*0, ω0, and a time vector that spans two periods of the waveform. - -``` ->> A = 1; D = @(n) -A*4j*sin(n*pi/2)./(n.^2*pi^2); ->> T0 = 2; omega0 = 2*pi/T0; t = (-T0:.001:T0); -``` - -Next, we set the dc portion of the signal. - ->> D0 = 0; x10 = D0\*ones(size(t)); - -To add the desired 10 harmonics, we enter a loop for 1 ≤ *n* ≤ 10 and add in the *Dn* and *D*−*n* terms. Although *x*(*t*) should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed using the real command. - -``` ->> for n = 1:10, ->> x10 = x10+real(D(n)*exp(1j*omega0*n*t)+D(-n)*exp(-1j*omega0*n*t)); ->> end -``` - -Lastly, we plot the resulting truncated Fourier series synthesis of *x*(*t*). - ->> plot(t,x10,'k'); xlabel('t'); ylabel('x\_{10}(t)'); - -Since the synthesized waveform shown in Fig. 6.19 closely matches the original waveform in Fig. 6.4, we have high confidence that the computed *Dn* are correct. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/082_6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/082_6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS.md deleted file mode 100644 index 99389e51be37fa82de5e9dabebd0437480e3f187..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/082_6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS.md +++ /dev/null @@ -1,120 +0,0 @@ -## **[6.4 LTIC SYSTEM](#page-12-0) RESPONSE TO PERIODIC INPUTS** - -A periodic signal can be expressed as a sum of everlasting exponentials (or sinusoids). We also know how to find the response of an LTIC system to an everlasting exponential. From this information, we can readily determine the response of an LTIC system to periodic inputs. A periodic signal *x*(*t*) with period *T*0 can be expressed as an exponential Fourier series - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -### 638 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -In Sec. 4.8, we showed that the response of an LTIC system with transfer function *H*(*s*) to an everlasting exponential input *ej*ω*t* is an everlasting exponential *H*(*j*ω)*ej*ω*t* . This input–output pair can be displayed as† - -$$ -\underbrace{e^{j\omega t}}_{\text{input}} \Longrightarrow \underbrace{H(j\omega)e^{j\omega t}}_{\text{output}} -$$ - -Therefore, from the linearity property, - -$$ -\underbrace{\sum_{n=-\infty}^{\infty} D_n e^{jn\omega_0 t}}_{\text{input } x(t)} \Longrightarrow \underbrace{\sum_{n=-\infty}^{\infty} D_n H(jn\omega_0) e^{jn\omega_0 t}}_{\text{response } y(t)} -$$ -(6.29) - -The response *y*(*t*) is obtained in the form of an exponential Fourier series and is therefore a periodic signal of the same period as that of the input. - -We shall demonstrate the utility of these results by the following example. - -### **EXAMPLE 6.12 Full-Wave Rectifier** - -A full-wave rectifier (Fig. 6.20a) is used to obtain a dc signal from a sinusoid sin *t*. The rectified signal *x*(*t*), depicted in Fig. 6.17, is applied to the input of a lowpass *RC* filter, which suppresses the time-varying component and yields a dc component with some residual ripple. Find the filter output *y*(*t*). Find also the dc output and the rms value of the ripple voltage. - -First, we shall find the Fourier series for the rectified signal *x*(*t*), whose period is *T*0 = π. Consequently, ω0 = 2, and - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{j2nt} -$$ - -where - -$$ -D_n = \frac{1}{\pi} \int_0^{\pi} \sin t e^{-j2nt} dt = \frac{2}{\pi (1 - 4n^2)} -$$ -(6.30) - -Therefore, - -$$ -x(t) = \sum_{n = -\infty}^{\infty} \frac{2}{\pi (1 - 4n^2)} e^{j2nt} -$$ - -† This result applies only to asymptotically stable systems. This is because when *s* = *j*ω, the integral on the right-hand side of Eq. (2.39) does not converge for unstable systems. Moreover, for marginally stable systems also, that integral does not converge in the ordinary sense, and *H*(*j*ω) cannot be obtained from *H*(*s*) by replacing *s* with *j*ω. - -**Figure 6.20 (a)** Full-wave rectifier with a lowpass filter and **(b)** its output. - -Next, we find the transfer function of the *RC* filter in Fig. 6.20a. This filter is identical to the *RC* circuit in Ex. 1.17 (Fig. 1.35) for which the differential equation relating the output (capacitor voltage) to the input *x*(*t*) was found to be [Eq. (1.31)]: - -$$ -(3D+1)y(t) = x(t) -$$ - -The transfer function *H*(*s*) for this system is found from Eq. (2.41) as - -$$ -H(s) = \frac{1}{3s+1} -$$ - -and - -$$ -H(j\omega) = \frac{1}{3j\omega + 1} \tag{6.31} -$$ - -From Eq. (6.29), the filter output *y*(*t*) can be expressed as (with ω0 = 2) - -$$ -y(t) = \sum_{n = -\infty}^{\infty} D_n H(jn\omega_0) e^{jn\omega_0 t} = \sum_{n = -\infty}^{\infty} D_n H(j2n) e^{j2nt} -$$ - -Substituting *Dn* and *H*(*j*2*n*) from Eqs. (6.30) and (6.31) in the foregoing equation, we obtain - -$$ -y(t) = \sum_{n = -\infty}^{\infty} \frac{2}{\pi (1 - 4n^2)(j6n + 1)} e^{j2nt} -$$ - -#### 640 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -Note that the output *y*(*t*) is also a periodic signal given by the exponential Fourier series on the right-hand side. The output is shown in Fig. 6.20b. - -The output Fourier series coefficient corresponding to *n* = 0 is the dc component of the output, given by 2/π. The remaining terms in the Fourier series constitute the unwanted component called the ripple. We can determine the rms value of the ripple voltage by using Eq. (6.27) to find the power of the ripple component. The power of the ripple is the power of all the components except the dc (*n* = 0). Note that *D*ˆ *n*, the exponential Fourier coefficient for the output *y*(*t*), is - -$$ -\hat{D}_n = \frac{2}{\pi (1 - 4n^2)(j6n + 1)} -$$ - -Therefore, from Eq. (6.28), we have - -$$ -P_{\text{right}} = 2 \sum_{n=1}^{\infty} |D_n|^2 = 2 \sum_{n=1}^{\infty} \left| \frac{2}{\pi (1 - 4n^2)(j6n + 1)} \right|^2 = \frac{8}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{(1 - 4n^2)^2 (36n^2 + 1)} -$$ - -Numerical computation of the right-hand side yields *P*ripple = 0.0025, and the ripple rms value = *P*ripple = 0.05. This shows that the rms ripple voltage is 5% of the amplitude of the input sinusoid. - -### WHY USE EXPONENTIALS? - -The exponential Fourier series is just another way of representing trigonometric Fourier series (or vice versa). The two forms carry identical information—no more, no less. The reasons for preferring the exponential form have already been mentioned: this form is more compact, and the expression for deriving the exponential coefficients is also more compact than those in the trigonometric series. Furthermore, the LTIC system response to exponential signals is also simpler (more compact) than the system response to sinusoids. In addition, the exponential form proves to be much easier than the trigonometric form to manipulate mathematically and otherwise handle in the area of signals as well as systems. Moreover, exponential representation proves much more convenient for analysis of complex *x*(*t*). For these reasons, in our future discussion we shall use the exponential form exclusively. - -A minor disadvantage of the exponential form is that it cannot be visualized as easily as sinusoids. For intuitive and qualitative understanding, the sinusoids have the edge over exponentials. Fortunately, this difficulty can be overcome readily because of the close connection between exponential and Fourier spectra. For the purpose of mathematical analysis, we shall continue to use exponential signals and spectra; but to understand the physical situation intuitively or qualitatively, we shall speak in terms of sinusoids and trigonometric spectra. Thus, although all mathematical manipulation will be in terms of exponential spectra, we shall now speak of exponential and sinusoids interchangeably when we discuss intuitive and qualitative insights in attempting to arrive at an understanding of physical situations. This is an important point; readers should make an extra effort to familiarize themselves with the two forms of spectra, their relationships, and their convertibility. - -### DUAL PERSONALITY OF A SIGNAL - -The discussion so far shows that a periodic signal has a dual personality—the time domain and the frequency domain. It can be described by its waveform or by its Fourier spectra. The timeand frequency-domain descriptions provide complementary insights into a signal. For in-depth perspective, we need to understand both these identities. It is important to learn to think of a signal from both perspectives. In the next chapter, we shall see that aperiodic signals also have this dual personality. Moreover, we shall show that even LTI systems have this dual personality, which offers complementary insights into the system behavior. - -### LIMITATIONS OF THE FOURIER SERIES METHOD OF ANALYSIS - -We have developed here a method of representing a periodic signal as a weighted sum of everlasting exponentials whose frequencies lie along the ω axis in the *s* plane. This representation (Fourier series) is valuable in many applications. However, as a tool for analyzing linear systems, it has serious limitations and consequently has limited utility for the following reasons: - -- 1. The Fourier series can be used only for periodic inputs. All practical inputs are aperiodic (remember that a periodic signal starts at *t* = −∞). -- 2. The Fourier methods can be applied readily to BIBO-stable (or asymptotically stable) systems. It cannot handle unstable or even marginally stable systems. - -The first limitation can be overcome by representing aperiodic signals in terms of everlasting exponentials. This representation can be achieved through the Fourier integral, which may be considered to be an extension of the Fourier series. We shall therefore use the Fourier series as a stepping-stone to the Fourier integral developed in the next chapter. The second limitation can be overcome by using exponentials *est*, where *s* is not restricted to the imaginary axis but is free to take on complex values. This generalization leads to the Laplace integral, discussed in Ch. 4 (the Laplace transform). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/083_6.5 GENERALIZED FOURIER SERIES - SIGNALS AS VECTORS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/083_6.5 GENERALIZED FOURIER SERIES - SIGNALS AS VECTORS.md deleted file mode 100644 index d509d6bdb389e9ee200bb7f66715e21d7ddd4980..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/083_6.5 GENERALIZED FOURIER SERIES - SIGNALS AS VECTORS.md +++ /dev/null @@ -1,707 +0,0 @@ -## **[6.5 GENERALIZED](#page-12-0) FOURIER SERIES: SIGNALS AS VECTORS** - -We now consider a very general approach to signal representation with far-reaching consequences.† There is a perfect analogy between signals and vectors; the analogy is so strong that the term *analogy* understates the reality. Signals are not just *like* vectors. Signals *are* vectors! A vector can be represented as a sum of its components in a variety of ways, depending on the choice of coordinate system. A signal can also be represented as a sum of its components in a variety of ways. Let us begin with some basic vector concepts and then apply these concepts to signals. - -† This section closely follows the material from the author's earlier book [10]. Omission of this section will not cause any discontinuity in understanding the rest of the book. Derivation of Fourier series through the signal-vector analogy provides an interesting insight into signal representation and other topics such as signal correlation, data truncation, and signal detection. - -### **[6.5-1 Component of a Vector](#page-12-0)** - -A vector is specified by its magnitude and its direction. We shall denote all vectors by boldface. For example, **x** is a certain vector with magnitude or length |**x**|. For the two vectors **x** and **y** shown in Fig. 6.21, we define their dot (inner or scalar) product as - -$$ -\mathbf{x} \cdot \mathbf{y} = |\mathbf{x}| |\mathbf{y}| \cos \theta -$$ - -where θ is the angle between these vectors. Using this definition, we can express |**x**|, the length of a vector **x**, as - -$$ -|\mathbf{x}|^2 = \mathbf{x} \cdot \mathbf{x} -$$ - -Let the component of **x** along **y** be *c***y** as depicted in Fig. 6.21. Geometrically, the component of **x** along **y** is the projection of **x** on **y** and is obtained by drawing a perpendicular from the tip of **x** on the vector **y**, as illustrated in Fig. 6.21. What is the mathematical significance of a component of a vector along another vector? As seen from Fig. 6.21, the vector **x** can be expressed in terms of vector **y** as - -$$ -\mathbf{x} = c\mathbf{y} + \mathbf{e} -$$ - -However, this is not the only way to express **x** in terms of **y**. From Fig. 6.22, which shows two of the infinite other possibilities, we have - -$$ -\mathbf{x} = c_1 \mathbf{y} + \mathbf{e}_1 = c_2 \mathbf{y} + \mathbf{e}_2 -$$ - -In each of these three representations, **x** is represented in terms of **y** plus another vector called the *error vector*. If we approximate **x** by *c***y**, - -$$ -\mathbf{x} \simeq c\mathbf{y} -$$ - -the error in the approximation is the vector **e** = **x** − *c***y**. Similarly, the errors in approximations in these drawings are **e**1 (Fig. 6.22a) and **e**2 (Fig. 6.22b). What is unique about the approximation in Fig. 6.21 is that the error vector is the smallest. We can now define mathematically the component of a vector **x** along vector **y** to be *c***y** where *c* is chosen to minimize the length of the error vector **e** = **x** − *c***y**. Now, the length of the component of **x** along **y** is |**x**| cos θ. But it is also *c*|**y**|, as seen from Fig. 6.21. Therefore, - -$$ -c|\mathbf{y}| = |\mathbf{x}| \cos \theta -$$ - -Multiplying both sides by |**y**| yields - -$$ -c|\mathbf{y}|^2 = |\mathbf{x}||\mathbf{y}|\cos\theta = \mathbf{x}\cdot\mathbf{y} -$$ - -**Figure 6.21** Component (projection) of a vector along another vector. - -**Figure 6.22** Approximation of a vector in terms of another vector. - -Therefore, - -$$ -c = \frac{\mathbf{x} \cdot \mathbf{y}}{\mathbf{y} \cdot \mathbf{y}} = \frac{1}{|\mathbf{y}|^2} \mathbf{x} \cdot \mathbf{y} -$$ - (6.32) - -From Fig. 6.21, it is apparent that when **x** and **y** are perpendicular, or orthogonal, then **x** has a zero component along **y**; consequently, *c* = 0. Keeping an eye on Eq. (6.32), we therefore define **x** and **y** to be *orthogonal* if the inner (scalar or dot) product of the two vectors is zero, that is, if - -**x** · **y** = 0 - -### **[6.5-2 Signal Comparison and Component of a Signal](#page-12-0)** - -The concept of a vector component and orthogonality can be extended to signals. Consider the problem of approximating a real signal *x*(*t*) in terms of another real signal *y*(*t*) over an interval (*t*1, *t*2): - -$$ -x(t) \simeq cy(t) \qquad t_1 < t < t_2 -$$ - -The error *e*(*t*) in this approximation is - -$$ -e(t) = \begin{cases} x(t) - cy(t) & t_1 < t < t_2 \\ 0 & \text{otherwise} \end{cases} -$$ - -We now select a criterion for the "best approximation." We know that the signal energy is one possible measure of a signal size. For best approximation, we shall use the criterion that minimizes the size or energy of the error signal *e*(*t*) over the interval (*t*1,*t*2). This energy *Ee* is given by - -$$ -E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt -$$ - -Note that the right-hand side is a definite integral with *t* as the dummy variable. Hence, *Ee* is a function of the parameter *c* (not *t*) and *Ee* is minimum for some choice of *c*. To minimize *Ee*, a necessary condition is - -$$ -\frac{dE_e}{dc} = 0 -$$ -$$ -\frac{d}{dc} \left[ \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt \right] = 0 -$$ - -*d* - -or - -#### 644 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -Expanding the squared term inside the integral, we obtain - -$$ -\frac{d}{dc} \left[ \int_{t_1}^{t_2} x^2(t) dt \right] - \frac{d}{dc} \left[ 2c \int_{t_1}^{t_2} x(t) y(t) dt \right] + \frac{d}{dc} \left[ c^2 \int_{t_1}^{t_2} y^2(t) dt \right] = 0 -$$ - -from which we get - -$$ --2\int_{t_1}^{t_2} x(t)y(t) dt + 2c \int_{t_1}^{t_2} y^2(t) dt = 0 -$$ - -$$ -c = \frac{\int_{t_1}^{t_2} x(t)y(t) dt}{\int_{t_1}^{t_2} y^2(t) dt} = \frac{1}{E_y} \int_{t_1}^{t_2} x(t)y(t) dt -$$ - (6.33) - -We observe a remarkable similarity between the behavior of vectors and signals, as indicated by Eqs. (6.32) and (6.33). It is evident from these two parallel expressions that *the area under the product of two signals corresponds to the inner (scalar or dot) product of two vectors*. In fact, the area under the product of *x*(*t*) and *y*(*t*) is called the *inner product* of *x*(*t*) and *y*(*t*), and is denoted by (*x*, *y*). The energy of a signal is the inner product of a signal with itself, and corresponds to the vector length square (which is the inner product of the vector with itself). - -To summarize our discussion, if a signal *x*(*t*) is approximated by another signal *y*(*t*) as - -*x*(*t*) *cy*(*t*) - -then the optimum value of *c* that minimizes the energy of the error signal in this approximation is given by Eq. (6.33). - -Taking our clue from vectors, we say that a signal *x*(*t*) contains a component *cy*(*t*), where *c* is given by Eq. (6.33). Note that in vector terminology, *cy*(*t*) is the projection of *x*(*t*) on *y*(*t*). Continuing with the analogy, we say that if the component of a signal *x*(*t*) of the form *y*(*t*) is zero (i.e., *c* = 0), the signals *x*(*t*) and *y*(*t*) are orthogonal over the interval (*t*1, *t*2). Therefore, we define the real signals *x*(*t*) and *y*(*t*) to be orthogonal over the interval (*t*1, *t*2) if† - -$$ -\int_{t_1}^{t_2} x(t)y(t) dt = 0 -$$ -\n(6.34) - -### **EXAMPLE 6.13 Sine-Wave Approximation of a Square Wave** - -For the square signal *x*(*t*) shown in Fig. 6.23, find the component in *x*(*t*) of the form sin*t*. In other words, approximate *x*(*t*) in terms of sin *t* - -*x*(*t*) *c*sin*t* 0 < *t* < 2π - -so that the energy of the error signal is minimum. - -† For complex signals, the definition is modified as in Eq. (6.37), in Sec. 6.5-3. - -**Figure 6.23** Approximation of a square wave in terms of a single sinusoid. - -In this case, - -$$ -y(t) = \sin t -$$ - and $E_y = \int_0^{2\pi} \sin^2(t) dt = \pi$ - -From Eq. (6.33), we find - -$$ -c = \frac{1}{\pi} \int_0^{2\pi} x(t) \sin t \, dt = \frac{1}{\pi} \left[ \int_0^{\pi} \sin t \, dt + \int_{\pi}^{2\pi} -\sin t \, dt \right] = \frac{4}{\pi} -$$ - -Thus, - -$$ -x(t) \simeq \frac{4}{\pi} \sin t -$$ - -represents the best approximation of *x*(*t*) by the function sin*t*, which will minimize the error energy. This sinusoidal component of *x*(*t*) is shaded in Fig. 6.23. By analogy with vectors, we say that the square function *x*(*t*) depicted in Fig. 6.23 has a component of signal sin*t* and that the magnitude of this component is 4/π. - -### **DR ILL 6.7 Sine Wave Approximation of a Ramp Function** - -Show that over an interval (−π < *t* < π), the "best" approximation of the signal *x*(*t*) = *t* in terms of the function sin*t* is 2 sin*t*. Verify that the error signal *e*(*t*) = *t* − 2 sin*t* is orthogonal to the signal sin*t* over the interval −π < *t* < π. Sketch the signals *t* and 2 sin*t* over the interval −π < *t* < π. - -### **[6.5-3 Extension to Complex Signals](#page-12-0)** - -So far we have restricted ourselves to real functions of *t*. To generalize the results to complex functions of *t*, consider again the problem of approximating a signal *x*(*t*) by a signal *y*(*t*) over an - -#### 646 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -interval (*t*1 < *t* < *t*2): - -$$ -x(t) \simeq cy(t) -$$ - -where *x*(*t*) and *y*(*t*) now can be complex functions of *t*. Recall that the energy *Ey* of the complex signal *y*(*t*) over an interval (*t*1, *t*2) is - -$$ -E_{y} = \int_{t_1}^{t_2} |y(t)|^2 dt -$$ - -In this case, both the coefficient *c* and the error - -$$ -e(t) = x(t) - cy(t) -$$ - -are complex (in general). For the "best" approximation, we choose *c* to minimize the energy *Ee* of the error signal *e*(*t*). Now, - -$$ -E_e = \int_{t_1}^{t_2} |x(t) - cy(t)|^2 dt -$$ -\n(6.35) - -Recall also that - -$$ -|u + v|^2 = (u + v)(u^* + v^*) = |u|^2 + |v|^2 + u^*v + uv^* \tag{6.36} -$$ - -After some manipulation, we can use this result to rearrange Eq. (6.35) as - -$$ -E_e = \int_{t_1}^{t_2} |x(t)|^2 dt - \left| \frac{1}{\sqrt{E_y}} \int_a^{t_2} x(t) y^*(t) dt \right|^2 + \left| c \sqrt{E_y} - \frac{1}{\sqrt{E_y}} \int_{t_1}^{t_2} x(t) y^*(t) dt \right|^2 -$$ - -Since the first two terms on the right-hand side are independent of *c*, it is clear that *Ee* is minimized by choosing *c* so that the third term on the right-hand side is zero. This yields - -$$ -c = \frac{1}{E_y} \int_{t_1}^{t_2} x(t) y^*(t) dt -$$ - -In light of this result, we need to redefine orthogonality for the complex case as follows: two complex functions *x*1(*t*) and *x*2(*t*) are orthogonal over an interval (*t*1 < *t* < *t*2) if - -$$ -\int_{t_1}^{t_2} x_1(t) x_2^*(t) dt = 0 \qquad \text{or} \qquad \int_{t_1}^{t_2} x_1^*(t) x_2(t) dt = 0 \tag{6.37} -$$ - -Either equality suffices. This is a general definition of orthogonality, which reduces to Eq. (6.34) when the functions are real. - -### **DR ILL 6.8 Complex Exponential Approximation of a Square Wave** - -Show that over an interval (0 < *t* < 2π ), the "best" approximation of the square signal *x*(*t*) in Fig. 6.23 in terms of the signal *ejt* is given by (2/*j*π ) *ejt*. Verify that the error signal *e*(*t*) = *x*(*t*)−(2/*j*π )*ejt* is orthogonal to the signal *ejt*. - -### ENERGY OF THE SUM OF ORTHOGONAL SIGNALS - -We know that the square of the length of a sum of two orthogonal vectors is equal to the sum of the squares of the lengths of the two vectors. Thus, if vectors **x** and **y** are orthogonal, and if **z** = **x**+**y**, then - -$$ -|\mathbf{z}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2 -$$ - -We have a similar result for signals. The energy of the sum of two orthogonal signals is equal to the sum of the energies of the two signals. Thus, if signals *x*(*t*) and *y*(*t*) are orthogonal over an interval (*t*1, *t*2), and if *z*(*t*) = *x*(*t*)+*y*(*t*), then - -$$ -E_z = E_x + E_y -$$ - -We now prove this result for complex signals, of which real signals are a special case. From Eq. (6.36), it follows that - -$$ -\int_{t_1}^{t_2} |x(t) + y(t)|^2 dt = \int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt + \int_{t_1}^{t_2} x(t)y^*(t) dt + \int_{t_1}^{t_2} x^*(t)y(t) dt -$$ - -= -$$ -\int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt -$$ - -The last result follows from the fact that because of orthogonality, the two integrals of the products *x*(*t*)*y*∗(*t*) and *x*∗(*t*)*y*(*t*) are zero [see Eq. (6.37)]. This result can be extended to the sum of any number of mutually orthogonal signals. - -### **[6.5-4 Signal Representation by an Orthogonal Signal Set](#page-12-0)** - -In this section we show a way of representing a signal as a sum of orthogonal signals. Here again we can benefit from the insight gained from a similar problem in vectors. We know that a vector can be represented as a sum of orthogonal vectors, which form the coordinate system of a vector space. The problem in signals is analogous, and the results for signals are parallel to those for vectors. So, let us review the case of vector representation. - -### ORTHOGONAL VECTOR SPACE - -Let us investigate a three-dimensional Cartesian vector space described by three mutually orthogonal vectors **x**1, **x**2, and **x**3, as illustrated in Fig. 6.24. First, we shall seek to approximate a three-dimensional vector **x** in terms of two mutually orthogonal vectors **x**1 and **x**2: - -$$ -\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 -$$ - -The error **e** in this approximation is - -$$ -\mathbf{e} = \mathbf{x} - (c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2) -$$ - -or - -$$ -\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + \mathbf{e} -$$ - -**Figure 6.24** Representation of a vector in three-dimensional space. - -As in the earlier geometrical argument, we see from Fig. 6.24 that the length of **e** is minimum when **e** is perpendicular to the **x**1–**x**2 plane, and *c*1**x**1 and *c*2**x**2 are the projections (components) of **x** on **x**1 and **x**2, respectively. Therefore, the constants *c*1 and *c*2 are given by Eq. (6.32). Observe that the error vector is orthogonal to both the vectors **x**1 and **x**2. - -Now, let us determine the "best" approximation to **x** in terms of all three mutually orthogonal vectors **x**1, **x**2, and **x**3: - -$$ -\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3 \tag{6.38} -$$ - -Figure 6.24 shows that a unique choice of *c*1, *c*2, and *c*3 exists, for which Eq. (6.38) is no longer an approximation but an equality - -$$ -\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3 -$$ - -In this case, *c*1**x**1, *c*2**x**2, and *c*3**x**3 are the projections (components) of **x** on **x**1,**x**2, and **x**3, respectively; that is, - -$$ -c_i = \frac{\mathbf{x} \cdot \mathbf{x}_i}{\mathbf{x}_i \cdot \mathbf{x}_i} = \frac{1}{|\mathbf{x}_i|^2} \mathbf{x} \cdot \mathbf{x}_i \qquad i = 1, 2, 3 -$$ - (6.39) - -Note that the error in the approximation is zero when **x** is approximated in terms of three mutually orthogonal vectors: **x**1, **x**2, and **x**3. The reason is that **x** is a three-dimensional vector, and the vectors **x**1, **x**2, and **x**3 represent a *complete set* of orthogonal vectors in three-dimensional space. Completeness here means that it is impossible to find another vector **x**4 in this space, which is orthogonal to all three vectors, **x**1,**x**2, and **x**3. Any vector in this space can then be represented (with zero error) in terms of these three vectors. Such vectors are known as *basis* vectors. If a set of vectors {**x***i*} is not complete, the error in the approximation will generally not be zero. Thus, in the three-dimensional case discussed earlier, it is generally not possible to represent a vector **x** in terms of only two basis vectors without an error. - -The choice of basis vectors is not unique. In fact, a set of basis vectors corresponds to a particular choice of coordinate system. Thus, a three-dimensional vector **x** may be represented in many different ways, depending on the coordinate system used. - -### ORTHOGONAL SIGNAL SPACE - -We start with real signals and then extend the discussion to complex signals. We proceed with our signal approximation problem, using clues and insights developed for vector approximation. As before, we define orthogonality of a real signal set *x*1(*t*), *x*2(*t*), ..., *xN*(*t*) over interval (*t*1,*t*2) as - -$$ -\int_{t_1}^{t_2} x_m(t) x_n(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases} -$$ - (6.40) - -If the energies *En* = 1 for all *n*, then the set is *normalized* and is called an *orthonormal set*. An orthogonal set can always be normalized by dividing *xn*(*t*) by √*En* for all *n*. - -Now, consider approximating a signal *x*(*t*) over the interval (*t*1, *t*2) by a set of *N* real, mutually orthogonal signals *x*1(*t*), *x*2(*t*),..., *xN*(*t*) as - -$$ -x(t) \simeq c_1 x_1(t) + c_2 x_2(t) + \dots + c_N x_N(t) \simeq \sum_{n=1}^N c_n x_n(t) -$$ - (6.41) - -In the approximation of Eq. (6.41), the error *e*(*t*) is - -$$ -e(t) = x(t) - \sum_{n=1}^{N} c_n x_n(t) -$$ - -and *Ee*, the error signal energy, is - -$$ -E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt -$$ - (6.42) - -According to our criterion for best approximation, we select the values of *ci* that minimize *Ee*. Hence, the necessary condition is ∂*Ee*/*dci* = 0 for *i* = 1, 2,...,*N*, that is, - -$$ -\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^N c_n x_n(t) \right]^2 dt = 0 -$$ - -When we expand the integrand, we find that all the cross-multiplication terms arising from the orthogonal signals are zero by virtue of orthogonality: that is, all terms of the form \$ *xm*(*t*)*xn*(*t*)*dt* with *m* = *n* vanish. Similarly, the derivative with respect to *ci* of all terms that do not contain *ci* is zero. For each *i*, this leaves only two nonzero terms: - -$$ -\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ -2c_i x(t) x_i(t) + c_i^2 x_i^2(t) \right] dt = 0 -$$ - -or - -$$ --2\int_{t_1}^{t_2} x(t)x_i(t) dt + 2c_i \int_{t_1}^{t_2} x_i^2(t) dt = 0 \qquad i = 1, 2, \dots, N -$$ - -Therefore, - -$$ -c_i = \frac{\int_{t_1}^{t_2} x(t) x_i(t) dt}{\int_{t_1}^{t_2} x_i^2(t) dt} = \frac{1}{E_i} \int_{t_1}^{t_2} x(t) x_i(t) dt \qquad i = 1, 2, ..., N -$$ - (6.43) - -A comparison of Eq. (6.43) with Eq. (6.39) forcefully brings out the analogy of signals with vectors. - -**Finality Property.** Equation (6.43) shows one interesting property of the coefficients of *c*1, *c*2, ..., *cN*: the optimum value of any coefficient in Eq. (6.41) is independent of the number of terms used in the approximation. For example, if we used only one term (*N* = 1) or two terms (*N* = 2) or any number of terms, the optimum value of the coefficient *c*1 would be the same [as given by Eq. (6.43)]. The advantage of this approximation of a signal *x*(*t*) by a set of mutually orthogonal signals is that we can continue to add terms to the approximation without disturbing the previous terms. This property of *finality* of the values of the coefficients is very important from a practical point of view.† - -### ENERGY OF THE ERROR SIGNAL - -When the coefficients *ci* in Eq. (6.41) are chosen according to Eq. (6.43), the error signal energy is minimized. This minimum value of *Ee* is given by Eq. (6.42): - -$$ -E_e = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt -$$ - -= -$$ -\int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 \int_{t_1}^{t_2} x_n^2(t) dt - 2 \sum_{n=1}^{N} c_n \int_{t_1}^{t_2} x(t) x_n(t) dt -$$ - -$$ -x(t_1) = a_0 + a_1t_1 -$$ - and $x(t_2) = a_0 + a_1t_2$ - -Solution of these equations yields the desired values of *a*0 and *a*1. For a three-point approximation, we must choose the polynomial *a*0 +*a*1*t* +*a*2*t* 2 with - -$$ -x(t_i) = a_0 + a_1t_i + a_2t_i^2 -$$ - $i = 1, 2, and 3$ - -The approximation improves with a larger number of points (higher-order polynomial), but the coefficients *a*0, *a*1, *a*2, ... do not have the finality property. Every time we increase the number of terms in the polynomial, we need to recalculate the coefficients. - -† Contrast this situation with a polynomial approximation of *x*(*t*). Suppose we wish to find a two-point approximation of *x*(*t*) by a polynomial in *t*; that is, the polynomial is to be equal to *x*(*t*) at two points *t*1 and *t*2. This can be done by choosing a first-order polynomial *a*0 +*a*1*t* with - -Substitution of Eqs. (6.40) and (6.43) in this equation yields - -$$ -E_e = \int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 E_n - 2 \sum_{n=1}^{N} c_n^2 E_n = \int_{t_1}^{t_2} x^2(t) dt - \sum_{n=1}^{N} c_n^2 E_n -$$ - (6.44) - -Observe that because the term *c*2 *kEk* is nonnegative, the error energy *Ee* generally decreases as *N*, the number of terms, is increased. Hence, it is possible that the error energy →0 as *N* → ∞. When this happens, the orthogonal signal set is said to be *complete*. In this case, Eq. (6.41) is no more an approximation but an equality - -$$ -x(t) = c_1 x_1(t) + c_2 x_2(t) + \dots + c_n x_n(t) + \dots = \sum_{n=1}^{\infty} c_n x_n(t) \qquad t_1 < t < t_2 \tag{6.45} -$$ - -where the coefficients *cn* are given by Eq. (6.43). Because the error signal energy approaches zero, it follows that the energy of *x*(*t*) is now equal to the sum of the energies of its orthogonal components *c*1*x*1(*t*), *c*2*x*2(*t*), *c*3*x*3(*t*), .... - -The series on the right-hand side of Eq. (6.45) is called the *generalized Fourier series* of *x*(*t*) with respect to the set {*xn*(*t*)}. When the set {*xn*(*t*)} is such that the error energy *Ee* → 0 as *N* → ∞ for every member of some particular class, we say that the set {*xn*(*t*)} is complete on (*t*1, *t*2) for that class of *x*(*t*), and the set {*xn*(*t*)} is called a set of *basis functions* or *basis signals*. Unless otherwise mentioned, in the future we shall consider only the class of energy signals. - -Thus, when the set {*xn*(*t*)} is complete, we have the equality of Eq. (6.45). One subtle point that must be understood clearly is the meaning of equality in Eq. (6.45). *The equality here is not an equality in the ordinary sense, but in the sense that the error energy, that is, the energy of the difference between the two sides of Eq. (6.45), approaches zero*. If the equality exists in the ordinary sense, the error energy is always zero, but the converse is not necessarily true. The error energy can approach zero even though *e*(*t*), the difference between the two sides, is nonzero at some isolated instants. The reason is that even if *e*(*t*) is nonzero at such instants, the area under *e*2(*t*) is still zero; thus the Fourier series on the right-hand side of Eq. (6.45) may differ from *x*(*t*) at a finite number of points. - -In Eq. (6.45), the energy of the left-hand side is *Ex*, and the energy of the right-hand side is the sum of the energies of all the orthogonal components.† Thus, - -$$ -\int_{t_1}^{t_2} x^2(t) dt = c_1^2 E_1 + c_2^2 E_2 + \dots = \sum_{n=1}^{\infty} c_n^2 E_n -$$ - (6.46) - -This is *Parseval's theorem* expressed for energy signals. In Eqs. (6.26) and (6.27), we have already encountered Parseval's theorem for power signals. Recall that the signal energy (area under the squared value of a signal) is analogous to the square of the length of a vector in the vector-signal analogy. In vector space, we know that the square of the length of a vector is equal to the sum of - -† Note that the energy of a signal *cx*(*t*) is *c*2*Ex*. - -### 652 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -the squares of the lengths of its orthogonal components. Parseval's theorem of Eq. (6.46) is the statement of this fact as it applies to signals. - -### GENERALIZATION TO COMPLEX SIGNALS - -The foregoing results can be generalized to complex signals as follows: a set of functions *x*1(*t*), *x*2(*t*), ..., *xN*(*t*) is mutually orthogonal over the interval (*t*1, *t*2) if - -$$ -\int_{t_1}^{t_2} x_m(t) x_n^*(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases} -$$ - -If this set is complete for a certain class of functions, then a function *x*(*t*) in this class can be expressed as - -$$ -x(t) = c_1 x_1(t) + c_2 x_2(t) + \cdots + c_i x_i(t) + \cdots -$$ - -where - -$$ -c_n = \frac{1}{E_n} \int_{t_1}^{t_2} x(t) x_n^*(t) dt -$$ -\n(6.47) - -### **EXAMPLE 6.14 Approximating a Square Wave with a Set of Harmonic Sine Waves** - -In Ex. 6.13, the square signal *x*(*t*) in Fig. 6.23 is approximated by a single sinusoid sin *t*. In this example, we approximate *x*(*t*) using the set of harmonic sine waves sin *t*, sin 2*t*, ..., sin *nt*, ..., and see how the approximation improves with the number of terms. - -To begin, we note that the set of harmonic sine waves sin *t*, sin 2*t*,..., sin *nt*,... is orthogonal over any interval of duration 2π. † The reader can verify this fact by showing that for any real number *a*, - -$$ -\int_{a}^{a+2\pi} \sin mt \sin nt dt = \begin{cases} 0 & m \neq n \\ \pi & m = n \end{cases} -$$ - (6.48) - -Using this set, we approximate *x*(*t*) as - -*x*(*t*) *c*1 sin *t* +*c*2 sin 2*t* +···+*cn* sin *Nt* - -† This sine set, along with the cosine set cos 0*t*, cos *t*, cos 2*t*,..., cos*nt*,..., forms a complete set. In this case, however, the coefficients *ci* corresponding to the cosine terms are zero. For this reason, we have omitted cosine terms in this example. This composite sine and cosine set is the basis set for the trigonometric Fourier series. - -**Figure 6.25** Approximation of a square wave by a sum of harmonic sinusoids. - -Therefore, - -$$ -x(t) \simeq \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \dots + \frac{1}{N} \sin Nt \right) -$$ - (6.49) - -Note that coefficients of terms sin*kt* are zero for even values of *k*. Figure 6.25 shows how the approximation improves as we increase the number of terms in the series. - -Let us investigate the error signal energy as *N* → ∞. From Eq. (6.44), - -$$ -E_e = \int_0^{2\pi} x^2(t) dt - \sum_{n=1}^{\infty} c_n^2 E_n -$$ - -Note that - -$$ -\int_0^{2\pi} x^2(t) dt = \int_0^{\pi} 1^2 dt + \int_{\pi}^{2\pi} -1^2 dt = 2\pi -$$ -$$ -c_n^2 = \begin{cases} \frac{16}{n^2 \pi^2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases} -$$ - -and from Eq. (6.48), - -$$ -E_n=\pi -$$ - -Therefore, - -$$ -E_e = 2\pi - \sum_{n=1,3,5,\dots}^{N} \frac{16}{n^2 \pi^2} \pi = 2\pi - \frac{16}{\pi} \sum_{n=1,3,5,\dots}^{N} \frac{1}{n^2} -$$ - -For a single-term approximation (*N* = 1), - -$$ -E_e = 2\pi - \frac{16}{\pi} = 1.1938 -$$ - -For a two-term approximation (*N* = 3), - -$$ -E_e = 2\pi - \frac{16}{\pi} \left( 1 + \frac{1}{9} \right) = 0.6243 -$$ - -Continuing this process, we compute the error energy *Ee* for various values of *N* as - -| N | 1 | 3 | 5 | 7 | 99 | $\infty$ | -|-------|--------|--------|--------|--------|---------|----------| -| $E_e$ | 1.1938 | 0.6243 | 0.4206 | 0.3166 | 0.02545 | 0 | - -Clearly, *x*(*t*) can be represented by the infinite series - -$$ -x(t) = \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \cdots \right) = \frac{4}{\pi} \sum_{n=1,3,5,\dots}^{\infty} \frac{1}{n} \sin nt -$$ - -The equality exists in the sense that the error signal energy → 0 as *N* → ∞. In this case, the error energy decreases rather slowly with *N*, indicating that the series converges slowly. This is to be expected because *x*(*t*) has jump discontinuities and consequently, according to discussion in Sec. 6.2-2, the series converges asymptotically as 1/*n*. - -### **DR ILL 6.9 Approximating a Ramp Signal with a Set of Harmonic Sine Waves** - -Approximate the signal *x*(*t*) = *t* − π (Fig. 6.26) over the interval (0, 2π ) in terms of the set of sinusoids {sin *nt*}, *n* = 0, 1, 2,..., used in Ex. 6.14. Find *Ee*, the error energy. Show that *Ee* → 0 as *N* → ∞. - -4π - -### **ANSWERS** - -%*N* - -1 - -*x*(*t*) −2 *n*=1 *N* sin *nt* and *Ee* = 2 3π3 −%*N n*=1 *n*2 *x*(*t*) *p p p* 2*p t* **Figure 6.26** Ramp signal for Drill 6.9. - -### SOME EXAMPLES OF GENERALIZED FOURIER SERIES - -Signals are vectors in every sense. Like a vector, a signal can be represented as a sum of its components in a variety of ways. Just as vector coordinate systems are formed by mutually orthogonal vectors (rectangular, cylindrical, spherical), we also have signal coordinate systems (basis signals) formed by a variety of sets of mutually orthogonal signals. There exist a large number of orthogonal signal sets that can be used as basis signals for generalized Fourier series. Some well-known signal sets are trigonometric (sinusoid) functions, exponential functions, Walsh functions, Bessel functions, Legendre polynomials, Laguerre functions, Jacobi polynomials, Hermite polynomials, and Chebyshev polynomials. The functions that concern us most in this book are the trigonometric and the exponential sets discussed earlier in this chapter. - -### LEGENDRE FOURIER SERIES - -A set of Legendre polynomials *Pn*(*t*) (*n* = 0, 1, 2, 3,...) forms a complete set of mutually orthogonal functions over an interval (−1 < *t* < 1). These polynomials can be defined by the Rodrigues formula: - -$$ -P_n(t) = \frac{1}{2^n n!} \frac{d^n}{dt^n} (t^2 - 1)^n \qquad n = 0, 1, 2, \dots -$$ - -It follows from this equation that - -$$ -P_0(t) = 1, P_1(t) = t, P_2(t) = \left(\frac{3}{2}t^2 - \frac{1}{2}\right), P_3(t) = \left(\frac{5}{2}t^3 - \frac{3}{2}t\right) -$$ -, and so on - -We may verify the orthogonality of these polynomials by showing that - -$$ -\int_{-1}^{1} P_m(t) P_n(t) dt = \begin{cases} 0 & m \neq n \\ \frac{2}{2m+1} & m = n \end{cases} -$$ - -We can express a function *x*(*t*) in terms of Legendre polynomials over an interval (−1 < *t* < 1) as - -$$ -x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots -$$ -\n(6.50) - -where - -$$ -c_r = \frac{\int_{-1}^{1} x(t)P_r(t) dt}{\int_{-1}^{1} P_r^2(t) dt} = \frac{2r+1}{2} \int_{-1}^{1} x(t)P_r(t) dt -$$ -\n(6.51) - -Note that although the series representation is valid over the interval (−1, 1), it can be extended to any interval by the appropriate time scaling (see Prob. 6.5-8). - -### **EXAMPLE 6.15 Legendre Fourier Series** - -Determine the Legendre Fourier series of the square signal shown in Fig. 6.27. - -**Figure 6.27** Square signal for Ex. 6.15. - -From Eq. (6.50), we know that the Legendre Fourier series takes the form - -$$ -x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots -$$ - -The coefficients *c*0, *c*1, *c*2,..., *cr* may be found from Eq. (6.51). We have - -$$ -x(t) = \begin{cases} 1 & \cdots -1 < t < 0 \\ -1 & \cdots 0 < t < 1 \end{cases} -$$ - -and - -$$ -c_0 = \frac{1}{2} \int_{-1}^{1} x(t) dt = 0 -$$ - -\n -$$ -c_1 = \frac{3}{2} \int_{-1}^{1} tx(t) dt = \frac{3}{2} \left( \int_{-1}^{0} t dt - \int_{0}^{1} t dt \right) = -\frac{3}{2} -$$ - -\n -$$ -c_2 = \frac{5}{2} \int_{-1}^{1} x(t) \left( \frac{3}{2} t^2 - \frac{1}{2} \right) dt = 0 -$$ - -This result follows immediately from the fact that the integrand is an odd function of *t*. In fact, this is true of all *cr* for even values of *r*, that is, - -$$ -c_0 = c_2 = c_4 = c_6 = \cdots = 0 -$$ - -Also, - -$$ -c_3 = \frac{7}{2} \int_{-1}^{1} x(t) \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt = \frac{7}{2} \left[ \int_{-1}^{0} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt - \int_{0}^{1} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt \right] = \frac{7}{8} -$$ - -In a similar way, coefficients *c*5, *c*7,... can be evaluated. We now have - -$$ -x(t) = -\frac{3}{2}t + \frac{7}{8}\left(\frac{5}{2}t^3 - \frac{3}{2}t\right) + \cdots -$$ - -### TRIGONOMETRIC FOURIER SERIES - -We have already proved [see Eqs. (6.4), (6.5), and (6.6)] that the trigonometric signal set - -{1, cosω0*t*, cos 2ω0*t*, ..., cos*n*ω0*t*, ...; sinω0*t*, sin 2ω0*t*, ..., sin*n*ω0*t*, ...} - -is orthogonal over any interval of duration *T*0, where *T*0 = 1/*f*0 is the period of the sinusoid of frequency *f*0. This is a complete set for a class of signals with finite energies [11, 12]. Therefore, we can express a signal *x*(*t*) by a trigonometric Fourier series over any interval of duration *T*0 seconds as - -$$ -x(t) = a_0 + a_1 \cos \omega_0 t + a_2 \cos 2\omega_0 t + \cdots -$$ -$$ -+ b_1 \sin \omega_0 t + b_2 \sin 2\omega_0 t + \cdots -$$ - -or - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \qquad t_1 < t < t_1 + T_0 -$$ - -where - -$$ -\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0} -$$ - -We can use Eq. (6.43) to determine the Fourier coefficients *a*0, *an*, and *bn*. Thus, - -$$ -a_n = \frac{\int_{t_1}^{t_1 + T_0} x(t) \cos n\omega_0 t \, dt}{\int_{t_1}^{t_1 + T_0} \cos^2 n\omega_0 t \, dt} -$$ -\n(6.52) - -The integral in the denominator of Eq. (6.52) has already been found to be *T*0/2 when *n* = 0 [Eq. (6.4) with *m* = *n*]. For *n* = 0, the denominator is *T*0. Hence, - -$$ -a_0 = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) dt \quad \text{and} \quad a_n = \frac{2}{T_0} \int_{t_1}^{t_1+T_0} x(t) \cos n\omega_0 t dt \qquad n = 1, 2, 3, \dots \tag{6.53} -$$ - -Similarly, we find that - -$$ -b_n = \frac{2}{T_0} \int_{t_1}^{t_1 + T_0} x(t) \sin n\omega_0 t \, dt \qquad n = 1, 2, 3, \dots \tag{6.54} -$$ - -Note that the Fourier series in Eq. (6.49) of Ex. 6.14 is indeed the trigonometric Fourier series with *T*0 = 2π and ω0 = 2π/*T*0. In this particular example, it is easy to verify from Eq. (6.53) that *an* = 0 for all *n*, including *n* = 0. Hence, the Fourier series in that example consisted only of sine terms. - -### EXPONENTIAL FOURIER SERIES - -As shown in the footnote on page 622, the set of exponentials *ejn*ω0*t* (*n* = 0,±1,±2,...) is a set of functions orthogonal over any interval of duration *T*0 = 2π/ω0. An arbitrary signal *x*(*t*) can now be expressed over an interval (*t*1,*t*1 +*T*0) as - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad t_1 < t < t_1 + T_0 -$$ - -where [see Eq. (6.47)] - -$$ -D_n = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) e^{-jn\omega_0 t} dt -$$ - -### WHY USE THE EXPONENTIAL SET? - -If *x*(*t*) can be represented in terms of hundreds of different orthogonal sets, why do we exclusively use the exponential (or trigonometric) set for the representation of signals or LTI systems? It so - -happens that the exponential signal is an eigenfunction of LTI systems. In other words, for an LTI system, only an exponential input *est* yields the response that is also an exponential of the same form, given by *H*(*s*)*est*. The same is true of the trigonometric set. This fact makes the use of exponential signals natural for LTI systems in the sense that the system analysis using exponentials as the basis signals is greatly simplified. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/084_6.6 NUMERICAL COMPUTATION OF D_n.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/084_6.6 NUMERICAL COMPUTATION OF D_n.md deleted file mode 100644 index bdf1445e2e0f7ce362d634b1fd4a89e78b2ef592..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/084_6.6 NUMERICAL COMPUTATION OF D_n.md +++ /dev/null @@ -1,67 +0,0 @@ -## **6.6 NUMERICAL [COMPUTATION OF](#page-12-0)** *Dn* - -We can compute *Dn* numerically by using the DFT (the discrete Fourier transform discussed in Sec. 8.5), which uses the samples of a periodic signal *x*(*t*) over one period. The sampling interval is *T* seconds. Hence, there are *N*0 =*T*0/*T* number of samples in one period *T*0. To find the relationship between *Dn* and the samples of *x*(*t*), consider Eq. (6.19) and write - -$$ -D_n = \frac{1}{T_0} \int_{T_0} x(t)e^{-jn\omega_0 t} dt -$$ - -= -$$ -\lim_{T \to 0} \frac{1}{N_0 T} \sum_{k=0}^{N_0 - 1} x(kT)e^{-jn\omega_0 kT} T -$$ - -= -$$ -\lim_{T \to 0} \frac{1}{N_0} \sum_{k=0}^{N_0 - 1} x(kT)e^{-jn\Omega_0 k} -$$ -(6.55) - -where *x*(*kT*) is the *k*th sample of *x*(*t*) and - -$$ -N_0 = \frac{T_0}{T} \quad \text{and} \quad \Omega_0 = \omega_0 T = \frac{2\pi}{N_0} -$$ - -In practice, it is impossible to make *T* → 0 in computing the right-hand side of Eq. (6.55). We can make *T* small, but not zero, which will cause the data to increase without limit. Thus, we shall ignore the limit on *T* in Eq. (6.55) with the implicit understanding that *T* is reasonably small. Nonzero *T* will result in some computational error, which is inevitable in any numerical evaluation of an integral. The error resulting from nonzero *T* is called the *aliasing error,* which is discussed in more detail in Ch. 8. Thus, we can express Eq. (6.55) as - -$$ -D_n \approx \frac{1}{N_0} \sum_{k=0}^{N_0 - 1} x(kT) e^{-jn\Omega_0 k} -$$ - (6.56) - -Since 0*N*0 = 2π, we know that *ejn*0(*k*+*N*0) = *ejn*0*k*, and it follows that - -$$ -D_{n+N_0}=D_n -$$ - -The periodicity property *Dn*+*N*0 = *Dn* means that beyond *n* = *N*0/2, the coefficients represent the values for negative *n*. For instance, when *N*0 = 32, *D*17 = *D*−15, *D*18 = *D*−14,...,*D*31 = *D*−1. The cycle repeats again from *n* = 32 on. - -We can use the efficient FFT (the *fast Fourier transform* discussed in Sec. 8.6) to compute the right-hand side of Eq. (6.56). We shall use MATLAB to implement the FFT algorithm. For - -### 660 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -this purpose, we need samples of *x*(*t*) over one period starting at *t* = 0. In this algorithm, it is also preferable (although not necessary) that *N*0 be a power of 2, (i.e., *N*0 = 2*m* where *m* is an integer). - -### **EXAMPLE 6.16 Numerical Computation of Fourier Spectra** - -Numerically compute and then plot the exponential Fourier spectra for the periodic signal in Fig. 6.2a (Ex. 6.1). - -The samples of *x*(*t*) start at *t* = 0 and the last (*N*0th) sample is at *t* = *T*0 − *T*. At the points of discontinuity, the sample value is taken as the average of the values of the function on two sides of the discontinuity. Thus, the sample at *t* = 0 is not 1 but (*e*−π/2 +1)/2 = 0.604. To determine *N*0, we require that *Dn* for *n* ≥ *N*0/2 be negligible. Because *x*(*t*) has a jump discontinuity, *Dn* decays rather slowly as 1/*n*. Hence, a choice of *N*0 = 200 is acceptable because the (*N*0/2)nd (100th) harmonic is about 1% of the fundamental. However, we also require *N*0 to be a power of 2. Hence, we shall take *N*0 = 256 = 28. - -First, the basic parameters are established. - ->> T\_0 = pi; N\_0 = 256; T = T\_0/N\_0; t = (0:T:T\*(N\_0-1))'; >> x = exp(-t/2); x(1) = (exp(-pi/2)+1)/2; - -Next, the DFT, computed by means of the fft function, is used to approximate the exponential Fourier spectra up to *n* = *N*0/2. To facilitate comparison with previous plots of *Dn*, we only plot the results over −5 ≤ *n* ≤ 5. - -``` ->> D_n = fft(x)/N_0; n = [-N_0/2:N_0/2-1]'; ->> clf; subplot(1,2,1); stem(n,abs(fftshift(D_n)),'.k'); ->> axis([-5 5 0 .6]); xlabel('n'); ylabel('|D_n|'); ->> subplot(1,2,2); stem(n,angle(fftshift(D_n)),'.k'); ->> axis([-5 5 -2 2]); xlabel('n'); ylabel('\angle D_n [rad]'); -``` - -As shown in Fig. 6.28, the resulting approximation is visually indistinguishable from the true Fourier series spectra shown in Fig. 6.12 or Fig. 6.13. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/085_6.7 MATLAB - FOURIER SERIES APPLICATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/085_6.7 MATLAB - FOURIER SERIES APPLICATIONS.md deleted file mode 100644 index 512e56df18c0c58181955e33ff458ce5e4ce2067..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/085_6.7 MATLAB - FOURIER SERIES APPLICATIONS.md +++ /dev/null @@ -1,190 +0,0 @@ -## **[6.7 MATLAB: FOURIER](#page-12-0) SERIES APPLICATIONS** - -Computational packages such as MATLAB simplify the Fourier-based analysis, design, and synthesis of periodic signals. MATLAB permits rapid and sophisticated calculations, which promote practical application and intuitive understanding of the Fourier series. - -### **[6.7-1 Periodic Functions and the Gibbs Phenomenon](#page-12-0)** - -It is sufficient to define any *T*0-periodic function over the interval (0 ≤ *t* < *T*0). For example, consider the 2π-periodic function given by - -$$ -x(t) = \begin{cases} t/A & 0 \le t < A \\ 1 & A \le t < \pi \\ 0 & \pi \le t < 2\pi \\ x(t + 2\pi) & \text{otherwise} \end{cases} -$$ - -Although similar to a square wave, *x*(*t*) has a linearly rising edge of width *A*, where (0 < *A* < π). As *A* → 0, *x*(*t*) approaches a square wave; as *A* → π, *x*(*t*) approaches a type of sawtooth wave. - -In MATLAB, the mod command helps represent periodic functions such as *x*(*t*). - ->> x = @(t,A) mod(t,2\*pi)/A.\*(mod(t,2\*pi)=A)&(mod(t,2\*pi)n* = *D*∗ *n*. Truncating the Fourier series at |*n*| = *N* yields the approximation - -$$ -x(t) \approx x_N(t) = D_0 + \sum_{n=1}^{N} \left( D_n e^{jnt} + D_n^* e^{-jnt} \right) -$$ -\n(6.57) - -For a user-specified *N*, program CH6MP1 uses Eq. (6.57) to compute *xN*(*t*) over (−π/4 ≤ *t* < 2π +π/4). - -``` -function [x_N,t] = CH6MP1(A,N); -% CH6MP1.m : Chapter 6, MATLAB Program 1 -% Function M-file approximates x(t) using Fourier series truncated at |n|=N -% INPUTS: A = width of rising edge -% N = largest harmonic of truncated Fourier series -% OUTPUTS: x_N = Nth harmonic truncated Fourier series -``` - -``` -% t = time vector for x_N -``` - -``` -% Define FS coefficients for signal x(t) -D = @(n) 1/(2*pi*n)*((exp(-1j*n*A)-1)/(n*A) + 1j*exp(-1j*n*pi)); -% Construct truncated FS approximation of x(t) using N harmonics -t = linspace(-pi/4,2*pi+pi/4,10000); % Time vector exceeds one period. -x_N = (2*pi-A)/(4*pi)*ones(size(t)); % Compute dc term -for n = 1:N, % Compute N remaining terms - x_N = x_N+real(D(n)*exp(1j*n*t) + conj(D(n))*exp(-1j*n*t)); -end -``` - -Although theoretically not required, the real command ensures that small computer round-off errors do not cause a complex-valued result. - -Using program CH6MP1 with *A* = π/2 and *N* = 20, Fig. 6.29 compares *x*(*t*) and *x*20(*t*). - ->> A = pi/2; [x\_20,t] = CH6MP1(A,20); >> plot(t,x\_20,'k',t,x(t,A),'k:'); axis([-pi/4,2\*pi+pi/4,-0.1,1.1]); >> xlabel('t'); ylabel('x\_{20}(t)'); - -As expected, the falling edge is accompanied by the overshoot that is characteristic of the Gibbs phenomenon. - -Increasing *N* to 100, as shown in Fig. 6.30, improves the approximation but does not reduce the overshoot. - -``` ->> [x_100,t] = CH6MP1(A,100); -``` - ->> plot(t,x\_100,'k',t,x(t,A),'k:'); axis([-pi/4,2\*pi+pi/4,-0.1,1.1]); - -``` ->> xlabel('t'); ylabel('x_{100}(t)'); -``` - -Reducing *A* to π/64 produces a curious result. For *N* = 20, both the rising and falling edges are accompanied by roughly 9% of overshoot, as shown in Fig. 6.31. As the number of terms is increased, overshoot persists only in the vicinity of jump discontinuities. For *xN*(*t*), increasing *N* decreases the overshoot near the rising edge but not near the falling edge. Remember that it is a - -**Figure 6.29** Comparison of *x*20(*t*) and *x*(*t*) when *A* = π/2. - -**Figure 6.30** Comparison of *x*100(*t*) and *x*(*t*) when *A* = π/2. - -**Figure 6.31** Comparison of *x*20(*t*) and *x*(*t*) when *A* = π/64. - -**Figure 6.32** Comparison of *x*100(*t*) and *x*(*t*) when *A* = π/64. - -true jump discontinuity that causes the Gibbs phenomenon. A continuous signal, no matter how sharply it rises, can always be represented by a Fourier series at every point within any small error by increasing *N*. This is not the case when a true jump discontinuity is present. Figure 6.32 illustrates this behavior using *N* = 100. - -### **[6.7-2 Optimization and Phase Spectra](#page-12-0)** - -Although magnitude spectra typically receive the most attention, phase spectra are critically important in some applications. Consider the problem of characterizing the frequency response of an unknown system. By applying sinusoids one at a time, the frequency response is empirically measured one point at a time. This process is tedious at best. Applying a superposition of many sinusoids, however, allows simultaneous measurement of many points of the frequency response. Such measurements can be taken by a spectrum analyzer equipped with a transfer function mode or by applying Fourier analysis techniques, which are discussed in later chapters. - -A multitone test signal *m*(*t*) is constructed as a superposition of *N* real sinusoids - -$$ -m(t) = \sum_{n=1}^{N} M_n \cos{(\omega_n t + \theta_n)} -$$ - -where *Mn* and θ*n* establish the relative magnitude and phase of each sinusoidal component. It is sensible to constrain all gains to be equal, *Mn* = *M* for all *n*. This ensures equal treatment at each point of the measured frequency response. Although the value *M* is normally chosen to set the desired signal power, we set *M* = 1 for convenience. - -While not required, it is also sensible to space the sinusoidal components uniformly in frequency. - -$$ -m(t) = \sum_{n=1}^{N} \cos(n\omega_0 t + \theta_n) -$$ -\n(6.58) - -Another sensible alternative, which spaces components logarithmically in frequency, is treated in Prob. 6.7-4. - -Equation (6.58) is now a truncated compact-form Fourier series with a flat magnitude spectrum. Frequency resolution and range are set by ω0 and *N*, respectively. For example, a 2 kHz range with a resolution of 100 Hz requires ω0 = 2π100 and *N* = 20. The only remaining unknowns are the θ*n*. - -While it is tempting to set θ*n* = 0 for all *n*, the results are quite unsatisfactory. MATLAB helps demonstrate the problem by using ω0 = 2π100 and *N* = 20 sinusoids, each with a peak-to-peak voltage of 1 volt. - -``` ->> m = @(theta,t,omega) sum(cos(omega*t+theta*ones(size(t)))); ->> N = 20; omega = 2*pi*100*[1:N]'; theta = zeros(size(omega)); ->> t = linspace(-0.01,0.01,10000); ->> plot(t,m(theta,t,omega),'k'); xlabel('t [sec]'); ylabel('m(t) [volts]'); -``` - -As shown in Fig. 6.33, θ*n* = 0 causes each sinusoid to constructively add. The resulting 20 volt peak can saturate system components, such as operational amplifiers operating with ±12 volt rails. To improve signal performance, the maximum amplitude of *m*(*t*) over *t* needs to be reduced. - -One way to reduce max*t*(|*m*(*t*)|) is to reduce *M*, the strength of each component. Unfortunately, this approach reduces the system's signal-to-noise ratio and ultimately degrades measurement quality. Therefore, reducing *M* is not a smart decision. The phases θ*n*, however, can be adjusted to reduce max*t*(|*m*(*t*)|) while preserving signal power. In fact, since θ*n* = 0 maximizes max*t*(|*m*(*t*)|), just about any other choice of θ*n* will improve the situation. Even a random choice should improve performance. - -**Figure 6.33** Test signal *m*(*t*) with θ*n* = 0. - -As with any computer, MATLAB cannot generate truly random numbers. Rather, it generates pseudo-random numbers. Pseudo-random numbers are deterministic sequences that appear to be random. The particular sequence of numbers that is realized depends entirely on the initial state of the pseudo-random number generator. Setting the generator's initial state to a known value allows a "random" experiment with reproducible results. The command rng(0) initializes the state of the pseudo-random number generator to a known condition of zero, and the MATLAB command rand(a,b) generates an a-by-b matrix of pseudo-random numbers that are uniformly distributed over the interval (0, 1). Radian phases occupy the wider interval (0, 2π ), so the results from rand need to be appropriately scaled. - ->> rng(0); theta\_rand0 = 2\*pi\*rand(N,1); - -Next, we recompute and plot *m*(*t*) using the randomly chosen θ*n*. - -``` ->> m_rand0 = m(theta_rand0,t,omega); -``` - -``` ->> plot(t,m_rand0,'k'); axis([-0.01,0.01,-10,10]); -``` - -``` ->> xlabel('t [sec]'); ylabel('m(t) [volts]'); -``` - ->> set(gca,'ytick',[min(m\_rand0),max(m\_rand0)]); grid on; - -For a vector input, the min and max commands return the minimum and maximum values of the vector. Using these values to set *y* axis tick marks makes it easy to identify the extreme values of the *m*(*t*). As seen from Fig. 6.34, the maximum amplitude is now 7.6307, which is significantly smaller than the maximum of 20 when θ*n* = 0. - -Randomly chosen phases suffer a fatal fault: there is little guarantee of optimal performance. For example, repeating the experiment with rng(5) produces a maximum magnitude of 8.2399 volts, as shown in Fig. 6.35. This value is significantly higher than the previous maximum of 7.6307 volts. Clearly, it is better to replace a random solution with an optimal solution. - -What constitutes "optimal"? Many choices exist, but desired signal criteria naturally suggest that optimal phases minimize the maximum magnitude of *m*(*t*) over all *t*. To find these optimal phases, MATLAB's fminsearch command is useful. First, the function to be minimized, called the objective function, is defined. - ->> maxmagm = @(theta,t,omega) max(abs(sum(cos(omega\*t+theta\*ones(size(t)))))); - -666 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -**Figure 6.34** Test signal *m*(*t*) with random θ*n* found by using rng(0). - -**Figure 6.35** Test signal *m*(*t*) with random θ*n* found by using rand('state',1). - -The anonymous function argument order is important; fminsearch uses the first input argument as the variable of minimization. To minimize over θ, as desired, θ must be the first argument of the objective function maxmagm. - -Next, the time vector is shortened to include only one period of *m*(*t*). - ->> t = linspace(0,0.01,401); - -A full period ensures that all values of *m*(*t*) are considered; the short length of t helps ensure that functions execute quickly. An initial value of θ is randomly chosen to begin the search. - -``` ->> rng(0); theta_init = 2*pi*rand(N,1); ->> theta_opt = fminsearch(maxmagm,theta_init,[],t,omega); -``` - -Notice that fminsearch finds the minimizer to maxmagm over θ by using an initial value theta\_init. Most numerical minimization techniques are capable of finding only local minima, and fminsearch is no exception. As a result, fminsearch does not always produce a unique solution. The empty square brackets indicate no special options are requested, and the remaining ordered arguments are secondary inputs for the objective function. Full format details for fminsearch are available from MATLAB's help facilities. - -**Figure 6.36** Test signal *m*(*t*) with optimized phases. - -Figure 6.36 shows the phase-optimized test signal. The maximum magnitude is reduced to a value of 5.3632 volts, which is a significant improvement over the original peak of 20 volts. - -Although the signals shown in Figs. 6.33 through 6.36 look different, they all possess the same magnitude spectra. The signals differ only in phase spectra. It is interesting to investigate the similarities and differences of these signals in ways other than graphs and mathematics. For example, is there an audible difference between the signals? For computers equipped with sound capability, the MATLAB sound command can be used to find out. - -``` ->> Fs = 8000; t = [0:1/Fs:2]; % Two second records at a sampling rate of 8kHz ->> sound(m(theta,t,omega)/20,Fs); % Play (scaled) m(t) constructed using zero phases -``` - -Since the sound command clips magnitudes that exceed 1, the input vector is scaled by 1/20 to avoid clipping and the resulting sound distortion. The signals using other phase assignments are created and played in a similar fashion. How well does the human ear discern the differences in phase spectra? If you are like most people, you will not be able to discern any differences in how these waveforms sound. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/086_6.8 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/086_6.8 SUMMARY.md deleted file mode 100644 index 756aa1db80af39988728e9569a28ac311158ccdd..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/086_6.8 SUMMARY.md +++ /dev/null @@ -1,17 +0,0 @@ -## **[6.8 SUMMARY](#page-12-0)** - -In this chapter we showed how a periodic signal can be represented as a sum of sinusoids or exponentials. If the frequency of a periodic signal is *f*0, then it can be expressed as a weighted sum of a sinusoid of frequency *f*0 and its harmonics (the trigonometric Fourier series). We can reconstruct the periodic signal from a knowledge of the amplitudes and phases of these sinusoidal components (amplitude and phase spectra). - -If a periodic signal *x*(*t*) has an even symmetry, its Fourier series contains only cosine terms (including dc). In contrast, if *x*(*t*) has an odd symmetry, its Fourier series contains only sine terms. If *x*(*t*) has neither type of symmetry, its Fourier series contains both sine and cosine terms. - -At points of discontinuity, the Fourier series for *x*(*t*) converges to the mean of the values of *x*(*t*) on either side of the discontinuity. For signals with discontinuities, the Fourier series converges in the mean and exhibits Gibbs phenomenon at the points of discontinuity. The amplitude spectrum of the Fourier series for a periodic signal *x*(*t*) with jump discontinuities decays slowly (as 1/*n*) with frequency. We need a large number of terms in the Fourier series to approximate *x*(*t*) within - -### 668 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -a given error. In contrast, the amplitude spectrum of a smoother periodic signal decays faster with frequency and we require a smaller number of terms in the series to approximate *x*(*t*) within a given error. - -A sinusoid can be expressed in terms of exponentials. Therefore, the Fourier series of a periodic signal can also be expressed as a sum of exponentials (the exponential Fourier series). The exponential form of the Fourier series and the expressions for the series coefficients are more compact than those of the trigonometric Fourier series. Also, the response of LTIC systems to an exponential input is much simpler than that for a sinusoidal input. Moreover, the exponential form of representation lends itself better to mathematical manipulations than does the trigonometric form. This includes the establishment of useful Fourier series properties that simplify work and help provide a more intuitive understanding of signals. For these reasons, the exponential form of the series is preferred in modern practice in the areas of signals and systems. - -The plots of amplitudes and angles of various exponential components of the Fourier series as functions of the frequency are the exponential Fourier spectra (amplitude and angle spectra) of the signal. Because a sinusoid cosω0*t* can be represented as a sum of two exponentials, *ej*ω0*t* and *e*−*j*ω0*t* , the frequencies in the exponential spectra range from ω = −∞ to ∞. By definition, frequency of a signal is always a positive quantity. Presence of a spectral component of a negative frequency −*n*ω0 merely indicates that the Fourier series contains terms of the form *e*−*jn*ω0*t* . The spectra of the trigonometric and exponential Fourier series are closely related, and one can be found by the inspection of the other. - -In Sec. 6.5 we discuss a method of representing signals by the generalized Fourier series, of which the trigonometric and exponential Fourier series are special cases. Signals are vectors in every sense. Just as a vector can be represented as a sum of its components in a variety of ways, depending on the choice of the coordinate system, a signal can be represented as a sum of its components in a variety of ways, of which the trigonometric and exponential Fourier series are only two examples. Just as we have vector coordinate systems formed by mutually orthogonal vectors, we also have signal coordinate systems (basis signals) formed by mutually orthogonal signals. Any signal in this signal space can be represented as a sum of the basis signals. Each set of basis signals yields a particular Fourier series representation of the signal. The signal is equal to its Fourier series, not in the ordinary sense, but in the special sense that the energy of the difference between the signal and its Fourier series approaches zero. This allows for the signal to differ from its Fourier series at some isolated points. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/087_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/087_REFERENCES.md deleted file mode 100644 index 465fd5f66a98c2052dc9915e26b58cbc4d6b0d6f..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/087_REFERENCES.md +++ /dev/null @@ -1,15 +0,0 @@ -### **[REFERENCES](#page-12-0)** - -- 1. Bell, E. T. *Men of Mathematics.* Simon & Schuster, New York, 1937. -- 2. Durant, W., and Durant, A. *The Age of Napoleon,* Part XI in *The Story of Civilization Series.* Simon & Schuster, New York, 1975. -- 3. Calinger, R. *Classics of Mathematics,* 4th ed. Moore Publishing, Oak Park, IL, 1982. -- 4. Lanczos, C. *Discourse on Fourier Series*. Oliver Boyd, London, 1966. -- 5. Körner, T. W. *Fourier Analysis.* Cambridge University Press, Cambridge, UK, 1989. -- 6. Guillemin, E. A. *Theory of Linear Physical Systems.* Wiley, New York, 1963. -- 7. Gibbs, W. J. *Nature,* vol. 59, p. 606, April 1899. -- 8. Bôcher, M. *Annals of Mathematics,* vol. 7, no. 2, 1906. - -- 9. Carslaw, H. S. *Bulletin of the American Mathematical Society,* vol. 31, pp. 420–424, October 1925. -- 10. Lathi, B. P. *Signals, Systems, and Communication.* Wiley, New York, 1965. -- 11. Walker P. L. *The Theory of Fourier Series and Integrals.* Wiley, New York, 1986. -- 12. Churchill, R. V., and Brown, J. W. *Fourier Series and Boundary Value Problems,* 3rd ed. McGraw-Hill, New York, 1978. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/088_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/088_PROBLEMS.md deleted file mode 100644 index 526bc2bce3f1e2882ebab7cb8cfa854c12a885ed..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/088_PROBLEMS.md +++ /dev/null @@ -1,851 +0,0 @@ -## **[PROBLEMS](#page-12-0)** - -- **6.1-1** For each of the periodic signals shown in Fig. P6.1-1, find the compact trigonometric Fourier series and sketch the amplitude and phase spectra. If either the sine or cosine terms are absent in the Fourier series, explain why. -- **6.1-2** (a) Find the trigonometric Fourier series for *y*(*t*) shown in Fig. P6.1-2. - - (b) The signal *y*(*t*) can be obtained by time reversal of *x*(*t*) shown in Fig. 6.2a. Use this fact to obtain the Fourier series for *y*(*t*) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a). - - (c) Show that, in general, time reversal of a periodic signal does not affect the amplitude spectrum, and the phase spectrum is also unchanged except for the change of sign. -- **6.1-3** (a) Find the trigonometric Fourier series for the periodic signal *y*(*t*) depicted in Fig. P6.1-3. - - (b) The signal *y*(*t*) can be obtained by time compression of *x*(*t*) shown in Fig. 6.2a by a factor 2. Use this fact to obtain the Fourier series for *y*(*t*) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a). - - (c) Show that, in general, time compression of a periodic signal by a factor *a* expands the Fourier spectra along the ω axis by the same factor *a*. In other words *C*0,*Cn*, and θ*n* remain unchanged, but the fundamental frequency is increased by the factor *a*, thus expanding the spectrum. Similarly, time expansion of a periodic signal by a factor *a* compresses its Fourier spectra along the ω axis by the factor *a*. -- **6.1-4** (a) Find the trigonometric Fourier series for the periodic signal *g*(*t*) in Fig. P6.1-4. Take advantage of the symmetry. - -- (b) Observe that *g*(*t*) is identical to *x*(*t*) in Fig. 6.4a left-shifted by 0.5 second. Use this fact to obtain the Fourier series for *g*(*t*) from the results in Ex. 6.2. Verify that the Fourier series thus obtained is identical to that found in part (a). -- (c) Show that, in general, a time shift of *T* seconds of a periodic signal does not affect the amplitude spectrum. However, the phase of the *n*th harmonic is increased or decreased *n*ω0*T* depending on whether the signal is advanced or delayed by *T* seconds. -- **6.1-5** Determine the trigonometric Fourier series coefficients *an* and *bn* for the following signals. In each case, also determine the signals' fundamental radian frequency ω0. No integration is required to solve this problem. - - (a) *x*a(*t*) = cos(3π*t*) - - (b) *x*b(*t*) = sin(7π*t*) - - (c) *x*c(*t*) = 2+4 cos(3π*t*)−2*j*sin(7π*t*) - - (d) *x*d(*t*) = (1+*j*)sin(3π*t*)+(2−*j*) cos(7π*t*) - - (e) *x*e(*t*) = sin(3π*t* +1)+2 cos(7π*t* −2) - - (f) *x*f(*t*) = sin(6π*t*)+2 cos(14π*t*) -- **6.1-6** If the two halves of one period of a periodic signal are identical in shape except that one is the negative of the other, the periodic signal is said to have a *half-wave symmetry*. If a periodic signal *x*(*t*) with a period *T*0 satisfies the half-wave symmetry condition, then - -$$ -x\left(t - \frac{T_0}{2}\right) = -x(t) -$$ - -In this case, show that all the even-numbered harmonics vanish and that the odd-numbered harmonic coefficients are given by - -$$ -a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t \, dt -$$ - -**Figure P6.1-2** - -(b) - -**Figure P6.1-6** - -and - -$$ -b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t \, dt -$$ - -Using these results, find the Fourier series for the periodic signals in Fig. P6.1-6. - -**6.1-7** Over a finite interval, a signal can be represented by more than one trigonometric (or exponential) Fourier series. For instance, if we wish to represent *x*(*t*) = *t* over an interval 0 < *t* < 1 by a Fourier series with fundamental frequency ω0 = 2, we can draw a pulse *x*(*t*) = *t* over the interval 0 < *t* < 1 and repeat the pulse every π seconds so that *T*0 = π and ω0 = 2 (Fig. P6.1-7a). If we want the fundamental frequency ω0 to be 4, we repeat the pulse every π/2 seconds. If we want the series to contain only cosine terms with ω0 = 2, we construct a pulse *x*(*t*) = |*t*| over −1 < *t* < 1, and repeat it every π seconds (Fig. P6.1-7b). The resulting signal is an even function with period π. Hence, its Fourier series will have only cosine terms with ω0 = 2. The resulting Fourier series represents *x*(*t*) = *t* over 0 < *t* < 1, as desired. We do not care what it represents outside this interval. - -Sketch the periodic signal *x*(*t*) such that *x*(*t*) = *t* for 0 < *t* < 1 and the Fourier series for *x*(*t*) satisfies the following conditions. - -- (a) ω0 = π/2 and contains all harmonics, but cosine terms only -- (b) ω0 = 2 and contains all harmonics, but sine terms only -- (c) ω0 = π/2 and contains all harmonics, which are exclusively neither sine nor cosine -- (d) ω0 = 1 and contains only odd harmonics and cosine terms -- (e) ω0 = π/2 and contains only odd harmonics and sine terms - -**Figure P6.1-7** - -(f) ω0 = 1 and contains only odd harmonics, which are exclusively neither sine nor cosine. - -[*Hint:* For parts (d), (e), and (f), you need to use half-wave symmetry discussed in Prob. 6.1-6. Cosine terms imply a possible dc component.] You are asked only to sketch the periodic signal *x*(*t*) satisfying the given conditions. Do not find the values of the Fourier coefficients. - -- **6.1-8** State with reasons whether the following signals are periodic or aperiodic. For periodic signals, find the period and state which harmonics are present in the series. - - (a) 3 sin *t* +2 sin 3*t* - - (b) 2+5 sin 4*t* +4 cos 7*t* - - (c) 2 sin 3*t* +7 cos π*t* - - (d) 7 cos π*t* +5 sin 2π*t* - - (e) 3 cos √2*t* +5 cos 2*t* - - (f) sin 5*t* 2 +3 cos 6*t* 5 +3 sin *t* 7 +30◦ *t* - -(g) -$$ -\sin 3t + \cos \frac{15}{4} -$$ - -- (h) (3 sin 2*t* +sin 5*t*)2 -- (i) (5 sin 2*t*)3 -- **6.3-1** For each of the periodic signals in Fig. P6.1-1, find exponential Fourier series and sketch the corresponding spectra. -- **6.3-2** A 2π-periodic signal *x*(*t*) is specified over one period as - -$$ -x(t) = \begin{cases} \frac{1}{A}t & 0 \le t < A \\ 1 & A \le t < \pi \\ 0 & \pi \le t < 2\pi \end{cases} -$$ - -Sketch *x*(*t*) over two periods from *t* = 0 to 4π. Show that the exponential Fourier series coefficients *Dn* for this series are given by - -$$ -D_n = \begin{cases} \frac{2\pi - A}{4\pi} & n = 0\\ \frac{1}{2\pi n} \left( \frac{e^{-jAn} - 1}{An} + je^{-jn\pi} \right) & n \neq 0 \end{cases} -$$ - -**6.3-3** A periodic signal *x*(*t*) is expressed by the following Fourier series: - -$$ -x(t) = 3\cos t + \sin\left(t - \frac{\pi}{6}\right) - 2\cos\left(t - \frac{\pi}{3}\right) -$$ - -- (a) Sketch the amplitude and phase spectra for the trigonometric series. -- (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra. -- (c) By inspection of spectra in part (b), write the exponential Fourier series for *x*(*t*). -- (d) Show that the series found in part (c) is equivalent to the trigonometric series for *x*(*t*). -- **6.3-4** The trigonometric Fourier series of a certain periodic signal is given by - -$$ -x(t) = 3 + \sqrt{3}\cos 2t + \sin 2t -$$ -$$ -+ \sin 3t - \frac{1}{2}\cos\left(5t + \frac{\pi}{3}\right) -$$ - -- (a) Sketch the trigonometric Fourier spectra. -- (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra. -- (c) By inspection of spectra in part (b), write the exponential Fourier series for *x*(*t*). -- (d) Show that the series found in part (c) is equivalent to the trigonometric series for *x*(*t*). -- **6.3-5** The exponential Fourier series of a certain function is given as - -$$ -x(t) = (2+j2)e^{-j3t} + j2e^{-jt} + 3 - j2e^{jt} + (2-j2)e^{j3t} -$$ - -- (a) Sketch the exponential Fourier spectra. -- (b) By inspection of the spectra in part (a), sketch the trigonometric Fourier spectra for *x*(*t*). Find the compact trigonometric Fourier series from these spectra. -- (c) Show that the trigonometric series found in part (b) is equivalent to the exponential series for *x*(*t*). -- (d) Find the signal bandwidth. -- **6.3-6** Figure P6.3-6 shows the trigonometric Fourier spectra of a periodic signal *x*(*t*). - - (a) By inspection of Fig. P6.3-6, find the trigonometric Fourier series representing *x*(*t*). - - (b) By inspection of Fig. P6.3-6, sketch the exponential Fourier spectra of *x*(*t*). - - (c) By inspection of the exponential Fourier spectra obtained in part (b), find the exponential Fourier series for *x*(*t*). - - (d) Show that the series found in parts (a) and (c) are equivalent. -- **6.3-7** Figure P6.3-7 shows the exponential Fourier spectra of a periodic signal *x*(*t*). - -- (a) By inspection of Fig. P6.3-7, find the exponential Fourier series representing *x*(*t*). -- (b) By inspection of Fig. P6.3-7, sketch the trigonometric Fourier spectra for *x*(*t*). -- (c) By inspection of the trigonometric Fourier spectra found in part (b), find the trigonometric Fourier series for *x*(*t*). -- (d) Show that the series found in parts (a) and (c) are equivalent. -- **6.3-8** Let periodic signal *x*(*t*) have exponential Fourier series spectrum *Dn*. Prove the following properties. - - (a) If *x*(*t*) has even symmetry, then *Dn* also has even symmetry. - - (b) If *x*(*t*) has odd symmetry, then *Dn* also has odd symmetry. - - (c) If *x*(*t*) is real, then *Dn* is conjugate symmetric (*Dn* = *D*∗ −*n*). - - (d) If *x*(*t*) is imaginary, then *Dn* is conjugate antisymmetric (*Dn* = −*D*∗ −*n*). -- **6.3-9** (a) Find the exponential Fourier series for the signal in Fig. P6.3-9a. - - (b) Using the results in part (a), find the Fourier series for the signal *x*ˆ(*t*) in Fig. P6.3-9b, - -**Figure P6.3-7** - -#### **Figure P6.3-9** - -which is a time-shifted version of the signal *x*(*t*). - -- (c) Using the results in part (a), find the Fourier series for the signal *x*˜(*t*) in Fig. P6.3-9c, which is a time-scaled version of the signal *x*(*t*). -- **6.3-10** A periodic signal *x*(*t*) is expressed as an exponential Fourier series - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} -$$ - -(a) Show that the exponential Fourier series for *x*ˆ(*t*) = *x*(*t* −*T*) is given by - -$$ -\hat{x}(t) = \sum_{n = -\infty}^{\infty} \hat{D}_n e^{jn\omega_0 t} -$$ - -in which - -$$ -|\tilde{D}_n| = |D_n| \quad \text{and} \quad \angle \tilde{D}_n = \angle D_n - n\omega_0 T -$$ - -This result shows that time shifting of a periodic signal by *T* seconds merely changes the phase spectrum by *n*ω0*T*. The amplitude spectrum is unchanged. - -(b) Show that the exponential Fourier series for *x*˜(*t*) = *x*(*at*) is given by - -$$ -\tilde{x}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn(a\omega_0)t} -$$ - -This result shows that time compression of a periodic signal by a factor *a* expands its Fourier spectra along the ω axis by the same factor *a*. Similarly, time expansion of a periodic signal by a factor *a* compresses its Fourier spectra along the ω axis by the factor *a*. Intuitively explain this result. - -**6.3-11** (a) The Fourier series for the periodic signal in Fig. 6.7a is given in Drill 6.1. Verify Parseval's theorem for this series, given that - -$$ -\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} -$$ - -- (b) If *x*(*t*) is approximated by the first *N* terms in this series, find *N* so that the power of the error signal is less than 1% of *Px*. -- **6.3-12** (a) The Fourier series for the periodic signal in Fig. 6.7b is given in Drill 6.1. Verify - -Problems 675 - -Parseval's theorem for this series, given that - -$$ -\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} -$$ - -- (b) If *x*(*t*) is approximated by the first *N* terms in this series, find *N* so that the power of the error signal is less than 10% of *Px*. -- **6.3-13** The signal *x*(*t*) in Fig. 6.17 is approximated by the first 2*N* + 1 terms (from *n* = −*N* to *N*) in its exponential Fourier series given in Drill 6.5. Determine the value of *N* if this (2*N* + 1)-term Fourier series power is to be no less than 99.75% of the power of *x*(*t*). -- **6.3-14** (a) A 2 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = 1 3 *x*1(−*t* −5) in terms of *X*1[*n*]. - - (b) A 2 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = cos(10*t*)*x*1(*t*) in terms of *X*1[*n*]. - - (c) A 3 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = *x*1(−*t*)−3*x*1(*t* +2) in terms of *X*1[*n*]. -- **6.3-15** A 2-periodic signal *x*(*t*) is defined as - -$$ -x(t) = \begin{cases} -t^2 - t + 0.25 & -1 \le t < 0\\ t^2 - t + 0.25 & 0 \le t < 1\\ x(t+2) & \forall t \end{cases} -$$ - -- (a) Plot *x*(*t*) over −2 ≤ *t* ≤ 2. -- (b) Determine *D*0, the dc content of *x*(*t*). -- (c) Similar to Ex. 6.11, use properties and not integration to determine *Dn* for *n* = 0. -- (d) Plot the magnitude spectrum |*Dn*| over a suitable range of *n*. -- (e) How does the magnitude spectrum |*Dn*| compare to that of signal *y*(*t*) = cos(π*t*)? Note similarities as well as major differences. -- **6.3-16** A 3-periodic signal *x*(*t*) is defined as - -$$ -x(t) = \begin{cases} \n|t| & -1 \le t \le 1\\ \n0 & 1 < |t| \le 1.5\\ \nx(t+3) & \forall t\n\end{cases} -$$ - -(a) Plot -$$ -x(t) -$$ - over $-3 \le t \le 3$ . - -- (b) Determine *D*0, the dc content of *x*(*t*). -- (c) Similar to Ex. 6.11, use properties and not integration to determine *Dn* for *n* = 0. -- (d) Plot the magnitude spectrum |*Dn*| over a suitable range of *n*. What is the most dominant frequency component of this signal? -- **6.4-1** Find the response of an LTIC system with transfer function - -$$ -H(s) = \frac{s}{s^2 + 2s + 3} -$$ - -to the periodic input shown in Fig. 6.2a. - -- **6.4-2** A periodic signal *x*(*t*) = 1 + 2 cos(5π*t*) + 3 sin(14π*t*) is applied to an LTIC system to produce output *y*(*t*). - - (a) Determine ω0, the fundamental radian frequency of *x*(*t*). - - (b) Determine *Dn*, the exponential Fourier series spectrum of *x*(*t*). - - (c) If the system is an ideal lowpass filter with cutoff frequency *f*c = 2 Hz, what is the output *y*(*t*)? - - (d) If the system is an ideal highpass filter with cutoff frequency *f*c = 2 Hz, what is the output *y*(*t*)? - - (e) If the system is an ideal bandpass filter with a 4 Hz passband centered at 4 Hz, what is the output *y*(*t*)? - - (f) If the system is an ideal bandstop filter with a 5 Hz stopband centered at 10 Hz, what is the output *y*(*t*)? - - (g) Describe the frequency response of a filter that, in response to *x*(*t*), would produce the output *y*(*t*) = 4 cos(5π*t*) −9 sin(14π*t*). -- **6.4-3** Consider a *T*0 = 1 periodic signal *x*(*t*) defined as - -$$ -x(t) = \begin{cases} 1 - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases} -$$ - -- (a) Sketch *x*(*t*) for −2 ≤ *t* ≤ 2. -- (b) Determine *Dn*, the exponential Fourier series spectrum of *x*(*t*). -- (c) If *x*(*t*) is applied to an ideal bandpass filter with a 1 Hz passband centered at 3 Hz, determine the output *y*(*t*). -- **6.4-4** (a) Find the exponential Fourier series for a signal *x*(*t*) = cos 5*t* sin 3*t*. You can do this without evaluating any integrals. - - (b) Sketch the Fourier spectra. - -(c) The signal *x*(*t*) is applied at the input of an LTIC system with frequency response, as shown in Fig. P6.4-4. Find the output *y*(*t*). - -**Figure P6.4-4** - -- **6.4-5** (a) Find the exponential Fourier series for a periodic signal *x*(*t*) shown in Fig. P6.4-5a. - - (b) The signal *x*(*t*) is applied at the input of an LTIC system shown in Fig. P6.4-5b. Find the expression for the output *y*(*t*). -- **6.4-6** A *T*-periodic τ/*T* duty-cycle square wave *p*(*t*) is defined as - -$$ -p(t) = \begin{cases} 1 & |t| < \frac{\tau}{2} \\ 0 & \frac{\tau}{2} < |t| < \frac{T}{2} \\ p(t+T) & \forall t \end{cases} -$$ - -where 0 <τ< *T*. Also consider the frequency response *H*(ω) of a lowpass communications channel with 10 rad/s bandwidth (e.g., |*H*(ω)| ≈ 0 for ω > 10). If *T* and τ are properly chosen, we can estimate *H*(ω) at points ω = *n*ω0 as *H*ˆ (*n*ω0) = 1 *P*0 *Yn*, where *Yn* is the exponential FS spectrum of the channel output *y*(*t*) in response to input *p*(*t*) and *P*0 is the dc component of *p*(*t*). - -- (a) Using direct integration, determine the exponential Fourier series coefficients *Pn* of signal *p*(*t*). -- (b) Determine a suitable value *T* so that *p*(*t*) applied to system *H*(ω) has 21 component frequencies over the system bandwidth 0 ≤ ω ≤ 10. -- (c) Assuming *T* is properly chosen, determine a suitable duty cycle τ/*T* so that *H*(*n*ω0) ≈ *Yn*. Carefully justify your result. -- (d) From the perspective of using *p*(*t*) to help measure the system frequency response *H*(ω), what happens if *T* is properly chosen but τ/*T* is chosen too small? -- (e) From the perspective of using *p*(*t*) to help measure the system frequency response *H*(ω), what happens if *T* is properly chosen but τ/*T* is chosen too large? -- **6.5-1** Derive Eq. (6.32) in an alternate way by observing that **e** = (**x**−*c***y**) and |**e**| 2 = (**x**−*c***y**)·(**x** − *c***y**) = |**x**| 2 +*c*2|**y**| 2 −2*c***x** · **y**. -- **6.5-2** A signal *x*(*t*) is approximated in terms of a signal *y*(*t*) over an interval (*t*1, *t*2): - -$$ -x(t) \simeq cy(t) \qquad t_1 < t < t_2 -$$ - -where *c* is chosen to minimize the error energy. - -**Figure P6.4-5** - -- (a) Show that *y*(*t*) and the error *e*(*t*) = *x*(*t*) − *cy*(*t*) are orthogonal over the interval (*t*1, *t*2). -- (b) If possible, explain the result in terms of a signal-vector analogy. -- (c) Verify this result for the square signal *x*(*t*) in Fig. 6.23 and its approximation in terms of signal sin *t*. -- **6.5-3** If *x*(*t*) and *y*(*t*) are orthogonal, then show that the energy of the signal *x*(*t*) + *y*(*t*) is identical to the energy of the signal *x*(*t*) − *y*(*t*) and is given by *Ex* + *Ey*. Explain this result by using the vector analogy. In general, show that for orthogonal signals *x*(*t*) and *y*(*t*) and for any pair of arbitrary real constants *c*1 and *c*2, the energies of *c*1*x*(*t*) + *c*2*y*(*t*) and *c*1*x*(*t*) − *c*2*y*(*t*) are both given by *c*2 1*Ex* +*c*2 2*Ey*. -- **6.5-4** (a) For the signals *x*(*t*) and *y*(*t*) depicted in Fig. P6.5-4, find the component of the form *y*(*t*) contained in *x*(*t*). In other words, find the optimum value of *c* in the approximation *x*(*t*) ≈ *cy*(*t*) so that the error signal energy is minimum. - - (b) Find the error signal *e*(*t*) and its energy *Ee*. Show that the error signal is orthogonal to *y*(*t*), and that *Ex* = *c*2*Ey* + *Ee*. Explain this result in terms of vectors. - -**Figure P6.5-4** - -- **6.5-5** For the signals *x*(*t*) and *y*(*t*) shown in Fig. P6.5-4, find the component of the form *x*(*t*) contained in *y*(*t*). In other words, find the optimum value of *c* in the approximation *y*(*t*) ≈ *cx*(*t*) so that the error signal energy is minimum. What is the error signal energy? -- **6.5-6** Represent the signal *x*(*t*) shown in Fig. P6.5-4a over the interval from 0 to 1 by a trigonometric Fourier series of fundamental frequency ω0 = 2π. Compute the error energy in the representation of *x*(*t*) by only the first *N* terms of this series for *N* = 1, 2, 3, and 4. - -- **6.5-7** Represent *x*(*t*) = *t* over the interval (0, 1) by a trigonometric Fourier series that has - - (a) ω0 = 2π and only sine terms - - (b) ω0 = π and only sine terms - - (c) ω0 = π and only cosine terms - -You may use a dc term in these series if necessary. - -- **6.5-8** In Ex. 6.15, we represented the function in Fig. 6.27 by Legendre polynomials. - - (a) Use the results in Ex. 6.15 to represent the signal *g*(*t*) in Fig. P6.5-8 by Legendre polynomials. - - (b) Compute the error energy for the approximations having one and two (nonzero) terms. - -**Figure P6.5-8** - -- **6.5-9** Walsh functions, which can take on only two amplitude values, form a complete set of orthonormal functions and are of great practical importance in digital applications because they can be easily generated by logic circuitry and because multiplication with these functions can be implemented by simply using a polarity-reversing switch. Figure P6.5-9 shows the first eight functions in this set. Represent *x*(*t*) in Fig. P6.5-4a over the interval (0, 1) by using a Walsh Fourier series with these eight basis functions. Compute the energy of *e*(*t*), the error in the approximation, using the first *N* nonzero terms in the series for *N* = 1, 2, 3, and 4. In Prob. 6.5-6 we found the trigonometric Fourier series for *x*(*t*). How does the Walsh series compare with the trigonometric series in Prob. 6.5-6 from the viewpoint of the error energy for a given *N*? -- **6.5-10** For the four-dimensional real space *R*4, the so-called Walsh basis is given by: φ1 = [1, 1, 1, 1], φ2 = [1, 1,−1,−1], φ3 = [1,−1,−1, 1], and φ4 = [1,−1, 1,−1]. Denoting elements *x* = [*x*1, *x*2, *x*3, *x*4] and *y* = [*y*1, *y*2, *y*3, *y*4] (*x*, *y* ∈ *R*4), we can define orthogonality as - -### **Figure P6.5-9** - -%4 *k*=1 *xky*∗ *k* = 0. In linear algebra terminology, orthogonality here means that the inner product of vectors *x* and *y* is zero. Lastly, define a vector *z* = [−4, 0, 1,−7]. - -- (a) Show that the Walsh basis functions are mutually orthogonal. This requires a total of six calculations. -- (b) Are the Walsh basis functions normal? That is, does the inner product of each Walsh basis function with itself evaluate to 1? -- (c) Determine the coefficients [*c*1, *c*2, *c*3, *c*4] to represent *z* using Walsh basis functions as *z*ˆ = %4 *k*=1 *ck*φ*k*. -- (d) Determine the best three-dimensional approximation *z*ˆ3*D* to *z* in terms of the Walsh basis functions. That is, your estimate can only be a linear combination of three functions from [φ1,φ2,φ3,φ4]. Evaluate the three-term sum to determine the four elements of vector *z*ˆ3*D*. -- **6.5-11** A function can be expanded in terms of many different types of basis functions, not just the complex exponentials of Fourier analysis. For example, Walsh functions are explored in Prob. 6.5-9. *Laguerre* polynomials *Lk*(*t*), which have support on the interval [0,∞), are another possible set of basis functions. The Laguerre expansion using any number of terms we choose. For the Laguerre expansion, we define orthogonality a little differently, as - -$$ -\int_0^\infty e^{-t}x(t)y(t)dt = 0. -$$ - -Notice the presence of the *e*−*t* term in the integral. Using this definition, Laguerre polynomials are orthonormal. - -(a) Show that *L*0(*t*) is normal. That is, show that - -$$ -\int_0^\infty e^{-t} L_0(t) L_0^*(t) dt = 1 -$$ - -(b) Show that *L*1(*t*) is normal. That is, show that - -$$ -\int_0^{\infty} e^{-t} L_1(t) L_1^*(t) dt = 1 -$$ - -- (c) Show that *L*0(*t*) is orthogonal to *L*1(*t*). -- (d) Compute the coefficient *c*0 that produces the best approximation *x*ˆ0(*t*) = *c*0*L*0(*t*) of the function *x*(*t*) = *e*−*t u*(*t*). -- (e) For the best estimate *x*ˆ1(*t*) = *c*0*L*0(*t*) + *c*1*L*1(*t*) of the function *x*(*t*) = *e*−*t u*(*t*), the coefficient *c*0 remains unchanged from part **()**(d) and *c*1 = 1 4 . Confirm that *c*1 = 1 4 and explain why the coefficient *c*0 does not change. -- **6.7-1** A periodic signal has ω0 = 2 3π and exponential Fourier series spectrum *Dn* = *j* cos (π*n*/10)(*u*[*n*+10] −*u*[*n*−11]). [*Hint:* Refer to Prob. 6.3-8 for some useful properties.] - - (a) Determine the period *T*0 of the corresponding signal *x*(*t*). - - (b) Is the time-domain signal real, imaginary, or neither? Justify your answer. - - (c) Is the time-domain signal even, odd, or neither? Justify your answer. - - (d) Use MATLAB to synthesize the timedomain signal *x*(*t*) and plot it over the interval [−*T*0,*T*0]. -- **6.7-2** Repeat Prob. 6.7-1 for a periodic signal with ω0 = 3 2π and *Dn* = 2 sin(π*n*/10)(*u*[*n* + 10] − *u*[*n*−11]). -- **6.7-3** Consider the (*T*0 = 1)-periodic signal *x*(*t*): - -$$ -x(t) = \begin{cases} 2t - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases}. -$$ - -- (a) Sketch *x*(*t*) for −2 ≤ *t* ≤ 2. -- (b) Using properties and minimal integration, determine *Dn*, the exponential Fourier spectrum of *x*(*t*). - -- (c) Verify the correctness of *Dn* by using MAT-LAB to synthesize *x*(*t*) with a suitable truncation of Eq. (6.19). -- (d) Suppose *x*(*t*) is applied to an ideal bandpass filter with passband between 2.5 and 3.5 Hz. Determine the filter output *y*(*t*). Simplify your answer. - -[*Hint:* Refer to Ex. 6.11.] - -**6.7-4** Section 6.7 discusses the construction of a phase-optimized multitone test signal with linearly spaced frequency components. This problem investigates a similar signal with logarithmically spaced frequency components. - -> A multitone test signal *m*(*t*) is constructed by using a superposition of *N* real sinusoids - -$$ -m(t) = \sum_{n=1}^{N} \cos(\omega_n t + \theta_n) -$$ - -where θ*n* establishes the relative phase of each sinusoidal component. - -- (a) Determine a suitable set of *N* = 10 frequencies ω*n* that logarithmically spans [(2π ) ≤ ω ≤ 100(2π )] yet still results in a periodic test signal *m*(*t*). Determine the period *T*0 of your signal. Using θ*n* = 0, plot the resulting (*T*0)-periodic signal over −*T*0/2 ≤ *t* ≤ *T*0/2. -- (b) Determine a suitable set of phases θ*n* that minimize the maximum magnitude of *m*(*t*). Plot the resulting signal and identify the maximum magnitude that results. -- (c) Many systems suffer from what is called one-over-*f* noise. The power of this undesirable noise is proportional to 1/*f* . Thus, low-frequency noise is stronger than high-frequency noise. What modifications to *m*(*t*) are appropriate for use in environments with 1/*f* noise? Justify your answer. - -# **[CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER TRANSFORM** - -We can analyze linear systems in many different ways by taking advantage of the property of linearity, whereby the input is expressed as a sum of simpler components. The system response to any complex input can be found by summing the system's response to these simpler components of the input. In time-domain analysis, we separated the input into impulse components. In the frequency-domain analysis in Ch. 4, we separated the input into exponentials of the form *est* (the Laplace transform), where the complex frequency *s* = σ + *j*ω. The Laplace transform, although very valuable for system analysis, proves somewhat awkward for signal analysis, where we prefer to represent signals in terms of exponentials *ej*ω*t* instead of *est*. This is accomplished by the Fourier transform. In a sense, the Fourier transform may be considered to be a special case of the Laplace transform with *s* = *j*ω. Although this view is true most of the time, it does not always hold because of the nature of convergence of the Laplace and Fourier integrals. - -In Ch. 6, we succeeded in representing periodic signals as a sum of (everlasting) sinusoids or exponentials of the form *ej*ω*t* . The Fourier integral developed in this chapter extends this spectral representation to aperiodic signals. - -## **7.1 APERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY THE FOURIER INTEGRAL** - -Applying a limiting process, we now show that an aperiodic signal can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal *x*(*t*) such as the one depicted in Fig. 7.1a by everlasting exponentials, let us construct a new periodic signal *xT*0 (*t*) formed by repeating the signal *x*(*t*) at intervals of *T*0 seconds, as illustrated in Fig. 7.1b. The period *T*0 is made long enough to avoid overlap between the repeating pulses. The periodic signal *xT*0 (*t*) can be represented by an exponential Fourier series. If we let *T*0 → ∞, the pulses in the periodic signal repeat after an infinite interval and, therefore, - -$$ -\lim_{T_0 \to \infty} x_{T_0}(t) = x(t) -$$ - -CHAPTER - -**7** - -**Figure 7.1** Construction of a periodic signal: (a) signal *x*(*t*) and (b) periodic extension of *x*(*t*). - -Thus, the Fourier series representing *xT*0 (*t*) will also represent *x*(*t*) in the limit *T*0 → ∞. The exponential Fourier series for *xT*0 (*t*) is given by - -$$ -x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} -$$ -\n(7.1) - -where ω0 = 2π *T*0 and - -$$ -D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x_{T_0}(t) e^{-jn\omega_0 t} dt -$$ -\n(7.2) - -Observe that integrating *xT*0 (*t*) over (−*T*0/2,*T*0/2) is the same as integrating *x*(*t*) over (−∞,∞). Therefore, Eq. (7.2) can be expressed as - -$$ -D_n = \frac{1}{T_0} \int_{-\infty}^{\infty} x(t) e^{-jn\omega_0 t} dt -$$ -\n(7.3) - -It is interesting to see how the nature of the spectrum changes as *T*0 increases. To understand this fascinating behavior, let us define *X*(ω), a continuous function of ω, as - -$$ -X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt -$$ -\n(7.4) - -A glance at Eqs. (7.3) and (7.4) shows that - -$$ -D_n = \frac{1}{T_0} X(n\omega_0) \tag{7.5} -$$ - -**Figure 7.2** Change in the Fourier spectrum when the period *T*0 in Fig. 7.1 is doubled. - -This means that the Fourier coefficients *Dn* are 1/*T*0 times the samples of *X*(ω) uniformly spaced at intervals of ω0, as depicted in Fig. 7.2a.† Therefore, (1/*T*0)*X*(ω) is the envelope for the coefficients *Dn*. We now let *T*0 → ∞ by doubling *T*0 repeatedly. Doubling *T*0 halves the fundamental frequency ω0 so that there are now twice as many components (samples) in the spectrum. However, by doubling *T*0, the envelope (1/*T*0)*X*(ω) is halved, as shown in Fig. 7.2b. If we continue this process of doubling *T*0 repeatedly, the spectrum progressively becomes denser while its magnitude becomes smaller. Note, however, that the relative shape of the envelope remains the same [proportional to *X*(ω) in Eq. (7.4)]. In the limit as *T*0 → ∞, ω0 → 0 and *Dn* → 0. This result makes for a spectrum so dense that the spectral components are spaced at zero (infinitesimal) intervals. At the same time, the amplitude of each component is zero (infinitesimal). We have *nothing of everything, yet we have something!* This paradox sounds like *Alice in Wonderland,* but as we shall see, these are the classic characteristics of a very familiar phenomenon.‡ - -Substitution of Eq. (7.5) in Eq. (7.1) yields - -$$ -x_{T_0}(t) = \sum_{n = -\infty}^{\infty} \frac{X(n\omega_0)}{T_0} e^{jn\omega_0 t} -$$ -\n(7.6) - -As *T*0 →∞, ω0 becomes infinitesimal (ω0 →0). Hence, we shall replace ω0 by a more appropriate notation, ω. In terms of this new notation, ω0 = 2π *T*0 becomes - -$$ -\Delta \omega = \frac{2\pi}{T_0} -$$ - -† For the sake of simplicity, we assume *Dn*, and therefore *X*(ω), in Fig. 7.2, to be real. The argument, however, is also valid for complex *Dn* [or *X*(ω)]. - -‡ If nothing else, the reader now has irrefutable proof of the proposition that 0% ownership of everything is better than 100% ownership of nothing. - -**Figure 7.3** The Fourier series becomes the Fourier integral in the limit as *T*0 → ∞. - -and Eq. (7.6) becomes - -$$ -x_{T_0}(t) = \sum_{n=-\infty}^{\infty} \left[ \frac{X(n\Delta\omega)\Delta\omega}{2\pi} \right] e^{(jn\Delta\omega)t} -$$ - -This equation shows that *xT*0 (*t*) can be expressed as a sum of everlasting exponentials of frequencies 0,±ω,±2ω,±3ω,... (the Fourier series). The amount of the component of frequency *n*ω is [*X*(*n*ω)ω]/2π. In the limit as *T*0 → ∞, ω → 0 and *xT*0 (*t*) → *x*(*t*). Therefore, - -$$ -x(t) = \lim_{T_0 \to \infty} x_{T_0}(t) = \lim_{\Delta \omega \to 0} \frac{1}{2\pi} \sum_{n = -\infty}^{\infty} X(n\Delta \omega) e^{(jn\Delta \omega)t} \Delta \omega -$$ -\n(7.7) - -The sum on the right-hand side of Eq. (7.7) can be viewed as the area under the function *X*(ω)*ej*ω*t* , as illustrated in Fig. 7.3. Therefore, - -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega -$$ - (7.8) - -The integral on the right-hand side is called the *Fourier integral*. We have now succeeded in representing an aperiodic signal *x*(*t*) by a Fourier integral (rather than a Fourier series).† This integral is basically a Fourier series (in the limit) with fundamental frequency ω → 0, as seen from Eq. (7.7). The amount of the exponential *ejn*ω*t* is *X*(*n*ω)ω/2π. Thus, the function *X*(ω) given by Eq. (7.4) acts as a spectral function. - -We call *X*(ω) the *direct* Fourier transform of *x*(*t*), and *x*(*t*) the *inverse* Fourier transform of *X*(ω). The same information is conveyed by the statement that *x*(*t*) and *X*(ω) are a Fourier transform pair. Symbolically, this statement is expressed as - -$$ -X(\omega) = \mathcal{F}[x(t)] -$$ - and $x(t) = \mathcal{F}^{-1}[X(\omega)]$ - -† This derivation should not be considered to be a rigorous proof of Eq. (7.8). The situation is not as simple as we have made it appear [1]. - -or - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -To recapitulate, - -$$ -X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt -$$ -\n(7.9) - -and - -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega -$$ -\n(7.10) - -It is helpful to keep in mind that the Fourier integral in Eq. (7.10) is of the nature of a Fourier series with fundamental frequency ω approaching zero [Eq. (7.7)]. Therefore, most of the discussion and properties of Fourier series apply to the Fourier transform as well. *The transform X*(ω) *is the frequency-domain specification of x*(*t*)*.* - -We can plot the spectrum *X*(ω) as a function of ω. Since *X*(ω) is complex, we have both amplitude and angle (or phase) spectra - -$$ -X(\omega) = |X(\omega)|e^{j\angle X(\omega)} -$$ - -in which |*X*(ω)| is the amplitude and *X*(ω) is the angle (or phase) of *X*(ω). According to Eq. (7.9), - -$$ -X(-\omega) = \int_{-\infty}^{\infty} x(t)e^{j\omega t}dt -$$ - -Taking the conjugates of both sides yields - -$$ -x^*(t) \Longleftrightarrow X^*(-\omega) \tag{7.11} -$$ - -This property is known as the *conjugation property.* Now, if *x*(*t*) is a real function of *t*, then *x*(*t*) = *x*∗(*t*), and from the conjugation property, we find that - -$$ -X(-\omega) = X^*(\omega) -$$ - -This is the *conjugate symmetry* property of the Fourier transform, applicable to real *x*(*t*). Therefore, for real *x*(*t*), - -$$ -|X(-\omega)| = |X(\omega)| \quad \text{and} \quad \angle X(-\omega) = -\angle X(\omega) \tag{7.12} -$$ - -Thus, for real *x*(*t*), the amplitude spectrum |*X*(ω)| is an even function, and the phase spectrum *X*(ω) is an odd function of ω. These results were derived earlier for the Fourier spectrum of a periodic signal [Eq. (6.22)] and should come as no surprise. - -### **EXAMPLE 7.1 Fourier Transform of a Causal Exponential** - -Find the Fourier transform of *e*−*atu*(*t*). - -By definition [Eq. (7.9)], - -$$ -X(\omega) = \int_{-\infty}^{\infty} e^{-at} u(t) e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt = \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty} -$$ - -But |*e*−*j*ω*t* | = 1. Therefore, as *t* → ∞, *e*−(*a*+*j*ω)*t* = *e*−*ate*−*j*ω*t* = ∞ if *a* < 0, but it is equal to 0 if *a* > 0. Therefore, - -$$ -X(\omega) = \frac{1}{a + j\omega} \qquad a > 0 -$$ - -Expressing *a*+*j*ω in the polar form as √ *a*2 +ω2 *ej*tan−1(ω/*a*) , we obtain - -$$ -X(\omega) = \frac{1}{\sqrt{a^2 + \omega^2}} e^{-j \tan^{-1}(\omega/a)} -$$ - -Therefore, - -$$ -|X(\omega)| = \frac{1}{\sqrt{a^2 + \omega^2}} \quad \text{and} \quad \angle X(\omega) = -\tan^{-1}\left(\frac{\omega}{a}\right) -$$ - -The amplitude spectrum |*X*(ω)| and the phase spectrum *X*(ω) are depicted in Fig. 7.4b. Observe that |*X*(ω)| is an even function of ω, and *X*(ω) is an odd function of ω, as expected. - -### EXISTENCE OF THE FOURIER TRANSFORM - -In Ex. 7.1 we observed that when *a* < 0, the Fourier integral for *e*−*atu*(*t*) does not converge. Hence, the Fourier transform for *e*−*atu*(*t*) does not exist if *a* < 0 (growing exponential). Clearly, not all signals are Fourier transformable. - -Because the Fourier transform is derived here as a limiting case of the Fourier series, it follows that the basic qualifications of the Fourier series, such as *equality in the mean* and convergence conditions in suitably modified form, apply to the Fourier transform as well. It can be shown that if *x*(*t*) has a finite energy, that is, if - -$$ -\int_{-\infty}^{\infty} |x(t)|^2 dt < \infty -$$ - -then the Fourier transform *X*(ω) is finite and converges to *x*(*t*) in the mean. This means, if we let - -$$ -\hat{x}(t) = \lim_{W \to \infty} \frac{1}{2\pi} \int_{-W}^{W} X(\omega) e^{j\omega t} d\omega -$$ - -then Eq. (7.10) implies - -$$ -\int_{-\infty}^{\infty} \left| x(t) - \hat{x}(t) \right|^2 dt = 0 \tag{7.13} -$$ - -In other words, *x*(*t*) and its Fourier integral [the right-hand side of Eq. (7.10)] can differ at some values of *t* without contradicting Eq. (7.13). We shall now discuss an alternate set of criteria due to Dirichlet for convergence of the Fourier transform. - -As with the Fourier series, if *x*(*t*) satisfies certain conditions (*Dirichlet conditions*), its Fourier transform is guaranteed to converge pointwise at all points where *x*(*t*) is continuous. Moreover, at the points of discontinuity, *x*(*t*) converges to the value midway between the two values of *x*(*t*) on either side of the discontinuity. The Dirichlet conditions are as follows: - -1. *x*(*t*) should be absolutely integrable, that is, - -$$ -\int_{-\infty}^{\infty} |x(t)| \, dt < \infty \tag{7.14} -$$ - -If this condition is satisfied, we see that the integral on the right-hand side of Eq. (7.9) is guaranteed to have a finite value. - -- 2. *x*(*t*) must have only a finite number of finite discontinuities within any finite interval. -- 3. *x*(*t*) must contain only a finite number of maxima and minima within any finite interval. - -We stress here that although the Dirichlet conditions are sufficient for the existence and pointwise convergence of the Fourier transform, they are not necessary. For example, we saw in Ex. 7.1 that a growing exponential, which violates Dirichlet's first condition in Eq. (7.14), does not have a Fourier transform. But the signal of the form (sin*at*)/*t*, which *does* violate this condition, does have a Fourier transform. - -Any signal that can be generated in practice satisfies the Dirichlet conditions and therefore has a Fourier transform. Thus, the physical existence of a signal is a sufficient condition for the existence of its transform. - -### LINEARITY OF THE FOURIER TRANSFORM - -The Fourier transform is linear; that is, if - -$$ -x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega) -$$ - -then - -$$ -a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(\omega) + a_2X_2(\omega) \tag{7.15} -$$ - -The proof is trivial and follows directly from Eq. (7.9). This result can be extended to any finite number of terms. It can be extended to an infinite number of terms only if the conditions required for interchangeability of the operations of summation and integration are satisfied. - -### **[7.1-1 Physical Appreciation of the Fourier Transform](#page-12-0)** - -In understanding any aspect of the Fourier transform, we should remember that Fourier representation is a way of expressing a signal in terms of everlasting sinusoids (or exponentials). The Fourier spectrum of a signal indicates the relative amplitudes and phases of sinusoids that are required to synthesize that signal. A periodic signal Fourier spectrum has finite amplitudes and exists at discrete frequencies (ω0 and its multiples). Such a spectrum is easy to visualize, but the spectrum of an aperiodic signal is not easy to visualize because it has a continuous spectrum. The continuous spectrum concept can be appreciated by considering an analogous, more tangible phenomenon. One familiar example of a continuous distribution is the loading of a beam. Consider a beam loaded with weights *D*1,*D*2,*D*3,...,*Dn* units at the uniformly spaced points *y*1, *y*2,..., *yn*, as shown in Fig. 7.5a. - -The total load *WT* on the beam is given by the sum of these loads at each of the *n* points: - -$$ -W_T = \sum_{i=1}^n D_i -$$ - -Consider now the case of a continuously loaded beam, as depicted in Fig. 7.5b. In this case, although there appears to be a load at every point, the load at any one point is zero. This does not mean that there is no load on the beam. A meaningful measure of load in this situation is not the load at a point, but rather the loading density per unit length at that point. Let *X*(*y*) be the loading density per unit length of beam. It then follows that the load over a beam length *y*(*y* → 0), at some point *y*, is *X*(*y*)*y*. To find the total load on the beam, we divide the beam into segments of interval *y*(*y* → 0). The load over the *n*th such segment of length *y* is *X*(*ny*)*y*. The total load *WT* is given by - -$$ -W_T = \lim_{\Delta y \to 0} \sum_{y_1}^{y_n} X(n\Delta y) \, \Delta y = \int_{y_1}^{y_n} X(y) \, dy -$$ - -The load now exists at every point, and *y* is now a continuous variable. In the case of discrete loading (Fig. 7.5a), the load exists only at *n* discrete points. At other points, there is no load. On the other hand, in the continuously loaded case, the load exists at every point, but at any specific - -**Figure 7.5** Weight-loading analogy for the Fourier transform. - -point *y*, the load is zero. The load over a small interval *y*, however, is [*X*(*ny*)]*y* (Fig. 7.5b). Thus, even though the load at a point *y* is zero, the relative load at that point is *X*(*y*). - -An exactly analogous situation exists in the case of a signal spectrum. When *x*(*t*) is periodic, the spectrum is discrete, and *x*(*t*) can be expressed as a sum of discrete exponentials with finite amplitudes: - -$$ -x(t) = \sum_{n} D_n e^{jn\omega_0 t} -$$ - -For an aperiodic signal, the spectrum becomes continuous; that is, the spectrum exists for every value of ω, but the amplitude of each component in the spectrum is zero. The meaningful measure here is not the amplitude of a component of some frequency but the spectral density per unit bandwidth. From Eq. (7.7), it is clear that *x*(*t*) is synthesized by adding exponentials of the form *ejn*ω*t* , in which the contribution by any one exponential component is zero. But the contribution by exponentials in an infinitesimal band ω located at ω = *n*ω is (1/2π )*X*(*n*ω)ω, and the addition of all these components yields *x*(*t*) in the integral form: - -$$ -x(t) = \lim_{\Delta\omega \to 0} \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} X(n\Delta\omega)e^{(jn\Delta\omega)t} \Delta\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{i\omega t} d\omega -$$ - -Thus, *n*ω approaches a continuous variable ω. The spectrum now exists at every ω. The contribution by components within a band *d*ω is (1/2π )*X*(ω)*d*ω = *X*(ω)*df* , where *df* is the bandwidth in hertz. Clearly, *X*(ω) is the *spectral density* per unit bandwidth (in hertz).† It also follows that even if the amplitude of any one component is infinitesimal, the relative amount of a component of frequency ω is *X*(ω). Although *X*(ω) is a spectral density, in practice, it is customarily called the *spectrum* of *x*(*t*) rather than the spectral density of *x*(*t*). Deferring to this convention, we shall call *X*(ω) the Fourier spectrum (or Fourier transform) of *x*(*t*). - -### A MARVELOUS BALANCING ACT - -An important point to remember here is that *x*(*t*) is represented (or synthesized) by exponentials or sinusoids that are everlasting (not causal). Such conceptualization leads to a rather fascinating picture when we try to visualize the synthesis of a timelimited pulse signal *x*(*t*) [Fig. 7.6] by the sinusoidal components in its Fourier spectrum. The signal *x*(*t*) exists only over an interval (*a*,*b*) and is zero outside this interval. The spectrum of *x*(*t*) contains an infinite number of exponentials (or sinusoids), which start at *t* = −∞ and continue forever. The amplitudes and phases of these components add up exactly to *x*(*t*) over the finite interval (*a*,*b*) and to zero everywhere outside this interval. Juggling the amplitudes and phases of an infinite number of components to achieve - -† To stress that the signal spectrum is a *density* function, we shall shade the plot of |*X*(ω)| (as in Fig. 7.4b). The representation of *X*(ω), however, will be a line plot, primarily to avoid visual confusion. - -such a perfect and delicate balance boggles the human imagination. Yet the Fourier transform accomplishes it routinely, without much thinking on our part. Indeed, we become so involved in mathematical manipulations that we fail to notice this marvel. - -## **[7.2 TRANSFORMS OF](#page-12-0) SOME USEFUL FUNCTIONS** - -For convenience, we now introduce a compact notation for the useful gate, triangle, and interpolation functions. - -### UNIT GATE FUNCTION - -We define a unit gate function rect(*x*) as a gate pulse of unit height and unit width, centered at the origin, as illustrated in Fig. 7.7a† : - -rect -$$ -(x) -$$ - = -$$ -\begin{cases} 0 & |x| > \frac{1}{2} \\ \frac{1}{2} & |x| = \frac{1}{2} \\ 1 & |x| < \frac{1}{2} \end{cases} -$$ - (7.16) - -The gate pulse in Fig. 7.7b is the unit gate pulse rect(*x*) expanded by a factor τ along the horizontal axis and therefore can be expressed as rect(*x*/τ ) (see Sec. 1.2-2). Observe that τ , the denominator of the argument of rect (*x*/τ ), indicates the width of the pulse. - -### UNIT TRIANGLE FUNCTION - -We define a unit triangle function (*x*) as a triangular pulse of unit height and unit width, centered at the origin, as shown in Fig. 7.8a - -$$ -\Delta(x) = \begin{cases} 0 & |x| \ge \frac{1}{2} \\ 1 - 2|x| & |x| < \frac{1}{2} \end{cases} -$$ -(7.17) - -**Figure 7.7** A gate pulse. - -† At |*x*| = 0.5, we require rect(*x*) = 0.5 because the inverse Fourier transform of a discontinuous signal converges to the mean of its two values at the discontinuity. - -**Figure 7.8** A triangle pulse. - -The pulse in Fig. 7.8b is (*x*/τ ). Observe that here, as for the gate pulse, the denominator τ of the argument of (*x*/τ ) indicates the pulse width. - -### INTERPOLATION FUNCTION SINC (*x*) - -The function sin*x*/*x* is the "sine over argument" function denoted by sinc (*x*). † This function plays an important role in signal processing. It is also known as the *filtering or interpolating function*. We define - -$$ -\operatorname{sinc}(x) = \frac{\sin x}{x} \tag{7.18} -$$ - -Inspection of Eq. (7.18) shows the following: - -- 1. sinc (*x*) is an even function of *x*. -- 2. sinc (*x*) = 0 when sin *x* = 0 except at *x* = 0, where it appears to be indeterminate. This means that sinc*x* = 0 for *x* = ±π,±2π,±3π,.... -- 3. Using L'Hôpital's rule, we find sinc (0) = 1. -- 4. sinc (*x*) is the product of an oscillating signal sin*x* (of period 2π) and a monotonically decreasing function 1/*x*. Therefore, sinc (*x*) exhibits damped oscillations of period 2π, with amplitude decreasing continuously as 1/*x*. - -Figure 7.9a shows sinc(*x*). Observe that sinc (*x*) = 0 for values of *x* that are positive and negative integer multiples of π. Figure 7.9b shows sinc (3ω/7). The argument 3ω/7 = π when ω = 7π/3. Therefore, the first zero of this function occurs at ω = 7π/3. - -### **DR ILL 7.1 Sketching Basic Functions** - -Sketch: **(a)** rect(*x*/8), **(b)** (ω/10), **(c)** sinc (3πω/2), and **(d)** sinc (*t*)rect(*t*/4π ). - -$$ -\operatorname{sinc}(x) = \frac{\sin \pi x}{\pi x} -$$ - -† sinc(*x*) is also denoted by Sa (*x*) in the literature. Some authors define sinc (*x*) as - -**Figure 7.9** A sinc pulse. - -### **EXAMPLE 7.2 Fourier Transform of a Rectangular Pulse** - -Find the Fourier transform of *x*(*t*) = rect(*t*/τ ) (Fig. 7.10a). - -$$ -X(\omega) = \int_{-\infty}^{\infty} \text{rect}\left(\frac{t}{\tau}\right) e^{-j\omega t} dt -$$ - -Since rect(*t*/τ ) = 1 for |*t*| < τ/2, and since it is zero for |*t*| > τ/2, - -$$ -X(\omega) = \int_{-\tau/2}^{\tau/2} e^{-j\omega t} dt -$$ - -= $-\frac{1}{j\omega} (e^{-j\omega \tau/2} - e^{j\omega \tau/2}) = \frac{2 \sin(\frac{\omega \tau}{2})}{\omega}$ -= $\tau \frac{\sin(\frac{\omega \tau}{2})}{(\frac{\omega \tau}{2})} = \tau \operatorname{sinc}(\frac{\omega \tau}{2})$ - -**Figure 7.10 (a)** A gate pulse *x*(*t*), **(b)** its Fourier spectrum *X*(ω), **(c)** its amplitude spectrum |*X*(ω)|, and **(d)** its phase spectrum *X*(ω). - -Therefore, - -$$ -\text{rect}\left(\frac{t}{\tau}\right) \Longleftrightarrow \tau \text{ sinc}\left(\frac{\omega\tau}{2}\right) \tag{7.19} -$$ - -Recall that sinc(*x*) = 0 when *x* = ±*n*π. Hence, sinc(ωτ /2) = 0 when ωτ/2 = ±*n*π; that is, when ω = ±2*n*π/τ ,(*n* = 1, 2, 3,. . .), as depicted in Fig. 7.10b. The Fourier transform *X*(ω) shown in Fig. 7.10b exhibits positive and negative values. A negative amplitude can be considered to be a positive amplitude with a phase of −π or π. We use this observation to plot the amplitude spectrum |*X*(ω)|=|sinc (ωτ /2)| (Fig. 7.10c) and the phase spectrum *X*(ω) (Fig. 7.10d). The phase spectrum, which is required to be an odd function of ω, may be drawn in several other ways because a negative sign can be accounted for by a phase of ±*n*π, where *n* is any odd integer. All such representations are equivalent. - -#### BANDWIDTH OF RECT *t τ* - -The spectrum *X*(ω) in Fig. 7.10 peaks at ω = 0 and decays at higher frequencies. Therefore, rect(*t*/τ ) is a lowpass signal with most of the signal energy in lower-frequency components. Strictly speaking, because the spectrum extends from 0 to ∞, the bandwidth is ∞. However, much of the spectrum is concentrated within the first lobe (from ω = 0 to ω = 2π/τ ). Therefore, a rough estimate of the bandwidth of a rectangular pulse of width τ seconds is 2π/τ rad/s, or 1/τ Hz.† Note the reciprocal relationship of the pulse width with its bandwidth. We shall observe later that this result is true, in general. - -† To compute bandwidth, we must consider the spectrum for positive values of ω only. See the discussion in Sec. 6.3. - -Find the Fourier transform of the unit impulse δ(*t*). - -Using the sampling property of the impulse [Eq. (1.11)], we obtain - -$$ -\mathcal{F}[\delta(t)] = \int_{-\infty}^{\infty} \delta(t)e^{-j\omega t}dt = 1 \quad \text{and} \quad \delta(t) \Longleftrightarrow 1 -$$ - -Figure 7.11 shows δ(*t*) and its spectrum. - -### **EXAMPLE 7.4 Inverse Fourier Transform of the Dirac Delta Function** - -Find the inverse Fourier transform of δ(ω). - -On the basis of Eq. (7.10) and the sampling property of the impulse function, - -$$ -\mathcal{F}^{-1}[\delta(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega) e^{j\omega t} d\omega = \frac{1}{2\pi} -$$ - -Therefore, - -$$ -\frac{1}{2\pi} \Longleftrightarrow \delta(\omega) \quad \text{and} \quad 1 \Longleftrightarrow 2\pi \delta(\omega) \tag{7.20} -$$ - -This result shows that the spectrum of a constant signal *x*(*t*) = 1 is an impulse 2πδ(ω), as illustrated in Fig. 7.12. - -The result [Eq. (7.20)] could have been anticipated on qualitative grounds. Recall that the Fourier transform of *x*(*t*) is a spectral representation of *x*(*t*) in terms of everlasting exponential components of the form *ej*ω*t* . Now, to represent a constant signal *x*(*t*) = 1, we need a single - -**Figure 7.12 (a)** A constant (dc) signal and **(b)** its Fourier spectrum. - -everlasting exponential *ej*ω*t* with ω = 0.† This results in a spectrum at a single frequency ω = 0. Another way of looking at the situation is that *x*(*t*) = 1 is a dc signal that has a single frequency ω = 0 (dc). - -If an impulse at ω = 0 is a spectrum of a dc signal, what does an impulse at ω = ω0 represent? We shall answer this question in the next example. - -### **EXAMPLE 7.5 Inverse Fourier Transform of a Shifted Dirac Delta Function** - -Find the inverse Fourier transform of δ(ω −ω0). - -Using the sampling property of the impulse function, we obtain - -$$ -\mathcal{F}^{-1}[\delta(\omega - \omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega = \frac{1}{2\pi} e^{j\omega_0 t} -$$ - -Therefore, - -$$ -\frac{1}{2\pi}e^{j\omega_0 t} \Longleftrightarrow \delta(\omega - \omega_0) \quad \text{and} \quad e^{j\omega_0 t} \Longleftrightarrow 2\pi\delta(\omega - \omega_0) \tag{7.21} -$$ - -This result shows that the spectrum of an everlasting exponential *ej*ω0*t* is a single impulse at ω = ω0. We reach the same conclusion by qualitative reasoning. To represent the everlasting - -† The constant multiplier 2π in the spectrum [*X*(ω) = 2πδ(ω)] may be a bit puzzling. Since 1 = *ej*ω*t* with ω = 0, it appears that the Fourier transform of *x*(*t*) = 1 should be an impulse of strength unity rather than 2π. Recall, however, that in the Fourier transform *x*(*t*) is synthesized by exponentials not of amplitude *X*(*n*ω)ω but of amplitude 1/2π times *X*(*n*ω)ω, as seen from Eq. (7.7). Had we used variable *f* (hertz) instead of ω, the spectrum would have been the unit impulse. - -exponential *ej*ω0*t* , we need a single everlasting exponential *ej*ω*t* with ω = ω0. Therefore, the spectrum consists of a single component at frequency ω = ω0. From Eq. (7.21) it follows that - -$$ -e^{-j\omega_0 t} \Longleftrightarrow 2\pi \delta(\omega + \omega_0) -$$ - -# **EXAMPLE 7.6 Fourier Transform of a Sinusoid** Find the Fourier transform of the everlasting sinusoid cos ω0*t* (Fig. 7.13a). *v*0 0 *v*0 *x*(*t*) cos *v*0*t X*(*v*) *p t v* 0 (a) (b) **Figure 7.13 (a)** A cosine signal and **(b)** its Fourier spectrum. - -Recall Euler's formula - -$$ -\cos \omega_0 t = \frac{1}{2} (e^{j\omega_0 t} + e^{-j\omega_0 t}) -$$ - -Applying Eq. (7.21), we obtain - -$$ -\cos \omega_0 t \Longleftrightarrow \pi [\delta(\omega + \omega_0) + \delta(\omega - \omega_0)] -$$ - -The spectrum of cos ω0*t* consists of two impulses at ω0 and −ω0, as shown in Fig. 7.13b. The result also follows from qualitative reasoning. An everlasting sinusoid cos ω0*t* can be synthesized by two everlasting exponentials, *ej*ω0*t* and *e*−*j*ω0*t* . Therefore, the Fourier spectrum consists of only two components of frequencies ω0 and −ω0. - -### **EXAMPLE 7.7 Fourier Transform of a Periodic Signal** - -Determine the Fourier transform of a periodic signal *x*(*t*) using its Fourier series representation. - -We can use a Fourier series to express a periodic signal as a sum of exponentials of the form *ejn*ω0*t* , whose Fourier transform is found in Eq. (7.21). Hence, we can readily find the Fourier transform of a periodic signal by using the linearity property in Eq. (7.15). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/089_7 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/089_7 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER TRANSFORM.md deleted file mode 100644 index 39a2e2fdb9718a2446013b2f13493eb827f3438e..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/089_7 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER TRANSFORM.md +++ /dev/null @@ -1,115 +0,0 @@ -### 696 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -The Fourier series of a periodic signal *x*(*t*) with period *T*0 is given by - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -Taking the Fourier transform of both sides, we obtain† - -$$ -X(\omega) = 2\pi \sum_{n=-\infty}^{\infty} D_n \delta(\omega - n\omega_0) -$$ - (7.22) - -**Figure 7.14 (a)** The uniform impulse train and **(b)** its Fourier transform. - -As shown in Eq. (6.24) from Ex. 6.9, the Fourier coefficients *Dn* for δ*T*0 (*t*) are constant *Dn* = 1/*T*0. From Eq. (7.22), the Fourier transform of δ*T*0 (*t*) is therefore - -$$ -X(\omega) = \frac{2\pi}{T_0} \sum_{n=-\infty}^{\infty} \delta(\omega - n\omega_0) = \omega_0 \delta_{\omega_0}(\omega), \quad \text{where } \omega_0 = \frac{2\pi}{T_0} -$$ - -The corresponding spectrum is shown in Fig. 7.14b. - -† We assume here that the linearity property can be extended to an infinite sum. - -### **EXAMPLE 7.9 Fourier Transform of the Unit Step Function** - -Find the Fourier transform of the unit step function *u*(*t*). - -Trying to find the Fourier transform of *u*(*t*) by direct integration leads to an indeterminate result because - -$$ -U(\omega) = \int_{-\infty}^{\infty} u(t)e^{-j\omega t}dt = \int_{0}^{\infty} e^{-j\omega t}dt = \left. \frac{-1}{j\omega}e^{-j\omega t} \right|_{0}^{\infty} -$$ - -The upper limit of *e*−*j*ω*t* as *t*→∞ yields an indeterminate answer. So we approach this problem by considering *u*(*t*) to be a decaying exponential *e*−*atu*(*t*) in the limit as *a* → 0 (Fig. 7.15a). Thus, - -$$ -u(t) = \lim_{a \to 0} e^{-at} u(t) -$$ - -and - -$$ -U(\omega) = \lim_{a \to 0} \mathcal{F}\lbrace e^{-at} u(t) \rbrace = \lim_{a \to 0} \frac{1}{a + j\omega} -$$ - -Expressing the right-hand side in terms of its real and imaginary parts yields - -$$ -U(\omega) = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} - j \frac{\omega}{a^2 + \omega^2} \right] = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} \right] + \frac{1}{j\omega} -$$ - -**Figure 7.15** Derivation of the Fourier transform of the step function. - -The function *a*/(*a*2 + ω2) has interesting properties. First, the area under this function (Fig. 7.15b) is π regardless of the value of *a*: - -$$ -\int_{-\infty}^{\infty} \frac{a}{a^2 + \omega^2} \, d\omega = \tan^{-1} \frac{\omega}{a} \bigg|_{-\infty}^{\infty} = \pi -$$ - -### 698 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Second, when *a* → 0, this function approaches zero for all ω = 0, and all its area (π) is concentrated at a single point ω = 0. Clearly, as *a* → 0, this function approaches an impulse of strength π. Thus, - -$$ -U(\omega) = \pi \delta(\omega) + \frac{1}{j\omega} \tag{7.23} -$$ - -Note that *u*(*t*) is not a "true" dc signal because it is not constant over the interval −∞ to ∞. To synthesize "true" dc, we require only one everlasting exponential with ω = 0 (impulse at ω = 0). The signal *u*(*t*) has a jump discontinuity at *t* = 0. It is impossible to synthesize such a signal with a single everlasting exponential *ej*ω*t* . To synthesize this signal from everlasting exponentials, we need, in addition to an impulse at ω = 0, all the frequency components, as indicated by the term 1/*j*ω in Eq. (7.23). - -### **EXAMPLE 7.10 Fourier Transform of the Sign Function** - -Find the Fourier transform of the sign function sgn(*t*) [pronounced *signum* (*t*)], depicted in Fig. 7.16. - -*t* sgn (*t*) 1 0 -1 **Figure 7.16** The signum function sgn(*t*). Observe that sgn(*t*)+1 = 2*u*(*t*) ⇒ sgn(*t*) = 2*u*(*t*)−1 Using Eqs. (7.20) and (7.23) and the linearity property, we obtain sgn(*t*) ⇐⇒ 2 *j*ω - -Table 7.1 provides many common Fourier transform pairs. - -| No. | x(t) | X(ω) | | -|-----|--------------------------|-------------------------------------------------------|--------------------| -| 1 | e−atu(t) | 1
a+jω | a > 0 | -| 2 | eatu(−t) | 1
a−jω | a > 0 | -| 3 | e−a t | 2a
a2 +ω2 | a > 0 | -| 4 | te−atu(t) | 1
(a+jω)2 | a > 0 | -| 5 | ne−atu(t)
t | n!
(a+jω)n+1 | a > 0 | -| 6 | δ(t) | 1 | | -| 7 | 1 | 2πδ(ω) | | -| 8 | ejω0t | 2πδ(ω −ω0) | | -| 9 | cos ω0t | π[δ(ω −ω0) +δ(ω +ω0)] | | -| 10 | sin ω0t | jπ[δ(ω +ω0)−δ(ω −ω0)] | | -| 11 | u(t) | 1
πδ(ω)+
jω | | -| 12 | sgnt | 2
jω | | -| 13 | cos ω0t u(t) | π
jω
2 [δ(ω −ω0) +δ(ω +ω0)] +
ω2
0 −ω2 | | -| 14 | sin ω0t u(t) | π
ω0
[δ(ω −ω0)−δ(ω +ω0)] +
ω2
2j
0 −ω2 | | -| 15 | e−atsin
ω0t u(t) | ω0
(a+jω)2 +ω2
0 | a > 0 | -| 16 | e−at cos
ω0t u(t) | a+jω
(a+jω)2 +ω2
0 | a > 0 | -| 17 | rect t

τ | τ sincωτ

2 | | -| 18 | W
π sinc(Wt) | rect ω

2W | | -| 19 | t

τ | τ
ωτ

sinc2
2
4 | | -| 20 | Wt
W
2π sinc2
2 | ω

2W | | -| 21 | "∞
δ(t −nT) | "∞
ω0
δ(ω −nω0) | 2π
ω0
=
T | -| 22 | n=−∞
2/2σ2
e−t | n=−∞
√2πe−σ2ω2/2
σ | | - -**TABLE 7.1** Select Fourier Transform Pairs - -### **DR ILL 7.2 Inverse Fourier Transform of a Rectangular Pulse** - -Show that the inverse Fourier transform of *X*(ω) illustrated in Fig. 7.17 is *x*(*t*) = (ω0/π )sinc (ω0*t*). Sketch *x*(*t*). - -### **DR ILL 7.3 Fourier Transform of a General Sinusoid** - -Show that cos(ω0*t* +θ ) ⇐⇒ π[δ(ω +ω0)*e*−*j*θ +δ(ω −ω0)*ej*θ ]. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/090_7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/090_7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS.md deleted file mode 100644 index 4a81801d85e704fc1372af68b41a4da8487a5a2e..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/090_7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS.md +++ /dev/null @@ -1,20 +0,0 @@ -### **[7.2-1 Connection Between the Fourier and Laplace Transforms](#page-12-0)** - -The general (bilateral) Laplace transform of a signal *x*(*t*), according to Eq. (4.1), is - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ - (7.24) - -Setting *s* = *j*ω in this equation yields - -$$ -X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt -$$ - -where *X*(*j*ω) = *X*(*s*)|*s*=*j*ω. But, the right-hand-side integral defines *X*(ω), the Fourier transform of *x*(*t*). Does this mean that the Fourier transform can be obtained from the corresponding Laplace transform by setting *s* = *j*ω? In other words, is it true that *X*(*j*ω) = *X*(ω)? Yes and no. Yes, it is true in most cases. For example, when *x*(*t*) = *e*−*atu*(*t*), its Laplace transform is 1/(*s* + *a*), and *X*(*j*ω) = 1/(*j*ω +*a*), which is equal to *X*(ω) (assuming *a* < 0). However, for the unit step function *u*(*t*), the Laplace transform is - -$$ -u(t) \Longleftrightarrow \frac{1}{s} \qquad \text{Re}\, s > 0 -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/091_7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/091_7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM.md deleted file mode 100644 index 19b4e74f167df8087408847494d0ff0b78665cbd..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/091_7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM.md +++ /dev/null @@ -1,608 +0,0 @@ -### 7.3 Some Properties of the Fourier Transform 701 - -The Fourier transform is given by - -$$ -u(t) \Longleftrightarrow \frac{1}{j\omega} + \pi \delta(\omega) -$$ - -Clearly, *X*(*j*ω) = *X*(ω) in this case. - -To understand this puzzle, consider the fact that we obtain *X*(*j*ω) by setting *s* = *j*ω in Eq. (7.24). This implies that the integral on the right-hand side of Eq. (7.24) converges for *s* = *j*ω, meaning that *s* = *j*ω (the imaginary axis) lies in the ROC for *X*(*s*). The general rule is that only when the ROC for *X*(*s*) includes the ω axis, does setting *s* = *j*ω in *X*(*s*) yield the Fourier transform *X*(ω), that is, *X*(*j*ω) = *X*(ω). This is the case of absolutely integrable *x*(*t*). If the ROC of *X*(*s*) excludes the ω axis, *X*(*j*ω) = *X*(ω). This is the case for exponentially growing *x*(*t*) and also *x*(*t*) that is constant or is oscillating with constant amplitude. - -The reason for this peculiar behavior has something to do with the nature of convergence of the Laplace and the Fourier integrals when *x*(*t*) is not absolutely integrable.† - -This discussion shows that although the Fourier transform may be considered as a special case of the Laplace transform, we need to circumscribe such a view. This fact can also be confirmed by noting that a periodic signal has the Fourier transform, but the Laplace transform does not exist. - -## **7.3 SOME [PROPERTIES OF THE](#page-12-0) FOURIER TRANSFORM** - -We now study some of the important properties of the Fourier transform and their implications as well as applications. We have already encountered two important properties, linearity [Eq. (7.15)] and the conjugation property [Eq. (7.11)]. - -Before embarking on this study, we shall explain an important and pervasive aspect of the Fourier transform: the time-frequency duality. - -† To explain this point, consider the unit step function and its transforms. Both the Laplace and the Fourier transform synthesize *x*(*t*), using everlasting exponentials of the form *est*. The frequency *s* can be anywhere in the complex plane for the Laplace transform, but it must be restricted to the ω axis in the case of the Fourier transform. The unit step function is readily synthesized in the Laplace transform by a relatively simple spectrum *X*(*s*) = 1/*s*, in which the frequencies *s* are chosen in the RHP [the region of convergence for *u*(*t*) is Re *s* > 0]. In the Fourier transform, however, we are restricted to values of *s* on the ω axis only. The function *u*(*t*) can still be synthesized by frequencies along the ω axis, but the spectrum is more complicated than it is when we are free to choose the frequencies in the RHP. In contrast, when *x*(*t*) is absolutely integrable, the region of convergence for the Laplace transform includes the ω axis, and we can synthesize *x*(*t*) by using frequencies along the ω axis in both transforms. This leads to *X*(*j*ω) = *X*(ω). - -We may explain this concept by an example of two countries, X and Y. Suppose these countries want to construct similar dams in their respective territories. Country X has financial resources but not much manpower. In contrast, Y has considerable manpower but few financial resources. The dams will still be constructed in both countries, although the methods used will be different. Country X will use expensive but efficient equipment to compensate for its lack of manpower, whereas Y will use the cheapest possible equipment in a labor-intensive approach to the project. Similarly, both Fourier and Laplace integrals converge for *u*(*t*), but the makeup of the components used to synthesize *u*(*t*) will be very different for two cases because of the constraints of the Fourier transform, which are not present for the Laplace transform. - -### TIME-FREQUENCY DUALITY IN THE TRANSFORM OPERATIONS - -Equations (7.9) and (7.10) show an interesting fact: the direct and the inverse transform operations are remarkably similar. These operations, required to go from *x*(*t*) to *X*(ω) and then from *X*(ω) to *x*(*t*), are depicted graphically in Fig. 7.18. The inverse transform equation can be obtained from the direct transform equation by replacing *x*(*t*) with *X*(ω), *t* with ω, and ω with *t*. In a similar way, we can obtain the direct from the inverse. There are only two minor differences in these operations: the factor 2π appears only in the inverse operator, and the exponential indices in the two operations have opposite signs. Otherwise the two equations are duals of each other.† - -This observation has far-reaching consequences in the study of the Fourier transform. It is the basis of the so-called duality of time and frequency. *The duality principle may be compared with a photograph and its negative. A photograph can be obtained from its negative, and by using an identical procedure, a negative can be obtained from the photograph*. For any result or relationship between *x*(*t*) and *X*(ω), there exists a dual result or relationship, obtained by interchanging the roles of *x*(*t*) and *X*(ω) in the original result (along with some minor modifications arising because of the factor 2π and a sign change). For example, the time-shifting property, to be proved later, states that if *x*(*t*) ⇐⇒ *X*(ω), then - -$$ -x(t-t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0} -$$ - -**Figure 7.18** A near symmetry between the direct and the inverse Fourier transforms. - -$$ -X(2\pi f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \quad \text{and} \quad x(t) = \int_{-\infty}^{\infty} X(2\pi f)e^{j2\pi ft} df -$$ - -This leaves only one significant difference, that of sign change in the exponential index. - -† Of the two differences, the former can be eliminated by change of variable from ω to *f* (in hertz). In this case ω = 2π*f* and *d*ω = 2π *df* . - -Therefore, the direct and the inverse transforms are given by - -The dual of this property (the frequency-shifting property) states that - -$$ -x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0) -$$ - -Observe the role reversal of time and frequency in these two equations (with the minor difference of the sign change in the exponential index). The value of this principle lies in the fact that *whenever we derive any result, we can be sure that it has a dual.* This possibility can give valuable insights about many unsuspected properties or results in signal processing. - -The properties of the Fourier transform are useful not only in deriving the direct and inverse transforms of many functions, but also in obtaining several valuable results in signal processing. The reader should not fail to observe the ever-present duality in this discussion. - -### LINEARITY - -The linearity property, already introduced as Eq. (7.15), states that if *x*1(*t*) ⇐⇒ *X*1(ω) and *x*2(*t*) ⇐⇒ *X*2(ω), then *a*1*x*1(*t*)+*a*2*x*2(*t*) ⇐⇒ *a*1*X*1(ω)+*a*2*X*2(ω). - -### CONJUGATION AND CONJUGATE SYMMETRY - -The conjugation property, which has already been introduced, states that if *x*(*t*) ⇐⇒ *X*(ω), then - -$$ -x^*(t) \Longleftrightarrow X^*(-\omega) -$$ - -From this property follows the conjugate symmetry property, also introduced earlier, which states that if *x*(*t*) is real, then - -$$ -X(-\omega) = X^*(\omega) -$$ - -### DUALITY - -The duality property states that if - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then - -$$ -X(t) \Longleftrightarrow 2\pi x(-\omega) \tag{7.25} -$$ - -**Proof.** From Eq. (7.10) we can write - -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(u)e^{iut} du -$$ - -Hence, - -$$ -2\pi x(-t) = \int_{-\infty}^{\infty} X(u)e^{-jut} du -$$ - -Changing *t* to ω yields Eq. (7.25). - -*t* - -From Eq. (7.19) we have - -1 - -$$ -\underbrace{\text{rect}\left(\frac{t}{\tau}\right)}_{x(t)} \Longleftrightarrow \underbrace{\tau \text{ sinc}\left(\frac{\omega \tau}{2}\right)}_{X(\omega)} -$$ - -Also, *X*(*t*) is the same as *X*(ω) with ω replaced by *t*, and *x*(−ω) is the same as *x*(*t*) with *t* replaced by −ω. Therefore, the duality property of Eq. (7.25) yields - -$$ -\underbrace{\tau \text{ sinc}\left(\frac{\tau t}{2}\right)}_{X(t)} \Longleftrightarrow \underbrace{2\pi \text{ rect}\left(\frac{-\omega}{\tau}\right)}_{2\pi x(-\omega)} = 2\pi \text{ rect}\left(\frac{\omega}{\tau}\right) -$$ - -In this result, we used the fact that rect(−*x*) = rect(*x*) because rect is an even function. Figure 7.19b shows this pair graphically. Observe the interchange of the roles of *t* and ω (with the minor adjustment of the factor 2π). This result appears as pair 18 in Table 7.1 (with τ/2 = *W*). - -As an interesting exercise, the reader should generate the dual of every pair in Table 7.1 by applying the duality property. - -## **DR ILL 7.4 Applying the Duality Property of the Fourier Transform** - -Apply the duality property to pairs 1, 3, and 9 (Table 7.1) to show that - -- **(a)** 1/(*jt* +*a*) ⇐⇒ 2π*ea*ω*u*(−ω) -- **(b)** 2*a*/(*t* 2 +*a*2) ⇐⇒ 2π*e*−*a*|ω| -- **(c)** δ(*t* +*t*0)+δ(*t* −*t*0) ⇐⇒ 2 cos *t*0ω - -### THE SCALING PROPERTY If - -*x*(*t*) ⇐⇒ *X*(ω) - -then, for any real constant *a*, - -$$ -x(at) \Longleftrightarrow \frac{1}{|a|}X\left(\frac{\omega}{a}\right) \tag{7.26} -$$ - -**Proof.** For a positive real constant *a*, - -$$ -\mathcal{F}[x(at)] = \int_{-\infty}^{\infty} x(at)e^{-j\omega t}dt = \frac{1}{a}\int_{-\infty}^{\infty} x(u)e^{(-j\omega/a)u}du = \frac{1}{a}X\left(\frac{\omega}{a}\right) -$$ - -Similarly, we can demonstrate that if *a* < 0, - -$$ -x(at) \Longleftrightarrow \frac{-1}{a}X\left(\frac{\omega}{a}\right) -$$ - -Hence follows Eq. (7.26). - -### SIGNIFICANCE OF THE SCALING PROPERTY - -The function *x*(*at*) represents the function *x*(*t*) compressed in time by a factor *a* (see Sec. 1.2-2). Similarly, a function *X*(ω/*a*) represents the function *X*(ω) expanded in frequency by the same factor *a*. *The scaling property states that time compression of a signal results in its spectral expansion, and time expansion of the signal results in its spectral compression*. Intuitively, compression in time by factor *a* means that the signal is varying faster by factor *a*. † To synthesize such a signal, the frequencies of its sinusoidal components must be increased by the factor *a*, implying that its frequency spectrum is expanded by the factor *a*. Similarly, a signal expanded in time varies more slowly; hence the frequencies of its components are lowered, implying that its frequency spectrum is compressed. For instance, the signal cos 2ω0*t* is the same as the signal cosω0*t* time-compressed by a factor of 2. Clearly, the spectrum of the former (impulse at ±2ω0) is an expanded version of the spectrum of the latter (impulse at ±ω0). The effect of this scaling is demonstrated in Fig. 7.20. - -† We are assuming *a* > 1, although the argument still holds if *a* < 1. In the latter case, compression becomes expansion by factor 1/*a*, and vice versa. - -**Figure 7.20** The scaling property of the Fourier transform. - -### RECIPROCITY OF SIGNAL DURATION AND ITS BANDWIDTH - -The scaling property implies that if *x*(*t*) is wider, its spectrum is narrower, and vice versa. Doubling the signal duration halves its bandwidth, and vice versa. This suggests that the bandwidth of a signal is inversely proportional to the signal duration or width (in seconds).† We have already verified this fact for the gate pulse, where we found that the bandwidth of a gate pulse of width τ seconds is 1/τ Hz. More discussion of this interesting topic can be found in the literature [2]. - -By letting *a* = −1 in Eq. (7.26), we obtain the *inversion (or reflection) property of time and frequency:* - -$$ -x(-t) \Longleftrightarrow X(-\omega) \tag{7.27} -$$ - -### **EXAMPLE 7.12 Fourier Transform Reflection Property** - -Using the reflection property of the Fourier transform and Table 7.1, find the Fourier transforms of *eatu*(−*t*) and *e*−*a*|*t*| . - -Application of Eq. (7.27) to pair 1 of Table 7.1 yields - -$$ -e^{at}u(-t) \Longleftrightarrow \frac{1}{a-j\omega} \qquad a > 0 -$$ - -Also, - -$$ -e^{-a|t|} = e^{-at}u(t) + e^{at}u(-t) -$$ - -† When a signal has infinite duration, we must consider its effective or equivalent duration. There is no unique definition of effective signal duration. One possible definition is given in Eq. (2.47). - -Therefore, - -$$ -e^{-a|t|} \Longleftrightarrow \frac{1}{a+j\omega} + \frac{1}{a-j\omega} = \frac{2a}{a^2 + \omega^2} \qquad a > 0 \tag{7.28} -$$ - -The signal *e*−*a*|*t*| and its spectrum are illustrated in Fig. 7.21. - -## THE TIME-SHIFTING PROPERTY If - -*x*(*t*) ⇐⇒ *X*(ω) - -then - -$$ -x(t - t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0} \tag{7.29} -$$ - -**Proof.** By definition, - -$$ -\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(t-t_0)e^{-j\omega t} dt -$$ - -Letting *t* −*t*0 = *u*, we have - -$$ -\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(u)e^{-j\omega(u+t_0)} du = e^{-j\omega t_0} \int_{-\infty}^{\infty} x(u)e^{-j\omega u} du = X(\omega)e^{-j\omega t_0} -$$ - -This result shows that *delaying a signal by t*0 *seconds does not change its amplitude spectrum. The phase spectrum, however, is changed by* −ω*t*0. - -### PHYSICAL EXPLANATION OF THE LINEAR PHASE - -Time delay in a signal causes a linear phase shift in its spectrum. This result can also be derived by heuristic reasoning. Imagine *x*(*t*) being synthesized by its Fourier components, which are sinusoids of certain amplitudes and phases. The delayed signal *x*(*t* − *t*0) can be synthesized by the same sinusoidal components, each delayed by *t*0 seconds. The amplitudes of the components remain unchanged. Therefore, the amplitude spectrum of *x*(*t* − *t*0) is identical to that of *x*(*t*). The time delay of *t*0 in each sinusoid, however, does change the phase of each component. Now, a sinusoid - -**Figure 7.22** Physical explanation of the time-shifting property. - -cosω*t* delayed by *t*0 is given by - -$$ -\cos \omega (t - t_0) = \cos (\omega t - \omega t_0) -$$ - -Therefore a time delay *t*0 in a sinusoid of frequency ω manifests as a phase delay of ω*t*0. This is a linear function of ω, meaning that higher-frequency components must undergo proportionately higher phase shifts to achieve the same time delay. This effect is depicted in Fig. 7.22 with two sinusoids, the frequency of the lower sinusoid being twice that of the upper. The same time delay *t*0 amounts to a phase shift of π/2 in the upper sinusoid and a phase shift of π in the lower sinusoid. This verifies the fact that *to achieve the same time delay, higher-frequency sinusoids must undergo proportionately higher phase shifts*. The principle of linear phase shift is very important, and we shall encounter it again in distortionless signal transmission and filtering applications. - -### **EXAMPLE 7.13 Fourier Transform Time-Shifting Property** - -Use the time-shifting property to find the Fourier transform of *e*−*a*|*t*−*t*0| . - -This function, shown in Fig. 7.23a, is a time-shifted version of *e*−*a*|*t*| (depicted in Fig. 7.21a). From Eqs. (7.28) and (7.29), we have - -$$ -e^{-a|t-t_0|} \Longleftrightarrow \frac{2a}{a^2 + \omega^2} e^{-j\omega t_0} -$$ - -The spectrum of *e*−*a*|*t*−*t*0| (Fig. 7.23b) is the same as that of *e*−*a*|*t*| (Fig. 7.21b), except for an added phase shift of −ω*t*0. - -### **EXAMPLE 7.14 Fourier Transform of a Time-Shifted Rectangular Pulse** - -Find the Fourier transform of the time-shifted rectangular pulse *x*(*t*) illustrated in Fig. 7.24a. - -The pulse *x*(*t*) is the gate pulse rect(*t*/τ ) in Fig. 7.10a delayed by 3τ/4 seconds. Hence, according to Eq. (7.29), its Fourier transform is the Fourier transform of rect(*t*/τ ) multiplied by *e*−*j*ω(3τ /4) . Therefore, - -$$ -X(\omega) = \tau \operatorname{sinc}\left(\frac{\omega \tau}{2}\right) e^{-j\omega(3\tau/4)} -$$ - -The amplitude spectrum |*X*(ω)| (depicted in Fig. 7.24b) of this pulse is the same as that indicated in Fig. 7.10c. But the phase spectrum has an added linear term −3ωτ/4. Hence, the phase spectrum of *x*(*t*) (Fig. 7.24a) is identical to that in Fig. 7.10d plus a linear term −3ωτ/4, as shown in Fig. 7.24c. - -### PHASE SPECTRUM USING PRINCIPAL VALUES - -There is an alternate way of spectral representation of *X*(ω). The phase angle computed on a calculator or by using a computer subroutine is generally the principal value (modulo 2π value) of the phase angle, which always lies in the range −π to π. For instance, the principal value of angle 3π/2 is −π/2, and so on. The principal value differs from the actual value by ±2π radians (and its integer multiples) in a way that ensures that the principal value remains within −π to π. Thus, the principal value will show jump discontinuities of ±2π whenever the actual - -### 710 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -phase crosses ±π. The phase plot in Fig. 7.24c is redrawn in Fig. 7.24d using the principal value for the phase. This phase pattern, which contains phase discontinuities of magnitudes 2π and π, becomes repetitive at intervals of ω = 8π/τ . - -## **DR ILL 7.5 Fourier Transform Time-Shifting Property** - -Use pair 18 of Table 7.1 and the time-shifting property to show that the Fourier transform of sinc [ω0(*t* − *T*)] is (π/ω0)rect(ω/2ω0)*e*−*j*ω*T* . Sketch the amplitude and phase spectra of the Fourier transform. - -### THE FREQUENCY-SHIFTING PROPERTY If - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then - -$$ -x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0) \tag{7.30} -$$ - -**Proof.** By definition, - -$$ -\mathcal{F}[x(t)e^{j\omega_0 t}] = \int_{-\infty}^{\infty} x(t)e^{j\omega_0 t}e^{-j\omega t} dt = \int_{-\infty}^{\infty} x(t)e^{-j(\omega - \omega_0)t} dt = X(\omega - \omega_0) -$$ - -According to this property, the multiplication of a signal by a factor *ej*ω0*t* shifts the spectrum of that signal by ω = ω0. Note the duality between the time-shifting and the frequency-shifting properties. - -Changing ω0 to −ω0 in Eq. (7.30) yields - -$$ -x(t)e^{-j\omega_0 t} \Longleftrightarrow X(\omega + \omega_0) \tag{7.31} -$$ - -Because *ej*ω0*t* is not a real function that can be generated, frequency shifting in practice is achieved by multiplying *x*(*t*) by a sinusoid. Observe that - -$$ -x(t)\cos \omega_0 t = \frac{1}{2} [x(t)e^{j\omega_0 t} + x(t)e^{-j\omega_0 t}] -$$ - -From Eqs. (7.30) and (7.31), it follows that - -$$ -x(t)\cos\omega_0 t \Longleftrightarrow \frac{1}{2}[X(\omega - \omega_0) + X(\omega + \omega_0)]\tag{7.32} -$$ - -This result shows that the multiplication of a signal *x*(*t*) by a sinusoid of frequency ω0 shifts the spectrum *X*(ω) by ±ω0, as depicted in Fig. 7.25. - -Multiplication of a sinusoid cosω0*t* by *x*(*t*) amounts to modulating the sinusoid amplitude. This type of modulation is known as *amplitude modulation*. The sinusoid cosω0*t* is called the *carrier,* the signal *x*(*t*) is the *modulating signal*, and the signal *x*(*t*) cosω0*t* is the *modulated signal*. Further discussion of modulation and demodulation appears in Sec. 7.7. - -To sketch a signal *x*(*t*) cos ω0*t*, we observe that - -$$ -x(t)\cos\omega_0 t = \begin{cases} x(t) & \text{when } \cos\omega_0 t = 1\\ -x(t) & \text{when } \cos\omega_0 t = -1 \end{cases} -$$ - -Therefore, *x*(*t*) cos ω0*t* touches *x*(*t*) when the sinusoid cos ω0*t* is at its positive peaks and touches −*x*(*t*) when cos ω0*t* is at its negative peaks. This means that *x*(*t*) and −*x*(*t*) act as envelopes for the signal *x*(*t*) cos ω0*t* (see Fig. 7.25). The signal −*x*(*t*) is a mirror image of *x*(*t*) about the horizontal axis. Figure 7.25 shows the signals *x*(*t*) and *x*(*t*) cos ω0*t* and their spectra. - -**Figure 7.25** Amplitude modulation of a signal causes spectral shifting. - -### **EXAMPLE 7.15 Spectral Shifting by Amplitude Modulation** - -Find and sketch the Fourier transform of the modulated signal *x*(*t*) cos 10*t* in which *x*(*t*) is a gate pulse rect(*t*/4), as illustrated in Fig. 7.26a. - -From pair 17 of Table 7.1, we find rect(*t*/4) ⇐⇒ 4 sinc (2ω), which is depicted in Fig. 7.26b. From Eq. (7.32) it follows that - -$$ -x(t)\cos 10t \Longleftrightarrow \frac{1}{2}[X(\omega+10) + X(\omega-10)] -$$ - -In this case, *X*(ω) = 4 sinc (2ω). Therefore, - -*x*(*t*) cos 10*t* ⇐⇒ 2 sinc [2(ω +10)] +2 sinc [2(ω −10)] - -The spectrum (Fig. 7.26c) of *x*(*t*) cos 10*t* is obtained by shifting *X*(ω) in Fig. 7.26b to the left by 10 and also to the right by 10, and then multiplying it by 0.5, as depicted in Fig. 7.26d. - -## **DR ILL 7.6 Fourier Transform of an Amplitude-Modulated Signal** - -Sketch signal *e*−|*t*| cos 10*t*. Find the Fourier transform of this signal and sketch its spectrum. **Answer:** *X*(ω) = 1 (ω−10)2+1 + 1 (ω+10)2+1 . See Fig. 7.21b for the spectrum of *e*−*a*|*t*| . - -## **DR ILL 7.7 Amplitude Modulation Using a Phase-Shifted Carrier** - -Show that - -``` -x(t) cos(ω0t +θ ) ⇐⇒ 1 - 2 - - X(ω −ω0)ejθ +X(ω +ω0)e−jθ -``` - -### APPLICATIONS OF MODULATION - -Modulation is used to shift signal spectra. Some of the situations that call for spectrum shifting are presented next. - -1. If several signals, all occupying the same frequency band, are transmitted simultaneously over the same transmission medium, they will all interfere; it will be impossible to separate or retrieve them at a receiver. For example, if all radio stations decide to broadcast audio signals simultaneously, a receiver will not be able to separate them. This problem is solved - -Old is gold, but sometimes it is fool's gold. - -by using modulation, whereby each radio station is assigned a distinct carrier frequency. Each station transmits a modulated signal. This procedure shifts the signal spectrum to its allocated band, which is not occupied by any other station. A radio receiver can pick up any station by tuning to the band of the desired station. The receiver must now demodulate the received signal (undo the effect of modulation). Demodulation therefore consists of another spectral shift required to restore the signal to its original band. Note that both modulation and demodulation implement spectral shifting; consequently, demodulation operation is similar to modulation (see Sec. 7.7). - -This method of transmitting several signals simultaneously over a channel by sharing its frequency band is known as *frequency-division multiplexing (FDM)*. - -2. For effective radiation of power over a radio link, the antenna size must be of the order of the wavelength of the signal to be radiated. Audio signal frequencies are so low (wavelengths are so large) that impracticably large antennas would be required for radiation. Here, shifting the spectrum to a higher frequency (a smaller wavelength) by modulation solves the problem. - -### CONVOLUTION - -The time-convolution property and its dual, the frequency-convolution property, state that if - -$$ -x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega) -$$ - -then - -$$ -x_1(t) * x_2(t) \Longleftrightarrow X_1(\omega) X_2(\omega) \quad \text{(time convolution)} \tag{7.33} -$$ - -### 7.3 Some Properties of the Fourier Transform 715 - -and - -$$ -x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi}X_1(\omega) * X_2(\omega) \quad \text{(frequency convolution)} \tag{7.34} -$$ - -**Proof.** By definition, - -$$ -\mathcal{F}|x_1(t) * x_2(t)| = \int_{-\infty}^{\infty} e^{-j\omega t} \left[ \int_{-\infty}^{\infty} x_1(\tau) x_2(t-\tau) d\tau \right] dt -$$ -$$ -= \int_{-\infty}^{\infty} x_1(\tau) \left[ \int_{-\infty}^{\infty} e^{-j\omega t} x_2(t-\tau) d\tau \right] d\tau -$$ - -The inner integral is the Fourier transform of *x*2(*t* − τ ), given by [time-shifting property in Eq. (7.29)] *X*2(ω)*e*−*j*ωτ . Hence, - -$$ -\mathcal{F}[x_1(t) * x_2(t)] = \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} X_2(\omega) d\tau = X_2(\omega) \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} d\tau = X_1(\omega) X_2(\omega) -$$ - -Let *H*(ω) be the Fourier transform of the unit impulse response *h*(*t*), that is, - -*h*(*t*) ⇐⇒ *H*(ω) - -Application of the time-convolution property to *y*(*t*) = *x*(*t*) ∗ *h*(*t*) yields [assuming that both *x*(*t*) and *h*(*t*) are Fourier transformable] - -$$ -Y(\omega) = X(\omega)H(\omega) \tag{7.35} -$$ - -The frequency-convolution property of Eq. (7.34) can be proved in exactly the same way by reversing the roles of *x*(*t*) and *X*(ω). - -### **EXAMPLE 7.16 Time-Convolution Property to Show the Time-Integration Property** - -Use the time-convolution property to show that if - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega) -$$ - -Because - -$$ -u(t-\tau) = \begin{cases} 1 & \tau \leq t \\ 0 & \tau > t \end{cases} -$$ - -it follows that - -$$ -x(t) * u(t) = \int_{-\infty}^{\infty} x(\tau)u(t-\tau) d\tau = \int_{-\infty}^{t} x(\tau) d\tau -$$ - -#### 716 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Now, from the time-convolution property [Eq. (7.33)], it follows that - -$$ -x(t) * u(t) = \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(\omega) \left[ \frac{1}{j\omega} + \pi \delta(\omega) \right] -$$ -$$ -= \frac{X(\omega)}{j\omega} + \pi X(0) \delta(\omega) -$$ - -In deriving the last result, we used Eq. (1.10). - -### **DR ILL 7.8 Fourier Transform Time-Convolution Property** - -Use the time-convolution property to show that: - -$$ -(a) x(t) * \delta(t) = x(t) -$$ - -**(b)** *e*−*atu*(*t*)∗*e*−*btu*(*t*) = 1 *b*−*a* [*e*−*at* −*e*−*bt*]*u*(*t*) - -## TIME DIFFERENTIATION AND TIME INTEGRATION If - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then† - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega) \quad \text{(time differentiation)} \tag{7.36} -$$ - -and - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega) \quad \text{(time integration)} \tag{7.37} -$$ - -**Proof.** Differentiation of both sides of Eq. (7.10) yields - -$$ -\frac{dx(t)}{dt} = \frac{1}{2\pi} \int_{-\infty}^{\infty} j\omega X(\omega) e^{j\omega t} d\omega -$$ - -$$ -\int_{-\infty}^{\infty} \left| \frac{dx(t)}{dt} \right| dt < \infty -$$ - -† Valid only if the transform of *dx*/*dt* exists. In other words, *dx*/*dt* must satisfy the Dirichlet conditions. The first Dirichlet condition implies - -We also require that *x*(*t*) → 0 as *t* → ±∞. Otherwise, *x*(*t*) has a dc component, which gets lost in differentiation, and there is no one-to-one relationship between *x*(*t*) and *dx*/*dt*. - -| Operation | x(t) | X(ω) | -|---------------------------------|---------------------|-----------------------------| -| Scalar multiplication | kx(t) | kX(ω) | -| Addition | x1(t)+x2(t) | X1(ω) +X2(ω) | -| Conjugation | x∗(t) | X∗(−ω) | -| Duality | X(t) | 2πx(−ω) | -| Scaling (a real) | x(at) | 1
ω

X
a
a | -| Time shifting | x(t −t0) | X(ω)e−jωt0 | -| Frequency shifting (ω0
real) | x(t)ejω0t | X(ω −ω0) | -| Time convolution | x1(t)∗x2(t) | X1(ω)X2(ω) | -| Frequency convolution | x1(t)x2(t) | 1
2π X1(ω)∗X2(ω) | -| Time differentiation | dnx(t)
dtn | (jω)nX(ω) | -| Time integration | # t
x(u)du
−∞ | X(ω)
+πX(0)δ(ω)
jω | - -**TABLE 7.2** Fourier Transform Properties - -This result shows that - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega) -$$ - -Repeated application of this property yields - -$$ -\frac{d^n x(t)}{dt^n} \Longleftrightarrow (j\omega)^n X(\omega) -$$ - -The time-integration property [Eq. (7.37)] has already been proved in Ex. 7.16. Table 7.2 summarizes the most important properties of the Fourier transform. - -### **EXAMPLE 7.17 Fourier Transform Time-Differentiation Property** - -Use the time-differentiation property to find the Fourier transform of the triangle pulse (*t*/τ ) illustrated in Fig. 7.27a. Verify the correctness of the spectrum by using it to synthesize a periodic replication of the original time-domain signal with τ = 1. - -**Figure 7.27** Finding the Fourier transform of a piecewise-linear signal using the time-differentiation property. - -To find the Fourier transform of this pulse, we differentiate the pulse successively, as illustrated in Fig. 7.27b and 7.27c. Because *dx*/*dt* is constant everywhere, its derivative, *d*2*x*/*dt*2, is zero everywhere. But *dx*/*dt* has jump discontinuities with a positive jump of 2/τ at *t* = ±τ/2, and a negative jump of 4/τ at *t* = 0. Recall that the derivative of a signal at a jump discontinuity is an impulse at that point of strength equal to the amount of jump. Hence, *d*2*x*/*dt*2, the derivative of *dx*/*dt*, consists of a sequence of impulses, as depicted in Fig. 7.27c; that is, - -$$ -\frac{d^2x(t)}{dt^2} = \frac{2}{\tau} \left[ \delta \left( t + \frac{\tau}{2} \right) - 2\delta(t) + \delta \left( t - \frac{\tau}{2} \right) \right] -$$ - -From the time-differentiation property [Eq. (7.36)], - -$$ -\frac{d^2x(t)}{dt^2} \Longleftrightarrow (j\omega)^2 X(\omega) = -\omega^2 X(\omega) -$$ - -Also, from the time-shifting property [Eq. (7.29)], - -$$ -\delta(t-t_0) \Longleftrightarrow e^{-j\omega t_0} -$$ - -Combining these results, we obtain - -$$ --\omega^2 X(\omega) = \frac{2}{\tau} \left[ e^{j(\omega \tau/2)} - 2 + e^{-j(\omega \tau/2)} \right] = \frac{4}{\tau} \left( \cos \frac{\omega \tau}{2} - 1 \right) = -\frac{8}{\tau} \sin^2 \left( \frac{\omega \tau}{4} \right) -$$ - -and - -$$ -X(\omega) = \frac{8}{\omega^2 \tau} \sin^2\left(\frac{\omega \tau}{4}\right) = \frac{\tau}{2} \left[ \frac{\sin\left(\frac{\omega \tau}{4}\right)}{\frac{\omega \tau}{4}} \right]^2 = \frac{\tau}{2} \text{sinc}^2\left(\frac{\omega \tau}{4}\right) -$$ - -The spectrum *X*(ω) is depicted in Fig. 7.27d. This procedure of finding the Fourier transform can be applied to any function *x*(*t*) made up of straight-line segments with *x*(*t*) → 0 as |*t*|→∞. The second derivative of such a signal yields a sequence of impulses whose Fourier transform can be found by inspection. This example suggests a numerical method of finding the Fourier transform of an arbitrary signal *x*(*t*) by approximating the signal by straight-line segments. - -### SYNTHESIZING A PERIODIC REPLICATION TO VERIFY SPECTRUM CORRECTNESS - -While a signal's spectrum *X*(ω) provides useful insight into signal character, it can be difficult to look at *X*(ω) and know that it is correct for a particular signal *x*(*t*). Is it obvious, for example, that *X*(ω) = τ 2 sinc2 (ωτ/4) is really the spectrum of a τ -duration rectangle function? Or is it possible that a mathematical error was made in the determination of *X*(ω)? It is difficult to be certain by simple inspection of the spectrum. - -The same uncertainties exist when we are looking at a periodic signal's Fourier series spectrum. In the Fourier series case, we can verify the correctness of a signal's spectrum by synthesizing *x*(*t*) with a truncated Fourier series; the synthesized signal will match the original only if the computed spectrum is correct. This is exactly the approach that was taken in Ex. 6.11. And since a truncated Fourier series involves a simple sum, tools like MATLAB make waveform synthesis relatively simple, at least in the case of the Fourier series. - -In the case of the Fourier transform, however, synthesis of *x*(*t*) using Eq. (7.10) requires integration, a task not well suited to numerical packages such as MATLAB. All is not lost, however. Consider Eq. (7.5). By scaling and sampling the spectrum *X*(ω) of an aperiodic signal *x*(*t*), we obtain the Fourier series coefficient of a signal that is the periodic replication of *x*(*t*). Similar to Ex. 6.11, we can then synthesize a periodic replication of *x*(*t*) with a truncated Fourier series to verify spectrum correctness. Let us demonstrate the idea for the current example with τ = 1. - -To begin, we represent *X*(ω) = τ 2 sinc2 (ωτ/4) using an anonymous function in MATLAB. Since MATLAB computes sinc(x) as (sin(π*x*))/π*x*, we must scale the input by 1/π to match the notation of sinc in this book. - ->> tau = 1; X = @(omega) tau/2\*(sinc(omega\*tau/(4\*pi))).^2; - -For our periodic replication, let us pick *T*0 = 2, which is comfortably wide enough to accommodate our (τ = 1)-width function without overlap. We use Eq. (7.5) to define the needed Fourier series coefficients *Dn*. - ->> TO = 2; omega0 = -$$ -2*pi/TO -$$ -; D = $\mathcal{Q}(n)$ X(n\*omega0)/TO; - -Let us use 25 harmonics to synthesize the periodic replication *x*25(*t*) of our triangular signal *x*(*t*). To begin waveform synthesis, we set the dc portion of the signal. - ->> t = (-T0:.001:T0); x25 = D(0)\*ones(size(t)); - -To add the desired 25 harmonics, we enter a loop for 1 ≤ *n* ≤ 25 and add in the *Dn* and *D*−*n* terms. Although the result should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed by using the real command. - ->> for n = 1:25, >> x25 = x25+real(D(n)\*exp(1j\*omega0\*n\*t)+D(-n)\*exp(-1j\*omega0\*n\*t)); >> end - -Lastly, we plot the resulting truncated Fourier series synthesis of *x*(*t*). - ->> plot(t,x25,'k'); xlabel('t'); ylabel('x\_{25}(t)'); - -Since the synthesized waveform shown in Fig. 7.28 closely matches a 2-periodic replication of the triangle wave in Fig. 7.27a, we have high confidence that both the computed *Dn* and, by extension, the Fourier spectrum *X*(ω) are correct. - -**Figure 7.28** Synthesizing a 2-periodic replication of *x*(*t*) using a truncated Fourier series. - -### **DR ILL 7.9 Fourier Transform Time-Differentiation Property** - -Use the time-differentiation property to find the Fourier transform of rect(*t*/τ ). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/092_7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/092_7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS.md deleted file mode 100644 index 41923204e4f234336c960fa8d500b40f5bdae268..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/092_7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS.md +++ /dev/null @@ -1,282 +0,0 @@ -## **7.4 SIGNAL [TRANSMISSION](#page-12-0) THROUGH LTIC SYSTEMS** - -If *x*(*t*) and *y*(*t*) are the input and output of an LTIC system with impulse response *h*(*t*), then, as demonstrated in Eq. (7.35), - -$$ -Y(\omega) = H(\omega)X(\omega) -$$ - -This equation does not apply to (asymptotically) unstable systems because *h*(*t*) for such systems is not Fourier transformable. It applies to BIBO-stable as well as most of the marginally stable systems.† Similarly, this equation does not apply if *x*(*t*) is not Fourier transformable. - -In Ch. 4, we saw that the Laplace transform is more versatile and capable of analyzing all kinds of LTIC systems whether stable, unstable, or marginally stable. Laplace transform can also handle exponentially growing inputs. In comparison to the Laplace transform, the Fourier transform in system analysis is not just clumsier, but also very restrictive. Hence, the Laplace transform is preferable to the Fourier transform in LTIC system analysis. We shall not belabor the application of the Fourier transform to LTIC system analysis. We consider just one example here. - -### **EXAMPLE 7.18 Fourier Transform to Determine the Zero-State Response** - -Use the Fourier transform to find the zero-state response of a stable LTIC system with frequency response - -$$ -H(s) = \frac{1}{s+2} -$$ - -and the input is *x*(*t*) = *e*−*t u*(*t*). Stability implies that the region of convergence of *H*(*s*) includes the ω axis. - -In this case, - -$$ -X(\omega) = \frac{1}{j\omega + 1} -$$ - -† For marginally stable systems, if the input *x*(*t*) contains a finite-amplitude sinusoid of the system's natural frequency, which leads to resonance, the output is not Fourier transformable. It does, however, apply to marginally stable systems if the input does not contain a finite-amplitude sinusoid of the system's natural frequency. - -### 722 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Moreover, because the system is stable, the frequency response *H*(*j*ω) = *H*(ω). Hence, - -$$ -H(\omega) = H(s)|_{s=j\omega} = \frac{1}{j\omega + 2} -$$ - -Therefore, - -$$ -Y(\omega) = H(\omega)X(\omega) = \frac{1}{(j\omega + 2)(j\omega + 1)} -$$ - -Expanding the right-hand side in partial fractions yields - -$$ -Y(\omega) = \frac{1}{j\omega + 1} - \frac{1}{j\omega + 2} -$$ - -and - -$$ -y(t) = (e^{-t} - e^{-2t})u(t) -$$ - -## **DR ILL 7.10 Fourier Transform to Determine the Zero-State Response** - -For the system in Ex. 7.18, show that the zero-input response to the input *et u*(−*t*) is *y*(*t*) = 1 3 [*et u*(−*t*)+*e*−2*t u*(*t*)]. [*Hint:* Use pair 2 (Table 7.1) to find the Fourier transform of *et u*(−*t*).] - -### HEURISTIC UNDERSTANDING OF LINEAR SYSTEM RESPONSE - -In finding the linear system response to arbitrary input, the time-domain method uses convolution integral and the frequency-domain method uses the Fourier integral. Despite the apparent dissimilarities of the two methods, their philosophies are amazingly similar. In the time-domain case, we express the input *x*(*t*) as a sum of its impulse components; in the frequency-domain case, the input is expressed as a sum of everlasting exponentials (or sinusoids). In the former case, the response *y*(*t*) obtained by summing the system's responses to impulse components results in the convolution integral; in the latter case, the response obtained by summing the system's response to everlasting exponential components results in the Fourier integral. These ideas can be expressed mathematically as follows: - -1. For the time-domain case, - -| δ(t)
⇒ h(t) | shows the system response
δ(t)
h(t)
to
is the impulse response | -|--------------------------------------------|------------------------------------------------------------------------------------------| -| = \$ ∞
x(t)
x(τ )δ(t
−τ )dτ
−∞ | expresses
x(t)
as a sum
of impulse components | -| = \$ ∞
y(t)
x(τ )h(t
−τ )dτ
−∞ | expresses
y(t)
as a sum of responses to
the impulse components of input
x(t) | - -2. For the frequency-domain case, - -$$ -e^{j\omega t} \implies H(\omega)e^{j\omega t} \qquad \text{shows the system response} -$$ - -\nto $e^{j\omega t}$ is $H(\omega)e^{j\omega t}$ -\n -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega \qquad \text{expresses } x(t) \text{ as a sum} -$$ - -\nof everyday exponential components -\n -$$ -y(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)H(\omega)e^{j\omega t} d\omega \qquad \text{expresses } y(t) \text{ as a sum of responses to} -$$ - -\nthe exponential components of input $x(t)$ - -The frequency-domain view sees a system in terms of its frequency response (system response to various sinusoidal components). It views a signal as a sum of various sinusoidal components. Transmission of an input signal through a (linear) system is viewed as transmission of various sinusoidal components of the input through the system. - -It was not by coincidence that we used the impulse function in time-domain analysis and the exponential *ej*ω*t* in studying the frequency domain. The two functions happen to be duals of each other. Thus, the Fourier transform of an impulse δ(*t* − τ ) is *e*−*j*ωτ , and the Fourier transform of *ej*ω0*t* is an impulse 2πδ(ω − ω0). This *time-frequency duality* is a constant theme in the Fourier transform and linear systems. - -### **[7.4-1 Signal Distortion During Transmission](#page-12-0)** - -For a system with frequency response *H*(ω), if *X*(ω) and *Y*(ω) are the spectra of the input and the output signals, respectively, then - -$$ -Y(\omega) = X(\omega)H(\omega) \tag{7.38} -$$ - -The transmission of the input signal *x*(*t*) through the system changes it into the output signal *y*(*t*). Equation (7.38) shows the nature of this change or modification. Here, *X*(ω) and *Y*(ω) are the spectra of the input and the output, respectively. Therefore, *H*(ω) is the spectral response of the system. The output spectrum is obtained by the input spectrum multiplied by the spectral response of the system. Equation (7.38), which clearly brings out the spectral shaping (or modification) of the signal by the system, can be expressed in polar form as - -$$ -|Y(\omega)|e^{j\angle Y(\omega)} = |X(\omega)||H(\omega)|e^{j[\angle X(\omega)+\angle H(j\omega)]} -$$ - -Therefore, - -$$ -|Y(\omega)| = |X(\omega)| |H(\omega)| \quad \text{and} \quad \angle Y(\omega) = \angle X(\omega) + \angle H(\omega) -$$ - -During transmission, the input signal amplitude spectrum |*X*(ω)| is changed to |*X*(ω)||*H*(ω)|. Similarly, the input signal phase spectrum *X*(ω) is changed to *X*(ω) + *H*(ω). An input signal spectral component of frequency ω is modified in amplitude by a factor |*H*(ω)| and is shifted in phase by an angle *H*(ω). Clearly, |*H*(ω)| is the amplitude response, and *H*(ω) is the phase response of the system. The plots of |*H*(ω)| and *H*(ω) as functions of ω show at a glance how the system modifies the amplitudes and phases of various sinusoidal inputs. This is the reason why *H*(ω) is also called the *frequency response* of the system. During transmission through the system, some frequency components may be boosted in amplitude, while others may be attenuated. The - -### 724 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -relative phases of the various components also change. In general, the output waveform will be different from the input waveform. - -### DISTORTIONLESS TRANSMISSION - -In several applications, such as signal amplification or message signal transmission over a communication channel, we require that the output waveform be a replica of the input waveform. In such cases we need to minimize the distortion caused by the amplifier or the communication channel. It is, therefore, of practical interest to determine the characteristics of a system that allows a signal to pass without distortion (*distortionless transmission*). - -Transmission is said to be distortionless if the input and the output have identical waveshapes within a multiplicative constant. A delayed output that retains the input waveform is also considered to be distortionless. Thus, in distortionless transmission, the input *x*(*t*) and the output *y*(*t*) satisfy the condition - -$$ -y(t) = G_0 x(t - t_d) -$$ - -The Fourier transform of this equation yields - -$$ -Y(\omega) = G_0 X(\omega) e^{-j\omega t_d} -$$ - -But - -$$ -Y(\omega) = X(\omega)H(\omega) -$$ - -Therefore, - -$$ -H(\omega) = G_0 e^{-j\omega t_d} -$$ - -This is the frequency response required of a system for distortionless transmission. From this equation, it follows that - -$$ -|H(\omega)| = G_0 \quad \text{and} \quad \angle H(\omega) = -\omega t_d \tag{7.39} -$$ - -This result shows that for distortionless transmission, the amplitude response |*H*(ω)| must be a constant, and the phase response *H*(ω) must be a linear function of ω with slope −*td*, where *td* is the delay of the output with respect to input (Fig. 7.29). - -### MEASURE OF TIME-DELAY VARIATION WITH FREQUENCY - -The gain |*H*(ω)| = *G*0 means that every spectral component is multiplied by a constant *G*0. Also, as seen in connection with Fig. 7.22, a linear phase *H*(ω) = −ω*td* means that every spectral - -**Figure 7.29** LTIC system frequency response for distortionless transmission. - -component is delayed by *td* seconds. This results in the output equal to *G*0 times the input delayed by *td* seconds. Because each spectral component is attenuated by the same factor (*G*0) and delayed by exactly the same amount (*td*), the output signal is an exact replica of the input (except for attenuating factor *G*0 and delay *td*). - -For distortionless transmission, we require a *linear phase* characteristic. The phase is not only a linear function of ω, it should also pass through the origin ω = 0. In practice, many systems have a phase characteristic that may be only approximately linear. A convenient way of judging phase linearity is to plot the slope of *H*(ω) as a function of frequency. This slope, which is constant for an ideal linear phase (ILP) system, is a function of ω in the general case and can be expressed as - -$$ -t_g(\omega) = -\frac{d}{d\omega} \angle H(\omega) -$$ -\n(7.40) - -If *tg*(ω) is constant, all the components are delayed by the same time interval *tg*. But if the slope is not constant, the time delay *tg* varies with frequency. This variation means that different frequency components undergo different amounts of time delay, and consequently, the output waveform will not be a replica of the input waveform. As we shall see, *tg*(ω) plays an important role in bandpass systems and is called the *group delay* or *envelope* delay. Observe that constant *td* [Eq. (7.39)] implies constant *tg*. Note that *H*(ω) = φ0 −ω*tg* also has a constant *tg*. Thus, constant group delay is a more relaxed condition. - -It is often thought (erroneously) that flatness of amplitude response |*H*(ω)| alone can guarantee signal quality. However, a system that has a flat amplitude response may yet distort a signal beyond recognition if the phase response is not linear (*td* not constant). - -### THE NATURE OF DISTORTION IN AUDIO AND VIDEO SIGNALS - -Generally speaking, the human ear can readily perceive amplitude distortion but is relatively insensitive to phase distortion. For the phase distortion to become noticeable, the variation in delay [variation in the slope of *H*(ω)] should be comparable to the signal duration (or the physically perceptible duration, in case the signal itself is long). In the case of audio signals, each spoken syllable can be considered to be an individual signal. The average duration of a spoken syllable is of a magnitude of the order of 0.01 to 0.1 second. Audio systems may have nonlinear phases, yet no noticeable signal distortion results because in practical audio systems, maximum variation in the slope of *H*(ω) is only a small fraction of a millisecond. This is the real truth underlying the statement that "the human ear is relatively insensitive to phase distortion" [3]. As a result, the manufacturers of audio equipment make available only |*H*(ω)|, the amplitude response characteristic of their systems. - -For video signals, in contrast, the situation is exactly the opposite. The human eye is sensitive to phase distortion but is relatively insensitive to amplitude distortion. Amplitude distortion in television signals manifests itself as a partial destruction of the relative half-tone values of the resulting picture, but this effect is not readily apparent to the human eye. Phase distortion (nonlinear phase), on the other hand, causes different time delays in different picture elements. The result is a smeared picture, and this effect is readily perceived by the human eye. Phase distortion is also very important in digital communication systems because the nonlinear phase characteristic of a channel causes pulse dispersion (spreading out), which in turn causes pulses to interfere with neighboring pulses. Such interference between pulses can cause an error in the pulse amplitude at the receiver: a binary **1** may read as **0**, and vice versa. - -### **[7.4-2 Bandpass Systems and Group Delay](#page-12-0)** - -The distortionless transmission conditions [Eq. (7.39)] can be relaxed slightly for bandpass systems. For lowpass systems, the phase characteristics not only should be linear over the band of interest but also should pass through the origin. For bandpass systems, the phase characteristics must be linear over the band of interest but need not pass through the origin. - -Consider an LTI system with amplitude and phase characteristics as shown in Fig. 7.30, where the amplitude spectrum is a constant *G*0 and the phase is φ0 − ω*tg* over a band 2*W* centered at frequency ω*c*. Over this band, we can describe *H*(ω) as† - -$$ -H(\omega) = G_0 e^{j(\phi_0 - \omega t_g)} \qquad \omega \ge 0 \tag{7.41} -$$ - -The phase of *H*(ω) in Eq. (7.41), shown dotted in Fig. 7.30b, is linear but does not pass through the origin. - -Consider a modulated input signal *z*(*t*) = *x*(*t*) cosω*ct*. This is a bandpass signal, whose spectrum is centered at ω = ω*c*. The signal cosω*ct* is the carrier, and the signal *x*(*t*), which is a lowpass signal of bandwidth *W* (see Fig. 7.25), is the *envelope* of *z*(*t*). ‡ We shall now show that the transmission of *z*(*t*) through *H*(ω) results in distortionless transmission of the envelope *x*(*t*). However, the carrier phase changes by φ0. To show this, consider an input *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* and the corresponding output *y*ˆ(*t*). From Eq. (7.30), *Z*ˆ(ω) = *X*(ω − ω*c*), and the corresponding output - -**Figure 7.30** Generalized linear phase characteristics. - -† Because the phase function is an odd function of ω, if *H*(ω) = φ0 − ω*tg* for ω ≥ 0, over the band 2*W* (centered at ω*c*), then *H*(ω) = −φ0 − ω*tg* for ω < 0 over the band 2*W* (centered at −ω*c*), as shown in Fig. 7.30a. - -‡ The envelope of a bandpass signal is well defined only when the bandwidth of the envelope is well below the carrier ω*c* (*W* ω*c*). - -spectrum *Y*ˆ(ω) is given by - -$$ -\hat{Y}(\omega) = H(\omega)\hat{Z}(\omega) = H(\omega)X(\omega - \omega_c) -$$ - -Recall that the bandwidth of *X*(ω) is *W* so that the bandwidth of *X*(ω −ω*c*) is 2*W*, centered at ω*c*. Over this range, *H*(ω) is given by Eq. (7.41). Hence, - -$$ -\hat{Y}(\omega) = G_0 X(\omega - \omega_c) e^{j(\phi_0 - \omega t_g)} = G_0 e^{j\phi_0} X(\omega - \omega_c) e^{-\omega t_g} -$$ - -Use of Eqs. (7.29) and (7.30) yields *y*ˆ(*t*) as - -$$ -\hat{y}(t) = G_0 e^{j\phi_0} x(t - t_g) e^{j\omega_c(t - t_g)} = G_0 x(t - t_g) e^{j[\omega_c(t - t_g) + \phi_0]} -$$ - -This is the system response to input *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* , which is a complex signal. We are really interested in finding the response to the input *z*(*t*) = *x*(*t*) cosω*ct*, which is the real part of *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* . Hence, we use Eq. (2.31) to obtain *y*(*t*), the system response to the input *z*(*t*) = *x*(*t*) cosω*ct*, as - -$$ -y(t) = G_0 x(t - t_g) \cos [\omega_c (t - t_g) + \phi_0)] -$$ -\n(7.42) - -where *tg*, the *group* (or *envelope*) delay, is the negative slope of *H*(ω) at ω*c*. † The output *y*(*t*) is basically the delayed input *z*(*t* − *tg*), except that the output carrier acquires an extra phase φ0. The output envelope *x*(*t* − *tg*) is the delayed version of the input envelope *x*(*t*) and is not affected by extra phase φ0 of the carrier. In a modulated signal, such as *x*(*t*) cosω*ct*, the information generally resides in the envelope *x*(*t*). Hence, the transmission is considered to be distortionless if the envelope *x*(*t*) remains undistorted. - -Most practical systems satisfy Eq. (7.41), at least over a very small band. Figure 7.30b shows a typical case in which this condition is satisfied for a small band *W* centered at frequency ω*c*. - -A system in Eq. (7.41) is said to have a *generalized linear phase* (GLP), as illustrated in Fig. 7.30. The ideal linear phase (ILP) characteristics is shown in Fig. 7.29. For distortionless transmission of bandpass signals, the system need satisfy Eq. (7.41) only over the bandwidth of the bandpass signal. - -**Caution.** Recall that the phase response associated with the amplitude response may have jump discontinuities when the amplitude response goes negative. Jump discontinuities also arise because of the use of the principal value for phase. Under such conditions, to compute the group delay [Eq. (7.40)], we should ignore the jump discontinuities. - -$$ -y(t) = G_o x(t - t_g) \cos \omega_c (t - t_{ph}) -$$ - -where *t*ph, called the *phase delay* at ω*c*, is given by *t*ph(ω*c*) = (ω*ctg* − φ0)/ω*c*. Generally, *t*ph varies with ω, and we can write - -$$ -t_{\rm ph}(\omega) = \frac{\omega t_g - \phi_0}{\omega} -$$ - -Recall also that *tg* itself may vary with ω. - -† Equation (7.42) can also be expressed as - -### **EXAMPLE 7.19 Distortionless Bandpass Transmission** - -**(a)** A signal *z*(*t*), shown in Fig. 7.31b, is given by - -$$ -z(t) = x(t) \cos \omega_c t -$$ - -where ω*c* = 2000π. The pulse *x*(*t*) (Fig. 7.31a) is a lowpass pulse of duration 0.1 second and has a bandwidth of about 10 Hz. This signal is passed through a filter whose frequency response is shown in Fig. 7.31c (shown only for positive ω). Find and sketch the filter output *y*(*t*). - -**(b)** Find the filter response if ω*c* = 4000π. - -**(a)** The spectrum *Z*(ω) is a narrow band of width 20 Hz, centered at frequency *f*0 = 1 kHz. The gain at the center frequency (1 kHz) is 2. The group delay, which is the negative of the slope of the phase plot, can be found by drawing tangents at ω*c*, as shown in Fig. 7.31c. The negative of the slope of the tangent represents *tg*, and the intercept along the vertical axis by the tangent represents φ0 at that frequency. From the tangents at ω*c*, we find *tg*, the group delay, as - -$$ -t_g = \frac{2.4\pi - 0.4\pi}{2000\pi} = 10^{-3} -$$ - -The vertical axis intercept is φ0 = −0.4π. Hence, by using Eq. (7.42) with gain *G*0 = 2, we obtain - -$$ -y(t) = 2x(t - t_g)\cos[\omega_c(t - t_g) - 0.4\pi] \qquad \omega_c = 2000\pi \quad t_g = 10^{-3} -$$ - -Figure 7.31d shows the output *y*(*t*), which consists of the modulated pulse envelope *x*(*t*) delayed by 1 ms and the phase of the carrier changed by −0.4π. The output shows no distortion of the envelope *x*(*t*), only the delay. The carrier phase change does not affect the shape of envelope. Hence, the transmission is considered distortionless. - -**(b)** Figure 7.31c shows that when ω*c* = 4000π, the slope of *H*(ω) is zero so that *tg* = 0. Also, the gain *G*0 = 1.5, and the intercept of the tangent with the vertical axis is φ0 = −3.1π. Hence, - -$$ -y(t) = 1.5x(t)\cos(\omega_c t - 3.1\pi) -$$ - -This, too, is a distortionless transmission for the same reasons as for case (a). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/093_7.5 IDEAL AND PRACTICAL FILTERS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/093_7.5 IDEAL AND PRACTICAL FILTERS.md deleted file mode 100644 index af4991185f496394be3a3586f2a541012ca91094..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/093_7.5 IDEAL AND PRACTICAL FILTERS.md +++ /dev/null @@ -1,78 +0,0 @@ -## **[7.5 IDEAL AND](#page-13-0) PRACTICAL FILTERS** - -Ideal filters allow distortionless transmission of a certain band of frequencies and completely suppress the remaining frequencies. The ideal lowpass filter (Fig. 7.32), for example, allows all components below ω = *W* rad/s to pass without distortion and suppresses all components above ω = *W*. Figure 7.33 illustrates ideal highpass and bandpass filter characteristics. - -The ideal lowpass filter in Fig. 7.32a has a linear phase of slope −*td*, which results in a time delay of *td* seconds for all its input components of frequencies below *W* rad/s. Therefore, if the input is a signal *x*(*t*) bandlimited to *W* rad/s, the output *y*(*t*) is *x*(*t*) delayed by *td*: that is, - -$$ -y(t) = x(t - t_d) -$$ - -The signal *x*(*t*) is transmitted by this system without distortion, but with time delay *td*. For this filter, |*H*(ω)| = rect(ω/2*W*) and *H*(ω) = *e*−*j*ω*td* so that - -**Figure 7.32** Ideal lowpass filter: **(a)** frequency response and **(b)** impulse response. - -**Figure 7.33** Ideal **(a)** highpass and **(b)** bandpass filter frequency responses. - -The unit impulse response *h*(*t*) of this filter is obtained from pair 18 (Table 7.1) and the time-shifting property - -$$ -h(t) = \mathcal{F}^{-1} \left[ \text{rect}\left(\frac{\omega}{2W}\right) e^{-j\omega t_d} \right] = \frac{W}{\pi} \operatorname{sinc}[W(t - t_d)] -$$ - -Recall that *h*(*t*) is the system response to impulse input δ(*t*), which is applied at *t* = 0. Figure 7.32b shows a curious fact: the response *h*(*t*) begins even before the input is applied (at *t* = 0). Clearly, the filter is noncausal and therefore physically unrealizable. Similarly, one can show that other ideal filters (such as the ideal highpass or ideal bandpass filters depicted in Fig. 7.33) are also physically unrealizable. - -For a physically realizable system, *h*(*t*) must be causal; that is, - -$$ -h(t) = 0 \qquad \text{for } t < 0 -$$ - -In the frequency domain, this condition is equivalent to the well-known *Paley–Wiener criterion,* which states that the necessary and sufficient condition for the amplitude response |*H*(ω)| to be realizable is† - -$$ -\int_{-\infty}^{\infty} \frac{|\ln|H(\omega)|}{1 + \omega^2} d\omega < \infty -$$ -\n(7.43) - -If *H*(ω) does not satisfy this condition, it is unrealizable. Note that if |*H*(ω)| = 0 over any finite band, |ln|*H*(ω)|| = ∞ over that band, and Eq. (7.43) is violated. If, however, *H*(ω) = 0 at a single frequency (or a set of discrete frequencies), the integral in Eq. (7.43) may still be finite even though the integrand is infinite at those discrete frequencies. Therefore, for a physically realizable system, *H*(ω) may be zero at some discrete frequencies, but it cannot be zero over any finite band. In addition, if |*H*(ω)| decays exponentially (or at a higher rate) with ω, the integral in Eq. (7.43) goes to infinity, and |*H*(ω)| cannot be realized. Clearly, |*H*(ω)| cannot decay too fast with ω. According to this criterion, ideal filter characteristics (Figs. 7.32 and 7.33) are unrealizable. - -The impulse response *h*(*t*) in Fig. 7.32 is not realizable. One practical approach to filter design is to cut off the tail of *h*(*t*) for *t* < 0. The resulting causal impulse response:*h*(*t*), given by - -$$ -h(t) = h(t)u(t) -$$ - -is physically realizable because it is causal (Fig. 7.34). If *td* is sufficiently large, :*h*(*t*) will be a close approximation of *h*(*t*), and the resulting filter *H* :(ω) will be a good approximation of an ideal filter. This close realization of the ideal filter is achieved because of the increased value of time delay *td*. This observation means that the price of close realization is higher delay in the output; this situation is common in noncausal systems. Of course, theoretically, a delay *td* = ∞ is needed to realize the ideal characteristics. But a glance at Fig. 7.32b shows that a delay *td* of three or four times π *W* will make :*h*(*t*) a reasonably close version of *h*(*t* − *td*). For instance, an audio filter is required to handle frequencies of up to 20 kHz (*W* = 40,000π). In this case, a *td* of about 10−4 - -$$ -\int_{-\infty}^{\infty} |H(\omega)|^2 d\omega < \infty -$$ - -Note that the Paley–Wiener criterion is a criterion for the realizability of the amplitude response |*H*(ω)|. - -† We are assuming that |*H*(ω)| is square integrable, that is, - -**Figure 7.34** Approximate realization of an ideal lowpass filter by truncation of its impulse response. - -(0.1 ms) would be a reasonable choice. The truncation operation [cutting the tail of *h*(*t*) to make it causal], however, creates some unsuspected problems. We discuss these problems and their cure in Sec. 7.8. - -In practice, we can realize a variety of filter characteristics that approach the ideal. Practical (realizable) filter characteristics are gradual, without jump discontinuities in amplitude response. - -### **DR ILL 7.11 The Unrealizable Gaussian Response** - -Show that a filter with Gaussian frequency response *H*(ω) = *e*−αω2 is unrealizable. Demonstrate this fact in two ways: first by showing that its impulse response is noncausal, and then by showing that |*H*(ω)| violates the Paley–Wiener criterion. [*Hint:* Use pair 22 in Table 7.1.] - -### THINKING IN THE TIME AND FREQUENCY DOMAINS: A TWO-DIMENSIONAL VIEW OF SIGNALS AND SYSTEMS - -Both signals and systems have dual personalities, the time domain and the frequency domain. For a deeper perspective, we should examine and understand both these identities because they offer complementary insights. An exponential signal, for instance, can be specified by its time-domain description such as *e*−2*t u*(*t*) or by its Fourier transform (its frequency-domain description) 1/(*j*ω +2). The time-domain description depicts the waveform of a signal. The frequency-domain description portrays its spectral composition [relative amplitudes of its sinusoidal (or exponential) components and their phases]. For the signal *e*−2*t* , for instance, the time-domain description portrays the exponentially decaying signal with a time constant 0.5. The frequency-domain description characterizes it as a lowpass signal, which can be synthesized by sinusoids with amplitudes decaying with frequency roughly as 1/ω. - -An LTIC system can also be described or specified in the time domain by its impulse response *h*(*t*) or in the frequency domain by its frequency response *H*(ω). In Sec. 2.6, we studied intuitive insights in the system behavior offered by the impulse response, which consists of characteristic modes of the system. By purely qualitative reasoning, we saw that the system responds well to signals that are similar to the characteristic modes and responds poorly to signals that are very different from those modes. We also saw that the shape of the impulse response *h*(*t*) determines the system time constant (speed of response), and pulse dispersion (spreading), which, in turn, determines the rate of pulse transmission. - -The frequency response *H*(ω) specifies the system response to exponential or sinusoidal input of various frequencies. This is precisely the filtering characteristic of the system. - -Experienced electrical engineers instinctively think in both domains (time and frequency) whenever possible. When they look at a signal, they consider its waveform, the signal width (duration), and the rate at which the waveform decays. This is basically a time-domain perspective. They also think of the signal in terms of its frequency spectrum, that is, in terms of its sinusoidal components and their relative amplitudes and phases, whether the spectrum is lowpass, bandpass, highpass, and so on. This is a frequency-domain perspective. Experienced electrical engineers think of a system in terms of its impulse response *h*(*t*). The width of *h*(*t*) indicates the time constant (response time): that is, how quickly the system is capable of responding to an input, and how much dispersion (spreading) it will cause. This is a time-domain perspective. From the frequency-domain perspective, these engineers view a system as a filter, which selectively transmits certain frequency components and suppresses the others [frequency response *H*(ω)]. Knowing the input signal spectrum and the frequency response of the system, they create a mental image of the output signal spectrum. This concept is precisely expressed by *Y*(ω) = *X*(ω)*H*(ω). - -We can analyze LTI systems by time-domain techniques or by frequency-domain techniques. Then why learn both? The reason is that the two domains offer complementary insights into system behavior. Some aspects are easily grasped in one domain; other aspects may be easier to see in the other domain. Both time-domain and frequency-domain methods are as essential for the study of signals and systems as two eyes are essential to a human being for correct visual perception of reality. A person can see with either eye, but for proper perception of three-dimensional reality, both eyes are essential. - -It is important to keep the two domains separate, and not to mix the entities in the two domains. If we are using the frequency domain to determine the system response, we must deal with all signals in terms of their spectra (Fourier transforms) and all systems in terms of their frequency responses. For example, to determine the system response *y*(*t*) to an input *x*(*t*), we must first convert the input signal into its frequency-domain description *X*(ω). The system description also must be in the frequency domain, that is, the frequency response *H*(ω). The output signal spectrum *Y*(ω) = *X*(ω)*H*(ω). Thus, the result (output) is also in the frequency domain. To determine the final answer *y*(*t*), we must take the inverse transform of *Y*(ω). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/094_7.6 SIGNAL ENERGY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/094_7.6 SIGNAL ENERGY.md deleted file mode 100644 index b87f2f257e1bce8f2e8dd24bc369953c31262e6c..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/094_7.6 SIGNAL ENERGY.md +++ /dev/null @@ -1,110 +0,0 @@ -## **[7.6 SIGNAL](#page-13-0) ENERGY** - -The signal energy *Ex* of a signal *x*(*t*) was defined in Ch. 1 as - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt -$$ -\n(7.44) - -Signal energy can be related to the signal spectrum *X*(ω) by substituting Eq. (7.10) in Eq. (7.44): - -$$ -E_x = \int_{-\infty}^{\infty} x(t)x^*(t) dt = \int_{-\infty}^{\infty} x(t) \left[ \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) e^{-j\omega t} d\omega \right] dt -$$ - -#### 734 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Here, we used the fact that *x*∗(*t*), being the conjugate of *x*(*t*), can be expressed as the conjugate of the right-hand side of Eq. (7.10). Now, interchanging the order of integration yields - -$$ -E_x = \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) \left[ \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \right] d\omega -$$ - -= $\frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) X^*(\omega) d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega$ - -Consequently, - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega \qquad (7.45) -$$ - -This is *Parseval's theorem* (for the Fourier transform). A similar result was obtained in Eqs. (6.26) and (6.27) for a periodic signal and its Fourier series. This result allows us to determine the signal energy from either the time-domain specification *x*(*t*) or the corresponding frequency-domain specification *X*(ω). - -The right-hand side of Eq. (7.45) can be interpreted to mean that the energy of a signal *x*(*t*) results from energies contributed by all the spectral components of the signal *x*(*t*). The total signal energy is the area under |*X*(ω)2| (divided by 2π). If we consider a small band ω (ω → 0), as illustrated in Fig. 7.35, the energy *Ex* of the spectral components in this band is the area of |*X*(ω)| 2 under this band (divided by 2π): - -$$ -\Delta E_x = \frac{1}{2\pi} |X(\omega)|^2 \,\Delta \omega = |X(\omega)|^2 \,\Delta f \qquad \frac{\Delta \omega}{2\pi} = \Delta f \,\mathrm{Hz} -$$ - -Therefore, the energy contributed by the components in this band of *f* (in hertz) is |*X*(ω)| 2*f* . The total signal energy is the sum of energies of all such bands and is indicated by the area under |*X*(ω)| 2 as in Eq. (7.45). Therefore, |*X*(ω)| 2 is the *energy spectral density* (per unit bandwidth in hertz). - -For real signals, *X*(ω) and *X*(−ω) are conjugates, and |*X*(ω)| 2 is an even function of ω because - -$$ -|X(\omega)|^2 = X(\omega)X^*(\omega) = X(\omega)X(-\omega) -$$ - -**Figure 7.35** Interpretation of energy spectral density of a signal. - -Therefore, the energy of real signal *x*(*t*) can be expressed as† - -$$ -E_x = \frac{1}{\pi} \int_0^\infty |X(\omega)|^2 d\omega \tag{7.46} -$$ - -The signal energy *Ex*, which results from contributions from all the frequency components from ω = 0 to ∞, is given by (1/π times) the area under |*X*(ω)| 2 from ω = 0 to ∞. It follows that the energy contributed by spectral components of frequencies between ω1 and ω2 is - -$$ -\Delta E_x = \frac{1}{\pi} \int_{\omega_1}^{\omega_2} |X(\omega)|^2 d\omega \tag{7.47} -$$ - -### **EXAMPLE 7.20 Signal Energy and Parseval's Theorem** - -Find the energy of signal *x*(*t*) = *e*−*atu*(*t*). Determine the frequency *W* (rad/s) so that the energy contributed by the spectral components of all the frequencies below *W* is 95% of the signal energy *Ex*. - -We have - -$$ -E_x = \int_{-\infty}^{\infty} x^2(t) \, dt = \int_0^{\infty} e^{-2at} \, dt = \frac{1}{2a} -$$ - -We can verify this result by Parseval's theorem. For this signal, - -$$ -X(\omega) = \frac{1}{j\omega + a} -$$ - -and - -$$ -E_x = \frac{1}{\pi} \int_0^{\infty} |X(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{\infty} \frac{1}{\omega^2 + a^2} d\omega = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^{\infty} = \frac{1}{2a} -$$ - -The band ω = 0 to ω = *W* contains 95% of the signal energy, that is, 0.95/2*a*. Therefore, from Eq. (7.47) with ω1 = 0 and ω2 = *W*, we obtain - -$$ -\frac{0.95}{2a} = \frac{1}{\pi} \int_0^W \frac{d\omega}{\omega^2 + a^2} = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^W = \frac{1}{\pi a} \tan^{-1} \frac{W}{a} -$$ - -or - -$$ -\frac{0.95\pi}{2} = \tan^{-1}\frac{W}{a} \implies W = 12.706a \text{ rad/s} -$$ - -† In Eq. (7.46), it is assumed that *X*(ω) does not contain an impulse at ω = 0. If such an impulse exists, it should be integrated separately with a multiplying factor of 1/2π rather than 1/π. - -### 736 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -This result indicates that the spectral components of *x*(*t*) in the band from 0 (dc) to 12.706*a* rad/s (2.02*a* Hz) contribute 95% of the total signal energy; all the remaining spectral components (in the band from 12.706*a* rad/s to ∞) contribute only 5% of the signal energy. - -### **DR ILL 7.12 Signal Energy and Parseval's Theorem** - -Use Parseval's theorem to show that the energy of the signal *x*(*t*) = 2*a*/(*t* 2 +*a*2) is 2π/*a*. [*Hint:* Find *X*(ω) using pair 3 of Table 7.1 and the duality property.] - -### THE ESSENTIAL BANDWIDTH OF A SIGNAL - -The spectra of all practical signals extend to infinity. However, because the energy of any practical signal is finite, the signal spectrum must approach 0 as ω → ∞. Most of the signal energy is contained within a certain band of *B* Hz, and the energy contributed by the components beyond *B* Hz is negligible. We can therefore suppress the signal spectrum beyond *B* Hz with little effect on the signal shape and energy. The bandwidth *B* is called the *essential bandwidth* of the signal. The criterion for selecting *B* depends on the error tolerance in a particular application. We may, for example, select *B* to be that band which contains 95% of the signal energy.† This figure may be higher or lower than 95%, depending on the precision needed. Using such a criterion, we can determine the essential bandwidth of a signal. The essential bandwidth *B* for the signal *e*−*atu*(*t*), using 95% energy criterion, was determined in Ex. 7.20 to be 2.02*a* Hz. - -Suppression of all the spectral components of *x*(*t*) beyond the essential bandwidth results in a signal *x*ˆ(*t*), which is a close approximation of *x*(*t*). If we use the 95% criterion for the essential bandwidth, the energy of the error (the difference) *x*(*t*)− ˆ*x*(*t*) is 5% of *Ex*. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/095_7.7 APPLICATION TO COMMUNICATIONS - AMPLITUDE MODULATION.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/095_7.7 APPLICATION TO COMMUNICATIONS - AMPLITUDE MODULATION.md deleted file mode 100644 index 537301e8ea6d085416e7b9d046a04ac603365a4b..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/095_7.7 APPLICATION TO COMMUNICATIONS - AMPLITUDE MODULATION.md +++ /dev/null @@ -1,242 +0,0 @@ -## **[7.7 APPLICATION TO](#page-13-0) COMMUNICATIONS: AMPLITUDE MODULATION** - -*Modulation* causes a spectral shift in a signal and is used to gain certain advantages mentioned in our discussion of the frequency-shifting property. Broadly speaking, there are two classes of modulation: amplitude (linear) modulation and angle (nonlinear) modulation. In this section, we shall discuss some practical forms of amplitude modulation. - -† For lowpass signals, the essential bandwidth may also be defined as a frequency at which the value of the amplitude spectrum is a small fraction (about 1%) of its peak value. In Ex. 7.20, for instance, the peak value, which occurs at ω = 0, is 1/*a*. - -### **[7.7-1 Double-Sideband, Suppressed-Carrier \(DSB-SC\) Modulation](#page-13-0)** - -In amplitude modulation, the amplitude *A* of the carrier *A*cos(ω*ct* + θ*c*) is varied in some manner with the *baseband* (message)† signal *m*(*t*) (known as the *modulating signal*). The frequency ω*c* and the phase θ*c* are constant. We can assume θ*c* = 0 without loss of generality. If the carrier amplitude *A* is made directly proportional to the modulating signal *m*(*t*), the modulated signal is *m*(*t*) cos ω*ct* (Fig. 7.36). As was indicated earlier [Eq. (7.32)], this type of modulation simply shifts the spectrum of *m*(*t*) to the carrier frequency (Fig. 7.36c). Thus, if - -**Figure 7.36** DSB-SC modulation. - -† The term *baseband* is used to designate the band of frequencies of the signal delivered by the source or the input transducer. - -then - -$$ -m(t)\cos\omega_c t \Longleftrightarrow \frac{1}{2}[M(\omega + \omega_c) + M(\omega - \omega_c)] \tag{7.48} -$$ - -Recall that *M*(ω − ω*c*) is *M*(ω)-shifted to the right by ω*c* and *M*(ω + ω*c*) is *M*(ω)-shifted to the left by ω*c*. Thus, the process of modulation shifts the spectrum of the modulating signal to the left and the right by ω*c*. Note also that if the bandwidth of *m*(*t*) is *B* Hz, then, as indicated in Fig. 7.36c, the bandwidth of the modulated signal is 2*B* Hz. We also observe that the modulated signal spectrum centered at ω*c* is composed of two parts: a portion that lies above ω*c*, known as the *upper sideband (USB)*, and a portion that lies below ω*c*, known as the *lower sideband (LSB)*. Similarly, the spectrum centered at −ω*c* has upper and lower sidebands. This form of modulation is called *double sideband (DSB)* modulation for the obvious reason. - -The relationship of *B* to ω*c* is of interest. Figure 7.36c shows that ω*c* ≥ 2π*B* to avoid the overlap of the spectra centered at ±ω*c*. If ω*c* < 2π*B*, the spectra overlap and the information of *m*(*t*) are lost in the process of modulation, a loss that makes it impossible to get back *m*(*t*) from the modulated signal *m*(*t*) cos ω*ct*. † - -### **EXAMPLE 7.21 Double-Sideband Suppressed-Carrier Modulation** - -For a baseband signal *m*(*t*) = cos ω*mt*, find the DSB-SC signal and sketch its spectrum. Identify the upper and lower sidebands. - -We shall work this problem in the frequency domain as well as the time domain to clarify the basic concepts of DSB-SC modulation. In the frequency-domain approach, we work with the signal spectra. The spectrum of the baseband signal *m*(*t*) = cos ω*mt* is given by - -$$ -M(\omega) = \pi \left[ \delta(\omega - \omega_m) + \delta(\omega + \omega_m) \right] -$$ - -The spectrum consists of two impulses located at ±ω*m*, as depicted in Fig. 7.37a. - -The DSB-SC (modulated) spectrum, as indicated by Eq. (7.48), is the baseband spectrum in Fig. 7.37a shifted to the right and the left by ω*c* (times 0.5), as depicted in Fig. 7.37b. This spectrum consists of impulses at ±(ω*c* − ω*m*) and ±(ω*c* + ω*m*). The spectrum beyond ω*c* is the upper sideband (USB), and the one below ω*c* is the lower sideband (LSB). Observe that the DSB-SC spectrum does not have as a component the carrier frequency ω*c*. This is why the term *double-sideband, suppressed carrier* (DSB-SC) is used for this type of modulation. - -† Practical factors may impose additional restrictions on ω*c*. For instance, in broadcast applications, a radiating antenna can radiate only a narrow band without distortion. This restriction implies that avoiding distortion caused by the radiating antenna calls for ω*c*/2π*B* 1. The broadcast band AM radio, for instance, with *B* = 5 kHz and the band of 550–1600 kHz for carrier frequency gives a ratio of ω*c*/2π*B* roughly in the range of 100–300. - -**Figure 7.37** An example of DSB-SC modulation. - -In the time-domain approach, we work directly with signals in the time domain. For the baseband signal *m*(*t*) = cos ω*mt*, the DSB-SC signal ϕDSB-SC(*t*) is - -$$ -\varphi_{\text{DSB-SC}}(t) = m(t) \cos \omega_c t -$$ - -= $\cos \omega_m t \cos \omega_c t$ -= $\frac{1}{2} [\cos (\omega_c + \omega_m)t + \cos (\omega_c - \omega_m)t]$ (7.49) - -This result shows that when the baseband (message) signal is a single sinusoid of frequency ω*m*, the modulated signal consists of two sinusoids: the component of frequency ω*c* + ω*m* (the upper sideband), and the component of frequency ω*c* − ω*m* (the lower sideband). Figure 7.37b illustrates precisely the spectrum of ϕDSB-SC(*t*). Thus, each component of frequency ω*m* in the modulating signal results in two components of frequencies ω*c* + ω*m* and ω*c* − ω*m* in the modulated signal. This being a DSB-SC (suppressed-carrier) modulation, there is no component of the carrier frequency ω*c* on the right-hand side of Eq. (7.49).† - -### DEMODULATION OF DSB-SC SIGNALS - -The DSB-SC modulation translates or shifts the frequency spectrum to the left and the right by ω*c* (i.e., at +ω*c* and −ω*c*), as seen from Eq. (7.48). To recover the original signal *m*(*t*) from - -† The term *suppressed carrier* does not necessarily mean absence of the spectrum at the carrier frequency. "Suppressed carrier" merely implies that there is no discrete component of the carrier frequency. Since no discrete component exists, the DSB-SC spectrum does not have impulses at ±ω*c*, which further implies that the modulated signal *m*(*t*) cos ω*ct* does not contain a term of the form *k* cos ω*ct* [assuming that *m*(*t*) has a zero mean value]. - -**Figure 7.38** Demodulation of DSB-SC: **(a)** demodulator and **(b)** spectrum of *e*(*t*). - -the modulated signal, we must retranslate the spectrum to its original position. The process of recovering the signal from the modulated signal (retranslating the spectrum to its original position) is referred to as *demodulation*, or *detection*. Observe that if the modulated signal spectrum in Fig. 7.36c is shifted to the left and to the right by ω*c* (and halved), we obtain the spectrum illustrated in Fig. 7.38b, which contains the desired baseband spectrum in addition to an unwanted spectrum at ±2ω*c*. The latter can be suppressed by a lowpass filter. Thus, demodulation, which is almost identical to modulation, consists of multiplication of the incoming modulated signal *m*(*t*) cos ω*ct* by a carrier cos ω*ct* followed by a lowpass filter, as depicted in Fig. 7.38a. We can verify this conclusion directly in the time domain by observing that the signal *e*(*t*) in Fig. 7.38a is - -$$ -e(t) = m(t)\cos^2\omega_c t = \frac{1}{2}[m(t) + m(t)\cos 2\omega_c t] -$$ - -Therefore, the Fourier transform of the signal *e*(*t*) is - -$$ -E(\omega) = \frac{1}{2}M(\omega) + \frac{1}{4}[M(\omega + 2\omega_c) + M(\omega - 2\omega_c)] -$$ - -Hence, *e*(*t*) consists of two components (1/2)*m*(*t*) and (1/2)*m*(*t*) cos 2ω*ct*, with their spectra, as illustrated in Fig. 7.38b. The spectrum of the second component, being a modulated signal with carrier frequency 2ω*c*, is centered at ±2ω*c*. Hence, this component is suppressed by the lowpass filter in Fig. 7.38a. The desired component (1/2)*M*(ω), being a lowpass spectrum (centered at ω = 0), passes through the filter unharmed, resulting in the output (1/2)*m*(*t*). - -A possible form of lowpass filter characteristics is depicted (dotted) in Fig. 7.38b. In this method of recovering the baseband signal, called *synchronous detection,* or *coherent detection,* we use a carrier of exactly the same frequency (and phase) as the carrier used for modulation. Thus, for demodulation, we need to generate a local carrier at the receiver in frequency and phase coherence (synchronism) with the carrier used at the modulator. We shall demonstrate in Ex. 7.22 that both phase and frequency synchronism are extremely critical. - -### **EXAMPLE 7.22 Frequency and Phase Incoherence in DSB-SC** - -Discuss the effect of lack of frequency and phase coherence (synchronism) between the carriers at the modulator (transmitter) and the demodulator (receiver) in DSB-SC. - -Let the modulator carrier be cos ω*ct* (Fig. 7.36a). For the demodulator in Fig. 7.38a, we shall consider two cases: with carrier cos(ω*ct*+θ ) (phase error of θ) and with carrier cos(ω*c*+ω)*t* (frequency error ω). - -**(a)** With the demodulator carrier cos(ω*ct* + θ ) (instead of cos ω*ct*) in Fig. 7.38a, the multiplier output is *e*(*t*) = *m*(*t*) cos ω*ct* cos(ω*ct* + θ ) instead of *m*(*t*) cos2ω*ct*. From the trigonometric identity, we obtain - -$$ -e(t) = m(t)\cos\omega_c t \cos(\omega_c t + \theta) -$$ - -= $\frac{1}{2}m(t)[\cos\theta + \cos(2\omega_c t + \theta)]$ - -The spectrum of the component (1/2)*m*(*t*) cos(2ω*ct* + θ ) is centered at ±2ω*c*. Consequently, it will be filtered out by the lowpass filter at the output. The component (1/2)*m*(*t*) cos θ is the signal *m*(*t*) multiplied by a constant (1/2) cos θ. The spectrum of this component is centered at ω = 0 (lowpass spectrum) and will pass through the lowpass filter at the output, yielding the output (1/2)*m*(*t*) cos θ. - -If θ is constant, the phase asynchronism merely yields an output that is attenuated (by a factor cos θ). Unfortunately, in practice, θ is often the phase difference between the carriers generated by two distant generators and varies randomly with time. This variation would result in an output whose gain varies randomly with time. - -**(b)** In the case of frequency error, the demodulator carrier is cos(ω*c*+ω)*t*. This situation is very similar to the phase error case in part (a) with θ replaced by (ω)*t*. Following the analysis in part (a), we can express the demodulator product *e*(*t*) as - -$$ -e(t) = m(t)\cos\omega_c t \cos(\omega_c + \Delta\omega)t -$$ - -= $\frac{1}{2}m(t)[\cos(\Delta\omega)t + \cos(2\omega_c + \Delta\omega)t]$ - -The spectrum of the component (1/2)*m*(*t*) cos(2ω*c* + ω)*t* is centered at ±(2ω*c* + ω). Consequently, this component will be filtered out by the lowpass filter at the output. The component (1/2)*m*(*t*) cos(ω)*t* is the signal *m*(*t*) multiplied by a low-frequency carrier of frequency ω. The spectrum of this component is centered at ±ω. In practice, the frequency error (ω) is usually very small. Hence, the signal (1/2)*m*(*t*) cos(ω)*t* (whose spectrum is centered at ±ω) is a lowpass signal and passes through the lowpass filter at the output, resulting in the output (1/2)*m*(*t*) cos(ω)*t*. The output is the desired signal *m*(*t*) multiplied by a very-low-frequency sinusoid cos(ω)*t*. The output in this case is not merely an attenuated replica of the desired signal *m*(*t*), but represents *m*(*t*) multiplied by a time-varying gain cos(ω)*t*. If, for instance, the transmitter and the receiver carrier frequencies differ just by 1 Hz, the output will be the desired signal *m*(*t*) multiplied by a time-varying signal whose gain goes from the maximum to 0 every half-second. This is like a restless child fiddling with the volume control knob of a receiver, going from maximum volume to zero volume every half-second. This kind of distortion (called the *beat effect*) is beyond repair. - -### **[7.7-2 Amplitude Modulation \(AM\)](#page-13-0)** - -For the suppressed-carrier scheme just discussed, a receiver must generate a carrier in frequency and phase synchronism with the carrier at a transmitter that may be located hundreds or thousands of miles away. This situation calls for a sophisticated receiver, which could be quite costly. The other alternative is for the transmitter to transmit a carrier *A* cosω*ct* [along with the modulated signal *m*(*t*) cosω*ct*] so that there is no need to generate a carrier at the receiver. In this case, the transmitter needs to transmit much larger power, a rather expensive procedure. In point-to-point communications, where there is one transmitter for each receiver, substantial complexity in the receiver system can be justified, provided there is a large enough saving in expensive high-power transmitting equipment. On the other hand, for a broadcast system with a multitude of receivers for each transmitter, it is more economical to have one expensive high-power transmitter and simpler, less expensive receivers. The second option (transmitting a carrier along with the modulated signal) is the obvious choice in this case. This is amplitude modulation (AM), in which the transmitted signal ϕAM (*t*) is given by - -$$ -\varphi_{AM}(t) = A\cos\omega_c t + m(t)\cos\omega_c t = [A + m(t)]\cos\omega_c t \tag{7.50} -$$ - -Recall that the DSB-SC signal is *m*(*t*) cos ω*ct*. From Eq. (7.50) it follows that the AM signal is identical to the DSB-SC signal with *A*+*m*(*t*) as the modulating signal [instead of *m*(*t*)]. Therefore, to sketch ϕAM (*t*), we sketch *A* + *m*(*t*) and −[*A* + *m*(*t*)] as the envelopes and fill in between with the sinusoid of the carrier frequency. Two cases are considered in Fig. 7.39. In the first case, *A* is large enough so that *A* + *m*(*t*) ≥ 0 (is nonnegative) for all values of *t*. In the second case, *A* is not large enough to satisfy this condition. In the first case, the envelope (Fig. 7.39d) has the same shape as *m*(*t*) (although riding on a dc of magnitude *A*). In the second case, the envelope shape is not *m*(*t*), for some parts get rectified (Fig. 7.39e). Thus, we can detect the desired signal *m*(*t*) by detecting the envelope in the first case. In the second case, such a detection is not possible. We shall see that envelope detection is an extremely simple and inexpensive operation, which does not require generation of a local carrier for the demodulation. But as just noted, the envelope of AM has the information about *m*(*t*) only if the AM signal [*A*+*m*(*t*)] cos ω*ct* satisfies the condition *A*+*m*(*t*) > 0 for all *t*. Thus, the condition for envelope detection of an AM signal is - -$$ -A + m(t) \ge 0 \qquad \text{for all } t \tag{7.51} -$$ - -If *mp* is the peak amplitude (positive or negative) of *m*(*t*), then Eq. (7.51) is equivalent to - -$$ -A\geq m_p -$$ - -Thus, the minimum carrier amplitude required for the viability of envelope detection is *mp*. This point is clearly illustrated in Fig. 7.39. - -We define the *modulation index* μ as - -$$ -\mu = \frac{m_p}{A} \tag{7.52} -$$ - -where *A* is the carrier amplitude. Note that *mp* is a constant of the signal *m*(*t*). Because *A* ≥ *mp* and because there is no upper bound on *A*, it follows that - -0 ≤ μ ≤ 1 - -**Figure 7.39** An AM signal **(a)** for two values of A **(b, c)** and the respective envelopes **(d, e)**. - -as the required condition for the viability of demodulation of AM by an envelope detector. - -When *A* < *mp*, Eq. (7.52) shows that μ > 1 (overmodulation, shown in Fig. 7.39e). In this case, the option of envelope detection is no longer viable. We then need to use synchronous demodulation. Note that synchronous demodulation can be used for any value of μ (see Prob. 7.7-7). The envelope detector, which is considerably simpler and less expensive than the synchronous detector, can be used only when μ ≤ 1. - -### **EXAMPLE 7.23 Amplitude Modulation** - -Sketch ϕAM (*t*) for modulation indices of μ = 0.5 (50% modulation) and μ = 1 (100% modulation), when *m*(*t*) = *B*cos ω*mt*. This case is referred to as *tone modulation* because the modulating signal is a pure sinusoid (or tone). - -### 744 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -In this case, *mp* = *B* and the modulation index according to Eq. (7.52) is - -$$ -\mu = \frac{B}{A} -$$ - -Hence, *B* = μ*A* and - -$$ -m(t) = B\cos\omega_m t = \mu A\cos\omega_m t -$$ - -Therefore, - -$$ -\varphi_{AM}(t) = [A + m(t)] \cos \omega_c t = A[1 + \mu \cos \omega_m t] \cos \omega_c t -$$ - -The modulated signals corresponding to μ = 0.5 and μ = 1 appear in Figs. 7.40a and 7.40b, respectively. - -### DEMODULATION OF AM: THE ENVELOPE DETECTOR - -The AM signal can be demodulated coherently by a locally generated carrier (see Prob. 7.7-7). Since, however, coherent, or synchronous, demodulation of AM (with μ ≤ 1) will defeat the very purpose of AM, it is rarely used in practice. We shall consider here one of the noncoherent methods of AM demodulation, *envelope detection*. † - -In an envelope detector, the output of the detector follows the envelope of the (modulated) input signal. The circuit illustrated in Fig. 7.41a functions as an envelope detector. During the positive cycle of the input signal, the diode conducts and the capacitor *C* charges up to the peak voltage of the input signal (Fig. 7.41b). As the input signal falls below this peak value, the diode is cut off, because the capacitor voltage (which is very nearly the peak voltage) is greater than the input signal voltage, a circumstance causing the diode to open. The capacitor now discharges through the resistor *R* at a slow rate (with a time constant *RC*). During the next positive cycle, - -† There are also other methods of noncoherent detection. The rectifier detector consists of a rectifier followed by a lowpass filter. This method is also simple and almost as inexpensive as the envelope detector [4]. The nonlinear detector, although simple and inexpensive, results in a distorted output. - -**Figure 7.41** Demodulation by means of envelope detector. - -the same drama repeats. When the input signal becomes greater than the capacitor voltage, the diode conducts again. The capacitor again charges to the peak value of this (new) cycle. As the input voltage falls below the new peak value, the diode cuts off again and the capacitor discharges slowly during the cutoff period, a process that changes the capacitor voltage very slightly. - -In this manner, during each positive cycle, the capacitor charges up to the peak voltage of the input signal and then decays slowly until the next positive cycle. Thus, the output voltage *vC*(*t*) follows closely the envelope of the input. The capacitor discharge between positive peaks, however, causes a ripple signal of frequency ω*c* in the output. This ripple can be reduced by increasing the time constant *RC* so that the capacitor discharges very little between the positive peaks (*RC* 1/ω*c*). Making *RC* too large, however, would make it impossible for the capacitor voltage to follow the envelope (see Fig. 7.41b). Thus, *RC* should be large in comparison to 1/ω*c* but small in comparison to 1/2π*B*, where *B* is the highest frequency in *m*(*t*). Incidentally, these two conditions also require that ω*c* 2π*B*, a condition necessary for a well-defined envelope. - -The envelope-detector output *vC*(*t*) is *A* + *m*(*t*) plus a ripple of frequency ω*c*. The dc term *A* can be blocked out by a capacitor or a simple *RC* highpass filter. The ripple is reduced further by - -### 746 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -another (lowpass) *RC* filter. In the case of audio signals, the speakers also act as lowpass filters, which further enhances suppression of the high-frequency ripple. - -### **[7.7-3 Single-Sideband Modulation \(SSB\)](#page-13-0)** - -Now consider the baseband spectrum *M*(ω) (Fig. 7.42a) and the spectrum of the DSB-SC modulated signal *m*(*t*) cos ω*ct* (Fig. 7.42b). The DSB spectrum in Fig. 7.42b has two sidebands: the upper and the lower (USB and LSB), both containing complete information on *M*(ω) [see Eq. (7.12)]. Clearly, it is redundant to transmit both sidebands, a process that requires twice the bandwidth of the baseband signal. A scheme in which only one sideband is transmitted is known - -**Figure 7.42** Spectra for single-sideband transmission: **(a)** baseband, **(b)** DSB, **(c)** USB, **(d)** LSB, and **(e)** synchronously demodulated signal. - -as *single-sideband (SSB) transmission,* which requires only half the bandwidth of the DSB signal. Thus, we transmit only the upper sidebands (Fig. 7.42c) or only the lower sidebands (Fig. 7.42d). - -An SSB signal can be coherently (synchronously) demodulated. For example, multiplication of a USB signal (Fig. 7.42c) by 2 cos ω*ct* shifts its spectrum to the left and to the right by ω*c*, yielding the spectrum in Fig. 7.42e. Lowpass filtering of this signal yields the desired baseband signal. The case is similar with an LSB signal. Hence, demodulation of SSB signals is identical to that of DSB-SC signals, and the synchronous demodulator in Fig. 7.38a can demodulate SSB signals. Note that we are talking of SSB signals without an additional carrier. Hence, they are suppressed-carrier signals (SSB-SC). - -### **EXAMPLE 7.24 Single-Sideband Modulation** - -Find the USB (upper sideband) and LSB (lower sideband) signals when *m*(*t*) = cos ω*mt*. Sketch their spectra, and show that these SSB signals can be demodulated using the synchronous demodulator in Fig. 7.38a. - -The DSB-SC signal for this case is - -$$ -\varphi_{\text{DSB-SC}}(t) = m(t)\cos\omega_c t = \cos\omega_m t \cos\omega_c t \qquad = \frac{1}{2} [\cos(\omega_c - \omega_m)t + \cos(\omega_c + \omega_m)t] -$$ - -As pointed out in Ex. 7.21, the terms (1/2) cos(ω*c* + ω*m*)*t* and (1/2) cos(ω*c* − ω*m*)*t* represent the upper and lower sidebands, respectively. The spectra of the upper and lower sidebands are given in Figs. 7.43a and 7.43b. Observe that these spectra can be obtained from the DSB-SC spectrum in Fig. 7.37b by using a proper filter to suppress the undesired sidebands. For instance, the USB signal in Fig. 7.43a can be obtained by passing the DSB-SC signal (Fig. 7.37b) through a highpass filter of cutoff frequency ω*c*. Similarly, the LSB signal in Fig. 7.43b can be obtained by passing the DSB-SC signal through a lowpass filter of cutoff frequency ω*c*. - -If we apply the LSB signal (1/2) cos(ω*c* − ω*m*)*t* to the synchronous demodulator in Fig. 7.38a, the multiplier output is - -$$ -e(t) = \frac{1}{2}\cos{(\omega_c - \omega_m)t}\cos{\omega_c t} = \frac{1}{4}[\cos{\omega_m t} + \cos{(2\omega_c - \omega_m)t}] -$$ - -The term (1/4) cos(2ω*c*−ω*m*)*t* is suppressed by the lowpass filter, producing the desired output (1/4) cos ω*mt* [which is *m*(*t*)/4]. The spectrum of this term is π[δ(ω+ω*m*)+δ(ω−ω*m*)]/4, as depicted in Fig. 7.43c. In the same way, we can show that the USB signal can be demodulated by the synchronous demodulator. - -In the frequency domain, demodulation (multiplication by cos ω*ct*) amounts to shifting the LSB spectrum (Fig. 7.43b) to the left and the right by ω*c* (times 0.5) and then suppressing the high frequency, as illustrated in Fig. 7.43c. The resulting spectrum represents the desired signal (1/4)*m*(*t*). - -### GENERATION OF SSB SIGNALS - -Two methods are commonly used to generate SSB signals. The *selective-filtering method* uses sharp cutoff filters to eliminate the undesired sideband, and the second method uses phase-shifting networks to achieve the same goal [4].† We shall consider here only the first method. - -Selective filtering is the most commonly used method of generating SSB signals. In this method, a DSB-SC signal is passed through a sharp cutoff filter to eliminate the undesired sideband. - -To obtain the USB, the filter should pass all components above ω*c* unattenuated and completely suppress all components below ω*c*. Such an operation requires an ideal filter, which is unrealizable. It can, however, be realized closely if there is some separation between the passband and the stopband. Fortunately, the voice signal provides this condition, because its spectrum shows little power content at the origin (Fig. 7.44). Moreover, articulation tests show that for speech signals, frequency components below 300 Hz are not important. In other words, we may suppress all speech components below 300 Hz without appreciably affecting intelligibility.‡ Thus, filtering of the unwanted sideband becomes relatively easy for speech signals because we have a 600 Hz transition region around the cutoff frequency ω*c*. For some signals, which have considerable power - -† Yet another method, known as Weaver's method, is also used to generate SSB signals. - -‡ Similarly, suppression of components of a speech signal above 3500 Hz causes no appreciable change in intelligibility. - -at low frequencies (around ω = 0), SSB techniques cause considerable distortion. Such is the case with video signals. Consequently, for video signals, instead of SSB, we use another technique, the *vestigial sideband (VSB)*, which is a compromise between SSB and DSB. It inherits the advantages of SSB and DSB but avoids their disadvantages at a cost of slightly increased bandwidth. VSB signals are relatively easy to generate, and their bandwidth is only slightly (typically 25%) greater than that of SSB signals. In VSB signals, instead of rejecting one sideband completely (as in SSB), we accept a gradual cutoff from one sideband [4]. - -### **[7.7-4 Frequency-Division Multiplexing](#page-13-0)** - -Signal multiplexing allows transmission of several signals on the same channel. Later, in Ch. 8 (Sec. 8.2-2), we shall discuss time-division multiplexing (TDM), where several signals time-share the same channel, such as a cable or an optical fiber. In frequency-division multiplexing (FDM), the use of modulation, as illustrated in Fig. 7.45, makes several signals share the band of the same channel. Each signal is modulated by a different carrier frequency. The various carriers are adequately separated to avoid overlap (or interference) between the spectra of various modulated signals. These carriers are referred to as *subcarriers*. Each signal may use a different kind of modulation, for example, DSB-SC, AM, SSB-SC, VSB-SC, or even other forms of modulation, not discussed here [such as FM (frequency modulation) or PM (phase modulation)]. The modulated-signal spectra may be separated by a small guard band to avoid interference and to facilitate signal separation at the receiver. - -When all the modulated spectra are added, we have a composite signal that may be considered to be a new baseband signal. Sometimes, this composite baseband signal may be used to further modulate a high-frequency (radio frequency, or RF) carrier for the purpose of transmission. - -At the receiver, the incoming signal is first demodulated by the RF carrier to retrieve the composite baseband, which is then bandpass-filtered to separate the modulated signals. Then each modulated signal is individually demodulated by an appropriate subcarrier to obtain all the basic baseband signals. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/096_7.8 DATA TRUNCATION - WINDOW FUNCTIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/096_7.8 DATA TRUNCATION - WINDOW FUNCTIONS.md deleted file mode 100644 index e95e5b274759cb061c02bae40b7d5850e999c7c8..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/096_7.8 DATA TRUNCATION - WINDOW FUNCTIONS.md +++ /dev/null @@ -1,73 +0,0 @@ -## **7.8 DATA [TRUNCATION: WINDOW](#page-13-0) FUNCTIONS** - -We often need to truncate data in diverse situations from numerical computations to filter design. For example, if we need to compute numerically the Fourier transform of some signal, say, *e*−*t u*(*t*), we will have to truncate the signal *e*−*t u*(*t*) beyond a sufficiently large value of *t* (typically five time constants and above). The reason is that in numerical computations, we have to deal with - -**Figure 7.45** Frequency-division multiplexing: **(a)** FDM spectrum **(b)** transmitter, and **(c)** receiver. - -data of finite duration. Similarly, the impulse response *h*(*t*) of an ideal lowpass filter is noncausal and approaches zero asymptotically as |*t*|→∞. For a practical design, we may want to truncate *h*(*t*) beyond a sufficiently large value of |*t*| to make *h*(*t*) causal and of finite duration. In signal sampling, to eliminate aliasing, we must use an antialiasing filter to truncate the signal spectrum beyond the half-sampling frequency ω*s*/2. Again, we may want to synthesize a periodic signal by adding the first *n* harmonics and truncating all the higher harmonics. These examples show that data truncation can occur in both time and frequency domains. On the surface, truncation appears to be a simple problem of cutting off the data at a point at which values are deemed to be sufficiently small. Unfortunately, this is not the case. Simple truncation can cause some unsuspected problems. - -### WINDOW FUNCTIONS - -Truncation operation may be regarded as multiplying a signal of a large width by a window function of a smaller (finite) width. Simple truncation amounts to using a *rectangular window wR*(*t*) (shown later in Fig. 7.48a) in which we assign unit weight to all the data within the window width (|*t*| < *T*/2), and assign zero weight to all the data lying outside the window (|*t*| > *T*/2). It is also possible to use a window in which the weight assigned to the data within the window may not be constant. In a *triangular window wT* (*t*), for example, the weight assigned to data decreases linearly over the window width (shown later in Fig. 7.48b). - -Consider a signal *x*(*t*) and a window function *w*(*t*). If *x*(*t*) ⇐⇒ *X*(ω) and *w*(*t*) ⇐⇒ *W*(ω), and if the windowed function *xw*(*t*) ⇐⇒ *Xw*(ω), then - -$$ -x_w(t) = x(t)w(t) -$$ - and $X_w(\omega) = \frac{1}{2\pi}X(\omega)*W(\omega)$ - -According to the width property of convolution, it follows that the width of *Xw*(ω) equals the sum of the widths of *X*(ω) and *W*(ω). Thus, truncation of a signal increases its bandwidth by the amount of bandwidth of *w*(*t*). Clearly, the truncation of a signal causes its spectrum to spread (or smear) by the amount of the bandwidth of *w*(*t*). Recall that the signal bandwidth is inversely proportional to the signal duration (width). Hence, the wider the window, the smaller its bandwidth, and the smaller the *spectral spreading*. This result is predictable because a wider window means that we are accepting more data (closer approximation), which should cause smaller distortion (smaller spectral spreading). Smaller window width (poorer approximation) causes more spectral spreading (more distortion). In addition, since *W*(ω) is really not strictly bandlimited and its spectrum → 0 only asymptotically, the spectrum of *Xw*(ω) → 0 asymptotically also at the same rate as that of *W*(ω), even if *X*(ω) is, in fact, strictly bandlimited. Thus, windowing causes the spectrum of *X*(ω) to spread into the band where it is supposed to be zero. This effect is called *leakage*. The following example clarifies these twin effects of spectral spreading and leakage. - -Let us consider *x*(*t*) = cos ω0*t* and a rectangular window *wR*(*t*) = rect(*t*/*T*), illustrated in Fig. 7.46b. The reason for selecting a sinusoid for *x*(*t*) is that its spectrum consists of spectral lines of zero width (Fig. 7.46a). Hence, this choice will make the effect of spectral spreading and leakage easily discernible. The spectrum of the truncated signal *xw*(*t*) is the convolution of the two impulses of *X*(ω) with the sinc spectrum of the window function. Because the convolution of any function with an impulse is the function itself (shifted at the location of the impulse), the resulting spectrum of the truncated signal is 1/2π times the two sinc pulses at ±ω0, as depicted in Fig. 7.46c (also see Fig. 7.26). Comparison of spectra *X*(ω) and *Xw*(ω) reveals the effects of truncation. These are: - -1. The spectral lines of *X*(ω) have zero width. But the truncated signal is spread out by 2π/*T* about each spectral line. The amount of spread is equal to the width of the mainlobe of the window spectrum. One effect of this *spectral spreading* (or smearing) is that if *x*(*t*) has two spectral components of frequencies differing by less than 4π/*T* rad/s (2/*T* Hz), they - -**Figure 7.46** Windowing and its effects. - -will be indistinguishable in the truncated signal. The result is loss of spectral resolution. We would like the spectral spreading [mainlobe width of *W*(ω)] to be as small as possible. - -2. In addition to the mainlobe spreading, the truncated signal has sidelobes, which decay slowly with frequency. The spectrum of *x*(*t*) is zero everywhere except at ±ω0. On the other hand, the truncated signal spectrum *Xw*(ω) is zero nowhere because of the sidelobes. These sidelobes decay asymptotically as 1/ω. Thus, the truncation causes spectral *leakage* in the band where the spectrum of the signal *x*(*t*) is zero. The peak *sidelobe* magnitude is 0.217 times the mainlobe magnitude (13.3 dB below the peak mainlobe magnitude). Also, the sidelobes decay at a rate 1/ω, which is −6 dB/octave (or −20 dB/decade). This is the sidelobe's *rolloff rate*. We want smaller sidelobes with a faster rate of decay (high rolloff rate). Figure 7.46d, which plots |*WR*(ω)| as a function of ω, clearly shows the mainlobe and sidelobe features, with the first sidelobe amplitude −13.3 dB below the mainlobe amplitude and the sidelobes decaying at a rate of −6 dB/octave (or −20 dB/decade). - -So far, we have discussed the effect on the signal spectrum of signal truncation (truncation in the time domain). Because of the time-frequency duality, the effect of spectral truncation (truncation in frequency domain) on the signal shape is similar. - -### REMEDIES FOR SIDE EFFECTS OF TRUNCATION - -For better results, we must try to minimize the twin side effects of truncations: spectral spreading (mainlobe width) and leakage (sidelobe). Let us consider each of these ills. - -- 1. The spectral spread (mainlobe width) of the truncated signal is equal to the bandwidth of the window function *w*(*t*). We know that the signal bandwidth is inversely proportional to the signal width (duration). Hence, to reduce the spectral spread (mainlobe width), we need to increase the window width. -- 2. To improve the leakage behavior, we must search for the cause of the slow decay of sidelobes. In Ch. 6, we saw that the Fourier spectrum decays as 1/ω for a signal with jump discontinuity, decays as 1/ω2 for a continuous signal whose first derivative is discontinuous, and so on.† Smoothness of a signal is measured by the number of continuous derivatives it possesses. The smoother the signal, the faster the decay of its spectrum. Thus, we can achieve a given leakage behavior by selecting a suitably smooth (tapered) window. -- 3. For a given window width, the remedies for the two effects are incompatible. If we try to improve one, the other deteriorates. For instance, among all the windows of a given width, the rectangular window has the smallest spectral spread (mainlobe width), but its sidelobes have high level and they decay slowly. A tapered (smooth) window of the same width has smaller and faster decaying sidelobes, but it has a wider mainlobe.‡ But we can compensate for the increased mainlobe width by widening the window. Thus, we can remedy both the side effects of truncation by selecting a suitably smooth window of sufficient width. - -There are several well-known tapered-window functions, such as Bartlett (triangular), Hanning (von Hann), Hamming, Blackman, and Kaiser, which truncate the data gradually. These - -† This result was demonstrated for periodic signals. However, it applies to aperiodic signals also. This is because we showed in the beginning of this chapter that if *xT*0 (*t*) is a periodic signal formed by periodic extension of an aperiodic signal *x*(*t*), then the spectrum of *xT*0 (*t*) is (1/*T*0 times) the samples of *X*(ω). Thus, - -what is true of the decay rate of the spectrum of *xT*0 (*t*) is also true of the rate of decay of *X*(ω). ‡ A tapered window yields a higher mainlobe width because the effective width of a tapered window is smaller than that of the rectangular window; see Sec. 2.6-2 [Eq. (2.47)] for the definition of effective width. Therefore, from the reciprocity of the signal width and its bandwidth, it follows that the rectangular window mainlobe is narrower than a tapered window. - -| No. | Window w(t) | Mainlobe
Width | Rolloff
Rate
Level (dB) | Peak
Sidelobe | -|-----|---------------------------------------------------------------------|-------------------|-------------------------------|------------------| -| 1 | t
Rectangular: rect
T | 4π
T | −6 | −13.3 | -| 2 | t
Bartlett:
2T | 8π
T | −12 | −26.5 | -| 3 | 2πt
!
Hanning: 0.5
1+cos
T | 8π
T | −18 | −31.5 | -| 4 | 2πt
Hamming: 0.54+0.46 cos
T | 8π
T | −6 | −42.7 | -| 5 | 2πt
4πt
Blackman: 0.42+0.5 cos
+0.08 cos
T
T | 12π
T | −18 | −58.1 | -| 6 |
2
t
I0
α
1−4
T
Kaiser:
0 ≤ α ≤ 10
I0(α) | 11.2π
T | −6 | −59.9 | -| | | | (α = 8.168) | | - -**TABLE 7.3** Some Window Functions and Their Characteristics - -**Figure 7.47 (a)** Hanning and **(b)** Hamming windows. - -windows offer different trade-offs with respect to spectral spread (mainlobe width), the peak sidelobe magnitude, and the leakage rolloff rate, as indicated in Table 7.3 [5, 6]. Observe that all windows are symmetrical about the origin (i.e., are even functions of *t*). Because of this feature, *W*(ω) is a real function of ω; that is, *W*(ω) is either 0 or π. Hence, the phase function of the truncated signal has a minimal amount of distortion. - -Figure 7.47 shows two well-known tapered-window functions, the von Hann (or Hanning) window *w*Han(*x*) and the Hamming window *w*Ham(*x*). We have intentionally used the independent variable *x* because windowing can be performed in the time domain as well as in the frequency domain, so *x* could be *t* or ω, depending on the application. - -There are hundreds of windows, all with different characteristics. But the choice depends on a particular application. The rectangular window has the narrowest mainlobe. The Bartlett (triangle) window (also called the Fejer or Cesaro) is inferior in all respects to the Hanning window. For this reason, it is rarely used in practice. Hanning is preferred over Hamming in spectral analysis because it has faster sidelobe decay. For filtering applications, on the other hand, the Hamming window is chosen because it has the smallest sidelobe magnitude for a given mainlobe width. The Hamming window is the most widely used general-purpose window. The Kaiser window, which uses *I*0(α), the modified zero-order Bessel function, is more versatile and adjustable. Selecting a proper value of α (0 ≤ α ≤ 10) allows the designer to tailor the window to suit a particular application. The parameter α controls the mainlobe-sidelobe trade-off. When α = 0, the Kaiser window is the rectangular window. For α = 5.4414, it is the Hamming window, and when α = 8.885, it is the Blackman window. As α increases, the mainlobe width increases and the sidelobe level decreases. - -### **[7.8-1 Using Windows in Filter Design](#page-13-0)** - -We shall design an ideal lowpass filter of bandwidth *W* rad/s, with frequency response *H*(ω), as shown in Fig. 7.48e or Fig. 7.48f. For this filter, the impulse response *h*(*t*) = (*W*/π )sinc (*Wt*) (Fig. 7.48c) is noncausal and, therefore, unrealizable. Truncation of *h*(*t*) by a suitable window (Fig. 7.48a) makes it realizable, although the resulting filter is now an approximation to the desired ideal filter.† We shall use a rectangular window *wR*(*t*) and a triangular (Bartlett) window *wT* (*t*) to truncate *h*(*t*), and then examine the resulting filters. The truncated impulse responses *hR*(*t*) = *h*(*t*)*wR*(*t*) and *hT* (*t*) = *h*(*t*)*wT* (*t*) are depicted in Fig. 7.48d. Hence, the windowed filter frequency response is the convolution of *H*(ω) with the Fourier transform of the window, as illustrated in Figs. 7.48e and 7.48f. We make the following observations. - -- 1. The windowed filter spectra show *spectral spreading* at the edges, and instead of a sudden switch there is a gradual transition from the passband to the stopband of the filter. The transition band is smaller (2π/*T* rad/s) for the rectangular case than for the triangular case (4π/*T* rad/s). -- 2. Although *H*(ω) is bandlimited, the windowed filters are not. But the stopband behavior of the triangular case is superior to that of the rectangular case. For the rectangular window, the leakage in the stopband decreases slowly (as 1/ω) in comparison to that of the triangular window (as 1/ω2). Moreover, the rectangular case has a higher peak sidelobe amplitude than that of the triangular window. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/097_7.9 MATLAB - FOURIER TRANSFORM TOPICS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/097_7.9 MATLAB - FOURIER TRANSFORM TOPICS.md deleted file mode 100644 index f3bcd37a42b8818601a90f236a9efd64bbf0c287..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/097_7.9 MATLAB - FOURIER TRANSFORM TOPICS.md +++ /dev/null @@ -1,204 +0,0 @@ -## **[7.9 MATLAB: FOURIER](#page-13-0) TRANSFORM TOPICS** - -MATLAB is useful for investigating a variety of Fourier transform topics. In this section, a rectangular pulse is used to investigate the scaling property, Parseval's theorem, essential bandwidth, and spectral sampling. Kaiser window functions are also investigated. - -† In addition to truncation, we need to delay the truncated function by *T*/2 to render it causal. However, the time delay only adds a linear phase to the spectrum without changing the amplitude spectrum. Thus, to simplify our discussion, we shall ignore the delay. - -**Figure 7.48** Window-based filter design. - -### **[7.9-1 The Sinc Function and the Scaling Property](#page-13-0)** - -As shown in Ex. 7.2, the Fourier transform of *x*(*t*) = rect(*t*/τ ) is *X*(ω) = τ sinc (ωτ/2). To represent *X*(ω) in MATLAB, a sinc function is first required. As an alternative to the signal processing toolbox function sinc, which computes sinc(*x*) as sin(π*x*)/π*x*, we create our own function that follows the conventions of this book and defines sinc(*x*) = sin(*x*)/*x*. - -function [y] = CH7MP1(x) % CH7MP1.m : Chapter 7, MATLAB Program 1 % Function M-file computes the sinc function, y = sin(x)/x. - -y(x==0) = 1; y(x~=0) = sin(x(x~=0))./x(x~=0); - -The computational simplicity of sinc (*x*) = sin(*x*)/*x* is somewhat deceptive: sin(0)/0 results in a divide-by-zero error. Thus, program CH7MP1 assigns sinc (0) = 1 and computes the remaining values according to the definition. Notice that CH7MP1 cannot be directly replaced by an anonymous function. Anonymous functions cannot have multiple lines or contain certain commands such as =, if, or for. M-files, however, can be used to define an anonymous function. For example, we can represent *X*(ω) as an anonymous function that is defined in terms of CH7MP1. - -``` ->> X = @(omega,tau) tau*CH7MP1(omega*tau/2); -``` - -Once we have defined *X*(ω), it is simple to investigate the effects of scaling the pulse width τ . Consider the three cases τ = 1.0, τ = 0.5, and τ = 2.0. - -``` ->> omega = linspace(-4*pi,4*pi,200); ->> plot(omega,X(omega,1),'k-',omega,X(omega,0.5),'k-.',omega,X(omega,2),'k--'); ->> grid; axis tight; xlabel('\omega'); ylabel('X(\omega)'); ->> legend('Baseline (\tau = 1)','Compressed (\tau = 0.5)',... ->> 'Expanded (\tau = 2.0)'); -``` - -Figure 7.49 confirms the reciprocal relationship between signal duration and spectral bandwidth: time compression causes spectral expansion, and time expansion causes spectral compression. Additionally, spectral amplitudes are directly related to signal energy. As a signal is compressed, signal energy and thus spectral magnitude decrease. The opposite effect occurs when the signal is expanded. - -**Figure 7.49** Spectra *X*(ω) = τ sinc (ωτ/2) for τ = 1.0, τ = 0.5, and τ = 2.0. - -### **[7.9-2 Parseval's Theorem and Essential Bandwidth](#page-13-0)** - -Parseval's theorem concisely relates energy between the time domain and the frequency domain: - -$$ -\int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega -$$ - -This too is easily verified with MATLAB. For example, a unit amplitude pulse *x*(*t*) with duration τ has energy *Ex* = τ . Thus, - -$$ -\int_{-\infty}^{\infty} |X(\omega)|^2 \, d\omega = 2\pi \, \tau -$$ - -Letting τ = 1, the energy of *X*(ω) is computed by using the quad function. - -``` ->> X_squared = @(omega, tau) (tau*CH7MP1(omega*tau/2)).^2; ->> quad(X_squared,-1e6,1e6,[],[],1) - ans = 6.2817 -``` - -Although not perfect, the result of the numerical integration is consistent with the expected value of 2π ≈ 6.2832. For quad, the first argument is the function to be integrated, the next two arguments are the limits of integration, the empty square brackets indicate default values for special options, and the last argument is the secondary input τ for the anonymous function X\_squared. Full format details for quad are available from MATLAB's help facilities. - -A more interesting problem involves computing a signal's essential bandwidth. Consider, for example, finding the essential bandwidth *W*, in radians per second, that contains fraction β of the energy of the square pulse *x*(*t*). That is, we want to find *W* such that - -$$ -\frac{1}{2\pi} \int_{-W}^{W} |X(\omega)|^2 d\omega = \beta \tau -$$ - -Program CH7MP2 uses a guess-and-check method to find *W*. - -``` -function [W,E_W] = CH7MP2(tau,beta,tol) -% CH7MP2.m : Chapter 7, MATLAB Program 2 -% Function M-file computes essential bandwidth W for square pulse. -% INPUTS: tau = pulse width -% beta = fraction of signal energy desired in W -% tol = tolerance of relative energy error -% OUTPUTS: W = essential bandwidth [rad/s] -% E_W = Energy contained in bandwidth W -W = 0; step = 2*pi/tau; % Initial guess and step values -X_squared = @(omega,tau) (tau*CH7MP1(omega*tau/2)).^2; -E = beta*tau; % Desired energy in W -relerr = (E-0)/E; % Initial relative error is 100 percent -while(abs(relerr) > tol), - if (relerr>0), % W too small, so... - W=W+step; % ... increase W by step - elseif (relerr<0), % W too large, so... - step = step/2; % ... decrease step and then W -``` - -``` -W = W-step; - end - E_W = 1/(2*pi)*quad(X_squared,-W,W,[],[],tau); - relerr = (E - E_W)/E; -end -``` - -Although this guess-and-check method is not the most efficient, it is relatively simple to understand: CH7MP2 sensibly adjusts *W* until the relative error is within tolerance. The number of iterations needed to converge to a solution depends on a variety of factors and is not known beforehand. The while command is ideal for such situations: - -``` -while expression, -``` - -*statements;* - -end - -While the *expression* is true, the *statements* are continually repeated. - -To demonstrate CH7MP2, consider the 90% essential bandwidth *W* for a pulse of 1 second duration. Typing [W,E\_W]=CH7MP2(1,0.9,0.001) returns an essential bandwidth *W* = 5.3014 that contains 89.97% of the energy. Reducing the error tolerance improves the estimate. CH7MP2(1,0.9,0.00005) returns an essential bandwidth *W* = 5.3321 that contains 90.00% of the energy. These essential bandwidth calculations are consistent with estimates presented after Ex. 7.2. - -### **[7.9-3 Spectral Sampling](#page-13-0)** - -Consider a signal with finite duration τ . A periodic signal *xT*0 (*t*) is constructed by repeating *x*(*t*) every *T*0 seconds, where *T*0 ≥ τ . From Eq. (7.5), we can write the Fourier series coefficients of *xT*0 (*t*) as *Dn* = (1/*T*0)*X*(*n*2π/*T*0). Put another way, the Fourier series coefficients are obtained by sampling the spectrum *X*(ω). - -By using spectral sampling, it is simple to determine the Fourier series coefficients for an arbitrary duty-cycle, square-pulse periodic signal. The square pulse *x*(*t*) = rect(*t*/τ ) has spectrum *X*(ω) = τ sinc(ωτ /2). Thus, the *n*th Fourier coefficient of the periodic extension *xT*0 (*t*) is *Dn* = (τ/*T*0)sinc (*n*πτ/*T*0). As in Ex. 6.4, τ = π and *T*0 = 2π provide a square-pulse periodic signal. The Fourier coefficients are determined by - -``` ->> tau = pi; T_0 = 2*pi; n = [0:10]; -``` - ->> D\_n = tau/T\_0\*MS7P1(n\*pi\*tau/T\_0); - -``` ->> stem(n,D_n); xlabel('n'); ylabel('D_n'); -``` - ->> axis([-0.5 10.5 -0.2 0.55]); - -The results, shown in Fig. 7.50, agree with Fig. 6.6b. Doubling the period to *T*0 = 4π effectively doubles the density of spectral samples and halves the spectral amplitude, as shown in Fig. 7.51. - -As *T*0 increases, the spectral sampling becomes progressively finer while the amplitude becomes infinitesimal. An evolution of the Fourier series toward the Fourier integral is seen by allowing the period *T*0 to become large. Figure 7.52 shows the result for *T*0 = 40π. - -If *T*0 = τ , the signal *xT*0 is a constant and the spectrum should concentrate energy at dc. In this case, the sinc function is sampled at the zero crossings and *Dn* = 0 for all *n* not equal to 0. Only the sample corresponding to *n* = 0 is nonzero, indicating a dc signal, as expected. It is a simple matter to modify the previous code to verify this case. - -**Figure 7.50** Fourier spectra for τ = π and *T*0 = 2π. - -**Figure 7.51** Fourier spectra for τ = π and *T*0 = 4π. - -**Figure 7.52** Fourier spectra for τ = π and *T*0 = 40π. - -### **[7.9-4 Kaiser Window Functions](#page-13-0)** - -A window function is useful only if it can be easily computed and applied to a signal. The Kaiser window, for example, is flexible but appears rather intimidating: - -$$ -w_K(t) = \begin{cases} \frac{I_0(\alpha\sqrt{1 - 4(t/T)^2})}{I_0(\alpha)} & |t| < T/2\\ 0 & \text{otherwise} \end{cases} -$$ - -Fortunately, the bark of a Kaiser window is worse than its bite! The function *I*0(*x*), a zero-order modified Bessel function of the first kind, can be computed according to - -$$ -I_0(x) = \sum_{k=0}^{\infty} \left(\frac{x^k}{2^k k!}\right)^2 -$$ - -or, more simply, by using the MATLAB function besseli(0,x). In fact, MATLAB supports a wide range of Bessel functions, including Bessel functions of the first and second kinds (besselj and bessely), modified Bessel functions of the first and second kinds (besseli and besselk), Hankel functions (besselh), and Airy functions (airy). - -Program CH7MP3 computes Kaiser windows at times *t* by using parameters *T* and α. - -``` -function [w_K] = CH7MP3(t,T,alpha) -% CH7MP3.m : Chapter 7, MATLAB Program 3 -% Function M-file computes a width-T Kaiser window using parameter alpha. -% Alpha can also be a string identifier: 'rectangular', 'Hamming', or -% 'Blackman'. -% INPUTS: t = independent variable of the window function -% T = window width -% alpha = Kaiser parameter or string identifier -% OUTPUTS: w_K = Kaiser window function -if strncmpi(alpha,'rectangular',1), - alpha = 0; -elseif strncmpi(alpha,'Hamming',3), - alpha = 5.4414; -elseif strncmpi(alpha,'Blackman',1), - alpha = 8.885; -elseif isa(alpha,'char') - disp('Unrecognized string identifier.'); return -end -w_K = zeros(size(t)); i = find(abs(t)**Figure 7.53** Special-case, unit-duration Kaiser windows. - -Figure 7.53 shows the three special-case, unit-duration Kaiser windows generated by - -``` ->> t = [-0.6:.001:0.6]; T = 1; -``` - -``` ->> plot(t,CH7MP3(t,T,'r'),'k-',t,CH7MP3(t,T,'ham'),'k-.',t,CH7MP3(t,T,'b'),'k--'); -``` - -``` ->> axis([-0.6 0.6 -.1 1.1]); xlabel('t'); ylabel('w_K(t)'); -``` - -``` ->> legend('Rectangular','Hamming','Blackman','Location','EastOutside'); -``` diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/098_7.10 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/098_7.10 SUMMARY.md deleted file mode 100644 index 34ba617620609862eda4e3cc5fc3d1ec64369a08..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/098_7.10 SUMMARY.md +++ /dev/null @@ -1,15 +0,0 @@ -## **[7.10 SUMMARY](#page-13-0)** - -In Ch. 6, we represented periodic signals as a sum of (everlasting) sinusoids or exponentials (Fourier series). In this chapter we extended this result to aperiodic signals, which are represented by the Fourier integral (instead of the Fourier series). An aperiodic signal *x*(*t*) may be regarded as a periodic signal with period *T*0 → ∞ so that the Fourier integral is basically a Fourier series with a fundamental frequency approaching zero. Therefore, for aperiodic signals, the Fourier spectra are continuous. This continuity means that a signal is represented as a sum of sinusoids (or exponentials) of all frequencies over a continuous frequency interval. The Fourier transform *X*(ω), therefore, is the spectral density (per unit bandwidth in hertz). - -An ever-present aspect of the Fourier transform is the duality between time and frequency, which also implies duality between the signal *x*(*t*) and its transform *X*(ω). This duality arises because of near-symmetrical equations for direct and inverse Fourier transforms. The duality principle has far-reaching consequences and yields many valuable insights into signal analysis. - -The scaling property of the Fourier transform leads to the conclusion that the signal bandwidth is inversely proportional to signal duration (signal width). Time shifting of a signal does not change its amplitude spectrum, but it does add a linear phase component to its spectrum. Multiplication of a signal by an exponential *ej*ω0*t* shifts the spectrum to the right by ω0. In practice, spectral shifting is achieved by multiplying a signal by a sinusoid such as cosω0*t* (rather than the exponential *ej*ω0*t* ). This process is known as amplitude modulation. Multiplication of two signals results in convolution of their spectra, whereas convolution of two signals results in multiplication of their spectra. - -For an LTIC system with the frequency response *H*(ω), the input and output spectra *X*(ω) and *Y*(ω) are related by the equation *Y*(ω) = *X*(ω)*H*(ω). This is valid only for asymptotically stable systems. It also applies to marginally stable systems if the input does not contain a finite-amplitude sinusoid of the natural frequency of the system. For asymptotically unstable systems, the frequency response *H*(ω) does not exist. For distortionless transmission of a signal through an LTIC system, the amplitude response |*H*(ω)| of the system must be constant, and the phase response *H*(ω) should be a linear function of ω over a band of interest. Ideal filters, which allow distortionless transmission of a certain band of frequencies and suppress all the remaining frequencies, are physically unrealizable (noncausal). In fact, it is impossible to build a physical system with zero gain [*H*(ω) = 0] over a finite band of frequencies. Such systems (which include ideal filters) can be realized only with infinite time delay in the response. - -The energy of a signal *x*(*t*) is equal to 1/2π times the area under |*X*(ω)2| (Parseval's theorem). The energy contributed by spectral components within a band *f* (in hertz) is given by |*X*(ω)| 2*f* . Therefore, |*X*(ω)| 2 is the energy spectral density per unit bandwidth (in hertz). - -The process of modulation shifts the signal spectrum to different frequencies. Modulation is used for many reasons: to transmit several messages simultaneously over the same channel for the sake of utilizing channel's high bandwidth, to effectively radiate power over a radio link, to shift a signal spectrum at higher frequencies to overcome the difficulties associated with signal processing at lower frequencies, and to effect the exchange of transmission bandwidth and transmission power required to transmit data at a certain rate. Broadly speaking, there are two types of modulation, amplitude and angle modulation. Each class has several subclasses. - -In practice, we often need to truncate data. Truncating is like viewing data through a window, which permits only certain portions of the data to be seen and hides (suppresses) the remainder. Abrupt truncation of data amounts to a rectangular window, which assigns a unit weight to data seen from the window and zero weight to the remaining data. Tapered windows, on the other hand, reduce the weight gradually from 1 to 0. Data truncation can cause some unsuspected problems. For example, in computation of the Fourier transform, windowing (data truncation) causes spectral spreading (spectral smearing) that is characteristic of the window function used. A rectangular window results in the least spreading, but it does so at the cost of a high and oscillatory spectral leakage outside the signal band, which decays slowly as 1/ω. In comparison to a rectangular window, tapered windows, in general, have larger spectral spreading (smearing), but the spectral leakage is smaller and decays faster with frequency. If we try to reduce spectral leakage by using a smoother window, the spectral spreading increases. Fortunately, spectral spreading can be reduced by increasing the window width. Therefore, we can achieve a given combination of spectral spread (transition bandwidth) and leakage characteristics by choosing a suitable tapered window function of a sufficiently long width *T*. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/099_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/099_REFERENCES.md deleted file mode 100644 index 66a9db6cce35e974a02c17f8ecff841f299b3804..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/099_REFERENCES.md +++ /dev/null @@ -1,8 +0,0 @@ -### **[REFERENCES](#page-13-0)** - -- 1. Churchill, R. V., and Brown, J. W. *Fourier Series and Boundary Value Problems,* 3rd ed. McGraw-Hill, New York, 1978. -- 2. Bracewell, R. N. *Fourier Transform and Its Applications,* rev. 2nd ed. McGraw-Hill, New York, 1986. -- 3. Guillemin, E. A. *Theory of Linear Physical Systems*. Wiley, New York, 1963. -- 4. Lathi, B. P. *Modern Digital and Analog Communication Systems,* 3rd ed. Oxford University Press, New York, 1998. -- 5. Hamming, R. W. *Digital Filters,* 2nd ed. Prentice-Hall, Englewood Cliffs, NJ, 1983. -- 6. Harris, F. J. On the use of windows for harmonic analysis with the discrete Fourier transform. *Proceedings of the IEEE,* vol. 66, no. 1, pp. 51–83, January 1978. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/100_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/100_PROBLEMS.md deleted file mode 100644 index 3a3cf0616c229948217aed1d11954f384f4e2a36..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/100_PROBLEMS.md +++ /dev/null @@ -1,440 +0,0 @@ -## **[PROBLEMS](#page-13-0)** - -- **7.1-1** Suppose signal *x*(*t*) = *t* 2 [*u*(*t*)−*u*(*t* −2)] has Fourier transform *X*(ω). Define a 3-periodic replication of *x*(*t*) as *y*(*t*) = %∞ *n*=−∞ 2*x*(*t* − 1 − 3*n*). Determine *Yk*, the Fourier series of *y*(*t*), in terms of the Fourier transform *X*(·). -- **7.1-2** Show that for a real *x*(*t*), Eq. (7.10) can be expressed as - -$$ -x(t) = \frac{1}{\pi} \int_0^\infty |X(\omega)| \cos[\omega t + \angle X(\omega)] d\omega -$$ - -This is the trigonometric form of the Fourier integral. Compare this with the compact trigonometric Fourier series. - -**7.1-3** Show that if *x*(*t*) is an even function of *t*, then - -$$ -X(\omega) = 2 \int_0^\infty x(t) \cos \omega t \, dt -$$ - -and if *x*(*t*) is an odd function of *t*, then - -$$ -X(\omega) = -2j \int_0^\infty x(t) \sin \omega t \, dt -$$ - -Hence, prove that if *x*(*t*) is a real and even function of *t*, then *X*(ω) is a real and even function of ω. In addition, if *x*(*t*) is a real and odd function of *t*, then *X*(ω) is an imaginary and odd function of ω. - -**7.1-4** A signal *x*(*t*) can be expressed as the sum of even and odd components (see Sec. 1.5-2): - -*x*(*t*) = *xe*(*t*) +*xo*(*t*) - -(a) If *x*(*t*) ⇐⇒ *X*(ω), show that for real *x*(*t*), - -*xe*(*t*) ⇐⇒ Re[*X*(ω)] - -and - -$$ -x_o(t) \Longleftrightarrow j \operatorname{Im}[X(\omega)] -$$ - -- (b) Verify these results by finding the Fourier transforms of the even and odd components of the following signals: **(i)** *u*(*t*) and **(ii)** *e*−*atu*(*t*). -- **7.1-5** Using Eq. (7.9), find the Fourier transforms of the signals *x*(*t*) in Fig. P7.1-5. -- **7.1-6** Using Eq. (7.9), find the Fourier transforms of the signals depicted in Fig. P7.1-6. -- **7.1-7** Use Eq. (7.10) to find the inverse Fourier transforms of the spectra in Fig. P7.1-7. - -**Figure P7.1-6** - -**Figure P7.1-8** - -- **7.1-8** Use Eq. (7.10) to find the inverse Fourier transforms of the spectra in Fig. P7.1-8. -- **7.1-9** If *x*(*t*) ⇐⇒ *X*(ω), then show that - -$$ -X(0) = \int_{-\infty}^{\infty} x(t) dt -$$ - -and - -$$ -x(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) d\omega -$$ - -Also show that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc}(x) dx = \int_{-\infty}^{\infty} \operatorname{sinc}^{2}(x) dx = \pi -$$ - -**Figure P7.2-4** - -- **7.2-1** Sketch the following functions: (a) rect(*t*/2) - - (b) (3ω/100) - - (c) rect((*t* −10)/8) - - (d) sinc(πω/5) - - (e) sinc ((ω/5)−2π) - - (f) sinc (*t*/5)rect(*t*/10π ) -- **7.2-2** Using Eq. (7.9), show that the Fourier transform of rect(*t* − 5) is sinc(ω/2)*e*−*j*5ω. Sketch the resulting amplitude and phase spectra. -- **7.2-3** Using Eq. (7.10), show that the inverse Fourier transform of rect((ω −10)/2π ) is sinc(π*t*) *ej*10*t* . -- **7.2-4** Find the inverse Fourier transform of *X*(ω) for the spectra illustrated in Fig. P7.2-4. [*Hint*: - -*X*(ω) = |*X*(ω)|*ej X*(ω). This problem illustrates how different phase spectra (both with the same amplitude spectrum) represent entirely different signals.] - -- **7.2-5** (a) Can you find the Fourier transform of *eatu*(*t*) when *a*>1 by setting *s* = *j*ω in the Laplace transform of *eatu*(*t*)? Explain. - - (b) Find the Laplace transform of *x*(*t*) shown in Fig. P7.2-5. Can you find the Fourier transform of *x*(*t*) by setting *s* = *j*ω in its Laplace transform? Explain. Verify your answer by finding the Fourier and the Laplace transforms of *x*(*t*). - -**Figure P7.2-5** - -- **7.3-1** Apply the duality property to the appropriate pair in Table 7.1 to show that - - (a) 1 2 [δ(*t*)+*j*/π*t*] ⇐⇒ *u*(ω) - - (b) δ(*t* +*T*) +δ(*t* −*T*) ⇐⇒ 2 cos *T*ω - - (c) δ(*t* +*T*)−δ(*t* −*T*) ⇐⇒ 2*j*sin *T*ω - -- **7.3-2** A signal *x*(*t*) has Fourier transform *X*(ω). Determine the Fourier transform *Y*(ω) in terms of *X*(ω) for each of the following signals *y*(*t*): (a) *y*(*t*) = 1 5 *x*(−2*t* +3) (b) *y*(*t*) = *ej*2*t x*∗(−3*t* −6) -- **7.3-3** A signal *x*(*t*) has Fourier transform *X*(ω). Determine the inverse Fourier transform *y*(*t*) in terms of *x*(*t*) for each of the following spectra *Y*(ω), - -(a) -$$ -Y(\omega) = \frac{4}{3}e^{-j2\omega/3}X(-\omega/3) -$$ - -(b) -$$ -Y(\omega) = \frac{1}{3}e^{j2(\omega - 2)}X^*\left(\frac{\omega - 2}{3}\right) -$$ - -**7.3-4** The Fourier transform of the triangular pulse *x*(*t*) in Fig. P7.3-4 is expressed as - -$$ -X(\omega) = \frac{1}{\omega^2} (e^{j\omega} - j\omega e^{j\omega} - 1) -$$ - -Use this information, and the time-shifting and time-scaling properties, to find the Fourier transforms of the signals *xi*(*t*)(*i* = 1, 2, 3, 4, 5) shown in Fig. P7.3-4. - -- **7.3-5** Using only the time-shifting property and Table 7.1, find the Fourier transforms of the signals depicted in Fig. P7.3-5. -- **7.3-6** Consider the fact that the τ -duration triangle function ω τ has inverse Fourier transform - -**Figure P7.3-4** - -2 0 234 0 - - -(a) (b) - -4 -3 - - -**Figure P7.3-7** - --4 -3 - - -τ 4π sinc2 *t*τ 4 . Use the duality property to determine the Fourier transform *Y*(ω) of signal *y*(*t*) = (*t*). - -**7.3-7** Use the time-shifting property to show that if *x*(*t*) ⇐⇒ *X*(ω), then - -$$ -x(t+T) + x(t-T) \Longleftrightarrow 2X(\omega)\cos T\omega -$$ - -This is the dual of Eq. (7.32). Use this result and Table 7.1 to find the Fourier transforms of the signals shown in Fig. P7.3-7. - -**7.3-8** Prove the following results, which are duals of each other: - -$$ -x(t)\sin \omega_0 t \Longleftrightarrow \frac{1}{2j}[X(\omega - \omega_0) - X(\omega + \omega_0)] -$$ - -$$ -\frac{1}{2j}[x(t+T) - x(t-T)] \Longleftrightarrow X(\omega)\sin T\omega -$$ - -Use the latter result and Table 7.1 to find the Fourier transform of the signal in Fig. P7.3-8. - -2 234 - -- **7.3-9** The signals in Fig. P7.3-9 are modulated signals with carrier cos 10*t*. Find the Fourier transforms of these signals by using the appropriate properties of the Fourier transform and Table 7.1. Sketch the amplitude and phase spectra for Figs. P7.3-9a and P7.3-9b. -- **7.3-10** Use the frequency-shifting property and Table 7.1 to find the inverse Fourier transform of the spectra depicted in Fig. P7.3-10. -- **7.3-11** Let *X*(ω) = rect(ω) be the Fourier transform of a signal *x*(*t*). - - (a) For *y*a(*t*) = *x*(*t*) ∗ *x*(*t*), sketch *Y*a(ω). - - (b) For *y*b(*t*) = *x*(*t*) ∗ *x*(*t*/2), sketch *Y*b(ω). - -(a) (b) - -**Figure P7.3-10** - -- (c) For *y*c(*t*) = 2*x*(*t*), sketch *Y*c(ω). -- (d) For *y*d(*t*) = *x*2(*t*), sketch *Y*d(ω). -- (e) For *y*e(*t*) = 1−*x*2(*t*), sketch *Y*e(ω). -- **7.3-12** Use the time-convolution property to prove pairs 2, 4, 13, and 14 in Table 2.1 (assume λ < 0 in pair 2, λ1 and λ2 < 0 in pair 4, λ1 < 0 and λ2 > 0 in pair 13, and λ1 and λ2 > 0 in pair 14). These restrictions are placed because of the Fourier transformability issue for the signals concerned. For pair 2, you need to apply the result in Eq. (1.10). -- **7.3-13** A signal *x*(*t*) is bandlimited to *B* Hz. Show that the signal *xn*(*t*) is bandlimited to *nB* Hz. -- **7.3-14** Find the Fourier transform of the signal in Fig. P7.3-5a by three different methods: - - (a) By direct integration using Eq. (7.9). - - (b) Using only pair 17 (Table 7.1) and the time-shifting property. - - (c) Using the time-differentiation and time-shifting properties, along with the fact that δ(*t*) ⇐⇒ 1. - -**7.3-15** (a) Prove the frequency-differentiation property (dual of the time-differentiation property): - -$$ --jtx(t) \Longleftrightarrow \frac{d}{d\omega}X(\omega) -$$ - -- (b) Use this property and pair 1 (Table 7.1) to determine the Fourier transform of *te*−*atu*(*t*). -- **7.3-16** Adapt the method of Ex. 7.17 and use the frequency-differentiation (see Prob. 7.3-15) and other properties to find the inverse Fourier transform *x*(*t*) of the triangular spectrum *X*(ω) = (ω/2). -- **7.3-17** Adapt the method of Ex. 7.17 and use the frequency-differentiation (see Prob. 7.3-15) and other properties to find the inverse Fourier transform *x*(*t*) of the spectrum *X*(ω) = π ω 2 rect ω 4 . -- **7.4-1** For a stable LTIC system with transfer function - -$$ -H(s) = \frac{1}{s+1} -$$ - -find the (zero-state) response if the input *x*(*t*) is - -- (a) *e*−2*t u*(*t*) -- (b) *e*−*t u*(*t*) -- (c) *et u*(−*t*) -- (d) *u*(*t*) -- **7.4-2** A stable LTIC system is specified by the frequency response - -$$ -H(\omega) = \frac{-1}{j\omega - 2} -$$ - -Find the impulse response of this system and show that this is a noncausal system. Find the (zero-state) response of this system if the input *x*(*t*) is - -- (a) *e*−*t u*(*t*) -- (b) *et u*(−*t*) -- **7.4-3** A periodic signal *x*(*t*) = 1 + 2 cos(5π*t*) + 3 sin(8π*t*) is applied to an LTIC system with impulse response *h*(*t*) = 8sinc(4*t*) cos(2π*t*) to produce output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Determine ω0, the fundamental radian frequency of *x*(*t*). - - (b) Determine *X*(ω), the Fourier transform of *x*(*t*). - - (c) Sketch the system's magnitude response |*H*(ω)| over −10π ≤ ω ≤ 10π. - - (d) Is the system *h*(*t*) distortionless? Explain. - - (e) Determine *y*(*t*). -- **7.4-4** A periodic delta train *x*(*t*) = %∞ *n*=−∞ δ(*t* − π*n*) is applied to an LTIC system with impulse response *h*(*t*) = sin(3*t*)sinc2 *t* π to produce zero-state output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Determine ω0, the fundamental radian frequency of *x*(*t*). - - (b) Determine *X*(ω), the Fourier transform of *x*(*t*). - - (c) Sketch the system's magnitude response |*H*(ω)| over −10π ≤ ω ≤ 10π. - - (d) Is the system *h*(*t*) distortionless? Explain. - - (e) Determine *y*(*t*). -- **7.4-5** Signals *x*1(*t*)=104rect(104*t*) and *x*2(*t*) = δ(*t*) are applied at the inputs of the ideal lowpass filters *H*1(ω) = rect(ω/40,000π ) and *H*2(ω) = rect(ω/20,000π ) (Fig. P7.4-5). The outputs *y*1(*t*) and *y*2(*t*) of these filters are multiplied to obtain the signal *y*(*t*) = *y*1(*t*)*y*2(*t*). (a) Sketch *X*1(ω) and *X*2(ω). - -- (b) Sketch *H*1(ω) and *H*2(ω). -- (c) Sketch *Y*1(ω) and *Y*2(ω). -- (d) Find the bandwidths of *y*1(*t*), *y*2(*t*), and *y*(*t*). - -- **7.4-6** A lowpass system time constant is often defined as the width of its unit impulse response *h*(*t*) (see Sec. 2.6-2). An input pulse *p*(*t*) to this system acts like an impulse of strength equal to the area of *p*(*t*) if the width of *p*(*t*) is much smaller than the system time constant, and provided *p*(*t*) is a lowpass pulse, implying that its spectrum is concentrated at low frequencies. Verify this behavior by considering a system whose unit impulse response is *h*(*t*) = rect(*t*/10−3). The input pulse is a triangle pulse *p*(*t*) = (*t*/10−6). Show that the system response to this pulse is very nearly the system response to the input *A*δ(*t*), where *A* is the area under the pulse *p*(*t*). -- **7.4-7** A lowpass system time constant is often defined as the width of its unit impulse response *h*(*t*) (see Sec. 2.6-2). An input pulse *p*(*t*) to this system passes practically without distortion if the width of *p*(*t*) is much greater than the system time constant, and provided *p*(*t*) is a lowpass pulse, implying that its spectrum is concentrated at low frequencies. Verify this behavior by considering a system whose unit impulse response is *h*(*t*) = rect(*t*/10−3). The input pulse is a triangle pulse *p*(*t*)=(*t*). Show that the system output to this pulse is very nearly *kp*(*t*), where *k* is the system gain to a dc signal, that is, *k* = *H*(0). -- **7.4-8** A causal signal *h*(*t*) has a Fourier transform *H*(ω). If *R*(ω) and *X*(ω) are the real and the imaginary parts of *H*(ω), that is, *H*(ω) = *R*(ω)+ *jX*(ω), then show that - -$$ -R(\omega) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{X(\omega)}{\omega - y} d\omega -$$ - -and - -$$ -X(\omega) = -\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{R(\omega)}{\omega - y} d\omega -$$ - -assuming that *h*(*t*) has no impulse at the origin. This pair of integrals defines the *Hilbert transform*. [*Hint:* Let *he*(*t*) and *ho*(*t*) be the even and odd components of *h*(*t*). Use the results in Prob. 7.1-4. See Fig. 1.24 for the relationship between *he*(*t*) and *ho*(*t*).] - -This problem states one of the important properties of causal systems: that the real and imaginary parts of the frequency response of a causal system are related. If one specifies the real part, the imaginary part cannot be specified independently. The imaginary part is predetermined by the real part, and vice versa. This result also leads to the conclusion that the magnitude and angle of *H*(ω) are related, provided all the poles and zeros of *H*(ω) lie in the LHP. - -**7.5-1** Consider a filter with the frequency response - -$$ -H(\omega) = e^{-(k\omega^2 + j\omega t_0)} -$$ - -Show that this filter is physically unrealizable by using the time-domain criterion [noncausal *h*(*t*)] and the frequency-domain (Paley–Wiener) criterion. Can this filter be made approximately realizable by choosing *t*0 sufficiently large? Use your own (reasonable) criterion of approximate realizability to determine *t*0. [*Hint:* Use pair 22 in Table 7.1.] - -**7.5-2** Show that a filter with frequency response - -$$ -H(\omega) = \frac{2(10^5)}{\omega^2 + 10^{10}} e^{-j\omega t_0} -$$ - -is unrealizable. Can this filter be made approximately realizable by choosing a sufficiently large *t*0? Use your own (reasonable) criterion of approximate realizability to determine *t*0. - -- **7.5-3** Determine whether the filters with the following frequency response *H*(ω) are physically realizable. If they are not realizable, can they be realized approximately by allowing a finite time delay in the response? - - (a) 10−6 sinc (10−6ω) - - (b) 10−4 (ω/40,000π) - - (c) 2π δ(ω) - -- **7.5-4** Consider signal *x*1(*t*), its Fourier transform *X*1(*f*), and several other signals, as shown in Fig. P7.5-4. Notice, spectra are drawn as a function of hertzian frequency *f* rather than radian frequency ω. - - (a) Accurately sketch *X*2(*f*), the Fourier transform of *x*2(*t*). - - (b) Accurately sketch *x*3(*t*), the inverse Fourier transform of *X*3(*f*). - - (c) The signal *x*4(*t*) = *x*1(*t*) + *x*2(*t*) is passed through an ideal lowpass filter with 3 Hz cutoff to produce output *y*4(*t*). Accurately sketch *y*4(*t*). - -**Figure P7.5-4** - -- **7.6-1** Define *x*(*t*) = 1 2π sinc(*t*/2) with Fourier transform *X*(ω) = rect(ω). Use Parseval's theorem to determine \$ ∞ −∞ sinc2(*t* −2)*dt*. -- **7.6-2** Show that the energy of a Gaussian pulse - -$$ -x(t) = \frac{1}{\sigma\sqrt{2\pi}}e^{-t^2/2\sigma^2} -$$ - -is 1/(2σ √π ). Verify this result by using Parseval's theorem to derive the energy *Ex* from *X*(ω). [*Hint:* See pair 22 in Table 7.1. Use the fact that \$ ∞ −∞ *e*−*x*2/2 *dx* = √2π.] - -**7.6-3** Use Parseval's theorem of Eq. (7.45) to show that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc}^2(kx) dx = \frac{\pi}{k} -$$ - -- **7.6-4** A lowpass signal *x*(*t*) is applied to a squaring device. The squarer output *x*2(*t*) is applied to a lowpass filter of bandwidth *f* (in hertz) (Fig. P7.6-4). Show that if *f* is very small (*f* → 0), then the filter output is a dc signal *y*(*t*) ≈ 2*Exf* . [*Hint:* If *x*2(*t*) ⇐⇒ *A*(ω), then show that *Y*(ω) ≈ [4π*A*(0)*f*]δ(ω) if *f* → 0. Now, show that *A*(0) = *Ex*.] -- **7.6-5** Generalize Parseval's theorem to show that for real, Fourier-transformable signals *x*1(*t*) and *x*2(*t*) - -$$ -\int_{-\infty}^{\infty} x_1(t) x_2(t) dt -$$ - -= $\frac{1}{2\pi} \int_{-\infty}^{\infty} X_1(-\omega) X_2(\omega) d\omega$ -= $\frac{1}{2\pi} \int_{-\infty}^{\infty} X_1(\omega) X_2(-\omega) d\omega$ - -**7.6-6** Show that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc} (Wt - m\pi) \operatorname{sinc} (Wt - n\pi) dt -$$ -$$ -= \begin{cases} 0 & m \neq n \\ \frac{\pi}{W} & m = n \end{cases} -$$ - -( )2 Lowpass filter *x*(*t*) *x y*(*t*) 2*Exf* 2(*t*) - -**Figure P7.6-4** - -[*Hint:* Recognize that - -$$ -\operatorname{sinc}(Wt - k\pi) = \operatorname{sinc}\left[W\left(t - \frac{k\pi}{W}\right)\right] \\ -\Longleftrightarrow \frac{\pi}{W} \operatorname{rect}\left(\frac{\omega}{2W}\right) e^{-jk\pi\omega/W} -$$ - -Use this fact and the result in Prob. 7.6-5.] - -- **7.6-7** (a) What does it mean to compute the 95% essential bandwidth *B* of a signal *x*(*t*) with Fourier transform *X*(ω)? - - (b) Determine the 95% essential bandwidth *B* of a signal with spectrum *X*(ω) = rect(ω). - - (c) Determine the 95% essential bandwidth *B* of a signal with spectrum *X*(ω) = (ω). -- **7.6-8** Using a 95% energy criterion, determine the essential bandwidth *B* of a signal that has a Fourier transform given by *X*(ω) = *e*−|ω| . -- **7.6-9** For the signal - -$$ -x(t) = \frac{2a}{t^2 + a^2} -$$ - -determine the essential bandwidth *B* (in hertz) of *x*(*t*) such that the energy contained in the spectral components of *x*(*t*) of frequencies below *B* Hz is 99% of the signal energy *Ex*. - -- **7.7-1** For each of the following baseband signals **(i)** *m*(*t*) = cos 1000*t*, **(ii)** *m*(*t*) = 2 cos 1000*t* + cos 2000*t*, and **(iii)** *m*(*t*) = cos 1000*t* cos 3000*t*: - - (a) Sketch the spectrum of *m*(*t*). - - (b) Sketch the spectrum of the DSB-SC signal *m*(*t*) cos 10,000*t*. - - (c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra. - - (d) Identify the frequencies in the baseband, and the corresponding frequencies in the DSB-SC, USB, and LSB spectra. Explain the nature of frequency shifting in each case. -- **7.7-2** A message signal *m*(*t*) with spectrum *M*(ω) = 1 1000 ω 2π6000 is to be transmitted using a communication system. Assume all single-sideband systems have suppressed carriers. - - (a) What is the hertzian bandwidth of *m*(*t*)? - -- (b) Sketch the spectrum of the transmitted signal if the communication system is DSB-SC with ω*c* = 2π 100,000. -- (c) Sketch the spectrum of the transmitted signal if the communication system is AM with ω*c* = 2π 100,000 and a modulation index of μ = 1. What is the corresponding carrier amplitude *A*? -- (d) Sketch the spectrum of the transmitted signal if the communication system is USB with ω*c* = 2π 100,000. -- (e) Sketch the spectrum of the transmitted signal if the communication system is LSB with ω*c* = 2π 100,000. -- (f) Suppose we want to transmit *m*(*t*) on each of an FDM system's four channels: DSB-SC at carrier ω1, AM (μ = 1) at carrier ω2, USB at carrier ω3, and LSB at carrier ω4. Determine carrier frequencies ω1 < ω2 < ω3 < ω4 so that the FDM spectrum begins at a frequency of 100,000 Hz with 5,000 Hz deadbands separating adjacent messages. What is the end hertzian frequency of the FDM signal? -- **7.7-3** You are asked to design a DSB-SC modulator to generate a modulated signal *km*(*t*) cosω*ct*, where *m*(*t*) is a signal bandlimited to *B* Hz (Fig. P7.7-3a). Figure P7.7-3b shows a DSB-SC modulator available in the stockroom. The bandpass filter is tuned to ω*c* and has a bandwidth of 2*B* Hz. The carrier generator available generates not cosω*ct*, but cos3 ω*ct*. - - (a) Explain whether you would be able to generate the desired signal using only this equipment. If so, what is the value of *k*? - -- (b) Determine the signal spectra at points *b* and *c*, and indicate the frequency bands occupied by these spectra. -- (c) What is the minimum usable value of ω*c*? -- (d) Would this scheme work if the carrier generator output were cos2 ω*ct*? Explain. -- (e) Would this scheme work if the carrier generator output were cos*n* ω*ct* for any integer *n* ≥ 2? -- **7.7-4** In practice, the analog multiplication operation is difficult and expensive. For this reason, in amplitude modulators, it is necessary to find some alternative to multiplication of *m*(*t*) with cosω*ct*. Fortunately, for this purpose, we can replace multiplication with a switching operation. A similar observation applies to demodulators. In the scheme depicted in Fig. P7.7-4a, the period of the rectangular periodic pulse *x*(*t*) - -**Figure P7.7-4** - -**Figure P7.7-3** - -shown in Fig. P7.7-4b is *T*0 = 2π/ω*c*. The bandpass filter is centered at ±ω*c* and has a bandwidth of 2*B* Hz. Note that multiplication by a square periodic pulse *x*(*t*) in Fig. P7.7-4b amounts to periodic on-off switching of *m*(*t*), which is bandlimited to *B* Hz. Such a switching operation is relatively simple and inexpensive. Show that this scheme can generate an amplitude-modulated signal *k* cos ω*ct*. Determine the value of *k*. Show that the same scheme can also be used for demodulation, provided the bandpass filter in Fig. P7.7-4a is replaced by a lowpass (or baseband) filter. - -**7.7-5** Figure P7.7-5a shows a scheme to transmit two signals *m*1(*t*) and *m*2(*t*) simultaneously on the same channel (without causing spectral interference). Such a scheme, which transmits more than one signal, is known as signal *multiplexing*. In this case, we transmit multiple signals by sharing an available spectral band on the channel; hence, this is an example of the *frequency-division* multiplexing. The signal at point *b* is the multiplexed signal, which now modulates a carrier of frequency 20,000 rad/s. The modulated signal at point *c* is now transmitted over the channel. - -- (a) Sketch the spectra at points *a*, *b*, and *c*. -- (b) What must be the minimum bandwidth of the channel? -- (c) Design a receiver to recover signals *m*1(*t*) and *m*2(*t*) from the modulated signal at point *c*. -- **7.7-6** The system shown in Fig. P7.7-6 is used for scrambling audio signals. The output *y*(*t*) is the scrambled version of the input *m*(*t*). - -**Figure P7.7-6** - -- (a) Find the spectrum of the scrambled signal *y*(*t*). -- (b) Suggest a method of descrambling *y*(*t*) to obtain *m*(*t*). - -A slightly modified version of this scrambler was first used commercially on the 25-mile radio-telephone circuit connecting Los Angeles and Santa Catalina Island. - -- **7.7-7** Figure P7.7-7 presents a scheme for coherent (synchronous) demodulation. Show that this scheme can demodulate the AM signal [*A* + *m*(*t*)] cos ω*ct* regardless of the value of *A*. -- **7.7-8** Sketch the AM signal [*A* + *m*(*t*)] cos ω*ct* for the periodic triangle signal *m*(*t*) illustrated in Fig. P7.7-8 corresponding to the following modulation indices: - - (a) μ = 0.5 - - (b) μ = 1 - - (c) μ = 2 - - (d) μ = ∞ - -How do you interpret the case μ = ∞? - -**7.9-1** Consider the signal *x*(*t*) defined as - -$$ -x(t) = \begin{cases} 1 - |t| & -\frac{1}{2} \le t \le \frac{1}{2} \\ 0 & \text{otherwise} \end{cases} -$$ - -- (a) Sketch the signal *x*(*t*) over −2 ≤ *t* ≤ 2. -- (b) Use time-differentiation and other Fourier transform properties to determine *X*(ω). The only integration you should use is to determine the dc component *X*(0). - -(c) Using MATLAB, verify the correctness of *X*(ω) by synthesizing a 3-periodic replication of the original time-domain signal *x*(*t*). - -[*Hint:* Follow the approach taken in Ex. 7.17.] - -- **7.9-2** Consider the signal *x*(*t*) = |*t*|rect *t*−1 3 . - - (a) Sketch the signal *x*(*t*) over −5 ≤ *t* ≤ 5. - - (b) Use time-differentiation and other Fourier transform properties to determine *X*(ω). The only integration you should use is to determine the dc component *X*(0). - - (c) Use MATLAB to plot the magnitude spectrum |*X*(ω)| and the phase spectrum *X*(ω) over suitable ranges of ω. - - (d) Using MATLAB, verify the correctness of *X*(ω) by synthesizing a 10-periodic replication of the original time-domain signal *x*(*t*). - -[*Hint:* Follow the approach taken in Ex. 7.17.] - -- **7.9-3** Consider the continuous-time aperiodic signal *x*(*t*) = rect(*t*) with Fourier transform *X*(ω) = sinc(ω/2). Furthermore, let *y*(*t*) = (1 − |*t* − 1|)(*u*(*t*) − *u*(*t* − 2)) with Fourier transform *Y*(ω). - - (a) Express the Fourier transform *Y*(ω) in terms of *X*(ω). - - (b) Suppose we create Fourier series coefficients *Vk* by sampling *Y*(ω) according to *Vk* = *Y*(2π*k*/3). Sketch the corresponding time-domain signal *v*(*t*) over a suitable range of time *t*. - - (c) Use MATLAB to synthesize and plot *v*(*t*) using the Fourier series coefficients *Vk* = - -*Y*(2π*k*/3). Verify that the synthesized waveform matches the result of part (b). - -- (d) Suppose we again create Fourier series coefficients *Vk* according to *Vk* = *Y*(2π*k*/3). Next, we upsample *Vk* by factor 2 to create *Wk*. Sketch the time domain signal *p*(*t*) that has Fourier series coefficients *Pk* = *Vk* +*Wk*. -- (e) Use MATLAB to synthesize and plot *p*(*t*) using the Fourier series coefficients *Pk* = *Vk* + *Wk* defined in part (d). Verify that the synthesized waveform matches the result of part (d). -- **7.9-4** Consider the signal *x*(*t*) = *e*−*atu*(*t*). Modify CH7MP2 to compute the following essential bandwidths. - - (a) Setting *a*=1, determine the essential bandwidth *W*1 that contains 95% of the signal energy. Compare this value with the theoretical value presented in Ex. 7.20. - - (b) Setting *a*=2, determine the essential bandwidth *W*2 that contains 90% of the signal energy. - - (c) Setting *a*=3, determine the essential bandwidth *W*3 that contains 75% of the signal energy. -- **7.9-5** A unit amplitude pulse with duration τ is defined as - -$$ -x(t) = \begin{cases} 1 & |t| \le \tau/2 \\ 0 & \text{otherwise} \end{cases} -$$ - -- (a) Determine the duration τ1 that results in a 95% essential bandwidth of 5 Hz. -- (b) Determine the duration τ2 that results in a 90% essential bandwidth of 10 Hz. -- (c) Determine the duration τ3 that results in a 75% essential bandwidth 20 Hz. -- **7.9-6** Consider the signal *x*(*t*) = *e*−*atu*(*t*). - -- (a) Determine the decay parameter *a*1 that results in a 95% essential bandwidth of 5 Hz. -- (b) Determine the decay parameter *a*2 that results in a 90% essential bandwidth of 10 Hz. -- (c) Determine the decay parameter *a*3 that results in a 75% essential bandwidth 20 Hz. -- **7.9-7** Use MATLAB to determine the 95, 90, and 75% essential bandwidths of a one-second triangle function with a peak amplitude of 1. Recall that a triangle function can be constructed by the convolution of two rectangular pulses. -- **7.9-8** A 1/3 duty-cycle square-pulse *T*0-periodic signal *x*(*t*) is described as - -$$ -x(t) = \begin{cases} 1 & -T_0/6 \le t \le T_0/6 \\ 0 & T_0/t \le |t| \le T_0/2 \\ x(t+T_0) & \forall t \end{cases} -$$ - -- (a) Use spectral sampling to determine the Fourier series coefficients *Dn* of *x*(*t*) for *T*0 = 2π. Evaluate and plot *Dn* for (0 ≤ *n* ≤ 10). -- (b) Use spectral sampling to determine the Fourier series coefficients *Dn* of *x*(*t*) for *T*0 = π. Evaluate and plot *Dn* for (0 ≤ *n* ≤ 10). How does this result compare with your answer to part (a)? What can be said about the relation of *T*0 to *Dn* for signal *x*(*t*), which has fixed duty cycle of 1/3? -- **7.9-9** Determine the Fourier transform of a Gaussian pulse defined as *x*(*t*) = *e*−*t* 2 . Plot both *x*(*t*) and *X*(ω). How do the two curves compare? [*Hint:* - -$$ -\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(t-a)^2/2}dt=1 -$$ - -for any real or imaginary *a*.] diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/101_8 SAMPLING - THE BRIDGE FROM CONTINUOUS TO DISCRETE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/101_8 SAMPLING - THE BRIDGE FROM CONTINUOUS TO DISCRETE.md deleted file mode 100644 index 4031ca6f1de5985ffb75216b028d2e56cf2137d2..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/101_8 SAMPLING - THE BRIDGE FROM CONTINUOUS TO DISCRETE.md +++ /dev/null @@ -1,5 +0,0 @@ -# **SAMPLING: THE BRIDGE FROM [CONTINUOUS TO](#page-13-0) DISCRETE** - -A continuous-time signal can be processed by applying its samples through a discrete-time system. For this purpose, it is important to maintain the signal sampling rate high enough to permit the reconstruction of the original signal from these samples without error (or with an error within a given tolerance). The necessary quantitative framework for this purpose is provided by the sampling theorem derived in Sec. 8.1. - -Sampling theory is the bridge between the continuous-time and discrete-time worlds. The information inherent in a sampled continuous-time signal is equivalent to that of a discrete-time signal. A sampled continuous-time signal is a sequence of impulses, while a discrete-time signal presents the same information as a sequence of numbers. These are basically two different ways of presenting the same data. Clearly, all the concepts in the analysis of sampled signals apply to discrete-time signals. We should not be surprised to see that the Fourier spectra of the two kinds of signal are also the same (within a multiplicative constant). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/102_8.1 THE SAMPLING THEOREM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/102_8.1 THE SAMPLING THEOREM.md deleted file mode 100644 index bcf9e0abbe6ae32886886d6f9fdc0d430ef83754..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/102_8.1 THE SAMPLING THEOREM.md +++ /dev/null @@ -1,168 +0,0 @@ -## **8.1 THE [SAMPLING](#page-13-0) THEOREM** - -We now show that a real signal whose spectrum is bandlimited to *B* Hz [*X*(ω) = 0 for |ω| > 2π*B*] can be reconstructed exactly (without any error) from its samples taken uniformly at a rate *fs* > 2*B* samples per second. In other words, the minimum sampling frequency is *fs* = 2*B* Hz.† - -To prove the sampling theorem, consider a signal *x*(*t*) (Fig. 8.1a) whose spectrum is bandlimited to *B* Hz (Fig. 8.1b).‡ For convenience, spectra are shown as functions of ω as well as of *f* (hertz). Sampling *x*(*t*) at a rate of *fs* Hz ( *fs* samples per second) can be accomplished by multiplying *x*(*t*) by an impulse train δ*T* (*t*) (Fig. 8.1c), consisting of unit impulses repeating periodically every *T* seconds, where *T* = 1/*fs*. The schematic of a sampler is shown in Fig. 8.1d. The resulting sampled signal *x*(*t*) is shown in Fig. 8.1e. The sampled signal consists of impulses - -CHAPTER - -**8** - -† The theorem stated here (and proved subsequently) applies to lowpass signals. A bandpass signal whose spectrum exists over a frequency band *fc* − (*B*/2) < |*f* | < *fc* + (*B*/2) has a bandwidth of *B* Hz. Such a signal is uniquely determined by 2*B* samples per second. In general, the sampling scheme is a bit more complex in this case. It uses two interlaced sampling trains, each at a rate of *B* samples per second. See, for example, [1]. ‡ The spectrum *X*(ω) in Fig. 8.1b is shown as real, for convenience. However, our arguments are valid for complex *X*(ω) as well. - -**Figure 8.1** Sampled signal and its Fourier spectrum. - -spaced every *T* seconds (the sampling interval). The *n*th impulse, located at *t* = *nT*, has a strength *x*(*nT*), the value of *x*(*t*) at *t* = *nT*. - -$$ -\bar{x}(t) = x(t)\delta_T(t) = \sum_n x(nT)\delta(t - nT) -$$ - -Because the impulse train δ*T* (*t*) is a periodic signal of period *T*, it can be expressed as a trigonometric Fourier series like that already obtained in Ex. 6.9 [Eq. (6.25)], - -$$ -\delta_T(t) = \frac{1}{T} [1 + 2\cos\omega_s t + 2\cos 2\omega_s t + 2\cos 3\omega_s t + \cdots] \qquad \omega_s = \frac{2\pi}{T} = 2\pi f_s -$$ - -Therefore, - -$$ -\bar{x}(t) = x(t)\delta_T(t) = \frac{1}{T}[x(t) + 2x(t)\cos\omega_s t + 2x(t)\cos 2\omega_s t + 2x(t)\cos 3\omega_s t + \cdots] -$$ -(8.1) - -To find *X*(ω), the Fourier transform of *x*(*t*), we take the Fourier transform of the right-hand side of Eq. (8.1), term by term. The transform of the first term in the brackets is *X*(ω). The transform - -#### 778 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -of the second term 2*x*(*t*) cos ω*st* is *X*(ω − ω*s*) + *X*(ω + ω*s*) [see Eq. (7.32)]. This represents spectrum *X*(ω) shifted to ω*s* and −ω*s*. Similarly, the transform of the third term 2*x*(*t*) cos 2ω*st* is *X*(ω − 2ω*s*) + *X*(ω + 2ω*s*), which represents the spectrum *X*(ω) shifted to 2ω*s* and −2ω*s*, and so on to infinity. This result means that the spectrum *X*(ω) consists of *X*(ω) repeating periodically with period ω*s* = 2π/*T* rad/s, or *fs* = 1/*T* Hz, as depicted in Fig. 8.1f. There is also a constant multiplier 1/*T* in Eq. (8.1). Therefore, - -$$ -\overline{X}(\omega) = \frac{1}{T} \sum_{n=-\infty}^{\infty} X(\omega - n\omega_s) -$$ -\n(8.2) - -If we are to reconstruct *x*(*t*) from *x*(*t*), we should be able to recover *X*(ω) from *X*(ω). This recovery is possible if there is no overlap between successive cycles of *X*(ω). Figure 8.1f indicates that this requires - -$$ -f_s > 2B \tag{8.3} -$$ - -Also, the sampling interval *T* = 1/*fs*. Therefore, - -$$ -T < \frac{1}{2B} -$$ - -Thus, as long as the sampling frequency *fs* is greater than twice the signal bandwidth *B* (in hertz), *X*(ω) consists of nonoverlapping repetitions of *X*(ω). Figure 8.1f shows that the gap between the two adjacent spectral repetitions is *fs* − 2*B* Hz, and *x*(*t*) can be recovered from its samples *x*(*t*) by passing the sampled signal *x*(*t*) through an ideal lowpass filter having a bandwidth of any value between *B* and *fs* − *B* Hz. The minimum sampling rate *fs* = 2*B* required to recover *x*(*t*) from its samples *x*(*t*) is called the *Nyquist rate* for *x*(*t*), and the corresponding sampling interval *T* = 1/2*B* is called the *Nyquist interval* for *x*(*t*). Samples of a signal taken at its Nyquist rate are the *Nyquist samples* of that signal. - -We are saying that the Nyquist rate 2*B* Hz is the minimum sampling rate required to preserve the information of *x*(*t*). This contradicts Eq. (8.3), where we showed that to preserve the information of *x*(*t*), the sampling rate *fs* needs to be greater than 2*B* Hz. Strictly speaking, Eq. (8.3) is the correct statement. However, if the spectrum *X*(ω) contains no impulse or its derivatives at the highest frequency *B* Hz, then the minimum sampling rate 2*B* Hz is adequate. In practice, it is rare to observe *X*(ω) with an impulse or its derivatives at the highest frequency. If the contrary situation were to occur, we should use Eq. (8.3).† - -† An interesting observation is that if the impulse is because of a cosine term, the sampling rate of 2*B* Hz is adequate. However, if the impulse is because of a sine term, then the rate must be greater than 2*B* Hz. This may be seen from the fact that samples of sin 2π*Bt* using *T* = 1/2*B* are all zero because sin 2π*BnT* = sinπ*n* = 0. But, samples of cos 2π*Bt* are cos 2π*BnT* = cosπ*n* = (−1)*n*. We can reconstruct cos 2π*Bt* from these samples. This peculiar behavior occurs because in the sampled signal spectrum corresponding to the signal cos 2π*Bt*, the impulses, which occur at frequencies (2*n* ± 1)*B* Hz (*n* = 0,±1,±2,. . .), interact constructively, whereas in the case of sin 2π*Bt*, the impulses, because of their opposite phases (*e*±*j*π/2), interact destructively and cancel out in the sampled signal spectrum. Hence, sin 2π*Bt* cannot be reconstructed from its samples at a rate 2*B* Hz. A similar situation exists for signal cos(2π*Bt*+θ ), which contains a component of the form sin 2π*Bt*. For this reason, it is advisable to maintain sampling rate above 2*B* Hz if a finite-amplitude component of a sinusoid of frequency *B* Hz is present in the signal. - -The sampling theorem proved here uses samples taken at uniform intervals. This condition is not necessary. Samples can be taken arbitrarily at any instants as long as the sampling instants are recorded and there are, on average, 2*B* samples per second [2]. The essence of the sampling theorem was known to mathematicians for a long time in the form of the *interpolation formula* [see later, Eq. (8.6)]. The origin of the sampling theorem was attributed by H. S. Black to Cauchy in 1841. The essential idea of the sampling theorem was rediscovered in the 1920s by Carson, Nyquist, and Hartley. - -### **EXAMPLE 8.1 Sampling at, Below, and Above the Nyquist Rate** - -In this example, we examine the effects of sampling a signal at the Nyquist rate, below the Nyquist rate (undersampling), and above the Nyquist rate (oversampling). Consider a signal *x*(*t*) = sinc2 (5π*t*) (Fig. 8.2a) whose spectrum is *X*(ω) = 0.2(ω/20π ) (Fig. 8.2b). The bandwidth of this signal is 5 Hz (10π rad/s). Consequently, the Nyquist rate is 10 Hz; that is, we must sample the signal at a rate no less than 10 samples/s. The Nyquist interval is *T* = 1/2*B* = 0.1 second. - -Recall that the sampled signal spectrum consists of (1/*T*)*X*(ω) = (0.2/*T*)(ω/20π ) repeating periodically with a period equal to the sampling frequency *fs* Hz. For the three sampling rates *fs* = 5 Hz (undersampling), 10 Hz (Nyquist rate), and 20 Hz (oversampling), we see that - -| fs
(Hz) | T
= 1
(s)
fs | TX(ω)
1 | Comments | -|------------|-----------------------|----------------|---------------| -| 5 | 0.2 | ω

20π | Undersampling | -| 10 | 0.1 | 2 ω

20π | Nyquist rate | -| 20 | 0.05 | 4 ω

20π | Oversampling | - -In the first case (undersampling), the sampling rate is 5 Hz (5 samples/s), and the spectrum (1/*T*)*X*(ω) repeats every 5 Hz (10π rad/s). The successive spectra overlap, as depicted in Fig. 8.2d, and the spectrum *X*(ω) are not recoverable from *X*(ω); that is, *x*(*t*) cannot be reconstructed from its samples *x*(*t*) in Fig. 8.2c. In the second case, we use the Nyquist sampling rate of 10 Hz (Fig. 8.2e). The spectrum *X*(ω) consists of back-to-back, nonoverlapping repetitions of (1/*T*)*X*(ω) repeating every 10 Hz. Hence, *X*(ω) can be recovered - -**Figure 8.2** Effects of undersampling and oversampling. - -from *X*(ω) using an ideal lowpass filter of bandwidth 5 Hz (Fig. 8.2f). Finally, in the last case of oversampling (sampling rate 20 Hz), the spectrum *X*(ω) consists of nonoverlapping repetitions of (1/*T*)*X*(ω) (repeating every 20 Hz) with empty bands between successive cycles (Fig. 8.2h). Hence, *X*(ω) can be recovered from *X*(ω) by using an ideal lowpass filter or even a practical lowpass filter (shown dashed in Fig. 8.2h).† - -### **DR ILL 8.1 Nyquist Sampling** - -Find the Nyquist rate and the Nyquist sampling interval for the signals sinc(100π*t*) and sinc(100π*t*)+sinc(50π*t*). - -### **ANSWERS** - -The Nyquist sampling interval is 0.01 s and the Nyquist sampling rate is 100 Hz for both signals. - -### FOR SKEPTICS ONLY - -Rare is the reader who, at first encounter, is not skeptical of the sampling theorem. It seems impossible that Nyquist samples can define the one and the only signal that passes through those sample values. We can easily picture infinite number of signals passing through a given set of samples. However, among all these (infinite number of) signals, only one has the minimum bandwidth *B* ≤ 1/2*T* Hz, where *T* is the sampling interval. See Prob. 8.2-15. - -To summarize, for a given set of samples taken at a rate *fs* Hz, there is only one signal of bandwidth *B* ≤ *fs*/2 that passes through those samples. All other signals that pass through those samples have bandwidth higher than *fs*/2, and the samples are sub-Nyquist rate samples for those signals. - -### **[8.1-1 Practical Sampling](#page-13-0)** - -In proving the sampling theorem, we assumed ideal samples obtained by multiplying a signal *x*(*t*) by an impulse train that is physically unrealizable. In practice, we multiply a signal *x*(*t*) by a train of pulses of finite width, depicted in Fig. 8.3c. The sampler is shown in Fig. 8.3d. The sampled signal *x*(*t*) is illustrated in Fig. 8.3e. We wonder whether it is possible to recover or reconstruct *x*(*t*) from this *x*(*t*). Surprisingly, the answer is affirmative, provided the sampling rate is not below the Nyquist rate. The signal *x*(*t*) can be recovered by lowpass filtering *x*(*t*) as if it were sampled by impulse train. - -† The filter should have a constant gain between 0 and 5 Hz and zero gain beyond 10 Hz. In practice, the gain beyond 10 Hz can be made negligibly small, but not zero. - -**Figure 8.3** Effect of practical sampling. - -The plausibility of this result becomes apparent when we consider the fact that reconstruction of *x*(*t*) requires the knowledge of the Nyquist sample values. This information is available or built into the sampled signal *x*(*t*) in Fig. 8.3e because the *n*th sampled pulse strength is *x*(*nT*). To prove the result analytically, we observe that the sampling pulse train *pT* (*t*) depicted in Fig. 8.3c, being a periodic signal, can be expressed as a trigonometric Fourier series - -$$ -p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n) \qquad \omega_s = \frac{2\pi}{T} -$$ - -Thus, - -$$ -\overline{x}(t) = x(t)p_T(t) = x(t)\left[C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n)\right] -$$ -$$ -= C_0 x(t) + \sum_{n=1}^{\infty} C_n x(t) \cos(n\omega_s t + \theta_n) -$$ - -The sampled signal *x*(*t*) consists of *C*0*x*(*t*), *C*1*x*(*t*) cos(ω*st* + θ1), *C*2*x*(*t*) cos(2ω*st* + θ2), ... . Note that the first term *C*0*x*(*t*) is the desired signal and all the other terms are modulated signals with spectra centered at ±ω*s*,±2ω*s*,±3ω*s*,..., as illustrated in Fig. 8.3f. Clearly the signal *x*(*t*) can be recovered by lowpass filtering of *x*(*t*), as shown in Fig. 8.3d. As before, it is necessary that ω*s* > 4π*B* (or *fs* > 2*B*). - -### **EXAMPLE 8.2 Practical Sampling** - -Demonstrate practical sampling by sampling signal *x*(*t*) = sinc2 (5π*t*) with the rectangular pulse sequence *pT* (*t*) illustrated in Fig. 8.4c. Sketch the original and sampled signals and their spectra, and discuss recovery of *x*(*t*) from its samples. - -**Figure 8.4** An example of practical sampling. - -#### 784 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -The signal *x*(*t*) and its spectrum are shown in Figs. 8.4a and 8.4b, respectively. - -The period of *pT* (*t*) is 0.1 second so that the fundamental frequency (which is the sampling frequency) is 10 Hz. Hence, ω*s* = 20π. The Fourier series for *pT* (*t*) can be expressed as - -$$ -p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos n\omega_s t -$$ - -Hence, - -$$ -\overline{x}(t) = x(t)p_T(t) -$$ - -= C1x(t) + C1x(t) cos 20 $\pi$ t + C2x(t) cos 40 $\pi$ t + C3x(t) cos 60 $\pi$ t + ... - -Use of Eq. (6.14) yields *C*0 = 1 4 and *Cn* = 2 *n*π sin *n*π 4 . Consequently, we have - -$$ -\overline{x}(t) = x(t)p_T(t) -$$ - -= $\frac{1}{4}x(t) + C_1x(t) \cos 20\pi t + C_2x(t) \cos 40\pi t + C_3x(t) \cos 60\pi t + \cdots$ - -and - -$$ -\overline{X}(\omega) = \frac{1}{4}X(\omega) + \frac{C_1}{2}[X(\omega - 20\pi) + X(\omega + 20\pi)] + \frac{C_2}{2}[X(\omega - 40\pi) + X(\omega + 40\pi)] + \frac{C_3}{2}[X(\omega - 60\pi) + X(\omega + 60\pi)] + \cdots -$$ - -where *Cn* = (2/*n*π )sin(*n*π/4). The sampled signal and its spectrum are shown in Figs. 8.4d and 8.4e, respectively. - -The spectrum *X*(ω) consists of *X*(ω) repeating periodically at the interval of 20π rad/s (10 Hz). Hence, there is no overlap between cycles, and *X*(ω) can be recovered by using an ideal lowpass filter of bandwidth 5 Hz. An ideal lowpass filter of unit gain (and bandwidth 5 Hz) will allow the first term on the right-hand side of the foregoing equation to pass fully and suppress all the other terms. Hence, the output *y*(*t*) is - -> *y*(*t*) = 1 4 *x*(*t*) - -### **DR ILL 8.2 The Role of Sampling Pulse Area** - -Show that the basic pulse *p*(*t*) used in the sampling pulse train in Fig. 8.4c cannot have zero area if we wish to reconstruct *x*(*t*) by lowpass-filtering the sampled signal. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/103_8.2 SIGNAL RECONSTRUCTION.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/103_8.2 SIGNAL RECONSTRUCTION.md deleted file mode 100644 index 5e472163ba2236d1b7bf9bc4d6d8710696e2e408..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/103_8.2 SIGNAL RECONSTRUCTION.md +++ /dev/null @@ -1,259 +0,0 @@ -## **8.2 SIGNAL [RECONSTRUCTION](#page-13-0)** - -The process of reconstructing a continuous-time signal *x*(*t*) from its samples is also known as *interpolation*. In Sec. 8.1, we saw that a signal *x*(*t*) bandlimited to *B* Hz can be reconstructed (interpolated) exactly from its samples if the sampling frequency *fs* exceeds 2*B* Hz or the sampling interval *T* is less than 1/2*B*. This reconstruction is accomplished by passing the sampled signal through an ideal lowpass filter of gain *T* and having a bandwidth of any value between *B* and *fs* −*B* Hz. From a practical viewpoint, a good choice is the middle value *fs*/2 = 1/2*T* Hz or π/*T* rad/s. This value allows for small deviations in the ideal filter characteristics on either side of the cutoff frequency. With this choice of cutoff frequency and gain *T*, the ideal lowpass filter required for signal reconstruction (or interpolation) is - -$$ -H(\omega) = T \operatorname{rect}\left(\frac{\omega}{2\pi f_s}\right) = T \operatorname{rect}\left(\frac{\omega T}{2\pi}\right) -$$ -\n(8.4) - -The interpolation process here is expressed in the frequency domain as a filtering operation. Now we shall examine this process from the time-domain viewpoint. - -### TIME-DOMAIN VIEW:ASIMPLE INTERPOLATION - -Consider the interpolation system shown in Fig. 8.5a. We start with a very simple interpolating filter, whose impulse response is rect(*t*/*T*), depicted in Fig. 8.5b. This is a gate pulse centered at the origin, having unit height, and width *T* (the sampling interval). We shall find the output of this filter when the input is the sampled signal *x*(*t*) consisting of an impulse train with the *n*th impulse at *t* = *nT* with strength *x*(*nT*). Each sample in *x*(*t*), being an impulse, produces at the output a gate pulse of height equal to the strength of the sample. For instance, the *n*th sample is an impulse of strength *x*(*nT*) located at *t* = *nT* and can be expressed as *x*(*nT*)δ(*t*−*nT*). When this impulse passes through the filter, it produces at the output a gate pulse of height *x*(*nT*), centered at *t* = *nT* (shaded in Fig. 8.5c). Each sample in *x*(*t*) will generate a corresponding gate pulse, resulting in the filter output that is a staircase approximation of *x*(*t*), shown dotted in Fig. 8.5c. This filter thus gives a crude form of interpolation. - -The frequency response of this filter *H*(ω) is the Fourier transform of the impulse response rect(*t*/*T*). Thus, - -$$ -h(t) = \text{rect}\left(\frac{t}{T}\right) \qquad \text{and} \qquad H(\omega) = T \text{sinc}\left(\frac{\omega T}{2}\right) \tag{8.5} -$$ - -The amplitude response |*H*(ω)| for this filter, illustrated in Fig. 8.5d, explains the reason for the crudeness of this interpolation. This filter, also known as the *zero-order hold* (ZOH) filter, is a poor form of the ideal lowpass filter (shaded in Fig. 8.5d) required for exact interpolation.† - -We can improve on the ZOH filter by using a *first-order hold* filter, which results in a linear interpolation instead of a staircase interpolation. A linear interpolator, whose impulse response is a triangle pulse (*t*/2*T*), results in an interpolation in which successive sample tops are connected by straight-line segments (see Prob. 8.2-3). - -† Figure 8.5b shows that the impulse response of this filter is noncausal, and this filter is not realizable. In practice, we make it realizable by delaying the impulse response by *T*/2. This merely delays the output of the filter by *T*/2. - -**Figure 8.5** Simple interpolation by means of a zero-order hold (ZOH) circuit. **(a)** ZOH interpolator. **(b)** Impulse response of a ZOH circuit. **(c)** Signal reconstruction by ZOH, as viewed in the time domain. **(d)** Frequency response of a ZOH. - -### TIME-DOMAIN VIEW: AN IDEAL INTERPOLATION - -The ideal interpolation filter frequency response obtained in Eq. (8.4) is illustrated in Fig. 8.6a. The impulse response of this filter, the inverse Fourier transform of *H*(ω) is - -$$ -h(t) = \text{sinc}\left(\frac{\pi t}{T}\right) -$$ - -For the Nyquist sampling rate, *T* = 1/2*B*, and - -$$ -h(t) = \text{sinc}\left(2\pi Bt\right) -$$ - -This *h*(*t*) is depicted in Fig. 8.6b. Observe the interesting fact that *h*(*t*) = 0 at all Nyquist sampling instants (*t* = ±*n*/2*B*) except at *t* = 0. When the sampled signal *x*¯(*t*) is applied at the input of this filter, the output is *x*(*t*). Each sample in *x*(*t*), being an impulse, generates a sinc pulse of height equal to the strength of the sample, as illustrated in Fig. 8.6c. The process is identical to that depicted in Fig. 8.5c, except that *h*(*t*) is a sinc pulse instead of a gate pulse. Addition of the sinc - -**Figure 8.6** Ideal interpolation for Nyquist sampling rate. - -pulses generated by all the samples results in *x*(*t*). The *n*th sample of the input *x*(*t*) is the impulse *x*(*nT*)δ(*t* −*nT*); the filter output of this impulse is *x*(*nT*)*h*(*t* −*nT*). Hence, the filter output to *x*(*t*), which is *x*(*t*), can now be expressed as a sum - -$$ -x(t) = \sum_{n} x(nT)h(t - nT) = \sum_{n} x(nT)\operatorname{sinc}\left[\frac{\pi}{T}(t - nT)\right] -$$ - -For the case of Nyquist sampling rate, *T* = 1/2*B*, this expression simplifies to - -$$ -x(t) = \sum_{n} x(nT) \operatorname{sinc}(2\pi Bt - n\pi) -$$ -\n(8.6) - -Equation (8.6) is the *interpolation formula*, which yields values of *x*(*t*) between samples as a weighted sum of all the sample values. - -## **EXAMPLE 8.3 Bandlimited Interpolation of the Kronecker Delta Function** - -Find a signal *x*(*t*) that is bandlimited to *B* Hz, and whose samples are - -$$ -x(0) = 1 -$$ - and $x(\pm T) = x(\pm 2T) = x(\pm 3T) = \cdots = 0$ - -### 788 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -where the sampling interval *T* is the Nyquist interval for *x*(*t*), that is, *T* = 1/2*B*. - -Because we are given the Nyquist sample values, we use the interpolation formula of Eq. (8.6) to construct *x*(*t*) from its samples. Since all but one of the Nyquist samples are zero, only one term (corresponding to *n* = 0) in the summation on the right-hand side of Eq. (8.6) survives. Thus, - -$$ -x(t) = \operatorname{sinc}\left(2\pi Bt\right) -$$ - -This signal is illustrated in Fig. 8.6b. Observe that this is the only signal that has a bandwidth *B* Hz and the sample values *x*(0) = 1 and *x*(*nT*) = 0(*n* = 0). No other signal satisfies these conditions. - -### **[8.2-1 Practical Difficulties in Signal Reconstruction](#page-13-0)** - -Consider the signal reconstruction procedure illustrated in Fig. 8.7a. If *x*(*t*) is sampled at the Nyquist rate *fs* = 2*B* Hz, the spectrum *X*(ω) consists of repetitions of *X*(ω) without any gap between successive cycles, as depicted in Fig. 8.7b. To recover *x*(*t*) from *x*(*t*), we need to pass the sampled signal *x*(*t*) through an ideal lowpass filter, shown dotted in Fig. 8.7b. As seen in Sec. 7.5, such a filter is unrealizable; it can be closely approximated only with infinite time delay in the response. In other words, we can recover the signal *x*(*t*) from its samples with infinite time delay. A practical solution to this problem is to sample the signal at a rate higher than the Nyquist rate (*fs* > 2*B* or ω*s* > 4π*B*). The result is *X*(ω), consisting of repetitions of *X*(ω) with a finite bandgap between successive cycles, as illustrated in Fig. 8.7c. Now, we can recover *X*(ω) from *X*(ω) using a lowpass filter with a gradual cutoff characteristic, shown dotted in Fig. 8.7c. But even in this case, if the unwanted spectrum is to be suppressed, the filter gain must be zero beyond some frequency (see Fig. 8.7c). According to the Paley–Wiener criterion [Eq. (7.43)], it is impossible to realize even this filter. The only advantage in this case is that the required filter can be closely approximated with a smaller time delay. All this means that it is impossible in practice to recover a bandlimited signal *x*(*t*) exactly from its samples, even if the sampling rate is higher than the Nyquist rate. However, as the sampling rate increases, the recovered signal approaches the desired signal more closely. - -### THE TREACHERY OF ALIASING - -There is another fundamental practical difficulty in reconstructing a signal from its samples. The sampling theorem was proved on the assumption that the signal *x*(*t*) is bandlimited. *All practical signals are timelimited*; that is, they are of finite duration or width. We can demonstrate (see Prob. 8.2-20) that a signal cannot be timelimited and bandlimited simultaneously. If a signal is timelimited, it cannot be bandlimited, and vice versa (but it can be simultaneously nontimelimited and nonbandlimited). Clearly, all practical signals, which are necessarily timelimited, are nonbandlimited, as shown in Fig. 8.8a; they have infinite bandwidth, and the spectrum *X*(ω) consists of overlapping cycles of *X*(ω) repeating every *fs* Hz (the sampling frequency), as - -**Figure 8.7 (a)** Signal reconstruction from its samples. **(b)** Spectrum of a signal sampled at the Nyquist rate. **(c)** Spectrum of a signal sampled above the Nyquist rate. - -illustrated in Fig. 8.8b.† Because of infinite bandwidth in this case, the spectral overlap is unavoidable, regardless of the sampling rate. Sampling at a higher rate reduces but does not eliminate overlapping between repeating spectral cycles. Because of the overlapping tails, *X*(ω) no longer has complete information about *X*(ω), and it is no longer possible, even theoretically, to recover *x*(*t*) exactly from the sampled signal *x*(*t*). If the sampled signal is passed through an ideal lowpass filter of cutoff frequency *fs*/2 Hz, the output is not *X*(ω) but *Xa*(ω) (Fig. 8.8c), which is a version of *X*(ω) distorted as a result of two separate causes: - -- 1. The loss of the tail of *X*(ω) beyond |*f* | > *fs*/2 Hz. -- 2. The reappearance of this tail inverted or folded onto the spectrum. Note that the spectra cross at frequency *fs*/2 = 1/2*T* Hz. This frequency is called the *folding* frequency. - -† Figure 8.8b shows that from the infinite number of repeating cycles, only the neighboring spectral cycles overlap. This is a somewhat simplified picture. In reality, all the cycles overlap and interact with every other cycle because of the infinite width of all practical signal spectra. Fortunately, all practical spectra also must decay at higher frequencies. This results in insignificant amount of interference from cycles other than the immediate neighbors. When such an assumption is not justified, aliasing computations become little more involved. - -**Figure 8.8** Aliasing effect. **(a)** Spectrum of a practical signal *x*(*t*). **(b)** Spectrum of sampled *x*(*t*). **(c)** Reconstructed signal spectrum. **(d)** Sampling scheme using anti-aliasing filter. **(e)** Sampled signal spectrum (dotted) and the reconstructed signal spectrum (solid) when anti-aliasing filter is used. - -The spectrum may be viewed as if the lost tail is folding back onto itself at the folding frequency. For instance, a component of frequency (*fs*/2) + *fz* shows up as or "impersonates" a component of lower frequency (*fs*/2) − *fz* in the reconstructed signal. Thus, the components of frequencies above *fs*/2 reappear as components of frequencies below *fs*/2. This tail inversion, known as *spectral folding* or *aliasing,* is shown shaded in Fig. 8.8b and also in Fig. 8.8c. In the process of aliasing, not only are we losing all the components of frequencies above the folding frequency *fs*/2 Hz, but these very components reappear (aliased) as lower-frequency components, as shown in Figs. 8.8b and 8.8c. Such aliasing destroys the integrity of the frequency components below the folding frequency *fs*/2, as depicted in Fig. 8.8c. - -The aliasing problem is analogous to that of an army with a platoon that has secretly defected to the enemy side. The platoon is, however, ostensibly loyal to the army. The army is in double jeopardy. First, the army has lost this platoon as a fighting force. In addition, during actual fighting, the army will have to contend with sabotage by the defectors and will have to find another loyal platoon to neutralize the defectors. Thus, the army has lost two platoons in nonproductive activity. - -### DEFECTORS ELIMINATED: THE ANTI-ALIASING FILTER - -If you were the commander of the betrayed army, the solution to the problem would be obvious. As soon as the commander got wind of the defection, he would incapacitate, by whatever means, the defecting platoon *before the fighting begins*. This way he loses only one (the defecting) platoon. This is a partial solution to the double jeopardy of betrayal and sabotage, a solution that partly rectifies the problem and cuts the losses to half. - -We follow exactly the same procedure. The potential defectors are all the frequency components beyond the folding frequency *fs*/2 = 1/2*T* Hz. We should eliminate (suppress) these components from *x*(*t*) *before sampling x*(*t*). Such suppression of higher frequencies can be accomplished by an ideal lowpass filter of cutoff *fs*/2 Hz, as shown in Fig. 8.8d. This is called the *anti-aliasing filter*. Figure 8.8d also shows that anti-aliasing filtering is performed before sampling. Figure 8.8e shows the sampled signal spectrum (dotted) and the reconstructed signal *Xaa*(ω) when an anti-aliasing scheme is used. An anti-aliasing filter essentially bandlimits the signal *x*(*t*) to *fs*/2 Hz. This way, we lose only the components beyond the folding frequency *fs*/2 Hz. These suppressed components now cannot reappear to corrupt the components of frequencies below the folding frequency. Clearly, use of an anti-aliasing filter results in the reconstructed signal spectrum *Xaa*(ω) = *X*(ω) for |*f* | < *fs*/2. Thus, although we lost the spectrum beyond *fs*/2 Hz, the spectrum for all the frequencies below *fs*/2 remains intact. The effective aliasing distortion is cut in half owing to elimination of folding. We stress again that the anti-aliasing operation must be performed *before the signal is sampled*. - -An anti-aliasing filter also helps to reduce noise. Noise, generally, has a wideband spectrum, and without anti-aliasing, the aliasing phenomenon itself will cause the noise lying outside the desired band to appear in the signal band. Anti-aliasing suppresses the entire noise spectrum beyond frequency *fs*/2. - -The anti-aliasing filter, being an ideal filter, is unrealizable. In practice, we use a steep cutoff filter, which leaves a sharply attenuated spectrum beyond the folding frequency *fs*/2. - -### SAMPLING FORCES NONBANDLIMITED SIGNALS TO APPEAR BANDLIMITED - -Figure 8.8b shows that the spectrum of a signal *x*(*t*) consists of overlapping cycles of *X*(ω). This means that *x*(*t*) are sub-Nyquist samples of *x*(*t*). However, we may also view the spectrum in Fig. 8.8b as the spectrum *Xa*(ω) (Fig. 8.8c), repeating periodically every *fs* Hz without overlap. The spectrum *Xa*(ω) is bandlimited to *fs*/2 Hz. Hence, these (sub-Nyquist) samples of *x*(*t*) are actually the Nyquist samples for signal *xa*(*t*). In conclusion, sampling a nonbandlimited signal *x*(*t*) at a rate *fs* Hz makes the samples appear to be the Nyquist samples of some signal *xa*(*t*), bandlimited to *fs*/2 Hz. In other words, sampling makes a nonbandlimited signal appear to be a bandlimited signal *xa*(*t*) with bandwidth *fs*/2 Hz. A similar conclusion applies if *x*(*t*) is bandlimited but sampled at a sub-Nyquist rate. - -### VERIFICATION OF ALIASING IN SINUSOIDS - -We showed in Fig. 8.8b how sampling a signal below the Nyquist rate causes aliasing, which makes a signal of higher frequency (*fs*/2) + *fz* Hz masquerade as a signal of lower frequency (*fs*/2) − *fz* Hz. Figure 8.8b demonstrates this result in the frequency domain. Let us now verify it in the time domain to gain a deeper appreciation of aliasing. - -We can prove our proposition by showing that samples of sinusoids of frequencies (ω*s*/2)+ω*z* and (ω*s*/2)−ω*z* are identical when the sampling frequency is *fs* = ω*s*/2π Hz. - -For a sinusoid *x*(*t*) = cos ω*t*, sampled at intervals of *T* seconds, *x*(*nT*), its *n*th sample (at *t* = *nT*) is - -*x*(*nT*) = cosω*nT n* integer - -Hence, samples of sinusoids of frequency ω = (ω*s*/2)±ω*z* are† - -$$ -x(nT) = \cos\left(\frac{\omega_s}{2} \pm \omega_z\right) nT = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT \mp \sin\left(\frac{\omega_s}{2}\right) nT \sin\omega_z nT -$$ - -Recognizing that ω*sT* = 2π*fsT* = 2π, and sin(ω*s*/2)*nT* = sinπ*n* = 0 for all integer *n*, we obtain - -$$ -x(nT) = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT -$$ - -Clearly, the samples of a sinusoid of frequency (*fs*/2)+*fz* are identical to the samples of a sinusoid (*fs*/2)−*fz*. ‡ For instance, when a sinusoid of frequency 100 Hz is sampled at a rate of 120 Hz, the apparent frequency of the sinusoid that results from reconstruction of the samples is 20 Hz. This follows from the fact that here, 100 = (*fs*/2)+*fz* = 60+*fz* so that *fz* = 40. Hence, (*fs*/2)−*fz* = 20. Such would precisely be the conclusion arrived at from Fig. 8.8b. - -† Here we have ignored the phase aspect of the sinusoid. Sampled versions of a sinusoid *x*(*t*) = cos(ω*t* + θ ) with two different frequencies (ω*s*/2) ± ω*z* have identical frequency, but the phase signs may be reversed - -depending on the value of ω*z*. ‡ The reader is encouraged to verify this result graphically by plotting the spectrum of a sinusoid of frequency (ω*s*/2) + ω*z* (impulses at ±[(ω*s*/2) + ω*z*]) and its periodic repetition at intervals ω*s*. Although the result is valid for all values of ω*z*, consider the case of ω*z* < ω*s*/2 to simplify the graphics. - -This discussion again shows that sampling a sinusoid of frequency *f* aliasing can be avoided if the sampling rate *fs* > 2*f* Hz. - -$$ -0 \le f < \frac{f_s}{2} \qquad \text{or} \qquad 0 \le \omega < \frac{\pi}{T} -$$ - -Violating this condition leads to aliasing, implying that the samples appear to be those of a lower-frequency signal. Because of this loss of identity, it is impossible to reconstruct the signal faithfully from its samples. - -### GENERAL CONDITION FOR ALIASING IN SINUSOIDS - -We can generalize the foregoing result by showing that samples of a sinusoid of frequency *f*0 are identical to those of a sinusoid of frequency *f*0 + *mfs* Hz (integer *m*), where *fs* is the sampling frequency. The samples of cos 2π(*f*0 +*mfs*)*t* are - -$$ -\cos 2\pi (f_0 + m f_s) nT = \cos (2\pi f_0 nT + 2\pi mn) = \cos 2\pi f_0 nT -$$ - -The result follows because *mn* is an integer and *fsT* = 1. This result shows that sinusoids of frequencies that differ by an integer multiple of *fs* result in identical set of samples. In other words, samples of sinusoids separated by frequency *fs* Hz are identical. This implies that samples of sinusoids in any frequency band of *fs* Hz are unique; that is, no two sinusoids in that band have the same samples (when sampled at a rate *fs* Hz). For instance, frequencies in the band from −*fs*/2 to *fs*/2 have unique samples (at the sampling rate *fs*). This band is called the *fundamental band*. Recall also that *fs*/2 is the folding frequency. - -From the discussion thus far, we conclude that if a continuous-time sinusoid of frequency *f* Hz is sampled at a rate of *fs* Hz (samples/s), the resulting samples would appear as samples of a continuous-time sinusoid of frequency *fa* in the fundamental band, where - -$$ -f_a = f - mf_s \qquad -\frac{f_s}{2} \le f_a < \frac{f_s}{2} \qquad m \text{ an integer} \tag{8.7} -$$ - -The frequency *fa* lies in the fundamental band from −*fs*/2 to *fs*/2. Figure 8.9a shows the plot of *fa* versus *f* , where *f* is the actual frequency and *fa* is the corresponding fundamental band frequency, whose samples are identical to those of the sinusoid of frequency *f* , when the sampling rate is *fs* Hz. - -Recall, however, that the sign change of a frequency does not alter the actual frequency of the waveform. This is because - -$$ -\cos(-\omega_a t + \theta) = \cos(\omega_a t - \theta) -$$ - -Clearly the *apparent frequency* of a sinusoid of frequency −*fa* is also *fa*. However, its phase undergoes a sign change. This means the apparent frequency of any sampled sinusoid lies in the range from 0 to *fs*/2 Hz. To summarize, if a continuous-time sinusoid of frequency *f* Hz is sampled at a rate of *fs* Hz (samples/second), the resulting samples would appear as samples of a continuous-time sinusoid of frequency |*fa*| that lies in the band from 0 to *fs*/2. According to Eq. (8.7), - -$$ -|f_a| = |f - mf_s| \qquad |f_a| \le \frac{f_s}{2} \qquad m \text{ an integer} -$$ - -**Figure 8.9** Apparent frequencies of a sampled sinusoid: **(a)** *fa* versus *f* and **(b)** |*fa*| versus *f* . - -The plot of the apparent frequency |*fa*| versus *f* is shown in Fig. 8.9b.† As expected, the apparent frequency |*fa*| of any sampled sinusoid, regardless of its frequency, is always in the range of 0 to *fs*/2 Hz. However, when *fa* is negative, the phase of the apparent sinusoid undergoes a sign change. The frequency belts in which such phase changes occur are shown shaded in Fig. 8.9b. - -Consider, for example, a sinusoid cos(2π*ft* +θ ) with *f* =8000 Hz sampled at a rate *fs* = 3000 Hz. Using Eq. (8.7), we obtain *fa* = 8000 − 3 × 3000 = −1000. Hence, |*fa*| = 1000. The samples would appear to have come from a sinusoid cos(2000π*t* − θ ). Observe the sign change of the phase because *fa* is negative.‡ - -In the light of the foregoing development, let us consider a sinusoid of frequency *f* = (*fs*/2)+*fz*, sampled at a rate of *fs* Hz. According to Eq. (8.7), - -$$ -f_a = \frac{f_s}{2} + f_z - (1 \times f_s) = -\frac{f_s}{2} + f_z -$$ - -Hence, the apparent frequency is |*fa*| = (*fs*/2) − *fz*, confirming our earlier result. However, the phase of the sinusoid will suffer a sign change because *fa* is negative. - -Figure 8.10 shows how samples of sinusoids of two different frequencies (sampled at the same rate) generate identical sets of samples. Both the sinusoids are sampled at a rate *fs* = 5 Hz (*T* = 0.2 second). The frequencies of the two sinusoids, 1 Hz (period 1) and 6 Hz (period 1/6), differ by *fs* = 5 Hz. - -† The plots in Figs. 8.9 and 5.17 are identical. This is because a sampled sinusoid is basically a discrete-time sinusoid. - -‡ For phase sign change, we are assuming that the signal has the form cos(2π*ft* + θ ). If the form is sin(2π*ft* + θ ), the rule changes slightly. It is left as an exercise for the reader to show that when *fa* < 0, this sinusoid appears as −sin(2π|*fa*|*t* − θ ). Thus, in addition to phase change, the amplitude also changes sign. - -**Figure 8.10** Demonstration of aliasing. - -The reason for aliasing can be clearly seen in Fig. 8.10. The root of the problem is the sampling rate, which may be adequate for the lower-frequency sinusoid but is clearly inadequate for the higher-frequency sinusoid. The figure clearly shows that between the successive samples of the higher-frequency sinusoid, there are wiggles, which are bypassed or ignored, and are unrepresented in the samples, indicating a sub-Nyquist rate of sampling. The frequency of the apparent signal *xa*(*t*) is always the lowest possible frequency that lies within the band |*f* | ≤ *fs*/2. Thus, the apparent frequency of the samples in this example is 1 Hz. If these samples are chosen to reconstruct a signal using a lowpass filter of bandwidth *fs*/2, we shall obtain a sinusoid of frequency 1 Hz. - -### **EXAMPLE 8.4 Apparent Frequency of Sampled Sinusoids** - -A continuous-time sinusoid cos(2π*ft* + θ ) is sampled at a rate *fs* = 1000 Hz. Determine the apparent (aliased) sinusoid of the resulting samples if the input signal frequency *f* is **(a)** 400 Hz, **(b)** 600 Hz, **(c)** 1000 Hz, and **(d)** 2400 Hz. - -The folding frequency is *fs*/2 = 500. Hence, sinusoids below 500 Hz (frequency within the fundamental band) will not be aliased and sinusoids of frequency above 500 Hz will be aliased. - -**(c)** Since *f* = 1000 Hz can be expressed as 1000 = 0 + 1000, we see that *fa* = 0. Hence, the aliased frequency is 0 Hz (dc), and there is no phase sign change. The apparent sinusoid is *y*(*t*) = cos(0π*t* ±θ ) = cos(θ ). This is a dc signal with constant sample values for all *n*. - -**(a)** Since *f* = 400 Hz is less than 500 Hz, there is no aliasing. The apparent sinusoid is cos(2π*ft* +θ ) with *f* = 400. - -**(b)** Since *f* = 600 Hz can be expressed as 600 = −400 + 1000, we see that *fa* = −400. Hence, the aliased frequency is 400 Hz and the phase changes sign. The apparent (aliased) sinusoid is cos(2π*ft* −θ ) with *f* = 400. - -### 796 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -**(d)** Here, *f* = 2400 Hz can be expressed as 2400 = 400 + (2 × 1000) so that *fa* = 400. Hence, the aliased frequency is 400 Hz and there is no sign change for the phase. The apparent sinusoid is cos(2π*ft* +θ ) with *f* = 400. - -We could have found these answers directly from Fig. 8.9b. For example, for case (b), we read |*fa*| = 400 corresponding to *f* = 600. Moreover, *f* = 600 lies in the shaded belt. Hence, there is a phase sign change. - -## **DR ILL 8.3 A Case of Identical Sampled Sinusoids** - -Show that samples of 90 Hz and 110 Hz sinusoids of the form cosω*t* are identical when sampled at a rate 200 Hz. - -### **DR ILL 8.4 Apparent Frequency of Sampled Sinusoids** - -A sinusoid of frequency *f*0 Hz is sampled at a rate of 100 Hz. Determine the apparent frequency of the samples if *f*0 is **(a)** 40 Hz, **(b)** 60 Hz, **(c)** 140 Hz, and **(d)** 160 Hz. - -### **ANSWERS** - -All four cases have an apparent frequency of 40 Hz. - -### **[8.2-2 Some Applications of the Sampling Theorem](#page-13-0)** - -The sampling theorem is very important in signal analysis, processing, and transmission because it allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing a continuous-time signal is therefore equivalent to processing a discrete sequence of numbers. Such processing leads us directly into the area of digital filtering. In the field of communication, the transmission of a continuous-time message reduces to the transmission of a sequence of numbers by means of pulse trains. The continuous-time signal *x*(*t*) is sampled, and sample values are used to modify certain parameters of a periodic pulse train. We may vary the amplitudes (Fig. 8.11b), widths (Fig. 8.11c), or positions (Fig. 8.11d) of the pulses in proportion to the sample values of the signal *x*(*t*). Accordingly, we may have *pulse-amplitude modulation* (PAM), *pulse-width modulation* (PWM), or *pulse-position modulation* (PPM). The most important form of pulse modulation today is *pulse-code modulation* (PCM), discussed in Sec. 8.3 in connection with Fig. 8.14b. In all these cases, instead of transmitting *x*(*t*), we transmit the corresponding pulse-modulated signal. At the receiver, we read the information of the pulse-modulated signal and reconstruct the analog signal *x*(*t*). - -**Figure 8.11** Pulse-modulated signals. **(a)** The signal. **(b)** The PAM signal. **(c)** The PWM (PDM) signal. **(d)** The PAM signal. - -One advantage of using pulse modulation is that it permits the simultaneous transmission of several signals on a time-sharing basis—*time-division multiplexing* (TDM). Because a pulse-modulated signal occupies only a part of the channel time, we can transmit several pulse-modulated signals on the same channel by interweaving them. Figure 8.12 shows the TDM of two PAM signals. In this manner, we can multiplex several signals on the same channel by reducing pulse widths.† - -Digital signals also offer an advantage in the area of communications, where signals must travel over distances. Transmission of digital signals is more rugged than that of analog signals because digital signals can withstand channel noise and distortion much better as long as the noise - -† Another method of transmitting several baseband signals simultaneously is frequency-division multiplexing (FDM) discussed in Sec. 7.7-4. In FDM, various signals are multiplexed by sharing the channel bandwidth. The spectrum of each message is shifted to a specific band not occupied by any other signal. The information of various signals is located in nonoverlapping frequency bands of the channel (Fig. 7.45). In a way, TDM and FDM are duals of each other. - -**Figure 8.12** Time-division multiplexing of two signals. - -**Figure 8.13** Digital signal transmission: **(a)** at the transmitter, **(b)** received distorted signal (without noise), **(c)** received distorted signal (with noise), and **(d)** regenerated signal at the receiver. - -and the distortion are within limits. An analog signal can be converted to digital binary form through sampling and quantization (rounding off), as explained in the next section. The digital (binary) message in Fig. 8.13a is distorted by the channel, as illustrated in Fig. 8.13b. Yet if the distortion remains within a limit, we can recover the data without error because we need only make a simple binary decision: Is the received pulse positive or negative? Figure 8.13c shows the same data with channel distortion and noise. Here again, the data can be recovered correctly as long as the distortion and the noise are within limits. Such is not the case with analog messages. Any distortion or noise, no matter how small, will distort the received signal. - -The greatest advantage of digital communication over the analog counterpart, however, is the viability of regenerative repeaters in the former. In an analog transmission system, a message signal grows progressively weaker as it travels along the channel (transmission path), whereas the channel noise and the signal distortion, being cumulative, become progressively stronger. Ultimately, the signal, overwhelmed by noise and distortion, is mutilated. Amplification is of little help because it enhances the signal and the noise in the same proportion. Consequently, the distance over which an analog message can be transmitted is limited by the transmitted power. If a transmission path is long enough, the channel distortion and noise will accumulate sufficiently to overwhelm even a digital signal. The trick is to set up repeaters along the transmission path at distances short enough to permit detection of signal pulses before the noise and distortion have a chance to accumulate sufficiently. At each repeater, the pulses are detected, and new, clean pulses are transmitted to the next repeater, which, in turn, duplicates the same process. If the noise and distortion remain within limits (which is possible because of the closely spaced repeaters), pulses can be detected correctly.† This way the digital messages can be transmitted over longer distances with greater reliability. In contrast, analog messages cannot be cleaned up periodically, and their transmission is therefore less reliable. The most significant error in digitized signals comes from quantizing (rounding off). This error, discussed in Sec. 8.3, can be reduced as much as desired by increasing the number of quantization levels, at the cost of an increased bandwidth of the transmission medium (channel). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/104_8.3 ANALOG-TO-DIGITAL (A - D) CONVERSION.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/104_8.3 ANALOG-TO-DIGITAL (A - D) CONVERSION.md deleted file mode 100644 index 570917e942f9eabb55109c511a1492f295b4d485..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/104_8.3 ANALOG-TO-DIGITAL (A - D) CONVERSION.md +++ /dev/null @@ -1,77 +0,0 @@ -## **[8.3 ANALOG-TO-DIGITAL](#page-13-0) (A/D) CONVERSION** - -The amplitude of an *analog* signal can take on any value over a continuous range. Hence, analog signal amplitude can take on an infinite number of values. In contrast, a *digital* signal amplitude can take on only a finite number of values. An analog signal can be converted into a digital signal by means of sampling and *quantizing* (rounding off). Sampling an analog signal alone will not yield a digital signal because a sample of analog signal can still take on any value in a continuous range. It is digitized by rounding off its value to one of the closest permissible numbers (or *quantized levels*), as illustrated in Fig. 8.14a, which represents one possible quantizing scheme. The amplitudes of the analog signal *x*(*t*) lie in the range (−*V*,*V*). This range is partitioned into *L* subintervals, each of magnitude = 2*V*/*L*. Next, each sample amplitude is approximated by the midpoint value of the subinterval in which the sample falls (see Fig. 8.14a for *L* = 16). It is clear that each sample is approximated to one of the *L* numbers. Thus, the signal is digitized with quantized samples taking on any one of the *L* values. This is an *L*-ary digital signal (see Sec. 1.3-2). Each sample can now be represented by one of *L* distinct pulses. - -From a practical viewpoint, dealing with a large number of distinct pulses is difficult. We prefer to use the smallest possible number of distinct pulses, the very smallest number being 2. A digital signal using only two symbols or values is the binary signal. A binary digital signal (a signal that can take on only two values) is very desirable because of its simplicity, economy, and ease of engineering. We can convert an *L*-ary signal into a binary signal by using pulse coding. Figure 8.14b shows one such code for the case of *L* = 16. This code, formed by binary representation of the 16 decimal digits from 0 to 15, is known as the *natural binary code (NBC)*. For *L* quantization levels, we need a minimum of *b* binary code digits, where 2*b* = *L* or *b* = log2 *L*. - -Each of the 16 levels is assigned one binary code word of four digits. Thus, each sample in this example is encoded by four binary digits. To transmit or digitally process the binary data, we need to assign a distinct electrical pulse to each of the two binary states. One possible way is to assign a negative pulse to a binary **0** and a positive pulse to a binary **1** so that each sample is now represented by a group of four binary pulses (pulse code), as depicted in Fig. 8.14b. The resulting binary signal is a digital signal obtained from the analog signal *x*(*t*) through A/D conversion. In communications jargon, such a signal is known as a pulse-code-modulated (PCM) signal. - -† The error in pulse detection can be made negligible. - -| (a) | -|-----| -| | -| | -| | - -| Digit | Binary equivalent | Pulse code waveform | -|-------|-------------------|---------------------| -| 0 | 0000 | | -| 1 | 0001 | | -| 2 | 0010 | | -| 3 | 0011 | | -| 4 | 0100 | | -| 5 | 0101 | | -| 6 | 0110 | | -| 7 | 0111 | | -| 8 | 1000 | | -| 9 | 1001 | | -| 10 | 1010 | | -| 11 | 1011 | | -| 12 | 1100 | | -| 13 | 1101 | | -| 14 | 1110 | | -| 15 | 1111 | | - -(b) - -**Figure 8.14** Analog-to-digital (A/D) conversion of a signal: **(a)** quantizing and **(b)** pulse coding. - -The convenient contraction of "*b*inary digi*t*" to *bit* has become an industry standard abbreviation. - -The audio signal bandwidth is about 15 kHz, but subjective tests show that signal articulation (intelligibility) is not affected if all the components above 3400 Hz are suppressed [3]. Since the objective in telephone communication is intelligibility rather than high fidelity, the components above 3400 Hz are eliminated by a lowpass filter.† The resulting signal is then sampled at a rate of 8000 samples/s (8 kHz). This rate is intentionally kept higher than the Nyquist sampling rate of 6.8 kHz to avoid unrealizable filters required for signal reconstruction. Each sample is finally quantized into 256 levels (*L* = 256), which requires a group of eight binary pulses to encode each sample (28 = 256). Thus, a digitized telephone signal consists of data amounting to 8 × 8000 = 64,000 or 64 kbit/s, requiring 64,000 binary pulses per second for its transmission. - -The compact disc (CD), a high-fidelity application of A/D conversion, requires the audio signal bandwidth of 20 kHz. Although the Nyquist sampling rate is only 40 kHz, an actual sampling rate of 44.1 kHz is used for the reason mentioned earlier. The signal is quantized into a rather large number of levels (*L* = 65,536) to reduce quantizing error. The binary-coded samples are now recorded on the CD. - -### A HISTORICAL NOTE - -The binary system of representing any number by using **1**s and **0**s was invented in India by Pingala (ca. 200 BCE). It was again worked out independently in the West by Gottfried Wilhelm Leibniz (1646–1716). He felt a spiritual significance in this discovery, reasoning that **1** representing unity was clearly a symbol for God, while **0** represented the nothingness. He reasoned that if all numbers can be represented merely by the use of **1** and **0**, this surely proves that God created the universe out of nothing! - -### **EXAMPLE 8.5 ADC Bit Number and Bit Rate** - -A signal *x*(*t*) bandlimited to 3 kHz is sampled at a rate 331 3% higher than the Nyquist rate. The maximum acceptable error in the sample amplitude (the maximum error due to quantization) is 0.5% of the peak amplitude *V*. The quantized samples are binary-coded. Find the required sampling rate, the number of bits required to encode each sample, and the bit rate of the resulting PCM signal. - -The Nyquist sampling rate is *f*Nyq = 2×3000 = 6000 Hz (samples/s). The actual sampling rate is *fA* = 6000×(11 3 ) = 8000 Hz. - -The quantization step is , and the maximum quantization error is ±/2, where = 2*V*/*L*. The maximum error due to quantization, /2, should be no greater than 0.5% of the - -†Components below 300 Hz may also be suppressed without affecting the articulation. - -### 802 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -signal peak amplitude *V*. Therefore, - -$$ -\frac{\Delta}{2} = \frac{V}{L} = \frac{0.5}{100}V \implies L = 200 -$$ - -For binary coding, *L* must be a power of 2. Hence, the next higher value of *L* that is a power of 2 is *L* = 256. Because log2 256 = 8, we need 8 bits to encode each sample. Therefore the bit rate of the PCM signal is - -8000×8 = 64,000 bits/s - -### **DR ILL 8.5 Bit Number and Bit Rate for ASCII** - -The American Standard Code for Information Interchange (ASCII) has 128 characters, which are binary-coded. A certain computer generates 100,000 characters per second. Show that - -- **(a)** 7 bits (binary digits) are required to encode each character -- **(b)** 700,000 bits/s are required to transmit the computer output. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/105_8.4 DUAL OF TIME SAMPLING - SPECTRAL SAMPLING.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/105_8.4 DUAL OF TIME SAMPLING - SPECTRAL SAMPLING.md deleted file mode 100644 index 6514cce10329be5f2ac98e16db778ff7ac884fa7..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/105_8.4 DUAL OF TIME SAMPLING - SPECTRAL SAMPLING.md +++ /dev/null @@ -1,82 +0,0 @@ -## **8.4 DUAL OF TIME [SAMPLING: SPECTRAL](#page-13-0) SAMPLING** - -As in other cases, the sampling theorem has its dual. In Sec. 8.1, we discussed the time-sampling theorem and showed that a signal bandlimited to *B* Hz can be reconstructed from the signal samples taken at a rate of *fs* > 2*B* samples/s. Note that the signal spectrum exists over the frequency range (in hertz) of −*B* to *B*. Therefore, 2*B* is the spectral width (not the bandwidth, which is *B*) of the signal. This fact means that a signal *x*(*t*) can be reconstructed from samples taken at a rate *fs* > the spectral width of *X*(ω) in hertz ( *fs* > 2*B*). - -We now prove the dual of the time-sampling theorem. This is the *spectral sampling theorem*, which applies to timelimited signals (the dual of bandlimited signals). A timelimited signal *x*(*t*) exists only over a finite interval of τ seconds, as shown in Fig. 8.15a. Generally, a timelimited signal is characterized by *x*(*t*) = 0 for *t* < *T*1 and *t* > *T*2 (assuming *T*2 > *T*1). The signal width or duration is τ = *T*2 −*T*1 seconds. - -The spectral sampling theorem states that the spectrum *X*(ω) of a signal *x*(*t*) timelimited to a duration of τ seconds can be reconstructed from the samples of *X*(ω) taken at a rate *R* samples/Hz, where *R* > τ (the signal width or duration) in seconds. - -Figure 8.15a shows a timelimited signal *x*(*t*) and its Fourier transform *X*(ω). Although *X*(ω) is complex in general, it is adequate for our line of reasoning to show *X*(ω) as a real function. - -$$ -X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt = \int_{0}^{\tau} x(t)e^{-j\omega t}dt -$$ -\n(8.8) - - - -**Figure 8.15** Periodic repetition of a signal amounts to sampling its spectrum. - -We now construct *xT*0 (*t*), a periodic signal formed by repeating *x*(*t*) every *T*0 seconds (*T*0 > τ ), as depicted in Fig. 8.15b. This periodic signal can be expressed by the exponential Fourier series - -$$ -x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -where (assuming *T*0 > τ ) - -$$ -D_n = \frac{1}{T_0} \int_0^{T_0} x(t) e^{-jn\omega_0 t} dt = \frac{1}{T_0} \int_0^{\tau} x(t) e^{-jn\omega_0 t} dt -$$ - -From Eq. (8.8), it follows that - -$$ -D_n = \frac{1}{T_0} X(n\omega_0) -$$ - -This result indicates that the coefficients of the Fourier series for *xT*0 (*t*) are (1/*T*0) times the sample values of the spectrum *X*(ω) taken at intervals of ω0. This means that the spectrum of the periodic signal *xT*0 (*t*) is the sampled spectrum *X*(ω), as illustrated in Fig. 8.15b. Now as long as *T*0 > τ , the successive cycles of *x*(*t*) appearing in *xT*0 (*t*) do not overlap, and *x*(*t*) can be recovered from *xT*0 (*t*). Such recovery implies indirectly that *X*(ω) can be reconstructed from its samples. These samples are separated by the fundamental frequency *f*0 = 1/*T*0 Hz of the periodic signal *xT*0 (*t*). Hence, the condition for recovery is *T*0 > τ ; that is, - -$$ -f_0 < \frac{1}{\tau} \, \mathrm{Hz} -$$ - -Therefore, to be able to reconstruct the spectrum *X*(ω) from the samples of *X*(ω), the samples should be taken at frequency intervals *f*0 < 1/τ Hz. If *R* is the sampling rate (samples/Hz), then - -$$ -R = \frac{1}{f_0} > \tau \text{ samples/Hz} -$$ - -#### 804 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -### SPECTRAL INTERPOLATION - -Consider a signal timelimited to τ seconds and centered at *Tc*. We now show that the spectrum *X*(ω) of *x*(*t*) can be reconstructed from the samples of *X*(ω). For this case, using the dual of the approach employed to derive the signal interpolation formula in Eq. (8.6), we obtain the spectral interpolation formula† - -$$ -X(\omega) = \sum_{n=-\infty}^{\infty} X(n\omega_0) \operatorname{sinc}\left(\frac{\omega T_0}{2} - n\pi\right) e^{-j(\omega - n\omega_0)T_c} \qquad \omega_0 = \frac{2\pi}{T_0} \qquad T_0 > \tau \tag{8.9} -$$ - -For the case in Fig. 8.15, *Tc* = *T*0/2. If the pulse *x*(*t*) were to be centered at the origin, then *Tc* = 0, and the exponential term at the extreme right in Eq. (8.9) would vanish. In such a case, Eq. (8.9) would be the exact dual of Eq. (8.6). - -### **EXAMPLE 8.6 Spectral Sampling and Interpolation** - -The spectrum *X*(ω) of a unit-duration signal *x*(*t*), centered at the origin, is sampled at the intervals of 1 Hz or 2π rad/s (the Nyquist rate). The samples are - -$$ -X(0) = 1 -$$ - and $X(\pm 2\pi n) = 0$ $n = 1, 2, 3, ...$ - -Find *x*(*t*). - -We use the interpolation formula Eq. (8.9) (with *Tc* = 0) to construct *X*(ω) from its samples. Since all but one of the Nyquist samples are zero, only one term (corresponding to *n* = 0) in the summation on the right-hand side of Eq. (8.9) survives. Thus, with *X*(0) = 1 and τ = *T*0 = 1, we obtain - -$$ -X(\omega) = \operatorname{sinc}\left(\frac{\omega}{2}\right) -$$ - and $x(t) = \operatorname{rect}(t)$ - -For a signal of unit duration, this is the only spectrum with the sample values *X*(0) = 1 and *X*(2π*n*) = 0(*n* = 0). No other spectrum satisfies these conditions. - -† This can be obtained by observing that the Fourier transform of *xT*0 (*t*) is 2π % *n Dn*δ(ω − *n*ω0) [see Eq. (7.22)]. We can recover *x*(*t*) from *xT*0 (*t*) by multiplying the latter with rect(*t* − *Tc*)/*T*0, whose Fourier transform is *T*0 sinc(ω*T*0/2)*e*−*j*ω*Tc* . Hence, *X*(ω) is 1/2π times the convolution of these two Fourier transforms, which yields Eq. (8.9). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/106_8.5 NUMERICAL COMPUTATION OF THE FOURIER TRANSFORM - THE DISCRETE FOURIER TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/106_8.5 NUMERICAL COMPUTATION OF THE FOURIER TRANSFORM - THE DISCRETE FOURIER TRANSFORM.md deleted file mode 100644 index 16acfa1c77df28649580a98b9ae453f20a3c6ac2..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/106_8.5 NUMERICAL COMPUTATION OF THE FOURIER TRANSFORM - THE DISCRETE FOURIER TRANSFORM.md +++ /dev/null @@ -1,552 +0,0 @@ -## **8.5 NUMERICAL [COMPUTATION OF THE](#page-13-0) FOURIER TRANSFORM: THE DISCRETE FOURIER TRANSFORM** - -Numerical computation of the Fourier transform of *x*(*t*) requires sample values of *x*(*t*) because a digital computer can work only with discrete data (sequence of numbers). Moreover, a computer can compute *X*(ω) only at some discrete values of ω [samples of *X*(ω)]. We therefore need to relate the samples of *X*(ω) to samples of *x*(*t*). This task can be accomplished by using the results of the two sampling theorems developed in Secs. 8.1 and 8.4. - -We begin with a timelimited signal *x*(*t*) (Fig. 8.16a) and its spectrum *X*(ω) (Fig. 8.16b). Since *x*(*t*) is timelimited, *X*(ω) is nonbandlimited. For convenience, we shall show all spectra as functions of the frequency variable *f* (in hertz) rather than ω. According to the sampling theorem, the spectrum *X*(ω) of the sampled signal *x*(*t*) consists of *X*(ω) repeating every *fs* Hz, where *fs* = 1/*T*, as depicted in Fig. 8.16d.† In the next step, the sampled signal in Fig. 8.16c is repeated periodically every *T*0 seconds, as illustrated in Fig. 8.16e. According to the spectral sampling theorem, such an operation results in sampling the spectrum at a rate of *T*0 samples/Hz. This sampling rate means that the samples are spaced at *f*0 = 1/*T*0 Hz, as depicted in Fig. 8.16f. - -The foregoing discussion shows that when a signal *x*(*t*) is sampled and then periodically repeated, the corresponding spectrum is also sampled and periodically repeated. Our goal is to relate the samples of *x*(*t*) to the samples of *X*(ω). - -### NUMBER OF SAMPLES - -One interesting observation from Figs. 8.16e and 8.16f is that *N*0, the number of samples of the signal in Fig. 8.16e in one period *T*0, is identical to *N* 0, the number of samples of the spectrum in Fig. 8.16f in one period *fs*. To see this, we notice that - -$$ -N_0 = \frac{T_0}{T} \quad N'_0 = \frac{f_s}{f_0} \quad f_s = \frac{1}{T} \quad \text{and} \quad f_0 = \frac{1}{T_0} -$$ -(8.10) - -Using these relations, we see that - -$$ -N_0 = \frac{T_0}{T} = \frac{f_s}{f_0} = N'_0 -$$ - -### ALIASING AND LEAKAGE IN NUMERICAL COMPUTATION - -Figure 8.16f shows the presence of aliasing in the samples of the spectrum *X*(ω). This aliasing error can be reduced as much as desired by increasing the sampling frequency *fs* (decreasing the sampling interval *T* = 1/*fs*). The aliasing can never be eliminated for timelimited *x*(*t*), however, because its spectrum *X*(ω) is nonbandlimited. Had we started with a signal having a bandlimited spectrum *X*(ω), there would be no aliasing in the spectrum in Fig. 8.16f. Unfortunately, such a signal is nontimelimited, and its repetition (in Fig. 8.16e) would result in signal overlapping (aliasing in the time domain). In this case, we shall have to contend with errors in signal - -† There is a multiplying constant 1/*T* for the spectrum in Fig. 8.16d [see Eq. (8.2)], but this is irrelevant to our discussion here. - -samples. In other words, in computing the direct or inverse Fourier transform numerically, we can reduce the error as much as we wish, but the error can never be eliminated. This is true of numerical computation of the direct and inverse Fourier transforms, regardless of the method used. For example, if we determine the Fourier transform by direct integration numerically, by using Eq. (7.9), there will be an error because the interval of integration *t* can never be made zero. Similar remarks apply to numerical computation of the inverse transform. Therefore, we should always keep in mind the nature of this error in our results. In our discussion (Fig. 8.16), we assumed *x*(*t*) to be a timelimited signal. If *x*(*t*) were not timelimited, we would need to timelimit it because numerical computations can work only with finite data. Furthermore, this data truncation causes error because of spectral spreading (smearing) and leakage, as discussed in Sec. 7.8. The leakage also causes aliasing. Leakage can be reduced by using a tapered window for signal truncation. But this choice increases spectral spreading or smearing. Spectral spreading can be reduced by increasing the window width (i.e., more data), which increases *T*0, and reduces *f*0 (increases *spectral* or *frequency resolution*). - -### PICKET FENCE EFFECT - -The numerical computation method yields only the uniform sample values of *X*(ω). The major peaks or valleys of *X*(ω) can lie between two samples and may remain hidden, giving a false picture of reality. Viewing samples is like viewing the signal and its spectrum through a "picket fence" with upright posts that are very wide and placed close together. What is hidden behind the pickets is much more than what we can see. Such misleading results can be avoided by using a sufficiently large *N*0, the number of samples, to increase resolution. We can also use zero padding (discussed later) or the spectral interpolation formula [Eq. (8.9)] to determine the values of *X*(ω) between samples. - -### POINTS OF DISCONTINUITY - -If *x*(*t*) or *X*(ω) has a jump discontinuity at a sampling point, the sample value should be taken as the average of the values on the two sides of the discontinuity because the Fourier representation at a point of discontinuity converges to the average value. - -### DERIVATION OF THE DISCRETE FOURIER TRANSFORM (DFT) - -If *x*(*nT*) and *X*(*r*ω0) are the *n*th and *r*th samples of *x*(*t*) and *X*(ω), respectively, then we define new variables *xn* and *Xr* as - -$$ -x_n = Tx(nT) = \frac{T_0}{N_0}x(nT) -$$ -\n(8.11) - -and - -$$ -X_r = X(r\omega_0) -$$ - -where - -$$ -\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0} -$$ - -We shall now show that *xn* and *Xr* are related by the following equations† : - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n} -$$ - (8.12) - -and - -$$ -x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{ir\Omega_0 n} -$$ -\n(8.13) - -where - -$$ -\Omega_0 = \omega_0 T = \frac{2\pi}{N_0} -$$ - -These equations define the direct and the inverse *discrete Fourier transforms,* with *Xr* the direct discrete Fourier transform (DFT) of *xn*, and *xn* the inverse discrete Fourier transform (IDFT) of *Xr*. The notation - -$$ -x_n \Longleftrightarrow X_r -$$ - -is also used to indicate that *xn* and *Xr* are a DFT pair. Remember that *xn* is *T*0/*N*0 times the *n*th sample of *x*(*t*) and *Xr* is the *r*th sample of *X*(ω). Knowing the sample values of *x*(*t*), we can use the DFT to compute the sample values of *X*(ω)—and vice versa. Note, however, that *xn* is a function of *n* (*n* = 0, 1, 2,...,*N*0 − 1) rather than of *t* and that *Xr* is a function of *r* (*r* = 0, 1, 2,...,*N*0 − 1) rather than of ω. Moreover, both *xn* and *Xr* are periodic sequences of period *N*0 (Figs. 8.16e, 8.16f). Such sequences are called *N*0*-periodic sequences*. The proof of the DFT relationships in Eqs. (8.12) and (8.13) follows directly from the results of the sampling theorem. The sampled signal *x*(*t*) (Fig. 8.16c) can be expressed as - -$$ -\overline{x}(t) = \sum_{n=0}^{N_0 - 1} x(nT)\delta(t - nT) -$$ - -Since δ (*t* −*nT*) ⇐⇒ *e*−*jn*ω*T* , applying the Fourier transform yields - -$$ -\overline{X}(\omega) = \sum_{n=0}^{N_0 - 1} x(nT) e^{-jn\omega T} -$$ - -But from Fig. 8.1f [or Eq. (8.2)], it is clear that over the interval |ω| ≤ ω*s*/2, *X*(ω), the Fourier transform of *x*(*t*) is *X*(ω)/*T*, assuming negligible aliasing. Hence, - -$$ -X(\omega) = T\overline{X}(\omega) = T\sum_{n=0}^{N_0 - 1} x(nT)e^{-jn\omega T} \qquad |\omega| \le \frac{\omega_s}{2} -$$ - -† In Eqs. (8.12) and (8.13), the summation is performed from 0 to *N*0 −1. It is shown in Sec. 9.1-2 [Eqs. (9.6) and (9.7)] that the summation may be performed over any successive *N*0 values of *n* or *r*. - -and - -$$ -X_r = X(r\omega_0) = T \sum_{n=0}^{N_0 - 1} x(nT)e^{-nkr\omega_0 T} -$$ -\n(8.14) - -If we let ω0*T* = 0, then from Eq. (8.10), - -$$ -\Omega_0 = \omega_0 T = 2\pi f_0 T = \frac{2\pi}{N_0} -$$ - -Also, from Eq. (8.11), - -$$ -Tx(nT) = x_n -$$ - -Therefore, Eq. (8.14) becomes - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -This is Eq. (8.12), which we set to prove. - -The inverse transform relationship of Eq. (8.13) can be derived by using a similar procedure with the roles of *t* and ω reversed, but here we shall use a more direct proof. To prove Eq. (8.13), we multiply both sides of Eq. (8.12) by *ejm*0*r* and sum over *r* as - -$$ -\sum_{r=0}^{N_0-1} X_r e^{jm\Omega_0 r} = \sum_{r=0}^{N_0-1} \left[ \sum_{n=0}^{N_0-1} x_n e^{-jr\Omega_0 n} \right] e^{jm\Omega_0 r} -$$ - -By interchanging the order of summation on the right-hand side, we have - -$$ -\sum_{r=0}^{N_0-1} X_r e^{jm\Omega_0 r} = \sum_{n=0}^{N_0-1} x_n \left[ \sum_{r=0}^{N_0-1} e^{j(m-n)\Omega_0 r} \right] -$$ - -As the footnote below readily shows, the inner sum on the right-hand side is zero for *n* = *m* and is *N*0 when *n* = *m*. † Thus, the outer sum will have only one nonzero term when *n* = *m*, and it is - -† We show that *N* - -$$ -\sum_{n=0}^{N_0-1} e^{jk\Omega_0 n} = \begin{cases} N_0 & k = 0, \pm N_0, \pm 2N_0, \dots \\ 0 & \text{otherwise} \end{cases} -$$ - (8.15) - -Recall that 0*N*0 = 2π. So *ejk*0*n* = 1 when *k* = 0,±*N*0,±2*N*0,.... Hence, the sum on the left-hand side of Eq. (8.15) is *N*0. To compute the sum for other values of *k*, we note that the sum on the left-hand side of Eq. (8.15) is a geometric progression with common ratio α = *ejk*0 . Therefore, (see Sec. B.8-3) - -$$ -\sum_{n=0}^{N_0-1} e^{jk\Omega_0 n} = \frac{e^{jk\Omega_0 N_0} - 1}{e^{jk\Omega_0} - 1} = 0 \qquad (e^{jk\Omega_0 N_0} = e^{j2\pi m} = 1) -$$ - -#### 810 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -*N*0*xn* = *N*0*xm*. Therefore, - -$$ -x_m = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{im\Omega_0 r} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -Because *Xr* is *N*0 periodic, we need to determine the values of *Xr* over any one period. It is customary to determine *Xr* over the range (0, *N*0 − 1), rather than over the range (−*N*0/2,(*N*0/2)−1).† - -## CHOICE OF *T* AND *T*0 - -In DFT computation, we first need to select suitable values for *N*0 and *T* or *T*0. For this purpose, we begin by deciding on *B*, the essential bandwidth (in hertz) of the signal. The sampling frequency *fs* must be at least 2*B*, that is, - -$$ -\frac{f_s}{2} \ge B -$$ - -Moreover, the sampling interval *T* = 1/*fs* [Eq. (8.10)], and - -$$ -T \le \frac{1}{2B} \tag{8.16} -$$ - -Once we pick *B*, we can choose *T* according to Eq. (8.16). Also, - -$$ -f_0 = \frac{1}{T_0} \tag{8.17} -$$ - -where *f*0 is the *frequency resolution* [separation between samples of *X*(ω)]. Hence, if *f*0 is given, we can pick *T*0 according to Eq. (8.17). Knowing *T*0 and *T*, we determine *N*0 from - -$$ -N_0 = \frac{T_0}{T} -$$ - -### ZERO PADDING - -Recall that observing *Xr* is like observing the spectrum *X*(ω) through a picket fence. If the frequency sampling interval *f*0 is not sufficiently small, we could miss out on some significant details and obtain a misleading picture. To obtain a higher number of samples, we need to reduce *f*0. Because *f*0 =1/*T*0, a higher number of samples requires us to increase the value of *T*0, the period of repetition for *x*(*t*). This option increases *N*0, the number of samples of *x*(*t*), by adding dummy samples of 0 value. This addition of dummy samples is known as *zero padding*. Thus, zero padding increases the number of samples and may help in getting a better idea of the spectrum *X*(ω) from its samples *Xr*. To continue with our picket fence analogy, zero padding is like using more, and narrower, pickets. - -† The DFT of Eq. (8.12) and the IDFT of Eq. (8.13) represent a transform in their own right, and they are exact. There is no approximation. However, *xn* and *Xr*, thus obtained, are only approximations to the actual samples of a signal *x*(*t*) and of its Fourier transform *X*(ω). - -### ZERO PADDING DOES NOT IMPROVE ACCURACY OR RESOLUTION - -Actually, we are not observing *X*(ω) through a picket fence. We are observing a distorted version of *X*(ω) resulting from the truncation of *x*(*t*). Hence, we should keep in mind that even if the fence were transparent, we would see a reality distorted by aliasing. Seeing through the picket fence just gives us an imperfect view of the imperfectly represented reality. Zero padding only allows us to look at more samples of that imperfect reality. It can never reduce the imperfection in what is behind the fence. The imperfection, which is caused by aliasing, can be lessened only by reducing the sampling interval *T*. Observe that reducing *T* also increases *N*0, the number of samples, and is like increasing the number of pickets while reducing their width. But in this case, the reality behind the fence is also better dressed and we see more of it. - -### **EXAMPLE 8.7 Number of Samples and Frequency Resolution** - -A signal *x*(*t*) has a duration of 2 ms and an essential bandwidth of 10 kHz. It is desirable to have a frequency resolution of 100 Hz in the DFT (*f*0 = 100). Determine *N*0. - -To have *f*0 = 100 Hz, the effective signal duration *T*0 must be - -$$ -T_0 = \frac{1}{f_0} = \frac{1}{100} = 10 -$$ - ms - -Since the signal duration is only 2 ms, we need zero padding over 8 ms. Also, *B* = 10,000. Hence, *fs* = 2*B* = 20,000 and *T* = 1/*fs* = 50 µs. Furthermore, - -$$ -N_0 = \frac{f_s}{f_0} = \frac{20,000}{100} = 200 -$$ - -The *fast Fourier transform* (FFT) algorithm (discussed later; see Sec. 8.6) is used to compute DFT, where it proves convenient (although not necessary) to select *N*0 as a power of 2; that is, *N*0 = 2*n* (*n*, integer). Let us choose *N*0 = 256. Increasing *N*0 from 200 to 256 can be used to reduce aliasing error (by reducing *T*), to improve resolution (by increasing *T*0 using zero padding), or a combination of both. - -**Reducing Aliasing Error.** We maintain the same *T*0 so that *f*0 = 100. Hence, - -$$ -f_s = N_0 f_0 = 256 \times 100 = 25,600 -$$ - and $T = \frac{1}{f_s} = 39 \,\mu s$ - -Thus, increasing *N*0 from 200 to 256 permits us to reduce the sampling interval *T* from 50 µs to 39 µs while maintaining the same frequency resolution (*f*0 = 100). - -**Improving Resolution.** Here, we maintain the same *T* = 50 µs, which yields - -$$ -T_0 = N_0 T = 256(50 \times 10^{-6}) = 12.8 -$$ - ms and $f_0 = \frac{1}{T_0} = 78.125$ Hz - -### 812 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -Thus, increasing *N*0 from 200 to 256 can improve the frequency resolution from 100 to 78.125 Hz while maintaining the same aliasing error (*T* = 50 µs). - -**Combination of Reducing Aliasing Error and Improving Resolution.** To simultaneously reduce alias error and improve resolution, we could choose *T* = 45 µs and *T*0 = 11.5 ms so that *f*0 = 86.96 Hz. Many other combinations exist as well. - -### **EXAMPLE 8.8 DFT to Compute the Fourier Transform of an Exponential** - -Use the DFT to compute (samples of) the Fourier transform of *e*−2*t u*(*t*). Plot the resulting Fourier spectra. - -We first determine *T* and *T*0. The Fourier transform of *e*−2*t u*(*t*) is 1/(*j*ω + 2). This lowpass signal is not bandlimited. In Sec. 7.6, we used the energy criterion to compute the essential bandwidth of a signal. Here, we shall present a simpler, but workable alternative to the energy criterion. The essential bandwidth of a signal will be taken as the frequency at which |*X*(ω)| drops to 1% of its peak value (see the footnote on page 736). In this case, the peak value occurs at ω = 0, where |*X*(0)| = 0.5. Observe that - -$$ -|X(\omega)| = \frac{1}{\sqrt{\omega^2 + 4}} \approx \frac{1}{\omega} \qquad \omega \gg 2 -$$ - -Also, 1% of the peak value is 0.01 × 0.5 = 0.005. Hence, the essential bandwidth *B* is at ω = 2π*B*, where - -$$ -|X(\omega)| \approx \frac{1}{2\pi B} = 0.005 \quad \Rightarrow \quad B = \frac{100}{\pi} \text{ Hz} -$$ - -and from Eq. (8.16), - -$$ -T \le \frac{1}{2B} = \frac{\pi}{200} = 0.015708 -$$ - -Had we used 1% energy criterion to determine the essential bandwidth, following the procedure in Ex. 7.20, we would have obtained *B* = 20.26 Hz, which is somewhat smaller than the value just obtained by using the 1% amplitude criterion. - -The second issue is to determine *T*0. Because the signal is not timelimited, we have to truncate it at *T*0 such that *x*(*T*0) 1. A reasonable choice would be *T*0 = 4 because *x*(4) = *e*−8 = 0.0003351. The result is *N*0 = *T*0/*T* = 254.6, which is not a power of 2. Hence, we choose *T*0 = 4, and *T* = 0.015625 = 1/64, yielding *N*0 = 256, which is a power of 2. - -Note that there is a great deal of flexibility in determining *T* and *T*0, depending on the accuracy desired and the computational capacity available. We could just as well have chosen *T* = 0.03125, yielding *N*0 = 128, although this choice would have given a slightly higher aliasing error. - -Because the signal has a jump discontinuity at *t* = 0, the first sample (at *t* = 0) is 0.5, the averages of the values on the two sides of the discontinuity. We compute *Xr* (the DFT) from the samples of *e*−2*t u*(*t*) according to Eq. (8.12). Note that *Xr* is the *r*th sample of *X*(ω), and these samples are spaced at *f*0 = 1/*T*0 = 0.25 Hz (ω0 = π/2 rad/s). - -Because *Xr* is *N*0 periodic, *Xr* = *X*(*r*+256) so that *X*256 = *X*0. Hence, we need to plot *Xr* over the range *r* = 0 to 255 (not 256). Moreover, because of this periodicity, *X*−*r* = *X*(−*r*+256), and the values of *Xr* over the range *r* = −127 to −1 are identical to those over the range *r* = 129 to 255. Thus, *X*−127 = *X*129, *X*−126 = *X*130,...,*X*−1 = *X*255. In addition, because of the property of conjugate symmetry of the Fourier transform, *X*−*r* = *X*∗ *r* , it follows that *X*−1 = *X*∗ 1 , *X*−2 = *X*∗ 2 ,...,*X*−128 = *X*∗ 128. Thus, we need *Xr* only over the range *r* = 0 to *N*0/2 (128 in this case). - -Figure 8.17 shows the computed plots of |*Xr*| and *Xr*. The exact spectra are depicted by continuous curves for comparison. Note the nearly perfect agreement between the two sets of spectra. We have depicted the plot of only the first 28 points rather than all 128 points, which would have made the figure very crowded, resulting in loss of clarity. The points are at the intervals of 1/*T*0 = 1/4 Hz or ω0 = 1.5708 rad/s. The 28 samples, therefore, exhibit the plots over the range ω = 0 to ω = 28(1.5708) ≈ 44 rad/s or 7 Hz. - -**Figure 8.17** Discrete Fourier transform of an exponential signal *e*−2*t u*(*t*). - -### 814 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -In this example, we knew *X*(ω) beforehand; hence we could make intelligent choices for *B* (or the sampling frequency *fs*). In practice, we generally do not know *X*(ω) beforehand. In fact, that is the very thing we are trying to determine. In such a case, we must make an intelligent guess for *B* or *fs* from circumstantial evidence. We should then continue reducing the value of *T* and recomputing the transform until the result stabilizes within the desired number of significant digits. - -### USING MATLAB TO COMPUTE AND PLOT THE RESULTS - -Let us now use MATLAB to confirm the results of this example. First, parameters are defined and MATLAB's fft command is used to compute the DFT. - -``` ->> T_0 = 4; N_0 = 256; T = T_0/N_0; t = (0:T:T*(N_0-1))'; ->> x = T*exp(-2*t); x(1) = x(1)/2; ->> X_r = fft(x); r = [-N_0/2:N_0/2-1]'; omega_r = r*2*pi/T_0; -``` - -The true Fourier transform is also computed for comparison. - ->> omega = linspace(-pi/T,pi/T,5001); X = 1./(j\*omega+2); - -For clarity, we display spectrum over a restricted frequency range. - -``` ->> subplot(1,2,1); stem(omega_r,fftshift(abs(X_r)),'k.'); -``` - -``` ->> line(omega,abs(X),'color',[0 0 0]); axis([-0.01 44 -0.01 0.51]); -``` - -- >> xlabel('\omega'); ylabel('|X(\omega)|'); -- >> subplot(1,2,2); stem(omega\_r,fftshift(angle(X\_r)),'k.'); -- >> line(omega,angle(X),'color',[0 0 0]); axis([-0.01 44 -pi/2-0.01 0.01]); -- >> xlabel('\omega'); ylabel('\angle X(\omega)'); - -The results, shown in Fig. 8.18, match the earlier results shown in Fig. 8.17. - -### **EXAMPLE 8.9 DFT to Compute the Fourier Transform of a Rectangular Pulse** - -Use the DFT to compute the Fourier transform of 8 rect(*t*). - -This gate function and its Fourier transform are illustrated in Figs. 8.19a and 8.19b. To determine the value of the sampling interval *T*, we must first decide on the essential bandwidth *B*. In Fig. 8.19b, we see that *X*(ω) decays rather slowly with ω. Hence, the essential bandwidth *B* is rather large. For instance, at *B* = 15.5 Hz (97.39 rad/s), *X*(ω) = −0.1643, which is about 2% of the peak at *X*(0). Hence, the essential bandwidth is well above 16 Hz if we use the 1% of the peak amplitude criterion for computing the essential bandwidth. However, we shall deliberately take *B* = 4 for two reasons: to show the effect of aliasing and because the use of *B* > 4 would give an enormous number of samples, which could not be conveniently displayed on the page without losing sight of the essentials. Thus, we shall intentionally accept approximation to graphically clarify the concepts of the DFT. - -The choice of *B* = 4 results in the sampling interval *T* = 1/2*B* = 1/8. Looking again at the spectrum in Fig. 8.19b, we see that the choice of the frequency resolution *f*0 = 1/4 Hz is reasonable. Such a choice gives us four samples in each lobe of *X*(ω). In this case *T*0 = 1/*f*0 = 4 seconds and *N*0 = *T*0/*T* = 32. The duration of *x*(*t*) is only 1 second. We must repeat it every 4 seconds (*T*0 = 4), as depicted in Fig. 8.19c, and take samples every 1/8 second. This choice yields 32 samples (*N*0 = 32). Also, - -$$ -x_n = Tx(nT) = \frac{1}{8}x(nT) -$$ - -Since *x*(*t*) = 8 rect(*t*), the values of *xn* are 1, 0, or 0.5 (at the points of discontinuity), as illustrated in Fig. 8.19c, where *xn* is depicted as a function of *t* as well as *n*, for convenience. - -In the derivation of the DFT, we assumed that *x*(*t*) begins at *t* = 0 (Fig. 8.16a), and then took *N*0 samples over the interval (0, *T*0). In the present case, however, *x*(*t*) begins at −1/2. This difficulty is easily resolved when we realize that the DFT obtained by this procedure is actually the DFT of *xn* repeating periodically every *T*0 seconds. Figure 8.19c clearly indicates that periodic repeating the segment of *xn* over the interval from −2 to 2 seconds yields the same signal as the periodic repeating the segment of *xn* over the interval from 0 to 4 seconds. Hence, the DFT of the samples taken from −2 to 2 seconds is the same as that of the samples taken from 0 to 4 seconds. Therefore, regardless of where *x*(*t*) starts, we can always take the samples of *x*(*t*) and its periodic extension over the interval from 0 to *T*0. In the present example, the 32 sample values are - -$$ -x_n = \begin{cases} 1 & 0 \le n \le 3 \text{ and } 29 \le n \le 31 \\ 0 & 5 \le n \le 27 \\ 0.5 & n = 4,28 \end{cases} -$$ - -816 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -**Figure 8.19** Discrete Fourier transform of a gate pulse. - -Observe that the last sample is at *t* = 31/8, not at 4, because the signal repetition starts at *t* = 4, and the sample at *t* = 4 is the same as the sample at *t* = 0. Now, *N*0 = 32 and 0 = 2π/32 = π/16. Therefore [see Eq. (8.12)], - -$$ -X_r = \sum_{n=0}^{31} x_n e^{-jr(\pi/16)n} -$$ - -Values of *Xr* are computed according to this equation and plotted in Fig. 8.19d. - -The samples *Xr* are separated by *f*0 = 1/*T*0 Hz. In this case *T*0 = 4, so the frequency resolution *f*0 is 1/4 Hz, as desired. The folding frequency *fs*/2 = *B* = 4 Hz corresponds to *r* = *N*0/2 = 16. Because *Xr* is *N*0 periodic (*N*0 = 32), the values of *Xr* for *r* = −16 to *n* = −1 are the same as those for *r* = 16 to *n* = 31. For instance, *X*17 = *X*−15, *X*18 = *X*−14, and so on. The DFT gives us the samples of the spectrum *X*(ω). - -For the sake of comparison, Fig. 8.19d also shows the shaded curve 8 sinc(ω/2), which is the Fourier transform of 8 rect(*t*). The values of *Xr* computed from the DFT equation show aliasing error, which is clearly seen by comparing the two superimposed plots. The error in *X*2 is just about 1.3%. However, the aliasing error increases rapidly with *r*. For instance, the error in *X*6 is about 12%, and the error in *X*10 is 33%. The error in *X*14 is a whopping 72%. The percent error increases rapidly near the folding frequency (*r* = 16) because *x*(*t*) has a jump discontinuity, which makes *X*(ω) decay slowly as 1/ω. Hence, near the folding frequency, the inverted tail (due to aliasing) is very nearly equal to *X*(ω) itself. Moreover, the final values are the difference between the exact and the folded values (which are very close to the exact values). Hence, the percent error near the folding frequency (*r* = 16 in this case) is very high, although the absolute error is very small. Clearly, for signals with jump discontinuities, the aliasing error near the folding frequency will always be high (in percentage terms), regardless of the choice of *N*0. To ensure a negligible aliasing error at any value *r*, we must make sure that *N*0 *r*. This observation is valid for all signals with jump discontinuities. - -### USING MATLAB TO COMPUTE AND PLOT THE RESULTS - -Once again, MATLAB lets us easily confirm the results of this example. First, parameters are defined and MATLAB's fft command is used to compute the DFT. - ->> T\_0 = 4; N\_0 = 32; T = T\_0/N\_0; >> x\_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]'; >> X\_r = fft(x\_n); r = [-N\_0/2:N\_0/2-1]'; omega\_r = r\*2\*pi/T\_0; - -The true Fourier transform is also computed for comparison. - -``` ->> omega = linspace(-pi/T,pi/T,5001); X = 8*sinc(omega/(2*pi)); -``` - -Since it is real, we can display the resulting spectrum using a single plot. - ->> clf; stem(omega\_r,fftshift(real(X\_r)),'k.'); >> line(omega,X,'color',[0 0 0]); - -### 818 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - ->> xlabel('\omega'); ylabel('X(\omega)'); axis tight - -The result, shown in Fig. 8.20, matches the earlier result shown in Fig. 8.19d. The DFT approximation does not perfectly follow the true Fourier transform, especially at high frequencies, because the parameter *B* is deliberately set too small. - -### **[8.5-1 Some Properties of the DFT](#page-13-0)** - -The discrete Fourier transform is basically the Fourier transform of a sampled signal repeated periodically. Hence, the properties derived earlier for the Fourier transform apply to the DFT as well. - -LINEARITY If *xn* ⇐⇒ *Xr* and *gn* ⇐⇒ *Gr*, then - -$$ -a_1x_n + a_2g_n \Longleftrightarrow a_1X_r + a_2G_r -$$ - -The proof is trivial. - -### CONJUGATE SYMMETRY - -From the conjugation property *x*∗(*t*) ⇐⇒ *X*∗(−ω), we have - -*x*∗ *n* ←→ *X*∗ −*r* - -From this equation and the time-reversal property, we obtain - -$$ -x_{-n}^* \longleftrightarrow X_r^* -$$ - -When *x*(*t*) is real, then the conjugate-symmetry property states that *X*∗(ω) = *X*(−ω). Hence, for real *xn*, - -$$ -X_r^* = X_{-r} -$$ - -Moreover, *Xr* is *N*0 periodic. Hence, - -$$ -X_r^* = X_{N_0-r} -$$ - -Because of this property, we need compute only half the DFTs for real *xn*. The other half are the conjugates. - -### TIME SHIFTING - -The time-shifting (circular shifting) property states† - -$$ -x_{n-k} \Longleftrightarrow X_r e^{-jr\Omega_0 k} -$$ - -**Proof.** We use Eq. (8.13) to find the inverse DFT of *Xre*−*jr*0*k* as - -$$ -\frac{1}{N_0} \sum_{r=0}^{N_0-1} X_r e^{-jr\Omega_0 k} e^{jr\omega_0 n} = \frac{1}{N_0} \sum_{r=0}^{N_0-1} X_r e^{jr\Omega_0 (n-k)} = x_{n-k} -$$ - -### FREQUENCY SHIFTING - -A dual of the time-shifting property, the frequency-shifting property states - -$$ -x_n e^{jn\Omega_0 m} \Longleftrightarrow X_{r-m} -$$ - -**Proof.** This proof is identical to that of the time-shifting property except that we start with Eq. (8.12). - -### CIRCULAR CONVOLUTION - -The circular (or periodic) convolution property states - -$$ -x_n \circledast g_n \Longleftrightarrow X_r G_r \tag{8.18} -$$ - -and - -$$ -x_n g_n \Longleftrightarrow \frac{1}{N_0} X_r \circledast G_r \tag{8.19} -$$ - -For two *N*0-periodic sequences *xn* and *gn*, circular (or periodic) convolution is defined by - -$$ -x_n \circledast g_n = \sum_{k=0}^{N_0 - 1} x_k g_{n-k} = \sum_{k=0}^{N_0 - 1} g_k x_{n-k} -$$ -\n(8.20) - -† Time shifting is also known as *circular shifting* because such a shift can be interpreted as a circular shift of the *N*0 samples in the first cycle 0 ≤ *n* ≤ *N*0 −1. - -**Figure 8.21** Graphical depictions of circular convolution. - -To prove Eq. (8.18), we find the DFT of the circular convolution *xn*-∗ *gn* as - -$$ -\sum_{n=0}^{N_0-1} \left( \sum_{k=0}^{N_0-1} x_k g_{n-k} \right) e^{-j r \omega_0 n} = \sum_{k=0}^{N_0-1} x_k \left( \sum_{n=0}^{N_0-1} g_{n-k} e^{-j r \omega_0 n} \right) -$$ -$$ -= \sum_{k=0}^{N_0-1} x_k (G_r e^{-j r \Omega_0 k}) = X_r G_r -$$ - -Equation (8.19) can be proved in the same way. - -For periodic sequences, the convolution can be visualized in terms of two sequences, with one sequence fixed and the other inverted and moved past the fixed sequence, one digit at a time. If the two sequences are *N*0 periodic, the same configuration will repeat after *N*0 shifts of the sequence. Clearly the convolution *xn*-∗ *gn* becomes *N*0 periodic. Such convolution can be conveniently visualized in terms of *N*0 sequences, as illustrated in Fig. 8.21, for the case of *N*0 = 4. The inner *N*0-point sequence *xn* is clockwise and fixed. The outer *N*0-point sequence *gn* is inverted so that it becomes counterclockwise. This sequence is now rotated clockwise 1 unit at a time. We multiply the overlapping numbers and add. For example, the value of *xn*-∗ *gn* at *n* = 0 (Fig. 8.21) is - -$$ -x_0g_0 + x_1g_3 + x_2g_2 + x_3g_1 -$$ - -and the value of *xn*-∗ *gn* at *n* = 1 is (Fig. 8.21) - -$$ -x_0g_1 + x_1g_0 + x_2g_3 + x_3g_2 -$$ - -and so on. - -### **[8.5-2 Some Applications of the DFT](#page-13-0)** - -The DFT is useful not only in the computation of direct and inverse Fourier transforms, but also in other applications such as convolution, correlation, and filtering. Use of the efficient FFT algorithm, discussed shortly (Sec. 8.6), makes it particularly appealing. - -### LINEAR CONVOLUTION - -Let *x*(*t*) and *g*(*t*) be the two signals to be convolved. In general, these signals may have different time durations. To convolve them by using their samples, they must be sampled at the same rate (not below the Nyquist rate of either signal). Let *xn* (0 ≤ *n* ≤ *N*1 − 1) and *gn* (0 ≤ *n* ≤ *N*2 − 1) be the corresponding discrete sequences representing these samples. Now, - -$$ -c(t) = x(t) * g(t) -$$ - -and if we define three sequences as *xn* = *Tx*(*nT*), *gn* = *Tg*(*nT*), and *cn* = *Tc*(*nT*), then† - -$$ -c_n = x_n * g_n -$$ - -where we define the linear convolution sum of two discrete sequences *xn* and *gn* as - -$$ -c_n = x_n * g_n = \sum_{k=-\infty}^{\infty} x_k g_{n-k} -$$ - -Because of the width property of the convolution, *cn* exists for 0≤ *n*≤ *N*1+*N*2−1. To be able to use the DFT circular convolution technique, we must make sure that the circular convolution will yield the same result as does linear convolution. In other words, the signal resulting from the circular convolution must have the same length (*N*1 + *N*2 − 1) as that of the signal resulting from linear convolution. This step can be accomplished by adding *N*2 − 1 dummy samples of zero value to *xn* and *N*1 −1 dummy samples of zero value to *gn* (zero padding). This procedure changes the length of both *xn* and *gn* to *N*1+*N*2 −1. The circular convolution now is identical to the linear convolution except that it repeats periodically with period *N*1 +*N*2 −1. A little reflection will show that in such a case the circular convolution procedure in Fig. 8.21 over one cycle (0 ≤ *n* ≤ *N*1 + *N*2 − 1) is identical to the linear convolution of the two sequences *xn* and *gn*. We can use the DFT to find the convolution *xn* ∗ *gn* in three steps, as follows: - -- 1. Find the DFTs *Xr* and *Gr* corresponding to suitably padded *xn* and *gn*. -- 2. Multiply *Xr* by *Gr*. -- 3. Find the IDFT of *XrGr*. This procedure of convolution, when implemented by the fast Fourier transform algorithm (discussed later), is known as *fast convolution*. - -### FILTERING - -We generally think of filtering in terms of a hardware-oriented solution (e.g., building a circuit with *RLC* components and operational amplifiers). However, filtering also has a software-oriented solution [a computer algorithm that yields the filtered output *y*(*t*) for a given input *x*(*t*)]. This goal can be conveniently accomplished by using the DFT. If *x*(*t*) is the signal to be filtered, then *Xr*, the DFT of *xn*, is found. The spectrum *Xr* is then shaped (filtered) as desired by multiplying *Xr* by *Hr*, where *Hr* are the samples of *H*(ω) for the filter [*Hr* = *H*(*r*ω0)]. Finally, we take the IDFT of *XrHr* to obtain the filtered output *yn*[*yn* =*Ty*(*nT*)]. This procedure is demonstrated in the following example. - -† We can show that *cn* = lim*T*→0 *xn* ∗ *gn*; [see 4]. Error is inherent in any numerical method used to compute convolution of continuous-time signals; since *T* = 0 in practice, there will be some error in this equation. - -The signal *x*(*t*) in Fig. 8.22a is passed through an ideal lowpass filter of frequency response *H*(ω) depicted in Fig. 8.22b. Use the DFT to find the sampled version of the filter output. - -**Figure 8.22** DFT solution for filtering *x*(*t*) through *H*(ω). - -We have already found the 32-point DFT of *x*(*t*) (see Fig. 8.19d). Next we multiply *Xr* by *Hr*. To find *Hr*, we recall using *f*0 = 1/4 in computing the 32-point DFT of *x*(*t*). Because *Xr* is 32-periodic, *Hr* must also be 32-periodic with samples separated by 1/4 Hz. This fact means that *Hr* must be repeated every 8 Hz or 16π rad/s (see Fig. 8.22c). The resulting 32 samples of *Hr* over (0 ≤ ω ≤ 16π ) are as follows: - -$$ -H_r = \begin{cases} 1 & 0 \le r \le 7 \\ 0 & 9 \le r \le 23 \\ 0.5 & r = 8,24 \end{cases} \text{ and } 25 \le r \le 31 -$$ - -We multiply *Xr* with *Hr*. The desired output signal samples *yn* are found by taking the inverse DFT of *XrHr*. The resulting output signal is illustrated in Fig. 8.22d. - -It is quite simple to verify the results of this filtering example using MATLAB. First, parameters are defined, and MATLAB's fft command is used to compute the DFT of *xn*. - ->> T\_0 = 4; N\_0 = 32; T = T\_0/N\_0; n = (0:N\_0-1); r = n; >> x\_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]'; X\_r = fft(x\_n); - -The DFT of the filter's output is just the product of the filter response *Hr* and the input DFT *Xr*. The output *yn* is obtained using the ifft command and then plotted. - -``` ->> H_r = [ones(1,8) 0.5 zeros(1,15) 0.5 ones(1,7)]'; ->> Y_r = H_r.*X_r; y_n = ifft(Y_r); ->> clf; stem(n,real(y_n),'k.'); ->> xlabel('n'); ylabel('y_n'); axis([0 31 -.1 1.1]); -``` - -The result, shown in Fig. 8.23, matches the earlier result shown in Fig. 8.22d. Recall, this DFT-based approach shows the samples *yn* of the filter output *y*(*t*) (sampled in this case at a rate *T* = 1 8 ) over 0 ≤ *n* ≤ *N*0 − 1 = 31 when the input pulse *x*(*t*) is periodically replicated to form samples *xn* (see Fig. 8.19c). - -**Figure 8.23** Using MATLAB and the DFT to determine filter output. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/107_8.6 THE FAST FOURIER TRANSFORM (FFT).md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/107_8.6 THE FAST FOURIER TRANSFORM (FFT).md deleted file mode 100644 index da8c8e3acf09bc3e1b1866d6c24635dfb84529da..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/107_8.6 THE FAST FOURIER TRANSFORM (FFT).md +++ /dev/null @@ -1,99 +0,0 @@ -## **8.6 THE FAST FOURIER [TRANSFORM](#page-13-0) (FFT)** - -The number of computations required in performing the DFT was dramatically reduced by an algorithm developed by Cooley and Tukey in 1965 [5]. This algorithm, known as the *fast Fourier transform* (FFT), reduces the number of computations from something on the order of *N*2 0 to *N*0 log*N*0. To compute one sample *Xr* from Eq. (8.12), we require *N*0 complex multiplications and *N*0 −1 complex additions. To compute *N*0 such values (*Xr* for *r* = 0, 1,...,*N*0 −1), we require a total of *N*2 0 complex multiplications and *N*0(*N*0 − 1) complex additions. For a large *N*0, these computations can be prohibitively time-consuming, even for a high-speed computer. The FFT algorithm is what made the use of Fourier transform accessible for digital signal processing. - -### HOW DOES THE FFT REDUCE THE NUMBER OF COMPUTATIONS? - -It is easy to understand the magic of the FFT. The secret is in the linearity of the Fourier transform and also of the DFT. Because of linearity, we can compute the Fourier transform of a signal *x*(*t*) as a sum of the Fourier transforms of segments of *x*(*t*) of shorter duration. The same principle applies to the computation of the DFT. Consider a signal of length *N*0 = 16 samples. As seen earlier, DFT computation of this sequence requires *N*2 0 = 256 multiplications and *N*0(*N*0 − 1) = 240 additions. We can split this sequence into two shorter sequences, each of length 8. To compute DFT of each of these segments, we need 64 multiplications and 56 additions. Thus, we need a total of 128 multiplications and 112 additions. Suppose, we split the original sequence in four segments of length 4 each. To compute the DFT of each segment, we require 16 multiplications and 12 additions. Hence, we need a total of 64 multiplications and 48 additions. If we split the sequence in eight segments of length 2 each, we need 4 multiplications and 2 additions for each segment, resulting in a total of 32 multiplications and 8 additions. Thus, we have been able to reduce the number of multiplications from 256 to 32 and the number of additions from 240 to 8. Moreover, some of these multiplications turn out to be multiplications by 1 or −1. All this fantastic economy in the number of computations is realized by the FFT without any approximation! The values obtained by the FFT are identical to those obtained by the DFT. In this example, we considered a relatively small value of *N*0 = 16. The reduction in the number of computations is much more dramatic for higher values of *N*0. - -The FFT algorithm is simplified if we choose *N*0 to be a power of 2, although such a choice is not essential. For convenience, we define - -$$ -W_{N_0} = e^{-(j2\pi/N_0)} = e^{-j\Omega_0} -$$ - -so that - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n W_{N_0}^{nr} \qquad 0 \le r \le N_0 - 1 \tag{8.21} -$$ - -and - -$$ -x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r W_{N_0}^{-nr} \qquad 0 \le n \le N_0 - 1 \tag{8.22} -$$ - -Although there are many variations of the Tukey–Cooley algorithm, these can be grouped into two basic types: *decimation in time* and *decimation in frequency*. - -### THE DECIMATION-IN-TIME ALGORITHM - -Here we divide the *N*0-point data sequence *xn* into two (*N*0/2)-point sequences consisting of evenand odd-numbered samples, respectively, as follows: - -$$ -\underbrace{x_0, x_2, x_4, \ldots, x_{N_0-2}}_{\text{sequence } g_n}, \underbrace{x_1, x_3, x_5, \ldots, x_{N_0-1}}_{\text{sequence } h_n} -$$ - -Then, from Eq. (8.21), - -$$ -X_r = \sum_{n=0}^{(N_0/2)-1} x_{2n} W_{N_0}^{2n} + \sum_{n=0}^{(N_0/2)-1} x_{2n+1} W_{N_0}^{(2n+1)r} -$$ - -Also, since - -$$ -W_{N_0/2} = W_{N_0}^2 -$$ - -we have - -$$ -X_r = \sum_{n=0}^{(N_0/2)-1} x_{2n} W_{N_0/2}^{nr} + W_{N_0}^r \sum_{n=0}^{(N_0/2)-1} x_{2n+1} W_{N_0/2}^{nr} -$$ - -= $G_r + W_{N_0}^r H_r$ 0 \le $r \le N_0 - 1$ (8.23) - -where *Gr* and *Hr* are the (*N*0/2)-point DFTs of the even- and odd-numbered sequences, *gn* and *hn*, respectively. Also, *Gr* and *Hr*, being the (*N*0/2)-point DFTs, are (*N*0/2) periodic. Hence, - -$$ -G_{r+(N_0/2)} = G_r \qquad \text{and} \qquad H_{r+(N_0/2)} = H_r \tag{8.24} -$$ - -Moreover, - -$$ -W_{N_0}^{r + (N_0/2)} = W_{N_0}^{N_0/2} W_{N_0}^r = e^{-j\pi} W_{N_0}^r = -W_{N_0}^r -$$ -\n(8.25) - -From Eqs. (8.23), (8.24), and (8.25), we obtain - -$$ -X_{r+(N_0/2)} = G_r - W_{N_0}^r H_r \tag{8.26} -$$ - -This property can be used to reduce the number of computations. We can compute the first *N*0/2 points (0 ≤ *n* ≤ (*N*0/2) − 1) of *Xr* by using Eq. (8.23) and the last *N*0/2 points by using Eq. (8.26) as - -$$ -X_r = G_r + W_{N_0}^r H_r \qquad 0 \le r \le \frac{N_0}{2} - 1 -$$ - -$$ -X_{r + (N_0/2)} = G_r - W_{N_0}^r H_r \qquad 0 \le r \le \frac{N_0}{2} - 1 -$$ - (8.27) - -**Figure 8.24** Butterfly signal flow graph. - -**Figure 8.25** Successive steps in an 8-point FFT. - -Thus, an *N*0-point DFT can be computed by combining the two (*N*0/2)-point DFTs, as in Eq. (8.27). These equations can be represented conveniently by the *signal flow* graph depicted in Fig. 8.24. This structure is known as a *butterfly*. Figure 8.25a shows the implementation of Eq. (8.24) for the case of *N*0 = 8. - -The next step is to compute the (*N*0/2)-point DFTs *Gr* and *Hr*. We repeat the same procedure by dividing *gn* and *hn* into two (*N*0/4)-point sequences corresponding to the even- and odd-numbered samples. Then we continue this process until we reach the one-point DFT. These steps for the case of *N*0 = 8 are shown in Figs. 8.25a, 8.25b, and 8.25c. Figure 8.25c shows that the two-point DFTs require no multiplication. - -To count the number of computations required in the first step, assume that *Gr* and *Hr* are known. Equation (8.27) clearly shows that to compute all the *N*0 points of the *Xr*, we require *N*0 complex additions and *N*0/2 complex multiplications† (corresponding to *Wr N*0 *Hr*). - -In the second step, to compute the (*N*0/2)-point DFT *Gr* from the (*N*0/4)-point DFT, we require *N*0/2 complex additions and *N*0/4 complex multiplications. We require an equal number of computations for *Hr*. Hence, in the second step, there are *N*0 complex additions and *N*0/2 complex multiplications. The number of computations required remains the same in each step. Since a total of log2*N*0 steps is needed to arrive at a one-point DFT, we require, conservatively, a total of *N*0 log2*N*0 complex additions and (*N*0/2)log2*N*0 complex multiplications, to compute the *N*0-point DFT. Actually, as Fig. 8.25c shows, many multiplications are multiplications by 1 or −1, which further reduces the number of computations. - -The procedure for obtaining IDFT is identical to that used to obtain the DFT except that *WN*0 = *ej*(2π/*N*0) instead of *e*−*j*(2π/*N*0) (in addition to the multiplier 1/*N*0). Another FFT algorithm, the *decimation-in-frequency* algorithm, is similar to the decimation-in-time algorithm. The only difference is that instead of dividing *xn* into two sequences of even- and odd-numbered samples, we divide *xn* into two sequences formed by the first *N*0/2 and the last *N*0/2 samples, proceeding in the same way until a single-point DFT is reached in log2*N*0 steps. The total number of computations in this algorithm is the same as that in the decimation-in-time algorithm. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/108_8.7 MATLAB - THE DISCRETE FOURIER TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/108_8.7 MATLAB - THE DISCRETE FOURIER TRANSFORM.md deleted file mode 100644 index 5d332b736f5b55fb6da664308c71e2fe6f9eb419..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/108_8.7 MATLAB - THE DISCRETE FOURIER TRANSFORM.md +++ /dev/null @@ -1,216 +0,0 @@ -## **[8.7 MATLAB: THE](#page-13-0) DISCRETE FOURIER TRANSFORM** - -As an idea, the discrete Fourier transform (DFT) has been known for hundreds of years. Practical computing devices, however, are responsible for bringing the DFT into common use. MATLAB is capable of DFT computations that would have been impractical just a few decades ago. - -### **[8.7-1 Computing the Discrete Fourier Transform](#page-13-0)** - -The MATLAB command fft(x) computes the DFT of a vector x that is defined over (0 ≤ *n* ≤ *N*0 −1) (Problem 8.7-1 considers how to scale the DFT to accommodate signals that do not begin at *n* = 0.) As its name suggests, the function fft uses the computationally more efficient fast Fourier transform algorithm when it is appropriate to do so. The inverse DFT is easily computed by using the ifft function. - -† Actually, *N*0/2 is a conservative figure because some multiplications corresponding to the cases of *Wr N*0 = 1,*j*, and so on, are eliminated. - -To illustrate MATLAB's DFT capabilities, consider 50 points of a 10 Hz sinusoid sampled at *fs* = 50 Hz and scaled by *T* = 1/*fs*. - -``` ->> T = 1/50; N_0 = 50; n = (0:N_0-1); ->> x = T*cos(2*pi*10*n*T); -``` - -In this case, the vector x contains exactly 10 cycles of the sinusoid. The fft command computes the DFT. - ->> X = fft(x); - -Since the DFT is both discrete and periodic, fft needs to return only the *N*0 discrete values contained in the single period (0 ≤ *f* < *fs*). - -While *Xr* can be plotted as a function of *r*, it is more convenient to plot the DFT as a function of frequency *f* . A frequency vector, in hertz, is created by using *N*0 and *T*. - ->> f = (0:N\_0-1)/(T\*N\_0); stem(f,abs(X),'k.'); >> axis([0 50 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); - -As expected, Fig. 8.26 shows content at a frequency of 10 Hz. Since the time-domain signal is real, *X*(*f*) is conjugate symmetric. Thus, content at 10 Hz implies equal content at −10 Hz. The content visible at 40 Hz is an alias of the −10 Hz content. - -Often, it is preferred to plot a DFT over the principal frequency range (−*fs*/2 ≤ *f* < *fs*/2). The MATLAB function fftshift properly rearranges the output of fft to accomplish this task. - -``` ->> stem(f-1/(T*2),fftshift(abs(X)),'k.'); ->> axis([-25 25 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); -``` - -When we use fftshift, the conjugate symmetry that accompanies the DFT of a real signal becomes apparent, as shown in Fig. 8.27. - -Since DFTs are generally complex-valued, the magnitude plots of Figs. 8.26 and 8.27 offer only half the picture; the signal's phase spectrum, shown in Fig. 8.28, completes it. - -``` ->> stem(f-1/(T*2),fftshift(angle(X)),'k.'); ->> axis([-25 25 -1.1*pi 1.1*pi]); xlabel('f [Hz]'); ylabel('\angle X(f)'); -``` - -**Figure 8.26** |*X*(*f*)| computed over (0 ≤ *f* < 50) by using fft. - -**Figure 8.27** |*X*(*f*)| displayed over (−25 ≤ *f* < 25) by using fftshift. - -**Figure 8.28** *X*(*f*) displayed over (−25≤*f* <25). - -Since the signal is real, the phase spectrum necessarily has odd symmetry. Additionally, the phase at ±10 Hz is zero, as expected for a zero-phase cosine function. More interesting, however, are the phase values found at the remaining frequencies. Does a simple cosine really have such complicated phase characteristics? The answer, of course, is no. The magnitude plot of Fig. 8.27 helps identify the problem: there is zero content at frequencies other than ±10 Hz. Phase computations are not reliable at points where the magnitude response is zero. One way to remedy this problem is to assign a phase of zero when the magnitude response is near or at zero. - -### **[8.7-2 Improving the Picture with Zero Padding](#page-13-0)** - -DFT magnitude and phase plots paint a picture of a signal's spectrum. At times, however, the picture can be somewhat misleading. Given a sampling frequency *fs* = 50 Hz and a sampling interval *T* = 1/*fs*, consider the signal - -$$ -y[n] = Te^{j2\pi \left(10\frac{1}{3}\right)n} -$$ - -This complex-valued, periodic signal contains a single positive frequency at 101 3 Hz. Let us compute the signal's DFT using 50 samples. - -``` ->> y = T*exp(j*2*pi*(10+1/3)*n*T); Y = fft(y); ->> stem(f-25,fftshift(abs(Y)),'k.'); ->> axis([-25 25 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|Y(f)|'); -``` - -**Figure 8.29** |*Y*(*f*)| using 50 data points. - -**Figure 8.30** |*Yzp*(*f*)| over 5 ≤ *f* ≤ 15 using 50 data points padded with 550 zeros. - -In this case, the vector y contains a noninteger number of cycles. Figure 8.29 shows the significant frequency leakage that results. Also notice that since *y*[*n*] is not real, the DFT is not conjugate symmetric. - -In this example, the discrete DFT frequencies do not include the actual 101 3 Hz frequency of the signal. Thus, it is difficult to determine the signal's frequency from Fig. 8.29. To improve the picture, the signal is zero-padded to 12 times its original length. - -``` ->> y_zp = [y,zeros(1,11*length(y))]; Y_zp = fft(y_zp); ->> f_zp = (0:12*N_0-1)/(T*12*N_0); ->> stem(f_zp-25,fftshift(abs(Y_zp)),'k.'); ->> axis([-25 25 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|Y_{zp}(f)|'); -``` - -Figure 8.30, zoomed in to 5 ≤ *f* ≤ 15, correctly shows the peak frequency at 101 3 Hz and better represents the signal's spectrum. - -It is important to keep in mind that zero padding does not increase the resolution or accuracy of the DFT. To return to the picket fence analogy, zero padding increases the number of pickets in our fence but cannot change what is behind the fence. More formally, the characteristics of the sinc function, such as main beam width and sidelobe levels, depend on the fixed width of the pulse, not on the number of zeros that follow. Adding zeros cannot change the characteristics of the sinc function and thus cannot change the resolution or accuracy of the DFT. Adding zeros simply allows the sinc function to be sampled more finely. - -### **[8.7-3 Quantization](#page-13-0)** - -A *B*-bit analog-to-digital converter (ADC) samples an analog signal and quantizes amplitudes by using 2*B* discrete levels. This quantization results in signal distortion that is particularly noticeable for small *B*. Typically, quantization is classified as symmetric or asymmetric and as either rounding or truncating. Let us investigate rounding-type quantizers. - -The quantized output *x*q of an asymmetric rounding converter is given as† - -$$ -x_{\rm q} = \frac{x_{\rm max}}{2^{B-1}}\lfloor \frac{x}{x_{\rm max}}2^{B-1} + \frac{1}{2}\rfloor -$$ - -The quantized output *x*q of a symmetric rounding converter is given as - -$$ -x_{\rm q} = \frac{x_{\rm max}}{2^{B-1}}\left(\lfloor \frac{x}{x_{\rm max}}2^{B-1}\rfloor + \frac{1}{2}\right) -$$ - -Program CH8MP1 quantizes a signal using one of these two rounding quantizer rules and also ensures no more than 2*B* output levels. - -``` -function [xq] = CH8MP1(x,xmax,B,method) -% CH8MP1.m : Chapter 8, MATLAB Program 1 -% Function M-file quantizes x over (-xmax,xmax) using 2^b levels. -% Uses rounding rule, supports symmetric and asymmetric quantization -% INPUTS: x = input signal -% xmax = maximum magnitude of signal to be quantized -% B = number of quantization bits -% method = default 'sym' for symmetrical, 'asym' for asymmetrical -% OUTPUTS: xq = quantized signal -if (nargin<3), - disp('Insufficient number of inputs.'); return -elseif (nargin==3), - method = 'sym'; -elseif (nargin>4), - disp('Too many inputs.'); return -end -x(abs(x)>xmax)=xmax*sign(x(abs(x)>xmax)); % Limit amplitude to xmax -switch lower(method) - case 'asym' - xq = xmax/(2^(B-1))*floor(x*2^(B-1)/xmax+1/2); - xq(xq>=xmax)=xmax*(1-2^(1-B)); % Ensure only 2^B levels - case 'sym' - xq = xmax/(2^(B-1))*(floor(x*2^(B-1)/xmax)+1/2); - xq(xq>=xmax)=xmax*(1-2^(1-B)/2); % Ensure only 2^B levels -``` - -† Large values of *x* may return quantized values *x*q outside the 2*B* allowable levels. In such cases, *x*q should be clamped to the nearest permitted level. - -``` -otherwise - disp('Unrecognized quantization method.'); return -end -``` - -Several MATLAB commands require discussion. First, the nargin function returns the number of input arguments. In this program, nargin is used to ensure that a correct number of inputs is supplied. If the number of inputs supplied is incorrect, an error message is displayed and the function terminates. If only three input arguments are detected, the quantization type is not explicitly specified and the program assigns the default symmetric method. - -As with many high-level languages such as C, MATLAB supports general switch/case structures† : - -``` -switch switch_expr, -case case_expr, - statements; -... -otherwise, - statements; -``` - -end - -CH8MP1 switches among cases of the string method. In this way, method-specific parameters are easily set. The command lower is used to convert a string to all lowercase characters. In this way, strings such as SYM, Sym, and sym are all indistinguishable. Similar to lower, the MATLAB command upper converts a string to all uppercase. - -The floor command rounds input values to the nearest integer toward minus infinity. Mathematically, it computes ·. To accommodate different types of rounding, MATLAB supplies three other rounding commands: ceil, round, and fix. The ceil command rounds input values to the nearest integers toward infinity, ( ·"); the round command rounds input values toward the nearest integer; the fix command rounds input values to the nearest integer toward zero. For example, if x = [-0.5 0.5];, floor(x) yields [-1 0], ceil(x) yields [0 1], round(x) yields [-1 1], and fix(x) yields [0 0]. Finally, CH8MP1 checks and, if necessary, corrects large values of *x*q that may be outside the allowable 2*B* levels. - -To verify operation, CH8MP1 is used to determine the transfer characteristics of a symmetric 3-bit quantizer operating over (−10,10). - -``` ->> x = (-10:.0001:10); xsq = CH8MP1(x,10,3,'sym'); -``` - -``` ->> plot(x,xsq,'k'); axis([-10 10 -10.5 10.5]); grid on; -``` - -``` ->> xlabel('Quantizer input'); ylabel('Quantizer output'); -``` - -Figure 8.31 shows the results. Clearly, the quantized output is limited to 2*B* = 8 levels. Zero is not a quantization level for symmetric quantizers, so half of the levels occur above zero and half of the levels occur below zero. In fact, *symmetric quantizers* get their name from the symmetry in quantization levels above and below zero. - -By changing the method in CH8MP1 from 'sym' to 'asym', we obtain the transfer characteristics of an asymmetric 3-bit quantizer, as shown in Fig. 8.32. Again, the quantized output is limited to 2*B* = 8 levels, and zero is now one of the included levels. With zero as a quantization - -† A functionally equivalent structure can be written by using if, elseif, and else statements. - -**Figure 8.31** Transfer characteristics of a symmetric 3-bit quantizer. - -**Figure 8.32** Transfer characteristics of an asymmetric 3-bit quantizer. - -level, we need one fewer quantization level above zero than there are levels below. Not surprisingly, *asymmetric quantizers* get their name from the asymmetry in quantization levels above and below zero. - -There is no doubt that quantization can change a signal. It follows that the spectrum of a quantized signal can also change. While these changes are difficult to characterize mathematically, they are easy to investigate by using MATLAB. Consider a 1 Hz cosine sampled at *fs* = 50 Hz over 1 second. - ->> x = cos(2\*pi\*n\*T); X = fft(x); T = 1/50; N\_0 = 50; n = (0:N\_0-1); - -Upon quantizing by means of a 2-bit asymmetric rounding quantizer, both the signal and spectrum are substantially changed. - -``` ->> xaq = CH8MP1(x,1,2,'asym'); Xaq = fft(xaq); ->> subplot(2,2,1); stem(n,x,'k'); axis([0 49 -1.1 1.1]); ->> xlabel('n');ylabel('x[n]'); ->> subplot(2,2,2); stem(f-25,fftshift(abs(X)),'k'); axis([-25,25 -1 26]) ->> xlabel('f');ylabel('|X(f)|'); ->> subplot(2,2,3); stem(n,xaq,'k');axis([0 49 -1.1 1.1]); -``` - -**Figure 8.33** Signal and spectrum effects of quantization. - -``` ->> xlabel('n');ylabel('x_{aq}[n]'); -``` - -``` ->> subplot(2,2,4); stem(f-25,fftshift(abs(fft(xaq))),'k'); axis([-25,25 -1 26]); -``` - -``` ->> xlabel('f');ylabel('|X_{aq}(f)|'); -``` - -The results are shown in Fig. 8.33. The original signal *x*[*n*] appears sinusoidal and has pure spectral content at ±1 Hz. The asymmetrically quantized signal *xaq*[*n*] is significantly distorted. The corresponding magnitude spectrum |*Xaq*(*f*)| is spread over a broad range of frequencies. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/109_8.8 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/109_8.8 SUMMARY.md deleted file mode 100644 index 03e6c07b69a5107b08f58e6c2e7e93a5a1a72160..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/109_8.8 SUMMARY.md +++ /dev/null @@ -1,9 +0,0 @@ -## **[8.8 SUMMARY](#page-13-0)** - -A signal bandlimited to *B* Hz can be reconstructed exactly from its samples if the sampling rate *fs* > 2*B* Hz (the sampling theorem). Such a reconstruction, although possible theoretically, poses practical problems such as the need for ideal filters, which are unrealizable or are realizable only with infinite delay. Therefore, in practice, there is always an error in reconstructing a signal from its samples. Moreover, practical signals are not bandlimited, which causes an additional error (aliasing error) in signal reconstruction from its samples. When a signal is sampled at a frequency *fs* Hz, samples of a sinusoid of frequency (*fs*/2) + *x* Hz appear as samples of a lower frequency (*fs*/2) − *x* Hz. This phenomenon, in which higher frequencies appear as lower frequencies, is known as aliasing. Aliasing error can be reduced by bandlimiting a signal to *fs*/2 Hz (half the sampling frequency). Such bandlimiting, done prior to sampling, is accomplished by an anti-aliasing filter that is an ideal lowpass filter of cutoff frequency *fs*/2 Hz. - -The sampling theorem is very important in signal analysis, processing, and transmission because it allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing a continuous-time signal is therefore equivalent to processing a discrete sequence of numbers. This leads us directly into the area of digital filtering (discrete-time systems). In the field of communication, the transmission of a continuous-time message reduces to the transmission of a sequence of numbers. This opens doors to many new techniques of communicating continuous-time signals by pulse trains. - -The dual of the sampling theorem states that for a signal timelimited to τ seconds, its spectrum *X*(ω) can be reconstructed from the samples of *X*(ω) taken at uniform intervals not greater than 1/τ Hz. In other words, the spectrum should be sampled at a rate not less than τ samples/Hz. - -To compute the direct or the inverse Fourier transform numerically, we need a relationship between the samples of *x*(*t*) and *X*(ω). The sampling theorem and its dual provide such a quantitative relationship in the form of a discrete Fourier transform (DFT). The DFT computations are greatly facilitated by a fast Fourier transform (FFT) algorithm, which reduces the number of computations from something on the order of *N*2 0 to *N*0 log*N*0. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/110_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/110_REFERENCES.md deleted file mode 100644 index 16f7d594d16e22fef901b7edfeddc33b793dc5e5..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/110_REFERENCES.md +++ /dev/null @@ -1,7 +0,0 @@ -### **[REFERENCES](#page-13-0)** - -- 1. Linden, D. A. A discussion of sampling theorem. *Proceedings of the IRE,* vol. 47, pp. 1219–1226, July 1959. -- 2. Siebert, W. M. *Circuits, Signals, and Systems*. MIT/McGraw-Hill, New York, 1986. -- 3. Bennett, W. R. *Introduction to Signal Transmission*. McGraw-Hill, New York, 1970. -- 4. Lathi, B. P. *Linear Systems and Signals*. Berkeley-Cambridge Press, Carmichael, CA, 1992. -- 5. Cooley, J. W., and Tukey, J. W. An algorithm for the machine calculation of complex Fourier series. *Mathematics of Computation*, vol. 19, pp. 297–301, April 1965. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/111_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/111_PROBLEMS.md deleted file mode 100644 index 7664299ea96800727112e533658fd273ffb675cf..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/111_PROBLEMS.md +++ /dev/null @@ -1,565 +0,0 @@ -## **[PROBLEMS](#page-13-0)** - -[*Note:* In many problems, the plots of spectra are shown as functions of frequency *f* Hz for convenience, although we have labeled them as functions of ω as *X*(ω), *Y*(ω), etc.] - -- **8.1-1** If *f*s is the Nyquist rate for signal *x*(*t*), determine the Nyquist rate for each of the following signals: - - (a) *y*a(*t*) = *d dt x*(*t*) - - (b) *y*b(*t*) = *x*(*t*) cos(2π*f*0*t*) - - (c) *y*c(*t*) = *x*(*t*+*a*)+*x*(*t*−*b*), for real constants *a* and *b* - - (d) *y*d(*t*) = *x*(*at*), for real *a* > 0 - -- **8.1-2** Figure P8.1-2 shows Fourier spectra of signals *x*1(*t*) and *x*2(*t*). Determine the Nyquist sampling rates for signals *x*1(*t*), *x*2(*t*), *x*2 1(*t*), *x*3 2(*t*), and *x*1(*t*)*x*2(*t*). -- **8.1-3** A signal *x*(*t*) has a bandwidth of *B* = 1000 Hz. For a positive integer *N*, what is the Nyquist rate for the signal *y*(*t*) = *xN*(*t*)? -- **8.1-4** Determine the Nyquist sampling rate and the Nyquist sampling interval for the signals: (a) sinc2(100π*t*) - - (b) 0.01 sinc2(100π*t*) - -**Figure P8.1-2** - -### 836 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -- (c) sinc(100π*t*)+3sinc2(60π*t*) -- (d) sinc(50π*t*)sinc(100π*t*) -- **8.1-5** (a) Sketch |*X*(ω)|, the amplitude spectrum of a signal *x*(*t*) = 3 cos 6π*t* + sin 18π*t* + 2 cos(28 − )π*t*, where is a very small number → 0. Determine the minimum sampling rate required to be able to reconstruct *x*(*t*) from these samples. - - (b) Sketch the amplitude spectrum of the sampled signal when the sampling rate is 25% above the Nyquist rate (show the spectrum over the frequency range ±50 Hz only). How would you reconstruct *x*(*t*) from these samples? -- **8.1-6** (a) Derive the sampling theorem by considering the fact that the sampled signal *x*(*t*) = *x*(*t*)δ*T* (*t*), and using the frequency-convolutionpropertyinEq.(7.34). - - (b) For a sampling train consisting of shifted unit impulses at instants *nT* + τ instead of at *nT* (for all positive and negative integer values of *n*), find the spectrum of the sampled signal. -- **8.1-7** A signal is bandlimited to 12 kHz. The band between 10 and 12 kHz has been so corrupted by excessive noise that the information in this band is nonrecoverable. Determine the minimum sampling rate for this signal so that the uncorrupted portion of the band can be recovered. If we were to filter out the corrupted - -spectrum prior to sampling, what would be the minimum sampling rate? - -- **8.1-8** A continuous-time signal *x*(*t*) = ((*t* − 1)/2) is sampled at three rates: 10, 2, and 1 Hz. Sketch the resulting sampled signals. Because *x*(*t*) is timelimited, its bandwidth is infinite. However, most of its energy is concentrated in a small band. Determine a reasonable minimum sampling rate that will allow reconstruction of this signal with a small error. The answer is not unique. Make a reasonable assumption of what you define as a "negligible" or "small" error. -- **8.1-9** (a) A signal *x*(*t*) = 5 sinc2 (5π*t*) + cos 20π*t* is sampled at a rate of 10 Hz. Find the spectrum of the sampled signal. Can *x*(*t*) be reconstructed by lowpass filtering the sampled signal? - - (b) Repeat part (a) for a sampling frequency of 20 Hz. Can you reconstruct the signal from this sampled signal? Explain. - - (c) If *x*(*t*) = 5 sinc2 (5π*t*) + sin 20π*t*, can you reconstruct *x*(*t*) from the samples of *x*(*t*) at a rate of 20 Hz? Explain your answer with spectral representation(s). - - (d) For *x*(*t*) = 5 sinc2 (5π*t*) + sin 20π*t*, can you reconstruct *x*(*t*) from the samples of *x*(*t*) at a rate of 21 Hz? Explain your answer with spectral representation(s). Comment on your results. -- **8.1-10** (a) The highest frequency in the spectrum *X*(ω) (Fig. P8.1-10a) of a bandpass signal *x*(*t*) - -**Figure P8.1-10** - -is 30 Hz. Hence, the minimum sampling frequency needed to sample *x*(*t*) is 60 Hz. Show the spectrum of the signal sampled at a rate of 60 Hz. Can you reconstruct *x*(*t*) from these samples? How? - -- (b) A certain busy student looks at *X*(ω), concludes that its bandwidth is really 10 Hz, and decides that the sampling rate 20 Hz is adequate for sampling *x*(*t*). Sketch the spectrum of the signal sampled at a rate of 20 Hz. Can *x*(*t*) be reconstructed from these samples? -- (c) The same student, using the same reasoning, looks at *Y*(ω) in Fig. P8.1-10b, the spectrum of another bandpass signal *y*(*t*), and concludes that the sampling rate of 20 Hz can be used to sample *y*(*t*). Sketch the spectrum of the signal *y*(*t*) sampled at a rate of 20 Hz. Can *y*(*t*) be reconstructed from these samples? -- **8.1-11** A signal *x*(*t*) whose spectrum *X*(ω), as shown in Fig. P8.1-11, is sampled at a frequency *fs* = *f*1 + *f*2 Hz. Find all the sample values of *x*(*t*) merely by inspection of *X*(ω). -- **8.1-12** As described in Sec. 8.1-1, practical sampling can be achieved by multiplying a signal *x*(*t*) by a periodic train of pulses *pT* (*t*). The pulse train *pT* (*t*) can be created by the periodic replication of some pulse *p*(*t*) as - -$$ -p_T(t) = \sum_{k=-\infty}^{\infty} p(t - kT) -$$ - -It is desired to sample a signal at a rate *f*s = 100 Hz, and two pulses are under consideration: - -$$ -p_a(t) = -\frac{1}{4}u(t) + \frac{5}{4}u(t - \frac{2T}{20}) + -$$ - -$$ --\frac{5}{4}u(t - \frac{3T}{20}) + \frac{1}{4}u(t - \frac{5T}{20}) -$$ - -and - -$$ -p_{\rm b}(t) = e^{-t/T} \left[ u(t) - u(t - 1.5T) \right] -$$ - -- (a) Plot *pT* (*t*) using *p*a(*t*) over 0 ≤ *t* ≤ 4*T*. -- (b) Plot *pT* (*t*) using *p*b(*t*) over 0 ≤ *t* ≤ 4*T*. -- (c) Which pulse, *p*a(*t*) or *p*b(*t*), is more suitable as a sampling pulse? Carefully explain your answer. -- **8.1-13** In digital data transmission over a communication channel, it is important to know the upper theoretical limit on the rate of digital pulses that can be transmitted over a channel of bandwidth *B* Hz. In digital transmission, the relative shape of the pulse is not important. We are interested in knowing only the amplitude represented by the pulse. For instance, in binary communication, we are interested in knowing whether the received pulse amplitude is 1 or −1 (positive or negative). Thus, each pulse represents one piece of information. Consider one independent amplitude value (not necessarily binary) as one piece of information. Show that 2*B* independent pieces of information per second can be transmitted correctly (assuming no noise) over a channel of bandwidth *B* Hz. This important principle in communication theory states that 1 Hz of bandwidth can transmit two independent pieces of information per second. It represents the upper rate of pulse transmission over a channel without any error in reception in the absence of noise. [*Hint:* According to the interpolation formula [Eq. (8.6)], a continuous-time signal of bandwidth *B* Hz can be constructed from 2*B* pieces of information/second.] -- **8.1-14** This example is one of those interesting situations leading to a curious result in the category of defying gravity. The sinc function can be recovered from its samples taken at extremely low frequencies in apparent defiance of the sampling theorem. - -Consider a sinc pulse *x*(*t*) = sinc(4π*t*) for which *X*(ω) = (1/4)rect(ω/8π ). The bandwidth of *x*(*t*) is *B* = 2 Hz, and its Nyquist rate is 4 Hz. - -**Figure P8.1-11** - -- (a) Sample *x*(*t*) at a rate 4 Hz and sketch the spectrum of the sampled signal. -- (b) To recover *x*(*t*) from its samples, we pass the sampled signal through an ideal lowpass filter of bandwidth *B* = 2 Hz and gain *G* = *T* = 1/4. Sketch this system and show that for this system *H*(ω)=(1/4)rect(ω/8π ). Show also that when the input is the sampled *x*(*t*) at a rate 4 Hz, the output of this system is indeed *x*(*t*), as expected. -- (c) Now sample *x*(*t*) at half the Nyquist rate, at 2 Hz. Apply this sampled signal at the input of the lowpass filter used in part (b). Find the output. -- (d) Repeat part (c) for the sampling rate 1 Hz. -- (e) Show that the output of the lowpass filter in part (b) is *x*(*t*) to the sampled *x*(*t*) if the sampling rate is 4/*N*, where *N* is any positive integer. This means that we can recover *x*(*t*) from its samples taken at an arbitrarily small rate by letting *N* → ∞. -- (f) The mystery may be clarified a bit by examining the problem in the time domain. Find the samples of *x*(*t*) when the sampling rate is 2/*N* (*N* integer). -- **8.2-1** A signal *x*(*t*) = sinc (200π*t*) is sampled (multiplied) by a periodic pulse train *pT* (*t*) represented in Fig. P8.2-1. Find and sketch the spectrum of the sampled signal. Explain whether you will be able to reconstruct *x*(*t*) from these samples. Find the filter output if the sampled signal is passed through an ideal lowpass filter of bandwidth 100 Hz and unit gain. What is the filter output if its bandwidth *B* Hz is between 100 and 150 Hz? What happens if the bandwidth exceeds 150 Hz? -- **8.2-2** Show that the circuit in Fig. P8.2-2 is a realization of the causal ZOH (zero-order hold) circuit. You can do this by showing that the unit impulse response *h*(*t*) of this circuit is indeed equal to that in Eq. (8.5) delayed by *T*/2 seconds to make it causal. - -### **Figure P8.2-2** - -- **8.2-3** (a) A first-order hold circuit (FOH) can also be used to reconstruct a signal *x*(*t*) from its samples. The impulse response of this circuit is *h*(*t*) = (*t*/2*T*), where *T* is the sampling interval. Consider a typical sampled signal *x*(*t*) and show that this circuit performs the linear interpolation. In other words, the filter output consists of sample tops connected by straight-line segments. Follow the procedure discussed in Sec. 8.2 (Fig. 8.5c). - - (b) Determine the frequency and magnitude responses of this filter, and compare it with **(i)** the ideal filter required for signal reconstruction and **(ii)** a ZOH circuit. - - (c) This filter, being noncausal, is unrealizable. By delaying its impulse response, the filter can be made realizable. What is the minimum delay required to make it realizable? How would this delay affect the reconstructed signal and the filter frequency response? - - (d) Show that the causal FOH circuit in part (c) can be realized by the ZOH circuit depicted in Fig. P8.2-2 followed by an identical filter in cascade. -- **8.2-4** Suppose signal *x*(*t*)=sin(2π*t*/8)(*u*(*t*)−*u*(*t* −8)) is sampled at a rate *fs* = 1 Hz to generate signal *x*[*n*]. - - (a) Sketch *x*(*t*) and *x*[*n*]. - - (b) Has aliasing occurred in sampling *x*(*t*) to produce *x*[*n*]? Explain. - - (c) Sketch the output *x*ˆ(*t*) produced when *x*[*n*] is applied to the causal ZOH reconstructor of Prob. 8.2-2. How does *x*ˆ(*t*) compare with *x*(*t*)? - -**Figure P8.2-1** - -- (d) Sketch the output *x*ˆ(*t*) produced when *x*[*n*] is applied to the FOH reconstructor of Prob.8.2-3.Howdoes*x*ˆ(*t*)comparewith*x*(*t*)? -- **8.2-5** Repeat Prob. 8.2-4 for the signal *x*(*t*) = cos(2π*t*/8)(*u*(*t*)−*u*(*t* −8)). -- **8.2-6** Is it possible to sample a physically realizable (nonzero) signal *x*(*t*) with a physically realizable system without aliasing? If possible, explain what conditions must be met. If not possible, explain why not. -- **8.2-7** In the text, for sampling purposes, we used timelimited narrow pulses such as impulses or rectangular pulses of width less than the sampling interval *T*. Show that it is not necessary to restrict the sampling pulse width. We can use sampling pulses of arbitrarily large duration and still be able to reconstruct the signal *x*(*t*) as long as the pulse rate is no less than the Nyquist rate for *x*(*t*). - -Consider *x*(*t*) to be bandlimited to *B* Hz. The sampling pulse to be used is an exponential *e*−*atu*(*t*). We multiply *x*(*t*) by a periodic train of exponential pulses of the form *e*−*atu*(*t*) spaced *T* seconds apart. Find the spectrum of the sampled signal, and show that *x*(*t*) can be reconstructed from this sampled signal provided the sampling rate is no less than 2*B* Hz or *T* < 1/2*B*. Explain how you would reconstruct *x*(*t*) from the sampled signal. - -- **8.2-8** In Ex. 8.2, the sampling of a signal *x*(*t*) was accomplished by multiplying the signal by a pulse train *pT* (*t*), resulting in the sampled signal depicted in Fig. 8.4d. This procedure is known as the *natural sampling*. Figure P8.2-8 shows the so-called *flat-top sampling* of the same signal *x*(*t*) = sinc2 (5π*t*). - - (a) Show that the signal *x*(*t*) can be recovered from flat-top samples if the sampling rate is no less than the Nyquist rate. - - (b) Explain how you would recover *x*(*t*) from the flat-top samples. - - (c) Find the expression for the sampled signal spectrum *X*(ω) and sketch it roughly. -- **8.2-9** A sinusoid of frequency *f*0 Hz is sampled at a rate *fs* = 20 Hz. Find the apparent frequency of the sampled signal if *f*0 is: - - (a) 8 Hz - - (b) 12 Hz - -(c) 20 Hz - -- (d) 22 Hz -- (e) 32 Hz -- **8.2-10** A sinusoid of unknown frequency *f*0 is sampled at a rate 60 Hz. The apparent frequency of the samples is 20 Hz. Determine *f*0 if it is known that *f*0 lies in the range: - - (a) 0–30 Hz - - (b) 30–60 Hz - - (c) 60–90 Hz - - (d) 90–120 Hz -- **8.2-11** A signal *x*(*t*) = 3 cos 6π*t*+cos 16π*t*+2 cos 20π*t* is sampled at a rate 25% above the Nyquist rate. Sketch the spectrum of the sampled signal. How would you reconstruct *x*(*t*) from these samples? If the sampling frequency is 25% below the Nyquist rate, what are the frequencies of the sinusoids present in the output of the filter with cutoff frequency equal to the folding frequency? Do not write the actual output; give just the frequencies of the sinusoids present in the output. -- **8.2-12** A complex signal *x*(*t*) has a spectrum given as - -$$ -X(\omega) = \begin{cases} \omega & 0 \le \omega \le 2\pi 10 \\ 0 & \text{otherwise} \end{cases} -$$ - -Let *x*(*t*) be sampled at rate *f*s = 24 Hz to produce signal *x*(*t*) with spectrum *X*(ω). - -- (a) Sketch *X*(ω). -- (b) Has aliasing occurred in sampling *x*(*t*) to produce *x*(*t*)? Explain. -- (c) Can *x*(*t*) be exactly recovered from *x*(*t*)? Explain. -- **8.2-13** Repeat Prob. 8.2-12 for the sampling rate *f*s = 16 Hz. - -- **8.2-14** Repeat Prob. 8.2-12 for the sampling rate *f*s = 8 Hz. -- **8.2-15** (a) Show that the signal *x*(*t*), reconstructed from its samples *x*(*nT*), using Eq. (8.6) has a bandwidth *B* ≤ 1/2*T* Hz. - - (b) Show that *x*(*t*) is the smallest bandwidth signal that passes through samples *x*(*nT*). [*Hint:* Use the *reductio ad absurdum* method.] -- **8.2-16** In digital communication systems, the efficient use of channel bandwidth is ensured by transmitting digital data encoded by means of bandlimited pulses. Unfortunately, bandlimited pulses are non-timelimited; that is, they have infinite duration, which causes pulses representing successive digits to interfere and cause errors in the reading of true pulse values. This difficulty can be resolved by shaping a pulse *p*(*t*) in such a way that it is bandlimited, yet causes zero interference at the sampling instants. To transmit *R* pulses per second, we require a minimum bandwidth *R*/2 Hz (see Prob. 8.1-13). The bandwidth of *p*(*t*) should be *R*/2 Hz, and its samples, in order to cause no interference at all other sampling instants, must satisfy the condition - -$$ -p(nT) = \begin{cases} 1 & n = 0 \quad T = \frac{1}{R} \\ 0 & n \neq 0 \end{cases} -$$ - -Because the pulse rate is *R* pulses per second, the sampling instants are located at intervals of 1/*R* seconds. Hence, the foregoing condition ensures that any given pulse will not interfere with the amplitude of any other pulse at its center. Find *p*(*t*). Is *p*(*t*) unique in the sense that no other pulse satisfies the given requirements? - -**8.2-17** The problem of pulse interference in digital data transmission was outlined in Prob. 8.2-16, where we found a pulse shape *p*(*t*) to eliminate the interference. Unfortunately, the pulse found is not only noncausal, (and unrealizable) but also has a serious drawback: because of its slow decay (as 1/*t*), it is prone to severe interference due to small parameter deviation. To make the pulse decay rapidly, Nyquist proposed relaxing the bandwidth requirement from *R*/2 Hz to *kR*/2 Hz with 1 ≤ *k* ≤ 2. The pulse must still have a property of noninterference with other pulses, for example, - -$$ -p(nT) = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases} \qquad T = \frac{1}{R} -$$ - -Show that this condition is satisfied only if the pulse spectrum *P*(ω) has an odd symmetry about the set of dotted axes, as shown in Fig. P8.2-17. The bandwidth of *P*(ω) is *kR*/2 Hz (1 ≤ *k* ≤ 2). - -**8.2-18** The Nyquist samples of a signal *x*(*t*) bandlimited to *B* Hz are - -$$ -x(nT) = \begin{cases} 1 & n = 0, 1 \\ 0 & \text{all } n \neq 0, 1 \end{cases} \qquad T = \frac{1}{2B} -$$ - -Show that - -$$ -x(t) = \frac{\operatorname{sinc}(2\pi Bt)}{1 - 2Bt} -$$ - -This pulse, known as the *duobinary pulse,* is used in digital transmission applications. - -**Figure P8.2-17** - -**8.2-19** A signal bandlimited to *B* Hz is sampled at a rate *fs* = 2*B* Hz. Show that - -$$ -\int_{-\infty}^{\infty} x(t) dt = T \sum_{-\infty}^{\infty} x(nT) -$$ -$$ -\int_{-\infty}^{\infty} |x(t)|^2 dt = T \sum_{-\infty}^{\infty} |x(nT)|^2 -$$ - -[*Hint:* Use the orthogonality property of the sinc function in Prob. 7.6-6.] - -- **8.2-20** Prove that a signal cannot be simultaneously timelimited and bandlimited. [*Hint:* Show that a contrary assumption leads to contradiction. Assume a signal to be simultaneously timelimited and bandlimited so that *X*(ω) = 0 for |ω| ≥ 2π*B*. In this case, *X*(ω) = *X*(ω)rect(ω/4π*B* ) for *B* > *B*. This fact means that *x*(*t*) is equal to *x*(*t*) ∗ 2*B* sinc (2π*B t*). The latter cannot be timelimited because the sinc function tail extends to infinity.] -- **8.3-1** Physically implementable digital systems, such as smartphones and computers, require that signals be both time-sampled and amplitude-quantized. - - (a) Why is time sampling necessary? When does time sampling result in unrecoverable changes to the signal? - - (b) Why is amplitude-quantization necessary? When does amplitude quantization result in unrecoverable changes to the signal? -- **8.3-2** Typical analog-to-digital converters (ADCs) operate over a range of input amplitudes [−*V*ref,*V*ref]. Why is it desirable to condition the input *x*(*t*) to an ADC so that its maximum magnitude is close to, but does not exceed, *V*ref? What happens if the maximum magnitude of *x*(*t*) is greater than *V*ref? What happens if the maximum magnitude of *x*(*t*) is much smaller than *V*ref? -- **8.3-3** A compact disc (CD) records audio signals digitally by means of a binary code. Assume an audio signal bandwidth of 15 kHz. - - (a) What is the Nyquist rate? - - (b) If the Nyquist samples are quantized into 65,536 levels (*L* = 65,536) and then binary-coded, what number of binary digits is required to encode a sample? - -- (c) Determine the number of binary digits per second (bits/s) required to encode the audio signal. -- (d) For practical reasons discussed in the text, signals are sampled at a rate well above the Nyquist rate. Practical CDs use 44,100 samples/s. If *L* = 65,536, determine the number of pulses per second required to encode the signal. -- **8.3-4** A TV signal (video and audio) has a bandwidth of 4.5 MHz. This signal is sampled, quantized, and binary-coded. - - (a) Determine the sampling rate if the signal is to be sampled at a rate 20% above the Nyquist rate. - - (b) If the samples are quantized into 1024 levels, what number of binary pulses is required to encode each sample? - - (c) Determine the binary pulse rate (bits/s) of the binary-coded signal. -- **8.3-5** (a) In a certain A/D scheme, there are 16 quantization levels. Give one possible binary code and one possible quaternary (4-ary) code. For the quaternary code, use **0, 1, 2,** and **3** as the four symbols. Use the minimum number of digits in your code. - - (b) To represent a given number of quantization levels *L*, we require a minimum of *bM* digits for an *M*-ary code. Show that the ratio of the number of digits in a binary code to the number of digits in a quaternary (4-ary) code is 2, that is, *b*2/*b*4 = 2. -- **8.3-6** Five telemetry signals, each of bandwidth 1 kHz, are quantized and binary-coded. These signals are time-division multiplexed (signal bits interleaved). Choose the number of quantization levels so that the maximum error in sample amplitudes is no greater than 0.2% of the peak signal amplitude. The signals must be sampled at least 20% above the Nyquist rate. Determine the data rate (bits per second) of the multiplexed signal. -- **8.4-1** A triangle function *x*(*t*) = (*t*/5) has spectrum *X*(ω). Sketch the corresponding time-domain signal *xT*0 (*t*) if *X*(ω) is sampled at the following rates: - - (a) *f*0 = 10 samples/Hz - - (b) *f*0 = 5 samples/Hz - -- (c) *f*0 = 4 samples/Hz -- (d) *f*0 = 2.5 samples/Hz -- **8.4-2** The Fourier transform of a signal *x*(*t*), bandlimited to *B* Hz, is *X*(ω). The signal *x*(*t*) is repeated periodically at intervals *T*, where *T* = 1.25/*B*. The resulting signal *y*(*t*) is - -$$ -y(t) = \sum_{-\infty}^{\infty} x(t - nT) -$$ - -Show that *y*(*t*) can be expressed as - -$$ -y(t) = C_0 + C_1 \cos(1.6\pi Bt + \theta_1) -$$ - -where - -$$ -C_0 = \frac{1}{T}X(0) -$$ - -$$ -C_1 = \frac{2}{T} \left| X\left(\frac{2\pi}{T}\right) \right| -$$ - -and - -$$ -\theta_1 = \angle X \left( \frac{2\pi}{T} \right) -$$ - -Recall that a bandlimited signal is not timelimited, and hence has infinite duration. The periodic repetitions are all overlapping. - -- **8.5-1** For a signal *x*(*t*) that is timelimited to 10 ms and has an essential bandwidth of 10 kHz, determine *N*0, the number of signal samples necessary to compute a power-of-2 FFT with a frequency resolution *f*0 of at least 50 Hz. Explain whether any zero padding is necessary. -- **8.5-2** To compute the DFT of signal *x*(*t*) in Fig. P8.5-2, write the sequence *xn* (for *n* = 0 to *N*0 − 1) if the frequency resolution *f*0 must be at least 0.25 Hz. Assume the essential bandwidth (the folding frequency) of *x*(*t*) to be at least 3 Hz. Do not compute the DFT; just write the appropriate sequence *xn*. -- **8.5-3** Suppose we want to sample a finite-duration signal *x*(*t*) that occupies 0 ≤ *t* ≤ *T*. - -- (a) Devise a way to use the DFT to help select a suitable sampling rate *f*s for signal *x*(*t*). [*Hint:* Consider the characteristics of an oversampled signal's DFT spectrum.] -- (b) Test the method you devised in part (a) using the signal *x*(*t*) = (*t*−1 2 ). Use MAT-LAB to compute any needed DFTs. What value *f*s seems reasonable for this signal? -- **8.5-4** Choose appropriate values for *N*0 and *T* and compute the DFT of the signal *e*−*t u*(*t*). Use two different criteria for determining the effective bandwidth of *e*−*t u*(*t*). As the bandwidth, use the frequency at which the amplitude response drops to 1% of its peak value (at ω = 0). Next, use the 99% energy criterion for determining the bandwidth (see Ex. 7.20). -- **8.5-5** Repeat Prob. 8.5-4 for the signal - -$$ -x(t) = \frac{2}{t^2 + 1} -$$ - -- **8.5-6** For the signals *x*(*t*) and *g*(*t*) represented in Fig. P8.5-6, write the appropriate sequences *xn* and *gn* necessary for the computation of the convolution of *x*(*t*) and *g*(*t*) using DFT. Use *T* = 1/8. -- **8.5-7** For this problem, interpret the *N*-point DFT as an *N*-periodic function of *r*. To stress this fact, we shall change the notation *Xr* to *X*(*r*). Are the following frequency-domain signals valid DFTs? Answer yes or no. For each valid DFT, determine the size *N* of the DFT and whether the time-domain signal is real. - - (a) *X*(*r*) = *j*−π - - (b) *X*(*r*) = sin(*r*/10) - - (c) *X*(*r*) = sin(π*r*/10) - - (d) *X*(*r*) = (1+*j*)/√2 *r* - - (e) *X*(*r*) = #*r* + π\$10 where #·\$10 denotes the modulo-*N* operation. -- **8.7-1** MATLAB's fft command computes the DFT of a vector x assuming the first sample occurs at time *n* = 0. Given that X = fft(x) has already - -been computed, derive a method to correct X to reflect an arbitrary starting time *n* = *n*0. - -- **8.7-2** Consider a complex signal composed of two closely spaced complex exponentials: *x*1[*n*] = *ej*2π*n*30/100 + *ej*2π*n*33/100. For each of the following cases, plot the length-*N* DFT magnitude as a function of frequency *fr*, where *fr* = *r*/*N*. - - (a) Compute and plot the DFT of *x*1[*n*] using 10 samples (0 ≤ *n* ≤ 9). From the plot, can both exponentials be identified? Explain. - - (b) Zero-pad the signal from part (a) with 490 zeros and then compute and plot the 500-point DFT. Does this improve the picture of the DFT? Explain. - - (c) Compute and plot the DFT of *x*1[*n*] using 100 samples (0 ≤ *n* ≤ 99). From the plot, can both exponentials be identified? Explain. - - (d) Zero-pad the signal from part (c) with 400 zeros and then compute and plot the 500-point DFT. Does this improve the picture of the DFT? Explain. -- **8.7-3** Repeat Prob. 8.7-2, using the complex signal *x*2[*n*] = *ej*2π*n*30/100 +*ej*2π*n*31.5/100. -- **8.7-4** Consider a complex signal composed of a dc term and two complex exponentials: *y*1[*n*] = 1 + *ej*2π*n*30/100 + 0.5 ∗ *ej*2π*n*43/100. For each of the following cases, plot the length-*N* DFT magnitude as a function of frequency *fr*, where *fr* = *r*/*N*. - - (a) Use MATLAB to compute and plot the DFT of *y*1[*n*] with 20 samples (0 ≤ *n*≤19). From the plot, can the two non-dc exponentials be identified? Given the amplitude relation between the two, the lower-frequency peak should be twice as large as the higher-frequency peak. Is this the case? Explain. - - (b) Zero-pad the signal from part (a) to a total length of 500. Does this improve locating the two non-dc exponential components? Is - -the lower-frequency peak twice as large as the higher-frequency peak? Explain. - -- (c) MATLAB's signal-processing toolbox function window allows window functions to be easily generated. Generate a length-20 Hanning window and apply it to *y*1[*n*]. Using this windowed function, repeat parts (a) and (b). Comment on whether the window function helps or hinders the analysis. -- **8.7-5** Repeat Prob. 8.7-4, using the complex signal *y*2[*n*] = 1+*ej*2π*n*30/100 +0.5*ej*2π*n*38/100. -- **8.7-6** This problem investigates the idea of zero padding applied in the frequency domain. When asked, plot the length-*N* DFT magnitude as a function of frequency *fr*, where *fr* = *r*/*N*. - - (a) In MATLAB, create a vector x that contains one period of the sinusoid *x*[*n*] = cos((π/2)*n*). Plot the result. How "sinusoidal" does the signal appear to be? - - (b) Use the fft command to compute the DFT X of vector x. Plot the magnitude of the DFT coefficients. Do they make sense? - - (c) Zero-pad the DFT vector to a total length of 100 by inserting the appropriate number of zeros in the middle of the vector X. Call this zero-padded DFT sequence Y. Why are zeros inserted in the middle rather than the end? Take the inverse DFT of Y and plot the result. What similarities exist between the new signal y and the original signal x? What are the differences between x and y? What is the effect of zero padding in the frequency domain? How is this type of zero padding similar to zero padding in the time domain? - - (d) Derive a general modification to the procedure of zero padding in the frequency domain to ensure that the amplitude of the resulting time-domain signal is left unchanged. - -- (e) Consider one period of a square wave described by the length-8 vector [1111 −1 −1 −1 −1]. Zero-pad the DFT of this vector to a length of 100, and call the result S. Scale S according to part (d), take the inverse DFT, and plot the result. Does the new time-domain signal *s*[*n*] look like a square wave? Explain. -- **8.7-7** The quantized output *x*q of a truncating asymmetric converter is given as - -$$ -\textstyle x_\mathrm{q} = \frac{x_\mathrm{max}}{2^{B-1}} \lfloor \frac{x}{x_\mathrm{max}} 2^{B-1} \frac{1}{2} \rfloor -$$ - -Any values outside the 2*B* allowable levels should be clamped to the nearest level. - -- (a) Similar to Fig. 8.31, plot the transfer characteristics for a 3-bit version of this quantizer. -- (b) Apply 3-bit truncating asymmetric quantization to a 1 Hz cosine sampled at *f*s = 50 Hz over 1 second. Plot the original signal - -*x*(*t*), the quantized signal *x*q(*t*), and the magnitude spectra of both. How does truncating asymmetric quantization compare to the results of asymmetric rounding quantization shown in Fig. 8.33? - -**8.7-8** The quantized output *x*q of a symmetric truncating converter is given as - -$$ -x_{\mathbf{q}} = \frac{x_{\max}}{2^{B-1}} \left( \lfloor \frac{x}{x_{\max}} 2^{B-1} - \frac{1}{2} \rfloor + \frac{1}{2} \right) -$$ - -Any values outside the 2*B* allowable levels should be clamped to the nearest level. - -- (a) Similar to Fig. 8.31, plot the transfer characteristics for a 3-bit version of this quantizer. -- (b) Apply 3-bit truncating symmetric quantization to a 1 Hz cosine sampled at *f*s = 50 Hz over 1 second. Plot the original signal *x*(*t*), the quantized signal *x*q(*t*), and the magnitude spectra of both. How does truncating symmetric quantization compare to the results of asymmetric rounding quantization shown in Fig. 8.33? - - - -# **FOURIER ANALYSIS OF [DISCRETE-TIME](#page-14-0) SIGNALS** - -In Chs. 6 and 7, we studied the ways of representing a continuous-time signal as a sum of sinusoids or exponentials. In this chapter we shall discuss similar development for discrete-time signals. Our approach is parallel to that used for continuous-time signals. We first represent a periodic *x*[*n*] as a Fourier series formed by a discrete-time exponential (or sinusoid) and its harmonics. Later we extend this representation to an aperiodic signal *x*[*n*] by considering *x*[*n*] as a limiting case of a periodic signal with the period approaching infinity. - -## **[9.1 DISCRETE-TIME](#page-14-0) FOURIER SERIES (DTFS)** - -A continuous-time sinusoid cosω*t* is a periodic signal regardless of the value of ω. Such is not the case for the discrete-time sinusoid cos*n* (or exponential *ejn*). A sinusoid cos*n* is periodic only if /2π is a rational number. This can be proved by observing that if this sinusoid is *N*0 periodic, then - -$$ -\cos\Omega(n+N_0)=\cos\Omega n -$$ - -This is possible only if - -*N*0 = 2π*m m* integer - -Here, both *m* and *N*0 are integers. Hence, /2π = *m*/*N*0 is a rational number. Thus, a sinusoid cos*n* (or exponential *ejn*) is periodic only if - -$$ -\frac{\Omega}{2\pi} = \frac{m}{N_0} -$$ - a rational number - -When this condition (/2π a rational number) is satisfied, the period *N*0 of the sinusoid cos*n* is given by - -$$ -N_0 = m\left(\frac{2\pi}{\Omega}\right) \tag{9.1} -$$ - -To compute *N*0, we must choose the smallest value of *m* that will make *m*(2π/) an integer. For example, if = 4π/17, then the smallest value of *m* that will make *m*(2π/) = *m*(17/2) an integer is 2. Therefore, - -$$ -N_0 = m\left(\frac{2\pi}{\Omega}\right) = 2\left(\frac{17}{2}\right) = 17 -$$ - -However, a sinusoid cos(0.8*n*) is not a periodic signal because 0.8/2π is not a rational number. - -### **[9.1-1 Periodic Signal Representation by](#page-14-0) Discrete-Time Fourier Series** - -A continuous-time periodic signal of period *T*0 can be represented as a trigonometric Fourier series consisting of a sinusoid of the fundamental frequency ω0 = 2π/*T*0, and all its harmonics. The exponential form of the Fourier series consists of exponentials *ej*0*t* , *e*±*j*ω0*t* , *e*±*j*2ω0*t* , *e*±*j*3ω0*t* ,... . - -A discrete-time periodic signal can be represented by a discrete-time Fourier series using a parallel development. Recall that a periodic signal *x*[*n*] with period *N*0 is characterized by the fact that - -$$ -x[n] = x[n+N_0] -$$ - -The smallest value of *N*0 for which this equation holds is the *fundamental period*. The *fundamental frequency* is 0 = 2π/*N*0 rad/sample. An *N*0-periodic signal *x*[*n*] can be represented by a discrete-time Fourier series made up of sinusoids of fundamental frequency 0 = 2π/*N*0 and its harmonics. As in the continuous-time case, we may use a trigonometric or an exponential form of the Fourier series. Because of its compactness and ease of mathematical manipulations, the exponential form is preferable to the trigonometric. For this reason, we shall bypass the trigonometric form and go directly to the exponential form of the discrete-time Fourier series. - -The exponential Fourier series consists of the exponentials *ej*0*n*, *e*±*j*0*n*, *e*±*j*20*n*, ..., *e*±*jn*0*n*, ..., and so on. There would be an infinite number of harmonics, except for the property proved in Sec. 5.5-1, that discrete-time exponentials whose frequencies are separated by 2π (or integer multiples of 2π) are identical because - -$$ -e^{i(\Omega \pm 2\pi m)n} = e^{i\Omega n} e^{\pm 2\pi mn} = e^{i\Omega n} \qquad m \text{ integer} -$$ - -The consequence of this result is that the *r*th harmonic is identical to the (*r* + *N*0)th harmonic. To demonstrate this, let *gn* denote the *n*th harmonic *ejn*0*n*. Then - -$$ -g_{r+N_0} = e^{j(r+N_0)\Omega_0 n} = e^{j(r\Omega_0 n + 2\pi n)} = e^{j r\Omega_0 n} = g_r -$$ - -and - -$$ -g_r = g_{r+N_0} = g_{r+2N_0} = \cdots = g_{r+mN_0} -$$ - *m* integer - -Thus, the first harmonic is identical to the (*N*0 +1)th harmonic, the second harmonic is identical to the (*N*0 +2)th harmonic, and so on. In other words, there are only *N*0 independent harmonics, and their frequencies range over an interval 2π (because the harmonics are separated by 0 = 2π/*N*0). This means that, unlike the continuous-time counterpart, the discrete-time Fourier series has only a finite number (*N*0) of terms. This result is consistent with our observation in Sec. 5.5-1 that all discrete-time signals are bandlimited to a band from −π to π. Because the harmonics are separated by 0 = 2π/*N*0, there can only be *N*0 harmonics in this band. We also saw that this band can be taken from 0 to 2π or any other contiguous band of width 2π. This means we may - - - -choose the *N*0 independent harmonics *ejr*0*n* over 0 ≤ *r* ≤ *N*0 −1, or over −1 ≤ *r* ≤ *N*0 −2, or over 1 ≤ *r* ≤ *N*0, or over any other suitable choice for that matter. Every one of these sets will have the same harmonics, although in different order. - -Let us consider the first choice, which corresponds to exponentials *ejr*0*n* for *r* = 0, 1, 2, ... , *N*0 −1. The Fourier series for an *N*0-periodic signal *x*[*n*] consists of only these *N*0 harmonics, and can be expressed as - -$$ -x[n] = \sum_{r=0}^{N_0 - 1} \mathcal{D}_r e^{jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -To compute coefficients *Dr*, we multiply both sides by *e*−*jm*0*n* and sum over *n* from *n* = 0 to (*N*0 −1). *N* - -$$ -\sum_{n=0}^{N_0-1} x[n]e^{-jm\Omega_0 n} = \sum_{n=0}^{N_0-1} \sum_{r=0}^{N_0-1} \mathcal{D}_r e^{j(r-m)\Omega_0 n} -$$ -(9.2) - -The right-hand sum, after interchanging the order of summation, results in - -$$ -\sum_{r=0}^{N_0-1} \mathcal{D}_r \left[ \sum_{n=0}^{N_0-1} e^{j(r-m)\Omega_0 n} \right] -$$ - -The inner sum, according to Eq. (8.15) in Sec. 8.5, is zero for all values of *r* = *m*. It is nonzero with a value *N*0 only when *r* = *m*. This fact means the outside sum has only one term *DmN*0 (corresponding to *r* = *m*). Therefore, the right-hand side of Eq. (9.2) is equal to *DmN*0, and - -$$ -\sum_{n=0}^{N_0-1} x[n]e^{-jm\Omega_0 n} = \mathcal{D}_m N_0 -$$ - -and - -$$ -\mathcal{D}_m = \frac{1}{N_0} \sum_{n=0}^{N_0 - 1} x[n] e^{-jm\Omega_0 n} -$$ - -We now have a discrete-time Fourier series (DTFS) representation of an *N*0-periodic signal *x*[*n*] as - -$$ -x[n] = \sum_{r=0}^{N_0 - 1} \mathcal{D}_r e^{jr\Omega_0 n} -$$ -\n(9.3) - -where - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n=0}^{N_0 - 1} x[n] e^{-j r \Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ -\n(9.4) - -Observe that DTFS Eqs. (9.3) and (9.4) are identical (within a scaling constant) to the DFT Eqs. (8.13) and (8.12).† Therefore, we can use the efficient FFT algorithm to compute the DTFS coefficients. - -† If we let *x*[*n*] = *N*0*xk* and *Dr* = *Xr*, Eqs. (9.3) and (9.4) are identical to Eqs. (8.13) and (8.12), respectively. - -## **[9.1-2 Fourier Spectra of a Periodic Signal](#page-14-0)** *x***[***n***]** - -The Fourier series consists of *N*0 components - -$$ -\mathcal{D}_0, \mathcal{D}_1 e^{j\Omega_0 n}, \mathcal{D}_2 e^{j2\Omega_0 n}, \dots, \mathcal{D}_{N_0-1} e^{j(N_0-1)\Omega_0 n} -$$ - -The frequencies of these components are 0, 0, 20, ..., (*N*0 − 1)0, where 0 = 2π/*N*0. The amount of the *r*th harmonic is *Dr*. We can plot this amount *Dr* (the Fourier coefficient) as a function of index *r* or frequency . Such a plot, called the *Fourier spectrum* of *x*[*n*], gives us, at a glance, the graphical picture of the amounts of various harmonics of *x*[*n*]. - -In general, the Fourier coefficients *Dr* are complex, and they can be represented in the polar form as - -$$ -\mathcal{D}_r = |\mathcal{D}_r|e^{j\angle{\mathcal{D}_r}} -$$ - -The plot of |*Dr*| versus is called the amplitude spectrum and that of *Dr* versus is called the angle (or phase) spectrum. These two plots together are the frequency spectra of *x*[*n*]. Knowing these spectra, we can reconstruct or synthesize *x*[*n*] according to Eq. (9.3). Therefore, the Fourier (or frequency) spectra, which are an alternative way of describing a periodic signal *x*[*n*], are in every way equivalent (in terms of the information) to the plot of *x*[*n*] as a function of *n*. The Fourier spectra of a signal constitute the *frequency-domain* description of *x*[*n*], in contrast to the time-domain description, where *x*[*n*] is specified as a function of index *n* (representing time). - -The results are very similar to the representation of a continuous-time periodic signal by an exponential Fourier series except that, generally, the continuous-time signal spectrum bandwidth is infinite and consists of an infinite number of exponential components (harmonics). The spectrum of the discrete-time periodic signal, in contrast, is bandlimited and has at most *N*0 components. - -### PERIODIC EXTENSION OF FOURIER SPECTRUM - -We now show that if φ[*r*] is an *N*0-periodic function of *r*, then - -$$ -\sum_{r=0}^{N_0-1} \phi[r] = \sum_{r=(N_0)} \phi[r] \tag{9.5} -$$ - -where *r* = #*N*0\$ indicates summation over any *N*0 consecutive values of *r*. Because φ[*r*] is *N*0 periodic, the same values repeat with period *N*0. Hence, the sum of any set of *N*0 consecutive values of φ[*r*] must be the same no matter the value of *r* at which we start summing. Basically, it represents the sum over one cycle. - -To apply this result to the DTFS, we observe that *e*−*jr*0*n* is *N*0 periodic because - -$$ -e^{-jr\Omega_0(n+N_0)} = e^{-jr\Omega_0 n}e^{-j2\pi r} = e^{-jr\Omega_0 n} -$$ - -Therefore, if *x*[*n*] is *N*0 periodic, *x*[*n*]*e*−*jr*0*n* is also *N*0 periodic. Hence, from Eq. (9.4), it follows that *Dr* is also *N*0 periodic, as is *Drejr*0*n*. Now, because of Eq. (9.5), we can express Eqs. (9.3) and (9.4) as - -$$ -x[n] = \sum_{r = \langle N_0 \rangle} \mathcal{D}_r e^{jr\Omega_0 n} \tag{9.6} -$$ - -9.1 Discrete-Time Fourier Series (DTFS) 849 - -and - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n = \langle N_0 \rangle} x[n] e^{-j r \Omega_0 n} \tag{9.7} -$$ - -If we plot *Dr* for all values of *r* (rather than only 0 ≤ *r* ≤ *N*0 − 1), then the spectrum *Dr* is *N*0 periodic. Moreover, Eq. (9.6) shows that *x*[*n*] can be synthesized not only by the *N*0 exponentials corresponding to 0 ≤ *r* ≤ *N*0 − 1, but also by any successive *N*0 exponentials in this spectrum, starting at any value of *r* (positive or negative). For this reason, it is customary to show the spectrum *Dr* for all values of *r* (not just over the interval 0 ≤ *r* ≤ *N*0 − 1). *Yet we must remember that to synthesize x*[*n*] *from this spectrum, we need to add only N*0 *consecutive components.* All these observations are consistent with our discussion in Ch. 5, where we showed that a sinusoid of a given frequency is equivalent to multitudes of sinusoids, all separated by integer multiple of 2π in frequency. - -Along the scale, *Dr* repeats every 2π intervals, and along the *r* scale, *Dr* repeats at intervals of *N*0. Equations (9.6) and (9.7) show that both *x*[*n*] and its spectrum *Dr* are *N*0 periodic and both have exactly the same number of components (*N*0) over one period. - -Equation (9.7) shows that *Dr* is complex in general, and *D*−*r* is the conjugate of *Dr* if *x*[*n*] is real. Thus, - -$$ -|\mathcal{D}_r| = |\mathcal{D}_{-r}| -$$ - and $\angle \mathcal{D}_r = -\angle \mathcal{D}_{-r}$ - -so that the amplitude spectrum |*Dr*| is an even function , and *Dr* is an odd function of *r* (or ). All these concepts will be clarified by the examples to follow. The first example is rather trivial and serves mainly to familiarize the reader with the basic concepts of DTFS. - -### **EXAMPLE 9.1 Discrete-Time Fourier Series of a Sinusoid** - -Find the discrete-time Fourier series (DTFS) for *x*[*n*] = sin 0.1π*n* (Fig. 9.1a). Sketch the amplitude and phase spectra. - -In this case, the sinusoid sin 0.1π*n* is periodic because /2π = 1/20 is a rational number and the period *N*0 is [see Eq. (9.1)] - -$$ -N_0 = m\left(\frac{2\pi}{\Omega}\right) = m\left(\frac{2\pi}{0.1\pi}\right) = 20m -$$ - -The smallest value of *m* that makes 20*m* an integer is *m* = 1. Therefore, the period *N*0 = 20 so that 0 = 2π/*N*0 = 0.1π, and from Eq. (9.6), - -$$ -x[n] = \sum_{r=\langle 20 \rangle} \mathcal{D}_r e^{j0.1\pi rn} -$$ - -where the sum is performed over any 20 consecutive values of *r*. We shall select the range −10 ≤ *r* < 10 (values of *r* from −10 to 9). This choice corresponds to synthesizing *x*[*n*] using - -**Figure 9.1** Discrete-time sinusoid sin 0.1π*n* and its Fourier spectra. - -the spectral components in the fundamental frequency range (−π ≤ <π). Thus, - -$$ -x[n] = \sum_{r=-10}^{9} \mathcal{D}_r e^{j0.1\pi rn} -$$ - -where, according to Eq. (9.7), - -$$ -\mathcal{D}_r = \frac{1}{20} \sum_{n=-10}^{9} \sin 0.1 \pi n e^{-j0.1 \pi r n} -$$ - -= -$$ -\frac{1}{20} \sum_{n=-10}^{9} \frac{1}{2j} (e^{j0.1 \pi n} - e^{-j0.1 \pi n}) e^{-j0.1 \pi r n} -$$ - -= -$$ -\frac{1}{40j} \left[ \sum_{n=-10}^{9} e^{j0.1 \pi n (1-r)} - \sum_{n=-10}^{9} e^{-j0.1 \pi n (1+r)} \right] -$$ - -In these sums, *r* takes on all values between −10 and 9. From Eq. (8.15), it follows that the first sum on the right-hand side is zero for all values of *r* except *r* = 1, when the sum is equal to *N*0 = 20. Similarly, the second sum is zero for all values of *r* except *r* = −1, when it is equal to *N*0 = 20. Therefore, - -$$ -\mathcal{D}_1 = \frac{1}{2j} \qquad \text{and} \qquad \mathcal{D}_{-1} = -\frac{1}{2j} -$$ - -and all other coefficients are zero. The corresponding Fourier series is given by - -$$ -x[n] = \sin 0.1\pi n = \frac{1}{2j} (e^{j0.1\pi n} - e^{-j0.1\pi n}) -$$ -\n(9.8) - -Here the fundamental frequency 0 = 0.1π, and there are only two nonzero components: - -$$ -\mathcal{D}_1 = \frac{1}{2j} = \frac{1}{2}e^{-j\pi/2} -$$ - and $\mathcal{D}_{-1} = -\frac{1}{2j} = \frac{1}{2}e^{j\pi/2}$ - -Therefore, - -$$ -|\mathcal{D}_1| = |\mathcal{D}_{-1}| = \frac{1}{2} \quad \text{and} \quad \angle \mathcal{D}_1 = -\frac{\pi}{2}, \ \angle \mathcal{D}_{-1} = \frac{\pi}{2} -$$ - -Sketches of *Dr* for the interval (−10 ≤ *r* < 10) appear in Figs. 9.1b and 9.1c. According to Eq. (9.8), there are only two components corresponding to *r* = 1 and −1. The remaining 18 coefficients are zero. The *r*th component *Dr* is the amplitude of the frequency *r*0 = 0.1*r*π. Therefore, the frequency interval corresponding to −10 ≤ *r* < 10 is −π ≤ <π, as depicted in Figs. 9.1b and 9.1c. This spectrum over the range −10 ≤ *r* < 10 (or −π ≤ < π) is sufficient to specify the frequency-domain description (Fourier series), and we can synthesize *x*[*n*] by adding these spectral components. Because of the periodicity property discussed in this section, the spectrum *Dr* is a periodic function of *r* with period *N*0 = 20. For this reason, we repeat the spectrum with period *N*0 = 20 (or = 2π), as illustrated in Figs. 9.1b and 9.1c, which are periodic extensions of the spectrum in the range −10 ≤ *r* < 10. Observe that the amplitude spectrum is an even function and the angle or phase spectrum is an odd function of *r* (or ), as expected. - -The result [Eq. (9.8)] is a trigonometric identity and could have been obtained immediately without the formality of finding the Fourier coefficients. We have intentionally chosen this trivial example to introduce the reader gently to the new concept of the discrete-time Fourier series and its periodic nature. The Fourier series is a way of expressing a periodic signal *x*[*n*] in terms of exponentials of the form *ejr*0*n* and its harmonics. The result in Eq. (9.8) is merely a statement of the (obvious) fact that sin 0.1π*n* can be expressed as a sum of two exponentials *ej*0.1π*n* and *e*−*j*0.1π*n*. - -Because of the periodicity of the discrete-time exponentials *ejr*0*n*, the Fourier series components can be selected in any range of length *N*0 = 20 (or = 2π). For example, if diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/112_9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/112_9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS.md deleted file mode 100644 index 46ef840ae8a2bdc9f55fc125d77f5d7daddba0eb..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/112_9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS.md +++ /dev/null @@ -1,121 +0,0 @@ -### 852 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -we select the frequency range 0 ≤ < 2π (or 0 ≤ *r* < 20), we obtain the Fourier series as - -$$ -x[n] = \sin 0.1\pi n = \frac{1}{2j} (e^{j0.1\pi n} - e^{j1.9\pi n}) -$$ - -This series is equivalent to that in Eq. (9.8) because the two exponentials *ej*1.9π*n* and *e*−*j*0.1π*n* are equivalent. This follows from the fact that *ej*1.9π*n* = *ej*1.9π*n* ×*e*−*j*2π*n* = *e*−*j*0.1π*n*. - -We could have selected the spectrum over any other range of width = 2π in Figs. 9.1b and 9.1c as a valid discrete-time Fourier series. The reader may verify this by proving that such a spectrum starting anywhere (and of width = 2π) is equivalent to the same two components on the right-hand side of Eq. (9.8). - -### **DR ILL 9.1 DTFS Spectra on Alternate Intervals** - -From the spectra in Fig. 9.1, write the Fourier series corresponding to the interval −10 ≥ *r* > −30 (or −π ≥ > −3π). Show that this Fourier is equivalent to that in Eq. (9.8). - -### **DR ILL 9.2 Discrete-Time Fourier Series of a Sum of Sinusoids** - -Find the period and the DTFS for - -*x*[*n*] = 4 cos 0.2π*n*+6 sin 0.5π*n* - -over the interval 0 ≤ *r* ≤ 19. Use Eq. (9.4) to compute *Dr*. - -### **ANSWERS** - -*N*0 = 20 and *x*[*n*] = 2*ej*0.2π*n* +(3*e*−*j*π/2)*ej*0.5π*n* +(3*ej*π/2)*ej*1.5π*n* +2*ej*1.8π*n* - -### **DR ILL 9.3 Fundamental Period of Discrete-Time Sinusoids** - -Find the fundamental periods *N*0, if any, for: **(a)** sin(301π*n*/4) and **(b)** cos 1.3*n*. - -### **ANSWERS** - -**(a)** *N*0 = 8, **(b)** *N*0 does not exist because the sinusoid is not periodic. - -Compute and plot the discrete-time Fourier series for the periodic sampled gate function shown in Fig. 9.2a. - -**Figure 9.2 (a)** Periodic sampled gate pulse and **(b)** its Fourier spectrum. - -In this case, *N*0 = 32 and 0 = 2π/32 = π/16. Therefore, - -$$ -x[n] = \sum_{r = \langle 32 \rangle} \mathcal{D}_r e^{jr(\pi/16)n} -$$ - -where - -$$ -\mathcal{D}_r = \frac{1}{32} \sum_{n=(32)} x[n] e^{-jr(\pi/16)n} -$$ - -For our convenience, we shall choose the interval −16 ≤ *n* ≤ 15 for this summation, although any other interval of the same width (32 points) would give the same result.† - -$$ -\mathcal{D}_r = \frac{1}{32} \sum_{n=-16}^{15} x[n] e^{-j r (\pi/16)n} -$$ - -† In this example we have used the same equations as those for the DFT in Ex. 8.9, within a scaling constant. In the present example, the values of *x*[*n*] at *n* = 4 and −4 are taken as 1 (full value), whereas in Ex. 8.9 these values are 0.5 (half the value). This is the reason for the slight difference in spectra in Figs. 9.2b and 8.19d. Unlike continuous-time signals, discontinuity is a meaningless concept in discrete-time signals. - -### 854 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -Now, *x*[*n*] = 1 for −4 ≤ *n* ≤ 4 and is zero for all other values of *n*. Therefore, - -$$ -\mathcal{D}_r = \frac{1}{32} \sum_{n=-4}^{4} e^{-jr(\pi/16)n} \tag{9.9} -$$ - -This is a geometric progression with a common ratio *e*−*j*(π/16)*r* . Therefore (see Sec. B.8-3),† - -$$ -\mathcal{D}_r = \frac{1}{32} \left[ \frac{e^{-j(5\pi r/16)} - e^{j(4\pi r/16)}}{e^{-j(\pi r/16)} - 1} \right] -$$ - -\n -$$ -= \left(\frac{1}{32}\right) \frac{e^{-j(0.5\pi r/16)} \left[e^{-j(4.5\pi r/16)} - e^{j(4.5\pi r/16)}\right]}{e^{-j(0.5\pi r/16)} \left[e^{-j(0.5\pi r/16)} - e^{j(0.5\pi r/16)}\right]} -$$ - -\n -$$ -= \left(\frac{1}{32}\right) \frac{\sin\left(\frac{4.5\pi r}{16}\right)}{\sin\left(\frac{0.5\pi r}{16}\right)} -$$ - -\n -$$ -= \left(\frac{1}{32}\right) \frac{\sin(4.5r\Omega_0)}{\sin(0.5r\Omega_0)} \qquad \Omega_0 = \frac{\pi}{16} -$$ - (9.10) - -This spectrum (with its periodic extension) is depicted in Fig. 9.2b. - -### DISCRETE-TIME FOURIER SERIES USING MATLAB - -Let us confirm our results by using MATLAB to directly compute the DTFS according to Eq. (9.4). - -``` ->> N_0 = 32; n = (0:N_0-1); Omega_0 = 2*pi/N_0; ->> x_n = [ones(1,5) zeros(1,23) ones(1,4)]; ->> for r = 0:N_0-1, ->> X_r(r+1) = sum(x_n.*exp(-j*r*Omega_0*n))/N_0; ->> end ->> r = n; stem(r,real(X_r),'k.'); ->> xlabel('r'); ylabel('X_r'); axis([0 31 -.1 0.3]); -``` - -The MATLAB result, shown in Fig. 9.3, matches Fig. 9.2b. Alternatively, scaling the FFT by *N*0 produces the exact same result (Fig. 9.3). - ->> X\_r = fft(x\_n)/N\_0; stem(r,real(X\_r),'k.'); >> xlabel('r'); ylabel('X\_r'); axis([0 31 -.1 0.3]); - -$$ -\frac{1}{32} \sum_{n=-4}^{4} x[n] = \frac{9}{32} -$$ - -Fortunately, the value of *D*0, as computed from Eq. (9.10), also happens to be 9/32. Hence, Eq. (9.10) is valid for all *r*. - -† Strictly speaking, the geometric progression sum formula applies only if the common ratio - -*e*−*j*(π/16)*r* = 1. When *r* = 0, this ratio is unity. Hence, Eq. (9.10) is valid for values of *r* = 0. For the case *r* = 0, the sum in Eq. (9.9) is given by - - diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/113_9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/113_9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL.md deleted file mode 100644 index 1272007079c213d8ee6fc10124c618e9280df737..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/113_9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL.md +++ /dev/null @@ -1,421 +0,0 @@ -## **9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL** - -In Sec. 9.1 we succeeded in representing periodic signals as a sum of (everlasting) exponentials. In this section we extend this representation to aperiodic signals. The procedure is identical conceptually to that used in Ch. 7 for continuous-time signals. - -Applying a limiting process, we now show that an aperiodic signal *x*[*n*] can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal *x*[*n*] such as the one illustrated in Fig. 9.4a by everlasting exponential signals, let us construct a new periodic signal *xN*0 [*n*] formed by repeating the signal *x*[*n*] every *N*0 units, as shown in Fig. 9.4b. The period *N*0 is made large enough to avoid overlap between the repeating cycles (*N*0 ≥ 2*N*+1). The periodic signal *xN*0 [*n*] can be represented by an exponential Fourier series. If we let *N*0 → ∞, the signal - -**Figure 9.4** Generation of a periodic signal by periodic extension of a signal *x*[*n*]. - -#### 856 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -*x*[*n*] repeats after an infinite interval, and therefore, - -$$ -\lim_{N_0 \to \infty} x_{N_0}[n] = x[n] -$$ - -Thus, the Fourier series representing *xN*0 [*n*] will also represent *x*[*n*] in the limit *N*0 → ∞. The exponential Fourier series for *xN*0 [*n*] is given by - -$$ -x_{N_0}[n] = \sum_{r = \langle N_0 \rangle} \mathcal{D}_r e^{jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ -\n(9.11) - -where - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n=-\infty}^{\infty} x[n] e^{-jr\Omega_0 n} \tag{9.12} -$$ - -The limits for the sum on the right-hand side of Eq. (9.12) should be from −*N* to *N*. But because *x*[*n*] = 0 for |*n*| > *N*, it does not matter if the limits are taken from −∞ to ∞. - -It is interesting to see how the nature of the spectrum changes as *N*0 increases. To understand this behavior, let us define *X*(), a continuous function of , as - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n} -$$ -\n(9.13) - -From this definition and Eq. (9.12), we have - -$$ -\mathcal{D}_r = \frac{1}{N_0} X(r\Omega_0) \tag{9.14} -$$ - -This result shows that the Fourier coefficients *Dr* are 1/*N*0 times the samples of *X*() taken every 0 rad/s.† Therefore, (1/*N*0)*X*() is the envelope for the coefficients *Dr*. We now let *N*0→∞ by doubling *N*0 repeatedly. Doubling *N*0 halves the fundamental frequency 0, with the result that the spacing between successive spectral components (harmonics) is halved, and there are now twice as many components (samples) in the spectrum. At the same time, by doubling *N*0, the envelope of the coefficients *Dr* is halved, as seen from Eq. (9.14). If we continue this process of doubling *N*0 repeatedly, the number of components doubles in each step; the spectrum progressively becomes denser, while its magnitude *Dr* becomes smaller. Note, however, that the relative shape of the envelope remains the same [proportional to *X*() in Eq. (9.13)]. In the limit, as *N*0 → ∞, the fundamental frequency 0 →0, and *Dr* →0. The separation between successive harmonics, which is 0, is approaching zero (infinitesimal), and the spectrum becomes so dense that it appears to be continuous. But as the number of harmonics increases indefinitely, the harmonic amplitudes *Dr* become vanishingly small (infinitesimal). We discussed an identical situation in Sec. 7.1. - -We follow the procedure in Sec. 7.1 and let *N*0 → ∞. According to Eq. (9.13), - -$$ -X(r\Omega_0) = \sum_{n=-\infty}^{\infty} x[n]e^{-jr\Omega_0 n} -$$ - -† For the sake of simplicity we assume *Dr* and therefore *X*() to be real. The argument, however, is also valid for complex *Dr* [or *X*()]. - -Using Eq. (9.14), we can express Eq. (9.11) as - -$$ -x_{N_0}[n] = \frac{1}{N_0} \sum_{r=N_0} X(r\Omega_0) e^{jr\Omega_0 n} = \sum_{r=N_0} X(r\Omega_0) e^{jr\Omega_0 n} \left(\frac{\Omega_0}{2\pi}\right) -$$ - -In the limit as *N*0 → ∞, 0 → 0 and *xN*0 [*n*] → *x*[*n*]. Therefore, - -$$ -x[n] = \lim_{\Omega_0 \to 0} \sum_{r = \langle N_0 \rangle} \left[ \frac{X(r\Omega_0)\Omega_0}{2\pi} \right] e^{jr\Omega_0 n} \tag{9.15} -$$ - -Because 0 is infinitesimal, it will be appropriate to replace 0 with an infinitesimal notation : - -$$ -\Delta \Omega = \frac{2\pi}{N_0} \tag{9.16} -$$ - -Equation (9.15) can be expressed as - -$$ -x[n] = \lim_{\Delta\Omega \to 0} \frac{1}{2\pi} \sum_{r=\langle N_0 \rangle} X(r\Delta\Omega) e^{jr\Delta\Omega n} \Delta\Omega \tag{9.17} -$$ - -The range *r* = #*N*0\$ implies the interval of *N*0 number of harmonics, which is *N*0 = 2π according to Eq. (9.16). In the limit, the right-hand side of Eq. (9.17) becomes the integral - -$$ -x[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{in\Omega} d\Omega -$$ -\n(9.18) - -where \$ 2π indicates integration over any continuous interval of 2π. The spectrum *X*() is given by [Eq. (9.13)] - -$$ -X(\Omega) = \sum_{n = -\infty}^{\infty} x[n]e^{-j\Omega n} -$$ -\n(9.19) - -The integral on the right-hand side of Eq. (9.18) is called the *Fourier integral*. We have now succeeded in representing an aperiodic signal *x*[*n*] by a Fourier integral (rather than a Fourier series). This integral is basically a Fourier series (in the limit) with fundamental frequency →0, as seen in Eq. (9.17). The amount of the exponential *ejrn* is *X*(*r*)/2π. Thus, the function *X*() given by Eq. (9.19) acts as a spectral function, which indicates the relative amounts of various exponential components of *x*[*n*]. - -We call *X*() the (direct) discrete-time Fourier transform (DTFT) of *x*[*n*], and *x*[*n*] the inverse discrete-time Fourier transform (IDTFT) of *X*(). This nomenclature can be represented as - -$$ -X(\Omega) = \text{DTFT}\{x[n]\} -$$ - and $x[n] = \text{IDTFT}\{X(\Omega)\}$ - -The same information is conveyed by the statement that *x*[*n*] and *X*() are a (discrete-time) Fourier transform pair. Symbolically, this is expressed as - -$$ -x[n] \Longleftrightarrow X(\Omega) -$$ - -The Fourier transform *X*() is the frequency-domain description of *x*[*n*]. - -### **9.2-1 Nature of Fourier Spectra** - -We now discuss several important features of the discrete-time Fourier transform and the spectra associated with it. - -### FOURIER SPECTRA ARE CONTINUOUS FUNCTIONS OF - -Although *x*[*n*] is a discrete-time signal, *X*(), its DTFT is a continuous function of for the simple reason that is a continuous variable, which can take any value over a continuous interval from −∞ to ∞. - -## FOURIER SPECTRA ARE PERIODIC FUNCTIONS OF WITH PERIOD 2π - -From Eq. (9.19), it follows that - -$$ -X(\Omega + 2\pi) = \sum_{n=-\infty}^{\infty} x[n]e^{-j(\Omega + 2\pi)n} = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n}e^{-j2\pi n} = X(\Omega) -$$ - -Clearly, the spectrum *X*() is a continuous, periodic function of with period 2π. We must remember, however, that to synthesize *x*[*n*], we need to use the spectrum over a frequency interval of only 2π, starting at any value of [see Eq. (9.18)]. As a matter of convenience, we shall choose this interval to be the fundamental frequency range (−π, π). It is, therefore, not necessary to show discrete-time-signal spectra beyond the fundamental range, although we often do so. - -The reason for the periodic behavior of *X*() was discussed in Ch. 5, where we showed that, in a basic sense, the discrete-time frequency is bandlimited to || ≤ π. However, all discrete-time sinusoids with frequencies separated by an integer multiple of 2π are identical. This is why the spectrum is 2π periodic. - -### CONJUGATE SYMMETRY OF *X*() - -From Eq. (9.19), we obtain the DTFT of *x*∗[*n*] as - -$$ -\text{DTFT}\{x^*[n]\} = \sum_{n=-\infty}^{\infty} x^*[n]e^{-j\Omega n} = X^*(-\Omega) -$$ - -In other words, - -$$ -x^*[n] \Longleftrightarrow X^*(-\Omega) \tag{9.20} -$$ - -For real *x*[*n*], Eq. (9.20) reduces to *x*[*n*] ⇐⇒ *X*∗(−), which implies that for real *x*[*n*] - -$$ -X(\Omega) = X^*(-\Omega) -$$ - -Therefore, for real *x*[*n*], *X*() and *X*(−) are conjugates. Since *X*() is generally complex, we have both amplitude and angle (or phase) spectra - -$$ -X(\Omega) = |X(\Omega)|e^{j\angle X(\Omega)} -$$ - -Because of conjugate symmetry of *X*(), it follows that for real *x*[*n*], - -$$ -|X(\Omega)| = |X(-\Omega)| \quad \text{and} \quad \angle X(\Omega) = -\angle X(-\Omega) -$$ - -Therefore, the amplitude spectrum |*X*()| is an even function of and the phase spectrum *X*() is an odd function of for real *x*[*n*]. - -### PHYSICAL APPRECIATION OF THE DISCRETE-TIME FOURIER TRANSFORM - -In understanding any aspect of the Fourier transform, we should remember that Fourier representation is a way of expressing a signal *x*[*n*] as a sum of everlasting exponentials (or sinusoids). The Fourier spectrum of a signal indicates the relative amplitudes and phases of the exponentials (or sinusoids) required to synthesize *x*[*n*]. - -A detailed explanation of the nature of such sums over a continuum of frequencies is provided in Sec. 7.1-1. - -### EXISTENCE OF THE DTFT - -Because |*e*−*jn*| = 1, from Eq. (9.19), it follows that the existence of *X*() is guaranteed if *x*[*n*] is absolutely summable; that is, - -$$ -\sum_{n=-\infty}^{\infty} |x[n]| < \infty \tag{9.21} -$$ - -This shows that the condition of absolute summability is a sufficient condition for the existence of the DTFT representation. This condition also guarantees its uniform convergence. The inequality - -$$ -\left[\sum_{n=-\infty}^{\infty} |x[n]| \right]^2 \ge \sum_{n=-\infty}^{\infty} |x[n]|^2 -$$ - -shows that the energy of an absolutely summable sequence is finite. However, not all finite-energy signals are absolutely summable. Signal *x*[*n*] = sinc (*n*) is such an example. For such signals, the DTFT converges, not uniformly, but in the mean.† - -To summarize, *X*() exists under a weaker condition - -$$ -\sum_{n=-\infty}^{\infty} |x[n]|^2 < \infty \tag{9.22} -$$ - -The DTFT under this condition is guaranteed to converge in the mean. Thus, the DTFT of the exponentially growing signal γ *nu*[*n*] does not exist when |γ | > 1 because the signal violates Eqs. (9.21) and (9.22). But the DTFT exists for the signal sinc(*n*), which violates Eq. (9.21) but does satisfy Eq. (9.22) (see later, Ex. 9.6). In addition, if the use of δ(), the continuous-time impulse function, is permitted, we can even find the DTFT of some signals that violate both Eq. (9.21) and Eq. (9.22). Such signals are not absolutely summable, nor do they have finite energy. For example, as seen from pairs 11 and 12 of Table 9.1, the DTFT of *x*[*n*] = 1 for all *n* and *x*[*n*] = *ej*0*n* exist, although they violate Eqs. (9.21) and (9.22). - -$$ -\lim_{M \to \infty} \int_{-\pi}^{\pi} \left| X(\Omega) - \sum_{n=-M}^{M} x[n] e^{-j\Omega n} \right|^2 d\Omega = 0 -$$ - -† This means - -| No. | x[n] | X() | | -|-----|--------------------------|--------------------------------------------------------------------------------------|-----------| -| 1 | δ[n−k] | e−jk | Integer k | -| 2 | γ nu[n] | ej
ej −γ | γ < 1 | -| 3 | −γ nu[−(n+1)] | ej
ej −γ | γ > 1 | -| 4 | γ n | 1−γ 2
1−2γ cos+γ 2 | γ < 1 | -| 5 | nγ nu[n] | γ ej
(ej −γ )2 | γ < 1 | -| 6 | γ n cos(0n+θ
)u[n] | ej[ej cos
θ −γ cos(0
−θ )]
ej2 −(2γ
cos0)ej +γ
2 | γ < 1 | -| 7 | u[n] −u[n− M] | sin(M/2)
e−j(M−1)/2
sin(/2) | | -| 8 | c
π sinc (cn) | "∞
−2πk
rect
2c
k=−∞ | c
≤ π | -| 9 | cn
c
2π sinc2
2 | "∞
−2πk
2c
k=−∞ | c
≤ π | -| 10 | u[n] | ej
+π "∞
δ(−2πk)
ej −1
k=−∞ | | -| 11 | 1
for all n | 2π "∞
δ(−2πk)
k=−∞ | | -| 12 | ej0n | 2π "∞
δ(−0
−2πk)
k=−∞ | | -| 13 | cos0n | π "∞
−2πk) +δ(+0
−2πk)
δ(−0
k=−∞ | | -| 14 | sin0n | jπ "∞
δ(+0
−2πk)−δ(−0
−2πk)
k=−∞ | | -| 15 | (cos0n)u[n] | ej2 −ej cos0
"∞
π
δ(−2πk−0)+δ(−2πk+0)
+1 +
ej2 −2ej cos0
2
k=−∞ | | -| 16 | (sin0n)u[n] | ej sin0
"∞
π
δ(−2πk−0)−δ(−2πk+0)
+1 +
ej2 −2ej cos0
2j
k=−∞ | | - -**TABLE 9.1** Select Discrete-Time Fourier Transform Pairs - -### **EXAMPLE 9.3 DTFT of a Causal Exponential** - -Find the DTFT of *x*[*n*] = γ *nu*[*n*]. - -Using the definition, the DTFT is - -$$ -X(\Omega) = \sum_{n=0}^{\infty} \gamma^n e^{-j\Omega n} = \sum_{n=0}^{\infty} (\gamma e^{-j\Omega})^n -$$ - -This is an infinite geometric series with a common ratio γ *e*−*j*. Therefore (see Sec. B.8-3), - -$$ -X(\Omega) = \frac{1}{1 - \gamma e^{-j\Omega}} -$$ - -provided |γ *e*−*j*| < 1. But because |*e*−*j*| = 1, this condition implies |γ | < 1. Therefore, - -$$ -X(\Omega) = \frac{1}{1 - \gamma e^{-j\Omega}} \qquad |\gamma| < 1 -$$ - -If |γ | > 1, *X*() does not converge. This result is in conformity with Eqs. (9.21) and (9.22). To determine magnitude and phase responses, we note that - -$$ -X(\Omega) = \frac{1}{1 - \gamma \cos \Omega + j\gamma \sin \Omega} \tag{9.23} -$$ - -so - -$$ -|X(\Omega)| = \frac{1}{\sqrt{(1 - \gamma \cos \Omega)^2 + (\gamma \sin \Omega)^2}} = \frac{1}{\sqrt{1 + \gamma^2 - 2\gamma \cos \Omega}} -$$ - -and - -$$ -\angle X(\Omega) = -\tan^{-1}\left[\frac{\gamma \sin \Omega}{1 - \gamma \cos \Omega}\right] -$$ - -Figure 9.5 shows *x*[*n*] = γ *nu*[*n*] and its spectra for γ = 0.8. Observe that the frequency spectra are continuous and periodic functions of with the period 2π. As explained earlier, we need to use the spectrum only over the frequency interval of 2π. We often select this interval to be the fundamental frequency range (−π,π). - -The amplitude spectrum |*X*()| is an even function and the phase spectrum *X*() is an odd function of . - -### **EXAMPLE 9.4 DTFT of an Anticausal Exponential** - -**Figure 9.6** Exponential γ *nu*[−(*n*+1)]. - -Using the definition, the DTFT is - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} \gamma^n u [-(n+1)] e^{-j\Omega n} = \sum_{n=-1}^{-\infty} (\gamma e^{-j\Omega})^n = \sum_{n=-1}^{-\infty} \left(\frac{1}{\gamma} e^{j\Omega}\right)^{-n} -$$ - -Setting *n* = −*m* yields - -$$ -x[n] = \sum_{m=1}^{\infty} \left(\frac{1}{\gamma} e^{j\Omega}\right)^m = \frac{1}{\gamma} e^{j\Omega} + \left(\frac{1}{\gamma} e^{j\Omega}\right)^2 + \left(\frac{1}{\gamma} e^{j\Omega}\right)^3 + \cdots -$$ - -This is a geometric series with a common ratio *ej*/γ . Therefore, from Sec. B.8-3, - -$$ -X(\Omega) = \frac{1}{\gamma e^{-j\Omega} - 1} = \frac{1}{(\gamma \cos \Omega - 1) - j\gamma \sin \Omega}, \qquad |\gamma| > 1 -$$ - -Therefore, - -$$ -|X(\Omega)| = \frac{1}{\sqrt{1 + \gamma^2 - 2\gamma \cos \Omega}} \quad \text{and} \quad \angle X(\Omega) = \tan^{-1} \left[ \frac{\gamma \sin \Omega}{\gamma \cos \Omega - 1} \right] -$$ - -Except for the change of sign, this Fourier transform (and the corresponding frequency spectra) is identical to that of *x*[*n*] = γ *nu*[*n*]. Yet there is no ambiguity in determining the IDTFT of *X*() = 1/(γ *e*−*j* −1) because of the restrictions on the value of γ in each case. If |γ | < 1, then the inverse transform is *x*[*n*]=−γ *nu*[*n*]. If |γ | > 1, it is *x*[*n*] = γ *n*[−(*n*+1)]. - -### **EXAMPLE 9.5 DTFT of a Rectangular Pulse** - -Find the DTFT of the discrete-time rectangular pulse illustrated in Fig. 9.7a. This pulse is also known as the 9-point rectangular window function. - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n} = \sum_{n=-(M-1)/2}^{(M-1)/2} (e^{-j\Omega})^n \qquad M=9 -$$ - -This is a geometric progression with a common ratio *e*−*j* and (see Sec. B.8-3) - -$$ -X(\Omega) = \frac{e^{-j[(M+1)/2]\Omega} - e^{j[(M-1)/2]\Omega}}{e^{-j\Omega} - 1} -$$ -$$ -= \frac{e^{-j\Omega/2} (e^{-j(M/2)\Omega} - e^{j(M/2)\Omega})}{e^{-j\Omega/2} (e^{-j\Omega/2} - e^{j\Omega/2})} -$$ - -$$ -=\frac{\sin\left(\frac{M}{2}\Omega\right)}{\sin\left(0.5\Omega\right)}\tag{9.24} -$$ - -$$ -=\frac{\sin(4.5\Omega)}{\sin(0.5\Omega)} \qquad \text{for } M=9 -$$ - (9.25) - -Figure 9.7b shows the spectrum *X*() for *M* = 9. - -**Figure 9.7 (a)** Discrete-time gate pulse and **(b)** its Fourier spectrum. - -### DISCRETE-TIME FOURIER TRANSFORM USING MATLAB - -Within a scale factor, the DTFS is identical to the DFT and, therefore, the FFT. That is, the DTFS is just the FFT scaled by 1 *N*0 . Combined with Eq. (9.14), we see that the DFT *Xr* of finite-duration signal *x*[*n*] (repeated with period *N*0 large enough to avoid overlap) is just samples of the DTFT *X*() taken at = *r*0. That is, the length-*N*0 DFT of signal *x*[*n*] yields *N*0 samples of its DTFT *X*() as - -$$ -X_r = X(r\Omega_0), \qquad \text{where } \Omega_0 = \frac{2\pi}{N_0} \tag{9.26} -$$ - -This relationship provides a way to use MATLAB's fft command to validate our DTFT calculations. By appropriately zero-padding *x*[*n*], we can obtain as many samples of *X*() as are desired. Let us demonstrate the process for the current example using *N*0 = 64. Notice that in taking the DFT, we modulo-*N*0 shift our rectangular pulse signal to occupy 0 ≤ *n* ≤ *N*0 −1. - -``` ->> Omega = linspace(0,2*pi,1000); -``` - -``` ->> X = sin(4.5*Omega)./sin(0.5*Omega); X(mod(Omega,2*pi)==0) = 4.5/0.5; -``` - -``` ->> N_0 = 64; M = 9; x = [ones(1,(M+1)/2) zeros(1,N_0-M) ones(1,(M-1)/2)]; -``` - -``` ->> Xr = fft(x); Omega_0 = 2*pi/N_0; r = 0:N_0-1; -``` - -``` ->> plot(Omega,abs(X),'k-',Omega_0*r,abs(Xr),'k.'); axis([0 2*pi 0 9.5]); -``` - ->> xlabel('\Omega'); ylabel('|X(\Omega)|'); - -As shown in Fig. 9.8, the FFT samples align exactly with our analytical DTFT result. - -### **EXAMPLE 9.6 Inverse DTFT of a Rectangular Spectrum** - -Find the inverse DTFT of the rectangular pulse spectrum described over the fundamental band (|| ≤ π) by *X*() = rect(/2*c*) for *c* ≤ π. Because of the periodicity property, *X*() repeats at the intervals of 2π, as shown in Fig. 9.9a. - -**Figure 9.9** Periodic gate spectrum and its inverse discrete-time Fourier transform. - -According to Eq. (9.18), - -$$ -x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\Omega) e^{jn\Omega} d\Omega = \frac{1}{2\pi} \int_{-\Omega_c}^{\Omega_c} e^{jn\Omega} d\Omega -$$ -$$ -= \frac{1}{j2\pi n} e^{jn\Omega} \Big|_{-\Omega_c}^{\Omega_c} = \frac{\sin(\Omega_c n)}{\pi n} = \frac{\Omega_c}{\pi} \text{sinc}(\Omega_c n) -$$ - -The signal *x*[*n*] is depicted in Fig. 9.9b (for the case *c* = π/4). - -### **DR ILL 9.4 Finding the DTFT** - -Find the DTFT and sketch the corresponding amplitude and phase spectra for - -(a) -$$ -x[n] = \gamma^{|k|} -$$ - with $|\gamma| < 1$ - -**(b)** -$$ -y[n] = \delta[n+1] - \delta[n-1] -$$ - -### **ANSWERS** - -(a) -$$ -X(\Omega) = \frac{1 - \gamma^2}{1 - 2\gamma \cos \Omega + \gamma^2} -$$ - -\n(b) $|Y(\Omega)| = 2|\sin \Omega|$ and $\angle Y(\omega) = (\pi/2)[1 - \text{sgn}(\sin \Omega)]$ - -## **9.2-2 Connection Between the DTFT and the** *z***-Transform** - -The connection between the (bilateral) *z*-transform and the DTFT is similar to that between the Laplace transform and the Fourier transform. The *z*-transform of *x*[*n*], according to Eq. (5.1), is - -$$ -X[z] = \sum_{n = -\infty}^{\infty} x[n]z^{-n} -$$ -\n(9.27) - -Setting *z* = *ej* in this equation yields - -$$ -X[e^{i\Omega}] = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n} -$$ - -The right-hand side sum defines *X*(), the DTFT of *x*[*n*]. Does this mean that the DTFT can be obtained from the corresponding *z*-transform by setting *z* = *ej*? In other words, is it true that *X*[*ej*] = *X*()? Yes, it is true in most cases. For example, when *x*[*n*] = *anu*[*n*], its *z*-transform is *z*/(*z* − *a*), and *X*[*ej*] = *ej*/(*ej* − *a*), which is equal to *X*() (assuming |*a*| < 1). However, for the unit step function *u*[*n*], the *z*-transform is *z*/(*z* − 1), and *X*[*ej*] = *ej*/(*ej* − 1). As seen from Table 9.1, pair 10, this is not equal to *X*() in this case. - -We obtained *X*[*ej*] by setting *z* = *ej* in Eq. (9.27). This implies that the sum on the right-hand side of Eq. (9.27) converges for *z* = *ej*, which means the unit circle (characterized by *z* = *ej*) lies in the region of convergence for *X*[*z*]. Hence, the general rule is that setting *z* = *ej* in *X*[*z*] yields the DTFT *X*() only when the ROC for *X*[*z*] includes the unit circle. This applies for all *x*[*n*] that are absolutely summable. If the ROC of *X*[*z*] excludes the unit circle, *X*[*ej*] = *X*(). This applies to all exponentially growing *x*[*n*] and also *x*[*n*], which either is constant or oscillates with constant amplitude. - -The reason for this peculiar behavior has something to do with the nature of convergence of the *z*-transform and the DTFT.† - -This discussion shows that although the DTFT may be considered to be a special case of the *z*-transform, we need to circumscribe such a view. This cautionary note is supported by the fact that a periodic signal has the DTFT, but its *z*-transform does not exist. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/114_9.3 PROPERTIES OF THE DTFT.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/114_9.3 PROPERTIES OF THE DTFT.md deleted file mode 100644 index 6b455c382fdc61390f7ea514bd3b5a2586bb7f69..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/114_9.3 PROPERTIES OF THE DTFT.md +++ /dev/null @@ -1,401 +0,0 @@ -## **[9.3 PROPERTIES OF THE](#page-14-0) DTFT** - -A close connection exists between the DTFT and the CTFT (continuous-time Fourier transform). For this reason, which Sec. 9.4 discusses, the properties of the DTFT are very similar to those of the CTFT, as the following discussion shows. - -### LINEARITY OF THE DTFT - -If - -*x*1[*n*] ⇐⇒ *X*1() and *x*2[*n*] ⇐⇒ *X*2() - -then - -$$ -a_1x_1[n]+a_2x_2[n] \Longleftrightarrow a_1X_1(\Omega)+a_2X_2(\Omega) -$$ - -The proof is trivial. The result can be extended to any finite sums. - -### CONJUGATE SYMMETRY OF *X*() - -In Eq. (9.20), we proved the *conjugation property* - -$$ -x^*[n] \Longleftrightarrow X^*(-\Omega) \tag{9.28} -$$ - -† To explain this point, consider the unit step function *u*[*n*] and its transforms. Both the *z*-transform and the DTFT synthesize *x*[*n*], using everlasting exponentials of the form *zn*. The value of *z* can be anywhere in the complex *z*-plane for the *z*-transform, but it must be restricted to the unit circle (*z* = *ej*) in the case of the DTFT. The unit step function is readily synthesized in the *z*-transform by a relatively simple spectrum *X*[*z*] = *z*/(*z* − 1), by choosing *z* outside the unit circle (the ROC for *u*[*n*] is |*z*| > 1). In the DTFT, however, we are restricted to values of *z* only on the unit circle (*z* = *ej*). The function *u*[*n*] can still be synthesized by values of *z* on the unit circle, but the spectrum is more complicated than when we are free to choose *z* anywhere, including the region outside the unit circle. In contrast, when *x*[*n*] is absolutely summable, the region of convergence for the *z*-transform includes the unit circle, and we can synthesize *x*[*n*] by using *z* along the unit circle in both the transforms. This leads to *X*[*ej*] = *X*(). - -### 868 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -We also showed that as a consequence of this, when *x*[*n*] is real, *X*() and *X*(−) are conjugates, that is, - -$$ -X(-\Omega) = X^*(\Omega) -$$ - -This is the *conjugate symmetry* property. Since *X*() is generally complex, we have both amplitude and angle (or phase) spectra - -$$ -X(\Omega) = |X(\Omega)|e^{j\angle X(\Omega)} -$$ - -Hence, for real *x*[*n*], it follows that - -$$ -|X(\Omega)| = |X(-\Omega)| \quad \text{and} \quad \angle X(\Omega) = -\angle X(-\Omega) -$$ - -Therefore, for real *x*[*n*], the amplitude spectrum |*X*()| is an even function of and the phase spectrum *X*() is an odd function of . - -### TIME AND FREQUENCY REVERSAL - -Also called the reflection property, the time and frequency reversal property states that - -$$ -x[-n] \Longleftrightarrow X(-\Omega) \tag{9.29} -$$ - -Demonstration of this property is straightforward. From Eq. (9.19), the DTFT of *x*[−*n*] is - -$$ -\text{DTFT}\{x[-n]\} = \sum_{n=-\infty}^{\infty} x[-n]e^{-j\Omega n} = \sum_{m=-\infty}^{\infty} x[m]e^{j\Omega m} = X(-\Omega) -$$ - -### **EXAMPLE 9.7 Using the Reflection Property** - -Use the time-frequency reversal property of Eq. (9.29) and pair 2 in Table 9.1 to derive pair 4 in Table 9.1. - -Pair 2 states that - -$$ -\gamma^n u[n] = \frac{e^{i\Omega}}{e^{i\Omega} - \gamma} \qquad |\gamma| < 1 -$$ - -Hence, from Eq. (9.29), - -$$ -\gamma^{-n}u[-n] = \frac{e^{-j\Omega}}{e^{-j\Omega} - \gamma} \qquad |\gamma| < 1 -$$ - -Moreover, γ |*n*| could be expressed as a sum of γ *n u*[*n*] and γ −*n u*[−*n*], except that the impulse at *n* = 0 is counted twice (once from each of the two exponentials). Hence, - -$$ -\gamma^{|n|} = \gamma^n u[n] + \gamma^{-n} u[-n] - \delta[n] -$$ - -Combining these results and invoking the linearity property, we can write - -$$ -\text{DTFT}\{\gamma^{|n|}\} = \frac{e^{j\Omega}}{e^{j\Omega} - \gamma} + \frac{e^{-j\Omega}}{e^{-j\Omega} - \gamma} - 1 = \frac{1 - \gamma^2}{1 - 2\gamma \cos \Omega + \gamma^2} \qquad |\gamma| < 1 -$$ - -which agrees with pair 4 in Table 9.1. - -### **DR ILL 9.5 Using the Reflection Property** - -In Table 9.1, derive pair 13 from pair 15 by using the time-reversal property of Eq. (9.29). - -### MULTIPLICATION BY *n*: FREQUENCY DIFFERENTIATION - -$$ -nx[n] \Longleftrightarrow j\frac{dX(\Omega)}{d\Omega} \tag{9.30} -$$ - -The result follows immediately by differentiating both sides of Eq. (9.19) with respect to . - -### **EXAMPLE 9.8 Using the Frequency-Differentiation Property** - -Use the frequency-differentiation property of Eq. (9.30) and pair 2 in Table 9.1 to derive pair 5 in Table 9.1. - -Pair 2 states that - -$$ -\gamma^n u[n] = \frac{e^{i\Omega}}{e^{i\Omega} - \gamma} \qquad |\gamma| < 1 -$$ - -Hence, from Eq. (9.30), - -$$ -n\gamma^{n}u[n] = j\frac{d}{d\Omega} \left\{ \frac{e^{j\Omega}}{e^{j\Omega} - \gamma} \right\} = \frac{\gamma e^{j\Omega}}{(e^{j\Omega} - \gamma)^{2}} \qquad |\gamma| < 1 -$$ - -which agrees with pair 5 in Table 9.1. - -### TIME-SHIFTING PROPERTY If - -*x*[*n*] ⇐⇒ *X*() - -then - -$$ -x[n-k] \Longleftrightarrow X(\Omega)e^{-jk\Omega} \qquad \text{for integer } k \tag{9.31} -$$ - -This property can be proved by direct substitution in the equation defining the direct transform. From Eq. (9.19), we obtain - -$$ -x[n-k] \Longleftrightarrow \sum_{n=-\infty}^{\infty} x[n-k]e^{-j\Omega n} = \sum_{m=-\infty}^{\infty} x[m]e^{-j\Omega[m+k]} -$$ -$$ -= e^{-j\Omega k} \sum_{n=-\infty}^{\infty} x[m]e^{-j\Omega m} = e^{-jk\Omega}X(\Omega) -$$ - -This result shows that *delaying a signal by k samples does not change its amplitude spectrum. The phase spectrum, however, is changed by* −*k*. This added phase is a linear function of with slope −*k*. - -### PHYSICAL EXPLANATION OF LINEAR PHASE - -Time delay in a signal causes a linear phase shift in its spectrum. The heuristic explanation of this result is exactly parallel to that for continuous-time signals given in Sec. 7.3 (see Fig. 7.22). - -### **EXAMPLE 9.9 Demonstrating Linear Phase** - -To demonstrate the linear phase associated with a time shift, find the DTFT of *x*[*n*] = (1/4)sinc (π(*n*−2)/4), shown in Fig. 9.10a. - -In Ex. 9.6, we found that - -$$ -\frac{1}{4}\operatorname{sinc}\left(\frac{\pi n}{4}\right) \Longleftrightarrow \sum_{m=-\infty}^{\infty} \operatorname{rect}\left(\frac{\Omega - 2\pi m}{\pi/2}\right) -$$ - -Use of the time-shifting property [Eq. (9.31)] yields (for integer *k*) - -$$ -\frac{1}{4}\operatorname{sinc}\left(\frac{\pi(n-2)}{4}\right) \Longleftrightarrow \sum_{m=-\infty}^{\infty} \operatorname{rect}\left(\frac{\Omega-2\pi m}{\pi/2}\right) e^{-j2\Omega} -$$ - -The spectrum of the shifted signal is shown in Fig. 9.10b. - -## **DR ILL 9.6 Using the Time-Shifting Property** - -Verify the result in Eq. (9.24) from pair 7 in Table 9.1 and the time-shifting property of the DTFT. - -# FREQUENCY-SHIFTING PROPERTY - -If - -$$ -x[n] \Longleftrightarrow X(\Omega) -$$ - -then - -$$ -x[n]e^{j\Omega_c n} \Longleftrightarrow X(\Omega - \Omega_c) \tag{9.32} -$$ - -This property is the dual of the time-shifting property. To prove the frequency-shifting property, we use Eq. (9.19) as - -$$ -x[n]e^{j\Omega_c n} \Longleftrightarrow \sum_{n=-\infty}^{\infty} x[n]e^{j\Omega_c n}e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x[n]e^{-j(\Omega-\Omega_c)n} = X(\Omega-\Omega_c) -$$ - -From this result, it follows that - -$$ -x[n]e^{-j\Omega_c n} \Longleftrightarrow X(\Omega + \Omega_c) -$$ - -### 872 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -Adding this pair to the pair in Eq. (9.32), we obtain - -$$ -x[n]\cos(\Omega_c n) \Longleftrightarrow \frac{1}{2}\{X(\Omega - \Omega_c) + X(\Omega + \Omega_c)\}\tag{9.33} -$$ - -This is the *modulation property*. - -Multiplying both sides of pair (9.32) by *ej*θ , we obtain - -$$ -x[n]e^{j(\Omega_c n + \theta)} \Longleftrightarrow X(\Omega - \Omega_c)e^{j\theta} -$$ - -Using this pair, we can generalize the modulation property as - -$$ -x[n]\cos\left(\Omega_c n + \theta\right) \Longleftrightarrow \frac{1}{2}\left\{X(\Omega - \Omega_c)e^{i\theta} + X(\Omega + \Omega_c)e^{-i\theta}\right\} -$$ - -### **EXAMPLE 9.10 Modulation Property** - -A signal *x*[*n*] = sinc (π*n*/4) modulates a carrier cos*cn*. Find and sketch the spectrum of the modulated signal *x*[*n*] cos*cn* for - -- **(a)** *c* = π/2 -- **(b)** *c* = 7π/8 = 0.875π - -**(a)** For *x*[*n*] = sinc (π*n*/4), we find (Table 9.1, pair 8) - -$$ -X(\Omega) = 4 \sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 2\pi m}{\pi/2}\right) -$$ - -Figure 9.11a shows the DTFT *X*(). From the modulation property of Eq. (9.33), we obtain - -$$ -x[n]\cos(0.5\pi n) \Longleftrightarrow 2\sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 0.5\pi - 2\pi m}{0.5\pi}\right) + \text{rect}\left(\frac{\Omega + 0.5\pi - 2\pi m}{0.5\pi}\right) -$$ - -Figure 9.11b shows half the *X*() shifted by π/2 and Fig. 9.11c shows half the *X*() shifted by −π/2. The spectrum of the modulated signal is obtained by adding these two shifted spectra and multiplying by half, as shown in Fig. 9.11d. - -**(b)** Figure 9.12a shows *X*(), which is the same as that in part (a). For *c* = 7π/8 = 0.875π, the modulation property of Eq. (9.33) yields - -$$ -x[n]\cos(0.875\pi n) \Longleftrightarrow 2\sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 0.875\pi - 2\pi m}{0.5\pi}\right) + \text{rect}\left(\frac{\Omega + 0.875\pi - 2\pi m}{0.5\pi}\right) -$$ - -**Figure 9.11** Instance of modulation for Ex. 9.10a. - -Figure 9.12b shows *X*() shifted by 7π/8 and Fig. 9.12c shows *X*() shifted by −7π/8. The spectrum of the modulated signal is obtained by adding these two shifted spectra and multiplying by half, as shown in Fig. 9.12d. In this case, the two shifted spectra overlap. Since the operation of modulation thus causes aliasing, it does not achieve the desired effect of spectral shifting. In this example, to realize spectral shifting without aliasing requires *c* ≤ 3π/4. - -### **DR ILL 9.7 Using the Frequency-Shifting Property** - -In Table 9.1, derive pairs 12 and 13 from pair 11 and the frequency-shifting/modulation property. - -## TIME- AND FREQUENCY-CONVOLUTION PROPERTY If - -*x*1[*n*] ⇐⇒ *X*1() and *x*2[*n*] ⇐⇒ *X*2() - -then - -$$ -x_1[n] * x_2[n] \Longleftrightarrow X_1(\Omega)X_2(\Omega) \tag{9.34} -$$ - -and - -$$ -x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi}X_1(\Omega)\circledast X_2(\Omega) \tag{9.35} -$$ - -where - -$$ -x_1[n] * x_2[n] = \sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m] -$$ - -For two continuous, periodic signals, we define the periodic convolution, denoted by symbol -∗ as† - -$$ -X_1(\Omega)\circledast X_2(\Omega) = \frac{1}{2\pi} \int_{2\pi} X_1(u)X_2(\Omega - u) du -$$ - -The convolution here is not the *linear* convolution used so far. This is a *periodic* (or *circular*) convolution applicable to the convolution of two continuous, periodic functions with the same period. The limit of integration in the convolution extends only to one period. - -Proof of the time-convolution property is identical to that given in Sec. 5.2 [Eq. (5.19)]. All we have to do is replace *z* with *ej*. To prove the frequency-convolution property of Eq. (9.35), we have - -$$ -x_1[n]x_2[n] \Longleftrightarrow \sum_{n=-\infty}^{\infty} x_1[n]x_2[n]e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x_2[n] \left[ \frac{1}{2\pi} \int_{2\pi} X_1(u)e^{-jnu} du \right] e^{-j\Omega n} -$$ - -Interchanging the order of summation and integration, we obtain - -$$ -x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi} \int_{2\pi} X_1(u) \left[ \sum_{n=-\infty}^{\infty} x_2[n] e^{-j(\Omega-u)n} \right] du = \frac{1}{2\pi} \int_{2\pi} X_1(u) X_2(\Omega-u) du -$$ - -### **EXAMPLE 9.11 DTFT of an Accumulator System** - -If -$$ -x[n] \leftrightarrow X(\Omega) -$$ -, then show that $\sum_{k=-\infty}^{n} x[k] \leftrightarrow \pi X(0) \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1} X(\Omega)$ . - -† In Eq. (8.20), we defined periodic convolution for two discrete, periodic sequences in a different way. Although we are using the same symbol -∗ for both discrete and continuous cases, the meaning will be clear from the context. - -### 876 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -To begin, we notice that - -$$ -x[n] * u[n] = \sum_{k=-\infty}^{\infty} x[k]u[n-k] = \sum_{k=-\infty}^{n} x[k] -$$ - -Applying the time-convolution property of Eq. (9.34) and pair 10 in Table 9.1, it follows that - -$$ -\sum_{k=-\infty}^{n} x[k] = x[n] * u[n] \Longleftrightarrow X(\Omega) \left(\pi \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1}\right) -$$ - -Because of 2π periodicity, *X*(0) = *X*(2π*k*). Moreover, *X*()δ( − 2π*k*) = *X*(2π*k*)δ( − 2π*k*) = *X*(0)δ(−2π*k*). Hence, - -$$ -\sum_{k=-\infty}^{n} x[k] \Longleftrightarrow \pi X(0) \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1} X(\Omega) -$$ - -### **DR ILL 9.8 Using the Frequency-Convolution Property** - -In Table 9.1, derive pair 9 from pair 8, assuming *c* ≤ π/2. Use the frequency-convolution property. - -PARSEVAL'S THEOREM If - -$$ -x[n] \Longleftrightarrow X(\Omega) -$$ - -then *Ex*, the energy of *x*[*n*], is given by - -$$ -E_x = \sum_{n=-\infty}^{\infty} |x[n]|^2 = \frac{1}{2\pi} \int_{2\pi} |X(\Omega)|^2 d\Omega -$$ -\n(9.36) - -To prove this property, we have from Eq. (9.28), - -$$ -X^*(\Omega) = \sum_{n=-\infty}^{\infty} x^*[n]e^{i\Omega n} -$$ - -Now, - -$$ -\sum_{n=-\infty}^{\infty} |x[n]|^2 = \sum_{n=-\infty}^{\infty} x^*[n]x[n] = \sum_{n=-\infty}^{\infty} x^*[n] \left[ \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{i\Omega n} d\Omega \right] -$$ -$$ -= \frac{1}{2\pi} \int_{2\pi} X(\Omega) \left[ \sum_{n=-\infty}^{\infty} x^*[n] e^{i\Omega n} \right] d\Omega -$$ -$$ -= \frac{1}{2\pi} \int_{2\pi} X(\Omega) X^*(\Omega) d\Omega = \frac{1}{2\pi} \int_{2\pi} |X(\Omega)|^2 d\Omega -$$ - -Table 9.2 summarizes Parseval's theorem and the other important properties of the DTFT. - -| Operation | x[n] | X() | -|-----------------------|-------------------------------|-----------------------------------------------| -| Linearity | a1x1[n] +a2x2[n] | a1X1()+a2X2() | -| Conjugation | x∗[n] | X∗(−) | -| Scalar multiplication | ax[n] | aX() | -| Multiplication by n | nx[n] | dX()
j
d | -| Time reversal | x[−n] | X(−) | -| Time shifting | x[n−k] | X()e−jk
k integer | -| Frequency shifting | x[n] ejcn | X(−c) | -| Time convolution | x1[n] ∗ x2[n] | X1()X2() | -| Frequency convolution | x1[n]x2[n] | #
1
X1[u]X2[−u]du
2π
2π | -| Parseval's theorem | = "∞
x[n] 2
Ex
n=−∞ | #
1
2 d
Ex
X()
=
2π
2π | - -**TABLE 9.2** Properties of the DTFT - -### **EXAMPLE 9.12 Using Parseval's Theorem to Find Signal Energy** - -Find the energy of *x*[*n*] = sinc (*cn*), assuming *c* < π. - -From pair 8, Table 9.1, the fundamental band spectrum of *x*[*n*] is - -$$ -\operatorname{sinc}(\Omega_c n) \Longleftrightarrow \frac{\pi}{\Omega_c} \operatorname{rect}\left(\frac{\Omega}{2\Omega_c}\right) \qquad |\Omega| \le \pi -$$ - -### 878 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -From Parseval's theorem [Eq. (9.36)], we have - -$$ -E_x = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{\pi^2}{\Omega_c^2} \left[ \text{rect}\left(\frac{\Omega}{2\Omega_c}\right) \right]^2 d\Omega -$$ - -Because rect(/2*c*) = 1 over || ≤ *c* and is zero otherwise, the preceding integral yields - -$$ -E_x = \frac{1}{2\pi} \left(\frac{\pi^2}{\Omega_c^2}\right) (2\Omega_c) = \frac{\pi}{\Omega_c} -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/115_9.4 LTI DISCRETE-TIME SYSTEM ANALYSIS BY DTFT.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/115_9.4 LTI DISCRETE-TIME SYSTEM ANALYSIS BY DTFT.md deleted file mode 100644 index 2adea6ce3484fc0167ac4c9cce228300d8dbbe81..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/115_9.4 LTI DISCRETE-TIME SYSTEM ANALYSIS BY DTFT.md +++ /dev/null @@ -1,193 +0,0 @@ -## **[9.4 LTI DISCRETE-TIME](#page-14-0) SYSTEM ANALYSIS BY DTFT** - -Consider a linear, time-invariant, discrete-time system with the unit impulse response *h*[*n*]. We shall find the (zero-state) system response *y*[*n*] for the input *x*[*n*]. Let - -*x*[*n*] ⇐⇒ *X*() *y*[*n*] ⇐⇒ *Y*() and *h*[*n*] ⇐⇒ *H*() - -Because *y*[*n*] = *x*[*n*] ∗ *h*[*n*], it follows from Eq. (9.34) that - -$$ -Y(\Omega) = X(\Omega)H(\Omega) \tag{9.37} -$$ - -This result is similar to that obtained for continuous-time systems. Let us examine the role of *H*(), the DTFT of the unit impulse response *h*[*n*]. - -Equation (9.37) holds for BIBO-stable systems and also for marginally stable systems if the input does not contain the system's natural mode(s). In other cases, the response grows with *n* and is not Fourier-transformable. Moreover, the input *x*[*n*] also has to be DTF-transformable. For cases where Eq. (9.37) does not apply, we use the *z*-transform for system analysis. - -Equation (9.37) shows that the output signal frequency spectrum is the product of the input signal frequency spectrum and the frequency response of the system. From this equation, we obtain - -|*Y*()|=|*X*()||*H*()| and *Y*() = *X*()+ *H*() - -This result shows that the output amplitude spectrum is the product of the input amplitude spectrum and the amplitude response of the system. The output phase spectrum is the sum of the input phase spectrum and the phase response of the system. - -We can also interpret Eq. (9.37) in terms of the frequency-domain viewpoint, which sees a system in terms of its frequency response (system response to various exponential or sinusoidal components). The frequency domain views a signal as a sum of various exponential or sinusoidal components. The transmission of a signal through a (linear) system is viewed as transmission of various exponential or sinusoidal components of the input signal through the system. This concept can be understood by displaying the input–output relationships by a directed arrow as follows: - -$$ -e^{i\Omega n} \Longrightarrow H(\Omega)e^{i\Omega n} -$$ - -which shows that the system response to *ejn* is *H*()*ejn*, and - -$$ -x[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{i\Omega n} d\Omega -$$ - -which shows *x*[*n*] as a sum of everlasting exponential components. Invoking the linearity property, we obtain - -$$ -y[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) H(\Omega) e^{i\Omega n} d\Omega -$$ - -which gives *y*[*n*] as a sum of responses to all input components and is equivalent to Eq. (9.37). Thus, *X*() is the input spectrum and *Y*() is the output spectrum, given by *X*()*H*(). - -### **EXAMPLE 9.13 LTID System Analysis by the DTFT** - -An LTID system is specified by the equation *y*[*n*] − 0.5*y*[*n* − 1] = *x*[*n*]. Find *H*(), the frequency response of this system. Determine the (zero-state) response *y*[*n*] if the input *x*[*n*] = (0.8)*nu*[*n*]. - -Let *x*[*n*] ⇐⇒ *X*() and *y*[*n*] ⇐⇒ *Y*(). Taking the DTFT of the system's difference equation yields - -$$ -(1 - 0.5e^{-j\Omega})Y(\Omega) = X(\Omega) -$$ - -According to Eq. (9.37), - -$$ -H(\Omega) = \frac{Y(\Omega)}{X(\Omega)} = \frac{1}{1 - e^{-j\Omega}} = \frac{e^{j\Omega}}{e^{j\Omega} - 0.5} -$$ - -Also, *x*[*n*] = (0.8)*nu*[*n*]. Hence, - -$$ -X(\Omega) = \frac{e^{i\Omega}}{e^{i\Omega} - 0.8} -$$ - -and - -$$ -Y(\Omega) = X(\Omega)H(\Omega) = \frac{2e^{i\Omega}}{(e^{i\Omega} - 0.8)(e^{i\Omega} - 0.5)} -$$ - -#### 880 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -We can express the right-hand side as a sum of two first-order terms (modified partial fraction expansion as discussed in Sec. B.5-6) as follows† : - -$$ -\frac{Y(\Omega)}{e^{i\Omega}} = \frac{e^{i\Omega}}{(e^{i\Omega} - 0.5)(e^{i\Omega} - 0.8)} = \frac{-\frac{5}{3}}{e^{i\Omega} - 0.5} + \frac{\frac{8}{3}}{e^{i\Omega} - 0.8} -$$ - -Consequently, - -$$ -Y(\Omega) = -\left(\frac{5}{3}\right) \frac{e^{i\Omega}}{e^{i\Omega} - 0.5} + \left(\frac{8}{3}\right) \frac{e^{i\Omega}}{e^{i\Omega} - 0.8} -$$ - -= -\left(\frac{5}{3}\right) \frac{1}{1 - 0.5e^{-i\Omega}} + \left(\frac{8}{3}\right) \frac{1}{1 - 0.8e^{-i\Omega}} - -From entry 2 of Table 9.1, the inverse DTFT of this equation is - -$$ -y[n] = \left[ -\frac{5}{3}(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] -$$ - -This example demonstrates the procedure for using the DTFT to determine an LTID system response. It is similar to the Fourier transform method in the analysis of LTIC systems. As in the case of the Fourier transform, this method can be used only if the system is asymptotically or BIBO-stable and if the input signal is DTF-transformable.‡ We shall not belabor this method further because it is clumsier and more restrictive than the *z*-transform method discussed in Ch. 5. - -### **[9.4-1 Distortionless Transmission](#page-14-0)** - -In several applications, digital signals are passed through LTI systems, and we require that the output waveform be a replica of the input waveform. As in the continuous-time case, transmission is said to be distortionless if the input *x*[*n*] and the output *y*[*n*] satisfy the condition - -$$ -y[n] = G_0 x[n - n_d] -$$ - -Here, *nd*, the delay (in samples), is assumed to be integer. Taking the Fourier transform yields - -$$ -Y(\Omega) = G_0 X(\Omega) e^{-j\Omega n_d} -$$ - -But - -$$ -Y(\Omega) = X(\Omega) H(\Omega) -$$ - -† Here, *Y*() is a function of variable *ej*. Hence, *x* = *ej* for the purpose of comparison with the expression in Sec. B.5-6. - -‡ It can also be applied to marginally stable systems if the input does not contain natural mode(s) of the system. - -**Figure 9.13** LTI system frequency response for distortionless transmission. - -Therefore, - -$$ -H(\Omega) = G_0 e^{-j\Omega n_d} -$$ - -This is the frequency response required for distortionless transmission. From this equation, it follows that - -$$ -|H(\Omega)| = G_0 \quad \text{and} \quad \angle H(\Omega) = -\Omega n_d \tag{9.38} -$$ - -Thus, for distortionless transmission, the amplitude response |*H*()| must be a constant, and the phase response *H*() must be a linear function of with slope −*nd*, where *nd* is the delay in the number of samples with respect to input (Fig. 9.13). These are precisely the characteristics of an ideal delay of *nd* samples with a gain of *G*0 [see Eq. (9.31)]. - -### MEASURE OF DELAY VARIATION - -For distortionless transmission, we require a *linear phase* characteristic. In practice, many systems have a phase characteristic that may be only approximately linear. A convenient way of judging phase linearity is to plot the slope of *H*() as a function of frequency. This slope is constant for the ideal linear phase (ILP) system, but it may vary with in the general case. The slope can be expressed as - -$$ -n_g(\Omega) = -\frac{d}{d\Omega} \angle H(\Omega) \tag{9.39} -$$ - -If *ng*() is constant, all the components are delayed by *ng* samples. But if the slope is not constant, the delay *ng* varies with frequency. This variation means that different frequency components undergo different amounts of delay, and consequently, the output waveform will not be a replica of the input waveform. As in the case of LTIC systems, *ng*(), as defined in Eq. (9.39), plays an important role in bandpass systems and is called the *group delay* or *envelope* delay. Observe that constant *nd* implies constant *ng*. Note that *H*() = φ0 − *nd* also has a constant *ng*. Thus, constant group delay is a more relaxed condition. - -### DISTORTIONLESS TRANSMISSION OVER BANDPASS SYSTEMS - -As in the case of continuous-time systems, the distortionless transmission conditions can be relaxed for discrete-time bandpass systems. For lowpass systems, the phase characteristic should not only be linear over the band of interest, it should also pass through the origin [Eq. (9.38)]. For bandpass systems, the phase characteristic should be linear over the band of interest, but it need not pass through the origin (*ng* should be constant). The amplitude response is required to - -#### 882 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -be constant over the passband. Thus, for distortionless transmission over a bandpass system, the frequency response for positive range of is of the form† - -$$ -H(\Omega) = G_0 e^{j(\phi_0 - \Omega n_g)} \qquad \Omega \ge 0 -$$ - -The proof is identical to that for the continuous-time case in Sec. 7.4-2 and will not be repeated. In using Eq. (9.39) to compute *ng*, we should ignore jump discontinuities in the phase function. - -### **[9.4-2 Ideal and Practical Filters](#page-14-0)** - -Ideal filters allow distortionless transmission of a certain band of frequencies and suppress all the remaining frequencies. The general ideal lowpass filter shown in Fig. 9.14 for || ≤ π allows all components below the cutoff frequency = *c* to pass without distortion and suppresses all components above *c*. Figure 9.15 illustrates ideal highpass and bandpass filter characteristics. - -The ideal lowpass filter in Fig. 9.14a has a linear phase of slope −*nd*, which results in a delay of *nd* samples for all its input components of frequencies below *c* rad/sample. Therefore, if the input is a signal *x*[*n*] bandlimited to *c*, the output *y*[*n*] is *x*[*n*] delayed by *nd*; that is, - -$$ -y[n] = x[n - n_d] -$$ - -The signal *x*[*n*] is transmitted by this system without distortion, but with delay of *nd* samples. For this filter, - -$$ -H(\Omega) = \sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 2\pi m}{2\Omega_c}\right) e^{-j\Omega n_d} -$$ - -The unit impulse response *h*[*n*] of this filter is obtained from pair 8 (Table 9.1) and the time-shifting property - -$$ -h[n] = \frac{\Omega_c}{\pi} \operatorname{sinc} \left[ \Omega_c (n - n_d) \right] -$$ - -Because *h*[*n*] is the system response to impulse input δ[*n*], which is applied at *n* = 0, it must be causal (i.e., it must not start before *n* = 0) for a realizable system. Figure 9.14b shows *h*[*n*] for - -**Figure 9.14** Ideal lowpass filter: its frequency response and impulse response. - -† Because the phase function is an odd function of , if *H*() = φ0 − *ng* for ≥ 0, over the band 2*W* (centered at *c*), then *H*() = −φ0 −*ng* for < 0 over the band 2*W* (centered at −*c*). - -**Figure 9.15** Ideal highpass and bandpass filter frequency response. - -**Figure 9.16** Approximate realization of an ideal lowpass filter by truncation of its impulse response. - -*c* = π/4 and *nd* = 12. This figure also shows that *h*[*n*] is noncausal, hence unrealizable. Similarly, one can show that other ideal filters (such as the ideal highpass or and bandpass filters depicted in Fig. 9.15) are also noncausal and therefore physically unrealizable. - -One practical approach to realize an ideal lowpass filter approximately is to truncate both tails (positive and negative) of *h*[*n*] so that it has a finite length and then delay sufficiently to make it causal (Fig. 9.16). We now synthesize a system with this truncated (and delayed) impulse response. For closer approximation, the truncating window has to be correspondingly wider. The delay required also increases correspondingly. Thus, the price of closer realization is higher delay in the output; this situation is common in noncausal systems. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/116_9.5 DTFT CONNECTION WITH THE CTFT.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/116_9.5 DTFT CONNECTION WITH THE CTFT.md deleted file mode 100644 index d340272b930a5106ae5dafe3b8fdc3254c562669..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/116_9.5 DTFT CONNECTION WITH THE CTFT.md +++ /dev/null @@ -1,94 +0,0 @@ -## **[9.5 DTFT CONNECTION WITH THE](#page-14-0) CTFT** - -Consider a continuous-time signal *xc*(*t*) (Fig. 9.17a) with the Fourier transform *Xc*(ω) bandlimited to *B* Hz (Fig. 9.17b). This signal is sampled with a sampling interval *T*. The sampling rate is at least equal to the Nyquist rate; that is, *T* ≤ 1/2*B*. The sampled signal *xc*(*t*) (Fig. 9.17c) can be - -**Figure 9.17** Connection between the DTFT and the Fourier transform. - -expressed as - -$$ -\bar{x}_c(t) = \sum_{n=-\infty}^{\infty} x_c(nT) \,\delta(t - nT) -$$ - -The continuous-time Fourier transform of the foregoing equation yields - -$$ -\overline{X}_c(\omega) = \sum_{n=-\infty}^{\infty} x_c(nT) e^{-jnT\omega} -$$ -\n(9.40) - -In Sec. 8.1 (Fig. 8.1f), we showed that *Xc*(ω) is *Xc*(ω)/*T* repeating periodically with a period ω*s* = 2π/*T*, as illustrated in Fig. 9.17d. Let us construct a discrete-time signal *x*[*n*] such that its *n*th sample value is equal to the value of the *n*th sample of *xc*(*t*), as depicted in Fig. 9.17e, that is, - -$$ -x[n] = x_c(nT) \tag{9.41} -$$ - -Now, *X*(), the DTFT of *x*[*n*], is given by - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-jn\Omega} = \sum_{n=-\infty}^{\infty} x_c(nT) e^{-jn\Omega} -$$ - -Comparison of this equation with Eq. (9.40) shows that letting ω*T* = in *Xc*(ω) yields *X*(), that is, - -$$ -X(\Omega) = X_c(\omega)|_{\omega T = \Omega} -$$ - -Alternately, *X*() can be obtained from *Xc*(ω) by replacing ω with /*T*, that is, - -$$ -X(\Omega) = \overline{X}_c \left(\frac{\Omega}{T}\right) \tag{9.42} -$$ - -Therefore, *X*() is identical to *Xc*(ω), frequency-scaled by factor *T*, as shown in Fig. 9.17f. Thus, ω = 2π/*T* in Fig. 9.17d corresponds to = 2π in Fig. 9.17f. - -### **[9.5-1 Use of DFT and FFT for Numerical Computation of the DTFT](#page-14-0)** - -The discrete Fourier transform (DFT), as discussed in Ch. 8, is a tool for computing the samples of the continuous-time Fourier transform (CTFT). Because of the close connection between CTFT and DTFT, as seen in Eq. (9.42), we can also use this same DFT to compute DTFT samples. - -In Ch. 8, Eqs. (8.12) and (8.13) relate an *N*0-point sequence *xn* to another *N*0-point sequence *Xr*. Changing the notation *xn* to *x*[*n*] in these equations, we obtain - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x[n] e^{-j r \Omega_0 n} -$$ -\n(9.43) - -and - -$$ -x[n] = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{jr\Omega_0 n} -$$ -\n(9.44) - -where 0 = 2π *N*0 . Comparing Eq. (9.19) with Eq. (9.43), we recognize that *Xr* is the sample of *X*() at = *r*0, that is, - -$$ -X_r = X(r\Omega_0) \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -Hence, DFT Eqs. (9.43) and (9.44) can be viewed to relate an *N*0-point sequence *x*[*n*] to the *N*0-point samples of corresponding *X*(). We can now use the efficient algorithm FFT (discussed in Ch. 8) to compute *Xr* from *x*[*n*], and vice versa. - -If *x*[*n*] is not timelimited, we can still find the approximate values of *Xr* by suitably windowing *x*[*n*]. To reduce the error, the window should be tapered and should have sufficient width to satisfy error specifications. In practice, the numerical computation of signals, which are generally non-timelimited, is performed in this manner because of the computational economy of the DFT, especially for signals of long duration. - -### COMPUTATION OF DISCRETE-TIME FOURIER SERIES (DTFS) - -The discrete-time Fourier series (DTFS) equations [(9.3) and (9.4)] are identical to the DFT equations [(8.13) and (8.12)] within a scaling constant *N*0. If we let *x*[*n*] = *N*0*xn* and *Dr* = *Xr* - -#### 886 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -in Eqs. (9.4) and (9.3), we obtain - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n} \qquad \text{and} \qquad x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{jr\Omega_0 n} -$$ - -This is precisely the DFT and IDFT of Eqs. (8.12) and (8.13). For instance, to compute the DTFS for the periodic signal in Fig. 9.2a, we use the values of *xn* = *x*[*n*]/*N*0 as - -$$ -x_n = \begin{cases} \frac{1}{32} & 0 \le n \le 4 \quad \text{and} \quad 28 \le n \le 31\\ 0 & 5 \le n \le 27 \end{cases} -$$ - -Numerical computations in modern digital signal processing are conveniently performed with the discrete Fourier transform, introduced in Sec. 8.5. The DFT computations can be very efficiently executed by using the fast Fourier transform (FFT) algorithm discussed in Sec. 8.6. The DFT is indeed the workhorse of modern digital signal processing. The discrete-time Fourier transform (DTFT) and the inverse discrete-time Fourier transform (IDTFT) can be computed by using the DFT. For an *N*0-point signal *x*[*n*], its DFT yields exactly *N*0 samples of *X*() at frequency intervals of 2π/*N*0. We can obtain a larger number of samples of *X*() by padding a sufficient number of zero-valued samples to *x*[*n*]. The *N*0-point DFT of *x*[*n*] gives exact values of the DTFT samples if *x*[*n*] has a finite length *N*0. If the length of *x*[*n*] is infinite, we need to use the appropriate window function to truncate *x*[*n*]. - -Because of the convolution property, we can use the DFT to compute the convolution of two signals *x*[*n*] and *h*[*n*], as discussed in Sec. 8.5. This procedure, known as fast convolution, requires padding both signals by a suitable number of zeros, to make the linear convolution of the two signals identical to the circular (or periodic) convolution of the padded signals. Large blocks of data may be processed by sectioning the data into smaller blocks and processing such smaller blocks in sequence. Such a procedure requires smaller memory and reduces the processing time [1]. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/117_9.6 GENERALIZATION OF THE DTFT TO THE z-TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/117_9.6 GENERALIZATION OF THE DTFT TO THE z-TRANSFORM.md deleted file mode 100644 index 82cc6599ba00a65574a6d71852fb4ae36ea8cd61..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/117_9.6 GENERALIZATION OF THE DTFT TO THE z-TRANSFORM.md +++ /dev/null @@ -1,103 +0,0 @@ -## **[9.6 GENERALIZATION OF THE](#page-14-0) DTFT TO THE** *z***-TRANSFORM** - -LTID systems can be analyzed by using the DTFT. This method, however, has the following limitations. - -- 1. Existence of the DTFT is guaranteed only for absolutely summable signals. The DTFT does not exist for exponentially or even linearly growing signals. This means that the DTFT method is applicable only for a limited class of inputs. -- 2. Moreover, this method can be applied only to asymptotically or BIBO-stable systems; it cannot be used for unstable or even marginally stable systems. - -These are serious limitations in the study of LTID system analysis. Actually, it is the first limitation that is also the cause of the second limitation. Because the DTFT is incapable of handling growing signals, it is incapable of handling unstable or marginally stable systems.† Our goal is, therefore, to extend the DTFT concept so that it can handle exponentially growing signals. - -We may wonder what causes this limitation on DTFT so that it is incapable of handling exponentially growing signals. Recall that in the DTFT, we are using sinusoids or exponentials of the form *ejn* to synthesize an arbitrary signal *x*[*n*]. These signals are sinusoids with constant amplitudes. They are incapable of synthesizing exponentially growing signals no matter how many such components we add. Our hope, therefore, lies in trying to synthesize *x*[*n*] by using exponentially growing sinusoids or exponentials. This goal can be accomplished by generalizing the frequency variable *j* to σ + *j*, that is, by using exponentials of the form *e*(σ+*j*)*n* instead of exponentials *ejn*. The procedure is almost identical to that used in extending the Fourier transform to the Laplace transform. - -Let us define a new variable *X*ˆ(*j*) = *X*(). Hence, - -$$ -\hat{X}(j\Omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\Omega n} -$$ -\n(9.45) - -and - -$$ -x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} \hat{X}(j\Omega) e^{j\Omega n} d\Omega -$$ - -Consider now the DTFT of *x*[*n*] *e*−σ*n* (σ real): - -$$ -\text{DTFT}\left\{x[n]e^{-\sigma n}\right\} = \sum_{n=-\infty}^{\infty} x[n]e^{-\sigma n}e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x[n]e^{-(\sigma+j\Omega)n} -$$ - -It follows from Eq. (9.45) that this sum is *X*ˆ(σ +*j*). Thus, - -$$ -\text{DTFT}\left\{x[n]e^{-\sigma n}\right\} = \sum_{n=-\infty}^{\infty} x[n]e^{-(\sigma+j\Omega)n} = \hat{X}(\sigma+j\Omega) \tag{9.46} -$$ - -Hence, the inverse DTFT of *X*ˆ(σ +*j*) is *x*[*n*] *e*−σ*n*. Therefore, - -$$ -x[n]e^{-\sigma n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \hat{X}(\sigma + j\Omega) e^{j\Omega n} d\Omega -$$ - -Multiplying both sides by *e*σ*n* yields - -$$ -x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} \hat{X}(\sigma + j\Omega) e^{(\sigma + j\Omega)n} d\Omega -$$ -\n(9.47) - -Let us define a new variable *z* as - -$$ -z = e^{\sigma + j\Omega} -$$ - so that $\ln z = \sigma + j\Omega$ and $\frac{1}{z}dz = j d\Omega$ - -†Recall that the output of an unstable system grows exponentially. Also, the output of a marginally stable system to characteristic mode input grows with time. - -**Figure 9.18** Contour of integration for the *z*-transform. - -Because *z* = *e*σ+*j* is complex, we can express it as *z* = *rej*, where *r* = *e*σ . Thus, *z* lies on a circle of radius *r*, and as varies from −π to π, *z* circumambulates along this circle, completing exactly one counterclockwise rotation, as illustrated in Fig. 9.18. Changing to variable *z* in Eq. (9.47) yields - -$$ -x[n] = \frac{1}{2\pi j} \oint \hat{X}(\ln z) z^{n-1} dz -$$ - (9.48) - -and from Eq. (9.46) we obtain - -$$ -\hat{X}(\ln z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} -$$ -\n(9.49) - -where the integral 6 indicates a contour integral around a circle of radius *r* in the counterclockwise direction. - -Equations (9.48) and (9.49) are the desired extensions. They are, however, in a clumsy form. For the sake of convenience, we make another notational change by observing that *X*ˆ(ln*z*) is a function of *z*. Let us denote it by a simpler notation *X*[*z*]. Thus, Eq. (9.48) becomes - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ -\n(9.50) - -and Eq. (9.49) becomes - -$$ -X[z] = \sum_{n=-\infty}^{\infty} x[n]z^{-n} -$$ -\n(9.51) - -This is the (bilateral) *z*-transform pair. Equation (9.50) expresses *x*[*n*] as a continuous sum of exponentials of the form *zn* = *e*(σ+*j*)*n* = *rn ejn*. Thus, by selecting a proper value for *r* (or σ), we can make the exponential grow (or decay) at any exponential rate we desire. - -If we let σ = 0, we have *z* = *ej* and - -$$ -X[z]|_{z=e^{j\Omega}} = \hat{X}(\ln z)\Big|_{z=e^{j\Omega}} = \hat{X}(j\Omega) = X(\Omega) -$$ - -Thus, the familiar DTFT is just a special case of the *z*-transform *X*[*z*] obtained by letting *z* = *ej* and assuming that the sum on the right-hand side of Eq. (9.51) converges when *z* = *ej*. This also implies that the ROC for *X*[*z*] includes the unit circle. - -## **[9.7 MATLAB: WORKING WITH THE](#page-14-0) DTFS AND THE DTFT** - -This section investigates various methods to compute the discrete-time Fourier series (DTFS). Performance of these methods is assessed by using MATLAB's stopwatch and profiling functions. Additionally, the discrete-time Fourier transform (DTFT) is applied to the important topic of finite impulse response (FIR) filter design. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/118_9.7 MATLAB - WORKING WITH THE DTFS AND THE DTFT.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/118_9.7 MATLAB - WORKING WITH THE DTFS AND THE DTFT.md deleted file mode 100644 index fec76cdf31e3ee60f6ed2de143afd98e67bddcde..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/118_9.7 MATLAB - WORKING WITH THE DTFS AND THE DTFT.md +++ /dev/null @@ -1,267 +0,0 @@ -### **[9.7-1 Computing the Discrete-Time Fourier Series](#page-14-0)** - -Within a scale factor, the DTFS is identical to the DFT. Thus, methods to compute the DFT can be readily used to compute the DTFS. Specifically, the DTFS is the DFT scaled by 1/*N*0. As an example, consider a 50 Hz sinusoid sampled at 1000 Hz over one-tenth of a second. - -``` ->> T = 1/1000; N_0 = 100; n = (0:N_0-1)'; ->> x = cos(2*pi*50*n*T); -``` - -The DTFS is obtained by scaling the DFT. - -``` ->> X = fft(x)/N_0; f = (0:N_0-1)/(T*N_0); ->> stem(f-1/(2*T),fftshift(abs(X)),'k.'); ->> axis([-500 500 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); -``` - -Figure 9.19 shows a peak magnitude of 0.5 at ±50 Hz. This result is consistent with Euler's representation - -$$ -\cos(2\pi 50nT) = \frac{1}{2}e^{j2\pi 50nT} + \frac{1}{2}e^{-j2\pi 50nT} -$$ - -Lacking the 1/*N*0 scale factor, the DFT would have a peak amplitude 100 times larger. - -The inverse DTFS is obtained by scaling the inverse DFT by *N*0. - -``` ->> x = real(ifft(X)*N_0); stem(n,x,'k.'); ->> axis([0 99 -1.1 1.1]); xlabel('n'); ylabel('x[n]'); -``` - -Figure 9.20 confirms that the sinusoid *x*[*n*] is properly recovered. Although the result is theoretically real, computer round-off errors produce a small imaginary component, which the real command removes. - -**Figure 9.19** DTFS computed by scaling the DFT. - -**Figure 9.20** Inverse DTFS computed by scaling the inverse DFT. - -Although MATLAB's fft command provides an efficient method to compute the DTFS, other important computational methods exist. A matrix-based approach is one popular way to implement Eq. (9.4). Although not as efficient as an FFT-based algorithm, matrix-based approaches provide insight into the DTFS and serve as an excellent model for solving similarly structured problems. - -To begin, define *WN*0 = *ej*0 , which is a constant for a given *N*0. Substituting *WN*0 into Eq. (9.4) yields - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n=0}^{N_0 - 1} x[n] W_{N_0}^{-nr} -$$ - -An inner product of two vectors computes *Dr*. - -$$ -\mathcal{D}_r = \frac{1}{N_0} \begin{bmatrix} 1 & W_{N_0}^{-r} & W_{N_0}^{-2r} & \dots & W_{N_0}^{-(N_0-1)r} \end{bmatrix} \begin{bmatrix} x[0] \\ x[1] \\ x[2] \\ \vdots \\ x[N_0-1] \end{bmatrix} -$$ - -Stacking the results for all *r* yields - -$$ -\begin{bmatrix}\n\begin{bmatrix}\n\mathcal{D}_0 \\ -\mathcal{D}_1 \\ -\mathcal{D}_2 \\ -\vdots \\ -\mathcal{D}_{N_0-1}\n\end{bmatrix} = \frac{1}{N_0} \begin{bmatrix}\n1 & 1 & 1 & \cdots & 1 \\ -1 & W_{N_0}^{-1} & W_{N_0}^{-2} & \cdots & W_{N_0}^{-(N_0-1)} \\ -1 & W_{N_0}^{-2} & W_{N_0}^{-4} & \cdots & W_{N_0}^{-2(N_0-1)} \\ -\vdots & \vdots & \vdots & \cdots & \vdots \\ -1 & W_{N_0}^{-(N_0-1)} & W_{N_0}^{-2(N_0-1)} & \cdots & W_{N_0}^{-(N_0-1)^2}\n\end{bmatrix} \begin{bmatrix}\nx[0] \\ -x[1] \\ -x[2] \\ -\vdots \\ -x[N_0-1]\n\end{bmatrix} -$$ - -In matrix notation, this equation is compactly written as - -$$ -\mathbf{D} = \frac{1}{N_0} \mathbf{W}_{N_0} \mathbf{x} -$$ - -Since it is also used to compute the DFT, matrix **W***N*0 is often called a DFT matrix. - -Let us create an anonymous function to compute the *N*0-by-*N*0 DFT matrix **W***N*0 . Although not used here, the signal-processing toolbox function dftmtx computes the same DFT matrix, although in a less obvious but more efficient fashion. - ->> W = @(N\_0) (exp(-j\*2\*pi/N\_0)).^((0:N\_0-1)'\*(0:N\_0-1)); - -While less efficient than FFT-based methods, the matrix approach correctly computes the DTFS. - -``` ->> X = W(N_0)*x/N_0; stem(f-1/(2*T),fftshift(abs(X)),'k.'); ->> axis([-500 500 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); -``` - -The resulting plot is indistinguishable from Fig. 9.19. Problem 9.7-1 investigates a matrix-based approach to compute Eq. (9.3), the inverse DTFS. - -### **[9.7-2 Measuring Code Performance](#page-14-0)** - -Writing efficient code is important, particularly if the code is frequently used, requires complicated operations, involves large data sets, or operates in real time. MATLAB provides several tools for assessing code performance. When properly used, the profile function provides detailed statistics that help assess code performance. MATLAB help thoroughly describes the use of the sophisticated profile command. - -A simpler method of assessing code efficiency is to measure execution time and compare it with a reference. The MATLAB command tic starts a stopwatch timer. The toc command reads the timer. Sandwiching instructions between tic and toc returns the elapsed time. For example, the execution time of the 100-point matrix-based DTFS computation is - -``` ->> tic; W(N_0)*x/N_0; toc - Elapsed time is 0.004417 seconds. -``` - -Different machines operate at different speeds with different operating systems and with different background tasks. Therefore, elapsed-time measurements can vary considerably from machine to machine and from execution to execution. For relatively simple and short events like the present case, execution times can be so brief that MATLAB may report unreliable times or fail to register an elapsed time at all. - -To increase the elapsed time and therefore the accuracy of the time measurement, a loop is used to repeat the calculation. - -``` ->> tic; for i=1:100, W(N_0)*x/N_0; end; toc - Elapsed time is 0.173388 seconds. -``` - -This elapsed time suggests that each 100-point DTFS calculation takes a little under 2 milliseconds. What exactly does this mean, however? Elapsed time is only meaningful relative to some reference. Let us see what difference occurs by precomputing the DFT matrix, rather than repeatedly using our anonymous function. - -``` ->> W100 = W(100); tic; for i=1:100, W100*x/N_0; end; toc - Elapsed time is 0.001199 seconds. -``` - -### 892 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -Amazingly, this small change makes a hundredfold change in our computational efficiency! Clearly, it is much better to precompute the DFT matrix. - -To provide another example, consider the time it takes to compute the same DTFS using the FFT-based approach. - ->> tic; for i=1:100, fft(x)/N\_0; end; toc Elapsed time is 0.000399 seconds. - -With this as a reference, our fastest matrix-based computations appear to be several times slower than the FFT-based computations. This difference becomes more dramatic as *N*0 is increased. Since the two methods provide identical results, there is little incentive to use the slower matrix-based approach, and the FFT-based algorithm is generally preferred. Even so, the FFT can exhibit curious behavior: adding a few data points, even the artificial samples introduced by zero padding, can dramatically increase or decrease execution times. The tic and toc commands illustrate this strange result. Consider computing the DTFS of 1015 random data points 100 times. - ->> x1 = rand(1015,1); tic; for i=1:100; fft(x1)/1015; end; T1 = toc T1 = 0.0067 - -Next, pad the sequence with four zeros. - ->> -$$ -x2 = [x1; zeros(4,1)] -$$ -; tic; for i=1:100; fft( $x2$ )/1019; end; T2 = toc -T2 = 0.0134 - -The ratio of the two elapsed times indicates that adding four points to an already long sequence increases the computation time by a factor of 2. Next, the sequence is zero-padded to a length of *N*0 = 1024. - ->> x3 = [x2;zeros(5,1)]; tic; for i=1:100; fft(x3)/1024; end; T3 = toc T3 = 0.0017 - -In this case, the added data decrease the original execution time by a factor of 4 and the second execution time by a factor of 8! These results are particularly surprising when it is realized that the lengths of y1, y2, and y3 differ by less than 1%. - -As it turns out, the efficiency of the fft command depends on the factorability of *N*0. With the factor command, 1015 = (5)(7)(29), 1019 is prime, and 1024 = (2)10. The most factorable length, 1024, results in the fastest execution, while the least factorable length, 1019, results in the slowest execution. To ensure the greatest factorability and fastest operation, vector lengths are ideally a power of 2. - -### **[9.7-3 FIR Filter Design by Frequency Sampling](#page-14-0)** - -Finite impulse response (FIR) digital filters are flexible, always stable, and relatively easy to implement. These qualities make FIR filters a popular choice among digital filter designers. The difference equation of a length-*N* causal FIR filter is conveniently expressed as - -$$ -y[n] = h_0 x[n] + h_1 x[n-1] + \dots + h_{N-1} x[n-(N-1)] = \sum_{k=0}^{N-1} h_k x[n-k] -$$ - -The filter coefficients, or tap weights as they are sometimes called, are expressed by using the variable *h* to emphasize that the coefficients themselves represent the impulse response of the filter. - -The filter's frequency response is - -$$ -H(\Omega) = \frac{Y(\Omega)}{X(\Omega)} = \sum_{k=0}^{N-1} h_k e^{-j\Omega k} -$$ - -Since *H*() is a 2π-periodic function of the continuous variable , it is sufficient to specify *H*() over a single period (0 ≤ < 2π ). - -In many filtering applications, the desired magnitude response |*Hd*()| is known but not the filter coefficients *h*[*n*]. The question, then, is one of determining the filter coefficients from the desired magnitude response. - -Consider the design of a lowpass filter with cutoff frequency *c* = π/4. An anonymous function represents the desired ideal frequency response. - ->> H\_d = @(Omega) (mod(Omega,2\*pi)2\*pi-pi/4); - -Since the inverse DTFT of *Hd*() is a sampled sinc function, it is impossible to perfectly achieve the desired response with a causal, finite-length FIR filter. A realizable FIR filter is necessarily an approximation, and an infinite number of possible solutions exist. Thought of another way, *Hd*() specifies an infinite number of points, but the FIR filter only has *N* unknown tap weights. In general, we expect a length-*N* filter to match only *N* points of the desired response over (0 ≤ < 2π ). Which frequencies should be chosen? - -A simple and sensible method is to select *N* frequencies uniformly spaced on the interval (0 ≤ < 2π ), (0, 2π/*N*, 4π/*N*, 6π/*N*,...,(*N* −1)2π/*N*). By choosing uniformly spaced frequency samples, the *N*-point inverse DFT can be used to determine the tap weights *h*[*n*]. Program CH9MP1 illustrates this procedure. - -``` -function [h] = CH9MP1(N,H_d); -% CH9MP1.m : Chapter 9, MATLAB Program 1 -% Function M-file designs a length-N FIR filter by sampling the desired -% magnitude response H_d. Phase response is left as zero. -% INPUTS: N = desired FIR filter length -% H_d = anonymous function that defines the desired magnitude response -% OUTPUTS: h = impulse response (FIR filter coefficients) -% Create N equally spaced frequency samples: -Omega = linspace(0,2*pi*(1-1/N),N)'; -% Sample the desired magnitude response and create h[n]: -H = 1.0*H_d(Omega); h = real(ifft(H)); -``` - -To complete the design, the filter length must be specified. Small values of *N* reduce the filter's complexity but also reduce the quality of the filter's response. Large values of *N* improve the approximation of *Hd*() but also increase complexity. A balance is needed. We choose an intermediate value of *N* = 21 and use CH9MP1 to design the filter. - ->> N = 21; h = CH9MP1(N,H\_d); - -To assess the filter quality, the frequency response is computed by means of program CH5MP1. - -``` ->> Omega = linspace(0,2*pi,1000); samples = linspace(0,2*pi*(1-1/N),N)'; -``` - -``` ->> H = CH5MP1(h,1,Omega); -``` - -``` ->> subplot(2,1,1); stem([0:N-1],h,'k.'); xlabel('n'); ylabel('h[n]'); -``` - -``` ->> subplot(2,1,2); -``` - ->> plot(samples,H\_d(samples),'k.',Omega,H\_d(Omega),'k:',Omega,abs(H),'k'); - -``` ->> axis([0 2*pi -0.1 1.6]); xlabel('\Omega'); ylabel('|H(\Omega)|'); -``` - ->> legend('Samples','Desired','Actual','Location','North'); - -As shown in Fig. 9.21, the filter's frequency response intersects the desired response at the sampled values of *Hd*(). The overall response, however, has significant ripple between sample points that renders the filter practically useless. Increasing the filter length does not alleviate the ripple problems. Figure 9.22 shows the case *N* = 41. - -To understand the poor behavior of filters designed with CH9MP1, remember that the impulse response of an ideal lowpass filter is a sinc function with the peak centered at zero. Thought of another way, the peak of the sinc is centered at *n* = 0 because the phase of *Hd*() is zero. Constrained to be causal, the impulse response of the designed filter still has a peak at *n* = 0 but cannot include values for negative *n*. As a result, the sinc function is split in an unnatural way with sharp discontinuities on both ends of *h*[*n*]. Sharp discontinuities in the time domain appear as high-frequency oscillations in the frequency domain, which is why *H*() has significant ripple. - -To improve the filter behavior, the peak of the sinc is moved to *n* = (*N* − 1)/2, the center of the length-*N* filter response. In this way, the peak is not split, no large discontinuities are present, and frequency response ripple is consequently reduced. From DFT properties, a cyclic shift of (*N* − 1)/2 in the time domain requires a scale factor of *e*−*j*(*N*−1)/2 in the frequency - -**Figure 9.21** Length-21 FIR lowpass filter using zero phase. - -**Figure 9.22** Length-41 FIR lowpass filter using zero phase. - -domain.† Notice that the scale factor *e*−*j*(*N*−1)/2 affects only phase, not magnitude, and results in a linear phase filter. Program CH9MP2 implements the procedure. - -``` -function [h] = CH9MP2(N,H_d); -% CH9MP2.m : Chapter 9, MATLAB Program 2 -% Function M-file designs a length-N FIR filter by sampling the desired -% magnitude response H_d. Phase is defined to shift h[n] by (N-1)/2. -% INPUTS: N = desired FIR filter length -% H_d = anonymous function that defines the desired magnitude response -% OUTPUTS: h = impulse response (FIR filter coefficients) -% Create N equally spaced frequency samples and use to sample H_d: -Omega = linspace(0,2*pi*(1-1/N),N)'; H = H_d(Omega); -% Define phase to shift h[n] by (N-1)/2: -H = H.*exp(-j*Omega*((N-1)/2)); -H(fix(N/2)+2:N,1) = H(fix(N/2)+2:N,1)*((-1)^(N-1)); -h = real(ifft(H)); -``` - -† Technically, the shift property requires (*N* −1)/2 to be an integer, which occurs only for odd-length filters. The next-to-last line of program CH9MP2 implements a correction factor, of sorts, required to accommodate the fractional shifts desired for even-length filters. The mathematical derivation of this correction is nontrivial and is not included here. Those hesitant to use this correction factor have an alternative: simply round (*N* − 1)/2 to the nearest integer. Although the rounded shift is slightly off-center for even-length filters, there is usually little or no appreciable difference in the characteristics of the filter. Even so, true centering is desirable because the resulting impulse response is symmetric, which can reduce by half the number of multiplies required to implement the filter. - -**Figure 9.23** Length-21 FIR lowpass filter using linear phase. - -Figure 9.23 shows the results for the *N* = 21 case using CH9MP2 to compute *h*[*n*]. As hoped, the impulse response looks like a sinc function with the peak centered at *n* = 10. Additionally, the frequency response ripple is greatly reduced. With CH9MP2, increasing *N* improves the quality of the filter, as shown in Fig. 9.24 for the case *N* = 41. While the magnitude response is needed to establish the general shape of the filter response, it is the proper selection of phase that ensures the acceptability of the filter's behavior. - -To illustrate the flexibility of the design method, consider a bandpass filter with passband (π/4 < || < π/2). - -``` ->> H_d = @(Omega) (mod(Omega,2*pi)>pi/4)&(mod(Omega,2*pi)> (mod(Omega,2*pi)>3*pi/2)&(mod(Omega,2*pi)<7*pi/4); -``` - -Figure 9.25 shows the results for *N* = 50. Notice that this even-length filter uses a fractional shift and is symmetric about *n* = 24.5. - -Although FIR filter design by means of frequency sampling is very flexible, it is not always appropriate. Extreme care is needed for filters, such as digital differentiators and Hilbert transformers, that require special phase characteristics for proper operation. Additionally, if frequency samples occur near jump discontinuities of *Hd*(), rounding errors may, in rare cases, disrupt the desired symmetry of the sampled magnitude response. Such cases are corrected by slightly adjusting the location of problematic jump discontinuities or by changing the value of *N*. - -**Figure 9.24** Length-41 FIR lowpass filter using linear phase. - -**Figure 9.25** Length-50 FIR bandpass filter using linear phase. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/119_9.8 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/119_9.8 SUMMARY.md deleted file mode 100644 index 21936ae8fc1718a52e9ba95dd2a34326f9b23975..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/119_9.8 SUMMARY.md +++ /dev/null @@ -1,13 +0,0 @@ -## **[9.8 SUMMARY](#page-14-0)** - -This chapter deals with the analysis and processing of discrete-time signals. For analysis, our approach is parallel to that used in continuous-time signals. We first represent a periodic *x*[*n*] as a Fourier series formed by a discrete-time exponential and its harmonics. Later we extend this representation to an aperiodic signal *x*[*n*] by considering *x*[*n*] to be a limiting case of a periodic signal with period approaching infinity. - -Periodic signals are represented by discrete-time Fourier series (DTFS); aperiodic signals are represented by the discrete-time Fourier integral. The development, although similar to that of continuous-time signals, also reveals some significant differences. The basic difference in the two cases arises because a continuous-time exponential *ej*ω*t* has a unique waveform for every value of ω in the range −∞ to ∞. In contrast, a discrete-time exponential *ejn* has a unique waveform only for values of in a continuous interval of 2π. Therefore, if 0 is the fundamental frequency, then at most 2π/0 exponentials in the Fourier series are independent. Consequently, the discrete-time exponential Fourier series has only *N*0 = 2π/0 terms. - -The discrete-time Fourier transform (DTFT) of an aperiodic signal is a continuous function of and is periodic with period 2π. We can synthesize *x*[*n*] from spectral components of *X*() in any band of width 2π. In a basic sense, the DTFT has a finite spectral width of 2π, which makes it bandlimited to π radians. - -Linear, time-invariant, discrete-time (LTID) systems can be analyzed by means of the DTFT if the input signals are DTF-transformable and if the system is stable. Analysis of unstable (or marginally stable) systems and/or exponentially growing inputs can be handled by the *z*-transform, which is a generalized DTFT. The relationship of the DTFT to the *z*-transform is similar to that of the Fourier transform to the Laplace transform. Whereas the *z*-transform is superior to the DTFT for analysis of LTID systems, the DTFT is preferable in signal analysis. - -If *H*() is the DTFT of the system's impulse response *h*[*n*], then |*H*()| is the amplitude response, and *H*() is the phase response of the system. Moreover, if *X*() and *Y*() are the DTFTs of the input *x*[*n*] and the corresponding output *y*[*n*], then *Y*() = *H*()*X*(). Therefore, the output spectrum is the product of the input spectrum and the system's frequency response. - -Because of the similarity between the DFT and DTFT relationships, numerical computations of the DTFT of finite-length signals can be handled by using the DFT and the FFT, introduced in Secs. 8.5 and 8.6. For signals of infinite length, we use a window of suitable length to truncate the signal so that the final results are within a given error tolerance. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/120_REFERENCE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/120_REFERENCE.md deleted file mode 100644 index 0960f506a61c7b2a899dac4ddeef69411482996b..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/120_REFERENCE.md +++ /dev/null @@ -1,48 +0,0 @@ -### **[REFERENCE](#page-14-0)** - -1. Mitra, S. K. *Digital Signal Processing: A Computer-Based Approach,* 2nd ed. McGraw-Hill, New York, 2001. - -## **[PROBLEMS](#page-14-0)** - -**9.1-1** Find the discrete-time Fourier series (DTFS) and sketch their spectra |*Dr*| and *Dr* for 0≤*r*≤*N*0−1 for the following periodic signal: - -*x*[*n*] = 4 cos 2.4π*n*+2 sin 3.2π*n* - -- **9.1-2** Repeat Prob. 9.1-1 for *x*[*n*] =cos 2.2π*n*cos 3.3π*n*. -- **9.1-3** Repeat Prob. 9.1-1 for *x*[*n*]=2 cos 3.2π(*n*−3). -- **9.1-4** Determine and sketch the DTFS spectrum *Dr* of a 7-periodic signal *x*[*n*] that over 0 ≤ *n* ≤ 6 is given by - -$$ -[0, 1, -2, 3, -4, 5, -6] -$$ - -How does the spectrum *Dr* change if *x*[*n*] is time reversed? - -- **9.1-5** Find the discrete-time Fourier series and the corresponding amplitude and phase spectra for the *x*[*n*] shown in Fig. P9.1-5. -- **9.1-6** Repeat Prob. 9.1-5 for the *x*[*n*] depicted in Fig. P9.1-6. -- **9.1-7** Repeat Prob. 9.1-5 for the *x*[*n*] illustrated in Fig. P9.1-7. -- **9.1-8** An *N*0-periodic signal *x*[*n*] is represented by its DTFS, as in Eq. (9.3). Prove Parseval's theorem - -**Figure P9.1-5** - -**Figure P9.1-6** - -**Figure P9.1-7** - -2 - -(for the DTFS), which states that - -$$ -\frac{1}{N_0}\sum_{n=\langle N_0\rangle} |x[n]|^2 = \sum_{r=\langle N_0\rangle} |\mathcal{D}_r| -$$ - -In the text [Eq. (9.36)], we obtain Parseval's theorem for the DTFT. [*Hint:* If *w* is complex, then |*w*| 2 = *ww*∗ and use Eq. (8.15).] - -- **9.1-9** Answer yes or no, and justify your answers with an appropriate example or proof. - - (a) Is a sum of aperiodic discrete-time sequences ever periodic? - - (b) Is a sum of periodic discrete-time sequences ever aperiodic? -- **9.2-1** Show that for a real *x*[*n*], Eq. (9.18) can be expressed as - -$$ -x[n] = \frac{1}{\pi} \int_0^{\pi} |X(\Omega)| \cos(\Omega n + \angle X(\Omega)) d\Omega \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/121_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/121_PROBLEMS.md deleted file mode 100644 index 44461426ee95c400529199e22dd5d97fcf4e5d14..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/121_PROBLEMS.md +++ /dev/null @@ -1,381 +0,0 @@ -$$ - -This is the trigonometric form of the DTFT. - -**9.2-2** A signal *x*[*n*] can be expressed as the sum of even and odd components (Sec. 1.5-1): - -$$ -x[n] = x_e[n] + x_o[n] -$$ - -(a) If *x*[*n*] ⇐⇒ *X*(), show that for real *x*[*n*], - -$$ -x_e[n] \Longleftrightarrow \operatorname{Re}[X(\Omega)] -$$ - -and - -$$ -x_o[n] \Longleftrightarrow j \operatorname{Im}[X(\Omega)] -$$ - -- (b) Verify these results by finding the DTFT of the even and odd components of the signal (0.8)*nu*[*n*]. -- **9.2-3** For the following signals, find the DTFT directly, using the definition in Eq. (9.19). Assume |γ | < 1. - - (a) δ[*n*] - -(b) -$$ -\delta[n-k] -$$ - -- (c) γ *nu*[*n*−1] -- (d) γ *nu*[*n*+1] -- (e) (−γ )*nu*[*n*] -- (f) γ |*n*| -- **9.2-4** Use Eq. (9.18) to find the inverse DTFT for the following spectra, given only over the interval || ≤ π. Assume *c* and 0 < π. (a) *ejk* integer *k* - - (b) cos*k* integer *k* - - (c) cos2(/2) - -(d) -$$ -\Delta \left( \frac{1}{2\Omega_c} \right) -$$ - -- (e) 2πδ(−0) -- (f) π[δ(−0) +δ(+0)] -- **9.2-5** (a) Determine and plot the DTFT *X*() of the triangular signal *x*[*n*] shown in Fig. P9.2-5. - - (b) Using Ex. 9.5 as a guide, use MATLAB and the FFT to validate the DTFT calculations and plot of part (a). - -**Figure P9.2-5** - -- **9.2-6** Using Eq. (9.18), show that the inverse DTFT of rect((−π/4)/π ) is 0.5 sinc(π*n*/2) *ej*π*n*/4. -- **9.2-7** Using Eq. (9.19), find the DTFT of the signals *x*[*n*] in Fig. P9.2-7. -- **9.2-8** Using Eq. (9.19), find the DTFT of the signals depicted in Fig. P9.2-8. -- **9.2-9** Use Eq. (9.18) to find the inverse DTFT of the spectra (shown only for || ≤ π) in Fig. P9.2-9. - -**Figure P9.2-9** - -**9.2-10** Use Eq. (9.18) to find the inverse DTFT of the spectra (shown only for || ≤ π) in Fig. P9.2-10. - -**9.2-11** Find the DTFT for the signals shown in Fig. P9.2-11. - -**Figure P9.2-10** - -(c) - -(d) - -**Figure P9.2-11** - -- **9.2-12** Find the inverse DTFT of *X*() (shown only for ||≤π) for the spectra illustrated in Fig. P9.2-12. [*Hint: X*() = |*X*()|*ej X*(). This problem illustrates how different phase spectra (both with the same amplitude spectrum) represent entirely different signals.] -- **9.2-13** (a) Show that time-expanded signal *xe*[*n*] in Eq. (3.2) can also be expressed as - -$$ -x_e[n] = \sum_{k=-\infty}^{\infty} x[k]\delta[n - Lk] -$$ - -- (b) Find the DTFT of *xe*[*n*] by finding the DTFT of the right-hand side of the equation in part (a). -- (c) Use the result in part (b) and Table 9.1 to find the DTFT of *z*[*n*], shown in Fig. P9.2-13. -- **9.2-14** (a) A glance at Eq. (9.18) shows that the inverse DTFT equation is identical to the inverse (continuous-time) Fourier transform Eq. (7.10) for a signal *x*(*t*) bandlimited to π rad/s. Hence, we should be able to use the continuous-time Fourier transform - -Table 7.1 to find DTFT pairs that correspond to continuous-time transform pairs for bandlimited signals. Use this fact to derive DTFT pairs 8, 9, 11, 12, 13, and 14 in Table 9.1 by means of the appropriate pairs in Table 7.1. - -- (b) Can this method be used to derive pairs 2, 3, 4, 5, 6, 7, 10, 15, and 16 in Table 9.1? Justify your answer with specific reason(s). -- **9.2-15** Are the following frequency-domain signals valid DTFT's? Answer yes or no, and justify your answers. - - (a) *X*() = +π - - (b) *X*() = *j*+π - - (c) *X*() = sin(10) - - (d) *X*() = sin(/10) - - (e) *X*() = δ() -- **9.3-1** Using only pairs 2 and 5 (Table 9.1) and the time-shifting property of Eq. (9.31), find the DTFT of the following signals, assuming |*a*| < 1. - -- (a) *u*[*n*] −*u*[*n*−9] -- (b) *an*−*mu*[*n*−*m*] -- (c) *an*−3(*u*[*n*] −*u*[*n*−10]) - -1 - -(b) - -**Figure P9.2-13** - -- (d) *an*−*mu*[*n*] -- (e) *anu*[*n*−*m*] -- (f) (*n*−*m*)*an*−*mu*[*n*−*m*] -- (g) (*n*−*m*)*anu*[*n*] -- (h) *nan*−*mu*[*n*−*m*] -- **9.3-2** The triangular pulse *x*[*n*] shown in Fig. P9.3-2a is given by - -$$ -X(\Omega) = \frac{4e^{j6\Omega} - 5e^{j5\Omega} + e^{j\Omega}}{(e^{j\Omega} - 1)^2} -$$ - -Use this information and the DTFT properties to find the DTFT of the signals *x*1[*n*], *x*2[*n*], *x*3[*n*], and *x*4[*n*] shown in Figs. P9.3-2b, P9.3-2c, P9.3-2d, and P9.3-2e, respectively. - -**9.3-3** Suppose signal *x*[*n*] = sinc2(π*n*/2) modulates a carrier cos(c*n*) to produce signal *y*[*n*] = *x*[*n*] cos(c*n*). Find and sketch the DTFT of: - -(a) *x*[*n*] - -- (b) *y*[*n*] for c = π/2 -- (c) *y*[*n*] for c = 3π/4 -- (d) *y*[*n*] for c = π - -**9.3-4** Show that periodic convolution *X*()-∗ *Y*() = 2π*X*() if - -$$ -X(\Omega) = \sum_{k=0}^{4} a_k e^{-jk\Omega} -$$ - -and - -$$ -Y(\Omega) = \frac{\sin(5\Omega/2)}{\sin(\Omega/2)} e^{-j2\Omega} -$$ - -where *ak* is a set of arbitrary constants. - -- **9.3-5** Using only pair 2 (Table 9.1) and properties of DTFT, find the DTFT of the following signals, assuming |*a*| < 1 and 0 < π. - - (a) *an* cos0*nu*[*n*] - - (b) *n*2*anu*[*n*] - - (c) (*n*−*k*)*a*2*nu*[*n*−*m*] -- **9.3-6** Use pair 10 in Table 9.1, and suitable properties of the DTFT, to derive pairs 11, 12, 13, 14, 15, and 16. -- **9.3-7** Use the time-shifting property to show that - -$$ -x[n+k]+x[n-k] \Longleftrightarrow 2X(\Omega)\cos k\Omega -$$ - -(e) - -**Figure P9.3-2** - -**Figure P9.3-7** - -Use this result to find the DTFT of the signals shown in Fig. P9.3-7. - -**9.3-8** Use the time-shifting property to show that - -$$ -x[n+k] - x[n-k] \Longleftrightarrow 2jX(\Omega) \sin k\Omega -$$ - -Use this result to find the DTFT of the signal shown in Fig. P9.3-8. - -**9.3-9** Suppose signal *x*[*n*] has spectrum *X*() that is bandlimited to π/2 rad/sample. Next, define signal *y*[*n*] as - -$$ -y[n] = \begin{cases} x[n] & n \text{ even} \\ 0 & n \text{ odd} \end{cases} -$$ - -Determine the spectrum of *Y*() in terms of *X*(). Sketch *Y*() if, over −π ≤ ≤ π, - -$$ -Y(\Omega) = \begin{cases} |2\Omega/\pi| & -\pi/2 \le \Omega \le \pi/2\\ 0 & \text{otherwise} \end{cases} -$$ - -**9.3-10** Repeat Prob. 9.3-9 if *y*[*n*] is instead defined as - -$$ -y[n] = \begin{cases} x[n] & n \text{ odd} \\ 0 & n \text{ even} \end{cases} -$$ - -- **9.3-11** Using only pair 2 in Table 9.1 and the convolution property, find the inverse DTFT of *X*() = *e*2*j*/(*ej* −γ )2. -- **9.3-12** In Table 9.1, you are given pair 1. From this information and using suitable properties of the DTFT, derive pairs 2, 3, 4, 5, 6, and 7 of Table 9.1. For example, starting with pair 1, derive pair 2. From pair 2, use suitable properties - -of the DTFT to derive pair 3. From pairs 2 and 3, derive pair 4, and so on. - -- **9.3-13** From the pair *ej*(0/2)*n* ⇐⇒ 2πδ( − (0/2)) over the fundamental band, and the frequency-convolution property, find the DTFT of *ej*0*n*. Assume 0 <π/2. -- **9.3-14** From the definition and properties of the DTFT, show that - -(a) -$$ -\sum_{n=-\infty}^{\infty} \operatorname{sinc}(\Omega_c n) = \frac{\pi}{\Omega_c} \quad \Omega_c < \pi -$$ - -\n(b) -$$ -\sum_{n=-\infty}^{\infty} (-1)^n \operatorname{sinc}(\Omega_c n) = 0 \quad \Omega_c < \pi -$$ - -\n(c) -$$ -\sum_{n=-\infty}^{\infty} \operatorname{sinc}^2(\Omega_c n) = \frac{\pi}{\Omega_c} \quad \Omega_c < \pi/2 -$$ - -\n(d) -$$ -\sum_{n=-\infty}^{\infty} (-1)^n \operatorname{sinc}^2(\Omega_c n) = 0 \quad \Omega_c < \pi/2 -$$ - -\n(e) -$$ -\int_{-\pi}^{\pi} \frac{\sin(M\Omega/2)}{\sin(\Omega/2)} = 2\pi \quad \text{odd } M -$$ - -$$ -\int_{-\pi}^{\infty} \frac{\sin(\alpha z/2)}{\sin(\alpha z/2)} \sin(\alpha z/2) -$$ - -(f) -$$ -\sum_{n=-\infty}^{\infty} |\sin(\alpha z/2)|^4 = 2\pi/3\Omega_c \quad \Omega_c < \pi/2 -$$ - -**9.3-15** Show that the energy of signal *xc*(*t*) specified in Eq. (9.41) is identical to *T* times the energy of the discrete-time signal *x*[*n*], assuming *xc*(*t*) is bandlimited to *B* ≤ 1/2*T* Hz. [*Hint:* Recall that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc} [\pi(t-m)] \operatorname{sinc} [\pi(t-n)] dt -$$ -$$ -= \begin{cases} 0 & m \neq n \\ 1 & m = n \end{cases} -$$ - -That is, sinc functions are orthogonal.] - -**9.4-1** Use the DTFT method to find the zero-state response *y*[*n*] of a causal system with frequency response - -$$ -H(\Omega) = \frac{e^{i\Omega} + 0.32}{e^{i2\Omega} + e^{i\Omega} + 0.16} -$$ - -and the input *x*[*n*] = (−0.5)*nu*[*n*]. - -**9.4-2** Repeat Prob. 9.4-1 for - -$$ -H(\Omega) = \frac{e^{i\Omega} + 0.32}{e^{i2\Omega} + e^{i\Omega} + 0.16} -$$ - -and input *x*[*n*] = *u*[*n*]. - -**9.4-3** Repeat Prob. 9.4-1 for - -$$ -H(\Omega) = \frac{e^{i\Omega}}{e^{i\Omega} - 0.5} -$$ - -and - -$$ -x[n] = 0.8nu[n] + 2(2)nu[-(n+1)] -$$ - -**9.4-4** Determine and sketch the magnitude and phase response for an LTID system specified by the equation - -$$ -y[n] + 0.5y[n-1] = x[n] - 0.9x[n-1] -$$ - -Determine the system output *y*[*n*] for the input *x*[*n*] = cos( π*n* 3 +0.5). - -**9.4-5** Repeat Prob. 9.4-4 if the LTID system is instead specified by the equation - -$$ -y[n] - 0.5y[n-1] = x[n] + 0.9x[n-1] -$$ - -**9.4-6** An accumulator system has the property that an input *x*[*n*] results in the output - -$$ -y[n] = \sum_{k=-\infty}^{n} x[k] -$$ - -- (a) Find the unit impulse response *h*[*n*] and the frequency response *H*() for the accumulator. -- (b) Use the results of part (a) to find the DTFT of *u*[*n*]. -- **9.4-7** A noncausal 7-point moving average is described by the equation - -$$ -y[n] = \frac{1}{7} \sum_{k=-3}^{3} x[n-k] -$$ - -- (a) Find and sketch the magnitude and phase responses of the system. -- (b) How can this system be made causal? Plot the magnitude and phase responses of the causal system, and comment on any differences from part (a). -- **9.4-8** An LTID system frequency response over || ≤ π is - -$$ -H(\Omega) = \text{rect}\bigg(\frac{\Omega}{\pi}\bigg)e^{-j2\Omega} -$$ - -Find the output *y*[*n*] of this system, if the input *x*[*n*] is given by - -- (a) sinc (π*n*/2) -- (b) sinc(π*n*) -- (c) sinc2 (π*n*/4) -- **9.4-9** (a) If *x*[*n*] ⇐⇒ *X*(), then, show that (−1)*nx*[*n*] ⇐⇒ *X*(−π ). - - (b) Sketch γ *nu*[*n*] and (−γ )*nu*[*n*] for γ = 0.8; see the spectra for γ *nu*[*n*] in Figs. 9.5b and 9.5c. From these spectra, sketch the spectra for (−γ )*nu*[*n*]. - - (c) An ideal lowpass filter of cutoff frequency *c* is specified by the frequency response *H*() = rect(/2*c*). Find its impulse response *h*[*n*]. Find the frequency response of a filter whose impulse response is (−1)*nh*[*n*]. Sketch the frequency response of this filter. What kind of filter is this? -- **9.4-10** An analog differentiator *y*(*t*) = *d dt x*(*t*) can be approximated using a backward difference system described as - -$$ -y[n] = \frac{x[n] - x[n-1]}{T} -$$ - -Find and sketch the magnitude and phase responses of this DT system. For what frequencies does the system most behave as a differentiator? For what frequencies does the system least behave as a differentiator? - -**9.4-11** A filter with impulse response *h*[*n*] is modified as shown in Fig. P9.4-11. Determine the resulting filter impulse response *h*1[*n*]. Find also the resulting filter frequency response *H*1() in terms of the frequency response *H*(). How are *H*() and *H*1() related? - -- **9.4-12** (a) Consider an LTID system *S*1, specified by a difference equation of the form of Eqs. (3.15) or (3.16) or (3.20) in Ch. 3. We construct another system *S*2 by replacing coefficients *ai* (*i*=0, 1, 2,...,*N*) by coefficients (−1)*i ai* and replacing all coefficients *bi* (*i* = 0, 1, 2,...,*N*) with coefficients (−1)*i bi*. How are the frequency responses of the two systems related? - - (b) If *S*1 represents a lowpass filter, what kind of filter is specified by *S*2? - - (c) What type of filter (lowpass, highpass, etc.) is specified by the difference equation - -$$ -y[n] - 0.8y[n-1] = x[n] -$$ - -What kind of filter is specified by the following difference equation? - -$$ -y[n] + 0.8y[n-1] = x[n] -$$ - -**9.4-13** (a) The system shown in Fig. P9.4-13 contains two identical LTID filters with frequency response *H*0() and corresponding impulse response *h*0[*n*]. It is easy to see that the system is linear. Show that this system is also time-invariant. Do this by finding the **Figure P9.4-11** - -response of the system to input δ[*n* − *k*] in terms of *h*0[*n*]. - -- (b) If *H*0() = rect(/2*W*) over the fundamental band, and *c* + *W* ≤ π, find *H*(), the frequency response of this system. What kind of filter is this? -- **9.5-1** Determine the DTFT of *x*[*n*] = sin(0*n*) from the CTFT of *x*c(*t*) = sin(ω0*t*). -- **9.5-2** A CT signal *x*(*t*), bandlimited to 25 kHz, is sampled at 50 kHz to produce - -$$ -x[n] = \delta[n+4] - 2\delta[n+2] + \delta[n+1] - 3\delta[n] - \delta[n-1] - 2\delta[n-2] - \delta[n-4] -$$ - -Determine the CTFT *X*(ω). - -- **9.7-1** This problem uses a matrix-based approach to investigate the computation of the inverse DTFS. - - (a) Implement Eq. (9.3), the inverse DTFS, using a matrix-based approach. - - (b) Compare the execution speed of the matrix-based approach to the IFFT-based approach for input vectors of sizes 10, 100, and 1000. - -**Figure P9.4-13** - -- (c) What is the result of multiplying the DFT matrix **W***N*0 by the inverse DTFS matrix? Discuss your result. -- **9.7-2** A stable, first-order highpass IIR digital filter has transfer function - -$$ -H[z] = \left(\frac{1+\alpha}{2}\right) \left(\frac{1-z^{-1}}{1-\alpha z^{-1}}\right) -$$ - -- (a) Derive an expression relating α to the 3 dB cutoff frequency *c*. -- (b) Test your expression from part (a) in the following manner. First, compute α to achieve a 3 dB cutoff frequency of 1 kHz, assuming a sampling rate of *Fs* = 5 kHz. Determine a difference equation description of the system, and verify that the system is stable. Next, compute and plot the magnitude response of the resulting filter. Verify that the filter is highpass and has the correct cutoff frequency. -- (c) Holding α constant, what happens to the cutoff frequency *c* as *Fs* is increased to 50 kHz? What happens to the cutoff frequency *fc* as *Fs* is increased to 50 kHz? -- (d) Is there a well-behaved inverse filter to *H*[*z*]? Explain. -- (e) Determine α for *c* = π/2. Comment on the resulting filter, particularly *h*[*n*]. -- **9.7-3** Using the frequency sampling method, design a length-35 linear phase FIR highstop filter that has cutoff frequency c = 2π/3. Plot the resulting filter's impulse response *h*[*n*] and magnitude response |*H*()|. -- **9.7-4** Using the frequency-sampling method, design a length-71 linear phase FIR bandstop filter that has stopband (π/3 < || < π/2). Plot the resulting filter's impulse response *h*[*n*] and magnitude response |*H*()|. -- **9.7-5** Figure P9.7-5 provides the desired magnitude response |*H*()| of a real filter. Mathematically, - -$$ -|H(\Omega)| = \begin{cases} 2\frac{4\Omega}{\pi} & 0 \leq \Omega < \frac{\pi}{4} \\ 2 - \frac{4\Omega}{\pi} & \frac{\pi}{4} \leq \Omega < \frac{\pi}{2} \\ 0 & \frac{\pi}{2} \leq \Omega \leq \pi \end{cases} -$$ - -Since the digital filter is real, |*H*()|=|*H*(−)| and |*H*()|=|*H*(+2π )| for all . - -- (a) Can a realizable filter have this exact magnitude response? Explain your answer. -- (b) Use the frequency-sampling method to design an FIR filter with this magnitude response (or a reasonable approximation). Use MATLAB to plot the magnitude response of your filter. - -- **9.7-6** A real FIR comb filter is needed that has magnitude response |*H*()|=[0, 3, 0, 3, 0, 3, 0, 3] for = [0,π/4,π/2, 3π/4,π, 5π/4, 3π/2, 7π/4], respectively. Provide the impulse response *h*[*n*] of a filter that accomplishes these specifications. -- **9.7-7** A permutation matrix **P** has a single one in each row and column with the remaining elements all zero. Permutation matrices are useful for reordering the elements of a vector; the operation **Px** reorders the elements of a column vector **x** based on the form of **P**. - - (a) Fully describe an *N*0 × *N*0 permutation matrix named **R***N*0 that reverses the order of the elements of a column vector **x**. - - (b) Given DFT matrix **W***N*0 , verify that (**W***N*0 )(**W***N*0 ) = **W**2 *N*0 produces a scaled permutation matrix. How does **W**2 *N*0 **x** reorder the elements of **x**? - - (c) What is the result of (**W**2 *N*0 )(**W**2 *N*0 )**x** = **W**4 *N*0 **x**? diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/122_10 STATE-SPACE ANALYSIS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/122_10 STATE-SPACE ANALYSIS.md deleted file mode 100644 index 33802ed0b25817698f30ed821dc4e6fbf3ee6a16..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/122_10 STATE-SPACE ANALYSIS.md +++ /dev/null @@ -1,65 +0,0 @@ -# **[STATE-SPACE](#page-14-0) ANALYSIS** - -In Sec. 1.10, basic notions of *state variables* were introduced. In this chapter, we shall discuss state variables in more depth. - -Most of this book deals with an external (input–output) description of systems. As noted in Ch. 1, such a description may be inadequate in some cases, and we need a systematic way of finding a system's *internal description*. State-space analysis of systems meets this need. In this method, we first select a set of key variables, called the *state variables,* in the system. Every possible signal or variable in the system at any instant *t* can be expressed in terms of the state variables and the input(s) at that instant *t*. If we know all the state variables as a function of *t*, we can determine every possible signal or variable in the system at any instant with a relatively simple relationship. The system description in this method consists of two parts: - -- 1. A set of equations relating the state variables to the inputs (*the state equation*). -- 2. A set of equations relating outputs to the state variables and the inputs (*the output equation*). - -The analysis procedure, therefore, consists of solving the state equation first, and then solving the output equation. The state-space description is capable of determining every possible system variable (or output) from knowledge of the input and the initial state (conditions) of the system. For this reason, it is an *internal description* of the system. - -By its nature, state variable analysis is eminently suited for multiple-input, multiple-output (MIMO) systems. A single-input, single output (SISO) system is a special case of MIMO systems. In addition, the state-space techniques are useful for several other reasons, mentioned in Sec. 1.10, and repeated here. - -- 1. The state equations of a system provide a mathematical model of great generality that can describe not just linear systems, but also nonlinear systems; not just time-invariant systems, but also time-varying parameter systems; not just SISO systems, but also MIMO systems. Indeed, state equations are ideally suited for analysis, synthesis, and optimization of MIMO systems. -- 2. Compact matrix notation along with powerful techniques of linear algebra greatly facilitates complex manipulations. Without such features, many important results of - -modern system theory would have been difficult to obtain. State equations can yield a great deal of information about a system even when they are not solved explicitly. - -- 3. State equations lend themselves readily to digital computer simulation of complex systems of high order, with or without nonlinearities, and with multiple inputs and outputs. -- 4. For second-order systems (*N* = 2), a graphical method called *phase-plane analysis* can be used on state equations, whether they are linear or nonlinear. - -## **[10.1 MATHEMATICAL](#page-14-0) PRELIMINARIES** - -This chapter requires some understanding of matrix algebra. Section B.6 introduces basic concepts of matrix algebra, but misses a few needed mathematical concepts, which we present next. - -### **[10.1-1 Derivatives and Integrals of a Matrix](#page-14-0)** - -Elements of a matrix need not be constants; they may be functions of a variable. For example, if - -$$ -\mathbf{A} = \begin{bmatrix} e^{-2t} & \sin t \\ e^t & e^{-t} + e^{-2t} \end{bmatrix} -$$ - (10.1) - -then the matrix elements are functions of *t*. Here, it is helpful to denote **A** by **A**(*t*). Next, we define the derivative and integral of **A**(*t*). - -The derivative of a matrix **A**(*t*) (with respect to *t*) is defined as a matrix whose *ij*th element is the derivative (with respect to *t*) of the *ij*th element of the matrix **A**. Thus, if - -$$ -\mathbf{A}(t) = [a_{ij}(t)]_{m \times n} -$$ - -then - -$$ -\frac{d}{dt}[\mathbf{A}(t)] = \left[\frac{d}{dt}a_{ij}(t)\right]_{m \times n} \quad \text{or} \quad \dot{\mathbf{A}}(t) = [\dot{a}_{ij}(t)]_{m \times n} -$$ - -Thus, the derivative of the matrix in Eq. (10.1) is given by - -$$ -\dot{\mathbf{A}}(t) = \begin{bmatrix} -2e^{-2t} & \cos t \\ e^t & -e^{-t} - 2e^{-2t} \end{bmatrix} -$$ - -Similarly, we define the integral of **A**(*t*) (with respect to *t*) as a matrix whose *ij*th element is the integral (with respect to *t*) of the *ij*th element of the matrix **A**: - -$$ -\int \mathbf{A}(t) dt = \left( \int a_{ij}(t) dt \right)_{m \times n} -$$ - -Thus, for the matrix **A** in Eq. (10.1), we have - -$$ -\int \mathbf{A}(t) dt = \begin{bmatrix} \int e^{-2t} dt & \int \sin dt \\ \int e^t dt & \int (e^{-t} + 2e^{-2t}) dt \end{bmatrix} -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/123_10.1 MATHEMATICAL PRELIMINARIES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/123_10.1 MATHEMATICAL PRELIMINARIES.md deleted file mode 100644 index ebdd0a338dc2db9299761ead52be846bafda9d21..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/123_10.1 MATHEMATICAL PRELIMINARIES.md +++ /dev/null @@ -1,217 +0,0 @@ -#### 910 CHAPTER 10 STATE-SPACE ANALYSIS - -Since differentiation is a linear operation, it is easy to show that - -$$ -\frac{d}{dt}(\mathbf{A} + \mathbf{B}) = \frac{d\mathbf{A}}{dt} + \frac{d\mathbf{B}}{dt} \quad \text{and} \quad \frac{d}{dt}(c\mathbf{A}) = c\frac{d\mathbf{A}}{dt} -$$ - -The derivative of a matrix product is given as - -$$ -\frac{d}{dt}(\mathbf{AB}) = \frac{d\mathbf{A}}{dt}\mathbf{B} + \mathbf{A}\frac{d\mathbf{B}}{dt} = \dot{\mathbf{A}}\mathbf{B} + \mathbf{A}\dot{\mathbf{B}}\tag{10.2} -$$ - -We can prove Eq. (10.2) as follows. Let **A** be an *m*×*n* matrix and **B** an *n*×*p* matrix. Then, if - -**C** = **AB** - -from Eq. (B.33), we have - -$$ -c_{ik} = \sum_{j=1}^{n} a_{ij} b_{jk} -$$ - -and - -$$ -\dot{c}_{ik} = \underbrace{\sum_{j=1}^{n} \dot{a}_{ij} b_{jk}}_{d_{ik}} + \underbrace{\sum_{j=1}^{n} a_{ij} \dot{b}_{jk}}_{e_{ik}} -$$ - or -$$ -\dot{c}_{ik} = d_{ik} + e_{ik} -$$ - (10.3) - -Equation (10.3) along with the multiplication rule clearly indicates that *dik* is the *ik*th element of matrix **AB**˙ and *eik* is the *ik*th element of matrix **AB**˙ . Equation (10.2) then follows. - -If we let **B** = **A**−1 in Eq. (10.2), we obtain - -$$ -\frac{d}{dt}(\mathbf{A}\mathbf{A}^{-1}) = \frac{d\mathbf{A}}{dt}\mathbf{A}^{-1} + \mathbf{A}\frac{d}{dt}\mathbf{A}^{-1} -$$ - -But since - -$$ -\frac{d}{dt}(\mathbf{A}\mathbf{A}^{-1}) = \frac{d}{dt}\mathbf{I} = 0 -$$ - -we have - -$$ -\frac{d}{dt}(\mathbf{A}^{-1}) = -\mathbf{A}^{-1}\frac{d\mathbf{A}}{dt}\mathbf{A}^{-1} -$$ - -### **[10.1-2 The Characteristic Equation of a Matrix:](#page-14-0) The Cayley–Hamilton Theorem** - -For an (*n*×*n*) square matrix **A**, any vector **x** (**x** = 0) that satisfies the equation - -$$ -\mathbf{A}\mathbf{x} = \lambda \mathbf{x} \tag{10.4} -$$ - -is an *eigenvector* (or *characteristic vector*), and λ is the corresponding *eigenvalue* (or *characteristic value*) of **A**. Equation (10.4) can be expressed as - -$$ -(\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = 0 \qquad \text{or} \qquad (\lambda \mathbf{I} - \mathbf{A})\mathbf{x} = 0 -$$ - -The solution for this set of homogeneous equations exists if and only if - -$$ -|\lambda \mathbf{I} - \mathbf{A}| = \begin{vmatrix} \lambda - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \lambda - a_{22} & \cdots & -a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & \lambda - a_{nn} \end{vmatrix} = 0 -$$ - (10.5) - -Equation (10.5) is known as the *characteristic equation* of matrix **A** and can be expressed as - -$$ -Q(\lambda) = |\lambda \mathbf{I} - \mathbf{A}| = \lambda^n + a_{n-1}\lambda^{n-1} + \dots + a_1\lambda + a_0\lambda^0 = 0 -$$ - (10.6) - -*Q*(λ) is called the *characteristic polynomial* of matrix **A**. The *n* zeros of the characteristic polynomial are the eigenvalues of **A** and, corresponding to each eigenvalue, there is an eigenvector that satisfies Eq. (10.4). - -The *Cayley–Hamilton theorem* states that every *n*×*n* matrix **A** satisfies its own characteristic equation. In other words, Eq. (10.6) is valid if λ is replaced by **A**: - -$$ -\mathbf{Q}(\mathbf{A}) = \mathbf{A}^n + a_{n-1}\mathbf{A}^{n-1} + \dots + a_1\mathbf{A} + a_0\mathbf{A}^0 = 0 -$$ - (10.7) - -### FUNCTIONS OF A MATRIX - -We now demonstrate the use of the Cayley–Hamilton theorem [Eq. (10.7)] to evaluate functions of an *n*×*n* square matrix **A**. - -Consider a function *f*(λ) in the form of an infinite power series: - -$$ -f(\lambda) = \alpha_0 + \alpha_1 \lambda + \alpha_2 \lambda_2^2 + \dots = \sum_{i=0}^{\infty} \alpha_i \lambda^i -$$ - (10.8) - -Since λ, being an eigenvalue (characteristic root) of **A**, satisfies the characteristic equation [Eq. (10.6)], we can write - -$$ -\lambda^{n} = -a_{n-1}\lambda^{n-1} - a_{n-2}\lambda^{n-2} - \dots - a_1\lambda - a_0 -$$ - (10.9) - -If we multiply both sides by λ, the left-hand side is λ*n*+1, and the right-hand side contains the terms λ*n*, λ*n*−1, ... , λ. Using Eq. (10.9), we substitute λ*n* in terms of λ*n*−1, λ*n*−2,..., λ so that the highest power on the right-hand side is reduced to *n* − 1. Continuing in this way, we see that λ*n*+*k* can be expressed in terms of λ*n*−1, λ*n*−2,...,λ for any *k*. Hence, the infinite series on the right-hand side of Eq. (10.8) can always be expressed in terms of λ*n*−1, λ*n*−2,...,λ and a constant as - -$$ -f(\lambda) = \beta_0 + \beta_1 \lambda + \beta_2 \lambda^2 + \dots + \beta_{n-1} \lambda^{n-1} -$$ - (10.10) - -If we assume that there are *n* distinct eigenvalues λ1, λ2, ... , λ*n*, then Eq. (10.10) holds for these *n* values of λ. The substitution of these values in Eq. (10.10) yields *n* simultaneous equations - -$$ -\begin{bmatrix} f(\lambda_1) \\ f(\lambda_2) \\ \vdots \\ f(\lambda_n) \end{bmatrix} = \begin{bmatrix} 1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1} \\ 1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{n-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & \lambda_n & \lambda_n^2 & \cdots & \lambda_n^{n-1} \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_{n-1} \end{bmatrix} -$$ - -### 912 CHAPTER 10 STATE-SPACE ANALYSIS - -Solving for the β coefficients yields - -$$ -\begin{bmatrix}\n\beta_0 \\ -\beta_1 \\ -\vdots \\ -\beta_{n-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1} \\ -1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{n-1} \\ -\vdots & \vdots & \vdots & \cdots & \vdots \\ -1 & \lambda_n & \lambda_n^2 & \cdots & \lambda_n^{n-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\nf(\lambda_1) \\ -f(\lambda_2) \\ -\vdots \\ -f(\lambda_n)\n\end{bmatrix} -$$ -\n(10.11) - -Since **A** also satisfies Eq. (10.9), we may advance a similar argument to show that if *f*(**A**) is a function of a square matrix **A** expressed as an infinite power series in **A**, then - -$$ -f(\mathbf{A}) = \alpha_0 \mathbf{I} + \alpha_1 \mathbf{A} + \alpha_2 \mathbf{A}^2 + \cdots = \sum_{i=0}^{\infty} \alpha_i \mathbf{A}^i -$$ - -and, as argued earlier, the right-hand side can be expressed by using terms of power less than or equal to *n*−1, - -$$ -f(\mathbf{A}) = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{n-1} \mathbf{A}^{n-1} = \sum_{i=0}^{n-1} \beta_i \mathbf{A}^i -$$ - (10.12) - -in which the coefficients β*i*s are found from Eq. (10.11). If some of the eigenvalues are repeated (multiple roots), the results are somewhat modified. - -We shall demonstrate the utility of this result with the following two examples. - -### **[10.1-3 Computation of an Exponential and a Power of a Matrix](#page-14-0)** - -Let us compute *e***A***t* defined by - -$$ -e^{\mathbf{A}t} = \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2 t^2}{2!} + \cdots + \frac{\mathbf{A}^n t^n}{n!} + \cdots = \sum_{k=0}^{\infty} \frac{\mathbf{A}^k t^k}{k!} -$$ - -From Eq. (10.12), we can express - -$$ -e^{\mathbf{A}t} = \sum_{i=1}^{n-1} \beta_i(\mathbf{A})^i -$$ - -in which the β*i*s are given by Eq. (10.11), with *f*(λ*i*) = *e*λ*it* . - -### **EXAMPLE 10.1 Computing the Exponential of a Matrix** - -Compute *e***A***t* for the case - -$$ -\mathbf{A} = \left[ \begin{array}{cc} 0 & 1 \\ -2 & -3 \end{array} \right] -$$ - -The characteristic equation is - -$$ -|\lambda \mathbf{I} - \mathbf{A}| = \begin{vmatrix} \lambda & -1 \\ 2 & \lambda + 3 \end{vmatrix} = \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0 -$$ - -Hence, the eigenvalues are λ1 = −1, λ2 = −2, and - -$$ -e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} -$$ - -in which - -$$ -\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & -2 \end{bmatrix}^{-1} \begin{bmatrix} e^{-t} \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^{-t} \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} 2e^{-t} - e^{-2t} \\ e^{-t} - e^{-2t} \end{bmatrix} -$$ - -and - -$$ -e^{\mathbf{A}t} = (2e^{-t} - e^{-2t}) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + (e^{-t} - e^{-2t}) \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} -$$ -$$ -= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} -$$ -(10.13) - -COMPUTATION OF *Ak* As Eq. (10.12) indicates, we can express **A***k* as - -$$ -\mathbf{A}^{k} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \cdots + \beta_{n-1} \mathbf{A}^{n-1} -$$ - -in which the β*i*s are given by Eq. (10.11) with *f*(λ*i*) = λ*k i* . For a completed example of the computation of **A***k* by this method, see Ex. 10.13. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/124_10.2 INTRODUCTION TO STATE SPACE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/124_10.2 INTRODUCTION TO STATE SPACE.md deleted file mode 100644 index e0b034682019a465f8543a25083e41e4240101ad..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/124_10.2 INTRODUCTION TO STATE SPACE.md +++ /dev/null @@ -1,179 +0,0 @@ -## **[10.2 INTRODUCTION TO](#page-14-0) STATE SPACE** - -From the discussion in Ch. 1, we know that to determine a system's response(s) at any instant *t*, we need to know the system's inputs during its entire past, from −∞ to *t*. If the inputs are known only for *t* > *t*0, we can still determine the system output(s) for any *t* > *t*0, provided we know certain initial conditions in the system at *t* = *t*0. These initial conditions collectively are called the *initial state* of the system (at *t* = *t*0). - -*The state variables q*1(*t*),*q*2(*t*),...,*qN*(*t*) are the minimum number of system variables such that their initial values at any instant *t*0 are sufficient to determine the behavior of the system for all time *t* ≥ *t*0 when the input(s) to the system is known for *t* ≥ *t*0. This statement implies that an output of a system at any instant is determined completely from a knowledge of the values of the system state and the input at that instant. - -Initial conditions of a system can be specified in many different ways. Consequently, the system state can also be specified in many different ways. This means that state variables are not unique. - -#### 914 CHAPTER 10 STATE-SPACE ANALYSIS - -This discussion is also valid for multiple-input, multiple-output (MIMO) systems, where every possible system output at any instant *t* is determined completely from a knowledge of the system state and the input(s) at the instant *t*. These ideas should become clear from the following example of an *RLC* circuit. - -## **EXAMPLE 10.2 State-Space Description and Output Equations of an** *RLC* **Circuit** - -Find a state-space description of the *RLC* circuit shown in Fig. 10.1. Verify that all possible system outputs at some instant *t* can be determined from knowledge of the system state and the input at that instant *t*. - -**Figure 10.1** Circuit for Ex. 10.2. - -It is known that inductor currents and capacitor voltages in an *RLC* circuit can be used as one possible choice of state variables. For this reason, we shall choose *q*1 (the capacitor voltage) and *q*2 (the inductor current) as our state variables. - -The node equation at the intermediate node is - -$$ -i_3 = i_1 - i_2 - q_2 -$$ - -but *i*3 = 0.2*q*˙1, *i*1 = 2(*x* −*q*1), *i*2 = 3*q*1. Hence, - -$$ -0.2\dot{q}_1 = 2(x - q_1) - 3q_1 - q_2 -$$ - -or - -$$ -\dot{q}_1 = -25q_1 - 5q_2 + 10x -$$ - -This is the first state equation. To obtain the second state equation, we sum the voltages in the extreme right loop formed by *C*, *L*, and the 2 resistor so that they are equal to zero: - -$$ --q_1 + \dot{q}_2 + 2q_2 = 0 -$$ - -$$ -\quad \text{or} \quad -$$ - -Thus, the two state equations are - -$$ -\dot{q}_1 = -25q_1 - 5q_2 + 10x -$$ - -$$ -\dot{q}_2 = q_1 - 2q_2 -$$ - -*q*˙2 = *q*1 −2*q*2 - -Every possible output can now be expressed as a linear combination of *q*1, *q*2, and *x*. From Fig. 10.1, we have - -$$ -v_1 = x - q_1 -$$ - -\n -$$ -i_1 = 2(x - q_1) -$$ - -\n -$$ -v_2 = q_1 -$$ - -\n -$$ -i_2 = 3q_1 -$$ - -\n -$$ -i_3 = i_1 - i_2 - q_2 = 2(x - q_1) - 3q_1 - q_2 = -5q_1 - q_2 + 2x -$$ - -\n -$$ -i_4 = q_2 -$$ - -\n -$$ -v_4 = 2i_4 = 2q_2 -$$ - -\n -$$ -v_3 = q_1 - v_4 = q_1 - 2q_2 -$$ - -This set of equations is known as the *output equation* of the system. It is clear from this set that every possible output at some instant *t* can be determined from knowledge of *q*1(*t*), *q*2(*t*), and *x*(*t*), the system state, and the input at the instant *t*. Once we have solved the state equations to obtain *q*1(*t*) and *q*2(*t*), we can determine every possible output for any given input *x*(*t*). - -For continuous-time systems, the state equations are *N* simultaneous first-order differential equations in *N* state variables *q*1, *q*2, ... , *qN* of the form - -$$ -\dot{q}_i = g_i(q_1, q_2, \dots, q_N, x_1, x_2, \dots, x_j) -$$ - $i = 1, 2, \dots, N$ - -where *x*1, *x*2, ... , *xj* are the *j* system inputs. For a linear system, these equations reduce to a simpler linear form - -$$ -\dot{q}_i = a_{i1}q_1 + a_{i2}q_2 + \dots + a_{iN}q_N + b_{i1}x_1 + b_{i2}x_2 + \dots + b_{ij}x_j \qquad i = 1, 2, \dots, N \qquad (10.14) -$$ - -If there are *k* outputs *y*1, *y*2,..., *yk*, the *k* output equations are of the form - -$$ -y_m = c_{m1}q_1 + c_{m2}q_2 + \dots + c_{mN}q_N + d_{m1}x_1 + d_{m2}x_2 + \dots + d_{mj}x_j \qquad m = 1, 2, \dots, k \quad (10.15) -$$ - -The *N* simultaneous first-order state equations are also known as the *normal-form* equations. - -These equations can be written more conveniently in matrix form: - -$$ -\begin{bmatrix}\n\dot{q}_1 \\ -\dot{q}_2 \\ -\vdots \\ -\dot{q}_N\n\end{bmatrix} = \begin{bmatrix}\na_{11} & a_{12} & \cdots & a_{1N} \\ -a_{21} & a_{22} & \cdots & a_{2N} \\ -\vdots & \vdots & \cdots & \vdots \\ -a_{N1} & a_{N2} & \cdots & a_{NN}\n\end{bmatrix} \begin{bmatrix}\nq_1 \\ -q_2 \\ -\vdots \\ -q_N\n\end{bmatrix} + \begin{bmatrix}\nb_{11} & b_{12} & \cdots & b_{1j} \\ -b_{21} & b_{22} & \cdots & b_{2j} \\ -\vdots & \vdots & \ddots & \vdots \\ -b_{N1} & b_{N2} & \cdots & b_{Nj}\n\end{bmatrix} \begin{bmatrix}\nx_1 \\ -x_2 \\ -\vdots \\ -x_j\n\end{bmatrix} -$$ -\n -$$ -\begin{bmatrix}\ny_1 \\ -y_2 \\ -\vdots \\ -y_k\n\end{bmatrix} = \begin{bmatrix}\nc_{11} & c_{12} & \cdots & c_{1N} \\ -c_{21} & c_{22} & \cdots & c_{2N} \\ -\vdots & \vdots & \ddots & \vdots \\ -c_{k1} & c_{k2} & \cdots & c_{kN}\n\end{bmatrix} \begin{bmatrix}\nq_1 \\ -q_2 \\ -\vdots \\ -q_N\n\end{bmatrix} + \begin{bmatrix}\nd_{11} & d_{12} & \cdots & d_{1j} \\ -d_{21} & d_{22} & \cdots & d_{2j} \\ -\vdots & \vdots & \cdots & \vdots \\ -d_{k1} & d_{k2} & \cdots & d_{kj}\n\end{bmatrix} \begin{bmatrix}\nx_1 \\ -x_2 \\ -\vdots \\ -x_j\n\end{bmatrix} -$$ - -or - -and - -**q**˙ = **Aq**+**Bx** (10.16) - -and - -$$ -y = Cq + Dx \tag{10.17} -$$ - -Equation (10.16) is the state equation and Eq. (10.17) is the output equation; **q, y,** and **x** are the state vector, the output vector, and the input vector, respectively. - -For discrete-time systems, the state equations are *N* simultaneous first-order difference equations. Discrete-time systems are discussed in Sec. 10.7. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/125_10.3 A SYSTEMATIC PROCEDURE TO DETERMINE STATE EQUATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/125_10.3 A SYSTEMATIC PROCEDURE TO DETERMINE STATE EQUATIONS.md deleted file mode 100644 index 0176e4dd868ae22b5bf5e0f537e0ea10ceedcfdc..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/125_10.3 A SYSTEMATIC PROCEDURE TO DETERMINE STATE EQUATIONS.md +++ /dev/null @@ -1,382 +0,0 @@ -## **[10.3 A SYSTEMATIC](#page-14-0) PROCEDURE TO DETERMINE STATE EQUATIONS** - -We shall discuss here a systematic procedure to determine the state-space description of linear time-invariant systems. In particular, we shall consider systems of two types: (1) *RLC* networks and (2) systems specified by block diagrams or *N*th-order transfer functions. - -### **[10.3-1 Electrical Circuits](#page-14-0)** - -The method used in Ex. 10.2 proves effective in most of the simple cases. The steps are as follows: - -- 1. Choose all independent capacitor voltages and inductor currents to be the state variables. -- 2. Choose a set of loop currents; express the state variables and their first derivatives in terms of these loop currents. -- 3. Write loop equations, and eliminate all variables other than state variables (and their first derivatives) from the equations derived in steps 2 and 3. - -### **EXAMPLE 10.3 State Equations of an** *RLC* **Circuit** - -Write the state equations for the network shown in Fig. 10.2. - -**Figure 10.2** Circuit for Ex. 10.3. - -**Step 1.** There is one inductor and one capacitor in the network. Therefore, we shall choose the inductor current *q*1 and the capacitor voltage *q*2 as the state variables. - -**Step 2.** The relationship between the loop currents and the state variables can be written by inspection: - -1 - -$$ -q_1 = i_2 \tag{10.18} -$$ - -$$ -\frac{1}{2}\dot{q}_2 = i_2 - i_3\tag{10.19} -$$ - -**Step 3.** The loop equations are - -$$ -4i_1 - 2i_2 = x \tag{10.20} -$$ - -$$ -2(i_2 - i_1) + \dot{q}_1 + q_2 = 0 \tag{10.21} -$$ - -$$ --q_2 + 3i_3 = 0 \tag{10.22} -$$ - -Now we eliminate *i*1, *i*2, and *i*3 from the state and loop equations as follows. From Eq. (10.21), we have - -$$ -\dot{q}_1 = 2(i_1 - i_2) - q_2 -$$ - -We can eliminate *i*1 and *i*2 from this equation by using Eqs. (10.18) and (10.20) to obtain - -$$ -\dot{q}_1 = -q_1 - q_2 + \frac{1}{2}x -$$ - -The substitution of Eqs. (10.18) and (10.22) in Eq. (10.19) yields - -$$ -\dot{q}_2 = 2q_1 - \frac{2}{3}q_2 -$$ - -These are the desired state equations. We can express them in matrix form as - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 2 & -\frac{2}{3} \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix} x -$$ -\n(10.23) - -The derivation of state equations from loop equations is facilitated considerably by choosing loops in such a way that only one loop current passes through each of the inductors or capacitors. - -### AN ALTERNATIVE PROCEDURE - -We can also determine the state equations by the following procedure. - -- 1. Choose all independent capacitor voltages and inductor currents to be the state variables. -- 2. Replace each capacitor by a voltage source equal to the capacitor voltage, and replace each inductor by a current source equal to the inductor current. This step will transform the *RLC* network into a network consisting only of resistors, current sources, and voltage sources. -- 3. Find the current through each capacitor and equate it to *Cq*˙*i*, where *qi* is the capacitor voltage. Similarly, find the voltage across each inductor and equate it to *Lq*˙*j*, where *qj* is the inductor current. - -### **EXAMPLE 10.4 Alternate Procedure to Determine State Equations** - -Use the three-step alternative procedure just outlined to write the state equations for the network in Fig. 10.2. - -In the network in Fig. 10.2, we replace the inductor by a current source of current *q*1 and the capacitor by a voltage source of voltage *q*2, as shown in Fig. 10.3. The resulting network consists of four resistors, two voltage sources, and one current source. - -**Figure 10.3** Equivalent circuit of the network in Fig. 10.2. - -We can determine the voltage *vL* across the inductor and the current *ic* through the capacitor by using the principle of superposition. This step can be accomplished by inspection. For example, *vL* has three components arising from three sources. To compute the component due to *x*, we assume that *q*1 =0 (open circuit) and *q*2 =0 (short circuit). Under these conditions, the entire network to the right of the 2 resistor is opened, and the component of *vL* due to *x* is the voltage across the 2 resistor. This voltage is clearly (1/2)*x*. Similarly, to find the component of *vL* due to *q*1, we short *x* and *q*2. The source *q*1 sees an equivalent resistor of 1 across it, and hence *vL* = −*q*1. Continuing the process, we find that the component of *vL* due to *q*2 is −*q*2. Hence, - -$$ -v_L = \dot{q}_1 = \frac{1}{2}x - q_1 - q_2 -$$ - -Using the same procedure, we find - -$$ -i_c = \frac{1}{2}\dot{q}_2 = q_1 - \frac{1}{3}q_2 -$$ - -These equations are identical to the state equations [Eq. (10.23)] obtained earlier.† - -### **[10.3-2 State Equations from a Transfer Function](#page-14-0)** - -It is relatively easy to determine the state equations of a system specified by its transfer function.‡ Consider, for example, a first-order system with the transfer function - -$$ -H(s) = \frac{1}{s+a} -$$ - -The system realization appears in Fig. 10.4. The integrator output *q* serves as a natural state variable since, in practical realization, initial conditions are placed on the integrator output. The - -† This procedure requires modification if the system contains all-capacitor and voltage-source tie sets or all-inductor and current-source cut sets. In the case of all-capacitor and voltage-source tie sets, all capacitor voltages cannot be independent. One capacitor voltage can be expressed in terms of the remaining capacitor voltages and the voltage source(s) in that tie set. Consequently, one of the capacitor voltages should not be used as a state variable, and that capacitor should not be replaced by a voltage source. Similarly, in all-inductor and current-source tie sets, one inductor should not be replaced by a current source. If there are all-capacitor tie sets or all-inductor cut sets only, no further complications occur. In all-capacitor voltage-source tie sets and/or all-inductor current-source cut sets, we have additional difficulties in that the terms involving derivatives of the input may occur. This problem can be solved by redefining the state variables. The final state variables will not be capacitor voltages and inductor currents. - -‡ We implicitly assume that the system is controllable and observable. This implies that there are no pole-zero cancellations in the transfer function. If such cancellations are present, the state variable description represents only the part of the system that is controllable and observable (the part of the system that is coupled to the input and the output). In other words, the internal description represented by the state equations is no better than the external description represented by the input–output equation. - -integrator input is naturally *q*˙. From Fig. 10.4, we have - -*q*˙ = −*aq*+*x* and *y* = *q* - -In Sec. 4.6 we saw that a given transfer function can be realized in several ways. Consequently, we should be able to obtain different state-space descriptions of the same system by using different realizations. This assertion will be clarified by the following example. - -### **EXAMPLE 10.5 State-Space Description from a Transfer Function** - -Consider a system specified by the transfer function - -$$ -H(s) = \underbrace{\frac{2s+10}{s^3+8s^2+19s+12}}_{\text{direct form}} = \underbrace{\left(\frac{2}{s+1}\right)\left(\frac{s+5}{s+3}\right)\left(\frac{1}{s+4}\right)}_{\text{cascade}} = \underbrace{\frac{\frac{4}{3}}{s+1} - \frac{2}{s+3} + \frac{\frac{2}{3}}{s+4}}_{\text{parallel}} -$$ - -The procedure developed in Sec. 4.6 allows us to realize *H*(*s*) as, among others, direct form II (DFII), transpose DFII (TDFII), cascade, and parallel. These realizations are depicted in Fig. 10.5. Determine state-space descriptions for each of these realizations. As mentioned earlier, the output of each integrator serves as a natural state variable. - -### **Direct Form II and Its Transpose** - -Here we shall realize the system using the canonical form (direct form II and its transpose) discussed in Sec. 4.6. If we choose the state variables to be the three integrator outputs *q*1, *q*2, and *q*3, then, according to Fig. 10.5a, - -$$ -\dot{q}_1 = q_2 \n\dot{q}_2 = q_3 \n\dot{q}_3 = -12q_1 - 19q_2 - 8q_3 + x -$$ - -**Figure 10.5 (a)** DFII, **(b)** TDFII, **(c)** cascade, and **(d)** parallel realizations of *H*(*s*). - -### 922 CHAPTER 10 STATE-SPACE ANALYSIS - -Also, the output *y* is given by - -$$ -y = 10q_1 + 2q_2 -$$ - -In matrix form, these state and output equations become - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \\ \dot{q}_3 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -12 & -19 & -8 \end{bmatrix}}_{\mathbf{A}} \underbrace{\begin{bmatrix} q_1 \\ q_2 \\ q_3 \end{bmatrix}}_{\mathbf{A}} + \underbrace{\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}}_{\mathbf{B}} x -$$ - -and - -$$ -\mathbf{y} = \underbrace{[10 \quad 2 \quad 0]}_{\text{C}} \begin{bmatrix} q_1 \\ q_2 \\ q_3 \end{bmatrix} -$$ - -We can readily verify the state equations of the DFII structure by using MATLAB's tf2ss command: - -``` ->> num = [2 10]; den = [1 8 19 12]; ->> [A,B,C,D] = tf2ss(num,den) - A = -8 -19 -12 - 100 - 010 - B= 1 - 0 - 0 - C = 0 2 10 - D= 0 -``` - -MATLAB's convention for labeling state variables *q*1,*q*2,...,*qn* in a block diagram, such as shown in Fig. 10.5a, is reversed. That is, MATLAB labels *q*1 as *qn*, *q*2 and *qn*−1, and so on. Keeping this in mind, we see that MATLAB indeed confirms our earlier results. - -It is also possible to determine the transfer function from the state-space representation using the ss2tf and tf commands: - -``` ->> [num,den] = ss2tf(A,B,C,D); H = tf(num,den) - H = - 2 s + 10 - ----------------------- - s^3 + 8 s^2 + 19 s + 12 -``` - -### **Transpose Direct Form II** - -We can also realize *H*(*s*) by using the transpose of the DFII form, as shown in Fig. 10.5b. If we label the output of the three integrators as the state variables *v*1, *v*2, and *v*3, then, according - -to Fig. 10.5b, - -$$ -\dot{v}_1 = -12v_3 + 10x -$$ - -\n -$$ -\dot{v}_2 = v_1 - 19v_3 + 2x -$$ - -\n -$$ -\dot{v}_3 = v_2 - 8v_3 -$$ - -and the output *y* is given by - -*y* = *v*3 - -The matrix form of these state and output equations become - -$$ -\begin{bmatrix} \dot{v}_1 \\ \dot{v}_2 \\ \dot{v}_3 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & 0 & -12 \\ 1 & 0 & -19 \\ 0 & 1 & -8 \end{bmatrix}}_{\hat{A}} \underbrace{\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}}_{+} + \underbrace{\begin{bmatrix} 10 \\ 2 \\ 0 \end{bmatrix}}_{\hat{B}} x -$$ - -and - -$$ -\mathbf{y} = \underbrace{[0 \quad 0 \quad 1]}_{\hat{\mathbf{c}}} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} -$$ - -Observe closely the relationship between the state-space descriptions of *H*(*s*) by means of the DFII and TDFII realizations. The **A** matrices in these two cases are the transpose of each other; also, the **B** of one is the transpose of **C** in the other, and vice versa. Hence, - -$$ -(\mathbf{A})^T = \hat{\mathbf{A}}, \qquad (\mathbf{B})^T = \hat{\mathbf{C}}, \qquad \text{and} \qquad (\mathbf{C})^T = \hat{\mathbf{B}} -$$ - -This is no coincidence. This duality relation is generally true [1]. - -### **Cascade Realization** - -The three integrator outputs *w*1, *w*2, and *w*3 in Fig. 10.5c are the state variables. Writing equations for the summer outputs yields - -$$ -\dot{w}_1 = -w_1 + x -$$ -, $\dot{w}_2 = 2w_1 - 3w_2$ , and $\dot{w}_3 = 5w_2 + \dot{w}_2 - 4w_3$ - -Since *w*˙ 2 = 2*w*1 −3*w*2, we see that *w*˙ 3 = 2*w*1 +2*w*2 −4*w*3. From Fig. 10.5c, we further see that *y* = *w*3. Put into matrix form, the state and output equations are therefore - -$$ -\begin{bmatrix} \dot{w}_1 \\ \dot{w}_2 \\ \dot{w}_3 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 2 & -3 & 0 \\ 2 & 2 & -4 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} x -$$ -$$ -y = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} -$$ - -and - -### 924 CHAPTER 10 STATE-SPACE ANALYSIS - -### **Parallel Realization (Diagonal Representation)** - -The three integrator outputs *z*1, *z*2, and *z*3 in Fig. 10.5d are the state variables. The state equations are - -> *z*˙1 = −*z*1 +*x z*˙2 = −3*z*2 +*x z*˙3 = −4*z*3 +*x* - -and the output equation is - -$$ -y = \frac{4}{3}z_1 - 2z_2 + \frac{2}{3}z_3 -$$ - -In matrix form, these equations are - -$$ -\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \\ \dot{z}_3 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -4 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} x -$$ -$$ -y = \begin{bmatrix} \frac{4}{3} & -2 & \frac{2}{3} \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} -$$ - -### A GENERAL CASE - -It is clear that a system has several state-space descriptions. Notable among these are the variables obtained from the DFII, its transpose, and the diagonalized variables (in the parallel realization). State equations in these forms can be written immediately by inspection of the transfer function. Consider the general *N*th-order transfer function - -$$ -H(s) = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} -$$ - -= -$$ -\frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{(s - \lambda_1)(s - \lambda_2) \dots (s - \lambda_N)} -$$ - -= -$$ -b_0 + \frac{k_1}{s - \lambda_1} + \frac{k_2}{s - \lambda_2} + \dots + \frac{k_N}{s - \lambda_N} -$$ - (10.25) - -The realizations of *H*(*s*) found by using direct form II [Eq. (10.24)] and the parallel form [Eq. (10.25)] appear in Figs. 10.6a and 10.6b, respectively. - -The *N* integrator outputs *q*1, *q*2, ... , *qN* in Fig. 10.6a are the state variables. By inspection of this figure, we obtain - -$$ -\begin{aligned}\n\dot{q}_1 &= q_2 \\ -\dot{q}_2 &= q_3 \\ -&\vdots \\ -\dot{q}_{N-1} &= q_N \\ -\dot{q}_N &= -a_N q_1 - a_{N-1} q_2 - \dots - a_2 q_{N-1} - a_1 q_N + x\n\end{aligned} -$$ - -**Figure 10.6 (a)** Direct form II and **(b)** parallel realizations for an *N*th-order LTIC system. - -and output *y* is - -$$ -y = b_N q_1 + b_{N-1} q_2 + \cdots + b_1 q_N + b_0 \dot{q}_N -$$ - -We can eliminate *q*˙*N* in this output equation by using the last state equation to yield - -$$ -y = (b_N - b_0 a_N)q_1 + (b_{N-1} - b_0 a_{N-1})q_2 + \dots + (b_1 - b_0 a_1)q_N + b_0 x -$$ - -= $\hat{b}_N q_1 + \hat{b}_{N-1} q_2 + \dots + \hat{b}_1 q_N + b_0 x$ - -where *b*ˆ*i* = *bi* −*b*0*ai*. In matrix form, we obtain - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \\ \vdots \\ \dot{q}_{N-1} \\ \dot{q}_N \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ -a_N & -a_{N-1} & -a_{N-2} & \cdots & -a_2 & -a_1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \\ \vdots \\ q_{N-1} \\ q_N \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix} x -$$ - -and - -$$ -y = [\hat{b}_N \quad \hat{b}_{N-1} \quad \cdots \quad \hat{b}_1] \begin{bmatrix} q_1 \\ q_2 \\ \vdots \\ q_N \end{bmatrix} + b_0 x -$$ - -In Fig. 10.6b, the *N* integrator outputs *z*1, *z*2, ... , *zN* are the state variables. By inspection of this figure, we obtain - -$$ -\dot{z}_1 = \lambda_1 z_1 + x -$$ - -\n -$$ -\dot{z}_2 = \lambda_2 z_2 + x -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -\dot{z}_N = \lambda_N z_N + x -$$ - -and - -$$ -y = k_1 z_1 + k_2 z_2 + \cdots + k_N z_N + b_0 x -$$ - -or - -$$ -\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \\ \vdots \\ \dot{z}_{N-1} \\ \dot{z}_N \end{bmatrix} = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_2 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \cdots & \lambda_{N-1} & 0 \\ 0 & 0 & \cdots & 0 & \lambda_N \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_{N-1} \\ z_N \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \\ 1 \end{bmatrix} x -$$ -(10.26) -$$ -y = \begin{bmatrix} k_1 & k_2 & \cdots & k_{N-1} & k_N \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_{N-1} \end{bmatrix} + b_0 x -$$ - -*zN* - -and - -Observe that the diagonalized form of the state matrix [Eq. (10.26)] has the transfer function poles as its diagonal elements. The presence of repeated poles in *H*(*s*) will modify the procedure slightly. The handling of these cases is discussed in Sec. 4.6. - -It is clear from the foregoing discussion that a state-space description is not unique. For any realization of *H*(*s*) obtained from integrators, scalar multipliers, and adders, a corresponding state-space description exists. Since there are uncountable possible realizations of *H*(*s*), there are uncountable possible state-space descriptions. - -The advantages and drawbacks of various types of realization were discussed in Sec. 4.6. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/126_10.4 SOLUTION OF STATE EQUATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/126_10.4 SOLUTION OF STATE EQUATIONS.md deleted file mode 100644 index ba5a9e04143c6b47f1836bcdb468b665d815c398..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/126_10.4 SOLUTION OF STATE EQUATIONS.md +++ /dev/null @@ -1,710 +0,0 @@ -## **[10.4 SOLUTION OF](#page-14-0) STATE EQUATIONS** - -The state equations of a linear system are *N* simultaneous linear differential equations of the first order. We studied the techniques of solving linear differential equations in Chs. 2 and 4. The same techniques can be applied to state equations without any modification. However, it is more convenient to carry out the solution in the framework of matrix notation. - -These equations can be solved in both the time and frequency domains (Laplace transform). The latter is relatively easier to deal with than the time-domain solution. For this reason, we shall first consider the Laplace transform solution. - -### **[10.4-1 Laplace Transform Solution of State Equations](#page-14-0)** - -The *i*th state equation [Eq. (10.14)] is of the form - -$$ -\dot{q}_i = a_{i1}q_1 + a_{i2}q_2 + \dots + a_{iN}q_N + b_{i1}x_1 + b_{i2}x_2 + \dots + b_{ij}x_j \tag{10.27} -$$ - -We shall take the Laplace transform of this equation. Let - -$$ -q_i(t) \Longleftrightarrow Q_i(s) -$$ - -so that - -$$ -\dot{q}_i(t) \Longleftrightarrow sQ_i(s) - q_i(0) -$$ - -Also, let - -$$ -x_i(t) \Longleftrightarrow X_i(s) -$$ - -The Laplace transform of Eq. (10.27) yields - -$$ -sQ_i(s) - q_i(0) = a_{i1}Q_1(s) + a_{i2}Q_2(s) + \cdots + a_{iN}Q_N(s) + b_{i1}X_1(s) + b_{i2}X_2(s) + \cdots + b_{ij}X_j(s) -$$ - -Taking the Laplace transforms of all *N* state equations, we obtain - -$$ -s\left[\begin{array}{c} Q_1(s) \\ Q_2(s) \\ \vdots \\ Q_N(s) \end{array}\right] - \left[\begin{array}{c} q_1(0) \\ q_2(0) \\ \vdots \\ q_N(0) \end{array}\right] = \left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1N} \\ a_{21} & a_{22} & \cdots & a_{2N} \\ \vdots & \vdots & \cdots & \vdots \\ a_{N1} & a_{N2} & \cdots & a_{NN} \end{array}\right] \left[\begin{array}{c} Q_1(s) \\ Q_2(s) \\ \vdots \\ Q_N(s) \end{array}\right] + \left[\begin{array}{cccc} q_1(0) \\ \vdots \\ q_N(0) \end{array}\right] + \left[\begin{array}{cccc} b_{11} & b_{12} & \cdots & b_{1j} \\ b_{21} & b_{22} & \cdots & b_{2j} \\ \vdots & \vdots & \cdots & \vdots \\ b_{N1} & b_{N2} & \cdots & b_{Nj} \end{array}\right] \left[\begin{array}{c} X_1(s) \\ X_2(s) \\ \vdots \\ X_j(s) \end{array}\right] -$$ - -Defining the vectors, as indicated, we have - -$$ -s\mathbf{Q}(s) - \mathbf{q}(0) = \mathbf{A}\mathbf{Q}(s) + \mathbf{B}\mathbf{X}(s) -$$ - -or - -$$ -s\mathbf{Q}(s) - \mathbf{A}\mathbf{Q}(s) = \mathbf{q}(0) + \mathbf{B}\mathbf{X}(s) -$$ - -and - -$$ -(s\mathbf{I} - \mathbf{A})\mathbf{Q}(s) = \mathbf{x}(0) + \mathbf{B}\mathbf{X}(s) -$$ - -where **I** is the *N* ×*N* identity matrix. Solving for **Q**(*s*), we have - -$$ -Q(s) = (sI - A)^{-1}[q(0) + BX(s)] -$$ - -= $\Phi(s)[q(0) + BX(s)]$ (10.28) - -where - -$$ -\mathbf{\Phi}(s) = (s\mathbf{I} - \mathbf{A})^{-1} -$$ - -Thus, from Eq. (10.28), - -$$ -\mathbf{Q}(s) = \mathbf{\Phi}(s)\mathbf{q}(0) + \mathbf{\Phi}(s)\mathbf{B}\mathbf{X}(s) -$$ - -and - -$$ -\mathbf{q}(t) = \underbrace{\mathcal{L}^{-1}[\Phi(s)]\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{\mathcal{L}^{-1}[\Phi(s)\mathbf{B}\mathbf{X}(s)]}_{\text{zero-state response}} -$$ -(10.29) - -Equation (10.29) gives the desired solution. Observe the two components of the solution. The first component yields **q**(*t*) when the input *x*(*t*) = 0. Hence, the first component is the zero-input response. In a similar manner, we see that the second component is the zero-state response. - -### **EXAMPLE 10.6 Laplace Transform Solution to State Equations** - -Using the Laplace transform, find the state vector **q**(*t*) for the system whose state equation is given by - -**q**˙ = **Aq**+**Bx** - -where - -$$ -\mathbf{A} = \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \qquad \mathbf{x}(t) = u(t) -$$ - -and the initial conditions are *q*1(0) = 2, *q*2(0) = 1. - -From Eq. (10.28), we have - -$$ -\mathbf{Q}(s) = \mathbf{\Phi}(s)[\mathbf{q}(0) + \mathbf{B}\mathbf{X}(s)] -$$ - -Let us first find (*s*). We have - -$$ -(s\mathbf{I} - \mathbf{A}) = s \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix} = \begin{bmatrix} s+12 & -\frac{2}{3} \\ 36 & s+1 \end{bmatrix} -$$ - -and - -$$ -\Phi(s) = (s\mathbf{I} - \mathbf{A})^{-1} = \begin{bmatrix} \frac{s+1}{(s+4)(s+9)} & \frac{2/3}{(s+4)(s+9)}\\ \frac{-36}{(s+4)(s+9)} & \frac{s+12}{(s+4)(s+9)} \end{bmatrix} -$$ - -Now, -$$ -q(0) -$$ - is given as - -$$ -\mathbf{q}(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} -$$ - -Also, *X*(*s*) = 1/*s*, and - -$$ -\mathbf{BX}(s) = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \frac{1}{s} = \begin{bmatrix} \frac{1}{3s} \\ \frac{1}{s} \end{bmatrix} -$$ - -Therefore, - -$$ -\mathbf{q}(0) + \mathbf{B}\mathbf{X}(s) = \begin{bmatrix} 2 + \frac{1}{3s} \\ 1 + \frac{1}{s} \end{bmatrix} = \begin{bmatrix} \frac{6s+1}{3s} \\ \frac{s+1}{s} \end{bmatrix} -$$ - -and - -$$ -Q(s) = \Phi(s) [q(0) + BX(s)] -$$ - -= -$$ -\begin{bmatrix} \frac{s+1}{(s+4)(s+9)} & \frac{2/3}{(s+4)(s+9)} \\ \frac{-36}{(s+4)(s+9)} & \frac{s+12}{(s+4)(s+9)} \end{bmatrix} \begin{bmatrix} \frac{6s+1}{3s} \\ \frac{s+1}{s} \end{bmatrix} -$$ - -= -$$ -\begin{bmatrix} \frac{2s^2+3s+1}{s(s+4)(s+9)} \\ \frac{s-59}{(s+4)(s+9)} \end{bmatrix} -$$ - -= -$$ -\begin{bmatrix} \frac{1/36}{s} - \frac{21/20}{s+4} + \frac{136/45}{s+9} \\ \frac{-63/5}{s+4} + \frac{68/5}{s+9} \end{bmatrix} -$$ - -The inverse Laplace transform of this equation yields - -$$ -\begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{36} - \frac{21}{20}e^{-4t} + \frac{136}{45}e^{-9t}\right)u(t) \\ \left(-\frac{63}{5}e^{-4t} + \frac{68}{5}e^{-9t}\right)u(t) \end{bmatrix} -$$ - -This result is readily confirmed using MATLAB and its symbolic toolbox. - ->> syms s >> A = [-12 2/3;-36 -1]; B = [1/3; 1]; q0 = [2;1]; X = 1/s; >> q = ilaplace(inv(s\*eye(2)-A)\*(q0+B\*X)) q = (136\*exp(-9\*t))/45 - (21\*exp(-4\*t))/20 + 1/36 (68\*exp(-9\*t))/5 - (63\*exp(-4\*t))/5 - -To create a plot of the state vector, we use MATLAB's subs command to substitute the symbolic variable *t* with a vector of desired values. - -### 930 CHAPTER 10 STATE-SPACE ANALYSIS - -``` ->> t = (0:.01:2); q = subs(q); q1 = q(1,:); q2 = q(2,:); ->> plot(t,q1,'k',t,q2,'k--'); xlabel('t'); ylabel('Amplitude'); ->> legend('q_1(t)','q_2(t)','Location','SE'); -``` - -The resulting plot is shown in Fig. 10.7. - -### THE OUTPUT - -The output equation is given by - -$$ -y = Cq + Dx -$$ - -and - -$$ -\mathbf{Y}(s) = \mathbf{C}\mathbf{Q}(s) + \mathbf{D}\mathbf{X}(s) -$$ - -Upon substituting Eq. (10.28) into this equation, we have - -$$ -\mathbf{Y}(s) = \mathbf{C}[\Phi(s)[\mathbf{q}(0) + \mathbf{B}\mathbf{x}\mathbf{X}(s)]] + \mathbf{D}\mathbf{X}(s) -$$ - -= -$$ -\underbrace{\mathbf{C}\Phi(s)\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{[\mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D}]\mathbf{X}(s)}_{\text{zero-state response}} -$$ -(10.30) - -The zero-state response [i.e., the response **Y**(*s*) when **q**(0) = **0**] is given by - -$$ -\mathbf{Y}(s) = [\mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D}]\mathbf{X}(s) -$$ - -Note that the transfer function of a system is defined under the zero-state condition [see Eq. (4.19)]. The matrix **C**(*s*)**B** + **D** is the *transfer function matrix* **H**(*s*) of the system, which relates the responses *y*1, *y*2, ... , *yk* to the inputs *x*1, *x*2, ... , *xj*: - -$$ -\mathbf{H}(s) = \mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D} \tag{10.31} -$$ - -and the zero-state response is - -$$ -\mathbf{Y}(s) = \mathbf{H}(s)\mathbf{X}(s) -$$ - -The matrix **H**(*s*) is a *k* × *j* matrix (*k* is the number of outputs and *j* is the number of inputs). The *ij*th element *Hij*(*s*) of *H*(*s*) is the transfer function that relates the output *yi*(*t*) to the input *xj*(*t*). - -### **EXAMPLE 10.7 Transfer Function Matrix from State-Space Description** - -Let us consider a system with a state equation - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} -$$ - -and an output equation - -$$ -\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} -$$ - -Determine the transfer function matrix of the system. - -In this case, - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \qquad \mathbf{D} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \tag{10.32} -$$ - -and - -$$ -\Phi(s) = (s\mathbf{I} - \mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 2 & s+3 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} -$$ -(10.33) - -Hence, the transfer function matrix **H**(*s*) is given by - -$$ -\mathbf{H}(s) = \mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D} -$$ -\n -$$ -= \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} -$$ -\n -$$ -= \begin{bmatrix} \frac{s+4}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{s+4}{s+2} & \frac{1}{s+2} \\ \frac{2(s-2)}{(s+1)(s+2)} & \frac{s^2+5s+2}{(s+1)(s+2)} \end{bmatrix} -$$ -\n(10.34) - -and the zero-state response is - -$$ -\mathbf{Y}(s) = \mathbf{H}(s)\mathbf{X}(s) -$$ - -Remember that the *ij*th element of the transfer function matrix in Eq. (10.34) represents the transfer function that relates the output *yi*(*t*) to the input *xj*(*t*). For instance, the transfer function that relates the output *y*3 to the input *x*2 is *H*32(*s*), where - -$$ -H_{32}(s) = \frac{s^2 + 5s + 2}{(s+1)(s+2)} -$$ - -We can readily verify the transfer function matrix using MATLAB and its symbolic toolbox functions. - ->> A = [0 1;-2 -3]; B = [1 0;1 1]; >> C = [1 0;1 1;0 2]; D = [0 0;1 0;0 1]; >> syms s; H = collect(simplify(C\*inv(s\*eye(2)-A)\*B+D)) H = [ (s + 4)/(s^2 + 3\*s + 2), 1/(s^2 + 3\*s + 2)] [ (s + 4)/(s + 2), 1/(s + 2)] [ (2\*s - 4)/(s^2 + 3\*s + 2), (s^2 + 5\*s + 2)/(s^2 + 3\*s + 2)] - -Transfer functions relating particular inputs to particular outputs, such as *H*32(*s*), can be obtained using the ss2tf and tf functions. - -``` ->> [num,den] = ss2tf(A,B,C,D,2); H_32 = tf(num(3,:),den) - H_32 = - s^2 + 5 s + 2 - ------------- - s^2 + 3 s + 2 -``` - -### CHARACTERISTIC ROOTS (EIGENVALUES) OF A MATRIX - -It is interesting to observe that the denominator of every transfer function in Eq. (10.34) is (*s* + 1)(*s* + 2) except for *H*21(*s*) and *H*22(*s*), where the factor (*s* + 1) is canceled. This is no coincidence. We see that the denominator of every element of (*s*) is |*s***I** − **A**| because (*s*) = (*s***I** − **A**)−1, and the inverse of a matrix has its determinant in the denominator. Since **C**, **B**, and **D** are matrices with constant elements, we see from Eq. (10.31) that the denominator of (*s*) will also be the denominator of **H**(*s*). Hence, the denominator of every element of **H**(*s*) is |*s***I** − **A**|, except for the possible cancellation of the common factors mentioned earlier. In other words, the zeros of the polynomial |*s***I**− **A**| are also the poles of all transfer functions of the system. *Therefore, the zeros of the polynomial* |*s***I**−**A**| *are the characteristic roots of the system.* Hence, the characteristic roots of the system are the roots of the equation - -$$ -|s\mathbf{I} - \mathbf{A}| = 0 \tag{10.35} -$$ - -Since |*s***I** − **A**| is an *N*th-order polynomial in *s* with *N* zeros λ1, λ2, ... , λ*N*, we can write Eq. (10.35) as - -$$ -|s\mathbf{I} - \mathbf{A}| = s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N = (s - \lambda_1)(s - \lambda_2) \cdot \dots (s - \lambda_N) = 0 -$$ - -For the system in Ex. 10.7, - -$$ -|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s & 0 \\ 0 & s \end{vmatrix} - \begin{vmatrix} 0 & 1 \\ -2 & -3 \end{vmatrix} = \begin{vmatrix} s & -1 \\ 2 & s + 3 \end{vmatrix} -$$ -$$ -= s^2 + 3s + 2 = (s + 1)(s + 2) -$$ - -Hence, - -$$ -\lambda_1 = -1 \quad \text{and} \quad \lambda_2 = -2 -$$ - -Equation (10.35) is known as the *characteristic equation of the matrix* **A**, and λ1, λ2, ... , λ*N* are the characteristic roots of **A**. The term *eigenvalue,* meaning "characteristic value" in German, is also commonly used in the literature. Thus, we have shown that the characteristic roots of a system are the eigenvalues (characteristic values) of the matrix **A**. - -At this point, the reader will recall that if λ1, λ2, ... , λ*N* are the poles of the transfer function, then the zero-input response is of the form - -$$ -y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \dots + c_N e^{\lambda_N t} -$$ - (10.36) - -This fact is also obvious from Eq. (10.30). The denominator of every element of the zero-input response matrix **C**(*s*)**q**(0) is |*s***I**−**A**| =(*s*−λ1)(*s*−λ2)···(*s*−λ*N*). Therefore, the partial fraction expansion and the subsequent inverse Laplace transform will yield a zero-input component of the form in Eq. (10.36). - -### **[10.4-2 Time-Domain Solution of State Equations](#page-14-0)** - -The state equation is - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x} \tag{10.37} -$$ - -We now show that the solution of the vector differential Eq. (10.37) is - -$$ -\mathbf{q}(t) = e^{\mathbf{A}t}\mathbf{q}(0) + \int_0^t e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{x}(\tau) d\tau -$$ - -Before proceeding further, we must define the matrix exponential *e***A***t* . An exponential of a matrix is defined by an infinite series identical to that used in defining an exponential of a scalar. We shall define - -$$ -e^{\mathbf{A}t} = \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2 t^2}{2!} + \frac{\mathbf{A}^3 t^3}{3!} + \dots + \frac{\mathbf{A}^n t^n}{n!} + \dots = \sum_{k=0}^{\infty} \frac{\mathbf{A}^k t^k}{k!} -$$ -(10.38) - -For example, if - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} -$$ - -then - -$$ -\mathbf{A}t = \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} t = \begin{bmatrix} 0 & t \\ 2t & t \end{bmatrix} -$$ - -and - -$$ -\frac{\mathbf{A}^2 t^2}{2!} = \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} \frac{t^2}{2} = \begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix} \frac{t^2}{2} = \begin{bmatrix} t^2 & \frac{t^2}{2} \\ t^2 & \frac{3t^2}{2} \end{bmatrix} -$$ - -and so on. - -#### 934 CHAPTER 10 STATE-SPACE ANALYSIS - -We can show that the infinite series in Eq. (10.38) is absolutely and uniformly convergent for all values of *t*. Consequently, it can be differentiated or integrated term by term. Thus, to find (*d*/*dt*)*e***A***t* , we differentiate the series on the right-hand side of Eq. (10.38) term by term: - -$$ -\frac{d}{dt}e^{\mathbf{A}t} = \mathbf{A} + \mathbf{A}^2t + \frac{\mathbf{A}^3t^2}{2!} + \frac{\mathbf{A}^4t^3}{3!} + \cdots -$$ - -\n -$$ -= \mathbf{A}\bigg[\mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots\bigg] = \mathbf{A}e^{\mathbf{A}t} -$$ - -\n -$$ -= \bigg[\mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots + \cdots\bigg]\mathbf{A} = e^{\mathbf{A}t}\mathbf{A} -$$ - -Hence, - -$$ -\frac{d}{dt}e^{\mathbf{A}t} = \mathbf{A}e^{\mathbf{A}t} = e^{\mathbf{A}t}\mathbf{A} -$$ - -Also note that from Eq. (10.38), it follows that - -*e***0** = **I** - -where **I** is just the identity matrix. If we premultiply or postmultiply the infinite series for *e***A***t* [Eq. (10.38)] by an infinite series for *e*−**A***t* , we find that - -$$ -(e^{-At})(e^{At}) = (e^{At})(e^{-At}) = I -$$ -\n(10.39) - -In Sec. 10.1-1, we showed that - -$$ -\frac{d}{dt}(\mathbf{U}\mathbf{V}) = \frac{d\mathbf{U}}{dt}\mathbf{V} + \mathbf{U}\frac{d\mathbf{V}}{dt} -$$ - -Using this relationship, we observe that - -$$ -\frac{d}{dt}[e^{-\mathbf{A}t}\mathbf{q}] = \left(\frac{d}{dt}e^{-\mathbf{A}t}\right)\mathbf{q} + e^{-\mathbf{A}t}\dot{\mathbf{q}} -$$ -$$ -= -e^{-\mathbf{A}t}\mathbf{A}\mathbf{q} + e^{-\mathbf{A}t}\dot{\mathbf{q}} -$$ -(10.40) - -We now premultiply both sides of Eq. (10.37) by *e*−**A***t* to yield - -$$ -e^{-At}\dot{\mathbf{q}} = e^{-At}\mathbf{A}\mathbf{q} + e^{-At}\mathbf{B}\mathbf{x} -$$ - -or - -$$ --e^{-At}\mathbf{A}\mathbf{q}+e^{-\mathbf{A}t}\dot{\mathbf{q}}=e^{-\mathbf{A}t}\mathbf{B}\mathbf{x} -$$ - -Substituting this result into Eq. (10.40) yields - -$$ -\frac{d}{dt}[e^{-At}\mathbf{q}] = e^{-At}\mathbf{B}\mathbf{x} -$$ - -The integration of both sides of this equation from 0 to *t* yields - -$$ -e^{-\mathbf{A}t}\mathbf{q}\big|_{0}^{t} = \int_{0}^{t} e^{-\mathbf{A}\tau} \mathbf{B} \mathbf{x}(\tau) d\tau -$$ - -or - -$$ -e^{-\mathbf{A}t}\mathbf{q}(t) - \mathbf{q}(0) = \int_0^t e^{-\mathbf{A}\tau} \mathbf{B}\mathbf{x}(\tau) d\tau -$$ - -Hence, - -$$ -e^{-\mathbf{A}t}\mathbf{q} = \mathbf{q}(0) + \int_0^t e^{-\mathbf{A}\tau} \mathbf{B} \mathbf{x}(\tau) d\tau -$$ - -Premultiplying this result by *e***A***t* and using Eq. (10.39), we have - -$$ -\mathbf{q}(t) = \underbrace{e^{\mathbf{A}t}\mathbf{q}(0)}_{\text{ZIR}} + \underbrace{\int_0^t e^{\mathbf{A}(t-\tau)}\mathbf{B}\mathbf{x}(\tau) d\tau}_{\text{ZSR}} -$$ -(10.41) - -This is the desired solution. The first term on the right-hand side represents *q*(*t*) when the input *x*(*t*) = 0. Hence, it is the zero-input component. The second term, by a similar argument, is seen to be the zero-state component. - -The results of Eq. (10.41) can be expressed more conveniently in terms of the matrix convolution. We can define the convolution of two matrices in a manner similar to the multiplication of two matrices, except that the multiplication of two elements is replaced by their convolution. For example, - -$$ -\begin{bmatrix} x_1 & x_2 \ x_3 & x_4 \end{bmatrix} * \begin{bmatrix} g_1 & g_2 \ g_3 & g_4 \end{bmatrix} = \begin{bmatrix} (x_1 * g_1 + x_2 * g_3) & (x_1 * g_2 + x_2 * g_4) \\ (x_3 * g_1 + x_4 * g_3) & (x_3 * g_2 + x_4 * g_4) \end{bmatrix} -$$ - -By using this definition of matrix convolution, we can express Eq. (10.41) as - -$$ -\mathbf{q}(t) = e^{\mathbf{A}t}\mathbf{q}(0) + e^{\mathbf{A}t} * \mathbf{B}\mathbf{x}(t) -$$ -\n(10.42) - -Note that the limits of the convolution integral [Eq. (10.41)] are from 0 to *t*. Hence, all the elements of *e***A***t* in the convolution term of Eq. (10.42) are implicitly assumed to be multiplied by *u*(*t*). - -The result of Eqs. (10.41) and (10.42) can be easily generalized for any initial value of *t*. It is left as an exercise for the reader to show that the solution of the state equation can be expressed as - -$$ -\mathbf{q}(t) = e^{\mathbf{A}(t-t_0)}\mathbf{q}(t_0) + \int_{t_0}^t e^{\mathbf{A}(t-\tau)} \mathbf{B}\mathbf{x}(\tau) d\tau -$$ - -## DETERMINING *eAt* - -The exponential *e***A***t* required in Eqs. (10.41) and (10.42) can be computed from the definition in Eq. (10.38). Unfortunately, this is an infinite series, and its computation can be quite laborious. Moreover, we may not be able to recognize the closed-form expression for the answer. There are several efficient methods of determining *e***A***t* in closed form. It was shown in Sec. 10.1-3 that for an *N* ×*N* matrix **A**, - -$$ -e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{N-1} \mathbf{A}^{N-1} -$$ - (10.43) - -where - -$$ -\begin{bmatrix}\n\beta_0 \\ -\beta_1 \\ -\vdots \\ -\beta_{N-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{N-1} \\ -1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{N-1} \\ -\vdots & \vdots & \ddots & \vdots \\ -1 & \lambda_N & \lambda_N^2 & \cdots & \lambda_N^{N-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\ne^{\lambda_1 t} \\ -ne^{\lambda_2 t} \\ -\vdots \\ -e^{\lambda_N t}\n\end{bmatrix} -$$ - -and λ1,λ2,...,λ*N* are the *N* characteristic values (eigenvalues) of **A**. - -We can also determine *e***A***t* by comparing Eqs. (10.41) and (10.29). It is clear that - -$$ -e^{\mathbf{A}t} = \mathcal{L}^{-1}[\Phi(s)] = \mathcal{L}^{-1}[(s\mathbf{I} - \mathbf{A})^{-1}] -$$ -\n(10.44) - -Thus, *e***A***t* and (*s*) are a Laplace transform pair. To be consistent with Laplace transform notation, *e***A***t* is often denoted by *φ*(*t*), *the state transition matrix* (STM): - -*e***A***t* = *φ*(*t*) - -### **EXAMPLE 10.8 Time-Domain Method to Solve State Equations** - -Use the time-domain method to solve Ex. 10.6. - -For this case, the characteristic roots are given by - -$$ -|\mathbf{sI} - \mathbf{A}| = \begin{vmatrix} s+12 & -\frac{2}{3} \\ 36 & s+1 \end{vmatrix} = s^2 + 13s + 36 = (s+4)(s+9) = 0 -$$ - -The roots are λ1 = −4 and λ2 = −9, so - -$$ -\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 1 & -9 \end{bmatrix}^{-1} \begin{bmatrix} e^{-4t} \\ e^{-9t} \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 9e^{-4t} - 4e^{-9t} \\ e^{-4t} - e^{-9t} \end{bmatrix} -$$ - -and - -$$ -e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} -$$ - -= $\left(\frac{9}{5}e^{-4t} - \frac{4}{5}e^{-9t}\right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \left(\frac{1}{5}e^{-4t} - \frac{1}{5}e^{-9t}\right) \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix}$ -= $\begin{bmatrix} \left(\frac{-3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right) & \frac{2}{15}(e^{-4t} - e^{-9t}) \\ \frac{36}{5}(-e^{-4t} + e^{-9t}) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right) \end{bmatrix}$ - -The zero-input response is given by [see Eq. (10.41)] - -$$ -e^{\mathbf{A}t}\mathbf{q}(0) = \begin{bmatrix} \left(-\frac{3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right) & \frac{2}{15}(e^{-4t} - e^{-9t})\\ \frac{36}{5}(-e^{-4t} + e^{-9t}) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right) \end{bmatrix} \begin{bmatrix} 2\\ 1 \end{bmatrix} -$$ -$$ -= \begin{bmatrix} \left(\frac{-16}{15}e^{-4t} + \frac{46}{15}e^{-9t}\right)u(t) \\ \left(\frac{-64}{5}e^{-4t} + \frac{69}{5}e^{-9t}\right)u(t) \end{bmatrix} -$$ - -Note here the presence of *u*(*t*), indicating that the response begins at *t* = 0. - -The zero-state component is *e***A***t* ∗**Bx** [see Eq. (10.42)], where - -$$ -\mathbf{B}\mathbf{x} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} u(t) = \begin{bmatrix} \frac{1}{3}u(t) \\ u(t) \end{bmatrix} -$$ - -and - -$$ -e^{\mathbf{A}t} * \mathbf{B} \mathbf{x}(t) = \begin{bmatrix} \left(\frac{-3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right)u(t) & \frac{2}{15}(e^{-4t} - e^{-9t})u(t) \\ \frac{36}{5}(-e^{-4t} + e^{-9t}u(t)) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right)u(t) \end{bmatrix} * \begin{bmatrix} \frac{1}{3}u(t) \\ u(t) \end{bmatrix} -$$ - -Note again the presence of the term *u*(*t*) in every element of *e***A***t* . This is the case because the limits of the convolution integral run from 0 to *t* [Eq. (10.41)]. Thus, - -$$ -e^{\mathbf{A}t} * \mathbf{Bx}(t) = \begin{bmatrix} \left(-\frac{3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right)u(t) * \frac{1}{3}u(t) & \frac{2}{15}(e^{-4t} - e^{-9t})u(t) * u(t) \\ \frac{36}{5}(-e^{-4t} + e^{-9t})u(t) * \frac{1}{3}u(t) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right)u(t) * u(t) \end{bmatrix} -$$ -$$ -= \begin{bmatrix} -\frac{1}{15}e^{-4t}u(t) * u(t) + \frac{2}{5}e^{-9t}u(t) * u(t) \\ -\frac{4}{5}e^{-4t}u(t) * u(t) + \frac{9}{5}e^{-9t}u(t) * u(t) \end{bmatrix} -$$ - -Substitution for the preceding convolution integrals from the convolution table (Table 2.1) yields - -$$ -e^{\mathbf{A}t} * \mathbf{Bx}(t) = \begin{bmatrix} -\frac{1}{60}(1 - e^{-4t})u(t) + \frac{2}{45}(1 - e^{-9t})u(t) \\ -\frac{1}{5}(1 - e^{-4t})u(t) + \frac{1}{5}(1 - e^{-9t})u(t) \end{bmatrix} -$$ -$$ -= \begin{bmatrix} (\frac{1}{36} + \frac{1}{60}e^{-4t} - \frac{2}{45}e^{-9t})u(t) \\ \frac{1}{5}(e^{-4t} - e^{-9t})u(t) \end{bmatrix} -$$ - -The sum of the two components now gives the desired solution for **q**(*t*): - -$$ -\mathbf{q}(t) = \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{36} - \frac{21}{20}e^{-4t} + \frac{136}{45}e^{-9t}\right)u(t) \\ \left(\frac{-63}{5}e^{-4t} + \frac{68}{5}e^{-9t}\right)u(t) \end{bmatrix} -$$ - -This result confirms the solution obtained by using the frequency-domain method [see Ex. 10.6]. Once the state variables *q*1 and *q*2 have been found for *t*≥0, all the remaining variables can be determined from the output equation. - -### THE OUTPUT - -The output equation is given by - -$$ -\mathbf{y}(t) = \mathbf{C}\mathbf{q}(t) + \mathbf{D}\mathbf{x}(t) -$$ - -The substitution of the solution for **q** [Eq. (10.42)] in this equation yields - -$$ -\mathbf{y}(t) = \mathbf{C} [e^{\mathbf{A}t} \mathbf{q}(0) + e^{\mathbf{A}t} * \mathbf{B} \mathbf{x}(t)] + \mathbf{D} \mathbf{x}(t) -$$ - -Since the elements of **B** are constants, - -$$ -e^{\mathbf{A}t} * \mathbf{B}\mathbf{x}(t) = e^{\mathbf{A}t}\mathbf{B} * \mathbf{x}(t) -$$ - -### 938 CHAPTER 10 STATE-SPACE ANALYSIS - -With this result, the output equation becomes - -$$ -\mathbf{y}(t) = \mathbf{C} [e^{\mathbf{A}t} \mathbf{q}(0) + e^{\mathbf{A}t} \mathbf{B} * \mathbf{x}(t)] + \mathbf{D} \mathbf{x}(t) -$$ - -Now recall that the convolution of *x*(*t*) with the unit impulse δ(*t*) yields *x*(*t*). Let us define a *j* × *j* diagonal matrix *δ*(*t*) such that all its diagonal terms are unit impulse functions. It is then obvious that - -$$ -\delta(t) * \mathbf{x}(t) = \mathbf{x}(t) -$$ - -and the output equation can be expressed as - -$$ -\mathbf{y}(t) = \mathbf{C}[e^{\mathbf{A}t}\mathbf{q}(0) + e^{\mathbf{A}t}\mathbf{B} * \mathbf{x}(t)] + \mathbf{D}\delta(t) * \mathbf{x}(t) -$$ -$$ -= \mathbf{C}e^{\mathbf{A}t}\mathbf{q}(0) + [\mathbf{C}e^{\mathbf{A}t}\mathbf{B} + \mathbf{D}\delta(t)] * \mathbf{x}(t) -$$ - -With the notation *φ*(*t*) for *e***A***t* , the output equation may be expressed as - -$$ -\mathbf{y}(t) = \underbrace{\mathbf{C}\boldsymbol{\phi}(t)\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{\left[\mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t)\right] * \mathbf{x}(t)}_{\text{zero-state response}} -$$ - -The zero-state response, that is, the response when **q**(0) = **0**, is - -$$ -\mathbf{y}(t) = [\mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t)] * \mathbf{x}(t) = \mathbf{h}(t) * \mathbf{x}(t) -$$ - -where - -$$ -\mathbf{h}(t) = \mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t) \tag{10.45} -$$ - -The matrix **h**(*t*) is a *k* × *j* matrix known as the *impulse response matrix*. The reason for this designation is obvious. The *ij*th element of **h**(*t*) is *hij*(*t*), which represents the zero-state response *yi* when the input *xj*(*t*) = δ(*t*) and when all other inputs (and all the initial conditions) are zero. Not surprisingly, vectors **h**(*t*) and **H**(*s*) form a Laplace transform pair, - -$$ -\mathcal{L}[\mathbf{h}(t)] = \mathbf{H}(s) -$$ - -### **EXAMPLE 10.9 State Transition Matrix by Inverse Laplace Transform** - -For the system described in Ex. 10.7, use Eq. (10.44) to determine *e***A***t* : - -$$ -\boldsymbol{\phi}(t) = e^{\mathbf{A}t} = \mathcal{L}^{-1} \boldsymbol{\Phi}(s) -$$ - -This problem was solved earlier with frequency-domain techniques. From Eq. (10.33), we have - -$$ -\begin{aligned} \n\phi(t) &= \mathcal{L}^{-1} \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)}\\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} \\ \n&= \mathcal{L}^{-1} \begin{bmatrix} \frac{2}{s+1} - \frac{1}{s+2} & \frac{1}{s+1} - \frac{1}{s+2} \\ \frac{-2}{s+1} + \frac{2}{s+2} & \frac{-1}{s+1} + \frac{2}{s+2} \end{bmatrix} \\ \n&= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} \n\end{aligned} -$$ - -The same result is obtained in Ex. 10.1 (Sec. 10.1-3) by using Eq. (10.43) [see Eq. (10.13)]. Also, *δ*(*t*) is a diagonal *j*×*j* or 2×2 matrix: - -$$ -\delta(t) = \begin{bmatrix} \delta(t) & 0\\ 0 & \delta(t) \end{bmatrix} -$$ - -Substituting the matrices *φ*(*t*), *δ*(*t*), **C**, **D**, and **B** [Eq. (10.32)] into Eq. (10.45), we have - -$$ -\mathbf{h}(t) = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \delta(t) & 0 \\ 0 & \delta(t) \end{bmatrix} -$$ -$$ -= \begin{bmatrix} 3e^{-t} - 2e^{-2t} & e^{-t} - e^{-2t} \\ \delta(t) + 2e^{-2t} & e^{-2t} \\ -6e^{-t} + 8e^{-2t} & \delta(t) - 2e^{-2t} + 4e^{-2t} \end{bmatrix} -$$ - -As the reader can verify, the Laplace transform of this equation yields the transfer function matrix **H**(*s*) in Eq. (10.34). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/127_10.5 LINEAR TRANSFORMATION OF A STATE VECTOR.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/127_10.5 LINEAR TRANSFORMATION OF A STATE VECTOR.md deleted file mode 100644 index acd24f2f468962c0faf212f7e0e5a8c2d375094b..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/127_10.5 LINEAR TRANSFORMATION OF A STATE VECTOR.md +++ /dev/null @@ -1,452 +0,0 @@ -## **10.5 LINEAR [TRANSFORMATION OF A](#page-14-0) STATE VECTOR** - -In Sec. 10.2 we saw that the state of a system can be specified in several ways. The sets of all possible state variables are related—in other words, if we are given one set of state variables, we should be able to relate it to any other set. We are particularly interested in a linear type of relationship. Let *q*1,*q*2,...,*qN* and *w*1,*w*2,...,*wN* be two different sets of state variables specifying the same system. Let these sets be related by linear equations as - -$$ -w_1 = p_{11}q_1 + p_{12}q_2 + \dots + p_{1N}q_N -$$ - -\n -$$ -w_2 = p_{21}q_1 + p_{22}q_2 + \dots + p_{2N}q_N -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -w_N = p_{N1}q_1 + p_{N2}q_2 + \dots + p_{NN}q_N -$$ - -or - -| ⎡
w1
w2
⎢ | ⎤
⎥ | ⎡
p11
p21
⎢ | p12
p22 | ···
··· | ⎤
p1N
p2N
⎥ | ⎡
⎤
q1
q2
⎢
⎥ | -|--------------------|------------------|----------------------|------------|------------|----------------------|------------------------------| -| ⎢
⎢
⎣ | ⎥
=
⎥
⎦ | ⎢
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⎦ | -| wN | | pN1 | pN2 | ··· | pNN | qN | -|

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q | - -Defining the vector **w** and matrix **P** as just shown, we obtain the compact matrix representation - -$$ -\mathbf{w} = \mathbf{P}\mathbf{q} \tag{10.46} -$$ - -and - -$$ -\mathbf{q} = \mathbf{P}^{-1}\mathbf{w} \tag{10.47} -$$ - -Thus, the state vector **q** is transformed into another state vector **w** through the linear transformation in Eq. (10.46). - -If we know **w**, we can determine **q** from **q** = **P**−1 **w**, provided **P**−1 exists. This is equivalent to saying that **P** is a nonsingular matrix† (|**P**| = 0). Thus, if **P** is a nonsingular matrix, the vector **w** defined by Eq. (10.46) is also a state vector. Consider the state equation of a system - -**q**˙ = **Aq**+**Bx** - -$$ -\mathbf{w} = \mathbf{P}\mathbf{q} -$$ -$$ -\mathbf{q} = \mathbf{P}^{-1}\mathbf{w} -$$ - -and - -then - -If - -Hence, the state equation now becomes - -$$ -\mathbf{P}^{-1}\dot{\mathbf{w}} = \mathbf{A}\mathbf{P}^{-1}\mathbf{w} + \mathbf{B}\mathbf{x} -$$ - -**q**˙ = **P**−1 - -**w**˙ - -or - -$$ -\dot{\mathbf{w}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1}\mathbf{w} + \mathbf{P}\mathbf{B}\mathbf{x} -$$ - -= $\hat{\mathbf{A}}\mathbf{w} + \hat{\mathbf{B}}\mathbf{x}$ (10.48) - -where - -$$ -\hat{\mathbf{A}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1} \qquad \text{and} \qquad \hat{\mathbf{B}} = \mathbf{P}\mathbf{B} \tag{10.49} -$$ - -Equation (10.48) is a state equation for the same system, but now it is expressed in terms of the state vector **w**. - -† This condition is equivalent to saying that all *N* equations in Eq. (10.46) are linearly independent; that is, none of the *N* equations can be expressed as a linear combination of the remaining equations. - -The output equation is also modified. Let the original output equation be - -$$ -y = Cq + Dx -$$ - -In terms of the new state variable **w**, this equation becomes - -$$ -\mathbf{y} = \mathbf{C}(\mathbf{P}^{-1}\mathbf{w}) + \mathbf{D}\mathbf{x} -$$ -$$ -= \hat{\mathbf{C}}\mathbf{w} + \mathbf{D}\mathbf{x} -$$ -$$ -\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} -$$ -(10.50) - -### **EXAMPLE 10.10 Linear Transformation of the State Vector** - -The state equations of a certain system are given by - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \end{bmatrix} x(t) -$$ - -Find the state equations for this system when the new state variables *w*1 and *w*2 are given as - -$$ -\begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} -$$ - (10.51) - -According to Eq. (10.48), the state equation for the state variable **w** is given by - -$$ -\dot{\mathbf{w}} = \hat{\mathbf{A}}\mathbf{w} + \hat{\mathbf{B}}\mathbf{x} -$$ - -where [see Eqs. (10.49) and (10.50)] - -$$ -\hat{\mathbf{A}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} -$$ -$$ -= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} -$$ -$$ -= \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix} -$$ - -and - -where - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} -$$ - -Therefore, - -$$ -\begin{bmatrix} \dot{w}_1 \\ \dot{w}_2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} + \begin{bmatrix} 3 \\ -1 \end{bmatrix} x(t) -$$ - -This is the desired state equation for the state vector **w**. The solution of this equation requires a knowledge of the initial state **w**(0). This can be obtained from the given initial state **q**(0) by using Eq. (10.51). - -We can obtain the same result with less effort using MATLAB. - -``` ->> A = [0 1;-2 -3]; B = [1; 2]; ->> P = [1 1;1 -1]; ->> Ahat = P*A*inv(P), Bhat = P*B - Ahat = -2 0 - 3 -1 - Bhat = 3 - -1 -``` - -### INVARIANCE OF EIGENVALUES - -We have seen that the poles of all possible transfer functions of a system are the eigenvalues of the matrix **A**. If we transform a state vector from **q** to **w**, the variables *w*1, *w*2, ... , *wN* are linear combinations of *q*1, *q*2, ... , *qN* and therefore may be considered to be outputs. Hence, the poles of the transfer functions relating *w*1, *w*2, ... , *wN* to the various inputs must also be the eigenvalues of matrix **A**. On the other hand, the system is also specified by Eq. (10.48). This means that the poles of the transfer functions must be the eigenvalues of **A**ˆ . Therefore, the eigenvalues of matrix **A** remain unchanged for the linear transformation of variables represented by Eq. (10.46), and the eigenvalues of matrix **A** and matrix **A**ˆ (**A**ˆ = **PAP**−1 ) are identical, implying that the characteristic equations of **A** and **A**ˆ are also identical. This result also can be proved alternately as follows. - -Consider the matrix **P**(*s***I**−**A**)**P**−1 . We have - -$$ -\mathbf{P}(s\mathbf{I} - \mathbf{A})\mathbf{P}^{-1} = \mathbf{P}s\mathbf{IP}^{-1} - \mathbf{P}\mathbf{A}\mathbf{P}^{-1} = s\mathbf{P}\mathbf{IP}^{-1} - \hat{\mathbf{A}} = s\mathbf{I} - \hat{\mathbf{A}} -$$ - -Taking the determinants of both sides, we obtain - -$$ -|\mathbf{P}||s\mathbf{I} - \mathbf{A}||\mathbf{P}^{-1}| = |s\mathbf{I} - \hat{\mathbf{A}}| -$$ - -The determinants |**P**| and |**P**−1 | are reciprocals of each other. Hence, - -$$ -|s\mathbf{I} - \mathbf{A}| = |s\mathbf{I} - \hat{\mathbf{A}}| -$$ - -This is the desired result. We have shown that the characteristic equations of **A** and **A**ˆ are identical. Hence, the eigenvalues of **A** and **A**ˆ are identical. - -In Ex. 10.10, matrix **A** is given as - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} -$$ - -The characteristic equation is - -$$ -|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s & -1 \\ 2 & s+3 \end{vmatrix} = s^2 + 3s + 2 = 0 -$$ - -Also, - -$$ -\hat{\mathbf{A}} = \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix} -$$ - -and - -$$ -|s\mathbf{I} - \hat{\mathbf{A}}| = \begin{bmatrix} s+2 & 0\\ -3 & s+1 \end{bmatrix} = s^2 + 3s + 2 = 0 -$$ - -This result verifies that the characteristic equations of **A** and **A**ˆ are identical. - -### **[10.5-1 Diagonalization of Matrix](#page-14-0) A** - -For several reasons, it is desirable to make matrix **A** diagonal. If **A** is not diagonal, we can transform the state variables such that the resulting matrix **A**ˆ is diagonal.† One can show that for any diagonal matrix **A**, the diagonal elements of this matrix must necessarily be λ1, λ2, ... , λ*N* (the eigenvalues) of the matrix. Consider the diagonal matrix **A**: - -$$ -\mathbf{A} = \begin{bmatrix} a_1 & 0 & 0 & \cdots & 0 \\ 0 & a_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & a_N \end{bmatrix} -$$ - -The characteristic equation is given by - -$$ -|\mathbf{sI} - \mathbf{A}| = \begin{bmatrix} (s - a_1) & 0 & 0 & \cdots & 0 \\ 0 & (s - a_2) & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & (s - a_N) \end{bmatrix} = 0 -$$ - -or - -$$ -(s - a_1)(s - a_2) \cdots (s - a_N) = 0 -$$ - -The nonzero (diagonal) elements of a diagonal matrix are therefore its eigenvalues λ1, λ2, ... , λ*N*. We shall denote the diagonal matrix by the symbol, **A**: - -$$ -\mathbf{\Lambda} = \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_N \end{bmatrix} -$$ - (10.52) - -Let us now consider the transformation of the state vector **A** such that the resulting matrix **A**ˆ is a diagonal matrix . - -† In this discussion we assume distinct eigenvalues. If the eigenvalues are not distinct, we can reduce the matrix to a modified diagonalized (Jordan) form. - -### 944 CHAPTER 10 STATE-SPACE ANALYSIS - -Consider the system - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x} -$$ - -We shall assume that λ1, λ2, ... , λ*N*, the eigenvalues of **A**, are distinct (no repeated roots). Let us transform the state vector **q** into the new state vector **z**, using the transformation - -$$ -z = Pq \tag{10.53} -$$ - -Then, after the development of Eq. (10.48), we have - -$$ -\dot{z} = PAP^{-1}z + PBx -$$ - -We desire the transformation to be such that **PAP**−1 is a diagonal matrix given by Eq. (10.52), or - -= **PAP**−1 - -$$ -\dot{\mathbf{z}} = \mathbf{\Lambda}\mathbf{z} + \mathbf{B}\mathbf{x} \tag{10.54} -$$ - -Hence, - -or - -$$ -\Lambda P = PA \tag{10.55} -$$ - -We know and **A**. Equation (10.55) therefore can be solved to determine **P**. - -### **EXAMPLE 10.11 Diagonal Form of the State Equations** - -Find the diagonalized form of the state equations for the system in Ex. 10.10. - -In this case, - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} -$$ - -We found λ1 = −1 and λ2 = −2. Hence, - -$$ -\mathbf{\Lambda} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} -$$ - -and Eq. (10.55) becomes - -$$ -\begin{bmatrix} -1 & 0 \ 0 & -2 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} 0 & 1 \ -2 & -3 \end{bmatrix} -$$ - -Equating the four elements on two sides, we obtain - -$$ --p_{11} = -2p_{12} -$$ - -\n -$$ --p_{12} = p_{11} - 3p_{12} -$$ - -\n -$$ --2p_{21} = -2p_{22} -$$ - -\n -$$ --2p_{22} = p_{21} - 3p_{22} -$$ - -The reader will immediately recognize that the first two equations are identical and that the last two equations are identical. Hence, two equations may be discarded, leaving us with only two equations [*p*11 = 2*p*12 and *p*21 = *p*22] and four unknowns. This observation means that there is no unique solution. There is, in fact, an infinite number of solutions. We can assign any value to *p*11 and *p*21 to yield one possible solution.† If *p*11 = *k*1 and *p*21 = *k*2, then we have *p*12 = *k*1/2 and *p*22 = *k*2: - -$$ -\mathbf{P} = \begin{bmatrix} k_1 & \frac{k_1}{2} \\ k_2 & k_2 \end{bmatrix} -$$ - -We may assign any values to *k*1 and *k*2. For convenience, let *k*1 = 2 and *k*2 = 1. This substitution yields - -$$ -\mathbf{P} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} -$$ - -The transformed variables [Eq. (10.53)] are - -$$ -\begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} = \begin{bmatrix} 2q_1 + q_2 \\ q_1 + q_2 \end{bmatrix} -$$ - -This expression relates the new state variables *z*1 and *z*2 to the original state variables *q*1 and *q*2. The system equation with **z** as the state vector is given by [see Eq. (10.54)] - -$$ -\dot{z} = \Lambda z + \hat{B}x -$$ - -where - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \end{bmatrix} -$$ -$$ -\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \begin{bmatrix} 4 \\ 4 \end{bmatrix}. -$$ - -*z*2 - -+ - -4 3 - -*x* (10.56) - -−1 0 0 −2 - -*z*˙1 *z*˙2 - -= - -Hence, - -† If, however, we want the state equations in diagonalized form, as in Eq. (10.26), where all the elements of -$$ -\hat{B} -$$ - matrix are unity, there is a unique solution. The reason is that the equation $\hat{B} = PB$ , where all the elements of $\hat{B}$ are unity, imposes additional constraints. In the present example, this condition will yield $p_{11} = 1/2$ , $p_{12} = 1/4$ , $p_{21} = 1/3$ , and $p_{22} = 1/3$ . The relationship between **z** and **q** is then - -$$ -z_1 = \frac{1}{2}q_1 + \frac{1}{4}q_2 -$$ - and $z_2 = \frac{1}{3}q_1 + \frac{1}{3}q_2$ - -or - -$$ -\begin{aligned}\n\dot{z}_1 &= -z_1 + 4x \\ -\dot{z}_2 &= -2z_2 + 3x\n\end{aligned} -$$ - -Note the distinctive nature of these state equations. Each state equation involves only one variable and therefore can be solved by itself. A general state equation has the derivative of one state variable equal to a linear combination of all state variables. Such is not the case with the diagonalized matrix . Each state variable *zi* is chosen so that it is uncoupled from the rest of the variables; hence, a system with *N* eigenvalues is split into *N* decoupled systems, each with an equation of the form - -$$ -\dot{z}_i = \lambda_i z_i + (\text{input terms}) -$$ - -This fact also can be readily seen from Fig. 10.8a, which is a realization of the system represented by Eq. (10.56). In contrast, consider the original state equations [see Ex. 10.10] - -$$ -\dot{q}_1 = q_2 + x(t) \n\dot{q}_2 = -2q_1 - 3q_2 + 2x(t) -$$ - -A realization for these equations is shown in Fig. 10.8b. It can be seen from Fig. 10.8a that the states *z*1 and *z*2 are decoupled, whereas the states *q*1 and *q*2 (Fig. 10.8b) are coupled. It should be remembered that Figs. 10.8a and 10.8b are realizations of the same system.† - -**Figure 10.8** Two realizations of the second-order system. - -† Here we have only a simulated state equation; the outputs are not shown. The outputs are linear combinations of state variables (and inputs). Hence, the output equation can be easily incorporated into these diagrams. - -### MATRIX DIAGONALIZATION VIA MATLAB - -The key to diagonalizing matrix **A** is to determine a matrix **P** that satisfies **P** = **PA** [Eq. (10.55)], where is a diagonal matrix of the eigenvalues of **A**. This problem is directly related to the classic eigenvalue problem, stated as - -### **AV** = **V** - -where **V** is a matrix of eigenvectors for **A**. If we can find **V**, we can take its inverse to determine **P**. That is, **P** = **V**−1 . This relationship is more fully developed in Sec. 10.8. - -MATLAB's built-in function eig can determine the eigenvectors of a matrix and, therefore, can help us determine a suitable matrix **P**. Let us demonstrate this approach for the current case. - -``` ->> A = [0 1;-2 -3]; B = [1; 2]; ->> [V, Lambda] = eig(A); ->> P = inv(V), Lambda, Bhat = P*B - P = 2.8284 1.4142 - 2.2361 2.2361 - Lambda = -1 0 - 0 -2 - Bhat = 5.6569 - 6.7082 -``` - -Therefore, - -$$ -\mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} 2.8284 & 1.4142 \\ 2.2361 & 2.2361 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} = \mathbf{P}\mathbf{q} -$$ - -and - -$$ -\dot{\mathbf{z}} = \begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \begin{bmatrix} 5.6569 \\ 6.7082 \end{bmatrix} x(t) = \Lambda \mathbf{z} + \hat{\mathbf{B}} \mathbf{x} -$$ - -Recall that neither **P** nor **B**ˆ are unique, which explains why the MATLAB output does not need to match our previous solution. Still, the MATLAB results do their job and successfully diagonalize matrix **A**. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/128_10.6 CONTROLLABILITY AND OBSERVABILITY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/128_10.6 CONTROLLABILITY AND OBSERVABILITY.md deleted file mode 100644 index c585c7c60376978d093d71a719d29ab5084e8b9d..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/128_10.6 CONTROLLABILITY AND OBSERVABILITY.md +++ /dev/null @@ -1,242 +0,0 @@ -## **[10.6 CONTROLLABILITY AND](#page-14-0) OBSERVABILITY** - -Consider a diagonalized state-space description of a system - -$$ -\dot{z} = \Lambda z + \dot{B}x \quad \text{and} \quad Y = \dot{C}z + Dx \tag{10.57} -$$ - -We shall assume that all *N* eigenvalues λ1, λ2, ... , λ*N* are distinct. The state equations in Eq. (10.57) are of the form - -$$ -\dot{z}_m = \lambda_m z_m + \hat{b}_{m1} x_1 + \hat{b}_{m2} x_2 + \cdots + \hat{b}_{mj} x_j \qquad m = 1, 2, \ldots, N -$$ - -### 948 CHAPTER 10 STATE-SPACE ANALYSIS - -If *b*ˆ*m*1, *b*ˆ*m*2, ... , *b*ˆ*mj* (the *m*th row in matrix **B**ˆ ) are all zero, then - -$$ -\dot{z}_m = \lambda_m z_m -$$ - -and the variable *zm* is uncontrollable because *zm* is not coupled to any of the inputs. Moreover, *zm* is decoupled from all the remaining (*N* − 1) state variables because of the diagonalized nature of the variables. Hence, there is no direct or indirect coupling of *zm* with any of the inputs, and the system is uncontrollable. In contrast, if at least one element in the *m*th row of **B**ˆ is nonzero, *zm* is coupled to at least one input and is therefore controllable. *Thus, a system with a diagonalized state [Eq. (10.57)] is completely controllable if and only if the matrix* **B**ˆ *has no row of zero elements*. - -The outputs [see Eq. (10.57)] are of the form - -$$ -y_i = \hat{c}_{i1}z_1 + \hat{c}_{i2}z_2 + \cdots + \hat{c}_{iN}z_N + \sum_{m=1}^j d_{im}x_m -$$ - $i = 1, 2, ..., k$ - -If *c*ˆ*im* =0, then the state *zm* will not appear in the expression for *yi*. Since all the states are decoupled because of the diagonalized nature of the equations, the state *zm* cannot be observed directly or indirectly (through other states) at the output *yi*. Hence, the *m*th mode *e*λ*mt* will not be observed at the output *yi*. If *c*ˆ1*m*, *c*ˆ2*m*, ... , *c*ˆ*km* (the *m*th column in matrix **C**ˆ ) are all zero, the state *zm* will not be observable at any of the *k* outputs, and the state *zm* is unobservable. In contrast, if at least one element in the *m*th column of **C**ˆ is nonzero, *zm* is observable at least at one output. *Thus, a system with diagonalized equations of the form in Eq. (10.57) is completely observable if and only if the matrix* **C**ˆ *has no column of zero elements*. In this discussion, we assumed distinct eigenvalues; for repeated eigenvalues, the modified criteria can be found in the literature [1, 2]. - -If the state-space description is not in diagonalized form, it may be converted into diagonalized form using the procedure in Ex. 10.11. It is also possible to test for controllability and observability even if the state-space description is in undiagonalized form [1, 2]. - -### **EXAMPLE 10.12 Controllability and Observability** - -Investigate the controllability and observability of the systems in Fig. 10.9. - -In both cases, the state variables are identified as the two integrator outputs, *q*1 and *q*2. The state equations for the system in Fig. 10.9a are - -$$ -\dot{q}_1 = q_1 + x \n\dot{q}_2 = q_1 - q_2 -$$ -\n(10.58) - -and - -$$ -y = \dot{q}_2 - q_2 = q_1 - 2q_2 -$$ - -**Figure 10.9** Systems for Ex. 10.12. - -Hence, - -$$ -\mathbf{A} = \begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 1 & -2 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ -$$ -|\mathbf{sI} - \mathbf{A}| = \begin{vmatrix} s-1 & 0 \\ -1 & s+1 \end{vmatrix} = (s-1)(s+1) -$$ - -Therefore, - -$$ -\lambda_1 = 1 \qquad \text{and} \qquad \lambda_2 = -1 -$$ - -and - -$$ -\mathbf{\Lambda} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} -$$ - -We shall now use the procedure in Sec. 10.5-1 to diagonalize this system. According to Eq. (10.55), we have - -$$ -\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & -1 \end{bmatrix} -$$ - -The solution of this equation yields - -$$ -p_{12} = 0 -$$ - and $-2p_{21} = p_{22}$ - -Choosing *p*11 = 1 and *p*21 = 1, we have - -$$ -\mathbf{P} = \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix} -$$ - -and - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} -$$ - -All the rows of **B**ˆ are nonzero. Hence, the system is controllable. Also, - -$$ -\mathbf{Y} = \mathbf{C}\mathbf{q} = \mathbf{C}\mathbf{P}^{-1}\mathbf{z} = \hat{\mathbf{C}}\mathbf{z} -$$ - -and - -$$ -\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} = \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 \end{bmatrix} -$$ - -The first column of **C**ˆ is zero. Hence, the mode *z*1 (corresponding to λ1 =1) is unobservable. The system is therefore controllable but not observable. We come to the same conclusion by realizing the system with the diagonalized state variables *z*1 and *z*2, whose state equations are - -$$ -\dot{z} = \Lambda z + Bx -$$ -$$ -y = \hat{C}z -$$ - -Using our previous calculations, we have - -$$ -\begin{aligned}\n\dot{z}_1 &= z_1 + x \\ -\dot{z}_2 &= -z_2 + x\n\end{aligned} -$$ - -and - -$$ -y\,{=}\,z_2 -$$ - -Figure 10.10a shows a realization of these equations. It is clear that each of the two modes is controllable, but the first mode (corresponding to λ = 1) is not observable at the output. - -The state equations for the system in Fig. 10.9b are - -$$ -\dot{q}_1 = -q_1 + x \n\dot{q}_2 = \dot{q}_1 - q_1 + q_2 = -2q_1 + q_2 + x -$$ -\n(10.59) - -and - -*y* = *q*2 - -Hence, - -$$ -\mathbf{A} = \begin{bmatrix} -1 & 0 \\ -2 & 1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ - -$$ -|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s+1 & 0 \\ -1 & s-1 \end{vmatrix} = (s+1)(s-1) -$$ - - $x_2 = 1$ , and - -so that λ1 = −1, λ2 = 1, and - -$$ -\Lambda = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} -$$ - -Diagonalizing the matrix, we have - -$$ -\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} -1 & 0 \ -2 & 1 \end{bmatrix} -$$ - -The solution of this equation yields *p*11 = −*p*12 and *p*22 = 0. Choosing *p*11 = −1 and *p*21 = 1, we obtain - -$$ -\mathbf{P} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} -$$ - -and - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \end{bmatrix} -$$ - -The first row of **B**ˆ is zero. Hence, the mode corresponding to λ1 = 1 is not controllable. However, since none of the columns of **C**ˆ vanish, both modes are observable at the output. Hence the system is observable but not controllable. - -We reach the same conclusion by realizing the system with the diagonalized state variables *z*1 and *z*2. The two state equations are - -$$ -\dot{z} = \Lambda z + \dot{B}x -$$ -$$ -y = \hat{C}z -$$ - -Using our previous calculations, we have - -$$ -\begin{aligned}\n\dot{z}_1 &= z_1\\ \n\dot{z}_2 &= -z_2 + x\n\end{aligned} -$$ - -and thus, - -*y* = *z*1 +*z*2 - -Figure 10.10b shows a realization of these equations. Clearly, each of the two modes is observable at the output, but the mode corresponding to λ1 = 1 is not controllable. - -### USING MATLAB TO DETERMINE CONTROLLABILITY AND OBSERVABILITY - -As demonstrated in Ex. 10.11, we can use MATLAB's eig function to determine the matrix **P** that will diagonalize **A**. We can then use **P** to determine **B**ˆ and **C**ˆ , from which we can determine the controllability and observability of a system. Let us demonstrate the process for the two present systems. - -First, let us use MATLAB to compute **B**ˆ and **C**ˆ for the system in Fig. 10.9a. - ->> A = [1 0;1 -1]; B = [1; 0]; C = [1 -2]; >> [V, Lambda] = eig(A); P=inv(V); Bhat = P\*B, Chat = C\*inv(P) Bhat = -0.5000 1.1180 Chat = -2 0 - -Since all the rows of **B**ˆ are nonzero, the system is controllable. However, one column of **C**ˆ is zero, so one mode is unobservable. - -Next, let us use MATLAB to compute **B**ˆ and **C**ˆ for the system in Fig. 10.9b. - ->> A = [-1 0;-2 1]; B = [1; 1]; C = [0 1]; >> [V, Lambda] = eig(A); P=inv(V); Bhat = P\*B, Chat = C\*inv(P) Bhat = 0 1.4142 Chat = 1.0000 0.7071 - -One of the rows of **B**ˆ is zero, so one mode is uncontrollable. Since all of the columns of **C**ˆ are nonzero, the system is observable. - -As expected, the MATLAB results confirm our earlier conclusions regarding the controllability and observability of the systems of Fig. 10.9. - -### **[10.6-1 Inadequacy of the Transfer Function Description of a System](#page-14-0)** - -Example 10.12 demonstrates the inadequacy of the transfer function to describe an LTI system in general. The systems in Figs. 10.9a and 10.9b both have the same transfer function - -$$ -H(s) = \frac{1}{s+1} -$$ - -Yet the two systems are very different. Their true nature is revealed in Figs. 10.10a and 10.10b, respectively. Both the systems are unstable, but their transfer function *H*(*s*) = 1/(*s* + 1) does not give any hint of it. Moreover, the systems are very different from the viewpoint of controllability and observability. The system in Fig. 10.9a is controllable but not observable, whereas the system in Fig. 10.9b is observable but not controllable. - -The transfer function description of a system looks at a system only from the input and output terminals. Consequently, the transfer function description can specify only the part of the system that is coupled to the input and the output terminals. From Figs. 10.10a and 10.10b, we see that in both cases only a part of the system that has a transfer function *H*(*s*) = 1/(*s* + 1) is coupled to the input and the output terminals. This is why both systems have the same transfer function *H*(*s*) = 1/(*s*+1). - -The state variable description [Eqs. (10.58) and (10.59)], on the other hand, contains all the information about these systems to describe them completely. The reason is that the state variable description is an internal description, not the external description obtained from the system behavior at external terminals. - -Apparently, the transfer function fails to describe these systems completely because the transfer functions of these systems have a common factor *s*−1 in the numerator and denominator; this common factor is canceled out in the systems in Fig. 10.9, with a consequent loss of the information. Such a situation occurs when a system is uncontrollable and/or unobservable. If a system is both controllable and observable (which is the case with most of the practical systems) the transfer function describes the system completely. In such a case, the internal and external descriptions are equivalent. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/129_10.7 STATE-SPACE ANALYSIS OF DISCRETE-TIME SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/129_10.7 STATE-SPACE ANALYSIS OF DISCRETE-TIME SYSTEMS.md deleted file mode 100644 index ae6369290315ea332b78d5ce14511f898dc101f1..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/129_10.7 STATE-SPACE ANALYSIS OF DISCRETE-TIME SYSTEMS.md +++ /dev/null @@ -1,414 +0,0 @@ -## **[10.7 STATE-SPACE](#page-15-0) ANALYSIS OF DISCRETE-TIME SYSTEMS** - -We have shown that an *N*th-order differential equation can be expressed in terms of *N* first-order differential equations. In the following analogous procedure, we show that a general *N*th-order difference equation can be expressed in terms of *N* first-order difference equations. - -Consider the *z*-transfer function - -$$ -H[z] = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N} -$$ - -The input *x*[*n*] and the output *y*[*n*] of this system are related by the difference equation - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] = (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] -$$ - -The DFII realization of this equation is illustrated in Fig. 10.11. - -Signals appearing at the outputs of *N* delay elements are denoted by *q*1[*n*], *q*2[*n*], ... , *qN*[*n*]. The input of the first delay is *qN*[*n* + 1]. We can now write *N* equations, one at the input of each delay: - -$$ -q_1[n+1] = q_2[n] -$$ - -\n -$$ -q_2[n+1] = q_3[n] -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -q_{N-1}[n+1] = q_N[n] -$$ - -\n -$$ -q_N[n+1] = -a_Nq_1[n] - a_{N-1}q_2[n] - \dots - a_1q_N[n] + x[n] -$$ -\n(10.60) - -and - -*y*[*n*] = *bNq*1[*n*] +*bN*−1*q*2[*n*]+···+*b*1*qN*[*n*] +*b*0*qN*+1[*n*] - -We can eliminate *qN*+1[*n*] from this equation by using the last equation in Eq. (10.60) to yield - -$$ -y[n] = (b_N - b_0 a_N)q_1[n] + (b_{N-1} - b_0 a_{N-1})q_2[n] + \dots + (b_1 - b_0 a_1)q_N[n] + b_0 x[n] -$$ - -= $\hat{b}_N q_1[n] + \hat{b}_{N-1} q_2[n] + \dots + \hat{b}_1 q_N[n] + b_0 x[n]$ (10.61) - -where *b*ˆ*i* = *bi* −*b*0*ai*. - -Equation (10.60) shows *N* first-order difference equations in *N* variables *q*1[*n*], *q*2[*n*], ... , *qN*[*n*]. These variables should immediately be recognized as state variables, since the specification of the initial values of these variables in Fig. 10.11 will uniquely determine the response *y*[*n*] for a given *x*[*n*]. Thus, Eq. (10.60) represents the state equations, and Eq. (10.61) is the output equation. In matrix form, we can write these equations as - -$$ -\begin{bmatrix} q_{1}[n+1] \\ q_{2}[n+1] \\ \vdots \\ q_{N-1}[n+1] \\ \hline q_{N}[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ -a_{N} & -a_{N-1} & -a_{N-2} & \cdots & -a_{2} & -a_{1} \end{bmatrix} \begin{bmatrix} q_{1}[n] \\ q_{2}[n] \\ \vdots \\ q_{N-1}[n] \\ \hline q_{N}[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix} x[n] \quad (10.62) -$$ - -and - -$$ -\mathbf{y}[n] = \underbrace{\begin{bmatrix} \hat{b}_N & \hat{b}_{N-1} & \cdots & \hat{b}_1 \end{bmatrix}}_{C} \begin{bmatrix} q_1[n] \\ q_2[n] \\ \vdots \\ q_N[n] \end{bmatrix} + \underbrace{b_0}_{D} x[n] \qquad (10.63) -$$ - -In general, - -$$ -\mathbf{q}[n+1] = \mathbf{A}\mathbf{q}[n] + \mathbf{B}\mathbf{x}[n] -$$ -$$ -\mathbf{y}[n] = \mathbf{C}\mathbf{q}[n] + \mathbf{D}\mathbf{x}[n] -$$ - -Here we have represented a discrete-time system with state equations for DFII form. There are several other possible representations, as discussed in Sec. 10.3. We may, for example, use the cascade, parallel, or transpose of DFII forms to realize the system, or we may use some linear transformation of the state vector to realize other forms. In all cases, the output of each delay element qualifies as a state variable. We then write the equation at the input of each delay element. The *N* equations thus obtained are the *N* state equations. - -### **[10.7-1 Solution in State Space](#page-15-0)** - -Consider the state equation - -$$ -\mathbf{q}[n+1] = \mathbf{A}\mathbf{q}[n] + \mathbf{B}\mathbf{x}[n] -$$ - -From this equation, it follows that - -$$ -\mathbf{q}[n] = \mathbf{A}\mathbf{q}[n-1] + \mathbf{B}\mathbf{x}[n-1] -$$ - -and - -$$ -q[n-1] = Aq[n-2] + Bx[n-2] -$$ - -\n -$$ -q[n-2] = Aq[n-3] + Bx[n-3] -$$ - -\n: -\n: -\n -$$ -q[1] = Aq[0] + Bx[0] -$$ - -#### 956 CHAPTER 10 STATE-SPACE ANALYSIS - -Substituting the expression for **q**[*n*−1] into that for **q**[*n*], we obtain - -$$ -\mathbf{q}[n] = \mathbf{A}^2 \mathbf{q}[n-2] + \mathbf{A} \mathbf{B} \mathbf{x}[n-2] + \mathbf{B} \mathbf{x}[n-1] -$$ - -Substituting the expression for **q**[*n*−2] in this equation, we obtain - -$$ -q[n] = A^{3}q[n-3] + A^{2}Bx[n-3] + ABx[n-2] + Bx[n-1] -$$ - -Continuing in this way, we obtain - -$$ -\mathbf{q}[n] = \mathbf{A}^n \mathbf{q}[0] + \mathbf{A}^{n-1} \mathbf{B} \mathbf{x}[0] + \mathbf{A}^{n-2} \mathbf{B} \mathbf{x}[1] + \dots + \mathbf{B} \mathbf{x}[n-1] -$$ - -= $\mathbf{A}^n \mathbf{q}[0] + \sum_{m=0}^{n-1} \mathbf{A}^{n-1-m} \mathbf{B} \mathbf{x}[m]$ - -The upper limit of this summation is nonnegative. Hence, *n* ≥ 1, and the summation is recognized as the convolution sum - -$$ -\mathbf{A}^{n-1}u[n-1]*\mathbf{B}\mathbf{x}[n] -$$ - -Consequently, - -$$ -\mathbf{q}[n] = \underbrace{\mathbf{A}^n \mathbf{q}[0]}_{\text{zero input}} + \underbrace{\mathbf{A}^{n-1} u[n-1] * \mathbf{B} \mathbf{x}[n]}_{\text{zero state}} -$$ -(10.64) - -and - -$$ -\mathbf{y}[n] = \mathbf{C}\mathbf{q} + \mathbf{D}\mathbf{x} -$$ - -= $\mathbf{C}\mathbf{A}^n \mathbf{q}[0] + \sum_{m=0}^{n-1} \mathbf{C}\mathbf{A}^{n-1-m} \mathbf{B}\mathbf{x}[m] + \mathbf{D}\mathbf{x}$ -= $\mathbf{C}\mathbf{A}^n \mathbf{q}[0] + \mathbf{C}\mathbf{A}^{n-1}u[n-1] * \mathbf{B}\mathbf{x}[n] + \mathbf{D}\mathbf{x}$ (10.65) - -In Sec. 10.1-3, we showed that - -$$ -\mathbf{A}^n = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{N-1} \mathbf{A}^{N-1} -$$ - (10.66) - -where (assuming *N* distinct eigenvalues of **A**) - -$$ -\begin{bmatrix}\n\beta_0 \\ -\beta_1 \\ -\vdots \\ -\beta_{N-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{N-1} \\ -1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{N-1} \\ -\vdots & \vdots & \vdots & \cdots & \vdots \\ -1 & \lambda_N & \lambda_N^2 & \cdots & \lambda_N^{N-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\n\lambda_1^n \\ -\lambda_2^n \\ -\vdots \\ -\lambda_N^n\n\end{bmatrix} -$$ -\n(10.67) - -and λ1, λ2, ... , λ*N* are the *N* eigenvalues of **A**. - -We can also determine **A***n* from the *z*-transform formula, which will be derived later, in Eq. (10.71): - -$$ -\mathbf{A}^n = \mathcal{Z}^{-1}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}] -$$ - -### **EXAMPLE 10.13 State-Space Analysis of a Discrete-Time System** - -Give a state-space description of the system in Fig. 10.12. Find the output *y*[*n*] if the input *x*[*n*] = *u*[*n*] and the initial conditions are *q*1[0] = 2 and *q*2[0] = 3. - -6 **Figure 10.12** System for Ex. 10.13. - -Recognizing that *q*2[*n*] = *q*1[*n*+1], the state equations are [see Eqs. (10.62) and (10.63)] - -$$ -\begin{bmatrix} q_1[n+1] \\ q_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & \frac{5}{6} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} x -$$ - -and - -$$ -y[n] = \begin{bmatrix} -1 & 5 \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} -$$ - -To find the solution [Eq. (10.65)], we must first determine **A***n* . The characteristic equation of **A** is - -$$ -|\lambda \mathbf{I} - \mathbf{A}| = \begin{vmatrix} \lambda & -1 \\ \frac{1}{6} & \lambda - \frac{5}{6} \end{vmatrix} = \lambda^2 - \frac{5}{6}\lambda + \frac{1}{6} = \left(\lambda - \frac{1}{3}\right)\left(\lambda - \frac{1}{2}\right) = 0 -$$ - -Hence, λ1 = 1/3 and λ2 = 1/2 are the eigenvalues of **A** and [see Eq. (10.66)] - -$$ -\mathbf{A}^n = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} -$$ - -where [see Eq. (10.67)] - -$$ -\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{3} \\ 1 & \frac{1}{2} \end{bmatrix}^{-1} \begin{bmatrix} \left(\frac{1}{3}\right)^n \\ \left(\frac{1}{2}\right)^n \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -6 & 6 \end{bmatrix} \begin{bmatrix} (3)^{-n} \\ (2)^{-n} \end{bmatrix} = \begin{bmatrix} 3(3)^{-n} - 2(2)^{-n} \\ -6(3)^{-n} + 6(2)^{-n} \end{bmatrix} -$$ - -and - -$$ -\mathbf{A}^{n} = [3(3)^{-n} - 2(2)^{-n}] \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + [-6(3)^{-n} + 6(2)^{-n}] \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & \frac{5}{6} \end{bmatrix} -$$ - -= -$$ -\begin{bmatrix} 3(3)^{-n} - 2(2)^{-n} & -6(3)^{-n} + 6(2)^{-n} \\ (3)^{-n} - (2)^{-n} & -2(3)^{-n} + 3(2)^{-n} \end{bmatrix} -$$ -(10.68) - -We can now determine the state vector **q**[*n*] from Eq. (10.64). Since we are interested in the output *y*[*n*], we shall use Eq. (10.65) directly. Note that - -$$ -CAn = [-1 \quad 5]An = [2(3)-n - 3(2)-n -4(3)-n + 9(2)-n] -$$ - -and the zero-input response is **CA***n* **q**[0], with - -$$ -\mathbf{q}[0] = \begin{bmatrix} 2 \\ 3 \end{bmatrix} -$$ - -Hence, the zero-input response is - -$$ -CAnq[0] = -8(3)-n + 21(2)-n -$$ - -The zero-state component is given by the convolution sum of **CA***n*−1 *u*[*n*−1] and **Bx**[*n*]. We can use the shifting property of the convolution sum [Eq. (3.32)] to obtain the zero-state component by finding the convolution sum of **CA***n u*[*n*] and **Bx**[*n*] and then replacing *n* with *n* − 1 in the result. We use this procedure because the convolution sums are listed in Table 3.1 for functions of the type *x*[*n*]*u*[*n*], rather than *x*[*n*]*u*[*n*−1]. - -$$ -\mathbf{CA}^{n}u[n] * \mathbf{B}x[n] = [2(3)^{-n} - 3(2)^{-n} - 4(3)^{-n} + 9(2)^{-n}] * \begin{bmatrix} 0 \\ u[n] \end{bmatrix} -$$ - -= -4(3)^{-n} \* u[n] + 9(2)^{-n} \* u[n] - -Using Table 3.1 (pair 4), we obtain - -$$ -\mathbf{CA}^{n}u[n] * \mathbf{B}x[n] = -4 \left[ \frac{1 - 3^{-(n+1)}}{1 - \frac{1}{3}} \right] u[n] + 9 \left[ \frac{1 - 2^{-(n+1)}}{1 - \frac{1}{2}} \right] u[n] -$$ -$$ -= [12 + 6(3^{-(n+1)}) - 18(2^{-(n+1)})]u[n] -$$ - -Now the desired (zero-state) response is obtained by replacing *n* by *n*−1. Hence, - -$$ -CAnu[n] * Bx[n-1] = [12+6(3)-n - 18(2)-n]u[n-1] -$$ - -It follows that - -$$ -y[n] = [-8(3)^{-n} + 21(2)^{-n}u[n] + [12 + 6(3)^{-n} - 18(2)^{-n}]u[n-1] -$$ - -This is the desired answer. We can simplify this answer by observing that 12 + 6(3)−*n* − 18(2)−*n* = 0 for *n* = 0. Hence, *u*[*n*−1] may be replaced by *u*[*n*], and - -$$ -y[n] = [12 - 2(3)^{-n} + 3(2)^{-n}]u[n] -$$ -\n(10.69) - -### USING MATLAB TO OBTAIN A GRAPHICAL SOLUTION - -MATLAB is equipped with tools to simulate digital systems, which makes it easy to obtain a graphical solution to the system. Let us use MATLAB simulation to determine the total system output over 0 ≤ *n* ≤ 25. - -- >> A = [0 1;-1/6 5/6]; B = [0; 1]; C = [-1 5]; D = 0; -- >> N = 25; n = (0:N); x = ones(1,N+1); q0 = [2;3]; -- >> sys = ss(A,B,C,D,-1); % Discrete-time state space model -- >> [y,q] = lsim(sys,x,n,q0); % Simulate output and state vector -- >> clf; stem(n,y,'k.'); xlabel('n'); ylabel('y[n]'); axis([-.5 25.5 11.5 13.5]); - -The MATLAB results, shown in Fig. 10.13, exactly align with the analytical solution derived earlier. Also notice that the zero-input and zero-state responses can be separately obtained using the same code and respectively setting either x or q0 to zero. - -## **10.7-2 The** *z***[-Transform Solution](#page-15-0)** - -The *z*-transform of Eq. (10.62) is given by - -$$ -z\mathbf{Q}[z] - z\mathbf{q}[0] = \mathbf{A}\mathbf{Q}[z] + \mathbf{B}\mathbf{X}[z] -$$ - -Therefore, - -$$ -(z\mathbf{I} - \mathbf{A})\mathbf{Q}[z] = z\mathbf{q}[0] + \mathbf{B}\mathbf{X}[z] -$$ - -and - -$$ -Q[z] = (zI - A)^{-1}zq[0] + (zI - A)^{-1}BX[z] -$$ -$$ -= (I - z^{-1}A)^{-1}q[0] + (zI - A)^{-1}BX[z] -$$ - -Hence, - -$$ -\mathbf{q}[n] = \underbrace{\mathcal{Z}^{-1}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}]\mathbf{q}[0]}_{\text{zero-input response}} + \underbrace{\mathcal{Z}^{-1}[(z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B}\mathbf{X}[z]]}_{\text{zero-state response}} -$$ -(10.70) - -#### 960 CHAPTER 10 STATE-SPACE ANALYSIS - -A comparison of Eq. (10.70) with Eq. (10.64) shows that - -$$ -\mathbf{A}^{n} = \mathcal{Z}^{-1}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}] -$$ - (10.71) - -The output equation is given by - -$$ -\mathbf{Y}[z] = \mathbf{C}\mathbf{Q}[z] + \mathbf{D}\mathbf{X}[z] -$$ - -= $\mathbf{C}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0] + (z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B}\mathbf{X}[z]] + \mathbf{D}\mathbf{X}[z]$ -= $\mathbf{C}(\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0] + [\mathbf{C}(z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B} + \mathbf{D}]\mathbf{X}[z]$ -= $\underbrace{\mathbf{C}(\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0]}_{\text{zero-input response}} + \underbrace{\mathbf{H}[z]\mathbf{X}[z]}_{\text{zero-state response}}$ (10.72) - -where - -$$ -\mathbf{H}[z] = \mathbf{C}(z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B} + \mathbf{D} -$$ - (10.73) - -Note that **H**[*z*] is the transfer function matrix of the system, and *Hij*[*z*], the *ij*th element of **H**[*z*], is the transfer function relating the output *yi*[*n*] to the input *xj*[*n*]. If we define **h**[*n*] as - -$$ -\mathbf{h}[n] = \mathcal{Z}^{-1}[\mathbf{H}[z]] -$$ - -then **h**[*n*] represents the unit impulse function response matrix of the system. Thus, *hij*[*n*], the *ij*th element of **h**[*n*], represents the zero-state response *yi*[*n*] when the input *xj*[*n*] = δ[*n*] and all other inputs are zero. - -### **EXAMPLE 10.14** *z***-Transform Solution to State Equations** - -Use the *z*-transform to find the response *y*[*n*] for the system in Ex. 10.13. - -According to Eq. (10.72), - -$$ -\mathbf{Y}[z] = [-1 \quad 5] \left[ \frac{1}{\frac{1}{6z}} \quad \frac{-\frac{1}{z}}{1 - \frac{z}{6z}} \right]^{-1} \left[ \frac{2}{3} \right] + [-1 \quad 5] \left[ \frac{z}{\frac{1}{6}} \quad \frac{-1}{z - \frac{5}{6}} \right]^{-1} \left[ \frac{0}{\frac{z}{z - 1}} \right] -$$ -\n -$$ -= [-1 \quad 5] \left[ \frac{\frac{z(6z - 5)}{6z^2 - 5z + 1}}{\frac{-z}{6z^2 - 5z + 1}} \quad \frac{\frac{6z}{6z^2 - 5z + 1}}{\frac{6z^2 - 5z + 1}{6z^2 - 5z + 1}} \right] \left[ \frac{2}{3} \right] + [-1 \quad 5] \left[ \frac{\frac{z}{(z - 1)(z^2 - \frac{5}{6}z + \frac{1}{6})}}{\frac{z^2}{(z - 1)(z^2 - \frac{5}{6}z + \frac{1}{6})}} \right] -$$ -\n -$$ -= \frac{13z^2 - 3z}{z^2 - \frac{5}{6}z + \frac{1}{6}} + \frac{(5z - 1)z}{(z - 1)(z^2 - \frac{5}{6}z + \frac{1}{6})} -$$ -\n -$$ -= \frac{-8z}{z - \frac{1}{3}} + \frac{21z}{z - \frac{1}{2}} + \frac{12z}{z - 1} + \frac{12z}{z - 1} + \frac{6z}{z - \frac{1}{3}} - \frac{18z}{z - \frac{1}{2}} -$$ - -Therefore, - -$$ -y[n] = \underbrace{[-8(3)^{-n} + 21(2)^{-n}}_{\text{zero-input response}} + \underbrace{12 + 6(3)^{-n} - 18(2)^{-n}}_{\text{zero-state response}}]u[n] -$$ - -### LINEAR TRANSFORMATION, CONTROLLABILITY, AND OBSERVABILITY - -The procedure for linear transformation is parallel to that in the continuous-time case (Sec. 10.5). If **w** is the transformed-state vector given by - -**w** = **Pq** - -then - -$$ -\mathbf{w}[n+1] = \mathbf{P}\mathbf{A}\mathbf{P}^{-1}\mathbf{w}[n] + \mathbf{P}\mathbf{B}\mathbf{x} -$$ - -and - -$$ -\mathbf{y}[n] = (\mathbf{C}\mathbf{P}^{-1})\mathbf{w} + \mathbf{D}\mathbf{x} -$$ - -Controllability and observability may be investigated by diagonalizing the matrix, as explained in Sec. 10.5-1. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/130_10.8 MATLAB - TOOLBOXES AND STATE-SPACE ANALYSIS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/130_10.8 MATLAB - TOOLBOXES AND STATE-SPACE ANALYSIS.md deleted file mode 100644 index 5409673a39102e7b09727007ed3d3ab14b42cf05..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/130_10.8 MATLAB - TOOLBOXES AND STATE-SPACE ANALYSIS.md +++ /dev/null @@ -1,353 +0,0 @@ -## **[10.8 MATLAB: TOOLBOXES AND](#page-15-0) STATE-SPACE ANALYSIS** - -The preceding MATLAB sections provide a comprehensive introduction to the basic MATLAB environment. However, MATLAB also offers a wide range of toolboxes that perform specialized tasks. Once installed, toolbox functions operate no differently from ordinary MATLAB functions. Although toolboxes are purchased at extra cost, they save time and offer the convenience of predefined functions. It would take significant effort to duplicate a toolbox's functionality by using custom user-defined programs. - -Three toolboxes are particularly appropriate in the study of signals and systems: the control system toolbox, the signal-processing toolbox, and the symbolic math toolbox. Functions from these toolboxes have been utilized throughout the text in the MATLAB examples as well as certain end-of-chapter problems. This section provides a more formal introduction to a selection of functions, both standard and toolbox, that are appropriate for state-space problems. - -## **10.8-1** *z***[-Transform Solutions to Discrete-Time, State-Space Systems](#page-15-0)** - -As with continuous-time systems, it is often more convenient to solve discrete-time systems in the transform domain rather than in the time domain. As given in Ex. 10.13, consider the state-space description of the system shown in Fig. 10.12. - -$$ -\begin{bmatrix} q_1[n+1] \\ q_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & \frac{5}{6} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} x[n] -$$ - -and - -$$ -y[n] = [-1 \quad 5] \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} -$$ - -We are interested in the output *y*[*n*] in response to the input *x*[*n*] = *u*[*n*] with initial conditions *q*1[0] = 2 and *q*2[0] = 3. - -To describe this system, the state matrices **A**, **B**, **C**, and **D** are first defined. - ->> A = [0 1;-1/6 5/6]; B = [0; 1]; C = [-1 5]; D = 0; - -Additionally, the vector of initial conditions is defined. - ->> q\_0 = [2;3]; - -In the transform domain, the solution to the state equation is - -$$ -\mathbf{Q}[z] = (\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0] + (z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B}\mathbf{X}[z] \tag{10.74} -$$ - -The solution is separated into two parts: the zero-input response and the zero-state response. - -MATLAB's symbolic toolbox makes possible a symbolic representation of Eq. (10.74). First, a symbolic variable *z* needs to be defined. - ->> z = sym('z'); - -The sym command is used to construct symbolic variables, objects, and numbers. Typing whos confirms that z is indeed a symbolic object. The syms command is a shorthand command for constructing symbolic objects. For example, syms z s is equivalent to the two instructions z = sym('z'); and s = sym('s');. - -Next, a symbolic expression for *X*[*z*] needs to be constructed for the unit step input, *x*[*n*] = *u*[*n*]. The *z*-transform is computed by means of the ztrans command. - ->> X = ztrans(sym('1')) X = z/(z-1) - -Several comments are in order. First, the ztrans command assumes a causal signal. For *n* ≥ 0, *u*[*n*] has a constant value of 1. Second, the argument of ztrans needs to be a symbolic expression, even if the expression is a constant. Thus, a symbolic one sym('1') is required. Also note that continuous-time systems use Laplace transforms rather than *z*-transforms. In such cases, the laplace command replaces the ztrans command. - -Construction of **Q**[*z*] is now trivial. - -$$ -\begin{aligned}\n&\text{Q} = \text{inv}\left(\text{eye}(2) - z^(-1) * A\right) * q_0 + \text{inv}\left(z * \text{eye}(2) - A\right) * B * X \\ -&\text{Q} = \\ -&\left(18 * z\right) / \left(6 * z^2 - 5 * z + 1\right) + \left(2 * z * \left(6 * z - 5\right)\right) / \left(6 * z^2 - 5 * z + 1\right) + \left(6 * z\right) / \left((z - 1) * \left(6 * z^2 - 5 * z + 1\right)\right) \\ -&\left(18 * z^2\right) / \left(6 * z^2 - 5 * z + 1\right) - \left(2 * z\right) / \left(6 * z^2 - 5 * z + 1\right) + \left(6 * z^2\right) / \left((z - 1) * \left(6 * z^2 - 5 * z + 1\right)\right)\n\end{aligned} -$$ - -Unfortunately, not all MATLAB functions work with symbolic objects. Still, the symbolic toolbox overloads many standard MATLAB functions, such as inv, to work with symbolic objects. Recall that overloaded functions have identical names but different behavior; proper function selection is typically determined by context. - -The expression Q is somewhat unwieldy. The simplify command uses various algebraic techniques to simplify the result. - -``` ->> Q = simplify(Q) - Q = -(2*z*(- 6*z^2 + 2*z + 1))/(6*z^3 - 11*z^2 + 6*z - 1) - (2*z*(9*z^2 - 7*z + 1))/(6*z^3 - 11*z^2 + 6*z - 1) -``` - -The resulting expression is mathematically equivalent to the original but notationally more compact. - -Since **D** = 0, the output *Y*[*z*] is given by *Y*[*z*] = **CQ**[*z*]. - -``` ->> Y = simplify(C*Q) - Y = (6*z*(13*z^2 - 11*z + 2))/(6*z^3 - 11*z^2 + 6*z - 1) -``` - -The corresponding time-domain expression is obtained by using the inverse *z*-transform command iztrans. - ->> y = iztrans(Y) y = 3\*(1/2)^n - 2\*(1/3)^n + 12 - -Like ztrans, the iztrans command assumes a causal signal, so the result implies multiplication by a unit step. That is, the system output is *y*[*n*] =(3(1/2)*n*−2(1/3)*n*+12)*u*[*n*], which is equivalent to Eq. (10.69) derived in Ex. 10.13. Continuous-time systems use inverse Laplace transforms rather than inverse *z*-transforms. In such cases, the ilaplace command therefore replaces the iztrans command. - -Following a similar procedure, it is a simple matter to compute the zero-input response *y*zir[*n*]: - ->> y\_zir = iztrans(simplify(C\*inv(eye(2)-z^(-1)\*A)\*q\_0)) y\_zir = 21\*(1/2)^n - 8\*(1/3)^n - -The zero-state response is given by - ->> y\_zsr=y- y\_zir y\_zsr = 6\*(1/3)^n - 18\*(1/2)^n + 12 - -Typing iztrans(simplify(C\*inv(z\*eye(2)-A)\*B\*X)) produces the same result. - -MATLAB plotting functions, such as plot and stem, do not directly support symbolic expressions. By using the subs command, however, it is easy to replace a symbolic variable with a vector of desired values. - -**Figure 10.14** Output *y*[*n*] computed by using the symbolic math toolbox. - -``` ->> n = [0:25]; stem(n,subs(y,n),'k.'); ->> xlabel('n'); ylabel('y[n]'); axis([-.5 25.5 11.5 13.5]); -``` - -Figure 10.14 shows the results, which are equivalent to the results obtained in Ex. 10.13. Although there are plotting commands in the symbolic math toolbox such as ezplot that plot symbolic expression, these plotting routines lack the flexibility needed to satisfactorily plot discrete-time functions. - -### **[10.8-2 Transfer Functions from State-Space Representations](#page-15-0)** - -A system's transfer function provides a wealth of useful information. From Eq. (10.73), the transfer function for the system described in Ex. 10.13 is - -``` ->> H = collect(simplify(C*inv(z*eye(2)-A)*B+D)) - H = (30*z - 6)/(6*z^2 - 5*z + 1) -``` - -It is also possible to determine the numerator and denominator transfer function coefficients from a state-space model by using the signal-processing toolbox function ss2tf. - ->> [num,den] = ss2tf(A,B,C,D) num = 0 5.0000 -1.0000 den = 1.0000 -0.8333 0.1667 - -The denominator of *H*[*z*] provides the characteristic polynomial - -γ 2 − 5 6γ + 1 6 - -Equivalently, the characteristic polynomial is the determinant of (*z***I**−**A**). - ->> syms gamma; char\_poly = subs(det(z\*eye(2)-A),z,gamma) char\_poly = gamma^2 - (5\*gamma)/6 + 1/6 - -Here, the subs command replaces the symbolic variable z with the desired symbolic variable gamma. - -The roots command does not accommodate symbolic expressions. Thus, the sym2poly command converts the symbolic expression into a polynomial coefficient vector suitable for the roots command. - ->> roots(sym2poly(char\_poly)) ans = 0.5000 0.3333 - -Taking the inverse *z*-transform of *H*[*z*] yields the impulse response *h*[*n*]. - ->> h = iztrans(H) h = 18\*(1/2)^n - 12\*(1/3)^n - 6\*kroneckerDelta(n, 0) - -As suggested by the characteristic roots, the characteristic modes of the system are (1/2)*n* and (1/3)*n*. Notice that the symbolic math toolbox represents δ[*n*] as kroneckerDelta(n, 0). In general, δ[*n* − *a*] is represented as kroneckerDelta(n-a, 0). This notation is frequently encountered. Consider, for example, delaying the input *x*[*n*] = *u*[*n*] by 2, *x*[*n* − 2] = *u*[*n* − 2]. In the transform domain, this is equivalent to *z*−2*X*[*z*]. Taking the inverse *z*-transform of *z*−2*X*[*z*] yields - -$$ -\Rightarrow \text{ 27 (--2)*X} = \text{ 27 (--2)*X} -$$ -\n -$$ -\text{ans} = 1 - \text{ 27 (--2)*X} = \text{ 27 (--2)*X} -$$ - -That is, MATLAB represents the delayed unit step *u*[*n*−2] as (−δ[*n*−1] −δ[*n*−0] +1)*u*[*n*]. The transfer function also permits convenient calculation of the zero-state response. - ->> -$$ -y\_zsr = iztrans(H*X) -$$ - -y\_ $zsr = 6*(1/3)^n - 18*(1/2)^n + 12$ - -The result agrees with previous calculations. - -### **[10.8-3 Controllability and Observability of Discrete-Time Systems](#page-15-0)** - -In their controllability and observability, discrete-time systems are analogous to continuous-time systems. For example, consider the LTID system described by the constant coefficient difference equation - -$$ -y[n] + \frac{5}{6}y[n-1] + \frac{1}{6}y[n-2] = x[n] + \frac{1}{2}x[n-1] -$$ - -Figure 10.15 illustrates the direct form II (DFII) realization of this system. The system input is *x*[*n*], the system output is *y*[*n*], and the outputs of the delay blocks are designated as state variables *q*1[*n*] and *q*2[*n*]. - -The corresponding state and output equations (see Prob. 10.7-1) are - -$$ -\mathbf{Q}[n+1] = \begin{bmatrix} q_1[n+1] \\ q_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & -\frac{5}{6} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} x[n] = \mathbf{A}\mathbf{Q}[n] + \mathbf{B}x[n] -$$ - -and - -$$ -y[n] = \begin{bmatrix} -\frac{1}{6} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + 1x[n] = \mathbf{CQ}[n] + \mathbf{D}x[n] -$$ - -To describe this system in MATLAB, the state matrices **A**, **B**, **C**, and **D** are first defined. - ->> A = -$$ -[0 \ 1; -1/6 \ -5/6] -$$ -; B = $[0; 1]$ ; C = $[-1/6 \ -1/3]$ ; D = 1; - -**Figure 10.15** Direct form II realization of *y*[*n*] + (5/6)*y*[*n* − 1] + (1/6)*y*[*n* − 2] = *x*[*n*] + (1/2)*x*[*n*−1]. - -To assess the controllability and observability of this system, the state matrix **A** needs to be diagonalized.† As shown in Eq. (10.55), this requires a transformation matrix **P** such that - -$$ -PA = AP -$$ - (10.75) - -where is a diagonal matrix containing the unique eigenvalues of **A**. Recall, the transformation matrix **P** is not unique. - -To determine a matrix **P**, it is helpful to review the eigenvalue problem. Mathematically, an eigendecomposition of **A** is expressed as - -$$ -AV = V\Lambda -$$ - -where **V** is a matrix of eigenvectors and is a diagonal matrix of eigenvalues. Pre- and post-multiplying both sides of this equation by **V**−1 yields - -$$ -\mathbf{V}^{-1}\mathbf{A}\mathbf{V}\mathbf{V}^{-1} = \mathbf{V}^{-1}\mathbf{V}\mathbf{\Lambda}\mathbf{V}^{-1} -$$ - -Simplification yields - -$$ -\mathbf{V}^{-1}\mathbf{A} = \mathbf{\Lambda}\mathbf{V}^{-1} \tag{10.76} -$$ - -Comparing Eqs. (10.75) and (10.76), we see that a suitable transformation matrix **P** is given by an inverse eigenvector matrix **V**−1 . - -The eig command is used to verify that **A** has the required distinct eigenvalues as well as compute the needed eigenvector matrix **V**. - -``` ->> [V,Lambda] = eig(A) - V = - 0.9487 -0.8944 - -0.3162 0.4472 - Lambda = - -0.3333 0 - 0 -0.5000 -``` - -Since the diagonal elements of Lambda are all unique, a transformation matrix **P** is given by - ->> P = inv(V); - -The transformed state matrices **A**ˆ = **PAP**−1 , **B**ˆ = **PB**, and **C**ˆ = **CP**−1 are easily computed by using transformation matrix **P**. Notice that matrix **D** is unaffected by state variable transformations. - -``` ->> Ahat = P*A*inv(P), Bhat = P*B, Chat = C*inv(P) - Ahat = - -0.3333 -0.0000 - 0.0000 -0.5000 -``` - -† This approach requires that the state matrix **A** have unique eigenvalues. Systems with repeated roots require that state matrix **A** be transformed into a modified diagonal form, also called the Jordan form. The MATLAB function jordan is used in these cases. - -``` -Bhat = - 6.3246 - 6.7082 -Chat = - -0.0527 -0.0000 -``` - -The proper operation of **P** is verified by the correct diagonalization of **A**, **A**ˆ = . Since no row of **B**ˆ is zero, the system is controllable. Since, however, at least one column of **C**ˆ is zero, the system is not observable. These characteristics are no coincidence. The DFII realization, which is more descriptively called the controller canonical form, is always controllable but not always observable. - -As a second example, consider the same system realized using the transposed direct form II structure (TDFII), as shown in Fig. 10.16. The system input is *x*[*n*], the system output is *y*[*n*], and the outputs of the delay blocks are designated as state variables *v*1[*n*] and *v*2[*n*]. - -The corresponding state and output equations (see Prob. 10.7-2) are - -$$ -\mathbf{V}[n+1] = \begin{bmatrix} v_1[n+1] \\ v_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{6} \\ 1 & -\frac{5}{6} \end{bmatrix} \begin{bmatrix} v_1[n] \\ v_2[n] \end{bmatrix} + \begin{bmatrix} -\frac{1}{6} \\ -\frac{1}{3} \end{bmatrix} x[n] = \mathbf{A}\mathbf{V}[n] + \mathbf{B}x[n] -$$ - -and - -$$ -y[n] = [0 \quad 1] \begin{bmatrix} v_1[n] \\ v_2[n] \end{bmatrix} + 1x[n] = \mathbf{CV}[n] + \mathbf{D}x[n] -$$ - -To describe this system in MATLAB, the state matrices **A**, **B**, **C**, and **D** are defined. - ->> A = -$$ -[0 -1/6; 1 -5/6] -$$ -; B = $[-1/6; -1/3]$ ; C = $[0 1]$ ; D = 1; - -To diagonalize **A**, a transformation matrix **P** is created. - ->> [V, Lambda] = eig(A) -\n -$$ -V = -$$ - 0.4472 0.3162 -\n0.8944 0.9487 -\nLambda = -\n-0.3333 0 -\n0 -0.5000 - -6 – **Figure 10.16** Transposed direct form II realization of *y*[*n*]+(5/6)*y*[*n*−1]+(1/6)*y*[*n*−2] = *x*[*n*]+(1/2)*x*[*n*−1]. - -The characteristic modes of a system do not depend on implementation, so the eigenvalues of the DFII and TDFII realizations are the same. However, the eigenvectors of the two realizations are quite different. Since the transformation matrix **P** depends on the eigenvectors, different realizations can possess different observability and controllability characteristics. - -Using transformation matrix **P**, the transformed state matrices **A**ˆ = **PAP**−1 , **B**ˆ = **PB**, and **C**ˆ = **CP**−1 are computed. - -``` ->> P = inv(V); ->> Ahat = P*A*inv(P), Bhat = P*B, Chat = C*inv(P) - Ahat = - -0.3333 0 - 0.0000 -0.5000 - Bhat = - -0.3727 - -0.0000 - Chat = - 0.8944 0.9487 -``` - -Again, the proper operation of **P** is verified by the correct diagonalization of **A**, **A**ˆ = . Since no column of **C**ˆ is zero, the system is observable. However, at least one row of **B**ˆ is zero, and therefore the system is not controllable. The TDFII realization, which is more descriptively called the observer canonical form, is always observable but not always controllable. It is interesting to note that the properties of controllability and observability are influenced by the particular realization of a system. - -### **[10.8-4 Matrix Exponentiation and the Matrix Exponential](#page-15-0)** - -Matrix exponentiation is important to many problems, including the solution of discrete-time state-space equations. Equation (10.64), for example, shows that the state response requires matrix exponentiation, **A***n* . For a square **A** and specific *n*, MATLAB happily returns **A***n* by using the ^ operator. From the system in Ex. 10.13 and *n* = 3, we have - -``` ->> A = [0 1;-1/6 5/6]; n = 3; A^n - ans = - -0.1389 0.5278 - -0.0880 0.3009 -``` - -The same result is also obtained by typing A\*A\*A. - -Often, it is useful to solve **A***n* symbolically. Noting **A***n* = *Z*−1[(**I** − *z*−1**A**)−1], the symbolic toolbox can produce a symbolic expression for **A***n* . - -``` ->> syms z n; An = simplify(iztrans(inv(eye(2)-z^(-1)*A))) - An = - [ 3*(1/3)^n - 2*(1/2)^n, 6*(1/2)^n - 6*(1/3)^n] - [ 1/3^n - 1/2^n, 3/2^n - 2/3^n] -``` - -Notice that this result is identical to Eq. (10.68), derived earlier. Substituting the case *n* = 3 into An provides a result that is identical to the one elicited by the previous A^n command. - ->> double(subs(An,n,3)) ans = -0.1389 0.5278 -0.0880 0.3009 - -For continuous-time systems, the matrix exponential *e***A***t* is commonly encountered. The expm command can compute the matrix exponential symbolically. Using the system from Ex. 10.8 yields - -``` ->> syms t; A = [-12 2/3;-36 -1]; eAt = simplify(expm(A*t)) - eAt = - [ -(exp(-9*t)*(3*exp(5*t) - 8))/5, (2*exp(-9*t)*(exp(5*t) - 1))/15] - [ -(36*exp(-9*t)*(exp(5*t) - 1))/5, (exp(-9*t)*(8*exp(5*t) - 3))/5] -``` - -This result is identical to the result computed in Ex. 10.8. Similar to the discrete-time case, an identical result is obtained by typing syms s; simplify(ilaplace(inv(s\*eye(2)-A))). - -For a specific *t*, the matrix exponential is also easy to compute, either through substitution or direct computation. Consider the case *t* = 3. - ->> double(subs(eAt,t,3)) ans = 1.0e-004 \* -0.0369 0.0082 -0.4424 0.0983 - -The command expm(A\*3) produces the same result. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/131_10.9 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/131_10.9 SUMMARY.md deleted file mode 100644 index 0860e5a2b179a899f7dd0340332a5111f918b75d..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/131_10.9 SUMMARY.md +++ /dev/null @@ -1,67 +0,0 @@ -## **[10.9 SUMMARY](#page-15-0)** - -An *N*th-order system can be described in terms of *N* key variables—the state variables of the system. The state variables are not unique; rather, they can be selected in a variety of ways. Every possible system output can be expressed as a linear combination of the state variables and the inputs. Therefore, the state variables describe the entire system, not merely the relationship between certain input(s) and output(s). For this reason, the state variable description is an internal description of the system. Such a description is therefore the most general system description, and it contains the information of the external descriptions, such as the impulse response and the transfer function. The state variable description can also be extended to time-varying parameter systems and nonlinear systems. An external description of a system may not characterize the system completely. - -The state equations of a system can be written directly from knowledge of the system structure, from the system equations, or from the block diagram representation of the system. State equations consist of a set of *N* first-order differential equations and can be solved by time-domain or frequency-domain (transform) methods. Suitable procedures exist to transform one given set of state variables into another. Because a set of state variables is not unique, we can have an infinite variety of state-space descriptions of the same system. The use of an appropriate transformation allows us to see clearly which of the system states are controllable and which are observable. - -### **[REFERENCES](#page-15-0)** - -- 1. Kailath, Thomas. *Linear Systems.* Prentice-Hall, Englewood Cliffs, NJ, 1980. -- 2. Zadeh, L., and C. Desoer. *Linear System Theory.* McGraw-Hill, New York, 1963. - -## **[PROBLEMS](#page-15-0)** - -- **10.1-1** Convert each of the following second-order differential equations into a set of two first-order differential equations (state equations). State which of the sets represent nonlinear equations. - - (a) *y*¨ +10*y*˙ +2*y* = *x* - -1 + 1 H \_ - -network in Fig. P10.2-2. - -network in Fig. P10.2-3. - -**Figure P10.2-1** - -1 2 F - -- (b) *y*¨ +2*eyy*˙ +log*y* = *x* -- (c) *y*¨ +φ1(*y*)*y*˙ +φ2(*y*)*y* = *x* -- **10.2-1** Write the state equations for the *RLC* network in Fig. P10.2-1. - -*x* 3 - -**10.2-2** Write the state and output equations for the - -**10.2-3** Write the state and output equations for the - -2 - -**10.2-4** Write the state and output equations for the electrical network in Fig. P10.2-4. - -**Figure P10.2-4** - -**Figure P10.2-2** - -**10.2-5** Write the state and output equations for the network in Fig. P10.2-5. - -**10.2-6** Write the state and output equations of the system shown in Fig. P10.2-6. - -**Figure P10.2-6** - -**10.2-7** Write the state and output equations of the system shown in Fig. P10.2-7. - -**Figure P10.2-7** - -**10.2-8** For a system specified by the transfer function - -$$ -H(s) = \frac{3s + 10}{s^2 + 7s + 12} -$$ - -write sets of state equations for DFII and its transpose, cascade, and parallel forms. Also write the corresponding output equations. - -**10.2-9** Repeat Prob. 10.2-8 for - -\n(a) - -\n \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/132_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/132_REFERENCES.md deleted file mode 100644 index c1667c6cccf32840a4dae0b02a97239271372439..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/132_REFERENCES.md +++ /dev/null @@ -1,244 +0,0 @@ -$$ -H(s) = \frac{4s}{(s+1)(s+2)^2} -$$ -\n(b) - -\n -$$ -H(s) = \frac{s^3 + 7s^2 + 12s}{(s+1)^3(s+2)} -$$ - -**10.3-1** Find the state vector **q**(*t*) by using the Laplace transform method if - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x} -$$ - -where - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 2 \\ -1 & -3 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \qquad x(t) = 0 -$$ - -**10.3-2** Repeat Prob. 10.3-1 for - -$$ -\mathbf{A} = \begin{bmatrix} -5 & -6 \\ 1 & 0 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 5 \\ 4 \end{bmatrix} \qquad \qquad x(t) = \sin 100t -$$ - -**10.3-3** Repeat Prob. 10.3-1 for - -$$ -\mathbf{A} = \begin{bmatrix} -2 & 0 \\ 1 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \qquad x(t) = u(t) -$$ - -**10.3-4** Repeat Prob. 10.3-1 for - -$$ -\mathbf{A} = \begin{bmatrix} -1 & 1 \\ 0 & -2 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \qquad \mathbf{x} = \begin{bmatrix} u(t) \\ \delta(t) \end{bmatrix} -$$ - -**10.3-5** Use the Laplace transform method to find the response *y* for - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x}(t) -$$ -$$ -y = \mathbf{C}\mathbf{q} + \mathbf{D}\mathbf{x}(t) -$$ - -where - -$$ -\mathbf{A} = \begin{bmatrix} -3 & 1 \\ -2 & 0 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} -$$ -$$ -\mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ - -and - -$$ -x(t) = u(t) \qquad \mathbf{q}(0) = \begin{bmatrix} 2 \\ 0 \end{bmatrix} -$$ - -**10.3-6** Repeat Prob. 10.3-5 for - -$$ -\mathbf{A} = \begin{bmatrix} -1 & 1 \\ -1 & -1 \end{bmatrix} \quad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\mathbf{C} = \begin{bmatrix} 1 & 1 \end{bmatrix} \quad \mathbf{D} = 1 -$$ -$$ -x(t) = u(t) \quad \mathbf{q}(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} -$$ - -! - -**10.3-7** The transfer function *H*(*s*) in Prob. 10.2-8 is realized as a cascade of *H*1(*s*) followed by *H*2(*s*), where - -$$ -H_1(s) = \frac{1}{s+3} -$$ - -$$ -H_2(s) = \frac{3s+10}{s+4} -$$ - -Let the outputs of these subsystems be state variables *q*1 and *q*2, respectively. Write the state equations and the output equation for this system and verify that **H**(*s*) = **C***φ*(*s*)**B**+**D**. - -- **10.3-8** Find the transfer function matrix **H**(*s*) for the system in Prob. 10.3-5. -- **10.3-9** Find the transfer function matrix **H**(*s*) for the system in Prob. 10.3-6. -- **10.3-10** Find the transfer function matrix **H**(*s*) for the system - -$$ -\dot{q} = Aq + Bx -$$ -$$ -y = Cq + Dx -$$ - -where - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -1 & -2 \end{bmatrix} \quad \mathbf{B} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \quad \mathbf{x} = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} -$$ - -$$ -\mathbf{C} = \begin{bmatrix} 1 & 2 \\ 4 & 1 \\ 1 & 1 \end{bmatrix} \qquad \mathbf{D} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{bmatrix} -$$ - -- **10.3-11** Repeat Prob. 10.3-1, using the time-domain method. -- **10.3-12** Repeat Prob. 10.3-2, using the time-domain method. -- **10.3-13** Repeat Prob. 10.3-3, using the time-domain method. -- **10.3-14** Repeat Prob. 10.3-4, using the time-domain method. -- **10.3-15** Repeat Prob. 10.3-5, using the time-domain method. -- **10.3-16** Repeat Prob. 10.3-6, using the time-domain method. -- **10.3-17** Find the unit impulse response matrix **h**(*t*) for the system in Prob. 10.3-7, using Eq. (10.45). -- **10.3-18** Find the unit impulse response matrix **h**(*t*) for the system in Prob. 10.3-6. -- **10.3-19** Find the unit impulse response matrix **h**(*t*) for the system in Prob. 10.3-10. -- **10.4-1** The state equations of a certain system are given as - -$$ -\dot{q}_1 = q_2 + 2x -$$ - -$$ -\dot{q}_2 = -q_1 - q_2 + x -$$ - -Define a new state vector **w** such that - -$$ -w_1 = q_2 -$$ - -$$ -w_2 = q_2 - q_1 -$$ - -Find the state equations of the system with **w** as the state vector. Determine the characteristic roots (eigenvalues) of the matrix **A** in the original and the transformed state equations. - -**10.4-2** The state equations of a certain system are - -$$ -\dot{q}_1 = q_2 \n\dot{q}_2 = -2q_1 - 3q_2 + 2x -$$ - -(a) Determine a new state vector **w** (in terms of vector **q**) such that the resulting state equations are in diagonalized form. - -(b) For output **y** given by - -$$ -y = Cq + Dx -$$ - -where - -$$ -\mathbf{C} = \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ - -determine the output **y** in terms of the new state vector **w**. - -### **10.4-3** Given a system - -$$ -\dot{\mathbf{q}} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & -3 \end{bmatrix} \mathbf{q} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} x -$$ - -determine a new state vector **w** such that the state equations are diagonalized. - -**10.4-4** The state equations of a certain system are given in diagonalized form as - -$$ -\dot{\mathbf{q}} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -2 \end{bmatrix} \mathbf{q} + \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} x -$$ - -The output equation is given by - -$$ -y = \begin{bmatrix} 1 & 3 & 1 \end{bmatrix} \mathbf{q} -$$ - -Determine the output *y* for - -$$ -\mathbf{q}(0) = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \qquad x(t) = u(t) -$$ - -**10.5-1** Write the state equations for the systems depicted in Fig. P10.5-1. Determine a new state vector **w** such that the resulting state equations are in diagonalized form. Write the output **y** in terms of **w**. Determine in each case whether the system is controllable and observable. - -### **Figure P10.5-1** - -**10.6-1** An LTI discrete-time system is specified by - - 2 - -! - -$$ -\mathbf{A} = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} \qquad \mathbf{D} = \begin{bmatrix} 1 \end{bmatrix} -$$ - -and - -- **q**(0) = 1 *x*[*n*] = *u*[*n*] -- (a) Find the output *y*[*n*], using the timedomain method. -- (b) Find the output *y*[*n*], using the frequencydomain method. -- **10.6-2** An LTI discrete-time system is specified by the difference equation - -$$ -y[n+2] + y[n+1] + 0.16y[n] -$$ - -= $x[n+1] + 0.32x[n]$ - -- (a) Show the DFII, its transpose, cascade, and parallel realizations of this system. -- (b) Write the state and the output equations from these realizations, using the output of each delay element as a state variable. -- **10.6-3** Repeat Prob. 10.6-2 for - -$$ -y[n+2] + y[n+1] - 6y[n] -$$ - -= 2x[n+2] + x[n+1] - -- **10.7-1** Verify the state and output equations for the LTID system shown in Fig. 10.15. -- **10.7-2** Verify the state and output equations for the LTID system shown in Fig. 10.16. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/133_INDEX.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/133_INDEX.md deleted file mode 100644 index f87b8f396011892a190bdd7a0fe8a62ab0b4db70..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/133_INDEX.md +++ /dev/null @@ -1,89 +0,0 @@ -# **[INDEX](#page-15-0)** - -Abscissa of convergence, 336 Accumulator systems, 259, 295, 519 discrete-time Fourier transform of, 875–76 Active circuits, 382–85 Adders, 388, 396, 399, 403 Addition of complex numbers, 11–12 of matrices, 38 of sinusoids, 18–20 Additivity, 97–98 Algebra of complex numbers, 5–15 matrix, 38–42 Aliasing, 536–38, 788–95, 805, 811–12, 817, 834 defined, 536 general condition for in sinusoids, 793–96 treachery of, 788–91 verification of in sinusoids, 792–93 Aliasing error, 659, 811 Amplitude, 16 Amplitude modulation, 711–13, 736–49, 762 Amplitude response, 413, 416–17, 421, 424, 435, 437–38, 440–42 Amplitude spectrum, 598, 607, 615–25, 667, 668, 707, 848, 870, 878 Analog filters, 261 Analog signals, 133 defined, 78 digital processing of, 547–54 properties of, 78 Analog systems, 109, 135, 261 Analog-to-digital (A/D) conversion, 799–802, 831 Analog-to-digital converters (ADC) bit number, 801–2 bit rate, 801–2 Angle modulation, 736, 763 Angles electronic calculators in computing, 8–11 principal value of, 9 Angular acceleration, 116 Angular position, 116 Angular velocity, 116 - -Anti-aliasing filters, 537, 791, 834 Anti-causal exponentials, 862–63 Anti-causal signals, 81 Aperiodic signals, 133 discrete-time Fourier integral and, 855–67 Fourier integral and, 680–89, 762 properties of, 78–82 Apparent frequency, 534–36, 792–96 *Ars Magna* (Cardano), 2–5 Associative property, 171, 283 Asymptotic stability. *See* Internal stability. Audio signals, 713–14, 725, 746 Automatic position control system, 406–8 Auxiliary conditions, 153 differential equation solution and, 161 Backward difference system, 258, 295, 519, 568–69 Bandlimited signals, 533, 788, 792, 802 Bandpass filters, 441–43, 542–44, 749, 882–83, 896 group delay and, 726–27 ideal, 730–31, 882–83 poles and zeros of *H*(*s*) and, 443 Bandstop filters, 441–42, 445, 545–46 Bandwidth, 628 continuous-time systems and, 208–10 data truncation and, 751–53 essential, 736, 758–59 Fourier transform and, 692, 706, 762–63 Bartlett window, 753–55 Baseband signals, 737–40, 746–47, 749 Basis signals, 651, 655, 668 Basis vectors, 648 Beat effect, 741 Bhaskar, 2 Bilateral Laplace transform, 330, 335–37, 445–55, 467 properties of, 451–55 Bilateral *z*-transform, 431, 490, 554–63 in discrete-time system analysis, 563 properties of, 559–60 Bilinear transformation, 569–70 Binary digital signals, 799 Black box, 95, 119, 120 - -Blackman window, 755, 761 Block diagrams, 386–88, 405, 407, 408, 519 Bôcher, M., 620–21 Bode plots, 419–35 constant of, 421 first-order pole and, 424–27 pole at the origin and, 422–23 second-order pole and, 426–35 Bombelli, Raphael, 3 Bonaparte, Napoleon, 347, 610–11 Bounded-input/bounded-output (BIBO) stability, 110, 135, 263 of continuous-time systems, 196–97, 199–203, 222–23 of discrete-time systems, 298–99, 301–4, 314, 526, 527 frequency response and, 412–13 internal stability relationship to, 199–203, 301–4 of the Laplace transform, 371–73 signal transmission and, 721 steady-state response and, 418 of the *z*-transform, 518 Butterfly signal flow graph, 825 Butterworth filters, 440, 551–54 cascaded second-order sections for Butterworth filter realization, 461–63 MATLAB on, 459–63 transformation of, 571–72 - -Canonic direct realization. *See* Direct form II realization Cardano, Gerolamo, 2–5 Cartesian form, 8–15 Cascade realization, 394, 526, 920, 923 Cascade systems, 190, 192, 372–73 Cascaded *RC* filters, 461–62 Causal exponentials, 861–62 Causal signals, 81, 83, 134 Causal sinusoidal input in continuous-time systems, 418–19 in discrete-time systems, 527 Causal systems, 104–6, 135, 263 properties of, 104–6 zero-state response and, 172, 283 Cayley–Hamilton theorem, 910–12 Characteristic equations of continuous-time systems, 153–55 of discrete-time systems, 271, 273, 309 of a matrix, 910–12, 933 Characteristic functions, 192 Characteristic modes of continuous-time systems, 153–55, 162–65, 167, 170, 196, 198–99, 203–6 of discrete-time systems, 271–74, 278–79, 297–301, 305, 313 Characteristic polynomials of continuous-time systems, 153–56, 164, 166, 202–3, 220 of discrete-time systems, 271, 274–75, 279, 303–4 of the Laplace transform, 371–72 of the *z*-transform, 518 - -Characteristic roots of continuous-time systems, 153–56, 162, 166, 198–203, 206, 209, 211–12, 214–17, 222–24 of discrete-time systems, 271, 273–75, 297, 299–301, 303–5, 309, 314 invariance of, 942–43 of a matrix, 911, 932–33 Characteristic values. *See* Characteristic roots Characteristic vectors, 910 Chebyshev filters, 440, 463–66 Circular convolution, 819–20, 821 Clearing fractions, 26–27, 32–33, 342–43 Closed Loop systems. *See* Feedback systems Coherent demodulation. *See* Synchronous demodulation Coefficients of Fourier series, computation, 595–98 Column vectors, 36 Commutative property of the convolution integral, 170, 173, 181, 191–92 of the convolution sum, 283 Compact disc (CD), 801 Compact form of Fourier series, 597–98, 599, 600, 604–7 Complex factors of *Q*(*x*), 29 Complex frequency, 89–91 Complex inputs, 177, 297 Complex numbers, 1–15, 54 algebra of, 5–15 arithmetical operations for, 12–15 conjugates of, 6–7 historical note, 1–5 logarithms of, 15 origins of, 2–5 standard forms of, 14–15 useful identities, 7–8 working with, 13–14 Complex poles, 395, 432, 497, 542 Complex roots, 154–56, 274–76 Complex signals, 94–95 Conjugate symmetry of the discrete Fourier transform, 818–19 of the discrete-time Fourier transform, 858–59, 867–68 of the Fourier transform, 684, 703 Conjugation, 684, 703 Constants, 54, 98, 100, 103, 130, 422 Constant-parameter systems. *See* Time-invariant systems Continuous functions, 858 Continuous-time filters, 455–63 Continuous-time Fourier transform (CTFT), 867, 884–85 Continuous-time signals, 107–8, 135 defined, 78 discrete-time systems and, 238 Fourier series and, 593–679 Fourier transform and, 678–769, 680–775 Continuous-time systems, 135, 150–236 analog systems compared with, 261 differential equations of, 161, 196, 213 - -discrete-time systems compared with, 261 external input, response to, 168–96 frequency response of, 412–18, 732–33 internal conditions, response to, 151–63 intuitive insights into, 189–90, 203–5 Laplace transform, 330–487 Periodic inputs and, 637–641 properties of, 107–8 signal transmission through, 721–29 stability of, 196–203, 222–23 state equations for, 915–16 Control systems, 404–12 analysis of, 406–12 design specifications, 411 step input and, 407–9 Controllability/observability, 123–24 of continuous-time systems, 197, 200–2, 223 of discrete-time systems, 303, 965–68 in state-space analysis, 947–53, 961 Convergence abscissa of, 336 of Fourier series, 613–14 to the mean, 613, 614 region of. *See* region of convergence Convolution, 507–9 with an impulse, 283 of the bilateral *z*-transform, 560 circular, 819–20, 821 discrete-time, 311–12 fast, 821, 886 frequency. *See* Frequency convolution of the Fourier transform, 714–16 linear, 821 periodic, 886 time. *See* Time convolution Convolution integral, 170–93, 222, 282, 288, 313, 722 explanation for use, 189–90 graphical understanding of, 178–90, 217–20 properties of, 170–72 Convolution sum, 282–86, 313 graphical procedure for, 288–93 properties of, 282–83 from a table, 285–86 Convolution table, 175–76 Cooley, J. W., 824 Corner frequency, 424 Cramer's rule, 23–25, 40, 51, 379, 385 Critically damped systems, 409, 410 Cubic equations, 2–3, 58 Custom filter function, 310–11 Cutoff frequency, 208, 209 - -Damping coefficient, 115–18 Dashpots linear, 115 torsional, 116 - -Data truncations, 749–55, 763 Decades, 422 Decibels, 421 Decimation-in-frequency algorithm, 824, 827 Decimation-in-time algorithm, 825–27 Decomposition, 99–100, 151 Delayed impulse, 168 Demodulation, 714 of amplitude modulation, 744–46 of DSB-SC signals, 739–41 synchronous, 743–44 Depressed cubic equation, 58 Derivative formulas, 56 Descartes, René, 2 Detection. *See* Demodulation Deterministic signals, 82, 134 Diagonal matrices, 37 Difference equations, 259–60, 265–70 causality condition in, 265–66 classical solution of, 298 differential equation kinship with, 260 frequency response, 532 order of, 260 recursive and non-recursive forms of, 259 recursive solution of, 266–70 sinusoidal response of difference equation systems, 528 *z*-transform solution of, 488, 510–19, 574 Differential equations, 161 classical solution of, 196 difference equation kinship with, 260 Laplace transform solution of, 346–48, 360–73 Differentiators digital, 256–58 ideal, 369–71, 373, 416–17 Digital differentiator example, 258–59 Digital filters, 108, 238, 261–62 Digital integrators, 258–59 Digital processing of analog signals, 547–53 Digital signals, 135, 797–99 advantages of, 261–62 binary, 799–801 defined, 78 *L*-ary, 799 properties of, 78 *See also* Analog-to-digital conversion Digital systems, 109, 135, 261 Dirac definition of an impulse, 88, 134 Dirac delta train, 696–97 Dirac, P.A.M., 86 Direct discrete Fourier transform (DFT), 808, 857 Direct form I (DFI) realization Laplace transform and, 390–91, 394 *z*-transform and, 521 *See also* Transposed direct form II realization - -Direct form II (DFII) realization, 920–25, 954, 965, 967 Laplace transform and, 391, 398 *z*-transform and, 520–22, 525 Direct Fourier transform, 683, 702–3, 762 Direct *z*-transform, 488–592 Dirichlet conditions, 612, 614, 686 Discrete Fourier transform (DFT), 659, 805–23, 827–34, 835 aliasing and leakage and, 805–6 applications of, 820–23 computing Fourier transform, 812–18 derivation of, 807–10 determining filter output, 822–23 direct, 808, 857 discrete-time Fourier transform and, 885–86, 898 inverse, 808, 835, 857 MATLAB on, 827–34 picket fence effect and, 807 points of discontinuity, 807 properties of, 818–20 zero padding and, 810–11, 829–30 Discrete-time complex exponentials, 252 Discrete-time convolution, 311–12 Discrete-time exponentials, 247–49 Discrete-time Fourier integral, 855–67 Discrete-time Fourier series (DTFS), 845–55 computation of, 885–86 MATLAB on, 889–97 of periodic gate function, 853–55 periodic signals and, 846–47, 898 of sinusoids, 849–52 Discrete-time Fourier transform (DTFT), 857–88 of accumulator systems, 875–76 of anti-causal exponentials, 862–63 of causal exponentials, 861–62 continuous-time Fourier transform and, 883–86 existence of, 859, 886 inverse, 886 linear time-invariant discrete-time system analysis by, 879–80 MATLAB on, 889–97 physical appreciation of, 859 properties of, 867–78 of rectangular pulses, 863–65 table of, 860 *z*-transform connection with, 866–67, 886–88, 898 Discrete-time signals, 78, 79, 107–8, 133, 237–53 defined, 78 Fourier analysis of, 845–907 inherently bandlimited, 533 size of, 238–40 useful models, 245–53 useful operations, 240–45 Discrete-time systems, 135, 237–329 classification of, 262–64 controllability/observability of, 303, 965–68 difference equations of, 259–60, 265–70, 298 - -discrete-time Fourier transform analysis of, 878–83 examples of, 253–65 external input, response to, 280–98 frequency response of, 526–38 internal conditions, response to, 270–76 intuitive insights into, 305–6 properties of, 107–8, 264–65 stability of, 263, 298–305, 314 state-space analysis of, 953–64 *z*-transform analysis of, 488–592 Distinct factors of *Q*(*x*), 27 Distortionless transmission, 724–28, 730, 763, 880–82 bandpass systems and, 726–27, 881–82 measure of delay variation, 881 Distributive property, 171, 283 Division of complex numbers, 12–14 Double-sideband, suppressed-carrier (DSB-SC) modulation, 737–41, 742, 746–49 Downsampling, 243–44 Duality, 703–4 Dynamic systems, 103–4, 134–35, 263 - -Eigenfunctions, 193 Eigenvalues. *See* Characteristic roots Eigenvectors, 910 Einstein, Albert, 348 Electrical systems, 95–96, 111–14 Laplace transform analysis of, 373–85, 467 state equations for, 916–19 Electromechanical systems, 118–19 Electronic calculators, 8–11 Energy signals, 67, 82, 134, 239–40 Energy spectral density, 734, 763 Envelope delay. *See* Group delay Envelope detector, 743–45 Equilibrium states, 196, 198 Error signals, 650–51 Error vectors, 642 Essential bandwidth, 736, 758–59 Euler, Leonhard, 2, 3 Euler's formula, 5–6, 45, 252 Even component of a signal, 93–95 Even functions, 92–93, 134 Everlasting exponentials continuous-time systems and, 189, 193–95, 222 discrete-time systems and, 296–97, 313 Fourier series and, 637, 638, 641 Fourier transform and, 687 Laplace transform and, 367–68, 412, 419 Everlasting signals, 81, 134 Exponential Fourier series, 621–37, 661, 803 periodic inputs and, 637–41 reasons for using, 640 symmetry effect on, 630–32 - -Exponential Fourier spectra, 624–32, 664, 667, 668 Exponential functions, 89–91, 134 Exponential input, 193, 296 Exponentials computation of matrix, 922–913 discrete-time, 247–49 discrete-time complex, 252 everlasting. *See* Everlasting exponentials matrix, 968–69 monotonic, 20–22, 90, 91, 134 sinusoid varying, 22–23, 90, 134 sinusoids expressed in, 20 *Exposition du système du monde* (Laplace), 346 External description of a system, 119–20, 135 External input continuous-time system response to, 168–96 discrete-time system response to, 280–98 External stability. *See* Bounded-input/bounded-output stability Fast convolution, 821, 886 Fast Fourier transform (FFT), 659, 811, 821, 824–27, 835 computations reduced by, 824 discrete-time Fourier series and, 847 discrete-time Fourier transform and, 885–86, 898 Feedback systems Laplace transform and, 386–88, 392–95, 399, 404–12 *z*-transform and, 521 Feedforward connections, 392–94, 403 Filtering discrete Fourier transform and, 821–23 MATLAB on, 308–10 selective, 748–49 time constant and, 207–8 Filters analog, 261 anti-aliasing, 537, 791, 834 bandpass, 441–43 bandstop, 441–42, 445, 545–46 Butterworth. *See* Butterworth Filters cascaded *RC*, 461–62 Chebyshev, 440, 463–66 continuous-time, 455–63 custom function, 310–11 digital, 108, 238, 261–62 finite impulse response, 524, 892–97 first-order hold, 785 frequency response of, 412–18 highpass, 443, 445, 542, 730–31, 882–83 Ideal. *See* Ideal filters impulse invariance criterion of, 548 infinite impulse response, 524, 565–74 lowpass, 439–41 lowpass. *See* Lowpass filters notch, 441–43, 540, 545–46 - -poles and zeros of *H*(*s*) and, 436–45 practical, 444–45, 882–83 sharp cutoff, 748 windows in design of, 755 zero-order hold, 785 Final value theorem, 359–61, 508 Finite impulse response (FIR) filters, 524, 892–97 Finite-duration signals, 333 Finite-memory systems, 104 First-order factors, method of, 497 First-order hold filters, 785 Folding frequency, 789–91, 793, 795, 817 For-loops, 216–18 Forced response difference equations and, 298 differential equations and, 198 Forward amplifiers, 405–6 Fourier integral, 722 aperiodic signal and, 680–89, 762 discrete-time, 855–67 Fourier series, 593–679 compact form of, 597–98, 599, 600, 604–7 computing the coefficients of, 595–98 discrete time. *See* Discrete-time Fourier series existence of, 612–13 exponential. *See* Exponential Fourier series generalized, 641–59, 668 Legendre, 656–57 limitations of analysis method, 641 trigonometric. *See* Trigonometric Fourier series waveshaping in, 615–17 Fourier spectrum, 598–607, 777 exponential, 624–32, 664, 667, 668 nature of, 858–59 of a periodic signal, 848–55 Fourier transform, 680–755, 778, 802–3 continuous-time, 867, 883–86 discrete. *See* Discrete Fourier transform discrete-time. *See* Discrete-time Fourier transform direct, 683, 702–3, 762 existence of, 685–86 fast. *See* fast Fourier transform interpolation and, 785 inverse, 683, 693–95, 699, 762, 786–87 physical appreciation of, 687–89 properties of, 701–21 useful functions of, 689–701 Fourier transform pairs, 683, 700 Fourier, Baron Jean-Baptiste-Joseph, 610–12 Fractions, 1–2 clearing, 26–27, 32–34, 342–43 partial. *See* Partial fractions Frequency apparent, 534–36, 793–94 complex, 89–91 - -Frequency (*continued*) corner, 424 cutoff, 208, 209 folding, 789–91, 793, 795, 817 fundamental, 594, 609–10, 846 negative, 626–28 neper, 91 radian, 16, 91, 594 reduction in range, 535 of sinusoids, 16 time delay variation with, 724–25 Frequency convolution of the bilateral Laplace transform, 452 of the discrete-time Fourier transform, 875–76 of the Fourier transform, 714–16 of the Laplace transform, 357 Frequency differentiation, 869 Frequency domain analysis, 368, 722–23, 848 of electrical networks, 374–78 of the Fourier series, 598, 601 two-dimensional view and, 732–33 *See also* Laplace transform Frequency inversion, 706 Frequency resolution, 807, 810–12, 815, 817 Frequency response, 724 Bode plots and, 419–22 of continuous-time systems, 412–18, 732–33 of discrete-time systems, 526–38 MATLAB on, 456–57, 531–32 periodic nature of, 532–36 from pole-zero location, 538–47 pole-zero plots and, 566–68 poles and zeros of *H*(*s*) and, 436–39 transfer function from, 435 Frequency reversal, 868–69 Frequency shifting of the bilateral Laplace transform, 451 of the discrete Fourier transform, 819 of the discrete-time Fourier transform, 871–74 of the Fourier transform, 711–13 of the Laplace transform, 353–54 Frequency spectra, 598, 601 Frequency-division multiplexing (FDM), 714, 749–50 Function M-files, 214–15 Functions characteristic, 193 continuous, 858 even, 92–93, 134 exponential, 89–91, 134 improper, 25–26, 34 interpolation, 690 MATLAB on, 126–33 odd, 92–95, 134 proper, 25–27 - -rational, 25–29, 338 singularity, 89 Fundamental band, 533, 534, 537, 793 Fundamental frequency, 594, 609–10, 846 Fundamental period, 79, 133, 239–40, 593, 595, 846 - -Gain enhancement by poles, 437–38 Gauss, Karl Friedrich, 3–4 Generalized Fourier series, 641–59, 668 Generalized linear phase (GLP), 726–27 Gibbs phenomenon, 619–21, 661–63 Gibbs, Josiah Willard, 620–21 Graphical interpretation of convolution integral, 178–90, 217–20 of convolution sum, 288–93 Greatest common factor of frequencies, 609–10 Group delay, 725–28, 881 - -#### *H*(*s*) - -filter design and, 436–45 realization of, 548–49 *See also* Transfer functions Half-wave symmetry, 608 Hamming window, 754–55, 761 Hanning window, 754–55, 761 Hardware realization, 64, 95, 133 Harmonic distortion, 634 Harmonically related frequencies, 609 Heaviside "cover-up" method, 27–30, 33–35, 341, 342–43, 497 Heaviside, Oliver, 347–48, 612 Highpass filters, 443, 445, 542, 745, 747, 882–83 Homogeneity, 97–98 - -Ideal delay, 369, 416 Ideal differentiators, 369–71, 373, 416–17 Ideal filters, 730–33, 763, 785, 791, 834, 882–83 Ideal integrators, 369, 370, 373, 400, 416–18 Ideal interpolation, 786–87 Ideal linear phase (ILP), 725, 727 Ideal masses, 114 Identity matrices, 37 Identity systems, 109, 192, 263 Imaginary numbers, 1–5 Impedance, 374–77, 379, 380, 382, 384, 387, 399 Improper functions, 25–26, 34 Impulse invariance criterion of filter design, 548 Impulse matching, 164–66 Impulse response matrix, 938 Indefinite integrals, 57 Indicator function. *See* Relational operators Inertia, moment of, 116–18 Infinite impulse response (IIR) filters, 524, 565–74 Information transmission rate, 209–10 - -Initial conditions, 97–100, 102, 122, 134, 335 at 0− and 0+, 363–64 continuous-time systems and, 158–61 generators of, 376–83 Initial value theorem, 359–61, 508 Input, 64 complex, 177, 297 exponential, 193, 296 external. *See* External input in linear systems, 97 multiple, 178, 287–88 ramp, 410–11 sinusoidal. *See* Sinusoidal input step, 407–10 Input–output description, 111–19 Instantaneous systems, 103–4, 134, 263 Integrals convolution. *See* Convolutional integral discrete-time Fourier, 855–67 Fourier. *See* Fourier integral indefinite, 57 of matrices, 909–10 Integrators digital, 258–59 ideal, 369, 370, 373, 400, 416–18 system realization and, 400 Integro-differential equations, 360–73, 466, 488 Interconnected systems continuous-time, 190–93 discrete-time, 294–97 Internal conditions continuous-time system response to, 151–63 discrete-time system response to, 270–76 Internal description of a system, 119–21, 135, 908 *See also* State-space description of a system Internal stability, 110, 135, 263 BIBO relationship to, 199–203, 301–4 of continuous-time systems, 196–203, 222–23 of discrete-time systems, 298–302, 305, 314, 526, 527 of the Laplace transform, 372 of the *z*-transform, 518 Interpolation, 785–88 of discrete-time signals, 243–44 ideal, 786–87 simple, 785–86 spectral, 804 Interpolation formula, 779, 787 Interpolation function, 690 Intuitive insights into continuous-time systems, 189–90, 203–12 into discrete-time systems, 305–6 into the Laplace transform, 367–68 Inverse continuous-time systems, 192–93 Inverse discrete Fourier transform (IDFT), 808, 827, 857 Inverse discrete-time Fourier transform (IDTFT), 886 of rectangular spectrum, 865–66 Inverse discrete-time systems, 294–95 Inverse Fourier transform, 683, 693–95, 699, 762, 786–87 Inverse Laplace transform, 333, 335, 445, 549 finding, 338–46 Inverse *z*-transform, 488–89, 491, 499, 500, 501, 510, 554, 555, 559 finding, 495 Inversion frequency, 706 matrix, 40–42 Invertible systems, 109–10, 135, 263 Irrational numbers, 1–2 Kaiser window, 755, 760–62 Kelvin, Lord, 348 Kennelly-Heaviside atmosphere layer, 348 Kirchhoff's laws, 95 current (KCL), 111, 213, 374 voltage (KVL), 111, 374 Kronecker delta functions, 245 bandlimited interpolation of, 787–88 *L*-ary digital signals, 799 L'Hôpital's rule, 58, 211, 690 Lagrange, Louis de, 347, 612, 613 Laplace transform, 167, 330–487, 721 bilateral. *See* Bilateral Laplace transform differential equation solutions and, 346–48, 360–73 electrical network analysis and, 373–85, 467 existence of, 336–37 Fourier transform connection with, 699–701, 866 intuitive interpretation of, 367–69 inverse, 549, 938–39 properties of, 349–62 stability of, 371–74 state equation solutions by, 927–33 system realization and, 388–404 unilateral, 333–36, 337, 338, 345, 360, 445, 467 *z*-transform connection with, 488, 489, 491, 563–65 Laplace transform pairs, 333 Laplace, Marquis Pierre-Simon de, 346–47, 611, 612, 613 Leakage, 751, 753–55, 763, 805–6 Left half plane (LHP), 91, 198–99, 202, 211, 223, 435 Left shift, 71, 73, 130, 134, 503, 509, 510, 512 Left-sided sequences, 555–56 Legendre Fourier series, 656–57 Leibniz, Gottfried Wilhelm, 801 Linear convolution, 821 Linear dashpots, 115 Linear phase distortionless transmission and, 725, 881 generalized, 726–27 ideal, 725, 727 - -Linear phase (*continued*) physical description of, 707–9 physical explanation of, 870–71 Linear springs, 114 Linear systems, 97–101, 134 heuristic understanding of, 722–23 response of, 98–100 Linear time-invariant continuous-time (LTIC) systems. *See* Continuous-time systems Linear time-invariant discrete-time (LTID) systems. *See* Discrete-time systems Linear time-invariant (LTI) systems, 103, 194–95 Linear time-invariant discrete-time (LTID) systems, 879–80 Linear time-varying systems, 103 Linear transformation of vectors, 36, 939–47, 961 Linearity of the bilateral Laplace transform, 451 of the bilateral *z*-transform, 559 concept of, 97–98 of the discrete Fourier transform, 818, 824 of the discrete-time Fourier transform, 867 of discrete-time systems, 262 of the Fourier transform, 686–87, 824 of the Laplace transform, 331–32 of the *z*-transform, 489 Log magnitude, 27, 422–24 Loop currents continuous-time systems and, 159–63, 175 Laplace transform and, 375 Lower sideband (LSB), 738–39, 747 Lowpass filters, 439–41, 540–42 ideal, 730, 784–85, 788–89, 882–83 poles and zeros of *H*(*s*) and, 436–45 M-files, 212–20 function, 214–15 script, 213–14, 218 Maclaurin series, 6, 55 Magnitude response. *See* Amplitude response Marginally stable systems continuous-time, 198–200, 203, 211, 222–24 discrete-time, 301–2, 304, 314 Laplace transform, 373 signal transmission and, 721 *z*-transform, 519 Mathematical models of systems, 95–96, 125 MATLAB on Butterworth filters, 459–63 calculator operations in, 43–45 on continuous-time filters, 455–63 on discrete Fourier transform, 827–34 on discrete-time Fourier series and transform, - -889–97 - -on discrete-time systems/signals, 306–12 - -elementary operations in, 42–53 on filtering, 308–10 Fourier series applications in, 661–67 Fourier transform topics in, 755–62 frequency response plots, 531–32 on functions, 126–33 impulse invariance, 553 impulse response and, 167 on infinite-impulse response filters, 565–74 M-files in, 212–20 matrix operations in, 49–53 multiple magnitude response curves, 544 partial fraction expansion in, 53 periodic functions, 661–63 phase spectrum, 664–67 polynomial roots and, 157 simple plotting in, 46–48 state-space analysis in, 961–69 vector operations in, 45–46 zero-input response and, 157–58 Matrices, 36–42 algebra of, 38–42 characteristic equation of, 909–10, 933 characteristic roots of, 932–33 computing exponential of, 912–13 definitions and properties of, 37–38 derivatives of, 909–10 diagonal, 37 diagonalization of, 943–44 equal, 37 functions of, 911–12 identity, 37 impulse response, 938 integrals of, 909–10 inversion of, 40–42 MATLAB operations, 49–53 nonsingular, 41 square, 36, 37, 41 state transition, 936 symmetric, 37 transpose of, 37–38 zero, 37 Matrix exponentials, 968–69 Matrix exponentiation, 968–69 Mechanical systems, 114–18 Memory, systems and, 104, 263 Memoryless systems. *See* Instantaneous systems Method of residues, 27 Michelson, Albert, 620–21 Minimum phase systems, 435, 436 Modified partial fractions, 35, 496 Modulation, 713–14, 736–49 amplitude, 711–13, 736, 742–46, 762 angle, 736, 763 of the discrete-time Fourier transform, 872 - -double-sideband, suppressed-carrier, 737–41, 742, 746–49 pulse-amplitude, 796 pulse-code, 796, 799 pulse-position, 796 pulse-width, 796 single-sideband, 746–49 Moment of inertia, 116–18 Monotonic exponentials, 20–22, 90, 91, 134 Multiple inputs, 178, 287–88 Multiple-input, multiple-output (MIMO) systems, 98, 125, 908 Multiplication bilateral *z*-transform and, 560 of complex numbers, 12–14 discrete-time Fourier transform and, 869 of a function by an impulse, 87 matrix, 38–40 scalar, 38, 400–1, 505 *z*-transform and, 506–7 Natural binary code (NBC), 799 Natural modes. *See* Characteristic modes Natural numbers, 1 Natural response difference equations and, 298 differential equations and, 196 Negative feedback, 406 Negative frequency, 626–28 Negative numbers, 1–3, 45 Neper frequency, 91 Neutral equilibrium, 197, 198 Newton, Sir Isaac, 2, 346–47 Noise, 66, 151, 371, 417, 791, 797–99 Nonanticipative systems. *See* Causal systems Non-bandlimited signals, 792 Noncausal signals, 81 Noncausal systems, 104–7, 135, 263 properties of, 104–6 reasons for studying, 106–7 Non-invertible systems, 109–10, 135, 263 Non-inverting amplifiers, 382 Nonlinear systems, 97–101, 134 Nonsingular matrices, 41 Non-uniqueness, 533 Normal-form equations, 915 Norton theorem, 375 Notch filters, 441–43, 540, 545–46. *See also* Bandstop filters Numerical integration, 131–33 Nyquist interval, 778, 779 Nyquist rate, 778–81, 788–89, 792, 795, 821 Nyquist samples, 778, 781, 782, 788, 792 - -Observability. *See* controllability/observability Octave, 422 - -Odd component of a signals, 93–95 Odd functions, 92–95, 134 Operational amplifiers, 382–83, 399, 467 Ordinary numbers, 1–5 Orthogonal signal space, 649–50 Orthogonal signals, 668 energy of the sum of, 647 signal representation by set, 647–59 Orthogonal vector space, 647–48 Orthogonality, 622 Orthonormal sets, 649 Oscillators, 203 Output, 64, 97 Output equations, 122, 124, 908, 930, 941 Overdamped systems, 409–10 Paley–Wiener criterion, 444, 731–32, 788 Parallel realization, 393–94, 525–26, 921, 924–25 Parallel systems, 190, 387 Parseval's theorem, 632, 651–52, 734–35, 755, 758–59, 876–78 Partial fractions expansion of, 25–35, 53 inverse transform by partial fraction expansion and tables, 495–98 Laplace transform and, 338–39, 341, 344, 362, 394, 395, 419, 454 modified, 35 *z*-transform, 499 Passbands, 441, 444–45, 748, 755 Peak time, 409–10 Percent overshoot (PO), 409–10 Periodic (circular) convolution, 819–20 of the discrete-time Fourier transform, 875 Periodic extension of the Fourier spectrum, 848–55 properties of, 80–81 Periodic functions Fourier spectra as, 858 MATLAB on, 661–63 Periodic gate function, 853–55 Periodic signals, 133, 637–40 discrete-time Fourier series and, 846–47 Fourier spectra of, 848–55 Fourier transform of, 695–96 properties of, 78–82 and trigonometric Fourier series, 593–612, 661 Periods fundamental, 79, 133, 239–40, 593, 595, 846 sinusoid, 16 Phase response, 413–25, 427–35, 439, 467 Phase spectrum, 598, 607, 617–18, 707, 848 MATLAB on, 664–67 using principal values, 709–10 - -Phase-plane analysis, 125, 909 Phasors, 18–20 Physical systems. *See* Causal systems Picket fence effect, 807 Pickoff nodes, 190, 254–55, 396 Pingala, 801 Pointwise convergent series, 613 Polar coordinates, 5–6 Polar form, 8–15 arithmetical operations in, 12–15 sinusoids and, 18 Pole-zero location, 538–47 Pole-zero plots, 566–68 Poles complex, 395, 432, 497, 542 controlling gain by, 540 first-order, 424–27 gain enhancement by, 437–38 *H*(*s*), filter design and, 436–45 at the origin, 422–23 repeated, 395, 525, 926 in the right half plane, 371, 435–36 second-order, 426–35 wall of, 439–41, 542 Polynomial expansion, 458–59 Polynomial roots, 157, 572 Positive feedback, 406 Power series, 55 Power signals, 67, 82, 134, 239–40. *See also* Signal power Power, determining, 68–69 matrix, 912–13 Powers, of complex numbers, 13–16 Practical filters, 730–33, 882–83 Preece, Sir William, 349 Prewarping, 570–71 Principal values of the angle, 9 phase spectrum using, 709–10 Proper functions, 25–27 Pulse-amplitude modulation (PAM), 796 Pulse-code modulation (PCM), 796, 799 Pulse dispersion, 209 Pulse-position modulation (PPM), 796 Pulse-width modulation (PWM), 796 Pupin, M., 348 Pythagoras, 2 - -Quadratic equations, 58 Quadratic factors, 29–30 for the Laplace transform, 341–42 for the *z*-transform, 497 Quantization, 799, 831–34 Quantized levels, 799 - -Radian frequency, 16, 91, 594 Random signals, 82, 134 Rational functions, 25–29, 338 Real numbers, 2–7, 43 Real time, 105–6 Rectangular pulses, 863–65 Rectangular spectrum, 865–66 Rectangular windows, 751, 753–55, 763 Reflection property, 868–69 Region of convergence (ROC) for continuous-time systems, 193 for finite-duration signals, 333 for the Laplace transform, 331–33, 337, 347, 448, 449, 454–55, 467 for the *z*-transform, 489–91, 555–58, 561 Relational operators, 128–29 Repeated factors of *Q*(*x*), 31–32 Repeated poles, 395, 525, 926 Repeated roots of continuous-time systems, 154–56, 195, 198, 202, 223 of discrete-time systems, 270, 273–74, 297, 301, 313–14 Resonance phenomenon, 163, 204, 205, 210–12, 305 Right half plane (RHP), 91, 198, 200–3, 223, 371, 435–36 Right shift, 71–72, 131, 134, 501–4, 509, 510 Right-sided sequences, 555–56 Rise time, 206–7, 405, 409–10, 411 *RLC* networks, 914, 916–18 RMS value, 68–69, 70 Rolloff rate, 753, 754 Roots complex, 154–56, 274–76 of complex numbers, 11–15 polynomial, 157, 572 repeated. *See* Repeated roots unrepeated, 198, 202, 223, 301, 314 Rotational systems, 116–19 Rotational mass. *See* Moment of inertia Row vectors, 36, 45, 48–50 - -Sales estimate example, 255–56 Sallen–Key circuit, 383, 384, 461–62, 463, 466 Sampled continuous-time sinusoids, 527–31 Sampling, 776–844 practical, 781–84 properties of, 87–88, 134 signal reconstruction and, 785–99 spectral, 759–60, 802–4 *See also* Discrete Fourier transform; Fast-Fourier transform Sampling interval, 550–54 Sampling rate, 243–44, 536–37 Sampling theorem, 537, 776–84, 834–35 applications of, 796–99 spectral, 802 Savings account example, 253–55 - -Scalar multiplication, 38, 400–1, 505, 509, 520 Scaling, 97–98, 130 of the Fourier transform, 705–6, 755, 757, 762 of the Laplace transform, 357 *See also* Time scaling Script M-files, 213–14, 216, 218 Selective-filtering method, 748–49 Sharp cutoff filters, 748 Shifting of the bilateral *z*-transform, 559 of the convolution integral, 171–72 of the convolution sum, 283 of discrete-time signals, 240 *See also* Frequency shifting; Time shifting Sideband, 746–49 sifting. *See* Sampling Signal distortion, 723–25 Signal energy, 65–66, 70, 131–33, 733–36, 757, 877–78. *See also* Energy signals Signal power, 65–67, 133. *See also* Power signals Signal reconstruction, 785–99. *See also* Interpolation Signal-to-noise power ratio, 66 Signal transmission, 721–29 Signals, 64–91, 133–34 analog. *See* Analog signals anti-causal, 81 aperiodic. *See* Aperiodic signals audio, 713–14, 725, 746 bandlimited, 533, 788, 792, 802 baseband, 737–40, 746–47, 749 basis, 651, 655, 668 causal, 81, 83, 134 classification of, 78–82, 133–34 comparison and components of, 643–45 complex, 94–95 continuous time. *See* continuous-time signals defined, 65 deterministic, 83, 134 digital. *See* Digital signals discrete time. *See* Discrete-time signals energy, 82, 134, 239–40 error, 650–51 even components of, 93–95 everlasting, 81, 134 finite-duration, 333 modulating, 711, 737–39 non-bandlimited, 792 noncausal, 81 odd components of, 93–95 orthogonal. *See* Orthogonal signals periodic. *See* Periodic signals phantoms of, 189 power, 82, 134, 239–40 random, 82, 134 size of, 64–70, 133 - -sketching, 20–23 time reversal of, 77 time limited, 802, 805, 807 two-dimensional view of, 732–33 useful models, 82–91 useful operations, 71–78 as vectors, 641–59 video, 725, 749 Sinc function, 757 Single-input, single-output (SISO) systems, 98, 125, 908 Single-sideband (SSB) modulation, 746–49 Singularity functions, 89 Sinusoidal input causal. *See* Causal sinusoidal input continuous-time systems and, 208 discrete-time systems and, 309 frequency response and, 413–17 steady-state response to causal sinusoidal input, 418–19 Sinusoids, 16–20, 89–91, 134 addition of, 18–20 apparent frequency of sampled, 795–96 compression and expansion, 76 continuous-time, 251–52, 533–37 discrete-time, 251, 527, 528, 533–37 discrete-time Fourier series of, 849–52 in exponential terms, 20 exponentially varying, 22–23, 80, 134 general condition for aliasing in, 793–96 power of a sum of two equal-frequency, 70 sampled continuous-time, 527–31 verification of aliasing in, 792–93 Sketching signals, 20–23 Sliding-tape method, 290–93 Software realization, 64, 95, 133 Spectral density, 688 Spectral folding. *See* Aliasing Spectral interpolation, 804 Spectral resolution, 807 Spectral sampling, 759–60, 802 Spectral sampling theorem, 802 Spectral spreading, 751–53, 755, 763, 807 Springs linear, 114 torsional, 116–17 Square matrices, 36, 37, 41 Square roots of negative numbers, 2–4 Stability BIBO. *See* Bounded-input/bounded-output stability of continuous-time systems, 196–203, 222–23 of discrete-time systems, 263, 298–305, 314 of the Laplace transform, 371–74 Internal. *See* Internal stability of the *z*-transform, 518–19 marginal. *See* marginally stable systems - -Stable equilibrium, 196–97 Stable systems, 110, 263 State equations, 122–25, 135, 908–9, 969 alternative procedure to determine, 918–19 diagonal form of, 944–47 solution of, 926–39 for the state vector, 941–42 systematic procedure for determining, 913–26 time-domain method to solve, 936–37 State transition matrix (STM), 936 State variables, 121–25, 135, 908, 969 State vectors, 927–30, 961 linear transformation of, 941–42 State-space analysis, 908–73 controllability/observability in, 947–53, 961 of discrete-time systems, 953–64 in MATLAB, 961–69 transfer function and, 920–24 transfer function matrix, 931–32 State-space description of a system, 121–25 Steady-state error, 409–11 Steady-state response in continuous-time systems, 418–19 in discrete-time systems, 527 Stem plots, 306–8 Step input, 407–10 Stiffness of linear springs, 114 of torsional springs, 116–17 Stopbands, 441, 444, 445, 456, 457, 459, 460, 463, 755 Subcarriers, 749 Subtraction of complex numbers, 11–12 Superposition, 98, 99, 100, 123, 134 continuous-time systems and, 168, 170, 178 discrete-time systems and, 287 Symmetric matrices, 37 Symmetry conjugate. *See* Conjugate symmetry exponential Fourier series and, 630–32 trigonometric Fourier series and, 607–8 Synchronous demodulation, 743–44, 747 System realization, 388–404, 519–25, 567 cascade, 394, 525–26, 919–20, 923 of complex conjugate poles, 395 direct. *See* Direct form I realization; Direct form II realization differences in performance, 525–26 hardware, 64, 95, 133 parallel. *See* Parallel realization software, 64, 95, 129 Systems, 95–133, 134–35 accumulator, 259, 295, 519 analog, 109, 135, 261 backward difference, 258, 295, 519, 568–69 BIBO stability, assessing, 110 cascade, 190, 192, 372, 373 - -causal. *See* causal systems causality, assessing, 105 classification of, 97–110, 134–35 continuous time. *See* Continuous-time systems control. *See* control systems critically damped, 409, 410 data for computing response, 96–97 defined, 64 digital, 78, 135, 261 discrete time. *See* discrete time systems dynamic, 103–4, 134–35, 263 electrical, 95–96, 111–14 electrical. *See* Electrical systems electromechanical, 118–19 feedback. *See* feedback systems finite-memory, 104 identity, 109, 192, 263 input–output description, 111–19 instantaneous, 103–4, 263 interconnected. *See* interconnected systems invertible, 109–10, 135, 263 linear. *See* Linear systems mathematical models of, 95–96, 125 mechanical, 114–18 memory and, 104, 263 minimum phase, 435, 436 multiple-input, multiple-output, 98, 125, 908 noncausal, 104–7, 263 non-invertible, 109–10, 135 nonlinear, 97–101, 134 overdamped, 409–10 parallel, 190, 387 phantoms of, 189 properties of, 264–65 rotational, 116–19 single-input, single-output, 98, 125, 908 stable, 110, 263 translational, 114–16 time invariant. *See* Time-invariant systems time varying. *See* Time-varying systems two-dimensional view of, 732–33 underdamped, 409 unstable, 110, 263 - -Tacoma Narrows Bridge failure, 212 Tapered windows, 753–54, 763, 807 Taylor series, 55 *Théorie analytique de la chaleur* (Fourier), 612 Thévenin's theorem, 375, 378, 379 Time constant of continuous-time systems, 205–10, 223 of the exponential, 21–22 filtering and, 207–9 information transmission rate and, 209–10 - -pulse dispersion and, 209 rise time and, 206–7 Time convolution of the bilateral Laplace transform, 452 of the discrete-time Fourier transform, 875–76 of the Fourier transform, 714–16 of the Laplace transform, 357 of the *z*-transform, 507–8 Time delay, variation with frequency, 724–25 Time differentiation of the bilateral Laplace transform, 451 of the Fourier transform, 716–18 of the Laplace transform, 354–56 Time integration of the bilateral Laplace transform, 451 of the Fourier transform, 716–18 of the Laplace transform, 356–57 Time inversion, 706 Time reversal, 134 of the bilateral Laplace transform, 452 of the bilateral *z*-transform, 560 of the convolution integral, 178, 181 described, 76–77 of the discrete-time Fourier transform, 868–69 of discrete-time signals, 242 of the *z*-transform, 506–7 Time scaling, 77 of the bilateral Laplace transform, 452 described, 73–74 Time shifting, 77, 79 of the bilateral Laplace transform, 451 of the convolution integral, 178 described, 71–73 of the discrete Fourier transform, 819 of the discrete-time Fourier transform, 870 of the Fourier transform, 707 of the Laplace transform, 349–51 of the *z*-transform, 501–5, 510 Time-division multiplexing (TDM), 749, 797 Time-domain analysis, 723 of continuous-time systems, 150–236 of discrete-time systems, 237–329 of the Fourier series, 598, 601 of interpolation, 785–88 state equation solution in, 933–39 two-dimensional view and, 732–33 Time-frequency duality, 702–3, 723, 753 Time invariant systems, 134 discrete-time, 262 linear. *See* Linear time-invariant systems properties of, 102–3 Time-varying systems, 134 discrete-time, 262 linear, 103 properties of, 102–3 - -Time-limited signals, 802, 805, 807 Torque, 116–18 Torsional dashpots, 116 Torsional springs, 116, 117 Total response of continuous-time systems, 195–96 of discrete-time systems, 297–98 *Traité de mécanique céleste* (Laplace), 346 Transfer functions, 522 analog filter realization with, 548–49 block diagrams and, 386–88 of continuous-time systems, 193–94, 222 of discrete-time systems, 296–97, 314, 514–15, 567–68 from the frequency response, 435 inadequacy for system description, 953 realization of, 389–99, 401, 524–25 state equations from, 916, 919–26 from state-space representations, 964–65 Translational systems, 114–16 Transpose of a matrix, 37–38 Transposed direct form II (TDFII) realization, 398, 967–69 state equations and, 920–24 *z*-transform and, 520–22, 525–26 Triangular windows, 751 Trigonometric Fourier series, 640, 652, 657–58, 667, 668 exponential, 621–37, 661 periodic signals and, 593–612, 661 sampling and, 777, 782 symmetry effect on, 607–8 Trigonometric identities, 55–56 Tukey, J. W., 824 - -Underdamped systems, 409 Uniformly convergent series, 613 Unilateral Laplace transform, 333–36, 337, 338, 345, 360, 445, 467 Unilateral *z*-transform, 489, 491, 492, 495, 554–55, 559 Uniqueness, 335 Unit delay, 517, 520, 521 Unit-gate function, 689 Unit-impulse function, 133 of discrete-time systems, 246–47, 280, 313 as a generalized function, 88–89 properties of, 86–89 Unit-impulse response of continuous-time systems, 163–68, 170, 189–93, 220–21, 222, 731 convolution with, 171 determining, 221 of discrete-time systems, 277–80, 286, 295, 313 Unit matrices, 37 Unit-step function, 84–86, 88–89 of discrete-time systems, 246–47 relational operators and, 128–30 - -### 988 Index - -Unit-triangle function, 689–90 Unrepeated roots, 198, 202, 223, 301, 314 Unstable equilibrium, 196–97 Unstable systems, 110, 263 Upper sideband (USB), 737–39, 746–48 Upsampling, 243–44 Vectors, 36–37, 641–59 basis, 648 characteristic, 910 column, 36 components of, 642–43 error, 642 MATLAB operations, 45–46 matrix multiplication by, 40 orthogonal space, 647–48 row, 36, 45, 48–50 signals as, 641–59 state, 927–30, 961 Vestigial sideband (VSB), 749 Video signals, 725, 749 Waveshaping, 615–17 Weber–Fechner law, 421 Width of the convolution integral, 172, 187 of the convolution sum, 283 Window functions, 749–55, 760–62 *z*-transform, 488–592 bilateral. *See* Bilateral z-transform difference equation solutions of, 488, 510–19, 574 direct, 488–592 discrete-time Fourier transform and, 866–67, 886–88, 898 existence of, 491–95 - -inverse. *See* inverse *z*-transform - -properties of, 501–9 stability of, 518–19 state-space analysis and, 956, 959–65 system realization and, 519–25, 567 time-reversal property, 506–7 time-shifting properties, 501–5 unilateral, 489, 491, 492, 495, 554–55, 559 *z*-domain differentiation property, 506 *z*-domain scaling property, 505 Zero matrices, 37 Zero padding, 810–11, 829–30 Zero-input response, 119, 123 of continuous-time systems, 151–63, 195–96, 203, 220–22 described, 98–100 of discrete-time systems, 270–76, 297–301, 309–11 insights into behavior of, 161–63 of the Laplace transform, 363, 368 in oscillators, 203 of the *z*-transform, 512–13 zero-state response independence from, 161 Zero-order hold (ZOH) filters, 785 Zero-state response, 119, 123 alternate interpretation, 515–18 causality and, 172–73 of continuous-time systems, 151, 161, 168–96, 221–22, 512–16 described, 98–101 of discrete-time systems, 280–98, 308–9, 311, 312, 313 of the Laplace transform, 358, 363, 366–67, 369, 370 zero-input response independence from, 161 Zeros controlling gain by, 540 filter design, 436–45 first-order, 424–27 gain suppression by, 439–40 at the origin, 422–23 - -second-order, 426–35 \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/archive/linear-systems-and-signals-3rd-edition-bp-lathi.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/archive/linear-systems-and-signals-3rd-edition-bp-lathi.md deleted file mode 100644 index b975d2b91bed17e1fe1f7a8e53872b25ea2ae986..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/archive/linear-systems-and-signals-3rd-edition-bp-lathi.md +++ /dev/null @@ -1,34732 +0,0 @@ -# LINEAR SYSTEMS AND SIGNALS - -### **THE OXFORD SERIES IN ELECTRICAL AND COMPUTER ENGINEERING** - -**Adel S. Sedra,** Series Editor - -Allen and Holberg, *CMOS Analog Circuit Design, 3rd edition* Boncelet, *Probability, Statistics, and Random Signals* Bobrow, *Elementary Linear Circuit Analysis, 2nd edition* Bobrow, *Fundamentals of Electrical Engineering, 2nd edition* Campbell, *Fabrication Engineering at the Micro- and Nanoscale, 4th edition* Chen, *Digital Signal Processing* Chen, *Linear System Theory and Design, 4th edition* Chen, *Signals and Systems, 3rd edition* Comer, *Digital Logic and State Machine Design, 3rd edition* Comer, *Microprocessor-Based System Design* Cooper and McGillem, *Probabilistic Methods of Signal and System Analysis, 3rd edition* Dimitrijev, *Principles of Semiconductor Device, 2nd edition* Dimitrijev, *Understanding Semiconductor Devices* Fortney, *Principles of Electronics: Analog & Digital* Franco, *Electric Circuits Fundamentals* Ghausi, *Electronic Devices and Circuits: Discrete and Integrated* Guru and Hiziroglu, ˘ *Electric Machinery and Transformers, 3rd edition* Houts, *Signal Analysis in Linear Systems* Jones, *Introduction to Optical Fiber Communication Systems* Krein, *Elements of Power Electronics, 2nd Edition* Kuo, *Digital Control Systems, 3rd edition* Lathi and Green, *Linear Systems and Signals, 3rd edition* Lathi and Ding, *Modern Digital and Analog Communication Systems, 5th edition* Lathi, *Signal Processing and Linear Systems* Martin, *Digital Integrated Circuit Design* Miner, *Lines and Electromagnetic Fields for Engineers* Mitra, *Signals and Systems* Parhami, *Computer Architecture* Parhami, *Computer Arithmetic, 2nd edition* Roberts and Sedra, *SPICE, 2nd edition* Roberts, Taenzler, and Burns, *An Introduction to Mixed-Signal IC Test and Measurement, 2nd edition* Roulston, *An Introduction to the Physics of Semiconductor Devices* Sadiku, *Elements of Electromagnetics, 7th edition* Santina, Stubberud, and Hostetter, *Digital Control System Design, 2nd edition* Sarma, *Introduction to Electrical Engineering* Schaumann, Xiao, and Van Valkenburg, *Design of Analog Filters, 3rd edition* Schwarz and Oldham, *Electrical Engineering: An Introduction, 2nd edition* Sedra and Smith, *Microelectronic Circuits, 7th edition* Stefani, Shahian, Savant, and Hostetter, *Design of Feedback Control Systems, 4th edition* Tsividis, *Operation and Modeling of the MOS Transistor, 3rd edition* Van Valkenburg, *Analog Filter Design* Warner and Grung, *Semiconductor Device Electronics* Wolovich, *Automatic Control Systems* Yariv and Yeh, Photonics: *Optical Electronics in Modern Communications, 6th edition* Zak, ˙ *Systems and Control* - -# LINEAR SYSTEMS AND SIGNALS - -THIRD EDITION - -**B. P. Lathi and R. A. Green** - -New York Oxford OXFORD UNIVERSITY PRESS 2018 - -Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide. - -Oxford New York Auckland Cape Town Dar es Salaam Hong Kong Karachi Kuala Lumpur Madrid Melbourne Mexico City Nairobi New Delhi Shanghai Taipei Toronto - -With offices in Argentina Austria Brazil Chile Czech Republic France Greece Guatemala Hungary Italy Japan Poland Portugal Singapore South Korea Switzerland Thailand Turkey Ukraine Vietnam - -Copyright c 2018 by Oxford University Press - -For titles covered by Section 112 of the US Higher Education Opportunity Act, please visit [www.oup.com/us/he](http://www.oup.com/us/he) for the latest information about pricing and alternate formats. - -Published by Oxford University Press. 198 Madison Avenue, New York, NY 10016 - -Oxford is a registered trademark of Oxford University Press. - -All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of Oxford University Press. - -Library of Congress Cataloging-in-Publication Data Names: Lathi, B. P. (Bhagwandas Pannalal), author. | Green, R. A. (Roger A.), author. Title: Linear systems and signals / B.P. Lathi and R.A. Green. Description: Third Edition. | New York : Oxford University Press, [2018] | Series: The Oxford Series in Electrical and Computer Engineering Identifiers: LCCN 2017034962 | ISBN 9780190200176 (hardcover : acid-free paper) Subjects: LCSH: Signal processing–Mathematics. | System analysis. | Linear time invariant systems. | Digital filters (Mathematics) Classification: LCC TK5102.5 L298 2017 | DDC 621.382/2–dc23 LC record available at - -ISBN 978–0–19–020017–6 - -Printing number: 9 8 7 6 5 4 3 2 1 - -Printed by R.R. Donnelly in the United States of America - -# **CONTENTS** - -[PREFACE](#page-16-0) xv - -### B BACKGROUND - -- [B.1 Complex Numbers](#page-20-0) 1 - - [B.1-1 A Historical Note](#page-20-0) 1 - - [B.1-2 Algebra of Complex Numbers](#page-24-0) 5 -- [B.2 Sinusoids](#page-35-0) 16 - - [B.2-1 Addition of Sinusoids](#page-37-0) 18 - - [B.2-2 Sinusoids in Terms of Exponentials](#page-39-0) 20 -- [B.3 Sketching Signals](#page-39-0) 20 - - [B.3-1 Monotonic Exponentials](#page-39-0) 20 - - [B.3-2 The Exponentially Varying Sinusoid](#page-41-0) 22 -- [B.4 Cramer's Rule](#page-42-0) 23 -- [B.5 Partial Fraction Expansion](#page-44-0) 25 - - [B.5-1 Method of Clearing Fractions](#page-45-0) 26 - - [B.5-2 The Heaviside "Cover-Up" Method](#page-46-0) 27 - - [B.5-3 Repeated Factors of](#page-50-0) *Q*(*x*) 31 - - [B.5-4 A Combination of Heaviside "Cover-Up" and Clearing Fractions](#page-51-0) 32 - - [B.5-5 Improper](#page-53-0) *F*(*x*) with *m* = *n* 34 - - [B.5-6 Modified Partial Fractions](#page-54-0) 35 -- [B.6 Vectors and Matrices](#page-55-0) 36 - - [B.6-1 Some Definitions and Properties](#page-56-0) 37 - - [B.6-2 Matrix Algebra](#page-57-0) 38 -- [B.7 MATLAB: Elementary Operations](#page-61-0) 42 - - [B.7-1 MATLAB Overview](#page-61-0) 42 - - [B.7-2 Calculator Operations](#page-62-0) 43 - - [B.7-3 Vector Operations](#page-64-0) 45 - - [B.7-4 Simple Plotting](#page-65-0) 46 - - [B.7-5 Element-by-Element Operations](#page-67-0) 48 - - [B.7-6 Matrix Operations](#page-68-0) 49 - - [B.7-7 Partial Fraction Expansions](#page-72-0) 53 -- [B.8 Appendix: Useful Mathematical Formulas](#page-73-0) 54 - - [B.8-1 Some Useful Constants](#page-73-0) 54 - -- [B.8-2 Complex Numbers](#page-73-0) 54 -- [B.8-3 Sums](#page-73-0) 54 -- [B.8-4 Taylor and Maclaurin Series](#page-74-0) 55 -- [B.8-5 Power Series](#page-74-0) 55 -- [B.8-6 Trigonometric Identities](#page-74-0) 55 -- [B.8-7 Common Derivative Formulas](#page-75-0) 56 -- [B.8-8 Indefinite Integrals](#page-76-0) 57 -- [B.8-9 L'Hôpital's Rule](#page-77-0) 58 -- [B.8-10 Solution of Quadratic and Cubic Equations](#page-77-0) 58 -- *[References](#page-77-0)* 58 *[Problems](#page-78-0)* 59 - -### 1 SIGNALS AND SYSTEMS - -- [1.1 Size of a Signal](#page-83-0) 64 - - [1.1-1 Signal Energy](#page-84-0) 65 - - [1.1-2 Signal Power](#page-84-0) 65 -- [1.2 Some Useful Signal Operations](#page-90-0) 71 - - [1.2-1 Time Shifting](#page-90-0) 71 - - [1.2-2 Time Scaling](#page-92-0) 73 - - [1.2-3 Time Reversal](#page-95-0) 76 - - [1.2-4 Combined Operations](#page-96-0) 77 -- [1.3 Classification of Signals](#page-97-0) 78 - - [1.3-1 Continuous-Time and Discrete-Time Signals](#page-97-0) 78 - - [1.3-2 Analog and Digital Signals](#page-97-0) 78 - - [1.3-3 Periodic and Aperiodic Signals](#page-98-0) 79 - - [1.3-4 Energy and Power Signals](#page-101-0) 82 - - [1.3-5 Deterministic and Random Signals](#page-101-0) 82 -- [1.4 Some Useful Signal Models](#page-101-0) 82 - - [1.4-1 The Unit Step Function](#page-102-0) *u*(*t*) 83 - - [1.4-2 The Unit Impulse Function](#page-105-0) δ(*t*) 86 - - [1.4-3 The Exponential Function](#page-108-0) *est* 89 -- [1.5 Even and Odd Functions](#page-111-0) 92 - - [1.5-1 Some Properties of Even and Odd Functions](#page-111-0) 92 - - [1.5-2 Even and Odd Components of a Signal](#page-112-0) 93 -- [1.6 Systems](#page-114-0) 95 -- [1.7 Classification of Systems](#page-116-0) 97 - - [1.7-1 Linear and Nonlinear Systems](#page-116-0) 97 - - [1.7-2 Time-Invariant and Time-Varying Systems](#page-121-0) 102 - - [1.7-3 Instantaneous and Dynamic Systems](#page-122-0) 103 - - [1.7-4 Causal and Noncausal Systems](#page-123-0) 104 - - [1.7-5 Continuous-Time and Discrete-Time Systems](#page-126-0) 107 - - [1.7-6 Analog and Digital Systems](#page-128-0) 109 - - [1.7-7 Invertible and Noninvertible Systems](#page-128-0) 109 - - [1.7-8 Stable and Unstable Systems](#page-129-0) 110 - -- [1.8 System Model: Input–Output Description](#page-130-0) 111 - - [1.8-1 Electrical Systems](#page-130-0) 111 - - [1.8-2 Mechanical Systems](#page-133-0) 114 - - [1.8-3 Electromechanical Systems](#page-137-0) 118 -- [1.9 Internal and External Descriptions of a System](#page-138-0) 119 -- [1.10 Internal Description: The State-Space Description](#page-140-0) 121 -- [1.11 MATLAB: Working with Functions](#page-145-0) 126 - - [1.11-1 Anonymous Functions](#page-145-0) 126 - - [1.11-2 Relational Operators and the Unit Step Function](#page-147-0) 128 - - [1.11-3 Visualizing Operations on the Independent Variable](#page-149-0) 130 - - [1.11-4 Numerical Integration and Estimating Signal Energy](#page-150-0) 131 -- [1.12 Summary](#page-152-0) 133 - -*[References](#page-154-0)* 135 *[Problems](#page-155-0)* 136 - -### 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -- [2.1 Introduction](#page-169-0) 150 -- [2.2 System Response to Internal Conditions: The Zero-Input Response](#page-170-0) 151 [2.2-1 Some Insights into the Zero-Input Behavior of a System](#page-180-0) 161 -- [2.3 The Unit Impulse Response](#page-182-0) *h*(*t*) 163 -- [2.4 System Response to External Input: The Zero-State Response](#page-187-0) 168 - - [2.4-1 The Convolution Integral](#page-189-0) 170 - - [2.4-2 Graphical Understanding of Convolution Operation](#page-197-0) 178 - - [2.4-3 Interconnected Systems](#page-209-0) 190 - - [2.4-4 A Very Special Function for LTIC Systems:](#page-212-0) - - The Everlasting Exponential *est* 193 - - [2.4-5 Total Response](#page-214-0) 195 -- [2.5 System Stability](#page-215-0) 196 - - [2.5-1 External \(BIBO\) Stability](#page-215-0) 196 - - [2.5-2 Internal \(Asymptotic\) Stability](#page-217-0) 198 - - [2.5-3 Relationship Between BIBO and Asymptotic Stability](#page-218-0) 199 -- [2.6 Intuitive Insights into System Behavior](#page-222-0) 203 - - [2.6-1 Dependence of System Behavior on Characteristic Modes](#page-222-0) 203 - - [2.6-2 Response Time of a System: The System Time Constant](#page-224-0) 205 - - [2.6-3 Time Constant and Rise Time of a System](#page-225-0) 206 - - [2.6-4 Time Constant and Filtering](#page-226-0) 207 - - [2.6-5 Time Constant and Pulse Dispersion \(Spreading\)](#page-228-0) 209 - - [2.6-6 Time Constant and Rate of Information Transmission](#page-228-0) 209 - - [2.6-7 The Resonance Phenomenon](#page-229-0) 210 -- [2.7 MATLAB: M-Files](#page-231-0) 212 - - [2.7-1 Script M-Files](#page-232-0) 213 - - [2.7-2 Function M-Files](#page-233-0) 214 - -- [2.7-3 For-Loops](#page-234-0) 215 -- [2.7-4 Graphical Understanding of Convolution](#page-236-0) 217 -- [2.8 Appendix: Determining the Impulse Response](#page-239-0) 220 -- [2.9 Summary](#page-240-0) 221 - -*[References](#page-242-0)* 223 *[Problems](#page-242-0)* 223 - -### 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -- [3.1 Introduction](#page-256-0) 237 - - [3.1-1 Size of a Discrete-Time Signal](#page-257-0) 238 -- [3.2 Useful Signal Operations](#page-259-0) 240 -- [3.3 Some Useful Discrete-Time Signal Models](#page-264-0) 245 - - [3.3-1 Discrete-Time Impulse Function](#page-264-0) δ[*n*] 245 - - [3.3-2 Discrete-Time Unit Step Function](#page-265-0) *u*[*n*] 246 - - [3.3-3 Discrete-Time Exponential](#page-266-0) γ *n* 247 - - [3.3-4 Discrete-Time Sinusoid cos](#page-270-0)(*n*+θ ) 251 - - [3.3-5 Discrete-Time Complex Exponential](#page-271-0) *ejn* 252 -- [3.4 Examples of Discrete-Time Systems](#page-272-0) 253 - - [3.4-1 Classification of Discrete-Time Systems](#page-281-0) 262 -- [3.5 Discrete-Time System Equations](#page-284-0) 265 - -[3.5-1 Recursive \(Iterative\) Solution of Difference Equation](#page-285-0) 266 - -- [3.6 System Response to Internal Conditions: The Zero-Input Response](#page-289-0) 270 -- [3.7 The Unit Impulse Response](#page-296-0) *h*[*n*] 277 - - [3.7-1 The Closed-Form Solution of](#page-297-0) *h*[*n*] 278 -- [3.8 System Response to External Input: The Zero-State Response](#page-299-0) 280 - - [3.8-1 Graphical Procedure for the Convolution Sum](#page-307-0) 288 - - [3.8-2 Interconnected Systems](#page-313-0) 294 - - [3.8-3 Total Response](#page-316-0) 297 -- [3.9 System Stability](#page-317-0) 298 - - [3.9-1 External \(BIBO\) Stability](#page-317-0) 298 - - [3.9-2 Internal \(Asymptotic\) Stability](#page-318-0) 299 - - [3.9-3 Relationship Between BIBO and Asymptotic Stability](#page-320-0) 301 -- [3.10 Intuitive Insights into System Behavior](#page-324-0) 305 -- [3.11 MATLAB: Discrete-Time Signals and Systems](#page-325-0) 306 - - [3.11-1 Discrete-Time Functions and Stem Plots](#page-325-0) 306 - - [3.11-2 System Responses Through Filtering](#page-327-0) 308 - - [3.11-3 A Custom Filter Function](#page-329-0) 310 - - [3.11-4 Discrete-Time Convolution](#page-330-0) 311 -- [3.12 Appendix: Impulse Response for a Special Case](#page-332-0) 313 -- [3.13 Summary](#page-332-0) 313 - -*[Problems](#page-333-0)* 314 - -### 4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM - -- [4.1 The Laplace Transform](#page-349-0) 330 - - [4.1-1 Finding the Inverse Transform](#page-357-0) 338 -- [4.2 Some Properties of the Laplace Transform](#page-368-0) 349 - - [4.2-1 Time Shifting](#page-368-0) 349 - - [4.2-2 Frequency Shifting](#page-372-0) 353 - - [4.2-3 The Time-Differentiation Property](#page-373-0) 354 - - [4.2-4 The Time-Integration Property](#page-375-0) 356 - - [4.2-5 The Scaling Property](#page-376-0) 357 - - [4.2-6 Time Convolution and Frequency Convolution](#page-376-0) 357 -- [4.3 Solution of Differential and Integro-Differential Equations](#page-379-0) 360 - - [4.3-1 Comments on Initial Conditions at 0](#page-382-0)− and at 0+ 363 - - [4.3-2 Zero-State Response](#page-385-0) 366 - - [4.3-3 Stability](#page-390-0) 371 - - [4.3-4 Inverse Systems](#page-392-0) 373 -- [4.4 Analysis of Electrical Networks: The Transformed Network](#page-392-0) 373 - - [4.4-1 Analysis of Active Circuits](#page-401-0) 382 -- [4.5 Block Diagrams](#page-405-0) 386 -- [4.6 System Realization](#page-407-0) 388 - - [4.6-1 Direct Form I Realization](#page-408-0) 389 - - [4.6-2 Direct Form II Realization](#page-409-0) 390 - - [4.6-3 Cascade and Parallel Realizations](#page-412-0) 393 - - [4.6-4 Transposed Realization](#page-415-0) 396 - - [4.6-5 Using Operational Amplifiers for System Realization](#page-418-0) 399 -- [4.7 Application to Feedback and Controls](#page-423-0) 404 - - [4.7-1 Analysis of a Simple Control System](#page-425-0) 406 -- [4.8 Frequency Response of an LTIC System](#page-431-0) 412 - - [4.8-1 Steady-State Response to Causal Sinusoidal Inputs](#page-437-0) 418 -- [4.9 Bode Plots](#page-438-0) 419 - - [4.9-1 Constant](#page-441-0) *Ka*1*a*2/*b*1*b*3 422 - - [4.9-2 Pole \(or Zero\) at the Origin](#page-441-0) 422 - - [4.9-3 First-Order Pole \(or Zero\)](#page-443-0) 424 - - [4.9-4 Second-Order Pole \(or Zero\)](#page-445-0) 426 - - [4.9-5 The Transfer Function from the Frequency Response](#page-454-0) 435 -- [4.10 Filter Design by Placement of Poles and Zeros of](#page-455-0) *H*(*s*) 436 - - [4.10-1 Dependence of Frequency Response on Poles](#page-455-0) and Zeros of *H*(*s*) 436 - - [4.10-2 Lowpass Filters](#page-458-0) 439 - - [4.10-3 Bandpass Filters](#page-460-0) 441 - - [4.10-4 Notch \(Bandstop\) Filters](#page-460-0) 441 - - [4.10-5 Practical Filters and Their Specifications](#page-463-0) 444 -- [4.11 The Bilateral Laplace Transform](#page-464-0) 445 - -- [4.11-1 Properties of the Bilateral Laplace Transform](#page-470-0) 451 -- [4.11-2 Using the Bilateral Transform for Linear System Analysis](#page-471-0) 452 -- [4.12 MATLAB: Continuous-Time Filters](#page-474-0) 455 - - [4.12-1 Frequency Response and Polynomial Evaluation](#page-475-0) 456 - - [4.12-2 Butterworth Filters and the](#page-478-0) Find Command 459 - - [4.12-3 Using Cascaded Second-Order Sections for Butterworth](#page-480-0) Filter Realization 461 - - [4.12-4 Chebyshev Filters](#page-482-0) 463 -- [4.13 Summary](#page-485-0) 466 - -*[References](#page-487-0)* 468 *[Problems](#page-487-0)* 468 - -### 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *z*-TRANSFORM - -- 5.1 The *z*[-Transform](#page-507-0) 488 - - [5.1-1 Inverse Transform by Partial Fraction Expansion and Tables](#page-514-0) 495 - - 5.1-2 Inverse *z*[-Transform by Power Series Expansion](#page-518-0) 499 -- [5.2 Some Properties of the](#page-520-0) *z*-Transform 501 - - [5.2-1 Time-Shifting Properties](#page-520-0) 501 - - 5.2-2 *z*[-Domain Scaling Property \(Multiplication by](#page-524-0) γ *n*) 505 - - 5.2-3 *z*[-Domain Differentiation Property \(Multiplication by](#page-525-0) *n*) 506 - - [5.2-4 Time-Reversal Property](#page-525-0) 506 - - [5.2-5 Convolution Property](#page-526-0) 507 -- 5.3 *z*[-Transform Solution of Linear Difference Equations](#page-529-0) 510 - - [5.3-1 Zero-State Response of LTID Systems: The Transfer Function](#page-533-0) 514 - - [5.3-2 Stability](#page-537-0) 518 - - [5.3-3 Inverse Systems](#page-538-0) 519 -- [5.4 System Realization](#page-538-0) 519 -- 5.5 Frequency Response of Discrete-Time Systems [526](#page-545-0) - - [5.5-1 The Periodic Nature of Frequency Response](#page-551-0) 532 - - [5.5-2 Aliasing and Sampling Rate](#page-555-0) 536 -- [5.6 Frequency Response from Pole-Zero Locations](#page-557-0) 538 -- [5.7 Digital Processing of Analog Signals](#page-566-0) 547 -- [5.8 The Bilateral](#page-573-0) *z*-Transform 554 - - [5.8-1 Properties of the Bilateral](#page-578-0) *z*-Transform 559 - - 5.8-2 Using the Bilateral *z*[-Transform for Analysis of LTID Systems](#page-579-0) 560 -- [5.9 Connecting the Laplace and](#page-582-0) *z*-Transforms 563 -- [5.10 MATLAB: Discrete-Time IIR Filters](#page-584-0) 565 - - [5.10-1 Frequency Response and Pole-Zero Plots](#page-585-0) 566 - - [5.10-2 Transformation Basics](#page-586-0) 567 - - [5.10-3 Transformation by First-Order Backward Difference](#page-587-0) 568 - - [5.10-4 Bilinear Transformation](#page-588-0) 569 - - [5.10-5 Bilinear Transformation with Prewarping](#page-589-0) 570 - - [5.10-6 Example: Butterworth Filter Transformation](#page-590-0) 571 - -[5.10-7 Problems Finding Polynomial Roots](#page-591-0) 572 - -[5.10-8 Using Cascaded Second-Order Sections to Improve Design](#page-591-0) 572 - -[5.11 Summary](#page-593-0) 574 - -*[References](#page-594-0)* 575 *[Problems](#page-594-0)* 575 - -### 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -- [6.1 Periodic Signal Representation by Trigonometric Fourier Series](#page-612-0) 593 - - [6.1-1 The Fourier Spectrum](#page-617-0) 598 - - [6.1-2 The Effect of Symmetry](#page-626-0) 607 - - [6.1-3 Determining the Fundamental Frequency and Period 609](#page-628-0) -- [6.2 Existence and Convergence of the Fourier Series](#page-631-0) 612 - - [6.2-1 Convergence of a Series](#page-632-0) 613 - - [6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping](#page-634-0) 615 -- [6.3 Exponential Fourier Series](#page-640-0) 621 - - [6.3-1 Exponential Fourier Spectra](#page-643-0) 624 - - [6.3-2 Parseval's Theorem](#page-651-0) 632 - - [6.3-3 Properties of the Fourier Series](#page-654-0) 635 -- [6.4 LTIC System Response to Periodic Inputs](#page-656-0) 637 -- [6.5 Generalized Fourier Series: Signals as Vectors](#page-660-0) 641 - - [6.5-1 Component of a Vector](#page-661-0) 642 - - [6.5-2 Signal Comparison and Component of a Signal](#page-662-0) 643 - - [6.5-3 Extension to Complex Signals](#page-664-0) 645 - - [6.5-4 Signal Representation by an Orthogonal Signal Set](#page-666-0) 647 -- [6.6 Numerical Computation of](#page-678-0) *Dn* 659 -- [6.7 MATLAB: Fourier Series Applications](#page-680-0) 661 - - [6.7-1 Periodic Functions and the Gibbs Phenomenon](#page-680-0) 661 - - [6.7-2 Optimization and Phase Spectra](#page-683-0) 664 -- [6.8 Summary](#page-686-0) 667 *[References](#page-687-0)* 668 *[Problems](#page-688-0)* 669 - -## 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -- [7.1 Aperiodic Signal Representation by the Fourier Integral](#page-699-0) 680 [7.1-1 Physical Appreciation of the Fourier Transform](#page-706-0) 687 -- [7.2 Transforms of Some Useful Functions](#page-708-0) 689 - - [7.2-1 Connection Between the Fourier and Laplace Transforms](#page-719-0) 700 -- [7.3 Some Properties of the Fourier Transform](#page-720-0) 701 -- [7.4 Signal Transmission Through LTIC Systems](#page-740-0) 721 - - [7.4-1 Signal Distortion During Transmission](#page-742-0) 723 - - [7.4-2 Bandpass Systems and Group Delay](#page-745-0) 726 - -### xii Contents - -- [7.5 Ideal and Practical Filters](#page-749-0) 730 -- [7.6 Signal Energy](#page-752-0) 733 -- [7.7 Application to Communications: Amplitude Modulation](#page-755-0) 736 - - [7.7-1 Double-Sideband, Suppressed-Carrier \(DSB-SC\) Modulation](#page-756-0) 737 - - [7.7-2 Amplitude Modulation \(AM\)](#page-761-0) 742 - - [7.7-3 Single-Sideband Modulation \(SSB\)](#page-765-0) 746 - - [7.7-4 Frequency-Division Multiplexing](#page-768-0) 749 -- [7.8 Data Truncation: Window Functions](#page-768-0) 749 - - [7.8-1 Using Windows in Filter Design](#page-774-0) 755 -- [7.9 MATLAB: Fourier Transform Topics](#page-774-0) 755 - - [7.9-1 The Sinc Function and the Scaling Property](#page-776-0) 757 - - [7.9-2 Parseval's Theorem and Essential Bandwidth](#page-777-0) 758 - - [7.9-3 Spectral Sampling](#page-778-0) 759 - - [7.9-4 Kaiser Window Functions](#page-779-0) 760 -- [7.10 Summary](#page-781-0) 762 - -*[References](#page-782-0)* 763 *[Problems](#page-783-0)* 764 - -### 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -- [8.1 The Sampling Theorem](#page-795-0) 776 [8.1-1 Practical Sampling](#page-800-0) 781 -- [8.2 Signal Reconstruction](#page-804-0) 785 - - [8.2-1 Practical Difficulties in Signal Reconstruction](#page-807-0) 788 - - [8.2-2 Some Applications of the Sampling Theorem](#page-815-0) 796 -- [8.3 Analog-to-Digital \(A/D\) Conversion](#page-818-0) 799 -- [8.4 Dual of Time Sampling: Spectral Sampling](#page-821-0) 802 -- [8.5 Numerical Computation of the Fourier Transform:](#page-824-0) The Discrete Fourier Transform 805 - - [8.5-1 Some Properties of the DFT](#page-837-0) 818 - - [8.5-2 Some Applications of the DFT](#page-839-0) 820 -- [8.6 The Fast Fourier Transform \(FFT\)](#page-843-0) 824 -- [8.7 MATLAB: The Discrete Fourier Transform](#page-846-0) 827 - - [8.7-1 Computing the Discrete Fourier Transform](#page-846-0) 827 - - [8.7-2 Improving the Picture with Zero Padding](#page-848-0) 829 - - [8.7-3 Quantization](#page-850-0) 831 -- [8.8 Summary](#page-853-0) 834 - -*[References](#page-854-0)* 835 *[Problems](#page-854-0)* 835 - -### 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -- [9.1 Discrete-Time Fourier Series \(DTFS\)](#page-864-0) 845 - - [9.1-1 Periodic Signal Representation by Discrete-Time Fourier Series](#page-865-0) 846 - - [9.1-2 Fourier Spectra of a Periodic Signal](#page-867-0) *x*[*n*] 848 -- [9.2 Aperiodic Signal Representation](#page-874-0) - -by Fourier Integral 855 - -- [9.2-1 Nature of Fourier Spectra](#page-877-0) 858 -- [9.2-2 Connection Between the DTFT and the](#page-885-0) *z*-Transform 866 -- [9.3 Properties of the DTFT](#page-886-0) 867 -- [9.4 LTI Discrete-Time System Analysis by DTFT](#page-897-0) 878 - - [9.4-1 Distortionless Transmission](#page-899-0) 880 - - [9.4-2 Ideal and Practical Filters](#page-901-0) 882 -- [9.5 DTFT Connection with the CTFT](#page-902-0) 883 - - [9.5-1 Use of DFT and FFT for Numerical Computation of the DTFT](#page-904-0) 885 -- [9.6 Generalization of the DTFT to the](#page-905-0) *z*-Transform 886 -- [9.7 MATLAB: Working with the DTFS and the DTFT](#page-908-0) 889 - - [9.7-1 Computing the Discrete-Time Fourier Series](#page-908-0) 889 - - [9.7-2 Measuring Code Performance](#page-910-0) 891 - - [9.7-3 FIR Filter Design by Frequency Sampling](#page-911-0) 892 -- [9.8 Summary](#page-917-0) 898 - -*[Reference](#page-917-0)* 898 *[Problems](#page-918-0)* 899 - -### 10 STATE-SPACE ANALYSIS - -- [10.1 Mathematical Preliminaries](#page-928-0) 909 - - [10.1-1 Derivatives and Integrals of a Matrix](#page-928-0) 909 - - [10.1-2 The Characteristic Equation of a Matrix:](#page-929-0) - - The Cayley–Hamilton Theorem 910 - - [10.1-3 Computation of an Exponential and a Power of a Matrix](#page-931-0) 912 -- [10.2 Introduction to State Space](#page-932-0) 913 -- [10.3 A Systematic Procedure to Determine State Equations](#page-935-0) 916 - - [10.3-1 Electrical Circuits](#page-935-0) 916 - - [10.3-2 State Equations from a Transfer Function](#page-938-0) 919 -- [10.4 Solution of State Equations](#page-945-0) 926 - - [10.4-1 Laplace Transform Solution of State Equations](#page-946-0) 927 - - [10.4-2 Time-Domain Solution of State Equations](#page-952-0) 933 -- [10.5 Linear Transformation of a State Vector](#page-958-0) 939 - - [10.5-1 Diagonalization of Matrix](#page-962-0) **A** 943 -- [10.6 Controllability and Observability](#page-966-0) 947 - - [10.6-1 Inadequacy of the Transfer Function Description of a System](#page-972-0) 953 - -[10.7 State-Space Analysis of Discrete-Time Systems](#page-972-0) 953 - -[10.7-1 Solution in State Space](#page-974-0) 955 - -10.7-2 The *z*[-Transform Solution](#page-978-0) 959 - -[10.8 MATLAB: Toolboxes and State-Space Analysis](#page-980-0) 961 - -10.8-1 *z*[-Transform Solutions to Discrete-Time, State-Space Systems](#page-980-0) 961 - -[10.8-2 Transfer Functions from State-Space Representations](#page-983-0) 964 - -[10.8-3 Controllability and Observability of Discrete-Time Systems](#page-984-0) 965 - -[10.8-4 Matrix Exponentiation and the Matrix Exponential](#page-987-0) 968 - -[10.9 Summary](#page-988-0) 969 - -*[References](#page-989-0)* 970 *[Problems](#page-989-0)* 970 - -[INDEX](#page-994-0) 975 - -# **[PREFACE](#page-6-0)** - -This book, *Linear Systems and Signals,* presents a comprehensive treatment of signals and linear systems at an introductory level. Following our preferred style, it emphasizes a physical appreciation of concepts through heuristic reasoning and the use of metaphors, analogies, and creative explanations. Such an approach is much different from a purely deductive technique that uses mere mathematical manipulation of symbols. There is a temptation to treat engineering subjects as a branch of applied mathematics. Such an approach is a perfect match to the public image of engineering as a dry and dull discipline. It ignores the physical meaning behind various derivations and deprives students of intuitive grasp and the enjoyable experience of logical uncovering of the subject matter. In this book, we use mathematics not so much to prove axiomatic theory as to support and enhance physical and intuitive understanding. Wherever possible, theoretical results are interpreted heuristically and are enhanced by carefully chosen examples and analogies. - -This third edition, which closely follows the organization of the second edition, has been refined in many ways. Discussions are streamlined, adding or trimming material as needed. Equation, example, and section labeling is simplified and improved. Computer examples are fully updated to reflect the most current version of MATLAB. Hundreds of added problems provide new opportunities to learn and understand topics. We have taken special care to improve the text without the topic creep and bloat that commonly occurs with each new edition of a text. - -## **NOTABLE FEATURES** - -The notable features of this book include the following. - -- 1. Intuitive and heuristic understanding of the concepts and physical meaning of mathematical results are emphasized throughout. Such an approach not only leads to deeper appreciation and easier comprehension of the concepts, but also makes learning enjoyable for students. -- 2. Often, students lack an adequate background in basic material such as complex numbers, sinusoids, hand-sketching of functions, Cramer's rule, partial fraction expansion, and matrix algebra. We include a background chapter that addresses these basic and pervasive topics in electrical engineering. Response by students has been unanimously enthusiastic. -- 3. There are hundreds of worked examples in addition to drills (usually with answers) for students to test their understanding. Additionally, there are over 900 end-of-chapter problems of varying difficulty. -- 4. Modern electrical engineering practice requires the use of computer calculation and simulation, most often using the software package MATLAB. Thus, we integrate - -MATLAB into many of the worked examples throughout the book. Additionally, each chapter concludes with a section devoted to learning and using MATLAB in the context and support of book topics. Problem sets also contain numerous computer problems. - -- 5. The discrete-time and continuous-time systems may be treated in sequence, or they may be integrated by using a parallel approach. -- 6. The summary at the end of each chapter proves helpful to students in summing up essential developments in the chapter. -- 7. There are several historical notes to enhance students' interest in the subject. This information introduces students to the historical background that influenced the development of electrical engineering. - -## **ORGANIZATION** - -The book may be conceived as divided into five parts: - -- 1. Introduction (Chs. B and 1). -- 2. Time-domain analysis of linear time-invariant (LTI) systems (Chs. 2 and 3). -- 3. Frequency-domain (transform) analysis of LTI systems (Chs. 4 and 5). -- 4. Signal analysis (Chs. 6, 7, 8, and 9). -- 5. State-space analysis of LTI systems (Ch. 10). - -The organization of the book permits much flexibility in teaching the continuous-time and discrete-time concepts. The natural sequence of chapters is meant to integrate continuous-time and discrete-time analysis. It is also possible to use a sequential approach in which all the continuous-time analysis is covered first (Chs. 1, 2, 4, 6, 7, and 8), followed by discrete-time analysis (Chs. 3, 5, and 9). - -## **SUGGESTIONS FOR USING THIS BOOK** - -The book can be readily tailored for a variety of courses spanning 30 to 45 lecture hours. Most of the material in the first eight chapters can be covered at a brisk pace in about 45 hours. The book can also be used for a 30-lecture-hour course by covering only analog material (Chs. 1, 2, 4, 6, 7, and possibly selected topics in Ch. 8). Alternately, one can also select Chs. 1 to 5 for courses purely devoted to systems analysis or transform techniques. To treat continuous- and discrete-time systems by using an integrated (or parallel) approach, the appropriate sequence of chapters is 1, 2, 3, 4, 5, 6, 7, and 8. For a sequential approach, where the continuous-time analysis is followed by discrete-time analysis, the proper chapter sequence is 1, 2, 4, 6, 7, 8, 3, 5, and possibly 9 (depending on the time available). - -## **MATLAB** - -MATLAB is a sophisticated language that serves as a powerful tool to better understand engineering topics, including control theory, filter design, and, of course, linear systems and signals. MATLAB's flexible programming structure promotes rapid development and analysis. Outstanding visualization capabilities provide unique insight into system behavior and signal character. - -As with any language, learning MATLAB is incremental and requires practice. This book provides two levels of exposure to MATLAB. First, MATLAB is integrated into many examples throughout the text to reinforce concepts and perform various computations. These examples utilize standard MATLAB functions as well as functions from the control system, signal-processing, and symbolic math toolboxes. MATLAB has many more toolboxes available, but these three are commonly available in most engineering departments. - -A second and deeper level of exposure to MATLAB is achieved by concluding each chapter with a separate MATLAB section. Taken together, these eleven sections provide a self-contained introduction to the MATLAB environment that allows even novice users to quickly gain MATLAB proficiency and competence. These sessions provide detailed instruction on how to use MATLAB to solve problems in linear systems and signals. Except for the very last chapter, special care has been taken to avoid the use of toolbox functions in the MATLAB sessions. Rather, readers are shown the process of developing their own code. In this way, those readers without toolbox access are not at a disadvantage. All of this book's MATLAB code is available for download at the OUP companion website [www.oup.com/us/lathi.](http://www.oup.com/us/lathi) - -## **CREDITS AND ACKNOWLEDGMENTS** - -The portraits of Gauss, Laplace, Heaviside, Fourier, and Michelson have been reprinted courtesy of the Smithsonian Institution Libraries. The likenesses of Cardano and Gibbs have been reprinted courtesy of the Library of Congress. The engraving of Napoleon has been reprinted courtesy of Bettmann/Corbis. The many fine cartoons throughout the text are the work of Joseph Coniglio, a former student of Dr. Lathi. - -Many individuals have helped us in the preparation of this book, as well as its earlier editions. We are grateful to each and every one for helpful suggestions and comments. Book writing is an obsessively time-consuming activity, which causes much hardship for an author's family. We both are grateful to our families for their enormous but invisible sacrifices. - -> *B. P. Lathi R. A. Green* - - - -# **[BACKGROUND](#page-6-0)** - -The topics discussed in this chapter are not entirely new to students taking this course. You have already studied many of these topics in earlier courses or are expected to know them from your previous training. Even so, this background material deserves a review because it is so pervasive in the area of signals and systems. Investing a little time in such a review will pay big dividends later. Furthermore, this material is useful not only for this course but also for several courses that follow. It will also be helpful later, as reference material in your professional career. - -## **[B.1 COMPLEX](#page-6-0) NUMBERS** - -*Complex numbers* are an extension of ordinary numbers and are an integral part of the modern number system. Complex numbers, particularly *imaginary numbers,* sometimes seem mysterious and unreal. This feeling of unreality derives from their unfamiliarity and novelty rather than their supposed nonexistence! Mathematicians blundered in calling these numbers "imaginary," for the term immediately prejudices perception. Had these numbers been called by some other name, they would have become demystified long ago, just as irrational numbers or negative numbers were. Many futile attempts have been made to ascribe some physical meaning to imaginary numbers. However, this effort is needless. In mathematics we assign symbols and operations any meaning we wish as long as internal consistency is maintained. The history of mathematics is full of entities that were unfamiliar and held in abhorrence until familiarity made them acceptable. This fact will become clear from the following historical note. - -### **[B.1-1 A Historical Note](#page-6-0)** - -Among early people the number system consisted only of natural numbers (positive integers) needed to express the number of children, cattle, and quivers of arrows. These people had no need for fractions. Whoever heard of two and one-half children or three and one-fourth cows! - -However, with the advent of agriculture, people needed to measure continuously varying quantities, such as the length of a field and the weight of a quantity of butter. The number system, therefore, was extended to include fractions. The ancient Egyptians and Babylonians knew how - -#### 2 CHAPTER B BACKGROUND - -to handle fractions, but *Pythagoras* discovered that some numbers (like the diagonal of a unit square) could not be expressed as a whole number or a fraction. Pythagoras, a number mystic, who regarded numbers as the essence and principle of all things in the universe, was so appalled at his discovery that he swore his followers to secrecy and imposed a death penalty for divulging this secret [1]. These numbers, however, were included in the number system by the time of Descartes, and they are now known as *irrational numbers*. - -Until recently, *negative numbers* were not a part of the number system. The concept of negative numbers must have appeared absurd to early man. However, the medieval Hindus had a clear understanding of the significance of positive and negative numbers [2, 3]. They were also the first to recognize the existence of absolute negative quantities [4]. The works of *Bhaskar* (1114–1185) on arithmetic (*L*¯*ilavat* ¯ ¯*i*) and algebra (*B*¯*ijaganit*) not only use the decimal system but also give rules for dealing with negative quantities. Bhaskar recognized that positive numbers have two square roots [5]. Much later, in Europe, the men who developed the banking system that arose in Florence and Venice during the late Renaissance (fifteenth century) are credited with introducing a crude form of negative numbers. The seemingly absurd subtraction of 7 from 5 seemed reasonable when bankers began to allow their clients to draw seven gold ducats while their deposit stood at five. All that was necessary for this purpose was to write the difference, 2, on the debit side of a ledger [6]. - -Thus, the number system was once again broadened (generalized) to include negative numbers. The acceptance of negative numbers made it possible to solve equations such as *x*+5=0, which had no solution before. Yet for equations such as *x*2 + 1 = 0, leading to *x*2 = −1, the solution could not be found in the real number system. It was therefore necessary to define a completely new kind of number with its square equal to −1. During the time of Descartes and Newton, imaginary (or complex) numbers came to be accepted as part of the number system, but they were still regarded as algebraic fiction. The Swiss mathematician *Leonhard Euler* introduced the notation *i* (for *imaginary*) around 1777 to represent √−1. Electrical engineers use the notation *j* instead of *i* to avoid confusion with the notation *i* often used for electrical current. Thus, - -$$ -j^2 = -1 \qquad \text{and} \qquad \sqrt{-1} = \pm j -$$ - -This notation allows us to determine the square root of any negative number. For example, - -$$ -\sqrt{-4} = \sqrt{4} \times \sqrt{-1} = \pm 2j -$$ - -When imaginary numbers are included in the number system, the resulting numbers are called *complex numbers*. - -### ORIGINS OF COMPLEX NUMBERS - -Ironically (and contrary to popular belief), it was not the solution of a quadratic equation, such as *x*2 + 1 = 0, but a cubic equation with real roots that made imaginary numbers plausible and acceptable to early mathematicians. They could dismiss √−1 as pure nonsense when it appeared as a solution to *x*2 + 1 = 0 because this equation has no real solution. But in 1545, *Gerolamo Cardano* of Milan published *Ars Magna* (The Great Art), the most important algebraic work of the Renaissance. In this book, he gave a method of solving a general cubic equation in which a root of a negative number appeared in an intermediate step. According to his method, the solution to a third-order equation† - -$$ -x^3 + ax + b = 0 -$$ - -is given by - -$$ -x = \sqrt[3]{-\frac{b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}} + \sqrt[3]{-\frac{b}{2} - \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}} -$$ - -For example, to find a solution of *x*3 + 6*x* − 20 = 0, we substitute *a* = 6,*b* = −20 in the foregoing equation to obtain - -$$ -x = \sqrt[3]{10 + \sqrt{108}} + \sqrt[3]{10 - \sqrt{108}} = \sqrt[3]{20.392} - \sqrt[3]{0.392} = 2 -$$ - -We can readily verify that 2 is indeed a solution of *x*3 + 6*x* − 20 = 0. But when Cardano tried to solve the equation *x*3 −15*x* −4 = 0 by this formula, his solution was - -$$ -x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}} -$$ - -What was Cardano to make of this equation in the year 1545? In those days, negative numbers were themselves suspect, and a square root of a negative number was doubly preposterous! Today, we know that - -$$ -(2 \pm j)^3 = 2 \pm j11 = 2 \pm \sqrt{-121} -$$ - -Therefore, Cardano's formula gives - -$$ -x = (2+j) + (2-j) = 4 -$$ - -We can readily verify that *x* = 4 is indeed a solution of *x*3 − 15*x* − 4 = 0. Cardano tried to explain halfheartedly the presence of √−121 but ultimately dismissed the whole enterprise as being "as subtle as it is useless." A generation later, however, *Raphael Bombelli* (1526–1573), after examining Cardano's results, proposed acceptance of imaginary numbers as a necessary vehicle that would transport the mathematician from the *real* cubic equation to its *real* solution. In other words, although we begin and end with real numbers, we seem compelled to move into an unfamiliar world of imaginaries to complete our journey. To mathematicians of the day, this proposal seemed incredibly strange [7]. Yet they could not dismiss the idea of imaginary numbers so easily because this concept yielded the real solution of an equation. It took two more centuries for the full importance of complex numbers to become evident in the works of Euler, Gauss, and Cauchy. Still, Bombelli deserves credit for recognizing that such numbers have a role to play in algebra [7]. - -† This equation is known as the *depressed cubic* equation. A general cubic equation - -*y*3 +*py*2 +*qy*+*r* = 0 - -can always be reduced to a depressed cubic form by substituting *y* = *x* − (*p*/3). Therefore, any general cubic equation can be solved if we know the solution to the depressed cubic. The depressed cubic was independently solved, first by *Scipione del Ferro* (1465–1526) and then by *Niccolo Fontana* (1499–1557). The latter is better known in the history of mathematics as *Tartaglia* ("Stammerer"). Cardano learned the secret of the depressed cubic solution from Tartaglia. He then showed that by using the substitution *y* = *x*−(*p*/3), a general cubic is reduced to a depressed cubic. - -#### 4 CHAPTER B BACKGROUND - -In 1799 the German mathematician *Karl Friedrich Gauss,* at the ripe age of 22, proved the fundamental theorem of algebra, namely that every algebraic equation in one unknown has a root in the form of a complex number. He showed that every equation of the *n*th order has exactly *n* solutions (roots), no more and no less. Gauss was also one of the first to give a coherent account of complex numbers and to interpret them as points in a complex plane. It is he who introduced the term *complex numbers* and paved the way for their general and systematic use. The number system was once again broadened or generalized to include imaginary numbers. Ordinary (or real) numbers became a special case of generalized (or complex) numbers. - -The utility of complex numbers can be understood readily by an analogy with two neighboring countries *X* and *Y*, as illustrated in Fig. B.1. If we want to travel from City *a* to City *b* (both in - -Gerolamo Cardano Karl Friedrich Gauss - -**Figure B.1** Use of complex numbers can reduce the work. - -Country *X*), the shortest route is through Country *Y*, although the journey begins and ends in Country *X*. We may, if we desire, perform this journey by an alternate route that lies exclusively in *X*, but this alternate route is longer. In mathematics we have a similar situation with real numbers (Country *X*) and complex numbers (Country *Y*). Most real-world problems start with real numbers, and the final results must also be in real numbers. But the derivation of results is considerably simplified by using complex numbers as an intermediary. It is also possible to solve any real-world problem by an alternate method, using real numbers exclusively, but such procedures would increase the work needlessly. - -### **[B.1-2 Algebra of Complex Numbers](#page-6-0)** - -A complex number (*a*,*b*) or *a* + *jb* can be represented graphically by a point whose Cartesian coordinates are (*a*,*b*) in a complex plane (Fig. B.2). Let us denote this complex number by *z* so that - -$$ -z = a + jb \tag{B.1} -$$ - -This representation is the Cartesian (or rectangular) form of complex number *z*. The numbers *a* and *b* (the abscissa and the ordinate) of *z* are the *real part* and the *imaginary part*, respectively, of *z*. They are also expressed as - -$$ -Re z = a \qquad \text{and} \qquad Im z = b -$$ - -Note that in this plane all real numbers lie on the horizontal axis, and all imaginary numbers lie on the vertical axis. - -Complex numbers may also be expressed in terms of polar coordinates. If (*r*, θ ) are the polar coordinates of a point *z* = *a*+*jb* (see Fig. B.2), then - -$$ -a = r \cos \theta -$$ - and $b = r \sin \theta$ - -Consequently, - -$$ -z = a + jb = r\cos\theta + jr\sin\theta = r(\cos\theta + j\sin\theta) -$$ - (B.2) - -*Euler's formula* states that - -$$ -e^{j\theta} = \cos\theta + j\sin\theta \tag{B.3} -$$ - -To prove Euler's formula, we use a Maclaurin series to expand *ej*θ , cos θ, and sin θ: - -$$ -e^{j\theta} = 1 + j\theta + \frac{(j\theta)^2}{2!} + \frac{(j\theta)^3}{3!} + \frac{(j\theta)^4}{4!} + \frac{(j\theta)^5}{5!} + \frac{(j\theta)^6}{6!} + \cdots -$$ - -\n -$$ -= 1 + j\theta - \frac{\theta^2}{2!} - j\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + j\frac{\theta^5}{5!} - \frac{\theta^6}{6!} - \cdots -$$ - -\n -$$ -\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} + \cdots -$$ - -\n -$$ -\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots -$$ - -Clearly, it follows that *ej*θ = cos θ +*j*sin θ. Using Eq. (B.3) in Eq. (B.2) yields - -$$ -z = re^{i\theta} \tag{B.4} -$$ - -This representation is the polar form of complex number *z*. - -Summarizing, a complex number can be expressed in rectangular form *a* + *jb* or polar form *rej*θ with - -$$ -a = r \cos \theta -$$ - -\n -$$ -b = r \sin \theta -$$ - and -$$ -r = \sqrt{a^2 + b^2} -$$ - -\n -$$ -\theta = \tan^{-1} \left(\frac{b}{a}\right) -$$ - (B.5) - -Observe that *r* is the distance of the point *z* from the origin. For this reason, *r* is also called the *magnitude* (or *absolute value*) of *z* and is denoted by |*z*|. Similarly, θ is called the angle of *z* and is denoted by *z*. Therefore, we can also write polar form of Eq. (B.4) as - -$$ -z = |z|e^{j\angle z} -$$ - where $|z| = r$ and $\angle z = \theta$ - -Using polar form, we see that the reciprocal of a complex number is given by - -$$ -\frac{1}{z} = \frac{1}{re^{j\theta}} = \frac{1}{r}e^{-j\theta} = \frac{1}{|z|}e^{-j\sqrt{z}} -$$ - -### CONJUGATE OF A COMPLEX NUMBER - -We define *z*∗, the *conjugate* of *z* = *a*+*jb*, as - -$$ -z^* = a - jb = re^{-j\theta} = |z|e^{-j\angle z} -$$ - (B.6) - -The graphical representations of a number *z* and its conjugate *z*∗ are depicted in Fig. B.2. Observe that *z*∗ is a mirror image of *z* about the horizontal axis. *To find the conjugate of any number, we need only replace j with* −*j in that number* (which is the same as changing the sign of its angle). - -The sum of a complex number and its conjugate is a real number equal to twice the real part of the number: - -$$ -z + z^* = (a + jb) + (a - jb) = 2a = 2 \operatorname{Re} z -$$ - -Thus, we see that the real part of complex number *z* can be computed as - -$$ -\text{Re}\,z = \frac{z + z^*}{2} \tag{B.7} -$$ - -Similarly, the imaginary part of complex number *z* can be computed as - -$$ -\operatorname{Im} z = \frac{z - z^*}{2j} \tag{B.8} -$$ - -The product of a complex number *z* and its conjugate is a real number |*z*| 2, the square of the magnitude of the number: - -$$ -zz^* = |z|e^{j\angle z}|z|e^{-j\angle z} = |z|^2 -$$ - (B.9) - -### UNDERSTANDING SOME USEFUL IDENTITIES - -In a complex plane, *rej*θ represents a point at a distance *r* from the origin and at an angle θ with the horizontal axis, as shown in Fig. B.3a. For example, the number −1 is at a unit distance from the origin and has an angle π or −π (more generally, π plus any integer multiple of 2π), as seen from Fig. B.3b. Therefore, - -$$ --1 = e^{j(\pi + 2\pi n)} \qquad n \text{ integer} -$$ - -The number 1, on the other hand, is also at a unit distance from the origin, but has an angle 0 (more generally, 0 plus any integer multiple of 2π). Therefore, - -$$ -1 = e^{j2\pi n} \qquad n \text{ integer} -$$ - (B.10) - -The number *j* is at a unit distance from the origin and its angle is π 2 (more generally, π 2 plus any integer multiple of 2π), as seen from Fig. B.3b. Therefore, - -$$ -j = e^{j(\frac{\pi}{2} + 2\pi n)} \qquad n \text{ integer} -$$ - -Similarly, - -$$ --j = e^{j(-\frac{\pi}{2} + 2\pi n)} \qquad n \text{ integer} -$$ - -Notice that the angle of any complex number is only known within an integer multiple of 2π. - -This discussion shows the usefulness of the graphic picture of *rej*θ . This picture is also helpful in several other applications. For example, to determine the limit of *e*(α+*j*ω)*t* as *t* → ∞, we note that - -*ej*ω*t* - -$$ -e^{(\alpha+j\omega)t} = e^{\alpha t} e^{j\omega t} -$$ -\n -$$ -\lim_{\rho \to 0} \frac{e^{j\theta}}{\log \rho} = \lim_{\rho \to 0} \frac{e^{j\theta}}{\log \rho} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} -$$ -\n -$$ -\lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} -$$ -\n -$$ -\lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{ -$$ - -**Figure B.3** Understanding some useful identities in terms of *rej*θ . - -### 8 CHAPTER B BACKGROUND - -Now the magnitude of *ej*ω*t* is unity regardless of the value of ω or *t* because *ej*ω*t* = *rej*θ with *r* = 1. Therefore, *e*α*t* determines the behavior of *e*(α+*j*ω)*t* as *t* → ∞ and - -$$ -\lim_{t \to \infty} e^{(\alpha + j\omega)t} = \lim_{t \to \infty} e^{\alpha t} e^{j\omega t} = \begin{cases} 0 & \alpha < 0 \\ \infty & \alpha > 0 \end{cases} -$$ - -In future discussions, you will find it very useful to remember *rej*θ as a number at a distance *r* from the origin and at an angle θ with the horizontal axis of the complex plane. - -### A WARNING ABOUT COMPUTING ANGLES WITH CALCULATORS - -From the Cartesian form *a* + *jb*, we can readily compute the polar form *rej*θ [see Eq. (B.5)]. Calculators provide ready conversion of rectangular into polar and vice versa. However, if a calculator computes an angle of a complex number by using an inverse tangent function θ = tan−1(*b*/*a*), proper attention must be paid to the quadrant in which the number is located. For instance, θ corresponding to the number −2 − *j*3 is tan−1(−3/−2). This result is not the same as tan−1(3/2). The former is −123.7◦, whereas the latter is 56.3◦. A calculator cannot make this distinction and can give a correct answer only for angles in the first and fourth quadrants.† A calculator will read tan−1(−3/−2) as tan−1(3/2), which is clearly wrong. When you are computing inverse trigonometric functions, if the angle appears in the second or third quadrant, the answer of the calculator is off by 180◦. The correct answer is obtained by adding or subtracting 180◦ to the value found with the calculator (either adding or subtracting yields the correct answer). For this reason, it is advisable to draw the point in the complex plane and determine the quadrant in which the point lies. This issue will be clarified by the following examples. - -### **EXAMPLE B.1 Cartesian to Polar Form** - -Express the following numbers in polar form: **(a)** 2+*j*3, **(b)** −2+*j*1, **(c)** −2−*j*3, and **(d)** 1−*j*3. - -**(a)** - -$$ -|z| = \sqrt{2^2 + 3^2} = \sqrt{13} -$$ - $\angle z = \tan^{-1}(\frac{3}{2}) = 56.3^{\circ}$ - -In this case the number is in the first quadrant, and a calculator will give the correct value of 56.3◦. Therefore (see Fig. B.4a), we can write - -$$ -2 + j3 = \sqrt{13} e^{j56.3^{\circ}} -$$ - -**(b)** - -$$ -|z| = \sqrt{(-2)^2 + 1^2} = \sqrt{5} -$$ - $\angle z = \tan^{-1}(\frac{1}{-2}) = 153.4^{\circ}$ - -In this case the angle is in the second quadrant (see Fig. B.4b), and therefore the answer given by the calculator, tan−1(1/−2) = −26.6◦, is off by 180◦. The correct answer is - -† Calculators with two-argument inverse tangent functions will correctly compute angles. - -(−26.6±180)◦ = 153.4◦ or −206.6◦. Both values are correct because they represent the same angle. It is a common practice to choose an angle whose numerical value is less than 180◦. Such a value is called the *principal value* of the angle, which in this case is 153.4◦. Therefore, - -$$ --2 + j1 = \sqrt{5}e^{j153.4^{\circ}} -$$ - -**(c)** - -$$ -|z| = \sqrt{(-2)^2 + (-3)^2} = \sqrt{13} -$$ - $\angle z = \tan^{-1}\left(\frac{-3}{-2}\right) = -123.7^{\circ}$ - -In this case the angle appears in the third quadrant (see Fig. B.4c), and therefore the answer obtained by the calculator (tan−1(−3/−2) = 56.3◦) is off by 180◦. The correct answer is (56.3 ± 180)◦ = 236.3◦ or −123.7◦. We choose the principal value −123.7◦ so that (see Fig. B.4c) - -$$ --2 - j3 = \sqrt{13}e^{-j123.7^{\circ}} -$$ - -**(d)** - -$$ -|z| = \sqrt{1^2 + (-3)^2} = \sqrt{10} -$$ - $\angle z = \tan^{-1}\left(\frac{-3}{1}\right) = -71.6^{\circ}$ - -In this case the angle appears in the fourth quadrant (see Fig. B.4d), and therefore the answer given by the calculator, tan−1(−3/1) = −71.6◦, is correct (see Fig. B.4d): - -$$ -1 - j3 = \sqrt{10}e^{-j71.6^{\circ}} -$$ - -### 10 CHAPTER B BACKGROUND - -We can easily verify these results using the MATLAB abs and angle commands. To obtain units of degrees, we must multiply the radian result of the angle command by 180 π . Furthermore, the angle command correctly computes angles for all four quadrants of √ the complex plane. To provide an example, let us use MATLAB to verify that −2 + *j*1 = 5*ej*153.4◦ = 2.2361*ej*153.4◦ . - ->> abs(-2+1j) ans = 2.2361 >> angle(-2+1j)\*180/pi ans = 153.4349 - -One can also use the cart2pol command to convert Cartesian to polar coordinates. Readers, particularly those who are unfamiliar with MATLAB, will benefit by reading the overview in Sec. B.7. - -### **EXAMPLE B.2 Polar to Cartesian Form** - -Represent the following numbers in the complex plane and express them in Cartesian form: **(a)** 2*ej*π/3, **(b)** 4*e*−*j*3π/4, **(c)** 2*ej*π/2, **(d)** 3*e*−*j*3π , **(e)** 2*ej*4π , and **(f)** 2*e*−*j*4π . - -**(a)** 2*ej*π/3 = 2(cos π/3+*j*sin π/3) = 1+*j* √3 (see Fig. B.5a) **(b)** 4*e*−*j*3π/4 = 4(cos 3π/4−*j*sin 3π/4) = −2 √2−*j*2 √2 (see Fig. B.5b) **(c)** 2*ej*π/2 = 2(cos π/2+*j*sin π/2) = 2(0+*j*1) = *j*2 (see Fig. B.5c) **(d)** 3*e*−*j*3π = 3(cos 3π −*j*sin 3π ) = 3(−1+*j*0) = −3 (see Fig. B.5d) **(e)** 2*ej*4π = 2(cos 4π +*j*sin 4π ) = 2(1+*j*0) = 2 (see Fig. B.5e) **(f)** 2*e*−*j*4π = 2(cos 4π −*j*sin 4π ) = 2(1−*j*0) = 2 (see Fig. B.5f) - -We can readily verify these results using MATLAB. First, we use the exp function to represent a number in polar form. Next, we use the real and imag commands to determine the real and imaginary components of that number. To provide an example, let us use MATLAB to verify the result of part (a): 2*ej*π/3 = 1+*j* √3 = 1+*j*1.7321. - -``` ->> real(2*exp(1j*pi/3)) - ans = 1.0000 ->> imag(2*exp(1j*pi/3)) - ans = 1.7321 -``` - -Since MATLAB defaults to Cartesian form, we could have verified the entire result in one step. - -``` ->> 2*exp(1j*pi/3) - ans = 1.0000 + 1.7321i -``` - -One can also use the pol2cart command to convert polar to Cartesian coordinates. - -### ARITHMETICAL OPERATIONS, POWERS, AND ROOTS OF COMPLEX NUMBERS - -To conveniently perform addition and subtraction, complex numbers should be expressed in Cartesian form. Thus, if - -$$ -z_1 = 3 + j4 = 5e^{j53.1^{\circ}} -$$ - -and - -$$ -z_2 = 2 + j3 = \sqrt{13}e^{j56.3^{\circ}} -$$ - -then - -$$ -z_1 + z_2 = (3 + j4) + (2 + j3) = 5 + j7 -$$ - -### 12 CHAPTER B BACKGROUND - -If *z*1 and *z*2 are given in polar form, we would need to convert them into Cartesian form for the purpose of adding (or subtracting). Multiplication and division, however, can be carried out in either Cartesian or polar form, although the latter proves to be much more convenient. This is because if *z*1 and *z*2 are expressed in polar form as - -$$ -z_1 = r_1 e^{j\theta_1} -$$ - and $z_2 = r_2 e^{j\theta_2}$ - -then - -$$ -z_1 z_2 = (r_1 e^{j\theta_1})(r_2 e^{j\theta_2}) = r_1 r_2 e^{j(\theta_1 + \theta_2)} -$$ - -and - -$$ -\frac{z_1}{z_2} = \frac{r_1 e^{j\theta_1}}{r_2 e^{j\theta_2}} = \frac{r_1}{r_2} e^{j(\theta_1 - \theta_2)} -$$ - -Moreover, - -$$ -z^n = (re^{j\theta})^n = r^n e^{jn\theta} -$$ - -and - -$$ -z^{1/n} = (re^{i\theta})^{1/n} = r^{1/n}e^{i\theta/n} -$$ - (B.11) - -This shows that the operations of multiplication, division, powers, and roots can be carried out with remarkable ease when the numbers are in polar form. - -Strictly speaking, there are *n* values for *z*1/*n* (the *n*th root of *z*). To find all the *n* roots, we reexamine Eq. (B.11): - -$$ -z^{1/n} = [re^{j\theta}]^{1/n} = [re^{j(\theta + 2\pi k)}]^{1/n} = r^{1/n}e^{j(\theta + 2\pi k)/n} \qquad k = 0, 1, 2, \dots, n-1 -$$ - (B.12) - -The value of *z*1/*n* given in Eq. (B.11) is the *principal value* of *z*1/*n*, obtained by taking the *n*th root of the principal value of *z*, which corresponds to the case *k* = 0 in Eq. (B.12). - -### **EXAMPLE B.3 Multiplication and Division of Complex Numbers** - -Using both polar and Cartesian forms, determine *z*1*z*2 and *z*1/*z*2 for the numbers - -$$ -z_1 = 3 + j4 = 5e^{j53.1^{\circ}} -$$ - and $z_2 = 2 + j3 = \sqrt{13}e^{j56.3^{\circ}}$ - -**Multiplication: Cartesian Form** - -$$ -z_1 z_2 = (3+j4)(2+j3) = (6-12) + j(8+9) = -6+j17 -$$ - -**Multiplication: Polar Form** - -$$ -z_1 z_2 = (5e^{j53.1^{\circ}})(\sqrt{13}e^{j56.3^{\circ}}) = 5\sqrt{13}e^{j109.4^{\circ}} -$$ - -**Division: Cartesian Form** - -$$ -\frac{z_1}{z_2} = \frac{3+j4}{2+j3} -$$ - -To eliminate the complex number in the denominator, we multiply both the numerator and the denominator of the right-hand side by 2−*j*3, the denominator's conjugate. This yields - -$$ -\frac{z_1}{z_2} = \frac{(3+j4)(2-j3)}{(2+j3)(2-j3)} = \frac{18-j1}{2^2+3^2} = \frac{18-j1}{13} = \frac{18}{13} - j\frac{1}{13} -$$ - -**Division: Polar Form** - -$$ -\frac{z_1}{z_2} = \frac{5e^{j53.1^{\circ}}}{\sqrt{13}e^{j56.3^{\circ}}} = \frac{5}{\sqrt{13}}e^{j(53.1^{\circ} - 56.3^{\circ})} = \frac{5}{\sqrt{13}}e^{-j3.2^{\circ}} -$$ - -It is clear from this example that multiplication and division are easier to accomplish in polar form than in Cartesian form. - -These results are also easily verified using MATLAB. To provide one example, let us use Cartesian forms in MATLAB to verify that *z*1*z*2 = −6+*j*17. - -``` ->> z1 = 3+4j; z2 = 2+3j; ->> z1*z2 - ans = -6.0000 + 17.0000i -``` - -As a second example, let us use polar forms in MATLAB to verify that *z*1/*z*2 = 1.3868*e*−*j*3.2◦ . Since MATLAB generally expects angles be represented in the natural units of radians, we must use appropriate conversion factors in moving between degrees and radians (and vice versa). - -``` ->> z1 = 5*exp(1j*53.1*pi/180); z2 = sqrt(13)*exp(1j*56.3*pi/180); ->> abs(z1/z2) - ans = 1.3868 ->> angle(z1/z2)*180/pi - ans = -3.2000 -``` - -### **EXAMPLE B.4 Working with Complex Numbers** - -For *z*1 = 2*ej*π/4 and *z*2 = 8*ej*π/3, find the following: **(a)** 2*z*1 −*z*2, **(b)** 1/*z*1, **(c)** *z*1/*z*2 2, and **(d)** √3 *z*2. - -**(a)** Since subtraction cannot be performed directly in polar form, we convert *z*1 and *z*2 to Cartesian form: - -$$ -z_1 = 2e^{j\pi/4} = 2\left(\cos\frac{\pi}{4} + j\sin\frac{\pi}{4}\right) = \sqrt{2} + j\sqrt{2} -$$ -$$ -z_2 = 8e^{j\pi/3} = 8\left(\cos\frac{\pi}{3} + j\sin\frac{\pi}{3}\right) = 4 + j4\sqrt{3} -$$ - -Therefore, - -$$ -2z_1 - z_2 = 2(\sqrt{2} + j\sqrt{2}) - (4 + j4\sqrt{3}) = (2\sqrt{2} - 4) + j(2\sqrt{2} - 4\sqrt{3}) = -1.17 - j4.1 -$$ - -**(b)** - -$$ -\frac{1}{z_1} = \frac{1}{2e^{j\pi/4}} = \frac{1}{2}e^{-j\pi/4} -$$ - -**(c)** - -$$ -\frac{z_1}{z_2^2} = \frac{2e^{j\pi/4}}{(8e^{j\pi/3})^2} = \frac{2e^{j\pi/4}}{64e^{j2\pi/3}} = \frac{1}{32}e^{j(\pi/4 - 2\pi/3)} = \frac{1}{32}e^{-j(5\pi/12)} -$$ - -**(d)** There are three cube roots of 8*ej*(π/3) = 8*ej*(π/3+2π*k*) , *k* = 0, 1, 2. - -$$ -\sqrt[3]{z_2} = z_2^{1/3} = \left[8e^{i(\pi/3 + 2\pi k)}\right]^{1/3} = 8^{1/3} \left(e^{i[(6\pi k + \pi)/3]}\right)^{1/3} = \begin{cases} 2e^{i\pi/9} & k = 0\\ 2e^{i7\pi/9} & k = 1\\ 2e^{i13\pi/9} & k = 2 \end{cases} -$$ - -The value corresponding to *k* = 0 is termed the *principal value*. - -### **EXAMPLE B.5 Standard Forms of Complex Numbers** - -Consider *X*(ω), a complex function of a real variable ω: - -$$ -X(\omega) = \frac{2 + j\omega}{3 + j4\omega} -$$ - -**(a)** Express *X*(ω) in Cartesian form, and find its real and imaginary parts. - -**(b)** Express *X*(ω) in polar form, and find its magnitude |*X*(ω)| and angle *X*(ω). - -$$ -X(\omega) = \frac{(2+j\omega)(3-j4\omega)}{(3+j4\omega)(3-j4\omega)} = \frac{(6+4\omega^2) - j5\omega}{9+16\omega^2} = \frac{6+4\omega^2}{9+16\omega^2} - j\frac{5\omega}{9+16\omega^2} -$$ - -This is the Cartesian form of *X*(ω). Clearly, the real and imaginary parts *Xr*(ω) and *Xi*(ω) are given by - -$$ -X_r(\omega) = \frac{6 + 4\omega^2}{9 + 16\omega^2} -$$ - and $X_i(\omega) = \frac{-5\omega}{9 + 16\omega^2}$ - -**(a)** To obtain the real and imaginary parts of *X*(ω), we must eliminate imaginary terms in the denominator of *X*(ω). This is readily done by multiplying both the numerator and the denominator of *X*(ω) by 3−*j*4ω, the conjugate of the denominator 3+*j*4ω so that - -$$ -(\mathbf{b}) -$$ - -$$ -X(\omega) = \frac{2 + j\omega}{3 + j4\omega} = \frac{\sqrt{4 + \omega^2} e^{j\tan^{-1}(\omega/2)}}{\sqrt{9 + 16\omega^2} e^{j\tan^{-1}(4\omega/3)}} = \sqrt{\frac{4 + \omega^2}{9 + 16\omega^2}} e^{j\tan^{-1}(\omega/2) - \tan^{-1}(4\omega/3)} -$$ - -This is the polar representation of *X*(ω). Observe that - -$$ -|X(\omega)| = \sqrt{\frac{4 + \omega^2}{9 + 16\omega^2}} \quad \text{and} \quad \angle X(\omega) = \tan^{-1}\left(\frac{\omega}{2}\right) - \tan^{-1}\left(\frac{4\omega}{3}\right) -$$ - -### LOGARITHMS OF COMPLEX NUMBERS - -To take the natural logarithm of a complex number *z*, we first express *z* in general polar form as - -$$ -z = re^{j\theta} = re^{j(\theta \pm 2\pi k)} -$$ - $k = 0, 1, 2, 3, ...$ - -Taking the natural logarithm, we see that - -$$ -\ln z = \ln \left( r e^{j(\theta \pm 2\pi k)} \right) = \ln r \pm j(\theta + 2\pi k) \qquad k = 0, 1, 2, 3, \dots -$$ - -The value of ln*z* for *k* = 0 is called the *principal value* of ln*z* and is denoted by Ln*z*. In this way, we see that - -$$ -\ln 1 = \ln(1e^{\pm j2\pi k}) = \pm j2\pi k \qquad k = 0, 1, 2, 3, \dots -$$ -$$ -\ln(-1) = \ln[1e^{\pm j\pi(2k+1)}] = \pm j(2k+1)\pi \qquad k = 0, 1, 2, 3, \dots -$$ -$$ -\ln j = \ln(e^{j\pi(1\pm 4k)/2}) = j\frac{\pi(1\pm 4k)}{2} \qquad k = 0, 1, 2, 3, \dots -$$ -$$ -j^j = e^{j\ln j} = e^{-\pi(1\pm 4k)/2} \qquad k = 0, 1, 2, 3, \dots -$$ - -In all of these cases, setting *k* = 0 yields the principal value of the expression. - -We can further our logarithm skills by noting that the familiar properties of logarithms hold for complex arguments. Therefore, we have - -$$ -log(z1z2) = log z1 + log z2 -$$ -$$ -log(z1/z2) = log z1 - log z2 -$$ -$$ -a(z1+z2) = az1 × az2 -$$ -$$ -zc = ecln z -$$ -$$ -az = ezln a -$$ - -## **[B.2 SINUSOIDS](#page-6-0)** - -Consider the sinusoid - -$$ -x(t) = C\cos(2\pi f_0 t + \theta) -$$ - (B.13) - -We know that - -$$ -\cos \varphi = \cos (\varphi + 2n\pi) -$$ - $n = 0, \pm 1, \pm 2, \pm 3, ...$ - -Therefore, cos ϕ repeats itself for every change of 2π in the angle ϕ. For the sinusoid in Eq. (B.13), the angle 2π*f*0*t*+θ changes by 2π when *t* changes by 1/*f*0. Clearly, this sinusoid repeats every 1/*f*0 seconds. As a result, there are *f*0 repetitions per second. This is the *frequency* of the sinusoid, and the repetition interval *T*0 given by - -$$ -T_0 = \frac{1}{f_0} -$$ - (B.14) - -is the *period*. For the sinusoid in Eq. (B.13), *C* is the *amplitude, f*0 is the *frequency* (in hertz), and θ is the phase. Let us consider two special cases of this sinusoid when θ = 0 and θ = −π/2 as follows: - -$$ -x(t) = C\cos 2\pi f_0 t \qquad (\theta = 0) -$$ - -and - -$$ -x(t) = C\cos(2\pi f_0 t - \pi/2) = C\sin 2\pi f_0 t \qquad (\theta = -\pi/2) -$$ - -The angle or phase can be expressed in units of degrees or radians. Although the radian is the proper unit, in this book we shall often use the degree unit because students generally have a better feel for the relative magnitudes of angles expressed in degrees rather than in radians. For example, we relate better to the angle 24◦ than to 0.419 radian. Remember, however, when in doubt, use the radian unit and, above all, be consistent. In other words, in a given problem or an expression, do not mix the two units. - -It is convenient to use the variable ω0 (*radian frequency*) to express 2π*f*0: - -$$ -\omega_0 = 2\pi f_0 \tag{B.15} -$$ - -With this notation, the sinusoid in Eq. (B.13) can be expressed as - -$$ -x(t) = C\cos{(\omega_0 t + \theta)} -$$ - -in which the period *T*0 and frequency ω0 are given by [see Eqs. (B.14) and (B.15)] - -$$ -T_0 = \frac{1}{\omega_0/2\pi} = \frac{2\pi}{\omega_0} \quad \text{and} \quad \omega_0 = \frac{2\pi}{T_0} -$$ - -Although we shall often refer to ω0 as the frequency of the signal cos(ω0*t*+θ ), it should be clearly understood that ω0 is the *radian frequency*; the *hertzian frequency* of this sinusoid is *f*0 = ω0/2π ). - -The signals *C*cos ω0*t* and *C*sin ω0*t* are illustrated in Figs. B.6a and B.6b, respectively. A general sinusoid *C*cos(ω0*t*+θ ) can be readily sketched by shifting the signal *C*cos ω0*t* in Fig. B.6a by the appropriate amount. Consider, for example, - -$$ -x(t) = C\cos{(\omega_0 t - 60^\circ)} -$$ - -**Figure B.6** Sketching a sinusoid. - -This signal can be obtained by shifting (delaying) the signal *C*cos ω0*t* (Fig. B.6a) to the right by a phase (angle) of 60◦. We know that a sinusoid undergoes a 360◦ change of phase (or angle) in one cycle. A quarter-cycle segment corresponds to a 90◦ change of angle. We therefore shift (delay) the signal in Fig. B.6a by two-thirds of a quarter-cycle segment to obtain *C*cos(ω0*t* − 60◦), as shown in Fig. B.6c. - -Observe that if we delay *C*cos ω0*t* in Fig. B.6a by a quarter-cycle (angle of 90◦ or π/2 radians), we obtain the signal *C*sin ω0*t*, depicted in Fig. B.6b. This verifies the well-known trigonometric identity - -$$ -C\cos{(\omega_0 t - \pi/2)} = C\sin{\omega_0 t} -$$ - -#### 18 CHAPTER B BACKGROUND - -Alternatively, if we advance *C*sin ω0*t* by a quarter-cycle, we obtain *C*cos ω0*t*. Therefore, - -$$ -C\sin(\omega_0 t + \pi/2) = C\cos\omega_0 t -$$ - -These observations mean that sin ω0*t* lags cos ω0*t* by 90◦(π/2 radians) and that cos ω0*t* leads sin ω0*t* by 90◦. - -### **[B.2-1 Addition of Sinusoids](#page-6-0)** - -Two sinusoids having the same frequency but different phases add to form a single sinusoid of the same frequency. This fact is readily seen from the well-known trigonometric identity - -*C*cos θ cos ω0*t* −*C*sin θ sin ω0*t* = *C*cos(ω0*t* +θ ) - -Setting *a* = *C*cos θ and *b* = −*C*sin θ, we see that - -$$ -a\cos\omega_0 t + b\sin\omega_0 t = C\cos(\omega_0 t + \theta) -$$ - (B.16) - -From trigonometry, we know that - -$$ -C = \sqrt{a^2 + b^2} \qquad \text{and} \qquad \theta = \tan^{-1}\left(\frac{-b}{a}\right) \tag{B.17} -$$ - -Equation (B.17) shows that *C* and θ are the magnitude and angle, respectively, of a complex number *a* − *jb*. In other words, *a* − *jb* = *Cej*θ . Hence, to find *C* and θ, we convert *a* − *jb* to polar form and the magnitude and the angle of the resulting polar number are *C* and θ, respectively. - -The process of adding two sinusoids with the same frequency can be clarified by using *phasors* to represent sinusoids. We represent the sinusoid *C*cos(ω0*t*+θ ) by a phasor of length *C* at an angle θ with the horizontal axis. Clearly, the sinusoid *a*cos ω0*t* is represented by a horizontal phasor of length *a*(θ = 0), while *b*sin ω0*t* = *b*cos(ω0*t* −π/2) is represented by a vertical phasor of length *b* at an angle −π/2 with the horizontal (Fig. B.7). Adding these two phasors results in a phasor of length *C* at an angle θ, as depicted in Fig. B.7. From this figure, we verify the values of *C* and θ found in Eq. (B.17). Proper care should be exercised in computing θ, as explained on page 8 ("A Warning About Computing Angles with Calculators"). - -**Figure B.7** Phasor addition of sinusoids. - -### **EXAMPLE B.6 Addition of Sinusoids** - -In the following cases, express *x*(*t*) as a single sinusoid: - -**(a)** *x*(*t*) = cos ω0*t* − √3 sin ω0*t* - -**(b)** *x*(*t*) = −3 cos ω0*t* +4 sin ω0*t* - -**(a)** In this case, *a* = 1 and *b* = −√3. Using Eq. (B.17) yields - -$$ -C = \sqrt{1^2 + (\sqrt{3})^2} = 2 -$$ - and $\theta = \tan^{-1}(\frac{\sqrt{3}}{1}) = 60^\circ$ - -Therefore, - -$$ -x(t) = 2\cos{(\omega_0 t + 60^\circ)} -$$ - -We can verify this result by drawing phasors corresponding to the two sinusoids. The sinusoid cos ω0*t* is represented by a phasor of unit length at a zero angle with the horizontal. The phasor sin ω0*t* is represented by a unit phasor at an angle of −90◦ with the horizontal. Therefore, − √3 sin ω0*t* is represented by a phasor of length √3 at 90◦ with the horizontal, as depicted in Fig. B.8a. The two phasors added yield a phasor of length 2 at 60◦ with the horizontal (also shown in Fig. B.8a). - -**Figure B.8** Phasor addition of sinusoids. - -Alternately, we note that *a*−*jb* = 1+*j* √3 = 2*ej*π/3. Hence, *C* = 2 and θ = π/3. Observe that a phase shift of ±π amounts to multiplication by −1. Therefore, *x*(*t*) can also be expressed alternatively as - -$$ -x(t) = -2\cos(\omega_0 t + 60^\circ \pm 180^\circ) = -2\cos(\omega_0 t - 120^\circ) = -2\cos(\omega_0 t + 240^\circ) -$$ - -In practice, the principal value, that is, −120◦, is preferred. - -**(b)** In this case, *a* = −3 and *b* = 4. Using Eq. (B.17) yields - -$$ -C = \sqrt{(-3)^2 + 4^2} = 5 -$$ - and $\theta = \tan^{-1}\left(\frac{-4}{-3}\right) = -126.9^{\circ}$ - -Observe that - -$$ -\tan^{-1}\left(\frac{-4}{-3}\right) \neq \tan^{-1}\left(\frac{4}{3}\right) = 53.1^{\circ} -$$ - -Therefore, - -$$ -x(t) = 5\cos\left(\omega_0 t - 126.9^\circ\right) -$$ - -This result is readily verified in the phasor diagram in Fig. B.8b. Alternately, *a*−*jb* = −3−*j*4 = 5*e*−*j*126.9◦ , a fact readily confirmed using MATLAB. - ->> C = abs(-3+4j) C=5 >> theta = angle(-3+4j)\*180/pi theta = 126.8699 - -``` -Hence, C = 5 and θ = −126.8699◦. -``` - -We can also perform the reverse operation, expressing *C*cos(ω0*t* +θ ) in terms of cos ω0*t* and sin ω0*t* by again using the trigonometric identity - -*C*cos(ω0*t* +θ ) = *C*cos θ cos ω0*t* −*C*sin θ sin ω0*t* - -For example, - -$$ -10\cos\left(\omega_0 t - 60^\circ\right) = 5\cos\omega_0 t + 5\sqrt{3}\sin\omega_0 t -$$ - -### **B.2-2 Sinusoids in Terms of Exponentials** - -From Eq. (B.3), we know that *ej*ϕ = cos ϕ + *j*sin ϕ and *e*−*j*ϕ = cos ϕ − *j*sin ϕ. Adding these two expressions and dividing by 2 provide an expression for cosine in terms of complex exponentials, while subtracting and scaling by 2*j* provide an expression for sine. That is, - -$$ -\cos \varphi = \frac{1}{2} (e^{j\varphi} + e^{-j\varphi}) -$$ - and $\sin \varphi = \frac{1}{2j} (e^{j\varphi} - e^{-j\varphi})$ (B.18) - -## **[B.3 SKETCHING](#page-6-0) SIGNALS** - -In this section, we discuss the sketching of a few useful signals, starting with exponentials. - -### **[B.3-1 Monotonic Exponentials](#page-6-0)** - -The signal *e*−*at* decays monotonically, and the signal *eat* grows monotonically with *t* (assuming *a* > 0), as depicted in Fig. B.9. For the sake of simplicity, we shall consider an exponential *e*−*at* starting at *t* = 0, as shown in Fig. B.10a. - -The signal *e*−*at* has a unit value at *t* = 0. At *t* = 1/*a*, the value drops to 1/*e* (about 37% of its initial value), as illustrated in Fig. B.10a. This time interval over which the exponential reduces by - -**Figure B.9** Monotonic exponentials. - -**Figure B.10** Sketching **(a)** *e*−*at* and **(b)** *e*−2*t* . - -a factor *e* (i.e., drops to about 37% of its value) is known as the *time constant* of the exponential. Therefore, the time constant of *e*−*at* is 1/*a*. Observe that the exponential is reduced to 37% of its initial value over any time interval of duration 1/*a*. This can be shown by considering any set of instants *t*1 and *t*2 separated by one time constant so that - -$$ -t_2 - t_1 = \frac{1}{a} -$$ - -Now the ratio of *e*−*at*2 to *e*−*at*1 is given by - -$$ -\frac{e^{-at_2}}{e^{-at_1}} = e^{-a(t_2 - t_1)} = \frac{1}{e} \approx 0.37 -$$ - -We can use this fact to sketch an exponential quickly. For example, consider - -$$ -x(t) = e^{-2t} -$$ - -The time constant in this case is 0.5. The value of *x*(*t*) at *t* = 0 is 1. At *t* = 0.5 (one time constant), it is 1/*e* (about 0.37). The value of *x*(*t*) continues to drop further by the factor 1/*e* (37%) over the next half-second interval (one time constant). Thus, *x*(*t*) at *t* = 1 is (1/*e*)2. Continuing in this manner, we see that *x*(*t*) = (1/*e*)3 at *t* = 1.5, and so on. A knowledge of the values of *x*(*t*) at *t* = 0, 0.5, 1, and 1.5 allows us to sketch the desired signal, as shown in Fig. B.10b.† - -For a monotonically growing exponential *eat*, the waveform increases by a factor *e* over each interval of 1/*a* seconds. - -### **[B.3-2 The Exponentially Varying Sinusoid](#page-6-0)** - -We now discuss sketching an exponentially varying sinusoid - -$$ -x(t) = Ae^{-at}\cos{(\omega_0 t + \theta)} -$$ - -Let us consider a specific example: - -$$ -x(t) = 4e^{-2t}\cos{(6t - 60^\circ)} -$$ - -We shall sketch 4*e*−2*t* and cos(6*t* −60◦) separately and then multiply them: - -- **(a) Sketching 4***e***−2***t* **.** This monotonically decaying exponential has a time constant of 0.5 second and an initial value of 4 at *t* = 0. Therefore, its values at *t* = 0.5, 1, 1.5, and 2 are 4/*e*, 4/*e*2, 4/*e*3, and 4/*e*4, or about 1.47, 0.54, 0.2, and 0.07, respectively. Using these values as a guide, we sketch 4*e*−2*t* , as illustrated in Fig. B.11a. -- **(b) Sketching cos***(***6***t* **− 60◦***)***.** The procedure for sketching cos(6*t* − 60◦) is discussed in Sec. B.2 (Fig. B.6c). Here, the period of the sinusoid is *T*0 = 2π/6 ≈ 1, and there is a phase delay of 60◦, or two-thirds of a quarter-cycle, which is equivalent to a delay of about (60/360)(1) ≈ 1/6 seconds (see Fig. B.11b). -- **(c) Sketching 4***e***−2***t* **cos***(***6***t* **− 60◦***)***.** We now multiply the waveforms in steps (a) and (b). This multiplication amounts to forcing the sinusoid 4 cos(6*t* −60◦) to decrease exponentially with a time constant of 0.5. The initial amplitude (at *t* = 0) is 4, decreasing to 4/*e* (=1.47) at *t* = 0.5, to 1.47/*e*(=0.54) at *t* = 1, and so on. This is depicted in Fig. B.11c. Note that when cos(6*t* −60◦) has a value of unity (peak amplitude), - -$$ -4e^{-2t}\cos{(6t - 60^\circ)} = 4e^{-2t} -$$ - -Therefore, 4*e*−2*t* cos(6*t*−60◦) touches 4*e*−2*t* at the instants at which the sinusoid cos(6*t* −60◦) is at its positive peaks. Clearly, 4*e*−2*t* is an envelope for positive amplitudes of 4*e*−2*t* cos(6*t* − 60◦). Similar argument shows that 4*e*−2*t* cos(6*t* − 60◦) touches −4*e*−2*t* at its negative peaks. Therefore, −4*e*−2*t* is an envelope for negative amplitudes of 4*e*−2*t* cos(6*t* − 60◦). Thus, to sketch 4*e*−2*t* cos(6*t* − 60◦), we first draw the envelopes 4*e*−2*t* and −4*e*−2*t* (the mirror image of 4*e*−2*t* about the horizontal axis), and then sketch the sinusoid cos(6*t* − 60◦), with these envelopes acting as constraints on the sinusoid's amplitude (see Fig. B.11c). - -In general, *Ke*−*at* cos(ω0*t* + θ ) can be sketched in this manner, with *Ke*−*at* and −*Ke*−*at* constraining the amplitude of cos(ω0*t* +θ ). - -† If we wish to refine the sketch further, we could consider intervals of half the time constant over which the signal decays by a factor 1/ √*e*. Thus, at *t* = 0.25, *x*(*t*) = 1/ √*e*, and at *t* = 0.75, *x*(*t*) = 1/*e* √*e*, and so on. - -**Figure B.11** Sketching an exponentially varying sinusoid. - -## **[B.4 CRAMER'S](#page-6-0) RULE** - -Cramer's rule offers a very convenient way to solve simultaneous linear equations. Consider a set of *n* linear simultaneous equations in *n* unknowns *x*1, *x*2,..., *xn*: - -$$ -a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = y_1 -$$ - -\n -$$ -a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = y_2 -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n = y_n -$$ - -\n(B.19) - -These equations can be expressed in matrix form as - -$$ -\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} -$$ - (B.20) - -We denote the matrix on the left-hand side formed by the elements *aij* as **A**. The determinant of **A** is denoted by |**A**|. If the determinant |**A**| is not zero, Eq. (B.19) has a unique solution given by Cramer's formula - -$$ -x_k = \frac{|\mathbf{D}_k|}{|\mathbf{A}|} \qquad k = 1, 2, \dots, n -$$ - (B.21) - -where |**D***k*| is obtained by replacing the *k*th column of |**A**| by the column on the right-hand side of Eq. (B.20) (with elements *y*1, *y*2,..., *yn*). - -We shall demonstrate the use of this rule with an example. - -### **EXAMPLE B.7 Using Cramer's Rule to Solve a System of Equations** - -Use Cramer's rule to solve the following simultaneous linear equations in three unknowns: - -$$ -2x_1 + x_2 + x_3 = 3 -$$ - -$$ -x_1 + 3x_2 - x_3 = 7 -$$ - -$$ -x_1 + x_2 + x_3 = 1 -$$ - -In matrix form, these equations can be expressed as - -| ⎡
2 | 1 | ⎤
1 | ⎡
x1 | ⎤ | ⎡
⎤
3 | -|--------|---|---------|---------|-----|-------------| -| 1
⎣ | 3 | −1
⎦ | x2
⎣ | ⎦ = | 7
⎣
⎦ | -| 1 | 1 | 1 | x3 | | 1 | - -Here, - -$$ -|\mathbf{A}| = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} = 4 -$$ - -Since |**A**| = 4 = 0, a unique solution exists for *x*1, *x*2, and *x*3. This solution is provided by Cramer's rule [Eq. (B.21)] as follows: - -$$ -x_1 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 3 & 1 & 1 \\ 7 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \frac{8}{4} = 2 -$$ - -$$ -x_2 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 2 & 3 & 1 \\ 1 & 7 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \frac{4}{4} = 1 -$$ - -$$ -x_3 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 2 & 1 & 3 \\ 1 & 3 & 7 \\ 1 & 1 & 1 \end{vmatrix} = \frac{-8}{4} = -2 -$$ - -MATLAB is well suited to compute Cramer's formula, so these results are easy to verify. To provide an example, let us verify that *x*1 = 2 using MATLAB's det command to compute the needed matrix determinants. - -``` ->> x1 = det([3 1 1;7 3 -1;1 1 1])/det([2 1 1;1 3 -1;1 1 1]) - x1 = 2.0000 -``` - -## **[B.5 PARTIAL](#page-6-0) FRACTION EXPANSION** - -In the analysis of linear time-invariant systems, we encounter functions that are ratios of two polynomials in a certain variable, say, *x*. Such functions are known as *rational functions*. A rational function *F*(*x*) can be expressed as - -$$ -F(x) = \frac{b_m x^m + b_{m-1} x^{m-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0} = \frac{P(x)}{Q(x)} -$$ -(B.22) - -The function *F*(*x*) is *improper* if *m* ≥ *n* and *proper* if *m* < *n*. † An improper function can always be separated into the sum of a polynomial in *x* and a proper function. Consider, for example, the function - -$$ -F(x) = \frac{2x^3 + 9x^2 + 11x + 2}{x^2 + 4x + 3} -$$ - -Because this is an improper function, we divide the numerator by the denominator until the remainder has a lower degree than the denominator. - -$$ -x^{2} + 4x + 3 \quad \begin{array}{c} 2x + 1 \\ 2x^{3} + 9x^{2} + 11x + 2 \\ 2x^{3} + 8x^{2} + 6x \\ x^{2} + 5x + 2 \\ x^{2} + 4x + 3 \\ x - 1 \end{array} -$$ - -† Some sources classify *F*(*x*) as strictly proper if *m* &lt; *n*, proper if *m* ≤ *n*, and improper if *m* &gt; *n*. - -### 26 CHAPTER B BACKGROUND - -Therefore, *F*(*x*) can be expressed as - -$$ -F(x) = \frac{2x^3 + 9x^2 + 11x + 2}{x^2 + 4x + 3} = \underbrace{2x + 1}_{\text{polynomial in } x} + \underbrace{\frac{x - 1}{x^2 + 4x + 3}}_{\text{proper function}} -$$ - -A proper function can be further expanded into partial fractions. The remaining discussion in this section is concerned with various ways of doing this. - -### **[B.5-1 Method of Clearing Fractions](#page-6-0)** - -A rational function can be written as a sum of appropriate partial fractions with unknown coefficients, which are determined by clearing fractions and equating the coefficients of similar powers on the two sides. This procedure is demonstrated by the following example. - -### **EXAMPLE B.8 Method of Clearing Fractions** - -Expand the following rational function *F*(*x*) into partial fractions: - -$$ -F(x) = \frac{x^3 + 3x^2 + 4x + 6}{(x+1)(x+2)(x+3)^2} -$$ - -This function can be expressed as a sum of partial fractions with denominators (*x* + 1), (*x* +2),(*x* +3), and (*x* +3)2, as follows: - -$$ -F(x) = \frac{x^3 + 3x^2 + 4x + 6}{(x+1)(x+2)(x+3)^2} = \frac{k_1}{x+1} + \frac{k_2}{x+2} + \frac{k_3}{x+3} + \frac{k_4}{(x+3)^2} -$$ - -To determine the unknowns *k*1, *k*2, *k*3, and *k*4, we clear fractions by multiplying both sides by (*x* +1)(*x* +2)(*x* +3)2 to obtain - -$$ -x^{3} + 3x^{2} + 4x + 6 = k_{1}(x^{3} + 8x^{2} + 21x + 18) + k_{2}(x^{3} + 7x^{2} + 15x + 9) -$$ - -+ $k_{3}(x^{3} + 6x^{2} + 11x + 6) + k_{4}(x^{2} + 3x + 2)$ -= $x^{3}(k_{1} + k_{2} + k_{3}) + x^{2}(8k_{1} + 7k_{2} + 6k_{3} + k_{4})$ -+ $x(21k_{1} + 15k_{2} + 11k_{3} + 3k_{4}) + (18k_{1} + 9k_{2} + 6k_{3} + 2k_{4})$ - -Equating coefficients of similar powers on both sides yields - -$$ -k_1 + k_2 + k_3 = 1 -$$ - -\n -$$ -8k_1 + 7k_2 + 6k_3 + k_4 = 3 -$$ - -\n -$$ -21k_1 + 15k_2 + 11k_3 + 3k_4 = 4 -$$ - -\n -$$ -18k_1 + 9k_2 + 6k_3 + 2k_4 = 6 -$$ - -Solution of these four simultaneous equations yields - -$$ -k_1 = 1 -$$ -, $k_2 = -2$ , $k_3 = 2$ , $k_4 = -3$ - -Therefore, - -$$ -F(x) = \frac{1}{x+1} - \frac{2}{x+2} + \frac{2}{x+3} - \frac{3}{(x+3)^2} -$$ - -Although this method is straightforward and applicable to all situations, it is not necessarily the most efficient. We now discuss other methods that can reduce numerical work considerably. - -### **[B.5-2 The Heaviside "Cover-Up" Method](#page-6-0)** - -### DISTINCT FACTORS OF *Q*(*x*) - -We shall first consider the partial fraction expansion of *F*(*x*) = *P*(*x*)/*Q*(*x*), in which all the factors of *Q*(*x*) are distinct (not repeated). Consider the proper function - -$$ -F(x) = \frac{b_m x^m + b_{m-1} x^{m-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0} \qquad m < n -$$ -$$ -= \frac{P(x)}{(x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n)} -$$ - -As seen in Ex. B.8, *F*(*x*) can be expressed as the sum of partial fractions - -$$ -F(x) = \frac{k_1}{x - \lambda_1} + \frac{k_2}{x - \lambda_2} + \dots + \frac{k_n}{x - \lambda_n} -$$ - (B.23) - -To determine the coefficient *k*1, we multiply both sides of Eq. (B.23) by *x*−λ1 and then let *x* = λ1. This yields - -$$ -(x - \lambda_1)F(x)|_{x = \lambda_1} = k_1 + \frac{k_2(x - \lambda_1)}{(x - \lambda_2)} + \frac{k_3(x - \lambda_1)}{(x - \lambda_3)} + \dots + \frac{k_n(x - \lambda_1)}{(x - \lambda_n)}\bigg|_{x = \lambda_1} -$$ - -On the right-hand side, all the terms except *k*1 vanish. Therefore, - -$$ -k_1 = (x - \lambda_1)F(x)|_{x = \lambda_1} -$$ - -Similarly, we can show that - -$$ -k_r = (x - \lambda_r)F(x)|_{x = \lambda_r} \qquad r = 1, 2, \dots, n -$$ - (B.24) - -This procedure also goes under the name *method of residues*. - -### **EXAMPLE B.9 Heaviside "Cover-Up" Method** - -Expand the following rational function *F*(*x*) into partial fractions: - -$$ -F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} = \frac{k_1}{x+1} + \frac{k_2}{x-2} + \frac{k_3}{x+3} -$$ - -To determine *k*1, we let *x* = −1 in (*x* + 1)*F*(*x*). Note that (*x* + 1)*F*(*x*) is obtained from *F*(*x*) by omitting the term (*x* + 1) from its denominator. Therefore, to compute *k*1 corresponding to the factor (*x* + 1), we cover up the term (*x* + 1) in the denominator of *F*(*x*) and then substitute *x* = −1 in the remaining expression. [Mentally conceal the term (*x* + 1) in *F*(*x*) with a finger and then let *x* = −1 in the remaining expression.] The steps in covering up the function - -$$ -F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} -$$ - -are as follows. - -**Step 1.** Cover up (conceal) the factor (*x* +1) from *F*(*x*): - -$$ -\frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} -$$ - -**Step 2.** Substitute *x* = −1 in the remaining expression to obtain *k*1: - -$$ -k_1 = \frac{2 - 9 - 11}{(-1 - 2)(-1 + 3)} = \frac{-18}{-6} = 3 -$$ - -Similarly, to compute *k*2, we cover up the factor (*x* − 2) in *F*(*x*) and let *x* = 2 in the remaining function, as follows: - -$$ -k_2 = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}\bigg|_{x=2} = \frac{8+18-11}{(2+1)(2+3)} = \frac{15}{15} = 1 -$$ - -and - -$$ -k_3 = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}\bigg|_{x=-3} = \frac{18 - 27 - 11}{(-3+1)(-3-2)} = \frac{-20}{10} = -2 -$$ - -Therefore, - -$$ -F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} = \frac{3}{x+1} + \frac{1}{x-2} - \frac{2}{x+3} -$$ - -### COMPLEX FACTORS OF *Q*(*x*) - -The procedure just given works regardless of whether the factors of *Q*(*x*) are real or complex. Consider, for example, - -$$ -F(x) = \frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)} -$$ - -= $\frac{k_1}{x+1} + \frac{k_2}{x+2-j3} + \frac{k_3}{x+2+j3}$ (B.25) - -where - -$$ -k_1 = \left[ \frac{4x^2 + 2x + 18}{(x+1)\ (x^2 + 4x + 13)} \right]_{x=-1} = 2 -$$ - -Similarly, - -$$ -k_2 = \left[ \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)} \right]_{x=-2+j3} = 1+j2 = \sqrt{5}e^{j63.43^{\circ}} -$$ -$$ -k_3 = \left[ \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)} \right]_{x=-2-j3} = 1-j2 = \sqrt{5}e^{-j63.43^{\circ}} -$$ - -Therefore, - -$$ -F(x) = \frac{2}{x+1} + \frac{\sqrt{5}e^{i63.43^{\circ}}}{x+2-j3} + \frac{\sqrt{5}e^{-i63.43^{\circ}}}{x+2+j3} -$$ - -The coefficients *k*2 and *k*3 corresponding to the complex-conjugate factors are also conjugates of each other. This is generally true when the coefficients of a rational function are real. In such a case, we need to compute only one of the coefficients. - -### QUADRATIC FACTORS - -Often we are required to combine the two terms arising from complex-conjugate factors into one quadratic factor. For example, *F*(*x*) in Eq. (B.25) can be expressed as - -$$ -F(x) = \frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{k_1}{x+1} + \frac{c_1x + c_2}{x^2 + 4x + 13} -$$ - -The coefficient *k*1 is found by the Heaviside method to be 2. Therefore, - -$$ -\frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{2}{x+1} + \frac{c_1x + c_2}{x^2 + 4x + 13} -$$ -(B.26) - -The values of *c*1 and *c*2 are determined by clearing fractions and equating the coefficients of similar powers of *x* on both sides of the resulting equation. Clearing fractions on both sides of Eq. (B.26) yields - -$$ -4x2 + 2x + 18 = 2(x2 + 4x + 13) + (c1x + c2)(x + 1) -$$ - -= (2+c1)x2 + (8+c1+c2)x + (26+c2) - -Equating terms of similar powers yields *c*1 = 2, *c*2 = −8, and - -$$ -\frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{2}{x+1} + \frac{2x - 8}{x^2 + 4x + 13} -$$ - -### SHORTCUTS - -The values of *c*1 and *c*2 in Eq. (B.26) can also be determined by using shortcuts. After computing *k*1 = 2 by the Heaviside method as before, we let *x* = 0 on both sides of Eq. (B.26) to eliminate *c*1. This gives us - -$$ -\frac{18}{13} = 2 + \frac{c_2}{13} \qquad \Rightarrow \qquad c_2 = -8 -$$ - -To determine *c*1, we multiply both sides of Eq. (B.26) by *x* and then let *x* → ∞. Remember that when *x* → ∞, only the terms of the highest power are significant. Therefore, - -$$ -4 = 2 + c_1 \qquad \Rightarrow \qquad c_1 = 2 -$$ - -In the procedure discussed here, we let *x* = 0 to determine *c*2 and then multiply both sides by *x* and let *x* → ∞ to determine *c*1. However, nothing is sacred about these values (*x* = 0 or *x* = ∞). We use them because they reduce the number of computations involved. We could just as well use other convenient values for *x*, such as *x* = 1. Consider the case - -$$ -F(x) = \frac{2x^2 + 4x + 5}{x(x^2 + 2x + 5)} = \frac{k}{x} + \frac{c_1x + c_2}{x^2 + 2x + 5} -$$ - -We find *k* = 1 by the Heaviside method in the usual manner. As a result, - -$$ -\frac{2x^2 + 4x + 5}{x(x^2 + 2x + 5)} = \frac{1}{x} + \frac{c_1x + c_2}{x^2 + 2x + 5} -$$ -(B.27) - -If we try letting *x* = 0 to determine *c*1 and *c*2, we obtain ∞ on both sides. So let us choose *x* = 1. This yields - -$$ -\frac{11}{8} = 1 + \frac{c_1 + c_2}{8} \qquad \text{or} \qquad c_1 + c_2 = 3 -$$ - -We can now choose some other value for *x*, such as *x* = 2, to obtain one more relationship to use in determining *c*1 and *c*2. In this case, however, a simple method is to multiply both sides of Eq. (B.27) by *x* and then let *x* → ∞. This yields - -$$ -2 = 1 + c_1 \qquad \Rightarrow \qquad c_1 = 1 -$$ - -Since *c*1 +*c*2 = 3, we see that *c*2 = 2 and therefore, - -$$ -F(x) = \frac{1}{x} + \frac{x+2}{x^2 + 2x + 5} -$$ - -## **[B.5-3 Repeated Factors of](#page-6-0)** *Q(x)* - -If a function *F*(*x*) has a repeated factor in its denominator, it has the form - -$$ -F(x) = \frac{P(x)}{(x - \lambda)^r (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_j)} -$$ - -Its partial fraction expansion is given by - -$$ -F(x) = \frac{a_0}{(x - \lambda)^r} + \frac{a_1}{(x - \lambda)^{r-1}} + \dots + \frac{a_{r-1}}{(x - \lambda)} -$$ - -+ -$$ -\frac{k_1}{x - \alpha_1} + \frac{k_2}{x - \alpha_2} + \dots + \frac{k_j}{x - \alpha_j} -$$ - (B.28) - -The coefficients *k*1, *k*2,..., *kj* corresponding to the unrepeated factors in this equation are determined by the Heaviside method, as before [Eq. (B.24)]. To find the coefficients *a*0,*a*1, *a*2,...,*ar*−1, we multiply both sides of Eq. (B.28) by (*x* −λ)*r* . This gives us - -$$ -(x - \lambda)^r F(x) = a_0 + a_1(x - \lambda) + a_2(x - \lambda)^2 + \dots + a_{r-1}(x - \lambda)^{r-1} + k_1 \frac{(x - \lambda)^r}{x - \alpha_1} + k_2 \frac{(x - \lambda)^r}{x - \alpha_2} + \dots + k_n \frac{(x - \lambda)^r}{x - \alpha_n} -$$ -(B.29) - -If we let *x* = λ on both sides of Eq. (B.29), we obtain - -$$ -(x - \lambda)^r F(x)|_{x = \lambda} = a_0 -$$ - -Therefore, *a*0 is obtained by concealing the factor (*x*−λ)*r* in *F*(*x*) and letting *x* =λ in the remaining expression (the Heaviside "cover-up" method). If we take the derivative (with respect to *x*) of both sides of Eq. (B.29), the right-hand side is *a*1+ terms containing a factor (*x*−λ) in their numerators. Letting *x* = λ on both sides of this equation, we obtain - -$$ -\frac{d}{dx}\left[ (x - \lambda)^r F(x) \right] \Big|_{x = \lambda} = a_1 -$$ - -Thus, *a*1 is obtained by concealing the factor (*x*−λ)*r* in *F*(*x*), taking the derivative of the remaining expression, and then letting *x* = λ. Continuing in this manner, we find - -$$ -a_j = \frac{1}{j!} \left. \frac{d^j}{dx^j} \left[ (x - \lambda)^r F(x) \right] \right|_{x = \lambda} -$$ - (B.30) - -Observe that (*x* − λ)*r F*(*x*) is obtained from *F*(*x*) by omitting the factor (*x* − λ)*r* from its denominator. Therefore, the coefficient *aj* is obtained by concealing the factor (*x* − λ)*r* in *F*(*x*), taking the *j*th derivative of the remaining expression, and then letting *x* = λ (while dividing by *j*!). - -### **EXAMPLE B.10 Partial Fraction Expansion with Repeated Factors** - -Expand *F*(*x*) into partial fractions if - -$$ -F(x) = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} -$$ - -The partial fractions are - -$$ -F(x) = \frac{a_0}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{k}{x+2} -$$ - -The coefficient *k* is obtained by concealing the factor (*x* + 2) in *F*(*x*) and then substituting *x* = −2 in the remaining expression: - -$$ -k = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}\bigg|_{x=-2} = 1 -$$ - -To find *a*0, we conceal the factor (*x* +1)3 in *F*(*x*) and let *x* = −1 in the remaining expression: - -$$ -a_0 = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}\bigg|_{x=-1} = 2 -$$ - -To find *a*1, we conceal the factor (*x* + 1)3 in *F*(*x*), take the derivative of the remaining expression, and then let *x* = −1: - -$$ -a_1 = \frac{d}{dx} \left[ \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} \right] \Big|_{x=-1} = 1 -$$ - -Similarly, - -$$ -a_2 = \frac{1}{2!} \frac{d^2}{dx^2} \left[ \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} \right] \Big|_{x=-1} = 3 -$$ - -Therefore, - -$$ -F(x) = \frac{2}{(x+1)^3} + \frac{1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2} -$$ - -### **[B.5-4 A Combination of Heaviside "Cover-Up" and Clearing Fractions](#page-6-0)** - -For multiple roots, especially of higher order, the Heaviside expansion method, which requires repeated differentiation, can become cumbersome. For a function that contains several repeated and unrepeated roots, a hybrid of the two procedures proves to be the best. The simpler coefficients are determined by the Heaviside method, and the remaining coefficients are found by clearing fractions or shortcuts, thus incorporating the best of the two methods. We demonstrate this procedure by solving Ex. B.10 once again by this method. - -In Ex. B.10, coefficients *k* and *a*0 are relatively simple to determine by the Heaviside expansion method. These values were found to be *k*1 = 1 and *a*0 = 2. Therefore, - -$$ -\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{1}{x+2} -$$ - -We now multiply both sides of this equation by (*x* +1)3(*x* +2) to clear the fractions. This yields - -$$ -4x3 + 16x2 + 23x + 13 = 2(x+2) + a1(x+1)(x+2) + a2(x+1)2(x+2) + (x+1)3 -$$ - -= (1+a2)x3 + (a1+4a2+3)x2 + (5+3a1+5a2)x + (4+2a1+2a2+1) - -Equating coefficients of the third and second powers of *x* on both sides, we obtain - -$$ -\begin{array}{ccc} 1 + a_2 = 4 \\ a_1 + 4a_2 + 3 = 16 \end{array} \implies \begin{array}{c} a_1 = 1 \\ a_2 = 3 \end{array} -$$ - -We may stop here if we wish because the two desired coefficients, *a*1 and *a*2, are now determined. However, equating the coefficients of the two remaining powers of *x* yields a convenient check on the answer. Equating the coefficients of the *x*1 and *x*0 terms, we obtain - -$$ -23 = 5 + 3a_1 + 5a_2 -$$ - -$$ -13 = 4 + 2a_1 + 2a_2 + 1 -$$ - -These equations are satisfied by the values *a*1 = 1 and *a*2 = 3, found earlier, providing an additional check for our answers. Therefore, - -$$ -F(x) = \frac{2}{(x+1)^3} + \frac{1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2} -$$ - -which agrees with the earlier result. - -### A COMBINATION OF HEAVISIDE "COVER-UP" AND SHORTCUTS - -In Ex. B.10, after determining the coefficients *a*0 = 2 and *k* = 1 by the Heaviside method as before, we have - -$$ -\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{1}{x+2} -$$ - -There are only two unknown coefficients, *a*1 and *a*2. If we multiply both sides of this equation by *x* and then let *x* → ∞, we can eliminate *a*1. This yields - -$$ -4 = a_2 + 1 \quad \Longrightarrow \quad a_2 = 3 -$$ - -Therefore, - -$$ -\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2} -$$ - -#### 34 CHAPTER B BACKGROUND - -There is now only one unknown *a*1, which can be readily found by setting *x* equal to any convenient value, say, *x* = 0. This yields - -$$ -\frac{13}{2} = 2 + a_1 + 3 + \frac{1}{2} \implies a_1 = 1 -$$ - -which agrees with our earlier answer. - -There are other possible shortcuts. For example, we can compute *a*0 (coefficient of the highest power of the repeated root), subtract this term from both sides, and then repeat the procedure. - -## **[B.5-5 Improper](#page-6-0)** *F(x)* **with** *m* **=** *n* - -A general method of handling an improper function is indicated in the beginning of this section. However, for the special case of when the numerator and denominator polynomials of *F*(*x*) have the same degree (*m* = *n*), the procedure is the same as that for a proper function. We can show that for - -$$ -F(x) = \frac{b_n x^n + b_{n-1} x^{n-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0} -$$ - -= $b_n + \frac{k_1}{x - \lambda_1} + \frac{k_2}{x - \lambda_2} + \dots + \frac{k_n}{x - \lambda_n}$ - -the coefficients *k*1, *k*2,..., *kn* are computed as if *F*(*x*) were proper. Thus, - -$$ -k_r = (x - \lambda_r)F(x)|_{x = \lambda_r} -$$ - -For quadratic or repeated factors, the appropriate procedures discussed in Secs. B.5-2 or B.5-3 should be used as if *F*(*x*) were proper. In other words, when *m* = *n*, the only difference between the proper and improper case is the appearance of an extra constant *bn* in the latter. Otherwise, the procedure remains the same. The proof is left as an exercise for the reader. - -### **EXAMPLE B.11 Partial Fraction Expansion of Improper Rational Function** - -Expand *F*(*x*) into partial fractions if - -$$ -F(x) = \frac{3x^2 + 9x - 20}{x^2 + x - 6} = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} -$$ - -Here, *m* = *n* = 2 with *bn* = *b*2 = 3. Therefore, - -$$ -F(x) = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} = 3 + \frac{k_1}{x - 2} + \frac{k_2}{x + 3} -$$ - -in which - -$$ -k_1 = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} \bigg|_{x=2} = \frac{12 + 18 - 20}{(2 + 3)} = \frac{10}{5} = 2 -$$ - -and - -$$ -k_2 = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)}\bigg|_{x=-3} = \frac{27 - 27 - 20}{(-3 - 2)} = \frac{-20}{-5} = 4 -$$ - -Therefore, - -$$ -F(x) = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} = 3 + \frac{2}{x - 2} + \frac{4}{x + 3} -$$ - -### **[B.5-6 Modified Partial Fractions](#page-6-0)** - -In finding the inverse *z*-transform (Ch. 5), we require partial fractions of the form *kx*/(*x* −λ*i*)*r* rather than *k*/(*x* −λ*i*)*r* . This can be achieved by expanding *F*(*x*)/*x* into partial fractions. Consider, for example, - -$$ -F(x) = \frac{5x^2 + 20x + 18}{(x+2)(x+3)^2} -$$ - -Dividing both sides by *x* yields - -$$ -\frac{F(x)}{x} = \frac{5x^2 + 20x + 18}{x(x+2)(x+3)^2} -$$ - -Expansion of the right-hand side into partial fractions as usual yields - -$$ -\frac{F(x)}{x} = \frac{5x^2 + 20x + 18}{x(x+2)(x+3)^2} = \frac{a_1}{x} + \frac{a_2}{x+2} + \frac{a_3}{(x+3)} + \frac{a_4}{(x+3)^2} -$$ - -Using the procedure discussed earlier, we find *a*1 = 1, *a*2 = 1, *a*3 = −2, and *a*4 = 1. Therefore, - -$$ -\frac{F(x)}{x} = \frac{1}{x} + \frac{1}{x+2} - \frac{2}{x+3} + \frac{1}{(x+3)^2} -$$ - -Now multiplying both sides by *x* yields - -$$ -F(x) = 1 + \frac{x}{x+2} - \frac{2x}{x+3} + \frac{x}{(x+3)^2} -$$ - -This expresses *F*(*x*) as the sum of partial fractions having the form *kx*/(*x* −λ*i*)*r* . - -## **[B.6 VECTORS AND](#page-6-0) MATRICES** - -An entity specified by *n* numbers in a certain order (ordered *n*-tuple) is an *n*-dimensional *vector*. Thus, an ordered *n*-tuple (*x*1, *x*2, ..., *xn*) represents an *n*-dimensional vector **x**. A vector may be represented as a row (*row vector*): - -$$ -\mathbf{x} = [x_1 \quad x_2 \quad \cdots \quad x_n] -$$ - -or as a column (*column vector*): - -$$ -\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} -$$ - -Simultaneous linear equations can be viewed as the transformation of one vector into another. Consider, for example, the *m* simultaneous linear equations - -$$ -y_1 = a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n -$$ - -\n -$$ -y_2 = a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -y_m = a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n -$$ - -\n(B.31) - -If we define two column vectors **x** and **y** as - -$$ -\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \quad \text{and} \quad \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix} -$$ - -then Eq. (B.31) may be viewed as the relationship or the function that transforms vector **x** into vector **y**. Such a transformation is called a *linear transformation* of vectors. To perform a linear transformation, we need to define the array of coefficients *aij* appearing in Eq. (B.31). This array is called a *matrix* and is denoted by **A** for convenience: - -$$ -\mathbf{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{array} \right] -$$ - -A matrix with *m* rows and *n* columns is called a matrix of order (*m*,*n*) or an (*m* × *n*) matrix. For the special case of *m* = *n*, the matrix is called a *square matrix* of order *n*. - -It should be stressed at this point that a matrix is not a number such as a determinant, but an array of numbers arranged in a particular order. It is convenient to abbreviate the representation of matrix **A** with the form (*aij*)*m*×*n*, implying a matrix of order *m* × *n* with *aij* as its *ij*th element. In practice, when the order *m* × *n* is understood or need not be specified, the notation can be abbreviated to (*aij*). Note that the first index *i* of *aij* indicates the row and the second index *j* indicates the column of the element *aij* in matrix **A**. - -Equation (B.31) may now be expressed in a matrix form as - -$$ -\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} -$$ - or $y = Ax$ (B.32) - -At this point, we have not defined the multiplication of a matrix by a vector. The quantity **Ax** is not meaningful until such an operation has been defined. - -### **[B.6-1 Some Definitions and Properties](#page-6-0)** - -A square matrix whose elements are zero everywhere except on the main diagonal is a *diagonal matrix*. An example of a diagonal matrix is - -| ⎡
2 | 0 | 0 | ⎤ | -|--------|---|---|---| -| 0
⎣ | 1 | 0 | ⎦ | -| 0 | 0 | 5 | | - -A diagonal matrix with unity for all its diagonal elements is called an *identity matrix* or a *unit matrix*, denoted by **I**. This is a square matrix: - -| | ⎡
1 | 0 | 0 | ··· | ⎤
0 | -|--------|-------------|---|---|-----|-------------| -| | 0
⎢ | 1 | 0 | ··· | 0
⎥ | -| I
= | ⎢
0
⎢ | 0 | 1 | ··· | ⎥
0
⎥ | -| | ⎢
⎢
⎣ | | | ··· | ⎥
⎥
⎦ | -| | 0 | 0 | 0 | ··· | 1 | - -The order of the unit matrix is sometimes indicated by a subscript. Thus, **I***n* represents the *n*×*n* unit matrix (or identity matrix). However, we shall omit the subscript since order is easily understood by context. - -A matrix having all its elements zero is a *zero matrix*. - -A square matrix **A** is a *symmetric matrix* if *aij* = *aji* (symmetry about the main diagonal). - -Two matrices of the same order are said to be *equal* if they are equal element by element. Thus, if - -$$ -\mathbf{A} = (a_{ij})_{m \times n} \quad \text{and} \quad \mathbf{B} = (b_{ij})_{m \times n} -$$ - -then **A** = **B** only if *aij* = *bij* for all *i* and *j*. - -If the rows and columns of an *m*×*n* matrix **A** are interchanged so that the elements in the *i*th row now become the elements of the *i*th column (for *i* = 1, 2,...,*m*), the resulting matrix is called the *transpose* of **A** and is denoted by **A***T* . It is evident that **A***T* is an *n*×*m* matrix. For example, if - -$$ -\mathbf{A} = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 3 \end{bmatrix}, \quad \text{then} \quad \mathbf{A}^T = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 2 & 3 \end{bmatrix} -$$ - -### 38 CHAPTER B BACKGROUND - -Using the abbreviated notation, if **A** = (*aij*)*m*×*n*, then **A***T* = (*aji*)*n*×*m*. Intuitively, further notice that (**A***T* )*T* = **A**. - -### **[B.6-2 Matrix Algebra](#page-6-0)** - -We shall now define matrix operations, such as addition, subtraction, multiplication, and division of matrices. The definitions should be formulated so that they are useful in the manipulation of matrices. - -### ADDITION OF MATRICES - -For two matrices **A** and **B**, both of the same order (*m*×*n*), - -$$ -\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \text{ and } \mathbf{B} = \begin{bmatrix} b_{11} & b_{12} & \cdots & b_{1n} \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ b_{m1} & b_{m2} & \cdots & b_{mn} \end{bmatrix} -$$ - -we define the sum **A**+**B** as - -$$ -\mathbf{A} + \mathbf{B} = \begin{bmatrix} (a_{11} + b_{11}) & (a_{12} + b_{12}) & \cdots & (a_{1n} + b_{1n}) \\ (a_{21} + b_{21}) & (a_{22} + b_{22}) & \cdots & (a_{2n} + b_{2n}) \\ \vdots & \vdots & \ddots & \vdots \\ (a_{m1} + b_{m1}) & (a_{m2} + b_{m2}) & \cdots & (a_{mn} + b_{mn}) \end{bmatrix} -$$ - -or - -$$ -\mathbf{A} + \mathbf{B} = (a_{ij} + b_{ij})_{m \times n} -$$ - -Note that two matrices can be added only if they are of the same order. - -### MULTIPLICATION OF A MATRIX BY A SCALAR - -We multiply a matrix **A** by a scalar *c* as follows: - -$$ -c\mathbf{A} = c \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} = \begin{bmatrix} ca_{11} & ca_{12} & \cdots & ca_{1n} \\ ca_{21} & ca_{22} & \cdots & ca_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ ca_{m1} & ca_{m2} & \cdots & ca_{mn} \end{bmatrix} = \mathbf{A}c -$$ - -Thus, we also observe that the scalar *c* and the matrix **A** commute: *c***A** = **A***c*. - -### MATRIX MULTIPLICATION - -We define the product - -$$ -AB = C -$$ - -in which *cij*, the element of **C** in the *i*th row and *j*th column, is found by adding the products of the elements of **A** in the *i*th row multiplied by the corresponding elements of **B** in the *j*th column. Thus, - -$$ -c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj} = \sum_{k=1}^{n} a_{ik}b_{kj} -$$ - (B.33) - -This result is expressed as follows: - -Note carefully that if this procedure is to work, the number of columns of **A** must be equal to the number of rows of **B**. In other words, **AB**, the product of matrices **A** and **B**, is defined only if the number of columns of **A** is equal to the number of rows of **B**. If this condition is not satisfied, the product **AB** is not defined and is meaningless. When the number of columns of **A** is equal to the number of rows of **B**, matrix **A** is said to be *conformable* to matrix **B** for the product **AB**. Observe that if **A** is an *m* × *n* matrix and **B** is an *n* × *p* matrix, **A** and **B** are conformable for the product, and **C** is an *m*×*p* matrix. - -We demonstrate the use of the rule in Eq. (B.33) with the following examples. - -$$ -\begin{bmatrix} 2 & 3 \\ 1 & 1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 1 & 2 \\ 2 & 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 9 & 5 & 7 \\ 3 & 4 & 2 & 3 \\ 5 & 10 & 4 & 7 \end{bmatrix} -$$ -$$ -\begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = 8 -$$ - -In both cases, the two matrices are conformable. However, if we interchange the order of the first matrices as follows: - -| 1 | 3 | 1 | !⎡
2 | 2 | 3 | ⎤ | -|---|---|---|---------|--------|---|---| -| | | | | 1
⎣ | 1 | ⎦ | -| 2 | 1 | 1 | 1 | 3 | 1 | | - -the matrices are no longer conformable for the product. It is evident that, in general, - -### **AB** = **BA** - -Indeed, **AB** may exist and **BA** may not exist, or vice versa, as in our examples. We shall see later that for some special matrices, **AB** = **BA**. When this is true, matrices **A** and **B** are said to *commute*. We re-emphasize that in general, matrices do not commute. - -### 40 CHAPTER B BACKGROUND - -In the matrix product **AB**, matrix **A** is said to be *postmultiplied* by **B** or matrix **B** is said to be *premultiplied* by **A**. We may also verify the following relationships: - -$$ -(A + B)C = AC + BC -$$ -$$ -C(A + B) = CA + CB -$$ - -We can verify that any matrix **A** premultiplied or postmultiplied by the identity matrix **I** remains unchanged: - -**AI** = **IA** = **A** - -Of course, we must make sure that the order of **I** is such that the matrices are conformable for the corresponding product. - -We give here, without proof, another important property of matrices: - -$$ -|\mathbf{A}\mathbf{B}| = |\mathbf{A}||\mathbf{B}| -$$ - -where |**A**| and |**B**| represent determinants of matrices **A** and **B**. - -### MULTIPLICATION OF A MATRIX BY A VECTOR - -Consider Eq. (B.32), which represents Eq. (B.31). The right-hand side of Eq. (B.32) is a product of the *m*×*n* matrix **A** and a vector **x**. If, for the time being, we treat the vector **x** as if it were an *n*×1 matrix, then the product **Ax**, according to the matrix multiplication rule, yields the right-hand side of Eq. (B.31). Thus, we may multiply a matrix by a vector by treating the vector as if it were an *n* × 1 matrix. Note that the constraint of conformability still applies. Thus, in this case, **xA** is not defined and is meaningless. - -### MATRIX INVERSION - -To define the inverse of a matrix, let us consider the set of equations represented by Eq. (B.32) when *m* = *n*: - -$$ -\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} -$$ - (B.34) - -We can solve this set of equations for *x*1, *x*2, ... , *xn* in terms of *y*1, *y*2, ... , *yn* by using Cramer's rule [see Eq. (B.21)]. This yields - -$$ -\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} \frac{|\mathbf{D}_{11}|}{|\mathbf{A}|} & \frac{|\mathbf{D}_{21}|}{|\mathbf{A}|} & \cdots & \frac{|\mathbf{D}_{n1}|}{|\mathbf{A}|} \\ \frac{|\mathbf{D}_{12}|}{|\mathbf{A}|} & \frac{|\mathbf{D}_{22}|}{|\mathbf{A}|} & \cdots & \frac{|\mathbf{D}_{n2}|}{|\mathbf{A}|} \\ \vdots & \vdots & \cdots & \vdots \\ \frac{|\mathbf{D}_{1n}|}{|\mathbf{A}|} & \frac{|\mathbf{D}_{2n}|}{|\mathbf{A}|} & \cdots & \frac{|\mathbf{D}_{nn}|}{|\mathbf{A}|} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} -$$ -(B.35) - -in which |**A**| is the determinant of the matrix **A** and |**D***ij*| is the *cofactor* of element *aij* in the matrix **A**. The cofactor of element *aij* is given by (−1)*i*+*j* times the determinant of the (*n* − 1) × (*n* − 1) matrix that is obtained when the *i*th row and the *j*th column in matrix **A** are deleted. - -We can express Eq. (B.34) in compact matrix form as - -$$ -y = Ax -$$ - (B.36) - -We now define **A**−1 , the inverse of a square matrix **A**, with the property - -> **A**−1 **A** = **I** (unit matrix) - -Then, premultiplying both sides of Eq. (B.36) by **A**−1 , we obtain - -$$ -\mathbf{A}^{-1}\mathbf{y} = \mathbf{A}^{-1}\mathbf{A}\mathbf{x} = \mathbf{I}\mathbf{x} = \mathbf{x} -$$ - -or - -$$ -\mathbf{x} = \mathbf{A}^{-1} \mathbf{y} \tag{B.37} -$$ - -A comparison of Eq. (B.37) with Eq. (B.35) shows that - -$$ -A^{-1} = \frac{1}{|A|} \begin{bmatrix} |D_{11}| & |D_{21}| & \cdots & |D_{n1}| \\ |D_{12}| & |D_{22}| & \cdots & |D_{n2}| \\ \vdots & \vdots & \cdots & \vdots \\ |D_{1n}| & |D_{2n}| & \cdots & |D_{nn}| \end{bmatrix} -$$ - -One of the conditions necessary for a unique solution of Eq. (B.34) is that the number of equations must equal the number of unknowns. This implies that the matrix **A** must be a square matrix. In addition, we observe from the solution as given in Eq. (B.35) that if the solution is to exist, |**A**| = 0.† Therefore, the inverse exists only for a square matrix and only under the condition that the determinant of the matrix be nonzero. A matrix whose determinant is nonzero is a *nonsingular* matrix. Thus, an inverse exists only for a nonsingular, square matrix. Since **A**−1 **A** = **I** = **AA**−1 , we further note that the matrices **A** and **A**−1 commute.‡ - -The operation of matrix division can be accomplished through matrix inversion. - -### **EXAMPLE B.12 Computing the Inverse of a Matrix** - -Let us find **A**−1 if - -**A** = ⎡ ⎣ 211 123 321 ⎤ ⎦ - -† These two conditions imply that the number of equations is equal to the number of unknowns and that all the equations are independent. - -‡ To prove **AA**−1 = **I**, notice first that we define **A**−1**A** = **I**. Thus, **IA** = **AI** = **A**(**A**−1**A**) = (**AA**−1)**A**. Subtracting (**AA**−1)**A**, we see that **IA**−(**AA**−1)**A** = 0 or (**I**−**AA**−1)**A** = 0. This requires **AA**−1 = **I**. - -Here, - -$$ -|\mathbf{D}_{11}| = -4, \t |\mathbf{D}_{12}| = 8, \t |\mathbf{D}_{13}| = -4 |\mathbf{D}_{21}| = 1, \t |\mathbf{D}_{22}| = -1, \t |\mathbf{D}_{23}| = -1 |\mathbf{D}_{31}| = 1, \t |\mathbf{D}_{32}| = -5, \t |\mathbf{D}_{33}| = 3 \text{and } |\mathbf{A}| = -4. \text{ Therefore, } \mathbf{A}^{-1} = -\frac{1}{4} \begin{bmatrix} -4 & 1 & 1 \\ 8 & -1 & -5 \\ -4 & -1 & 3 \end{bmatrix} -$$ - -## **[B.7 MATLAB: ELEMENTARY](#page-6-0) OPERATIONS** - -### **[B.7-1 MATLAB Overview](#page-6-0)** - -Although MATLAB´l (a registered trademark of The MathWorks, Inc.) is easy to use, it can be intimidating to new users. Over the years, MATLAB has evolved into a sophisticated computational package with thousands of functions and thousands of pages of documentation. This section provides a brief introduction to the software environment. - -When MATLAB is first launched, its command window appears. When MATLAB is ready to accept an instruction or input, a command prompt (>>) is displayed in the command window. Nearly all MATLAB activity is initiated at the command prompt. - -Entering instructions at the command prompt generally results in the creation of an object or objects. Many classes of objects are possible, including functions and strings, but usually objects are just data. Objects are placed in what is called the MATLAB workspace. If not visible, the workspace can be viewed in a separate window by typing workspace at the command prompt. The workspace provides important information about each object, including the object's name, size, and class. - -Another way to view the workspace is the whos command. When whos is typed at the command prompt, a summary of the workspace is printed in the command window. The who command is a short version of whos that reports only the names of workspace objects. - -Several functions exist to remove unnecessary data and help free system resources. To remove specific variables from the workspace, the clear command is typed, followed by the names of the variables to be removed. Just typing clear removes all objects from the workspace. Additionally, the clc command clears the command window, and the clf command clears the current figure window. - -Often, important data and objects created in one session need to be saved for future use. The save command, followed by the desired filename, saves the entire workspace to a file, which has the .mat extension. It is also possible to selectively save objects by typing save followed by the filename and then the names of the objects to be saved. The load command followed by the filename is used to load the data and objects contained in a MATLAB data file (.mat file). - -Although MATLAB does not automatically save workspace data from one session to the next, lines entered at the command prompt are recorded in the command history. Previous command lines can be viewed, copied, and executed directly from the command history window. From the command window, pressing the up or down arrow key scrolls through previous commands and redisplays them at the command prompt. Typing the first few characters and then pressing the arrow keys scrolls through the previous commands that start with the same characters. The arrow keys allow command sequences to be repeated without retyping. - -Perhaps the most important and useful command for new users is help. To learn more about a function, simply type help followed by the function name. Helpful text is then displayed in the command window. The obvious shortcoming of help is that the function name must first be known. This is especially limiting for MATLAB beginners. Fortunately, help screens often conclude by referencing related or similar functions. These references are an excellent way to learn new MATLAB commands. Typing help help, for example, displays detailed information on the help command itself and also provides reference to relevant functions, such as the lookfor command. The lookfor command helps locate MATLAB functions based on a keyword search. Simply type lookfor followed by a single keyword, and MATLAB searches for functions that contain that keyword. - -MATLAB also has comprehensive HTML-based help. The HTML help is accessed by using MATLAB's integrated help browser, which also functions as a standard web browser. The HTML help facility includes a function and topic index as well as full text-searching capabilities. Since HTML documents can contain graphics and special characters, HTML help can provide more information than the command-line help. After a little practice, it is easy to find information in MATLAB. - -When MATLAB graphics are created, the print command can save figures in a common file format such as postscript, encapsulated postscript, JPEG, or TIFF. The format of displayed data, such as the number of digits displayed, is selected by using the format command. MATLAB help provides the necessary details for both these functions. When a MATLAB session is complete, the exit command terminates MATLAB. - -### **[B.7-2 Calculator Operations](#page-6-0)** - -MATLAB can function as a simple calculator, working as easily with complex numbers as with real numbers. Scalar addition, subtraction, multiplication, division, and exponentiation are accomplished using the traditional operator symbols +, -, \*, /, and ^. Since MATLAB predefines i = j = √−1, a complex constant is readily created using Cartesian coordinates. For example, - ->> z = -3-4j z = -3.0000 - 4.0000i - -assigns the complex constant −3−*j*4 to the variable *z*. - -The real and imaginary components of *z* are extracted by using the real and imag operators. In MATLAB, the input to a function is placed parenthetically following the function name. - -$$ -\Rightarrow \quad z\_real = real(z); \ z\_imag = imag(z); -$$ - -When a command is terminated with a semicolon, the statement is evaluated but the results are not displayed to the screen. This feature is useful when one is computing intermediate results, and it allows multiple instructions on a single line. Although not displayed, the results z\_real = -3 and z\_imag = -4 are calculated and available for additional operations such as computing |*z*|. - -There are many ways to compute the modulus, or magnitude, of a complex quantity. Trigonometry confirms that *z* = −3 − *j*4, which corresponds to a 3-4-5 triangle, has modulus |*z*| = |−3 − *j*4| = (−3)2 +(−4)2 = 5. The MATLAB sqrt command provides one way to compute the required square root. - ->> z\_mag = sqrt(z\_real^2 + z\_imag^2) z\_mag = 5 - -In MATLAB, most commands, including sqrt, accept inputs in a variety of forms, including constants, variables, functions, expressions, and combinations thereof. - -The same result is also obtained by computing |*z*| = √*zz*∗. In this case, complex conjugation is performed by using the conj command. - ->> z\_mag = sqrt(z\*conj(z)) z\_mag = 5 - -More simply, MATLAB computes absolute values directly by using the abs command. - ->> z\_mag = abs(z) z\_mag = 5 - -In addition to magnitude, polar notation requires phase information. The angle command provides the angle of a complex number. - ->> z\_rad = angle(z) z\_rad = -2.2143 - -MATLAB expects and returns angles in a radian measure. Angles expressed in degrees require an appropriate conversion factor. - ->> z\_deg = angle(z)\*180/pi z\_deg = -126.8699 - -Notice, MATLAB predefines the variable pi = π. - -It is also possible to obtain the angle of *z* using a two-argument arc-tangent function, atan2. - -``` ->> z_rad = atan2(z_imag,z_real) - z_rad = -2.2143 -``` - -Unlike a single-argument arctangent function, the two-argument arctangent function ensures that the angle reflects the proper quadrant. MATLAB supports a full complement of trigonometric functions: standard trigonometric functions cos, sin, tan; reciprocal trigonometric functions sec, csc, cot; inverse trigonometric functions acos, asin, atan, asec, acsc, acot; and hyperbolic variations cosh, sinh, tanh, sech, csch, coth, acosh, asinh, atanh, asech, acsch, and acoth. Of course, MATLAB comfortably supports complex arguments for any trigonometric function. As with the angle command, MATLAB trigonometric functions utilize units of radians. - -The concept of trigonometric functions with complex-valued arguments is rather intriguing. The results can contradict what is often taught in introductory mathematics courses. For example, a common claim is that |cos(*x*)| ≤ 1. While this is true for real *x*, it is not necessarily true for complex *x*. This is readily verified by example using MATLAB and the cos function. - -``` ->> cos(1j) - ans = 1.5431 -``` - -Problem B.1-19 investigates these ideas further. - -Similarly, the claim that it is impossible to take the logarithm of a negative number is false. For example, the principal value of ln(−1) is *j*π, a fact easily verified by means of Euler's equation. In MATLAB, base-10 and base-*e* logarithms are computed by using the log10 and log commands, respectively. - ->> log(-1) ans = 0 + 3.1416i - -### **[B.7-3 Vector Operations](#page-6-0)** - -The power of MATLAB becomes apparent when vector arguments replace scalar arguments. Rather than computing one value at a time, a single expression computes many values. Typically, vectors are classified as row vectors or column vectors. For now, we consider the creation of row vectors with evenly spaced, real elements. To create such a vector, the notation a:b:c is used, where a is the initial value, b designates the step size, and c is the termination value. For example, 0:2:11 creates the length-6 vector of even-valued integers ranging from 0 to 10. - ->> k = 0:2:11 k = 0 2 4 6 8 10 - -In this case, the termination value does not appear as an element of the vector. Negative and noninteger step sizes are also permissible. - ->> k = 11:-10/3:0 k = 11.0000 7.6667 4.3333 1.0000 - -If a step size is not specified, a value of 1 is assumed. - ->> k = 0:11 k = 0 1 2 3 4 5 6 7 8 9 10 11 - -Vector notation provides the basis for solving a wide variety of problems. - -For example, consider finding the three cube roots of minus one, *w*3 = −1 = *ej*(π+2π*k*) for integer *k*. Taking the cube root of each side yields *w* = *ej*(π/3+2π*k*/3) . To find the three unique solutions, use any three consecutive integer values of *k* and MATLAB's exp function. - ->> k = 0:2; w = exp(1j\*(pi/3 + 2\*pi\*k/3)) w = 0.5000 + 0.8660i -1.0000 + 0.0000i 0.5000 - 0.8660i - -The solutions, particularly *w* = −1, are easy to verify. - -Finding the 100 unique roots of *w*100 = −1 is just as simple. - ->> -$$ -k = 0.99 -$$ -; w = exp(1j\*(pi/100 + 2\*pi\*k/100)); - -A semicolon concludes the final instruction to suppress the inconvenient display of all 100 solutions. To view a particular solution, the user must use an index to specify desired elements. MATLAB indices are integers that increase from a starting value of 1. For example, the fifth element of *w* is extracted using an index of 5.† - ->> w(5) ans = 0.9603 + 0.2790i - -Notice that this solution corresponds to *k* = 4. The independent variable of a function, in this case *k*, rarely serves as the index. Since *k* is also a vector, it can likewise be indexed. In this way, we can verify that the fifth value of *k* is indeed 4. - ->> k(5) ans = 4 - -It is also possible to use a vector index to access multiple values. For example, index vector 98:100 identifies the last three solutions corresponding to *k* = [97, 98, 99]. - ->> w(98:100) ans = 0.9877 - 0.1564i 0.9956 - 0.0941i 0.9995 - 0.0314i - -Vector representations provide the foundation to rapidly create and explore various signals. Consider the simple 10 Hz sinusoid described by *f*(*t*) = sin(2π10*t* + π/6). Two cycles of this sinusoid are included in the interval 0 ≤ *t* <0.2. A vector *t* is used to uniformly represent 500 points over this interval. - ->> t = 0:0.2/500:0.2-0.2/500; - -Next, the function *f*(*t*) is evaluated at these points. - -``` ->> f = sin(2*pi*10*t+pi/6) -``` - -The value of *f*(*t*) at *t* = 0 is the first element of the vector and is thus obtained by using an index of 1. - ->> f(1) ans = 0.5000 - -Unfortunately, MATLAB's indexing syntax conflicts with standard equation notation.‡ That is, the MATLAB indexing command f(1) is not the same as the standard notation *f*(1) = *f*(*t*)|*t*=1. Care must be taken to avoid confusion; remember that the index parameter rarely reflects the independent variable of a function. - -## **[B.7-4 Simple Plotting](#page-6-0)** - -MATLAB's plot command provides a convenient way to visualize data, such as graphing *f*(*t*) against the independent variable *t*. - -``` ->> plot(t,f); -``` - -† Some other programming languages, such as C, begin indexing at 0. Careful attention is warranted. - -‡ MATLAB anonymous functions, considered in Sec. 1.11, are an important and useful exception. - -**Figure B.12** *f*(*t*) = sin(2π10*t* +π/6). - -Axis labels are added using the xlabel and ylabel commands, where the desired string must be enclosed by single quotation marks. The result is shown in Fig. B.12. - ->> xlabel('t'); ylabel('f(t)') - -The title command is used to add a title above the current axis. - -By default, MATLAB connects data points with solid lines. Plotting discrete points, such as the 100 unique roots of *w*100 = −1, is accommodated by supplying the plot command with an additional string argument. For example, the string 'o' tells MATLAB to mark each data point with a circle rather than connecting points with lines. A full description of the supported plot options is available from MATLAB's help facilities. - -``` ->> plot(real(w),imag(w),'o'); ->> xlabel('Re(w)'); ylabel('Im(w)'); axis equal -``` - -The axis equal command ensures that the scale used for the horizontal axis is equal to the scale used for the vertical axis. Without axis equal, the plot would appear elliptical rather than circular. Figure B.13 illustrates that the 100 unique roots of *w*100 = −1 lie equally spaced on the unit circle, a fact not easily discerned from the raw numerical data. - -MATLAB also includes many specialized plotting functions. For example, MATLAB commands semilogx, semilogy, and loglog operate like the plot command but use base-10 logarithmic scales for the horizontal axis, vertical axis, and the horizontal and vertical axes, - -**Figure B.13** Unique roots of *w*100 = −1. - -### 48 CHAPTER B BACKGROUND - -respectively. Monochrome and color images can be displayed by using the image command, and contour plots are easily created with the contour command. Furthermore, a variety of three-dimensional plotting routines are available, such as plot3, contour3, mesh, and surf. Information about these instructions, including examples and related functions, is available from MATLAB help. - -### **[B.7-5 Element-by-Element Operations](#page-6-0)** - -Suppose a new function *h*(*t*) is desired that forces an exponential envelope on the sinusoid *f*(*t*), *h*(*t*) = *f*(*t*)*g*(*t*), where *g*(*t*) = *e*−10*t* . First, row vector *g*(*t*) is created. - ->> g = exp(-10\*t); - -Given MATLAB's vector representation of *g*(*t*) and *f*(*t*), computing *h*(*t*) requires some form of vector multiplication. There are three standard ways to multiply vectors: inner product, outer product, and element-by-element product. As a matrix-oriented language, MATLAB defines the standard multiplication operator \* according to the rules of matrix algebra: the multiplicand must be conformable to the multiplier. A 1 × *N* row vector times an *N* × 1 column vector results in the scalar-valued inner product. An *N* × 1 column vector times a 1 × *M* row vector results in the outer product, which is an *N* × *M* matrix. Matrix algebra prohibits multiplication of two row vectors or multiplication of two column vectors. Thus, the \* operator is not used to perform element-by-element multiplication.† - -Element-by-element operations require vectors to have the same dimensions. An error occurs if element-by-element operations are attempted between row and column vectors. In such cases, one vector must first be transposed to ensure both vector operands have the same dimensions. In MATLAB, most element-by-element operations are preceded by a period. For example, element-by-element multiplication, division, and exponentiation are accomplished using .\*, ./, and .^, respectively. Vector addition and subtraction are intrinsically element-by-element operations and require no period. Intuitively, we know *h*(*t*) should be the same size as both *g*(*t*) and *f*(*t*). Thus, *h*(*t*) is computed using element-by-element multiplication. - -The plot command accommodates multiple curves and also allows modification of line properties. This facilitates side-by-side comparison of different functions, such as *h*(*t*) and *f*(*t*). Line characteristics are specified by using options that follow each vector pair and are enclosed in single quotes. - ->> plot(t,f,'-k',t,h,':k'); >> xlabel('t'); ylabel('Amplitude'); >> legend('f(t)','h(t)'); - -Here, '-k' instructs MATLAB to plot *f*(*t*) using a solid black line, while ':k' instructs MATLAB to use a dotted black line to plot *h*(*t*). A legend and axis labels complete the plot, as shown in - -&gt;> h = f.\*g; - -† While grossly inefficient, element-by-element multiplication can be accomplished by extracting the main diagonal from the outer product of two *N*-length vectors. - -**Figure B.14** Graphical comparison of *f*(*t*) and *h*(*t*). - -Fig. B.14. It is also possible, although more cumbersome, to use pull down menus to modify line properties and to add labels and legends directly in the figure window. - -### **[B.7-6 Matrix Operations](#page-6-0)** - -Many applications require more than row vectors with evenly spaced elements; row vectors, column vectors, and matrices with arbitrary elements are typically needed. - -MATLAB provides several functions to generate common, useful matrices. Given integers m, n, and vector x, the function eye(m) creates the *m*×*m* identity matrix; the function ones(m,n) creates the *m* × *n* matrix of all ones; the function zeros(m,n) creates the *m* × *n* matrix of all zeros; and the function diag(x) uses vector x to create a diagonal matrix. The creation of general matrices and vectors, however, requires each individual element to be specified. - -Vectors and matrices can be input spreadsheet style by using MATLAB's array editor. This graphical approach is rather cumbersome and is not often used. A more direct method is preferable. - -Consider a simple row vector **r**, - -$$ -\mathbf{r} = [1 \ 0 \ 0] -$$ - -The MATLAB notation a:b:c cannot create this row vector. Rather, square brackets are used to create **r**. - ->> r = [1 0 0] r=1 0 0 - -Square brackets enclose elements of the vector, and spaces or commas are used to separate row elements. - -Next, consider the 3×2 matrix **A**, - -$$ -\mathbf{A} = \left[ \begin{array}{cc} 2 & 3 \\ 4 & 5 \\ 0 & 6 \end{array} \right] -$$ - -Matrix **A** can be viewed as a three-high stack of two-element row vectors. With a semicolon to separate rows, square brackets are used to create the matrix. - ->> A = [2 3;4 5;0 6] A=2 3 4 5 0 6 - -Each row vector needs to have the same length to create a sensible matrix. - -In addition to enclosing string arguments, a single quote performs the complex conjugate transpose operation. In this way, row vectors become column vectors and vice versa. For example, a column vector **c** is easily created by transposing row vector **r**. - ->> c = r' c=1 0 0 - -Since vector **r** is real, the complex-conjugate transpose is just the transpose. Had **r** been complex, the simple transpose could have been accomplished by either r.' or (conj(r))'. - -More formally, square brackets are referred to as a concatenation operator. A concatenation combines or connects smaller pieces into a larger whole. Concatenations can involve simple numbers, such as the six-element concatenation used to create the 3×2 matrix **A**. It is also possible to concatenate larger objects, such as vectors and matrices. For example, vector **c** and matrix **A** can be concatenated to form a 3×3 matrix **B**. - ->> B = [c A] B=1 2 3 045 006 - -Errors will occur if the component dimensions do not sensibly match; a 2×2 matrix would not be concatenated with a 3×3 matrix, for example. - -Elements of a matrix are indexed much like vectors, except two indices are typically used to specify row and column.† Element (1, 2) of matrix **B**, for example, is 2. - ->> B(1,2) ans = 2 - -Indices can likewise be vectors. For example, vector indices allow us to extract the elements common to the first two rows and last two columns of matrix **B**. - -``` ->> B(1:2,2:3) - ans = 2 3 - 4 5 -``` - -† Matrix elements can also be accessed by means of a single index, which enumerates along columns. Formally, the element from row *m* and column *n* of an *M* × *N* matrix may be obtained with a single index (*n*−1)*M* +*m*. For example, element (1, 2) of matrix **B** is accessed by using the index (2−1)3+1 = 4. That is, B(4) yields 2. - -One indexing technique is particularly useful and deserves special attention. A colon can be used to specify all elements along a specified dimension. For example, B(2,:) selects all column elements along the second row of **B**. - ->> B(2,:) ans = 0 4 5 - -Now that we understand basic vector and matrix creation, we turn our attention to using these tools on real problems. Consider solving a set of three linear simultaneous equations in three unknowns. - -$$ -x_1 - 2x_2 + 3x_3 = 1 -$$ -$$ --\sqrt{3}x_1 + x_2 - \sqrt{5}x_3 = \pi -$$ -$$ -3x_1 - \sqrt{7}x_2 + x_3 = e -$$ - -This system of equations is represented in matrix form according to **Ax** = **y**, where - -$$ -\mathbf{A} = \begin{bmatrix} 1 & -2 & 3 \\ -\sqrt{3} & 1 & -\sqrt{5} \\ 3 & -\sqrt{7} & 1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \text{ and } \mathbf{y} = \begin{bmatrix} 1 \\ \pi \\ e \end{bmatrix} -$$ - -Although Cramer's rule can be used to solve **Ax** = **y**, it is more convenient to solve by multiplying both sides by the matrix inverse of **A**. That is, **x** = **A**−1**Ax** = **A**−1**y**. Solving for **x** by hand or by calculator would be tedious at best, so MATLAB is used. We first create **A** and **y**. - ->> A = [1 -2 3;-sqrt(3) 1 -sqrt(5);3 -sqrt(7) 1]; y = [1;pi;exp(1)]; - -The vector solution is found by using MATLAB's inv function. - ->> -$$ -x = inv(A)*y -$$ - - $x = -1.9999$ - $-3.8998$ - $-1.5999$ - -It is also possible to use MATLAB's left divide operator x=A\y to find the same solution. The left divide is generally more computationally efficient than matrix inverses. As with matrix multiplication, left division requires that the two arguments be conformable. - -Of course, Cramer's rule can be used to compute individual solutions, such as *x*1, by using vector indexing, concatenation, and MATLAB's det command to compute determinants. - ->> x1 = det([y,A(:,2:3)])/det(A) x1 = -1.9999 - -Another nice application of matrices is the simultaneous creation of a family of curves. Consider *h*α(*t*) = *e*−α*t*sin(2π10*t* + π/6) over 0 ≤ *t* ≤ 0.2. Figure B.14 shows *h*α(*t*) for α = 0 and α = 10. Let's investigate the family of curves *h*α(*t*) for α = [0, 1,..., 10]. - -An inefficient way to solve this problem is create *h*α(*t*) for each α of interest. This requires 11 individual cases. Instead, a matrix approach allows all 11 curves to be computed simultaneously. First, a vector is created that contains the desired values of α. - ->> alpha = (0:10); - -By using a sampling interval of one millisecond, *t* = 0.001, a time vector is also created. - ->> t = (0:0.001:0.2)'; - -The result is a length-201 column vector. By replicating the time vector for each of the 11 curves required, a time matrix T is created. This replication can be accomplished by using an outer product between t and a 1×11 vector of ones.† - ->> T = t\*ones(1,11); - -The result is a 201 × 11 matrix that has identical columns. Right multiplying T by a diagonal matrix created from α, columns of T can be individually scaled and the final result is computed. - ->> H = exp(-T\*diag(alpha)).\*sin(2\*pi\*10\*T+pi/6); - -Here, H is a 201 × 11 matrix, where each column corresponds to a different value of α. That is, **H** = [**h**0,**h**1,...,**h**10], where **h**α are column vectors. As shown in Fig. B.15, the 11 desired curves are simultaneously displayed by using MATLAB's plot command, which allows matrix arguments. - ->> plot(t,H); xlabel('t'); ylabel('h(t)'); - -This example illustrates an important technique called vectorization, which increases execution efficiency for interpretive languages such as MATLAB. Algorithm vectorization uses matrix and - -**Figure B.15** *h*α(*t*) for α = [0, 1,..., 10]. - -† The repmat command provides a more flexible method to replicate or tile objects. Equivalently, T = repmat(t,1,11). - -vector operations to avoid manual repetition and loop structures. It takes practice and effort to become proficient at vectorization, but the worthwhile result is efficient, compact code.† - -### **[B.7-7 Partial Fraction Expansions](#page-6-0)** - -There are a wide variety of techniques and shortcuts to compute the partial fraction expansion of rational function *F*(*x*) = *B*(*x*)/*A*(*x*), but few are more simple than the MATLAB residue command. The basic form of this command is - ->> [R,P,K] = residue(B,A) - -The two input vectors B and A specify the polynomial coefficients of the numerator and denominator, respectively. These vectors are ordered in descending powers of the independent variable. Three vectors are output. The vector R contains the coefficients of each partial fraction, and vector P contains the corresponding roots of each partial fraction. For a root repeated *r* times, the *r* partial fractions are ordered in ascending powers. When the rational function is not proper, the vector K contains the direct terms, which are ordered in descending powers of the independent variable. - -To demonstrate the power of the residue command, consider finding the partial fraction expansion of - -$$ -F(x) = \frac{x^5 + \pi}{(x + \sqrt{2})(x - \sqrt{2})^3} = \frac{x^5 + \pi}{x^4 - \sqrt{8x^3 + \sqrt{32x - 4}}} -$$ - -By hand, the partial fraction expansion of *F*(*x*) is difficult to compute. MATLAB, however, makes short work of the expansion. - ->> [R,P,K] = residue([10000 pi],[1 -sqrt(8) 0 sqrt(32) -4]); R.', P.', K R = 7.8888 5.9713 3.1107 0.1112 P = 1.4142 1.4142 1.4142 -1.4142 K = 1.0000 2.8284 - -Written in standard form, the partial fraction expansion of *F*(*x*) is - -$$ -F(x) = x + 2.8284 + \frac{7.8888}{x - \sqrt{2}} + \frac{5.9713}{(x - \sqrt{2})^2} + \frac{3.1107}{(x - \sqrt{2})^3} + \frac{0.1112}{x + \sqrt{2}} -$$ - -The signal–processing toolbox function residuez is similar to the residue command and offers more convenient expansion of certain rational functions, such as those commonly encountered in the study of discrete-time systems. Additional information about the residue and residuez commands is available from MATLAB's help facilities. - -† The benefits of vectorization are less pronounced in recent versions of MATLAB. - -## **[B.8 APPENDIX: USEFUL](#page-6-0) MATHEMATICAL FORMULAS** - -We conclude this chapter with a selection of useful mathematical facts. - -### **[B.8-1 Some Useful Constants](#page-6-0)** - -π ≈ 3.1415926535 *e* ≈ 2.7182818284 1 *e* ≈ 0.3678794411 log10 2 ≈ 0.30103 log10 3 ≈ 0.47712 - -### **[B.8-2 Complex Numbers](#page-7-0)** - -$$ -e^{\pm j\pi/2} = \pm j -$$ - -\n -$$ -e^{\pm j n\pi} = \begin{cases} 1 & n \text{ even} \\ -1 & n \text{ odd} \end{cases} -$$ - -\n -$$ -e^{\pm j\theta} = \cos \theta \pm j \sin \theta -$$ - -\n -$$ -a + jb = re^{j\theta} \qquad r = \sqrt{a^2 + b^2}, \theta = \tan^{-1} \left(\frac{b}{a}\right) -$$ - -\n -$$ -(re^{j\theta})^k = r^k e^{jk\theta} -$$ - -\n -$$ -(r_1 e^{j\theta_1})(r_2 e^{j\theta_2}) = r_1 r_2 e^{j(\theta_1 + \theta_2)} -$$ - -### **[B.8-3 Sums](#page-7-0)** - -$$ -\sum_{k=m}^{n} r^{k} = \frac{r^{n+1} - r^{m}}{r - 1} \qquad r \neq 1 -$$ -\n -$$ -\sum_{k=0}^{n} k = \frac{n(n+1)}{2} -$$ -\n -$$ -\sum_{k=0}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6} -$$ -\n -$$ -\sum_{k=0}^{n} kr^{k} = \frac{r + [n(r-1) - 1]r^{n+1}}{(r-1)^{2}} \qquad r \neq 1 -$$ -\n -$$ -\sum_{k=0}^{n} k^{2}r^{k} = \frac{r[(1+r)(1-r^{n}) - 2n(1-r)r^{n} - n^{2}(1-r)^{2}r^{n}]}{(1-r)^{3}} \qquad r \neq 1 -$$ - -### **[B.8-4 Taylor and Maclaurin Series](#page-7-0)** - -$$ -f(x) = f(a) + \frac{(x-a)}{1!}\dot{f}(a) + \frac{(x-a)^2}{2!}\ddot{f}(a) + \dots = \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!}f^{(k)}(a) -$$ -$$ -f(x) = f(0) + \frac{x}{1!}\dot{f}(0) + \frac{x^2}{2!}\ddot{f}(0) + \dots = \sum_{k=0}^{\infty} \frac{x^k}{k!}f^{(k)}(0) -$$ - -### **[B.8-5 Power Series](#page-7-0)** - -$$ -e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots + \frac{x^{n}}{n!} + \dots -$$ -\n -$$ -\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots -$$ -\n -$$ -\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - \dots -$$ -\n -$$ -\tan x = x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \frac{17x^{7}}{315} + \dots \qquad x^{2} < \pi^{2}/4 -$$ -\n -$$ -\tanh x = x - \frac{x^{3}}{3} + \frac{2x^{5}}{15} - \frac{17x^{7}}{315} + \dots \qquad x^{2} < \pi^{2}/4 -$$ -\n -$$ -(1 + x)^{n} = 1 + nx + \frac{n(n-1)}{2!}x^{2} + \frac{n(n-1)(n-2)}{3!}x^{3} + \dots + \binom{n}{k}x^{k} + \dots + x^{n} -$$ -\n -$$ -(1 + x)^{n} \approx 1 + nx \qquad |x| \ll 1 -$$ -\n -$$ -\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \dots \qquad |x| < 1 -$$ - -### **[B.8-6 Trigonometric Identities](#page-7-0)** - -$$ -e^{\pm jx} = \cos x \pm j \sin x -$$ - -\n -$$ -\cos x = \frac{1}{2} [e^{jx} + e^{-jx}] -$$ - -\n -$$ -\sin x = \frac{1}{2j} [e^{jx} - e^{-jx}] -$$ - -\n -$$ -\cos (x \pm \frac{\pi}{2}) = \pm \sin x -$$ - -\n -$$ -\sin (x \pm \frac{\pi}{2}) = \pm \cos x -$$ - -\n -$$ -2 \sin x \cos x = \sin 2x -$$ - -\n -$$ -\sin^2 x + \cos^2 x = 1 -$$ - -\n -$$ -\cos^2 x - \sin^2 x = \cos 2x -$$ - -\n -$$ -\cos^2 x = \frac{1}{2} (1 + \cos 2x) -$$ - -\n -$$ -\sin^2 x = \frac{1}{2} (1 - \cos 2x) -$$ - - -$$ -\cos^3 x = \frac{1}{4} (3 \cos x + \cos 3x) -$$ - -\n -$$ -\sin^3 x = \frac{1}{4} (3 \sin x - \sin 3x) -$$ - -\n -$$ -\sin (x \pm y) = \sin x \cos y \pm \cos x \sin y -$$ - -\n -$$ -\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y -$$ - -\n -$$ -\tan (x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} -$$ - -\n -$$ -\sin x \sin y = \frac{1}{2} [\cos (x - y) - \cos (x + y)] -$$ - -\n -$$ -\cos x \cos y = \frac{1}{2} [\cos (x - y) + \cos (x + y)] -$$ - -\n -$$ -\sin x \cos y = \frac{1}{2} [\sin (x - y) + \sin (x + y)] -$$ - -\n -$$ -\sin x \cos x + b \sin x = C \cos (x + \theta) \qquad C = \sqrt{a^2 + b^2}, \theta = \tan^{-1} (\frac{-b}{a}) -$$ - -**[B.8-7 Common Derivative Formulas](#page-7-0)** - -$$ -\frac{d}{dx}f(u) = \frac{d}{du}f(u)\frac{du}{dx} -$$ -\n -$$ -\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} -$$ -\n -$$ -\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} -$$ -\n -$$ -\frac{dx^n}{dx} = nx^{n-1} -$$ -\n -$$ -\frac{d}{dx}\ln(ax) = \frac{1}{x} -$$ -\n -$$ -\frac{d}{dx}\log(ax) = \frac{\log e}{x} -$$ -\n -$$ -\frac{d}{dx}e^{bx} = be^{bx} -$$ -\n -$$ -\frac{d}{dx}a^{bx} = b(\ln a)a^{bx} -$$ -\n -$$ -\frac{d}{dx}\sin ax = a\cos ax -$$ -\n -$$ -\frac{d}{dx}\cos ax = -a\sin ax -$$ -\n -$$ -\frac{d}{dx}\tan ax = \frac{a}{\cos^2 ax} -$$ -\n -$$ -\frac{d}{dx}(\sin^{-1}ax) = \frac{a}{\sqrt{1 - a^2x^2}} -$$ -\n -$$ -\frac{d}{dx}(\cos^{-1}ax) = \frac{-a}{\sqrt{1 - a^2x^2}} -$$ -\n -$$ -\frac{d}{dx}(\tan^{-1}ax) = \frac{a}{1 + a^2x^2} -$$ - -### **[B.8-8 Indefinite Integrals](#page-7-0)** - -$$ -\int u dv = uv - \int v du -$$ - -\n -$$ -\int f(x)\dot{g}(x) dx = f(x)g(x) - \int f(x)g(x) dx -$$ - -\n -$$ -\int \sin ax dx = -\frac{1}{a} \cos ax \qquad \int \cos ax dx = \frac{1}{a} \sin ax -$$ - -\n -$$ -\int \sin^2 ax dx = \frac{x}{2} - \frac{\sin 2ax}{4a} \qquad \int \cos^2 ax dx = \frac{x}{2} + \frac{\sin 2ax}{4a} -$$ - -\n -$$ -\int x \sin ax dx = \frac{1}{a^2} (\sin ax - ax \cos ax) -$$ - -\n -$$ -\int x \cos ax dx = \frac{1}{a^2} (\cos ax + ax \sin ax) -$$ - -\n -$$ -\int x^2 \sin ax dx = \frac{1}{a^3} (2ax \sin ax + 2 \cos ax - a^2x^2 \cos ax) -$$ - -\n -$$ -\int x^2 \cos ax dx = \frac{1}{a^3} (2ax \cos ax - 2 \sin ax + a^2x^2 \sin ax) -$$ - -\n -$$ -\int \sin ax \sin bx dx = \frac{\sin (a - b)x}{2(a - b)} - \frac{\sin (a + b)x}{2(a + b)} \qquad a^2 \neq b^2 -$$ - -\n -$$ -\int \sin ax \cos bx dx = -\left[ \frac{\cos (a - b)x}{2(a - b)} + \frac{\cos (a + b)x}{2(a + b)} \right] \qquad a^2 \neq b^2 -$$ - -\n -$$ -\int \cos ax \cos bx dx = \frac{\sin (a - b)x}{2(a - b)} + \frac{\sin (a + b)x}{2(a + b)} \qquad a^2 \neq b^2 -$$ - -\n -$$ -\int e^{ax} dx = \frac{e^{ax}}{a^2} (ax - 1) -$$ - -\n -$$ -\int x^2 e^{ax} dx = \frac{e^{ax}}{a^2} (a^2x^2 - 2ax + 2) -$$ - -\n -$$ -\int e^{ax} \sin bx dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) -$$ - -\n -$$ -\int e^{ax} \cos bx dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) -$$ - -\n -$$ -\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \ln(x -$$ - -### **[B.8-9 L'Hôpital's Rule](#page-7-0)** - -If lim *f*(*x*)/*g*(*x*) results in the indeterministic form 0/0 or ∞/∞, then - -$$ -\lim \frac{f(x)}{g(x)} = \lim \frac{\dot{f}(x)}{\dot{g}(x)} -$$ - -### **[B.8-10 Solution of Quadratic and Cubic Equations](#page-7-0)** - -Any *quadratic* equation can be reduced to the form - -$$ -ax^2 + bx + c = 0 -$$ - -The solution of this equation is provided by - -$$ -x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} -$$ - -A general *cubic* equation - -$$ -y^3 + py^2 + qy + r = 0 -$$ - -may be reduced to the *depressed cubic* form - -$$ -x^3 + ax + b = 0 -$$ - -by substituting - -$$ -y = x - \frac{p}{3} -$$ - -This yields - -$$ -a = \frac{1}{3}(3q - p^2) \qquad b = \frac{1}{27}(2p^3 - 9pq + 27r) -$$ - -Now let - -$$ -A = \sqrt[3]{-\frac{b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}} \qquad B = \sqrt[3]{-\frac{b}{2} - \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}} -$$ - -The solution of the depressed cubic is - -$$ -x = A + B -$$ -, $x = -\frac{A+B}{2} + \frac{A-B}{2}\sqrt{-3}$ , $x = -\frac{A+B}{2} - \frac{A-B}{2}\sqrt{-3}$ - -and - -$$ -y = x - \frac{p}{3} -$$ - -### **[REFERENCES](#page-7-0)** - -- 1. Asimov, Isaac. *Asimov on Numbers.* Bell Publishing, New York, 1982. -- 2. Calinger, R., ed. *Classics of Mathematics.* Moore Publishing, Oak Park, IL, 1982. -- 3. Hogben, Lancelot. *Mathematics in the Making.* Doubleday, New York, 1960. - -- 4. Cajori, Florian. *A History of Mathematics,* 4th ed. Chelsea, New York, 1985. -- 5. Encyclopaedia Britannica. *Micropaedia* IV, 15th ed., vol. 11, p. 1043. Chicago, 1982. -- 6. Singh, Jagjit. *Great Ideas of Modern Mathematics.* Dover, New York, 1959. -- 7. Dunham, William. *Journey Through Genius.* Wiley, New York, 1990. - -## **[PROBLEMS](#page-7-0)** - -- **B.1-1** Given a complex number *w* = *x* + *jy*, the complex conjugate of *w* is defined in rectangular coordinates as *w*∗ =*x*−*jy*. Use this fact to derive complex conjugation in polar form. -- **B.1-2** Express the following numbers in polar form: - - (a) *w*a = 1+*j* - - (b) *w*b = 1+*ej* - - (c) *w*c = −4+*j*3 - - (d) *w*d = (1+*j*)(−4+*j*3) - - (e) *w*e = *ej*π/4 +2*e*−*j*π/4 - -(f) -$$ -w_f = \frac{1+j}{2j} -$$ - -(g) -$$ -w_g = (1+j)/(-4+j3) -$$ - -(h) -$$ -w_h = \frac{1-j}{\sin(j)} -$$ - -- **B.1-3** Express the following numbers in Cartesian (rectangular) form: - - (a) *w*a = *j*+*ej* - - (b) *w*b = 3*ej*π/4 - - (c) *w*c = 1/*ej* - - (d) *w*d = (1+*j*)(−4+*j*3) - -(e) -$$ -w_e = e^{j\pi/4} + 2e^{-j\pi/4} -$$ - -- (f) *w*f = *ej* +1 -- (g) *w*g = 1/2*j* -- (h) *w*h = *j j j* (*j* raised to the *j* raised to the *j*) -- **B.1-4** Showing all work and simplifying your answer, determine the real part of the following numbers: - - (a) *w*a = 1 *j* (*j*−5*e*2−3*j* ) - - (b) *w*b = (1+*j*)ln(1+*j*) -- **B.1-5** Showing all work and simplifying your answer, determine the imaginary part of the following numbers: - - (a) *w*a = −*jej*π/4 - -(b) -$$ -w_b = 1 - 2je^{2-4j} -$$ - -- (c) *w*c = tan(*j*) -- **B.1-6** For complex constant *w*, prove: - - (a) Re(*w*) = (*w* +*w*∗)/2 - - (b) Im(*w*) = (*w* −*w*∗)/2*j* - -- **B.1-7** Given *w* = *x* −*jy*, determine: (a) Re(*ew*) - - (b) Im(*ew*) -- **B.1-8** For arbitrary complex constants *w*1 and *w*2, prove or disprove the following: - - (a) Re(*jw*1) = −Im(*w*1) - - (b) Im(*jw*1) = Re(*w*1) - - (c) Re(*w*1)+Re(*w*2) = Re(*w*1 +*w*2) - - (d) Im(*w*1)+Im(*w*2) = Im(*w*1 +*w*2) - - (e) Re(*w*1)Re(*w*2) = Re(*w*1*w*2) - - (f) Im(*w*1)/Im(*w*2) = Im(*w*1/*w*2) -- **B.1-9** Given *w*1 = 3+*j*4 and *w*2 = 2*ej*π/4. - - (a) Express *w*1 in standard polar form. - - (b) Express *w*2 in standard rectangular form. - - (c) Determine |*w*1| 2 and |*w*2| 2. - - (d) Express *w*1 + *w*2 in standard rectangular form. - - (e) Express *w*1 −*w*2 in standard polar form. - - (f) Express *w*1*w*2 in standard rectangular form. - - (g) Express *w*1/*w*2 in standard polar form. -- **B.1-10** Repeat Prob. B.1-9 using *w*1 = (3 + *j*4)2 and *w*2 = 2.5*je*−*j*40π . -- **B.1-11** Repeat Prob. B.1-9 using *w*1 = *j* + *e*π/4 and *w*2 = cos(*j*). -- **B.1-12** Using the complex plane: - - (a) Evaluate and locate the distinct solutions to (*w*) 4 = −1. - - (b) Evaluate and locate the distinct solutions to (*w* −(1+*j*2))5 = (32/ √2)(1+*j*). - - (c) Sketch the solution to |*w* −2*j*| = 3. - - (d) Graph *w*(*t*) = (1+*t*)*ejt* for (−10 ≤ *t* ≤ 10). -- **B.1-13** The distinct solutions to (*w* −*w*1) *n* = *w*2 lie on a circle in the complex plane, as shown in Fig. PB.1-13. One solution is located on the real axis at √3 + 1 = 2.732, and one solution is located on the imaginary axis at √3−1=0.732. Determine *w*1, *w*2, and *n*. - -- **B.1-14** Find the distinct solutions to each of the following. Use MATLAB to graph each solution set in the complex plane. - - (a) *w*3 = − 8 27 - - (b) (*w* +1)8 = 1 - - (c) *w*2 +*j* = 0 - - (d) 16(*w* −1)4 +81 = 0 - - (e) (*w* +2*j*)3 = −8 - - (f) (*j*−*w*)1.5 = 2+*j*2 - - (g) (*w* −1)2.5 = *j*4 √ 2 -- **B.1-15** If *j* = √−1, what is √*j*? -- **B.1-16** Find all the values of ln(−*e*), expressing your answer in Cartesian form. -- **B.1-17** Determine all values of log10(−1), expressing your answer in Cartesian form. Notice that the logarithm has base 10, not *e*. -- **B.1-18** Express the following in standard rectangular coordinates: - - (a) *w*a = ln(1/(1+*j*)) - - (b) *w*b = cos(1+*j*) - - (c) *w*c = (1−*j*)*j* -- **B.1-19** By constraining *w* to be purely imaginary, show that the equation cos(*w*) = 2 can be represented as a standard quadratic equation. Solve this equation for *w*. -- **B.1-20** Certain integrals, although expressed in relatively simple form, are quite difficult to solve. For example, \$ *e*−*x*2 *dx* cannot be evaluated in terms of elementary functions; most calculators that perform integration cannot handle this indefinite integral. Fortunately, you are smarter than most calculators. - -- (a) Express *e*−*x*2 using a Taylor series expansion. -- (b) Using your series expansion for *e*−*x*2 , determine \$ *e*−*x*2 *dx*. -- (c) Using a suitably truncated series, evaluate the definite integral \$ 1 0 *e*−*x*2 *dx*. -- **B.1-21** Repeat Prob. B.1-20 for \$ *e*−*x*3 *dx*. -- **B.1-22** Repeat Prob. B.1-20 for \$ cos *x*2 *dx*. -- **B.1-23** For each function, determine a suitable series expansion. - - (a) *f*a(*x*) = (2−*x*2)−1 - - (b) *f*b(*x*) = (0.5)*x* -- **B.1-24** Consider the function *f*(*x*) = 1+*x*+*x*2+*x*3. - - (a) Express *f*(*x*) using a Taylor series with expansion point of *a* = 1. Explicitly write out every term. [*Hint:* See Sec. B.8-4.] - - (b) Describe a good reason why you might want to express a function that is already a simple polynomial using such a series. -- **B.1-25** Determine the Maclaurin series expansion of each of the following. [*Hint:* See Sec. B.8-4.] (a) *f*a(*x*) = 2*x* - - (b) *f*b(*x*) = 1 3 *x* -- **B.2-1** Determine the fundamental period *T*0, frequency *f*0, and radian frequency ω0 for the following sinusoids: - - (a) cos(5π*t* +3) - - (b) 7 sin 2*t*−π 3 -- **B.2-2** Determine an expression for a sinusoid that oscillates 15 times per second, that has a value of -1 at *t* = 0, and whose peak amplitude is 3. Use MATLAB to plot the signal over 0 ≤ *t* ≤ 1. -- **B.2-3** Let *x*1(*t*) = 2 cos(3*t* + 1) and *x*2(*t*) = −3 cos (3*t* −2). - - (a) Determine *a*1 and *b*1 so that *x*1(*t*) = *a*1 cos(3*t*)+*b*1 sin(3*t*). - - (b) Determine *a*2 and *b*2 so that *x*2(*t*) = *a*2 cos(3*t*)+*b*2 sin(3*t*). - - (c) Determine *C* and θ so that *x*1(*t*) + *x*2(*t*) = *C*cos(3*t* +θ ). -- **B.2-4** In addition to the traditional sine and cosine functions, there are the *hyperbolic* sine and cosine functions, which are defined by sinh(*w*) = (*ew* −*e*−*w*)/2 and cosh(*w*) = (*ew* +*e*−*w*)/2. In general, the argument is a complex constant *w* = *x* +*jy*. - -- (a) Show that cosh(*w*) = cosh(*x*) cos(*y*) + *j*sinh(*x*)sin(*y*). -- (b) Determine a similar expression for sinh(*w*) in rectangular form that only uses functions of real arguments, such as sin(*x*), cosh(*y*), and so on. -- **B.2-5** Use Euler's identity to solve or prove the following: - - (a) Find real, positive constants *c* and φ for all real *t* such that 2.5 cos(3*t*) − 1.5 sin(3*t* + π/3) = *c* cos(3*t* + φ). Sketch the resulting sinusoid. - - (b) Prove that cos(θ ± φ) = cos(θ ) cos(φ) ∓ sin(θ )sin(φ). - - (c) Given real constants *a*, *b*, and α, complex constant *w*, and the fact that - -$$ -\int_a^b e^{wx} dx = \frac{1}{w} (e^{wb} - e^{wa}) -$$ - -evaluate the integral - -$$ -\int_a^b e^{wx} \sin(\alpha x) dx -$$ - -- **B.2-6** A particularly boring stretch of interstate highway has a posted speed limit of 70 mph. A highway engineer wants to install "rumble bars" (raised ridges on the side of the road) so that cars traveling the speed limit will produce quarter-second bursts of 1 kHz sound every second, a strategy that is particularly effective at startling sleepy drivers awake. Provide design specifications for the engineer. -- **B.3-1** By hand, accurately sketch the following signals over (0 ≤ *t* ≤ 1): - - (a) *x*a(*t*) = *e*−*t* - - (b) *x*b(*t*) = sin(2π5*t*) - -(c) -$$ -x_c(t) = e^{-t} \sin(2\pi 5t) -$$ - -- **B.3-2** In 1950, the human population was approximately 2.5 billion people. Assuming a doubling time of 40 years, formulate an exponential model for human population in the form *p*(*t*) = *aebt*, where *t* is measured in years. Sketch *p*(*t*) over the interval 1950 ≤ *t* ≤ 2100. According to this model, in what year can we expect the population to reach the estimated 15 billion carrying capacity of the earth? -- **B.3-3** Determine an expression for an exponentially decaying sinusoid that oscillates three - -times per second and whose amplitude envelope decreases by 50% every 2 seconds. Use MATLAB to plot the signal over −2 ≤ *t* ≤ 2. - -**B.3-4** By hand, sketch the following against independent variable *t*: - -(a) -$$ -x_a(t) = \text{Re}\left(2e^{(-1+j2\pi)t}\right) -$$ - -(b) -$$ -x_b(t) = \text{Im}\left(3 - e^{(1-j2\pi)t}\right) -$$ - -(c) $x_c(t) = 3 - \text{Im}\left(e^{(1-j2\pi)t}\right)$ - -**B.4-1** Consider the following system of equations: - -$$ -\left[\begin{array}{cc} -1 & 2 \\ 3 & -4 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{c} 3 \\ -1 \end{array}\right] -$$ - -Expressing all answers in rational form (ratio of integers), use Cramer's rule to determine *x*1 and *x*2. Perform all calculations by hand, including matrix determinants. - -**B.4-2** Consider the following system of equations: - -| ⎡
1 | 2 | ⎤
0 | ⎡
⎤
x1 | | ⎡
⎤
7 | -|--------|---|--------|--------------|-----|-------------| -| 0
⎣ | 3 | 4
⎦ | x2
⎣ | ⎦ = | 8
⎣
⎦ | -| 5 | 0 | 6 | x3 | | 9 | - -Expressing all answers in rational form (ratio of integers), use Cramer's rule to determine *x*1, *x*2, and *x*3. Perform all calculations by hand, including matrix determinants. - -**B.4-3** Consider the following system of equations. - -$$ -x_1 + x_2 + x_3 = 1 -$$ - -$$ -x_1 + 2x_2 + 3x_3 = 3 -$$ - -$$ -x_1 - x_2 = -3 -$$ - -Use Cramer's rule to determine *x*1, *x*2, and *x*3. Matrix determinants can be computed by using MATLAB's det command. - -**B.5-1** Determine the constants *a*0, *a*1, and *a*2 of the partial fraction expansion - -$$ -F(s) = \frac{s}{(s+1)^3} -$$ - -= $\frac{a_0}{(s+1)^3} + \frac{a_1}{(s+1)^2} + \frac{a_2}{(s+1)}$ - -- **B.5-2** Compute by hand the partial fraction expansions of the following rational functions: - - (a) *H*a(*s*) = *s*2+5*s*+6 *s*3+*s*2+*s*+1 , which has denominator poles at *s* = ±*j* and *s* = −1 - -(b) -$$ -H_b(s) = \frac{1}{H_1(s)} = \frac{s^3 + s^2 + s + 1}{s^2 + 5s + 6} -$$ - -(c) -$$ -H_c(s) = \frac{1}{(s+1)^2(s^2+1)} -$$ - -(d) $H_d(s) = \frac{s^2+5s+6}{3s^2+2s+1}$ - -- **B.5-3** Compute by hand the partial fraction expansions of the following rational functions: (a) *F*a(*x*) = (*x*−1)(*x*−2) (*x*−3)2 (b) *F*b(*x*) = (*x*−1)2 (3*x*−1)(2*x*−1) (c) *F*c(*x*) = (*x*−1)2 (3*x*−1)2(2*x*−1) (d) *F*d(*x*) = *x*2−5*x*+6 2*x*2+8*x*+6 (e) *F*e(*x*) = 2*x*2−3*x*−11 *x*2−*x*−2 (f) *F*f(*x*) = 3+2*x*2 −3+2*x*+*x*2 - - (g) *F*g(*x*) = *x*3+2*x*2+3*x*+4 *x*2+1 (h) *F*h(*x*) = 1+2*x*+3*x*2 *x*2+5*x*+6 (i) *F*i(*x*) = 3*x*3−*x*2+14*x*+4 *x*2+4 (j) *F*j(*x*) = 2*x*−1−1+2*x x*−5+6*x*−1 (k) *F*k(*x*) = 3 −5*x*2−9*x*+23 -- *x*2+*x*−2 **B.6-1** A system of equations in terms of unknowns *x*1 and *x*2 and arbitrary constants *a*, *b*, *c*, *d*, *e*, and *f* is given by - -$$ -ax_1 + bx_2 = c -$$ - -$$ -dx_1 + ex_2 = f -$$ - -- (a) Represent this system of equations in matrix form. -- (b) Identify specific constants *a*, *b*, *c*, *d*, *e*, and *f* such that *x*1 = 3 and *x*2 = −2. Are the constants you selected unique? -- (c) Identify nonzero constants *a*, *b*, *c*, *d*, *e*, and *f* such that no solutions *x*1 and *x*2 exist. -- (d) Identify nonzero constants *a*, *b*, *c*, *d*, *e*, and *f* such that an infinite number of solutions *x*1 and *x*2 exist. -- **B.6-2** Using a matrix approach, solve the following system of equations: - -$$ -x_1 + x_2 + x_3 + x_4 = 4 -$$ - -\n -$$ -x_1 + x_2 + x_3 - x_4 = 2 -$$ - -\n -$$ -x_1 + x_2 - x_3 - x_4 = 0 -$$ - -\n -$$ -x_1 - x_2 - x_3 - x_4 = -2 -$$ - -**B.6-3** Using a matrix approach, solve the following system of equations: - -$$ -x_1 + x_2 + x_3 + x_4 = 1 -$$ -$$ -x_1 - 2x_2 + 3x_3 = 2 -$$ -$$ -x_1 - x_3 + 7x_4 = 3 -$$ -$$ --2x_2 + 3x_3 - 4x_4 = 4 -$$ - -**B.6-4** A signal *f*(*t*) = *a*cos(3*t*) + *b*sin(3*t*) reaches a peak amplitude of 5 at *t* = 1.8799 and has a zero crossing at *t* = 0.3091. Use a matrix-based approach to determine the constants *a* and *b*. - -### **B.6-5** Define - -$$ -\mathbf{x} = \begin{bmatrix} 1 & 3 \\ -2 & 4 \end{bmatrix}, \quad \mathbf{y} = \begin{bmatrix} -5 \\ 2 \end{bmatrix}, -$$ - -and -$$ -\mathbf{z} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} -$$ - -By hand, calculate the following: - -- (a) **f**a = **y***T* **y** (b) **f**b = **yy***T* (c) **f**c = **xy** (d) **f**d = **x***T* **y** (e) **f**e = **y***T* **x** (f) **f**f = **xz** (g) **f**g = **zxz** (h) **f**h = **x***T* −**z** -- **B.7-1** Use MATLAB to produce the plots requested in Prob. B.3-4. -- **B.7-2** Use MATLAB to plot the function *x*(*t*) = *t*sin(2π*t*) over 0 ≤ *t* ≤ 10 using 501 equally spaced points. What is the maximum value of *x*(*t*) over this range of *t*? -- **B.7-3** Use MATLAB to plot *x*(*t*) = cos(*t*)sin(20*t*) over a suitable range of *t*. -- **B.7-4** Use MATLAB to plot *x*(*t*) = %10 *k*=1 cos(2π*kt*) over a suitable range of *t*. The MATLAB command sum may prove useful. -- **B.7-5** When a bell is struck with a mallet, it produces a ringing sound. Write an equation that approximates the sound produced by a small, light bell. Carefully identify your assumptions. How does your equation change if the bell is large and heavy? You can assess the quality of your models by using the MATLAB sound command to listen to your "bell." - -- **B.7-6** You are working on a digital quadrature amplitude modulation (QAM) communication receiver. The QAM receiver requires a pair of quadrature signals: cos*n* and sin*n*. These can be simultaneously generated by following a simple procedure: (1) choose a point *w* on the unit circle, (2) multiply *w* by itself and store the result, (3) multiply *w* by the last result and store, and (4) repeat step 3. - - (a) Show that this method can generate the desired pair of quadrature sinusoids. - - (b) Determine a suitable value of *w* so that good-quality, periodic, 2π × 100,000 rad/s signals can be generated. How much time is available for the processing unit to compute each sample? - - (c) Simulate this procedure by using MATLAB and report your results. - - (d) Identify as many assumptions and limitations to this technique as possible. For example, can your system operate correctly for an indefinite period of time? -- **B.7-7** Using MATLAB's residue command, - - (a) Verify the results of Prob. B.5-2a. - - (b) Verify the results of Prob. B.5-2b. - - (c) Verify the results of Prob. B.5-2c. - - (d) Verify the results of Prob. B.5-2d. -- **B.7-8** Using MATLAB's residue command, - - (a) Verify the results of Prob. B.5-3a. - - (b) Verify the results of Prob. B.5-3b. - - (c) Verify the results of Prob. B.5-3c. - - (d) Verify the results of Prob. B.5-3d. - - (e) Verify the results of Prob. B.5-3e. - - (f) Verify the results of Prob. B.5-3f. - - (g) Verify the results of Prob. B.5-3g. - - (h) Verify the results of Prob. B.5-3h. - - (i) Verify the results of Prob. B.5-3i. - - (j) Verify the results of Prob. B.5-3j. - - (k) Verify the results of Prob. B.5-3k. -- **B.7-9** Determine the original length-3 vectors a and b need to produce the MATLAB output: - - >> [r,p,k] = residue(b,a) r = 0 + 2.0000i 0 - 2.0000i - -$$ -p = 3 -$$ - --3 -$$ -k = 0 + 1.0000i -$$ - -**B.7-10** Let *N* = [*n*7,*n*6,*n*5,...,*n*2,*n*1] represent the seven digits of your phone number. Construct a rational function according to - -$$ -H_N(s) = \frac{n_7s^2 + n_6s + n_5 + n_4s^{-1}}{n_3s^2 + n_2s + n_1} -$$ - -Use MATLAB's residue command to compute the partial fraction expansion of *HN*(*s*). - -- **B.7-11** When plotted in the complex plane for −π ≤ ω ≤ π, the function *f*(ω) = cos(ω) + *j*0.1 sin(2ω) results in a so-called Lissajous figure that resembles a two-bladed propeller. - - (a) In MATLAB, create two row vectors fr and fi corresponding to the real and imaginary portions of *f*(ω), respectively, over a suitable number *N* samples of ω. Plot the real portion against the imaginary portion and verify the figure resembles a propeller. - - (b) Let complex constant *w* = *x* + *jy* be represented in vector form - -$$ -\mathbf{w} = \left[ \begin{array}{c} x \\ y \end{array} \right] -$$ - -Consider the 2×2 rotational matrix **R**: - -$$ -\mathbf{R} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} -$$ - -Show that **Rw** rotates vector **w** by θ radians. - -- (c) Create a rotational matrix R corresponding to 10◦ and multiply it by the 2×*N* matrix f = [fr;fi];. Plot the result to verify that the "propeller" has indeed rotated counterclockwise. -- (d) Given the matrix R determined in part (c), what is the effect of performing RRf? How about RRRf? Generalize the result. -- (e) Investigate the behavior of multiplying *f*(ω) by the function *ej*θ . - - - -# **[SIGNALS AND](#page-7-0) SYSTEMS** - -In this chapter we shall discuss basic aspects of signals and systems. We shall also introduce fundamental concepts and qualitative explanations of the hows and whys of systems theory, thus building a solid foundation for understanding the quantitative analysis in the remainder of the book. For simplicity, the focus of this chapter is on continuous-time signals and systems. Chapter 3 presents the same ideas for discrete-time signals and systems. - -### SIGNALS - -A *signal* is a set of data or information. Examples include a telephone or a television signal, monthly sales of a corporation, or daily closing prices of a stock market (e.g., the Dow Jones averages). In all these examples, the signals are functions of the independent variable *time*. This is not always the case, however. When an electrical charge is distributed over a body, for instance, the signal is the charge density, a function of *space* rather than time. In this book we deal almost exclusively with signals that are functions of time. The discussion, however, applies equally well to other independent variables. - -## SYSTEMS - -Signals may be processed further by *systems,* which may modify them or extract additional information from them. For example, an anti-aircraft gun operator may want to know the future location of a hostile moving target that is being tracked by his radar. Knowing the radar signal, he knows the past location and velocity of the target. By properly processing the radar signal (the input), he can approximately estimate the future location of the target. Thus, a system is an entity that *processes* a set of signals (*inputs*) to yield another set of signals (*outputs*). A system may be made up of physical components, as in electrical, mechanical, or hydraulic systems (hardware realization), or it may be an algorithm that computes an output from an input signal (software realization). - -## **[1.1 SIZE OF A](#page-7-0) SIGNAL** - -The size of any entity is a number that indicates the largeness or strength of that entity. Generally speaking, the signal amplitude varies with time. How can a signal that exists over a certain time interval with varying amplitude be measured by one number that will indicate the signal size or signal strength? Such a measure must consider not only the signal amplitude, but also its duration. For instance, if we are to devise a single number *V* as a measure of the size of a human being, we must consider not only his or her width (girth), but also the height. If we make a simplifying assumption that the shape of a person is a cylinder of variable radius *r* (which varies with the height *h*), then one possible measure of the size of a person of height *H* is the person's volume *V*, given by - -$$ -V = \pi \int_0^H r^2(h) \, dh -$$ - -### **[1.1-1 Signal Energy](#page-7-0)** - -Arguing in this manner, we may consider the area under a signal *x*(*t*) as a possible measure of its size, because it takes account not only of the amplitude but also of the duration. However, this will be a defective measure because even for a large signal *x*(*t*), its positive and negative areas could cancel each other, indicating a signal of small size. This difficulty can be corrected by defining the signal size as the area under |*x*(*t*)| 2, which is always positive. We call this measure the *signal energy Ex*, defined as - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt -$$ - (1.1) - -This definition simplifies for a real-valued signal *x*(*t*) to *Ex* = \$ ∞ −∞ *x*2(*t*)*dt*. There are also other possible measures of signal size, such as the area under |*x*(*t*)|. The energy measure, however, is not only more tractable mathematically but is also more meaningful (as shown later) in the sense that it is indicative of the energy that can be extracted from the signal. - -### **[1.1-2 Signal Power](#page-7-0)** - -Signal energy must be finite for it to be a meaningful measure of signal size. A necessary condition for the energy to be finite is that the signal amplitude → 0 as |*t*|→∞ (Fig. 1.1a). Otherwise the integral in Eq. (1.1) will not converge. - -When the amplitude of *x*(*t*) does not → 0 as |*t*|→∞ (Fig. 1.1b), the signal energy is infinite. A more meaningful measure of the signal size in such a case would be the time average of the energy, if it exists. This measure is called the *power* of the signal. For a signal *x*(*t*), we define its power *Px* as - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |x(t)|^2 dt -$$ - (1.2) - -This definition simplifies for a real-valued signal *x*(*t*) to *Px* = lim*T*→∞ 1 *T* \$ *T*/2 −*T*/2 *x*2(*t*)*dt*. Observe that the signal power *Px* is the time average (mean) of the signal magnitude squared, that is, the *mean-square* value of |*x*(*t*)|. Indeed, the square root of *Px* is the familiar *rms* (root-mean-square) value of *x*(*t*). - -Generally, the mean of an entity averaged over a large time interval approaching infinity exists if the entity either is periodic or has a statistical regularity. If such a condition is not satisfied, the average may not exist. For instance, a ramp signal *x*(*t*) = *t* increases indefinitely as |*t*|→∞, and neither the energy nor the power exists for this signal. However, the unit step function, which is not periodic nor has statistical regularity, does have a finite power. - -**Figure 1.1** Examples of signals: **(a)** a signal with finite energy and **(b)** a signal with finite power. - -When *x*(*t*) is periodic, |*x*(*t*)| 2 is also periodic. Hence, the power of *x*(*t*) can be computed from Eq. (1.2) by averaging |*x*(*t*)| 2 over one period. - -**Comments.** The signal energy as defined in Eq. (1.1) does not indicate the actual energy (in the conventional sense) of the signal because the signal energy depends not only on the signal, but also on the load. It can, however, be interpreted as the energy dissipated in a normalized load of a 1 ohm resistor if a voltage *x*(*t*) were to be applied across the 1 ohm resistor [or if a current *x*(*t*) were to be passed through the 1 ohm resistor]. The measure of "energy" is therefore indicative of the energy capability of the signal, not the actual energy. For this reason the concepts of conservation of energy should not be applied to this "signal energy." Parallel observation applies to "signal power" defined in Eq. (1.2). These measures are but convenient indicators of the signal size, which prove useful in many applications. For instance, if we approximate a signal *x*(*t*) by another signal *g*(*t*), the error in the approximation is *e*(*t*) = *x*(*t*) − *g*(*t*). The energy (or power) of *e*(*t*) is a convenient indicator of the goodness of the approximation. It provides us with a quantitative measure of determining the closeness of the approximation. In communication systems, during transmission over a channel, message signals are corrupted by unwanted signals (noise). The quality of the received signal is judged by the relative sizes of the desired signal and the unwanted signal (noise). In this case the ratio of the message signal and noise signal powers (signal-to-noise power ratio) is a good indication of the received signal quality. - -**Units of Energy and Power.** Equation (1.1) is not correct dimensionally. This is because here we are using the term *energy* not in its conventional sense, but to indicate the signal size. The same observation applies to Eq. (1.2) for power. The units of energy and power, as defined here, depend on the nature of the signal *x*(*t*). If *x*(*t*) is a voltage signal, its energy *Ex* has units of volts squared-seconds (V2 s), and its power *Px* has units of volts squared. If *x*(*t*) is a current signal, these units will be amperes squared-seconds (A2 s) and amperes squared, respectively. - -**Figure 1.2** Signals for Ex. 1.1 - -In Fig. 1.2a, the signal amplitude → 0 as |*t*|→∞. Therefore the suitable measure for this signal is its energy *Ex* given by - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-1}^{0} (2)^2 dt + \int_{0}^{\infty} 4e^{-t} dt = 4 + 4 = 8 -$$ - -In Fig. 1.2b, the signal magnitude does not → 0 as |*t*|→∞. However, it is periodic, and therefore its power exists. We can use Eq. (1.2) to determine its power. We can simplify the procedure for periodic signals by observing that a periodic signal repeats regularly each period (2 seconds in this case). Therefore, averaging |*x*(*t*)| 2 over an infinitely large interval is identical to averaging this quantity over one period (2 seconds in this case). Thus - -$$ -P_x = \frac{1}{2} \int_{-1}^{1} |x(t)|^2 dt = \frac{1}{2} \int_{-1}^{1} t^2 dt = \frac{1}{3} -$$ - -Recall that the signal power is the square of its rms value. Therefore, the rms value of this signal is 1/ √3. - -### **EXAMPLE 1.2 Determining Power and RMS Value** - -Determine the power and the rms value of - -- **(a)** *x*(*t*) = *C* cos(ω0*t* +θ ) -- **(b)** *x*(*t*) = *C*1 cos(ω1*t* +θ1)+*C*2 cos(ω2*t* +θ2) ω1 = ω2 -- **(c)** *x*(*t*) = *Dej*ω0*t* - -**(a)** This is a periodic signal with period *T*0 = 2π/ω0. The suitable measure of this signal is its power. Because it is a periodic signal, we may compute its power by averaging its energy over one period *T*0 = 2π/ω0. However, for the sake of demonstration, we shall use Eq. (1.2) to solve this problem by averaging over an infinitely large time interval. - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C^2 \cos^2 (\omega_0 t + \theta) dt = \lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} [1 + \cos (2\omega_0 t + 2\theta)] dt -$$ - -= -$$ -\lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} dt + \lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} \cos (2\omega_0 t + 2\theta) dt -$$ - -The first term on the right-hand side is equal to *C*2/2. The second term, however, is zero because the integral appearing in this term represents the area under a sinusoid over a very large time interval *T* with *T* → ∞. This area is at most equal to the area of half the cycle because of cancellations of the positive and negative areas of a sinusoid. The second term is this area multiplied by *C*2/2*T* with *T* → ∞. Clearly this term is zero, and - -$$ -P_x = \frac{C^2}{2} -$$ - -This shows that a sinusoid of amplitude *C* has a power *C*2/2 regardless of the value of its frequency ω0 (ω0 = 0) and phase θ. The rms value is *C*/ √2. If the signal frequency is zero (dc or a constant signal of amplitude *C*), the reader can show that the power is *C*2. - -**(b)** In Ch. 6, we shall show that a sum of two sinusoids may or may not be periodic, depending on whether the ratio ω1/ω2 is a rational number. Therefore, the period of this signal is not known. Hence, its power will be determined by averaging its energy over *T* seconds with *T* → ∞. Thus, - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} [C_1 \cos(\omega_1 t + \theta_1) + C_2 \cos(\omega_2 t + \theta_2)]^2 dt -$$ - -= -$$ -\lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C_1^2 \cos^2(\omega_1 t + \theta_1) dt + \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C_2^2 \cos^2(\omega_2 t + \theta_2) dt -$$ - -+ -$$ -\lim_{T \to \infty} \frac{2C_1 C_2}{T} \int_{-T/2}^{T/2} \cos(\omega_1 t + \theta_1) \cos(\omega_2 t + \theta_2) dt -$$ - -The first and second integrals on the right-hand side are the powers of the two sinusoids, which are *C*1 2/2 and *C*2 2/2, as found in part (a). The third term, the product of two sinusoids, can be expressed as a sum of two sinusoids cos[(ω1+ω2)*t*+(θ1+θ2)] and cos[(ω1−ω2)*t*+(θ1−θ2)], respectively. Now, arguing as in part (a), we see that the third term is zero. Hence, we have† - -$$ -P_x = \frac{C_1^2}{2} + \frac{C_2}{2} -$$ - -2 - -and the rms value is (*C*1 2 +*C*2 2)/2. - -We can readily extend this result to a sum of any number of sinusoids with distinct frequencies. Thus, if - -$$ -x(t) = \sum_{n=1}^{\infty} C_n \cos{(\omega_n t + \theta_n)} -$$ - -assuming that none of the two sinusoids have identical frequencies and ω*n* = 0, then - -$$ -P_x = \frac{1}{2} \sum_{n=1}^{\infty} C_n^2 -$$ - -If *x*(*t*) also has a dc term, as - -$$ -x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(\omega_n t + \theta_n) -$$ - -then - -$$ -P_x = C_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} C_n^2 -$$ - (1.3) - -**(c)** In this case the signal is complex, and we use Eq. (1.2) to compute the power. - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |De^{j\omega_0 t}|^2 dt -$$ - -Recall that |*ej*ω0*t* | = 1 so that |*Dej*ω0*t* | 2 = |*D*| 2, and - -$$ -P_x = |D|^2 \tag{1.4} -$$ - -The rms value is |*D*|. - -**Comment.** In part (b) of Ex. 1.2, we have shown that the power of the sum of two sinusoids is equal to the sum of the powers of the sinusoids. It may appear that the power of *x*1(*t*) + *x*2(*t*) - -† This is true only if ω1 = ω2. If ω1 = ω2, the integrand of the third term contains a constant cos(θ1 − θ2), and the third term → 2*C*1*C*2 cos(θ1 −θ2) as *T* → ∞. - -is *Px*1 + *Px*2 . Unfortunately, this conclusion is not true in general. It is true only under a certain condition (orthogonality), discussed later (Sec. 6.5-3). - -### **DR ILL 1.1 Computing Energy, Power, and RMS Value** - -Show that the energies of the signals in Figs. 1.3a, 1.3b, 1.3c, and 1.3d are 4, 1, 4/3, and 4/3, respectively. Observe that doubling a signal quadruples the energy, and time-shifting a signal has no effect on the energy. Show also that the power of the signal in Fig. 1.3e is 0.4323. What is the rms value of signal in Fig. 1.3e? - -## **DR ILL 1.2 Computing Power over a Period** - -Redo Ex. 1.1a to find the power of a sinusoid *C* cos(ω0*t* + θ ) by averaging the signal energy over one period *T*0 = 2π/ω0 (rather than averaging over the infinitely large interval). Show also that the power of a dc signal *x*(*t*) = *C*0 is *C*2 0, and its rms value is *C*0. - -## **DR ILL 1.3 Power of a Sum of Two Equal-Frequency Sinusoids** - -Show that if ω1 = ω2, the power of *x*(*t*) = *C*1 cos(ω1*t* +θ1)+*C*2 cos(ω2*t* +θ2) is [*C*1 2 +*C*2 2 + 2*C*1*C*2 cos(θ1 −θ2)]/2, which is not equal to the Ex. 1.2b result of (*C*1 2 +*C*2 2)/2. - -## **1.2 SOME USEFUL SIGNAL [OPERATIONS](#page-7-0)** - -We discuss here three useful signal operations: shifting, scaling, and inversion. Since the independent variable in our signal description is time, these operations are discussed as *time shifting, time scaling,* and *time reversal* (inversion). However, this discussion is valid for functions having independent variables other than time (e.g., frequency or distance). - -### **[1.2-1 Time Shifting](#page-7-0)** - -Consider a signal *x*(*t*) (Fig. 1.4a) and the same signal delayed by *T* seconds (Fig. 1.4b), which we shall denote by φ(*t*). Whatever happens in *x*(*t*) (Fig. 1.4a) at some instant *t* also happens in φ(*t*) (Fig. 1.4b) *T* seconds later at the instant *t* +*T*. Therefore - -$$ -\phi(t+T) = x(t) \qquad \text{and} \qquad \phi(t) = x(t-T) -$$ - -Therefore, to time-shift a signal by *T*, we replace *t* with *t* − *T*. Thus *x*(*t* − *T*) represents *x*(*t*) time-shifted by *T* seconds. If *T* is positive, the shift is to the right (delay), as in Fig. 1.4b. If *T* is negative, the shift is to the left (advance), as in Fig. 1.4c. Clearly, *x*(*t* − 2) is *x*(*t*) delayed (right-shifted) by 2 seconds, and *x*(*t* + 2) is *x*(*t*) advanced (left-shifted) by 2 seconds. - -**Figure 1.4** Time-shifting a signal. - -### **EXAMPLE 1.3 Time Shifting** - -An exponential function *x*(*t*) = *e*−2*t* shown in Fig. 1.5a is delayed by 1 second. Sketch and mathematically describe the delayed function. Repeat the problem with *x*(*t*) advanced by 1 second. - -**Figure 1.5 (a)** Signal *x*(*t*). **(b)** Signal *x*(*t*) delayed by 1 second. **(c)** Signal *x*(*t*) advanced by 1 second. - -The function *x*(*t*) can be described mathematically as - -$$ -x(t) = \begin{cases} e^{-2t} & t \ge 0\\ 0 & t < 0 \end{cases} \tag{1.5} -$$ - -Let *xd*(*t*) represent the function *x*(*t*) delayed (right-shifted) by 1 second, as illustrated in Fig. 1.5b. This function is *x*(*t* − 1); its mathematical description can be obtained from *x*(*t*) by replacing *t* with *t* −1 in Eq. (1.5). Thus, - -$$ -x_d(t) = x(t-1) = \begin{cases} e^{-2(t-1)} & t-1 \ge 0 \text{ or } t \ge 1\\ 0 & t-1 < 0 \text{ or } t < 1 \end{cases} -$$ - -Let *xa*(*t*) represent the function *x*(*t*) advanced (left-shifted) by 1 second, as depicted in Fig. 1.5c. This function is *x*(*t* + 1); its mathematical description can be obtained from *x*(*t*) by replacing *t* with *t* +1 in Eq. (1.5). Thus, - -$$ -x_a(t) = x(t+1) = \begin{cases} e^{-2(t+1)} & t+1 \ge 0 \text{ or } t \ge -1\\ 0 & t+1 < 0 \text{ or } t < -1 \end{cases} -$$ - -### **DR ILL 1.4 Working with Time Delay and Time Advance** - -Write a mathematical description of the signal *x*3(*t*) in Fig. 1.3c. Next, delay this signal by 2 seconds. Sketch the delayed signal. Show that this delayed signal *xd*(*t*) can be described mathematically as *xd*(*t*) = 2(*t* − 2) for 2 ≤ *t* ≤ 3, and equal to 0 otherwise. Now repeat the procedure with the signal advanced (left-shifted) by 1 second. Show that this advanced signal *xa*(*t*) can be described as *xa*(*t*) = 2(*t* +1) for −1 ≤ *t* ≤ 0, and 0 otherwise. - -### **[1.2-2 Time Scaling](#page-7-0)** - -The compression or expansion of a signal in time is known as *time scaling*. Consider the signal *x*(*t*) of Fig. 1.6a. The signal φ(*t*) in Fig. 1.6b is *x*(*t*) compressed in time by a factor of 2. Therefore, whatever happens in *x*(*t*) at some instant *t* also happens to φ(*t*) at the instant *t*/2 so that - -$$ -\phi\left(\frac{t}{2}\right) = x(t) -$$ - and $\phi(t) = x(2t)$ - -Observe that because *x*(*t*) = 0 at *t* = *T*1 and *T*2, we must have φ(*t*) = 0 at *t* = *T*1/2 and *T*2/2, as shown in Fig. 1.6b. If *x*(*t*) were recorded on a tape and played back at twice the normal recording speed, we would obtain *x*(2*t*). In general, if *x*(*t*) is compressed in time by a factor *a* (*a* > 1), the resulting signal φ(*t*) is given by - -$$ -\phi(t) = x(at) -$$ - -Using a similar argument, we can show that *x*(*t*) expanded (slowed down) in time by a factor *a* (*a* > 1) is given by - -$$ -\phi(t) = x \left(\frac{t}{a}\right) -$$ - -Figure 1.6c shows *x*(*t*/2), which is *x*(*t*) expanded in time by a factor of 2. Observe that in a time-scaling operation, the origin *t* = 0 is the anchor point, which remains unchanged under the scaling operation because at *t* = 0, *x*(*t*) = *x*(*at*) = *x*(0). - -In summary, to time-scale a signal by a factor *a*, we replace *t* with *at*. If *a* > 1, the scaling results in compression, and if *a* < 1, the scaling results in expansion. - -### **EXAMPLE 1.4 Continuous Time-Scaling Operation** - -Figure 1.7a shows a signal *x*(*t*). Sketch and describe mathematically this signal time-compressed by factor 3. Repeat the problem for the same signal time-expanded by factor 2. - -The signal *x*(*t*) can be described as - -$$ -x(t) = \begin{cases} 2 & -1.5 \le t < 0 \\ 2e^{-t/2} & 0 \le t < 3 \\ 0 & \text{otherwise} \end{cases} -$$ - (1.6) - -Figure 1.7b shows *xc*(*t*), which is *x*(*t*) time-compressed by factor 3; consequently, it can be described mathematically as *x*(3*t*), which is obtained by replacing *t* with 3*t* in the right-hand side of Eq. (1.6). Thus, - -$$ -x_c(t) = x(3t) = \begin{cases} 2 & -1.5 \le 3t < 0 \text{ or } -0.5 \le t < 0 \\ 2e^{-3t/2} & 0 \le 3t < 3 \text{ or } 0 \le t < 1 \\ 0 & \text{otherwise} \end{cases} -$$ - -Observe that the instants *t* = −1.5 and 3 in *x*(*t*) correspond to the instants *t* = −0.5, and 1 in the compressed signal *x*(3*t*). - -Figure 1.7c shows *xe*(*t*), which is *x*(*t*) time-expanded by factor 2; consequently, it can be described mathematically as *x*(*t*/2), which is obtained by replacing *t* with *t*/2 in *x*(*t*). Thus, - -$$ -x_e(t) = x\left(\frac{t}{2}\right) = \begin{cases} 2 & -1.5 \le \frac{t}{2} < 0 \text{ or } -3 \le t < 0 \\ 2e^{-t/4} & 0 \le \frac{t}{2} < 3 \text{ or } 0 \le t < 6 \\ 0 & \text{otherwise} \end{cases} -$$ - -Observe that the instants *t* = −1.5 and 3 in *x*(*t*) correspond to the instants *t* = −3 and 6 in the expanded signal *x*(*t*/2). - -### **DR ILL 1.5 Compression and Expansion of Sinusoids** - -Show that the time compression by an integer factor *n* (*n* > 1) of a sinusoid results in a sinusoid of the same amplitude and phase, but with the frequency increased *n*-fold. Similarly, the time expansion by an integer factor *n* (*n* > 1) of a sinusoid results in a sinusoid of the same amplitude and phase, but with the frequency reduced by a factor *n*. Verify your conclusion by sketching a sinusoid sin 2*t* and the same sinusoid compressed by a factor 3 and expanded by a factor 2. - -### **[1.2-3 Time Reversal](#page-7-0)** - -Consider the signal *x*(*t*) in Fig. 1.8a. We can view *x*(*t*) as a rigid wire frame hinged at the vertical axis. To time-reverse *x*(*t*), we rotate this frame 180◦ about the vertical axis. This time reversal [the reflection of *x*(*t*) about the vertical axis] gives us the signal φ(*t*) (Fig. 1.8b). Observe that whatever happens in Fig. 1.8a at some instant *t* also happens in Fig. 1.8b at the instant −*t*, and vice versa. Therefore, - -φ(*t*) = *x*(−*t*) - -Thus, to time-reverse a signal we replace *t* with −*t*, and the time reversal of signal *x*(*t*) results in a signal *x*(−*t*). We must remember that the reversal is performed about the vertical axis, which acts as an anchor or a hinge. Recall also that the reversal of *x*(*t*) about the horizontal axis results in −*x*(*t*). - -**Figure 1.8** Time reversal of a signal. - -### **EXAMPLE 1.5 Time Reversal of a Signal** - -For the signal *x*(*t*) illustrated in Fig. 1.9a, sketch *x*(−*t*), which is time-reversed *x*(*t*). - -The instants −1 and −5 in *x*(*t*) are mapped into instants 1 and 5 in *x*(−*t*). Because *x*(*t*) = *et*/2, we have *x*(−*t*) = *e*−*t*/2. The signal *x*(−*t*) is depicted in Fig. 1.9b. We can describe *x*(*t*) and *x*(−*t*) as - -$$ -x(t) = \begin{cases} e^{t/2} & -1 \ge t > -5\\ 0 & \text{otherwise} \end{cases} -$$ - -and its time-reversed version *x*(−*t*) is obtained by replacing *t* with −*t* in *x*(*t*) as - -*x*(−*t*) = *e*−*t*/2 −1 ≥ −*t* > −5 or 1 ≤ *t* < 5 0 otherwise - -### **[1.2-4 Combined Operations](#page-7-0)** - -Certain complex operations require simultaneous use of more than one of the operations just described. The most general operation involving all the three operations is *x*(*at* − *b*), which is realized in two possible sequences of operation: - -- 1. Time-shift *x*(*t*) by *b* to obtain *x*(*t*−*b*). Now time-scale the shifted signal *x*(*t*−*b*) by *a* [i.e., replace *t* with *at*] to obtain *x*(*at* −*b*). -- 2. Time-scale *x*(*t*) by *a* to obtain *x*(*at*). Now time-shift *x*(*at*) by *b*/*a* [i.e., replace *t* with *t* − (*b*/*a*)] to obtain *x*[*a*(*t* − *b*/*a*)] = *x*(*at* − *b*). In either case, if *a* is negative, time scaling involves time reversal. - -For example, the signal *x*(2*t*−6) can be obtained in two ways. We can delay *x*(*t*) by 6 to obtain *x*(*t* − 6), and then time-compress this signal by factor 2 (replace *t* with 2*t*) to obtain *x*(2*t* − 6). Alternately, we can first time-compress *x*(*t*) by factor 2 to obtain *x*(2*t*), then delay this signal by 3 (replace *t* with *t* −3) to obtain *x*(2*t* −6). - -## **[1.3 CLASSIFICATION OF](#page-7-0) SIGNALS** - -Classification helps us better understand and utilize the items around us. Cars, for example, are classified as sports, offroad, family, and so forth. Knowing you have a sports car is useful in deciding whether to drive on a highway or on a dirt road. Knowing you want to drive up a mountain, you would probably choose an offroad vehicle over a family sedan. Similarly, there are several classes of signals. Some signal classes are more suitable for certain applications than others. Further, different signal classes often require different mathematical tools. Here we shall consider only the following classes of signals, which are suitable for the scope of this book: - -- 1. Continuous-time and discrete-time signals -- 2. Analog and digital signals -- 3. Periodic and aperiodic signals -- 4. Energy and power signals -- 5. Deterministic and probabilistic signals - -### **[1.3-1 Continuous-Time and Discrete-Time Signals](#page-7-0)** - -A signal that is specified for a continuum of values of time *t* (Fig. 1.10a) is a *continuous-time signal,* and a signal that is specified only at discrete values of *t* (Fig. 1.10b) is a *discrete-time signal*. Telephone and video camera outputs are continuous-time signals, whereas the quarterly gross national product (GNP), monthly sales of a corporation, and stock market daily averages are discrete-time signals. - -### **[1.3-2 Analog and Digital Signals](#page-7-0)** - -The concept of continuous time is often confused with that of analog. The two are not the same. The same is true of the concepts of discrete time and digital. A signal whose amplitude can take on any value in a continuous range is an *analog signal*. This means that an analog signal amplitude can take on an infinite number of values. A *digital signal,* on the other hand, is one whose amplitude can take on only a finite number of values. Signals associated with a digital computer are digital because they take on only two values (binary signals). A digital signal whose amplitudes can take on *M* values is an *M*-ary signal of which binary (*M* = 2) is a special case. The terms *continuous time* and *discrete time* qualify the nature of a signal along the time (horizontal) axis. The terms *analog* and *digital,* on the other hand, qualify the nature of the signal amplitude (vertical axis). Figure 1.11 shows examples of signals of various types. It is clear that analog is not necessarily continuous-time and digital need not be discrete-time. Figure 1.11c shows an example of an analog discrete-time signal. An analog signal can be converted into a digital signal [analog-to-digital (A/D) conversion] through quantization (rounding off ), as explained in Sec. 8.3. - -**Figure 1.10 (a)** Continuous-time and **(b)** discrete-time signals. - -### **[1.3-3 Periodic and Aperiodic Signals](#page-7-0)** - -A signal *x*(*t*) is said to be *periodic* if for some positive constant *T*0 - -$$ -x(t) = x(t + T_0) \qquad \text{for all } t \tag{1.7} -$$ - -The *smallest* value of *T*0 that satisfies the periodicity condition of Eq. (1.7) is the *fundamental period* of *x*(*t*). The signals in Figs. 1.2b and 1.3e are periodic signals with periods 2 and 1, respectively. A signal is *aperiodic* if it is not periodic. Signals in Figs. 1.2a, 1.3a, 1.3b, 1.3c, and 1.3d are all aperiodic. - -By definition, a periodic signal *x*(*t*) remains unchanged when time-shifted by one period. For this reason, a periodic signal must start at *t* = −∞: if it started at some finite instant, say, *t* = 0, the time-shifted signal *x*(*t* + *T*0) would start at *t* = −*T*0 and *x*(*t* + *T*0) would not be the same as 80 CHAPTER 1 SIGNALS AND SYSTEMS - -**Figure 1.11** Examples of signals: **(a)** analog, continuous time; **(b)** digital, continuous time; **(c)** analog, discrete time; and **(d)** digital, discrete time. - -**Figure 1.12** A periodic signal of period *T*0. - -*x*(*t*). Therefore, a *periodic signal, by definition, must start at t* = −∞ *and continue forever, as illustrated in Fig. 1.12.* - -Another important property of a periodic signal *x*(*t*) is that *x*(*t*) can be generated by *periodic extension* of any segment of *x*(*t*) of duration *T*0 (the period). As a result, we can generate *x*(*t*) from any segment of *x*(*t*) having a duration of one period by placing this segment and the reproduction thereof end to end ad infinitum on either side. Figure 1.13 shows a periodic signal *x*(*t*) of period *T*0 = 6. The shaded portion of Fig. 1.13a shows a segment of *x*(*t*) starting at *t* = −1 and having a duration of one period (6 seconds). This segment, when repeated forever in either direction, results in the periodic signal *x*(*t*). Figure 1.13b shows another shaded segment of *x*(*t*) of duration *T*0 starting at *t* = 0. Again, we see that this segment, when repeated forever on either side, results in *x*(*t*). The reader can verify that this construction is possible with any segment of *x*(*t*) starting at any instant as long as the segment duration is one period. - -**Figure 1.13** Generation of a periodic signal by periodic extension of its segment of one-period duration. - -An additional useful property of a periodic signal *x*(*t*) of period *T*0 is that the area under *x*(*t*) over any interval of duration *T*0 is the same; that is, for any real numbers *a* and *b*, - -$$ -\int_{a}^{a+T_0} x(t) dt = \int_{b}^{b+T_0} x(t) dt -$$ - -This result follows from the fact that a periodic signal takes the same values at the intervals of *T*0. Hence, the values over any segment of duration *T*0 are repeated in any other interval of the same duration. For convenience, the area under *x*(*t*) over any interval of duration *T*0 will be denoted by - -$$ -\int_{T_0} x(t) \, dt -$$ - -It is helpful to label signals that start at *t* = −∞ and continue forever as *everlasting* signals. Thus, an everlasting signal exists over the entire interval −∞ < *t* < ∞. The signals in Figs. 1.1b and 1.2b are examples of everlasting signals. Clearly, a periodic signal, by definition, is an everlasting signal. - -A signal that does not start before *t* = 0 is a *causal* signal. In other words, *x*(*t*) is a causal signal if - -$$ -x(t) = 0 \qquad t < 0 -$$ - -The signals in Figs. 1.3a–1.3c are causal signals. A signal that starts before *t* = 0 is a *noncausal* signal. All the signals in Figs. 1.1 and 1.2 are noncausal. Observe that an everlasting signal is always noncausal but a noncausal signal is not necessarily everlasting. The everlasting signal in Fig. 1.2b is noncausal; however, the noncausal signal in Fig. 1.2a is not everlasting. A signal that is zero for all *t* ≥ 0 is called an *anti-causal* signal. - -**Comment.** A true everlasting signal cannot be generated in practice for obvious reasons. Why should we bother to postulate such a signal? In later chapters we shall see that certain signals - -### 82 CHAPTER 1 SIGNALS AND SYSTEMS - -(e.g., an impulse and an everlasting sinusoid) that cannot be generated in practice *do* serve a very useful purpose in the study of signals and systems. - -### **1.3-4 Energy and Power Signals** - -A signal with finite energy is an *energy signal,* and a signal with finite and nonzero power is a *power signal*. The signals in Figs. 1.2a and 1.2b are examples of energy and power signals, respectively. Observe that power is the time average of energy. Since the averaging is over an infinitely large interval, a signal with finite energy has zero power, and a signal with finite power has infinite energy. Therefore, a signal cannot be both an energy signal and a power signal. If it is one, it cannot be the other. On the other hand, there are signals that are neither energy nor power signals. The ramp signal is one such case. - -**Comments.** All practical signals have finite energies and are therefore energy signals. A power signal must necessarily have infinite duration; otherwise, its power, which is its energy averaged over an infinitely large interval, will not approach a (nonzero) limit. Clearly, it is impossible to generate a true power signal in practice because such a signal has infinite duration and infinite energy. - -Also, because of periodic repetition, periodic signals for which the area under |*x*(*t*)| 2 over one period is finite are power signals; however, not all power signals are periodic. - -### **DR ILL 1.6 Neither Energy nor Power** - -Show that an everlasting exponential *e*−*at* is neither an energy nor a power signal for any real value of *a*. However, if *a* is imaginary, it is a power signal with power *Px* = 1 regardless of the value of *a*. - -### **[1.3-5 Deterministic and Random Signals](#page-7-0)** - -A signal whose physical description is known completely, in either a mathematical form or a graphical form, is a *deterministic signal*. A signal whose values cannot be predicted precisely but are known only in terms of probabilistic description, such as mean value or mean-squared value, is a *random signal*. In this book we shall exclusively deal with deterministic signals. Random signals are beyond the scope of this study. - -## **[1.4 SOME](#page-7-0) USEFUL SIGNAL MODELS** - -In the area of signals and systems, the step, the impulse, and the exponential functions play very important roles. Not only do they serve as a basis for representing other signals, but their use can simplify many aspects of the signals and systems. - -## **[1.4-1 The Unit Step Function](#page-7-0)** *u(t)* - -In much of our discussion, the signals begin at *t* = 0 (causal signals). Such signals can be conveniently described in terms of unit step function *u*(*t*) shown in Fig. 1.14a. This function is defined by - -$$ -u(t) = \begin{cases} 1 & t \ge 0 \\ 0 & t < 0 \end{cases} -$$ - (1.8) - -If we want a signal to start at *t* = 0 (so that it has a value of zero for *t* < 0), we need only multiply the signal by *u*(*t*). For instance, the signal *e*−*at* represents an everlasting exponential that starts at *t* = −∞. The causal form of this exponential (Fig. 1.14b) can be described as *e*−*atu*(*t*). - -The unit step function also proves very useful in specifying a function with different mathematical descriptions over different intervals. Examples of such functions appear in Fig. 1.7. These functions have different mathematical descriptions over different segments of time, as seen from Eqs. (1.5) and (1.6). Such a description often proves clumsy and inconvenient in mathematical treatment. We can use the unit step function to describe such functions by a single expression that is valid for all *t*. - -Consider, for example, the rectangular pulse depicted in Fig. 1.15a. We can express such a pulse in terms of familiar step functions by observing that the pulse *x*(*t*) can be expressed as the sum of the two delayed unit step functions, as shown in Fig. 1.15b. The unit step function *u*(*t*) delayed by *T* seconds is *u*(*t* −*T*). From Fig. 1.15b, it is clear that - -$$ -x(t) = u(t-2) - u(t-4) -$$ - -**Figure 1.14 (a)** Unit step function *u*(*t*). **(b)** Exponential *e*−*atu*(*t*). - -**Figure 1.15** Representation of a rectangular pulse by step functions. - -### **EXAMPLE 1.6 Describing a Triangle Function with the Unit Step** - -Use the unit step function to describe the signal in Fig. 1.16a. - -**Figure 1.16** Representation of a signal defined interval by interval. - -$$ -x_1(t) = t[u(t) - u(t-2)] -$$ - -The signal *x*2(*t*) can be obtained by multiplying another ramp by the gate pulse illustrated in Fig. 1.16c. This ramp has a slope −2; hence it can be described by −2*t* + *c*. Now, because the ramp has a zero value at *t* = 3, the constant *c* = 6, and the ramp can be described by −2(*t*−3). Also, the gate pulse in Fig. 1.16c is *u*(*t* −2)−*u*(*t* −3). Therefore, - -$$ -x_2(t) = -2(t-3)[u(t-2) - u(t-3)] -$$ - -The signal illustrated in Fig. 1.16a can be conveniently handled by breaking it up into the two components *x*1(*t*) and *x*2(*t*), depicted in Figs. 1.16b and 1.16c, respectively. Here, *x*1(*t*) can be obtained by multiplying the ramp *t* by the gate pulse *u*(*t*) − *u*(*t* − 2), as shown in Fig. 1.16b. Therefore, - -and - -$$ -x(t) = x_1(t) + x_2(t) -$$ - -= t[u(t) - u(t-2)] - 2(t-3)[u(t-2) - u(t-3)] -= tu(t) - 3(t-2)u(t-2) + 2(t-3)u(t-3) - -### **EXAMPLE 1.7 Describing a Piecewise Function with the Unit Step** - -Describe the signal in Fig. 1.7a by a single expression valid for all *t*. - -Over the interval from −1.5 to 0, the signal can be described by a constant 2, and over the interval from 0 to 3, it can be described by 2*e*−*t*/2. Therefore, - -$$ -x(t) = 2[u(t+1.5) - u(t)] + 2e^{-t/2}[u(t) - u(t-3)] -$$ - -constant part -$$ -= 2u(t+1.5) - 2(1 - e^{-t/2})u(t) - 2e^{-t/2}u(t-3) -$$ - -Compare this expression with the expression for the same function found in Eq. (1.6). - -### **DR ILL 1.7 Using Reflected Unit Step Functions** - -Show that the signals depicted in Figs. 1.17a and 1.17b can be described as *u*(−*t*) and *e*−*atu*(−*t*), respectively. - -### **DR ILL 1.8 Describing a Piecewise Function with the Unit Step** - -Show that the signal shown in Fig. 1.18 can be described as - -$$ -x(t) = (t-1)u(t-1) - (t-2)u(t-2) - u(t-4) -$$ - -## **[1.4-2 The Unit Impulse Function](#page-7-0)** *δ(t)* - -The unit impulse function δ(*t*) is one of the most important functions in the study of signals and systems. This function was first defined in two parts by P. A. M. Dirac as - -$$ -\delta(t) = 0 \quad t \neq 0 \quad \text{and} \quad \int_{-\infty}^{\infty} \delta(t) dt = 1 \tag{1.9} -$$ - -**Figure 1.18** Signal for Drill 1.8. - -We can visualize an impulse as a tall, narrow, rectangular pulse of unit area, as illustrated in Fig. 1.19b. The width of this rectangular pulse is a very small value → 0. Consequently, its height is a very large value 1/ → ∞. The unit impulse therefore can be regarded as a rectangular pulse with a width that has become infinitesimally small, a height that has become infinitely large, and an overall area that has been maintained at unity. Thus δ(*t*) = 0 everywhere except at *t* = 0, where it is undefined. For this reason, a unit impulse is represented by the spearlike symbol in Fig. 1.19a. - -Other pulses, such as the exponential, triangular, or Gaussian types, may also be used in impulse approximation. The important feature of the unit impulse function is not its shape but the fact that its effective duration (pulse width) approaches zero while its area remains at unity. For example, the exponential pulse α*e*−α*t u*(*t*) in Fig. 1.20a becomes taller and narrower as α increases. - -**Figure 1.19** A unit impulse and its - -**Figure 1.20** Other possible approximations to a unit impulse. - -In the limit as α → ∞, the pulse height → ∞, and its width or duration → 0. Yet, the area under the pulse is unity regardless of the value of α because - -$$ -\int_0^\infty \alpha e^{-\alpha t} dt = 1 -$$ - -The pulses in Figs. 1.20b and 1.20c behave in a similar fashion. Clearly, the exact impulse function cannot be generated in practice; it can only be approached. - -From Eq. (1.9), it follows that the function *k*δ(*t*) = 0 for all *t* = 0, and its area is *k*. Thus, *k*δ(*t*) is an impulse function whose area is *k* (in contrast to the unit impulse function, whose area is 1). - -### MULTIPLICATION OF A FUNCTION BY AN IMPULSE - -Let us now consider what happens when we multiply the unit impulse δ(*t*) by a function φ(*t*) that is known to be continuous at *t* = 0. Since the impulse has nonzero value only at *t* = 0, and the value of φ(*t*) at *t* = 0 is φ(0), we obtain - -$$ -\phi(t)\delta(t) = \phi(0)\delta(t) -$$ - -Thus, multiplication of a continuous-time function φ(*t*) with an unit impulse located at *t* = 0 results in an impulse, which is located at *t* = 0 and has strength φ(0) [the value of φ(*t*) at the location of the impulse]. Use of exactly the same argument leads to the generalization of this result, stating that provided φ(*t*) is continuous at *t* = *T*,φ(*t*) multiplied by an impulse δ(*t* − *T*) (impulse located at *t* = *T*) results in an impulse located at *t* = *T* and having strength φ(*T*) [the value of φ(*t*) at the location of the impulse]. - -$$ -\phi(t)\delta(t-T) = \phi(T)\delta(t-T) \tag{1.10} -$$ - -### SAMPLING PROPERTY OF THE UNIT IMPULSE FUNCTION - -From Eq. (1.10) it follows that - -$$ -\int_{-\infty}^{\infty} \phi(t)\delta(t-T) dt = \phi(T) \int_{-\infty}^{\infty} \delta(t) dt = \phi(T) -$$ -\n(1.11) - -provided φ(*t*) is continuous at *t* = *T*. This result means that *the area under the product of a function with an impulse* δ(*t* − *T*) *is equal to the value of that function at the instant at which the unit impulse is located.* This property is very important and useful and is known as the *sampling* or *sifting property* of the unit impulse. - -### UNIT IMPULSE AS A GENERALIZED FUNCTION - -The definition of the unit impulse function given in Eq. (1.9) is not mathematically rigorous, which leads to serious difficulties. First, the impulse function does not define a unique function: for example, it can be shown that δ(*t*)+δ(˙ *t*) also satisfies Eq. (1.9) [1]. Moreover, δ(*t*) is not even a true function in the ordinary sense. An ordinary function is specified by its values for all time *t*. The impulse function is zero everywhere except at *t* = 0, and at this, the only interesting part of its range, it is undefined. These difficulties are resolved by defining the impulse as a generalized function rather than an ordinary function. A *generalized function* is defined by its effect on other functions instead of by its value at every instant of time. - -In this approach the impulse function is defined by the sampling property [Eq. (1.11)]. We say nothing about what the impulse function is or what it looks like. Instead, the impulse function is defined in terms of its effect on a test function φ(*t*). We define a unit impulse as a function for which the area under its product with a function φ(*t*) is equal to the value of the function φ(*t*) at the instant at which the impulse is located. It is assumed that φ(*t*) is continuous at the location of the impulse. Recall that the sampling property [Eq. (1.11)] is the consequence of the classical (Dirac) definition of the unit impulse in Eq. (1.9). In contrast, *the sampling property [Eq. (1.11)] defines the impulse function in the generalized function approach.* - -We now present an interesting application of the generalized function definition of an impulse. Because the unit step function *u*(*t*) is discontinuous at *t* = 0, its derivative *du*/*dt* does not exist at *t* = 0 in the ordinary sense. We now show that this derivative *does* exist in the generalized sense, and it is, in fact, δ(*t*). As a proof, let us evaluate the integral of (*du*/*dt*)φ(*t*), using integration by parts: - -$$ -\int_{-\infty}^{\infty} \frac{du(t)}{dt} \phi(t) dt = u(t) \phi(t) \Big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} u(t) \dot{\phi}(t) dt -$$ -$$ -= \phi(\infty) - 0 - \int_{0}^{\infty} \dot{\phi}(t) dt -$$ -$$ -= \phi(\infty) - \phi(t) \Big|_{0}^{\infty} = \phi(0) -$$ - -This result shows that *du*/*dt* satisfies the sampling property of δ(*t*). Therefore it is an impulse δ(*t*) in the generalized sense—that is, - -$$ -\frac{du(t)}{dt} = \delta(t) \tag{1.12} -$$ - -Consequently, - -$$ -\int_{-\infty}^{t} \delta(\tau) d\tau = u(t) -$$ - -These results can also be obtained graphically from Fig. 1.19b. We observe that the area from −∞ to *t* under the limiting form of δ(*t*) in Fig. 1.19b is zero if *t* < −/2 and unity if *t* ≥ /2 with → 0. Consequently, - -$$ -\int_{-\infty}^{t} \delta(\tau) d\tau = \begin{cases} 0 & t < 0 \\ 1 & t \ge 0 \end{cases} -$$ -$$ -= u(t) -$$ - -This result shows that the unit step function can be obtained by integrating the unit impulse function. Similarly the unit ramp function *x*(*t*) = *tu*(*t*) can be obtained by integrating the unit step function. We may continue with unit parabolic function *t* 2/2 obtained by integrating the unit ramp, and so on. On the other side, we have derivatives of impulse function, which can be defined as generalized functions (see Prob. 1.4-12). All these functions, derived from the unit impulse function (successive derivatives and integrals), are called *singularity functions*. † - -### **DR ILL 1.9 Simplifying Expressions Containing the Unit Impulse** - -Show that - -(a) -$$ -(t^3 + 3)\delta(t) = 3\delta(t) -$$ - -\n(b) $\left[\sin\left(t^2 - \frac{\pi}{2}\right)\right] \delta(t) = -\delta(t)$ -\n(c) $e^{-2t}\delta(t) = \delta(t)$ - -(d) -$$ -\frac{\omega^2 + 1}{\omega^2 + 9} \delta(\omega - 1) = \frac{1}{5} \delta(\omega - 1) -$$ - -## **DR ILL 1.10 Simplifying Integrals Containing the Unit Impulse** - -Show that - -(a) -$$ -\int_{-\infty}^{\infty} \delta(t) e^{-j\omega t} dt = 1 -$$ - -\n(b) -$$ -\int_{-\infty}^{\infty} \delta(t-2) \cos\left(\frac{\pi t}{4}\right) dt = 0 -$$ - -\n(c) -$$ -\int_{-\infty}^{\infty} e^{-2(x-t)} \delta(2-t) dt = e^{-2(x-2)} -$$ - -## **[1.4-3 The Exponential Function](#page-7-0)** *est* - -Another important function in the area of signals and systems is the exponential signal *est*, where *s* is complex in general, given by - -``` -s = σ +jω -``` - -† Singularity functions were defined by late Prof. S. J. Mason as follows. A singularity is a point at which a function does not possess a derivative. Each of the singularity functions (or if not the function itself, then the function differentiated a finite number of times) has a singular point at the origin and is zero elsewhere [2]. - -Therefore, - -$$ -e^{st} = e^{(\sigma + j\omega)t} = e^{\sigma t} e^{j\omega t} = e^{\sigma t} (\cos \omega t + j \sin \omega t) -$$ - (1.13) - -Since *s*∗ = σ −*j*ω (the conjugate of *s*), then - -$$ -e^{s^*t} = e^{(\sigma - j\omega)t} = e^{\sigma t}e^{-j\omega t} = e^{\sigma t}(\cos \omega t - j\sin \omega t) -$$ - -and - -$$ -e^{\sigma t} \cos \omega t = \frac{1}{2} (e^{st} + e^{s^* t}) -$$ -\n(1.14) - -A comparison of Eq. (1.13) with Euler's formula shows that *est* is a generalization of the function *ej*ω*t* , where the frequency variable *j*ω is generalized to a complex variable *s* = σ + *j*ω. For this reason, we designate the variable *s* as the *complex frequency*. In fact, function *est* encompasses a large class of functions. The following functions are either special cases of or can be expressed in terms of *est*: - -- 1. A constant *k* = *ke*0*t* (*s* = 0) -- 2. A monotonic exponential *e*σ*t* (ω = 0, *s* = σ ) -- 3. A sinusoid cos ω*t* (σ = 0, *s* = ±*j*ω) -- 4. An exponentially varying sinusoid *e*σ*t* cos ω*t* (*s* = σ ±*j*ω) - -These functions are illustrated in Fig. 1.21. - -The complex frequency *s* can be conveniently represented on a *complex frequency plane* (*s* plane), as depicted in Fig. 1.22. The horizontal axis is the real axis (σ axis), and the vertical axis is the imaginary axis (ω axis). The absolute value of the imaginary part of *s* is |ω| (the - -**Figure 1.21** Sinusoids of complex frequency σ +*j*ω. - -**Figure 1.22** Complex frequency plane. - -*radian* frequency), which indicates the frequency of oscillation of *est*; the real part σ (the *neper* frequency) gives information about the rate of increase or decrease of the amplitude of *est*. For signals whose complex frequencies lie on the real axis (σ axis, where ω = 0), the frequency of oscillation is zero. Consequently these signals are monotonically increasing or decreasing exponentials (Fig. 1.21a). For signals whose frequencies lie on the imaginary axis (ω axis, where σ = 0), *e*σ*t* = 1. Therefore, these signals are conventional sinusoids with constant amplitude (Fig. 1.21b). The case *s* = 0 (σ = ω = 0) corresponds to a constant (dc) signal because *e*0*t* = 1. For the signals illustrated in Figs. 1.21c and 1.21d, both σ and ω are nonzero; the frequency *s* is complex and does not lie on either axis. The signal in Fig. 1.21c decays exponentially. Therefore, σ is negative, and *s* lies to the left of the imaginary axis. In contrast, the signal in Fig. 1.21d *grows* exponentially. Therefore, σ is positive, and *s* lies to the right of the imaginary axis. Thus the *s* plane (Fig. 1.21) can be separated into two parts: the *left half-plane* (LHP) corresponding to exponentially decaying signals and the *right half-plane* (RHP) corresponding to exponentially growing signals. The imaginary axis separates the two regions and corresponds to signals of constant amplitude. - -An exponentially growing sinusoid *e*2*t* cos 5*t*, for example, can be expressed as a linear combination of exponentials *e*(2+*j*5)*t* and *e*(2−*j*5)*t* with complex frequencies 2 + *j*5 and 2−*j*5, respectively, which lie in the RHP. An exponentially decaying sinusoid *e*−2*t* cos 5*t* can be expressed as a linear combination of exponentials *e*(−2+*j*5)*t* and *e*(−2−*j*5)*t* with complex frequencies −2 + *j*5 and −2 − *j*5, respectively, which lie in the LHP. A constant-amplitude sinusoid cos 5*t* can be expressed as a linear combination of exponentials *ej*5*t* and *e*−*j*5*t* with complex frequencies ±*j*5, which lie on the imaginary axis. Observe that the monotonic exponentials *e*±2*t* are also generalized sinusoids with complex frequencies ±2. - -## **[1.5 EVEN AND](#page-7-0) ODD FUNCTIONS** - -A function *xe*(*t*) is said to be an *even function* of *t* if it is symmetrical about the vertical axis. A function *xo*(*t*) is said to be an *odd function* of *t* if it is antisymmetrical about the vertical axis. Mathematically expressed, these symmetry conditions require - -$$ -x_e(t) = x_e(-t) -$$ - and $x_o(t) = -x_o(-t)$ (1.15) - -An even function has the same value at the instants *t* and −*t* for all values of *t*. On the other hand, the value of an odd function at the instant *t* is the negative of its value at the instant −*t*. An example even signal and an example odd signal are shown in Figs. 1.23a and 1.23b, respectively. - -### **[1.5-1 Some Properties of Even and Odd Functions](#page-7-0)** - -Even and odd functions have the following properties: - -even function ×odd function = odd function odd function ×odd function = even function even function ×even function = even function - -The proofs are trivial and follow directly from the definition of odd and even functions [Eq. (1.15)]. - -### AREA - -Because of the symmetries of even and odd functions about the vertical axis, it follows from Eq. (1.15) [or Fig. 1.23] that - -$$ -\int_{-a}^{a} x_e(t) dt = 2 \int_{0}^{a} x_e(t) dt \quad \text{and} \quad \int_{-a}^{a} x_o(t) dt = 0 \quad (1.16) -$$ - -These results are valid under the assumption that there is no impulse (or its derivatives) at the origin. The proof of these statements is obvious from the plots of even and odd functions. Formal proofs, left as an exercise for the reader, can be accomplished by using the definitions in Eq. (1.15). - -Because of their properties, study of odd and even functions proves useful in many applications, as will become evident in later chapters. - -### **[1.5-2 Even and Odd Components of a Signal](#page-7-0)** - -Every signal *x*(*t*) can be expressed as a sum of even and odd components because - -$$ -x(t) = \underbrace{\frac{1}{2}[x(t) + x(-t)]}_{\text{even}} + \underbrace{\frac{1}{2}[x(t) - x(-t)]}_{\text{odd}} -$$ -(1.17) - -From the definitions in Eq. (1.15), we can clearly see that the first component on the right-hand side is an even function, while the second component is odd. This is apparent from the fact that replacing *t* by −*t* in the first component yields the same function. The same maneuver in the second component yields the negative of that component. - -### **EXAMPLE 1.8 Finding the Even and Odd Components of a Signal** - -Find and sketch the even and odd components of *x*(*t*) = *e*−*atu*(*t*). - -Based on Eq. (1.17), we can express *x*(*t*) as a sum of the even component *xe*(*t*) and the odd component *xo*(*t*) as - -$$ -x(t) = x_e(t) + x_o(t) -$$ - -where - -$$ -x_e(t) = \frac{1}{2} [e^{-at}u(t) + e^{at}u(-t)] \quad \text{and} \quad x_o(t) = \frac{1}{2} [e^{-at}u(t) - e^{at}u(-t)] -$$ - -The function *e*−*atu*(*t*) and its even and odd components are illustrated in Fig. 1.24. - -### **EXAMPLE 1.9 Finding the Even and Odd Components of a Complex Signal** - -Find the even and odd components of *ejt*. - -From Eq. (1.17), - -*ejt* = *xe*(*t*)+*xo*(*t*) - -where - -*xe*(*t*) = 1 2 [*ejt* +*e*−*jt*] = cos *t* and *xo*(*t*) = 1 2 [*ejt* −*e*−*jt*] = *j*sin *t* - -### A MODIFICATION FOR COMPLEX SIGNALS - -While a complex signal can be decomposed into even and odd components, it is more common to decompose complex signals using conjugate symmetries. A complex signal *x*(*t*) is said to be *conjugate-symmetric* if *x*(*t*) = *x*∗(−*t*). A conjugate-symmetric signal is even in the real part and odd in the imaginary part. Thus, a real conjugate-symmetric signal is an even signal. A signal is *conjugate-antisymmetric* if *x*(*t*) = −*x*∗(−*t*). A conjugate-antisymmetric signal is odd in the real part and even in the imaginary part. A real conjugate-antisymmetric signal is an odd signal. Any signal *x*(*t*) can be decomposed into a conjugate-symmetric portion *xcs*(*t*) plus a conjugate-antisymmetric portion *xca*(*t*). That is, - -$$ -x(t) = x_{cs}(t) + x_{ca}(t) -$$ - -where - -$$ -x_{cs}(t) = \frac{x(t) + x^*(-t)}{2} -$$ - and $x_{ca}(t) = \frac{x(t) - x^*(-t)}{2}$ - -The proof is similar to the one for decomposing a signal into even and odd components. As we shall see in later chapters, conjugate symmetries commonly occur in real-world signals and their transforms. - -## **[1.6 SYSTEMS](#page-7-0)** - -As mentioned in Sec. 1.1, systems are used to process signals to allow modification or extraction of additional information from the signals. A system may consist of physical components (hardware realization) or of an algorithm that computes the output signal from the input signal (software realization). - -Roughly speaking, a physical system consists of interconnected components, which are characterized by their terminal (input–output) relationships. In addition, a system is governed by laws of interconnection. For example, in electrical systems, the terminal relationships are the familiar voltage-current relationships for the resistors, capacitors, inductors, transformers, transistors, and so on, as well as the laws of interconnection (i.e., Kirchhoff's laws). We use these laws to derive mathematical equations relating the outputs to the inputs. These equations then represent a *mathematical model* of the system. - -A system can be conveniently illustrated by a "black box" with one set of accessible terminals where the input variables *x*1(*t*), *x*2(*t*), ..., *xj*(*t*) are applied and another set of accessible terminals where the output variables *y*1(*t*), *y*2(*t*),..., *yk*(*t*) are observed (Fig. 1.25). - -The study of systems consists of three major areas: mathematical modeling, analysis, and design. Although we shall be dealing with mathematical modeling, our main concern is with - -| x1(t) | y1(t) | | -|--------|--------|-----------------------------------------| -| x2(t) | y2(t) | | -| •
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• | | -| xj(t) | yk(t) | Figure 1.25 Representation of a system. | - -analysis and design. The major portion of this book is devoted to the analysis problem—how to determine the system outputs for the given inputs and a given mathematical model of the system (or rules governing the system). To a lesser extent, we will also consider the problem of design or synthesis—how to construct a system that will produce a desired set of outputs for the given inputs. - -### DATA NEEDED TO COMPUTE SYSTEM RESPONSE - -To understand what data we need to compute a system response, consider a simple *RC* circuit with a current source *x*(*t*) as its input (Fig. 1.26). - -The output voltage *y*(*t*) is given by - -$$ -y(t) = Rx(t) + \frac{1}{C} \int_{-\infty}^{t} x(\tau) d\tau -$$ -\n(1.18) - -The limits of the integral on the right-hand side are from −∞ to *t* because this integral represents the capacitor charge due to the current *x*(*t*) flowing in the capacitor, and this charge is the result of the current flowing in the capacitor from −∞. Now, Eq. (1.18) can be expressed as - -$$ -y(t) = Rx(t) + \frac{1}{C} \int_{-\infty}^{0} x(\tau) d\tau + \frac{1}{C} \int_{0}^{t} x(\tau) d\tau -$$ - -The middle term on the right-hand side is *vC*(0), the capacitor voltage at *t* = 0. Therefore, - -$$ -y(t) = v_C(0) + Rx(t) + \frac{1}{C} \int_0^t x(\tau) d\tau \qquad t \ge 0 -$$ - -This equation can be readily generalized as - -$$ -y(t) = v_C(t_0) + Rx(t) + \frac{1}{C} \int_{t_0}^t x(\tau) d\tau \qquad t \ge t_0 -$$ -\n(1.19) - -From Eq. (1.18), the output voltage *y*(*t*) at an instant *t* can be computed if we know the input current flowing in the capacitor throughout its entire past (−∞ to *t*). Alternatively, if we know the input current *x*(*t*) from some moment *t*0 onward, then, using Eq. (1.19), we can still calculate *y*(*t*) for *t* ≥ *t*0 from a knowledge of the input current, provided we know *vC*(*t*0), the initial capacitor voltage (voltage at *t*0). Thus *vC*(*t*0) contains all the relevant information about the circuit's entire - -**Figure 1.26** Example of a simple electrical system. - -past (−∞ to *t*0) that we need to compute *y*(*t*) for *t* ≥ *t*0. Therefore, the response of a system at *t* ≥ *t*0 can be determined from its input(s) during the interval *t*0 to *t* and from certain *initial conditions* at *t* = *t*0. - -In the preceding example, we needed only one initial condition. However, in more complex systems, several initial conditions may be necessary. We know, for example, that in passive *RLC* networks, the initial values of all inductor currents and all capacitor voltages† are needed to determine the outputs at any instant *t* ≥ 0 if the inputs are given over the interval [0,*t*]. - -## **[1.7 CLASSIFICATION OF](#page-7-0) SYSTEMS** - -Systems may be classified broadly in the following categories: - -- 1. Linear and nonlinear systems -- 2. Constant-parameter and time-varying-parameter systems -- 3. Instantaneous (memoryless) and dynamic (with memory) systems -- 4. Causal and noncausal systems -- 5. Continuous-time and discrete-time systems -- 6. Analog and digital systems -- 7. Invertible and noninvertible systems -- 8. Stable and unstable systems - -Other classifications, such as deterministic and probabilistic systems, are beyond the scope of this text and are not considered. - -### **[1.7-1 Linear and Nonlinear Systems](#page-7-0)** - -### THE CONCEPT OF LINEARITY - -A system whose output is proportional to its input is an *example* of a linear system. But linearity implies more than this; it also implies the *additivity property:* that is, if several inputs are acting on a system, then the total effect on the system due to all these inputs can be determined by considering one input at a time while assuming all the other inputs to be zero. The total effect is then the sum of all the component effects. This property may be expressed as follows: for a linear system, if an input *x*1 acting alone has an effect *y*1, and if another input *x*2, also acting alone, has an effect *y*2, then, with both inputs acting on the system, the total effect will be *y*1 +*y*2. Thus, if - -$$ -x_1 \longrightarrow y_1 -$$ - and $x_2 \longrightarrow y_2$ - -then for all *x*1 and *x*2 - -*x*1 +*x*2 −→ *y*1 +*y*2 (1.20) - -In addition, a linear system must satisfy the *homogeneity* or scaling property, which states that for arbitrary real or imaginary number *k*, if an input is increased *k*-fold, the effect also increases *k*-fold. Thus, if - -*x* −→ *y* - -† Strictly speaking, this means independent inductor currents and capacitor voltages. - -### 98 CHAPTER 1 SIGNALS AND SYSTEMS - -then for all real or imaginary *k* - -$$ -kx \longrightarrow ky \tag{1.21} -$$ - -Thus, linearity implies two properties: homogeneity (scaling) and additivity.† Both these properties can be combined into one property (*superposition*), which is expressed as follows: If - -$$ -x_1 \longrightarrow y_1 -$$ - and $x_2 \longrightarrow y_2$ - -then for all inputs *x*1 and *x*2 and all constants *k*1 and *k*2, - -$$ -k_1x_1 + k_2x_2 \longrightarrow k_1y_1 + k_2y_2 \tag{1.22} -$$ - -There is another useful way to view the linearity condition described in Eq. (1.22): the response of a linear system is unchanged whether the operations of summing and scaling precede the system (sum and scale act on inputs) or follow the system (sum and scale act on outputs). *Thus, linearity implies commutability between a system and the operations of summing and scaling.* It may appear that additivity implies homogeneity. Unfortunately, homogeneity does not always follow from additivity. Drill 1.11 demonstrates such a case. - -### **DR ILL 1.11 Additivity but Not Homogeneity** - -Show that a system with the input *x*(*t*) and the output *y*(*t*) related by *y*(*t*) = Re{*x*(*t*)} satisfies the additivity property but violates the homogeneity property. Hence, such a system is not linear. [*Hint:* Show that Eq. (1.21) is not satisfied when *k* is complex.] - -### RESPONSE OF A LINEAR SYSTEM - -For the sake of simplicity, we discuss only *single-input, single-output* (*SISO*) systems. But the discussion can be readily extended to *multiple-input, multiple-output* (*MIMO*) systems. - -A system's output for *t* ≥ 0 is the result of two independent causes: the initial conditions of the system (or the system state) at *t* = 0 and the input *x*(*t*) for *t* ≥ 0. If a system is to be linear, the output must be the sum of the two components resulting from these two causes: first, the *zero-input response* (ZIR) that results only from the initial conditions at *t* = 0 with the input *x*(*t*) = 0 for *t* ≥ 0, and then the *zero-state response* (ZSR) that results only from the input *x*(*t*) for *t* ≥ 0 when the initial conditions (at *t* = 0) are assumed to be zero. When all the appropriate initial conditions are zero, the system is said to be in *zero state*. The system output is zero when the input is zero only if the system is in zero state. - -In summary, a linear system response can be expressed as the sum of the zero-input and zero-state responses: - -total response = zero-input response + zero-state response - -† A linear system must also satisfy the additional condition of *smoothness,* where small changes in the system's inputs must result in small changes in its outputs [3]. - -This property of linear systems, which permits the separation of an output into components resulting from the initial conditions and from the input, is called the *decomposition property*. For the *RC* circuit of Fig. 1.26, the response *y*(*t*) was found to be [see Eq. (1.19) with *t*0 = 0] - -$$ -y(t) = \underbrace{v_C(0)}_{\text{ZIR}} + \underbrace{Rx(t) + \frac{1}{C} \int_0^t x(\tau) d\tau}_{\text{ZSR}} \tag{1.23} -$$ - -From Eq. (1.23), it is clear that if the input *x*(*t*) = 0 for *t* ≥ 0, the output *y*(*t*) = *vC*(0). Hence *vC*(0) is the zero-input response of the response *y*(*t*). Similarly, if the system state (the voltage *vC* in this case) is zero at *t* = 0, the output is given by the second component on the right-hand side of Eq. (1.23). Clearly this is the zero-state response of the response *y*(*t*). - -In addition to the decomposition property, linearity implies that both the zero-input and zero-state components must obey the principle of superposition with respect to each of their respective causes. For example, if we increase the initial condition *k*-fold, the zero-input response must also increase *k*-fold. Similarly, if we increase the input *k*-fold, the zero-state response must also increase *k*-fold. These facts can be readily verified from Eq. (1.23) for the *RC* circuit in Fig. 1.26. For instance, if we double the initial condition *vC*(0), the zero-input response doubles; if we double the input *x*(*t*), the zero-state response doubles. - -## **EXAMPLE 1.10 Linearity of Constant-Coefficient Linear Differential Equations** - -Show that the system described by the equation - -$$ -\frac{dy(t)}{dt} + 3y(t) = x(t) -$$ - (1.24) - -is linear. - -Let the system response to the inputs *x*1(*t*) and *x*2(*t*) be *y*1(*t*) and *y*2(*t*), respectively. Then - -$$ -\frac{dy_1(t)}{dt} + 3y_1(t) = x_1(t) \qquad \text{and} \qquad \frac{dy_2(t)}{dt} + 3y_2(t) = x_2(t) -$$ - -Multiplying the first equation by *k*1, the second by *k*2, and adding them yield - -$$ -\frac{d}{dt}[k_1y_1(t) + k_2y_2(t)] + 3[k_1y_1(t) + k_2y_2(t)] = k_1x_1(t) + k_2x_2(t) -$$ - -But this equation is the system equation [Eq. (1.24)] with - -*x*(*t*) = *k*1*x*1(*t*)+*k*2*x*2(*t*) and *y*(*t*) = *k*1*y*1(*t*)+*k*2*y*2(*t*) - -Therefore, when the input is *k*1*x*1(*t*) + *k*2*x*2(*t*), the system response is *k*1*y*1(*t*) + *k*2*y*2(*t*). Consequently, the system is linear. Using this argument, we can readily generalize the result to - -### 100 CHAPTER 1 SIGNALS AND SYSTEMS - -show that a system described by a differential equation of the form - -$$ -a_0 \frac{d^N y(t)}{dt^N} + a_1 \frac{d^{N-1} y(t)}{dt^{N-1}} + \dots + a_N y(t) = b_{N-M} \frac{d^M x(t)}{dt^M} + \dots + b_{N-1} \frac{dx(t)}{dt} + b_N x(t) \quad (1.25) -$$ - -is a linear system. The coefficients *ai* and *bi* in this equation can be constants or functions of time. Although here we proved only zero-state linearity, it can be shown that such systems are also zero-input linear and have the decomposition property. - -## **DR ILL 1.12 Linearity of a Differential Equation with Time-Varying Parameters** - -Show that the system described by the following equation is linear: - -$$ -\frac{dy(t)}{dt} + t^2 y(t) = (2t + 3)x(t) -$$ - -### **DR ILL 1.13 A Nonlinear Differential Equation** - -Show that the system described by the following equation is nonlinear: - -$$ -y(t)\frac{dy(t)}{dt} + 3y(t) = x(t) -$$ - -### MORE COMMENTS ON LINEAR SYSTEMS - -Almost all systems observed in practice become nonlinear when large enough signals are applied to them. However, it is possible to approximate most of the nonlinear systems by linear systems for small-signal analysis. The analysis of nonlinear systems is generally difficult. Nonlinearities can arise in so many ways that describing them with a common mathematical form is impossible. Not only is each system a category in itself, but even for a given system, changes in initial conditions or input amplitudes may change the nature of the problem. On the other hand, the superposition property of linear systems is a powerful unifying principle that allows for a general solution. The superposition property (linearity) greatly simplifies the analysis of linear systems. Because of the decomposition property, we can evaluate separately the two components of the output. The zero-input response can be computed by assuming the input to be zero, and the zero-state response can be computed by assuming zero initial conditions. Moreover, if we express an input *x*(*t*) as a sum of simpler functions, - -$$ -x(t) = a_1 x_1(t) + a_2 x_2(t) + \cdots + a_m x_m(t) -$$ - -then, by virtue of linearity, the response *y*(*t*) is given by - -$$ -y(t) = a_1 y_1(t) + a_2 y_2(t) + \cdots + a_m y_m(t) -$$ - -where *yk*(*t*) is the zero-state response to an input *xk*(*t*). This apparently trivial observation has profound implications. As we shall see repeatedly in later chapters, it proves extremely useful and opens new avenues for analyzing linear systems. - -For example, consider an arbitrary input *x*(*t*) such as the one shown in Fig. 1.27a. We can approximate *x*(*t*) with a sum of rectangular pulses of width *t* and of varying heights. The approximation improves as *t* → 0, when the rectangular pulses become impulses spaced *t* seconds apart (with *t* → 0).† Thus, an arbitrary input can be replaced by a weighted sum of impulses spaced *t* (*t* → 0) seconds apart. Therefore, if we know the system response to a unit impulse, we can immediately determine the system response to an arbitrary input *x*(*t*) by adding the system response to each impulse component of *x*(*t*). A similar situation is depicted in Fig. 1.27b, where *x*(*t*) is approximated by a sum of step functions of varying magnitude and spaced *t* seconds apart. The approximation improves as *t* becomes smaller. Therefore, if we know the system response to a unit step input, we can compute the system response to any arbitrary input *x*(*t*) with relative ease. Time-domain analysis of linear systems (discussed in Ch. 2) uses this approach. - -Chapters 4, 5, 6, and 7 employ the same approach but instead use sinusoids or exponentials as the basic signal components. We show that any arbitrary input signal can be expressed as a weighted sum of sinusoids (or exponentials) having various frequencies. Thus a knowledge of the system response to a sinusoid enables us to determine the system response to an arbitrary input *x*(*t*). - -**Figure 1.27** Signal representation in terms of impulse and step components. - -† Here, the discussion of a rectangular pulse approaching an impulse at *t* → 0 is somewhat imprecise. It is explained in Sec. 2.4 with more rigor. - -### **[1.7-2 Time-Invariant and Time-Varying Systems](#page-7-0)** - -Systems whose parameters do not change with time are *time-invariant* (also *constant-parameter*) systems. For such a system, if the input is delayed by *T* seconds, the output is the same as before but delayed by *T* (assuming initial conditions are also delayed by *T*). This property is expressed graphically in Fig. 1.28. We can also illustrate this property, as shown in Fig. 1.29. We can delay the output *y*(*t*) of a system *S* by applying the output *y*(*t*) to a *T* second delay (Fig. 1.29a). If the system is time invariant, then the delayed output *y*(*t*−*T*) can also be obtained by first delaying the input *x*(*t*) before applying it to the system, as shown in Fig. 1.29b. In other words, the system *S* and the time delay commute if the system *S* is time invariant. This would not be true for time-varying systems. Consider, for instance, a time-varying system specified by *y*(*t*) = *e*−*t x*(*t*). The output for such a system in Fig. 1.29a is *e*−(*t*−*T*) *x*(*t* − *T*). In contrast, the output for the system in Fig. 1.29b is *e*−*t x*(*t* −*T*). - -**Figure 1.28** Time-invariance property. - -**Figure 1.29** Illustration of timeinvariance property. - -It is possible to verify that the system in Fig. 1.26 is a time-invariant system. Networks composed of *RLC* elements and other commonly used active elements such as transistors are time-invariant systems. A system with an input–output relationship described by a linear differential equation of the form given in Ex. 1.10 [Eq. (1.25)] is a linear time-invariant (LTI) system when the coefficients *ai* and *bi* of such equation are constants. If these coefficients are functions of time, then the system is a linear *time-varying* system. - -The system described in Drill 1.12 is linear time varying. Another familiar example of a time-varying system is the carbon microphone, in which the resistance *R* is a function of the mechanical pressure generated by sound waves on the carbon granules of the microphone. The output current from the microphone is thus modulated by the sound waves, as desired. - -### **EXAMPLE 1.11 Assessing System Time Invariance** - -Determine the time invariance of the following systems: **(a)** *y*(*t*)=*x*(*t*)*u*(*t*) and **(b)** *y*(*t*)= *d dt x*(*t*). - -**(a)** In this case, the output equals the input for *t* ≥ 0 and is otherwise zero. Clearly, the input is being modified by a time-dependent function, so the system is likely time variant. We can prove that the system is not time invariant through a counterexample. Letting *x*1(*t*) = δ(*t*+1), we see that *y*1(*t*) = 0. However, *x*2(*t*) = *x*1(*t*−2) = δ(*t*−1) produces an output of *y*2(*t*) = δ(*t* − 1), which does equal *y*1(*t* − 2) = 0 as time-invariance would require. Thus, *y*(*t*) = *x*(*t*)*u*(*t*) is a time variant system. - -**(b)** Although it appears that *x*(*t*) is being modified by a time-dependent function, this is not the case. The output of this system is simply the slope of the input. If the input is delayed, so too is the output. Applying input *x*(*t*) to the system produces output *y*(*t*) = *d dt x*(*t*); delaying this output by *T* produces *y*(*t* − *T*) = *d d*(*t*−*T*) *x*(*t* − *T*) = *d dt x*(*t* − *T*). This is just the output of the system to a delayed input *x*(*t* − *T*). Since the *T*-delayed output of the system to input *x*(*t*) equals the output of the system to the *T*-delayed input *x*(*t* −*T*), the system is time invariant. - -### **DR ILL 1.14 A Time-Variant System** - -Show that a system described by the following equation is a time-varying-parameter system: - -$$ -y(t) = (\sin t)x(t-2) -$$ - -[*Hint:* Show that the system fails to satisfy the time-invariance property.] - -### **[1.7-3 Instantaneous and Dynamic Systems](#page-7-0)** - -As observed earlier, a system's output at any instant *t* generally depends on the entire past input. However, in a special class of systems, the output at any instant *t* depends only on its input at that - -### 104 CHAPTER 1 SIGNALS AND SYSTEMS - -instant. In resistive networks, for example, any output of the network at some instant *t* depends only on the input at the instant *t*. In these systems, past history is irrelevant in determining the response. Such systems are said to be *instantaneous* or *memoryless* systems. More precisely, a system is said to be instantaneous (or memoryless) if its output at any instant *t* depends, at most, on the strength of its input(s) at the same instant *t*, and not on any past or future values of the input(s). Otherwise, the system is said to be *dynamic* (or a system with memory). A system whose response at *t* is completely determined by the input signals over the past *T* seconds [interval from (*t*−*T*) to *t*] is a *finite-memory system* with a memory of *T* seconds. Networks containing inductive and capacitive elements generally have infinite memory because the response of such networks at any instant *t* is determined by their inputs over the entire past (−∞,*t*). This is true for the *RC* circuit of Fig. 1.26. - -### **EXAMPLE 1.12 Assessing System Memory** - -Determine whether the following systems are memoryless: **(a)** *y*(*t* − 1) = 2*x*(*t* − 1), **(b)** *y*(*t*) = *d dt x*(*t*), and **(c)** *y*(*t*) = (*t* −1)*x*(*t*). - -**(a)** In this case, the output at time *t* −1 is just twice the input at the same time *t* −1. Since the output at a particular time depends only on the strength of the input at the same time, the system is memoryless. - -**(b)** Although it appears that the output *y*(*t*) at time *t* depends on the input *x*(*t*) at the same time *t*, we know that the slope (derivative) of *x*(*t*) cannot be determined solely from a single point. There must be some memory, even if infinitesimally small, involved. This is confirmed by using the fundamental theorem of calculus to express the system as - -$$ -y(t) = \lim_{T \to 0} \frac{x(t) - x(t - T)}{T} -$$ - -Since the output at a particular time depends on more than just the input at the same time, the system is not memoryless. - -**(c)** The output *y*(*t*) at time *t* is just the input *x*(*t*) at the same time *t* multiplied by the (time-dependent) coefficient *t* − 1. Since the output at a particular time depends only on the strength of the input at the same time, the system is memoryless. - -### **[1.7-4 Causal and Noncausal Systems](#page-7-0)** - -A *causal* (also known as a *physical* or *nonanticipative*) system is one for which the output at any instant *t*0 depends only on the value of the input *x*(*t*) for *t* ≤ *t*0. In other words, the value of the output at the present instant depends only on the past and present values of the input *x*(*t*), not on its future values. To put it simply, in a causal system the output cannot start before the input is applied. If the response starts before the input, it means that the system knows the input in the - -**Figure 1.30** Input–output of a noncausal system and the causal output achieved by delay. - -future and acts on this knowledge before the input is applied. A system that violates the condition of causality is called a *noncausal* (or *anticipative*) system. - -Any practical system that operates in real time† must necessarily be causal. We do not yet know how to build a system that can respond to future inputs (inputs not yet applied). A noncausal system is a prophetic system that knows the future input and acts on it in the present. Thus, if we apply an input starting at *t* = 0 to a noncausal system, the output would begin even before *t* = 0. For example, consider the system specified by - -$$ -y(t) = x(t-2) + x(t+2) -$$ -\n(1.26) - -For the input *x*(*t*) illustrated in Fig. 1.30a, the output *y*(*t*), as computed from Eq. (1.26) (shown in Fig. 1.30b), starts even before the input is applied. Equation (1.26) shows that *y*(*t*), the output at *t*, is given by the sum of the input values 2 seconds before and 2 seconds after *t* (at *t* − 2 and *t* + 2, respectively). But if we are operating the system in real time at *t*, we do not know what the value of the input will be 2 seconds later. Thus it is impossible to implement this system in real time. For this reason, noncausal systems are unrealizable in *real time*. - -### **EXAMPLE 1.13 Assessing System Causality** - -Determine whether the following systems are causal: **(a)** *y*(*t*) = *x*(−*t*), **(b)** *y*(*t*) = *x*(*t* + 1), and **(c)** *y*(*t* +1) = *x*(*t*). - -† In real-time operations, the response to an input is essentially simultaneous (contemporaneous) with the input itself. - -**(a)** Here, the output is a reflection of the input. We can easily use a counterexample to disprove the causality of this system. The input *x*(*t*) = δ(*t* − 1), which is nonzero at *t* = 1, produces an output *y*(*t*) = δ(*t* + 1), which is nonzero at *t* = −1, a time 2 seconds earlier than the input! Clearly the system is not causal. - -**(b)** In this case, the output at time *t* depends on the input at future time of *t* + 1. Clearly the system is not causal. - -**(c)** In this case, the output at time *t* + 1 depends on the input one second in the past, at time *t*. Since the output does not depend on future values of the input, the system is causal. - -### WHY STUDY NONCAUSAL SYSTEMS? - -The foregoing discussion may suggest that noncausal systems have no practical purpose. This is not the case; they are valuable in the study of systems for several reasons. First, noncausal systems *are* realizable when the independent variable is other than "time" (e.g., *space*). Consider, for example, an electric charge of density *q*(*x*) placed along the *x* axis for *x* ≥ 0. This charge density produces an electric field *E*(*x*) that is present at every point on the *x* axis from *x* = −∞ to ∞. In this case the input [i.e., the charge density *q*(*x*)] starts at *x* = 0, but its output [the electric field *E*(*x*)] begins before *x* = 0. Clearly, this space-charge system is noncausal. This discussion shows that only temporal systems (systems with time as independent variable) must be causal to be realizable. The terms "before" and "after" have a special connection to causality only when the independent variable is time. This connection is lost for variables other than time. Nontemporal systems, such as those occurring in optics, can be noncausal and still realizable. - -Moreover, even for temporal systems, such as those used for signal processing, the study of noncausal systems is important. In such systems we may have all input data prerecorded. This often happens with speech, geophysical, and meteorological signals, and with space probes. In such cases, the input's future values are available to us. For example, suppose we had a set of input signal records available for the system described by Eq. (1.26). We can then compute *y*(*t*) since, for any *t*, we need only refer to the records to find the input's value 2 seconds before and 2 seconds after *t*. Thus, noncausal systems can be realized, although not in real time. We may therefore be able to realize a noncausal system, provided we are willing to accept a time delay in the output. Consider a system whose output *y*ˆ(*t*) is the same as *y*(*t*) in Eq. (1.26) delayed by 2 seconds (Fig. 1.30c), so that - -$$ -\hat{y}(t) = y(t-2) = x(t-4) + x(t) -$$ - -Here the value of the output *y*ˆ at any instant *t* is the sum of the values of the input *x* at *t* and at the instant 4 seconds earlier [at (*t* − 4)]. In this case, the output at any instant *t* does not depend on future values of the input, and the system is causal. The output of this system, which is *y*ˆ(*t*), is identical to that in Eq. (1.26) or Fig. 1.30b except for a delay of 2 seconds. Thus, a noncausal system may be realized or satisfactorily approximated in real time by using a causal system with a delay. - -A third reason for studying noncausal systems is that they provide an upper bound on the performance of causal systems. For example, if we wish to design a filter for separating a signal from noise, then the optimum filter is invariably a noncausal system. Although unrealizable, this - -Noncausal systems are realizable with time delay! - -noncausal system's performance acts as the upper limit on what can be achieved and gives us a standard for evaluating the performance of causal filters. - -At first glance, noncausal systems may seem to be inscrutable. Actually, there is nothing mysterious about these systems and their approximate realization through physical systems with delay. If we want to know what will happen one year from now, we have two choices: go to a prophet (an unrealizable person) who can give the answers instantly, or go to a wise man and allow him a delay of one year to give us the answer! If the wise man is truly wise, he may even be able, by studying trends, to shrewdly guess the future very closely with a delay of less than a year. Such is the case with noncausal systems—nothing more and nothing less. - -### **DR ILL 1.15 A Noncausal System** - -Show that a system described by the following equation is noncausal: - -$$ -y(t) = \int_{t-5}^{t+5} x(\tau) d\tau -$$ - -Show that this system can be realized physically if we accept a delay of 5 seconds in the output. - -### **[1.7-5 Continuous-Time and Discrete-Time Systems](#page-7-0)** - -Signals defined or specified over a continuous range of time are *continuous-time signals,* denoted by symbols *x*(*t*), *y*(*t*), and so on. Systems whose inputs and outputs are continuous-time signals are *continuous-time systems*. On the other hand, signals defined only at discrete instants of time *t*0, *t*1, *t*2,...,*tn*,... are *discrete-time signals,* denoted by the symbols *x*(*tn*), *y*(*tn*), and so on, where *n* is some integer. Systems whose inputs and outputs are discrete-time signals are *discrete-time systems*. A digital computer is a familiar example of this type of system. In practice, discrete-time signals can arise from sampling continuous-time signals. For example, when the sampling is - -#### 108 CHAPTER 1 SIGNALS AND SYSTEMS - -uniform, the discrete instants *t*0, *t*1, *t*2, ... are uniformly spaced so that - -$$ -t_{k+1} - t_k = T \qquad \text{for all } k -$$ - -In such case, the discrete-time signals represented by the samples of continuous-time signals *x*(*t*), *y*(*t*), and so on can be expressed as *x*(*nT*), *y*(*nT*), and so on; for convenience, we further simplify this notation to *x*[*n*], *y*[*n*], ..., where it is understood that *x*[*n*] = *x*(*nT*) and that *n* is some integer. A typical discrete-time signal is shown in Fig. 1.31. A discrete-time signal may also be viewed as a sequence of numbers ..., *x*[−1], *x*[0], *x*[1], *x*[2], .... Thus, a discrete-time system may be seen as processing a sequence of numbers *x*[*n*] and yielding as an output another sequence of numbers *y*[*n*]. - -Discrete-time signals arise naturally in situations that are inherently discrete time, such as population studies, amortization problems, national income models, and radar tracking. They may also arise as a result of sampling continuous-time signals in sampled data systems, digital filtering, and the like. Digital filtering is a particularly interesting application in which continuous-time signals are processed by using discrete-time systems, as shown in Fig. 1.32. A continuous-time signal *x*(*t*) is first sampled to convert it into a discrete-time signal *x*[*n*], which then is processed by the discrete-time system to yield a discrete-time output *y*[*n*]. A continuous-time signal *y*(*t*) is finally constructed from *y*[*n*]. In this manner, we can process a continuous-time signal with an appropriate discrete-time system such as a digital computer. Because discrete-time systems have several significant advantages over continuous-time systems, there is an accelerating trend toward processing continuous-time signals with discrete-time systems. - -**Figure 1.32** Processing continuous-time signals by discrete-time systems. - -### **[1.7-6 Analog and Digital Systems](#page-7-0)** - -Analog and digital signals are discussed in Sec. 1.3-2. A system whose input and output signals are analog is an *analog system;* a system whose input and output signals are digital is a *digital system*. A digital computer is an example of a digital (binary) system. Observe that a digital computer is a digital as well as a discrete-time system. - -### **[1.7-7 Invertible and Noninvertible Systems](#page-7-0)** - -A system *S* performs certain operation(s) on input signal(s). If we can obtain the input *x*(*t*) back from the corresponding output *y*(*t*) by some operation, the system *S* is said to be *invertible*. When several different inputs result in the same output (as in a rectifier), it is impossible to obtain the input from the output, and the system is *noninvertible*. Therefore, for an invertible system, it is essential that every input have a unique output so that there is a one-to-one mapping between an input and the corresponding output. The system that achieves the inverse operation [of obtaining *x*(*t*) from *y*(*t*)] is the *inverse system* for *S*. For instance, if *S* is an ideal integrator, then its inverse system is an ideal differentiator. Consider a system *S* connected in tandem with its inverse *Si*, as shown in Fig. 1.33. The input *x*(*t*) to this tandem system results in signal *y*(*t*) at the output of *S*, and the signal *y*(*t*), which now acts as an input to *Si*, yields back the signal *x*(*t*) at the output of *Si*. Thus, *Si* undoes the operation of *S* on *x*(*t*), yielding back *x*(*t*). A system whose output is equal to the input (for all possible inputs) is an *identity* system. Cascading a system with its inverse system, as shown in Fig. 1.33, results in an identity system. - -In contrast, a rectifier, specified by an equation *y*(*t*) = |*x*(*t*)|, is noninvertible because the rectification operation cannot be undone. - -Inverse systems are very important in signal processing. In many applications, the signals are distorted during the processing, and it is necessary to undo the distortion. For instance, in transmission of data over a communication channel, the signals are distorted owing to non-ideal frequency response and finite bandwidth of a channel. It is necessary to restore the signal as closely as possible to its original shape. Such equalization is also used in audio systems and photographic systems. - -inverse results in an identity system. - -### **EXAMPLE 1.14 Assessing System Invertibility** - -Determine whether the following systems are invertible: **(a)** *y*(*t*) = *x*(−*t*), **(b)** *y*(*t*) = *tx*(*t*), and **(c)** *y*(*t*) = *d dt x*(*t*). - -**(a)** Here, the output is a reflection of the input, which does not cause any loss to the input. The input can, in fact, be exactly recovered by simply reflecting the output [*x*(*t*) = *y*(−*t*)], which is to say that a reflecting system is its own inverse. Thus, *y*(*t*) = *x*(−*t*) is an invertible system. - -### 110 CHAPTER 1 SIGNALS AND SYSTEMS - -**(b)** In this case, one might be tempted to recover the input from the output as *x*(*t*) = 1 *t y*(*t*). This approach works almost everywhere, except at *t* = 0 where the input value *x*(0) cannot be recovered. Due to this single lost point, the system *y*(*t*) = *tx*(*t*) is not invertible. - -**(c)** Differentiation eliminates any dc component. For example, the inputs *x*1(*t*) = 1 and *x*2(*t*) = 2 both produce the same output *y*(*t*) = 0. Given only *y*(*t*) = 0, it is impossible to know if the original input was *x*1(*t*) = 1, *x*2(*t*) = 2, or something else entirely. Since unique inputs do produce unique outputs, we know that *y*(*t*) = *d dt x*(*t*) is not an invertible system. - -### **[1.7-8 Stable and Unstable Systems](#page-7-0)** - -Systems can also be classified as *stable* or *unstable* systems. Stability can be *internal* or *external*. If every *bounded input* applied at the input terminal results in a *bounded output,* the system is said to be stable *externally*. External stability can be ascertained by measurements at the external terminals (input and output) of the system. This type of stability is also known as the stability in the BIBO (bounded-input/bounded-output) sense. The concept of internal stability is postponed to Ch. 2 because it requires some understanding of internal system behavior, introduced in that chapter. - -### **EXAMPLE 1.15 Assessing System BIBO Stability** - -Determine whether the following systems are BIBO-stable: **(a)** *y*(*t*) = *x*2(*t*), **(b)** *y*(*t*) = *tx*(*t*), and **(c)** *y*(*t*) = *d dt x*(*t*). - -**(a)** This system squares an input to produce the output. If the input is bounded, which is to say that |*x*(*t*)| ≤ *Mx* < ∞ for all *t*, then we see that - -$$ -|y(t)| = |x^2(t)| = |x(t)|^2 \le M_x^2 < \infty -$$ - -Since the output amplitude is guaranteed to be bounded for any bounded-amplitude input, the system *y*(*t*) = *x*2(*t*) is BIBO-stable. - -**(b)** We can prove that *y*(*t*) = *tx*(*t*) is not BIBO-stable with a simple example. The bounded-amplitude input *x*(*t*) = *u*(*t*) produces the output *y*(*t*) = *tu*(*t*) whose amplitude grows to infinity as *t* → ∞. Thus, *y*(*t*) = *tx*(*t*) is a BIBO-unstable system. - -**(c)** We can prove that *y*(*t*) = *d dt x*(*t*) is not BIBO-stable with an example. The bounded-amplitude input *x*(*t*) = *u*(*t*) produces the output *y*(*t*) = δ(*t*) whose amplitude is infinite at *t* = 0. Thus, *y*(*t*) = *d dt x*(*t*) is a BIBO-unstable system. - -### **DR ILL 1.16 A Noninvertible BIBO-Stable System** - -Show that a system described by the equation *y*(*t*) = *x*2(*t*) is noninvertible but BIBO-stable. - -## **1.8 SYSTEM [MODEL: INPUT–OUTPUT](#page-8-0) DESCRIPTION** - -A system description in terms of the measurements at the input and output terminals is called the *input–output description*. As mentioned earlier, systems theory encompasses a variety of systems, such as electrical, mechanical, hydraulic, acoustic, electromechanical, and chemical, as well as social, political, economic, and biological. The first step in analyzing any system is the construction of a system model, which is a mathematical expression or a rule that satisfactorily approximates the dynamical behavior of the system. In this chapter we shall consider only continuous-time systems. Modeling of discrete-time systems is discussed in Ch. 3. - -### **[1.8-1 Electrical Systems](#page-8-0)** - -To construct a system model, we must study the relationships between different variables in the system. In electrical systems, for example, we must determine a satisfactory model for the voltage-current relationship of each element, such as Ohm's law for a resistor. In addition, we must determine the various constraints on voltages and currents when several electrical elements are interconnected. These are the laws of interconnection—the well-known Kirchhoff laws for voltage and current (KVL and KCL). From all these equations, we eliminate unwanted variables to obtain equation(s) relating the desired output variable(s) to the input(s). The following examples demonstrate the procedure of deriving input–output relationships for some LTI electrical systems. - -### 112 CHAPTER 1 SIGNALS AND SYSTEMS - -By using the voltage-current laws of each element (inductor, resistor, and capacitor), we can express this equation as - -$$ -\frac{dy(t)}{dt} + 3y(t) + 2\int_{-\infty}^{t} y(\tau) d\tau = x(t) -$$ -\n(1.27) - -Differentiating both sides of this equation, we obtain - -$$ -\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt} -$$ -\n(1.28) - -This differential equation is the input–output relationship between the output *y*(*t*) and the input *x*(*t*). - -It proves convenient to use a compact notation *D* for the differential operator *d*/*dt*. This notation can be repeatedly applied. Thus, - -$$ -\frac{dy(t)}{dt} \equiv Dy(t), \qquad \frac{d^2y(t)}{dt^2} \equiv D^2y(t), \qquad \dots, \qquad \frac{d^Ny(t)}{dt^N} \equiv D^Ny(t) -$$ - -With this notation, Eq. (1.28) can be expressed as - -$$ -(D2 + 3D + 2)y(t) = Dx(t) -$$ -\n(1.29) - -The differential operator is the inverse of the integral operator, so we can use the operator 1/*D* to represent integration.† - -$$ -\int_{-\infty}^{t} y(\tau) d\tau \equiv \frac{1}{D} y(t) -$$ - -$$ -\frac{d}{dt} \left[ \int_{-\infty}^{t} y(\tau) d\tau \right] = y(t) -$$ - -† Use of operator 1/*D* for integration generates some subtle mathematical difficulties because the operators *D* and 1/*D* do not commute. For instance, we know that *D*(1/*D*) = 1 because - -However, (1/*D*)*D* is not necessarily unity. Use of Cramer's rule in solving simultaneous integro-differential equations will always result in cancellation of operators 1/*D* and *D*. This procedure may yield erroneous results when the factor *D* occurs in the numerator as well as in the denominator. This happens, for instance, in circuits with all-inductor loops or all-capacitor cut sets. To eliminate this problem, avoid the integral operation in system equations so that the resulting equations are differential rather than integro-differential. In electrical circuits, this can be done by using charge (instead of current) variables in loops containing capacitors and choosing current variables for loops without capacitors. In the literature this problem of commutativity of *D* and 1/*D* is largely ignored. As mentioned earlier, such a procedure gives erroneous results only in special systems, such as the circuits with all-inductor loops or all-capacitor cut sets. Fortunately such systems constitute a very small fraction of the systems we deal with. For further discussion of this topic and a correct method of handling problems involving integrals, see [4]. - -Consequently, Eq. (1.27) can be expressed as - -$$ -\left(D+3+\frac{2}{D}\right)y(t) = x(t) -$$ - -Multiplying both sides by *D* to differentiate the expression, we obtain - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -which is identical to Eq. (1.29). - -Recall that Eq. (1.29) is not an algebraic equation, and *D*2+3*D*+2 is not an algebraic term that multiplies *y*(*t*); it is an operator that operates on *y*(*t*). It means that we must perform the following operations on *y*(*t*): take the second derivative of *y*(*t*) and add to it 3 times the first derivative of *y*(*t*) and 2 times *y*(*t*). Clearly, a polynomial in *D* multiplied by *y*(*t*) represents a certain differential operation on *y*(*t*). - -### **EXAMPLE 1.17 Input–Output Equation of a Series** *RC* **Circuit** - -Using operator notation, find the equation relating input to output for the series *RC* circuit of Fig. 1.35 if the input is the voltage *x*(*t*) and output is - -- **(a)** the loop current *i*(*t*) -- **(b)** the capacitor voltage *y*(*t*) - -**(a)** The loop equation for the circuit is - -$$ -R i(t) + \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau = x(t) -$$ - -or - -$$ -15i(t) + 5 \int_{-\infty}^{t} i(\tau) d\tau = x(t) -$$ - -With operator notation, this equation can be expressed as - -$$ -15 i(t) + \frac{5}{D} i(t) = x(t) -$$ -\n(1.30) - -**(b)** Multiplying both sides of Eq. (1.30) by *D* (i.e., differentiating the equation), we obtain - -(15*D*+5)*i*(*t*) = *Dx*(*t*) - -Using the fact that *i*(*t*) = *C dy*(*t*) *dt* = 1 5*Dy*(*t*), simple substitution yields - -$$ -(3D+1)y(t) = x(t) -$$ -\n(1.31) - -## **DR ILL 1.17 Input–Output Equation of a Series** *RLC* **Circuit with Inductor Voltage as Output** - -If the inductor voltage *vL*(*t*) is taken as the output, show that the *RLC* circuit in Fig. 1.34 has an input–output equation of (*D*2 +3*D*+2)*vL*(*t*) = *D*2*x*(*t*). - -## **DR ILL 1.18 Input–Output Equation of a Series** *RC* **Circuit with Capacitor Voltage as Output** - -If the capacitor voltage *vC*(*t*) is taken as the output, show that the *RLC* circuit in Fig. 1.34 has an input–output equation of (*D*2 +3*D*+2)*vC*(*t*) = 2*x*(*t*). - -### **[1.8-2 Mechanical Systems](#page-8-0)** - -Planar motion can be resolved into translational (rectilinear) motion and rotational (torsional) motion. Translational motion will be considered first. We shall restrict ourselves to motions in one dimension. - -### TRANSLATIONAL SYSTEMS - -The basic elements used in modeling translational systems are ideal masses, linear springs, and dashpots providing viscous damping. The laws of various mechanical elements are now discussed. - -For a *mass M* (Fig. 1.36a), a force *x*(*t*) causes a motion *y*(*t*) and acceleration *y*¨(*t*). From Newton's law of motion, - -$$ -x(t) = M\ddot{y}(t) = M\frac{d^2y(t)}{dt^2} = MD^2y(t) -$$ - -The force *x*(*t*) required to stretch (or compress) a *linear spring* (Fig. 1.36b) by an amount *y*(*t*) is given by - -$$ -x(t) = Ky(t) -$$ - -where *K* is the *stiffness* of the spring. - -**Figure 1.36** Some elements in translational mechanical systems. - -For *a linear dashpot* (Fig. 1.36c), which operates by virtue of viscous friction, the force moving the dashpot is proportional to the relative velocity *y*˙(*t*) of one surface with respect to the other. Thus - -$$ -x(t) = B\dot{y}(t) = B\frac{dy(t)}{dt} = BDy(t) -$$ - -where *B* is the *damping coefficient* of the dashpot or the viscous friction. - -### **EXAMPLE 1.18 Input–Output Equation for a Translational Mechanical System** - -Find the input–output relationship for the translational mechanical system shown in Fig. 1.37a or its equivalent in Fig. 1.37b. The input is the force *x*(*t*), and the output is the mass position *y*(*t*). - -### 116 CHAPTER 1 SIGNALS AND SYSTEMS - -In mechanical systems it is helpful to draw a free-body diagram of each junction, which is a point at which two or more elements are connected. In Fig. 1.37, the point representing the mass is a junction. The displacement of the mass is denoted by *y*(*t*). The spring is also stretched by the amount *y*(*t*), and therefore it exerts a force −*Ky*(*t*) on the mass. The dashpot exerts a force −*By*˙(*t*) on the mass, as shown in the free-body diagram (Fig. 1.37c). By Newton's second law, the net force must be *My*¨(*t*). Therefore, - -$$ -M\ddot{y}(t) = -B\dot{y}(t) - Ky(t) + x(t) -$$ - -or - -$$ -(MD2 + BD + K)y(t) = x(t) -$$ - -### ROTATIONAL SYSTEMS - -In rotational systems, the motion of a body may be defined as its motion about a certain axis. The variables used to describe rotational motion are torque (in place of force), angular position (in place of linear position), angular velocity (in place of linear velocity), and angular acceleration (in place of linear acceleration). The system elements are *rotational mass* or *moment of inertia* (in place of mass) and *torsional springs* and *torsional dashpots* (in place of linear springs and dashpots). The terminal equations for these elements are analogous to the corresponding equations for translational elements. If *J* is the moment of inertia (or rotational mass) of a rotating body about a certain axis, then the external torque required for this motion is equal to *J* (rotational mass) times the angular acceleration. If θ (*t*) is the angular position of the body, θ (¨ *t*) is its angular acceleration, and - -torque = -$$ -J\ddot{\theta}(t) = J\frac{d^2\theta(t)}{dt^2} = JD^2\theta(t) -$$ - -Similarly, if *K* is the stiffness of a torsional spring (per unit angular twist), and θ is the angular displacement of one terminal of the spring with respect to the other, then - -torque = -$$ -K\theta(t) -$$ - -Finally, the torque due to viscous damping of a torsional dashpot with damping coefficient *B* is - -torque = -$$ -B\dot{\theta}(t) -$$ - = $BD\theta(t)$ - -### **EXAMPLE 1.19 Input–Output Equation for Aircraft Roll Angle** - -The attitude of an aircraft can be controlled by three sets of surfaces (shown shaded in Fig. 1.38): elevators, rudder, and ailerons. By manipulating these surfaces, one can set the aircraft on a desired flight path. The roll angle ϕ(*t*) can be controlled by deflecting in the opposite direction the two aileron surfaces as shown in Fig. 1.38. Assuming only rolling motion, find the equation relating the roll angle ϕ(*t*) to the input (deflection) θ (*t*). - -**Figure 1.38** Attitude control of an airplane. - -*J* - -The aileron surfaces generate a torque about the roll axis proportional to the aileron deflection angle θ (*t*). Let this torque be *c*θ (*t*), where *c* is the constant of proportionality. Air friction dissipates the torque *B*ϕ(˙ *t*). The torque available for rolling motion is then *c*θ (*t*) − *B*ϕ(˙ *t*). If *J* is the moment of inertia of the plane about the *x* axis (roll axis), then - -net torque = -$$ -J\ddot{\varphi}(t) = c\theta(t) - B\dot{\varphi}(t) -$$ - -and - -$$ -J\frac{d^2\varphi(t)}{dt^2} + B\frac{d\varphi(t)}{dt} = c\theta(t) \qquad \text{or} \qquad (JD^2 + BD)\varphi(t) = c\theta(t) -$$ - -This is the desired equation relating the output (roll angle ϕ(*t*)) to the input (aileron angle θ (*t*)). - -The roll velocity ω(*t*) is ϕ(˙ *t*). If the desired output is the roll velocity ω(*t*) rather than the roll angle ϕ(*t*), then the input–output equation would be - -$$ -\frac{d\omega(t)}{dt} + B\omega(t) = c\theta(t) \qquad \text{or} \qquad (JD + B)\omega(t) = c\theta(t) -$$ - -## **DR ILL 1.19 Input–Output Equation of a Rotational Mechanical System** - -Torque *T* (*t*) is applied to the rotational mechanical system shown in Fig. 1.39a. The torsional spring stiffness is *K*; the rotational mass (the cylinder's moment of inertia about the shaft) is *J*; the viscous damping coefficient between the cylinder and the ground is *B*. Find the equation relating the output angle θ (*t*) to the input torque *T* (*t*). [*Hint:* A free-body diagram is shown in Fig. 1.39b.] - -### **[1.8-3 Electromechanical Systems](#page-8-0)** - -A wide variety of electromechanical systems is used to convert electrical signals into mechanical motion (mechanical energy) and vice versa. Here we consider a rather simple example of an armature-controlled dc motor driven by a current source *x*(*t*), as shown in Fig. 1.40a. The torque *T* (*t*) generated in the motor is proportional to the armature current *x*(*t*). Therefore, - -$$ -\mathcal{T}(t) = K_T x(t) -$$ - -where *KT* is a constant of the motor. This torque drives a mechanical load whose free-body diagram is shown in Fig. 1.40b. The viscous damping (with coefficient *B*) dissipates a torque *B*θ (˙ *t*). If *J* is the moment of inertia of the load (including the rotor of the motor), then the net torque *T* (*t*)−*B*θ (˙ *t*) must be equal to *J*θ (¨ *t*): - -$$ -J\ddot{\theta}(t) = \mathcal{T}(t) - B\dot{\theta}(t) -$$ - -Thus, - -$$ -(JD2 + BD)\theta(t) = \mathcal{T}(t) = K_T x(t) -$$ - -which in conventional form can be expressed as - -$$ -J\frac{d^2\theta(t)}{dt^2} + B\frac{d\theta(t)}{dt} = K_T x(t) -$$ -\n(1.32) - -**Figure 1.40** Armature-controlled dc motor. - -## **1.9 INTERNAL AND EXTERNAL [DESCRIPTIONS OF A](#page-8-0) SYSTEM** - -The input–output relationship of a system is an *external description* of that system. We have found an external description (not the *internal description*) of systems in all the examples discussed so far. This may puzzle the reader because in each of these cases, we derived the input–output relationship by analyzing the internal structure of that system. Why is this not an internal description? What makes a description internal? Although it is true that we did find the input–output description by internal analysis of the system, we did so strictly for convenience. We could have obtained the input–output description by making observations at the external (input and output) terminals, for example, by measuring the output for certain inputs, such as an impulse or a sinusoid. A description that can be obtained from measurements at the external terminals (even when the rest of the system is sealed inside an inaccessible black box) is an external description. Clearly, the input–output description is an external description. What, then, is an internal description? An internal description is capable of providing complete information about all possible signals in the system. An external description may not give such complete information. An external description can always be found from an internal description, but the converse is not necessarily true. We shall now give an example to clarify the distinction between an external and an internal description. - -Let the circuit in Fig. 1.41a with the input *x*(*t*) and the output *y*(*t*) be enclosed inside a "black box" with only the input and the output terminals accessible. To determine its external description, let us apply a known voltage *x*(*t*) at the input terminals and measure the resulting output voltage *y*(*t*). - -Let us also assume that there is some initial charge *Q*0 present on the capacitor. The output voltage will generally depend on both, the input *x*(*t*) and the initial charge *Q*0. To compute the output resulting because of the charge *Q*0, assume the input *x*(*t*) = 0 (short across the input). In this case, the currents in the two 2 resistors in the upper and the lower branches at the output terminals are equal and opposite because of the balanced nature of the circuit. Clearly, the capacitor charge results in zero voltage at the output.† - -† The output voltage *y*(*t*) resulting because of the capacitor charge [assuming *x*(*t*) = 0] is the zero-input response, which, as argued above, is zero. The output component due to the input *x*(*t*) (assuming zero initial capacitor charge) is the zero-state response. Complete analysis of this problem is given later in Ex. 1.21. - -**Figure 1.41** A system that cannot be described by external measurements. - -Now, to compute the output *y*(*t*) resulting from the input voltage *x*(*t*), we assume zero initial capacitor charge (short across the capacitor terminals). The current *i*(*t*) (Fig. 1.41a), in this case, divides equally between the two parallel branches because the circuit is balanced. Thus, the voltage across the capacitor continues to remain zero. Therefore, for the purpose of computing the current *i*(*t*), the capacitor may be removed or replaced by a short. The resulting circuit is equivalent to that shown in Fig. 1.41b, which shows that the input *x*(*t*) sees a load of 5, and - -$$ -i(t) = \frac{1}{5}x(t) -$$ - -Also, because *y*(*t*) = 2*i*(*t*), - -$$ -y(t) = \frac{2}{5}x(t) -$$ - -This is the total response. Clearly, for the external description, the capacitor does not exist. No external measurement or external observation can detect the presence of the capacitor. Furthermore, if the circuit is enclosed inside a "black box" so that only the external terminals are accessible, it is impossible to determine the currents (or voltages) inside the circuit from external measurements or observations. An internal description, however, can provide every possible signal inside the system. In Ex. 1.21, we shall find the internal description of this system and show that it is capable of determining every possible signal in the system. - -For most systems, the external and internal descriptions are equivalent, but there are a few exceptions, as in the present case, where the external description gives an inadequate picture of the system. This happens when the system is *uncontrollable* and/or *unobservable*. - -Figure 1.42 shows structural representations of simple uncontrollable and unobservable systems. In Fig. 1.42a, we note that part of the system (subsystem *S*2) inside the box cannot be controlled by the input *x*(*t*). In Fig. 1.42b, some of the system outputs (those in subsystem *S*2) cannot be observed from the output terminals. If we try to describe either of these systems by applying an external input *x*(*t*) and then measuring the output *y*(*t*), the measurement will not characterize the complete system but only the part of the system (here *S*1) that is both controllable - -**Figure 1.42** Structures of uncontrollable and unobservable systems. - -and observable (linked to both the input and output). Such systems are undesirable in practice and should be avoided in any system design. The system in Fig. 1.41a can be shown to be neither controllable nor observable. It can be represented structurally as a combination of the systems in Figs. 1.42a and 1.42b. - -## **1.10 INTERNAL [DESCRIPTION: THE](#page-8-0) STATE-SPACE DESCRIPTION** - -We shall now introduce the *state-space* description of a linear system, which is an internal description of a system. In this approach, we identify certain key variables, called the *state variables,* of the system. These variables have the property that every possible signal in the system can be expressed as a linear combination of these state variables. For example, we can show that every possible signal in a passive *RLC* circuit can be expressed as a linear combination of independent capacitor voltages and inductor currents, which, therefore, are state variables for the circuit. - -To illustrate this point, consider the network in Fig. 1.43. We identify two state variables: the capacitor voltage *q*1 and the inductor current *q*2. If the values of *q*1, *q*2, and the input *x*(*t*) are known at some instant *t*, we can demonstrate that every possible signal (current or voltage) in the circuit can be determined at *t*. For example, if *q*1 = 10, *q*2 = 1, and the input *x* = 20 at some instant, the remaining voltages and currents at that instant will be - -$$ -i_1 = (x - q_1)/1 = 20 - 10 = 10 \text{A} -$$ - -\n -$$ -v_1 = x - q_1 = 20 - 10 = 10 \text{V} -$$ - -\n -$$ -v_2 = q_1 = 10 \text{V} -$$ - -\n -$$ -i_2 = q_1/2 = 5 \text{A} -$$ - -\n -$$ -i_C = i_1 - i_2 - q_2 = 10 - 5 - 1 = 4 \text{A} -$$ - -\n -$$ -i_3 = q_2 = 1 \text{A} -$$ - -\n -$$ -v_3 = 5q_2 = 5 \text{V} -$$ - -\n -$$ -v_L = q_1 - v_3 = 10 - 5 = 5 \text{V} -$$ - -\n(1.33) - -Thus all signals in this circuit are determined. Clearly, state variables consist of the *key variables* in a system; a knowledge of the state variables allows one to determine every possible output of the system. Note that the *state-variable description is an internal description* of a system because it is capable of describing all possible signals in the system. - -**Figure 1.43** Choosing suitable initial conditions in a network. - -### **EXAMPLE 1.20 State-Space Description of a System** - -This example illustrates how state equations may be natural and easier to determine than other descriptions, such as loop or node equations. Consider again the network in Fig. 1.43 with *q*1 and *q*2 as the state variables and write the state equations. - -This can be done by simple inspection of Fig. 1.43. Since *q*˙1 is the current through the capacitor, - -$$ -\dot{q}_1 = i_C = i_1 - i_2 - q_2 -$$ - -= $(x - q_1) - 0.5 q_1 - q_2$ -= $-1.5 q_1 - q_2 + x$ - -Also 2*q*˙2, the voltage across the inductor, is given by - -$$ -2\dot{q}_2 = q_1 - v_3 -$$ -$$ -= q_1 - 5q_2 -$$ - -or - -$$ -\dot{q}_2 = 0.5 q_1 - 2.5 q_2 -$$ - -Thus, the state equations are - -$$ -\dot{q}_1 = -1.5q_1 - q_2 + x \n\dot{q}_2 = 0.5q_1 - 2.5q_2 -$$ -\n(1.34) - -This is a set of two simultaneous first-order differential equations. This set of equations comprises the *state equations*. Once these equations have been solved for *q*1 and *q*2, everything else in the circuit can be determined by using Eq. (1.33), which are known as the *output equations*. Thus, in this approach, we have two sets of equations, the state equations and the output equations. Once we have solved the state equations, all possible outputs can be obtained - -from the output equations. In the input–output description, an *N*th-order system is described by an *N*th-order equation. In the state-variable approach, the same system is described by *N* simultaneous first-order state equations.† - -### **EXAMPLE 1.21 Controllability and Observability** - -Investigate the nature of state equations and the issue of controllability and observability for the circuit in Fig. 1.41a. - -This circuit has only one capacitor and no inductors. Hence, there is only one state variable, the capacitor voltage *q*(*t*). Since *C* = 1 F, the capacitor current is *q*˙. There are two sources in this circuit: the input *x*(*t*) and the capacitor voltage *q*(*t*). The response due to *x*(*t*), assuming *q*(*t*) = 0, is the zero-state response, which can be found from Fig. 1.44a, where we have shorted the capacitor [*q*(*t*) = 0]. The response due to *q*(*t*) assuming *x*(*t*) = 0, is the zero-input response, which can be found from Fig. 1.44b, where we have shorted *x*(*t*) to ensure *x*(*t*) = 0. It is now trivial to find both the components. - -Figure 1.44a shows zero-state currents in every branch. It is clear that the input *x*(*t*) sees an effective resistance of 5 , and, hence, the current through *x*(*t*) is *x*/5 A, which divides in the two parallel branches, resulting in the current *x*/10 through each branch. - -Examining the circuit in Fig. 1.44b for the zero-input response, we note that the capacitor voltage is *q* and the current is *q*˙. We also observe that the capacitor sees two loops in parallel, each with resistance 4 and current *q*˙/2. Interestingly, the 3 branch is effectively shorted because the circuit is balanced, and thus the voltage across the terminals *cd* is zero. The total current in any branch is the sum of the currents in that branch in Figs. 1.44a and 1.44b (principle of superposition). - -Branch - -\n -$$ -c = \frac{x}{10} + \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} + \frac{\dot{q}}{2}\right) -$$ -\n -$$ -cb = \frac{x}{10} - \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} - \frac{\dot{q}}{2}\right) -$$ -\n -$$ -ad = \frac{x}{10} - \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} - \frac{\dot{q}}{2}\right) -$$ -\n -$$ -bd = \frac{x}{10} + \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} + \frac{\dot{q}}{2}\right) -$$ -\n -$$ -ec = \frac{x}{5} \quad 3\left(\frac{x}{5}\right) -$$ -\n -$$ -ed = \frac{x}{5} \quad x -$$ -\n(1.35) - -† This assumes the system to be controllable and observable. If it is not, the input–output description equation will be of an order lower than the corresponding number of state equations. - -**Figure 1.44** Analysis of a system that is neither controllable nor observable. - -To find the state equation, we note that the current in branch *ca* is (*x*/10)+ ˙*q*/2 and the current in branch *cb* is (*x*/10)− ˙*q*/2. Hence, the equation around the loop *acba* is - -$$ -q = 2\left[-\frac{x}{10} - \frac{\dot{q}}{2}\right] + 2\left[\frac{x}{10} - \frac{\dot{q}}{2}\right] = -2\dot{q} -$$ - -$$ -\dot{q} = -0.5q -$$ - (1.36) - -or - -This is the desired state equation. - -Substitution of *q*˙ = −0.5*q* in Eq. (1.35) shows that every possible current and voltage in the circuit can be expressed in terms of the state variable *q* and the input *x*, as desired. Hence, the set of Eq. (1.35) is the output equation for this circuit. Once we have solved the state equation [Eq. (1.36)] for *q*, we can determine every possible output in the circuit. - -The output *y*(*t*) is given by - -$$ -y(t) = 2\left[\frac{x}{10} - \frac{\dot{q}}{2}\right] + 2\left[\frac{x}{10} + \frac{\dot{q}}{2}\right] = \frac{2}{5}x(t) -$$ - (1.37) - -A little examination of the state and the output equations indicates the nature of this system. Equation (1.36) shows that the state *q*(*t*) is independent of the input *x*(*t*); hence the system state *q* cannot be controlled by the input. Moreover, Eq. (1.37) shows that the output *y*(*t*) does not depend on the state *q*(*t*). Thus, the system state cannot be observed from the output terminals. Hence, the system is neither controllable nor observable. Such is not the case of other systems examined earlier. Consider, for example, the circuit in Fig. 1.43. The state equations [Eq. (1.34)] show that the states are influenced by the input directly or indirectly. Hence, the system is controllable. Moreover, as Eq. (1.33) shows, every possible output is expressed in terms of the state variables and the input. Hence, the states are also observable. - -State-space techniques are useful not just because of their ability to provide internal system description, but for several other reasons, including the following. - -- 1. State equations of a system provide a mathematical model of great generality that can describe not just linear systems, but also nonlinear systems; not just time-invariant systems, but also time-varying parameter systems; not just SISO (single-input/single-output) systems, but also multiple-input/multiple-output (MIMO) systems. Indeed, state equations are ideally suited for the analysis, synthesis, and optimization of MIMO systems. -- 2. Compact matrix notation and the powerful techniques of linear algebra greatly facilitate complex manipulations. Without such features, many important results of the modern system theory would have been difficult to obtain. State equations can yield a great deal of information about a system even when they are not solved explicitly. -- 3. State equations lend themselves readily to digital computer simulation of complex systems of high order, with or without nonlinearities, and with multiple inputs and outputs. -- 4. For second-order systems (*N* = 2), a graphical method called *phase-plane analysis* can be used on state equations, whether they are linear or nonlinear. - -The real benefits of the state-space approach, however, are realized for highly complex systems of large order. Much of the book is devoted to introduction of the basic concepts of linear systems analysis, which must necessarily begin with simpler systems without using the state-space approach. Chapter 10 deals with the state-space analysis of linear, time-invariant, continuous-time, and discrete-time systems. - -### **DR ILL 1.20 State Equations for a Series** *RLC* **Circuit** - -Write the state equations for the series *RLC* circuit shown in Fig. 1.45, using the inductor current *q*1(*t*) and the capacitor voltage *q*2(*t*) as state variables. Express every voltage and current in this circuit as a linear combination of *q*1, *q*2, and *x*. - -### **ANSWERS** - -*q*1 = −3*q*1 −*q*2 +*x* and *q*2 = 2*q*1. - -**Figure 1.45** Circuit for Drill 1.20. - -## **[1.11 MATLAB: WORKING WITH](#page-8-0) FUNCTIONS** - -Working with functions is fundamental to signals and systems applications. MATLAB provides several methods of defining and evaluating functions. An understanding and proficient use of these methods are therefore necessary and beneficial. - -### **[1.11-1 Anonymous Functions](#page-8-0)** - -Many simple functions are most conveniently represented by using MATLAB anonymous functions. An anonymous function provides a symbolic representation of a function defined in terms of MATLAB operators, functions, or other anonymous functions. For example, consider defining the exponentially damped sinusoid *f*(*t*) = *e*−*t* cos(2π*t*). - -``` ->> f = @(t) exp(-t).*cos(2*pi*t); -``` - -In this context, the @ symbol identifies the expression as an anonymous function, which is assigned a name of f. Parentheses following the @ symbol are used to identify the function's independent variables (input arguments), which in this case is the single time variable t. Input arguments, such as t, are local to the anonymous function and are not related to any workspace variables with the same names. - -Once defined, *f*(*t*) can be evaluated simply by passing the input values of interest. For example, - ->> t = 0; f(t) ans = 1 - -evaluates *f*(*t*) at *t* = 0, confirming the expected result of unity. The same result is obtained by passing *t* = 0 directly. - ->> f(0) ans = 1 - -Vector inputs allow the evaluation of multiple values simultaneously. Consider the task of plotting *f*(*t*) over the interval (−2 ≤ *t* ≤ 2). Gross function behavior is clear: *f*(*t*) should oscillate four times with a decaying envelope. Since accurate hand sketches are cumbersome, MATLAB-generated plots are an attractive alternative. As the following example illustrates, care must be taken to ensure reliable results. - -Suppose vector t is chosen to include only the integers contained in (−2 ≤ *t* ≤ 2), namely, [−2,−1, 0, 1, 2]. - ->> t = (-2:2); - -This vector input is evaluated to form a vector output. - ->> f(t) ans = 7.3891 2.7183 1.0000 0.3679 0.1353 The plot command graphs the result, which is shown in Fig. 1.46. - -``` ->> plot(t,f(t)); ->> xlabel('t'); ylabel('f(t)'); grid; -``` - -Grid lines, added by using the grid command, aid feature identification. Unfortunately, the plot does not illustrate the expected oscillatory behavior. More points are required to adequately represent *f*(*t*). - -The question, then, is how many points is enough?† If too few points are chosen, information is lost. If too many points are chosen, memory and time are wasted. A balance is needed. For oscillatory functions, plotting 20 to 200 points per oscillation is normally adequate. For the present case, t is chosen to give 100 points per oscillation. - ->> t = (-2:0.01:2); - -Again, the function is evaluated and plotted. - -**Figure 1.46** *f*(*t*) = *e*−*t* cos(2π*t*) for t = (-2:2). - -**Figure 1.47** *f*(*t*) = *e*−*t* cos(2π*t*) for t = (-2:0.01:2). - -† Sampling theory, presented later, formally addresses important aspects of this question. - ->> plot(t,f(t)); >> xlabel('t'); ylabel('f(t)'); grid; - -The result, shown in Fig. 1.47, is an accurate depiction of *f*(*t*). - -### **[1.11-2 Relational Operators and the Unit Step Function](#page-8-0)** - -The unit step function *u*(*t*) arises naturally in many practical situations. For example, a unit step can model the act of turning on a system. With the help of relational operators, anonymous functions can represent the unit step function. - -In MATLAB, a relational operator compares two items. If the comparison is true, a logical true (1) is returned. If the comparison is false, a logical false (0) is returned. Sometimes called indicator functions, relational operators indicates whether a condition is true. Six relational operators are available: <, >, <=, >=, ==, and ~=. - -The unit step function is readily defined using the >= relational operator. - ->> u = @(t) 1.0.\*(t>=0); - -Any function with a jump discontinuity, such as the unit step, is difficult to plot. Consider plotting *u*(*t*) by using t = (-2:2). - ->> t = (-2:2); plot(t,u(t)); >> xlabel('t'); ylabel('u(t)'); - -Two significant problems are apparent in the resulting plot, shown in Fig. 1.48. First, MATLAB automatically scales plot axes to tightly bound the data. In this case, this normally desirable feature obscures most of the plot. Second, MATLAB connects plot data with lines, making a true jump discontinuity difficult to achieve. The coarse resolution of vector t emphasizes the effect by showing an erroneous sloping line between *t* = −1 and *t* = 0. - -The first problem is corrected by vertically enlarging the bounding box with the axis command. The second problem is reduced, but not eliminated, by adding points to vector t. - -**Figure 1.48** *u*(*t*) for t = (-2:2). - -**Figure 1.49** *u*(*t*) for t = (-2:0.01:2) with axis modification. - ->> t = (-2:0.01:2); plot(t,u(t)); >> xlabel('t'); ylabel('u(t)'); >> axis([-2 2 -0.1 1.1]); - -The four-element vector argument of axis specifies *x* axis minimum, *x* axis maximum, *y* axis minimum, and *y* axis maximum, respectively. The improved results are shown in Fig. 1.49. - -Relational operators can be combined using logical AND, logical OR, and logical negation: &, |, and ~, respectively. For example, (t>0)&(t<1) and ~((t<=0)|(t>=1)) both test if 0 < *t* < 1. To demonstrate, consider defining and plotting the unit pulse *p*(*t*) = *u*(*t*) − *u*(*t* − 1), as shown in Fig. 1.50: - ->> p = @(t) 1.0.\*((t>=0)&(t<1)); >> t = (-1:0.01:2); plot(t,p(t)); >> xlabel('t'); ylabel('p(t) = u(t)-u(t-1)'); >> axis([-1 2 -.1 1.1]); - -Since anonymous functions can be constructed using other anonymous functions, we could have used our previously defined unit step anonymous function to define *p*(*t*) as p = @(t) u(t)-u(t-1);. - -**Figure 1.50** *p*(*t*) = *u*(*t*)−*u*(*t* −1) over (−1 ≤ *t* ≤ 2). - -### 130 CHAPTER 1 SIGNALS AND SYSTEMS - -For scalar operands, MATLAB also supports two short-circuit logical constructs. A short-circuit logical AND is performed by using &&, and a short-circuit logical OR is performed by using ||. Short-circuit logical operators are often more efficient than traditional logical operators because they test the second portion of the expression only when necessary. That is, when scalar expression A is found false in (A&&B), scalar expression B is not evaluated, since a false result is already guaranteed. Similarly, scalar expression B is not evaluated when scalar expression A is found true in (A||B), since a true result is already guaranteed. - -### **[1.11-3 Visualizing Operations on the Independent Variable](#page-8-0)** - -Two operations on a function's independent variable are commonly encountered: shifting and scaling. Anonymous functions are well suited to investigate both operations. - -Consider *g*(*t*) = *f*(*t*)*u*(*t*) = *e*−*t* cos(2π*t*)*u*(*t*), a causal version of *f*(*t*). MATLAB easily multiplies anonymous functions. Thus, we create *g*(*t*) by multiplying our anonymous functions for *f*(*t*) and *u*(*t*). † - ->> g = @(t) f(t).\*u(t); - -A combined shifting and scaling operation is represented by *g*(*at* + *b*), where *a* and *b* are arbitrary real constants. As an example, consider plotting *g*(2*t* +1) over (−2 ≤ *t* ≤ 2). With *a* = 2, the function is compressed by a factor of 2, resulting in twice the oscillations per unit *t*. Adding the condition *b* > 0 shifts the waveform to the left. Given anonymous function g, an accurate plot is nearly trivial to obtain. - -``` ->> t = (-2:0.01:2); ->> plot(t,g(2*t+1)); xlabel('t'); ylabel('g(2t+1)'); grid; -``` - -Figure 1.51 confirms the expected waveform compression and left shift. As a final check, realize that function *g*(·) turns on when the input argument is zero. Therefore, *g*(2*t* + 1) should turn on when 2*t* +1 = 0 or at *t* = −0.5, a fact again confirmed by Fig. 1.51. - -**Figure 1.51** *g*(2*t* +1) over (−2 ≤ *t* ≤ 2). - -† Although we define g in terms of f and u, the function g will not change if we later change either f or u unless we subsequently redefine g as well. - -**Figure 1.52** *g*(−*t* +1) over (−2 ≤ *t* ≤ 2). - -**Figure 1.53** *h*(*t*) = *g*(2*t* +1)+*g*(−*t* +1) over (−2 ≤ *t* ≤ 2). - -Next, consider plotting *g*(−*t* + 1) over (−2 ≤ *t* ≤ 2). Since *a* < 0, the waveform will be reflected. Adding the condition *b* > 0 shifts the final waveform to the right. - ->> plot(t,g(-t+1)); xlabel('t'); ylabel('g(-t+1)'); grid; - -Figure 1.52 confirms both the reflection and the right shift. - -Up to this point, Figs. 1.51 and 1.52 could be reasonably sketched by hand. Consider plotting the more complicated function *h*(*t*) = *g*(2*t* + 1) + *g*(−*t* + 1) over (−2 ≤ *t* ≤ 2) (Fig. 1.53); an accurate hand sketch would be quite difficult. With MATLAB, the work is much less burdensome. - ->> plot(t,g(2\*t+1)+g(-t+1)); xlabel('t'); ylabel('h(t)'); grid; - -### **[1.11-4 Numerical Integration and Estimating Signal Energy](#page-8-0)** - -Interesting signals often have nontrivial mathematical representations. Computing signal energy, which involves integrating the square of these expressions, can be a daunting task. Fortunately, many difficult integrals can be accurately estimated by means of numerical integration techniques. - -### 132 CHAPTER 1 SIGNALS AND SYSTEMS - -Even if the integration appears simple, numerical integration provides a good way to verify analytical results. - -To start, consider the simple signal *x*(*t*) = *e*−*t* (*u*(*t*)−*u*(*t*−1)). The energy of *x*(*t*) is expressed as *Ex* = \$ ∞ −∞ |*x*(*t*)| 2 *dt* = \$ 1 0 *e*−2*t dt*. Integrating yields *Ex* = 0.5(1 − *e*−2) ≈ 0.4323. The energy integral can also be evaluated numerically. Figure 1.27 helps illustrate the simple method of rectangular approximation: evaluate the integrand at points uniformly separated by *t*, multiply each by *t* to compute rectangle areas, and then sum over all rectangles. First, we create function *x*(*t*). - ->> x = @(t) exp(-t).\*((t>=0)&(t<1)); - -With *t* = 0.01, a suitable time vector is created. - ->> t = (0:0.01:1); - -The final result is computed by using the sum command. - ->> E\_x = sum(x(t).\*x(t)\*0.01) E\_x = 0.4367 - -The result is not perfect, but at 1% relative error it is close. By reducing *t*, the approximation is improved. For example, *t* = 0.001 yields E\_x = 0.4328, or 0.1% relative error. - -Although simple to visualize, rectangular approximation is not the best numerical integration technique. The MATLAB function quad implements a better numerical integration technique called recursive adaptive Simpson quadrature.† To operate, quad requires a function describing the integrand, the lower limit of integration, and the upper limit of integration. Notice that no *t* needs to be specified. - -To use quad to estimate *Ex*, the integrand must first be described. - ->> x\_squared = @(t) x(t).\*x(t); - -Estimating *Ex* immediately follows. - ->> E\_x = quad(x\_squared,0,1) E\_x = 0.4323 - -In this case, the relative error is −0.0026%. - -The same techniques can be used to estimate the energy of more complex signals. Consider *g*(*t*), defined previously. Energy is expressed as *Eg* = \$ ∞ 0 *e*−2*t* cos2 (2π*t*)*dt*. A closed-form solution exists, but it takes some effort. MATLAB provides an answer more quickly. - ->> g\_squared = @(t) g(t).\*g(t); - -† A comprehensive treatment of numerical integration is outside the scope of this text. Details of this particular method are not important for the current discussion; it is sufficient to say that it is better than the rectangular approximation. - -Although the upper limit of integration is infinity, the exponentially decaying envelope ensures *g*(*t*) is effectively zero well before *t* = 100. Thus, an upper limit of *t* = 100 is used along with *t* = 0.001. - ->> t = (0:0.001:100); >> E\_g = sum(g\_squared(t)\*0.001) E\_g = 0.2567 - -A slightly better approximation is obtained with the quad function. - ->> E\_g = quad(g\_squared,0,100) E\_g = 0.2562 - -### **DR ILL 1.21 Computing Signal Energy with MATLAB** - -Use MATLAB to confirm that the energy of signal *h*(*t*), defined previously as *h*(*t*) = *g*(2*t* + 1)+*g*(−*t* +1), is *Eh* = 0.3768. - -## **[1.12 SUMMARY](#page-8-0)** - -A *signal* is a set of data or information. A *system* processes input signals to modify them or extract additional information from them to produce output signals (response). A system may be made up of physical components (hardware realization), or it may be an algorithm that computes an output signal from an input signal (software realization). - -A convenient measure of the size of a signal is its energy, if it is finite. If the signal energy is infinite, the appropriate measure is its power, if it exists. The signal power is the time average of its energy (averaged over the entire time interval from −∞ to ∞). For periodic signals, the time averaging need be performed over only one period in view of the periodic repetition of the signal. Signal power is also equal to the mean squared value of the signal (averaged over the entire time interval from *t* = −∞ to ∞). - -Signals can be classified in several ways. - -- 1. A *continuous-time signal* is specified for a continuum of values of the independent variable (such as time *t*). A *discrete-time signal* is specified only at a finite or a countable set of time instants. -- 2. An *analog signal* is a signal whose amplitude can take on any value over a continuum. On the other hand, a signal whose amplitudes can take on only a finite number of values is a *digital signal*. The terms *discrete-time* and *continuous-time* qualify the nature of a signal along the time axis (horizontal axis). The terms *analog* and *digital,* on the other hand, qualify the nature of the signal amplitude (vertical axis). -- 3. A *periodic signal x*(*t*) is defined by the fact that *x*(*t*) = *x*(*t* +*T*0) for some *T*0. The smallest positive value of *T*0 for which this relationship is satisfied is called the *fundamental period*. A periodic signal remains unchanged when shifted by an integer multiple of its period. A periodic signal *x*(*t*) can be generated by a periodic extension of any contiguous segment of *x*(*t*) of duration *T*0. Finally, a periodic signal, by definition, must exist over the entire time interval −∞ < *t* < ∞. A signal is *aperiodic* if it is not periodic. - -- 4. An *everlasting signal* starts at *t* = −∞ and continues forever to *t* = ∞. Hence, periodic signals are everlasting signals. A *causal signal* is a signal that is zero for *t* < 0. -- 5. A signal with finite energy is an *energy signal*. Similarly a signal with a finite and nonzero power (mean-square value) is a *power signal*. A signal can be either an energy signal or a power signal, but not both. However, there are signals that are neither energy nor power signals. -- 6. A signal whose physical description is known completely in a mathematical or graphical form is a *deterministic signal*. A *random signal* is known only in terms of its probabilistic description such as mean value or mean-square value, rather than by its mathematical or graphical form. - -A signal *x*(*t*) delayed by *T* seconds (right-shifted) can be expressed as *x*(*t* − *T*); on the other hand, *x*(*t*) advanced by *T* (left-shifted) is *x*(*t* + *T*). A signal *x*(*t*) time-compressed by a factor *a*(*a* > 1) is expressed as *x*(*at*); on the other hand, the same signal time-expanded by factor *a*(*a* > 1) is *x*(*t*/*a*). The signal *x*(*t*) when time-reversed can be expressed as *x*(−*t*). - -The unit step function *u*(*t*) is very useful in representing causal signals and signals with different mathematical descriptions over different intervals. - -In the classical (Dirac) definition, the unit impulse function δ(*t*) is characterized by unit area and is concentrated at a single instant *t* = 0. The impulse function has a sampling (or sifting) property, which states that the area under the product of a function with a unit impulse is equal to the value of that function at the instant at which the impulse is located (assuming the function to be continuous at the impulse location). In the modern approach, the impulse function is viewed as a generalized function and is defined by the sampling property. - -The exponential function *est*, where *s* is complex, encompasses a large class of signals that includes a constant, a monotonic exponential, a sinusoid, and an exponentially varying sinusoid. - -A real signal that is symmetrical about the vertical axis (*t* = 0) is an *even* function of time, and a real signal that is antisymmetrical about the vertical axis is an *odd* function of time. The product of an even function and an odd function is an odd function. However, the product of an even function and an even function or an odd function and an odd function is an even function. The area under an odd function from *t* = −*a* to *a* is always zero regardless of the value of *a*. On the other hand, the area under an even function from *t* = −*a* to *a* is two times the area under the same function from *t* = 0 to *a* (or from *t* = −*a* to 0). Every signal can be expressed as a sum of odd and even functions of time. - -A system processes input signals to produce output signals (response). The input is the cause, and the output is its effect. In general, the output is affected by two causes: the internal conditions of the system (such as the initial conditions) and the external input. - -Systems can be classified in several ways. - -- 1. Linear systems are characterized by the linearity property, which implies superposition; if several causes (such as various inputs and initial conditions) are acting on a linear system, the total output (response) is the sum of the responses from each cause, assuming that all the remaining causes are absent. A system is nonlinear if superposition does not hold. -- 2. In time-invariant systems, system parameters do not change with time. The parameters of time-varying-parameter systems change with time. -- 3. For memoryless (or instantaneous) systems, the system response at any instant *t* depends only on the value of the input at *t*. For systems with memory (also known as dynamic - -systems), the system response at any instant *t* depends not only on the present value of the input, but also on the past values of the input (values before *t*). - -- 4. In contrast, if a system response at *t* also depends on the future values of the input (values of input beyond *t*), the system is noncausal. In causal systems, the response does not depend on the future values of the input. Because of the dependence of the response on the future values of input, the effect (response) of noncausal systems occurs before the cause. When the independent variable is time (temporal systems), the noncausal systems are prophetic systems, and therefore, unrealizable, although close approximation is possible with some time delay in the response. Noncausal systems with independent variables other than time (e.g., space) are realizable. -- 5. Systems whose inputs and outputs are continuous-time signals are continuous-time systems; systems whose inputs and outputs are discrete-time signals are discrete-time systems. If a continuous-time signal is sampled, the resulting signal is a discrete-time signal. We can process a continuous-time signal by processing the samples of the signal with a discrete-time system. -- 6. Systems whose inputs and outputs are analog signals are analog systems; those whose inputs and outputs are digital signals are digital systems. -- 7. If we can obtain the input *x*(*t*) back from the output *y*(*t*) of a system *S* by some operation, the system *S* is said to be invertible. Otherwise the system is noninvertible. -- 8. A system is stable if bounded input produces bounded output. This defines external stability because it can be ascertained from measurements at the external terminals of the system. External stability is also known as the stability in the BIBO (bounded-input/bounded-output) sense. Internal stability, discussed later in Ch. 2, is measured in terms of the internal behavior of the system. - -The system model derived from a knowledge of the internal structure of the system is its internal description. In contrast, an external description is a representation of a system as seen from its input and output terminals; it can be obtained by applying a known input and measuring the resulting output. In the majority of practical systems, an external description of a system so obtained is equivalent to its internal description. At times, however, the external description fails to describe the system adequately. Such is the case with the so-called uncontrollable or unobservable systems. - -A system may also be described in terms of certain set of key variables called state variables. In this description, an *N*th-order system can be characterized by a set of *N* simultaneous first-order differential equations in *N* state variables. State equations of a system represent an internal description of that system. - -### **[REFERENCES](#page-8-0)** - -- 1. Papoulis, A., *The Fourier Integral and Its Applications*. McGraw-Hill, New York, 1962. -- 2. Mason, S. J., *Electronic Circuits, Signals, and Systems*. Wiley, New York, 1960. -- 3. Kailath, T., *Linear Systems*. Prentice-Hall, Englewood Cliffs, NJ, 1980. -- 4. Lathi, B. P., *Signals and Systems*. Berkeley-Cambridge Press, Carmichael, CA, 1987. - -## **[PROBLEMS](#page-8-0)** - -- **1.1-1** Find the energies of the signals illustrated in Fig. P1.1-1. Comment on the effect on energy of sign change, time shifting, or doubling of the signal. What is the effect on the energy if the signal is multiplied by *k*? -- **1.1-2** Repeat Prob. 1.1-1 for the signals in Fig. P1.1-2. -- **1.1-3** (a) Find the energies of the pair of signals *x*(*t*) and *y*(*t*) depicted in Figs. P1.1-3a and P1.1-3b. Sketch and find the energies of signals *x*(*t*) + *y*(*t*) and *x*(*t*) − *y*(*t*). Can you make any observation from these results? - - (b) Repeat part (a) for the signal pair illustrated in Fig. P1.1-3c. Is your observation in part (a) still valid? - -- **1.1-4** Find the power of the periodic signal *x*(*t*) shown in Fig. P1.1-4. Find also the powers and the rms values of: - - (a) −*x*(*t*) - - (b) 2*x*(*t*) - - (c) *cx*(*t*) - - Comment. -- **1.1-5** By original design, a system outputs a 10-volt pulse that is 3 seconds in duration. It is desired to upgrade the square-pulse output with a "soft-start" pulse that steps up to 10 volts in 1-volt increments spaced every 20 milliseconds. Determine the signal duration *T* so that the "soft-start" pulse has the same signal energy as the original square pulse. - -**Figure P1.1-2** - --8 - -*t* 3 - -- **1.1-6** Determine the power and the rms value for each of the following signals: - - (a) 5+10 cos(100*t* +π/3) - - (b) 10 cos(100*t* +π/3)+16 sin(150*t* +π/5) - - (c) (10+2 sin 3*t*) cos 10*t* - - (d) 10 cos 5*t* cos 10*t* - - (e) 10 sin 5*t* cos 10*t* - - (f) *ej*α*t* cosω0*t* -- **1.1-7** Figure P1.1-7 shows a periodic 50% duty cycle dc-offset sawtooth wave *x*(*t*) with peak amplitude *A*. Determine the energy and power of *x*(*t*). -- **1.1-8** Two periodic signals that differ only by a 90-degree phase shift are considered to be quadrature signals. For example, cos(2π*t*) and sin(2π*t*) are quadrature signals. Another pair of quadrature signals is *x*(*t*) = sgn[cos(2π*t*)] and - -*y*(*t*) = sgn[sin(2π*t*)], where sgn is the sign (or signum) function. - -- (a) Plot *x*(*t*) and determine its power *Px* and energy *Ex*. -- (b) Plot *y*(*t*) and determine its power *Py* and energy *Ey*. -- (c) Consider the complex function *f*(*t*) = *x*(*t*)+ *jy*(*t*). Determine the power and energy of *f*(*t*). -- (d) When real functions *x*(*t*) and *y*(*t*) are combined as *f*(*t*) = *x*(*t*) + *jy*(*t*), is it generally true that *Ef* = *Ex* + *Ey* and *Pf* = *Px* + *Py*? Prove your answer. -- **1.1-9** There are many useful properties related to signal energy. Prove each of the following statements. In each case, let energy signal *x*1(*t*) have energy *E*[*x*1(*t*)], let energy signal *x*2(*t*) have - -**Figure P1.1-7** - -energy *E*[*x*2(*t*)], and let *T* be a nonzero, finite, real-valued constant. - -- (a) Prove *E*[*Tx*1(*t*)] = *T*2*E*[*x*1(*t*)]. That is, amplitude scaling a signal by constant *T* scales the signal energy by *T*2. -- (b) Prove *E*[*x*1(*t*)] = *E*[*x*1(*t* − *T*)]. That is, shifting a signal does not affect its energy. -- (c) If (*x*1(*t*) = 0) ⇒ (*x*2(*t*) = 0) and (*x*2(*t*) = 0) ⇒ (*x*1(*t*) = 0), then prove *E*[*x*1(*t*) + *x*2(*t*)] = *E*[*x*1(*t*)] + *E*[*x*2(*t*)]. That is, the energy of the sum of two nonoverlapping signals is the sum of the two individual energies. -- (d) Prove *E*[*x*1(*Tt*)] = (1/|*T*|)*E*[*x*1(*t*)]. That is, time-scaling a signal by *T* reciprocally scales the signal energy by 1/|*T*|. -- **1.1-10** Consider the signal *x*(*t*) shown in Fig. P1.1-10. Outside the interval shown, *x*(*t*) is zero. Determine the signal energy *E*[*x*(*t*)]. [*Hint:* Use the results of Prob. 1.1-9.] - -**Figure P1.1-10** - -**1.1-11** (a) Show that the power of a signal - -$$ -x(t) = \sum_{k=m}^{n} D_k e^{j\omega_k t} -$$ - -$$ -P_x = \sum_{k=m}^{n} |D_k|^2 -$$ - -assuming all frequencies to be distinct, that is, ω*i* = ω*k* for all *i* = *k*. - -- (b) Use the result in part (a) to determine the power of each of the signals in Prob. 1.1-6. -- **1.1-12** A binary signal *x*(*t*) = 0 for *t* < 0. For positive time, *x*(*t*) toggles between one and zero as follows: one for 1 second, zero for 1 second, one for 1 second, zero for 2 seconds, one for 1 second, zero for 3 seconds, and so forth. That is, the "on" time is always 1 second, but the "off" time successively increases by 1 second between each toggle. A portion of *x*(*t*) is shown in Fig. P1.1-12. Determine the energy and power of *x*(*t*). -- **1.2-1** For the signal *x*(*t*) depicted in Fig. P1.2-1, sketch the signals - - (a) *x*(−*t*) - -is - -- (b) *x*(*t* +6) -- (c) *x*(3*t*) -- (d) *x*(*t*/2) -- **1.2-2** For the signal *x*(*t*) illustrated in Fig. P1.2-2, sketch - - (a) *x*(*t* −4) - - (b) *x*(*t*/1.5) - - (c) *x*(−*t*) - - (d) *x*(2*t* −4) - - (e) *x*(2−*t*) -- **1.2-3** In Fig. P1.2-3, express signals *x*1(*t*), *x*2(*t*), *x*3(*t*), *x*4(*t*), and *x*5(*t*) in terms of signal *x*(*t*) and its time-shifted, time-scaled, or time-reversed versions. -- **1.2-4** For an energy signal *x*(*t*) with energy *Ex*, show that the energy of any one of the signals −*x*(*t*), *x*(−*t*), and *x*(*t* − *T*) is *Ex*. Show also that the energy of *x*(*at*) as well as *x*(*at* − *b*) is *Ex*/*a*, but the energy of *ax*(*t*) is *a*2*Ex*. This shows that time - -#### **Figure P1.2-3** - -inversion and time shifting do not affect signal energy. On the other hand, time compression of a signal (*a* > 1) reduces the energy, and time expansion of a signal (*a* < 1) increases the energy. What is the effect on signal energy if the signal is multiplied by a constant *a*? - -**1.2-5** Define 2*x*(−3*t* + 1) = *t*[*u*(−*t* − 1) − *u*(−*t* + 1)], where *u*(*t*) is the unit step function. (a) Plot 2*x*(−3*t* +1) over a suitable range of *t*. - -- (b) Plot *x*(*t*) over a suitable range of *t*. -- **1.2-6** Consider the signal *x*(*t*) = 2−*tu*(*t*) , where *u*(*t*) is the unit step function. - - (a) Accurately sketch *x*(*t*) over (−1 ≤ *t* ≤ 1). - - (b) Accurately sketch *y*(*t*) = 0.5*x*(1 − 2*t*) over (−1 ≤ *t* ≤ 1). -- **1.2-7** Define signals *y*(*t*) and *z*(*t*) as in Fig. P1.2-7. - - (a) Determine constants *a*, *b*, and *c* to produce *z*(*t*) = *ax*(*bt* +*c*) in Fig. P1.2-7. - -- **Figure P1.2-7** -- (b) Determine and sketch a signal *v*(*t*) such that *z*(*t*) = \$ *t* −∞ *v*(τ )*d*τ . -- **1.3-1** Think of a real-world signal that is a personally relevant and interesting. Describe the signal and then classify it according to the six following characteristics: - - (a) continuous-time or discrete-time - - (b) analog or digital - - (c) periodic or aperiodic - - (d) energy or power - - (e) causal or noncausal - - (f) deterministic or random - -If possible, think of a second real-world signal that has the opposite six characteristics of your first signal. If such a second signal is not possible, carefully explain why that is the case. - -**1.3-2** Define signal *y*(*t*) = %∞ *k*=−∞ *x*(0.5*t* − 10*k*), where - -$$ -x(t) = \begin{cases} e^{-2t} & t \ge 1 \\ 0 & t < 1 \end{cases} -$$ - -- (a) Determine the constant *a* such that the signal *x*(−2*t* +*a*) is borderline anticausal. -- (b) Is the signal *y*(*t*) periodic? If so, determine the period *Ty*. If not, explain why *y*(*t*) is not periodic. -- **1.3-3** Determine whether each of the following statements is true or false. If the statement is false, demonstrate this by proof or example. - - (a) Every continuous-time signal is an analog signal. - - (b) Every discrete-time signal is a digital signal. - - (c) If a signal is not an energy signal, then it must be a power signal and vice versa. - - (d) An energy signal must be of finite duration. - - (e) A power signal cannot be causal. - - (f) A periodic signal cannot be anticausal. -- **1.3-4** Determine whether each of the following statements is true or false. If the statement is - -false, demonstrate by proof or example why the statement is false. - -- (a) Every bounded periodic signal is a power signal. -- (b) Every bounded power signal is a periodic signal. -- (c) If an energy signal *x*(*t*) has energy *E*, then the energy of *x*(*at*) is *E*/*a*. Assume *a* is real and positive. -- (d) If a power signal *x*(*t*) has power *P*, then the power of *x*(*at*) is *P*/*a*. Assume *a* is real and positive. -- **1.3-5** Given *x*1(*t*) = cos(*t*), *x*2(*t*) = sin(π*t*), and *x*3(*t*) = *x*1(*t*)+*x*2(*t*). - - (a) Determine the fundamental periods *T*1 and *T*2 of signals *x*1(*t*) and *x*2(*t*). - - (b) Show that *x*3(*t*) is not periodic, which requires *T*3 = *k*1*T*1 = *k*2*T*2 for some integers *k*1 and *k*2. - - (c) Determine the powers *Px*1 , *Px*2 , and *Px*3 of signals *x*1(*t*), *x*2(*t*), and *x*3(*t*). -- **1.3-6** For any constant ω, is the function *f*(*t*) = sin(ω*t*) a periodic function of the independent variable *t*? Justify your answer. -- **1.3-7** The signal shown in Fig. P1.3-7 is defined as - -$$ -x(t) = \begin{cases} t & 0 \le t < 1 \\ 0.5 + 0.5 \cos(2\pi t) & 1 \le t < 2 \\ 3 - t & 2 \le t < 3 \\ 0 & \text{otherwise} \end{cases} -$$ - -The energy of *x*(*t*) is *E* ≈ 1.0417. - -- (a) What is the energy of *y*1(*t*) = (1/3)*x*(2*t*)? -- (b) A periodic signal *y*2(*t*) is defined as - -$$ -y_2(t) = \begin{cases} x(t) & 0 \le t < 4\\ y_2(t+4) & \forall t \end{cases} -$$ - -What is the power of *y*2(*t*)? - -(c) What is the power of *y*3(*t*) = (1/3)*y*2(2*t*)? - -**Figure P1.3-7** - -- **1.3-8** Let *y*1(*t*) = *y*2(*t*) = *t* 2 over 0 ≤ *t* ≤ 1. Notice, this statement does not require *y*1(*t*) = *y*2(*t*) for all *t*. - - (a) Define *y*1(*t*) as an even, periodic signal with period *T*1 = 2. Sketch *y*1(*t*) and determine its power. - - (b) Design an odd, periodic signal *y*2(*t*) with period *T*2 = 3 and power equal to unity. Fully describe *y*2(*t*) and sketch the signal over at least one full period. [*Hint:* There are an infinite number of possible solutions to this problem—you need to find only one of them!] - - (c) We can create a complex-valued function *y*3(*t*) = *y*1(*t*) + *jy*2(*t*). Determine whether this signal is periodic. If yes, determine the period *T*3. If no, justify why the signal is not periodic. - - (d) Determine the power of *y*3(*t*) defined in part (c). The power of a complex-valued function *z*(*t*) is - -$$ -P = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} z(\tau) z^*(\tau) d\tau -$$ - -- **1.4-1** Sketch the following signals: - - (a) *u*(*t* −5)−*u*(*t* −7) - - (b) *u*(*t* −5) +*u*(*t* −7) - - (c) *t* 2[*u*(*t* −1)−*u*(*t* −2)] - - (d) (*t* −4)[*u*(*t* −2) −*u*(*t* −4)] -- **1.4-2** Express each of the signals in Fig. P1.4-2 by a single expression valid for all *t*. - -**1.4-3** Letting *w*(*t*) = *t*[*u*(*t*)−*u*(*t* −1)], define the periodic signal *x*(*t*) as - -$$ -x(t) = \sum_{k=-\infty}^{\infty} w(2t + 2k) - 0.5w(2t + 2k - 1) -$$ - -- (a) Sketch *w*(*t*) and *x*(*t*). What is the fundamental period *T*0 of signal *x*(*t*)? -- (b) Sketch *y*(*t*) = *d dt x*(1−0.5*t*). -- (c) Determine the energy *Ez* and power *Pz* of the signal *z*(*t*) = *x*(0.5 − 1.5*t*)[*u*(*t*)−*u*(*t* −1)]. Sketching *z*(*t*) should help. -- **1.4-4** Define signal *x*(*t*) = *u*(*t*−1)−*u*(*t*−2.5)−2δ(*t*− 4)+δ(*t* −6). - - (a) Sketch *y*(*t*) = \$ *t* −∞ *x*(τ )*d*τ . - - (b) Describe a simple change that can be made to the right-most delta function in *x*(*t*) so that *y*(*t*) = \$ *t* −∞ *x*(τ )*d*τ has finite energy. - - (c) Sketch *z*(*t*) = \$ ∞ *t x*(τ )*d*τ . - - (d) Determine real constants *A* and *B* so that *w*(*t*) = *x t*−*A B* has a region of support [−2, 2]. -- **1.4-5** Simplify the following expressions: - -(a) -$$ -\left(\frac{\sin t}{t^2 + 2}\right) \delta(t) -$$ - -\n(b) $\left(\frac{j\omega + 2}{\omega^2 + 9}\right) \delta(\omega)$ - -(c) [*e*−*t* cos(3*t* −60◦)]δ(*t*) - -(d) -$$ -\left(\frac{\sin\left[\frac{\pi}{2}(t-2)\right]}{t^2+4}\right)\delta(1-t) -$$ - -(e) -$$ -\left(\frac{1}{j\omega+2}\right)\delta(\omega+3) -$$ - -(f) -$$ -\left(\frac{\sin k\omega}{\omega}\right)\delta(\omega) -$$ - -[*Hint:* Use Eq. (1.10). For part (f) use L'Hôpital's rule.] - -**1.4-6** Evaluate the following integrals: - -(a) -$$ -\int_{-\infty}^{\infty} \delta(\tau) x(t-\tau) d\tau -$$ - -\n(b) -$$ -\int_{-\infty}^{\infty} x(\tau) \delta(t-\tau) d\tau -$$ - -\n(c) -$$ -\int_{-\infty}^{\infty} \delta(t) e^{-j\omega t} dt -$$ - -\n(d) -$$ -\int_{-\infty}^{\infty} \delta(2t-3) \sin \pi t dt -$$ - -\n(e) -$$ -\int_{-\infty}^{\infty} \delta(t+3) e^{-t} dt -$$ - -\n(f) -$$ -\int_{-\infty}^{\infty} (t^3+4) \delta(1-t) dt -$$ - -\n(g) -$$ -\int_{-\infty}^{\infty} x(2-t) \delta(3-t) dt -$$ - -\n(h) -$$ -\int_{-\infty}^{\infty} e^{(x-1)} \cos \left[ \frac{\pi}{2} (x-5) \right] \delta(x-3) dx -$$ - -**1.4-7** For real and positive constant *a*, evaluate the following integral: - -$$ -\int_{-\infty}^{\infty} \delta(at) \, dt -$$ - -- **1.4-8** (a) Find and sketch *dx*/*dt* for the signal *x*(*t*) shown in Fig. P1.2-2. - - (b) Find and sketch *d*2*x*/*dt*2 for the signal *x*1(*t*) depicted in Fig. P1.4-2a. -- **1.4-9** Find and sketch \$ *t* −∞ *x*(*t*)*dt* for the signals *x*(*t*) illustrated in Fig. P1.4-9. -- **1.4-10** Using the generalized function definition of impulse [Eq. (1.11) with *T* = 0], show that δ(*t*) is an even function of *t*. -- **1.4-11** Using the generalized function definition of impulse [Eq. (1.11) with *T* = 0], show that - -$$ -\delta(at) = \frac{1}{|a|} \delta(t) -$$ - -**1.4-12** Show that - -**Figure P1.4-9** - -$$ -\int_{-\infty}^{\infty} \dot{\delta}(t)\phi(t) dt = -\dot{\phi}(0) -$$ - -where φ(*t*) and φ(˙ *t*) are continuous at *t* = 0, and φ(*t*) → 0 as *t* → ±∞. This integral defines δ(˙ *t*) as a generalized function. [*Hint:* Use integration by parts.] - -- **1.4-13** A sinusoid *e*σ*t* cos ω*t* can be expressed as a sum of exponentials *est* and *e*−*st* [Eq. (1.14)] with complex frequencies *s* = σ +*j*ω and *s* = σ −*j*ω. Locate in the complex plane the frequencies of the following sinusoids: - - (a) cos 3*t* - - (b) *e*−3*t* cos 3*t* - - (c) *e*2*t* cos 3*t* - - (d) *e*−2*t* - - (e) *e*2*t* - - (f) 5 -- **1.5-1** Find and sketch the odd and the even components of the following: - - (a) *u*(*t*) - - (b) *tu*(*t*) - - (c) sinω0*t* - - (d) cosω0*t* - - (e) cos(ω0*t* +θ ) - - (f) sinω0*tu*(*t*) - - (g) cosω0*tu*(*t*) -- **1.5-2** Define *x*(*t*) = 2*u*(*t*+1)−*u*(*t*−2)−*u*(*t*−3). - - (a) Letting *xo*(*t*) designate the odd portion of *x*(*t*), accurately sketch *xo*(1−2*t*). - -- (b) Letting *xe*(*t*) designate the even portion of *x*(*t*), accurately sketch *xe*(2+*t*/3). -- **1.5-3** (a) Determine even and odd components of the signal *x*(*t*) = *e*−2*t u*(*t*). - - (b) Show that the energy of *x*(*t*) is the sum of energies of its odd and even components found in part (a). - - (c) Generalize the result in part (b) for any finite energy signal. -- **1.5-4** (a) If *xe*(*t*) and *xo*(*t*) are even and the odd components of a real signal *x*(*t*), then show that - -$$ -\int_{-\infty}^{\infty} x_e(t)x_o(t) dt = 0 -$$ - -(b) Show that - -$$ -\int_{-\infty}^{\infty} x(t) dt = \int_{-\infty}^{\infty} x_e(t) dt -$$ - -- **1.5-5** An aperiodic signal is defined as *x*(*t*) = sin(π*t*)*u*(*t*), where *u*(*t*) is the continuous-time step function. Is the odd portion of this signal, *xo*(*t*), periodic? Justify your answer. -- **1.5-6** An aperiodic signal is defined as *x*(*t*) = cos(π*t*)*u*(*t*), where *u*(*t*) is the continuous-time step function. Is the even portion of this signal, *xe*(*t*), periodic? Justify your answer. -- **1.5-7** Consider the signal *x*(*t*) shown in Fig. P1.5-7. - -- (a) Determine and carefully sketch *v*(*t*) = 3*x*(−(1/2)(*t* +1)). -- (b) Determine the energy and power of *v*(*t*). -- (c) Determine and carefully sketch the even portion of *v*(*t*), *ve*(*t*). -- (d) Let *a* = 2 and *b* = 3; sketch *v*(*at* + *b*), *v*(*at*) +*b*, *av*(*t* +*b*), and *av*(*t*)+*b*. -- (e) Let *a* = −3 and *b* = −2; sketch *v*(*at* + *b*), *v*(*at*)+*b*, *av*(*t* +*b*), and *av*(*t*) +*b*. -- **1.5-8** Consider the signal *y*(*t*) = (1/5)*x*(−2*t* − 3) shown in Fig. P1.5-8. - -**Figure P1.5-8** - -- (a) Does *y*(*t*) have an odd portion, *yo*(*t*)? If so, determine and carefully sketch *yo*(*t*). Otherwise, explain why no odd portion exists. -- (b) Determine and carefully sketch the original signal *x*(*t*). -- **1.5-9** Consider the signal −(1/2)*x*(−3*t* + 2) shown in Fig. P1.5-9. - -### **Figure P1.5-9** - -- (a) Determine and carefully sketch the original signal *x*(*t*). -- (b) Determine and carefully sketch the even portion of the original signal *x*(*t*). -- (c) Determine and carefully sketch the odd portion of the original signal *x*(*t*). -- **1.5-10** The conjugate symmetric (or Hermitian) portion of a signal is defined as *wcs*(*t*) = (*w*(*t*) + *w*∗(−*t*))/2. Show that the real portion of *wcs*(*t*) is even and that the imaginary portion of *wcs*(*t*) is odd. -- **1.5-11** The conjugate antisymmetric (or skew-Hermitian) portion of a signal is defined as *wca*(*t*) = (*w*(*t*) − *w*∗(−*t*))/2. Show that the real portion of *wca*(*t*) is odd and that the imaginary portion of *wca*(*t*) is even. - -**1.5-12** Define *w*(*t*) = *ej*(*t*+π/4) . - -- (a) Referring to the definition in Prob. 1.5-10, determine *wcs*(*t*). Express your simplified answer in standard rectangular form. -- (b) Referring to the definition in Prob. 1.5-11, determine *wca*(*t*). Express your simplified answer in standard polar form. - -### 144 CHAPTER 1 SIGNALS AND SYSTEMS - -**1.5-13** Figure P1.5-13 plots a complex signal *w*(*t*) in the complex plane over the time range (0 ≤ *t* ≤ 1). The time *t* =0 corresponds with the origin, while the time *t* = 1 corresponds with the point (2, 1). - -**Figure P1.5-13** - -- (a) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is an even signal. -- (b) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is an odd signal. -- (c) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is a conjugate symmetric signal. [*Hint:* See Prob. 1.5-10.] -- (d) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is a conjugate antisymmetric signal. [*Hint:* See Prob. 1.5-11.] -- (e) In the complex plane, plot as much of *w*(3*t*) as possible. -- **1.5-14** Define complex signal *x*(*t*) = *t* 2(1 + *j*) over interval (1 ≤ *t* ≤ 2). The remaining portion is defined such that *x*(*t*) is a minimum-energy, skew-Hermitian signal. - - (a) Fully describe *x*(*t*) for all *t*. - - (b) Sketch *y*(*t*) = Re{*x*(*t*)} versus the independent variable *t*. - - (c) Sketch *z*(*t*) = Re{*jx*(−2*t* + 1)} versus the independent variable *t*. - - (d) Determine the energy and power of *x*(*t*). - -[*Hint:* See Prob. 1.5-11 for a definition of skew-Hermitian signals.] - -- **1.6-1** Write the input–output relationship for an ideal integrator. Determine the zero-input and zero-state components of the response. -- **1.6-2** A force *x*(*t*) acts on a ball of mass *M* (Fig. P1.6-2). Show that the velocity *v*(*t*) of the ball at any instant *t* > 0 can be determined if we know the force *x*(*t*) over the interval from 0 to *t* and the ball's initial velocity *v*(0). - -### **Figure P1.6-2** - -- **1.6-3** From your personal experience, provide an example of: - - (a) a single-input, single-output (SISO) system - - (b) a multiple-input, single-output (MISO) system - - (c) a single-input, multiple-output (SIMO) system - - (d) a multiple-input, multiple-output (MIMO) system -- **1.7-1** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), determine which of the systems are linear and which are nonlinear. - -(a) -$$ -\frac{dy(t)}{dt} + 2y(t) = x^2(t) -$$ - -\n(b) -$$ -\frac{dy(t)}{dt} + 3ty(t) = t^2x(t) -$$ - -\n(c) -$$ -3y(t) + 2 = x(t) -$$ - -\n(d) -$$ -\frac{dy(t)}{dt} + y^2(t) = x(t) -$$ - -\n(e) -$$ -\left(\frac{dy(t)}{dt}\right)^2 + 2y(t) = x(t) -$$ - -\n(f) -$$ -\frac{dy(t)}{dt} + (\sin t)y(t) = \frac{dx(t)}{dt} + 2x(t) -$$ - -(g) -$$ -\frac{dy(t)}{dt} + 2y(t) = x(t) \frac{dx(t)}{dt} -$$ - -(h) -$$ -y(t) = \int_0^t x(\tau) d\tau -$$ - -−∞ **1.7-2** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), explain with reasons which of the systems are time-invariant parameter systems and which are time-varying-parameter systems. - -(a) -$$ -y(t) = x(t-2) -$$ - -\n(b) $y(t) = x(-t)$ -\n(c) $y(t) = x(at)$ -\n(d) $y(t) = tx(t-2)$ -\n(e) $y(t) = \int_{-5}^{5} x(\tau) d\tau$ -\n(f) $y(t) = \left(\frac{dx(t)}{dt}\right)^2$ - -- **1.7-3** Two inputs, temperature *T*(*t*) and wind speed *V*(*t*), produce an output, wind chill *W*(*t*), according to *W*(*t*) = 35.74 + 0.6215*T*(*t*) − 35.75{*V*(*t*)} 0.16 + 0.4275*T*(*t*){*V*(*t*)} 0.16. The independent variable here is time, *t*. Answer the following questions yes or no, and provide mathematical justification for each answer. - - (a) Is this system BIBO-stable? - - (b) Is the system memoryless? - - (c) Is the system causal? - - (d) For simplicity, let the wind speed be constant, *V*(*t*) = *kV* . Thus, *W*(*t*) = *k*1 + *k*2*T*(*t*) for some constants *k*1 and *k*2. Is this simplified system linear? - - (e) For simplicity, let the temperature be constant, *T*(*t*) = *kT* . Thus, *W*(*t*) = *k*3 + *k*4 {*V*(*t*)} 0.16 for some constants *k*3 and *k*4. Is this simplified system linear? -- **1.7-4** Input voltage *x*(*t*) applied to an inverting op-amp follower circuit produces output *y*(*t*) according to - -$$ -y(t + t_{p}) = \begin{cases} -V_{ref} & x(t) > V_{ref} \\ V_{ref} & x(t) < -V_{ref} \\ -x(t) & \text{otherwise} \end{cases} -$$ - -where op-amp reference voltage *V*ref and propagation delay *t*p are both positive constants. Answer the following questions yes or no, and provide mathematical justification for each answer. - -- (a) Is this system BIBO-stable? -- (b) Is the system causal? -- (c) Is the system invertible? -- (d) Is the system linear? -- (e) Is the system memoryless? -- (f) Is the system time invariant? -- **1.7-5** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t+1) = \begin{cases} -2x(t) & \text{when } x(t) \ge 0\\ 0 & \text{otherwise} \end{cases} -$$ - -**1.7-6** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t-1) = \begin{cases} x(t-1) & \text{when } \frac{d}{dt}x(t) \ge 0\\ x(t-2) & \text{otherwise} \end{cases} -$$ - -- **1.7-7** Repeat Prob. 1.7-4 for a system that multiplies a given input by a ramp function, *r*(*t*) = *tu*(*t*). That is, *y*(*t*) = *x*(*t*)*r*(*t*). -- **1.7-8** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t) = \frac{d}{dt}x(t-1) -$$ - -**1.7-9** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t) = \begin{cases} x(t) & \text{if } x(t) > 0\\ 0 & \text{if } x(t) \le 0 \end{cases} -$$ - -**1.7-10** A continuous-time system is given by - -$$ -y(t) = 0.5 \int_{-\infty}^{\infty} x(\tau) [\delta(t-\tau) - \delta(t+\tau)] d\tau -$$ - -Recall that δ(*t*) designates the Dirac delta function. - -- (a) Explain what this system does. -- (b) Is the system BIBO-stable? Justify your answer. -- (c) Is the system linear? Justify your answer. -- (d) Is the system memoryless? Justify your answer. -- (e) Is the system causal? Justify your answer. -- (f) Is the system time invariant? Justify your answer. -- **1.7-11** For a certain LTI system with the input *x*(*t*), the output *y*(*t*) and the two initial conditions *q*1(0) and *q*2(0), the following observations were made: - -| x(t) | q1(0) | q2(0) | y
(t) | -|------|-------|-------|--------------------| -| 0 | 1 | −1 | e−t
u(t) | -| 0 | 2 | 1 | e−t
(3t +2)u(t) | -| u(t) | −1 | −1 | 2u(t) | - -Determine *y*(*t*) when both the initial conditions are zero and the input *x*(*t*) is as shown in Fig. P1.7-11. [*Hint:* There are three causes: the input and each of the two initial conditions. Because of the linearity property, if a cause is increased by a factor *k*, the response to that cause also increases by the same factor *k*. Moreover, if causes are added, the corresponding responses add.] - -**Figure P1.7-11** - -**1.7-12** A system is specified by its input–output relationship as - -$$ -y(t) = \frac{x^2(t)}{dx(t)/dt} -$$ - -Show that the system satisfies the homogeneity property but not the additivity property. - -**1.7-13** Show that the circuit in Fig. P1.7-13 is zero-state linear but not zero-input linear. Assume all diodes to have identical (matched) characteristics. The output is the current *y*(*t*). - -**Figure P1.7-13** - -**1.7-14** The inductor *L* and the capacitor *C* in Fig. P1.7-14 are nonlinear, which makes the circuit nonlinear. The remaining three elements are linear. Show that the output *y*(*t*) of this nonlinear circuit satisfies the linearity conditions with respect to the input *x*(*t*) and the initial conditions (all the initial inductor currents and capacitor voltages). - -**1.7-15** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), determine which are causal and which are noncausal. - -(a) -$$ -y(t) = x(t-2) -$$ - -(b) $y(t) = x(-t)$ - -(c) *y*(*t*) = *x*(*at*) *a* > 1 - -(d) *y*(*t*) = *x*(*at*) *a* < 1 - -**1.7-16** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), determine which are invertible and which are noninvertible. For the invertible systems, find the input–output relationship of the inverse system. - -(a) -$$ -y(t) = \int_{-\infty}^{t} x(\tau) d\tau -$$ - -\n(b) $y(t) = x^{n}(t), x(t)$ real, *n* integer -\n(c) $y(t) = \frac{dx(t)}{dt}$ - -$$ -(c) \ y(t) = \frac{t}{dt} -$$ - -(d) -$$ -y(t) = x(3t - 6) -$$ - -(e) -$$ -y(t) = \cos [x(t)] -$$ - -(f) -$$ -y(t) = e^{x(t)} -$$ -, $x(t)$ real - -- **1.7-17** Figure P1.7-17 displays an input *x*1(*t*) to a linear time-invariant (LTI) system *H*, the corresponding output *y*1(*t*), and a second input *x*2(*t*). - - (a) Bill suggests that *x*2(*t*) = 2*x*1(3*t*)−*x*1(*t*−1). Is Bill correct? If yes, prove it. If not, correct his error. - - (b) Bill wants to know the output *y*2(*t*) in response to the input *x*2(*t*). Provide him with an expression for *y*2(*t*) in terms of *y*1(*t*). Use MATLAB to plot *y*2(*t*). - -- **1.7-18** A linear time-invariant system *H* acts on input *x*(*t*) = *u*(*t* − 0.5) − *u*(*t* − 1.5) to produce output *y*(*t*) = *H* {*x*(*t*)} = 0.5*u*(*t*) + 0.5*u*(*t* − 1) −*u*(*t* −2). - - (a) Is it possible that the system is causal? Explain your answer. If not causal, determine the shift necessary to make the system causal. - - (b) Is it possible that the system is memoryless? Explain your answer. - - (c) Suppose the output *y*(*t*) is applied to an identical system *H* to produce output - -$$ -z(t) = H\{y(t)\} = H\{H\{x(t)\}\} -$$ - -If possible, determine and sketch *z*(*t*). If not possible, explain why *z*(*t*) cannot be determined using the information given. - -**1.8-1** For the circuit depicted in Fig. P1.8-1, find the differential equations relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*). - -**1.8-2** For the circuit depicted in Fig. P1.8-2, find the differential equations relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*). - -**Figure P1.8-2** - -- **1.8-3** A simplified (one-dimensional) model of an automobile suspension system is shown in Fig. P1.8-3. In this case, the input is not a force but a displacement *x*(*t*) (the road contour). Find the differential equation relating the output *y*(*t*) (auto body displacement) to the input *x*(*t*) (the road contour). -- **1.8-4** A field-controlled dc motor is shown in Fig. P1.8-4. Its armature current *ia* is maintained constant. The torque generated by this motor is proportional to the field current *if* (torque= *Kf if*). Find the differential equation relating the output position θ to the input voltage *x*(*t*). The motor and load together have a moment of inertia *J*. -- **1.8-5** Water flows into a tank at a rate of *qi* units/s and flows out through the outflow valve at a rate of *q*0 units/s (Fig. P1.8-5). Determine the equation relating the outflow *q*0 to the input *qi*. The outflow rate is proportional to the head *h*. Thus *q*0 = *Rh*, where *R* is the valve resistance. Determine also the differential equation relating the head *h* to the input *qi*. [Hint: The net inflow of water in time Δ*t* is (*qi* − *q*0)Δ*t*. This inflow is also *A*Δ*h*, where *A* is the cross section of the tank.] -- **1.8-6** Consider the circuit shown in Fig. P1.8-6, with input voltage *x*(*t*) and output currents *y*1(*t*), *y*2(*t*), and *y*3(*t*). - - (a) What is the order of this system? Explain your answer. - - (b) Determine the matrix representation for this system. - - (c) Use Cramer's rule to determine the output current *y*3(*t*) for the input voltage *x*(*t*) = [2− | cos(*t*)|]*u*(*t* −1). -- **1.10-1** Write state equations for the parallel *RLC* circuit in Fig. P1.8-2. Use the capacitor voltage *q*1 and the inductor current *q*2 as your state variables. - -Show that every possible current or voltage in the circuit can be expressed in terms of *q*1, *q*2 and the input *x*(*t*). - -**1.10-2** Write state equations for the third-order circuit shown in Fig. P1.10-2, using the inductor currents *q*1, *q*2 and the capacitor voltage *q*3 as state variables. Show that every possible voltage or current in this circuit can be expressed as a linear combination of *q*1, *q*2, *q*3, and the input *x*(*t*). Also, at some instant *t*, it was found that - -*q*1 = 5, *q*2 = 1, *q*3 = 2, and *x* = 10. Determine the voltage across and the current through every element in this circuit. - -- **1.11-1** Provide MATLAB code and output that plots the odd portion *xo*(*t*) of the function *x*(*t*) = 2−*t* cos(2π*t*)*u*(*t*−π ) over a suitable-length interval using a suitable number of points. -- **1.11-2** Provide MATLAB code and output that plots the even portion *xe*(*t*) of the function *x*(*t*) = 2−*t*/2 cos(4π*t*)*u*(*t* − 0.5) over a suitable *t* using *t* = 0.002 second between points. -- **1.11-3** Define *x*(*t*) = *et*(1+*j*2π )*u*(−*t*) and *y*(*t*) = Re\* 2*x* −5−*t* 2 +. - - (a) Use MATLAB to plot Re{*x*(*t*)} versus Im{*x*(*at*)} for *a* = 0.5, 1, and 2 and −10 ≤ - -### **Figure P1.10-2** - -*t* ≤ 10. How important is the scale factor *a* on the shape of the resulting figure? - -- (b) Use MATLAB to plot *y*(*t*) over −10 ≤ *t* ≤ 10. Analytically determine the time *t*0 where *y*(*t*) has a jump discontinuity. Verify your calculation of *t*0 using the plot of *y*(*t*). -- (c) Use MATLAB and numerical integration to compute the energy *Ex* of signal *x*(*t*). -- (d) Use MATLAB and numerical integration to compute the energy *Ey* of signal *y*(*t*). -- **1.11-4** Consider the signal *x*(*t*) = *u*( *t* 2 + 1) − *u*(*t* − 1) − δ( *t* 2 ). Define *y*(*t*) = \$ *t*−3 −∞ *x*(τ )*d*τ and *z*(*t*) = \$ ∞ *t x*(τ )*d*τ . - - (a) Using MATLAB, accurately plot *y*(*t*). - - (b) Using MATLAB, accurately plot *z*(*t*). - - (c) Using MATLAB, accurately plot *w*(*t*) = *d dt y*(*t*) +*z*(*t*) . - - - -# **TIME-DOMAIN ANALYSIS OF [CONTINUOUS-TIME](#page-8-0) SYSTEMS** - -In this book we consider two methods of analysis of linear time-invariant (LTI) systems: the time-domain method and the frequency-domain method. In this chapter we discuss the *time-domain analysis* of linear, time-invariant, continuous-time (LTIC) systems. - -## **[2.1 INTRODUCTION](#page-8-0)** - -For the purpose of analysis, we shall consider *linear differential systems*. This is the class of LTIC systems introduced in Ch. 1, for which the input *x*(*t*) and the output *y*(*t*) are related by linear differential equations of the form - -$$ -\frac{d^{N}y(t)}{dt^{N}} + a_{1} \frac{d^{N-1}y(t)}{dt^{N-1}} + \dots + a_{N-1} \frac{dy(t)}{dt} + a_{N}y(t) -$$ -\n -$$ -= b_{N-M} \frac{d^{M}x(t)}{dt^{M}} + b_{N-M+1} \frac{d^{M-1}x(t)}{dt^{M-1}} + \dots + b_{N-1} \frac{dx(t)}{dt} + b_{N}x(t) \tag{2.1} -$$ - -where all the coefficients *ai* and *bi* are constants. Using operator notation *D* to represent *d*/*dt*, we can express this equation as - -$$ -(DN + a1DN-1 + ··· + aN-1D + aN)y(t) -$$ - -= (bN-MDM + bN-M+1DM-1 + ··· + bN-1D + bN)x(t) - -or - -$$ -Q(D)y(t) = P(D)x(t) -$$ -\n(2.2) - -where the polynomials *Q*(*D*) and *P*(*D*) are - -$$ -Q(D) = DN + a1DN-1 + \dots + aN-1D + aN -$$ - -$$ -P(D) = bN-MDM + bN-M+1DM-1 + \dots + bN-1D + bN -$$ - -Theoretically the powers *M* and *N* in the foregoing equations can take on any value. However, practical considerations make *M* > *N* undesirable for two reasons. In Sec. 4.3-3, we shall show that an LTIC system specified by Eq. (2.1) acts as an (*M* − *N*)th-order differentiator. A differentiator represents an unstable system because a bounded input like the step input results in an unbounded output, δ(*t*). Second, noise is enhanced by a differentiator. Noise is a wideband signal containing components of all frequencies from 0 to a very high frequency approaching ∞. † Hence, noise contains a significant amount of rapidly varying components. We know that the derivative of any rapidly varying signal is high. Therefore, any system specified by Eq. (2.1) in which *M* > *N* will magnify the high-frequency components of noise through differentiation. It is entirely possible for noise to be magnified so much that it swamps the desired system output even if the noise signal at the system's input is tolerably small. Hence, practical systems generally use *M* ≤ *N*. For the rest of this text we assume implicitly that *M* ≤ *N*. For the sake of generality, we shall assume *M* = *N* in Eq. (2.1). - -In Ch. 1, we demonstrated that a system described by Eq. (2.2) is linear. Therefore, its response can be expressed as the sum of two components: the zero-input response and the zero-state response (decomposition property).‡ Therefore, - -total response = zero-input response + zero-state response - -The zero-input response is the system output when the input *x*(*t*) = 0, and thus it is the result of internal system conditions (such as energy storages, initial conditions) alone. It is independent of the external input *x*(*t*). In contrast, the zero-state response is the system output to the external input *x*(*t*) when the system is in zero state, meaning the absence of all internal energy storages: that is, all initial conditions are zero. - -## **2.2 SYSTEM RESPONSE TO INTERNAL [CONDITIONS:](#page-8-0) THE ZERO-INPUT RESPONSE** - -The zero-input response *y*0(*t*) is the solution of Eq. (2.2) when the input *x*(*t*) = 0 so that - -$$ -Q(D)y_0(t) = 0 -$$ - -$$ -Q(D)y_0(t) = 0 -$$ - -If *y*(*t*) is the zero-state response, then *y*(*t*) is the solution of - -$$ -Q(D)y(t) = P(D)x(t) -$$ - -subject to zero initial conditions (zero-state). Adding these two equations, we have - -$$ -Q(D)[y_0(t) + y(t)] = P(D)x(t) -$$ - -Clearly, *y*0(*t*)+*y*(*t*) is the general solution of Eq. (2.2). - -† Noise is any undesirable signal, natural or manufactured, that interferes with the desired signals in the system. Some of the sources of noise are the electromagnetic radiation from stars, the random motion of electrons in system components, interference from nearby radio and television stations, transients produced by automobile ignition systems, and fluorescent lighting. - -‡ We can verify readily that the system described by Eq. (2.2) has the decomposition property. If *y*0(*t*) is the zero-input response, then, by definition, - -or - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)y0(t) = 0 -$$ -\n(2.3) - -A solution to this equation can be obtained systematically [1]. However, we will take a shortcut by using heuristic reasoning. Equation (2.3) shows that a linear combination of *y*0(*t*) and its *N* successive derivatives is zero, not at *some* values of *t*, but for all *t*. Such a result is possible *if and only if y*0(*t*) and all its *N* successive derivatives are of the same form. Otherwise their sum can never add to zero for all values of *t*. We know that only an exponential function *e*λ*t* has this property. So let us assume that - -$$ -y_0(t) = ce^{\lambda t} -$$ - -is a solution to Eq. (2.3). Then - -$$ -Dy_0(t) = \frac{dy_0(t)}{dt} = c\lambda e^{\lambda t} -$$ -$$ -D^2y_0(t) = \frac{d^2y_0(t)}{dt^2} = c\lambda^2 e^{\lambda t} -$$ -$$ -\vdots -$$ -$$ -D^Ny_0(t) = \frac{d^Ny_0(t)}{dt^N} = c\lambda^N e^{\lambda t} -$$ - -Substituting these results in Eq. (2.3), we obtain - -$$ -c(\lambda^N + a_1 \lambda^{N-1} + \dots + a_{N-1} \lambda + a_N)e^{\lambda t} = 0 -$$ - -For a nontrivial solution of this equation, - -$$ -\lambda^{N} + a_{1}\lambda^{N-1} + \dots + a_{N-1}\lambda + a_{N} = 0 -$$ -\n(2.4) - -This result means that *ce*λ*t* is indeed a solution of Eq. (2.3), provided λ satisfies Eq. (2.4). Note that the polynomial in Eq. (2.4) is identical to the polynomial *Q*(*D*) in Eq. (2.3), with λ replacing *D*. Therefore, Eq. (2.4) can be expressed as - -$$ -Q(\lambda) = 0 -$$ - -Expressing *Q*(λ) in factorized form, we obtain - -$$ -Q(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_N) = 0 -$$ -\n(2.5) - -Clearly, λ has *N* solutions: λ1, λ2, ..., λ*N*, assuming that all λ*i* are distinct. Consequently, Eq. (2.3) has *N* possible solutions: *c*1*e*λ1*t* , *c*2*e*λ2*t* , ..., *cNe*λ*Nt* , with *c*1, *c*2,..., *cN* as arbitrary constants. We - -can readily show that a general solution is given by the sum of these *N* solutions† so that - -$$ -y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \dots + c_N e^{\lambda_N t} -$$ - (2.6) - -where *c*1, *c*2, ..., *cN* are arbitrary constants determined by *N* constraints (the auxiliary conditions) on the solution. - -Observe that the polynomial *Q*(λ), which is characteristic of the system, has nothing to do with the input. For this reason the polynomial *Q*(λ) is called the *characteristic polynomial* of the system. The equation - -$$ -Q(\lambda) = 0 -$$ - -is called the *characteristic equation* of the system. Equation (2.5) clearly indicates that λ1, λ2, ..., λ*N* are the roots of the characteristic equation; consequently, they are called the *characteristic roots* of the system. The terms *characteristic values, eigenvalues,* and *natural frequencies* are also used for characteristic roots.‡ The exponentials *e*λ*it* (*i* = 1, 2,...,*n*) in the zero-input response are the *characteristic modes* (also known as *natural modes* or simply as *modes*) of the system. There is a characteristic mode for each characteristic root of the system, and the *zero-input response is a linear combination of the characteristic modes of the system*. - -An LTIC system's characteristic modes comprise its single most important attribute. Characteristic modes not only determine the zero-input response but also play an important role in determining the zero-state response. In other words, the entire behavior of a system is dictated primarily by its characteristic modes. In the rest of this chapter we shall see the pervasive presence of characteristic modes in every aspect of system behavior. - -### REPEATED ROOTS - -The solution of Eq. (2.3) as given in Eq. (2.6) assumes that the *N* characteristic roots λ1, λ2, ..., λ*N* are distinct. If there are repeated roots (same root occurring more than once), the form of the solution is modified slightly. By direct substitution we can show that the solution of the equation - -$$ -(D - \lambda)^2 y_0(t) = 0 -$$ - -is given by - -$$ -y_0(t) = (c_1 + c_2 t)e^{\lambda t} -$$ - -† To prove this assertion, assume that *y*1(*t*), *y*2(*t*), ..., *yN*(*t*) are all solutions of Eq. (2.3). Then - -$$ -Q(D)y_1(t) = 0 -$$ - -\n -$$ -Q(D)y_2(t) = 0 -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -Q(D)y_N(t) = 0 -$$ - -Multiplying these equations by *c*1, *c*2, ..., *cN*, respectively, and adding them together yield - -$$ -Q(D)[c_1y_1(t) + c_2y_2(t) + \cdots + c_Ny_n(t)] = 0 -$$ - -This result shows that *c*1*y*1(*t*) + *c*2*y*2(*t*) +···+ *cNyn*(*t*) is also a solution of the homogeneous equation [Eq. (2.3)]. - -‡ *Eigenvalue* is German for "characteristic value." - -#### 154 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -In this case the root λ repeats twice. Observe that the characteristic modes in this case are *e*λ*t* and *te*λ*t* . Continuing this pattern, we can show that for the differential equation - -$$ -(D - \lambda)^r y_0(t) = 0 -$$ - -the characteristic modes are *e*λ*t* , *te*λ*t* , *t* 2*e*λ*t* , ..., *t r*−1*e*λ*t* , and that the solution is - -$$ -y_0(t) = (c_1 + c_2t + \dots + c_rt^{r-1})e^{\lambda t} -$$ - -Consequently, for a system with the characteristic polynomial - -$$ -Q(\lambda) = (\lambda - \lambda_1)^r (\lambda - \lambda_{r+1}) \cdots (\lambda - \lambda_N) -$$ - -the characteristic modes are *e*λ1*t* , *te*λ1*t* , ..., *t r*−1*e*λ1*t* , *e*λ*r*+1*t* , ..., *e*λ*Nt* and the solution is - -$$ -y_0(t) = (c_1 + c_2t + \dots + c_rt^{r-1})e^{\lambda_1 t} + c_{r+1}e^{\lambda_{r+1} t} + \dots + c_N e^{\lambda_N t} -$$ - -### COMPLEX ROOTS - -The procedure for handling complex roots is the same as that for real roots. For complex roots, the usual procedure leads to complex characteristic modes and the complex form of solution. However, it is possible to avoid the complex form altogether by selecting a real-form of solution, as described next. - -For a real system, complex roots must occur in pairs of conjugates if the coefficients of the characteristic polynomial *Q*(λ) are to be real. Therefore, if α + *j*β is a characteristic root, α − *j*β must also be a characteristic root. The zero-input response corresponding to this pair of complex conjugate roots is - -$$ -y_0(t) = c_1 e^{(\alpha + j\beta)t} + c_2 e^{(\alpha - j\beta)t} -$$ -\n(2.7) - -For a real system, the response *y*0(*t*) must also be real. This is possible only if *c*1 and *c*2 are conjugates. Let - -$$ -c_1 = \frac{c}{2}e^{j\theta} \qquad \text{and} \qquad c_2 = \frac{c}{2}e^{-j\theta} -$$ - -This yields - -$$ -y_0(t) = \frac{c}{2}e^{i\theta}e^{(\alpha+j\beta)t} + \frac{c}{2}e^{-j\theta}e^{(\alpha-j\beta)t} -$$ - -= -$$ -\frac{c}{2}e^{\alpha t}[e^{i(\beta t + \theta)} + e^{-j(\beta t + \theta)}] -$$ - -= -$$ -ce^{\alpha t}\cos(\beta t + \theta) -$$ - (2.8) - -Therefore, the zero-input response corresponding to complex conjugate roots α ± *j*β can be expressed in a complex form [Eq. (2.7)] or a real form [Eq. (2.8)]. - -### **EXAMPLE 2.1 Finding the Zero-Input Response** - -Find *y*0(*t*), the zero-input response of the response for an LTIC system described by - -- **(a)** the simple-root system (*D*2 + 3*D* + 2)*y*(*t*) = *Dx*(*t*) with initial conditions *y*0(0) = 0 and *y*˙0(0) = −5. -- **(b)** the repeated-root system (*D*2 + 6*D* + 9)*y*(*t*) = (3*D* + 5)*x*(*t*) with initial conditions *y*0(0) = 3 and *y*˙0(0) = −7. -- **(c)** the complex-root system (*D*2 + 4*D* + 40)*y*(*t*) = (*D* + 2)*x*(*t*) with initial conditions *y*0(0) = 2 and *y*˙0(0) = 16.78. - -**(a)** Note that *y*0(*t*), being the zero-input response (*x*(*t*) = 0), is the solution of (*D*2 +3*D*+ 2)*y*0(*t*) = 0. The characteristic polynomial of the system is λ2 + 3λ + 2. The characteristic equation of the system is therefore λ2 + 3λ + 2 = (λ + 1)(λ + 2) = 0. The characteristic roots of the system are λ1 = −1 and λ2 = −2, and the characteristic modes of the system are *e*−*t* and *e*−2*t* . Consequently, the zero-input response is - -$$ -y_0(t) = c_1 e^{-t} + c_2 e^{-2t} -$$ - -Differentiating this expression, we obtain - -$$ -\dot{y}_0(t) = -c_1 e^{-t} - 2c_2 e^{-2t} -$$ - -To determine the constants *c*1 and *c*2, we set *t* = 0 in the equations for *y*0(*t*) and *y*˙0(*t*) and substitute the initial conditions *y*0(0) = 0 and *y*˙0(0) = −5, yielding - -$$ -0 = c_1 + c_2 -$$ - -$$ --5 = -c_1 - 2c_2 -$$ - -Solving these two simultaneous equations in two unknowns for *c*1 and *c*2 yields - -$$ -c_1 = -5 \qquad \text{and} \qquad c_2 = 5 -$$ - -Therefore, - -$$ -y_0(t) = -5e^{-t} + 5e^{-2t} -$$ - (2.9) - -This is the zero-input response of *y*(*t*). Because *y*0(*t*) is present at *t* = 0−, we are justified in assuming that it exists for *t* ≥ 0.† - -**(b)** The characteristic polynomial is λ2 + 6λ + 9 = (λ + 3)2, and its characteristic roots are λ1 = −3, λ2 = −3 (repeated roots). Consequently, the characteristic modes of the system are *e*−3*t* and *te*−3*t* . The zero-input response, being a linear combination of the characteristic modes, is given by - -$$ -y_0(t) = (c_1 + c_2 t)e^{-3t} -$$ - -† *y*0(*t*) may be present even before *t* = 0−. However, we can be sure of its presence only from *t* = 0− onward. - -We can find the arbitrary constants *c*1 and *c*2 from the initial conditions *y*0(0) = 3 and *y*˙0(0) = −7 following the procedure in part (a). The reader can show that *c*1 = 3 and *c*2 = 2. Hence, - -$$ -y_0(t) = (3+2t)e^{-3t} \qquad t \ge 0 -$$ - -**(c)** The characteristic polynomial is λ2 + 4λ + 40 = (λ + 2 − *j*6)(λ + 2 + *j*6). The characteristic roots are −2 ± *j*6.† The solution can be written either in the complex form [Eq. (2.7)] or in the real form [Eq. (2.8)]. The complex form is *y*0(*t*) = *c*1*e*λ1*t* + *c*2*e*λ2*t* , where λ1 = −2+*j*6 and λ2 = −2−*j*6. Since α = −2 and β =6, the real-form solution is [see Eq. (2.8)] - -$$ -y_0(t) = ce^{-2t}\cos(6t+\theta) -$$ - -Differentiating this expression, we obtain - -$$ -\dot{y}_0(t) = -2ce^{-2t}\cos{(6t + \theta)} - 6ce^{-2t}\sin{(6t + \theta)} -$$ - -To determine the constants *c* and θ, we set *t* = 0 in the equations for *y*0(*t*) and *y*˙0(*t*) and substitute the initial conditions *y*0(0) = 2 and *y*˙0(0) = 16.78, yielding - -$$ -2 = c \cos \theta -$$ - -16.78 = $-2c \cos \theta - 6c \sin \theta$ - -Solution of these two simultaneous equations in two unknowns *c*cos θ and *c*sinθ yields - -*c*cos θ = 2 and *c*sinθ = −3.463 - -Squaring and then adding these two equations yield - -$$ -c^{2} = (2)^{2} + (-3.464)^{2} = 16 \Longrightarrow c = 4 -$$ - -Next, dividing *c*sinθ = −3.463 by *c*cos θ = 2 yields - -$$ -\tan \theta = \frac{-3.463}{2} -$$ - -and - -$$ -\theta = \tan^{-1} \left( \frac{-3.463}{2} \right) = -\frac{\pi}{3} -$$ - -Therefore, - -$$ -y_0(t) = 4e^{-2t}\cos\left(6t - \frac{\pi}{3}\right) -$$ - -For the plot of *y*0(*t*), refer again to Fig. B.11c. - -$$ -\lambda^2 + 4\lambda + 40 = (\lambda^2 + 4\lambda + 4) + 36 = (\lambda + 2)^2 + (6)^2 = (\lambda + 2 - j6)(\lambda + 2 + j6) -$$ - -† The complex conjugate roots of a second-order polynomial can be determined by using the formula in Sec. B.8-10 or by expressing the polynomial as a sum of two squares. The latter can be accomplished by completing the square with the first two terms, as follows: - -### **EXAMPLE 2.2 Using MATLAB to Find Polynomial Roots** - -Find the roots λ1 and λ2 of the polynomial λ2 + 4λ + *k* for three values of *k*: **(a)** *k* = 3, **(b)** *k* = 4, and **(c)** *k* = 40. - -**(a)** >> r = roots([1 4 3]).' r = -3 -1 For *k* = 3, the polynomial roots are therefore λ1 = −3 and λ2 = −1. **(b)** >> r = roots([1 4 4]).' r = -2 -2 For *k* = 4, the polynomial roots are therefore λ1 = λ2 = −2. **(c)** >> r = roots([1 4 40]).' r = -2.00+6.00i -2.00-6.00i For *k* = 40, the polynomial roots are therefore λ1 = −2+*j*6 and λ2 = −2−*j*6. - -### **EXAMPLE 2.3 Using MATLAB to Find the Zero-Input Response** - -Consider an LTIC system specified by the differential equation - -$$ -(D^2 + 4D + k)y(t) = (3D + 5)x(t) -$$ - -Using initial conditions *y*0(0) = 3 and *y*˙0(0) = −7, apply MATLAB's dsolve command to determine the zero-input response when: **(a)** *k* = 3, **(b)** *k* = 4, and **(c)** *k* = 40. - -**(a)** - ->> y\_0 = dsolve('D2y+4\*Dy+3\*y=0','y(0)=3','Dy(0)=-7','t') y\_0 = 1/exp(t) + 2/exp(3\*t) - -For *k* = 3, the zero-input response is therefore *y*0(*t*) = *e*−*t* +2*e*−3*t* . **(b)** - ->> y\_0 = dsolve('D2y+4\*Dy+4\*y=0','y(0)=3','Dy(0)=-7','t') y\_0 = 3/exp(2\*t) - t/exp(2\*t) - -For *k* = 4, the zero-input response is therefore *y*0(*t*) = 3*e*−2*t* −*te*−2*t* . **(c)** - ->> -$$ -y_0 = dsolve('D2y+4*Dy+40*y=0', 'y(0)=3', 'Dy(0)=-7', 't') -$$ - -\n $y_0 = (3*cos(6*t))/exp(2*t) - sin(6*t)/(6*exp(2*t))$ - -For *k* = 40, the zero-input response is therefore *y*0(*t*) = 3*e*−2*t* cos(6*t*)− 1 6 *e*−2*t*sin(6*t*). - -### **DR ILL 2.1 Finding the Zero-Input Response of a First-Order System** - -Find the zero-input response of an LTIC system described by (*D* + 5)*y*(*t*) = *x*(*t*) if the initial condition is *y*(0) = 5. - -### **ANSWER** - -*y*0(*t*) = 5*e*−5*t t* ≥ 0 - -## **DR ILL 2.2 Finding the Zero-Input Response of a Second-Order System** - -Letting *y*0(0) = 1 and *y*˙0(0) = 4, solve - -(*D*2 +2*D*)*y*0(*t*) = 0 - -**ANSWER** - -*y*0(*t*) = 3−2*e*−2*t t* ≥ 0 - -### PRACTICAL INITIAL CONDITIONS AND THE MEANING OF 0− AND 0+ - -In Ex. 2.1 the initial conditions *y*0(0) and *y*˙0(0) were supplied. In practical problems, we must derive such conditions from the physical situation. For instance, in an *RLC* circuit, we may be given the conditions (initial capacitor voltages, initial inductor currents, etc.). - -From this information, we need to derive *y*0(0), *y*˙0(0), ... for the desired variable as demonstrated in the next example. - -In much of our discussion, the input is assumed to start at *t* = 0, unless otherwise mentioned. Hence, *t* = 0 is the reference point. The conditions immediately before *t* = 0 (just before the input is applied) are the conditions at *t* = 0−, and those immediately after *t* = 0 (just after the input is applied) are the conditions at *t* = 0+ (compare this with the historical time frames BCE and CE). In practice, we are likely to know the initial conditions at *t* = 0− rather than at *t* = 0+. The two sets of conditions are generally different, although in some cases they may be identical. - -The total response *y*(*t*) consists of two components: the zero-input response *y*0(*t*) [response due to the initial conditions alone with *x*(*t*) = 0] and the zero-state response resulting from the input alone with all initial conditions zero. At *t* = 0−, the total response *y*(*t*) consists solely of the zero-input response *y*0(*t*) because the input has not started yet. Hence the initial conditions on *y*(*t*) are identical to those of *y*0(*t*). Thus, *y*(0−) = *y*0(0−), *y*˙(0−) = ˙*y*0(0−), and so on. Moreover, *y*0(*t*) is the response due to initial conditions alone and does not depend on the input *x*(*t*). Hence, application of the input at *t* = 0 does not affect *y*0(*t*). This means the initial conditions on *y*0(*t*) at *t* = 0− and 0+ are identical; that is, *y*0(0−), *y*˙0(0−), ... are identical to *y*0(0+), *y*˙0(0+), ..., respectively. It is clear that for *y*0(*t*), there is no distinction between the initial conditions at *t* = 0−, 0, and 0+. They are all the same. But this is not the case with the total response *y*(*t*), which consists of both the zero-input and zero-state responses. Thus, in general, *y*(0−) = *y*(0+), *y*˙(0−) = ˙*y*(0+), and so on. - -### **EXAMPLE 2.4 Consideration of Initial Conditions** - -A voltage *x*(*t*) = 10*e*−3*t u*(*t*) is applied at the input of the *RLC* circuit illustrated in Fig. 2.2a. Find the loop current *y*(*t*) for *t* ≥ 0 if the initial inductor current is zero [*y*(0−) = 0] and the initial capacitor voltage is 5 volts [*vC*(0−) = 5]. - -The differential (loop) equation relating *y*(*t*) to *x*(*t*) was derived in Eq. (1.29) as - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -The zero-state component of *y*(*t*) resulting from the input *x*(*t*), assuming that all initial conditions are zero, that is, *y*(0−) = *vC*(0−) = 0, will be obtained later in Ex. 2.9. In this example we shall find the zero-input reponse *y*0(*t*). For this purpose, we need two initial conditions, *y*0(0) and *y*˙0(0). These conditions can be derived from the given initial conditions, *y*(0−) = 0 and *vC*(0−) = 5, as follows. Recall that *y*0(*t*) is the loop current when the input terminals are shorted so that the input *x*(*t*) = 0 (zero-input), as depicted in Fig. 2.2b. We now compute *y*0(0) and *y*˙0(0), the values of the loop current and its derivative at *t* = 0, from the initial values of the inductor current and the capacitor voltage. Remember that the inductor current cannot change instantaneously in the absence of an impulsive voltage. Similarly, the capacitor voltage cannot change instantaneously in the absence of an impulsive current. Therefore, when the input terminals are shorted at *t* = 0, the inductor current is still zero and the capacitor voltage is still 5 volts. Thus, - -*y*0(0) = 0 - -**Figure 2.1** Circuits for Ex. 2.4. - -To determine *y*˙0(0), we use the loop equation for the circuit in Fig. 2.2b. Because the voltage across the inductor is *L*(*dy*0/*dt*) or *y*˙0(*t*), this equation can be written as follows: - -$$ -\dot{y}_0(t) + 3y_0(t) + v_C(t) = 0 -$$ - -Setting *t* = 0, we obtain - -$$ -\dot{y}_0(0) + 3y_0(0) + v_C(0) = 0 -$$ - -But *y*0(0) = 0 and *vC*(0) = 5. Consequently, - -*y*˙0(0) = −5 - -Therefore, the desired initial conditions are - -$$ -y_0(0) = 0 -$$ - and $\dot{y}_0(0) = -5$ - -Thus, the problem reduces to finding *y*0(*t*), the zero-input component of *y*(*t*) of the system specified by the equation (*D*2 +3*D*+2)*y*(*t*) = *Dx*(*t*), when the initial conditions are *y*0(0) = 0 and *y*˙0(0) = −5. We have already solved this problem in Ex. 2.1a, where we found - -$$ -y_0(t) = -5e^{-t} + 5e^{-2t} \qquad t \ge 0 -$$ - -This is the zero-input component of the loop current *y*(*t*). - -It is interesting to find the initial conditions at *t* = 0− and 0+ for the total response *y*(*t*). Let us compare *y*(0−) and *y*˙(0−) with *y*(0+) and *y*˙(0+). The two pairs can be compared by writing the loop equation for the circuit in Fig. 2.2a at *t* = 0− and *t* = 0+. The only difference between the two situations is that at *t* = 0−, the input *x*(*t*) = 0, whereas at *t* = 0+, the input *x*(*t*) = 10 [because *x*(*t*) = 10*e*−3*t* ]. Hence, the two loop equations are - -$$ -\dot{y}(0^-) + 3y(0^-) + v_C(0^-) = 0 -$$ - -$$ -\dot{y}(0^+) + 3y(0^+) + v_C(0^+) = 10 -$$ - -The loop current *y*(0+) = *y*(0−) = 0 because it cannot change instantaneously in the absence of impulsive voltage. The same is true of the capacitor voltage. Hence, *vC*(0+) = *vC*(0−) = 5. Substituting these values in the foregoing equations, we obtain *y*˙(0−) = −5 and *y*˙(0+) = 5. Thus, - -*y*(0−) = 0, *y*˙(0−) = −5 and *y*(0+) = 0, *y*˙(0+) = 5 (2.10) - -### **DR ILL 2.3 Zero-Input Response of an** *RC* **Circuit** - -In the circuit in Fig. 2.2a, the inductance *L* = 0 and the initial capacitor voltage *vC*(0) = 30 volts. Show that the zero-input component of the loop current is given by *y*0(*t*) = −10*e*−2*t*/3 for *t* ≥ 0. - -### INDEPENDENCE OF THE ZERO-INPUT AND ZERO-STATE RESPONSES - -In Ex. 2.4 we computed the zero-input component without using the input *x*(*t*). The zero-state response can be computed from the knowledge of the input *x*(*t*) alone; the initial conditions are assumed to be zero (system in zero state). The two components of the system response (the zero-input and zero-state responses) are independent of each other. *The two worlds of zero-input response and zero-state response coexist side by side, neither one knowing or caring what the other is doing. For each component, the other is totally irrelevant.* - -### ROLE OF AUXILIARY CONDITIONS IN SOLUTION OF DIFFERENTIAL EQUATIONS - -The solution of a differential equation requires additional pieces of information (the *auxiliary conditions*). Why? We now show heuristically why a differential equation does not, in general, have a unique solution unless some additional constraints (or conditions) on the solution are known. - -Differentiation operation is not invertible unless one piece of information about *y*(*t*) is given. To get back *y*(*t*) from *dy*/*dt*, we must know one piece of information, such as *y*(0). Thus, differentiation is an irreversible (noninvertible) operation during which certain information is lost. To invert this operation, one piece of information about *y*(*t*) must be provided to restore the original *y*(*t*). Using a similar argument, we can show that, given *d*2*y*/*dt*2, we can determine *y*(*t*) uniquely only if two additional pieces of information (constraints) about *y*(*t*) are given. In general, to determine *y*(*t*) uniquely from its *N*th derivative, we need *N* additional pieces of information (constraints) about *y*(*t*). These constraints are also called *auxiliary conditions*. When these conditions are given at *t* = 0, they are called *initial conditions*. - -### **[2.2-1 Some Insights into the Zero-Input Behavior of a System](#page-8-0)** - -By definition, the zero-input response is the system response to its internal conditions, assuming that its input is zero. Understanding this phenomenon provides interesting insight into system behavior. If a system is disturbed momentarily from its rest position and if the disturbance is then - -#### 162 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -removed, the system will not come back to rest instantaneously. In general, it will come back to rest over a period of time and only through a special type of motion that is characteristic of the system.† For example, if we press on an automobile fender momentarily and then release it at *t* = 0, there is no external force on the automobile for *t* > 0.‡ The auto body will eventually come back to its rest (equilibrium) position, but not through any arbitrary motion. It must do so by using only a form of response that is sustainable by the system on its own without any external source, since the input is zero. Only characteristic modes satisfy this condition. *The system uses a proper combination of characteristic modes to come back to the rest position while satisfying appropriate boundary (or initial) conditions.* - -If the shock absorbers of the automobile are in good condition (high damping coefficient), the characteristic modes will be monotonically decaying exponentials, and the auto body will come to rest rapidly without oscillation. In contrast, for poor shock absorbers (low damping coefficients), the characteristic modes will be exponentially decaying sinusoids, and the body will come to rest through oscillatory motion. When a series *RC* circuit with an initial charge on the capacitor is shorted, the capacitor will start to discharge exponentially through the resistor. This response of the *RC* circuit is caused entirely by its internal conditions and is sustained by this system without the aid of any external input. The exponential current waveform is therefore the characteristic mode of the *RC* circuit. - -Mathematically we know that *any combination of characteristic modes can be sustained by the system alone without requiring an external input*. This fact can be readily verified for the series *RL* circuit shown in Fig. 2.2. The loop equation for this system is - -$$ -(D+2)y(t) = x(t) -$$ - -It has a single characteristic root λ = −2, and the characteristic mode is *e*−2*t* . We now verify that a loop current *y*(*t*) = *ce*−2*t* can be sustained through this circuit without any input voltage. The input voltage *x*(*t*) required to drive a loop current *y*(*t*) = *ce*−2*t* is given by - -$$ -x(t) = L\frac{dy(t)}{dt} + Ry(t) -$$ - -= $\frac{d}{dt}(ce^{-2t}) + 2ce^{-2t}$ -= $-2ce^{-2t} + 2ce^{-2t} = 0$ - -**Figure 2.2** Modes always get a free ride. - -† This assumes that the system will eventually come back to its original rest (or equilibrium) position. - -‡ We ignore the force of gravity, which merely causes a constant displacement of the auto body without affecting the other motion. - -Clearly, the loop current *y*(*t*) = *ce*−2*t* is sustained by the *RL* circuit on its own, without the necessity of an external input. - -### THE RESONANCE PHENOMENON - -We have seen that any signal consisting of a system's characteristic mode is sustained by the system on its own; the system offers no obstacle to such signals. Imagine what would happen if we were to drive the system with an external input that is one of its characteristic modes. This would be like pouring gasoline on a fire in a dry forest or hiring a child to eat ice cream. A child would gladly do the job without pay. Think what would happen if he were paid by the amount of ice cream he ate! He would work overtime. He would work day and night, until he became sick. The same thing happens with a system driven by an input of the form of characteristic mode. The system response grows without limit, until it burns out.† We call this behavior the *resonance phenomenon*. An intelligent discussion of this important phenomenon requires an understanding of the zero-state response; for this reason we postpone this topic until Sec. 2.6-7. - -## **2.3 THE UNIT IMPULSE [RESPONSE](#page-8-0)** *h(t)* - -In Ch. 1 we explained how a system response to an input *x*(*t*) may be found by breaking this input into narrow rectangular pulses, as illustrated earlier in Fig. 1.27a, and then summing the system response to all the components. The rectangular pulses become impulses in the limit as their widths approach zero. Therefore, the system response is the sum of its responses to various impulse components. This discussion shows that if we know the system response to an impulse input, we can determine the system response to an arbitrary input *x*(*t*). We now discuss a method of determining *h*(*t*), the unit impulse response of an LTIC system described by the *N*th-order differential equation [Eq. (2.1)] - -$$ -\frac{d^N y(t)}{dt^N} + a_1 \frac{d^{N-1} y(t)}{dt^{N-1}} + \dots + a_{N-1} \frac{dy(t)}{dt} + a_N y(t) -$$ - -= $b_{N-M} \frac{d^M x(t)}{dt^M} + b_{N-M+1} \frac{d^{M-1} x(t)}{dt^{M-1}} + \dots + b_{N-1} \frac{dx(t)}{dt} + b_N x(t)$ - -Recall that noise considerations restrict practical systems to *M* ≤ *N*. Under this constraint, the most general case is *M* = *N*. Therefore, Eq. (2.1) can be expressed as - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)y(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)x(t) -$$ - (2.11) - -Before deriving the general expression for the unit impulse response *h*(*t*), it is illuminating to understand qualitatively the nature of *h*(*t*). The impulse response *h*(*t*) is the system response to an impulse input δ(*t*) applied at *t* = 0 with all the initial conditions zero at *t* = 0−. An impulse input δ(*t*) is like lightning, which strikes instantaneously and then vanishes. But in its wake, in that single moment, objects that have been struck are rearranged. Similarly, an impulse input δ(*t*) appears momentarily at *t* = 0, and then it is gone forever. But in that moment it generates energy storages; that is, it creates nonzero initial conditions instantaneously within the system at - -† In practice, the system in resonance is more likely to go in saturation because of high amplitude levels. - -*t* = 0+. Although the impulse input δ(*t*) vanishes for *t* > 0 so that the system has no input after the impulse has been applied, the system will still have a response generated by these newly created initial conditions. The impulse response *h*(*t*), therefore, must consist of the system's characteristic modes for *t* ≥ 0+. As a result, - -*h*(*t*) = characteristic mode terms *t* ≥ 0+ - -This response is valid for *t* > 0. But what happens at *t* = 0? At a single moment *t* = 0, there can at most be an impulse,† so the form of the complete response *h*(*t*) is - -$$ -h(t) = A_0 \delta(t) + \text{characteristic mode terms} \qquad t \ge 0 \tag{2.12} -$$ - -because *h*(*t*) is the unit impulse response. Setting *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*) in Eq. (2.11) yields - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)h(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)\delta(t) -$$ - -In this equation we substitute *h*(*t*) from Eq. (2.12) and compare the coefficients of similar impulsive terms on both sides. The highest order of the derivative of impulse on both sides is *N*, with its coefficient value as *A*0 on the left-hand side and *b*0 on the right-hand side. The two values must be matched. Therefore, *A*0 = *b*0 and - -$$ -h(t) = b_0 \delta(t) + \text{characteristic modes} \tag{2.13} -$$ - -In Eq. (2.11), if *M* < *N*, *b*0 = 0. Hence, the impulse term *b*0δ(*t*) exists only if *M* = *N*. The unknown coefficients of the *N* characteristic modes in *h*(*t*) in Eq. (2.13) can be determined by using the technique of impulse matching, as explained in the following example. - -### **EXAMPLE 2.5 Impulse Response via Impulse Matching** - -Find the impulse response *h*(*t*) for a system specified by - -$$ -(D2 + 5D + 6)y(t) = (D + 1)x(t) -$$ -\n(2.14) - -In this case, *b*0 = 0. Hence, *h*(*t*) consists of only the characteristic modes. The characteristic polynomial is λ2 + 5λ + 6 = (λ + 2)(λ + 3). The roots are −2 and −3. Hence, the impulse - -† It might be possible for the derivatives of δ(*t*) to appear at the origin. However, if *M* ≤ *N*, it is impossible for *h*(*t*) to have any derivatives of δ(*t*). This conclusion follows from Eq. (2.11) with *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*). The coefficients of the impulse and all its derivatives must be matched on both sides of this equation. If *h*(*t*) contains δ(1) (*t*), the first derivative of δ(*t*), the left-hand side of Eq. (2.11) will contain a term δ(*N*+1) (*t*). But the highest-order derivative term on the right-hand side is δ(*N*) (*t*). Therefore, the two sides cannot match. Similar arguments can be made against the presence of the impulse's higher-order derivatives in *h*(*t*). - -response *h*(*t*) is - -$$ -h(t) = (c_1 e^{-2t} + c_2 e^{-3t}) u(t) -$$ -\n(2.15) - -Letting *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*) in Eq. (2.14), we obtain - -$$ -\ddot{h}(t) + 5\dot{h}(t) + 6h(t) = \dot{\delta}(t) + \delta(t) -$$ -\n(2.16) - -Recall that initial conditions *h*(0−) and *h*˙(0−) are both zero. But the application of an impulse at *t* = 0 creates new initial conditions at *t* = 0+. Let *h*(0+) = *K*1 and *h*˙(0+) = *K*2. These jump discontinuities in *h*(*t*) and *h*˙(*t*) at *t* = 0 result in impulse terms *h*˙(0) = *K*1δ(*t*) and *h*¨(0) = *K*1δ(˙ *t*) + *K*2δ(*t*) on the left-hand side. Matching the coefficients of impulse terms on both sides of Eq. (2.16) yields - -$$ -5K_1 + K_2 = 1 -$$ -, $K_1 = 1$ $\implies$ $K_1 = 1, K_2 = -4$ - -We now use these values *h*(0+) = *K*1 = 1 and *h*˙(0+) = *K*2 = −4 in Eq. (2.15) to find *c*1 and *c*2. Setting *t* = 0+ in Eq. (2.15), we obtain *c*1 + *c*2 = 1. Also setting *t* = 0+ in *h*˙(*t*), we obtain −2*c*1 −3*c*1 = −4. These two simultaneous equations yield *c*1 = −1 and *c*2 = 2. Therefore, - -$$ -h(t) = (-e^{-2t} + 2e^{-3t})u(t) -$$ - -Although the method used in this example is relatively simple, we can simplify it still further by using a modified version of impulse matching. - -### SIMPLIFIED IMPULSE MATCHING METHOD - -The alternate technique we present now allows us to reduce the procedure to a simple routine to determine *h*(*t*). To avoid the needless distraction, the proof for this procedure is placed in Sec. 2.8. There, we show that for an LTIC system specified by Eq. (2.11), the unit impulse response *h*(*t*) is given by - -$$ -h(t) = b_0 \delta(t) + [P(D)y_n(t)]u(t) -$$ -\n(2.17) - -where *yn*(*t*) is a linear combination of the characteristic modes of the system subject to the following initial conditions: - -$$ -y_n(0) = \dot{y}_n(0) = \ddot{y}_n(0) = \dots = y_n^{(N-2)}(0) = 0 -$$ - and $y_n^{(N-1)}(0) = 1$ (2.18) - -where *y*(*k*) *n* (0) is the value of the *k*th derivative of *yn*(*t*) at *t* = 0. We can express this set of conditions for various values of *N* (the system order) as follows: - -$$ -N = 1 : y_n(0) = 1 -$$ - -\n -$$ -N = 2 : y_n(0) = 0, \dot{y}_n(0) = 1 -$$ - -\n -$$ -N = 3 : y_n(0) = \dot{y}_n(0) = 0, \ddot{y}_n(0) = 1 -$$ - -and so on. - -As stated earlier, if the order of *P*(*D*) is less than the order of *Q*(*D*), that is, if *M* < *N*, then *b*0 = 0, and the impulse term *b*0δ(*t*) in *h*(*t*) is zero. - -### **EXAMPLE 2.6 Impulse Response via Simplified Impulse Matching** - -Determine the unit impulse response *h*(*t*) for a system specified by the equation - -$$ -(D2 + 3D + 2) y(t) = Dx(t) -$$ - (2.19) - -This is a second-order system (*N* = 2) having the characteristic polynomial - -$$ -(\lambda^2 + 3\lambda + 2) = (\lambda + 1)(\lambda + 2) -$$ - -The characteristic roots of this system are λ = −1 and λ = −2. Therefore, - -$$ -y_n(t) = c_1 e^{-t} + c_2 e^{-2t} -$$ -\n(2.20) - -Differentiation of this equation yields - -$$ -\dot{y}_n(t) = -c_1 e^{-t} - 2c_2 e^{-2t} \tag{2.21} -$$ - -The initial conditions are [see Eq. (2.18)] - -$$ -\dot{y}_n(0) = 1 \qquad \text{and} \qquad y_n(0) = 0 -$$ - -Setting *t* = 0 in Eqs. (2.20) and (2.21), and substituting the initial conditions just given, we obtain - -$$ -0 = c_1 + c_2 -$$ - -$$ -1 = -c_1 - 2c_2 -$$ - -Solution of these two simultaneous equations yields - -$$ -c_1 = 1 \qquad \text{and} \qquad c_2 = -1 -$$ - -Therefore, - -$$ -y_n(t) = e^{-t} - e^{-2t} -$$ - -Moreover, according to Eq. (2.19), *P*(*D*) = *D* so that - -$$ -P(D)y_n(t) = Dy_n(t) = \dot{y}_n(t) = -e^{-t} + 2e^{-2t} -$$ - -Also in this case, *b*0 = 0 [the second-order term is absent in *P*(*D*)]. Therefore, - -$$ -h(t) = [P(D)y_n(t)]u(t) = (-e^{-t} + 2e^{-2t})u(t) -$$ - -**Comment.** In the above discussion, we have assumed *M* ≤ *N*, as specified by Eq. (2.11). Section 2.8 shows that the expression for *h*(*t*) applicable to all possible values of *M* and *N* is given by - -$$ -h(t) = P(D)[y_n(t)u(t)] -$$ - -where *yn*(*t*) is a linear combination of the characteristic modes of the system subject to initial conditions [Eq. (2.18)]. This expression reduces to Eq. (2.17) when *M* ≤ *N*. - -Determination of the impulse response *h*(*t*) using the procedures in this section is relatively simple. However, in Ch. 4 we shall discuss another, even simpler method using the Laplace transform. As the next example demonstrates, it is also possible to find *h*(*t*) using functions from MATLAB's symbolic math toolbox. - -### **EXAMPLE 2.7 Using MATLAB to Find the Impulse Response** - -Determine the impulse response *h*(*t*) for an LTIC system specified by the differential equation - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -This is a second-order system with *b*0 = 0. First we find the zero-input component for initial conditions *y*(0−) = 0, and *y*˙(0−) = 1. Since *P*(*D*) = *D*, the zero-input response is differentiated and the impulse response immediately follows as *h*(*t*) = 0δ(*t*)+ [*Dyn*(*t*)]*u*(*t*). - ->> y\_n = dsolve('D2y+3\*Dy+2\*y=0','y(0)=0','Dy(0)=1','t'); h = diff(y\_n) h = 2/exp(2\*t) - 1/exp(t) - -Therefore, *h*(*t*) = (2*e*−2*t* −*e*−*t* )*u*(*t*). - -### **DR ILL 2.4 Finding the Impulse Response** - -Determine the unit impulse response of LTIC systems described by the following equations: - -- **(a)** (*D*+2)*y*(*t*) = (3*D*+5)*x*(*t*) -- **(b)** *D*(*D*+2)*y*(*t*) = (*D*+4)*x*(*t*) -- **(c)** (*D*2 +2*D*+1)*y*(*t*) = *Dx*(*t*) - -### **ANSWERS** - -- **(a)** 3δ(*t*)−*e*−2*t u*(*t*) -- **(b)** (2−*e*−2*t* )*u*(*t*) -- **(c)** (1−*t*)*e*−*t u*(*t*) - -### SYSTEM RESPONSE TO DELAYED IMPULSE - -If *h*(*t*) is the response of an LTIC system to the input δ(*t*), then *h*(*t*−*T*) is the response of this same system to the input δ(*t* − *T*). This conclusion follows from the time-invariance property of LTIC systems. Thus, by knowing the unit impulse response *h*(*t*), we can determine the system response to a delayed impulse δ(*t* − *T*). Next, we put this result to good use in finding an LTIC system's zero-state response. - -## **2.4 SYSTEM [RESPONSE TO](#page-8-0) EXTERNAL INPUT: THE ZERO-STATE RESPONSE** - -This section is devoted to the determination of the zero-state response of an LTIC system. This is the system response *y*(*t*) to an input *x*(*t*) when the system is in the zero state, that is, when all initial conditions are zero. *We shall assume that the systems discussed in this section are in the zero state unless mentioned otherwise.* Under these conditions, the zero-state response will be the total response of the system. - -We shall use the superposition property for finding the system response to an arbitrary input *x*(*t*). Let us define a basic pulse *p*(*t*) of unit height and width τ , starting at *t* = 0 as illustrated in Fig. 2.3a. Figure 2.3b shows an input *x*(*t*) as a sum of narrow rectangular pulses. The pulse starting at *t* = *n*τ in Fig. 2.3b has a height *x*(*n*τ ) and can be expressed as *x*(*n*τ )*p*(*t*−*n*τ ). Now, *x*(*t*) is the sum of all such pulses. Hence, - -$$ -x(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau) p(t - n\Delta \tau) = \lim_{\Delta \tau \to 0} \sum_{\tau} \left[ \frac{x(n\Delta \tau)}{\Delta \tau} \right] p(t - n\Delta \tau) \Delta \tau -$$ - -The term [*x*(*n*τ )/τ ]*p*(*t* − *n*τ ) represents a pulse *p*(*t* − *n*τ ) with height *x*(*n*τ )/τ . As τ → 0, the height of this strip → ∞, but its area remains *x*(*n*τ ). Hence, this strip approaches an impulse *x*(*n*τ )δ(*t* −*n*τ ) as τ → 0 (Fig. 2.3e). Therefore, - -$$ -x(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau) \delta(t - n\Delta \tau) \Delta \tau -$$ - (2.22) - -To find the response for this input *x*(*t*), we consider the input and the corresponding output pairs, as shown in Figs. 2.3c–2.3f and also shown by directed arrow notation as follows: - -input -$$ -\Rightarrow -$$ - output -\n -$$ -\delta(t) \Longrightarrow h(t) -$$ -\n -$$ -\delta(t - n\Delta\tau) \Longrightarrow h(t - n\Delta\tau) -$$ -\n -$$ -[x(n\Delta\tau)\Delta\tau]\delta(t - n\Delta\tau) \Longrightarrow [x(n\Delta\tau)\Delta\tau]h(t - n\Delta\tau) -$$ -\n -$$ -\lim_{\Delta\tau \to 0} \sum_{\tau} x(n\Delta\tau)\delta(t - n\Delta\tau) \Delta\tau \Longrightarrow \lim_{\Delta\tau \to 0} \sum_{\tau} x(n\Delta\tau)h(t - n\Delta\tau) \Delta\tau -$$ -\n -$$ -x(t) \quad [\text{see Eq. (2.22)}] -$$ - -**Figure 2.3** Finding the system response to an arbitrary input *x*(*t*). - -Therefore,† - -$$ -y(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau)h(t - n\Delta \tau)\Delta \tau -$$ - -= -$$ -\int_{-\infty}^{\infty} x(\tau)h(t - \tau) d\tau -$$ - (2.23) - -This is the result we seek. We have obtained the system response *y*(*t*) to an arbitrary input *x*(*t*) in terms of the unit impulse response *h*(*t*). Knowing *h*(*t*), we can determine the response *y*(*t*) to any input. *Observe once again the all-pervasive nature of the system's characteristic modes. The system response to any input is determined by the impulse response, which, in turn, is made up of characteristic modes of the system.* - -It is important to keep in mind the assumptions used in deriving Eq. (2.23). We assumed a linear time-invariant (LTI) system. Linearity allowed us to use the principle of superposition, and time invariance made it possible to express the system's response to δ(*t* −*n*τ ) as *h*(*t* −*n*τ ). - -### **[2.4-1 The Convolution Integral](#page-8-0)** - -The zero-state response *y*(*t*) obtained in Eq. (2.23) is given by an integral that occurs frequently in the physical sciences, engineering, and mathematics. For this reason this integral is given a special name: the *convolution integral*. The convolution integral of two functions *x*1(*t*) and *x*2(*t*) is denoted symbolically by *x*1(*t*) ∗ *x*2(*t*) and is defined as - -$$ -x_1(t) * x_2(t) \equiv \int_{-\infty}^{\infty} x_1(\tau) x_2(t - \tau) d\tau -$$ - (2.24) - -Some important properties of the convolution integral follow. - -### THE COMMUTATIVE PROPERTY - -Convolution operation is commutative; that is, *x*1(*t*) ∗ *x*2(*t*) = *x*2(*t*) ∗ *x*1(*t*). This property can be proved by a change of variable. In Eq. (2.24), if we let *z* = *t* − τ so that τ = *t* − *z* and *d*τ = −*dz*, we obtain - -$$ -x_1(t) * x_2(t) = -\int_{-\infty}^{-\infty} x_2(z) x_1(t-z) dz -$$ - -= -$$ -\int_{-\infty}^{\infty} x_2(z) x_1(t-z) dz -$$ - -= -$$ -x_2(t) * x_1(t) -$$ - (2.25) - -$$ -y(t) = \int_{-\infty}^{\infty} x(\tau)h(t, \tau) d\tau -$$ - -where *h*(*t*, τ ) is the system response at instant *t* to a unit impulse input located at τ . - - - -† In deriving this result we have assumed a time-invariant system. If the system is time-varying, then the system response to the input δ(*t*−*n*Δτ ) cannot be expressed as *h*(*t*−*n*Δτ ) but instead has the form *h*(*t*,*n*Δτ ). Use of this form modifies Eq. (2.23) to - -### THE DISTRIBUTIVE PROPERTY - -According to the distributive property, - -$$ -x_1(t) * [x_2(t) + x_3(t)] = x_1(t) * x_2(t) + x_1(t) * x_3(t) -$$ -\n(2.26) - -### THE ASSOCIATIVE PROPERTY - -According to the associative property, - -$$ -x_1(t) * [x_2(t) * x_3(t)] = [x_1(t) * x_2(t)] * x_3(t) -$$ -\n(2.27) - -The proofs of Eqs. (2.26) and (2.27) follow directly from the definition of the convolution integral. They are left as an exercise for the reader. - -THE SHIFT PROPERTY If - -$$ -x_1(t) * x_2(t) = c(t) -$$ - -then - -$$ -x_1(t) * x_2(t - T) = x_1(t - T) * x_2(t) = c(t - T) -$$ - -More generally, we see that - -$$ -x_1(t - T_1) * x_2(t - T_2) = c(t - T_1 - T_2) -$$ -\n(2.28) - -**Proof.** We are given - -$$ -x_1(t) * x_2(t) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t - \tau) d\tau = c(t) -$$ - -Therefore, - -$$ -x_1(t) * x_2(t-T) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t-T-\tau) d\tau -$$ -$$ -= c(t-T) -$$ - -The equally simple proof of Eq. (2.28) follows a similar approach. - -### CONVOLUTION WITH AN IMPULSE - -Convolution of a function *x*(*t*) with a unit impulse results in the function *x*(*t*) itself. By definition of convolution, - -$$ -x(t) * \delta(t) = \int_{-\infty}^{\infty} x(\tau) \delta(t - \tau) d\tau -$$ - -Because δ(*t* −τ ) is an impulse located at τ = *t*, according to the sampling property of the impulse [Eq. (1.11)], the integral here is just the value of *x*(τ ) at τ = *t*, that is, *x*(*t*). Therefore, - -$$ -x(t) * \delta(t) = x(t) -$$ - -Actually this result was derived earlier [Eq. (2.22)]. - -#### 172 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -### THE WIDTH PROPERTY - -If the durations (widths) of *x*1(*t*) and *x*2(*t*) are finite, given by *T*1 and *T*2, respectively, then the duration (width) of *x*1(*t*) ∗ *x*2(*t*) is *T*1 + *T*2 (Fig. 2.4). The proof of this property follows readily from the graphical considerations discussed later in Sec. 2.4-2. - -**Figure 2.4** Width property of convolution. - -### ZERO-STATE RESPONSE AND CAUSALITY - -The (zero-state) response *y*(*t*) of an LTIC system is - -$$ -y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau -$$ -\n(2.29) - -In deriving Eq. (2.29), we assumed the system to be linear and time-invariant. There were no other restrictions either on the system or on the input signal *x*(*t*). Since, in practice, most systems are causal, their response cannot begin before the input. Furthermore, most inputs are also causal, which means they start at *t* = 0. - -Causality restriction on both signals and systems further simplifies the limits of integration in Eq. (2.29). By definition, the response of a causal system cannot begin before its input begins. Consequently, the causal system's response to a unit impulse δ(*t*) (which is located at *t* = 0) cannot begin before *t* = 0. Therefore, a *causal system's unit impulse response h*(*t*) *is a causal signal*. - -It is important to remember that the integration in Eq. (2.29) is performed with respect to τ (not *t*). If the input *x*(*t*) is causal, *x*(τ ) = 0 for τ < 0. Therefore, *x*(τ ) = 0 for τ < 0, as illustrated in Fig. 2.5a. Similarly, if *h*(*t*) is causal, *h*(*t* − τ ) = 0 for *t* − τ < 0; that is, for τ > *t*, as depicted in Fig. 2.5a. Therefore, the product *x*(τ )*h*(*t* − τ ) = 0 everywhere except over the nonshaded interval 0 ≤ τ ≤ *t* shown in Fig. 2.5a (assuming *t* ≥ 0). Observe that if *t* is negative, *x*(τ )*h*(*t* − τ ) = 0 for all τ , as shown in Fig. 2.5b. Therefore, Eq. (2.29) reduces to - -$$ -y(t) = x(t) * h(t) = \begin{cases} \int_0^t x(\tau)h(t-\tau) d\tau & t \ge 0\\ 0 & t < 0 \end{cases} -$$ - (2.30) - -The lower limit of integration in Eq. (2.30) is taken as 0− to avoid the difficulty in integration that can arise if *x*(*t*) contains an impulse at the origin. This result shows that if *x*(*t*) and *h*(*t*) are both causal, the response *y*(*t*) is also causal. - -Because of the convolution's commutative property [Eq. (2.25)], we can also express Eq. (2.30) as [assuming causal *x*(*t*) and *h*(*t*)] - -$$ -y(t) = \begin{cases} \int_0^t h(\tau)x(t-\tau)d\tau & t \ge 0\\ 0 & t < 0 \end{cases} -$$ - -Hereafter, the lower limit of 0− will be implied even when we write it as 0. As in Eq. (2.30), this result assumes that both the input and the system are causal. - -### **EXAMPLE 2.8 Computing the Zero-State Response** - -For an LTIC system with the unit impulse response *h*(*t*) = *e*−2*t u*(*t*), determine the response *y*(*t*) for the input - -$$ -x(t) = e^{-t}u(t) -$$ - -Here both *x*(*t*) and *h*(*t*) are causal (Fig. 2.6). Hence, from Eq. (2.30), we obtain - -$$ -y(t) = \int_0^t x(\tau)h(t-\tau) d\tau \qquad t \ge 0 -$$ - -Because *x*(*t*) = *e*−*t u*(*t*) and *h*(*t*) = *e*−2*t u*(*t*), - -$$ -x(\tau) = e^{-\tau} u(\tau) -$$ - and $h(t - \tau) = e^{-2(t - \tau)} u(t - \tau)$ - -Remember that the integration is performed with respect to τ (not *t*), and the region of integration is 0 ≤ τ ≤ *t*. Hence, τ ≥ 0 and *t* − τ ≥ 0. Therefore, *u*(τ ) = 1 and *u*(*t* − τ ) = 1; consequently, - -$$ -y(t) = \int_0^t e^{-\tau} e^{-2(t-\tau)} d\tau \qquad t \ge 0 -$$ - -Because this integration is with respect to τ , we can pull *e*−2*t* outside the integral, giving us - -$$ -y(t) = e^{-2t} \int_0^t e^{\tau} d\tau = e^{-2t} (e^t - 1) = e^{-t} - e^{-2t} \qquad t \ge 0 -$$ - -Moreover, *y*(*t*) = 0 when *t* < 0 [see Eq. (2.30)]. Therefore, - -$$ -y(t) = (e^{-t} - e^{-2t})u(t) -$$ - -The response is depicted in Fig. 2.6c. - -### **DR ILL 2.5 Computing the Zero-State Response** - -For an LTIC system with the impulse response *h*(*t*) = 6*e*−*t u*(*t*), determine the system response to the input: **(a)** 2*u*(*t*) and **(b)** 3*e*−3*t u*(*t*). - -### **ANSWERS** - -- **(a)** 12(1−*e*−*t* )*u*(*t*) -- **(b)** 9(*e*−*t* −*e*−3*t* )*u*(*t*) - -### **DR ILL 2.6 Zero-State Response with Resonance** - -Repeat Drill 2.5 for the input *x*(*t*) = *e*−*t u*(*t*). - -### **ANSWER** - -6*te*−*t u*(*t*) - -### THE CONVOLUTION TABLE - -The task of convolution is considerably simplified by a ready-made convolution table (Table 2.1). This table, which lists several pairs of signals and their convolution, can conveniently determine *y*(*t*), a system response to an input *x*(*t*), without performing the tedious job of integration. For instance, we could have readily found the convolution in Ex. 2.8 by using pair 4 (with λ1 = −1 and λ2 = −2) to be (*e*−*t* −*e*−2*t* )*u*(*t*). The following example demonstrates the utility of this table. - -### **EXAMPLE 2.9 Convolution by Tables** - -Use Table 2.1 to compute the loop current *y*(*t*) of the *RLC* circuit in Ex. 2.4 for the input *x*(*t*) = 10*e*−3*t u*(*t*) when all the initial conditions are zero. - -The loop equation for this circuit [see Ex. 1.16 or Eq. (1.29)] is - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -The impulse response *h*(*t*) for this system, as obtained in Ex. 2.6, is - -$$ -h(t) = (2e^{-2t} - e^{-t})u(t) -$$ - -The input is *x*(*t*) = 10*e*−3*t u*(*t*), and the response *y*(*t*) is - -$$ -y(t) = x(t) * h(t) = 10e^{-3t}u(t) * [2e^{-2t} - e^{-t}]u(t) -$$ - -Using the distributive property of the convolution [Eq. (2.26)], we obtain - -$$ -y(t) = 10e^{-3t}u(t) * 2e^{-2t}u(t) - 10e^{-3t}u(t) * e^{-t}u(t) -$$ - -= 20[e-3tu(t) \* e-2tu(t)] - 10[e-3tu(t) \* e-tu(t)] - -Now the use of pair 4 in Table 2.1 yields - -$$ -y(t) = \frac{20}{-3 - (-2)} [e^{-3t} - e^{-2t}] u(t) - \frac{10}{-3 - (-1)} [e^{-3t} - e^{-t}] u(t) -$$ - -= $-20(e^{-3t} - e^{-2t}) u(t) + 5(e^{-3t} - e^{-t}) u(t)$ -= $(-5e^{-t} + 20e^{-2t} - 15e^{-3t}) u(t)$ - -| No. | x1(t) | x2(t) | x1(t)
∗ x2(t)
= x2(t)
∗ x1(t) | -|-----|-------------------------|--------------------|-----------------------------------------------------------------------------------| -| 1 | x(t) | δ(t −T) | x(t −T) | -| 2 | eλt
u(t) | u(t) | 1−eλt
u(t)
−λ | -| 3 | u(t) | u(t) | tu(t) | -| 4 | eλ1t
u(t) | eλ2t
u(t) | eλ1t −eλ2t
u(t)
λ1
= λ2
λ1
−λ2 | -| 5 | eλt
u(t) | eλt
u(t) | teλt
u(t) | -| 6 | teλt
u(t) | eλt
u(t) | 1
2eλt
t
u(t)
2 | -| 7 | Nu(t)
t | eλt
u(t) | "N
N! eλt
N−k
N!t
λN+1 u(t) −
u(t)
λk+1(N
−k)!
k=0 | -| 8 | Mu(t)
t | Nu(t)
t | M!N!
M+N+1u(t)
t
(M +N +1)! | -| 9 | teλ1t
u(t) | eλ2t
u(t) | eλ2t −eλ1t +(λ1
−λ2)teλ1t
u(t)
−λ2)2
(λ1 | -| 10 | Meλt
t
u(t) | Neλt
t
u(t) | M!N!
M+N+1eλt
t
u(t)
(N + M +1)! | -| 11 | Meλ1t
t
u(t) | Neλ2t
t
u(t) | "M
(−1)kM!(N
M−keλ1t
+k)!t
−λ2)N+k+1 u(t)
k!(M −k)!(λ1
k=0 | -| | λ1
= λ2 | | "N
(−1)kN!(M
N−keλ2t
+k)!t
−λ1)M+k+1 u(t)
+
k!(N −k)!(λ2
k=0 | -| 12 | e−αt cos(βt
+θ )u(t) | eλt
u(t) | cos(θ −φ)eλt −e−αt cos(βt
+θ −φ)
u(t)

(α +λ)2 +β2 | -| | | | φ = tan−1[−β/(α
+λ)] | -| 13 | eλ1t
u(t) | eλ2t
u(−t) | eλ1t
u(t)+eλ2t
u(−t)
Reλ2
> Reλ1
λ2
−λ1 | -| 14 | eλ1t
u(−t) | eλ2t
u(−t) | eλ1t −eλ2t
u(−t)
λ2
−λ1 | - -**TABLE 2.1** Select Convolution Integrals - -### **DR ILL 2.7 Convolution by Tables** - -Use Table 2.1 to show *e*−2*t u*(*t*) ∗ (1−*e*−*t* )*u*(*t*) = 1 2 −*e*−*t* + 1 2 *e*−2*t u*(*t*). - -### **DR ILL 2.8 Zero-State Response by Convolution Table** - -Rework Drills 2.5 and 2.6 using Table 2.1. - -### **DR ILL 2.9 Another Zero-State Response by Convolution Table** - -For an LTIC system with the unit impulse response *h*(*t*) = *e*−2*t u*(*t*), determine the zero-state response *y*(*t*) if the input *x*(*t*) = sin 3*t u*(*t*). [*Hint:* Use pair 12 from Table 2.1.] - -### **ANSWER** - -1 13 [3*e*−2*t* + √13 cos(3*t* −146.32◦)]*u*(*t*) or 1 13 [3*e*−2*t* − √13 cos(3*t* +33.68◦)]*u*(*t*) - -### RESPONSE TO COMPLEX INPUTS - -The LTIC system response discussed so far applies to general input signals, real or complex. However, if the system is real, that is, if *h*(*t*) is real, then we shall show that the real part of the input generates the real part of the output, and a similar conclusion applies to the imaginary part. - -If the input is *x*(*t*) = *xr*(*t*) + *jxi*(*t*), where *xr*(*t*) and *xi*(*t*) are the real and imaginary parts of *x*(*t*), then for real *h*(*t*) - -$$ -y(t) = h(t) * [x_r(t) + jx_i(t)] = h(t) * x_r(t) + jh(t) * x_i(t) = y_r(t) + jy_i(t) -$$ - -where *yr*(*t*) and *yi*(*t*) are the real and the imaginary parts of *y*(*t*). Using the right-directed-arrow notation to indicate a pair of the input and the corresponding output, the foregoing result can be expressed as follows. If - -$$ -x(t) = x_r(t) + jx_i(t) \implies y(t) = y_r(t) + jy_i(t) -$$ - -then - -$$ -x_r(t) \implies y_r(t) \quad \text{and} \quad x_i(t) \implies y_i(t) \tag{2.31} -$$ - -### MULTIPLE INPUTS - -Multiple inputs to LTI systems can be treated by applying the superposition principle. Each input is considered separately, with all other inputs assumed to be zero. The sum of all these individual system responses constitutes the total system output when all the inputs are applied simultaneously. - -### **[2.4-2 Graphical Understanding of Convolution Operation](#page-8-0)** - -The convolution operation can be grasped readily through a graphical interpretation of the convolution integral. Such an understanding is helpful in evaluating the convolution integral of more complex signals. In addition, graphical convolution allows us to grasp visually or mentally the convolution integral's result, which can be of great help in sampling, filtering, and many other problems. Finally, many signals have no exact mathematical description, so they can be described only graphically. If two such signals are to be convolved, we have no choice but to perform their convolution graphically. - -We shall now explain the convolution operation by convolving the signals *x*(*t*) and *g*(*t*), illustrated in Figs. 2.7a and 2.7b, respectively. If *c*(*t*) is the convolution of *x*(*t*) with *g*(*t*), then - -$$ -c(t) = \int_{-\infty}^{\infty} x(\tau)g(t-\tau) d\tau -$$ - -One of the crucial points to remember here is that this integration is performed with respect to τ so that *t* is just a parameter (like a constant). This consideration is especially important when we sketch the graphical representations of the functions *x*(τ ) and *g*(*t*−τ ). Both these functions should be sketched as functions of τ , not of *t*. - -The function *x*(τ ) is identical to *x*(*t*), with τ replacing *t* (Fig. 2.7c). Therefore, *x*(*t*) and *x*(τ ) will have the same graphical representations. Similar remarks apply to *g*(*t*) and *g*(τ ) (Fig. 2.7d). - -To appreciate what *g*(*t* − τ ) looks like, let us start with the function *g*(τ ) (Fig. 2.7d). Time reversal of this function (reflection about the vertical axis τ = 0) yields *g*(−τ ) (Fig. 2.7e). Let us denote this function by φ(τ ): - -$$ -\phi(\tau) = g(-\tau) -$$ - -Now φ(τ ) shifted by *t* seconds is φ(τ −*t*), given by - -$$ -\phi(\tau - t) = g[-(\tau - t)] = g(t - \tau) -$$ - -Therefore, we first time-reverse *g*(τ ) to obtain *g*(−τ ) and then time-shift *g*(−τ ) by *t* to obtain *g*(*t* − τ ). For positive *t*, the shift is to the right (Fig. 2.7f); for negative *t*, the shift is to the left (Figs. 2.7g, 2.7h). - -The preceding discussion gives us a graphical interpretation of the functions *x*(τ ) and *g*(*t*−τ ). The convolution *c*(*t*) is the area under the product of these two functions. Thus, to compute *c*(*t*) at some positive instant *t* = *t*1, we first obtain *g*(−τ ) by inverting *g*(τ ) about the vertical axis. Next, we right-shift or delay *g*(−τ ) by *t*1 to obtain *g*(*t*1 − τ ) (Fig. 2.7f), and then we multiply this function by *x*(τ ), giving us the product *x*(τ )*g*(*t*1 − τ ) (shaded portion in Fig. 2.7f). The area *A*1 under this product is *c*(*t*1), the value of *c*(*t*) at *t* = *t*1. We can therefore plot *c*(*t*1) = *A*1 on a curve describing *c*(*t*), as shown in Fig. 2.7i. The area under the product *x*(τ )*g*(−τ ) in Fig. 2.7e is *c*(0), the value of the convolution for *t* = 0 (at the origin). - -**Figure 2.7** Graphical explanation of the convolution operation. - -### 180 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -A similar procedure is followed in computing the value of *c*(*t*) at *t* = *t*2, where *t*2 is negative (Fig. 2.7g). In this case, the function *g*(−τ ) is shifted by a negative amount (that is, left-shifted) to obtain *g*(*t*2−τ ). Multiplication of this function with *x*(τ ) yields the product *x*(τ )*g*(*t*2−τ ). The area under this product is *c*(*t*2) = *A*2, giving us another point on the curve *c*(*t*) at *t* = *t*2 (Fig. 2.7i). This procedure can be repeated for all values of *t*, from −∞ to ∞. The result will be a curve describing *c*(*t*) for all time *t*. Note that when *t* ≤ −3, *x*(τ ) and *g*(*t*−τ ) do not overlap (see Fig. 2.7h); therefore, *c*(*t*) = 0 for *t* ≤ −3. - -### SUMMARY OF THE GRAPHICAL PROCEDURE - -The procedure for graphical convolution can be summarized as follows: - -- 1. Keep the function *x*(τ ) fixed. -- 2. Visualize the function *g*(τ ) as a rigid wire frame, and rotate (or invert) this frame about the vertical axis (τ = 0) to obtain *g*(−τ ). -- 3. Shift the inverted frame along the τ axis by *t*0 seconds. The shifted frame now represents *g*(*t*0 −τ ). -- 4. The area under the product of *x*(τ ) and *g*(*t*0 − τ ) (the shifted frame) is *c*(*t*0), the value of the convolution at *t* = *t*0. -- 5. Repeat this procedure, shifting the frame by different values (positive and negative) to obtain *c*(*t*) for all values of *t*. - -The graphical procedure discussed here appears very complicated and discouraging at first reading. Indeed, some people claim that convolution has driven many electrical engineering undergraduates to contemplate theology either for salvation or as an alternative career (*IEEE Spectrum,* March 1991, p. 60). Actually, the bark of convolution is worse than its bite. In graphical convolution, we need to determine the area under the product *x*(τ )*g*(*t* − τ ) for all values of *t* from −∞ to ∞. However, a mathematical description of *x*(τ )*g*(*t* − τ ) is generally valid over a range - -Convolution: Its bark is worse than its bite! - -of *t*. Therefore, repeating the procedure for every value of *t* amounts to repeating it only a few times for different ranges of *t*. - -We can also use the commutative property of convolution to our advantage by computing *x*(*t*) ∗ *g*(*t*) or *g*(*t*) ∗ *x*(*t*), whichever is simpler. As a rule of thumb, *convolution computations are simplified if we choose to invert (time-reverse) the simpler of the two functions*. For example, if the mathematical description of *g*(*t*) is simpler than that of *x*(*t*), then *x*(*t*) ∗ *g*(*t*) will be easier to compute than *g*(*t*) ∗ *x*(*t*). In contrast, if the mathematical description of *x*(*t*) is simpler, the reverse will be true. - -We shall demonstrate graphical convolution with the following examples. Let us start by using this graphical method to rework Ex. 2.8. - -### **EXAMPLE 2.10 Graphical Convolution of Two Causal Functions** - -Determine graphically *y*(*t*) = *x*(*t*) ∗ *h*(*t*) for *x*(*t*) = *e*−*t u*(*t*) and *h*(*t*) = *e*−2*t u*(*t*). - -In Figs. 2.8a and 2.8b we have *x*(*t*) and *h*(*t*), respectively; and Fig. 2.8c shows *x*(τ ) and *h*(−τ ) as functions of τ . The function *h*(*t* −τ ) is now obtained by shifting *h*(−τ ) by *t*. If *t* is positive, the shift is to the right (delay); if *t* is negative, the shift is to the left (advance). Figure 2.8d shows that for negative *t*, *h*(*t* −τ ) [obtained by left-shifting *h*(−τ )] does not overlap *x*(τ ), and the product *x*(τ )*h*(*t* −τ ) = 0, so that - -$$ -y(t) = 0 \qquad t < 0 -$$ - -Figure 2.8e shows the situation for *t* ≥ 0. Here *x*(τ ) and *h*(*t* −τ ) do overlap, but the product is nonzero only over the interval 0 ≤ τ ≤ *t* (shaded interval). Therefore, - -$$ -y(t) = \int_0^t x(\tau)h(t-\tau) d\tau \qquad t \ge 0 -$$ - -All we need to do now is substitute correct expressions for *x*(τ ) and *h*(*t* − τ ) in this integral. From Figs. 2.8a and 2.8b, it is clear that the segments of *x*(*t*) and *g*(*t*) to be used in this convolution (Fig. 2.8e) are described by - -$$ -x(t) = e^{-t} \qquad \text{and} \qquad h(t) = e^{-2t} -$$ - -Therefore, - -$$ -x(\tau) = e^{-\tau} -$$ - and $h(t - \tau) = e^{-2(t - \tau)}$ - -Consequently, - -$$ -y(t) = \int_0^t e^{-\tau} e^{-2(t-\tau)} d\tau = e^{-2t} \int_0^t e^{\tau} d\tau = e^{-t} - e^{-2t} \qquad t \ge 0 -$$ - -Moreover, *y*(*t*) = 0 for *t* < 0 so that - -$$ -y(t) = (e^{-t} - e^{-2t})u(t) -$$ - -## **EXAMPLE 2.11 Graphical Convolution: Causal Function and Two-Sided Function** - -Find *c*(*t*) = *x*(*t*) ∗ *g*(*t*) for the signals depicted in Figs. 2.9a and 2.9b. - -Since *x*(*t*) is simpler than *g*(*t*), it is easier to evaluate *g*(*t*) ∗ *x*(*t*) than *x*(*t*) ∗ *g*(*t*). However, we shall intentionally take the more difficult route and evaluate *x*(*t*) ∗ *g*(*t*). - -From *x*(*t*) and *g*(*t*) (Figs. 2.9a and 2.9b, respectively), observe that *g*(*t*) is composed of two segments. As a result, it can be described as - -$$ -g(t) = \begin{cases} 2e^{-t} & \text{segment A} \\ -2e^{2t} & \text{segment B} \end{cases} -$$ - -Therefore, - -$$ -g(t - \tau) = \begin{cases} 2e^{-(t-\tau)} & \text{segment A} \\ -2e^{2(t-\tau)} & \text{segment B} \end{cases} -$$ - -The segment of *x*(*t*) that is used in convolution is *x*(*t*) = 1 so that *x*(τ ) = 1. Figure 2.9c shows *x*(τ ) and *g*(−τ ). - -To compute *c*(*t*) for *t* ≥ 0, we right-shift *g*(−τ ) to obtain *g*(*t*−τ ), as illustrated in Fig. 2.9d. Clearly, *g*(*t* − τ ) overlaps with *x*(τ ) over the shaded interval, that is, over the range τ ≥ 0; segment A overlaps with *x*(τ ) over the interval (0,*t*), while segment B overlaps with *x*(τ ) over (*t*,∞). Remembering that *x*(τ ) = 1, we have - -$$ -c(t) = \int_0^\infty x(\tau)g(t-\tau) d\tau -$$ - -= -$$ -\int_0^t 2e^{-(t-\tau)} d\tau + \int_t^\infty -2e^{2(t-\tau)} d\tau -$$ - -= -$$ -2(1 - e^{-t}) - 1 = 1 - 2e^{-t} \qquad t \ge 0 -$$ - -Figure 2.9e shows the situation for *t* < 0. Here the overlap is over the shaded interval, that is, over the range τ ≥ 0, where only the segment B of *g*(*t*) is involved. Therefore, - -$$ -c(t) = \int_0^\infty x(\tau)g(t-\tau) d\tau = \int_0^\infty -2e^{2(t-\tau)} d\tau = -e^{2t} \qquad t \le 0 -$$ - -Therefore, - -$$ -c(t) = \begin{cases} 1 - 2e^{-t} & t \ge 0\\ -e^{2t} & t \le 0 \end{cases} -$$ - -Figure 2.9f shows a plot of *c*(*t*). - -## **EXAMPLE 2.12 Graphical Convolution of Two Finite-Duration Functions** - -Find *x*(*t*) ∗ *g*(*t*) for the functions *x*(*t*) and *g*(*t*) shown in Figs. 2.10a and 2.10b. - -Here, *x*(*t*) has a simpler mathematical description than that of *g*(*t*), so it is preferable to time-reverse *x*(*t*). Hence, we shall determine *g*(*t*) ∗ *x*(*t*) rather than *x*(*t*) ∗ *g*(*t*). Thus, - -$$ -c(t) = g(t) * x(t) = \int_{-\infty}^{\infty} g(\tau) x(t - \tau) d\tau -$$ - -First, we determine the expressions for the segments of *x*(*t*) and *g*(*t*) used in finding *c*(*t*). According to Figs. 2.10a and 2.10b, these segments can be expressed as - -$$ -x(t) = 1 -$$ - and $g(t) = \frac{1}{3}t$ - -so that - -$$ -x(t - \tau) = 1 \qquad \text{and} \qquad g(\tau) = \frac{1}{3}\tau -$$ - -Figure 2.10c shows *g*(τ ) and *x*(−τ ), whereas Fig. 2.10d shows *g*(τ ) and *x*(*t* − τ ), which is *x*(−τ ) shifted by *t*. Because the edges of *x*(−τ ) are at τ = −1 and 1, the edges of *x*(*t* − τ ) are at −1 + *t* and 1 + *t*. The two functions overlap over the interval (0, 1 + *t*) (shaded interval) so that - -$$ -c(t) = \int_0^{1+t} g(\tau) x(t-\tau) d\tau = \int_0^{1+t} \frac{1}{3} \tau d\tau = \frac{1}{6} (t+1)^2 \qquad -1 \le t \le 1 \qquad (2.32) -$$ - -This situation, depicted in Fig. 2.10d, is valid only for −1 ≤ *t* ≤ 1. For *t* ≥ 1 but ≤ 2, the situation is as illustrated in Fig. 2.10e. The two functions overlap only over the range −1 + *t* to 1 + *t* (shaded interval). Note that the expressions for *g*(τ ) and *x*(*t* − τ ) do not change; only the range of integration changes. Therefore, - -$$ -c(t) = \int_{-1+t}^{1+t} \frac{1}{3}\tau \,d\tau = \frac{2}{3}t \qquad \qquad 1 \le t \le 2 \tag{2.33} -$$ - -Also note that the expressions in Eqs. (2.32) and (2.33) both apply at *t* = 1, the transition point between their respective ranges. We can readily verify that both expressions yield a value of 2/3 at *t* = 1 so that *c*(1) = 2/3. The continuity of *c*(*t*) at transition points indicates a high probability of a correct answer. Continuity of *c*(*t*) at transition points is assured as long as *x*(*t*) and *g*(*t*) contain no impulse functions. - -For *t* ≥ 2 but ≤ 4, the situation is as shown in Fig. 2.10f. The functions *g*(τ ) and *x*(*t* − τ ) overlap over the interval from −1+*t* to 3 (shaded interval) so that - -$$ -c(t) = \int_{-1+t}^{3} \frac{1}{3}\tau \,d\tau = -\frac{1}{6}(t^2 - 2t - 8) \qquad 2 \le t \le 4 \qquad (2.34) -$$ - -Both Eqs. (2.33) and (2.34) apply at the transition point *t* = 2. We can readily verify that *c*(2) = 4/3 when either of these expressions is used. - -For *t* ≥ 4, *x*(*t* − τ ) has been shifted so far to the right that it no longer overlaps with *g*(τ ) as depicted in Fig. 2.10g. Consequently, - -$$ -c(t) = 0 \qquad \qquad t \ge 4 -$$ - -We now turn our attention to negative values of *t*. We have already determined *c*(*t*) up to *t* = −1. For *t* < −1, there is no overlap between the two functions, as illustrated in Fig. 2.10h, so that - -$$ -c(t) = 0 \qquad \qquad t \le -1 -$$ - -Combining our results, we see that - -$$ -c(t) = \begin{cases} \frac{1}{6}(t+1)^2 & -1 \le t < 1 \\ \frac{2}{3}t & 1 \le t < 2 \\ -\frac{1}{6}(t^2 - 2t - 8) & 2 \le t < 4 \\ 0 & \text{otherwise} \end{cases} -$$ - -Figure 2.10i plots *c*(*t*) according to this expression. - -### THE WIDTH OF CONVOLVED FUNCTIONS - -The widths (durations) of *x*(*t*), *g*(*t*), and *c*(*t*) in Ex. 2.12 (Fig. 2.10) are 2, 3, and 5, respectively. Note that the width of *c*(*t*) in this case is the sum of the widths of *x*(*t*) and *g*(*t*). This observation is not a coincidence. Using the concept of graphical convolution, we can readily see that if *x*(*t*) and *g*(*t*) have the finite widths of *T*1 and *T*2 respectively, then the width of *c*(*t*) is equal to *T*1 +*T*2. The reason is that the time it takes for a signal of width (duration) *T*1 to completely pass another signal of width (duration) *T*2 so that they become non-overlapping is *T*1+*T*2. When the two signals become non-overlapping, the convolution goes to zero. - -### **DR ILL 2.10 Interchanging Convolution Order** - -Rework Ex. 2.11 by evaluating *g*(*t*) ∗ *x*(*t*). - -### **DR ILL 2.11 Showing Commutability Using Two Causal Signals** - -Use graphical convolution to show that *x*(*t*) ∗ *g*(*t*) = *g*(*t*) ∗ *x*(*t*) = *c*(*t*) in Fig. 2.11. - -## **DR ILL 2.12 Showing Commutability Using a Causal Signal and an Anticausal Signal** - -## **DR ILL 2.13 Showing Commutability Using Shifted Signals** - -Repeat Drill 2.11 for the functions in Fig. 2.13. - -### THE PHANTOM OF THE SIGNALS AND SYSTEMS OPERA - -In the study of signals and systems we often come across some signals such as an impulse, which cannot be generated in practice and have never been sighted by anyone.† One wonders why we even consider such idealized signals. The answer should be clear from our discussion so far in this chapter. Even if the impulse function has no physical existence, we can compute the system response *h*(*t*) to this phantom input according to the procedure in Sec. 2.3, and knowing *h*(*t*), we can compute the system response to any arbitrary input. The concept of impulse response, therefore, provides an effective intermediary for computing system response to an arbitrary input. In addition, the impulse response *h*(*t*) itself provides a great deal of information and insight about the system behavior. In Sec. 2.6 we show that the knowledge of impulse response provides much valuable information, such as the response time, pulse dispersion, and filtering properties of the system. Many other useful insights about the system behavior can be obtained by inspection of *h*(*t*). - -Similarly, in frequency-domain analysis (discussed in later chapters), we use an *everlasting exponential* (or *sinusoid)* to determine system response. An everlasting exponential (or sinusoid), too, is a phantom, which nobody has ever seen and which has no physical existence. But it provides another effective intermediary for computing the system response to an arbitrary input. Moreover, the system response to everlasting exponential (or sinusoid) provides valuable information and insight regarding the system's behavior. Clearly, idealized impulses and everlasting sinusoids are friendly and helpful spirits. - -Interestingly, the unit impulse and the everlasting exponential (or sinusoid) are the dual of each other in the time-frequency duality, to be studied in Ch. 7. Actually, the time-domain and the frequency-domain methods of analysis are the dual of each other. - -### WHY CONVOLUTION? AN INTUITIVE EXPLANATION OF SYSTEM RESPONSE - -On the surface, it appears rather strange that the response of linear systems (those gentlest of the gentle systems) should be given by such a tortuous operation of convolution, where one signal is fixed and the other is inverted and shifted. To understand this odd behavior, consider a hypothetical impulse response *h*(*t*) that decays linearly with time (Fig. 2.14a). This response is strongest at *t* =0, the moment the impulse is applied, and it decays linearly at future instants so that one second later (at *t* = 1 and beyond), it ceases to exist. This means that the closer the impulse input is to an instant *t*, the stronger is its response at *t*. - -Now consider the input *x*(*t*) shown in Fig. 2.14b. To compute the system response, we break the input into rectangular pulses and approximate these pulses with impulses. Generally, the response of a causal system at some instant *t* will be determined by all the impulse components of the input before *t*. Each of these impulse components will have different weight in determining the response at the instant *t*, depending on its proximity to *t*. As seen earlier, the closer the impulse is to *t*, the stronger is its influence at *t*. The impulse at *t* has the greatest weight (unity) in determining - -† The late Prof. S. J. Mason, the inventor of signal flow graph techniques, used to tell a story of a student frustrated with the impulse function. The student said, "The unit impulse is a thing that is so small you can't see it, except at one place (the origin), where it is so big you can't see it. In other words, you can't see it at all; at least I can't!" [2]. - -**Figure 2.14** Intuitive explanation of convolution. - -the response at *t*. The weight decreases linearly for all impulses before *t* until the instant *t* − 1. The input before *t* − 1 has no influence (zero weight). Thus, to determine the system response at *t*, we must assign a linearly decreasing weight to impulses occurring before *t*, as shown in Fig. 2.14b. This weighting function is precisely the function *h*(*t* − τ ). The system response at *t* is then determined not by the input *x*(τ ) but by the weighted input *x*(τ )*h*(*t* − τ ), and the summation of all these weighted inputs is the convolution integral. - -### **[2.4-3 Interconnected Systems](#page-8-0)** - -A larger, more complex system can often be viewed as the interconnection of several smaller subsystems, each of which is easier to characterize. Knowing the characterizations of these subsystems, it becomes simpler to analyze such large systems. We shall consider here two basic interconnections, cascade and parallel. Figure 2.15a shows *S*1 and *S*2, two LTIC subsystems connected in parallel, and Fig. 2.15b shows the same two systems connected in cascade. - -In Fig. 2.15a, the device depicted by the symbol inside a circle represents an adder, which adds signals at its inputs. Also the junction from which two (or more) branches radiate out is called the *pickoff node*. Every branch that radiates out from the pickoff node carries the same signal (the signal at the junction). In Fig. 2.15a, for instance, the junction at which the input is applied is a pickoff node from which two branches radiate out, each of which carries the input signal at the node. - -Let the impulse response of *S*1 and *S*2 be *h*1(*t*) and *h*2(*t*), respectively. Further assume that interconnecting these systems, as shown in Fig. 2.15, does not load them. This means that the impulse response of either of these systems remains unchanged whether observed when these systems are unconnected or when they are interconnected. - -To find *hp*(*t*), the impulse response of the parallel system *Sp* in Fig. 2.15a, we apply an impulse at the input of *Sp*. This results in the signal δ(*t*) at the inputs of *S*1 and *S*2, leading to their outputs *h*1(*t*) and *h*2(*t*), respectively. These signals are added by the adder to yield *h*1(*t*) + *h*2(*t*) as the output of *Sp*: - -$$ -h_p(t) = h_1(t) + h_2(t) -$$ - -To find *hc*(*t*), the impulse response of the cascade system *Sc* in Fig. 2.15b, we apply the input δ(*t*) at the input of *Sc*, which is also the input to *S*1. Hence, the output of *S*1 is *h*1(*t*), which now acts - -**Figure 2.15** Interconnected systems. - -as the input to *S*2. The response of *S*2 to input *h*1(*t*) is *h*1(*t*) ∗ *h*2(*t*). Therefore, - -$$ -h_c(t) = h_1(t) * h_2(t) -$$ - -Because of the commutative property of convolution, it follows that interchanging the systems *S*1 and *S*2, as shown in Fig. 2.15c, results in the same impulse response *h*1(*t*) ∗ *h*2(*t*). This means that when several LTIC systems are cascaded, the order of systems does not affect the impulse response of the composite system. In other words, linear operations, performed in cascade, commute. The order in which they are performed is not important, at least theoretically.† - -†Change of order, however, could affect performance because of physical limitations and sensitivities to changes in the subsystems involved. - -#### 192 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -We shall give here another interesting application of the commutative property of LTIC systems. Figure 2.15d shows a cascade of two LTIC systems: a system *S* with impulse response *h*(*t*), followed by an ideal integrator. Figure 2.15e shows a cascade of the same two systems in reverse order; an ideal integrator followed by *S*. In Fig. 2.15d, if the input *x*(*t*) to *S* yields the output *y*(*t*), then the output of the system of Fig. 2.15d is the integral of *y*(*t*). In Fig. 2.15e, the output of the integrator is the integral of *x*(*t*). The output in Fig. 2.15e is identical to the output in Fig. 2.15d. Hence, it follows that if an LTIC system response to input *x*(*t*) is *y*(*t*), then the response of the same system to the integral of *x*(*t*) is the integral of *y*(*t*). In other words, - -if -$$ -x(t) \Longrightarrow y(t) -$$ - then $\int_{-\infty}^{t} x(\tau) d\tau \Longrightarrow \int_{-\infty}^{t} y(\tau) d\tau$ - -Replacing the ideal integrator with an ideal differentiator in Figs. 2.15d and 2.15e, and following a similar argument, we conclude that - -if -$$ -x(t) \Longrightarrow y(t) -$$ - then $\frac{dx(t)}{dt} \Longrightarrow \frac{dy(t)}{dt}$ - -If we let *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*) in Fig. 2.15e, we find that *g*(*t*), the unit step response of an LTIC system with impulse *h*(*t*), is given by - -$$ -g(t) = \int_{-\infty}^{t} h(\tau) d\tau -$$ -\n(2.35) - -We can also show that the system response to δ(˙ *t*) is *dh*(*t*)/*dt*. These results can be extended to other singularity functions. For example, the unit ramp response of an LTIC system is the integral of its unit step response, and so on. - -### INVERSE SYSTEMS - -In Fig. 2.15b, if *S*1 and *S*2 are inverse systems with impulse response *h*(*t*) and *hi*(*t*), respectively, then the impulse response of the cascade of these systems is *h*(*t*) ∗ *hi*(*t*). But, the cascade of a system with its inverse is an identity system, whose output is the same as the input. In other words, the unit impulse response of the cascade of inverse systems is also an unit impulse δ(*t*). Hence, - -$$ -h(t) * h_i(t) = \delta(t) \tag{2.36} -$$ - -We shall give an interesting application of the commutative property. As seen from Eq. (2.36), a cascade of inverse systems is an identity system. Moreover, in a cascade of several LTIC subsystems, changing the order of the subsystems in any manner does not affect the impulse response of the cascade system. Using these facts, we observe that the two systems, shown in Fig. 2.15f, are equivalent. We can compute the response of the cascade system on the right-hand side, by computing the response of the system inside the dotted box to the input *x*˙(*t*). The impulse response of the dotted box is *g*(*t*), the integral of *h*(*t*), as given in Eq. (2.35). Hence, it follows that - -$$ -y(t) = x(t) * h(t) = \dot{x}(t) * g(t) -$$ -\n(2.37) - -Recall that *g*(*t*) is the unit step response of the system. Hence, an LTIC response can also be obtained as a convolution of *x*˙(*t*) (the derivative of the input) with the unit step response of the system. This result can be readily extended to higher derivatives of the input. An LTIC system response is the convolution of the *n*th derivative of the input with the *n*th integral of the impulse response. - -### **[2.4-4 A Very Special Function for LTIC Systems:](#page-8-0) The Everlasting Exponential** *est* - -There is a very special connection of LTIC systems with the everlasting exponential function *est*, where *s* is a complex variable, in general. We now show that the LTIC system's (zero-state) response to everlasting exponential input *est* is also the same everlasting exponential (within a multiplicative constant). Moreover, no other function can make the same claim. Such an input for which the system response is also of the same form is called the *characteristic function* (also *eigenfunction*) of the system. Because a sinusoid is a form of exponential (*s* = ±*j*ω), everlasting sinusoid is also a characteristic function of an LTIC system. Note that we are talking here of an everlasting exponential (or sinusoid), which starts at *t* = −∞. - -If *h*(*t*) is the system's unit impulse response, then system response *y*(*t*) to an everlasting exponential *est* is given by - -$$ -y(t) = h(t) * e^{st} = \int_{-\infty}^{\infty} h(\tau) e^{s(t-\tau)} d\tau = e^{st} \int_{-\infty}^{\infty} h(\tau) e^{-s\tau} d\tau -$$ - -The integral on the right-most side is a function of a complex variable *s* and a constant with respect to *t*. Let us denote this term by *H*(*s*), which is also complex, in general. Thus, - -$$ -y(t) = H(s)e^{st} -$$ -\n(2.38) - -where - -$$ -H(s) = \int_{-\infty}^{\infty} h(\tau) e^{-s\tau} d\tau -$$ - (2.39) - -Equation (2.38) is valid only for the values of *s* for which *H*(*s*) exists, that is, if \$ ∞ −∞ *h*(τ )*e*−*s*τ *d*τ exists (or converges). The region in the *s* plane for which this integral converges is called the *region of convergence* for *H*(*s*). Further elaboration of the region of convergence is presented in Ch. 4. - -For a given *s*, note that *H*(*s*) is a constant. Thus, the input and the output are the same (within a multiplicative constant) for the everlasting exponential signal. - -*H*(*s*), which is called the *transfer function* of the system, is a function of complex variable *s*. An alternate definition of the transfer function *H*(*s*) of an LTIC system, as seen from Eq. (2.38), is - -$$ -H(s) = \frac{\text{output signal}}{\text{input signal}} \bigg|_{\text{input} = \text{everlasting exponential } e^{st}} \tag{2.40} -$$ - -The transfer function is defined for, and is meaningful to, LTIC systems only. It does not exist for nonlinear or time-varying systems, in general. - -We repeat again that this discussion is about the everlasting exponential, which starts at *t* = −∞, not the causal exponential *estu*(*t*), which starts at *t* = 0. - -#### 194 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -For a system specified by Eq. (2.2), the transfer function is given by - -$$ -H(s) = \frac{P(s)}{Q(s)}\tag{2.41} -$$ - -This follows readily by considering an everlasting input *x*(*t*) = *est*. According to Eq. (2.38), the output is *y*(*t*) = *H*(*s*)*est*. Substitution of this *x*(*t*) and *y*(*t*) in Eq. (2.2) yields - -$$ -H(s)[Q(D)e^{st}] = P(D)e^{st} -$$ - -Moreover, - -$$ -D^r e^{st} = \frac{d^r e^{st}}{dt^r} = s^r e^{st} -$$ - -Hence, - -$$ -P(D)e^{st} = P(s)e^{st} \qquad \text{and} \qquad Q(D)e^{st} = Q(s)e^{st} -$$ - -Consequently, - -$$ -H(s) = \frac{P(s)}{Q(s)} -$$ - -### **DR ILL 2.14 Ideal Integrator and Differentiator Transfer Functions** - -Show that the transfer function of an ideal integrator is *H*(*s*) = 1/*s* and that of an ideal differentiator is *H*(*s*) = *s*. Find the answer in two ways: using Eq. (2.39) and using Eq. (2.41). [*Hint:* Find *h*(*t*) for the ideal integrator and differentiator. You also may need to use the result in Prob. 1.4-12.] - -### A FUNDAMENTAL PROPERTY OF LTI SYSTEMS - -We can show that Eq. (2.38) is a fundamental property of LTI systems and it follows directly as a consequence of linearity and time invariance. To show this let us assume that the response of an LTI system to an everlasting exponential *est* is *y*(*s*,*t*). If we define - -$$ -H(s,t) = \frac{y(s,t)}{e^{st}} -$$ - -then - -$$ -y(s,t) = H(s,t) e^{st} -$$ - -Because of the time-invariance property, the system response to input *es*(*t*−*T*) is *H*(*s*,*t* − *T*) *es*(*t*−*T*) , that is, - -$$ -y(s, t - T) = H(s, t - T) e^{s(t - T)} -$$ -\n(2.42) - -The delayed input *es*(*t*−*T*) represents the input *est* multiplied by a constant *e*−*sT* . Hence, according to the linearity property, the system response to *es*(*t*−*T*) must be *y*(*s*,*t*) *e*−*sT* . Hence, - -$$ -y(s,t-T) = y(s,t) e^{-sT} = H(s,t) e^{s(t-T)} -$$ - -Comparison of this result with Eq. (2.42) shows that - -$$ -H(s,t) = H(s,t-T) \qquad \text{for all } T -$$ - -This means *H*(*s*,*t*) is independent of *t*, and we can express *H*(*s*,*t*) = *H*(*s*). Hence, - -$$ -y(s,t) = H(s) e^{st} -$$ - -### **[2.4-5 Total Response](#page-8-0)** - -Assuming distinct roots, the total response of a linear system can be expressed as the sum of its zero-input response (ZIR) and its zero-state response (ZSR): - -total response = -$$ -\underbrace{\sum_{k=1}^{N} c_k e^{\lambda_k t}}_{\text{ZIR}} + \underbrace{x(t) * h(t)}_{\text{ZSR}} -$$ - -For repeated roots, the zero-input component should be appropriately modified. - -For the series *RLC* circuit in Ex. 2.4 with the input *x*(*t*) = 10*e*−3*t u*(*t*) and the initial conditions *y*(0−) = 0, *vC*(0−) = 5, we determined the zero-input response in Ex. 2.1a [Eq. (2.9)]. We found the zero-state response in Ex. 2.9. From the results in Exs. 2.1a and 2.9, we obtain - -total current = -$$ -\underbrace{(-5e^{-t} + 5e^{-2t})}_{\text{zero-input current}} + \underbrace{(-5e^{-t} + 20e^{-2t} - 15e^{-3t})}_{\text{zero-state current}} \qquad t \ge 0 -$$ - (2.43) - -Figure 2.16a shows the zero-input, zero-state, and total responses. - -**Figure 2.16** Total response and its components. - -### NATURAL AND FORCED RESPONSE - -For the *RLC* circuit in Ex. 2.4, the characteristic modes were found to be *e*−*t* and *e*−2*t* . As we expected, the zero-input response is composed exclusively of characteristic modes. Note, however, that even the zero-state response [Eq. (2.43)] contains characteristic mode terms. This observation is generally true of LTIC systems. We can now lump together all the characteristic mode terms in the total response, giving us a component known as the *natural response yn*(*t*). The remainder, consisting entirely of noncharacteristic mode terms, is known as the *forced response y*φ(*t*). The total response of the *RLC* circuit in Ex. 2.4 can be expressed in terms of natural and forced components by regrouping the terms in Eq. (2.43) as - -total current = -$$ -\underbrace{(-10e^{-t} + 25e^{-2t})}_{\text{natural response } y_n(t)} + \underbrace{(-15e^{-3t})}_{\text{forced response } y_\phi(t)} \qquad t \ge 0 -$$ - (2.44) - -Figure 2.16b shows the natural, forced, and total responses. - -The classical solution to a differential equation includes the natural (also called the *homogeneous* or *complementary*) solution and the forced (also known as the *particular*) solution; traditional courses on differential equations provide simplified procedures to determine these components. Unfortunately, the classical solution lacks the engineering intuition and utility afforded by the zero-input and zero-state responses. The classical approach cannot separate the responses arising from internal conditions and external input. While the natural and forced solutions can be obtained from the zero-input and zero-state responses, the converse is not true. Further, the classical method is unable to express the system response to an input *x*(*t*) as an explicit function of *x*(*t*). In fact, the classical method is restricted to a certain class of inputs and cannot handle arbitrary inputs, as can the method to determine the zero-state response. For these (and other) reasons, we do not further detail the classical solution of differential equations. - -## **[2.5 SYSTEM](#page-8-0) STABILITY** - -Stability is an important system property. Two types of system stability are generally considered: external (BIBO) stability and internal (asymptotic) stability. Let us consider both stability types in turn. - -### **[2.5-1 External \(BIBO\) Stability](#page-8-0)** - -To understand the intuitive basis for the BIBO (bounded-input/bounded-output) stability of a system introduced in Sec. 1.7, let us examine the stability concept as applied to a right circular cone. Such a cone can be made to stand forever on its circular base, on its apex, or on its side. For this reason, these three states of the cone are said to be *equilibrium states*. Qualitatively, however, the three states show very different behavior. If the cone, standing on its circular base, were to be disturbed slightly and then left to itself, it would eventually return to its original equilibrium position. In such a case, the cone is said to be in *stable equilibrium*. In contrast, if the cone stands on its apex, then the slightest disturbance will cause the cone to move farther and farther away from its equilibrium state. The cone in this case is said to be in an *unstable equilibrium*. The cone lying on its side, if disturbed, will neither go back to the original state nor continue to move farther away from the original state. Thus it is said to be in a *neutral equilibrium*. Clearly, when a system is in stable equilibrium, application of a small disturbance (input) produces a small response. In contrast, when the system is in unstable equilibrium, even a minuscule disturbance (input) produces an unbounded response. The BIBO-stability definition can be understood in the light of this concept. If every bounded input produces bounded output, the system is (BIBO) stable.† In contrast, if even one bounded input results in unbounded response, the system is (BIBO) unstable. - -For an LTIC system, - -$$ -y(t) = h(t) * x(t) = \int_{-\infty}^{\infty} h(\tau) x(t - \tau) d\tau -$$ - -Therefore, - -$$ -|y(t)| \le \int_{-\infty}^{\infty} |h(\tau)| |x(t - \tau)| d\tau -$$ - -Moreover, if *x*(*t*) is bounded, then |*x*(*t* −τ )| < *K*1 < ∞, and - -$$ -|y(t)| \leq K_1 \int_{-\infty}^{\infty} |h(\tau)| d\tau -$$ - -Hence for BIBO stability, - -$$ -\int_{-\infty}^{\infty} |h(\tau)| d\tau < \infty \tag{2.45} -$$ - -This is a sufficient condition for BIBO stability. We can show that this is also a necessary condition (see Prob. 2.5-7). Therefore, for an LTIC system, if its impulse response *h*(*t*) is absolutely integrable, the system is (BIBO) stable. Otherwise it is (BIBO) unstable. In addition, we shall show in Ch. 4 that a necessary (but not sufficient) condition for an LTIC system described by Eq. (2.1) to be BIBO-stable is *M* ≤ *N*. If *M* > *N*, the system is unstable. This is one of the reasons to avoid systems with *M* > *N*. - -Because the BIBO stability of a system can be ascertained by measurements at the external terminals (input and output), this is an external stability criterion. It is no coincidence that the BIBO criterion in Eq. (2.45) is in terms of the impulse response, which is an external description of the system. - -As observed in Sec. 1.9, the internal behavior of a system is not always ascertainable from the external terminals. Therefore, external (BIBO) stability may not be a correct indication of internal stability. Indeed, some systems that appear stable by the BIBO criterion may be internally unstable. This is like a room on fire inside a house: no trace of fire is visible from outside, but the entire house will be burned to ashes. - -The BIBO stability is meaningful only for systems in which the internal and the external description are equivalent (controllable and observable systems). Fortunately, most practical systems fall into this category, and whenever we apply this criterion, we implicitly assume that the system, in fact, belongs to this category. Internal stability is all-inclusive, and external stability can always be determined from internal stability. For this reason, we now investigate the internal stability criterion. - -† The system is assumed to be in zero state. - -### **[2.5-2 Internal \(Asymptotic\) Stability](#page-8-0)** - -Because of the great variety of possible system behaviors, there are several definitions of internal stability in the literature. Here we shall consider a definition that is suitable for causal, linear, time-invariant (LTI) systems. - -If, in the absence of an external input, a system remains in a particular state (or condition) indefinitely, then that state is said to be an *equilibrium state* of the system. For an LTI system, zero state, in which all initial conditions are zero, is an equilibrium state. Now suppose an LTI system is in zero state and we change this state by creating small nonzero initial conditions (small disturbance). These initial conditions will generate signals consisting of characteristic modes in the system. By analogy with the cone, if the system is stable, it should eventually return to zero state. In other words, when left to itself, every mode in a stable system arising as a result of nonzero initial conditions should approach 0 as *t*→∞. However, if even one of the modes grows with time, the system will never return to zero state, and the system would be identified as unstable. In the borderline case, some modes neither decay to zero nor grow indefinitely, while all the remaining modes decay to zero. This case is like the neutral equilibrium in the cone. Such a system is said to be *marginally* stable. Internal stability is also called *asymptotic* stability or stability in the sense of *Lyapunov*. - -For a system characterized by Eq. (2.1), we can restate the internal stability criterion in terms of the location of the *N* characteristic roots λ1, λ2, ..., λ*N* of the system in a complex plane. The characteristic modes are of the form *e*λ*kt* or *t r e*λ*kt* . The locations of various roots in the complex plane and the corresponding modes are shown in Fig. 2.17. These modes → 0 as *t* → ∞ if Re λ*k* < 0. In contrast, the modes → ∞ as *t* → ∞ if Reλ*k* > 0.† - -From Fig. 2.17, we see that a system is (asymptotically) stable if all its characteristic roots lie in the LHP, that is, if Reλ*k* < 0 for all *k*. If even a single characteristic root lies in the RHP, the system is (asymptotically) unstable. Modes due to roots on the imaginary axis (λ = ±*j*ω0) are of the form *e*±*j*ω0*t* . Hence, if some roots are on the imaginary axis, and all the remaining roots are in the LHP, the system is marginally stable (assuming that the roots on the imaginary axis are not repeated). If the imaginary axis roots are repeated, the characteristic modes are of the form *t r e*±*j*ω*kt* , which *do* grow with time indefinitely. Hence, the system is unstable. Figure 2.18 shows stability regions in the complex plane. - -To summarize: - -- 1. An LTIC system is asymptotically stable if, and only if, all the characteristic roots are in the LHP. The roots may be simple (unrepeated) or repeated. -- 2. An LTIC system is unstable if, and only if, one or both of the following conditions exist: (i) at least one root is in the RHP; (ii) there are repeated roots on the imaginary axis. -- 3. An LTIC system is marginally stable if, and only if, there are no roots in the RHP, and there are some unrepeated roots on the imaginary axis. - -$$ -\lim_{t \to \infty} e^{\lambda t} = \lim_{t \to \infty} e^{(\alpha + j\beta)t} = \lim_{t \to \infty} e^{\alpha t} e^{j\beta t} = \begin{cases} 0 & \alpha < 0 \\ \infty & \alpha > 0 \end{cases} -$$ - -This conclusion is also valid for the terms of the form *t r e*λ*t* . - -† This may be seen from the fact that if α and β are the real and the imaginary parts of a root λ, then - -**Figure 2.17** Location of characteristic roots and the corresponding characteristic modes. - -### **[2.5-3 Relationship Between BIBO and Asymptotic Stability](#page-8-0)** - -External stability is determined by applying an external input with zero initial conditions, while internal stability is determined by applying the nonzero initial conditions and no external input. This is why these stabilities are also called the *zero-state stability* and the *zero-input stability*, respectively. - -Recall that *h*(*t*), the impulse response of an LTIC system, is a linear combination of the system characteristic modes. For an LTIC system, specified by Eq. (2.1), we can readily show that when a characteristic root λ*k* is in the LHP, the corresponding mode *e*λ*kt* is absolutely integrable. In - -contrast, if λ*k* is in the RHP or on the imaginary axis, *e*λ*kt* is not absolutely integrable.† This means that an asymptotically stable system is BIBO-stable. Moreover, a marginally stable or asymptotically unstable system is BIBO-unstable. The converse is not necessarily true; that is, BIBO stability does not necessarily inform us about the internal stability of the system. For instance, if a system is uncontrollable and/or unobservable, some modes of the system are invisible and/or uncontrollable from the external terminals [3]. Hence, the stability picture portrayed by the external description is of questionable value. BIBO (external) stability cannot assure internal (asymptotic) stability, as the following example shows. - -### **EXAMPLE 2.13 A BIBO-Stable but Asymptotically Unstable System** - -An LTID system consists of two subsystems *S*1 and *S*2 in cascade (Fig. 2.19). The impulse response of these systems are *h*1(*t*) and *h*2(*t*), respectively, given by - -$$ -h_1(t) = \delta(t) - 2e^{-t}u(t) -$$ - and $h_2(t) = e^t u(t)$ - -Comment on the BIBO and asymptotic stability of the composite system. - -$$ -\int_{-\infty}^{\infty} |e^{\lambda \tau} u(\tau)| d\tau = \int_{0}^{\infty} e^{\alpha \tau} d\tau = \begin{cases} -1/\alpha & \alpha < 0\\ \infty & \alpha \ge 0 \end{cases} -$$ - -This conclusion is also valid when the integrand is of the form |*t ke*λ*t u*(*t*)|. - -† Consider a mode of the form *e*λ*t* , where λ = α +*j*β. Hence, *e*λ*t* = *e*α*t ej*β*t* and |*e*λ*t* | = *e*α*t* . Therefore, - -The composite system impulse response *h*(*t*) is given by - -$$ -h(t) = h_1(t) * h_2(t) = h_2(t) * h_1(t) = e^t u(t) * [\delta(t) - 2e^{-t} u(t)] -$$ - -= $e^t u(t) - 2 \left[ \frac{e^t - e^{-t}}{2} \right] u(t)$ -= $e^{-t} u(t)$ - -If the composite cascade system were to be enclosed in a black box with only the input and the output terminals accessible, any measurement from these external terminals would show that the impulse response of the system is *e*−*t u*(*t*), without any hint of the dangerously unstable system the system is harboring within. - -The composite system is BIBO-stable because its impulse response, *e*−*t u*(*t*), is absolutely integrable. Observe, however, the subsystem *S*2 has a characteristic root 1, which lies in the RHP. Hence, *S*2 is asymptotically unstable. Eventually, *S*2 will burn out (or saturate) because of the unbounded characteristic response generated by intended or unintended initial conditions, no matter how small. We shall show in Ex. 10.12 that this composite system is observable, but not controllable. If the positions of *S*1 and *S*2 were interchanged (*S*2 followed by *S*1), the system is still BIBO-stable, but asymptotically unstable. In this case, the analysis in Ex. 10.12 shows that the composite system is controllable, but not observable. - -This example shows that BIBO stability does not always imply asymptotic stability. However, asymptotic stability always implies BIBO stability. - -Fortunately, uncontrollable and/or unobservable systems are not commonly observed in practice. Henceforth, in determining system stability, we shall assume that unless otherwise mentioned, the internal and the external descriptions of a system are equivalent, implying that the system is controllable and observable. - -### **EXAMPLE 2.14 Investigating Asymptotic and BIBO Stability** - -Investigate the asymptotic and the BIBO stability of LTIC system described by the following equations, assuming that the equations are internal system descriptions: - -- **(a)** (*D*+1)(*D*2 +4*D*+8)*y*(*t*) = (*D*−3)*x*(*t*) -- **(b)** (*D*−1)(*D*2 +4*D*+8)*y*(*t*) = (*D*+2)*x*(*t*) -- **(c)** (*D*+2)(*D*2 +4)*y*(*t*) = (*D*2 +*D*+1)*x*(*t*) -- **(d)** (*D*+1)(*D*2 +4)2*y*(*t*) = (*D*2 +2*D*+8)*x*(*t*) - -The characteristic polynomials of these systems are - -- **(a)** (λ+1)(λ2 +4λ+8) = (λ+1)(λ+2−*j*2)(λ+2+*j*2) -- **(b)** (λ−1)(λ2 +4λ+8) = (λ−1)(λ+2−*j*2)(λ+2+*j*2) -- **(c)** (λ+2)(λ2 +4) = (λ+2)(λ−*j*2)(λ+*j*2) -- **(d)** (λ+1)(λ2 +4)2 = (λ+2)(λ−*j*2)2(λ+*j*2)2 - -Consequently, the characteristic roots of the systems are (see Fig. 2.20): - -- **(a)** −1, −2±*j*2 -- **(b)** 1, −2±*j*2 -- **(c)** −2, ±*j*2 -- **(d)** −1, ±*j*2, ±*j*2 - -System (a) is asymptotically stable (all roots in LHP), system (b) is unstable (one root in RHP), system (c) is marginally stable (unrepeated roots on imaginary axis) and no roots in RHP, and system (d) is unstable (repeated roots on the imaginary axis). BIBO stability is readily determined from the asymptotic stability. System (a) is BIBO-stable, system (b) is BIBO-unstable, system (c) is BIBO-unstable, and system (d) is BIBO-unstable. We have assumed that these systems are controllable and observable. - -### **DR ILL 2.15 Assessing Stability by Characteristic Roots** - -For each case, plot the characteristic roots and determine asymptotic and BIBO stabilities. Assume the equations reflect internal descriptions. - -- **(a)** *D*(*D*+2)*y*(*t*) = 3*x*(*t*) -- **(b)** *D*2(*D*+3)*y*(*t*) = (*D*+5)*x*(*t*) -- **(c)** (*D*+1)(*D*+2)*y*(*t*) = (2*D*+3)*x*(*t*) -- **(d)** (*D*2 +1)(*D*2 +9)*y*(*t*) = (*D*2 +2*D*+4)*x*(*t*) -- **(e)** (*D*+1)(*D*2 −4*D*+9)*y*(*t*) = (*D*+7)*x*(*t*) - -### **ANSWERS** - -- **(a)** Marginally stable, but BIBO-unstable -- **(b)** Unstable in both senses -- **(c)** Stable in both senses -- **(d)** Marginally stable, but BIBO-unstable -- **(e)** Unstable in both senses. - -### IMPLICATIONS OF STABILITY - -All practical signal-processing systems must be asymptotically stable. Unstable systems are useless from the viewpoint of signal processing because any set of intended or unintended initial conditions leads to an unbounded response that either destroys the system or (more likely) leads it to some saturation conditions that change the nature of the system. Even if the discernible initial conditions are zero, stray voltages or thermal noise signals generated within the system will act as initial conditions. Because of exponential growth of a mode or modes in unstable systems, a stray signal, no matter how small, will eventually cause an unbounded output. - -Marginally stable systems, though BIBO unstable, do have one important application in the oscillator, which is a system that generates a signal on its own without the application of an external input. Consequently, the oscillator output is a zero-input response. If such a response is to be a sinusoid of frequency ω0, the system should be marginally stable with characteristic roots at ±*j*ω0. Thus, to design an oscillator of frequency ω0, we should pick a system with the characteristic polynomial (λ−*j*ω0)(λ+*j*ω0) = λ2 +ω0 2. A system described by the differential equation - -$$ -(D2 + \omega_02)y(t) = x(t) -$$ - -will do the job. However, practical oscillators are invariably realized using nonlinear systems. - -## **2.6 INTUITIVE [INSIGHTS INTO](#page-8-0) SYSTEM BEHAVIOR** - -This section attempts to provide an understanding of what determines system behavior. Because of its intuitive nature, the discussion is more or less qualitative. We shall now show that the most important attributes of a system are its characteristic roots or characteristic modes because they determine not only the zero-input response but also the entire behavior of the system. - -### **[2.6-1 Dependence of System Behavior on Characteristic Modes](#page-8-0)** - -Recall that the zero-input response of a system consists of the system's characteristic modes. For a stable system, these characteristic modes decay exponentially and eventually vanish. This behavior may give the impression that these modes do not substantially affect system behavior in general and system response in particular. This impression is totally wrong! We shall now see that the system's characteristic modes leave their imprint on every aspect of the system behavior. *We may compare the system's characteristic modes (or roots) to a seed that eventually dissolves in the* - -### 204 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -*ground; however, the plant that springs from it is totally determined by the seed. The imprint of the seed exists on every cell of the plant.* - -To understand this interesting phenomenon, recall that the characteristic modes of a system are very special to that system because it can sustain these signals without the application of an external input. In other words, the system offers a free ride and ready access to these signals. Now imagine what would happen if we actually drove the system with an input having the form of a characteristic mode! We would expect the system to respond strongly (this is, in fact, the resonance phenomenon discussed later in this section). If the input is not exactly a characteristic mode but is close to such a mode, we would still expect the system response to be strong. However, if the input is very different from any of the characteristic modes, we would expect the system to respond poorly. We shall now show that these intuitive deductions are indeed true. - -Intuition can cut the math jungle instantly! - -We have devised a measure of similarity of signals later (see in Ch. 6). Here we shall take a simpler approach. Let us restrict the system's inputs to exponentials of the form *e*ζ *t* , where ζ is generally a complex number. The similarity of two exponential signals *e*ζ *t* and *e*λ*t* will then be measured by the closeness of ζ and λ. If the difference ζ − λ is small, the signals are similar; if ζ −λ is large, the signals are dissimilar. - -Now consider a first-order system with a single characteristic mode *e*λ*t* and the input *e*ζ *t* . The impulse response of this system is then given by *Ae*λ*t* , where the exact value of *A* is not important for this qualitative discussion. The system response *y*(*t*) is given by - -$$ -y(t) = h(t) * x(t) = Ae^{\lambda t}u(t) * e^{\zeta t}u(t) -$$ - -From the convolution table (Table 2.1), we obtain - -$$ -y(t) = \frac{A}{\zeta - \lambda} [e^{\zeta t} - e^{\lambda t}] u(t) -$$ -\n(2.46) - -Clearly, if the input *e*ζ *t* is similar to *e*λ*t* , ζ −λ is small and the system response is large. *The closer the input x*(*t*) *to the characteristic mode, the stronger the system response.* In contrast, if the input is very different from the natural mode, ζ − λ is large and the system responds poorly. This is precisely what we set out to prove. - -We have proved the foregoing assertion for a single-mode (first-order) system. It can be generalized to an *N*th-order system, which has *N* characteristic modes. The impulse response *h*(*t*) of such a system is a linear combination of its *N* modes. Therefore, if *x*(*t*) is similar to any one of the modes, the corresponding response will be high; if it is similar to none of the modes, the response will be small. Clearly, the characteristic modes are very influential in determining system response to a given input. - -It would be tempting to conclude on the basis of Eq. (2.46) that if the input is identical to the characteristic mode, so that ζ = λ, then the response goes to infinity. Remember, however, that if ζ = λ, the numerator on the right-hand side of Eq. (2.46) also goes to zero. We shall study this interesting behavior (resonance phenomenon) later in this section. - -We now show that *mere inspection of the impulse response h*(*t*) *(which is composed of characteristic modes) reveals a great deal about the system behavior*. - -### **[2.6-2 Response Time of a System: The System Time Constant](#page-8-0)** - -Like human beings, systems have a certain response time. In other words, when an input (stimulus) is applied to a system, a certain amount of time elapses before the system fully responds to that input. This time lag or response time is called the system *time constant*. As we shall see, a system's time constant is equal to the width of its impulse response *h*(*t*). - -An input δ(*t*) to a system is instantaneous (zero duration), but its response *h*(*t*) has a duration *Th*. Therefore, the system requires a time *Th* to respond fully to this input, and we are justified in viewing *Th* as the system's response time or time constant. We arrive at the same conclusion via another argument. The output is a convolution of the input with *h*(*t*). If an input is a pulse of width *Tx*, then the output pulse width is *Tx* + *Th* according to the width property of convolution. This conclusion shows that the system requires *Th* seconds to respond fully to any input. *The system time constant indicates how fast the system is. A system with a smaller time constant is a faster system that responds quickly to an input. A system with a relatively large time constant is a sluggish system that cannot respond well to rapidly varying signals.* - -Strictly speaking, the duration of the impulse response *h*(*t*) is ∞ because the characteristic modes approach zero asymptotically as *t* → ∞. However, beyond some value of *t*, *h*(*t*) becomes negligible. It is therefore necessary to use some suitable measure of the impulse response's effective width. - -There is no single satisfactory definition of effective signal duration (or width) applicable to every situation. For the situation depicted in Fig. 2.21, a reasonable definition of the duration *h*(*t*) would be *Th*, the width of the rectangular pulse *h*ˆ(*t*). This rectangular pulse *h*ˆ(*t*) has an area identical to that of *h*(*t*) and a height identical to that of *h*(*t*) at some suitable instant *t* = *t*0. In Fig. 2.21, *t*0 is chosen as the instant at which *h*(*t*) is maximum. According to this definition,† - -$$ -T_h h(t_0) = \int_{-\infty}^{\infty} h(t) dt -$$ - -† This definition is satisfactory when *h*(*t*) is a single, mostly positive (or mostly negative) pulse. Such systems are lowpass systems. This definition should not be applied indiscriminately to all systems. - -or - -$$ -T_h = \frac{\int_{-\infty}^{\infty} h(t) dt}{h(t_0)} -$$ -\n(2.47) - -Now if a system has a single mode - -*h*(*t*) = *Ae*λ*t u*(*t*) - -with λ negative and real, then *h*(*t*) is maximum at *t* = 0 with value *h*(0) = *A*. Therefore, according to Eq. (2.47), - -$$ -T_h = \frac{1}{A} \int_0^\infty A e^{\lambda t} dt = -\frac{1}{\lambda} -$$ - -Thus, the time constant in this case is simply the (negative of the) reciprocal of the system's characteristic root. For the multimode case, *h*(*t*) is a weighted sum of the system's characteristic modes, and *Th* is a weighted average of the time constants associated with the *N* modes of the system. - -### **[2.6-3 Time Constant and Rise Time of a System](#page-8-0)** - -Rise time of a system, defined as the time required for the unit step response to rise from 10% to 90% of its steady-state value, is an indication of the speed of response.† The system time constant may also be viewed from a perspective of rise time. The unit step response *y*(*t*) of a system is the convolution of *u*(*t*) with *h*(*t*). Let the impulse response *h*(*t*) be a rectangular pulse of width *Th*, as shown in Fig. 2.22. This assumption simplifies the discussion, yet gives satisfactory results for qualitative discussion. The result of this convolution is illustrated in Fig. 2.22. Note that the output does not rise from zero to a final value instantaneously as the input rises; instead, the output takes *Th* seconds to accomplish this. Hence, the rise time *Tr* of the system is equal to the system time constant - -$$ -T_r = T_h -$$ - -This result and Fig. 2.22 show clearly that a system generally does not respond to an input instantaneously. Instead, it takes time *Th* for the system to respond fully. - -† Because of varying definitions of rise time, the reader may find different results in the literature. The qualitative and intuitive nature of this discussion should always be kept in mind. - -**Figure 2.22** Rise time of a system. - -### **[2.6-4 Time Constant and Filtering](#page-8-0)** - -A larger time constant implies a sluggish system because the system takes longer to respond fully to an input. Such a system cannot respond effectively to rapid variations in the input. In contrast, a smaller time constant indicates that a system is capable of responding to rapid variations in the input. Thus, there is a direct connection between a system's time constant and its filtering properties. - -A high-frequency sinusoid varies rapidly with time. A system with a large time constant will not be able to respond well to this input. Therefore, such a system will suppress rapidly varying (high-frequency) sinusoids and other high-frequency signals, thereby acting as a lowpass filter (a filter allowing the transmission of low-frequency signals only). We shall now show that a system - -**Figure 2.23** Time constant and filtering. - -with a time constant *Th* acts as a lowpass filter having a cutoff frequency of *fc* = 1/*Th* hertz, so that sinusoids with frequencies below *fc* Hz are transmitted reasonably well, while those with frequencies above *fc* Hz are suppressed. - -To demonstrate this fact, let us determine the system response to a sinusoidal input *x*(*t*) by convolving this input with the effective impulse response *h*(*t*) in Fig. 2.23a. From Figs. 2.23b and 2.23c we see the process of convolution of *h*(*t*) with the sinusoidal inputs of two different frequencies. The sinusoid in Fig. 2.23b has a relatively high frequency, while the frequency of the sinusoid in Fig. 2.23c is low. Recall that the convolution of *x*(*t*) and *h*(*t*) is equal to the area under the product *x*(τ )*h*(*t* − τ ). This area is shown shaded in Figs. 2.23b and 2.23c for the two cases. For the high-frequency sinusoid, it is clear from Fig. 2.23b that the area under *x*(τ )*h*(*t* − τ ) is very small because its positive and negative areas nearly cancel each other out. In this case the output *y*(*t*) remains periodic but has a rather small amplitude. This happens when the period of the sinusoid is much smaller than the system time constant *Th*. In contrast, for the low-frequency sinusoid, the period of the sinusoid is larger than *Th*, rendering the partial cancellation of area under *x*(τ )*h*(*t* −τ ) less effective. Consequently, the output *y*(*t*) is much larger, as depicted in Fig. 2.23c. - -Between these two possible extremes in system behavior, a transition point occurs when the period of the sinusoid is equal to the system time constant *Th*. The frequency at which this transition occurs is known as the *cutoff frequency fc* of the system. Because *Th* is the period of cutoff frequency *fc*, - -$$ -f_c = \frac{1}{T_h} -$$ - -The frequency *fc* is also known as the bandwidth of the system because the system transmits or passes sinusoidal components with frequencies below *fc* while attenuating components with frequencies above *fc*. Of course, the transition in system behavior is gradual. There is no dramatic change in system behavior at *fc* = 1/*Th*. Moreover, these results are based on an idealized (rectangular pulse) impulse response; in practice these results will vary somewhat, depending on the exact shape of *h*(*t*). Remember that the "feel" of general system behavior is more important than exact system response for this qualitative discussion. - -Since the system time constant is equal to its rise time, we have - -$$ -T_r = \frac{1}{f_c} \qquad \text{or} \qquad f_c = \frac{1}{T_r} \tag{2.48} -$$ - -Thus, a system's bandwidth is inversely proportional to its rise time. Although Eq. (2.48) was derived for an idealized (rectangular) impulse response, its implications are valid for lowpass LTIC systems, in general. For a general case, we can show that [1] - -$$ -f_c = \frac{k}{T_r} -$$ - -where the exact value of *k* depends on the nature of *h*(*t*). An experienced engineer often can estimate quickly the bandwidth of an unknown system by simply observing the system response to a step input on an oscilloscope. - -### **[2.6-5 Time Constant and Pulse Dispersion \(Spreading\)](#page-8-0)** - -In general, the transmission of a pulse through a system causes pulse dispersion (or spreading). Therefore, the output pulse is generally wider than the input pulse. This system behavior can have serious consequences in communication systems in which information is transmitted by pulse amplitudes. Dispersion (or spreading) causes interference or overlap with neighboring pulses, thereby distorting pulse amplitudes and introducing errors in the received information. - -Earlier we saw that if an input *x*(*t*) is a pulse of width *Tx*, then *Ty*, the width of the output *y*(*t*), is - -$$ -T_{y}=T_{x}+T_{h} -$$ - -This result shows that an input pulse spreads out (disperses) as it passes through a system. Since *Th* is also the system's time constant or rise time, the amount of spread in the pulse is equal to the time constant (or rise time) of the system. - -### **[2.6-6 Time Constant and Rate of Information Transmission](#page-8-0)** - -In pulse communications systems, which convey information through pulse amplitudes, the rate of information transmission is proportional to the rate of pulse transmission. We shall demonstrate that to avoid the destruction of information caused by dispersion of pulses during their transmission through the channel (transmission medium), the rate of information transmission should not exceed the bandwidth of the communications channel. - -Since an input pulse spreads out by *Th* seconds, the consecutive pulses should be spaced *Th* seconds apart to avoid interference between pulses. Thus, the rate of pulse transmission should not exceed 1/*Th* pulses/second. But 1/*Th* = *fc*, the channel's bandwidth, so that we can transmit pulses through a communications channel at a rate of *fc* pulses per second and still avoid significant interference between the pulses. The rate of information transmission is therefore proportional to the channel's bandwidth (or to the reciprocal of its time constant).† - -The discussion of Secs. 2.6-2, 2.6-3, 2.6-4, 2.6-5, and 2.6-6) shows that the system time constant determines much of a system's behavior—its filtering characteristics, rise time, pulse dispersion, and so on. In turn, the time constant is determined by the system's characteristic roots. Clearly the characteristic roots and their relative amounts in the impulse response *h*(*t*) determine the behavior of a system. - -### **EXAMPLE 2.15 Intuitive Insights into Lowpass System Behavior** - -Find the time constant *Th*, rise time *Tr*, and cutoff frequency *fc* for a lowpass system that has impulse response *h*(*t*) = *te*−*t u*(*t*). Determine the maximum rate that pulses of 1 second - -† Theoretically, a channel of bandwidth *fc* can transmit correctly up to 2*fc* pulse amplitudes per second [4]. Our derivation here, being very simple and qualitative, yields only half the theoretical limit. In practice it is not easy to attain the upper theoretical limit. - -#### 210 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -duration can be transmitted through the system so that interference is essentially avoided between adjacent pulses at the system output. - -The system impulse response *h*(*t*) = *te*−*t u*(*t*), which looks similar to the impulse response of Fig. 2.21, has a peak value of *e*−1 = 0.3679 at a time *t*0 = 1. According to Eq. (2.47) and using integration by parts, the system time constant is therefore - -$$ -T_h = \frac{\int_0^\infty t e^{-t} dt}{e^{-1}} = e^1 \left( -t e^{-t} \Big|_0^\infty + \int_0^\infty e^{-t} dt \right) = e^1 \left( 0 - e^{-t} \Big|_0^\infty \right) = e^1(1) = 2.7183 -$$ - -Thus, - -$$ -T_h = 2.7183 -$$ - s, $T_r = T_h = 2.7183$ s, and $f_c = \frac{1}{T_h} = 0.3679$ Hz - -Due to its lowpass nature, this system will spread an input pulse of 1 second to an output with width - -*Ty* = *Tx* +*Th* = 1+2.7183 = 3.7183 s - -To avoid interference between pulses at the output, the pulse transmission rate should be no more than the reciprocal of the output pulse width. That is, - -> maximum pulse transmission rate = 1 3.7183 = 0.2689 pulse/s - -By narrowing the input pulses, the pulse transmission rate could increase up to *fc* = 0.3679 pulse/s. - -### **[2.6-7 The Resonance Phenomenon](#page-8-0)** - -Finally, we come to the fascinating phenomenon of resonance. As we have already mentioned several times, this phenomenon is observed when the input signal is identical or is very close to a characteristic mode of the system. For the sake of simplicity and clarity, we consider a first-order system having only a single mode, *e*λ*t* . Let the impulse response of this system be† - -$$ -h(t) = Ae^{\lambda t} -$$ - -and let the input be - -$$ -x(t) = e^{(\lambda - \epsilon)t} -$$ - -The system response *y*(*t*) is then given by - -$$ -y(t) = Ae^{\lambda t} * e^{(\lambda - \epsilon)t} -$$ - -† For convenience, we omit multiplying *x*(*t*) and *h*(*t*) by *u*(*t*). Throughout this discussion, we assume that they are causal. - -### 2.6 Intuitive Insights into System Behavior 211 - -From the convolution table we obtain - -$$ -y(t) = \frac{A}{\epsilon} \left[ e^{\lambda t} - e^{(\lambda - \epsilon)t} \right] = A e^{\lambda t} \left( \frac{1 - e^{-\epsilon t}}{\epsilon} \right) -$$ - (2.49) - -Now, as → 0, both the numerator and the denominator of the term in the parentheses approach zero. Applying L'Hôpital's rule to this term yields - -$$ -\lim_{\epsilon \to 0} y(t) = A t e^{\lambda t} -$$ - -Clearly, the response does not go to infinity as → 0, but it acquires a factor *t*, which approaches ∞ as *t* → ∞. If λ has a negative real part (so that it lies in the LHP), *e*λ*t* decays faster than *t* and *y*(*t*) → 0 as *t* → ∞. The resonance phenomenon in this case is present, but its manifestation is aborted by the signal's own exponential decay. - -This discussion shows that*resonance is a cumulative phenomenon,* not instantaneous. It builds up linearly with *t*. † When the mode decays exponentially, the signal decays too fast for resonance to counteract the decay; as a result, the signal vanishes before resonance has a chance to build it up. However, if the mode were to decay at a rate less than 1/*t*, we should see the resonance phenomenon clearly. This specific condition would be possible if Re λ ≥ 0. For instance, when Re λ = 0 so that λ lies on the imaginary axis of the complex plane (λ = *j*ω), the output becomes - -$$ -y(t) = A t e^{j\omega t} -$$ - -Here, the response does go to infinity linearly with *t*. - -For a real system, if λ = *j*ω is a root, λ∗ = −*j*ω must also be a root; the impulse response is of the form *Aej*ω*t* + *Ae*−*j*ω*t* = 2*A*cos ω*t*. The response of this system to input *A*cosω*t* is 2*A*cosω*t* ∗ cosω*t*. The reader can show that this convolution contains a term of the form *At* cos ω*t*. The resonance phenomenon is clearly visible. The system response to its characteristic mode increases linearly with time, eventually reaching ∞, as indicated in Fig. 2.24. - -Recall that when λ = *j*ω, the system is marginally stable. As we have indicated, the full effect of resonance cannot be seen for an asymptotically stable system; only in a marginally stable system does the resonance phenomenon boost the system's response to infinity when the system's input - -**Figure 2.24** Buildup of system response in resonance. - -† If the characteristic root in question repeats *r* times, resonance effect increases as *t r*−1. However, *t r*−1*e*λ*t* →0 as *t* → ∞ for any value of *r*, provided Re λ < 0 (λ in the LHP). - -### 212 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -is a characteristic mode. But even in an asymptotically stable system, we see a manifestation of resonance if its characteristic roots are close to the imaginary axis so that Re λ is a small, negative value. We can show that when the characteristic roots of a system are σ ± *j*ω0, then the system response to the input *ej*ω0*t* or the sinusoid cosω0*t* is very large for small σ. † The system response drops off rapidly as the input signal frequency moves away from ω0. This frequency-selective behavior can be studied more profitably after an understanding of frequency-domain analysis has been acquired. For this reason we postpone full discussion of this subject until Ch. 4. - -### IMPORTANCE OF THE RESONANCE PHENOMENON - -The resonance phenomenon is very important because it allows us to design frequency-selective systems by choosing their characteristic roots properly. Lowpass, bandpass, highpass, and bandstop filters are all examples of frequency-selective networks. In mechanical systems, the inadvertent presence of resonance can cause signals of such tremendous magnitude that the system may fall apart. A musical note (periodic vibrations) of proper frequency can shatter glass if the frequency is matched to the characteristic root of the glass, which acts as a mechanical system. Similarly, a company of soldiers marching in step across a bridge amounts to applying a periodic force to the bridge. If the frequency of this input force happens to be nearer to a characteristic root of the bridge, the bridge may respond (vibrate) violently and collapse, even though it would have been strong enough to carry many soldiers marching out of step. A case in point is the Tacoma Narrows Bridge failure of 1940. This bridge was opened to traffic in July 1940. Within four months of opening (on November 7, 1940), it collapsed in a mild gale, not because of the wind's brute force but because the frequencies of wind-generated vortices, which matched the natural frequencies (characteristic roots) of the bridge, caused resonance. - -Because of the great damage that may occur, mechanical resonance is generally to be avoided, especially in structures or vibrating mechanisms. If an engine with periodic force (such as piston motion) is mounted on a platform, the platform with its mass and springs should be designed so that their characteristic roots are not close to the engine's frequency of vibration. Proper design of this platform can not only avoid resonance, but also attenuate vibrations if the system roots are placed far away from the frequency of vibration. - -## **[2.7 MATLAB: M-FILES](#page-8-0)** - -M-files are stored sequences of MATLAB commands and help simplify complicated tasks. There are two types of M-file: script and function. Both types are simple text files and require a .m filename extension. - -Although M-files can be created by using any text editor, MATLAB's built-in editor is the preferable choice because of its special features. As with any program, comments improve the readability of an M-file. Comments begin with the % character and continue through the end of the line. - -An M-file is executed by simply typing the filename (without the .m extension). To execute, M-files need to be located in the current directory or any other directory in the MATLAB path. New directories are easily added to the MATLAB path by using the addpath command. - -† This follows directly from Eq. (2.49) with λ = σ +*j*ω0 and = σ. - -### **[2.7-1 Script M-Files](#page-8-0)** - -Script files, the simplest type of M-file, consist of a series of MATLAB commands. Script files record and automate a series of steps, and they are easy to modify. To demonstrate the utility of a script file, consider the operational amplifier circuit shown in Fig. 2.25. - -The system's characteristic modes define the circuit's behavior and provide insight regarding system behavior. Using ideal, infinite gain difference amplifier characteristics, we first derive the differential equation that relates output *y*(*t*) to input *x*(*t*). Kirchhoff's current law (KCL) at the node shared by *R*1 and *R*3 provides - -$$ -\frac{x(t) - v(t)}{R_3} + \frac{y(t) - v(t)}{R_2} + \frac{0 - v(t)}{R_1} - C_2 \dot{v}(t) = 0 -$$ - -KCL at the inverting input of the op amp gives - -$$ -\frac{v(t)}{R_1} + C_1 \dot{y}(t) = 0 -$$ - -Combining and simplifying the KCL equations yield - -$$ -\ddot{y}(t) + \frac{1}{C_2} \left\{ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right\} \dot{y}(t) + \frac{1}{R_1 R_2 C_1 C_2} y(t) = -\frac{1}{R_1 R_3 C_1 C_2} x(t) -$$ - -which is the desired constant coefficient differential equation. Thus, the characteristic equation is given by - -$$ -\lambda^2 + \frac{1}{C_2} \left\{ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right\} \lambda + \frac{1}{R_1 R_2 C_1 C_2} = (a_0 \lambda^2 + a_1 \lambda + a_2) = 0 \tag{2.50} -$$ - -The roots λ1 and λ2 of Eq. (2.50) establish the nature of the characteristic modes *e*λ1*t* and *e*λ2*t* . - -As a first case, assign nominal component values of *R*1 = *R*2 = *R*3 = 10 k and *C*1 = *C*2 = 1 µF. A series of MATLAB commands allows convenient computation of the roots λ = [λ1;λ2]. Although λ can be determined using the quadratic equation, MATLAB's roots command is more convenient. The roots command requires an input vector that contains the polynomial coefficients in descending order. Even if a coefficient is zero, it must still be included in the vector. - -**Figure 2.25** Operation-amplifier circuit. - -``` -% CH2MP1.m : Chapter 2, MATLAB Program 1 -% Script M-file determines characteristic roots of op-amp circuit. -% Set component values: -R = [1e4, 1e4, 1e4]; C = [1e-6, 1e-6]; -% Determine coefficients for characteristic equation: -A = [1, (1/R(1)+1/R(2)+1/R(3))/C(2), 1/(R(1)*R(2)*C(1)*C(2))]; -% Determine characteristic roots: -lambda = roots(A); -``` - -A script file is created by placing these commands in a text file, which in this case is named CH2MP1.m. While comment lines improve program clarity, their removal does not affect program functionality. The program is executed by typing - -``` ->> CH2MP1 -``` - -After execution, all the resulting variables are available in the workspace. For example, to view the characteristic roots, type - ->> lambda lambda = -261.8034 -38.1966 - -Thus, the characteristic modes are simple decaying exponentials: *e*−261.8034*t* and *e*−38.1966*t* . - -Script files permit simple or incremental changes, thereby saving significant effort. Consider what happens when capacitor *C*1 is changed from 1.0 µF to 1.0 nF. Changing CH2MP1.m so that C = [1e-9, 1e-6] allows computation of the new characteristic roots: - -``` ->> CH2MP1 ->> lambda - lambda = 1.0e+003 * - -0.1500 + 3.1587i - -0.1500 - 3.1587i -``` - -Perhaps surprisingly, the characteristic modes are now complex exponentials capable of supporting oscillations. The imaginary portion of λ dictates an oscillation rate of 3158.7 rad/s or about 503 Hz. The real portion dictates the rate of decay. The time expected to reduce the amplitude to 25% is approximately *t* = ln 0.25/Re(λ) ≈ 0.01 second. - -### **[2.7-2 Function M-Files](#page-8-0)** - -It is inconvenient to modify and save a script file each time a change of parameters is desired. Function M-files provide a sensible alternative. Unlike script M-files, function M-files can accept input arguments as well as return outputs. Functions truly extend the MATLAB language in ways that script files cannot. - -Syntactically, a function M-file is identical to a script M-file except for the first line. The general form of the first line is - -function [*output1, ..., outputN*] = filename(*input1, ..., inputM*) - -For example, consider modification of CH2MP1.m to make function CH2MP2.m. Component values are passed to the function as two separate inputs: a length-3 vector of resistor values and a length-2 vector of capacitor values. The characteristic roots are returned as a 2×1 complex vector. - -``` -function [lambda] = CH2MP2(R,C) -% CH2MP2.m : Chapter 2, MATLAB Program 2 -% Function M-file finds characteristic roots of op-amp circuit. -% INPUTS: R = length-3 vector of resistances -% C = length-2 vector of capacitances -% OUTPUTS: lambda = characteristic roots -% Determine coefficients for characteristic equation: -A = [1, (1/R(1)+1/R(2)+1/R(3))/C(2), 1/(R(1)*R(2)*C(1)*C(2))]; -% Determine characteristic roots: -lambda = roots(A); -``` - -As with script M-files, function M-files execute by typing the name at the command prompt. However, inputs must also be included. For example, CH2MP2 easily confirms the oscillatory modes of the preceding example. - -``` ->> lambda = CH2MP2([1e4, 1e4, 1e4],[1e-9, 1e-6]) - lambda = 1.0e+003 * - -0.1500 + 3.1587i - -0.1500 - 3.1587i -``` - -Although scripts and functions have similarities, they also have distinct differences that are worth pointing out. Scripts operate on workspace data; either functions must be supplied data through inputs or they must create their own data. Unless passed as an output, variables and data created by functions remain local to the function; variables or data generated by scripts are global and are added to the workspace. To emphasize this point, consider polynomial coefficient vector A, which is created and used in both CH2MP1.m and CH2MP2.m. Following execution of function CH2MP2, the variable A is not added to the workspace. Following execution of script CH2MP1, however, A is available in the workspace. Recall, the workspace is easily viewed by typing either who or whos. - -### **[2.7-3 For-Loops](#page-9-0)** - -Real resistors and capacitors never exactly equal their nominal values. Suppose that the circuit components are measured as *R*1 = 10.322 k, *R*2 = 9.952 k, *R*3 = 10.115 k, *C*1 = 1.120 nF, and *C*2 = 1.320 µF. These values are consistent with the 10 and 25% tolerance resistor and capacitor values commonly and readily available. CH2MP2.m uses these component values to calculate the new values of λ. - -### 216 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -``` ->> lambda = CH2MP2([10322,9592,10115],[1.12e-9, 1.32e-6]) - lambda = 1.0e+003 * - -0.1136 + 2.6113i - -0.1136 - 2.6113i -``` - -Now the natural modes oscillate at 2611.3 rad/s or about 416 Hz. Decay to 25% amplitude is expected in *t* = ln 0.25/(−113.6) ≈ 0.012 second. These values, which differ significantly from the nominal values of 503 Hz and *t* ≈ 0.01 second, warrant a more formal investigation of the effect of component variations on the locations of the characteristic roots. - -It is sensible to look at three values for each component: the nominal value, a low value, and a high value. Low and high values are based on component tolerances. For example, a 10% 1 k resistor could have an expected low value of 1000(1 − 0.1) = 900 and an expected high value of 1000(1+0.1) = 1100 . For the five passive components in the design, 35 = 243 permutations are possible. - -Using either CH2MP1.m or CH2MP2.m to solve each of the 243 cases would be very tedious and boring. For-loops help automate repetitive tasks such as this. In MATLAB, the general structure of a for statement is - -``` -for variable = expression, statement, ..., statement, end -``` - -Five nested for-loops, one for each passive component, are required for the present example. - -``` -% CH2MP3.m : Chapter 2, MATLAB Program 3 -% Script M-file determines characteristic roots over a range of component - values. -% Pre-allocate memory for all computed roots: -lambda = zeros(2,243); -% Initialize index to identify each permutation: -p=0; -for R1 = 1e4*[0.9,1.0,1.1], - for R2 = 1e4*[0.9,1.0,1.1], - for R3 = 1e4*[0.9,1.0,1.1], - for C1 = 1e-9*[0.75,1.0,1.25], - for C2 = 1e-6*[0.75,1.0,1.25], - p = p+1; - lambda(:,p) = CH2MP2([R1 R2 R3],[C1 C2]); - end - end - end - end -end -plot(real(lambda(:)),imag(lambda(:)),'kx',... - real(lambda(:,1)),imag(lambda(:,1)),'kv',... - real(lambda(:,end)),imag(lambda(:,end)),'k^') -xlabel('Real'),ylabel('Imaginary') -legend('Char. Roots','Min. Val. Roots','Max. Val. Roots','Location','West'); -``` - -**Figure 2.26** Effect of component values on characteristic root locations. - -The command lambda = zeros(2,243) preallocates a 2×243 array to store the computed roots. When necessary, MATLAB performs dynamic memory allocation, so this command is not strictly necessary. However, preallocation significantly improves script execution speed. Notice also that it would be nearly useless to call script CH2MP1 from within the nested loop; script file parameters cannot be changed during execution. - -The plot instruction is quite long. Long commands can be broken across several lines by terminating intermediate lines with three dots (...). The three dots tell MATLAB to continue the present command to the next line. Black x's locate roots of each permutation. The command lambda(:) vectorizes the 2 × 243 matrix lambda into a 486 × 1 vector. This is necessary in this case to ensure that a proper legend is generated. Because of loop order, permutation *p* = 1 corresponds to the case of all components at the smallest values and permutation *p* = 243 corresponds to the case of all components at the largest values. This information is used to separately highlight the minimum and maximum cases using down-triangles () and up-triangles (), respectively. In addition to terminating each for loop, end is used to indicate the final index along a particular dimension, which eliminates the need to remember the particular size of a variable. An overloaded function, such as end, serves multiple uses and is typically interpreted based on context. - -The graphical results provided by CH2MP3 are shown in Fig. 2.26. Between extremes, root oscillations vary from 365 to 745 Hz and decay times to 25% amplitude vary from 6.2 to 12.7 ms. Clearly, this circuit's behavior is quite sensitive to ordinary component variations. - -### **[2.7-4 Graphical Understanding of Convolution](#page-9-0)** - -MATLAB graphics effectively illustrate the convolution process. Consider the case of *y*(*t*) = *x*(*t*)∗ *h*(*t*), where *x*(*t*) = 1.5 sin(π*t*)(*u*(*t*) − *u*(*t* − 1)) and *h*(*t*) = 1.5(*u*(*t*) − *u*(*t* − 1.5)) − *u*(*t* − 2) + *u*(*t* − 2.5). Program CH2MP4 steps through the convolution over the time interval (−0.25 ≤ *t* ≤ 3.75). - -``` -% CH2MP4.m : Chapter 2, MATLAB Program 4 -% Script M-file graphically demonstrates the convolution process. -figure(1) % Create figure window and make visible on screen -u = @(t) 1.0*(t>=0); -x = @(t) 1.5*sin(pi*t).*(u(t)-u(t-1)); -h = @(t) 1.5*(u(t)-u(t-1.5))-u(t-2)+u(t-2.5); -dtau = 0.005; tau = -1:dtau:4; -ti = 0; tvec = -.25:.1:3.75; -y = NaN*zeros(1,length(tvec)); % Pre-allocate memory -for t = tvec, - ti = ti+1; % Time index - xh = x(t-tau).*h(tau); lxh = length(xh); - y(ti) = sum(xh.*dtau); % Trapezoidal approximation of convolution integral - subplot(2,1,1),plot(tau,h(tau),'k-',tau,x(t-tau),'k--',t,0,'ok'); - axis([tau(1) tau(end) -2.0 2.5]); - patch([tau(1:end-1);tau(1:end-1);tau(2:end);tau(2:end)],... - [zeros(1,lxh-1);xh(1:end-1);xh(2:end);zeros(1,lxh-1)],... - [.8 .8 .8],'edgecolor','none'); - xlabel('\tau'); title('h(\tau) [solid], x(t-\tau) [dashed], h(\tau)x(t-\tau) [gray]'); - c = get(gca,'children'); set(gca,'children',[c(2);c(3);c(4);c(1)]); - subplot(2,1,2),plot(tvec,y,'k',tvec(ti),y(ti),'ok'); - xlabel('t'); ylabel('y(t) = \int h(\tau)x(t-\tau) d\tau'); - axis([tau(1) tau(end) -1.0 2.0]); grid; - drawnow; -end -``` - -At each step, the program plots *h*(τ ), *x*(*t* − τ ), and shades the area *h*(τ )*x*(*t* − τ ) gray. This gray area, which reflects the integral of *h*(τ )*x*(*t* − τ ), is also the desired result, *y*(*t*). Figures 2.27, 2.28, and 2.29 display the convolution process at times *t* of 0.75, 2.25, and 2.85 seconds, respectively. These figures help illustrate how the regions of integration change with time. Figure 2.27 has limits of integration from 0 to (*t* = 0.75). Figure 2.28 has two regions of integration, with limits (*t* −1 = 1.25) to 1.5 and 2.0 to (*t* = 2.25). The last plot, Fig. 2.29, has limits from 2.0 to 2.5. - -Several comments regarding CH2MP4 are in order. The command figure(1) opens the first figure window and, more important, makes sure it is visible. Anonymous functions are used to represent the functions *u*(*t*), *x*(*t*), and *h*(*t*). NaN, standing for not-a-number, usually results from operations such as 0/0 or ∞−∞. MATLAB refuses to plot NaN values, so preallocating *y*(*t*) with NaNs ensures that MATLAB displays only values of *y*(*t*) that have been computed. As its name suggests, length returns the length of the input vector. The subplot(a,b,c) command partitions the current figure window into an a-by-b matrix of axes and selects axes c for use. Subplots facilitate graphical comparison by allowing multiple axes in a single figure window. The patch command is used to create the gray-shaded area for *h*(τ )*x*(*t* − τ ). In CH2MP4, the get and set commands are used to reorder plot objects so that the gray area does not obscure other lines. Details of the patch, get, and set commands, as used in CH2MP4, are somewhat advanced and are not pursued here.† MATLAB also prints most Greek letters if the Greek name is preceded by a backslash (\) character. For example, \tau in the xlabel command produces the symbol τ in the plot's axis label. Similarly, an integral sign is produced by \int. Finally, the drawnow - -† Interested students should consult the MATLAB help facilities for further information. Actually, the get and set commands are extremely powerful and can help modify plots in almost any conceivable way. - -**Figure 2.27** Graphical convolution at step *t* = 0.75 second. - -**Figure 2.28** Graphical convolution at step *t* = 2.25 seconds. - -command forces MATLAB to update the graphics window for each loop iteration. Although slow, this creates an animation-like effect. Replacing drawnow with the pause command allows users to manually step through the convolution process. The pause command still forces the graphics window to update, but the program will not continue until a key is pressed. - -**Figure 2.29** Graphical convolution at step *t* = 2.85 seconds. - -## **[2.8 APPENDIX: DETERMINING THE](#page-9-0) IMPULSE RESPONSE** - -In Eq. (2.13), we showed that for an LTIC system *S* specified by Eq. (2.11), the unit impulse response *h*(*t*) can be expressed as - -$$ -h(t) = b_0 \delta(t) + \text{characteristic modes} \tag{2.51} -$$ - -To determine the characteristic mode terms in Eq. (2.51), let us consider a system *S*0 whose input *x*(*t*) and the corresponding output *w*(*t*) are related by - -$$ -Q(D)w(t) = x(t) \tag{2.52} -$$ - -Observe that both the systems *S* and *S*0 have the same characteristic polynomial; namely, *Q*(λ), and, consequently, the same characteristic modes. Moreover, *S*0 is the same as *S* with *P*(*D*)=1, that is, *b*0 = 0. Therefore, according to Eq. (2.51), the impulse response of *S*0 consists of characteristic mode terms only without an impulse at *t* = 0. Let us denote this impulse response of *S*0 by *yn*(*t*). Observe that *yn*(*t*) consists of characteristic modes of *S* and therefore may be viewed as a zero-input response of *S*. Now *yn*(*t*) is the response of *S*0 to input δ(*t*). Therefore, according to Eq. (2.52), - -$$ -Q(D)y_n(t) = \delta(t) -$$ - -or - -$$ -(D^N + a_1 D^{N-1} + \dots + a_{N1} D + a_N) y_n(t) = \delta(t) -$$ - -or - -$$ -y_n^{(N)}(t) + a_1 y_n^{(N-1)}(t) + \dots + a_{N-1} y_n^{(1)}(t) + a_N y_n(t) = \delta(t) -$$ - -where *y*(*k*) *n* (*t*) represents the *k*th derivative of *yn*(*t*). The right-hand side contains a single impulse term, δ(*t*). This is possible only if *y*(*N*−1) *n* (*t*) has a unit jump discontinuity at *t* = 0, so that *y*(*N*) *n* (*t*) = δ(*t*). Moreover, the lower-order terms cannot have any jump discontinuity because this would mean the presence of the derivatives of δ(*t*). Therefore *yn*(0) = *y*(1) *n* (0) =···= *y*(*N*−2) *n* (0) = 0 (no discontinuity at *t* = 0), and the *N* initial conditions on *yn*(*t*) are - -$$ -y_n(0) = y_n^{(1)}(0) = \dots = y_n^{(N-2)}(0) = 0 -$$ - and $y_n^{(N-1)}(0) = 1$ (2.53) - -This discussion means that *yn*(*t*) is the zero-input response of the system *S* subject to initial conditions [Eq. (2.53)]. - -We now show that for the same input *x*(*t*) to both systems, *S* and *S*0, their respective outputs *y*(*t*) and *w*(*t*) are related by - -$$ -y(t) = P(D)w(t) -$$ -\n(2.54) - -To prove this result, we operate on both sides of Eq. (2.52) by *P*(*D*) to obtain - -$$ -Q(D)P(D)w(t) = P(D)x(t) -$$ - -Comparison of this equation with Eq. (2.2) leads immediately to Eq. (2.54). - -Now if the input *x*(*t*) = δ(*t*), the output of *S*0 is *yn*(*t*), and the output of *S*, according to Eq. (2.54), is *P*(*D*)*yn*(*t*). This output is *h*(*t*), the unit impulse response of *S*. Note, however, that because it is an impulse response of a causal system *S*0, the function *yn*(*t*) is causal. To incorporate this fact we must represent this function as *yn*(*t*)*u*(*t*). Now it follows that *h*(*t*), the unit impulse response of the system *S*, is given by - -$$ -h(t) = P(D)[y_n(t)u(t)] -$$ -\n -$$ -(2.55) -$$ - -where *yn*(*t*) is a linear combination of the characteristic modes of the system subject to initial conditions (2.53). - -The right-hand side of Eq. (2.55) is a linear combination of the derivatives of *yn*(*t*)*u*(*t*). Evaluating these derivatives is clumsy and inconvenient because of the presence of *u*(*t*). The derivatives will generate an impulse and its derivatives at the origin. Fortunately when *M* ≤ *N* [Eq. (2.11)], we can avoid this difficulty by using the observation in Eq. (2.51), which asserts that at *t* = 0 (the origin), *h*(*t*) = *b*0δ(*t*). Therefore, we need not bother to find *h*(*t*) at the origin. This simplification means that instead of deriving *P*(*D*)[*yn*(*t*)*u*(*t*)], we can derive *P*(*D*)*yn*(*t*) and add to it the term *b*0δ(*t*) so that - -$$ -h(t) = b_0 \delta(t) + P(D) y_n(t) \qquad t \ge 0 -$$ - -= $b_0 \delta(t) + [P(D) y_n(t)] u(t)$ - -This expression is valid when *M* ≤ *N* [the form given in Eq. (2.11)]. When *M* > *N*, Eq. (2.55) should be used. - -## **[2.9 SUMMARY](#page-9-0)** - -This chapter discusses time-domain analysis of LTIC systems. The total response of a linear system is a sum of the zero-input response and zero-state response. The zero-input response is the system - -#### 222 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -response generated only by the internal conditions (initial conditions) of the system, assuming that the external input is zero; hence the adjective "zero-input." The zero-state response is the system response generated by the external input, assuming that all initial conditions are zero, that is, when the system is in zero state. - -Every system can sustain certain forms of response on its own with no external input (zero input). These forms are intrinsic characteristics of the system; that is, they do not depend on any external input. For this reason they are called characteristic modes of the system. Needless to say, the zero-input response is made up of characteristic modes chosen in a combination required to satisfy the initial conditions of the system. For an *N*th-order system, there are *N* distinct modes. - -The unit impulse function is an idealized mathematical model of a signal that cannot be generated in practice.† Nevertheless, introduction of such a signal as an intermediary is very helpful in analysis of signals and systems. The unit impulse response of a system is a combination of the characteristic modes of the system‡ because the impulse δ(*t*) = 0 for *t* > 0. Therefore, the system response for *t* > 0 must necessarily be a zero-input response, which, as seen earlier, is a combination of characteristic modes. - -The zero-state response (response due to external input) of a linear system can be obtained by breaking the input into simpler components and then adding the responses to all the components. In this chapter we represent an arbitrary input *x*(*t*) as a sum of narrow rectangular pulses [staircase approximation of *x*(*t*)]. In the limit as the pulse width → 0, the rectangular pulse components approach impulses. Knowing the impulse response of the system, we can find the system response to all the impulse components and add them to yield the system response to the input *x*(*t*). The sum of the responses to the impulse components is in the form of an integral, known as the convolution integral. The system response is obtained as the convolution of the input *x*(*t*) with the system's impulse response *h*(*t*). Therefore, the knowledge of the system's impulse response allows us to determine the system response to any arbitrary input. - -LTIC systems have a very special relationship to the everlasting exponential signal *est* because the response of an LTIC system to such an input signal is the same signal within a multiplicative constant. The response of an LTIC system to the everlasting exponential input *est* is *H*(*s*)*est*, where *H*(*s*) is the transfer function of the system. - -If every bounded input results in a bounded output, the system is stable in the bounded-input/bounded-output (BIBO) sense. An LTIC system is BIBO-stable if and only if its impulse response is absolutely integrable. Otherwise, it is BIBO-unstable. BIBO stability is a stability seen from external terminals of the system. Hence, it is also called external stability or zero-state stability. - -In contrast, internal stability (or the zero-input stability) examines the system stability from inside. When some initial conditions are applied to a system in zero state, then, if the system eventually returns to zero state, the system is said to be stable in the asymptotic or Lyapunov sense. If the system's response increases without bound, it is unstable. If the system does not go to zero state and the response does not increase indefinitely, the system is marginally stable. The internal stability criterion, in terms of the location of a system's characteristic roots, can be summarized as follows: - -† However, it can be closely approximated by a narrow pulse of unit area and having a width that is much smaller than the time constant of an LTIC system in which it is used. - -‡ There is the possibility of an impulse in addition to the characteristic modes. - -- 1. An LTIC system is asymptotically stable if, and only if, all the characteristic roots are in the LHP. The roots may be repeated or unrepeated. -- 2. An LTIC system is unstable if, and only if, either one or both of the following conditions exist: (i) at least one root is in the RHP; (ii) there are repeated roots on the imaginary axis. -- 3. An LTIC system is marginally stable if, and only if, there are no roots in the RHP, and there are some unrepeated roots on the imaginary axis. - -It is possible for a system to be externally (BIBO) stable but internally unstable. When a system is controllable and observable, its external and internal descriptions are equivalent. Hence, external (BIBO) and internal (asymptotic) stabilities are equivalent and provide the same information. Such a BIBO-stable system is also asymptotically stable, and vice versa. Similarly, a BIBO-unstable system is either marginally stable or asymptotically unstable system. - -The characteristic behavior of a system is extremely important because it determines not only the system response to internal conditions (zero-input behavior), but also the system response to external inputs (zero-state behavior) and the system stability. The system response to external inputs is determined by the impulse response, which itself is made up of characteristic modes. The width of the impulse response is called the time constant of the system, which indicates how fast the system can respond to an input. The time constant plays an important role in determining such diverse system behaviors as the response time and filtering properties of the system, dispersion of pulses, and the rate of pulse transmission through the system. - -### **[REFERENCES](#page-9-0)** - -- 1. Lathi, B. P., *Signals and Systems*. Berkeley-Cambridge Press, Carmichael, CA, 1987. -- 2. Mason, S. J., *Electronic Circuits, Signals, and Systems*. Wiley, New York, 1960. -- 3. Kailath, T., *Linear System*. Prentice-Hall, Englewood Cliffs, NJ, 1980. -- 4. Lathi, B. P., *Modern Digital and Analog Communication Systems,* 3rd ed. Oxford University Press, New York, 1998. - -## **[PROBLEMS](#page-9-0)** - -- **2.2-1** Determine the constants *c*1, *c*2, λ1, and λ2 for each of the following second-order systems, which have zero-input responses of the form *y*zir(*t*) = *c*1*e*λ1*t* +*c*2*e*λ2*t* . - - (a) *y*¨(*t*) + 2*y*˙(*t*) + 5*y*(*t*) = ¨*x*(*t*) − 5*x*(*t*) with *y*zir(0) = 2 and *y*˙zir(0) = 0. - - (b) *y*¨(*t*) + 2*y*˙(*t*) + 5*y*(*t*) = ¨*x*(*t*) − 5*x*(*t*) with *y*zir(0) = 4 and *y*˙zir(0) = −1. - - (c) *d*2 *dt*2 *y*(*t*) + 2 *d dt y*(*t*) = *x*(*t*) with *y*zir(0) = 1 and *y*˙zir(0) = 2. - - (d) (*D*2 +2*D*+10){*y*(*t*)} = (*D*5 −*D*){*x*(*t*)} with *y*zir(0) = ˙*y*zir(0) = 1. - - (e) (*D*2 + 7 2*D* + 3 2 ){*y*(*t*)} = (*D* + 2){*x*(*t*)} with *y*zir(0) = 3 and *y*¨zir(0) = −8. [*Caution:* The - -second IC is given in terms of the second derivative, not the first derivative]. - -- (f) 13*y*(*t*) + 4 *d dt y*(*t*) + *d*2 *dt*2 *y*(*t*) = 2*x*(*t*) − 4 *d dt x*(*t*) with *y*zir(0) = 3 and *y*¨zir(0) = −15. [*Caution:* The second IC is given in terms of the second derivative, not the first derivative]. -- **2.2-2** Consider a linear time-invariant system with input *x*(*t*) and output *y*(*t*) that is described by the differential equation - -$$ -(D+1)(D2-1) \{y(t)\} = (D5-1) \{x(t)\} -$$ - -Furthermore, assume *y*(0) = ˙*y*(0) = ¨*y*(0) = 1. - -### 224 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -- (a) What is the order of this system? -- (b) What are the characteristic roots of this system? -- (c) Determine the zero-input response *y*zir(*t*). Simplify your answer. -- **2.2-3** A real LTIC system with input *x*(*t*) and output *y*(*t*) is described by the following constant-coefficient linear differential equation: - -$$ -(D3 + 9D) \{y(t)\} = (2D3 + 1) \{x(t)\}. -$$ - -- (a) What is the characteristic equation of this system? -- (b) What are the characteristic modes of this system? -- (c) Assuming *y*zir(0) = 4, *y*˙zir(0) = −18, and *y*¨zir(0) = 0, determine this system's zero-input response *y*zir(*t*). Simplify *y*zir(*t*) to include only real terms (i.e., no *j*'s should appear in your answer). -- **2.2-4** An LTIC system is specified by the equation - -$$ -(D2 + 5D + 6)y(t) = (D + 1)x(t) -$$ - -- (a) Find the characteristic polynomial, characteristic equation, characteristic roots, and characteristic modes of this system. -- (b) Find *y*0(*t*), the zero-input component of the response *y*(*t*) for *t* ≥ 0, if the initial conditions are *y*0(0−) = 2 and *y*˙0(0−) = −1. -- **2.2-5** Repeat Prob. 2.2-4 for - -$$ -(D^2 + 4D + 4)y(t) = Dx(t) -$$ - -and *y*0(0−) = 3, *y*˙0(0−) = −4. - -**2.2-6** Repeat Prob. 2.2-4 for - -$$ -D(D+1)y(t) = (D+2)x(t) -$$ - -and *y*0(0−) = ˙*y*0(0−) = 1. - -**2.2-7** Repeat Prob. 2.2-4 for - -$$ -(D^2 + 9)y(t) = (3D + 2)x(t) -$$ - -and *y*0(0−) = 0, *y*˙0(0−) = 6. - -**2.2-8** Repeat Prob. 2.2-4 for - -$$ -(D2 + 4D + 13)y(t) = 4(D+2)x(t) -$$ - -with -$$ -y_0(0^-) = 5 -$$ -, $\dot{y}_0(0^-) = 15.98$ . - -**2.2-9** Repeat Prob. 2.2-4 for - -$$ -D^2(D+1)y(t) = (D^2+2)x(t) -$$ - -with *y*0(0−) = 4, *y*˙0(0−) = 3, and *y*¨0(0−) = −1. - -**2.2-10** Repeat Prob. 2.2-4 for - -$$ -(D+1)(D^2 + 5D + 6)y(t) = Dx(t) -$$ - -with *y*0(0−) = 2, *y*˙0(0−) = −1, and *y*¨0(0−) = 5. - -- **2.2-11** A system is described by a constant-coefficient linear differential equation and has zero-input response given by *y*0(*t*) = 2*e*−*t* +3. - - (a) Is it possible for the system's characteristic equation to be λ + 1 = 0? Justify your answer. - - (b) Is it possible for the system's characteristic equation to be √3(λ2 +λ) = 0? Justify your answer. - - (c) Is it possible for the system's characteristic equation to be λ(λ + 1)2 = 0? Justify your answer. -- **2.2-12** Consider the circuit of Fig. P2.2-12. Using operator notation, this system can be described as (*D*+*a*1){*y*(*t*)} = (*b*0*D*+*b*1){*x*(*t*)}. - - (a) Determine the constants *a*1, *b*0, and *b*1 in terms of the system components *R*, *Rf* , and *C*. - - (b) Assume that *R* = 300 k, *Rf* = 1.2 M, and *C* = 5 µF. What is the zero-input response *y*0(*t*) of this system, assuming *vC*(0) = 1 V? - -**2.3-1** Determine the characteristic equation, characteristic modes, and impulse response *h*(*t*) for each of the following real LTIC systems. Since the systems are real, express each *h*(*t*) using only real terms (i.e., no *j*'s should appear in your answers). - -Problems 225 - -- (a) (*D*2 +1){*y*(*t*)} = 2*D*{*x*(*t*)} -- (b) (*D*3 +*D*){*y*(*t*)} = (2*D*3 +1){*x*(*t*)} - -(c) -$$ -\frac{d^2}{dt^2}y(t) + 2\frac{d}{dt}y(t) + 5y(t) = 8x(t) -$$ - -**2.3-2** Find the unit impulse response of a system specified by the equation - -$$ -(D2 + 4D + 3)y(t) = (D + 5)x(t) -$$ - -**2.3-3** Repeat Prob. 2.3-2 for - -$$ -(D2 + 5D + 6)y(t) = (D2 + 7D + 11)x(t) -$$ - -**2.3-4** Repeat Prob. 2.3-2 for the first-order allpass filter specified by the equation - -$$ -(D+1)y(t) = -(D-1)x(t) -$$ - -**2.3-5** Find the unit impulse response of an LTIC system specified by the equation - -(*D*2 +6*D*+9)*y*(*t*) = (2*D*+9)*x*(*t*) - -- **2.3-6** Determine and plot the unit impulse response *h*(*t*) of the op-amp circuit of Fig. P2.2-12, assuming that *R* = 300 k, *Rf* = 1.2 M, and *C* = 5 µF. -- **2.3-7** A causal LTIC system with input *x*(*t*) and output *y*(*t*) is described by the constant coefficient integral equation - -$$ -y(t) + \int 3y(t) dt + \int \int 2y(t) dt = -$$ - -$$ -\int \int x(t) dt - \int \int \int x(t) dt. -$$ - -- (a) Express this system as a constant coefficient linear differential equation in standard operator form. -- (b) Determine the characteristic modes of this system. - -**Figure P2.4-1** - -- (c) Determine the impulse response *h*(*t*) of this system. -- **2.4-1** Let *f*(*t*) = *h*1(*t*)∗*h*2(*t*), where *h*1(*t*) and *h*2(*t*) are shown in Fig. P2.4-1. In the following, use the graphical convolution procedure where you flip and shift *h*2(*t*). - - (a) Plot *h*1(τ ) and *h*2(*t* − τ ) as functions of τ . Clearly label the plots, including necessary function parameterizations. - - (b) Determine the (piecewise) regions of *f*(*t*) and set up the corresponding integrals that describe *f*(*t*) in those regions. Do not evaluate the integrals, only set them up! - - (c) Determine *f*(1), which is *f*(*t*) evaluated at *t* = 1. Provide a number, not a formula. -- **2.4-2** Consider signals *h*(*t*) = *u*(*t* + 3) − 2*u*(*t* + 1) + *u*(*t* − 1) and *x*(*t*) = cos(*t*) *u*(*t* − π/2)− *u*(*t* −3π/2) . Let *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Determine the last time *t*last that *y*(*t*) is nonzero. That is, find the smallest value *t*last such that *y*(*t*) = 0 for all *t* > *t*last. - - (b) Determine the approximate time *t*max where *y*(*t*) is a maximum. -- **2.4-3** Consider signals *h*(*t*) = −*u*(*t* + 2) + 3*u*(*t* − 1) − 2*u*(*t* − 5 2 ) and *x*(*t*) = sin(*t*)[*u*(*t* + 2π ) −*u*(*t* +π )]. Determine the approximate time *t*min where *y*(*t*)=*x*(*t*)∗*h*(*t*) is a minimum. Note, the minimum value of *y*(*t*) = 0! -- **2.4-4** An LTIC system has impulse response *h*(*t*) = 3*u*(*t* − 2). For input *x*(*t*) shown in Fig. P2.4-4, use the graphical convolution procedure to determine *y*zsr(*t*) = *h*(*t*) ∗ *x*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *x*(*t*) and explicitly show all integration steps—even if apparently trivial! - -**Figure P2.4-4** - -**2.4-5** Suppose an LTIC system has impulse response *h*(*t*) and input *x*(*t*) = *u*(*t*). Figure P2.4-5 shows *x*(*t*) and *h*(*t* + 1), respectively. Be careful! Figure P2.4-5 shows *h*(*t* +1), not *h*(*t*). - -**Figure P2.4-5** - -- (a) Is system *h*(*t*) causal? Mathematically justify your answer. -- (b) Use the graphical convolution procedure to determine *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *x*(*t*) and explicitly show all integration steps. - -**2.4-6** Repeat Prob. 2.4-5 using the signals of Fig. P2.4-6, rather than those of Fig. P2.4-5. Be careful! Figure P2.4-6 shows *h*(*t* −1), not *h*(*t*). - -**Figure P2.4-6** - -- **2.4-7** Suppose an LTIC system has impulse response *h*(*t*) and input *x*(*t*), both shown in Fig. P2.4-7. Use the graphical convolution procedure to determine *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *x*(*t*) and explicitly show all integration steps. -- **2.4-8** An LTIC system has impulse response *h*(*t*), as shown in Fig. P2.4-8. Let *t* have units of seconds. Let the input be *x*(*t*) = *u*(−*t* − 2) and designate the output as *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Use the graphical convolution procedure where *h*(*t*) is flipped and shifted to determine *y*zsr(*t*). Accurately plot your result. - - (b) Use the graphical convolution procedure where *x*(*t*) is flipped and shifted to determine *y*zsr(*t*). Accurately plot your result. -- **2.4-9** If *c*(*t*) = *x*(*t*) ∗ *g*(*t*), then show that *Ac* = *AxAg*, where *Ax*,*Ag*, and *Ac* are the areas under *x*(*t*), *g*(*t*), and *c*(*t*), respectively. Verify this *area property* of convolution in Exs. 2.10 and 2.12. - -**Figure P2.4-7** - -**2.4-10** If *x*(*t*) ∗ *g*(*t*) = *c*(*t*), then show that *x*(*at*) ∗ *g*(*at*) = |1/*a*|*c*(*at*). This *time-scaling property* of convolution states that if both *x*(*t*) and *g*(*t*) are time-scaled by *a*, their convolution is also time-scaled by *a* (and multiplied by |1/*a*|). - -**Figure P2.4-8** - -- **2.4-11** Show that the convolution of an odd and an even function is an odd function and the convolution of two odd or two even functions is an even function. [*Hint:* Use the time-scaling property of convolution in Prob. 2.4-10.] -- **2.4-12** Suppose an LTIC system has impulse response *h*(*t*) = (1 − *t*)[*u*(*t*) − *u*(*t* − 1)] and input *x*(*t*) = *u*(−*t* −1)+*u*(*t* −1). Use the graphical convolution procedure to determine *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *h*(*t*), explicitly show all integration steps, and simplify your answer. -- **2.4-13** Using direct integration, find *e*−*atu*(*t*) ∗ *e*−*btu*(*t*). -- **2.4-14** Using direct integration, find *u*(*t*) ∗ *u*(*t*), *e*−*atu*(*t*) ∗ *e*−*atu*(*t*), and *tu*(*t*) ∗ *u*(*t*). -- **2.4-15** Using direct integration, find sin *t u*(*t*) ∗ *u*(*t*) and cos *t u*(*t*) ∗ *u*(*t*). -- **2.4-16** The unit impulse response of an LTIC system is - -$$ -h(t) = e^{-t}u(t) -$$ - -Find this system's (zero-state) response *y*(*t*) if the input *x*(*t*) is: - -$$ -(a) u(t) -$$ - -$$ -(b) e^{-t}u(t) -$$ - -$$ -(c) e^{-2t}u(t) -$$ - -(d) sin 3*t u*(*t*) - -Use the convolution table (Table 2.1) to find your answers. - -**2.4-17** Repeat Prob. 2.4-16 for - -$$ -h(t) = [2e^{-3t} - e^{-2t}]u(t) -$$ - -and if the input -$$ -x(t) -$$ - is: -\n(a) $u(t)$ -\n(b) $e^{-t}u(t)$ -\n(c) $e^{-2t}u(t)$ - -**2.4-18** Repeat Prob. 2.4-16 for - -$$ -h(t) = (1 - 2t)e^{-2t}u(t) -$$ - -and input *x*(*t*) = *u*(*t*). - -**2.4-19** Repeat Prob. 2.4-16 for - -$$ -h(t) = 4e^{-2t}\cos 3t u(t) -$$ - -and each of the following inputs *x*(*t*): - -- (a) *u*(*t*) -- (b) *e*−*t u*(*t*) -- **2.4-20** Repeat Prob. 2.4-16 for - -$$ -h(t) = e^{-t}u(t) -$$ - -and each of the following inputs *x*(*t*): - -- (a) *e*−2*t u*(*t*) -- (b) *e*−2(*t*−3) *u*(*t*) -- (c) *e*−2*t u*(*t* −3) -- (d) The gate pulse depicted in Fig. P2.4-20—and provide a sketch of *y*(*t*). - -**Figure P2.4-20** - -**2.4-21** A first-order allpass filter impulse response is given by - -$$ -h(t) = -\delta(t) + 2e^{-t}u(t) -$$ - -- (a) Find the zero-state response of this filter for the input *et u*(−*t*). -- (b) Sketch the input and the corresponding zero-state response. -- **2.4-22** Figure P2.4-22 shows the input *x*(*t*) and the impulse response *h*(*t*) for an LTIC system. Let the output be *y*(*t*). - - (a) By inspection of *x*(*t*) and *h*(*t*), find *y*(−1), *y*(0), *y*(1), *y*(2), *y*(3), *y*(4), *y*(5), and - -### **Figure P2.4-22** - -*y*(6). Thus, by merely examining *x*(*t*) and *h*(*t*), you are required to see what the result of convolution yields at *t* = −1, 0, 1, 2, 3, 4, 5, and 6. - -- (b) Find the system response to the input *x*(*t*). -- **2.4-23** The zero-state response of an LTIC system to an input *x*(*t*) = 2*e*−2*t u*(*t*) is *y*(*t*) = [4*e*−2*t* + 6*e*−3*t* ]*u*(*t*). Find the impulse response of the system. [*Hint:* We have not yet developed a method of finding *h*(*t*) from the knowledge of the input and the corresponding output. Knowing the form of *x*(*t*) and *y*(*t*), you will have to make the best guess of the general form of *h*(*t*).] -- **2.4-24** Sketch the functions *x*(*t*) = 1/(*t* 2 +1) and *u*(*t*). Now find *x*(*t*) ∗ *u*(*t*) and sketch the result. -- **2.4-25** Figure P2.4-25 shows *x*(*t*) and *g*(*t*). Find and sketch *c*(*t*) = *x*(*t*) ∗ *g*(*t*). -- **2.4-26** Find and sketch *c*(*t*) = *x*(*t*) ∗ *g*(*t*) for the functions depicted in Fig. P2.4-26. -- **2.4-27** Find and sketch *c*(*t*) = *x*1(*t*) ∗ *x*2(*t*) for the pairs of functions illustrated in Fig. P2.4-27. -- **2.4-28** Use Eq. (2.37) to find the convolution of *x*(*t*) and *w*(*t*), shown in Fig. P2.4-28. - -- **2.4-29** Determine *H*(*s*), the transfer function of an ideal time delay of *T* seconds. Find your answer by two methods: using Eq. (2.39) and using Eq. (2.40). -- **2.4-30** Determine *y*(*t*) = *x*(*t*) ∗ *h*(*t*) for the signals depicted in Fig. P2.4-30. -- **2.4-31** Two linear time-invariant systems, each with impulse response *h*(*t*), are connected in cascade. Refer to Fig. P2.4-31. Given input *x*(*t*) = *u*(*t*), determine *y*(1). That is, determine the step response at time *t* = 1 for the cascaded system shown. -- **2.4-32** Consider the electric circuit shown in Fig. P2.4-32. - - (a) Determine the differential equation that relates the input *x*(*t*) to output *y*(*t*). Recall that *iC*(*t*) = *CdvC*(*t*) *dt* and *vL*(*t*) = *LdiL*(*t*) *dt* . - - (b) Find the characteristic equation for this circuit, and express the root(s) of the characteristic equation in terms of *L* and *C*. - - (c) Determine the zero-input response given an initial capacitor voltage of one volt and an initial inductor current of zero amps. That is, find *y*0(*t*) given *vC*(0) = 1 V and - -**Figure P2.4-26** - -Problems 229 - -**Figure P2.4-27** - -*iL*(0) = 0 A. [*Hint:* The coefficient(s) in *y*0(*t*) are independent of *L* and *C*.] - -- (d) Plot *y*0(*t*) for *t* ≥ 0. Does the zero-input response, which is caused solely by initial conditions, ever "die out"? -- (e) Determine the total response *y*(*t*) to the input *x*(*t*) = *e*−*t u*(*t*). Assume an initial inductor current of *iL*(0−) = 0 A, an initial capacitor voltage of *vC*(0−) = 1 V, *L* = 1 H, and *C* = 1 F. - -**Figure P2.4-33** - -- **2.4-33** Two LTIC systems have impulse response functions given by *h*1(*t*) = (1−*t*)[*u*(*t*)−*u*(*t*−1)] and *h*2(*t*) = *t*[*u*(*t* +2)−*u*(*t* −2)]. - - (a) Carefully sketch the functions *h*1(*t*) and *h*2(*t*). - - (b) Assume that the two systems are connected in parallel, as shown in Fig. P2.4-33a. Carefully plot the equivalent impulse response function, *hp*(*t*). - - (c) Assume that the two systems are connected in cascade, as shown in Fig. P2.4-33b. Carefully plot the equivalent impulse response function, *hs*(*t*). - -- **2.4-34** Consider the circuit shown in Fig. P2.4-34. - - (a) Find the output *y*(*t*) given an initial capacitor voltage of *y*(0) = 2 volts and an input *x*(*t*) = *u*(*t*). - - (b) Given an input *x*(*t*) = *u*(*t* − 1), determine the initial capacitor voltage *y*(0) so that the output *y*(*t*) is 0.5 volt at *t* = 2 seconds. - -**Figure P2.4-34** - -**2.4-35** An analog signal is given by *x*(*t*) = *t*[*u*(*t*) − *u*(*t* − 1)], as shown in Fig. P2.4-35. Determine and plot *y*(*t*) = *x*(*t*) ∗ *x*(2*t*). - -**Figure P2.4-35** - -- **2.4-36** Consider the electric circuit shown in Fig. P2.4-36. - - (a) Determine the differential equation that relates the input current *x*(*t*) to output current *y*(*t*). Recall that - -$$ -v_L(t) = L \frac{di_L(t)}{dt} -$$ - -- (b) Find the characteristic equation for this circuit, and express the root(s) of the characteristic equation in terms of *L*1, *L*2, and *R*. -- (c) Determine the zero-input response given initial inductor currents of one ampere each. That is, find *y*0(*t*) given *iL*1 (0) = *iL*2 (0) = 1 A. - -**Figure P2.4-36** - -**Figure P2.4-38** - -- **2.4-37** An LTI system has step response given by *g*(*t*) = *e*−*t u*(*t*) − *e*−2*t u*(*t*). Determine the output of this system *y*(*t*) given an input *x*(*t*) = δ(*t* −π ) −cos( √3)*u*(*t*). -- **2.4-38** The periodic signal *x*(*t*) shown in Fig. P2.4-38 is input to a system with impulse response function *h*(*t*) = *t*[*u*(*t*) − *u*(*t* − 1.5)], also shown in Fig. P2.4-38. Use convolution to determine the output *y*(*t*) of this system. Plot *y*(*t*) over (−3 ≤ *t* ≤ 3). -- **2.4-39** Consider the electric circuit shown in Fig. P2.4-39. - - (a) Determine the differential equation relating input *x*(*t*) to output *y*(*t*). - - (b) Determine the output *y*(*t*) in response to the input *x*(*t*) = 4*te*−3*t*/2*u*(*t*). Assume component values of *R* = 1 , *C*1 = 1 F, and *C*2 = 2 F, and initial capacitor voltages of *VC*1 = 2 V and *VC*2 = 1 V. - -**Figure P2.4-39** - -- **2.4-40** An LTIC system has impulse response *h*(*t*) = 3*e*−|*t*| . - - (a) Is the system causal? Mathematically justify your answer. - - (b) Determine the zero-state response of this system if the input is *x*(*t*) = *u*(2−*t*). -- **2.4-41** A cardiovascular researcher is attempting to model the human heart. He has recorded ventricular pressure, which he believes corresponds to the heart's impulse response - -function *h*(*t*), as shown in Fig. P2.4-41. Comment on the function *h*(*t*) shown in Fig. P2.4-41. Can you establish any system properties, such as causality or stability? Do the data suggest any reason to suspect that the measurement is not a true impulse response? - -- **2.4-42** Consider an integrator system, *y*(*t*) = \$ *t* −∞ *x*(τ )*d*τ . - - (a) What is the unit impulse response *h*i(*t*) of this system? - - (b) If two such integrators are put in parallel, what is the resulting impulse response *h*p(*t*)? - - (c) If two such integrators are put in series, what is the resulting impulse response *h*s(*t*)? -- **2.4-43** The autocorrelation of a function *x*(*t*) is given by *rxx*(*t*) = \$ ∞ −∞ *x*(τ )*x*(τ − *t*)*d*τ . This equation is computed in a manner nearly identical to convolution. - - (a) Show *rxx*(*t*) = *x*(*t*) ∗ *x*(−*t*). - - (b) Determine and plot *rxx*(*t*) for the signal *x*(*t*) depicted in Fig. P2.4-43. [*Hint: rxx*(*t*) = *rxx*(−*t*).] - -**Figure P2.4-43** - -**2.4-44** Consider the circuit shown in Fig. P2.4-44. This circuit functions as an integrator. Assume ideal op-amp behavior and recall that - -*dVC*(*t*) - -- (a) Determine the differential equation that relates the input *x*(*t*) to the output *y*(*t*). -- (b) This circuit does not behave well at dc. Demonstrate this by computing the zero-state response *y*(*t*) for a unit step input *x*(*t*) = *u*(*t*). -- **2.4-45** Derive the result in Eq. (2.37) in another way. As mentioned in Ch. 1 (Fig. 1.27b), it is possible to express an input in terms of its step components, as shown in Fig. P2.4-45. Find the system response as a sum of the responses to the step components of the input. - -**2.4-46** Show that an LTIC system response to an everlasting sinusoid cosω0*t* is given by - -$$ -y(t) = |H(j\omega_0)| \cos [\omega_0 t + \angle H(j\omega_0)] -$$ - -where - -$$ -H(j\omega) = \int_{-\infty}^{\infty} h(t)e^{-j\omega t} dt -$$ - -assuming the integral on the right-hand side exists. - -**2.4-47** A line charge is located along the *x* axis with a charge density *Q*(*x*) coulombs per meter. Show that the electric field *E*(*x*) produced by this line charge at a point *x* is given by - -$$ -E(x) = Q(x) * h(x) -$$ - -where *h*(*x*) = 1/4π *x*2. [*Hint:* The charge over an interval τ located at τ = *n*τ is *Q*(*n*τ )τ . Also by Coulomb's law, the electric field *E*(*r*) at a distance *r* from a charge *q* coulombs is given by *E*(*r*) = *q*/4π *r*2.] - -- **2.4-48** A system is called complex if a real-valued input can produce a complex-valued output. Suppose a linear time-invariant complex system has impulse response *h*(*t*) = *j*[*u*(−*t* + 2) − *u*(−*t*)]. - - (a) Is this system causal? Explain. - - (b) Use convolution to determine the zero-state response *y*1(*t*) of this system in response to the unit-duration pulse *x*1(*t*) = *u*(*t*) − *u*(*t* − 1). - - (c) Using the result from part (a), determine the zero-state response *y*2(*t*) in response to *x*2(*t*) = 2*u*(*t* −1)−*u*(*t* −2)−*u*(*t* −3). -- **2.5-1** Explain, with reasons, whether the LTIC systems described by the following equations are (i) stable or unstable in the BIBO sense; (ii) asymptotically stable, unstable, or marginally stable. Assume that the systems are controllable and observable. - - (a) (*D*2 +8*D*+12)*y*(*t*) = (*D*−1)*x*(*t*) - - (b) *D*(*D*2 +3*D*+2)*y*(*t*) = (*D*+5)*x*(*t*) - - (c) *D*2(*D*2 +2)*y*(*t*) = *x*(*t*) - - (d) (*D*+1)(*D*2 −6*D*+5)*y*(*t*) = (3*D*+1)*x*(*t*) -- **2.5-2** Repeat Prob. 2.5-1 for the following: - - (a) (*D*+1)(*D*2 +2*D*+5)2*y*(*t*) = *x*(*t*) - - (b) (*D*+1)(*D*2 +9)*y*(*t*) = (2*D*+9)*x*(*t*) - - (c) (*D*+1)(*D*2 +9)2*y*(*t*) = (2*D*+9)*x*(*t*) - - (d) (*D*2 +1)(*D*2 +4)(*D*2 +9)*y*(*t*) = 3*Dx*(*t*) -- **2.5-3** Consider an LTIC system with unit impulse response *h*(*t*) = *et* 2 3 cos( 3 2 *t*) + 1 3 sin(π*t*) *u*(123 − *t*). Is this system BIBO-stable? Mathematically justify your answer. - -- **2.5-4** Consider an LTIC system with unit impulse response *h*(*t*) = 1 *t u*(*t* −*T*). - - (a) Determine, if possible, the value(s) of *T* for which this system is causal. - - (b) Determine, if possible, the value(s) of *T* for which this system is BIBO-stable. Justify all answers mathematically. -- **2.5-5** You are given the choice of a system that is guaranteed internally stable or a system that is guaranteed externally stable. Which do you choose? Why? -- **2.5-6** For a certain LTIC system, the impulse response *h*(*t*) = *u*(*t*). - - (a) Determine the characteristic root(s) of this system. - - (b) Is this system asymptotically or marginally stable, or is it unstable? - - (c) Is this system BIBO-stable? - - (d) What can this system be used for? -- **2.5-7** In Sec. 2.5 we demonstrated that for an LTIC system, the condition of Eq. (2.45) is sufficient for BIBO stability. Show that this is also a necessary condition for BIBO stability in such systems. In other words, show that if Eq. (2.45) is not satisfied, then there exists a bounded input that produces an unbounded output. [Hint: Assume that a system exists for which *h*(*t*) violates Eq. (2.45) and yet produces an output that is bounded for every bounded input. Establish the contradiction in this statement by considering an input *x*(*t*) defined by *x*(*t*1−τ )=1 when *h*(τ ) ≥ 0 and *x*(*t*1 − τ ) = −1 when *h*(τ ) < 0, where *t*1 is some fixed instant.] -- **2.5-8** An analog LTIC system with impulse response function *h*(*t*) = *u*(*t* + 2) − *u*(*t* − 2) is presented with an input *x*(*t*) = *t*(*u*(*t*) −*u*(*t* −2)). - - (a) Determine and plot the system output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (b) Is this system stable? Is this system causal? Justify your answers. -- **2.5-9** A system has an impulse response function shaped like a rectangular pulse, *h*(*t*) = *u*(*t*) − *u*(*t* − 1). Is the system stable? Is the system causal? -- **2.5-10** A continuous-time LTI system has impulse response function *h*(*t*) =%∞ *i*=0(0.5)*i* δ(*t*−*i*). - - (a) Is the system causal? Prove your answer. - - (b) Is the system stable? Prove your answer. - -- **2.6-1** Data at a rate of 1 million pulses per second are to be transmitted over a certain communications channel. The unit step response *g*(*t*) for this channel is shown in Fig. P2.6-1. - - (a) Can this channel transmit data at the required rate? Explain your answer. - - (b) Can an audio signal consisting of components with frequencies up to 15 kHz be transmitted over this channel with reasonable fidelity? - -**Figure P2.6-1** - -- **2.6-2** Determine a frequency ω that will cause the input *x*(*t*) = cos(ω*t*) to produce a strong response when applied to the system described by (*D*2 + 2*D* + 13/4){*y*(*t*)} = *x*(*t*). Carefully explain your choice. -- **2.6-3** Figure P2.6-3 shows the impulse response *h*(*t*) of a lowpass LTIC system. Determine the peak amplitude *A* and time constant *Th* so that rectangular impulse response *h*ˆ(*t*) is an appropriate approximation of *h*(*t*). The two graphs of Fig. P2.6-3 are not necessarily drawn to the same scale. - -**Figure P2.6-3** - -- **2.6-4** A certain communication channel has a bandwidth of 10 kHz. A pulse of 0.5 ms duration is transmitted over this channel. - - (a) Determine the width (duration) of the received pulse. - - (b) Find the maximum rate at which these pulses can be transmitted over this channel without interference between the successive pulses. - -- **2.6-5** A first-order LTIC system has a characteristic root λ = −104. - - (a) Determine *Tr*, the rise time of its unit step input response. - - (b) Determine the bandwidth of this system. - - (c) Determine the rate at which the information pulses can be transmitted through this system. -- **2.6-6** A lowpass system with a 6 MHz cutoff frequency needs to transmit data pulse that are 500 6 ns wide. Determine a suitable transmission rate *F*rate (pulses/s) for this system. -- **2.6-7** Sketch an impulse response *h*(*t*) of a non-causal LP system that has an approximate cutoff frequency of 5 kHz. Since many solutions are possible, be sure to properly justify your answer. -- **2.6-8** Two LTIC transmission channels are available: the first has impulse response *h*1(*t*) = *u*(*t*) − *u*(*t* − 1) and the second has impulse response *h*2(*t*) = δ(*t*) + 0.5δ(*t* − 1) + 0.25δ(*t* − 2). Explain which channel is better suited for the transmission of high-speed digital data (pulses). -- **2.6-9** Consider a linear time-invariant system with impulse response *h*(*t*) shown in Fig. P2.6-9. Outside the interval shown, *h*(*t*) = 0. - -### **Figure P2.6-9** - -- (a) What is the rise time *Tr* of this system? Remember, rise time is the time between the application of a unit step and the moment at which the system has "fully" responded. -- (b) Suppose *h*(*t*) represents the response of a communication channel. What conditions might cause the channel to have such an impulse response? What is the maximum - -average number of pulses per unit time that can be transmitted without causing interference? Justify your answer. - -- (c) Determine the system output *y*(*t*) = *x*(*t*) ∗ *h*(*t*) for *x*(*t*) = [*u*(*t* − 2) − *u*(*t*)]. Accurately sketch *y*(*t*) over (0 ≤ *t* ≤ 10). -- **2.6-10** A lowpass LTIC system has impulse response *h*(*t*) = −*te*−*t u*(*t*). - - (a) Accurately sketch *h*(*t*). - - (b) Describe a rectangular impulse response *h*ˆ(*t*) as an appropriate approximation of *h*(*t*). What is the approximate cutoff frequency of this system? -- **2.6-11** A lowpass LTIC system has impulse response *h*(*t*), as shown in Fig. P2.4-8. - - (a) As discussed in Sec. 2.6-2, determine a rectangular approximation *h*ˆ(*t*) to *h*(*t*). - - (b) Using *h*ˆ(*t*), what is the time constant *Th* of this lowpass system? - - (c) Using *h*ˆ(*t*), what is the approximate radian cutoff frequency ω*c* of this lowpass system? - - (d) Assuming a frequency ω0 ω*c*, what is the system response *y*(*t*) to the input *x*(*t*) = sin(ω0*t* +π/3)? -- **2.6-12** A first CT lowpass system with time constant *T*1 = 4 µs is put in series with a second CT lowpass system with time constant *T*2 = 2 µs. Make an educated sketch of the overall impulse response function *h*series(*t*). What is the time constant *T*series of the overall series-connected system? -- **2.7-1** An LTIC system with input *x*(*t*) and output *y*(*t*) is described by the following constant coefficient linear differential equation: - -$$ -(D4 - 16) \{y(t)\} = (D - 2) \{x(t)\}. -$$ - -- (a) What are the 4 characteristic roots of this system (λ1, λ2, λ3, and λ4)? Determine the roots by hand and then verify your answers using MATLAB's roots command. -- (b) From Eq. (2.17), computing *h*(*t*) requires a signal *y*˜*n*(*t*) = %4 *k*=1 *cke*λ*kt* . First, determine a matrix representation of the system of equations needed to solve for the four coefficients *ck*. Second, write MATLAB code that computes the length-4 column vector of coefficients *ck*. - -**2.7-2** Define *x*(*t*) = 2*u*(*t* + 2 3 ) − 2*u*(*t*). Further, define the periodic signal *h*1(*t*) as - -$$ -h_1(t) = \begin{cases} t & 0 \le t < 1 \\ h_1(t+1) & \forall t \end{cases} -$$ - -Lastly, define the aperiodic signal *h*2(*t*) in terms of *h*1(*t*) as - -$$ -h_2(t) = h_1(t)[u(t-1) - u(t-2)] -$$ - -- (a) Use MATLAB to plot *x*(*t*), *h*1(*t*), and *h*2(*t*) over the interval −2.5 ≤ *t* ≤ 3.5. -- (b) Using the graphical convolution procedure, compute *y*2(*t*) = *x*(*t*) ∗ *h*2(*t*). -- (c) Compute by hand and then MATLAB plot *y*1(*t*) = *x*(*t*) ∗ *h*1(*t*). Modify program CH2MP4.m in Sec. 2.7-4 to validate your analytical result. -- **2.7-3** Consider the circuit shown in Fig. P2.7-3. Assume ideal op-amp behavior and recall that - -$$ -i_C(t) = C \frac{dV_C(t)}{dt} -$$ - -Without a feedback resistor *Rf* , the circuit functions as an integrator and is unstable, particularly at dc. A feedback resistor *Rf* corrects this problem and results in a stable circuit that functions as a "lossy" integrator. - -**Figure P2.7-3** - -- (a) Determine the differential equation that relates the input *x*(*t*) to the output *y*(*t*). What is the corresponding characteristic equation? -- (b) To demonstrate that this "lossy" integrator is well behaved at dc, determine the zero-state - -response *y*(*t*) given a unit step input *x*(*t*) = *u*(*t*). - -- (c) Investigate the effect of 10% resistor and 25% capacitor tolerances on the system's characteristic root(s). -- **2.7-4** Consider the electric circuit shown in Fig. P2.7-4. Let *C*1 = *C*2 = 10 µF, *R*1 = *R*2 = 100 k, and *R*3 = 50 k. - - (a) Determine the corresponding differential equation describing this circuit. Is the circuit BIBO-stable? - - (b) Determine the zero-input response *y*0(*t*) if the output of each op amp initially reads one volt. - -- (c) Determine the zero-state response *y*(*t*) to a step input *x*(*t*) = *u*(*t*). -- (d) Investigate the effect of 10% resistor and 25% capacitor tolerances on the system's characteristic roots. -- **2.7-5** Input *x*(*t*)=3[*u*(*t*)−*u*(*t*−1)]+2[*u*(*t*−2)−*u*(*t*− 3)] is applied to a lowpass LTIC system with impulse response *h*(*t*) = (4 − *t*)[*u*(*t*) − *u*(*t* − 2)] to produce output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). Modify program CH2MP4.m in Sec. 2.7-4 to perform the graphical convolution procedure to produce a plot of *y*(*t*). - - - -# **TIME-DOMAIN ANALYSIS OF [DISCRETE-TIME](#page-9-0) SYSTEMS** - -In this chapter we introduce the basic concepts of discrete-time signals and systems. Furthermore, we explore the time-domain analysis of linear, time-invariant, discrete-time (LTID) systems. We show how to compute the zero-input response, determine the unit impulse response, and use convolution to evaluate the zero-state response. - -## **[3.1 INTRODUCTION](#page-9-0)** - -A *discrete-time signal* is basically a sequence of numbers. Such signals arise naturally in inherently discrete-time situations such as population studies, amortization problems, national income models, and radar tracking. They may also arise as a result of sampling continuous-time signals in sampled data systems and digital filtering. Such signals can be denoted by *x*[*n*], *y*[*n*], and so on, where the variable *n* takes integer values, and *x*[*n*] denotes the *n*th number in the sequence labeled *x*. In this notation, the discrete-time variable *n* is enclosed in square brackets instead of parentheses, which we have reserved for enclosing continuous-time variables, such as *t*. - -Systems whose inputs and outputs are discrete-time signals are called *discrete-time systems*. A digital computer is a familiar example of this type of system. A discrete-time signal is a sequence of numbers, and a discrete-time system processes a sequence of numbers *x*[*n*] to yield another sequence *y*[*n*] as the output.† - -A discrete-time signal, when obtained by uniform sampling of a continuous-time signal *x*(*t*), can also be expressed as *x*(*nT*), where *T* is the sampling interval and *n*, the discrete variable taking on integer values. Thus, *x*(*nT*) denotes the value of the signal *x*(*t*) at *t* = *nT*. The signal *x*(*nT*) is a sequence of numbers (sample values), and hence, by definition, is a discrete-time signal. Such a signal can also be denoted by the customary discrete-time notation *x*[*n*], where *x*[*n*] = *x*(*nT*). A typical discrete-time signal is depicted in Fig. 3.1, which shows both forms of notation. By way of an example, a continuous-time exponential *x*(*t*) = *e*−*t* , when sampled every *T* = 0.1 seconds, results in a discrete-time signal *x*(*nT*) given by - -$$ -x(nT) = e^{-nT} = e^{-0.1n} -$$ - -† There may be more than one input and more than one output. - -**Figure 3.2** Processing a continuous-time signal by means of a discrete-time system. - -Clearly, this signal is a function of *n* and may be expressed as *x*[*n*]. Such representation is more convenient and will be followed throughout this book, even for signals resulting from sampling continuous-time signals. - -Digital filters can process continuous-time signals by discrete-time systems, using appropriate interfaces at the input and the output, as illustrated in Fig. 3.2. A continuous-time signal *x*(*t*) is first sampled to convert it into a discrete-time signal *x*[*n*], which is then processed by a discrete-time system to yield the output *y*[*n*]. A continuous-time signal *y*(*t*) is finally constructed from *y*[*n*]. We shall use the notations C/D and D/C for conversion from continuous to discrete time and from discrete to continuous time. By using the interfaces in this manner, we can use an appropriate discrete-time system to process a continuous-time signal. As we shall see later in our discussion, discrete-time systems have several advantages over continuous-time systems. For this reason, there is an accelerating trend toward processing continuous-time signals with discrete-time systems. - -### **[3.1-1 Size of a Discrete-Time Signal](#page-9-0)** - -Arguing along the lines similar to those used for continuous-time signals, the size of a discrete-time signal *x*[*n*] will be measured by its energy *Ex*, defined by - -$$ -E_x = \sum_{n=-\infty}^{\infty} |x[n]|^2 -$$ -\n(3.1) - -This definition is valid for real or complex *x*[*n*]. For this measure to be meaningful, the energy of a signal must be finite. A necessary condition for the energy to be finite is that the signal amplitude must → 0 as |*n*|→∞. Otherwise the sum in Eq. (3.1) will not converge. If *Ex* is finite, the signal is called an *energy signal*. - -In some cases, for instance, when the amplitude of *x*[*n*] does not → 0 as |*n*|→∞, then the signal energy is infinite, and a more meaningful measure of the signal in such a case would be the time average of the energy (if it exists), which is the signal power *Px*, defined by - -$$ -P_{x} = \lim_{N \to \infty} \frac{1}{2N + 1} \sum_{-N}^{N} |x[n]|^{2} -$$ - -In this equation, the sum is divided by 2*N* + 1 because there are 2*N* + 1 samples in the interval from −*N* to *N*. For periodic signals, the time averaging need be performed over only one period in view of the periodic repetition of the signal. If *Px* is finite and nonzero, the signal is called a *power signal*. As in the continuous-time case, a discrete-time signal can either be an energy signal or a power signal, but cannot be both at the same time. Some signals are neither energy nor power signals. - -### **EXAMPLE 3.1 Computing DT Energy and Power** - -Find the energy of the signal *x*[*n*] = *n*(*u*[*n*] − *u*[*n* − 6]), shown in Fig. 3.3a and the power for the periodic signal *y*[*n*] in Fig. 3.3b. - -By definition, - -$$ -E_x = \sum_{n=0}^{5} n^2 = 55 -$$ - -A periodic signal *x*[*n*] with period *N*0 is characterized by the fact that - -$$ -x[n] = x[n+N_0] -$$ - -The smallest value of *N*0 for which the preceding equation holds is the *fundamental period*. Such a signal is called *N*0 *periodic*. Figure 3.3b shows an example of a periodic signal *y*[*n*] of period *N*0 = 6 because each period contains 6 samples. Note that if the first sample is taken at *n* = 0, the last sample is at *n* = *N*0 − 1 = 5, not at *n* = *N*0 = 6. Because the signal *y*[*n*] is periodic, its power *Py* can be found by averaging its energy over one period. Averaging the energy over one period, we obtain - -$$ -P_{y} = \frac{1}{6} \sum_{n=0}^{5} n^{2} = \frac{55}{6} -$$ - -**Figure 3.3 (a)** Energy and **(b)** power computations for a signal. - -### **DR ILL 3.1 DT Signal Classification: Energy, Power, and Neither** - -Show that the signal *x*[*n*] = *anu*[*n*] is an energy signal of energy *Ex* = 1/(1 − |*a*| 2) if |*a*| < 1, that it is a power signal of power *Px* = 0.5 if |*a*| = 1, and that it is neither an energy signal nor a power signal if |*a*| > 1. - -## **3.2 USEFUL SIGNAL [OPERATIONS](#page-9-0)** - -Signal operations for *shifting,* and *scaling,* as discussed for continuous-time signals also apply, with some modifications, to discrete-time signals. - -### SHIFTING - -Consider a signal *x*[*n*] (Fig. 3.4a) and the same signal delayed (right-shifted) by 5 units (Fig. 3.4b), which we shall denote by *xs*[*n*]. † Using the argument employed for a similar operation in continuous-time signals (Sec. 1.2), we obtain - -$$ -x_s[n] = x[n-5] -$$ - -Therefore, to shift a sequence by *M* units (*M* integer), we replace *n* with *n* − *M*. Thus *x*[*n* − *M*] represents *x*[*n*] shifted by *M* units. If *M* is positive, the shift is to the right (delay). If *M* is negative, the shift is to the left (advance). Accordingly, *x*[*n* − 5] is *x*[*n*] delayed (right-shifted) by 5 units, and *x*[*n*+5] is *x*[*n*] advanced (left-shifted) by 5 units. - -† The terms "delay" and "advance" are meaningful only when the independent variable is time. For other independent variables, such as frequency or distance, it is more appropriate to refer to the "right shift" and "left shift" of a sequence. - -**Figure 3.4** Shifting and time reversal of a signal. - -## **DR ILL 3.2 Left-Shift Operation** - -Show that *x*[*n*] in Fig. 3.4a left-shifted by 3 units can be expressed as 0.729(0.9)*n* for 0 ≤ *n* ≤ 7, and zero otherwise. Sketch the shifted signal. - -### **DR ILL 3.3 Right-Shift Operation** - -Show that *x*[−*k* − *n*] can be obtained from *x*[*n*] by first right-shifting *x*[*n*] by *k* units and then time-reversing this shifted signal. - -### TIME REVERSAL - -To time-reverse *x*[*n*] in Fig. 3.4a, we rotate *x*[*n*] about the vertical axis to obtain the time-reversed signal *xr*[*n*] shown in Fig. 3.4c. Using the argument employed for a similar operation in continuous-time signals (Sec. 1.2), we obtain - -$$ -x_r[n] = x[-n] -$$ - -Therefore, to time-reverse a signal, we replace *n* with −*n* so that *x*[−*n*] is the time-reversed *x*[*n*]. For example, if *x*[*n*] = (0.9)*n* for 3 ≤ *n* ≤ 10, then *xr*[*n*] = (0.9)−*n* for 3 ≤ −*n* ≤ 10; that is, −3 ≥ *n* ≥ −10, as shown in Fig. 3.4c. - -The origin *n* = 0 is the anchor point, which remains unchanged under time-reversal operation because at *n* = 0, *x*[*n*] = *x*[−*n*] = *x*[0]. Note that while the reversal of *x*[*n*] about the vertical axis is *x*[−*n*], the reversal of *x*[*n*] about the horizontal axis is −*x*[*n*]. - -### **EXAMPLE 3.2 Time Reversal and Shifting** - -In the convolution operation, discussed later, we need to find the function *x*[*k* −*n*] from *x*[*n*]. - -This can be done in two steps: (i) time-reverse the signal *x*[*n*] to obtain *x*[−*n*]; (ii) now, right-shift *x*[−*n*] by *k*. Recall that right-shifting is accomplished by replacing *n* with *n* − *k*. Hence, right-shifting *x*[−*n*] by *k* units is *x*[−(*n* − *k*)] = *x*[*k* − *n*]. Figure 3.4d shows *x*[5 − *n*], obtained this way. We first time-reverse *x*[*n*] to obtain *x*[−*n*] in Fig. 3.4c. Next, we shift *x*[−*n*] by *k* = 5 to obtain *x*[*k* −*n*] = *x*[5−*n*], as shown in Fig. 3.4d. - -In this particular example, the order of the two operations employed is interchangeable. We can first left-shift *x*[*k*] to obtain *x*[*n* + 5]. Next, we time-reverse *x*[*n* + 5] to obtain *x*[−*n* + 5] = *x*[5 − *n*]. The reader is encouraged to verify that this procedure yields the same result, as in Fig. 3.4d. - -### **DR ILL 3.4 Time Reversal** - -Sketch the signal *x*[*n*] = *e*−0.5*n* for −3 ≤ *n* ≤ 2, and zero otherwise. Sketch the corresponding time-reversed signal and show that it can be expressed as *xr*[*n*] = *e*0.5*n* for −2 ≤ *n* ≤ 3. - -### SAMPLING RATE ALTERATION: DOWNSAMPLING, UPSAMPLING, AND INTERPOLATION - -Alteration of the sampling rate is somewhat similar to time-scaling in continuous-time signals. Consider a signal *x*[*n*] compressed by factor *M*. Compressing a signal *x*[*n*] by factor *M* yields *xd*[*n*] given by - -$$ -x_d[n] = x[Mn] -$$ - -Because of the restriction that discrete-time signals are defined only for integer values of the argument, we must restrict *M* to integer values. The values of *x*[*Mn*] at *n* = 0, 1, 2, 3,... are *x*[0], *x*[*M*], *x*[2*M*], *x*[3*M*], ... . This means *x*[*Mn*] selects every *M*th sample of *x*[*n*] and deletes all the samples in between. It reduces the number of samples by factor *M*. If *x*[*n*] is obtained by sampling a continuous-time signal, this operation implies reducing the sampling rate by factor *M*. For this reason, this operation is commonly called *downsampling*. Figure 3.5a shows a signal *x*[*n*] and Fig. 3.5b shows the signal *x*[2*n*], which is obtained by deleting odd-numbered samples of *x*[*n*]. † - -In the continuous-time case, time compression merely speeds up the signal without loss of any data. In contrast, downsampling *x*[*n*] generally causes loss of data. Under certain conditions—for example, if *x*[*n*] is the result of oversampling some continuous-time signal—then *xd*[*n*] may still retain the complete information about *x*[*n*]. - -An *interpolated* signal is generated in two steps; first, we expand *x*[*n*] by an integer factor *L* to obtain the expanded signal *xe*[*n*], as - -$$ -x_e[n] = \begin{cases} x[n/L] & n = 0, \pm L \pm 2L, \dots, \\ 0 & \text{otherwise} \end{cases} \tag{3.2} -$$ - -To understand this expression, consider a simple case of expanding *x*[*n*] by a factor 2 (*L* = 2). When *n* is odd, *n*/2 is noninteger, and *xe*[*n*] = 0. That is, *xe*[1] = *xe*[3] = *xe*[5],... are all zero, as depicted in Fig. 3.5c. Moreover, *n*/2 is integer for even *n*, and the values of *xe*[*n*] = *x*[*n*/2] for *n* = 0, 2, 4, 6,..., are *x*[0], *x*[1], *x*[2], *x*[3], ... , as shown in Fig. 3.5c. In general, for *n* = 0, 1, 2,..., *xe*[*n*] is given by the sequence - -$$ -x[0], \underbrace{0,0,\ldots,0,0}_{L-1 \text{ zeros}}, x[1], \underbrace{0,0,\ldots,0,0}_{L-1 \text{ zeros}}, x[2], \underbrace{0,0,\ldots,0,0}_{L-1 \text{ zeros}},\ldots -$$ - -Thus, the sampling rate of *xe*[*n*] is *L* times that of *x*[*n*]. Hence, this operation is commonly called *upsampling*. The upsampled signal *xe*[*n*] contains all the data of *x*[*n*], although in an expanded form. - -In the expanded signal in Fig. 3.5c, the missing (zero-valued) odd-numbered samples can be reconstructed from the non-zero-valued samples by using some suitable interpolation formula. Figure 3.5d shows such an interpolated signal *xi*[*n*], where the missing samples are constructed by using an interpolating filter. The optimum interpolating filter is usually an ideal lowpass - -† Odd-numbered samples of *x*[*n*] can be retained (and even-numbered samples deleted) by using the transformation *xd*[*n*] = *x*[2*n*+1]. - -**Figure 3.5** Compression (downsampling) and expansion (upsampling, interpolation) of a signal. - -filter, which is realizable only approximately. In practice, we may use an interpolation that is nonoptimum but realizable. The process of filtering to interpolate the zero-valued samples is called *interpolation*. Since the interpolated data are computed from the existing data, interpolation does not result in gain of information. While further discussion of interpolation is beyond our scope, Drill 3.5 and Prob. 3.11-10 introduce the idea of linear interpolation. - -### **DR ILL 3.5 Expansion and Interpolation** - -A signal *x*[*n*] is expanded by factor 2 to obtain signal *x*[*n*/2]. The odd-numbered samples (*n* odd) in this signal have zero value. Show that the linearly interpolated odd-numbered samples are given by *xi*[*n*] = (1/2){*x*[*n*−1] +*x*[*n*+1]}. - -## **3.3 SOME USEFUL [DISCRETE-TIME](#page-9-0) SIGNAL MODELS** - -We now discuss some important discrete-time signal models that are encountered frequently in the study of discrete-time signals and systems. - -## **[3.3-1 Discrete-Time Impulse Function](#page-9-0)** *δ***[***n***]** - -The discrete-time counterpart of the continuous-time impulse function δ(*t*) is δ[*n*], a Kronecker delta function, defined by - -$$ -\delta[n] = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases} -$$ - -This function, also called the unit impulse sequence, is shown in Fig. 3.6a. The shifted impulse sequence δ[*n* − *m*] is depicted in Fig. 3.6b. Unlike its continuous-time counterpart δ(*t*) (the Dirac delta), the Kronecker delta is a very simple function, requiring no special esoteric knowledge of distribution theory. - -**Figure 3.6** Discrete-time impulse function: **(a)** unit impulse sequence and **(b)** shifted impulse sequence. - -## **[3.3-2 Discrete-Time Unit Step Function](#page-9-0)** *u***[***n***]** - -The discrete-time counterpart of the unit step function *u*(*t*) is *u*[*n*] (Fig. 3.7a), defined by - -$$ -u[n] = \begin{cases} 1 & \text{for } n \ge 0 \\ 0 & \text{for } n < 0 \end{cases} -$$ - -If we want a signal to start at *n* = 0 (so that it has a zero value for all *n* < 0), we need only multiply the signal by *u*[*n*]. - -**Figure 3.7 (a)** A discrete-time unit step function *u*[*n*] and **(b)** its application. - -### **EXAMPLE 3.3 Describing Signals with Unit Step and Unit Impulse Functions** - -Describe the signal *x*[*n*] shown in Fig. 3.7b by a single expression valid for all *n*. - -The signal *x*[*n*] can be broken into three components: (1) a ramp component *x*1[*n*] from *n* = 0 to 4, (2) a scaled step component *x*2[*n*] from *n* = 5 to 10, and (3) an impulse component *x*3[*n*] represented by the negative spike at *n* = 8. Let us consider each one separately. - -We express *x*1[*n*] = *n*(*u*[*n*]−*u*[*n*−5]) to account for the signal from *n* = 0 to 4. Assuming that the spike at *n* = 8 does not exist, we can express *x*2[*n*] = 4(*u*[*n*−5] −*u*[*n*−11]) to account for the signal from *n* = 5 to 10. Once these two components have been added, the only part that is unaccounted for is a spike of amplitude −2 at *n* = 8, which can be represented by - -There are many different ways of viewing *x*[*n*]. Although each way of viewing yields a different expression, they are all equivalent. We shall consider here just one possible expression. - -*x*3[*n*]=−2δ[*n*−8]. Hence, - -$$ -x[n] = x_1[n] + x_2[n] + x_3[n] -$$ - -= $n(u[n] - u[n-5]) + 4(u[n-5] - u[n-11]) - 2\delta[n-8]$ for all *n* - -We stress again that the expression is valid for all values of *n*. The reader can find several other equivalent expressions for *x*[*n*]. For example, one may consider a scaled step function from *n* = 0 to 10, subtract a ramp over the range *n* = 0 to 3, and subtract the spike. You can also play with breaking *n* into different ranges for your expression. - -## **[3.3-3 Discrete-Time Exponential](#page-9-0)** *γ n* - -A continuous-time exponential *e*λ*t* can be expressed in an alternate form as - -$$ -e^{\lambda t} = \gamma^t \qquad (\gamma = e^{\lambda} \text{ or } \lambda = \ln \gamma) -$$ - -For example, *e*−0.3*t* = (0.7408)*t* because *e*−0.3 = 0.7408. Conversely, 4*t* = *e*1.386*t* because *e*1.386 = 4, that is, ln 4 = 1.386. In the study of continuous-time signals and systems, we prefer the form *e*λ*t* rather than γ *t* . In contrast, the exponential form γ *n* is preferable in the study of discrete-time signals and systems, as will become apparent later. The discrete-time exponential γ *n* can also be expressed by using a natural base, as - -$$ -e^{\lambda n} = \gamma^n \qquad (\gamma = e^{\lambda} \text{ or } \lambda = \ln \gamma) -$$ - -Because of unfamiliarity with exponentials with bases other than *e*, exponentials of the form γ *n* may seem inconvenient and confusing at first. The reader is urged to plot some exponentials to acquire a sense of these functions. Also observe that γ −*n* = 1 γ *n* . - -### **DR ILL 3.6 Equivalent Forms of DT Exponentials** - -**(a)** Show that (i) (0.25)−*n* = 4*n*, (ii) 4−*n* = (0.25)*n*, (iii) *e*2*t* = (7.389)*t* , (iv) *e*−2*t* = (0.1353)*t* = (7.389)−*t* , (v) *e*3*n* = (20.086)*n*, and (vi) *e*−1.5*n* = (0.2231)*n* = (4.4817)−*n*. **(b)** Show that (i) 2*n* = *e*0.693*n*, (ii) (0.5)*n* = *e*−0.693*n*, and (iii) (0.8)−*n* = *e*0.2231*n*. - -**Nature of** *γ n***.** The signal *e*λ*n* grows exponentially with *n* if Reλ > 0 (λ in the RHP), and decays exponentially if Reλ < 0 (λ in the LHP). It is constant or oscillates with constant amplitude if Reλ = 0 (λ on the imaginary axis). Clearly, the location of λ in the complex plane indicates whether the signal *e*λ*n* will grow exponentially, decay exponentially, or oscillate with constant - -**Figure 3.8** The λ plane, the γ plane, and their mapping. - -amplitude (Fig. 3.8a). A constant signal (λ = 0) is also an oscillation with zero frequency. We now find a similar criterion for determining the nature of γ *n* from the location of γ in the complex plane. - -Figure 3.8a shows a complex plane (λ plane). Consider a signal *ejn*. In this case, λ = *j* lies on the imaginary axis (Fig. 3.8a), and therefore is a constant-amplitude oscillating signal. This signal *ejn* can be expressed as γ *n*, where γ = *ej*. Because the magnitude of *ej* is unity, |γ | = 1. Hence, when λ lies on the imaginary axis, the corresponding γ lies on a circle of unit radius, centered at the origin (the *unit circle* illustrated in Fig. 3.8b). Therefore, a signal γ *n* oscillates with constant amplitude if γ lies on the unit circle. Thus, the imaginary axis in the λ plane maps into the unit circle in the γ plane. - -Next consider the signal *e*λ*n*, where λ lies in the left half-plane in Fig. 3.8a. This means λ = *a* + *jb*, where *a* is negative (*a* < 0). In this case, the signal decays exponentially. This signal can be expressed as γ *n*, where - -$$ -\gamma = e^{\lambda} = e^{a+jb} = e^a e^{jb} -$$ - -and - -$$ -|\gamma| = |e^a| \, |e^{ib}| = e^a -$$ - because $|e^{ib}| = 1$ - -Also, *a* is negative (*a* < 0). Hence, |γ | = *ea* < 1. This result means that the corresponding γ lies inside the unit circle. Therefore, a signal γ *n* decays exponentially if γ lies within the unit circle (Fig. 3.8b). If, in the preceding case we select *a* to be positive (λ in the right half-plane), then |γ | > 1, and γ lies outside the unit circle. Therefore, a signal γ *n* grows exponentially if γ lies outside the unit circle (Fig. 3.8b). - -To summarize, the imaginary axis in the λ plane maps into the unit circle in the γ plane. The left half-plane in the λ plane maps into the inside of the unit circle and the right half of the λ plane maps into the outside of the unit circle in the γ plane, as depicted in Fig. 3.8. - -**Figure 3.9** Discrete-time exponentials γ *n*. - -Plots of (0.8)*n* and (−0.8)*n* appear in Figs. 3.9a and 3.9b, respectively. Plots of (0.5)*n* and (1.1)*n* appear in Figs. 3.9c and 3.9d, respectively. These plots verify our earlier conclusions about the location of γ and the nature of signal growth. Observe that a signal (−|γ |)*n* alternates sign successively (is positive for even values of *n* and negative for odd values of *n*, as depicted in Fig. 3.9b). Also, the exponential (0.5)*n* decays faster than (0.8)*n* because 0.5 is closer to the origin than 0.8. The exponential (0.5)*n* can also be expressed as 2−*n* because (0.5)−1 = 2. - -### **DR ILL 3.7 Sketching DT Exponentials** - -Sketch the following signals: **(a)** (1)*n*, **(b)** (−1)*n*, **(c)** (0.5)*n*, **(d)** (−0.5)*n*, **(e)** (0.5)−*n*, **(f)** 2−*n*, and **(g)** (−2)*n*. Express these exponentials as γ *n*, and plot γ in the complex plane for each case. Verify that γ *n* decays exponentially with *n* if γ lies inside the unit circle and that γ *n* grows with *n* if γ is outside the unit circle. If γ is on the unit circle, γ *n* is constant or oscillates with a constant amplitude. - -### 250 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Accurately hand-sketching DT signals can be tedious and difficult. As the next example shows, MATLAB is particularly well suited to plot DT signals, including exponentials. - -### **EXAMPLE 3.4 Plotting DT Exponentials with MATLAB** - -Use MATLAB to plot the following discrete-time signals over (0 ≤ *n* ≤ 8): **(a)** *xa*[*n*] = (0.8)*n*, **(b)** *xb*[*n*] = (−0.8)*n*, **(c)** *xc*[*n*] = (0.5)*n*, and **(d)** *xd*[*n*] = (1.1)*n*. - -To begin, we use anonymous functions to represent each of the four signals. Next, we plot these functions over the desired range of *n*. The results, shown in Fig. 3.10, match the earlier Fig. 3.9 plots of the same signals. - -``` ->> n = (0:8); x_a = @(n) (0.8).^n; x_b = @(n) (-0.8).^(n); ->> x_c = @(n) (0.5).^n; x_d = @(n) (1.1).^n; ->> subplot(2,2,1); stem(n,x_a(n),'k'); ylabel('x_a[n]'); xlabel('n'); ->> subplot(2,2,2); stem(n,x_b(n),'k'); ylabel('x_b[n]'); xlabel('n'); ->> subplot(2,2,3); stem(n,x_c(n),'k'); ylabel('x_c[n]'); xlabel('n'); ->> subplot(2,2,4); stem(n,x_d(n),'k'); ylabel('x_d[n]'); xlabel('n'); -``` - -## **[3.3-4 Discrete-Time Sinusoid](#page-9-0) cos***(n* **+***θ )* - -A general discrete-time sinusoid can be expressed as *C*cos(*n*+θ ), where *C* is the *amplitude,* and θ is the *phase* in radians. Also, *n* is an angle in radians. Hence, the dimensions of the frequency are *radians per sample*. This sinusoid may also be expressed as - -$$ -C\cos\left(\Omega n + \theta\right) = C\cos\left(2\pi\mathcal{F}n + \theta\right) -$$ - -where *F* = /2π. Therefore, the dimensions of the discrete-time frequency *F* are (radians/2π) per sample, which is equal to *cycles per sample*. This means if *N*0 is the period (samples/cycle) of the sinusoid, then the frequency of the sinusoid *F* = 1/*N*0 (samples/cycle). - -Figure 3.11 shows a discrete-time sinusoid cos( π 12 *n* + π 4 ). For this case, the frequency is = π/12 radians/sample. Alternately, the frequency is *F* = 1/24 cycles/sample. In other words, there are 24 samples in one cycle of the sinusoid. - -Because cos(−*x*) = cos(*x*), - -$$ -\cos\left(-\Omega n + \theta\right) = \cos\left(\Omega n - \theta\right) -$$ - -This shows that both cos(*n*+θ ) and cos(−*n*+θ ) have the same frequency (). Therefore, *the frequency of* cos(*n*+θ ) *is* ||. - -**Figure 3.11** A discrete-time sinusoid cos( π 12 *n*+ π 4 ). - -### SAMPLED CONTINUOUS-TIME SINUSOID YIELDS A DISCRETE-TIME SINUSOID - -A continuous-time sinusoid cosω*t* sampled every *T* seconds yields a discrete-time sequence whose *n*th element (at *t* = *nT*) is cosω*nT*. Thus, the sampled signal *x*[*n*] is given by - -$$ -x[n] = \cos \omega nT = \cos \Omega n \quad \text{where } \Omega = \omega T -$$ - -### 252 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Thus, a continuous-time sinusoid cosω*t* sampled every *T* seconds yields a discrete-time sinusoid cos*n*, where = ω*T*. † - -## **[3.3-5 Discrete-Time Complex Exponential](#page-9-0)** *ejn* - -Using Euler's formula, we can express an exponential *ejn* in terms of sinusoids as - -*ejn* = (cos*n*+*j*sin*n*) and *e*−*jn* = (cos*n*−*j*sin*n*) - -These equations show that *the frequency of both ejn and e*−*jn is* (radians/sample). Therefore, the frequency of *ejn* is ||. - -Observe that for *r* = 1 and θ = *n*, - -$$ -e^{j\Omega n}=re^{j\theta} -$$ - -This equation shows that the magnitude and angle of *ejn* are 1 and *n*, respectively. In the complex plane, *ejn* is a point on a unit circle at an angle *n*. - -### **EXAMPLE 3.5 Plotting a DT Sinusoid with MATLAB** - -Using MATLAB, plot the discrete-time sinusoid *x*[*n*] = cos π 12 *n*+ π 4 . - -We represent the desired sinusoid using an anonymous function. Next, we plot this function over the desired range of *n*. The result, shown in Fig. 3.12, matches the plot of the same signal shown in Fig. 3.11. - ->> n = (-30:30); x = @(n) cos(n\*pi/12+pi/4); >> clf; stem(n,x(n),'k'); ylabel('x[n]'); xlabel('n'); - -† Superficially, it may appear that a discrete-time sinusoid is a continuous-time sinusoid's cousin in a striped suit. However, some of the properties of discrete-time sinusoids are very different from those of continuous-time sinusoids. For instance, not every discrete-time sinusoid is periodic. A sinusoid cos*n* is periodic only if is a rational multiple of 2π. Also, discrete-time sinusoids are bandlimited to = π. Any sinusoid with ≥ π can always be expressed as a sinusoid of some frequency ≤ π. These peculiar properties are the direct consequence of the fact that the period of a discrete-time sinusoid must be an integer. These topics are discussed in Chs. 5 and 9. - - - -## **[3.4 EXAMPLES OF](#page-9-0) DISCRETE-TIME SYSTEMS** - -We shall give here four examples of discrete-time systems. In the first two examples, the signals are inherently of the discrete-time variety. In the third and fourth examples, a continuous-time signal is processed by a discrete-time system, as illustrated in Fig. 3.2, by discretizing the signal through sampling. - -### **EXAMPLE 3.6 Savings Account** - -A person makes a deposit (the input) in a bank regularly at an interval of *T* (say, 1 month). The bank pays a certain interest on the account balance during the period *T* and mails out a periodic statement of the account balance (the output) to the depositor. Find the equation relating the output *y*[*n*] (the balance) to the input *x*[*n*] (the deposit). - -In this case, the signals are inherently discrete time. Let - -*x*[*n*] = deposit made at the *n*th discrete instant - -*y*[*n*] = account balance at the *n*th instant computed - -immediately after receipt of the *n*th deposit *x*[*n*] - -*r* = interest per dollar per period *T* - -The balance *y*[*n*] is the sum of (i) the previous balance *y*[*n* − 1], (ii) the interest on *y*[*n* − 1] during the period *T*, and (iii) the deposit *x*[*n*] - -$$ -y[n] = y[n-1] + ry[n-1] + x[n] -$$ - -= (1+r)y[n-1] + x[n] - -or - -$$ -y[n] - ay[n-1] = x[n] \qquad a = 1+r -$$ -\n(3.3) - -In this example the deposit *x*[*n*] is the input (cause) and the balance *y*[*n*] is the output (effect). - -### 254 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -A withdrawal from the account is a negative deposit. Therefore, this formulation can handle deposits as well as withdrawals. It also applies to a loan payment problem with the initial value *y*[0]=−*M*, where *M* is the amount of the loan. A loan is an initial deposit with a negative value. Alternately, we may treat a loan of *M* dollars taken at *n* = 0 as an input of −*M* at *n* = 0 (see Prob. 3.8-23). - -We can express Eq. (3.3) in an alternate form. The choice of index *n* in Eq. (3.3) is completely arbitrary, so we can substitute *n*+1 for *n* to obtain - -$$ -y[n+1] - ay[n] = x[n+1] -$$ -\n(3.4) - -We also could have obtained Eq. (3.4) directly by realizing that *y*[*n*+1], the balance at instant (*n* + 1), is the sum of *y*[*n*] plus *ry*[*n*] (the interest on *y*[*n*]) plus the deposit (input) *x*[*n* + 1] at instant (*n*+1). - -The difference equation in Eq. (3.3) uses delays, whereas the form in Eq. (3.4) uses advances. Thus, Eq. (3.3) is said to be in *delay form* and Eq. (3.4) is said to be in *advance form*. The delay form is more natural because operation of delay is causal, hence realizable. In contrast, advance operation, being noncausal, is unrealizable. We use the advance form primarily for its mathematical convenience over the delay form.† - -We shall now represent this system in a block diagram form, which is basically a road map to a hardware (or software) realization of the system. For this purpose, the causal (realizable) delay form in Eq. (3.3) will be used. There are three basic operations in this equation: *addition, scalar multiplication,* and *delay*. Figure 3.13 shows their schematic representation. In addition, we also have a *pickoff* node (Fig. 3.13d), which is used to provide multiple copies of a signal at its input. - -**Figure 3.13** Schematic representations of basic operations on sequences. - -† Use of the advance form results in discrete-time system equations that are identical in form to those for continuous-time systems. This will become apparent later. In transform analysis, advance form leads to the more convenient variable *z* instead of the clumsy *z*−1 that arises from delay form. - -**Figure 3.14** Realization of the savings account system. - -Figure 3.14 shows in block diagram form a system represented by Eq. (3.3). To understand this realization, it is helpful to rewrite Eq. (3.3) as *y*[*n*] = *ay*[*n* − 1] + *x*[*n*] (*a* = 1 + *r*). Now, assume that the output *y*[*n*] is available at the pickoff node *N*. Unit delay of *y*[*n*] results in *y*[*n* − 1], which is multiplied by a scalar of value *a* to yield *ay*[*n* − 1]. Next, we generate *y*[*n*] by adding the input *x*[*n*] and *ay*[*n* − 1]. † Observe that node *N* is a pickoff node, from which two copies of the output signal flow out: one as the feedback signal and the other as the output signal. - -### **EXAMPLE 3.7 Sales Estimate** - -During semester *n*, *x*[*n*] students enroll in a course requiring a certain textbook while the publisher sells *y*[*n*] new copies of the same book. On the average, one-quarter of students with books in salable condition resell the texts at the end of the semester, and the book life is three semesters. Write the equation relating *y*[*n*], the new books sold by the publisher, to *x*[*n*], the number of students enrolled in the *n*th semester, assuming that every student buys a book. - -In the *n*th semester, the total books *x*[*n*] sold to students must be equal to *y*[*n*] (new books from the publisher) plus the used books from students enrolled in the preceding two semesters (because the book life is only three semesters). There are *y*[*n* − 1] new books sold in semester (*n* − 1), and one-quarter of these books, that is, (1/4)*y*[*n* − 1], will be resold in the *n*th semester. Also, *y*[*n* − 2] new books are sold in semester *n* − 2, and one-quarter of these, that is, (1/4)*y*[*n* − 2], will be resold in semester (*n* − 1). Again, a quarter of these, that is, (1/16)*y*[*n* − 2], will be resold in the *n*th semester. Therefore, *x*[*n*] must be equal to the sum of *y*[*n*], (1/4)*y*[*n*−1], and (1/16)*y*[*n*−2]. - -$$ -y[n] + \frac{1}{4}y[n-1] + \frac{1}{16}y[n-2] = x[n] -$$ -\n(3.5) - -Equation (3.5) can also be expressed in an alternative form by realizing that this equation is valid for any value of *n*. Therefore, replacing *n* by *n*+2, we obtain - -$$ -y[n+2] + \frac{1}{4}y[n+1] + \frac{1}{16}y[n] = x[n+2] -$$ -\n(3.6) - -This is the alternative form of Eq. (3.5). - -† A unit delay represents 1 unit of time delay. In this example, 1 unit of delay in the output corresponds to period *T* for the actual output. - -### 256 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -To facilitate a realization of a system with this input–output equation, we rewrite the delay-form Eq. (3.5) as *y*[*n*]=−1 4 *y*[*n* − 1] − 1 16 *y*[*n* − 2] + *x*[*n*]. Figure 3.15 shows a corresponding hardware realization using two unit delays in cascade.† - -**Figure 3.15** Realization of the system representing sales estimate in Ex. 3.7. - -### **EXAMPLE 3.8 Digital Differentiator** - -Design a discrete-time system, like the one in Fig. 3.2, to differentiate continuous-time signals. This differentiator is used in an audio system having an input signal bandwidth below 20 kHz. - -In this case, the output *y*(*t*) is required to be the derivative of the input *x*(*t*). The discrete-time processor (system) *G* processes the samples of *x*(*t*) to produce the discrete-time output *y*[*n*]. Let *x*[*n*] and *y*[*n*] represent the samples *T* seconds apart of the signals *x*(*t*) and *y*(*t*), respectively, that is, - -$$ -x[n] = x(nT) \qquad \text{and} \qquad y[n] = y(nT) \tag{3.7} -$$ - -The signals *x*[*n*] and *y*[*n*] are the input and the output for the discrete-time system *G*. Now, we require that - -$$ -y(t) = \frac{dx(t)}{dt} -$$ - -Therefore, at *t* = *nT* (see Fig. 3.16a), - -$$ -y(nT) = \frac{dx(t)}{dt}\bigg|_{t=nT} = \lim_{T \to 0} \frac{1}{T} [x(nT) - x[(n - 1)T]] -$$ - -† The comments in the preceding footnote apply here also. Although 1 unit of delay in this example is one semester, we need not use this value in the hardware realization. Any value other than one semester results in a time-scaled output. - -**Figure 3.16** Digital differentiator and its realization. - -By using the notation in Eq. (3.7), the foregoing equation can be expressed as - -$$ -y[n] = \lim_{T \to 0} \frac{1}{T} \{x[n] - x[n-1]\} -$$ - -This is the input–output relationship for *G* required to achieve our objective. In practice, the sampling interval *T* cannot be zero. Assuming *T* to be sufficiently small, the equation just given can be expressed as - -$$ -y[n] = \frac{1}{T} \{x[n] - x[n-1]\} -$$ -\n(3.8) - -The approximation improves as *T* approaches 0. A discrete-time processor *G* to realize Eq. (3.8) is shown inside the shaded box in Fig. 3.16b. The system in Fig. 3.16b acts as a differentiator. This example shows how a continuous-time signal can be processed by a - -### 258 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -discrete-time system. The considerations for determining the sampling interval *T* are discussed in Chs. 5 and 8, where it is shown that to process frequencies below 20 kHz, the proper choice is - -$$ -T \le \frac{1}{2 \times \text{highest frequency}} = \frac{1}{40,000} = 25 \,\mu\text{s} -$$ - -To see how well this method works, let us consider the differentiator in Fig. 3.16b with a ramp input *x*(*t*) = *t*, depicted in Fig. 3.16c. If the system were to act as a differentiator, then the output *y*(*t*) of the system should be the unit step function *u*(*t*). Let us investigate how the system performs this particular operation and how well the system achieves the objective. - -The samples of the input *x*(*t*) = *t* at the interval of *T* seconds act as the input to the discrete-time system *G*. These samples, denoted by a compact notation *x*[*n*], are, therefore, - -$$ -x[n] = x(t)|_{t=nT} = t|_{t=nT} \qquad t \ge 0 -$$ - -= nT \qquad n \ge 0 - -Figure 3.16d shows the sampled signal *x*[*n*]. This signal acts as an input to the discrete-time system *G*. Figure 3.16b shows that the operation of *G* consists of subtracting a sample from the preceding (delayed) sample and then multiplying the difference with 1/*T*. From Fig. 3.16d, it is clear that the difference between the successive samples is a constant *nT* −(*n*−1)*T* = *T* for all samples, except for the sample at *n* = 0 (because there is no preceding sample at *n* = 0). The output of *G* is 1/*T* times the difference *T*, which is unity for all values of *n*, except *n* = 0, where it is zero. Therefore, the output *y*[*n*] of *G* consists of samples of unit values for *n* ≥ 1, as illustrated in Fig. 3.16e. The D/C (discrete-time to continuous-time) converter converts these samples into a continuous-time signal *y*(*t*), as shown in Fig. 3.16f. Ideally, the output should have been *y*(*t*) = *u*(*t*). This deviation from the ideal is due to our use of a nonzero sampling interval *T*. As *T* approaches zero, the output *y*(*t*) approaches the desired output *u*(*t*). - -The digital differentiator in Eq. (3.8) is an example of what is known as the *backward difference* system. The reason for calling it so is obvious from Fig. 3.16a. To compute the derivative of *y*(*t*), we are using the difference between the present sample value and the preceding (backward) sample value. If we use the difference between the next (forward) sample at *t* = (*n* + 1)*T* and the present sample at *t* = *nT*, we obtain a forward difference form of differentiator as - -$$ -y[n] = \frac{1}{T} \{x[n+1] - x[n]\} -$$ -\n(3.9) - -### **EXAMPLE 3.9 Digital Integrator** - -Design a digital integrator along the same lines as the digital differentiator in Ex. 3.8. - -For an integrator, the input *x*(*t*) and the output *y*(*t*) are related by - -$$ -y(t) = \int_{-\infty}^{t} x(\tau) d\tau -$$ - -### 3.4 Examples of Discrete-Time Systems 259 - -Therefore, at *t* = *nT* (see Fig. 3.16a), - -$$ -y(nT) = \lim_{T \to 0} \sum_{k=-\infty}^{n} x(kT)T -$$ - -Using the usual notation *x*(*kT*) = *x*[*k*], *y*(*nT*) = *y*[*n*], and so on, this equation can be expressed as - -$$ -y[n] = \lim_{T \to 0} T \sum_{k=-\infty}^{n} x[k] -$$ - -Assuming that *T* is small enough to justify the assumption *T* → 0, we have - -$$ -y[n] = T \sum_{k=-\infty}^{n} x[k] -$$ - (3.10) - -This equation represents an example of *accumulator* system. This digital integrator equation can be expressed in an alternate form. From Eq. (3.10), it follows that - -$$ -y[n] - y[n-1] = Tx[n] -$$ -\n(3.11) - -This is an alternate description for the digital integrator. Equations (3.10) and (3.11) are equivalent; the one can be derived from the other. Observe that the form of Eq. (3.11) is similar to that of Eq. (3.3). Hence, the block diagram representation of a digital integrator in the form of Eq. (3.11) is identical to that in Fig. 3.14 with *a* = 1 and the input multiplied by *T*. - -## RECURSIVE AND NONRECURSIVE FORMS OF DIFFERENCE EQUATION - -If Eq. (3.11) expresses Eq. (3.10) in another form, what is the difference between these two forms? Which form is preferable? To answer these questions, let us examine how the output is computed by each of these forms. In Eq. (3.10), the output *y*[*n*] at any instant *n* is computed by adding all the past input values till *n*. This can mean a large number of additions. In contrast, Eq. (3.11) can be expressed as *y*[*n*] = *y*[*n*−1] +*Tx*[*n*]. Hence, computation of *y*[*n*] involves addition of only two values: the preceding output value *y*[*n* − 1] and the present input value *x*[*n*]. The computations are done recursively by using the preceding output values. For example, if the input starts at *n* = 0, we first compute *y*[0]. Then we use the computed value *y*[0] to compute *y*[1]. Knowing *y*[1], we compute *y*[2], and so on. The computations are recursive. This is why the form of Eq. (3.11) is called *recursive* form and the form of Eq. (3.10) is called *nonrecursive* form. Clearly, "recursive" and "nonrecursive" describe two different ways of presenting the same information. Equations (3.3), (3.5), and (3.11) are examples of recursive form, and Eqs. (3.8) and (3.10) are examples of nonrecursive form. - -### KINSHIP OF DIFFERENCE EQUATIONS TO DIFFERENTIAL EQUATIONS - -We now show that a digitized version of a differential equation results in a difference equation. Let us consider a simple first-order differential equation - -$$ -\frac{dy(t)}{dt} + cy(t) = x(t) -$$ -\n(3.12) - -Consider uniform samples of *x*(*t*) at intervals of *T* seconds. As usual, we use the notation *x*[*n*] to denote *x*(*nT*), the *n*th sample of *x*(*t*). Similarly, *y*[*n*] denotes *y*[*nT*], the *n*th sample of *y*(*t*). From the basic definition of a derivative, we can express Eq. (3.12) at *t* = *nT* as - -$$ -\lim_{T \to 0} \frac{y[n] - y[n-1]}{T} + cy[n] = x[n] -$$ - -Clearing the fractions and rearranging the terms yield (assuming nonzero, but very small *T*) - -$$ -y[n] + \alpha y[n-1] = \beta x[n] -$$ -\n(3.13) - -where - -$$ -\alpha = \frac{-1}{1 + cT} \quad \text{and} \quad \beta = \frac{T}{1 + cT} -$$ - -We can also express Eq. (3.13) in advance form as - -$$ -y[n+1] + \alpha y[n] = \beta x[n+1] -$$ - -It is clear that a differential equation can be approximated by a difference equation of the same order. In this way, we can approximate an *n*th-order differential equation by a difference equation of *n*th order. Indeed, a digital computer solves differential equations by using an equivalent difference equation, which can be solved by means of simple operations of addition, multiplication, and shifting. Recall that a computer can perform only these simple operations. It must necessarily approximate complex operation like differentiation and integration in terms of such simple operations. The approximation can be made as close to the exact answer as possible by choosing sufficiently small value for *T*. - -At this stage, we have not developed tools required to choose a suitable value of the sampling interval *T*. This subject is discussed in Ch. 5 and also in Ch. 8. In Sec. 5.7, we shall discuss a systematic procedure (impulse invariance method) for finding a discrete-time system with which to realize an *N*th-order LTIC system. - -### ORDER OF A DIFFERENCE EQUATION - -Equations (3.3), (3.5), (3.9), (3.11), and (3.13) are examples of difference equations. The highest-order difference of the output signal or the input signal, whichever is higher, represents the *order* of the difference equation. Hence, Eqs. (3.3), (3.9), (3.11), and (3.13) are first-order difference equations, whereas Eq. (3.5) is of the second order. - -### **DR ILL 3.8 Digital Integrator Design** - -Design a digital integrator in Ex. 3.9 using the fact that for an integrator, the output *y*(*t*) and the input *x*(*t*) are related by *dy*(*t*)/*dt* = *x*(*t*). Approximation (similar to that in Ex. 3.8) of this equation at *t* = *nT* yields the recursive form in Eq. (3.11). - -### ANALOG, DIGITAL, CONTINUOUS-TIME, AND DISCRETE-TIME SYSTEMS - -The basic difference between continuous-time systems and analog systems, as also between discrete-time and digital systems, is fully explained in Secs. 1.7-5 and 1.7-6.† Historically, discrete-time systems have been realized with digital computers, where continuous-time signals are processed through digitized samples rather than unquantized samples. Therefore, the terms *digital filters* and *discrete-time systems* are used synonymously in the literature. This distinction is irrelevant in the analysis of discrete-time systems. For this reason, we follow this loose convention in this book, where the term *digital filter* implies a *discrete-time system,* and *analog filter* means *continuous-time system*. Moreover, the terms C/D (continuous-to-discrete-time ) and D/C will occasionally be used interchangeably with terms A/D (analog-to-digital) and D/A, respectively. - -### ADVANTAGES OF DIGITAL SIGNAL PROCESSING - -- 1. Digital systems operation can tolerate considerable variation in signal values, and hence are less sensitive to changes in the component parameter values due to temperature variation, aging, and other factors. This results in greater degree of precision and stability. Since digital systems are binary circuits, their accuracy can be increased by using more complex circuitry to increase word length, subject to cost limitations. -- 2. Digital systems do not require any factory adjustment and can be easily duplicated in volume without having to worry about precise component values. They can be fully integrated, and even highly complex systems can be placed on a single chip by using *VLSI* (very-large-scale integrated) circuits. -- 3. Digital filters are more flexible. Their characteristics can be easily altered simply by changing the program. Digital hardware implementation permits the use of microprocessors, miniprocessors, digital switching, and large-scale integrated circuits. -- 4. A greater variety of filters can be realized by digital systems. -- 5. Digital signals can be stored easily and inexpensively on various media (e.g., magnetic, optical, and solid state) without deterioration of signal quality. It is also possible (and increasingly popular) to search and select information from distant electronic storehouses, such as the cloud. -- 6. Digital signals can be coded to yield extremely low error rates and high fidelity, as well as privacy. Also, more sophisticated signal-processing algorithms can be used to process digital signals. - -† The terms *discrete-time* and *continuous-time* qualify the nature of a signal along the time axis (horizontal axis). The terms *analog* and *digital,* in contrast, qualify the nature of the signal amplitude (vertical axis). - -### 262 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -- 7. Digital filters can be easily time-shared and therefore can serve a number of inputs simultaneously. Moreover, it is easier and more efficient to multiplex several digital signals on the same channel. -- 8. Reproduction with digital messages is extremely reliable without deterioration. Analog messages such as photocopies and films, for example, lose quality at each successive stage of reproduction and have to be transported physically from one distant place to another, often at relatively high cost. - -One must weigh these advantages against such disadvantages as increased system complexity due to use of A/D and D/A interfaces, limited range of frequencies available in practice (affordable rates are gigahertz or less), and use of more power than is needed for the passive analog circuits. Digital systems use power-consuming active devices. - -### **[3.4-1 Classification of Discrete-Time Systems](#page-9-0)** - -Before examining the nature of discrete-time system equations, let us consider the concepts of linearity, time invariance (or shift invariance), and causality, which apply to discrete-time systems also. - -### LINEARITY AND TIME INVARIANCE - -For discrete-time systems, the definition of *linearity* is identical to that for continuous-time systems, as given in Eq. (1.22). We can show that the systems in Exs. 3.6, 3.7, 3.8, and 3.9 are all linear. - -Time invariance (or *shift invariance*) for discrete-time systems is also defined in a way similar to that for continuous-time systems. Systems whose parameters do not change with time (with *n*) are *time-invariant* or shift-invariant (also *constant-parameter*) systems. For such a system, if the input is delayed by *k* units or samples, the output is the same as before but delayed by *k* samples (assuming the initial conditions also are delayed by *k*). The systems in Exs. 3.6, 3.7, 3.8, and 3.9 are time-invariant because the coefficients in the system equations are constants (independent of *n*). If these coefficients were functions of *n* (time), then the systems would be linear *time-varying* systems. Consider, for example, a system described by - -$$ -y[n] = e^{-n}x[n] -$$ - -For this system, let a signal *x*1[*n*] yield the output *y*1[*n*], and another input *x*2[*n*] yield the output *y*2[*n*]. Then - -$$ -y_1[n] = e^{-n}x_1[n] -$$ - and $y_2[n] = e^{-n}x_2[n]$ - -If we let *x*2[*n*] = *x*1[*n*−*N*0], then - -$$ -y_2[n] = e^{-n}x_2[n] = e^{-n}x_1[n - N_0] \neq y_1[n - N_0] -$$ - -Clearly, this is a time-varying parameter system. - -### CAUSAL AND NONCAUSAL SYSTEMS - -A *causal* (also known as a *physical* or *nonanticipative*) system is one for which the output at any instant *n* = *k* depends only on the value of the input *x*[*n*] for *n* ≤ *k*. In other words, the value of the output at the present instant depends only on the past and present values of the input *x*[*n*], not on its future values. As we shall see, the systems in Exs. 3.6, 3.7, 3.8, and 3.9 are all causal. - -### INVERTIBLE AND NONINVERTIBLE SYSTEMS - -A discrete-time system *S* is invertible if an inverse system *Si* exists such that the cascade of *S* and *Si* results in an *identity* system. An identity system is defined as one whose output is identical to the input. In other words, for an invertible system, the input can be uniquely determined from the corresponding output. For every input there is a unique output. When a signal is processed through such a system, its input can be reconstructed from the corresponding output. There is no loss of information when a signal is processed through an invertible system. - -A cascade of a unit delay with a unit advance results in an identity system because the output of such a cascaded system is identical to the input. Clearly, the inverse of an ideal unit delay is ideal unit advance, which is a noncausal (and unrealizable) system. In contrast, a compressor *y*[*n*] = *x*[*Mn*] is not invertible because this operation loses all but every *M*th sample of the input, and, generally, the input cannot be reconstructed. Similarly, operations, such as *y*[*n*] = cos *x*[*n*] or *y*[*n*]=|*x*[*n*]|, are not invertible. - -### **DR ILL 3.9 Invertibility** - -Show that a system specified by equation *y*[*n*] = *ax*[*n*] + *b* is invertible but that the system *y*[*n*]=|*x*[*n*]|2 is noninvertible. - -### STABLE AND UNSTABLE SYSTEMS - -The concept of stability is similar to that in continuous-time systems. Stability can be *internal* or *external*. If every *bounded input* applied at the input terminal results in a *bounded output,* the system is said to be stable *externally*. External stability can be ascertained by measurements at the external terminals of the system. This type of stability is also known as the stability in the BIBO (bounded-input/bounded-output) sense. Both internal and external stability are discussed in greater detail in Sec. 3.9. - -### MEMORYLESS SYSTEMS AND SYSTEMS WITH MEMORY - -The concepts of memoryless (or instantaneous) systems and those with memory (or dynamic) are identical to the corresponding concepts of the continuous-time case. A system is memoryless if its response at any instant *n* depends at most on the input at the same instant *n*. The output at any instant of a system with memory generally depends on the past, present, and future values of the input. For example, *y*[*n*] =sin*x*[*n*] is an example of instantaneous system, and *y*[*n*]−*y*[*n*−1] =*x*[*n*] is an example of a dynamic system or a system with memory. - -### **EXAMPLE 3.10 Investigating DT System Properties** - -Consider a DT system described as *y*[*n* + 1] = *x*[*n* + 1]*x*[*n*]. Determine whether the system is **(a)** linear, **(b)** time-invariant, **(c)** causal, **(d)** invertible, **(e)** BIBO-stable, and **(f)** memoryless. - -Let us delay the input–output equation by one to obtain the equivalent but more convenient representation of *y*[*n*] = *x*[*n*]*x*[*n*−1]. - -**(a)** Linearity requires both homogeneity and additivity. Let us first investigate homogeneity. Assuming *x*[*n*] -⇒ *y*[*n*], we see that - -$$ -ax[n] \Longrightarrow (ax[n])(ax[n-1]) = a^2y[n] \neq ay[n] -$$ - -Thus, the system does not satisfy the homogeneity property. - -The system also does not satisfy the additivity property. Assuming *x*1[*n*] -⇒ *y*1[*n*] and *x*2[*n*] -⇒ *y*2[*n*], we see that input *x*[*n*] = *x*1[*n*] +*x*2[*n*] produces output *y*[*n*] as - -$$ -y[n] = (x_1[n] + x_2[n])(x_1[n-1] + x_2[n-1]) -$$ - -= $x_1[n]x_1[n-1] + x_2[n]x_2[n-1] + x_1[n]x_2[n-1] + x_2[n]x_1[n-1]$ -= $y_1[n] + y_2[n] + x_1[n]x_2[n-1] + x_2[n]x_1[n-1]$ - $\neq y_1[n] + y_2[n]$ - -Clearly, additivity is not satisfied. - -Since the system does not satisfy both the homogeneity and additivity properties, we conclude that the system is not linear. - -**(b)** To be time-invariant, a shift in any input should cause a corresponding shift in respective output. Assume that *x*[*n*] -⇒ *y*[*n*]. Applying a delay version of this input to the system yields - -$$ -x[n - N] \Longrightarrow x[n - N]x[n - 1 - N] = x[(n - N)]x[(n - N) - 1] = y[n - N] -$$ - -Since shifting an input causes a corresponding shift in the output, we conclude that the system is time-invariant. - -**(c)** To be causal, an output value cannot depend on any future input values. The output *y* at time *n* depends on the input *x* at present and past times *n* and *n*−1. Since the current output does not depend on future input values, the system is causal. - -**(d)** For a system to be invertible, every input must generate a unique output, which allows exact recovery of the input from the output. Consider two inputs to this system: *x*1[*n*] = 1 and *x*2[*n*]=−1. Both inputs generate the same output: *y*1[*n*] = *y*2[*n*] = 1. Since unique inputs do not always generate unique outputs, we conclude that the system is not invertible. - -**(e)** To be BIBO-stable, any bounded input must generate a bounded output. A bounded input satisfies |*x*[*n*]| ≤ *Mx* < ∞ for all *n*. Given this condition, the system output magnitude behaves as - -> |*y*[*n*]| = |*x*[*n*]*x*[*n*−1]| = |*x*[*n*]||*x*[*n*−1]| ≤ *M*2 *x* < ∞ - -Since any bounded input is guaranteed to produce a bounded output, it follows that the system is BIBO-stable. - -**(f)** To be memoryless, a system's output can only depend on the strength of the current input. Since the output *y* at time *n* depends on the input *x* not only at present time *n* but also on past time *n*−1, we see that the system is not memoryless. - -## **[3.5 DISCRETE-TIME](#page-9-0) SYSTEM EQUATIONS** - -In this section we discuss time-domain analysis of LTID (linear, time-invariant, discrete-time systems). With minor differences, the procedure is parallel to that for continuous-time systems. - -### DIFFERENCE EQUATIONS - -Equations (3.3), (3.5), (3.8), and (3.13) are examples of difference equations. Equations (3.3), (3.8), and (3.13) are first-order difference equations, and Eq. (3.5) is a second-order difference equation. All these equations are linear, with constant (not time-varying) coefficients.† Before giving a general form of an *N*th-order linear difference equation, we recall that a difference equation can be written in two forms: the first form uses delay terms such as *y*[*n* − 1], *y*[*n* − 2], *x*[*n*−1], *x*[*n*−2], and so on; and the alternate form uses advance terms such as *y*[*n*+1], *y*[*n*+2], and so on. Although the delay form is more natural, we shall often prefer the advance form, not just for the general notational convenience, but also for resulting notational uniformity with the operator form for differential equations. This facilitates the commonality of the solutions and concepts for continuous-time and discrete-time systems. - -We start here with a general difference equation, written in advance form as - -$$ -y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n] = -$$ - -\n -$$ -b_{N-M}x[n+M] + b_{N-M+1}x[n+M-1] + \cdots + b_{N-1}x[n+1] + b_Nx[n] -$$ - (3.14) - -This is a linear difference equation whose order is max(*N*,*M*). We have assumed the coefficient of *y*[*n* + *N*] to be unity (*a*0 = 1) without loss of generality. If *a*0 = 1, we can divide the equation throughout by *a*0 to normalize the equation to have *a*0 = 1. - -### CAUSALITY CONDITION - -For a causal system, the output cannot depend on future input values. This means that when the system equation is in the advance form of Eq. (3.14), causality requires *M* ≤ *N*. If *M* were to be greater than *N*, then *y*[*n*+*N*], the output at *n*+*N* would depend on *x*[*n*+ *M*], which is the input at the later instant *n*+ *M*. For a general causal case, *M* = *N*, and Eq. (3.14) can be expressed as - -$$ -y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n] = -$$ - -\n -$$ -b_0x[n+N] + b_1x[n+N-1] + \cdots + b_{N-1}x[n+1] + b_Nx[n] -$$ -\n(3.15) - -† Equations such as (3.3), (3.5), (3.8), and (3.13) are considered to be linear according to the classical definition of linearity. Some authors label such equations as *incrementally linear*. We prefer the classical definition. It is just a matter of individual choice and makes no difference in the final results. - -where some of the coefficients on either side can be zero. In this *N*th-order equation, *a*0, the coefficient of *y*[*n*+*N*], is normalized to unity. Equation (3.15) is valid for all values of *n*. Therefore, it is still valid if we replace *n* by *n* − *N* throughout the equation [see Eqs. (3.3) and (3.4)]. Such replacement yields a delay-form alternative: - -$$ -y[n] + a_1y[n-1] + \dots + a_{N-1}y[n-N+1] + a_Ny[n-N] = -$$ - -\n -$$ -b_0x[n] + b_1x[n-1] + \dots + b_{N-1}x[n-N+1] + b_Nx[n-N] -$$ - (3.16) - -### **[3.5-1 Recursive \(Iterative\) Solution of Difference Equation](#page-9-0)** - -Equation (3.16) can be expressed as - -$$ -y[n] = -a_1y[n-1] - a_2y[n-2] - \cdots - a_Ny[n-N] + b_0x[n] + b_1x[n-1] + \cdots + b_Nx[n-N] -$$ -(3.17) - -In Eq. (3.17), *y*[*n*] is computed from 2*N* + 1 pieces of information; the preceding *N* values of the output: *y*[*n* − 1], *y*[*n* − 2], ... , *y*[*n* − *N*], and the preceding *N* values of the input: *x*[*n* − 1], *x*[*n* − 2], ... , *x*[*n* − *N*], and the present value of the input *x*[*n*]. Initially, to compute *y*[0], the *N* initial conditions *y*[−1], *y*[−2], ... , *y*[−*N*] serve as the preceding *N* output values. Hence, knowing the *N* initial conditions and the input, we can determine recursively the entire output *y*[0], *y*[1], *y*[2], *y*[3], ... , one value at a time. For instance, to find *y*[0] we set *n* = 0 in Eq. (3.17). The left-hand side is *y*[0], and the right-hand side is expressed in terms of *N* initial conditions *y*[−1], *y*[−2], ... , *y*[−*N*] and the input *x*[0] if *x*[*n*] is causal (because of causality, other input terms *x*[−*n*] = 0). Similarly, knowing *y*[0] and the input, we can compute *y*[1] by setting *n* = 1 in Eq. (3.17). Knowing *y*[0] and *y*[1], we find *y*[2], and so on. Thus, we can use this recursive procedure to find the complete response *y*[0], *y*[1], *y*[2], .... For this reason, this equation is classed as a recursive form. This method basically reflects the manner in which a computer would solve a recursive difference equation, given the input and initial conditions. Equation (3.17) [or Eq. (3.16)] is nonrecursive if all the *N* − 1 coefficients *ai* = 0 (*i* = 1, 2,...,*N* − 1). In this case, it can be seen that *y*[*n*] is computed only from the input values and without using any previous outputs. Generally speaking, the recursive procedure applies only to equations in the recursive form. The recursive (iterative) procedure is demonstrated by the following examples. - -### **EXAMPLE 3.11 Iterative Solution to a First-Order Difference Equation** - -Solve iteratively - -$$ -y[n] - 0.5y[n-1] = x[n] -$$ - -with initial condition *y*[−1] = 16 and causal input *x*[*n*] = *n*2*u*[*n*]. This equation can be expressed as - -*y*[*n*] = 0.5*y*[*n*−1] +*x*[*n*] (3.18) - -If we set *n* = 0 in Eq. (3.18), we obtain - -$$ -y[0] = 0.5y[-1] + x[0] -$$ - -= 0.5(16) + 0 = 8 - -Now, setting *n* = 1 in Eq. (3.18) and using the value *y*[0] = 8 (computed in the first step) and *x*[1] = (1)2 = 1, we obtain - -$$ -y[1] = 0.5(8) + (1)^2 = 5 -$$ - -Next, setting *n* = 2 in Eq. (3.18) and using the value *y*[1] = 5 (computed in the previous step) and *x*[2] = (2)2, we obtain - -$$ -y[2] = 0.5(5) + (2)^2 = 6.5 -$$ - -Continuing in this way iteratively, we obtain - -$$ -y[3] = 0.5(6.5) + (3)^{2} = 12.25 -$$ - -\n -$$ -y[4] = 0.5(12.25) + (4)^{2} = 22.125 -$$ - -\n -$$ -\vdots -$$ - -The output *y*[*n*] is depicted in Fig. 3.17. - -We now present one more example of iterative solution—this time for a second-order equation. The iterative method can be applied to a difference equation in delay form or advance form. In Ex. 3.11 we considered the former. Let us now apply the iterative method to the advance form. - -### **EXAMPLE 3.12 Iterative Solution to a Second-Order Difference Equation** - -Solve iteratively - -$$ -y[n+2] - y[n+1] + 0.24y[n] = x[n+2] - 2x[n+1] -$$ - -### 268 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -with initial conditions *y*[−1] = 2, *y*[−2] = 1 and a causal input *x*[*n*] = *nu*[*n*]. The system equation can be expressed as - -$$ -y[n+2] = y[n+1] - 0.24y[n] + x[n+2] - 2x[n+1] -$$ -\n(3.19) - -Setting *n* = −2 in Eq. (3.19) and then substituting *y*[−1] = 2, *y*[−2] = 1, *x*[0] = *x*[−1] = 0, we obtain - -$$ -y[0] = 2 - 0.24(1) + 0 - 0 = 1.76 -$$ - -Setting *n* = −1 in Eq. (3.19) and then substituting *y*[0] = 1.76, *y*[−1] = 2, *x*[1] = 1, *x*[0] = 0, we obtain - -$$ -y[1] = 1.76 - 0.24(2) + 1 - 0 = 2.28 -$$ - -Setting *n* = 0 in Eq. (3.19) and then substituting *y*[0] = 1.76, *y*[1] = 2.28, *x*[2] = 2, and *x*[1] = 1 yield - -$$ -y[2] = 2.28 - 0.24(1.76) + 2 - 2(1) = 1.8576 -$$ - -and so on. - -With MATLAB, we can readily verify and extend these recursive calculations. - -``` ->> n = -2:5; y = [1,2,zeros(1,length(n)-2)]; x = [0,0,n(3:end)]; ->> for k = 1:length(n)-2, ->> y(k+2) = y(k+1)-0.24*y(k)+x(k+2)-2*x(k+1); ->> end ->> n,y - n = -2 -1 0 1 2 3 4 5 - y = 1.0000 2.0000 1.7600 2.2800 1.8576 0.3104 -2.1354 -5.2099 -``` - -Note carefully the recursive nature of the computations. From the *N* initial conditions (and the input), we obtained *y*[0] first. Then, using this value of *y*[0] and the preceding *N* − 1 initial conditions (along with the input), we find *y*[1]. Next, using *y*[0], *y*[1] along with the past *N* − 2 initial conditions and input, we obtained *y*[2], and so on. This method is general and can be applied to a recursive difference equation of any order. It is interesting that the hardware realization of Eq. (3.18) depicted in Fig. 3.14 (with *a* = 0.5) generates the solution precisely in this (iterative) fashion. - -### **DR ILL 3.10 Iterative Solution to a Difference Equation** - -Using the iterative method, find the first three terms of *y*[*n*] for - -*y*[*n*+1] −2*y*[*n*] = *x*[*n*] - -The initial condition is *y*[−1] = 10 and the input *x*[*n*] = 2 starting at *n* = 0. - -**ANSWER** *y*[0] = 20, *y*[1] = 42, and *y*[2] = 86 - -We shall see in the future that the solution of a difference equation obtained in this direct (iterative) way is useful in many situations. Despite the many uses of this method, a closed-form solution of a difference equation is far more useful in the study of system behavior and its dependence on the input and various system parameters. For this reason we shall develop a systematic procedure to analyze discrete-time systems along lines similar to those used for continuous-time systems. - -### OPERATOR NOTATION - -In difference equations, it is convenient to use operator notation similar to that used in differential equations for the sake of compactness. In continuous-time systems, we used the operator *D* to denote the operation of differentiation. For discrete-time systems, we shall use the operator *E* to denote the operation for advancing a sequence by one time unit. Thus, - -$$ -Ex[n] \equiv x[n+1] -$$ - -$$ -E^{2}x[n] \equiv x[n+2] -$$ - -$$ -\vdots -$$ - -$$ -E^{N}x[n] \equiv x[n+N] -$$ - -Let us use this advance operator notation to represent several systems investigated earlier. The first-order difference equation of a savings account is [see Eq. (3.4)] - -$$ -y[n+1] - ay[n] = x[n+1] -$$ - -Using the operator notation, we can express this equation as - -$$ -E\mathbf{y}[n] - a\mathbf{y}[n] = Ex[n] \qquad \text{or} \qquad (E - a)\mathbf{y}[n] = Ex[n] -$$ - -Similarly, the second-order book sales estimate described by Eq. (3.6) as - -$$ -y[n+2] + \frac{1}{4}y[n+1] + \frac{1}{16}y[n] = x[n+2] -$$ - -can be expressed in operator notation as - -$$ -(E^2 + \frac{1}{4}E + \frac{1}{16})y[n] = E^2x[n] -$$ - -The general *N*th-order advance-form difference equation of Eq. (3.15) can be expressed as - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] = (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] -$$ - -or - -$$ -Q[E]y[n] = P[E]x[n] \tag{3.20} -$$ - -where *Q*[*E*] and *P*[*E*] are *N*th-order polynomial operators - -$$ -Q[E] = EN + a1EN-1 + \dots + aN-1E + aN -$$ - -$$ -P[E] = b0EN + b1EN-1 + \dots + bN-1E + bN -$$ - -### RESPONSE OF LINEAR DISCRETE-TIME SYSTEMS - -Following the procedure used for continuous-time systems, we can show that Eq. (3.20) is a linear equation (with constant coefficients). A system described by such an equation is a linear, time-invariant, discrete-time (LTID) system. We can verify, as in the case of LTIC systems (see the footnote on page 151), that the general solution of Eq. (3.20) consists of zero-input and zero-state components. - -## **3.6 SYSTEM RESPONSE TO INTERNAL [CONDITIONS:](#page-9-0) THE ZERO-INPUT RESPONSE** - -The zero-input response *y*0[*n*] is the solution of Eq. (3.20) with *x*[*n*] = 0; that is, - -$$ -Q[E]y_0[n] = 0 -$$ - -or - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y0[n] = 0 -$$ -\n(3.21) - -Although we can solve this equation systematically, even a cursory examination points to the solution. This equation states that a linear combination of *y*0[*n*] and advanced *y*0[*n*] is zero, *not for some values of n, but for all n.* Such a situation is possible *if and only if y*0[*n*] and advanced *y*0[*n*] have the same form. Only an exponential function γ *n* has this property, as the following equation indicates: - -$$ -E^k\{\gamma^n\} = \gamma^{n+k} = \gamma^k\gamma^n -$$ - -This expression shows that γ *n* advanced by *k* units is a constant (γ *k*) times γ *n*. Therefore, the solution of Eq. (3.21) must be of the form† - -$$ -y_0[n] = c\gamma^n \tag{3.22} -$$ - -To determine *c* and γ , we substitute this solution in Eq. (3.21). Since *Eky*0[*n*] = *y*0[*n*+*k*] = *c*γ *n*+*k*, this produces - -$$ -c(\gamma^{N} + a_1 \gamma^{N-1} + \cdots + a_{N-1} \gamma + a_N) \gamma^{n} = 0 -$$ - -For a nontrivial solution of this equation, - -$$ -\gamma^{N} + a_{1}\gamma^{n-1} + \dots + a_{N-1}\gamma + a_{N} = 0 -$$ -\n(3.23) - -or - -*Q*[γ ] = 0 - -Our solution *c*γ *n* [Eq. (3.22)] is correct, provided γ satisfies Eq. (3.23). Now, *Q*[γ ] is an *N*th-order polynomial and can be expressed in the factored form (assuming all distinct roots): - -$$ -(\gamma - \gamma_1)(\gamma - \gamma_2) \cdots (\gamma - \gamma_N) = 0 -$$ - -Clearly, γ has *N* solutions γ1, γ2, ... , γ*N* and, therefore, Eq. (3.21) also has *N* solutions *c*1γ *n* 1 , *c*2γ *n* 2 , ... , *cn*γ *n N*. In such a case, we have shown that the general solution is a linear combination - -† A signal of the form *nm*γ *n* also satisfies this requirement under certain conditions (repeated roots), discussed later. - -#### 3.6 System Response to Internal Conditions: The Zero-Input Response 271 - -of the *N* solutions (see the footnote on page 153). Thus, - -$$ -y_0[n] = c_1 \gamma_1^n + c_2 \gamma_2^n + \cdots + c_n \gamma_N^n -$$ - -where γ1, γ2, ... , γ*n* are the roots of Eq. (3.23) and *c*1, *c*2, ... , *cn* are arbitrary constants determined from *N* auxiliary conditions, generally given in the form of initial conditions. The polynomial *Q*[γ ] is called the *characteristic polynomial* of the system, and *Q*[γ ] = 0 [Eq. (3.23)] is the *characteristic equation* of the system. Moreover, γ1, γ2, ... , γ*N*, the roots of the characteristic equation, are called *characteristic roots* or *characteristic values* (also *eigenvalues*) of the system. The exponentials γ *n i* (*i* = 1, 2,...,*N*) are the *characteristic modes* or *natural modes* of the system. A characteristic mode corresponds to each characteristic root of the system, and the *zero-input response is a linear combination of the characteristic modes of the system*. - -### **EXAMPLE 3.13 Zero-Input Response of a Second-Order System with Real Roots** - -The LTID system described by the difference equation - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -has input *x*[*n*] = 4−*nu*[*n*] and initial conditions *y*[−1] = 0 and *y*[−2] = 25/4. Determine the zero-input response *y*0[*n*]. The zero-state response of this system is considered later, in Ex. 3.21. - -The system equation in operator notation is - -$$ -(E^2 - 0.6E - 0.16)y[n] = 5E^2x[n] -$$ - -The characteristic polynomial is - -$$ -\gamma^2 - 0.6\gamma - 0.16 = (\gamma + 0.2)(\gamma - 0.8) -$$ - -The characteristic equation is - -$$ -(\gamma + 0.2)(\gamma - 0.8) = 0 -$$ - -The characteristic roots are γ1 = −0.2 and γ2 = 0.8. The zero-input response is - -$$ -y_0[n] = c_1(-0.2)^n + c_2(0.8)^n \tag{3.24} -$$ - -To determine arbitrary constants *c*1 and *c*2, we set *n* = −1 and −2 in Eq. (3.24), then substitute *y*0[−1] = 0 and *y*0[−2] = 25/4 to obtain† - -$$ -\begin{array}{c}\n0 = -5c_1 + \frac{5}{4}c_2 \\ -\frac{25}{4} = 25c_1 + \frac{25}{16}c_2\n\end{array}\n\right\} \quad \Longrightarrow \quad c_1 = \frac{1}{5} -$$ -\n -$$ -c_2 = \frac{4}{5} -$$ - -† The initial conditions *y*[−1] and *y*[−2] are the conditions given on the total response. But because the input does not start until *n* = 0, the zero-state response is zero for *n* < 0. Hence, at *n* = −1 and −2 the total response consists of the zero-input component only so that *y*[−1] = *y*0[−1] and *y*[−2] = *y*0[−2]. - -### 272 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Therefore, - -$$ -y_0[n] = \frac{1}{5}(-0.2)^n + \frac{4}{5}(0.8)^n -$$ - $n \ge 0$ - -The reader can verify this solution by computing the first few terms using the iterative method (see Exs. 3.11 and 3.12). - -### **DR ILL 3.11 Zero-Input Response of First-Order Systems** - -Find and sketch the zero-input response for the systems described by the following equations: - -- **(a)** *y*[*n*+1] −0.8*y*[*n*] = 3*x*[*n*+1] -- **(b)** *y*[*n*+1] +0.8*y*[*n*] = 3*x*[*n*+1] - -In each case the initial condition is *y*[−1] = 10. Verify the solutions by computing the first three terms using the iterative method. - -### **ANSWERS** - -- **(a)** 8(0.8)*n* -- **(b)** −8(−0.8)*n* - -### **DR ILL 3.12 Zero-Input Response of a Second-Order System with Real Roots** - -Find the zero-input response of a system described by the equation - -$$ -y[n] + 0.3y[n-1] - 0.1y[n-2] = x[n] + 2x[n-1] -$$ - -The initial conditions are *y*0[−1] = 1 and *y*0[−2] = 33. Verify the solution by computing the first three terms iteratively. - -### **ANSWER** - -*y*0[*n*] = (0.2)*n* +2(−0.5)*n* - -Section 3.5-1 introduced the method of recursion to solve difference equations. As the next example illustrates, the zero-input response can likewise be found through recursion. Since it does not provide a closed-form solution, recursion is generally not the preferred method of solving difference equations. - -### REPEATED ROOTS - -So far we have assumed the system to have *N* distinct characteristic roots γ1, γ2, ... , γ*N* with corresponding characteristic modes γ *n* 1 , γ *n* 2 , ... , γ *n N*. If two or more roots coincide (repeated roots), the form of characteristic modes is modified. Direct substitution shows that if a root γ repeats *r* times (root of multiplicity *r*), the corresponding characteristic modes for this root are γ *n*, *n*γ *n*, *n*2γ *n*, ... , *nr*−1γ *n*. Thus, if the characteristic equation of a system is - -$$ -Q[\gamma]=(\gamma-\gamma_1)^r(\gamma-\gamma_{r+1})(\gamma-\gamma_{r+2})\cdots(\gamma-\gamma_N) -$$ - -then the zero-input response of the system is - -$$ -y_0[n] = (c_1 + c_2n + c_3n^2 + \dots + c_rn^{r-1})\gamma_1^n + c_{r+1}\gamma_{r+1}^n + c_{r+2}\gamma_{r+2}^n + \dots + c_n\gamma_N^n -$$ - -### **EXAMPLE 3.15 Zero-Input Response of a Second-Order System with Repeated Roots** - -Consider a second-order difference equation with repeated roots: - -$$ -(E2 + 6E + 9)y[n] = (2E2 + 6E)x[n] -$$ - -Determine the zero-input response *y*0[*n*] if the initial conditions are *y*0[−1]=−1/3 and *y*0[−2]=−2/9. - -The characteristic polynomial is γ 2 +6γ +9 = (γ +3)2, and we have a repeated characteristic root at γ = −3. The characteristic modes are (−3)*n* and *n*(−3)*n*. Hence, the zero-input response is - -> *y*0[*n*] = (*c*1 +*c*2*n*)(−3) *n* - -Although we can determine the constants *c*1 and *c*2 from the initial conditions following a procedure similar to Ex. 3.13, we instead use MATLAB to perform the needed calculations. - ->> c = inv([(-3)^(-1) -1\*(-3)^(-1);(-3)^(-2) -2\*(-3)^(-2)])\*[-1/3;-2/9] c=4 3 - -Thus, the zero-input response is - -$$ -y_0[n] = (4+3n)(-3)^n -$$ - $n \ge 0$ - -### COMPLEX ROOTS - -As in the case of continuous-time systems, the complex roots of a discrete-time system will occur in pairs of conjugates if the system equation coefficients are real. Complex roots can be treated exactly as we would treat real roots. However, just as in the case of continuous-time systems, we can also use the real form of solution as an alternative. - -First we express the complex conjugate roots γ and γ ∗ in polar form. If |γ | is the magnitude and β is the angle of γ , then - -$$ -\gamma = |\gamma|e^{i\beta} -$$ - and $\gamma^* = |\gamma|e^{-j\beta}$ - -The zero-input response is given by - -$$ -y_0[n] = c_1 \gamma^n + c_2(\gamma^*)^n = c_1 |\gamma|^n e^{i\beta n} + c_2 |\gamma|^n e^{-j\beta n} -$$ - -For a real system, *c*1 and *c*2 must be conjugates so that *y*0[*n*] is a real function of *n*. Let - -$$ -c_1 = \frac{c}{2}e^{j\theta} \qquad \text{and} \qquad c_2 = \frac{c}{2}e^{-j\theta} -$$ - -Then - -$$ -y_0[n] = \frac{c}{2} |\gamma|^n \left[ e^{j(\beta n + \theta)} + e^{-j(\beta n + \theta)} \right] = c |\gamma|^n \cos(\beta n + \theta) -$$ -\n(3.25) - -where *c* and θ are arbitrary constants determined from the auxiliary conditions. This is the solution in real form, which avoids dealing with complex numbers. - -### **EXAMPLE 3.16 Zero-Input Response of a Second-Order System with Complex Roots** - -Consider a second-order difference equation with complex-conjugate roots: - -$$ -(E2 - 1.56E + 0.81)y[n] = (E + 3)x[n] -$$ - -Determine the zero-input response *y*0[*n*] if the initial conditions are *y*0[−1] = 2 and *y*0[−2] = 1. - -The characteristic polynomial is (γ 2 − 1.56γ + 0.81) = (γ − 0.78 − *j*0.45)(γ − 0.78 + *j*0.45). The characteristic roots are 0.78 ± *j*0.45; that is, 0.9*e*±*j*(π/6) . We could immediately write the solution as - -$$ -y_0[n] = c(0.9)^n e^{j\pi n/6} + c^*(0.9)^n e^{-j\pi n/6} -$$ - -Setting *n* = −1 and −2 and using the initial conditions *y*0[−1] = 2 and *y*0[−2] = 1, we find *c* = 1.1550−*j*0.2025 = 1.1726*e*−*j*0.1735 and *c*∗ = 1.1550+*j*0.2025 = 1.1726*ej*0.1735. - ->> gamma = roots([1 -1.56 0.81]); >> c = inv([gamma(1)^(-1) gamma(2)^(-1);gamma(1)^(-2) gamma(2)^(-2)])\*[2;1] c = 1.1550 - 0.2025i 1.1550 + 0.2025i - -Alternately, we could also find the unknown coefficient by using the real form of the solution, as given in Eq. (3.25). In the present case, the roots are 0.9*e*±*j*(π/6) . Hence, |γ | = 0.9 and β = π/6, and the zero-input response, according to Eq. (3.25), is given by - -$$ -y_0[n] = c(0.9)^n \cos\left(\frac{\pi}{6}n + \theta\right) -$$ - -To determine the constants *c* and θ, we set *n* = −1 and −2 in this equation and substitute the initial conditions *y*0[−1] = 2 and *y*0[−2] = 1 to obtain - -$$ -2 = \frac{c}{0.9} \cos\left(-\frac{\pi}{6} + \theta\right) = \frac{c}{0.9} \left[\frac{\sqrt{3}}{2} \cos\theta + \frac{1}{2} \sin\theta\right] -$$ -$$ -1 = \frac{c}{(0.9)^2} \cos\left(-\frac{\pi}{3} + \theta\right) = \frac{c}{0.81} \left[\frac{1}{2} \cos\theta + \frac{\sqrt{3}}{2} \sin\theta\right] -$$ - -or - -$$ -\frac{\sqrt{3}}{1.8}c\cos\theta + \frac{1}{1.8}c\sin\theta = 2 -$$ -$$ -\frac{1}{1.62}c\cos\theta + \frac{\sqrt{3}}{1.62}c\sin\theta = 1 -$$ - -These are two simultaneous equations in two unknowns *c*cos θ and *c*sin θ. Solution of these equations yields - -$$ -c \cos \theta = 2.308 -$$ - -$$ -c \sin \theta = -0.397 -$$ - -Dividing *c*sin θ by *c*cos θ yields - -$$ -\tan \theta = \frac{-0.397}{2.308} = \frac{-0.172}{1} -$$ -$$ -\theta = \tan^{-1}(-0.172) = -0.17 \text{ rad} -$$ - -Substituting θ = −0.17 radian in *c*cos θ = 2.308 yields *c* = 2.34 and - -$$ -y_0[n] = 2.34(0.9)^n \cos\left(\frac{\pi}{6}n - 0.17\right) -$$ - $n \ge 0$ - -Observe that here we have used radian units for both β and θ. We also could have used the degree unit, although this practice is not recommended. The important consideration is to be consistent and to use the same units for both β and θ. - -### **DR ILL 3.13 Zero-Input Response of a Second-Order System with Complex Roots** - -Find the zero-input response of a system described by the equation - -$$ -y[n] + 4y[n-2] = 2x[n] -$$ - -The initial conditions are *y*0[−1]=−1/(2 √2) and *y*0[−2] = 1/(4 √ 2). Verify the solution by computing the first three terms iteratively. - -**ANSWER** - -*y*0[*n*] = (2)*n* cos π 2 *n*− 3π 4 - -# **3.7 THE UNIT IMPULSE [RESPONSE](#page-9-0)** *h***[***n***]** - -Consider an *n*th-order system specified by the equation - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] = (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] -$$ - -or - -$$ -Q[E]y[n] = P[E]x[n] -$$ - -The unit impulse response *h*[*n*] is the solution of this equation for the input δ[*n*] with all the initial conditions zero; that is, - -$$ -Q[E]h[n] = P[E]\delta[n] \tag{3.26} -$$ - -subject to initial conditions - -$$ -h[-1] = h[-2] = \cdots = h[-N] = 0 -$$ - -Equation (3.26) can be solved to determine *h*[*n*] iteratively or in a closed form. The following example demonstrates the iterative solution. - -### **EXAMPLE 3.17 Iterative Determination of the Impulse Response** - -Iteratively compute the first two values of the impulse response *h*[*n*] of a system described by the equation - -*y*[*n*] −0.6*y*[*n*−1] −0.16*y*[*n*−2] = 5*x*[*n*] - -To determine the unit impulse response, we let the input *x*[*n*] = δ[*n*] and the output *y*[*n*] = *h*[*n*] in the system's difference equation to obtain - -$$ -h[n] - 0.6h[n-1] - 0.16h[n-2] = 5\delta[n] -$$ - -subject to zero initial state; that is, *h*[−1] = *h*[−2] = 0. Setting *n* = 0 in this equation yields - -*h*[0] −0.6(0)−0.16(0) = 5(1) ⇒ *h*[0] = 5 - -Setting *n* = 1 in the same equation and using *h*[0] = 5, we obtain - -*h*[1] −0.6(5)−0.16(0) = 5(0) ⇒ *h*[1] = 3 - -Continuing this way, we can determine any number of terms of *h*[*n*]. Unfortunately, such a solution does not yield a closed-form expression for *h*[*n*]. Nevertheless, determining a few values of *h*[*n*] can be useful in determining the closed-form solution, as the following development shows. - -## **[3.7-1 The Closed-Form Solution of](#page-9-0)** *h***[***n***]** - -Recall that *h*[*n*] is the system response to input δ[*n*], which is zero for *n* > 0. We know that when the input is zero, only the characteristic modes can be sustained by the system. Therefore, *h*[*n*] must be made up of characteristic modes for *n* > 0. At *n* = 0, it may have some nonzero value *A*0 so that a general form of *h*[*n*] can be expressed as† - -$$ -h[n] = A_0 \delta[n] + y_c[n]u[n] -$$ -\n(3.27) - -where *yc*[*n*] is a linear combination of the characteristic modes. We now substitute Eq. (3.27) in Eq. (3.26) to obtain *Q*[*E*](*A*0δ[*n*] +*yc*[*n*]*u*[*n*]) = *P*[*E*]δ[*n*]. Because *yc*[*n*] is made up of characteristic modes, *Q*[*E*]*yc*[*n*]*u*[*n*] = 0, and we obtain *A*0*Q*[*E*]δ[*n*] = *P*[*E*]δ[*n*], that is, - -$$ -A_0 (\delta[n+N] + a_1 \delta[n+N-1] + \cdots + a_N \delta[n]) = b_0 \delta[n+N] + \cdots + b_N \delta[n] -$$ - -Setting *n* = 0 in this equation and using the fact that δ[*m*] = 0 for all *m* = 0, and δ[0] = 1, we obtain - -$$ -A_0 a_N = b_N \quad \Longrightarrow \quad A_0 = \frac{b_N}{a_N} \tag{3.28} -$$ - -Hence,‡ - -$$ -h[n] = \frac{b_N}{a_N} \delta[n] + y_c[n]u[n] \tag{3.29} -$$ - -The *N* unknown coefficients in *yc*[*n*] (on the right-hand side) can be determined from a knowledge of *N* values of *h*[*n*]. Fortunately, it is a straightforward task to determine values of *h*[*n*] iteratively, as demonstrated in Ex. 3.17. We compute *N* values *h*[0], *h*[1], *h*[2], ... , *h*[*N* −1] iteratively. Now, setting *n* = 0, 1, 2, ... , *N* −1 in Eq. (3.29), we can determine the *N* unknowns in *yc*[*n*]. This point will become clear in the following example. - -### **EXAMPLE 3.18 Closed-Form Determination of the Impulse Response** - -Determine the unit impulse response *h*[*n*] for a system in Ex. 3.17 specified by the equation - -$$ -y[n] - 0.6y[n-1] - 0.16y[n-2] = 5x[n] -$$ - -† We assume that the term *yc*[*n*] consists of characteristic modes for *n* &gt; 0 only. To reflect this behavior, the characteristic terms should be expressed in the form γ *n j u*[*n* − 1]. But because *u*[*n* − 1] = *u*[*n*] − δ[*n*], *cj*γ *n j u*[*n* − 1] = *cj*γ *n j u*[*n*] − *cj*δ[*n*], and *yc*[*n*] can be expressed in terms of exponentials γ *n j u*[*n*] (which start at *n* = 0), plus an impulse at *n* = 0. - -‡ If *aN* = 0, then *A*0 cannot be determined by Eq. (3.28). In such a case, we show in Sec. 3.12 that *h*[*n*] is of the form *A*0δ[*n*] + *A*1δ[*n* − 1] + *yc*[*n*]*u*[*n*]. We have here *N* + 2 unknowns, which can be determined from *N* +2 values *h*[0],*h*[1],...,*h*[*N* +1] found iteratively. - -This equation can be expressed in the advance form as - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -or in advance operator form as - -$$ -(E^2 - 0.6E - 0.16)y[n] = 5E^2x[n] -$$ - -The characteristic polynomial is - -$$ -\gamma^2 - 0.6\gamma - 0.16 = (\gamma + 0.2)(\gamma - 0.8) -$$ - -The characteristic modes are (−0.2)*n* and (0.8)*n*. Therefore, - -$$ -y_c[n] = c_1(-0.2)^n + c_2(0.8)^n -$$ - -Inspecting the system difference equation, we see that *aN* = −0.16 and *bN* = 0. Therefore, according to Eq. (3.29), - -$$ -h[n] = [c_1(-0.2)^n + c_2(0.8)^n]u[n] -$$ - -To determine *c*1 and *c*2, we need to find two values of *h*[*n*] iteratively. From Ex. 3.17, we know that *h*[0] = 5 and *h*[1] = 3. Setting *n* = 0 and 1 in our expression for *h*[*n*] and using the fact that *h*[0] = 5 and *h*[1] = 3, we obtain - -$$ -\begin{array}{c}\n5 = c_1 + c_2 \\ -3 = -0.2c_1 + 0.8c_2\n\end{array}\n\right\} \implies c_1 = 1\n\begin{array}{c}\nc_1 = 1 \\ -c_2 = 4\n\end{array} -$$ - -Therefore, - -$$ -h[n] = [(-0.2)^n + 4(0.8)^n]u[n] -$$ - -### **DR ILL 3.14 Closed-Form Determination of the Impulse Response** - -Find *h*[*n*], the unit impulse response of the LTID systems specified by the following equations: - -- **(a)** *y*[*n*+1] −*y*[*n*] = *x*[*n*] -- **(b)** *y*[*n*] −5*y*[*n*−1] +6*y*[*n*−2] = 8*x*[*n*−1] −19*x*[*n*−2] -- **(c)** *y*[*n*+2] −4*y*[*n*+1] +4*y*[*n*] = 2*x*[*n*+2] −2*x*[*n*+1] -- **(d)** *y*[*n*] = 2*x*[*n*] −2*x*[*n*−1] - -### **ANSWERS** - -- **(a)** *h*[*n*] = *u*[*n*−1] -- **(b)** *h*[*n*]=−19 6 δ[*n*] + 3 2 (2)*n* + 5 3 (3)*n u*[*n*] -- **(c)** *h*[*n*] = (2+*n*)2*nu*[*n*] -- **(d)** *h*[*n*] = 2δ[*n*] −2δ[*n*−1] - -### **EXAMPLE 3.19 Filtering Perspective of the Unit Impulse Response** - -Use the MATLAB filter command to solve Ex. 3.18. - -There are several ways to find the impulse response using MATLAB. In this method, we first specify the unit impulse function, which will serve as our input. Vectors a and b are created to specify the system. The filter command is then used to determine the impulse response. In fact, this method can be used to determine the zero-state response for any input. - -**Comment.** Although it is relatively simple to determine the impulse response *h*[*n*] by using the procedure in this section, in Ch. 5 we shall discuss the much simpler method of the *z*-transform. - -## **3.8 SYSTEM [RESPONSE TO](#page-9-0) EXTERNAL INPUT: THE ZERO-STATE RESPONSE** - -The zero-state response *y*[*n*] is the system response to an input *x*[*n*] when the system is in the zero state. In this section we shall assume that systems are in the zero state unless mentioned otherwise, so that the zero-state response will be the total response of the system. Here we follow the procedure parallel to that used in the continuous-time case by expressing an arbitrary input *x*[*n*] as a sum of impulse components. A signal *x*[*n*] in Fig. 3.20a can be expressed as a sum of impulse components, such as those depicted in Figs. 3.20b–3.20f. The component of *x*[*n*] at *n* = *m* is *x*[*m*]δ[*n*−*m*], and *x*[*n*] is the sum of all these components summed from *m* = −∞ to ∞. - -Therefore, - -$$ -x[n] = x[0]\delta[n] + x[1]\delta[n-1] + x[2]\delta[n-2] + \cdots -$$ - -+ x[-1]\delta[n+1] + x[-2]\delta[n+2] + \cdots -= -$$ -\sum_{m=-\infty}^{\infty} x[m]\delta[n-m] -$$ -(3.30) - -### 282 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -For a linear system, if we know the system response to impulse δ[*n*], we can obtain the system response to any arbitrary input by summing the system response to various impulse components. Let *h*[*n*] be the system response to impulse input δ[*n*]. We shall use the notation - -$$ -x[n] \Longrightarrow y[n] -$$ - -to indicate the input and the corresponding response of the system. Thus, if - -$$ -\delta[n] \Longrightarrow h[n] -$$ - -then because of time invariance - -$$ -\delta[n-m] \Longrightarrow h[n-m] -$$ - -and because of linearity - -$$ -x[m]\delta[n-m] \Longrightarrow x[m]h[n-m] -$$ - -and again because of linearity - -$$ -\underbrace{\sum_{m=-\infty}^{\infty} x[m]\delta[n-m]}_{x[n]} \quad \Longrightarrow \quad \underbrace{\sum_{m=-\infty}^{\infty} x[m]h[n-m]}_{y[n]} -$$ - -The left-hand side is *x*[*n*] [see Eq. (3.30)], and the right-hand side is the system response *y*[*n*] to input *x*[*n*]. Therefore,† - -$$ -y[n] = \sum_{m = -\infty}^{\infty} x[m]h[n-m] -$$ -\n(3.31) - -The summation on the right-hand side is known as the *convolution sum* of *x*[*n*] and *h*[*n*], and is represented symbolically by *x*[*n*] ∗ *h*[*n*] - -$$ -x[n] * h[n] = \sum_{m=-\infty}^{\infty} x[m]h[n-m] -$$ - -### PROPERTIES OF THE CONVOLUTION SUM - -The structure of the convolution sum is similar to that of the convolution integral. Moreover, the properties of the convolution sum are similar to those of the convolution integral. We shall enumerate these properties here without proof. The proofs are similar to those for the convolution integral and may be derived by the reader. - -$$ -y[n] = \sum_{m=-\infty}^{\infty} x[m]h[n,m] -$$ - -† In deriving this result, we have assumed a time-invariant system. The system response to input δ[*n* − *m*] for a time-varying system cannot be expressed as *h*[*n* − *m*]; instead, it has the form *h*[*n*, *m*]. Using this form, Eq. (3.31) is modified as follows: - -**The Commutative Property.** - -$$ -x_1[n] * x_2[n] = x_2[n] * x_1[n] -$$ - -**The Distributive Property.** - -$$ -x_1[n] * (x_2[n] + x_3[n]) = x_1[n] * x_2[n] + x_1[n] * x_3[n] -$$ - -**The Associative Property.** - -$$ -x_1[n] * (x_2[n] * x_3[n]) = (x_1[n] * x_2[n]) * x_3[n] -$$ - -**The Shifting Property.** If - -$$ -x_1[n] * x_2[n] = c[n] -$$ - -then - -$$ -x_1[n-m]*x_2[n-p] = c[n-m-p] -$$ -\n(3.32) - -**The Convolution with an Impulse.** - -$$ -x[n] * \delta[n] = x[n] -$$ - -**The Width Property.** If *x*1[*n*] and *x*2[*n*] have finite widths of *W*1 and *W*2, respectively, then the width of *x*1[*n*] ∗ *x*2[*n*] is *W*1 + *W*2. The width of a signal is 1 less than the number of its elements (length). Thus the signal in Fig. 3.22h has six elements (length of 6) but a width of only 5. Alternately, the property may be stated in terms of lengths as follows: if *x*1[*n*] and *x*2[*n*] have finite lengths of *L*1 and *L*2 elements, respectively, then the length of *x*1[*n*] ∗ *x*2[*n*] is *L*1 + *L*2 − 1 elements. - -### CAUSALITY AND ZERO-STATE RESPONSE - -In deriving Eq. (3.31), we assumed the system to be linear and time-invariant. There were no other restrictions on either the input signal or the system. In our applications, almost all the input signals are causal, and a majority of the systems are also causal. These restrictions further simplify the limits of the sum in Eq. (3.31). If the input *x*[*n*] is causal, *x*[*m*] = 0 for *m* < 0. Similarly, if the system is causal (i.e., if *h*[*n*] is causal), then *h*[*x*] = 0 for negative *x* so that *h*[*n* − *m*] = 0 when *m* > *n*. Therefore, if *x*[*n*] and *h*[*n*] are both causal, the product *x*[*m*]*h*[*n*−*m*] = 0 for *m* < 0 and for *m* > *n*, and it is nonzero only for the range 0 ≤ *m* ≤ *n*. Therefore, Eq. (3.31) in this case reduces to - -$$ -y[n] = \sum_{m=0}^{n} x[m]h[n-m] -$$ -\n(3.33) - -We shall evaluate the convolution sum first by an analytical method and later with graphical aid. - -### **EXAMPLE 3.20 Convolution of Causal Signals** - -Determine *c*[*n*] = *x*[*n*] ∗ *g*[*n*] for - -$$ -x[n] = (0.8)^n u[n] -$$ - and $g[n] = (0.3)^n u[n]$ - -We have - -$$ -c[n] = \sum_{m=-\infty}^{\infty} x[m]g[n-m] -$$ - -Note that - -$$ -x[m] = (0.8)^m u[m] -$$ - and $g[n-m] = (0.3)^{n-m} u[n-m]$ - -Both *x*[*n*] and *g*[*n*] are causal. Therefore [see Eq. (3.33)], - -$$ -c[n] = \sum_{m=0}^{n} x[m]g[n-m] = \sum_{m=0}^{n} (0.8)^m u[m] (0.3)^{n-m} u[n-m] -$$ - -In this summation, *m* lies between 0 and *n* (0 ≤ *m* ≤ *n*). Therefore, if *n* ≥ 0, then both *m* and *n*−*m* ≥ 0 so that *u*[*m*] = *u*[*n*−*m*] = 1. If *n* < 0, *m* is negative because *m* lies between 0 and *n*, and *u*[*m*] = 0. Therefore, - -$$ -c[n] = \begin{cases} \sum_{m=0}^{n} (0.8)^m (0.3)^{n-m} & n \ge 0\\ 0 & n < 0 \end{cases} -$$ - -or - -$$ -c[n] = (0.3)^n \sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m u[n] -$$ - -This is a geometric progression with common ratio (0.8/0.3). From Sec. B.8-3 we have - -$$ -c[n] = (0.3)^n \frac{(0.8)^{n+1} - (0.3)^{n+1}}{(0.3)^n (0.8 - 0.3)} u[n] -$$ - -= 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n] - -### **DR ILL 3.15 Convolution of Causal Signals** - -Show that (0.8)*nu*[*n*] ∗ *u*[*n*] = 5[1−(0.8)*n*+1]*u*[*n*]. - -### CONVOLUTION SUM FROM A TABLE - -Just as in the continuous-time case, we have prepared a table (Table 3.1) from which convolution sums may be determined directly for a variety of signal pairs. For example, the convolution in Ex. 3.20 can be read directly from this table (pair 4) as - -$$ -(0.8)^n u[n] * (0.3)^n u[n] = \frac{(0.8)^{n+1} - (0.3)^{n+1}}{0.8 - 0.3} u[n] = 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n] -$$ - -We shall demonstrate the use of the convolution table in the following example. - -| No. | x1[n] | x2[n] | x1[n]
∗ x2[n]
= x2[n]
∗ x1[n] | -|-----|-----------------------------|---------------|---------------------------------------------------------------------------------------------------------| -| 1 | δ[n−k] | x[n] | x[n−k] | -| 2 | γ nu[n] | u[n] | 1−γ n+1
!
u[n]
1−γ | -| 3 | u[n] | u[n] | (n+1)u[n] | -| 4 | γ n
1 u[n] | γ n
2 u[n] | γ n+1
−γ n+1
1
2
u[n]
γ1
= γ2
γ1
−γ2 | -| 5 | u[n] | nu[n] | n(n+1)
u[n]
2 | -| 6 | γ nu[n] | nu[n] | γ (γ n −1)
!
+n(1−γ )
u[n]
(1−γ )2 | -| 7 | nu[n] | nu[n] | 1
6 n(n−1)(n+1)u[n] | -| 8 | γ nu[n] | γ nu[n] | (n+1)γ nu[n] | -| 9 | nγ n
1 u[n] | γ n
2 u[n] | !
γ1γ2
γ1
−γ2
γ n
2 −γ n
nγ n
u[n]
γ1
= γ2
1 +
1
−γ2)2
(γ1
γ2 | -| 10 | n cos(βn+θ
γ1
)u[n] | nu[n]
γ2 | 1
n+1 cos[β(n+1)+θ−φ]− γ2
n+1 cos(θ−φ)]u[n]
R[ γ1 | -| | | | 1/2
R =
2 +
2 −2 γ1 γ2
γ1
γ2
cosβ | -| | | | !
( γ1 sinβ)
φ = tan−1
( γ1 cosβ − γ2 ) | -| 11 | γ n
1 u[−(n+1)] | γ n
2 u[n] | γ2
γ1
γ n
γ n
2 u[n] +
1 u[−(n+1)]
γ1 > γ2
γ1
−γ2
γ1
−γ2 | - -**TABLE 3.1** Select Convolution Sums - -### **EXAMPLE 3.21 Convolution by Tables** - -Using Table 3.1, find the (zero-state) response *y*[*n*] of an LTID system described by the equation - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -if the input *x*[*n*] = 4−*nu*[*n*]. - -The input can be expressed as *x*[*n*] = 4−*nu*[*n*] = (1/4)*nu*[*n*] = (0.25)*nu*[*n*]. The unit impulse response of this system, obtained in Ex. 3.18, is - -$$ -h[n] = [(-0.2)^n + 4(0.8)^n]u[n] -$$ - -Therefore, - -$$ -y[n] = x[n] * h[n] -$$ - -= (0.25)nu[n] \* [(-0.2)nu[n] + 4(0.8)nu[n] -= (0.25)nu[n] \* (-0.2)nu[n] + (0.25)nu[n] \* 4(0.8)nu[n] - -We use pair 4 (Table 3.1) to find the foregoing convolution sums. - -$$ -y[n] = \left[\frac{(0.25)^{n+1} - (-0.2)^{n+1}}{0.25 - (-0.2)} + 4 \frac{(0.25)^{n+1} - (0.8)^{n+1}}{0.25 - 0.8}\right] u[n] -$$ - -= $(2.22[(0.25)^{n+1} - (-0.2)^{n+1}] - 7.27[(0.25)^{n+1} - (0.8)^{n+1}])u[n]$ -= $[-5.05(0.25)^{n+1} - 2.22(-0.2)^{n+1} + 7.27(0.8)^{n+1}]u[n]$ - -Recognizing that - -$$ -\gamma^{n+1} = \gamma(\gamma)^n -$$ - -we can express *y*[*n*] as - -$$ -y[n] = [-1.26(0.25)^{n} + 0.444(-0.2)^{n} + 5.81(0.8)^{n}]u[n] -$$ - -= [-1.26(4)-n + 0.444(-0.2)n + 5.81(0.8)n]u[n] - -### **DR ILL 3.16 Convolution by Tables** - -Use Table 3.1 to show that - -(a) -$$ -(0.8)^{n+1}u[n] * u[n] = 4[1 - 0.8(0.8)^n]u[n] -$$ - -**(b)** *n*3−*nu*[*n*] ∗ (0.2)*nu*[*n*] = 15 4 (0.2)*n* − 1− 2 3 *n* 3−*n u*[*n*] - -(c) -$$ -e^{-n}u[n] * 2^{-n}u[n] = \frac{2}{2-e} \left[ e^{-n} - \frac{e}{2}2^{-n} \right] u[n] -$$ - -### **EXAMPLE 3.22 Filtering Perspective of the Zero-State Response** - -Use the MATLAB filter command to compute and sketch the zero-state response for the system described by (*E*2 +0.5*E* −1)*y*[*n*] = (2*E*2 +6*E*)*x*[*n*] and the input *x*[*n*] = 4−*nu*[*n*]. - -We solve this problem using the same approach as Ex. 3.19. Although the input is bounded and quickly decays to zero, the system itself is unstable and an unbounded output results. - ->> n = (0:11); x = @(n) 4.^(-n).\*(n>=0); >> a = [1 0.5 -1]; b = [2 6 0]; y = filter(b,a,x(n)); >> clf; stem(n,y,'k'); xlabel('n'); ylabel('y[n]'); axis([-0.5 11.5 -20 25]); - -### RESPONSE TO COMPLEX INPUTS - -As in the case of real continuous-time systems, we can show that for an LTID system with real *h*[*n*], if the input and the output are expressed in terms of their real and imaginary parts, then the real part of the input generates the real part of the response and the imaginary part of the input generates the imaginary part. Thus, if - -$$ -x[n] = x_r[n] + jx_i[n] -$$ - and $y[n] = y_r[n] + jy_i[n]$ - -using the right-directed arrow to indicate the input–output pair, we can show that - -$$ -x_r[n] \Longrightarrow y_r[n] -$$ - and $x_i[n] \Longrightarrow y_i[n]$ (3.34) - -The proof is similar to that used to derive Eq. (2.31) for LTIC systems. - -### MULTIPLE INPUTS - -Multiple inputs to LTI systems can be treated by applying the superposition principle. Each input is considered separately, with all other inputs assumed to be zero. The sum of all these individual system responses constitutes the total system output when all the inputs are applied simultaneously. - -### **DR ILL 3.17 Response to Multiple Inputs** - -Show that the system described by *y*[*n*] − 0.6*y*[*n* − 1] − 0.16*y*[*n* − 2] = 5*x*[*n*] responds to input *x*[*n*] = δ[*n*] +4−*nu*[*n*] with output *y*[*n*] = [−1.26(4)−*n* +1.444(−0.2)*n* +9.81(0.8)*n*]*u*[*n*]. [*Hint:* Use the results of Exs. 3.18 and 3.21.] - -### **[3.8-1 Graphical Procedure for the Convolution Sum](#page-9-0)** - -The steps in evaluating the convolution sum are parallel to those followed in evaluating the convolution integral. The convolution sum of causal signals *x*[*n*] and *g*[*n*] is given by - -$$ -c[n] = \sum_{m=0}^{n} x[m]g[n-m] -$$ - -We first plot *x*[*m*] and *g*[*n* − *m*] as functions of *m* (not *n*), because the summation is over *m*. Functions *x*[*m*] and *g*[*m*] are the same as *x*[*n*] and *g*[*n*], plotted, respectively, as functions of *m* (see Fig. 3.22). The convolution operation can be performed as follows: - -- 1. Invert *g*[*m*] about the vertical axis (*m* = 0) to obtain *g*[−*m*] (Fig. 3.22d). Figure 3.22e shows both *x*[*m*] and *g*[−*m*]. -- 2. Shift *g*[−*m*] by *n* units to obtain *g*[*n* − *m*]. For *n* > 0, the shift is to the right (delay); for *n* < 0, the shift is to the left (advance). Figure 3.22f shows *g*[*n* − *m*] for *n* > 0; for *n* < 0, see Fig. 3.22g. -- 3. Next we multiply *x*[*m*] and *g*[*n*−*m*] and add all the products to obtain *c*[*n*]. The procedure is repeated for each value of *n* over the range −∞ to ∞. - -We shall demonstrate by an example the graphical procedure for finding the convolution sum. Although both the functions in this example are causal, this procedure is applicable to the general case. - -### **EXAMPLE 3.23 Graphical Procedure for the Convolution Sum** - -Find *c*[*n*] = *x*[*n*] ∗*g*[*n*], where *x*[*n*] and *g*[*n*] are depicted in Figs. 3.22a and 3.22b, respectively. - -We are given - -*x*[*n*] = (0.8) *n* and *g*[*n*] = (0.3) *n* - -Therefore, - -*x*[*m*] = (0.8) *m* and *g*[*n*−*m*] = (0.3) *n*−*m* - -**Figure 3.22** Graphical procedure to convolve *x*[*n*] and *g*[*n*]. - -### 290 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Figure 3.22f shows the general situation for *n* ≥ 0. The two functions *x*[*m*] and *g*[*n*−*m*] overlap over the interval 0 ≤ *m* ≤ *n*. Therefore, - -$$ -c[n] = \sum_{m=0}^{n} x[m]g[n-m] -$$ - -= $\sum_{m=0}^{n} (0.8)^m (0.3)^{n-m}$ -= $(0.3)^n \sum_{m=0}^{n} \left(\frac{0.8}{0.3}\right)^m$ -= $2[(0.8)^{n+1} - (0.3)^{n+1}]$ $n \ge 0$ (see Sec. B.8-3) - -For *n* < 0, there is no overlap between *x*[*m*] and *g*[*n*−*m*], as shown in Fig. 3.22g, so that - -*c*[*n*] = 0 *n* < 0 - -Combining pieces, we see that - -$$ -c[n] = 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n] -$$ - -which agrees with the result found earlier in Ex. 3.20. - -### **DR ILL 3.18 Graphical Procedure for the Convolution Sum** - -Find (0.8)*nu*[*n*] ∗ *u*[*n*] graphically and sketch the result. - -### **ANSWER** - -5(1−(0.8)*n*+1)*u*[*n*] - -### AN ALTERNATIVE FORM OF GRAPHICAL PROCEDURE: THE SLIDING-TAPE METHOD - -This algorithm is convenient when the sequences *x*[*n*] and *g*[*n*] are short or when they are available only in graphical form. The algorithm is basically the same as the graphical procedure in Fig. 3.22. The only difference is that instead of presenting the data as graphical plots, we display it as a sequence of numbers on tapes. Otherwise the procedure is the same, as will become clear in the following example. - -### **EXAMPLE 3.24 Sliding-Tape Method for the Convolution Sum** - -Use the sliding-tape method to convolve the two sequences *x*[*n*] and *g*[*n*] depicted in Figs. 3.23a and 3.23b, respectively. - -In this procedure we write the sequences *x*[*n*] and*g*[*n*] in the slots of two tapes: *x* tape and *g* tape (Fig. 3.23c). Now leave the *x* tape stationary (to correspond to *x*[*m*]). The *g*[−*m*] tape is obtained by inverting the *g*[*m*] tape about the origin (*m* = 0) so that the slots corresponding to *x*[0] and *g*[0] remain aligned (Fig. 3.23d). We now shift the inverted tape by *n* slots, multiply values on two tapes in adjacent slots, and add all the products to find *c*[*n*]. Figures 3.23d–3.23i show the cases for *n* = 0–5. Figures 3.23j, 3.23k, and 3.23l show the cases for *n* = −1,−2, and −3, respectively. - -For the case of *n* = 0, for example (Fig. 3.23d), - -$$ -c[0] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) = -3 -$$ - -For *n* = 1 (Fig. 3.23e), - -$$ -c[1] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) = -2 -$$ - -Similarly, - -$$ -c[2] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) = 0 -$$ - -\n -$$ -c[3] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) = 3 -$$ - -\n -$$ -c[4] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) + (4 \times 1) = 7 -$$ - -\n -$$ -c[5] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) + (4 \times 1) = 7 -$$ - -Figure 3.23i shows that *c*[*n*] = 7 for *n* ≥ 4. - -Similarly, we compute *c*[*n*] for negative *n* by sliding the tape backward, one slot at a time, as shown in the plots corresponding to *n* = −1, −2, and −3, respectively (Figs. 3.23j, 3.23k, and 3.23l). - -$$ -c[-1] = (-2 \times 1) + (-1 \times 1) = -3 -$$ - -\n -$$ -c[-2] = (-2 \times 1) = -2 -$$ - -\n -$$ -c[-3] = 0 -$$ - -Figure 3.23l shows that *c*[*n*] = 0 for *n* ≤ 3. Figure 3.23m shows the plot of *c*[*n*]. - -Rotate the *g* tape about the vertical axis as shown in (d) - -### **DR ILL 3.19 Sliding-Tape Method for the Convolution Sum** - -Use the graphical procedure of Ex. 3.24 (sliding-tape technique) to show that *x*[*n*] ∗ *g*[*n*] = *c*[*n*] in Fig. 3.24. Verify the width property of convolution. - -## **EXAMPLE 3.25 Convolution of Two Finite-Duration Signals Using MATLAB** - -For the signals *x*[*n*] and *g*[*n*] depicted in Fig. 3.24, use MATLAB to compute and plot *c*[*n*] = *x*[*n*] ∗ *g*[*n*]. - -### **[3.8-2 Interconnected Systems](#page-9-0)** - -As with continuous-time case, we can determine the impulse response of systems connected in parallel (Fig. 3.26a) and cascade (Figs. 3.26b, 3.26c). We can use arguments identical to those used for the continuous-time systems in Sec. 2.4-3 to show that if two LTID systems *S*1 and *S*2 with impulse responses *h*1[*n*] and *h*2[*n*], respectively, are connected in parallel, the composite parallel system impulse response is *h*1[*n*] + *h*2[*n*]. Similarly, if these systems are connected in cascade, the impulse response of the composite system is *h*1[*n*] ∗ *h*2[*n*]. Moreover, because *h*1[*n*] ∗ *h*2[*n*] = *h*2[*n*] ∗ *h*1[*n*], linear systems commute. Their orders can be interchanged without affecting the composite system behavior. - -**Figure 3.26** Interconnected systems. - -### INVERSE SYSTEMS - -If the two systems in cascade are the inverse of each other, with impulse responses *h*[*n*] and *hi*[*n*], respectively, then the impulse response of the cascade of these systems is *h*[*n*] ∗ *hi*[*n*]. But, the cascade of a system with its inverse is an identity system, whose output is the same as the input. Hence, the unit impulse response of an identity system is δ[*n*]. Consequently, - -$$ -h[n] * h_i[n] = \delta[n] -$$ - -As an example, we show that an accumulator system and a backward difference system are the inverse of each other. An accumulator system is specified by† - -$$ -y[n] = \sum_{k=-\infty}^{n} x[k] \tag{3.35} -$$ - -The backward difference system is specified by - -$$ -y[n] = x[n] - x[n-1] \tag{3.36} -$$ - -From Eq. (3.35), we find *h*acc[*n*], the impulse response of the accumulator, as - -$$ -h_{\text{acc}}[n] = \sum_{k=-\infty}^{n} \delta[k] = u[n] -$$ - -Similarly, from Eq. (3.36), *h*bdf[*n*], the impulse response of the backward difference system is given by - -$$ -h_{\text{bdf}}[n] = \delta[n] - \delta[n-1] -$$ - -We can verify that - -$$ -h_{\text{acc}} * h_{\text{bdf}} = u[n] * {\delta[n] - \delta[n-1]} = u[n] - u[n-1] = \delta[n] -$$ - -Roughly speaking, a discrete-time accumulator is analogous to a continuous-time integrator, and a backward difference system is analogous to a differentiator. We have already encountered examples of these systems in Exs. 3.8 and 3.9 (digital differentiator and integrator). - -### SYSTEM RESPONSE TO %*n k*=−∞ *x*[*k*] - -Figure 3.26d shows a cascade of two LTID systems: a system *S* with impulse response *h*[*n*], followed by an accumulator. Figure 3.26e shows a cascade of the same two systems in reverse order: an accumulator followed by *S*. In Fig. 3.26d, if the input *x*[*n*] to *S* results in the output *y*[*n*], then the output of the system in Fig. 3.26d is the %*y*[*k*]. In Fig. 3.26e, the output of the accumulator is the sum %*x*[*k*]. Because the output of the system in Fig. 3.26e is identical to that of system Fig. 3.26d, it follows that - -if -$$ -x[n] \Longrightarrow y[n] -$$ -, then $\sum_{k=-\infty}^{n} x[k] \Longrightarrow \sum_{k=-\infty}^{n} y[k]$ - -† Equations (3.35) and (3.36) are identical to Eqs. (3.10) and (3.8), respectively, with *T* = 1. - -#### 296 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -If we let *x*[*n*] = δ[*n*] and *y*[*n*] = *h*[*n*], we find that *g*[*n*], the unit step response of an LTID system with impulse response *h*[*n*], is given by - -$$ -g[n] = \sum_{k=-\infty}^{n} h[k] \tag{3.37} -$$ - -The reader can readily prove the inverse relationship - -$$ -h[n] = g[n] - g[n-1] -$$ - -### A VERY SPECIAL FUNCTION FOR LTID SYSTEMS: THE EVERLASTING EXPONENTIAL *zn* - -In Sec. 2.4-4, we showed that there exists one signal for which the response of an LTIC system is the same as the input within a multiplicative constant. The response of an LTIC system to an everlasting exponential input *est* is *H*(*s*)*est*, where *H*(*s*) is the system transfer function. We now show that for an LTID system, the same role is played by an everlasting exponential *zn*. The system response *y*[*n*] in this case is given by - -$$ -y[n] = h[n] * zn -$$ - -= -$$ -\sum_{m=-\infty}^{\infty} h[m] z^{n-m} -$$ - -= -$$ -zn \sum_{m=-\infty}^{\infty} h[m] z^{-m} -$$ - -For causal *h*[*n*], the limits on the sum on the right-hand side would range from 0 to ∞. In any case, this sum is a function of *z*. Assuming that this sum converges, let us denote it by *H*[*z*]. Thus, - -$$ -y[n] = H[z]z^n \tag{3.38} -$$ - -where - -$$ -H[z] = \sum_{m=-\infty}^{\infty} h[m]z^{-m} -$$ -\n(3.39) - -Equation (3.38) is valid only for values of *z* for which the sum on the right-hand side of Eq. (3.39) exists (converges). Note that *H*[*z*] is a constant for a given *z*. Thus, the input and the output are the same (within a multiplicative constant) for the everlasting exponential input *zn*. - -*H*[*z*], which is called the *transfer function* of the system, is a function of the complex variable *z*. An alternate definition of the transfer function *H*[*z*] of an LTID system from Eq. (3.38) is - -$$ -H[z] = \frac{\text{output signal}}{\text{input signal}} \bigg|_{\text{input} = \text{everlasting exponential } z^n} -$$ - (3.40) - -The transfer function is defined for, and is meaningful to, LTID systems only. It does not exist for nonlinear or time-varying systems in general. - -We repeat again that in this discussion we are talking of the everlasting exponential, which starts at *n* = −∞, not the causal exponential *znu*[*n*], which starts at *n* = 0. - -For a system specified by Eq. (3.20), the transfer function is given by - -$$ -H[z] = \frac{P[z]}{Q[z]} \tag{3.41} -$$ - -This follows readily by considering an everlasting input *x*[*n*] = *zn*. According to Eq. (3.40), the output is *y*[*n*] = *H*[*z*]*zn*. Substitution of this *x*[*n*] and *y*[*n*] in Eq. (3.20) yields - -$$ -H[z]\{Q[E]z^n\} = P[E]z^n -$$ - -Moreover, - -$$ -E^k z^n = z^{n+k} = z^k z^n -$$ - -Hence, - -$$ -P[E]z^n = P[z]z^n \qquad \text{and} \qquad Q[E]z^n = Q[z]z^n -$$ - -Consequently, - -$$ -H[z] = \frac{P[z]}{Q[z]} -$$ - -### **DR ILL 3.20 DT System Transfer Function** - -Show that the transfer function of the digital differentiator in Ex. 3.8 (big shaded block in Fig. 3.16b) is given by *H*[*z*] = (*z*−1)/*Tz*, and the transfer function of an unit delay, specified by *y*[*n*] = *x*[*n*−1], is given by 1/*z*. - -### **[3.8-3 Total Response](#page-9-0)** - -The total response of an LTID system can be expressed as a sum of the zero-input and zero-state responses: - -total response = -$$ -\underbrace{\sum_{j=1}^{N} c_j \gamma_j^n}_{\text{ZIR}} + \underbrace{x[n] * h[n]}_{\text{ZSR}} -$$ - -In this expression, the zero-input response should be appropriately modified for the case of repeated roots. We have developed procedures to determine these two components. From the system equation, we find the characteristic roots and characteristic modes. The zero-input response is a linear combination of the characteristic modes. From the system equation, we also determine *h*[*n*], the impulse response, as discussed in Sec. 3.7. Knowing *h*[*n*] and the input *x*[*n*], we find the zero-state response as the convolution of *x*[*n*] and *h*[*n*]. The arbitrary constants *c*1, *c*2,..., *cn* in the zero-input response are determined from the *n* initial conditions. For the system described by the equation - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -#### 298 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -with initial conditions *y*[−1] = 0, *y*[−2] = 25/4 and input *x*[*n*] = (4)−*nu*[*n*], we have determined the two components of the response in Exs. 3.13 and 3.21, respectively. From the results in these examples, the total response for *n* ≥ 0 is - -total response = -$$ -\underbrace{0.2(-0.2)^n + 0.8(0.8)^n}_{\text{ZIR}} + \underbrace{0.444(-0.2)^n + 5.81(0.8)^n - 1.26(4)^{-n}}_{\text{ZSR}} -$$ -(3.42) - -### NATURAL AND FORCED RESPONSE - -The characteristic modes of this system are (−0.2)*n* and (0.8)*n*. The zero-input response is made up of characteristic modes exclusively, as expected, but the characteristic modes also appear in the zero-state response. When all the characteristic mode terms in the total response are lumped together, the resulting component is the *natural response*. The remaining part of the total response that is made up of noncharacteristic modes is the *forced response*. For the present case, Eq. (3.42) yields - -total response = -$$ -\underbrace{0.644(-0.2)^n + 6.61(0.8)^n}_{\text{natural response}} + \underbrace{-1.26(4)^{-n}}_{\text{forced response}} \qquad n \ge 0 -$$ - -Just like differential equations, the classical solution to difference equations includes the natural and forced responses, a decomposition that lacks the engineering intuition and utility afforded by the zero-input and zero-state responses. The classical approach cannot separate the responses arising from internal conditions and external input. While the natural and forced solutions can be obtained from the zero-input and zero-state responses, the converse is not true. Further, the classical method is unable to express the system response to an input *x*[*n*] as an explicit function of *x*[*n*]. In fact, the classical method is restricted to a certain class of inputs and cannot handle arbitrary inputs as can the method to determine the zero-state response. For these (and other) reasons, we do not further detail the classical approach and its direct calculation of the forced and natural responses. - -## **[3.9 SYSTEM](#page-9-0) STABILITY** - -The concepts and criteria for the BIBO (external) stability and internal (asymptotic) stability for discrete-time systems are identical to those corresponding to continuous-time systems. The comments in Sec. 2.5 for LTIC systems concerning the distinction between external and internal stability are also valid for LTID systems. Let us begin with external (BIBO) stability. - -### **[3.9-1 External \(BIBO\) Stability](#page-9-0)** - -Recall that - -$$ -y[n] = h[n] * x[n] = \sum_{m=-\infty}^{\infty} h[m]x[n-m] -$$ - -and - -$$ -|y[n]| = \left|\sum_{m=-\infty}^{\infty} h[m]x[n-m]\right| \le \sum_{m=-\infty}^{\infty} |h[m]| |x[n-m]| -$$ - -If *x*[*n*] is bounded, then |*x*[*n*−*m*]| < *K*1 < ∞, and - -$$ -|y[n]| \leq K_1 \sum_{m=-\infty}^{\infty} |h[m]| -$$ - -Clearly the output is bounded if the summation on the right-hand side is bounded; that is, if - -$$ -\sum_{n=-\infty}^{\infty} |h[n]| < K_2 < \infty \tag{3.43} -$$ - -This is a sufficient condition for BIBO stability. We can show that this is also a necessary condition (see Prob. 3.9-1). Therefore, if the impulse response *h*[*n*] of an LTID system is absolutely summable, the system is (BIBO) stable. Otherwise it is unstable. - -All the comments about the nature of external and internal stability in Ch. 2 apply to discrete-time case. We shall not elaborate them further. - -### **[3.9-2 Internal \(Asymptotic\) Stability](#page-9-0)** - -For LTID systems, as in the case of LTIC systems, internal stability, called asymptotical stability or stability in the sense of Lyapunov (also the zero-input stability), is defined in terms of the zero-input response of a system. - -For an LTID system specified by a difference equation in the form of Eq. (3.15) [or Eq. (3.20)], the zero-input response consists of the characteristic modes of the system. The mode corresponding to a characteristic root γ is γ *n*. To be more general, let γ be complex so that - -$$ -\gamma = |\gamma|e^{i\beta} -$$ - and $\gamma^n = |\gamma|^n e^{i\beta n}$ - -Since the magnitude of *ej*β*n* is always unity regardless of the value of *n*, the magnitude of γ *n* is |γ | *n*. Therefore, - -if -$$ -|\gamma| < 1 -$$ -, then $\gamma^n \to 0$ as $n \to \infty$ -if $|\gamma| > 1$ , then $\gamma^n \to \infty$ as $n \to \infty$ -and if $|\gamma| = 1$ , then $|\gamma|^n = 1$ for all *n* - -The characteristic modes corresponding to characteristic roots at various locations in the complex plane appear in Fig. 3.27. - -These results can be grasped more effectively in terms of the location of characteristic roots in the complex plane. Figure 3.28 shows a circle of unit radius, centered at the origin in a complex plane. Our discussion shows that if all characteristic roots of the system lie inside the *unit circle*, |γ*i*| < 1 for all *i* and the system is asymptotically stable. On the other hand, even if one characteristic root lies outside the unit circle, the system is unstable. If none of the characteristic - -**Figure 3.27** Characteristic roots locations and the corresponding characteristic modes. - -(d) - -(a) - -(b) - -roots lie outside the unit circle, but some simple (unrepeated) roots lie on the circle itself, the system is marginally stable. If two or more characteristic roots coincide on the unit circle (repeated roots), the system is unstable. The reason is that for repeated roots, the zero-input response is of the form *nr*−1γ *n*, and if |γ | = 1, then |*nr*−1γ *n*| = *nr*−1 → ∞ as *n* → ∞. † Note, however, that repeated roots inside the unit circle do not cause instability. - -**Figure 3.28** Characteristic root locations and system stability. - -To summarize: - -- 1. An LTID system is asymptotically stable if, and only if, all the characteristic roots are inside the unit circle. The roots may be simple or repeated. -- 2. An LTID system is unstable if, and only if, either one or both of the following conditions exist: (i) at least one root is outside the unit circle; (ii) there are repeated roots on the unit circle. -- 3. An LTID system is marginally stable if and only if there are no roots outside the unit circle and there are some unrepeated roots on the unit circle. - -### **[3.9-3 Relationship Between BIBO and Asymptotic Stability](#page-9-0)** - -For LTID systems, the relation between the two types of stability is similar to those in LTIC systems. For a system specified by Eq. (3.15), we can readily show that if a characteristic root γ*k* - -† If the development of discrete-time systems is parallel to that of continuous-time systems, we wonder why the parallel breaks down here. Why, for instance, are LHP and RHP not the regions demarcating stability and instability? The reason lies in the form of the characteristic modes. In continuous-time systems, we chose the form of characteristic mode as *e*λ*it* . In discrete-time systems, for computational convenience, we choose the form to be γ *n i* . Had we chosen this form to be *e*λ*in* where γ*i* = *e*λ*i* , then the LHP and RHP (for the location of λ*i*) again would demarcate stability and instability. The reason is that if γ = *e*λ, |γ | = 1 implies |*e*λ| = 1, and therefore λ = *j*ω. This shows that the unit circle in γ plane maps into the imaginary axis in the λ plane. - -### 302 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -is inside the unit circle, the corresponding mode γ *n k* is absolutely summable. In contrast, if γ*k* lies outside the unit circle, or on the unit circle, γ *n k* is not absolutely summable.† - -This means that an asymptotically stable system is BIBO-stable. Moreover, a marginally stable or asymptotically unstable system is BIBO-unstable. The converse is not necessarily true. The stability picture portrayed by the external description is of questionable value. BIBO (external) stability cannot ensure internal (asymptotic) stability, as the following example shows. - -### **EXAMPLE 3.26 A BIBO-Stable but Asymptotically Unstable System** - -An LTID systems consists of two subsystems *S*1 and *S*2 in cascade (Fig. 3.29). The impulse response of these systems are *h*1[*n*] and *h*2[*n*], respectively, given by - -> *h*1[*n*] = 4δ[*n*] −3(0.5) *n u*[*n*] and *h*2[*n*] = 2*n u*[*n*] - -Investigate the BIBO and asymptotic stability of the composite system. - -The composite system impulse response *h*[*n*] is given by - -$$ -h[n] = h_1[n] * h_2[n] = h_2[n] * h_1[n] = 2nu[n] * (4\delta[n] - 3(0.5)nu[n]) -$$ - -= 4(2)nu[n] - 3\left[\frac{2^{n+1} - (0.5)n+1}{2 - 0.5}\right]u[n] -= (0.5)nu[n] - -If the composite cascade system were to be enclosed in a black box with only the input and the output terminals accessible, any measurement from these external terminals would show that the impulse response of the system is (0.5)*nu*[*n*], without any hint of the unstable system sheltered inside the composite system. - -† This conclusion follows from the fact that (see Sec. B.8-3) - -$$ -\sum_{n=-\infty}^{\infty} |\gamma_k^n| u[n] = \sum_{n=0}^{\infty} |\gamma_k|^n = \frac{1}{1 - |\gamma_k|} \qquad |\gamma_k| < 1 -$$ - -Moreover, if |γ | ≥ 1, the sum diverges and goes to ∞. These conclusions are valid also for the modes of the form *nr* γ *n k* . - -The composite system is BIBO-stable because its impulse response (0.5)*nu*[*n*] is absolutely summable. However, the system *S*2 is asymptotically unstable because its characteristic root, 2, lies outside the unit circle. This system will eventually burn out (or saturate) because of the unbounded characteristic response generated by intended or unintended initial conditions, no matter how small. - -The system is asymptotically unstable, though BIBO-stable. This example shows that BIBO stability does not necessarily ensure asymptotic stability when a system is uncontrollable, unobservable, or both. The internal and the external descriptions of a system are equivalent only when the system is controllable and observable. In such a case, BIBO stability means the system is asymptotically stable, and vice versa. - -Fortunately, uncontrollable or unobservable systems are not common in practice. Henceforth, in determining system stability, we shall assume that unless otherwise mentioned, the internal and the external descriptions of the system are equivalent, implying that the system is controllable and observable. - -### **EXAMPLE 3.27 Investigating Asymptotic and BIBO Stability** - -Determine the internal and external stability of systems specified by the following equations. In each case plot the characteristic roots in the complex plane. - -- **(a)** *y*[*n*+2] +2.5*y*[*n*+1] +*y*[*n*] = *x*[*n*+1] −2*x*[*n*] -- **(b)** *y*[*n*] −*y*[*n*−1] +0.21*y*[*n*−2] = 2*x*[*n*−1] +3*x*[*n*−2] -- **(c)** *y*[*n*+3] +2*y*[*n*+2] + 3 2 *y*[*n*+1] + 1 2 *y*[*n*] = *x*[*n*+1] -- **(d)** (*E*2 −*E* +1)2*y*[*n*] = (3*E* +1)*x*[*n*] - -**(a)** The characteristic polynomial is - -$$ -\gamma^2 + 2.5\gamma + 1 = (\gamma + 0.5)(\gamma + 2) -$$ - -The characteristic roots are −0.5 and −2. Because | − 2| > 1 (−2 lies outside the unit circle), the system is BIBO-unstable and also asymptotically unstable (Fig. 3.30a). - -**(b)** The characteristic polynomial is - -$$ -\gamma^2 - \gamma + 0.21 = (\gamma - 0.3)(\gamma - 0.7) -$$ - -The characteristic roots are 0.3 and 0.7, both of which lie inside the unit circle. The system is BIBO-stable and asymptotically stable (Fig. 3.30b). - -**(c)** The characteristic polynomial is - -$$ -\gamma^{3} + 2\gamma^{2} + \frac{3}{2}\gamma + \frac{1}{2} = (\gamma + 1)(\gamma^{2} + \gamma + \frac{1}{2}) = (\gamma + 1)(\gamma + 0.5 - j0.5)(\gamma + 0.5 + j0.5) -$$ - -### 304 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -The characteristic roots are −1, −0.5 ± *j*0.5 (Fig. 3.30c). One of the characteristic roots is on the unit circle and the remaining two roots are inside the unit circle. The system is BIBO-unstable but marginally stable. - -**(d)** The characteristic polynomial is - -$$ -(\gamma^2 - \gamma + 1)^2 = \left(\gamma - \frac{1}{2} - j\frac{\sqrt{3}}{2}\right)^2 \left(\gamma - \frac{1}{2} + j\frac{\sqrt{3}}{2}\right)^2 -$$ - -The characteristic roots are (1/2)±*j*( √3/2) = 1*e*±*j*(π/3) repeated twice, and they lie on the unit circle (Fig. 3.30d). The system is BIBO-unstable and asymptotically unstable. - -### **DR ILL 3.21 Assessing Stability by Characteristic Roots** - -Using the complex plane, locate the characteristic roots of the following systems, and use the characteristic root locations to determine external and internal stability of each system. - -**(a)** (*E* +1)(*E*2 +6*E* +25)*y*[*n*] = 3*Ex*[*n*] - -**(b)** (*E* −1)2(*E* +0.5)*y*[*n*] = (*E*2 +2*E* +3)*x*[*n*] - -### **ANSWERS** - -Both systems are BIBO-and asymptotically unstable. - -## **[3.10 INTUITIVE](#page-9-0) INSIGHTS INTO SYSTEM BEHAVIOR** - -The intuitive insights into the behavior of continuous-time systems and their qualitative proofs, discussed in Sec. 2.6, also apply to discrete-time systems. For this reason, we shall merely mention here without discussion some of the insights presented in Sec. 2.6. - -The system's entire (zero-input and zero-state) behavior is strongly influenced by the characteristic roots (or modes) of the system. The system responds strongly to input signals similar to its characteristic modes and poorly to inputs very different from its characteristic modes. In fact, when the input is a characteristic mode of the system, the response goes to infinity, provided the mode is a nondecaying signal. This is the resonance phenomenon. The width of an impulse response *h*[*n*] indicates the response time (time required to respond fully to an input) of the system. It is the time constant of the system.† Discrete-time pulses are generally dispersed when passed through a discrete-time system. The amount of dispersion (or spreading out) is equal to the system time constant (or width of *h*[*n*]). The system time constant also determines the rate at which the system can transmit information. A smaller time constant corresponds to a higher rate of information transmission, and vice versa. We keep in mind that concepts such as time constant and pulse dispersion only coarsely illustrate system behavior. Let us illustrate these ideas with an example. - -### **EXAMPLE 3.28 Intuitive Insights into Lowpass DT System Behavior** - -Determine the time constant, rise time, pulse dispersion, and filter characteristics of a lowpass DT system with impulse response *h*[*n*] = 2(0.6)*nu*[*n*]. - -† This part of the discussion applies to systems with impulse response *h*[*n*] that is a mostly positive (or mostly negative) pulse. - -### 306 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Since *h*[*n*] resembles a single, mostly positive pulse, we know that the DT system is lowpass. Similar to the CT case shown in Sec. 2.6, we can determine the time constant *Th* as the width of a rectangle that approximates *h*[*n*]. This rectangle possesses the same peak height and total sum (area), as does *h*[*n*]. The peak of *h*[*n*] is 2, and the total sum (area) is - -$$ -\sum_{n=0}^{\infty} 2(0.6)^n = 2\frac{1-0}{1-0.6} = 5 -$$ - -Since the width of a DT signal is 1 less than its length, we see that the time constant *Th* (rectangle width) is - -$$ -T_h = \text{rectangle width} = \frac{\text{area}}{\text{height}} - 1 = \frac{5}{2} - 1 = 1.5 \text{ samples} -$$ - -Since time constant, rise time, pulse dispersion are all given by the same value, we see that - -time constant = rise time = pulse dispersion = *Th* = 1.5 samples - -The approximate cutoff frequency of our DT system can be determined as the frequency of a DT sinusoid whose period equals the length of the rectangle approximation to *h*[*n*]. That is, - -cutoff frequency = -$$ -\frac{1}{T_h + 1} = \frac{2}{5} -$$ - cycles/sample - -Equivalently, we can express the cutoff frequency as 4π/5 radians/sample. - -Notice that *Th* is not an integer and thus lacks a clear physical meaning for our DT system. How, for example, can it take 1.5 samples for our DT system to fully respond to an input? We can put our minds at ease by remembering the approximate nature of *Th*, which is meant to provide only a rough understanding of system behavior. - -## **[3.11 MATLAB: DISCRETE-TIME](#page-9-0) SIGNALS AND SYSTEMS** - -MATLAB is naturally and ideally suited to discrete-time signals and systems. Many special functions are available for discrete-time data operations, including the stem, filter, and conv commands. In this section, we investigate and apply these and other commands. - -### **[3.11-1 Discrete-Time Functions and Stem Plots](#page-9-0)** - -Consider the discrete-time function *f*[*n*] = *e*−*n*/5 cos(π*n*/5)*u*[*n*]. In MATLAB, there are many ways to represent *f*[*n*] including M-files or, for particular *n*, explicit command line evaluation. In this example, however, we use an anonymous function. - ->> f = @(n) exp(-n/5).\*cos(pi\*n/5).\*(n>=0); - -A true discrete-time function is undefined (or zero) for noninteger *n*. Although anonymous function f is intended as a discrete-time function, its present construction does not restrict *n* to be integer, and it can therefore be misused. For example, MATLAB dutifully returns 0.8606 to f(0.5) when a NaN (not-a-number) or zero is more appropriate. The user is responsible for appropriate function use. - -Next, consider plotting the discrete-time function *f*[*n*] over (−10 ≤ *n* ≤ 10). The stem command simplifies this task. - -``` ->> n = (-10:10)'; ->> stem(n,f(n),'k'); ->> xlabel('n'); ylabel('f[n]'); -``` - -Here, stem operates much like the plot command: dependent variable f(n) is plotted against independent variable n with black lines. The stem command emphasizes the discrete-time nature of the data, as Fig. 3.31 illustrates. - -For discrete-time functions, the operations of shifting, inversion, and scaling can have surprising results. Compare *f*[−2*n*] with *f*[−2*n* + 1]. Contrary to the continuous case, the second is not a shifted version of the first. We can use separate subplots, each over (−10 ≤ *n* ≤ 10), to help illustrate this fact. Notice that unlike the plot command, the stem command cannot simultaneously plot multiple functions on a single axis; overlapping stem lines would make such plots difficult to read anyway. - -``` ->> subplot(2,1,1); stem(n,f(-2*n),'k'); ylabel('f[-2n]'); ->> subplot(2,1,2); stem(n,f(-2*n+1),'k'); ylabel('f[-2n+1]'); xlabel('n'); -``` - -The results are shown in Fig. 3.32. Interestingly, the original function *f*[*n*] can be recovered by interleaving samples of *f*[−2*n*] and *f*[−2*n*+1] and then time-reflecting the result. - -Care must always be taken to ensure that MATLAB performs the desired computations. Our anonymous function f is a case in point: although it correctly downsamples, it does not properly upsample (see Prob. 3.11-2). MATLAB does what it is told, but it is not always told how to do everything correctly! - -**Figure 3.31** *f*[*n*] over (−10 ≤ *n* ≤ 10). - -**Figure 3.32** *f*[−2*n*] and *f*[−2*n*+1] over (−10 ≤ *n* ≤ 10). - -### **[3.11-2 System Responses Through Filtering](#page-9-0)** - -MATLAB's filter command provides an efficient way to evaluate the system response of a constant coefficient linear difference equation represented in delay form as - -$$ -\sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{N} b_k x[n-k] -$$ -\n(3.44) - -In the simplest form, filter requires three input arguments: a length-(*N* + 1) vector of feedforward coefficients [*b*0,*b*1,...,*bN*], a length-(*N* + 1) vector of feedback coefficients [*a*0,*a*1,...,*aN*], and an input vector.† Since no initial conditions are specified, the output corresponds to the system's zero-state response. - -To serve as an example, consider a system described by *y*[*n*]−*y*[*n*−1]+*y*[*n*−2] = *x*[*n*]. When *x*[*n*] = δ[*n*], the zero-state response is equal to the impulse response *h*[*n*], which we compute over (0 ≤ *n* ≤ 30). - ->> b = [1 0 0]; a = [1 -1 1]; >> n = (0:30)'; delta = @(n) 1.0.\*(n==0); >> h = filter(b,a,delta(n)); >> clf; stem(n,h,'k'); axis([-.5 30.5 -1.1 1.1]); >> xlabel('n'); ylabel('h[n]'); - -† It is important to pay close attention to the inevitable notational differences found throughout engineering documents. In MATLAB help documents, coefficient subscripts begin at 1 rather than 0 to better conform with MATLAB indexing conventions. That is, MATLAB labels *a*0 as a(1), *b*0 as b(1), and so forth. - -**Figure 3.33** *h*[*n*] for *y*[*n*] −*y*[*n*−1] +*y*[*n*−2] = *x*[*n*]. - -**Figure 3.34** Resonant zero-state response *y*[*n*] for *x*[*n*] = cos(2π*n*/6)*u*[*n*]. - -As shown in Fig. 3.33, *h*[*n*] appears to be (*N*0 = 6)-periodic for *n* ≥ 0. Since periodic signals are not absolutely summable, %∞ *n*=−∞ |*h*[*n*]| is not finite and the system is not BIBO-stable. Furthermore, the sinusoidal input *x*[*n*] = cos(2π*n*/6)*u*[*n*], which is (*N*0 = 6)-periodic for *n* ≥ 0, should generate a resonant zero-state response. - -``` ->> x = @(n) cos(2*pi*n/6).*(n>=0); ->> y = filter(b,a,x(n)); ->> stem(n,y,'k'); xlabel('n'); ylabel('y[n]'); -``` - -The response's linear envelope, shown in Fig. 3.34, confirms a resonant response. The characteristic equation of the system is γ 2 − γ + 1, which has roots γ = *e*±*j*π/3. Since the input *x*[*n*] = cos(2π*n*/6)*u*[*n*] = (1/2)(*ej*π*n*/3 + *e*−*j*π*n*/3)*u*[*n*] coincides with the characteristic roots, a resonant response is guaranteed. - -By adding initial conditions, the filter command can also compute a system's zero-input response and total response. Continuing the preceding example, consider finding the zero-input response for *y*[−1] = 1 and *y*[−2] = 2 over (0 ≤ *n* ≤ 30). - -``` ->> z_i = filtic(b,a,[1 2]); ->> y_0 = filter(b,a,zeros(size(n)),z_i); ->> stem(n,y_0,'k'); xlabel('n'); ylabel('y_{0} [n]'); ->> axis([-0.5 30.5 -2.1 2.1]); -``` - -**Figure 3.35** Zero-input response *y*0[*n*] for *y*[−1] = 1 and *y*[−2] = 2. - -There are many physical ways to implement a particular equation. MATLAB implements Eq. (3.44) by using the popular direct form II transposed structure.† Consequently, initial conditions must be compatible with this implementation structure. The signal-processing toolbox function filtic converts the traditional *y*[−1], *y*[−2], ..., *y*[−*N*] initial conditions for use with the filter command. An input of zero is created with the zeros command. The dimensions of this zero input are made to match the vector n by using the size command. Finally, \_{ } forces subscript text in the graphics window, and ^{ } forces superscript text. The results are shown in Fig. 3.35. - -Given *y*[−1] = 1 and *y*[−2] = 2 and an input *x*[*n*] = cos(2π*n*/6)*u*[*n*], the total response is easy to obtain with the filter command. - -$$ -\Rightarrow y\_total = filter(b,a,x(n),z_i); -$$ - -Summing the zero-state and zero-input response gives the same result. Computing the total absolute error provides a check. - -``` ->> sum(abs(y_total-(y + y_0))) - ans = 1.8430e-014 -``` - -Within computer round-off, both methods return the same sequence. - -### **[3.11-3 A Custom Filter Function](#page-9-0)** - -The filtic command is available only if the signal-processing toolbox is installed. To accommodate installations without the signal-processing toolbox and to help develop your MATLAB skills, consider writing a function similar in syntax to filter that directly uses the ICs *y*[−1], *y*[−2], ..., *y*[−*N*]. Normalizing *a*0 = 1 and solving Eq. (3.44) for *y*[*n*] yield - -$$ -y[n] = \sum_{k=0}^{N} b_k x[n-k] - \sum_{k=1}^{N} a_k y[n-k] -$$ - -This recursive form provides a good basis for our custom filter function. - -† Implementation structures, such as direct form II transposed, are discussed in Ch. 4. - -``` -function [y] = CH3MP1(b,a,x,yi); -% CH3MP1.m : Chapter 3, MATLAB Program 1 -% Function M-file filters data x to create y -% INPUTS: b = vector of feedforward coefficients -% a = vector of feedback coefficients -% x = input data vector -% yi = vector of initial conditions [y[-1], y[-2], ...] -% OUTPUTS: y = vector of filtered output data -yi = flipud(yi(:)); % Properly format IC's. -y = [yi;zeros(length(x),1)]; % Preinitialize y, beginning with IC's. -x = [zeros(length(yi),1);x(:)]; % Append x with zeros to match size of y. -b = b/a(1);a = a/a(1); % Normalize coefficients. -for n = length(yi)+1:length(y), - for nb = 0:length(b)-1, - y(n) = y(n) + b(nb+1)*x(n-nb); % Feedforward terms. - end - for na = 1:length(a)-1, - y(n) = y(n) - a(na+1)*y(n-na); % Feedback terms. - end -end -y = y(length(yi)+1:end); % Strip off IC's for final output. -``` - -Most instructions in CH3MP1 have been discussed; now we turn to the flipud instruction. The flip up-down command flipud reverses the order of elements in a column vector. Although not used here, the flip left-right command fliplr reverses the order of elements in a row vector. Note that typing help *filename* displays the first contiguous set of comment lines in an M-file. Thus, it is good programming practice to document M-files, as in CH3MP1, with an initial block of clear comment lines. - -As an exercise, the reader should verify that CH3MP1 correctly computes the impulse response *h*[*n*], the zero-state response *y*[*n*], the zero-input response *y*0[*n*], and the total response *y*[*n*]+*y*0[*n*]. - -### **[3.11-4 Discrete-Time Convolution](#page-9-0)** - -Convolution of two finite-duration discrete-time signals is accomplished by using the conv command. For example, the discrete-time convolution of two length-4 rectangular pulses, *g*[*n*] = (*u*[*n*]−*u*[*n*−4])∗(*u*[*n*]−*u*[*n*−4]), is a length-(4+4−1=7) triangle. Representing *u*[*n*]−*u*[*n*−4] by the vector [1, 1, 1, 1], the convolution is computed by - ->> conv([1 1 1 1],[1 1 1 1]) ans = 1 2 3 4 3 2 1 - -Notice that (*u*[*n*+4] −*u*[*n*]) ∗ (*u*[*n*] −*u*[*n*−4]) is also computed by conv([1 1 1 1],[1 1 1 1]) and obviously yields the same result. The difference between these two cases is the regions of support: (0 ≤ *n* ≤ 6) for the first and (−4 ≤ *n* ≤ 2) for the second. Although the conv command - -### 312 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -does not compute the region of support, it is relatively easy to obtain. If vector w begins at *n* = *nw* and vector v begins at *n* = *nv*, then conv(w,v) begins at *n* = *nw* +*nv*. - -In general, the conv command cannot properly convolve infinite-duration signals. This is not too surprising, since computers themselves cannot store an infinite-duration signal. For special cases, however, conv can correctly compute a portion of such convolution problems. Consider the common case of convolving two causal signals. By passing the first *N* samples of each, conv returns a length-(2*N* − 1) sequence. The first *N* samples of this sequence are valid; the remaining *N* −1 samples are not. - -To illustrate this point, reconsider the zero-state response *y*[*n*] over (0 ≤ *n* ≤ 30) for system *y*[*n*]−*y*[*n*−1]+*y*[*n*−2] = *x*[*n*] given input *x*[*n*] = cos(2π*n*/6)*u*[*n*]. The results obtained by using a filtering approach are shown in Fig. 3.34. - -The response can also be computed using convolution according to *y*[*n*] = *h*[*n*] ∗ *x*[*n*]. The impulse response of this system is† - -$$ -h[n] = \left\{ \cos\left(\frac{\pi n}{3}\right) + \frac{1}{\sqrt{3}} \sin\left(\frac{\pi n}{3}\right) \right\} u[n] -$$ - -Both *h*[*n*] and *x*[*n*] are causal and have infinite duration, so conv can be used to obtain a portion of the convolution. - -``` ->> u = @(n) 1.0.*(n>=0); h = @(n) (cos(pi*n/3)+sin(pi*n/3)/sqrt(3)).*u(n); ->> y = conv(h(n),x(n)); ->> stem([0:60],y,'k'); xlabel('n'); ylabel('y[n]'); -``` - -The conv output is fully displayed in Fig. 3.36. As expected, the results are correct over (0 ≤ *n* ≤ 30). The remaining values are clearly incorrect; the output envelope should continue to grow, not decay. Normally, these incorrect values are not displayed. - ->> stem(n,y(1:31),'k'); xlabel('n'); ylabel('y[n]'); - -The resulting plot is identical to Fig. 3.34. - -**Figure 3.36** *y*[*n*] for *x*[*n*] = cos(2π*n*/6)*u*[*n*] computed with conv. - -† Techniques to analytically determine *h*[*n*] are presented in Ch. 5. - -## **[3.12 APPENDIX: IMPULSE](#page-9-0) RESPONSE FOR A SPECIAL CASE** - -When *aN* =0, *A*0 =*bN*/*aN* becomes indeterminate, and the procedure needs to be modified slightly. When *aN* = 0, *Q*[*E*] can be expressed as *EQ*ˆ [*E*], and Eq. (3.26) can be expressed as - -$$ -E\tilde{Q}[E]h[n] = P[E]\delta[n] = P[E]\{E\delta[n-1]\} = EP[E]\delta[n-1] -$$ - -Hence, - -$$ -\hat{Q}[E]h[n] = P[E]\delta[n-1] -$$ - -In this case the input vanishes not for *n* ≥ 1, but for *n* ≥ 2. Therefore, the response consists not only of the zero-input term and an impulse *A*0δ[*n*] (at *n* = 0), but also of an impulse *A*1δ[*n*−1] (at *n* = 1). Therefore, - -$$ -h[n] = A_0 \delta[n] + A_1 \delta[n-1] + y_c[n]u[n] -$$ - -We can determine the unknowns *A*0, *A*1, and the *N* − 1 coefficients in *yc*[*n*] from the *N* + 1 number of initial values *h*[0], *h*[1], ... , *h*[*N*], determined as usual from the iterative solution of the equation *Q*[*E*]*h*[*n*] = *P*[*E*]δ[*n*]. † Similarly, if *aN* = *aN*−1 = 0, we need to use the form *h*[*n*] = *A*0δ[*n*]+*A*1δ[*n*−1]+*A*2δ[*n*−2]+*yc*[*n*]*u*[*n*]. The *N* +1 unknown constants are determined from the *N* +1 values *h*[0], *h*[1], ... , *h*[*N*], determined iteratively, and so on. - -## **[3.13 SUMMARY](#page-9-0)** - -This chapter discusses time-domain analysis of LTID (linear, time-invariant, discrete-time) systems. The analysis is parallel to that of LTIC systems, with some minor differences. Discrete-time systems are described by difference equations. For an *N*th-order system, *N* auxiliary conditions must be specified for a unique solution. Characteristic modes are discrete-time exponentials of the form γ *n* corresponding to an unrepeated root γ , and the modes are of the form *ni* γ *n* corresponding to a repeated root γ . - -The unit impulse function δ[*n*] is a sequence of a single number of unit value at *n* = 0. The unit impulse response *h*[*n*] of a discrete-time system is a linear combination of its characteristic modes.‡ - -The zero-state response (response due to external input) of a linear system is obtained by breaking the input into impulse components and then adding the system responses to all the impulse components. The sum of the system responses to the impulse components is in the form of a sum, known as the convolution sum, whose structure and properties are similar to the convolution integral. The system response is obtained as the convolution sum of the input *x*[*n*] with the system's impulse response *h*[*n*]. Therefore, the knowledge of the system's impulse response allows us to determine the system response to any arbitrary input. - -LTID systems have a very special relationship to the everlasting exponential signal *zn* because the response of an LTID system to such an input signal is the same signal within a multiplicative - -† *Q*ˆ [γ ] is now an (*N* −1)-order polynomial. Hence there are only *N* −1 unknowns in *yc*[*n*]. ‡ There is a possibility of an impulse δ[*n*] in addition to characteristic modes. - -### 314 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -constant. The response of an LTID system to the everlasting exponential input *zn* is *H*[*z*]*zn*, where *H*[*z*] is the transfer function of the system. - -The external stability criterion, the bounded-input/bounded-output (BIBO) stability criterion, states that a system is stable if and only if every bounded input produces a bounded output. Otherwise the system is unstable. - -The internal stability criterion can be stated in terms of the location of characteristic roots of the system as follows: - -- 1. An LTID system is asymptotically stable if and only if all the characteristic roots are inside the unit circle. The roots may be repeated or unrepeated. -- 2. An LTID system is unstable if and only if either one or both of the following conditions exist: (i) at least one root is outside the unit circle; (ii) there are repeated roots on the unit circle. -- 3. An LTID system is marginally stable if and only if there are no roots outside the unit circle and some unrepeated roots on the unit circle. - -An asymptotically stable system is always BIBO-stable. The converse is not necessarily true. - -## **[PROBLEMS](#page-9-0)** - -- **3.1-1** Find the energy of the signals depicted in Fig. P3.1-1. -- **3.1-2** Find the power of the signals illustrated in Fig. P3.1-2. -- **3.1-3** Show that the power of a signal *Dej*(2π/*N*0)*n* is |*D*| 2. Hence, show that the power of a signal *x*[*n*] =%*N*0−1 *r*=0 *Drejr*(2π/*N*0)*n* is *Px* =%*N*0−1 *r*=0 |*Dr*| 2. Use the fact that - -$$ -\sum_{k=0}^{N_0-1} e^{j(r-m)2\pi k/N_0} = \begin{cases} N_0 & r=m\\ 0 & \text{otherwise} \end{cases} -$$ - -- **3.1-4** (a) Determine even and odd components of the signal *x*[*n*] = (0.8)*nu*[*n*]. - - (b) Show that the energy of *x*[*n*] is the sum of energies of its odd and even components found in part (a). - - (c) Generalize the result in part (b) for any finite energy signal. -- **3.1-5** (a) If *xe*[*n*] and *xo*[*n*] are the even and the odd components of causal energy signal *x*[*n*], then determine *Exe* and *Ex*0 , and show that *Exe* +*Ex*0 = *Ex*. - - (b) Show that the cross-energy of *xe* and *xo* is zero, that is, - -$$ -\sum_{n=-\infty}^{\infty} x_e[n]x_o[n] = 0 -$$ - -**3.1-6** Define - -$$ -x[n] = \begin{cases} \left(\frac{1}{3}\right)^n & n \ge 0\\ A^n & n < 0 \end{cases} -$$ - -- (a) Determine the energy *Ex* and power *Px* of *x*[*n*] if *A* = 1 2 . -- (b) Determine the energy *Ex* and power *Px* of *x*[*n*] if *A* = 1. -- (c) Determine the energy *Ex* and power *Px* of *x*[*n*] if *A* = 2. -- **3.1-7** Determine the energy *Ex* and power *Px* of the complex DT signal *x*[*n*] = Re\* 3(*ej*π/4)*n* + -- **3.2-1** If the energy of a signal *x*[*n*] is *Ex*, then find the energy of the following: - - (a) *x*[−*n*] - - (b) *x*[*n*−*m*] - - (c) *x*[*m*−*n*] - - (d) *Kx*[*n*] (*m* integer and *K* constant) -- **3.2-2** If the power of a periodic signal *x*[*n*] is *Px*, find and comment on the powers and the rms values of the following: - - (a) −*x*[*n*] - - (b) *x*[−*n*] - - (c) *x*[*n*−*m*] (*m* integer) - - (d) *cx*[*n*] - - (e) *x*[*m*−*n*] (*m* integer) - -Problems 315 - -**Figure P3.1-2** - -- **3.2-3** Letting ↓ identify *n* = 0, define the nonzero values of signal *x*[*n*] as [−1, 2,−3, 4,−5, 4,−3, ↓ 2,−1]. - - (a) Using vector form, represent signal *y*[*n*] = *x*[−3*n* + 2]. Be sure to identify the *n* = 0 element. - - (b) Using vector form, represent signal *z*[*n*] = *x*[*n*/2 − 3]. Be sure to identify the *n* = 0 element. -- **3.2-4** Letting ↓ identify *n* = 0, define the nonzero values of signal *x*[*n*] as [−1, ↓ 2, - - −3, 4,−5, 4,−3, 2,−1]. (a) Determine the energy *Ex* and power *Px* of the signal *x*[*n*]. - - (b) Using vector form, represent signal *y*[*n*] = *x*[2(*n* + 2)]. Be sure to identify the *n* = 0 element. - - (c) Using vector form, represent signal *z*[*n*] = *x*[−*n*−6 3 ]. Be sure to identify the *n* = 0 element. -- **3.2-5** Letting ↓ identify the *n* = 0 value, describe a 4-periodic signal *w*[*n*] using vector notation as - - [· · · , 1, 2, 3, 4, ↓ 1, 2, 3, 4, 1, 2, 3, 4,···]. - - (a) Determine the energy *Ex* and power *Px* of signal *x*[*n*] = *w*[2*n*]. - - (b) Determine the energy *Ey* and power *Py* of signal *y*[*n*] = *w*[2− *n* 3 ]. -- **3.2-6** Let DT signal *x*[*n*] have values [1, 2, 3, 4, 5, 6] for 0 ≤ *n* ≤ 5 and let DT signal *y*[*n*] have values [5, 0, 0, 3, 0, 0, 1] for 0 ≤ *n* ≤ 6. Both signals are zero outside the ranges given. Further, define a 6-periodic replication of *y*[*n*] as *y*˜[*n*] = %∞ *k*=−∞ *y*[*n*−6*k*] . - - (a) Determine the energy *Ex* and power *Px* of signal *x*[*n*]. - - (b) Determine the smallest-magnitude integers *N*1, *N*2, and *N*3 such that *y*[*n*] = *x*[ *N*1*n N*2 +*N*3]. - - (c) Determine the energy *Ey*˜ and power *Py*˜ of *y*˜[*n*]. -- **3.2-7** For the signal shown in Fig. P3.1-1b, sketch the following signals: - - (a) *x*[−*n*] - - (b) *x*[*n*+6] - - (c) *x*[*n*−6] - - (d) *x*[3*n*] - - (e) *x* ' *n* ( - - 3 (f) *x*[3−*n*] - -- **3.2-8** Repeat Prob. 3.2-7 for the signal depicted in Fig. P3.1-1c. -- **3.2-9** Letting ↓ identify *n* = 0, consider a DT signal *x*[*n*] whose nonzero values are given as *x*[*n*]=[1, −3, 2, 2, 3, −2, −1, 1, 2, −3, ↓ 3, 3, −2, 1, −3, 2, 3, −1]. Accurately sketch *y*[*n*] = *x*[−1 − 2*n*] and *z*[*n*] = *x*[−2 + *n*/3] over −5 ≤ *n* ≤ 4. -- **3.2-10** Define - -$$ -x[n] = \begin{cases} \left(\frac{1}{2}\right)^n & n \ge 0\\ 0 & n < 0 \end{cases} -$$ - -Determine and locate the two largest non-zero values of: - -- (a) *y*a[*n*] = *x*[2*n*] -- (b) *y*b[*n*] = *x*[*n*/3] -- (c) *y*c[*n*] = *x*[3*n*+1] -- (d) *y*d[*n*] = *x*[−2*n*+5] -- (e) *y*e[*n*] = *x*[−(*n*+8)/2] -- **3.3-1** Sketch, and find the power of, the following signals: - - (a) (1)*n* - - (b) (−1)*n* - - (c) *u*[*n*] - - (d) (−1)*nu*[*n*] - - (e) cos'π 3 *n*+ π 6 ( -- **3.3-2** Show that - - (a) δ[*n*] +δ[*n*−1] = *u*[*n*] −*u*[*n*−2] - - (b) 2*n*−1 sin π*n* 3 *u*[*n*]= 1 2 2*n* sin π*n* 3 *u*[*n*−1] - - (c) *n*(*n*−1)γ *nu*[*n*] = *n*(*n*−1)γ *nu*[*n*−2] - - (d) (*u*[*n*] +(−1)*nu*[*n*])sinπ*n* = 0 for all *n* - -(e) -$$ -(u[n] + (-1)^{n+1}u[n])\cos(\frac{\pi n}{2}) = 0 -$$ - for all *n* - -- **3.3-3** Sketch the following signals: - - (a) *u*[*n*−2] −*u*[*n*−6] - - (b) *n*{*u*[*n*] −*u*[*n*−7]} - - (c) (*n*−2){*u*[*n*−2] −*u*[*n*−6]} - - (d) (−*n*+8){*u*[*n*−6] −*u*[*n*−9]} - - (e) (*n*−2){*u*[*n*−2]−*u*[*n*−6]}+(−*n*+8){*u*[*n*− 6] −*u*[*n*−9]} -- **3.3-4** Describe each of the signals in Fig. P3.1-1 by a single expression valid for all *n*. -- **3.3-5** Why are DT signals of the form *zn* so important to the study of LTID systems? -- **3.3-6** Explain the similarities and differences between the Kronecker delta function δ[*n*] and the Dirac delta function δ(*t*). - -- **3.3-7** The following signals are in the form *e*λ*n*. Express them in the form γ *n*: - - (a) *e*−0.5*n* - - (b) *e*0.5*n* - - (c) *e*−*j*π*n* - - (d) *ej*π*n* - -In each case show the locations of λ and γ in the complex plane. Verify that an exponential is growing if γ lies outside the unit circle (or if λ lies in the RHP), is decaying if γ lies within the unit circle (or if λ lies in the LHP), and has a constant amplitude if γ lies on the unit circle (or if λ lies on the imaginary axis). - -- **3.3-8** Express the following signals, which are in the form *e*λ*n*, in the form γ *n*: - - (a) *e*−(1+*j*π )*n* - - (b) *e*−(1−*j*π )*n* - - (c) *e*(1+*j*π )*n* - - (d) *e*(1−*j*π )*n* - - (e) *e*−[1+*j*(π/3)]*n* - - (f) *e*[1−*j*(π/3)]*n* -- **3.3-9** The concepts of even and odd functions for discrete-time signals are identical to those of the continuous-time signals discussed in Sec. 1.5. Using these concepts, find and sketch the odd and the even components of the following: - - (a) *u*[*n*] - - (b) *nu*[*n*] - - (c) sinπ*n* 4 - - (d) cosπ*n* 4 -- **3.4-1** A cash register output *y*[*n*] represents the total cost of *n* items rung up by a cashier. The input *x*[*n*] is the cost of the *n*th item. - - (a) Write the difference equation relating *y*[*n*] to *x*[*n*]. - - (b) Realize this system using a time-delay element. -- **3.4-2** Let *p*[*n*] be the population of a certain country at the beginning of the *n*th year. The birth and death rates of the population during any year are 3.3 and 1.3%, respectively. If *i*[*n*] is the total number of immigrants entering the country during the *n*th year, write the difference equation relating *p*[*n* + 1], *p*[*n*], and *i*[*n*]. Assume that the immigrants enter the country throughout the year at a uniform rate. - -- **3.4-3** A moving average is used to detect a trend of a rapidly fluctuating variable, such as the stock market average. A variable may fluctuate (up and down) daily, masking its long-term (secular) trend. We can discern the long-term trend by smoothing or averaging the past *N* values of the variable. For the stock market average, we may consider a 5-day moving average *y*[*n*] to be the mean of the past 5 days' market closing values *x*[*n*], *x*[*n*−1],..., *x*[*n*−4]. - - (a) Write the difference equation relating *y*[*n*] to the input *x*[*n*]. - - (b) Use time-delay elements to realize the 5-day moving-average filter. -- **3.4-4** The digital integrator in Ex. 3.9 is specified by - -$$ -y[n] - y[n-1] = Tx[n] -$$ - -If an input *u*[*n*] is applied to such an integrator, show that the output is (*n* + 1)*Tu*[*n*], which approaches the desired ramp *nTu*[*n*] as *T* → 0. - -**3.4-5** Approximate the following second-order differential equation with a difference equation. - -$$ -\frac{d^2y(t)}{dt^2} + a_1 \frac{dy(t)}{dt} + a_0 y(t) = x(t) -$$ - -- **3.4-6** Letting ↓ identify *n* = 0, define the nonzero values of signal *g*[*n*] in vector form as [1, 2, 3, 4, 5, 4, 3, ↓ 2, 1]. The impulse response of an LTID system is defined in terms of *g*[*n*] as *h*[*n*] = *g*[−2*n*−1]. - - (a) Express the nonzero values of *h*[*n*] in vector form, taking care to identify the *n* = 0 point. - - (b) Write a constant-coefficient linear difference equation (input *x*[*n*] and output *y*[*n*]) that has impulse response *h*[*n*]. - - (c) Show that the system is both linear and time-invariant. - - (d) Determine, if possible, whether the system is BIBO-stable. - - (e) Determine, if possible, whether the system is memoryless. - - (f) Determine, if possible, whether the system is causal. -- **3.4-7** An LTID system has an impulse response function *h*[*n*] = *u*[−(5−*n*)/3]. - - (a) Using an accurate sketch or vector representation, graphically depict *h*[*n*]. - -- (b) Determine, if possible, whether the system is BIBO-stable. -- (c) Determine, if possible, whether the system is memoryless. -- (d) Determine, if possible, whether the system is causal. -- **3.4-8** The voltage at the *n*th node of a resistive ladder in Fig. P3.4-8 is *v*[*n*], (*n* = 0, 1, 2,...,*N*). Show that *v*[*n*] satisfies the second-order difference equation - -$$ -v[n+2] - Av[n+1] + v[n] = 0 \quad A = 2 + \frac{1}{a} -$$ - -[*Hint:* Consider the node equation at the *n*th node with voltage *v*[*n*].] - -- **3.4-9** Determine whether each of the following statements is true or false. If the statement is false, demonstrate by proof or example why the statement is false. If the statement is true, explain why. - - (a) A discrete-time signal with finite power cannot be an energy signal. - -- (b) A discrete-time signal with infinite energy must be a power signal. -- (c) The system described by *y*[*n*] = (*n*+1)*x*[*n*] is causal. -- (d) The system described by *y*[*n* − 1] = *x*[*n*] is causal. -- (e) If an energy signal *x*[*n*] has energy *E*, then the energy of *x*[*an*] is *E*/|*a*|. -- **3.4-10** A linear time-invariant system produces output *y*1[*n*] in response to input *x*1[*n*], as shown in Fig. P3.4-10. Determine and sketch the output *y*2[*n*] that results when input *x*2[*n*] is applied to the same system. -- **3.4-11** A system is described by - -$$ -y[n] = \frac{1}{2} \sum_{k=-\infty}^{\infty} x[k] (\delta[n-k] + \delta[n+k]) -$$ - -- (a) Explain what this system does. -- (b) Is the system BIBO-stable? Justify your answer. -- (c) Is the system linear? Justify your answer. - -**Figure P3.4-8** - -**Figure P3.4-10** - -- (d) Is the system memoryless? Justify your answer. -- (e) Is the system causal? Justify your answer. -- (f) Is the system time-invariant? Justify your answer. -- **3.4-12** A discrete-time system is given by - -$$ -y[n+1] = \frac{x[n]}{x[n+1]} -$$ - -- (a) Is the system BIBO-stable? Justify your answer. -- (b) Is the system memoryless? Justify your answer. -- (c) Is the system causal? Justify your answer. -- **3.4-13** Explain why the continuous-time system *y*(*t*) = *x*(2*t*) is always invertible and yet the corresponding discrete-time system *y*[*n*] = *x*[2*n*] is not invertible. -- **3.4-14** Consider the input–output relationships of two similar discrete-time systems: - -$$ -y_1[n] = \sin\left(\frac{\pi}{2}n + 1\right)x[n] -$$ - -and - -$$ -y_2[n] = \sin\left(\frac{\pi}{2}(n+1)\right) x[n] -$$ - -Explain why *x*[*n*] can be recovered from *y*1[*n*] yet *x*[*n*] cannot be recovered from *y*2[*n*]. - -- **3.4-15** Consider a system that multiplies a given input by a ramp function, *r*[*n*]. That is, *y*[*n*] = *x*[*n*]*r*[*n*]. - - (a) Is the system BIBO-stable? Justify your answer. - - (b) Is the system linear? Justify your answer. - - (c) Is the system memoryless? Justify your answer. - - (d) Is the system causal? Justify your answer. - - (e) Is the system time-invariant? Justify your answer. -- **3.4-16** A jet-powered car is filmed using a camera operating at 60 frames per second. Let variable *n* designate the film frame, where *n* = 0 corresponds to engine ignition (film before ignition is discarded). By analyzing each frame of the film, it is possible to determine the car position *x*[*n*], measured in meters, from the original starting position *x*[0] = 0. - -From physics, we know that velocity is the time derivative of position: - -$$ -v(t) = \frac{d}{dt}x(t) -$$ - -Furthermore, we know that acceleration is the time derivative of velocity: - -$$ -a(t) = \frac{d}{dt}v(t) -$$ - -We can estimate the car velocity from the film data by using a simple difference equation *v*[*n*] = *k*(*x*[*n*] −*x*[*n*−1]). - -- (a) Determine the appropriate constant *k* to ensure *v*[*n*] has units of meters per second. -- (b) Determine a standard-form constant coefficient difference equation that outputs an estimate of acceleration, *a*[*n*], using an input of position, *x*[*n*]. Identify the advantages and shortcomings of estimating acceleration *a*(*t*) with *a*[*n*]. What is the impulse response *h*[*n*] for this system? -- **3.5-1** An LTID system is described by a constant coefficient linear difference equation 2*y*[*n*] + 2*y*[*n*−1] = *x*[*n*−1]. - - (a) Express this system in standard advance operator form. - - (b) Using recursion, determine the first 5 values of the system impulse response *h*[*n*]. - - (c) Using recursion, determine the first 5 values of the system zero-state response to input *x*[*n*] = 2*u*[*n*]. - - (d) Using recursion, determine for (0 ≤ *n* ≤ 4) the system zero-input response if *y*[−1] = 1. -- **3.5-2** Solve recursively (first three terms only): - - (a) *y*[*n*+1] −0.5*y*[*n*] = 0, with *y*[−1] = 10 - - (b) *y*[*n* + 1] + 2*y*[*n*] = *x*[*n* + 1], with *x*[*n*] = *e*−*nu*[*n*] and *y*[−1] = 0 -- **3.5-3** Solve the following equation recursively (first three terms only): - -$$ -y[n] - 0.6y[n-1] - 0.16y[n-2] = 0 -$$ - -with - -$$ -y[-1] = -25, y[-2] = 0. -$$ - -**3.5-4** Solve recursively the second-order difference Eq. (3.6) for sales estimate (first three terms only), assuming *y*[−1] = *y*[−2] = 0 and *x*[*n*] = 100*u*[*n*]. - -**3.5-5** Solve the following equation recursively (first three terms only): - -$$ -y[n+2] + 3y[n+1] + 2y[n] = x[n+2] + 3x[n+1] + 3x[n] -$$ - -with *x*[*n*] = (3)*nu*[*n*], *y*[−1] = 3, and *y*[−2] = 2 - -**3.5-6** Repeat Prob. 3.5-5 for - -$$ -y[n] + 2y[n-1] + y[n-2] = 2x[n] - x[n-1] -$$ - -with *x*[*n*] = (3)−*nu*[*n*], *y*[−1] = 2, and *y*[−2] = 3. - -- **3.6-1** Given *y*0[−1] = 3 and *y*0[−2]=−1, determine the closed-form expression of the zero-input response *y*0[*n*] of an LTID system described by the equation *y*[*n*]+ 1 6 *y*[*n*−1]− 1 6 *y*[*n*−2] = 1 3 *x*[*n*] +2 3 *x*[*n*−2]. -- **3.6-2** Solve - -$$ -y[n+2] + 3y[n+1] + 2y[n] = 0 -$$ - -if -$$ -y[-1] = 0 -$$ - and $y[-2] = 1$ . - -**3.6-3** Solve - -$$ -y[n+2] + 2y[n+1] + y[n] = 0 -$$ - -if -$$ -y[-1] = 1 -$$ - and $y[-2] = 1$ . - -**3.6-4** Solve - -$$ -y[n+2] - 2y[n+1] + 2y[n] = 0 -$$ - -if *y*[−1] = 1 and *y*[−2] = 0. - -**3.6-5** For the general *N*th-order difference Eq. (3.16), letting - -*a*1 = *a*2 =···= *aN*−1 = 0 - -results in a general causal *N*th-order LTI *nonrecursive* difference equation - -$$ -y[n] = b_0 x[n] + b_1 x[n-1] + \cdots + b_N x[n-N] -$$ - -Show that the characteristic roots for this system are zero—hence, that the zero-input response is zero. Consequently, the total response consists of the zero-state component only. - -**3.6-6** Leonardo Pisano Fibonacci, a famous thirteenthcentury mathematician, generated the sequence of integers - -$$ -\{0,1,1,2,3,5,8,13,21,34,\dots\} -$$ - -while addressing, oddly enough, a problem involving rabbit reproduction. An element of the Fibonacci sequence is the sum of the previous two. - -- (a) Find the constant-coefficient difference equation whose zero-input response *f*[*n*] with auxiliary conditions *f*[1] = 0 and *f*[2] = 1 is a Fibonacci sequence. Given *f*[*n*] is the system output, what is the system input? -- (b) What are the characteristic roots of this system? Is the system stable? -- (c) Designating 0 and 1 as the first and second Fibonacci numbers, determine the fiftieth Fibonacci number. Determine the one thousandth Fibonacci number. -- **3.6-7** Find *v*[*n*], the voltage at the *n*th node of the resistive ladder depicted in Fig. P3.4-8, if *V* = 100 volts and *a* = 2. [*Hint* 1: Consider the node equation at the *n*th node with voltage *v*[*n*]. *Hint* 2: See Prob. 3.4-8 for the equation for *v*[*n*]. The auxiliary conditions are *v*[0] = 100 and *v*[*N*] = 0.] -- **3.6-8** Consider the discrete-time system *y*[*n*] + *y*[*n* − 1] + 0.25*y*[*n* − 2] = √3*x*[*n* − 8]. Find the zero input response, *y*0[*n*], if *y*0[−1] = 1 and *y*0[1] = 1. -- **3.6-9** Provide a standard-form polynomial *Q*(*X*) such that *Q*(*E*){*y*[*n*]} = *x*[*n*] corresponds to a marginally stable third-order LTID system and *Q*(*D*){*y*(*t*)} = *x*(*t*) corresponds to a stable third-order LTIC system. -- **3.7-1** Find the unit impulse response *h*[*n*] of systems specified by the following equations: (a) *y*[*n*+1] +2*y*[*n*] = *x*[*n*] (b) *y*[*n*] +2*y*[*n*−1] = *x*[*n*] -- **3.7-2** Determine the unit impulse response *h*[*n*] of the following systems. In each case, use recursion to verify the *n* = 3 value of the closed-form expression of *h*[*n*]. - - (a) (*E*2 +1){*y*[*n*]} = (*E* +0.5){*x*[*n*]} - - (b) *y*[*n*] −*y*[*n*−1] +0.25*y*[*n*−2] = *x*[*n*] - - (c) *y*[*n*] − 1 6 *y*[*n*−1] − 1 6 *y*[*n*−2] = 1 3 *x*[*n*−2] - -- (d) *y*[*n*] + 1 6 *y*[*n*−1] − 1 6 *y*[*n*−2] = 1 3 *x*[*n*] -- (e) *y*[*n*] + 1 4 *y*[*n*−2] = *x*[*n*] -- (f) (*E*2 − 4 9 ){*y*[*n*]} = (*E*2 +1){*x*[*n*]} -- (g) (*E*2 − 1 4 )(*E* + 1 2 ){*y*[*n*]} = *E*3{*x*[*n*]} -- (h) (*E* − 1 2 )2{*y*[*n*]} = *x*[*n*] -- **3.7-3** Consider a DT system with input *x*[*n*] and output *y*[*n*] described by the difference equation - -$$ -4y[n+1] + y[n-1] = 8x[n+1] + 8x[n] -$$ - -- (a) What is the order of this system? -- (b) Determine the characteristic mode(s) of the system. -- (c) Determine a closed-form expression for the system's impulse response *h*[*n*]. -- **3.7-4** Repeat Prob. 3.7-3 for a system described by the difference equation - -$$ -y[n+3] - \frac{3}{10}y[n+2] - \frac{1}{10}y[n+1] = 2x[n+1] -$$ - -**3.7-5** Repeat Prob. 3.7-1 for - -$$ -(E2 - 6E + 9)y[n] = Ex[n] -$$ - -**3.7-6** Repeat Prob. 3.7-1 for - -$$ -y[n] - 6y[n-1] + 25y[n-2] = 2x[n] - 4x[n-1] -$$ - -**3.7-7** (a) For the general *N*th-order difference Eq. (3.16), letting - -$$ -a_0 = a_1 = a_2 = \cdots = a_{N-1} = 0 -$$ - -results in a general causal *N*th-order LTI *nonrecursive* difference equation - -$$ -y[n] = \sum_{i=0}^{N} b_i x[n-i] -$$ - -Find the impulse response *h*[*n*] for this system. [*Hint:* The characteristic equation for this case is γ *n* = 0. Hence, all the characteristic roots are zero. In this case, *yc*[*n*] = 0, and the approach in Sec. 3.7 does not work. Use a direct method to find *h*[*n*] by realizing that *h*[*n*] is the response to unit impulse input.] - -(b) Find the impulse response of a nonrecursive LTID system described by the equation - -$$ -y[n] = 3x[n] - 5x[n-1] - 2x[n-3] -$$ - -Observe that the impulse response has only a finite (*N*) number of nonzero elements. For this reason, such systems are called *finite-impulse response* (FIR) systems. For a general recursive case [Eq. (3.20)], the impulse response has an infinite number of nonzero elements, and such systems are called *infinite-impulse response* (IIR) systems. - -**3.8-1** The convolution *y*[*n*] = 5 2*n u*[*n*+5] ∗ (3*nu*[−*n*−2]) can be represented as - -$$ -y[n] = \begin{cases} C_1(\gamma_1)^n & n < N \\ C_2(\gamma_2)^n & n \ge N \end{cases} -$$ - -Using the graphical convolution procedure, determine constants *C*1, *C*2, γ1, γ2, and *N*. - -- **3.8-2** Use the graphical convolution procedure to determine the following: - - (a) *y*a[*n*] = *u*[*n*] ∗ (*u*[*n* − 5] − *u*[*n* − 9] + (0.5)(*n*−8) *u*[*n*−9]) - - (b) *y*b[*n*] = ( 1 2 )|*n*| ∗ *u*[−*n*+5] -- **3.8-3** Let *x*[*n*] = (0.5)*n* (*u*[*n*+4] −*u*[*n*−4]) be input into an LTID system with an impulse response given by - -$$ -h[n] = \begin{cases} 2 & \text{[(}n \text{ mod } 6) < 4 \text{]} \text{ and } [n \ge 0] \\ 0 & \text{otherwise} \end{cases} -$$ - -Recall, (*n* mod *p*) is the remainder of the division *n*/*p*. The system is described according to the difference equation *y*[*n*]−*y*[*n*−6] = 2*x*[*n*]+ 2*x*[*n*−1] +2*x*[*n*−2] +2*x*[*n*−3]. - -- (a) Determine the six characteristic roots (γ1 through γ6) of the system. -- (b) Determine the value of *y*[10], the zero-state output of system *h*[*n*] in response to *x*[*n*] at time *n* = 10. Express your result in decimal form to at least three decimal places (e.g., *y*[10] = 3.142). -- **3.8-4** An LTID system has impulse response *h*[*n*] = (0.5)(*n*+3) (*u*[*n*] −*u*[*n*+6]). A 6-periodic DT - -input signal *x*[*n*] is given by - -$$ -x[n] = \begin{cases} 1 & n = 0, \pm 3, \pm 6, \pm 9, \pm 12, \dots \\ 2 & n = 1, 1 \pm 6, 1 \pm 12, \dots \\ 3 & n = 2, 2 \pm 6, 2 \pm 12, \dots \\ 0 & \text{otherwise} \end{cases} -$$ - -- (a) Is system *h*[*n*] causal? Mathematically justify your answer. -- (b) Determine the value of *y*[12], the zero-state output of system *h*[*n*] in response to *x*[*n*] at time *n* = 12. Express your result in decimal form to at least three decimal places (e.g., *y*[12] = 1.234). -- **3.8-5** Find the (zero-state) response *y*[*n*] of an LTID system whose unit impulse response is - -$$ -h[n] = (-2)^n u[n-1] -$$ - -and the input is *x*[*n*] = *e*−*nu*[*n* + 1]. Find your answer by computing the convolution sum and also by using Table 3.1. - -**3.8-6** Find the (zero-state) response *y*[*n*] of an LTID system if the input is *x*[*n*] = 3*n*−1*u*[*n*+2], and - -$$ -h[n] = \frac{1}{2} [\delta[n-2] - (-2)^{n+1}] u[n-3] -$$ - -**3.8-7** Find the (zero-state) response *y*[*n*] of an LTID system if the input *x*[*n*] = (3)*n*+2*u*[*n*+1], and - -$$ -h[n] = [(2)^{n-2} + 3(-5)^{n+2}]u[n-1] -$$ - -**3.8-8** Find the (zero-state) response *y*[*n*] of an LTID system if the input *x*[*n*] = (3)−*n*+2*u*[*n*+3], and - -$$ -h[n] = 3(n-2)(2)^{n-3}u[n-4] -$$ - -**3.8-9** Find the (zero-state) response *y*[*n*] of an LTID system if its input *x*[*n*] = (2)*nu*[*n*−1], and - -$$ -h[n] = (3)^n \cos\left(\frac{\pi}{3}n - 0.5\right) u[n] -$$ - -Find your answer using only Table 3.1. - -- **3.8-10** Consider an LTID system ("system 1") described by (*E* − 1 2 ){*y*[*n*]} = *x*[*n*]. - - (a) Determine the impulse response *h*1[*n*] for system 1. Simplify your answer. - - (b) Determine the step response *s*[*n*] for system 1 (the step response is the output in response to a unit step input). Simplify your answer. - -- (c) Determine the impulse response *h*cascade[*n*] of system 1 cascaded with an LTID system with impulse response *h*2[*n*]=−3*u*[*n*−13]. Simplify your answer. -- **3.8-11** Derive the results in entries 1, 2, and 3 in Table 3.1. [*Hint:* You may need to use the information in Sec. B.8-3.] -- **3.8-12** Derive the results in entries 4, 5, and 6 in Table 3.1. -- **3.8-13** Derive the results in entries 7 and 8 in Table 3.1. [*Hint:* You may need to use the information in Sec. B.8-3.] -- **3.8-14** Derive the results in entries 9 and 11 in Table 3.1. [*Hint:* You may need to use the information in Sec. B.8-3.] -- **3.8-15** Find the total response of a system specified by the equation - -$$ -y[n+1] + 2y[n] = x[n+1] -$$ - -if *y*[−1] = 10, and the input *x*[*n*] = *e*−*nu*[*n*]. - -- **3.8-16** Find an LTID system (zero-state) response if its impulse response *h*[*n*] = (0.5)*nu*[*n*], and the input *x*[*n*] is (a) 2*nu*[*n*] (b) 2*n*−3*u*[*n*] - - (c) 2*nu*[*n*−2] - -[*Hint:* You may need to use the convolution shift property of Eq. (3.32).] - -**3.8-17** For a system specified by equation - -$$ -y[n] = x[n] - 2x[n-1] -$$ - -Find the system response to input *x*[*n*] = *u*[*n*]. What is the order of the system? What type of system (recursive or nonrecursive) is this? Is the knowledge of initial condition(s) necessary to find the system response? Explain. - -- **3.8-18** (a) A discrete-time LTI system is shown in Fig. P3.8-18. Express the overall impulse response of the system, *h*[*n*], in terms of *h*1[*n*], *h*2[*n*], *h*3[*n*], *h*4[*n*], and *h*5[*n*]. - - (b) Two LTID systems in cascade have impulse response *h*1[*n*] and *h*2[*n*], respectively. Show that if *h*1[*n*] = (0.9)*nu*[*n*] − 0.5(0.9)*n*−1*u*[*n* − 1] and *h*2[*n*] = (0.5)*nu*[*n*] − 0.9(0.5)*n*−1*u*[*n* − 1], the cascade system is an identity system. - -**Figure P3.8-18** - -**3.8-19** (a) Show that for a causal system, Eq. (3.37) can also be expressed as - -$$ -g[n] = \sum_{k=0}^{n} h[n-k] -$$ - -- (b) How would the expressions in part (a) change if the system is not causal? -- **3.8-20** An LTID system with input *x*[*n*] and output *y*[*n*] has impulse response *h*[*n*] = 2(*u*[*n* + 2] − *u*[*n* − 3]). - - (a) Write a constant-coefficient linear difference equation that has the given impulse response. [*Hint:* First express *h*[*n*] in terms of delta functions δ[*n*].] - - (b) Using graphical convolution, determine the zero-state output of this system in response to the anticausal input *x*[*n*] = 2*nu*[−*n*]. A simplified closed-form solution is required. -- **3.8-21** Consider three LTID systems: system 1 has impulse response *h*1[*n*]=[ ↓ 2, −3, 4], system 2 has impulse response *h*2[*n*]=[ ↓ 0, 0, −6, − 9, 3], and system 3 is an identity system (output equals input). - - (a) Determine the overall impulse response *h*[*n*] if system 1 is connected in cascade with a parallel connection of systems 2 and 3. - - (b) For input *x*[*n*] = *u*[−*n*], determine the zero-state response *y*zsr[*n*] of system 2. -- **3.8-22** In the savings account problem described in Ex. 3.6, a person deposits \$500 at the beginning of every month, starting at *n* = 0 with the exception at *n* = 4, when instead of depositing \$500, she withdraws \$1000. Find *y*[*n*] if the interest rate is 1% per month (*r* = 0.01). - -**3.8-23** To pay off a loan of *M* dollars in *N* number of payments using a fixed monthly payment of *P* dollars, show that - -$$ -P = \frac{rM}{1 - (1 + r)^{-N}} -$$ - -where *r* is the interest rate per dollar per month. [*Hint:* This problem can be modeled by Eq. (3.3) with the payments of *P* dollars starting at *n* = 1. The problem can be approached in two ways. First, consider the loan as the initial condition *y*0[0]=−*M*, and the input *x*[*n*] = *Pu*[*n* − 1]. The loan balance is the sum of the zero-input component (due to the initial condition) and the zero-state component *h*[*n*] ∗ *x*[*n*]. Second, consider the loan as an input −*M* at *n* = 0 along with the input due to payments. The loan balance is now exclusively a zero-state component *h*[*n*] ∗ *x*[*n*]. Because the loan is paid off in *N* payments, set *y*[*N*] = 0.] - -- **3.8-24** A person receives an automobile loan of \$10,000 from a bank at the interest rate of 1.5% per month. His monthly payment is \$500, with the first payment due one month after he receives the loan. Compute the number of payments required to pay off the loan. Note that the last payment may not be exactly \$500. [*Hint:* Follow the procedure in Prob. 3.8-23 to determine the balance *y*[*n*]. To determine *N*, the number of payments, set *y*[*N*] = 0. In general, *N* will not be an integer. The number of payments *K* is the largest integer ≤ *N*. The residual payment is |*y*[*K*]|.] -- **3.8-25** Letting ↓ identify the *n* = 0 values, use the sliding-tape method to determine the following: - - (a) *y*a = [ ↓ 2, 3,−2,−3] ∗ [−10, ↓ 0,−5] (b) *y*b = [2, ↓ −1, 3,−2] ∗ [−1,−4, 1, ↓ −2] (c) *y*c = [ ↓ 0, 0, 3, 2, 1, 2, 3]∗[2, 3,−2, ↓ 1] - - (d) *y*d = [5, 0, 0, ↓ −2, 8] ∗ [−1, 1, ↓ 3, 3,−2, 3] - - (e) *y*e = ([1, ↓ −1]∗[ ↓ 1,−1]) ∗ ([ ↓ 1,−1]∗[1, ↓ −1]) - - (f) *y*f = ([2, ↓ −1]∗[ ↓ 1,−2]) ∗ ([ ↓ 1,−2]∗[2, ↓ −1]) Outside the values shown, assume all signals are zero. - -### 324 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -- **3.8-26** Let ↓ identify the *n* = 0 and consider DT signals *x*[*n*] and *h*[*n*] whose nonzero values are given as *x*[*n*]=[1, 2, 3, ↓ 4, 5] and *h*[*n*] = [−2, −1, 1, ↓ 2]. Use DT convolution (any method) to determine *y*[*n*] = (2*x*[*n* − 30]) ∗ −3 2 *h*[*n*−10] . Express your result in vector notation, making sure to indicate the time index of the leftmost (nonzero) element. -- **3.8-27** Using the sliding-tape algorithm, show that (a) *u*[*n*] ∗ *u*[*n*] = (*n*+1)*u*[*n*] (b) (*u*[*n*] −*u*[*n*−*m*])∗*u*[*n*] = (*n*+1)*u*[*n*]−(*n*− *m*+1)*u*[*n*−*m*] -- **3.8-28** Using the sliding-tape algorithm, find *x*[*n*] ∗*g*[*n*] for the signals shown in Fig. P3.8-28. - -- **3.8-29** Repeat Prob. 3.8-28 for the signals shown in Fig. P3.8-29. -- **3.8-30** Repeat Prob. 3.8-28 for the signals shown in Fig. P3.8-30. -- **3.8-31** Letting ↓ identify *n* = 0, define the nonzero values of signal *x*[*n*] as [1, 2, ↓ 2]. Similarly, define the non-zero values of signal *y*[*n*] as [3, 4, 6, 6, 11, ↓ 2,−2]. Using the sliding-tape algorithm as the basis for your work, determine the signal *h*[*n*] so that *y*[*n*] = *x*[*n*] ∗ *h*[*n*]. -- **3.8-32** The convolution sum in Eq. (3.33) can be expressed in a matrix form as *y* = *Hx*, where *y* is a column vector containing *y*[0], *y*[1],... *y*[*n*]; *x* is a column vector containing *x*[0], *x*[1],... *x*[*n*]; - -**Figure P3.8-30** - -and *H* is a lower triangular matrix defined as - -| | ⎡
h[0] | 0 | 0 | | 0 | ⎤ | -|--------|-----------|--------|---|---------|------|--------| -| | h[1]
⎢ | h[0] | 0 | | 0 | ⎥ | -| H
= | ⎢
⎢ | | | | | ⎥
⎥ | -| | ⎣
h[n] | h[n−1] | | ···
| h[0] | ⎦ | -| | | | | | | | - -Knowing *h*[*n*] and the output *y*[*n*], we can determine the input *x*[*n*] according to *x* = *H*−1*y*. This operation is the reverse of convolution and is known as *deconvolution*. Moreover, knowing *x*[*n*] and *y*[*n*], we can determine *h*[*n*]. This can be done by expressing the foregoing matrix equation as *n* + 1 simultaneous equations in terms of *n* + 1 unknowns *h*[0], *h*[1], ... , *h*[*n*]. These equations can readily be solved iteratively. Thus, we can synthesize a system that yields a certain output *y*[*n*] for a given input *x*[*n*]. - -- (a) Design a system (i.e., determine *h*[*n*]) that will yield the output sequence (8, 12, 14, 15, 15.5, 15.75, ...) for the input sequence (1, 1, 1, 1, 1, 1, ...). -- (b) For a system with the impulse response sequence (1, 2, 4, ...), the output sequence was (1, 7/3, 43/9, ...). Determine the input sequence. -- **3.8-33** A second-order LTID system has zero-input response - -$$ -y_0[n] = [3, 2\frac{1}{3}, 2\frac{1}{9}, 2\frac{1}{27}, \dots] -$$ - -= -$$ -\sum_{k=0}^{\infty} \left\{ 2 + \left(\frac{1}{3}\right)^k \right\} \delta[n-k] -$$ - -- (a) Determine the characteristic equation of this system, *a*0γ 2 +*a*1γ +*a*2 = 0. -- (b) Find a bounded, causal input with infinite duration that would cause a strong response from this system. Justify your choice. - -**Figure P3.8-34** - -- (c) Find a bounded, causal input with infinite duration that would cause a weak response from this system. Justify your choice. -- **3.8-34** An LTID filter has an impulse response function given by *h*1[*n*] = δ[*n* + 2] − δ[*n* − 2]. A second LTID system has an impulse response function given by *h*2[*n*] = *n*(*u*[*n*+4] −*u*[*n*−4]). - - (a) Carefully sketch the functions *h*1[*n*] and *h*2[*n*] over (−10 ≤ *n* ≤ 10). - - (b) Assume that the two systems are connected in parallel, as shown in Fig. P3.8-34a. Determine the impulse response *hp*[*n*] for the parallel system in terms of *h*1[*n*] and *h*2[*n*]. Sketch *hp*[*n*] over (−10 ≤ *n* ≤ 10). - - (c) Assume that the two systems are connected in cascade, as shown in Fig. P3.8-34b. Determine the impulse response *hs*[*n*] for the cascade system in terms of *h*1[*n*] and *h*2[*n*]. Sketch *hs*[*n*] over (−10 ≤ *n* ≤ 10). -- **3.8-35** This problem investigates an interesting application of discrete-time convolution: the expansion of certain polynomial expressions. - - (a) By hand, expand (*z*3+*z*2+*z*+1)2. Compare the coefficients to [1, 1, 1, 1]∗[1, 1, 1, 1]. - - (b) Formulate a relationship between discretetime convolution and the expansion of constant-coefficient polynomial expressions. - - (c) Use convolution to expand (*z*−4 − 2*z*−3 + 3*z*−2)4. - - (d) Use convolution to expand (*z*5 +2*z*4 +3*z*2 + 5)2(*z*−4 −5*z*−2 +13). -- **3.8-36** Joe likes coffee, and he drinks his coffee according to a very particular routine. He begins by adding two teaspoons of sugar to his mug, which he then fills to the brim with hot coffee. He drinks 2/3 of the mug's contents, adds another two teaspoons of sugar, and tops the mug off with steaming hot coffee. This refill procedure - -continues, sometimes for many, many cups of coffee. Joe has noted that his coffee tends to taste sweeter with the number of refills. - -Let independent variable *n* designate the coffee refill number. In this way, *n* = 0 indicates the first cup of coffee, *n* = 1 is the first refill, and so forth. Let *x*[*n*] represent the sugar (measured in teaspoons) added into the system (a coffee mug) on refill *n*. Let *y*[*n*] designate the amount of sugar (again, teaspoons) contained in the mug on refill *n*. - -- (a) The sugar (teaspoons) in Joe's coffee can be represented using a standard second-order constant coefficient difference equation *y*[*n*] + *a*1*y*[*n* − 1] + *a*2*y*[*n* − 2] = *b*0*x*[*n*] + *b*1*x*[*n* − 1] + *b*2*x*[*n* − 2]. Determine the constants *a*1, *a*2, *b*0, *b*1, and *b*2. -- (b) Determine *x*[*n*], the driving function to this system. -- (c) Solve the difference equation for *y*[*n*]. This requires finding the total solution. Joe always starts with a clean mug from the dishwasher, so *y*[−1] (the sugar content before the first cup) is zero. -- (d) Determine the steady-state value of *y*[*n*]. That is, what is *y*[*n*] as *n* → ∞? If possible, suggest a way of modifying *x*[*n*] so that the sugar content of Joe's coffee remains a constant for all nonnegative *n*. -- **3.8-37** A system is called complex if a real-valued input can produce a complex-valued output. Consider a causal complex system described by a first-order constant coefficient linear difference equation: - -(*jE* +0.5)*y*[*n*] = (−5*E*)*x*[*n*] - -- (a) Determine the impulse response function *h*[*n*] for this system. -- (b) Given input *x*[*n*] = *u*[*n* − 5] and initial condition *y*0[−1] = *j*, determine the system's total output *y*[*n*] for *n* ≥ 0. -- **3.8-38** A discrete-time LTI system has impulse response function *h*[*n*] = *n*(*u*[*n* − 2] − *u*[*n*+2]). - - (a) Carefully sketch the function *h*[*n*] over (−5 ≤ *n* ≤ 5). - - (b) Determine the difference equation representation of this system, using *y*[*n*] to designate the output and *x*[*n*] to designate the input. - -- **3.8-39** Consider three discrete-time signals: *x*[*n*], *y*[*n*], and *z*[*n*]. Denoting convolution as ∗, identify the expression(s) that is(are) equivalent to *x*[*n*](*y*[*n*] ∗ *z*[*n*]): - - (a) (*x*[*n*] ∗ *y*[*n*])*z*[*n*] (b) (*x*[*n*]*y*[*n*]) ∗ (*x*[*n*]*z*[*n*]) - - (c) (*x*[*n*]*y*[*n*]) ∗ *z*[*n*] - - (d) none of the above - - Justify your answer! -- **3.8-40** A causal system with input *x*[*n*] and output *y*[*n*] is described by - -$$ -y[n] - ny[n-1] = x[n] -$$ - -- (a) By recursion, determine the first six nonzero values of *h*[*n*], the response to *x*[*n*] = δ[*n*]. Do you think this system is BIBO-stable? Why? -- (b) Compute *y*R[4] recursively from *y*R[*n*] − *ny*R[*n*−1] = *x*[*n*], assuming all initial conditions are zero and *x*[*n*] = *u*[*n*]. The subscript R is only used to emphasize a recursive solution. -- (c) Define *y*C[*n*] = *x*[*n*]∗*h*[*n*]. Using *x*[*n*] = *u*[*n*] and *h*[*n*] from part (a), compute *y*C[4]. The subscript C is only used to emphasize a convolution solution. -- (d) In this chapter, both recursion and convolution are presented as potential methods to compute the zero-state response (ZSR) of a discrete-time system. Comparing parts (b) and (c), we see that *y*R[4] = *y*C[4]. Why are the two results not the same? Which method, if any, yields the correct ZSR value? -- **3.9-1** In Sec. 3.9-1 we showed that for BIBO stability in an LTID system, it is sufficient for its impulse response *h*[*n*] to satisfy Eq. (3.43). Show that this is also a necessary condition for the system to be BIBO-stable. In other words, show that if Eq. (3.43) is not satisfied, there exists a bounded input that produces unbounded output. [*Hint:* Assume that a system exists for which *h*[*n*] violates Eq. (3.43), yet its output is bounded for every bounded input. Establish the contradiction in this statement by considering an input *x*[*n*] defined by *x*[*n*1 − *m*] = 1 when *h*[*m*] > 0 and *x*[*n*1−*m*]=−1 when *h*[*m*]<0, where *n*1 is some fixed integer.] - -- **3.9-2** Each of the following equations specifies an LTID system. Determine whether each of these systems is BIBO-stable or -unstable. Determine also whether each is asymptotically stable, unstable, or marginally stable. - - (a) *y*[*n* + 2] + 0.6*y*[*n* + 1] − 0.16*y*[*n*] = *x*[*n* + 1] −2*x*[*n*] - - (b) *y*[*n*] + 3*y*[*n* − 1] + 2*y*[*n* − 2] = *x*[*n* − 1] + 2*x*[*n*−2] - - (c) (*E* −1)2 *E* + 1 2 *y*[*n*] = *x*[*n*] - - (d) *y*[*n*] +2*y*[*n*−1] +0.96*y*[*n*−2] = *x*[*n*] - - (e) *y*[*n*]+*y*[*n*−1]−2*y*[*n*−2] = *x*[*n*]+2*x*[*n*−1] - - (f) (*E*2 −1)(*E*2 +1)*y*[*n*] = *x*[*n*] -- **3.9-3** Consider two LTIC systems in cascade, as illustrated in Fig. 3.29. The impulse response of the system *S*1 is *h*1[*n*] = 2*n u*[*n*] and the impulse response of the system *S*2 is *h*2[*n*] = δ[*n*] − 2δ[*n*−1]. Is the cascaded system asymptotically stable or unstable? Determine the BIBO stability of the composite system. -- **3.9-4** Figure P3.9-4 locates the characteristic roots of ten causal, LTID systems, labeled A through J. Each system has only two roots and is described using operator notation as *Q*(*E*)*y*[*n*] = *P*(*E*)*x*[*n*]. All plots are drawn to scale, with the unit circle shown for reference. For each of the following parts, identify all the answers that are correct. - - (a) Identify all systems that are unstable. - - (b) Assuming all systems have *P*(*E*) = *E*2, identify all systems that are real. Recall that a real system always generates a real-valued response to a real-valued input. - - (c) Identify all systems that support oscillatory natural modes. - -- (d) Identify all systems that have at least one mode whose envelop decays at a rate of 2−*n*. -- (e) Identify all systems that have only one mode. -- **3.9-5** A discrete-time LTI system has impulse response given by - -$$ -h[n] = \delta[n] + \left(\frac{1}{3}\right)^n u[n-1] -$$ - -- (a) Is the system stable? Is the system causal? Justify your answers. -- (b) Plot the signal *x*[*n*] = *u*[*n*−3] −*u*[*n*+3]. -- (c) Determine the system's zero-state response *y*[*n*] to the input *x*[*n*] = *u*[*n* − 3] − *u*[*n* + 3]. Plot *y*[*n*] over (−10 ≤ *n* ≤ 10). -- **3.9-6** An LTID system has an impulse response given by - -$$ -h[n] = \left(\frac{1}{2}\right)^{|n|} -$$ - -- (a) Is the system causal? Justify your answer. -- (b) Compute %∞ *n*=−∞ |*h*[*n*]|. Is this system BIBO-stable? -- (c) Compute the energy and power of input signal *x*[*n*] = 3*u*[*n*−5]. -- (d) Using input *x*[*n*] = 3*u*[*n* − 5], determine the zero-state response of this system at time *n* = 10. That is, determine *y*zsr[10]. -- **3.10-1** Determine a constant coefficient linear difference equation that describes a system for which the input *x*[*n*] = 2( 1 3 )*nu*[−*n* − 4] causes resonance. -- **3.10-2** If one exists, determine a real input *x*[*n*] that will cause resonance in the causal LTID system described by (*E*2 +1){*y*[*n*]} = (*E*+0.5){*x*[*n*]}. If no such input exists, explain why not. - -**Figure P3.9-4** - -- **3.10-3** Consider two lowpass LTID systems, one with infinite-duration impulse response *h*1[*n*] = −(0.5)*nu*[*n*] and the other with finite-duration impulse response *h*2[*n*] = 2(*u*[*n*] − *u*[*n* − 4]). Which system (1, 2, both, or neither) would more efficiently transmit a binary communication signal? Carefully justify your result. -- **3.11-1** Write a MATLAB program that recursively computes and then plots the solution to *y*[*n*] − 1 3 *y*[*n* − 1] + 1 2 *y*[*n* − 2] = *x*[*n*] for (0 ≤ *n* ≤ 100) given *x*[*n*] = δ[*n*] + *u*[*n* − 50] and *y*[−2] = *y*[−1] = 2. -- **3.11-2** Consider the discrete-time function *f*[*n*] = *e*−*n*/5 cos(π*n*/5)*u*[*n*]. Section 3.11 uses anonymous functions in describing DT signals. - - f = @(n) exp(-n/5).\*cos(pi\*n/5).\*(n>=0); - -While this anonymous function operates correctly for a downsampling operation such as f[2n], it does not operate correctly for an upsampling operation, such as f[n/2]. Modify the anonymous function f so that it also correctly accommodates upsampling operations. Test your code by computing and plotting f(n/2) over (−10 ≤ *n* ≤ 10). - -- **3.11-3** Write MATLAB code to compute and plot the DT convolutions of Prob. 3.8-25. -- **3.11-4** An indecisive student contemplates whether he should stay home or take his final exam, which is being held 2 miles away. Starting at home, the student travels half the distance to the exam location before changing his mind. The student turns around and travels half the distance between his current location and his home before changing his mind again. This process of changing direction and traveling half the remaining distance continues until the student either reaches a destination or dies from exhaustion. - - (a) Determine a suitable difference equation description of this system. - - (b) Use MATLAB to simulate the difference equation in part (a). Where does the student end up as *n* → ∞? How does your answer change if the student goes two-thirds the way each time, rather than halfway? - - (c) Determine a closed-form solution to the equation in part (a). Use this solution to verify the results in part (b). - -**3.11-5** The cross-correlation function between *x*[*n*] and *y*[*n*] is given as - -$$ -r_{xy}[k] = \sum_{n=-\infty}^{\infty} x[n]y[n-k] -$$ - -Notice that *rxy*[*k*] is quite similar to the convolution sum. The independent variable *k* corresponds to the relative *shift* between the two inputs. - -- (a) Express *rxy*[*k*] in terms of convolution. Is *rxy*[*k*] = *ryx*[*k*]? -- (b) Cross-correlation is said to indicate similarity between two signals. Do you agree? Why or why not? -- (c) If *x*[*n*] and *y*[*n*] are both finite duration, MATLAB's conv command is well suited to compute *rxy*[*k*]. Write a MATLAB function that computes the cross-correlation function using the conv command. Four vectors are passed to the function (x, y, nx, and ny) corresponding to the inputs *x*[*n*], *y*[*n*], and their respective time vectors. Notice that x and y are not necessarily the same length. Two outputs should be created (rxy and k) corresponding to *rxy*[*k*] and its shift vector. -- (d) Test your code from part (c) using *x*[*n*] = *u*[*n* − 5] − *u*[*n* − 10] over (0 ≤ *n* = *nx* ≤ 20) and *y*[*n*] = *u*[−*n*−15]−*u*[−*n*−10]+δ[*n*− 2] over (−20 ≤ *n* = *ny* ≤ 10). Plot the result rxy as a function of the shift vector k. What shift *k* gives the largest magnitude of *rxy*[*k*]? Does this make sense? -- **3.11-6** Suppose a vector x exists in the MATLAB workspace, corresponding to a finite-duration DT signal *x*[*n*] - - (a) Write a MATLAB function that, when passed vector x, computes and returns Ex, the energy of *x*[*n*]. - - (b) Write a MATLAB function that, when passed vector x, computes and returns Px, the power of *x*[*n*]. Assume that *x*[*n*] is periodic and that vector x contains data for an integer number of periods of *x*[*n*]. -- **3.11-7** A causal *N*-point max filter assigns *y*[*n*] to the maximum of {*x*[*n*],..., *x*[*n*−(*N* −1)]}. - - (a) Write a MATLAB function that performs *N*-point max filtering on a length-*M* input vector x. The two function inputs are vector x and scalar N. To create the length-*M* output - -vector y, initially pad the input vector with *N* − 1 zeros. The MATLAB command max may be helpful. - -- (b) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4, *N* = 8, and *N* = 12. Comment on the filter behavior. -- **3.11-8** A causal *N*-point min filter assigns *y*[*n*] to the minimum of {*x*[*n*],..., *x*[*n*−(*N* −1)]}. - - (a) Write a MATLAB function that performs *N*-point min filtering on a length-*M* input vector x. The two function inputs are vector x and scalar N. To create the length-*M* output vector y, initially pad the input vector with *N* − 1 zeros. The MATLAB command min may be helpful. - - (b) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4,*N* = 8, and *N* = 12. Comment on the filter behavior. -- **3.11-9** A causal *N*-point median filter assigns *y*[*n*] to the median of {*x*[*n*],..., *x*[*n*−(*N* −1)]}. The median is found by sorting sequence {*x*[*n*],..., *x*[*n* − (*N* − 1)]} and choosing the middle value (odd *N*) or the average of the two middle values (even *N*). - - (a) Write a MATLAB function that performs *N*-point median filtering on a length-*M* input vector x. The two function inputs are vector x and scalar N. To create the length-*M* output vector y, initially pad the input vector with *N* − 1 zeros. The MATLAB command sort or median may be helpful. - - (b) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4, *N* = 8, and *N* = 12. Comment on the filter behavior. -- **3.11-10** Recall that *y*[*n*] = *x*[*n*/*N*] represents an upsample by *N* operation. An interpolation filter replaces the inserted zeros with more realistic values. A linear interpolation filter has impulse response - -$$ -h[n] = \sum_{k=-(N-1)}^{N-1} \left(1 - \left|\frac{k}{N}\right|\right) \delta(n-k) -$$ - -- (a) Determine a constant coefficient difference equation that has impulse response *h*[*n*]. -- (b) The impulse response *h*[*n*] is noncausal. What is the smallest time shift necessary to make the filter causal? What is the effect of this shift on the behavior of the filter? -- (c) Write a MATLAB function that will compute the parameters necessary to implement an interpolation filter using MATLAB's filter command. That is, your function should output filter vectors b and a given an input scalar *N*. -- (d) Test your filter and MATLAB code. To do this, create *x*[*n*] = cos(*n*) for (0 ≤ *n* ≤ 9). Upsample *x*[*n*] by *N* = 10 to create a new signal *xup*[*n*]. Design the corresponding *N* = 10 linear interpolation filter, filter *xup*[*n*] to produce *y*[*n*], and plot the results. -- **3.11-11** A causal *N*-point moving-average filter has impulse response *h*[*n*] = (*u*[*n*] − *u*[*n* − *N*])/*N*. - - (a) Determine a constant-coefficient difference equation that has impulse response *h*[*n*]. - - (b) Write a MATLAB function that will compute the parameters necessary to implement an *N*-point moving-average filter using MATLAB's filter command. That is, your function should output filter vectors b and a given a scalar input *N*. - - (c) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4, *N* = 8, and *N* = 12. Comment on the filter behavior. - - (d) Problem 3.11-10 introduces linear interpolation filters, for use following an upsample by *N* operation. Within a scale factor, show that a cascade of two *N*-point moving-average filters is equivalent to the linear interpolation filter. What is the scale factor difference? Test this idea with MAT-LAB. Create *x*[*n*] = cos(*n*) for (0 ≤ *n* ≤ 9). Upsample *x*[*n*] by *N* = 10 to create a new signal *xup*[*n*]. Design an *N* = 10 moving-average filter. Filter *xup*[*n*] twice and scale to produce *y*[*n*]. Plot the results. Does the output from the cascaded pair of moving-average filters linearly interpolate the upsampled data? - -# **[CONTINUOUS-TIME](#page-10-0) SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM** - -Because of the linearity (superposition) property of linear time-invariant systems, we can find the response of these systems by breaking the input *x*(*t*) into several components and then summing the system response to all the components of *x*(*t*). We have already used this procedure in time-domain analysis, in which the input *x*(*t*) is broken into impulsive components. In the *frequency-domain analysis* developed in this chapter, we break up the input *x*(*t*) into exponentials of the form *est*, where the parameter *s* is the complex frequency of the signal *est*, as explained in Sec. 1.4-3. This method offers an insight into the system behavior complementary to that seen in the time-domain analysis. In fact, the time-domain and the frequency-domain methods are duals of each other. - -The tool that makes it possible to represent arbitrary input *x*(*t*) in terms of exponential components is the *Laplace transform*, which is discussed in the following section. - -## **4.1 THE LAPLACE [TRANSFORM](#page-10-0)** - -For a signal *x*(*t*), its Laplace transform *X*(*s*) is defined by - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ -\n(4.1) - -The signal *x*(*t*) is said to be the *inverse Laplace transform* of *X*(*s*). It can be shown that - -$$ -x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds -$$ -\n(4.2) - -where *c* is a constant chosen to ensure the convergence of the integral in Eq. (4.1), as explained later. See also [1]. - -This pair of equations is known as the *bilateral Laplace transform pair*, where *X*(*s*) is the direct Laplace transform of *x*(*t*) and *x*(*t*) is the inverse Laplace transform of *X*(*s*). Symbolically, - -> *X*(*s*) = *L*[*x*(*t*)] and *x*(*t*) = *L*−1 [*X*(*s*)] - -Note that - -CHAPTER - -**4** - -$$ -\mathcal{L}^{-1}\{\mathcal{L}[x(t)]\} = x(t) \qquad \text{and} \qquad \mathcal{L}\{\mathcal{L}^{-1}[X(s)]\} = X(s) -$$ - -It is also common practice to use a bidirectional arrow to indicate a Laplace transform pair, as follows: - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -The Laplace transform, defined in this way, can handle signals existing over the entire time interval from −∞ to ∞ (causal and noncausal signals). For this reason it is called the *bilateral* (or *two-sided*) Laplace transform. Later we shall consider a special case—the *unilateral* or *one-sided* Laplace transform—which can handle only causal signals. - -### LINEARITY OF THE LAPLACE TRANSFORM - -We now prove that the Laplace transform is a linear operator by showing that the principle of superposition holds, implying that if - -$$ -x_1(t) \Longleftrightarrow X_1(s) -$$ - and $x_2(t) \Longleftrightarrow X_2(s)$ - -then - -$$ -a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(s) + a_2X_2(s) -$$ - -The proof is simple. By definition, - -$$ -\mathcal{L}[a_1x_1(t) + a_2x_2(t)] = \int_{-\infty}^{\infty} [a_1x_1(t) + a_2x_2(t)]e^{-st} dt -$$ - -\n -$$ -= a_1 \int_{-\infty}^{\infty} x_1(t)e^{-st} dt + a_2 \int_{-\infty}^{\infty} x_2(t)e^{-st} dt -$$ - -\n -$$ -= a_1X_1(s) + a_2X_2(s) -$$ -\n(4.3) - -This result can be extended to any finite sum. - -### THE REGION OF CONVERGENCE (ROC) - -The *region of convergence* (ROC), also called the region of existence, for the Laplace transform, *X*(*s*), is the set of values of *s* (the region in the complex plane) for which the integral in Eq. (4.1) converges. This concept will become clear in the following example. - -### **EXAMPLE 4.1 Laplace Transform and ROC of a Causal Exponential** - -For a signal *x*(*t*) = *e*−*atu*(*t*), find the Laplace transform *X*(*s*) and its ROC. - -By definition, - -$$ -X(s) = \int_{-\infty}^{\infty} e^{-at} u(t) e^{-st} dt -$$ - -Because *u*(*t*) = 0 for *t* < 0 and *u*(*t*) = 1 for *t* ≥ 0, - -$$ -X(s) = \int_0^\infty e^{-at} e^{-st} dt = \int_0^\infty e^{-(s+a)t} dt = -\frac{1}{s+a} e^{-(s+a)t} \Big|_0^\infty \tag{4.4} -$$ - -### 332 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Note that *s* is complex and as *t* → ∞, the term *e*−(*s*+*a*)*t* does not necessarily vanish. Here we recall that for a complex number *z* = α +*j*β, - -$$ -e^{-zt} = e^{-(\alpha+j\beta)t} = e^{-\alpha t}e^{-j\beta t} -$$ - -Now |*e*−*j*β*t* | = 1 regardless of the value of β*t*. Therefore, as *t* → ∞, *e*−*zt* → 0 only if α > 0, and *e*−*zt* → ∞ if α < 0. Thus, - -$$ -\lim_{t \to \infty} e^{-zt} = \begin{cases} 0 & \text{Re } z > 0 \\ \infty & \text{Re } z < 0 \end{cases} -$$ - (4.5) - -Clearly, - -$$ -\lim_{t \to \infty} e^{-(s+a)t} = \begin{cases} 0 & \text{Re}(s+a) > 0\\ \infty & \text{Re}(s+a) < 0 \end{cases} -$$ - -Use of this result in Eq. (4.4) yields - -$$ -X(s) = \frac{1}{s+a} \qquad \text{Re}(s+a) > 0 -$$ -$$ -e^{-at}u(t) \Longleftrightarrow \frac{1}{s+a} \qquad \text{Re } s > -a \tag{4.6} -$$ - -or - -The ROC of -$$ -X(s) -$$ - is Re $s > -a$ , as shown in the shaded area in Fig. 4.1a. This fact means that the integral defining $X(s)$ in Eq. (4.4) exists only for the values of $s$ in the shaded region in Fig. 4.1a. For other values of $s$ , the integral in Eq. (4.4) does not converge. For this reason, the shaded region is called the *ROC* (or the *region of existence*) for $X(s)$ . - -**Figure 4.1** Signals **(a)** *e*−*atu*(*t*) and **(b)** −*e*−*atu*(−*t*) have the same Laplace transform but different regions of convergence. - -### REGION OF CONVERGENCE FOR FINITE-DURATION SIGNALS - -A finite-duration signal *xf*(*t*) is a signal that is nonzero only for *t*1 ≤ *t* ≤ *t*2, where both *t*1 and *t*2 are finite numbers and *t*2 > *t*1. For a finite-duration, absolutely integrable signal, the ROC is the entire *s* plane. This is clear from the fact that if *xf*(*t*) is absolutely integrable and a finite-duration signal, then *x*(*t*)*e*−σ*t* is also absolutely integrable for any value of σ because the integration is over the finite range of *t* only. Hence, the Laplace transform of such a signal converges for every value of *s*. This means that the ROC of a general signal *x*(*t*) remains unaffected by the addition of any absolutely integrable, finite-duration signal *xf*(*t*) to *x*(*t*). In other words, if *R* represents the ROC of a signal *x*(*t*), then the ROC of a signal *x*(*t*)+*xf*(*t*) is also *R*. - -### ROLE OF THE REGION OF CONVERGENCE - -The ROC is required for evaluating the inverse Laplace transform *x*(*t*) from *X*(*s*), as defined by Eq. (4.2). The operation of finding the inverse transform requires an integration in the complex plane, which needs some explanation. The path of integration is along *c*+*j*ω, with ω varying from −∞ to ∞. † Moreover, the path of integration must lie in the ROC (or existence) for *X*(*s*). For the signal *e*−*atu*(*t*), this is possible if *c* > −*a*. One possible path of integration is shown (dotted) in Fig. 4.1a. Thus, to obtain *x*(*t*) from *X*(*s*), the integration in Eq. (4.2) is performed along this path. When we integrate [1/(*s* + *a*)]*est* along this path, the result is *e*−*atu*(*t*). Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of Laplace transforms (Table 4.1), where the Laplace transform pairs are tabulated for a variety of signals. To find the inverse Laplace transform of, say, 1/(*s* + *a*), instead of using the complex integral of Eq. (4.2), we look up the table and find the inverse Laplace transform to be *e*−*atu*(*t*) (assuming that the ROC is Re *s* > −*a*). Although the table given here is rather short, it comprises the functions of most practical interest. A more comprehensive table appears in Doetsch [2]. - -### THE UNILATERAL LAPLACE TRANSFORM - -To understand the need for defining unilateral transform, let us find the Laplace transform of signal *x*(*t*) illustrated in Fig. 4.1b: - -$$ -x(t) = -e^{-at}u(-t) -$$ - -The Laplace transform of this signal is - -$$ -X(s) = \int_{-\infty}^{\infty} -e^{-at}u(-t)e^{-st}dt -$$ - -Because *u*(−*t*) = 1 for *t* < 0 and *u*(−*t*) = 0 for *t* > 0, - -$$ -X(s) = \int_{-\infty}^{0} -e^{-at}e^{-st} dt = -\int_{-\infty}^{0} e^{-(s+a)t} dt = \frac{1}{s+a}e^{-(s+a)t} \Big|_{-\infty}^{0} -$$ - -† The discussion about the path of convergence is rather complicated, requiring the concepts of contour integration and understanding of the theory of complex variables. For this reason, the discussion here is somewhat simplified. - -| No. | x(t) | X(s) | -|-----|-------------------------------------------------------|-----------------------------------------------------------| -| 1 | δ(t) | 1 | -| 2 | u(t) | 1
s | -| 3 | tu(t) | 1
s2 | -| 4 | nu(t)
t | n!
sn+1 | -| 5 | eλt
u(t) | 1
s−λ | -| 6 | teλt
u(t) | 1
(s−λ)2 | -| 7 | neλt
t
u(t) | n!
(s−λ)n+1 | -| 8a | cos bt u(t) | s
s2 +b2 | -| 8b | sin bt u(t) | b
s2 +b2 | -| 9a | e−at cos
bt u(t) | s+a
(s+a)2 +b2 | -| 9b | e−atsin
bt u(t) | b
(s+a)2 +b2 | -| 10a | re−at cos(bt
+θ )u(t) | (r cos θ )s +(ar cos θ −brsin θ )
s2 +2as
+(a2 +b2) | -| 10b | re−at cos(bt
+θ )u(t) | 0.5rejθ
0.5re−jθ
s+a−jb +
s +a+jb | -| 10c | re−at cos(bt
+θ )u(t) | As+B
s2 +2as+c | -| |
A2c+B2 −2ABa
r =
c−a2 | | -| | Aa−B
θ = tan−1
√
c−a2
A | | -| | b = √
c−a2 | | -| 10d | e−at
!
B−Aa
Acos bt +
sin bt
u(t)
b | As+B
s2 +2as+c | -| | b = √
c−a2 | | - -**TABLE 4.1** Select (Unilateral) Laplace Transform Pairs - -Equation (4.5) shows that - -lim *t*→−∞*e*−(*s*+*a*)*t* = 0 Re (*s*+*a*) < 0 - -Hence, - -$$ -X(s) = \frac{1}{s+a} \qquad \text{Re } s < -a -$$ - -The signal −*e*−*atu*(−*t*) and its ROC (Re *s* < −*a*) are depicted in Fig. 4.1b. Note that the Laplace transforms for the signals *e*−*atu*(*t*) and −*e*−*atu*(−*t*) are identical except for their regions of convergence. Therefore, for a given *X*(*s*), there may be more than one inverse transform, depending on the ROC. In other words, unless the ROC is specified, there is no one-to-one correspondence between *X*(*s*) and *x*(*t*). This fact increases the complexity in using the Laplace transform. The complexity is the result of trying to handle causal as well as noncausal signals. If we restrict all our signals to the causal type, such an ambiguity does not arise. There is only one inverse transform of *X*(*s*) = 1/(*s* + *a*), namely, *e*−*atu*(*t*). To find *x*(*t*) from *X*(*s*), we need not even specify the ROC. In summary, if all signals are restricted to the causal type, then, for a given *X*(*s*), there is only one inverse transform *x*(*t*). † - -The unilateral Laplace transform is a special case of the bilateral Laplace transform in which all signals are restricted to being causal; consequently, the limits of integration for the integral in Eq. (4.1) can be taken from 0 to ∞. Therefore, the unilateral Laplace transform *X*(*s*) of a signal *x*(*t*) is defined as - -$$ -X(s) = \int_{0^{-}}^{\infty} x(t)e^{-st} dt -$$ - (4.7) - -We choose 0− (rather than 0+ used in some texts) as the lower limit of integration. This convention not only ensures inclusion of an impulse function at *t* = 0, but also allows us to use initial conditions at 0− (rather than at 0+) in the solution of differential equations via the Laplace transform. In practice, we are likely to know the initial conditions before the input is applied (at 0−), not after the input is applied (at 0+). Indeed, the very meaning of the term "initial conditions" implies conditions at *t* = 0− (conditions before the input is applied). Detailed analysis of desirability of using *t* = 0− appears in Sec. 4.3. - -The unilateral Laplace transform simplifies the system analysis problem considerably because of its *uniqueness property*, which says that for a given *X*(*s*), there is a unique inverse transform. But there is a price for this simplification: we cannot analyze noncausal systems or use noncausal inputs. However, in most practical problems, this restriction is of little consequence. For this reason, we shall first consider the unilateral Laplace transform and its application to system analysis. (The bilateral Laplace transform is discussed later, in Sec. 4.11.) - -Basically there is no difference between the unilateral and the bilateral Laplace transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at *t* = 0 (causal signals). Therefore, the expression [Eq. (4.2)] for the inverse Laplace transform remains unchanged. In practice, the term *Laplace transform* means *the unilateral Laplace transform*. - -† Actually, *X*(*s*) specifies *x*(*t*) within a null function *n*(*t*), which has the property that the area under |*n*(*t*)| 2 is zero over any finite interval 0 to *t* (*t* > 0) (Lerch's theorem). For example, if two functions are identical everywhere except at finite number of points, they differ by a null function. - -### EXISTENCE OF THE LAPLACE TRANSFORM - -The variable *s* in the Laplace transform is complex in general, and it can be expressed as *s* =σ +*j*ω. By definition, - -$$ -X(s) = \int_{0^-}^{\infty} x(t)e^{-st} dt = \int_{0^-}^{\infty} [x(t)e^{-\sigma t}]e^{-j\omega t} dt -$$ - -Because |*ej*ω*t* | = 1, the integral on the right-hand side of this equation converges if - -$$ -\int_{0^{-}}^{\infty} \left| x(t)e^{-\sigma t} \right| dt < \infty \tag{4.8} -$$ - -Hence the existence of the Laplace transform is guaranteed if the integral in Eq. (4.8) is finite for some value of σ. Any signal that grows no faster than an exponential signal *Me*σ0*t* for some *M* and σ0 satisfies the condition of Eq. (4.8). Thus, if for some *M* and σ0, - -$$ -|x(t)| \le Me^{\sigma_0 t} \tag{4.9} -$$ - -we can choose σ>σ0 to satisfy Eq. (4.8).† The signal *et* 2 , in contrast, grows at a rate faster than *e*σ0*t* , and consequently is not Laplace–transformable.‡ Fortunately such signals (which are not Laplace–transformable) are of little consequence from either a practical or a theoretical viewpoint. If σ0 is the smallest value of σ for which the integral in Eq. (4.8) is finite, σ0 is called the *abscissa of convergence* and the ROC of *X*(*s*) is Re *s* > σ0. The abscissa of convergence for *e*−*atu*(*t*) is −*a* (the ROC is Re *s* > −*a*). - -### **EXAMPLE 4.2 Bilateral Laplace Transform of Common Causal Signals** - -Determine the Laplace transform of the following: **(a)** δ(*t*), **(b)** *u*(*t*), and **(c)** cos ω0*t u*(*t*). - -**(a)** - -$$ -\mathcal{L}[\delta(t)] = \int_{0^-}^{\infty} \delta(t) e^{-st} dt -$$ - -Using the sampling property [Eq. (1.11) with *T* = 0], we obtain - -$$ -\mathcal{L}[\delta(t)] = 1 \qquad \text{for all } s -$$ - -that is, - -$$ -\delta(t) \Longleftrightarrow 1 \qquad \text{for all } s -$$ - -† The condition of Eq. (4.9) is sufficient but not necessary for the existence of the Laplace transform. For example, *x*(*t*) = 1/ √*t* is infinite at *t* = 0, and Eq. (4.9) cannot be satisfied; but the transform of 1/ √*t* exists and is given by √π/*s*. - -‡ However, if we consider a truncated (finite-duration) signal *et* 2 , the Laplace transform exists. - -**(b)** To find the Laplace transform of *u*(*t*), recall that *u*(*t*) = 1 for *t* ≥ 0. Therefore, - -$$ -\mathcal{L}[u(t)] = \int_{0^{-}}^{\infty} u(t)e^{-st} dt = \int_{0^{-}}^{\infty} e^{-st} dt = -\frac{1}{s}e^{-st} \Big|_{0^{-}}^{\infty} -$$ - -= $\frac{1}{s}$ Re $s > 0$ - -We also could have obtained this result from Eq. (4.6) by letting *a* = 0. **(c)** Because cos ω0*t u*(*t*) = 1 2 [*ej*ω0*t* +*e*−*j*ω0*t* ]*u*(*t*), we know that - -$$ -\mathcal{L}[\cos \omega_0 t u(t)] = \frac{1}{2} \mathcal{L}[e^{j\omega_0 t} u(t) + e^{-j\omega_0 t} u(t)] -$$ - -From Eq. (4.6), it follows that - -$$ -\mathcal{L}[\cos \omega_0 t u(t)] = \frac{1}{2} \left[ \frac{1}{s - j\omega_0} + \frac{1}{s + j\omega_0} \right] \qquad \text{Re}\,(s \pm j\omega) = \text{Re}\,s > 0 -$$ -\n -$$ -= \frac{s}{s^2 + \omega_0^2} \qquad \text{Re}\,s > 0 \tag{4.10} -$$ - -For the unilateral Laplace transform, there is a unique inverse transform of *X*(*s*); consequently, there is no need to specify the ROC explicitly. For this reason, we shall generally ignore any mention of the ROC for unilateral transforms. Recall, also, that in the unilateral Laplace transform it is understood that every signal *x*(*t*) is zero for *t* < 0, and it is appropriate to indicate this fact by multiplying the signal by *u*(*t*). - -### **DR ILL 4.1 Bilateral Laplace Transform of Gate Functions** - -By direct integration, find the Laplace transform *X*(*s*) and the region of convergence of *X*(*s*) for the gate functions shown in Fig. 4.2. - -**Figure 4.2** Gate functions for Drill 4.1. - -### **ANSWERS** - -(a) -$$ -\frac{1}{s}(1-e^{-2s}) -$$ - for all *s* -\n(b) $\frac{1}{s}(1-e^{-2s})e^{-2s}$ for all *s* - -### **[4.1-1 Finding the Inverse Transform](#page-10-0)** - -Finding the inverse Laplace transform by using Eq. (4.2) requires integration in the complex plane, a subject beyond the scope of this book (but see, e.g., [3]). For our purpose, we can find the inverse transforms from Table 4.1. All we need is to express *X*(*s*) as a sum of simpler functions of the forms listed in the table. Most of the transforms *X*(*s*) of practical interest are *rational functions*, that is, ratios of polynomials in *s*. Such functions can be expressed as a sum of simpler functions by using partial fraction expansion (see Sec. B.5). - -Values of *s* for which *X*(*s*) = 0 are called the *zeros* of *X*(*s*); the values of *s* for which *X*(*s*)→∞ are called the *poles* of *X*(*s*). If *X*(*s*) is a rational function of the form *P*(*s*)/*Q*(*s*), the roots of *P*(*s*) are the zeros and the roots of *Q*(*s*) are the poles of *X*(*s*). - -### **EXAMPLE 4.3 Inverse Unilateral Laplace Transform** - -Find the inverse unilateral Laplace transforms of - -(a) -$$ -\frac{7s-6}{s^2-s-6} -$$ - -\n(b) -$$ -\frac{2s^2+5}{s^2+3s+2} -$$ - -\n(c) -$$ -\frac{6(s+34)}{s(s^2+10s+34)} -$$ - -\n(d) -$$ -\frac{8s+10}{(s+1)(s+2)^3} -$$ - -In no case is the inverse transform of these functions directly available in Table 4.1. Rather, we need to expand these functions into partial fractions, as discussed in Sec. B.5-1. Today, it is very easy to find partial fractions via software such as MATLAB. However, just as the availability of a calculator does not obviate the need for learning the mechanics of arithmetical operations (addition, multiplication, etc.), the widespread availability of computers does not eliminate the need to learn the mechanics of partial fraction expansion. - -$$ -\left( \mathbf{a}\right) -$$ - -$$ -X(s) = \frac{7s - 6}{(s + 2)(s - 3)} = \frac{k_1}{s + 2} + \frac{k_2}{s - 3} -$$ - -To determine *k*1, corresponding to the term (*s* + 2), we cover up (conceal) the term (*s* + 2) in *X*(*s*) and substitute *s* = −2 (the value of *s* that makes *s* + 2 = 0) in the remaining expression (see Sec. B.5-2): - -$$ -k_1 = \frac{7s - 6}{(s + 2)(s - 3)}\bigg|_{s = -2} = \frac{-14 - 6}{-2 - 3} = 4 -$$ - -Similarly, to determine *k*2 corresponding to the term (*s* − 3), we cover up the term (*s* − 3) in *X*(*s*) and substitute *s* = 3 in the remaining expression - -$$ -k_2 = \frac{7s - 6}{(s + 2)(s - 3)}\bigg|_{s=3} = \frac{21 - 6}{3 + 2} = 3 -$$ - -Therefore, - -$$ -X(s) = \frac{7s - 6}{(s + 2)(s - 3)} = \frac{4}{s + 2} + \frac{3}{s - 3} -$$ -(4.11) - -### CHECKING THE ANSWER - -It is easy to make a mistake in partial fraction computations. Fortunately it is simple to check the answer by recognizing that *X*(*s*) and its partial fractions must be equal for every value of *s* if the partial fractions are correct. Let us verify this assertion in Eq. (4.11) for some convenient value, say, *s* = 0. Substitution of *s* = 0 in Eq. (4.11) yields† - -$$ -1 = 2 - 1 = 1 -$$ - -We can now be sure of our answer with a high margin of confidence. Using pair 5 of Table 4.1 in Eq. (4.11), we obtain - -$$ -x(t) = \mathcal{L}^{-1}\left(\frac{4}{s+2} + \frac{3}{s-3}\right) = (4e^{-2t} + 3e^{3t})u(t) -$$ - -**(b)** - -$$ -X(s) = \frac{2s^2 + 5}{s^2 + 3s + 2} = \frac{2s^2 + 5}{(s+1)(s+2)} -$$ - -Observe that *X*(*s*) is an improper function with *M* = *N*. In such a case, we can express *X*(*s*) as a sum of the coefficient of the highest power in the numerator plus partial fractions corresponding to the poles of *X*(*s*) (see Sec. B.5-5). In the present case, the coefficient of the highest power in the numerator is 2. Therefore, - -$$ -X(s) = 2 + \frac{k_1}{s+1} + \frac{k_2}{s+2} -$$ - -where - -$$ -k_1 = \frac{2s^2 + 5}{(s+1)(s+2)}\bigg|_{s=-1} = \frac{2+5}{-1+2} = 7 -$$ - -and - -$$ -k_2 = \frac{2s^2 + 5}{(s+1)(s+2)}\bigg|_{s=-2} = \frac{8+5}{-2+1} = -13 -$$ - -Therefore, - -$$ -X(s) = 2 + \frac{7}{s+1} - \frac{13}{s+2} -$$ - -From Table 4.1, pairs 1 and 5, we obtain - -$$ -x(t) = 2\delta(t) + (7e^{-t} - 13e^{-2t})u(t) -$$ - -† Because *X*(*s*) = ∞ at its poles, we should avoid the pole values (−2 and 3 in the present case) for checking. The answers may check even if partial fractions are wrong. This situation can occur when two or more errors cancel their effects. But the chances of this problem arising for randomly selected values of *s* are extremely small. - -**(c)** - -$$ -X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{6(s+34)}{s(s+5-j3)(s+5+j3)} -$$ -$$ -= \frac{k_1}{s} + \frac{k_2}{s+5-j3} + \frac{k_2^*}{s+5+j3} -$$ - -Note that the coefficients (*k*2 and *k*∗ 2) of the conjugate terms must also be conjugate (see Sec. B.5). Now - -$$ -k_1 = \frac{6(s+34)}{s(s^2+10s+34)}\bigg|_{s=0} = \frac{6 \times 34}{34} = 6 -$$ - -$$ -k_2 = \frac{6(s+34)}{s(s+5-j3)(s+5+j3)}\bigg|_{s=-5+j3} = \frac{29+j3}{-3-j5} = -3+j4 -$$ - -Therefore, - -$$ -k_2^* = -3 - j4 -$$ - -To use pair 10b of Table 4.1, we need to express *k*2 and *k*∗ 2 in polar form. - -$$ --3 + j4 = (\sqrt{3^2 + 4^2}) e^{j \tan^{-1}(4/4)} = 5 e^{j \tan^{-1}(4/4)} -$$ - -Observe that tan−1(4/−3) = tan−1(−4/3). This fact is evident in Fig. 4.3. For further discussion of this topic, see Ex. B.1. - -**Figure 4.3** Visualizing tan−1(−4/3) = tan−1(4/−3). - -From Fig. 4.3, we observe that - -$$ -k_2 = -3 + j4 = 5e^{j126.9^{\circ}} -$$ - -so - -$$ -k_2^* = 5e^{-j126.9^\circ} -$$ - -Therefore, - -$$ -X(s) = \frac{6}{s} + \frac{5e^{j126.9^{\circ}}}{s+5-j3} + \frac{5e^{-j126.9^{\circ}}}{s+5+j3} -$$ - -From Table 4.1 (pairs 2 and 10b), we obtain - -$$ -x(t) = [6 + 10e^{-5t}\cos(3t + 126.9^\circ)]u(t) -$$ - -### ALTERNATIVE METHOD USING QUADRATIC FACTORS - -The foregoing procedure involves considerable manipulation of complex numbers. Pair 10c (Table 4.1) indicates that the inverse transform of quadratic terms (with complex conjugate poles) can be found directly without having to find first-order partial fractions. We discussed such a procedure in Sec. B.5-2. For this purpose, we shall express *X*(*s*) as - -$$ -X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{k_1}{s} + \frac{As+B}{s^2+10s+34} -$$ - -We have already determined that *k*1 = 6 by the (Heaviside) "cover-up" method. Therefore, - -$$ -\frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{As+B}{s^2+10s+34} -$$ - -Clearing the fractions by multiplying both sides by *s*(*s*2 +10*s*+34) yields - -$$ -6(s+34) = (6+A)s2 + (60+B)s + 204 -$$ - -Now, equating the coefficients of *s*2 and *s* on both sides yields - -$$ -A = -6 \qquad \text{and} \qquad B = -54 -$$ - -and - -$$ -X(s) = \frac{6}{s} + \frac{-6s - 54}{s^2 + 10s + 34} -$$ - -We now use pairs 2 and 10c to find the inverse Laplace transform. The parameters for pair 10c are *A* = −6, *B* = −54, *a* = 5, *c* = 34, *b* = √ *c*−*a*2 = 3, and - -$$ -r = \sqrt{\frac{A^2c + B^2 - 2ABA}{c - a^2}} = 10 \qquad \theta = \tan^{-1} \frac{Aa - B}{A\sqrt{c - a^2}} = 126.9^{\circ} -$$ - -Therefore, - -$$ -x(t) = [6 + 10e^{-5t}\cos(3t + 126.9^\circ)]u(t) -$$ - -which agrees with the earlier result. - -### SHORTCUTS - -The partial fractions with quadratic terms also can be obtained by using shortcuts. We have - -$$ -X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{As+B}{s^2+10s+34} -$$ - -### 342 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -We can determine *A* by eliminating *B* on the right-hand side. This step can be accomplished by multiplying both sides of the equation for *X*(*s*) by *s* and then letting *s*→ ∞. This procedure yields - -$$ -0 = 6 + A \quad \Longrightarrow \quad A = -6 -$$ - -Therefore, - -$$ -\frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{-6s+16}{s^2+10s+34} -$$ - -To find *B*, we let *s* take on any convenient value, say, *s* = 1, in this equation to obtain - -$$ -\frac{210}{45} = 6 + \frac{B - 6}{45} \implies B = -54 -$$ - -a result that agrees with the answer found earlier. - -**(d)** - -$$ -X(s) = \frac{8s+10}{(s+1)(s+2)^3} = \frac{k_1}{s+1} + \frac{a_0}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2} -$$ - -where - -$$ -k_1 = \frac{8s + 10}{(s+1)(s+2)^3}\Big|_{s=-1} = 2 -$$ - -\n -$$ -a_0 = \frac{8s + 10}{(s+1)(s+2)^3}\Big|_{s=-2} = 6 -$$ - -\n -$$ -a_1 = \left\{\frac{d}{ds}\left[\frac{8s + 10}{(s+1)(s+2)^3}\right]\right\}_{s=-2} = -2 -$$ - -\n -$$ -a_2 = \frac{1}{2}\left\{\frac{d^2}{ds^2}\left[\frac{8s + 10}{(s+1)(s+2)^3}\right]\right\}_{s=-2} = -2 -$$ - -Therefore, - -$$ -X(s) = \frac{2}{s+1} + \frac{6}{(s+2)^3} - \frac{2}{(s+2)^2} - \frac{2}{s+2} -$$ - -and - -$$ -x(t) = [2e^{-t} + (3t^2 - 2t - 2)e^{-2t}]u(t) -$$ - -### ALTERNATIVE METHOD:AHYBRID OF HEAVISIDE AND CLEARING FRACTIONS - -In this method, the simpler coefficients *k*1 and *a*0 are determined by the Heaviside "cover-up" procedure, as discussed earlier. To determine the remaining coefficients, we use the clearing-fraction method. Using the values *k*1 = 2 and *a*0 = 6 obtained earlier by the Heaviside "cover-up" method, we have - -$$ -\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2} -$$ - -We now clear fractions by multiplying both sides of the equation by (*s* + 1)(*s* + 2)3. This procedure yields† - -$$ -8s + 10 = 2(s + 2)3 + 6(s + 1) + a1(s + 1)(s + 2) + a2(s + 1)(s + 2)2 -$$ - -= (2 + a2)s3 + (12 + a1 + 5a2)s2 + (30 + 3a1 + 8a2)s + (22 + 2a1 + 4a2) - -Equating coefficients of *s*3 and *s*2 on both sides, we obtain - -$$ -0 = (2 + a_2) \implies a_2 = -2 0 = 12 + a_1 + 5a_2 = 2 + a_1 \implies a_1 = -2 -$$ - -We can stop here if we wish, since the two desired coefficients *a*1 and *a*2 have already been found. However, equating the coefficients of *s*1 and *s*0 serves as a check on our answers. This step yields - -$$ -8 = 30 + 3a_1 + 8a_2 -$$ - -$$ -10 = 22 + 2a_1 + 4a_2 -$$ - -Substitution of *a*1 = *a*2 = −2, obtained earlier, satisfies these equations. This step confirms the correctness of our answers. - -### ANOTHER ALTERNATIVE:AHYBRID OF HEAVISIDE AND SHORTCUTS - -In this method, the simpler coefficients *k*1 and *a*0 are determined by the Heaviside "cover-up" procedure, as discussed earlier. The usual shortcuts are then used to determine the remaining coefficients. Using the values *k*1 = 2 and *a*0 = 6, determined earlier by the Heaviside method, we have - -$$ -\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2} -$$ - -There are two unknowns, *a*1 and *a*2. If we multiply both sides by *s* and then let *s* → ∞, we eliminate *a*1. This procedure yields - -0 = 2+*a*2 ⇒ *a*2 = −2 - -Therefore, - -$$ -\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} - \frac{2}{s+2} -$$ - -There is now only one unknown, *a*1. This value can be determined readily by setting *s* equal to any convenient value, say, *s* = 0. This step yields - -$$ -\frac{10}{8} = 2 + \frac{3}{4} + \frac{a_1}{4} - 1 \implies a_1 = -2 -$$ - -† We could have cleared fractions without finding *k*1 and *a*0. This alternative, however, proves more laborious because it increases the number of unknowns to 4. By predetermining *k*1 and *a*0, we reduce the unknowns to 2. Moreover, this method provides a convenient check on the solution. This hybrid procedure achieves the best of both methods. - -### **EXAMPLE 4.4 Inverse Laplace Transform with MATLAB** - -Using the MATLAB residue command, determine the inverse Laplace transform of each of the following functions: - -(a) -$$ -X_a(s) = \frac{2s^2 + 5}{s^2 + 3s + 2} -$$ - -\n(b) $X_b(s) = \frac{2s^2 + 7s + 4}{(s+1)(s+2)^2}$ -\n(c) $X_c(s) = \frac{8s^2 + 21s + 19}{(s+2)(s^2 + s + 7)}$ - -In each case, we use the MATLAB residue command to perform the necessary partial fraction expansions. The inverse Laplace transform follows using Table 4.1. - -``` -(a) ->> num = [2 0 5]; den = [1 3 2]; ->> [r, p, k] = residue(num,den) - r = -13 - 7 - p = -2 -``` - -Therefore, *Xa*(*s*) = −13/(*s*+2) +7/(*s*+1) +2 and *xa*(*t*) = (−13*e*−2*t* +7*e*−*t* )*u*(*t*)+2δ(*t*). - -``` -(b) -``` - --1 k= 2 - -``` ->> num = [2 7 4]; den = [conv([1 1],conv([1 2],[1 2]))]; ->> [r, p, k] = residue(num,den) - r= 3 - 2 - -1 - p = -2 - -2 - -1 - k = [] -``` - -Therefore, *Xb*(*s*) = 3/(*s*+2) +2/(*s*+2)2 −1/(*s*+1) and *xb*(*t*) = (3*e*−2*t* +2*te*−2*t* −*e*−*t* )*u*(*t*). - -**(c)** In this case, a few calculations are needed beyond the results of the residue command so that pair 10b of Table 4.1 can be utilized. - ->> num = [8 21 19]; den = [conv([1 2],[1 1 7])]; >> [r, p, k]= residue(num,den) - -``` -r = 3.5000-0.48113i - 3.5000+0.48113i - 1.0000 - p = -0.5000+2.5981i - -0.5000-2.5981i - -2.0000 - k = [] ->> ang = angle(r), mag = abs(r) - ang = -0.13661 - 0.13661 - 0 - mag = 3.5329 - 3.5329 - 1.0000 -``` - -Thus, - -$$ -X_c(s) = \frac{1}{s+2} + \frac{3.5329e^{-j0.13661}}{s+0.5-j2.5981} + \frac{3.5329e^{j0.13661}}{s+0.5+j2.5981} -$$ - -and - -$$ -x_c(t) = [e^{-2t} + 1.7665e^{-0.5t}\cos(2.5981t - 0.1366)]u(t). -$$ - -## **EXAMPLE 4.5 Symbolic Laplace and Inverse Laplace Transforms with MATLAB** - -Using MATLAB's symbolic math toolbox, determine the following: - -- **(a)** the direct unilateral Laplace transform of *xa*(*t*) = sin(*at*)+cos(*bt*) -- **(b)** the inverse unilateral Laplace transform of *Xb*(*s*) = *as*2/(*s*2 +*b*2) - -**(a)** Here, we use the sym command to symbolically define our variables and expression for *xa*(*t*), and then we use the laplace command to compute the (unilateral) Laplace transform. - ->> syms a b t; x\_a = sin(a\*t)+cos(b\*t); >> X\_a = laplace(x\_a); X\_a = a/(a^2 + s^2) + s/(b^2 + s^2) - -Therefore, *Xa*(*s*) = *a s*2+*a*2 + *s s*2+*b*2 . It is also easy to use MATLAB to determine *Xa*(*s*) in standard rational form. - ->> X\_a = collect(X\_a) X\_a = (a^2\*s + a\*b^2 + a\*s^2 + s^3)/(s^4 + (a^2 + b^2)\*s^2 + a^2\*b^2) - -Thus, we also see that *Xa*(*s*) = *s*3+*as*2+*a*2*s*+*ab*2 *s*4+(*a*2+*b*2)*s*2+*a*2*b*2 - -**(b)** A similar approach is taken for the inverse Laplace transform, except that the ilaplace command is used rather than the laplace command. - -``` ->> syms a b s; X_b = (a*s^2)/(s^2+b^2); ->> x_b = ilaplace(X_b) - x_b = a*dirac(t) - a*b*sin(b*t) -``` - -Therefore, *xb*(*t*) = *a*δ(*t*)−*ab*sin(*bt*)*u*(*t*). - -### **DR ILL 4.2 Laplace Transform** - -Show that the Laplace transform of 10*e*−3*t* cos (4*t* + 53.13◦) is (6*s* − 14)/(*s*2 + 6*s* + 25). Use Table 4.1. - -### **DR ILL 4.3 Inverse Laplace Transform** - -Find the inverse Laplace transform of the following: - -(a) -$$ -\frac{s+17}{s^2+4s-5} -$$ - -\n(b) -$$ -\frac{3s-5}{(s+1)(s^2+2s+5)} -$$ - -\n(c) -$$ -\frac{16s+43}{(s-2)(s+3)^2} -$$ - -### **ANSWERS** - -(a) -$$ -(3e^t - 2e^{-5t})u(t) -$$ - -- **(b)** −2*e*−*t* + 5 2 *e*−*t* cos(2*t* −36.87◦) *u*(*t*) -- **(c)** [3*e*2*t* +(*t* −3)*e*−3*t* ]*u*(*t*) - -## A HISTORICAL NOTE: MARQUIS PIERRE-SIMON DE LAPLACE (1749–1827) - -The Laplace transform is named after the great French mathematician and astronomer Laplace, who first presented the transform and its applications to differential equations in a paper published in 1779. - -Laplace developed the foundations of potential theory and made important contributions to special functions, probability theory, astronomy, and celestial mechanics. In his *Exposition du système du monde* (1796), Laplace formulated a nebular hypothesis of cosmic origin and tried to explain the universe as a pure mechanism. In his *Traité de mécanique céleste* (*celestial mechanics*), which completed the work of Newton, Laplace used mathematics and physics to subject the solar system and all heavenly bodies to the laws of motion and the principle of gravitation. Newton had - -Pierre-Simon de Laplace and Oliver Heaviside - -been unable to explain the irregularities of some heavenly bodies; in desperation, he concluded that God himself must intervene now and then to prevent such catastrophes as Jupiter eventually falling into the sun (and the moon into the earth), as predicted by Newton's calculations. Laplace proposed to show that these irregularities would correct themselves periodically and that a little patience—in Jupiter's case, 929 years—would see everything returning automatically to order; thus there was no reason why the solar and the stellar systems could not continue to operate by the laws of Newton and Laplace to the end of time [4]. - -Laplace presented a copy of *Mécanique céleste* to Napoleon, who, after reading the book, took Laplace to task for not including God in his scheme: "You have written this huge book on the system of the world without once mentioning the author of the universe." "Sire," Laplace retorted, "I had no need of that hypothesis." Napoleon was not amused, and when he reported this reply to another great mathematician-astronomer, Louis de Lagrange, the latter remarked, "Ah, but that is a fine hypothesis. It explains so many things" [5]. - -Napoleon, following his policy of honoring and promoting scientists, made Laplace the minister of the interior. To Napoleon's dismay, however, the new appointee attempted to bring "the spirit of infinitesimals" into administration, and so Laplace was transferred hastily to the Senate. - -### OLIVER HEAVISIDE (1850–1925) - -Although Laplace published his transform method to solve differential equations in 1779, the method did not catch on until a century later. It was rediscovered independently in a rather awkward form by an eccentric British engineer, Oliver Heaviside (1850–1925), one of the tragic figures in the history of science and engineering. Despite his prolific contributions to electrical engineering, he was severely criticized during his lifetime and was neglected later to the point that - -### 348 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -hardly a textbook today mentions his name or credits him with contributions. Nevertheless, his studies had a major impact on many aspects of modern electrical engineering. It was Heaviside who made transatlantic communication possible by inventing cable loading, but few mention him as a pioneer or an innovator in telephony. It was Heaviside who suggested the use of inductive cable loading, but the credit is given to M. Pupin, who was not even responsible for building the first loading coil.† In addition, Heaviside was [6]: - -- The first to find a solution to the distortionless transmission line. -- The innovator of lowpass filters. -- The first to write Maxwell's equations in modern form. -- The codiscoverer of rate energy transfer by an electromagnetic field. -- An early champion of the now-common phasor analysis. -- An important contributor to the development of vector analysis. In fact, he essentially created the subject independently of Gibbs [7]. -- An originator of the use of operational mathematics used to solve linear integro-differential equations, which eventually led to rediscovery of the ignored Laplace transform. -- The first to theorize (along with Kennelly of Harvard) that a conducting layer (the Kennelly–Heaviside layer) of atmosphere exists, which allows radio waves to follow earth's curvature instead of traveling off into space in a straight line. -- The first to posit that an electrical charge would increase in mass as its velocity increases, an anticipation of an aspect of Einstein's special theory of relativity [8]. He also forecast the possibility of superconductivity. - -Heaviside was a self-made, self-educated man. Although his formal education ended with elementary school, he eventually became a pragmatically successful mathematical physicist. He began his career as a telegrapher, but increasing deafness forced him to retire at the age of 24. He then devoted himself to the study of electricity. His creative work was disdained by many professional mathematicians because of his lack of formal education and his unorthodox methods. - -Heaviside had the misfortune to be criticized both by mathematicians, who faulted him for lack of rigor, and by men of practice, who faulted him for using too much mathematics and thereby confusing students. Many mathematicians, trying to find solutions to the distortionless transmission line, failed because no rigorous tools were available at the time. Heaviside succeeded because he used mathematics not with rigor, but with insight and intuition. Using his much maligned operational method, Heaviside successfully attacked problems that the rigid mathematicians could not solve, problems such as the flow-of-heat in a body of spatially varying conductivity. Heaviside brilliantly used this method in 1895 to demonstrate a fatal flaw in Lord Kelvin's determination of the geological age of the earth by secular cooling; he used the same flow-of-heat theory as for his cable analysis. Yet the mathematicians of the Royal Society remained unmoved and were not the least impressed by the fact that Heaviside had found the answer to problems no one else could solve. Many mathematicians who examined his work dismissed it - -† Heaviside developed the theory for cable loading, George Campbell built the first loading coil, and the telephone circuits using Campbell's coils were in operation before Pupin published his paper. In the legal fight over the patent, however, Pupin won the battle: he was a shrewd self-promoter, and Campbell had poor legal support. - -with contempt, asserting that his methods were either complete nonsense or a rehash of known ideas [6]. - -Sir William Preece, the chief engineer of the British Post Office, a savage critic of Heaviside, ridiculed Heaviside's work as too theoretical and, therefore, leading to faulty conclusions. Heaviside's work on transmission lines and loading was dismissed by the British Post Office and might have remained hidden, had not Lord Kelvin himself publicly expressed admiration for it [6]. - -Heaviside's operational calculus may be formally inaccurate, but in fact it anticipated the operational methods developed in more recent years [9]. Although his method was not fully understood, it provided correct results. When Heaviside was attacked for the vague meaning of his operational calculus, his pragmatic reply was, "Shall I refuse my dinner because I do not fully understand the process of digestion?" - -Heaviside lived as a bachelor hermit, often in near-squalid conditions, and died largely unnoticed, in poverty. His life demonstrates the persistent arrogance and snobbishness of the intellectual establishment, which does not respect creativity unless it is presented in the strict language of the establishment. - -## **4.2 SOME [PROPERTIES OF THE](#page-10-0) LAPLACE TRANSFORM** - -Properties of the Laplace transform are useful not only in the derivation of the Laplace transform of functions but also in the solutions of linear integro-differential equations. A glance at Eqs. (4.2) and (4.1) shows that there is a certain measure of symmetry in going from *x*(*t*) to *X*(*s*), and vice versa. This symmetry or duality is also carried over to the properties of the Laplace transform. This fact will be evident in the following development. - -We are already familiar with two properties: linearity [Eq. (4.3)] and the uniqueness property of the Laplace transform discussed earlier. - -### **[4.2-1 Time Shifting](#page-10-0)** - -The time-shifting property states that if - -*x*(*t*) ⇐⇒ *X*(*s*) - -then for *t*0 ≥ 0 - -$$ -x(t - t_0) \Longleftrightarrow X(s)e^{-st_0} \tag{4.12} -$$ - -Observe that *x*(*t*) starts at *t* = 0, and, therefore, *x*(*t* − *t*0) starts at *t* = *t*0. This fact is implicit, but is not explicitly indicated in Eq. (4.12). This often leads to inadvertent errors. To avoid such a pitfall, we should restate the property as follows. If - -$$ -x(t)u(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -x(t-t_0)u(t-t_0) \Longleftrightarrow X(s)e^{-st_0} \qquad t_0 \ge 0 -$$ - -**Proof.** - -$$ -\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_0^\infty x(t-t_0)u(t-t_0)e^{-st}dt -$$ - -Setting *t* −*t*0 = τ , we obtain - -$$ -\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_{-t_0}^{\infty} x(\tau)u(\tau)e^{-s(\tau+t_0)}d\tau -$$ - -Because *u*(τ ) = 0 for τ < 0 and *u*(τ ) = 1 for τ ≥ 0, the limits of integration can be taken from 0 to ∞. Thus, - -$$ -\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_0^\infty x(\tau)e^{-s(\tau+t_0)}d\tau -$$ -$$ -= e^{-st_0} \int_0^\infty x(\tau)e^{-s\tau}d\tau -$$ -$$ -= X(s)e^{-st_0} -$$ - -Note that *x*(*t* − *t*0)*u*(*t* − *t*0) is the signal *x*(*t*)*u*(*t*) delayed by *t*0 seconds. The time-shifting property states that *delaying a signal by t*0 *seconds amounts to multiplying its transform e*−*st*0 . - -This property of the unilateral Laplace transform holds only for positive *t*0 because if *t*0 were negative, the signal *x*(*t* −*t*0)*u*(*t* −*t*0) may not be causal. - -We can readily verify this property in Drill 4.1. If the signal in Fig. 4.2a is *x*(*t*)*u*(*t*), then the signal in Fig. 4.2b is *x*(*t* − 2)*u*(*t* − 2). The Laplace transform for the pulse in Fig. 4.2a is (1/*s*)(1−*e*−2*s* ). Therefore, the Laplace transform for the pulse in Fig. 4.2b is (1/*s*)(1−*e*−2*s* )*e*−2*s* . - -The time-shifting property proves very convenient in finding the Laplace transform of functions with different descriptions over different intervals, as the following example demonstrates. - -### **EXAMPLE 4.6 Laplace Transform and the Time-Shifting Property** - -Find the Laplace transform of *x*(*t*) depicted in Fig. 4.4a. - -Describing mathematically a function such as the one in Fig. 4.4a is discussed in Sec. 1.4. The function *x*(*t*) in Fig. 4.4a can be described as a sum of two components shown in Fig. 4.4b. The equation for the first component is *t*−1 over 1 ≤ *t* ≤ 2 so that this component can be described by (*t* −1)[*u*(*t* −1)−*u*(*t* −2)]. The second component can be described by *u*(*t* −2)−*u*(*t* −4). Therefore, - -$$ -x(t) = (t-1)[u(t-1) - u(t-2)] + [u(t-2) - u(t-4)] -$$ - -= $(t-1)u(t-1) - (t-1)u(t-2) + u(t-2) - u(t-4)$ (4.13) - -**Figure 4.4** Finding a piecewise representation of a signal *x*(*t*). - -The first term on the right-hand side is the signal *tu*(*t*) delayed by 1 second. Also, the third and fourth terms are the signal *u*(*t*) delayed by 2 and 4 seconds, respectively. The second term, however, cannot be interpreted as a delayed version of any entry in Table 4.1. For this reason, we rearrange it as - -$$ -(t-1)u(t-2) = (t-2+1)u(t-2) = (t-2)u(t-2) + u(t-2) -$$ - -We have now expressed the second term in the desired form as *tu*(*t*) delayed by 2 seconds plus *u*(*t*) delayed by 2 seconds. With this result, Eq. (4.13) can be expressed as - -$$ -x(t) = (t-1)u(t-1) - (t-2)u(t-2) - u(t-4) -$$ - -Application of the time-shifting property to *tu*(*t*) ⇐⇒ 1/*s*2 yields - -$$ -(t-1)u(t-1) \Longleftrightarrow \frac{1}{s^2}e^{-s} \qquad \text{and} \qquad (t-2)u(t-2) \Longleftrightarrow \frac{1}{s^2}e^{-2s} -$$ - -Also - -$$ -u(t) \Longleftrightarrow \frac{1}{s} -$$ - and $u(t-4) \Longleftrightarrow \frac{1}{s}e^{-4s}$ - -Therefore, - -$$ -X(s) = \frac{1}{s^2}e^{-s} - \frac{1}{s^2}e^{-2s} - \frac{1}{s}e^{-4s} -$$ - -### **EXAMPLE 4.7 Inverse Laplace Transform and the Time-Shifting Property** - -Find the inverse Laplace transform of - -$$ -X(s) = \frac{s+3+5e^{-2s}}{(s+1)(s+2)} -$$ - -### 352 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Observe the exponential term *e*−2*s* in the numerator of *X*(*s*), indicating time delay. In such a case, we should separate *X*(*s*) into terms with and without a delay factor, as - -$$ -X(s) = \underbrace{\frac{s+3}{(s+1)(s+2)}}_{X_1(s)} + \underbrace{\frac{5e^{-2s}}{(s+1)(s+2)}}_{X_2(s)e^{-2s}} -$$ - -where - -$$ -X_1(s) = \frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{1}{s+2} -$$ -$$ -X_2(s) = \frac{5}{(s+1)(s+2)} = \frac{5}{s+1} - \frac{5}{s+2} -$$ - -Therefore, - -$$ -x_1(t) = (2e^{-t} - e^{-2t})u(t) -$$ - -$$ -x_2(t) = 5(e^{-t} - e^{-2t})u(t) -$$ - -Also, because - -$$ -X(s) = X_1(s) + X_2(s)e^{-2s} -$$ - -we can write - -$$ -x(t) = x_1(t) + x_2(t-2) -$$ - -= $(2e^{-t} - e^{-2t})u(t) + 5[e^{-(t-2)} - e^{-2(t-2)}]u(t-2)$ - -## **DR ILL 4.4 Laplace Transform and the Time-Shifting Property** - -Find the Laplace transform of the signal illustrated in Fig. 4.5. - -## **DR ILL 4.5 Inverse Laplace Transform and the Time-Shifting Property** - -Find the inverse Laplace transform of *X*(*s*) = 3*e*−2*s* (*s*−1)(*s*+2) . - -**ANSWER** *et*−2 −*e*−2(*t*−2) *u*(*t* −2) - -### **[4.2-2 Frequency Shifting](#page-10-0)** - -The frequency-shifting property states that if - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -x(t)e^{s_0t} \Longleftrightarrow X(s-s_0) \tag{4.14} -$$ - -Observe the symmetry (or duality) between this property and the time-shifting property of Eq. (4.12). - -**Proof.** - -$$ -\mathcal{L}[x(t)e^{s_0t}] = \int_{0^-}^{\infty} x(t)e^{s_0t}e^{-st} dt = \int_{0^-}^{\infty} x(t)e^{-(s-s_0)t} dt = X(s-s_0) -$$ - -### **EXAMPLE 4.8 Frequency-Shifting Property** - -Derive pair 9a in Table 4.1 from pair 8a and the frequency-shifting property. - -Pair 8a is - -$$ -\cos btu(t) \Longleftrightarrow \frac{s}{s^2 + b^2} -$$ - -From the frequency-shifting property [Eq. (4.14)] with *s*0 = −*a*, we obtain - -$$ -e^{-at}\cos btu(t) \Longleftrightarrow \frac{s+a}{(s+a)^2 + b^2} -$$ - -### **DR ILL 4.6 Frequency-Shifting Property** - -Derive pair 6 in Table 4.1 from pair 3 and the frequency-shifting property. - -### 354 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -We are now ready to consider the two most important properties of the Laplace transform: time differentiation and time integration. - -### **[4.2-3 The Time-Differentiation Property](#page-10-0)** - -The time-differentiation property states that if† - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow sX(s) - x(0^-) -$$ - -Repeating this property a second time (differentiating twice) yields - -$$ -\frac{d^2x(t)}{dt^2} \Longleftrightarrow s^2X(s) - sx(0^-) - \dot{x}(0^-) -$$ - -Repeated differentiation yields - -$$ -\frac{d^n x(t)}{dt^n} \Longleftrightarrow s^n X(s) - s^{n-1} x(0^-) - s^{n-2} \dot{x}(0^-) - \dots - x^{(n-1)}(0^-) -$$ - -= $s^n X(s) - \sum_{k=1}^n s^{n-k} x^{(k-1)}(0^-)$ (4.15) - -where *x*(*r*) (0−) is *dr x*/*dtr* at *t* = 0−. - -**Proof.** - -$$ -\mathcal{L}\left[\frac{dx(t)}{dt}\right] = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ - -Integrating by parts, we obtain - -$$ -\mathcal{L}\left[\frac{dx(t)}{dt}\right] = x(t)e^{-st}\Big|_{0^{-}}^{\infty} + s\int_{0^{-}}^{\infty} x(t)e^{-st}dt -$$ - -For the Laplace integral to converge [i.e., for *X*(*s*) to exist], it is necessary that *x*(*t*)*e*−*st* → 0 as *t* → ∞ for the values of *s* in the ROC for *X*(*s*). Thus, - -$$ -\mathcal{L}\left[\frac{dx(t)}{dt}\right] = -x(0^-) + sX(s) -$$ - -Repeated application of this procedure yields Eq. (4.15). - -$$ -tx(t) \Longleftrightarrow -\frac{d}{ds}X(s) -$$ - -† The dual of the time-differentiation property is the frequency-differentiation property, which states that - -Find the Laplace transform of the signal *x*(*t*) in Fig. 4.6a by using Table 4.1 and the time-differentiation and time-shifting properties of the Laplace transform. - -Figures 4.6b and 4.6c show the first two derivatives of *x*(*t*). Recall that the derivative at a point of jump discontinuity is an impulse of strength equal to the amount of jump [see Eq. (1.12)]. Therefore, - -$$ -\frac{d^2x(t)}{dt^2} = \delta(t) - 3\delta(t-2) + 2\delta(t-3) -$$ - -#### 356 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -The Laplace transform of this equation yields - -$$ -\mathcal{L}\left(\frac{d^2x(t)}{dt^2}\right) = \mathcal{L}\left[\delta(t) - 3\delta(t-2) + 2\delta(t-3)\right] -$$ - -Using the time-differentiation property of Eq. (4.15), the time-shifting property of Eq. (4.12), and the facts that *x*(0−) = ˙*x*(0−) = 0, and δ(*t*) ⇐⇒ 1, we obtain - -$$ -s^2 X(s) - 0 - 0 = 1 - 3e^{-2s} + 2e^{-3s} -$$ - -Therefore, - -$$ -X(s) = \frac{1}{s^2} (1 - 3e^{-2s} + 2e^{-3s}) -$$ - -which confirms the earlier result in Drill 4.4. - -### **[4.2-4 The Time-Integration Property](#page-10-0)** - -The time-integration property states that if† - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -\int_{0^{-}}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} \quad \text{and} \quad \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} + \frac{\int_{-\infty}^{0^{-}} x(\tau) d\tau}{s} \quad (4.16) -$$ - -**Proof.** To prove the first part of Eq. (4.16), we define - -$$ -g(t) = \int_{0^-}^{t} x(\tau) d\tau -$$ - -so that - -$$ -\frac{d}{dt}g(t) = x(t) \qquad \text{and} \qquad g(0^-) = 0 -$$ - -Now, if - -$$ -g(t) \Longleftrightarrow G(s) -$$ - -then - -$$ -X(s) = \mathcal{L}\left[\frac{d}{dt}g(t)\right] = sG(s) - g(0^{-}) = sG(s) -$$ - -$$ -\frac{x(t)}{t} \Longleftrightarrow \int_{s}^{\infty} X(z) dz -$$ - -† The dual of the time-integration property is the frequency-integration property, which states that - -### 4.2 Some Properties of the Laplace Transform 357 - -Therefore, - -$$ -G(s) = \frac{X(s)}{s} -$$ - -or - -$$ -\int_{0^-}^t x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} -$$ - -To prove the second part of Eq. (4.16), observe that - -$$ -\int_{-\infty}^{t} x(\tau) d\tau = \int_{-\infty}^{0^-} x(\tau) d\tau + \int_{0^-}^{t} x(\tau) d\tau -$$ - -Note that the first term on the right-hand side is a constant for *t* ≥ 0. Taking the Laplace transform of the foregoing equation and using the first part of Eq. (4.16), we obtain - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{\int_{-\infty}^{0^-} x(\tau) d\tau}{s} + \frac{X(s)}{s} -$$ - -### **[4.2-5 The Scaling Property](#page-10-0)** - -The scaling property states that if - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then for *a* > 0 - -$$ -x(at) \Longleftrightarrow \frac{1}{a}X\left(\frac{s}{a}\right) -$$ - -The proof is given in Ch. 7. Note that *a* is restricted to positive values because if *x*(*t*) is causal, then *x*(*at*) is anticausal (is zero for *t* ≥ 0) for negative *a*, and anticausal signals are not permitted in the (unilateral) Laplace transform. - -Recall that *x*(*at*) is the signal *x*(*t*) time-compressed by the factor *a*, and *X*( *s a* ) is *X*(*s*) expanded along the *s* scale by the same factor *a* (see Sec. 1.2-2). The scaling property states that *time compression of a signal by a factor a causes expansion of its Laplace transform in the s scale by the same factor. Similarly, time expansion x*(*t*) *causes compression of X*(*s*) *in the s scale by the same factor.* - -### **[4.2-6 Time Convolution and Frequency Convolution](#page-10-0)** - -Another pair of properties states that if - -$$ -x_1(t) \Longleftrightarrow X_1(s) -$$ - and $x_2(t) \Longleftrightarrow X_2(s)$ - -then (*time-convolution property*) - -$$ -x_1(t) * x_2(t) \Longleftrightarrow X_1(s)X_2(s) \tag{4.17} -$$ - -### 358 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -and (*frequency-convolution property*) - -$$ -x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi j}[X_1(s) * X_2(s)] -$$ - -Observe the symmetry (or duality) between the two properties. Proofs of these properties are postponed to Ch. 7. - -Equation (2.39) indicates that *H*(*s*), the transfer function of an LTIC system, is the Laplace transform of the system's impulse response *h*(*t*); that is, - -$$ -h(t) \Longleftrightarrow H(s) -$$ - -If the system is causal, *h*(*t*) is causal, and, according to Eq. (2.39), *H*(*s*) is the unilateral Laplace transform of *h*(*t*). Similarly, if the system is noncausal, *h*(*t*) is noncausal, and *H*(*s*) is the bilateral transform of *h*(*t*). - -We can apply the time-convolution property to the LTIC input–output relationship *y*(*t*) = *x*(*t*) ∗ *h*(*t*) to obtain - -$$ -Y(s) = X(s)H(s) \tag{4.18} -$$ - -The response *y*(*t*) is the zero-state response of the LTIC system to the input *x*(*t*). From Eq. (4.18), it follows that - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{\mathcal{L}[zero-state response]}{\mathcal{L}[input]} -$$ -(4.19) - -This may be considered an alternate definition of the LTIC system transfer function *H*(*s*). It is the *ratio of the transform of zero-state response to the transform of the input.* - -### **EXAMPLE 4.10 Time-Convolution Property** - -Use the time-convolution property of the Laplace transform to determine *c*(*t*)=*eatu*(*t*)∗*ebtu*(*t*). - -From Eq. (4.17), it follows that - -$$ -C(s) = \frac{1}{(s-a)(s-b)} = \frac{1}{a-b} \left[ \frac{1}{s-a} - \frac{1}{s-b} \right] -$$ - -The inverse transform of this equation yields - -$$ -c(t) = \frac{1}{a-b}(e^{at} - e^{bt})u(t) -$$ - -### INITIAL AND FINAL VALUES - -In certain applications, it is desirable to know the values of *x*(*t*) as *t* → 0 and *t* → ∞ [initial and final values of *x*(*t*)] from the knowledge of its Laplace transform *X*(*s*). Initial and final value theorems provide such information. - -*The initial value theorem* states that if *x*(*t*) and its derivative *dx*/*dt* are both Laplace transformable, then - -$$ -x(0^+) = \lim_{s \to \infty} sX(s) \tag{4.20} -$$ - -provided the limit on the right-hand side of Eq. (4.20) exists. - -*The final value theorem* states that if both *x*(*t*) and *dx*/*dt* are Laplace transformable, then - -$$ -\lim_{t \to \infty} x(t) = \lim_{s \to 0} sX(s) \tag{4.21} -$$ - -provided *sX*(*s*) has no poles in the RHP or on the imaginary axis. To prove these theorems, we begin by setting *n* = 1 in Eq. (4.15). Using the definition of the Laplace transform, we see that - -$$ -sX(s) - x(0^{-}) = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ - -= $\int_{0^{-}}^{0^{+}} \frac{dx(t)}{dt} e^{-st} dt + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$ -= $x(t) \Big|_{0^{-}}^{0^{+}} + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$ -= $x(0^{+}) - x(0^{-}) + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$ - -Therefore, - -$$ -sX(s) = x(0^+) + \int_{0^+}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ - -and - -$$ -\lim_{s \to \infty} sX(s) = x(0^+) + \lim_{s \to \infty} \int_{0^+}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ -$$ -= x(0^+) + \int_{0^+}^{\infty} \frac{dx(t)}{dt} \left( \lim_{s \to \infty} e^{-st} \right) dt -$$ -$$ -= x(0^+) -$$ - -**Comment.** The initial value theorem applies only if *X*(*s*) is strictly proper (*M* < *N*), because for *M* ≥ *N*, lim*s*→∞ *sX*(*s*) does not exist, and the theorem does not apply. In such a case, we can still find the answer by using long division to express *X*(*s*) as a polynomial in *s* plus a strictly proper fraction, where *M* < *N*. For example, by using long division, we can express - -$$ -\frac{s^3 + 3s^2 + s + 1}{s^2 + 2s + 1} = (s + 1) - \frac{2s}{s^2 + 2s + 1} -$$ - -The inverse transform of the polynomial in *s* is in terms of δ(*t*), and its derivatives, which are zero at *t* = 0+. In the foregoing case, the inverse transform of *s* + 1 is δ(˙ *t*) + δ(*t*). Hence, the desired - -### 360 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -*x*(0+) is the value of the remainder (strictly proper) fraction, for which the initial value theorem applies. In the present case, - -$$ -x(0^+) = \lim_{s \to \infty} \frac{-2s^2}{s^2 + 2s + 1} = -2 -$$ - -To prove the final value theorem, we let *n* = 1 and *s* → 0 in Eq. (4.15) to obtain - -$$ -\lim_{s \to 0} [sX(s) - x(0^{-})] = \lim_{s \to 0} \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} dt -$$ -$$ -= x(t)|_{0^{-}}^{\infty} = \lim_{t \to \infty} x(t) - x(0^{-}) -$$ - -a deduction that leads to the desired result, Eq. (4.21). - -**Comment.** The final value theorem applies only if the poles of *X*(*s*) are in the LHP (including *s* = 0). If *X*(*s*) has a pole in the RHP, *x*(*t*) contains an exponentially growing term and *x*(∞) does not exist. If there is a pole on the imaginary axis, then *x*(*t*) contains an oscillating term and *x*(∞) does not exist. However, if there is a pole at the origin, then *x*(*t*) contains a constant term, and hence, *x*(∞) exists and is a constant. - -### **EXAMPLE 4.11 Initial and Final Values** - -Determine the initial and final values of *y*(*t*) if its Laplace transform *Y*(*s*) is given by - -$$ -Y(s) = \frac{10(2s+3)}{s(s^2+2s+5)} -$$ - -Equations (4.20) and (4.21) yield - -$$ -y(0^+) = \lim_{s \to \infty} sY(s) = \lim_{s \to \infty} \frac{10(2s+3)}{(s^2+2s+5)} = 0 -$$ - -$$ -y(\infty) = \lim_{s \to 0} sY(s) = \lim_{s \to 0} \frac{10(2s+3)}{(s^2+2s+5)} = 6 -$$ - -Table 4.2 summarizes the most important unilateral Laplace transform properties. - -## **4.3 SOLUTION OF DIFFERENTIAL AND [INTEGRO-DIFFERENTIAL](#page-10-0) EQUATIONS** - -The time-differentiation property of the Laplace transform has set the stage for solving linear differential (or integro-differential) equations with constant coefficients. Because *dky*/*dtk* ⇐⇒ *skY*(*s*), the Laplace transform of a differential equation is an algebraic equation that can be readily - -| Operation | x(t) | X(s) | -|------------------------------|------------------------|-------------------------------------------------------| -| Addition | x1(t)+x2(t) | X1(s)+X2(s) | -| Scalar multiplication | kx(t) | kX(s) | -| Time differentiation | dx(t)
dt | sX(s)−x(0−) | -| | d2x(t)
dt2 | s2X(s)−sx(0−)−
˙x(0−) | -| | d3x(t)
dt3 | s3X(s)−s2x(0−)−sx˙(0−)−
¨x(0−) | -| | dnx(t)
dtn | %n
snX(s)−
sn−kx(k−1)
(0−)
k=1 | -| Time integration | # t
x(τ )dτ
0− | 1
X(s)
s | -| | # t
x(τ )dτ | # 0−
1
1
X(s)+
x(t)dt
s
s | -| Time shifting | −∞
x(t −t0)u(t −t0) | −∞
X(s)e−st0
t0
≥ 0 | -| Frequency shifting | x(t)es0t | X(s−s0) | -| Frequency
differentiation | −tx(t) | dX(s)
ds | -| Frequency integration | x(t)
t | # ∞
X(z)dz
s | -| Scaling | x(at),a ≥ 0 | s
1
X
a
a | -| Time convolution | x1(t) ∗ x2(t) | X1(s)X2(s) | -| Frequency convolution | x1(t)x2(t) | 1
X1(s) ∗ X2(s)
2πj | -| Initial value | x(0+) | sX(s)
(n > m)
lim | -| Final value | x(∞) | s→∞
sX(s)
[poles of sX(s) in LHP]
lim
s→0 | - -**TABLE 4.2** Unilateral Laplace Transform Properties - -solved for *Y*(*s*). Next we take the inverse Laplace transform of *Y*(*s*) to find the desired solution *y*(*t*). The following examples demonstrate the Laplace transform procedure for solving linear differential equations with constant coefficients. - -## **EXAMPLE 4.12 Laplace Transform to Solve a Second-Order Linear Differential Equation** - -Solve the second-order linear differential equation - -$$ -(D2 + 5D + 6)y(t) = (D + 1)x(t) -$$ - -for the initial conditions *y*(0−) = 2 and *y*˙(0−) = 1 and the input *x*(*t*) = *e*−4*t u*(*t*). 362 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -The equation is - -$$ -\frac{d^2y(t)}{dt^2} + 5\frac{dy(t)}{dt} + 6y(t) = \frac{dx(t)}{dt} + x(t) -$$ -\n(4.22) - -Let - -$$ -y(t) \Longleftrightarrow Y(s) -$$ - -Then from Eq. (4.15), - -$$ -\frac{dy(t)}{dt} \Longleftrightarrow sY(s) - y(0^-) = sY(s) - 2 -$$ - -and - -$$ -\frac{d^2y(t)}{dt^2} \Longleftrightarrow s^2Y(s) - sy(0^-) - \dot{y}(0^-) = s^2Y(s) - 2s - 1 -$$ - -Moreover, for *x*(*t*) = *e*−4*t u*(*t*), - -$$ -X(s) = \frac{1}{s+4} \quad \text{and} \quad \frac{dx(t)}{dt} \Longleftrightarrow sX(s) - x(0^-) = \frac{s}{s+4} - 0 = \frac{s}{s+4} -$$ - -Taking the Laplace transform of Eq. (4.22), we obtain - -$$ -[s^{2}Y(s) - 2s - 1] + 5[sY(s) - 2] + 6Y(s) = \frac{s}{s+4} + \frac{1}{s+4} -$$ - -Collecting all the terms of *Y*(*s*) and the remaining terms separately on the left-hand side, we obtain - -$$ -(s2 + 5s + 6)Y(s) - (2s + 11) = \frac{s+1}{s+4} -$$ - (4.23) - -Therefore, - -$$ -(s2 + 5s + 6)Y(s) = (2s + 11) + \frac{s+1}{s+4} = \frac{2s2 + 20s + 45}{s+4} -$$ - -and - -$$ -Y(s) = \frac{2s^2 + 20s + 45}{(s^2 + 5s + 6)(s + 4)} = \frac{2s^2 + 20s + 45}{(s + 2)(s + 3)(s + 4)} -$$ - -Expanding the right-hand side into partial fractions yields - -$$ -Y(s) = \frac{13/2}{s+2} - \frac{3}{s+3} - \frac{3/2}{s+4} -$$ - -The inverse Laplace transform of this equation yields - -$$ -y(t) = \left(\frac{13}{2}e^{-2t} - 3e^{-3t} - \frac{3}{2}e^{-4t}\right)u(t) -$$ -\n(4.24) - -Example 4.12 demonstrates the ease with which the Laplace transform can solve linear differential equations with constant coefficients. The method is general and can solve a linear differential equation with constant coefficients of any order. - -### ZERO-INPUT AND ZERO-STATE COMPONENTS OF RESPONSE - -The Laplace transform method gives the total response, which includes zero-input and zero-state components. It is possible to separate the two components if we so desire. The initial condition terms in the response give rise to the zero-input response. For instance, in Ex. 4.12, the terms attributable to initial conditions *y*(0−) = 2 and *y*˙(0−) = 1 in Eq. (4.23) generate the zero-input response. These initial condition terms are −(2*s* + 11), as seen in Eq. (4.23). The terms on the right-hand side are exclusively due to the input. Equation (4.23) is reproduced below with the proper labeling of the terms - -$$ -(s2 + 5s + 6) Y(s) - (2s + 11) = \frac{s+1}{s+4} -$$ - -so that - -$$ -(s2 + 5s + 6) Y(s) = \underbrace{(2s + 11)}_{initial condition terms} + \underbrace{\underbrace{s + 1}_{s + 4}}_{input terms} -$$ - -Therefore, - -$$ -Y(s) = \underbrace{\frac{2s+11}{s^2+5s+6}}_{\text{ZIR}} + \underbrace{\frac{s+1}{(s+4)(s^2+5s+6)}}_{\text{ZSR}} \\ -= \left[ \frac{7}{s+2} - \frac{5}{s+3} \right] + \left[ \frac{-1/2}{s+2} + \frac{2}{s+3} - \frac{3/2}{s+4} \right] -$$ - -Taking the inverse transform of this equation yields - -$$ -y(t) = \underbrace{\left(7e^{-2t} - 5e^{-3t}\right)u(t)}_{\text{ZIR}} + \underbrace{\left(-\frac{1}{2}e^{-2t} + 2e^{-3t} - \frac{3}{2}e^{-4t}\right)u(t)}_{\text{ZSR}} -$$ - -### **[4.3-1 Comments on Initial Conditions at](#page-10-0) 0− and at 0+** - -The initial conditions in Ex. 4.12 are *y*(0−) = 2 and *y*˙(0−) = 1. If we let *t* = 0 in the total response in Eq. (4.24), we find *y*(0) = 2 and *y*˙(0) = 2, which is at odds with the given initial conditions. Why? Because the initial conditions are given at *t* = 0− (just before the input is applied), when only the zero-input response is present. The zero-state response is the result of the input *x*(*t*) applied at *t* = 0. Hence, this component does not exist at *t* = 0−. Consequently, the initial conditions at *t* = 0− are satisfied by the zero-input response, not by the total response. We can readily verify in this example that the zero-input response does indeed satisfy the given initial conditions at *t* = 0−. It is the total response that satisfies the initial conditions at *t* = 0+, which are generally different from the initial conditions at 0−. - -There also exists a *L*+ version of the Laplace transform, which uses the initial conditions at *t* = 0+ rather than at 0− (as in our present *L*− version). The *L*+ version, which was in vogue till the early 1960s, is identical to the *L*− version except the limits of Laplace integral [Eq. (4.7)] are from 0+ to ∞. Hence, by definition, the origin *t* = 0 is excluded from the domain. This version, still used in some math books, has some serious difficulties. For instance, the Laplace transform of δ(*t*) is zero because δ(*t*) = 0 for *t* ≥ 0+. Moreover, this approach is rather clumsy in the theoretical study - -### 364 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -of linear systems because the response obtained cannot be separated into zero-input and zero-state components. As we know, the zero-state component represents the system response as an explicit function of the input, and without knowing this component, it is not possible to assess the effect of the input on the system response in a general way. The *L*+ version can separate the response in terms of the natural and the forced components, which are not as interesting as the zero-input and the zero-state components. Note that we can always determine the natural and the forced components from the zero-input and the zero-state components [e.g., Eq. (2.44) from Eq. (2.43)], but the converse is not true. Because of these and some other problems, electrical engineers (wisely) started discarding the *L*+ version in the early 1960s. - -It is interesting to note the time-domain duals of these two Laplace versions. The classical method is the dual of the *L*+ method, and the convolution (zero-input/zero-state) method is the dual of the *L*− method. The first pair uses the initial conditions at 0+, and the second pair uses those at *t* = 0−. The first pair (the classical method and the *L*+ version) is awkward in the theoretical study of linear system analysis. It was no coincidence that the *L*− version was adopted immediately after the introduction to the electrical engineering community of state-space analysis (which uses zero-input/zero-state separation of the output). - -## **DR ILL 4.7 Laplace Transform to Solve a Second-Order Linear Differential Equation** - -Solve - -$$ -\frac{d^2y(t)}{dt^2} + 4\frac{dy(t)}{dt} + 3y(t) = 2\frac{dx(t)}{dt} + x(t) -$$ - -for the input *x*(*t*) = *u*(*t*). The initial conditions are *y*(0−) = 1 and *y*˙(0−) = 2. - -### **ANSWER** - -*y*(*t*) = 1 3 (1+9*e*−*t* −7*e*−3*t* )*u*(*t*) - -### **EXAMPLE 4.13 Laplace Transform to Solve an Electric Circuit** - -In the circuit of Fig. 4.7a, the switch is in the closed position for a long time before *t* = 0, when it is opened instantaneously. Find the inductor current *y*(*t*) for *t* ≥ 0. - -When the switch is in the closed position (for a long time), the inductor current is 2 amperes and the capacitor voltage is 10 volts. When the switch is opened, the circuit is equivalent to that depicted in Fig. 4.7b, with the initial inductor current *y*(0−) = 2 and the initial capacitor voltage *vC*(0−) = 10. The input voltage is 10 volts, starting at *t* = 0, and, therefore, can be represented by 10*u*(*t*). - -**Figure 4.7** Analysis of a network with a switching action. - -The loop equation of the circuit in Fig. 4.7b is - -$$ -\frac{dy(t)}{dt} + 2y(t) + 5 \int_{-\infty}^{t} y(\tau) d\tau = 10u(t) -$$ -\n(4.25) - -If - -*y*(*t*) ⇐⇒ *Y*(*s*) - -then - -$$ -\frac{dy(t)}{dt} \Longleftrightarrow sY(s) - y(0^-) = sY(s) - 2 -$$ - -and [see Eq. (4.16)] - -$$ -\int_{-\infty}^{t} y(\tau) d\tau \Longleftrightarrow \frac{Y(s)}{s} + \frac{\int_{-\infty}^{0} y(\tau) d\tau}{s} -$$ - -Because *y*(*t*) is the capacitor current, the integral \$ 0− −∞ *y*(τ )*d*τ is *qC*(0−), the capacitor charge at *t* = 0−, which is given by *C* times the capacitor voltage at *t* = 0−. Therefore, - -$$ -\int_{-\infty}^{0^-} y(\tau) d\tau = q_C(0^-) = C v_C(0^-) = \frac{1}{5}(10) = 2 -$$ - -and - -$$ -\int_{-\infty}^{t} y(\tau) d\tau \Longleftrightarrow \frac{Y(s)}{s} + \frac{2}{s} -$$ - -#### 366 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Using these results, the Laplace transform of Eq. (4.25) is - -$$ -sY(s) - 2 + 2Y(s) + \frac{5Y(s)}{s} + \frac{10}{s} = \frac{10}{s} -$$ - -or - -$$ -\[s+2+\frac{5}{s}\]Y(s) = 2 -$$ - -and - -$$ -Y(s) = \frac{2s}{s^2 + 2s + 5} -$$ - -To find the inverse Laplace transform of *Y*(*s*), we use pair 10c (Table 4.1) with values *A* = 2, *B* = 0, *a* = 1, and *c* = 5. This yields - -$$ -r = \sqrt{\frac{20}{4}} = \sqrt{5} -$$ -, $b = \sqrt{c - a^2} = 2$ and $\theta = \tan^{-1}(\frac{2}{4}) = 26.6^{\circ}$ - -Therefore, - -$$ -y(t) = \sqrt{5}e^{-t}\cos(2t + 26.6^{\circ})u(t) -$$ - -This response is shown in Fig. 4.7c. - -**Comment.** In our discussion so far, we have multiplied input signals by *u*(*t*), implying that the signals are zero prior to *t* = 0. This is needlessly restrictive. These signals can have any arbitrary value prior to *t* = 0. As long as the initial conditions at *t* = 0 are specified, we need only the knowledge of the input for *t* ≥ 0 to compute the response for *t* ≥ 0. Some authors use the notation 1(*t*) to denote a function that is equal to *u*(*t*) for *t* ≥ 0 and that has arbitrary value for negative *t*. We have abstained from this usage to avoid needless confusion caused by the introduction of a new function, which is very similar to *u*(*t*). - -### **[4.3-2 Zero-State Response](#page-10-0)** - -Consider an *N*th-order LTIC system specified by the equation - -$$ -Q(D)y(t) = P(D)x(t) -$$ - -or - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)y(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)x(t) -$$ - (4.26) - -We shall now find the general expression for the zero-state response of an LTIC system. Zero-state response *y*(*t*), by definition, is the system response to an input when the system is initially relaxed (in zero state). Therefore, *y*(*t*) satisfies Eq. (4.26) with zero initial conditions - -$$ -y(0^-) -$$ - = $\dot{y}(0^-)$ = $\ddot{y}(0^-)$ = $\dots$ = $y^{(N-1)}(0^-)$ = 0 - -Moreover, the input *x*(*t*) is causal so that - -$$ -x(0^-) = \dot{x}(0^-) = \ddot{x}(0^-) = \cdots = x^{(N-1)}(0^-) = 0 -$$ - -Let - -$$ -y(t) \Longleftrightarrow Y(s) -$$ - and $x(t) \Longleftrightarrow X(s)$ - -Because of zero initial conditions, - -$$ -D^{r}y(t) = \frac{d^{r}}{dt^{r}}y(t) \Longleftrightarrow s^{r}Y(s) -$$ -$$ -D^{k}x(t) = \frac{d^{k}}{dt^{k}}x(t) \Longleftrightarrow s^{k}X(s) -$$ - -Therefore, the Laplace transform of Eq. (4.26) yields - -$$ -(sN + a1sN-1 + \dots + aN-1s + aN)Y(s) = (b0sN + b1sN-1 + \dots + bN-1s + bN)X(s) -$$ - -or - -$$ -Y(s) = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} X(s) = \frac{P(s)}{Q(s)} X(s) -$$ - -But we have shown in Eq. (4.18) that *Y*(*s*) = *H*(*s*)*X*(*s*). Consequently, - -$$ -H(s) = \frac{P(s)}{Q(s)}\tag{4.27} -$$ - -This is the transfer function of a linear differential system specified in Eq. (4.26). The same result has been derived earlier in Eq. (2.41) using an alternate (time-domain) approach. - -We have shown that *Y*(*s*), the Laplace transform of the zero-state response *y*(*t*), is the product of *X*(*s*) and *H*(*s*), where *X*(*s*) is the Laplace transform of the input *x*(*t*) and *H*(*s*) is the system transfer function [relating the particular output *y*(*t*) to the input *x*(*t*)]. - -### INTUITIVE INTERPRETATION OF THE LAPLACE TRANSFORM - -So far we have treated the Laplace transform as a machine that converts linear integro-differential equations into algebraic equations. There is no physical understanding of how this is accomplished or what it means. We now discuss a more intuitive interpretation and meaning of the Laplace transform. - -In Ch. 2, Eq. (2.38), we showed that LTI system response to an everlasting exponential *est* is *H*(*s*)*est*. If we could express every signal as a linear combination of everlasting exponentials of the form *est*, we could readily obtain the system response to any input. For example, if - -$$ -x(t) = \sum_{k=1}^{K} X(s_i) e^{s_i t} -$$ - -the response of an LTIC system to such input *x*(*t*) is given by - -$$ -y(t) = \sum_{k=1}^{K} X(s_i) H(s_i) e^{s_i t} -$$ - -### 368 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Unfortunately, the class of signals that can be expressed in this form is very small. However, we can express almost all signals of practical utility as a sum of everlasting exponentials over a continuum of frequencies. This is precisely what the Laplace transform in Eq. (4.2) does. - -$$ -x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds -$$ -\n(4.28) - -Invoking the linearity property of the Laplace transform, we can find the system response *y*(*t*) to input *x*(*t*) in Eq. (4.28) as† - -$$ -y(t) = \frac{1}{2\pi j} \int_{c'-j\infty}^{c'+j\infty} X(s)H(s)e^{st} ds = \mathcal{L}^{-1}X(s)H(s) -$$ -(4.29) - -Clearly, - -*Y*(*s*) = *X*(*s*)*H*(*s*) - -We can now represent the transformed version of the system, as depicted in Fig. 4.8a. The input *X*(*s*) is the Laplace transform of *x*(*t*), and the output *Y*(*s*) is the Laplace transform of (the zero-input response) *y*(*t*). The system is described by the transfer function *H*(*s*). The output *Y*(*s*) is the product *X*(*s*)*H*(*s*). - -Recall that *s* is the complex frequency of *est*. This explains why the Laplace transform method is also called the *frequency-domain* method. Note that *X*(*s*),*Y*(*s*), and *H*(*s*) are the frequency-domain representations of *x*(*t*), *y*(*t*), and *h*(*t*), respectively. We may view the boxes marked *L* and *L*−1 in Fig. 4.8a as the interfaces that convert the time-domain entities into the corresponding frequency-domain entities, and vice versa. All real-life signals begin in the time domain, and the final answers must also be in the time domain. First, we convert the time-domain input(s) into the frequency-domain counterparts. The problem itself is solved in the frequency domain, resulting in the answer *Y*(*s*), also in the frequency domain. Finally, we convert *Y*(*s*) to *y*(*t*). Solving the problem is relatively simpler in the frequency domain than in the time domain. Henceforth, we shall omit the explicit representation of the interface boxes *L* and *L*−1, representing signals and systems in the frequency domain, as shown in Fig. 4.8b. - -**Figure 4.8** Alternate interpretation of the Laplace transform. - -† Recall that *H*(*s*) has its own region of validity. Hence, the limits of integration for the integral in Eq. (4.28) are modified in Eq. (4.29) to accommodate the region of existence (validity) of *X*(*s*) as well as *H*(*s*). - -### **EXAMPLE 4.14 Laplace Transform to Find the Zero-State Response** - -Find the response *y*(*t*) of an LTIC system described by the equation - -$$ -\frac{d^2y(t)}{dt^2} + 5\frac{dy(t)}{dt} + 6y(t) = \frac{dx(t)}{dt} + x(t) -$$ - -if the input *x*(*t*) = 3*e*−5*t u*(*t*) and all the initial conditions are zero; that is, the system is in the zero state. - -The system equation is - -$$ -\underbrace{(D^2 + 5D + 6)}_{Q(D)} y(t) = \underbrace{(D + 1)}_{P(D)} x(t) -$$ - -Therefore, - -$$ -H(s) = \frac{P(s)}{Q(s)} = \frac{s+1}{s^2 + 5s + 6} -$$ - -Also, - -$$ -X(s) = \mathcal{L}[3e^{-5t}u(t)] = \frac{3}{s+5} -$$ - -and - -$$ -Y(s) = X(s)H(s) = \frac{3(s+1)}{(s+5)(s^2+5s+6)} -$$ - -= -$$ -\frac{3(s+1)}{(s+5)(s+2)(s+3)} = \frac{-2}{s+5} - \frac{1}{s+2} + \frac{3}{s+3} -$$ - -The inverse Laplace transform of this equation is - -$$ -y(t) = (-2e^{-5t} - e^{-2t} + 3e^{-3t})u(t) -$$ - -### **EXAMPLE 4.15 Laplace Transform to Find System Transfer Functions** - -Show that the transfer function of: - -- **(a)** an ideal delay of *T* seconds is *e*−*sT* -- **(b)** an ideal differentiator is *s* -- **(c)** an ideal integrator is 1/*s* - -**(a) Ideal Delay.** For an ideal delay of *T* seconds, the input *x*(*t*) and output *y*(*t*) are related by - -*y*(*t*) = *x*(*t* −*T*) and *Y*(*s*) = *X*(*s*)*e*−*sT* [see Eq. (4.12)] - -### 370 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Therefore, - -$$ -H(s) = \frac{Y(s)}{X(s)} = e^{-sT} -$$ -\n(4.30) - -**(b) Ideal Differentiator.** For an ideal differentiator, the input *x*(*t*) and the output *y*(*t*) are related by - -$$ -y(t) = \frac{dx(t)}{dt} -$$ - -The Laplace transform of this equation yields - -$$ -Y(s) = sX(s) \qquad [x(0^-) = 0 \text{ for a causal signal}] -$$ - -and - -$$ -H(s) = \frac{Y(s)}{X(s)} = s -$$ -\n(4.31) - -**(c) Ideal Integrator.** For an ideal integrator with zero initial state, that is, *y*(0−) = 0, - -$$ -y(t) = \int_0^t x(\tau) d\tau \quad \text{and} \quad Y(s) = \frac{1}{s}X(s) -$$ - -Therefore, - -$$ -H(s) = \frac{1}{s} \tag{4.32} -$$ - -### **DR ILL 4.8 Differential Equation and Zero-State Response from a System Transfer Function** - -For an LTIC system with transfer function, - -$$ -H(s) = \frac{s+5}{s^2 + 4s + 3} -$$ - -- **(a)** Describe the differential equation relating the input *x*(*t*) and output *y*(*t*). -- **(b)** Find the system response *y*(*t*) to the input *x*(*t*) = *e*−2*t u*(*t*) if the system is initially in zero state. - -**ANSWERS** - -(a) -$$ -\frac{d^2y(t)}{dt^2} + 4\frac{dy(t)}{dt} + 3y(t) = \frac{dx(t)}{dt} + 5x(t) -$$ - -**(b)** -$$ -y(t) = (2e^{-t} - 3e^{-2t} + e^{-3t})u(t) -$$ - -### **[4.3-3 Stability](#page-10-0)** - -Equation (4.27) shows that the denominator of *H*(*s*) is *Q*(*s*), which is apparently identical to the characteristic polynomial *Q*(λ) defined in Ch. 2. Does this mean that the denominator of *H*(*s*) is the characteristic polynomial of the system? This may or may not be the case, since if *P*(*s*) and *Q*(*s*) in Eq. (4.27) have any common factors, they cancel out, and the effective denominator of *H*(*s*) is not necessarily equal to *Q*(*s*). Recall also that the system transfer function *H*(*s*), like *h*(*t*), is defined in terms of measurements at the external terminals. Consequently, *H*(*s*) and *h*(*t*) are both external descriptions of the system. In contrast, the characteristic polynomial *Q*(*s*) is an internal description. Clearly, we can determine only external stability, that is, BIBO stability, from *H*(*s*). If all the poles of *H*(*s*) are in LHP, all the terms in *h*(*t*) are decaying exponentials, and *h*(*t*) is absolutely integrable [see Eq. (2.45)].† Consequently, the system is BIBO-stable. Otherwise the system is BIBO-unstable. - -Beware of right half-plane poles! - -So far, we have assumed that *H*(*s*) is a proper function, that is, *M* ≤ *N*. We now show that if *H*(*s*) is improper, that is, if *M* > *N*, the system is BIBO-unstable. In such a case, using long division, we obtain *H*(*s*) = *R*(*s*) + *H* (*s*), where *R*(*s*) is an (*M* − *N*)th-order polynomial and *H* (*s*) is a proper transfer function. For example, - -$$ -H(s) = \frac{s^3 + 4s^2 + 4s + 5}{s^2 + 3s + 2} = s + \frac{s^2 + 2s + 5}{s^2 + 3s + 2} -$$ - -As shown in Eq. (4.31), the term *s* is the transfer function of an ideal differentiator. If we apply step function (bounded input) to this system, the output will contain an impulse (unbounded output). Clearly, the system is BIBO-unstable. Moreover, such a system greatly amplifies noise because differentiation enhances higher frequencies, which generally predominate in a noise signal. These - -† Values of *s* for which *H*(*s*) is ∞ are the *poles* of *H*(*s*). Thus, poles of *H*(*s*) are the values of *s* for which the denominator of *H*(*s*) is zero. - -### 372 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -are two good reasons to avoid improper systems (*M* > *N*). In our future discussion, we shall implicitly assume that the systems are proper, unless stated otherwise. - -If *P*(*s*) and *Q*(*s*) do not have common factors, then the denominator of *H*(*s*) is identical to *Q*(*s*), the characteristic polynomial of the system. In this case, we can determine internal stability by using the criterion described in Sec. 2.5. Thus, if *P*(*s*) and *Q*(*s*) have no common factors, the asymptotic stability criterion in Sec. 2.5 can be restated in terms of the poles of the transfer function of a system, as follows: - -- 1. An LTIC system is asymptotically stable if and only if all the poles of its transfer function *H*(*s*) are in the LHP. The poles may be simple or repeated. -- 2. An LTIC system is unstable if and only if either one or both of the following conditions exist: (i) at least one pole of *H*(*s*) is in the RHP; (ii) there are repeated poles of *H*(*s*) on the imaginary axis. -- 3. An LTIC system is marginally stable if and only if there are no poles of *H*(*s*) in the RHP and some unrepeated poles on the imaginary axis. - -The locations of zeros of *H*(*s*) have no role in determining the system stability. - -### **EXAMPLE 4.16 BIBO and Asymptotic Stability** - -Figure 4.9a shows a cascade connection of two LTIC systems *S*1 followed by *S*2. The transfer functions of these systems are *H*1(*s*) = 1/(*s* − 1) and *H*2(*s*) = (*s* − 1)/(*s* + 1), respectively. Determine the BIBO and asymptotic stability of the composite (cascade) system. - -**Figure 4.9** Distinction between BIBO and asymptotic stability. - -If the impulse responses of *S*1 and *S*2 are *h*1(*t*) and *h*2(*t*), respectively, then the impulse response of the cascade system is *h*(*t*) = *h*1(*t*)∗*h*2(*t*). Hence, *H*(*s*) = *H*1(*s*)*H*2(*s*). In the present case, - -$$ -H(s) = \left(\frac{1}{s-1}\right)\left(\frac{s-1}{s+1}\right) = \frac{1}{s+1} -$$ - -The pole of *S*1 at *s* = 1 cancels with the zero at *s* = 1 of *S*2. This results in a composite system having a single pole at *s* = −1. If the composite cascade system were to be enclosed inside a black box with only the input and the output terminals accessible, any measurement from these external terminals would show that the transfer function of the system is 1/(*s*+1), without any hint of the fact that the system is housing an unstable system (Fig. 4.9b). - -The impulse response of the cascade system is *h*(*t*) = *e*−*t u*(*t*), which is absolutely integrable. Consequently, the system is BIBO-stable. - -To determine the asymptotic stability, we note that *S*1 has one characteristic root at 1, and *S*2 also has one root at −1. Recall that the two systems are independent (one does not load the other), and the characteristic modes generated in each subsystem are independent of the other. Clearly, the mode *et* will not be eliminated by the presence of *S*2. Hence, the composite system has two characteristic roots, located at ±1, and the system is asymptotically unstable, though BIBO-stable. - -Interchanging the positions of *S*1 and *S*2 makes no difference in this conclusion. This example shows that BIBO stability can be misleading. If a system is asymptotically unstable, it will destroy itself (or, more likely, lead to saturation condition) because of unchecked growth of the response due to intended or unintended stray initial conditions. BIBO stability is not going to save the system. Control systems are often compensated to realize certain desirable characteristics. One should never try to stabilize an unstable system by canceling its RHP pole(s) with RHP zero(s). Such a misguided attempt will fail, not because of the practical impossibility of exact cancellation but for the more fundamental reason, as just explained. - -### **DR ILL 4.9 BIBO and Asymptotic Stability** - -Show that an ideal integrator is marginally stable but BIBO-unstable. - -### **[4.3-4 Inverse Systems](#page-10-0)** - -If *H*(*s*) is the transfer function of a system *S*, then *Si*, its inverse system has a transfer function *Hi*(*s*) given by - -$$ -H_i(s) = \frac{1}{H(s)} -$$ - -This follows from the fact the cascade of *S* with its inverse system *Si* is an identity system, with impulse response δ(*t*), implying *H*(*s*)*Hi*(*s*) = 1. For example, an ideal integrator and its inverse, an ideal differentiator, have transfer functions 1/*s* and *s*, respectively, leading to *H*(*s*)*Hi*(*s*) = 1. - -## **4.4 ANALYSIS OF ELECTRICAL NETWORKS: THE TRANSFORMED NETWORK** - -Example 4.12 shows how electrical networks may be analyzed by writing the integro-differential equation(s) of the system and then solving these equations by the Laplace transform. We now show that it is also possible to analyze electrical networks directly without having to write the - -### 374 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -integro-differential equations. This procedure is considerably simpler because it permits us to treat an electrical network as if it were a resistive network. For this purpose, we need to represent a network in the "frequency domain" where all the voltages and currents are represented by their Laplace transforms. - -For the sake of simplicity, let us first discuss the case with zero initial conditions. If *v*(*t*) and *i*(*t*) are the voltage across and the current through an inductor of *L* henries, then - -$$ -v(t) = L \frac{di(t)}{dt} -$$ - -The Laplace transform of this equation (assuming zero initial current) is - -$$ -V(s) = LsI(s) -$$ - -Similarly, for a capacitor of *C* farads, the voltage-current relationship is *i*(*t*) = *C*(*dv*/*dt*) and its Laplace transform, assuming zero initial capacitor voltage, yields *I*(*s*) = *CsV*(*s*); that is, - -$$ -V(s) = \frac{1}{Cs}I(s) -$$ - -For a resistor of *R* ohms, the voltage-current relationship is *v*(*t*) = *Ri*(*t*), and its Laplace transform is - -$$ -V(s) = RI(s) -$$ - -Thus, in the "frequency domain," the voltage-current relationships of an inductor and a capacitor are algebraic; these elements behave like resistors of "resistance" *Ls* and 1/*Cs*, respectively. The generalized "resistance" of an element is called its *impedance* and is given by the ratio *V*(*s*)/*I*(*s*) for the element (under zero initial conditions). The impedances of a resistor of *R* ohms, an inductor of *L* henries, and a capacitance of *C* farads are *R*, *Ls*, and 1/*Cs*, respectively. - -Also, the interconnection constraints (Kirchhoff's laws) remain valid for voltages and currents in the frequency domain. To demonstrate this point, let *vj*(*t*) (*j* = 1, 2,..., *k*) be the voltages across *k* elements in a loop and let *ij*(*t*)(*j* = 1, 2,...,*m*) be the *j* currents entering a node. Then - -$$ -\sum_{j=1}^{k} v_j(t) = 0 \quad \text{and} \quad \sum_{j=1}^{m} i_j(t) = 0 -$$ - -Now if - -$$ -v_j(t) \Longleftrightarrow V_j(s) -$$ - and $i_j(t) \Longleftrightarrow I_j(s)$ - -then - -$$ -\sum_{j=1}^{k} V_j(s) = 0 \quad \text{and} \quad \sum_{j=1}^{m} I_j(s) = 0 -$$ - -This result shows that if we represent all the voltages and currents in an electrical network by their Laplace transforms, we can treat the network as if it consisted of the "resistances" *R*, *Ls*, and 1/*Cs* corresponding to a resistor *R*, an inductor *L*, and a capacitor *C*, respectively. The system equations (loop or node) are now algebraic. Moreover, the simplification techniques that have been developed for resistive circuits—equivalent series and parallel impedances, voltage and current divider rules, Thévenin and Norton theorems—can be applied to general electrical networks. The following examples demonstrate these concepts. - -**Figure 4.10 (a)** A circuit and **(b)** its transformed version. - -In the first step, we represent the circuit in the frequency domain, as illustrated in Fig. 4.10b. All the voltages and currents are represented by their Laplace transforms. The voltage 10*u*(*t*) is represented by 10/*s* and the (unknown) current *i*(*t*) is represented by its Laplace transform *I*(*s*). All the circuit elements are represented by their respective impedances. The inductor of 1 henry is represented by *s*, the capacitor of 1/2 farad is represented by 2/*s*, and the resistor of 3 ohms is represented by 3. We now consider the frequency-domain representation of voltages and currents. The voltage across any element is *I*(*s*) times its impedance. Therefore, the total voltage drop in the loop is *I*(*s*) times the total loop impedance, and it must be equal to *V*(*s*), (transform of) the input voltage. The total impedance in the loop is - -$$ -Z(s) = s + 3 + \frac{2}{s} = \frac{s^2 + 3s + 2}{s} -$$ - -The input"voltage" is *V*(*s*) = 10/*s*. Therefore, the "loop current" *I*(*s*) is - -$$ -I(s) = \frac{V(s)}{Z(s)} = \frac{10/s}{(s^2 + 3s + 2)/s} = \frac{10}{s^2 + 3s + 2} = \frac{10}{(s+1)(s+2)} = \frac{10}{s+1} - \frac{10}{s+2} -$$ - -The inverse transform of this equation yields the desired result: - -$$ -i(t) = 10(e^{-t} - e^{-2t})u(t) -$$ - -### 376 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -### INITIAL CONDITION GENERATORS - -The discussion in which we assumed zero initial conditions can be readily extended to the case of nonzero initial conditions because the initial condition in a capacitor or an inductor can be represented by an equivalent source. We now show that a capacitor *C* with an initial voltage *v*(0−) (Fig. 4.11a) can be represented in the frequency domain by an uncharged capacitor of impedance 1/*Cs* in series with a voltage source of value *v*(0−)/*s* (Fig. 4.11b) or as the same uncharged capacitor in parallel with a current source of value *Cv*(0−) (Fig. 4.11c). Similarly, an inductor *L* with an initial current *i*(0−) (Fig. 4.11d) can be represented in the frequency domain by an inductor of impedance *Ls* in series with a voltage source of value *Li*(0−) (Fig. 4.11e) or by the same inductor in parallel with a current source of value *i*(0−)/*s* (Fig. 4.11f). - -**Figure 4.11** Initial condition generators for a capacitor and an inductor. - -To prove this point, consider the terminal relationship of the capacitor in Fig. 4.11a: - -$$ -i(t) = C \frac{dv(t)}{dt} -$$ - -The Laplace transform of this equation yields - -$$ -I(s) = C[sV(s) - v(0^-)] -$$ - -### 4.4 Analysis of Electrical Networks: The Transformed Network 377 - -This equation can be rearranged as - -$$ -V(s) = \frac{1}{Cs}I(s) + \frac{v(0^{-})}{s} -$$ -\n(4.33) - -Observe that *V*(*s*) is the voltage (in the frequency domain) across the charged capacitor and *I*(*s*)/*Cs* is the voltage across the same capacitor without any charge. Therefore, the charged capacitor can be represented by the uncharged capacitor in series with a voltage source of value *v*(0−)/*s*, as depicted in Fig. 4.11b. Equation (4.33) can also be rearranged as - -$$ -V(s) = \frac{1}{Cs}[I(s) + Cv(0^{-})] -$$ - -This equation shows that the charged capacitor voltage *V*(*s*) is equal to the uncharged capacitor voltage caused by a current *I*(*s*) + *Cv*(0−). This result is reflected precisely in Fig. 4.11c, where the current through the uncharged capacitor is *I*(*s*)+*Cv*(0−). † - -For the inductor in Fig. 4.11d, the terminal equation is - -$$ -v(t) = L \frac{di(t)}{dt} -$$ - -and - -$$ -V(s) = L[sI(s) - i(0^{-})] = LsI(s) - Li(0^{-}) -$$ -\n(4.34) - -This expression is consistent with Fig. 4.11e. We can rearrange Eq. (4.34) as - -$$ -V(s) = Ls \left[ I(s) - \frac{i(0^{-})}{s} \right] -$$ - -This expression is consistent with Fig. 4.11f. - -Let us rework Ex. 4.13 using these concepts. Figure 4.12a shows the circuit in Fig. 4.7b with the initial conditions *y*(0−) = 2 and *vC*(0−) = 10. Figure 4.12b shows the frequency-domain representation (transformed circuit) of the circuit in Fig. 4.12a. The resistor is represented by its impedance 2; the inductor with initial current of 2 amperes is represented according to the arrangement in Fig. 4.11e with a series voltage source *Ly*(0−) = 2. The capacitor with initial voltage of 10 volts is represented according to the arrangement in Fig. 4.11b with a series voltage source *v*(0−)/*s* = 10/*s*. Note that the impedance of the inductor is *s* and that of the capacitor is 5/*s*. The input of 10*u*(*t*) is represented by its Laplace transform 10/*s*. - -The total voltage in the loop is (10/*s*) + 2 − (10/*s*) = 2, and the loop impedance is (*s*+2+(5/*s*)). Therefore, - -$$ -Y(s) = \frac{2}{s + 2 + 5/s} = \frac{2s}{s^2 + 2s + 5} -$$ - -which confirms our earlier result in Ex. 4.13. - -† In the time domain, a charged capacitor *C* with initial voltage *v*(0−) can be represented as the same capacitor uncharged in series with a voltage source *v*(0−)*u*(*t*), or in parallel with a current source *Cv*(0−)δ(*t*). Similarly, an inductor *L* with initial current *i*(0−) can be represented by the same inductor with zero initial current in series with a voltage source *Li*(0−)δ(*t*) or with a parallel current source *i*(0−)*u*(*t*). - -**Figure 4.12** A circuit and its transformed version with initial-condition generators. - -**Figure 4.13** Using initial condition generators and Thévenin equivalent representation. - -Inspection of this circuit shows that when the switch is closed and the steady-state conditions are reached, the capacitor voltage *vC* = 16 volts, and the inductor current *y*2 = 4 amperes. Therefore, when the switch is opened (at *t* = 0), the initial conditions are *vC*(0−) = 16 and *y*2(0−) = 4. Figure 4.13b shows the transformed version of the circuit in Fig. 4.13a. We have used equivalent sources to account for the initial conditions. The initial capacitor voltage of 16 volts is represented by a series voltage of 16/*s* and the initial inductor current of 4 amperes is represented by a source of value *Ly*2(0−) = 2. - -From Fig. 4.13b, the loop equations can be written directly in the frequency domain as - -$$ -\frac{Y_1(s)}{s} + \frac{1}{5}[Y_1(s) - Y_2(s)] = \frac{4}{s} -$$ -$$ --\frac{1}{5}Y_1(s) + \frac{6}{5}Y_2(s) + \frac{s}{2}Y_2(s) = 2 -$$ -$$ -\begin{bmatrix} \frac{1}{s} + \frac{1}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{6}{5} + \frac{s}{2} \end{bmatrix} \begin{bmatrix} Y_1(s) \\ Y_2(s) \end{bmatrix} = \begin{bmatrix} \frac{4}{5} \\ \frac{5}{2} \end{bmatrix} -$$ - -Application of Cramer's rule to this equation yields - -$$ -Y_1(s) = \frac{24(s+2)}{s^2 + 7s + 12} = \frac{24(s+2)}{(s+3)(s+4)} = \frac{-24}{s+3} + \frac{48}{s+4} -$$ - -and - -$$ -y_1(t) = (-24e^{-3t} + 48e^{-4t})u(t) -$$ - -Similarly, we obtain - -$$ -Y_2(s) = \frac{4(s+7)}{s^2 + 7s + 12} = \frac{16}{s+3} - \frac{12}{s+4} -$$ - -and - -$$ -y_2(t) = (16e^{-3t} - 12e^{-4t})u(t) -$$ - -We also could have used Thévenin's theorem to compute *Y*1(*s*) and *Y*2(*s*) by replacing the circuit to the right of the capacitor (right of terminals *ab*) with its Thévenin equivalent, as shown in Fig. 4.13c. Figure 4.13b shows that the Thévenin impedance *Z*(*s*) and the Thévenin source *V*(*s*) are - -$$ -Z(s) = \frac{\frac{1}{5} \left(\frac{s}{2} + 1\right)}{\frac{1}{5} + \frac{s}{2} + 1} = \frac{s+2}{5s+12} -$$ -$$ -V(s) = \frac{-\frac{1}{5}}{\frac{1}{5} + \frac{s}{2} + 1} = \frac{-4}{5s+12} -$$ - -According to Fig. 4.13c, the current *Y*1(*s*) is given by - -$$ -Y_1(s) = \frac{\frac{4}{s} - V(s)}{\frac{1}{s} + Z(s)} = \frac{24(s+2)}{s^2 + 7s + 12} -$$ - -which confirms the earlier result. We may determine *Y*2(*s*) in a similar manner. - -### **EXAMPLE 4.19 Transformed Analysis of a Coupled Inductive Network** - -The switch in the circuit in Fig. 4.14a is at position a for a long time before *t* = 0, when it is moved instantaneously to position b. Determine the current *y*1(*t*) and the output voltage *v*0(*t*) for *t* ≥ 0. - -Just before switching, the values of the loop currents are 2 and 1, respectively, that is, *y*1(0−) = 2 and *y*2(0−) = 1. - -The equivalent circuits for two types of inductive coupling are illustrated in Figs. 4.14b and 4.14c. For our situation, the circuit in Fig. 4.14c applies. Figure 4.14d shows the transformed version of the circuit in Fig. 4.14a after switching. Note that the inductors *L*1 + *M*, *L*2 + *M*, and −*M* are 3, 4, and −1 henries with impedances 3*s*, 4*s*, and −*s* respectively. The initial condition voltages in the three branches are (*L*1 + *M*)*y*1(0−) = 6, (*L*2 + *M*)*y*2(0−) = 4, and −*M*[*y*1(0−)−*y*2(0−)]=−1, respectively. The two loop equations of the circuit are - -$$ -(2s+3)Y1(s) + (s-1)Y2(s) = \frac{10}{s} + 5 -$$ -$$ -(s-1)Y1(s) + (3s+2)Y2(s) = 5 -$$ - -or - -$$ -\begin{bmatrix} 2s+3 & s-1 \ s-1 & 3s+2 \end{bmatrix} \begin{bmatrix} Y_1(s) \\ Y_2(s) \end{bmatrix} \begin{bmatrix} \frac{5s+10}{s} \\ 5 \end{bmatrix} -$$ - -Solving for *Y*1(*s*), we obtain - -$$ -Y_1(s) = \frac{2s^2 + 9s + 4}{s(s^2 + 3s + 1)} = \frac{4}{s} - \frac{1}{s + 0.382} - \frac{1}{s + 2.618} -$$ - -Therefore, - -$$ -y_1(t) = (4 - e^{-0.382t} - e^{-2.618t})u(t) -$$ - -Similarly, - -$$ -Y_2(s) = \frac{s^2 + 2s + 2}{s(s^2 + 3s + 1)} = \frac{2}{s} - \frac{1.618}{s + 0.382} + \frac{0.618}{s + 2.618} -$$ - -and - -$$ -y_2(t) = (2 - 1.618e^{-0.382t} + 0.618e^{-2.618t})u(t) -$$ - -The output voltage is therefore - -$$ -v_0(t) = y_2(t) = (2 - 1.618e^{-0.382t} + 0.618e^{-2.618t})u(t) -$$ - -## **DR ILL 4.10 Transformed Analysis of an** *RLC* **Circuit with a Switch** - -For the *RLC* circuit in Fig. 4.15, the input is switched on at *t* = 0. The initial conditions are *y*(0−) = 2 amperes and *vC*(0−) = 50 volts. Find the loop current *y*(*t*) and the capacitor voltage *vC*(*t*) for *t* ≥ 0. - -### **ANSWERS** - - - -### **[4.4-1 Analysis of Active Circuits](#page-10-0)** - -Although we have considered examples of only passive networks so far, the circuit analysis procedure using the Laplace transform is also applicable to active circuits. All that is needed is to replace the active elements with their mathematical models (or equivalent circuits) and proceed as before. - -The operational amplifier (depicted by the triangular symbol in Fig. 4.16a) is a well-known element in modern electronic circuits. The terminals with the positive and the negative signs correspond to noninverting and inverting terminals, respectively. This means that the polarity of the output voltage *v*2 is the same as that of the input voltage at the terminal marked by the positive sign (noninverting). The opposite is true for the inverting terminal, marked by the negative sign. - -Figure 4.16b shows the model (equivalent circuit) of the operational amplifier (op amp) in Fig. 4.16a. A typical op amp has a very large gain. The output voltage *v*2 = −*Av*1, where *A* is typically 105 to 106. The input impedance is very high, of the order of 1012 , and the output impedance is very low (50–100 ). For most applications, we are justified in assuming the gain *A* and the input impedance to be infinite and the output impedance to be zero. For this reason we see an ideal voltage source at the output. - -Consider now the operational amplifier with resistors *Ra* and *Rb* connected, as shown in Fig. 4.16c. This configuration is known as the *noninverting amplifier*. Observe that the input polarities in this configuration are inverted in comparison to those in Fig. 4.16a. We now show that the output voltage *v*2 and the input voltage *v*1 in this case are related by - -$$ -v_2 = Kv_1, \qquad \text{where } K = 1 + \frac{R_b}{R_a} -$$ - -First, we recognize that because the input impedance and the gain of the operational amplifier approach infinity, the input current *ix* and the input voltage *vx* in Fig. 4.16c are infinitesimal and may be taken as zero. The dependent source in this case is *Avx* instead of −*Avx* because of the input polarity inversion. The dependent source *Avx* (see Fig. 4.16b) at the output will generate current *io*, as illustrated in Fig. 4.16c. Now - -$$ -v_2 = (R_b + R_a)i_o -$$ - -**Figure 4.16** Operational amplifier and its equivalent circuit. - -and also - -$$ -v_1 = v_x + R_a i_o = R_a i_o -$$ - -Therefore, - -$$ -\frac{v_2}{v_1} = \frac{R_b + R_a}{R_a} = 1 + \frac{R_b}{R_a} = K -$$ - -or - -$$ -v_2(t) = Kv_1(t) -$$ - -The equivalent circuit of the noninverting amplifier is depicted in Fig. 4.16d. - -### **EXAMPLE 4.20 Transform Analysis of a Sallen–Key Circuit** - -The circuit in Fig. 4.17a is called the *Sallen–Key* circuit, which is frequently used in filter design. Find the transfer function *H*(*s*) relating the output voltage *vo*(*t*) to the input voltage *vi*(*t*). - -**Figure 4.17 (a)** Sallen–Key circuit and **(b)** its equivalent. - -We are required to find - -$$ -H(s) = \frac{V_o(s)}{V_i(s)} -$$ - -assuming all initial conditions to be zero. - -Figure 4.17b shows the transformed version of the circuit in Fig. 4.17a. The noninverting amplifier is replaced by its equivalent circuit. All the voltages are replaced by their Laplace transforms, and all the circuit elements are shown by their impedances. All the initial conditions are assumed to be zero, as required for determining *H*(*s*). - -We shall use node analysis to derive the result. There are two unknown node voltages, *Va*(*s*) and *Vb*(*s*), requiring two node equations. - -At node *a*, *IR*1 (*s*), the current in *R*1 (leaving the node *a*), is [*Va*(*s*) − *Vi*(*s*)]/*R*1. Similarly, *IR*2 (*s*), the current in *R*2 (leaving the node *a*), is [*Va*(*s*) − *Vb*(*s*)]/*R*2, and *IC*1 (*s*), the current in capacitor *C*1 (leaving the node *a*), is [*Va*(*s*)−*Vo*(*s*)]*C*1*s* = [*Va*(*s*)−*KVb*(*s*)]*C*1*s*. - -The sum of all the three currents is zero. Therefore, - -$$ -\frac{V_a(s) - V_i(s)}{R_1} + \frac{V_a(s) - V_b(s)}{R_2} + [V_a(s) - KV_b(s)]C_1s = 0 -$$ - -or - -$$ -\left(\frac{1}{R_1} + \frac{1}{R_2} + C_1 s\right) V_a(s) - \left(\frac{1}{R_2} + KC_1 s\right) V_b(s) = \frac{1}{R_1} V_i(s) -$$ - -Similarly, the node equation at node *b* yields - -$$ -\frac{V_b(s) - V_a(s)}{R_2} + C_2 s V_b(s) = 0 -$$ - -or - -$$ --\frac{1}{R_2}V_a(s) + \left(\frac{1}{R_2} + C_2s\right)V_b(s) = 0 -$$ - -The two node equations in two unknown node voltages *Va*(*s*) and *Vb*(*s*) can be expressed in matrix form as - -$$ -\begin{bmatrix} G_1 + G_2 + C_1 s & -(G_2 + KC_1 s) \ -G_2 & (G_2 + C_2 s) \end{bmatrix} \begin{bmatrix} V_a(s) \ V_b(s) \end{bmatrix} = \begin{bmatrix} G_1 V_i(s) \ 0 \end{bmatrix} -$$ - -where - -$$ -G_1 = \frac{1}{R_1} \quad \text{and} \quad G_2 = \frac{1}{R_2} -$$ - -Application of Cramer's rule yields - -$$ -\frac{V_b(s)}{V_i(s)} = \frac{G_1 G_2}{C_1 C_2 s^2 + [G_1 C_2 + G_2 C_2 + G_2 C_1 (1 - K)]s + G_1 G_2} -$$ - -= -$$ -\frac{\omega_0^2}{s^2 + 2\alpha s + \omega_0^2} -$$ - -where - -$$ -K = 1 + \frac{R_b}{R_a} \quad \text{and} \quad \omega_0^2 = \frac{G_1 G_2}{C_1 C_2} = \frac{1}{R_1 R_2 C_1 C_2} -$$ -$$ -2\alpha = \frac{G_1 C_2 + G_2 C_2 + G_2 C_1 (1 - K)}{C_1 C_2} = \frac{1}{R_1 C_1} + \frac{1}{R_2 C_1} + \frac{1}{R_2 C_2} (1 - K) -$$ - -Now - -$$ -V_o(s) = KV_b(s) -$$ - -Therefore, - -$$ -H(s) = \frac{V_o(s)}{V_i(s)} = K \frac{V_b(s)}{V_i(s)} = \frac{K\omega_0^2}{s^2 + 2\alpha s + \omega_0} -$$ - -2 - -## **4.5 BLOCK [DIAGRAMS](#page-10-0)** - -Large systems may consist of an enormous number of components or elements. As anyone who has seen the circuit diagram of a radio or a television receiver can appreciate, analyzing such systems all at once could be next to impossible. In such cases, it is convenient to represent a system by suitably interconnected subsystems, each of which can be readily analyzed. Each subsystem can be characterized in terms of its input–output relationships. A linear system can be characterized by its transfer function *H*(*s*). Figure 4.18a shows a block diagram of a system with a transfer function *H*(*s*) and its input and output *X*(*s*) and *Y*(*s*), respectively. - -Subsystems may be interconnected by using cascade, parallel, and feedback interconnections (Figs. 4.18b, 4.18c, 4.18d), the three elementary types. When transfer functions appear in cascade, as depicted in Fig. 4.18b, then, as shown earlier, the transfer function of the overall system is the product of the two transfer functions. This result can also be proved by observing that in Fig. 4.18b - -$$ -\frac{Y(s)}{X(s)} = \frac{W(s)}{X(s)} \frac{Y(s)}{W(s)} = H_1(s)H_2(s) -$$ - -**Figure 4.18** Elementary connections of blocks and their equivalents. - -We can extend this result to any number of transfer functions in cascade. It follows from this discussion that the subsystems in cascade can be interchanged without affecting the overall transfer function. This commutation property of LTI systems follows directly from the commutative (and associative) property of convolution. We have already proved this property in Sec. 2.4-3. Every possible ordering of the subsystems yields the same overall transfer function. However, there may be practical consequences (such as sensitivity to parameter variation) affecting the behavior of different ordering. - -Similarly, when two transfer functions, *H*1(*s*) and *H*2(*s*), appear in parallel, as illustrated in Fig. 4.18c, the overall transfer function is given by *H*1(*s*) + *H*2(*s*), the sum of the two transfer functions. The proof is trivial. This result can be extended to any number of systems in parallel. - -When the output is fed back to the input, as shown in Fig. 4.18d, the overall transfer function *Y*(*s*)/*X*(*s*) can be computed as follows. The inputs to the adder are *X*(*s*) and −*H*(*s*)*Y*(*s*). Therefore, *E*(*s*), the output of the adder, is - -$$ -E(s) = X(s) - H(s)Y(s) -$$ - -But - -$$ -Y(s) = G(s)E(s) -$$ - -= $G(s)[X(s) - H(s)Y(s)]$ - -Therefore, - -$$ -Y(s)[1 + G(s)H(s)] = G(s)X(s) -$$ - -so that - -$$ -\frac{Y(s)}{X(s)} = \frac{G(s)}{1 + G(s)H(s)} -$$ -(4.35) - -Therefore, the feedback loop can be replaced by a single block with the transfer function shown in Eq. (4.35) (see Fig. 4.18d). - -In deriving these equations, we implicitly assume that when the output of one subsystem is connected to the input of another subsystem, the latter does not load the former. For example, the transfer function *H*1(*s*) in Fig. 4.18b is computed by assuming that the second subsystem *H*2(*s*) was not connected. This is the same as assuming that *H*2(*s*) does not load *H*1(*s*). In other words, the input–output relationship of *H*1(*s*) will remain unchanged regardless of whether *H*2(*s*) is connected. Many modern circuits use op amps with high input impedances, so this assumption is justified. When such an assumption is not valid, *H*1(*s*) must be computed under operating conditions [i.e., with *H*2(*s*) connected]. - -### **EXAMPLE 4.21 Transfer Functions of Feedback Systems Using MATLAB** - -Consider the feedback system of Fig. 4.18d with *G*(*s*) = *K*/(*s*(*s* + 8)) and *H*(*s*) = 1. Use MATLAB to determine the transfer function for each of the following cases: **(a)** *K* = 7, **(b)** *K* = 16, and **(c)** *K* = 80. - -We solve these cases using the control system toolbox function feedback. - -``` -(a) ->> H = tf(1,1); K = 7; G = tf([0 0 K],[1 8 0]); TFa = feedback(G,H) - Ha = - 7 - ------------- - s^2 + 8 s + 7 -Thus, Ha(s) = 7/(s2 +8s+7). -(b) ->> H = tf(1,1); K = 16; G = tf([0 0 K],[1 8 0]); TFb = feedback(G,H) - Hb = - 16 - -------------- - s^2 + 8 s + 16 -Thus, Hb(s) = 16/(s2 +8s+16). -(c) ->> H = tf(1,1); K = 80; G = tf([0 0 K],[1 8 0]); TFc = feedback(G,H) - Hc = - 80 - -------------- - s^2 + 8 s + 80 -Thus, Hc(s) = 80/(s2 +8s+80). -``` - -## **4.6 SYSTEM [REALIZATION](#page-10-0)** - -We now develop a systematic method for realization (or implementation) of an arbitrary *N*th-order transfer function. The most general transfer function with *M* = *N* is given by - -$$ -H(s) = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} -$$ -(4.36) - -Since realization is basically a synthesis problem, there is no unique way of realizing a system. A given transfer function can be realized in many different ways. A transfer function *H*(*s*) can be realized by using integrators or differentiators along with adders and multipliers. We avoid use of differentiators for practical reasons discussed in Secs. 2.1 and 4.3-3. Hence, in our implementation, we shall use integrators along with scalar multipliers and adders. We are already familiar with representation of all these elements except the integrator. The integrator can be represented by a box with integral sign (time-domain representation, Fig. 4.19a) or by a box with transfer function 1/*s* (frequency-domain representation, Fig. 4.19b). - -**Figure 4.19 (a)** Time-domain and **(b)** frequency-domain representations of an integrator. - -### **[4.6-1 Direct Form I Realization](#page-10-0)** - -Rather than realize the general *N*th-order system described by Eq. (4.36), we begin with a specific case of the following third-order system and then extend the results to the *N*th-order case: - -$$ -H(s) = \frac{b_0 s^3 + b_1 s^2 + b_2 s + b_3}{s^3 + a_1 s^2 + a_2 s + a_3} = \frac{b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \frac{b_3}{s^3}}{1 + \frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}} -$$ - -We can express *H*(*s*) as - -$$ -H(s) = \underbrace{\left(b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \frac{b_3}{s^3}\right)}_{H_1(s)} \underbrace{\left(\frac{1}{1 + \frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}\right)}_{H_2(s)}} -$$ - -We can realize *H*(*s*) as a cascade of transfer function *H*1(*s*) followed by *H*2(*s*), as depicted in Fig. 4.20a, where the output of *H*1(*s*) is denoted by *W*(*s*). Because of the commutative property of LTI system transfer functions in cascade, we can also realize *H*(*s*) as a cascade of *H*2(*s*) followed by *H*1(*s*), as illustrated in Fig. 4.20b, where the (intermediate) output of *H*2(*s*) is denoted by *V*(*s*). - -**Figure 4.20** Realization of a transfer function in two steps. - -The output of *H*1(*s*) in Fig. 4.20a is given by *W*(*s*) = *H*1(*s*)*X*(*s*). Hence, - -$$ -W(s) = \left(b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \frac{b_3}{s^3}\right)X(s) -$$ -\n(4.37) - -Also, the output *Y*(*s*) and the input *W*(*s*) of *H*2(*s*) in Fig. 4.20a are related by *Y*(*s*) = *H*2(*s*)*W*(*s*). Hence, - -$$ -W(s) = \left(1 + \frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}\right)Y(s) -$$ -\n(4.38) - -**Figure 4.21** Direct form I realization of an LTIC system: **(a)** third-order and **(b)** *N*th-order. - -We shall first realize *H*1(*s*). Equation (4.37) shows that the output *W*(*s*) can be synthesized by adding the input *b*0*X*(*s*) to *b*1(*X*(*s*)/*s*),*b*2(*X*(*s*)/*s*2), and *b*3(*X*(*s*)/*s*3). Because the transfer function of an integrator is 1/*s*, the signals *X*(*s*)/*s*,*X*(*s*)/*s*2, and *X*(*s*)/*s*3 can be obtained by successive integration of the input *x*(*t*). The left-half section of Fig. 4.21a shows how *W*(*s*) can be synthesized from *X*(*s*), according to Eq. (4.37). Hence, this section represents a realization of *H*1(*s*). - -To complete the picture, we shall realize *H*2(*s*), which is specified by Eq. (4.38). We can rearrange Eq. (4.38) as - -$$ -Y(s) = W(s) - \left(\frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}\right) Y(s) -$$ -\n(4.39) - -Hence, to obtain *Y*(*s*), we subtract *a*1*Y*(*s*)/*s*, *a*2*Y*(*s*)/*s*2, and *a*3*Y*(*s*)/*s*3 from *W*(*s*). We have already obtained *W*(*s*) from the first step [output of *H*1(*s*)]. To obtain signals *Y*(*s*)/*s*, *Y*(*s*)/*s*2, and *Y*(*s*)/*s*3, we assume that we already have the desired output *Y*(*s*). Successive integration of *Y*(*s*) yields the needed signals *Y*(*s*)/*s*, *Y*(*s*)/*s*2, and *Y*(*s*)/*s*3. We now synthesize the final output *Y*(*s*) according to Eq. (4.39), as seen in the right-half section of Fig. 4.21a.† The left-half section in Fig. 4.21a represents *H*1(*s*) and the right-half is *H*2(*s*). We can generalize this procedure, known as the *direct form I* (DFI) realization, for any value of *N*. This procedure requires 2*N* integrators to realize an *N*th-order transfer function, as shown in Fig. 4.21b. - -### **[4.6-2 Direct Form II Realization](#page-10-0)** - -In the direct form I, we realize *H*(*s*) by implementing *H*1(*s*) followed by *H*2(*s*), as shown in Fig. 4.20a. We can also realize *H*(*s*), as shown in Fig. 4.20b, where *H*2(*s*) is followed by *H*1(*s*). - -† It may seem odd that we first assumed the existence of *Y*(*s*), integrated it successively, and then in turn generated *Y*(*s*) from *W*(*s*) and the three successive integrals of signal *Y*(*s*). This procedure poses a dilemma similar to "Which came first, the chicken or the egg?" The problem here is satisfactorily resolved by writing the expression for *Y*(*s*) at the output of the right-hand adder (at the top) in Fig. 4.21a and verifying that this expression is indeed the same as Eq. (4.38). - -**Figure 4.22** Direct form II realization of an *N*th-order LTIC system. - -This procedure is known as the *direct form II* realization. Figure 4.22a shows direct form II realization, where we have interchanged sections representing *H*1(*s*) and *H*2(*s*) in Fig. 4.21b. The output of *H*2(*s*) in this case is denoted by *V*(*s*). ‡ - -An interesting observation in Fig. 4.22a is that the input signal to both the chains of integrators is *V*(*s*). Clearly, the outputs of integrators in the left-side chain are identical to the corresponding outputs of the right-side integrator chain, thus making the right-side chain redundant. We can eliminate this chain and obtain the required signals from the left-side chain, as shown in Fig. 4.22b. This implementation halves the number of integrators to *N*, and, thus, is more efficient in hardware utilization than either Figs. 4.21b or 4.22a. This is the *direct form II* (DFII) realization. - -An *N*th-order differential equation with *N* = *M* has a property that its implementation requires a minimum of *N* integrators. A realization is *canonic* if the number of integrators used in the realization is equal to the order of the transfer function realized. Thus, canonic realization has no redundant integrators. The DFII form in Fig. 4.22b is a canonic realization, and is also called the *direct canonic* form. Note that the DFI is noncanonic. - -The direct form I realization (Fig. 4.22b) implements zeros first [the left-half section represented by *H*1(*s*)] followed by realization of poles [the right-half section represented by *H*2(*s*)] of *H*(*s*). In contrast, canonic direct implements poles first followed by zeros. Although both these realizations result in the same transfer function, they generally behave differently from the viewpoint of sensitivity to parameter variations. - -$$ -V(s) = X(s) - \left(\frac{a_1}{s} + \frac{a_2}{s^2} + \dots + \frac{a_N}{s^N}\right) V(s) -$$ - -and - -$$ -Y(s) = \left(b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \dots + \frac{b_N}{s^N}\right) V(s) -$$ - -‡ The reader can show that the equations relating *X*(*s*),*V*(*s*), and *Y*(*s*) in Fig. 4.22a are - -### **EXAMPLE 4.22 Canonic Direct Form Realizations** - -Find the canonic direct form realization of the following transfer functions: - -(a) -$$ -\frac{5}{s+7} -$$ - -\n(b) $\frac{s}{s+7}$ -\n(c) $\frac{s+5}{s+7}$ -\n(d) $\frac{4s+28}{s^2+6s+5}$ - -All four of these transfer functions are special cases of *H*(*s*) in Eq. (4.36). - -**(a)** The transfer function 5/(*s*+7) is of the first order (*N* = 1); therefore, we need only one integrator for its realization. The feedback and feedforward coefficients are - -$$ -a_1 = 7 -$$ - and $b_0 = 0$ , $b_1 = 5$ - -The realization is depicted in Fig. 4.23a. Because *N* = 1, there is a single feedback connection from the output of the integrator to the input adder with coefficient *a*1 = 7. For *N* = 1, generally, there are *N* + 1 = 2 feedforward connections. However, in this case, *b*0 = 0, and there is only one feedforward connection with coefficient *b*1 = 5 from the output of the integrator to the output adder. Because there is only one input signal to the output adder, we can do away with the adder, as shown in Fig. 4.23a. - -**(b)** - -$$ -H(s) = \frac{s}{s+7} -$$ - -In this first-order transfer function, *b*1 = 0. The realization is shown in Fig. 4.23b. Because there is only one signal to be added at the output adder, we can discard the adder. - -**(c)** - -$$ -H(s) = \frac{s+5}{s+7} -$$ - -The realization appears in Fig. 4.23c. Here *H*(*s*) is a first-order transfer function with *a*1 = 7 and *b*0 = 1, *b*1 = 5. There is a single feedback connection (with coefficient 7) from the integrator output to the input adder. There are two feedforward connections (Fig. 4.23c).† - -$$ -H(s) = 1 - \frac{2}{s+7} -$$ - -We now realize *H*(*s*) as a parallel combination of two transfer functions, as indicated by this equation. - -† When *M* = *N* (as in this case), *H*(*s*) can also be realized in another way by recognizing that - -$$ -f_{\rm{max}} -$$ - -**(d)** - -$$ -H(s) = \frac{4s + 28}{s^2 + 6s + 5} -$$ - -This is a second-order system with *b*0 = 0, *b*1 = 4, *b*2 = 28, *a*1 = 6, and *a*2 = 5. Figure 4.23d shows a realization with two feedback connections and two feedforward connections. - -## **DR ILL 4.11 Canonic Direct Form Realization** - -Give the canonic direct realization of - -$$ -H(s) = \frac{2s}{s^2 + 6s + 25} -$$ - -### **[4.6-3 Cascade and Parallel Realizations](#page-10-0)** - -An *N*th-order transfer function *H*(*s*) can be expressed as a product or a sum of *N* first-order transfer functions. Accordingly, we can also realize *H*(*s*) as a cascade (series) or parallel form of these *N* first-order transfer functions. Consider, for instance, the transfer function in part (d) of Ex. 4.22. - -$$ -H(s) = \frac{4s + 28}{s^2 + 6s + 5} -$$ - -We can express *H*(*s*) as - -$$ -H(s) = \frac{4s + 28}{(s+1)(s+5)} = \underbrace{\left(\frac{4s + 28}{s+1}\right)}_{H_1(s)} \underbrace{\left(\frac{1}{s+5}\right)}_{H_2(s)} -$$ - -We can also express *H*(*s*) as a sum of partial fractions as - -$$ -H(s) = \frac{4s + 28}{(s+1)(s+5)} = \underbrace{\frac{6}{s+1}}_{H_3(s)} - \underbrace{\frac{2}{s+5}}_{H_4(s)} -$$ - -These equations give us the option of realizing *H*(*s*) as a cascade of *H*1(*s*) and *H*2(*s*), as shown in Fig. 4.24a, or a parallel of *H*3(*s*) and *H*4(*s*), as depicted in Fig. 4.24b. Each of the first-order transfer functions in Fig. 4.24 can be implemented by using canonic direct realizations, discussed earlier. - -This discussion by no means exhausts all the possibilities. In the cascade form alone, there are different ways of grouping the factors in the numerator and the denominator of *H*(*s*), and each grouping can be realized in DFI or canonic direct form. Accordingly, several cascade forms are possible. In Sec. 4.6-4, we shall discuss yet another form that essentially doubles the numbers of realizations discussed so far. - -From a practical viewpoint, parallel and cascade forms are preferable because parallel and certain cascade forms are numerically less sensitive than canonic direct form to small parameter variations in the system. Qualitatively, this difference can be explained by the fact that in a canonic realization all the coefficients interact with each other, and a change in any coefficient will be magnified through its repeated influence from feedback and feedforward connections. In a parallel realization, in contrast, the change in a coefficient will affect only a localized segment; the case with a cascade realization is similar. - -In the examples of cascade and parallel realization, we have separated *H*(*s*) into first-order factors. For *H*(*s*) of higher orders, we could group *H*(*s*) into factors, not all of which are necessarily of the first order. For example, if *H*(*s*) is a third-order transfer function, we could realize this function as a cascade (or a parallel) combination of a first-order and a second-order factor. - -**Figure 4.24** Realization of (4*s* +28)/[(*s*+1)(*s*+5)]: **(a)** cascade form and **(b)** parallel form. - -### REALIZATION OF COMPLEX CONJUGATE POLES - -The complex poles in *H*(*s*) should be realized as a second-order (quadratic) factor because we cannot implement multiplication by complex numbers. Consider, for example, - -$$ -H(s) = \frac{10s + 50}{(s+3)(s^2+4s+13)} -$$ - -= -$$ -\frac{10s + 50}{(s+3)(s+2-j3)(s+2+j3)} -$$ - -= -$$ -\frac{2}{s+3} - \frac{1+j2}{s+2-j3} - \frac{1-j2}{s+2+j3} -$$ - -We cannot realize first-order transfer functions individually with the poles −2 ± *j*3 because they require multiplication by complex numbers in the feedback and the feedforward paths. Therefore, we need to combine the conjugate poles and realize them as a second-order transfer function.† In the present example, we can create a cascade realization from *H*(*s*) expressed in product form as - -$$ -H(s) = \left(\frac{10}{s+3}\right) \left(\frac{s+5}{s^2+4s+13}\right) -$$ - -Similarly, we can create a parallel realization from *H*(*s*) expressed in sum form as - -$$ -H(s) = \frac{2}{s+3} - \frac{2s-8}{s^2+4s+13} -$$ - -### REALIZATION OF REPEATED POLES - -When repeated poles occur, the procedure for canonic and cascade realization is exactly the same as before. For a parallel realization, however, the procedure requires special handling, as explained in Ex. 4.23. - -### **EXAMPLE 4.23 Parallel Realization** - -Determine the parallel realization of - -$$ -H(s) = \frac{7s^2 + 37s + 51}{(s+2)(s+3)^2} = \frac{5}{s+2} + \frac{2}{s+3} - \frac{3}{(s+3)^2} -$$ - -This third-order transfer function should require no more than three integrators. But if we try to realize each of the three partial fractions separately, we require four integrators because of the one second-order term. This difficulty can be avoided by observing that the terms 1/(*s*+3) and - -† It is possible to realize complex, conjugate poles indirectly by using a cascade of two first-order transfer functions and feedback. A transfer function with poles −*a* ± *jb* can be realized by using a cascade of two identical first-order transfer functions, each having a pole at −*a* (see Prob. 4.6-15). - -**Figure 4.25** Parallel realization of (7*s*2 +37*s* +51)/((*s* +2)(*s* +3)2). - -1/(*s* + 3)2 can be realized with a cascade of two subsystems, each having a transfer function 1/(*s* + 3), as shown in Fig. 4.25. Each of the three first-order transfer functions in Fig. 4.25 may now be realized as in Fig. 4.23. - -### **DR ILL 4.12 Canonic, Cascade, and Parallel Realizations** - -Find the canonic, cascade, and parallel realization of - -$$ -H(s) = \frac{s+3}{s^2 + 7s + 10} = \left(\frac{s+3}{s+2}\right)\left(\frac{1}{s+5}\right) -$$ - -### **[4.6-4 Transposed Realization](#page-10-0)** - -Two realizations are said to be *equivalent* if they have the same transfer function. A simple way to generate an equivalent realization from a given realization is to use its *transpose*. To generate a transpose of any realization, we change the given realization as follows: - -- 1. Reverse all the arrow directions without changing the scalar multiplier values. -- 2. Replace pickoff nodes by adders and vice versa. -- 3. Replace the input *X*(*s*) with the output *Y*(*s*) and vice versa. - -Figure 4.26a shows the transposed version of the canonic direct form realization in Fig. 4.22b found according to the rules just listed. Figure 4.26b is Fig. 4.26a reoriented in the conventional form so that the input *X*(*s*) appears at the left and the output *Y*(*s*) appears at the right. Observe that this realization is also canonic. - -Rather than prove the theorem on equivalence of the transposed realizations, we shall verify that the transfer function of the realization in Fig. 4.26b is identical to that in Eq. (4.36). - -**Figure 4.26** Realization of an *N*th-order LTI transfer function in the transposed form. - -Figure 4.26b shows that *Y*(*s*) is being fed back through *N* paths. The fed-back signal appearing at the input of the top adder is - -$$ -\left(\frac{-a_1}{s} + \frac{-a_2}{s^2} + \cdots + \frac{-a_{N-1}}{s^{N-1}} + \frac{-a_N}{s^N}\right)Y(s) -$$ - -The signal *X*(*s*), fed to the top adder through *N* +1 forward paths, contributes - -$$ -\left(b_0 + \frac{b_1}{s} + \cdots + \frac{b_{N-1}}{s^{N-1}} + \frac{b_N}{s^N}\right)X(s) -$$ - -The output *Y*(*s*) is equal to the sum of these two signals (feed forward and feed back). Hence, - -$$ -Y(s) = \left(\frac{-a_1}{s} + \frac{-a_2}{s^2} + \dots + \frac{-a_{N-1}}{s^{N-1}} + \frac{-a_N}{s^N}\right)Y(s) + \left(b_0 + \frac{b_1}{s} + \dots + \frac{b_{N-1}}{s^{N-1}} + \frac{b_N}{s^N}\right)X(s) -$$ - -Transporting all the *Y*(*s*) terms to the left side and multiplying throughout by *sN*, we obtain - -$$ -(sN + a1sN-1 + \dots + aN-1s + aN)Y(s) = (b0sN + b1sN-1 + \dots + bN-1s + bN)X(s) -$$ - -Consequently, - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} -$$ - -Hence, the transfer function *H*(*s*) is identical to that in Eq. (4.36). - -We have essentially doubled the number of possible realizations. Every realization that was found earlier has a transpose. Note that the transpose of a transpose results in the same realization. - -### **EXAMPLE 4.24 Transposed Realizations** - -Find the transpose canonic direct realizations for parts (a) and (d) of Ex. 4.22 (Figs. 4.23c and 4.23d). The transfer functions are: - -(a) -$$ -\frac{s+5}{s+7} -$$ - -\n(b) $\frac{4s+28}{s^2+6s+5}$ - -Both these realizations are special cases of the one in Fig. 4.26b. - -**(a)** In this case, *N* = 1 with *a*1 = 7,*b*0 = 1,*b*1 = 5. The desired realization can be obtained by transposing Fig. 4.23c. However, we already have the general model of the transposed realization in Fig. 4.26b. The desired solution is a special case of Fig. 4.26b with *N* = 1 and *a*1 = 7,*b*0 = 1,*b*1 = 5, as shown in Fig. 4.27a. - -**(b)** In this case, *N* = 2 with *b*0 = 0, *b*1 = 4, *b*2 = 28, *a*1 = 6, *a*2 = 5. Using the model of Fig. 4.26b, we obtain the desired realization, as shown in Fig. 4.27b. - -**Figure 4.27** Transposed canonic direct form realizations of **(a)** (*s*+5)/(*s*+7) and **(b)** (4*s*+28)/(*s*2 +6*s*+5). - -### **DR ILL 4.13 Transposed Realizations** - -Find the transposed DFI and transposed canonic direct (TDFII) realizations of *H*(*s*) in Drill 4.11 - -### **[4.6-5 Using Operational Amplifiers for System Realization](#page-10-0)** - -In this section, we discuss practical implementation of the realizations described in Sec. 4.6-4. Earlier we saw that the basic elements required for the synthesis of an LTIC system (or a given transfer function) are (scalar) multipliers, integrators, and adders. All these elements can be realized by operational amplifier (op-amp) circuits. - -### OPERATIONAL AMPLIFIER CIRCUITS - -Figure 4.28 shows an op-amp circuit in the frequency domain (the transformed circuit). Because the input impedance of the op amp is infinite (very high), all the current *I*(*s*) flows in the feedback path, as illustrated. Moreover *Vx*(*s*), the voltage at the input of the op amp, is zero (very small) because of the infinite (very large) gain of the op amp. Therefore, for all practical purposes, - -$$ -Y(s) = -I(s)Z_f(s) -$$ - -Moreover, because *vx* ≈ 0, - -$$ -I(s) = \frac{X(s)}{Z(s)} -$$ - -Substitution of the second equation in the first yields - -$$ -Y(s) = -\frac{Z_f(s)}{Z(s)}X(s) -$$ - -Therefore, the op-amp circuit in Fig. 4.28 has the transfer function - -$$ -H(s) = -\frac{Z_f(s)}{Z(s)} -$$ - -By properly choosing *Z*(*s*) and *Zf*(*s*), we can obtain a variety of transfer functions, as the following development shows. - -**Figure 4.28** A basic inverting configuration op-amp circuit. - -### THE SCALAR MULTIPLIER - -If we use a resistor *Rf* in the feedback and a resistor *R* at the input (Fig. 4.29a), then *Zf*(*s*) = *Rf* , *Z*(*s*) = *R*, and - -$$ -H(s) = -\frac{R_f}{R} -$$ - -The system acts as a scalar multiplier (or an amplifier) with a negative gain *Rf* /*R*. A positive gain can be obtained by using two such multipliers in cascade or by using a single noninverting amplifier, as depicted in Fig. 4.16c. Figure 4.29a also shows the compact symbol used in circuit diagrams for a scalar multiplier. - -### THE INTEGRATOR - -If we use a capacitor *C* in the feedback and a resistor *R* at the input (Fig. 4.29b), then *Zf*(*s*) = 1/*Cs*, *Z*(*s*) = *R*, and - -$$ -H(s) = \left(-\frac{1}{RC}\right)\frac{1}{s} -$$ - -The system acts as an ideal integrator with a gain −1/*RC*. Figure 4.29b also shows the compact symbol used in circuit diagrams for an integrator. - -**Figure 4.29 (a)** Op-amp inverting amplifier. **(b)** Integrator. - -**Figure 4.30** Op-amp summing and amplifying circuit. - -### THE ADDER - -Consider now the circuit in Fig. 4.30a with *r* inputs *X*1(*s*), *X*2(*s*), ... , *Xr*(*s*). As usual, the input voltage *Vx*(*s*) 0 because the op-amp gain → ∞. Moreover, the current going into the op amp is very small ( 0) because the input impedance → ∞. Therefore, the total current in the feedback resistor *Rf* is *I*1(*s*)+*I*2(*s*)+···+*Ir*(*s*). Moreover, because *Vx*(*s*) = 0, - -$$ -I_j(s) = \frac{X_j(s)}{R_j} -$$ - $j = 1, 2, ..., r$ - -Also, - -$$ -Y(s) = -R_f[I_1(s) + I_2(s) + \dots + I_r(s)] -$$ - -= -$$ --\left[\frac{R_f}{R_1}X_1(s) + \frac{R_f}{R_2}X_2(s) + \dots + \frac{R_f}{R_r}X_r(s)\right] -$$ - -= $k_1X_1(s) + k_2X_2(s) + \dots + k_rX_r(s)$ - -where - -$$ -k_i = \frac{-R_f}{R_i} -$$ - -Clearly, the circuit in Fig. 4.30 serves an adder and an amplifier with any desired gain for each of the input signals. Figure 4.30b shows the compact symbol used in circuit diagrams for an adder with *r* inputs. - -### **EXAMPLE 4.25 Op-Amp Realization** - -Use op-amp circuits to realize the canonic direct form of the transfer function - -$$ -H(s) = \frac{2s+5}{s^2+4s+10} -$$ - -**Figure 4.31** Op-amp realization of a second-order transfer function (2*s*+5)/(*s*2 +4*s* +10). - -The basic canonic realization is shown in Fig. 4.31a. The same realization with horizontal reorientation is shown in Fig. 4.31b. Signals at various points are also indicated in the realization. For convenience, we denote the output of the last integrator by *W*(*s*). Consequently, the signals at the inputs of the two integrators are *sW*(*s*) and *s*2*W*(*s*), as shown in Figs. 4.31a and 4.31b. Op-amp elements (multipliers, integrators, and adders) change the polarity of the output signals. To incorporate this fact, we modify the canonic realization in Fig. 4.31b to that depicted in Fig. 4.31c. In Fig. 4.31b, the successive outputs of the adder and the integrators are *s*2*W*(*s*),*sW*(*s*), and *W*(*s*), respectively. Because of polarity reversals in op-amp circuits, these outputs are −*s*2*W*(*s*),*sW*(*s*), and −*W*(*s*), respectively, in Fig. 4.31c. This polarity reversal requires corresponding modifications in the signs of feedback and feedforward gains. According to Fig. 4.31b, - -$$ -s^2W(s) = X(s) - 4sW(s) - 10W(s) -$$ - -Therefore, - -$$ --s^2W(s) = -X(s) + 4sW(s) + 10W(s) -$$ - -Because the adder gains are always negative (see Fig. 4.30b), we rewrite the foregoing equation as - -$$ --s2W(s) = -1[X(s)] - 4[-sW(s)] - 10[-W(s)] -$$ - -Figure 4.31c shows the implementation of this equation. The hardware realization appears in Fig. 4.31d. Both integrators have a unity gain, which requires *RC* = 1. We have used *R* = 100 k and *C* = 10 µF. The gain of 10 in the outer feedback path is obtained in the adder by choosing the feedback resistor of the adder to be 100 k and an input resistor of 10 k. Similarly, the gain of 4 in the inner feedback path is obtained by using the corresponding input resistor of 25 k. The gains of 2 and 5, required in the feedforward connections, are obtained by using a feedback resistor of 100 k and input resistors of 50 and 20 k, respectively.† - -The op-amp realization in Fig. 4.31 is not necessarily the one that uses the fewest op amps. This example is given just to illustrate a systematic procedure for designing an op-amp circuit of an arbitrary transfer function. There are more efficient circuits (such as Sallen–Key or biquad) that use fewer op amps to realize a second-order transfer function. - -### **DR ILL 4.14 Transfer Functions of Op-Amp Circuits** - -Show that the transfer functions of the op-amp circuits in Figs. 4.32a and 4.32b are *H*1(*s*) and *H*2(*s*), respectively, where - -$$ -H_1(s) = \frac{-R_f}{R} \left( \frac{a}{s+a} \right) \qquad a = \frac{1}{R_f C_f} -$$ - -$$ -H_2(s) = -\frac{C}{C_f} \left( \frac{s+b}{s+a} \right) \qquad a = \frac{1}{R_f C_f} \qquad b = \frac{1}{RC} -$$ - -† It is possible to avoid the two inverting op amps (with gain −1) in Fig. 4.31d by adding signal *sW*(*s*) to the input and output adders directly, using the noninverting amplifier configuration in Fig. 4.16d. - - - -## **[4.7 APPLICATION TO](#page-10-0) FEEDBACK AND CONTROLS** - -Generally, systems are designed to produce a desired output *y*(*t*) for a given input *x*(*t*). Using the given performance criteria, we can design a system, as shown in Fig. 4.33a. Ideally, such an open-loop system should yield the desired output. In practice, however, the system characteristics change with time, as a result of aging or replacement of some components, or because of changes in the operating environment. Such variations cause changes in the output for the same input. Clearly, this is undesirable in precision systems. - -**Figure 4.33 (a)** Open-loop and **(b)** closed-loop (feedback) systems. - -A possible solution to this problem is to add a signal component to the input that is not a predetermined function of time but will change to counteract the effects of changing system characteristics and the environment. In short, we must provide a correction at the system input to account for the undesired changes just mentioned. Yet since these changes are generally unpredictable, it is not clear how to preprogram appropriate corrections to the input. However, the difference between the actual output and the desired output gives an indication of the suitable correction to be applied to the system input. It may be possible to counteract the variations by feeding the output (or some function of output) back to the input. - -We unconsciously apply this principle in daily life. Consider an example of marketing a certain product. The optimum price of the product is the value that maximizes the profit of a merchant. The output in this case is the profit, and the input is the price of the item. The output (profit) can be controlled (within limits) by varying the input (price). The merchant may price the product too high initially, in which case, he will sell too few items, reducing the profit. Using feedback of the profit (output), he adjusts the price (input), to maximize his profit. If there is a sudden or unexpected change in the business environment, such as a strike-imposed shutdown of a large factory in town, the demand for the item goes down, thus reducing his output (profit). He adjusts his input (reduces price) using the feedback of the output (profit) in a way that will optimize his profit in the changed circumstances. If the town suddenly becomes more prosperous because a new factory opens, he will increase the price to maximize the profit. Thus, by continuous feedback of the output to the input, he realizes his goal of maximum profit (optimum output) in any given circumstances. We observe thousands of examples of feedback systems around us in everyday life. Most social, economical, educational, and political processes are, in fact, feedback processes. A block diagram of such a system, called the *feedback* or *closed-loop* system, is shown in Fig. 4.33b. - -A feedback system can address the problems arising because of unwanted disturbances such as random-noise signals in electronic systems, a gust of wind affecting a tracking antenna, a meteorite hitting a spacecraft, and the rolling motion of antiaircraft gun platforms mounted on ships or moving tanks. Feedback may also be used to reduce nonlinearities in a system or to control its rise time (or bandwidth). Feedback is used to achieve, with a given system, the desired objective within a given tolerance, despite partial ignorance of the system and the environment. A feedback system, thus, has an ability for supervision and self-correction in the face of changes in the system parameters and external disturbances (change in the environment). - -Consider the feedback amplifier in Fig. 4.34. Let the forward amplifier gain *G* = 10,000. One-hundredth of the output is fed back to the input (*H* = 0.01). The gain *T* of the feedback amplifier is obtained by [see Eq. (4.35)] - -$$ -T = \frac{G}{1 + GH} = \frac{10,000}{1 + 100} = 99.01 -$$ - -Suppose that because of aging or replacement of some transistors, the gain *G* of the forward amplifier changes from 10,000 to 20,000. The new gain of the feedback amplifier is given by - -$$ -T = \frac{G}{1 + GH} = \frac{20,000}{1 + 200} = 99.5 -$$ - -Surprisingly, 100% variation in the forward gain *G* causes only 0.5% variation in the feedback amplifier gain *T*. Such reduced sensitivity to parameter variations is a must in precision amplifiers. In this example, we reduced the sensitivity of gain to parameter variations at the cost of forward - -*H* **Figure 4.34** Effects of negative and positive feedback. - -#### 406 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -gain, which is reduced from 10,000 to 99. There is no dearth of forward gain (obtained by cascading stages). But low sensitivity is extremely precious in precision systems. - -Now, consider what happens when we add (instead of subtract) the signal fed back to the input. Such addition means the sign on the feedback connection is + instead of − (which is same as changing the sign of *H* in Fig. 4.34). Consequently, - -$$ -T = \frac{G}{1 - GH} -$$ - -If we let *G* = 10,000 as before and *H* = 0.9×10−4, then - -$$ -T = \frac{10,000}{1 - 0.9(10^{4})(10^{-4})} = 100,000 -$$ - -Suppose that because of aging or replacement of some transistors, the gain of the forward amplifier changes to 11,000. The new gain of the feedback amplifier is - -$$ -T = \frac{11,000}{1 - 0.9(11,000)(10^{-4})} = 1,100,000 -$$ - -Observe that in this case, a mere 10% increase in the forward gain *G* caused 1000% increase in the gain *T* (from 100,000 to 1,100,000). Clearly, the amplifier is very sensitive to parameter variations. This behavior is exactly opposite of what was observed earlier, when the signal fed back was subtracted from the input. - -What is the difference between the two situations? Crudely speaking, the former case is called the *negative feedback* and the latter is the *positive feedback*. The positive feedback increases system gain but tends to make the system more sensitive to parameter variations. It can also lead to instability. In our example, if *G* were to be 111,111, then *GH* = 1, *T* = ∞, and the system would become unstable because the signal fed back was exactly equal to the input signal itself, since *GH* = 1. Hence, once a signal has been applied, no matter how small and how short in duration, it comes back to reinforce the input undiminished, which further passes to the output, and is fed back again and again and again. In essence, the signal perpetuates itself forever. This perpetuation, even when the input ceases to exist, is precisely the symptom of instability. - -Generally speaking, a feedback system cannot be described in black and white terms, such as positive or negative. Usually *H* is a frequency-dependent component, more accurately represented by *H*(*s*); hence it varies with frequency. Consequently, what was negative feedback at lower frequencies can turn into positive feedback at higher frequencies and may give rise to instability. This is one of the serious aspects of feedback systems, which warrants a designer's careful attention. - -### **[4.7-1 Analysis of a Simple Control System](#page-10-0)** - -Figure 4.35a represents an automatic position control system, which can be used to control the angular position of a heavy object (e.g., a tracking antenna, an anti-aircraft gun mount, or the position of a ship). The input θ*i* is the desired angular position of the object, which can be set at any given value. The actual angular position θ*o* of the object (the output) is measured by a potentiometer whose wiper is mounted on the output shaft. The difference between the input θ*i* - -(set at the desired output position) and the output θ*o* (actual position) is amplified; the amplified output, which is proportional to θ*i* − θ*o*, is applied to the motor input. If θ*i* − θ*o* = 0 (the output being equal to the desired angle), there is no input to the motor, and the motor stops. But if θ*o* = θ*i*, there will be a nonzero input to the motor, which will turn the shaft until θ*o* = θ*i*. It is evident that by setting the input potentiometer at a desired position in this system, we can control the angular position of a heavy remote object. - -The block diagram of this system is shown in Fig. 4.35b. The amplifier gain is *K*, where *K* is adjustable. Let the motor (with load) transfer function that relates the output angle θ*o* to the motor input voltage be *G*(*s*) [for a starting point, see Eq. (1.32)]. This feedback arrangement is identical to that in Fig. 4.18d with *H*(*s*) = 1. Hence, *T*(*s*), the (closed-loop) system transfer function relating the output θ*o* to the input θ*i*, is - -$$ -\frac{\Theta_o(s)}{\Theta_i(s)} = T(s) = \frac{KG(s)}{1 + KG(s)} -$$ - -From this equation, we shall investigate the behavior of the automatic position control system in Fig. 4.35a for a step and a ramp input. - -### STEP INPUT - -If we desire to change the angular position of the object instantaneously, we need to apply a step input. We may then want to know how long the system takes to position itself at the desired angle, whether it reaches the desired angle, and whether it reaches the desired position smoothly (monotonically) or oscillates about the final position. If the system oscillates, we may want to know how long it takes for the oscillations to settle down. All these questions can be readily answered by finding the output θ*o*(*t*) when the input θ*i*(*t*) = *u*(*t*). A step input implies instantaneous change in the angle. This input would be one of the most difficult to follow; if the system can perform well for this input, it is likely to give a good account of itself under most other expected situations. This is why we test control systems for a step input. - -For the step input θ*i*(*t*) = *u*(*t*), *i*(*s*) = 1/*s* and - -$$ -\Theta_o(s) = \frac{1}{s}T(s) = \frac{KG(s)}{s[1+KG(s)]} -$$ - -Let the motor (with load) transfer function relating the load angle θ*o*(*t*) to the motor input voltage be *G*(*s*) = 1/(*s*(*s*+8)). This yields - -$$ -\Theta_o(s) = \frac{\frac{K}{s(s+8)}}{s\left[1 + \frac{K}{s(s+8)}\right]} = \frac{K}{s(s^2 + 8s + K)} -$$ - -Let us investigate the system behavior for three different values of gain *K*. For *K* = 7, - -$$ -\Theta_o(s) = \frac{7}{s(s^2 + 8s + 7)} = \frac{7}{s(s+1)(s+7)} = \frac{1}{s} - \frac{\frac{7}{6}}{s+1} + \frac{\frac{1}{6}}{s+7} -$$ - -**Figure 4.35 (a)** An automatic position control system. **(b)** Its block diagram. **(c)** The unit step response. **(d)** The unit ramp response. - -and - -$$ -\theta_o(t) = \left(1 - \frac{7}{6}e^{-t} + \frac{1}{6}e^{-7t}\right)u(t) -$$ - -This response, illustrated in Fig. 4.35c, shows that the system reaches the desired angle, but at a rather leisurely pace. To speed up the response let us increase the gain to, say, 80. - -For *K* = 80, - -$$ -\Theta_o(s) = \frac{80}{s(s^2 + 8s + 80)} = \frac{80}{s(s + 4 - j8)(s + 4 + j8)} -$$ -$$ -= \frac{1}{s} + \frac{\frac{\sqrt{5}}{4}e^{j153^\circ}}{s + 4 - j8} + \frac{\frac{\sqrt{5}}{4}e^{-j153^\circ}}{s + 4 + j8} -$$ - -and - -$$ -\theta_o(t) = \left[1 + \frac{\sqrt{5}}{2}e^{-4t}\cos{(8t + 153^\circ)}\right]u(t) -$$ - -This response, also depicted in Fig. 4.35c, achieves the goal of reaching the final position at a faster rate than that in the earlier case (*K* = 7). Unfortunately the improvement is achieved at the cost of ringing (oscillations) with high overshoot. In the present case, the *percent overshoot* (PO) is 21%. The response reaches its peak value at *peak time tp* = 0.393 second. The *rise time*, defined as the time required for the response to rise from 10% to 90% of its steady-state value, indicates the speed of response.† In the present case *tr* = 0.175 second. The steady-state value of the response is unity so that the *steady-state error* is zero. Theoretically it takes infinite time for the response to reach the desired value of unity. In practice, however, we may consider the response to have settled to the final value if it closely approaches the final value. A widely accepted measure of closeness is within 2% of the final value. The time required for the response to reach and stay within 2% of the final value is called the settling time *ts*. ‡ In Fig. 4.35c, we find *ts* ≈ 1 second (when *K* = 80). A good system has a small overshoot, small *tr* and *ts* and a small steady-state error. - -A large overshoot, as in the present case, may be unacceptable in many applications. Let us try to determine *K* (the gain) that yields the fastest response without oscillations. Complex characteristic roots lead to oscillations; to avoid oscillations, the characteristic roots should be real. In the present case, the characteristic polynomial is *s*2 +8*s*+*K*. For *K* > 16, the characteristic roots are complex; for *K* < 16, the roots are real. The fastest response without oscillations is obtained by choosing *K* = 16. We now consider this case. - -For *K* = 16, - -$$ -\Theta_o(s) = \frac{16}{s(s^2 + 8s + 16)} = \frac{16}{s(s+4)^2} = \frac{1}{s} - \frac{1}{s+4} - \frac{4}{(s+4)^2} -$$ - -and - -$$ -\theta_o(t) = [1 - (4t + 1)e^{-4t}]u(t) -$$ - -This response also appears in Fig. 4.35c. The system with *K* > 16 is said to be *underdamped* (oscillatory response), whereas the system with *K* < 16 is said to be *overdamped*. For *K* = 16, the system is said to be *critically damped*. - -† *Delay time td*, defined as the time required for the response to reach 50% of its steady-state value, is another indication of speed. For the present case, *td* = 0.141 second. - -‡ Typical percentage values used are 2 to 5% for *ts*. - -### 410 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -There is a trade-off between undesirable overshoot and rise time. Reducing overshoots leads to higher rise time (sluggish system). In practice, a small overshoot, which is still faster than the critical damping, may be acceptable. Note that percent overshoot PO and peak time *tp* are meaningless for the overdamped or critically damped cases. In addition to adjusting gain *K*, we may need to augment the system with some type of compensator if the specifications on overshoot and the speed of response are too stringent. - -### RAMP INPUT - -If the anti-aircraft gun in Fig. 4.35a is tracking an enemy plane moving with a uniform velocity, the gun-position angle must increase linearly with *t*. Hence, the input in this case is a ramp; that is, θ*i*(*t*) = *tu*(*t*). Let us find the response of the system to this input when *K* = 80. In this case, *i*(*s*) = 1/*s*2, and - -$$ -\Theta_o(s) = \frac{80}{s^2(s^2 + 8s + 80)} = -\frac{0.1}{s} + \frac{1}{s^2} + \frac{0.1(s - 2)}{s^2 + 8s + 80} -$$ - -Use of Table 4.1 yields - -$$ -\theta_o(t) = \left[ -0.1 + t + \frac{1}{8}e^{-8t}\cos\left(8t + 36.87^\circ\right) \right] u(t) -$$ - -This response, sketched in Fig. 4.35d, shows that there is a steady-state error *er* = 0.1 radian. In many cases such a small steady-state error may be tolerable. If, however, a zero steady-state error to a ramp input is required, this system in its present form is unsatisfactory. We must add some form of compensator to the system. - -### **EXAMPLE 4.26 Step and Ramp Responses of Feedback Systems Using MATLAB** - -Using the feedback system of Fig. 4.18d with *G*(*s*) = *K*/(*s*(*s* + 8)) and *H*(*s*) = 1, determine the step response for each of the following cases: **(a)** *K* = 7, **(b)** *K* = 16, and **(c)** *K* = 80. Additionally, find the unit ramp response when **(d)** *K* = 80. - -Example 4.21 computes the transfer functions of these feedback systems in a simple way. In this example, the conv command is used to demonstrate polynomial multiplication of the two denominator factors of *G*(*s*). Step responses are computed by using the step command. **(a–c)** - ->> H = tf(1,1); K = 7; G = tf([K],conv([1 0],[1 8])); Ha = feedback(G,H); - -&gt;> H = tf(1,1); K = 16; G = tf([K],conv([1 0],[1 8])); Hb = feedback(G,H); - -&gt;> H = tf(1,1); K = 80; G = tf([K],conv([1 0],[1 8])); Hc = feedback(G,H); - -&gt;> clf; step(Ha,'k-',Hb,'k--',Hc,'k-.'); - -&gt;> legend('K = 7','K = 16','K = 80','Location','best'); - -**Figure 4.36** Step responses for Ex. 4.26. - -**(d)** The unit ramp response is equivalent to the integral of the unit step response. We can obtain the ramp response by taking the step response of the system in cascade with an integrator. To help highlight waveform detail, we compute the ramp response over the short time interval of 0 ≤ *t* ≤ 1.5. - ->> t = 0:.001:1.5; Hd = series(Hc,tf([1],[1 0])); >> step(Hd,'k-',t); title('Unit Ramp Response'); - -### DESIGN SPECIFICATIONS - -Now the reader has some idea of the various specifications a control system might require. Generally, a control system is designed to meet given transient specifications, steady-state error specifications, and sensitivity specifications. Transient specifications include overshoot, rise time, and settling time of the response to step input. The steady-state error is the difference between - -### 412 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -the desired response and the actual response to a test input in steady state. The system should also satisfy a specified sensitivity specifications to some system parameter variations, or to certain disturbances. Above all, the system must remain stable under operating conditions. Discussion of design procedures used to realize given specifications is beyond the scope of this book. - -## **4.8 FREQUENCY [RESPONSE OF AN](#page-10-0) LTIC SYSTEM** - -Filtering is an important area of signal processing. Filtering characteristics of a system are indicated by its response to sinusoids of various frequencies varying from 0 to ∞. Such characteristics are called the frequency response of the system. In this section, we shall find the frequency response of LTIC systems. - -In Sec. 2.4-4 we showed that an LTIC system response to an everlasting exponential input *x*(*t*) = *est* is also an everlasting exponential *H*(*s*)*est*. As before, we use an arrow directed from the input to the output to represent an input–output pair: - -$$ -e^{st} \Longrightarrow H(s)e^{st} \tag{4.40} -$$ - -Setting *s* = *j*ω in this relationship yields - -$$ -e^{j\omega t} \Longrightarrow H(j\omega)e^{j\omega t} \tag{4.41} -$$ - -Noting that cosω*t* is the real part of *ej*ω*t* , use of Eq. (2.31) yields - -$$ -\cos \omega t \Longrightarrow \text{Re}[H(j\omega)e^{j\omega t}] \tag{4.42} -$$ - -We can express *H*(*j*ω) in the polar form as - -$$ -H(j\omega) = |H(j\omega)|e^{j\angle H(j\omega)} -$$ - -With this result, Eq. (4.42) becomes - -$$ -\cos \omega t \Longrightarrow |H(j\omega)| \cos [\omega t + \angle H(j\omega)] -$$ - -In other words, the system response *y*(*t*) to a sinusoidal input cosω*t* is given by - -$$ -y(t) = |H(j\omega)| \cos[\omega t + \angle H(j\omega)] -$$ - -Using a similar argument, we can show that the system response to a sinusoid cos(ω*t* +θ ) is - -$$ -y(t) = |H(j\omega)|\cos[\omega t + \theta + \angle H(j\omega)]\tag{4.43} -$$ - -This result is valid only for BIBO-stable systems. The frequency response is meaningless for BIBO-unstable systems. This follows from the fact that the frequency response in Eq. (4.41) is obtained by setting *s* = *j*ω in Eq. (4.40). But, as shown in Sec. 2.4-4 [Eqs. (2.38) and (2.39)], Eq. (4.40) applies only for the values of *s* for which *H*(*s*) exists. For BIBO-unstable systems, the ROC for *H*(*s*) does not include the ω axis where *s* = *j*ω [see Eq. (4.10)]. This means that *H*(*s*) when *s* = *j*ω is meaningless for BIBO-unstable systems.† - -Equation (4.43) shows that for a sinusoidal input of radian frequency ω, the system response is also a sinusoid of the same frequency ω. *The amplitude of the output sinusoid is* |*H*(*j*ω)| *times the input amplitude, and the phase of the output sinusoid is shifted by H*(*j*ω) *with respect to the input phase* (see later Fig. 4.38 in Ex. 4.27). For instance, a certain system with |*H*(*j*10)| = 3 and *H*(*j*10) = −30◦ amplifies a sinusoid of frequency ω = 10 by a factor of 3 and delays its phase by 30◦. The system response to an input 5cos(10*t* + 50◦) is 3 × 5 cos(10*t* + 50◦ − 30◦) = 15 cos(10*t* +20◦). - -Clearly |*H*(*j*ω)| is the amplitude *gain* of the system, and a plot of |*H*(*j*ω)| versus ω shows the amplitude gain as a function of frequency ω. We shall call |*H*(*j*ω)| the *amplitude response*. It also goes under the name *magnitude response*. ‡ Similarly, *H*(*j*ω) is the *phase response*, and a plot of *H*(*j*ω) versus ω shows how the system modifies or changes the phase of the input sinusoid. Plots of the magnitude response |*H*(*j*ω)| and phase response *H*(*j*ω) show at a glance how a system responds to sinusoids of various frequencies. Observe that *H*(*j*ω) has the information of |*H*(*j*ω)| and *H*(*j*ω) and is therefore termed the *frequency response* of the system. Clearly, the frequency response of a system represents its filtering characteristics. - -### **EXAMPLE 4.27 Frequency Response** - -Find the frequency response (amplitude and phase responses) of a system whose transfer function is - -$$ -H(s) = \frac{s+0.1}{s+5} -$$ - -Also, find the system response *y*(*t*) if the input *x*(*t*) is - -**(a)** cos 2*t* - -**(b)** cos(10*t* −50◦) - -In this case, - -$$ -H(j\omega) = \frac{j\omega + 0.1}{j\omega + 5} -$$ - -† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cosω0*t* or *eat* cosω0*t* (*a* > 0). Hence, the response of such a system to a sinusoid cosω*t* will contain not just the sinusoid of frequency ω, but also nondecaying natural modes, rendering the concept of frequency response meaningless. - -‡ Strictly speaking, |*H*(ω)| is magnitude response. There is a fine distinction between amplitude and magnitude. Amplitude *A* can be positive and negative. In contrast, the magnitude |*A*| is always nonnegative. We refrain from relying on this useful distinction between amplitude and magnitude in the interest of avoiding proliferation of essentially similar entities. This is also why we shall use the "amplitude" (instead of "magnitude") spectrum for |*H*(ω)|. - -### 414 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Therefore, - -$$ -|H(j\omega)| = \frac{\sqrt{\omega^2 + 0.01}}{\sqrt{\omega^2 + 25}} \quad \text{and} \quad \angle H(j\omega) = \tan^{-1}\left(\frac{\omega}{0.1}\right) - \tan^{-1}\left(\frac{\omega}{5}\right) -$$ - -Both the amplitude and the phase response are depicted in Fig. 4.38a as functions of ω. These plots furnish the complete information about the frequency response of the system to sinusoidal inputs. - -**(a)** For the input *x*(*t*) = cos 2*t*, ω = 2, and - -$$ -|H(j2)| = \frac{\sqrt{(2)^2 + 0.01}}{\sqrt{(2)^2 + 25}} = 0.372 -$$ - -\n -$$ -\angle H(j2) = \tan^{-1}\left(\frac{2}{0.1}\right) - \tan^{-1}\left(\frac{2}{5}\right) = 87.1^{\circ} - 21.8^{\circ} = 65.3^{\circ} -$$ - -**Figure 4.38** Responses for the system of Ex. 4.27. - -We also could have read these values directly from the frequency response plots in Fig. 4.38a corresponding to ω = 2. This result means that for a sinusoidal input with frequency ω = 2, the amplitude gain of the system is 0.372, and the phase shift is 65.3◦. In other words, the output amplitude is 0.372 times the input amplitude, and the phase of the output is shifted with respect to that of the input by 65.3◦. Therefore, the system response to the input cos 2*t* is - -$$ -y(t) = 0.372 \cos(2t + 65.3^{\circ}) -$$ - -The input cos 2*t* and the corresponding system response 0.372cos(2*t* + 65.3◦) are illustrated in Fig. 4.38b. - -**(b)** For the input cos(10*t* − 50◦), instead of computing the values |*H*(*j*ω)| and *H*(*j*ω) as in part (a), we shall read them directly from the frequency response plots in Fig. 4.38a corresponding to ω = 10. These are - -$$ -|H(j10)| = 0.894 -$$ - and $\angle H(j10) = 26^{\circ}$ - -Therefore, for a sinusoidal input of frequency ω = 10, the output sinusoid amplitude is 0.894 times the input amplitude, and the output sinusoid is shifted with respect to the input sinusoid by 26◦. Therefore, the system response *y*(*t*) to an input cos(10*t* −50◦) is - -$$ -y(t) = 0.894 \cos (10t - 50^\circ + 26^\circ) = 0.894 \cos (10t - 24^\circ) -$$ - -If the input were sin(10*t* − 50◦), the response would be 0.894 sin(10*t* − 50◦ + 26◦) = 0.894 sin(10*t* −24◦). - -The frequency response plots in Fig. 4.38a show that the system has highpass filtering characteristics; it responds well to sinusoids of higher frequencies (ω well above 5), and suppresses sinusoids of lower frequencies (ω well below 5). - -### PLOTTING FREQUENCY RESPONSE WITH MATLAB - -It is simple to use MATLAB to create magnitude and phase response plots. Here, we consider two methods. In the first method, we use an anonymous function to define the transfer function *H*(*s*) and then obtain the frequency response plots by substituting *j*ω for *s*. - -``` ->> H = @(s) (s+0.1)./(s+5); omega = 0:.01:20; ->> subplot(1,2,1); plot(omega,abs(H(1j*omega)),'k-'); ->> subplot(1,2,2); plot(omega,angle(H(1j*omega))*180/pi,'k-'); -``` - -In the second method, we define vectors that contain the numerator and denominator coefficients of *H*(*s*) and then use the freqs command to compute frequency response. - -``` ->> B = [1 0.1]; A = [1 5]; H = freqs(B,A,omega); omega = 0:.01:20; ->> subplot(1,2,1); plot(omega,abs(H),'k-'); ->> subplot(1,2,2); plot(omega,angle(H)*180/pi,'k-'); -``` - -Both approaches generate plots that match Fig. 4.38a. - -### **EXAMPLE 4.28 Frequency Responses of Delay, Differentiator, and Integrator Systems** - -Find and sketch the frequency responses (magnitude and phase) for **(a)** an ideal delay of *T* seconds, **(b)** an ideal differentiator, and **(c)** an ideal integrator. - -**(a) Ideal delay of** *T* **seconds.** The transfer function of an ideal delay is [see Eq. (4.30)] - -$$ -H(s) = e^{-sT} -$$ - -Therefore, - -$$ -H(j\omega) = e^{-j\omega T} -$$ - -Consequently, - -$$ -|H(j\omega)| = 1 \quad \text{and} \quad \angle H(j\omega) = -\omega T -$$ - -These amplitude and phase responses are shown in Fig. 4.39a. The amplitude response is constant (unity) for all frequencies. The phase shift increases linearly with frequency with a slope of −*T*. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal delay of *T* seconds, the output is cosω(*t* − *T*). The output sinusoid amplitude is the same as that of the input for all values of ω. Therefore, the amplitude response (gain) is unity for all frequencies. Moreover, the output cosω(*t* − *T*) = cos(ω*t* − ω*T*) has a phase shift −ω*T* with respect to the input cosω*t*. Therefore, the phase response is linearly proportional to the frequency ω with a slope −*T*. - -**(b) An ideal differentiator.** The transfer function of an ideal differentiator is [see Eq. (4.31)] - -*H*(*s*) = *s* - -Therefore, - -*H*(*j*ω) = *j*ω = ω*ej*π/2 - -Consequently, - -$$ -|H(j\omega)| = \omega -$$ - and $\angle H(j\omega) = \frac{\pi}{2}$ - -These amplitude and phase responses are depicted in Fig. 4.39b. The amplitude response increases linearly with frequency, and phase response is constant (π/2) for all frequencies. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal differentiator, the output is −ωsin ω*t* = ωcos[ω*t* + (π/2)]. Therefore, the output sinusoid amplitude is ω times the input amplitude; that is, the amplitude response (gain) increases linearly with frequency ω. Moreover, the output sinusoid undergoes a phase shift π/2 with respect to the input cosω*t*. Therefore, the phase response is constant (π/2) with frequency. - -**Figure 4.39** Frequency response of an ideal **(a)** delay, **(b)** differentiator, and **(c)** integrator. - -In an ideal differentiator, the amplitude response (gain) is proportional to frequency [|*H*(*j*ω)| = ω] so that the higher-frequency components are enhanced (see Fig. 4.39b). All practical signals are contaminated with noise, which, by its nature, is a broadband (rapidly varying) signal containing components of very high frequencies. A differentiator can increase the noise disproportionately to the point of drowning out the desired signal. This is why ideal differentiators are avoided in practice. - -**(c) An ideal integrator.** The transfer function of an ideal integrator is [see Eq. (4.32)] - -$$ -H(s) = \frac{1}{s} -$$ - -Therefore, - -$$ -H(j\omega) = \frac{1}{j\omega} = \frac{-j}{\omega} = \frac{1}{\omega}e^{-j\pi/2} -$$ - -Consequently, - -$$ -|H(j\omega)| = \frac{1}{\omega} \quad \text{and} \quad \angle H(j\omega) = -\frac{\pi}{2} -$$ - -These amplitude and phase responses are illustrated in Fig. 4.39c. The amplitude response is inversely proportional to frequency, and the phase shift is constant (−π/2) with frequency. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal integrator, the output is (1/ω)sin ω*t* = (1/ω) cos[ω*t* −(π/2)]. Therefore, the amplitude response is inversely proportional to ω, and the phase response is constant (−π/2) - -### 418 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -with frequency.† Because its gain is 1/ω, the ideal integrator suppresses higher-frequency components but enhances lower-frequency components with ω < 1. Consequently, noise signals (if they do not contain an appreciable amount of very-low-frequency components) are suppressed (smoothed out) by an integrator. - -### **DR ILL 4.15 Sinusoidal Response of an LTIC System** - -Find the response of an LTIC system specified by - -$$ -\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt} + 5x(t) -$$ - -if the input is a sinusoid 20 sin(3*t* +35◦). - -### **ANSWER** - -10.23 sin(3*t* −61.91◦) - -### **[4.8-1 Steady-State Response to Causal Sinusoidal Inputs](#page-10-0)** - -So far we have discussed the LTIC system response to everlasting sinusoidal inputs (starting at *t* = −∞). In practice, we are more interested in causal sinusoidal inputs (sinusoids starting at *t* = 0). Consider the input *ej*ω*t u*(*t*), which starts at *t* = 0 rather than at *t* = −∞. In this case *X*(*s*) = 1/(*s* + *j*ω). Moreover, according to Eq. (4.27), *H*(*s*) = *P*(*s*)/*Q*(*s*), where *Q*(*s*) is the - -† A puzzling aspect of this result is that in deriving the transfer function of the integrator in Eq. (4.32), we have assumed that the input starts at *t* = 0. In contrast, in deriving its frequency response, we assume that the everlasting exponential input *ej*ω*t* starts at *t* = −∞. There appears to be a fundamental contradiction between the everlasting input, which starts at *t* = −∞, and the integrator, which opens its gates only at *t* = 0. Of what use is everlasting input, since the integrator starts integrating at *t* = 0? The answer is that the integrator gates are always open, and integration begins whenever the input starts. We restricted the input to start at *t* = 0 in deriving Eq. (4.32) because we were finding the transfer function using the unilateral transform, where the inputs begin at *t* = 0. So the integrator starting to integrate at *t* = 0 is restricted because of the limitations of the unilateral transform method, not because of the limitations of the integrator itself. If we were to find the integrator transfer function using Eq. (2.40), where there is no such restriction on the input, we would still find the transfer function of an integrator as 1/*s*. Similarly, even if we were to use the bilateral Laplace transform, where *t* starts at −∞, we would find the transfer function of an integrator to be 1/*s*. The transfer function of a system is the property of the system and does not depend on the method used to find it. - -characteristic polynomial given by *Q*(*s*) = (*s*−λ1)(*s*−λ2)··· (*s*−λ*N*). † Hence, - -$$ -Y(s) = X(s)H(s) = \frac{P(s)}{(s - \lambda_1)(s - \lambda_2) \cdots (s - \lambda_N)(s - j\omega)} -$$ - -In the partial fraction expansion of the right-hand side, let the coefficients corresponding to the *N* terms (*s* − λ1), (*s* − λ2), ... , (*s* − λ*N*) be *k*1, *k*2, ... , *kN*. The coefficient corresponding to the last term (*s*−*j*ω) is *P*(*s*)/*Q*(*s*)|*s*=*j*ω = *H*(*j*ω). Hence, - -$$ -Y(s) = \sum_{i=1}^{n} \frac{k_i}{s - \lambda_i} + \frac{H(j\omega)}{s - j\omega} -$$ - -and - -$$ -y(t) = \underbrace{\sum_{i=1}^{n} k_i e^{\lambda_i t} u(t)}_{\text{transient component } y_{\text{tr}}(t)} + \underbrace{H(j\omega)e^{j\omega t} u(t)}_{\text{steady-state component } y_{\text{ss}}(t)} -$$ - -For an asymptotically stable system, the characteristic mode terms *e*λ*it* decay with time, and, therefore, constitute the so-called *transient* component of the response. The last term *H*(*j*ω)*ej*ω*t* persists forever, and is the *steady-state* component of the response given by - -$$ -y_{ss}(t) = H(j\omega)e^{j\omega t}u(t) -$$ - -This result also explains why an everlasting exponential input *ej*ω*t* results in the total response *H*(*j*ω)*ej*ω*t* for BIBO systems. Because the input started at *t* = −∞, at any finite time the decaying transient component has long vanished, leaving only the steady-state component. Hence, the total response appears to be *H*(*j*ω)*ej*ω*t* . - -From the argument that led to Eq. (4.43), it follows that for a causal sinusoidal input cosω*t*, the steady-state response *yss*(*t*) is given by - -$$ -y_{ss}(t) = |H(j\omega)| \cos[\omega t + \angle H(j\omega)]u(t) -$$ - -In summary, |*H*(*j*ω)| cos[ω*t* + *H*(*j*ω)] is the total response to everlasting sinusoid cosω*t*. In contrast, it is the steady-state response to the same input applied at *t* = 0. - -## **[4.9 BODE](#page-10-0) PLOTS** - -Sketching frequency response plots (|*H*(*j*ω)| and *H*(*j*ω) versus ω) is considerably facilitated by the use of logarithmic scales. The amplitude and phase response plots as a function of ω on a logarithmic scale are known as *Bode plots*. By using the asymptotic behavior of the amplitude and the phase responses, we can sketch these plots with remarkable ease, even for higher-order transfer functions. - -† For simplicity, we have assumed nonrepeating characteristic roots. The procedure is readily modified for repeated roots, and the same conclusion results. - -#### 420 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Let us consider a system with the transfer function - -$$ -H(s) = \frac{K(s+a_1)(s+a_2)}{s(s+b_1)(s^2+b_2s+b_3)} -$$ -\n(4.44) - -where the second-order factor (*s*2 + *b*2*s* + *b*3) is assumed to have complex conjugate roots.† We shall rearrange Eq. (4.44) in the form - -$$ -H(s) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(\frac{s}{a_1} + 1\right)\left(\frac{s}{a_2} + 1\right)}{s\left(\frac{s}{b_1} + 1\right)\left(\frac{s^2}{b_3} + \frac{b_2}{b_3} s + 1\right)} -$$ - -and - -$$ -H(j\omega) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(1 + \frac{j\omega}{a_1}\right)\left(1 + \frac{j\omega}{a_2}\right)}{j\omega\left(1 + \frac{j\omega}{b_1}\right)\left[1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right]} -$$ - -This equation shows that *H*(*j*ω) is a complex function of ω. The amplitude response |*H*(*j*ω)| and the phase response *H*(*j*ω) are given by - -$$ -|H(j\omega)| = \left| \frac{Ka_1a_2}{b_1b_3} \right| \frac{\left| 1 + \frac{j\omega}{a_1} \right| \left| 1 + \frac{j\omega}{a_2} \right|}{|j\omega| \left| 1 + \frac{j\omega}{b_1} \right| \left| 1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3} \right|} -$$ -(4.45) - -and - -$$ -\angle H(j\omega) = \angle \left(\frac{Ka_1a_2}{b_1b_3}\right) + \angle \left(1 + \frac{j\omega}{a_1}\right) + \angle \left(1 + \frac{j\omega}{a_2}\right) -$$ -$$ -- \angle j\omega - \angle \left(1 + \frac{j\omega}{b_1}\right) - \angle \left[1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right] -$$ -(4.46) - -From Eq. (4.46) we see that the phase function consists of the addition of terms of four kinds: (i) the phase of a constant, (ii) the phase of *j*ω, which is 90◦ for all values of ω, (iii) the phase for the first-order term of the form 1+*j*ω/*a*, and (iv) the phase of the second-order term - -$$ -\[1+\frac{jb_2\omega}{b_3}+\frac{(j\omega)^2}{b_3}\] -$$ - -We can plot these basic phase functions for ω in the range 0 to ∞ and then, using these plots, we can construct the phase function of any transfer function by properly adding these basic responses. Note that if a particular term is in the numerator, its phase is added, but if the term is in the - -† Coefficients *a*1, *a*2 and *b*1, *b*2, *b*3 used in this section are not to be confused with those used in the representation of *N*th-order LTIC system equations given earlier [Eqs. (2.1) or (4.26)]. - -denominator, its phase is subtracted. This makes it easy to plot the phase function *H*(*j*ω) as a function of ω. Computation of |*H*(*j*ω)|, unlike that of the phase function, however, involves the multiplication and division of various terms. This is a formidable task, especially when we have to plot this function for the entire range of ω (0 to ∞). - -We know that a log operation converts multiplication and division to addition and subtraction. So, instead of plotting |*H*(*j*ω)|, why not plot log |*H*(*j*ω)| to simplify our task? We can take advantage of the fact that logarithmic units are desirable in several applications, where the variables considered have a very large range of variation. This is particularly true in frequency response plots, where we may have to plot frequency response over a range from a very low frequency, near 0, to a very high frequency, in the range of 1010 or higher. A plot on a linear scale of frequencies for such a large range will bury much of the useful information at lower frequencies. Also, the amplitude response may have a very large dynamic range from a low of 10−6 to a high of 106 . A linear plot would be unsuitable for such a situation. Therefore, logarithmic plots not only simplify our task of plotting, but, fortunately, they are also desirable in this situation. - -There is another important reason for using logarithmic scale. The Weber–Fechner law (first observed by Weber in 1834) states that human senses (sight, touch, hearing, etc.) generally respond in a logarithmic way. For instance, when we hear sound at two different power levels, we judge one sound twice as loud when the ratio of the two sound powers is 10. Human senses respond to equal ratios of power, not equal increments in power [10]. This is clearly a logarithmic response.† - -The logarithmic unit is the *decibel* and is equal to 20 times the logarithm of the quantity (log to the base 10). Therefore, 20log10 |*H*(*j*ω)| is simply the log amplitude in decibels (dB).‡ Thus, instead of plotting |*H*(*j*ω)|, we shall plot 20log10 |*H*(*j*ω)| as a function of ω. These plots (log amplitude and phase) are called *Bode plots*. For the transfer function in Eq. (4.45), the *log amplitude* is - -$$ -20\log|H(j\omega)| = 20\log\left|\frac{Ka_1a_2}{b_1b_3}\right| + 20\log\left|1 + \frac{j\omega}{a_1}\right| + 20\log\left|1 + \frac{j\omega}{a_2}\right| - 20\log|j\omega| -$$ -$$ --20\log\left|1 + \frac{j\omega}{b_1}\right| - 20\log\left|1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right| \tag{4.47} -$$ - -The term 20log(*Ka*1*a*2/*b*1*b*3) is a constant. We observe that the log amplitude is a sum of four basic terms corresponding to a constant, a pole or zero at the origin (20log|*j*ω|), a first-order pole or zero (20log|1+*j*ω/*a*|), and complex-conjugate poles or zeros (20log|1+*j*ω*b*2/*b*3 +(*j*ω)2/*b*3|). - -† Observe that the frequencies of musical notes are spaced logarithmically (not linearly). The octave is a ratio of 2. The frequencies of the same note in the successive octaves have a ratio of 2. On the Western musical scale, there are 12 distinct notes in each octave. The frequency of each note is about 6% higher than the frequency of the preceding note. Thus, the successive notes are separated not by some constant frequency, but by constant ratio of 1.06. - -‡ Originally, the unit *bel* (after the inventor of telephone, Alexander Graham Bell) was introduced to represent power ratio as log10 *P*2/*P*1 bels. A tenth of this unit is a decibel, as in 10 log10 *P*2/*P*1 decibels. Since the power ratio of two signals is proportional to the amplitude ratio squared, or |*H*(*j*ω)| 2, we have 10 log10 *P*2/*P*1 = 10 log10 |*H*(*j*ω)| 2 = 20 log10 |*H*(*j*ω)| dB. - -### 422 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -We can sketch these four basic terms as functions of ω and use them to construct the log-amplitude plot of any desired transfer function. Let us discuss each of the terms. - -## **[4.9-1 Constant](#page-10-0)** *Ka***1***a***2***/b***1***b***3** - -The log amplitude of the constant *Ka*1*a*2/*b*1*b*2 term is also a constant, 20log|*Ka*1*a*2/*b*1*b*3|. The phase contribution from this term is zero for positive value and π for negative value of the constant (complex constants can have different phases). - -### **[4.9-2 Pole \(or Zero\) at the Origin](#page-10-0)** - -LOG MAGNITUDE - -A pole at the origin gives rise to the term −20log|*j*ω|, which can be expressed as - -$$ --20\log|j\omega| = -20\log\omega -$$ - -This function can be plotted as a function of ω. However, we can effect further simplification by using the logarithmic scale for the variable ω itself. Let us define a new variable *u* such that - -*u* = logω - -Hence, - -$$ --20\log\omega = -20u -$$ - -The log-amplitude function −20*u* is plotted as a function of *u* in Fig. 4.40a. This is a straight line with a slope of −20. It crosses the *u* axis at *u* = 0. The ω-scale (*u* = logω) also appears in Fig. 4.40a. Semilog graphs can be conveniently used for plotting, and we can directly plot ω on semilog paper. A ratio of 10 is a *decade*, and a ratio of 2 is known as an *octave*. Furthermore, a decade along the ω scale is equivalent to 1 unit along the *u* scale. We can also show that a ratio of 2 (an octave) along the ω scale equals to 0.3010 (which is log10 2) along the *u* scale.† - -$$ -u_2 - u_1 = \log_{10} \omega_2 - \log_{10} \omega_1 = \log_{10} (\omega_2/\omega_1) -$$ - -Thus, if - -(ω2/ω1) = 10 (which is a decade) - -then - -*u*2 −*u*1 = log10 10 = 1 - -and if - -(ω2/ω1) = 2 (which is an octave) - -then - -$$ -u_2 - u_1 = \log_{10} 2 = 0.3010 -$$ - -† This point can be shown as follows. Let ω1 and ω2 along the ω scale correspond to *u*1 and *u*2 along the *u* scale so that logω1 = *u*1 and logω2 = *u*2. Then - -**Figure 4.40 (a)** Amplitude and **(b)** phase responses of a pole or a zero at the origin. - -Note that equal increments in *u* are equivalent to equal ratios on the ω scale. Thus, 1 unit along the *u* scale is the same as one decade along the ω scale. This means that the amplitude plot has a slope of −20 dB/decade or −20(0.3010) = −6.02 dB/octave (commonly stated as −6 dB/octave). Moreover, the amplitude plot crosses the ω axis at ω = 1, since *u* = log10ω = 0 when ω = 1. - -For the case of a zero at the origin, the log-amplitude term is 20 log ω. This is a straight line passing through ω = 1 and having a slope of 20 dB/decade (or 6 dB/octave). This plot is a mirror image about the ω axis of the plot for a pole at the origin and is shown dashed in Fig. 4.40a. - -### PHASE - -The phase function corresponding to the pole at the origin is − *j*ω [see Eq. (4.46)]. Thus, - -$$ -\angle H(j\omega) = -\angle j\omega = -90^{\circ} -$$ - -The phase is constant (−90◦) for all values of ω, as depicted in Fig. 4.40b. For a zero at the origin, the phase is *j*ω = 90◦. This is a mirror image of the phase plot for a pole at the origin and is shown dashed in Fig. 4.40b. - -### **[4.9-3 First-Order Pole \(or Zero\)](#page-10-0)** - -### THE LOG MAGNITUDE - -The log amplitude of a first-order pole at −*a* is −20log|1+*j*ω/*a*|. Let us investigate the asymptotic behavior of this function for extreme values of ω (ω *a* and ω *a*). - -**(a)** For ω *a*, - -$$ --20\log\left|1+\frac{j\omega}{a}\right|\approx-20\log 1=0 -$$ - -Hence, the log-amplitude function → 0 asymptotically for ω *a* (Fig. 4.41a). - -**(a)** For the other extreme case, where ω *a*, - -$$ --20\log\left|1+\frac{j\omega}{a}\right| \approx -20\log\left(\frac{\omega}{a}\right) = -20\log\omega + 20\log a = -20u + 20\log a -$$ - -This represents a straight line (when plotted as a function of *u*, the log of ω) with a slope of −20 dB/decade (or −6 dB/octave). When ω = *a*, the log amplitude is zero. Hence, this line crosses the ω axis at ω = *a*, as illustrated in Fig. 4.41a. Note that the asymptotes in (a) and (b) meet at ω = *a*. - -The exact log amplitude for this pole is - -$$ --20\log\left|1+\frac{j\omega}{a}\right| = -20\log\left(1+\frac{\omega^2}{a^2}\right)^{1/2} = -10\log\left(1+\frac{\omega^2}{a^2}\right) -$$ - -This exact log magnitude function also appears in Fig. 4.41a. Observe that the actual and the asymptotic plots are very close. A maximum error of 3 dB occurs at ω = *a*. This frequency is known as the *corner frequency* or *break frequency*. The error everywhere else is less than 3 dB. A plot of the error as a function of ω is shown in Fig. 4.42a. This figure shows that the error at 1 octave above or below the corner frequency is 1 dB and the error at 2 octaves above or below the corner frequency is 0.3 dB. The actual plot can be obtained by adding the error to the asymptotic plot. - -The amplitude response for a zero at −*a* (shown dotted in Fig. 4.41a) is identical to that of the pole at −*a* with a sign change and therefore is the mirror image (about the 0 dB line) of the amplitude plot for a pole at −*a*. - -### PHASE - -The phase for the first-order pole at −*a* is - -$$ -\angle H(j\omega) = -\angle \left(1 + \frac{j\omega}{a}\right) = -\tan^{-1}\left(\frac{\omega}{a}\right) -$$ - -Let us investigate the asymptotic behavior of this function. For ω *a*, - -$$ --\tan^{-1}\left(\frac{\omega}{a}\right) \approx 0 -$$ - -and, for ω *a*, - -$$ --\tan^{-1}\left(\frac{\omega}{a}\right) \approx -90^{\circ} -$$ - -**Figure 4.41 (a)** Amplitude and **(b)** phase responses of a first-order pole or zero at *s* = −*a*. - -The actual plot along with the asymptotes is depicted in Fig. 4.41b. In this case, we use a three-line segment asymptotic plot for greater accuracy. The asymptotes are a phase angle of 0◦ for ω ≤ *a*/10, a phase angle of −90◦ for ω ≥ 10*a*, and a straight line with a slope −45◦/decade connecting these two asymptotes (from ω = *a*/10 to 10*a*) crossing the ω axis at ω = *a*/10. It can be seen from Fig. 4.41b that the asymptotes are very close to the curve and the maximum error is 5.7◦. Figure 4.42b plots the error as a function of ω; the actual plot can be obtained by adding the error to the asymptotic plot. - -**Figure 4.42** Errors in asymptotic approximation of a first-order pole at *s* = −*a*. - -The phase for a zero at −*a* (shown dotted in Fig. 4.41b) is identical to that of the pole at −*a* with a sign change, and therefore is the mirror image (about the 0◦ line) of the phase plot for a pole at −*a*. - -### **[4.9-4 Second-Order Pole \(or Zero\)](#page-10-0)** - -Let us consider the second-order pole in Eq. (4.44). The denominator term is *s*2 + *b*2*s* + *b*3. We shall introduce the often-used standard form *s*2 + 2ζω*ns* + ω2 *n* instead of *s*2 + *b*2*s* + *b*3. With this form, the log amplitude function for the second-order term in Eq. (4.47) becomes - -$$ --20\log\left|1+2j\zeta\frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right| -$$ - -and the phase function is - -$$ --\angle \left[1+2j\zeta \frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right] -$$ -\n(4.48) - -### THE LOG MAGNITUDE - -The log amplitude is given by - -log amplitude = -$$ --20 \log \left| 1 + 2j\zeta \left( \frac{\omega}{\omega_n} \right) + \left( \frac{j\omega}{\omega_n} \right)^2 \right| -$$ - (4.49) - -For ω ω*n*, the log amplitude becomes - -$$ -log amplitude \approx -20 log 1 = 0 -$$ - -For ω ω*n*, the log amplitude is - -$$ -\log amplitude \approx -20 \log \left| \left( -\frac{\omega}{\omega_n} \right)^2 \right| = -40 \log \left( \frac{\omega}{\omega_n} \right) -$$ - -= -40 log $\omega$ - 40 log $\omega_n$ = -40u - 40 log $\omega_n$ (4.50) - -The two asymptotes are zero for ω<ω*n* and −40*u*−40logω*n* for ω>ω*n*. The second asymptote is a straight line with a slope of −40 dB/decade (or −12 dB/octave) when plotted against the log ω scale. It begins at ω = ω*n* [see Eq. (4.50)]. The asymptotes are depicted in Fig. 4.43a. The exact log amplitude is given by [see Eq. (4.49)] - -log amplitude = -$$ --20 \log \left\{ \left[ 1 - \left( \frac{\omega}{\omega_n} \right)^2 \right]^2 + 4 \zeta^2 \left( \frac{\omega}{\omega_n} \right)^2 \right\}^{1/2} -$$ - (4.51) - -The log amplitude in this case involves a parameter ζ , resulting in a different plot for each value of ζ . For complex-conjugate poles,† ζ < 1. Hence, we must sketch a family of curves for a number of values of ζ in the range 0 to 1. This is illustrated in Fig. 4.43a. The error between the actual plot and the asymptotes is shown in Fig. 4.44. The actual plot can be obtained by adding the error to the asymptotic plot. - -For second-order zeros (complex-conjugate zeros), the plots are mirror images (about the 0 dB line) of the plots depicted in Fig. 4.43a. Note the resonance phenomenon of the complex-conjugate poles. This phenomenon is barely noticeable for ζ > 0.707 but becomes pronounced as ζ → 0. - -### PHASE - -The phase function for second-order poles, as apparent in Eq. (4.48), is - -$$ -\angle H(j\omega) = -\tan^{-1}\left[\frac{2\zeta\left(\frac{\omega}{\omega_n}\right)}{1 - \left(\frac{\omega}{\omega_n}\right)^2}\right] -$$ -(4.52) - -For ω ω*n*, - -$$ -\angle H(j\omega) \approx 0 -$$ - -† For ζ ≥ 1, the two poles in the second-order factor are no longer complex but real, and each of these two real poles can be dealt with as a separate first-order factor. - -**Figure 4.43** Amplitude and phase response of a second-order pole. - -For ω ω*n*, - -$$ -\angle H(j\omega) \simeq -180^\circ -$$ - -Hence, the phase → −180◦ as ω → ∞. As in the case of amplitude, we also have a family of phase plots for various values of ζ , as illustrated in Fig. 4.43b. A convenient asymptote for the phase of complex-conjugate poles is a step function that is 0◦ for ω<ω*n* and −180◦ for ω>ω*n*. - -**Figure 4.44** Errors in the asymptotic approximation of a second-order pole. - -Error plots for such an asymptote are shown in Fig. 4.44 for various values of ζ . The exact phase is the asymptotic value plus the error. - -For complex-conjugate zeros, the amplitude and phase plots are mirror images of those for complex conjugate-poles. - -We shall demonstrate the application of these techniques with two examples. - -### **EXAMPLE 4.29 Bode Plots for Second-Order Transfer Function with Real Roots** - -Sketch Bode plots for the transfer function - -$$ -H(s) = \frac{20s(s+100)}{(s+2)(s+10)} -$$ - -### MAGNITUDE PLOT - -First, we write the transfer function in normalized form - -$$ -H(s) = \frac{20 \times 100}{2 \times 10} \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)} = 100 \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)} -$$ - -Here, the constant term is 100; that is, 40 dB (20 log 100 = 40). This term can be added to the plot by simply relabeling the horizontal axis (from which the asymptotes begin) as the 40 dB line (see Fig. 4.45a). Such a step implies shifting the horizontal axis upward by 40 dB. This is precisely what is desired. - -In addition, we have two first-order poles at −2 and −10, one zero at the origin, and one zero at −100. - -**Step 1.** For each of these terms, we draw an asymptotic plot as follows (shown in Fig. 4.45a by dashed lines): - -- **(a)** For the zero at the origin, draw a straight line with a slope of 20 dB/decade passing through ω = 1. -- **(b)** For the pole at −2, draw a straight line with a slope of −20 dB/decade (for ω > 2) beginning at the corner frequency ω = 2. -- **(c)** For the pole at −10, draw a straight line with a slope of −20 dB/decade beginning at the corner frequency ω = 10. -- **(d)** For the zero at −100, draw a straight line with a slope of 20 dB/decade beginning at the corner frequency ω = 100. -- **Step 2.** Add all the asymptotes, as depicted in Fig. 4.45a by solid line segments. -- **Step 3.** Apply the following corrections (see Fig. 4.42a): - - **(a)** The correction at ω = 1 because of the corner frequency at ω = 2 is −1 dB. The correction at ω = 1 because of the corner frequencies at ω = 10 and ω = 100 is quite small (see Fig. 4.42a) and may be ignored. Hence, the net correction at ω = 1 is −1 dB. - -**Figure 4.45 (a)** Amplitude and **(b)** phase responses of the second-order system. - -- **(b)** The correction at ω = 2 because of the corner frequency at ω = 2 is −3 dB, and the correction because of the corner frequency at ω = 10 is −0.17 dB. The correction because of the corner frequency ω = 100 can be safely ignored. Hence the net correction at ω = 2 is −3.17 dB. -- **(c)** The correction at ω = 10 because of the corner frequency at ω = 10 is −3 dB, and the correction because of the corner frequency at ω = 2 is −0.17 dB. The correction because of ω = 100 can be ignored. Hence the net correction at ω = 10 is −3.17 dB. - -- **(d)** The correction at ω = 100 because of the corner frequency at ω = 100 is 3 dB, and the corrections because of the other corner frequencies may be ignored. -- **(e)** In addition to the corrections at corner frequencies, we may consider corrections at intermediate points for more accurate plots. For instance, the corrections at ω = 4 because of corner frequencies at ω = 2 and 10 are −1 and about −0.65, totaling −1.65 dB. In the same way, the corrections at ω = 5 because of corner frequencies at ω = 2 and 10 are −0.65 and −1, totaling −1.65 dB. - -With these corrections, the resulting amplitude plot is illustrated in Fig. 4.45a. - -### PHASE PLOT - -We draw the asymptotes corresponding to each of the four factors: - -- **(a)** The zero at the origin causes a 90◦ phase shift. -- **(b)** The pole at *s* = −2 has an asymptote with a zero value for −∞ <ω< 0.2 and a slope of −45◦/decade beginning at ω = 0.2 and going up to ω = 20. The asymptotic value for ω > 20 is −90◦. -- **(c)** The pole at *s* = −10 has an asymptote with a zero value for −∞ <ω< 1 and a slope of −45◦/decade beginning at ω = 1 and going up to ω = 100. The asymptotic value for ω > 100 is −90◦. -- **(d)** The zero at *s* = −100 has an asymptote with a zero value for −∞ <ω< 10 and a slope of 45◦/decade beginning at ω = 10 and going up to ω = 1000. The asymptotic value for ω > 1000 is 90◦. All the asymptotes are added, as shown in Fig. 4.45b. The appropriate corrections are applied from Fig. 4.42b, and the exact phase plot is depicted in Fig. 4.45b. - -### **EXAMPLE 4.30 Bode Plots for Second-Order Transfer Function with Complex Poles** - -Sketch the amplitude and phase response (Bode plots) for the transfer function - -$$ -H(s) = \frac{10(s+100)}{s^2 + 2s + 100} = 10 \frac{1 + \frac{s}{100}}{1 + \frac{s}{50} + \frac{s^2}{100}} -$$ - -### MAGNITUDE PLOT - -Here, the constant term is 10: that is, 20 dB(20 log 10 = 20). To add this term, we simply label the horizontal axis (from which the asymptotes begin) as the 20 dB line, as before (see Fig. 4.46a). - -**Figure 4.46 (a)** Amplitude and **(b)** phase responses of the second-order system. - -In addition, we have a real zero at *s* = −100 and a pair of complex conjugate poles. When we express the second-order factor in standard form, - -$$ -s^2 + 2s + 100 = s^2 + 2\zeta \omega_n s + \omega_n^2 -$$ - -### 434 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -we have - -$$ -\omega_n = 10 \quad \text{and} \quad \zeta = 0.1 -$$ - -**Step 1.** Draw an asymptote of −40 dB/decade (−12 dB/octave) starting at ω = 10 for the complex conjugate poles, and draw another asymptote of 20 dB/decade starting at ω = 100 for the (real) zero. - -**Step 2.** Add both asymptotes. - -**Step 3.** Apply the correction at ω = 100, where the correction because of the corner frequency ω = 100 is 3 dB. The correction because of the corner frequency ω = 10, as seen from Fig. 4.44a for ζ = 0.1, can be safely ignored. Next, the correction at ω = 10 because of the corner frequency ω = 10 is 13.90 dB (see Fig. 4.44a for ζ = 0.1). The correction because of the real zero at −100 can be safely ignored at ω = 10. We may find corrections at a few more points. The resulting plot is illustrated in Fig. 4.46a. - -### PHASE PLOT - -The asymptote for the complex conjugate poles is a step function with a jump of −180◦ at ω = 10. The asymptote for the zero at *s* = −100 is zero for ω ≤ 10 and is a straight line with a slope of 45◦/decade, starting at ω = 10 and going to ω = 1000. For ω ≥ 1000, the asymptote is 90◦. The two asymptotes add to give the sawtooth shown in Fig. 4.46b. We now apply the corrections from Figs. 4.42b and 4.44b to obtain the exact plot. - -**Figure 4.47** MATLAB-generated Bode plots for Ex. 4.30. - -### BODE PLOTS WITH MATLAB - -Bode plots make it relatively simple to hand-draw straight-line approximations to a system's magnitude and frequency responses. To produce exact Bode plots, we turn to MATLAB and its bode command. - ->> bode(tf([10 1000],[1 2 100]),'k-'); - -The resulting MATLAB plots, shown in Fig. 4.47, match the plots shown in Fig. 4.46. - -**Comment.** These two examples demonstrate that actual frequency response plots are very close to asymptotic plots, which are so easy to construct. Thus, by mere inspection of *H*(*s*) and its poles and zeros, one can rapidly construct a mental image of the frequency response of a system. This is the principal virtue of Bode plots. - -### POLES AND ZEROS IN THE RIGHT HALF-PLANE - -In our discussion so far, we have assumed the poles and zeros of the transfer function to be in the left half-plane. What if some of the poles and/or zeros of *H*(*s*) lie in the RHP? If there is a pole in the RHP, the system is unstable. Such systems are useless for any signal-processing application. For this reason, we shall consider only the case of the RHP zero. The term corresponding to RHP zero at *s* = *a* is (*s*/*a*) −1, and the corresponding frequency response is (*j*ω/*a*) −1. The amplitude response is - -$$ -\left|\frac{j\omega}{a} - 1\right| = \left(\frac{\omega^2}{a^2} + 1\right)^{1/2} -$$ - -This shows that the amplitude response of an RHP zero at *s* = *a* is identical to that of an LHP zero or *s* = −*a*. Therefore, the log amplitude plots remain unchanged whether the zeros are in the LHP or the RHP. However, the phase corresponding to the RHP zero at *s* = *a* is - -$$ -\angle \left(\frac{j\omega}{a} - 1\right) = \angle -\left(1 - \frac{j\omega}{a}\right) = \pi + \tan^{-1}\left(\frac{-\omega}{a}\right) = \pi - \tan^{-1}\left(\frac{\omega}{a}\right) -$$ - -whereas the phase corresponding to the LHP zero at *s* = −*a* is tan−1(ω/*a*). - -The complex-conjugate zeros in the RHP give rise to a term *s*2−2ζω*ns*+ω2 *n*, which is identical to the term *s*2 +2ζω*ns*+ω2 *n* with a sign change in ζ . Hence, from Eqs. (4.51) and (4.52), it follows that the amplitudes are identical, but the phases are of opposite signs for the two terms. - -Systems whose poles and zeros are restricted to the LHP are classified as *minimum phase* systems. Minimum phase systems are particularly desirable because the system *and its inverse* are both stable. - -### **[4.9-5 The Transfer Function from the Frequency Response](#page-10-0)** - -In the preceding section we were given the transfer function of a system. From a knowledge of the transfer function, we developed techniques for determining the system response to sinusoidal inputs. We can also reverse the procedure to determine the transfer function of a minimum phase - -### 436 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -system from the system's response to sinusoids. This application has significant practical utility. If we are given a system in a black box with only the input and output terminals available, the transfer function has to be determined by experimental measurements at the input and output terminals. The frequency response to sinusoidal inputs is one of the possibilities that is very attractive because the measurements involved are so simple. One needs only to apply a sinusoidal signal at the input and observe the output. We find the amplitude gain |*H*(*j*ω)| and the output phase shift *H*(*j*ω) (with respect to the input sinusoid) for various values of ω over the entire range from 0 to ∞. This information yields the frequency response plots (Bode plots) when plotted against log ω. From these plots we determine the appropriate asymptotes by taking advantage of the fact that the slopes of all asymptotes must be multiples of ±20 dB/decade if the transfer function is a rational function (function that is a ratio of two polynomials in *s*). From the asymptotes, the corner frequencies are obtained. Corner frequencies determine the poles and zeros of the transfer function. Because of the ambiguity about the location of zeros since LHP and RHP zeros (zeros at *s* = ±*a*) have identical magnitudes, this procedure works only for minimum phase systems. - -## **4.10 FILTER DESIGN BY [PLACEMENT OF](#page-10-0) POLES AND ZEROS OF** *H(s)* - -In this section we explore the strong dependence of frequency response on the location of poles and zeros of *H*(*s*). This dependence points to a simple intuitive procedure to filter design. - -### **[4.10-1 Dependence of Frequency Response on Poles](#page-10-0) and Zeros of** *H(s)* - -Frequency response of a system is basically the information about the filtering capability of the system. A system transfer function can be expressed as - -$$ -H(s) = \frac{P(s)}{Q(s)} = b_0 \frac{(s - z_1)(s - z_2) \cdots (s - z_N)}{(s - \lambda_1)(s - \lambda_2) \cdots (s - \lambda_N)} -$$ - -where *z*1, *z*2, ... , *zN* are λ1, λ2, ... , λ*N* are the poles of *H*(*s*). Now the value of the transfer function *H*(*s*) at some frequency *s* = *p* is - -$$ -H(s)|_{s=p} = b_0 \frac{(p-z_1)(p-z_2)\cdots(p-z_N)}{(p-\lambda_1)(p-\lambda_2)\cdots(p-\lambda_N)} -$$ -(4.53) - -This equation consists of factors of the form *p*−*zi* and *p*−λ*i*. The factor *p*−*zi* is a complex number represented by a vector drawn from point *z* to the point *p* in the complex plane, as illustrated in Fig. 4.48a. The length of this line segment is |*p* − *zi*|, the magnitude of *p* − *zi*. The angle of this directed line segment (with the horizontal axis) is (*p* − *zi*). To compute *H*(*s*) at *s* = *p*, we draw line segments from all poles and zeros of *H*(*s*) to the point *p*, as shown in Fig. 4.48b. The vector connecting a zero *zi* to the point *p* is *p* − *zi*. Let the length of this vector be *ri*, and let its angle with the horizontal axis be φ*i*. Then *p*−*zi* = *riej*φ*i* . Similarly, the vector connecting a pole λ*i* to the point *p* is *p* − λ*i* = *diej*θ*i* , where *di* and θ*i* are the length and the angle (with the horizontal axis), - -**Figure 4.48** Vector representations of **(a)** complex numbers and **(b)** factors of *H*(*s*). - -respectively, of the vector *p*−λ*i*. Now from Eq. (4.53) it follows that - -$$ -H(s)|_{s=p} = b_0 \frac{(r_1 e^{j\phi_1})(r_2 e^{j\phi_2}) \cdots (r_N e^{j\phi_N})}{(d_1 e^{j\theta_1})(d_2 e^{j\theta_2}) \cdots (d_N e^{j\theta_N})} -$$ - -= $b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} e^{j[(\phi_1 + \phi_2 + \cdots + \phi_N) - (\theta_1 + \theta_2 + \cdots + \theta_N)]}$ - -Therefore - -$$ -|H(s)|_{s=p} = b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} = b_0 \frac{\text{product of distances of zeros to } p}{\text{product of distances of poles to } p} -$$ -(4.54) - -and - -$$ -\angle H(s)|_{s=p} = (\phi_1 + \phi_2 + \dots + \phi_N) - (\theta_1 + \theta_2 + \dots + \theta_N) -$$ - -= sum of angles of zeros to $p$ – sum of angles of poles to $p$ (4.55) - -Here, we have assumed positive *b*0. If *b*0 is negative, there is an additional phase π. Using this procedure, we can determine *H*(*s*) for any value of *s*. To compute the frequency response *H*(*j*ω), we use *s* = *j*ω (a point on the imaginary axis), connect all poles and zeros to the point *j*ω, and determine |*H*(*j*ω)| and *H*(*j*ω) from Eqs. (4.54) and (4.55). We repeat this procedure for all values of ω from 0 to ∞ to obtain the frequency response. - -### GAIN ENHANCEMENT BY A POLE - -To understand the effect of poles and zeros on the frequency response, consider a hypothetical case of a single pole −α + *j*ω0, as depicted in Fig. 4.49a. To find the amplitude response |*H*(*j*ω)| for a certain value of ω, we connect the pole to the point *j*ω (Fig. 4.49a). If the length of this line is *d*, then |*H*(*j*ω)| is proportional to 1/*d*, - -$$ -|H(j\omega)| = \frac{K}{d} \tag{4.56} -$$ - -where the exact value of constant *K* is not important at this point. As ω increases from zero, *d* decreases progressively until ω reaches the value ω0. As ω increases beyond ω0, *d* increases - -**Figure 4.49** The role of poles and zeros in determining the frequency response of an LTIC system. - -progressively. Therefore, according to Eq. (4.56), the amplitude response |*H*(*j*ω)| increases from ω = 0 until ω = ω0, and it decreases continuously as ω increases beyond ω0, as illustrated in Fig. 4.49b. Therefore, a pole at −α + *j*ω0 results in a frequency-selective behavior that enhances the gain at the frequency ω0 (resonance). Moreover, as the pole moves closer to the imaginary axis (as α is reduced), this enhancement (resonance) becomes more pronounced. This is because α, the distance between the pole and *j*ω0 (*d* corresponding to *j*ω0), becomes smaller, which increases the gain *K*/*d*. In the extreme case, when α = 0 (pole on the imaginary axis), the gain at ω0 goes to infinity. Repeated poles further enhance the frequency-selective effect. To summarize, we can enhance a gain at a frequency ω0 by placing a pole opposite the point *j*ω0. The closer the pole is to *j*ω0, the higher is the gain at ω0, and the gain variation is more rapid (more frequency selective) in the vicinity of frequency ω0. Note that a pole must be placed in the LHP for stability. - -Here we have considered the effect of a single complex pole on the system gain. For a real system, a complex pole −α + *j*ω0 must accompany its conjugate −α − *j*ω0. We can readily show that the presence of the conjugate pole does not appreciably change the frequency-selective behavior in the vicinity of ω0. This is because the gain in this case is *K*/*dd* , where *d* is the distance of a point *j*ω from the conjugate pole −α − *j*ω0. Because the conjugate pole is far from *j*ω0, there is no dramatic change in the length *d* as ω varies in the vicinity of ω0. There is a gradual increase in the value of *d* as ω increases, which leaves the frequency-selective behavior as it was originally, with only minor changes. - -### GAIN SUPPRESSION BY A ZERO - -Using the same argument, we observe that zeros at −α ± *j*ω0 (Fig. 4.49d) will have exactly the opposite effect of suppressing the gain in the vicinity of ω0, as shown in Fig. 4.49e). A zero on the imaginary axis at *j*ω0 will totally suppress the gain (zero gain) at frequency ω0. Repeated zeros will further enhance the effect. Also, a closely placed pair of a pole and a zero (dipole) tend to cancel out each other's influence on the frequency response. Clearly, a proper placement of poles and zeros can yield a variety of frequency-selective behavior. We can use these observations to design lowpass, highpass, bandpass, and bandstop (or notch) filters. - -Phase response can also be computed graphically. In Fig. 4.49a, angles formed by the complex conjugate poles −α±*j*ω0 at ω =0 (the origin) are equal and opposite. As ω increases from 0 up, the angle θ1 (due to the pole −α +*j*ω0), which has a negative value at ω = 0, is reduced in magnitude; the angle θ2 because of the pole −α − *j*ω0, which has a positive value at ω = 0, increases in magnitude. As a result, θ1 + θ2, the sum of the two angles, increases continuously, approaching a value π as ω → ∞. The resulting phase response *H*(*j*ω) = −(θ1 +θ2) is illustrated in Fig. 4.49c. Similar arguments apply to zeros at −α ± *j*ω0. The resulting phase response *H*(*j*ω) = (φ1 + φ2) is depicted in Fig. 4.49f. - -We now focus on simple filters, using the intuitive insights gained in this discussion. The discussion is essentially qualitative. - -### **[4.10-2 Lowpass Filters](#page-10-0)** - -A typical lowpass filter has a maximum gain at ω = 0. Because a pole enhances the gain at frequencies in its vicinity, we need to place a pole (or poles) on the real axis opposite the origin (*j*ω = 0), as shown in Fig. 4.50a. The transfer function of this system is - -$$ -H(s) = \frac{\omega_c}{s + \omega_c} -$$ - -We have chosen the numerator of *H*(*s*) to be ω*c* to normalize the dc gain *H*(0) to unity. If *d* is the distance from the pole −ω*c* to a point *j*ω (Fig. 4.50a), then - -$$ -|H(j\omega)| = \frac{\omega_c}{d} -$$ - -with *H*(0) = 1. As ω increases, *d* increases and |*H*(*j*ω)| decreases monotonically with ω, as illustrated in Fig. 4.50d with label *N* = 1. This is clearly a lowpass filter with gain enhanced in the vicinity of ω = 0. - -### WALL OF POLES - -An ideal lowpass filter characteristic (shaded in Fig. 4.50d) has a constant gain of unity up to frequency ω*c*. Then the gain drops suddenly to 0 for ω>ω*c*. To achieve the ideal lowpass - -**Figure 4.50** Pole-zero configuration and the amplitude response of a lowpass (Butterworth) filter. - -characteristic, we need enhanced gain over the entire frequency band from 0 to ω*c*. We know that to enhance a gain at any frequency ω, we need to place a pole opposite ω. To achieve an enhanced gain for all frequencies over the band (0 to ω*c*), we need to place a pole opposite every frequency in this band. In other words, we need a *continuous wall of poles* facing the imaginary axis opposite the frequency band 0 to ω*c* (and from 0 to −ω*c* for conjugate poles), as depicted in Fig. 4.50b. At this point, the optimum shape of this wall is not obvious because our arguments are qualitative and intuitive. Yet, it is certain that to have enhanced gain (constant gain) at every frequency over this range, we need an infinite number of poles on this wall. We can show that for a maximally flat† response over the frequency range (0 to ω*c*), the wall is a semicircle with an infinite number of poles uniformly distributed along the wall [11]. In practice, we compromise by using a finite number (*N*) of poles with less-than-ideal characteristics. Figure 4.50c shows the pole configuration for a fifth-order (*N* = 5) filter. The amplitude response for various values of *N* is illustrated in Fig. 4.50d. As *N* → ∞, the filter response approaches the ideal. This family of filters is known as the *Butterworth* filters. There are also other families. In *Chebyshev* filters, the wall shape is a semiellipse rather than a semicircle. The characteristics of a Chebyshev filter are inferior to those of Butterworth over the passband (0,ω*c*), where the characteristics show a rippling effect - -† Maximally flat amplitude response means the first 2*N* − 1 derivatives of |*H*(*j*ω)| with respect to ω are zero at ω = 0. - -instead of the maximally flat response of Butterworth. But in the stopband (ω>ω*c*), Chebyshev behavior is superior in the sense that Chebyshev filter gain drops faster than that of the Butterworth. - -### **[4.10-3 Bandpass Filters](#page-10-0)** - -The shaded characteristic in Fig. 4.51b shows the ideal bandpass filter gain. In the bandpass filter, the gain is enhanced over the entire passband. Our earlier discussion indicates that this can be realized by a wall of poles opposite the imaginary axis in front of the passband centered at ω0. (There is also a wall of conjugate poles opposite −ω0.) Ideally, an infinite number of poles is required. In practice, we compromise by using a finite number of poles and accepting less-than-ideal characteristics (Fig. 4.51). - -### **[4.10-4 Notch \(Bandstop\) Filters](#page-10-0)** - -An ideal notch filter amplitude response (shaded in Fig. 4.52b) is a complement of the amplitude response of an ideal bandpass filter. Its gain is zero over a small band centered at some frequency ω0 and is unity over the remaining frequencies. Realization of such a characteristic requires an infinite number of poles and zeros. Let us consider a practical second-order notch filter to obtain zero gain at a frequency ω = ω0. For this purpose, we must have zeros at ±*j*ω0. The requirement of unity gain at ω = ∞ requires the number of poles to be equal to the number of zeros (*M* = *N*). This ensures that for very large values of ω, the product of the distances of poles from ω will be equal to the product of the distances of zeros from ω. Moreover, unity gain at ω = 0 requires a pole and the corresponding zero to be equidistant from the origin. For example, if we use two (complex-conjugate) zeros, we must have two poles; the distance from the origin of the poles and of the zeros should be the same. This requirement can be met by placing the two conjugate poles on the semicircle of radius ω0, as depicted in Fig. 4.52a. The poles can be anywhere on the semicircle to satisfy the equidistance condition. Let the two conjugate poles be at angles ±θ with respect to the negative real axis. Recall that a pole and a zero in the same vicinity tend to cancel out - -**Figure 4.51 (a)** Pole-zero configuration and **(b)** the amplitude response of a bandpass filter. - -**Figure 4.52 (a)** Pole-zero configuration and **(b)** the amplitude response of a bandstop (notch) filter. - -each other's influences. Therefore, placing poles closer to zeros (selecting θ closer to π/2) results in a rapid recovery of the gain from value 0 to 1 as we move away from ω0 in either direction. Figure 4.52b shows the gain |*H*(*j*ω)| for three different values of θ. - -### **EXAMPLE 4.31 Notch Filter Design** - -Design a second-order notch filter to suppress 60 Hz hum in a radio receiver. - -We use the poles and zeros in Fig. 4.52a with ω0 = 120π. The zeros are at *s* = ±*j*ω0. The two poles are at −ω0 cos θ ±*j*ω0 sin θ. The filter transfer function is (with ω0 = 120π) - -$$ -H(s) = \frac{(s - j\omega_0)(s + j\omega_0)}{(s + \omega_0 \cos \theta + j\omega_0 \sin \theta)(s + \omega_0 \cos \theta - j\omega_0 \sin \theta)} -$$ - -= -$$ -\frac{s^2 + \omega_0^2}{s^2 + (2\omega_0 \cos \theta)s + \omega_0^2} = \frac{s^2 + 142122.3}{s^2 + (753.98 \cos \theta)s + 142122.3} -$$ - -and - -$$ -|H(j\omega)| = \frac{-\omega^2 + 142122.3}{\sqrt{(-\omega^2 + 142122.3)^2 + (753.98\omega\cos\theta)^2}} -$$ - -The closer the poles are to the zeros (the closer θ is to π/2), the faster the gain recovery from 0 to 1 on either side of ω0 = 120π. Figure 4.52b shows the amplitude response for three different values of θ. This example is a case of very simple design. To achieve zero gain over a band, we need an infinite number of poles as well as an infinite number of zeros. - -MATLAB easily computes and plots the magnitude response curves of Fig. 4.52b. To illustrate, let us plot the magnitude response using θ = 60◦ over a frequency range of 0 ≤ *f* ≤ 150 Hz. The result, shown in Fig. 4.53, matches the θ = 60◦ case of Fig. 4.52b. - ->> f = (0:.01:150); omega0 = 2\*pi\*60; theta = 60\*pi/180; - -- >> H = @(s) (s.^2+omega0^2)./(s.^2+2\*omega0\*cos(theta)\*s+omega0^2); -- >> plot(f,abs(H(1j\*2\*pi\*f)),'k-'); -- >> xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); - -## **DR ILL 4.16 Magnitude Response from Pole-Zero Plots** - -Use the qualitative method of sketching the frequency response to show that the system with the pole-zero configuration in Fig. 4.54a is a highpass filter and the configuration in Fig. 4.54b is a bandpass filter. - -### **[4.10-5 Practical Filters and Their Specifications](#page-10-0)** - -For ideal filters, everything is black and white; the gains are either zero or unity over certain bands. As we saw earlier, real life does not permit such a worldview. Things have to be gray or shades of gray. In practice, we can realize a variety of filter characteristics that can only approach ideal characteristics. - -An ideal filter has a passband (unity gain) and a stopband (zero gain) with a sudden transition from the passband to the stopband. There is no transition band. For practical (or realizable) filters, on the other hand, the transition from the passband to the stopband (or vice versa) is gradual and takes place over a finite band of frequencies. Moreover, for realizable filters, the gain cannot be zero over a finite band (Paley–Wiener condition). As a result, there can be no true stopband for practical filters. We therefore define a *stopband* to be a band over which the gain is below some small number *Gs*, as illustrated in Fig. 4.55. Similarly, we define a *passband* to be a band over which the gain is between 1 and some number *Gp* (*Gp* < 1), as shown in Fig. 4.55. We have selected the passband gain of unity for convenience. It could be any constant. Usually the gains are specified in terms of decibels. This is simply 20 times the log (to base 10) of the gain. Thus, - -$$ -G(\text{dB}) = 20\log_{10} G -$$ - -A gain of unity is 0 dB and a gain of √ 2 is 3.01 dB, usually approximated by 3 dB. Sometimes the specification may be in terms of attenuation, which is the negative of the gain in dB. Thus, a gain of 1/ √2, that is, 0.707, is −3 dB, but is an attenuation of 3 dB. - -**Figure 4.55** Passband, stopband, and transition band in filters of various types. - -In a typical design procedure, *Gp* (*minimum passband gain*) and *Gs* (*maximum stopband gain*) are specified. Figure 4.55 shows the passband, the stopband, and the transition band for typical lowpass, bandpass, highpass, and bandstop filters. Fortunately, the highpass, bandpass, and bandstop filters can be obtained from a basic lowpass filter by simple frequency transformations. For example, replacing *s* with ω*c*/*s* in the lowpass filter transfer function results in a highpass filter. Similarly, other frequency transformations yield the bandpass and bandstop filters. Hence, it is necessary to develop a design procedure only for a basic lowpass filter. Then, by using appropriate transformations, we can design filters of other types. The design procedures are beyond our scope here and will not be discussed. The interested reader is referred to [1]. - -## **4.11 THE BILATERAL LAPLACE [TRANSFORM](#page-10-0)** - -Situations involving noncausal signals and/or systems cannot be handled by the (unilateral) Laplace transform discussed so far. These cases can be analyzed by the *bilateral* (or *two-sided*) Laplace transform defined by - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ - -and *x*(*t*) can be obtained from *X*(*s*) by the inverse transformation - -$$ -x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds -$$ - -Observe that the unilateral Laplace transform discussed so far is a special case of the bilateral Laplace transform, where the signals are restricted to the causal type. Basically, the two transforms are the same. For this reason we use the same notation for the bilateral Laplace transform. - -Earlier we showed that the Laplace transforms of *e*−*atu*(*t*) and of −*e*−*atu*(−*t*) are identical. The only difference is in their regions of convergence (ROC). The ROC for the former is Re*s* > −*a*; that for the latter is Re *s* < −*a*, as illustrated in Fig. 4.1. Clearly, the inverse Laplace transform of *X*(*s*) is not unique unless the ROC is specified. If we restrict all our signals to the causal type, however, this ambiguity does not arise. The inverse transform of 1/(*s*+*a*) is *e*−*atu*(*t*). Thus, in the unilateral Laplace transform, we can ignore the ROC in determining the inverse transform of *X*(*s*). - -We now show that any bilateral transform can be expressed in terms of two unilateral transforms. It is, therefore, possible to evaluate bilateral transforms from a table of unilateral transforms. - -Consider the function *x*(*t*) appearing in Fig. 4.56a. We separate *x*(*t*) into two components, *x*1(*t*) and *x*2(*t*), representing the positive time (*causal*) component and the negative time (*anticausal*) component of *x*(*t*), respectively (Figs. 4.56b and 4.56c): - -$$ -x_1(t) = x(t)u(t) -$$ - and $x_2(t) = x(t)u(-t)$ - -**Figure 4.56** Expressing a signal as a sum of causal and anticausal components. - -The bilateral Laplace transform of *x*(*t*) is given by - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ - -= -$$ -\int_{-\infty}^{0^-} x_2(t)e^{-st} dt + \int_{0^-}^{\infty} x_1(t)e^{-st} dt -$$ - -= -$$ -X_2(s) + X_1(s) -$$ - (4.57) - -where *X*1(*s*) is the Laplace transform of the causal component *x*1(*t*), and *X*2(*s*) is the Laplace transform of the anticausal component *x*2(*t*). Consider *X*2(*s*), given by - -$$ -X_2(s) = \int_{-\infty}^{0^-} x_2(t)e^{-st} dt = \int_{0^+}^{\infty} x_2(-t)e^{st} dt -$$ - -Therefore, - -$$ -X_2(-s) = \int_{0^+}^{\infty} x_2(-t)e^{-st} dt -$$ - -If *x*(*t*) has any impulse or its derivative(s) at the origin, they are included in *x*1(*t*). Consequently, *x*2(*t*) = 0 at the origin; that is, *x*2(0) = 0. Hence, the lower limit on the integration in the preceding equation can be taken as 0− instead of 0+. Therefore, - -$$ -X_2(-s) = \int_{0^-}^{\infty} x_2(-t)e^{-st} dt -$$ - -Because *x*2(−*t*) is causal (Fig. 4.56d), *X*2(−*s*) can be found from the unilateral transform table. Changing the sign of *s* in *X*2(−*s*) yields *X*2(*s*). - -To summarize, the bilateral transform *X*(*s*) in Eq. (4.57) can be computed from the unilateral transforms in two steps: - -- 1. Split *x*(*t*) into its causal and anticausal components, *x*1(*t*) and *x*2(*t*), respectively. -- 2. Since the signals *x*1(*t*) and *x*2(−*t*) are both causal, take the (unilateral) Laplace transform of *x*1(*t*) and add to it the (unilateral) Laplace transform of *x*2(−*t*), with *s* replaced by −*s*. This procedure gives the (bilateral) Laplace transform of *x*(*t*). - -Since *x*1(*t*) and *x*2(−*t*) are both causal, *X*1(*s*) and *X*2(−*s*) are both unilateral Laplace transforms. Let σ*c*1 and σ*c*2 be the abscissas of convergence of *X*1(*s*) and *X*2(−*s*), respectively. This statement implies that *X*1(*s*) exists for all *s* with Re *s* > σ*c*1, and *X*2(−*s*) exists for all *s* with Re*s* > σ*c*2. Therefore, *X*2(*s*) exists for all *s* with Re *s* < −σ*c*2. † Therefore, *X*(*s*) = *X*1(*s*) + *X*2(*s*) exists for all *s* such that - -$$ -\sigma_{c1} < \text{Re}\,s < -\sigma_{c2} -$$ - -The regions of convergence of *X*1(*s*), *X*2(*s*), and *X*(*s*) are shown in Fig. 4.57. Because *X*(*s*) is finite for all values of *s* lying in the strip of convergence (σ*c*1 < Re *s* < −σ*c*2), poles of *X*(*s*) must lie outside this strip. The poles of *X*(*s*) arising from the causal component *x*1(*t*) lie to the left of the *strip* (region) *of convergence*, and those arising from its anticausal component *x*2(*t*) lie to its right (see Fig. 4.57). This fact is of crucial importance in finding the inverse bilateral transform. - -This result can be generalized to left-sided and right-sided signals. We define a signal *x*(*t*) as a *right-sided* signal if *x*(*t*) = 0 for *t* < *T*1 for some finite positive or negative number *T*1. A causal signal is always a right-sided signal, but the converse is not necessarily true. A signal is said to *left-sided* if it is zero for *t* > *T*2 for some finite, positive, or negative number *T*2. An anticausal signal is always a left-sided signal, but the converse is not necessarily true. A *two-sided* signal is of infinite duration on both positive and negative sides of *t* and is neither right-sided nor left-sided. - -We can show that the conclusions for ROC for causal signals also hold for right-sided signals, and those for anticausal signals hold for left-sided signals. In other words, if *x*(*t*) is causal or - -† For instance, if *x*(*t*) exists for all *t* &gt; 10, then *x*(−*t*), its time-inverted form, exists for *t* &lt; −10. - -**Figure 4.57** Regions of convergence for causal, anticausal, and combined signals. - -right-sided, the poles of *X*(*s*) lie to the left of the ROC, and if *x*(*t*) is anticausal or left-sided, the poles of *X*(*s*) lie to the right of the ROC. - -To prove this generalization, we observe that a right-sided signal can be expressed as *x*(*t*) + *xf*(*t*), where *x*(*t*) is a causal signal and *xf*(*t*) is some finite-duration signal. The ROC of any finite-duration signal is the entire *s*-plane (no finite poles). Hence, the ROC of the right-sided signal *x*(*t*) + *xf*(*t*) is the region common to the ROCs of *x*(*t*) and *xf*(*t*), which is same as the ROC for *x*(*t*). This proves the generalization for right-sided signals. We can use a similar argument to generalize the result for left-sided signals. Let us find the bilateral Laplace transform of - -$$ -x(t) = e^{bt}u(-t) + e^{at}u(t) -$$ -\n(4.58) - -We already know the Laplace transform of the causal component - -$$ -e^{at}u(t) \Longleftrightarrow \frac{1}{s-a} \qquad \text{Re}\,s > a \tag{4.59} -$$ - -For the anticausal component, *x*2(*t*) = *ebtu*(−*t*), we have - -$$ -x_2(-t) = e^{-bt}u(t) \Longleftrightarrow \frac{1}{s+b} \qquad \text{Re}\, s > -b -$$ - -so that - -$$ -X_2(s) = \frac{1}{-s+b} = \frac{-1}{s-b} \qquad \text{Re}\, s < b -$$ - -Therefore, - -$$ -e^{bt}u(-t) \Longleftrightarrow \frac{-1}{s-b} \qquad \text{Re}\,s < b \tag{4.60} -$$ - -and the Laplace transform of *x*(*t*) in Eq. (4.58) is - -$$ -X(s) = -\frac{1}{s-b} + \frac{1}{s-a} -$$ - Res > a and Res < b -= -$$ -\frac{a-b}{(s-b)(s-a)} -$$ - Res < b -(4.61) - -Figure 4.58 shows *x*(*t*) and the ROC of *X*(*s*) for various values of *a* and *b*. Equation (4.61) indicates that the ROC of *X*(*s*) does not exist if *a* > *b*, which is precisely the case in Fig. 4.58f. Observe that the poles of *X*(*s*) are outside (on the edges) of the ROC. The poles of *X*(*s*) because of the anticausal component of *x*(*t*) lie to the right of the ROC, and those due to the causal component of *x*(*t*) lie to its left. - -When *X*(*s*) is expressed as a sum of several terms, the ROC for *X*(*s*) is the intersection of (region common to) the ROCs of all the terms. In general, if *x*(*t*) = %*k i*=1 *xi*(*t*), then the ROC for *X*(*s*) is the intersection of the ROCs (region common to all ROCs) for the transforms *X*1(*s*), *X*2(*s*), ... , *Xk*(*s*). - -**Figure 4.58** Various two exponential signals and their regions of convergence. - -### **EXAMPLE 4.32 Inverse Bilateral Laplace Transform** - -Find the inverse bilateral Laplace transform of - -$$ -X(s) = \frac{-3}{(s+2)(s-1)} -$$ - -if the ROC is **(a)** −2 < Re *s* < 1, **(b)** Re *s* > 1, and **(c)** Re *s* < −2. - -**(a)** - -$$ -X(s) = \frac{1}{s+2} - \frac{1}{s-1} -$$ - -Now, *X*(*s*) has poles at −2 and 1. The strip of convergence is −2 < Re *s* < 1. The pole at −2, being to the left of the strip of convergence, corresponds to a causal signal. The pole at 1, being to the right of the strip of convergence, corresponds to an anticausal signal. Equations (4.59) and (4.60) yield - -$$ -x(t) = e^{-2t}u(t) + e^t u(-t) -$$ - -**(b)** Both poles lie to the left of the ROC, so both poles correspond to causal signals. Therefore, - -)*u*(*t*) - -*x*(*t*) = (*e*−2*t* −*et* - -**Figure 4.59** Three possible inverse transforms of −3/((*s* +2)(*s*−1)). - -**(c)** Both poles lie to the right of the region of convergence, so both poles correspond to anticausal signals, and - -$$ -x(t) = (-e^{-2t} + e^t)u(-t) -$$ - -Figure 4.59 shows the three inverse transforms corresponding to the same *X*(*s*) but with different regions of convergence. - -### **[4.11-1 Properties of the Bilateral Laplace Transform](#page-11-0)** - -Properties of the bilateral Laplace transform are similar to those of the unilateral transform. We shall merely state the properties here without proofs. Let the ROC of *X*(*s*) be *a* < Re *s* < *b*. Similarly, let the ROC of *Xi*(*s*) be *ai* < Re *s* < *bi* for (*i* = 1, 2). - -LINEARITY - -$$ -a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(s) + a_2X_2(s) -$$ - -The ROC for *a*1*X*1(*s*) + *a*2*X*2(*s*) is the region common to (intersection of) the ROCs for *X*1(*s*) and *X*2(*s*). - -TIME SHIFT - -$$ -x(t-T) \Longleftrightarrow X(s)e^{-sT} -$$ - -The ROC for *X*(*s*)*e*−*sT* is identical to the ROC for *X*(*s*). - -FREQUENCY SHIFT - -$$ -x(t)e^{s_0t} \Longleftrightarrow X(s-s_0) -$$ - -The ROC for *X*(*s*−*s*0) is *a*+*c* < Re *s* < *b*+*c*, where *c* = Re *s*0. - -TIME DIFFERENTIATION - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow sX(s) -$$ - -The ROC for *sX*(*s*) contains the ROC for *X*(*s*) and may be larger than that of *X*(*s*) under certain conditions [e.g., if *X*(*s*) has a first-order pole at *s* = 0, it is canceled by the factor *s* in *sX*(*s*)]. - -TIME INTEGRATION - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(s)/s -$$ - -The ROC for *sX*(*s*) is max (*a*, 0) < Re *s* < *b*. - -TIME SCALING - -$$ -x(\beta t) \Longleftrightarrow \frac{1}{|\beta|}X\left(\frac{s}{\beta}\right) -$$ - -The ROC for *X*(*s*/β) is β*a* < Re *s* < β*b*. For β > 1, *x*(β*t*) represents time compression and the corresponding ROC expands by factor β. For 0 >β> 1, *x*(β*t*) represents time expansion and the corresponding ROC is compressed by factor β. - -TIME CONVOLUTION - -$$ -x_1(t) * x_2(t) \Longleftrightarrow X_1(s)X_2(s) -$$ - -The ROC for *X*1(*s*)*X*2(*s*) is the region common to (intersection of ) the ROCs for *X*1(*s*) and *X*2(*s*). - -FREQUENCY CONVOLUTION - -$$ -x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty} X_1(w)X_2(s-w) dw -$$ - -The ROC for *X*1(*s*) ∗ *X*2(*s*) is *a*1 +*a*2 < Re *s* < *b*1 +*b*2. - -TIME REVERSAL - -$$ -x(-t) \Longleftrightarrow X(-s) -$$ - -The ROC for *X*(−*s*) is −*b* < Re *s* < −*a*. - -### **[4.11-2 Using the Bilateral Transform for Linear System Analysis](#page-11-0)** - -Since the bilateral Laplace transform can handle noncausal signals, we can analyze noncausal LTIC systems using the bilateral Laplace transform. We have shown that the (zero-state) output *y*(*t*) is given by - -$$ -y(t) = \mathcal{L}^{-1}[X(s)H(s)] -$$ - -This expression is valid only if *X*(*s*)*H*(*s*) exists. The ROC of *X*(*s*)*H*(*s*) is the region in which both *X*(*s*) and *H*(*s*) exist. In other words, the ROC of *X*(*s*)*H*(*s*) is the region common to the regions of convergence of both *X*(*s*) and *H*(*s*). These ideas are clarified in the following examples. - -Find the current *y*(*t*) for the *RC* circuit in Fig. 4.60a if the voltage *x*(*t*) is - -$$ -x(t) = e^t u(t) + e^{2t} u(-t) -$$ - -**Figure 4.60** Response of a circuit to a noncausal input. - -The transfer function *H*(*s*) of the circuit is given by - -$$ -H(s) = \frac{s}{s+1} \qquad \text{Re}\, s > -1 -$$ - -Because *h*(*t*) is a causal function, the ROC of *H*(*s*) is Re *s* > −1. Next, the bilateral Laplace transform of *x*(*t*) is given by - -$$ -X(s) = \frac{1}{s-1} - \frac{1}{s-2} = \frac{-1}{(s-1)(s-2)} \qquad 1 < \text{Re } s < 2 -$$ - -The response *y*(*t*) is the inverse transform of *X*(*s*)*H*(*s*): - -$$ -y(t) = \mathcal{L}^{-1} \left[ \frac{-s}{(s+1)(s-1)(s-2)} \right] = \mathcal{L}^{-1} \left[ \frac{1}{6} \frac{1}{s+1} + \frac{1}{2} \frac{1}{s-1} - \frac{2}{3} \frac{1}{s-2} \right] -$$ - -The ROC of *X*(*s*)*H*(*s*) is that ROC common to both *X*(*s*) and *H*(*s*). This is 1 < Re *s* < 2. The poles *s* = ±1 lie to the left of the ROC and, therefore, correspond to causal signals; the pole - -### 454 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -*s* = 2 lies to the right of the ROC and thus represents an anticausal signal. Hence, - -$$ -y(t) = \frac{1}{6}e^{-t}u(t) + \frac{1}{2}e^{t}u(t) + \frac{2}{3}e^{2t}u(-t) -$$ - -Figure 4.60c shows *y*(*t*). Note that in this example, if - -$$ -x(t) = e^{-4t}u(t) + e^{-2t}u(-t) -$$ - -then the ROC of *X*(*s*) is −4 < Re *s* < −2. Here no region of convergence exists for *X*(*s*)*H*(*s*). Hence, the response *y*(*t*) goes to infinity. - -### **EXAMPLE 4.34 Response of a Noncausal System** - -Find the response *y*(*t*) of a noncausal system with the transfer function - -$$ -H(s) = \frac{-1}{s-1} \qquad \text{Re}\, s < 1 -$$ - -to the input *x*(*t*) = *e*−2*t u*(*t*). - -We have - -$$ -X(s) = \frac{1}{s+2} \qquad \text{Re}\, s > -2 -$$ - -and - -$$ -Y(s) = X(s)H(s) = \frac{-1}{(s-1)(s+2)} -$$ - -The ROC of *X*(*s*)*H*(*s*) is the region −2 < Re *s* < 1. By partial fraction expansion, - -$$ -Y(s) = \frac{-1/3}{s-1} + \frac{1/3}{s+2} \qquad -2 < \text{Re}\, s < 1 -$$ - -and - -$$ -y(t) = \frac{1}{3} [e^t u(-t) + e^{-2t} u(t)] -$$ - -Note that the pole of *H*(*s*) lies in the RHP at 1. Yet the system is not unstable. The pole(s) in the RHP may indicate instability or noncausality, depending on its location with respect to the region of convergence of *H*(*s*). For example, if *H*(*s*) = −1/(*s*−1) with Re *s* > 1, the system is causal and unstable, with *h*(*t*) = −*et u*(*t*). In contrast, if *H*(*s*) = −1/(*s* − 1) with Re *s* < 1, the system is noncausal and stable, with *h*(*t*) = *et u*(−*t*). - -### **EXAMPLE 4.35 System Response to a Noncausal Input** - -Find the response *y*(*t*) of a system with the transfer function - -$$ -H(s) = \frac{1}{s+5} \qquad \text{Re}\, s > -5 -$$ - -and the input - -$$ -x(t) = e^{-t}u(t) + e^{-2t}u(-t) -$$ - -The input *x*(*t*) is of the type depicted in Fig. 4.58f, and the region of convergence for *X*(*s*) does not exist. In this case, we must determine separately the system response to each of the two input components, *x*1(*t*) = *e*−*t u*(*t*) and *x*2(*t*) = *e*−2*t u*(−*t*). - -$$ -X_1(s) = \frac{1}{s+1} \qquad \text{Re}\, s > -1 -$$ -\n -$$ -X_2(s) = \frac{-1}{s+2} \qquad \text{Re}\, s < -2 -$$ - -If *y*1(*t*) and *y*2(*t*) are the system responses to *x*1(*t*) and *x*2(*t*), respectively, then - -$$ -Y_1(s) = \frac{1}{(s+1)(s+5)} = \frac{1/4}{s+1} - \frac{1/4}{s+5} -$$ - Re $s > -1$ - -so that - -$$ -y_1(t) = \frac{1}{4}(e^{-t} - e^{-5t})u(t) -$$ - -and - -$$ -Y_2(s) = \frac{-1}{(s+2)(s+5)} = \frac{-1/3}{s+2} + \frac{1/3}{s+5} \qquad -5 < \text{Re}\,s < -2 -$$ - -so that - -$$ -y_2(t) = \frac{1}{3} [e^{-2t}u(-t) + e^{-5t}u(t)] -$$ - -Therefore, - -$$ -y(t) = y_1(t) + y_2(t) = \frac{1}{3}e^{-2t}u(-t) + \left(\frac{1}{4}e^{-t} + \frac{1}{12}e^{-5t}\right)u(t) -$$ - -## **[4.12 MATLAB: CONTINUOUS-TIME](#page-11-0) FILTERS** - -Continuous-time filters are essential to many if not most engineering systems, and MATLAB is an excellent assistant for filter design and analysis. Although a comprehensive treatment of continuous-time filter techniques is outside the scope of this book, quality filters can be designed and realized with minimal additional theory. - -A simple yet practical example demonstrates basic filtering concepts. Telephone voice signals are often lowpass-filtered to eliminate frequencies above a cutoff of 3 kHz, or ω*c* = 3000(2π ) ≈ 18,850 rad/s. Filtering maintains satisfactory speech quality and reduces signal bandwidth, thereby increasing the phone company's call capacity. How, then, do we design and realize an acceptable 3 kHz lowpass filter? - -### **[4.12-1 Frequency Response and Polynomial Evaluation](#page-11-0)** - -Magnitude response plots help assess a filter's performance and quality. The magnitude response of an ideal filter is a brick-wall function with unity passband gain and perfect stopband attenuation. For a lowpass filter with cutoff frequency ω*c*, the ideal magnitude response is - -$$ -|H_{\text{ideal}}(j\omega)| = \begin{cases} 1 & |\omega| \le \omega_c \\ 0 & |\omega| > \omega_c \end{cases} -$$ - -Unfortunately, ideal filters cannot be implemented in practice. Realizable filters require compromises, although good designs will closely approximate the desired brick-wall response. - -A realizable LTIC system often has a rational transfer function that is represented in the *s*-domain as - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{B(s)}{A(s)} = \frac{\sum_{k=0}^{M} b_{k+N-M} s^{M-k}}{\sum_{k=0}^{N} a_k s^{N-k}} -$$ - -Frequency response *H*(*j*ω) is obtained by letting *s* = *j*ω, where frequency ω is in radians per second. - -MATLAB is ideally suited to evaluate frequency response functions. Defining a length-(*N* + 1) coefficient vector **A** = [*a*0,*a*1,...,*aN*] and a length-(*M* + 1) coefficient vector **B** = [*bN*−*M*,*bN*−*M*+1,..., *bN*], program CH4MP1 computes *H*(*j*ω) for each frequency in the input vector *ω*. - -``` -function [H] = CH4MP1(B,A,omega); -% CH4MP1.m : Chapter 4, MATLAB Program 1 -% Function M-file computes frequency response for LTIC system -% INPUTS: B = vector of feedforward coefficients -% A = vector of feedback coefficients -% omega = vector of frequencies [rad/s]. -% OUTPUTS: H = frequency response -``` - -``` -H = polyval(B,j*omega)./polyval(A,j*omega); -``` - -The function polyval efficiently evaluates simple polynomials and makes the program nearly trivial. For example, when A is the vector of coefficients [*a*0,*a*1,...,*aN*], polyval (A,j\*omega) computes - -$$ -\sum_{k=0}^N a_k (j\omega)^{N-k} -$$ - -for each value of the frequency vector omega. It is also possible to compute frequency responses by using the signal-processing toolbox function freqs. - -### DESIGN AND EVALUATION OF A SIMPLE *RC* FILTER - -One of the simplest lowpass filters is realized by using an *RC* circuit, as shown in Fig. 4.61. This one-pole system has transfer function *HRC*(*s*) = (*RCs* + 1)−1 and magnitude response |*HRC*(*j*ω)|=|(*j*ω*RC* + 1)−1| = 1/ 1+(*RC*ω)2. Independent of component values *R* and *C*, this circuit has many desirable characteristics, such as unity gain at ω = 0 and magnitude response that monotonically decreases to zero as ω → ∞. - -Components *R* and *C* are chosen to set the desired 3 kHz cutoff frequency. For many filter types, the cutoff frequency corresponds to the half-power point, or |*HRC*(*j*ω*c*)| = 1/ √2. Assign *C* a realistic capacitance of 1 nF, then the required resistance is computed by *R* = 1/ *C*2ω2 *c* = 1/ (10−9)2(2π3000)2. - -``` ->> omega_c = 2*pi*3000; C = 1e-9; R = 1/sqrt(C^2*omega_c^2) - R = 5.3052e+004 -``` - -The root of this first-order *RC* filter is directly related to the cutoff frequency, λ = −1/*RC* = −18,850 = −ω*c*. - -To evaluate the *RC* filter performance, the magnitude response is plotted over the mostly audible frequency range (0 ≤ *f* ≤ 20 kHz). - -``` ->> f = linspace(0,20000,200); Hmag_RC = abs(CH4MP1([1],[R*C 1],f*2*pi)); -``` - -``` ->> plot(f,abs(f*2*pi)<=omega_c,'k-',f,Hmag_RC,'k--'); -``` - -``` ->> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); -``` - -``` ->> legend('Ideal','First-order RC','location','best'); -``` - -The linspace(X1,X2,N) command generates an N-length vector of linearly spaced points between X1 and X2. - -As shown in Fig. 4.62, the first-order *RC* response is indeed lowpass with a half-power cutoff frequency equal to 3 kHz. It rather poorly approximates the desired brick-wall response: the passband is not very flat, and stopband attenuation increases very slowly to less than 20 dB at 20 kHz. - -**Figure 4.62** Magnitude response |*HRC*(*j*2π*f*)| of a first-order *RC* filter. - -**Figure 4.63** A cascaded *RC* filter. - -### A CASCADED *RC* FILTER AND POLYNOMIAL EXPANSION - -A first-order *RC* filter is destined for poor performance; one pole is simply insufficient to obtain good results. A cascade of *RC* circuits increases the number of poles and improves the filter response. To simplify the analysis and prevent loading between stages, we employ op-amp followers to buffer the output of each stage, as shown in Fig. 4.63. A cascade of *N* stages results in an *N*th-order filter with transfer function given by - -$$ -H_{\text{cascade}}(s) = [H_{RC}(s)]^N = (RCs + 1)^{-N} -$$ - -Upon choosing a cascade of 10 stages and *C* = 1 nF, a 3 kHz cutoff frequency is obtained by setting *R* = √ 21/10 −1/(*C*ω*c*) = √ 21/10 −1/(6π(10)−6). - ->> R = sqrt(2^(1/10)-1)/(C\*omega\_c) R = 1.4213e+004 - -This cascaded filter has a 10th-order pole at λ = −1/*RC* and no finite zeros. To compute the magnitude response, polynomial coefficient vectors **A** and **B** are needed. Setting **B** = [1] ensures there are no finite zeros or, equivalently, that all zeros are at infinity. The poly command, which expands a vector of roots into a corresponding vector of polynomial coefficients, is used to obtain **A**. - -- >> B = 1; A = poly(-1/(R\*C)\*ones(10,1));A = A/A(end); -- >> Hmag\_cascade = abs(CH4MP1(B,A,f\*2\*pi)); -- >> plot(f,abs(f\*2\*pi)<=omega\_c,'k-',f,Hmag\_cascade,'k--'); -- >> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); -- >> legend('Ideal','Tenth-order RC cascade','location','best'); - -Notice that scaling a polynomial by a constant does not change its roots. Conversely, the roots of a polynomial specify a polynomial within a scale factor. The command A = A/A(end) properly scales the denominator polynomial to ensure unity gain at ω = 0. - -The magnitude response plot of the tenth-order *RC* cascade is shown in Fig. 4.64. Compared with the simple *RC* response of Fig. 4.62, the passband remains relatively unchanged, but stopband attenuation is greatly improved to over 60 dB at 20 kHz. - -**Figure 4.64** Magnitude response |*H*cascade(*j*2π*f*)| of a tenth-order *RC* cascade. - -## **[4.12-2 Butterworth Filters and the](#page-11-0)** Find **Command** - -The pole location of a first-order lowpass filter is necessarily fixed by the cutoff frequency. There is little reason, however, to place all the poles of a 10th-order filter at one location. Better pole placement will improve our filter's magnitude response. One strategy, discussed in Sec. 4.10, is to place a wall of poles opposite the passband frequencies. A semicircular wall of poles leads to the Butterworth family of filters, and a semi-elliptical shape leads to the Chebyshev family of filters. Butterworth filters are considered first. - -To begin, notice that a transfer function *H*(*s*) with real coefficients has a squared magnitude response given by |*H*(*j*ω)| 2 = *H*(*j*ω)*H*∗(*j*ω) = *H*(*j*ω)*H*(−*j*ω) = *H*(*s*)*H*(−*s*)|*s*=*j*ω. Thus, half the poles of |*H*(*j*ω)| 2 correspond to the filter *H*(*s*) and the other half correspond to *H*(−*s*). Filters that are both stable and causal require *H*(*s*) to include only left-half-plane poles. - -The squared magnitude response of a Butterworth filter is - -$$ -|H_{\text{BW}}(j\omega)|^2 = \frac{1}{1 + (j\omega/j\omega_c)^{2N}} -$$ - -This function has the same appealing characteristics as the first-order *RC* filter: a gain that is unity at ω = 0 and monotonically decreases to zero as ω → ∞. By construction, the half-power gain - -occurs at ω*c*. Perhaps most importantly, however, the first 2*N* − 1 derivatives of |*H*BW(*j*ω)| with respect to ω are zero at ω = 0. Put another way, the passband is constrained to be very flat for low frequencies. For this reason, Butterworth filters are sometimes called maximally flat filters. - -As discussed in Sec. B.7, the roots of minus 1 must lie equally spaced on a circle centered at the origin. Thus, the 2*N* poles of |*H*BW(*j*ω)| 2 naturally lie equally spaced on a circle of radius ω*c* centered at the origin. Figure 4.65 displays the 20 poles corresponding to the case *N* = 10 and ω*c* = 3000(2π ) rad/s. An *N*th-order Butterworth filter that is both causal and stable uses the *N* left-half-plane poles of |*H*BW(*j*ω)| 2. - -To design a 10th-order Butterworth filter, we first compute the 20 poles of |*H*BW(*j*ω)| 2: - ->> N=10; poles = roots([(1j\*omega\_c)^(-2\*N),zeros(1,2\*N-1),1]); - -The find command is a powerful and useful function that returns the indices of a vector's nonzero elements. Combined with relational operators, the find command allows us to extract the 10 left-half-plane roots that correspond to the poles of our Butterworth filter. - ->> BW\_poles = poles(find(real(poles)<0)); - -To compute the magnitude response, these roots are converted to coefficient vector **A**. - -``` ->> A = poly(BW_poles); A = A/A(end); Hmag_BW = abs(CH4MP1(B,A,f*2*pi)); -``` - -``` ->> plot(f,abs(f*2*pi)<=omega_c,'k-',f,Hmag_BW,'k--'); -``` - -``` ->> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|'); -``` - -``` ->> legend('Ideal','Tenth-order Butterworth','location','best'); -``` - -The magnitude response plot of the Butterworth filter is shown in Fig. 4.66. The Butterworth response closely approximates the brick-wall function and provides excellent filter characteristics: flat passband, rapid transition to the stopband, and excellent stopband attenuation (>40 dB at 5 kHz). - -**Figure 4.66** Magnitude response |*H*BW(*j*2π*f*)| of a tenth-order Butterworth filter. - -### **[4.12-3 Using Cascaded Second-Order Sections](#page-11-0) for Butterworth Filter Realization** - -For our *RC* filters, realization preceded design. For our Butterworth filter, however, design has preceded realization. For our Butterworth filter to be useful, we must be able to implement it. - -Since the transfer function *H*BW(*s*) is known, the differential equation is also known. Therefore, it is possible to try to implement the design by using op-amp integrators, summers, and scalar multipliers. Unfortunately, this approach will not work well. To understand why, consider the denominator coefficients *a*0 = 1.766×10−43 and *a*10 = 1. The smallest coefficient is 43 orders of magnitude smaller than the largest coefficient! It is practically impossible to accurately realize such a broad range in scale values. To understand this, skeptics should try to find realistic resistors such that *Rf* /*R* = 1.766×10−43. Additionally, small component variations will cause large changes in actual pole location. - -A better approach is to cascade five second-order sections, where each section implements one complex conjugate pair of poles. By pairing poles in complex conjugate pairs, each of the resulting second-order sections has real coefficients. With this approach, the smallest coefficients are only about nine orders of magnitude smaller than the largest coefficients. Furthermore, pole placement is typically less sensitive to component variations for cascaded structures. - -The Sallen–Key circuit shown in Fig. 4.67 provides a good way to realize a pair of complex-conjugate poles.† The transfer function of this circuit is - -$$ -H_{SK}(s) = \frac{\frac{1}{R_1R_2C_1C_2}}{s^2 + \left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1}\right)s + \frac{1}{R_1R_2C_1C_2}} = \frac{\omega_0^2}{s^2 + \left(\frac{\omega_0}{Q}\right)s + \omega_0^2} -$$ - -Geometrically, ω0 is the distance from the origin to the poles and *Q* = 1/2 cosψ, where ψ is the angle between the negative real axis and the pole. Termed the "quality factor" of a circuit, *Q* - -† A more general version of the Sallen–Key circuit has a resistor *Ra* from the negative terminal to ground and a resistor *Rb* between the negative terminal and the output. In Fig. 4.67, *Ra* = ∞ and *Rb* = 0. - -provides a measure of the peakedness of the response. High-*Q* filters have poles close to the ω axis, which boost the magnitude response near those frequencies. - -Although many ways exist to determine suitable component values, a simple method is to assign *R*1 a realistic value and then let *R*2 = *R*1, *C*1 = 2*Q*/ω0*R*1, and *C*2 = 1/2*Q*ω0*R*2. Butterworth poles are a distance ω*c* from the origin, so ω0 = ω*c*. For our 10th-order Butterworth filter, the angles ψ are regularly spaced at 9, 27, 45, 63, and 81 degrees. MATLAB program CH4MP2 automates the task of computing component values and magnitude responses for each stage. - -``` -% CH4MP2.m : Chapter 4, MATLAB Program 2 -% Script M-file computes Sallen-Key component values and magnitude -% responses for each of the five cascaded second-order filter sections. -omega_0 = 3000*2*pi; % Filter cut-off frequency -psi = [9 27 45 63 81]*pi/180; % Butterworth pole angles -f = linspace(0,6000,200); % Frequency range for magnitude response calculations -Hmag_SK = zeros(5,200); % Pre-allocate array for magnitude responses -for stage = 1:5, - Q = 1/(2*cos(psi(stage))); % Compute Q for current stage - % Compute and display filter components to the screen: - disp(['Stage ',num2str(stage),... - ' (Q = ',num2str(Q),... - '): R1 = R2 = ',num2str(56000),... - ', C1 = ',num2str(2*Q/(omega_0*56000)),... - ', C2 = ',num2str(1/(2*Q*omega_0*56000))]); - B = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute filter coefficients - Hmag_SK(stage,:) = abs(CH4MP1(B,A,2*pi*f)); % Compute magnitude response -end -plot(f,Hmag_SK,'k',f,prod(Hmag_SK),'k:') -xlabel('f [Hz]'); ylabel('Magnitude Response') -``` - -The disp command displays a character string to the screen. Character strings must be enclosed in single quotation marks. The num2str command converts numbers to character strings and facilitates the formatted display of information. The prod command multiplies along the columns of a matrix; it computes the total magnitude response as the product of the magnitude responses of the five stages. - -Executing the program produces the following output: - -``` ->> CH4MP2 - Stage 1 (Q = 0.50623): R1 = R2 = 56000, C1 = 9.5916e-10, C2 = 9.3569e-10 - Stage 2 (Q = 0.56116): R1 = R2 = 56000, C1 = 1.0632e-09, C2 = 8.441e-10 - Stage 3 (Q = 0.70711): R1 = R2 = 56000, C1 = 1.3398e-09, C2 = 6.6988e-10 - Stage 4 (Q = 1.1013): R1 = R2 = 56000, C1 = 2.0867e-09, C2 = 4.3009e-10 - Stage 5 (Q = 3.1962): R1 = R2 = 56000, C1 = 6.0559e-09, C2 = 1.482e-10 -``` - -**Figure 4.68** Magnitude responses for Sallen–Key filter stages. - -Since all the component values are practical, this filter is possible to implement. Figure 4.68 displays the magnitude responses for all five stages (solid lines). The total response (dotted line) confirms a 10th-order Butterworth response. Stage 5, which has the largest *Q* and implements the pair of conjugate poles nearest the ω axis, is the most peaked response. Stage 1, which has the smallest *Q* and implements the pair of conjugate poles furthest from the ω axis, is the least peaked response. In practice, it is best to order high-*Q* stages last; this reduces the risk that the high gains will saturate the filter hardware. - -### **[4.12-4 Chebyshev Filters](#page-11-0)** - -Like an order-*N* Butterworth lowpass filter (LPF), an order-*N* Chebyshev LPF is an all-pole filter that possesses many desirable characteristics. Compared with an equal-order Butterworth filter, the Chebyshev filter achieves better stopband attenuation and reduced transition bandwidth by allowing an adjustable amount of ripple within the passband. - -The squared magnitude response of a Chebyshev filter is - -$$ -|H_{\rm C}(j\omega)|^2 = \frac{1}{1 + \epsilon^2 C_N^2(\omega/\omega_c)} -$$ - -where controls the passband ripple, *CN*(ω/ω*c*) is a degree-*N* Chebyshev polynomial, and ω*c* is the radian cutoff frequency. Several characteristics of Chebyshev LPFs are noteworthy: - -• An order-*N* Chebyshev LPF is equi-ripple in the passband (|ω| ≤ ω*c*), has a total of *N* maxima and minima over (0 ≤ ω ≤ ω*c*), and is monotonic decreasing in the stopband (|ω| > ω*c*). - -### 464 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -- In the passband, the maximum gain is 1 and the minimum gain is 1/ √1+ 2. For odd-valued *N*, |*H*(*j*0)| = 1. For even-valued *N*, |*H*C(*j*0)| = 1/ √1+ 2. -- Ripple is controlled by setting = √ 10*R*/10 −1, where *R* is the allowable passband ripple expressed in decibels. Reducing adversely affects filter performance (see Prob. 4.12-10). -- Unlike Butterworth filters, the cutoff frequency ω*c* rarely specifies the 3 dB point. For = 1, |*H*C(*j*ω*c*)| 2 = 1/(1+ 2) = 0.5. The cutoff frequency ω*c* simply indicates the frequency after which |*H*C(*j*ω)| < 1/ √ 1+ 2. - -The Chebyshev polynomial *CN*(*x*) is defined as - -$$ -C_N(x) = \cos[N\cos^{-1}(x)] = \cosh[N\cosh^{-1}(x)] -$$ - -In this form, it is difficult to verify that *CN*(*x*) is a degree-*N* polynomial in *x*. A recursive form of *CN*(*x*) makes this fact more clear (see Prob. 4.12-13). - -$$ -C_N(x) = 2xC_{N-1}(x) - C_{N-2}(x) -$$ - -With *C*0(*x*) = 1 and *C*1(*x*) = *x*, the recursive form shows that any *CN* is a linear combination of degree-*N* polynomials and is therefore a degree-*N* polynomial itself. For *N* ≥ 2, MATLAB program CH4MP3 generates the (*N* +1) coefficients of Chebyshev polynomial *CN*(*x*). - -``` -function [C_N] = CH4MP3(N); -% CH4MP3.m : Chapter 4, MATLAB Program 3 -% Function M-file computes Chebyshev polynomial coefficients -% using the recursion relation C_N(x) = 2xC_{N-1}(x) - C_{N-2}(x) -% INPUTS: N = degree of Chebyshev polynomial -% OUTPUTS: C_N = vector of Chebyshev polynomial coefficients -C_Nm2 = 1; C_Nm1 = [1 0]; % Initial polynomial coefficients: -for t = 2:N; - C_N = 2*conv([1 0],C_Nm1)-[zeros(1,length(C_Nm1)-length(C_Nm2)+1),C_Nm2]; - C_Nm2 = C_Nm1; C_Nm1 = C_N; -``` - -``` -end -``` - -``` -As examples, consider C2(x) = 2xC1(x) − C0(x) = 2x(x) − 1 = 2x2 − 1 and C3(x) = 2xC2(x) − -C1(x) = 2x(2x2 −1)−x = 4x3 −3x. CH4MP3 easily confirms these cases. -``` - -``` ->> CH4MP3(2) - ans = 2 0 -1 ->> CH4MP3(3) - ans = 4 0 -3 0 -``` - -Since *CN*(ω/ω*c*) is a degree-*N* polynomial, |*H*C(*j*ω)| 2 is an all-pole rational function with 2*N* finite poles. Similar to the Butterworth case, the *N* poles specifying a causal and stable Chebyshev filter can be found by selecting the *N* left-half-plane roots of 1+ 2*C*2 *N*[*s*/(*j*ω*c*)]. - -Root locations and dc gain are sufficient to specify a Chebyshev filter for a given *N* and . To demonstrate, consider the design of an order-8 Chebyshev filter with cutoff frequency *fc* = 1 kHz and allowable passband ripple *R* = 1 dB. First, filter parameters are specified. - -``` ->> omega_c = 2*pi*1000; R = 1; N = 8; ->> epsilon = sqrt(10^(R/10)-1); -``` - -The coefficients of *CN*[*s*/(*j*ω*c*)] are obtained with the help of CH4MP3, and then the coefficients of [1+ 2*C*2 *N*(*s*/(*j*ω*c*))] are computed by using convolution to perform polynomial multiplication. - -``` ->> CN = CH4MP3(N).*((1/(1j*omega_c)).^[N:-1:0]); ->> CP = epsilon^2*conv(CN,CN); CP(end) = CP(end)+1; -``` - -Next, the polynomial roots are found, and the left-half-plane poles are retained and plotted. - -``` ->> poles = roots(CP); i = find(real(poles)<0); C_poles = poles(i); ->> plot(real(C_poles),imag(C_poles),'kx'); axis equal; ->> axis(omega_c*[-1.1 1.1 -1.1 1.1]); ->> xlabel('Real'); ylabel('Imaginary'); -``` - -As shown in Fig. 4.69, the roots of a Chebyshev filter lie on an ellipse† (see Prob. 4.12-14). - -**Figure 4.69** Pole-zero plot for an order-8 Chebyshev LPF with *fc* = 1 kHz and *R* = 1 dB. - -To compute the filter's magnitude response, the poles are expanded into a polynomial, the dc gain is set based on the even value of *N*, and CH4MP1 is used. - -``` ->> A = poly(C_poles); B = A(end)/sqrt(1+epsilon^2); -``` - ->> omega = linspace(0,2\*pi\*2000,2001); H\_C = CH4MP1(B,A,omega); - -``` ->> plot(omega/2/pi,abs(H_C),'k'); axis([0 2000 0 1.1]); -``` - -``` ->> xlabel('f [Hz]'); ylabel('|H_C(j2\pi f)|'); -``` - -† E. A. Guillemin demonstrates a wonderful relationship between the Chebyshev ellipse and the Butterworth circle in his book *Synthesis of Passive Networks* (Wiley, New York, 1957). - -**Figure 4.70** Magnitude responses for an order-8 Chebyshev LPF with *fc* = 1 kHz and *R* = 1 dB. - -As seen in Fig. 4.70, the magnitude response exhibits correct Chebyshev filter characteristics: passband ripples are equal in height and never exceed *R* = 1 dB; there are a total of *N* = 8 maxima and minima in the passband; and the gain rapidly and monotonically decreases after the cutoff frequency of *fc* = 1 kHz. - -For higher-order filters, polynomial rooting may not provide reliable results. Fortunately, Chebyshev roots can also be determined analytically. For - -$$ -\phi_k = \frac{2k+1}{2N}\pi \quad \text{and} \quad \xi = \frac{1}{N}\sinh^{-1}\left(\frac{1}{\epsilon}\right) -$$ - -the Chebyshev poles are - -$$ -p_k = \omega_c \sinh(\xi) \sin(\phi_k) + j\omega_c \cosh(\xi) \cos(\phi_k) -$$ - -Continuing the same example, the poles are recomputed and again plotted. The result is identical to Fig. 4.69. - -``` ->> k = [1:N]; xi = 1/N*asinh(1/epsilon); phi = (k*2-1)/(2*N)*pi; ->> C_poles = omega_c*(-sinh(xi)*sin(phi)+1j*cosh(xi)*cos(phi)); ->> plot(real(C_poles),imag(C_poles),'kx'); axis equal; ->> axis(omega_c*[-1.1 1.1 -1.1 1.1]); ->> xlabel('Real'); ylabel('Imaginary'); -``` - -As in the case of high-order Butterworth filters, a cascade of second-order filter sections facilitates practical implementation of Chebyshev filters. Problems 4.12-5 and 4.12-8 use second-order Sallen–Key circuit stages to investigate such implementations. - -## **[4.13 SUMMARY](#page-11-0)** - -This chapter discusses analysis of LTIC (linear, time-invariant, continuous-time) systems by the Laplace transform, which transforms integro-differential equations of such systems into algebraic equations. Therefore solving these integro-differential equations reduces to solving algebraic equations. The Laplace transform method cannot be used for time-varying-parameter systems or for nonlinear systems in general. - -The transfer function *H*(*s*) of an LTIC system is the Laplace transform of its impulse response. It may also be defined as a ratio of the Laplace transform of the output to the Laplace transform of the input when all initial conditions are zero (system in zero state). If *X*(*s*) is the Laplace transform of the input *x*(*t*) and *Y*(*s*) is the Laplace transform of the corresponding output *y*(*t*) (when all initial conditions are zero), then *Y*(*s*) = *X*(*s*)*H*(*s*). For an LTIC system described by an *N*th-order differential equation *Q*(*D*)*y*(*t*) = *P*(*D*)*x*(*t*), the transfer function *H*(*s*) = *P*(*s*)/*Q*(*s*). Like the impulse response *h*(*t*), the transfer function *H*(*s*) is also an external description of the system. - -Electrical circuit analysis can also be carried out by using a transformed circuit method, in which all signals (voltages and currents) are represented by their Laplace transforms, all elements by their impedances (or admittances), and initial conditions by their equivalent sources (initial condition generators). In this method, a network can be analyzed as if it were a resistive circuit. - -Large systems can be depicted by suitably interconnected subsystems represented by blocks. Each subsystem, being a smaller system, can be readily analyzed and represented by its input–output relationship, such as its transfer function. Analysis of large systems can be carried out with the knowledge of input–output relationships of its subsystems and the nature of interconnection of various subsystems. - -LTIC systems can be realized by scalar multipliers, adders, and integrators. A given transfer function can be synthesized in many different ways, such as canonic, cascade, and parallel. Moreover, every realization has a transpose, which also has the same transfer function. In practice, all the building blocks (scalar multipliers, adders, and integrators) can be obtained from operational amplifiers. - -The system response to an everlasting exponential *est* is also an everlasting exponential *H*(*s*)*est*. Consequently, the system response to an everlasting exponential *ej*ω*t* is *H*(*j*ω) *ej*ω*t* . Hence, *H*(*j*ω) is the frequency response of the system. For a sinusoidal input of unit amplitude and having frequency ω, the system response is also a sinusoid of the same frequency (ω) with amplitude |*H*(*j*ω)|, and its phase is shifted by *H*(*j*ω) with respect to the input sinusoid. For this reason |*H*(*j*ω)| is called the amplitude response (gain) and *H*(*j*ω) is called the phase response of the system. Amplitude and phase response of a system indicate the filtering characteristics of the system. The general nature of the filtering characteristics of a system can be quickly determined from a knowledge of the location of poles and zeros of the system transfer function. - -Most of the input signals and practical systems are causal. Consequently we are required most of the time to deal with causal signals. When all signals must be causal, the Laplace transform analysis is greatly simplified; the region of convergence of a signal becomes irrelevant to the analysis process. This special case of the Laplace transform (which is restricted to causal signals) is called the unilateral Laplace transform. Much of the chapter deals with this variety of Laplace transform. Section 4.11 discusses the general Laplace transform (the bilateral Laplace transform), which can handle causal and noncausal signals and systems. In the bilateral transform, the inverse transform of *X*(*s*) is not unique but depends on the region of convergence of *X*(*s*). Thus, the region of convergence plays a very crucial role in the bilateral Laplace transform. - -### 468 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -### **[REFERENCES](#page-11-0)** - -- 1. Lathi, B. P. *Signal Processing and Linear Systems*, 1st ed. Oxford University Press, New York, 1998. -- 2. Doetsch, G. *Introduction to the Theory and Applications of the Laplace Transformation with a Table of Laplace Transformations.* Springer-Verlag, New York, 1974. -- 3. LePage, W. R. *Complex Variables and the Laplace Transforms for Engineers*. McGraw-Hill, New York, 1961. -- 4. Durant, Will, and Ariel Durant. *The Age of Napoleon*, Part XI in *The Story of Civilization Series*. Simon & Schuster, New York, 1975. -- 5. Bell, E. T. *Men of Mathematics*. Simon & Schuster, New York, 1937. -- 6. Nahin, P. J. "Oliver Heaviside: Genius and Curmudgeon." *IEEE Spectrum*, vol. 20, pp. 63–69, July 1983. -- 7. Berkey, D. *Calculus*, 2nd ed. Saunders, Philadelphia, 1988. -- 8. Encyclopaedia Britannica. *Micropaedia IV*, 15th ed., p. 981, Chicago, 1982. -- 9. Churchill, R. V. *Operational Mathematics*, 2nd ed. McGraw-Hill, New York, 1958. -- 10. Truxal, J. G. *The Age of Electronic Messages*. McGraw-Hill, New York, 1990. -- 11. Van Valkenberg, M. *Analog Filter Design*. Oxford University Press, New York, 1982. - -## **[PROBLEMS](#page-11-0)** - -- **4.1-1** By direct integration [Eq. (4.1)] find the Laplace transforms and the region of convergence of the following functions: - - (a) *u*(*t*)−*u*(*t* −1) - - (b) *te*−*t u*(*t*) - - (c) *t* cos ω0*t u*(*t*) - - (d) (*e*2*t* −2*e*−*t* )*u*(*t*) - - (e) cos ω1*t* cos ω2*t u*(*t*) - - (f) cosh(*at*)*u*(*t*) - - (g) sinh(*at*)*u*(*t*) - - (h) *e*−2*t* cos(5*t* +θ )*u*(*t*) -- **4.1-2** By direct integration [Eq. (4.1)] find the Laplace transforms and the region of convergence of the following functions: - -(a) -$$ -e^{-2t}u(t-5) + \delta(t-1) -$$ - -(b) -$$ -\pi e^{3t} u(t+5) - \delta(2t) -$$ - -(c) -$$ -\sum_{k=0}^{\infty} \delta(t - kT), T > 0 -$$ - -- **4.1-3** By direct integration find the Laplace transforms of the signals shown in Fig. P4.1-3. -- **4.1-4** Find the inverse (unilateral) Laplace transforms of the following functions: - -(a) -$$ -\frac{2s+5}{s^2+5s+6} -$$ - -\n(b) -$$ -\frac{3s+5}{s^2+4s+13} -$$ - -\n(c) -$$ -\frac{(s+1)^2}{s^2-s-6} -$$ - -\n(d) -$$ -\frac{5}{s^2(s+2)} -$$ - -**Figure P4.1-3** - -(e) -$$ -\frac{2s+1}{(s+1)(s^2+2s+2)} -$$ - -\n(f) -$$ -\frac{s+2}{s(s+1)^2} -$$ - -\n(g) -$$ -\frac{1}{(s+1)(s+2)^4} -$$ - -\n(h) -$$ -\frac{s+1}{s(s+2)^2(s^2+4s+5)} -$$ - -(i) -$$ -\frac{s(s+2)^2(s^2+4s+5)}{(s+1)^2(s^2+2s+5)} -$$ - -- **4.2-1** Suppose a CT signal *x*(*t*)=2[*u*(*t* −2)−*u*(*t* +1)] has a transform *X*(*s*). - - (a) If *Y*a(*s*) = *e*−5*s sX s* + 1 2 , determine and sketch the corresponding signal *y*a(*t*). - - (b) If *Y*b(*s*) = 2−*s sX*(*s* −2), determine and sketch the corresponding signal *y*b(*t*). -- **4.2-2** Find the Laplace transforms of the following functions using only Table 4.1 and the time-shifting property (if needed) of the unilateral Laplace transform: - - (a) *u*(*t*) −*u*(*t* −1) - - (b) *e*−(*t*−τ )*u*(*t* −τ ) - - (c) *e*−(*t*−τ )*u*(*t*) - - (d) *e*−*t u*(*t* −τ ) - - (e) *te*−*t u*(*t* −τ ) - - (f) sin[ω0(*t* −τ )]*u*(*t* −τ ) - - (g) sin[ω0(*t* −τ )]*u*(*t*) - - (h) sin ω0*t u*(*t* −τ ) - - (i) *t*sin(*t*)*u*(*t*) - - (j) (1−*t*) cos(*t* −1)*u*(*t* −1) -- **4.2-3** Using only Table 4.1 and the time-shifting property, determine the Laplace transform of the signals in Fig. P4.1-3. [*Hint:* See Sec. 1.4 for discussion of expressing such signals analytically.] -- **4.2-4** Prove the frequency-differentiation property, −*tx*(*t*) ⇐⇒ *d dsX*(*s*). This property holds for both the unilateral and bilateral Laplace transforms. -- **4.2-5** Consider the signal *x*(*t*) = *te*−2(*t*−3) *u*(*t*−2). - - (a) Determine the *unilateral* Laplace transform *X*u(*s*) = *L*u {*x*(*t*)}. - -- (b) Determine the *bilateral* Laplace transform *X*(*s*) = *L*{*x*(*t*)}. -- **4.2-6** Consider the signals *x*(*t*) and *y*(*t*), as shown in Fig. P4.2-6. - - (a) Using the definition, compute *X*(*s*), the bilateral Laplace transform of *x*(*t*). - - (b) Using Laplace transform properties, express *Y*(*s*), the bilateral Laplace transform of *y*(*t*), as a function of *X*(*s*), the bilateral Laplace transform of *x*(*t*). Simplify as much as possible without substituting your answer from part (a). -- **4.2-7** Find the inverse Laplace transforms of the following functions: - -(a) -$$ -\frac{(2s+5)e^{-2s}}{s^2+5s+6} -$$ - -(b) -$$ -\frac{se^{-3s}+2}{s^2+2s+2} -$$ - -(c) -$$ -\frac{e^{-(s-1)}+3}{2(2s+5)} -$$ - -$$ -\begin{array}{c}\n\text{(c)} \quad s^2 - 2s + 5 \\ -\text{(d)} \quad \frac{e^{-s} + e^{-2s} + 1}{s^2 + 3s + 2}\n\end{array} -$$ - -- **4.2-8** Using ROC σ > 0, determine the inverse Laplace transform of *X*(*s*) = *s*−1 *d ds e*−2*s s* . -- **4.2-9** The Laplace transform of a causal periodic signal can be determined from the knowledge of the Laplace transform of its first cycle (period). - - (a) If the Laplace transform of *x*(*t*) in Fig. P4.2-9a is *X*(*s*), then show that *G*(*s*), the Laplace transform of *g*(*t*) (Fig. P4.2-9b), is - -$$ -G(s) = \frac{X(s)}{1 - e^{-sT_0}} \qquad \text{Re}\, s > 0 -$$ - -- (b) Use this result to find the Laplace transform of the signal *p*(*t*) illustrated in Fig. P4.2-9c. -- **4.2-10** Starting only with the fact that δ(*t*) ⇐⇒ 1, build pairs 2 through 10b in Table 4.1, using various properties of the Laplace transform. - -#### **Figure P4.2-9** - -- **4.2-11** (a) Find the Laplace transform of the pulses in Fig. 4.2 by using only the time-different iation property, the time-shifting property, and the fact that δ(*t*) ⇐⇒ 1. - - (b) In Ex. 4.9, the Laplace transform of *x*(*t*) is found by finding the Laplace transform of *d*2*x*/*dt*2. Find the Laplace transform of *x*(*t*) in that example by finding the Laplace transform of *dx*/*dt* and using Table 4.1, if necessary. -- **4.2-12** Determine the inverse unilateral Laplace transform of - -$$ -X(s) = \frac{1}{e^{s+3}} \frac{s^2}{(s+1)(s+2)} -$$ - -- **4.2-13** Since 13 is such a lucky number, determine the inverse Laplace transform of *X*(*s*) = 1/(*s* +1)13 given region of convergence σ > −1. [*Hint:* What is the *n*th derivative of 1/(*s* +*a*)?] -- **4.2-14** It is difficult to compute the Laplace transform *X*(*s*) of signal - -$$ -x(t) = \frac{1}{t}u(t) -$$ - -by using direct integration. Instead, properties provide a simpler method. - -- (a) Use Laplace transform properties to express the Laplace transform of *tx*(*t*) in terms of the unknown quantity *X*(*s*). -- (b) Use the definition to determine the Laplace transform of *y*(*t*) = *tx*(*t*). - -- (c) Solve for *X*(*s*) by using the two pieces from **()**(a) and **()**(b). Simplify your answer. -- **4.3-1** Use the Laplace transform to solve the following differential equations: - - (a) (*D*2 + 3*D* + 2)*y*(*t*) = *Dx*(*t*) if *y*(0−) = *y*˙(0−) = 0 and *x*(*t*) = *u*(*t*) - - (b) (*D*2 + 4*D* + 4)*y*(*t*) = (*D* + 1)*x*(*t*) if *y*(0−) = 2, *y*˙(0−) = 1 and *x*(*t*) = *e*−*t u*(*t*) - - (c) (*D*2 +6*D*+25)*y*(*t*) = (*D*+2)*x*(*t*) if *y*(0−) = *y*˙(0−) = 1 and *x*(*t*) = 25*u*(*t*) -- **4.3-2** Solve the differential equations in Prob. 4.3-1 using the Laplace transform. In each case determine the zero-input and zero-state components of the solution. -- **4.3-3** Consider a causal LTIC system described by the differential equation - -$$ -2\dot{y}(t) + 6y(t) = \dot{x}(t) - 4x(t) -$$ - -- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*(0−) = −3. -- (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *e*δ(*t* −π ). -- **4.3-4** Consider a causal LTIC system described by the differential equation - -$$ -\ddot{y}(t) + 3\dot{y}(t) + 2y(t) = 2\dot{x}(t) - x(t) -$$ - -- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*˙(0−) = 2 and *y*(0−) = −3. -- (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *u*(*t*). -- **4.3-5** Solve the following simultaneous differential equations using the Laplace transform, assuming all initial conditions to be zero and the input *x*(*t*) = *u*(*t*): - - (a) (*D*+3)*y*1(*t*)−2*y*2(*t*) = *x*(*t*) −2*y*1(*t*) +(2*D*+4)*y*2(*t*) = 0 - - (b) (*D*+2)*y*1(*t*)−(*D*+1)*y*2(*t*) = 0 −(*D*+1)*y*1(*t*) +(2*D*+1)*y*2(*t*) = *x*(*t*) - -Determine the transfer functions relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*). - -- **4.3-6** Consider a causal LTIC system described by *y*˙(*t*) +2*y*(*t*) = ˙*x*(*t*). - - (a) Determine the transfer function *H*(*s*) for this system. - - (b) Using your result from part (a), determine the impulse response *h*(*t*) for this system. - - (c) Using Laplace transform techniques, determine the output *y*(*t*) if the input is *x*(*t*) = *e*−*t u*(*t*) and *y*(0−) = √ 2. -- **4.3-7** Repeat Prob. 4.3-6 for a causal LTIC system described by 3*y*(*t*) + ˙*y*(*t*)+ ˙*x*(*t*) = 0. -- **4.3-8** For the circuit in Fig. P4.3-8, the switch is in the open position for a long time before *t* = 0, when it is closed instantaneously. - - (a) Write loop equations (in time domain) for *t* ≥ 0. - - (b) Solve for *y*1(*t*) and *y*2(*t*) by taking the Laplace transform of loop equations found in part (a). - -**4.3-9** For each of the systems described by the following differential equations, find the system transfer function: - -(a) -$$ -\frac{d^2y(t)}{dt^2} + 11\frac{dy(t)}{dt} + 24y(t) = 5\frac{dx(t)}{dt} + 3x(t) -$$ - -\n(b) -$$ -\frac{d^3y(t)}{dt^3} + 6\frac{d^2y(t)}{dt^2} - 11\frac{dy(t)}{dt} + 6y(t) -$$ -$$ -= 3\frac{d^2x(t)}{dt^2} + 7\frac{dx(t)}{dt} + 5x(t) -$$ - -\n(c) -$$ -\frac{d^4y(t)}{dt^4} + 4\frac{dy(t)}{dt} = 3\frac{dx(t)}{dt} + 2x(t) -$$ - -\n(d) -$$ -\frac{d^2y(t)}{dt^2} - y(t) = \frac{dx(t)}{dt} - x(t) -$$ - -**4.3-10** For each of the systems specified by the following transfer functions, find the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable: - -(a) -$$ -H(s) = \frac{s+5}{s^2 + 3s + 8} -$$ - -\n(b) $H(s) = \frac{s^2 + 3s + 5}{s^3 + 8s^2 + 5s + 7}$ -\n(c) $H(s) = \frac{5s^2 + 7s + 2}{s^2 - 2s + 5}$ - -**4.3-11** For a system with transfer function - -$$ -H(s) = \frac{2s+3}{s^2+2s+5} -$$ - -- (a) Find the (zero-state) response for inputs *x*1(*t*) = 10*u*(*t*) and *x*2(*t*) = *u*(*t* −5). -- (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable. -- **4.3-12** For a system with transfer function - -$$ -H(s) = \frac{s}{s^2 + 9} -$$ - -- (a) Find the (zero-state) response if the input *x*(*t*) = (1−*e*−*t* )*u*(*t*) -- (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable. -- **4.3-13** Consider a system with transfer function - -$$ -H(s) = \frac{s+5}{s^2 + 5s + 6} -$$ - -Find the (zero-state) response for the following inputs: - -- (a) *x*a(*t*) = *e*−3*t u*(*t*) -- (b) *x*b(*t*) = *e*−4*t u*(*t*) -- (c) *x*c(*t*) = *e*−4(*t*−5) *u*(*t* −5) -- (d) *x*d(*t*) = *e*−4(*t*−5) *u*(*t*) -- (e) *x*e(*t*) = *e*−4*t u*(*t* −5) - -Assuming that the system *H*(*s*) is controllable and observable, - -- (f) write the differential equation relating the output *y*(*t*) to the input *x*(*t*). -- **4.3-14** An LTI system has a step response given by *s*(*t*) = *e*−*t u*(*t*) − *e*−2*t u*(*t*). Determine the output of this system *y*(*t*) given an input *x*(*t*) = δ(*t* − π )−cos( √3)*u*(*t*). -- **4.3-15** For an LTIC system with zero initial conditions (system initially in zero state), if an input *x*(*t*) produces an output *y*(*t*), then using the Laplace transform, show the following: - - (a) The input *dx*/*dt* produces an output *dy*/*dt*. - - (b) The input \$ *t* 0 *x*(τ )*d*τ produces an output \$ *t* 0 *y*(τ )*d*τ . Hence, show that the unit step response of a system is an integral of the impulse response; that is, \$ *t* 0 *h*(τ )*d*τ . -- **4.3-16** Discuss asymptotic and BIBO stabilities for the systems described by the following transfer functions, assuming that the systems are controllable and observable: - -(a) -$$ -\frac{(s+5)}{s^2+3s+2} -$$ - -(b) -$$ -\frac{s+5}{s^2(s+2)} -$$ - -$$ -(c) \frac{s(s+2)}{s+5} -$$ - -(d) -$$ -\frac{s+5}{s(s+2)} -$$ - -(e) -$$ -\frac{s+5}{s^2-2s+3} -$$ - -- **4.3-17** Repeat Prob. 4.3-16 for systems described by the following differential equations. Systems may be uncontrollable and/or unobservable. - - (a) (*D*2 +3*D*+2)*y*(*t*) = (*D*+3)*x*(*t*) - - (b) (*D*2 +3*D*+2)*y*(*t*) = (*D*+1)*x*(*t*) - - (c) (*D*2 +*D*−2)*y*(*t*) = (*D*−1)*x*(*t*) - - (d) (*D*2 −3*D*+2)*y*(*t*) = (*D*−1)*x*(*t*) -- **4.4-1** The circuit shown in Fig. P4.4-1 has system function given by *H*(*s*) = 1 1+*RCs*. Let *R* = 2 and *C* = 3 and use Laplace transform techniques to solve the following. - - (a) Find the output *y*(*t*) given an initial capacitor voltage of *y*(0−) = 3 and an input *x*(*t*) = *u*(*t*). - - (b) Given an input *x*(*t*) = *u*(*t* − 3), determine the initial capacitor voltage *y*(0−) so that the output *y*(*t*) is 1 volt at *t* = 6 seconds. - -- **4.4-2** Consider the circuit shown in Fig. P4.4-2. Use Laplace transform techniques to solve the following. - - (a) Determine the standard-form, constantcoefficient differential equation description of this circuit. - - (b) Letting *R* = *C* = 1, determine the total response *y*(*t*) to input *x*(*t*) = 3*e*−*t u*(*t*) and initial capacitor voltage of *vC*(0−) = 5. - -**4.4-3** Find the zero-state response *y*(*t*) of the network in Fig. P4.4-3 if the input voltage *x*(*t*) = *te*−*t u*(*t*). Find the transfer function relating the output *Y*(*s*) to the input *X*(*s*). From the transfer function, write the differential equation relating *y*(*t*) to *x*(*t*). - -### **Figure P4.4-3** - -**4.4-4** The switch in the circuit of Fig. P4.4-4 is closed for a long time and then opened instantaneously at *t* = 0. Find and sketch the current *y*(*t*). - -#### **Figure P4.4-4** - -**4.4-5** Find the current *y*(*t*) for the parallel resonant circuit in Fig. P4.4-5 if the input is: (a) *x*(*t*) = *A*cos ω0*t u*(*t*) (b) *x*(*t*) = *A*sin ω0*t u*(*t*) Assume all initial conditions to be zero and, in both cases, ω2 0 = 1/*LC*. - -**Figure P4.4-5** - -**Figure P4.4-6** - -- **4.4-6** Find the loop currents *y*1(*t*) and *y*2(*t*) for *t* ≥ 0 in the circuit of Fig. P4.4-6a for the input *x*(*t*) in Fig. P4.4-6b. -- **4.4-7** For the network in Fig. P4.4-7, the switch is in a closed position for a long time before *t* = 0, when it is opened instantaneously. Find *y*1(*t*) and *vs*(*t*) for *t* ≥ 0. - -#### **Figure P4.4-7** - -- **4.4-8** Find the output voltage *v*0(*t*) for *t* ≥ 0 for the circuit in Fig. P4.4-8, if the input *x*(*t*) = 100*u*(*t*). The system is in the zero state initially. -- **4.4-9** Find the output voltage *y*(*t*) for the network in Fig. P4.4-9 for the initial conditions *iL*(0) = 1 A and *vC*(0) = 3 V. -- **4.4-10** For the network in Fig. P4.4-10, the switch is in position *a* for a long time and then is moved to position *b* instantaneously at *t* = 0. Determine the current *y*(*t*) for *t* > 0. - -**4.4-11** Consider the circuit of Fig. P4.4-11. - -- (a) Using transform-domain techniques, determine the system's standard-form transfer function *H*(*s*). -- (b) Using transform-domain techniques and letting *R* = *L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*t u*(*t* −1). - -**Figure P4.4-10** - -(c) Using transform-domain techniques and letting *R* = 2*L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*t u*(*t* −1). - -**4.4-12** Show that the transfer function that relates the output voltage *y*(*t*) to the input voltage *x*(*t*) for the op-amp circuit in Fig. P4.4-12a is given by - -$$ -H(s) = \frac{Ka}{s+a} \quad \text{where} -$$ - -$$ -K = 1 + \frac{R_b}{R_a} \quad \text{and} \quad a = \frac{1}{RC} -$$ - -and that the transfer function for the circuit in Fig. P4.4-12b is given by - -$$ -H(s) = \frac{Ks}{s+a} -$$ - -**4.4-13** For the second-order op-amp circuit in Fig. P4.4-13, show that the transfer function *H*(*s*) relating the output voltage *y*(*t*) to the input - -### **Figure P4.4-14** - -voltage *x*(*t*) is given by - -$$ -H(s) = \frac{-s}{s^2 + 8s + 12} -$$ - -- **4.4-14** Consider the op-amp circuit of Fig. P4.4-14. - - (a) Determine the standard-form transfer function *H*(*s*) of this system. - - (b) Determine the standard-form constant coefficient linear differential equation description of this circuit. - -- (c) Using transform-domain techniques, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*2*t u*(*t* +1). -- (d) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0− capacitor voltage (first op-amp output voltage) is 3 volts. -- **4.4-15** We desire the op-amp circuit of Fig. P4.4-15 to behave as *y*˙(*t*)−1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*). - - (a) Determine resistors *R*1, *R*2, and *R*3 so that the circuit's input–output behavior follows - -the desired differential equation of *y*˙(*t*) − 1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*). - -- (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltage (first op-amp output voltage) is 2 volts. -- (c) Using transform-domain techniques, determine the impulse response *h*(*t*) of this circuit. -- (d) Determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *u*(*t* −2). -- **4.4-16** We desire the op-amp circuit of Fig. P4.4-16 to behave as \$ \$ *y*(*t*)+ 2 5 \$ *y*(*t*)+ 1 5 *y*(*t*) = \$ \$ *x*(*t*)− \$ *x*(*t*) - - (a) Determine the resistors *R*1, *R*2, and *R*3 to produce the desired behavior. - - (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltages (first two op-amp outputs) are each 1 volt. - -**4.4-17** (a) Using the initial and final value theorems, find the initial and final value of the zero-state response of a system with the transfer function - -$$ -H(s) = \frac{6s^2 + 3s + 10}{2s^2 + 6s + 5} -$$ - -and input *x*(*t*) = *u*(*t*). - -(b) Repeat part (a) for the input *x*(*t*) = *e*−*t u*(*t*). - -(c) Find y(0+) and y( -$$ -\infty -$$ -) if $Y(s) = \frac{s^2 + 5s + 6}{s^2 + 3s + 2}$ - -. - -- (d) Find *y*(0+) and *y*(∞) if *Y*(*s*) = *s*3 +4*s*2 +10*s*+7 *s*2 +2*s* +3 . -- **4.5-1** Consider two LTIC systems. The first has transfer function *H*1(*s*) = 2*s s*+1 , and the second has transfer function *H*2(*s*) = 1 *se*3(*s*−1) . - -**Figure P4.5-2** - -- (a) Determine the overall impulse response *h*s(*t*) if the two systems are connected in series. -- (b) Determine the overall impulse response *h*p(*t*) if the two systems are connected in parallel. -- **4.5-2** Figure P4.5-2a shows two resistive ladder segments. The transfer function of each segment (ratio of output to input voltage) is 1/2. Figure P4.5-2b shows these two segments connected in cascade. - - (a) Is the transfer function (ratio of output to input voltage) of this cascaded network (1/2)(1/2) = 1/4? - - (b) If your answer is affirmative, verify the answer by direct computation of the transfer function. Does this computation confirm the earlier value 1/4? If not, why? - - (c) Repeat the problem with *R*3 = *R*4 = 20 k. Does this result suggest the answer to the problem in part (b)? -- **4.5-3** In communication channels, transmitted signal is propagated simultaneously by several paths of varying lengths. This causes the signal to reach the destination with varying time delays and varying gains. Such a system generally distorts the received signal. For error-free communication, it is necessary to undo this distortion as - -For simplicity, let us assume that a signal is propagated by two paths whose time delays differ by τ seconds. The channel over the intended path has a delay of *T* seconds and unity gain. The signal over the unintended path has a delay of *T* + τ seconds and gain *a*. Such a channel can be modeled, as shown in Fig. P4.5-3. Find the inverse system transfer function to correct the delay distortion and show that the inverse system can be realized by a feedback system. The inverse system should be causal to be realizable. [*Hint:* We want to correct only the distortion caused by the relative delay τ seconds. For distortionless transmission, the signal may be delayed. What is important is to maintain the shape of *x*(*t*). Thus, a received signal of the form *c x*(*t* −*T*) is considered to be distortionless.] - -**4.5-4** Discuss BIBO stability of the feedback systems depicted in Fig. P4.5-4. For the system in - -**Figure P4.5-4** - -Fig. P4.5-4b, consider three cases: **(a)** *K* = 10, **(b)** *K* = 50, and **(c)** *K* = 48. - -**4.6-1** Realize - -$$ -H(s) = \frac{s(s+2)}{(s+1)(s+3)(s+4)} -$$ - -by canonic direct, series, and parallel forms. - -- **4.6-2** Realize the transfer function in Prob. 4.6-1 by using the transposed form of the realizations found in Prob. 4.6-1. -- **4.6-3** Repeat Prob. 4.6-1 for (a) *H*(*s*) = 3*s*(*s* +2) (*s* +1)(*s*2 +2*s*+2) (b) *H*(*s*) = 2*s* −4 (*s* +2)(*s*2 +4) -- **4.6-4** Realize the transfer functions in Prob. 4.6-3 by using the transposed form of the realizations found in Prob. 4.6-3. -- **4.6-5** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{2s+3}{5s(s+2)^2(s+3)} -$$ - -- **4.6-6** Realize the transfer function in Prob. 4.6-5 by using the transposed form of the realizations found in Prob. 4.6-5. -- **4.6-7** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{s(s+1)(s+2)}{(s+5)(s+6)(s+8)} -$$ - -- **4.6-8** Realize the transfer function in Prob. 4.6-7 by using the transposed form of the realizations found in Prob. 4.6-7. -- **4.6-9** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{s^3}{(s+1)^2(s+2)(s+3)} -$$ - -- **4.6-10** Realize the transfer function in Prob. 4.6-9 by using the transposed form of the realizations found in Prob. 4.6-9. -- **4.6-11** Repeat Prob. 4.6-1 for - -$$ -H(s) = \frac{s^3}{(s+1)(s^2+4s+13)} -$$ - -**4.6-12** Realize the transfer function in Prob. 4.6-11 by using the transposed form of the realizations found in Prob. 4.6-11. - -- **4.6-13** Draw a TDFII block realization of a causal LTIC system with transfer function *H*(*s*) = (*s*−2*j*)(*s*+2*j*) (*s*−*j*)(*s*+*j*)(*s*+2) . Give two reasons why TDFII tends to be a good structure. -- **4.6-14** Consider a causal LTIC system with transfer function *H*(*s*) = (*s*−2*j*)(*s*+2*j*)(*s*−3*j*)(*s*+3*j*) 9(*s*+1)(*s*+2)(*s*+1−*j*)(*s*+1+*j*) . - - (a) Realize *H*(*s*) using a single fourth-order real TDFII structure. Is this block realization unique? Explain. - - (b) Realize *H*(*s*) using a cascade of secondorder real DFII structures. Is this block realization unique? Explain. - - (c) Realize *H*(*s*) using a parallel connection of second-order real DFI structures. Is this block realization unique? Explain. -- **4.6-15** In this problem we show how a pair of complex conjugate poles may be realized by using a cascade of two first-order transfer functions and feedback. Show that the transfer functions of the block diagrams in Figs. P4.6-15a and P4.6-15b are: (a) - -$$ -H_a(s) = \frac{1}{(s+a)^2 + b^2} -$$ - -= -$$ -\frac{1}{s^2 + 2as + (a^2 + b^2)} -$$ - -(b) - -$$ -Hb(s) = \frac{s+a}{(s+a)^2 + b^2} -$$ - -= -$$ -\frac{s+a}{s^2 + 2as + (a^2 + b^2)} -$$ - -Hence, show that the transfer function of the block diagram in Fig. P4.6-15c is (c) - -$$ -H_c(s) = \frac{As + B}{(s+a)^2 + b^2} -$$ - -= -$$ -\frac{As + B}{s^2 + 2as + (a^2 + b^2)} -$$ - -**4.6-16** Show op-amp realizations of the following transfer functions: - -(a) -$$ -\frac{-10}{s+5} -$$ - -\n(b) $\frac{10}{s+5}$ -\n(c) $\frac{s+2}{s+5}$ - -(c) - -*b*2 - -**4.6-17** Show two different op-amp circuit realizations of the transfer function - -$$ -H(s) = \frac{s+2}{s+5} = 1 - \frac{3}{s+5} -$$ - -**4.6-18** Show an op-amp canonic direct realization of the transfer function - -$$ -H(s) = \frac{3s + 7}{s^2 + 4s + 10} -$$ - -**4.6-19** Show an op-amp canonic direct realization of the transfer function - -$$ -H(s) = \frac{s^2 + 5s + 2}{s^2 + 4s + 13} -$$ - -**4.6-20** Consider a system described by a constantcoefficient linear differential equation as *d dt y*(*t*) - -+ 2*y*(*t*) = *x*(*t*) − 3 *d dt x*(*t*). Draw an op-amp realization of this system if resistors and inductors are available but not capacitors. Would using inductors rather than capacitors in this circuit pose any problem? Explain. - -- **4.7-1** Feedback can be used to increase (or decrease) the system bandwidth. Consider the system in Fig. P4.7-1a with transfer function *G*(*s*) = ω*c*/(*s* +ω*c*). - - (a) Show that the 3 dB bandwidth of this system is ω*c* and the dc gain is unity; that is, |*H*(*j*0)| = 1. - - (b) To increase the bandwidth of this system, we use negative feedback with *H*(*s*) = 9, as depicted in Fig. P4.7-1b. Show that the 3 dB bandwidth of this system is 10ω*c*. What is the dc gain? - -- (c) To decrease the bandwidth of this system, we use positive feedback with *H*(*s*) = −0.9, as illustrated in Fig. P4.7-1c. Show that the 3 dB bandwidth of this system is ω*c*/10. What is the dc gain? -- (d) The system gain at dc times its 3 dB bandwidth is the *gain-bandwidth product* of a system. Show that this product is the same for all the three systems in Fig. P4.7-1. This result shows that if we increase the bandwidth, the gain decreases and vice versa. -- **4.8-1** Suppose an engineer builds a controllable, observable LTIC system with transfer function *H*(*s*) = *s*2+4 2*s*2+4*s*+4 . - - (a) By direct calculation, compute the magnitude response at frequencies ω = 0, 1, 2, 3, 5, 10, and ∞. Use these calculations to roughly sketch the magnitude response over 0 ≤ ω ≤ 10. - - (b) To test the system, the engineer connects a signal generator to the system in hopes to measure the magnitude response using a standard oscilloscope. What type of signal should the engineer input into the system to make the measurements? How should the engineer make the measurements? Provide sufficient detail to fully justify your answers. - - (c) Suppose the engineer accidentally constructs the system *H*−1(*s*)= 1 *H*(*s*) = 2*s*2+4*s*+4 *s*2+4 . What impact will this mistake have on his tests? -- **4.8-2** For an LTIC system described by the transfer function - -$$ -H(s) = \frac{s+2}{s^2 + 5s + 4} -$$ - -find the response to the following everlasting sinusoidal inputs: - -- (a) 5 cos(2*t* +30◦) -- (b) 10 sin(2*t* +45◦) -- (c) 10 cos(3*t* +40◦) - -Observe that these are everlasting sinusoids. - -**4.8-3** For an LTIC system described by the transfer function - -$$ -H(s) = \frac{s+3}{(s+2)^2} -$$ - -find the steady-state system response to the following inputs: - -- (a) 10*u*(*t*) -- (b) cos(2*t* +60◦)*u*(*t*) -- (c) sin(3*t* −45◦)*u*(*t*) -- (d) *ej*3*t u*(*t*) -- **4.8-4** For an allpass filter specified by the transfer function - -$$ -H(s) = \frac{-(s-10)}{s+10} -$$ - -find the system response to the following (everlasting) inputs: - -- (a) *ej*ω*t* -- (b) cos(ω*t* +θ ) -- (c) cos *t* -- (d) sin 2*t* -- (e) cos 10*t* -- (f) cos 100*t* - -Comment on the filter response. - -- **4.8-5** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.8-5. The dc response of this system is minus 1, *H*(*j*0) = −1. - - (a) Letting *H*(*s*) = *k*(*s*2 +*b*1*s*+*b*2)/(*s*2 +*a*1*s*+ *a*2), determine the constants *k*, *b*1, *b*2, *a*1, and *a*2. - - (b) What is the output *y*(*t*) of this system in response to the input *x*(*t*) = 4 + cos(*t*/2 + π/3)? - -### **Figure P4.8-5** - -- **4.8-6** Consider a CT system described by (*D*+1)(*D*+ 2){*y*(*t*)} = *x*(*t* − 1). Notice that this differential equation is in terms of *x*(*t* −1), not *x*(*t*)! - - (a) Determine the output *y*(*t*) given input *x*(*t*) = 1. - -- (b) Determine the output *y*(*t*) given input *x*(*t*) = cos(*t*). -- **4.8-7** An LTIC system has transfer function *H*(*s*) = 4*s s*2+2*s*+37 = 4*s* (*s*+1+6*j*)(*s*+1−6*j*). Determine the steady-state output in response to input *x*(*t*) = 1 3 *ej*(6*t*+π/3) *u*(6*t* +π/3). -- **4.9-1** Suppose a real first-order lowpass system *H*(*s*) has unity gain in the passband, one finite pole at *s* = −2, and one finite zero at an unspecified location. - - (a) Determine the location of the system zero so that the filter achieves 40 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response. - - (b) Determine the location of the system zero so that the filter achieves 30 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response. -- **4.9-2** Repeat Prob. 4.9-1 for a highpass rather than a lowpass system. -- **4.9-3** Repeat Prob. 4.9-1 for a second-order system that has a pair of repeated poles and a pair of repeated zeros. -- **4.9-4** Sketch Bode plots for the following transfer functions: - -(a) -$$ -\frac{s(s+100)}{(s+2)(s+20)} -$$ - -(b) -$$ -\frac{(s+10)(s+20)}{s^2(s+100)} -$$ - -(c) -$$ -\frac{(s+10)(s+200)}{(s+20)^2(s+1000)} -$$ - -$$ -4.9-5 \quad \text{Repeat Prob. } 4.9-4 \text{ for} -$$ - -(a) -$$ -\frac{s^2}{(s+1)(s^2+4s+16)} -$$ - -\n(b) -$$ -\frac{s}{(s+1)(s^2+14.14s+100)} -$$ - -\n(c) -$$ -\frac{(s+10)}{s(s^2+14.14s+100)} -$$ - -- **4.9-6** Using the lowest order possible, determine a system function *H*(*s*) with real-valued roots that matches the frequency response in Fig. P4.9-6. Verify your answer with MATLAB. -- **4.9-7** A graduate student recently implemented an analog phase lock loop (PLL) as part of his thesis. His PLL consists of four basic components: a phase/frequency detector, a charge pump, a loop filter, and a voltage-controlled oscillator. This problem considers only the loop filter, which is shown in Fig. P4.9-7a. The loop filter input is the current *x*(*t*), and the output is the voltage *y*(*t*). - - (a) Derive the loop filter's transfer function *H*(*s*). Express *H*(*s*) in standard form. - - (b) Figure P4.9-7b provides four possible frequency response plots, labeled A through D. Each log-log plot is drawn to the same scale, and line slopes are either 20 dB/decade, 0 dB/decade, or −20 dB/decade. Clearly identify which plot(s), if any, could represent the loop filter. - - (c) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for low-frequency inputs? - -**Figure P4.9-6** - -(d) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for high-frequency inputs? - -**Figure P4.9-7** - -- **4.10-1** A causal LTIC system *H*(*s*) = 2(*s*−4*j*)(*s*+4*j*) (*s*+1+2*j*)(*s*+1−2*j*) has input *x*(*t*) = −1 + 2 cos(2*t*) − 3 sin(4*t* + π/3) + 4 cos(10*t*). Below, perform accurate calculations at ω = 0, ±2, ±4, and ±10. - - (a) Using the graphical method of Sec. 4.10-1, accurately sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - - (b) Using the graphical method of Sec. 4.10-1, accurately sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10. - - (c) Approximate the system output *y*(*t*) in response to the input *x*(*t*). -- **4.10-2** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.10-2. The dc response of this system is minus 2, *H*(*j*0) = −2. - - (a) Letting *H*(*s*) = *k s*2+*b*1*s*+*b*2 *s*2+*a*1*s*+*a*2 , determine the constants *k*, *b*1, *b*2, *a*1, and *a*2. - -- (b) Using the graphical method of Sec. 4.10-1, hand-sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB. -- (c) Using the graphical method of Sec. 4.10-1, hand-sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB. -- (d) What is the output *y*(*t*) in response to input *x*(*t*) = −3+cos(3*t*+π/3)−sin(4*t*−π/8)? - -**Figure P4.10-2** - -**4.10-3** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of an LTIC system described by the transfer function - -$$ -H(s) = \frac{s^2 - 2s + 50}{s^2 + 2s + 50} -$$ - -= -$$ -\frac{(s - 1 - j7)(s - 1 + j7)}{(s + 1 - j7)(s + 1 + j7)} -$$ - -What kind of filter is this? - -**4.10-4** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of LTIC systems whose pole-zero plots are shown in Fig. P4.10-4. - -**Figure P4.10-4** - -- **4.10-5** A causal LTIC system *H*(*s*) = (*s*−3*j*)(*s*+3*j*) 3(*s*+2+*j*)(*s*+2−*j*) has input *x*(*t*) = cos(*t*) + sin(3*t* + π/3) + cos(100*t*). - - (a) Using the graphical method of Sec. 4.10-1, sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - - (b) Determine the system output *y*(*t*) in response to the input *x*(*t*). - - (c) Suppose we create a second causal system with transfer function *H*2(*s*) = *H*(−*s*). Sketch this system's pole/zero plot. What is the response *y*2(*t*) of system *H*2(*s*) to input *x*(*t*)? -- **4.10-6** Design a second-order bandpass filter with center frequency ω = 10. The gain should be zero at ω = 0 and at ω = ∞. Select poles at −*a* ± *j*10. Leave your answer in terms of *a*. Explain the influence of *a* on the frequency response. -- **4.10-7** The LTIC system described by *H*(*s*) = (*s*−1)/(*s* +1) has unity magnitude response |*H*(*j*ω)| = 1. Positive Pat claims that the output *y*(*t*) of this system is equal the input *x*(*t*), since the system is allpass. Cynical Cynthia doesn't think so. "This is *signals and systems* class," she complains. "It *has* to be more complicated!" Who is correct, Pat or Cynthia? Justify your answer. -- **4.10-8** Two students, Amy and Jeff, disagree about an analog system function given by *H*1(*s*) = *s*. Sensible Jeff claims the system has a zero at *s* = 0. Rebellious Amy, however, notes that the system function can be rewritten as *H*1(*s*) = 1/*s*−1 and claims that this implies a system pole - -at *s* = ∞. Who is correct? Why? What are the poles and zeros of the system *H*2(*s*) = 1/*s*? - -- **4.10-9** A rational transfer function *H*(*s*) is often used to represent an analog filter. Why must *H*(*s*) be strictly proper for lowpass and bandpass filters? Why must *H*(*s*) be proper for highpass and bandstop filters? -- **4.10-10** For a given filter order *N*, why is the stopband attenuation rate of an all-pole lowpass filter better than filters with finite zeros? -- **4.10-11** Is it possible, with real coefficients ([*k*,*b*1,*b*2,*a*1,*a*2] ∈ *R*), for a system - -$$ -H(s) = k \frac{s^2 + b_1 s + b_2}{s^2 + a_1 s + a_2} -$$ - -to function as a lowpass filter? Explain your answer. - -- **4.10-12** Nick recently built a simple second-order Butterworth lowpass filter for his home stereo. Although the system performs fairly well, Nick is an overachiever and hopes to improve the system performance. Unfortunately, Nick is lazy and doesn't want to design another filter. Thinking "Twice the filtering gives twice the performance," he suggests filtering the audio signal not once but twice with a cascade of two identical filters. His overworked, underpaid signals professor is skeptical and states, "If you are using *identical* filters, it makes no difference whether you filter once or twice!" Who is correct? Why? -- **4.10-13** An LTIC system impulse response is given by *h*(*t*) = *u*(*t*)−*u*(*t* −1). - - (a) Determine the transfer function *H*(*s*). Using *H*(*s*), determine and plot the magnitude response |*H*(*j*ω)|. Which type of filter most accurately describes the behavior of this system: lowpass, highpass, bandpass, or bandstop? - - (b) What are the poles and zeros of *H*(*s*)? Explain your answer. - - (c) Can you determine the impulse response of the inverse system? If so, provide it. If not, suggest a method that could be used to approximate the impulse response of the inverse system. -- **4.10-14** An ideal lowpass filter *H*LP(*s*) has magnitude response that is unity for low frequencies - -and zero for high frequencies. An ideal highpass filter *H*HP(*s*) has an opposite magnitude response: zero for low frequencies and unity for high frequencies. A student suggests a possible lowpass-to-highpass filter transformation: *H*HP(*s*) = 1 − *H*LP(*s*). In general, will this transformation work? Explain your answer. - -- **4.10-15** An LTIC system has a rational transfer function *H*(*s*). When appropriate, assume that all initial conditions are zero. - - (a) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = cos(100π*t*)*u*(*t*)? Explain. - - (b) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = sin(50π*t*)*u*(*t*)? Explain. - - (c) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = cos(100π*t*)? Explain. - - (d) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = sin(50π*t*)? Explain. -- **4.11-1** Find the ROC, if it exists, of the (bilateral) Laplace transform of the following signals: - - (a) *etu*(*t*) (b) *e*−*tu*(*t*) - -(c) -$$ -\frac{1}{1+t^2} -$$ - -(d) -$$ -\frac{1}{1+e^t} -$$ - -(e) $e^{-kt^2}$ - -- **4.11-2** Using the definition and direct integration, find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals. If the Laplace transform does not exist, carefully explain why. - - (a) *x*a(*t*) = *e*(−1−*j*)*t u*(1−*t*) - - (b) *x*b(*t*) = *j* (*t*+1) *u*(−*t* −1) - - (c) *x*c(*t*) = *ej*π/3*u*(2−*t*) +*j*δ(*t* −5) - - (d) *x*d(*t*) = 1+1 = 2 - - (e) *x*e(*t*) = 3*u*(−*t*) +*e*−2*t* [*u*(*t*)−*u*(*t* −10)] - - (f) *x*f(*t*) = *et*−2*u*(1−*t*) +*e*−2*t u*(*t* +1) -- **4.11-3** Determine the bilateral Laplace transform *X*(*s*) of the signal - -$$ -x(t) = \left[e^t u(-t)\right] * \left[t\cos(2t)u(t)\right] -$$ - -**4.11-4** A signal has bilateral Laplace transform *X*(*s*) = (*s*+1) (*s*−2)(*s*−3) but unknown region of convergence. What ROC results in the smallest maximum amplitude of *x*(*t*)? Justify your answer. - -- **4.11-5** Find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals: - - (a) *e*−|*t*| - - (b) *e*−|*t*| cos *t* - -(c) -$$ -e^t u(t) + e^{2t} u(-t) -$$ - -(d) $e^{-tu(t)}$ - -$$ -\begin{array}{cc}\n(e) & e^{-tu(-t)} \\ -(e) & e^{-tu(-t)}\n\end{array} -$$ - -- (f) cosω0*t u*(*t*) +*et u*(−*t*) -- **4.11-6** Find the inverse (bilateral) Laplace transforms of the following functions: - -(a) -$$ -\frac{2s+5}{(s+2)(s+3)} -$$ - -\n -$$ --3 < \sigma < -2 -$$ - -\n(b) -$$ -\frac{2s-5}{(s-2)(s-3)} -$$ - -\n -$$ -2 < \sigma < 3 -$$ - -\n(c) -$$ -\frac{2s+3}{(s+1)(s+2)} -$$ - -\n -$$ -\sigma > -1 -$$ - -\n(d) -$$ -\frac{2s+3}{(s+1)(s+2)} -$$ - -\n -$$ -\sigma < -2 -$$ - -(d) -$$ -\frac{2s+6}{(s+1)(s+2)} \quad \sigma < -2 -$$ - -(e) -$$ -\frac{3s^2 - 2s - 17}{(s+1)(s+3)(s-5)} \quad -1 < \sigma < 5 -$$ - -**4.11-7** Find - -$$ -\mathcal{L}^{-1}\left[\frac{2s^2 - 2s - 6}{(s+1)(s-1)(s+2)}\right] -$$ - -if the ROC is - -- (a) Re *s* > 1 -- (b) Re *s* < −2 -- (c) −1 < Re*s* < 1 -- (d) −2 < Re*s* < −1 -- **4.11-8** For a causal LTIC system having a transfer function *H*(*s*) = 1/(*s*+1), find the output *y*(*t*) if the input *x*(*t*) is given by (a) *e*−|*t*|/2 (b) *et u*(*t*)+*e*2*t u*(−*t*) (c) *e*−*t*/2*u*(*t*) +*e*−*t*/4*u*(−*t*) (d) *e*2*t u*(*t*)+*et u*(−*t*) (e) *e*−*t*/4*u*(*t*) +*e*−*t*/2*u*(−*t*) - - (f) *e*−3*t u*(*t*)+*e*−2*t u*(−*t*) -- **4.11-9** The autocorrelation function *rxx*(*t*) of a signal *x*(*t*) is given by - -$$ -r_{xx}(t) = \int_{-\infty}^{\infty} x(\tau) x(\tau + t) d\tau -$$ - -Derive an expression for *Rxx*(*s*) = *L*(*rxx*(*t*)) in terms of *X*(*s*), where *X*(*s*) = *L*(*x*(*t*)). - -**4.11-10** Determine the inverse Laplace transform of - -$$ -X(s) = \frac{2}{s} + \frac{s}{2} -$$ - -given that the region of convergence is σ < 0. - -**4.11-11** An absolutely integrable signal *x*(*t*) has a pole at *s* = π. It is possible that other poles may be present. Recall that an absolutely integrable signal satisfies - -$$ -\int_{-\infty}^{\infty} |x(t)| dt < \infty -$$ - -- (a) Can *x*(*t*) be left-sided? Explain. -- (b) Can *x*(*t*) be right-sided? Explain. -- (c) Can *x*(*t*) be two-sided? Explain. -- (d) Can *x*(*t*) be of finite duration? Explain. -- **4.11-12** Using ROC σ < 0, determine the inverse Laplace transform of *X*(*s*) = *s d ds e*−2*s s* . [*Hint:* Use Laplace transform properties to avoid tedious calculus.] -- **4.11-13** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal - -$$ -X(s) = \frac{2}{e^s} + \frac{1}{s} \left[ e^s \frac{4}{\frac{s}{3} + 2} \right] -$$ - -where the ROC is −6 < Re{*s*} < 0. - -**4.11-14** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal - -$$ -X(s) = \frac{d^7}{ds^7} \left[ \frac{e^{-4s}}{(s+2)(s+3)} \right] -$$ - -where the ROC is −3 < Re{*s*} < −2. - -**4.11-15** Using the definition, compute the bilateral Laplace transform, including the region of convergence (ROC), of the following complexvalued functions: - -(a) -$$ -x_1(t) = (j + e^{jt})u(t) -$$ - -- (b) *x*2(*t*) = *j* cosh(*t*)*u*(−*t*) -- (c) *x*3(*t*) = *ej*( π 4 ) *u*(−*t* +1)+*j*δ(*t* −5) -- (d) *x*4(*t*) = *j t u*(−*t*) +δ(*t* −π ) - -**4.11-16** A bounded-amplitude signal *x*(*t*) has bilateral Laplace transform *X*(*s*) given by - -$$ -X(s) = \frac{s2^s}{(s-1)(s+1)} -$$ - -- (a) Determine the corresponding region of convergence. -- (b) Determine the time-domain signal *x*(*t*). -- **4.12-1** Express the polynomial *C*20(*x*) in standard form. That is, determine the coefficients % *ak* of *C*20(*x*)= 20 *k*=0 *akx*20−*k*. -- **4.12-2** Consider an LTIC system with - -$$ -H(s) = \frac{1}{s^3 + 4s^2 + 8s + 8} = \frac{1}{(s^2 + 2s + 4)(s + 2)} -$$ - -= -$$ -\frac{1}{(s - 2e^{j2\pi/3})(s - 2e^{-j2\pi/3})(s + 2)}. -$$ - -- (a) Write MATLAB code that accurately plots the system magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. -- (b) Write MATLAB code that accurately plots the system phase response over −10 ≤ ω ≤ 10. -- (c) Determine the max value *y*max of output *y*(*t*) in response to input *x*(*t*) = 2−sin(2*t*+π/3). -- (d) Draw a parallel representation of this system using real DFI structures of order 2 or less. [*Hint:* Use MATLAB to perform a partial fraction expansion of *H*(*s*).] -- **4.12-3** Consider the op-amp circuit of Fig. P4.12-3. Further, let *RC* = 1. - - (a) From Fig. P4.12-3, determine the (simplified, standard form, rational) transfer function *H*(*s*). - - (b) Use MATLAB to accurately plot |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - - (c) Determine the output of this system in response to *x*(*t*) = cos(10*t*)−1. - - (d) The circuit of Fig. P4.12-3 contains two capacitors. Suppose one capacitor must be a 25% tolerance part, while the other must be a 10% tolerance part. If the goal is to preserve the original magnitude response, should you use the 25% tolerance capacitor with the first op-amp or the second op-amp? Justify your answer with appropriate MAT-LAB simulations. - -**Figure P4.12-3** - -- **4.12-4** Design an order-12 Butterworth lowpass filter with a cutoff frequency of ω*c* = 2π5000 by completing the following. - - (a) Locate and plot the filter's poles and zeros in the complex plane. Plot the corresponding magnitude response |*H*LP(*j*ω)| to verify proper design. - - (b) Setting all resistor values to 100,000, determine the capacitor values to implement the filter using a cascade of six second-order Sallen–Key circuit sections. The form of a Sallen–Key stage is shown in Fig. P4.12-4. On a single plot, plot the magnitude response of each section as well as the overall magnitude response. Identify the poles that correspond to each section's magnitude response curve. Are the capacitor values realistic? -- **4.12-5** Rather than a Butterworth filter, repeat Prob. 4.12-4 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each Sallen–Key stage is constrained to have unity gain at dc, an overall gain error of 1/ √ 1+ 2 is acceptable. -- **4.12-6** An analog lowpass filter with cutoff frequency ω*c* can be transformed into a highpass filter - -with cutoff frequency ω*c* by using an *RC*–*CR* transformation rule: each resistor *Ri* is replaced by a capacitor *C i* = 1/*Ri*ω*c* and each capacitor *Ci* is replaced by a resistor *R i* = 1/*Ci*ω*c*. - -Use this rule to design an order-8 Butterworth highpass filter with ω*c* = 2π4000 by completing the following. - -- (a) Design an order-8 Butterworth lowpass filter with ω*c* = 2π4000 by using four second-order Sallen–Key circuit stages, the form of which is shown in Fig. P4.12-4. Give resistor and capacitor values for each stage. Choose the resistors so that the *RC*–*CR* transformation will result in 1 nF capacitors. At this point, are the component values realistic? -- (b) Draw an *RC*–*CR* transformed Sallen–Key circuit stage. Determine the transfer function *H*(*s*) of the transformed stage in terms of the variables *R* 1, *R* 2, *C* 1, and *C* 2. -- (c) Transform the LPF designed in part (a) by using an *RC*–*CR* transformation. Give the resistor and capacitor values for each stage. Are the component values realistic? Using *H*(*s*) derived in part (b), plot the - -magnitude response of each section as well - -as the overall magnitude response. Does the overall response look like a highpass Butterworth filter? - -Plot the HPF system poles and zeros in the complex *s* plane. How do these locations compare with those of the Butterworth LPF? - -- **4.12-7** Repeat Prob. 4.12-6, using ω*c* = 2π1500 and an order-16 filter. That is, eight second-order stages need to be designed. -- **4.12-8** Rather than a Butterworth filter, repeat Prob. 4.12-6 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each transformed Sallen–Key stage is constrained to have unity gain at ω = ∞, an overall gain error of 1/ √ 1+ 2 is acceptable. -- **4.12-9** The MATLAB signal-processing toolbox function butter helps design analog Butterworth filters. Use MATLAB help to learn how butter works. For each of the following cases, design the filter, plot the filter's poles and zeros in the complex *s* plane, and plot the decibel magnitude response 20log10 |*H*(*j*ω)|: - - (a) Design a sixth-order analog lowpass filter with ω*c* = 2π3500. - - (b) Design a sixth-order analog highpass filter with ω*c* = 2π3500. - - (c) Design a sixth-order analog bandpass filter with a passband between 2 and 4 kHz. - - (d) Design a sixth-order analog bandstop filter with a stopband between 2 and 4 kHz. -- **4.12-10** The MATLAB signal-processing toolbox function cheby1 helps design analog Chebyshev - -type I filters. A Chebyshev type I filter has a passband ripple and a smooth stopband. Setting the passband ripple to *Rp* = 3 dB, repeat Prob. 4.12-9 using the cheby1 command. With all other parameters held constant, what is the general effect of reducing *Rp*, the allowable passband ripple? - -- **4.12-11** The MATLAB signal-processing toolbox function cheby2 helps design analog Chebyshev type II filters. A Chebyshev type II filter has a smooth passband and ripple in the stopband. Setting the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the cheby2 command. With all other parameters held constant, what is the general effect of increasing *Rs*, the minimum stopband attenuation? -- **4.12-12** The MATLAB signal-processing toolbox function ellip helps design analog elliptic filters. An elliptic filter has ripple in both the passband and the stopband. Setting the passband ripple to *Rp* = 3 dB and the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the ellip command. -- **4.12-13** Using the definition *CN*(*x*)=cosh(*N* cosh−1(*x*)), prove the recursive relation *CN*(*x*) = 2*xCN*−1(*x*) −*CN*−2(*x*). -- **4.12-14** Prove that the poles of a Chebyshev filter, which are located at *pk* = ω*c* sinh(ξ )sin(φ*k*) + *j*ω*c* cosh(ξ ) cos(φ*k*), lie on an ellipse. [*Hint:* The equation of an ellipse in the *x*–*y* plane is (*x*/*a*) 2+ (*y*/*b*) 2 = 1, where constants *a* and *b* define the major and minor axes of the ellipse.] - -# **[DISCRETE-TIME](#page-11-0) SYSTEM ANALYSIS USING THE** *z***-TRANSFORM** - -The counterpart of the Laplace transform for discrete-time systems is the *z*-transform. The Laplace transform converts integro-differential equations into algebraic equations. In the same way, the *z*-transforms changes difference equations into algebraic equations, thereby simplifying the analysis of discrete-time systems. The *z*-transform method of analysis of discrete-time systems parallels the Laplace transform method of analysis of continuous-time systems, with some minor differences. In fact, we shall see that *the z-transform is the Laplace transform in disguise*. - -The behavior of discrete-time systems is similar to that of continuous-time systems (with some differences). The frequency-domain analysis of discrete-time systems is based on the fact (proved in Sec. 3.8-2) that the response of a linear, time-invariant, discrete-time (LTID) system to an everlasting exponential *zn* is the same exponential (within a multiplicative constant) given by *H*[*z*]*zn*. We then express an input *x*[*n*] as a sum of (everlasting) exponentials of the form *zn*. The system response to *x*[*n*] is then found as a sum of the system's responses to all these exponential components. The tool that allows us to represent an arbitrary input *x*[*n*] as a sum of (everlasting) exponentials of the form *zn* is the *z*-transform. - -## **5.1 THE** *z***[-TRANSFORM](#page-11-0)** - -We define *X*[*z*], the direct *z*-transform of *x*[*n*], as - -$$ -X[z] = \sum_{n = -\infty}^{\infty} x[n]z^{-n} -$$ - (5.1) - -where *z* is a complex variable. The signal *x*[*n*], which is the inverse *z*-transform of *X*[*z*], can be obtained from *X*[*z*] by using the following inverse *z*-transformation: - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ -\n(5.2) - -The symbol 6 indicates an integration in counterclockwise direction around a closed path in the complex plane (see Fig. 5.1). We derive this *z*-transform pair later, in Ch. 9, as an extension of the discrete-time Fourier transform pair. - -CHAPTER - -**5** - -As in the case of the Laplace transform, we need not worry about this integral at this point because inverse *z*-transforms of many signals of engineering interest can be found in a *z*-transform table. The direct and inverse *z*-transforms can be expressed symbolically as - -$$ -X[z] = \mathcal{Z}\{x[n]\} \qquad \text{and} \qquad x[n] = \mathcal{Z}^{-1}\{X[z]\} -$$ - -or simply as - -*x*[*n*] ⇐⇒ *X*[*z*] - -Note that - -*Z*−1 [*Z*{*x*[*n*]}] = *x*[*n*] and *Z*[*Z*−1 {*X*[*z*]}] = *X*[*z*] - -### LINEARITY OF THE *z*-TRANSFORM - -Like the Laplace transform, the *z*-transform is a linear operator. If - -*x*1[*n*] ⇐⇒ *X*1[*z*] and *x*2[*n*] ⇐⇒ *X*2[*z*] - -then - -*a*1*x*1[*n*] +*a*2*x*2[*n*] ⇐⇒ *a*1*X*1[*z*] +*a*2*X*2[*z*] - -The proof is trivial and follows from the definition of the *z*-transform. This result can be extended to finite sums. - -### THE UNILATERAL *z*-TRANSFORM - -For the same reasons discussed in Ch. 4, we find it convenient to consider the unilateral *z*-transform. As seen for the Laplace case, the bilateral transform has some complications because of the non-uniqueness of the inverse transform. In contrast, the unilateral transform has a unique inverse. This fact simplifies the analysis problem considerably, but at a price: the unilateral version can handle only causal signals and systems. Fortunately, most of the practical cases are causal. The more general *bilateral z-transform* is discussed later, in Sec. 5.8. In practice, the term *z-transform* generally means *the unilateral z-transform*. - -In a basic sense, there is no difference between the unilateral and the bilateral *z*-transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at *n* = 0 (causal signals). Hence, the definition of the unilateral transform is the same as that of the bilateral [Eq. (5.1)], except that the limits of the sum are from 0 to ∞: - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - (5.3) - -The expression for the inverse *z*-transform in Eq. (5.2) remains valid for the unilateral case also. - -## THE REGION OF CONVERGENCE (ROC) OF *X*[*z*] - -The sum in Eq. (5.1) [or Eq. (5.3)] defining the direct *z*-transform *X*[*z*] may not converge (exist) for all values of *z*. The values of *z* (the region in the complex plane) for which the sum in Eq. (5.1) converges (or exists) are called the *region of existence,* or more commonly the *region of convergence* (ROC), for *X*[*z*]. This concept will become clear in the following example. - -### **EXAMPLE 5.1 Bilateral** *z***-Transform of a Causal Exponential** - -Find the *z*-transform and the corresponding ROC for the signal γ *nu*[*n*]. - -By definition, - -$$ -X[z] = \sum_{n=0}^{\infty} \gamma^n u[n] z^{-n} -$$ - -Since *u*[*n*] = 1 for all *n* ≥ 0, - -$$ -X[z] = \sum_{n=0}^{\infty} \left(\frac{\gamma}{z}\right)^n = 1 + \left(\frac{\gamma}{z}\right) + \left(\frac{\gamma}{z}\right)^2 + \left(\frac{\gamma}{z}\right)^3 + \dots + \dots -$$ - (5.4) - -It is helpful to remember the geometric progression and its sum [see Sec. B.8-3]: - -$$ -1 + x + x2 + x3 + \dots = \frac{1}{1 - x} \quad \text{if} \quad |x| < 1 -$$ - -Applying this relationship to Eq. (5.4) yields - -$$ -X[z] = \frac{1}{1 - \frac{\gamma}{z}} \qquad \left| \frac{\gamma}{z} \right| < 1 -$$ -\n -$$ -= \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.5} -$$ - -Observe that *X*[*z*] exists only for |*z*| > |γ |. For |*z*| < |γ |, the sum in Eq. (5.4) does not converge; it goes to infinity. Therefore, the ROC of *X*[*z*] is the shaded region outside the circle of radius |γ |, centered at the origin, in the *z*-plane, as depicted in Fig. 5.1b. - -**Figure 5.1** γ *nu*[*n*] and the region of convergence of its *z*-transform. - -Later in Eq. (5.52), we show that the *z*-transform of another signal, −γ *nu*[−(*n* + 1)], is also *z*/(*z* − γ ). However, the ROC in this case is |*z*| < |γ |. Clearly, the inverse *z*-transform of *z*/(*z* − γ ) is not unique. However, if we restrict the inverse transform to be causal, then the inverse transform is unique, namely, γ *nu*[*n*]. - -The ROC is required for evaluating *x*[*n*] from *X*[*z*], according to Eq. (5.2). The integral in Eq. (5.2) is a contour integral, implying integration in a counterclockwise direction along a closed path centered at the origin and satisfying the condition |*z*| > |γ |. Thus, any circular path centered at the origin and with a radius greater than |γ | (Fig. 5.1b) will suffice. We can show that the integral in Eq. (5.2) along any such path (with a radius greater than |γ |) yields the same result, namely, *x*[*n*]. † Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of *z*-transforms (Table 5.1), where *z*-transform pairs are tabulated for a variety of signals. To find the inverse *z*-transform of say, *z*/(*z* − γ ), instead of using the complex integration in Eq. (5.2), we consult the table and find the inverse *z*-transform of *z*/(*z*−γ ) as γ *nu*[*n*]. Because of the uniqueness property of the unilateral *z*-transform, there is only one inverse for each *X*[*z*]. Although the table given here is rather short, it comprises the functions of most practical interest. - -The situation of the *z*-transform regarding the uniqueness of the inverse transform is parallel to that of the Laplace transform. For the bilateral case, the inverse *z*-transform is not unique unless the ROC is specified. For the unilateral case, the inverse transform is unique; the region of convergence need not be specified to determine the inverse *z*-transform. For this reason, we shall ignore the ROC in the unilateral *z*-transform Table 5.1. - -## EXISTENCE OF THE *z*-TRANSFORM - -By definition, - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} = \sum_{n=0}^{\infty} \frac{x[n]}{z^n} -$$ - -The existence of the *z*-transform is guaranteed if - -$$ -|X[z]| \leq \sum_{n=0}^{\infty} \frac{|x[n]|}{|z|^n} < \infty -$$ - -for some |*z*|. Any signal *x*[*n*] that grows no faster than an exponential signal *rn* 0, for some *r*0, satisfies this condition. Thus, if - -$$ -|x[n]| \le r_0^n \qquad \text{for some } r_0 \tag{5.6} -$$ - -† Indeed, the path need not even be circular. It can have any odd shape, as long as it encloses the pole(s) of *X*[*z*] and the path of integration is counterclockwise. - -| No. | x[n] | X[z] | -|-----|----------------------------------------------------|-----------------------------------------------------------------| -| 1 | δ[n−k] | z−k | -| 2 | u[n] | z
z−1 | -| 3 | nu[n] | z
(z−1)2 | -| 4 | n2u[n] | z(z+1)
(z−1)3 | -| 5 | n3u[n] | z(z2 +4z+1)
(z−1)4 | -| 6 | γ nu[n] | z
z−γ | -| 7 | γ n−1u[n−1] | 1
z−γ | -| 8 | nγ nu[n] | γ z
(z−γ )2 | -| 9 | n2γ
nu[n] | γ z(z+γ )
(z−γ )3 | -| 10 | n(n−1)(n−2)···(n−m+1)
γ nu[n]
γ mm! | z
(z−γ )m+1 | -| 11a | n cos
γ
βn u[n] | z(z− γ cos β)
z2 −(2 γ
2
cos β)z+ γ | -| 11b | n sin
γ
βn u[n] | z γ sin β
z2 −(2 γ
2
cos β)z+ γ | -| 12a | n cos(βn+θ
r γ
)u[n] | rz[z cos θ − γ cos(β −θ )]
z2 −(2 γ
2
cos β)z+ γ | -| 12b | n cos(βn+θ
γ = γ ejβ
r γ
)u[n] | (0.5rejθ )z
(0.5re−jθ )z
+
z−γ
z−γ ∗ | -| 12c | n cos(βn+θ
r γ
)u[n] | z(Az+B)
z2 +2az+ γ
2
| -| |
A2 γ
2 +B2 −2AaB

r =
2 −a2
γ | | -| | β = cos−1 −a
γ | | -| | Aa−B
θ = tan−1

2 −a2
A
γ | | - -**TABLE 5.1** Select (Unilateral) *z*-Transform Pairs - -$$ -|X[z]| \le \sum_{n=0}^{\infty} \left(\frac{r_0}{|z|}\right)^n = \frac{1}{1 - \frac{r_0}{|z|}} \qquad |z| > r_0 -$$ - -Therefore, *X*[*z*] exists for |*z*| > *r*0. Almost all practical signals satisfy Eq. (5.6) and are therefore *z*-transformable. Some signal models (e.g., γ *n*2 ) grow faster than the exponential signal *rn* 0 (for any *r*0) and do not satisfy Eq. (5.6) and therefore are not *z*-transformable. Fortunately, such signals are of little practical or theoretical interest. Even such signals over a finite interval are *z*-transformable. - -then - -This geometric sum simplifies [see Sec. B.8-3] to - -$$ -X[z] = \frac{1}{1 - \frac{1}{z}} \qquad \left| \frac{1}{z} \right| < 1 -$$ -\n -$$ -= \frac{z}{z - 1} \qquad |z| > 1 -$$ - -Therefore, - -$$ -u[n] \Longleftrightarrow \frac{z}{z-1} \qquad |z| > 1 -$$ - -**(c)** Recall that cos β*n* = (*ej*β*n* +*e*−*j*β*n*)/2. Moreover, according to Eq. (5.5), - -$$ -e^{\pm j\beta n}u[n] \Longleftrightarrow \frac{z}{z - e^{\pm j\beta}} \qquad |z| > |e^{\pm j\beta}| = 1 -$$ - -Therefore, - -$$ -X[z] = \frac{1}{2} \left[ \frac{z}{z - e^{j\beta}} + \frac{z}{z - e^{-j\beta}} \right] = \frac{z(z - \cos \beta)}{z^2 - 2z \cos \beta + 1} \qquad |z| > 1 -$$ - -**(d)** Here *x*[0] = *x*[1] = *x*[2] = *x*[3] = *x*[4] = 1 and *x*[5] = *x*[6]=···= 0. Therefore, according to Eq. (5.7), - -$$ -X[z] = 1 + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + \frac{1}{z^4} = \frac{z^4 + z^3 + z^2 + z + 1}{z^4} -$$ - for all $z \neq 0$ - -We can also express this result in a more compact form by summing the geometric progression on the right-hand side of the foregoing equation. From the result in Sec. B.8-3 with *r* =1/*z*,*m*= 0, and *n* = 4, we obtain - -$$ -X[z] = \frac{\left(\frac{1}{z}\right)^5 - \left(\frac{1}{z}\right)^0}{\frac{1}{z} - 1} = \frac{z}{z - 1} (1 - z^{-5}) -$$ - -### **DR ILL 5.1 Bilateral** *z***-Transform** - -- **(a)** Find the *z*-transform of a signal shown in Fig. 5.3. -- **(b)** Use pair 12a (Table 5.1) to find the *z*-transform of *x*[*n*] = 20.65( √2)*n* cos[(π/4)*n* − 1.415]*u*[*n*]. - - - -### **[5.1-1 Inverse Transform by Partial Fraction Expansion and Tables](#page-11-0)** - -As in the Laplace transform, we shall avoid the integration in the complex plane required to find the inverse *z*-transform [Eq. (5.2)] by using the (unilateral) transform table (Table 5.1). Many of the transforms *X*[*z*] of practical interest are rational functions (ratio of polynomials in *z*), which can be expressed as a sum of partial fractions, whose inverse transforms can be readily found in a table of transform. The partial fraction method works because for every transformable *x*[*n*] defined for *n* ≥ 0, there is a corresponding unique *X*[*z*] defined for |*z*| > *r*0 (where *r*0 is some constant), and vice versa. - -### **EXAMPLE 5.3 Inverse** *z***-Transform by Partial Fraction Expansion** - -Find the inverse *z*-transforms of - -(a) -$$ -\frac{8z-19}{(z-2)(z-3)} -$$ - -\n(b) -$$ -\frac{z(2z^2-11z+12)}{(z-1)(z-2)^3} -$$ - -\n(c) -$$ -\frac{2z(3z+17)}{(z-1)(z^2-6z+25)} -$$ - -**(a)** Expanding *X*[*z*] into partial fractions yields - -$$ -X[z] = \frac{8z - 19}{(z - 2)(z - 3)} = \frac{3}{z - 2} + \frac{5}{z - 3} -$$ - -### 496 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -From Table 5.1, pair 7, we obtain - -$$ -x[n] = [3(2)^{n-1} + 5(3)^{n-1}]u[n-1] -$$ -\n(5.8) - -If we expand rational *X*[*z*] into partial fractions directly, we shall always obtain an answer that is multiplied by *u*[*n* − 1] because of the nature of pair 7 in Table 5.1. This form is rather awkward as well as inconvenient. We prefer the form that contains *u*[*n*] rather than *u*[*n*−1]. A glance at Table 5.1 shows that the *z*-transform of every signal that is multiplied by *u*[*n*] has a factor *z* in the numerator. This observation suggests that we expand *X*[*z*] into *modified partial fractions*, where each term has a factor *z* in the numerator. This goal can be accomplished by expanding *X*[*z*]/*z* into partial fractions and then multiplying both sides by *z*. We shall demonstrate this procedure by reworking part (a). For this case, - -$$ -\frac{X[z]}{z} = \frac{8z - 19}{z(z - 2)(z - 3)} = \frac{(-19/6)}{z} + \frac{(3/2)}{z - 2} + \frac{(5/3)}{z - 3} -$$ - -Multiplying both sides by *z* yields - -$$ -X[z] = -\frac{19}{6} + \frac{3}{2} \left( \frac{z}{z-2} \right) + \frac{5}{3} \left( \frac{z}{z-3} \right) -$$ - -From pairs 1 and 6 in Table 5.1, it follows that - -$$ -x[n] = -\frac{19}{6}\delta[n] + \left[\frac{3}{2}(2)^n + \frac{5}{3}(3)^n\right]u[n] \tag{5.9} -$$ - -The reader can verify that this answer is equivalent to that in Eq. (5.8) by computing *x*[*n*] in both cases for *n* = 0, 1, 2, 3,..., and comparing the results. The form in Eq. (5.9) is more convenient than that in Eq. (5.8). For this reason, we shall always expand *X*[*z*]/*z* rather than *X*[*z*] into partial fractions and then multiply both sides by *z* to obtain modified partial fractions of *X*[*z*], which have a factor *z* in the numerator. - -**(b)** - -$$ -X[z] = \frac{z(2z^2 - 11z + 12)}{(z - 1)(z - 2)^3} -$$ - -and - -$$ -\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{k}{z - 1} + \frac{a_0}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)} -$$ - -where - -$$ -k = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=1} = -3 -$$ - -$$ -a_0 = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=2} = -2 -$$ - -Therefore, - -$$ -\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{-3}{z - 1} - \frac{2}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)} -$$ -(5.10) - -We can determine *a*1 and *a*2 by clearing fractions. Or we may use a shortcut. For example, to determine *a*2, we multiply both sides of Eq. (5.10) by *z* and let *z* → ∞. This yields - -$$ -0 = -3 - 0 + 0 + a_2 \implies a_2 = 3 -$$ - -This result leaves only one unknown, *a*1, which is readily determined by letting *z* take any convenient value, say, *z* = 0, on both sides of Eq. (5.10). This produces - -$$ -\frac{12}{8} = 3 + \frac{1}{4} + \frac{a_1}{4} - \frac{3}{2} -$$ - -which yields *a*1 = −1. Therefore, - -$$ -\frac{X[z]}{z} = \frac{-3}{z-1} - \frac{2}{(z-2)^3} - \frac{1}{(z-2)^2} + \frac{3}{z-2} -$$ - -and - -$$ -X[z] = -3\frac{z}{z-1} - 2\frac{z}{(z-2)^3} - \frac{z}{(z-2)^2} + 3\frac{z}{z-2} -$$ - -Now the use of Table 5.1, pairs 6 and 10, yields - -$$ -x[n] = \left[ -3 - 2\frac{n(n-1)}{8} (2)^n - \frac{n}{2} (2)^n + 3(2)^n \right] u[n] -$$ - -= -$$ -- \left[ 3 + \frac{1}{4} (n^2 + n - 12) 2^n \right] u[n] -$$ - -**(c) Complex Poles.** - -$$ -X[z] = \frac{2z(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2z(3z+17)}{(z-1)(z-3-j4)(z-3+j4)} -$$ - -The poles of *X*[*z*] are 1, 3 + *j*4, and 3 − *j*4. Whenever there are complex-conjugate poles, the problem can be worked out in two ways. In the first method we expand *X*[*z*] into (modified) first-order partial fractions. In the second method, rather than obtain one factor corresponding to each complex-conjugate pole, we obtain quadratic factors corresponding to each pair of complex-conjugate poles. This procedure is explained next. - -MENT-ORDER FACTORS - -\n -$$ -\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2(3z+17)}{(z-1)(z-3-j4)(z-3+j4)} -$$ - -We find the partial fraction of *X*[*z*]/*z* using the Heaviside "cover-up" method: - -$$ -\frac{X[z]}{z} = \frac{2}{z-1} + \frac{1.6e^{-j2.246}}{z-3-j4} + \frac{1.6e^{j2.246}}{z-3+j4} -$$ - -and - -$$ -X[z] = 2\frac{z}{z-1} + (1.6e^{-j2.246})\frac{z}{z-3-j4} + (1.6e^{j2.246})\frac{z}{z-3+j4} -$$ - -The inverse transform of the first term on the right-hand side is 2*u*[*n*]. The inverse transform of the remaining two terms (complex conjugate poles) can be obtained from pair 12b (Table 5.1) by identifying *r*/2 = 1.6, θ = −2.246 rad, γ = 3 + *j*4 = 5*ej*0.927, so that |γ | = 5, β = 0.927. Therefore, - -$$ -x[n] = [2 + 3.2(5)n \cos(0.927n - 2.246)]u[n] -$$ - -### METHOD OF QUADRATIC FACTORS - -$$ -\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{Az+B}{z^2-6z+25} -$$ - -Multiplying both sides by *z* and letting *z* → ∞, we find - -$$ -0 = 2 + A \Longrightarrow A = -2 -$$ - -and - -$$ -\frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{-2z+B}{z^2-6z+25} -$$ - -To find *B*, we let *z* take any convenient value, say, *z* = 0. This step yields - -$$ -\frac{-34}{25} = -2 + \frac{B}{25} \Longrightarrow B = 16 -$$ - -Therefore, - -$$ -\frac{X[z]}{z} = \frac{2}{z-1} + \frac{-2z+16}{z^2-6z+25} -$$ - -and - -$$ -X[z] = \frac{2z}{z-1} + \frac{z(-2z+16)}{z^2 - 6z + 25} -$$ - -We now use pair 12c, where we identify *A* = −2, *B* = 16, |γ | = 5, and *a* = −3. Therefore, - -$$ -r = \sqrt{\frac{100 + 256 - 192}{25 - 9}} = 3.2 -$$ -, $\beta = \cos^{-1}\left(\frac{3}{5}\right) = 0.927$ rad - -and - -$$ -\theta = \tan^{-1}\left(\frac{-10}{-8}\right) = -2.246 \,\text{rad} -$$ - -so that - -$$ -x[n] = [2 + 3.2(5)n \cos (0.927n - 2.246)]u[n] -$$ - -### **DR ILL 5.2 Inverse** *z***-Transform by Partial Fraction Expansion** - -Find the inverse *z*-transform of the following functions: - -(a) -$$ -\frac{z(2z-1)}{(z-1)(z+0.5)} -$$ - -\n(b) -$$ -\frac{1}{(z-1)(z+0.5)} -$$ - -\n(c) -$$ -\frac{9}{(z+2)(z-0.5)^2} -$$ - -\n(d) -$$ -\frac{5z(z-1)}{z^2-1.6z+0.8} -$$ - -\n[*Hint:* $\sqrt{0.8} = 2/\sqrt{5}$ .] -\n**ANSWERS** -\n(a) $\left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]$ -\n(b) $-2\delta[n] + \left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]$ - -- **(c)** 18δ[*n*]−[0.72(−2)*n* +17.28(0.5)*n* −14.4*n*(0.5)*n*]*u*[*n*] -- **(d)** 5 √5 2 √ 2 5 *n* cos(0.464*n*+0.464)*u*[*n*] - -## **5.1-2 Inverse** *z***[-Transform by Power Series Expansion](#page-11-0)** - -By definition, - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - -= $x[0] + \frac{x[1]}{z} + \frac{x[2]}{z^2} + \frac{x[3]}{z^3} + \cdots$ -= $x[0]z^0 + x[1]z^{-1} + x[2]z^{-2} + x[3]z^{-3} + \cdots$ - -This result is a power series in *z*−1. Therefore, if we can expand *X*[*z*] into the power series in *z*−1, the coefficients of this power series can be identified as *x*[0], *x*[1], *x*[2], *x*[3], .... A rational *X*[*z*] can be expanded into a power series of *z*−1 by dividing its numerator by the denominator. Consider, for example, - -$$ -X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = \frac{7z^3 - 2z^2}{z^3 - 1.7z^2 + 0.8z - 0.1} -$$ - -### 500 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -To obtain a series expansion in powers of *z*−1, we divide the numerator by the denominator as follows: - -$$ -z^{3}-1.7z^{2}+0.8z-0.1\overline{)7z^{3}-2z^{2}} -$$ - -$$ -\underline{7z^{3}-11.9z^{2}+5.60z-0.7} -$$ - -$$ -\underline{7z^{3}-11.9z^{2}+5.60z-0.7} -$$ - -$$ -\underline{9.9z^{2}-5.60z+0.7} -$$ - -$$ -\underline{9.9z^{2}-16.83z+7.92-0.99z^{-1}} -$$ - -$$ -\underline{11.23z-7.22+0.99z^{-1}} -$$ - -$$ -\underline{11.23z-19.09+8.98z^{-1}} -$$ - -$$ -\underline{11.87-7.99z^{-1}} -$$ - -Thus, - -$$ -X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = 7 + 9.9z^{-1} + 11.23z^{-2} + 11.87z^{-3} + \cdots -$$ - -Therefore, - -$$ -x[0] = 7, x[1] = 9.9, x[2] = 11.23, x[3] = 11.87, \dots -$$ - -Although this procedure yields *x*[*n*] directly, it does not provide a closed-form solution. For this reason, it is not very useful unless we want to know only the first few terms of the sequence *x*[*n*]. - -### **DR ILL 5.3 Inverse** *z***-Transform by Long Division** - -Using long division to find the power series in *z*−1, show that the inverse *z*-transform of *z*/(*z*−0.5) is (0.5)*nu*[*n*] or (2)−*nu*[*n*]. - -## RELATIONSHIP BETWEEN *h*[*n*] AND *H*[*z*] - -For an LTID system, if *h*[*n*] is its unit impulse response, then from Eq. (3.39), where we defined *H*[*z*], the system transfer function, we write - -$$ -H[z] = \sum_{n=-\infty}^{\infty} h[n]z^{-n} -$$ - (5.11) - -For causal systems, the limits on the sum are from *n* = 0 to ∞. This equation shows that the transfer function *H*[*z*] is the *z*-transform of the impulse response *h*[*n*] of an LTID system; that is, - -$$ -h[n] \Longleftrightarrow H[z] -$$ - -This important result relates the time-domain specification *h*[*n*] of a system to *H*[*z*], the frequency-domain specification of a system. The result is parallel to that for LTIC systems. - -### **DR ILL 5.4 Impulse Response by Inverse** *z***-Transform** - -Redo Drill 3.14 by taking the inverse *z*-transform of *H*[*z*], as given by Eq. (3.41). - -## **5.2 SOME [PROPERTIES OF THE](#page-11-0)** *z***-TRANSFORM** - -The *z*-transform properties are useful in the derivation of *z*-transforms of many functions and also in the solution of linear difference equations with constant coefficients. Here we consider a few important properties of the *z*-transform. - -In our discussion, the variable *n* appearing in signals, such as *x*[*n*] and *y*[*n*], may or may not stand for time. However, in most applications of our interest, *n* is proportional to time. For this reason, we shall loosely refer to the variable *n* as time. - -### **[5.2-1 Time-Shifting Properties](#page-11-0)** - -In the following discussion of the shift property, we deal with shifted signals *x*[*n*]*u*[*n*], *x*[*n*−*k*]*u*[*n*− *k*], *x*[*n*−*k*]*u*[*n*], and *x*[*n*+*k*]*u*[*n*]. Unless we physically understand the meaning of such shifts, our understanding of the shift property remains mechanical rather than intuitive or heuristic. For this reason, using a hypothetical signal *x*[*n*], we have illustrated various shifted signals for *k* = 1 in Fig. 5.4. - -## RIGHT SHIFT (DELAY) If - -*x*[*n*]*u*[*n*] ⇐⇒ *X*[*z*] - -then - -$$ -x[n-1]u[n-1] \Longleftrightarrow \frac{1}{z}X[z] \tag{5.12} -$$ - -In general, - -$$ -x[n-m]u[n-m] \Longleftrightarrow \frac{1}{z^m}X[z] -$$ -\n(5.13) - -Moreover, - -$$ -x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1] -$$ -\n(5.14) - -Repeated application of this property yields - -$$ -x[n-2]u[n] \Longleftrightarrow \frac{1}{z} \left[ \frac{1}{z} X[z] + x[-1] \right] + x[-2] = \frac{1}{z^2} X[z] + \frac{1}{z} x[-1] + x[-2] -$$ - -In general, for integer value of *m*, - -$$ -x[n-m]u[n] \Longleftrightarrow z^{-m}X[z] + z^{-m}\sum_{n=1}^{m}x[-n]z^{n} -$$ -\n(5.15) - -A look at Eqs. (5.12) and (5.14) shows that they are identical except for the extra term *x*[−1] in Eq. (5.14). We see from Figs. 5.4c and 5.4d that *x*[*n* − 1]*u*[*n*] is the same as *x*[*n* − 1]*u*[*n* − 1] plus *x*[−1]δ[*n*]. Hence, the difference between their transforms is *x*[−1]. - -### 5.2 Some Properties of the *z*-Transform 503 - -**Proof.** For the integer value of *m*, - -$$ -\mathcal{Z}{x[n-m]u[n-m]} = \sum_{n=0}^{\infty} x[n-m]u[n-m]z^{-n} -$$ - -Recall that *x*[*n* − *m*]*u*[*n* − *m*] = 0 for *n* < *m* so that the limits on the summation on the right-hand side can be taken from *n* = *m* to ∞. Therefore, - -$$ -\mathcal{Z}\{x[n-m]u[n-m]\} = \sum_{n=m}^{\infty} x[n-m]z^{-n} -$$ -$$ -= \sum_{r=0}^{\infty} x[r]z^{-(r+m)} -$$ -$$ -= \frac{1}{z^m} \sum_{r=0}^{\infty} x[r]z^{-r} = \frac{1}{z^m} X[z] -$$ - -To prove Eq. (5.15), we have - -$$ -\mathcal{Z}\{x[n-m]u[n]\} = \sum_{n=0}^{\infty} x[n-m]z^{-n} = \sum_{r=-m}^{\infty} x[r]z^{-(r+m)} -$$ -$$ -= z^{-m} \left[ \sum_{r=-m}^{-1} x[r]z^{-r} + \sum_{r=0}^{\infty} x[r]z^{-r} \right] -$$ -$$ -= z^{-m} \sum_{n=1}^{m} x[-n]z^{n} + z^{-m}X[z] -$$ - -LEFT SHIFT (ADVANCE) If - -$$ -x[n]u[n] \Longleftrightarrow X[z] -$$ - -then - -$$ -x[n+1]u[n] \Longleftrightarrow zX[z] - zx[0] -$$ - -Repeated application of this property yields - -$$ -x[n+2]u[n] \Longleftrightarrow z\{z(X[z] - zx[0]) - x[1]\} = z^2 X[z] - z^2 x[0] - zx[1] -$$ - -and for the integer value of *m*, - -$$ -x[n+m]u[n] \Longleftrightarrow z^m X[z] - z^m \sum_{n=0}^{m-1} x[n]z^{-n} -$$ -\n(5.16) - -### 504 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -**Proof.** By definition, - -$$ -\mathcal{Z}\{x[n+m]u[n]\} = \sum_{n=0}^{\infty} x[n+m]z^{-n} -$$ -$$ -= \sum_{r=m}^{\infty} x[r]z^{-(r-m)} -$$ -$$ -= z^m \sum_{r=m}^{\infty} x[r]z^{-r} -$$ -$$ -= z^m \left[ \sum_{r=0}^{\infty} x[r]z^{-r} - \sum_{r=0}^{m-1} x[r]z^{-r} \right] -$$ -$$ -= z^m X[z] - z^m \sum_{r=0}^{m-1} x[r]z^{-r} -$$ - -### **EXAMPLE 5.4** *z***-Transform Using the Right-Shift Property** - -The signal *x*[*n*] can be expressed as a product of *n* and a gate pulse *u*[*n*] − *u*[*n* − 6]. Therefore, - -$$ -x[n] = n\{u[n] - u[n-6]\} = nu[n] - nu[n-6] -$$ - -We cannot find the *z*-transform of *nu*[*n* − 6] directly by using the right-shift property [Eq. (5.13)]. So we rearrange it in terms of (*n*−6)*u*[*n*−6] as follows: - -$$ -x[n] = nu[n] - (n - 6 + 6)u[n - 6] -$$ - -= $nu[n] - (n - 6)u[n - 6] - 6u[n - 6]$ - -We can now find the *z*-transform of the bracketed term by using the right-shift property [Eq. (5.13)]. Because *u*[*n*] ⇐⇒ *z*/(*z*−1), - -$$ -u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{z-1} = \frac{1}{z^5(z-1)} -$$ - -Also, because *nu*[*n*] ⇐⇒ *z*/(*z*−1)2, - -$$ -(n-6)u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{(z-1)^2} = \frac{1}{z^5(z-1)^2} -$$ - -Therefore, - -$$ -X[z] = \frac{z}{(z-1)^2} - \frac{1}{z^5(z-1)^2} - \frac{6}{z^5(z-1)} = \frac{z^6 - 6z + 5}{z^5(z-1)^2} -$$ - -### **DR ILL 5.5** *z***-Transform Using the Right-Shift Property** - -Using only the fact that *u*[*n*] ⇐⇒ *z*/(*z*−1) and the right-shift property [Eq. (5.13)], find the *z*-transforms of the signals in Figs. 5.2 and 5.3. - -### **ANSWERS** - -See Ex. 5.2d and Drill 5.1a. - -## **5.2-2** *z***[-Domain Scaling Property \(Multiplication by](#page-11-0)** *γ n***)** - -Scaling in the *z*-domain is equivalent to multiplying a time-domain signal by an exponential. That is, if - -$$ -x[n]u[n] \Longleftrightarrow X[z] -$$ - -then - -$$ -\gamma^n x[n]u[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right] \tag{5.17} -$$ - -**Proof.** - -$$ -\mathcal{Z}\{\gamma^{n}x[n]u[n]\} = \sum_{n=0}^{\infty} \gamma^{n}x[n]z^{-n} = \sum_{n=0}^{\infty} x[n]\left(\frac{z}{\gamma}\right)^{-n} = X\left[\frac{z}{\gamma}\right] -$$ - -### **DR ILL 5.6 Using the** *z***-Domain Scaling Property** - -Use Eq. (5.17) to derive pairs 6 and 8 in Table 5.1 from pairs 2 and 3, respectively. - -## **5.2-3** *z***[-Domain Differentiation Property \(Multiplication by](#page-11-0)** *n***)** - -Multiplying a signal by *n* in the time domain produces differentiation in the *z*-domain. That is, if - -$$ -x[n]u[n] \Longleftrightarrow X[z] -$$ - -then - -$$ -nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z] \tag{5.18} -$$ - -**Proof.** - -$$ --z\frac{d}{dz}X[z] = -z\frac{d}{dz}\sum_{n=0}^{\infty}x[n]z^{-n} -$$ -$$ -= -z\sum_{n=0}^{\infty} -nx[n]z^{-n-1} -$$ -$$ -= \sum_{n=0}^{\infty}nx[n]z^{-n} = \mathcal{Z}\{nx[n]u[n]\} -$$ - -### **DR ILL 5.7 Using the** *z***-Domain Differentiation Property** - -Use Eq. (5.18) to derive pairs 3 and 4 in Table 5.1 from pair 2. Similarly, derive pairs 8 and 9 from pair 6. - -### **[5.2-4 Time-Reversal Property](#page-11-0)** - -If - -$$ -x[n] \Longleftrightarrow X[z] -$$ - -then† - -$$ -x[-n] \Longleftrightarrow X[1/z] -$$ - -**Proof.** - -$$ -\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[-n]z^{-n} -$$ - -$$ -x^*[-n] \Longleftrightarrow X^*[1/z^*] -$$ - -† For complex signal *x*[*n*], the time-reversal property is modified as follows: - -Changing the sign of the dummy variable *n* yields - -$$ -\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[n]z^n -$$ -$$ -= \sum_{n=-\infty}^{\infty} x[n](1/z)^{-n} -$$ -$$ -= X[1/z] -$$ - -The region of convergence is also inverted; that is, if the ROC of *x*[*n*] is |*z*| > |γ |, then the ROC of *x*[−*n*] is |*z*| < 1/|γ |. - -### **DR ILL 5.8 Using the Time-Reversal Property** - -Use the time-reversal property and pair 2 in Table 5.1 to show that *u*[−*n*] ⇐⇒ −1/(*z*−1) with the ROC |*z*| < 1. - -### **[5.2-5 Convolution Property](#page-11-0)** - -The time-convolution property states that if‡ - -$$ -x_1[n] \Longleftrightarrow X_1[z] -$$ - and $x_2[n] \Longleftrightarrow X_2[z]$ , - -then (*time convolution*) - -$$ -x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z] \tag{5.19} -$$ - -**Proof.** This property applies to causal as well as noncausal sequences. We shall prove it for the more general case of noncausal sequences, where the convolution sum ranges from −∞ to ∞. - -We have - -$$ -\mathcal{Z}\{x_1[n] * x_2[n]\} = \mathcal{Z}\left[\sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m]\right] -$$ -$$ -= \sum_{n=-\infty}^{\infty} z^{-n} \sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m] -$$ - -‡ There is also the frequency-convolution property, which states that - -$$ -x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi j} \oint X_1[u]X_2\left[\frac{z}{u}\right]u^{-1} du -$$ - -### 508 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -Interchanging the order of summation, we have - -$$ -\mathcal{Z}[x_1[n]*x_2[n]] = \sum_{m=-\infty}^{\infty} x_1[m] \sum_{n=-\infty}^{\infty} x_2[n-m]z^{-n} -$$ -$$ -= \sum_{m=-\infty}^{\infty} x_1[m] \sum_{r=-\infty}^{\infty} x_2[r]z^{-(r+m)} -$$ -$$ -= \sum_{m=-\infty}^{\infty} x_1[m]z^{-m} \sum_{r=-\infty}^{\infty} x_2[r]z^{-r} -$$ -$$ -= X_1[z]X_2[z] -$$ - -### LTID SYSTEM RESPONSE - -It is interesting to apply the time-convolution property to the LTID input–output equation *y*[*n*] = *x*[*n*] ∗ *h*[*n*]. Since *h*[*n*] ⇐⇒ *H*[*z*], it follows from Eq. (5.19) that - -$$ -Y[z] = X[z]H[z] \tag{5.20} -$$ - -### **DR ILL 5.9 Using the Convolution Property** - -Use the time-convolution property and appropriate pairs in Table 5.1 to show that *u*[*n*] ∗ *u*[*n*−1] = *nu*[*n*]. - -### INITIAL AND FINAL VALUES - -For a causal *x*[*n*], the initial value theorem states that - -$$ -x[0] = \lim_{z \to \infty} X[z] -$$ - -This result follows immediately from Eq. (5.7). - -If (*z*−1)*X*[*z*] has no poles outside the unit circle, then the final value theorem states that - -$$ -\lim_{N \to \infty} x[N] = \lim_{z \to 1} (z - 1)X[z] -$$ - -This can be shown from the fact that - -$$ -x[n] - x[n-1] \Longleftrightarrow \left\{1 - \frac{1}{z}\right\} X[z] = \frac{(z-1)X[z]}{z} -$$ - -and - -$$ -\frac{(z-1)X[z]}{z} = \sum_{n=-\infty}^{\infty} \{x[n] - x[n-1]\}z^{-n} -$$ - -$$ -\lim_{z \to 1} \frac{(z-1)X[z]}{z} = \lim_{z \to 1} (z-1)X[z] = \lim_{z \to 1} \lim_{N \to \infty} \sum_{n=-\infty}^{N} \{x[n] - x[n-1]\}z^{-n} = \lim_{N \to \infty} x[N] -$$ - -All these properties of the *z*-transform are listed in Table 5.2. - -| Operation | x[n] | X[z] | -|-----------------------|--------------------|-------------------------------------------------------------| -| Addition | x1[n] +x2[n] | X1[z] +X2[z] | -| Scalar multiplication | ax[n] | aX[z] | -| Right shifting | x[n−m]u[n−m] | 1
zm X[z] | -| | x[n−m]u[n] | "m
1
1
n
zm X[z] +
x[−n]z
zm
n=1 | -| | x[n−1]u[n] | 1
X[z] +x[−1]
z | -| | x[n−2]u[n] | 1
1
z2 X[z] +
x[−1] +x[−2]
z | -| | x[n−3]u[n] | 1
1
1
z3 X[z] +
z2 x[−1] +
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Poles of (z −1)X[z]
z→1 | -| | | inside the unit circle | - -**TABLE 5.2** *z*-Transform Properties - -and - -## **5.3** *z***-TRANSFORM [SOLUTION OF](#page-11-0) LINEAR DIFFERENCE EQUATIONS** - -The time-shifting (left-shift or right-shift) property has set the stage for solving linear difference equations with constant coefficients. As in the case of the Laplace transform with differential equations, the *z*-transform converts difference equations into algebraic equations that are readily solved to find the solution in the *z* domain. Taking the inverse *z*-transform of the *z*-domain solution yields the desired time-domain solution. The following examples demonstrate the procedure. - -### **EXAMPLE 5.5** *z***-Transform Solution of a Linear Difference Equation** - -Solve - -$$ -y[n+2] - 5y[n+1] + 6y[n] = 3x[n+1] + 5x[n] -$$ - -if the initial conditions are *y*[−1] = 11/6, *y*[−2] = 37/36, and the input *x*[*n*] = (2)−*nu*[*n*]. - -As we shall see, difference equations can be solved by using the right-shift or the leftshift property. Because the difference equation here is in advance form, the use of the left-shift property in Eq. (5.16) may seem appropriate for its solution. Unfortunately, this left-shift property requires a knowledge of auxiliary conditions *y*[0], *y*[1], ... , *y*[*N* − 1] rather than of the initial conditions *y*[−1], *y*[−2], ... , *y*[−*n*], which are generally given. This difficulty can be overcome by expressing the difference equation in delay form (obtained by replacing *n* with *n*−2) and then using the right-shift property.† The resulting delay-form difference equation is - -$$ -y[n] - 5y[n-1] + 6y[n-2] = 3x[n-1] + 5x[n-2] -$$ -\n(5.21) - -We now use the right-shift property to take the *z*-transform of this equation. But before proceeding, we must be clear about the meaning of a term like *y*[*n* − 1] here. Does it mean *y*[*n* − 1]*u*[*n* − 1] or *y*[*n* − 1]*u*[*n*]? In any equation, we must have some time reference *n* = 0, and every term is referenced from this instant. Hence, *y*[*n*−*k*] means *y*[*n*−*k*]*u*[*n*]. Remember also that although we are considering the situation for *n* ≥ 0, *y*[*n*] is present even before *n* = 0 (in the form of initial conditions). Now - -$$ -y[n]u[n] \Longleftrightarrow Y[z] -$$ - -\n -$$ -y[n-1]u[n] \Longleftrightarrow \frac{1}{z}Y[z] + y[-1] = \frac{1}{z}Y[z] + \frac{11}{6} -$$ - -\n -$$ -y[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}Y[z] + \frac{1}{z}y[-1] + y[-2] = \frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36} -$$ - -Noting that for causal input *x*[*n*], - -$$ -x[-1] = x[-2] = \cdot \cdot \cdot = x[-n] = 0 -$$ - -† Another approach is to find *y*[0], *y*[1], *y*[2], ... , *y*[*N* − 1] from *y*[−1], *y*[−2], ... , *y*[−*n*] iteratively, as in Sec. 3.5-1, and then apply the left-shift property to the advance-form difference equation. - -We obtain - -$$ -x[n] = (2)^{-n}u[n] = (2^{-1})^{n}u[n] = (0.5)^{n}u[n] \Longleftrightarrow \frac{z}{z - 0.5} -$$ - -$$ -x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1] = \frac{1}{z}\frac{z}{z-0.5} + 0 = \frac{1}{z-0.5} -$$ -$$ -x[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}X[z] + \frac{1}{z}x[-1] + x[-2] = \frac{1}{z^2}X[z] + 0 + 0 = \frac{1}{z(z-0.5)} -$$ - -In general, - -$$ -x[n - r]u[n] \Longleftrightarrow \frac{1}{z^r}X[z] -$$ - -Taking the *z*-transform of Eq. (5.21) and substituting the foregoing results, we obtain - -$$ -Y[z] - 5\left[\frac{1}{z}Y[z] + \frac{11}{6}\right] + 6\left[\frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36}\right] = \frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)} -$$ - -or - -1− 5 *z* + 6 *z*2 *Y*[*z*] − 3− 11 *z* = 3 *z*−0.5 + 5 *z*(*z*−0.5) (5.22) - -from which we obtain - -$$ -(z2 - 5z + 6)Y[z] = \frac{z(3z2 - 9.5z + 10.5)}{(z - 0.5)} -$$ - -so that - -$$ -Y[z] = \frac{z(3z^2 - 9.5z + 10.5)}{(z - 0.5)(z^2 - 5z + 6)} -$$ - -and - -$$ -\frac{Y[z]}{z} = \frac{3z^2 - 9.5z + 10.5}{(z - 0.5)(z - 2)(z - 3)} = \frac{(26/15)}{z - 0.5} - \frac{(7/3)}{z - 2} + \frac{(18/5)}{z - 3} -$$ - -Therefore, - -$$ -Y[z] = \frac{26}{15} \left(\frac{z}{z - 0.5}\right) - \frac{7}{3} \left(\frac{z}{z - 2}\right) + \frac{18}{5} \left(\frac{z}{z - 3}\right) -$$ -$$ -y[n] = \left[\frac{26}{15} (0.5)^n - \frac{7}{3} (2)^n + \frac{18}{5} (3)^n\right] u[n] \tag{5.23} -$$ - -and - -This example demonstrates the ease with which linear difference equations with constant coefficients can be solved by the *z*-transform. This method is general: it can be used to solve a single difference equation or a set of simultaneous difference equations of any order as long as the equations are linear with constant coefficients. - -### **Comment.** - -Sometimes, instead of initial conditions *y*[−1], *y*[−2], ... , *y*[−*n*], auxiliary conditions *y*[0], *y*[1], ... , *y*[*N* − 1] are given to solve a difference equation. In this case, the equation can be solved by expressing it in the advance form and then using the left-shift property (see Drill 5.11). - -### **DR ILL 5.10** *z***-Transform Solution of a Linear Difference Equation** - -Solve the following equation if the initial conditions *y*[−1] = 2, *y*[−2] = 0, and the input *x*[*n*] = *u*[*n*]: - -$$ -y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n] -$$ - -### **ANSWER** - -*y*[*n*] = 12−15 1 2 *n* + 14 3 1 3 *n u*[*n*] - -## **DR ILL 5.11 Difference Equation Solution Using** *y*[**0**]**,** *y*[**1**]**,** ... **,** *y*[*N* −**1**] - -Solve the following equation if the auxiliary conditions are *y*[0] = 1, *y*[1] = 2, and the input *x*[*n*] = *u*[*n*]: - -*y*[*n*] +3*y*[*n*−1] +2*y*[*n*−2] = *x*[*n*−1] +3*x*[*n*−2] - -### **ANSWER** - -*y*[*n*] = 2 3 +2(−1)*n* − 5 3 (−2)*n u*[*n*] - -### ZERO-INPUT AND ZERO-STATE COMPONENTS - -In Ex. 5.5 we found the total solution of the difference equation. It is relatively easy to separate the solution into zero-input and zero-state components. All we have to do is to separate the response into terms arising from the input and terms arising from initial conditions (IC). We can separate the response in Eq. (5.22) as follows: - -$$ -\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] - \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} = \underbrace{\frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)}}_{\text{input terms}} -$$ - -Therefore, - -$$ -\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] = \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} + \underbrace{\frac{(3z + 5)}{z(z - 0.5)}}_{\text{input terms}} -$$ - -Multiplying both sides by *z*2 yields - -$$ -(z2 - 5z + 6)Y[z] = \underbrace{z(3z - 11)}_{\text{IC terms}} + \underbrace{\frac{z(3z + 5)}{z - 0.5}}_{\text{input terms}} -$$ - -and - -$$ -Y[z] = \underbrace{\frac{z(3z-11)}{z^2 - 5z + 6}}_{\text{zero-input response}} + \underbrace{\frac{z(3z+5)}{(z-0.5)(z^2 - 5z + 6)}}_{\text{zero-state response}} -$$ - -We expand both terms on the right-hand side into modified partial fractions to yield - -$$ -Y[z] = \underbrace{\left[5\left(\frac{z}{z-2}\right) - 2\left(\frac{z}{z-3}\right)\right]}_{\text{zero-input response}} + \underbrace{\left[\frac{26}{15}\left(\frac{z}{z-0.5}\right) - \frac{22}{3}\left(\frac{z}{z-2}\right) + \frac{28}{5}\left(\frac{z}{z-3}\right)\right]}_{\text{zero-state response}} -$$ - -and - -$$ -y[n] = \underbrace{(5(2)^n - 2(3)^n) u[n]}_{\text{zero-input response}} + \underbrace{\left(\frac{26}{15}(0.5)^n - \frac{22}{3}(2)^n + \frac{28}{5}(3)^n\right) u[n]}_{\text{zero-state response}} -$$ - -= -$$ -\left[-\frac{7}{3}(2)^n + \frac{18}{5}(3)^n + \frac{26}{15}(0.5)^n\right] u[n] -$$ - -which agrees with the result in Eq. (5.23). - -### **DR ILL 5.12 Separating Zero-Input and Zero-State Responses** - -Solve - -$$ -y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n] -$$ - -if the initial conditions are *y*[−1] = 2, *y*[−2] = 0, and the input *x*[*n*] = *u*[*n*]. Separate the response into zero-input and zero-state responses. - -### **ANSWER** - -$$ -y[n] = \underbrace{\left(3\left(\frac{1}{2}\right)^n - \frac{4}{3}\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-input response}} + \underbrace{\left(12 - 18\left(\frac{1}{2}\right)^n + 6\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-state response}} -$$ -\n -$$ -= \left[12 - 15\left(\frac{1}{2}\right)^n + \frac{14}{3}\left(\frac{1}{3}\right)^n\right]u[n] -$$ - -### **[5.3-1 Zero-State Response of LTID Systems: The Transfer Function](#page-11-0)** - -Consider an *N*th-order LTID system specified by the difference equation - -$$ -Q[E]y[n] = P[E]x[n] -$$ - -or - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] -$$ - -= (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] - -or - -$$ -y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n] -$$ - -= $b_0x[n+N] + \cdots + b_{N-1}x[n+1] + b_Nx[n]$ (5.24) - -We now derive the general expression for the zero-state response: that is, the system response to input *x*[*n*] when all the initial conditions *y*[−1] = *y*[−2]=···= *y*[−*N*] = 0 (zero state). The input *x*[*n*] is assumed to be causal so that *x*[−1] = *x*[−2]=···= *x*[−*N*] = 0. - -Equation (5.24) can be expressed in delay form as - -$$ -y[n] + a_1y[n-1] + \dots + a_Ny[n-N] = b_0x[n] + b_1x[n-1] + \dots + b_Nx[n-N] -$$ - (5.25) - -Because *y*[−*r*] = *x*[−*r*] = 0 for *r* = 1, 2,...,*N*, - -$$ -y[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}Y[z] -$$ - -$$ -x[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}X[z] \qquad m=1,2,\ldots,N -$$ - -Now the *z*-transform of Eq. (5.25) is given by - -$$ -\left(1+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots+\frac{a_N}{z^N}\right)Y[z] = \left(b_0+\frac{b_1}{z}+\frac{b_2}{z^2}+\cdots+\frac{b_N}{z^N}\right)X[z] -$$ - -Multiplication of both sides by *zN* yields - -$$ -(zN + a1zN-1 + \dots + aN-1z + aN)Y[z] -$$ - -= (b0zN + b1zN-1 + \dots + bN-1z + bN)X[z] - -Therefore, - -$$ -Y[z] = \left(\frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}\right) X[z] -$$ - -= $\frac{P[z]}{Q[z]} X[z]$ - -We have shown in Eq. (5.20) that *Y*[*z*] = *X*[*z*]*H*[*z*]. Hence, it follows that - -$$ -H[z] = \frac{P[z]}{Q[z]} = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N} -$$ -(5.26) - -As in the case of LTIC systems, this result leads to an alternative definition of the LTID system transfer function as the ratio of *Y*[*z*] to *X*[*z*] (assuming all initial conditions zero). - -$$ -H[z] \equiv \frac{Y[z]}{X[z]} = \frac{\mathcal{Z}[zero-state response]}{\mathcal{Z}[input]} -$$ - -### ALTERNATE INTERPRETATION OF THE *z*-TRANSFORM - -So far we have treated the *z*-transform as a machine that converts linear difference equations into algebraic equations. There is no physical understanding of how this is accomplished or what it means. We now discuss more intuitive interpretation and meaning of the *z*-transform. - -In Ch. 3, Eq. (3.38), we showed that the LTID system response to an everlasting exponential *zn* is *H*[*z*]*zn*. If we could express every discrete-time signal as a linear combination of everlasting exponentials of the form *zn*, we could readily obtain the system response to any input. For example, if - -$$ -x[n] = \sum_{k=1}^{K} X[z_k] z_k^n -$$ -\n(5.27) - -the response of an LTID system to this input is given by - -$$ -y[n] = \sum_{k=1}^{K} X[z_k]H[z_k]z_k^n -$$ - -Unfortunately, a very small class of signals can be expressed in the form of Eq. (5.27). However, we can express almost all signals of practical utility as a sum of everlasting exponentials over a continuum of values of *z*. This is precisely what the *z*-transform in Eq. (5.2) does. - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ -\n(5.28) - -Invoking the linearity property of the *z*-transform, we can find the system response *y*[*n*] to input *x*[*n*] in Eq. (5.28) as† - -$$ -y[n] = \frac{1}{2\pi j} \oint X[z]H[z]z^{n-1} dz = \mathcal{Z}^{-1}{X[z]H[z]} -$$ - -Clearly, - -*Y*[*z*] = *X*[*z*]*H*[*z*] - -This viewpoint of finding the response of LTID system is illustrated in Fig. 5.6a. Just as in continuous-time systems, we can model discrete-time systems in the transformed manner by - -† In computing *y*[*n*], the contour along which the integration is performed is modified to consider the ROC of *X*[*z*] as well as *H*[*z*]. We ignore this consideration in this intuitive discussion. - -**Figure 5.6** The transformed representation of an LTID system. - -representing all signals by their *z*-transforms and all system components (or elements) by their transfer functions, as shown in Fig. 5.6b. - -The result *Y*[*z*] = *H*[*z*]*X*[*z*] greatly facilitates derivation of the system response to a given input. We shall demonstrate this assertion by an example. - -### **EXAMPLE 5.6 Transfer Function to Find the Zero-State Response** - -Find the response *y*[*n*] of an LTID system described by the difference equation - -$$ -y[n+2] + y[n+1] + 0.16y[n] = x[n+1] + 0.32x[n] -$$ - -or - -$$ -(E2 + E + 0.16)y[n] = (E + 0.32)x[n] -$$ - -for the input *x*[*n*] = (−2)−*nu*[*n*] and with all the initial conditions zero (system in the zero state). - -From the difference equation, we find - -$$ -H[z] = \frac{P[z]}{Q[z]} = \frac{z + 0.32}{z^2 + z + 0.16} -$$ - -For the input *x*[*n*] = (−2)−*nu*[*n*]=[(−2)−1] *nu*(*n*) = (−0.5)*nu*[*n*], - -$$ -X[z] = \frac{z}{z+0.5} -$$ - -and - -$$ -Y[z] = X[z]H[z] = \frac{z(z+0.32)}{(z^2 + z + 0.16)(z+0.5)} -$$ - -Therefore, - -$$ -\frac{Y[z]}{z} = \frac{(z+0.32)}{(z^2+z+0.16)(z+0.5)} = \frac{(z+0.32)}{(z+0.2)(z+0.8)(z+0.5)} -$$ -$$ -= \frac{2/3}{z+0.2} - \frac{8/3}{z+0.8} + \frac{2}{z+0.5} -$$ - -so that - -$$ -Y[z] = \frac{2}{3} \left( \frac{z}{z+0.2} \right) - \frac{8}{3} \left( \frac{z}{z+0.8} \right) + 2 \left( \frac{z}{z+0.5} \right) -$$ - -and - -$$ -y[n] = \left[\frac{2}{3}(-0.2)^n - \frac{8}{3}(-0.8)^n + 2(-0.5)^n\right]u[n] -$$ - -### **EXAMPLE 5.7 Transfer Function of a Unit Delay** - -Show that the transfer function of a unit delay is 1/*z*. - -If the input to the unit delay is *x*[*n*]*u*[*n*], then its output (Fig. 5.7) is given by - -*y*[*n*] = *x*[*n*−1]*u*[*n*−1] - -The *z*-transform of this equation yields [see Eq. (5.12)] - -$$ -Y[z] = \frac{1}{z}X[z] = H[z]X[z] -$$ - -It follows that the transfer function of the unit delay is - -$$ -H[z] = \frac{1}{z} -$$ - -*x*[*n*]*u*[*n*] *X*[*z*] *x*[*n* - 1]*u*[*n* - 1] *Y*[*z*] *X*[*z*] 1 *z* 1 *z* **Figure 5.7** Ideal unit delay and its transfer function. - -### **DR ILL 5.13 Transfer Function to Find Zero-State Response and Difference Equation** - -A discrete-time system is described by the following transfer function: - -$$ -H[z] = \frac{z - 0.5}{(z + 0.5)(z - 1)} -$$ - -- **(a)** Find the system response to input *x*[*n*] = 3−(*n*+1) *u*[*n*] if all initial conditions are zero. -- **(b)** Write the difference equation relating the output *y*[*n*] to input *x*[*n*] for this system. - -### **ANSWERS** - -- **(a)** *y*[*n*] = 1 3 1 2 −0.8(−0.5)*n* +0.3 1 3 *n u*[*n*] -- **(b)** *y*[*n*+2] −0.5*y*[*n*+1] −0.5*y*[*n*] = *x*[*n*+1] −0.5*x*[*n*] - -### **[5.3-2 Stability](#page-11-0)** - -Equation (5.26) shows that the denominator of *H*[*z*] is *Q*[*z*], which is apparently identical to the characteristic polynomial *Q*[γ ] defined in Ch. 3. Does this mean that the denominator of *H*[*z*] is the characteristic polynomial of the system? This may or may not be the case: if *P*[*z*] and *Q*[*z*] in Eq. (5.26) have any common factors, they cancel out, and the effective denominator of *H*[*z*] is not necessarily equal to *Q*[*z*]. Recall also that the system transfer function *H*[*z*], like *h*[*n*], is defined in terms of measurements at the external terminals. Consequently, *H*[*z*] and *h*[*n*] are both external descriptions of the system. In contrast, the characteristic polynomial *Q*[*z*] is an internal description. Clearly, we can determine only external stability, that is, BIBO stability, from *H*[*z*]. If all the poles of *H*[*z*] are within the unit circle, all the terms in *h*[*n*] are decaying exponentials, and as shown in Sec. 3.9, *h*[*n*] is absolutely summable. Consequently, the system is BIBO-stable. Otherwise the system is BIBO-unstable. - -If *P*[*z*] and *Q*[*z*] do not have common factors, then the denominator of *H*[*z*] is identical to *Q*[*z*]. † The poles of *H*[*z*] are the characteristic roots of the system. We can now determine internal stability. The internal stability criterion in Sec. 3.9-2 can be restated in terms of the poles of *H*[*z*], as follows. - -- 1. An LTID system is asymptotically stable if and only if all the poles of its transfer function *H*[*z*] are within the unit circle. The poles may be repeated or simple. -- 2. An LTID system is unstable if and only if either one or both of the following conditions exist: (i) at least one pole of *H*[*z*] is outside the unit circle; (ii) there are repeated poles of *H*[*z*] on the unit circle. - -† There is no way of determining whether any common factors in *P*[*z*] and *Q*[*z*] were canceled out. This is because in our derivation of *H*[*z*], we generally get the final result after the cancellations have been effected. When we use internal description of the system to derive *Q*[*z*], however, we find pure *Q*[*z*] unaffected by any common factor in *P*[*z*]. - -3. An LTID system is marginally stable if and only if there are no poles of *H*[*z*] outside the unit circle, and there are some simple poles on the unit circle. - -### **DR ILL 5.14 Transfer Function to Determine Stability** - -Show that an *accumulator* whose impulse response is *h*[*n*] = *u*[*n*] is marginally stable but BIBO-unstable. - -### **[5.3-3 Inverse Systems](#page-11-0)** - -If *H*[*z*] is the transfer function of a system *S*, then *Si*, its inverse system, has a transfer function *Hi*[*z*] given by - -$$ -H_i[z] = \frac{1}{H[z]} -$$ - -This follows from the fact the inverse system *Si* undoes the operation of *S*. Hence, if *H*[*z*] is placed in cascade with *Hi*[*z*], the transfer function of the composite system (identity system) is unity. For example, an *accumulator* whose transfer function is *H*[*z*] = *z*/(*z* − 1) and a *backward difference system* whose transfer function is *Hi*[*z*] = (*z*−1)/*z* are inverse of each other. Similarly if - -$$ -H[z] = \frac{z - 0.4}{z - 0.7} -$$ - -its inverse system transfer function is - -$$ -H_i[z] = \frac{z - 0.7}{z - 0.4} -$$ - -as required by the property *H*[*z*]*Hi*[*z*] = 1. Hence, it follows that - -$$ -h[n] * h_i[n] = \delta[n] -$$ - -### **DR ILL 5.15 Inverse Systems** - -Find the impulse responses of an accumulator and a first-order backward difference system. Show that the convolution of the two impulse responses yields δ[*n*]. - -## **5.4 SYSTEM [REALIZATION](#page-11-0)** - -Because of the similarity between LTIC and LTID systems, conventions for block diagrams and rules of interconnection for LTID are identical to those for continuous-time (LTIC) systems. It is not necessary to rederive these relationships. We shall merely restate them to refresh the reader's memory. - -The block diagram representations of the basic operations, such as an adder, a scalar multiplier, unit delay, and pickoff points, is shown in Fig. 3.13. In our development, the unit delay, which is represented by a box marked D in Fig. 3.13, will be represented by its transfer function 1/*z*. All the signals will also be represented in terms of their *z*-transforms. Thus, the input and the output will be labeled *X*[*z*] and *Y*[*z*], respectively. - -When two systems with transfer functions *H*1[*z*] and *H*2[*z*] are connected in cascade (as in Fig. 4.18b), the transfer function of the composite system is *H*1[*z*]*H*2[*z*]. If the same two systems are connected in parallel (as in Fig. 4.18c), the transfer function of the composite - -**Figure 5.8** Realization of an *N*th-order causal LTID system transfer function by using **(a)** DFI, **(b)** canonic direct (DFII), and **(c)** the transpose form of DFII. - -system is *H*1[*z*] + *H*2[*z*]. For a feedback system (as in Fig. 4.18d), the transfer function is *G*[*z*]/(1+*G*[*z*]*H*[*z*]). - -We now consider a systematic method for realization (or simulation) of an arbitrary *N*th-order LTID transfer function. Since realization is basically a synthesis problem, there is no unique way of realizing a system. A given transfer function can be realized in many different ways. We present here the two forms of *direct realization*. Each of these forms can be executed in several other ways, such as cascade and parallel. Furthermore, a system can be realized by the transposed version of any known realization of that system. This artifice doubles the number of system realizations. A transfer function *H*[*z*] can be realized by using time delays along with adders and multipliers. - -We shall consider a realization of a general *N*th-order causal LTID system, whose transfer function is given by - -$$ -H[z] = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N} -$$ -(5.29) - -This equation is identical to the transfer function of a general *N*th-order proper LTIC system given in Eq. (4.36). The only difference is that the variable *z* in the former is replaced by the variable *s* in the latter. Hence, the procedure for realizing an LTID transfer function is identical to that for the LTIC transfer function with the basic element 1/*s* (integrator) replaced by the element 1/*z* (unit delay). The reader is encouraged to follow the steps in Sec. 4.6 and rederive the results for the LTID transfer function in Eq. (5.29). Here we shall merely reproduce the realizations from Sec. 4.6 with integrators (1/*s*) replaced by unit delays (1/*z*). - -The direct form I (DFI) is shown in Fig. 5.8a, the canonic direct form (DFII) is shown in Fig. 5.8b and the transpose of canonic direct is shown in Fig. 5.8c. The DFII and its transpose are canonic because they require *N* delays, which is the minimum number needed to implement the *N*th-order LTID transfer function in Eq. (5.29). In contrast, the form DFI is a noncanonic because it generally requires 2*N* delays. The DFII realization in Fig. 5.8b is also called a *canonic direct* form. - -### **EXAMPLE 5.8 Canonical Realizations of Transfer Functions** - -Find the canonic direct and the transposed canonic direct realizations of the following transfer functions: **(a)** 2 *z*+5 , **(b)** 4*z*+28 *z*+1 , **(c)** *z z*+7 , and **(d)** 4*z*+28 *z*2 +6*z*+5 . - -All four of these transfer functions are special cases of *H*[*z*] in Eq. (5.29). **(a)** - -$$ -H[z] = \frac{2}{z+5} -$$ - -For this case, the transfer function is of the first order (*N* = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are - -$$ -a_1 = 5 -$$ - and $b_0 = 0$ , $b_1 = 2$ - -### 522 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -We use Fig. 5.8 as our model and reduce it to the case of *N* = 1. Figure 5.9a shows the canonic direct (DFII) form, and Fig. 5.9b its transpose. The two realizations are almost the same. The minor difference is that in the DFII form, the gain 2 is provided at the output, and in the transpose, the same gain is provided at the input. - -**Figure 5.9** Realization of transfer function 2/(*z*+5): **(a)** canonic direct form and **(b)** its transpose. - -In a similar way, we realize the remaining transfer functions. **(b)** - -$$ -H[z] = \frac{4z + 28}{z + 1} -$$ - -In this case also, the transfer function is of the first order (*N* = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are - -$$ -a_1 = 1 -$$ - and $b_0 = 4$ , $b_1 = 28$ - -Figure 5.10 illustrates the canonic direct and its transpose for this case.† - -**Figure 5.10** Realization of (4*z* +28)/(*z* +1): **(a)** canonic direct form and **(b)** its transpose. - -$$ -H[z] = \frac{4z + 28}{z+1} = 4 + \frac{24}{z+1} -$$ - -Hence, this transfer function can also be realized as two transfer functions in parallel. - -† Transfer functions with *N* = *M* may also be expressed as a sum of a constant and a strictly proper transfer function. For example, - -$$ -H[z] = \frac{z}{z+7} -$$ - -Here *N* = 1 and *b*0 = 1,*b*1 = 0 and *a*1 = 7. Figure 5.11 shows the direct and the transposed realizations. Observe that the realizations are almost alike. - -**Figure 5.11** Realization of *z*/(*z* +7): **(a)** canonic direct form and **(b)** its transpose. - -**(d)** - -$$ -H[z] = \frac{4z + 28}{z^2 + 6z + 5} -$$ - -This is a second-order system (*N* = 2) with *b*0 = 0, *b*1 = 4, *b*2 = 28, *a*1 = 6, *a*2 = 5. Figure 5.12 shows the canonic direct and transposed canonic direct realizations. - -**Figure 5.12** Realization of (4*z* +28)/(*z*2 +6*z*+5): **(a)** canonic direct form and **(b)** its transpose. - -**(c)** - -### **DR ILL 5.16 Realization of a Second-Order Transfer Function** - -Realize the transfer function - -$$ -H[z] = \frac{2z}{z^2 + 6z + 25} -$$ - -### REALIZATION OF FINITE IMPULSE RESPONSE (FIR) FILTERS - -So far we have been quite general in our development of realization techniques. They can be applied to infinite impulse response (IIR) or FIR filters. For FIR filters, the coefficients *ai* = 0 for all *i* = 0.† Hence, FIR filters can be readily implemented by means of the schemes developed so far by eliminating all branches with *ai* coefficients. The condition *ai* = 0 implies that all the poles of a FIR filter are at *z* = 0. - -### **EXAMPLE 5.9 Realization of an FIR Filter** - -Realize *H*[*z*] = (*z*3 +4*z*2 +5*z*+2)/*z*3 using canonic direct and transposed forms. - -† This statement is true for all *i* = 0 because *a*0 is assumed to be unity. - -For *H*[*z*], *b*0 = 1, *b*1 = 4, *b*2 = 5, and *b*3 = 2. Hence, we obtain the canonic direct realization, shown in Fig. 5.13a. We have shown the horizontal orientation because it is easier to see that this filter is basically a tapped delay line. That is why this structure is also known as a *tapped delay line* or *transversal filter*. Figure 5.13b shows the corresponding transposed implementation. - -## CASCADE AND PARALLEL REALIZATIONS, COMPLEX AND REPEATED POLES - -The considerations and observations for cascade and parallel realizations as well as complex and multiple poles are identical to those discussed for LTIC systems in Sec. 4.6-3. - -## **DR ILL 5.17 Cascade and Parallel Realizations of a Transfer Function** - -Find canonic direct realizations of the following transfer function by using the cascade and parallel forms. The specific cascade decomposition is as follows: - -$$ -H[z] = \frac{z+3}{z^2 + 7z + 10} = \left(\frac{z+3}{z+2}\right)\left(\frac{1}{z+5}\right) -$$ - -### DO ALL REALIZATIONS LEAD TO THE SAME PERFORMANCE? - -For a given transfer function, we have presented here several possible different realizations (DFI, canonic form DFII, and its transpose). There are also cascade and parallel versions, and there are many possible grouping of the factors in the numerator and the denominator of *H*[*z*], leading to different realizations. We can also use various combinations of these forms in implementing different subsections of a system. Moreover, the transpose of each version doubles the number. However, this discussion by no means exhausts all the possibilities. Transforming variables affords limitless potential realizations of the same transfer function. - -Theoretically, all these realizations are equivalent; that is, they lead to the same transfer function. This, however, is true only when we implement them with infinite precision. In practice, finite wordlength restriction causes each realization to behave differently in terms of sensitivity to parameter variation, stability, frequency response distortion error, and so on. These effects are serious for higher-order transfer functions, which require correspondingly higher numbers of delay elements. The finite wordlength errors that plague these implementations are coefficient quantization, overflow errors, and round-off errors. From a practical viewpoint, parallel and cascade forms using low-order filters minimize the effects of finite wordlength. Parallel and certain cascade forms are numerically less sensitive than the canonic direct form to small parameter variations in the system. In the canonic direct form structure with large *N*, a small change in a filter coefficient due to parameter quantization results in a large change in the location of the poles and the zeros of the system. Qualitatively, this difference can be explained by the fact that - -### 526 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -in a direct form (or its transpose), all the coefficients interact with each other, and a change in any coefficient will be magnified through its repeated influence from feedback and feedforward connections. In a parallel realization, in contrast, a change in a coefficient will affect only a localized segment; the case of a cascade realization is similar. For this reason, the most popular technique for minimizing finite wordlength effects is to design filters by using cascade or parallel forms employing low-order filters. In practice, high-order filters are realized by using multiple second-order sections in cascade, because second-order filters not only are easier to design but are less susceptible to coefficient quantization and round-off errors, and their implementations allow easier data word scaling to reduce the potential overflow effects of data word-size growth. A cascaded system using second-order building blocks usually requires fewer multiplications for a given filter frequency response [1]. - -There are several ways to pair the poles and zeros of an *N*th-order *H*[*z*] into a cascade of second-order sections, and several ways to order the resulting sections. Quantizing error will be different for each combination. Although several papers published provide guidelines in predicting and minimizing finite wordlength errors, it is advisable to resort to computer simulation of the filter design. This way, one can vary filter hardware characteristic, such as coefficient wordlengths, accumulator register sizes, sequencing of cascaded sections, and input signal sets. Such an approach is both reliable and economical [1]. - -## **5.5 FREQUENCY RESPONSE OF [DISCRETE-TIME](#page-11-0) SYSTEMS** - -For (asymptotically or BIBO-stable) continuous-time systems, we showed that the system response to an input *ej*ω*t* is *H*(*j*ω)*ej*ω*t* and that the response to an input cos ω*t* is |*H*(*j*ω)| cos[ω*t* + *H*(*j*ω)]. Similar results hold for discrete-time systems. We now show that for an (asymptotically or BIBO-stable) LTID system, the system response to an input *ejn* is *H*[*ej*]*ejn* and the response to an input cos *n* is |*H*[*ej*]| cos(*n*+ *H*[*ej*]). - -The proof is similar to the one used for continuous-time systems. In Sec. 3.8-2, we showed that an LTID system response to an (everlasting) exponential *zn* is also an (everlasting) exponential *H*[*z*]*zn*. This result is valid only for values of *z* for which *H*[*z*], as defined in Eq. (5.11), exists (converges). As usual, we represent this input–output relationship by a directed arrow notation as - -$$ -z^n \Longrightarrow H[z]z^n \tag{5.30} -$$ - -Setting *z* = *ej* in this relationship yields - -$$ -e^{i\Omega n} \Longrightarrow H[e^{i\Omega}]e^{i\Omega n} \tag{5.31} -$$ - -Noting that cos *n* is the real part of *ejn*, use of Eq. (3.34) yields - -$$ -\cos \Omega n \Longrightarrow \text{Re}\{H[e^{i\Omega}]e^{i\Omega n}\}\tag{5.32} -$$ - -Expressing *H*[*ej*] in the polar form - -$$ -H[e^{i\Omega}] = |H[e^{i\Omega}]|e^{i\angle H[e^{i\Omega}]} -$$ - -Eq. (5.32) can be expressed as - -$$ -\cos \Omega n \Longrightarrow |H[e^{i\Omega}]|\cos(\Omega n + \angle H[e^{i\Omega}]) -$$ - -In other words, the system response *y*[*n*] to a sinusoidal input cos *n* is given by - -$$ -y[n] = |H[e^{j\Omega}]|\cos(\Omega n + \angle H[e^{j\Omega}]) -$$ - -Following the same argument, the system response to a sinusoid cos(*n*+θ ) is - -$$ -y[n] = |H[e^{j\Omega}]\cos(\Omega n + \theta + \angle H[e^{j\Omega}])\tag{5.33} -$$ - -This result is valid only for BIBO-stable or asymptotically stable systems. The frequency response is meaningless for BIBO-unstable systems (which include marginally stable and asymptotically unstable systems). This follows from the fact that the frequency response in Eq. (5.31) is obtained by setting *z*=*ej* in Eq. (5.30). But, as shown in Sec. 3.8-2 [Eqs. (3.38) and (3.39)], the relationship of Eq. (5.30) applies only for values of *z* for which *H*[*z*] exists. For BIBO-unstable systems, the ROC for *H*[*z*] does not include the unit circle where *z* = *ej*. This means, for BIBO-unstable systems, that *H*[*z*] is meaningless when *z* = *ej*. † - -This important result shows that the response of an asymptotically or BIBO-stable LTID system to a discrete-time sinusoidal input of frequency is also a discrete-time sinusoid of the same frequency. *The amplitude of the output sinusoid is* |*H*[*ej*]| *times the input amplitude, and the phase of the output sinusoid is shifted by H*[*ej*] *with respect to the input phase*. Clearly, |*H*[*ej*]| is the amplitude gain, and a plot of |*H*[*ej*]| versus is the amplitude response of the discrete-time system. Similarly, *H*[*ej*] is the phase response of the system, and a plot of *H*[*ej*] versus shows how the system modifies or shifts the phase of the input sinusoid. Note that *H*[*ej*] incorporates the information of both amplitude and phase responses and therefore is called the *frequency responses* of the system. - -### STEADY-STATE RESPONSE TO CAUSAL SINUSOIDAL INPUT - -As in the case of continuous-time systems, we can show that the response of an LTID system to a causal sinusoidal input cos *n u*[*n*] is *y*[*n*] in Eq. (5.33), plus a natural component consisting of the characteristic modes (see Prob. 5.5-9). For a stable system, all the modes decay exponentially, and only the sinusoidal component in Eq. (5.33) persists. For this reason, this component is called the sinusoidal *steady-state* response of the system. Thus, *yss*[*n*], the steady-state response of a system to a causal sinusoidal input cos *n u*[*n*], is - -$$ -y_{ss}[n] = |H[e^{j\Omega}]\cos{(\Omega n + \angle H[e^{j\Omega}])}u[n] -$$ - -### SYSTEM RESPONSE TO SAMPLED CONTINUOUS-TIME SINUSOIDS - -So far we have considered the response of a discrete-time system to a discrete-time sinusoid cos *n* (or exponential *ejn*). In practice, the input may be a sampled continuous-time sinusoid cos ω*t* (or an exponential *ej*ω*t* ). When a sinusoid cos ω*t* is sampled with sampling interval *T*, the resulting - -† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cos0*n* or γ *n* cos0*n* (γ > 1). Hence, the response of such a system to a sinusoid cos*n* will contain not just the sinusoid of frequency but also nondecaying natural modes, rendering the concept of frequency response meaningless. Alternately, we can argue that when *z*=*ej*, a BIBO-unstable system violates the dominance condition |γ*i*| &lt; |*ej*| for all *i*, where γ*i* represents *i*th characteristic root of the system. - -### 528 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -signal is a discrete-time sinusoid cos ω*nT*, obtained by setting *t* = *nT* in cosω*t*. Therefore, all the results developed in this section apply if we substitute ω*T* for : - -$$ -\Omega = \omega T \tag{5.34} -$$ - -### **EXAMPLE 5.10 Sinusoidal Response of a Difference Equation System** - -For a system specified by the equation - -$$ -y[n+1] - 0.8y[n] = x[n+1] -$$ - -find the system response to the inputs - -(a) -$$ -1^n = 1 -$$ - -(b) $\cos\left[\frac{\pi}{6}n - 0.2\right]$ - -**(c)** a sampled sinusoid cos 1500*t* with sampling interval *T* = 0.001 - -The system equation can be expressed as - -$$ -(E - 0.8)y[n] = Ex[n] -$$ - -Therefore, the transfer function of the system is - -$$ -H[z] = \frac{z}{z - 0.8} = \frac{1}{1 - 0.8z^{-1}} -$$ - -The frequency response is - -$$ -H[e^{i\Omega}] = \frac{1}{1 - 0.8e^{-i\Omega}} = \frac{1}{(1 - 0.8\cos\Omega) + i0.8\sin\Omega} -$$ - -Therefore, - -$$ -|H[e^{i\Omega}]| = \frac{1}{\sqrt{(1 - 0.8 \cos \Omega)^2 + (0.8 \sin \Omega)^2}} = \frac{1}{\sqrt{1.64 - 1.6 \cos \Omega}} -$$ -(5.35) - -and - -$$ -\angle H[e^{i\Omega}] = -\tan^{-1}\left[\frac{0.8\sin\Omega}{1 - 0.8\cos\Omega}\right] -$$ -\n(5.36) - -The amplitude response |*H*[*ej*]| can also be obtained by observing that |*H*| 2 = *HH*∗. Since our system is real, we therefore see that - -$$ -|H[e^{i\Omega}]|^{2} = H[e^{i\Omega}]H^{*}[e^{i\Omega}] = H[e^{i\Omega}]H[e^{-i\Omega}] -$$ -\n(5.37) - -Substituting for *H*[*ej*], it follows that - -$$ -|H[e^{i\Omega}]|^2 = \left(\frac{1}{1 - 0.8e^{-i\Omega}}\right)\left(\frac{1}{1 - 0.8e^{i\Omega}}\right) = \frac{1}{1.64 - 1.6\cos\Omega} -$$ - -which matches the result found earlier. - -Figure 5.14 shows plots of amplitude and phase response as functions of . We now compute the amplitude and the phase response for the various inputs. - -**Figure 5.14** Frequency response of the LTID system. - -**(a)** Since 1*n* = (*ej*)*n* with = 0, the amplitude response is *H*[*ej*0]. From Eq. (5.35) we obtain - -$$ -H[e^{i0}] = \frac{1}{\sqrt{1.64 - 1.6 \cos(0)}} = \frac{1}{\sqrt{0.04}} = 5 = 5 \angle 0 -$$ - -Therefore, - -$$ -|H[e^{j0}]| = 5 \quad \text{and} \quad \angle H[e^{j0}] = 0 -$$ - -These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = 0. Therefore, the system response to input 1 is - -$$ -y[n] = 5(1^n) = 5 \qquad \text{for all } n -$$ - -**(b)** For *x*[*n*] = cos[(π/6)*n*−0.2], = π/6. According to Eqs. (5.35) and (5.36), - -$$ -|H[e^{j\pi/6}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos \frac{\pi}{6}}} = 1.983 -$$ - -$$ -\angle H[e^{j\pi/6}] = -\tan^{-1} \left[ \frac{0.8 \sin \frac{\pi}{6}}{1 - 0.8 \cos \frac{\pi}{6}} \right] = -0.916 \text{ rad} -$$ - -These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = π/6. Therefore, - -$$ -y[n] = 1.983 \cos\left(\frac{\pi}{6}n - 0.2 - 0.916\right) = 1.983 \cos\left(\frac{\pi}{6}n - 1.116\right) -$$ - -Figure 5.15 shows the input *x*[*n*] and the corresponding system response. - -**Figure 5.15** Sinusoidal input and the corresponding output of the LTID system. - -**(c)** A sinusoid cos 1500*t* sampled every *T* seconds (*t* = *nT*) results in a discrete-time sinusoid - -$$ -x[n] = \cos 1500nT -$$ - -For *T* = 0.001, the input is - -*x*[*n*] = cos(1.5*n*) - -In this case, = 1.5. According to Eqs. (5.35) and (5.36), - -$$ -|H[e^{j1.5}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos(1.5)}} = 0.809 -$$ - -$$ -\angle H[e^{j1.5}] = -\tan^{-1} \left[ \frac{0.8 \sin(1.5)}{1 - 0.8 \cos(1.5)} \right] = -0.702 \text{ rad} -$$ - -These values also could be read directly from Fig. 5.14 corresponding to = 1.5. Therefore, - -$$ -y[n] = 0.809 \cos(1.5n - 0.702) -$$ - -### FREQUENCY RESPONSE PLOTS USING MATLAB - -MATLAB makes it easy to compute and plot magnitude and phase responses directly using a system's transfer function. As the following code demonstrates, there is no need to derive separate expressions for the magnitude and phase responses. - ->> Omega = linspace(-pi,pi,400); H = @(z) z./(z-0.8); - ->> subplot(1,2,1); plot(Omega,abs(H(exp(1j\*Omega))),'k'); axis tight; - ->> xlabel('\Omega'); ylabel('|H[e^{j \Omega}]|'); - ->> subplot(1,2,2); plot(Omega,angle(H(exp(1j\*Omega))\*180/pi),'k'); axis tight; - -``` ->> xlabel('\Omega'); ylabel('\angle H[e^{j \Omega}] [deg]'); -``` - -The resulting plots, shown in Fig. 5.16, confirm the earlier results of Fig. 5.14. - -### 532 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -**Comment:** Figures 5.14 and 5.16 show amplitude and phase response plots as functions of . These plots as well as Eqs. (5.35) and (5.36) indicate that the frequency response of a discrete-time system is a continuous (rather than discrete) function of frequency . There is no contradiction here. This behavior is merely an indication of the fact that the frequency variable is continuous (takes on all possible values) and therefore the system response exists at every value of . - -### **DR ILL 5.18 Frequency Response of Difference Equation** - -For a system specified by the equation - -$$ -y[n+1] - 0.5y[n] = x[n] -$$ - -find the amplitude and the phase response. Find the system response to sinusoidal input cos[1000*t* −(π/3)] sampled every *T* = 0.5 ms. - -### **ANSWER** - -$$ -|H[e^{i\Omega}]| = \frac{1}{\sqrt{1.25 - \cos \Omega}} -$$ - -\n -$$ -\angle H[e^{i\Omega}] = -\tan^{-1} \left[ \frac{\sin \Omega}{\cos \Omega - 0.5} \right] -$$ - -\n -$$ -y[n] = 1.639 \cos \left( 0.5n - \frac{\pi}{3} - 0.904 \right) = 1.639 \cos (0.5n - 1.951) -$$ - -### **DR ILL 5.19 Frequency Response of an Ideal Delay System** - -Show that for an ideal delay (*H*[*z*] = 1/*z*), the amplitude response |*H*[*ej*]| = 1, and the phase response *H*[*ej*]=−. Thus, a pure time delay does not affect the amplitude gain of sinusoidal input, but it causes a phase shift (delay) of radians in a discrete sinusoid of frequency . Thus, for an ideal delay, the phase shift of the output sinusoid is proportional to the frequency of the input sinusoid (linear phase shift). - -### **[5.5-1 The Periodic Nature of Frequency Response](#page-11-0)** - -In Ex. 5.10 and Fig. 5.14, we saw that the frequency response *H*[*ej*] is a periodic function of . This is not a coincidence. Unlike continuous-time systems, all LTID systems have periodic frequency response. This is seen clearly from the nature of the expression of the frequency response of an LTID system. Because *e*±*j*2π*m* = 1 for all integer values of *m* [see Eq. (B.10)], - -$$ -H[e^{j\Omega}] = H[e^{j(\Omega + 2\pi m)}] \qquad m \text{ integer} -$$ - -Therefore, the frequency response *H*[*ej*] is a periodic function of with a period 2π. This is the mathematical explanation of the periodic behavior. The physical explanation that follows provides a much better insight into the periodic behavior. - -### NON-UNIQUENESS OF DISCRETE-TIME SINUSOID WAVEFORMS - -A continuous-time sinusoid cos ω*t* has a unique waveform for every real value of ω in the range 0 to ∞. Increasing ω results in a sinusoid of ever-increasing frequency. Such is not the case for the discrete-time sinusoid cos *n* because - -$$ -\cos[(\Omega \pm 2\pi m)n] = \cos \Omega n \qquad m \text{ integer} -$$ - -and - -$$ -e^{j(\Omega \pm 2\pi m)n} = e^{j\Omega n} \qquad m \text{ integer} -$$ - -This shows that the discrete-time sinusoids cos *n* (and exponentials *ejn*) separated by values of in integral multiples of 2π are identical. The reason for the periodic nature of the frequency response of an LTID system is now clear. Since the sinusoids (or exponentials) with frequencies separated by interval 2π are identical, the system response to such sinusoids is also identical and, hence, is periodic with period 2π. - -This discussion shows that the discrete-time sinusoid cos *n* has a unique waveform only for the values of in the range −π to π. This band is called the *fundamental band*. Every frequency , no matter how large, is identical to some frequency, *a*, in the fundamental band (−π ≤ *a* < π), where - -$$ -\Omega_a = \Omega - 2\pi m \qquad -\pi \le \Omega_a < \pi \quad \text{and} \quad m \text{ integer} \tag{5.38} -$$ - -The integer *m* can be positive or negative. We use Eq. (5.38) to plot the fundamental band frequency *a* versus the frequency of a sinusoid (Fig. 5.17a). The frequency *a* is modulo 2π value of . - -All these conclusions are also valid for exponential *ejn*. - -### ALL DISCRETE-TIME SIGNALS ARE INHERENTLY BANDLIMITED - -This discussion leads to the surprising conclusion that all discrete-time signals are inherently bandlimited, with frequencies lying in the range −π to π radians per sample. In terms of frequency *F* = /2π, where *F* is in cycles per sample, all frequencies *F* separated by an integer number are identical. For instance, all discrete-time sinusoids of frequencies 0.3, 1.3, 2.3, ... cycles per sample are identical. The fundamental range of frequencies is −0.5 to 0.5 cycles per sample. - -Any discrete-time sinusoid of frequency beyond the fundamental band, when plotted, appears and behaves, in every way, like a sinusoid having its frequency in the fundamental band. It is impossible to distinguish between the two signals. Thus, in a basic sense, discrete-time frequencies beyond || = π or |*F*| = 1/2 do not exist. Yet, in a "mathematical" sense, we must admit the existence of sinusoids of frequencies beyond = π. What does this mean? - -**Figure 5.17 (a)** Actual frequency versus **(b)** apparent frequency. - -### A MAN NAMED ROBERT - -To give an analogy, consider a fictitious person Mr. Robert Thompson. His mother calls him Robby; his acquaintances call him Bob, his close friends call him by his nickname, Shorty. Yet, Robert, Robby, Bob, and Shorty are one and the same person. However, we cannot say that only Mr. Robert Thompson exists, or only Robby exists, or only Shorty exists, or only Bob exists. All these four persons exist, although they are one and the same individual. In a same way, we cannot say that the frequency π/2 exists and frequency 5π/2 does not exist; they are both the same entity, called by different names. - -It is in this sense that we have to admit the existence of frequencies beyond the fundamental band. Indeed, mathematical expressions in the frequency domain automatically cater to this need by their built-in periodicity. As seen earlier, the very structure of the frequency response is 2π-periodic. We shall also see later, in Ch. 9, that discrete-time signal spectra are also 2π-periodic. - -Admitting the existence of frequencies beyond π also serves mathematical and computational convenience in digital signal-processing applications. Values of frequencies beyond π may also originate naturally in the process of sampling continuous-time sinusoids. Because there is no upper limit on the value of ω, there is no upper limit on the value of the resulting discrete-time frequency = ω*T* either.† - -The highest possible frequency is π and the lowest frequency is 0 (dc or constant). Clearly, the high frequencies are those in the vicinity of = (2*m* + 1)π and the low frequencies are those in the vicinity of = 2π*m* for all positive or negative integer values of *m*. - -† However, if goes beyond π, the resulting aliasing reduces the apparent frequency to *a* < π. - -### FURTHER REDUCTION IN THE FREQUENCY RANGE - -Because cos(−*n* + θ ) = cos(*n* − θ ), a frequency in the range −π to 0 is identical to the frequency (of the same magnitude) in the range 0 to π (but with a change in phase sign). Consequently the *apparent frequency* for a discrete-time sinusoid of any frequency is equal to some value in the range 0 to π. Thus, cos(8.7π*n* + θ ) = cos(0.7π*n* + θ ), and the apparent frequency is 0.7π. Similarly, - -$$ -\cos(9.6\pi n + \theta) = \cos(-0.4\pi n + \theta) = \cos(0.4\pi n - \theta) -$$ - -Hence, the frequency 9.6π is identical (in every respect) to frequency −0.4π, which, in turn, is equal (within the sign of its phase) to frequency 0.4π. In this case, the apparent frequency reduces to |*a*| = 0.4π. We can generalize the result to say that the apparent frequency of a discrete-time sinusoid is |*a*|, as found from Eq. (5.38), and if *a* <0, there is a phase reversal. Figure 5.17b plots versus the apparent frequency |*a*|. The shaded bands represent the ranges of for which there is a phase reversal, when represented in terms of |*a*|. For example, the apparent frequency for both the sinusoids cos(2.4π + θ ) and cos(3.6π + θ ) is |*a*| = 0.4π, as seen from Fig. 5.17b. But 2.4π is in a clear band and 3.6π is in a shaded band. Hence, these sinusoids appear as cos(0.4π +θ ) and cos(0.4π −θ ), respectively. - -Although every discrete-time sinusoid can be expressed as having frequency in the range from 0 to π, we generally use the frequency range from −π to π instead of 0 to π for two reasons. First, exponential representation of sinusoids with frequencies in the range 0 to π requires a frequency range −π to π. Second, even when we are using a trigonometric representation, we generally need the frequency range −π to π to have exact identity (without phase reversal) of a higher-frequency sinusoid. - -For certain practical advantages, in place of the range −π to π, we often use other contiguous ranges of width 2π. The range 0 to 2π, for instance, is used in many applications. It is left as an exercise for the reader to show that the frequencies in the range from π to 2π are identical to those in the range from −π to 0. - -### **EXAMPLE 5.11 Apparent Frequency** - -Express the following signals in terms of their apparent frequencies: **(a)** cos(0.5π*n* + θ ), **(b)** cos(1.6π*n*+θ ), **(c)** sin(1.6π*n*+θ ), **(d)** cos(2.3π*n*+θ ), and **(e)** cos(34.699*n*+θ ). - -**(a)** =0.5π is in the reduced range already. This is also apparent from Fig. 5.17a or 5.17b. Because *a* = 0.5π, there is no phase reversal, and the apparent sinusoid is cos(0.5π*n*+θ ). - -**(b)** We express 1.6π = −0.4π + 2π so that *a* = −0.4π and |*a*| = 0.4. Also, *a* is negative, implying sign change for the phase. Hence, the apparent sinusoid is cos(0.4π*n*−θ ). This fact is also apparent from Fig. 5.17b. - -**(c)** We first convert the sine form to cosine form as sin(1.6π*n*+θ ) = cos(1.6π*n* − (π/2) + θ ). In part **(b)**, we found *a* = −0.4π. Hence, the apparent sinusoid is cos(0.4π*n* + (π/2)−θ ) = −sin(0.4π*n*−θ ). In this case, both the phase and the amplitude change signs. - -**(d)** 2.3π = 0.3π +2π so that *a* = 0.3π. Hence, the apparent sinusoid is cos(0.3π*n*+θ ). - -**(e)** We have 34.699 = −3+6(2π ). Hence, *a* = −3, and the apparent frequency |*a*| = 3 rad/sample. Because *a* is negative, there is a sign change of the phase. Hence, the apparent sinusoid is cos(3*n*−θ ). - -### **DR ILL 5.20 Apparent Frequency** - -Show that the sinusoids having frequencies of **(a)** 2π, **(b)** 3π, **(c)** 5π, **(d)** 3.2π, **(e)** 22.1327, and **(f)** π + 2 can be expressed, respectively, as sinusoids of frequencies **(a)** 0, **(b)** π, **(c)** π, **(d)** 0.8π, **(e)** 3, and **(f)** π −2. Show that in cases (d), (e), and (f), phase changes sign. - -### **[5.5-2 Aliasing and Sampling Rate](#page-11-0)** - -The non-uniqueness of discrete-time sinusoids and the periodic repetition of the same waveforms at intervals of 2π may seem innocuous, but in reality it leads to a serious problem for processing continuous-time signals by digital filters. A continuous-time sinusoid cosω*t* sampled every *T* seconds (*t* = *nT*) results in a discrete-time sinusoid cosω*nT*, which is cos*n* with = ω*T*. The discrete-time sinusoids cos*n* have unique waveforms only for the values of frequencies in the range <π or ω*T* < π. Therefore, samples of continuous-time sinusoids of two (or more) different frequencies can generate the same discrete-time signal, as shown in Fig. 5.18. *This phenomenon is known as aliasing because through sampling, two entirely different analog sinusoids take on the same "discrete-time" identity*. † - -Aliasing causes ambiguity in digital signal processing, which makes it impossible to determine the true frequency of the sampled signal. Consider, for instance, digitally processing - -**Figure 5.18** Demonstration of the aliasing effect. - -† Figure 5.18 shows samples of two sinusoids cos 12π*t* and cos 2π*t* taken every 0.2 second. The corresponding discrete-time frequencies ( = ω*T* = 0.2ω) are cos 2.4π and cos 0.4π. The apparent frequency of 2.4π is 0.4π, identical to the discrete-time frequency corresponding to the lower sinusoid. This shows that the samples of both these continuous-time sinusoids at 0.2-second intervals are identical, as verified from Fig. 5.18. - -a continuous-time signal that contains two distinct components of frequencies ω1 and ω2. The samples of these components appear as discrete-time sinusoids of frequencies 1 = ω1*T* and 2 = ω2*T*. If 1 and 2 happen to differ by an integer multiple of 2π (if ω2 − ω1 = 2*k*π/*T*), the two frequencies will be read as the same (lower of the two) frequency by the digital processor.‡ As a result, the higher-frequency component ω2 not only is lost for good (by losing its identity to ω1), but also it reincarnates as a component of frequency ω1, thus distorting the true amplitude of the original component of frequency ω1. Hence, the resulting processed signal will be distorted. Clearly, aliasing is highly undesirable and should be avoided. To avoid aliasing, the frequencies of the continuous-time sinusoids to be processed should be kept within the fundamental band ω*T* ≤ π or ω ≤ π/*T*. Under this condition the question of ambiguity or aliasing does not arise because any continuous-time sinusoid of frequency in this range has a unique waveform when it is sampled. Therefore, if ω*h* is the highest frequency to be processed, then, to avoid aliasing, - -$$ -\omega_h < \frac{\pi}{T} -$$ - -If *fh* is the highest frequency in hertz, *fh* = ω*h*/2π, and we avoid aliasing if - -$$ -f_h < \frac{1}{2T} \qquad \text{or} \qquad T < \frac{1}{2f_h} \tag{5.39} -$$ - -This shows that discrete-time signal processing places the limit on the highest frequency *fh* that can be processed for a given value of the sampling interval *T*. Fortunately, we can process a signal of any frequency (without aliasing) by choosing a suitably small value of *T*. Since the sampling frequency *fs* is the reciprocal of the sampling interval *T*, we can also express Eq. (5.39) as - -$$ -f_s = \frac{1}{T} > 2f_h -$$ - or $f_h < \frac{f_s}{2}$ (5.40) - -This result is a special case of the well-known *sampling theorem* (to be proved in Ch. 8). It states that for a discrete-time system to process a continuous-time sinusoid, the sampling rate must be greater than twice the frequency (in hertz) of the sinusoid. In short, *a sampled sinusoid must have a minimum of two samples per cycle*. † For sampling rates below this minimum value, the output signal will be aliased, which means it will be mistaken for a sinusoid of lower frequency. - -### ANTI-ALIASING FILTER - -If the sampling rate fails to satisfy Eq. (5.40), aliasing occurs, causing the frequencies beyond *fs*/2 Hz to masquerade as lower frequencies to corrupt the spectrum at frequencies below *fs*/2. To avoid such a corruption, a signal to be sampled is passed through an *anti-aliasing* filter of bandwidth *fs*/2 prior to sampling. This operation ensures the condition of Eq. (5.40). The drawback of such a filter is that we lose the spectral components of the signal beyond frequency *fs*/2, which is preferable to the aliasing corruption of the signal at frequencies below *fs*/2. Chapter 8 presents a detailed analysis of the aliasing problem. - -‡ In the case shown in Fig. 5.18, ω1 = 12π, ω2 = 2π, and *T* = 0.2. Hence, ω2 − ω1 = 10π*T* = 2π, and the two frequencies are read as the same frequency = 0.4π by the digital processor. - -† Strictly speaking, we must have more than two samples per cycle. - -### **EXAMPLE 5.12 Maximum Sampling Interval** - -Determine the maximum sampling interval *T* that can be used in a discrete-time oscillator that generates a sinusoid of 50 kHz. - -Here the highest significant frequency *fh* = 50 kHz. Therefore from Eq. (5.39), - -$$ -T < \frac{1}{2f_h} = 10\,\mu\,\mathrm{s} -$$ - -The sampling interval must be less than 10µs. The sampling frequency is *fs* = 1/*T* > 100 kHz. - -### **EXAMPLE 5.13 Maximum Frequency Without Aliasing** - -A discrete-time amplifier uses a sampling interval *T* = 25µs. What is the highest frequency of a signal that can be processed with this amplifier without aliasing? - -From Eq. (5.39) - -$$ -f_h < \frac{1}{2T} = 20 \, \text{kHz} -$$ - -## **5.6 FREQUENCY [RESPONSE FROM](#page-11-0) POLE-ZERO LOCATIONS** - -The frequency responses (amplitude and phase responses) of a system are determined by pole-zero locations of the transfer function *H*[*z*]. Just as in continuous-time systems, it is possible to determine quickly the amplitude and the phase response and to obtain physical insight into the filter characteristics of a discrete-time system by using a graphical technique. The general *N*th-order transfer function *H*[*z*] in Eq. (5.26) can be expressed in factored form as - -$$ -H[z] = b_0 \frac{(z - z_1)(z - z_2) \cdots (z - z_N)}{(z - \gamma_1)(z - \gamma_2) \cdots (z - \gamma_N)} -$$ - -We can compute *H*[*z*] graphically by using the concepts discussed in Sec. 4.10. The directed line segment from *zi* to *z* in the complex plane (Fig. 5.19a) represents the complex number *z* − *zi*. The length of this segment is |*z*−*zi*| and its angle with the horizontal axis is (*z*−*zi*). - -To compute the frequency response *H*[*ej*] we evaluate *H*[*z*] at *z* = *ej*. But for *z* = *ej*, |*z*| = 1 and *z* = so that *z* = *ej* represents a point on the unit circle at an angle with the horizontal. We now connect all zeros (*z*1, *z*2,...,*zN*) and all poles (γ1, γ2, ... , γ*N*) to the point *ej*, as indicated in - -**Figure 5.19** Vector representations of **(a)** complex numbers and **(b)** factors of *H*[*z*]. - -Fig. 5.19b. Let *r*1, *r*2, ... , *rN* be the lengths and φ1, φ2, ... , φ*N* be the angles, respectively, of the straight lines connecting *z*1, *z*2, ... , *zN* to the point *ej*. Similarly, let *d*1, *d*2, ... , *dN* be the lengths and θ1, θ2, ... , θ*N* be the angles, respectively, of the lines connecting γ1, γ2, ... , γ*N* to *ej*. Then - -$$ -H[e^{j\Omega}] = H[z]|_{z=e^{j\Omega}} = b_0 \frac{(r_1 e^{j\phi_1})(r_2 e^{j\phi_2}) \cdots (r_N e^{j\phi_N})}{(d_1 e^{j\theta_1})(d_2 e^{j\theta_2}) \cdots (d_N e^{j\theta_N})} -$$ - -= $b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} e^{j[(\phi_1 + \phi_2 + \cdots + \phi_N) - (\theta_1 + \theta_2 + \cdots + \theta_N)]}$ - -Therefore (assuming *b*0 > 0), - -$$ -|H[e^{j\Omega}]| = b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} = b_0 \frac{\text{product of the distances of zeros to } e^{j\Omega}}{\text{product of distances of poles to } e^{j\Omega}} -$$ -(5.41) - -and - -$$ -\angle H[e^{i\Omega}] = (\phi_1 + \phi_2 + \dots + \phi_N) - (\theta_1 + \theta_2 + \dots + \theta_N) -$$ - -= sum of zero angles to $e^{i\Omega}$ - sum of pole angles to $e^{i\Omega}$ - -In this manner, we can compute the frequency response *H*[*ej*] for any value of by selecting the point on the unit circle at an angle . This point is *ej*. To compute the frequency response *H*[*ej*], we connect all poles and zeros to this point and use the foregoing equations to determine |*H*[*ej*]| and *H*[*ej*]. We repeat this procedure for all values of from 0 to π to obtain the frequency response. - -### CONTROLLING GAIN BY PLACEMENT OF POLES AND ZEROS - -The nature of the influence of pole and zero locations on the frequency response is similar to that observed in continuous-time systems, with minor differences. In place of the imaginary axis of the continuous-time systems, we have the unit circle in the discrete-time case. The nearer the pole (or zero) is to a point *ej* (on the unit circle) representing some frequency , the more influence that pole (or zero) wields on the amplitude response at that frequency because the length of the vector joining that pole (or zero) to the point *ej* is small. The proximity of a pole (or a zero) has a similar effect on the phase response. From Eq. (5.41), it is clear that to enhance the amplitude response at a frequency , we should place a pole as close as possible to the point *ej* (which is on the unit circle).† Similarly, to suppress the amplitude response at a frequency , we should place a zero as close as possible to the point *ej* on the unit circle. Placing repeated poles or zeros will further enhance their influence. - -Total suppression of signal transmission at any frequency can be achieved by placing a zero on the unit circle at a point corresponding to that frequency. This observation is used in the notch (bandstop) filter design. - -Placing a pole or a zero at the origin does not influence the amplitude response because the length of the vector connecting the origin to any point on the unit circle is unity. However, a pole (or a zero) at the origin adds angle − (or ) to *H*[*ej*]. Hence, the phase spectrum − (or ) is a linear function of frequency and therefore represents a pure time delay (or time advance) of *T* seconds (see Drill 5.19). Therefore, a pole (a zero) at the origin causes a time delay (or a time advance) of *T* seconds in the response. There is no change in the amplitude response. - -For a stable system, all the poles must be located inside the unit circle. The zeros may lie anywhere. Also, for a physically realizable system, *H*[*z*] must be a proper fraction, that is, *N* ≥ *M*. If, to achieve a certain amplitude response, we require *M* > *N*, we can still make the system realizable by placing a sufficient number of poles at the origin to make *N* = *M*. This will not change the amplitude response, but it will increase the time delay of the response. - -In general, a pole at a point has the opposite effect of a zero at that point. Placing a zero closer to a pole tends to cancel the effect of that pole on the frequency response. - -### LOWPASS FILTERS - -A lowpass filter generally has a maximum gain at or near = 0, which corresponds to point *ej*0 = 1 on the unit circle. Clearly, placing a pole inside the unit circle near the point *z* = 1 (Fig. 5.20a) would result in a lowpass response.‡ The corresponding amplitude and phase response appear in Fig. 5.20a. For smaller values of , the point *ej* (a point on the unit circle at an angle ) is closer to the pole, and consequently the gain is higher. As increases, the distance of the point *ej* from the pole increases. Consequently the gain decreases, resulting in a lowpass characteristic. Placing a zero at the origin does not change the amplitude response but it does modify the phase response, as illustrated in Fig. 5.20b. Placing a zero at *z* = −1, however, changes both the amplitude and the phase response (Fig. 5.20c). The point *z* = −1 corresponds to frequency - -† The closest we can place a pole is on the unit circle at the point representing . This choice would lead to infinite gain, but should be avoided because it will render the system marginally stable (BIBO-unstable). The closer the point to the unit circle, the more sensitive the system gain to parameter variations. - -‡ Placing the pole at *z* = 1 results in maximum (infinite) gain but renders the system BIBO-unstable, hence should be avoided. - -**Figure 5.20** Various pole-zero configurations and the corresponding frequency responses. - -### 542 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - - = π (*z* = *ej* = *ej*π = −1). Consequently, the amplitude response now becomes more attenuated at higher frequencies, with a zero gain at = π. We can approach ideal lowpass characteristics by using more poles staggered near *z* = 1 (but within the unit circle). Figure 5.20d shows a third-order lowpass filter with three poles near *z* = 1 and a third-order zero at *z* = −1, with corresponding amplitude and phase response. For an ideal lowpass filter, we need an enhanced gain at every frequency in the band (0, *c*). This can be achieved by placing a continuous wall of poles (requiring an infinite number of poles) opposite this band. - -### HIGHPASS FILTERS - -A highpass filter has a small gain at lower frequencies and a high gain at higher frequencies. Such a characteristic can be realized by placing a pole or poles near *z* = −1 because we want the gain at = π to be the highest. Placing a zero at *z* = 1 further enhances suppression of gain at lower frequencies. Figure 5.20e shows a possible pole-zero configuration of the third-order highpass filter with corresponding amplitude and phase responses. - -In the following two examples, we shall realize analog filters by using digital processors and suitable interface devices (C/D and D/C), as shown in Fig. 3.2. At this point, we shall examine the design of a digital processor with the transfer function *H*[*z*] for the purpose of realizing bandpass and bandstop filters in the following examples. - -As Fig. 3.2 shows, the C/D device samples the continuous-time input *x*(*t*) to yield a discrete-time signal *x*[*n*], which serves as the input to *H*[*z*]. The output *y*[*n*] of *H*[*z*] is converted to a continuous-time signal *y*(*t*) by a D/C device. We also saw in Eq. (5.34) that a continuous-time sinusoid of frequency ω, when sampled, results in a discrete-time sinusoid = ω*T*. - -### **EXAMPLE 5.14 Bandpass Filter by Pole-Zero Placement** - -By trial and error, design a tuned (bandpass) analog filter with zero transmission at 0 Hz and also at the highest frequency *fh* = 500 Hz. The resonant frequency is required to be 125 Hz. - -Because *fh* = 500, we require *T* < 1/1000 [see Eq. (5.39)]. Let us select *T* = 10−3. † Recall that the analog frequencies ω correspond to digital frequencies = ω*T*. Hence, analog frequencies ω = 0 and 1000π correspond to = 0 and π, respectively. The gain is required to be zero at these frequencies. Hence, we need to place zeros at *ej* corresponding to = 0 and = π. For = 0, *z* = *ej* = 1; for = π, *ej* = −1. Hence, there must be zeros at *z* = ±1. Moreover, we need enhanced response at the resonant frequency ω = 250π, which corresponds to = π/4, which, in turn, corresponds to *z* = *ej* = *ej*π/4. Therefore, to enhance the frequency response at ω = 250π, we place a pole in the vicinity of *ej*π/4. Because this is a complex pole, we also need its conjugate near *e*−*j*π/4, as indicated in Fig. 5.21a. Let us choose these poles γ1 and γ2 as - -γ1 = |γ |*ej*π/4 and γ2 = |γ |*e*−*j*π/4 - -† Strictly speaking, we need *T* < 0.001. However, we shall show in Ch. 8 that if the input does not contain a finite amplitude component of 500 Hz, *T* = 0.001 is adequate. Generally, practical signals satisfy this condition. - -where |γ | < 1 for stability. The closer γ is to the unit circle, the more sharply peaked is the response around ω = 250π. We also have zeros at ±1. Hence, - -$$ -H[z] = K \frac{(z-1)(z+1)}{(z-|\gamma|e^{j\pi/4})(z-|\gamma|e^{-j\pi/4})} = K \frac{z^2 - 1}{z^2 - \sqrt{2}|\gamma|z+|\gamma|^2} -$$ - -For convenience, we shall choose *K* = 1. The amplitude response is given by - -$$ -|H[e^{i\Omega}]| = \frac{|e^{i2\Omega} - 1|}{|e^{i\Omega} - |\gamma|e^{i\pi/4}||e^{i\Omega} - |\gamma|e^{-i\pi/4}|} -$$ - -Now, by using Eq. (5.37), we obtain - -$$ -|H[e^{i\Omega}]|^2 = \frac{2(1 - \cos 2\Omega)}{\left[1 + |\gamma|^2 - 2|\gamma|\cos\left(\Omega - \frac{\pi}{4}\right)\right] \left[1 + |\gamma|^2 - 2|\gamma|\cos\left(\Omega + \frac{\pi}{4}\right)\right]} -$$ - -**Figure 5.21** Designing a bandpass filter. - -### 544 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -Figure 5.21b shows the amplitude response as a function of ω, as well as = ω*T* = 10−3ω for values of |γ | = 0.83, 0.96, and 1. As expected, the gain is zero at ω = 0 and at 500 Hz (ω = 1000π). The gain peaks at about 125 Hz (ω = 250π). The resonance (peaking) becomes pronounced as |γ | approaches 1. Figure 5.21c shows a canonical realization of this filter, which follows from the transfer function *H*[*z*]. - -## MULTIPLE MAGNITUDE RESPONSE CURVES USING MATLAB - -By defining an anonymous function of the two variables *z* and γ in MATLAB, it is straightforward to duplicate the three magnitude response curves in Fig. 5.21b, corresponding to the cases γ = 0.83, 0.96, and 1. - -``` ->> Omega = linspace(0,pi,400); ->> H = @(z,gamma_m) (z.^2-1)./(z.^2-sqrt(2)*gamma_m*z+gamma_m^2); ->> plot(Omega,abs(H(exp(1j*Omega),0.83)),... ->> Omega,abs(H(exp(1j*Omega),0.96)),... ->> Omega,abs(H(exp(1j*Omega),0.99))); ->> text(.27*pi,35,'|\gamma|=1'); ->> text(.28*pi,25.5,'|\gamma|=0.96'); ->> text(.35*pi,6.41,'|\gamma|=0.83'); ->> set(gca,'xtick',0:pi/4:pi,'ytick',[0 6.41,25.5]); ->> axis([0 pi 0 40]); xlabel('\Omega'); ylabel('|H[e^{j \Omega}]|'); -``` - -The result, shown in Fig. 5.22, confirms the earlier result of Fig. 5.21b. Phase response curves can be generated with minor modification to the MATLAB code. - -**Figure 5.22** MATLAB-generated magnitude response curves for Ex. 5.14. - -### **EXAMPLE 5.15 Bandstop Filter by Pole-Zero Placement** - -Design a second-order notch filter to have zero transmission at 250 Hz and a sharp recovery of gain to unity on both sides of 250 Hz. The highest significant frequency to be processed is *fh* = 400 Hz. - -In this case, *T* < 1/2*fh* = 1.25 × 10−3. Let us choose *T* = 10−3. For the frequency 250 Hz, = 2π(250)*T* = π/2. Thus, the frequency 250 Hz is represented by a point *ej* = *ej*π/2 = *j* on the unit circle, as depicted in Fig. 5.23a. Since we need zero transmission at this frequency, we must place a zero at *z* = *ej*π/2 = *j* and its conjugate at *z* = *e*−*j*π/2 = −*j*. We also require a sharp recovery of gain on both sides of frequency 250 Hz. To accomplish this goal, we place two poles close to the two zeros, to cancel out the effect of the two zeros as we move away from the point *j* (corresponding to frequency 250 Hz). For this reason, let us use poles at ±*ja* with *a* < 1 for stability. The closer the poles are to zeros (the closer the *a* to 1), the faster is the gain recovery on either side of 250 Hz. The resulting transfer function is - -$$ -H[z] = K \frac{(z-j)(z+j)}{(z-ja)(z+ja)} = K \frac{z^2+1}{z^2+a^2} -$$ - -The dc gain (gain at = 0, or *z* = 1 ) of this filter is - -$$ -H[1] = K \frac{2}{1 + a^2} -$$ - -Because we require a dc gain of unity, we must select *K* = (1+*a*2)/2. The transfer function is therefore - -$$ -H[z] = \frac{(1+a^2)(z^2+1)}{2(z^2+a^2)} -$$ - -and according to Eq. (5.37), - -$$ -|H[e^{j\Omega}]|^{2} = \frac{(1+a^{2})^{2}}{4} \frac{(e^{j2\Omega}+1)(e^{-j2\Omega}+1)}{(e^{j2\Omega}+a^{2})(e^{-j2\Omega}+a^{2})} -$$ -$$ -= \frac{(1+a^{2})^{2}(1+\cos 2\Omega)}{2(1+a^{4}+2a^{2}\cos 2\Omega)} -$$ - -Figure 5.23b shows |*H*[*ej*]| for values of *a* = 0.3, 0.6, and 0.95. Figure 5.23c shows a realization of this filter. - -### **DR ILL 5.21 Highpass Filter by Pole-Zero Placement** - -Use the graphical argument to show that a filter with transfer function - -$$ -H[z] = \frac{z - 0.9}{z} -$$ - -acts like a highpass filter. Make a rough sketch of the amplitude response. - -## **5.7 DIGITAL [PROCESSING OF](#page-11-0) ANALOG SIGNALS** - -An analog (meaning continuous-time) signal can be processed digitally by sampling the analog signal and processing the samples by a digital (meaning discrete-time) processor. The output of the processor is then converted back to analog signal, as shown in Fig. 5.24a. We saw some simple cases of such processing in Exs. 3.8, 3.9, 5.14, and 5.15. In this section, we shall derive a criterion for designing such a digital processor for a general LTIC system. - -Suppose that we wish to realize an equivalent of an analog system with transfer function *Ha*(*s*), shown in Fig. 5.24b. Let the digital processor transfer function in Fig. 5.24a that realizes this desired *Ha*(*s*) be *H*[*z*]. In other words, we wish to make the two systems in Fig. 5.24 equivalent (at least approximately). - -By "equivalence" we mean that for a given input *x*(*t*), the systems in Fig. 5.24 yield the same output *y*(*t*). Therefore, *y*(*nT*), the samples of the output in Fig. 5.24b, are identical to *y*[*n*], the output of *H*[*z*] in Fig. 5.24a. - -**Figure 5.24** Analog filter realization with a digital filter. - -### 548 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -For the sake of generality, we are assuming a noncausal system. The argument and the results are also valid for causal systems. The output *y*(*t*) of the system in Fig. 5.24b is - -$$ -y(t) = \int_{-\infty}^{\infty} x(\tau)h_a(t-\tau) d\tau = \lim_{\Delta \tau \to 0} \sum_{m=-\infty}^{\infty} x(m\Delta \tau)h_a(t-m\Delta \tau) \Delta \tau -$$ - -For our purpose, it is convenient to use the notation *T* for τ . Assuming *T* (the sampling interval) to be small enough, such a change of notation yields - -$$ -y(t) = T \sum_{m=-\infty}^{\infty} x(mT)h_a(t - mT) -$$ - -The response at the *n*th sampling instant is *y*(*nT*) obtained by setting *t* = *nT* in the equation is - -$$ -y(nT) = T \sum_{m=-\infty}^{\infty} x(mT)h_a[(n-m)T] -$$ -\n(5.42) - -In Fig. 5.24a, the input to *H*[*z*] is *x*(*nT*) = *x*[*n*]. If *h*[*n*] is the unit impulse response of *H*[*z*], then *y*[*n*], the output of *H*[*z*], is given by - -$$ -y[n] = \sum_{m = -\infty}^{\infty} x[m]h[n-m] -$$ -\n(5.43) - -If the two systems are to be equivalent, *y*(*nT*) in Eq. (5.42) must be equal to *y*[*n*] in Eq. (5.43). Therefore, - -$$ -h[n] = Th_a(nT) \tag{5.44} -$$ - -This is the time-domain criterion for equivalence of the two systems.† According to this criterion, *h*[*n*], the unit impulse response of *H*[*z*] in Fig. 5.24a, should be *T* times the samples of *ha*(*t*), the unit impulse response of the system in Fig. 5.24b. This is known as the *impulse invariance criterion* of filter design. - -Strictly speaking, this realization guarantees the output equivalence only at the sampling instants, that is, *y*(*nT*) = *y*[*n*], and that also requires the assumption that *T* → 0. Clearly, this criterion leads to an approximate realization of *Ha*(*s*). However, it can be shown that when the frequency response of |*Ha*(*j*ω)| is bandlimited, the realization is exact [2], provided the sampling rate is high enough to avoid any aliasing (*T* < 1/2*fh*). - -### REALIZATION OF RATIONAL *H*(*s*) - -If we wish to realize an analog filter with transfer function - -$$ -H_a(s) = \frac{c}{s - \lambda} -$$ - -† Because *T* is a constant, some authors ignore the factor *T*, which yields a simplified criterion *h*[*n*] = *ha*(*nT*). Ignoring *T* merely scales the amplitude response of the resulting filter. - -The impulse response *h*(*t*), given by the inverse Laplace transform of *Ha*(*s*), is - -$$ -h_a(t) = ce^{\lambda t}u(t) -$$ - -The corresponding digital filter unit impulse response *h*[*n*], per Eq. (5.44), is - -$$ -h[n] = Th_a(nT) = Tce^{n\lambda T} -$$ - -Figure 5.25 shows *ha*(*t*) and *h*[*n*]. The corresponding *H*[*z*], the *z*-transform of *h*[*n*], as found from Table 5.1, is - -$$ -H[z] = \frac{Tcz}{z - e^{\lambda T}} -$$ -\n(5.45) - -**Figure 5.25** Impulse response for analog and digital systems in the impulse invariance method of filter design. - -| No. | Ha(s) | ha(t) | h[n] | H[z] | -|-----|------------------|-------------------------------|-----------------------------|-----------------------------------------------------------------| -| 1 | K | Kδ(t) | TKδ[n] | TK | -| 2 | 1
s | u(t) | Tu[n] | Tz
z−1 | -| 3 | 1
s2 | t | nT2 | T2z
(z−1)2 | -| 4 | 1
s3 | 2
t
2 | k2T3
2 | T3z(z
+1)
2(z−1)3 | -| 5 | 1
s −λ | eλt | TeλnT | Tz
z−eλT | -| 6 | 1
(s −λ)2 | teλt | nT2eλnT | T2zeλT
(z−eλT )2 | -| 7 | As+B
s2+2as+c | Tre−at cos(bt+θ
) | Tre−anT cos(bnT+θ
) | Trz[z cos θ −e−aT cos(bT−θ
)]
z2−(2e−aT cos
bT)z+e−2aT | -| | r = |
A2c+B2 −2ABa
,
c−a2 | b = √
θ = tan−1
c−a2, | Aa−B
√
c−a2
A | - -**TABLE 5.3** Select Impulse-Invariance Pairs - -The procedure of finding *H*[*z*] can be systematized for any *N*th-order system. First we express an *N*th-order analog transfer function *Ha*(*s*) as a sum of partial fractions as‡ - -$$ -H_a(s) = \sum_{i=1}^n \frac{c_i}{s - \lambda_i} -$$ - -Then the corresponding *H*[*z*] is given by - -$$ -H[z] = T \sum_{i=1}^{n} \frac{c_i z}{z - e^{\lambda_i T}} -$$ - -This transfer function can be readily realized, as explained in Sec. 5.4. Table 5.3 lists several pairs of *Ha*(*s*) and their corresponding *H*[*z*]. For instance, to realize a digital integrator, we examine its *Ha*(*s*) = 1/*s*. From Table 5.3, corresponding to *Ha*(*s*) = 1/*s* (pair 2), we find *H*[*z*] = *Tz*/(*z* − 1). This is exactly the result we obtained in Ex. 3.9 using another approach. - -Note that the frequency response *Ha*(*j*ω) of a practical analog filter cannot be bandlimited. Consequently, all these realizations are approximate. - -### CHOOSING THE SAMPLING INTERVAL *T* - -The impulse-invariance criterion (5.44) was derived under the assumption that *T* → 0. Such an assumption is neither practical nor necessary for satisfactory design. Avoiding of aliasing is the most important consideration for the choice of *T*. In Eq. (5.39), we showed that for a sampling interval *T* seconds, the highest frequency that can be sampled without aliasing is 1/2*T* Hz or π/*T* radians per second. This implies that *Ha*(*j*ω), the frequency response of the analog filter in Fig. 5.24b should not have spectral components beyond frequency π/*T* radians per second. In other words, to avoid aliasing, the frequency response of the system *Ha*(*s*) must be bandlimited to π/*T* radians per second. We shall see later in Ch. 7 that frequency response of a realizable LTIC system cannot be bandlimited; that is, the response generally exists for all frequencies up to ∞. Therefore, it is impossible to digitally realize an LTIC system exactly without aliasing. The saving grace is that the frequency response of every realizable LTIC system decays with frequency. This allows for a compromise in digitally realizing an LTIC system with an acceptable level of aliasing. The smaller the value of *T*, the smaller the aliasing, and the better the approximation. Since it is impossible to make |*Ha*(*j*ω)| zero, we are satisfied with making it negligible beyond the frequency π/*T*. As a rule of thumb [3], we choose *T* such that |*Ha*(*j*ω)| at the frequency ω = π/*T* is less than a certain fraction (often taken as 1%) of the peak value of |*Ha*(*j*ω)|. This ensures that aliasing is negligible. The peak |*Ha*(*j*ω)| usually occurs at ω = 0 for lowpass filters and at the band center frequency ω*c* for bandpass filters. - -‡ Assuming *Ha*(*s*) has simple poles. For repeated poles, the form changes accordingly. Entry 6 in Table 5.3 is suitable for repeated poles. - -### **EXAMPLE 5.16 Butterworth Filter Design by the Impulse-Invariance Method** - -Design a digital filter to realize a first-order lowpass Butterworth filter with the transfer function - -$$ -H_a(s) = \frac{\omega_c}{s + \omega_c} \qquad \omega_c = 10^5 \tag{5.46} -$$ - -For this filter, we find the corresponding *H*[*z*] according to Eq. (5.45) (or pair 5 in Table 5.3) as - -$$ -H[z] = \frac{\omega_c T z}{z - e^{-\omega_c T}} -$$ -\n(5.47) - -Next, we select the value of *T* by means of the criterion according to which the gain at ω = π/*T* drops to 1% of the maximum filter gain. However, this choice results in such a good design that aliasing is imperceptible. The resulting amplitude response is so close to the desired response that we can hardly notice the aliasing effect in our plot. For the sake of demonstrating the aliasing effect, we shall deliberately select a 10% criterion (instead of 1%). We have - -$$ -|H_a(j\omega)| = \left|\frac{\omega_c}{\sqrt{\omega^2 + \omega_c^2}}\right| -$$ - -In this case |*Ha*(*j*ω)|max = 1, which occurs at ω = 0. Use of 10% criterion leads to |*Ha*(π/*T*)| = 0.1. Observe that - -$$ -|H_a(j\omega)| \approx \frac{\omega_c}{\omega} \qquad \omega \gg \omega_c -$$ - -Hence, - -$$ -|H_a(\pi/T)| \approx \frac{\omega_c}{\pi/T} = 0.1 \quad \Longrightarrow \quad \pi/T = 10\omega_c = 10^6 -$$ - -Thus, the 10% criterion yields *T* = 10−6π. The 1% criterion would have given *T* = 10−7π. Substitution of *T* = 10−6π in Eq. (5.47) yields - -$$ -H[z] = \frac{0.3142z}{z - 0.7304} -$$ -\n(5.48) - -A canonical realization of this filter is shown in Fig. 5.26a. - -To find the frequency response of this digital filter, we rewrite *H*[*z*] as - -$$ -H[z] = \frac{0.3142}{1 - 0.7304z^{-1}} -$$ - -Therefore, - -$$ -H[e^{j\omega T}] = \frac{0.3142}{1 - 0.7304e^{-j\omega T}} = \frac{0.3142}{(1 - 0.7304 \cos \omega T) + j0.7304 \sin \omega T} -$$ - -**Figure 5.26** An example of filter design by the impulse-invariance method: **(a)** filter realization, **(b)** amplitude response, and **(c)** phase response. - -The corresponding magnitude response is - -$$ -|H[e^{j\omega T}]| = \frac{0.3142}{\sqrt{(1 - 0.7304 \cos \omega T)^2 + (0.7304 \sin \omega T)^2}} -$$ -$$ -= \frac{0.3142}{\sqrt{1.533 - 1.4608 \cos \omega T}} -$$ -(5.49) - -### 5.7 Digital Processing of Analog Signals 553 - -and the phase response is - -$$ -\angle H[e^{j\omega T}] = -\tan^{-1}\left(\frac{0.7304 \sin \omega T}{1 - 0.7304 \cos \omega T}\right) -$$ - (5.50) - -This frequency response differs from the desired response *Ha*(*j*ω) because aliasing causes frequencies above π/*T* to appear as frequencies below π/*T*. This generally results in increased gain for frequencies below π/*T*. For instance, the realized filter gain at ω = 0 is *H*[*ej*0] = *H*[1]. This value, as obtained from Eq. (5.48), is 1.1654 instead of the desired value 1. We can partly compensate for this distortion by multiplying *H*[*z*] or *H*[*ej*ω*T* ] by a normalizing constant *K* = *Ha*(0)/*H*[1] = 1/1.1654 = 0.858. This forces the resulting gain of *H*[*ej*ω*T* ] to be equal to 1 at ω = 0. The normalized *Hn*[*z*] = 0.858*H*[*z*] = 0.858(0.1π*z*/(*z*−0.7304)). The amplitude response in Eq. (5.49) is multiplied by *K* = 0.858 and plotted in Fig. 5.26b over the frequency range 0 ≤ ω ≤ π/*T* = 106. The multiplying constant *K* has no effect on the phase response in Eq. (5.50), which is shown in Fig. 5.26c. - -Also, the desired frequency response, according to Eq. (5.46) with ω*c* = 105, is - -$$ -H_a(j\omega) = \frac{\omega_c}{j\omega + \omega_c} = \frac{10^5}{j\omega + 10^5} -$$ - -Therefore, - -$$ -|H_a(j\omega)| = \frac{10^5}{\sqrt{\omega^2 + 10^{10}}} \quad \text{and} \quad \angle H_a(j\omega) = -\tan^{-1}\frac{\omega}{10^5} -$$ - -This desired amplitude and phase response are plotted (dotted) in Figs. 5.26b and 5.26c for comparison with realized digital filter response. Observe that the amplitude response behavior of the analog and the digital filter is very close over the range ω ≤ ω*c* = 105. However, for higher frequencies, there is considerable aliasing, especially in the phase spectrum. Had we used the 1% rule, the realized frequency response would have been closer over another decade of the frequency range. - -### IMPULSE INVARIANCE BY MATLAB - -We can readily use the MATLAB impinvar command to confirm our digital filter designed by the impulse-invariance method. - -``` ->> omegac = 10^5; Ba = [omegac]; Aa = [1 omegac]; Fs = 10^6/pi; ->> [B,A] = impinvar(Ba,Aa,Fs) - B = 0.3142 - A = 1.0000 -0.7304 -``` - -This confirms our earlier result of Eq. (5.48) that the digital filter transfer function is - -$$ -H[z] = \frac{0.3142z}{z - 0.7304} -$$ - -### **DR ILL 5.22 Filter Design by the Impulse-Invariance Method** - -Design a digital filter to realize an analog transfer function - -$$ -H_a(s) = \frac{20}{s+20} -$$ - -**ANSWER** - -*H*[*z*] = 20*Tz z*−*e*−20*T* with *T* = π 2000 - -## **5.8 THE BILATERAL** *z***[-TRANSFORM](#page-11-0)** - -Situations involving noncausal signals or systems cannot be handled by the (unilateral) *z*-transform discussed so far. Such cases can be analyzed by the *bilateral* (or two-sided) *z*-transform defined in Eq. (5.1) as - -$$ -X[z] = \sum_{n=-\infty}^{\infty} x[n]z^{-n} -$$ - -As in Eq. (5.2), the inverse *z*-transform is given by - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ - -These equations define the bilateral *z*-transform. Earlier, we showed that - -$$ -\gamma^n u[n] \Longleftrightarrow \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.51} -$$ - -In contrast, the *z*-transform of the signal −γ *nu*[−(*n*+1)], illustrated in Fig. 5.27a, is - -$$ -\mathcal{Z}\{-\gamma^n u[-(n+1)]\} = \sum_{-\infty}^{-1} -\gamma^n z^{-n} = \sum_{-\infty}^{-1} -\left(\frac{\gamma}{z}\right)^n -$$ -$$ -= -\left[\frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right] -$$ -$$ -= 1 - \left[1 + \frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right] -$$ -$$ -= 1 - \frac{1}{1 - \frac{z}{\gamma}} \qquad \left|\frac{z}{\gamma}\right| < 1 -$$ -$$ -= \frac{z}{z - \gamma} \qquad |z| < |\gamma| -$$ - -**Figure 5.27 (a)** −γ *nu*[−(*n*+1)] and **(b)** the region of convergence (ROC) of its *z*-transform. - -Therefore, - -$$ -\mathcal{Z}\{-\gamma^{n}u[-(n+1)]\} = \frac{z}{z-\gamma} \qquad |z| < |\gamma| \tag{5.52} -$$ - -A comparison of Eqs. (5.51) and (5.52) shows that the *z*-transform of γ *nu*[*n*] is identical to that of −γ *nu*[−(*n* + 1)]. The regions of convergence, however, are different. In the former case, *X*[*z*] converges for |*z*| > |γ |; in the latter, *X*[*z*] converges for |*z*| < |γ | (see Fig. 5.27b). Clearly, the inverse transform of *X*[*z*] is not unique unless the region of convergence is specified. If we add the restriction that all our signals be causal, however, this ambiguity does not arise. The inverse transform of *z*/(*z* − γ ) is γ *nu*[*n*] even without specifying the ROC. Thus, in the unilateral transform, we can ignore the ROC in determining the inverse *z*-transform of *X*[*z*]. - -As in the case of the bilateral Laplace transform, if *x*[*n*] = %*k i*=1 *xi*[*n*], then the ROC for *X*[*z*] is the intersection of the ROCs (region common to all ROCs) for the transforms *X*1[*z*],*X*2[*z*],...,*Xk*[*z*]. - -The preceding results lead to the conclusion (similar to that for the Laplace transform) that if *z* = β is the largest magnitude pole for a causal sequence, its ROC is |*z*| > |β|. If *z* = α is the smallest magnitude nonzero pole for an anticausal sequence, its ROC is |*z*| < |α|. - -### REGION OF CONVERGENCE FOR LEFT-SIDED AND RIGHT-SIDED SEQUENCES - -Let us first consider a finite duration sequence *xf*[*n*], defined as a sequence that is nonzero for *N*1 ≤ *n* ≤ *N*2, where both *N*1 and *N*2 are finite numbers and *N*2 > *N*1. Also, - -$$ -X_f[z] = \sum_{n=N_1}^{N_2} x_f[n]z^{-n} -$$ - -For example, if *N*1 = −2 and *N*2 = 1, then - -$$ -X_f[z] = x_f[-2]z^2 + x_f[-1]z + x_f[0] + \frac{x_f[1]}{z} -$$ - -Assuming all the elements in *xf*[*n*] are finite, we observe that *Xf*[*z*] has two poles at *z* = ∞ because of terms *xf*[−2]*z*2 +*xf*[−1]*z* and one pole at *z* = 0 because of term *xf*[1]/*z*. Thus, a finite-duration sequence could have poles at *z* = 0 and *z* = ∞. Observe that *Xf*[*z*] converges for all values of *z* except possibly *z* = 0 and *z* = ∞. - -This means that the ROC of a general signal *x*[*n*] + *xf*[*n*] is the same as the ROC of *x*[*n*] with the possible exception of *z* = 0 and *z* = ∞. - -A *right-sided* sequence is zero for *n* < *N*2 < ∞ and a left-sided sequence is zero for *n* > *N*1 > −∞. A causal sequence is always a right-sided sequence, but the converse is not necessarily true. An anticausal sequence is always a left-sided sequence, but the converse is not necessarily true. A *two-sided* sequence is of infinite duration and is neither right-sided nor left-sided. - -A right-sided sequence *xr*[*n*] can be expressed as *xr*[*n*] = *xc*[*n*] +*xf*[*n*], where *xc*[*n*] is a causal signal and *xf*[*n*] is a finite-duration signal. Therefore, the ROC for *xr*[*n*] is the same as the ROC for *xc*[*n*] except possibly *z* = ∞. If *z* = β is the largest magnitude pole for a right-sided sequence *xr*[*n*], its ROC is |β| < |*z*|≤∞. Similarly, a left-sided sequence can be expressed as *xl*[*n*] = *xa*[*n*]+*xf*[*n*], where *xa*[*n*] is an anticausal sequence and *xf*[*n*] is a finite-duration signal. Therefore, the ROC for *xl*[*n*] is the same as the ROC for *xa*[*n*] except possibly *z*=0. Thus, if*z*=α is the smallest magnitude nonzero pole for a left-sided sequence, its ROC is 0 ≤ |*z*| < |α|. - -### **EXAMPLE 5.17 Bilateral** *z***-Transform** - -Determine the bilateral *z*-transform of - -$$ -x[n] = \underbrace{(0.9)^n u[n]}_{x_1[n]} + \underbrace{(1.2)^n u[-(n+1)]}_{x_2[n]} -$$ - -From the results in Eqs. (5.51) and (5.52), we have - -$$ -X_1[z] = \frac{z}{z - 0.9} \qquad |z| > 0.9 -$$ - -$$ -X_2[z] = \frac{-z}{z - 1.2} \qquad |z| < 1.2 -$$ - -The common region where both *X*1[*z*] and *X*2[*z*] converge is 0.9 < |*z*| < 1.2 (Fig. 5.28b). Hence, - -$$ -X[z] = X_1[z] + X_2[z] -$$ - -= $\frac{z}{z - 0.9} - \frac{z}{z - 1.2}$ -= $\frac{-0.3z}{(z - 0.9)(z - 1.2)}$ 0.9 < |z| < 1.2 - -The sequence *x*[*n*] and the ROC of *X*[*z*] are depicted in Fig. 5.28. - -### **EXAMPLE 5.18 Inverse Bilateral** *z***-Transform** - -Find the inverse bilateral *z*-transform of - -$$ -X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)} -$$ - -if the ROC is **(a)** |*z*| > 2, **(b)** |*z*| < 0.8, and **(c)** 0.8 < |*z*| < 2. - -**(a)** - -$$ -\frac{X[z]}{z} = \frac{-(z+0.4)}{(z-0.8)(z-2)} = \frac{1}{z-0.8} - \frac{2}{z-2} -$$ - -and - -$$ -X[z] = \frac{z}{z - 0.8} - 2\frac{z}{z - 2} -$$ - -Since the ROC is |*z*| > 2, both terms correspond to causal sequences and - -$$ -x[n] = [(0.8)^n - 2(2)^n]u[n] -$$ - -This sequence appears in Fig. 5.29a. - -**(b)** In this case, |*z*| < 0.8, which is less than the magnitudes of both poles. Hence, both terms correspond to anticausal sequences, and - -$$ -x[n] = [-(0.8)^n + 2(2)^n]u[-(n+1)] -$$ - -This sequence appears in Fig. 5.29b. - -**(c)** In this case, 0.8 < |*z*| < 2; the part of *X*[*z*] corresponding to the pole at 0.8 is a causal sequence, and the part corresponding to the pole at 2 is an anticausal sequence: - -$$ -x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)] -$$ - -This sequence appears in Fig. 5.29c. - -### **DR ILL 5.23 Inverse Bilateral** *z***-Transform** - -Find the inverse bilateral *z*-transform of - -$$ -X[z] = \frac{z}{z^2 + \frac{5}{6}z + \frac{1}{6}} \qquad \frac{1}{2} > |z| > \frac{1}{3} -$$ - -### **ANSWER** - - − 1 3 *n u*[*n*] +6 − 1 2 *n u*[−(*n*+1)] - -INVERSE TRANSFORM BY EXPANSION OF *X*[*z*] IN POWER SERIES OF *z* We have - -$$ -X[z] = \sum_{n} x[n]z^{-n} -$$ - -For an anticausal sequence, which exists only for *n* ≤ −1, this equation becomes - -$$ -X[z] = x[-1]z + x[-2]z^{2} + x[-3]z^{3} + \cdots -$$ - -We can find the inverse *z*-transform of *X*[*z*] by dividing the numerator polynomial by the denominator polynomial, both in ascending powers of *z*, to obtain a polynomial in ascending powers of *z*. Thus, to find the inverse transform of *z*/(*z* − 0.5) (when the ROC is |*z*| < 0.5), we divide *z* by −0.5+*z* to obtain −2*z*−4*z*2−8*z*3−· · ·. Hence, *x*[−1]=−2, *x*[−2]=−4, *x*[−3]=−8, and so on. - -## **[5.8-1 Properties of the Bilateral](#page-11-0)** *z***-Transform** - -Properties of the bilateral *z*-transform are similar to those of the unilateral transform. We shall merely state the properties here, without proofs, for *xi*[*n*] ⇐⇒ *Xi*[*z*]. - -LINEARITY - -$$ -a_1x_1[n]+a_2x_2[n] \Longleftrightarrow a_1X_1[z]+a_2X_2[z] -$$ - -The ROC for *a*1*X*1[*z*] + *a*2*X*2[*z*] is the region common to (intersection of) the ROCs for *X*1[*z*] and *X*2[*z*]. - -SHIFT - -*x*[*n*−*m*] ⇐⇒ 1 *zm X*[*z*] *m* is positive or negative integer - -The ROC for *X*[*z*]/*zm* is the ROC for *X*[*z*] except for the addition or deletion of *z* = 0 or *z* = ∞ caused by the factor 1/*zm*. - -CONVOLUTION - -$$ -x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z] -$$ - -The ROC for *X*1[*z*]*X*2[*z*] is the region common to (intersection of) the ROCs for *X*1[*z*] and *X*2[*z*]. - -MULTIPLICATION BY γ *n* - -$$ -\gamma^n x[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right] -$$ - -If the ROC for *X*[*z*] is |γ1| < |*z*| < |γ2|, then the ROC for *X*[*z*/γ ] is |γ γ1| < |*z*| < |γ γ2|, indicating that the ROC is scaled by the factor |γ |. - -MULTIPLICATION BY *n* - -$$ -nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z] -$$ - -The ROC for −*z*(*dX*/*dz*) is the same as the ROC for *X*[*z*]. - -TIME REVERSAL - -$$ -x[-n] \Longleftrightarrow X[1/z] -$$ - -If the ROC for *X*[*z*] is |γ1| < |*z*| < |γ2|, then the ROC for *X*[1/*z*] is 1/|γ1| > |*z*| > |1/γ2|. - -COMPLEX CONJUGATION - -*x*∗[*n*] ⇐⇒ *X*∗[*z* ∗] - -The ROC for *X*∗[*z*∗] is the same as the ROC for *X*[*z*]. - -## **5.8-2 Using the Bilateral** *z***[-Transform for Analysis of LTID Systems](#page-11-0)** - -Because the bilateral *z*-transform can handle noncausal signals, we can use this transform to analyze noncausal linear systems. The zero-state response *y*[*n*] is given by - -$$ -y[n] = \mathcal{Z}^{-1}\{X[z]H[z]\} -$$ - -provided *X*[*z*]*H*[*z*] exists. The ROC of *X*[*z*]*H*[*z*] is the region in which both *X*[*z*] and *H*[*z*] exist, which means that the region is the common part of the ROC of both *X*[*z*] and *H*[*z*]. - -### **EXAMPLE 5.19 Zero-State Response by Bilateral** *z***-Transform** - -For a causal system specified by the transfer function - -$$ -H[z] = \frac{z}{z - 0.5} -$$ - -find the zero-state response to input - -$$ -x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)] -$$ - -$$ -X[z] = \frac{z}{z - 0.8} - \frac{2z}{z - 2} = \frac{-z(z + 0.4)}{(z - 0.8)(z - 2)} -$$ - -The ROC corresponding to the causal term is |*z*| > 0.8, and that corresponding to the anticausal term is |*z*| < 2. Hence, the ROC for *X*[*z*] is the common region, given by 0.8 < |*z*| < 2. Hence, - -$$ -X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)} \qquad 0.8 < |z| < 2 -$$ - -Therefore, - -$$ -Y[z] = X[z]H[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)} -$$ - -Since the system is causal, the ROC of *H*[*z*] is |*z*| > 0.5. The ROC of *X*[*z*] is 0.8 < |*z*| < 2. The common region of convergence for *X*[*z*] and *H*[*z*] is 0.8 < |*z*| < 2. Therefore, - -$$ -Y[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)} \qquad 0.8 < |z| < 2 -$$ - -Expanding *Y*[*z*] into modified partial fractions yields - -$$ -Y[z] = -\frac{z}{z - 0.5} + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) - \frac{8}{3} \left( \frac{z}{z - 2} \right) \qquad 0.8 < |z| < 2 -$$ - -Since the ROC extends outward from the pole at 0.8, both poles at 0.5 and 0.8 correspond to causal sequence. The ROC extends inward from the pole at 2. Hence, the pole at 2 corresponds to anticausal sequence. Therefore, - -$$ -y[n] = \left[ -(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] + \frac{8}{3}(2)^n u[-(n+1)] -$$ - -### **EXAMPLE 5.20 Zero-State Response for an Input with No** *z***-Transform** - -For the system in Ex. 5.19, find the zero-state response to input - -$$ -x[n] = \underbrace{(0.8)^n u[n]}_{x_1[n]} + \underbrace{(0.6)^n u[-(n+1)]}_{x_2[n]} -$$ - -The *z*-transforms of the causal and anticausal components *x*1[*n*] and *x*2[*n*] of the output are - -$$ -X_1[z] = \frac{z}{z - 0.8} \qquad |z| > 0.8 -$$ - -$$ -X_2[z] = \frac{-z}{z - 0.6} \qquad |z| < 0.6 -$$ - -Observe that a common ROC for *X*1[*z*] and *X*2[*z*] does not exist. Therefore, *X*[*z*] does not exist. In such a case we take advantage of the superposition principle and find *y*1[*n*] and *y*2[*n*], the system responses to *x*1[*n*] and *x*2[*n*], separately. The desired response *y*[*n*] is the sum of *y*1[*n*] and *y*2[*n*]. Now - -$$ -H[z] = \frac{z}{z - 0.5} -$$ - $|z| > 0.5$ -\n -$$ -Y_1[z] = X_1[z]H[z] = \frac{z^2}{(z - 0.5)(z - 0.8)} -$$ - $|z| > 0.8$ -\n -$$ -Y_2[z] = X_2[z]H[z] = \frac{-z^2}{(z - 0.5)(z - 0.6)} -$$ - $0.5 < |z| < 0.6$ - -Expanding *Y*1[*z*] and *Y*2[*z*] into modified partial fractions yields - -$$ -Y_1[z] = -\frac{5}{3} \left( \frac{z}{z - 0.5} \right) + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) \qquad |z| > 0.8 -$$ - -$$ -Y_2[z] = 5 \left( \frac{z}{z - 0.5} \right) - 6 \left( \frac{z}{z - 0.6} \right) \qquad 0.5 < |z| < 0.6 -$$ - -Therefore, - -$$ -y_1[n] = \left[ -\frac{5}{3}(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] -$$ - -$$ -y_2[n] = 5(0.5)^n u[n] + 6(0.6)^n u[-(n+1)] -$$ - -and - -$$ -y[n] = y_1[n] + y_2[n] = \left[\frac{10}{3}(0.5)^n + \frac{8}{3}(0.8)^n\right]u[n] + 6(0.6)^nu[-(n+1)] -$$ - -### **DR ILL 5.24 Zero-State Response by Bilateral** *z***-Transform** - -For the causal system in Ex. 5.19, find the zero-state response to input - -$$ -x[n] = \left(\frac{1}{4}\right)^n u[n] + 5(3)^n u[-(n+1)] -$$ - -**ANSWER** - -− 1 4 *n* +3 1 2 *n u*[*n*] +6(3)*nu*[−(*n*+1)] - -## **[5.9 CONNECTING THE](#page-11-0) LAPLACE AND** *z***-TRANSFORMS** - -We now show that discrete-time systems also can be analyzed by means of the Laplace transform. In fact, we shall see that *the z-transform is the Laplace transform in disguise* and that discrete-time systems can be analyzed as if they were continuous-time systems. - -So far we have considered the discrete-time signal as a sequence of numbers and not as an electrical signal (voltage or current). Similarly, we considered a discrete-time system as a mechanism that processes a sequence of numbers (input) to yield another sequence of numbers (output). The system was built by using delays (along with adders and multipliers) that delay sequences of numbers. A digital computer is a perfect example: every signal is a sequence of numbers, and the processing involves delaying sequences of numbers (along with addition and multiplication). - -Now suppose we have a discrete-time system with transfer function *H*[*z*] and input *x*[*n*]. Consider a continuous-time signal *x*(*t*) such that its *n*th sample value is *x*[*n*], as shown in Fig. 5.30.† Let the sampled signal be *x*(*t*), consisting of impulses spaced *T* seconds apart with the *n*th impulse of strength *x*[*n*]. Thus, - -$$ -\bar{x}(t) = \sum_{n=0}^{\infty} x[n]\delta(t - nT) -$$ - -Figure 5.30 shows *x*[*n*] and the corresponding *x*(*t*). The signal *x*[*n*] is applied to the input of a discrete-time system with transfer function *H*[*z*], which is generally made up of delays, adders, and scalar multipliers. Hence, processing *x*[*n*] through *H*[*z*] amounts to operating on the sequence *x*[*n*] by means of delays, adders, and scalar multipliers. Suppose for *x*(*t*) samples, we perform operations identical to those performed on the samples of *x*[*n*] by *H*[*z*]. For this purpose, we need a continuous-time system with transfer function *H*(*s*) that is identical in structure to the discrete-time system *H*[*z*] except that the delays in *H*[*z*] are replaced by elements that delay continuous-time signals (such as voltages or currents). There is no other difference between realizations of *H*[*z*] and *H*(*s*). If a continuous-time impulse δ(*t*) is applied to such a delay of *T* seconds, the output will be δ(*t* −*T*). The continuous-time transfer function of such a delay is *e*−*sT* [see Eq. (4.30)]. Hence, the delay elements with transfer function 1/*z* in the realization of *H*[*z*] will be replaced by the delay elements with transfer function *e*−*sT* in the realization of the corresponding *H*(*s*). This is the same - -† We can construct such *x*(*t*) from the sample values, as will be explained in Ch. 8. - -**Figure 5.30** Connection between the Laplace transform and the *z*-transform. - -as *z* being replaced by *esT* . Therefore, *H*(*s*) = *H*[*esT* ]. Let us now apply *x*[*n*] to the input of *H*[*z*] and apply *x*(*t*) at the input of *H*[*esT* ]. Whatever operations are performed by the discrete-time system *H*[*z*] on *x*[*n*] (Fig. 5.30a) are also performed by the corresponding continuous-time system *H*[*esT* ] on the impulse sequence *x*(*t*) (Fig. 5.30b). The delaying of a sequence in *H*[*z*] would amount to delaying of an impulse train in *H*[*esT* ]. Adding and multiplying operations are the same in both cases. In other words, one-to-one correspondence of the two systems is preserved in every aspect. Therefore if *y*[*n*] is the output of the discrete-time system in Fig. 5.30a, then *y*(*t*), the output of the continuous-time system in Fig. 5.30b, would be a sequence of impulse whose *n*th impulse strength is *y*[*n*]. Thus, - -$$ -\bar{y}(t) = \sum_{n=0}^{\infty} y[n]\delta(t - nT) -$$ - -The system in Fig. 5.30b, being a continuous-time system, can be analyzed via the Laplace transform. If - -$$ -\overline{x}(t) \Longleftrightarrow \overline{X}(s) \quad \text{and} \quad \overline{y}(t) \Longleftrightarrow \overline{Y}(s) -$$ - -then - -$$ -\overline{Y}(s) = H[e^{sT}]\overline{X}(s) -$$ -\n(5.53) - -Also, - -$$ -\overline{X}(s) = \mathcal{L}\left[\sum_{n=0}^{\infty} x[n]\delta(t - nT)\right] -$$ - -Now because the Laplace transform of δ(*t* −*nT*) is *e*−*snT* , - -$$ -\overline{X}(s) = \sum_{n=0}^{\infty} x[n]e^{-snT} \quad \text{and} \quad \overline{Y}(s) = \sum_{n=0}^{\infty} y[n]e^{-snT} -$$ - -Substitution of these expressions into Eq. (5.53) yields - -$$ -\sum_{n=0}^{\infty} y[n]e^{-snT} = H[e^{sT}]\left[\sum_{n=0}^{\infty} x[n]e^{-snT}\right] -$$ - -By introducing a new variable *z* = *esT* , this equation can be expressed as - -$$ -\sum_{n=0}^{\infty} y[n]z^{-n} = H[z] \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - -or - -$$ -Y[z] = H[z]X[z] -$$ - -where - -$$ -X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} -$$ - and $Y[z] = \sum_{n=0}^{\infty} y[n]z^{-n}$ - -It is clear from this discussion that the *z*-transform can be considered to be the Laplace transform with a change of variable *z* = *esT* or *s* = (1/*T*)ln*z*. Note that the transformation *z* = *esT* transforms the imaginary axis in the *s* plane (*s* = *j*ω) into a unit circle in the *z* plane (*z* = *esT* = *ej*ω*T* , or |*z*| = 1). The LHP and RHP in the *s*-plane map into the inside and the outside, respectively, of the unit circle in the *z* plane. - -## **[5.10 MATLAB: DISCRETE-TIME](#page-11-0) IIR FILTERS** - -Recent technological advancements have dramatically increased the popularity of discrete-time filters. Unlike their continuous-time counterparts, the performance of discrete-time filters is not affected by component variations, temperature, humidity, or age. Furthermore, digital hardware is easily reprogrammed, which allows convenient change of device function. For example, certain digital hearing aids are individually programmed to match the required response of a user. - -Typically, discrete-time filters are categorized as infinite-impulse response (IIR) or finite-impulse response (FIR). A popular method to obtain a discrete-time IIR filter is by transformation of a corresponding continuous-time filter design. MATLAB greatly assists this process. Although discrete-time IIR filter design is the emphasis of this section, methods for discrete-time FIR filter design are considered in Sec. 9.7. - -### **[5.10-1 Frequency Response and Pole-Zero Plots](#page-11-0)** - -Frequency response and pole-zero plots help characterize filter behavior. Similar to continuous-time systems, rational transfer functions for realizable LTID systems are represented in the *z*-domain as - -$$ -H[z] = \frac{Y[z]}{X[z]} = \frac{B[z]}{A[z]} = \frac{\sum_{k=0}^{N} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} = \frac{\sum_{k=0}^{N} b_k z^{N-k}}{\sum_{k=0}^{N} a_k z^{N-k}} -$$ -(5.54) - -When only the first (*N*1 + 1) numerator coefficients are nonzero and only the first (*N*2 + 1) denominator coefficients are nonzero, Eq. (5.54) simplifies to - -$$ -H[z] = \frac{Y[z]}{X[z]} = \frac{B[z]}{A[z]} = \frac{\sum_{k=0}^{N_1} b_k z^{-k}}{\sum_{k=0}^{N_2} a_k z^{-k}} = \frac{\sum_{k=0}^{N_1} b_k z^{N_1 - k}}{\sum_{k=0}^{N_2} a_k z^{N_2 - k}} z^{N_2 - N_1} -$$ -(5.55) - -The form of Eq. (5.55) has many advantages. It can be more efficient than Eq. (5.54); it still works when *N*1 = *N*2 = *N*; and it more closely conforms to the notation of built-in MATLAB discrete-time signal-processing functions. - -The right-hand side of Eq. (5.55) is a form that is convenient for MATLAB computations. The frequency response *H*[*ej*] is obtained by letting *z* = *ej*, where has units of radians. Often, = ω*T*, where ω is the continuous-time frequency in radians per second and *T* is the sampling period in seconds. Defining length-(*N*2 + 1) coefficient vector **A** = [*a*0,*a*1,...,*aN*2 ] and length-(*N*1+1) coefficient vector **B** = [*b*0,*b*1,...,*bN*1 ], program CH5MP1 computes *H*[*ej*] by using Eq. (5.55) for each frequency in the input vector . - -``` -function [H] = CH5MP1(B,A,Omega); -% CH5MP1.m : Chapter 5, MATLAB Program 1 -% Function M-file computes frequency response for LTID systems -% INPUTS: B = vector of feedforward coefficients -% A = vector of feedback coefficients -% Omega = vector of frequencies [rad], typically -pi<=Omega<=pi -% OUTPUTS: H = frequency response -N_1 = length(B)-1; N_2 = length(A)-1; -H = polyval(B,exp(1j*Omega))./polyval(A,exp(1j*Omega)).*... - exp(1j*Omega*(N_2-N_1)); -``` - -Note that owing to MATLAB's indexing scheme, A(k) corresponds to coefficient *ak*−1 and B(k) corresponds to coefficient *bk*−1. It is also possible to use the signal-processing toolbox function freqz to evaluate the frequency response of a system described by Eq. (5.55). Under special circumstances, the control system toolbox function bode can also be used. - -Program CH5MP2 computes and plots the poles and zeros of an LTID system described by Eq. (5.55), again using vectors **B** and **A**. - -``` -function [p,z] = CH5MP2(B,A); -% CH5MP2.m : Chapter 5, MATLAB Program 2 -% Function M-file computes and plots poles and zeros for LTID systems -% INPUTS: B = vector of feedforward coefficients -% A = vector of feedback coefficients -N_1 = length(B)-1; N_2 = length(A)-1; -p = roots([A,zeros(1,N_1-N_2)]); z = roots([B,zeros(1,N_2-N_1)]); -ucirc = exp(1j*linspace(0,2*pi,200)); % Compute unit circle for plot -plot(real(p),imag(p),'xk',real(z),imag(z),'ok',real(ucirc),imag(ucirc),'k:'); -xlabel('Real'); ylabel('Imag'); -ax = axis; dx = 0.05*(ax(2)-ax(1)); dy = 0.05*(ax(4)-ax(3)); -axis(ax+[-dx,dx,-dy,dy]); axis equal; -``` - -The right-hand side of Eq. (5.55) helps explain how the roots are computed. When *N*1 = *N*2, the term *zN*2−*N*1 implies additional roots at the origin. If *N*1 > *N*2, the roots are poles, which are added by concatenating A with zeros(N\_1-N\_2,1); since *N*2 − *N*1 ≤ 0, zeros(N\_2-N\_1,1) produces the empty set and B is unchanged. If *N*2 > *N*1, the roots are zeros, which are added by concatenating B with zeros(N\_2-N\_1,1); since *N*1 − *N*2 ≤ 0, zeros(N\_1-N\_2,1) produces the empty set and A is unchanged. Poles and zeros are indicated with black x's and o's, respectively. For visual reference, the unit circle is also plotted. The last two lines in CH5MP2 expand the plot axis box so that root locations are not obscured and also ensure that the real and imaginary axes are drawn to the same scale. - -### **[5.10-2 Transformation Basics](#page-11-0)** - -Transformation of a continuous-time filter to a discrete-time filter begins with the desired continuous-time transfer function - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{B(s)}{A(s)} = \frac{\sum_{k=0}^{M} b_{k+N-M} s^{M-k}}{\sum_{k=0}^{N} a_k s^{N-k}} -$$ - -As a matter of convenience, *H*(*s*) is represented in factored form as - -$$ -H(s) = \frac{b_{N-M}}{a_0} \frac{\prod_{k=1}^{M} (s - z_k)}{\prod_{k=1}^{N} (s - p_k)} -$$ -(5.56) - -where *zk* and *pk* are the system poles and zeros, respectively. - -A mapping rule converts the rational function *H*(*s*) to a rational function *H*[*z*]. Requiring that the result be rational ensures that the system realization can proceed with only delay, sum, and multiplier blocks. There are many possible mapping rules. For obvious reasons, good transformations tend to map the ω axis to the unit circle, ω = 0 to *z* = 1, ω = ∞ to *z* = −1, and the left half-plane to the interior of the unit circle. Put another way, sinusoids map to sinusoids, zero frequency maps to zero frequency, high frequency maps to high frequency, and stable systems map to stable systems. - -### 568 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -Section 5.9 suggests that the *z*-transform can be considered to be a Laplace transform with a change of variable *z* = *esT* or *s* = (1/*T*)ln*z*, where *T* is the sampling interval. It is tempting, therefore, to convert a continuous-time filter to a discrete-time filter by substituting *s* = (1/*T*)ln*z* into *H*(*s*), or *H*[*z*] = *H*(*s*)|*s*=(1/*T*)ln*z*. This approach is impractical, however, since the resulting *H*[*z*] is not rational and therefore cannot be implemented by using standard blocks. Although not considered here, the so-called matched-*z* transformation relies on the relationship *z* = *esT* to transform system poles and zeros, so the connection is not completely without merit. - -### **[5.10-3 Transformation by First-Order Backward Difference](#page-11-0)** - -Consider the transfer function *H*(*s*) = *Y*(*s*)/*X*(*s*) = *s*, which corresponds to the first-order continuous-time differentiator - -$$ -y(t) = \frac{d}{dt}x(t) -$$ - -An approximation that resembles the fundamental theorem of calculus is the first-order backward difference - -$$ -y(t) = \frac{x(t) - x(t - T)}{T} -$$ - -For sampling interval *T* and *t* = *nT*, the corresponding discrete-time approximation is - -$$ -y[n] = \frac{x[n] - x[n-1]}{T} -$$ - -which has transfer function - -$$ -H[z] = Y[z]/X[z] = \frac{1 - z^{-1}}{T} -$$ - -This implies a transformation rule that uses the change of variable *s* = (1−*z*−1)/*T* or *z* = 1/(1−*sT*). This transformation rule is appealing since the resulting *H*[*z*] is rational and has the same number of poles and zeros as *H*(*s*). Section 3.4 discusses this transformation strategy in a different way in describing the kinship of difference equations to differential equations. - -After some algebra, substituting *s* = (1−*z*−1)/*T* into Eq. (5.56) yields - -$$ -H[z] = \left(\frac{b_{N-M} \prod_{k=1}^{M} (1/T - z_k)}{a_0 \prod_{k=1}^{N} (1/T - p_k)}\right) \frac{\prod_{k=1}^{M} \left(1 - \frac{1}{1 - Tz_k} z^{-1}\right)}{\prod_{k=1}^{N} \left(1 - \frac{1}{1 - Tp_k} z^{-1}\right)} -$$ -(5.57) - -The discrete-time system has *M* zeros at 1/(1−*Tzk*) and *N* poles at 1/(1−*Tpk*). This transformation rule preserves system stability but does not map the ω axis to the unit circle (see Prob. 5.7-10). - -MATLAB program CH5MP3 uses the first-order backward difference method of Eq. (5.57) to convert a continuous-time filter described by coefficient vectors **A** = [*a*0,*a*1,...,*aN*] and **B** = [*bN*−*M*,*bN*−*M*+1,...,*bN*] into a discrete-time filter. The form of the discrete-time filter follows Eq. (5.55). - -function [Bd,Ad] = CH5MP3(B,A,T); % CH5MP3.m : Chapter 5, MATLAB Program 3 % Function M-file first-order backward difference transformation - -``` -% of a continuous-time filter described by B and A into a discrete-time filter. -% INPUTS: B = vector of continuous-time filter feedforward coefficients -% A = vector of continuous-time filter feedback coefficients -% T = sampling interval -% OUTPUTS: Bd = vector of discrete-time filter feedforward coefficients -% Ad = vector of discrete-time filter feedback coefficients -z = roots(B); p = roots(A); % s-domain roots -gain = B(1)/A(1)*prod(1/T-z)/prod(1/T-p); -zd = 1./(1-T*z); pd = 1./(1-T*p); % z-domain roots -Bd = gain*poly(zd); Ad = poly(pd); -``` - -### **[5.10-4 Bilinear Transformation](#page-11-0)** - -The bilinear transformation is based on a better approximation than first-order backward differences. Again, consider the continuous-time integrator - -$$ -y(t) = \frac{d}{dt}x(t) -$$ - -Represent signal *x*(*t*) as - -$$ -x(t) = \int_{t-T}^{t} \frac{d}{d\tau} x(\tau) d\tau + x(t-T) -$$ - -Letting *t* = *nT* and replacing the integral with a trapezoidal approximation yield - -$$ -x(nT) = \frac{T}{2} \left[ \frac{d}{dt} x(nT) + \frac{d}{dt} x(nT - T) \right] + x(nT - T) -$$ - -Substituting *y*(*t*) for (*d*/*dt*)*x*(*t*), the equivalent discrete-time system is - -$$ -x[n] = \frac{T}{2}(y[n] + y[n-1]) + x[n-1] -$$ - -From *z*-transforms, the transfer function is - -$$ -H[z] = \frac{Y[z]}{X[z]} = \frac{2(1 - z^{-1})}{T(1 + z^{-1})} -$$ - -The implied change of variable *s* = 2(1−*z*−1)/*T*(1+*z*−1) or *z* = (1+*sT*/2)/(1−*sT*/2) is called the bilinear transformation. Not only does the bilinear transformation result in a rational function *H*[*z*], the ω axis is correctly mapped to the unit circle (see Prob. 5.6-18a). - -After some algebra, substituting *s* = 2(1−*z*−1)/*T*(1+*z*−1) into Eq. (5.56) yields - -$$ -H[z] = \left(\frac{b_{N-M} \prod_{k=1}^{M} (2/T - z_k)}{a_0 \prod_{k=1}^{N} (2/T - p_k)}\right) \frac{\prod_{k=1}^{M} \left(1 - \frac{1 + z_k T/2}{1 - z_k T/2} z^{-1}\right)}{\prod_{k=1}^{N} \left(1 - \frac{1 + p_k T/2}{1 - p_k T/2} z^{-1}\right)} (1 + z^{-1})^{N-M} \tag{5.58} -$$ - -In addition to the *M* zeros at (1+*zkT*/2)/(1−*zkT*/2) and *N* poles at (1+*pkT*/2)/(1−*pkT*/2), there are *N*−*M* zeros at minus 1. Since practical continuous-time filters require *M* ≤ *N* for stability, the number of added zeros is thankfully always nonnegative. - -MATLAB program CH5MP4 converts a continuous-time filter described by coefficient vectors **A** = [*a*0,*a*1,...,*aN*] and **B** = [*bN*−*M*,*bN*−*M*+1,...,*bN*] into a discrete-time filter by using the bilinear transformation of Eq. (5.58). The form of the discrete-time filter follows Eq. (5.55). If available, it is also possible to use the signal-processing toolbox function bilinear to perform the bilinear transformation. - -``` -function [Bd,Ad] = CH5MP4(B,A,T); -% CH5MP4.m : Chapter 5, MATLAB Program 4 -% Function M-file bilinear transformation of a continuous-time filter -% described by vectors B and A into a discrete-time filter. -% Length of B must not exceed A. -% INPUTS: B = vector of continuous-time filter feedforward coefficients -% A = vector of continuous-time filter feedback coefficients -% T = sampling interval -% OUTPUTS: Bd = vector of discrete-time filter feedforward coefficients -% Ad = vector of discrete-time filter feedback coefficients -if (length(B)>length(A)), - disp('Numerator order must not exceed denominator order.'); - return -end -z = roots(B); p = roots(A); % s-domain roots -gain = real(B(1)/A(1)*prod(2/T-z)/prod(2/T-p)); -zd = (1+z*T/2)./(1-z*T/2); pd = (1+p*T/2)./(1-p*T/2); % z-domain roots -Bd = gain*poly([zd;-ones(length(A)-length(B),1)]); Ad = poly(pd); -As with most high-level languages, MATLAB supports general if-structures: -``` - -``` -if expression, - statements; -elseif expression, - statements; -else, - statements; -end -``` - -In the program CH5MP4, the if statement tests *M* > *N*. When true, an error message is displayed and the return command terminates program execution to prevent errors. - -### **[5.10-5 Bilinear Transformation with Prewarping](#page-11-0)** - -The bilinear transformation maps the entire infinite-length ω axis onto the finite-length unit circle (*z* = *ej*) according to ω = (2/*T*)tan(/2) (see Prob. 5.6-18b). Equivalently, = 2arctan(ω*T*/2). The nonlinearity of the tangent function causes a frequency compression, commonly called frequency warping, that distorts the transformation. - -To illustrate the warping effect, consider the bilinear transformation of a continuous-time lowpass filter with cutoff frequency ω*c* = 2π3000 rad/s. If the target digital system uses a sampling rate of 10 kHz, then *T* = 1/(10,000) and ω*c* maps to *c* = 2arctan(ω*cT*/2) = 1.5116. Thus, the transformed cutoff frequency is short of the desired *c* = ω*cT* = 0.6π = 1.8850. - -Cutoff frequencies are important and need to be as accurate as possible. By adjusting the parameter *T* used in the bilinear transform, one continuous-time frequency can be exactly mapped to one discrete-time frequency; the process is called prewarping. Continuing the last example, adjusting *T* = (2/ω*c*)tan(*c*/2) ≈ 1/6848 achieves the appropriate prewarping to ensure ω*c* = 2π3000 maps to *c* = 0.6π. - -### **[5.10-6 Example: Butterworth Filter Transformation](#page-11-0)** - -To illustrate the transformation techniques, consider a continuous-time 10th-order Butterworth lowpass filter with cutoff frequency ω*c* = 2π3000, as designed in Sec. 4.12. First, we determine continuous-time coefficient vectors **A** and **B**. - -``` ->> omega_c = 2*pi*3000; N=10; ->> poles = roots([(1j*omega_c)^(-2*N),zeros(1,2*N-1),1]); ->> poles = poles(find(poles<0)); ->> B = 1; A = poly(poles); A = A/A(end); -``` - -Programs CH5MP3 and CH5MP4 are used to perform first-order forward difference and bilinear transformations, respectively. - -``` ->> Omega = linspace(0,pi,200); T = 1/10000; Omega_c = omega_c*T; ->> [B1,A1] = CH5MP3(B,A,T); % First-order backward difference transformation ->> [B2,A2] = CH5MP4(B,A,T); % Bilinear transformation ->> [B3,A3] = CH5MP4(B,A,2/omega_c*tan(Omega_c/2)); % Bilinear with prewarping -``` - -Magnitude responses are computed using CH5MP1 and then plotted. - -``` ->> H1mag = abs(CH5MP1(B1,A1,Omega)); ->> H2mag = abs(CH5MP1(B2,A2,Omega)); ->> H3mag = abs(CH5MP1(B3,A3,Omega)); ->> plot(Omega,(Omega<=Omega_c),'k',Omega,H1mag,'k-.',... ->> Omega,H2mag,'k--',Omega,H3mag,'k:'); ->> axis([0 pi -.05 1.5]); ->> xlabel('\Omega [rad]'); ylabel('Magnitude Response'); ->> legend('Ideal','FOBD','BLT','Prewarp BLT','location','best'); -``` - -The result of each transformation method is shown in Fig. 5.31, where FOBD and BLT stand for first-order backward difference and bilinear transformation, respectively. - -Although the first-order backward difference results in a lowpass filter, the method causes significant distortion that makes the resulting filter unacceptable with regard to cutoff frequency. The bilinear transformation is better, but, as predicted, the cutoff frequency falls short of the desired value. Bilinear transformation with prewarping properly locates the cutoff frequency and produces a very acceptable filter response. - -**Figure 5.31** Comparison of various transformation techniques. - -### **[5.10-7 Problems Finding Polynomial Roots](#page-12-0)** - -Numerically, it is difficult to accurately determine the roots of a polynomial. Consider, for example, a simple polynomial that has a root at minus 1 repeated four times, (*s* + 1)4 = *s*4 + 4*s*3 +6*s*2 +4*s*+1. The MATLAB roots command returns a surprising result: - -``` ->> roots([1464 1])' - ans = -1.0002 -1.0000-0.0002i -1.0000+0.0002i -0.9998 -``` - -Even for this low-degree polynomial, MATLAB does not return the true roots. - -The problem worsens as polynomial degree increases. The bilinear transformation of the 10th-order Butterworth filter, for example, should have 10 zeros at minus 1. Figure 5.32 shows that the zeros, computed by CH5MP2 with the roots command, are not correctly located. - -When possible, programs should avoid root computations that may limit accuracy. For example, results from the transformation programs CH5MP3 and CH5MP4 are more accurate if the true transfer function poles and zeros are passed directly as inputs rather than the polynomial coefficient vectors. When roots must be computed, result accuracy should always be verified. - -### **[5.10-8 Using Cascaded Second-Order Sections to Improve Design](#page-12-0)** - -The dynamic range of high-degree polynomial coefficients is often large. Adding the difficulties associated with factoring a high-degree polynomial, it is little surprise that high-order designs are difficult. - -As with continuous-time filters, performance is improved by using a cascade of second-order sections to design and realize a discrete-time filter. Cascades of second-order sections are also more robust to the coefficient quantization that occurs when discrete-time filters are implemented on fixed-point digital hardware. - -To illustrate the performance possible with a cascade of second-order sections, consider a 180th-order transformed Butterworth discrete-time filter with cutoff frequency *c* = 0.6π ≈ 1.8850. Program CH5MP5 completes this design, taking care to initially locate poles and zeros without root computations. - -**Figure 5.32** Pole-zero plot computed by using roots. - -``` -% CH5MP5.m : Chapter 5, MATLAB Program 5 -% Script M-file designs a 180th-order Butterworth lowpass discrete-time filter -% with cutoff Omega_c = 0.6*pi using 90 cascaded second-order filter sections. -omega_0 = 1; % Use normalized cutoff frequency for analog prototype -psi = [0.5:1:90]*pi/180; % Butterworth pole angles -Omega_c = 0.6*pi; % Discrete-time cutoff frequency -Omega = linspace(0,pi,1000); % Frequency range for magnitude response -Hmag = zeros(90,1000); p = zeros(1,180); z = zeros(1,180); % Pre-allocation -for stage = 1:90, - Q = 1/(2*cos(psi(stage))); % Compute Q for stage - B = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute stage coefficients - [B1,A1] = CH5MP4(B,A,2/omega_0*tan(0.6*pi/2)); % Transform stage to DT - p(stage*2-1:stage*2) = roots(A1); % Compute z-domain poles for stage - z(stage*2-1:stage*2) = roots(B1); % Compute z-domain zeros for stage - Hmag(stage,:) = abs(CH5MP1(B1,A1,Omega)); % Compute stage mag response -end -ucirc = exp(j*linspace(0,2*pi,200)); % Compute unit circle for pole-zero plot -figure; -plot(real(p),imag(p),'kx',real(z),imag(z),'ok',real(ucirc),imag(ucirc),'k:'); -axis equal; xlabel('Real'); ylabel('Imag'); -figure; plot(Omega,prod(Hmag),'k'); axis([0 pi -0.05 1.05]); -xlabel('\Omega [rad]'); ylabel('Magnitude Response'); -``` - -The figure command preceding each plot command opens a separate window for each plot. - -The filter's pole-zero plot is shown in Fig. 5.33, along with the unit circle, for reference. All 180 zeros of the cascaded design are properly located at minus 1. The wall of poles provides an amazing approximation to the desired brick-wall response, as shown by the magnitude response in Fig. 5.34. It is virtually impossible to realize such high-order designs with continuous-time filters, which adds another reason for the popularity of discrete-time filters. Still, the design is not - -**Figure 5.33** Pole-zero plot for 180th-order discrete-time Butterworth filter. - -**Figure 5.34** Magnitude response for a 180th-order discrete-time Butterworth filter. - -trivial; even functions from the MATLAB signal-processing toolbox fail to properly design such a high-order discrete-time Butterworth filter. - -## **[5.11 SUMMARY](#page-12-0)** - -In this chapter we discussed the analysis of linear, time-invariant, discrete-time (LTID) systems by means of the *z*-transform. The *z*-transform changes the difference equations of LTID systems into algebraic equations. Therefore, solving these difference equations reduces to solving algebraic equations. - -The transfer function *H*[*z*] of an LTID system is equal to the ratio of the *z*-transform of the output to the *z*-transform of the input when all initial conditions are zero. Therefore, if *X*[*z*] is the *z*-transform of the input *x*[*n*] and *Y*[*z*] is the *z*-transform of the corresponding output *y*[*n*] (when all initial conditions are zero), then *Y*[*z*] = *H*[*z*]*X*[*z*]. For an LTID system specified by the difference equation *Q*[*E*]*y*[*n*] = *P*[*E*]*x*[*n*], the transfer function *H*[*z*] = *P*[*z*]/*Q*[*z*]. Moreover, *H*[*z*] is the *z*-transform of the system impulse response *h*[*n*]. We showed in Ch. 3 that the system response to an everlasting exponential *zn* is *H*[*z*]*zn*. - -We may also view the *z*-transform as a tool that expresses a signal *x*[*n*] as a sum of exponentials of the form *zn* over a continuum of the values of *z*. Using the fact that an LTID system response to *zn* is *H*[*z*]*zn*, we find the system response to *x*[*n*] as a sum of the system's responses to all the components of the form *zn* over the continuum of values of *z*. - -LTID systems can be realized by scalar multipliers, adders, and time delays. A given transfer function can be synthesized in many different ways. We discussed canonical, transposed canonical, cascade, and parallel forms of realization. The realization procedure is identical to that for continuous-time systems with 1/*s* (integrator) replaced by 1/*z* (unit delay). - -The majority of the input signals and practical systems are causal. Consequently, we are required to deal with causal signals most of the time. Restricting all signals to the causal type greatly simplifies *z*-transform analysis; the ROC of a signal becomes irrelevant to the analysis process. This special case of *z*-transform (which is restricted to causal signals) is called the unilateral *z*-transform. Much of the chapter deals with this transform. Section 5.8 discusses the general variety of the *z*-transform (bilateral *z*-transform), which can handle causal and noncausal signals and systems. In the bilateral transform, the inverse transform of *X*[*z*] is not unique, but depends on the ROC of *X*[*z*]. Thus, the ROC plays a crucial role in the bilateral *z*-transform. - -In Sec. 5.9, we showed that discrete-time systems can be analyzed by the Laplace transform as if they were continuous-time systems. In fact, we showed that the *z*-transform is the Laplace transform with a change in variable. - -### **[REFERENCES](#page-12-0)** - -- 1. Lyons, R. G. *Understanding Digital Signal Processing*. Addison-Wesley, Reading, MA, 1997. -- 2. Oppenheim, A. V., and R. W. Schafer. *Discrete-Time Signal Processing*, 2nd ed. Prentice-Hall, Upper Saddle River, NJ, 1999. -- 3. Mitra, S. K. *Digital Signal Processing*, 2nd ed. McGraw-Hill, New York, 2001. - -## **[PROBLEMS](#page-12-0)** - -- **5.1-1** Using the definition, compute the *z*-transform of *x*[*n*] = (−1)*n*(*u*[*n*] − *u*[*n* − 8]). Sketch the poles and zeros of *X*[*z*] in the *z* plane. No calculator is needed to do this problem! -- **5.1-2** Determine the unilateral *z*-transform *X*[*z*] of the signal *x*[*n*] shown in Fig. P5.1-2. As the picture suggests, *x*[*n*]=−3 for all *n* ≥ 9 and *x*[*n*] = 0 for all *n* < 3. -- **5.1-3** (a) A causal signal has *z*-transform given by *X*[*z*] = *z*2 *z*3−1 . Determine the time-domain signal *x*[*n*] and sketch *x*[*n*] over −4 ≤ *n* ≤ 11. [*Hint:* No complex arithmetic is needed to solve this problem!] - -**Figure P5.1-2** - -### 576 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -- (b) Consider the causal semiperiodic signal *y*[*n*] shown in Fig. P5.1-3. Notice, *y*[*n*] continually repeats the sequence [1, 2, 3] for *n* ≥ 0. Determine the unilateral *z*-transform *Y*[*z*] of this signal. If possible, express your result as a rational function in standard form. -- **5.1-4** Using the definition of the *z*-transform, find the *z*-transform and the ROC for each of the following signals. - - (a) *u*[*n*− *m*] - - (b) γ *n* sinπ*n u*[*n*] - - (c) γ *n* cosπ*n u*[*n*] - -**Figure P5.1-3** - -(d) -$$ -\gamma^n \sin \frac{\pi n}{2} u[n] -$$ - -\n(e) $\gamma^n \cos \frac{\pi n}{2} u[n]$ -\n(f) $\sum_{k=0}^{\infty} 2^{2k} \delta[n-2k]$ -\n(g) $\gamma^{n-1} u[n-1]$ - -(h) *n*γ *n u*[*n*] - -$$ -(i) \; n \, u[n] -$$ - -(j) -$$ -\frac{\gamma^n}{n!}u[n] -$$ - -(k) -$$ -[2^{n-1} - (-2)^{n-1}]u[n] -$$ - -(l) $\frac{(\ln \alpha)^n}{n!}u[n]$ - -- **5.1-5** Showing all work, evaluate \$∞ *n*=0 *n*(−3/2)−*n*. -- **5.1-6** Using only the *z*-transforms of Table 5.1, determine the *z*-transform of each of the following signals. - -(a) *u*[*n*] −*u*[*n*−2] (b) γ *n*−2*u*[*n*− 2] (c) 2*n*+1*u*[*n*− 1] +*en*−1*u*[*n*] (d) 2−*n* cosπ 3 *n u*[*n*−1] (e) *n*γ *nu*[*n*− 1] (f) *n*(*n*−1)(*n*−2)2*n*−3*u*[*n*−*m*] for *m* = 0, 1, 2, 3 (g) (−1)*nnu*[*n*] - -(h) -$$ -\sum_{k=0}^{\infty} k\delta(n-2k+1) -$$ - -**5.1-7** Find the inverse unilateral *z*-transform of each of the following: (a) *z*(*z* −4) - -(a) -$$ -z^2 - 5z + 6 -$$ - -\n(b) $\frac{z-4}{z^2 - 5z + 6}$ -\n(c) $\frac{(e^{-2} - 2)z}{(z - e^{-2})(z - 2)}$ -\n(d) $\frac{(z - 1)^2}{z^3}$ -\n(e) $\frac{z(2z + 3)}{(z - 1)(z^2 - 5z + 6)}$ -\n(f) $\frac{z(-5z + 22)}{(z + 1)(z - 2)^2}$ -\n(g) $\frac{z(1.4z + 0.08)}{(z - 0.2)(z - 0.8)^2}$ - -(h) -$$ -\frac{z(z-2)}{z^2 - z + 1} -$$ - -\n(i) -$$ -\frac{2z^2 - 0.3z + 0.25}{z^2 + 0.6z + 0.25} -$$ - -\n(j) -$$ -\frac{2z(3z-23)}{(z-1)(z^2 - 6z + 25)} -$$ - -\n(k) -$$ -\frac{z(3.83z + 11.34)}{(z-2)(z^2 - 5z + 25)} -$$ - -\n(k) -$$ -\frac{z^2(-2z^2 + 8z - 7)}{(z^2 - 2z^2 + 8z - 7)} -$$ - -$$ -(1) \frac{z^2(-2z^2+8z-7)}{(z-1)(z-2)^3} -$$ - -**5.1-8** (a) Expanding *X*[*z*] as a power series in *z*−1, find the first three terms of *x*[*n*] if - -$$ -X[z] = \frac{2z^3 + 13z^2 + z}{z^3 + 7z^2 + 2z + 1} -$$ - -(b) Extend the procedure used in part (a) to find the first four terms of *x*[*n*] if - -$$ -X[z] = \frac{2z^4 + 16z^3 + 17z^2 + 3z}{z^3 + 7z^2 + 2z + 1} -$$ - -- **5.1-9** A right-sided signal *x*[*n*] has *z*-transform given by *X*[*z*] = *z*6+2*z*5+3*z*4+4*z*3 *z*4−1 . Using a power series expansion of *X*[*z*], determine *x*[*n*] over −5 ≤ *n* ≤ 5. -- **5.1-10** Find *x*[*n*] by expanding - -$$ -X[z] = \frac{\gamma z}{(z - \gamma)^2} -$$ - -as a power series in *z*−1. - -- **5.1-11** (a) In Table 5.1, if the numerator and the denominator powers of *X*[*z*] are *M* and *N*, respectively, explain why in some cases *N* − *M* = 0, while in others *N* − *M* = 1 or *N* − *M* = *m* (*m* any positive integer). - - (b) Without actually finding the *z*-transform, state what is *N* − *M* for *X*[*z*] corresponding to *x*[*n*] = γ *nu*[*n*−4]. -- **5.2-1** For a discrete-time signal shown in Fig. P5.2-1, show that - -$$ -X[z] = \frac{1 - z^{-m}}{1 - z^{-1}} -$$ - -Find your answer by using the definition in Eq. (5.1) and by using Table 5.1 and an appropriate property of the *z*-transform. - -**Figure P5.2-1** - -- **5.2-2** Determine the unilateral *z*-transform of signal *x*[*n*] = (1−*n*) cos π 2 (*n*−1) *u*[*n*−1]. -- **5.2-3** Suppose a DT signal *x*[*n*] = 2(*u*[*n*−10] −*u*[*n*− 6]) has a transform *X*(*z*). Define *Y*(*z*) = 1 2*z*−3 *d dzX*(2*z*). Using graphic plot or vector notation, determine the corresponding signal *y*[*n*]. -- **5.2-4** Suppose a DT signal *x*[*n*] = 3(*u*[*n*] −*u*[*n*−5]) has a transform *X*(*z*). Define *Y*(*z*) = 2*z*−4 *d dzX z* 2 . Using graphic plot or vector notation, determine the corresponding signal *y*[*n*]. -- **5.2-5** Find the *z*-transform of the signal illustrated in Fig. P5.2-5. Solve this problem in two ways, as in Exs. 5.2d and 5.4. Verify that the two answers are equivalent. -- **5.2-6** Using *z*-transform techniques and properties (no time-domain convolution sum!), determine the convolution *y*[*n*] =( 1 2 )*nu*[*n*−3]∗( 1 3 )*n*−6*u*[*n*−4]. Express your answer in the form *y*[*n*] = *c*1γ *n*−*N*1 1 *u*[*n* − *N*1] + *c*2γ *n*−*N*2 2 *u*[*n* − *N*2], making sure to clearly identify the constants *c*1, *c*2, γ1, γ2, *N*1, and *N*2. - -**Figure P5.2-5** - -**5.2-7** Determine the inverse unilateral *z*-transform *x*[*n*] of the signal - -$$ -X[z] = \frac{d^7}{dz^7} \left[ \frac{z^{-4}}{(z - \frac{1}{2})(z + 3)} \right] -$$ - -- **5.2-8** Using only the fact that γ *nu*[*n*]⇐⇒*z*/(*z*−γ ) and properties of the *z*-transform, find the *z*-transform of each of the following: - - (a) *n*2*u*[*n*] - - (b) *n*2γ *nu*[*n*] - - (c) *n*3*u*[*n*] - - (d) *an*[*u*[*n*] −*u*[*n*−*m*]] - - (e) *ne*−2*nu*[*n*−*m*] - - (f) (*n*−2)(0.5)*n*−3 *u*[*n*−4] -- **5.2-9** Using only pair 1 in Table 5.1 and appropriate properties of the *z*-transform, derive iteratively pairs 2 through 9. In other words, first derive pair 2. Then use pair 2 (and pair 1, if needed) to derive pair 3, and so on. -- **5.2-10** Find the *z*-transform of cos(π*n*/4)*u*[*n*] using only pairs 1 and 11b in Table 5.1 and a suitable property of the *z*-transform. -- **5.2-11** Apply the time-reversal property to pair 6 of Table 5.1 to show that γ *nu*[−(*n* + 1)] ⇐⇒ −*z*/(*z*−γ ) and the ROC is given by |*z*| < |γ |. -- **5.2-12** (a) If *x*[*n*] ⇐⇒ *X*[*z*], then show that (−1)*nx*[*n*] ⇐⇒ *X*[−*z*]. - - (b) Use this result to show that (−γ )*nu*[*n*] ⇐⇒ *z*/(*z* +γ ). - - (c) Use these results to find the *z*-transforms of *x*i[*n*]=[2*n*−1 −(−2)*n*−1]*u*[*n*] and *x*−ii[*n*] = γ *n* cosπ*n u*[*n*] -- **5.2-13** (a) If *x*[*n*] ⇐⇒ *X*[*z*], then show that - -$$ -\sum_{k=0}^{n} x[k] \Longleftrightarrow \frac{zX[z]}{z-1} -$$ - -### 578 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM - -- (b) Use this result to derive pair 2 from pair 1 in Table 5.1. -- **5.2-14** A number of causal time-domain functions are shown in Fig. P5.2-14. List the function of time that corresponds to each of the following functions of *z*. Few or no calculations are necessary! Be careful, the graphs may be scaled differently. - -(a) -$$ -\frac{z^2}{(z-0.75)^2} -$$ - -\n(b) $\frac{z^2 - 0.9z/\sqrt{2}}{z^2 - 0.9\sqrt{2}z + 0.81}$ -\n(c) $\sum_{k=1}^{4} z^{-2k}$ - -$$ -k=0 -$$ -\n(d) - -\n -$$ -\frac{z^{-5}}{1-z^{-1}} -$$ - -(e) -$$ -\frac{z^2}{z^4 - 1} -$$ - -(c) 0.75z - -(f) -$$ -\frac{1}{(z-0.75)^2} -$$ - -\n(g) $\frac{z^2 - z/\sqrt{2}}{z-\sqrt{2}}$ - -(g) -$$ -\frac{z^2 - \sqrt{2}z + 1}{z^2 - 5z^{-5} + 4z^{-6}} -$$ - -\n(h) -$$ -\frac{z^{-1} - 5z^{-5} + 4z^{-6}}{5(1 - z^{-1})^2} -$$ - -\n(i) -$$ -\frac{z}{z - 1.1} -$$ - -(j) -$$ -\frac{0.25z^{-1}}{(1-z^{-1})(1-0.75z^{-1})} -$$ - -- **5.2-15** Suppose we upsample a causal signal *x*[*n*] by factor *N* to produce signal *y*[*n*]. Express *Y*(*z*) in terms of *X*(*z*), taking care to mathematically justify your result. -- **5.3-1** Using *z*-transform techniques, find the output *y*[*n*] of an LTID system specified by the equation *y*[*n*]− 1 3 *y*[*n*−1] = *x*[*n*−1] if the initial condition is *y*[−1] = 2 and the input is *x*[*n*]=−*u*[*n*]. -- **5.3-2** Consider an LTID system *y*[*n*] − *y*[*n* − 2] = *x*[*n*] with *y*[−1] = 0, *y*[−2] = 1, and *x*[*n*] = *u*[*n*]. - - (a) Determine *Y*[*z*], expressed as a rational function in standard factored form. - - (b) Use *Y*[*z*] from part (a) to solve for the system output *y*[*n*]. -- **5.3-3** Solve Prob. 3.8-23 by the *z*-transform method. -- **5.3-4** Consider a DT system with transfer function *H*[*z*] = 2*z*−2 *z*−0.5 . Assuming the system is both controllable and observable, determine the ZIR *y*zir[*n*] given *y*[−1] = 1. -- **5.3-5** (a) Solve - -$$ -y[n+1] + 2y[n] = x[n+1] -$$ - -when -$$ -y[0] = 1 -$$ - and $x[n] = e^{-(n-1)}u[n]$ - -- (b) Find the zero-input and the zero-state components of the response. -- **5.3-6** Consider a LTID system that is described by the difference equation *y*[*n*] − 1 4 *y*[*n* − 2] = *x*[*n*−1]. - -- (a) Use transform-domain techniques to determine the zero-state response *y*zsr[*n*] to input *x*[*n*] = 3*u*[*n*−5]. -- (b) Use transform-domain techniques to determine the zero-input response *y*zir[*n*] given *y*zir[−2] = *y*zir[−1] = 1. -- **5.3-7** (a) Find the output *y*[*n*] of an LTID system specified by the equation - -$$ -2y[n+2] - 3y[n+1] + y[n] -$$ - -= $4x[n+2] - 3x[n+1]$ - -for input *x*[*n*] = (4)−*nu*[*n*] and initial conditions *y*[−1] = 0 and *y*[−2] = 1. - -- (b) Find the zero-input and the zero-state components of the response. -- (c) Find the transient and the steady-state components of the response. -- **5.3-8** Solve Prob. 5.3-7 if initial conditions *y*[−1] and *y*[−2] are instead replaced with auxiliary conditions *y*[0] = 3/2 and *y*[1] = 35/4. -- **5.3-9** (a) Solve - -$$ -4y[n+2] + 4y[n+1] + y[n] = x[n+1] -$$ - -with *y*[−1] = 0, *y*[−2] = 1, and *x*[*n*] = *u*[*n*]. - -- (b) Find the zero-input and the zero-state components of the response. -- (c) Find the transient and the steady-state components of the response. -- **5.3-10** Solve - -$$ -y[n+2] - 3y[n+1] + 2y[n] = x[n+1] -$$ - -if *y*[−1] = 2, *y*[−2] = 3, and *x*[*n*] = (3)*nu*[*n*]. - -**5.3-11** Solve - -$$ -y[n+2] - 2y[n+1] + 2y[n] = x[n] -$$ - -with *y*[−1] = 1, *y*[−2] = 0, and *x*[*n*] = *u*[*n*]. - -- **5.3-12** Consider a causal LTID system described as *H*(*z*) = 21(*z*2+1) 16(*z*2+ 1 4 *z*− 3 8 ) . - - (a) Determine the standard delay-form difference equation description of this system. - - (b) Using transform-domain techniques, determine the system impulse response *h*[*n*]. - - (c) Using transform-domain techniques, determine *y*zir[*n*] given *y*[−1] = 16 and *y*[−2] = 8. - -- **5.3-13** Consider a causal LTID system described as *y*[*n*] − 5 6 *y*[*n* − 1] + 1 6 *y*[*n* − 2] = 3 2 *x*[*n* − 1] + 3 2 *x*[*n*−2]. - - (a) Determine the (standard-form) system transfer function *H*(*z*) and sketch the system pole-zero plot. - - (b) Using transform-domain techniques, determine *y*zir[*n*] given *y*[−1] = 2 and *y*[−2]=−2. -- **5.3-14** Solve - -$$ -y[n] + 2y[n-1] + 2y[n-2] -$$ - -= $x[n-1] + 2x[n-2]$ - -with *y*[0] = 0, *y*[1] = 1, and *x*[*n*] = *enu*[*n*]. - -- **5.3-15** A system with impulse response *h*[*n*] = 2(1/3)*nu*[*n* − 1] produces an output *y*[*n*] = (−2)*nu*[*n* − 1]. Determine the corresponding input *x*[*n*]. -- **5.3-16** A professor recently received an unexpected \$10 (a futile bribe attached to a test). Being the savvy investor that she is, the professor decides to invest the \$10 into a savings account that earns 0.5% interest compounded monthly (6.17% APY). Furthermore, she decides to supplement this initial investment with an additional \$5 deposit made every month, beginning the month immediately following her initial investment. - - (a) Model the professor's savings account as a constant coefficient linear difference equation. Designate *y*[*n*] as the account balance at month *n*, where *n* = 0 corresponds to the first month that interest is awarded (and that her \$5 deposits begin). - - (b) Determine a closed-form solution for *y*[*n*]. That is, you should express *y*[*n*] as a function only of *n*. - - (c) If we consider the professor's bank account as a system, what is the system impulse response *h*[*n*]? What is the system transfer function *H*[*z*]? - - (d) Explain this fact: if the input to the professor's bank account is the everlasting exponential *x*[*n*] = 1*n* = 1, then the output is **not** *y*[*n*] = 1*nH*[1] = *H*[1]. -- **5.3-17** Sally deposits \$100 into her savings account on the first day of every month except for each December, when she uses her money to buy - -holiday gifts. Define *b*[*m*] as the balance in Sally's account on the first day of month *m*. Assume Sally opens her account in January (*m* = 0), continues making monthly payments forever (except each December!), and that her monthly interest rate is 1%. Sally's account balance satisfies a simple difference equation *b*[*m*] = (1.01)*b*[*m* − 1] + *p*[*m*], where *p*[*m*] designates Sally's monthly deposits. Determine a closed-form expression for *b*[*m*] that is only a function of the month *m*. - -- **5.3-18** For each impulse response, determine the number of system poles, whether the poles are real or complex, and whether the system is BIBO-stable. - - (a) *h*1[*n*] = (−1+(0.5)*n*)*u*[*n*] - - (b) *h*2[*n*] = (*j*)*n*(*u*[*n*] −*u*[*n*−10]) -- **5.3-19** Find the following sums: - -(a) -$$ -\sum_{k=0}^{n} k -$$ - -(b) -$$ -\sum_{k=0}^{n} k^2 -$$ - -*k*=0 [*Hint:* Consider a system whose output *y*[*n*] is the desired sum. Examine the relationship between *y*[*n*] and *y*[*n* − 1]. Note also that *y*[0] = 0.] - -### **5.3-20** Find the following sum: - -$$ -\sum_{k=0}^{n} k^3 -$$ - -[*Hint:* See the hint for Prob. 5.3-19.] - -**5.3-21** Find the following sum: - -$$ -\sum_{k=0}^{n} ka^k \qquad a \neq 1 -$$ - -[*Hint:* See the hint for Prob. 5.3-19.] - -- **5.3-22** Redo Prob. 5.3-19 using the result in Prob. 5.2-13a. -- **5.3-23** Redo Prob. 5.3-20 using the result in Prob. 5.2-13a. -- **5.3-24** Redo Prob. 5.3-21 using the result in Prob. 5.2-13a. - -**5.3-25** (a) Find the zero-state response of an LTID system with transfer function - -$$ -H[z] = \frac{z}{(z+0.2)(z-0.8)} -$$ - -and the input *x*[*n*] = *e*(*n*+1) *u*[*n*]. - -(b) Write the difference equation relating the output *y*[*n*] to input *x*[*n*]. - -**5.3-26** Repeat Prob. 5.3-25 for *x*[*n*] = *u*[*n*] and - -$$ -H[z] = \frac{2z+3}{(z-2)(z-3)} -$$ - -**5.3-27** Repeat Prob. 5.3-25 for - -$$ -H[z] = \frac{6(5z - 1)}{6z^2 - 5z + 1} -$$ - -- and the input *x*[*n*] is (a) (4)−*nu*[*n*] (b) (4)−(*n*−2) *u*[*n*−2] (c) (4)−(*n*−2) *u*[*n*] (d) (4)−*nu*[*n*−2] -- **5.3-28** Repeat Prob. 5.3-25 for *x*[*n*] = *u*[*n*] and - -$$ -H[z] = \frac{2z - 1}{z^2 - 1.6z + 0.8} -$$ - -- **5.3-29** Find the transfer functions corresponding to each of the systems specified by difference equations in Probs. 5.3-5, 5.3-7, 5.3-9, and 5.3-14. -- **5.3-30** Find *h*[*n*], the unit impulse response of the systems described by the following equations: - - (a) *y*[*n*] +3*y*[*n*−1] +2*y*[*n*−2] = *x*[*n*] +3*x*[*n*− 1] +3*x*[*n*−2] - - (b) *y*[*n* + 2] + 2*y*[*n* + 1] + *y*[*n*] = 2*x*[*n* + 2] − *x*[*n*+1] - - (c) *y*[*n*]−*y*[*n*−1]+0.5*y*[*n*−2] = *x*[*n*]+2*x*[*n*− 1] -- **5.3-31** Find *h*[*n*], the unit impulse response of the systems in Probs. 5.3-25, 5.3-26, and 5.3-28. -- **5.3-32** A system has impulse response *h*[*n*] = *u*[*n* − 3]. - - (a) Determine the impulse response of the inverse system *h*−1[*n*]. - - (b) Is the inverse stable? Is the inverse causal? - - (c) Your boss asks you to implement *h*−1[*n*] to the best of your ability. Describe your realizable design, taking care to identify any deficiencies. - -**5.4-1** A system has impulse response given by - -$$ -h[n] = \left[ \left( \frac{1+j}{\sqrt{8}} \right)^n + \left( \frac{1-j}{\sqrt{8}} \right)^n \right] u[n] -$$ - -This system can be implemented according to Fig. P5.4-1. - -**Figure P5.4-1** - -- (a) Determine the coefficients *A*1 and *A*2 to implement *h*[*n*] using the structure shown in Fig. P5.4-1. -- (b) What is the zero-state response *y*0[*n*] of this system, given a shifted unit step input *x*[*n*] = *u*[*n*+3]? -- **5.4-2** (a) Show the canonic direct form, a cascade, and a parallel realization of - -$$ -H[z] = \frac{z(3z - 1.8)}{z^2 - z + 0.16} -$$ - -- (b) Find the transpose of the realizations obtained in part (a). -- **5.4-3** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{5z + 2.2}{z^2 + z + 0.16} -$$ - -**5.4-4** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{3.8z - 1.1}{(z - 0.2)(z^2 - 0.6z + 0.25)} -$$ - -**5.4-5** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{z(1.6z - 1.8)}{(z - 0.2)(z^2 + z + 0.5)} -$$ - -**5.4-6** Repeat Prob. 5.4-2 for - -$$ -H[z] = \frac{z(2z^2 + 1.3z + 0.96)}{(z + 0.5)(z - 0.4)^2} -$$ - -**5.4-7** Consider the LTID system shown in Fig. P5.4-7. - -$$ -x[n] -$$ - $\rightarrow$ $z^{-1}$ $\rightarrow$ $z^{-1}$ $\rightarrow$ $z^{-1}$ $\rightarrow$ $y[n]$ - -- **Figure P5.4-7** -- (a) Determine the standard delay-form difference equation description of this system. -- (b) Determine the impulse response *h*[*n*] of this system. -- (c) Is this realization canonical? Explain. -- (d) Is this system stable? Explain. -- (e) Is this system causal? Explain. -- **5.4-8** Realize a system whose transfer function is - -$$ -H[z] = \frac{2z^4 + z^3 + 0.8z^2 + 2z + 8}{z^4} -$$ - -**5.4-9** Realize a system whose transfer function is given by - -$$ -H[z] = \sum_{n=0}^{6} nz^{-n} -$$ - -**5.4-10** Consider the LTID system shown in Fig. P5.4-10, where parameter *c* is an arbitrary, real constant. - -**Figure P5.4-10** - -- (a) Determine the system transfer function *H*[*z*], expressed in standard rational form. -- (b) Determine all system poles and all system zeros. -- (c) Is the system of Fig. P5.4-10 canonical? Explain. -- (d) What constraints, if any, exist on parameter *c* to ensure that the system is stable? -- **5.4-11** This problem demonstrates the enormous number of ways of implementing even a relatively low-order transfer function. A second-order transfer function has two real zeros and two real poles. Discuss various ways of realizing - -such a transfer function. Consider canonic direct, cascade, parallel, and the corresponding transposed forms. Note also that interchange of cascaded sections yields a different realization. - -**5.4-12** Consider a digital audio system: an input analog-to-digital converter (ADC) is used to collect input samples at a CD-quality rate *Fs* = 44 kHz. Input samples are processed with a digital filter to generate output samples, which are sent to a digital-to-analog converter (DAC) at the same rate *Fs*. Every sample interval *T* = 1/*Fs*, the digital processor executes the following MATLAB-compatible code: - -``` -% Read input sample from the ADC -x = read_ADC; -% Process input and... -mem(1) = x - mem(3)*9/16; -% ...compute output sample -y = mem(1)*7/16 - mem(3)*7/16; -% Send output sample to the DAC -write_DAC = y; -% Update memory for next iteration -mem(3) = mem(2); -mem(2) = mem(1); -``` - -- (a) Does the code implement DFI, DFII, TDFI, or TDFII? Support your answer by drawing the appropriate block diagram labeled in a manner that is consistent with the code. -- (b) Determine the transfer function *H*[*z*] of this system. -- (c) What is the basic filtering function of this system: LP, HP, BP, or BS? Justify your answer. -- (d) Determine the transfer function *H*−1[*z*] of the inverse system to *H*[*z*] and draw its DFI block implementation. How well will the inverse system operate? -- **5.4-13** Repeat Prob. 5.4-12 but instead use the code: - -``` -% Read input sample from the ADC -x = read_ADC; -% Compute output sample -y = x*7/32+mem(1); -% Send output sample to the DAC -write_DAC = y; -% Update memory for next iteration -mem(1) = mem(2); -mem(2) = x*7/32 + y*9/16; -``` - -- **5.5-1** A CT sinusoid *x*(*t*) = cos(ω*t*) is sampled at a greater-than-Nyquist rate *F*s = 1000 Hz to produce a DT sinusoid *x*(*t*) = cos(*n*). Determine the analog frequency ω if (a) = π 4 - - (b) = 2π 3 - - (c) = 7 8 -- **5.5-2** Find the amplitude and phase response of the digital filters depicted in Fig. P5.5-2. -- **5.5-3** A causal LTID system *H*(*z*) = 21(*z*−*j*)(*z*+*j*) 16(*z*− 1 2 )(*z*+ 3 4 ) has a periodic input *x*[*n*] that toggles between the values 1 and 2. That is, *x*[*n*]=[..., 1, 2, 1, ↓ 2 , 1, 2, 1, ...], where *x*[0] = 2. - - (a) Plot the magnitude response |*H*(*ej*)| over −2π ≤ ≤ 2π. - - (b) Plot the phase response *H*(*ej*) over −2π ≤ ≤ 2π. - - (c) Determine the system output *y*[*n*] in response to the periodic input *x*[*n*]. - -**5.5-4** A causal LTID system *H*(*z*) = −7(*z*+1) 32(*z*−*j* 3 4 )(*z*+*j* 3 4 ) has a periodic input *x*[*n*] that cycles through the 4 values 3, 2, 1, and 2. That is, *x*[*n*] = [..., 3, 2, 1, 2, ↓ 3, 2, 1, 2, ...], where *x*[0] = 3. - -- (a) Plot the magnitude response |*H*(*ej*)| over −2π ≤ ≤ 2π. -- (b) Plot the phase response *H*(*ej*) over −2π ≤ ≤ 2π. -- (c) Determine the system output *y*[*n*] in response to the periodic input *x*[*n*]. -- **5.5-5** Find the amplitude and the phase response of the filters shown in Fig. P5.5-5. [*Hint:* Express *H*[*ej*] as *e*−*j*2.5*Ha*[*ej*].] -- **5.5-6** Find the frequency response for the moving-average system in Prob. 3.4-3. The input–output equation of this system is given by - -$$ -y[n] = \frac{1}{5} \sum_{k=0}^{4} x[n-k] -$$ - -**5.5-7** (a) Input–output relationships of two filters are described by - -(i) *y*[*n*]=−0.9*y*[*n*−1] +*x*[*n*] - -(ii) *y*[*n*] = 0.9*y*[*n*−1] +*x*[*n*] - -For each case, find the transfer function, the amplitude response, and the phase response. Sketch the amplitude response, and state the type (highpass, lowpass, etc.) of each filter. - -- (b) Find the response of each of these filters to a sinusoid *x*[*n*] = cos*n* for = 0.01π and 0.99π. In general, show that the gain (amplitude response) of filter (i) at frequency 0 is the same as the gain of filter (ii) at frequency π −0. -- **5.5-8** For an LTID system specified by the equation - -$$ -y[n+1] - 0.5y[n] = x[n+1] + 0.8x[n] -$$ - -(a) Find the amplitude and the phase response. - -- (b) Find the system response *y*[*n*] for the input *x*[*n*] = cos(0.5*k* −(π/3)). -- **5.5-9** For an asymptotically stable LTID system, show that the steady-state response to input *ejnu*[*n*] is *H*[*ej*]*ejnu*[*n*]. The steady-state response is that part of the response which does not decay with time and persists forever. -- **5.5-10** Express the following signals in terms of apparent frequencies: - - (a) cos(0.8π*n*+θ ) - - (b) sin(1.2π*n*+θ ) - - (c) cos(6.9*n*+θ ) - -**Figure P5.5-2** - -(b) - -**Figure P5.5-5** - -- (d) cos(2.8π*n*+θ ) +2 sin(3.7π*n*+θ ) -- (e) sinc(π*n*/2) -- (f) sinc(3π*n*/2) -- (g) sinc(2π*n*) -- **5.5-11** Show that cos √ (0.6π*n* + (π/6)) + 3 cos(1.4π*n* + (π/3)) = 2 cos(0.6π*n* − (π/6)). -- **5.5-12** (a) A digital filter has the sampling interval *T* = 50µs. Determine the highest frequency that can be processed by this filter without aliasing. - - (b) If the highest frequency to be processed is 50 kHz, determine the minimum value of the sampling frequency *Fs* and the maximum value of the sampling interval *T* that can be used. -- **5.5-13** Consider the discrete-time system represented by - -$$ -y[n] = \sum_{k=0}^{\infty} (0.5)^k x[n-k] -$$ - -- (a) Determine and plot the magnitude response |*H*[*ej*]| of the system. -- (b) Determine and plot the phase response *H*[*ej*] of the system. -- (c) Find an efficient block representation that implements this system. -- **5.6-1** Pole-zero configurations of certain filters are shown in Fig. P5.6-1. Sketch roughly the amplitude response of these filters. -- **5.6-2** Figure P5.6-2 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[−1]=−1. - -**Figure P5.6-2** - -- (a) Determine the five constants *b*0, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *b*0*z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−2π ≤ ≤ 0). -- (c) Determine the output *y*[*n*] of this system if the input is *x*[*n*] = sin π*n* 2 . -- **5.6-3** Repeat Prob. 5.6-2 if the zero at *z* = 1 is moved to *z* = −1 and *H*[1]=−1 is specified rather than *H*[−1]=−1. -- **5.6-4** Figure P5.6-4 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[−1] = 1. - - (a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *k z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . - -**Figure P5.6-4** - -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−π ≤ ≤ π ). -- (c) A signal *x*(*t*) = cos(100π*t*) + cos(500π*t*) is sampled at a greater than Nyquist rate *F*s Hz and then input into the above LTID system to produce DT output *y*[*n*] = β cos(0*n* + θ ). Determine *F*s and 0. You do not need to find constants β and θ. -- (d) Is the inpulse response *h*[*n*] of this system absolutely summable? Justify your answer. -- **5.6-5** Figure P5.6-5 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[1]=−1. - -(a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *k z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . - -**Figure P5.6-5** - -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−π ≤ ≤ π ). -- (c) A signal *x*(*t*) = cos(2π*ft*) is sampled at a rate *F*s = 1 kHz and then input into the above LTID system to produce DT output *y*[*n*]. Determine, if possible, the frequency or frequencies *f* that will produce zero output, *y*[*n*] = 0. -- **5.6-6** The system *y*[*n*] − *y*[*n* − 1] = *x*[*n*] − *x*[*n* − 1] is an all-pass system that has zero phase response. Is there any difference between this system and the system *y*[*n*] = *x*[*n*]? Justify your answer. -- **5.6-7** Figure P5.6-7 displays the pole-zero plot of a second-order real, causal LTID system that has a repeated zero and *H*[1] = 4. The solid circle is the unit circle. - -### **Figure P5.6-7** - -- (a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*2 that specify the transfer function *H*[*z*] = *k z*2+*b*1*z*+*b*2 *z*2+*a*1*z*+*a*2 . -- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response |*H*[*ej*]| over the range (−π ≤ ≤ π ). -- (c) Determine the steady-state output *y*ss[*n*] of this system if the input is *x*[*n*] = cos( 3π*n* 4 )*u*[*n*]. -- (d) State whether this system is LP, HP, BP, BS, or other. If the digital system operates at *F*s = 8 kHz, what is the approximate hertzian cutoff frequency (or frequencies) of this system? -- **5.6-8** The magnitude and phase responses of a real, stable, LTI system are shown in Fig. P5.6-8. - - (a) What type of system is this: lowpass, highpass, bandpass, or bandstop? - - (b) What is the output of this system in response to - -$$ -x_1[n] = 2\sin\left(\frac{\pi}{2}n + \frac{\pi}{4}\right) -$$ - -(c) What is the output of this system in response to - -$$ -x_2[n] = \cos\left(\frac{7\pi}{4}n\right) -$$ - -- **5.6-9** Consider an LTID system with system function *H*[*z*] = *b*0 *z*2+1 *z*2−9/16 . - - (a) Determine the constant *b*0 so that the system frequency response at = −π is −1. - - (b) Accurately sketch the system poles and zeros. - - (c) Using the locations of the system poles and zeros, sketch |*H*[*ej*]| over 0 ≤ ≤ 2π. - -- (d) Determine the response *y*[*n*] to the input *x*[*n*] = (−1+*j*)+*j n* +(1−*j*)sin(π*n*+1). -- (e) Draw an appropriate block diagram representation of this system. -- **5.6-10** Do Prob. 5.10-3 by graphical procedure. Do the sketches approximately, without using MATLAB. -- **5.6-11** Do Prob. 5.10-8 by graphical procedure. Do the sketches approximately, without using MATLAB. -- **5.6-12** (a) Realize a digital filter whose transfer function is given by - -$$ -H[z] = K \frac{z+1}{z-a} -$$ - -- (b) Sketch the amplitude response of this filter, assuming |*a*| < 1. -- (c) The amplitude response of this lowpass filter is maximum at = 0. The 3 dB bandwidth is the frequency at which the amplitude response drops to 0.707 (or 1/ √2) times its maximum value. Determine the 3 dB bandwidth of this filter when *a* = 0.2. -- **5.6-13** Design a digital notch filter to reject frequency 5000 Hz completely and to have a sharp recovery on either side of 5000 Hz to a gain of unity. The highest frequency to be processed is 20 kHz (*Fh* = 20,000). [*Hint:* See Ex. 5.15. The zeros should be at *e*±*j*ω*T* for ω corresponding to 5000 Hz, and the poles are at *ae*±*j*ω*T* with *a* < 1. Leave your answer in terms of *a*. Realize this filter using the canonical form. Find the amplitude response of the filter.] - -**5.6-14** Consider the desired DT system magnitude response |*H*[*ej*]| in Fig. P5.6-14. - -**Figure P5.6-14** - -- (a) Is the filter LP, HP, BP, BS, or other? Explain. -- (b) Sketch the pole-zero plot of a 2nd-order system that behaves as a reasonable approximation of Fig. P5.6-14. What is the coefficient *b*0? -- **5.6-15** Show that a first-order LTID system with a pole at *z* = *r* and a zero at *z* = 1/*r* (*r* ≤ 1) is an allpass filter. In other words, show that the amplitude response |*H*[*ej*]| of a system with the transfer function - -$$ -H[z] = \frac{z - \frac{1}{r}}{z - r} \qquad r \le 1 -$$ - -is constant with frequency. This is a first-order allpass filter. [*Hint:* Show that the ratio of the distances of any point on the unit circle from the zero (at *z* = 1/*r*) and the pole (at *z* = *r*) is a constant 1/*r*.] - -Generalize this result to show that an LTID system with two poles at *z* = *re*±*j*θ and two zeros at *z* = (1/*r*)*e*±*j*θ (*r* ≤ 1) is an allpass filter. In other words, show that the amplitude response of a system with the transfer function - -$$ -H[z] = \frac{\left(z - \frac{1}{r}e^{j\theta}\right)\left(z - \frac{1}{r}e^{-j\theta}\right)}{(z - re^{j\theta})(z - re^{-j\theta})} -$$ -$$ -= \frac{z^2 - \left(\frac{2}{r}\cos\theta\right)z + \frac{1}{r^2}}{z^2 - (2r\cos\theta)z + r^2} \qquad r \le 1 -$$ - -is constant with frequency. - -**5.6-16** (a) If *h*1[*n*] and *h*2[*n*], the impulse responses of two LTID systems are related by *h*2[*n*] = (−1)*nh*1[*n*], then show that - -$$ -H_2[e^{j\Omega}] = H_1[e^{j(\Omega \pm \pi)}] -$$ - -How is the frequency response spectrum *H*2[*ej*] related to the *H*1[*ej*]? - -- (b) If *H*1[*z*] represents an ideal lowpass filter with cutoff frequency *c*, sketch *H*2[*ej*]. What type of filter is *H*2[*ej*]? -- **5.6-17** Mappings such as the bilinear transformation are useful in the conversion of continuous-time filters to discrete-time filters. Another useful type of transformation is one that converts a discrete-time filter into a different type of discrete-time filter. Consider a transformation that replaces *z* with −*z*. - - (a) Show that this transformation converts lowpass filters into highpass filters and highpass filters into lowpass filters. - - (b) If the original filter is an FIR filter with impulse response *h*[*n*], what is the impulse response of the transformed filter? -- **5.6-18** The bilinear transformation is defined by the rule *s* = 2(1−*z*−1)/*T*(1+*z*−1). - - (a) Show that this transformation maps the ω axis in the *s* plane to the unit circle *z* = *ej* in the *z* plane. - - (b) Show that this transformation maps to 2 arctan(ω*T*/2). -- **5.7-1** In Ch. 3, we used another approximation to find a digital system to realize an analog system. We showed that an analog system specified by Eq. (3.12) can be realized by using the digital system specified by Eq. (3.13). Compare that solution with the one resulting from the - -impulse-invariance method. Show that one result is a close approximation of the other and that the approximation improves as *T* → 0. - -- **5.7-2** A CT system has impulse response *h*ct(*t*) = *e*−*t u*(*t*). Draw the DFI realization of the corresponding DT system designed by the impulse-invariance method with *T* = 0.1. -- **5.7-3** (a) Using the impulse-invariance criterion, design a digital filter to realize an analog filter with transfer function - -$$ -H_a(s) = \frac{7s + 20}{2(s^2 + 7s + 10)} -$$ - -- (b) Show a canonical and a parallel realization of the filter. Use a 1% criterion for the choice of *T*. -- **5.7-4** Use the impulse-invariance criterion to design a digital filter to realize the second-order analog Butterworth filter with transfer function - -$$ -H_a(s) = \frac{1}{s^2 + \sqrt{2}s + 1} -$$ - -Use a 1% criterion for the choice of *T*. - -- **5.7-5** Design a digital integrator using the impulse-invariance method. Find and give a rough sketch of the amplitude response, and compare it with that of the ideal integrator. If this integrator is used primarily for integrating audio signals (whose bandwidth is 20 kHz), determine a suitable value for *T*. -- **5.7-6** An oscillator by definition is a source (no input) that generates a sinusoid of a certain frequency ω0. Therefore, an oscillator is a system whose zero-input response is a sinusoid of the desired frequency. Find the transfer function of a digital oscillator to oscillate at 10 kHz by the methods described in parts (a) and (b). In both methods, select *T* so that there are 10 samples in each cycle of the sinusoid. - - (a) Choose *H*[*z*] directly so that its zero-input response is a discrete-time sinusoid of frequency =ω*T* corresponding to 10 kHz. - - (b) Choose *Ha*(*s*) whose zero-input response is an analog sinusoid of 10 kHz. Now use the impulse invariance method to determine *H*[*z*]. - - (c) Show a canonical realization of the oscillator. - -- **5.7-7** A variant of the impulse invariance method is the *step-invariance* method of digital filter synthesis. In this method, for a given *Ha*(*s*), we design *H*[*z*] in Fig. 5.24a such that *y*(*nT*) in Fig. 5.24b is identical to *y*[*n*] in Fig. 5.24a when *x*(*t*) = *u*(*t*). - - (a) Show that, in general, - -$$ -H[z] = \frac{z-1}{z} \mathcal{Z} \left[ \left( \mathcal{L}^{-1} \frac{H_a(s)}{s} \right)_{t=kT} \right] -$$ - -(b) Use this method to design *H*[*z*] for - -$$ -H_a(s) = \frac{\omega_c}{s + \omega_c} -$$ - -- (c) Use the step-invariance method to synthesize a discrete-time integrator and compare its amplitude response with that of the ideal integrator. -- **5.7-8** Use the *ramp-invariance* method to synthesize a discrete-time differentiator and integrator. In this method, for a given *Ha*(*s*), we design *H*[*z*] such that *y*(*nT*) in Fig. 5.24b is identical to *y*[*n*] in Fig. 5.24a when *x*(*t*) = *tu*(*t*). -- **5.7-9** In an impulse-invariance design, show that if *Ha*(*s*) is a transfer function of a stable system, the corresponding *H*[*z*] is also a transfer function of a stable system. -- **5.7-10** First-order backward differences provide the transformation rule *s* = (1−*z*−1)/*T*. - - (a) Show that this transformation maps the ω axis in the *s* plane to a circle of radius 1/2 centered at (1/2, 0) in the *z* plane. - - (b) Show that this transformation maps the left-half *s* plane to the interior of the unit circle in the *z* plane, which ensures that stability is preserved. -- **5.8-1** Find the *z*-transform (if it exists) and the corresponding ROC for each of the following signals: - - (a) (0.8)*nu*[*n*] +2*nu*[−(*n*+1)] - -(b) -$$ -2^n u[n] - 3^n u[-(n+1)] -$$ - -(c) -$$ -(-2)^{n+3}u[-n] + \sum_{k=0}^{\infty} (0.5)^{k-1} \delta(n-2k) -$$ - -(d) -$$ -(0.8)^n u[n] + (0.9)^n u[-(n+1)] -$$ - -(e) -$$ -[(0.8)^n + 3(0.4)^n]u[-(n+1)] -$$ - -(f) [(0.8)*n* +3(0.4)*n*]*u*[*n*] - -(g) -$$ -(0.8)^n u[n] + 3(0.4)^n u[-(n+1)] -$$ - -$$ -(0.6) \quad u[n] + 5(0.4) \quad u[ -$$ - -(h) -$$ -(0.5)^{|n|} -$$ - -$$ -(i) \t n u[-(n+1)] -$$ - -- **5.8-2** Using the definition, compute the bilateral *z*-transform *X*(*z*) of (a) *x*[*n*] = 3*nu*[−*n*] (b) *x*[*n*] = ( 1 3 )*nu*[*n*] Express your answers in standard rational form. -- **5.8-3** Determine the inverse *z*-transform *x*[*n*] of *X*[*z*] = *z*2− 1 3 *z* (*z*−1)(*z*+2) with ROC 1 &lt; |*z*| &lt; 2. - -**5.8-4** Find the inverse -$$ -z -$$ --transform of - -$$ -X[z] = \frac{(e^{-2} - 2)z}{(z - e^{-2})(z - 2)} -$$ - -when the ROC is, -\n(a) -$$ -|z| > 2 -$$ - -\n(b) $e^{-2} < |z| < 2$ -\n(c) $|z| < e^{-2}$ - -**5.8-5** Use partial fraction expansions, *z*-transform tables, and a region of convergence (|*z*| < 1/2) to determine the inverse *z*-transform of - -$$ -X(z) = \frac{1}{(2z+1)(z+1)(z+\frac{1}{2})} -$$ - -- **5.8-6** Using *z*-transform techniques and properties (no time-domain convolution sum!), determine the convolution *y*[*n*] = ( 1 3 )*n*−3*u*[*n* − 2] ∗ (2)*nu*[−*n*]. Express your answer in the form *y*[*n*] = *c*1γ *n* 1 *u*[*n* + *N*1] + *c*2γ *n* 2 *u*[−*n* + *N*2], making sure to clearly identify the constants *c*1, *c*2, γ1, γ2, *N*1, and *N*2. -- **5.8-7** Using partial fraction expansions, *z*-transform tables, and the fact that *h*[*n*] is stable, determine the inverse *z*-transform of - -$$ -H[z] = \frac{z^4 + z^3}{(z - 2)(z + \frac{1}{2})}. -$$ - -**5.8-8** Consider the system - -$$ -H[z] = \frac{z(z - \frac{1}{2})}{(z^3 - \frac{27}{8})} -$$ - -- (a) Draw the pole-zero diagram for *H*[*z*] and identify all possible regions of convergence. -- (b) Draw the pole-zero diagram for *H*−1[*z*] and identify all possible regions of convergence. -- **5.8-9** A discrete-time signal *x*[*n*] has a rational *z*-transform that contains a pole at *z* = 0.5. Given *x*1[*n*] = (1/3)*nx*[*n*] is absolutely summable and - -Problems 589 - -*x*2[*n*] = (1/4)*nx*[*n*] is **not** absolutely summable, determine whether *x*[*n*] is left-sided, right-sided, or two-sided. Justify your answer! - -- **5.8-10** Let *x*[*n*] be an absolutely summable signal with rational *z*-transform *X*[*z*]. *X*[*z*] is known to have a pole at *z* = (0.75+0.75*j*), and other poles may be present. Recall that an absolutely summable signal satisfies %∞ −∞ |*x*[*n*]| < ∞. - - (a) Can *x*[*n*] be left-sided? Explain. - - (b) Can *x*[*n*] be right-sided? Explain. - - (c) Can *x*[*n*] be two-sided? Explain. - - (d) Can *x*[*n*] be of finite duration? Explain. -- **5.8-11** Consider a causal system that has transfer function - -$$ -H[z] = \frac{z - 0.5}{z + 0.5} -$$ - -When appropriate, assume initial conditions of zero. - -- (a) Determine the output *y*1[*n*] of this system in response to *x*1[*n*] = (3/4)*nu*[*n*]. -- (b) Determine the output *y*2[*n*] of this system in response to *x*2[*n*] = (3/4)*n*. -- (c) Determine the output *y*3[*n*] of this system in response to *x*3[*n*] = (3/4)*nu*[−*n*−1]. -- **5.8-12** Let *x*[*n*] =(−1)*nu*[*n*−*n*0]+α*nu*[−*n*]. Determine the constraints on the complex number α and the integer *n*0 so that the *z*-transform *X*[*z*] exists with region of convergence 1 < |*z*| < 2. -- **5.8-13** Using the definition, compute the bilateral *z*-transform, including the region of convergence (ROC), of the following complex-valued functions: - - (a) *x*1[*n*] = (−*j*)−*nu*[−*n*] +δ[−*n*] - - (b) *x*2[*n*] = (*j*)*n* cos(*n*+1)*u*[*n*] - - (c) *x*3[*n*] = *j*sinh(*n*)*u*[−*n*+1] - - (d) *x*4[*n*] = %0 *k*=−∞(2*j*)*n*δ[*n*−2*k*] -- **5.8-14** Use partial fraction expansions, *z*-transform tables, and a region of convergence (0.5 < |*z*| < 2) to determine the inverse *z*-transform of - - (a) *X*1[*z*] = 1 1+ 13 6 *z*−1 + 1 6 *z*−2 − 1 3 *z*−3 1 - -(b) -$$ -X_2[z] = \frac{1}{z^{-3}(2 - z^{-1})(1 + 2z^{-1})} -$$ - -**5.8-15** Use partial fraction expansions, *z*-transform tables, and the fact that the systems are stable to determine the inverse *z*-transform of - -(a) -$$ -H_1[z] = \frac{z^{-1}}{(z - \frac{1}{2})(1 + \frac{1}{2}z^{-1})} -$$ - -(b) -$$ -H_2[z] = \frac{z+1}{z^3(z-2)(z+\frac{1}{2})} -$$ - -**5.8-16** By inserting *N* − 1 zeros between every sample of a unit step, we obtain a signal - -$$ -h[n] = \sum_{k=0}^{\infty} \delta[n - Nk] -$$ - -Determine *H*[*z*], the bilateral *z*-transform of *h*[*n*]. Identify the number and location(s) of the poles of *H*[*z*]. - -- **5.8-17** Using transform-domain techniques, determine the zero-state response *y*zsr[*n*] of LTID system *y*[*n*] − 1 4 *y*[*n* − 2] = *x*[*n*] to the noncausal input *x*[*n*] = 2*nu*[2−*n*]. -- **5.8-18** Determine the zero-state response of a system having a transfer function - -$$ -H[z] = \frac{z}{(z+0.2)(z-0.8)} \qquad |z| > 0.8 -$$ - -and an input *x*[*n*] given by - -- (a) *x*[*n*] = *enu*[*n*] -- (b) *x*[*n*] = 2*nu*[−(*n*+1)] -- (c) *x*[*n*] = *enu*[*n*] +2*nu*[−(*n*+1)] -- (d) *x*[*n*] = 2*nu*[*n*] +*u*[−(*n*+1)] -- (e) *x*[*n*] = *e*−2*nu*[−(*n*+1)] -- **5.8-19** The discrete cross-correlation between real signal *x*[*n*] and real signal *y*[*n*] is - -$$ -c_{xy}[n] = \sum_{k=-\infty}^{\infty} x[k]y[k-n] -$$ - -Let signal *x*[*n*] have *z*-transform *X*[*z*] with ROC *Rx*, and let signal *y*[*n*] have *z*-transform *Y*[*z*] with ROC *Ry*. Determine *CXY* [*z*] (the bilateral *z*-transform of *cxy*[*n*]) in terms of the *z*-transforms of *x*[*n*] and *y*[*n*]. - -**5.8-20** Transform properties can be very useful. The accumulation property states - -$$ -\sum_{k=-\infty}^{n} x[k] \Longleftrightarrow \frac{z}{z-1} X[z] -$$ - -This property is the DT version of the Laplace transform "integration in time" property. Given *x*[*n*] ⇐⇒ *X*[*z*], prove the accumulation property. [*Hint:* Polynomial long division can be helpful.] **5.10-1** Use MATLAB to generate pole-zero plots for the causal systems with the following transfer functions: - -(a) -$$ -H_a[z] = \frac{z^4 - \sqrt{2}z^2 + 1}{z^4 + 0.4096} -$$ - -(b) $H_b[z] = \frac{-3z^{-1} + \frac{3}{4}z^{-3}}{3 + \frac{3}{8}z^{-2} + 2z^{-4}}$ - -3+ 3 2 *z*−2+2*z*−4 In each case, determine whether the system is - -- stable. -- **5.10-2** For each of the following stable LTID systems, use MATLAB to generate magnitude and phase response plots over −π ≤ <π. (a) *H*a[*z*] = cos(*z*) *z*−0.5 - - (b) *H*b[*z*] = *z*3 sin(*z*−1) -- **5.10-3** Consider an LTID system described by the difference equation 4*y*[*n* + 2] − *y*[*n*] = *x*[*n* + 2] +*x*[*n*]. - - (a) Plot the pole-zero diagram for this system. - - (b) Plot the system's magnitude response |*H*[*ej*]| over −π ≤ ≤ π. - - (c) What type of system is this: lowpass, highpass, bandpass, or bandstop? - - (d) Is this system stable? Justify your answer. - - (e) Is this system real? Justify your answer. - - (f) If the system input is of the form *x*[*n*] = cos(*n*), what is the greatest possible amplitude of the output? Justify your answer. - - (g) Draw an efficient, causal implementation of this system using only add, scale, and delay blocks. -- **5.10-4** Consider an LTID system with system function *H*(*z*) = *b*0 *z*2+1 *z*2−4/9 . - - (a) Determine the constant *b*0 so that the system frequency response at = −π is −1. - - (b) Plot the pole-zero diagram for this system. - - (c) Plot the system's magnitude response |*H*[*ej*]| over −π ≤ ≤ π. - - (d) Plot the system's phase response *H*[*ej*] over −π ≤ ≤ π. - - (e) What type of system is this: lowpass, highpass, bandpass, or bandstop? - - (f) Determine the response *y*[*n*] to the input *x*[*n*] = (−1 − *j*) + (−*j*)*n* + (1 − *j*) cos(π*n* + 1 3 ). - - (g) Draw a TDFII block diagram representation of this system. - -**5.10-5** Consider the LTID system shown in Fig. P5.10-5, where parameters *c*1 and *c*2 are arbitrary constants. - -### **Figure P5.10-5** - -- (a) Determine the system function *H*[*z*], expressed in standard rational form. -- (b) What is the order *N* of this system? -- (c) Determine the *N* poles and *N* zeros of this system. Use MATLAB to create the corresponding pole-zero plot. -- (d) What constraints, if any, exist on parameters *c*1 and *c*2 to ensure that the system is stable? -- (e) Determine *c*1 and *c*2 so that this system functions as an LPF with narrow passband. Use MATLAB to generate the corresponding magnitude response |*H*[*ej*]| over −π ≤ ≤ π. -- (f) Ms. Zeroine, the heroine of DT systems, believes that if *x*[*n*] = 0, then *y*[*n*] = 0 also. Is Ms. Zeroine correct? Fully justify your answer. -- (g) Dr. Strange suggests that by setting *c*1 = −1 and *c*2 = −2, the system will act as a highpass filter. Is Dr. Strange right? Fully justify your answer. -- **5.10-6** One interesting and useful application of discrete systems is the implementation of complex (rather than real) systems. A complex system is one in which a real-valued input can produce a complex-valued output. Complex systems that are described by constant coefficient difference equations require at least one complex-valued coefficient, and they are capable of operating on complex-valued inputs. Consider the complex discrete-time system - -$$ -H[z] = \frac{z^2 - j}{z - 0.9e^{j3\pi/4}} -$$ - -(a) Determine and plot the system zeros and poles. - -- (b) Sketch the magnitude response |*H*[*ej*ω]| of this system over −2π ≤ ω ≤ 2π. Comment on the system's behavior. -- **5.10-7** Consider the complex system - -$$ -H[z] = \frac{z^4 - 1}{2(z^2 + 0.81j)} -$$ - -Refer to Prob. 5.10-6 for an introduction to complex systems. - -- (a) Plot the pole-zero diagram for *H*[*z*]. -- (b) Plot the system's magnitude response |*H*[*ej*]| over −π ≤ ≤ π. -- (c) Explain why *H*[*z*] is a noncausal system. Do not give a general definition of causality; specifically identify what makes this system noncausal. -- (d) One way to make this system causal is to add two poles to *H*[*z*]. That is, - -$$ -H_{\text{causal}}[z] = H[z] \frac{1}{(z-a)(z-b)} -$$ - -Find poles *a* and *b* such that |*H*causal[*ej*]| = |*H*[*ej*]|. - -- (e) Draw an efficient block implementation of *H*causal[*z*]. -- **5.10-8** A discrete-time LTI system is shown in Fig. P5.10-8. - - (a) Determine the difference equation that describes this system. - - (b) Determine the magnitude response |*H*[*ej*]| for this system and simplify your answer. Plot the magnitude response over −π ≤ ≤ π. What type of standard filter (lowpass, highpass, bandpass, or bandstop) best describes this system? - - (c) Determine the impulse response *h*[*n*] of this system. - -**Figure P5.10-8** - -- **5.10-9** Determine the impulse response *h*[*n*] for the system shown in Fig. P5.10-9. Is the system stable? Is the system causal? -- **5.10-10** An LTID filter has an impulse response function given by *h*[*n*] = δ[*n* − 1] + δ[*n* + 1]. Determine and carefully sketch the magnitude response |*H*[*ej*]| over the range −π ≤ ≤ π. For this range of frequencies, is this filter lowpass, highpass, bandpass, or bandstop? -- **5.10-11** A causal, stable discrete system has the rather strange transfer function *H*[*z*] = cos(*z*−1). - - (a) Write MATLAB code that will compute and plot the magnitude response of this system over an appropriate range of digital frequencies . Comment on the system. - - (b) Determine the impulse response *h*[*n*]. Plot *h*[*n*] over (0 ≤ *n* ≤ 10). - - (c) Determine a difference equation description for an FIR filter that closely approximates the system *H*[*z*] = cos(*z*−1). To verify proper behavior, plot the FIR filter's magnitude response and compare it with the magnitude response computed in part (a). -- **5.10-12** The MATLAB signal-processing toolbox function butter helps design digital Butterworth filters. Use MATLAB help to learn how butter works. For each of the following cases, design the filter, plot the filter's poles and zeros in the complex *z* plane, and plot the decibel magnitude response 20log10 |*H*[*ej*]|. - - (a) Design an eighth-order digital lowpass filter with *c* = π/3. - - (b) Design an eighth-order digital highpass filter with *c* = π/3. - - (c) Design an eighth-order digital bandpass filter with passband between 5π/24 and 11π/24. - - (d) Design an eighth-order digital bandstop filter with stopband between 5π/24 and 11π/24. -- **5.10-13** The MATLAB signal-processing toolbox function cheby1 helps design digital Chebyshev - -type I filters. A Chebyshev type I filter has passband ripple and smooth stopband. Setting the passband ripple to *Rp* = 3 dB, repeat Prob. 5.10-12 using the cheby1 command. With all other parameters held constant, what is the general effect of reducing *Rp*, the allowable passband ripple? - -**5.10-14** The MATLAB signal-processing toolbox function cheby2 helps design digital Chebyshev type II filters. A Chebyshev type II filter has smooth passband and ripple in the stopband. Setting the stopband ripple *Rs* = 20 dB down, repeat Prob. 5.10-12 using the cheby2 command. With all other parameters held constant, what is the general effect of increasing *Rs*, the minimum stopband attenuation? - -**5.10-15** The MATLAB signal-processing toolbox function ellip helps design digital elliptic filters. An elliptic filter has ripple in both the passband and the stopband. Setting the passband ripple to *Rp* = 3 dB and the stopband ripple *Rs* = 20 dB down, repeat Prob. 5.10-12 using the ellip command. - - - -# **[CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER SERIES** - -Electrical engineers instinctively think of signals in terms of their frequency spectra and think of systems in terms of their frequency response. Most teenagers know about the audible portion of audio signals having a bandwidth of about 20 kHz and the need for good-quality speakers to respond up to 20 kHz. This is basically thinking in the frequency domain. In Chs. 4 and 5 we discussed extensively the frequency-domain representation of systems and their spectral response (system response to signals of various frequencies). In Chs. 6 through 9, we discuss spectral representation of signals, where signals are expressed as a sum of sinusoids or exponentials. Actually, we touched on this topic in Chs. 4 and 5. Recall that the Laplace transform of a continuous-time signal is its spectral representation in terms of exponentials (or sinusoids) of complex frequencies. Similarly the *z*-transform of a discrete-time signal is its spectral representation in terms of discrete-time exponentials. However, in the earlier chapters we were concerned mainly with system representation; the spectral representation of signals was incidental to the system analysis. Spectral analysis of signals is an important topic in its own right, and now we turn to this subject. - -In this chapter we show that a periodic signal can be represented as a sum of sinusoids (or exponentials) of various frequencies. These results are extended to aperiodic signals in Ch. 7 and to discrete-time signals in Ch. 9. The fascinating subject of sampling of continuous-time signals is discussed in Ch. 8, leading to A/D (analog-to-digital) and D/A conversion. Chapter 8 forms the bridge between the continuous-time and the discrete-time worlds. - -## **6.1 PERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY TRIGONOMETRIC FOURIER SERIES** - -As seen in Sec. 1.3-3 [Eq. (1.7)], a periodic signal *x*(*t*) with period *T*0 (Fig. 6.1) has the property - -$$ -x(t) = x(t + T_0) \qquad \text{for all } t -$$ - -The *smallest* value of *T*0 that satisfies this periodicity condition is the *fundamental period* of *x*(*t*). As argued in Sec. 1.3-3, this equation implies that *x*(*t*) starts at −∞ and continues to ∞. Moreover, the area under a periodic signal *x*(*t*) over any interval of duration *T*0 is the same; that is, for any - -**Figure 6.1** A periodic signal of period *T*0. - -real numbers *a* and *b* - -$$ -\int_{a}^{a+T_0} x(t) dt = \int_{b}^{b+T_0} x(t) dt -$$ - -This result follows from the fact that a periodic signal takes the same values at intervals of *T*0. Hence, the values over any segment of duration *T*0 are repeated in any other interval of the same duration. For convenience, the area under *x*(*t*) over any interval of duration *T*0 will be denoted by - -$$ -\int_{T_0} x(t) \, dt -$$ - -The frequency of a sinusoid cos 2π*f*0*t* or sin 2π*f*0*t* is *f*0, and the period is *T*0 = 1/*f*0. These sinusoids can also be expressed as cosω0*t* or sinω0*t*, where ω0 = 2π*f*0 is the *radian frequency*, although for brevity, it is often referred to as frequency (see Sec. B.2). A sinusoid of frequency *nf*0 is said to be the *nth harmonic* of the sinusoid of frequency *f*0. - -Let us consider a signal *x*(*t*) made up of a sines and cosines of frequency ω0 and all of its harmonics (including the zeroth harmonic; i.e., dc) with arbitrary amplitudes† : - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.1} -$$ - -The frequency ω0 is called the *fundamental frequency.* - -We now prove an extremely important property: *x*(*t*) in Eq. (6.1) is a periodic signal with the same period as that of the fundamental, regardless of the values of the amplitudes *an* and *bn*. Note that the period *T*0 of the fundamental satisfies - -$$ -T_0 = \frac{1}{f_0} = \frac{2\pi}{\omega_0} \quad \text{and} \quad \omega_0 T_0 = 2\pi \tag{6.2} -$$ - -† In Eq. (6.1), the constant term *a*0 corresponds to the cosine term for *n* = 0 because cos(0 × ω0)*t* = 1. However, sin(0×ω0)*t* = 0. Hence, the sine term for *n* = 0 is nonexistent. - -To prove the periodicity of *x*(*t*), all we need is to show that *x*(*t*) = *x*(*t* +*T*0). From Eq. (6.1), - -$$ -x(t+T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 (t+T_0) + b_n \sin n\omega_0 (t+T_0) -$$ - -= $a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + n\omega_0 T_0) + b_n \sin(n\omega_0 t + n\omega_0 T_0)$ - -From Eq. (6.2), we have *n*ω0*T*0 = 2π*n*, and - -$$ -x(t + T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + 2\pi n) + b_n \sin(n\omega_0 t + 2\pi n) -$$ - -= $a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t = x(t)$ - -We could also infer this result intuitively. In one fundamental period *T*0, the *n*th harmonic executes *n* complete cycles. Hence, every sinusoid on the right-hand side of Eq. (6.1) executes a complete number of cycles in one fundamental period *T*0. Therefore, at *t* = *T*0, every sinusoid starts as if it were the origin and repeats the same drama over the next *T*0 seconds, and so on, ad infinitum. Hence, the sum of such harmonics results in a periodic signal of period *T*0. - -This result shows that any combination of sinusoids of frequencies 0, *f*0, 2*f*0, ..., *kf*0 is a periodic signal of period *T*0 = 1/*f*0 regardless of the values of amplitudes *ak* and *bk* of these sinusoids. By changing the values of *ak* and *bk* in Eq. (6.1), we can construct a variety of periodic signals, all of the same period *T*0 (*T*0 = 1/*f*0 = 2π/ω0). - -The converse of this result is also true. We shall show in Sec. 6.5-4 that *a periodic signal x*(*t*) *with a period T*0 *can be expressed as a sum of a sinusoid of frequency f*0 *(f*0 = 1/*T*0*) and all its harmonics, as shown in Eq. (6.1)*. † The infinite series on the right-hand side of Eq. (6.1) is known as the *trigonometric Fourier series* of a periodic signal *x*(*t*). - -### COMPUTING THE COEFFICIENTS OF A FOURIER SERIES - -To determine the coefficients of a Fourier series, consider an integral *I* defined by - -$$ -I = \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt -$$ - -where \$ *T*0 stands for integration over any contiguous interval of *T*0 seconds. By using a trigonometric identity (see Sec. B.8-6), this integral can be expressed as - -$$ -I = \frac{1}{2} \left[ \int_{T_0} \cos(n+m)\omega_0 t \, dt + \int_{T_0} \cos(n-m)\omega_0 t \, dt \right] \tag{6.3} -$$ - -† Strictly speaking, this statement applies only if a periodic signal *x*(*t*) is a continuous function of *t*. However, Sec. 6.5-4 shows that it can be applied even for discontinuous signals, if we interpret the equality in Eq. (6.1) in the mean-square sense instead of in the ordinary sense. This means that the power of the difference between the periodic signal *x*(*t*) and its Fourier series on the right-hand side of Eq. (6.1) approaches zero as the number of terms in the series approaches infinity. - -#### 596 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -Because cos ω0*t* executes one complete cycle during any interval of duration *T*0, cos(*n* + *m*)ω0*t* executes (*n* + *m*) complete cycles during any interval of duration *T*0. Therefore, the first integral in Eq. (6.3), which represents the area under *n*+*m* complete cycles of a sinusoid, equals zero. The same argument shows that the second integral in Eq. (6.3) is also zero, except when *n* = *m*. Hence, *I* in Eq. (6.3) is zero for all *n* = *m*. When *n* = *m*, the first integral in Eq. (6.3) is still zero, but the second integral yields - -$$ -I = \frac{1}{2} \int_{T_0} dt = \frac{T_0}{2} -$$ - -Thus, - -$$ -\int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & m = n \neq 0 \end{cases} \tag{6.4} -$$ - -Using similar arguments, we can show that - -$$ -\int_{T_0} \sin n\omega_0 t \sin m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & n = m \neq 0 \end{cases} \tag{6.5} -$$ - -and - -$$ -\int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt = 0 \qquad \text{for all } n \text{ and } m \tag{6.6} -$$ - -To determine *a*0 in Eq. (6.1), we integrate both sides of Eq. (6.1) over one period *T*0 to yield - -$$ -\int_{T_0} x(t) dt = a_0 \int_{T_0} dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t dt + b_n \int_{T_0} \sin n\omega_0 t dt \right] -$$ - -Recall that *T*0 is the period of a sinusoid of frequency ω0. Therefore, functions cos *n*ω0*t* and sin *n*ω0*t* execute *n* complete cycles over any interval of *T*0 seconds so that the area under these functions over an interval *T*0 is zero, and the last two integrals on the right-hand side of the foregoing equation are zero. This yields - -$$ -\int_{T_0} x(t) dt = a_0 \int_{T_0} dt = a_0 T_0 \quad \text{and} \quad a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt -$$ - -Next we multiply both sides of Eq. (6.1) by cos*m*ω0*t* and integrate the resulting equation over an interval *T*0: - -$$ -\int_{T_0} x(t) \cos m\omega_0 t \, dt = a_0 \int_{T_0} \cos m\omega_0 t \, dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt + b_n \int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt \right] -$$ - -The first integral on the right-hand side is zero because it is an area under *m* integral number of cycles of a sinusoid. Also, the last integral on the right-hand side vanishes because of Eq. (6.6). This leaves only the middle integral, which is also zero for all *n* = *m* because of Eq. (6.4). But *n* takes on all values from 1 to ∞, including *m*. When *n* = *m*, this integral is *T*0/2, according to Eq. (6.4). Therefore, from the infinite number of terms on the right-hand side, only one term survives to yield *anT*0/2 = *amT*0/2 (recall that *n* = *m*). Therefore, - -$$ -\int_{T_0} x(t) \cos m\omega_0 t \, dt = \frac{a_m T_0}{2} \qquad \text{and} \qquad a_m = \frac{2}{T_0} \int_{T_0} x(t) \cos m\omega_0 t \, dt -$$ - -Similarly, by multiplying both sides of Eq. (6.1) by sin *n*ω0*t* and then integrating over an interval *T*0, we obtain - -$$ -b_m = \frac{2}{T_0} \int_{T_0} x(t) \sin m\omega_0 t \, dt -$$ - -To sum up our discussion, which applies to real or complex *x*(*t*), we have shown that a periodic signal *x*(*t*) with period *T*0 can be expressed as a sum of a sinusoid of period *T*0 and its harmonics: - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.7} -$$ - -where ω0 = 2π*f*0 = 2π *T*0 and - -$$ -a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{T_0} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{T_0} x(t) \sin n\omega_0 t dt \tag{6.8} -$$ - -### COMPACT FORM OF FOURIER SERIES - -The results derived so far are general and apply whether *x*(*t*) is a real or a complex function of *t*. However, when *x*(*t*) is real, coefficients *an* and *bn* are real for all *n*, and the trigonometric Fourier series can be expressed in a *compact form,* using the results in Eq. (B.16): - -$$ -x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \theta_n) -$$ -\n(6.9) - -where *Cn* and θ*n* are related to *an* and *bn*, as [see Eq. (B.17)] - -$$ -C_0 = a_0 -$$ -, $C_n = \sqrt{a_n^2 + b_n^2}$ , and $\theta_n = \tan^{-1} \left( \frac{-b_n}{a_n} \right)$ (6.10) - -These results are summarized in Table 6.1. - -The compact form in Eq. (6.9) uses the cosine form. We could just as well have used the sine form, with terms sin(*n*ω0*t* + θ*n*) instead of cos(*n*ω0*t* + θ*n*). The literature overwhelmingly favors the cosine form, for no apparent reason except possibly that the cosine phasor is represented by the horizontal axis, which happens to be the reference axis in phasor representation. - -Equation (6.8) shows that *a*0 (or *C*0) is the average value of *x*(*t*) (averaged over one period). This value can often be determined by inspection of *x*(*t*). - -Because *an* and *bn* are real, *Cn* and θ*n* are also real. In the following discussion of trigonometric Fourier series, we shall assume real *x*(*t*), unless mentioned otherwise. - -| Series Form | Coefficient Computation | Conversion Formulas | -|--------------------------------------------------------|----------------------------------------------------|-------------------------------| -| Trigonometric | #
1
a0
f(t)dt
=
T0
T0 | a0
= C0
= D0 | -| "∞
f(t) = a0+
an
cosnω0t+bn
sinnω0t
n=1 | #
2
an
f(t) cosnω0t dt
=
T0
T0 | = Cnejθn =
an−jbn
2Dn | -| | #
2
bn
f(t)sinnω0t dt
=
T0 | = Cne−jθn =
an+jbn
2D−n | -| Compact trigonometric | T0
C0
= a0 | C0
= D0 | -| "∞
f(t) = C0
Cn
cos(nω0t +θn)
+ | =
2
2 +bn
Cn
an | Cn
= 2 Dn
n ≥ 1 | -| n=1 | −bn
= tan−1
θn
an | θn
= Dn | -| Exponential | | | -| f(t) = "∞
Dnejnω0t
n=−∞ | #
1
f(t)e−jnω0t
Dn
dt
=
T0
T0 | | - -**TABLE 6.1** Fourier Series Representation of a Periodic Signal of Period *T*0 (ω0 = 2π/*T*0) - -### **[6.1-1 The Fourier Spectrum](#page-12-0)** - -The compact trigonometric Fourier series in Eq. (6.9) indicates that a periodic signal *x*(*t*) can be expressed as a sum of sinusoids of frequencies 0 (dc), ω0, 2ω0, ..., *n*ω0, ..., whose amplitudes are *C*0, *C*1, *C*2, ..., *Cn*, ..., and whose phases are 0, θ1, θ2, ..., θ*n*, ..., respectively. We can readily plot amplitude *Cn* versus *n* (*the amplitude spectrum*) and θ*n* versus *n* (the *phase spectrum*).† Because *n* is proportional to the frequency *n*ω0, these plots are scaled plots of *Cn* versus ω and θ*n* versus ω. The two plots together are the *frequency spectra* of *x*(*t*). These spectra show at a glance the frequency contents of the signal *x*(*t*) with their amplitudes and phases. Knowing these spectra, we can reconstruct or synthesize the signal *x*(*t*) according to Eq. (6.9). Therefore, frequency spectra, which are an alternative way of describing a periodic signal *x*(*t*), are in every way equivalent to the plot of *x*(*t*) as a function of *t*. The frequency spectra of a signal constitute the *frequency-domain description* of *x*(*t*), in contrast to the *time-domain description,* where *x*(*t*) is specified as a function of time. - -In computing θ*n*, the phase of the *n*th harmonic from Eq. (6.10), the quadrant in which θ*n* lies should be determined from the signs of *an* and *bn*. For example, if *an* = −1 and *bn* = 1, θ*n* lies in the third quadrant, and - -$$ -\theta_n = \tan^{-1}\left(\frac{-1}{-1}\right) = -135^\circ -$$ - -Observe that - -$$ -\tan^{-1}\left(\frac{-1}{-1}\right) \neq \tan^{-1}(1) = 45^{\circ} -$$ - -† The amplitude *Cn*, by definition here, is nonnegative. Some authors define amplitude *An* that can take positive or negative values and magnitude *Cn* = |*An*| that can only be nonnegative. Thus, what we call amplitude spectrum becomes magnitude spectrum. The distinction between amplitude and magnitude, although useful, is avoided in this book in the interest of keeping definitions of essentially similar entities to a minimum. - -Although *Cn*, the amplitude of the *n*th harmonic as defined in Eq. (6.10), is positive, we shall find it convenient to allow *Cn* to take on negative values when *bn* = 0. This will become clear in later examples. - -## **EXAMPLE 6.1 Compact Trigonometric Fourier Series of Periodic Exponential Wave** - -Find the compact trigonometric Fourier series for the periodic signal *x*(*t*) shown in Fig. 6.2a. Sketch the amplitude and phase spectra for *x*(*t*). - -**Figure 6.2 (a)** A periodic signal and **(b, c)** its Fourier spectra. - -In this case the period *T*0 = π and the fundamental frequency *f*0 = 1/*T*0 = 1/π Hz, and - -$$ -\omega_0 = \frac{2\pi}{T_0} = 2 \,\text{rad/s} -$$ - -Therefore, - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos 2nt + b_n \sin 2nt -$$ - -where - -$$ -a_0 = \frac{1}{\pi} \int_{T_0} x(t) dt -$$ - -In this example the obvious choice for the interval of integration is from 0 to π. Hence, - -$$ -a_0 = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} dt = 0.504 -$$ - -$$ -a_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \cos 2nt dt = 0.504 \left(\frac{2}{1 + 16n^2}\right) -$$ - -and - -$$ -b_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \sin 2nt \, dt = 0.504 \left( \frac{8n}{1 + 16n^2} \right) -$$ - -Therefore, - -$$ -x(t) = 0.504 \left[ 1 + \sum_{n=1}^{\infty} \frac{2}{1 + 16n^2} (\cos 2nt + 4n \sin 2nt) \right] -$$ - -Also from Eq. (6.10), - -$$ -C_0 = a_0 = 0.504 -$$ - -\n -$$ -C_n = \sqrt{a_n^2 + b_n^2} = 0.504 \sqrt{\frac{4}{(1 + 16n^2)^2} + \frac{64n^2}{(1 + 16n^2)^2}} = 0.504 \left(\frac{2}{\sqrt{1 + 16n^2}}\right) -$$ - -\n -$$ -\theta_n = \tan^{-1}\left(\frac{-b_n}{a_n}\right) = \tan^{-1}(-4n) = -\tan^{-1} 4n -$$ - -Amplitude and phases of the dc and the first seven harmonics are computed from the above equations as - -| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | -|----|-------|---------|---------|---------|---------|---------|---------|---------| -| Cn | 0.504 | 0.244 | 0.125 | 0.084 | 0.063 | 0.0504 | 0.042 | 0.036 | -| θn | 0◦ | −75.96◦ | −82.87◦ | −85.24◦ | −86.42◦ | −87.14◦ | −87.61◦ | −87.95◦ | - -We can use these numerical values to express *x*(*t*) as - -$$ -x(t) = 0.504 + 0.504 \sum_{n=1}^{\infty} \frac{2}{\sqrt{1 + 16n^2}} \cos(2nt - \tan^{-1} 4n) -$$ - -= 0.504 + 0.244 cos (2t - 75.96°) + 0.125 cos (4t - 82.87°) -+ 0.084 cos (6t - 85.24°) + 0.063 cos (8t - 86.42°) + ... (6.11) - -### PLOTTING FOURIER SERIES SPECTRA USING MATLAB - -MATLAB is well suited to compute and plot Fourier series spectra. The results in Fig. 6.3, which plot *Cn* and θ*n* as functions of *n*, match Figs. 6.2b and 6.2c, which plot *Cn* and θ*n* as functions of ω = *n*ω0 = 2*n*. Plots of *an* and *bn* are similarly simple to generate. - -The amplitude and phase spectra for *x*(*t*), in Figs. 6.2b and 6.2c, tell us at a glance the frequency composition of *x*(*t*), that is, the amplitudes and phases of various sinusoidal components of *x*(*t*). Knowing the frequency spectra, we can reconstruct *x*(*t*), as shown on the right-hand side of Eq. (6.11). Therefore the frequency spectra (Figs. 6.2b, 6.2c) provide an alternative description—the frequency-domain description of *x*(*t*). The time-domain description of *x*(*t*) is shown in Fig. 6.2a. *A signal, therefore, has a dual identity: the time-domain identity x*(*t*) *and the frequency-domain identity (Fourier spectra). The two identities complement each other; taken together, they provide a better understanding of a signal.* - -An interesting aspect of Fourier series is that whenever there is a jump discontinuity in *x*(*t*), the series at the point of discontinuity converges to an average of the left-hand and right-hand limits of *x*(*t*) at the instant of discontinuity.† In the present example, for instance, *x*(*t*) is discontinuous at *t* = 0 with *x*(0+) = 1 and *x*(0−) = *x*(π ) = *e*−π/2 = 0.208. The corresponding Fourier series converges to a value (1 + 0.208)/2 = 0.604 at *t* = 0. This is easily verified from Eq. (6.11) by setting *t* = 0. - -† This behavior of the Fourier series is dictated by its convergence in the mean, discussed later in Secs. 6.2 and 6.5. - -### 602 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -### **EXAMPLE 6.2 Compact Trigonometric Fourier Series of a Periodic Triangle Wave** - -Find the compact trigonometric Fourier series for the triangular periodic signal *x*(*t*) shown in Fig. 6.4a, and sketch the amplitude and phase spectra for *x*(*t*). - -**Figure 6.4 (a)** A triangular periodic signal and **(b, c)** its Fourier spectra. - -In this case the period *T*0 = 2. Hence, - -$$ -\omega_0 = \frac{2\pi}{2} = \pi -$$ - -and - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\pi t + b_n \sin n\pi t -$$ - -where - -$$ -x(t) = \begin{cases} 2At & |t| < \frac{1}{2} \\ 2A(1-t) & \frac{1}{2} < t < \frac{3}{2} \end{cases} -$$ - -Here it will be advantageous to choose the interval of integration from −1/2 to 3/2 rather than 0 to 2. - -A glance at Fig. 6.4a shows that the average value (dc) of *x*(*t*) is zero so that *a*0 = 0. Also, - -$$ -a_n = \frac{2}{2} \int_{-1/2}^{3/2} x(t) \cos n\pi t dt -$$ - -= -$$ -\int_{-1/2}^{1/2} 2A t \cos n\pi t dt + \int_{1/2}^{3/2} 2A(1-t) \cos n\pi t dt -$$ - -Detailed evaluation of these integrals shows that both have a value of zero. Therefore *an* = 0. Next, - -$$ -b_n = \int_{-1/2}^{1/2} 2At \sin n\pi t \, dt + \int_{1/2}^{3/2} 2A(1-t) \sin n\pi t \, dt -$$ - -Detailed evaluation of these integrals yields, in turn, - -$$ -b_n = \frac{8A}{n^2 \pi^2} \sin\left(\frac{n\pi}{2}\right) = \begin{cases} 0 & n \text{ even} \\ \frac{8A}{n^2 \pi^2} & n = 1, 5, 9, 13, \dots \\ -\frac{8A}{n^2 \pi^2} & n = 3, 7, 11, 15, \dots \end{cases} -$$ - -Therefore, - -$$ -x(t) = \frac{8A}{\pi^2} \left[ \sin \pi t - \frac{1}{9} \sin 3\pi t + \frac{1}{25} \sin 5\pi t - \frac{1}{49} \sin 7\pi t + \dots \right] -$$ - (6.12) - -To plot Fourier spectra, the series must be converted into compact trigonometric form as in Eq. (6.9). In this case this is readily done by converting sine terms into cosine terms with a suitable phase shift. For example, - -$$ -\sin kt = \cos (kt - 90^\circ) \qquad \text{and} \qquad -\sin kt = \cos (kt + 90^\circ) -$$ - -By using these identities, Eq. (6.12) can be expressed as - -$$ -x(t) = \frac{8A}{\pi^2} \bigg[ \cos(\pi t - 90^\circ) + \frac{1}{9} \cos(3\pi t + 90^\circ) + \frac{1}{25} \cos(5\pi t - 90^\circ) + \frac{1}{49} \cos(7\pi t + 90^\circ) + \cdots \bigg] -$$ - -In this series all the even harmonics are missing. The phases of the odd harmonics alternate from −90◦ to 90◦. Figure 6.4 shows amplitude and phase spectra for *x*(*t*). - -## **EXAMPLE 6.3 Converting a Trigonometric FS to a Compact Trigonometric FS** - -A periodic signal *x*(*t*) is represented by a trigonometric Fourier series - -$$ -x(t) = 2 + 3\cos 2t + 4\sin 2t + 2\sin (3t + 30^\circ) - \cos (7t + 150^\circ) -$$ - -Express this series as a compact trigonometric Fourier series, and sketch amplitude and phase spectra for *x*(*t*). - -In compact trigonometric Fourier series, the sine and cosine terms of the same frequency are combined into a single term and all terms are expressed as cosine terms with positive amplitudes. Using Eqs. (6.9) and (6.10), we have - -$$ -3\cos 2t + 4\sin 2t = 5\cos (2t - 53.13^{\circ}) -$$ - -Also, - -$$ -\sin(3t + 30^{\circ}) = \cos(3t + 30^{\circ} - 90^{\circ}) = \cos(3t - 60^{\circ}) -$$ - -and - -$$ --\cos(7t + 150^{\circ}) = \cos(7t + 150^{\circ} - 180^{\circ}) = \cos(7t - 30^{\circ}) -$$ - -Therefore, - -$$ -x(t) = 2 + 5\cos(2t - 53.13^{\circ}) + 2\cos(3t - 60^{\circ}) + \cos(7t - 30^{\circ}) -$$ - -**Figure 6.5** Fourier spectra of the signal. - -In this case only four components (including dc) are present. The amplitude of dc is 2. The remaining three components are of frequencies ω = 2, 3, and 7 with amplitudes 5, 2, and 1 and phases −53.13◦, −60◦, and −30◦, respectively. The amplitude and phase spectra for this signal are shown in Figs. 6.5a and 6.5b, respectively. - -### **EXAMPLE 6.4 Compact Trigonometric Fourier Series of a Periodic Square Wave** - -Find the compact trigonometric Fourier series for the square-pulse periodic signal shown in Fig. 6.6a and sketch its Fourier spectrum. - -**Figure 6.6 (a)** A square pulse periodic signal and **(b)** its Fourier spectrum. - -Here the period is *T*0 = 2π and ω0 = 2π/*T*0 = 1. Therefore, - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos nt + b_n \sin nt -$$ - -where - -$$ -a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt -$$ - -From Fig. 6.6a, it is clear that a proper choice of region of integration is from −π to π. But since *x*(*t*) = 1 only over (−π/2, π/2), and *x*(*t*) = 0 over the remaining segment, - -$$ -a_0 = \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} dt = \frac{1}{2} -$$ - -We could have found *a*0, the average value of *x*(*t*), to be 1/2 merely by inspection of *x*(*t*) in Fig. 6.6a. Also, - -$$ -a_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos nt \, dt = \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right) -$$ - -= -$$ -\begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n = 1, 5, 9, 13, \dots \\ -\frac{2}{\pi n} & n = 3, 7, 11, 15, \dots \end{cases} -$$ - -$$ -b_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \sin nt \, dt = 0 -$$ - -Therefore - -$$ -x(t) = \frac{1}{2} + \frac{2}{\pi} \left( \cos t - \frac{1}{3} \cos 3t + \frac{1}{5} \cos 5t - \frac{1}{7} \cos 7t + \cdots \right) -$$ -(6.13) - -Observe that *bn* = 0 and all the sine terms are zero. Only the cosine terms appear in the trigonometric series. The series is therefore already in the compact form except that the amplitudes of alternating harmonics are negative. Now by definition, amplitudes *Cn* are positive [see Eq. (6.10)]. The negative sign can be accommodated by associating a proper phase, as seen from the trigonometric identity† - -−cos *x* = cos(*x* −π ) - -Using this fact, we can express the series in Eq. (6.13) as - -$$ -x(t) = \frac{1}{2} + \frac{2}{\pi} \left[ \cos \omega_0 t + \frac{1}{3} \cos (3\omega_0 t - \pi) + \frac{1}{5} \cos 5\omega_0 t + \frac{1}{7} \cos (7\omega_0 t - \pi) + \frac{1}{9} \cos 9\omega_0 t + \cdots \right] -$$ - -† Because cos(*x*±π ) = −cos *x*, we could have chosen the phase π or −π. In fact, cos(*x*±*N*π ) = −cos *x* for any odd integral value of *N*. Therefore the phase can be chosen as ±*N*π, where *N* is any convenient odd integer. - -This is the desired form of the compact trigonometric Fourier series. The amplitudes are - -$$ -C_0 = \frac{1}{2} -$$ - and $C_n = \begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n \text{ odd} \end{cases}$ - -The phases are - -$$ -\theta_n = \begin{cases} 0 & \text{for all } n \neq 3, 7, 11, 15, \dots \\ -\pi & n = 3, 7, 11, 15, \dots \end{cases} -$$ - -We might use these values to plot amplitude and phase spectra. However, we can simplify our task in this special case if we allow amplitude *Cn* to take on negative values. If this is allowed, we do not need a phase of −π to account for the sign as seen from Eq. (6.13). This means that phases of all components are zero, and we can discard the phase spectrum and manage with only the amplitude spectrum, as shown in Fig. 6.6b. Observe that there is no loss of information in doing so and that the amplitude spectrum in Fig. 6.6b has the complete information about the Fourier series in Eq. (6.13). *Therefore, whenever all sine terms vanish (bn* = 0*), it is convenient to allow Cn to take on negative values*. This permits the spectral information to be conveyed by a single spectrum.† - -Let us investigate the behavior of the series at the points of discontinuities. For the discontinuity at *t* =π/2, the values of *x*(*t*) on either sides of the discontinuity are *x*((π/2)−)=1 and *x*((π/2)+) = 0. We can verify by setting *t* = π/2 in Eq. (6.13) that *x*(π/2) = 0.5, which is a value midway between the values of *x*(*t*) on either side of the discontinuity at *t* = π/2. - -### **[6.1-2 The Effect of Symmetry](#page-12-0)** - -The Fourier series for the signal *x*(*t*) in Fig. 6.2a (Ex. 6.1) consists of sine and cosine terms, but the series for the signal *x*(*t*) in Fig. 6.4a (Ex. 6.2) consists of sine terms only, and the series for the signal *x*(*t*) in Fig. 6.6a (Ex. 6.4) consists of cosine terms only. This is no accident. We can show that the Fourier series of any even periodic function *x*(*t*) consists of cosine terms only and the series for any odd periodic function *x*(*t*) consists of sine terms only. Moreover, because of symmetry (even or odd), the information of one period of *x*(*t*) is implicit in only half the period, as seen in Figs. 6.4a and 6.6a. In these cases, knowing the signal over a half-period and knowing the kind of symmetry (even or odd), we can determine the signal waveform over a complete period. For this reason, the Fourier coefficients in these cases can be computed by integrating over only half the period rather than a complete period. To prove this result, recall that - -$$ -a_0 = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \sin n\omega_0 t dt -$$ - -Recall also that cos*n*ω0*t* is an even function and sin*n*ω0*t* is an odd function of *t*. If *x*(*t*) is an even function of *t*, then *x*(*t*) cos*n*ω0*t* is also an even function and *x*(*t*)sin *n*ω0*t* is an odd function of *t* - -† Here, the distinction between amplitude *An* and magnitude *Cn* = |*An*| would have been useful. But, for the reasons mentioned in the footnote on page 598, we refrain from this distinction formally. - -#### 608 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -(see Sec. 1.5-1). Therefore, following from Eq. (1.16), - -$$ -a_0 = \frac{2}{T_0} \int_0^{T_0/2} x(t) dt, \quad a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = 0 \tag{6.14} -$$ - -Similarly, if *x*(*t*) is an odd function of *t*, then *x*(*t*) cos *n*ω0*t* is an odd function of *t* and *x*(*t*)sin *n*ω0*t* is an even function of *t*. Therefore, - -$$ -a_n = 0 -$$ - and $b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t dt$ (6.15) - -Observe that because of symmetry, the integration required to compute the coefficients need be performed over only half the period. - -If a periodic signal *x*(*t*) shifted by half the period remains unchanged except for a sign—that is, if - -$$ -x\left(t - \frac{T_0}{2}\right) = -x(t) -$$ - -then the signal is said to have a *half-wave* symmetry. It can be shown that for a signal with a half-wave symmetry, all the even-numbered harmonics vanish (see Prob. 6.1-6). The signal in Fig. 6.4a is an example of such a symmetry. The signal in Fig. 6.6a also has this symmetry, although it is not obvious owing to a dc component. If we subtract the dc component of 0.5 from this signal, the remaining signal has half-wave symmetry. For this reason, this signal has only odd harmonics and a dc component of 0.5. - -### **DR ILL 6.1 Compact Trigonometric Fourier Series** - -Find the compact trigonometric Fourier series for periodic signals shown in Fig. 6.7. Sketch their amplitude and phase spectra. Allow *Cn* to take on negative values if *bn* = 0 so that the phase spectrum can be eliminated. [*Hint:* Use Eqs. (6.14) and (6.15) for appropriate symmetry conditions.] - -### **ANSWERS** - -(a) -$$ -x(t) = \frac{1}{3} - \frac{4}{\pi^2} \left( \cos \pi t - \frac{1}{4} \cos 2\pi t + \frac{1}{9} \cos 3\pi t - \frac{1}{16} \cos 4\pi t + \cdots \right) -$$ - -\n -$$ -= \frac{1}{3} + \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos n\pi t -$$ -\n(b) $x(t) = \frac{2A}{\pi} \left[ \sin \pi t - \frac{1}{2} \sin 2\pi t + \frac{1}{3} \sin 3\pi t - \frac{1}{4} \sin 4\pi t + \cdots \right]$ -\n -$$ -= \frac{2A}{\pi} \left[ \cos (\pi t - 90^\circ) + \frac{1}{2} \cos (2\pi t + 90^\circ) + \frac{1}{3} \cos (3\pi t - 90^\circ) + \frac{1}{4} \cos (4\pi t + 90^\circ) + \cdots \right] -$$ - - - -### **[6.1-3 Determining the Fundamental Frequency and Period](#page-12-0)** - -We have seen that every periodic signal can be expressed as a sum of sinusoids of a fundamental frequency ω0 and its harmonics. One may ask whether a sum of sinusoids of *any* frequencies represents a periodic signal. If so, how does one determine the period? Consider the following three functions: - -$$ -x_1(t) = 2 + 7\cos(\frac{1}{2}t + \theta_1) + 3\cos(\frac{2}{3}t + \theta_2) + 5\cos(\frac{7}{6}t + \theta_3) -$$ - -\n -$$ -x_2(t) = 2\cos(2t + \theta_1) + 5\sin(\pi t + \theta_2) -$$ - -\n -$$ -x_3(t) = 3\sin(3\sqrt{2}t + \theta) + 7\cos(6\sqrt{2}t + \phi) -$$ - -Recall that every frequency in a periodic signal is an integer multiple of the fundamental frequency ω0. Therefore the ratio of any two frequencies is of the form *m*/*n*, where *m* and *n* are integers. This means that the ratio of any two frequencies is a rational number. When the ratio of two frequencies is a rational number, the frequencies are said to be *harmonically* related. - -The largest number of which all the frequencies are integer multiples is the fundamental frequency. In other words, the fundamental frequency is the *greatest common factor* (GCF) of all the frequencies in the series. The frequencies in the spectrum of *x*1(*t*) are 1/2, 2/3, and 7/6 (we do not consider dc). The ratios of the successive frequencies are 3:4 and 4:7, respectively. Because both these numbers are rational, all the three frequencies in the spectrum are harmonically related, and the signal *x*1(*t*) is periodic. The GCF, that is, the greatest number of which 1/2, 2/3, and 7/6 are integer multiples, is 1/6.† Moreover, 3(1/6) = 1/2, 4(1/6) = 2/3, and 7(1/6) = 7/6. Therefore the fundamental frequency is 1/6, and the three frequencies in the spectrum are the third, fourth, and seventh harmonics. Observe that the fundamental frequency component is absent in this Fourier series. - -† The greatest common factor of *a*1/*b*1, *a*2/*b*2, ..., *am*/*bm* is the ratio of the GCF of the numerators set (*a*1,*a*2,...,*am*) to the LCM (least common multiple) of the denominator set (*b*1,*b*2,...,*bm*). For instance, for the set (2/3, 6/7, 2), the GCF of the numerator set (2, 6, 2) is 2; the LCM of the denominator set (3, 7, 1) is 21. Therefore, 2/21 is the largest number of which 2/3, 6/7, and 2 are integer multiples. - -#### 610 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -The signal *x*2(*t*) is not periodic because the ratio of two frequencies in the spectrum is 2/π, which is not a rational number. The signal *x*3(*t*) is periodic because the ratio of frequencies 3 √ 2 and 6√ 2 is 1/2, a rational number. The greatest common factor of 3√2 and 6√2 is 3√2. Therefore, the fundamental frequency ω0 = 3 √2, and the period - -$$ -T_0 = \frac{2\pi}{(3\sqrt{2})} = \frac{\sqrt{2}}{3}\pi -$$ - -### **DR ILL 6.2 Determining Periodicity, Fundamental Frequency, and Harmonic Content** - -Determine whether the signal - -$$ -x(t) = \cos\left(\frac{2}{3}t + 30^{\circ}\right) + \sin\left(\frac{4}{5}t + 45^{\circ}\right) -$$ - -is periodic. If it is periodic, find the fundamental frequency and the period. What harmonics are present in *x*(*t*)? - -### **ANSWERS** - -Periodic with ω0 =2/15 and period *T*0 =15π. Signal *x*(*t*) contains the fifth and sixth harmonics. - -## A HISTORICAL NOTE: BARON JEAN-BAPTISTE-JOSEPH FOURIER (1768–1830) - -The Fourier series and integral comprise a most beautiful and fruitful development, which serves as an indispensable instrument in the treatment of many problems in mathematics, science, and engineering. Maxwell was so taken by the beauty of the Fourier series that he called it a great mathematical poem. In electrical engineering, it is central to the areas of communication, signal processing, and several other fields, including antennas, but its initial reception by the scientific world was not enthusiastic. In fact, Fourier could not get his results published as a paper. - -Fourier, a tailor's son, was orphaned at age 8 and educated at a local military college (run by Benedictine monks), where he excelled in mathematics. The Benedictines prevailed upon the young genius to choose the priesthood as his vocation, but revolution broke out before he could take his vows. Fourier joined the people's party. But in its early days, the French Revolution, like most such upheavals, liquidated a large segment of the intelligentsia, including prominent scientists such as Lavoisier. Observing this trend, many intellectuals decided to leave France to save themselves from a rapidly rising tide of barbarism. Fourier, despite his early enthusiasm for the Revolution, narrowly escaped the guillotine twice. It was to the everlasting credit of Napoleon that he stopped the persecution of the intelligentsia and founded new schools to replenish their ranks. The 26-year-old Fourier was appointed chair of mathematics at the newly created École Normale in 1794 [1]. - -Jean-Baptiste-Joseph Fourier and Napoleon - -Napoleon was the first modern ruler with a scientific education, and he was one of the rare persons who are equally comfortable with soldiers and scientists. The age of Napoleon was one of the most fruitful in the history of science. Napoleon liked to sign himself as "member of *Institut de France*" (a fraternity of scientists), and he once expressed to Laplace his regret that "force of circumstances has led me so far from the career of a scientist" [2]. Many great figures in science and mathematics, including Fourier and Laplace, were honored and promoted by Napoleon. In 1798 he took a group of scientists, artists, and scholars—Fourier among them—on his Egyptian expedition, with the promise of an exciting and historic union of adventure and research. Fourier proved to be a capable administrator of the newly formed Institut d'Égypte, which, incidentally, was responsible for the discovery of the Rosetta Stone. The inscription on this stone in two languages and three scripts (hieroglyphic, demotic, and Greek) enabled Thomas Young and Jean-François Champollion, a protégé of Fourier, to invent a method of translating hieroglyphic writings of ancient Egypt—the only significant result of Napoleon's Egyptian expedition. - -Back in France in 1801, Fourier briefly served in his former position as professor of mathematics at the École Polytechnique in Paris. In 1802 Napoleon appointed him the prefect of Isère (with its headquarters in Grenoble), a position in which Fourier served with distinction. Fourier was named Baron of the Empire by Napoleon in 1809. Later, when Napoleon was exiled to Elba, his route was to take him through Grenoble. Fourier had the route changed to avoid meeting Napoleon, which would have displeased Fourier's new master, King Louis XVIII. Within a year, Napoleon escaped from Elba and returned to France. At Grenoble, Fourier was brought before him in chains. Napoleon scolded Fourier for his ungrateful behavior but reappointed him the prefect of Rhône at Lyons. Within four months Napoleon was defeated at Waterloo and was exiled to St. Helena, where he died in 1821. Fourier once again was in disgrace as a Bonapartist and had to pawn his possessions to keep himself alive. But through the intercession of a former student, who was now a prefect of Paris, he was appointed director of the statistical bureau of the Seine, a position that allowed him ample time for scholarly pursuits. Later, in 1827, he was elected to the powerful position of perpetual secretary of the Paris Academy of Science, a section of the Institut de France [3]. - -While serving as the prefect of Grenoble, Fourier carried on his elaborate investigation of the propagation of heat in solid bodies, which led him to the Fourier series and the Fourier integral. On December 21, 1807, he announced these results in a prize paper on the theory of heat. Fourier claimed that an arbitrary function (continuous or with discontinuities) defined in a finite interval by an arbitrarily capricious graph can always be expressed as a sum of sinusoids (Fourier series). The judges, who included the great French mathematicians Laplace, Lagrange, Legendre, Monge, and LaCroix, admitted the novelty and importance of Fourier's work but criticized it for lack of mathematical rigor and generality. Lagrange thought it incredible that a sum of sines and cosines could add up to anything but an infinitely differentiable function. Moreover, one of the properties of an infinitely differentiable function is that if we know its behavior over an arbitrarily small interval, we can determine its behavior over the entire range (the Taylor–Maclaurin series). Such a function is far from an arbitrary or a capriciously drawn graph [4]. Laplace had additional reason to criticize Fourier's work. Laplace and his students had already approached the problem of heat conduction from a different angle, and Laplace was reluctant to accept the superiority of Fourier's method [5]. Fourier thought the criticism unjustified but was unable to prove his claim because the tools required for operations with infinite series were not available at the time. However, posterity has proved Fourier to be closer to the truth than his critics. This is the classic conflict between pure mathematicians and physicists or engineers, as we saw earlier (Ch. 4) in the life of Oliver Heaviside. In 1829 Dirichlet proved Fourier's claim concerning capriciously drawn functions with a few restrictions (Dirichlet conditions). - -Although three of the four judges were in favor of publication, Fourier's paper was rejected because of vehement opposition by Lagrange. Fifteen years later, after several attempts and disappointments, Fourier published the results in expanded form as a text, *Théorie analytique de la chaleur,* which is now a classic. - -## **[6.2 EXISTENCE AND](#page-12-0) CONVERGENCE OF THE FOURIER SERIES** - -For the existence of the Fourier series, coefficients *a*0,*an*, and *bn* in Eq. (6.8) must be finite. It follows from Eq. (6.8) that the existence of these coefficients is guaranteed if *x*(*t*) is absolutely integrable over one period; that is, - -$$ -\int_{T_0} |x(t)| \, dt < \infty \tag{6.16} -$$ - -However, existence, by itself, does not inform us about the nature and the manner in which the series converges. We shall first discuss the notion of convergence. - -### **[6.2-1 Convergence of a Series](#page-12-0)** - -The key to many puzzles lies in the nature of the convergence of the Fourier series. Convergence of infinite series is a complex problem. It took mathematicians several decades to understand the convergence aspect of the Fourier series. We shall barely scratch the surface here. - -Nothing annoys a student more than the discussion of convergence. "Have we not proved," they ask, "that a periodic signal *x*(*t*) can be expressed as a Fourier series"? Then why spoil the fun by this annoying discussion? All we have shown so far is that a signal represented by a Fourier series in Eq. (6.1) is periodic. We have not proved the converse, that every periodic signal can be expressed as a Fourier series. This issue will be tackled later, in Sec. 6.5-4, where it will be shown that a periodic signal can be represented by a Fourier series, as in Eq. (6.1), where the equality of the two sides of the equation is not in the ordinary sense, but in the mean-square sense (explained later in this discussion). But the astute reader should have been skeptical of the claims of the Fourier series to represent discontinuous functions in Figs. 6.2a and 6.6a. If *x*(*t*) has a jump discontinuity, say, at *t* = 0, then *x*(0+), *x*(0), and *x*(0−) are generally different. How could a series consisting of the sum of continuous functions of the smoothest type (sinusoids) add to one value at *t* = 0− and a different value at *t* = 0 and yet another value at *t* = 0+? The demand is impossible to satisfy unless the math involved executes some spectacular acrobatics. How does a Fourier series act under such conditions? Precisely for this reason, the great mathematicians Lagrange and Laplace, two of the judges examining Fourier's paper, were skeptical of Fourier's claims and voted against publication of the paper that later became a classic. - -There are also other issues. In any practical application, we can use only a finite number of terms in a series. If, with a fixed number of terms, the series guarantees convergence within an arbitrarily small error at every value of *t*, such a series is highly desirable and is called a *uniformly convergent* series. If a series converges at every value of *t*, but to guarantee convergence within a given error requires a different number of terms at different *t*, then the series is still convergent, but less desirable. It goes under the name *pointwise convergent* series. - -Finally, we have the case of a series that refuses to converge at some *t*, no matter how many terms are added. But the series may *converge in the mean*; that is, the energy of the difference between *x*(*t*) and the corresponding finite term series approaches zero as the number of terms approaches infinity.† To explain this concept, let us consider representation of a function *x*(*t*) by an infinite series - -$$ -x(t) = \sum_{n=1}^{\infty} z_n(t) -$$ - -Let the partial sum of the first *N* terms of the series on the right-hand side be denoted by *xN*(*t*), that is, - -$$ -x_N(t) = \sum_{n=1}^N z_n(t) -$$ - -† The behavior is called "convergence in the mean" because minimizing the error energy over a certain interval is equivalent to minimizing the mean-square value of the error over the same interval. - -#### 614 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -If we approximate *x*(*t*) by *xN*(*t*) (the partial sum of the first *N* terms of the series), the *error* in the approximation is the difference *x*(*t*) − *xN*(*t*). The series converges *in the mean* to *x*(*t*) in the interval (0, *T*0) if - -$$ -\int_0^{T_0} |x(t) - x_N(t)|^2 dt \to 0 \quad \text{as} \quad N \to \infty -$$ - -Hence, the energy of the error *x*(*t*)−*xN*(*t*) approaches zero as *N* → ∞. This form of convergence does not require the series to be equal to *x*(*t*) for all *t*. It just requires the energy of the difference (area under |*x*(*t*)−*xN*(*t*)| 2) to vanish as *N* → ∞. Superficially it may appear that if the energy of a signal over an interval is zero, the signal (the error) must be zero everywhere. This is not true. The signal energy can be zero even if there are nonzero values at a finite number of isolated points. This is because although the signal is nonzero at a point (and zero everywhere else), the area under its square is still zero. Thus, a series that converges in the mean to *x*(*t*) need not converge to *x*(*t*) at a finite number of points. This is precisely what happens to the Fourier series when *x*(*t*) has jump discontinuities. This is also what makes Fourier series convergence compatible with the Gibbs phenomenon, to be discussed later in this section. - -There is a simple criterion for ensuring that a periodic signal *x*(*t*) has a Fourier series that converges in the mean. The Fourier series for *x*(*t*) converges to *x*(*t*) in the mean if *x*(*t*) has a finite energy over one period, that is, - -$$ -\int_{T_0} |x(t)|^2 dt < \infty \tag{6.17} -$$ - -Thus, the periodic signal *x*(*t*), having a finite energy over one period, guarantees the convergence in the mean of its Fourier series. In all the examples discussed so far, Eq. (6.17) is satisfied; hence the corresponding Fourier series converges in the mean. Equation (6.17), like Eq. (6.16), guarantees that the Fourier coefficients are finite. - -We shall now discuss an alternate set of criteria, due to Dirichlet, for convergence of the Fourier series. - -### DIRICHLET CONDITIONS - -Dirichlet showed that if *x*(*t*) satisfies certain conditions (*Dirichlet conditions*), its Fourier series is guaranteed to converge pointwise at all points where *x*(*t*) is continuous. Moreover, at the points of discontinuities, *x*(*t*) converges to the value midway between the two values of *x*(*t*) on either side of the discontinuity. These conditions are: - -- 1. The function *x*(*t*) must be absolutely integrable; that is, it must satisfy Eq. (6.16). -- 2. The function *x*(*t*) must have only a finite number of finite discontinuities in one period. -- 3. The function *x*(*t*) must contain only a finite number of maxima and minima in one period. - -All practical signals, including those in Exs. 6.1, 6.2, 6.3, and 6.4, satisfy these conditions. - -### **[6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping](#page-12-0)** - -The trigonometric Fourier series of a signal *x*(*t*) shows explicitly the sinusoidal components of *x*(*t*). We can synthesize *x*(*t*) by adding the sinusoids in the spectrum of *x*(*t*). Let us synthesize the square-pulse periodic signal *x*(*t*) of Fig. 6.6a by adding successive harmonics in its spectrum step by step and observing the similarity of the resulting signal to *x*(*t*). The Fourier series for this function as found in Eq. (6.13) is - -$$ -x(t) = \frac{1}{2} + \frac{2}{\pi} \left( \cos t - \frac{1}{3} \cos 3t + \frac{1}{5} \cos 5t - \frac{1}{7} \cos 7t + \cdots \right) -$$ - -We start the synthesis with only the first term in the series (*n* = 0), a constant 1/2 (dc); this is a gross approximation of the square wave, as shown in Fig. 6.8a. In the next step we add the dc (*n* = 0) and the first harmonic (fundamental), which results in a signal shown in Fig. 6.8b. Observe that the synthesized signal somewhat resembles *x*(*t*). It is a smoothed-out version of *x*(*t*). The sharp corners in *x*(*t*) are not reproduced in this signal because sharp corners mean rapid changes, and their reproduction requires rapidly varying (i.e., higher-frequency) components, which are excluded. Figure 6.8c shows the sum of dc, first, and third harmonics (even harmonics are absent). As we increase the number of harmonics progressively, as shown in Figs. 6.8d (sum up to the fifth harmonic) and 6.8e (sum up to the nineteenth harmonic), the edges of the pulses become sharper and the signal resembles *x*(*t*) more closely. - -### ASYMPTOTIC RATE OF AMPLITUDE SPECTRUM DECAY - -Figure 6.8 brings out one interesting aspect of the Fourier series. Lower frequencies in the Fourier series affect the large-scale behavior of *x*(*t*), whereas the higher frequencies determine the fine structure such as rapid wiggling. Hence, sharp changes in *x*(*t*), being a part of fine structure, necessitate higher frequencies in the Fourier series. The sharper the change [the higher the time derivative *x*˙(*t*)], the higher are the frequencies needed in the series. - -The amplitude spectrum indicates the amounts (amplitudes) of various frequency components of *x*(*t*). If *x*(*t*) is a smooth function, its variations are less rapid. Synthesis of such a function requires predominantly lower-frequency sinusoids and relatively small amounts of rapidly varying (higher-frequency) sinusoids. The amplitude spectrum of such a function would decay swiftly with frequency. To synthesize such a function, we require fewer terms in the Fourier series for a good approximation. On the other hand, a signal with sharp changes, such as jump discontinuities, contains rapid variations, and its synthesis requires a relatively large amount of high-frequency components. The amplitude spectrum of such a signal would decay slowly with frequency, and to synthesize such a function, we require many terms in its Fourier series for a good approximation. The square wave *x*(*t*) is a discontinuous function with jump discontinuities, and therefore its amplitude spectrum decays rather slowly, as 1/*n* [see Eq. (6.13)]. On the other hand, the triangular-pulse periodic signal in Fig. 6.4a is smoother because it is a continuous function (no jump discontinuities). Its spectrum decays rapidly with frequency as 1/*n*2 [see Eq. (6.12)]. - -We can show that if the first *k* − 1 derivatives of a periodic signal *x*(*t*) are continuous and the *k*th derivative is discontinuous, then its amplitude spectrum *Cn* decays with frequency at least as rapidly as 1/*nk*+1 [6]. This result provides a simple and useful means for predicting the asymptotic - -**Figure 6.8** Synthesis of a square-pulse periodic signal by successive addition of its harmonics. - -rate of convergence of the Fourier series. In the case of the square-wave signal (Fig. 6.6a), the zeroth derivative of the signal (the signal itself) is discontinuous so that *k* = 0. For the triangular periodic signal in Fig. 6.4a, the first derivative is discontinuous; that is, *k* = 1. For this reason, the spectra of these signals decay as 1/*n* and 1/*n*2, respectively. - -## **EXAMPLE 6.5 Square-Wave Synthesis by Truncated Fourier Series Using MATLAB** - -Use MATLAB to synthesize and plot the square wave of Fig. 6.8a using a Fourier series that is truncated to the 19th harmonic. The result should match Fig. 6.8e. - -To synthesize the waveform, we use the Fourier series of Eq. (6.13). - -``` ->> x = @(t) 1.0*(mod(t+pi/2,2*pi)<=pi); ->> t = linspace(-2*pi,2*pi,10001); ->> x19 = 0.5*ones(size(t)); ->> for n=1:19, x19 = x19+2/(pi*n)*sin(pi*n/2)*cos(n*t); end ->> plot(t,x19,'k-'); axis([-2*pi 2*pi -0.2 1.2]); ->> xlabel('t'); ylabel('x_{19}(t)'); -``` - -As expected, the result of Fig. 6.9 matches Fig. 6.8e. - -### PHASE SPECTRUM: THE WOMAN BEHIND A SUCCESSFUL MAN - -The role of the amplitude spectrum in shaping the waveform *x*(*t*) is quite clear. However, the role of the phase spectrum in shaping this waveform is less obvious. Yet, the phase spectrum, like the woman behind a successful man,† plays an equally important role in waveshaping. We can explain this role by considering a signal *x*(*t*) that has rapid changes, such as jump discontinuities. To synthesize an instantaneous change at a jump discontinuity, the phases of the various sinusoidal components in its spectrum must be such that all (or most) of the harmonic components will have - -† Or, to keep up with the times, the man behind a successful woman. - -#### 618 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -one sign before the discontinuity and the opposite sign after the discontinuity. This will result in a sharp change in *x*(*t*) at the point of discontinuity. We can verify this fact in any waveform with jump discontinuity. Consider, for example, the sawtooth waveform in Fig. 6.7b. This waveform has a discontinuity at *t* = 1. The Fourier series for this waveform, as given in Drill 6.1b, is - -$$ -x(t) = \frac{2A}{\pi} \left[ \cos(\pi t - 90^\circ) + \frac{1}{2} \cos(2\pi t + 90^\circ) + \frac{1}{3} \cos(3\pi t - 90^\circ) + \frac{1}{4} \cos(4\pi t + 90^\circ) + \cdots \right] -$$ - -Figure 6.10 shows the first three components of this series. The phases of all the (infinite) components are such that all the components are positive just before *t* = 1 and turn negative just after *t* = 1, the point of discontinuity. The same behavior is also observed at *t* = −1, where a similar discontinuity occurs. This sign change in all the harmonics adds up to produce very nearly a jump discontinuity. The role of the phase spectrum is crucial in achieving a sharp change in the waveform. If we ignore the phase spectrum when trying to reconstruct this signal, the result will be a smeared and spread-out waveform. In general, the phase spectrum is just as crucial as the amplitude spectrum in determining the waveform. *The synthesis of any signal x*(*t*) *is achieved by using a proper combination of amplitudes and phases of various sinusoids. This unique combination is the Fourier spectrum of x*(*t*)*.* - -**Figure 6.10** Role of the phase spectrum in shaping a periodic signal. - -### FOURIER SYNTHESIS OF DISCONTINUOUS FUNCTIONS: THE GIBBS PHENOMENON - -Figure 6.8 showed the square function *x*(*t*) and its approximation by a truncated trigonometric Fourier series that includes only the first *N* harmonics for *N* = 1, 3, 5, and 19. The plot of the truncated series approximates closely the function *x*(*t*) as *N* increases, and we expect that the series will converge exactly to *x*(*t*) as *N* → ∞. Yet the curious fact, as seen from Fig. 6.8, is that even for large *N*, the truncated series exhibits an oscillatory behavior and an overshoot approaching a value of about 9% in the vicinity of the discontinuity at the nearest peak of oscillation.† Regardless of the value of *N*, the overshoot remains at about 9%. Such strange behavior certainly would undermine anyone's faith in the Fourier series. In fact, this behavior puzzled many scholars at the turn of the century. Josiah Willard Gibbs, an eminent mathematical physicist who was the inventor of vector analysis, gave a mathematical explanation of this behavior (now called the *Gibbs phenomenon*). - -We can reconcile the apparent aberration in the behavior of the Fourier series by observing from Fig. 6.8 that the frequency of oscillation of the synthesized signal is *Nf*0, so the width of the spike with 9% overshoot is approximately 1/2*Nf*0. As we increase *N*, the frequency of oscillation increases and the spike width 1/2*Nf*0 diminishes. As *N* → ∞, the error power → 0 because the error consists mostly of the spikes, whose widths → 0. Therefore, as *N* → ∞, the corresponding Fourier series differs from *x*(*t*) by about 9% at the immediate left and right of the points of discontinuity, and yet the error power →0. The reason for all this confusion is that in this case, the Fourier series converges in the mean. When this happens, all we promise is that the error energy (over one period) → 0 as *N* → ∞. Thus, the series may differ from *x*(*t*) at some points and yet have the error signal power zero, as verified earlier. Note that the series, in this case, also converges pointwise at all points except the points of discontinuity. It is precisely at the discontinuities that the series differs from *x*(*t*) by 9%.‡ - -When we use only the first *N* terms in the Fourier series to synthesize a signal, we are abruptly terminating the series, giving a unit weight to the first *N* harmonics and zero weight to all the remaining harmonics beyond *N*. This abrupt termination of the series causes the Gibbs phenomenon in synthesis of discontinuous functions. Section 7.8 offers more discussion on the Gibbs phenomenon, its ramifications, and cure. - -The Gibbs phenomenon is present only when there is a jump discontinuity in *x*(*t*). When a continuous function *x*(*t*) is synthesized by using the first *N* terms of the Fourier series, the synthesized function approaches *x*(*t*) for all *t* as *N* → ∞. No Gibbs phenomenon appears. This can be seen in Fig. 6.11, which shows one cycle of a continuous periodic signal being synthesized from the first 19 harmonics. Compare the similar situation for a discontinuous signal in Fig. 6.8. - -### **DR ILL 6.3 Rate of Spectral Decay** - -By inspection of signals in Figs. 6.2a, 6.7a, and 6.7b, determine the asymptotic rate of decay of their amplitude spectra. - -† There is also an undershoot of 9% at the other side [at *t* = (π/2)+] of the discontinuity. - -‡ Actually, at discontinuities, the series converges to a value midway between the values on either side of the discontinuity. The 9% overshoot occurs at *t* = (π/2)− and 9% undershoot occurs at *t* = (π/2)+. - -**Figure 6.11** Fourier synthesis of a continuous signal using first 19 harmonics. - -### **ANSWERS** - -1/*n*, 1/*n*2, and 1/*n*, respectively. - -### A HISTORICAL NOTE ON THE GIBBS PHENOMENON - -Normally speaking, troublesome functions with strange behavior are invented by mathematicians; we rarely see such oddities in practice. In the case of the Gibbs phenomenon, however, the tables were turned. A rather puzzling behavior was observed in a mundane object, a mechanical wave synthesizer, and then well-known mathematicians of the day were dispatched on the scent of it to discover its hideout. - -Albert Michelson (of Michelson–Morley fame) was an intense, practical man who developed ingenious physical instruments of extraordinary precision, mostly in the field of optics. His harmonic analyzer, developed in 1898, could compute the first 80 coefficients of the Fourier series of a signal *x*(*t*) specified by any graphical description. The instrument could also be used as a harmonic synthesizer, which could plot a function *x*(*t*) generated by summing the first 80 harmonics (Fourier components) of arbitrary amplitudes and phases. This analyzer, therefore, had the ability of self-checking its operation by analyzing a signal *x*(*t*) and then adding the resulting 80 components to see whether the sum yielded a close approximation of *x*(*t*). - -Michelson found that the instrument checked very well with most of signals analyzed. However, when he tried a discontinuous function, such as a square wave,† a curious behavior was observed. The sum of 80 components showed oscillatory behavior (ringing), with an overshoot of 9% in the vicinity of the points of discontinuity. Moreover, this behavior was a constant feature regardless of the number of terms added. A larger number of terms made the oscillations proportionately faster, but regardless of the number of terms added, the overshoot remained 9%. This puzzling behavior caused Michelson to suspect some mechanical defect in his synthesizer. He wrote about his observation in a letter to *Nature* (December 1898). Josiah Willard Gibbs, who was a professor at Yale, investigated and clarified this behavior for a sawtooth periodic signal in a letter to *Nature* [7]. Later, in 1906, Bôcher generalized the result for any function with discontinuity [8]. - -† Actually, it was a periodic sawtooth signal. - -Albert Michelson and Josiah Willard Gibbs - -It was Bôcher who gave the name *Gibbs phenomenon* to this behavior. Gibbs showed that the peculiar behavior in the synthesis of a square wave was inherent in the behavior of the Fourier series because of nonuniform convergence at the points of discontinuity. - -This, however, is not the end of the story. Both Bôcher and Gibbs were under the impression that this property had remained undiscovered until Gibbs's work published in 1899. It is now known that what is called the Gibbs phenomenon had been observed in 1848 by Wilbraham of Trinity College, Cambridge, who clearly saw the behavior of the sum of the Fourier series components in the periodic sawtooth signal later investigated by Gibbs [9]. Apparently, this work was not known to most people, including Gibbs and Bôcher. - -## **[6.3 EXPONENTIAL](#page-12-0) FOURIER SERIES** - -By using Euler's equality, we can express cos *n*ω0*t* and sin *n*ω0*t* in terms of exponentials *ejn*ω0*t* and *e*−*jn*ω0*t* . Clearly, we should be able to express the trigonometric Fourier series in Eq. (6.7) in terms of exponentials of the form *ejn*ω0*t* with the index *n* taking on all integer values from −∞ to ∞, including zero. Derivation of the exponential Fourier series from the results already derived for the trigonometric Fourier series is straightforward, involving conversion of sinusoids to exponentials. We shall, however, derive them here independently, without using the prior results of the trigonometric series. - -This discussion shows that the *exponential Fourier series* for a periodic signal *x*(*t*) can be expressed as - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} -$$ - -#### 622 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -To derive the coefficients *Dn*, we multiply both sides of this equation by *e*−*jm*ω0*t* (*m* integer) and integrate over one period. This yields - -$$ -\int_{T_0} x(t) e^{-jm\omega_0 t} dt = \sum_{n=-\infty}^{\infty} D_n \int_{T_0} e^{j(n-m)\omega_0 t} dt -$$ - -To simplify this expression, we use the *orthogonality* property of exponentials, which states that† - -$$ -\int_{T_0} e^{jn\omega_0 t} e^{-jm\omega_0 t} dt = \begin{cases} 0 & m \neq n \\ T_0 & m = n \end{cases} -$$ -\n(6.18) - -Thus, - -$$ -\int_{T_0} x(t) e^{-jm\omega_0 t} dt = D_m T_0 -$$ - -from which we obtain - -$$ -D_m = \frac{1}{T_0} \int_{T_0} x(t) e^{-jm\omega_0 t} dt. -$$ - -To summarize, the exponential Fourier series can be expressed as - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \text{where} \qquad D_n = \frac{1}{T_0} \int_{T_0} x(t) e^{-jn\omega_0 t} dt \tag{6.19} -$$ - -Observe the compactness of Eq. (6.19) and compare it with the trigonometric Fourier series expression. Such a comparison demonstrates very clearly the principal virtue of the exponential Fourier series. First, the form of the series is most compact. Second, the mathematical expression for deriving the coefficients of the series is also compact. It is much more convenient to handle the exponential series than the trigonometric one. For these reasons we shall use the exponential (rather than trigonometric) representation of signals in the rest of the book. - -We can now relate *Dn* to trigonometric series coefficients *an* and *bn*. Setting *n*=0 in Eq. (6.19), we obtain - -$$ -D_0=a_0 -$$ - -Moreover, for *n* = 0, - -$$ -D_n = \frac{1}{T_0} \int_{T_0} x(t) \cos n\omega_0 t \, dt - \frac{j}{T_0} \int_{T_0} x(t) \sin n\omega_0 t \, dt = \frac{1}{2} (a_n - jb_n) \tag{6.20} -$$ - -$$ -\int_{T_0} e^{j(n-m)\omega_0 t} dt = \int_{T_0} \cos{(n-m)\omega_0 t} dt + j \int_{T_0} \sin{(n-m)\omega_0 t} dt -$$ - -Both the integrals on the right-hand side represent area under *n* − *m* number of cycles. Because *n* − *m* is an integer, both the areas are zero. Hence, Eq. (6.18) follows. - -† We can readily prove this property as follows. For the case of *m* = *n*, the integrand in Eq. (6.18) is unity and the integral is *T*0. When *m* = *n*, the integral on the left-hand side of Eq. (6.18) can be expressed as - -and - -$$ -D_{-n} = \frac{1}{T_0} \int_{T_0} x(t) \cos n\omega_0 t \, dt + \frac{j}{T_0} \int_{T_0} x(t) \sin n\omega_0 t \, dt = \frac{1}{2} (a_n + jb_n) \tag{6.21} -$$ - -These results are valid for general *x*(*t*), real or complex. When *x*(*t*) is real, *an* and *bn* are real, and Eqs. (6.20) and (6.21) show that *Dn* and *D*−*n* are conjugates. - -$$ -D_{-n}=D_n^* -$$ - -Moreover, from Eq. (6.10), we observe that - -$$ -a_n - jb_n = \sqrt{a_n^2 + b_n^2} \, e^{j \tan^{-1} \left( \frac{-b_n}{a_n} \right)} = C_n e^{j \theta_n} -$$ - -Hence, - -$$ -D_0=a_0=C_0 -$$ - -and - -$$ -D_n = \frac{1}{2} C_n e^{j\theta_n} \qquad D_{-n} = \frac{1}{2} C_n e^{-j\theta_n} -$$ - -Therefore, for *n* = 0, - -$$ -|D_n| = |D_{-n}| = \frac{1}{2}C_n, \quad \angle D_n = \theta_n, \quad \text{and} \angle D_{-n} = -\theta_n \tag{6.22} -$$ - -Note that |*Dn*| are the amplitudes and *Dn* are the angles of various exponential components. From Eq. (6.22) it follows that when *x*(*t*) is real, the amplitude spectrum (|*Dn*| versus ω) is an even function of ω and the angle spectrum ( *Dn* versus ω) is an odd function of ω. For complex *x*(*t*), *Dn* and *D*−*n* are generally not conjugates. - -### **EXAMPLE 6.6 Exponential Fourier Series of Periodic Exponential Wave** - -Find the exponential Fourier series for the signal of Fig. 6.2a from Ex. 6.1. - -In this case *T*0 = π, ω0 = 2π/*T*0 = 2, and - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{j2nt} -$$ - -where - -$$ -D_n = \frac{1}{T_0} \int_{T_0} x(t) e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-(1/2 + j2n)t} dt -$$ - -= $\frac{-1}{\pi (\frac{1}{2} + j2n)} e^{-(1/2 + j2n)t} \Big|_0^{\pi} = \frac{0.504}{1 + j4n}$ - -$$ -x(t) = 0.504 \sum_{n=-\infty}^{\infty} \frac{1}{1+j4n} e^{j2nt} -$$ - -= 0.504 $\left[ 1 + \frac{1}{1+j4} e^{j2t} + \frac{1}{1+j8} e^{j4t} + \frac{1}{1+j12} e^{j6t} + \cdots + \frac{1}{1-j4} e^{-j2t} + \frac{1}{1-j8} e^{-j4t} + \frac{1}{1-j12} e^{-j6t} + \cdots \right]$ - -Observe that the coefficients *Dn* are complex. Moreover, *Dn* and *D*−*n* are conjugates, as expected. - -### **[6.3-1 Exponential Fourier Spectra](#page-12-0)** - -In exponential spectra, we plot coefficients *Dn* as a function of ω. But since *Dn* is complex in general, we need both parts of one of two sets of plots: the real and the imaginary parts of *Dn*, or the magnitude and the angle of *Dn*. We prefer the latter because of its close connection to the amplitudes and phases of corresponding components of the trigonometric Fourier series. We therefore plot |*Dn*| versus ω and *Dn* versus ω. This requires that the coefficients *Dn* be expressed in polar form as |*Dn*|*ej Dn* , where |*Dn*| are the amplitudes and *Dn* are the angles of various exponential components. Equation (6.22) shows that for real *x*(*t*), the amplitude spectrum (|*Dn*| versus ω) is an even function of ω and the angle spectrum ( *Dn* versus ω) is an odd function of ω. - -For the series in Ex. 6.6, for instance, - -$$ -D_0 = 0.504 -$$ - -\n -$$ -D_1 = \frac{0.504}{1+j4} = 0.122e^{-j75.96^\circ} \implies |D_1| = 0.122, \angle D_1 = -75.96^\circ -$$ - -\n -$$ -D_{-1} = \frac{0.504}{1-j4} = 0.122e^{j75.96^\circ} \implies |D_{-1}| = 0.122, \angle D_{-1} = 75.96^\circ -$$ - -and - -$$ -D_2 = \frac{0.504}{1+j8} = 0.0625e^{-j82.87^\circ} \implies |D_2| = 0.0625, \ \angle D_2 = -82.87^\circ -$$ - -$$ -D_{-2} = \frac{0.504}{1-j8} = 0.0625e^{j82.87^\circ} \implies |D_{-2}| = 0.0625, \ \angle D_{-2} = 82.87^\circ -$$ - -and so on. Note that *Dn* and *D*−*n* are conjugates, as expected [see Eq. (6.22)]. - -Figure 6.12 shows the frequency spectra (amplitude and angle) of the exponential Fourier series for the periodic signal *x*(*t*) in Fig. 6.2a. - -We notice some interesting features of these spectra. First, the spectra exist for positive as well as negative values of ω (the frequency). Second, the amplitude spectrum is an even function of ω and the angle spectrum is an odd function of ω. - -and - -**Figure 6.12** Exponential Fourier spectra for the signal in Fig. 6.2a. - -At times it may appear that the phase spectrum of a real periodic signal fails to satisfy the odd symmetry: for example, when *Dk* = *D*−*k* = −10. In this case, *Dk* = 10*ej*π , and therefore, *D*−*k* = 10*e*−*j*π . Recall that *e*±*j*π = −1. Here, although *Dk* = *D*−*k*, their phases should be taken as π and −π. - -### **EXAMPLE 6.7 Plotting Fourier Series Spectra with MATLAB** - -Using MATLAB and the results of Ex. 6.6, compute and plot the exponential Fourier spectra for the periodic signal *x*(*t*) shown in Fig. 6.2a. The result should match Fig. 6.12. - -The expression for *Dn* is derived in Ex. 6.6. - -``` ->> clf; n = (-5:5); D_n = 0.504./(1+4j*n); -``` - -- >> subplot(1,2,1); stem(n,abs(D\_n),'.k'); -- >> xlabel('n'); ylabel('|D\_n|'); -- >> subplot(1,2,2); stem(n,angle(D\_n),'.k'); -- >> xlabel('n'); ylabel('\angle D\_n [rad]'); - -Except that it is plotted as a function of *n* rather than ω, the result in Fig. 6.13 matches Fig. 6.12. - -### WHAT IS A NEGATIVE FREQUENCY? - -The existence of the spectrum at negative frequencies is somewhat disturbing because, by definition, the frequency (number of repetitions per second) is a positive quantity. How do we interpret a negative frequency? We can use a trigonometric identity to express a sinusoid of a negative frequency −ω0 as - -$$ -\cos(-\omega_0 t + \theta) = \cos(\omega_0 t - \theta) -$$ - -This equation clearly shows that the frequency of a sinusoid cos(ω0*t* + θ ) is |ω0|, which is a positive quantity. The same conclusion is reached by observing that - -$$ -e^{\pm j\omega_0 t} = \cos \omega_0 t \pm j \sin \omega_0 t -$$ - -Thus, the frequency of exponentials *e*±*j*ω0*t* is indeed |ω0|. How do we then interpret the spectral plots for negative values of ω? A more satisfying way of looking at the situation is to say that *exponential spectra are a graphical representation of coefficients Dn as a function of* ω*. Existence of the spectrum at* ω = −*n*ω0 *is merely an indication that an exponential component e*−*jn*ω0*t exists in the series*. We know that a sinusoid of frequency *n*ω0 can be expressed in terms of a pair of exponentials *ejn*ω0*t* and *e*−*jn*ω0*t* . - -We see a close connection between the exponential spectra in Fig. 6.12 and the spectra of the corresponding trigonometric Fourier series for *x*(*t*) (Figs. 6.2b, 6.2c). Equation (6.22) explains the reason for the close connection, for real *x*(*t*), between the trigonometric spectra (*Cn* and θ*n*) with exponential spectra (|*Dn*| and *Dn*). The dc components *D*0 and *C*0 are identical in both spectra. Moreover, the exponential amplitude spectrum |*Dn*| is half the trigonometric amplitude spectrum *Cn* for *n* ≥ 1. The exponential angle spectrum *Dn* is identical to the trigonometric phase spectrum θ*n* for *n* ≥ 0. We can therefore produce the exponential spectra merely by inspection of trigonometric spectra, and vice versa. The following example demonstrates this feature. - -### **EXAMPLE 6.8 Relating Exponential to Trigonometric Fourier Series Spectra** - -The trigonometric Fourier spectra of a certain periodic signal *x*(*t*) are shown in Fig. 6.14a. After inspecting these spectra, sketch the corresponding exponential Fourier spectra and verify your results analytically. - -**Figure 6.14** Fourier series spectra for Ex. 6.8. - -The trigonometric spectral components exist at frequencies 0, 3, 6, and 9. The exponential spectral components exist at 0, 3, 6, 9, and −3, −6, −9. Consider first the amplitude spectrum. The dc component remains unchanged: that is, *D*0 = *C*0 = 16. Now |*Dn*| is an even function of ω and |*Dn*|=|*D*−*n*| = *Cn*/2. Thus, all the remaining spectrum |*Dn*| for positive *n* is half the trigonometric amplitude spectrum *Cn*, and the spectrum |*Dn*| for negative *n* is a reflection about the vertical axis of the spectrum for positive *n*, as shown in Fig. 6.14b. - -The angle spectrum is *Dn* = θ*n* for positive *n* and is −θ*n* for negative *n*, as depicted in Fig. 6.14b. We shall now verify that both sets of spectra represent the same signal. - -Signal *x*(*t*), whose trigonometric spectra are shown in Fig. 6.14a, has four spectral components of frequencies 0, 3, 6, and 9. The dc component is 16. The amplitude and the phase of the component of frequency 3 are 12 and −π/4, respectively. Therefore, this component can be expressed as 12cos(3*t* − π/4). Proceeding in this manner, we can write the Fourier series for *x*(*t*) as - -$$ -x(t) = 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right) -$$ - -Consider now the exponential spectra in Fig. 6.14b. They contain components of frequencies 0 (dc), ±3, ±6, and ±9. The dc component is *D*0 = 16. The component *ej*3*t* (frequency 3) has magnitude 6 and angle −π/4. Therefore, this component strength is 6*e*−*j*π/4, and it can be expressed as (6*e*−*j*π/4)*ej*3*t* . Similarly, the component of frequency −3 is (6*ej*π/4)*e*−*j*3*t* . Proceeding in this manner, *x*ˆ(*t*), the signal corresponding to the spectra in Fig. 6.14b, is - -$$ -\hat{x}(t) = 16 + \left[6e^{-j\pi/4}e^{j3t} + 6e^{j\pi/4}e^{-j3t}\right] + \left[4e^{-j\pi/2}e^{j6t} + 4e^{j\pi/2}e^{-j6t}\right] -$$ -\n -$$ -+ \left[2e^{-j\pi/4}e^{j9t} + 2e^{j\pi/4}e^{-j9t}\right] -$$ -\n -$$ -= 16 + 6\left[e^{j(3t - \pi/4)} + e^{-j(3t - \pi/4)}\right] + 4\left[e^{j(6t - \pi/2)} + e^{-j(6t - \pi/2)}\right] -$$ -\n -$$ -+ 2\left[e^{j(9t - \pi/4)} + e^{-j(9t - \pi/4)}\right] -$$ -\n -$$ -= 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right) -$$ - -Clearly both sets of spectra represent the same periodic signal. - -### BANDWIDTH OF A SIGNAL - -The difference between the highest and the lowest frequencies of the spectral components of a signal is the *bandwidth* of the signal. The bandwidth of the signal whose exponential spectra are shown in Fig. 6.14b is 9 (in radians). The highest and lowest frequencies are 9 and 0, respectively. Note that the component of frequency 12 has zero amplitude and is nonexistent. Moreover, the lowest frequency is 0, not −9. Recall that the frequencies (in the conventional sense) of the spectral components at ω = −3, −6, and −9 in reality are 3, 6, and 9.† The bandwidth can be more readily seen from the trigonometric spectra in Fig. 6.14a. - -### **EXAMPLE 6.9 Fourier Series Spectra of an Impulse Train** - -Find the exponential Fourier series and sketch the corresponding spectra for the impulse train δ*T*0 (*t*) depicted in Fig. 6.15a. From this result, sketch the trigonometric spectrum and write the trigonometric Fourier series for δ*T*0 (*t*). - -† Some authors *do* define bandwidth as the difference between the highest and the lowest (negative) frequency in the exponential spectrum. The bandwidth according to this definition is twice that defined here. In reality, this phrasing defines not the signal bandwidth but the *spectral width* (width of the exponential spectrum of the signal). - -The unit impulse train shown in Fig. 6.15a can be expressed as - -$$ -\sum_{n=-\infty}^{\infty} \delta(t - nT_0) -$$ - -Following Papoulis, we shall denote this function as δ*T*0 (*t*) for the sake of notational brevity. The exponential Fourier series is given by - -$$ -\delta_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ -\n(6.23) - -where - -$$ -D_n = \frac{1}{T_0} \int_{T_0} \delta_{T_0}(t) e^{-jn\omega_0 t} dt -$$ - -Choosing the interval of integration (−*T*0/2,*T*0/2) and recognizing that over this interval δ*T*0 (*t*) = δ(*t*), we get - -$$ -D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} \delta(t) e^{-jn\omega_0 t} dt -$$ - -### 630 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -In this integral, the impulse is located at *t* = 0. From the sampling property of Eq. (1.11), the integral on the right-hand side is the value of *e*−*jn*ω0*t* at *t* = 0 (where the impulse is located). Therefore, - -$$ -D_n = \frac{1}{T_0} \tag{6.24} -$$ - -From this result, we see that the exponential spectrum is constant for all frequencies, as shown in Fig. 6.15b. The spectrum, being real, requires only the amplitude plot. All phases are zero. - -Substituting *Dn* = 1 *T*0 into Eq. (6.23) yields the desired exponential Fourier series - -$$ -\delta_{T_0}(t) = \frac{1}{T_0} \sum_{n = -\infty}^{\infty} e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -To sketch the trigonometric spectrum, we use Eq. (6.22) to obtain - -$$ -C_0 = D_0 = \frac{1}{T_0} -$$ - -\n -$$ -C_n = 2|D_n| = \frac{2}{T_0} \qquad n = 1, 2, 3, ... -$$ - -\n -$$ -\theta_n = 0 -$$ - -Figure 6.15c shows the trigonometric Fourier spectrum. From this spectrum we can express δ*T*0 (*t*) as - -$$ -\delta_{T_0}(t) = \frac{1}{T_0} \left[ 1 + 2(\cos \omega_0 t + \cos 2\omega_0 t + \cos 3\omega_0 t + \cdots) \right] \qquad \omega_0 = \frac{2\pi}{T_0} \tag{6.25} -$$ - -### EFFECT OF SYMMETRY IN EXPONENTIAL FOURIER SERIES - -When *x*(*t*) has an even symmetry, *bn* = 0, and from Eq. (6.20), *Dn* = *an*/2, which is real (positive or negative). Hence, *Dn* can only be 0 or ±π. Moreover, we may compute *Dn* = *an*/2 by using Eq. (6.14), which requires integration over a half-period only. Similarly, when *x*(*t*) has an odd symmetry, *an* = 0, and *Dn* = −*jbn*/2 is imaginary (positive or negative). Hence, *Dn* can only be 0 or ±π/2. Moreover, we may compute *Dn* = −*jbn*/2 by using Eq. (6.15), which requires integration over a half-period only. Note, however, that in the exponential case, we are using the symmetry property indirectly by finding the trigonometric coefficients. We cannot apply it directly in finding *Dn* from Eq. (6.19) since the function *ejn*ω0*t* is neither even nor odd. - -## **DR ILL 6.4 Relating Trigonometric to Exponential Fourier Series Spectra** - -The exponential Fourier spectra of a certain periodic signal *x*(*t*) are shown in Fig. 6.16. Determine and sketch the trigonometric Fourier spectra of *x*(*t*) by inspection of Fig. 6.16. Now write the (compact) trigonometric Fourier series for *x*(*t*). - -## **DR ILL 6.5 Fourier Series Spectrum of a Full-Wave Rectified Sine Wave** - -Find the exponential Fourier series and sketch the corresponding Fourier spectrum *Dn* versus ω for the full-wave rectified sine wave depicted in Fig. 6.17. - -### **ANSWER** - -### **DR ILL 6.6 Exponential Fourier Series and Spectra** - -Find the exponential Fourier series and sketch the corresponding Fourier spectra for the periodic signals shown in Fig. 6.7. - -#### **ANSWERS (a)** *x*(*t*) = 1 3 + 2 π2 "∞ *n*=−∞(*n*=0) (−1)*n n*2 *ejn*π*t* **(b)** *x*(*t*) = *jA* π "∞ *n*=−∞(*n*=0) (−1)*n n ejn*π*t* - -### **[6.3-2 Parseval's Theorem](#page-12-0)** - -The trigonometric Fourier series of a periodic signal *x*(*t*) is given by - -$$ -x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \theta_n) -$$ - -Every term on the right-hand side of this equation is a power signal. As shown in Ex. 1.2, Eq. (1.3), the power of *x*(*t*) is equal to the sum of the powers of all the sinusoidal components on the right-hand side. - -$$ -P_x = C_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} C_n^2 -$$ -\n(6.26) - -This result is one form of *Parseval's theorem,* as applied to power signals. It states that the power of a periodic signal is equal to the sum of the powers of its Fourier components. - -We can apply the same argument to the exponential Fourier series (see Prob. 1.1-11). The power of a periodic signal *x*(*t*) can be expressed as a sum of the powers of its exponential components. In Eq. (1.4), we showed that the power of an exponential *Dej*ω0*t* is |*D*2|. We can use this result to express the power of a periodic signal *x*(*t*) in terms of its exponential Fourier series coefficients as - -$$ -P_x = \sum_{n=-\infty}^{\infty} |D_n|^2 \tag{6.27} -$$ - -For a real *x*(*t*), |*D*−*n*|=|*Dn*|. Therefore, - -$$ -P_x = D_0^2 + 2\sum_{n=1}^{\infty} |D_n|^2 -$$ -\n(6.28) - -### **EXAMPLE 6.10 Harmonic Distortion of Clipped Sinusoid** - -The input signal to an audio amplifier of gain 100 is given by *x*(*t*) = 0.1 cosω0*t*. Hence, the output is a sinusoid 10 cos ω0*t*. However, the amplifier, being nonlinear at higher amplitude levels, clips all amplitudes beyond ±8 volts, as shown in Fig. 6.18a. We shall determine the harmonic distortion incurred in this operation. - -**Figure 6.18 (a)** A clipped sinusoid cos ω0*t*. **(b)** The distortion component *xd*(*t*) of the signal in (a). - -The output *y*(*t*) is the clipped signal in Fig. 6.18a. The distortion signal *yd*(*t*), shown in Fig. 6.18b, is the difference between the undistorted sinusoid 10cosω0*t* and the output signal *y*(*t*). The signal *yd*(*t*), whose period is *T*0 [the same as that of *y*(*t*)], can be described over the first cycle as - -$$ -y_d(t) = \begin{cases} 10 \cos \omega_0 t - 8 & |t| \le 0.1024T_0 \\ 10 \cos \omega_0 t + 8 & \frac{T_0}{2} - 0.1024T_0 \le |t| \le \frac{T_0}{2} + 0.1024T_0 \\ 0 & \text{everywhere else} \end{cases} -$$ - -Observe that *yd*(*t*) is an even function of *t* and its mean value is zero. Hence, *a*0 = *C*0 = 0, and *bn* = 0. Thus, *Cn* = *an* and the Fourier series for *yd*(*t*) can be expressed as - -$$ -y_d(t) = \sum_{n=1}^{\infty} C_n \cos n\omega_0 t -$$ - -As usual, we can compute the coefficients *Cn* (which is equal to *an*) by integrating *yd*(*t*) cos *n*ω0*t* over one cycle (and then dividing by 2/*T*0). Because *yd*(*t*) has even symmetry, we can find *an* by integrating the expression over a half-cycle only using Eq. (6.14). The - -#### 634 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -straightforward evaluation of the appropriate integral yields† - -$$ -C_n = \begin{cases} \frac{20}{\pi} \left[ \frac{\sin\left[0.6435(n+1)\right]}{n+1} + \frac{\sin\left[0.6435(n-1)\right]}{n-1} \right] - \frac{32}{\pi} \left[ \frac{\sin\left(0.6435n\right)}{n} \right] & n \text{ odd} \\ 0 & n \text{ even} \end{cases} -$$ - -Computing the coefficients *C*1, *C*2, *C*3, ... from this expression, we can write - -*yd*(*t*) = 1.04 cos ω0*t* +0.733 cos 3ω0*t* +0.311 cos 5ω0*t* +··· - -### COMPUTING HARMONIC DISTORTION - -We can compute the amount of harmonic distortion in the output signal by computing the power of the distortion component *yd*(*t*). Because *yd*(*t*) is an even function of *t* and because the energy in the first half-cycle is identical to the energy in the second half-cycle, we can compute the power by averaging the energy over a quarter-cycle. Thus, - -$$ -P_{y_d} = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} y_d^2(t) dt = \frac{1}{T_0/4} \int_0^{T_0/4} y_d^2(t) dt -$$ - -= $\frac{4}{T_0} \int_0^{0.1024T_0} (10 \cos \omega_0 t - 8)^2 dt = 0.865$ - -The power of the desired signal 10cos ω0*t* is (10)2/2 = 50. Hence, the total harmonic distortion is‡ - -$$ -D_{\text{tot}} = \frac{0.865}{50} \times 100 = 1.73\% -$$ - -The power of the third harmonic components of *yd*(*t*) is (0.733)2/2 = 0.2686. The third harmonic distortion is - -$$ -D_3 = \frac{0.2686}{50} \times 100 = 0.5372\% -$$ - -$$ -C_n = a_n = \frac{8}{T_0} \int_0^{0.1024T_0} [10 \cos \omega_0 t - 8] \cos n\omega_0 t dt -$$ - -† In addition, *yd*(*t*) exhibits half-wave symmetry (see Prob. 6.1-6), where the second half-cycle is the negative of the first. Because of this property, all the even harmonics vanish, and the odd harmonics can be computed by integrating the appropriate expressions over the first half-cycle only (from −*T*0/4 to *T*0/4) and doubling the resulting values. Moreover, because of even symmetry, we can integrate the appropriate expressions over 0 to *T*0/4 (instead of from −*T*0/4 to *T*0/4) and double the resulting values. In essence, this allows us to compute *Cn* by integrating the expression over the quarter-cycle only and then quadrupling the resulting values. Thus, - -‡ In the literature, the harmonic distortion often refers to the rms distortion rather than the power distortion. The rms values are the square-root values of the corresponding powers. Thus, the third harmonic distortion in this sense is √(0.2686/50) × 100 = 7.33%. Alternately, we may also compute this value directly from the amplitudes of the third harmonic 0.733 and that of the fundamental as 10. The ratio of the rms values is (0.733/ √2) : (10/ √ 2) = 0.0733 and the percentage distortion is 7.33%. - -### **[6.3-3 Properties of the Fourier Series](#page-12-0)** - -As with the Laplace and *z*-transforms, the Fourier series has a variety of properties that can simplify work and help provide a more intuitive understanding of signals. Table 6.2 provides the most important properties of the Fourier series for a periodic signal *x*(*t*) and its spectrum *Dn*. Properties that involve two signals require that the two signals have a common fundamental frequency ω0. While not given here, the proofs of these properties are straightforward and parallel the proofs of the Fourier transform properties given in Ch. 7. - -To demonstrate the utility of Fourier series properties, let us consider an example where we use a selection of properties to simplify the work of finding a piecewise polynomial signal's spectrum. - -| Operation | x(t) | Dn | -|-----------------------|------------------------------|----------------| -| Scalar multiplication | kx(t) | kDn | -| Addition | x1(t)+x2(t) | D1,n
+D2,n | -| | x1(t), x2(t) require same ω0 | | -| Conjugation | x∗(t) | D∗
−n | -| Reversal | x(−t) | D−n | -| Time shifting | x(t −t0) | Dne−jnω0t0 | -| Frequency shifting | x(t)ejn0ω0t | Dn−n0 | -| Frequency convolution | x1(t)x2(t) | D1,n
∗ D2,n | -| | x1(t), x2(t) require same ω0 | | -| Time differentiation | dkx(t)
dtk | (jnω0)kDn | - -**TABLE 6.2** Selected Fourier Series Properties - -### **EXAMPLE 6.11 Using Fourier Series Properties** - -Use properties rather than integration to compute the exponential Fourier series coefficients *Dn* of the triangular signal *x*(*t*) shown in Fig. 6.4. Verify the correctness of *Dn* for *A* = 1 by synthesizing *x*(*t*) with a suitable truncation of Eq. (6.19). - -From Fig. 6.4, we see that *x*(*t*) is a piecewise linear function that is *T*0 = 2 periodic. To compute *Dn* directly using Eq. (6.19) would therefore require tedious integration by parts. Fortunately, we can compute *Dn* without integration by instead using Fourier series properties. First, however, we must compute the dc component *D*0 separately from other *Dn*. By simple inspection of Fig. 6.4, we see that *x*(*t*) has no dc component, so *D*0 = 0. - -To determine the remaining *Dn*, we begin by noting that *x*(*t*) has a constant slope of either 2*A* or −2*A*. Thus, differentiating *x*(*t*) once yields a square wave with amplitudes ±2*A*. Here, differentiation reduces *x*(*t*) from a piecewise linear to a piecewise constant function, - -thereby eliminating the need for integration by parts.† Differentiating *x*(*t*) twice yields a pair of shifted impulse trains, weighted by ±4*A*. This second differentiation eliminates the need for any integration whatsoever, since the Fourier series coefficients of an impulse train are known to be 1 *T*0 (see Ex. 6.9). - -Stated mathematically, we see that - -$$ -\frac{d^2}{dt^2}x(t) = 4A\delta_2(t + \frac{1}{2}) - 4A\delta_2(t - \frac{1}{2}) -$$ - -Transforming this expression and using the Fourier series properties of scalar multiplication, addition, frequency shifting, and time differentiation yield (for *n* = 0) - -$$ -(in\pi)^2 D_n = 4A \frac{e^{jn\omega_0/2}}{2} - 4A \frac{e^{-jn\omega_0/2}}{2} -$$ - -Substituting ω0 = 2π/*T*0 = π and solving for *Dn* yield - -$$ -D_n = 4A \frac{e^{in\pi/2} - e^{-in\pi/2}}{-2n^2\pi^2} -$$ - -Combining with the dc component and simplifying expressions, the final result is - -$$ -D_n = \begin{cases} 0 & n = 0\\ \frac{-A4j\sin(n\pi/2)}{n^2\pi^2} & n \neq 0 \end{cases} -$$ - -Overall, this is a neat way to determine the signal's spectrum. Through the selective use of properties, we have determined *Dn* without any integration. With a little care, this basic approach can yield the spectrum of *any piecewise polynomial periodic function*. Since all periodic functions of practical interest to engineers can, to an arbitrary level of accuracy, be represented as piecewise polynomial functions, we can always find their spectra without integration except for the dc term *D*0. - -### USING A TRUNCATED FOURIER SERIES TO VERIFY SPECTRUM CORRECTNESS - -While a signal's spectrum *Dn* provides useful insight into signal character, it can be difficult to look at *Dn* and know that it is correct for a particular signal *x*(*t*). For example, is it at all obvious that *Dn* = 4*Aj*sin(*n*π/2) −*n*2π2 is the spectrum for the triangle wave of Fig. 6.4? Probably not. - -There is an easy way, however, to verify a signal's spectrum: synthesize *x*(*t*) using a suitable truncation of Eq. (6.19). If the synthesized signal closely matches the original, we can be relatively certain that the spectrum *Dn* is correct. What is a suitable truncation? Well, it depends. All significant *Dn* terms need to be included, but not so many as to make the reconstruction impractical to compute. Since in the present case *Dn* decays at a rate 1/*n*2, a good approximation is possible with a relatively few number of terms; a 10-harmonic truncation should be just fine. Let us use MATLAB to synthesize *x*(*t*) using a 10-harmonic truncated Fourier series. - -† Differentiation also destroys the dc component of the signal, providing further justification as to why the dc component needs to be separately computed. - -To begin, we define *A*, *Dn*, *T*0, ω0, and a time vector that spans two periods of the waveform. - -``` ->> A = 1; D = @(n) -A*4j*sin(n*pi/2)./(n.^2*pi^2); ->> T0 = 2; omega0 = 2*pi/T0; t = (-T0:.001:T0); -``` - -Next, we set the dc portion of the signal. - ->> D0 = 0; x10 = D0\*ones(size(t)); - -To add the desired 10 harmonics, we enter a loop for 1 ≤ *n* ≤ 10 and add in the *Dn* and *D*−*n* terms. Although *x*(*t*) should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed using the real command. - -``` ->> for n = 1:10, ->> x10 = x10+real(D(n)*exp(1j*omega0*n*t)+D(-n)*exp(-1j*omega0*n*t)); ->> end -``` - -Lastly, we plot the resulting truncated Fourier series synthesis of *x*(*t*). - ->> plot(t,x10,'k'); xlabel('t'); ylabel('x\_{10}(t)'); - -Since the synthesized waveform shown in Fig. 6.19 closely matches the original waveform in Fig. 6.4, we have high confidence that the computed *Dn* are correct. - -## **[6.4 LTIC SYSTEM](#page-12-0) RESPONSE TO PERIODIC INPUTS** - -A periodic signal can be expressed as a sum of everlasting exponentials (or sinusoids). We also know how to find the response of an LTIC system to an everlasting exponential. From this information, we can readily determine the response of an LTIC system to periodic inputs. A periodic signal *x*(*t*) with period *T*0 can be expressed as an exponential Fourier series - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -### 638 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -In Sec. 4.8, we showed that the response of an LTIC system with transfer function *H*(*s*) to an everlasting exponential input *ej*ω*t* is an everlasting exponential *H*(*j*ω)*ej*ω*t* . This input–output pair can be displayed as† - -$$ -\underbrace{e^{j\omega t}}_{\text{input}} \Longrightarrow \underbrace{H(j\omega)e^{j\omega t}}_{\text{output}} -$$ - -Therefore, from the linearity property, - -$$ -\underbrace{\sum_{n=-\infty}^{\infty} D_n e^{jn\omega_0 t}}_{\text{input } x(t)} \Longrightarrow \underbrace{\sum_{n=-\infty}^{\infty} D_n H(jn\omega_0) e^{jn\omega_0 t}}_{\text{response } y(t)} -$$ -(6.29) - -The response *y*(*t*) is obtained in the form of an exponential Fourier series and is therefore a periodic signal of the same period as that of the input. - -We shall demonstrate the utility of these results by the following example. - -### **EXAMPLE 6.12 Full-Wave Rectifier** - -A full-wave rectifier (Fig. 6.20a) is used to obtain a dc signal from a sinusoid sin *t*. The rectified signal *x*(*t*), depicted in Fig. 6.17, is applied to the input of a lowpass *RC* filter, which suppresses the time-varying component and yields a dc component with some residual ripple. Find the filter output *y*(*t*). Find also the dc output and the rms value of the ripple voltage. - -First, we shall find the Fourier series for the rectified signal *x*(*t*), whose period is *T*0 = π. Consequently, ω0 = 2, and - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{j2nt} -$$ - -where - -$$ -D_n = \frac{1}{\pi} \int_0^{\pi} \sin t e^{-j2nt} dt = \frac{2}{\pi (1 - 4n^2)} -$$ -(6.30) - -Therefore, - -$$ -x(t) = \sum_{n = -\infty}^{\infty} \frac{2}{\pi (1 - 4n^2)} e^{j2nt} -$$ - -† This result applies only to asymptotically stable systems. This is because when *s* = *j*ω, the integral on the right-hand side of Eq. (2.39) does not converge for unstable systems. Moreover, for marginally stable systems also, that integral does not converge in the ordinary sense, and *H*(*j*ω) cannot be obtained from *H*(*s*) by replacing *s* with *j*ω. - -**Figure 6.20 (a)** Full-wave rectifier with a lowpass filter and **(b)** its output. - -Next, we find the transfer function of the *RC* filter in Fig. 6.20a. This filter is identical to the *RC* circuit in Ex. 1.17 (Fig. 1.35) for which the differential equation relating the output (capacitor voltage) to the input *x*(*t*) was found to be [Eq. (1.31)]: - -$$ -(3D+1)y(t) = x(t) -$$ - -The transfer function *H*(*s*) for this system is found from Eq. (2.41) as - -$$ -H(s) = \frac{1}{3s+1} -$$ - -and - -$$ -H(j\omega) = \frac{1}{3j\omega + 1} \tag{6.31} -$$ - -From Eq. (6.29), the filter output *y*(*t*) can be expressed as (with ω0 = 2) - -$$ -y(t) = \sum_{n = -\infty}^{\infty} D_n H(jn\omega_0) e^{jn\omega_0 t} = \sum_{n = -\infty}^{\infty} D_n H(j2n) e^{j2nt} -$$ - -Substituting *Dn* and *H*(*j*2*n*) from Eqs. (6.30) and (6.31) in the foregoing equation, we obtain - -$$ -y(t) = \sum_{n = -\infty}^{\infty} \frac{2}{\pi (1 - 4n^2)(j6n + 1)} e^{j2nt} -$$ - -#### 640 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -Note that the output *y*(*t*) is also a periodic signal given by the exponential Fourier series on the right-hand side. The output is shown in Fig. 6.20b. - -The output Fourier series coefficient corresponding to *n* = 0 is the dc component of the output, given by 2/π. The remaining terms in the Fourier series constitute the unwanted component called the ripple. We can determine the rms value of the ripple voltage by using Eq. (6.27) to find the power of the ripple component. The power of the ripple is the power of all the components except the dc (*n* = 0). Note that *D*ˆ *n*, the exponential Fourier coefficient for the output *y*(*t*), is - -$$ -\hat{D}_n = \frac{2}{\pi (1 - 4n^2)(j6n + 1)} -$$ - -Therefore, from Eq. (6.28), we have - -$$ -P_{\text{right}} = 2 \sum_{n=1}^{\infty} |D_n|^2 = 2 \sum_{n=1}^{\infty} \left| \frac{2}{\pi (1 - 4n^2)(j6n + 1)} \right|^2 = \frac{8}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{(1 - 4n^2)^2 (36n^2 + 1)} -$$ - -Numerical computation of the right-hand side yields *P*ripple = 0.0025, and the ripple rms value = *P*ripple = 0.05. This shows that the rms ripple voltage is 5% of the amplitude of the input sinusoid. - -### WHY USE EXPONENTIALS? - -The exponential Fourier series is just another way of representing trigonometric Fourier series (or vice versa). The two forms carry identical information—no more, no less. The reasons for preferring the exponential form have already been mentioned: this form is more compact, and the expression for deriving the exponential coefficients is also more compact than those in the trigonometric series. Furthermore, the LTIC system response to exponential signals is also simpler (more compact) than the system response to sinusoids. In addition, the exponential form proves to be much easier than the trigonometric form to manipulate mathematically and otherwise handle in the area of signals as well as systems. Moreover, exponential representation proves much more convenient for analysis of complex *x*(*t*). For these reasons, in our future discussion we shall use the exponential form exclusively. - -A minor disadvantage of the exponential form is that it cannot be visualized as easily as sinusoids. For intuitive and qualitative understanding, the sinusoids have the edge over exponentials. Fortunately, this difficulty can be overcome readily because of the close connection between exponential and Fourier spectra. For the purpose of mathematical analysis, we shall continue to use exponential signals and spectra; but to understand the physical situation intuitively or qualitatively, we shall speak in terms of sinusoids and trigonometric spectra. Thus, although all mathematical manipulation will be in terms of exponential spectra, we shall now speak of exponential and sinusoids interchangeably when we discuss intuitive and qualitative insights in attempting to arrive at an understanding of physical situations. This is an important point; readers should make an extra effort to familiarize themselves with the two forms of spectra, their relationships, and their convertibility. - -### DUAL PERSONALITY OF A SIGNAL - -The discussion so far shows that a periodic signal has a dual personality—the time domain and the frequency domain. It can be described by its waveform or by its Fourier spectra. The timeand frequency-domain descriptions provide complementary insights into a signal. For in-depth perspective, we need to understand both these identities. It is important to learn to think of a signal from both perspectives. In the next chapter, we shall see that aperiodic signals also have this dual personality. Moreover, we shall show that even LTI systems have this dual personality, which offers complementary insights into the system behavior. - -### LIMITATIONS OF THE FOURIER SERIES METHOD OF ANALYSIS - -We have developed here a method of representing a periodic signal as a weighted sum of everlasting exponentials whose frequencies lie along the ω axis in the *s* plane. This representation (Fourier series) is valuable in many applications. However, as a tool for analyzing linear systems, it has serious limitations and consequently has limited utility for the following reasons: - -- 1. The Fourier series can be used only for periodic inputs. All practical inputs are aperiodic (remember that a periodic signal starts at *t* = −∞). -- 2. The Fourier methods can be applied readily to BIBO-stable (or asymptotically stable) systems. It cannot handle unstable or even marginally stable systems. - -The first limitation can be overcome by representing aperiodic signals in terms of everlasting exponentials. This representation can be achieved through the Fourier integral, which may be considered to be an extension of the Fourier series. We shall therefore use the Fourier series as a stepping-stone to the Fourier integral developed in the next chapter. The second limitation can be overcome by using exponentials *est*, where *s* is not restricted to the imaginary axis but is free to take on complex values. This generalization leads to the Laplace integral, discussed in Ch. 4 (the Laplace transform). - -## **[6.5 GENERALIZED](#page-12-0) FOURIER SERIES: SIGNALS AS VECTORS** - -We now consider a very general approach to signal representation with far-reaching consequences.† There is a perfect analogy between signals and vectors; the analogy is so strong that the term *analogy* understates the reality. Signals are not just *like* vectors. Signals *are* vectors! A vector can be represented as a sum of its components in a variety of ways, depending on the choice of coordinate system. A signal can also be represented as a sum of its components in a variety of ways. Let us begin with some basic vector concepts and then apply these concepts to signals. - -† This section closely follows the material from the author's earlier book [10]. Omission of this section will not cause any discontinuity in understanding the rest of the book. Derivation of Fourier series through the signal-vector analogy provides an interesting insight into signal representation and other topics such as signal correlation, data truncation, and signal detection. - -### **[6.5-1 Component of a Vector](#page-12-0)** - -A vector is specified by its magnitude and its direction. We shall denote all vectors by boldface. For example, **x** is a certain vector with magnitude or length |**x**|. For the two vectors **x** and **y** shown in Fig. 6.21, we define their dot (inner or scalar) product as - -$$ -\mathbf{x} \cdot \mathbf{y} = |\mathbf{x}| |\mathbf{y}| \cos \theta -$$ - -where θ is the angle between these vectors. Using this definition, we can express |**x**|, the length of a vector **x**, as - -$$ -|\mathbf{x}|^2 = \mathbf{x} \cdot \mathbf{x} -$$ - -Let the component of **x** along **y** be *c***y** as depicted in Fig. 6.21. Geometrically, the component of **x** along **y** is the projection of **x** on **y** and is obtained by drawing a perpendicular from the tip of **x** on the vector **y**, as illustrated in Fig. 6.21. What is the mathematical significance of a component of a vector along another vector? As seen from Fig. 6.21, the vector **x** can be expressed in terms of vector **y** as - -$$ -\mathbf{x} = c\mathbf{y} + \mathbf{e} -$$ - -However, this is not the only way to express **x** in terms of **y**. From Fig. 6.22, which shows two of the infinite other possibilities, we have - -$$ -\mathbf{x} = c_1 \mathbf{y} + \mathbf{e}_1 = c_2 \mathbf{y} + \mathbf{e}_2 -$$ - -In each of these three representations, **x** is represented in terms of **y** plus another vector called the *error vector*. If we approximate **x** by *c***y**, - -$$ -\mathbf{x} \simeq c\mathbf{y} -$$ - -the error in the approximation is the vector **e** = **x** − *c***y**. Similarly, the errors in approximations in these drawings are **e**1 (Fig. 6.22a) and **e**2 (Fig. 6.22b). What is unique about the approximation in Fig. 6.21 is that the error vector is the smallest. We can now define mathematically the component of a vector **x** along vector **y** to be *c***y** where *c* is chosen to minimize the length of the error vector **e** = **x** − *c***y**. Now, the length of the component of **x** along **y** is |**x**| cos θ. But it is also *c*|**y**|, as seen from Fig. 6.21. Therefore, - -$$ -c|\mathbf{y}| = |\mathbf{x}| \cos \theta -$$ - -Multiplying both sides by |**y**| yields - -$$ -c|\mathbf{y}|^2 = |\mathbf{x}||\mathbf{y}|\cos\theta = \mathbf{x}\cdot\mathbf{y} -$$ - -**Figure 6.21** Component (projection) of a vector along another vector. - -**Figure 6.22** Approximation of a vector in terms of another vector. - -Therefore, - -$$ -c = \frac{\mathbf{x} \cdot \mathbf{y}}{\mathbf{y} \cdot \mathbf{y}} = \frac{1}{|\mathbf{y}|^2} \mathbf{x} \cdot \mathbf{y} -$$ - (6.32) - -From Fig. 6.21, it is apparent that when **x** and **y** are perpendicular, or orthogonal, then **x** has a zero component along **y**; consequently, *c* = 0. Keeping an eye on Eq. (6.32), we therefore define **x** and **y** to be *orthogonal* if the inner (scalar or dot) product of the two vectors is zero, that is, if - -**x** · **y** = 0 - -### **[6.5-2 Signal Comparison and Component of a Signal](#page-12-0)** - -The concept of a vector component and orthogonality can be extended to signals. Consider the problem of approximating a real signal *x*(*t*) in terms of another real signal *y*(*t*) over an interval (*t*1, *t*2): - -$$ -x(t) \simeq cy(t) \qquad t_1 < t < t_2 -$$ - -The error *e*(*t*) in this approximation is - -$$ -e(t) = \begin{cases} x(t) - cy(t) & t_1 < t < t_2 \\ 0 & \text{otherwise} \end{cases} -$$ - -We now select a criterion for the "best approximation." We know that the signal energy is one possible measure of a signal size. For best approximation, we shall use the criterion that minimizes the size or energy of the error signal *e*(*t*) over the interval (*t*1,*t*2). This energy *Ee* is given by - -$$ -E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt -$$ - -Note that the right-hand side is a definite integral with *t* as the dummy variable. Hence, *Ee* is a function of the parameter *c* (not *t*) and *Ee* is minimum for some choice of *c*. To minimize *Ee*, a necessary condition is - -$$ -\frac{dE_e}{dc} = 0 -$$ -$$ -\frac{d}{dc} \left[ \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt \right] = 0 -$$ - -*d* - -or - -#### 644 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -Expanding the squared term inside the integral, we obtain - -$$ -\frac{d}{dc} \left[ \int_{t_1}^{t_2} x^2(t) dt \right] - \frac{d}{dc} \left[ 2c \int_{t_1}^{t_2} x(t) y(t) dt \right] + \frac{d}{dc} \left[ c^2 \int_{t_1}^{t_2} y^2(t) dt \right] = 0 -$$ - -from which we get - -$$ --2\int_{t_1}^{t_2} x(t)y(t) dt + 2c \int_{t_1}^{t_2} y^2(t) dt = 0 -$$ - -$$ -c = \frac{\int_{t_1}^{t_2} x(t)y(t) dt}{\int_{t_1}^{t_2} y^2(t) dt} = \frac{1}{E_y} \int_{t_1}^{t_2} x(t)y(t) dt -$$ - (6.33) - -We observe a remarkable similarity between the behavior of vectors and signals, as indicated by Eqs. (6.32) and (6.33). It is evident from these two parallel expressions that *the area under the product of two signals corresponds to the inner (scalar or dot) product of two vectors*. In fact, the area under the product of *x*(*t*) and *y*(*t*) is called the *inner product* of *x*(*t*) and *y*(*t*), and is denoted by (*x*, *y*). The energy of a signal is the inner product of a signal with itself, and corresponds to the vector length square (which is the inner product of the vector with itself). - -To summarize our discussion, if a signal *x*(*t*) is approximated by another signal *y*(*t*) as - -*x*(*t*) *cy*(*t*) - -then the optimum value of *c* that minimizes the energy of the error signal in this approximation is given by Eq. (6.33). - -Taking our clue from vectors, we say that a signal *x*(*t*) contains a component *cy*(*t*), where *c* is given by Eq. (6.33). Note that in vector terminology, *cy*(*t*) is the projection of *x*(*t*) on *y*(*t*). Continuing with the analogy, we say that if the component of a signal *x*(*t*) of the form *y*(*t*) is zero (i.e., *c* = 0), the signals *x*(*t*) and *y*(*t*) are orthogonal over the interval (*t*1, *t*2). Therefore, we define the real signals *x*(*t*) and *y*(*t*) to be orthogonal over the interval (*t*1, *t*2) if† - -$$ -\int_{t_1}^{t_2} x(t)y(t) dt = 0 -$$ -\n(6.34) - -### **EXAMPLE 6.13 Sine-Wave Approximation of a Square Wave** - -For the square signal *x*(*t*) shown in Fig. 6.23, find the component in *x*(*t*) of the form sin*t*. In other words, approximate *x*(*t*) in terms of sin *t* - -*x*(*t*) *c*sin*t* 0 < *t* < 2π - -so that the energy of the error signal is minimum. - -† For complex signals, the definition is modified as in Eq. (6.37), in Sec. 6.5-3. - -**Figure 6.23** Approximation of a square wave in terms of a single sinusoid. - -In this case, - -$$ -y(t) = \sin t -$$ - and $E_y = \int_0^{2\pi} \sin^2(t) dt = \pi$ - -From Eq. (6.33), we find - -$$ -c = \frac{1}{\pi} \int_0^{2\pi} x(t) \sin t \, dt = \frac{1}{\pi} \left[ \int_0^{\pi} \sin t \, dt + \int_{\pi}^{2\pi} -\sin t \, dt \right] = \frac{4}{\pi} -$$ - -Thus, - -$$ -x(t) \simeq \frac{4}{\pi} \sin t -$$ - -represents the best approximation of *x*(*t*) by the function sin*t*, which will minimize the error energy. This sinusoidal component of *x*(*t*) is shaded in Fig. 6.23. By analogy with vectors, we say that the square function *x*(*t*) depicted in Fig. 6.23 has a component of signal sin*t* and that the magnitude of this component is 4/π. - -### **DR ILL 6.7 Sine Wave Approximation of a Ramp Function** - -Show that over an interval (−π < *t* < π), the "best" approximation of the signal *x*(*t*) = *t* in terms of the function sin*t* is 2 sin*t*. Verify that the error signal *e*(*t*) = *t* − 2 sin*t* is orthogonal to the signal sin*t* over the interval −π < *t* < π. Sketch the signals *t* and 2 sin*t* over the interval −π < *t* < π. - -### **[6.5-3 Extension to Complex Signals](#page-12-0)** - -So far we have restricted ourselves to real functions of *t*. To generalize the results to complex functions of *t*, consider again the problem of approximating a signal *x*(*t*) by a signal *y*(*t*) over an - -#### 646 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -interval (*t*1 < *t* < *t*2): - -$$ -x(t) \simeq cy(t) -$$ - -where *x*(*t*) and *y*(*t*) now can be complex functions of *t*. Recall that the energy *Ey* of the complex signal *y*(*t*) over an interval (*t*1, *t*2) is - -$$ -E_{y} = \int_{t_1}^{t_2} |y(t)|^2 dt -$$ - -In this case, both the coefficient *c* and the error - -$$ -e(t) = x(t) - cy(t) -$$ - -are complex (in general). For the "best" approximation, we choose *c* to minimize the energy *Ee* of the error signal *e*(*t*). Now, - -$$ -E_e = \int_{t_1}^{t_2} |x(t) - cy(t)|^2 dt -$$ -\n(6.35) - -Recall also that - -$$ -|u + v|^2 = (u + v)(u^* + v^*) = |u|^2 + |v|^2 + u^*v + uv^* \tag{6.36} -$$ - -After some manipulation, we can use this result to rearrange Eq. (6.35) as - -$$ -E_e = \int_{t_1}^{t_2} |x(t)|^2 dt - \left| \frac{1}{\sqrt{E_y}} \int_a^{t_2} x(t) y^*(t) dt \right|^2 + \left| c \sqrt{E_y} - \frac{1}{\sqrt{E_y}} \int_{t_1}^{t_2} x(t) y^*(t) dt \right|^2 -$$ - -Since the first two terms on the right-hand side are independent of *c*, it is clear that *Ee* is minimized by choosing *c* so that the third term on the right-hand side is zero. This yields - -$$ -c = \frac{1}{E_y} \int_{t_1}^{t_2} x(t) y^*(t) dt -$$ - -In light of this result, we need to redefine orthogonality for the complex case as follows: two complex functions *x*1(*t*) and *x*2(*t*) are orthogonal over an interval (*t*1 < *t* < *t*2) if - -$$ -\int_{t_1}^{t_2} x_1(t) x_2^*(t) dt = 0 \qquad \text{or} \qquad \int_{t_1}^{t_2} x_1^*(t) x_2(t) dt = 0 \tag{6.37} -$$ - -Either equality suffices. This is a general definition of orthogonality, which reduces to Eq. (6.34) when the functions are real. - -### **DR ILL 6.8 Complex Exponential Approximation of a Square Wave** - -Show that over an interval (0 < *t* < 2π ), the "best" approximation of the square signal *x*(*t*) in Fig. 6.23 in terms of the signal *ejt* is given by (2/*j*π ) *ejt*. Verify that the error signal *e*(*t*) = *x*(*t*)−(2/*j*π )*ejt* is orthogonal to the signal *ejt*. - -### ENERGY OF THE SUM OF ORTHOGONAL SIGNALS - -We know that the square of the length of a sum of two orthogonal vectors is equal to the sum of the squares of the lengths of the two vectors. Thus, if vectors **x** and **y** are orthogonal, and if **z** = **x**+**y**, then - -$$ -|\mathbf{z}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2 -$$ - -We have a similar result for signals. The energy of the sum of two orthogonal signals is equal to the sum of the energies of the two signals. Thus, if signals *x*(*t*) and *y*(*t*) are orthogonal over an interval (*t*1, *t*2), and if *z*(*t*) = *x*(*t*)+*y*(*t*), then - -$$ -E_z = E_x + E_y -$$ - -We now prove this result for complex signals, of which real signals are a special case. From Eq. (6.36), it follows that - -$$ -\int_{t_1}^{t_2} |x(t) + y(t)|^2 dt = \int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt + \int_{t_1}^{t_2} x(t)y^*(t) dt + \int_{t_1}^{t_2} x^*(t)y(t) dt -$$ - -= -$$ -\int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt -$$ - -The last result follows from the fact that because of orthogonality, the two integrals of the products *x*(*t*)*y*∗(*t*) and *x*∗(*t*)*y*(*t*) are zero [see Eq. (6.37)]. This result can be extended to the sum of any number of mutually orthogonal signals. - -### **[6.5-4 Signal Representation by an Orthogonal Signal Set](#page-12-0)** - -In this section we show a way of representing a signal as a sum of orthogonal signals. Here again we can benefit from the insight gained from a similar problem in vectors. We know that a vector can be represented as a sum of orthogonal vectors, which form the coordinate system of a vector space. The problem in signals is analogous, and the results for signals are parallel to those for vectors. So, let us review the case of vector representation. - -### ORTHOGONAL VECTOR SPACE - -Let us investigate a three-dimensional Cartesian vector space described by three mutually orthogonal vectors **x**1, **x**2, and **x**3, as illustrated in Fig. 6.24. First, we shall seek to approximate a three-dimensional vector **x** in terms of two mutually orthogonal vectors **x**1 and **x**2: - -$$ -\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 -$$ - -The error **e** in this approximation is - -$$ -\mathbf{e} = \mathbf{x} - (c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2) -$$ - -or - -$$ -\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + \mathbf{e} -$$ - -**Figure 6.24** Representation of a vector in three-dimensional space. - -As in the earlier geometrical argument, we see from Fig. 6.24 that the length of **e** is minimum when **e** is perpendicular to the **x**1–**x**2 plane, and *c*1**x**1 and *c*2**x**2 are the projections (components) of **x** on **x**1 and **x**2, respectively. Therefore, the constants *c*1 and *c*2 are given by Eq. (6.32). Observe that the error vector is orthogonal to both the vectors **x**1 and **x**2. - -Now, let us determine the "best" approximation to **x** in terms of all three mutually orthogonal vectors **x**1, **x**2, and **x**3: - -$$ -\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3 \tag{6.38} -$$ - -Figure 6.24 shows that a unique choice of *c*1, *c*2, and *c*3 exists, for which Eq. (6.38) is no longer an approximation but an equality - -$$ -\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3 -$$ - -In this case, *c*1**x**1, *c*2**x**2, and *c*3**x**3 are the projections (components) of **x** on **x**1,**x**2, and **x**3, respectively; that is, - -$$ -c_i = \frac{\mathbf{x} \cdot \mathbf{x}_i}{\mathbf{x}_i \cdot \mathbf{x}_i} = \frac{1}{|\mathbf{x}_i|^2} \mathbf{x} \cdot \mathbf{x}_i \qquad i = 1, 2, 3 -$$ - (6.39) - -Note that the error in the approximation is zero when **x** is approximated in terms of three mutually orthogonal vectors: **x**1, **x**2, and **x**3. The reason is that **x** is a three-dimensional vector, and the vectors **x**1, **x**2, and **x**3 represent a *complete set* of orthogonal vectors in three-dimensional space. Completeness here means that it is impossible to find another vector **x**4 in this space, which is orthogonal to all three vectors, **x**1,**x**2, and **x**3. Any vector in this space can then be represented (with zero error) in terms of these three vectors. Such vectors are known as *basis* vectors. If a set of vectors {**x***i*} is not complete, the error in the approximation will generally not be zero. Thus, in the three-dimensional case discussed earlier, it is generally not possible to represent a vector **x** in terms of only two basis vectors without an error. - -The choice of basis vectors is not unique. In fact, a set of basis vectors corresponds to a particular choice of coordinate system. Thus, a three-dimensional vector **x** may be represented in many different ways, depending on the coordinate system used. - -### ORTHOGONAL SIGNAL SPACE - -We start with real signals and then extend the discussion to complex signals. We proceed with our signal approximation problem, using clues and insights developed for vector approximation. As before, we define orthogonality of a real signal set *x*1(*t*), *x*2(*t*), ..., *xN*(*t*) over interval (*t*1,*t*2) as - -$$ -\int_{t_1}^{t_2} x_m(t) x_n(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases} -$$ - (6.40) - -If the energies *En* = 1 for all *n*, then the set is *normalized* and is called an *orthonormal set*. An orthogonal set can always be normalized by dividing *xn*(*t*) by √*En* for all *n*. - -Now, consider approximating a signal *x*(*t*) over the interval (*t*1, *t*2) by a set of *N* real, mutually orthogonal signals *x*1(*t*), *x*2(*t*),..., *xN*(*t*) as - -$$ -x(t) \simeq c_1 x_1(t) + c_2 x_2(t) + \dots + c_N x_N(t) \simeq \sum_{n=1}^N c_n x_n(t) -$$ - (6.41) - -In the approximation of Eq. (6.41), the error *e*(*t*) is - -$$ -e(t) = x(t) - \sum_{n=1}^{N} c_n x_n(t) -$$ - -and *Ee*, the error signal energy, is - -$$ -E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt -$$ - (6.42) - -According to our criterion for best approximation, we select the values of *ci* that minimize *Ee*. Hence, the necessary condition is ∂*Ee*/*dci* = 0 for *i* = 1, 2,...,*N*, that is, - -$$ -\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^N c_n x_n(t) \right]^2 dt = 0 -$$ - -When we expand the integrand, we find that all the cross-multiplication terms arising from the orthogonal signals are zero by virtue of orthogonality: that is, all terms of the form \$ *xm*(*t*)*xn*(*t*)*dt* with *m* = *n* vanish. Similarly, the derivative with respect to *ci* of all terms that do not contain *ci* is zero. For each *i*, this leaves only two nonzero terms: - -$$ -\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ -2c_i x(t) x_i(t) + c_i^2 x_i^2(t) \right] dt = 0 -$$ - -or - -$$ --2\int_{t_1}^{t_2} x(t)x_i(t) dt + 2c_i \int_{t_1}^{t_2} x_i^2(t) dt = 0 \qquad i = 1, 2, \dots, N -$$ - -Therefore, - -$$ -c_i = \frac{\int_{t_1}^{t_2} x(t) x_i(t) dt}{\int_{t_1}^{t_2} x_i^2(t) dt} = \frac{1}{E_i} \int_{t_1}^{t_2} x(t) x_i(t) dt \qquad i = 1, 2, ..., N -$$ - (6.43) - -A comparison of Eq. (6.43) with Eq. (6.39) forcefully brings out the analogy of signals with vectors. - -**Finality Property.** Equation (6.43) shows one interesting property of the coefficients of *c*1, *c*2, ..., *cN*: the optimum value of any coefficient in Eq. (6.41) is independent of the number of terms used in the approximation. For example, if we used only one term (*N* = 1) or two terms (*N* = 2) or any number of terms, the optimum value of the coefficient *c*1 would be the same [as given by Eq. (6.43)]. The advantage of this approximation of a signal *x*(*t*) by a set of mutually orthogonal signals is that we can continue to add terms to the approximation without disturbing the previous terms. This property of *finality* of the values of the coefficients is very important from a practical point of view.† - -### ENERGY OF THE ERROR SIGNAL - -When the coefficients *ci* in Eq. (6.41) are chosen according to Eq. (6.43), the error signal energy is minimized. This minimum value of *Ee* is given by Eq. (6.42): - -$$ -E_e = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt -$$ - -= -$$ -\int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 \int_{t_1}^{t_2} x_n^2(t) dt - 2 \sum_{n=1}^{N} c_n \int_{t_1}^{t_2} x(t) x_n(t) dt -$$ - -$$ -x(t_1) = a_0 + a_1t_1 -$$ - and $x(t_2) = a_0 + a_1t_2$ - -Solution of these equations yields the desired values of *a*0 and *a*1. For a three-point approximation, we must choose the polynomial *a*0 +*a*1*t* +*a*2*t* 2 with - -$$ -x(t_i) = a_0 + a_1t_i + a_2t_i^2 -$$ - $i = 1, 2, and 3$ - -The approximation improves with a larger number of points (higher-order polynomial), but the coefficients *a*0, *a*1, *a*2, ... do not have the finality property. Every time we increase the number of terms in the polynomial, we need to recalculate the coefficients. - -† Contrast this situation with a polynomial approximation of *x*(*t*). Suppose we wish to find a two-point approximation of *x*(*t*) by a polynomial in *t*; that is, the polynomial is to be equal to *x*(*t*) at two points *t*1 and *t*2. This can be done by choosing a first-order polynomial *a*0 +*a*1*t* with - -Substitution of Eqs. (6.40) and (6.43) in this equation yields - -$$ -E_e = \int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 E_n - 2 \sum_{n=1}^{N} c_n^2 E_n = \int_{t_1}^{t_2} x^2(t) dt - \sum_{n=1}^{N} c_n^2 E_n -$$ - (6.44) - -Observe that because the term *c*2 *kEk* is nonnegative, the error energy *Ee* generally decreases as *N*, the number of terms, is increased. Hence, it is possible that the error energy →0 as *N* → ∞. When this happens, the orthogonal signal set is said to be *complete*. In this case, Eq. (6.41) is no more an approximation but an equality - -$$ -x(t) = c_1 x_1(t) + c_2 x_2(t) + \dots + c_n x_n(t) + \dots = \sum_{n=1}^{\infty} c_n x_n(t) \qquad t_1 < t < t_2 \tag{6.45} -$$ - -where the coefficients *cn* are given by Eq. (6.43). Because the error signal energy approaches zero, it follows that the energy of *x*(*t*) is now equal to the sum of the energies of its orthogonal components *c*1*x*1(*t*), *c*2*x*2(*t*), *c*3*x*3(*t*), .... - -The series on the right-hand side of Eq. (6.45) is called the *generalized Fourier series* of *x*(*t*) with respect to the set {*xn*(*t*)}. When the set {*xn*(*t*)} is such that the error energy *Ee* → 0 as *N* → ∞ for every member of some particular class, we say that the set {*xn*(*t*)} is complete on (*t*1, *t*2) for that class of *x*(*t*), and the set {*xn*(*t*)} is called a set of *basis functions* or *basis signals*. Unless otherwise mentioned, in the future we shall consider only the class of energy signals. - -Thus, when the set {*xn*(*t*)} is complete, we have the equality of Eq. (6.45). One subtle point that must be understood clearly is the meaning of equality in Eq. (6.45). *The equality here is not an equality in the ordinary sense, but in the sense that the error energy, that is, the energy of the difference between the two sides of Eq. (6.45), approaches zero*. If the equality exists in the ordinary sense, the error energy is always zero, but the converse is not necessarily true. The error energy can approach zero even though *e*(*t*), the difference between the two sides, is nonzero at some isolated instants. The reason is that even if *e*(*t*) is nonzero at such instants, the area under *e*2(*t*) is still zero; thus the Fourier series on the right-hand side of Eq. (6.45) may differ from *x*(*t*) at a finite number of points. - -In Eq. (6.45), the energy of the left-hand side is *Ex*, and the energy of the right-hand side is the sum of the energies of all the orthogonal components.† Thus, - -$$ -\int_{t_1}^{t_2} x^2(t) dt = c_1^2 E_1 + c_2^2 E_2 + \dots = \sum_{n=1}^{\infty} c_n^2 E_n -$$ - (6.46) - -This is *Parseval's theorem* expressed for energy signals. In Eqs. (6.26) and (6.27), we have already encountered Parseval's theorem for power signals. Recall that the signal energy (area under the squared value of a signal) is analogous to the square of the length of a vector in the vector-signal analogy. In vector space, we know that the square of the length of a vector is equal to the sum of - -† Note that the energy of a signal *cx*(*t*) is *c*2*Ex*. - -### 652 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -the squares of the lengths of its orthogonal components. Parseval's theorem of Eq. (6.46) is the statement of this fact as it applies to signals. - -### GENERALIZATION TO COMPLEX SIGNALS - -The foregoing results can be generalized to complex signals as follows: a set of functions *x*1(*t*), *x*2(*t*), ..., *xN*(*t*) is mutually orthogonal over the interval (*t*1, *t*2) if - -$$ -\int_{t_1}^{t_2} x_m(t) x_n^*(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases} -$$ - -If this set is complete for a certain class of functions, then a function *x*(*t*) in this class can be expressed as - -$$ -x(t) = c_1 x_1(t) + c_2 x_2(t) + \cdots + c_i x_i(t) + \cdots -$$ - -where - -$$ -c_n = \frac{1}{E_n} \int_{t_1}^{t_2} x(t) x_n^*(t) dt -$$ -\n(6.47) - -### **EXAMPLE 6.14 Approximating a Square Wave with a Set of Harmonic Sine Waves** - -In Ex. 6.13, the square signal *x*(*t*) in Fig. 6.23 is approximated by a single sinusoid sin *t*. In this example, we approximate *x*(*t*) using the set of harmonic sine waves sin *t*, sin 2*t*, ..., sin *nt*, ..., and see how the approximation improves with the number of terms. - -To begin, we note that the set of harmonic sine waves sin *t*, sin 2*t*,..., sin *nt*,... is orthogonal over any interval of duration 2π. † The reader can verify this fact by showing that for any real number *a*, - -$$ -\int_{a}^{a+2\pi} \sin mt \sin nt dt = \begin{cases} 0 & m \neq n \\ \pi & m = n \end{cases} -$$ - (6.48) - -Using this set, we approximate *x*(*t*) as - -*x*(*t*) *c*1 sin *t* +*c*2 sin 2*t* +···+*cn* sin *Nt* - -† This sine set, along with the cosine set cos 0*t*, cos *t*, cos 2*t*,..., cos*nt*,..., forms a complete set. In this case, however, the coefficients *ci* corresponding to the cosine terms are zero. For this reason, we have omitted cosine terms in this example. This composite sine and cosine set is the basis set for the trigonometric Fourier series. - -**Figure 6.25** Approximation of a square wave by a sum of harmonic sinusoids. - -Therefore, - -$$ -x(t) \simeq \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \dots + \frac{1}{N} \sin Nt \right) -$$ - (6.49) - -Note that coefficients of terms sin*kt* are zero for even values of *k*. Figure 6.25 shows how the approximation improves as we increase the number of terms in the series. - -Let us investigate the error signal energy as *N* → ∞. From Eq. (6.44), - -$$ -E_e = \int_0^{2\pi} x^2(t) dt - \sum_{n=1}^{\infty} c_n^2 E_n -$$ - -Note that - -$$ -\int_0^{2\pi} x^2(t) dt = \int_0^{\pi} 1^2 dt + \int_{\pi}^{2\pi} -1^2 dt = 2\pi -$$ -$$ -c_n^2 = \begin{cases} \frac{16}{n^2 \pi^2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases} -$$ - -and from Eq. (6.48), - -$$ -E_n=\pi -$$ - -Therefore, - -$$ -E_e = 2\pi - \sum_{n=1,3,5,\dots}^{N} \frac{16}{n^2 \pi^2} \pi = 2\pi - \frac{16}{\pi} \sum_{n=1,3,5,\dots}^{N} \frac{1}{n^2} -$$ - -For a single-term approximation (*N* = 1), - -$$ -E_e = 2\pi - \frac{16}{\pi} = 1.1938 -$$ - -For a two-term approximation (*N* = 3), - -$$ -E_e = 2\pi - \frac{16}{\pi} \left( 1 + \frac{1}{9} \right) = 0.6243 -$$ - -Continuing this process, we compute the error energy *Ee* for various values of *N* as - -| N | 1 | 3 | 5 | 7 | 99 | $\infty$ | -|-------|--------|--------|--------|--------|---------|----------| -| $E_e$ | 1.1938 | 0.6243 | 0.4206 | 0.3166 | 0.02545 | 0 | - -Clearly, *x*(*t*) can be represented by the infinite series - -$$ -x(t) = \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \cdots \right) = \frac{4}{\pi} \sum_{n=1,3,5,\dots}^{\infty} \frac{1}{n} \sin nt -$$ - -The equality exists in the sense that the error signal energy → 0 as *N* → ∞. In this case, the error energy decreases rather slowly with *N*, indicating that the series converges slowly. This is to be expected because *x*(*t*) has jump discontinuities and consequently, according to discussion in Sec. 6.2-2, the series converges asymptotically as 1/*n*. - -### **DR ILL 6.9 Approximating a Ramp Signal with a Set of Harmonic Sine Waves** - -Approximate the signal *x*(*t*) = *t* − π (Fig. 6.26) over the interval (0, 2π ) in terms of the set of sinusoids {sin *nt*}, *n* = 0, 1, 2,..., used in Ex. 6.14. Find *Ee*, the error energy. Show that *Ee* → 0 as *N* → ∞. - -4π - -### **ANSWERS** - -%*N* - -1 - -*x*(*t*) −2 *n*=1 *N* sin *nt* and *Ee* = 2 3π3 −%*N n*=1 *n*2 *x*(*t*) *p p p* 2*p t* **Figure 6.26** Ramp signal for Drill 6.9. - -### SOME EXAMPLES OF GENERALIZED FOURIER SERIES - -Signals are vectors in every sense. Like a vector, a signal can be represented as a sum of its components in a variety of ways. Just as vector coordinate systems are formed by mutually orthogonal vectors (rectangular, cylindrical, spherical), we also have signal coordinate systems (basis signals) formed by a variety of sets of mutually orthogonal signals. There exist a large number of orthogonal signal sets that can be used as basis signals for generalized Fourier series. Some well-known signal sets are trigonometric (sinusoid) functions, exponential functions, Walsh functions, Bessel functions, Legendre polynomials, Laguerre functions, Jacobi polynomials, Hermite polynomials, and Chebyshev polynomials. The functions that concern us most in this book are the trigonometric and the exponential sets discussed earlier in this chapter. - -### LEGENDRE FOURIER SERIES - -A set of Legendre polynomials *Pn*(*t*) (*n* = 0, 1, 2, 3,...) forms a complete set of mutually orthogonal functions over an interval (−1 < *t* < 1). These polynomials can be defined by the Rodrigues formula: - -$$ -P_n(t) = \frac{1}{2^n n!} \frac{d^n}{dt^n} (t^2 - 1)^n \qquad n = 0, 1, 2, \dots -$$ - -It follows from this equation that - -$$ -P_0(t) = 1, P_1(t) = t, P_2(t) = \left(\frac{3}{2}t^2 - \frac{1}{2}\right), P_3(t) = \left(\frac{5}{2}t^3 - \frac{3}{2}t\right) -$$ -, and so on - -We may verify the orthogonality of these polynomials by showing that - -$$ -\int_{-1}^{1} P_m(t) P_n(t) dt = \begin{cases} 0 & m \neq n \\ \frac{2}{2m+1} & m = n \end{cases} -$$ - -We can express a function *x*(*t*) in terms of Legendre polynomials over an interval (−1 < *t* < 1) as - -$$ -x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots -$$ -\n(6.50) - -where - -$$ -c_r = \frac{\int_{-1}^{1} x(t)P_r(t) dt}{\int_{-1}^{1} P_r^2(t) dt} = \frac{2r+1}{2} \int_{-1}^{1} x(t)P_r(t) dt -$$ -\n(6.51) - -Note that although the series representation is valid over the interval (−1, 1), it can be extended to any interval by the appropriate time scaling (see Prob. 6.5-8). - -### **EXAMPLE 6.15 Legendre Fourier Series** - -Determine the Legendre Fourier series of the square signal shown in Fig. 6.27. - -**Figure 6.27** Square signal for Ex. 6.15. - -From Eq. (6.50), we know that the Legendre Fourier series takes the form - -$$ -x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots -$$ - -The coefficients *c*0, *c*1, *c*2,..., *cr* may be found from Eq. (6.51). We have - -$$ -x(t) = \begin{cases} 1 & \cdots -1 < t < 0 \\ -1 & \cdots 0 < t < 1 \end{cases} -$$ - -and - -$$ -c_0 = \frac{1}{2} \int_{-1}^{1} x(t) dt = 0 -$$ - -\n -$$ -c_1 = \frac{3}{2} \int_{-1}^{1} tx(t) dt = \frac{3}{2} \left( \int_{-1}^{0} t dt - \int_{0}^{1} t dt \right) = -\frac{3}{2} -$$ - -\n -$$ -c_2 = \frac{5}{2} \int_{-1}^{1} x(t) \left( \frac{3}{2} t^2 - \frac{1}{2} \right) dt = 0 -$$ - -This result follows immediately from the fact that the integrand is an odd function of *t*. In fact, this is true of all *cr* for even values of *r*, that is, - -$$ -c_0 = c_2 = c_4 = c_6 = \cdots = 0 -$$ - -Also, - -$$ -c_3 = \frac{7}{2} \int_{-1}^{1} x(t) \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt = \frac{7}{2} \left[ \int_{-1}^{0} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt - \int_{0}^{1} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt \right] = \frac{7}{8} -$$ - -In a similar way, coefficients *c*5, *c*7,... can be evaluated. We now have - -$$ -x(t) = -\frac{3}{2}t + \frac{7}{8}\left(\frac{5}{2}t^3 - \frac{3}{2}t\right) + \cdots -$$ - -### TRIGONOMETRIC FOURIER SERIES - -We have already proved [see Eqs. (6.4), (6.5), and (6.6)] that the trigonometric signal set - -{1, cosω0*t*, cos 2ω0*t*, ..., cos*n*ω0*t*, ...; sinω0*t*, sin 2ω0*t*, ..., sin*n*ω0*t*, ...} - -is orthogonal over any interval of duration *T*0, where *T*0 = 1/*f*0 is the period of the sinusoid of frequency *f*0. This is a complete set for a class of signals with finite energies [11, 12]. Therefore, we can express a signal *x*(*t*) by a trigonometric Fourier series over any interval of duration *T*0 seconds as - -$$ -x(t) = a_0 + a_1 \cos \omega_0 t + a_2 \cos 2\omega_0 t + \cdots -$$ -$$ -+ b_1 \sin \omega_0 t + b_2 \sin 2\omega_0 t + \cdots -$$ - -or - -$$ -x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \qquad t_1 < t < t_1 + T_0 -$$ - -where - -$$ -\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0} -$$ - -We can use Eq. (6.43) to determine the Fourier coefficients *a*0, *an*, and *bn*. Thus, - -$$ -a_n = \frac{\int_{t_1}^{t_1 + T_0} x(t) \cos n\omega_0 t \, dt}{\int_{t_1}^{t_1 + T_0} \cos^2 n\omega_0 t \, dt} -$$ -\n(6.52) - -The integral in the denominator of Eq. (6.52) has already been found to be *T*0/2 when *n* = 0 [Eq. (6.4) with *m* = *n*]. For *n* = 0, the denominator is *T*0. Hence, - -$$ -a_0 = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) dt \quad \text{and} \quad a_n = \frac{2}{T_0} \int_{t_1}^{t_1+T_0} x(t) \cos n\omega_0 t dt \qquad n = 1, 2, 3, \dots \tag{6.53} -$$ - -Similarly, we find that - -$$ -b_n = \frac{2}{T_0} \int_{t_1}^{t_1 + T_0} x(t) \sin n\omega_0 t \, dt \qquad n = 1, 2, 3, \dots \tag{6.54} -$$ - -Note that the Fourier series in Eq. (6.49) of Ex. 6.14 is indeed the trigonometric Fourier series with *T*0 = 2π and ω0 = 2π/*T*0. In this particular example, it is easy to verify from Eq. (6.53) that *an* = 0 for all *n*, including *n* = 0. Hence, the Fourier series in that example consisted only of sine terms. - -### EXPONENTIAL FOURIER SERIES - -As shown in the footnote on page 622, the set of exponentials *ejn*ω0*t* (*n* = 0,±1,±2,...) is a set of functions orthogonal over any interval of duration *T*0 = 2π/ω0. An arbitrary signal *x*(*t*) can now be expressed over an interval (*t*1,*t*1 +*T*0) as - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad t_1 < t < t_1 + T_0 -$$ - -where [see Eq. (6.47)] - -$$ -D_n = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) e^{-jn\omega_0 t} dt -$$ - -### WHY USE THE EXPONENTIAL SET? - -If *x*(*t*) can be represented in terms of hundreds of different orthogonal sets, why do we exclusively use the exponential (or trigonometric) set for the representation of signals or LTI systems? It so - -happens that the exponential signal is an eigenfunction of LTI systems. In other words, for an LTI system, only an exponential input *est* yields the response that is also an exponential of the same form, given by *H*(*s*)*est*. The same is true of the trigonometric set. This fact makes the use of exponential signals natural for LTI systems in the sense that the system analysis using exponentials as the basis signals is greatly simplified. - -## **6.6 NUMERICAL [COMPUTATION OF](#page-12-0)** *Dn* - -We can compute *Dn* numerically by using the DFT (the discrete Fourier transform discussed in Sec. 8.5), which uses the samples of a periodic signal *x*(*t*) over one period. The sampling interval is *T* seconds. Hence, there are *N*0 =*T*0/*T* number of samples in one period *T*0. To find the relationship between *Dn* and the samples of *x*(*t*), consider Eq. (6.19) and write - -$$ -D_n = \frac{1}{T_0} \int_{T_0} x(t)e^{-jn\omega_0 t} dt -$$ - -= -$$ -\lim_{T \to 0} \frac{1}{N_0 T} \sum_{k=0}^{N_0 - 1} x(kT)e^{-jn\omega_0 kT} T -$$ - -= -$$ -\lim_{T \to 0} \frac{1}{N_0} \sum_{k=0}^{N_0 - 1} x(kT)e^{-jn\Omega_0 k} -$$ -(6.55) - -where *x*(*kT*) is the *k*th sample of *x*(*t*) and - -$$ -N_0 = \frac{T_0}{T} \quad \text{and} \quad \Omega_0 = \omega_0 T = \frac{2\pi}{N_0} -$$ - -In practice, it is impossible to make *T* → 0 in computing the right-hand side of Eq. (6.55). We can make *T* small, but not zero, which will cause the data to increase without limit. Thus, we shall ignore the limit on *T* in Eq. (6.55) with the implicit understanding that *T* is reasonably small. Nonzero *T* will result in some computational error, which is inevitable in any numerical evaluation of an integral. The error resulting from nonzero *T* is called the *aliasing error,* which is discussed in more detail in Ch. 8. Thus, we can express Eq. (6.55) as - -$$ -D_n \approx \frac{1}{N_0} \sum_{k=0}^{N_0 - 1} x(kT) e^{-jn\Omega_0 k} -$$ - (6.56) - -Since 0*N*0 = 2π, we know that *ejn*0(*k*+*N*0) = *ejn*0*k*, and it follows that - -$$ -D_{n+N_0}=D_n -$$ - -The periodicity property *Dn*+*N*0 = *Dn* means that beyond *n* = *N*0/2, the coefficients represent the values for negative *n*. For instance, when *N*0 = 32, *D*17 = *D*−15, *D*18 = *D*−14,...,*D*31 = *D*−1. The cycle repeats again from *n* = 32 on. - -We can use the efficient FFT (the *fast Fourier transform* discussed in Sec. 8.6) to compute the right-hand side of Eq. (6.56). We shall use MATLAB to implement the FFT algorithm. For - -### 660 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -this purpose, we need samples of *x*(*t*) over one period starting at *t* = 0. In this algorithm, it is also preferable (although not necessary) that *N*0 be a power of 2, (i.e., *N*0 = 2*m* where *m* is an integer). - -### **EXAMPLE 6.16 Numerical Computation of Fourier Spectra** - -Numerically compute and then plot the exponential Fourier spectra for the periodic signal in Fig. 6.2a (Ex. 6.1). - -The samples of *x*(*t*) start at *t* = 0 and the last (*N*0th) sample is at *t* = *T*0 − *T*. At the points of discontinuity, the sample value is taken as the average of the values of the function on two sides of the discontinuity. Thus, the sample at *t* = 0 is not 1 but (*e*−π/2 +1)/2 = 0.604. To determine *N*0, we require that *Dn* for *n* ≥ *N*0/2 be negligible. Because *x*(*t*) has a jump discontinuity, *Dn* decays rather slowly as 1/*n*. Hence, a choice of *N*0 = 200 is acceptable because the (*N*0/2)nd (100th) harmonic is about 1% of the fundamental. However, we also require *N*0 to be a power of 2. Hence, we shall take *N*0 = 256 = 28. - -First, the basic parameters are established. - ->> T\_0 = pi; N\_0 = 256; T = T\_0/N\_0; t = (0:T:T\*(N\_0-1))'; >> x = exp(-t/2); x(1) = (exp(-pi/2)+1)/2; - -Next, the DFT, computed by means of the fft function, is used to approximate the exponential Fourier spectra up to *n* = *N*0/2. To facilitate comparison with previous plots of *Dn*, we only plot the results over −5 ≤ *n* ≤ 5. - -``` ->> D_n = fft(x)/N_0; n = [-N_0/2:N_0/2-1]'; ->> clf; subplot(1,2,1); stem(n,abs(fftshift(D_n)),'.k'); ->> axis([-5 5 0 .6]); xlabel('n'); ylabel('|D_n|'); ->> subplot(1,2,2); stem(n,angle(fftshift(D_n)),'.k'); ->> axis([-5 5 -2 2]); xlabel('n'); ylabel('\angle D_n [rad]'); -``` - -As shown in Fig. 6.28, the resulting approximation is visually indistinguishable from the true Fourier series spectra shown in Fig. 6.12 or Fig. 6.13. - -## **[6.7 MATLAB: FOURIER](#page-12-0) SERIES APPLICATIONS** - -Computational packages such as MATLAB simplify the Fourier-based analysis, design, and synthesis of periodic signals. MATLAB permits rapid and sophisticated calculations, which promote practical application and intuitive understanding of the Fourier series. - -### **[6.7-1 Periodic Functions and the Gibbs Phenomenon](#page-12-0)** - -It is sufficient to define any *T*0-periodic function over the interval (0 ≤ *t* < *T*0). For example, consider the 2π-periodic function given by - -$$ -x(t) = \begin{cases} t/A & 0 \le t < A \\ 1 & A \le t < \pi \\ 0 & \pi \le t < 2\pi \\ x(t + 2\pi) & \text{otherwise} \end{cases} -$$ - -Although similar to a square wave, *x*(*t*) has a linearly rising edge of width *A*, where (0 < *A* < π). As *A* → 0, *x*(*t*) approaches a square wave; as *A* → π, *x*(*t*) approaches a type of sawtooth wave. - -In MATLAB, the mod command helps represent periodic functions such as *x*(*t*). - ->> x = @(t,A) mod(t,2\*pi)/A.\*(mod(t,2\*pi)=A)&(mod(t,2\*pi)n* = *D*∗ *n*. Truncating the Fourier series at |*n*| = *N* yields the approximation - -$$ -x(t) \approx x_N(t) = D_0 + \sum_{n=1}^{N} \left( D_n e^{jnt} + D_n^* e^{-jnt} \right) -$$ -\n(6.57) - -For a user-specified *N*, program CH6MP1 uses Eq. (6.57) to compute *xN*(*t*) over (−π/4 ≤ *t* < 2π +π/4). - -``` -function [x_N,t] = CH6MP1(A,N); -% CH6MP1.m : Chapter 6, MATLAB Program 1 -% Function M-file approximates x(t) using Fourier series truncated at |n|=N -% INPUTS: A = width of rising edge -% N = largest harmonic of truncated Fourier series -% OUTPUTS: x_N = Nth harmonic truncated Fourier series -``` - -``` -% t = time vector for x_N -``` - -``` -% Define FS coefficients for signal x(t) -D = @(n) 1/(2*pi*n)*((exp(-1j*n*A)-1)/(n*A) + 1j*exp(-1j*n*pi)); -% Construct truncated FS approximation of x(t) using N harmonics -t = linspace(-pi/4,2*pi+pi/4,10000); % Time vector exceeds one period. -x_N = (2*pi-A)/(4*pi)*ones(size(t)); % Compute dc term -for n = 1:N, % Compute N remaining terms - x_N = x_N+real(D(n)*exp(1j*n*t) + conj(D(n))*exp(-1j*n*t)); -end -``` - -Although theoretically not required, the real command ensures that small computer round-off errors do not cause a complex-valued result. - -Using program CH6MP1 with *A* = π/2 and *N* = 20, Fig. 6.29 compares *x*(*t*) and *x*20(*t*). - ->> A = pi/2; [x\_20,t] = CH6MP1(A,20); >> plot(t,x\_20,'k',t,x(t,A),'k:'); axis([-pi/4,2\*pi+pi/4,-0.1,1.1]); >> xlabel('t'); ylabel('x\_{20}(t)'); - -As expected, the falling edge is accompanied by the overshoot that is characteristic of the Gibbs phenomenon. - -Increasing *N* to 100, as shown in Fig. 6.30, improves the approximation but does not reduce the overshoot. - -``` ->> [x_100,t] = CH6MP1(A,100); -``` - ->> plot(t,x\_100,'k',t,x(t,A),'k:'); axis([-pi/4,2\*pi+pi/4,-0.1,1.1]); - -``` ->> xlabel('t'); ylabel('x_{100}(t)'); -``` - -Reducing *A* to π/64 produces a curious result. For *N* = 20, both the rising and falling edges are accompanied by roughly 9% of overshoot, as shown in Fig. 6.31. As the number of terms is increased, overshoot persists only in the vicinity of jump discontinuities. For *xN*(*t*), increasing *N* decreases the overshoot near the rising edge but not near the falling edge. Remember that it is a - -**Figure 6.29** Comparison of *x*20(*t*) and *x*(*t*) when *A* = π/2. - -**Figure 6.30** Comparison of *x*100(*t*) and *x*(*t*) when *A* = π/2. - -**Figure 6.31** Comparison of *x*20(*t*) and *x*(*t*) when *A* = π/64. - -**Figure 6.32** Comparison of *x*100(*t*) and *x*(*t*) when *A* = π/64. - -true jump discontinuity that causes the Gibbs phenomenon. A continuous signal, no matter how sharply it rises, can always be represented by a Fourier series at every point within any small error by increasing *N*. This is not the case when a true jump discontinuity is present. Figure 6.32 illustrates this behavior using *N* = 100. - -### **[6.7-2 Optimization and Phase Spectra](#page-12-0)** - -Although magnitude spectra typically receive the most attention, phase spectra are critically important in some applications. Consider the problem of characterizing the frequency response of an unknown system. By applying sinusoids one at a time, the frequency response is empirically measured one point at a time. This process is tedious at best. Applying a superposition of many sinusoids, however, allows simultaneous measurement of many points of the frequency response. Such measurements can be taken by a spectrum analyzer equipped with a transfer function mode or by applying Fourier analysis techniques, which are discussed in later chapters. - -A multitone test signal *m*(*t*) is constructed as a superposition of *N* real sinusoids - -$$ -m(t) = \sum_{n=1}^{N} M_n \cos{(\omega_n t + \theta_n)} -$$ - -where *Mn* and θ*n* establish the relative magnitude and phase of each sinusoidal component. It is sensible to constrain all gains to be equal, *Mn* = *M* for all *n*. This ensures equal treatment at each point of the measured frequency response. Although the value *M* is normally chosen to set the desired signal power, we set *M* = 1 for convenience. - -While not required, it is also sensible to space the sinusoidal components uniformly in frequency. - -$$ -m(t) = \sum_{n=1}^{N} \cos(n\omega_0 t + \theta_n) -$$ -\n(6.58) - -Another sensible alternative, which spaces components logarithmically in frequency, is treated in Prob. 6.7-4. - -Equation (6.58) is now a truncated compact-form Fourier series with a flat magnitude spectrum. Frequency resolution and range are set by ω0 and *N*, respectively. For example, a 2 kHz range with a resolution of 100 Hz requires ω0 = 2π100 and *N* = 20. The only remaining unknowns are the θ*n*. - -While it is tempting to set θ*n* = 0 for all *n*, the results are quite unsatisfactory. MATLAB helps demonstrate the problem by using ω0 = 2π100 and *N* = 20 sinusoids, each with a peak-to-peak voltage of 1 volt. - -``` ->> m = @(theta,t,omega) sum(cos(omega*t+theta*ones(size(t)))); ->> N = 20; omega = 2*pi*100*[1:N]'; theta = zeros(size(omega)); ->> t = linspace(-0.01,0.01,10000); ->> plot(t,m(theta,t,omega),'k'); xlabel('t [sec]'); ylabel('m(t) [volts]'); -``` - -As shown in Fig. 6.33, θ*n* = 0 causes each sinusoid to constructively add. The resulting 20 volt peak can saturate system components, such as operational amplifiers operating with ±12 volt rails. To improve signal performance, the maximum amplitude of *m*(*t*) over *t* needs to be reduced. - -One way to reduce max*t*(|*m*(*t*)|) is to reduce *M*, the strength of each component. Unfortunately, this approach reduces the system's signal-to-noise ratio and ultimately degrades measurement quality. Therefore, reducing *M* is not a smart decision. The phases θ*n*, however, can be adjusted to reduce max*t*(|*m*(*t*)|) while preserving signal power. In fact, since θ*n* = 0 maximizes max*t*(|*m*(*t*)|), just about any other choice of θ*n* will improve the situation. Even a random choice should improve performance. - -**Figure 6.33** Test signal *m*(*t*) with θ*n* = 0. - -As with any computer, MATLAB cannot generate truly random numbers. Rather, it generates pseudo-random numbers. Pseudo-random numbers are deterministic sequences that appear to be random. The particular sequence of numbers that is realized depends entirely on the initial state of the pseudo-random number generator. Setting the generator's initial state to a known value allows a "random" experiment with reproducible results. The command rng(0) initializes the state of the pseudo-random number generator to a known condition of zero, and the MATLAB command rand(a,b) generates an a-by-b matrix of pseudo-random numbers that are uniformly distributed over the interval (0, 1). Radian phases occupy the wider interval (0, 2π ), so the results from rand need to be appropriately scaled. - ->> rng(0); theta\_rand0 = 2\*pi\*rand(N,1); - -Next, we recompute and plot *m*(*t*) using the randomly chosen θ*n*. - -``` ->> m_rand0 = m(theta_rand0,t,omega); -``` - -``` ->> plot(t,m_rand0,'k'); axis([-0.01,0.01,-10,10]); -``` - -``` ->> xlabel('t [sec]'); ylabel('m(t) [volts]'); -``` - ->> set(gca,'ytick',[min(m\_rand0),max(m\_rand0)]); grid on; - -For a vector input, the min and max commands return the minimum and maximum values of the vector. Using these values to set *y* axis tick marks makes it easy to identify the extreme values of the *m*(*t*). As seen from Fig. 6.34, the maximum amplitude is now 7.6307, which is significantly smaller than the maximum of 20 when θ*n* = 0. - -Randomly chosen phases suffer a fatal fault: there is little guarantee of optimal performance. For example, repeating the experiment with rng(5) produces a maximum magnitude of 8.2399 volts, as shown in Fig. 6.35. This value is significantly higher than the previous maximum of 7.6307 volts. Clearly, it is better to replace a random solution with an optimal solution. - -What constitutes "optimal"? Many choices exist, but desired signal criteria naturally suggest that optimal phases minimize the maximum magnitude of *m*(*t*) over all *t*. To find these optimal phases, MATLAB's fminsearch command is useful. First, the function to be minimized, called the objective function, is defined. - ->> maxmagm = @(theta,t,omega) max(abs(sum(cos(omega\*t+theta\*ones(size(t)))))); - -666 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -**Figure 6.34** Test signal *m*(*t*) with random θ*n* found by using rng(0). - -**Figure 6.35** Test signal *m*(*t*) with random θ*n* found by using rand('state',1). - -The anonymous function argument order is important; fminsearch uses the first input argument as the variable of minimization. To minimize over θ, as desired, θ must be the first argument of the objective function maxmagm. - -Next, the time vector is shortened to include only one period of *m*(*t*). - ->> t = linspace(0,0.01,401); - -A full period ensures that all values of *m*(*t*) are considered; the short length of t helps ensure that functions execute quickly. An initial value of θ is randomly chosen to begin the search. - -``` ->> rng(0); theta_init = 2*pi*rand(N,1); ->> theta_opt = fminsearch(maxmagm,theta_init,[],t,omega); -``` - -Notice that fminsearch finds the minimizer to maxmagm over θ by using an initial value theta\_init. Most numerical minimization techniques are capable of finding only local minima, and fminsearch is no exception. As a result, fminsearch does not always produce a unique solution. The empty square brackets indicate no special options are requested, and the remaining ordered arguments are secondary inputs for the objective function. Full format details for fminsearch are available from MATLAB's help facilities. - -**Figure 6.36** Test signal *m*(*t*) with optimized phases. - -Figure 6.36 shows the phase-optimized test signal. The maximum magnitude is reduced to a value of 5.3632 volts, which is a significant improvement over the original peak of 20 volts. - -Although the signals shown in Figs. 6.33 through 6.36 look different, they all possess the same magnitude spectra. The signals differ only in phase spectra. It is interesting to investigate the similarities and differences of these signals in ways other than graphs and mathematics. For example, is there an audible difference between the signals? For computers equipped with sound capability, the MATLAB sound command can be used to find out. - -``` ->> Fs = 8000; t = [0:1/Fs:2]; % Two second records at a sampling rate of 8kHz ->> sound(m(theta,t,omega)/20,Fs); % Play (scaled) m(t) constructed using zero phases -``` - -Since the sound command clips magnitudes that exceed 1, the input vector is scaled by 1/20 to avoid clipping and the resulting sound distortion. The signals using other phase assignments are created and played in a similar fashion. How well does the human ear discern the differences in phase spectra? If you are like most people, you will not be able to discern any differences in how these waveforms sound. - -## **[6.8 SUMMARY](#page-12-0)** - -In this chapter we showed how a periodic signal can be represented as a sum of sinusoids or exponentials. If the frequency of a periodic signal is *f*0, then it can be expressed as a weighted sum of a sinusoid of frequency *f*0 and its harmonics (the trigonometric Fourier series). We can reconstruct the periodic signal from a knowledge of the amplitudes and phases of these sinusoidal components (amplitude and phase spectra). - -If a periodic signal *x*(*t*) has an even symmetry, its Fourier series contains only cosine terms (including dc). In contrast, if *x*(*t*) has an odd symmetry, its Fourier series contains only sine terms. If *x*(*t*) has neither type of symmetry, its Fourier series contains both sine and cosine terms. - -At points of discontinuity, the Fourier series for *x*(*t*) converges to the mean of the values of *x*(*t*) on either side of the discontinuity. For signals with discontinuities, the Fourier series converges in the mean and exhibits Gibbs phenomenon at the points of discontinuity. The amplitude spectrum of the Fourier series for a periodic signal *x*(*t*) with jump discontinuities decays slowly (as 1/*n*) with frequency. We need a large number of terms in the Fourier series to approximate *x*(*t*) within - -### 668 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES - -a given error. In contrast, the amplitude spectrum of a smoother periodic signal decays faster with frequency and we require a smaller number of terms in the series to approximate *x*(*t*) within a given error. - -A sinusoid can be expressed in terms of exponentials. Therefore, the Fourier series of a periodic signal can also be expressed as a sum of exponentials (the exponential Fourier series). The exponential form of the Fourier series and the expressions for the series coefficients are more compact than those of the trigonometric Fourier series. Also, the response of LTIC systems to an exponential input is much simpler than that for a sinusoidal input. Moreover, the exponential form of representation lends itself better to mathematical manipulations than does the trigonometric form. This includes the establishment of useful Fourier series properties that simplify work and help provide a more intuitive understanding of signals. For these reasons, the exponential form of the series is preferred in modern practice in the areas of signals and systems. - -The plots of amplitudes and angles of various exponential components of the Fourier series as functions of the frequency are the exponential Fourier spectra (amplitude and angle spectra) of the signal. Because a sinusoid cosω0*t* can be represented as a sum of two exponentials, *ej*ω0*t* and *e*−*j*ω0*t* , the frequencies in the exponential spectra range from ω = −∞ to ∞. By definition, frequency of a signal is always a positive quantity. Presence of a spectral component of a negative frequency −*n*ω0 merely indicates that the Fourier series contains terms of the form *e*−*jn*ω0*t* . The spectra of the trigonometric and exponential Fourier series are closely related, and one can be found by the inspection of the other. - -In Sec. 6.5 we discuss a method of representing signals by the generalized Fourier series, of which the trigonometric and exponential Fourier series are special cases. Signals are vectors in every sense. Just as a vector can be represented as a sum of its components in a variety of ways, depending on the choice of the coordinate system, a signal can be represented as a sum of its components in a variety of ways, of which the trigonometric and exponential Fourier series are only two examples. Just as we have vector coordinate systems formed by mutually orthogonal vectors, we also have signal coordinate systems (basis signals) formed by mutually orthogonal signals. Any signal in this signal space can be represented as a sum of the basis signals. Each set of basis signals yields a particular Fourier series representation of the signal. The signal is equal to its Fourier series, not in the ordinary sense, but in the special sense that the energy of the difference between the signal and its Fourier series approaches zero. This allows for the signal to differ from its Fourier series at some isolated points. - -### **[REFERENCES](#page-12-0)** - -- 1. Bell, E. T. *Men of Mathematics.* Simon & Schuster, New York, 1937. -- 2. Durant, W., and Durant, A. *The Age of Napoleon,* Part XI in *The Story of Civilization Series.* Simon & Schuster, New York, 1975. -- 3. Calinger, R. *Classics of Mathematics,* 4th ed. Moore Publishing, Oak Park, IL, 1982. -- 4. Lanczos, C. *Discourse on Fourier Series*. Oliver Boyd, London, 1966. -- 5. Körner, T. W. *Fourier Analysis.* Cambridge University Press, Cambridge, UK, 1989. -- 6. Guillemin, E. A. *Theory of Linear Physical Systems.* Wiley, New York, 1963. -- 7. Gibbs, W. J. *Nature,* vol. 59, p. 606, April 1899. -- 8. Bôcher, M. *Annals of Mathematics,* vol. 7, no. 2, 1906. - -- 9. Carslaw, H. S. *Bulletin of the American Mathematical Society,* vol. 31, pp. 420–424, October 1925. -- 10. Lathi, B. P. *Signals, Systems, and Communication.* Wiley, New York, 1965. -- 11. Walker P. L. *The Theory of Fourier Series and Integrals.* Wiley, New York, 1986. -- 12. Churchill, R. V., and Brown, J. W. *Fourier Series and Boundary Value Problems,* 3rd ed. McGraw-Hill, New York, 1978. - -## **[PROBLEMS](#page-12-0)** - -- **6.1-1** For each of the periodic signals shown in Fig. P6.1-1, find the compact trigonometric Fourier series and sketch the amplitude and phase spectra. If either the sine or cosine terms are absent in the Fourier series, explain why. -- **6.1-2** (a) Find the trigonometric Fourier series for *y*(*t*) shown in Fig. P6.1-2. - - (b) The signal *y*(*t*) can be obtained by time reversal of *x*(*t*) shown in Fig. 6.2a. Use this fact to obtain the Fourier series for *y*(*t*) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a). - - (c) Show that, in general, time reversal of a periodic signal does not affect the amplitude spectrum, and the phase spectrum is also unchanged except for the change of sign. -- **6.1-3** (a) Find the trigonometric Fourier series for the periodic signal *y*(*t*) depicted in Fig. P6.1-3. - - (b) The signal *y*(*t*) can be obtained by time compression of *x*(*t*) shown in Fig. 6.2a by a factor 2. Use this fact to obtain the Fourier series for *y*(*t*) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a). - - (c) Show that, in general, time compression of a periodic signal by a factor *a* expands the Fourier spectra along the ω axis by the same factor *a*. In other words *C*0,*Cn*, and θ*n* remain unchanged, but the fundamental frequency is increased by the factor *a*, thus expanding the spectrum. Similarly, time expansion of a periodic signal by a factor *a* compresses its Fourier spectra along the ω axis by the factor *a*. -- **6.1-4** (a) Find the trigonometric Fourier series for the periodic signal *g*(*t*) in Fig. P6.1-4. Take advantage of the symmetry. - -- (b) Observe that *g*(*t*) is identical to *x*(*t*) in Fig. 6.4a left-shifted by 0.5 second. Use this fact to obtain the Fourier series for *g*(*t*) from the results in Ex. 6.2. Verify that the Fourier series thus obtained is identical to that found in part (a). -- (c) Show that, in general, a time shift of *T* seconds of a periodic signal does not affect the amplitude spectrum. However, the phase of the *n*th harmonic is increased or decreased *n*ω0*T* depending on whether the signal is advanced or delayed by *T* seconds. -- **6.1-5** Determine the trigonometric Fourier series coefficients *an* and *bn* for the following signals. In each case, also determine the signals' fundamental radian frequency ω0. No integration is required to solve this problem. - - (a) *x*a(*t*) = cos(3π*t*) - - (b) *x*b(*t*) = sin(7π*t*) - - (c) *x*c(*t*) = 2+4 cos(3π*t*)−2*j*sin(7π*t*) - - (d) *x*d(*t*) = (1+*j*)sin(3π*t*)+(2−*j*) cos(7π*t*) - - (e) *x*e(*t*) = sin(3π*t* +1)+2 cos(7π*t* −2) - - (f) *x*f(*t*) = sin(6π*t*)+2 cos(14π*t*) -- **6.1-6** If the two halves of one period of a periodic signal are identical in shape except that one is the negative of the other, the periodic signal is said to have a *half-wave symmetry*. If a periodic signal *x*(*t*) with a period *T*0 satisfies the half-wave symmetry condition, then - -$$ -x\left(t - \frac{T_0}{2}\right) = -x(t) -$$ - -In this case, show that all the even-numbered harmonics vanish and that the odd-numbered harmonic coefficients are given by - -$$ -a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t \, dt -$$ - -**Figure P6.1-2** - -(b) - -**Figure P6.1-6** - -and - -$$ -b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t \, dt -$$ - -Using these results, find the Fourier series for the periodic signals in Fig. P6.1-6. - -**6.1-7** Over a finite interval, a signal can be represented by more than one trigonometric (or exponential) Fourier series. For instance, if we wish to represent *x*(*t*) = *t* over an interval 0 < *t* < 1 by a Fourier series with fundamental frequency ω0 = 2, we can draw a pulse *x*(*t*) = *t* over the interval 0 < *t* < 1 and repeat the pulse every π seconds so that *T*0 = π and ω0 = 2 (Fig. P6.1-7a). If we want the fundamental frequency ω0 to be 4, we repeat the pulse every π/2 seconds. If we want the series to contain only cosine terms with ω0 = 2, we construct a pulse *x*(*t*) = |*t*| over −1 < *t* < 1, and repeat it every π seconds (Fig. P6.1-7b). The resulting signal is an even function with period π. Hence, its Fourier series will have only cosine terms with ω0 = 2. The resulting Fourier series represents *x*(*t*) = *t* over 0 < *t* < 1, as desired. We do not care what it represents outside this interval. - -Sketch the periodic signal *x*(*t*) such that *x*(*t*) = *t* for 0 < *t* < 1 and the Fourier series for *x*(*t*) satisfies the following conditions. - -- (a) ω0 = π/2 and contains all harmonics, but cosine terms only -- (b) ω0 = 2 and contains all harmonics, but sine terms only -- (c) ω0 = π/2 and contains all harmonics, which are exclusively neither sine nor cosine -- (d) ω0 = 1 and contains only odd harmonics and cosine terms -- (e) ω0 = π/2 and contains only odd harmonics and sine terms - -**Figure P6.1-7** - -(f) ω0 = 1 and contains only odd harmonics, which are exclusively neither sine nor cosine. - -[*Hint:* For parts (d), (e), and (f), you need to use half-wave symmetry discussed in Prob. 6.1-6. Cosine terms imply a possible dc component.] You are asked only to sketch the periodic signal *x*(*t*) satisfying the given conditions. Do not find the values of the Fourier coefficients. - -- **6.1-8** State with reasons whether the following signals are periodic or aperiodic. For periodic signals, find the period and state which harmonics are present in the series. - - (a) 3 sin *t* +2 sin 3*t* - - (b) 2+5 sin 4*t* +4 cos 7*t* - - (c) 2 sin 3*t* +7 cos π*t* - - (d) 7 cos π*t* +5 sin 2π*t* - - (e) 3 cos √2*t* +5 cos 2*t* - - (f) sin 5*t* 2 +3 cos 6*t* 5 +3 sin *t* 7 +30◦ *t* - -(g) -$$ -\sin 3t + \cos \frac{15}{4} -$$ - -- (h) (3 sin 2*t* +sin 5*t*)2 -- (i) (5 sin 2*t*)3 -- **6.3-1** For each of the periodic signals in Fig. P6.1-1, find exponential Fourier series and sketch the corresponding spectra. -- **6.3-2** A 2π-periodic signal *x*(*t*) is specified over one period as - -$$ -x(t) = \begin{cases} \frac{1}{A}t & 0 \le t < A \\ 1 & A \le t < \pi \\ 0 & \pi \le t < 2\pi \end{cases} -$$ - -Sketch *x*(*t*) over two periods from *t* = 0 to 4π. Show that the exponential Fourier series coefficients *Dn* for this series are given by - -$$ -D_n = \begin{cases} \frac{2\pi - A}{4\pi} & n = 0\\ \frac{1}{2\pi n} \left( \frac{e^{-jAn} - 1}{An} + je^{-jn\pi} \right) & n \neq 0 \end{cases} -$$ - -**6.3-3** A periodic signal *x*(*t*) is expressed by the following Fourier series: - -$$ -x(t) = 3\cos t + \sin\left(t - \frac{\pi}{6}\right) - 2\cos\left(t - \frac{\pi}{3}\right) -$$ - -- (a) Sketch the amplitude and phase spectra for the trigonometric series. -- (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra. -- (c) By inspection of spectra in part (b), write the exponential Fourier series for *x*(*t*). -- (d) Show that the series found in part (c) is equivalent to the trigonometric series for *x*(*t*). -- **6.3-4** The trigonometric Fourier series of a certain periodic signal is given by - -$$ -x(t) = 3 + \sqrt{3}\cos 2t + \sin 2t -$$ -$$ -+ \sin 3t - \frac{1}{2}\cos\left(5t + \frac{\pi}{3}\right) -$$ - -- (a) Sketch the trigonometric Fourier spectra. -- (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra. -- (c) By inspection of spectra in part (b), write the exponential Fourier series for *x*(*t*). -- (d) Show that the series found in part (c) is equivalent to the trigonometric series for *x*(*t*). -- **6.3-5** The exponential Fourier series of a certain function is given as - -$$ -x(t) = (2+j2)e^{-j3t} + j2e^{-jt} + 3 - j2e^{jt} + (2-j2)e^{j3t} -$$ - -- (a) Sketch the exponential Fourier spectra. -- (b) By inspection of the spectra in part (a), sketch the trigonometric Fourier spectra for *x*(*t*). Find the compact trigonometric Fourier series from these spectra. -- (c) Show that the trigonometric series found in part (b) is equivalent to the exponential series for *x*(*t*). -- (d) Find the signal bandwidth. -- **6.3-6** Figure P6.3-6 shows the trigonometric Fourier spectra of a periodic signal *x*(*t*). - - (a) By inspection of Fig. P6.3-6, find the trigonometric Fourier series representing *x*(*t*). - - (b) By inspection of Fig. P6.3-6, sketch the exponential Fourier spectra of *x*(*t*). - - (c) By inspection of the exponential Fourier spectra obtained in part (b), find the exponential Fourier series for *x*(*t*). - - (d) Show that the series found in parts (a) and (c) are equivalent. -- **6.3-7** Figure P6.3-7 shows the exponential Fourier spectra of a periodic signal *x*(*t*). - -- (a) By inspection of Fig. P6.3-7, find the exponential Fourier series representing *x*(*t*). -- (b) By inspection of Fig. P6.3-7, sketch the trigonometric Fourier spectra for *x*(*t*). -- (c) By inspection of the trigonometric Fourier spectra found in part (b), find the trigonometric Fourier series for *x*(*t*). -- (d) Show that the series found in parts (a) and (c) are equivalent. -- **6.3-8** Let periodic signal *x*(*t*) have exponential Fourier series spectrum *Dn*. Prove the following properties. - - (a) If *x*(*t*) has even symmetry, then *Dn* also has even symmetry. - - (b) If *x*(*t*) has odd symmetry, then *Dn* also has odd symmetry. - - (c) If *x*(*t*) is real, then *Dn* is conjugate symmetric (*Dn* = *D*∗ −*n*). - - (d) If *x*(*t*) is imaginary, then *Dn* is conjugate antisymmetric (*Dn* = −*D*∗ −*n*). -- **6.3-9** (a) Find the exponential Fourier series for the signal in Fig. P6.3-9a. - - (b) Using the results in part (a), find the Fourier series for the signal *x*ˆ(*t*) in Fig. P6.3-9b, - -**Figure P6.3-7** - -#### **Figure P6.3-9** - -which is a time-shifted version of the signal *x*(*t*). - -- (c) Using the results in part (a), find the Fourier series for the signal *x*˜(*t*) in Fig. P6.3-9c, which is a time-scaled version of the signal *x*(*t*). -- **6.3-10** A periodic signal *x*(*t*) is expressed as an exponential Fourier series - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} -$$ - -(a) Show that the exponential Fourier series for *x*ˆ(*t*) = *x*(*t* −*T*) is given by - -$$ -\hat{x}(t) = \sum_{n = -\infty}^{\infty} \hat{D}_n e^{jn\omega_0 t} -$$ - -in which - -$$ -|\tilde{D}_n| = |D_n| \quad \text{and} \quad \angle \tilde{D}_n = \angle D_n - n\omega_0 T -$$ - -This result shows that time shifting of a periodic signal by *T* seconds merely changes the phase spectrum by *n*ω0*T*. The amplitude spectrum is unchanged. - -(b) Show that the exponential Fourier series for *x*˜(*t*) = *x*(*at*) is given by - -$$ -\tilde{x}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn(a\omega_0)t} -$$ - -This result shows that time compression of a periodic signal by a factor *a* expands its Fourier spectra along the ω axis by the same factor *a*. Similarly, time expansion of a periodic signal by a factor *a* compresses its Fourier spectra along the ω axis by the factor *a*. Intuitively explain this result. - -**6.3-11** (a) The Fourier series for the periodic signal in Fig. 6.7a is given in Drill 6.1. Verify Parseval's theorem for this series, given that - -$$ -\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} -$$ - -- (b) If *x*(*t*) is approximated by the first *N* terms in this series, find *N* so that the power of the error signal is less than 1% of *Px*. -- **6.3-12** (a) The Fourier series for the periodic signal in Fig. 6.7b is given in Drill 6.1. Verify - -Problems 675 - -Parseval's theorem for this series, given that - -$$ -\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} -$$ - -- (b) If *x*(*t*) is approximated by the first *N* terms in this series, find *N* so that the power of the error signal is less than 10% of *Px*. -- **6.3-13** The signal *x*(*t*) in Fig. 6.17 is approximated by the first 2*N* + 1 terms (from *n* = −*N* to *N*) in its exponential Fourier series given in Drill 6.5. Determine the value of *N* if this (2*N* + 1)-term Fourier series power is to be no less than 99.75% of the power of *x*(*t*). -- **6.3-14** (a) A 2 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = 1 3 *x*1(−*t* −5) in terms of *X*1[*n*]. - - (b) A 2 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = cos(10*t*)*x*1(*t*) in terms of *X*1[*n*]. - - (c) A 3 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = *x*1(−*t*)−3*x*1(*t* +2) in terms of *X*1[*n*]. -- **6.3-15** A 2-periodic signal *x*(*t*) is defined as - -$$ -x(t) = \begin{cases} -t^2 - t + 0.25 & -1 \le t < 0\\ t^2 - t + 0.25 & 0 \le t < 1\\ x(t+2) & \forall t \end{cases} -$$ - -- (a) Plot *x*(*t*) over −2 ≤ *t* ≤ 2. -- (b) Determine *D*0, the dc content of *x*(*t*). -- (c) Similar to Ex. 6.11, use properties and not integration to determine *Dn* for *n* = 0. -- (d) Plot the magnitude spectrum |*Dn*| over a suitable range of *n*. -- (e) How does the magnitude spectrum |*Dn*| compare to that of signal *y*(*t*) = cos(π*t*)? Note similarities as well as major differences. -- **6.3-16** A 3-periodic signal *x*(*t*) is defined as - -$$ -x(t) = \begin{cases} \n|t| & -1 \le t \le 1\\ \n0 & 1 < |t| \le 1.5\\ \nx(t+3) & \forall t\n\end{cases} -$$ - -(a) Plot -$$ -x(t) -$$ - over $-3 \le t \le 3$ . - -- (b) Determine *D*0, the dc content of *x*(*t*). -- (c) Similar to Ex. 6.11, use properties and not integration to determine *Dn* for *n* = 0. -- (d) Plot the magnitude spectrum |*Dn*| over a suitable range of *n*. What is the most dominant frequency component of this signal? -- **6.4-1** Find the response of an LTIC system with transfer function - -$$ -H(s) = \frac{s}{s^2 + 2s + 3} -$$ - -to the periodic input shown in Fig. 6.2a. - -- **6.4-2** A periodic signal *x*(*t*) = 1 + 2 cos(5π*t*) + 3 sin(14π*t*) is applied to an LTIC system to produce output *y*(*t*). - - (a) Determine ω0, the fundamental radian frequency of *x*(*t*). - - (b) Determine *Dn*, the exponential Fourier series spectrum of *x*(*t*). - - (c) If the system is an ideal lowpass filter with cutoff frequency *f*c = 2 Hz, what is the output *y*(*t*)? - - (d) If the system is an ideal highpass filter with cutoff frequency *f*c = 2 Hz, what is the output *y*(*t*)? - - (e) If the system is an ideal bandpass filter with a 4 Hz passband centered at 4 Hz, what is the output *y*(*t*)? - - (f) If the system is an ideal bandstop filter with a 5 Hz stopband centered at 10 Hz, what is the output *y*(*t*)? - - (g) Describe the frequency response of a filter that, in response to *x*(*t*), would produce the output *y*(*t*) = 4 cos(5π*t*) −9 sin(14π*t*). -- **6.4-3** Consider a *T*0 = 1 periodic signal *x*(*t*) defined as - -$$ -x(t) = \begin{cases} 1 - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases} -$$ - -- (a) Sketch *x*(*t*) for −2 ≤ *t* ≤ 2. -- (b) Determine *Dn*, the exponential Fourier series spectrum of *x*(*t*). -- (c) If *x*(*t*) is applied to an ideal bandpass filter with a 1 Hz passband centered at 3 Hz, determine the output *y*(*t*). -- **6.4-4** (a) Find the exponential Fourier series for a signal *x*(*t*) = cos 5*t* sin 3*t*. You can do this without evaluating any integrals. - - (b) Sketch the Fourier spectra. - -(c) The signal *x*(*t*) is applied at the input of an LTIC system with frequency response, as shown in Fig. P6.4-4. Find the output *y*(*t*). - -**Figure P6.4-4** - -- **6.4-5** (a) Find the exponential Fourier series for a periodic signal *x*(*t*) shown in Fig. P6.4-5a. - - (b) The signal *x*(*t*) is applied at the input of an LTIC system shown in Fig. P6.4-5b. Find the expression for the output *y*(*t*). -- **6.4-6** A *T*-periodic τ/*T* duty-cycle square wave *p*(*t*) is defined as - -$$ -p(t) = \begin{cases} 1 & |t| < \frac{\tau}{2} \\ 0 & \frac{\tau}{2} < |t| < \frac{T}{2} \\ p(t+T) & \forall t \end{cases} -$$ - -where 0 <τ< *T*. Also consider the frequency response *H*(ω) of a lowpass communications channel with 10 rad/s bandwidth (e.g., |*H*(ω)| ≈ 0 for ω > 10). If *T* and τ are properly chosen, we can estimate *H*(ω) at points ω = *n*ω0 as *H*ˆ (*n*ω0) = 1 *P*0 *Yn*, where *Yn* is the exponential FS spectrum of the channel output *y*(*t*) in response to input *p*(*t*) and *P*0 is the dc component of *p*(*t*). - -- (a) Using direct integration, determine the exponential Fourier series coefficients *Pn* of signal *p*(*t*). -- (b) Determine a suitable value *T* so that *p*(*t*) applied to system *H*(ω) has 21 component frequencies over the system bandwidth 0 ≤ ω ≤ 10. -- (c) Assuming *T* is properly chosen, determine a suitable duty cycle τ/*T* so that *H*(*n*ω0) ≈ *Yn*. Carefully justify your result. -- (d) From the perspective of using *p*(*t*) to help measure the system frequency response *H*(ω), what happens if *T* is properly chosen but τ/*T* is chosen too small? -- (e) From the perspective of using *p*(*t*) to help measure the system frequency response *H*(ω), what happens if *T* is properly chosen but τ/*T* is chosen too large? -- **6.5-1** Derive Eq. (6.32) in an alternate way by observing that **e** = (**x**−*c***y**) and |**e**| 2 = (**x**−*c***y**)·(**x** − *c***y**) = |**x**| 2 +*c*2|**y**| 2 −2*c***x** · **y**. -- **6.5-2** A signal *x*(*t*) is approximated in terms of a signal *y*(*t*) over an interval (*t*1, *t*2): - -$$ -x(t) \simeq cy(t) \qquad t_1 < t < t_2 -$$ - -where *c* is chosen to minimize the error energy. - -**Figure P6.4-5** - -- (a) Show that *y*(*t*) and the error *e*(*t*) = *x*(*t*) − *cy*(*t*) are orthogonal over the interval (*t*1, *t*2). -- (b) If possible, explain the result in terms of a signal-vector analogy. -- (c) Verify this result for the square signal *x*(*t*) in Fig. 6.23 and its approximation in terms of signal sin *t*. -- **6.5-3** If *x*(*t*) and *y*(*t*) are orthogonal, then show that the energy of the signal *x*(*t*) + *y*(*t*) is identical to the energy of the signal *x*(*t*) − *y*(*t*) and is given by *Ex* + *Ey*. Explain this result by using the vector analogy. In general, show that for orthogonal signals *x*(*t*) and *y*(*t*) and for any pair of arbitrary real constants *c*1 and *c*2, the energies of *c*1*x*(*t*) + *c*2*y*(*t*) and *c*1*x*(*t*) − *c*2*y*(*t*) are both given by *c*2 1*Ex* +*c*2 2*Ey*. -- **6.5-4** (a) For the signals *x*(*t*) and *y*(*t*) depicted in Fig. P6.5-4, find the component of the form *y*(*t*) contained in *x*(*t*). In other words, find the optimum value of *c* in the approximation *x*(*t*) ≈ *cy*(*t*) so that the error signal energy is minimum. - - (b) Find the error signal *e*(*t*) and its energy *Ee*. Show that the error signal is orthogonal to *y*(*t*), and that *Ex* = *c*2*Ey* + *Ee*. Explain this result in terms of vectors. - -**Figure P6.5-4** - -- **6.5-5** For the signals *x*(*t*) and *y*(*t*) shown in Fig. P6.5-4, find the component of the form *x*(*t*) contained in *y*(*t*). In other words, find the optimum value of *c* in the approximation *y*(*t*) ≈ *cx*(*t*) so that the error signal energy is minimum. What is the error signal energy? -- **6.5-6** Represent the signal *x*(*t*) shown in Fig. P6.5-4a over the interval from 0 to 1 by a trigonometric Fourier series of fundamental frequency ω0 = 2π. Compute the error energy in the representation of *x*(*t*) by only the first *N* terms of this series for *N* = 1, 2, 3, and 4. - -- **6.5-7** Represent *x*(*t*) = *t* over the interval (0, 1) by a trigonometric Fourier series that has - - (a) ω0 = 2π and only sine terms - - (b) ω0 = π and only sine terms - - (c) ω0 = π and only cosine terms - -You may use a dc term in these series if necessary. - -- **6.5-8** In Ex. 6.15, we represented the function in Fig. 6.27 by Legendre polynomials. - - (a) Use the results in Ex. 6.15 to represent the signal *g*(*t*) in Fig. P6.5-8 by Legendre polynomials. - - (b) Compute the error energy for the approximations having one and two (nonzero) terms. - -**Figure P6.5-8** - -- **6.5-9** Walsh functions, which can take on only two amplitude values, form a complete set of orthonormal functions and are of great practical importance in digital applications because they can be easily generated by logic circuitry and because multiplication with these functions can be implemented by simply using a polarity-reversing switch. Figure P6.5-9 shows the first eight functions in this set. Represent *x*(*t*) in Fig. P6.5-4a over the interval (0, 1) by using a Walsh Fourier series with these eight basis functions. Compute the energy of *e*(*t*), the error in the approximation, using the first *N* nonzero terms in the series for *N* = 1, 2, 3, and 4. In Prob. 6.5-6 we found the trigonometric Fourier series for *x*(*t*). How does the Walsh series compare with the trigonometric series in Prob. 6.5-6 from the viewpoint of the error energy for a given *N*? -- **6.5-10** For the four-dimensional real space *R*4, the so-called Walsh basis is given by: φ1 = [1, 1, 1, 1], φ2 = [1, 1,−1,−1], φ3 = [1,−1,−1, 1], and φ4 = [1,−1, 1,−1]. Denoting elements *x* = [*x*1, *x*2, *x*3, *x*4] and *y* = [*y*1, *y*2, *y*3, *y*4] (*x*, *y* ∈ *R*4), we can define orthogonality as - -### **Figure P6.5-9** - -%4 *k*=1 *xky*∗ *k* = 0. In linear algebra terminology, orthogonality here means that the inner product of vectors *x* and *y* is zero. Lastly, define a vector *z* = [−4, 0, 1,−7]. - -- (a) Show that the Walsh basis functions are mutually orthogonal. This requires a total of six calculations. -- (b) Are the Walsh basis functions normal? That is, does the inner product of each Walsh basis function with itself evaluate to 1? -- (c) Determine the coefficients [*c*1, *c*2, *c*3, *c*4] to represent *z* using Walsh basis functions as *z*ˆ = %4 *k*=1 *ck*φ*k*. -- (d) Determine the best three-dimensional approximation *z*ˆ3*D* to *z* in terms of the Walsh basis functions. That is, your estimate can only be a linear combination of three functions from [φ1,φ2,φ3,φ4]. Evaluate the three-term sum to determine the four elements of vector *z*ˆ3*D*. -- **6.5-11** A function can be expanded in terms of many different types of basis functions, not just the complex exponentials of Fourier analysis. For example, Walsh functions are explored in Prob. 6.5-9. *Laguerre* polynomials *Lk*(*t*), which have support on the interval [0,∞), are another possible set of basis functions. The Laguerre expansion using any number of terms we choose. For the Laguerre expansion, we define orthogonality a little differently, as - -$$ -\int_0^\infty e^{-t}x(t)y(t)dt = 0. -$$ - -Notice the presence of the *e*−*t* term in the integral. Using this definition, Laguerre polynomials are orthonormal. - -(a) Show that *L*0(*t*) is normal. That is, show that - -$$ -\int_0^\infty e^{-t} L_0(t) L_0^*(t) dt = 1 -$$ - -(b) Show that *L*1(*t*) is normal. That is, show that - -$$ -\int_0^{\infty} e^{-t} L_1(t) L_1^*(t) dt = 1 -$$ - -- (c) Show that *L*0(*t*) is orthogonal to *L*1(*t*). -- (d) Compute the coefficient *c*0 that produces the best approximation *x*ˆ0(*t*) = *c*0*L*0(*t*) of the function *x*(*t*) = *e*−*t u*(*t*). -- (e) For the best estimate *x*ˆ1(*t*) = *c*0*L*0(*t*) + *c*1*L*1(*t*) of the function *x*(*t*) = *e*−*t u*(*t*), the coefficient *c*0 remains unchanged from part **()**(d) and *c*1 = 1 4 . Confirm that *c*1 = 1 4 and explain why the coefficient *c*0 does not change. -- **6.7-1** A periodic signal has ω0 = 2 3π and exponential Fourier series spectrum *Dn* = *j* cos (π*n*/10)(*u*[*n*+10] −*u*[*n*−11]). [*Hint:* Refer to Prob. 6.3-8 for some useful properties.] - - (a) Determine the period *T*0 of the corresponding signal *x*(*t*). - - (b) Is the time-domain signal real, imaginary, or neither? Justify your answer. - - (c) Is the time-domain signal even, odd, or neither? Justify your answer. - - (d) Use MATLAB to synthesize the timedomain signal *x*(*t*) and plot it over the interval [−*T*0,*T*0]. -- **6.7-2** Repeat Prob. 6.7-1 for a periodic signal with ω0 = 3 2π and *Dn* = 2 sin(π*n*/10)(*u*[*n* + 10] − *u*[*n*−11]). -- **6.7-3** Consider the (*T*0 = 1)-periodic signal *x*(*t*): - -$$ -x(t) = \begin{cases} 2t - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases}. -$$ - -- (a) Sketch *x*(*t*) for −2 ≤ *t* ≤ 2. -- (b) Using properties and minimal integration, determine *Dn*, the exponential Fourier spectrum of *x*(*t*). - -- (c) Verify the correctness of *Dn* by using MAT-LAB to synthesize *x*(*t*) with a suitable truncation of Eq. (6.19). -- (d) Suppose *x*(*t*) is applied to an ideal bandpass filter with passband between 2.5 and 3.5 Hz. Determine the filter output *y*(*t*). Simplify your answer. - -[*Hint:* Refer to Ex. 6.11.] - -**6.7-4** Section 6.7 discusses the construction of a phase-optimized multitone test signal with linearly spaced frequency components. This problem investigates a similar signal with logarithmically spaced frequency components. - -> A multitone test signal *m*(*t*) is constructed by using a superposition of *N* real sinusoids - -$$ -m(t) = \sum_{n=1}^{N} \cos(\omega_n t + \theta_n) -$$ - -where θ*n* establishes the relative phase of each sinusoidal component. - -- (a) Determine a suitable set of *N* = 10 frequencies ω*n* that logarithmically spans [(2π ) ≤ ω ≤ 100(2π )] yet still results in a periodic test signal *m*(*t*). Determine the period *T*0 of your signal. Using θ*n* = 0, plot the resulting (*T*0)-periodic signal over −*T*0/2 ≤ *t* ≤ *T*0/2. -- (b) Determine a suitable set of phases θ*n* that minimize the maximum magnitude of *m*(*t*). Plot the resulting signal and identify the maximum magnitude that results. -- (c) Many systems suffer from what is called one-over-*f* noise. The power of this undesirable noise is proportional to 1/*f* . Thus, low-frequency noise is stronger than high-frequency noise. What modifications to *m*(*t*) are appropriate for use in environments with 1/*f* noise? Justify your answer. - -# **[CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER TRANSFORM** - -We can analyze linear systems in many different ways by taking advantage of the property of linearity, whereby the input is expressed as a sum of simpler components. The system response to any complex input can be found by summing the system's response to these simpler components of the input. In time-domain analysis, we separated the input into impulse components. In the frequency-domain analysis in Ch. 4, we separated the input into exponentials of the form *est* (the Laplace transform), where the complex frequency *s* = σ + *j*ω. The Laplace transform, although very valuable for system analysis, proves somewhat awkward for signal analysis, where we prefer to represent signals in terms of exponentials *ej*ω*t* instead of *est*. This is accomplished by the Fourier transform. In a sense, the Fourier transform may be considered to be a special case of the Laplace transform with *s* = *j*ω. Although this view is true most of the time, it does not always hold because of the nature of convergence of the Laplace and Fourier integrals. - -In Ch. 6, we succeeded in representing periodic signals as a sum of (everlasting) sinusoids or exponentials of the form *ej*ω*t* . The Fourier integral developed in this chapter extends this spectral representation to aperiodic signals. - -## **7.1 APERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY THE FOURIER INTEGRAL** - -Applying a limiting process, we now show that an aperiodic signal can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal *x*(*t*) such as the one depicted in Fig. 7.1a by everlasting exponentials, let us construct a new periodic signal *xT*0 (*t*) formed by repeating the signal *x*(*t*) at intervals of *T*0 seconds, as illustrated in Fig. 7.1b. The period *T*0 is made long enough to avoid overlap between the repeating pulses. The periodic signal *xT*0 (*t*) can be represented by an exponential Fourier series. If we let *T*0 → ∞, the pulses in the periodic signal repeat after an infinite interval and, therefore, - -$$ -\lim_{T_0 \to \infty} x_{T_0}(t) = x(t) -$$ - -CHAPTER - -**7** - -**Figure 7.1** Construction of a periodic signal: (a) signal *x*(*t*) and (b) periodic extension of *x*(*t*). - -Thus, the Fourier series representing *xT*0 (*t*) will also represent *x*(*t*) in the limit *T*0 → ∞. The exponential Fourier series for *xT*0 (*t*) is given by - -$$ -x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} -$$ -\n(7.1) - -where ω0 = 2π *T*0 and - -$$ -D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x_{T_0}(t) e^{-jn\omega_0 t} dt -$$ -\n(7.2) - -Observe that integrating *xT*0 (*t*) over (−*T*0/2,*T*0/2) is the same as integrating *x*(*t*) over (−∞,∞). Therefore, Eq. (7.2) can be expressed as - -$$ -D_n = \frac{1}{T_0} \int_{-\infty}^{\infty} x(t) e^{-jn\omega_0 t} dt -$$ -\n(7.3) - -It is interesting to see how the nature of the spectrum changes as *T*0 increases. To understand this fascinating behavior, let us define *X*(ω), a continuous function of ω, as - -$$ -X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt -$$ -\n(7.4) - -A glance at Eqs. (7.3) and (7.4) shows that - -$$ -D_n = \frac{1}{T_0} X(n\omega_0) \tag{7.5} -$$ - -**Figure 7.2** Change in the Fourier spectrum when the period *T*0 in Fig. 7.1 is doubled. - -This means that the Fourier coefficients *Dn* are 1/*T*0 times the samples of *X*(ω) uniformly spaced at intervals of ω0, as depicted in Fig. 7.2a.† Therefore, (1/*T*0)*X*(ω) is the envelope for the coefficients *Dn*. We now let *T*0 → ∞ by doubling *T*0 repeatedly. Doubling *T*0 halves the fundamental frequency ω0 so that there are now twice as many components (samples) in the spectrum. However, by doubling *T*0, the envelope (1/*T*0)*X*(ω) is halved, as shown in Fig. 7.2b. If we continue this process of doubling *T*0 repeatedly, the spectrum progressively becomes denser while its magnitude becomes smaller. Note, however, that the relative shape of the envelope remains the same [proportional to *X*(ω) in Eq. (7.4)]. In the limit as *T*0 → ∞, ω0 → 0 and *Dn* → 0. This result makes for a spectrum so dense that the spectral components are spaced at zero (infinitesimal) intervals. At the same time, the amplitude of each component is zero (infinitesimal). We have *nothing of everything, yet we have something!* This paradox sounds like *Alice in Wonderland,* but as we shall see, these are the classic characteristics of a very familiar phenomenon.‡ - -Substitution of Eq. (7.5) in Eq. (7.1) yields - -$$ -x_{T_0}(t) = \sum_{n = -\infty}^{\infty} \frac{X(n\omega_0)}{T_0} e^{jn\omega_0 t} -$$ -\n(7.6) - -As *T*0 →∞, ω0 becomes infinitesimal (ω0 →0). Hence, we shall replace ω0 by a more appropriate notation, ω. In terms of this new notation, ω0 = 2π *T*0 becomes - -$$ -\Delta \omega = \frac{2\pi}{T_0} -$$ - -† For the sake of simplicity, we assume *Dn*, and therefore *X*(ω), in Fig. 7.2, to be real. The argument, however, is also valid for complex *Dn* [or *X*(ω)]. - -‡ If nothing else, the reader now has irrefutable proof of the proposition that 0% ownership of everything is better than 100% ownership of nothing. - -**Figure 7.3** The Fourier series becomes the Fourier integral in the limit as *T*0 → ∞. - -and Eq. (7.6) becomes - -$$ -x_{T_0}(t) = \sum_{n=-\infty}^{\infty} \left[ \frac{X(n\Delta\omega)\Delta\omega}{2\pi} \right] e^{(jn\Delta\omega)t} -$$ - -This equation shows that *xT*0 (*t*) can be expressed as a sum of everlasting exponentials of frequencies 0,±ω,±2ω,±3ω,... (the Fourier series). The amount of the component of frequency *n*ω is [*X*(*n*ω)ω]/2π. In the limit as *T*0 → ∞, ω → 0 and *xT*0 (*t*) → *x*(*t*). Therefore, - -$$ -x(t) = \lim_{T_0 \to \infty} x_{T_0}(t) = \lim_{\Delta \omega \to 0} \frac{1}{2\pi} \sum_{n = -\infty}^{\infty} X(n\Delta \omega) e^{(jn\Delta \omega)t} \Delta \omega -$$ -\n(7.7) - -The sum on the right-hand side of Eq. (7.7) can be viewed as the area under the function *X*(ω)*ej*ω*t* , as illustrated in Fig. 7.3. Therefore, - -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega -$$ - (7.8) - -The integral on the right-hand side is called the *Fourier integral*. We have now succeeded in representing an aperiodic signal *x*(*t*) by a Fourier integral (rather than a Fourier series).† This integral is basically a Fourier series (in the limit) with fundamental frequency ω → 0, as seen from Eq. (7.7). The amount of the exponential *ejn*ω*t* is *X*(*n*ω)ω/2π. Thus, the function *X*(ω) given by Eq. (7.4) acts as a spectral function. - -We call *X*(ω) the *direct* Fourier transform of *x*(*t*), and *x*(*t*) the *inverse* Fourier transform of *X*(ω). The same information is conveyed by the statement that *x*(*t*) and *X*(ω) are a Fourier transform pair. Symbolically, this statement is expressed as - -$$ -X(\omega) = \mathcal{F}[x(t)] -$$ - and $x(t) = \mathcal{F}^{-1}[X(\omega)]$ - -† This derivation should not be considered to be a rigorous proof of Eq. (7.8). The situation is not as simple as we have made it appear [1]. - -or - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -To recapitulate, - -$$ -X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt -$$ -\n(7.9) - -and - -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega -$$ -\n(7.10) - -It is helpful to keep in mind that the Fourier integral in Eq. (7.10) is of the nature of a Fourier series with fundamental frequency ω approaching zero [Eq. (7.7)]. Therefore, most of the discussion and properties of Fourier series apply to the Fourier transform as well. *The transform X*(ω) *is the frequency-domain specification of x*(*t*)*.* - -We can plot the spectrum *X*(ω) as a function of ω. Since *X*(ω) is complex, we have both amplitude and angle (or phase) spectra - -$$ -X(\omega) = |X(\omega)|e^{j\angle X(\omega)} -$$ - -in which |*X*(ω)| is the amplitude and *X*(ω) is the angle (or phase) of *X*(ω). According to Eq. (7.9), - -$$ -X(-\omega) = \int_{-\infty}^{\infty} x(t)e^{j\omega t}dt -$$ - -Taking the conjugates of both sides yields - -$$ -x^*(t) \Longleftrightarrow X^*(-\omega) \tag{7.11} -$$ - -This property is known as the *conjugation property.* Now, if *x*(*t*) is a real function of *t*, then *x*(*t*) = *x*∗(*t*), and from the conjugation property, we find that - -$$ -X(-\omega) = X^*(\omega) -$$ - -This is the *conjugate symmetry* property of the Fourier transform, applicable to real *x*(*t*). Therefore, for real *x*(*t*), - -$$ -|X(-\omega)| = |X(\omega)| \quad \text{and} \quad \angle X(-\omega) = -\angle X(\omega) \tag{7.12} -$$ - -Thus, for real *x*(*t*), the amplitude spectrum |*X*(ω)| is an even function, and the phase spectrum *X*(ω) is an odd function of ω. These results were derived earlier for the Fourier spectrum of a periodic signal [Eq. (6.22)] and should come as no surprise. - -### **EXAMPLE 7.1 Fourier Transform of a Causal Exponential** - -Find the Fourier transform of *e*−*atu*(*t*). - -By definition [Eq. (7.9)], - -$$ -X(\omega) = \int_{-\infty}^{\infty} e^{-at} u(t) e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt = \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty} -$$ - -But |*e*−*j*ω*t* | = 1. Therefore, as *t* → ∞, *e*−(*a*+*j*ω)*t* = *e*−*ate*−*j*ω*t* = ∞ if *a* < 0, but it is equal to 0 if *a* > 0. Therefore, - -$$ -X(\omega) = \frac{1}{a + j\omega} \qquad a > 0 -$$ - -Expressing *a*+*j*ω in the polar form as √ *a*2 +ω2 *ej*tan−1(ω/*a*) , we obtain - -$$ -X(\omega) = \frac{1}{\sqrt{a^2 + \omega^2}} e^{-j \tan^{-1}(\omega/a)} -$$ - -Therefore, - -$$ -|X(\omega)| = \frac{1}{\sqrt{a^2 + \omega^2}} \quad \text{and} \quad \angle X(\omega) = -\tan^{-1}\left(\frac{\omega}{a}\right) -$$ - -The amplitude spectrum |*X*(ω)| and the phase spectrum *X*(ω) are depicted in Fig. 7.4b. Observe that |*X*(ω)| is an even function of ω, and *X*(ω) is an odd function of ω, as expected. - -### EXISTENCE OF THE FOURIER TRANSFORM - -In Ex. 7.1 we observed that when *a* < 0, the Fourier integral for *e*−*atu*(*t*) does not converge. Hence, the Fourier transform for *e*−*atu*(*t*) does not exist if *a* < 0 (growing exponential). Clearly, not all signals are Fourier transformable. - -Because the Fourier transform is derived here as a limiting case of the Fourier series, it follows that the basic qualifications of the Fourier series, such as *equality in the mean* and convergence conditions in suitably modified form, apply to the Fourier transform as well. It can be shown that if *x*(*t*) has a finite energy, that is, if - -$$ -\int_{-\infty}^{\infty} |x(t)|^2 dt < \infty -$$ - -then the Fourier transform *X*(ω) is finite and converges to *x*(*t*) in the mean. This means, if we let - -$$ -\hat{x}(t) = \lim_{W \to \infty} \frac{1}{2\pi} \int_{-W}^{W} X(\omega) e^{j\omega t} d\omega -$$ - -then Eq. (7.10) implies - -$$ -\int_{-\infty}^{\infty} \left| x(t) - \hat{x}(t) \right|^2 dt = 0 \tag{7.13} -$$ - -In other words, *x*(*t*) and its Fourier integral [the right-hand side of Eq. (7.10)] can differ at some values of *t* without contradicting Eq. (7.13). We shall now discuss an alternate set of criteria due to Dirichlet for convergence of the Fourier transform. - -As with the Fourier series, if *x*(*t*) satisfies certain conditions (*Dirichlet conditions*), its Fourier transform is guaranteed to converge pointwise at all points where *x*(*t*) is continuous. Moreover, at the points of discontinuity, *x*(*t*) converges to the value midway between the two values of *x*(*t*) on either side of the discontinuity. The Dirichlet conditions are as follows: - -1. *x*(*t*) should be absolutely integrable, that is, - -$$ -\int_{-\infty}^{\infty} |x(t)| \, dt < \infty \tag{7.14} -$$ - -If this condition is satisfied, we see that the integral on the right-hand side of Eq. (7.9) is guaranteed to have a finite value. - -- 2. *x*(*t*) must have only a finite number of finite discontinuities within any finite interval. -- 3. *x*(*t*) must contain only a finite number of maxima and minima within any finite interval. - -We stress here that although the Dirichlet conditions are sufficient for the existence and pointwise convergence of the Fourier transform, they are not necessary. For example, we saw in Ex. 7.1 that a growing exponential, which violates Dirichlet's first condition in Eq. (7.14), does not have a Fourier transform. But the signal of the form (sin*at*)/*t*, which *does* violate this condition, does have a Fourier transform. - -Any signal that can be generated in practice satisfies the Dirichlet conditions and therefore has a Fourier transform. Thus, the physical existence of a signal is a sufficient condition for the existence of its transform. - -### LINEARITY OF THE FOURIER TRANSFORM - -The Fourier transform is linear; that is, if - -$$ -x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega) -$$ - -then - -$$ -a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(\omega) + a_2X_2(\omega) \tag{7.15} -$$ - -The proof is trivial and follows directly from Eq. (7.9). This result can be extended to any finite number of terms. It can be extended to an infinite number of terms only if the conditions required for interchangeability of the operations of summation and integration are satisfied. - -### **[7.1-1 Physical Appreciation of the Fourier Transform](#page-12-0)** - -In understanding any aspect of the Fourier transform, we should remember that Fourier representation is a way of expressing a signal in terms of everlasting sinusoids (or exponentials). The Fourier spectrum of a signal indicates the relative amplitudes and phases of sinusoids that are required to synthesize that signal. A periodic signal Fourier spectrum has finite amplitudes and exists at discrete frequencies (ω0 and its multiples). Such a spectrum is easy to visualize, but the spectrum of an aperiodic signal is not easy to visualize because it has a continuous spectrum. The continuous spectrum concept can be appreciated by considering an analogous, more tangible phenomenon. One familiar example of a continuous distribution is the loading of a beam. Consider a beam loaded with weights *D*1,*D*2,*D*3,...,*Dn* units at the uniformly spaced points *y*1, *y*2,..., *yn*, as shown in Fig. 7.5a. - -The total load *WT* on the beam is given by the sum of these loads at each of the *n* points: - -$$ -W_T = \sum_{i=1}^n D_i -$$ - -Consider now the case of a continuously loaded beam, as depicted in Fig. 7.5b. In this case, although there appears to be a load at every point, the load at any one point is zero. This does not mean that there is no load on the beam. A meaningful measure of load in this situation is not the load at a point, but rather the loading density per unit length at that point. Let *X*(*y*) be the loading density per unit length of beam. It then follows that the load over a beam length *y*(*y* → 0), at some point *y*, is *X*(*y*)*y*. To find the total load on the beam, we divide the beam into segments of interval *y*(*y* → 0). The load over the *n*th such segment of length *y* is *X*(*ny*)*y*. The total load *WT* is given by - -$$ -W_T = \lim_{\Delta y \to 0} \sum_{y_1}^{y_n} X(n\Delta y) \, \Delta y = \int_{y_1}^{y_n} X(y) \, dy -$$ - -The load now exists at every point, and *y* is now a continuous variable. In the case of discrete loading (Fig. 7.5a), the load exists only at *n* discrete points. At other points, there is no load. On the other hand, in the continuously loaded case, the load exists at every point, but at any specific - -**Figure 7.5** Weight-loading analogy for the Fourier transform. - -point *y*, the load is zero. The load over a small interval *y*, however, is [*X*(*ny*)]*y* (Fig. 7.5b). Thus, even though the load at a point *y* is zero, the relative load at that point is *X*(*y*). - -An exactly analogous situation exists in the case of a signal spectrum. When *x*(*t*) is periodic, the spectrum is discrete, and *x*(*t*) can be expressed as a sum of discrete exponentials with finite amplitudes: - -$$ -x(t) = \sum_{n} D_n e^{jn\omega_0 t} -$$ - -For an aperiodic signal, the spectrum becomes continuous; that is, the spectrum exists for every value of ω, but the amplitude of each component in the spectrum is zero. The meaningful measure here is not the amplitude of a component of some frequency but the spectral density per unit bandwidth. From Eq. (7.7), it is clear that *x*(*t*) is synthesized by adding exponentials of the form *ejn*ω*t* , in which the contribution by any one exponential component is zero. But the contribution by exponentials in an infinitesimal band ω located at ω = *n*ω is (1/2π )*X*(*n*ω)ω, and the addition of all these components yields *x*(*t*) in the integral form: - -$$ -x(t) = \lim_{\Delta\omega \to 0} \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} X(n\Delta\omega)e^{(jn\Delta\omega)t} \Delta\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{i\omega t} d\omega -$$ - -Thus, *n*ω approaches a continuous variable ω. The spectrum now exists at every ω. The contribution by components within a band *d*ω is (1/2π )*X*(ω)*d*ω = *X*(ω)*df* , where *df* is the bandwidth in hertz. Clearly, *X*(ω) is the *spectral density* per unit bandwidth (in hertz).† It also follows that even if the amplitude of any one component is infinitesimal, the relative amount of a component of frequency ω is *X*(ω). Although *X*(ω) is a spectral density, in practice, it is customarily called the *spectrum* of *x*(*t*) rather than the spectral density of *x*(*t*). Deferring to this convention, we shall call *X*(ω) the Fourier spectrum (or Fourier transform) of *x*(*t*). - -### A MARVELOUS BALANCING ACT - -An important point to remember here is that *x*(*t*) is represented (or synthesized) by exponentials or sinusoids that are everlasting (not causal). Such conceptualization leads to a rather fascinating picture when we try to visualize the synthesis of a timelimited pulse signal *x*(*t*) [Fig. 7.6] by the sinusoidal components in its Fourier spectrum. The signal *x*(*t*) exists only over an interval (*a*,*b*) and is zero outside this interval. The spectrum of *x*(*t*) contains an infinite number of exponentials (or sinusoids), which start at *t* = −∞ and continue forever. The amplitudes and phases of these components add up exactly to *x*(*t*) over the finite interval (*a*,*b*) and to zero everywhere outside this interval. Juggling the amplitudes and phases of an infinite number of components to achieve - -† To stress that the signal spectrum is a *density* function, we shall shade the plot of |*X*(ω)| (as in Fig. 7.4b). The representation of *X*(ω), however, will be a line plot, primarily to avoid visual confusion. - -such a perfect and delicate balance boggles the human imagination. Yet the Fourier transform accomplishes it routinely, without much thinking on our part. Indeed, we become so involved in mathematical manipulations that we fail to notice this marvel. - -## **[7.2 TRANSFORMS OF](#page-12-0) SOME USEFUL FUNCTIONS** - -For convenience, we now introduce a compact notation for the useful gate, triangle, and interpolation functions. - -### UNIT GATE FUNCTION - -We define a unit gate function rect(*x*) as a gate pulse of unit height and unit width, centered at the origin, as illustrated in Fig. 7.7a† : - -rect -$$ -(x) -$$ - = -$$ -\begin{cases} 0 & |x| > \frac{1}{2} \\ \frac{1}{2} & |x| = \frac{1}{2} \\ 1 & |x| < \frac{1}{2} \end{cases} -$$ - (7.16) - -The gate pulse in Fig. 7.7b is the unit gate pulse rect(*x*) expanded by a factor τ along the horizontal axis and therefore can be expressed as rect(*x*/τ ) (see Sec. 1.2-2). Observe that τ , the denominator of the argument of rect (*x*/τ ), indicates the width of the pulse. - -### UNIT TRIANGLE FUNCTION - -We define a unit triangle function (*x*) as a triangular pulse of unit height and unit width, centered at the origin, as shown in Fig. 7.8a - -$$ -\Delta(x) = \begin{cases} 0 & |x| \ge \frac{1}{2} \\ 1 - 2|x| & |x| < \frac{1}{2} \end{cases} -$$ -(7.17) - -**Figure 7.7** A gate pulse. - -† At |*x*| = 0.5, we require rect(*x*) = 0.5 because the inverse Fourier transform of a discontinuous signal converges to the mean of its two values at the discontinuity. - -**Figure 7.8** A triangle pulse. - -The pulse in Fig. 7.8b is (*x*/τ ). Observe that here, as for the gate pulse, the denominator τ of the argument of (*x*/τ ) indicates the pulse width. - -### INTERPOLATION FUNCTION SINC (*x*) - -The function sin*x*/*x* is the "sine over argument" function denoted by sinc (*x*). † This function plays an important role in signal processing. It is also known as the *filtering or interpolating function*. We define - -$$ -\operatorname{sinc}(x) = \frac{\sin x}{x} \tag{7.18} -$$ - -Inspection of Eq. (7.18) shows the following: - -- 1. sinc (*x*) is an even function of *x*. -- 2. sinc (*x*) = 0 when sin *x* = 0 except at *x* = 0, where it appears to be indeterminate. This means that sinc*x* = 0 for *x* = ±π,±2π,±3π,.... -- 3. Using L'Hôpital's rule, we find sinc (0) = 1. -- 4. sinc (*x*) is the product of an oscillating signal sin*x* (of period 2π) and a monotonically decreasing function 1/*x*. Therefore, sinc (*x*) exhibits damped oscillations of period 2π, with amplitude decreasing continuously as 1/*x*. - -Figure 7.9a shows sinc(*x*). Observe that sinc (*x*) = 0 for values of *x* that are positive and negative integer multiples of π. Figure 7.9b shows sinc (3ω/7). The argument 3ω/7 = π when ω = 7π/3. Therefore, the first zero of this function occurs at ω = 7π/3. - -### **DR ILL 7.1 Sketching Basic Functions** - -Sketch: **(a)** rect(*x*/8), **(b)** (ω/10), **(c)** sinc (3πω/2), and **(d)** sinc (*t*)rect(*t*/4π ). - -$$ -\operatorname{sinc}(x) = \frac{\sin \pi x}{\pi x} -$$ - -† sinc(*x*) is also denoted by Sa (*x*) in the literature. Some authors define sinc (*x*) as - -**Figure 7.9** A sinc pulse. - -### **EXAMPLE 7.2 Fourier Transform of a Rectangular Pulse** - -Find the Fourier transform of *x*(*t*) = rect(*t*/τ ) (Fig. 7.10a). - -$$ -X(\omega) = \int_{-\infty}^{\infty} \text{rect}\left(\frac{t}{\tau}\right) e^{-j\omega t} dt -$$ - -Since rect(*t*/τ ) = 1 for |*t*| < τ/2, and since it is zero for |*t*| > τ/2, - -$$ -X(\omega) = \int_{-\tau/2}^{\tau/2} e^{-j\omega t} dt -$$ - -= $-\frac{1}{j\omega} (e^{-j\omega \tau/2} - e^{j\omega \tau/2}) = \frac{2 \sin(\frac{\omega \tau}{2})}{\omega}$ -= $\tau \frac{\sin(\frac{\omega \tau}{2})}{(\frac{\omega \tau}{2})} = \tau \operatorname{sinc}(\frac{\omega \tau}{2})$ - -**Figure 7.10 (a)** A gate pulse *x*(*t*), **(b)** its Fourier spectrum *X*(ω), **(c)** its amplitude spectrum |*X*(ω)|, and **(d)** its phase spectrum *X*(ω). - -Therefore, - -$$ -\text{rect}\left(\frac{t}{\tau}\right) \Longleftrightarrow \tau \text{ sinc}\left(\frac{\omega\tau}{2}\right) \tag{7.19} -$$ - -Recall that sinc(*x*) = 0 when *x* = ±*n*π. Hence, sinc(ωτ /2) = 0 when ωτ/2 = ±*n*π; that is, when ω = ±2*n*π/τ ,(*n* = 1, 2, 3,. . .), as depicted in Fig. 7.10b. The Fourier transform *X*(ω) shown in Fig. 7.10b exhibits positive and negative values. A negative amplitude can be considered to be a positive amplitude with a phase of −π or π. We use this observation to plot the amplitude spectrum |*X*(ω)|=|sinc (ωτ /2)| (Fig. 7.10c) and the phase spectrum *X*(ω) (Fig. 7.10d). The phase spectrum, which is required to be an odd function of ω, may be drawn in several other ways because a negative sign can be accounted for by a phase of ±*n*π, where *n* is any odd integer. All such representations are equivalent. - -#### BANDWIDTH OF RECT *t τ* - -The spectrum *X*(ω) in Fig. 7.10 peaks at ω = 0 and decays at higher frequencies. Therefore, rect(*t*/τ ) is a lowpass signal with most of the signal energy in lower-frequency components. Strictly speaking, because the spectrum extends from 0 to ∞, the bandwidth is ∞. However, much of the spectrum is concentrated within the first lobe (from ω = 0 to ω = 2π/τ ). Therefore, a rough estimate of the bandwidth of a rectangular pulse of width τ seconds is 2π/τ rad/s, or 1/τ Hz.† Note the reciprocal relationship of the pulse width with its bandwidth. We shall observe later that this result is true, in general. - -† To compute bandwidth, we must consider the spectrum for positive values of ω only. See the discussion in Sec. 6.3. - -Find the Fourier transform of the unit impulse δ(*t*). - -Using the sampling property of the impulse [Eq. (1.11)], we obtain - -$$ -\mathcal{F}[\delta(t)] = \int_{-\infty}^{\infty} \delta(t)e^{-j\omega t}dt = 1 \quad \text{and} \quad \delta(t) \Longleftrightarrow 1 -$$ - -Figure 7.11 shows δ(*t*) and its spectrum. - -### **EXAMPLE 7.4 Inverse Fourier Transform of the Dirac Delta Function** - -Find the inverse Fourier transform of δ(ω). - -On the basis of Eq. (7.10) and the sampling property of the impulse function, - -$$ -\mathcal{F}^{-1}[\delta(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega) e^{j\omega t} d\omega = \frac{1}{2\pi} -$$ - -Therefore, - -$$ -\frac{1}{2\pi} \Longleftrightarrow \delta(\omega) \quad \text{and} \quad 1 \Longleftrightarrow 2\pi \delta(\omega) \tag{7.20} -$$ - -This result shows that the spectrum of a constant signal *x*(*t*) = 1 is an impulse 2πδ(ω), as illustrated in Fig. 7.12. - -The result [Eq. (7.20)] could have been anticipated on qualitative grounds. Recall that the Fourier transform of *x*(*t*) is a spectral representation of *x*(*t*) in terms of everlasting exponential components of the form *ej*ω*t* . Now, to represent a constant signal *x*(*t*) = 1, we need a single - -**Figure 7.12 (a)** A constant (dc) signal and **(b)** its Fourier spectrum. - -everlasting exponential *ej*ω*t* with ω = 0.† This results in a spectrum at a single frequency ω = 0. Another way of looking at the situation is that *x*(*t*) = 1 is a dc signal that has a single frequency ω = 0 (dc). - -If an impulse at ω = 0 is a spectrum of a dc signal, what does an impulse at ω = ω0 represent? We shall answer this question in the next example. - -### **EXAMPLE 7.5 Inverse Fourier Transform of a Shifted Dirac Delta Function** - -Find the inverse Fourier transform of δ(ω −ω0). - -Using the sampling property of the impulse function, we obtain - -$$ -\mathcal{F}^{-1}[\delta(\omega - \omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega = \frac{1}{2\pi} e^{j\omega_0 t} -$$ - -Therefore, - -$$ -\frac{1}{2\pi}e^{j\omega_0 t} \Longleftrightarrow \delta(\omega - \omega_0) \quad \text{and} \quad e^{j\omega_0 t} \Longleftrightarrow 2\pi\delta(\omega - \omega_0) \tag{7.21} -$$ - -This result shows that the spectrum of an everlasting exponential *ej*ω0*t* is a single impulse at ω = ω0. We reach the same conclusion by qualitative reasoning. To represent the everlasting - -† The constant multiplier 2π in the spectrum [*X*(ω) = 2πδ(ω)] may be a bit puzzling. Since 1 = *ej*ω*t* with ω = 0, it appears that the Fourier transform of *x*(*t*) = 1 should be an impulse of strength unity rather than 2π. Recall, however, that in the Fourier transform *x*(*t*) is synthesized by exponentials not of amplitude *X*(*n*ω)ω but of amplitude 1/2π times *X*(*n*ω)ω, as seen from Eq. (7.7). Had we used variable *f* (hertz) instead of ω, the spectrum would have been the unit impulse. - -exponential *ej*ω0*t* , we need a single everlasting exponential *ej*ω*t* with ω = ω0. Therefore, the spectrum consists of a single component at frequency ω = ω0. From Eq. (7.21) it follows that - -$$ -e^{-j\omega_0 t} \Longleftrightarrow 2\pi \delta(\omega + \omega_0) -$$ - -# **EXAMPLE 7.6 Fourier Transform of a Sinusoid** Find the Fourier transform of the everlasting sinusoid cos ω0*t* (Fig. 7.13a). *v*0 0 *v*0 *x*(*t*) cos *v*0*t X*(*v*) *p t v* 0 (a) (b) **Figure 7.13 (a)** A cosine signal and **(b)** its Fourier spectrum. - -Recall Euler's formula - -$$ -\cos \omega_0 t = \frac{1}{2} (e^{j\omega_0 t} + e^{-j\omega_0 t}) -$$ - -Applying Eq. (7.21), we obtain - -$$ -\cos \omega_0 t \Longleftrightarrow \pi [\delta(\omega + \omega_0) + \delta(\omega - \omega_0)] -$$ - -The spectrum of cos ω0*t* consists of two impulses at ω0 and −ω0, as shown in Fig. 7.13b. The result also follows from qualitative reasoning. An everlasting sinusoid cos ω0*t* can be synthesized by two everlasting exponentials, *ej*ω0*t* and *e*−*j*ω0*t* . Therefore, the Fourier spectrum consists of only two components of frequencies ω0 and −ω0. - -### **EXAMPLE 7.7 Fourier Transform of a Periodic Signal** - -Determine the Fourier transform of a periodic signal *x*(*t*) using its Fourier series representation. - -We can use a Fourier series to express a periodic signal as a sum of exponentials of the form *ejn*ω0*t* , whose Fourier transform is found in Eq. (7.21). Hence, we can readily find the Fourier transform of a periodic signal by using the linearity property in Eq. (7.15). - -### 696 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -The Fourier series of a periodic signal *x*(*t*) with period *T*0 is given by - -$$ -x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -Taking the Fourier transform of both sides, we obtain† - -$$ -X(\omega) = 2\pi \sum_{n=-\infty}^{\infty} D_n \delta(\omega - n\omega_0) -$$ - (7.22) - -**Figure 7.14 (a)** The uniform impulse train and **(b)** its Fourier transform. - -As shown in Eq. (6.24) from Ex. 6.9, the Fourier coefficients *Dn* for δ*T*0 (*t*) are constant *Dn* = 1/*T*0. From Eq. (7.22), the Fourier transform of δ*T*0 (*t*) is therefore - -$$ -X(\omega) = \frac{2\pi}{T_0} \sum_{n=-\infty}^{\infty} \delta(\omega - n\omega_0) = \omega_0 \delta_{\omega_0}(\omega), \quad \text{where } \omega_0 = \frac{2\pi}{T_0} -$$ - -The corresponding spectrum is shown in Fig. 7.14b. - -† We assume here that the linearity property can be extended to an infinite sum. - -### **EXAMPLE 7.9 Fourier Transform of the Unit Step Function** - -Find the Fourier transform of the unit step function *u*(*t*). - -Trying to find the Fourier transform of *u*(*t*) by direct integration leads to an indeterminate result because - -$$ -U(\omega) = \int_{-\infty}^{\infty} u(t)e^{-j\omega t}dt = \int_{0}^{\infty} e^{-j\omega t}dt = \left. \frac{-1}{j\omega}e^{-j\omega t} \right|_{0}^{\infty} -$$ - -The upper limit of *e*−*j*ω*t* as *t*→∞ yields an indeterminate answer. So we approach this problem by considering *u*(*t*) to be a decaying exponential *e*−*atu*(*t*) in the limit as *a* → 0 (Fig. 7.15a). Thus, - -$$ -u(t) = \lim_{a \to 0} e^{-at} u(t) -$$ - -and - -$$ -U(\omega) = \lim_{a \to 0} \mathcal{F}\lbrace e^{-at} u(t) \rbrace = \lim_{a \to 0} \frac{1}{a + j\omega} -$$ - -Expressing the right-hand side in terms of its real and imaginary parts yields - -$$ -U(\omega) = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} - j \frac{\omega}{a^2 + \omega^2} \right] = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} \right] + \frac{1}{j\omega} -$$ - -**Figure 7.15** Derivation of the Fourier transform of the step function. - -The function *a*/(*a*2 + ω2) has interesting properties. First, the area under this function (Fig. 7.15b) is π regardless of the value of *a*: - -$$ -\int_{-\infty}^{\infty} \frac{a}{a^2 + \omega^2} \, d\omega = \tan^{-1} \frac{\omega}{a} \bigg|_{-\infty}^{\infty} = \pi -$$ - -### 698 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Second, when *a* → 0, this function approaches zero for all ω = 0, and all its area (π) is concentrated at a single point ω = 0. Clearly, as *a* → 0, this function approaches an impulse of strength π. Thus, - -$$ -U(\omega) = \pi \delta(\omega) + \frac{1}{j\omega} \tag{7.23} -$$ - -Note that *u*(*t*) is not a "true" dc signal because it is not constant over the interval −∞ to ∞. To synthesize "true" dc, we require only one everlasting exponential with ω = 0 (impulse at ω = 0). The signal *u*(*t*) has a jump discontinuity at *t* = 0. It is impossible to synthesize such a signal with a single everlasting exponential *ej*ω*t* . To synthesize this signal from everlasting exponentials, we need, in addition to an impulse at ω = 0, all the frequency components, as indicated by the term 1/*j*ω in Eq. (7.23). - -### **EXAMPLE 7.10 Fourier Transform of the Sign Function** - -Find the Fourier transform of the sign function sgn(*t*) [pronounced *signum* (*t*)], depicted in Fig. 7.16. - -*t* sgn (*t*) 1 0 -1 **Figure 7.16** The signum function sgn(*t*). Observe that sgn(*t*)+1 = 2*u*(*t*) ⇒ sgn(*t*) = 2*u*(*t*)−1 Using Eqs. (7.20) and (7.23) and the linearity property, we obtain sgn(*t*) ⇐⇒ 2 *j*ω - -Table 7.1 provides many common Fourier transform pairs. - -| No. | x(t) | X(ω) | | -|-----|--------------------------|-------------------------------------------------------|--------------------| -| 1 | e−atu(t) | 1
a+jω | a > 0 | -| 2 | eatu(−t) | 1
a−jω | a > 0 | -| 3 | e−a t | 2a
a2 +ω2 | a > 0 | -| 4 | te−atu(t) | 1
(a+jω)2 | a > 0 | -| 5 | ne−atu(t)
t | n!
(a+jω)n+1 | a > 0 | -| 6 | δ(t) | 1 | | -| 7 | 1 | 2πδ(ω) | | -| 8 | ejω0t | 2πδ(ω −ω0) | | -| 9 | cos ω0t | π[δ(ω −ω0) +δ(ω +ω0)] | | -| 10 | sin ω0t | jπ[δ(ω +ω0)−δ(ω −ω0)] | | -| 11 | u(t) | 1
πδ(ω)+
jω | | -| 12 | sgnt | 2
jω | | -| 13 | cos ω0t u(t) | π
jω
2 [δ(ω −ω0) +δ(ω +ω0)] +
ω2
0 −ω2 | | -| 14 | sin ω0t u(t) | π
ω0
[δ(ω −ω0)−δ(ω +ω0)] +
ω2
2j
0 −ω2 | | -| 15 | e−atsin
ω0t u(t) | ω0
(a+jω)2 +ω2
0 | a > 0 | -| 16 | e−at cos
ω0t u(t) | a+jω
(a+jω)2 +ω2
0 | a > 0 | -| 17 | rect t

τ | τ sincωτ

2 | | -| 18 | W
π sinc(Wt) | rect ω

2W | | -| 19 | t

τ | τ
ωτ

sinc2
2
4 | | -| 20 | Wt
W
2π sinc2
2 | ω

2W | | -| 21 | "∞
δ(t −nT) | "∞
ω0
δ(ω −nω0) | 2π
ω0
=
T | -| 22 | n=−∞
2/2σ2
e−t | n=−∞
√2πe−σ2ω2/2
σ | | - -**TABLE 7.1** Select Fourier Transform Pairs - -### **DR ILL 7.2 Inverse Fourier Transform of a Rectangular Pulse** - -Show that the inverse Fourier transform of *X*(ω) illustrated in Fig. 7.17 is *x*(*t*) = (ω0/π )sinc (ω0*t*). Sketch *x*(*t*). - -### **DR ILL 7.3 Fourier Transform of a General Sinusoid** - -Show that cos(ω0*t* +θ ) ⇐⇒ π[δ(ω +ω0)*e*−*j*θ +δ(ω −ω0)*ej*θ ]. - -### **[7.2-1 Connection Between the Fourier and Laplace Transforms](#page-12-0)** - -The general (bilateral) Laplace transform of a signal *x*(*t*), according to Eq. (4.1), is - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ - (7.24) - -Setting *s* = *j*ω in this equation yields - -$$ -X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt -$$ - -where *X*(*j*ω) = *X*(*s*)|*s*=*j*ω. But, the right-hand-side integral defines *X*(ω), the Fourier transform of *x*(*t*). Does this mean that the Fourier transform can be obtained from the corresponding Laplace transform by setting *s* = *j*ω? In other words, is it true that *X*(*j*ω) = *X*(ω)? Yes and no. Yes, it is true in most cases. For example, when *x*(*t*) = *e*−*atu*(*t*), its Laplace transform is 1/(*s* + *a*), and *X*(*j*ω) = 1/(*j*ω +*a*), which is equal to *X*(ω) (assuming *a* < 0). However, for the unit step function *u*(*t*), the Laplace transform is - -$$ -u(t) \Longleftrightarrow \frac{1}{s} \qquad \text{Re}\, s > 0 -$$ - -### 7.3 Some Properties of the Fourier Transform 701 - -The Fourier transform is given by - -$$ -u(t) \Longleftrightarrow \frac{1}{j\omega} + \pi \delta(\omega) -$$ - -Clearly, *X*(*j*ω) = *X*(ω) in this case. - -To understand this puzzle, consider the fact that we obtain *X*(*j*ω) by setting *s* = *j*ω in Eq. (7.24). This implies that the integral on the right-hand side of Eq. (7.24) converges for *s* = *j*ω, meaning that *s* = *j*ω (the imaginary axis) lies in the ROC for *X*(*s*). The general rule is that only when the ROC for *X*(*s*) includes the ω axis, does setting *s* = *j*ω in *X*(*s*) yield the Fourier transform *X*(ω), that is, *X*(*j*ω) = *X*(ω). This is the case of absolutely integrable *x*(*t*). If the ROC of *X*(*s*) excludes the ω axis, *X*(*j*ω) = *X*(ω). This is the case for exponentially growing *x*(*t*) and also *x*(*t*) that is constant or is oscillating with constant amplitude. - -The reason for this peculiar behavior has something to do with the nature of convergence of the Laplace and the Fourier integrals when *x*(*t*) is not absolutely integrable.† - -This discussion shows that although the Fourier transform may be considered as a special case of the Laplace transform, we need to circumscribe such a view. This fact can also be confirmed by noting that a periodic signal has the Fourier transform, but the Laplace transform does not exist. - -## **7.3 SOME [PROPERTIES OF THE](#page-12-0) FOURIER TRANSFORM** - -We now study some of the important properties of the Fourier transform and their implications as well as applications. We have already encountered two important properties, linearity [Eq. (7.15)] and the conjugation property [Eq. (7.11)]. - -Before embarking on this study, we shall explain an important and pervasive aspect of the Fourier transform: the time-frequency duality. - -† To explain this point, consider the unit step function and its transforms. Both the Laplace and the Fourier transform synthesize *x*(*t*), using everlasting exponentials of the form *est*. The frequency *s* can be anywhere in the complex plane for the Laplace transform, but it must be restricted to the ω axis in the case of the Fourier transform. The unit step function is readily synthesized in the Laplace transform by a relatively simple spectrum *X*(*s*) = 1/*s*, in which the frequencies *s* are chosen in the RHP [the region of convergence for *u*(*t*) is Re *s* > 0]. In the Fourier transform, however, we are restricted to values of *s* on the ω axis only. The function *u*(*t*) can still be synthesized by frequencies along the ω axis, but the spectrum is more complicated than it is when we are free to choose the frequencies in the RHP. In contrast, when *x*(*t*) is absolutely integrable, the region of convergence for the Laplace transform includes the ω axis, and we can synthesize *x*(*t*) by using frequencies along the ω axis in both transforms. This leads to *X*(*j*ω) = *X*(ω). - -We may explain this concept by an example of two countries, X and Y. Suppose these countries want to construct similar dams in their respective territories. Country X has financial resources but not much manpower. In contrast, Y has considerable manpower but few financial resources. The dams will still be constructed in both countries, although the methods used will be different. Country X will use expensive but efficient equipment to compensate for its lack of manpower, whereas Y will use the cheapest possible equipment in a labor-intensive approach to the project. Similarly, both Fourier and Laplace integrals converge for *u*(*t*), but the makeup of the components used to synthesize *u*(*t*) will be very different for two cases because of the constraints of the Fourier transform, which are not present for the Laplace transform. - -### TIME-FREQUENCY DUALITY IN THE TRANSFORM OPERATIONS - -Equations (7.9) and (7.10) show an interesting fact: the direct and the inverse transform operations are remarkably similar. These operations, required to go from *x*(*t*) to *X*(ω) and then from *X*(ω) to *x*(*t*), are depicted graphically in Fig. 7.18. The inverse transform equation can be obtained from the direct transform equation by replacing *x*(*t*) with *X*(ω), *t* with ω, and ω with *t*. In a similar way, we can obtain the direct from the inverse. There are only two minor differences in these operations: the factor 2π appears only in the inverse operator, and the exponential indices in the two operations have opposite signs. Otherwise the two equations are duals of each other.† - -This observation has far-reaching consequences in the study of the Fourier transform. It is the basis of the so-called duality of time and frequency. *The duality principle may be compared with a photograph and its negative. A photograph can be obtained from its negative, and by using an identical procedure, a negative can be obtained from the photograph*. For any result or relationship between *x*(*t*) and *X*(ω), there exists a dual result or relationship, obtained by interchanging the roles of *x*(*t*) and *X*(ω) in the original result (along with some minor modifications arising because of the factor 2π and a sign change). For example, the time-shifting property, to be proved later, states that if *x*(*t*) ⇐⇒ *X*(ω), then - -$$ -x(t-t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0} -$$ - -**Figure 7.18** A near symmetry between the direct and the inverse Fourier transforms. - -$$ -X(2\pi f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \quad \text{and} \quad x(t) = \int_{-\infty}^{\infty} X(2\pi f)e^{j2\pi ft} df -$$ - -This leaves only one significant difference, that of sign change in the exponential index. - -† Of the two differences, the former can be eliminated by change of variable from ω to *f* (in hertz). In this case ω = 2π*f* and *d*ω = 2π *df* . - -Therefore, the direct and the inverse transforms are given by - -The dual of this property (the frequency-shifting property) states that - -$$ -x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0) -$$ - -Observe the role reversal of time and frequency in these two equations (with the minor difference of the sign change in the exponential index). The value of this principle lies in the fact that *whenever we derive any result, we can be sure that it has a dual.* This possibility can give valuable insights about many unsuspected properties or results in signal processing. - -The properties of the Fourier transform are useful not only in deriving the direct and inverse transforms of many functions, but also in obtaining several valuable results in signal processing. The reader should not fail to observe the ever-present duality in this discussion. - -### LINEARITY - -The linearity property, already introduced as Eq. (7.15), states that if *x*1(*t*) ⇐⇒ *X*1(ω) and *x*2(*t*) ⇐⇒ *X*2(ω), then *a*1*x*1(*t*)+*a*2*x*2(*t*) ⇐⇒ *a*1*X*1(ω)+*a*2*X*2(ω). - -### CONJUGATION AND CONJUGATE SYMMETRY - -The conjugation property, which has already been introduced, states that if *x*(*t*) ⇐⇒ *X*(ω), then - -$$ -x^*(t) \Longleftrightarrow X^*(-\omega) -$$ - -From this property follows the conjugate symmetry property, also introduced earlier, which states that if *x*(*t*) is real, then - -$$ -X(-\omega) = X^*(\omega) -$$ - -### DUALITY - -The duality property states that if - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then - -$$ -X(t) \Longleftrightarrow 2\pi x(-\omega) \tag{7.25} -$$ - -**Proof.** From Eq. (7.10) we can write - -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(u)e^{iut} du -$$ - -Hence, - -$$ -2\pi x(-t) = \int_{-\infty}^{\infty} X(u)e^{-jut} du -$$ - -Changing *t* to ω yields Eq. (7.25). - -*t* - -From Eq. (7.19) we have - -1 - -$$ -\underbrace{\text{rect}\left(\frac{t}{\tau}\right)}_{x(t)} \Longleftrightarrow \underbrace{\tau \text{ sinc}\left(\frac{\omega \tau}{2}\right)}_{X(\omega)} -$$ - -Also, *X*(*t*) is the same as *X*(ω) with ω replaced by *t*, and *x*(−ω) is the same as *x*(*t*) with *t* replaced by −ω. Therefore, the duality property of Eq. (7.25) yields - -$$ -\underbrace{\tau \text{ sinc}\left(\frac{\tau t}{2}\right)}_{X(t)} \Longleftrightarrow \underbrace{2\pi \text{ rect}\left(\frac{-\omega}{\tau}\right)}_{2\pi x(-\omega)} = 2\pi \text{ rect}\left(\frac{\omega}{\tau}\right) -$$ - -In this result, we used the fact that rect(−*x*) = rect(*x*) because rect is an even function. Figure 7.19b shows this pair graphically. Observe the interchange of the roles of *t* and ω (with the minor adjustment of the factor 2π). This result appears as pair 18 in Table 7.1 (with τ/2 = *W*). - -As an interesting exercise, the reader should generate the dual of every pair in Table 7.1 by applying the duality property. - -## **DR ILL 7.4 Applying the Duality Property of the Fourier Transform** - -Apply the duality property to pairs 1, 3, and 9 (Table 7.1) to show that - -- **(a)** 1/(*jt* +*a*) ⇐⇒ 2π*ea*ω*u*(−ω) -- **(b)** 2*a*/(*t* 2 +*a*2) ⇐⇒ 2π*e*−*a*|ω| -- **(c)** δ(*t* +*t*0)+δ(*t* −*t*0) ⇐⇒ 2 cos *t*0ω - -### THE SCALING PROPERTY If - -*x*(*t*) ⇐⇒ *X*(ω) - -then, for any real constant *a*, - -$$ -x(at) \Longleftrightarrow \frac{1}{|a|}X\left(\frac{\omega}{a}\right) \tag{7.26} -$$ - -**Proof.** For a positive real constant *a*, - -$$ -\mathcal{F}[x(at)] = \int_{-\infty}^{\infty} x(at)e^{-j\omega t}dt = \frac{1}{a}\int_{-\infty}^{\infty} x(u)e^{(-j\omega/a)u}du = \frac{1}{a}X\left(\frac{\omega}{a}\right) -$$ - -Similarly, we can demonstrate that if *a* < 0, - -$$ -x(at) \Longleftrightarrow \frac{-1}{a}X\left(\frac{\omega}{a}\right) -$$ - -Hence follows Eq. (7.26). - -### SIGNIFICANCE OF THE SCALING PROPERTY - -The function *x*(*at*) represents the function *x*(*t*) compressed in time by a factor *a* (see Sec. 1.2-2). Similarly, a function *X*(ω/*a*) represents the function *X*(ω) expanded in frequency by the same factor *a*. *The scaling property states that time compression of a signal results in its spectral expansion, and time expansion of the signal results in its spectral compression*. Intuitively, compression in time by factor *a* means that the signal is varying faster by factor *a*. † To synthesize such a signal, the frequencies of its sinusoidal components must be increased by the factor *a*, implying that its frequency spectrum is expanded by the factor *a*. Similarly, a signal expanded in time varies more slowly; hence the frequencies of its components are lowered, implying that its frequency spectrum is compressed. For instance, the signal cos 2ω0*t* is the same as the signal cosω0*t* time-compressed by a factor of 2. Clearly, the spectrum of the former (impulse at ±2ω0) is an expanded version of the spectrum of the latter (impulse at ±ω0). The effect of this scaling is demonstrated in Fig. 7.20. - -† We are assuming *a* > 1, although the argument still holds if *a* < 1. In the latter case, compression becomes expansion by factor 1/*a*, and vice versa. - -**Figure 7.20** The scaling property of the Fourier transform. - -### RECIPROCITY OF SIGNAL DURATION AND ITS BANDWIDTH - -The scaling property implies that if *x*(*t*) is wider, its spectrum is narrower, and vice versa. Doubling the signal duration halves its bandwidth, and vice versa. This suggests that the bandwidth of a signal is inversely proportional to the signal duration or width (in seconds).† We have already verified this fact for the gate pulse, where we found that the bandwidth of a gate pulse of width τ seconds is 1/τ Hz. More discussion of this interesting topic can be found in the literature [2]. - -By letting *a* = −1 in Eq. (7.26), we obtain the *inversion (or reflection) property of time and frequency:* - -$$ -x(-t) \Longleftrightarrow X(-\omega) \tag{7.27} -$$ - -### **EXAMPLE 7.12 Fourier Transform Reflection Property** - -Using the reflection property of the Fourier transform and Table 7.1, find the Fourier transforms of *eatu*(−*t*) and *e*−*a*|*t*| . - -Application of Eq. (7.27) to pair 1 of Table 7.1 yields - -$$ -e^{at}u(-t) \Longleftrightarrow \frac{1}{a-j\omega} \qquad a > 0 -$$ - -Also, - -$$ -e^{-a|t|} = e^{-at}u(t) + e^{at}u(-t) -$$ - -† When a signal has infinite duration, we must consider its effective or equivalent duration. There is no unique definition of effective signal duration. One possible definition is given in Eq. (2.47). - -Therefore, - -$$ -e^{-a|t|} \Longleftrightarrow \frac{1}{a+j\omega} + \frac{1}{a-j\omega} = \frac{2a}{a^2 + \omega^2} \qquad a > 0 \tag{7.28} -$$ - -The signal *e*−*a*|*t*| and its spectrum are illustrated in Fig. 7.21. - -## THE TIME-SHIFTING PROPERTY If - -*x*(*t*) ⇐⇒ *X*(ω) - -then - -$$ -x(t - t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0} \tag{7.29} -$$ - -**Proof.** By definition, - -$$ -\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(t-t_0)e^{-j\omega t} dt -$$ - -Letting *t* −*t*0 = *u*, we have - -$$ -\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(u)e^{-j\omega(u+t_0)} du = e^{-j\omega t_0} \int_{-\infty}^{\infty} x(u)e^{-j\omega u} du = X(\omega)e^{-j\omega t_0} -$$ - -This result shows that *delaying a signal by t*0 *seconds does not change its amplitude spectrum. The phase spectrum, however, is changed by* −ω*t*0. - -### PHYSICAL EXPLANATION OF THE LINEAR PHASE - -Time delay in a signal causes a linear phase shift in its spectrum. This result can also be derived by heuristic reasoning. Imagine *x*(*t*) being synthesized by its Fourier components, which are sinusoids of certain amplitudes and phases. The delayed signal *x*(*t* − *t*0) can be synthesized by the same sinusoidal components, each delayed by *t*0 seconds. The amplitudes of the components remain unchanged. Therefore, the amplitude spectrum of *x*(*t* − *t*0) is identical to that of *x*(*t*). The time delay of *t*0 in each sinusoid, however, does change the phase of each component. Now, a sinusoid - -**Figure 7.22** Physical explanation of the time-shifting property. - -cosω*t* delayed by *t*0 is given by - -$$ -\cos \omega (t - t_0) = \cos (\omega t - \omega t_0) -$$ - -Therefore a time delay *t*0 in a sinusoid of frequency ω manifests as a phase delay of ω*t*0. This is a linear function of ω, meaning that higher-frequency components must undergo proportionately higher phase shifts to achieve the same time delay. This effect is depicted in Fig. 7.22 with two sinusoids, the frequency of the lower sinusoid being twice that of the upper. The same time delay *t*0 amounts to a phase shift of π/2 in the upper sinusoid and a phase shift of π in the lower sinusoid. This verifies the fact that *to achieve the same time delay, higher-frequency sinusoids must undergo proportionately higher phase shifts*. The principle of linear phase shift is very important, and we shall encounter it again in distortionless signal transmission and filtering applications. - -### **EXAMPLE 7.13 Fourier Transform Time-Shifting Property** - -Use the time-shifting property to find the Fourier transform of *e*−*a*|*t*−*t*0| . - -This function, shown in Fig. 7.23a, is a time-shifted version of *e*−*a*|*t*| (depicted in Fig. 7.21a). From Eqs. (7.28) and (7.29), we have - -$$ -e^{-a|t-t_0|} \Longleftrightarrow \frac{2a}{a^2 + \omega^2} e^{-j\omega t_0} -$$ - -The spectrum of *e*−*a*|*t*−*t*0| (Fig. 7.23b) is the same as that of *e*−*a*|*t*| (Fig. 7.21b), except for an added phase shift of −ω*t*0. - -### **EXAMPLE 7.14 Fourier Transform of a Time-Shifted Rectangular Pulse** - -Find the Fourier transform of the time-shifted rectangular pulse *x*(*t*) illustrated in Fig. 7.24a. - -The pulse *x*(*t*) is the gate pulse rect(*t*/τ ) in Fig. 7.10a delayed by 3τ/4 seconds. Hence, according to Eq. (7.29), its Fourier transform is the Fourier transform of rect(*t*/τ ) multiplied by *e*−*j*ω(3τ /4) . Therefore, - -$$ -X(\omega) = \tau \operatorname{sinc}\left(\frac{\omega \tau}{2}\right) e^{-j\omega(3\tau/4)} -$$ - -The amplitude spectrum |*X*(ω)| (depicted in Fig. 7.24b) of this pulse is the same as that indicated in Fig. 7.10c. But the phase spectrum has an added linear term −3ωτ/4. Hence, the phase spectrum of *x*(*t*) (Fig. 7.24a) is identical to that in Fig. 7.10d plus a linear term −3ωτ/4, as shown in Fig. 7.24c. - -### PHASE SPECTRUM USING PRINCIPAL VALUES - -There is an alternate way of spectral representation of *X*(ω). The phase angle computed on a calculator or by using a computer subroutine is generally the principal value (modulo 2π value) of the phase angle, which always lies in the range −π to π. For instance, the principal value of angle 3π/2 is −π/2, and so on. The principal value differs from the actual value by ±2π radians (and its integer multiples) in a way that ensures that the principal value remains within −π to π. Thus, the principal value will show jump discontinuities of ±2π whenever the actual - -### 710 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -phase crosses ±π. The phase plot in Fig. 7.24c is redrawn in Fig. 7.24d using the principal value for the phase. This phase pattern, which contains phase discontinuities of magnitudes 2π and π, becomes repetitive at intervals of ω = 8π/τ . - -## **DR ILL 7.5 Fourier Transform Time-Shifting Property** - -Use pair 18 of Table 7.1 and the time-shifting property to show that the Fourier transform of sinc [ω0(*t* − *T*)] is (π/ω0)rect(ω/2ω0)*e*−*j*ω*T* . Sketch the amplitude and phase spectra of the Fourier transform. - -### THE FREQUENCY-SHIFTING PROPERTY If - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then - -$$ -x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0) \tag{7.30} -$$ - -**Proof.** By definition, - -$$ -\mathcal{F}[x(t)e^{j\omega_0 t}] = \int_{-\infty}^{\infty} x(t)e^{j\omega_0 t}e^{-j\omega t} dt = \int_{-\infty}^{\infty} x(t)e^{-j(\omega - \omega_0)t} dt = X(\omega - \omega_0) -$$ - -According to this property, the multiplication of a signal by a factor *ej*ω0*t* shifts the spectrum of that signal by ω = ω0. Note the duality between the time-shifting and the frequency-shifting properties. - -Changing ω0 to −ω0 in Eq. (7.30) yields - -$$ -x(t)e^{-j\omega_0 t} \Longleftrightarrow X(\omega + \omega_0) \tag{7.31} -$$ - -Because *ej*ω0*t* is not a real function that can be generated, frequency shifting in practice is achieved by multiplying *x*(*t*) by a sinusoid. Observe that - -$$ -x(t)\cos \omega_0 t = \frac{1}{2} [x(t)e^{j\omega_0 t} + x(t)e^{-j\omega_0 t}] -$$ - -From Eqs. (7.30) and (7.31), it follows that - -$$ -x(t)\cos\omega_0 t \Longleftrightarrow \frac{1}{2}[X(\omega - \omega_0) + X(\omega + \omega_0)]\tag{7.32} -$$ - -This result shows that the multiplication of a signal *x*(*t*) by a sinusoid of frequency ω0 shifts the spectrum *X*(ω) by ±ω0, as depicted in Fig. 7.25. - -Multiplication of a sinusoid cosω0*t* by *x*(*t*) amounts to modulating the sinusoid amplitude. This type of modulation is known as *amplitude modulation*. The sinusoid cosω0*t* is called the *carrier,* the signal *x*(*t*) is the *modulating signal*, and the signal *x*(*t*) cosω0*t* is the *modulated signal*. Further discussion of modulation and demodulation appears in Sec. 7.7. - -To sketch a signal *x*(*t*) cos ω0*t*, we observe that - -$$ -x(t)\cos\omega_0 t = \begin{cases} x(t) & \text{when } \cos\omega_0 t = 1\\ -x(t) & \text{when } \cos\omega_0 t = -1 \end{cases} -$$ - -Therefore, *x*(*t*) cos ω0*t* touches *x*(*t*) when the sinusoid cos ω0*t* is at its positive peaks and touches −*x*(*t*) when cos ω0*t* is at its negative peaks. This means that *x*(*t*) and −*x*(*t*) act as envelopes for the signal *x*(*t*) cos ω0*t* (see Fig. 7.25). The signal −*x*(*t*) is a mirror image of *x*(*t*) about the horizontal axis. Figure 7.25 shows the signals *x*(*t*) and *x*(*t*) cos ω0*t* and their spectra. - -**Figure 7.25** Amplitude modulation of a signal causes spectral shifting. - -### **EXAMPLE 7.15 Spectral Shifting by Amplitude Modulation** - -Find and sketch the Fourier transform of the modulated signal *x*(*t*) cos 10*t* in which *x*(*t*) is a gate pulse rect(*t*/4), as illustrated in Fig. 7.26a. - -From pair 17 of Table 7.1, we find rect(*t*/4) ⇐⇒ 4 sinc (2ω), which is depicted in Fig. 7.26b. From Eq. (7.32) it follows that - -$$ -x(t)\cos 10t \Longleftrightarrow \frac{1}{2}[X(\omega+10) + X(\omega-10)] -$$ - -In this case, *X*(ω) = 4 sinc (2ω). Therefore, - -*x*(*t*) cos 10*t* ⇐⇒ 2 sinc [2(ω +10)] +2 sinc [2(ω −10)] - -The spectrum (Fig. 7.26c) of *x*(*t*) cos 10*t* is obtained by shifting *X*(ω) in Fig. 7.26b to the left by 10 and also to the right by 10, and then multiplying it by 0.5, as depicted in Fig. 7.26d. - -## **DR ILL 7.6 Fourier Transform of an Amplitude-Modulated Signal** - -Sketch signal *e*−|*t*| cos 10*t*. Find the Fourier transform of this signal and sketch its spectrum. **Answer:** *X*(ω) = 1 (ω−10)2+1 + 1 (ω+10)2+1 . See Fig. 7.21b for the spectrum of *e*−*a*|*t*| . - -## **DR ILL 7.7 Amplitude Modulation Using a Phase-Shifted Carrier** - -Show that - -``` -x(t) cos(ω0t +θ ) ⇐⇒ 1 - 2 - - X(ω −ω0)ejθ +X(ω +ω0)e−jθ -``` - -### APPLICATIONS OF MODULATION - -Modulation is used to shift signal spectra. Some of the situations that call for spectrum shifting are presented next. - -1. If several signals, all occupying the same frequency band, are transmitted simultaneously over the same transmission medium, they will all interfere; it will be impossible to separate or retrieve them at a receiver. For example, if all radio stations decide to broadcast audio signals simultaneously, a receiver will not be able to separate them. This problem is solved - -Old is gold, but sometimes it is fool's gold. - -by using modulation, whereby each radio station is assigned a distinct carrier frequency. Each station transmits a modulated signal. This procedure shifts the signal spectrum to its allocated band, which is not occupied by any other station. A radio receiver can pick up any station by tuning to the band of the desired station. The receiver must now demodulate the received signal (undo the effect of modulation). Demodulation therefore consists of another spectral shift required to restore the signal to its original band. Note that both modulation and demodulation implement spectral shifting; consequently, demodulation operation is similar to modulation (see Sec. 7.7). - -This method of transmitting several signals simultaneously over a channel by sharing its frequency band is known as *frequency-division multiplexing (FDM)*. - -2. For effective radiation of power over a radio link, the antenna size must be of the order of the wavelength of the signal to be radiated. Audio signal frequencies are so low (wavelengths are so large) that impracticably large antennas would be required for radiation. Here, shifting the spectrum to a higher frequency (a smaller wavelength) by modulation solves the problem. - -### CONVOLUTION - -The time-convolution property and its dual, the frequency-convolution property, state that if - -$$ -x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega) -$$ - -then - -$$ -x_1(t) * x_2(t) \Longleftrightarrow X_1(\omega) X_2(\omega) \quad \text{(time convolution)} \tag{7.33} -$$ - -### 7.3 Some Properties of the Fourier Transform 715 - -and - -$$ -x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi}X_1(\omega) * X_2(\omega) \quad \text{(frequency convolution)} \tag{7.34} -$$ - -**Proof.** By definition, - -$$ -\mathcal{F}|x_1(t) * x_2(t)| = \int_{-\infty}^{\infty} e^{-j\omega t} \left[ \int_{-\infty}^{\infty} x_1(\tau) x_2(t-\tau) d\tau \right] dt -$$ -$$ -= \int_{-\infty}^{\infty} x_1(\tau) \left[ \int_{-\infty}^{\infty} e^{-j\omega t} x_2(t-\tau) d\tau \right] d\tau -$$ - -The inner integral is the Fourier transform of *x*2(*t* − τ ), given by [time-shifting property in Eq. (7.29)] *X*2(ω)*e*−*j*ωτ . Hence, - -$$ -\mathcal{F}[x_1(t) * x_2(t)] = \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} X_2(\omega) d\tau = X_2(\omega) \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} d\tau = X_1(\omega) X_2(\omega) -$$ - -Let *H*(ω) be the Fourier transform of the unit impulse response *h*(*t*), that is, - -*h*(*t*) ⇐⇒ *H*(ω) - -Application of the time-convolution property to *y*(*t*) = *x*(*t*) ∗ *h*(*t*) yields [assuming that both *x*(*t*) and *h*(*t*) are Fourier transformable] - -$$ -Y(\omega) = X(\omega)H(\omega) \tag{7.35} -$$ - -The frequency-convolution property of Eq. (7.34) can be proved in exactly the same way by reversing the roles of *x*(*t*) and *X*(ω). - -### **EXAMPLE 7.16 Time-Convolution Property to Show the Time-Integration Property** - -Use the time-convolution property to show that if - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega) -$$ - -Because - -$$ -u(t-\tau) = \begin{cases} 1 & \tau \leq t \\ 0 & \tau > t \end{cases} -$$ - -it follows that - -$$ -x(t) * u(t) = \int_{-\infty}^{\infty} x(\tau)u(t-\tau) d\tau = \int_{-\infty}^{t} x(\tau) d\tau -$$ - -#### 716 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Now, from the time-convolution property [Eq. (7.33)], it follows that - -$$ -x(t) * u(t) = \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(\omega) \left[ \frac{1}{j\omega} + \pi \delta(\omega) \right] -$$ -$$ -= \frac{X(\omega)}{j\omega} + \pi X(0) \delta(\omega) -$$ - -In deriving the last result, we used Eq. (1.10). - -### **DR ILL 7.8 Fourier Transform Time-Convolution Property** - -Use the time-convolution property to show that: - -$$ -(a) x(t) * \delta(t) = x(t) -$$ - -**(b)** *e*−*atu*(*t*)∗*e*−*btu*(*t*) = 1 *b*−*a* [*e*−*at* −*e*−*bt*]*u*(*t*) - -## TIME DIFFERENTIATION AND TIME INTEGRATION If - -$$ -x(t) \Longleftrightarrow X(\omega) -$$ - -then† - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega) \quad \text{(time differentiation)} \tag{7.36} -$$ - -and - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega) \quad \text{(time integration)} \tag{7.37} -$$ - -**Proof.** Differentiation of both sides of Eq. (7.10) yields - -$$ -\frac{dx(t)}{dt} = \frac{1}{2\pi} \int_{-\infty}^{\infty} j\omega X(\omega) e^{j\omega t} d\omega -$$ - -$$ -\int_{-\infty}^{\infty} \left| \frac{dx(t)}{dt} \right| dt < \infty -$$ - -† Valid only if the transform of *dx*/*dt* exists. In other words, *dx*/*dt* must satisfy the Dirichlet conditions. The first Dirichlet condition implies - -We also require that *x*(*t*) → 0 as *t* → ±∞. Otherwise, *x*(*t*) has a dc component, which gets lost in differentiation, and there is no one-to-one relationship between *x*(*t*) and *dx*/*dt*. - -| Operation | x(t) | X(ω) | -|---------------------------------|---------------------|-----------------------------| -| Scalar multiplication | kx(t) | kX(ω) | -| Addition | x1(t)+x2(t) | X1(ω) +X2(ω) | -| Conjugation | x∗(t) | X∗(−ω) | -| Duality | X(t) | 2πx(−ω) | -| Scaling (a real) | x(at) | 1
ω

X
a
a | -| Time shifting | x(t −t0) | X(ω)e−jωt0 | -| Frequency shifting (ω0
real) | x(t)ejω0t | X(ω −ω0) | -| Time convolution | x1(t)∗x2(t) | X1(ω)X2(ω) | -| Frequency convolution | x1(t)x2(t) | 1
2π X1(ω)∗X2(ω) | -| Time differentiation | dnx(t)
dtn | (jω)nX(ω) | -| Time integration | # t
x(u)du
−∞ | X(ω)
+πX(0)δ(ω)
jω | - -**TABLE 7.2** Fourier Transform Properties - -This result shows that - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega) -$$ - -Repeated application of this property yields - -$$ -\frac{d^n x(t)}{dt^n} \Longleftrightarrow (j\omega)^n X(\omega) -$$ - -The time-integration property [Eq. (7.37)] has already been proved in Ex. 7.16. Table 7.2 summarizes the most important properties of the Fourier transform. - -### **EXAMPLE 7.17 Fourier Transform Time-Differentiation Property** - -Use the time-differentiation property to find the Fourier transform of the triangle pulse (*t*/τ ) illustrated in Fig. 7.27a. Verify the correctness of the spectrum by using it to synthesize a periodic replication of the original time-domain signal with τ = 1. - -**Figure 7.27** Finding the Fourier transform of a piecewise-linear signal using the time-differentiation property. - -To find the Fourier transform of this pulse, we differentiate the pulse successively, as illustrated in Fig. 7.27b and 7.27c. Because *dx*/*dt* is constant everywhere, its derivative, *d*2*x*/*dt*2, is zero everywhere. But *dx*/*dt* has jump discontinuities with a positive jump of 2/τ at *t* = ±τ/2, and a negative jump of 4/τ at *t* = 0. Recall that the derivative of a signal at a jump discontinuity is an impulse at that point of strength equal to the amount of jump. Hence, *d*2*x*/*dt*2, the derivative of *dx*/*dt*, consists of a sequence of impulses, as depicted in Fig. 7.27c; that is, - -$$ -\frac{d^2x(t)}{dt^2} = \frac{2}{\tau} \left[ \delta \left( t + \frac{\tau}{2} \right) - 2\delta(t) + \delta \left( t - \frac{\tau}{2} \right) \right] -$$ - -From the time-differentiation property [Eq. (7.36)], - -$$ -\frac{d^2x(t)}{dt^2} \Longleftrightarrow (j\omega)^2 X(\omega) = -\omega^2 X(\omega) -$$ - -Also, from the time-shifting property [Eq. (7.29)], - -$$ -\delta(t-t_0) \Longleftrightarrow e^{-j\omega t_0} -$$ - -Combining these results, we obtain - -$$ --\omega^2 X(\omega) = \frac{2}{\tau} \left[ e^{j(\omega \tau/2)} - 2 + e^{-j(\omega \tau/2)} \right] = \frac{4}{\tau} \left( \cos \frac{\omega \tau}{2} - 1 \right) = -\frac{8}{\tau} \sin^2 \left( \frac{\omega \tau}{4} \right) -$$ - -and - -$$ -X(\omega) = \frac{8}{\omega^2 \tau} \sin^2\left(\frac{\omega \tau}{4}\right) = \frac{\tau}{2} \left[ \frac{\sin\left(\frac{\omega \tau}{4}\right)}{\frac{\omega \tau}{4}} \right]^2 = \frac{\tau}{2} \text{sinc}^2\left(\frac{\omega \tau}{4}\right) -$$ - -The spectrum *X*(ω) is depicted in Fig. 7.27d. This procedure of finding the Fourier transform can be applied to any function *x*(*t*) made up of straight-line segments with *x*(*t*) → 0 as |*t*|→∞. The second derivative of such a signal yields a sequence of impulses whose Fourier transform can be found by inspection. This example suggests a numerical method of finding the Fourier transform of an arbitrary signal *x*(*t*) by approximating the signal by straight-line segments. - -### SYNTHESIZING A PERIODIC REPLICATION TO VERIFY SPECTRUM CORRECTNESS - -While a signal's spectrum *X*(ω) provides useful insight into signal character, it can be difficult to look at *X*(ω) and know that it is correct for a particular signal *x*(*t*). Is it obvious, for example, that *X*(ω) = τ 2 sinc2 (ωτ/4) is really the spectrum of a τ -duration rectangle function? Or is it possible that a mathematical error was made in the determination of *X*(ω)? It is difficult to be certain by simple inspection of the spectrum. - -The same uncertainties exist when we are looking at a periodic signal's Fourier series spectrum. In the Fourier series case, we can verify the correctness of a signal's spectrum by synthesizing *x*(*t*) with a truncated Fourier series; the synthesized signal will match the original only if the computed spectrum is correct. This is exactly the approach that was taken in Ex. 6.11. And since a truncated Fourier series involves a simple sum, tools like MATLAB make waveform synthesis relatively simple, at least in the case of the Fourier series. - -In the case of the Fourier transform, however, synthesis of *x*(*t*) using Eq. (7.10) requires integration, a task not well suited to numerical packages such as MATLAB. All is not lost, however. Consider Eq. (7.5). By scaling and sampling the spectrum *X*(ω) of an aperiodic signal *x*(*t*), we obtain the Fourier series coefficient of a signal that is the periodic replication of *x*(*t*). Similar to Ex. 6.11, we can then synthesize a periodic replication of *x*(*t*) with a truncated Fourier series to verify spectrum correctness. Let us demonstrate the idea for the current example with τ = 1. - -To begin, we represent *X*(ω) = τ 2 sinc2 (ωτ/4) using an anonymous function in MATLAB. Since MATLAB computes sinc(x) as (sin(π*x*))/π*x*, we must scale the input by 1/π to match the notation of sinc in this book. - ->> tau = 1; X = @(omega) tau/2\*(sinc(omega\*tau/(4\*pi))).^2; - -For our periodic replication, let us pick *T*0 = 2, which is comfortably wide enough to accommodate our (τ = 1)-width function without overlap. We use Eq. (7.5) to define the needed Fourier series coefficients *Dn*. - ->> TO = 2; omega0 = -$$ -2*pi/TO -$$ -; D = $\mathcal{Q}(n)$ X(n\*omega0)/TO; - -Let us use 25 harmonics to synthesize the periodic replication *x*25(*t*) of our triangular signal *x*(*t*). To begin waveform synthesis, we set the dc portion of the signal. - ->> t = (-T0:.001:T0); x25 = D(0)\*ones(size(t)); - -To add the desired 25 harmonics, we enter a loop for 1 ≤ *n* ≤ 25 and add in the *Dn* and *D*−*n* terms. Although the result should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed by using the real command. - ->> for n = 1:25, >> x25 = x25+real(D(n)\*exp(1j\*omega0\*n\*t)+D(-n)\*exp(-1j\*omega0\*n\*t)); >> end - -Lastly, we plot the resulting truncated Fourier series synthesis of *x*(*t*). - ->> plot(t,x25,'k'); xlabel('t'); ylabel('x\_{25}(t)'); - -Since the synthesized waveform shown in Fig. 7.28 closely matches a 2-periodic replication of the triangle wave in Fig. 7.27a, we have high confidence that both the computed *Dn* and, by extension, the Fourier spectrum *X*(ω) are correct. - -**Figure 7.28** Synthesizing a 2-periodic replication of *x*(*t*) using a truncated Fourier series. - -### **DR ILL 7.9 Fourier Transform Time-Differentiation Property** - -Use the time-differentiation property to find the Fourier transform of rect(*t*/τ ). - -## **7.4 SIGNAL [TRANSMISSION](#page-12-0) THROUGH LTIC SYSTEMS** - -If *x*(*t*) and *y*(*t*) are the input and output of an LTIC system with impulse response *h*(*t*), then, as demonstrated in Eq. (7.35), - -$$ -Y(\omega) = H(\omega)X(\omega) -$$ - -This equation does not apply to (asymptotically) unstable systems because *h*(*t*) for such systems is not Fourier transformable. It applies to BIBO-stable as well as most of the marginally stable systems.† Similarly, this equation does not apply if *x*(*t*) is not Fourier transformable. - -In Ch. 4, we saw that the Laplace transform is more versatile and capable of analyzing all kinds of LTIC systems whether stable, unstable, or marginally stable. Laplace transform can also handle exponentially growing inputs. In comparison to the Laplace transform, the Fourier transform in system analysis is not just clumsier, but also very restrictive. Hence, the Laplace transform is preferable to the Fourier transform in LTIC system analysis. We shall not belabor the application of the Fourier transform to LTIC system analysis. We consider just one example here. - -### **EXAMPLE 7.18 Fourier Transform to Determine the Zero-State Response** - -Use the Fourier transform to find the zero-state response of a stable LTIC system with frequency response - -$$ -H(s) = \frac{1}{s+2} -$$ - -and the input is *x*(*t*) = *e*−*t u*(*t*). Stability implies that the region of convergence of *H*(*s*) includes the ω axis. - -In this case, - -$$ -X(\omega) = \frac{1}{j\omega + 1} -$$ - -† For marginally stable systems, if the input *x*(*t*) contains a finite-amplitude sinusoid of the system's natural frequency, which leads to resonance, the output is not Fourier transformable. It does, however, apply to marginally stable systems if the input does not contain a finite-amplitude sinusoid of the system's natural frequency. - -### 722 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Moreover, because the system is stable, the frequency response *H*(*j*ω) = *H*(ω). Hence, - -$$ -H(\omega) = H(s)|_{s=j\omega} = \frac{1}{j\omega + 2} -$$ - -Therefore, - -$$ -Y(\omega) = H(\omega)X(\omega) = \frac{1}{(j\omega + 2)(j\omega + 1)} -$$ - -Expanding the right-hand side in partial fractions yields - -$$ -Y(\omega) = \frac{1}{j\omega + 1} - \frac{1}{j\omega + 2} -$$ - -and - -$$ -y(t) = (e^{-t} - e^{-2t})u(t) -$$ - -## **DR ILL 7.10 Fourier Transform to Determine the Zero-State Response** - -For the system in Ex. 7.18, show that the zero-input response to the input *et u*(−*t*) is *y*(*t*) = 1 3 [*et u*(−*t*)+*e*−2*t u*(*t*)]. [*Hint:* Use pair 2 (Table 7.1) to find the Fourier transform of *et u*(−*t*).] - -### HEURISTIC UNDERSTANDING OF LINEAR SYSTEM RESPONSE - -In finding the linear system response to arbitrary input, the time-domain method uses convolution integral and the frequency-domain method uses the Fourier integral. Despite the apparent dissimilarities of the two methods, their philosophies are amazingly similar. In the time-domain case, we express the input *x*(*t*) as a sum of its impulse components; in the frequency-domain case, the input is expressed as a sum of everlasting exponentials (or sinusoids). In the former case, the response *y*(*t*) obtained by summing the system's responses to impulse components results in the convolution integral; in the latter case, the response obtained by summing the system's response to everlasting exponential components results in the Fourier integral. These ideas can be expressed mathematically as follows: - -1. For the time-domain case, - -| δ(t)
⇒ h(t) | shows the system response
δ(t)
h(t)
to
is the impulse response | -|--------------------------------------------|------------------------------------------------------------------------------------------| -| = \$ ∞
x(t)
x(τ )δ(t
−τ )dτ
−∞ | expresses
x(t)
as a sum
of impulse components | -| = \$ ∞
y(t)
x(τ )h(t
−τ )dτ
−∞ | expresses
y(t)
as a sum of responses to
the impulse components of input
x(t) | - -2. For the frequency-domain case, - -$$ -e^{j\omega t} \implies H(\omega)e^{j\omega t} \qquad \text{shows the system response} -$$ - -\nto $e^{j\omega t}$ is $H(\omega)e^{j\omega t}$ -\n -$$ -x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega \qquad \text{expresses } x(t) \text{ as a sum} -$$ - -\nof everyday exponential components -\n -$$ -y(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)H(\omega)e^{j\omega t} d\omega \qquad \text{expresses } y(t) \text{ as a sum of responses to} -$$ - -\nthe exponential components of input $x(t)$ - -The frequency-domain view sees a system in terms of its frequency response (system response to various sinusoidal components). It views a signal as a sum of various sinusoidal components. Transmission of an input signal through a (linear) system is viewed as transmission of various sinusoidal components of the input through the system. - -It was not by coincidence that we used the impulse function in time-domain analysis and the exponential *ej*ω*t* in studying the frequency domain. The two functions happen to be duals of each other. Thus, the Fourier transform of an impulse δ(*t* − τ ) is *e*−*j*ωτ , and the Fourier transform of *ej*ω0*t* is an impulse 2πδ(ω − ω0). This *time-frequency duality* is a constant theme in the Fourier transform and linear systems. - -### **[7.4-1 Signal Distortion During Transmission](#page-12-0)** - -For a system with frequency response *H*(ω), if *X*(ω) and *Y*(ω) are the spectra of the input and the output signals, respectively, then - -$$ -Y(\omega) = X(\omega)H(\omega) \tag{7.38} -$$ - -The transmission of the input signal *x*(*t*) through the system changes it into the output signal *y*(*t*). Equation (7.38) shows the nature of this change or modification. Here, *X*(ω) and *Y*(ω) are the spectra of the input and the output, respectively. Therefore, *H*(ω) is the spectral response of the system. The output spectrum is obtained by the input spectrum multiplied by the spectral response of the system. Equation (7.38), which clearly brings out the spectral shaping (or modification) of the signal by the system, can be expressed in polar form as - -$$ -|Y(\omega)|e^{j\angle Y(\omega)} = |X(\omega)||H(\omega)|e^{j[\angle X(\omega)+\angle H(j\omega)]} -$$ - -Therefore, - -$$ -|Y(\omega)| = |X(\omega)| |H(\omega)| \quad \text{and} \quad \angle Y(\omega) = \angle X(\omega) + \angle H(\omega) -$$ - -During transmission, the input signal amplitude spectrum |*X*(ω)| is changed to |*X*(ω)||*H*(ω)|. Similarly, the input signal phase spectrum *X*(ω) is changed to *X*(ω) + *H*(ω). An input signal spectral component of frequency ω is modified in amplitude by a factor |*H*(ω)| and is shifted in phase by an angle *H*(ω). Clearly, |*H*(ω)| is the amplitude response, and *H*(ω) is the phase response of the system. The plots of |*H*(ω)| and *H*(ω) as functions of ω show at a glance how the system modifies the amplitudes and phases of various sinusoidal inputs. This is the reason why *H*(ω) is also called the *frequency response* of the system. During transmission through the system, some frequency components may be boosted in amplitude, while others may be attenuated. The - -### 724 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -relative phases of the various components also change. In general, the output waveform will be different from the input waveform. - -### DISTORTIONLESS TRANSMISSION - -In several applications, such as signal amplification or message signal transmission over a communication channel, we require that the output waveform be a replica of the input waveform. In such cases we need to minimize the distortion caused by the amplifier or the communication channel. It is, therefore, of practical interest to determine the characteristics of a system that allows a signal to pass without distortion (*distortionless transmission*). - -Transmission is said to be distortionless if the input and the output have identical waveshapes within a multiplicative constant. A delayed output that retains the input waveform is also considered to be distortionless. Thus, in distortionless transmission, the input *x*(*t*) and the output *y*(*t*) satisfy the condition - -$$ -y(t) = G_0 x(t - t_d) -$$ - -The Fourier transform of this equation yields - -$$ -Y(\omega) = G_0 X(\omega) e^{-j\omega t_d} -$$ - -But - -$$ -Y(\omega) = X(\omega)H(\omega) -$$ - -Therefore, - -$$ -H(\omega) = G_0 e^{-j\omega t_d} -$$ - -This is the frequency response required of a system for distortionless transmission. From this equation, it follows that - -$$ -|H(\omega)| = G_0 \quad \text{and} \quad \angle H(\omega) = -\omega t_d \tag{7.39} -$$ - -This result shows that for distortionless transmission, the amplitude response |*H*(ω)| must be a constant, and the phase response *H*(ω) must be a linear function of ω with slope −*td*, where *td* is the delay of the output with respect to input (Fig. 7.29). - -### MEASURE OF TIME-DELAY VARIATION WITH FREQUENCY - -The gain |*H*(ω)| = *G*0 means that every spectral component is multiplied by a constant *G*0. Also, as seen in connection with Fig. 7.22, a linear phase *H*(ω) = −ω*td* means that every spectral - -**Figure 7.29** LTIC system frequency response for distortionless transmission. - -component is delayed by *td* seconds. This results in the output equal to *G*0 times the input delayed by *td* seconds. Because each spectral component is attenuated by the same factor (*G*0) and delayed by exactly the same amount (*td*), the output signal is an exact replica of the input (except for attenuating factor *G*0 and delay *td*). - -For distortionless transmission, we require a *linear phase* characteristic. The phase is not only a linear function of ω, it should also pass through the origin ω = 0. In practice, many systems have a phase characteristic that may be only approximately linear. A convenient way of judging phase linearity is to plot the slope of *H*(ω) as a function of frequency. This slope, which is constant for an ideal linear phase (ILP) system, is a function of ω in the general case and can be expressed as - -$$ -t_g(\omega) = -\frac{d}{d\omega} \angle H(\omega) -$$ -\n(7.40) - -If *tg*(ω) is constant, all the components are delayed by the same time interval *tg*. But if the slope is not constant, the time delay *tg* varies with frequency. This variation means that different frequency components undergo different amounts of time delay, and consequently, the output waveform will not be a replica of the input waveform. As we shall see, *tg*(ω) plays an important role in bandpass systems and is called the *group delay* or *envelope* delay. Observe that constant *td* [Eq. (7.39)] implies constant *tg*. Note that *H*(ω) = φ0 −ω*tg* also has a constant *tg*. Thus, constant group delay is a more relaxed condition. - -It is often thought (erroneously) that flatness of amplitude response |*H*(ω)| alone can guarantee signal quality. However, a system that has a flat amplitude response may yet distort a signal beyond recognition if the phase response is not linear (*td* not constant). - -### THE NATURE OF DISTORTION IN AUDIO AND VIDEO SIGNALS - -Generally speaking, the human ear can readily perceive amplitude distortion but is relatively insensitive to phase distortion. For the phase distortion to become noticeable, the variation in delay [variation in the slope of *H*(ω)] should be comparable to the signal duration (or the physically perceptible duration, in case the signal itself is long). In the case of audio signals, each spoken syllable can be considered to be an individual signal. The average duration of a spoken syllable is of a magnitude of the order of 0.01 to 0.1 second. Audio systems may have nonlinear phases, yet no noticeable signal distortion results because in practical audio systems, maximum variation in the slope of *H*(ω) is only a small fraction of a millisecond. This is the real truth underlying the statement that "the human ear is relatively insensitive to phase distortion" [3]. As a result, the manufacturers of audio equipment make available only |*H*(ω)|, the amplitude response characteristic of their systems. - -For video signals, in contrast, the situation is exactly the opposite. The human eye is sensitive to phase distortion but is relatively insensitive to amplitude distortion. Amplitude distortion in television signals manifests itself as a partial destruction of the relative half-tone values of the resulting picture, but this effect is not readily apparent to the human eye. Phase distortion (nonlinear phase), on the other hand, causes different time delays in different picture elements. The result is a smeared picture, and this effect is readily perceived by the human eye. Phase distortion is also very important in digital communication systems because the nonlinear phase characteristic of a channel causes pulse dispersion (spreading out), which in turn causes pulses to interfere with neighboring pulses. Such interference between pulses can cause an error in the pulse amplitude at the receiver: a binary **1** may read as **0**, and vice versa. - -### **[7.4-2 Bandpass Systems and Group Delay](#page-12-0)** - -The distortionless transmission conditions [Eq. (7.39)] can be relaxed slightly for bandpass systems. For lowpass systems, the phase characteristics not only should be linear over the band of interest but also should pass through the origin. For bandpass systems, the phase characteristics must be linear over the band of interest but need not pass through the origin. - -Consider an LTI system with amplitude and phase characteristics as shown in Fig. 7.30, where the amplitude spectrum is a constant *G*0 and the phase is φ0 − ω*tg* over a band 2*W* centered at frequency ω*c*. Over this band, we can describe *H*(ω) as† - -$$ -H(\omega) = G_0 e^{j(\phi_0 - \omega t_g)} \qquad \omega \ge 0 \tag{7.41} -$$ - -The phase of *H*(ω) in Eq. (7.41), shown dotted in Fig. 7.30b, is linear but does not pass through the origin. - -Consider a modulated input signal *z*(*t*) = *x*(*t*) cosω*ct*. This is a bandpass signal, whose spectrum is centered at ω = ω*c*. The signal cosω*ct* is the carrier, and the signal *x*(*t*), which is a lowpass signal of bandwidth *W* (see Fig. 7.25), is the *envelope* of *z*(*t*). ‡ We shall now show that the transmission of *z*(*t*) through *H*(ω) results in distortionless transmission of the envelope *x*(*t*). However, the carrier phase changes by φ0. To show this, consider an input *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* and the corresponding output *y*ˆ(*t*). From Eq. (7.30), *Z*ˆ(ω) = *X*(ω − ω*c*), and the corresponding output - -**Figure 7.30** Generalized linear phase characteristics. - -† Because the phase function is an odd function of ω, if *H*(ω) = φ0 − ω*tg* for ω ≥ 0, over the band 2*W* (centered at ω*c*), then *H*(ω) = −φ0 − ω*tg* for ω < 0 over the band 2*W* (centered at −ω*c*), as shown in Fig. 7.30a. - -‡ The envelope of a bandpass signal is well defined only when the bandwidth of the envelope is well below the carrier ω*c* (*W* ω*c*). - -spectrum *Y*ˆ(ω) is given by - -$$ -\hat{Y}(\omega) = H(\omega)\hat{Z}(\omega) = H(\omega)X(\omega - \omega_c) -$$ - -Recall that the bandwidth of *X*(ω) is *W* so that the bandwidth of *X*(ω −ω*c*) is 2*W*, centered at ω*c*. Over this range, *H*(ω) is given by Eq. (7.41). Hence, - -$$ -\hat{Y}(\omega) = G_0 X(\omega - \omega_c) e^{j(\phi_0 - \omega t_g)} = G_0 e^{j\phi_0} X(\omega - \omega_c) e^{-\omega t_g} -$$ - -Use of Eqs. (7.29) and (7.30) yields *y*ˆ(*t*) as - -$$ -\hat{y}(t) = G_0 e^{j\phi_0} x(t - t_g) e^{j\omega_c(t - t_g)} = G_0 x(t - t_g) e^{j[\omega_c(t - t_g) + \phi_0]} -$$ - -This is the system response to input *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* , which is a complex signal. We are really interested in finding the response to the input *z*(*t*) = *x*(*t*) cosω*ct*, which is the real part of *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* . Hence, we use Eq. (2.31) to obtain *y*(*t*), the system response to the input *z*(*t*) = *x*(*t*) cosω*ct*, as - -$$ -y(t) = G_0 x(t - t_g) \cos [\omega_c (t - t_g) + \phi_0)] -$$ -\n(7.42) - -where *tg*, the *group* (or *envelope*) delay, is the negative slope of *H*(ω) at ω*c*. † The output *y*(*t*) is basically the delayed input *z*(*t* − *tg*), except that the output carrier acquires an extra phase φ0. The output envelope *x*(*t* − *tg*) is the delayed version of the input envelope *x*(*t*) and is not affected by extra phase φ0 of the carrier. In a modulated signal, such as *x*(*t*) cosω*ct*, the information generally resides in the envelope *x*(*t*). Hence, the transmission is considered to be distortionless if the envelope *x*(*t*) remains undistorted. - -Most practical systems satisfy Eq. (7.41), at least over a very small band. Figure 7.30b shows a typical case in which this condition is satisfied for a small band *W* centered at frequency ω*c*. - -A system in Eq. (7.41) is said to have a *generalized linear phase* (GLP), as illustrated in Fig. 7.30. The ideal linear phase (ILP) characteristics is shown in Fig. 7.29. For distortionless transmission of bandpass signals, the system need satisfy Eq. (7.41) only over the bandwidth of the bandpass signal. - -**Caution.** Recall that the phase response associated with the amplitude response may have jump discontinuities when the amplitude response goes negative. Jump discontinuities also arise because of the use of the principal value for phase. Under such conditions, to compute the group delay [Eq. (7.40)], we should ignore the jump discontinuities. - -$$ -y(t) = G_o x(t - t_g) \cos \omega_c (t - t_{ph}) -$$ - -where *t*ph, called the *phase delay* at ω*c*, is given by *t*ph(ω*c*) = (ω*ctg* − φ0)/ω*c*. Generally, *t*ph varies with ω, and we can write - -$$ -t_{\rm ph}(\omega) = \frac{\omega t_g - \phi_0}{\omega} -$$ - -Recall also that *tg* itself may vary with ω. - -† Equation (7.42) can also be expressed as - -### **EXAMPLE 7.19 Distortionless Bandpass Transmission** - -**(a)** A signal *z*(*t*), shown in Fig. 7.31b, is given by - -$$ -z(t) = x(t) \cos \omega_c t -$$ - -where ω*c* = 2000π. The pulse *x*(*t*) (Fig. 7.31a) is a lowpass pulse of duration 0.1 second and has a bandwidth of about 10 Hz. This signal is passed through a filter whose frequency response is shown in Fig. 7.31c (shown only for positive ω). Find and sketch the filter output *y*(*t*). - -**(b)** Find the filter response if ω*c* = 4000π. - -**(a)** The spectrum *Z*(ω) is a narrow band of width 20 Hz, centered at frequency *f*0 = 1 kHz. The gain at the center frequency (1 kHz) is 2. The group delay, which is the negative of the slope of the phase plot, can be found by drawing tangents at ω*c*, as shown in Fig. 7.31c. The negative of the slope of the tangent represents *tg*, and the intercept along the vertical axis by the tangent represents φ0 at that frequency. From the tangents at ω*c*, we find *tg*, the group delay, as - -$$ -t_g = \frac{2.4\pi - 0.4\pi}{2000\pi} = 10^{-3} -$$ - -The vertical axis intercept is φ0 = −0.4π. Hence, by using Eq. (7.42) with gain *G*0 = 2, we obtain - -$$ -y(t) = 2x(t - t_g)\cos[\omega_c(t - t_g) - 0.4\pi] \qquad \omega_c = 2000\pi \quad t_g = 10^{-3} -$$ - -Figure 7.31d shows the output *y*(*t*), which consists of the modulated pulse envelope *x*(*t*) delayed by 1 ms and the phase of the carrier changed by −0.4π. The output shows no distortion of the envelope *x*(*t*), only the delay. The carrier phase change does not affect the shape of envelope. Hence, the transmission is considered distortionless. - -**(b)** Figure 7.31c shows that when ω*c* = 4000π, the slope of *H*(ω) is zero so that *tg* = 0. Also, the gain *G*0 = 1.5, and the intercept of the tangent with the vertical axis is φ0 = −3.1π. Hence, - -$$ -y(t) = 1.5x(t)\cos(\omega_c t - 3.1\pi) -$$ - -This, too, is a distortionless transmission for the same reasons as for case (a). - -## **[7.5 IDEAL AND](#page-13-0) PRACTICAL FILTERS** - -Ideal filters allow distortionless transmission of a certain band of frequencies and completely suppress the remaining frequencies. The ideal lowpass filter (Fig. 7.32), for example, allows all components below ω = *W* rad/s to pass without distortion and suppresses all components above ω = *W*. Figure 7.33 illustrates ideal highpass and bandpass filter characteristics. - -The ideal lowpass filter in Fig. 7.32a has a linear phase of slope −*td*, which results in a time delay of *td* seconds for all its input components of frequencies below *W* rad/s. Therefore, if the input is a signal *x*(*t*) bandlimited to *W* rad/s, the output *y*(*t*) is *x*(*t*) delayed by *td*: that is, - -$$ -y(t) = x(t - t_d) -$$ - -The signal *x*(*t*) is transmitted by this system without distortion, but with time delay *td*. For this filter, |*H*(ω)| = rect(ω/2*W*) and *H*(ω) = *e*−*j*ω*td* so that - -**Figure 7.32** Ideal lowpass filter: **(a)** frequency response and **(b)** impulse response. - -**Figure 7.33** Ideal **(a)** highpass and **(b)** bandpass filter frequency responses. - -The unit impulse response *h*(*t*) of this filter is obtained from pair 18 (Table 7.1) and the time-shifting property - -$$ -h(t) = \mathcal{F}^{-1} \left[ \text{rect}\left(\frac{\omega}{2W}\right) e^{-j\omega t_d} \right] = \frac{W}{\pi} \operatorname{sinc}[W(t - t_d)] -$$ - -Recall that *h*(*t*) is the system response to impulse input δ(*t*), which is applied at *t* = 0. Figure 7.32b shows a curious fact: the response *h*(*t*) begins even before the input is applied (at *t* = 0). Clearly, the filter is noncausal and therefore physically unrealizable. Similarly, one can show that other ideal filters (such as the ideal highpass or ideal bandpass filters depicted in Fig. 7.33) are also physically unrealizable. - -For a physically realizable system, *h*(*t*) must be causal; that is, - -$$ -h(t) = 0 \qquad \text{for } t < 0 -$$ - -In the frequency domain, this condition is equivalent to the well-known *Paley–Wiener criterion,* which states that the necessary and sufficient condition for the amplitude response |*H*(ω)| to be realizable is† - -$$ -\int_{-\infty}^{\infty} \frac{|\ln|H(\omega)|}{1 + \omega^2} d\omega < \infty -$$ -\n(7.43) - -If *H*(ω) does not satisfy this condition, it is unrealizable. Note that if |*H*(ω)| = 0 over any finite band, |ln|*H*(ω)|| = ∞ over that band, and Eq. (7.43) is violated. If, however, *H*(ω) = 0 at a single frequency (or a set of discrete frequencies), the integral in Eq. (7.43) may still be finite even though the integrand is infinite at those discrete frequencies. Therefore, for a physically realizable system, *H*(ω) may be zero at some discrete frequencies, but it cannot be zero over any finite band. In addition, if |*H*(ω)| decays exponentially (or at a higher rate) with ω, the integral in Eq. (7.43) goes to infinity, and |*H*(ω)| cannot be realized. Clearly, |*H*(ω)| cannot decay too fast with ω. According to this criterion, ideal filter characteristics (Figs. 7.32 and 7.33) are unrealizable. - -The impulse response *h*(*t*) in Fig. 7.32 is not realizable. One practical approach to filter design is to cut off the tail of *h*(*t*) for *t* < 0. The resulting causal impulse response:*h*(*t*), given by - -$$ -h(t) = h(t)u(t) -$$ - -is physically realizable because it is causal (Fig. 7.34). If *td* is sufficiently large, :*h*(*t*) will be a close approximation of *h*(*t*), and the resulting filter *H* :(ω) will be a good approximation of an ideal filter. This close realization of the ideal filter is achieved because of the increased value of time delay *td*. This observation means that the price of close realization is higher delay in the output; this situation is common in noncausal systems. Of course, theoretically, a delay *td* = ∞ is needed to realize the ideal characteristics. But a glance at Fig. 7.32b shows that a delay *td* of three or four times π *W* will make :*h*(*t*) a reasonably close version of *h*(*t* − *td*). For instance, an audio filter is required to handle frequencies of up to 20 kHz (*W* = 40,000π). In this case, a *td* of about 10−4 - -$$ -\int_{-\infty}^{\infty} |H(\omega)|^2 d\omega < \infty -$$ - -Note that the Paley–Wiener criterion is a criterion for the realizability of the amplitude response |*H*(ω)|. - -† We are assuming that |*H*(ω)| is square integrable, that is, - -**Figure 7.34** Approximate realization of an ideal lowpass filter by truncation of its impulse response. - -(0.1 ms) would be a reasonable choice. The truncation operation [cutting the tail of *h*(*t*) to make it causal], however, creates some unsuspected problems. We discuss these problems and their cure in Sec. 7.8. - -In practice, we can realize a variety of filter characteristics that approach the ideal. Practical (realizable) filter characteristics are gradual, without jump discontinuities in amplitude response. - -### **DR ILL 7.11 The Unrealizable Gaussian Response** - -Show that a filter with Gaussian frequency response *H*(ω) = *e*−αω2 is unrealizable. Demonstrate this fact in two ways: first by showing that its impulse response is noncausal, and then by showing that |*H*(ω)| violates the Paley–Wiener criterion. [*Hint:* Use pair 22 in Table 7.1.] - -### THINKING IN THE TIME AND FREQUENCY DOMAINS: A TWO-DIMENSIONAL VIEW OF SIGNALS AND SYSTEMS - -Both signals and systems have dual personalities, the time domain and the frequency domain. For a deeper perspective, we should examine and understand both these identities because they offer complementary insights. An exponential signal, for instance, can be specified by its time-domain description such as *e*−2*t u*(*t*) or by its Fourier transform (its frequency-domain description) 1/(*j*ω +2). The time-domain description depicts the waveform of a signal. The frequency-domain description portrays its spectral composition [relative amplitudes of its sinusoidal (or exponential) components and their phases]. For the signal *e*−2*t* , for instance, the time-domain description portrays the exponentially decaying signal with a time constant 0.5. The frequency-domain description characterizes it as a lowpass signal, which can be synthesized by sinusoids with amplitudes decaying with frequency roughly as 1/ω. - -An LTIC system can also be described or specified in the time domain by its impulse response *h*(*t*) or in the frequency domain by its frequency response *H*(ω). In Sec. 2.6, we studied intuitive insights in the system behavior offered by the impulse response, which consists of characteristic modes of the system. By purely qualitative reasoning, we saw that the system responds well to signals that are similar to the characteristic modes and responds poorly to signals that are very different from those modes. We also saw that the shape of the impulse response *h*(*t*) determines the system time constant (speed of response), and pulse dispersion (spreading), which, in turn, determines the rate of pulse transmission. - -The frequency response *H*(ω) specifies the system response to exponential or sinusoidal input of various frequencies. This is precisely the filtering characteristic of the system. - -Experienced electrical engineers instinctively think in both domains (time and frequency) whenever possible. When they look at a signal, they consider its waveform, the signal width (duration), and the rate at which the waveform decays. This is basically a time-domain perspective. They also think of the signal in terms of its frequency spectrum, that is, in terms of its sinusoidal components and their relative amplitudes and phases, whether the spectrum is lowpass, bandpass, highpass, and so on. This is a frequency-domain perspective. Experienced electrical engineers think of a system in terms of its impulse response *h*(*t*). The width of *h*(*t*) indicates the time constant (response time): that is, how quickly the system is capable of responding to an input, and how much dispersion (spreading) it will cause. This is a time-domain perspective. From the frequency-domain perspective, these engineers view a system as a filter, which selectively transmits certain frequency components and suppresses the others [frequency response *H*(ω)]. Knowing the input signal spectrum and the frequency response of the system, they create a mental image of the output signal spectrum. This concept is precisely expressed by *Y*(ω) = *X*(ω)*H*(ω). - -We can analyze LTI systems by time-domain techniques or by frequency-domain techniques. Then why learn both? The reason is that the two domains offer complementary insights into system behavior. Some aspects are easily grasped in one domain; other aspects may be easier to see in the other domain. Both time-domain and frequency-domain methods are as essential for the study of signals and systems as two eyes are essential to a human being for correct visual perception of reality. A person can see with either eye, but for proper perception of three-dimensional reality, both eyes are essential. - -It is important to keep the two domains separate, and not to mix the entities in the two domains. If we are using the frequency domain to determine the system response, we must deal with all signals in terms of their spectra (Fourier transforms) and all systems in terms of their frequency responses. For example, to determine the system response *y*(*t*) to an input *x*(*t*), we must first convert the input signal into its frequency-domain description *X*(ω). The system description also must be in the frequency domain, that is, the frequency response *H*(ω). The output signal spectrum *Y*(ω) = *X*(ω)*H*(ω). Thus, the result (output) is also in the frequency domain. To determine the final answer *y*(*t*), we must take the inverse transform of *Y*(ω). - -## **[7.6 SIGNAL](#page-13-0) ENERGY** - -The signal energy *Ex* of a signal *x*(*t*) was defined in Ch. 1 as - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt -$$ -\n(7.44) - -Signal energy can be related to the signal spectrum *X*(ω) by substituting Eq. (7.10) in Eq. (7.44): - -$$ -E_x = \int_{-\infty}^{\infty} x(t)x^*(t) dt = \int_{-\infty}^{\infty} x(t) \left[ \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) e^{-j\omega t} d\omega \right] dt -$$ - -#### 734 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -Here, we used the fact that *x*∗(*t*), being the conjugate of *x*(*t*), can be expressed as the conjugate of the right-hand side of Eq. (7.10). Now, interchanging the order of integration yields - -$$ -E_x = \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) \left[ \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \right] d\omega -$$ - -= $\frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) X^*(\omega) d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega$ - -Consequently, - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega \qquad (7.45) -$$ - -This is *Parseval's theorem* (for the Fourier transform). A similar result was obtained in Eqs. (6.26) and (6.27) for a periodic signal and its Fourier series. This result allows us to determine the signal energy from either the time-domain specification *x*(*t*) or the corresponding frequency-domain specification *X*(ω). - -The right-hand side of Eq. (7.45) can be interpreted to mean that the energy of a signal *x*(*t*) results from energies contributed by all the spectral components of the signal *x*(*t*). The total signal energy is the area under |*X*(ω)2| (divided by 2π). If we consider a small band ω (ω → 0), as illustrated in Fig. 7.35, the energy *Ex* of the spectral components in this band is the area of |*X*(ω)| 2 under this band (divided by 2π): - -$$ -\Delta E_x = \frac{1}{2\pi} |X(\omega)|^2 \,\Delta \omega = |X(\omega)|^2 \,\Delta f \qquad \frac{\Delta \omega}{2\pi} = \Delta f \,\mathrm{Hz} -$$ - -Therefore, the energy contributed by the components in this band of *f* (in hertz) is |*X*(ω)| 2*f* . The total signal energy is the sum of energies of all such bands and is indicated by the area under |*X*(ω)| 2 as in Eq. (7.45). Therefore, |*X*(ω)| 2 is the *energy spectral density* (per unit bandwidth in hertz). - -For real signals, *X*(ω) and *X*(−ω) are conjugates, and |*X*(ω)| 2 is an even function of ω because - -$$ -|X(\omega)|^2 = X(\omega)X^*(\omega) = X(\omega)X(-\omega) -$$ - -**Figure 7.35** Interpretation of energy spectral density of a signal. - -Therefore, the energy of real signal *x*(*t*) can be expressed as† - -$$ -E_x = \frac{1}{\pi} \int_0^\infty |X(\omega)|^2 d\omega \tag{7.46} -$$ - -The signal energy *Ex*, which results from contributions from all the frequency components from ω = 0 to ∞, is given by (1/π times) the area under |*X*(ω)| 2 from ω = 0 to ∞. It follows that the energy contributed by spectral components of frequencies between ω1 and ω2 is - -$$ -\Delta E_x = \frac{1}{\pi} \int_{\omega_1}^{\omega_2} |X(\omega)|^2 d\omega \tag{7.47} -$$ - -### **EXAMPLE 7.20 Signal Energy and Parseval's Theorem** - -Find the energy of signal *x*(*t*) = *e*−*atu*(*t*). Determine the frequency *W* (rad/s) so that the energy contributed by the spectral components of all the frequencies below *W* is 95% of the signal energy *Ex*. - -We have - -$$ -E_x = \int_{-\infty}^{\infty} x^2(t) \, dt = \int_0^{\infty} e^{-2at} \, dt = \frac{1}{2a} -$$ - -We can verify this result by Parseval's theorem. For this signal, - -$$ -X(\omega) = \frac{1}{j\omega + a} -$$ - -and - -$$ -E_x = \frac{1}{\pi} \int_0^{\infty} |X(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{\infty} \frac{1}{\omega^2 + a^2} d\omega = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^{\infty} = \frac{1}{2a} -$$ - -The band ω = 0 to ω = *W* contains 95% of the signal energy, that is, 0.95/2*a*. Therefore, from Eq. (7.47) with ω1 = 0 and ω2 = *W*, we obtain - -$$ -\frac{0.95}{2a} = \frac{1}{\pi} \int_0^W \frac{d\omega}{\omega^2 + a^2} = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^W = \frac{1}{\pi a} \tan^{-1} \frac{W}{a} -$$ - -or - -$$ -\frac{0.95\pi}{2} = \tan^{-1}\frac{W}{a} \implies W = 12.706a \text{ rad/s} -$$ - -† In Eq. (7.46), it is assumed that *X*(ω) does not contain an impulse at ω = 0. If such an impulse exists, it should be integrated separately with a multiplying factor of 1/2π rather than 1/π. - -### 736 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -This result indicates that the spectral components of *x*(*t*) in the band from 0 (dc) to 12.706*a* rad/s (2.02*a* Hz) contribute 95% of the total signal energy; all the remaining spectral components (in the band from 12.706*a* rad/s to ∞) contribute only 5% of the signal energy. - -### **DR ILL 7.12 Signal Energy and Parseval's Theorem** - -Use Parseval's theorem to show that the energy of the signal *x*(*t*) = 2*a*/(*t* 2 +*a*2) is 2π/*a*. [*Hint:* Find *X*(ω) using pair 3 of Table 7.1 and the duality property.] - -### THE ESSENTIAL BANDWIDTH OF A SIGNAL - -The spectra of all practical signals extend to infinity. However, because the energy of any practical signal is finite, the signal spectrum must approach 0 as ω → ∞. Most of the signal energy is contained within a certain band of *B* Hz, and the energy contributed by the components beyond *B* Hz is negligible. We can therefore suppress the signal spectrum beyond *B* Hz with little effect on the signal shape and energy. The bandwidth *B* is called the *essential bandwidth* of the signal. The criterion for selecting *B* depends on the error tolerance in a particular application. We may, for example, select *B* to be that band which contains 95% of the signal energy.† This figure may be higher or lower than 95%, depending on the precision needed. Using such a criterion, we can determine the essential bandwidth of a signal. The essential bandwidth *B* for the signal *e*−*atu*(*t*), using 95% energy criterion, was determined in Ex. 7.20 to be 2.02*a* Hz. - -Suppression of all the spectral components of *x*(*t*) beyond the essential bandwidth results in a signal *x*ˆ(*t*), which is a close approximation of *x*(*t*). If we use the 95% criterion for the essential bandwidth, the energy of the error (the difference) *x*(*t*)− ˆ*x*(*t*) is 5% of *Ex*. - -## **[7.7 APPLICATION TO](#page-13-0) COMMUNICATIONS: AMPLITUDE MODULATION** - -*Modulation* causes a spectral shift in a signal and is used to gain certain advantages mentioned in our discussion of the frequency-shifting property. Broadly speaking, there are two classes of modulation: amplitude (linear) modulation and angle (nonlinear) modulation. In this section, we shall discuss some practical forms of amplitude modulation. - -† For lowpass signals, the essential bandwidth may also be defined as a frequency at which the value of the amplitude spectrum is a small fraction (about 1%) of its peak value. In Ex. 7.20, for instance, the peak value, which occurs at ω = 0, is 1/*a*. - -### **[7.7-1 Double-Sideband, Suppressed-Carrier \(DSB-SC\) Modulation](#page-13-0)** - -In amplitude modulation, the amplitude *A* of the carrier *A*cos(ω*ct* + θ*c*) is varied in some manner with the *baseband* (message)† signal *m*(*t*) (known as the *modulating signal*). The frequency ω*c* and the phase θ*c* are constant. We can assume θ*c* = 0 without loss of generality. If the carrier amplitude *A* is made directly proportional to the modulating signal *m*(*t*), the modulated signal is *m*(*t*) cos ω*ct* (Fig. 7.36). As was indicated earlier [Eq. (7.32)], this type of modulation simply shifts the spectrum of *m*(*t*) to the carrier frequency (Fig. 7.36c). Thus, if - -**Figure 7.36** DSB-SC modulation. - -† The term *baseband* is used to designate the band of frequencies of the signal delivered by the source or the input transducer. - -then - -$$ -m(t)\cos\omega_c t \Longleftrightarrow \frac{1}{2}[M(\omega + \omega_c) + M(\omega - \omega_c)] \tag{7.48} -$$ - -Recall that *M*(ω − ω*c*) is *M*(ω)-shifted to the right by ω*c* and *M*(ω + ω*c*) is *M*(ω)-shifted to the left by ω*c*. Thus, the process of modulation shifts the spectrum of the modulating signal to the left and the right by ω*c*. Note also that if the bandwidth of *m*(*t*) is *B* Hz, then, as indicated in Fig. 7.36c, the bandwidth of the modulated signal is 2*B* Hz. We also observe that the modulated signal spectrum centered at ω*c* is composed of two parts: a portion that lies above ω*c*, known as the *upper sideband (USB)*, and a portion that lies below ω*c*, known as the *lower sideband (LSB)*. Similarly, the spectrum centered at −ω*c* has upper and lower sidebands. This form of modulation is called *double sideband (DSB)* modulation for the obvious reason. - -The relationship of *B* to ω*c* is of interest. Figure 7.36c shows that ω*c* ≥ 2π*B* to avoid the overlap of the spectra centered at ±ω*c*. If ω*c* < 2π*B*, the spectra overlap and the information of *m*(*t*) are lost in the process of modulation, a loss that makes it impossible to get back *m*(*t*) from the modulated signal *m*(*t*) cos ω*ct*. † - -### **EXAMPLE 7.21 Double-Sideband Suppressed-Carrier Modulation** - -For a baseband signal *m*(*t*) = cos ω*mt*, find the DSB-SC signal and sketch its spectrum. Identify the upper and lower sidebands. - -We shall work this problem in the frequency domain as well as the time domain to clarify the basic concepts of DSB-SC modulation. In the frequency-domain approach, we work with the signal spectra. The spectrum of the baseband signal *m*(*t*) = cos ω*mt* is given by - -$$ -M(\omega) = \pi \left[ \delta(\omega - \omega_m) + \delta(\omega + \omega_m) \right] -$$ - -The spectrum consists of two impulses located at ±ω*m*, as depicted in Fig. 7.37a. - -The DSB-SC (modulated) spectrum, as indicated by Eq. (7.48), is the baseband spectrum in Fig. 7.37a shifted to the right and the left by ω*c* (times 0.5), as depicted in Fig. 7.37b. This spectrum consists of impulses at ±(ω*c* − ω*m*) and ±(ω*c* + ω*m*). The spectrum beyond ω*c* is the upper sideband (USB), and the one below ω*c* is the lower sideband (LSB). Observe that the DSB-SC spectrum does not have as a component the carrier frequency ω*c*. This is why the term *double-sideband, suppressed carrier* (DSB-SC) is used for this type of modulation. - -† Practical factors may impose additional restrictions on ω*c*. For instance, in broadcast applications, a radiating antenna can radiate only a narrow band without distortion. This restriction implies that avoiding distortion caused by the radiating antenna calls for ω*c*/2π*B* 1. The broadcast band AM radio, for instance, with *B* = 5 kHz and the band of 550–1600 kHz for carrier frequency gives a ratio of ω*c*/2π*B* roughly in the range of 100–300. - -**Figure 7.37** An example of DSB-SC modulation. - -In the time-domain approach, we work directly with signals in the time domain. For the baseband signal *m*(*t*) = cos ω*mt*, the DSB-SC signal ϕDSB-SC(*t*) is - -$$ -\varphi_{\text{DSB-SC}}(t) = m(t) \cos \omega_c t -$$ - -= $\cos \omega_m t \cos \omega_c t$ -= $\frac{1}{2} [\cos (\omega_c + \omega_m)t + \cos (\omega_c - \omega_m)t]$ (7.49) - -This result shows that when the baseband (message) signal is a single sinusoid of frequency ω*m*, the modulated signal consists of two sinusoids: the component of frequency ω*c* + ω*m* (the upper sideband), and the component of frequency ω*c* − ω*m* (the lower sideband). Figure 7.37b illustrates precisely the spectrum of ϕDSB-SC(*t*). Thus, each component of frequency ω*m* in the modulating signal results in two components of frequencies ω*c* + ω*m* and ω*c* − ω*m* in the modulated signal. This being a DSB-SC (suppressed-carrier) modulation, there is no component of the carrier frequency ω*c* on the right-hand side of Eq. (7.49).† - -### DEMODULATION OF DSB-SC SIGNALS - -The DSB-SC modulation translates or shifts the frequency spectrum to the left and the right by ω*c* (i.e., at +ω*c* and −ω*c*), as seen from Eq. (7.48). To recover the original signal *m*(*t*) from - -† The term *suppressed carrier* does not necessarily mean absence of the spectrum at the carrier frequency. "Suppressed carrier" merely implies that there is no discrete component of the carrier frequency. Since no discrete component exists, the DSB-SC spectrum does not have impulses at ±ω*c*, which further implies that the modulated signal *m*(*t*) cos ω*ct* does not contain a term of the form *k* cos ω*ct* [assuming that *m*(*t*) has a zero mean value]. - -**Figure 7.38** Demodulation of DSB-SC: **(a)** demodulator and **(b)** spectrum of *e*(*t*). - -the modulated signal, we must retranslate the spectrum to its original position. The process of recovering the signal from the modulated signal (retranslating the spectrum to its original position) is referred to as *demodulation*, or *detection*. Observe that if the modulated signal spectrum in Fig. 7.36c is shifted to the left and to the right by ω*c* (and halved), we obtain the spectrum illustrated in Fig. 7.38b, which contains the desired baseband spectrum in addition to an unwanted spectrum at ±2ω*c*. The latter can be suppressed by a lowpass filter. Thus, demodulation, which is almost identical to modulation, consists of multiplication of the incoming modulated signal *m*(*t*) cos ω*ct* by a carrier cos ω*ct* followed by a lowpass filter, as depicted in Fig. 7.38a. We can verify this conclusion directly in the time domain by observing that the signal *e*(*t*) in Fig. 7.38a is - -$$ -e(t) = m(t)\cos^2\omega_c t = \frac{1}{2}[m(t) + m(t)\cos 2\omega_c t] -$$ - -Therefore, the Fourier transform of the signal *e*(*t*) is - -$$ -E(\omega) = \frac{1}{2}M(\omega) + \frac{1}{4}[M(\omega + 2\omega_c) + M(\omega - 2\omega_c)] -$$ - -Hence, *e*(*t*) consists of two components (1/2)*m*(*t*) and (1/2)*m*(*t*) cos 2ω*ct*, with their spectra, as illustrated in Fig. 7.38b. The spectrum of the second component, being a modulated signal with carrier frequency 2ω*c*, is centered at ±2ω*c*. Hence, this component is suppressed by the lowpass filter in Fig. 7.38a. The desired component (1/2)*M*(ω), being a lowpass spectrum (centered at ω = 0), passes through the filter unharmed, resulting in the output (1/2)*m*(*t*). - -A possible form of lowpass filter characteristics is depicted (dotted) in Fig. 7.38b. In this method of recovering the baseband signal, called *synchronous detection,* or *coherent detection,* we use a carrier of exactly the same frequency (and phase) as the carrier used for modulation. Thus, for demodulation, we need to generate a local carrier at the receiver in frequency and phase coherence (synchronism) with the carrier used at the modulator. We shall demonstrate in Ex. 7.22 that both phase and frequency synchronism are extremely critical. - -### **EXAMPLE 7.22 Frequency and Phase Incoherence in DSB-SC** - -Discuss the effect of lack of frequency and phase coherence (synchronism) between the carriers at the modulator (transmitter) and the demodulator (receiver) in DSB-SC. - -Let the modulator carrier be cos ω*ct* (Fig. 7.36a). For the demodulator in Fig. 7.38a, we shall consider two cases: with carrier cos(ω*ct*+θ ) (phase error of θ) and with carrier cos(ω*c*+ω)*t* (frequency error ω). - -**(a)** With the demodulator carrier cos(ω*ct* + θ ) (instead of cos ω*ct*) in Fig. 7.38a, the multiplier output is *e*(*t*) = *m*(*t*) cos ω*ct* cos(ω*ct* + θ ) instead of *m*(*t*) cos2ω*ct*. From the trigonometric identity, we obtain - -$$ -e(t) = m(t)\cos\omega_c t \cos(\omega_c t + \theta) -$$ - -= $\frac{1}{2}m(t)[\cos\theta + \cos(2\omega_c t + \theta)]$ - -The spectrum of the component (1/2)*m*(*t*) cos(2ω*ct* + θ ) is centered at ±2ω*c*. Consequently, it will be filtered out by the lowpass filter at the output. The component (1/2)*m*(*t*) cos θ is the signal *m*(*t*) multiplied by a constant (1/2) cos θ. The spectrum of this component is centered at ω = 0 (lowpass spectrum) and will pass through the lowpass filter at the output, yielding the output (1/2)*m*(*t*) cos θ. - -If θ is constant, the phase asynchronism merely yields an output that is attenuated (by a factor cos θ). Unfortunately, in practice, θ is often the phase difference between the carriers generated by two distant generators and varies randomly with time. This variation would result in an output whose gain varies randomly with time. - -**(b)** In the case of frequency error, the demodulator carrier is cos(ω*c*+ω)*t*. This situation is very similar to the phase error case in part (a) with θ replaced by (ω)*t*. Following the analysis in part (a), we can express the demodulator product *e*(*t*) as - -$$ -e(t) = m(t)\cos\omega_c t \cos(\omega_c + \Delta\omega)t -$$ - -= $\frac{1}{2}m(t)[\cos(\Delta\omega)t + \cos(2\omega_c + \Delta\omega)t]$ - -The spectrum of the component (1/2)*m*(*t*) cos(2ω*c* + ω)*t* is centered at ±(2ω*c* + ω). Consequently, this component will be filtered out by the lowpass filter at the output. The component (1/2)*m*(*t*) cos(ω)*t* is the signal *m*(*t*) multiplied by a low-frequency carrier of frequency ω. The spectrum of this component is centered at ±ω. In practice, the frequency error (ω) is usually very small. Hence, the signal (1/2)*m*(*t*) cos(ω)*t* (whose spectrum is centered at ±ω) is a lowpass signal and passes through the lowpass filter at the output, resulting in the output (1/2)*m*(*t*) cos(ω)*t*. The output is the desired signal *m*(*t*) multiplied by a very-low-frequency sinusoid cos(ω)*t*. The output in this case is not merely an attenuated replica of the desired signal *m*(*t*), but represents *m*(*t*) multiplied by a time-varying gain cos(ω)*t*. If, for instance, the transmitter and the receiver carrier frequencies differ just by 1 Hz, the output will be the desired signal *m*(*t*) multiplied by a time-varying signal whose gain goes from the maximum to 0 every half-second. This is like a restless child fiddling with the volume control knob of a receiver, going from maximum volume to zero volume every half-second. This kind of distortion (called the *beat effect*) is beyond repair. - -### **[7.7-2 Amplitude Modulation \(AM\)](#page-13-0)** - -For the suppressed-carrier scheme just discussed, a receiver must generate a carrier in frequency and phase synchronism with the carrier at a transmitter that may be located hundreds or thousands of miles away. This situation calls for a sophisticated receiver, which could be quite costly. The other alternative is for the transmitter to transmit a carrier *A* cosω*ct* [along with the modulated signal *m*(*t*) cosω*ct*] so that there is no need to generate a carrier at the receiver. In this case, the transmitter needs to transmit much larger power, a rather expensive procedure. In point-to-point communications, where there is one transmitter for each receiver, substantial complexity in the receiver system can be justified, provided there is a large enough saving in expensive high-power transmitting equipment. On the other hand, for a broadcast system with a multitude of receivers for each transmitter, it is more economical to have one expensive high-power transmitter and simpler, less expensive receivers. The second option (transmitting a carrier along with the modulated signal) is the obvious choice in this case. This is amplitude modulation (AM), in which the transmitted signal ϕAM (*t*) is given by - -$$ -\varphi_{AM}(t) = A\cos\omega_c t + m(t)\cos\omega_c t = [A + m(t)]\cos\omega_c t \tag{7.50} -$$ - -Recall that the DSB-SC signal is *m*(*t*) cos ω*ct*. From Eq. (7.50) it follows that the AM signal is identical to the DSB-SC signal with *A*+*m*(*t*) as the modulating signal [instead of *m*(*t*)]. Therefore, to sketch ϕAM (*t*), we sketch *A* + *m*(*t*) and −[*A* + *m*(*t*)] as the envelopes and fill in between with the sinusoid of the carrier frequency. Two cases are considered in Fig. 7.39. In the first case, *A* is large enough so that *A* + *m*(*t*) ≥ 0 (is nonnegative) for all values of *t*. In the second case, *A* is not large enough to satisfy this condition. In the first case, the envelope (Fig. 7.39d) has the same shape as *m*(*t*) (although riding on a dc of magnitude *A*). In the second case, the envelope shape is not *m*(*t*), for some parts get rectified (Fig. 7.39e). Thus, we can detect the desired signal *m*(*t*) by detecting the envelope in the first case. In the second case, such a detection is not possible. We shall see that envelope detection is an extremely simple and inexpensive operation, which does not require generation of a local carrier for the demodulation. But as just noted, the envelope of AM has the information about *m*(*t*) only if the AM signal [*A*+*m*(*t*)] cos ω*ct* satisfies the condition *A*+*m*(*t*) > 0 for all *t*. Thus, the condition for envelope detection of an AM signal is - -$$ -A + m(t) \ge 0 \qquad \text{for all } t \tag{7.51} -$$ - -If *mp* is the peak amplitude (positive or negative) of *m*(*t*), then Eq. (7.51) is equivalent to - -$$ -A\geq m_p -$$ - -Thus, the minimum carrier amplitude required for the viability of envelope detection is *mp*. This point is clearly illustrated in Fig. 7.39. - -We define the *modulation index* μ as - -$$ -\mu = \frac{m_p}{A} \tag{7.52} -$$ - -where *A* is the carrier amplitude. Note that *mp* is a constant of the signal *m*(*t*). Because *A* ≥ *mp* and because there is no upper bound on *A*, it follows that - -0 ≤ μ ≤ 1 - -**Figure 7.39** An AM signal **(a)** for two values of A **(b, c)** and the respective envelopes **(d, e)**. - -as the required condition for the viability of demodulation of AM by an envelope detector. - -When *A* < *mp*, Eq. (7.52) shows that μ > 1 (overmodulation, shown in Fig. 7.39e). In this case, the option of envelope detection is no longer viable. We then need to use synchronous demodulation. Note that synchronous demodulation can be used for any value of μ (see Prob. 7.7-7). The envelope detector, which is considerably simpler and less expensive than the synchronous detector, can be used only when μ ≤ 1. - -### **EXAMPLE 7.23 Amplitude Modulation** - -Sketch ϕAM (*t*) for modulation indices of μ = 0.5 (50% modulation) and μ = 1 (100% modulation), when *m*(*t*) = *B*cos ω*mt*. This case is referred to as *tone modulation* because the modulating signal is a pure sinusoid (or tone). - -### 744 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -In this case, *mp* = *B* and the modulation index according to Eq. (7.52) is - -$$ -\mu = \frac{B}{A} -$$ - -Hence, *B* = μ*A* and - -$$ -m(t) = B\cos\omega_m t = \mu A\cos\omega_m t -$$ - -Therefore, - -$$ -\varphi_{AM}(t) = [A + m(t)] \cos \omega_c t = A[1 + \mu \cos \omega_m t] \cos \omega_c t -$$ - -The modulated signals corresponding to μ = 0.5 and μ = 1 appear in Figs. 7.40a and 7.40b, respectively. - -### DEMODULATION OF AM: THE ENVELOPE DETECTOR - -The AM signal can be demodulated coherently by a locally generated carrier (see Prob. 7.7-7). Since, however, coherent, or synchronous, demodulation of AM (with μ ≤ 1) will defeat the very purpose of AM, it is rarely used in practice. We shall consider here one of the noncoherent methods of AM demodulation, *envelope detection*. † - -In an envelope detector, the output of the detector follows the envelope of the (modulated) input signal. The circuit illustrated in Fig. 7.41a functions as an envelope detector. During the positive cycle of the input signal, the diode conducts and the capacitor *C* charges up to the peak voltage of the input signal (Fig. 7.41b). As the input signal falls below this peak value, the diode is cut off, because the capacitor voltage (which is very nearly the peak voltage) is greater than the input signal voltage, a circumstance causing the diode to open. The capacitor now discharges through the resistor *R* at a slow rate (with a time constant *RC*). During the next positive cycle, - -† There are also other methods of noncoherent detection. The rectifier detector consists of a rectifier followed by a lowpass filter. This method is also simple and almost as inexpensive as the envelope detector [4]. The nonlinear detector, although simple and inexpensive, results in a distorted output. - -**Figure 7.41** Demodulation by means of envelope detector. - -the same drama repeats. When the input signal becomes greater than the capacitor voltage, the diode conducts again. The capacitor again charges to the peak value of this (new) cycle. As the input voltage falls below the new peak value, the diode cuts off again and the capacitor discharges slowly during the cutoff period, a process that changes the capacitor voltage very slightly. - -In this manner, during each positive cycle, the capacitor charges up to the peak voltage of the input signal and then decays slowly until the next positive cycle. Thus, the output voltage *vC*(*t*) follows closely the envelope of the input. The capacitor discharge between positive peaks, however, causes a ripple signal of frequency ω*c* in the output. This ripple can be reduced by increasing the time constant *RC* so that the capacitor discharges very little between the positive peaks (*RC* 1/ω*c*). Making *RC* too large, however, would make it impossible for the capacitor voltage to follow the envelope (see Fig. 7.41b). Thus, *RC* should be large in comparison to 1/ω*c* but small in comparison to 1/2π*B*, where *B* is the highest frequency in *m*(*t*). Incidentally, these two conditions also require that ω*c* 2π*B*, a condition necessary for a well-defined envelope. - -The envelope-detector output *vC*(*t*) is *A* + *m*(*t*) plus a ripple of frequency ω*c*. The dc term *A* can be blocked out by a capacitor or a simple *RC* highpass filter. The ripple is reduced further by - -### 746 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM - -another (lowpass) *RC* filter. In the case of audio signals, the speakers also act as lowpass filters, which further enhances suppression of the high-frequency ripple. - -### **[7.7-3 Single-Sideband Modulation \(SSB\)](#page-13-0)** - -Now consider the baseband spectrum *M*(ω) (Fig. 7.42a) and the spectrum of the DSB-SC modulated signal *m*(*t*) cos ω*ct* (Fig. 7.42b). The DSB spectrum in Fig. 7.42b has two sidebands: the upper and the lower (USB and LSB), both containing complete information on *M*(ω) [see Eq. (7.12)]. Clearly, it is redundant to transmit both sidebands, a process that requires twice the bandwidth of the baseband signal. A scheme in which only one sideband is transmitted is known - -**Figure 7.42** Spectra for single-sideband transmission: **(a)** baseband, **(b)** DSB, **(c)** USB, **(d)** LSB, and **(e)** synchronously demodulated signal. - -as *single-sideband (SSB) transmission,* which requires only half the bandwidth of the DSB signal. Thus, we transmit only the upper sidebands (Fig. 7.42c) or only the lower sidebands (Fig. 7.42d). - -An SSB signal can be coherently (synchronously) demodulated. For example, multiplication of a USB signal (Fig. 7.42c) by 2 cos ω*ct* shifts its spectrum to the left and to the right by ω*c*, yielding the spectrum in Fig. 7.42e. Lowpass filtering of this signal yields the desired baseband signal. The case is similar with an LSB signal. Hence, demodulation of SSB signals is identical to that of DSB-SC signals, and the synchronous demodulator in Fig. 7.38a can demodulate SSB signals. Note that we are talking of SSB signals without an additional carrier. Hence, they are suppressed-carrier signals (SSB-SC). - -### **EXAMPLE 7.24 Single-Sideband Modulation** - -Find the USB (upper sideband) and LSB (lower sideband) signals when *m*(*t*) = cos ω*mt*. Sketch their spectra, and show that these SSB signals can be demodulated using the synchronous demodulator in Fig. 7.38a. - -The DSB-SC signal for this case is - -$$ -\varphi_{\text{DSB-SC}}(t) = m(t)\cos\omega_c t = \cos\omega_m t \cos\omega_c t \qquad = \frac{1}{2} [\cos(\omega_c - \omega_m)t + \cos(\omega_c + \omega_m)t] -$$ - -As pointed out in Ex. 7.21, the terms (1/2) cos(ω*c* + ω*m*)*t* and (1/2) cos(ω*c* − ω*m*)*t* represent the upper and lower sidebands, respectively. The spectra of the upper and lower sidebands are given in Figs. 7.43a and 7.43b. Observe that these spectra can be obtained from the DSB-SC spectrum in Fig. 7.37b by using a proper filter to suppress the undesired sidebands. For instance, the USB signal in Fig. 7.43a can be obtained by passing the DSB-SC signal (Fig. 7.37b) through a highpass filter of cutoff frequency ω*c*. Similarly, the LSB signal in Fig. 7.43b can be obtained by passing the DSB-SC signal through a lowpass filter of cutoff frequency ω*c*. - -If we apply the LSB signal (1/2) cos(ω*c* − ω*m*)*t* to the synchronous demodulator in Fig. 7.38a, the multiplier output is - -$$ -e(t) = \frac{1}{2}\cos{(\omega_c - \omega_m)t}\cos{\omega_c t} = \frac{1}{4}[\cos{\omega_m t} + \cos{(2\omega_c - \omega_m)t}] -$$ - -The term (1/4) cos(2ω*c*−ω*m*)*t* is suppressed by the lowpass filter, producing the desired output (1/4) cos ω*mt* [which is *m*(*t*)/4]. The spectrum of this term is π[δ(ω+ω*m*)+δ(ω−ω*m*)]/4, as depicted in Fig. 7.43c. In the same way, we can show that the USB signal can be demodulated by the synchronous demodulator. - -In the frequency domain, demodulation (multiplication by cos ω*ct*) amounts to shifting the LSB spectrum (Fig. 7.43b) to the left and the right by ω*c* (times 0.5) and then suppressing the high frequency, as illustrated in Fig. 7.43c. The resulting spectrum represents the desired signal (1/4)*m*(*t*). - -### GENERATION OF SSB SIGNALS - -Two methods are commonly used to generate SSB signals. The *selective-filtering method* uses sharp cutoff filters to eliminate the undesired sideband, and the second method uses phase-shifting networks to achieve the same goal [4].† We shall consider here only the first method. - -Selective filtering is the most commonly used method of generating SSB signals. In this method, a DSB-SC signal is passed through a sharp cutoff filter to eliminate the undesired sideband. - -To obtain the USB, the filter should pass all components above ω*c* unattenuated and completely suppress all components below ω*c*. Such an operation requires an ideal filter, which is unrealizable. It can, however, be realized closely if there is some separation between the passband and the stopband. Fortunately, the voice signal provides this condition, because its spectrum shows little power content at the origin (Fig. 7.44). Moreover, articulation tests show that for speech signals, frequency components below 300 Hz are not important. In other words, we may suppress all speech components below 300 Hz without appreciably affecting intelligibility.‡ Thus, filtering of the unwanted sideband becomes relatively easy for speech signals because we have a 600 Hz transition region around the cutoff frequency ω*c*. For some signals, which have considerable power - -† Yet another method, known as Weaver's method, is also used to generate SSB signals. - -‡ Similarly, suppression of components of a speech signal above 3500 Hz causes no appreciable change in intelligibility. - -at low frequencies (around ω = 0), SSB techniques cause considerable distortion. Such is the case with video signals. Consequently, for video signals, instead of SSB, we use another technique, the *vestigial sideband (VSB)*, which is a compromise between SSB and DSB. It inherits the advantages of SSB and DSB but avoids their disadvantages at a cost of slightly increased bandwidth. VSB signals are relatively easy to generate, and their bandwidth is only slightly (typically 25%) greater than that of SSB signals. In VSB signals, instead of rejecting one sideband completely (as in SSB), we accept a gradual cutoff from one sideband [4]. - -### **[7.7-4 Frequency-Division Multiplexing](#page-13-0)** - -Signal multiplexing allows transmission of several signals on the same channel. Later, in Ch. 8 (Sec. 8.2-2), we shall discuss time-division multiplexing (TDM), where several signals time-share the same channel, such as a cable or an optical fiber. In frequency-division multiplexing (FDM), the use of modulation, as illustrated in Fig. 7.45, makes several signals share the band of the same channel. Each signal is modulated by a different carrier frequency. The various carriers are adequately separated to avoid overlap (or interference) between the spectra of various modulated signals. These carriers are referred to as *subcarriers*. Each signal may use a different kind of modulation, for example, DSB-SC, AM, SSB-SC, VSB-SC, or even other forms of modulation, not discussed here [such as FM (frequency modulation) or PM (phase modulation)]. The modulated-signal spectra may be separated by a small guard band to avoid interference and to facilitate signal separation at the receiver. - -When all the modulated spectra are added, we have a composite signal that may be considered to be a new baseband signal. Sometimes, this composite baseband signal may be used to further modulate a high-frequency (radio frequency, or RF) carrier for the purpose of transmission. - -At the receiver, the incoming signal is first demodulated by the RF carrier to retrieve the composite baseband, which is then bandpass-filtered to separate the modulated signals. Then each modulated signal is individually demodulated by an appropriate subcarrier to obtain all the basic baseband signals. - -## **7.8 DATA [TRUNCATION: WINDOW](#page-13-0) FUNCTIONS** - -We often need to truncate data in diverse situations from numerical computations to filter design. For example, if we need to compute numerically the Fourier transform of some signal, say, *e*−*t u*(*t*), we will have to truncate the signal *e*−*t u*(*t*) beyond a sufficiently large value of *t* (typically five time constants and above). The reason is that in numerical computations, we have to deal with - -**Figure 7.45** Frequency-division multiplexing: **(a)** FDM spectrum **(b)** transmitter, and **(c)** receiver. - -data of finite duration. Similarly, the impulse response *h*(*t*) of an ideal lowpass filter is noncausal and approaches zero asymptotically as |*t*|→∞. For a practical design, we may want to truncate *h*(*t*) beyond a sufficiently large value of |*t*| to make *h*(*t*) causal and of finite duration. In signal sampling, to eliminate aliasing, we must use an antialiasing filter to truncate the signal spectrum beyond the half-sampling frequency ω*s*/2. Again, we may want to synthesize a periodic signal by adding the first *n* harmonics and truncating all the higher harmonics. These examples show that data truncation can occur in both time and frequency domains. On the surface, truncation appears to be a simple problem of cutting off the data at a point at which values are deemed to be sufficiently small. Unfortunately, this is not the case. Simple truncation can cause some unsuspected problems. - -### WINDOW FUNCTIONS - -Truncation operation may be regarded as multiplying a signal of a large width by a window function of a smaller (finite) width. Simple truncation amounts to using a *rectangular window wR*(*t*) (shown later in Fig. 7.48a) in which we assign unit weight to all the data within the window width (|*t*| < *T*/2), and assign zero weight to all the data lying outside the window (|*t*| > *T*/2). It is also possible to use a window in which the weight assigned to the data within the window may not be constant. In a *triangular window wT* (*t*), for example, the weight assigned to data decreases linearly over the window width (shown later in Fig. 7.48b). - -Consider a signal *x*(*t*) and a window function *w*(*t*). If *x*(*t*) ⇐⇒ *X*(ω) and *w*(*t*) ⇐⇒ *W*(ω), and if the windowed function *xw*(*t*) ⇐⇒ *Xw*(ω), then - -$$ -x_w(t) = x(t)w(t) -$$ - and $X_w(\omega) = \frac{1}{2\pi}X(\omega)*W(\omega)$ - -According to the width property of convolution, it follows that the width of *Xw*(ω) equals the sum of the widths of *X*(ω) and *W*(ω). Thus, truncation of a signal increases its bandwidth by the amount of bandwidth of *w*(*t*). Clearly, the truncation of a signal causes its spectrum to spread (or smear) by the amount of the bandwidth of *w*(*t*). Recall that the signal bandwidth is inversely proportional to the signal duration (width). Hence, the wider the window, the smaller its bandwidth, and the smaller the *spectral spreading*. This result is predictable because a wider window means that we are accepting more data (closer approximation), which should cause smaller distortion (smaller spectral spreading). Smaller window width (poorer approximation) causes more spectral spreading (more distortion). In addition, since *W*(ω) is really not strictly bandlimited and its spectrum → 0 only asymptotically, the spectrum of *Xw*(ω) → 0 asymptotically also at the same rate as that of *W*(ω), even if *X*(ω) is, in fact, strictly bandlimited. Thus, windowing causes the spectrum of *X*(ω) to spread into the band where it is supposed to be zero. This effect is called *leakage*. The following example clarifies these twin effects of spectral spreading and leakage. - -Let us consider *x*(*t*) = cos ω0*t* and a rectangular window *wR*(*t*) = rect(*t*/*T*), illustrated in Fig. 7.46b. The reason for selecting a sinusoid for *x*(*t*) is that its spectrum consists of spectral lines of zero width (Fig. 7.46a). Hence, this choice will make the effect of spectral spreading and leakage easily discernible. The spectrum of the truncated signal *xw*(*t*) is the convolution of the two impulses of *X*(ω) with the sinc spectrum of the window function. Because the convolution of any function with an impulse is the function itself (shifted at the location of the impulse), the resulting spectrum of the truncated signal is 1/2π times the two sinc pulses at ±ω0, as depicted in Fig. 7.46c (also see Fig. 7.26). Comparison of spectra *X*(ω) and *Xw*(ω) reveals the effects of truncation. These are: - -1. The spectral lines of *X*(ω) have zero width. But the truncated signal is spread out by 2π/*T* about each spectral line. The amount of spread is equal to the width of the mainlobe of the window spectrum. One effect of this *spectral spreading* (or smearing) is that if *x*(*t*) has two spectral components of frequencies differing by less than 4π/*T* rad/s (2/*T* Hz), they - -**Figure 7.46** Windowing and its effects. - -will be indistinguishable in the truncated signal. The result is loss of spectral resolution. We would like the spectral spreading [mainlobe width of *W*(ω)] to be as small as possible. - -2. In addition to the mainlobe spreading, the truncated signal has sidelobes, which decay slowly with frequency. The spectrum of *x*(*t*) is zero everywhere except at ±ω0. On the other hand, the truncated signal spectrum *Xw*(ω) is zero nowhere because of the sidelobes. These sidelobes decay asymptotically as 1/ω. Thus, the truncation causes spectral *leakage* in the band where the spectrum of the signal *x*(*t*) is zero. The peak *sidelobe* magnitude is 0.217 times the mainlobe magnitude (13.3 dB below the peak mainlobe magnitude). Also, the sidelobes decay at a rate 1/ω, which is −6 dB/octave (or −20 dB/decade). This is the sidelobe's *rolloff rate*. We want smaller sidelobes with a faster rate of decay (high rolloff rate). Figure 7.46d, which plots |*WR*(ω)| as a function of ω, clearly shows the mainlobe and sidelobe features, with the first sidelobe amplitude −13.3 dB below the mainlobe amplitude and the sidelobes decaying at a rate of −6 dB/octave (or −20 dB/decade). - -So far, we have discussed the effect on the signal spectrum of signal truncation (truncation in the time domain). Because of the time-frequency duality, the effect of spectral truncation (truncation in frequency domain) on the signal shape is similar. - -### REMEDIES FOR SIDE EFFECTS OF TRUNCATION - -For better results, we must try to minimize the twin side effects of truncations: spectral spreading (mainlobe width) and leakage (sidelobe). Let us consider each of these ills. - -- 1. The spectral spread (mainlobe width) of the truncated signal is equal to the bandwidth of the window function *w*(*t*). We know that the signal bandwidth is inversely proportional to the signal width (duration). Hence, to reduce the spectral spread (mainlobe width), we need to increase the window width. -- 2. To improve the leakage behavior, we must search for the cause of the slow decay of sidelobes. In Ch. 6, we saw that the Fourier spectrum decays as 1/ω for a signal with jump discontinuity, decays as 1/ω2 for a continuous signal whose first derivative is discontinuous, and so on.† Smoothness of a signal is measured by the number of continuous derivatives it possesses. The smoother the signal, the faster the decay of its spectrum. Thus, we can achieve a given leakage behavior by selecting a suitably smooth (tapered) window. -- 3. For a given window width, the remedies for the two effects are incompatible. If we try to improve one, the other deteriorates. For instance, among all the windows of a given width, the rectangular window has the smallest spectral spread (mainlobe width), but its sidelobes have high level and they decay slowly. A tapered (smooth) window of the same width has smaller and faster decaying sidelobes, but it has a wider mainlobe.‡ But we can compensate for the increased mainlobe width by widening the window. Thus, we can remedy both the side effects of truncation by selecting a suitably smooth window of sufficient width. - -There are several well-known tapered-window functions, such as Bartlett (triangular), Hanning (von Hann), Hamming, Blackman, and Kaiser, which truncate the data gradually. These - -† This result was demonstrated for periodic signals. However, it applies to aperiodic signals also. This is because we showed in the beginning of this chapter that if *xT*0 (*t*) is a periodic signal formed by periodic extension of an aperiodic signal *x*(*t*), then the spectrum of *xT*0 (*t*) is (1/*T*0 times) the samples of *X*(ω). Thus, - -what is true of the decay rate of the spectrum of *xT*0 (*t*) is also true of the rate of decay of *X*(ω). ‡ A tapered window yields a higher mainlobe width because the effective width of a tapered window is smaller than that of the rectangular window; see Sec. 2.6-2 [Eq. (2.47)] for the definition of effective width. Therefore, from the reciprocity of the signal width and its bandwidth, it follows that the rectangular window mainlobe is narrower than a tapered window. - -| No. | Window w(t) | Mainlobe
Width | Rolloff
Rate
Level (dB) | Peak
Sidelobe | -|-----|---------------------------------------------------------------------|-------------------|-------------------------------|------------------| -| 1 | t
Rectangular: rect
T | 4π
T | −6 | −13.3 | -| 2 | t
Bartlett:
2T | 8π
T | −12 | −26.5 | -| 3 | 2πt
!
Hanning: 0.5
1+cos
T | 8π
T | −18 | −31.5 | -| 4 | 2πt
Hamming: 0.54+0.46 cos
T | 8π
T | −6 | −42.7 | -| 5 | 2πt
4πt
Blackman: 0.42+0.5 cos
+0.08 cos
T
T | 12π
T | −18 | −58.1 | -| 6 |
2
t
I0
α
1−4
T
Kaiser:
0 ≤ α ≤ 10
I0(α) | 11.2π
T | −6 | −59.9 | -| | | | (α = 8.168) | | - -**TABLE 7.3** Some Window Functions and Their Characteristics - -**Figure 7.47 (a)** Hanning and **(b)** Hamming windows. - -windows offer different trade-offs with respect to spectral spread (mainlobe width), the peak sidelobe magnitude, and the leakage rolloff rate, as indicated in Table 7.3 [5, 6]. Observe that all windows are symmetrical about the origin (i.e., are even functions of *t*). Because of this feature, *W*(ω) is a real function of ω; that is, *W*(ω) is either 0 or π. Hence, the phase function of the truncated signal has a minimal amount of distortion. - -Figure 7.47 shows two well-known tapered-window functions, the von Hann (or Hanning) window *w*Han(*x*) and the Hamming window *w*Ham(*x*). We have intentionally used the independent variable *x* because windowing can be performed in the time domain as well as in the frequency domain, so *x* could be *t* or ω, depending on the application. - -There are hundreds of windows, all with different characteristics. But the choice depends on a particular application. The rectangular window has the narrowest mainlobe. The Bartlett (triangle) window (also called the Fejer or Cesaro) is inferior in all respects to the Hanning window. For this reason, it is rarely used in practice. Hanning is preferred over Hamming in spectral analysis because it has faster sidelobe decay. For filtering applications, on the other hand, the Hamming window is chosen because it has the smallest sidelobe magnitude for a given mainlobe width. The Hamming window is the most widely used general-purpose window. The Kaiser window, which uses *I*0(α), the modified zero-order Bessel function, is more versatile and adjustable. Selecting a proper value of α (0 ≤ α ≤ 10) allows the designer to tailor the window to suit a particular application. The parameter α controls the mainlobe-sidelobe trade-off. When α = 0, the Kaiser window is the rectangular window. For α = 5.4414, it is the Hamming window, and when α = 8.885, it is the Blackman window. As α increases, the mainlobe width increases and the sidelobe level decreases. - -### **[7.8-1 Using Windows in Filter Design](#page-13-0)** - -We shall design an ideal lowpass filter of bandwidth *W* rad/s, with frequency response *H*(ω), as shown in Fig. 7.48e or Fig. 7.48f. For this filter, the impulse response *h*(*t*) = (*W*/π )sinc (*Wt*) (Fig. 7.48c) is noncausal and, therefore, unrealizable. Truncation of *h*(*t*) by a suitable window (Fig. 7.48a) makes it realizable, although the resulting filter is now an approximation to the desired ideal filter.† We shall use a rectangular window *wR*(*t*) and a triangular (Bartlett) window *wT* (*t*) to truncate *h*(*t*), and then examine the resulting filters. The truncated impulse responses *hR*(*t*) = *h*(*t*)*wR*(*t*) and *hT* (*t*) = *h*(*t*)*wT* (*t*) are depicted in Fig. 7.48d. Hence, the windowed filter frequency response is the convolution of *H*(ω) with the Fourier transform of the window, as illustrated in Figs. 7.48e and 7.48f. We make the following observations. - -- 1. The windowed filter spectra show *spectral spreading* at the edges, and instead of a sudden switch there is a gradual transition from the passband to the stopband of the filter. The transition band is smaller (2π/*T* rad/s) for the rectangular case than for the triangular case (4π/*T* rad/s). -- 2. Although *H*(ω) is bandlimited, the windowed filters are not. But the stopband behavior of the triangular case is superior to that of the rectangular case. For the rectangular window, the leakage in the stopband decreases slowly (as 1/ω) in comparison to that of the triangular window (as 1/ω2). Moreover, the rectangular case has a higher peak sidelobe amplitude than that of the triangular window. - -## **[7.9 MATLAB: FOURIER](#page-13-0) TRANSFORM TOPICS** - -MATLAB is useful for investigating a variety of Fourier transform topics. In this section, a rectangular pulse is used to investigate the scaling property, Parseval's theorem, essential bandwidth, and spectral sampling. Kaiser window functions are also investigated. - -† In addition to truncation, we need to delay the truncated function by *T*/2 to render it causal. However, the time delay only adds a linear phase to the spectrum without changing the amplitude spectrum. Thus, to simplify our discussion, we shall ignore the delay. - -**Figure 7.48** Window-based filter design. - -### **[7.9-1 The Sinc Function and the Scaling Property](#page-13-0)** - -As shown in Ex. 7.2, the Fourier transform of *x*(*t*) = rect(*t*/τ ) is *X*(ω) = τ sinc (ωτ/2). To represent *X*(ω) in MATLAB, a sinc function is first required. As an alternative to the signal processing toolbox function sinc, which computes sinc(*x*) as sin(π*x*)/π*x*, we create our own function that follows the conventions of this book and defines sinc(*x*) = sin(*x*)/*x*. - -function [y] = CH7MP1(x) % CH7MP1.m : Chapter 7, MATLAB Program 1 % Function M-file computes the sinc function, y = sin(x)/x. - -y(x==0) = 1; y(x~=0) = sin(x(x~=0))./x(x~=0); - -The computational simplicity of sinc (*x*) = sin(*x*)/*x* is somewhat deceptive: sin(0)/0 results in a divide-by-zero error. Thus, program CH7MP1 assigns sinc (0) = 1 and computes the remaining values according to the definition. Notice that CH7MP1 cannot be directly replaced by an anonymous function. Anonymous functions cannot have multiple lines or contain certain commands such as =, if, or for. M-files, however, can be used to define an anonymous function. For example, we can represent *X*(ω) as an anonymous function that is defined in terms of CH7MP1. - -``` ->> X = @(omega,tau) tau*CH7MP1(omega*tau/2); -``` - -Once we have defined *X*(ω), it is simple to investigate the effects of scaling the pulse width τ . Consider the three cases τ = 1.0, τ = 0.5, and τ = 2.0. - -``` ->> omega = linspace(-4*pi,4*pi,200); ->> plot(omega,X(omega,1),'k-',omega,X(omega,0.5),'k-.',omega,X(omega,2),'k--'); ->> grid; axis tight; xlabel('\omega'); ylabel('X(\omega)'); ->> legend('Baseline (\tau = 1)','Compressed (\tau = 0.5)',... ->> 'Expanded (\tau = 2.0)'); -``` - -Figure 7.49 confirms the reciprocal relationship between signal duration and spectral bandwidth: time compression causes spectral expansion, and time expansion causes spectral compression. Additionally, spectral amplitudes are directly related to signal energy. As a signal is compressed, signal energy and thus spectral magnitude decrease. The opposite effect occurs when the signal is expanded. - -**Figure 7.49** Spectra *X*(ω) = τ sinc (ωτ/2) for τ = 1.0, τ = 0.5, and τ = 2.0. - -### **[7.9-2 Parseval's Theorem and Essential Bandwidth](#page-13-0)** - -Parseval's theorem concisely relates energy between the time domain and the frequency domain: - -$$ -\int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega -$$ - -This too is easily verified with MATLAB. For example, a unit amplitude pulse *x*(*t*) with duration τ has energy *Ex* = τ . Thus, - -$$ -\int_{-\infty}^{\infty} |X(\omega)|^2 \, d\omega = 2\pi \, \tau -$$ - -Letting τ = 1, the energy of *X*(ω) is computed by using the quad function. - -``` ->> X_squared = @(omega, tau) (tau*CH7MP1(omega*tau/2)).^2; ->> quad(X_squared,-1e6,1e6,[],[],1) - ans = 6.2817 -``` - -Although not perfect, the result of the numerical integration is consistent with the expected value of 2π ≈ 6.2832. For quad, the first argument is the function to be integrated, the next two arguments are the limits of integration, the empty square brackets indicate default values for special options, and the last argument is the secondary input τ for the anonymous function X\_squared. Full format details for quad are available from MATLAB's help facilities. - -A more interesting problem involves computing a signal's essential bandwidth. Consider, for example, finding the essential bandwidth *W*, in radians per second, that contains fraction β of the energy of the square pulse *x*(*t*). That is, we want to find *W* such that - -$$ -\frac{1}{2\pi} \int_{-W}^{W} |X(\omega)|^2 d\omega = \beta \tau -$$ - -Program CH7MP2 uses a guess-and-check method to find *W*. - -``` -function [W,E_W] = CH7MP2(tau,beta,tol) -% CH7MP2.m : Chapter 7, MATLAB Program 2 -% Function M-file computes essential bandwidth W for square pulse. -% INPUTS: tau = pulse width -% beta = fraction of signal energy desired in W -% tol = tolerance of relative energy error -% OUTPUTS: W = essential bandwidth [rad/s] -% E_W = Energy contained in bandwidth W -W = 0; step = 2*pi/tau; % Initial guess and step values -X_squared = @(omega,tau) (tau*CH7MP1(omega*tau/2)).^2; -E = beta*tau; % Desired energy in W -relerr = (E-0)/E; % Initial relative error is 100 percent -while(abs(relerr) > tol), - if (relerr>0), % W too small, so... - W=W+step; % ... increase W by step - elseif (relerr<0), % W too large, so... - step = step/2; % ... decrease step and then W -``` - -``` -W = W-step; - end - E_W = 1/(2*pi)*quad(X_squared,-W,W,[],[],tau); - relerr = (E - E_W)/E; -end -``` - -Although this guess-and-check method is not the most efficient, it is relatively simple to understand: CH7MP2 sensibly adjusts *W* until the relative error is within tolerance. The number of iterations needed to converge to a solution depends on a variety of factors and is not known beforehand. The while command is ideal for such situations: - -``` -while expression, -``` - -*statements;* - -end - -While the *expression* is true, the *statements* are continually repeated. - -To demonstrate CH7MP2, consider the 90% essential bandwidth *W* for a pulse of 1 second duration. Typing [W,E\_W]=CH7MP2(1,0.9,0.001) returns an essential bandwidth *W* = 5.3014 that contains 89.97% of the energy. Reducing the error tolerance improves the estimate. CH7MP2(1,0.9,0.00005) returns an essential bandwidth *W* = 5.3321 that contains 90.00% of the energy. These essential bandwidth calculations are consistent with estimates presented after Ex. 7.2. - -### **[7.9-3 Spectral Sampling](#page-13-0)** - -Consider a signal with finite duration τ . A periodic signal *xT*0 (*t*) is constructed by repeating *x*(*t*) every *T*0 seconds, where *T*0 ≥ τ . From Eq. (7.5), we can write the Fourier series coefficients of *xT*0 (*t*) as *Dn* = (1/*T*0)*X*(*n*2π/*T*0). Put another way, the Fourier series coefficients are obtained by sampling the spectrum *X*(ω). - -By using spectral sampling, it is simple to determine the Fourier series coefficients for an arbitrary duty-cycle, square-pulse periodic signal. The square pulse *x*(*t*) = rect(*t*/τ ) has spectrum *X*(ω) = τ sinc(ωτ /2). Thus, the *n*th Fourier coefficient of the periodic extension *xT*0 (*t*) is *Dn* = (τ/*T*0)sinc (*n*πτ/*T*0). As in Ex. 6.4, τ = π and *T*0 = 2π provide a square-pulse periodic signal. The Fourier coefficients are determined by - -``` ->> tau = pi; T_0 = 2*pi; n = [0:10]; -``` - ->> D\_n = tau/T\_0\*MS7P1(n\*pi\*tau/T\_0); - -``` ->> stem(n,D_n); xlabel('n'); ylabel('D_n'); -``` - ->> axis([-0.5 10.5 -0.2 0.55]); - -The results, shown in Fig. 7.50, agree with Fig. 6.6b. Doubling the period to *T*0 = 4π effectively doubles the density of spectral samples and halves the spectral amplitude, as shown in Fig. 7.51. - -As *T*0 increases, the spectral sampling becomes progressively finer while the amplitude becomes infinitesimal. An evolution of the Fourier series toward the Fourier integral is seen by allowing the period *T*0 to become large. Figure 7.52 shows the result for *T*0 = 40π. - -If *T*0 = τ , the signal *xT*0 is a constant and the spectrum should concentrate energy at dc. In this case, the sinc function is sampled at the zero crossings and *Dn* = 0 for all *n* not equal to 0. Only the sample corresponding to *n* = 0 is nonzero, indicating a dc signal, as expected. It is a simple matter to modify the previous code to verify this case. - -**Figure 7.50** Fourier spectra for τ = π and *T*0 = 2π. - -**Figure 7.51** Fourier spectra for τ = π and *T*0 = 4π. - -**Figure 7.52** Fourier spectra for τ = π and *T*0 = 40π. - -### **[7.9-4 Kaiser Window Functions](#page-13-0)** - -A window function is useful only if it can be easily computed and applied to a signal. The Kaiser window, for example, is flexible but appears rather intimidating: - -$$ -w_K(t) = \begin{cases} \frac{I_0(\alpha\sqrt{1 - 4(t/T)^2})}{I_0(\alpha)} & |t| < T/2\\ 0 & \text{otherwise} \end{cases} -$$ - -Fortunately, the bark of a Kaiser window is worse than its bite! The function *I*0(*x*), a zero-order modified Bessel function of the first kind, can be computed according to - -$$ -I_0(x) = \sum_{k=0}^{\infty} \left(\frac{x^k}{2^k k!}\right)^2 -$$ - -or, more simply, by using the MATLAB function besseli(0,x). In fact, MATLAB supports a wide range of Bessel functions, including Bessel functions of the first and second kinds (besselj and bessely), modified Bessel functions of the first and second kinds (besseli and besselk), Hankel functions (besselh), and Airy functions (airy). - -Program CH7MP3 computes Kaiser windows at times *t* by using parameters *T* and α. - -``` -function [w_K] = CH7MP3(t,T,alpha) -% CH7MP3.m : Chapter 7, MATLAB Program 3 -% Function M-file computes a width-T Kaiser window using parameter alpha. -% Alpha can also be a string identifier: 'rectangular', 'Hamming', or -% 'Blackman'. -% INPUTS: t = independent variable of the window function -% T = window width -% alpha = Kaiser parameter or string identifier -% OUTPUTS: w_K = Kaiser window function -if strncmpi(alpha,'rectangular',1), - alpha = 0; -elseif strncmpi(alpha,'Hamming',3), - alpha = 5.4414; -elseif strncmpi(alpha,'Blackman',1), - alpha = 8.885; -elseif isa(alpha,'char') - disp('Unrecognized string identifier.'); return -end -w_K = zeros(size(t)); i = find(abs(t)**Figure 7.53** Special-case, unit-duration Kaiser windows. - -Figure 7.53 shows the three special-case, unit-duration Kaiser windows generated by - -``` ->> t = [-0.6:.001:0.6]; T = 1; -``` - -``` ->> plot(t,CH7MP3(t,T,'r'),'k-',t,CH7MP3(t,T,'ham'),'k-.',t,CH7MP3(t,T,'b'),'k--'); -``` - -``` ->> axis([-0.6 0.6 -.1 1.1]); xlabel('t'); ylabel('w_K(t)'); -``` - -``` ->> legend('Rectangular','Hamming','Blackman','Location','EastOutside'); -``` - -## **[7.10 SUMMARY](#page-13-0)** - -In Ch. 6, we represented periodic signals as a sum of (everlasting) sinusoids or exponentials (Fourier series). In this chapter we extended this result to aperiodic signals, which are represented by the Fourier integral (instead of the Fourier series). An aperiodic signal *x*(*t*) may be regarded as a periodic signal with period *T*0 → ∞ so that the Fourier integral is basically a Fourier series with a fundamental frequency approaching zero. Therefore, for aperiodic signals, the Fourier spectra are continuous. This continuity means that a signal is represented as a sum of sinusoids (or exponentials) of all frequencies over a continuous frequency interval. The Fourier transform *X*(ω), therefore, is the spectral density (per unit bandwidth in hertz). - -An ever-present aspect of the Fourier transform is the duality between time and frequency, which also implies duality between the signal *x*(*t*) and its transform *X*(ω). This duality arises because of near-symmetrical equations for direct and inverse Fourier transforms. The duality principle has far-reaching consequences and yields many valuable insights into signal analysis. - -The scaling property of the Fourier transform leads to the conclusion that the signal bandwidth is inversely proportional to signal duration (signal width). Time shifting of a signal does not change its amplitude spectrum, but it does add a linear phase component to its spectrum. Multiplication of a signal by an exponential *ej*ω0*t* shifts the spectrum to the right by ω0. In practice, spectral shifting is achieved by multiplying a signal by a sinusoid such as cosω0*t* (rather than the exponential *ej*ω0*t* ). This process is known as amplitude modulation. Multiplication of two signals results in convolution of their spectra, whereas convolution of two signals results in multiplication of their spectra. - -For an LTIC system with the frequency response *H*(ω), the input and output spectra *X*(ω) and *Y*(ω) are related by the equation *Y*(ω) = *X*(ω)*H*(ω). This is valid only for asymptotically stable systems. It also applies to marginally stable systems if the input does not contain a finite-amplitude sinusoid of the natural frequency of the system. For asymptotically unstable systems, the frequency response *H*(ω) does not exist. For distortionless transmission of a signal through an LTIC system, the amplitude response |*H*(ω)| of the system must be constant, and the phase response *H*(ω) should be a linear function of ω over a band of interest. Ideal filters, which allow distortionless transmission of a certain band of frequencies and suppress all the remaining frequencies, are physically unrealizable (noncausal). In fact, it is impossible to build a physical system with zero gain [*H*(ω) = 0] over a finite band of frequencies. Such systems (which include ideal filters) can be realized only with infinite time delay in the response. - -The energy of a signal *x*(*t*) is equal to 1/2π times the area under |*X*(ω)2| (Parseval's theorem). The energy contributed by spectral components within a band *f* (in hertz) is given by |*X*(ω)| 2*f* . Therefore, |*X*(ω)| 2 is the energy spectral density per unit bandwidth (in hertz). - -The process of modulation shifts the signal spectrum to different frequencies. Modulation is used for many reasons: to transmit several messages simultaneously over the same channel for the sake of utilizing channel's high bandwidth, to effectively radiate power over a radio link, to shift a signal spectrum at higher frequencies to overcome the difficulties associated with signal processing at lower frequencies, and to effect the exchange of transmission bandwidth and transmission power required to transmit data at a certain rate. Broadly speaking, there are two types of modulation, amplitude and angle modulation. Each class has several subclasses. - -In practice, we often need to truncate data. Truncating is like viewing data through a window, which permits only certain portions of the data to be seen and hides (suppresses) the remainder. Abrupt truncation of data amounts to a rectangular window, which assigns a unit weight to data seen from the window and zero weight to the remaining data. Tapered windows, on the other hand, reduce the weight gradually from 1 to 0. Data truncation can cause some unsuspected problems. For example, in computation of the Fourier transform, windowing (data truncation) causes spectral spreading (spectral smearing) that is characteristic of the window function used. A rectangular window results in the least spreading, but it does so at the cost of a high and oscillatory spectral leakage outside the signal band, which decays slowly as 1/ω. In comparison to a rectangular window, tapered windows, in general, have larger spectral spreading (smearing), but the spectral leakage is smaller and decays faster with frequency. If we try to reduce spectral leakage by using a smoother window, the spectral spreading increases. Fortunately, spectral spreading can be reduced by increasing the window width. Therefore, we can achieve a given combination of spectral spread (transition bandwidth) and leakage characteristics by choosing a suitable tapered window function of a sufficiently long width *T*. - -### **[REFERENCES](#page-13-0)** - -- 1. Churchill, R. V., and Brown, J. W. *Fourier Series and Boundary Value Problems,* 3rd ed. McGraw-Hill, New York, 1978. -- 2. Bracewell, R. N. *Fourier Transform and Its Applications,* rev. 2nd ed. McGraw-Hill, New York, 1986. -- 3. Guillemin, E. A. *Theory of Linear Physical Systems*. Wiley, New York, 1963. -- 4. Lathi, B. P. *Modern Digital and Analog Communication Systems,* 3rd ed. Oxford University Press, New York, 1998. -- 5. Hamming, R. W. *Digital Filters,* 2nd ed. Prentice-Hall, Englewood Cliffs, NJ, 1983. -- 6. Harris, F. J. On the use of windows for harmonic analysis with the discrete Fourier transform. *Proceedings of the IEEE,* vol. 66, no. 1, pp. 51–83, January 1978. - -## **[PROBLEMS](#page-13-0)** - -- **7.1-1** Suppose signal *x*(*t*) = *t* 2 [*u*(*t*)−*u*(*t* −2)] has Fourier transform *X*(ω). Define a 3-periodic replication of *x*(*t*) as *y*(*t*) = %∞ *n*=−∞ 2*x*(*t* − 1 − 3*n*). Determine *Yk*, the Fourier series of *y*(*t*), in terms of the Fourier transform *X*(·). -- **7.1-2** Show that for a real *x*(*t*), Eq. (7.10) can be expressed as - -$$ -x(t) = \frac{1}{\pi} \int_0^\infty |X(\omega)| \cos[\omega t + \angle X(\omega)] d\omega -$$ - -This is the trigonometric form of the Fourier integral. Compare this with the compact trigonometric Fourier series. - -**7.1-3** Show that if *x*(*t*) is an even function of *t*, then - -$$ -X(\omega) = 2 \int_0^\infty x(t) \cos \omega t \, dt -$$ - -and if *x*(*t*) is an odd function of *t*, then - -$$ -X(\omega) = -2j \int_0^\infty x(t) \sin \omega t \, dt -$$ - -Hence, prove that if *x*(*t*) is a real and even function of *t*, then *X*(ω) is a real and even function of ω. In addition, if *x*(*t*) is a real and odd function of *t*, then *X*(ω) is an imaginary and odd function of ω. - -**7.1-4** A signal *x*(*t*) can be expressed as the sum of even and odd components (see Sec. 1.5-2): - -*x*(*t*) = *xe*(*t*) +*xo*(*t*) - -(a) If *x*(*t*) ⇐⇒ *X*(ω), show that for real *x*(*t*), - -*xe*(*t*) ⇐⇒ Re[*X*(ω)] - -and - -$$ -x_o(t) \Longleftrightarrow j \operatorname{Im}[X(\omega)] -$$ - -- (b) Verify these results by finding the Fourier transforms of the even and odd components of the following signals: **(i)** *u*(*t*) and **(ii)** *e*−*atu*(*t*). -- **7.1-5** Using Eq. (7.9), find the Fourier transforms of the signals *x*(*t*) in Fig. P7.1-5. -- **7.1-6** Using Eq. (7.9), find the Fourier transforms of the signals depicted in Fig. P7.1-6. -- **7.1-7** Use Eq. (7.10) to find the inverse Fourier transforms of the spectra in Fig. P7.1-7. - -**Figure P7.1-6** - -**Figure P7.1-8** - -- **7.1-8** Use Eq. (7.10) to find the inverse Fourier transforms of the spectra in Fig. P7.1-8. -- **7.1-9** If *x*(*t*) ⇐⇒ *X*(ω), then show that - -$$ -X(0) = \int_{-\infty}^{\infty} x(t) dt -$$ - -and - -$$ -x(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) d\omega -$$ - -Also show that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc}(x) dx = \int_{-\infty}^{\infty} \operatorname{sinc}^{2}(x) dx = \pi -$$ - -**Figure P7.2-4** - -- **7.2-1** Sketch the following functions: (a) rect(*t*/2) - - (b) (3ω/100) - - (c) rect((*t* −10)/8) - - (d) sinc(πω/5) - - (e) sinc ((ω/5)−2π) - - (f) sinc (*t*/5)rect(*t*/10π ) -- **7.2-2** Using Eq. (7.9), show that the Fourier transform of rect(*t* − 5) is sinc(ω/2)*e*−*j*5ω. Sketch the resulting amplitude and phase spectra. -- **7.2-3** Using Eq. (7.10), show that the inverse Fourier transform of rect((ω −10)/2π ) is sinc(π*t*) *ej*10*t* . -- **7.2-4** Find the inverse Fourier transform of *X*(ω) for the spectra illustrated in Fig. P7.2-4. [*Hint*: - -*X*(ω) = |*X*(ω)|*ej X*(ω). This problem illustrates how different phase spectra (both with the same amplitude spectrum) represent entirely different signals.] - -- **7.2-5** (a) Can you find the Fourier transform of *eatu*(*t*) when *a*>1 by setting *s* = *j*ω in the Laplace transform of *eatu*(*t*)? Explain. - - (b) Find the Laplace transform of *x*(*t*) shown in Fig. P7.2-5. Can you find the Fourier transform of *x*(*t*) by setting *s* = *j*ω in its Laplace transform? Explain. Verify your answer by finding the Fourier and the Laplace transforms of *x*(*t*). - -**Figure P7.2-5** - -- **7.3-1** Apply the duality property to the appropriate pair in Table 7.1 to show that - - (a) 1 2 [δ(*t*)+*j*/π*t*] ⇐⇒ *u*(ω) - - (b) δ(*t* +*T*) +δ(*t* −*T*) ⇐⇒ 2 cos *T*ω - - (c) δ(*t* +*T*)−δ(*t* −*T*) ⇐⇒ 2*j*sin *T*ω - -- **7.3-2** A signal *x*(*t*) has Fourier transform *X*(ω). Determine the Fourier transform *Y*(ω) in terms of *X*(ω) for each of the following signals *y*(*t*): (a) *y*(*t*) = 1 5 *x*(−2*t* +3) (b) *y*(*t*) = *ej*2*t x*∗(−3*t* −6) -- **7.3-3** A signal *x*(*t*) has Fourier transform *X*(ω). Determine the inverse Fourier transform *y*(*t*) in terms of *x*(*t*) for each of the following spectra *Y*(ω), - -(a) -$$ -Y(\omega) = \frac{4}{3}e^{-j2\omega/3}X(-\omega/3) -$$ - -(b) -$$ -Y(\omega) = \frac{1}{3}e^{j2(\omega - 2)}X^*\left(\frac{\omega - 2}{3}\right) -$$ - -**7.3-4** The Fourier transform of the triangular pulse *x*(*t*) in Fig. P7.3-4 is expressed as - -$$ -X(\omega) = \frac{1}{\omega^2} (e^{j\omega} - j\omega e^{j\omega} - 1) -$$ - -Use this information, and the time-shifting and time-scaling properties, to find the Fourier transforms of the signals *xi*(*t*)(*i* = 1, 2, 3, 4, 5) shown in Fig. P7.3-4. - -- **7.3-5** Using only the time-shifting property and Table 7.1, find the Fourier transforms of the signals depicted in Fig. P7.3-5. -- **7.3-6** Consider the fact that the τ -duration triangle function ω τ has inverse Fourier transform - -**Figure P7.3-4** - -2 0 234 0 - - -(a) (b) - -4 -3 - - -**Figure P7.3-7** - --4 -3 - - -τ 4π sinc2 *t*τ 4 . Use the duality property to determine the Fourier transform *Y*(ω) of signal *y*(*t*) = (*t*). - -**7.3-7** Use the time-shifting property to show that if *x*(*t*) ⇐⇒ *X*(ω), then - -$$ -x(t+T) + x(t-T) \Longleftrightarrow 2X(\omega)\cos T\omega -$$ - -This is the dual of Eq. (7.32). Use this result and Table 7.1 to find the Fourier transforms of the signals shown in Fig. P7.3-7. - -**7.3-8** Prove the following results, which are duals of each other: - -$$ -x(t)\sin \omega_0 t \Longleftrightarrow \frac{1}{2j}[X(\omega - \omega_0) - X(\omega + \omega_0)] -$$ - -$$ -\frac{1}{2j}[x(t+T) - x(t-T)] \Longleftrightarrow X(\omega)\sin T\omega -$$ - -Use the latter result and Table 7.1 to find the Fourier transform of the signal in Fig. P7.3-8. - -2 234 - -- **7.3-9** The signals in Fig. P7.3-9 are modulated signals with carrier cos 10*t*. Find the Fourier transforms of these signals by using the appropriate properties of the Fourier transform and Table 7.1. Sketch the amplitude and phase spectra for Figs. P7.3-9a and P7.3-9b. -- **7.3-10** Use the frequency-shifting property and Table 7.1 to find the inverse Fourier transform of the spectra depicted in Fig. P7.3-10. -- **7.3-11** Let *X*(ω) = rect(ω) be the Fourier transform of a signal *x*(*t*). - - (a) For *y*a(*t*) = *x*(*t*) ∗ *x*(*t*), sketch *Y*a(ω). - - (b) For *y*b(*t*) = *x*(*t*) ∗ *x*(*t*/2), sketch *Y*b(ω). - -(a) (b) - -**Figure P7.3-10** - -- (c) For *y*c(*t*) = 2*x*(*t*), sketch *Y*c(ω). -- (d) For *y*d(*t*) = *x*2(*t*), sketch *Y*d(ω). -- (e) For *y*e(*t*) = 1−*x*2(*t*), sketch *Y*e(ω). -- **7.3-12** Use the time-convolution property to prove pairs 2, 4, 13, and 14 in Table 2.1 (assume λ < 0 in pair 2, λ1 and λ2 < 0 in pair 4, λ1 < 0 and λ2 > 0 in pair 13, and λ1 and λ2 > 0 in pair 14). These restrictions are placed because of the Fourier transformability issue for the signals concerned. For pair 2, you need to apply the result in Eq. (1.10). -- **7.3-13** A signal *x*(*t*) is bandlimited to *B* Hz. Show that the signal *xn*(*t*) is bandlimited to *nB* Hz. -- **7.3-14** Find the Fourier transform of the signal in Fig. P7.3-5a by three different methods: - - (a) By direct integration using Eq. (7.9). - - (b) Using only pair 17 (Table 7.1) and the time-shifting property. - - (c) Using the time-differentiation and time-shifting properties, along with the fact that δ(*t*) ⇐⇒ 1. - -**7.3-15** (a) Prove the frequency-differentiation property (dual of the time-differentiation property): - -$$ --jtx(t) \Longleftrightarrow \frac{d}{d\omega}X(\omega) -$$ - -- (b) Use this property and pair 1 (Table 7.1) to determine the Fourier transform of *te*−*atu*(*t*). -- **7.3-16** Adapt the method of Ex. 7.17 and use the frequency-differentiation (see Prob. 7.3-15) and other properties to find the inverse Fourier transform *x*(*t*) of the triangular spectrum *X*(ω) = (ω/2). -- **7.3-17** Adapt the method of Ex. 7.17 and use the frequency-differentiation (see Prob. 7.3-15) and other properties to find the inverse Fourier transform *x*(*t*) of the spectrum *X*(ω) = π ω 2 rect ω 4 . -- **7.4-1** For a stable LTIC system with transfer function - -$$ -H(s) = \frac{1}{s+1} -$$ - -find the (zero-state) response if the input *x*(*t*) is - -- (a) *e*−2*t u*(*t*) -- (b) *e*−*t u*(*t*) -- (c) *et u*(−*t*) -- (d) *u*(*t*) -- **7.4-2** A stable LTIC system is specified by the frequency response - -$$ -H(\omega) = \frac{-1}{j\omega - 2} -$$ - -Find the impulse response of this system and show that this is a noncausal system. Find the (zero-state) response of this system if the input *x*(*t*) is - -- (a) *e*−*t u*(*t*) -- (b) *et u*(−*t*) -- **7.4-3** A periodic signal *x*(*t*) = 1 + 2 cos(5π*t*) + 3 sin(8π*t*) is applied to an LTIC system with impulse response *h*(*t*) = 8sinc(4*t*) cos(2π*t*) to produce output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Determine ω0, the fundamental radian frequency of *x*(*t*). - - (b) Determine *X*(ω), the Fourier transform of *x*(*t*). - - (c) Sketch the system's magnitude response |*H*(ω)| over −10π ≤ ω ≤ 10π. - - (d) Is the system *h*(*t*) distortionless? Explain. - - (e) Determine *y*(*t*). -- **7.4-4** A periodic delta train *x*(*t*) = %∞ *n*=−∞ δ(*t* − π*n*) is applied to an LTIC system with impulse response *h*(*t*) = sin(3*t*)sinc2 *t* π to produce zero-state output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Determine ω0, the fundamental radian frequency of *x*(*t*). - - (b) Determine *X*(ω), the Fourier transform of *x*(*t*). - - (c) Sketch the system's magnitude response |*H*(ω)| over −10π ≤ ω ≤ 10π. - - (d) Is the system *h*(*t*) distortionless? Explain. - - (e) Determine *y*(*t*). -- **7.4-5** Signals *x*1(*t*)=104rect(104*t*) and *x*2(*t*) = δ(*t*) are applied at the inputs of the ideal lowpass filters *H*1(ω) = rect(ω/40,000π ) and *H*2(ω) = rect(ω/20,000π ) (Fig. P7.4-5). The outputs *y*1(*t*) and *y*2(*t*) of these filters are multiplied to obtain the signal *y*(*t*) = *y*1(*t*)*y*2(*t*). (a) Sketch *X*1(ω) and *X*2(ω). - -- (b) Sketch *H*1(ω) and *H*2(ω). -- (c) Sketch *Y*1(ω) and *Y*2(ω). -- (d) Find the bandwidths of *y*1(*t*), *y*2(*t*), and *y*(*t*). - -- **7.4-6** A lowpass system time constant is often defined as the width of its unit impulse response *h*(*t*) (see Sec. 2.6-2). An input pulse *p*(*t*) to this system acts like an impulse of strength equal to the area of *p*(*t*) if the width of *p*(*t*) is much smaller than the system time constant, and provided *p*(*t*) is a lowpass pulse, implying that its spectrum is concentrated at low frequencies. Verify this behavior by considering a system whose unit impulse response is *h*(*t*) = rect(*t*/10−3). The input pulse is a triangle pulse *p*(*t*) = (*t*/10−6). Show that the system response to this pulse is very nearly the system response to the input *A*δ(*t*), where *A* is the area under the pulse *p*(*t*). -- **7.4-7** A lowpass system time constant is often defined as the width of its unit impulse response *h*(*t*) (see Sec. 2.6-2). An input pulse *p*(*t*) to this system passes practically without distortion if the width of *p*(*t*) is much greater than the system time constant, and provided *p*(*t*) is a lowpass pulse, implying that its spectrum is concentrated at low frequencies. Verify this behavior by considering a system whose unit impulse response is *h*(*t*) = rect(*t*/10−3). The input pulse is a triangle pulse *p*(*t*)=(*t*). Show that the system output to this pulse is very nearly *kp*(*t*), where *k* is the system gain to a dc signal, that is, *k* = *H*(0). -- **7.4-8** A causal signal *h*(*t*) has a Fourier transform *H*(ω). If *R*(ω) and *X*(ω) are the real and the imaginary parts of *H*(ω), that is, *H*(ω) = *R*(ω)+ *jX*(ω), then show that - -$$ -R(\omega) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{X(\omega)}{\omega - y} d\omega -$$ - -and - -$$ -X(\omega) = -\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{R(\omega)}{\omega - y} d\omega -$$ - -assuming that *h*(*t*) has no impulse at the origin. This pair of integrals defines the *Hilbert transform*. [*Hint:* Let *he*(*t*) and *ho*(*t*) be the even and odd components of *h*(*t*). Use the results in Prob. 7.1-4. See Fig. 1.24 for the relationship between *he*(*t*) and *ho*(*t*).] - -This problem states one of the important properties of causal systems: that the real and imaginary parts of the frequency response of a causal system are related. If one specifies the real part, the imaginary part cannot be specified independently. The imaginary part is predetermined by the real part, and vice versa. This result also leads to the conclusion that the magnitude and angle of *H*(ω) are related, provided all the poles and zeros of *H*(ω) lie in the LHP. - -**7.5-1** Consider a filter with the frequency response - -$$ -H(\omega) = e^{-(k\omega^2 + j\omega t_0)} -$$ - -Show that this filter is physically unrealizable by using the time-domain criterion [noncausal *h*(*t*)] and the frequency-domain (Paley–Wiener) criterion. Can this filter be made approximately realizable by choosing *t*0 sufficiently large? Use your own (reasonable) criterion of approximate realizability to determine *t*0. [*Hint:* Use pair 22 in Table 7.1.] - -**7.5-2** Show that a filter with frequency response - -$$ -H(\omega) = \frac{2(10^5)}{\omega^2 + 10^{10}} e^{-j\omega t_0} -$$ - -is unrealizable. Can this filter be made approximately realizable by choosing a sufficiently large *t*0? Use your own (reasonable) criterion of approximate realizability to determine *t*0. - -- **7.5-3** Determine whether the filters with the following frequency response *H*(ω) are physically realizable. If they are not realizable, can they be realized approximately by allowing a finite time delay in the response? - - (a) 10−6 sinc (10−6ω) - - (b) 10−4 (ω/40,000π) - - (c) 2π δ(ω) - -- **7.5-4** Consider signal *x*1(*t*), its Fourier transform *X*1(*f*), and several other signals, as shown in Fig. P7.5-4. Notice, spectra are drawn as a function of hertzian frequency *f* rather than radian frequency ω. - - (a) Accurately sketch *X*2(*f*), the Fourier transform of *x*2(*t*). - - (b) Accurately sketch *x*3(*t*), the inverse Fourier transform of *X*3(*f*). - - (c) The signal *x*4(*t*) = *x*1(*t*) + *x*2(*t*) is passed through an ideal lowpass filter with 3 Hz cutoff to produce output *y*4(*t*). Accurately sketch *y*4(*t*). - -**Figure P7.5-4** - -- **7.6-1** Define *x*(*t*) = 1 2π sinc(*t*/2) with Fourier transform *X*(ω) = rect(ω). Use Parseval's theorem to determine \$ ∞ −∞ sinc2(*t* −2)*dt*. -- **7.6-2** Show that the energy of a Gaussian pulse - -$$ -x(t) = \frac{1}{\sigma\sqrt{2\pi}}e^{-t^2/2\sigma^2} -$$ - -is 1/(2σ √π ). Verify this result by using Parseval's theorem to derive the energy *Ex* from *X*(ω). [*Hint:* See pair 22 in Table 7.1. Use the fact that \$ ∞ −∞ *e*−*x*2/2 *dx* = √2π.] - -**7.6-3** Use Parseval's theorem of Eq. (7.45) to show that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc}^2(kx) dx = \frac{\pi}{k} -$$ - -- **7.6-4** A lowpass signal *x*(*t*) is applied to a squaring device. The squarer output *x*2(*t*) is applied to a lowpass filter of bandwidth *f* (in hertz) (Fig. P7.6-4). Show that if *f* is very small (*f* → 0), then the filter output is a dc signal *y*(*t*) ≈ 2*Exf* . [*Hint:* If *x*2(*t*) ⇐⇒ *A*(ω), then show that *Y*(ω) ≈ [4π*A*(0)*f*]δ(ω) if *f* → 0. Now, show that *A*(0) = *Ex*.] -- **7.6-5** Generalize Parseval's theorem to show that for real, Fourier-transformable signals *x*1(*t*) and *x*2(*t*) - -$$ -\int_{-\infty}^{\infty} x_1(t) x_2(t) dt -$$ - -= $\frac{1}{2\pi} \int_{-\infty}^{\infty} X_1(-\omega) X_2(\omega) d\omega$ -= $\frac{1}{2\pi} \int_{-\infty}^{\infty} X_1(\omega) X_2(-\omega) d\omega$ - -**7.6-6** Show that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc} (Wt - m\pi) \operatorname{sinc} (Wt - n\pi) dt -$$ -$$ -= \begin{cases} 0 & m \neq n \\ \frac{\pi}{W} & m = n \end{cases} -$$ - -( )2 Lowpass filter *x*(*t*) *x y*(*t*) 2*Exf* 2(*t*) - -**Figure P7.6-4** - -[*Hint:* Recognize that - -$$ -\operatorname{sinc}(Wt - k\pi) = \operatorname{sinc}\left[W\left(t - \frac{k\pi}{W}\right)\right] \\ -\Longleftrightarrow \frac{\pi}{W} \operatorname{rect}\left(\frac{\omega}{2W}\right) e^{-jk\pi\omega/W} -$$ - -Use this fact and the result in Prob. 7.6-5.] - -- **7.6-7** (a) What does it mean to compute the 95% essential bandwidth *B* of a signal *x*(*t*) with Fourier transform *X*(ω)? - - (b) Determine the 95% essential bandwidth *B* of a signal with spectrum *X*(ω) = rect(ω). - - (c) Determine the 95% essential bandwidth *B* of a signal with spectrum *X*(ω) = (ω). -- **7.6-8** Using a 95% energy criterion, determine the essential bandwidth *B* of a signal that has a Fourier transform given by *X*(ω) = *e*−|ω| . -- **7.6-9** For the signal - -$$ -x(t) = \frac{2a}{t^2 + a^2} -$$ - -determine the essential bandwidth *B* (in hertz) of *x*(*t*) such that the energy contained in the spectral components of *x*(*t*) of frequencies below *B* Hz is 99% of the signal energy *Ex*. - -- **7.7-1** For each of the following baseband signals **(i)** *m*(*t*) = cos 1000*t*, **(ii)** *m*(*t*) = 2 cos 1000*t* + cos 2000*t*, and **(iii)** *m*(*t*) = cos 1000*t* cos 3000*t*: - - (a) Sketch the spectrum of *m*(*t*). - - (b) Sketch the spectrum of the DSB-SC signal *m*(*t*) cos 10,000*t*. - - (c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra. - - (d) Identify the frequencies in the baseband, and the corresponding frequencies in the DSB-SC, USB, and LSB spectra. Explain the nature of frequency shifting in each case. -- **7.7-2** A message signal *m*(*t*) with spectrum *M*(ω) = 1 1000 ω 2π6000 is to be transmitted using a communication system. Assume all single-sideband systems have suppressed carriers. - - (a) What is the hertzian bandwidth of *m*(*t*)? - -- (b) Sketch the spectrum of the transmitted signal if the communication system is DSB-SC with ω*c* = 2π 100,000. -- (c) Sketch the spectrum of the transmitted signal if the communication system is AM with ω*c* = 2π 100,000 and a modulation index of μ = 1. What is the corresponding carrier amplitude *A*? -- (d) Sketch the spectrum of the transmitted signal if the communication system is USB with ω*c* = 2π 100,000. -- (e) Sketch the spectrum of the transmitted signal if the communication system is LSB with ω*c* = 2π 100,000. -- (f) Suppose we want to transmit *m*(*t*) on each of an FDM system's four channels: DSB-SC at carrier ω1, AM (μ = 1) at carrier ω2, USB at carrier ω3, and LSB at carrier ω4. Determine carrier frequencies ω1 < ω2 < ω3 < ω4 so that the FDM spectrum begins at a frequency of 100,000 Hz with 5,000 Hz deadbands separating adjacent messages. What is the end hertzian frequency of the FDM signal? -- **7.7-3** You are asked to design a DSB-SC modulator to generate a modulated signal *km*(*t*) cosω*ct*, where *m*(*t*) is a signal bandlimited to *B* Hz (Fig. P7.7-3a). Figure P7.7-3b shows a DSB-SC modulator available in the stockroom. The bandpass filter is tuned to ω*c* and has a bandwidth of 2*B* Hz. The carrier generator available generates not cosω*ct*, but cos3 ω*ct*. - - (a) Explain whether you would be able to generate the desired signal using only this equipment. If so, what is the value of *k*? - -- (b) Determine the signal spectra at points *b* and *c*, and indicate the frequency bands occupied by these spectra. -- (c) What is the minimum usable value of ω*c*? -- (d) Would this scheme work if the carrier generator output were cos2 ω*ct*? Explain. -- (e) Would this scheme work if the carrier generator output were cos*n* ω*ct* for any integer *n* ≥ 2? -- **7.7-4** In practice, the analog multiplication operation is difficult and expensive. For this reason, in amplitude modulators, it is necessary to find some alternative to multiplication of *m*(*t*) with cosω*ct*. Fortunately, for this purpose, we can replace multiplication with a switching operation. A similar observation applies to demodulators. In the scheme depicted in Fig. P7.7-4a, the period of the rectangular periodic pulse *x*(*t*) - -**Figure P7.7-4** - -**Figure P7.7-3** - -shown in Fig. P7.7-4b is *T*0 = 2π/ω*c*. The bandpass filter is centered at ±ω*c* and has a bandwidth of 2*B* Hz. Note that multiplication by a square periodic pulse *x*(*t*) in Fig. P7.7-4b amounts to periodic on-off switching of *m*(*t*), which is bandlimited to *B* Hz. Such a switching operation is relatively simple and inexpensive. Show that this scheme can generate an amplitude-modulated signal *k* cos ω*ct*. Determine the value of *k*. Show that the same scheme can also be used for demodulation, provided the bandpass filter in Fig. P7.7-4a is replaced by a lowpass (or baseband) filter. - -**7.7-5** Figure P7.7-5a shows a scheme to transmit two signals *m*1(*t*) and *m*2(*t*) simultaneously on the same channel (without causing spectral interference). Such a scheme, which transmits more than one signal, is known as signal *multiplexing*. In this case, we transmit multiple signals by sharing an available spectral band on the channel; hence, this is an example of the *frequency-division* multiplexing. The signal at point *b* is the multiplexed signal, which now modulates a carrier of frequency 20,000 rad/s. The modulated signal at point *c* is now transmitted over the channel. - -- (a) Sketch the spectra at points *a*, *b*, and *c*. -- (b) What must be the minimum bandwidth of the channel? -- (c) Design a receiver to recover signals *m*1(*t*) and *m*2(*t*) from the modulated signal at point *c*. -- **7.7-6** The system shown in Fig. P7.7-6 is used for scrambling audio signals. The output *y*(*t*) is the scrambled version of the input *m*(*t*). - -**Figure P7.7-6** - -- (a) Find the spectrum of the scrambled signal *y*(*t*). -- (b) Suggest a method of descrambling *y*(*t*) to obtain *m*(*t*). - -A slightly modified version of this scrambler was first used commercially on the 25-mile radio-telephone circuit connecting Los Angeles and Santa Catalina Island. - -- **7.7-7** Figure P7.7-7 presents a scheme for coherent (synchronous) demodulation. Show that this scheme can demodulate the AM signal [*A* + *m*(*t*)] cos ω*ct* regardless of the value of *A*. -- **7.7-8** Sketch the AM signal [*A* + *m*(*t*)] cos ω*ct* for the periodic triangle signal *m*(*t*) illustrated in Fig. P7.7-8 corresponding to the following modulation indices: - - (a) μ = 0.5 - - (b) μ = 1 - - (c) μ = 2 - - (d) μ = ∞ - -How do you interpret the case μ = ∞? - -**7.9-1** Consider the signal *x*(*t*) defined as - -$$ -x(t) = \begin{cases} 1 - |t| & -\frac{1}{2} \le t \le \frac{1}{2} \\ 0 & \text{otherwise} \end{cases} -$$ - -- (a) Sketch the signal *x*(*t*) over −2 ≤ *t* ≤ 2. -- (b) Use time-differentiation and other Fourier transform properties to determine *X*(ω). The only integration you should use is to determine the dc component *X*(0). - -(c) Using MATLAB, verify the correctness of *X*(ω) by synthesizing a 3-periodic replication of the original time-domain signal *x*(*t*). - -[*Hint:* Follow the approach taken in Ex. 7.17.] - -- **7.9-2** Consider the signal *x*(*t*) = |*t*|rect *t*−1 3 . - - (a) Sketch the signal *x*(*t*) over −5 ≤ *t* ≤ 5. - - (b) Use time-differentiation and other Fourier transform properties to determine *X*(ω). The only integration you should use is to determine the dc component *X*(0). - - (c) Use MATLAB to plot the magnitude spectrum |*X*(ω)| and the phase spectrum *X*(ω) over suitable ranges of ω. - - (d) Using MATLAB, verify the correctness of *X*(ω) by synthesizing a 10-periodic replication of the original time-domain signal *x*(*t*). - -[*Hint:* Follow the approach taken in Ex. 7.17.] - -- **7.9-3** Consider the continuous-time aperiodic signal *x*(*t*) = rect(*t*) with Fourier transform *X*(ω) = sinc(ω/2). Furthermore, let *y*(*t*) = (1 − |*t* − 1|)(*u*(*t*) − *u*(*t* − 2)) with Fourier transform *Y*(ω). - - (a) Express the Fourier transform *Y*(ω) in terms of *X*(ω). - - (b) Suppose we create Fourier series coefficients *Vk* by sampling *Y*(ω) according to *Vk* = *Y*(2π*k*/3). Sketch the corresponding time-domain signal *v*(*t*) over a suitable range of time *t*. - - (c) Use MATLAB to synthesize and plot *v*(*t*) using the Fourier series coefficients *Vk* = - -*Y*(2π*k*/3). Verify that the synthesized waveform matches the result of part (b). - -- (d) Suppose we again create Fourier series coefficients *Vk* according to *Vk* = *Y*(2π*k*/3). Next, we upsample *Vk* by factor 2 to create *Wk*. Sketch the time domain signal *p*(*t*) that has Fourier series coefficients *Pk* = *Vk* +*Wk*. -- (e) Use MATLAB to synthesize and plot *p*(*t*) using the Fourier series coefficients *Pk* = *Vk* + *Wk* defined in part (d). Verify that the synthesized waveform matches the result of part (d). -- **7.9-4** Consider the signal *x*(*t*) = *e*−*atu*(*t*). Modify CH7MP2 to compute the following essential bandwidths. - - (a) Setting *a*=1, determine the essential bandwidth *W*1 that contains 95% of the signal energy. Compare this value with the theoretical value presented in Ex. 7.20. - - (b) Setting *a*=2, determine the essential bandwidth *W*2 that contains 90% of the signal energy. - - (c) Setting *a*=3, determine the essential bandwidth *W*3 that contains 75% of the signal energy. -- **7.9-5** A unit amplitude pulse with duration τ is defined as - -$$ -x(t) = \begin{cases} 1 & |t| \le \tau/2 \\ 0 & \text{otherwise} \end{cases} -$$ - -- (a) Determine the duration τ1 that results in a 95% essential bandwidth of 5 Hz. -- (b) Determine the duration τ2 that results in a 90% essential bandwidth of 10 Hz. -- (c) Determine the duration τ3 that results in a 75% essential bandwidth 20 Hz. -- **7.9-6** Consider the signal *x*(*t*) = *e*−*atu*(*t*). - -- (a) Determine the decay parameter *a*1 that results in a 95% essential bandwidth of 5 Hz. -- (b) Determine the decay parameter *a*2 that results in a 90% essential bandwidth of 10 Hz. -- (c) Determine the decay parameter *a*3 that results in a 75% essential bandwidth 20 Hz. -- **7.9-7** Use MATLAB to determine the 95, 90, and 75% essential bandwidths of a one-second triangle function with a peak amplitude of 1. Recall that a triangle function can be constructed by the convolution of two rectangular pulses. -- **7.9-8** A 1/3 duty-cycle square-pulse *T*0-periodic signal *x*(*t*) is described as - -$$ -x(t) = \begin{cases} 1 & -T_0/6 \le t \le T_0/6 \\ 0 & T_0/t \le |t| \le T_0/2 \\ x(t+T_0) & \forall t \end{cases} -$$ - -- (a) Use spectral sampling to determine the Fourier series coefficients *Dn* of *x*(*t*) for *T*0 = 2π. Evaluate and plot *Dn* for (0 ≤ *n* ≤ 10). -- (b) Use spectral sampling to determine the Fourier series coefficients *Dn* of *x*(*t*) for *T*0 = π. Evaluate and plot *Dn* for (0 ≤ *n* ≤ 10). How does this result compare with your answer to part (a)? What can be said about the relation of *T*0 to *Dn* for signal *x*(*t*), which has fixed duty cycle of 1/3? -- **7.9-9** Determine the Fourier transform of a Gaussian pulse defined as *x*(*t*) = *e*−*t* 2 . Plot both *x*(*t*) and *X*(ω). How do the two curves compare? [*Hint:* - -$$ -\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(t-a)^2/2}dt=1 -$$ - -for any real or imaginary *a*.] - -# **SAMPLING: THE BRIDGE FROM [CONTINUOUS TO](#page-13-0) DISCRETE** - -A continuous-time signal can be processed by applying its samples through a discrete-time system. For this purpose, it is important to maintain the signal sampling rate high enough to permit the reconstruction of the original signal from these samples without error (or with an error within a given tolerance). The necessary quantitative framework for this purpose is provided by the sampling theorem derived in Sec. 8.1. - -Sampling theory is the bridge between the continuous-time and discrete-time worlds. The information inherent in a sampled continuous-time signal is equivalent to that of a discrete-time signal. A sampled continuous-time signal is a sequence of impulses, while a discrete-time signal presents the same information as a sequence of numbers. These are basically two different ways of presenting the same data. Clearly, all the concepts in the analysis of sampled signals apply to discrete-time signals. We should not be surprised to see that the Fourier spectra of the two kinds of signal are also the same (within a multiplicative constant). - -## **8.1 THE [SAMPLING](#page-13-0) THEOREM** - -We now show that a real signal whose spectrum is bandlimited to *B* Hz [*X*(ω) = 0 for |ω| > 2π*B*] can be reconstructed exactly (without any error) from its samples taken uniformly at a rate *fs* > 2*B* samples per second. In other words, the minimum sampling frequency is *fs* = 2*B* Hz.† - -To prove the sampling theorem, consider a signal *x*(*t*) (Fig. 8.1a) whose spectrum is bandlimited to *B* Hz (Fig. 8.1b).‡ For convenience, spectra are shown as functions of ω as well as of *f* (hertz). Sampling *x*(*t*) at a rate of *fs* Hz ( *fs* samples per second) can be accomplished by multiplying *x*(*t*) by an impulse train δ*T* (*t*) (Fig. 8.1c), consisting of unit impulses repeating periodically every *T* seconds, where *T* = 1/*fs*. The schematic of a sampler is shown in Fig. 8.1d. The resulting sampled signal *x*(*t*) is shown in Fig. 8.1e. The sampled signal consists of impulses - -CHAPTER - -**8** - -† The theorem stated here (and proved subsequently) applies to lowpass signals. A bandpass signal whose spectrum exists over a frequency band *fc* − (*B*/2) < |*f* | < *fc* + (*B*/2) has a bandwidth of *B* Hz. Such a signal is uniquely determined by 2*B* samples per second. In general, the sampling scheme is a bit more complex in this case. It uses two interlaced sampling trains, each at a rate of *B* samples per second. See, for example, [1]. ‡ The spectrum *X*(ω) in Fig. 8.1b is shown as real, for convenience. However, our arguments are valid for complex *X*(ω) as well. - -**Figure 8.1** Sampled signal and its Fourier spectrum. - -spaced every *T* seconds (the sampling interval). The *n*th impulse, located at *t* = *nT*, has a strength *x*(*nT*), the value of *x*(*t*) at *t* = *nT*. - -$$ -\bar{x}(t) = x(t)\delta_T(t) = \sum_n x(nT)\delta(t - nT) -$$ - -Because the impulse train δ*T* (*t*) is a periodic signal of period *T*, it can be expressed as a trigonometric Fourier series like that already obtained in Ex. 6.9 [Eq. (6.25)], - -$$ -\delta_T(t) = \frac{1}{T} [1 + 2\cos\omega_s t + 2\cos 2\omega_s t + 2\cos 3\omega_s t + \cdots] \qquad \omega_s = \frac{2\pi}{T} = 2\pi f_s -$$ - -Therefore, - -$$ -\bar{x}(t) = x(t)\delta_T(t) = \frac{1}{T}[x(t) + 2x(t)\cos\omega_s t + 2x(t)\cos 2\omega_s t + 2x(t)\cos 3\omega_s t + \cdots] -$$ -(8.1) - -To find *X*(ω), the Fourier transform of *x*(*t*), we take the Fourier transform of the right-hand side of Eq. (8.1), term by term. The transform of the first term in the brackets is *X*(ω). The transform - -#### 778 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -of the second term 2*x*(*t*) cos ω*st* is *X*(ω − ω*s*) + *X*(ω + ω*s*) [see Eq. (7.32)]. This represents spectrum *X*(ω) shifted to ω*s* and −ω*s*. Similarly, the transform of the third term 2*x*(*t*) cos 2ω*st* is *X*(ω − 2ω*s*) + *X*(ω + 2ω*s*), which represents the spectrum *X*(ω) shifted to 2ω*s* and −2ω*s*, and so on to infinity. This result means that the spectrum *X*(ω) consists of *X*(ω) repeating periodically with period ω*s* = 2π/*T* rad/s, or *fs* = 1/*T* Hz, as depicted in Fig. 8.1f. There is also a constant multiplier 1/*T* in Eq. (8.1). Therefore, - -$$ -\overline{X}(\omega) = \frac{1}{T} \sum_{n=-\infty}^{\infty} X(\omega - n\omega_s) -$$ -\n(8.2) - -If we are to reconstruct *x*(*t*) from *x*(*t*), we should be able to recover *X*(ω) from *X*(ω). This recovery is possible if there is no overlap between successive cycles of *X*(ω). Figure 8.1f indicates that this requires - -$$ -f_s > 2B \tag{8.3} -$$ - -Also, the sampling interval *T* = 1/*fs*. Therefore, - -$$ -T < \frac{1}{2B} -$$ - -Thus, as long as the sampling frequency *fs* is greater than twice the signal bandwidth *B* (in hertz), *X*(ω) consists of nonoverlapping repetitions of *X*(ω). Figure 8.1f shows that the gap between the two adjacent spectral repetitions is *fs* − 2*B* Hz, and *x*(*t*) can be recovered from its samples *x*(*t*) by passing the sampled signal *x*(*t*) through an ideal lowpass filter having a bandwidth of any value between *B* and *fs* − *B* Hz. The minimum sampling rate *fs* = 2*B* required to recover *x*(*t*) from its samples *x*(*t*) is called the *Nyquist rate* for *x*(*t*), and the corresponding sampling interval *T* = 1/2*B* is called the *Nyquist interval* for *x*(*t*). Samples of a signal taken at its Nyquist rate are the *Nyquist samples* of that signal. - -We are saying that the Nyquist rate 2*B* Hz is the minimum sampling rate required to preserve the information of *x*(*t*). This contradicts Eq. (8.3), where we showed that to preserve the information of *x*(*t*), the sampling rate *fs* needs to be greater than 2*B* Hz. Strictly speaking, Eq. (8.3) is the correct statement. However, if the spectrum *X*(ω) contains no impulse or its derivatives at the highest frequency *B* Hz, then the minimum sampling rate 2*B* Hz is adequate. In practice, it is rare to observe *X*(ω) with an impulse or its derivatives at the highest frequency. If the contrary situation were to occur, we should use Eq. (8.3).† - -† An interesting observation is that if the impulse is because of a cosine term, the sampling rate of 2*B* Hz is adequate. However, if the impulse is because of a sine term, then the rate must be greater than 2*B* Hz. This may be seen from the fact that samples of sin 2π*Bt* using *T* = 1/2*B* are all zero because sin 2π*BnT* = sinπ*n* = 0. But, samples of cos 2π*Bt* are cos 2π*BnT* = cosπ*n* = (−1)*n*. We can reconstruct cos 2π*Bt* from these samples. This peculiar behavior occurs because in the sampled signal spectrum corresponding to the signal cos 2π*Bt*, the impulses, which occur at frequencies (2*n* ± 1)*B* Hz (*n* = 0,±1,±2,. . .), interact constructively, whereas in the case of sin 2π*Bt*, the impulses, because of their opposite phases (*e*±*j*π/2), interact destructively and cancel out in the sampled signal spectrum. Hence, sin 2π*Bt* cannot be reconstructed from its samples at a rate 2*B* Hz. A similar situation exists for signal cos(2π*Bt*+θ ), which contains a component of the form sin 2π*Bt*. For this reason, it is advisable to maintain sampling rate above 2*B* Hz if a finite-amplitude component of a sinusoid of frequency *B* Hz is present in the signal. - -The sampling theorem proved here uses samples taken at uniform intervals. This condition is not necessary. Samples can be taken arbitrarily at any instants as long as the sampling instants are recorded and there are, on average, 2*B* samples per second [2]. The essence of the sampling theorem was known to mathematicians for a long time in the form of the *interpolation formula* [see later, Eq. (8.6)]. The origin of the sampling theorem was attributed by H. S. Black to Cauchy in 1841. The essential idea of the sampling theorem was rediscovered in the 1920s by Carson, Nyquist, and Hartley. - -### **EXAMPLE 8.1 Sampling at, Below, and Above the Nyquist Rate** - -In this example, we examine the effects of sampling a signal at the Nyquist rate, below the Nyquist rate (undersampling), and above the Nyquist rate (oversampling). Consider a signal *x*(*t*) = sinc2 (5π*t*) (Fig. 8.2a) whose spectrum is *X*(ω) = 0.2(ω/20π ) (Fig. 8.2b). The bandwidth of this signal is 5 Hz (10π rad/s). Consequently, the Nyquist rate is 10 Hz; that is, we must sample the signal at a rate no less than 10 samples/s. The Nyquist interval is *T* = 1/2*B* = 0.1 second. - -Recall that the sampled signal spectrum consists of (1/*T*)*X*(ω) = (0.2/*T*)(ω/20π ) repeating periodically with a period equal to the sampling frequency *fs* Hz. For the three sampling rates *fs* = 5 Hz (undersampling), 10 Hz (Nyquist rate), and 20 Hz (oversampling), we see that - -| fs
(Hz) | T
= 1
(s)
fs | TX(ω)
1 | Comments | -|------------|-----------------------|----------------|---------------| -| 5 | 0.2 | ω

20π | Undersampling | -| 10 | 0.1 | 2 ω

20π | Nyquist rate | -| 20 | 0.05 | 4 ω

20π | Oversampling | - -In the first case (undersampling), the sampling rate is 5 Hz (5 samples/s), and the spectrum (1/*T*)*X*(ω) repeats every 5 Hz (10π rad/s). The successive spectra overlap, as depicted in Fig. 8.2d, and the spectrum *X*(ω) are not recoverable from *X*(ω); that is, *x*(*t*) cannot be reconstructed from its samples *x*(*t*) in Fig. 8.2c. In the second case, we use the Nyquist sampling rate of 10 Hz (Fig. 8.2e). The spectrum *X*(ω) consists of back-to-back, nonoverlapping repetitions of (1/*T*)*X*(ω) repeating every 10 Hz. Hence, *X*(ω) can be recovered - -**Figure 8.2** Effects of undersampling and oversampling. - -from *X*(ω) using an ideal lowpass filter of bandwidth 5 Hz (Fig. 8.2f). Finally, in the last case of oversampling (sampling rate 20 Hz), the spectrum *X*(ω) consists of nonoverlapping repetitions of (1/*T*)*X*(ω) (repeating every 20 Hz) with empty bands between successive cycles (Fig. 8.2h). Hence, *X*(ω) can be recovered from *X*(ω) by using an ideal lowpass filter or even a practical lowpass filter (shown dashed in Fig. 8.2h).† - -### **DR ILL 8.1 Nyquist Sampling** - -Find the Nyquist rate and the Nyquist sampling interval for the signals sinc(100π*t*) and sinc(100π*t*)+sinc(50π*t*). - -### **ANSWERS** - -The Nyquist sampling interval is 0.01 s and the Nyquist sampling rate is 100 Hz for both signals. - -### FOR SKEPTICS ONLY - -Rare is the reader who, at first encounter, is not skeptical of the sampling theorem. It seems impossible that Nyquist samples can define the one and the only signal that passes through those sample values. We can easily picture infinite number of signals passing through a given set of samples. However, among all these (infinite number of) signals, only one has the minimum bandwidth *B* ≤ 1/2*T* Hz, where *T* is the sampling interval. See Prob. 8.2-15. - -To summarize, for a given set of samples taken at a rate *fs* Hz, there is only one signal of bandwidth *B* ≤ *fs*/2 that passes through those samples. All other signals that pass through those samples have bandwidth higher than *fs*/2, and the samples are sub-Nyquist rate samples for those signals. - -### **[8.1-1 Practical Sampling](#page-13-0)** - -In proving the sampling theorem, we assumed ideal samples obtained by multiplying a signal *x*(*t*) by an impulse train that is physically unrealizable. In practice, we multiply a signal *x*(*t*) by a train of pulses of finite width, depicted in Fig. 8.3c. The sampler is shown in Fig. 8.3d. The sampled signal *x*(*t*) is illustrated in Fig. 8.3e. We wonder whether it is possible to recover or reconstruct *x*(*t*) from this *x*(*t*). Surprisingly, the answer is affirmative, provided the sampling rate is not below the Nyquist rate. The signal *x*(*t*) can be recovered by lowpass filtering *x*(*t*) as if it were sampled by impulse train. - -† The filter should have a constant gain between 0 and 5 Hz and zero gain beyond 10 Hz. In practice, the gain beyond 10 Hz can be made negligibly small, but not zero. - -**Figure 8.3** Effect of practical sampling. - -The plausibility of this result becomes apparent when we consider the fact that reconstruction of *x*(*t*) requires the knowledge of the Nyquist sample values. This information is available or built into the sampled signal *x*(*t*) in Fig. 8.3e because the *n*th sampled pulse strength is *x*(*nT*). To prove the result analytically, we observe that the sampling pulse train *pT* (*t*) depicted in Fig. 8.3c, being a periodic signal, can be expressed as a trigonometric Fourier series - -$$ -p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n) \qquad \omega_s = \frac{2\pi}{T} -$$ - -Thus, - -$$ -\overline{x}(t) = x(t)p_T(t) = x(t)\left[C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n)\right] -$$ -$$ -= C_0 x(t) + \sum_{n=1}^{\infty} C_n x(t) \cos(n\omega_s t + \theta_n) -$$ - -The sampled signal *x*(*t*) consists of *C*0*x*(*t*), *C*1*x*(*t*) cos(ω*st* + θ1), *C*2*x*(*t*) cos(2ω*st* + θ2), ... . Note that the first term *C*0*x*(*t*) is the desired signal and all the other terms are modulated signals with spectra centered at ±ω*s*,±2ω*s*,±3ω*s*,..., as illustrated in Fig. 8.3f. Clearly the signal *x*(*t*) can be recovered by lowpass filtering of *x*(*t*), as shown in Fig. 8.3d. As before, it is necessary that ω*s* > 4π*B* (or *fs* > 2*B*). - -### **EXAMPLE 8.2 Practical Sampling** - -Demonstrate practical sampling by sampling signal *x*(*t*) = sinc2 (5π*t*) with the rectangular pulse sequence *pT* (*t*) illustrated in Fig. 8.4c. Sketch the original and sampled signals and their spectra, and discuss recovery of *x*(*t*) from its samples. - -**Figure 8.4** An example of practical sampling. - -#### 784 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -The signal *x*(*t*) and its spectrum are shown in Figs. 8.4a and 8.4b, respectively. - -The period of *pT* (*t*) is 0.1 second so that the fundamental frequency (which is the sampling frequency) is 10 Hz. Hence, ω*s* = 20π. The Fourier series for *pT* (*t*) can be expressed as - -$$ -p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos n\omega_s t -$$ - -Hence, - -$$ -\overline{x}(t) = x(t)p_T(t) -$$ - -= C1x(t) + C1x(t) cos 20 $\pi$ t + C2x(t) cos 40 $\pi$ t + C3x(t) cos 60 $\pi$ t + ... - -Use of Eq. (6.14) yields *C*0 = 1 4 and *Cn* = 2 *n*π sin *n*π 4 . Consequently, we have - -$$ -\overline{x}(t) = x(t)p_T(t) -$$ - -= $\frac{1}{4}x(t) + C_1x(t) \cos 20\pi t + C_2x(t) \cos 40\pi t + C_3x(t) \cos 60\pi t + \cdots$ - -and - -$$ -\overline{X}(\omega) = \frac{1}{4}X(\omega) + \frac{C_1}{2}[X(\omega - 20\pi) + X(\omega + 20\pi)] + \frac{C_2}{2}[X(\omega - 40\pi) + X(\omega + 40\pi)] + \frac{C_3}{2}[X(\omega - 60\pi) + X(\omega + 60\pi)] + \cdots -$$ - -where *Cn* = (2/*n*π )sin(*n*π/4). The sampled signal and its spectrum are shown in Figs. 8.4d and 8.4e, respectively. - -The spectrum *X*(ω) consists of *X*(ω) repeating periodically at the interval of 20π rad/s (10 Hz). Hence, there is no overlap between cycles, and *X*(ω) can be recovered by using an ideal lowpass filter of bandwidth 5 Hz. An ideal lowpass filter of unit gain (and bandwidth 5 Hz) will allow the first term on the right-hand side of the foregoing equation to pass fully and suppress all the other terms. Hence, the output *y*(*t*) is - -> *y*(*t*) = 1 4 *x*(*t*) - -### **DR ILL 8.2 The Role of Sampling Pulse Area** - -Show that the basic pulse *p*(*t*) used in the sampling pulse train in Fig. 8.4c cannot have zero area if we wish to reconstruct *x*(*t*) by lowpass-filtering the sampled signal. - -## **8.2 SIGNAL [RECONSTRUCTION](#page-13-0)** - -The process of reconstructing a continuous-time signal *x*(*t*) from its samples is also known as *interpolation*. In Sec. 8.1, we saw that a signal *x*(*t*) bandlimited to *B* Hz can be reconstructed (interpolated) exactly from its samples if the sampling frequency *fs* exceeds 2*B* Hz or the sampling interval *T* is less than 1/2*B*. This reconstruction is accomplished by passing the sampled signal through an ideal lowpass filter of gain *T* and having a bandwidth of any value between *B* and *fs* −*B* Hz. From a practical viewpoint, a good choice is the middle value *fs*/2 = 1/2*T* Hz or π/*T* rad/s. This value allows for small deviations in the ideal filter characteristics on either side of the cutoff frequency. With this choice of cutoff frequency and gain *T*, the ideal lowpass filter required for signal reconstruction (or interpolation) is - -$$ -H(\omega) = T \operatorname{rect}\left(\frac{\omega}{2\pi f_s}\right) = T \operatorname{rect}\left(\frac{\omega T}{2\pi}\right) -$$ -\n(8.4) - -The interpolation process here is expressed in the frequency domain as a filtering operation. Now we shall examine this process from the time-domain viewpoint. - -### TIME-DOMAIN VIEW:ASIMPLE INTERPOLATION - -Consider the interpolation system shown in Fig. 8.5a. We start with a very simple interpolating filter, whose impulse response is rect(*t*/*T*), depicted in Fig. 8.5b. This is a gate pulse centered at the origin, having unit height, and width *T* (the sampling interval). We shall find the output of this filter when the input is the sampled signal *x*(*t*) consisting of an impulse train with the *n*th impulse at *t* = *nT* with strength *x*(*nT*). Each sample in *x*(*t*), being an impulse, produces at the output a gate pulse of height equal to the strength of the sample. For instance, the *n*th sample is an impulse of strength *x*(*nT*) located at *t* = *nT* and can be expressed as *x*(*nT*)δ(*t*−*nT*). When this impulse passes through the filter, it produces at the output a gate pulse of height *x*(*nT*), centered at *t* = *nT* (shaded in Fig. 8.5c). Each sample in *x*(*t*) will generate a corresponding gate pulse, resulting in the filter output that is a staircase approximation of *x*(*t*), shown dotted in Fig. 8.5c. This filter thus gives a crude form of interpolation. - -The frequency response of this filter *H*(ω) is the Fourier transform of the impulse response rect(*t*/*T*). Thus, - -$$ -h(t) = \text{rect}\left(\frac{t}{T}\right) \qquad \text{and} \qquad H(\omega) = T \text{sinc}\left(\frac{\omega T}{2}\right) \tag{8.5} -$$ - -The amplitude response |*H*(ω)| for this filter, illustrated in Fig. 8.5d, explains the reason for the crudeness of this interpolation. This filter, also known as the *zero-order hold* (ZOH) filter, is a poor form of the ideal lowpass filter (shaded in Fig. 8.5d) required for exact interpolation.† - -We can improve on the ZOH filter by using a *first-order hold* filter, which results in a linear interpolation instead of a staircase interpolation. A linear interpolator, whose impulse response is a triangle pulse (*t*/2*T*), results in an interpolation in which successive sample tops are connected by straight-line segments (see Prob. 8.2-3). - -† Figure 8.5b shows that the impulse response of this filter is noncausal, and this filter is not realizable. In practice, we make it realizable by delaying the impulse response by *T*/2. This merely delays the output of the filter by *T*/2. - -**Figure 8.5** Simple interpolation by means of a zero-order hold (ZOH) circuit. **(a)** ZOH interpolator. **(b)** Impulse response of a ZOH circuit. **(c)** Signal reconstruction by ZOH, as viewed in the time domain. **(d)** Frequency response of a ZOH. - -### TIME-DOMAIN VIEW: AN IDEAL INTERPOLATION - -The ideal interpolation filter frequency response obtained in Eq. (8.4) is illustrated in Fig. 8.6a. The impulse response of this filter, the inverse Fourier transform of *H*(ω) is - -$$ -h(t) = \text{sinc}\left(\frac{\pi t}{T}\right) -$$ - -For the Nyquist sampling rate, *T* = 1/2*B*, and - -$$ -h(t) = \text{sinc}\left(2\pi Bt\right) -$$ - -This *h*(*t*) is depicted in Fig. 8.6b. Observe the interesting fact that *h*(*t*) = 0 at all Nyquist sampling instants (*t* = ±*n*/2*B*) except at *t* = 0. When the sampled signal *x*¯(*t*) is applied at the input of this filter, the output is *x*(*t*). Each sample in *x*(*t*), being an impulse, generates a sinc pulse of height equal to the strength of the sample, as illustrated in Fig. 8.6c. The process is identical to that depicted in Fig. 8.5c, except that *h*(*t*) is a sinc pulse instead of a gate pulse. Addition of the sinc - -**Figure 8.6** Ideal interpolation for Nyquist sampling rate. - -pulses generated by all the samples results in *x*(*t*). The *n*th sample of the input *x*(*t*) is the impulse *x*(*nT*)δ(*t* −*nT*); the filter output of this impulse is *x*(*nT*)*h*(*t* −*nT*). Hence, the filter output to *x*(*t*), which is *x*(*t*), can now be expressed as a sum - -$$ -x(t) = \sum_{n} x(nT)h(t - nT) = \sum_{n} x(nT)\operatorname{sinc}\left[\frac{\pi}{T}(t - nT)\right] -$$ - -For the case of Nyquist sampling rate, *T* = 1/2*B*, this expression simplifies to - -$$ -x(t) = \sum_{n} x(nT) \operatorname{sinc}(2\pi Bt - n\pi) -$$ -\n(8.6) - -Equation (8.6) is the *interpolation formula*, which yields values of *x*(*t*) between samples as a weighted sum of all the sample values. - -## **EXAMPLE 8.3 Bandlimited Interpolation of the Kronecker Delta Function** - -Find a signal *x*(*t*) that is bandlimited to *B* Hz, and whose samples are - -$$ -x(0) = 1 -$$ - and $x(\pm T) = x(\pm 2T) = x(\pm 3T) = \cdots = 0$ - -### 788 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -where the sampling interval *T* is the Nyquist interval for *x*(*t*), that is, *T* = 1/2*B*. - -Because we are given the Nyquist sample values, we use the interpolation formula of Eq. (8.6) to construct *x*(*t*) from its samples. Since all but one of the Nyquist samples are zero, only one term (corresponding to *n* = 0) in the summation on the right-hand side of Eq. (8.6) survives. Thus, - -$$ -x(t) = \operatorname{sinc}\left(2\pi Bt\right) -$$ - -This signal is illustrated in Fig. 8.6b. Observe that this is the only signal that has a bandwidth *B* Hz and the sample values *x*(0) = 1 and *x*(*nT*) = 0(*n* = 0). No other signal satisfies these conditions. - -### **[8.2-1 Practical Difficulties in Signal Reconstruction](#page-13-0)** - -Consider the signal reconstruction procedure illustrated in Fig. 8.7a. If *x*(*t*) is sampled at the Nyquist rate *fs* = 2*B* Hz, the spectrum *X*(ω) consists of repetitions of *X*(ω) without any gap between successive cycles, as depicted in Fig. 8.7b. To recover *x*(*t*) from *x*(*t*), we need to pass the sampled signal *x*(*t*) through an ideal lowpass filter, shown dotted in Fig. 8.7b. As seen in Sec. 7.5, such a filter is unrealizable; it can be closely approximated only with infinite time delay in the response. In other words, we can recover the signal *x*(*t*) from its samples with infinite time delay. A practical solution to this problem is to sample the signal at a rate higher than the Nyquist rate (*fs* > 2*B* or ω*s* > 4π*B*). The result is *X*(ω), consisting of repetitions of *X*(ω) with a finite bandgap between successive cycles, as illustrated in Fig. 8.7c. Now, we can recover *X*(ω) from *X*(ω) using a lowpass filter with a gradual cutoff characteristic, shown dotted in Fig. 8.7c. But even in this case, if the unwanted spectrum is to be suppressed, the filter gain must be zero beyond some frequency (see Fig. 8.7c). According to the Paley–Wiener criterion [Eq. (7.43)], it is impossible to realize even this filter. The only advantage in this case is that the required filter can be closely approximated with a smaller time delay. All this means that it is impossible in practice to recover a bandlimited signal *x*(*t*) exactly from its samples, even if the sampling rate is higher than the Nyquist rate. However, as the sampling rate increases, the recovered signal approaches the desired signal more closely. - -### THE TREACHERY OF ALIASING - -There is another fundamental practical difficulty in reconstructing a signal from its samples. The sampling theorem was proved on the assumption that the signal *x*(*t*) is bandlimited. *All practical signals are timelimited*; that is, they are of finite duration or width. We can demonstrate (see Prob. 8.2-20) that a signal cannot be timelimited and bandlimited simultaneously. If a signal is timelimited, it cannot be bandlimited, and vice versa (but it can be simultaneously nontimelimited and nonbandlimited). Clearly, all practical signals, which are necessarily timelimited, are nonbandlimited, as shown in Fig. 8.8a; they have infinite bandwidth, and the spectrum *X*(ω) consists of overlapping cycles of *X*(ω) repeating every *fs* Hz (the sampling frequency), as - -**Figure 8.7 (a)** Signal reconstruction from its samples. **(b)** Spectrum of a signal sampled at the Nyquist rate. **(c)** Spectrum of a signal sampled above the Nyquist rate. - -illustrated in Fig. 8.8b.† Because of infinite bandwidth in this case, the spectral overlap is unavoidable, regardless of the sampling rate. Sampling at a higher rate reduces but does not eliminate overlapping between repeating spectral cycles. Because of the overlapping tails, *X*(ω) no longer has complete information about *X*(ω), and it is no longer possible, even theoretically, to recover *x*(*t*) exactly from the sampled signal *x*(*t*). If the sampled signal is passed through an ideal lowpass filter of cutoff frequency *fs*/2 Hz, the output is not *X*(ω) but *Xa*(ω) (Fig. 8.8c), which is a version of *X*(ω) distorted as a result of two separate causes: - -- 1. The loss of the tail of *X*(ω) beyond |*f* | > *fs*/2 Hz. -- 2. The reappearance of this tail inverted or folded onto the spectrum. Note that the spectra cross at frequency *fs*/2 = 1/2*T* Hz. This frequency is called the *folding* frequency. - -† Figure 8.8b shows that from the infinite number of repeating cycles, only the neighboring spectral cycles overlap. This is a somewhat simplified picture. In reality, all the cycles overlap and interact with every other cycle because of the infinite width of all practical signal spectra. Fortunately, all practical spectra also must decay at higher frequencies. This results in insignificant amount of interference from cycles other than the immediate neighbors. When such an assumption is not justified, aliasing computations become little more involved. - -**Figure 8.8** Aliasing effect. **(a)** Spectrum of a practical signal *x*(*t*). **(b)** Spectrum of sampled *x*(*t*). **(c)** Reconstructed signal spectrum. **(d)** Sampling scheme using anti-aliasing filter. **(e)** Sampled signal spectrum (dotted) and the reconstructed signal spectrum (solid) when anti-aliasing filter is used. - -The spectrum may be viewed as if the lost tail is folding back onto itself at the folding frequency. For instance, a component of frequency (*fs*/2) + *fz* shows up as or "impersonates" a component of lower frequency (*fs*/2) − *fz* in the reconstructed signal. Thus, the components of frequencies above *fs*/2 reappear as components of frequencies below *fs*/2. This tail inversion, known as *spectral folding* or *aliasing,* is shown shaded in Fig. 8.8b and also in Fig. 8.8c. In the process of aliasing, not only are we losing all the components of frequencies above the folding frequency *fs*/2 Hz, but these very components reappear (aliased) as lower-frequency components, as shown in Figs. 8.8b and 8.8c. Such aliasing destroys the integrity of the frequency components below the folding frequency *fs*/2, as depicted in Fig. 8.8c. - -The aliasing problem is analogous to that of an army with a platoon that has secretly defected to the enemy side. The platoon is, however, ostensibly loyal to the army. The army is in double jeopardy. First, the army has lost this platoon as a fighting force. In addition, during actual fighting, the army will have to contend with sabotage by the defectors and will have to find another loyal platoon to neutralize the defectors. Thus, the army has lost two platoons in nonproductive activity. - -### DEFECTORS ELIMINATED: THE ANTI-ALIASING FILTER - -If you were the commander of the betrayed army, the solution to the problem would be obvious. As soon as the commander got wind of the defection, he would incapacitate, by whatever means, the defecting platoon *before the fighting begins*. This way he loses only one (the defecting) platoon. This is a partial solution to the double jeopardy of betrayal and sabotage, a solution that partly rectifies the problem and cuts the losses to half. - -We follow exactly the same procedure. The potential defectors are all the frequency components beyond the folding frequency *fs*/2 = 1/2*T* Hz. We should eliminate (suppress) these components from *x*(*t*) *before sampling x*(*t*). Such suppression of higher frequencies can be accomplished by an ideal lowpass filter of cutoff *fs*/2 Hz, as shown in Fig. 8.8d. This is called the *anti-aliasing filter*. Figure 8.8d also shows that anti-aliasing filtering is performed before sampling. Figure 8.8e shows the sampled signal spectrum (dotted) and the reconstructed signal *Xaa*(ω) when an anti-aliasing scheme is used. An anti-aliasing filter essentially bandlimits the signal *x*(*t*) to *fs*/2 Hz. This way, we lose only the components beyond the folding frequency *fs*/2 Hz. These suppressed components now cannot reappear to corrupt the components of frequencies below the folding frequency. Clearly, use of an anti-aliasing filter results in the reconstructed signal spectrum *Xaa*(ω) = *X*(ω) for |*f* | < *fs*/2. Thus, although we lost the spectrum beyond *fs*/2 Hz, the spectrum for all the frequencies below *fs*/2 remains intact. The effective aliasing distortion is cut in half owing to elimination of folding. We stress again that the anti-aliasing operation must be performed *before the signal is sampled*. - -An anti-aliasing filter also helps to reduce noise. Noise, generally, has a wideband spectrum, and without anti-aliasing, the aliasing phenomenon itself will cause the noise lying outside the desired band to appear in the signal band. Anti-aliasing suppresses the entire noise spectrum beyond frequency *fs*/2. - -The anti-aliasing filter, being an ideal filter, is unrealizable. In practice, we use a steep cutoff filter, which leaves a sharply attenuated spectrum beyond the folding frequency *fs*/2. - -### SAMPLING FORCES NONBANDLIMITED SIGNALS TO APPEAR BANDLIMITED - -Figure 8.8b shows that the spectrum of a signal *x*(*t*) consists of overlapping cycles of *X*(ω). This means that *x*(*t*) are sub-Nyquist samples of *x*(*t*). However, we may also view the spectrum in Fig. 8.8b as the spectrum *Xa*(ω) (Fig. 8.8c), repeating periodically every *fs* Hz without overlap. The spectrum *Xa*(ω) is bandlimited to *fs*/2 Hz. Hence, these (sub-Nyquist) samples of *x*(*t*) are actually the Nyquist samples for signal *xa*(*t*). In conclusion, sampling a nonbandlimited signal *x*(*t*) at a rate *fs* Hz makes the samples appear to be the Nyquist samples of some signal *xa*(*t*), bandlimited to *fs*/2 Hz. In other words, sampling makes a nonbandlimited signal appear to be a bandlimited signal *xa*(*t*) with bandwidth *fs*/2 Hz. A similar conclusion applies if *x*(*t*) is bandlimited but sampled at a sub-Nyquist rate. - -### VERIFICATION OF ALIASING IN SINUSOIDS - -We showed in Fig. 8.8b how sampling a signal below the Nyquist rate causes aliasing, which makes a signal of higher frequency (*fs*/2) + *fz* Hz masquerade as a signal of lower frequency (*fs*/2) − *fz* Hz. Figure 8.8b demonstrates this result in the frequency domain. Let us now verify it in the time domain to gain a deeper appreciation of aliasing. - -We can prove our proposition by showing that samples of sinusoids of frequencies (ω*s*/2)+ω*z* and (ω*s*/2)−ω*z* are identical when the sampling frequency is *fs* = ω*s*/2π Hz. - -For a sinusoid *x*(*t*) = cos ω*t*, sampled at intervals of *T* seconds, *x*(*nT*), its *n*th sample (at *t* = *nT*) is - -*x*(*nT*) = cosω*nT n* integer - -Hence, samples of sinusoids of frequency ω = (ω*s*/2)±ω*z* are† - -$$ -x(nT) = \cos\left(\frac{\omega_s}{2} \pm \omega_z\right) nT = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT \mp \sin\left(\frac{\omega_s}{2}\right) nT \sin\omega_z nT -$$ - -Recognizing that ω*sT* = 2π*fsT* = 2π, and sin(ω*s*/2)*nT* = sinπ*n* = 0 for all integer *n*, we obtain - -$$ -x(nT) = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT -$$ - -Clearly, the samples of a sinusoid of frequency (*fs*/2)+*fz* are identical to the samples of a sinusoid (*fs*/2)−*fz*. ‡ For instance, when a sinusoid of frequency 100 Hz is sampled at a rate of 120 Hz, the apparent frequency of the sinusoid that results from reconstruction of the samples is 20 Hz. This follows from the fact that here, 100 = (*fs*/2)+*fz* = 60+*fz* so that *fz* = 40. Hence, (*fs*/2)−*fz* = 20. Such would precisely be the conclusion arrived at from Fig. 8.8b. - -† Here we have ignored the phase aspect of the sinusoid. Sampled versions of a sinusoid *x*(*t*) = cos(ω*t* + θ ) with two different frequencies (ω*s*/2) ± ω*z* have identical frequency, but the phase signs may be reversed - -depending on the value of ω*z*. ‡ The reader is encouraged to verify this result graphically by plotting the spectrum of a sinusoid of frequency (ω*s*/2) + ω*z* (impulses at ±[(ω*s*/2) + ω*z*]) and its periodic repetition at intervals ω*s*. Although the result is valid for all values of ω*z*, consider the case of ω*z* < ω*s*/2 to simplify the graphics. - -This discussion again shows that sampling a sinusoid of frequency *f* aliasing can be avoided if the sampling rate *fs* > 2*f* Hz. - -$$ -0 \le f < \frac{f_s}{2} \qquad \text{or} \qquad 0 \le \omega < \frac{\pi}{T} -$$ - -Violating this condition leads to aliasing, implying that the samples appear to be those of a lower-frequency signal. Because of this loss of identity, it is impossible to reconstruct the signal faithfully from its samples. - -### GENERAL CONDITION FOR ALIASING IN SINUSOIDS - -We can generalize the foregoing result by showing that samples of a sinusoid of frequency *f*0 are identical to those of a sinusoid of frequency *f*0 + *mfs* Hz (integer *m*), where *fs* is the sampling frequency. The samples of cos 2π(*f*0 +*mfs*)*t* are - -$$ -\cos 2\pi (f_0 + m f_s) nT = \cos (2\pi f_0 nT + 2\pi mn) = \cos 2\pi f_0 nT -$$ - -The result follows because *mn* is an integer and *fsT* = 1. This result shows that sinusoids of frequencies that differ by an integer multiple of *fs* result in identical set of samples. In other words, samples of sinusoids separated by frequency *fs* Hz are identical. This implies that samples of sinusoids in any frequency band of *fs* Hz are unique; that is, no two sinusoids in that band have the same samples (when sampled at a rate *fs* Hz). For instance, frequencies in the band from −*fs*/2 to *fs*/2 have unique samples (at the sampling rate *fs*). This band is called the *fundamental band*. Recall also that *fs*/2 is the folding frequency. - -From the discussion thus far, we conclude that if a continuous-time sinusoid of frequency *f* Hz is sampled at a rate of *fs* Hz (samples/s), the resulting samples would appear as samples of a continuous-time sinusoid of frequency *fa* in the fundamental band, where - -$$ -f_a = f - mf_s \qquad -\frac{f_s}{2} \le f_a < \frac{f_s}{2} \qquad m \text{ an integer} \tag{8.7} -$$ - -The frequency *fa* lies in the fundamental band from −*fs*/2 to *fs*/2. Figure 8.9a shows the plot of *fa* versus *f* , where *f* is the actual frequency and *fa* is the corresponding fundamental band frequency, whose samples are identical to those of the sinusoid of frequency *f* , when the sampling rate is *fs* Hz. - -Recall, however, that the sign change of a frequency does not alter the actual frequency of the waveform. This is because - -$$ -\cos(-\omega_a t + \theta) = \cos(\omega_a t - \theta) -$$ - -Clearly the *apparent frequency* of a sinusoid of frequency −*fa* is also *fa*. However, its phase undergoes a sign change. This means the apparent frequency of any sampled sinusoid lies in the range from 0 to *fs*/2 Hz. To summarize, if a continuous-time sinusoid of frequency *f* Hz is sampled at a rate of *fs* Hz (samples/second), the resulting samples would appear as samples of a continuous-time sinusoid of frequency |*fa*| that lies in the band from 0 to *fs*/2. According to Eq. (8.7), - -$$ -|f_a| = |f - mf_s| \qquad |f_a| \le \frac{f_s}{2} \qquad m \text{ an integer} -$$ - -**Figure 8.9** Apparent frequencies of a sampled sinusoid: **(a)** *fa* versus *f* and **(b)** |*fa*| versus *f* . - -The plot of the apparent frequency |*fa*| versus *f* is shown in Fig. 8.9b.† As expected, the apparent frequency |*fa*| of any sampled sinusoid, regardless of its frequency, is always in the range of 0 to *fs*/2 Hz. However, when *fa* is negative, the phase of the apparent sinusoid undergoes a sign change. The frequency belts in which such phase changes occur are shown shaded in Fig. 8.9b. - -Consider, for example, a sinusoid cos(2π*ft* +θ ) with *f* =8000 Hz sampled at a rate *fs* = 3000 Hz. Using Eq. (8.7), we obtain *fa* = 8000 − 3 × 3000 = −1000. Hence, |*fa*| = 1000. The samples would appear to have come from a sinusoid cos(2000π*t* − θ ). Observe the sign change of the phase because *fa* is negative.‡ - -In the light of the foregoing development, let us consider a sinusoid of frequency *f* = (*fs*/2)+*fz*, sampled at a rate of *fs* Hz. According to Eq. (8.7), - -$$ -f_a = \frac{f_s}{2} + f_z - (1 \times f_s) = -\frac{f_s}{2} + f_z -$$ - -Hence, the apparent frequency is |*fa*| = (*fs*/2) − *fz*, confirming our earlier result. However, the phase of the sinusoid will suffer a sign change because *fa* is negative. - -Figure 8.10 shows how samples of sinusoids of two different frequencies (sampled at the same rate) generate identical sets of samples. Both the sinusoids are sampled at a rate *fs* = 5 Hz (*T* = 0.2 second). The frequencies of the two sinusoids, 1 Hz (period 1) and 6 Hz (period 1/6), differ by *fs* = 5 Hz. - -† The plots in Figs. 8.9 and 5.17 are identical. This is because a sampled sinusoid is basically a discrete-time sinusoid. - -‡ For phase sign change, we are assuming that the signal has the form cos(2π*ft* + θ ). If the form is sin(2π*ft* + θ ), the rule changes slightly. It is left as an exercise for the reader to show that when *fa* < 0, this sinusoid appears as −sin(2π|*fa*|*t* − θ ). Thus, in addition to phase change, the amplitude also changes sign. - -**Figure 8.10** Demonstration of aliasing. - -The reason for aliasing can be clearly seen in Fig. 8.10. The root of the problem is the sampling rate, which may be adequate for the lower-frequency sinusoid but is clearly inadequate for the higher-frequency sinusoid. The figure clearly shows that between the successive samples of the higher-frequency sinusoid, there are wiggles, which are bypassed or ignored, and are unrepresented in the samples, indicating a sub-Nyquist rate of sampling. The frequency of the apparent signal *xa*(*t*) is always the lowest possible frequency that lies within the band |*f* | ≤ *fs*/2. Thus, the apparent frequency of the samples in this example is 1 Hz. If these samples are chosen to reconstruct a signal using a lowpass filter of bandwidth *fs*/2, we shall obtain a sinusoid of frequency 1 Hz. - -### **EXAMPLE 8.4 Apparent Frequency of Sampled Sinusoids** - -A continuous-time sinusoid cos(2π*ft* + θ ) is sampled at a rate *fs* = 1000 Hz. Determine the apparent (aliased) sinusoid of the resulting samples if the input signal frequency *f* is **(a)** 400 Hz, **(b)** 600 Hz, **(c)** 1000 Hz, and **(d)** 2400 Hz. - -The folding frequency is *fs*/2 = 500. Hence, sinusoids below 500 Hz (frequency within the fundamental band) will not be aliased and sinusoids of frequency above 500 Hz will be aliased. - -**(c)** Since *f* = 1000 Hz can be expressed as 1000 = 0 + 1000, we see that *fa* = 0. Hence, the aliased frequency is 0 Hz (dc), and there is no phase sign change. The apparent sinusoid is *y*(*t*) = cos(0π*t* ±θ ) = cos(θ ). This is a dc signal with constant sample values for all *n*. - -**(a)** Since *f* = 400 Hz is less than 500 Hz, there is no aliasing. The apparent sinusoid is cos(2π*ft* +θ ) with *f* = 400. - -**(b)** Since *f* = 600 Hz can be expressed as 600 = −400 + 1000, we see that *fa* = −400. Hence, the aliased frequency is 400 Hz and the phase changes sign. The apparent (aliased) sinusoid is cos(2π*ft* −θ ) with *f* = 400. - -### 796 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -**(d)** Here, *f* = 2400 Hz can be expressed as 2400 = 400 + (2 × 1000) so that *fa* = 400. Hence, the aliased frequency is 400 Hz and there is no sign change for the phase. The apparent sinusoid is cos(2π*ft* +θ ) with *f* = 400. - -We could have found these answers directly from Fig. 8.9b. For example, for case (b), we read |*fa*| = 400 corresponding to *f* = 600. Moreover, *f* = 600 lies in the shaded belt. Hence, there is a phase sign change. - -## **DR ILL 8.3 A Case of Identical Sampled Sinusoids** - -Show that samples of 90 Hz and 110 Hz sinusoids of the form cosω*t* are identical when sampled at a rate 200 Hz. - -### **DR ILL 8.4 Apparent Frequency of Sampled Sinusoids** - -A sinusoid of frequency *f*0 Hz is sampled at a rate of 100 Hz. Determine the apparent frequency of the samples if *f*0 is **(a)** 40 Hz, **(b)** 60 Hz, **(c)** 140 Hz, and **(d)** 160 Hz. - -### **ANSWERS** - -All four cases have an apparent frequency of 40 Hz. - -### **[8.2-2 Some Applications of the Sampling Theorem](#page-13-0)** - -The sampling theorem is very important in signal analysis, processing, and transmission because it allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing a continuous-time signal is therefore equivalent to processing a discrete sequence of numbers. Such processing leads us directly into the area of digital filtering. In the field of communication, the transmission of a continuous-time message reduces to the transmission of a sequence of numbers by means of pulse trains. The continuous-time signal *x*(*t*) is sampled, and sample values are used to modify certain parameters of a periodic pulse train. We may vary the amplitudes (Fig. 8.11b), widths (Fig. 8.11c), or positions (Fig. 8.11d) of the pulses in proportion to the sample values of the signal *x*(*t*). Accordingly, we may have *pulse-amplitude modulation* (PAM), *pulse-width modulation* (PWM), or *pulse-position modulation* (PPM). The most important form of pulse modulation today is *pulse-code modulation* (PCM), discussed in Sec. 8.3 in connection with Fig. 8.14b. In all these cases, instead of transmitting *x*(*t*), we transmit the corresponding pulse-modulated signal. At the receiver, we read the information of the pulse-modulated signal and reconstruct the analog signal *x*(*t*). - -**Figure 8.11** Pulse-modulated signals. **(a)** The signal. **(b)** The PAM signal. **(c)** The PWM (PDM) signal. **(d)** The PAM signal. - -One advantage of using pulse modulation is that it permits the simultaneous transmission of several signals on a time-sharing basis—*time-division multiplexing* (TDM). Because a pulse-modulated signal occupies only a part of the channel time, we can transmit several pulse-modulated signals on the same channel by interweaving them. Figure 8.12 shows the TDM of two PAM signals. In this manner, we can multiplex several signals on the same channel by reducing pulse widths.† - -Digital signals also offer an advantage in the area of communications, where signals must travel over distances. Transmission of digital signals is more rugged than that of analog signals because digital signals can withstand channel noise and distortion much better as long as the noise - -† Another method of transmitting several baseband signals simultaneously is frequency-division multiplexing (FDM) discussed in Sec. 7.7-4. In FDM, various signals are multiplexed by sharing the channel bandwidth. The spectrum of each message is shifted to a specific band not occupied by any other signal. The information of various signals is located in nonoverlapping frequency bands of the channel (Fig. 7.45). In a way, TDM and FDM are duals of each other. - -**Figure 8.12** Time-division multiplexing of two signals. - -**Figure 8.13** Digital signal transmission: **(a)** at the transmitter, **(b)** received distorted signal (without noise), **(c)** received distorted signal (with noise), and **(d)** regenerated signal at the receiver. - -and the distortion are within limits. An analog signal can be converted to digital binary form through sampling and quantization (rounding off), as explained in the next section. The digital (binary) message in Fig. 8.13a is distorted by the channel, as illustrated in Fig. 8.13b. Yet if the distortion remains within a limit, we can recover the data without error because we need only make a simple binary decision: Is the received pulse positive or negative? Figure 8.13c shows the same data with channel distortion and noise. Here again, the data can be recovered correctly as long as the distortion and the noise are within limits. Such is not the case with analog messages. Any distortion or noise, no matter how small, will distort the received signal. - -The greatest advantage of digital communication over the analog counterpart, however, is the viability of regenerative repeaters in the former. In an analog transmission system, a message signal grows progressively weaker as it travels along the channel (transmission path), whereas the channel noise and the signal distortion, being cumulative, become progressively stronger. Ultimately, the signal, overwhelmed by noise and distortion, is mutilated. Amplification is of little help because it enhances the signal and the noise in the same proportion. Consequently, the distance over which an analog message can be transmitted is limited by the transmitted power. If a transmission path is long enough, the channel distortion and noise will accumulate sufficiently to overwhelm even a digital signal. The trick is to set up repeaters along the transmission path at distances short enough to permit detection of signal pulses before the noise and distortion have a chance to accumulate sufficiently. At each repeater, the pulses are detected, and new, clean pulses are transmitted to the next repeater, which, in turn, duplicates the same process. If the noise and distortion remain within limits (which is possible because of the closely spaced repeaters), pulses can be detected correctly.† This way the digital messages can be transmitted over longer distances with greater reliability. In contrast, analog messages cannot be cleaned up periodically, and their transmission is therefore less reliable. The most significant error in digitized signals comes from quantizing (rounding off). This error, discussed in Sec. 8.3, can be reduced as much as desired by increasing the number of quantization levels, at the cost of an increased bandwidth of the transmission medium (channel). - -## **[8.3 ANALOG-TO-DIGITAL](#page-13-0) (A/D) CONVERSION** - -The amplitude of an *analog* signal can take on any value over a continuous range. Hence, analog signal amplitude can take on an infinite number of values. In contrast, a *digital* signal amplitude can take on only a finite number of values. An analog signal can be converted into a digital signal by means of sampling and *quantizing* (rounding off). Sampling an analog signal alone will not yield a digital signal because a sample of analog signal can still take on any value in a continuous range. It is digitized by rounding off its value to one of the closest permissible numbers (or *quantized levels*), as illustrated in Fig. 8.14a, which represents one possible quantizing scheme. The amplitudes of the analog signal *x*(*t*) lie in the range (−*V*,*V*). This range is partitioned into *L* subintervals, each of magnitude = 2*V*/*L*. Next, each sample amplitude is approximated by the midpoint value of the subinterval in which the sample falls (see Fig. 8.14a for *L* = 16). It is clear that each sample is approximated to one of the *L* numbers. Thus, the signal is digitized with quantized samples taking on any one of the *L* values. This is an *L*-ary digital signal (see Sec. 1.3-2). Each sample can now be represented by one of *L* distinct pulses. - -From a practical viewpoint, dealing with a large number of distinct pulses is difficult. We prefer to use the smallest possible number of distinct pulses, the very smallest number being 2. A digital signal using only two symbols or values is the binary signal. A binary digital signal (a signal that can take on only two values) is very desirable because of its simplicity, economy, and ease of engineering. We can convert an *L*-ary signal into a binary signal by using pulse coding. Figure 8.14b shows one such code for the case of *L* = 16. This code, formed by binary representation of the 16 decimal digits from 0 to 15, is known as the *natural binary code (NBC)*. For *L* quantization levels, we need a minimum of *b* binary code digits, where 2*b* = *L* or *b* = log2 *L*. - -Each of the 16 levels is assigned one binary code word of four digits. Thus, each sample in this example is encoded by four binary digits. To transmit or digitally process the binary data, we need to assign a distinct electrical pulse to each of the two binary states. One possible way is to assign a negative pulse to a binary **0** and a positive pulse to a binary **1** so that each sample is now represented by a group of four binary pulses (pulse code), as depicted in Fig. 8.14b. The resulting binary signal is a digital signal obtained from the analog signal *x*(*t*) through A/D conversion. In communications jargon, such a signal is known as a pulse-code-modulated (PCM) signal. - -† The error in pulse detection can be made negligible. - -| (a) | -|-----| -| | -| | -| | - -| Digit | Binary equivalent | Pulse code waveform | -|-------|-------------------|---------------------| -| 0 | 0000 | | -| 1 | 0001 | | -| 2 | 0010 | | -| 3 | 0011 | | -| 4 | 0100 | | -| 5 | 0101 | | -| 6 | 0110 | | -| 7 | 0111 | | -| 8 | 1000 | | -| 9 | 1001 | | -| 10 | 1010 | | -| 11 | 1011 | | -| 12 | 1100 | | -| 13 | 1101 | | -| 14 | 1110 | | -| 15 | 1111 | | - -(b) - -**Figure 8.14** Analog-to-digital (A/D) conversion of a signal: **(a)** quantizing and **(b)** pulse coding. - -The convenient contraction of "*b*inary digi*t*" to *bit* has become an industry standard abbreviation. - -The audio signal bandwidth is about 15 kHz, but subjective tests show that signal articulation (intelligibility) is not affected if all the components above 3400 Hz are suppressed [3]. Since the objective in telephone communication is intelligibility rather than high fidelity, the components above 3400 Hz are eliminated by a lowpass filter.† The resulting signal is then sampled at a rate of 8000 samples/s (8 kHz). This rate is intentionally kept higher than the Nyquist sampling rate of 6.8 kHz to avoid unrealizable filters required for signal reconstruction. Each sample is finally quantized into 256 levels (*L* = 256), which requires a group of eight binary pulses to encode each sample (28 = 256). Thus, a digitized telephone signal consists of data amounting to 8 × 8000 = 64,000 or 64 kbit/s, requiring 64,000 binary pulses per second for its transmission. - -The compact disc (CD), a high-fidelity application of A/D conversion, requires the audio signal bandwidth of 20 kHz. Although the Nyquist sampling rate is only 40 kHz, an actual sampling rate of 44.1 kHz is used for the reason mentioned earlier. The signal is quantized into a rather large number of levels (*L* = 65,536) to reduce quantizing error. The binary-coded samples are now recorded on the CD. - -### A HISTORICAL NOTE - -The binary system of representing any number by using **1**s and **0**s was invented in India by Pingala (ca. 200 BCE). It was again worked out independently in the West by Gottfried Wilhelm Leibniz (1646–1716). He felt a spiritual significance in this discovery, reasoning that **1** representing unity was clearly a symbol for God, while **0** represented the nothingness. He reasoned that if all numbers can be represented merely by the use of **1** and **0**, this surely proves that God created the universe out of nothing! - -### **EXAMPLE 8.5 ADC Bit Number and Bit Rate** - -A signal *x*(*t*) bandlimited to 3 kHz is sampled at a rate 331 3% higher than the Nyquist rate. The maximum acceptable error in the sample amplitude (the maximum error due to quantization) is 0.5% of the peak amplitude *V*. The quantized samples are binary-coded. Find the required sampling rate, the number of bits required to encode each sample, and the bit rate of the resulting PCM signal. - -The Nyquist sampling rate is *f*Nyq = 2×3000 = 6000 Hz (samples/s). The actual sampling rate is *fA* = 6000×(11 3 ) = 8000 Hz. - -The quantization step is , and the maximum quantization error is ±/2, where = 2*V*/*L*. The maximum error due to quantization, /2, should be no greater than 0.5% of the - -†Components below 300 Hz may also be suppressed without affecting the articulation. - -### 802 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -signal peak amplitude *V*. Therefore, - -$$ -\frac{\Delta}{2} = \frac{V}{L} = \frac{0.5}{100}V \implies L = 200 -$$ - -For binary coding, *L* must be a power of 2. Hence, the next higher value of *L* that is a power of 2 is *L* = 256. Because log2 256 = 8, we need 8 bits to encode each sample. Therefore the bit rate of the PCM signal is - -8000×8 = 64,000 bits/s - -### **DR ILL 8.5 Bit Number and Bit Rate for ASCII** - -The American Standard Code for Information Interchange (ASCII) has 128 characters, which are binary-coded. A certain computer generates 100,000 characters per second. Show that - -- **(a)** 7 bits (binary digits) are required to encode each character -- **(b)** 700,000 bits/s are required to transmit the computer output. - -## **8.4 DUAL OF TIME [SAMPLING: SPECTRAL](#page-13-0) SAMPLING** - -As in other cases, the sampling theorem has its dual. In Sec. 8.1, we discussed the time-sampling theorem and showed that a signal bandlimited to *B* Hz can be reconstructed from the signal samples taken at a rate of *fs* > 2*B* samples/s. Note that the signal spectrum exists over the frequency range (in hertz) of −*B* to *B*. Therefore, 2*B* is the spectral width (not the bandwidth, which is *B*) of the signal. This fact means that a signal *x*(*t*) can be reconstructed from samples taken at a rate *fs* > the spectral width of *X*(ω) in hertz ( *fs* > 2*B*). - -We now prove the dual of the time-sampling theorem. This is the *spectral sampling theorem*, which applies to timelimited signals (the dual of bandlimited signals). A timelimited signal *x*(*t*) exists only over a finite interval of τ seconds, as shown in Fig. 8.15a. Generally, a timelimited signal is characterized by *x*(*t*) = 0 for *t* < *T*1 and *t* > *T*2 (assuming *T*2 > *T*1). The signal width or duration is τ = *T*2 −*T*1 seconds. - -The spectral sampling theorem states that the spectrum *X*(ω) of a signal *x*(*t*) timelimited to a duration of τ seconds can be reconstructed from the samples of *X*(ω) taken at a rate *R* samples/Hz, where *R* > τ (the signal width or duration) in seconds. - -Figure 8.15a shows a timelimited signal *x*(*t*) and its Fourier transform *X*(ω). Although *X*(ω) is complex in general, it is adequate for our line of reasoning to show *X*(ω) as a real function. - -$$ -X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt = \int_{0}^{\tau} x(t)e^{-j\omega t}dt -$$ -\n(8.8) - - - -**Figure 8.15** Periodic repetition of a signal amounts to sampling its spectrum. - -We now construct *xT*0 (*t*), a periodic signal formed by repeating *x*(*t*) every *T*0 seconds (*T*0 > τ ), as depicted in Fig. 8.15b. This periodic signal can be expressed by the exponential Fourier series - -$$ -x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0} -$$ - -where (assuming *T*0 > τ ) - -$$ -D_n = \frac{1}{T_0} \int_0^{T_0} x(t) e^{-jn\omega_0 t} dt = \frac{1}{T_0} \int_0^{\tau} x(t) e^{-jn\omega_0 t} dt -$$ - -From Eq. (8.8), it follows that - -$$ -D_n = \frac{1}{T_0} X(n\omega_0) -$$ - -This result indicates that the coefficients of the Fourier series for *xT*0 (*t*) are (1/*T*0) times the sample values of the spectrum *X*(ω) taken at intervals of ω0. This means that the spectrum of the periodic signal *xT*0 (*t*) is the sampled spectrum *X*(ω), as illustrated in Fig. 8.15b. Now as long as *T*0 > τ , the successive cycles of *x*(*t*) appearing in *xT*0 (*t*) do not overlap, and *x*(*t*) can be recovered from *xT*0 (*t*). Such recovery implies indirectly that *X*(ω) can be reconstructed from its samples. These samples are separated by the fundamental frequency *f*0 = 1/*T*0 Hz of the periodic signal *xT*0 (*t*). Hence, the condition for recovery is *T*0 > τ ; that is, - -$$ -f_0 < \frac{1}{\tau} \, \mathrm{Hz} -$$ - -Therefore, to be able to reconstruct the spectrum *X*(ω) from the samples of *X*(ω), the samples should be taken at frequency intervals *f*0 < 1/τ Hz. If *R* is the sampling rate (samples/Hz), then - -$$ -R = \frac{1}{f_0} > \tau \text{ samples/Hz} -$$ - -#### 804 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -### SPECTRAL INTERPOLATION - -Consider a signal timelimited to τ seconds and centered at *Tc*. We now show that the spectrum *X*(ω) of *x*(*t*) can be reconstructed from the samples of *X*(ω). For this case, using the dual of the approach employed to derive the signal interpolation formula in Eq. (8.6), we obtain the spectral interpolation formula† - -$$ -X(\omega) = \sum_{n=-\infty}^{\infty} X(n\omega_0) \operatorname{sinc}\left(\frac{\omega T_0}{2} - n\pi\right) e^{-j(\omega - n\omega_0)T_c} \qquad \omega_0 = \frac{2\pi}{T_0} \qquad T_0 > \tau \tag{8.9} -$$ - -For the case in Fig. 8.15, *Tc* = *T*0/2. If the pulse *x*(*t*) were to be centered at the origin, then *Tc* = 0, and the exponential term at the extreme right in Eq. (8.9) would vanish. In such a case, Eq. (8.9) would be the exact dual of Eq. (8.6). - -### **EXAMPLE 8.6 Spectral Sampling and Interpolation** - -The spectrum *X*(ω) of a unit-duration signal *x*(*t*), centered at the origin, is sampled at the intervals of 1 Hz or 2π rad/s (the Nyquist rate). The samples are - -$$ -X(0) = 1 -$$ - and $X(\pm 2\pi n) = 0$ $n = 1, 2, 3, ...$ - -Find *x*(*t*). - -We use the interpolation formula Eq. (8.9) (with *Tc* = 0) to construct *X*(ω) from its samples. Since all but one of the Nyquist samples are zero, only one term (corresponding to *n* = 0) in the summation on the right-hand side of Eq. (8.9) survives. Thus, with *X*(0) = 1 and τ = *T*0 = 1, we obtain - -$$ -X(\omega) = \operatorname{sinc}\left(\frac{\omega}{2}\right) -$$ - and $x(t) = \operatorname{rect}(t)$ - -For a signal of unit duration, this is the only spectrum with the sample values *X*(0) = 1 and *X*(2π*n*) = 0(*n* = 0). No other spectrum satisfies these conditions. - -† This can be obtained by observing that the Fourier transform of *xT*0 (*t*) is 2π % *n Dn*δ(ω − *n*ω0) [see Eq. (7.22)]. We can recover *x*(*t*) from *xT*0 (*t*) by multiplying the latter with rect(*t* − *Tc*)/*T*0, whose Fourier transform is *T*0 sinc(ω*T*0/2)*e*−*j*ω*Tc* . Hence, *X*(ω) is 1/2π times the convolution of these two Fourier transforms, which yields Eq. (8.9). - -## **8.5 NUMERICAL [COMPUTATION OF THE](#page-13-0) FOURIER TRANSFORM: THE DISCRETE FOURIER TRANSFORM** - -Numerical computation of the Fourier transform of *x*(*t*) requires sample values of *x*(*t*) because a digital computer can work only with discrete data (sequence of numbers). Moreover, a computer can compute *X*(ω) only at some discrete values of ω [samples of *X*(ω)]. We therefore need to relate the samples of *X*(ω) to samples of *x*(*t*). This task can be accomplished by using the results of the two sampling theorems developed in Secs. 8.1 and 8.4. - -We begin with a timelimited signal *x*(*t*) (Fig. 8.16a) and its spectrum *X*(ω) (Fig. 8.16b). Since *x*(*t*) is timelimited, *X*(ω) is nonbandlimited. For convenience, we shall show all spectra as functions of the frequency variable *f* (in hertz) rather than ω. According to the sampling theorem, the spectrum *X*(ω) of the sampled signal *x*(*t*) consists of *X*(ω) repeating every *fs* Hz, where *fs* = 1/*T*, as depicted in Fig. 8.16d.† In the next step, the sampled signal in Fig. 8.16c is repeated periodically every *T*0 seconds, as illustrated in Fig. 8.16e. According to the spectral sampling theorem, such an operation results in sampling the spectrum at a rate of *T*0 samples/Hz. This sampling rate means that the samples are spaced at *f*0 = 1/*T*0 Hz, as depicted in Fig. 8.16f. - -The foregoing discussion shows that when a signal *x*(*t*) is sampled and then periodically repeated, the corresponding spectrum is also sampled and periodically repeated. Our goal is to relate the samples of *x*(*t*) to the samples of *X*(ω). - -### NUMBER OF SAMPLES - -One interesting observation from Figs. 8.16e and 8.16f is that *N*0, the number of samples of the signal in Fig. 8.16e in one period *T*0, is identical to *N* 0, the number of samples of the spectrum in Fig. 8.16f in one period *fs*. To see this, we notice that - -$$ -N_0 = \frac{T_0}{T} \quad N'_0 = \frac{f_s}{f_0} \quad f_s = \frac{1}{T} \quad \text{and} \quad f_0 = \frac{1}{T_0} -$$ -(8.10) - -Using these relations, we see that - -$$ -N_0 = \frac{T_0}{T} = \frac{f_s}{f_0} = N'_0 -$$ - -### ALIASING AND LEAKAGE IN NUMERICAL COMPUTATION - -Figure 8.16f shows the presence of aliasing in the samples of the spectrum *X*(ω). This aliasing error can be reduced as much as desired by increasing the sampling frequency *fs* (decreasing the sampling interval *T* = 1/*fs*). The aliasing can never be eliminated for timelimited *x*(*t*), however, because its spectrum *X*(ω) is nonbandlimited. Had we started with a signal having a bandlimited spectrum *X*(ω), there would be no aliasing in the spectrum in Fig. 8.16f. Unfortunately, such a signal is nontimelimited, and its repetition (in Fig. 8.16e) would result in signal overlapping (aliasing in the time domain). In this case, we shall have to contend with errors in signal - -† There is a multiplying constant 1/*T* for the spectrum in Fig. 8.16d [see Eq. (8.2)], but this is irrelevant to our discussion here. - -samples. In other words, in computing the direct or inverse Fourier transform numerically, we can reduce the error as much as we wish, but the error can never be eliminated. This is true of numerical computation of the direct and inverse Fourier transforms, regardless of the method used. For example, if we determine the Fourier transform by direct integration numerically, by using Eq. (7.9), there will be an error because the interval of integration *t* can never be made zero. Similar remarks apply to numerical computation of the inverse transform. Therefore, we should always keep in mind the nature of this error in our results. In our discussion (Fig. 8.16), we assumed *x*(*t*) to be a timelimited signal. If *x*(*t*) were not timelimited, we would need to timelimit it because numerical computations can work only with finite data. Furthermore, this data truncation causes error because of spectral spreading (smearing) and leakage, as discussed in Sec. 7.8. The leakage also causes aliasing. Leakage can be reduced by using a tapered window for signal truncation. But this choice increases spectral spreading or smearing. Spectral spreading can be reduced by increasing the window width (i.e., more data), which increases *T*0, and reduces *f*0 (increases *spectral* or *frequency resolution*). - -### PICKET FENCE EFFECT - -The numerical computation method yields only the uniform sample values of *X*(ω). The major peaks or valleys of *X*(ω) can lie between two samples and may remain hidden, giving a false picture of reality. Viewing samples is like viewing the signal and its spectrum through a "picket fence" with upright posts that are very wide and placed close together. What is hidden behind the pickets is much more than what we can see. Such misleading results can be avoided by using a sufficiently large *N*0, the number of samples, to increase resolution. We can also use zero padding (discussed later) or the spectral interpolation formula [Eq. (8.9)] to determine the values of *X*(ω) between samples. - -### POINTS OF DISCONTINUITY - -If *x*(*t*) or *X*(ω) has a jump discontinuity at a sampling point, the sample value should be taken as the average of the values on the two sides of the discontinuity because the Fourier representation at a point of discontinuity converges to the average value. - -### DERIVATION OF THE DISCRETE FOURIER TRANSFORM (DFT) - -If *x*(*nT*) and *X*(*r*ω0) are the *n*th and *r*th samples of *x*(*t*) and *X*(ω), respectively, then we define new variables *xn* and *Xr* as - -$$ -x_n = Tx(nT) = \frac{T_0}{N_0}x(nT) -$$ -\n(8.11) - -and - -$$ -X_r = X(r\omega_0) -$$ - -where - -$$ -\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0} -$$ - -We shall now show that *xn* and *Xr* are related by the following equations† : - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n} -$$ - (8.12) - -and - -$$ -x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{ir\Omega_0 n} -$$ -\n(8.13) - -where - -$$ -\Omega_0 = \omega_0 T = \frac{2\pi}{N_0} -$$ - -These equations define the direct and the inverse *discrete Fourier transforms,* with *Xr* the direct discrete Fourier transform (DFT) of *xn*, and *xn* the inverse discrete Fourier transform (IDFT) of *Xr*. The notation - -$$ -x_n \Longleftrightarrow X_r -$$ - -is also used to indicate that *xn* and *Xr* are a DFT pair. Remember that *xn* is *T*0/*N*0 times the *n*th sample of *x*(*t*) and *Xr* is the *r*th sample of *X*(ω). Knowing the sample values of *x*(*t*), we can use the DFT to compute the sample values of *X*(ω)—and vice versa. Note, however, that *xn* is a function of *n* (*n* = 0, 1, 2,...,*N*0 − 1) rather than of *t* and that *Xr* is a function of *r* (*r* = 0, 1, 2,...,*N*0 − 1) rather than of ω. Moreover, both *xn* and *Xr* are periodic sequences of period *N*0 (Figs. 8.16e, 8.16f). Such sequences are called *N*0*-periodic sequences*. The proof of the DFT relationships in Eqs. (8.12) and (8.13) follows directly from the results of the sampling theorem. The sampled signal *x*(*t*) (Fig. 8.16c) can be expressed as - -$$ -\overline{x}(t) = \sum_{n=0}^{N_0 - 1} x(nT)\delta(t - nT) -$$ - -Since δ (*t* −*nT*) ⇐⇒ *e*−*jn*ω*T* , applying the Fourier transform yields - -$$ -\overline{X}(\omega) = \sum_{n=0}^{N_0 - 1} x(nT) e^{-jn\omega T} -$$ - -But from Fig. 8.1f [or Eq. (8.2)], it is clear that over the interval |ω| ≤ ω*s*/2, *X*(ω), the Fourier transform of *x*(*t*) is *X*(ω)/*T*, assuming negligible aliasing. Hence, - -$$ -X(\omega) = T\overline{X}(\omega) = T\sum_{n=0}^{N_0 - 1} x(nT)e^{-jn\omega T} \qquad |\omega| \le \frac{\omega_s}{2} -$$ - -† In Eqs. (8.12) and (8.13), the summation is performed from 0 to *N*0 −1. It is shown in Sec. 9.1-2 [Eqs. (9.6) and (9.7)] that the summation may be performed over any successive *N*0 values of *n* or *r*. - -and - -$$ -X_r = X(r\omega_0) = T \sum_{n=0}^{N_0 - 1} x(nT)e^{-nkr\omega_0 T} -$$ -\n(8.14) - -If we let ω0*T* = 0, then from Eq. (8.10), - -$$ -\Omega_0 = \omega_0 T = 2\pi f_0 T = \frac{2\pi}{N_0} -$$ - -Also, from Eq. (8.11), - -$$ -Tx(nT) = x_n -$$ - -Therefore, Eq. (8.14) becomes - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -This is Eq. (8.12), which we set to prove. - -The inverse transform relationship of Eq. (8.13) can be derived by using a similar procedure with the roles of *t* and ω reversed, but here we shall use a more direct proof. To prove Eq. (8.13), we multiply both sides of Eq. (8.12) by *ejm*0*r* and sum over *r* as - -$$ -\sum_{r=0}^{N_0-1} X_r e^{jm\Omega_0 r} = \sum_{r=0}^{N_0-1} \left[ \sum_{n=0}^{N_0-1} x_n e^{-jr\Omega_0 n} \right] e^{jm\Omega_0 r} -$$ - -By interchanging the order of summation on the right-hand side, we have - -$$ -\sum_{r=0}^{N_0-1} X_r e^{jm\Omega_0 r} = \sum_{n=0}^{N_0-1} x_n \left[ \sum_{r=0}^{N_0-1} e^{j(m-n)\Omega_0 r} \right] -$$ - -As the footnote below readily shows, the inner sum on the right-hand side is zero for *n* = *m* and is *N*0 when *n* = *m*. † Thus, the outer sum will have only one nonzero term when *n* = *m*, and it is - -† We show that *N* - -$$ -\sum_{n=0}^{N_0-1} e^{jk\Omega_0 n} = \begin{cases} N_0 & k = 0, \pm N_0, \pm 2N_0, \dots \\ 0 & \text{otherwise} \end{cases} -$$ - (8.15) - -Recall that 0*N*0 = 2π. So *ejk*0*n* = 1 when *k* = 0,±*N*0,±2*N*0,.... Hence, the sum on the left-hand side of Eq. (8.15) is *N*0. To compute the sum for other values of *k*, we note that the sum on the left-hand side of Eq. (8.15) is a geometric progression with common ratio α = *ejk*0 . Therefore, (see Sec. B.8-3) - -$$ -\sum_{n=0}^{N_0-1} e^{jk\Omega_0 n} = \frac{e^{jk\Omega_0 N_0} - 1}{e^{jk\Omega_0} - 1} = 0 \qquad (e^{jk\Omega_0 N_0} = e^{j2\pi m} = 1) -$$ - -#### 810 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -*N*0*xn* = *N*0*xm*. Therefore, - -$$ -x_m = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{im\Omega_0 r} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -Because *Xr* is *N*0 periodic, we need to determine the values of *Xr* over any one period. It is customary to determine *Xr* over the range (0, *N*0 − 1), rather than over the range (−*N*0/2,(*N*0/2)−1).† - -## CHOICE OF *T* AND *T*0 - -In DFT computation, we first need to select suitable values for *N*0 and *T* or *T*0. For this purpose, we begin by deciding on *B*, the essential bandwidth (in hertz) of the signal. The sampling frequency *fs* must be at least 2*B*, that is, - -$$ -\frac{f_s}{2} \ge B -$$ - -Moreover, the sampling interval *T* = 1/*fs* [Eq. (8.10)], and - -$$ -T \le \frac{1}{2B} \tag{8.16} -$$ - -Once we pick *B*, we can choose *T* according to Eq. (8.16). Also, - -$$ -f_0 = \frac{1}{T_0} \tag{8.17} -$$ - -where *f*0 is the *frequency resolution* [separation between samples of *X*(ω)]. Hence, if *f*0 is given, we can pick *T*0 according to Eq. (8.17). Knowing *T*0 and *T*, we determine *N*0 from - -$$ -N_0 = \frac{T_0}{T} -$$ - -### ZERO PADDING - -Recall that observing *Xr* is like observing the spectrum *X*(ω) through a picket fence. If the frequency sampling interval *f*0 is not sufficiently small, we could miss out on some significant details and obtain a misleading picture. To obtain a higher number of samples, we need to reduce *f*0. Because *f*0 =1/*T*0, a higher number of samples requires us to increase the value of *T*0, the period of repetition for *x*(*t*). This option increases *N*0, the number of samples of *x*(*t*), by adding dummy samples of 0 value. This addition of dummy samples is known as *zero padding*. Thus, zero padding increases the number of samples and may help in getting a better idea of the spectrum *X*(ω) from its samples *Xr*. To continue with our picket fence analogy, zero padding is like using more, and narrower, pickets. - -† The DFT of Eq. (8.12) and the IDFT of Eq. (8.13) represent a transform in their own right, and they are exact. There is no approximation. However, *xn* and *Xr*, thus obtained, are only approximations to the actual samples of a signal *x*(*t*) and of its Fourier transform *X*(ω). - -### ZERO PADDING DOES NOT IMPROVE ACCURACY OR RESOLUTION - -Actually, we are not observing *X*(ω) through a picket fence. We are observing a distorted version of *X*(ω) resulting from the truncation of *x*(*t*). Hence, we should keep in mind that even if the fence were transparent, we would see a reality distorted by aliasing. Seeing through the picket fence just gives us an imperfect view of the imperfectly represented reality. Zero padding only allows us to look at more samples of that imperfect reality. It can never reduce the imperfection in what is behind the fence. The imperfection, which is caused by aliasing, can be lessened only by reducing the sampling interval *T*. Observe that reducing *T* also increases *N*0, the number of samples, and is like increasing the number of pickets while reducing their width. But in this case, the reality behind the fence is also better dressed and we see more of it. - -### **EXAMPLE 8.7 Number of Samples and Frequency Resolution** - -A signal *x*(*t*) has a duration of 2 ms and an essential bandwidth of 10 kHz. It is desirable to have a frequency resolution of 100 Hz in the DFT (*f*0 = 100). Determine *N*0. - -To have *f*0 = 100 Hz, the effective signal duration *T*0 must be - -$$ -T_0 = \frac{1}{f_0} = \frac{1}{100} = 10 -$$ - ms - -Since the signal duration is only 2 ms, we need zero padding over 8 ms. Also, *B* = 10,000. Hence, *fs* = 2*B* = 20,000 and *T* = 1/*fs* = 50 µs. Furthermore, - -$$ -N_0 = \frac{f_s}{f_0} = \frac{20,000}{100} = 200 -$$ - -The *fast Fourier transform* (FFT) algorithm (discussed later; see Sec. 8.6) is used to compute DFT, where it proves convenient (although not necessary) to select *N*0 as a power of 2; that is, *N*0 = 2*n* (*n*, integer). Let us choose *N*0 = 256. Increasing *N*0 from 200 to 256 can be used to reduce aliasing error (by reducing *T*), to improve resolution (by increasing *T*0 using zero padding), or a combination of both. - -**Reducing Aliasing Error.** We maintain the same *T*0 so that *f*0 = 100. Hence, - -$$ -f_s = N_0 f_0 = 256 \times 100 = 25,600 -$$ - and $T = \frac{1}{f_s} = 39 \,\mu s$ - -Thus, increasing *N*0 from 200 to 256 permits us to reduce the sampling interval *T* from 50 µs to 39 µs while maintaining the same frequency resolution (*f*0 = 100). - -**Improving Resolution.** Here, we maintain the same *T* = 50 µs, which yields - -$$ -T_0 = N_0 T = 256(50 \times 10^{-6}) = 12.8 -$$ - ms and $f_0 = \frac{1}{T_0} = 78.125$ Hz - -### 812 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -Thus, increasing *N*0 from 200 to 256 can improve the frequency resolution from 100 to 78.125 Hz while maintaining the same aliasing error (*T* = 50 µs). - -**Combination of Reducing Aliasing Error and Improving Resolution.** To simultaneously reduce alias error and improve resolution, we could choose *T* = 45 µs and *T*0 = 11.5 ms so that *f*0 = 86.96 Hz. Many other combinations exist as well. - -### **EXAMPLE 8.8 DFT to Compute the Fourier Transform of an Exponential** - -Use the DFT to compute (samples of) the Fourier transform of *e*−2*t u*(*t*). Plot the resulting Fourier spectra. - -We first determine *T* and *T*0. The Fourier transform of *e*−2*t u*(*t*) is 1/(*j*ω + 2). This lowpass signal is not bandlimited. In Sec. 7.6, we used the energy criterion to compute the essential bandwidth of a signal. Here, we shall present a simpler, but workable alternative to the energy criterion. The essential bandwidth of a signal will be taken as the frequency at which |*X*(ω)| drops to 1% of its peak value (see the footnote on page 736). In this case, the peak value occurs at ω = 0, where |*X*(0)| = 0.5. Observe that - -$$ -|X(\omega)| = \frac{1}{\sqrt{\omega^2 + 4}} \approx \frac{1}{\omega} \qquad \omega \gg 2 -$$ - -Also, 1% of the peak value is 0.01 × 0.5 = 0.005. Hence, the essential bandwidth *B* is at ω = 2π*B*, where - -$$ -|X(\omega)| \approx \frac{1}{2\pi B} = 0.005 \quad \Rightarrow \quad B = \frac{100}{\pi} \text{ Hz} -$$ - -and from Eq. (8.16), - -$$ -T \le \frac{1}{2B} = \frac{\pi}{200} = 0.015708 -$$ - -Had we used 1% energy criterion to determine the essential bandwidth, following the procedure in Ex. 7.20, we would have obtained *B* = 20.26 Hz, which is somewhat smaller than the value just obtained by using the 1% amplitude criterion. - -The second issue is to determine *T*0. Because the signal is not timelimited, we have to truncate it at *T*0 such that *x*(*T*0) 1. A reasonable choice would be *T*0 = 4 because *x*(4) = *e*−8 = 0.0003351. The result is *N*0 = *T*0/*T* = 254.6, which is not a power of 2. Hence, we choose *T*0 = 4, and *T* = 0.015625 = 1/64, yielding *N*0 = 256, which is a power of 2. - -Note that there is a great deal of flexibility in determining *T* and *T*0, depending on the accuracy desired and the computational capacity available. We could just as well have chosen *T* = 0.03125, yielding *N*0 = 128, although this choice would have given a slightly higher aliasing error. - -Because the signal has a jump discontinuity at *t* = 0, the first sample (at *t* = 0) is 0.5, the averages of the values on the two sides of the discontinuity. We compute *Xr* (the DFT) from the samples of *e*−2*t u*(*t*) according to Eq. (8.12). Note that *Xr* is the *r*th sample of *X*(ω), and these samples are spaced at *f*0 = 1/*T*0 = 0.25 Hz (ω0 = π/2 rad/s). - -Because *Xr* is *N*0 periodic, *Xr* = *X*(*r*+256) so that *X*256 = *X*0. Hence, we need to plot *Xr* over the range *r* = 0 to 255 (not 256). Moreover, because of this periodicity, *X*−*r* = *X*(−*r*+256), and the values of *Xr* over the range *r* = −127 to −1 are identical to those over the range *r* = 129 to 255. Thus, *X*−127 = *X*129, *X*−126 = *X*130,...,*X*−1 = *X*255. In addition, because of the property of conjugate symmetry of the Fourier transform, *X*−*r* = *X*∗ *r* , it follows that *X*−1 = *X*∗ 1 , *X*−2 = *X*∗ 2 ,...,*X*−128 = *X*∗ 128. Thus, we need *Xr* only over the range *r* = 0 to *N*0/2 (128 in this case). - -Figure 8.17 shows the computed plots of |*Xr*| and *Xr*. The exact spectra are depicted by continuous curves for comparison. Note the nearly perfect agreement between the two sets of spectra. We have depicted the plot of only the first 28 points rather than all 128 points, which would have made the figure very crowded, resulting in loss of clarity. The points are at the intervals of 1/*T*0 = 1/4 Hz or ω0 = 1.5708 rad/s. The 28 samples, therefore, exhibit the plots over the range ω = 0 to ω = 28(1.5708) ≈ 44 rad/s or 7 Hz. - -**Figure 8.17** Discrete Fourier transform of an exponential signal *e*−2*t u*(*t*). - -### 814 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -In this example, we knew *X*(ω) beforehand; hence we could make intelligent choices for *B* (or the sampling frequency *fs*). In practice, we generally do not know *X*(ω) beforehand. In fact, that is the very thing we are trying to determine. In such a case, we must make an intelligent guess for *B* or *fs* from circumstantial evidence. We should then continue reducing the value of *T* and recomputing the transform until the result stabilizes within the desired number of significant digits. - -### USING MATLAB TO COMPUTE AND PLOT THE RESULTS - -Let us now use MATLAB to confirm the results of this example. First, parameters are defined and MATLAB's fft command is used to compute the DFT. - -``` ->> T_0 = 4; N_0 = 256; T = T_0/N_0; t = (0:T:T*(N_0-1))'; ->> x = T*exp(-2*t); x(1) = x(1)/2; ->> X_r = fft(x); r = [-N_0/2:N_0/2-1]'; omega_r = r*2*pi/T_0; -``` - -The true Fourier transform is also computed for comparison. - ->> omega = linspace(-pi/T,pi/T,5001); X = 1./(j\*omega+2); - -For clarity, we display spectrum over a restricted frequency range. - -``` ->> subplot(1,2,1); stem(omega_r,fftshift(abs(X_r)),'k.'); -``` - -``` ->> line(omega,abs(X),'color',[0 0 0]); axis([-0.01 44 -0.01 0.51]); -``` - -- >> xlabel('\omega'); ylabel('|X(\omega)|'); -- >> subplot(1,2,2); stem(omega\_r,fftshift(angle(X\_r)),'k.'); -- >> line(omega,angle(X),'color',[0 0 0]); axis([-0.01 44 -pi/2-0.01 0.01]); -- >> xlabel('\omega'); ylabel('\angle X(\omega)'); - -The results, shown in Fig. 8.18, match the earlier results shown in Fig. 8.17. - -### **EXAMPLE 8.9 DFT to Compute the Fourier Transform of a Rectangular Pulse** - -Use the DFT to compute the Fourier transform of 8 rect(*t*). - -This gate function and its Fourier transform are illustrated in Figs. 8.19a and 8.19b. To determine the value of the sampling interval *T*, we must first decide on the essential bandwidth *B*. In Fig. 8.19b, we see that *X*(ω) decays rather slowly with ω. Hence, the essential bandwidth *B* is rather large. For instance, at *B* = 15.5 Hz (97.39 rad/s), *X*(ω) = −0.1643, which is about 2% of the peak at *X*(0). Hence, the essential bandwidth is well above 16 Hz if we use the 1% of the peak amplitude criterion for computing the essential bandwidth. However, we shall deliberately take *B* = 4 for two reasons: to show the effect of aliasing and because the use of *B* > 4 would give an enormous number of samples, which could not be conveniently displayed on the page without losing sight of the essentials. Thus, we shall intentionally accept approximation to graphically clarify the concepts of the DFT. - -The choice of *B* = 4 results in the sampling interval *T* = 1/2*B* = 1/8. Looking again at the spectrum in Fig. 8.19b, we see that the choice of the frequency resolution *f*0 = 1/4 Hz is reasonable. Such a choice gives us four samples in each lobe of *X*(ω). In this case *T*0 = 1/*f*0 = 4 seconds and *N*0 = *T*0/*T* = 32. The duration of *x*(*t*) is only 1 second. We must repeat it every 4 seconds (*T*0 = 4), as depicted in Fig. 8.19c, and take samples every 1/8 second. This choice yields 32 samples (*N*0 = 32). Also, - -$$ -x_n = Tx(nT) = \frac{1}{8}x(nT) -$$ - -Since *x*(*t*) = 8 rect(*t*), the values of *xn* are 1, 0, or 0.5 (at the points of discontinuity), as illustrated in Fig. 8.19c, where *xn* is depicted as a function of *t* as well as *n*, for convenience. - -In the derivation of the DFT, we assumed that *x*(*t*) begins at *t* = 0 (Fig. 8.16a), and then took *N*0 samples over the interval (0, *T*0). In the present case, however, *x*(*t*) begins at −1/2. This difficulty is easily resolved when we realize that the DFT obtained by this procedure is actually the DFT of *xn* repeating periodically every *T*0 seconds. Figure 8.19c clearly indicates that periodic repeating the segment of *xn* over the interval from −2 to 2 seconds yields the same signal as the periodic repeating the segment of *xn* over the interval from 0 to 4 seconds. Hence, the DFT of the samples taken from −2 to 2 seconds is the same as that of the samples taken from 0 to 4 seconds. Therefore, regardless of where *x*(*t*) starts, we can always take the samples of *x*(*t*) and its periodic extension over the interval from 0 to *T*0. In the present example, the 32 sample values are - -$$ -x_n = \begin{cases} 1 & 0 \le n \le 3 \text{ and } 29 \le n \le 31 \\ 0 & 5 \le n \le 27 \\ 0.5 & n = 4,28 \end{cases} -$$ - -816 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -**Figure 8.19** Discrete Fourier transform of a gate pulse. - -Observe that the last sample is at *t* = 31/8, not at 4, because the signal repetition starts at *t* = 4, and the sample at *t* = 4 is the same as the sample at *t* = 0. Now, *N*0 = 32 and 0 = 2π/32 = π/16. Therefore [see Eq. (8.12)], - -$$ -X_r = \sum_{n=0}^{31} x_n e^{-jr(\pi/16)n} -$$ - -Values of *Xr* are computed according to this equation and plotted in Fig. 8.19d. - -The samples *Xr* are separated by *f*0 = 1/*T*0 Hz. In this case *T*0 = 4, so the frequency resolution *f*0 is 1/4 Hz, as desired. The folding frequency *fs*/2 = *B* = 4 Hz corresponds to *r* = *N*0/2 = 16. Because *Xr* is *N*0 periodic (*N*0 = 32), the values of *Xr* for *r* = −16 to *n* = −1 are the same as those for *r* = 16 to *n* = 31. For instance, *X*17 = *X*−15, *X*18 = *X*−14, and so on. The DFT gives us the samples of the spectrum *X*(ω). - -For the sake of comparison, Fig. 8.19d also shows the shaded curve 8 sinc(ω/2), which is the Fourier transform of 8 rect(*t*). The values of *Xr* computed from the DFT equation show aliasing error, which is clearly seen by comparing the two superimposed plots. The error in *X*2 is just about 1.3%. However, the aliasing error increases rapidly with *r*. For instance, the error in *X*6 is about 12%, and the error in *X*10 is 33%. The error in *X*14 is a whopping 72%. The percent error increases rapidly near the folding frequency (*r* = 16) because *x*(*t*) has a jump discontinuity, which makes *X*(ω) decay slowly as 1/ω. Hence, near the folding frequency, the inverted tail (due to aliasing) is very nearly equal to *X*(ω) itself. Moreover, the final values are the difference between the exact and the folded values (which are very close to the exact values). Hence, the percent error near the folding frequency (*r* = 16 in this case) is very high, although the absolute error is very small. Clearly, for signals with jump discontinuities, the aliasing error near the folding frequency will always be high (in percentage terms), regardless of the choice of *N*0. To ensure a negligible aliasing error at any value *r*, we must make sure that *N*0 *r*. This observation is valid for all signals with jump discontinuities. - -### USING MATLAB TO COMPUTE AND PLOT THE RESULTS - -Once again, MATLAB lets us easily confirm the results of this example. First, parameters are defined and MATLAB's fft command is used to compute the DFT. - ->> T\_0 = 4; N\_0 = 32; T = T\_0/N\_0; >> x\_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]'; >> X\_r = fft(x\_n); r = [-N\_0/2:N\_0/2-1]'; omega\_r = r\*2\*pi/T\_0; - -The true Fourier transform is also computed for comparison. - -``` ->> omega = linspace(-pi/T,pi/T,5001); X = 8*sinc(omega/(2*pi)); -``` - -Since it is real, we can display the resulting spectrum using a single plot. - ->> clf; stem(omega\_r,fftshift(real(X\_r)),'k.'); >> line(omega,X,'color',[0 0 0]); - -### 818 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - ->> xlabel('\omega'); ylabel('X(\omega)'); axis tight - -The result, shown in Fig. 8.20, matches the earlier result shown in Fig. 8.19d. The DFT approximation does not perfectly follow the true Fourier transform, especially at high frequencies, because the parameter *B* is deliberately set too small. - -### **[8.5-1 Some Properties of the DFT](#page-13-0)** - -The discrete Fourier transform is basically the Fourier transform of a sampled signal repeated periodically. Hence, the properties derived earlier for the Fourier transform apply to the DFT as well. - -LINEARITY If *xn* ⇐⇒ *Xr* and *gn* ⇐⇒ *Gr*, then - -$$ -a_1x_n + a_2g_n \Longleftrightarrow a_1X_r + a_2G_r -$$ - -The proof is trivial. - -### CONJUGATE SYMMETRY - -From the conjugation property *x*∗(*t*) ⇐⇒ *X*∗(−ω), we have - -*x*∗ *n* ←→ *X*∗ −*r* - -From this equation and the time-reversal property, we obtain - -$$ -x_{-n}^* \longleftrightarrow X_r^* -$$ - -When *x*(*t*) is real, then the conjugate-symmetry property states that *X*∗(ω) = *X*(−ω). Hence, for real *xn*, - -$$ -X_r^* = X_{-r} -$$ - -Moreover, *Xr* is *N*0 periodic. Hence, - -$$ -X_r^* = X_{N_0-r} -$$ - -Because of this property, we need compute only half the DFTs for real *xn*. The other half are the conjugates. - -### TIME SHIFTING - -The time-shifting (circular shifting) property states† - -$$ -x_{n-k} \Longleftrightarrow X_r e^{-jr\Omega_0 k} -$$ - -**Proof.** We use Eq. (8.13) to find the inverse DFT of *Xre*−*jr*0*k* as - -$$ -\frac{1}{N_0} \sum_{r=0}^{N_0-1} X_r e^{-jr\Omega_0 k} e^{jr\omega_0 n} = \frac{1}{N_0} \sum_{r=0}^{N_0-1} X_r e^{jr\Omega_0 (n-k)} = x_{n-k} -$$ - -### FREQUENCY SHIFTING - -A dual of the time-shifting property, the frequency-shifting property states - -$$ -x_n e^{jn\Omega_0 m} \Longleftrightarrow X_{r-m} -$$ - -**Proof.** This proof is identical to that of the time-shifting property except that we start with Eq. (8.12). - -### CIRCULAR CONVOLUTION - -The circular (or periodic) convolution property states - -$$ -x_n \circledast g_n \Longleftrightarrow X_r G_r \tag{8.18} -$$ - -and - -$$ -x_n g_n \Longleftrightarrow \frac{1}{N_0} X_r \circledast G_r \tag{8.19} -$$ - -For two *N*0-periodic sequences *xn* and *gn*, circular (or periodic) convolution is defined by - -$$ -x_n \circledast g_n = \sum_{k=0}^{N_0 - 1} x_k g_{n-k} = \sum_{k=0}^{N_0 - 1} g_k x_{n-k} -$$ -\n(8.20) - -† Time shifting is also known as *circular shifting* because such a shift can be interpreted as a circular shift of the *N*0 samples in the first cycle 0 ≤ *n* ≤ *N*0 −1. - -**Figure 8.21** Graphical depictions of circular convolution. - -To prove Eq. (8.18), we find the DFT of the circular convolution *xn*-∗ *gn* as - -$$ -\sum_{n=0}^{N_0-1} \left( \sum_{k=0}^{N_0-1} x_k g_{n-k} \right) e^{-j r \omega_0 n} = \sum_{k=0}^{N_0-1} x_k \left( \sum_{n=0}^{N_0-1} g_{n-k} e^{-j r \omega_0 n} \right) -$$ -$$ -= \sum_{k=0}^{N_0-1} x_k (G_r e^{-j r \Omega_0 k}) = X_r G_r -$$ - -Equation (8.19) can be proved in the same way. - -For periodic sequences, the convolution can be visualized in terms of two sequences, with one sequence fixed and the other inverted and moved past the fixed sequence, one digit at a time. If the two sequences are *N*0 periodic, the same configuration will repeat after *N*0 shifts of the sequence. Clearly the convolution *xn*-∗ *gn* becomes *N*0 periodic. Such convolution can be conveniently visualized in terms of *N*0 sequences, as illustrated in Fig. 8.21, for the case of *N*0 = 4. The inner *N*0-point sequence *xn* is clockwise and fixed. The outer *N*0-point sequence *gn* is inverted so that it becomes counterclockwise. This sequence is now rotated clockwise 1 unit at a time. We multiply the overlapping numbers and add. For example, the value of *xn*-∗ *gn* at *n* = 0 (Fig. 8.21) is - -$$ -x_0g_0 + x_1g_3 + x_2g_2 + x_3g_1 -$$ - -and the value of *xn*-∗ *gn* at *n* = 1 is (Fig. 8.21) - -$$ -x_0g_1 + x_1g_0 + x_2g_3 + x_3g_2 -$$ - -and so on. - -### **[8.5-2 Some Applications of the DFT](#page-13-0)** - -The DFT is useful not only in the computation of direct and inverse Fourier transforms, but also in other applications such as convolution, correlation, and filtering. Use of the efficient FFT algorithm, discussed shortly (Sec. 8.6), makes it particularly appealing. - -### LINEAR CONVOLUTION - -Let *x*(*t*) and *g*(*t*) be the two signals to be convolved. In general, these signals may have different time durations. To convolve them by using their samples, they must be sampled at the same rate (not below the Nyquist rate of either signal). Let *xn* (0 ≤ *n* ≤ *N*1 − 1) and *gn* (0 ≤ *n* ≤ *N*2 − 1) be the corresponding discrete sequences representing these samples. Now, - -$$ -c(t) = x(t) * g(t) -$$ - -and if we define three sequences as *xn* = *Tx*(*nT*), *gn* = *Tg*(*nT*), and *cn* = *Tc*(*nT*), then† - -$$ -c_n = x_n * g_n -$$ - -where we define the linear convolution sum of two discrete sequences *xn* and *gn* as - -$$ -c_n = x_n * g_n = \sum_{k=-\infty}^{\infty} x_k g_{n-k} -$$ - -Because of the width property of the convolution, *cn* exists for 0≤ *n*≤ *N*1+*N*2−1. To be able to use the DFT circular convolution technique, we must make sure that the circular convolution will yield the same result as does linear convolution. In other words, the signal resulting from the circular convolution must have the same length (*N*1 + *N*2 − 1) as that of the signal resulting from linear convolution. This step can be accomplished by adding *N*2 − 1 dummy samples of zero value to *xn* and *N*1 −1 dummy samples of zero value to *gn* (zero padding). This procedure changes the length of both *xn* and *gn* to *N*1+*N*2 −1. The circular convolution now is identical to the linear convolution except that it repeats periodically with period *N*1 +*N*2 −1. A little reflection will show that in such a case the circular convolution procedure in Fig. 8.21 over one cycle (0 ≤ *n* ≤ *N*1 + *N*2 − 1) is identical to the linear convolution of the two sequences *xn* and *gn*. We can use the DFT to find the convolution *xn* ∗ *gn* in three steps, as follows: - -- 1. Find the DFTs *Xr* and *Gr* corresponding to suitably padded *xn* and *gn*. -- 2. Multiply *Xr* by *Gr*. -- 3. Find the IDFT of *XrGr*. This procedure of convolution, when implemented by the fast Fourier transform algorithm (discussed later), is known as *fast convolution*. - -### FILTERING - -We generally think of filtering in terms of a hardware-oriented solution (e.g., building a circuit with *RLC* components and operational amplifiers). However, filtering also has a software-oriented solution [a computer algorithm that yields the filtered output *y*(*t*) for a given input *x*(*t*)]. This goal can be conveniently accomplished by using the DFT. If *x*(*t*) is the signal to be filtered, then *Xr*, the DFT of *xn*, is found. The spectrum *Xr* is then shaped (filtered) as desired by multiplying *Xr* by *Hr*, where *Hr* are the samples of *H*(ω) for the filter [*Hr* = *H*(*r*ω0)]. Finally, we take the IDFT of *XrHr* to obtain the filtered output *yn*[*yn* =*Ty*(*nT*)]. This procedure is demonstrated in the following example. - -† We can show that *cn* = lim*T*→0 *xn* ∗ *gn*; [see 4]. Error is inherent in any numerical method used to compute convolution of continuous-time signals; since *T* = 0 in practice, there will be some error in this equation. - -The signal *x*(*t*) in Fig. 8.22a is passed through an ideal lowpass filter of frequency response *H*(ω) depicted in Fig. 8.22b. Use the DFT to find the sampled version of the filter output. - -**Figure 8.22** DFT solution for filtering *x*(*t*) through *H*(ω). - -We have already found the 32-point DFT of *x*(*t*) (see Fig. 8.19d). Next we multiply *Xr* by *Hr*. To find *Hr*, we recall using *f*0 = 1/4 in computing the 32-point DFT of *x*(*t*). Because *Xr* is 32-periodic, *Hr* must also be 32-periodic with samples separated by 1/4 Hz. This fact means that *Hr* must be repeated every 8 Hz or 16π rad/s (see Fig. 8.22c). The resulting 32 samples of *Hr* over (0 ≤ ω ≤ 16π ) are as follows: - -$$ -H_r = \begin{cases} 1 & 0 \le r \le 7 \\ 0 & 9 \le r \le 23 \\ 0.5 & r = 8,24 \end{cases} \text{ and } 25 \le r \le 31 -$$ - -We multiply *Xr* with *Hr*. The desired output signal samples *yn* are found by taking the inverse DFT of *XrHr*. The resulting output signal is illustrated in Fig. 8.22d. - -It is quite simple to verify the results of this filtering example using MATLAB. First, parameters are defined, and MATLAB's fft command is used to compute the DFT of *xn*. - ->> T\_0 = 4; N\_0 = 32; T = T\_0/N\_0; n = (0:N\_0-1); r = n; >> x\_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]'; X\_r = fft(x\_n); - -The DFT of the filter's output is just the product of the filter response *Hr* and the input DFT *Xr*. The output *yn* is obtained using the ifft command and then plotted. - -``` ->> H_r = [ones(1,8) 0.5 zeros(1,15) 0.5 ones(1,7)]'; ->> Y_r = H_r.*X_r; y_n = ifft(Y_r); ->> clf; stem(n,real(y_n),'k.'); ->> xlabel('n'); ylabel('y_n'); axis([0 31 -.1 1.1]); -``` - -The result, shown in Fig. 8.23, matches the earlier result shown in Fig. 8.22d. Recall, this DFT-based approach shows the samples *yn* of the filter output *y*(*t*) (sampled in this case at a rate *T* = 1 8 ) over 0 ≤ *n* ≤ *N*0 − 1 = 31 when the input pulse *x*(*t*) is periodically replicated to form samples *xn* (see Fig. 8.19c). - -**Figure 8.23** Using MATLAB and the DFT to determine filter output. - -## **8.6 THE FAST FOURIER [TRANSFORM](#page-13-0) (FFT)** - -The number of computations required in performing the DFT was dramatically reduced by an algorithm developed by Cooley and Tukey in 1965 [5]. This algorithm, known as the *fast Fourier transform* (FFT), reduces the number of computations from something on the order of *N*2 0 to *N*0 log*N*0. To compute one sample *Xr* from Eq. (8.12), we require *N*0 complex multiplications and *N*0 −1 complex additions. To compute *N*0 such values (*Xr* for *r* = 0, 1,...,*N*0 −1), we require a total of *N*2 0 complex multiplications and *N*0(*N*0 − 1) complex additions. For a large *N*0, these computations can be prohibitively time-consuming, even for a high-speed computer. The FFT algorithm is what made the use of Fourier transform accessible for digital signal processing. - -### HOW DOES THE FFT REDUCE THE NUMBER OF COMPUTATIONS? - -It is easy to understand the magic of the FFT. The secret is in the linearity of the Fourier transform and also of the DFT. Because of linearity, we can compute the Fourier transform of a signal *x*(*t*) as a sum of the Fourier transforms of segments of *x*(*t*) of shorter duration. The same principle applies to the computation of the DFT. Consider a signal of length *N*0 = 16 samples. As seen earlier, DFT computation of this sequence requires *N*2 0 = 256 multiplications and *N*0(*N*0 − 1) = 240 additions. We can split this sequence into two shorter sequences, each of length 8. To compute DFT of each of these segments, we need 64 multiplications and 56 additions. Thus, we need a total of 128 multiplications and 112 additions. Suppose, we split the original sequence in four segments of length 4 each. To compute the DFT of each segment, we require 16 multiplications and 12 additions. Hence, we need a total of 64 multiplications and 48 additions. If we split the sequence in eight segments of length 2 each, we need 4 multiplications and 2 additions for each segment, resulting in a total of 32 multiplications and 8 additions. Thus, we have been able to reduce the number of multiplications from 256 to 32 and the number of additions from 240 to 8. Moreover, some of these multiplications turn out to be multiplications by 1 or −1. All this fantastic economy in the number of computations is realized by the FFT without any approximation! The values obtained by the FFT are identical to those obtained by the DFT. In this example, we considered a relatively small value of *N*0 = 16. The reduction in the number of computations is much more dramatic for higher values of *N*0. - -The FFT algorithm is simplified if we choose *N*0 to be a power of 2, although such a choice is not essential. For convenience, we define - -$$ -W_{N_0} = e^{-(j2\pi/N_0)} = e^{-j\Omega_0} -$$ - -so that - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n W_{N_0}^{nr} \qquad 0 \le r \le N_0 - 1 \tag{8.21} -$$ - -and - -$$ -x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r W_{N_0}^{-nr} \qquad 0 \le n \le N_0 - 1 \tag{8.22} -$$ - -Although there are many variations of the Tukey–Cooley algorithm, these can be grouped into two basic types: *decimation in time* and *decimation in frequency*. - -### THE DECIMATION-IN-TIME ALGORITHM - -Here we divide the *N*0-point data sequence *xn* into two (*N*0/2)-point sequences consisting of evenand odd-numbered samples, respectively, as follows: - -$$ -\underbrace{x_0, x_2, x_4, \ldots, x_{N_0-2}}_{\text{sequence } g_n}, \underbrace{x_1, x_3, x_5, \ldots, x_{N_0-1}}_{\text{sequence } h_n} -$$ - -Then, from Eq. (8.21), - -$$ -X_r = \sum_{n=0}^{(N_0/2)-1} x_{2n} W_{N_0}^{2n} + \sum_{n=0}^{(N_0/2)-1} x_{2n+1} W_{N_0}^{(2n+1)r} -$$ - -Also, since - -$$ -W_{N_0/2} = W_{N_0}^2 -$$ - -we have - -$$ -X_r = \sum_{n=0}^{(N_0/2)-1} x_{2n} W_{N_0/2}^{nr} + W_{N_0}^r \sum_{n=0}^{(N_0/2)-1} x_{2n+1} W_{N_0/2}^{nr} -$$ - -= $G_r + W_{N_0}^r H_r$ 0 \le $r \le N_0 - 1$ (8.23) - -where *Gr* and *Hr* are the (*N*0/2)-point DFTs of the even- and odd-numbered sequences, *gn* and *hn*, respectively. Also, *Gr* and *Hr*, being the (*N*0/2)-point DFTs, are (*N*0/2) periodic. Hence, - -$$ -G_{r+(N_0/2)} = G_r \qquad \text{and} \qquad H_{r+(N_0/2)} = H_r \tag{8.24} -$$ - -Moreover, - -$$ -W_{N_0}^{r + (N_0/2)} = W_{N_0}^{N_0/2} W_{N_0}^r = e^{-j\pi} W_{N_0}^r = -W_{N_0}^r -$$ -\n(8.25) - -From Eqs. (8.23), (8.24), and (8.25), we obtain - -$$ -X_{r+(N_0/2)} = G_r - W_{N_0}^r H_r \tag{8.26} -$$ - -This property can be used to reduce the number of computations. We can compute the first *N*0/2 points (0 ≤ *n* ≤ (*N*0/2) − 1) of *Xr* by using Eq. (8.23) and the last *N*0/2 points by using Eq. (8.26) as - -$$ -X_r = G_r + W_{N_0}^r H_r \qquad 0 \le r \le \frac{N_0}{2} - 1 -$$ - -$$ -X_{r + (N_0/2)} = G_r - W_{N_0}^r H_r \qquad 0 \le r \le \frac{N_0}{2} - 1 -$$ - (8.27) - -**Figure 8.24** Butterfly signal flow graph. - -**Figure 8.25** Successive steps in an 8-point FFT. - -Thus, an *N*0-point DFT can be computed by combining the two (*N*0/2)-point DFTs, as in Eq. (8.27). These equations can be represented conveniently by the *signal flow* graph depicted in Fig. 8.24. This structure is known as a *butterfly*. Figure 8.25a shows the implementation of Eq. (8.24) for the case of *N*0 = 8. - -The next step is to compute the (*N*0/2)-point DFTs *Gr* and *Hr*. We repeat the same procedure by dividing *gn* and *hn* into two (*N*0/4)-point sequences corresponding to the even- and odd-numbered samples. Then we continue this process until we reach the one-point DFT. These steps for the case of *N*0 = 8 are shown in Figs. 8.25a, 8.25b, and 8.25c. Figure 8.25c shows that the two-point DFTs require no multiplication. - -To count the number of computations required in the first step, assume that *Gr* and *Hr* are known. Equation (8.27) clearly shows that to compute all the *N*0 points of the *Xr*, we require *N*0 complex additions and *N*0/2 complex multiplications† (corresponding to *Wr N*0 *Hr*). - -In the second step, to compute the (*N*0/2)-point DFT *Gr* from the (*N*0/4)-point DFT, we require *N*0/2 complex additions and *N*0/4 complex multiplications. We require an equal number of computations for *Hr*. Hence, in the second step, there are *N*0 complex additions and *N*0/2 complex multiplications. The number of computations required remains the same in each step. Since a total of log2*N*0 steps is needed to arrive at a one-point DFT, we require, conservatively, a total of *N*0 log2*N*0 complex additions and (*N*0/2)log2*N*0 complex multiplications, to compute the *N*0-point DFT. Actually, as Fig. 8.25c shows, many multiplications are multiplications by 1 or −1, which further reduces the number of computations. - -The procedure for obtaining IDFT is identical to that used to obtain the DFT except that *WN*0 = *ej*(2π/*N*0) instead of *e*−*j*(2π/*N*0) (in addition to the multiplier 1/*N*0). Another FFT algorithm, the *decimation-in-frequency* algorithm, is similar to the decimation-in-time algorithm. The only difference is that instead of dividing *xn* into two sequences of even- and odd-numbered samples, we divide *xn* into two sequences formed by the first *N*0/2 and the last *N*0/2 samples, proceeding in the same way until a single-point DFT is reached in log2*N*0 steps. The total number of computations in this algorithm is the same as that in the decimation-in-time algorithm. - -## **[8.7 MATLAB: THE](#page-13-0) DISCRETE FOURIER TRANSFORM** - -As an idea, the discrete Fourier transform (DFT) has been known for hundreds of years. Practical computing devices, however, are responsible for bringing the DFT into common use. MATLAB is capable of DFT computations that would have been impractical just a few decades ago. - -### **[8.7-1 Computing the Discrete Fourier Transform](#page-13-0)** - -The MATLAB command fft(x) computes the DFT of a vector x that is defined over (0 ≤ *n* ≤ *N*0 −1) (Problem 8.7-1 considers how to scale the DFT to accommodate signals that do not begin at *n* = 0.) As its name suggests, the function fft uses the computationally more efficient fast Fourier transform algorithm when it is appropriate to do so. The inverse DFT is easily computed by using the ifft function. - -† Actually, *N*0/2 is a conservative figure because some multiplications corresponding to the cases of *Wr N*0 = 1,*j*, and so on, are eliminated. - -To illustrate MATLAB's DFT capabilities, consider 50 points of a 10 Hz sinusoid sampled at *fs* = 50 Hz and scaled by *T* = 1/*fs*. - -``` ->> T = 1/50; N_0 = 50; n = (0:N_0-1); ->> x = T*cos(2*pi*10*n*T); -``` - -In this case, the vector x contains exactly 10 cycles of the sinusoid. The fft command computes the DFT. - ->> X = fft(x); - -Since the DFT is both discrete and periodic, fft needs to return only the *N*0 discrete values contained in the single period (0 ≤ *f* < *fs*). - -While *Xr* can be plotted as a function of *r*, it is more convenient to plot the DFT as a function of frequency *f* . A frequency vector, in hertz, is created by using *N*0 and *T*. - ->> f = (0:N\_0-1)/(T\*N\_0); stem(f,abs(X),'k.'); >> axis([0 50 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); - -As expected, Fig. 8.26 shows content at a frequency of 10 Hz. Since the time-domain signal is real, *X*(*f*) is conjugate symmetric. Thus, content at 10 Hz implies equal content at −10 Hz. The content visible at 40 Hz is an alias of the −10 Hz content. - -Often, it is preferred to plot a DFT over the principal frequency range (−*fs*/2 ≤ *f* < *fs*/2). The MATLAB function fftshift properly rearranges the output of fft to accomplish this task. - -``` ->> stem(f-1/(T*2),fftshift(abs(X)),'k.'); ->> axis([-25 25 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); -``` - -When we use fftshift, the conjugate symmetry that accompanies the DFT of a real signal becomes apparent, as shown in Fig. 8.27. - -Since DFTs are generally complex-valued, the magnitude plots of Figs. 8.26 and 8.27 offer only half the picture; the signal's phase spectrum, shown in Fig. 8.28, completes it. - -``` ->> stem(f-1/(T*2),fftshift(angle(X)),'k.'); ->> axis([-25 25 -1.1*pi 1.1*pi]); xlabel('f [Hz]'); ylabel('\angle X(f)'); -``` - -**Figure 8.26** |*X*(*f*)| computed over (0 ≤ *f* < 50) by using fft. - -**Figure 8.27** |*X*(*f*)| displayed over (−25 ≤ *f* < 25) by using fftshift. - -**Figure 8.28** *X*(*f*) displayed over (−25≤*f* <25). - -Since the signal is real, the phase spectrum necessarily has odd symmetry. Additionally, the phase at ±10 Hz is zero, as expected for a zero-phase cosine function. More interesting, however, are the phase values found at the remaining frequencies. Does a simple cosine really have such complicated phase characteristics? The answer, of course, is no. The magnitude plot of Fig. 8.27 helps identify the problem: there is zero content at frequencies other than ±10 Hz. Phase computations are not reliable at points where the magnitude response is zero. One way to remedy this problem is to assign a phase of zero when the magnitude response is near or at zero. - -### **[8.7-2 Improving the Picture with Zero Padding](#page-13-0)** - -DFT magnitude and phase plots paint a picture of a signal's spectrum. At times, however, the picture can be somewhat misleading. Given a sampling frequency *fs* = 50 Hz and a sampling interval *T* = 1/*fs*, consider the signal - -$$ -y[n] = Te^{j2\pi \left(10\frac{1}{3}\right)n} -$$ - -This complex-valued, periodic signal contains a single positive frequency at 101 3 Hz. Let us compute the signal's DFT using 50 samples. - -``` ->> y = T*exp(j*2*pi*(10+1/3)*n*T); Y = fft(y); ->> stem(f-25,fftshift(abs(Y)),'k.'); ->> axis([-25 25 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|Y(f)|'); -``` - -**Figure 8.29** |*Y*(*f*)| using 50 data points. - -**Figure 8.30** |*Yzp*(*f*)| over 5 ≤ *f* ≤ 15 using 50 data points padded with 550 zeros. - -In this case, the vector y contains a noninteger number of cycles. Figure 8.29 shows the significant frequency leakage that results. Also notice that since *y*[*n*] is not real, the DFT is not conjugate symmetric. - -In this example, the discrete DFT frequencies do not include the actual 101 3 Hz frequency of the signal. Thus, it is difficult to determine the signal's frequency from Fig. 8.29. To improve the picture, the signal is zero-padded to 12 times its original length. - -``` ->> y_zp = [y,zeros(1,11*length(y))]; Y_zp = fft(y_zp); ->> f_zp = (0:12*N_0-1)/(T*12*N_0); ->> stem(f_zp-25,fftshift(abs(Y_zp)),'k.'); ->> axis([-25 25 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|Y_{zp}(f)|'); -``` - -Figure 8.30, zoomed in to 5 ≤ *f* ≤ 15, correctly shows the peak frequency at 101 3 Hz and better represents the signal's spectrum. - -It is important to keep in mind that zero padding does not increase the resolution or accuracy of the DFT. To return to the picket fence analogy, zero padding increases the number of pickets in our fence but cannot change what is behind the fence. More formally, the characteristics of the sinc function, such as main beam width and sidelobe levels, depend on the fixed width of the pulse, not on the number of zeros that follow. Adding zeros cannot change the characteristics of the sinc function and thus cannot change the resolution or accuracy of the DFT. Adding zeros simply allows the sinc function to be sampled more finely. - -### **[8.7-3 Quantization](#page-13-0)** - -A *B*-bit analog-to-digital converter (ADC) samples an analog signal and quantizes amplitudes by using 2*B* discrete levels. This quantization results in signal distortion that is particularly noticeable for small *B*. Typically, quantization is classified as symmetric or asymmetric and as either rounding or truncating. Let us investigate rounding-type quantizers. - -The quantized output *x*q of an asymmetric rounding converter is given as† - -$$ -x_{\rm q} = \frac{x_{\rm max}}{2^{B-1}}\lfloor \frac{x}{x_{\rm max}}2^{B-1} + \frac{1}{2}\rfloor -$$ - -The quantized output *x*q of a symmetric rounding converter is given as - -$$ -x_{\rm q} = \frac{x_{\rm max}}{2^{B-1}}\left(\lfloor \frac{x}{x_{\rm max}}2^{B-1}\rfloor + \frac{1}{2}\right) -$$ - -Program CH8MP1 quantizes a signal using one of these two rounding quantizer rules and also ensures no more than 2*B* output levels. - -``` -function [xq] = CH8MP1(x,xmax,B,method) -% CH8MP1.m : Chapter 8, MATLAB Program 1 -% Function M-file quantizes x over (-xmax,xmax) using 2^b levels. -% Uses rounding rule, supports symmetric and asymmetric quantization -% INPUTS: x = input signal -% xmax = maximum magnitude of signal to be quantized -% B = number of quantization bits -% method = default 'sym' for symmetrical, 'asym' for asymmetrical -% OUTPUTS: xq = quantized signal -if (nargin<3), - disp('Insufficient number of inputs.'); return -elseif (nargin==3), - method = 'sym'; -elseif (nargin>4), - disp('Too many inputs.'); return -end -x(abs(x)>xmax)=xmax*sign(x(abs(x)>xmax)); % Limit amplitude to xmax -switch lower(method) - case 'asym' - xq = xmax/(2^(B-1))*floor(x*2^(B-1)/xmax+1/2); - xq(xq>=xmax)=xmax*(1-2^(1-B)); % Ensure only 2^B levels - case 'sym' - xq = xmax/(2^(B-1))*(floor(x*2^(B-1)/xmax)+1/2); - xq(xq>=xmax)=xmax*(1-2^(1-B)/2); % Ensure only 2^B levels -``` - -† Large values of *x* may return quantized values *x*q outside the 2*B* allowable levels. In such cases, *x*q should be clamped to the nearest permitted level. - -``` -otherwise - disp('Unrecognized quantization method.'); return -end -``` - -Several MATLAB commands require discussion. First, the nargin function returns the number of input arguments. In this program, nargin is used to ensure that a correct number of inputs is supplied. If the number of inputs supplied is incorrect, an error message is displayed and the function terminates. If only three input arguments are detected, the quantization type is not explicitly specified and the program assigns the default symmetric method. - -As with many high-level languages such as C, MATLAB supports general switch/case structures† : - -``` -switch switch_expr, -case case_expr, - statements; -... -otherwise, - statements; -``` - -end - -CH8MP1 switches among cases of the string method. In this way, method-specific parameters are easily set. The command lower is used to convert a string to all lowercase characters. In this way, strings such as SYM, Sym, and sym are all indistinguishable. Similar to lower, the MATLAB command upper converts a string to all uppercase. - -The floor command rounds input values to the nearest integer toward minus infinity. Mathematically, it computes ·. To accommodate different types of rounding, MATLAB supplies three other rounding commands: ceil, round, and fix. The ceil command rounds input values to the nearest integers toward infinity, ( ·"); the round command rounds input values toward the nearest integer; the fix command rounds input values to the nearest integer toward zero. For example, if x = [-0.5 0.5];, floor(x) yields [-1 0], ceil(x) yields [0 1], round(x) yields [-1 1], and fix(x) yields [0 0]. Finally, CH8MP1 checks and, if necessary, corrects large values of *x*q that may be outside the allowable 2*B* levels. - -To verify operation, CH8MP1 is used to determine the transfer characteristics of a symmetric 3-bit quantizer operating over (−10,10). - -``` ->> x = (-10:.0001:10); xsq = CH8MP1(x,10,3,'sym'); -``` - -``` ->> plot(x,xsq,'k'); axis([-10 10 -10.5 10.5]); grid on; -``` - -``` ->> xlabel('Quantizer input'); ylabel('Quantizer output'); -``` - -Figure 8.31 shows the results. Clearly, the quantized output is limited to 2*B* = 8 levels. Zero is not a quantization level for symmetric quantizers, so half of the levels occur above zero and half of the levels occur below zero. In fact, *symmetric quantizers* get their name from the symmetry in quantization levels above and below zero. - -By changing the method in CH8MP1 from 'sym' to 'asym', we obtain the transfer characteristics of an asymmetric 3-bit quantizer, as shown in Fig. 8.32. Again, the quantized output is limited to 2*B* = 8 levels, and zero is now one of the included levels. With zero as a quantization - -† A functionally equivalent structure can be written by using if, elseif, and else statements. - -**Figure 8.31** Transfer characteristics of a symmetric 3-bit quantizer. - -**Figure 8.32** Transfer characteristics of an asymmetric 3-bit quantizer. - -level, we need one fewer quantization level above zero than there are levels below. Not surprisingly, *asymmetric quantizers* get their name from the asymmetry in quantization levels above and below zero. - -There is no doubt that quantization can change a signal. It follows that the spectrum of a quantized signal can also change. While these changes are difficult to characterize mathematically, they are easy to investigate by using MATLAB. Consider a 1 Hz cosine sampled at *fs* = 50 Hz over 1 second. - ->> x = cos(2\*pi\*n\*T); X = fft(x); T = 1/50; N\_0 = 50; n = (0:N\_0-1); - -Upon quantizing by means of a 2-bit asymmetric rounding quantizer, both the signal and spectrum are substantially changed. - -``` ->> xaq = CH8MP1(x,1,2,'asym'); Xaq = fft(xaq); ->> subplot(2,2,1); stem(n,x,'k'); axis([0 49 -1.1 1.1]); ->> xlabel('n');ylabel('x[n]'); ->> subplot(2,2,2); stem(f-25,fftshift(abs(X)),'k'); axis([-25,25 -1 26]) ->> xlabel('f');ylabel('|X(f)|'); ->> subplot(2,2,3); stem(n,xaq,'k');axis([0 49 -1.1 1.1]); -``` - -**Figure 8.33** Signal and spectrum effects of quantization. - -``` ->> xlabel('n');ylabel('x_{aq}[n]'); -``` - -``` ->> subplot(2,2,4); stem(f-25,fftshift(abs(fft(xaq))),'k'); axis([-25,25 -1 26]); -``` - -``` ->> xlabel('f');ylabel('|X_{aq}(f)|'); -``` - -The results are shown in Fig. 8.33. The original signal *x*[*n*] appears sinusoidal and has pure spectral content at ±1 Hz. The asymmetrically quantized signal *xaq*[*n*] is significantly distorted. The corresponding magnitude spectrum |*Xaq*(*f*)| is spread over a broad range of frequencies. - -## **[8.8 SUMMARY](#page-13-0)** - -A signal bandlimited to *B* Hz can be reconstructed exactly from its samples if the sampling rate *fs* > 2*B* Hz (the sampling theorem). Such a reconstruction, although possible theoretically, poses practical problems such as the need for ideal filters, which are unrealizable or are realizable only with infinite delay. Therefore, in practice, there is always an error in reconstructing a signal from its samples. Moreover, practical signals are not bandlimited, which causes an additional error (aliasing error) in signal reconstruction from its samples. When a signal is sampled at a frequency *fs* Hz, samples of a sinusoid of frequency (*fs*/2) + *x* Hz appear as samples of a lower frequency (*fs*/2) − *x* Hz. This phenomenon, in which higher frequencies appear as lower frequencies, is known as aliasing. Aliasing error can be reduced by bandlimiting a signal to *fs*/2 Hz (half the sampling frequency). Such bandlimiting, done prior to sampling, is accomplished by an anti-aliasing filter that is an ideal lowpass filter of cutoff frequency *fs*/2 Hz. - -The sampling theorem is very important in signal analysis, processing, and transmission because it allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing a continuous-time signal is therefore equivalent to processing a discrete sequence of numbers. This leads us directly into the area of digital filtering (discrete-time systems). In the field of communication, the transmission of a continuous-time message reduces to the transmission of a sequence of numbers. This opens doors to many new techniques of communicating continuous-time signals by pulse trains. - -The dual of the sampling theorem states that for a signal timelimited to τ seconds, its spectrum *X*(ω) can be reconstructed from the samples of *X*(ω) taken at uniform intervals not greater than 1/τ Hz. In other words, the spectrum should be sampled at a rate not less than τ samples/Hz. - -To compute the direct or the inverse Fourier transform numerically, we need a relationship between the samples of *x*(*t*) and *X*(ω). The sampling theorem and its dual provide such a quantitative relationship in the form of a discrete Fourier transform (DFT). The DFT computations are greatly facilitated by a fast Fourier transform (FFT) algorithm, which reduces the number of computations from something on the order of *N*2 0 to *N*0 log*N*0. - -### **[REFERENCES](#page-13-0)** - -- 1. Linden, D. A. A discussion of sampling theorem. *Proceedings of the IRE,* vol. 47, pp. 1219–1226, July 1959. -- 2. Siebert, W. M. *Circuits, Signals, and Systems*. MIT/McGraw-Hill, New York, 1986. -- 3. Bennett, W. R. *Introduction to Signal Transmission*. McGraw-Hill, New York, 1970. -- 4. Lathi, B. P. *Linear Systems and Signals*. Berkeley-Cambridge Press, Carmichael, CA, 1992. -- 5. Cooley, J. W., and Tukey, J. W. An algorithm for the machine calculation of complex Fourier series. *Mathematics of Computation*, vol. 19, pp. 297–301, April 1965. - -## **[PROBLEMS](#page-13-0)** - -[*Note:* In many problems, the plots of spectra are shown as functions of frequency *f* Hz for convenience, although we have labeled them as functions of ω as *X*(ω), *Y*(ω), etc.] - -- **8.1-1** If *f*s is the Nyquist rate for signal *x*(*t*), determine the Nyquist rate for each of the following signals: - - (a) *y*a(*t*) = *d dt x*(*t*) - - (b) *y*b(*t*) = *x*(*t*) cos(2π*f*0*t*) - - (c) *y*c(*t*) = *x*(*t*+*a*)+*x*(*t*−*b*), for real constants *a* and *b* - - (d) *y*d(*t*) = *x*(*at*), for real *a* > 0 - -- **8.1-2** Figure P8.1-2 shows Fourier spectra of signals *x*1(*t*) and *x*2(*t*). Determine the Nyquist sampling rates for signals *x*1(*t*), *x*2(*t*), *x*2 1(*t*), *x*3 2(*t*), and *x*1(*t*)*x*2(*t*). -- **8.1-3** A signal *x*(*t*) has a bandwidth of *B* = 1000 Hz. For a positive integer *N*, what is the Nyquist rate for the signal *y*(*t*) = *xN*(*t*)? -- **8.1-4** Determine the Nyquist sampling rate and the Nyquist sampling interval for the signals: (a) sinc2(100π*t*) - - (b) 0.01 sinc2(100π*t*) - -**Figure P8.1-2** - -### 836 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE - -- (c) sinc(100π*t*)+3sinc2(60π*t*) -- (d) sinc(50π*t*)sinc(100π*t*) -- **8.1-5** (a) Sketch |*X*(ω)|, the amplitude spectrum of a signal *x*(*t*) = 3 cos 6π*t* + sin 18π*t* + 2 cos(28 − )π*t*, where is a very small number → 0. Determine the minimum sampling rate required to be able to reconstruct *x*(*t*) from these samples. - - (b) Sketch the amplitude spectrum of the sampled signal when the sampling rate is 25% above the Nyquist rate (show the spectrum over the frequency range ±50 Hz only). How would you reconstruct *x*(*t*) from these samples? -- **8.1-6** (a) Derive the sampling theorem by considering the fact that the sampled signal *x*(*t*) = *x*(*t*)δ*T* (*t*), and using the frequency-convolutionpropertyinEq.(7.34). - - (b) For a sampling train consisting of shifted unit impulses at instants *nT* + τ instead of at *nT* (for all positive and negative integer values of *n*), find the spectrum of the sampled signal. -- **8.1-7** A signal is bandlimited to 12 kHz. The band between 10 and 12 kHz has been so corrupted by excessive noise that the information in this band is nonrecoverable. Determine the minimum sampling rate for this signal so that the uncorrupted portion of the band can be recovered. If we were to filter out the corrupted - -spectrum prior to sampling, what would be the minimum sampling rate? - -- **8.1-8** A continuous-time signal *x*(*t*) = ((*t* − 1)/2) is sampled at three rates: 10, 2, and 1 Hz. Sketch the resulting sampled signals. Because *x*(*t*) is timelimited, its bandwidth is infinite. However, most of its energy is concentrated in a small band. Determine a reasonable minimum sampling rate that will allow reconstruction of this signal with a small error. The answer is not unique. Make a reasonable assumption of what you define as a "negligible" or "small" error. -- **8.1-9** (a) A signal *x*(*t*) = 5 sinc2 (5π*t*) + cos 20π*t* is sampled at a rate of 10 Hz. Find the spectrum of the sampled signal. Can *x*(*t*) be reconstructed by lowpass filtering the sampled signal? - - (b) Repeat part (a) for a sampling frequency of 20 Hz. Can you reconstruct the signal from this sampled signal? Explain. - - (c) If *x*(*t*) = 5 sinc2 (5π*t*) + sin 20π*t*, can you reconstruct *x*(*t*) from the samples of *x*(*t*) at a rate of 20 Hz? Explain your answer with spectral representation(s). - - (d) For *x*(*t*) = 5 sinc2 (5π*t*) + sin 20π*t*, can you reconstruct *x*(*t*) from the samples of *x*(*t*) at a rate of 21 Hz? Explain your answer with spectral representation(s). Comment on your results. -- **8.1-10** (a) The highest frequency in the spectrum *X*(ω) (Fig. P8.1-10a) of a bandpass signal *x*(*t*) - -**Figure P8.1-10** - -is 30 Hz. Hence, the minimum sampling frequency needed to sample *x*(*t*) is 60 Hz. Show the spectrum of the signal sampled at a rate of 60 Hz. Can you reconstruct *x*(*t*) from these samples? How? - -- (b) A certain busy student looks at *X*(ω), concludes that its bandwidth is really 10 Hz, and decides that the sampling rate 20 Hz is adequate for sampling *x*(*t*). Sketch the spectrum of the signal sampled at a rate of 20 Hz. Can *x*(*t*) be reconstructed from these samples? -- (c) The same student, using the same reasoning, looks at *Y*(ω) in Fig. P8.1-10b, the spectrum of another bandpass signal *y*(*t*), and concludes that the sampling rate of 20 Hz can be used to sample *y*(*t*). Sketch the spectrum of the signal *y*(*t*) sampled at a rate of 20 Hz. Can *y*(*t*) be reconstructed from these samples? -- **8.1-11** A signal *x*(*t*) whose spectrum *X*(ω), as shown in Fig. P8.1-11, is sampled at a frequency *fs* = *f*1 + *f*2 Hz. Find all the sample values of *x*(*t*) merely by inspection of *X*(ω). -- **8.1-12** As described in Sec. 8.1-1, practical sampling can be achieved by multiplying a signal *x*(*t*) by a periodic train of pulses *pT* (*t*). The pulse train *pT* (*t*) can be created by the periodic replication of some pulse *p*(*t*) as - -$$ -p_T(t) = \sum_{k=-\infty}^{\infty} p(t - kT) -$$ - -It is desired to sample a signal at a rate *f*s = 100 Hz, and two pulses are under consideration: - -$$ -p_a(t) = -\frac{1}{4}u(t) + \frac{5}{4}u(t - \frac{2T}{20}) + -$$ - -$$ --\frac{5}{4}u(t - \frac{3T}{20}) + \frac{1}{4}u(t - \frac{5T}{20}) -$$ - -and - -$$ -p_{\rm b}(t) = e^{-t/T} \left[ u(t) - u(t - 1.5T) \right] -$$ - -- (a) Plot *pT* (*t*) using *p*a(*t*) over 0 ≤ *t* ≤ 4*T*. -- (b) Plot *pT* (*t*) using *p*b(*t*) over 0 ≤ *t* ≤ 4*T*. -- (c) Which pulse, *p*a(*t*) or *p*b(*t*), is more suitable as a sampling pulse? Carefully explain your answer. -- **8.1-13** In digital data transmission over a communication channel, it is important to know the upper theoretical limit on the rate of digital pulses that can be transmitted over a channel of bandwidth *B* Hz. In digital transmission, the relative shape of the pulse is not important. We are interested in knowing only the amplitude represented by the pulse. For instance, in binary communication, we are interested in knowing whether the received pulse amplitude is 1 or −1 (positive or negative). Thus, each pulse represents one piece of information. Consider one independent amplitude value (not necessarily binary) as one piece of information. Show that 2*B* independent pieces of information per second can be transmitted correctly (assuming no noise) over a channel of bandwidth *B* Hz. This important principle in communication theory states that 1 Hz of bandwidth can transmit two independent pieces of information per second. It represents the upper rate of pulse transmission over a channel without any error in reception in the absence of noise. [*Hint:* According to the interpolation formula [Eq. (8.6)], a continuous-time signal of bandwidth *B* Hz can be constructed from 2*B* pieces of information/second.] -- **8.1-14** This example is one of those interesting situations leading to a curious result in the category of defying gravity. The sinc function can be recovered from its samples taken at extremely low frequencies in apparent defiance of the sampling theorem. - -Consider a sinc pulse *x*(*t*) = sinc(4π*t*) for which *X*(ω) = (1/4)rect(ω/8π ). The bandwidth of *x*(*t*) is *B* = 2 Hz, and its Nyquist rate is 4 Hz. - -**Figure P8.1-11** - -- (a) Sample *x*(*t*) at a rate 4 Hz and sketch the spectrum of the sampled signal. -- (b) To recover *x*(*t*) from its samples, we pass the sampled signal through an ideal lowpass filter of bandwidth *B* = 2 Hz and gain *G* = *T* = 1/4. Sketch this system and show that for this system *H*(ω)=(1/4)rect(ω/8π ). Show also that when the input is the sampled *x*(*t*) at a rate 4 Hz, the output of this system is indeed *x*(*t*), as expected. -- (c) Now sample *x*(*t*) at half the Nyquist rate, at 2 Hz. Apply this sampled signal at the input of the lowpass filter used in part (b). Find the output. -- (d) Repeat part (c) for the sampling rate 1 Hz. -- (e) Show that the output of the lowpass filter in part (b) is *x*(*t*) to the sampled *x*(*t*) if the sampling rate is 4/*N*, where *N* is any positive integer. This means that we can recover *x*(*t*) from its samples taken at an arbitrarily small rate by letting *N* → ∞. -- (f) The mystery may be clarified a bit by examining the problem in the time domain. Find the samples of *x*(*t*) when the sampling rate is 2/*N* (*N* integer). -- **8.2-1** A signal *x*(*t*) = sinc (200π*t*) is sampled (multiplied) by a periodic pulse train *pT* (*t*) represented in Fig. P8.2-1. Find and sketch the spectrum of the sampled signal. Explain whether you will be able to reconstruct *x*(*t*) from these samples. Find the filter output if the sampled signal is passed through an ideal lowpass filter of bandwidth 100 Hz and unit gain. What is the filter output if its bandwidth *B* Hz is between 100 and 150 Hz? What happens if the bandwidth exceeds 150 Hz? -- **8.2-2** Show that the circuit in Fig. P8.2-2 is a realization of the causal ZOH (zero-order hold) circuit. You can do this by showing that the unit impulse response *h*(*t*) of this circuit is indeed equal to that in Eq. (8.5) delayed by *T*/2 seconds to make it causal. - -### **Figure P8.2-2** - -- **8.2-3** (a) A first-order hold circuit (FOH) can also be used to reconstruct a signal *x*(*t*) from its samples. The impulse response of this circuit is *h*(*t*) = (*t*/2*T*), where *T* is the sampling interval. Consider a typical sampled signal *x*(*t*) and show that this circuit performs the linear interpolation. In other words, the filter output consists of sample tops connected by straight-line segments. Follow the procedure discussed in Sec. 8.2 (Fig. 8.5c). - - (b) Determine the frequency and magnitude responses of this filter, and compare it with **(i)** the ideal filter required for signal reconstruction and **(ii)** a ZOH circuit. - - (c) This filter, being noncausal, is unrealizable. By delaying its impulse response, the filter can be made realizable. What is the minimum delay required to make it realizable? How would this delay affect the reconstructed signal and the filter frequency response? - - (d) Show that the causal FOH circuit in part (c) can be realized by the ZOH circuit depicted in Fig. P8.2-2 followed by an identical filter in cascade. -- **8.2-4** Suppose signal *x*(*t*)=sin(2π*t*/8)(*u*(*t*)−*u*(*t* −8)) is sampled at a rate *fs* = 1 Hz to generate signal *x*[*n*]. - - (a) Sketch *x*(*t*) and *x*[*n*]. - - (b) Has aliasing occurred in sampling *x*(*t*) to produce *x*[*n*]? Explain. - - (c) Sketch the output *x*ˆ(*t*) produced when *x*[*n*] is applied to the causal ZOH reconstructor of Prob. 8.2-2. How does *x*ˆ(*t*) compare with *x*(*t*)? - -**Figure P8.2-1** - -- (d) Sketch the output *x*ˆ(*t*) produced when *x*[*n*] is applied to the FOH reconstructor of Prob.8.2-3.Howdoes*x*ˆ(*t*)comparewith*x*(*t*)? -- **8.2-5** Repeat Prob. 8.2-4 for the signal *x*(*t*) = cos(2π*t*/8)(*u*(*t*)−*u*(*t* −8)). -- **8.2-6** Is it possible to sample a physically realizable (nonzero) signal *x*(*t*) with a physically realizable system without aliasing? If possible, explain what conditions must be met. If not possible, explain why not. -- **8.2-7** In the text, for sampling purposes, we used timelimited narrow pulses such as impulses or rectangular pulses of width less than the sampling interval *T*. Show that it is not necessary to restrict the sampling pulse width. We can use sampling pulses of arbitrarily large duration and still be able to reconstruct the signal *x*(*t*) as long as the pulse rate is no less than the Nyquist rate for *x*(*t*). - -Consider *x*(*t*) to be bandlimited to *B* Hz. The sampling pulse to be used is an exponential *e*−*atu*(*t*). We multiply *x*(*t*) by a periodic train of exponential pulses of the form *e*−*atu*(*t*) spaced *T* seconds apart. Find the spectrum of the sampled signal, and show that *x*(*t*) can be reconstructed from this sampled signal provided the sampling rate is no less than 2*B* Hz or *T* < 1/2*B*. Explain how you would reconstruct *x*(*t*) from the sampled signal. - -- **8.2-8** In Ex. 8.2, the sampling of a signal *x*(*t*) was accomplished by multiplying the signal by a pulse train *pT* (*t*), resulting in the sampled signal depicted in Fig. 8.4d. This procedure is known as the *natural sampling*. Figure P8.2-8 shows the so-called *flat-top sampling* of the same signal *x*(*t*) = sinc2 (5π*t*). - - (a) Show that the signal *x*(*t*) can be recovered from flat-top samples if the sampling rate is no less than the Nyquist rate. - - (b) Explain how you would recover *x*(*t*) from the flat-top samples. - - (c) Find the expression for the sampled signal spectrum *X*(ω) and sketch it roughly. -- **8.2-9** A sinusoid of frequency *f*0 Hz is sampled at a rate *fs* = 20 Hz. Find the apparent frequency of the sampled signal if *f*0 is: - - (a) 8 Hz - - (b) 12 Hz - -(c) 20 Hz - -- (d) 22 Hz -- (e) 32 Hz -- **8.2-10** A sinusoid of unknown frequency *f*0 is sampled at a rate 60 Hz. The apparent frequency of the samples is 20 Hz. Determine *f*0 if it is known that *f*0 lies in the range: - - (a) 0–30 Hz - - (b) 30–60 Hz - - (c) 60–90 Hz - - (d) 90–120 Hz -- **8.2-11** A signal *x*(*t*) = 3 cos 6π*t*+cos 16π*t*+2 cos 20π*t* is sampled at a rate 25% above the Nyquist rate. Sketch the spectrum of the sampled signal. How would you reconstruct *x*(*t*) from these samples? If the sampling frequency is 25% below the Nyquist rate, what are the frequencies of the sinusoids present in the output of the filter with cutoff frequency equal to the folding frequency? Do not write the actual output; give just the frequencies of the sinusoids present in the output. -- **8.2-12** A complex signal *x*(*t*) has a spectrum given as - -$$ -X(\omega) = \begin{cases} \omega & 0 \le \omega \le 2\pi 10 \\ 0 & \text{otherwise} \end{cases} -$$ - -Let *x*(*t*) be sampled at rate *f*s = 24 Hz to produce signal *x*(*t*) with spectrum *X*(ω). - -- (a) Sketch *X*(ω). -- (b) Has aliasing occurred in sampling *x*(*t*) to produce *x*(*t*)? Explain. -- (c) Can *x*(*t*) be exactly recovered from *x*(*t*)? Explain. -- **8.2-13** Repeat Prob. 8.2-12 for the sampling rate *f*s = 16 Hz. - -- **8.2-14** Repeat Prob. 8.2-12 for the sampling rate *f*s = 8 Hz. -- **8.2-15** (a) Show that the signal *x*(*t*), reconstructed from its samples *x*(*nT*), using Eq. (8.6) has a bandwidth *B* ≤ 1/2*T* Hz. - - (b) Show that *x*(*t*) is the smallest bandwidth signal that passes through samples *x*(*nT*). [*Hint:* Use the *reductio ad absurdum* method.] -- **8.2-16** In digital communication systems, the efficient use of channel bandwidth is ensured by transmitting digital data encoded by means of bandlimited pulses. Unfortunately, bandlimited pulses are non-timelimited; that is, they have infinite duration, which causes pulses representing successive digits to interfere and cause errors in the reading of true pulse values. This difficulty can be resolved by shaping a pulse *p*(*t*) in such a way that it is bandlimited, yet causes zero interference at the sampling instants. To transmit *R* pulses per second, we require a minimum bandwidth *R*/2 Hz (see Prob. 8.1-13). The bandwidth of *p*(*t*) should be *R*/2 Hz, and its samples, in order to cause no interference at all other sampling instants, must satisfy the condition - -$$ -p(nT) = \begin{cases} 1 & n = 0 \quad T = \frac{1}{R} \\ 0 & n \neq 0 \end{cases} -$$ - -Because the pulse rate is *R* pulses per second, the sampling instants are located at intervals of 1/*R* seconds. Hence, the foregoing condition ensures that any given pulse will not interfere with the amplitude of any other pulse at its center. Find *p*(*t*). Is *p*(*t*) unique in the sense that no other pulse satisfies the given requirements? - -**8.2-17** The problem of pulse interference in digital data transmission was outlined in Prob. 8.2-16, where we found a pulse shape *p*(*t*) to eliminate the interference. Unfortunately, the pulse found is not only noncausal, (and unrealizable) but also has a serious drawback: because of its slow decay (as 1/*t*), it is prone to severe interference due to small parameter deviation. To make the pulse decay rapidly, Nyquist proposed relaxing the bandwidth requirement from *R*/2 Hz to *kR*/2 Hz with 1 ≤ *k* ≤ 2. The pulse must still have a property of noninterference with other pulses, for example, - -$$ -p(nT) = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases} \qquad T = \frac{1}{R} -$$ - -Show that this condition is satisfied only if the pulse spectrum *P*(ω) has an odd symmetry about the set of dotted axes, as shown in Fig. P8.2-17. The bandwidth of *P*(ω) is *kR*/2 Hz (1 ≤ *k* ≤ 2). - -**8.2-18** The Nyquist samples of a signal *x*(*t*) bandlimited to *B* Hz are - -$$ -x(nT) = \begin{cases} 1 & n = 0, 1 \\ 0 & \text{all } n \neq 0, 1 \end{cases} \qquad T = \frac{1}{2B} -$$ - -Show that - -$$ -x(t) = \frac{\operatorname{sinc}(2\pi Bt)}{1 - 2Bt} -$$ - -This pulse, known as the *duobinary pulse,* is used in digital transmission applications. - -**Figure P8.2-17** - -**8.2-19** A signal bandlimited to *B* Hz is sampled at a rate *fs* = 2*B* Hz. Show that - -$$ -\int_{-\infty}^{\infty} x(t) dt = T \sum_{-\infty}^{\infty} x(nT) -$$ -$$ -\int_{-\infty}^{\infty} |x(t)|^2 dt = T \sum_{-\infty}^{\infty} |x(nT)|^2 -$$ - -[*Hint:* Use the orthogonality property of the sinc function in Prob. 7.6-6.] - -- **8.2-20** Prove that a signal cannot be simultaneously timelimited and bandlimited. [*Hint:* Show that a contrary assumption leads to contradiction. Assume a signal to be simultaneously timelimited and bandlimited so that *X*(ω) = 0 for |ω| ≥ 2π*B*. In this case, *X*(ω) = *X*(ω)rect(ω/4π*B* ) for *B* > *B*. This fact means that *x*(*t*) is equal to *x*(*t*) ∗ 2*B* sinc (2π*B t*). The latter cannot be timelimited because the sinc function tail extends to infinity.] -- **8.3-1** Physically implementable digital systems, such as smartphones and computers, require that signals be both time-sampled and amplitude-quantized. - - (a) Why is time sampling necessary? When does time sampling result in unrecoverable changes to the signal? - - (b) Why is amplitude-quantization necessary? When does amplitude quantization result in unrecoverable changes to the signal? -- **8.3-2** Typical analog-to-digital converters (ADCs) operate over a range of input amplitudes [−*V*ref,*V*ref]. Why is it desirable to condition the input *x*(*t*) to an ADC so that its maximum magnitude is close to, but does not exceed, *V*ref? What happens if the maximum magnitude of *x*(*t*) is greater than *V*ref? What happens if the maximum magnitude of *x*(*t*) is much smaller than *V*ref? -- **8.3-3** A compact disc (CD) records audio signals digitally by means of a binary code. Assume an audio signal bandwidth of 15 kHz. - - (a) What is the Nyquist rate? - - (b) If the Nyquist samples are quantized into 65,536 levels (*L* = 65,536) and then binary-coded, what number of binary digits is required to encode a sample? - -- (c) Determine the number of binary digits per second (bits/s) required to encode the audio signal. -- (d) For practical reasons discussed in the text, signals are sampled at a rate well above the Nyquist rate. Practical CDs use 44,100 samples/s. If *L* = 65,536, determine the number of pulses per second required to encode the signal. -- **8.3-4** A TV signal (video and audio) has a bandwidth of 4.5 MHz. This signal is sampled, quantized, and binary-coded. - - (a) Determine the sampling rate if the signal is to be sampled at a rate 20% above the Nyquist rate. - - (b) If the samples are quantized into 1024 levels, what number of binary pulses is required to encode each sample? - - (c) Determine the binary pulse rate (bits/s) of the binary-coded signal. -- **8.3-5** (a) In a certain A/D scheme, there are 16 quantization levels. Give one possible binary code and one possible quaternary (4-ary) code. For the quaternary code, use **0, 1, 2,** and **3** as the four symbols. Use the minimum number of digits in your code. - - (b) To represent a given number of quantization levels *L*, we require a minimum of *bM* digits for an *M*-ary code. Show that the ratio of the number of digits in a binary code to the number of digits in a quaternary (4-ary) code is 2, that is, *b*2/*b*4 = 2. -- **8.3-6** Five telemetry signals, each of bandwidth 1 kHz, are quantized and binary-coded. These signals are time-division multiplexed (signal bits interleaved). Choose the number of quantization levels so that the maximum error in sample amplitudes is no greater than 0.2% of the peak signal amplitude. The signals must be sampled at least 20% above the Nyquist rate. Determine the data rate (bits per second) of the multiplexed signal. -- **8.4-1** A triangle function *x*(*t*) = (*t*/5) has spectrum *X*(ω). Sketch the corresponding time-domain signal *xT*0 (*t*) if *X*(ω) is sampled at the following rates: - - (a) *f*0 = 10 samples/Hz - - (b) *f*0 = 5 samples/Hz - -- (c) *f*0 = 4 samples/Hz -- (d) *f*0 = 2.5 samples/Hz -- **8.4-2** The Fourier transform of a signal *x*(*t*), bandlimited to *B* Hz, is *X*(ω). The signal *x*(*t*) is repeated periodically at intervals *T*, where *T* = 1.25/*B*. The resulting signal *y*(*t*) is - -$$ -y(t) = \sum_{-\infty}^{\infty} x(t - nT) -$$ - -Show that *y*(*t*) can be expressed as - -$$ -y(t) = C_0 + C_1 \cos(1.6\pi Bt + \theta_1) -$$ - -where - -$$ -C_0 = \frac{1}{T}X(0) -$$ - -$$ -C_1 = \frac{2}{T} \left| X\left(\frac{2\pi}{T}\right) \right| -$$ - -and - -$$ -\theta_1 = \angle X \left( \frac{2\pi}{T} \right) -$$ - -Recall that a bandlimited signal is not timelimited, and hence has infinite duration. The periodic repetitions are all overlapping. - -- **8.5-1** For a signal *x*(*t*) that is timelimited to 10 ms and has an essential bandwidth of 10 kHz, determine *N*0, the number of signal samples necessary to compute a power-of-2 FFT with a frequency resolution *f*0 of at least 50 Hz. Explain whether any zero padding is necessary. -- **8.5-2** To compute the DFT of signal *x*(*t*) in Fig. P8.5-2, write the sequence *xn* (for *n* = 0 to *N*0 − 1) if the frequency resolution *f*0 must be at least 0.25 Hz. Assume the essential bandwidth (the folding frequency) of *x*(*t*) to be at least 3 Hz. Do not compute the DFT; just write the appropriate sequence *xn*. -- **8.5-3** Suppose we want to sample a finite-duration signal *x*(*t*) that occupies 0 ≤ *t* ≤ *T*. - -- (a) Devise a way to use the DFT to help select a suitable sampling rate *f*s for signal *x*(*t*). [*Hint:* Consider the characteristics of an oversampled signal's DFT spectrum.] -- (b) Test the method you devised in part (a) using the signal *x*(*t*) = (*t*−1 2 ). Use MAT-LAB to compute any needed DFTs. What value *f*s seems reasonable for this signal? -- **8.5-4** Choose appropriate values for *N*0 and *T* and compute the DFT of the signal *e*−*t u*(*t*). Use two different criteria for determining the effective bandwidth of *e*−*t u*(*t*). As the bandwidth, use the frequency at which the amplitude response drops to 1% of its peak value (at ω = 0). Next, use the 99% energy criterion for determining the bandwidth (see Ex. 7.20). -- **8.5-5** Repeat Prob. 8.5-4 for the signal - -$$ -x(t) = \frac{2}{t^2 + 1} -$$ - -- **8.5-6** For the signals *x*(*t*) and *g*(*t*) represented in Fig. P8.5-6, write the appropriate sequences *xn* and *gn* necessary for the computation of the convolution of *x*(*t*) and *g*(*t*) using DFT. Use *T* = 1/8. -- **8.5-7** For this problem, interpret the *N*-point DFT as an *N*-periodic function of *r*. To stress this fact, we shall change the notation *Xr* to *X*(*r*). Are the following frequency-domain signals valid DFTs? Answer yes or no. For each valid DFT, determine the size *N* of the DFT and whether the time-domain signal is real. - - (a) *X*(*r*) = *j*−π - - (b) *X*(*r*) = sin(*r*/10) - - (c) *X*(*r*) = sin(π*r*/10) - - (d) *X*(*r*) = (1+*j*)/√2 *r* - - (e) *X*(*r*) = #*r* + π\$10 where #·\$10 denotes the modulo-*N* operation. -- **8.7-1** MATLAB's fft command computes the DFT of a vector x assuming the first sample occurs at time *n* = 0. Given that X = fft(x) has already - -been computed, derive a method to correct X to reflect an arbitrary starting time *n* = *n*0. - -- **8.7-2** Consider a complex signal composed of two closely spaced complex exponentials: *x*1[*n*] = *ej*2π*n*30/100 + *ej*2π*n*33/100. For each of the following cases, plot the length-*N* DFT magnitude as a function of frequency *fr*, where *fr* = *r*/*N*. - - (a) Compute and plot the DFT of *x*1[*n*] using 10 samples (0 ≤ *n* ≤ 9). From the plot, can both exponentials be identified? Explain. - - (b) Zero-pad the signal from part (a) with 490 zeros and then compute and plot the 500-point DFT. Does this improve the picture of the DFT? Explain. - - (c) Compute and plot the DFT of *x*1[*n*] using 100 samples (0 ≤ *n* ≤ 99). From the plot, can both exponentials be identified? Explain. - - (d) Zero-pad the signal from part (c) with 400 zeros and then compute and plot the 500-point DFT. Does this improve the picture of the DFT? Explain. -- **8.7-3** Repeat Prob. 8.7-2, using the complex signal *x*2[*n*] = *ej*2π*n*30/100 +*ej*2π*n*31.5/100. -- **8.7-4** Consider a complex signal composed of a dc term and two complex exponentials: *y*1[*n*] = 1 + *ej*2π*n*30/100 + 0.5 ∗ *ej*2π*n*43/100. For each of the following cases, plot the length-*N* DFT magnitude as a function of frequency *fr*, where *fr* = *r*/*N*. - - (a) Use MATLAB to compute and plot the DFT of *y*1[*n*] with 20 samples (0 ≤ *n*≤19). From the plot, can the two non-dc exponentials be identified? Given the amplitude relation between the two, the lower-frequency peak should be twice as large as the higher-frequency peak. Is this the case? Explain. - - (b) Zero-pad the signal from part (a) to a total length of 500. Does this improve locating the two non-dc exponential components? Is - -the lower-frequency peak twice as large as the higher-frequency peak? Explain. - -- (c) MATLAB's signal-processing toolbox function window allows window functions to be easily generated. Generate a length-20 Hanning window and apply it to *y*1[*n*]. Using this windowed function, repeat parts (a) and (b). Comment on whether the window function helps or hinders the analysis. -- **8.7-5** Repeat Prob. 8.7-4, using the complex signal *y*2[*n*] = 1+*ej*2π*n*30/100 +0.5*ej*2π*n*38/100. -- **8.7-6** This problem investigates the idea of zero padding applied in the frequency domain. When asked, plot the length-*N* DFT magnitude as a function of frequency *fr*, where *fr* = *r*/*N*. - - (a) In MATLAB, create a vector x that contains one period of the sinusoid *x*[*n*] = cos((π/2)*n*). Plot the result. How "sinusoidal" does the signal appear to be? - - (b) Use the fft command to compute the DFT X of vector x. Plot the magnitude of the DFT coefficients. Do they make sense? - - (c) Zero-pad the DFT vector to a total length of 100 by inserting the appropriate number of zeros in the middle of the vector X. Call this zero-padded DFT sequence Y. Why are zeros inserted in the middle rather than the end? Take the inverse DFT of Y and plot the result. What similarities exist between the new signal y and the original signal x? What are the differences between x and y? What is the effect of zero padding in the frequency domain? How is this type of zero padding similar to zero padding in the time domain? - - (d) Derive a general modification to the procedure of zero padding in the frequency domain to ensure that the amplitude of the resulting time-domain signal is left unchanged. - -- (e) Consider one period of a square wave described by the length-8 vector [1111 −1 −1 −1 −1]. Zero-pad the DFT of this vector to a length of 100, and call the result S. Scale S according to part (d), take the inverse DFT, and plot the result. Does the new time-domain signal *s*[*n*] look like a square wave? Explain. -- **8.7-7** The quantized output *x*q of a truncating asymmetric converter is given as - -$$ -\textstyle x_\mathrm{q} = \frac{x_\mathrm{max}}{2^{B-1}} \lfloor \frac{x}{x_\mathrm{max}} 2^{B-1} \frac{1}{2} \rfloor -$$ - -Any values outside the 2*B* allowable levels should be clamped to the nearest level. - -- (a) Similar to Fig. 8.31, plot the transfer characteristics for a 3-bit version of this quantizer. -- (b) Apply 3-bit truncating asymmetric quantization to a 1 Hz cosine sampled at *f*s = 50 Hz over 1 second. Plot the original signal - -*x*(*t*), the quantized signal *x*q(*t*), and the magnitude spectra of both. How does truncating asymmetric quantization compare to the results of asymmetric rounding quantization shown in Fig. 8.33? - -**8.7-8** The quantized output *x*q of a symmetric truncating converter is given as - -$$ -x_{\mathbf{q}} = \frac{x_{\max}}{2^{B-1}} \left( \lfloor \frac{x}{x_{\max}} 2^{B-1} - \frac{1}{2} \rfloor + \frac{1}{2} \right) -$$ - -Any values outside the 2*B* allowable levels should be clamped to the nearest level. - -- (a) Similar to Fig. 8.31, plot the transfer characteristics for a 3-bit version of this quantizer. -- (b) Apply 3-bit truncating symmetric quantization to a 1 Hz cosine sampled at *f*s = 50 Hz over 1 second. Plot the original signal *x*(*t*), the quantized signal *x*q(*t*), and the magnitude spectra of both. How does truncating symmetric quantization compare to the results of asymmetric rounding quantization shown in Fig. 8.33? - - - -# **FOURIER ANALYSIS OF [DISCRETE-TIME](#page-14-0) SIGNALS** - -In Chs. 6 and 7, we studied the ways of representing a continuous-time signal as a sum of sinusoids or exponentials. In this chapter we shall discuss similar development for discrete-time signals. Our approach is parallel to that used for continuous-time signals. We first represent a periodic *x*[*n*] as a Fourier series formed by a discrete-time exponential (or sinusoid) and its harmonics. Later we extend this representation to an aperiodic signal *x*[*n*] by considering *x*[*n*] as a limiting case of a periodic signal with the period approaching infinity. - -## **[9.1 DISCRETE-TIME](#page-14-0) FOURIER SERIES (DTFS)** - -A continuous-time sinusoid cosω*t* is a periodic signal regardless of the value of ω. Such is not the case for the discrete-time sinusoid cos*n* (or exponential *ejn*). A sinusoid cos*n* is periodic only if /2π is a rational number. This can be proved by observing that if this sinusoid is *N*0 periodic, then - -$$ -\cos\Omega(n+N_0)=\cos\Omega n -$$ - -This is possible only if - -*N*0 = 2π*m m* integer - -Here, both *m* and *N*0 are integers. Hence, /2π = *m*/*N*0 is a rational number. Thus, a sinusoid cos*n* (or exponential *ejn*) is periodic only if - -$$ -\frac{\Omega}{2\pi} = \frac{m}{N_0} -$$ - a rational number - -When this condition (/2π a rational number) is satisfied, the period *N*0 of the sinusoid cos*n* is given by - -$$ -N_0 = m\left(\frac{2\pi}{\Omega}\right) \tag{9.1} -$$ - -To compute *N*0, we must choose the smallest value of *m* that will make *m*(2π/) an integer. For example, if = 4π/17, then the smallest value of *m* that will make *m*(2π/) = *m*(17/2) an integer is 2. Therefore, - -$$ -N_0 = m\left(\frac{2\pi}{\Omega}\right) = 2\left(\frac{17}{2}\right) = 17 -$$ - -However, a sinusoid cos(0.8*n*) is not a periodic signal because 0.8/2π is not a rational number. - -### **[9.1-1 Periodic Signal Representation by](#page-14-0) Discrete-Time Fourier Series** - -A continuous-time periodic signal of period *T*0 can be represented as a trigonometric Fourier series consisting of a sinusoid of the fundamental frequency ω0 = 2π/*T*0, and all its harmonics. The exponential form of the Fourier series consists of exponentials *ej*0*t* , *e*±*j*ω0*t* , *e*±*j*2ω0*t* , *e*±*j*3ω0*t* ,... . - -A discrete-time periodic signal can be represented by a discrete-time Fourier series using a parallel development. Recall that a periodic signal *x*[*n*] with period *N*0 is characterized by the fact that - -$$ -x[n] = x[n+N_0] -$$ - -The smallest value of *N*0 for which this equation holds is the *fundamental period*. The *fundamental frequency* is 0 = 2π/*N*0 rad/sample. An *N*0-periodic signal *x*[*n*] can be represented by a discrete-time Fourier series made up of sinusoids of fundamental frequency 0 = 2π/*N*0 and its harmonics. As in the continuous-time case, we may use a trigonometric or an exponential form of the Fourier series. Because of its compactness and ease of mathematical manipulations, the exponential form is preferable to the trigonometric. For this reason, we shall bypass the trigonometric form and go directly to the exponential form of the discrete-time Fourier series. - -The exponential Fourier series consists of the exponentials *ej*0*n*, *e*±*j*0*n*, *e*±*j*20*n*, ..., *e*±*jn*0*n*, ..., and so on. There would be an infinite number of harmonics, except for the property proved in Sec. 5.5-1, that discrete-time exponentials whose frequencies are separated by 2π (or integer multiples of 2π) are identical because - -$$ -e^{i(\Omega \pm 2\pi m)n} = e^{i\Omega n} e^{\pm 2\pi mn} = e^{i\Omega n} \qquad m \text{ integer} -$$ - -The consequence of this result is that the *r*th harmonic is identical to the (*r* + *N*0)th harmonic. To demonstrate this, let *gn* denote the *n*th harmonic *ejn*0*n*. Then - -$$ -g_{r+N_0} = e^{j(r+N_0)\Omega_0 n} = e^{j(r\Omega_0 n + 2\pi n)} = e^{j r\Omega_0 n} = g_r -$$ - -and - -$$ -g_r = g_{r+N_0} = g_{r+2N_0} = \cdots = g_{r+mN_0} -$$ - *m* integer - -Thus, the first harmonic is identical to the (*N*0 +1)th harmonic, the second harmonic is identical to the (*N*0 +2)th harmonic, and so on. In other words, there are only *N*0 independent harmonics, and their frequencies range over an interval 2π (because the harmonics are separated by 0 = 2π/*N*0). This means that, unlike the continuous-time counterpart, the discrete-time Fourier series has only a finite number (*N*0) of terms. This result is consistent with our observation in Sec. 5.5-1 that all discrete-time signals are bandlimited to a band from −π to π. Because the harmonics are separated by 0 = 2π/*N*0, there can only be *N*0 harmonics in this band. We also saw that this band can be taken from 0 to 2π or any other contiguous band of width 2π. This means we may - - - -choose the *N*0 independent harmonics *ejr*0*n* over 0 ≤ *r* ≤ *N*0 −1, or over −1 ≤ *r* ≤ *N*0 −2, or over 1 ≤ *r* ≤ *N*0, or over any other suitable choice for that matter. Every one of these sets will have the same harmonics, although in different order. - -Let us consider the first choice, which corresponds to exponentials *ejr*0*n* for *r* = 0, 1, 2, ... , *N*0 −1. The Fourier series for an *N*0-periodic signal *x*[*n*] consists of only these *N*0 harmonics, and can be expressed as - -$$ -x[n] = \sum_{r=0}^{N_0 - 1} \mathcal{D}_r e^{jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -To compute coefficients *Dr*, we multiply both sides by *e*−*jm*0*n* and sum over *n* from *n* = 0 to (*N*0 −1). *N* - -$$ -\sum_{n=0}^{N_0-1} x[n]e^{-jm\Omega_0 n} = \sum_{n=0}^{N_0-1} \sum_{r=0}^{N_0-1} \mathcal{D}_r e^{j(r-m)\Omega_0 n} -$$ -(9.2) - -The right-hand sum, after interchanging the order of summation, results in - -$$ -\sum_{r=0}^{N_0-1} \mathcal{D}_r \left[ \sum_{n=0}^{N_0-1} e^{j(r-m)\Omega_0 n} \right] -$$ - -The inner sum, according to Eq. (8.15) in Sec. 8.5, is zero for all values of *r* = *m*. It is nonzero with a value *N*0 only when *r* = *m*. This fact means the outside sum has only one term *DmN*0 (corresponding to *r* = *m*). Therefore, the right-hand side of Eq. (9.2) is equal to *DmN*0, and - -$$ -\sum_{n=0}^{N_0-1} x[n]e^{-jm\Omega_0 n} = \mathcal{D}_m N_0 -$$ - -and - -$$ -\mathcal{D}_m = \frac{1}{N_0} \sum_{n=0}^{N_0 - 1} x[n] e^{-jm\Omega_0 n} -$$ - -We now have a discrete-time Fourier series (DTFS) representation of an *N*0-periodic signal *x*[*n*] as - -$$ -x[n] = \sum_{r=0}^{N_0 - 1} \mathcal{D}_r e^{jr\Omega_0 n} -$$ -\n(9.3) - -where - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n=0}^{N_0 - 1} x[n] e^{-j r \Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ -\n(9.4) - -Observe that DTFS Eqs. (9.3) and (9.4) are identical (within a scaling constant) to the DFT Eqs. (8.13) and (8.12).† Therefore, we can use the efficient FFT algorithm to compute the DTFS coefficients. - -† If we let *x*[*n*] = *N*0*xk* and *Dr* = *Xr*, Eqs. (9.3) and (9.4) are identical to Eqs. (8.13) and (8.12), respectively. - -## **[9.1-2 Fourier Spectra of a Periodic Signal](#page-14-0)** *x***[***n***]** - -The Fourier series consists of *N*0 components - -$$ -\mathcal{D}_0, \mathcal{D}_1 e^{j\Omega_0 n}, \mathcal{D}_2 e^{j2\Omega_0 n}, \dots, \mathcal{D}_{N_0-1} e^{j(N_0-1)\Omega_0 n} -$$ - -The frequencies of these components are 0, 0, 20, ..., (*N*0 − 1)0, where 0 = 2π/*N*0. The amount of the *r*th harmonic is *Dr*. We can plot this amount *Dr* (the Fourier coefficient) as a function of index *r* or frequency . Such a plot, called the *Fourier spectrum* of *x*[*n*], gives us, at a glance, the graphical picture of the amounts of various harmonics of *x*[*n*]. - -In general, the Fourier coefficients *Dr* are complex, and they can be represented in the polar form as - -$$ -\mathcal{D}_r = |\mathcal{D}_r|e^{j\angle{\mathcal{D}_r}} -$$ - -The plot of |*Dr*| versus is called the amplitude spectrum and that of *Dr* versus is called the angle (or phase) spectrum. These two plots together are the frequency spectra of *x*[*n*]. Knowing these spectra, we can reconstruct or synthesize *x*[*n*] according to Eq. (9.3). Therefore, the Fourier (or frequency) spectra, which are an alternative way of describing a periodic signal *x*[*n*], are in every way equivalent (in terms of the information) to the plot of *x*[*n*] as a function of *n*. The Fourier spectra of a signal constitute the *frequency-domain* description of *x*[*n*], in contrast to the time-domain description, where *x*[*n*] is specified as a function of index *n* (representing time). - -The results are very similar to the representation of a continuous-time periodic signal by an exponential Fourier series except that, generally, the continuous-time signal spectrum bandwidth is infinite and consists of an infinite number of exponential components (harmonics). The spectrum of the discrete-time periodic signal, in contrast, is bandlimited and has at most *N*0 components. - -### PERIODIC EXTENSION OF FOURIER SPECTRUM - -We now show that if φ[*r*] is an *N*0-periodic function of *r*, then - -$$ -\sum_{r=0}^{N_0-1} \phi[r] = \sum_{r=(N_0)} \phi[r] \tag{9.5} -$$ - -where *r* = #*N*0\$ indicates summation over any *N*0 consecutive values of *r*. Because φ[*r*] is *N*0 periodic, the same values repeat with period *N*0. Hence, the sum of any set of *N*0 consecutive values of φ[*r*] must be the same no matter the value of *r* at which we start summing. Basically, it represents the sum over one cycle. - -To apply this result to the DTFS, we observe that *e*−*jr*0*n* is *N*0 periodic because - -$$ -e^{-jr\Omega_0(n+N_0)} = e^{-jr\Omega_0 n}e^{-j2\pi r} = e^{-jr\Omega_0 n} -$$ - -Therefore, if *x*[*n*] is *N*0 periodic, *x*[*n*]*e*−*jr*0*n* is also *N*0 periodic. Hence, from Eq. (9.4), it follows that *Dr* is also *N*0 periodic, as is *Drejr*0*n*. Now, because of Eq. (9.5), we can express Eqs. (9.3) and (9.4) as - -$$ -x[n] = \sum_{r = \langle N_0 \rangle} \mathcal{D}_r e^{jr\Omega_0 n} \tag{9.6} -$$ - -9.1 Discrete-Time Fourier Series (DTFS) 849 - -and - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n = \langle N_0 \rangle} x[n] e^{-j r \Omega_0 n} \tag{9.7} -$$ - -If we plot *Dr* for all values of *r* (rather than only 0 ≤ *r* ≤ *N*0 − 1), then the spectrum *Dr* is *N*0 periodic. Moreover, Eq. (9.6) shows that *x*[*n*] can be synthesized not only by the *N*0 exponentials corresponding to 0 ≤ *r* ≤ *N*0 − 1, but also by any successive *N*0 exponentials in this spectrum, starting at any value of *r* (positive or negative). For this reason, it is customary to show the spectrum *Dr* for all values of *r* (not just over the interval 0 ≤ *r* ≤ *N*0 − 1). *Yet we must remember that to synthesize x*[*n*] *from this spectrum, we need to add only N*0 *consecutive components.* All these observations are consistent with our discussion in Ch. 5, where we showed that a sinusoid of a given frequency is equivalent to multitudes of sinusoids, all separated by integer multiple of 2π in frequency. - -Along the scale, *Dr* repeats every 2π intervals, and along the *r* scale, *Dr* repeats at intervals of *N*0. Equations (9.6) and (9.7) show that both *x*[*n*] and its spectrum *Dr* are *N*0 periodic and both have exactly the same number of components (*N*0) over one period. - -Equation (9.7) shows that *Dr* is complex in general, and *D*−*r* is the conjugate of *Dr* if *x*[*n*] is real. Thus, - -$$ -|\mathcal{D}_r| = |\mathcal{D}_{-r}| -$$ - and $\angle \mathcal{D}_r = -\angle \mathcal{D}_{-r}$ - -so that the amplitude spectrum |*Dr*| is an even function , and *Dr* is an odd function of *r* (or ). All these concepts will be clarified by the examples to follow. The first example is rather trivial and serves mainly to familiarize the reader with the basic concepts of DTFS. - -### **EXAMPLE 9.1 Discrete-Time Fourier Series of a Sinusoid** - -Find the discrete-time Fourier series (DTFS) for *x*[*n*] = sin 0.1π*n* (Fig. 9.1a). Sketch the amplitude and phase spectra. - -In this case, the sinusoid sin 0.1π*n* is periodic because /2π = 1/20 is a rational number and the period *N*0 is [see Eq. (9.1)] - -$$ -N_0 = m\left(\frac{2\pi}{\Omega}\right) = m\left(\frac{2\pi}{0.1\pi}\right) = 20m -$$ - -The smallest value of *m* that makes 20*m* an integer is *m* = 1. Therefore, the period *N*0 = 20 so that 0 = 2π/*N*0 = 0.1π, and from Eq. (9.6), - -$$ -x[n] = \sum_{r=\langle 20 \rangle} \mathcal{D}_r e^{j0.1\pi rn} -$$ - -where the sum is performed over any 20 consecutive values of *r*. We shall select the range −10 ≤ *r* < 10 (values of *r* from −10 to 9). This choice corresponds to synthesizing *x*[*n*] using - -**Figure 9.1** Discrete-time sinusoid sin 0.1π*n* and its Fourier spectra. - -the spectral components in the fundamental frequency range (−π ≤ <π). Thus, - -$$ -x[n] = \sum_{r=-10}^{9} \mathcal{D}_r e^{j0.1\pi rn} -$$ - -where, according to Eq. (9.7), - -$$ -\mathcal{D}_r = \frac{1}{20} \sum_{n=-10}^{9} \sin 0.1 \pi n e^{-j0.1 \pi r n} -$$ - -= -$$ -\frac{1}{20} \sum_{n=-10}^{9} \frac{1}{2j} (e^{j0.1 \pi n} - e^{-j0.1 \pi n}) e^{-j0.1 \pi r n} -$$ - -= -$$ -\frac{1}{40j} \left[ \sum_{n=-10}^{9} e^{j0.1 \pi n (1-r)} - \sum_{n=-10}^{9} e^{-j0.1 \pi n (1+r)} \right] -$$ - -In these sums, *r* takes on all values between −10 and 9. From Eq. (8.15), it follows that the first sum on the right-hand side is zero for all values of *r* except *r* = 1, when the sum is equal to *N*0 = 20. Similarly, the second sum is zero for all values of *r* except *r* = −1, when it is equal to *N*0 = 20. Therefore, - -$$ -\mathcal{D}_1 = \frac{1}{2j} \qquad \text{and} \qquad \mathcal{D}_{-1} = -\frac{1}{2j} -$$ - -and all other coefficients are zero. The corresponding Fourier series is given by - -$$ -x[n] = \sin 0.1\pi n = \frac{1}{2j} (e^{j0.1\pi n} - e^{-j0.1\pi n}) -$$ -\n(9.8) - -Here the fundamental frequency 0 = 0.1π, and there are only two nonzero components: - -$$ -\mathcal{D}_1 = \frac{1}{2j} = \frac{1}{2}e^{-j\pi/2} -$$ - and $\mathcal{D}_{-1} = -\frac{1}{2j} = \frac{1}{2}e^{j\pi/2}$ - -Therefore, - -$$ -|\mathcal{D}_1| = |\mathcal{D}_{-1}| = \frac{1}{2} \quad \text{and} \quad \angle \mathcal{D}_1 = -\frac{\pi}{2}, \ \angle \mathcal{D}_{-1} = \frac{\pi}{2} -$$ - -Sketches of *Dr* for the interval (−10 ≤ *r* < 10) appear in Figs. 9.1b and 9.1c. According to Eq. (9.8), there are only two components corresponding to *r* = 1 and −1. The remaining 18 coefficients are zero. The *r*th component *Dr* is the amplitude of the frequency *r*0 = 0.1*r*π. Therefore, the frequency interval corresponding to −10 ≤ *r* < 10 is −π ≤ <π, as depicted in Figs. 9.1b and 9.1c. This spectrum over the range −10 ≤ *r* < 10 (or −π ≤ < π) is sufficient to specify the frequency-domain description (Fourier series), and we can synthesize *x*[*n*] by adding these spectral components. Because of the periodicity property discussed in this section, the spectrum *Dr* is a periodic function of *r* with period *N*0 = 20. For this reason, we repeat the spectrum with period *N*0 = 20 (or = 2π), as illustrated in Figs. 9.1b and 9.1c, which are periodic extensions of the spectrum in the range −10 ≤ *r* < 10. Observe that the amplitude spectrum is an even function and the angle or phase spectrum is an odd function of *r* (or ), as expected. - -The result [Eq. (9.8)] is a trigonometric identity and could have been obtained immediately without the formality of finding the Fourier coefficients. We have intentionally chosen this trivial example to introduce the reader gently to the new concept of the discrete-time Fourier series and its periodic nature. The Fourier series is a way of expressing a periodic signal *x*[*n*] in terms of exponentials of the form *ejr*0*n* and its harmonics. The result in Eq. (9.8) is merely a statement of the (obvious) fact that sin 0.1π*n* can be expressed as a sum of two exponentials *ej*0.1π*n* and *e*−*j*0.1π*n*. - -Because of the periodicity of the discrete-time exponentials *ejr*0*n*, the Fourier series components can be selected in any range of length *N*0 = 20 (or = 2π). For example, if - -### 852 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -we select the frequency range 0 ≤ < 2π (or 0 ≤ *r* < 20), we obtain the Fourier series as - -$$ -x[n] = \sin 0.1\pi n = \frac{1}{2j} (e^{j0.1\pi n} - e^{j1.9\pi n}) -$$ - -This series is equivalent to that in Eq. (9.8) because the two exponentials *ej*1.9π*n* and *e*−*j*0.1π*n* are equivalent. This follows from the fact that *ej*1.9π*n* = *ej*1.9π*n* ×*e*−*j*2π*n* = *e*−*j*0.1π*n*. - -We could have selected the spectrum over any other range of width = 2π in Figs. 9.1b and 9.1c as a valid discrete-time Fourier series. The reader may verify this by proving that such a spectrum starting anywhere (and of width = 2π) is equivalent to the same two components on the right-hand side of Eq. (9.8). - -### **DR ILL 9.1 DTFS Spectra on Alternate Intervals** - -From the spectra in Fig. 9.1, write the Fourier series corresponding to the interval −10 ≥ *r* > −30 (or −π ≥ > −3π). Show that this Fourier is equivalent to that in Eq. (9.8). - -### **DR ILL 9.2 Discrete-Time Fourier Series of a Sum of Sinusoids** - -Find the period and the DTFS for - -*x*[*n*] = 4 cos 0.2π*n*+6 sin 0.5π*n* - -over the interval 0 ≤ *r* ≤ 19. Use Eq. (9.4) to compute *Dr*. - -### **ANSWERS** - -*N*0 = 20 and *x*[*n*] = 2*ej*0.2π*n* +(3*e*−*j*π/2)*ej*0.5π*n* +(3*ej*π/2)*ej*1.5π*n* +2*ej*1.8π*n* - -### **DR ILL 9.3 Fundamental Period of Discrete-Time Sinusoids** - -Find the fundamental periods *N*0, if any, for: **(a)** sin(301π*n*/4) and **(b)** cos 1.3*n*. - -### **ANSWERS** - -**(a)** *N*0 = 8, **(b)** *N*0 does not exist because the sinusoid is not periodic. - -Compute and plot the discrete-time Fourier series for the periodic sampled gate function shown in Fig. 9.2a. - -**Figure 9.2 (a)** Periodic sampled gate pulse and **(b)** its Fourier spectrum. - -In this case, *N*0 = 32 and 0 = 2π/32 = π/16. Therefore, - -$$ -x[n] = \sum_{r = \langle 32 \rangle} \mathcal{D}_r e^{jr(\pi/16)n} -$$ - -where - -$$ -\mathcal{D}_r = \frac{1}{32} \sum_{n=(32)} x[n] e^{-jr(\pi/16)n} -$$ - -For our convenience, we shall choose the interval −16 ≤ *n* ≤ 15 for this summation, although any other interval of the same width (32 points) would give the same result.† - -$$ -\mathcal{D}_r = \frac{1}{32} \sum_{n=-16}^{15} x[n] e^{-j r (\pi/16)n} -$$ - -† In this example we have used the same equations as those for the DFT in Ex. 8.9, within a scaling constant. In the present example, the values of *x*[*n*] at *n* = 4 and −4 are taken as 1 (full value), whereas in Ex. 8.9 these values are 0.5 (half the value). This is the reason for the slight difference in spectra in Figs. 9.2b and 8.19d. Unlike continuous-time signals, discontinuity is a meaningless concept in discrete-time signals. - -### 854 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -Now, *x*[*n*] = 1 for −4 ≤ *n* ≤ 4 and is zero for all other values of *n*. Therefore, - -$$ -\mathcal{D}_r = \frac{1}{32} \sum_{n=-4}^{4} e^{-jr(\pi/16)n} \tag{9.9} -$$ - -This is a geometric progression with a common ratio *e*−*j*(π/16)*r* . Therefore (see Sec. B.8-3),† - -$$ -\mathcal{D}_r = \frac{1}{32} \left[ \frac{e^{-j(5\pi r/16)} - e^{j(4\pi r/16)}}{e^{-j(\pi r/16)} - 1} \right] -$$ - -\n -$$ -= \left(\frac{1}{32}\right) \frac{e^{-j(0.5\pi r/16)} \left[e^{-j(4.5\pi r/16)} - e^{j(4.5\pi r/16)}\right]}{e^{-j(0.5\pi r/16)} \left[e^{-j(0.5\pi r/16)} - e^{j(0.5\pi r/16)}\right]} -$$ - -\n -$$ -= \left(\frac{1}{32}\right) \frac{\sin\left(\frac{4.5\pi r}{16}\right)}{\sin\left(\frac{0.5\pi r}{16}\right)} -$$ - -\n -$$ -= \left(\frac{1}{32}\right) \frac{\sin(4.5r\Omega_0)}{\sin(0.5r\Omega_0)} \qquad \Omega_0 = \frac{\pi}{16} -$$ - (9.10) - -This spectrum (with its periodic extension) is depicted in Fig. 9.2b. - -### DISCRETE-TIME FOURIER SERIES USING MATLAB - -Let us confirm our results by using MATLAB to directly compute the DTFS according to Eq. (9.4). - -``` ->> N_0 = 32; n = (0:N_0-1); Omega_0 = 2*pi/N_0; ->> x_n = [ones(1,5) zeros(1,23) ones(1,4)]; ->> for r = 0:N_0-1, ->> X_r(r+1) = sum(x_n.*exp(-j*r*Omega_0*n))/N_0; ->> end ->> r = n; stem(r,real(X_r),'k.'); ->> xlabel('r'); ylabel('X_r'); axis([0 31 -.1 0.3]); -``` - -The MATLAB result, shown in Fig. 9.3, matches Fig. 9.2b. Alternatively, scaling the FFT by *N*0 produces the exact same result (Fig. 9.3). - ->> X\_r = fft(x\_n)/N\_0; stem(r,real(X\_r),'k.'); >> xlabel('r'); ylabel('X\_r'); axis([0 31 -.1 0.3]); - -$$ -\frac{1}{32} \sum_{n=-4}^{4} x[n] = \frac{9}{32} -$$ - -Fortunately, the value of *D*0, as computed from Eq. (9.10), also happens to be 9/32. Hence, Eq. (9.10) is valid for all *r*. - -† Strictly speaking, the geometric progression sum formula applies only if the common ratio - -*e*−*j*(π/16)*r* = 1. When *r* = 0, this ratio is unity. Hence, Eq. (9.10) is valid for values of *r* = 0. For the case *r* = 0, the sum in Eq. (9.9) is given by - - - -## **9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL** - -In Sec. 9.1 we succeeded in representing periodic signals as a sum of (everlasting) exponentials. In this section we extend this representation to aperiodic signals. The procedure is identical conceptually to that used in Ch. 7 for continuous-time signals. - -Applying a limiting process, we now show that an aperiodic signal *x*[*n*] can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal *x*[*n*] such as the one illustrated in Fig. 9.4a by everlasting exponential signals, let us construct a new periodic signal *xN*0 [*n*] formed by repeating the signal *x*[*n*] every *N*0 units, as shown in Fig. 9.4b. The period *N*0 is made large enough to avoid overlap between the repeating cycles (*N*0 ≥ 2*N*+1). The periodic signal *xN*0 [*n*] can be represented by an exponential Fourier series. If we let *N*0 → ∞, the signal - -**Figure 9.4** Generation of a periodic signal by periodic extension of a signal *x*[*n*]. - -#### 856 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -*x*[*n*] repeats after an infinite interval, and therefore, - -$$ -\lim_{N_0 \to \infty} x_{N_0}[n] = x[n] -$$ - -Thus, the Fourier series representing *xN*0 [*n*] will also represent *x*[*n*] in the limit *N*0 → ∞. The exponential Fourier series for *xN*0 [*n*] is given by - -$$ -x_{N_0}[n] = \sum_{r = \langle N_0 \rangle} \mathcal{D}_r e^{jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ -\n(9.11) - -where - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n=-\infty}^{\infty} x[n] e^{-jr\Omega_0 n} \tag{9.12} -$$ - -The limits for the sum on the right-hand side of Eq. (9.12) should be from −*N* to *N*. But because *x*[*n*] = 0 for |*n*| > *N*, it does not matter if the limits are taken from −∞ to ∞. - -It is interesting to see how the nature of the spectrum changes as *N*0 increases. To understand this behavior, let us define *X*(), a continuous function of , as - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n} -$$ -\n(9.13) - -From this definition and Eq. (9.12), we have - -$$ -\mathcal{D}_r = \frac{1}{N_0} X(r\Omega_0) \tag{9.14} -$$ - -This result shows that the Fourier coefficients *Dr* are 1/*N*0 times the samples of *X*() taken every 0 rad/s.† Therefore, (1/*N*0)*X*() is the envelope for the coefficients *Dr*. We now let *N*0→∞ by doubling *N*0 repeatedly. Doubling *N*0 halves the fundamental frequency 0, with the result that the spacing between successive spectral components (harmonics) is halved, and there are now twice as many components (samples) in the spectrum. At the same time, by doubling *N*0, the envelope of the coefficients *Dr* is halved, as seen from Eq. (9.14). If we continue this process of doubling *N*0 repeatedly, the number of components doubles in each step; the spectrum progressively becomes denser, while its magnitude *Dr* becomes smaller. Note, however, that the relative shape of the envelope remains the same [proportional to *X*() in Eq. (9.13)]. In the limit, as *N*0 → ∞, the fundamental frequency 0 →0, and *Dr* →0. The separation between successive harmonics, which is 0, is approaching zero (infinitesimal), and the spectrum becomes so dense that it appears to be continuous. But as the number of harmonics increases indefinitely, the harmonic amplitudes *Dr* become vanishingly small (infinitesimal). We discussed an identical situation in Sec. 7.1. - -We follow the procedure in Sec. 7.1 and let *N*0 → ∞. According to Eq. (9.13), - -$$ -X(r\Omega_0) = \sum_{n=-\infty}^{\infty} x[n]e^{-jr\Omega_0 n} -$$ - -† For the sake of simplicity we assume *Dr* and therefore *X*() to be real. The argument, however, is also valid for complex *Dr* [or *X*()]. - -Using Eq. (9.14), we can express Eq. (9.11) as - -$$ -x_{N_0}[n] = \frac{1}{N_0} \sum_{r=N_0} X(r\Omega_0) e^{jr\Omega_0 n} = \sum_{r=N_0} X(r\Omega_0) e^{jr\Omega_0 n} \left(\frac{\Omega_0}{2\pi}\right) -$$ - -In the limit as *N*0 → ∞, 0 → 0 and *xN*0 [*n*] → *x*[*n*]. Therefore, - -$$ -x[n] = \lim_{\Omega_0 \to 0} \sum_{r = \langle N_0 \rangle} \left[ \frac{X(r\Omega_0)\Omega_0}{2\pi} \right] e^{jr\Omega_0 n} \tag{9.15} -$$ - -Because 0 is infinitesimal, it will be appropriate to replace 0 with an infinitesimal notation : - -$$ -\Delta \Omega = \frac{2\pi}{N_0} \tag{9.16} -$$ - -Equation (9.15) can be expressed as - -$$ -x[n] = \lim_{\Delta\Omega \to 0} \frac{1}{2\pi} \sum_{r=\langle N_0 \rangle} X(r\Delta\Omega) e^{jr\Delta\Omega n} \Delta\Omega \tag{9.17} -$$ - -The range *r* = #*N*0\$ implies the interval of *N*0 number of harmonics, which is *N*0 = 2π according to Eq. (9.16). In the limit, the right-hand side of Eq. (9.17) becomes the integral - -$$ -x[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{in\Omega} d\Omega -$$ -\n(9.18) - -where \$ 2π indicates integration over any continuous interval of 2π. The spectrum *X*() is given by [Eq. (9.13)] - -$$ -X(\Omega) = \sum_{n = -\infty}^{\infty} x[n]e^{-j\Omega n} -$$ -\n(9.19) - -The integral on the right-hand side of Eq. (9.18) is called the *Fourier integral*. We have now succeeded in representing an aperiodic signal *x*[*n*] by a Fourier integral (rather than a Fourier series). This integral is basically a Fourier series (in the limit) with fundamental frequency →0, as seen in Eq. (9.17). The amount of the exponential *ejrn* is *X*(*r*)/2π. Thus, the function *X*() given by Eq. (9.19) acts as a spectral function, which indicates the relative amounts of various exponential components of *x*[*n*]. - -We call *X*() the (direct) discrete-time Fourier transform (DTFT) of *x*[*n*], and *x*[*n*] the inverse discrete-time Fourier transform (IDTFT) of *X*(). This nomenclature can be represented as - -$$ -X(\Omega) = \text{DTFT}\{x[n]\} -$$ - and $x[n] = \text{IDTFT}\{X(\Omega)\}$ - -The same information is conveyed by the statement that *x*[*n*] and *X*() are a (discrete-time) Fourier transform pair. Symbolically, this is expressed as - -$$ -x[n] \Longleftrightarrow X(\Omega) -$$ - -The Fourier transform *X*() is the frequency-domain description of *x*[*n*]. - -### **9.2-1 Nature of Fourier Spectra** - -We now discuss several important features of the discrete-time Fourier transform and the spectra associated with it. - -### FOURIER SPECTRA ARE CONTINUOUS FUNCTIONS OF - -Although *x*[*n*] is a discrete-time signal, *X*(), its DTFT is a continuous function of for the simple reason that is a continuous variable, which can take any value over a continuous interval from −∞ to ∞. - -## FOURIER SPECTRA ARE PERIODIC FUNCTIONS OF WITH PERIOD 2π - -From Eq. (9.19), it follows that - -$$ -X(\Omega + 2\pi) = \sum_{n=-\infty}^{\infty} x[n]e^{-j(\Omega + 2\pi)n} = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n}e^{-j2\pi n} = X(\Omega) -$$ - -Clearly, the spectrum *X*() is a continuous, periodic function of with period 2π. We must remember, however, that to synthesize *x*[*n*], we need to use the spectrum over a frequency interval of only 2π, starting at any value of [see Eq. (9.18)]. As a matter of convenience, we shall choose this interval to be the fundamental frequency range (−π, π). It is, therefore, not necessary to show discrete-time-signal spectra beyond the fundamental range, although we often do so. - -The reason for the periodic behavior of *X*() was discussed in Ch. 5, where we showed that, in a basic sense, the discrete-time frequency is bandlimited to || ≤ π. However, all discrete-time sinusoids with frequencies separated by an integer multiple of 2π are identical. This is why the spectrum is 2π periodic. - -### CONJUGATE SYMMETRY OF *X*() - -From Eq. (9.19), we obtain the DTFT of *x*∗[*n*] as - -$$ -\text{DTFT}\{x^*[n]\} = \sum_{n=-\infty}^{\infty} x^*[n]e^{-j\Omega n} = X^*(-\Omega) -$$ - -In other words, - -$$ -x^*[n] \Longleftrightarrow X^*(-\Omega) \tag{9.20} -$$ - -For real *x*[*n*], Eq. (9.20) reduces to *x*[*n*] ⇐⇒ *X*∗(−), which implies that for real *x*[*n*] - -$$ -X(\Omega) = X^*(-\Omega) -$$ - -Therefore, for real *x*[*n*], *X*() and *X*(−) are conjugates. Since *X*() is generally complex, we have both amplitude and angle (or phase) spectra - -$$ -X(\Omega) = |X(\Omega)|e^{j\angle X(\Omega)} -$$ - -Because of conjugate symmetry of *X*(), it follows that for real *x*[*n*], - -$$ -|X(\Omega)| = |X(-\Omega)| \quad \text{and} \quad \angle X(\Omega) = -\angle X(-\Omega) -$$ - -Therefore, the amplitude spectrum |*X*()| is an even function of and the phase spectrum *X*() is an odd function of for real *x*[*n*]. - -### PHYSICAL APPRECIATION OF THE DISCRETE-TIME FOURIER TRANSFORM - -In understanding any aspect of the Fourier transform, we should remember that Fourier representation is a way of expressing a signal *x*[*n*] as a sum of everlasting exponentials (or sinusoids). The Fourier spectrum of a signal indicates the relative amplitudes and phases of the exponentials (or sinusoids) required to synthesize *x*[*n*]. - -A detailed explanation of the nature of such sums over a continuum of frequencies is provided in Sec. 7.1-1. - -### EXISTENCE OF THE DTFT - -Because |*e*−*jn*| = 1, from Eq. (9.19), it follows that the existence of *X*() is guaranteed if *x*[*n*] is absolutely summable; that is, - -$$ -\sum_{n=-\infty}^{\infty} |x[n]| < \infty \tag{9.21} -$$ - -This shows that the condition of absolute summability is a sufficient condition for the existence of the DTFT representation. This condition also guarantees its uniform convergence. The inequality - -$$ -\left[\sum_{n=-\infty}^{\infty} |x[n]| \right]^2 \ge \sum_{n=-\infty}^{\infty} |x[n]|^2 -$$ - -shows that the energy of an absolutely summable sequence is finite. However, not all finite-energy signals are absolutely summable. Signal *x*[*n*] = sinc (*n*) is such an example. For such signals, the DTFT converges, not uniformly, but in the mean.† - -To summarize, *X*() exists under a weaker condition - -$$ -\sum_{n=-\infty}^{\infty} |x[n]|^2 < \infty \tag{9.22} -$$ - -The DTFT under this condition is guaranteed to converge in the mean. Thus, the DTFT of the exponentially growing signal γ *nu*[*n*] does not exist when |γ | > 1 because the signal violates Eqs. (9.21) and (9.22). But the DTFT exists for the signal sinc(*n*), which violates Eq. (9.21) but does satisfy Eq. (9.22) (see later, Ex. 9.6). In addition, if the use of δ(), the continuous-time impulse function, is permitted, we can even find the DTFT of some signals that violate both Eq. (9.21) and Eq. (9.22). Such signals are not absolutely summable, nor do they have finite energy. For example, as seen from pairs 11 and 12 of Table 9.1, the DTFT of *x*[*n*] = 1 for all *n* and *x*[*n*] = *ej*0*n* exist, although they violate Eqs. (9.21) and (9.22). - -$$ -\lim_{M \to \infty} \int_{-\pi}^{\pi} \left| X(\Omega) - \sum_{n=-M}^{M} x[n] e^{-j\Omega n} \right|^2 d\Omega = 0 -$$ - -† This means - -| No. | x[n] | X() | | -|-----|--------------------------|--------------------------------------------------------------------------------------|-----------| -| 1 | δ[n−k] | e−jk | Integer k | -| 2 | γ nu[n] | ej
ej −γ | γ < 1 | -| 3 | −γ nu[−(n+1)] | ej
ej −γ | γ > 1 | -| 4 | γ n | 1−γ 2
1−2γ cos+γ 2 | γ < 1 | -| 5 | nγ nu[n] | γ ej
(ej −γ )2 | γ < 1 | -| 6 | γ n cos(0n+θ
)u[n] | ej[ej cos
θ −γ cos(0
−θ )]
ej2 −(2γ
cos0)ej +γ
2 | γ < 1 | -| 7 | u[n] −u[n− M] | sin(M/2)
e−j(M−1)/2
sin(/2) | | -| 8 | c
π sinc (cn) | "∞
−2πk
rect
2c
k=−∞ | c
≤ π | -| 9 | cn
c
2π sinc2
2 | "∞
−2πk
2c
k=−∞ | c
≤ π | -| 10 | u[n] | ej
+π "∞
δ(−2πk)
ej −1
k=−∞ | | -| 11 | 1
for all n | 2π "∞
δ(−2πk)
k=−∞ | | -| 12 | ej0n | 2π "∞
δ(−0
−2πk)
k=−∞ | | -| 13 | cos0n | π "∞
−2πk) +δ(+0
−2πk)
δ(−0
k=−∞ | | -| 14 | sin0n | jπ "∞
δ(+0
−2πk)−δ(−0
−2πk)
k=−∞ | | -| 15 | (cos0n)u[n] | ej2 −ej cos0
"∞
π
δ(−2πk−0)+δ(−2πk+0)
+1 +
ej2 −2ej cos0
2
k=−∞ | | -| 16 | (sin0n)u[n] | ej sin0
"∞
π
δ(−2πk−0)−δ(−2πk+0)
+1 +
ej2 −2ej cos0
2j
k=−∞ | | - -**TABLE 9.1** Select Discrete-Time Fourier Transform Pairs - -### **EXAMPLE 9.3 DTFT of a Causal Exponential** - -Find the DTFT of *x*[*n*] = γ *nu*[*n*]. - -Using the definition, the DTFT is - -$$ -X(\Omega) = \sum_{n=0}^{\infty} \gamma^n e^{-j\Omega n} = \sum_{n=0}^{\infty} (\gamma e^{-j\Omega})^n -$$ - -This is an infinite geometric series with a common ratio γ *e*−*j*. Therefore (see Sec. B.8-3), - -$$ -X(\Omega) = \frac{1}{1 - \gamma e^{-j\Omega}} -$$ - -provided |γ *e*−*j*| < 1. But because |*e*−*j*| = 1, this condition implies |γ | < 1. Therefore, - -$$ -X(\Omega) = \frac{1}{1 - \gamma e^{-j\Omega}} \qquad |\gamma| < 1 -$$ - -If |γ | > 1, *X*() does not converge. This result is in conformity with Eqs. (9.21) and (9.22). To determine magnitude and phase responses, we note that - -$$ -X(\Omega) = \frac{1}{1 - \gamma \cos \Omega + j\gamma \sin \Omega} \tag{9.23} -$$ - -so - -$$ -|X(\Omega)| = \frac{1}{\sqrt{(1 - \gamma \cos \Omega)^2 + (\gamma \sin \Omega)^2}} = \frac{1}{\sqrt{1 + \gamma^2 - 2\gamma \cos \Omega}} -$$ - -and - -$$ -\angle X(\Omega) = -\tan^{-1}\left[\frac{\gamma \sin \Omega}{1 - \gamma \cos \Omega}\right] -$$ - -Figure 9.5 shows *x*[*n*] = γ *nu*[*n*] and its spectra for γ = 0.8. Observe that the frequency spectra are continuous and periodic functions of with the period 2π. As explained earlier, we need to use the spectrum only over the frequency interval of 2π. We often select this interval to be the fundamental frequency range (−π,π). - -The amplitude spectrum |*X*()| is an even function and the phase spectrum *X*() is an odd function of . - -### **EXAMPLE 9.4 DTFT of an Anticausal Exponential** - -**Figure 9.6** Exponential γ *nu*[−(*n*+1)]. - -Using the definition, the DTFT is - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} \gamma^n u [-(n+1)] e^{-j\Omega n} = \sum_{n=-1}^{-\infty} (\gamma e^{-j\Omega})^n = \sum_{n=-1}^{-\infty} \left(\frac{1}{\gamma} e^{j\Omega}\right)^{-n} -$$ - -Setting *n* = −*m* yields - -$$ -x[n] = \sum_{m=1}^{\infty} \left(\frac{1}{\gamma} e^{j\Omega}\right)^m = \frac{1}{\gamma} e^{j\Omega} + \left(\frac{1}{\gamma} e^{j\Omega}\right)^2 + \left(\frac{1}{\gamma} e^{j\Omega}\right)^3 + \cdots -$$ - -This is a geometric series with a common ratio *ej*/γ . Therefore, from Sec. B.8-3, - -$$ -X(\Omega) = \frac{1}{\gamma e^{-j\Omega} - 1} = \frac{1}{(\gamma \cos \Omega - 1) - j\gamma \sin \Omega}, \qquad |\gamma| > 1 -$$ - -Therefore, - -$$ -|X(\Omega)| = \frac{1}{\sqrt{1 + \gamma^2 - 2\gamma \cos \Omega}} \quad \text{and} \quad \angle X(\Omega) = \tan^{-1} \left[ \frac{\gamma \sin \Omega}{\gamma \cos \Omega - 1} \right] -$$ - -Except for the change of sign, this Fourier transform (and the corresponding frequency spectra) is identical to that of *x*[*n*] = γ *nu*[*n*]. Yet there is no ambiguity in determining the IDTFT of *X*() = 1/(γ *e*−*j* −1) because of the restrictions on the value of γ in each case. If |γ | < 1, then the inverse transform is *x*[*n*]=−γ *nu*[*n*]. If |γ | > 1, it is *x*[*n*] = γ *n*[−(*n*+1)]. - -### **EXAMPLE 9.5 DTFT of a Rectangular Pulse** - -Find the DTFT of the discrete-time rectangular pulse illustrated in Fig. 9.7a. This pulse is also known as the 9-point rectangular window function. - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n} = \sum_{n=-(M-1)/2}^{(M-1)/2} (e^{-j\Omega})^n \qquad M=9 -$$ - -This is a geometric progression with a common ratio *e*−*j* and (see Sec. B.8-3) - -$$ -X(\Omega) = \frac{e^{-j[(M+1)/2]\Omega} - e^{j[(M-1)/2]\Omega}}{e^{-j\Omega} - 1} -$$ -$$ -= \frac{e^{-j\Omega/2} (e^{-j(M/2)\Omega} - e^{j(M/2)\Omega})}{e^{-j\Omega/2} (e^{-j\Omega/2} - e^{j\Omega/2})} -$$ - -$$ -=\frac{\sin\left(\frac{M}{2}\Omega\right)}{\sin\left(0.5\Omega\right)}\tag{9.24} -$$ - -$$ -=\frac{\sin(4.5\Omega)}{\sin(0.5\Omega)} \qquad \text{for } M=9 -$$ - (9.25) - -Figure 9.7b shows the spectrum *X*() for *M* = 9. - -**Figure 9.7 (a)** Discrete-time gate pulse and **(b)** its Fourier spectrum. - -### DISCRETE-TIME FOURIER TRANSFORM USING MATLAB - -Within a scale factor, the DTFS is identical to the DFT and, therefore, the FFT. That is, the DTFS is just the FFT scaled by 1 *N*0 . Combined with Eq. (9.14), we see that the DFT *Xr* of finite-duration signal *x*[*n*] (repeated with period *N*0 large enough to avoid overlap) is just samples of the DTFT *X*() taken at = *r*0. That is, the length-*N*0 DFT of signal *x*[*n*] yields *N*0 samples of its DTFT *X*() as - -$$ -X_r = X(r\Omega_0), \qquad \text{where } \Omega_0 = \frac{2\pi}{N_0} \tag{9.26} -$$ - -This relationship provides a way to use MATLAB's fft command to validate our DTFT calculations. By appropriately zero-padding *x*[*n*], we can obtain as many samples of *X*() as are desired. Let us demonstrate the process for the current example using *N*0 = 64. Notice that in taking the DFT, we modulo-*N*0 shift our rectangular pulse signal to occupy 0 ≤ *n* ≤ *N*0 −1. - -``` ->> Omega = linspace(0,2*pi,1000); -``` - -``` ->> X = sin(4.5*Omega)./sin(0.5*Omega); X(mod(Omega,2*pi)==0) = 4.5/0.5; -``` - -``` ->> N_0 = 64; M = 9; x = [ones(1,(M+1)/2) zeros(1,N_0-M) ones(1,(M-1)/2)]; -``` - -``` ->> Xr = fft(x); Omega_0 = 2*pi/N_0; r = 0:N_0-1; -``` - -``` ->> plot(Omega,abs(X),'k-',Omega_0*r,abs(Xr),'k.'); axis([0 2*pi 0 9.5]); -``` - ->> xlabel('\Omega'); ylabel('|X(\Omega)|'); - -As shown in Fig. 9.8, the FFT samples align exactly with our analytical DTFT result. - -### **EXAMPLE 9.6 Inverse DTFT of a Rectangular Spectrum** - -Find the inverse DTFT of the rectangular pulse spectrum described over the fundamental band (|| ≤ π) by *X*() = rect(/2*c*) for *c* ≤ π. Because of the periodicity property, *X*() repeats at the intervals of 2π, as shown in Fig. 9.9a. - -**Figure 9.9** Periodic gate spectrum and its inverse discrete-time Fourier transform. - -According to Eq. (9.18), - -$$ -x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\Omega) e^{jn\Omega} d\Omega = \frac{1}{2\pi} \int_{-\Omega_c}^{\Omega_c} e^{jn\Omega} d\Omega -$$ -$$ -= \frac{1}{j2\pi n} e^{jn\Omega} \Big|_{-\Omega_c}^{\Omega_c} = \frac{\sin(\Omega_c n)}{\pi n} = \frac{\Omega_c}{\pi} \text{sinc}(\Omega_c n) -$$ - -The signal *x*[*n*] is depicted in Fig. 9.9b (for the case *c* = π/4). - -### **DR ILL 9.4 Finding the DTFT** - -Find the DTFT and sketch the corresponding amplitude and phase spectra for - -(a) -$$ -x[n] = \gamma^{|k|} -$$ - with $|\gamma| < 1$ - -**(b)** -$$ -y[n] = \delta[n+1] - \delta[n-1] -$$ - -### **ANSWERS** - -(a) -$$ -X(\Omega) = \frac{1 - \gamma^2}{1 - 2\gamma \cos \Omega + \gamma^2} -$$ - -\n(b) $|Y(\Omega)| = 2|\sin \Omega|$ and $\angle Y(\omega) = (\pi/2)[1 - \text{sgn}(\sin \Omega)]$ - -## **9.2-2 Connection Between the DTFT and the** *z***-Transform** - -The connection between the (bilateral) *z*-transform and the DTFT is similar to that between the Laplace transform and the Fourier transform. The *z*-transform of *x*[*n*], according to Eq. (5.1), is - -$$ -X[z] = \sum_{n = -\infty}^{\infty} x[n]z^{-n} -$$ -\n(9.27) - -Setting *z* = *ej* in this equation yields - -$$ -X[e^{i\Omega}] = \sum_{n=-\infty}^{\infty} x[n]e^{-j\Omega n} -$$ - -The right-hand side sum defines *X*(), the DTFT of *x*[*n*]. Does this mean that the DTFT can be obtained from the corresponding *z*-transform by setting *z* = *ej*? In other words, is it true that *X*[*ej*] = *X*()? Yes, it is true in most cases. For example, when *x*[*n*] = *anu*[*n*], its *z*-transform is *z*/(*z* − *a*), and *X*[*ej*] = *ej*/(*ej* − *a*), which is equal to *X*() (assuming |*a*| < 1). However, for the unit step function *u*[*n*], the *z*-transform is *z*/(*z* − 1), and *X*[*ej*] = *ej*/(*ej* − 1). As seen from Table 9.1, pair 10, this is not equal to *X*() in this case. - -We obtained *X*[*ej*] by setting *z* = *ej* in Eq. (9.27). This implies that the sum on the right-hand side of Eq. (9.27) converges for *z* = *ej*, which means the unit circle (characterized by *z* = *ej*) lies in the region of convergence for *X*[*z*]. Hence, the general rule is that setting *z* = *ej* in *X*[*z*] yields the DTFT *X*() only when the ROC for *X*[*z*] includes the unit circle. This applies for all *x*[*n*] that are absolutely summable. If the ROC of *X*[*z*] excludes the unit circle, *X*[*ej*] = *X*(). This applies to all exponentially growing *x*[*n*] and also *x*[*n*], which either is constant or oscillates with constant amplitude. - -The reason for this peculiar behavior has something to do with the nature of convergence of the *z*-transform and the DTFT.† - -This discussion shows that although the DTFT may be considered to be a special case of the *z*-transform, we need to circumscribe such a view. This cautionary note is supported by the fact that a periodic signal has the DTFT, but its *z*-transform does not exist. - -## **[9.3 PROPERTIES OF THE](#page-14-0) DTFT** - -A close connection exists between the DTFT and the CTFT (continuous-time Fourier transform). For this reason, which Sec. 9.4 discusses, the properties of the DTFT are very similar to those of the CTFT, as the following discussion shows. - -### LINEARITY OF THE DTFT - -If - -*x*1[*n*] ⇐⇒ *X*1() and *x*2[*n*] ⇐⇒ *X*2() - -then - -$$ -a_1x_1[n]+a_2x_2[n] \Longleftrightarrow a_1X_1(\Omega)+a_2X_2(\Omega) -$$ - -The proof is trivial. The result can be extended to any finite sums. - -### CONJUGATE SYMMETRY OF *X*() - -In Eq. (9.20), we proved the *conjugation property* - -$$ -x^*[n] \Longleftrightarrow X^*(-\Omega) \tag{9.28} -$$ - -† To explain this point, consider the unit step function *u*[*n*] and its transforms. Both the *z*-transform and the DTFT synthesize *x*[*n*], using everlasting exponentials of the form *zn*. The value of *z* can be anywhere in the complex *z*-plane for the *z*-transform, but it must be restricted to the unit circle (*z* = *ej*) in the case of the DTFT. The unit step function is readily synthesized in the *z*-transform by a relatively simple spectrum *X*[*z*] = *z*/(*z* − 1), by choosing *z* outside the unit circle (the ROC for *u*[*n*] is |*z*| > 1). In the DTFT, however, we are restricted to values of *z* only on the unit circle (*z* = *ej*). The function *u*[*n*] can still be synthesized by values of *z* on the unit circle, but the spectrum is more complicated than when we are free to choose *z* anywhere, including the region outside the unit circle. In contrast, when *x*[*n*] is absolutely summable, the region of convergence for the *z*-transform includes the unit circle, and we can synthesize *x*[*n*] by using *z* along the unit circle in both the transforms. This leads to *X*[*ej*] = *X*(). - -### 868 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -We also showed that as a consequence of this, when *x*[*n*] is real, *X*() and *X*(−) are conjugates, that is, - -$$ -X(-\Omega) = X^*(\Omega) -$$ - -This is the *conjugate symmetry* property. Since *X*() is generally complex, we have both amplitude and angle (or phase) spectra - -$$ -X(\Omega) = |X(\Omega)|e^{j\angle X(\Omega)} -$$ - -Hence, for real *x*[*n*], it follows that - -$$ -|X(\Omega)| = |X(-\Omega)| \quad \text{and} \quad \angle X(\Omega) = -\angle X(-\Omega) -$$ - -Therefore, for real *x*[*n*], the amplitude spectrum |*X*()| is an even function of and the phase spectrum *X*() is an odd function of . - -### TIME AND FREQUENCY REVERSAL - -Also called the reflection property, the time and frequency reversal property states that - -$$ -x[-n] \Longleftrightarrow X(-\Omega) \tag{9.29} -$$ - -Demonstration of this property is straightforward. From Eq. (9.19), the DTFT of *x*[−*n*] is - -$$ -\text{DTFT}\{x[-n]\} = \sum_{n=-\infty}^{\infty} x[-n]e^{-j\Omega n} = \sum_{m=-\infty}^{\infty} x[m]e^{j\Omega m} = X(-\Omega) -$$ - -### **EXAMPLE 9.7 Using the Reflection Property** - -Use the time-frequency reversal property of Eq. (9.29) and pair 2 in Table 9.1 to derive pair 4 in Table 9.1. - -Pair 2 states that - -$$ -\gamma^n u[n] = \frac{e^{i\Omega}}{e^{i\Omega} - \gamma} \qquad |\gamma| < 1 -$$ - -Hence, from Eq. (9.29), - -$$ -\gamma^{-n}u[-n] = \frac{e^{-j\Omega}}{e^{-j\Omega} - \gamma} \qquad |\gamma| < 1 -$$ - -Moreover, γ |*n*| could be expressed as a sum of γ *n u*[*n*] and γ −*n u*[−*n*], except that the impulse at *n* = 0 is counted twice (once from each of the two exponentials). Hence, - -$$ -\gamma^{|n|} = \gamma^n u[n] + \gamma^{-n} u[-n] - \delta[n] -$$ - -Combining these results and invoking the linearity property, we can write - -$$ -\text{DTFT}\{\gamma^{|n|}\} = \frac{e^{j\Omega}}{e^{j\Omega} - \gamma} + \frac{e^{-j\Omega}}{e^{-j\Omega} - \gamma} - 1 = \frac{1 - \gamma^2}{1 - 2\gamma \cos \Omega + \gamma^2} \qquad |\gamma| < 1 -$$ - -which agrees with pair 4 in Table 9.1. - -### **DR ILL 9.5 Using the Reflection Property** - -In Table 9.1, derive pair 13 from pair 15 by using the time-reversal property of Eq. (9.29). - -### MULTIPLICATION BY *n*: FREQUENCY DIFFERENTIATION - -$$ -nx[n] \Longleftrightarrow j\frac{dX(\Omega)}{d\Omega} \tag{9.30} -$$ - -The result follows immediately by differentiating both sides of Eq. (9.19) with respect to . - -### **EXAMPLE 9.8 Using the Frequency-Differentiation Property** - -Use the frequency-differentiation property of Eq. (9.30) and pair 2 in Table 9.1 to derive pair 5 in Table 9.1. - -Pair 2 states that - -$$ -\gamma^n u[n] = \frac{e^{i\Omega}}{e^{i\Omega} - \gamma} \qquad |\gamma| < 1 -$$ - -Hence, from Eq. (9.30), - -$$ -n\gamma^{n}u[n] = j\frac{d}{d\Omega} \left\{ \frac{e^{j\Omega}}{e^{j\Omega} - \gamma} \right\} = \frac{\gamma e^{j\Omega}}{(e^{j\Omega} - \gamma)^{2}} \qquad |\gamma| < 1 -$$ - -which agrees with pair 5 in Table 9.1. - -### TIME-SHIFTING PROPERTY If - -*x*[*n*] ⇐⇒ *X*() - -then - -$$ -x[n-k] \Longleftrightarrow X(\Omega)e^{-jk\Omega} \qquad \text{for integer } k \tag{9.31} -$$ - -This property can be proved by direct substitution in the equation defining the direct transform. From Eq. (9.19), we obtain - -$$ -x[n-k] \Longleftrightarrow \sum_{n=-\infty}^{\infty} x[n-k]e^{-j\Omega n} = \sum_{m=-\infty}^{\infty} x[m]e^{-j\Omega[m+k]} -$$ -$$ -= e^{-j\Omega k} \sum_{n=-\infty}^{\infty} x[m]e^{-j\Omega m} = e^{-jk\Omega}X(\Omega) -$$ - -This result shows that *delaying a signal by k samples does not change its amplitude spectrum. The phase spectrum, however, is changed by* −*k*. This added phase is a linear function of with slope −*k*. - -### PHYSICAL EXPLANATION OF LINEAR PHASE - -Time delay in a signal causes a linear phase shift in its spectrum. The heuristic explanation of this result is exactly parallel to that for continuous-time signals given in Sec. 7.3 (see Fig. 7.22). - -### **EXAMPLE 9.9 Demonstrating Linear Phase** - -To demonstrate the linear phase associated with a time shift, find the DTFT of *x*[*n*] = (1/4)sinc (π(*n*−2)/4), shown in Fig. 9.10a. - -In Ex. 9.6, we found that - -$$ -\frac{1}{4}\operatorname{sinc}\left(\frac{\pi n}{4}\right) \Longleftrightarrow \sum_{m=-\infty}^{\infty} \operatorname{rect}\left(\frac{\Omega - 2\pi m}{\pi/2}\right) -$$ - -Use of the time-shifting property [Eq. (9.31)] yields (for integer *k*) - -$$ -\frac{1}{4}\operatorname{sinc}\left(\frac{\pi(n-2)}{4}\right) \Longleftrightarrow \sum_{m=-\infty}^{\infty} \operatorname{rect}\left(\frac{\Omega-2\pi m}{\pi/2}\right) e^{-j2\Omega} -$$ - -The spectrum of the shifted signal is shown in Fig. 9.10b. - -## **DR ILL 9.6 Using the Time-Shifting Property** - -Verify the result in Eq. (9.24) from pair 7 in Table 9.1 and the time-shifting property of the DTFT. - -# FREQUENCY-SHIFTING PROPERTY - -If - -$$ -x[n] \Longleftrightarrow X(\Omega) -$$ - -then - -$$ -x[n]e^{j\Omega_c n} \Longleftrightarrow X(\Omega - \Omega_c) \tag{9.32} -$$ - -This property is the dual of the time-shifting property. To prove the frequency-shifting property, we use Eq. (9.19) as - -$$ -x[n]e^{j\Omega_c n} \Longleftrightarrow \sum_{n=-\infty}^{\infty} x[n]e^{j\Omega_c n}e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x[n]e^{-j(\Omega-\Omega_c)n} = X(\Omega-\Omega_c) -$$ - -From this result, it follows that - -$$ -x[n]e^{-j\Omega_c n} \Longleftrightarrow X(\Omega + \Omega_c) -$$ - -### 872 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -Adding this pair to the pair in Eq. (9.32), we obtain - -$$ -x[n]\cos(\Omega_c n) \Longleftrightarrow \frac{1}{2}\{X(\Omega - \Omega_c) + X(\Omega + \Omega_c)\}\tag{9.33} -$$ - -This is the *modulation property*. - -Multiplying both sides of pair (9.32) by *ej*θ , we obtain - -$$ -x[n]e^{j(\Omega_c n + \theta)} \Longleftrightarrow X(\Omega - \Omega_c)e^{j\theta} -$$ - -Using this pair, we can generalize the modulation property as - -$$ -x[n]\cos\left(\Omega_c n + \theta\right) \Longleftrightarrow \frac{1}{2}\left\{X(\Omega - \Omega_c)e^{i\theta} + X(\Omega + \Omega_c)e^{-i\theta}\right\} -$$ - -### **EXAMPLE 9.10 Modulation Property** - -A signal *x*[*n*] = sinc (π*n*/4) modulates a carrier cos*cn*. Find and sketch the spectrum of the modulated signal *x*[*n*] cos*cn* for - -- **(a)** *c* = π/2 -- **(b)** *c* = 7π/8 = 0.875π - -**(a)** For *x*[*n*] = sinc (π*n*/4), we find (Table 9.1, pair 8) - -$$ -X(\Omega) = 4 \sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 2\pi m}{\pi/2}\right) -$$ - -Figure 9.11a shows the DTFT *X*(). From the modulation property of Eq. (9.33), we obtain - -$$ -x[n]\cos(0.5\pi n) \Longleftrightarrow 2\sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 0.5\pi - 2\pi m}{0.5\pi}\right) + \text{rect}\left(\frac{\Omega + 0.5\pi - 2\pi m}{0.5\pi}\right) -$$ - -Figure 9.11b shows half the *X*() shifted by π/2 and Fig. 9.11c shows half the *X*() shifted by −π/2. The spectrum of the modulated signal is obtained by adding these two shifted spectra and multiplying by half, as shown in Fig. 9.11d. - -**(b)** Figure 9.12a shows *X*(), which is the same as that in part (a). For *c* = 7π/8 = 0.875π, the modulation property of Eq. (9.33) yields - -$$ -x[n]\cos(0.875\pi n) \Longleftrightarrow 2\sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 0.875\pi - 2\pi m}{0.5\pi}\right) + \text{rect}\left(\frac{\Omega + 0.875\pi - 2\pi m}{0.5\pi}\right) -$$ - -**Figure 9.11** Instance of modulation for Ex. 9.10a. - -Figure 9.12b shows *X*() shifted by 7π/8 and Fig. 9.12c shows *X*() shifted by −7π/8. The spectrum of the modulated signal is obtained by adding these two shifted spectra and multiplying by half, as shown in Fig. 9.12d. In this case, the two shifted spectra overlap. Since the operation of modulation thus causes aliasing, it does not achieve the desired effect of spectral shifting. In this example, to realize spectral shifting without aliasing requires *c* ≤ 3π/4. - -### **DR ILL 9.7 Using the Frequency-Shifting Property** - -In Table 9.1, derive pairs 12 and 13 from pair 11 and the frequency-shifting/modulation property. - -## TIME- AND FREQUENCY-CONVOLUTION PROPERTY If - -*x*1[*n*] ⇐⇒ *X*1() and *x*2[*n*] ⇐⇒ *X*2() - -then - -$$ -x_1[n] * x_2[n] \Longleftrightarrow X_1(\Omega)X_2(\Omega) \tag{9.34} -$$ - -and - -$$ -x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi}X_1(\Omega)\circledast X_2(\Omega) \tag{9.35} -$$ - -where - -$$ -x_1[n] * x_2[n] = \sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m] -$$ - -For two continuous, periodic signals, we define the periodic convolution, denoted by symbol -∗ as† - -$$ -X_1(\Omega)\circledast X_2(\Omega) = \frac{1}{2\pi} \int_{2\pi} X_1(u)X_2(\Omega - u) du -$$ - -The convolution here is not the *linear* convolution used so far. This is a *periodic* (or *circular*) convolution applicable to the convolution of two continuous, periodic functions with the same period. The limit of integration in the convolution extends only to one period. - -Proof of the time-convolution property is identical to that given in Sec. 5.2 [Eq. (5.19)]. All we have to do is replace *z* with *ej*. To prove the frequency-convolution property of Eq. (9.35), we have - -$$ -x_1[n]x_2[n] \Longleftrightarrow \sum_{n=-\infty}^{\infty} x_1[n]x_2[n]e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x_2[n] \left[ \frac{1}{2\pi} \int_{2\pi} X_1(u)e^{-jnu} du \right] e^{-j\Omega n} -$$ - -Interchanging the order of summation and integration, we obtain - -$$ -x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi} \int_{2\pi} X_1(u) \left[ \sum_{n=-\infty}^{\infty} x_2[n] e^{-j(\Omega-u)n} \right] du = \frac{1}{2\pi} \int_{2\pi} X_1(u) X_2(\Omega-u) du -$$ - -### **EXAMPLE 9.11 DTFT of an Accumulator System** - -If -$$ -x[n] \leftrightarrow X(\Omega) -$$ -, then show that $\sum_{k=-\infty}^{n} x[k] \leftrightarrow \pi X(0) \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1} X(\Omega)$ . - -† In Eq. (8.20), we defined periodic convolution for two discrete, periodic sequences in a different way. Although we are using the same symbol -∗ for both discrete and continuous cases, the meaning will be clear from the context. - -### 876 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -To begin, we notice that - -$$ -x[n] * u[n] = \sum_{k=-\infty}^{\infty} x[k]u[n-k] = \sum_{k=-\infty}^{n} x[k] -$$ - -Applying the time-convolution property of Eq. (9.34) and pair 10 in Table 9.1, it follows that - -$$ -\sum_{k=-\infty}^{n} x[k] = x[n] * u[n] \Longleftrightarrow X(\Omega) \left(\pi \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1}\right) -$$ - -Because of 2π periodicity, *X*(0) = *X*(2π*k*). Moreover, *X*()δ( − 2π*k*) = *X*(2π*k*)δ( − 2π*k*) = *X*(0)δ(−2π*k*). Hence, - -$$ -\sum_{k=-\infty}^{n} x[k] \Longleftrightarrow \pi X(0) \sum_{k=-\infty}^{\infty} \delta(\Omega - 2\pi k) + \frac{e^{i\Omega}}{e^{i\Omega} - 1} X(\Omega) -$$ - -### **DR ILL 9.8 Using the Frequency-Convolution Property** - -In Table 9.1, derive pair 9 from pair 8, assuming *c* ≤ π/2. Use the frequency-convolution property. - -PARSEVAL'S THEOREM If - -$$ -x[n] \Longleftrightarrow X(\Omega) -$$ - -then *Ex*, the energy of *x*[*n*], is given by - -$$ -E_x = \sum_{n=-\infty}^{\infty} |x[n]|^2 = \frac{1}{2\pi} \int_{2\pi} |X(\Omega)|^2 d\Omega -$$ -\n(9.36) - -To prove this property, we have from Eq. (9.28), - -$$ -X^*(\Omega) = \sum_{n=-\infty}^{\infty} x^*[n]e^{i\Omega n} -$$ - -Now, - -$$ -\sum_{n=-\infty}^{\infty} |x[n]|^2 = \sum_{n=-\infty}^{\infty} x^*[n]x[n] = \sum_{n=-\infty}^{\infty} x^*[n] \left[ \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{i\Omega n} d\Omega \right] -$$ -$$ -= \frac{1}{2\pi} \int_{2\pi} X(\Omega) \left[ \sum_{n=-\infty}^{\infty} x^*[n] e^{i\Omega n} \right] d\Omega -$$ -$$ -= \frac{1}{2\pi} \int_{2\pi} X(\Omega) X^*(\Omega) d\Omega = \frac{1}{2\pi} \int_{2\pi} |X(\Omega)|^2 d\Omega -$$ - -Table 9.2 summarizes Parseval's theorem and the other important properties of the DTFT. - -| Operation | x[n] | X() | -|-----------------------|-------------------------------|-----------------------------------------------| -| Linearity | a1x1[n] +a2x2[n] | a1X1()+a2X2() | -| Conjugation | x∗[n] | X∗(−) | -| Scalar multiplication | ax[n] | aX() | -| Multiplication by n | nx[n] | dX()
j
d | -| Time reversal | x[−n] | X(−) | -| Time shifting | x[n−k] | X()e−jk
k integer | -| Frequency shifting | x[n] ejcn | X(−c) | -| Time convolution | x1[n] ∗ x2[n] | X1()X2() | -| Frequency convolution | x1[n]x2[n] | #
1
X1[u]X2[−u]du
2π
2π | -| Parseval's theorem | = "∞
x[n] 2
Ex
n=−∞ | #
1
2 d
Ex
X()
=
2π
2π | - -**TABLE 9.2** Properties of the DTFT - -### **EXAMPLE 9.12 Using Parseval's Theorem to Find Signal Energy** - -Find the energy of *x*[*n*] = sinc (*cn*), assuming *c* < π. - -From pair 8, Table 9.1, the fundamental band spectrum of *x*[*n*] is - -$$ -\operatorname{sinc}(\Omega_c n) \Longleftrightarrow \frac{\pi}{\Omega_c} \operatorname{rect}\left(\frac{\Omega}{2\Omega_c}\right) \qquad |\Omega| \le \pi -$$ - -### 878 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -From Parseval's theorem [Eq. (9.36)], we have - -$$ -E_x = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{\pi^2}{\Omega_c^2} \left[ \text{rect}\left(\frac{\Omega}{2\Omega_c}\right) \right]^2 d\Omega -$$ - -Because rect(/2*c*) = 1 over || ≤ *c* and is zero otherwise, the preceding integral yields - -$$ -E_x = \frac{1}{2\pi} \left(\frac{\pi^2}{\Omega_c^2}\right) (2\Omega_c) = \frac{\pi}{\Omega_c} -$$ - -## **[9.4 LTI DISCRETE-TIME](#page-14-0) SYSTEM ANALYSIS BY DTFT** - -Consider a linear, time-invariant, discrete-time system with the unit impulse response *h*[*n*]. We shall find the (zero-state) system response *y*[*n*] for the input *x*[*n*]. Let - -*x*[*n*] ⇐⇒ *X*() *y*[*n*] ⇐⇒ *Y*() and *h*[*n*] ⇐⇒ *H*() - -Because *y*[*n*] = *x*[*n*] ∗ *h*[*n*], it follows from Eq. (9.34) that - -$$ -Y(\Omega) = X(\Omega)H(\Omega) \tag{9.37} -$$ - -This result is similar to that obtained for continuous-time systems. Let us examine the role of *H*(), the DTFT of the unit impulse response *h*[*n*]. - -Equation (9.37) holds for BIBO-stable systems and also for marginally stable systems if the input does not contain the system's natural mode(s). In other cases, the response grows with *n* and is not Fourier-transformable. Moreover, the input *x*[*n*] also has to be DTF-transformable. For cases where Eq. (9.37) does not apply, we use the *z*-transform for system analysis. - -Equation (9.37) shows that the output signal frequency spectrum is the product of the input signal frequency spectrum and the frequency response of the system. From this equation, we obtain - -|*Y*()|=|*X*()||*H*()| and *Y*() = *X*()+ *H*() - -This result shows that the output amplitude spectrum is the product of the input amplitude spectrum and the amplitude response of the system. The output phase spectrum is the sum of the input phase spectrum and the phase response of the system. - -We can also interpret Eq. (9.37) in terms of the frequency-domain viewpoint, which sees a system in terms of its frequency response (system response to various exponential or sinusoidal components). The frequency domain views a signal as a sum of various exponential or sinusoidal components. The transmission of a signal through a (linear) system is viewed as transmission of various exponential or sinusoidal components of the input signal through the system. This concept can be understood by displaying the input–output relationships by a directed arrow as follows: - -$$ -e^{i\Omega n} \Longrightarrow H(\Omega)e^{i\Omega n} -$$ - -which shows that the system response to *ejn* is *H*()*ejn*, and - -$$ -x[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{i\Omega n} d\Omega -$$ - -which shows *x*[*n*] as a sum of everlasting exponential components. Invoking the linearity property, we obtain - -$$ -y[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) H(\Omega) e^{i\Omega n} d\Omega -$$ - -which gives *y*[*n*] as a sum of responses to all input components and is equivalent to Eq. (9.37). Thus, *X*() is the input spectrum and *Y*() is the output spectrum, given by *X*()*H*(). - -### **EXAMPLE 9.13 LTID System Analysis by the DTFT** - -An LTID system is specified by the equation *y*[*n*] − 0.5*y*[*n* − 1] = *x*[*n*]. Find *H*(), the frequency response of this system. Determine the (zero-state) response *y*[*n*] if the input *x*[*n*] = (0.8)*nu*[*n*]. - -Let *x*[*n*] ⇐⇒ *X*() and *y*[*n*] ⇐⇒ *Y*(). Taking the DTFT of the system's difference equation yields - -$$ -(1 - 0.5e^{-j\Omega})Y(\Omega) = X(\Omega) -$$ - -According to Eq. (9.37), - -$$ -H(\Omega) = \frac{Y(\Omega)}{X(\Omega)} = \frac{1}{1 - e^{-j\Omega}} = \frac{e^{j\Omega}}{e^{j\Omega} - 0.5} -$$ - -Also, *x*[*n*] = (0.8)*nu*[*n*]. Hence, - -$$ -X(\Omega) = \frac{e^{i\Omega}}{e^{i\Omega} - 0.8} -$$ - -and - -$$ -Y(\Omega) = X(\Omega)H(\Omega) = \frac{2e^{i\Omega}}{(e^{i\Omega} - 0.8)(e^{i\Omega} - 0.5)} -$$ - -#### 880 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -We can express the right-hand side as a sum of two first-order terms (modified partial fraction expansion as discussed in Sec. B.5-6) as follows† : - -$$ -\frac{Y(\Omega)}{e^{i\Omega}} = \frac{e^{i\Omega}}{(e^{i\Omega} - 0.5)(e^{i\Omega} - 0.8)} = \frac{-\frac{5}{3}}{e^{i\Omega} - 0.5} + \frac{\frac{8}{3}}{e^{i\Omega} - 0.8} -$$ - -Consequently, - -$$ -Y(\Omega) = -\left(\frac{5}{3}\right) \frac{e^{i\Omega}}{e^{i\Omega} - 0.5} + \left(\frac{8}{3}\right) \frac{e^{i\Omega}}{e^{i\Omega} - 0.8} -$$ - -= -\left(\frac{5}{3}\right) \frac{1}{1 - 0.5e^{-i\Omega}} + \left(\frac{8}{3}\right) \frac{1}{1 - 0.8e^{-i\Omega}} - -From entry 2 of Table 9.1, the inverse DTFT of this equation is - -$$ -y[n] = \left[ -\frac{5}{3}(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] -$$ - -This example demonstrates the procedure for using the DTFT to determine an LTID system response. It is similar to the Fourier transform method in the analysis of LTIC systems. As in the case of the Fourier transform, this method can be used only if the system is asymptotically or BIBO-stable and if the input signal is DTF-transformable.‡ We shall not belabor this method further because it is clumsier and more restrictive than the *z*-transform method discussed in Ch. 5. - -### **[9.4-1 Distortionless Transmission](#page-14-0)** - -In several applications, digital signals are passed through LTI systems, and we require that the output waveform be a replica of the input waveform. As in the continuous-time case, transmission is said to be distortionless if the input *x*[*n*] and the output *y*[*n*] satisfy the condition - -$$ -y[n] = G_0 x[n - n_d] -$$ - -Here, *nd*, the delay (in samples), is assumed to be integer. Taking the Fourier transform yields - -$$ -Y(\Omega) = G_0 X(\Omega) e^{-j\Omega n_d} -$$ - -But - -$$ -Y(\Omega) = X(\Omega) H(\Omega) -$$ - -† Here, *Y*() is a function of variable *ej*. Hence, *x* = *ej* for the purpose of comparison with the expression in Sec. B.5-6. - -‡ It can also be applied to marginally stable systems if the input does not contain natural mode(s) of the system. - -**Figure 9.13** LTI system frequency response for distortionless transmission. - -Therefore, - -$$ -H(\Omega) = G_0 e^{-j\Omega n_d} -$$ - -This is the frequency response required for distortionless transmission. From this equation, it follows that - -$$ -|H(\Omega)| = G_0 \quad \text{and} \quad \angle H(\Omega) = -\Omega n_d \tag{9.38} -$$ - -Thus, for distortionless transmission, the amplitude response |*H*()| must be a constant, and the phase response *H*() must be a linear function of with slope −*nd*, where *nd* is the delay in the number of samples with respect to input (Fig. 9.13). These are precisely the characteristics of an ideal delay of *nd* samples with a gain of *G*0 [see Eq. (9.31)]. - -### MEASURE OF DELAY VARIATION - -For distortionless transmission, we require a *linear phase* characteristic. In practice, many systems have a phase characteristic that may be only approximately linear. A convenient way of judging phase linearity is to plot the slope of *H*() as a function of frequency. This slope is constant for the ideal linear phase (ILP) system, but it may vary with in the general case. The slope can be expressed as - -$$ -n_g(\Omega) = -\frac{d}{d\Omega} \angle H(\Omega) \tag{9.39} -$$ - -If *ng*() is constant, all the components are delayed by *ng* samples. But if the slope is not constant, the delay *ng* varies with frequency. This variation means that different frequency components undergo different amounts of delay, and consequently, the output waveform will not be a replica of the input waveform. As in the case of LTIC systems, *ng*(), as defined in Eq. (9.39), plays an important role in bandpass systems and is called the *group delay* or *envelope* delay. Observe that constant *nd* implies constant *ng*. Note that *H*() = φ0 − *nd* also has a constant *ng*. Thus, constant group delay is a more relaxed condition. - -### DISTORTIONLESS TRANSMISSION OVER BANDPASS SYSTEMS - -As in the case of continuous-time systems, the distortionless transmission conditions can be relaxed for discrete-time bandpass systems. For lowpass systems, the phase characteristic should not only be linear over the band of interest, it should also pass through the origin [Eq. (9.38)]. For bandpass systems, the phase characteristic should be linear over the band of interest, but it need not pass through the origin (*ng* should be constant). The amplitude response is required to - -#### 882 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -be constant over the passband. Thus, for distortionless transmission over a bandpass system, the frequency response for positive range of is of the form† - -$$ -H(\Omega) = G_0 e^{j(\phi_0 - \Omega n_g)} \qquad \Omega \ge 0 -$$ - -The proof is identical to that for the continuous-time case in Sec. 7.4-2 and will not be repeated. In using Eq. (9.39) to compute *ng*, we should ignore jump discontinuities in the phase function. - -### **[9.4-2 Ideal and Practical Filters](#page-14-0)** - -Ideal filters allow distortionless transmission of a certain band of frequencies and suppress all the remaining frequencies. The general ideal lowpass filter shown in Fig. 9.14 for || ≤ π allows all components below the cutoff frequency = *c* to pass without distortion and suppresses all components above *c*. Figure 9.15 illustrates ideal highpass and bandpass filter characteristics. - -The ideal lowpass filter in Fig. 9.14a has a linear phase of slope −*nd*, which results in a delay of *nd* samples for all its input components of frequencies below *c* rad/sample. Therefore, if the input is a signal *x*[*n*] bandlimited to *c*, the output *y*[*n*] is *x*[*n*] delayed by *nd*; that is, - -$$ -y[n] = x[n - n_d] -$$ - -The signal *x*[*n*] is transmitted by this system without distortion, but with delay of *nd* samples. For this filter, - -$$ -H(\Omega) = \sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 2\pi m}{2\Omega_c}\right) e^{-j\Omega n_d} -$$ - -The unit impulse response *h*[*n*] of this filter is obtained from pair 8 (Table 9.1) and the time-shifting property - -$$ -h[n] = \frac{\Omega_c}{\pi} \operatorname{sinc} \left[ \Omega_c (n - n_d) \right] -$$ - -Because *h*[*n*] is the system response to impulse input δ[*n*], which is applied at *n* = 0, it must be causal (i.e., it must not start before *n* = 0) for a realizable system. Figure 9.14b shows *h*[*n*] for - -**Figure 9.14** Ideal lowpass filter: its frequency response and impulse response. - -† Because the phase function is an odd function of , if *H*() = φ0 − *ng* for ≥ 0, over the band 2*W* (centered at *c*), then *H*() = −φ0 −*ng* for < 0 over the band 2*W* (centered at −*c*). - -**Figure 9.15** Ideal highpass and bandpass filter frequency response. - -**Figure 9.16** Approximate realization of an ideal lowpass filter by truncation of its impulse response. - -*c* = π/4 and *nd* = 12. This figure also shows that *h*[*n*] is noncausal, hence unrealizable. Similarly, one can show that other ideal filters (such as the ideal highpass or and bandpass filters depicted in Fig. 9.15) are also noncausal and therefore physically unrealizable. - -One practical approach to realize an ideal lowpass filter approximately is to truncate both tails (positive and negative) of *h*[*n*] so that it has a finite length and then delay sufficiently to make it causal (Fig. 9.16). We now synthesize a system with this truncated (and delayed) impulse response. For closer approximation, the truncating window has to be correspondingly wider. The delay required also increases correspondingly. Thus, the price of closer realization is higher delay in the output; this situation is common in noncausal systems. - -## **[9.5 DTFT CONNECTION WITH THE](#page-14-0) CTFT** - -Consider a continuous-time signal *xc*(*t*) (Fig. 9.17a) with the Fourier transform *Xc*(ω) bandlimited to *B* Hz (Fig. 9.17b). This signal is sampled with a sampling interval *T*. The sampling rate is at least equal to the Nyquist rate; that is, *T* ≤ 1/2*B*. The sampled signal *xc*(*t*) (Fig. 9.17c) can be - -**Figure 9.17** Connection between the DTFT and the Fourier transform. - -expressed as - -$$ -\bar{x}_c(t) = \sum_{n=-\infty}^{\infty} x_c(nT) \,\delta(t - nT) -$$ - -The continuous-time Fourier transform of the foregoing equation yields - -$$ -\overline{X}_c(\omega) = \sum_{n=-\infty}^{\infty} x_c(nT) e^{-jnT\omega} -$$ -\n(9.40) - -In Sec. 8.1 (Fig. 8.1f), we showed that *Xc*(ω) is *Xc*(ω)/*T* repeating periodically with a period ω*s* = 2π/*T*, as illustrated in Fig. 9.17d. Let us construct a discrete-time signal *x*[*n*] such that its *n*th sample value is equal to the value of the *n*th sample of *xc*(*t*), as depicted in Fig. 9.17e, that is, - -$$ -x[n] = x_c(nT) \tag{9.41} -$$ - -Now, *X*(), the DTFT of *x*[*n*], is given by - -$$ -X(\Omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-jn\Omega} = \sum_{n=-\infty}^{\infty} x_c(nT) e^{-jn\Omega} -$$ - -Comparison of this equation with Eq. (9.40) shows that letting ω*T* = in *Xc*(ω) yields *X*(), that is, - -$$ -X(\Omega) = X_c(\omega)|_{\omega T = \Omega} -$$ - -Alternately, *X*() can be obtained from *Xc*(ω) by replacing ω with /*T*, that is, - -$$ -X(\Omega) = \overline{X}_c \left(\frac{\Omega}{T}\right) \tag{9.42} -$$ - -Therefore, *X*() is identical to *Xc*(ω), frequency-scaled by factor *T*, as shown in Fig. 9.17f. Thus, ω = 2π/*T* in Fig. 9.17d corresponds to = 2π in Fig. 9.17f. - -### **[9.5-1 Use of DFT and FFT for Numerical Computation of the DTFT](#page-14-0)** - -The discrete Fourier transform (DFT), as discussed in Ch. 8, is a tool for computing the samples of the continuous-time Fourier transform (CTFT). Because of the close connection between CTFT and DTFT, as seen in Eq. (9.42), we can also use this same DFT to compute DTFT samples. - -In Ch. 8, Eqs. (8.12) and (8.13) relate an *N*0-point sequence *xn* to another *N*0-point sequence *Xr*. Changing the notation *xn* to *x*[*n*] in these equations, we obtain - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x[n] e^{-j r \Omega_0 n} -$$ -\n(9.43) - -and - -$$ -x[n] = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{jr\Omega_0 n} -$$ -\n(9.44) - -where 0 = 2π *N*0 . Comparing Eq. (9.19) with Eq. (9.43), we recognize that *Xr* is the sample of *X*() at = *r*0, that is, - -$$ -X_r = X(r\Omega_0) \qquad \Omega_0 = \frac{2\pi}{N_0} -$$ - -Hence, DFT Eqs. (9.43) and (9.44) can be viewed to relate an *N*0-point sequence *x*[*n*] to the *N*0-point samples of corresponding *X*(). We can now use the efficient algorithm FFT (discussed in Ch. 8) to compute *Xr* from *x*[*n*], and vice versa. - -If *x*[*n*] is not timelimited, we can still find the approximate values of *Xr* by suitably windowing *x*[*n*]. To reduce the error, the window should be tapered and should have sufficient width to satisfy error specifications. In practice, the numerical computation of signals, which are generally non-timelimited, is performed in this manner because of the computational economy of the DFT, especially for signals of long duration. - -### COMPUTATION OF DISCRETE-TIME FOURIER SERIES (DTFS) - -The discrete-time Fourier series (DTFS) equations [(9.3) and (9.4)] are identical to the DFT equations [(8.13) and (8.12)] within a scaling constant *N*0. If we let *x*[*n*] = *N*0*xn* and *Dr* = *Xr* - -#### 886 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -in Eqs. (9.4) and (9.3), we obtain - -$$ -X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n} \qquad \text{and} \qquad x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{jr\Omega_0 n} -$$ - -This is precisely the DFT and IDFT of Eqs. (8.12) and (8.13). For instance, to compute the DTFS for the periodic signal in Fig. 9.2a, we use the values of *xn* = *x*[*n*]/*N*0 as - -$$ -x_n = \begin{cases} \frac{1}{32} & 0 \le n \le 4 \quad \text{and} \quad 28 \le n \le 31\\ 0 & 5 \le n \le 27 \end{cases} -$$ - -Numerical computations in modern digital signal processing are conveniently performed with the discrete Fourier transform, introduced in Sec. 8.5. The DFT computations can be very efficiently executed by using the fast Fourier transform (FFT) algorithm discussed in Sec. 8.6. The DFT is indeed the workhorse of modern digital signal processing. The discrete-time Fourier transform (DTFT) and the inverse discrete-time Fourier transform (IDTFT) can be computed by using the DFT. For an *N*0-point signal *x*[*n*], its DFT yields exactly *N*0 samples of *X*() at frequency intervals of 2π/*N*0. We can obtain a larger number of samples of *X*() by padding a sufficient number of zero-valued samples to *x*[*n*]. The *N*0-point DFT of *x*[*n*] gives exact values of the DTFT samples if *x*[*n*] has a finite length *N*0. If the length of *x*[*n*] is infinite, we need to use the appropriate window function to truncate *x*[*n*]. - -Because of the convolution property, we can use the DFT to compute the convolution of two signals *x*[*n*] and *h*[*n*], as discussed in Sec. 8.5. This procedure, known as fast convolution, requires padding both signals by a suitable number of zeros, to make the linear convolution of the two signals identical to the circular (or periodic) convolution of the padded signals. Large blocks of data may be processed by sectioning the data into smaller blocks and processing such smaller blocks in sequence. Such a procedure requires smaller memory and reduces the processing time [1]. - -## **[9.6 GENERALIZATION OF THE](#page-14-0) DTFT TO THE** *z***-TRANSFORM** - -LTID systems can be analyzed by using the DTFT. This method, however, has the following limitations. - -- 1. Existence of the DTFT is guaranteed only for absolutely summable signals. The DTFT does not exist for exponentially or even linearly growing signals. This means that the DTFT method is applicable only for a limited class of inputs. -- 2. Moreover, this method can be applied only to asymptotically or BIBO-stable systems; it cannot be used for unstable or even marginally stable systems. - -These are serious limitations in the study of LTID system analysis. Actually, it is the first limitation that is also the cause of the second limitation. Because the DTFT is incapable of handling growing signals, it is incapable of handling unstable or marginally stable systems.† Our goal is, therefore, to extend the DTFT concept so that it can handle exponentially growing signals. - -We may wonder what causes this limitation on DTFT so that it is incapable of handling exponentially growing signals. Recall that in the DTFT, we are using sinusoids or exponentials of the form *ejn* to synthesize an arbitrary signal *x*[*n*]. These signals are sinusoids with constant amplitudes. They are incapable of synthesizing exponentially growing signals no matter how many such components we add. Our hope, therefore, lies in trying to synthesize *x*[*n*] by using exponentially growing sinusoids or exponentials. This goal can be accomplished by generalizing the frequency variable *j* to σ + *j*, that is, by using exponentials of the form *e*(σ+*j*)*n* instead of exponentials *ejn*. The procedure is almost identical to that used in extending the Fourier transform to the Laplace transform. - -Let us define a new variable *X*ˆ(*j*) = *X*(). Hence, - -$$ -\hat{X}(j\Omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\Omega n} -$$ -\n(9.45) - -and - -$$ -x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} \hat{X}(j\Omega) e^{j\Omega n} d\Omega -$$ - -Consider now the DTFT of *x*[*n*] *e*−σ*n* (σ real): - -$$ -\text{DTFT}\left\{x[n]e^{-\sigma n}\right\} = \sum_{n=-\infty}^{\infty} x[n]e^{-\sigma n}e^{-j\Omega n} = \sum_{n=-\infty}^{\infty} x[n]e^{-(\sigma+j\Omega)n} -$$ - -It follows from Eq. (9.45) that this sum is *X*ˆ(σ +*j*). Thus, - -$$ -\text{DTFT}\left\{x[n]e^{-\sigma n}\right\} = \sum_{n=-\infty}^{\infty} x[n]e^{-(\sigma+j\Omega)n} = \hat{X}(\sigma+j\Omega) \tag{9.46} -$$ - -Hence, the inverse DTFT of *X*ˆ(σ +*j*) is *x*[*n*] *e*−σ*n*. Therefore, - -$$ -x[n]e^{-\sigma n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \hat{X}(\sigma + j\Omega) e^{j\Omega n} d\Omega -$$ - -Multiplying both sides by *e*σ*n* yields - -$$ -x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} \hat{X}(\sigma + j\Omega) e^{(\sigma + j\Omega)n} d\Omega -$$ -\n(9.47) - -Let us define a new variable *z* as - -$$ -z = e^{\sigma + j\Omega} -$$ - so that $\ln z = \sigma + j\Omega$ and $\frac{1}{z}dz = j d\Omega$ - -†Recall that the output of an unstable system grows exponentially. Also, the output of a marginally stable system to characteristic mode input grows with time. - -**Figure 9.18** Contour of integration for the *z*-transform. - -Because *z* = *e*σ+*j* is complex, we can express it as *z* = *rej*, where *r* = *e*σ . Thus, *z* lies on a circle of radius *r*, and as varies from −π to π, *z* circumambulates along this circle, completing exactly one counterclockwise rotation, as illustrated in Fig. 9.18. Changing to variable *z* in Eq. (9.47) yields - -$$ -x[n] = \frac{1}{2\pi j} \oint \hat{X}(\ln z) z^{n-1} dz -$$ - (9.48) - -and from Eq. (9.46) we obtain - -$$ -\hat{X}(\ln z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} -$$ -\n(9.49) - -where the integral 6 indicates a contour integral around a circle of radius *r* in the counterclockwise direction. - -Equations (9.48) and (9.49) are the desired extensions. They are, however, in a clumsy form. For the sake of convenience, we make another notational change by observing that *X*ˆ(ln*z*) is a function of *z*. Let us denote it by a simpler notation *X*[*z*]. Thus, Eq. (9.48) becomes - -$$ -x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz -$$ -\n(9.50) - -and Eq. (9.49) becomes - -$$ -X[z] = \sum_{n=-\infty}^{\infty} x[n]z^{-n} -$$ -\n(9.51) - -This is the (bilateral) *z*-transform pair. Equation (9.50) expresses *x*[*n*] as a continuous sum of exponentials of the form *zn* = *e*(σ+*j*)*n* = *rn ejn*. Thus, by selecting a proper value for *r* (or σ), we can make the exponential grow (or decay) at any exponential rate we desire. - -If we let σ = 0, we have *z* = *ej* and - -$$ -X[z]|_{z=e^{j\Omega}} = \hat{X}(\ln z)\Big|_{z=e^{j\Omega}} = \hat{X}(j\Omega) = X(\Omega) -$$ - -Thus, the familiar DTFT is just a special case of the *z*-transform *X*[*z*] obtained by letting *z* = *ej* and assuming that the sum on the right-hand side of Eq. (9.51) converges when *z* = *ej*. This also implies that the ROC for *X*[*z*] includes the unit circle. - -## **[9.7 MATLAB: WORKING WITH THE](#page-14-0) DTFS AND THE DTFT** - -This section investigates various methods to compute the discrete-time Fourier series (DTFS). Performance of these methods is assessed by using MATLAB's stopwatch and profiling functions. Additionally, the discrete-time Fourier transform (DTFT) is applied to the important topic of finite impulse response (FIR) filter design. - -### **[9.7-1 Computing the Discrete-Time Fourier Series](#page-14-0)** - -Within a scale factor, the DTFS is identical to the DFT. Thus, methods to compute the DFT can be readily used to compute the DTFS. Specifically, the DTFS is the DFT scaled by 1/*N*0. As an example, consider a 50 Hz sinusoid sampled at 1000 Hz over one-tenth of a second. - -``` ->> T = 1/1000; N_0 = 100; n = (0:N_0-1)'; ->> x = cos(2*pi*50*n*T); -``` - -The DTFS is obtained by scaling the DFT. - -``` ->> X = fft(x)/N_0; f = (0:N_0-1)/(T*N_0); ->> stem(f-1/(2*T),fftshift(abs(X)),'k.'); ->> axis([-500 500 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); -``` - -Figure 9.19 shows a peak magnitude of 0.5 at ±50 Hz. This result is consistent with Euler's representation - -$$ -\cos(2\pi 50nT) = \frac{1}{2}e^{j2\pi 50nT} + \frac{1}{2}e^{-j2\pi 50nT} -$$ - -Lacking the 1/*N*0 scale factor, the DFT would have a peak amplitude 100 times larger. - -The inverse DTFS is obtained by scaling the inverse DFT by *N*0. - -``` ->> x = real(ifft(X)*N_0); stem(n,x,'k.'); ->> axis([0 99 -1.1 1.1]); xlabel('n'); ylabel('x[n]'); -``` - -Figure 9.20 confirms that the sinusoid *x*[*n*] is properly recovered. Although the result is theoretically real, computer round-off errors produce a small imaginary component, which the real command removes. - -**Figure 9.19** DTFS computed by scaling the DFT. - -**Figure 9.20** Inverse DTFS computed by scaling the inverse DFT. - -Although MATLAB's fft command provides an efficient method to compute the DTFS, other important computational methods exist. A matrix-based approach is one popular way to implement Eq. (9.4). Although not as efficient as an FFT-based algorithm, matrix-based approaches provide insight into the DTFS and serve as an excellent model for solving similarly structured problems. - -To begin, define *WN*0 = *ej*0 , which is a constant for a given *N*0. Substituting *WN*0 into Eq. (9.4) yields - -$$ -\mathcal{D}_r = \frac{1}{N_0} \sum_{n=0}^{N_0 - 1} x[n] W_{N_0}^{-nr} -$$ - -An inner product of two vectors computes *Dr*. - -$$ -\mathcal{D}_r = \frac{1}{N_0} \begin{bmatrix} 1 & W_{N_0}^{-r} & W_{N_0}^{-2r} & \dots & W_{N_0}^{-(N_0-1)r} \end{bmatrix} \begin{bmatrix} x[0] \\ x[1] \\ x[2] \\ \vdots \\ x[N_0-1] \end{bmatrix} -$$ - -Stacking the results for all *r* yields - -$$ -\begin{bmatrix}\n\begin{bmatrix}\n\mathcal{D}_0 \\ -\mathcal{D}_1 \\ -\mathcal{D}_2 \\ -\vdots \\ -\mathcal{D}_{N_0-1}\n\end{bmatrix} = \frac{1}{N_0} \begin{bmatrix}\n1 & 1 & 1 & \cdots & 1 \\ -1 & W_{N_0}^{-1} & W_{N_0}^{-2} & \cdots & W_{N_0}^{-(N_0-1)} \\ -1 & W_{N_0}^{-2} & W_{N_0}^{-4} & \cdots & W_{N_0}^{-2(N_0-1)} \\ -\vdots & \vdots & \vdots & \cdots & \vdots \\ -1 & W_{N_0}^{-(N_0-1)} & W_{N_0}^{-2(N_0-1)} & \cdots & W_{N_0}^{-(N_0-1)^2}\n\end{bmatrix} \begin{bmatrix}\nx[0] \\ -x[1] \\ -x[2] \\ -\vdots \\ -x[N_0-1]\n\end{bmatrix} -$$ - -In matrix notation, this equation is compactly written as - -$$ -\mathbf{D} = \frac{1}{N_0} \mathbf{W}_{N_0} \mathbf{x} -$$ - -Since it is also used to compute the DFT, matrix **W***N*0 is often called a DFT matrix. - -Let us create an anonymous function to compute the *N*0-by-*N*0 DFT matrix **W***N*0 . Although not used here, the signal-processing toolbox function dftmtx computes the same DFT matrix, although in a less obvious but more efficient fashion. - ->> W = @(N\_0) (exp(-j\*2\*pi/N\_0)).^((0:N\_0-1)'\*(0:N\_0-1)); - -While less efficient than FFT-based methods, the matrix approach correctly computes the DTFS. - -``` ->> X = W(N_0)*x/N_0; stem(f-1/(2*T),fftshift(abs(X)),'k.'); ->> axis([-500 500 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|'); -``` - -The resulting plot is indistinguishable from Fig. 9.19. Problem 9.7-1 investigates a matrix-based approach to compute Eq. (9.3), the inverse DTFS. - -### **[9.7-2 Measuring Code Performance](#page-14-0)** - -Writing efficient code is important, particularly if the code is frequently used, requires complicated operations, involves large data sets, or operates in real time. MATLAB provides several tools for assessing code performance. When properly used, the profile function provides detailed statistics that help assess code performance. MATLAB help thoroughly describes the use of the sophisticated profile command. - -A simpler method of assessing code efficiency is to measure execution time and compare it with a reference. The MATLAB command tic starts a stopwatch timer. The toc command reads the timer. Sandwiching instructions between tic and toc returns the elapsed time. For example, the execution time of the 100-point matrix-based DTFS computation is - -``` ->> tic; W(N_0)*x/N_0; toc - Elapsed time is 0.004417 seconds. -``` - -Different machines operate at different speeds with different operating systems and with different background tasks. Therefore, elapsed-time measurements can vary considerably from machine to machine and from execution to execution. For relatively simple and short events like the present case, execution times can be so brief that MATLAB may report unreliable times or fail to register an elapsed time at all. - -To increase the elapsed time and therefore the accuracy of the time measurement, a loop is used to repeat the calculation. - -``` ->> tic; for i=1:100, W(N_0)*x/N_0; end; toc - Elapsed time is 0.173388 seconds. -``` - -This elapsed time suggests that each 100-point DTFS calculation takes a little under 2 milliseconds. What exactly does this mean, however? Elapsed time is only meaningful relative to some reference. Let us see what difference occurs by precomputing the DFT matrix, rather than repeatedly using our anonymous function. - -``` ->> W100 = W(100); tic; for i=1:100, W100*x/N_0; end; toc - Elapsed time is 0.001199 seconds. -``` - -### 892 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS - -Amazingly, this small change makes a hundredfold change in our computational efficiency! Clearly, it is much better to precompute the DFT matrix. - -To provide another example, consider the time it takes to compute the same DTFS using the FFT-based approach. - ->> tic; for i=1:100, fft(x)/N\_0; end; toc Elapsed time is 0.000399 seconds. - -With this as a reference, our fastest matrix-based computations appear to be several times slower than the FFT-based computations. This difference becomes more dramatic as *N*0 is increased. Since the two methods provide identical results, there is little incentive to use the slower matrix-based approach, and the FFT-based algorithm is generally preferred. Even so, the FFT can exhibit curious behavior: adding a few data points, even the artificial samples introduced by zero padding, can dramatically increase or decrease execution times. The tic and toc commands illustrate this strange result. Consider computing the DTFS of 1015 random data points 100 times. - ->> x1 = rand(1015,1); tic; for i=1:100; fft(x1)/1015; end; T1 = toc T1 = 0.0067 - -Next, pad the sequence with four zeros. - ->> -$$ -x2 = [x1; zeros(4,1)] -$$ -; tic; for i=1:100; fft( $x2$ )/1019; end; T2 = toc -T2 = 0.0134 - -The ratio of the two elapsed times indicates that adding four points to an already long sequence increases the computation time by a factor of 2. Next, the sequence is zero-padded to a length of *N*0 = 1024. - ->> x3 = [x2;zeros(5,1)]; tic; for i=1:100; fft(x3)/1024; end; T3 = toc T3 = 0.0017 - -In this case, the added data decrease the original execution time by a factor of 4 and the second execution time by a factor of 8! These results are particularly surprising when it is realized that the lengths of y1, y2, and y3 differ by less than 1%. - -As it turns out, the efficiency of the fft command depends on the factorability of *N*0. With the factor command, 1015 = (5)(7)(29), 1019 is prime, and 1024 = (2)10. The most factorable length, 1024, results in the fastest execution, while the least factorable length, 1019, results in the slowest execution. To ensure the greatest factorability and fastest operation, vector lengths are ideally a power of 2. - -### **[9.7-3 FIR Filter Design by Frequency Sampling](#page-14-0)** - -Finite impulse response (FIR) digital filters are flexible, always stable, and relatively easy to implement. These qualities make FIR filters a popular choice among digital filter designers. The difference equation of a length-*N* causal FIR filter is conveniently expressed as - -$$ -y[n] = h_0 x[n] + h_1 x[n-1] + \dots + h_{N-1} x[n-(N-1)] = \sum_{k=0}^{N-1} h_k x[n-k] -$$ - -The filter coefficients, or tap weights as they are sometimes called, are expressed by using the variable *h* to emphasize that the coefficients themselves represent the impulse response of the filter. - -The filter's frequency response is - -$$ -H(\Omega) = \frac{Y(\Omega)}{X(\Omega)} = \sum_{k=0}^{N-1} h_k e^{-j\Omega k} -$$ - -Since *H*() is a 2π-periodic function of the continuous variable , it is sufficient to specify *H*() over a single period (0 ≤ < 2π ). - -In many filtering applications, the desired magnitude response |*Hd*()| is known but not the filter coefficients *h*[*n*]. The question, then, is one of determining the filter coefficients from the desired magnitude response. - -Consider the design of a lowpass filter with cutoff frequency *c* = π/4. An anonymous function represents the desired ideal frequency response. - ->> H\_d = @(Omega) (mod(Omega,2\*pi)2\*pi-pi/4); - -Since the inverse DTFT of *Hd*() is a sampled sinc function, it is impossible to perfectly achieve the desired response with a causal, finite-length FIR filter. A realizable FIR filter is necessarily an approximation, and an infinite number of possible solutions exist. Thought of another way, *Hd*() specifies an infinite number of points, but the FIR filter only has *N* unknown tap weights. In general, we expect a length-*N* filter to match only *N* points of the desired response over (0 ≤ < 2π ). Which frequencies should be chosen? - -A simple and sensible method is to select *N* frequencies uniformly spaced on the interval (0 ≤ < 2π ), (0, 2π/*N*, 4π/*N*, 6π/*N*,...,(*N* −1)2π/*N*). By choosing uniformly spaced frequency samples, the *N*-point inverse DFT can be used to determine the tap weights *h*[*n*]. Program CH9MP1 illustrates this procedure. - -``` -function [h] = CH9MP1(N,H_d); -% CH9MP1.m : Chapter 9, MATLAB Program 1 -% Function M-file designs a length-N FIR filter by sampling the desired -% magnitude response H_d. Phase response is left as zero. -% INPUTS: N = desired FIR filter length -% H_d = anonymous function that defines the desired magnitude response -% OUTPUTS: h = impulse response (FIR filter coefficients) -% Create N equally spaced frequency samples: -Omega = linspace(0,2*pi*(1-1/N),N)'; -% Sample the desired magnitude response and create h[n]: -H = 1.0*H_d(Omega); h = real(ifft(H)); -``` - -To complete the design, the filter length must be specified. Small values of *N* reduce the filter's complexity but also reduce the quality of the filter's response. Large values of *N* improve the approximation of *Hd*() but also increase complexity. A balance is needed. We choose an intermediate value of *N* = 21 and use CH9MP1 to design the filter. - ->> N = 21; h = CH9MP1(N,H\_d); - -To assess the filter quality, the frequency response is computed by means of program CH5MP1. - -``` ->> Omega = linspace(0,2*pi,1000); samples = linspace(0,2*pi*(1-1/N),N)'; -``` - -``` ->> H = CH5MP1(h,1,Omega); -``` - -``` ->> subplot(2,1,1); stem([0:N-1],h,'k.'); xlabel('n'); ylabel('h[n]'); -``` - -``` ->> subplot(2,1,2); -``` - ->> plot(samples,H\_d(samples),'k.',Omega,H\_d(Omega),'k:',Omega,abs(H),'k'); - -``` ->> axis([0 2*pi -0.1 1.6]); xlabel('\Omega'); ylabel('|H(\Omega)|'); -``` - ->> legend('Samples','Desired','Actual','Location','North'); - -As shown in Fig. 9.21, the filter's frequency response intersects the desired response at the sampled values of *Hd*(). The overall response, however, has significant ripple between sample points that renders the filter practically useless. Increasing the filter length does not alleviate the ripple problems. Figure 9.22 shows the case *N* = 41. - -To understand the poor behavior of filters designed with CH9MP1, remember that the impulse response of an ideal lowpass filter is a sinc function with the peak centered at zero. Thought of another way, the peak of the sinc is centered at *n* = 0 because the phase of *Hd*() is zero. Constrained to be causal, the impulse response of the designed filter still has a peak at *n* = 0 but cannot include values for negative *n*. As a result, the sinc function is split in an unnatural way with sharp discontinuities on both ends of *h*[*n*]. Sharp discontinuities in the time domain appear as high-frequency oscillations in the frequency domain, which is why *H*() has significant ripple. - -To improve the filter behavior, the peak of the sinc is moved to *n* = (*N* − 1)/2, the center of the length-*N* filter response. In this way, the peak is not split, no large discontinuities are present, and frequency response ripple is consequently reduced. From DFT properties, a cyclic shift of (*N* − 1)/2 in the time domain requires a scale factor of *e*−*j*(*N*−1)/2 in the frequency - -**Figure 9.21** Length-21 FIR lowpass filter using zero phase. - -**Figure 9.22** Length-41 FIR lowpass filter using zero phase. - -domain.† Notice that the scale factor *e*−*j*(*N*−1)/2 affects only phase, not magnitude, and results in a linear phase filter. Program CH9MP2 implements the procedure. - -``` -function [h] = CH9MP2(N,H_d); -% CH9MP2.m : Chapter 9, MATLAB Program 2 -% Function M-file designs a length-N FIR filter by sampling the desired -% magnitude response H_d. Phase is defined to shift h[n] by (N-1)/2. -% INPUTS: N = desired FIR filter length -% H_d = anonymous function that defines the desired magnitude response -% OUTPUTS: h = impulse response (FIR filter coefficients) -% Create N equally spaced frequency samples and use to sample H_d: -Omega = linspace(0,2*pi*(1-1/N),N)'; H = H_d(Omega); -% Define phase to shift h[n] by (N-1)/2: -H = H.*exp(-j*Omega*((N-1)/2)); -H(fix(N/2)+2:N,1) = H(fix(N/2)+2:N,1)*((-1)^(N-1)); -h = real(ifft(H)); -``` - -† Technically, the shift property requires (*N* −1)/2 to be an integer, which occurs only for odd-length filters. The next-to-last line of program CH9MP2 implements a correction factor, of sorts, required to accommodate the fractional shifts desired for even-length filters. The mathematical derivation of this correction is nontrivial and is not included here. Those hesitant to use this correction factor have an alternative: simply round (*N* − 1)/2 to the nearest integer. Although the rounded shift is slightly off-center for even-length filters, there is usually little or no appreciable difference in the characteristics of the filter. Even so, true centering is desirable because the resulting impulse response is symmetric, which can reduce by half the number of multiplies required to implement the filter. - -**Figure 9.23** Length-21 FIR lowpass filter using linear phase. - -Figure 9.23 shows the results for the *N* = 21 case using CH9MP2 to compute *h*[*n*]. As hoped, the impulse response looks like a sinc function with the peak centered at *n* = 10. Additionally, the frequency response ripple is greatly reduced. With CH9MP2, increasing *N* improves the quality of the filter, as shown in Fig. 9.24 for the case *N* = 41. While the magnitude response is needed to establish the general shape of the filter response, it is the proper selection of phase that ensures the acceptability of the filter's behavior. - -To illustrate the flexibility of the design method, consider a bandpass filter with passband (π/4 < || < π/2). - -``` ->> H_d = @(Omega) (mod(Omega,2*pi)>pi/4)&(mod(Omega,2*pi)> (mod(Omega,2*pi)>3*pi/2)&(mod(Omega,2*pi)<7*pi/4); -``` - -Figure 9.25 shows the results for *N* = 50. Notice that this even-length filter uses a fractional shift and is symmetric about *n* = 24.5. - -Although FIR filter design by means of frequency sampling is very flexible, it is not always appropriate. Extreme care is needed for filters, such as digital differentiators and Hilbert transformers, that require special phase characteristics for proper operation. Additionally, if frequency samples occur near jump discontinuities of *Hd*(), rounding errors may, in rare cases, disrupt the desired symmetry of the sampled magnitude response. Such cases are corrected by slightly adjusting the location of problematic jump discontinuities or by changing the value of *N*. - -**Figure 9.24** Length-41 FIR lowpass filter using linear phase. - -**Figure 9.25** Length-50 FIR bandpass filter using linear phase. - -## **[9.8 SUMMARY](#page-14-0)** - -This chapter deals with the analysis and processing of discrete-time signals. For analysis, our approach is parallel to that used in continuous-time signals. We first represent a periodic *x*[*n*] as a Fourier series formed by a discrete-time exponential and its harmonics. Later we extend this representation to an aperiodic signal *x*[*n*] by considering *x*[*n*] to be a limiting case of a periodic signal with period approaching infinity. - -Periodic signals are represented by discrete-time Fourier series (DTFS); aperiodic signals are represented by the discrete-time Fourier integral. The development, although similar to that of continuous-time signals, also reveals some significant differences. The basic difference in the two cases arises because a continuous-time exponential *ej*ω*t* has a unique waveform for every value of ω in the range −∞ to ∞. In contrast, a discrete-time exponential *ejn* has a unique waveform only for values of in a continuous interval of 2π. Therefore, if 0 is the fundamental frequency, then at most 2π/0 exponentials in the Fourier series are independent. Consequently, the discrete-time exponential Fourier series has only *N*0 = 2π/0 terms. - -The discrete-time Fourier transform (DTFT) of an aperiodic signal is a continuous function of and is periodic with period 2π. We can synthesize *x*[*n*] from spectral components of *X*() in any band of width 2π. In a basic sense, the DTFT has a finite spectral width of 2π, which makes it bandlimited to π radians. - -Linear, time-invariant, discrete-time (LTID) systems can be analyzed by means of the DTFT if the input signals are DTF-transformable and if the system is stable. Analysis of unstable (or marginally stable) systems and/or exponentially growing inputs can be handled by the *z*-transform, which is a generalized DTFT. The relationship of the DTFT to the *z*-transform is similar to that of the Fourier transform to the Laplace transform. Whereas the *z*-transform is superior to the DTFT for analysis of LTID systems, the DTFT is preferable in signal analysis. - -If *H*() is the DTFT of the system's impulse response *h*[*n*], then |*H*()| is the amplitude response, and *H*() is the phase response of the system. Moreover, if *X*() and *Y*() are the DTFTs of the input *x*[*n*] and the corresponding output *y*[*n*], then *Y*() = *H*()*X*(). Therefore, the output spectrum is the product of the input spectrum and the system's frequency response. - -Because of the similarity between the DFT and DTFT relationships, numerical computations of the DTFT of finite-length signals can be handled by using the DFT and the FFT, introduced in Secs. 8.5 and 8.6. For signals of infinite length, we use a window of suitable length to truncate the signal so that the final results are within a given error tolerance. - -### **[REFERENCE](#page-14-0)** - -1. Mitra, S. K. *Digital Signal Processing: A Computer-Based Approach,* 2nd ed. McGraw-Hill, New York, 2001. - -## **[PROBLEMS](#page-14-0)** - -**9.1-1** Find the discrete-time Fourier series (DTFS) and sketch their spectra |*Dr*| and *Dr* for 0≤*r*≤*N*0−1 for the following periodic signal: - -*x*[*n*] = 4 cos 2.4π*n*+2 sin 3.2π*n* - -- **9.1-2** Repeat Prob. 9.1-1 for *x*[*n*] =cos 2.2π*n*cos 3.3π*n*. -- **9.1-3** Repeat Prob. 9.1-1 for *x*[*n*]=2 cos 3.2π(*n*−3). -- **9.1-4** Determine and sketch the DTFS spectrum *Dr* of a 7-periodic signal *x*[*n*] that over 0 ≤ *n* ≤ 6 is given by - -$$ -[0, 1, -2, 3, -4, 5, -6] -$$ - -How does the spectrum *Dr* change if *x*[*n*] is time reversed? - -- **9.1-5** Find the discrete-time Fourier series and the corresponding amplitude and phase spectra for the *x*[*n*] shown in Fig. P9.1-5. -- **9.1-6** Repeat Prob. 9.1-5 for the *x*[*n*] depicted in Fig. P9.1-6. -- **9.1-7** Repeat Prob. 9.1-5 for the *x*[*n*] illustrated in Fig. P9.1-7. -- **9.1-8** An *N*0-periodic signal *x*[*n*] is represented by its DTFS, as in Eq. (9.3). Prove Parseval's theorem - -**Figure P9.1-5** - -**Figure P9.1-6** - -**Figure P9.1-7** - -2 - -(for the DTFS), which states that - -$$ -\frac{1}{N_0}\sum_{n=\langle N_0\rangle} |x[n]|^2 = \sum_{r=\langle N_0\rangle} |\mathcal{D}_r| -$$ - -In the text [Eq. (9.36)], we obtain Parseval's theorem for the DTFT. [*Hint:* If *w* is complex, then |*w*| 2 = *ww*∗ and use Eq. (8.15).] - -- **9.1-9** Answer yes or no, and justify your answers with an appropriate example or proof. - - (a) Is a sum of aperiodic discrete-time sequences ever periodic? - - (b) Is a sum of periodic discrete-time sequences ever aperiodic? -- **9.2-1** Show that for a real *x*[*n*], Eq. (9.18) can be expressed as - -$$ -x[n] = \frac{1}{\pi} \int_0^{\pi} |X(\Omega)| \cos(\Omega n + \angle X(\Omega)) d\Omega -$$ - -This is the trigonometric form of the DTFT. - -**9.2-2** A signal *x*[*n*] can be expressed as the sum of even and odd components (Sec. 1.5-1): - -$$ -x[n] = x_e[n] + x_o[n] -$$ - -(a) If *x*[*n*] ⇐⇒ *X*(), show that for real *x*[*n*], - -$$ -x_e[n] \Longleftrightarrow \operatorname{Re}[X(\Omega)] -$$ - -and - -$$ -x_o[n] \Longleftrightarrow j \operatorname{Im}[X(\Omega)] -$$ - -- (b) Verify these results by finding the DTFT of the even and odd components of the signal (0.8)*nu*[*n*]. -- **9.2-3** For the following signals, find the DTFT directly, using the definition in Eq. (9.19). Assume |γ | < 1. - - (a) δ[*n*] - -(b) -$$ -\delta[n-k] -$$ - -- (c) γ *nu*[*n*−1] -- (d) γ *nu*[*n*+1] -- (e) (−γ )*nu*[*n*] -- (f) γ |*n*| -- **9.2-4** Use Eq. (9.18) to find the inverse DTFT for the following spectra, given only over the interval || ≤ π. Assume *c* and 0 < π. (a) *ejk* integer *k* - - (b) cos*k* integer *k* - - (c) cos2(/2) - -(d) -$$ -\Delta \left( \frac{1}{2\Omega_c} \right) -$$ - -- (e) 2πδ(−0) -- (f) π[δ(−0) +δ(+0)] -- **9.2-5** (a) Determine and plot the DTFT *X*() of the triangular signal *x*[*n*] shown in Fig. P9.2-5. - - (b) Using Ex. 9.5 as a guide, use MATLAB and the FFT to validate the DTFT calculations and plot of part (a). - -**Figure P9.2-5** - -- **9.2-6** Using Eq. (9.18), show that the inverse DTFT of rect((−π/4)/π ) is 0.5 sinc(π*n*/2) *ej*π*n*/4. -- **9.2-7** Using Eq. (9.19), find the DTFT of the signals *x*[*n*] in Fig. P9.2-7. -- **9.2-8** Using Eq. (9.19), find the DTFT of the signals depicted in Fig. P9.2-8. -- **9.2-9** Use Eq. (9.18) to find the inverse DTFT of the spectra (shown only for || ≤ π) in Fig. P9.2-9. - -**Figure P9.2-9** - -**9.2-10** Use Eq. (9.18) to find the inverse DTFT of the spectra (shown only for || ≤ π) in Fig. P9.2-10. - -**9.2-11** Find the DTFT for the signals shown in Fig. P9.2-11. - -**Figure P9.2-10** - -(c) - -(d) - -**Figure P9.2-11** - -- **9.2-12** Find the inverse DTFT of *X*() (shown only for ||≤π) for the spectra illustrated in Fig. P9.2-12. [*Hint: X*() = |*X*()|*ej X*(). This problem illustrates how different phase spectra (both with the same amplitude spectrum) represent entirely different signals.] -- **9.2-13** (a) Show that time-expanded signal *xe*[*n*] in Eq. (3.2) can also be expressed as - -$$ -x_e[n] = \sum_{k=-\infty}^{\infty} x[k]\delta[n - Lk] -$$ - -- (b) Find the DTFT of *xe*[*n*] by finding the DTFT of the right-hand side of the equation in part (a). -- (c) Use the result in part (b) and Table 9.1 to find the DTFT of *z*[*n*], shown in Fig. P9.2-13. -- **9.2-14** (a) A glance at Eq. (9.18) shows that the inverse DTFT equation is identical to the inverse (continuous-time) Fourier transform Eq. (7.10) for a signal *x*(*t*) bandlimited to π rad/s. Hence, we should be able to use the continuous-time Fourier transform - -Table 7.1 to find DTFT pairs that correspond to continuous-time transform pairs for bandlimited signals. Use this fact to derive DTFT pairs 8, 9, 11, 12, 13, and 14 in Table 9.1 by means of the appropriate pairs in Table 7.1. - -- (b) Can this method be used to derive pairs 2, 3, 4, 5, 6, 7, 10, 15, and 16 in Table 9.1? Justify your answer with specific reason(s). -- **9.2-15** Are the following frequency-domain signals valid DTFT's? Answer yes or no, and justify your answers. - - (a) *X*() = +π - - (b) *X*() = *j*+π - - (c) *X*() = sin(10) - - (d) *X*() = sin(/10) - - (e) *X*() = δ() -- **9.3-1** Using only pairs 2 and 5 (Table 9.1) and the time-shifting property of Eq. (9.31), find the DTFT of the following signals, assuming |*a*| < 1. - -- (a) *u*[*n*] −*u*[*n*−9] -- (b) *an*−*mu*[*n*−*m*] -- (c) *an*−3(*u*[*n*] −*u*[*n*−10]) - -1 - -(b) - -**Figure P9.2-13** - -- (d) *an*−*mu*[*n*] -- (e) *anu*[*n*−*m*] -- (f) (*n*−*m*)*an*−*mu*[*n*−*m*] -- (g) (*n*−*m*)*anu*[*n*] -- (h) *nan*−*mu*[*n*−*m*] -- **9.3-2** The triangular pulse *x*[*n*] shown in Fig. P9.3-2a is given by - -$$ -X(\Omega) = \frac{4e^{j6\Omega} - 5e^{j5\Omega} + e^{j\Omega}}{(e^{j\Omega} - 1)^2} -$$ - -Use this information and the DTFT properties to find the DTFT of the signals *x*1[*n*], *x*2[*n*], *x*3[*n*], and *x*4[*n*] shown in Figs. P9.3-2b, P9.3-2c, P9.3-2d, and P9.3-2e, respectively. - -**9.3-3** Suppose signal *x*[*n*] = sinc2(π*n*/2) modulates a carrier cos(c*n*) to produce signal *y*[*n*] = *x*[*n*] cos(c*n*). Find and sketch the DTFT of: - -(a) *x*[*n*] - -- (b) *y*[*n*] for c = π/2 -- (c) *y*[*n*] for c = 3π/4 -- (d) *y*[*n*] for c = π - -**9.3-4** Show that periodic convolution *X*()-∗ *Y*() = 2π*X*() if - -$$ -X(\Omega) = \sum_{k=0}^{4} a_k e^{-jk\Omega} -$$ - -and - -$$ -Y(\Omega) = \frac{\sin(5\Omega/2)}{\sin(\Omega/2)} e^{-j2\Omega} -$$ - -where *ak* is a set of arbitrary constants. - -- **9.3-5** Using only pair 2 (Table 9.1) and properties of DTFT, find the DTFT of the following signals, assuming |*a*| < 1 and 0 < π. - - (a) *an* cos0*nu*[*n*] - - (b) *n*2*anu*[*n*] - - (c) (*n*−*k*)*a*2*nu*[*n*−*m*] -- **9.3-6** Use pair 10 in Table 9.1, and suitable properties of the DTFT, to derive pairs 11, 12, 13, 14, 15, and 16. -- **9.3-7** Use the time-shifting property to show that - -$$ -x[n+k]+x[n-k] \Longleftrightarrow 2X(\Omega)\cos k\Omega -$$ - -(e) - -**Figure P9.3-2** - -**Figure P9.3-7** - -Use this result to find the DTFT of the signals shown in Fig. P9.3-7. - -**9.3-8** Use the time-shifting property to show that - -$$ -x[n+k] - x[n-k] \Longleftrightarrow 2jX(\Omega) \sin k\Omega -$$ - -Use this result to find the DTFT of the signal shown in Fig. P9.3-8. - -**9.3-9** Suppose signal *x*[*n*] has spectrum *X*() that is bandlimited to π/2 rad/sample. Next, define signal *y*[*n*] as - -$$ -y[n] = \begin{cases} x[n] & n \text{ even} \\ 0 & n \text{ odd} \end{cases} -$$ - -Determine the spectrum of *Y*() in terms of *X*(). Sketch *Y*() if, over −π ≤ ≤ π, - -$$ -Y(\Omega) = \begin{cases} |2\Omega/\pi| & -\pi/2 \le \Omega \le \pi/2\\ 0 & \text{otherwise} \end{cases} -$$ - -**9.3-10** Repeat Prob. 9.3-9 if *y*[*n*] is instead defined as - -$$ -y[n] = \begin{cases} x[n] & n \text{ odd} \\ 0 & n \text{ even} \end{cases} -$$ - -- **9.3-11** Using only pair 2 in Table 9.1 and the convolution property, find the inverse DTFT of *X*() = *e*2*j*/(*ej* −γ )2. -- **9.3-12** In Table 9.1, you are given pair 1. From this information and using suitable properties of the DTFT, derive pairs 2, 3, 4, 5, 6, and 7 of Table 9.1. For example, starting with pair 1, derive pair 2. From pair 2, use suitable properties - -of the DTFT to derive pair 3. From pairs 2 and 3, derive pair 4, and so on. - -- **9.3-13** From the pair *ej*(0/2)*n* ⇐⇒ 2πδ( − (0/2)) over the fundamental band, and the frequency-convolution property, find the DTFT of *ej*0*n*. Assume 0 <π/2. -- **9.3-14** From the definition and properties of the DTFT, show that - -(a) -$$ -\sum_{n=-\infty}^{\infty} \operatorname{sinc}(\Omega_c n) = \frac{\pi}{\Omega_c} \quad \Omega_c < \pi -$$ - -\n(b) -$$ -\sum_{n=-\infty}^{\infty} (-1)^n \operatorname{sinc}(\Omega_c n) = 0 \quad \Omega_c < \pi -$$ - -\n(c) -$$ -\sum_{n=-\infty}^{\infty} \operatorname{sinc}^2(\Omega_c n) = \frac{\pi}{\Omega_c} \quad \Omega_c < \pi/2 -$$ - -\n(d) -$$ -\sum_{n=-\infty}^{\infty} (-1)^n \operatorname{sinc}^2(\Omega_c n) = 0 \quad \Omega_c < \pi/2 -$$ - -\n(e) -$$ -\int_{-\pi}^{\pi} \frac{\sin(M\Omega/2)}{\sin(\Omega/2)} = 2\pi \quad \text{odd } M -$$ - -$$ -\int_{-\pi}^{\infty} \frac{\sin(\alpha z/2)}{\sin(\alpha z/2)} \sin(\alpha z/2) -$$ - -(f) -$$ -\sum_{n=-\infty}^{\infty} |\sin(\alpha z/2)|^4 = 2\pi/3\Omega_c \quad \Omega_c < \pi/2 -$$ - -**9.3-15** Show that the energy of signal *xc*(*t*) specified in Eq. (9.41) is identical to *T* times the energy of the discrete-time signal *x*[*n*], assuming *xc*(*t*) is bandlimited to *B* ≤ 1/2*T* Hz. [*Hint:* Recall that - -$$ -\int_{-\infty}^{\infty} \operatorname{sinc} [\pi(t-m)] \operatorname{sinc} [\pi(t-n)] dt -$$ -$$ -= \begin{cases} 0 & m \neq n \\ 1 & m = n \end{cases} -$$ - -That is, sinc functions are orthogonal.] - -**9.4-1** Use the DTFT method to find the zero-state response *y*[*n*] of a causal system with frequency response - -$$ -H(\Omega) = \frac{e^{i\Omega} + 0.32}{e^{i2\Omega} + e^{i\Omega} + 0.16} -$$ - -and the input *x*[*n*] = (−0.5)*nu*[*n*]. - -**9.4-2** Repeat Prob. 9.4-1 for - -$$ -H(\Omega) = \frac{e^{i\Omega} + 0.32}{e^{i2\Omega} + e^{i\Omega} + 0.16} -$$ - -and input *x*[*n*] = *u*[*n*]. - -**9.4-3** Repeat Prob. 9.4-1 for - -$$ -H(\Omega) = \frac{e^{i\Omega}}{e^{i\Omega} - 0.5} -$$ - -and - -$$ -x[n] = 0.8nu[n] + 2(2)nu[-(n+1)] -$$ - -**9.4-4** Determine and sketch the magnitude and phase response for an LTID system specified by the equation - -$$ -y[n] + 0.5y[n-1] = x[n] - 0.9x[n-1] -$$ - -Determine the system output *y*[*n*] for the input *x*[*n*] = cos( π*n* 3 +0.5). - -**9.4-5** Repeat Prob. 9.4-4 if the LTID system is instead specified by the equation - -$$ -y[n] - 0.5y[n-1] = x[n] + 0.9x[n-1] -$$ - -**9.4-6** An accumulator system has the property that an input *x*[*n*] results in the output - -$$ -y[n] = \sum_{k=-\infty}^{n} x[k] -$$ - -- (a) Find the unit impulse response *h*[*n*] and the frequency response *H*() for the accumulator. -- (b) Use the results of part (a) to find the DTFT of *u*[*n*]. -- **9.4-7** A noncausal 7-point moving average is described by the equation - -$$ -y[n] = \frac{1}{7} \sum_{k=-3}^{3} x[n-k] -$$ - -- (a) Find and sketch the magnitude and phase responses of the system. -- (b) How can this system be made causal? Plot the magnitude and phase responses of the causal system, and comment on any differences from part (a). -- **9.4-8** An LTID system frequency response over || ≤ π is - -$$ -H(\Omega) = \text{rect}\bigg(\frac{\Omega}{\pi}\bigg)e^{-j2\Omega} -$$ - -Find the output *y*[*n*] of this system, if the input *x*[*n*] is given by - -- (a) sinc (π*n*/2) -- (b) sinc(π*n*) -- (c) sinc2 (π*n*/4) -- **9.4-9** (a) If *x*[*n*] ⇐⇒ *X*(), then, show that (−1)*nx*[*n*] ⇐⇒ *X*(−π ). - - (b) Sketch γ *nu*[*n*] and (−γ )*nu*[*n*] for γ = 0.8; see the spectra for γ *nu*[*n*] in Figs. 9.5b and 9.5c. From these spectra, sketch the spectra for (−γ )*nu*[*n*]. - - (c) An ideal lowpass filter of cutoff frequency *c* is specified by the frequency response *H*() = rect(/2*c*). Find its impulse response *h*[*n*]. Find the frequency response of a filter whose impulse response is (−1)*nh*[*n*]. Sketch the frequency response of this filter. What kind of filter is this? -- **9.4-10** An analog differentiator *y*(*t*) = *d dt x*(*t*) can be approximated using a backward difference system described as - -$$ -y[n] = \frac{x[n] - x[n-1]}{T} -$$ - -Find and sketch the magnitude and phase responses of this DT system. For what frequencies does the system most behave as a differentiator? For what frequencies does the system least behave as a differentiator? - -**9.4-11** A filter with impulse response *h*[*n*] is modified as shown in Fig. P9.4-11. Determine the resulting filter impulse response *h*1[*n*]. Find also the resulting filter frequency response *H*1() in terms of the frequency response *H*(). How are *H*() and *H*1() related? - -- **9.4-12** (a) Consider an LTID system *S*1, specified by a difference equation of the form of Eqs. (3.15) or (3.16) or (3.20) in Ch. 3. We construct another system *S*2 by replacing coefficients *ai* (*i*=0, 1, 2,...,*N*) by coefficients (−1)*i ai* and replacing all coefficients *bi* (*i* = 0, 1, 2,...,*N*) with coefficients (−1)*i bi*. How are the frequency responses of the two systems related? - - (b) If *S*1 represents a lowpass filter, what kind of filter is specified by *S*2? - - (c) What type of filter (lowpass, highpass, etc.) is specified by the difference equation - -$$ -y[n] - 0.8y[n-1] = x[n] -$$ - -What kind of filter is specified by the following difference equation? - -$$ -y[n] + 0.8y[n-1] = x[n] -$$ - -**9.4-13** (a) The system shown in Fig. P9.4-13 contains two identical LTID filters with frequency response *H*0() and corresponding impulse response *h*0[*n*]. It is easy to see that the system is linear. Show that this system is also time-invariant. Do this by finding the **Figure P9.4-11** - -response of the system to input δ[*n* − *k*] in terms of *h*0[*n*]. - -- (b) If *H*0() = rect(/2*W*) over the fundamental band, and *c* + *W* ≤ π, find *H*(), the frequency response of this system. What kind of filter is this? -- **9.5-1** Determine the DTFT of *x*[*n*] = sin(0*n*) from the CTFT of *x*c(*t*) = sin(ω0*t*). -- **9.5-2** A CT signal *x*(*t*), bandlimited to 25 kHz, is sampled at 50 kHz to produce - -$$ -x[n] = \delta[n+4] - 2\delta[n+2] + \delta[n+1] - 3\delta[n] - \delta[n-1] - 2\delta[n-2] - \delta[n-4] -$$ - -Determine the CTFT *X*(ω). - -- **9.7-1** This problem uses a matrix-based approach to investigate the computation of the inverse DTFS. - - (a) Implement Eq. (9.3), the inverse DTFS, using a matrix-based approach. - - (b) Compare the execution speed of the matrix-based approach to the IFFT-based approach for input vectors of sizes 10, 100, and 1000. - -**Figure P9.4-13** - -- (c) What is the result of multiplying the DFT matrix **W***N*0 by the inverse DTFS matrix? Discuss your result. -- **9.7-2** A stable, first-order highpass IIR digital filter has transfer function - -$$ -H[z] = \left(\frac{1+\alpha}{2}\right) \left(\frac{1-z^{-1}}{1-\alpha z^{-1}}\right) -$$ - -- (a) Derive an expression relating α to the 3 dB cutoff frequency *c*. -- (b) Test your expression from part (a) in the following manner. First, compute α to achieve a 3 dB cutoff frequency of 1 kHz, assuming a sampling rate of *Fs* = 5 kHz. Determine a difference equation description of the system, and verify that the system is stable. Next, compute and plot the magnitude response of the resulting filter. Verify that the filter is highpass and has the correct cutoff frequency. -- (c) Holding α constant, what happens to the cutoff frequency *c* as *Fs* is increased to 50 kHz? What happens to the cutoff frequency *fc* as *Fs* is increased to 50 kHz? -- (d) Is there a well-behaved inverse filter to *H*[*z*]? Explain. -- (e) Determine α for *c* = π/2. Comment on the resulting filter, particularly *h*[*n*]. -- **9.7-3** Using the frequency sampling method, design a length-35 linear phase FIR highstop filter that has cutoff frequency c = 2π/3. Plot the resulting filter's impulse response *h*[*n*] and magnitude response |*H*()|. -- **9.7-4** Using the frequency-sampling method, design a length-71 linear phase FIR bandstop filter that has stopband (π/3 < || < π/2). Plot the resulting filter's impulse response *h*[*n*] and magnitude response |*H*()|. -- **9.7-5** Figure P9.7-5 provides the desired magnitude response |*H*()| of a real filter. Mathematically, - -$$ -|H(\Omega)| = \begin{cases} 2\frac{4\Omega}{\pi} & 0 \leq \Omega < \frac{\pi}{4} \\ 2 - \frac{4\Omega}{\pi} & \frac{\pi}{4} \leq \Omega < \frac{\pi}{2} \\ 0 & \frac{\pi}{2} \leq \Omega \leq \pi \end{cases} -$$ - -Since the digital filter is real, |*H*()|=|*H*(−)| and |*H*()|=|*H*(+2π )| for all . - -- (a) Can a realizable filter have this exact magnitude response? Explain your answer. -- (b) Use the frequency-sampling method to design an FIR filter with this magnitude response (or a reasonable approximation). Use MATLAB to plot the magnitude response of your filter. - -- **9.7-6** A real FIR comb filter is needed that has magnitude response |*H*()|=[0, 3, 0, 3, 0, 3, 0, 3] for = [0,π/4,π/2, 3π/4,π, 5π/4, 3π/2, 7π/4], respectively. Provide the impulse response *h*[*n*] of a filter that accomplishes these specifications. -- **9.7-7** A permutation matrix **P** has a single one in each row and column with the remaining elements all zero. Permutation matrices are useful for reordering the elements of a vector; the operation **Px** reorders the elements of a column vector **x** based on the form of **P**. - - (a) Fully describe an *N*0 × *N*0 permutation matrix named **R***N*0 that reverses the order of the elements of a column vector **x**. - - (b) Given DFT matrix **W***N*0 , verify that (**W***N*0 )(**W***N*0 ) = **W**2 *N*0 produces a scaled permutation matrix. How does **W**2 *N*0 **x** reorder the elements of **x**? - - (c) What is the result of (**W**2 *N*0 )(**W**2 *N*0 )**x** = **W**4 *N*0 **x**? - -# **[STATE-SPACE](#page-14-0) ANALYSIS** - -In Sec. 1.10, basic notions of *state variables* were introduced. In this chapter, we shall discuss state variables in more depth. - -Most of this book deals with an external (input–output) description of systems. As noted in Ch. 1, such a description may be inadequate in some cases, and we need a systematic way of finding a system's *internal description*. State-space analysis of systems meets this need. In this method, we first select a set of key variables, called the *state variables,* in the system. Every possible signal or variable in the system at any instant *t* can be expressed in terms of the state variables and the input(s) at that instant *t*. If we know all the state variables as a function of *t*, we can determine every possible signal or variable in the system at any instant with a relatively simple relationship. The system description in this method consists of two parts: - -- 1. A set of equations relating the state variables to the inputs (*the state equation*). -- 2. A set of equations relating outputs to the state variables and the inputs (*the output equation*). - -The analysis procedure, therefore, consists of solving the state equation first, and then solving the output equation. The state-space description is capable of determining every possible system variable (or output) from knowledge of the input and the initial state (conditions) of the system. For this reason, it is an *internal description* of the system. - -By its nature, state variable analysis is eminently suited for multiple-input, multiple-output (MIMO) systems. A single-input, single output (SISO) system is a special case of MIMO systems. In addition, the state-space techniques are useful for several other reasons, mentioned in Sec. 1.10, and repeated here. - -- 1. The state equations of a system provide a mathematical model of great generality that can describe not just linear systems, but also nonlinear systems; not just time-invariant systems, but also time-varying parameter systems; not just SISO systems, but also MIMO systems. Indeed, state equations are ideally suited for analysis, synthesis, and optimization of MIMO systems. -- 2. Compact matrix notation along with powerful techniques of linear algebra greatly facilitates complex manipulations. Without such features, many important results of - -modern system theory would have been difficult to obtain. State equations can yield a great deal of information about a system even when they are not solved explicitly. - -- 3. State equations lend themselves readily to digital computer simulation of complex systems of high order, with or without nonlinearities, and with multiple inputs and outputs. -- 4. For second-order systems (*N* = 2), a graphical method called *phase-plane analysis* can be used on state equations, whether they are linear or nonlinear. - -## **[10.1 MATHEMATICAL](#page-14-0) PRELIMINARIES** - -This chapter requires some understanding of matrix algebra. Section B.6 introduces basic concepts of matrix algebra, but misses a few needed mathematical concepts, which we present next. - -### **[10.1-1 Derivatives and Integrals of a Matrix](#page-14-0)** - -Elements of a matrix need not be constants; they may be functions of a variable. For example, if - -$$ -\mathbf{A} = \begin{bmatrix} e^{-2t} & \sin t \\ e^t & e^{-t} + e^{-2t} \end{bmatrix} -$$ - (10.1) - -then the matrix elements are functions of *t*. Here, it is helpful to denote **A** by **A**(*t*). Next, we define the derivative and integral of **A**(*t*). - -The derivative of a matrix **A**(*t*) (with respect to *t*) is defined as a matrix whose *ij*th element is the derivative (with respect to *t*) of the *ij*th element of the matrix **A**. Thus, if - -$$ -\mathbf{A}(t) = [a_{ij}(t)]_{m \times n} -$$ - -then - -$$ -\frac{d}{dt}[\mathbf{A}(t)] = \left[\frac{d}{dt}a_{ij}(t)\right]_{m \times n} \quad \text{or} \quad \dot{\mathbf{A}}(t) = [\dot{a}_{ij}(t)]_{m \times n} -$$ - -Thus, the derivative of the matrix in Eq. (10.1) is given by - -$$ -\dot{\mathbf{A}}(t) = \begin{bmatrix} -2e^{-2t} & \cos t \\ e^t & -e^{-t} - 2e^{-2t} \end{bmatrix} -$$ - -Similarly, we define the integral of **A**(*t*) (with respect to *t*) as a matrix whose *ij*th element is the integral (with respect to *t*) of the *ij*th element of the matrix **A**: - -$$ -\int \mathbf{A}(t) dt = \left( \int a_{ij}(t) dt \right)_{m \times n} -$$ - -Thus, for the matrix **A** in Eq. (10.1), we have - -$$ -\int \mathbf{A}(t) dt = \begin{bmatrix} \int e^{-2t} dt & \int \sin dt \\ \int e^t dt & \int (e^{-t} + 2e^{-2t}) dt \end{bmatrix} -$$ - -#### 910 CHAPTER 10 STATE-SPACE ANALYSIS - -Since differentiation is a linear operation, it is easy to show that - -$$ -\frac{d}{dt}(\mathbf{A} + \mathbf{B}) = \frac{d\mathbf{A}}{dt} + \frac{d\mathbf{B}}{dt} \quad \text{and} \quad \frac{d}{dt}(c\mathbf{A}) = c\frac{d\mathbf{A}}{dt} -$$ - -The derivative of a matrix product is given as - -$$ -\frac{d}{dt}(\mathbf{AB}) = \frac{d\mathbf{A}}{dt}\mathbf{B} + \mathbf{A}\frac{d\mathbf{B}}{dt} = \dot{\mathbf{A}}\mathbf{B} + \mathbf{A}\dot{\mathbf{B}}\tag{10.2} -$$ - -We can prove Eq. (10.2) as follows. Let **A** be an *m*×*n* matrix and **B** an *n*×*p* matrix. Then, if - -**C** = **AB** - -from Eq. (B.33), we have - -$$ -c_{ik} = \sum_{j=1}^{n} a_{ij} b_{jk} -$$ - -and - -$$ -\dot{c}_{ik} = \underbrace{\sum_{j=1}^{n} \dot{a}_{ij} b_{jk}}_{d_{ik}} + \underbrace{\sum_{j=1}^{n} a_{ij} \dot{b}_{jk}}_{e_{ik}} -$$ - or -$$ -\dot{c}_{ik} = d_{ik} + e_{ik} -$$ - (10.3) - -Equation (10.3) along with the multiplication rule clearly indicates that *dik* is the *ik*th element of matrix **AB**˙ and *eik* is the *ik*th element of matrix **AB**˙ . Equation (10.2) then follows. - -If we let **B** = **A**−1 in Eq. (10.2), we obtain - -$$ -\frac{d}{dt}(\mathbf{A}\mathbf{A}^{-1}) = \frac{d\mathbf{A}}{dt}\mathbf{A}^{-1} + \mathbf{A}\frac{d}{dt}\mathbf{A}^{-1} -$$ - -But since - -$$ -\frac{d}{dt}(\mathbf{A}\mathbf{A}^{-1}) = \frac{d}{dt}\mathbf{I} = 0 -$$ - -we have - -$$ -\frac{d}{dt}(\mathbf{A}^{-1}) = -\mathbf{A}^{-1}\frac{d\mathbf{A}}{dt}\mathbf{A}^{-1} -$$ - -### **[10.1-2 The Characteristic Equation of a Matrix:](#page-14-0) The Cayley–Hamilton Theorem** - -For an (*n*×*n*) square matrix **A**, any vector **x** (**x** = 0) that satisfies the equation - -$$ -\mathbf{A}\mathbf{x} = \lambda \mathbf{x} \tag{10.4} -$$ - -is an *eigenvector* (or *characteristic vector*), and λ is the corresponding *eigenvalue* (or *characteristic value*) of **A**. Equation (10.4) can be expressed as - -$$ -(\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = 0 \qquad \text{or} \qquad (\lambda \mathbf{I} - \mathbf{A})\mathbf{x} = 0 -$$ - -The solution for this set of homogeneous equations exists if and only if - -$$ -|\lambda \mathbf{I} - \mathbf{A}| = \begin{vmatrix} \lambda - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \lambda - a_{22} & \cdots & -a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & \lambda - a_{nn} \end{vmatrix} = 0 -$$ - (10.5) - -Equation (10.5) is known as the *characteristic equation* of matrix **A** and can be expressed as - -$$ -Q(\lambda) = |\lambda \mathbf{I} - \mathbf{A}| = \lambda^n + a_{n-1}\lambda^{n-1} + \dots + a_1\lambda + a_0\lambda^0 = 0 -$$ - (10.6) - -*Q*(λ) is called the *characteristic polynomial* of matrix **A**. The *n* zeros of the characteristic polynomial are the eigenvalues of **A** and, corresponding to each eigenvalue, there is an eigenvector that satisfies Eq. (10.4). - -The *Cayley–Hamilton theorem* states that every *n*×*n* matrix **A** satisfies its own characteristic equation. In other words, Eq. (10.6) is valid if λ is replaced by **A**: - -$$ -\mathbf{Q}(\mathbf{A}) = \mathbf{A}^n + a_{n-1}\mathbf{A}^{n-1} + \dots + a_1\mathbf{A} + a_0\mathbf{A}^0 = 0 -$$ - (10.7) - -### FUNCTIONS OF A MATRIX - -We now demonstrate the use of the Cayley–Hamilton theorem [Eq. (10.7)] to evaluate functions of an *n*×*n* square matrix **A**. - -Consider a function *f*(λ) in the form of an infinite power series: - -$$ -f(\lambda) = \alpha_0 + \alpha_1 \lambda + \alpha_2 \lambda_2^2 + \dots = \sum_{i=0}^{\infty} \alpha_i \lambda^i -$$ - (10.8) - -Since λ, being an eigenvalue (characteristic root) of **A**, satisfies the characteristic equation [Eq. (10.6)], we can write - -$$ -\lambda^{n} = -a_{n-1}\lambda^{n-1} - a_{n-2}\lambda^{n-2} - \dots - a_1\lambda - a_0 -$$ - (10.9) - -If we multiply both sides by λ, the left-hand side is λ*n*+1, and the right-hand side contains the terms λ*n*, λ*n*−1, ... , λ. Using Eq. (10.9), we substitute λ*n* in terms of λ*n*−1, λ*n*−2,..., λ so that the highest power on the right-hand side is reduced to *n* − 1. Continuing in this way, we see that λ*n*+*k* can be expressed in terms of λ*n*−1, λ*n*−2,...,λ for any *k*. Hence, the infinite series on the right-hand side of Eq. (10.8) can always be expressed in terms of λ*n*−1, λ*n*−2,...,λ and a constant as - -$$ -f(\lambda) = \beta_0 + \beta_1 \lambda + \beta_2 \lambda^2 + \dots + \beta_{n-1} \lambda^{n-1} -$$ - (10.10) - -If we assume that there are *n* distinct eigenvalues λ1, λ2, ... , λ*n*, then Eq. (10.10) holds for these *n* values of λ. The substitution of these values in Eq. (10.10) yields *n* simultaneous equations - -$$ -\begin{bmatrix} f(\lambda_1) \\ f(\lambda_2) \\ \vdots \\ f(\lambda_n) \end{bmatrix} = \begin{bmatrix} 1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1} \\ 1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{n-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & \lambda_n & \lambda_n^2 & \cdots & \lambda_n^{n-1} \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_{n-1} \end{bmatrix} -$$ - -### 912 CHAPTER 10 STATE-SPACE ANALYSIS - -Solving for the β coefficients yields - -$$ -\begin{bmatrix}\n\beta_0 \\ -\beta_1 \\ -\vdots \\ -\beta_{n-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{n-1} \\ -1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{n-1} \\ -\vdots & \vdots & \vdots & \cdots & \vdots \\ -1 & \lambda_n & \lambda_n^2 & \cdots & \lambda_n^{n-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\nf(\lambda_1) \\ -f(\lambda_2) \\ -\vdots \\ -f(\lambda_n)\n\end{bmatrix} -$$ -\n(10.11) - -Since **A** also satisfies Eq. (10.9), we may advance a similar argument to show that if *f*(**A**) is a function of a square matrix **A** expressed as an infinite power series in **A**, then - -$$ -f(\mathbf{A}) = \alpha_0 \mathbf{I} + \alpha_1 \mathbf{A} + \alpha_2 \mathbf{A}^2 + \cdots = \sum_{i=0}^{\infty} \alpha_i \mathbf{A}^i -$$ - -and, as argued earlier, the right-hand side can be expressed by using terms of power less than or equal to *n*−1, - -$$ -f(\mathbf{A}) = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{n-1} \mathbf{A}^{n-1} = \sum_{i=0}^{n-1} \beta_i \mathbf{A}^i -$$ - (10.12) - -in which the coefficients β*i*s are found from Eq. (10.11). If some of the eigenvalues are repeated (multiple roots), the results are somewhat modified. - -We shall demonstrate the utility of this result with the following two examples. - -### **[10.1-3 Computation of an Exponential and a Power of a Matrix](#page-14-0)** - -Let us compute *e***A***t* defined by - -$$ -e^{\mathbf{A}t} = \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2 t^2}{2!} + \cdots + \frac{\mathbf{A}^n t^n}{n!} + \cdots = \sum_{k=0}^{\infty} \frac{\mathbf{A}^k t^k}{k!} -$$ - -From Eq. (10.12), we can express - -$$ -e^{\mathbf{A}t} = \sum_{i=1}^{n-1} \beta_i(\mathbf{A})^i -$$ - -in which the β*i*s are given by Eq. (10.11), with *f*(λ*i*) = *e*λ*it* . - -### **EXAMPLE 10.1 Computing the Exponential of a Matrix** - -Compute *e***A***t* for the case - -$$ -\mathbf{A} = \left[ \begin{array}{cc} 0 & 1 \\ -2 & -3 \end{array} \right] -$$ - -The characteristic equation is - -$$ -|\lambda \mathbf{I} - \mathbf{A}| = \begin{vmatrix} \lambda & -1 \\ 2 & \lambda + 3 \end{vmatrix} = \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0 -$$ - -Hence, the eigenvalues are λ1 = −1, λ2 = −2, and - -$$ -e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} -$$ - -in which - -$$ -\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & -2 \end{bmatrix}^{-1} \begin{bmatrix} e^{-t} \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^{-t} \\ e^{-2t} \end{bmatrix} = \begin{bmatrix} 2e^{-t} - e^{-2t} \\ e^{-t} - e^{-2t} \end{bmatrix} -$$ - -and - -$$ -e^{\mathbf{A}t} = (2e^{-t} - e^{-2t}) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + (e^{-t} - e^{-2t}) \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} -$$ -$$ -= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} -$$ -(10.13) - -COMPUTATION OF *Ak* As Eq. (10.12) indicates, we can express **A***k* as - -$$ -\mathbf{A}^{k} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \cdots + \beta_{n-1} \mathbf{A}^{n-1} -$$ - -in which the β*i*s are given by Eq. (10.11) with *f*(λ*i*) = λ*k i* . For a completed example of the computation of **A***k* by this method, see Ex. 10.13. - -## **[10.2 INTRODUCTION TO](#page-14-0) STATE SPACE** - -From the discussion in Ch. 1, we know that to determine a system's response(s) at any instant *t*, we need to know the system's inputs during its entire past, from −∞ to *t*. If the inputs are known only for *t* > *t*0, we can still determine the system output(s) for any *t* > *t*0, provided we know certain initial conditions in the system at *t* = *t*0. These initial conditions collectively are called the *initial state* of the system (at *t* = *t*0). - -*The state variables q*1(*t*),*q*2(*t*),...,*qN*(*t*) are the minimum number of system variables such that their initial values at any instant *t*0 are sufficient to determine the behavior of the system for all time *t* ≥ *t*0 when the input(s) to the system is known for *t* ≥ *t*0. This statement implies that an output of a system at any instant is determined completely from a knowledge of the values of the system state and the input at that instant. - -Initial conditions of a system can be specified in many different ways. Consequently, the system state can also be specified in many different ways. This means that state variables are not unique. - -#### 914 CHAPTER 10 STATE-SPACE ANALYSIS - -This discussion is also valid for multiple-input, multiple-output (MIMO) systems, where every possible system output at any instant *t* is determined completely from a knowledge of the system state and the input(s) at the instant *t*. These ideas should become clear from the following example of an *RLC* circuit. - -## **EXAMPLE 10.2 State-Space Description and Output Equations of an** *RLC* **Circuit** - -Find a state-space description of the *RLC* circuit shown in Fig. 10.1. Verify that all possible system outputs at some instant *t* can be determined from knowledge of the system state and the input at that instant *t*. - -**Figure 10.1** Circuit for Ex. 10.2. - -It is known that inductor currents and capacitor voltages in an *RLC* circuit can be used as one possible choice of state variables. For this reason, we shall choose *q*1 (the capacitor voltage) and *q*2 (the inductor current) as our state variables. - -The node equation at the intermediate node is - -$$ -i_3 = i_1 - i_2 - q_2 -$$ - -but *i*3 = 0.2*q*˙1, *i*1 = 2(*x* −*q*1), *i*2 = 3*q*1. Hence, - -$$ -0.2\dot{q}_1 = 2(x - q_1) - 3q_1 - q_2 -$$ - -or - -$$ -\dot{q}_1 = -25q_1 - 5q_2 + 10x -$$ - -This is the first state equation. To obtain the second state equation, we sum the voltages in the extreme right loop formed by *C*, *L*, and the 2 resistor so that they are equal to zero: - -$$ --q_1 + \dot{q}_2 + 2q_2 = 0 -$$ - -$$ -\quad \text{or} \quad -$$ - -Thus, the two state equations are - -$$ -\dot{q}_1 = -25q_1 - 5q_2 + 10x -$$ - -$$ -\dot{q}_2 = q_1 - 2q_2 -$$ - -*q*˙2 = *q*1 −2*q*2 - -Every possible output can now be expressed as a linear combination of *q*1, *q*2, and *x*. From Fig. 10.1, we have - -$$ -v_1 = x - q_1 -$$ - -\n -$$ -i_1 = 2(x - q_1) -$$ - -\n -$$ -v_2 = q_1 -$$ - -\n -$$ -i_2 = 3q_1 -$$ - -\n -$$ -i_3 = i_1 - i_2 - q_2 = 2(x - q_1) - 3q_1 - q_2 = -5q_1 - q_2 + 2x -$$ - -\n -$$ -i_4 = q_2 -$$ - -\n -$$ -v_4 = 2i_4 = 2q_2 -$$ - -\n -$$ -v_3 = q_1 - v_4 = q_1 - 2q_2 -$$ - -This set of equations is known as the *output equation* of the system. It is clear from this set that every possible output at some instant *t* can be determined from knowledge of *q*1(*t*), *q*2(*t*), and *x*(*t*), the system state, and the input at the instant *t*. Once we have solved the state equations to obtain *q*1(*t*) and *q*2(*t*), we can determine every possible output for any given input *x*(*t*). - -For continuous-time systems, the state equations are *N* simultaneous first-order differential equations in *N* state variables *q*1, *q*2, ... , *qN* of the form - -$$ -\dot{q}_i = g_i(q_1, q_2, \dots, q_N, x_1, x_2, \dots, x_j) -$$ - $i = 1, 2, \dots, N$ - -where *x*1, *x*2, ... , *xj* are the *j* system inputs. For a linear system, these equations reduce to a simpler linear form - -$$ -\dot{q}_i = a_{i1}q_1 + a_{i2}q_2 + \dots + a_{iN}q_N + b_{i1}x_1 + b_{i2}x_2 + \dots + b_{ij}x_j \qquad i = 1, 2, \dots, N \qquad (10.14) -$$ - -If there are *k* outputs *y*1, *y*2,..., *yk*, the *k* output equations are of the form - -$$ -y_m = c_{m1}q_1 + c_{m2}q_2 + \dots + c_{mN}q_N + d_{m1}x_1 + d_{m2}x_2 + \dots + d_{mj}x_j \qquad m = 1, 2, \dots, k \quad (10.15) -$$ - -The *N* simultaneous first-order state equations are also known as the *normal-form* equations. - -These equations can be written more conveniently in matrix form: - -$$ -\begin{bmatrix}\n\dot{q}_1 \\ -\dot{q}_2 \\ -\vdots \\ -\dot{q}_N\n\end{bmatrix} = \begin{bmatrix}\na_{11} & a_{12} & \cdots & a_{1N} \\ -a_{21} & a_{22} & \cdots & a_{2N} \\ -\vdots & \vdots & \cdots & \vdots \\ -a_{N1} & a_{N2} & \cdots & a_{NN}\n\end{bmatrix} \begin{bmatrix}\nq_1 \\ -q_2 \\ -\vdots \\ -q_N\n\end{bmatrix} + \begin{bmatrix}\nb_{11} & b_{12} & \cdots & b_{1j} \\ -b_{21} & b_{22} & \cdots & b_{2j} \\ -\vdots & \vdots & \ddots & \vdots \\ -b_{N1} & b_{N2} & \cdots & b_{Nj}\n\end{bmatrix} \begin{bmatrix}\nx_1 \\ -x_2 \\ -\vdots \\ -x_j\n\end{bmatrix} -$$ -\n -$$ -\begin{bmatrix}\ny_1 \\ -y_2 \\ -\vdots \\ -y_k\n\end{bmatrix} = \begin{bmatrix}\nc_{11} & c_{12} & \cdots & c_{1N} \\ -c_{21} & c_{22} & \cdots & c_{2N} \\ -\vdots & \vdots & \ddots & \vdots \\ -c_{k1} & c_{k2} & \cdots & c_{kN}\n\end{bmatrix} \begin{bmatrix}\nq_1 \\ -q_2 \\ -\vdots \\ -q_N\n\end{bmatrix} + \begin{bmatrix}\nd_{11} & d_{12} & \cdots & d_{1j} \\ -d_{21} & d_{22} & \cdots & d_{2j} \\ -\vdots & \vdots & \cdots & \vdots \\ -d_{k1} & d_{k2} & \cdots & d_{kj}\n\end{bmatrix} \begin{bmatrix}\nx_1 \\ -x_2 \\ -\vdots \\ -x_j\n\end{bmatrix} -$$ - -or - -and - -**q**˙ = **Aq**+**Bx** (10.16) - -and - -$$ -y = Cq + Dx \tag{10.17} -$$ - -Equation (10.16) is the state equation and Eq. (10.17) is the output equation; **q, y,** and **x** are the state vector, the output vector, and the input vector, respectively. - -For discrete-time systems, the state equations are *N* simultaneous first-order difference equations. Discrete-time systems are discussed in Sec. 10.7. - -## **[10.3 A SYSTEMATIC](#page-14-0) PROCEDURE TO DETERMINE STATE EQUATIONS** - -We shall discuss here a systematic procedure to determine the state-space description of linear time-invariant systems. In particular, we shall consider systems of two types: (1) *RLC* networks and (2) systems specified by block diagrams or *N*th-order transfer functions. - -### **[10.3-1 Electrical Circuits](#page-14-0)** - -The method used in Ex. 10.2 proves effective in most of the simple cases. The steps are as follows: - -- 1. Choose all independent capacitor voltages and inductor currents to be the state variables. -- 2. Choose a set of loop currents; express the state variables and their first derivatives in terms of these loop currents. -- 3. Write loop equations, and eliminate all variables other than state variables (and their first derivatives) from the equations derived in steps 2 and 3. - -### **EXAMPLE 10.3 State Equations of an** *RLC* **Circuit** - -Write the state equations for the network shown in Fig. 10.2. - -**Figure 10.2** Circuit for Ex. 10.3. - -**Step 1.** There is one inductor and one capacitor in the network. Therefore, we shall choose the inductor current *q*1 and the capacitor voltage *q*2 as the state variables. - -**Step 2.** The relationship between the loop currents and the state variables can be written by inspection: - -1 - -$$ -q_1 = i_2 \tag{10.18} -$$ - -$$ -\frac{1}{2}\dot{q}_2 = i_2 - i_3\tag{10.19} -$$ - -**Step 3.** The loop equations are - -$$ -4i_1 - 2i_2 = x \tag{10.20} -$$ - -$$ -2(i_2 - i_1) + \dot{q}_1 + q_2 = 0 \tag{10.21} -$$ - -$$ --q_2 + 3i_3 = 0 \tag{10.22} -$$ - -Now we eliminate *i*1, *i*2, and *i*3 from the state and loop equations as follows. From Eq. (10.21), we have - -$$ -\dot{q}_1 = 2(i_1 - i_2) - q_2 -$$ - -We can eliminate *i*1 and *i*2 from this equation by using Eqs. (10.18) and (10.20) to obtain - -$$ -\dot{q}_1 = -q_1 - q_2 + \frac{1}{2}x -$$ - -The substitution of Eqs. (10.18) and (10.22) in Eq. (10.19) yields - -$$ -\dot{q}_2 = 2q_1 - \frac{2}{3}q_2 -$$ - -These are the desired state equations. We can express them in matrix form as - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 2 & -\frac{2}{3} \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix} x -$$ -\n(10.23) - -The derivation of state equations from loop equations is facilitated considerably by choosing loops in such a way that only one loop current passes through each of the inductors or capacitors. - -### AN ALTERNATIVE PROCEDURE - -We can also determine the state equations by the following procedure. - -- 1. Choose all independent capacitor voltages and inductor currents to be the state variables. -- 2. Replace each capacitor by a voltage source equal to the capacitor voltage, and replace each inductor by a current source equal to the inductor current. This step will transform the *RLC* network into a network consisting only of resistors, current sources, and voltage sources. -- 3. Find the current through each capacitor and equate it to *Cq*˙*i*, where *qi* is the capacitor voltage. Similarly, find the voltage across each inductor and equate it to *Lq*˙*j*, where *qj* is the inductor current. - -### **EXAMPLE 10.4 Alternate Procedure to Determine State Equations** - -Use the three-step alternative procedure just outlined to write the state equations for the network in Fig. 10.2. - -In the network in Fig. 10.2, we replace the inductor by a current source of current *q*1 and the capacitor by a voltage source of voltage *q*2, as shown in Fig. 10.3. The resulting network consists of four resistors, two voltage sources, and one current source. - -**Figure 10.3** Equivalent circuit of the network in Fig. 10.2. - -We can determine the voltage *vL* across the inductor and the current *ic* through the capacitor by using the principle of superposition. This step can be accomplished by inspection. For example, *vL* has three components arising from three sources. To compute the component due to *x*, we assume that *q*1 =0 (open circuit) and *q*2 =0 (short circuit). Under these conditions, the entire network to the right of the 2 resistor is opened, and the component of *vL* due to *x* is the voltage across the 2 resistor. This voltage is clearly (1/2)*x*. Similarly, to find the component of *vL* due to *q*1, we short *x* and *q*2. The source *q*1 sees an equivalent resistor of 1 across it, and hence *vL* = −*q*1. Continuing the process, we find that the component of *vL* due to *q*2 is −*q*2. Hence, - -$$ -v_L = \dot{q}_1 = \frac{1}{2}x - q_1 - q_2 -$$ - -Using the same procedure, we find - -$$ -i_c = \frac{1}{2}\dot{q}_2 = q_1 - \frac{1}{3}q_2 -$$ - -These equations are identical to the state equations [Eq. (10.23)] obtained earlier.† - -### **[10.3-2 State Equations from a Transfer Function](#page-14-0)** - -It is relatively easy to determine the state equations of a system specified by its transfer function.‡ Consider, for example, a first-order system with the transfer function - -$$ -H(s) = \frac{1}{s+a} -$$ - -The system realization appears in Fig. 10.4. The integrator output *q* serves as a natural state variable since, in practical realization, initial conditions are placed on the integrator output. The - -† This procedure requires modification if the system contains all-capacitor and voltage-source tie sets or all-inductor and current-source cut sets. In the case of all-capacitor and voltage-source tie sets, all capacitor voltages cannot be independent. One capacitor voltage can be expressed in terms of the remaining capacitor voltages and the voltage source(s) in that tie set. Consequently, one of the capacitor voltages should not be used as a state variable, and that capacitor should not be replaced by a voltage source. Similarly, in all-inductor and current-source tie sets, one inductor should not be replaced by a current source. If there are all-capacitor tie sets or all-inductor cut sets only, no further complications occur. In all-capacitor voltage-source tie sets and/or all-inductor current-source cut sets, we have additional difficulties in that the terms involving derivatives of the input may occur. This problem can be solved by redefining the state variables. The final state variables will not be capacitor voltages and inductor currents. - -‡ We implicitly assume that the system is controllable and observable. This implies that there are no pole-zero cancellations in the transfer function. If such cancellations are present, the state variable description represents only the part of the system that is controllable and observable (the part of the system that is coupled to the input and the output). In other words, the internal description represented by the state equations is no better than the external description represented by the input–output equation. - -integrator input is naturally *q*˙. From Fig. 10.4, we have - -*q*˙ = −*aq*+*x* and *y* = *q* - -In Sec. 4.6 we saw that a given transfer function can be realized in several ways. Consequently, we should be able to obtain different state-space descriptions of the same system by using different realizations. This assertion will be clarified by the following example. - -### **EXAMPLE 10.5 State-Space Description from a Transfer Function** - -Consider a system specified by the transfer function - -$$ -H(s) = \underbrace{\frac{2s+10}{s^3+8s^2+19s+12}}_{\text{direct form}} = \underbrace{\left(\frac{2}{s+1}\right)\left(\frac{s+5}{s+3}\right)\left(\frac{1}{s+4}\right)}_{\text{cascade}} = \underbrace{\frac{\frac{4}{3}}{s+1} - \frac{2}{s+3} + \frac{\frac{2}{3}}{s+4}}_{\text{parallel}} -$$ - -The procedure developed in Sec. 4.6 allows us to realize *H*(*s*) as, among others, direct form II (DFII), transpose DFII (TDFII), cascade, and parallel. These realizations are depicted in Fig. 10.5. Determine state-space descriptions for each of these realizations. As mentioned earlier, the output of each integrator serves as a natural state variable. - -### **Direct Form II and Its Transpose** - -Here we shall realize the system using the canonical form (direct form II and its transpose) discussed in Sec. 4.6. If we choose the state variables to be the three integrator outputs *q*1, *q*2, and *q*3, then, according to Fig. 10.5a, - -$$ -\dot{q}_1 = q_2 \n\dot{q}_2 = q_3 \n\dot{q}_3 = -12q_1 - 19q_2 - 8q_3 + x -$$ - -**Figure 10.5 (a)** DFII, **(b)** TDFII, **(c)** cascade, and **(d)** parallel realizations of *H*(*s*). - -### 922 CHAPTER 10 STATE-SPACE ANALYSIS - -Also, the output *y* is given by - -$$ -y = 10q_1 + 2q_2 -$$ - -In matrix form, these state and output equations become - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \\ \dot{q}_3 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -12 & -19 & -8 \end{bmatrix}}_{\mathbf{A}} \underbrace{\begin{bmatrix} q_1 \\ q_2 \\ q_3 \end{bmatrix}}_{\mathbf{A}} + \underbrace{\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}}_{\mathbf{B}} x -$$ - -and - -$$ -\mathbf{y} = \underbrace{[10 \quad 2 \quad 0]}_{\text{C}} \begin{bmatrix} q_1 \\ q_2 \\ q_3 \end{bmatrix} -$$ - -We can readily verify the state equations of the DFII structure by using MATLAB's tf2ss command: - -``` ->> num = [2 10]; den = [1 8 19 12]; ->> [A,B,C,D] = tf2ss(num,den) - A = -8 -19 -12 - 100 - 010 - B= 1 - 0 - 0 - C = 0 2 10 - D= 0 -``` - -MATLAB's convention for labeling state variables *q*1,*q*2,...,*qn* in a block diagram, such as shown in Fig. 10.5a, is reversed. That is, MATLAB labels *q*1 as *qn*, *q*2 and *qn*−1, and so on. Keeping this in mind, we see that MATLAB indeed confirms our earlier results. - -It is also possible to determine the transfer function from the state-space representation using the ss2tf and tf commands: - -``` ->> [num,den] = ss2tf(A,B,C,D); H = tf(num,den) - H = - 2 s + 10 - ----------------------- - s^3 + 8 s^2 + 19 s + 12 -``` - -### **Transpose Direct Form II** - -We can also realize *H*(*s*) by using the transpose of the DFII form, as shown in Fig. 10.5b. If we label the output of the three integrators as the state variables *v*1, *v*2, and *v*3, then, according - -to Fig. 10.5b, - -$$ -\dot{v}_1 = -12v_3 + 10x -$$ - -\n -$$ -\dot{v}_2 = v_1 - 19v_3 + 2x -$$ - -\n -$$ -\dot{v}_3 = v_2 - 8v_3 -$$ - -and the output *y* is given by - -*y* = *v*3 - -The matrix form of these state and output equations become - -$$ -\begin{bmatrix} \dot{v}_1 \\ \dot{v}_2 \\ \dot{v}_3 \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & 0 & -12 \\ 1 & 0 & -19 \\ 0 & 1 & -8 \end{bmatrix}}_{\hat{A}} \underbrace{\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}}_{+} + \underbrace{\begin{bmatrix} 10 \\ 2 \\ 0 \end{bmatrix}}_{\hat{B}} x -$$ - -and - -$$ -\mathbf{y} = \underbrace{[0 \quad 0 \quad 1]}_{\hat{\mathbf{c}}} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} -$$ - -Observe closely the relationship between the state-space descriptions of *H*(*s*) by means of the DFII and TDFII realizations. The **A** matrices in these two cases are the transpose of each other; also, the **B** of one is the transpose of **C** in the other, and vice versa. Hence, - -$$ -(\mathbf{A})^T = \hat{\mathbf{A}}, \qquad (\mathbf{B})^T = \hat{\mathbf{C}}, \qquad \text{and} \qquad (\mathbf{C})^T = \hat{\mathbf{B}} -$$ - -This is no coincidence. This duality relation is generally true [1]. - -### **Cascade Realization** - -The three integrator outputs *w*1, *w*2, and *w*3 in Fig. 10.5c are the state variables. Writing equations for the summer outputs yields - -$$ -\dot{w}_1 = -w_1 + x -$$ -, $\dot{w}_2 = 2w_1 - 3w_2$ , and $\dot{w}_3 = 5w_2 + \dot{w}_2 - 4w_3$ - -Since *w*˙ 2 = 2*w*1 −3*w*2, we see that *w*˙ 3 = 2*w*1 +2*w*2 −4*w*3. From Fig. 10.5c, we further see that *y* = *w*3. Put into matrix form, the state and output equations are therefore - -$$ -\begin{bmatrix} \dot{w}_1 \\ \dot{w}_2 \\ \dot{w}_3 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 2 & -3 & 0 \\ 2 & 2 & -4 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} x -$$ -$$ -y = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} -$$ - -and - -### 924 CHAPTER 10 STATE-SPACE ANALYSIS - -### **Parallel Realization (Diagonal Representation)** - -The three integrator outputs *z*1, *z*2, and *z*3 in Fig. 10.5d are the state variables. The state equations are - -> *z*˙1 = −*z*1 +*x z*˙2 = −3*z*2 +*x z*˙3 = −4*z*3 +*x* - -and the output equation is - -$$ -y = \frac{4}{3}z_1 - 2z_2 + \frac{2}{3}z_3 -$$ - -In matrix form, these equations are - -$$ -\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \\ \dot{z}_3 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -4 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} x -$$ -$$ -y = \begin{bmatrix} \frac{4}{3} & -2 & \frac{2}{3} \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ z_3 \end{bmatrix} -$$ - -### A GENERAL CASE - -It is clear that a system has several state-space descriptions. Notable among these are the variables obtained from the DFII, its transpose, and the diagonalized variables (in the parallel realization). State equations in these forms can be written immediately by inspection of the transfer function. Consider the general *N*th-order transfer function - -$$ -H(s) = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} -$$ - -= -$$ -\frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{(s - \lambda_1)(s - \lambda_2) \dots (s - \lambda_N)} -$$ - -= -$$ -b_0 + \frac{k_1}{s - \lambda_1} + \frac{k_2}{s - \lambda_2} + \dots + \frac{k_N}{s - \lambda_N} -$$ - (10.25) - -The realizations of *H*(*s*) found by using direct form II [Eq. (10.24)] and the parallel form [Eq. (10.25)] appear in Figs. 10.6a and 10.6b, respectively. - -The *N* integrator outputs *q*1, *q*2, ... , *qN* in Fig. 10.6a are the state variables. By inspection of this figure, we obtain - -$$ -\begin{aligned}\n\dot{q}_1 &= q_2 \\ -\dot{q}_2 &= q_3 \\ -&\vdots \\ -\dot{q}_{N-1} &= q_N \\ -\dot{q}_N &= -a_N q_1 - a_{N-1} q_2 - \dots - a_2 q_{N-1} - a_1 q_N + x\n\end{aligned} -$$ - -**Figure 10.6 (a)** Direct form II and **(b)** parallel realizations for an *N*th-order LTIC system. - -and output *y* is - -$$ -y = b_N q_1 + b_{N-1} q_2 + \cdots + b_1 q_N + b_0 \dot{q}_N -$$ - -We can eliminate *q*˙*N* in this output equation by using the last state equation to yield - -$$ -y = (b_N - b_0 a_N)q_1 + (b_{N-1} - b_0 a_{N-1})q_2 + \dots + (b_1 - b_0 a_1)q_N + b_0 x -$$ - -= $\hat{b}_N q_1 + \hat{b}_{N-1} q_2 + \dots + \hat{b}_1 q_N + b_0 x$ - -where *b*ˆ*i* = *bi* −*b*0*ai*. In matrix form, we obtain - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \\ \vdots \\ \dot{q}_{N-1} \\ \dot{q}_N \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ -a_N & -a_{N-1} & -a_{N-2} & \cdots & -a_2 & -a_1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \\ \vdots \\ q_{N-1} \\ q_N \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix} x -$$ - -and - -$$ -y = [\hat{b}_N \quad \hat{b}_{N-1} \quad \cdots \quad \hat{b}_1] \begin{bmatrix} q_1 \\ q_2 \\ \vdots \\ q_N \end{bmatrix} + b_0 x -$$ - -In Fig. 10.6b, the *N* integrator outputs *z*1, *z*2, ... , *zN* are the state variables. By inspection of this figure, we obtain - -$$ -\dot{z}_1 = \lambda_1 z_1 + x -$$ - -\n -$$ -\dot{z}_2 = \lambda_2 z_2 + x -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -\dot{z}_N = \lambda_N z_N + x -$$ - -and - -$$ -y = k_1 z_1 + k_2 z_2 + \cdots + k_N z_N + b_0 x -$$ - -or - -$$ -\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \\ \vdots \\ \dot{z}_{N-1} \\ \dot{z}_N \end{bmatrix} = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 & 0 \\ 0 & \lambda_2 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \cdots & \lambda_{N-1} & 0 \\ 0 & 0 & \cdots & 0 & \lambda_N \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_{N-1} \\ z_N \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \\ 1 \end{bmatrix} x -$$ -(10.26) -$$ -y = \begin{bmatrix} k_1 & k_2 & \cdots & k_{N-1} & k_N \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_{N-1} \end{bmatrix} + b_0 x -$$ - -*zN* - -and - -Observe that the diagonalized form of the state matrix [Eq. (10.26)] has the transfer function poles as its diagonal elements. The presence of repeated poles in *H*(*s*) will modify the procedure slightly. The handling of these cases is discussed in Sec. 4.6. - -It is clear from the foregoing discussion that a state-space description is not unique. For any realization of *H*(*s*) obtained from integrators, scalar multipliers, and adders, a corresponding state-space description exists. Since there are uncountable possible realizations of *H*(*s*), there are uncountable possible state-space descriptions. - -The advantages and drawbacks of various types of realization were discussed in Sec. 4.6. - -## **[10.4 SOLUTION OF](#page-14-0) STATE EQUATIONS** - -The state equations of a linear system are *N* simultaneous linear differential equations of the first order. We studied the techniques of solving linear differential equations in Chs. 2 and 4. The same techniques can be applied to state equations without any modification. However, it is more convenient to carry out the solution in the framework of matrix notation. - -These equations can be solved in both the time and frequency domains (Laplace transform). The latter is relatively easier to deal with than the time-domain solution. For this reason, we shall first consider the Laplace transform solution. - -### **[10.4-1 Laplace Transform Solution of State Equations](#page-14-0)** - -The *i*th state equation [Eq. (10.14)] is of the form - -$$ -\dot{q}_i = a_{i1}q_1 + a_{i2}q_2 + \dots + a_{iN}q_N + b_{i1}x_1 + b_{i2}x_2 + \dots + b_{ij}x_j \tag{10.27} -$$ - -We shall take the Laplace transform of this equation. Let - -$$ -q_i(t) \Longleftrightarrow Q_i(s) -$$ - -so that - -$$ -\dot{q}_i(t) \Longleftrightarrow sQ_i(s) - q_i(0) -$$ - -Also, let - -$$ -x_i(t) \Longleftrightarrow X_i(s) -$$ - -The Laplace transform of Eq. (10.27) yields - -$$ -sQ_i(s) - q_i(0) = a_{i1}Q_1(s) + a_{i2}Q_2(s) + \cdots + a_{iN}Q_N(s) + b_{i1}X_1(s) + b_{i2}X_2(s) + \cdots + b_{ij}X_j(s) -$$ - -Taking the Laplace transforms of all *N* state equations, we obtain - -$$ -s\left[\begin{array}{c} Q_1(s) \\ Q_2(s) \\ \vdots \\ Q_N(s) \end{array}\right] - \left[\begin{array}{c} q_1(0) \\ q_2(0) \\ \vdots \\ q_N(0) \end{array}\right] = \left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1N} \\ a_{21} & a_{22} & \cdots & a_{2N} \\ \vdots & \vdots & \cdots & \vdots \\ a_{N1} & a_{N2} & \cdots & a_{NN} \end{array}\right] \left[\begin{array}{c} Q_1(s) \\ Q_2(s) \\ \vdots \\ Q_N(s) \end{array}\right] + \left[\begin{array}{cccc} q_1(0) \\ \vdots \\ q_N(0) \end{array}\right] + \left[\begin{array}{cccc} b_{11} & b_{12} & \cdots & b_{1j} \\ b_{21} & b_{22} & \cdots & b_{2j} \\ \vdots & \vdots & \cdots & \vdots \\ b_{N1} & b_{N2} & \cdots & b_{Nj} \end{array}\right] \left[\begin{array}{c} X_1(s) \\ X_2(s) \\ \vdots \\ X_j(s) \end{array}\right] -$$ - -Defining the vectors, as indicated, we have - -$$ -s\mathbf{Q}(s) - \mathbf{q}(0) = \mathbf{A}\mathbf{Q}(s) + \mathbf{B}\mathbf{X}(s) -$$ - -or - -$$ -s\mathbf{Q}(s) - \mathbf{A}\mathbf{Q}(s) = \mathbf{q}(0) + \mathbf{B}\mathbf{X}(s) -$$ - -and - -$$ -(s\mathbf{I} - \mathbf{A})\mathbf{Q}(s) = \mathbf{x}(0) + \mathbf{B}\mathbf{X}(s) -$$ - -where **I** is the *N* ×*N* identity matrix. Solving for **Q**(*s*), we have - -$$ -Q(s) = (sI - A)^{-1}[q(0) + BX(s)] -$$ - -= $\Phi(s)[q(0) + BX(s)]$ (10.28) - -where - -$$ -\mathbf{\Phi}(s) = (s\mathbf{I} - \mathbf{A})^{-1} -$$ - -Thus, from Eq. (10.28), - -$$ -\mathbf{Q}(s) = \mathbf{\Phi}(s)\mathbf{q}(0) + \mathbf{\Phi}(s)\mathbf{B}\mathbf{X}(s) -$$ - -and - -$$ -\mathbf{q}(t) = \underbrace{\mathcal{L}^{-1}[\Phi(s)]\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{\mathcal{L}^{-1}[\Phi(s)\mathbf{B}\mathbf{X}(s)]}_{\text{zero-state response}} -$$ -(10.29) - -Equation (10.29) gives the desired solution. Observe the two components of the solution. The first component yields **q**(*t*) when the input *x*(*t*) = 0. Hence, the first component is the zero-input response. In a similar manner, we see that the second component is the zero-state response. - -### **EXAMPLE 10.6 Laplace Transform Solution to State Equations** - -Using the Laplace transform, find the state vector **q**(*t*) for the system whose state equation is given by - -**q**˙ = **Aq**+**Bx** - -where - -$$ -\mathbf{A} = \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \qquad \mathbf{x}(t) = u(t) -$$ - -and the initial conditions are *q*1(0) = 2, *q*2(0) = 1. - -From Eq. (10.28), we have - -$$ -\mathbf{Q}(s) = \mathbf{\Phi}(s)[\mathbf{q}(0) + \mathbf{B}\mathbf{X}(s)] -$$ - -Let us first find (*s*). We have - -$$ -(s\mathbf{I} - \mathbf{A}) = s \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix} = \begin{bmatrix} s+12 & -\frac{2}{3} \\ 36 & s+1 \end{bmatrix} -$$ - -and - -$$ -\Phi(s) = (s\mathbf{I} - \mathbf{A})^{-1} = \begin{bmatrix} \frac{s+1}{(s+4)(s+9)} & \frac{2/3}{(s+4)(s+9)}\\ \frac{-36}{(s+4)(s+9)} & \frac{s+12}{(s+4)(s+9)} \end{bmatrix} -$$ - -Now, -$$ -q(0) -$$ - is given as - -$$ -\mathbf{q}(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} -$$ - -Also, *X*(*s*) = 1/*s*, and - -$$ -\mathbf{BX}(s) = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \frac{1}{s} = \begin{bmatrix} \frac{1}{3s} \\ \frac{1}{s} \end{bmatrix} -$$ - -Therefore, - -$$ -\mathbf{q}(0) + \mathbf{B}\mathbf{X}(s) = \begin{bmatrix} 2 + \frac{1}{3s} \\ 1 + \frac{1}{s} \end{bmatrix} = \begin{bmatrix} \frac{6s+1}{3s} \\ \frac{s+1}{s} \end{bmatrix} -$$ - -and - -$$ -Q(s) = \Phi(s) [q(0) + BX(s)] -$$ - -= -$$ -\begin{bmatrix} \frac{s+1}{(s+4)(s+9)} & \frac{2/3}{(s+4)(s+9)} \\ \frac{-36}{(s+4)(s+9)} & \frac{s+12}{(s+4)(s+9)} \end{bmatrix} \begin{bmatrix} \frac{6s+1}{3s} \\ \frac{s+1}{s} \end{bmatrix} -$$ - -= -$$ -\begin{bmatrix} \frac{2s^2+3s+1}{s(s+4)(s+9)} \\ \frac{s-59}{(s+4)(s+9)} \end{bmatrix} -$$ - -= -$$ -\begin{bmatrix} \frac{1/36}{s} - \frac{21/20}{s+4} + \frac{136/45}{s+9} \\ \frac{-63/5}{s+4} + \frac{68/5}{s+9} \end{bmatrix} -$$ - -The inverse Laplace transform of this equation yields - -$$ -\begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{36} - \frac{21}{20}e^{-4t} + \frac{136}{45}e^{-9t}\right)u(t) \\ \left(-\frac{63}{5}e^{-4t} + \frac{68}{5}e^{-9t}\right)u(t) \end{bmatrix} -$$ - -This result is readily confirmed using MATLAB and its symbolic toolbox. - ->> syms s >> A = [-12 2/3;-36 -1]; B = [1/3; 1]; q0 = [2;1]; X = 1/s; >> q = ilaplace(inv(s\*eye(2)-A)\*(q0+B\*X)) q = (136\*exp(-9\*t))/45 - (21\*exp(-4\*t))/20 + 1/36 (68\*exp(-9\*t))/5 - (63\*exp(-4\*t))/5 - -To create a plot of the state vector, we use MATLAB's subs command to substitute the symbolic variable *t* with a vector of desired values. - -### 930 CHAPTER 10 STATE-SPACE ANALYSIS - -``` ->> t = (0:.01:2); q = subs(q); q1 = q(1,:); q2 = q(2,:); ->> plot(t,q1,'k',t,q2,'k--'); xlabel('t'); ylabel('Amplitude'); ->> legend('q_1(t)','q_2(t)','Location','SE'); -``` - -The resulting plot is shown in Fig. 10.7. - -### THE OUTPUT - -The output equation is given by - -$$ -y = Cq + Dx -$$ - -and - -$$ -\mathbf{Y}(s) = \mathbf{C}\mathbf{Q}(s) + \mathbf{D}\mathbf{X}(s) -$$ - -Upon substituting Eq. (10.28) into this equation, we have - -$$ -\mathbf{Y}(s) = \mathbf{C}[\Phi(s)[\mathbf{q}(0) + \mathbf{B}\mathbf{x}\mathbf{X}(s)]] + \mathbf{D}\mathbf{X}(s) -$$ - -= -$$ -\underbrace{\mathbf{C}\Phi(s)\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{[\mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D}]\mathbf{X}(s)}_{\text{zero-state response}} -$$ -(10.30) - -The zero-state response [i.e., the response **Y**(*s*) when **q**(0) = **0**] is given by - -$$ -\mathbf{Y}(s) = [\mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D}]\mathbf{X}(s) -$$ - -Note that the transfer function of a system is defined under the zero-state condition [see Eq. (4.19)]. The matrix **C**(*s*)**B** + **D** is the *transfer function matrix* **H**(*s*) of the system, which relates the responses *y*1, *y*2, ... , *yk* to the inputs *x*1, *x*2, ... , *xj*: - -$$ -\mathbf{H}(s) = \mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D} \tag{10.31} -$$ - -and the zero-state response is - -$$ -\mathbf{Y}(s) = \mathbf{H}(s)\mathbf{X}(s) -$$ - -The matrix **H**(*s*) is a *k* × *j* matrix (*k* is the number of outputs and *j* is the number of inputs). The *ij*th element *Hij*(*s*) of *H*(*s*) is the transfer function that relates the output *yi*(*t*) to the input *xj*(*t*). - -### **EXAMPLE 10.7 Transfer Function Matrix from State-Space Description** - -Let us consider a system with a state equation - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} -$$ - -and an output equation - -$$ -\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} -$$ - -Determine the transfer function matrix of the system. - -In this case, - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \qquad \mathbf{D} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \tag{10.32} -$$ - -and - -$$ -\Phi(s) = (s\mathbf{I} - \mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 2 & s+3 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} -$$ -(10.33) - -Hence, the transfer function matrix **H**(*s*) is given by - -$$ -\mathbf{H}(s) = \mathbf{C}\Phi(s)\mathbf{B} + \mathbf{D} -$$ -\n -$$ -= \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} -$$ -\n -$$ -= \begin{bmatrix} \frac{s+4}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)} \\ \frac{s+4}{s+2} & \frac{1}{s+2} \\ \frac{2(s-2)}{(s+1)(s+2)} & \frac{s^2+5s+2}{(s+1)(s+2)} \end{bmatrix} -$$ -\n(10.34) - -and the zero-state response is - -$$ -\mathbf{Y}(s) = \mathbf{H}(s)\mathbf{X}(s) -$$ - -Remember that the *ij*th element of the transfer function matrix in Eq. (10.34) represents the transfer function that relates the output *yi*(*t*) to the input *xj*(*t*). For instance, the transfer function that relates the output *y*3 to the input *x*2 is *H*32(*s*), where - -$$ -H_{32}(s) = \frac{s^2 + 5s + 2}{(s+1)(s+2)} -$$ - -We can readily verify the transfer function matrix using MATLAB and its symbolic toolbox functions. - ->> A = [0 1;-2 -3]; B = [1 0;1 1]; >> C = [1 0;1 1;0 2]; D = [0 0;1 0;0 1]; >> syms s; H = collect(simplify(C\*inv(s\*eye(2)-A)\*B+D)) H = [ (s + 4)/(s^2 + 3\*s + 2), 1/(s^2 + 3\*s + 2)] [ (s + 4)/(s + 2), 1/(s + 2)] [ (2\*s - 4)/(s^2 + 3\*s + 2), (s^2 + 5\*s + 2)/(s^2 + 3\*s + 2)] - -Transfer functions relating particular inputs to particular outputs, such as *H*32(*s*), can be obtained using the ss2tf and tf functions. - -``` ->> [num,den] = ss2tf(A,B,C,D,2); H_32 = tf(num(3,:),den) - H_32 = - s^2 + 5 s + 2 - ------------- - s^2 + 3 s + 2 -``` - -### CHARACTERISTIC ROOTS (EIGENVALUES) OF A MATRIX - -It is interesting to observe that the denominator of every transfer function in Eq. (10.34) is (*s* + 1)(*s* + 2) except for *H*21(*s*) and *H*22(*s*), where the factor (*s* + 1) is canceled. This is no coincidence. We see that the denominator of every element of (*s*) is |*s***I** − **A**| because (*s*) = (*s***I** − **A**)−1, and the inverse of a matrix has its determinant in the denominator. Since **C**, **B**, and **D** are matrices with constant elements, we see from Eq. (10.31) that the denominator of (*s*) will also be the denominator of **H**(*s*). Hence, the denominator of every element of **H**(*s*) is |*s***I** − **A**|, except for the possible cancellation of the common factors mentioned earlier. In other words, the zeros of the polynomial |*s***I**− **A**| are also the poles of all transfer functions of the system. *Therefore, the zeros of the polynomial* |*s***I**−**A**| *are the characteristic roots of the system.* Hence, the characteristic roots of the system are the roots of the equation - -$$ -|s\mathbf{I} - \mathbf{A}| = 0 \tag{10.35} -$$ - -Since |*s***I** − **A**| is an *N*th-order polynomial in *s* with *N* zeros λ1, λ2, ... , λ*N*, we can write Eq. (10.35) as - -$$ -|s\mathbf{I} - \mathbf{A}| = s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N = (s - \lambda_1)(s - \lambda_2) \cdot \dots (s - \lambda_N) = 0 -$$ - -For the system in Ex. 10.7, - -$$ -|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s & 0 \\ 0 & s \end{vmatrix} - \begin{vmatrix} 0 & 1 \\ -2 & -3 \end{vmatrix} = \begin{vmatrix} s & -1 \\ 2 & s + 3 \end{vmatrix} -$$ -$$ -= s^2 + 3s + 2 = (s + 1)(s + 2) -$$ - -Hence, - -$$ -\lambda_1 = -1 \quad \text{and} \quad \lambda_2 = -2 -$$ - -Equation (10.35) is known as the *characteristic equation of the matrix* **A**, and λ1, λ2, ... , λ*N* are the characteristic roots of **A**. The term *eigenvalue,* meaning "characteristic value" in German, is also commonly used in the literature. Thus, we have shown that the characteristic roots of a system are the eigenvalues (characteristic values) of the matrix **A**. - -At this point, the reader will recall that if λ1, λ2, ... , λ*N* are the poles of the transfer function, then the zero-input response is of the form - -$$ -y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \dots + c_N e^{\lambda_N t} -$$ - (10.36) - -This fact is also obvious from Eq. (10.30). The denominator of every element of the zero-input response matrix **C**(*s*)**q**(0) is |*s***I**−**A**| =(*s*−λ1)(*s*−λ2)···(*s*−λ*N*). Therefore, the partial fraction expansion and the subsequent inverse Laplace transform will yield a zero-input component of the form in Eq. (10.36). - -### **[10.4-2 Time-Domain Solution of State Equations](#page-14-0)** - -The state equation is - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x} \tag{10.37} -$$ - -We now show that the solution of the vector differential Eq. (10.37) is - -$$ -\mathbf{q}(t) = e^{\mathbf{A}t}\mathbf{q}(0) + \int_0^t e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{x}(\tau) d\tau -$$ - -Before proceeding further, we must define the matrix exponential *e***A***t* . An exponential of a matrix is defined by an infinite series identical to that used in defining an exponential of a scalar. We shall define - -$$ -e^{\mathbf{A}t} = \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2 t^2}{2!} + \frac{\mathbf{A}^3 t^3}{3!} + \dots + \frac{\mathbf{A}^n t^n}{n!} + \dots = \sum_{k=0}^{\infty} \frac{\mathbf{A}^k t^k}{k!} -$$ -(10.38) - -For example, if - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} -$$ - -then - -$$ -\mathbf{A}t = \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} t = \begin{bmatrix} 0 & t \\ 2t & t \end{bmatrix} -$$ - -and - -$$ -\frac{\mathbf{A}^2 t^2}{2!} = \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} \frac{t^2}{2} = \begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix} \frac{t^2}{2} = \begin{bmatrix} t^2 & \frac{t^2}{2} \\ t^2 & \frac{3t^2}{2} \end{bmatrix} -$$ - -and so on. - -#### 934 CHAPTER 10 STATE-SPACE ANALYSIS - -We can show that the infinite series in Eq. (10.38) is absolutely and uniformly convergent for all values of *t*. Consequently, it can be differentiated or integrated term by term. Thus, to find (*d*/*dt*)*e***A***t* , we differentiate the series on the right-hand side of Eq. (10.38) term by term: - -$$ -\frac{d}{dt}e^{\mathbf{A}t} = \mathbf{A} + \mathbf{A}^2t + \frac{\mathbf{A}^3t^2}{2!} + \frac{\mathbf{A}^4t^3}{3!} + \cdots -$$ - -\n -$$ -= \mathbf{A}\bigg[\mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots\bigg] = \mathbf{A}e^{\mathbf{A}t} -$$ - -\n -$$ -= \bigg[\mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots + \cdots\bigg]\mathbf{A} = e^{\mathbf{A}t}\mathbf{A} -$$ - -Hence, - -$$ -\frac{d}{dt}e^{\mathbf{A}t} = \mathbf{A}e^{\mathbf{A}t} = e^{\mathbf{A}t}\mathbf{A} -$$ - -Also note that from Eq. (10.38), it follows that - -*e***0** = **I** - -where **I** is just the identity matrix. If we premultiply or postmultiply the infinite series for *e***A***t* [Eq. (10.38)] by an infinite series for *e*−**A***t* , we find that - -$$ -(e^{-At})(e^{At}) = (e^{At})(e^{-At}) = I -$$ -\n(10.39) - -In Sec. 10.1-1, we showed that - -$$ -\frac{d}{dt}(\mathbf{U}\mathbf{V}) = \frac{d\mathbf{U}}{dt}\mathbf{V} + \mathbf{U}\frac{d\mathbf{V}}{dt} -$$ - -Using this relationship, we observe that - -$$ -\frac{d}{dt}[e^{-\mathbf{A}t}\mathbf{q}] = \left(\frac{d}{dt}e^{-\mathbf{A}t}\right)\mathbf{q} + e^{-\mathbf{A}t}\dot{\mathbf{q}} -$$ -$$ -= -e^{-\mathbf{A}t}\mathbf{A}\mathbf{q} + e^{-\mathbf{A}t}\dot{\mathbf{q}} -$$ -(10.40) - -We now premultiply both sides of Eq. (10.37) by *e*−**A***t* to yield - -$$ -e^{-At}\dot{\mathbf{q}} = e^{-At}\mathbf{A}\mathbf{q} + e^{-At}\mathbf{B}\mathbf{x} -$$ - -or - -$$ --e^{-At}\mathbf{A}\mathbf{q}+e^{-\mathbf{A}t}\dot{\mathbf{q}}=e^{-\mathbf{A}t}\mathbf{B}\mathbf{x} -$$ - -Substituting this result into Eq. (10.40) yields - -$$ -\frac{d}{dt}[e^{-At}\mathbf{q}] = e^{-At}\mathbf{B}\mathbf{x} -$$ - -The integration of both sides of this equation from 0 to *t* yields - -$$ -e^{-\mathbf{A}t}\mathbf{q}\big|_{0}^{t} = \int_{0}^{t} e^{-\mathbf{A}\tau} \mathbf{B} \mathbf{x}(\tau) d\tau -$$ - -or - -$$ -e^{-\mathbf{A}t}\mathbf{q}(t) - \mathbf{q}(0) = \int_0^t e^{-\mathbf{A}\tau} \mathbf{B}\mathbf{x}(\tau) d\tau -$$ - -Hence, - -$$ -e^{-\mathbf{A}t}\mathbf{q} = \mathbf{q}(0) + \int_0^t e^{-\mathbf{A}\tau} \mathbf{B} \mathbf{x}(\tau) d\tau -$$ - -Premultiplying this result by *e***A***t* and using Eq. (10.39), we have - -$$ -\mathbf{q}(t) = \underbrace{e^{\mathbf{A}t}\mathbf{q}(0)}_{\text{ZIR}} + \underbrace{\int_0^t e^{\mathbf{A}(t-\tau)}\mathbf{B}\mathbf{x}(\tau) d\tau}_{\text{ZSR}} -$$ -(10.41) - -This is the desired solution. The first term on the right-hand side represents *q*(*t*) when the input *x*(*t*) = 0. Hence, it is the zero-input component. The second term, by a similar argument, is seen to be the zero-state component. - -The results of Eq. (10.41) can be expressed more conveniently in terms of the matrix convolution. We can define the convolution of two matrices in a manner similar to the multiplication of two matrices, except that the multiplication of two elements is replaced by their convolution. For example, - -$$ -\begin{bmatrix} x_1 & x_2 \ x_3 & x_4 \end{bmatrix} * \begin{bmatrix} g_1 & g_2 \ g_3 & g_4 \end{bmatrix} = \begin{bmatrix} (x_1 * g_1 + x_2 * g_3) & (x_1 * g_2 + x_2 * g_4) \\ (x_3 * g_1 + x_4 * g_3) & (x_3 * g_2 + x_4 * g_4) \end{bmatrix} -$$ - -By using this definition of matrix convolution, we can express Eq. (10.41) as - -$$ -\mathbf{q}(t) = e^{\mathbf{A}t}\mathbf{q}(0) + e^{\mathbf{A}t} * \mathbf{B}\mathbf{x}(t) -$$ -\n(10.42) - -Note that the limits of the convolution integral [Eq. (10.41)] are from 0 to *t*. Hence, all the elements of *e***A***t* in the convolution term of Eq. (10.42) are implicitly assumed to be multiplied by *u*(*t*). - -The result of Eqs. (10.41) and (10.42) can be easily generalized for any initial value of *t*. It is left as an exercise for the reader to show that the solution of the state equation can be expressed as - -$$ -\mathbf{q}(t) = e^{\mathbf{A}(t-t_0)}\mathbf{q}(t_0) + \int_{t_0}^t e^{\mathbf{A}(t-\tau)} \mathbf{B}\mathbf{x}(\tau) d\tau -$$ - -## DETERMINING *eAt* - -The exponential *e***A***t* required in Eqs. (10.41) and (10.42) can be computed from the definition in Eq. (10.38). Unfortunately, this is an infinite series, and its computation can be quite laborious. Moreover, we may not be able to recognize the closed-form expression for the answer. There are several efficient methods of determining *e***A***t* in closed form. It was shown in Sec. 10.1-3 that for an *N* ×*N* matrix **A**, - -$$ -e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{N-1} \mathbf{A}^{N-1} -$$ - (10.43) - -where - -$$ -\begin{bmatrix}\n\beta_0 \\ -\beta_1 \\ -\vdots \\ -\beta_{N-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{N-1} \\ -1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{N-1} \\ -\vdots & \vdots & \ddots & \vdots \\ -1 & \lambda_N & \lambda_N^2 & \cdots & \lambda_N^{N-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\ne^{\lambda_1 t} \\ -ne^{\lambda_2 t} \\ -\vdots \\ -e^{\lambda_N t}\n\end{bmatrix} -$$ - -and λ1,λ2,...,λ*N* are the *N* characteristic values (eigenvalues) of **A**. - -We can also determine *e***A***t* by comparing Eqs. (10.41) and (10.29). It is clear that - -$$ -e^{\mathbf{A}t} = \mathcal{L}^{-1}[\Phi(s)] = \mathcal{L}^{-1}[(s\mathbf{I} - \mathbf{A})^{-1}] -$$ -\n(10.44) - -Thus, *e***A***t* and (*s*) are a Laplace transform pair. To be consistent with Laplace transform notation, *e***A***t* is often denoted by *φ*(*t*), *the state transition matrix* (STM): - -*e***A***t* = *φ*(*t*) - -### **EXAMPLE 10.8 Time-Domain Method to Solve State Equations** - -Use the time-domain method to solve Ex. 10.6. - -For this case, the characteristic roots are given by - -$$ -|\mathbf{sI} - \mathbf{A}| = \begin{vmatrix} s+12 & -\frac{2}{3} \\ 36 & s+1 \end{vmatrix} = s^2 + 13s + 36 = (s+4)(s+9) = 0 -$$ - -The roots are λ1 = −4 and λ2 = −9, so - -$$ -\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 1 & -9 \end{bmatrix}^{-1} \begin{bmatrix} e^{-4t} \\ e^{-9t} \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 9e^{-4t} - 4e^{-9t} \\ e^{-4t} - e^{-9t} \end{bmatrix} -$$ - -and - -$$ -e^{\mathbf{A}t} = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} -$$ - -= $\left(\frac{9}{5}e^{-4t} - \frac{4}{5}e^{-9t}\right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \left(\frac{1}{5}e^{-4t} - \frac{1}{5}e^{-9t}\right) \begin{bmatrix} -12 & \frac{2}{3} \\ -36 & -1 \end{bmatrix}$ -= $\begin{bmatrix} \left(\frac{-3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right) & \frac{2}{15}(e^{-4t} - e^{-9t}) \\ \frac{36}{5}(-e^{-4t} + e^{-9t}) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right) \end{bmatrix}$ - -The zero-input response is given by [see Eq. (10.41)] - -$$ -e^{\mathbf{A}t}\mathbf{q}(0) = \begin{bmatrix} \left(-\frac{3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right) & \frac{2}{15}(e^{-4t} - e^{-9t})\\ \frac{36}{5}(-e^{-4t} + e^{-9t}) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right) \end{bmatrix} \begin{bmatrix} 2\\ 1 \end{bmatrix} -$$ -$$ -= \begin{bmatrix} \left(\frac{-16}{15}e^{-4t} + \frac{46}{15}e^{-9t}\right)u(t) \\ \left(\frac{-64}{5}e^{-4t} + \frac{69}{5}e^{-9t}\right)u(t) \end{bmatrix} -$$ - -Note here the presence of *u*(*t*), indicating that the response begins at *t* = 0. - -The zero-state component is *e***A***t* ∗**Bx** [see Eq. (10.42)], where - -$$ -\mathbf{B}\mathbf{x} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} u(t) = \begin{bmatrix} \frac{1}{3}u(t) \\ u(t) \end{bmatrix} -$$ - -and - -$$ -e^{\mathbf{A}t} * \mathbf{B} \mathbf{x}(t) = \begin{bmatrix} \left(\frac{-3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right)u(t) & \frac{2}{15}(e^{-4t} - e^{-9t})u(t) \\ \frac{36}{5}(-e^{-4t} + e^{-9t}u(t)) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right)u(t) \end{bmatrix} * \begin{bmatrix} \frac{1}{3}u(t) \\ u(t) \end{bmatrix} -$$ - -Note again the presence of the term *u*(*t*) in every element of *e***A***t* . This is the case because the limits of the convolution integral run from 0 to *t* [Eq. (10.41)]. Thus, - -$$ -e^{\mathbf{A}t} * \mathbf{Bx}(t) = \begin{bmatrix} \left(-\frac{3}{5}e^{-4t} + \frac{8}{5}e^{-9t}\right)u(t) * \frac{1}{3}u(t) & \frac{2}{15}(e^{-4t} - e^{-9t})u(t) * u(t) \\ \frac{36}{5}(-e^{-4t} + e^{-9t})u(t) * \frac{1}{3}u(t) & \left(\frac{8}{5}e^{-4t} - \frac{3}{5}e^{-9t}\right)u(t) * u(t) \end{bmatrix} -$$ -$$ -= \begin{bmatrix} -\frac{1}{15}e^{-4t}u(t) * u(t) + \frac{2}{5}e^{-9t}u(t) * u(t) \\ -\frac{4}{5}e^{-4t}u(t) * u(t) + \frac{9}{5}e^{-9t}u(t) * u(t) \end{bmatrix} -$$ - -Substitution for the preceding convolution integrals from the convolution table (Table 2.1) yields - -$$ -e^{\mathbf{A}t} * \mathbf{Bx}(t) = \begin{bmatrix} -\frac{1}{60}(1 - e^{-4t})u(t) + \frac{2}{45}(1 - e^{-9t})u(t) \\ -\frac{1}{5}(1 - e^{-4t})u(t) + \frac{1}{5}(1 - e^{-9t})u(t) \end{bmatrix} -$$ -$$ -= \begin{bmatrix} (\frac{1}{36} + \frac{1}{60}e^{-4t} - \frac{2}{45}e^{-9t})u(t) \\ \frac{1}{5}(e^{-4t} - e^{-9t})u(t) \end{bmatrix} -$$ - -The sum of the two components now gives the desired solution for **q**(*t*): - -$$ -\mathbf{q}(t) = \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{36} - \frac{21}{20}e^{-4t} + \frac{136}{45}e^{-9t}\right)u(t) \\ \left(\frac{-63}{5}e^{-4t} + \frac{68}{5}e^{-9t}\right)u(t) \end{bmatrix} -$$ - -This result confirms the solution obtained by using the frequency-domain method [see Ex. 10.6]. Once the state variables *q*1 and *q*2 have been found for *t*≥0, all the remaining variables can be determined from the output equation. - -### THE OUTPUT - -The output equation is given by - -$$ -\mathbf{y}(t) = \mathbf{C}\mathbf{q}(t) + \mathbf{D}\mathbf{x}(t) -$$ - -The substitution of the solution for **q** [Eq. (10.42)] in this equation yields - -$$ -\mathbf{y}(t) = \mathbf{C} [e^{\mathbf{A}t} \mathbf{q}(0) + e^{\mathbf{A}t} * \mathbf{B} \mathbf{x}(t)] + \mathbf{D} \mathbf{x}(t) -$$ - -Since the elements of **B** are constants, - -$$ -e^{\mathbf{A}t} * \mathbf{B}\mathbf{x}(t) = e^{\mathbf{A}t}\mathbf{B} * \mathbf{x}(t) -$$ - -### 938 CHAPTER 10 STATE-SPACE ANALYSIS - -With this result, the output equation becomes - -$$ -\mathbf{y}(t) = \mathbf{C} [e^{\mathbf{A}t} \mathbf{q}(0) + e^{\mathbf{A}t} \mathbf{B} * \mathbf{x}(t)] + \mathbf{D} \mathbf{x}(t) -$$ - -Now recall that the convolution of *x*(*t*) with the unit impulse δ(*t*) yields *x*(*t*). Let us define a *j* × *j* diagonal matrix *δ*(*t*) such that all its diagonal terms are unit impulse functions. It is then obvious that - -$$ -\delta(t) * \mathbf{x}(t) = \mathbf{x}(t) -$$ - -and the output equation can be expressed as - -$$ -\mathbf{y}(t) = \mathbf{C}[e^{\mathbf{A}t}\mathbf{q}(0) + e^{\mathbf{A}t}\mathbf{B} * \mathbf{x}(t)] + \mathbf{D}\delta(t) * \mathbf{x}(t) -$$ -$$ -= \mathbf{C}e^{\mathbf{A}t}\mathbf{q}(0) + [\mathbf{C}e^{\mathbf{A}t}\mathbf{B} + \mathbf{D}\delta(t)] * \mathbf{x}(t) -$$ - -With the notation *φ*(*t*) for *e***A***t* , the output equation may be expressed as - -$$ -\mathbf{y}(t) = \underbrace{\mathbf{C}\boldsymbol{\phi}(t)\mathbf{q}(0)}_{\text{zero-input response}} + \underbrace{\left[\mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t)\right] * \mathbf{x}(t)}_{\text{zero-state response}} -$$ - -The zero-state response, that is, the response when **q**(0) = **0**, is - -$$ -\mathbf{y}(t) = [\mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t)] * \mathbf{x}(t) = \mathbf{h}(t) * \mathbf{x}(t) -$$ - -where - -$$ -\mathbf{h}(t) = \mathbf{C}\boldsymbol{\phi}(t)\mathbf{B} + \mathbf{D}\boldsymbol{\delta}(t) \tag{10.45} -$$ - -The matrix **h**(*t*) is a *k* × *j* matrix known as the *impulse response matrix*. The reason for this designation is obvious. The *ij*th element of **h**(*t*) is *hij*(*t*), which represents the zero-state response *yi* when the input *xj*(*t*) = δ(*t*) and when all other inputs (and all the initial conditions) are zero. Not surprisingly, vectors **h**(*t*) and **H**(*s*) form a Laplace transform pair, - -$$ -\mathcal{L}[\mathbf{h}(t)] = \mathbf{H}(s) -$$ - -### **EXAMPLE 10.9 State Transition Matrix by Inverse Laplace Transform** - -For the system described in Ex. 10.7, use Eq. (10.44) to determine *e***A***t* : - -$$ -\boldsymbol{\phi}(t) = e^{\mathbf{A}t} = \mathcal{L}^{-1} \boldsymbol{\Phi}(s) -$$ - -This problem was solved earlier with frequency-domain techniques. From Eq. (10.33), we have - -$$ -\begin{aligned} \n\phi(t) &= \mathcal{L}^{-1} \begin{bmatrix} \frac{s+3}{(s+1)(s+2)} & \frac{1}{(s+1)(s+2)}\\ \frac{-2}{(s+1)(s+2)} & \frac{s}{(s+1)(s+2)} \end{bmatrix} \\ \n&= \mathcal{L}^{-1} \begin{bmatrix} \frac{2}{s+1} - \frac{1}{s+2} & \frac{1}{s+1} - \frac{1}{s+2} \\ \frac{-2}{s+1} + \frac{2}{s+2} & \frac{-1}{s+1} + \frac{2}{s+2} \end{bmatrix} \\ \n&= \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} \n\end{aligned} -$$ - -The same result is obtained in Ex. 10.1 (Sec. 10.1-3) by using Eq. (10.43) [see Eq. (10.13)]. Also, *δ*(*t*) is a diagonal *j*×*j* or 2×2 matrix: - -$$ -\delta(t) = \begin{bmatrix} \delta(t) & 0\\ 0 & \delta(t) \end{bmatrix} -$$ - -Substituting the matrices *φ*(*t*), *δ*(*t*), **C**, **D**, and **B** [Eq. (10.32)] into Eq. (10.45), we have - -$$ -\mathbf{h}(t) = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \delta(t) & 0 \\ 0 & \delta(t) \end{bmatrix} -$$ -$$ -= \begin{bmatrix} 3e^{-t} - 2e^{-2t} & e^{-t} - e^{-2t} \\ \delta(t) + 2e^{-2t} & e^{-2t} \\ -6e^{-t} + 8e^{-2t} & \delta(t) - 2e^{-2t} + 4e^{-2t} \end{bmatrix} -$$ - -As the reader can verify, the Laplace transform of this equation yields the transfer function matrix **H**(*s*) in Eq. (10.34). - -## **10.5 LINEAR [TRANSFORMATION OF A](#page-14-0) STATE VECTOR** - -In Sec. 10.2 we saw that the state of a system can be specified in several ways. The sets of all possible state variables are related—in other words, if we are given one set of state variables, we should be able to relate it to any other set. We are particularly interested in a linear type of relationship. Let *q*1,*q*2,...,*qN* and *w*1,*w*2,...,*wN* be two different sets of state variables specifying the same system. Let these sets be related by linear equations as - -$$ -w_1 = p_{11}q_1 + p_{12}q_2 + \dots + p_{1N}q_N -$$ - -\n -$$ -w_2 = p_{21}q_1 + p_{22}q_2 + \dots + p_{2N}q_N -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -w_N = p_{N1}q_1 + p_{N2}q_2 + \dots + p_{NN}q_N -$$ - -or - -| ⎡
w1
w2
⎢ | ⎤
⎥ | ⎡
p11
p21
⎢ | p12
p22 | ···
··· | ⎤
p1N
p2N
⎥ | ⎡
⎤
q1
q2
⎢
⎥ | -|--------------------|------------------|----------------------|------------|------------|----------------------|------------------------------| -| ⎢
⎢
⎣ | ⎥
=
⎥
⎦ | ⎢
⎢
⎣ | | | ⎥
⎥
⎦ | ⎢
⎥
⎢
⎥
⎣
⎦ | -| wN | | pN1 | pN2 | ··· | pNN | qN | -|

w | | | |
P | |

q | - -Defining the vector **w** and matrix **P** as just shown, we obtain the compact matrix representation - -$$ -\mathbf{w} = \mathbf{P}\mathbf{q} \tag{10.46} -$$ - -and - -$$ -\mathbf{q} = \mathbf{P}^{-1}\mathbf{w} \tag{10.47} -$$ - -Thus, the state vector **q** is transformed into another state vector **w** through the linear transformation in Eq. (10.46). - -If we know **w**, we can determine **q** from **q** = **P**−1 **w**, provided **P**−1 exists. This is equivalent to saying that **P** is a nonsingular matrix† (|**P**| = 0). Thus, if **P** is a nonsingular matrix, the vector **w** defined by Eq. (10.46) is also a state vector. Consider the state equation of a system - -**q**˙ = **Aq**+**Bx** - -$$ -\mathbf{w} = \mathbf{P}\mathbf{q} -$$ -$$ -\mathbf{q} = \mathbf{P}^{-1}\mathbf{w} -$$ - -and - -then - -If - -Hence, the state equation now becomes - -$$ -\mathbf{P}^{-1}\dot{\mathbf{w}} = \mathbf{A}\mathbf{P}^{-1}\mathbf{w} + \mathbf{B}\mathbf{x} -$$ - -**q**˙ = **P**−1 - -**w**˙ - -or - -$$ -\dot{\mathbf{w}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1}\mathbf{w} + \mathbf{P}\mathbf{B}\mathbf{x} -$$ - -= $\hat{\mathbf{A}}\mathbf{w} + \hat{\mathbf{B}}\mathbf{x}$ (10.48) - -where - -$$ -\hat{\mathbf{A}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1} \qquad \text{and} \qquad \hat{\mathbf{B}} = \mathbf{P}\mathbf{B} \tag{10.49} -$$ - -Equation (10.48) is a state equation for the same system, but now it is expressed in terms of the state vector **w**. - -† This condition is equivalent to saying that all *N* equations in Eq. (10.46) are linearly independent; that is, none of the *N* equations can be expressed as a linear combination of the remaining equations. - -The output equation is also modified. Let the original output equation be - -$$ -y = Cq + Dx -$$ - -In terms of the new state variable **w**, this equation becomes - -$$ -\mathbf{y} = \mathbf{C}(\mathbf{P}^{-1}\mathbf{w}) + \mathbf{D}\mathbf{x} -$$ -$$ -= \hat{\mathbf{C}}\mathbf{w} + \mathbf{D}\mathbf{x} -$$ -$$ -\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} -$$ -(10.50) - -### **EXAMPLE 10.10 Linear Transformation of the State Vector** - -The state equations of a certain system are given by - -$$ -\begin{bmatrix} \dot{q}_1 \\ \dot{q}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \end{bmatrix} x(t) -$$ - -Find the state equations for this system when the new state variables *w*1 and *w*2 are given as - -$$ -\begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} -$$ - (10.51) - -According to Eq. (10.48), the state equation for the state variable **w** is given by - -$$ -\dot{\mathbf{w}} = \hat{\mathbf{A}}\mathbf{w} + \hat{\mathbf{B}}\mathbf{x} -$$ - -where [see Eqs. (10.49) and (10.50)] - -$$ -\hat{\mathbf{A}} = \mathbf{P}\mathbf{A}\mathbf{P}^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} -$$ -$$ -= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} -$$ -$$ -= \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix} -$$ - -and - -where - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} -$$ - -Therefore, - -$$ -\begin{bmatrix} \dot{w}_1 \\ \dot{w}_2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} + \begin{bmatrix} 3 \\ -1 \end{bmatrix} x(t) -$$ - -This is the desired state equation for the state vector **w**. The solution of this equation requires a knowledge of the initial state **w**(0). This can be obtained from the given initial state **q**(0) by using Eq. (10.51). - -We can obtain the same result with less effort using MATLAB. - -``` ->> A = [0 1;-2 -3]; B = [1; 2]; ->> P = [1 1;1 -1]; ->> Ahat = P*A*inv(P), Bhat = P*B - Ahat = -2 0 - 3 -1 - Bhat = 3 - -1 -``` - -### INVARIANCE OF EIGENVALUES - -We have seen that the poles of all possible transfer functions of a system are the eigenvalues of the matrix **A**. If we transform a state vector from **q** to **w**, the variables *w*1, *w*2, ... , *wN* are linear combinations of *q*1, *q*2, ... , *qN* and therefore may be considered to be outputs. Hence, the poles of the transfer functions relating *w*1, *w*2, ... , *wN* to the various inputs must also be the eigenvalues of matrix **A**. On the other hand, the system is also specified by Eq. (10.48). This means that the poles of the transfer functions must be the eigenvalues of **A**ˆ . Therefore, the eigenvalues of matrix **A** remain unchanged for the linear transformation of variables represented by Eq. (10.46), and the eigenvalues of matrix **A** and matrix **A**ˆ (**A**ˆ = **PAP**−1 ) are identical, implying that the characteristic equations of **A** and **A**ˆ are also identical. This result also can be proved alternately as follows. - -Consider the matrix **P**(*s***I**−**A**)**P**−1 . We have - -$$ -\mathbf{P}(s\mathbf{I} - \mathbf{A})\mathbf{P}^{-1} = \mathbf{P}s\mathbf{IP}^{-1} - \mathbf{P}\mathbf{A}\mathbf{P}^{-1} = s\mathbf{P}\mathbf{IP}^{-1} - \hat{\mathbf{A}} = s\mathbf{I} - \hat{\mathbf{A}} -$$ - -Taking the determinants of both sides, we obtain - -$$ -|\mathbf{P}||s\mathbf{I} - \mathbf{A}||\mathbf{P}^{-1}| = |s\mathbf{I} - \hat{\mathbf{A}}| -$$ - -The determinants |**P**| and |**P**−1 | are reciprocals of each other. Hence, - -$$ -|s\mathbf{I} - \mathbf{A}| = |s\mathbf{I} - \hat{\mathbf{A}}| -$$ - -This is the desired result. We have shown that the characteristic equations of **A** and **A**ˆ are identical. Hence, the eigenvalues of **A** and **A**ˆ are identical. - -In Ex. 10.10, matrix **A** is given as - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} -$$ - -The characteristic equation is - -$$ -|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s & -1 \\ 2 & s+3 \end{vmatrix} = s^2 + 3s + 2 = 0 -$$ - -Also, - -$$ -\hat{\mathbf{A}} = \begin{bmatrix} -2 & 0 \\ 3 & -1 \end{bmatrix} -$$ - -and - -$$ -|s\mathbf{I} - \hat{\mathbf{A}}| = \begin{bmatrix} s+2 & 0\\ -3 & s+1 \end{bmatrix} = s^2 + 3s + 2 = 0 -$$ - -This result verifies that the characteristic equations of **A** and **A**ˆ are identical. - -### **[10.5-1 Diagonalization of Matrix](#page-14-0) A** - -For several reasons, it is desirable to make matrix **A** diagonal. If **A** is not diagonal, we can transform the state variables such that the resulting matrix **A**ˆ is diagonal.† One can show that for any diagonal matrix **A**, the diagonal elements of this matrix must necessarily be λ1, λ2, ... , λ*N* (the eigenvalues) of the matrix. Consider the diagonal matrix **A**: - -$$ -\mathbf{A} = \begin{bmatrix} a_1 & 0 & 0 & \cdots & 0 \\ 0 & a_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & a_N \end{bmatrix} -$$ - -The characteristic equation is given by - -$$ -|\mathbf{sI} - \mathbf{A}| = \begin{bmatrix} (s - a_1) & 0 & 0 & \cdots & 0 \\ 0 & (s - a_2) & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & (s - a_N) \end{bmatrix} = 0 -$$ - -or - -$$ -(s - a_1)(s - a_2) \cdots (s - a_N) = 0 -$$ - -The nonzero (diagonal) elements of a diagonal matrix are therefore its eigenvalues λ1, λ2, ... , λ*N*. We shall denote the diagonal matrix by the symbol, **A**: - -$$ -\mathbf{\Lambda} = \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_N \end{bmatrix} -$$ - (10.52) - -Let us now consider the transformation of the state vector **A** such that the resulting matrix **A**ˆ is a diagonal matrix . - -† In this discussion we assume distinct eigenvalues. If the eigenvalues are not distinct, we can reduce the matrix to a modified diagonalized (Jordan) form. - -### 944 CHAPTER 10 STATE-SPACE ANALYSIS - -Consider the system - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x} -$$ - -We shall assume that λ1, λ2, ... , λ*N*, the eigenvalues of **A**, are distinct (no repeated roots). Let us transform the state vector **q** into the new state vector **z**, using the transformation - -$$ -z = Pq \tag{10.53} -$$ - -Then, after the development of Eq. (10.48), we have - -$$ -\dot{z} = PAP^{-1}z + PBx -$$ - -We desire the transformation to be such that **PAP**−1 is a diagonal matrix given by Eq. (10.52), or - -= **PAP**−1 - -$$ -\dot{\mathbf{z}} = \mathbf{\Lambda}\mathbf{z} + \mathbf{B}\mathbf{x} \tag{10.54} -$$ - -Hence, - -or - -$$ -\Lambda P = PA \tag{10.55} -$$ - -We know and **A**. Equation (10.55) therefore can be solved to determine **P**. - -### **EXAMPLE 10.11 Diagonal Form of the State Equations** - -Find the diagonalized form of the state equations for the system in Ex. 10.10. - -In this case, - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} -$$ - -We found λ1 = −1 and λ2 = −2. Hence, - -$$ -\mathbf{\Lambda} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} -$$ - -and Eq. (10.55) becomes - -$$ -\begin{bmatrix} -1 & 0 \ 0 & -2 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} 0 & 1 \ -2 & -3 \end{bmatrix} -$$ - -Equating the four elements on two sides, we obtain - -$$ --p_{11} = -2p_{12} -$$ - -\n -$$ --p_{12} = p_{11} - 3p_{12} -$$ - -\n -$$ --2p_{21} = -2p_{22} -$$ - -\n -$$ --2p_{22} = p_{21} - 3p_{22} -$$ - -The reader will immediately recognize that the first two equations are identical and that the last two equations are identical. Hence, two equations may be discarded, leaving us with only two equations [*p*11 = 2*p*12 and *p*21 = *p*22] and four unknowns. This observation means that there is no unique solution. There is, in fact, an infinite number of solutions. We can assign any value to *p*11 and *p*21 to yield one possible solution.† If *p*11 = *k*1 and *p*21 = *k*2, then we have *p*12 = *k*1/2 and *p*22 = *k*2: - -$$ -\mathbf{P} = \begin{bmatrix} k_1 & \frac{k_1}{2} \\ k_2 & k_2 \end{bmatrix} -$$ - -We may assign any values to *k*1 and *k*2. For convenience, let *k*1 = 2 and *k*2 = 1. This substitution yields - -$$ -\mathbf{P} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} -$$ - -The transformed variables [Eq. (10.53)] are - -$$ -\begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} = \begin{bmatrix} 2q_1 + q_2 \\ q_1 + q_2 \end{bmatrix} -$$ - -This expression relates the new state variables *z*1 and *z*2 to the original state variables *q*1 and *q*2. The system equation with **z** as the state vector is given by [see Eq. (10.54)] - -$$ -\dot{z} = \Lambda z + \hat{B}x -$$ - -where - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 3 \end{bmatrix} -$$ -$$ -\begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \begin{bmatrix} 4 \\ 4 \end{bmatrix}. -$$ - -*z*2 - -+ - -4 3 - -*x* (10.56) - -−1 0 0 −2 - -*z*˙1 *z*˙2 - -= - -Hence, - -† If, however, we want the state equations in diagonalized form, as in Eq. (10.26), where all the elements of -$$ -\hat{B} -$$ - matrix are unity, there is a unique solution. The reason is that the equation $\hat{B} = PB$ , where all the elements of $\hat{B}$ are unity, imposes additional constraints. In the present example, this condition will yield $p_{11} = 1/2$ , $p_{12} = 1/4$ , $p_{21} = 1/3$ , and $p_{22} = 1/3$ . The relationship between **z** and **q** is then - -$$ -z_1 = \frac{1}{2}q_1 + \frac{1}{4}q_2 -$$ - and $z_2 = \frac{1}{3}q_1 + \frac{1}{3}q_2$ - -or - -$$ -\begin{aligned}\n\dot{z}_1 &= -z_1 + 4x \\ -\dot{z}_2 &= -2z_2 + 3x\n\end{aligned} -$$ - -Note the distinctive nature of these state equations. Each state equation involves only one variable and therefore can be solved by itself. A general state equation has the derivative of one state variable equal to a linear combination of all state variables. Such is not the case with the diagonalized matrix . Each state variable *zi* is chosen so that it is uncoupled from the rest of the variables; hence, a system with *N* eigenvalues is split into *N* decoupled systems, each with an equation of the form - -$$ -\dot{z}_i = \lambda_i z_i + (\text{input terms}) -$$ - -This fact also can be readily seen from Fig. 10.8a, which is a realization of the system represented by Eq. (10.56). In contrast, consider the original state equations [see Ex. 10.10] - -$$ -\dot{q}_1 = q_2 + x(t) \n\dot{q}_2 = -2q_1 - 3q_2 + 2x(t) -$$ - -A realization for these equations is shown in Fig. 10.8b. It can be seen from Fig. 10.8a that the states *z*1 and *z*2 are decoupled, whereas the states *q*1 and *q*2 (Fig. 10.8b) are coupled. It should be remembered that Figs. 10.8a and 10.8b are realizations of the same system.† - -**Figure 10.8** Two realizations of the second-order system. - -† Here we have only a simulated state equation; the outputs are not shown. The outputs are linear combinations of state variables (and inputs). Hence, the output equation can be easily incorporated into these diagrams. - -### MATRIX DIAGONALIZATION VIA MATLAB - -The key to diagonalizing matrix **A** is to determine a matrix **P** that satisfies **P** = **PA** [Eq. (10.55)], where is a diagonal matrix of the eigenvalues of **A**. This problem is directly related to the classic eigenvalue problem, stated as - -### **AV** = **V** - -where **V** is a matrix of eigenvectors for **A**. If we can find **V**, we can take its inverse to determine **P**. That is, **P** = **V**−1 . This relationship is more fully developed in Sec. 10.8. - -MATLAB's built-in function eig can determine the eigenvectors of a matrix and, therefore, can help us determine a suitable matrix **P**. Let us demonstrate this approach for the current case. - -``` ->> A = [0 1;-2 -3]; B = [1; 2]; ->> [V, Lambda] = eig(A); ->> P = inv(V), Lambda, Bhat = P*B - P = 2.8284 1.4142 - 2.2361 2.2361 - Lambda = -1 0 - 0 -2 - Bhat = 5.6569 - 6.7082 -``` - -Therefore, - -$$ -\mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} 2.8284 & 1.4142 \\ 2.2361 & 2.2361 \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix} = \mathbf{P}\mathbf{q} -$$ - -and - -$$ -\dot{\mathbf{z}} = \begin{bmatrix} \dot{z}_1 \\ \dot{z}_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} + \begin{bmatrix} 5.6569 \\ 6.7082 \end{bmatrix} x(t) = \Lambda \mathbf{z} + \hat{\mathbf{B}} \mathbf{x} -$$ - -Recall that neither **P** nor **B**ˆ are unique, which explains why the MATLAB output does not need to match our previous solution. Still, the MATLAB results do their job and successfully diagonalize matrix **A**. - -## **[10.6 CONTROLLABILITY AND](#page-14-0) OBSERVABILITY** - -Consider a diagonalized state-space description of a system - -$$ -\dot{z} = \Lambda z + \dot{B}x \quad \text{and} \quad Y = \dot{C}z + Dx \tag{10.57} -$$ - -We shall assume that all *N* eigenvalues λ1, λ2, ... , λ*N* are distinct. The state equations in Eq. (10.57) are of the form - -$$ -\dot{z}_m = \lambda_m z_m + \hat{b}_{m1} x_1 + \hat{b}_{m2} x_2 + \cdots + \hat{b}_{mj} x_j \qquad m = 1, 2, \ldots, N -$$ - -### 948 CHAPTER 10 STATE-SPACE ANALYSIS - -If *b*ˆ*m*1, *b*ˆ*m*2, ... , *b*ˆ*mj* (the *m*th row in matrix **B**ˆ ) are all zero, then - -$$ -\dot{z}_m = \lambda_m z_m -$$ - -and the variable *zm* is uncontrollable because *zm* is not coupled to any of the inputs. Moreover, *zm* is decoupled from all the remaining (*N* − 1) state variables because of the diagonalized nature of the variables. Hence, there is no direct or indirect coupling of *zm* with any of the inputs, and the system is uncontrollable. In contrast, if at least one element in the *m*th row of **B**ˆ is nonzero, *zm* is coupled to at least one input and is therefore controllable. *Thus, a system with a diagonalized state [Eq. (10.57)] is completely controllable if and only if the matrix* **B**ˆ *has no row of zero elements*. - -The outputs [see Eq. (10.57)] are of the form - -$$ -y_i = \hat{c}_{i1}z_1 + \hat{c}_{i2}z_2 + \cdots + \hat{c}_{iN}z_N + \sum_{m=1}^j d_{im}x_m -$$ - $i = 1, 2, ..., k$ - -If *c*ˆ*im* =0, then the state *zm* will not appear in the expression for *yi*. Since all the states are decoupled because of the diagonalized nature of the equations, the state *zm* cannot be observed directly or indirectly (through other states) at the output *yi*. Hence, the *m*th mode *e*λ*mt* will not be observed at the output *yi*. If *c*ˆ1*m*, *c*ˆ2*m*, ... , *c*ˆ*km* (the *m*th column in matrix **C**ˆ ) are all zero, the state *zm* will not be observable at any of the *k* outputs, and the state *zm* is unobservable. In contrast, if at least one element in the *m*th column of **C**ˆ is nonzero, *zm* is observable at least at one output. *Thus, a system with diagonalized equations of the form in Eq. (10.57) is completely observable if and only if the matrix* **C**ˆ *has no column of zero elements*. In this discussion, we assumed distinct eigenvalues; for repeated eigenvalues, the modified criteria can be found in the literature [1, 2]. - -If the state-space description is not in diagonalized form, it may be converted into diagonalized form using the procedure in Ex. 10.11. It is also possible to test for controllability and observability even if the state-space description is in undiagonalized form [1, 2]. - -### **EXAMPLE 10.12 Controllability and Observability** - -Investigate the controllability and observability of the systems in Fig. 10.9. - -In both cases, the state variables are identified as the two integrator outputs, *q*1 and *q*2. The state equations for the system in Fig. 10.9a are - -$$ -\dot{q}_1 = q_1 + x \n\dot{q}_2 = q_1 - q_2 -$$ -\n(10.58) - -and - -$$ -y = \dot{q}_2 - q_2 = q_1 - 2q_2 -$$ - -**Figure 10.9** Systems for Ex. 10.12. - -Hence, - -$$ -\mathbf{A} = \begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 1 & -2 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ -$$ -|\mathbf{sI} - \mathbf{A}| = \begin{vmatrix} s-1 & 0 \\ -1 & s+1 \end{vmatrix} = (s-1)(s+1) -$$ - -Therefore, - -$$ -\lambda_1 = 1 \qquad \text{and} \qquad \lambda_2 = -1 -$$ - -and - -$$ -\mathbf{\Lambda} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} -$$ - -We shall now use the procedure in Sec. 10.5-1 to diagonalize this system. According to Eq. (10.55), we have - -$$ -\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} 1 & 0 \ 1 & -1 \end{bmatrix} -$$ - -The solution of this equation yields - -$$ -p_{12} = 0 -$$ - and $-2p_{21} = p_{22}$ - -Choosing *p*11 = 1 and *p*21 = 1, we have - -$$ -\mathbf{P} = \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix} -$$ - -and - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} -$$ - -All the rows of **B**ˆ are nonzero. Hence, the system is controllable. Also, - -$$ -\mathbf{Y} = \mathbf{C}\mathbf{q} = \mathbf{C}\mathbf{P}^{-1}\mathbf{z} = \hat{\mathbf{C}}\mathbf{z} -$$ - -and - -$$ -\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} = \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & -2 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 \end{bmatrix} -$$ - -The first column of **C**ˆ is zero. Hence, the mode *z*1 (corresponding to λ1 =1) is unobservable. The system is therefore controllable but not observable. We come to the same conclusion by realizing the system with the diagonalized state variables *z*1 and *z*2, whose state equations are - -$$ -\dot{z} = \Lambda z + Bx -$$ -$$ -y = \hat{C}z -$$ - -Using our previous calculations, we have - -$$ -\begin{aligned}\n\dot{z}_1 &= z_1 + x \\ -\dot{z}_2 &= -z_2 + x\n\end{aligned} -$$ - -and - -$$ -y\,{=}\,z_2 -$$ - -Figure 10.10a shows a realization of these equations. It is clear that each of the two modes is controllable, but the first mode (corresponding to λ = 1) is not observable at the output. - -The state equations for the system in Fig. 10.9b are - -$$ -\dot{q}_1 = -q_1 + x \n\dot{q}_2 = \dot{q}_1 - q_1 + q_2 = -2q_1 + q_2 + x -$$ -\n(10.59) - -and - -*y* = *q*2 - -Hence, - -$$ -\mathbf{A} = \begin{bmatrix} -1 & 0 \\ -2 & 1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ - -$$ -|s\mathbf{I} - \mathbf{A}| = \begin{vmatrix} s+1 & 0 \\ -1 & s-1 \end{vmatrix} = (s+1)(s-1) -$$ - - $x_2 = 1$ , and - -so that λ1 = −1, λ2 = 1, and - -$$ -\Lambda = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} -$$ - -Diagonalizing the matrix, we have - -$$ -\begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} p_{11} & p_{12} \ p_{21} & p_{22} \end{bmatrix} \begin{bmatrix} -1 & 0 \ -2 & 1 \end{bmatrix} -$$ - -The solution of this equation yields *p*11 = −*p*12 and *p*22 = 0. Choosing *p*11 = −1 and *p*21 = 1, we obtain - -$$ -\mathbf{P} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} -$$ - -and - -$$ -\hat{\mathbf{B}} = \mathbf{P}\mathbf{B} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}^{-1} = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \end{bmatrix} -$$ - -The first row of **B**ˆ is zero. Hence, the mode corresponding to λ1 = 1 is not controllable. However, since none of the columns of **C**ˆ vanish, both modes are observable at the output. Hence the system is observable but not controllable. - -We reach the same conclusion by realizing the system with the diagonalized state variables *z*1 and *z*2. The two state equations are - -$$ -\dot{z} = \Lambda z + \dot{B}x -$$ -$$ -y = \hat{C}z -$$ - -Using our previous calculations, we have - -$$ -\begin{aligned}\n\dot{z}_1 &= z_1\\ \n\dot{z}_2 &= -z_2 + x\n\end{aligned} -$$ - -and thus, - -*y* = *z*1 +*z*2 - -Figure 10.10b shows a realization of these equations. Clearly, each of the two modes is observable at the output, but the mode corresponding to λ1 = 1 is not controllable. - -### USING MATLAB TO DETERMINE CONTROLLABILITY AND OBSERVABILITY - -As demonstrated in Ex. 10.11, we can use MATLAB's eig function to determine the matrix **P** that will diagonalize **A**. We can then use **P** to determine **B**ˆ and **C**ˆ , from which we can determine the controllability and observability of a system. Let us demonstrate the process for the two present systems. - -First, let us use MATLAB to compute **B**ˆ and **C**ˆ for the system in Fig. 10.9a. - ->> A = [1 0;1 -1]; B = [1; 0]; C = [1 -2]; >> [V, Lambda] = eig(A); P=inv(V); Bhat = P\*B, Chat = C\*inv(P) Bhat = -0.5000 1.1180 Chat = -2 0 - -Since all the rows of **B**ˆ are nonzero, the system is controllable. However, one column of **C**ˆ is zero, so one mode is unobservable. - -Next, let us use MATLAB to compute **B**ˆ and **C**ˆ for the system in Fig. 10.9b. - ->> A = [-1 0;-2 1]; B = [1; 1]; C = [0 1]; >> [V, Lambda] = eig(A); P=inv(V); Bhat = P\*B, Chat = C\*inv(P) Bhat = 0 1.4142 Chat = 1.0000 0.7071 - -One of the rows of **B**ˆ is zero, so one mode is uncontrollable. Since all of the columns of **C**ˆ are nonzero, the system is observable. - -As expected, the MATLAB results confirm our earlier conclusions regarding the controllability and observability of the systems of Fig. 10.9. - -### **[10.6-1 Inadequacy of the Transfer Function Description of a System](#page-14-0)** - -Example 10.12 demonstrates the inadequacy of the transfer function to describe an LTI system in general. The systems in Figs. 10.9a and 10.9b both have the same transfer function - -$$ -H(s) = \frac{1}{s+1} -$$ - -Yet the two systems are very different. Their true nature is revealed in Figs. 10.10a and 10.10b, respectively. Both the systems are unstable, but their transfer function *H*(*s*) = 1/(*s* + 1) does not give any hint of it. Moreover, the systems are very different from the viewpoint of controllability and observability. The system in Fig. 10.9a is controllable but not observable, whereas the system in Fig. 10.9b is observable but not controllable. - -The transfer function description of a system looks at a system only from the input and output terminals. Consequently, the transfer function description can specify only the part of the system that is coupled to the input and the output terminals. From Figs. 10.10a and 10.10b, we see that in both cases only a part of the system that has a transfer function *H*(*s*) = 1/(*s* + 1) is coupled to the input and the output terminals. This is why both systems have the same transfer function *H*(*s*) = 1/(*s*+1). - -The state variable description [Eqs. (10.58) and (10.59)], on the other hand, contains all the information about these systems to describe them completely. The reason is that the state variable description is an internal description, not the external description obtained from the system behavior at external terminals. - -Apparently, the transfer function fails to describe these systems completely because the transfer functions of these systems have a common factor *s*−1 in the numerator and denominator; this common factor is canceled out in the systems in Fig. 10.9, with a consequent loss of the information. Such a situation occurs when a system is uncontrollable and/or unobservable. If a system is both controllable and observable (which is the case with most of the practical systems) the transfer function describes the system completely. In such a case, the internal and external descriptions are equivalent. - -## **[10.7 STATE-SPACE](#page-15-0) ANALYSIS OF DISCRETE-TIME SYSTEMS** - -We have shown that an *N*th-order differential equation can be expressed in terms of *N* first-order differential equations. In the following analogous procedure, we show that a general *N*th-order difference equation can be expressed in terms of *N* first-order difference equations. - -Consider the *z*-transfer function - -$$ -H[z] = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N} -$$ - -The input *x*[*n*] and the output *y*[*n*] of this system are related by the difference equation - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] = (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] -$$ - -The DFII realization of this equation is illustrated in Fig. 10.11. - -Signals appearing at the outputs of *N* delay elements are denoted by *q*1[*n*], *q*2[*n*], ... , *qN*[*n*]. The input of the first delay is *qN*[*n* + 1]. We can now write *N* equations, one at the input of each delay: - -$$ -q_1[n+1] = q_2[n] -$$ - -\n -$$ -q_2[n+1] = q_3[n] -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -q_{N-1}[n+1] = q_N[n] -$$ - -\n -$$ -q_N[n+1] = -a_Nq_1[n] - a_{N-1}q_2[n] - \dots - a_1q_N[n] + x[n] -$$ -\n(10.60) - -and - -*y*[*n*] = *bNq*1[*n*] +*bN*−1*q*2[*n*]+···+*b*1*qN*[*n*] +*b*0*qN*+1[*n*] - -We can eliminate *qN*+1[*n*] from this equation by using the last equation in Eq. (10.60) to yield - -$$ -y[n] = (b_N - b_0 a_N)q_1[n] + (b_{N-1} - b_0 a_{N-1})q_2[n] + \dots + (b_1 - b_0 a_1)q_N[n] + b_0 x[n] -$$ - -= $\hat{b}_N q_1[n] + \hat{b}_{N-1} q_2[n] + \dots + \hat{b}_1 q_N[n] + b_0 x[n]$ (10.61) - -where *b*ˆ*i* = *bi* −*b*0*ai*. - -Equation (10.60) shows *N* first-order difference equations in *N* variables *q*1[*n*], *q*2[*n*], ... , *qN*[*n*]. These variables should immediately be recognized as state variables, since the specification of the initial values of these variables in Fig. 10.11 will uniquely determine the response *y*[*n*] for a given *x*[*n*]. Thus, Eq. (10.60) represents the state equations, and Eq. (10.61) is the output equation. In matrix form, we can write these equations as - -$$ -\begin{bmatrix} q_{1}[n+1] \\ q_{2}[n+1] \\ \vdots \\ q_{N-1}[n+1] \\ \hline q_{N}[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ -a_{N} & -a_{N-1} & -a_{N-2} & \cdots & -a_{2} & -a_{1} \end{bmatrix} \begin{bmatrix} q_{1}[n] \\ q_{2}[n] \\ \vdots \\ q_{N-1}[n] \\ \hline q_{N}[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix} x[n] \quad (10.62) -$$ - -and - -$$ -\mathbf{y}[n] = \underbrace{\begin{bmatrix} \hat{b}_N & \hat{b}_{N-1} & \cdots & \hat{b}_1 \end{bmatrix}}_{C} \begin{bmatrix} q_1[n] \\ q_2[n] \\ \vdots \\ q_N[n] \end{bmatrix} + \underbrace{b_0}_{D} x[n] \qquad (10.63) -$$ - -In general, - -$$ -\mathbf{q}[n+1] = \mathbf{A}\mathbf{q}[n] + \mathbf{B}\mathbf{x}[n] -$$ -$$ -\mathbf{y}[n] = \mathbf{C}\mathbf{q}[n] + \mathbf{D}\mathbf{x}[n] -$$ - -Here we have represented a discrete-time system with state equations for DFII form. There are several other possible representations, as discussed in Sec. 10.3. We may, for example, use the cascade, parallel, or transpose of DFII forms to realize the system, or we may use some linear transformation of the state vector to realize other forms. In all cases, the output of each delay element qualifies as a state variable. We then write the equation at the input of each delay element. The *N* equations thus obtained are the *N* state equations. - -### **[10.7-1 Solution in State Space](#page-15-0)** - -Consider the state equation - -$$ -\mathbf{q}[n+1] = \mathbf{A}\mathbf{q}[n] + \mathbf{B}\mathbf{x}[n] -$$ - -From this equation, it follows that - -$$ -\mathbf{q}[n] = \mathbf{A}\mathbf{q}[n-1] + \mathbf{B}\mathbf{x}[n-1] -$$ - -and - -$$ -q[n-1] = Aq[n-2] + Bx[n-2] -$$ - -\n -$$ -q[n-2] = Aq[n-3] + Bx[n-3] -$$ - -\n: -\n: -\n -$$ -q[1] = Aq[0] + Bx[0] -$$ - -#### 956 CHAPTER 10 STATE-SPACE ANALYSIS - -Substituting the expression for **q**[*n*−1] into that for **q**[*n*], we obtain - -$$ -\mathbf{q}[n] = \mathbf{A}^2 \mathbf{q}[n-2] + \mathbf{A} \mathbf{B} \mathbf{x}[n-2] + \mathbf{B} \mathbf{x}[n-1] -$$ - -Substituting the expression for **q**[*n*−2] in this equation, we obtain - -$$ -q[n] = A^{3}q[n-3] + A^{2}Bx[n-3] + ABx[n-2] + Bx[n-1] -$$ - -Continuing in this way, we obtain - -$$ -\mathbf{q}[n] = \mathbf{A}^n \mathbf{q}[0] + \mathbf{A}^{n-1} \mathbf{B} \mathbf{x}[0] + \mathbf{A}^{n-2} \mathbf{B} \mathbf{x}[1] + \dots + \mathbf{B} \mathbf{x}[n-1] -$$ - -= $\mathbf{A}^n \mathbf{q}[0] + \sum_{m=0}^{n-1} \mathbf{A}^{n-1-m} \mathbf{B} \mathbf{x}[m]$ - -The upper limit of this summation is nonnegative. Hence, *n* ≥ 1, and the summation is recognized as the convolution sum - -$$ -\mathbf{A}^{n-1}u[n-1]*\mathbf{B}\mathbf{x}[n] -$$ - -Consequently, - -$$ -\mathbf{q}[n] = \underbrace{\mathbf{A}^n \mathbf{q}[0]}_{\text{zero input}} + \underbrace{\mathbf{A}^{n-1} u[n-1] * \mathbf{B} \mathbf{x}[n]}_{\text{zero state}} -$$ -(10.64) - -and - -$$ -\mathbf{y}[n] = \mathbf{C}\mathbf{q} + \mathbf{D}\mathbf{x} -$$ - -= $\mathbf{C}\mathbf{A}^n \mathbf{q}[0] + \sum_{m=0}^{n-1} \mathbf{C}\mathbf{A}^{n-1-m} \mathbf{B}\mathbf{x}[m] + \mathbf{D}\mathbf{x}$ -= $\mathbf{C}\mathbf{A}^n \mathbf{q}[0] + \mathbf{C}\mathbf{A}^{n-1}u[n-1] * \mathbf{B}\mathbf{x}[n] + \mathbf{D}\mathbf{x}$ (10.65) - -In Sec. 10.1-3, we showed that - -$$ -\mathbf{A}^n = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} + \beta_2 \mathbf{A}^2 + \dots + \beta_{N-1} \mathbf{A}^{N-1} -$$ - (10.66) - -where (assuming *N* distinct eigenvalues of **A**) - -$$ -\begin{bmatrix}\n\beta_0 \\ -\beta_1 \\ -\vdots \\ -\beta_{N-1}\n\end{bmatrix} = \begin{bmatrix}\n1 & \lambda_1 & \lambda_1^2 & \cdots & \lambda_1^{N-1} \\ -1 & \lambda_2 & \lambda_2^2 & \cdots & \lambda_2^{N-1} \\ -\vdots & \vdots & \vdots & \cdots & \vdots \\ -1 & \lambda_N & \lambda_N^2 & \cdots & \lambda_N^{N-1}\n\end{bmatrix}^{-1} \begin{bmatrix}\n\lambda_1^n \\ -\lambda_2^n \\ -\vdots \\ -\lambda_N^n\n\end{bmatrix} -$$ -\n(10.67) - -and λ1, λ2, ... , λ*N* are the *N* eigenvalues of **A**. - -We can also determine **A***n* from the *z*-transform formula, which will be derived later, in Eq. (10.71): - -$$ -\mathbf{A}^n = \mathcal{Z}^{-1}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}] -$$ - -### **EXAMPLE 10.13 State-Space Analysis of a Discrete-Time System** - -Give a state-space description of the system in Fig. 10.12. Find the output *y*[*n*] if the input *x*[*n*] = *u*[*n*] and the initial conditions are *q*1[0] = 2 and *q*2[0] = 3. - -6 **Figure 10.12** System for Ex. 10.13. - -Recognizing that *q*2[*n*] = *q*1[*n*+1], the state equations are [see Eqs. (10.62) and (10.63)] - -$$ -\begin{bmatrix} q_1[n+1] \\ q_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & \frac{5}{6} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} x -$$ - -and - -$$ -y[n] = \begin{bmatrix} -1 & 5 \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} -$$ - -To find the solution [Eq. (10.65)], we must first determine **A***n* . The characteristic equation of **A** is - -$$ -|\lambda \mathbf{I} - \mathbf{A}| = \begin{vmatrix} \lambda & -1 \\ \frac{1}{6} & \lambda - \frac{5}{6} \end{vmatrix} = \lambda^2 - \frac{5}{6}\lambda + \frac{1}{6} = \left(\lambda - \frac{1}{3}\right)\left(\lambda - \frac{1}{2}\right) = 0 -$$ - -Hence, λ1 = 1/3 and λ2 = 1/2 are the eigenvalues of **A** and [see Eq. (10.66)] - -$$ -\mathbf{A}^n = \beta_0 \mathbf{I} + \beta_1 \mathbf{A} -$$ - -where [see Eq. (10.67)] - -$$ -\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{3} \\ 1 & \frac{1}{2} \end{bmatrix}^{-1} \begin{bmatrix} \left(\frac{1}{3}\right)^n \\ \left(\frac{1}{2}\right)^n \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -6 & 6 \end{bmatrix} \begin{bmatrix} (3)^{-n} \\ (2)^{-n} \end{bmatrix} = \begin{bmatrix} 3(3)^{-n} - 2(2)^{-n} \\ -6(3)^{-n} + 6(2)^{-n} \end{bmatrix} -$$ - -and - -$$ -\mathbf{A}^{n} = [3(3)^{-n} - 2(2)^{-n}] \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + [-6(3)^{-n} + 6(2)^{-n}] \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & \frac{5}{6} \end{bmatrix} -$$ - -= -$$ -\begin{bmatrix} 3(3)^{-n} - 2(2)^{-n} & -6(3)^{-n} + 6(2)^{-n} \\ (3)^{-n} - (2)^{-n} & -2(3)^{-n} + 3(2)^{-n} \end{bmatrix} -$$ -(10.68) - -We can now determine the state vector **q**[*n*] from Eq. (10.64). Since we are interested in the output *y*[*n*], we shall use Eq. (10.65) directly. Note that - -$$ -CAn = [-1 \quad 5]An = [2(3)-n - 3(2)-n -4(3)-n + 9(2)-n] -$$ - -and the zero-input response is **CA***n* **q**[0], with - -$$ -\mathbf{q}[0] = \begin{bmatrix} 2 \\ 3 \end{bmatrix} -$$ - -Hence, the zero-input response is - -$$ -CAnq[0] = -8(3)-n + 21(2)-n -$$ - -The zero-state component is given by the convolution sum of **CA***n*−1 *u*[*n*−1] and **Bx**[*n*]. We can use the shifting property of the convolution sum [Eq. (3.32)] to obtain the zero-state component by finding the convolution sum of **CA***n u*[*n*] and **Bx**[*n*] and then replacing *n* with *n* − 1 in the result. We use this procedure because the convolution sums are listed in Table 3.1 for functions of the type *x*[*n*]*u*[*n*], rather than *x*[*n*]*u*[*n*−1]. - -$$ -\mathbf{CA}^{n}u[n] * \mathbf{B}x[n] = [2(3)^{-n} - 3(2)^{-n} - 4(3)^{-n} + 9(2)^{-n}] * \begin{bmatrix} 0 \\ u[n] \end{bmatrix} -$$ - -= -4(3)^{-n} \* u[n] + 9(2)^{-n} \* u[n] - -Using Table 3.1 (pair 4), we obtain - -$$ -\mathbf{CA}^{n}u[n] * \mathbf{B}x[n] = -4 \left[ \frac{1 - 3^{-(n+1)}}{1 - \frac{1}{3}} \right] u[n] + 9 \left[ \frac{1 - 2^{-(n+1)}}{1 - \frac{1}{2}} \right] u[n] -$$ -$$ -= [12 + 6(3^{-(n+1)}) - 18(2^{-(n+1)})]u[n] -$$ - -Now the desired (zero-state) response is obtained by replacing *n* by *n*−1. Hence, - -$$ -CAnu[n] * Bx[n-1] = [12+6(3)-n - 18(2)-n]u[n-1] -$$ - -It follows that - -$$ -y[n] = [-8(3)^{-n} + 21(2)^{-n}u[n] + [12 + 6(3)^{-n} - 18(2)^{-n}]u[n-1] -$$ - -This is the desired answer. We can simplify this answer by observing that 12 + 6(3)−*n* − 18(2)−*n* = 0 for *n* = 0. Hence, *u*[*n*−1] may be replaced by *u*[*n*], and - -$$ -y[n] = [12 - 2(3)^{-n} + 3(2)^{-n}]u[n] -$$ -\n(10.69) - -### USING MATLAB TO OBTAIN A GRAPHICAL SOLUTION - -MATLAB is equipped with tools to simulate digital systems, which makes it easy to obtain a graphical solution to the system. Let us use MATLAB simulation to determine the total system output over 0 ≤ *n* ≤ 25. - -- >> A = [0 1;-1/6 5/6]; B = [0; 1]; C = [-1 5]; D = 0; -- >> N = 25; n = (0:N); x = ones(1,N+1); q0 = [2;3]; -- >> sys = ss(A,B,C,D,-1); % Discrete-time state space model -- >> [y,q] = lsim(sys,x,n,q0); % Simulate output and state vector -- >> clf; stem(n,y,'k.'); xlabel('n'); ylabel('y[n]'); axis([-.5 25.5 11.5 13.5]); - -The MATLAB results, shown in Fig. 10.13, exactly align with the analytical solution derived earlier. Also notice that the zero-input and zero-state responses can be separately obtained using the same code and respectively setting either x or q0 to zero. - -## **10.7-2 The** *z***[-Transform Solution](#page-15-0)** - -The *z*-transform of Eq. (10.62) is given by - -$$ -z\mathbf{Q}[z] - z\mathbf{q}[0] = \mathbf{A}\mathbf{Q}[z] + \mathbf{B}\mathbf{X}[z] -$$ - -Therefore, - -$$ -(z\mathbf{I} - \mathbf{A})\mathbf{Q}[z] = z\mathbf{q}[0] + \mathbf{B}\mathbf{X}[z] -$$ - -and - -$$ -Q[z] = (zI - A)^{-1}zq[0] + (zI - A)^{-1}BX[z] -$$ -$$ -= (I - z^{-1}A)^{-1}q[0] + (zI - A)^{-1}BX[z] -$$ - -Hence, - -$$ -\mathbf{q}[n] = \underbrace{\mathcal{Z}^{-1}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}]\mathbf{q}[0]}_{\text{zero-input response}} + \underbrace{\mathcal{Z}^{-1}[(z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B}\mathbf{X}[z]]}_{\text{zero-state response}} -$$ -(10.70) - -#### 960 CHAPTER 10 STATE-SPACE ANALYSIS - -A comparison of Eq. (10.70) with Eq. (10.64) shows that - -$$ -\mathbf{A}^{n} = \mathcal{Z}^{-1}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}] -$$ - (10.71) - -The output equation is given by - -$$ -\mathbf{Y}[z] = \mathbf{C}\mathbf{Q}[z] + \mathbf{D}\mathbf{X}[z] -$$ - -= $\mathbf{C}[(\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0] + (z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B}\mathbf{X}[z]] + \mathbf{D}\mathbf{X}[z]$ -= $\mathbf{C}(\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0] + [\mathbf{C}(z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B} + \mathbf{D}]\mathbf{X}[z]$ -= $\underbrace{\mathbf{C}(\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0]}_{\text{zero-input response}} + \underbrace{\mathbf{H}[z]\mathbf{X}[z]}_{\text{zero-state response}}$ (10.72) - -where - -$$ -\mathbf{H}[z] = \mathbf{C}(z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B} + \mathbf{D} -$$ - (10.73) - -Note that **H**[*z*] is the transfer function matrix of the system, and *Hij*[*z*], the *ij*th element of **H**[*z*], is the transfer function relating the output *yi*[*n*] to the input *xj*[*n*]. If we define **h**[*n*] as - -$$ -\mathbf{h}[n] = \mathcal{Z}^{-1}[\mathbf{H}[z]] -$$ - -then **h**[*n*] represents the unit impulse function response matrix of the system. Thus, *hij*[*n*], the *ij*th element of **h**[*n*], represents the zero-state response *yi*[*n*] when the input *xj*[*n*] = δ[*n*] and all other inputs are zero. - -### **EXAMPLE 10.14** *z***-Transform Solution to State Equations** - -Use the *z*-transform to find the response *y*[*n*] for the system in Ex. 10.13. - -According to Eq. (10.72), - -$$ -\mathbf{Y}[z] = [-1 \quad 5] \left[ \frac{1}{\frac{1}{6z}} \quad \frac{-\frac{1}{z}}{1 - \frac{z}{6z}} \right]^{-1} \left[ \frac{2}{3} \right] + [-1 \quad 5] \left[ \frac{z}{\frac{1}{6}} \quad \frac{-1}{z - \frac{5}{6}} \right]^{-1} \left[ \frac{0}{\frac{z}{z - 1}} \right] -$$ -\n -$$ -= [-1 \quad 5] \left[ \frac{\frac{z(6z - 5)}{6z^2 - 5z + 1}}{\frac{-z}{6z^2 - 5z + 1}} \quad \frac{\frac{6z}{6z^2 - 5z + 1}}{\frac{6z^2 - 5z + 1}{6z^2 - 5z + 1}} \right] \left[ \frac{2}{3} \right] + [-1 \quad 5] \left[ \frac{\frac{z}{(z - 1)(z^2 - \frac{5}{6}z + \frac{1}{6})}}{\frac{z^2}{(z - 1)(z^2 - \frac{5}{6}z + \frac{1}{6})}} \right] -$$ -\n -$$ -= \frac{13z^2 - 3z}{z^2 - \frac{5}{6}z + \frac{1}{6}} + \frac{(5z - 1)z}{(z - 1)(z^2 - \frac{5}{6}z + \frac{1}{6})} -$$ -\n -$$ -= \frac{-8z}{z - \frac{1}{3}} + \frac{21z}{z - \frac{1}{2}} + \frac{12z}{z - 1} + \frac{12z}{z - 1} + \frac{6z}{z - \frac{1}{3}} - \frac{18z}{z - \frac{1}{2}} -$$ - -Therefore, - -$$ -y[n] = \underbrace{[-8(3)^{-n} + 21(2)^{-n}}_{\text{zero-input response}} + \underbrace{12 + 6(3)^{-n} - 18(2)^{-n}}_{\text{zero-state response}}]u[n] -$$ - -### LINEAR TRANSFORMATION, CONTROLLABILITY, AND OBSERVABILITY - -The procedure for linear transformation is parallel to that in the continuous-time case (Sec. 10.5). If **w** is the transformed-state vector given by - -**w** = **Pq** - -then - -$$ -\mathbf{w}[n+1] = \mathbf{P}\mathbf{A}\mathbf{P}^{-1}\mathbf{w}[n] + \mathbf{P}\mathbf{B}\mathbf{x} -$$ - -and - -$$ -\mathbf{y}[n] = (\mathbf{C}\mathbf{P}^{-1})\mathbf{w} + \mathbf{D}\mathbf{x} -$$ - -Controllability and observability may be investigated by diagonalizing the matrix, as explained in Sec. 10.5-1. - -## **[10.8 MATLAB: TOOLBOXES AND](#page-15-0) STATE-SPACE ANALYSIS** - -The preceding MATLAB sections provide a comprehensive introduction to the basic MATLAB environment. However, MATLAB also offers a wide range of toolboxes that perform specialized tasks. Once installed, toolbox functions operate no differently from ordinary MATLAB functions. Although toolboxes are purchased at extra cost, they save time and offer the convenience of predefined functions. It would take significant effort to duplicate a toolbox's functionality by using custom user-defined programs. - -Three toolboxes are particularly appropriate in the study of signals and systems: the control system toolbox, the signal-processing toolbox, and the symbolic math toolbox. Functions from these toolboxes have been utilized throughout the text in the MATLAB examples as well as certain end-of-chapter problems. This section provides a more formal introduction to a selection of functions, both standard and toolbox, that are appropriate for state-space problems. - -## **10.8-1** *z***[-Transform Solutions to Discrete-Time, State-Space Systems](#page-15-0)** - -As with continuous-time systems, it is often more convenient to solve discrete-time systems in the transform domain rather than in the time domain. As given in Ex. 10.13, consider the state-space description of the system shown in Fig. 10.12. - -$$ -\begin{bmatrix} q_1[n+1] \\ q_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & \frac{5}{6} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} x[n] -$$ - -and - -$$ -y[n] = [-1 \quad 5] \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} -$$ - -We are interested in the output *y*[*n*] in response to the input *x*[*n*] = *u*[*n*] with initial conditions *q*1[0] = 2 and *q*2[0] = 3. - -To describe this system, the state matrices **A**, **B**, **C**, and **D** are first defined. - ->> A = [0 1;-1/6 5/6]; B = [0; 1]; C = [-1 5]; D = 0; - -Additionally, the vector of initial conditions is defined. - ->> q\_0 = [2;3]; - -In the transform domain, the solution to the state equation is - -$$ -\mathbf{Q}[z] = (\mathbf{I} - z^{-1}\mathbf{A})^{-1}\mathbf{q}[0] + (z\mathbf{I} - \mathbf{A})^{-1}\mathbf{B}\mathbf{X}[z] \tag{10.74} -$$ - -The solution is separated into two parts: the zero-input response and the zero-state response. - -MATLAB's symbolic toolbox makes possible a symbolic representation of Eq. (10.74). First, a symbolic variable *z* needs to be defined. - ->> z = sym('z'); - -The sym command is used to construct symbolic variables, objects, and numbers. Typing whos confirms that z is indeed a symbolic object. The syms command is a shorthand command for constructing symbolic objects. For example, syms z s is equivalent to the two instructions z = sym('z'); and s = sym('s');. - -Next, a symbolic expression for *X*[*z*] needs to be constructed for the unit step input, *x*[*n*] = *u*[*n*]. The *z*-transform is computed by means of the ztrans command. - ->> X = ztrans(sym('1')) X = z/(z-1) - -Several comments are in order. First, the ztrans command assumes a causal signal. For *n* ≥ 0, *u*[*n*] has a constant value of 1. Second, the argument of ztrans needs to be a symbolic expression, even if the expression is a constant. Thus, a symbolic one sym('1') is required. Also note that continuous-time systems use Laplace transforms rather than *z*-transforms. In such cases, the laplace command replaces the ztrans command. - -Construction of **Q**[*z*] is now trivial. - -$$ -\begin{aligned}\n&\text{Q} = \text{inv}\left(\text{eye}(2) - z^(-1) * A\right) * q_0 + \text{inv}\left(z * \text{eye}(2) - A\right) * B * X \\ -&\text{Q} = \\ -&\left(18 * z\right) / \left(6 * z^2 - 5 * z + 1\right) + \left(2 * z * \left(6 * z - 5\right)\right) / \left(6 * z^2 - 5 * z + 1\right) + \left(6 * z\right) / \left((z - 1) * \left(6 * z^2 - 5 * z + 1\right)\right) \\ -&\left(18 * z^2\right) / \left(6 * z^2 - 5 * z + 1\right) - \left(2 * z\right) / \left(6 * z^2 - 5 * z + 1\right) + \left(6 * z^2\right) / \left((z - 1) * \left(6 * z^2 - 5 * z + 1\right)\right)\n\end{aligned} -$$ - -Unfortunately, not all MATLAB functions work with symbolic objects. Still, the symbolic toolbox overloads many standard MATLAB functions, such as inv, to work with symbolic objects. Recall that overloaded functions have identical names but different behavior; proper function selection is typically determined by context. - -The expression Q is somewhat unwieldy. The simplify command uses various algebraic techniques to simplify the result. - -``` ->> Q = simplify(Q) - Q = -(2*z*(- 6*z^2 + 2*z + 1))/(6*z^3 - 11*z^2 + 6*z - 1) - (2*z*(9*z^2 - 7*z + 1))/(6*z^3 - 11*z^2 + 6*z - 1) -``` - -The resulting expression is mathematically equivalent to the original but notationally more compact. - -Since **D** = 0, the output *Y*[*z*] is given by *Y*[*z*] = **CQ**[*z*]. - -``` ->> Y = simplify(C*Q) - Y = (6*z*(13*z^2 - 11*z + 2))/(6*z^3 - 11*z^2 + 6*z - 1) -``` - -The corresponding time-domain expression is obtained by using the inverse *z*-transform command iztrans. - ->> y = iztrans(Y) y = 3\*(1/2)^n - 2\*(1/3)^n + 12 - -Like ztrans, the iztrans command assumes a causal signal, so the result implies multiplication by a unit step. That is, the system output is *y*[*n*] =(3(1/2)*n*−2(1/3)*n*+12)*u*[*n*], which is equivalent to Eq. (10.69) derived in Ex. 10.13. Continuous-time systems use inverse Laplace transforms rather than inverse *z*-transforms. In such cases, the ilaplace command therefore replaces the iztrans command. - -Following a similar procedure, it is a simple matter to compute the zero-input response *y*zir[*n*]: - ->> y\_zir = iztrans(simplify(C\*inv(eye(2)-z^(-1)\*A)\*q\_0)) y\_zir = 21\*(1/2)^n - 8\*(1/3)^n - -The zero-state response is given by - ->> y\_zsr=y- y\_zir y\_zsr = 6\*(1/3)^n - 18\*(1/2)^n + 12 - -Typing iztrans(simplify(C\*inv(z\*eye(2)-A)\*B\*X)) produces the same result. - -MATLAB plotting functions, such as plot and stem, do not directly support symbolic expressions. By using the subs command, however, it is easy to replace a symbolic variable with a vector of desired values. - -**Figure 10.14** Output *y*[*n*] computed by using the symbolic math toolbox. - -``` ->> n = [0:25]; stem(n,subs(y,n),'k.'); ->> xlabel('n'); ylabel('y[n]'); axis([-.5 25.5 11.5 13.5]); -``` - -Figure 10.14 shows the results, which are equivalent to the results obtained in Ex. 10.13. Although there are plotting commands in the symbolic math toolbox such as ezplot that plot symbolic expression, these plotting routines lack the flexibility needed to satisfactorily plot discrete-time functions. - -### **[10.8-2 Transfer Functions from State-Space Representations](#page-15-0)** - -A system's transfer function provides a wealth of useful information. From Eq. (10.73), the transfer function for the system described in Ex. 10.13 is - -``` ->> H = collect(simplify(C*inv(z*eye(2)-A)*B+D)) - H = (30*z - 6)/(6*z^2 - 5*z + 1) -``` - -It is also possible to determine the numerator and denominator transfer function coefficients from a state-space model by using the signal-processing toolbox function ss2tf. - ->> [num,den] = ss2tf(A,B,C,D) num = 0 5.0000 -1.0000 den = 1.0000 -0.8333 0.1667 - -The denominator of *H*[*z*] provides the characteristic polynomial - -γ 2 − 5 6γ + 1 6 - -Equivalently, the characteristic polynomial is the determinant of (*z***I**−**A**). - ->> syms gamma; char\_poly = subs(det(z\*eye(2)-A),z,gamma) char\_poly = gamma^2 - (5\*gamma)/6 + 1/6 - -Here, the subs command replaces the symbolic variable z with the desired symbolic variable gamma. - -The roots command does not accommodate symbolic expressions. Thus, the sym2poly command converts the symbolic expression into a polynomial coefficient vector suitable for the roots command. - ->> roots(sym2poly(char\_poly)) ans = 0.5000 0.3333 - -Taking the inverse *z*-transform of *H*[*z*] yields the impulse response *h*[*n*]. - ->> h = iztrans(H) h = 18\*(1/2)^n - 12\*(1/3)^n - 6\*kroneckerDelta(n, 0) - -As suggested by the characteristic roots, the characteristic modes of the system are (1/2)*n* and (1/3)*n*. Notice that the symbolic math toolbox represents δ[*n*] as kroneckerDelta(n, 0). In general, δ[*n* − *a*] is represented as kroneckerDelta(n-a, 0). This notation is frequently encountered. Consider, for example, delaying the input *x*[*n*] = *u*[*n*] by 2, *x*[*n* − 2] = *u*[*n* − 2]. In the transform domain, this is equivalent to *z*−2*X*[*z*]. Taking the inverse *z*-transform of *z*−2*X*[*z*] yields - -$$ -\Rightarrow \text{ 27 (--2)*X} = \text{ 27 (--2)*X} -$$ -\n -$$ -\text{ans} = 1 - \text{ 27 (--2)*X} = \text{ 27 (--2)*X} -$$ - -That is, MATLAB represents the delayed unit step *u*[*n*−2] as (−δ[*n*−1] −δ[*n*−0] +1)*u*[*n*]. The transfer function also permits convenient calculation of the zero-state response. - ->> -$$ -y\_zsr = iztrans(H*X) -$$ - -y\_ $zsr = 6*(1/3)^n - 18*(1/2)^n + 12$ - -The result agrees with previous calculations. - -### **[10.8-3 Controllability and Observability of Discrete-Time Systems](#page-15-0)** - -In their controllability and observability, discrete-time systems are analogous to continuous-time systems. For example, consider the LTID system described by the constant coefficient difference equation - -$$ -y[n] + \frac{5}{6}y[n-1] + \frac{1}{6}y[n-2] = x[n] + \frac{1}{2}x[n-1] -$$ - -Figure 10.15 illustrates the direct form II (DFII) realization of this system. The system input is *x*[*n*], the system output is *y*[*n*], and the outputs of the delay blocks are designated as state variables *q*1[*n*] and *q*2[*n*]. - -The corresponding state and output equations (see Prob. 10.7-1) are - -$$ -\mathbf{Q}[n+1] = \begin{bmatrix} q_1[n+1] \\ q_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{1}{6} & -\frac{5}{6} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} x[n] = \mathbf{A}\mathbf{Q}[n] + \mathbf{B}x[n] -$$ - -and - -$$ -y[n] = \begin{bmatrix} -\frac{1}{6} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} q_1[n] \\ q_2[n] \end{bmatrix} + 1x[n] = \mathbf{CQ}[n] + \mathbf{D}x[n] -$$ - -To describe this system in MATLAB, the state matrices **A**, **B**, **C**, and **D** are first defined. - ->> A = -$$ -[0 \ 1; -1/6 \ -5/6] -$$ -; B = $[0; 1]$ ; C = $[-1/6 \ -1/3]$ ; D = 1; - -**Figure 10.15** Direct form II realization of *y*[*n*] + (5/6)*y*[*n* − 1] + (1/6)*y*[*n* − 2] = *x*[*n*] + (1/2)*x*[*n*−1]. - -To assess the controllability and observability of this system, the state matrix **A** needs to be diagonalized.† As shown in Eq. (10.55), this requires a transformation matrix **P** such that - -$$ -PA = AP -$$ - (10.75) - -where is a diagonal matrix containing the unique eigenvalues of **A**. Recall, the transformation matrix **P** is not unique. - -To determine a matrix **P**, it is helpful to review the eigenvalue problem. Mathematically, an eigendecomposition of **A** is expressed as - -$$ -AV = V\Lambda -$$ - -where **V** is a matrix of eigenvectors and is a diagonal matrix of eigenvalues. Pre- and post-multiplying both sides of this equation by **V**−1 yields - -$$ -\mathbf{V}^{-1}\mathbf{A}\mathbf{V}\mathbf{V}^{-1} = \mathbf{V}^{-1}\mathbf{V}\mathbf{\Lambda}\mathbf{V}^{-1} -$$ - -Simplification yields - -$$ -\mathbf{V}^{-1}\mathbf{A} = \mathbf{\Lambda}\mathbf{V}^{-1} \tag{10.76} -$$ - -Comparing Eqs. (10.75) and (10.76), we see that a suitable transformation matrix **P** is given by an inverse eigenvector matrix **V**−1 . - -The eig command is used to verify that **A** has the required distinct eigenvalues as well as compute the needed eigenvector matrix **V**. - -``` ->> [V,Lambda] = eig(A) - V = - 0.9487 -0.8944 - -0.3162 0.4472 - Lambda = - -0.3333 0 - 0 -0.5000 -``` - -Since the diagonal elements of Lambda are all unique, a transformation matrix **P** is given by - ->> P = inv(V); - -The transformed state matrices **A**ˆ = **PAP**−1 , **B**ˆ = **PB**, and **C**ˆ = **CP**−1 are easily computed by using transformation matrix **P**. Notice that matrix **D** is unaffected by state variable transformations. - -``` ->> Ahat = P*A*inv(P), Bhat = P*B, Chat = C*inv(P) - Ahat = - -0.3333 -0.0000 - 0.0000 -0.5000 -``` - -† This approach requires that the state matrix **A** have unique eigenvalues. Systems with repeated roots require that state matrix **A** be transformed into a modified diagonal form, also called the Jordan form. The MATLAB function jordan is used in these cases. - -``` -Bhat = - 6.3246 - 6.7082 -Chat = - -0.0527 -0.0000 -``` - -The proper operation of **P** is verified by the correct diagonalization of **A**, **A**ˆ = . Since no row of **B**ˆ is zero, the system is controllable. Since, however, at least one column of **C**ˆ is zero, the system is not observable. These characteristics are no coincidence. The DFII realization, which is more descriptively called the controller canonical form, is always controllable but not always observable. - -As a second example, consider the same system realized using the transposed direct form II structure (TDFII), as shown in Fig. 10.16. The system input is *x*[*n*], the system output is *y*[*n*], and the outputs of the delay blocks are designated as state variables *v*1[*n*] and *v*2[*n*]. - -The corresponding state and output equations (see Prob. 10.7-2) are - -$$ -\mathbf{V}[n+1] = \begin{bmatrix} v_1[n+1] \\ v_2[n+1] \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{6} \\ 1 & -\frac{5}{6} \end{bmatrix} \begin{bmatrix} v_1[n] \\ v_2[n] \end{bmatrix} + \begin{bmatrix} -\frac{1}{6} \\ -\frac{1}{3} \end{bmatrix} x[n] = \mathbf{A}\mathbf{V}[n] + \mathbf{B}x[n] -$$ - -and - -$$ -y[n] = [0 \quad 1] \begin{bmatrix} v_1[n] \\ v_2[n] \end{bmatrix} + 1x[n] = \mathbf{CV}[n] + \mathbf{D}x[n] -$$ - -To describe this system in MATLAB, the state matrices **A**, **B**, **C**, and **D** are defined. - ->> A = -$$ -[0 -1/6; 1 -5/6] -$$ -; B = $[-1/6; -1/3]$ ; C = $[0 1]$ ; D = 1; - -To diagonalize **A**, a transformation matrix **P** is created. - ->> [V, Lambda] = eig(A) -\n -$$ -V = -$$ - 0.4472 0.3162 -\n0.8944 0.9487 -\nLambda = -\n-0.3333 0 -\n0 -0.5000 - -6 – **Figure 10.16** Transposed direct form II realization of *y*[*n*]+(5/6)*y*[*n*−1]+(1/6)*y*[*n*−2] = *x*[*n*]+(1/2)*x*[*n*−1]. - -The characteristic modes of a system do not depend on implementation, so the eigenvalues of the DFII and TDFII realizations are the same. However, the eigenvectors of the two realizations are quite different. Since the transformation matrix **P** depends on the eigenvectors, different realizations can possess different observability and controllability characteristics. - -Using transformation matrix **P**, the transformed state matrices **A**ˆ = **PAP**−1 , **B**ˆ = **PB**, and **C**ˆ = **CP**−1 are computed. - -``` ->> P = inv(V); ->> Ahat = P*A*inv(P), Bhat = P*B, Chat = C*inv(P) - Ahat = - -0.3333 0 - 0.0000 -0.5000 - Bhat = - -0.3727 - -0.0000 - Chat = - 0.8944 0.9487 -``` - -Again, the proper operation of **P** is verified by the correct diagonalization of **A**, **A**ˆ = . Since no column of **C**ˆ is zero, the system is observable. However, at least one row of **B**ˆ is zero, and therefore the system is not controllable. The TDFII realization, which is more descriptively called the observer canonical form, is always observable but not always controllable. It is interesting to note that the properties of controllability and observability are influenced by the particular realization of a system. - -### **[10.8-4 Matrix Exponentiation and the Matrix Exponential](#page-15-0)** - -Matrix exponentiation is important to many problems, including the solution of discrete-time state-space equations. Equation (10.64), for example, shows that the state response requires matrix exponentiation, **A***n* . For a square **A** and specific *n*, MATLAB happily returns **A***n* by using the ^ operator. From the system in Ex. 10.13 and *n* = 3, we have - -``` ->> A = [0 1;-1/6 5/6]; n = 3; A^n - ans = - -0.1389 0.5278 - -0.0880 0.3009 -``` - -The same result is also obtained by typing A\*A\*A. - -Often, it is useful to solve **A***n* symbolically. Noting **A***n* = *Z*−1[(**I** − *z*−1**A**)−1], the symbolic toolbox can produce a symbolic expression for **A***n* . - -``` ->> syms z n; An = simplify(iztrans(inv(eye(2)-z^(-1)*A))) - An = - [ 3*(1/3)^n - 2*(1/2)^n, 6*(1/2)^n - 6*(1/3)^n] - [ 1/3^n - 1/2^n, 3/2^n - 2/3^n] -``` - -Notice that this result is identical to Eq. (10.68), derived earlier. Substituting the case *n* = 3 into An provides a result that is identical to the one elicited by the previous A^n command. - ->> double(subs(An,n,3)) ans = -0.1389 0.5278 -0.0880 0.3009 - -For continuous-time systems, the matrix exponential *e***A***t* is commonly encountered. The expm command can compute the matrix exponential symbolically. Using the system from Ex. 10.8 yields - -``` ->> syms t; A = [-12 2/3;-36 -1]; eAt = simplify(expm(A*t)) - eAt = - [ -(exp(-9*t)*(3*exp(5*t) - 8))/5, (2*exp(-9*t)*(exp(5*t) - 1))/15] - [ -(36*exp(-9*t)*(exp(5*t) - 1))/5, (exp(-9*t)*(8*exp(5*t) - 3))/5] -``` - -This result is identical to the result computed in Ex. 10.8. Similar to the discrete-time case, an identical result is obtained by typing syms s; simplify(ilaplace(inv(s\*eye(2)-A))). - -For a specific *t*, the matrix exponential is also easy to compute, either through substitution or direct computation. Consider the case *t* = 3. - ->> double(subs(eAt,t,3)) ans = 1.0e-004 \* -0.0369 0.0082 -0.4424 0.0983 - -The command expm(A\*3) produces the same result. - -## **[10.9 SUMMARY](#page-15-0)** - -An *N*th-order system can be described in terms of *N* key variables—the state variables of the system. The state variables are not unique; rather, they can be selected in a variety of ways. Every possible system output can be expressed as a linear combination of the state variables and the inputs. Therefore, the state variables describe the entire system, not merely the relationship between certain input(s) and output(s). For this reason, the state variable description is an internal description of the system. Such a description is therefore the most general system description, and it contains the information of the external descriptions, such as the impulse response and the transfer function. The state variable description can also be extended to time-varying parameter systems and nonlinear systems. An external description of a system may not characterize the system completely. - -The state equations of a system can be written directly from knowledge of the system structure, from the system equations, or from the block diagram representation of the system. State equations consist of a set of *N* first-order differential equations and can be solved by time-domain or frequency-domain (transform) methods. Suitable procedures exist to transform one given set of state variables into another. Because a set of state variables is not unique, we can have an infinite variety of state-space descriptions of the same system. The use of an appropriate transformation allows us to see clearly which of the system states are controllable and which are observable. - -### **[REFERENCES](#page-15-0)** - -- 1. Kailath, Thomas. *Linear Systems.* Prentice-Hall, Englewood Cliffs, NJ, 1980. -- 2. Zadeh, L., and C. Desoer. *Linear System Theory.* McGraw-Hill, New York, 1963. - -## **[PROBLEMS](#page-15-0)** - -- **10.1-1** Convert each of the following second-order differential equations into a set of two first-order differential equations (state equations). State which of the sets represent nonlinear equations. - - (a) *y*¨ +10*y*˙ +2*y* = *x* - -1 + 1 H \_ - -network in Fig. P10.2-2. - -network in Fig. P10.2-3. - -**Figure P10.2-1** - -1 2 F - -- (b) *y*¨ +2*eyy*˙ +log*y* = *x* -- (c) *y*¨ +φ1(*y*)*y*˙ +φ2(*y*)*y* = *x* -- **10.2-1** Write the state equations for the *RLC* network in Fig. P10.2-1. - -*x* 3 - -**10.2-2** Write the state and output equations for the - -**10.2-3** Write the state and output equations for the - -2 - -**10.2-4** Write the state and output equations for the electrical network in Fig. P10.2-4. - -**Figure P10.2-4** - -**Figure P10.2-2** - -**10.2-5** Write the state and output equations for the network in Fig. P10.2-5. - -**10.2-6** Write the state and output equations of the system shown in Fig. P10.2-6. - -**Figure P10.2-6** - -**10.2-7** Write the state and output equations of the system shown in Fig. P10.2-7. - -**Figure P10.2-7** - -**10.2-8** For a system specified by the transfer function - -$$ -H(s) = \frac{3s + 10}{s^2 + 7s + 12} -$$ - -write sets of state equations for DFII and its transpose, cascade, and parallel forms. Also write the corresponding output equations. - -**10.2-9** Repeat Prob. 10.2-8 for - -\n(a) - -\n -$$ -H(s) = \frac{4s}{(s+1)(s+2)^2} -$$ -\n(b) - -\n -$$ -H(s) = \frac{s^3 + 7s^2 + 12s}{(s+1)^3(s+2)} -$$ - -**10.3-1** Find the state vector **q**(*t*) by using the Laplace transform method if - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x} -$$ - -where - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 2 \\ -1 & -3 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \qquad x(t) = 0 -$$ - -**10.3-2** Repeat Prob. 10.3-1 for - -$$ -\mathbf{A} = \begin{bmatrix} -5 & -6 \\ 1 & 0 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 5 \\ 4 \end{bmatrix} \qquad \qquad x(t) = \sin 100t -$$ - -**10.3-3** Repeat Prob. 10.3-1 for - -$$ -\mathbf{A} = \begin{bmatrix} -2 & 0 \\ 1 & -1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \qquad x(t) = u(t) -$$ - -**10.3-4** Repeat Prob. 10.3-1 for - -$$ -\mathbf{A} = \begin{bmatrix} -1 & 1 \\ 0 & -2 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} -$$ -$$ -\mathbf{q}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \qquad \mathbf{x} = \begin{bmatrix} u(t) \\ \delta(t) \end{bmatrix} -$$ - -**10.3-5** Use the Laplace transform method to find the response *y* for - -$$ -\dot{\mathbf{q}} = \mathbf{A}\mathbf{q} + \mathbf{B}\mathbf{x}(t) -$$ -$$ -y = \mathbf{C}\mathbf{q} + \mathbf{D}\mathbf{x}(t) -$$ - -where - -$$ -\mathbf{A} = \begin{bmatrix} -3 & 1 \\ -2 & 0 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} -$$ -$$ -\mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ - -and - -$$ -x(t) = u(t) \qquad \mathbf{q}(0) = \begin{bmatrix} 2 \\ 0 \end{bmatrix} -$$ - -**10.3-6** Repeat Prob. 10.3-5 for - -$$ -\mathbf{A} = \begin{bmatrix} -1 & 1 \\ -1 & -1 \end{bmatrix} \quad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\mathbf{C} = \begin{bmatrix} 1 & 1 \end{bmatrix} \quad \mathbf{D} = 1 -$$ -$$ -x(t) = u(t) \quad \mathbf{q}(0) = \begin{bmatrix} 2 \\ 1 \end{bmatrix} -$$ - -! - -**10.3-7** The transfer function *H*(*s*) in Prob. 10.2-8 is realized as a cascade of *H*1(*s*) followed by *H*2(*s*), where - -$$ -H_1(s) = \frac{1}{s+3} -$$ - -$$ -H_2(s) = \frac{3s+10}{s+4} -$$ - -Let the outputs of these subsystems be state variables *q*1 and *q*2, respectively. Write the state equations and the output equation for this system and verify that **H**(*s*) = **C***φ*(*s*)**B**+**D**. - -- **10.3-8** Find the transfer function matrix **H**(*s*) for the system in Prob. 10.3-5. -- **10.3-9** Find the transfer function matrix **H**(*s*) for the system in Prob. 10.3-6. -- **10.3-10** Find the transfer function matrix **H**(*s*) for the system - -$$ -\dot{q} = Aq + Bx -$$ -$$ -y = Cq + Dx -$$ - -where - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -1 & -2 \end{bmatrix} \quad \mathbf{B} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \quad \mathbf{x} = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} -$$ - -$$ -\mathbf{C} = \begin{bmatrix} 1 & 2 \\ 4 & 1 \\ 1 & 1 \end{bmatrix} \qquad \mathbf{D} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{bmatrix} -$$ - -- **10.3-11** Repeat Prob. 10.3-1, using the time-domain method. -- **10.3-12** Repeat Prob. 10.3-2, using the time-domain method. -- **10.3-13** Repeat Prob. 10.3-3, using the time-domain method. -- **10.3-14** Repeat Prob. 10.3-4, using the time-domain method. -- **10.3-15** Repeat Prob. 10.3-5, using the time-domain method. -- **10.3-16** Repeat Prob. 10.3-6, using the time-domain method. -- **10.3-17** Find the unit impulse response matrix **h**(*t*) for the system in Prob. 10.3-7, using Eq. (10.45). -- **10.3-18** Find the unit impulse response matrix **h**(*t*) for the system in Prob. 10.3-6. -- **10.3-19** Find the unit impulse response matrix **h**(*t*) for the system in Prob. 10.3-10. -- **10.4-1** The state equations of a certain system are given as - -$$ -\dot{q}_1 = q_2 + 2x -$$ - -$$ -\dot{q}_2 = -q_1 - q_2 + x -$$ - -Define a new state vector **w** such that - -$$ -w_1 = q_2 -$$ - -$$ -w_2 = q_2 - q_1 -$$ - -Find the state equations of the system with **w** as the state vector. Determine the characteristic roots (eigenvalues) of the matrix **A** in the original and the transformed state equations. - -**10.4-2** The state equations of a certain system are - -$$ -\dot{q}_1 = q_2 \n\dot{q}_2 = -2q_1 - 3q_2 + 2x -$$ - -(a) Determine a new state vector **w** (in terms of vector **q**) such that the resulting state equations are in diagonalized form. - -(b) For output **y** given by - -$$ -y = Cq + Dx -$$ - -where - -$$ -\mathbf{C} = \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix} \qquad \mathbf{D} = 0 -$$ - -determine the output **y** in terms of the new state vector **w**. - -### **10.4-3** Given a system - -$$ -\dot{\mathbf{q}} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & -3 \end{bmatrix} \mathbf{q} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} x -$$ - -determine a new state vector **w** such that the state equations are diagonalized. - -**10.4-4** The state equations of a certain system are given in diagonalized form as - -$$ -\dot{\mathbf{q}} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -2 \end{bmatrix} \mathbf{q} + \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} x -$$ - -The output equation is given by - -$$ -y = \begin{bmatrix} 1 & 3 & 1 \end{bmatrix} \mathbf{q} -$$ - -Determine the output *y* for - -$$ -\mathbf{q}(0) = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \qquad x(t) = u(t) -$$ - -**10.5-1** Write the state equations for the systems depicted in Fig. P10.5-1. Determine a new state vector **w** such that the resulting state equations are in diagonalized form. Write the output **y** in terms of **w**. Determine in each case whether the system is controllable and observable. - -### **Figure P10.5-1** - -**10.6-1** An LTI discrete-time system is specified by - - 2 - -! - -$$ -\mathbf{A} = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \qquad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} -$$ -$$ -\mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} \qquad \mathbf{D} = \begin{bmatrix} 1 \end{bmatrix} -$$ - -and - -- **q**(0) = 1 *x*[*n*] = *u*[*n*] -- (a) Find the output *y*[*n*], using the timedomain method. -- (b) Find the output *y*[*n*], using the frequencydomain method. -- **10.6-2** An LTI discrete-time system is specified by the difference equation - -$$ -y[n+2] + y[n+1] + 0.16y[n] -$$ - -= $x[n+1] + 0.32x[n]$ - -- (a) Show the DFII, its transpose, cascade, and parallel realizations of this system. -- (b) Write the state and the output equations from these realizations, using the output of each delay element as a state variable. -- **10.6-3** Repeat Prob. 10.6-2 for - -$$ -y[n+2] + y[n+1] - 6y[n] -$$ - -= 2x[n+2] + x[n+1] - -- **10.7-1** Verify the state and output equations for the LTID system shown in Fig. 10.15. -- **10.7-2** Verify the state and output equations for the LTID system shown in Fig. 10.16. - -# **[INDEX](#page-15-0)** - -Abscissa of convergence, 336 Accumulator systems, 259, 295, 519 discrete-time Fourier transform of, 875–76 Active circuits, 382–85 Adders, 388, 396, 399, 403 Addition of complex numbers, 11–12 of matrices, 38 of sinusoids, 18–20 Additivity, 97–98 Algebra of complex numbers, 5–15 matrix, 38–42 Aliasing, 536–38, 788–95, 805, 811–12, 817, 834 defined, 536 general condition for in sinusoids, 793–96 treachery of, 788–91 verification of in sinusoids, 792–93 Aliasing error, 659, 811 Amplitude, 16 Amplitude modulation, 711–13, 736–49, 762 Amplitude response, 413, 416–17, 421, 424, 435, 437–38, 440–42 Amplitude spectrum, 598, 607, 615–25, 667, 668, 707, 848, 870, 878 Analog filters, 261 Analog signals, 133 defined, 78 digital processing of, 547–54 properties of, 78 Analog systems, 109, 135, 261 Analog-to-digital (A/D) conversion, 799–802, 831 Analog-to-digital converters (ADC) bit number, 801–2 bit rate, 801–2 Angle modulation, 736, 763 Angles electronic calculators in computing, 8–11 principal value of, 9 Angular acceleration, 116 Angular position, 116 Angular velocity, 116 - -Anti-aliasing filters, 537, 791, 834 Anti-causal exponentials, 862–63 Anti-causal signals, 81 Aperiodic signals, 133 discrete-time Fourier integral and, 855–67 Fourier integral and, 680–89, 762 properties of, 78–82 Apparent frequency, 534–36, 792–96 *Ars Magna* (Cardano), 2–5 Associative property, 171, 283 Asymptotic stability. *See* Internal stability. Audio signals, 713–14, 725, 746 Automatic position control system, 406–8 Auxiliary conditions, 153 differential equation solution and, 161 Backward difference system, 258, 295, 519, 568–69 Bandlimited signals, 533, 788, 792, 802 Bandpass filters, 441–43, 542–44, 749, 882–83, 896 group delay and, 726–27 ideal, 730–31, 882–83 poles and zeros of *H*(*s*) and, 443 Bandstop filters, 441–42, 445, 545–46 Bandwidth, 628 continuous-time systems and, 208–10 data truncation and, 751–53 essential, 736, 758–59 Fourier transform and, 692, 706, 762–63 Bartlett window, 753–55 Baseband signals, 737–40, 746–47, 749 Basis signals, 651, 655, 668 Basis vectors, 648 Beat effect, 741 Bhaskar, 2 Bilateral Laplace transform, 330, 335–37, 445–55, 467 properties of, 451–55 Bilateral *z*-transform, 431, 490, 554–63 in discrete-time system analysis, 563 properties of, 559–60 Bilinear transformation, 569–70 Binary digital signals, 799 Black box, 95, 119, 120 - -Blackman window, 755, 761 Block diagrams, 386–88, 405, 407, 408, 519 Bôcher, M., 620–21 Bode plots, 419–35 constant of, 421 first-order pole and, 424–27 pole at the origin and, 422–23 second-order pole and, 426–35 Bombelli, Raphael, 3 Bonaparte, Napoleon, 347, 610–11 Bounded-input/bounded-output (BIBO) stability, 110, 135, 263 of continuous-time systems, 196–97, 199–203, 222–23 of discrete-time systems, 298–99, 301–4, 314, 526, 527 frequency response and, 412–13 internal stability relationship to, 199–203, 301–4 of the Laplace transform, 371–73 signal transmission and, 721 steady-state response and, 418 of the *z*-transform, 518 Butterfly signal flow graph, 825 Butterworth filters, 440, 551–54 cascaded second-order sections for Butterworth filter realization, 461–63 MATLAB on, 459–63 transformation of, 571–72 - -Canonic direct realization. *See* Direct form II realization Cardano, Gerolamo, 2–5 Cartesian form, 8–15 Cascade realization, 394, 526, 920, 923 Cascade systems, 190, 192, 372–73 Cascaded *RC* filters, 461–62 Causal exponentials, 861–62 Causal signals, 81, 83, 134 Causal sinusoidal input in continuous-time systems, 418–19 in discrete-time systems, 527 Causal systems, 104–6, 135, 263 properties of, 104–6 zero-state response and, 172, 283 Cayley–Hamilton theorem, 910–12 Characteristic equations of continuous-time systems, 153–55 of discrete-time systems, 271, 273, 309 of a matrix, 910–12, 933 Characteristic functions, 192 Characteristic modes of continuous-time systems, 153–55, 162–65, 167, 170, 196, 198–99, 203–6 of discrete-time systems, 271–74, 278–79, 297–301, 305, 313 Characteristic polynomials of continuous-time systems, 153–56, 164, 166, 202–3, 220 of discrete-time systems, 271, 274–75, 279, 303–4 of the Laplace transform, 371–72 of the *z*-transform, 518 - -Characteristic roots of continuous-time systems, 153–56, 162, 166, 198–203, 206, 209, 211–12, 214–17, 222–24 of discrete-time systems, 271, 273–75, 297, 299–301, 303–5, 309, 314 invariance of, 942–43 of a matrix, 911, 932–33 Characteristic values. *See* Characteristic roots Characteristic vectors, 910 Chebyshev filters, 440, 463–66 Circular convolution, 819–20, 821 Clearing fractions, 26–27, 32–33, 342–43 Closed Loop systems. *See* Feedback systems Coherent demodulation. *See* Synchronous demodulation Coefficients of Fourier series, computation, 595–98 Column vectors, 36 Commutative property of the convolution integral, 170, 173, 181, 191–92 of the convolution sum, 283 Compact disc (CD), 801 Compact form of Fourier series, 597–98, 599, 600, 604–7 Complex factors of *Q*(*x*), 29 Complex frequency, 89–91 Complex inputs, 177, 297 Complex numbers, 1–15, 54 algebra of, 5–15 arithmetical operations for, 12–15 conjugates of, 6–7 historical note, 1–5 logarithms of, 15 origins of, 2–5 standard forms of, 14–15 useful identities, 7–8 working with, 13–14 Complex poles, 395, 432, 497, 542 Complex roots, 154–56, 274–76 Complex signals, 94–95 Conjugate symmetry of the discrete Fourier transform, 818–19 of the discrete-time Fourier transform, 858–59, 867–68 of the Fourier transform, 684, 703 Conjugation, 684, 703 Constants, 54, 98, 100, 103, 130, 422 Constant-parameter systems. *See* Time-invariant systems Continuous functions, 858 Continuous-time filters, 455–63 Continuous-time Fourier transform (CTFT), 867, 884–85 Continuous-time signals, 107–8, 135 defined, 78 discrete-time systems and, 238 Fourier series and, 593–679 Fourier transform and, 678–769, 680–775 Continuous-time systems, 135, 150–236 analog systems compared with, 261 differential equations of, 161, 196, 213 - -discrete-time systems compared with, 261 external input, response to, 168–96 frequency response of, 412–18, 732–33 internal conditions, response to, 151–63 intuitive insights into, 189–90, 203–5 Laplace transform, 330–487 Periodic inputs and, 637–641 properties of, 107–8 signal transmission through, 721–29 stability of, 196–203, 222–23 state equations for, 915–16 Control systems, 404–12 analysis of, 406–12 design specifications, 411 step input and, 407–9 Controllability/observability, 123–24 of continuous-time systems, 197, 200–2, 223 of discrete-time systems, 303, 965–68 in state-space analysis, 947–53, 961 Convergence abscissa of, 336 of Fourier series, 613–14 to the mean, 613, 614 region of. *See* region of convergence Convolution, 507–9 with an impulse, 283 of the bilateral *z*-transform, 560 circular, 819–20, 821 discrete-time, 311–12 fast, 821, 886 frequency. *See* Frequency convolution of the Fourier transform, 714–16 linear, 821 periodic, 886 time. *See* Time convolution Convolution integral, 170–93, 222, 282, 288, 313, 722 explanation for use, 189–90 graphical understanding of, 178–90, 217–20 properties of, 170–72 Convolution sum, 282–86, 313 graphical procedure for, 288–93 properties of, 282–83 from a table, 285–86 Convolution table, 175–76 Cooley, J. W., 824 Corner frequency, 424 Cramer's rule, 23–25, 40, 51, 379, 385 Critically damped systems, 409, 410 Cubic equations, 2–3, 58 Custom filter function, 310–11 Cutoff frequency, 208, 209 - -Damping coefficient, 115–18 Dashpots linear, 115 torsional, 116 - -Data truncations, 749–55, 763 Decades, 422 Decibels, 421 Decimation-in-frequency algorithm, 824, 827 Decimation-in-time algorithm, 825–27 Decomposition, 99–100, 151 Delayed impulse, 168 Demodulation, 714 of amplitude modulation, 744–46 of DSB-SC signals, 739–41 synchronous, 743–44 Depressed cubic equation, 58 Derivative formulas, 56 Descartes, René, 2 Detection. *See* Demodulation Deterministic signals, 82, 134 Diagonal matrices, 37 Difference equations, 259–60, 265–70 causality condition in, 265–66 classical solution of, 298 differential equation kinship with, 260 frequency response, 532 order of, 260 recursive and non-recursive forms of, 259 recursive solution of, 266–70 sinusoidal response of difference equation systems, 528 *z*-transform solution of, 488, 510–19, 574 Differential equations, 161 classical solution of, 196 difference equation kinship with, 260 Laplace transform solution of, 346–48, 360–73 Differentiators digital, 256–58 ideal, 369–71, 373, 416–17 Digital differentiator example, 258–59 Digital filters, 108, 238, 261–62 Digital integrators, 258–59 Digital processing of analog signals, 547–53 Digital signals, 135, 797–99 advantages of, 261–62 binary, 799–801 defined, 78 *L*-ary, 799 properties of, 78 *See also* Analog-to-digital conversion Digital systems, 109, 135, 261 Dirac definition of an impulse, 88, 134 Dirac delta train, 696–97 Dirac, P.A.M., 86 Direct discrete Fourier transform (DFT), 808, 857 Direct form I (DFI) realization Laplace transform and, 390–91, 394 *z*-transform and, 521 *See also* Transposed direct form II realization - -Direct form II (DFII) realization, 920–25, 954, 965, 967 Laplace transform and, 391, 398 *z*-transform and, 520–22, 525 Direct Fourier transform, 683, 702–3, 762 Direct *z*-transform, 488–592 Dirichlet conditions, 612, 614, 686 Discrete Fourier transform (DFT), 659, 805–23, 827–34, 835 aliasing and leakage and, 805–6 applications of, 820–23 computing Fourier transform, 812–18 derivation of, 807–10 determining filter output, 822–23 direct, 808, 857 discrete-time Fourier transform and, 885–86, 898 inverse, 808, 835, 857 MATLAB on, 827–34 picket fence effect and, 807 points of discontinuity, 807 properties of, 818–20 zero padding and, 810–11, 829–30 Discrete-time complex exponentials, 252 Discrete-time convolution, 311–12 Discrete-time exponentials, 247–49 Discrete-time Fourier integral, 855–67 Discrete-time Fourier series (DTFS), 845–55 computation of, 885–86 MATLAB on, 889–97 of periodic gate function, 853–55 periodic signals and, 846–47, 898 of sinusoids, 849–52 Discrete-time Fourier transform (DTFT), 857–88 of accumulator systems, 875–76 of anti-causal exponentials, 862–63 of causal exponentials, 861–62 continuous-time Fourier transform and, 883–86 existence of, 859, 886 inverse, 886 linear time-invariant discrete-time system analysis by, 879–80 MATLAB on, 889–97 physical appreciation of, 859 properties of, 867–78 of rectangular pulses, 863–65 table of, 860 *z*-transform connection with, 866–67, 886–88, 898 Discrete-time signals, 78, 79, 107–8, 133, 237–53 defined, 78 Fourier analysis of, 845–907 inherently bandlimited, 533 size of, 238–40 useful models, 245–53 useful operations, 240–45 Discrete-time systems, 135, 237–329 classification of, 262–64 controllability/observability of, 303, 965–68 difference equations of, 259–60, 265–70, 298 - -discrete-time Fourier transform analysis of, 878–83 examples of, 253–65 external input, response to, 280–98 frequency response of, 526–38 internal conditions, response to, 270–76 intuitive insights into, 305–6 properties of, 107–8, 264–65 stability of, 263, 298–305, 314 state-space analysis of, 953–64 *z*-transform analysis of, 488–592 Distinct factors of *Q*(*x*), 27 Distortionless transmission, 724–28, 730, 763, 880–82 bandpass systems and, 726–27, 881–82 measure of delay variation, 881 Distributive property, 171, 283 Division of complex numbers, 12–14 Double-sideband, suppressed-carrier (DSB-SC) modulation, 737–41, 742, 746–49 Downsampling, 243–44 Duality, 703–4 Dynamic systems, 103–4, 134–35, 263 - -Eigenfunctions, 193 Eigenvalues. *See* Characteristic roots Eigenvectors, 910 Einstein, Albert, 348 Electrical systems, 95–96, 111–14 Laplace transform analysis of, 373–85, 467 state equations for, 916–19 Electromechanical systems, 118–19 Electronic calculators, 8–11 Energy signals, 67, 82, 134, 239–40 Energy spectral density, 734, 763 Envelope delay. *See* Group delay Envelope detector, 743–45 Equilibrium states, 196, 198 Error signals, 650–51 Error vectors, 642 Essential bandwidth, 736, 758–59 Euler, Leonhard, 2, 3 Euler's formula, 5–6, 45, 252 Even component of a signal, 93–95 Even functions, 92–93, 134 Everlasting exponentials continuous-time systems and, 189, 193–95, 222 discrete-time systems and, 296–97, 313 Fourier series and, 637, 638, 641 Fourier transform and, 687 Laplace transform and, 367–68, 412, 419 Everlasting signals, 81, 134 Exponential Fourier series, 621–37, 661, 803 periodic inputs and, 637–41 reasons for using, 640 symmetry effect on, 630–32 - -Exponential Fourier spectra, 624–32, 664, 667, 668 Exponential functions, 89–91, 134 Exponential input, 193, 296 Exponentials computation of matrix, 922–913 discrete-time, 247–49 discrete-time complex, 252 everlasting. *See* Everlasting exponentials matrix, 968–69 monotonic, 20–22, 90, 91, 134 sinusoid varying, 22–23, 90, 134 sinusoids expressed in, 20 *Exposition du système du monde* (Laplace), 346 External description of a system, 119–20, 135 External input continuous-time system response to, 168–96 discrete-time system response to, 280–98 External stability. *See* Bounded-input/bounded-output stability Fast convolution, 821, 886 Fast Fourier transform (FFT), 659, 811, 821, 824–27, 835 computations reduced by, 824 discrete-time Fourier series and, 847 discrete-time Fourier transform and, 885–86, 898 Feedback systems Laplace transform and, 386–88, 392–95, 399, 404–12 *z*-transform and, 521 Feedforward connections, 392–94, 403 Filtering discrete Fourier transform and, 821–23 MATLAB on, 308–10 selective, 748–49 time constant and, 207–8 Filters analog, 261 anti-aliasing, 537, 791, 834 bandpass, 441–43 bandstop, 441–42, 445, 545–46 Butterworth. *See* Butterworth Filters cascaded *RC*, 461–62 Chebyshev, 440, 463–66 continuous-time, 455–63 custom function, 310–11 digital, 108, 238, 261–62 finite impulse response, 524, 892–97 first-order hold, 785 frequency response of, 412–18 highpass, 443, 445, 542, 730–31, 882–83 Ideal. *See* Ideal filters impulse invariance criterion of, 548 infinite impulse response, 524, 565–74 lowpass, 439–41 lowpass. *See* Lowpass filters notch, 441–43, 540, 545–46 - -poles and zeros of *H*(*s*) and, 436–45 practical, 444–45, 882–83 sharp cutoff, 748 windows in design of, 755 zero-order hold, 785 Final value theorem, 359–61, 508 Finite impulse response (FIR) filters, 524, 892–97 Finite-duration signals, 333 Finite-memory systems, 104 First-order factors, method of, 497 First-order hold filters, 785 Folding frequency, 789–91, 793, 795, 817 For-loops, 216–18 Forced response difference equations and, 298 differential equations and, 198 Forward amplifiers, 405–6 Fourier integral, 722 aperiodic signal and, 680–89, 762 discrete-time, 855–67 Fourier series, 593–679 compact form of, 597–98, 599, 600, 604–7 computing the coefficients of, 595–98 discrete time. *See* Discrete-time Fourier series existence of, 612–13 exponential. *See* Exponential Fourier series generalized, 641–59, 668 Legendre, 656–57 limitations of analysis method, 641 trigonometric. *See* Trigonometric Fourier series waveshaping in, 615–17 Fourier spectrum, 598–607, 777 exponential, 624–32, 664, 667, 668 nature of, 858–59 of a periodic signal, 848–55 Fourier transform, 680–755, 778, 802–3 continuous-time, 867, 883–86 discrete. *See* Discrete Fourier transform discrete-time. *See* Discrete-time Fourier transform direct, 683, 702–3, 762 existence of, 685–86 fast. *See* fast Fourier transform interpolation and, 785 inverse, 683, 693–95, 699, 762, 786–87 physical appreciation of, 687–89 properties of, 701–21 useful functions of, 689–701 Fourier transform pairs, 683, 700 Fourier, Baron Jean-Baptiste-Joseph, 610–12 Fractions, 1–2 clearing, 26–27, 32–34, 342–43 partial. *See* Partial fractions Frequency apparent, 534–36, 793–94 complex, 89–91 - -Frequency (*continued*) corner, 424 cutoff, 208, 209 folding, 789–91, 793, 795, 817 fundamental, 594, 609–10, 846 negative, 626–28 neper, 91 radian, 16, 91, 594 reduction in range, 535 of sinusoids, 16 time delay variation with, 724–25 Frequency convolution of the bilateral Laplace transform, 452 of the discrete-time Fourier transform, 875–76 of the Fourier transform, 714–16 of the Laplace transform, 357 Frequency differentiation, 869 Frequency domain analysis, 368, 722–23, 848 of electrical networks, 374–78 of the Fourier series, 598, 601 two-dimensional view and, 732–33 *See also* Laplace transform Frequency inversion, 706 Frequency resolution, 807, 810–12, 815, 817 Frequency response, 724 Bode plots and, 419–22 of continuous-time systems, 412–18, 732–33 of discrete-time systems, 526–38 MATLAB on, 456–57, 531–32 periodic nature of, 532–36 from pole-zero location, 538–47 pole-zero plots and, 566–68 poles and zeros of *H*(*s*) and, 436–39 transfer function from, 435 Frequency reversal, 868–69 Frequency shifting of the bilateral Laplace transform, 451 of the discrete Fourier transform, 819 of the discrete-time Fourier transform, 871–74 of the Fourier transform, 711–13 of the Laplace transform, 353–54 Frequency spectra, 598, 601 Frequency-division multiplexing (FDM), 714, 749–50 Function M-files, 214–15 Functions characteristic, 193 continuous, 858 even, 92–93, 134 exponential, 89–91, 134 improper, 25–26, 34 interpolation, 690 MATLAB on, 126–33 odd, 92–95, 134 proper, 25–27 - -rational, 25–29, 338 singularity, 89 Fundamental band, 533, 534, 537, 793 Fundamental frequency, 594, 609–10, 846 Fundamental period, 79, 133, 239–40, 593, 595, 846 - -Gain enhancement by poles, 437–38 Gauss, Karl Friedrich, 3–4 Generalized Fourier series, 641–59, 668 Generalized linear phase (GLP), 726–27 Gibbs phenomenon, 619–21, 661–63 Gibbs, Josiah Willard, 620–21 Graphical interpretation of convolution integral, 178–90, 217–20 of convolution sum, 288–93 Greatest common factor of frequencies, 609–10 Group delay, 725–28, 881 - -#### *H*(*s*) - -filter design and, 436–45 realization of, 548–49 *See also* Transfer functions Half-wave symmetry, 608 Hamming window, 754–55, 761 Hanning window, 754–55, 761 Hardware realization, 64, 95, 133 Harmonic distortion, 634 Harmonically related frequencies, 609 Heaviside "cover-up" method, 27–30, 33–35, 341, 342–43, 497 Heaviside, Oliver, 347–48, 612 Highpass filters, 443, 445, 542, 745, 747, 882–83 Homogeneity, 97–98 - -Ideal delay, 369, 416 Ideal differentiators, 369–71, 373, 416–17 Ideal filters, 730–33, 763, 785, 791, 834, 882–83 Ideal integrators, 369, 370, 373, 400, 416–18 Ideal interpolation, 786–87 Ideal linear phase (ILP), 725, 727 Ideal masses, 114 Identity matrices, 37 Identity systems, 109, 192, 263 Imaginary numbers, 1–5 Impedance, 374–77, 379, 380, 382, 384, 387, 399 Improper functions, 25–26, 34 Impulse invariance criterion of filter design, 548 Impulse matching, 164–66 Impulse response matrix, 938 Indefinite integrals, 57 Indicator function. *See* Relational operators Inertia, moment of, 116–18 Infinite impulse response (IIR) filters, 524, 565–74 Information transmission rate, 209–10 - -Initial conditions, 97–100, 102, 122, 134, 335 at 0− and 0+, 363–64 continuous-time systems and, 158–61 generators of, 376–83 Initial value theorem, 359–61, 508 Input, 64 complex, 177, 297 exponential, 193, 296 external. *See* External input in linear systems, 97 multiple, 178, 287–88 ramp, 410–11 sinusoidal. *See* Sinusoidal input step, 407–10 Input–output description, 111–19 Instantaneous systems, 103–4, 134, 263 Integrals convolution. *See* Convolutional integral discrete-time Fourier, 855–67 Fourier. *See* Fourier integral indefinite, 57 of matrices, 909–10 Integrators digital, 258–59 ideal, 369, 370, 373, 400, 416–18 system realization and, 400 Integro-differential equations, 360–73, 466, 488 Interconnected systems continuous-time, 190–93 discrete-time, 294–97 Internal conditions continuous-time system response to, 151–63 discrete-time system response to, 270–76 Internal description of a system, 119–21, 135, 908 *See also* State-space description of a system Internal stability, 110, 135, 263 BIBO relationship to, 199–203, 301–4 of continuous-time systems, 196–203, 222–23 of discrete-time systems, 298–302, 305, 314, 526, 527 of the Laplace transform, 372 of the *z*-transform, 518 Interpolation, 785–88 of discrete-time signals, 243–44 ideal, 786–87 simple, 785–86 spectral, 804 Interpolation formula, 779, 787 Interpolation function, 690 Intuitive insights into continuous-time systems, 189–90, 203–12 into discrete-time systems, 305–6 into the Laplace transform, 367–68 Inverse continuous-time systems, 192–93 Inverse discrete Fourier transform (IDFT), 808, 827, 857 Inverse discrete-time Fourier transform (IDTFT), 886 of rectangular spectrum, 865–66 Inverse discrete-time systems, 294–95 Inverse Fourier transform, 683, 693–95, 699, 762, 786–87 Inverse Laplace transform, 333, 335, 445, 549 finding, 338–46 Inverse *z*-transform, 488–89, 491, 499, 500, 501, 510, 554, 555, 559 finding, 495 Inversion frequency, 706 matrix, 40–42 Invertible systems, 109–10, 135, 263 Irrational numbers, 1–2 Kaiser window, 755, 760–62 Kelvin, Lord, 348 Kennelly-Heaviside atmosphere layer, 348 Kirchhoff's laws, 95 current (KCL), 111, 213, 374 voltage (KVL), 111, 374 Kronecker delta functions, 245 bandlimited interpolation of, 787–88 *L*-ary digital signals, 799 L'Hôpital's rule, 58, 211, 690 Lagrange, Louis de, 347, 612, 613 Laplace transform, 167, 330–487, 721 bilateral. *See* Bilateral Laplace transform differential equation solutions and, 346–48, 360–73 electrical network analysis and, 373–85, 467 existence of, 336–37 Fourier transform connection with, 699–701, 866 intuitive interpretation of, 367–69 inverse, 549, 938–39 properties of, 349–62 stability of, 371–74 state equation solutions by, 927–33 system realization and, 388–404 unilateral, 333–36, 337, 338, 345, 360, 445, 467 *z*-transform connection with, 488, 489, 491, 563–65 Laplace transform pairs, 333 Laplace, Marquis Pierre-Simon de, 346–47, 611, 612, 613 Leakage, 751, 753–55, 763, 805–6 Left half plane (LHP), 91, 198–99, 202, 211, 223, 435 Left shift, 71, 73, 130, 134, 503, 509, 510, 512 Left-sided sequences, 555–56 Legendre Fourier series, 656–57 Leibniz, Gottfried Wilhelm, 801 Linear convolution, 821 Linear dashpots, 115 Linear phase distortionless transmission and, 725, 881 generalized, 726–27 ideal, 725, 727 - -Linear phase (*continued*) physical description of, 707–9 physical explanation of, 870–71 Linear springs, 114 Linear systems, 97–101, 134 heuristic understanding of, 722–23 response of, 98–100 Linear time-invariant continuous-time (LTIC) systems. *See* Continuous-time systems Linear time-invariant discrete-time (LTID) systems. *See* Discrete-time systems Linear time-invariant (LTI) systems, 103, 194–95 Linear time-invariant discrete-time (LTID) systems, 879–80 Linear time-varying systems, 103 Linear transformation of vectors, 36, 939–47, 961 Linearity of the bilateral Laplace transform, 451 of the bilateral *z*-transform, 559 concept of, 97–98 of the discrete Fourier transform, 818, 824 of the discrete-time Fourier transform, 867 of discrete-time systems, 262 of the Fourier transform, 686–87, 824 of the Laplace transform, 331–32 of the *z*-transform, 489 Log magnitude, 27, 422–24 Loop currents continuous-time systems and, 159–63, 175 Laplace transform and, 375 Lower sideband (LSB), 738–39, 747 Lowpass filters, 439–41, 540–42 ideal, 730, 784–85, 788–89, 882–83 poles and zeros of *H*(*s*) and, 436–45 M-files, 212–20 function, 214–15 script, 213–14, 218 Maclaurin series, 6, 55 Magnitude response. *See* Amplitude response Marginally stable systems continuous-time, 198–200, 203, 211, 222–24 discrete-time, 301–2, 304, 314 Laplace transform, 373 signal transmission and, 721 *z*-transform, 519 Mathematical models of systems, 95–96, 125 MATLAB on Butterworth filters, 459–63 calculator operations in, 43–45 on continuous-time filters, 455–63 on discrete Fourier transform, 827–34 on discrete-time Fourier series and transform, - -889–97 - -on discrete-time systems/signals, 306–12 - -elementary operations in, 42–53 on filtering, 308–10 Fourier series applications in, 661–67 Fourier transform topics in, 755–62 frequency response plots, 531–32 on functions, 126–33 impulse invariance, 553 impulse response and, 167 on infinite-impulse response filters, 565–74 M-files in, 212–20 matrix operations in, 49–53 multiple magnitude response curves, 544 partial fraction expansion in, 53 periodic functions, 661–63 phase spectrum, 664–67 polynomial roots and, 157 simple plotting in, 46–48 state-space analysis in, 961–69 vector operations in, 45–46 zero-input response and, 157–58 Matrices, 36–42 algebra of, 38–42 characteristic equation of, 909–10, 933 characteristic roots of, 932–33 computing exponential of, 912–13 definitions and properties of, 37–38 derivatives of, 909–10 diagonal, 37 diagonalization of, 943–44 equal, 37 functions of, 911–12 identity, 37 impulse response, 938 integrals of, 909–10 inversion of, 40–42 MATLAB operations, 49–53 nonsingular, 41 square, 36, 37, 41 state transition, 936 symmetric, 37 transpose of, 37–38 zero, 37 Matrix exponentials, 968–69 Matrix exponentiation, 968–69 Mechanical systems, 114–18 Memory, systems and, 104, 263 Memoryless systems. *See* Instantaneous systems Method of residues, 27 Michelson, Albert, 620–21 Minimum phase systems, 435, 436 Modified partial fractions, 35, 496 Modulation, 713–14, 736–49 amplitude, 711–13, 736, 742–46, 762 angle, 736, 763 of the discrete-time Fourier transform, 872 - -double-sideband, suppressed-carrier, 737–41, 742, 746–49 pulse-amplitude, 796 pulse-code, 796, 799 pulse-position, 796 pulse-width, 796 single-sideband, 746–49 Moment of inertia, 116–18 Monotonic exponentials, 20–22, 90, 91, 134 Multiple inputs, 178, 287–88 Multiple-input, multiple-output (MIMO) systems, 98, 125, 908 Multiplication bilateral *z*-transform and, 560 of complex numbers, 12–14 discrete-time Fourier transform and, 869 of a function by an impulse, 87 matrix, 38–40 scalar, 38, 400–1, 505 *z*-transform and, 506–7 Natural binary code (NBC), 799 Natural modes. *See* Characteristic modes Natural numbers, 1 Natural response difference equations and, 298 differential equations and, 196 Negative feedback, 406 Negative frequency, 626–28 Negative numbers, 1–3, 45 Neper frequency, 91 Neutral equilibrium, 197, 198 Newton, Sir Isaac, 2, 346–47 Noise, 66, 151, 371, 417, 791, 797–99 Nonanticipative systems. *See* Causal systems Non-bandlimited signals, 792 Noncausal signals, 81 Noncausal systems, 104–7, 135, 263 properties of, 104–6 reasons for studying, 106–7 Non-invertible systems, 109–10, 135, 263 Non-inverting amplifiers, 382 Nonlinear systems, 97–101, 134 Nonsingular matrices, 41 Non-uniqueness, 533 Normal-form equations, 915 Norton theorem, 375 Notch filters, 441–43, 540, 545–46. *See also* Bandstop filters Numerical integration, 131–33 Nyquist interval, 778, 779 Nyquist rate, 778–81, 788–89, 792, 795, 821 Nyquist samples, 778, 781, 782, 788, 792 - -Observability. *See* controllability/observability Octave, 422 - -Odd component of a signals, 93–95 Odd functions, 92–95, 134 Operational amplifiers, 382–83, 399, 467 Ordinary numbers, 1–5 Orthogonal signal space, 649–50 Orthogonal signals, 668 energy of the sum of, 647 signal representation by set, 647–59 Orthogonal vector space, 647–48 Orthogonality, 622 Orthonormal sets, 649 Oscillators, 203 Output, 64, 97 Output equations, 122, 124, 908, 930, 941 Overdamped systems, 409–10 Paley–Wiener criterion, 444, 731–32, 788 Parallel realization, 393–94, 525–26, 921, 924–25 Parallel systems, 190, 387 Parseval's theorem, 632, 651–52, 734–35, 755, 758–59, 876–78 Partial fractions expansion of, 25–35, 53 inverse transform by partial fraction expansion and tables, 495–98 Laplace transform and, 338–39, 341, 344, 362, 394, 395, 419, 454 modified, 35 *z*-transform, 499 Passbands, 441, 444–45, 748, 755 Peak time, 409–10 Percent overshoot (PO), 409–10 Periodic (circular) convolution, 819–20 of the discrete-time Fourier transform, 875 Periodic extension of the Fourier spectrum, 848–55 properties of, 80–81 Periodic functions Fourier spectra as, 858 MATLAB on, 661–63 Periodic gate function, 853–55 Periodic signals, 133, 637–40 discrete-time Fourier series and, 846–47 Fourier spectra of, 848–55 Fourier transform of, 695–96 properties of, 78–82 and trigonometric Fourier series, 593–612, 661 Periods fundamental, 79, 133, 239–40, 593, 595, 846 sinusoid, 16 Phase response, 413–25, 427–35, 439, 467 Phase spectrum, 598, 607, 617–18, 707, 848 MATLAB on, 664–67 using principal values, 709–10 - -Phase-plane analysis, 125, 909 Phasors, 18–20 Physical systems. *See* Causal systems Picket fence effect, 807 Pickoff nodes, 190, 254–55, 396 Pingala, 801 Pointwise convergent series, 613 Polar coordinates, 5–6 Polar form, 8–15 arithmetical operations in, 12–15 sinusoids and, 18 Pole-zero location, 538–47 Pole-zero plots, 566–68 Poles complex, 395, 432, 497, 542 controlling gain by, 540 first-order, 424–27 gain enhancement by, 437–38 *H*(*s*), filter design and, 436–45 at the origin, 422–23 repeated, 395, 525, 926 in the right half plane, 371, 435–36 second-order, 426–35 wall of, 439–41, 542 Polynomial expansion, 458–59 Polynomial roots, 157, 572 Positive feedback, 406 Power series, 55 Power signals, 67, 82, 134, 239–40. *See also* Signal power Power, determining, 68–69 matrix, 912–13 Powers, of complex numbers, 13–16 Practical filters, 730–33, 882–83 Preece, Sir William, 349 Prewarping, 570–71 Principal values of the angle, 9 phase spectrum using, 709–10 Proper functions, 25–27 Pulse-amplitude modulation (PAM), 796 Pulse-code modulation (PCM), 796, 799 Pulse dispersion, 209 Pulse-position modulation (PPM), 796 Pulse-width modulation (PWM), 796 Pupin, M., 348 Pythagoras, 2 - -Quadratic equations, 58 Quadratic factors, 29–30 for the Laplace transform, 341–42 for the *z*-transform, 497 Quantization, 799, 831–34 Quantized levels, 799 - -Radian frequency, 16, 91, 594 Random signals, 82, 134 Rational functions, 25–29, 338 Real numbers, 2–7, 43 Real time, 105–6 Rectangular pulses, 863–65 Rectangular spectrum, 865–66 Rectangular windows, 751, 753–55, 763 Reflection property, 868–69 Region of convergence (ROC) for continuous-time systems, 193 for finite-duration signals, 333 for the Laplace transform, 331–33, 337, 347, 448, 449, 454–55, 467 for the *z*-transform, 489–91, 555–58, 561 Relational operators, 128–29 Repeated factors of *Q*(*x*), 31–32 Repeated poles, 395, 525, 926 Repeated roots of continuous-time systems, 154–56, 195, 198, 202, 223 of discrete-time systems, 270, 273–74, 297, 301, 313–14 Resonance phenomenon, 163, 204, 205, 210–12, 305 Right half plane (RHP), 91, 198, 200–3, 223, 371, 435–36 Right shift, 71–72, 131, 134, 501–4, 509, 510 Right-sided sequences, 555–56 Rise time, 206–7, 405, 409–10, 411 *RLC* networks, 914, 916–18 RMS value, 68–69, 70 Rolloff rate, 753, 754 Roots complex, 154–56, 274–76 of complex numbers, 11–15 polynomial, 157, 572 repeated. *See* Repeated roots unrepeated, 198, 202, 223, 301, 314 Rotational systems, 116–19 Rotational mass. *See* Moment of inertia Row vectors, 36, 45, 48–50 - -Sales estimate example, 255–56 Sallen–Key circuit, 383, 384, 461–62, 463, 466 Sampled continuous-time sinusoids, 527–31 Sampling, 776–844 practical, 781–84 properties of, 87–88, 134 signal reconstruction and, 785–99 spectral, 759–60, 802–4 *See also* Discrete Fourier transform; Fast-Fourier transform Sampling interval, 550–54 Sampling rate, 243–44, 536–37 Sampling theorem, 537, 776–84, 834–35 applications of, 796–99 spectral, 802 Savings account example, 253–55 - -Scalar multiplication, 38, 400–1, 505, 509, 520 Scaling, 97–98, 130 of the Fourier transform, 705–6, 755, 757, 762 of the Laplace transform, 357 *See also* Time scaling Script M-files, 213–14, 216, 218 Selective-filtering method, 748–49 Sharp cutoff filters, 748 Shifting of the bilateral *z*-transform, 559 of the convolution integral, 171–72 of the convolution sum, 283 of discrete-time signals, 240 *See also* Frequency shifting; Time shifting Sideband, 746–49 sifting. *See* Sampling Signal distortion, 723–25 Signal energy, 65–66, 70, 131–33, 733–36, 757, 877–78. *See also* Energy signals Signal power, 65–67, 133. *See also* Power signals Signal reconstruction, 785–99. *See also* Interpolation Signal-to-noise power ratio, 66 Signal transmission, 721–29 Signals, 64–91, 133–34 analog. *See* Analog signals anti-causal, 81 aperiodic. *See* Aperiodic signals audio, 713–14, 725, 746 bandlimited, 533, 788, 792, 802 baseband, 737–40, 746–47, 749 basis, 651, 655, 668 causal, 81, 83, 134 classification of, 78–82, 133–34 comparison and components of, 643–45 complex, 94–95 continuous time. *See* continuous-time signals defined, 65 deterministic, 83, 134 digital. *See* Digital signals discrete time. *See* Discrete-time signals energy, 82, 134, 239–40 error, 650–51 even components of, 93–95 everlasting, 81, 134 finite-duration, 333 modulating, 711, 737–39 non-bandlimited, 792 noncausal, 81 odd components of, 93–95 orthogonal. *See* Orthogonal signals periodic. *See* Periodic signals phantoms of, 189 power, 82, 134, 239–40 random, 82, 134 size of, 64–70, 133 - -sketching, 20–23 time reversal of, 77 time limited, 802, 805, 807 two-dimensional view of, 732–33 useful models, 82–91 useful operations, 71–78 as vectors, 641–59 video, 725, 749 Sinc function, 757 Single-input, single-output (SISO) systems, 98, 125, 908 Single-sideband (SSB) modulation, 746–49 Singularity functions, 89 Sinusoidal input causal. *See* Causal sinusoidal input continuous-time systems and, 208 discrete-time systems and, 309 frequency response and, 413–17 steady-state response to causal sinusoidal input, 418–19 Sinusoids, 16–20, 89–91, 134 addition of, 18–20 apparent frequency of sampled, 795–96 compression and expansion, 76 continuous-time, 251–52, 533–37 discrete-time, 251, 527, 528, 533–37 discrete-time Fourier series of, 849–52 in exponential terms, 20 exponentially varying, 22–23, 80, 134 general condition for aliasing in, 793–96 power of a sum of two equal-frequency, 70 sampled continuous-time, 527–31 verification of aliasing in, 792–93 Sketching signals, 20–23 Sliding-tape method, 290–93 Software realization, 64, 95, 133 Spectral density, 688 Spectral folding. *See* Aliasing Spectral interpolation, 804 Spectral resolution, 807 Spectral sampling, 759–60, 802 Spectral sampling theorem, 802 Spectral spreading, 751–53, 755, 763, 807 Springs linear, 114 torsional, 116–17 Square matrices, 36, 37, 41 Square roots of negative numbers, 2–4 Stability BIBO. *See* Bounded-input/bounded-output stability of continuous-time systems, 196–203, 222–23 of discrete-time systems, 263, 298–305, 314 of the Laplace transform, 371–74 Internal. *See* Internal stability of the *z*-transform, 518–19 marginal. *See* marginally stable systems - -Stable equilibrium, 196–97 Stable systems, 110, 263 State equations, 122–25, 135, 908–9, 969 alternative procedure to determine, 918–19 diagonal form of, 944–47 solution of, 926–39 for the state vector, 941–42 systematic procedure for determining, 913–26 time-domain method to solve, 936–37 State transition matrix (STM), 936 State variables, 121–25, 135, 908, 969 State vectors, 927–30, 961 linear transformation of, 941–42 State-space analysis, 908–73 controllability/observability in, 947–53, 961 of discrete-time systems, 953–64 in MATLAB, 961–69 transfer function and, 920–24 transfer function matrix, 931–32 State-space description of a system, 121–25 Steady-state error, 409–11 Steady-state response in continuous-time systems, 418–19 in discrete-time systems, 527 Stem plots, 306–8 Step input, 407–10 Stiffness of linear springs, 114 of torsional springs, 116–17 Stopbands, 441, 444, 445, 456, 457, 459, 460, 463, 755 Subcarriers, 749 Subtraction of complex numbers, 11–12 Superposition, 98, 99, 100, 123, 134 continuous-time systems and, 168, 170, 178 discrete-time systems and, 287 Symmetric matrices, 37 Symmetry conjugate. *See* Conjugate symmetry exponential Fourier series and, 630–32 trigonometric Fourier series and, 607–8 Synchronous demodulation, 743–44, 747 System realization, 388–404, 519–25, 567 cascade, 394, 525–26, 919–20, 923 of complex conjugate poles, 395 direct. *See* Direct form I realization; Direct form II realization differences in performance, 525–26 hardware, 64, 95, 133 parallel. *See* Parallel realization software, 64, 95, 129 Systems, 95–133, 134–35 accumulator, 259, 295, 519 analog, 109, 135, 261 backward difference, 258, 295, 519, 568–69 BIBO stability, assessing, 110 cascade, 190, 192, 372, 373 - -causal. *See* causal systems causality, assessing, 105 classification of, 97–110, 134–35 continuous time. *See* Continuous-time systems control. *See* control systems critically damped, 409, 410 data for computing response, 96–97 defined, 64 digital, 78, 135, 261 discrete time. *See* discrete time systems dynamic, 103–4, 134–35, 263 electrical, 95–96, 111–14 electrical. *See* Electrical systems electromechanical, 118–19 feedback. *See* feedback systems finite-memory, 104 identity, 109, 192, 263 input–output description, 111–19 instantaneous, 103–4, 263 interconnected. *See* interconnected systems invertible, 109–10, 135, 263 linear. *See* Linear systems mathematical models of, 95–96, 125 mechanical, 114–18 memory and, 104, 263 minimum phase, 435, 436 multiple-input, multiple-output, 98, 125, 908 noncausal, 104–7, 263 non-invertible, 109–10, 135 nonlinear, 97–101, 134 overdamped, 409–10 parallel, 190, 387 phantoms of, 189 properties of, 264–65 rotational, 116–19 single-input, single-output, 98, 125, 908 stable, 110, 263 translational, 114–16 time invariant. *See* Time-invariant systems time varying. *See* Time-varying systems two-dimensional view of, 732–33 underdamped, 409 unstable, 110, 263 - -Tacoma Narrows Bridge failure, 212 Tapered windows, 753–54, 763, 807 Taylor series, 55 *Théorie analytique de la chaleur* (Fourier), 612 Thévenin's theorem, 375, 378, 379 Time constant of continuous-time systems, 205–10, 223 of the exponential, 21–22 filtering and, 207–9 information transmission rate and, 209–10 - -pulse dispersion and, 209 rise time and, 206–7 Time convolution of the bilateral Laplace transform, 452 of the discrete-time Fourier transform, 875–76 of the Fourier transform, 714–16 of the Laplace transform, 357 of the *z*-transform, 507–8 Time delay, variation with frequency, 724–25 Time differentiation of the bilateral Laplace transform, 451 of the Fourier transform, 716–18 of the Laplace transform, 354–56 Time integration of the bilateral Laplace transform, 451 of the Fourier transform, 716–18 of the Laplace transform, 356–57 Time inversion, 706 Time reversal, 134 of the bilateral Laplace transform, 452 of the bilateral *z*-transform, 560 of the convolution integral, 178, 181 described, 76–77 of the discrete-time Fourier transform, 868–69 of discrete-time signals, 242 of the *z*-transform, 506–7 Time scaling, 77 of the bilateral Laplace transform, 452 described, 73–74 Time shifting, 77, 79 of the bilateral Laplace transform, 451 of the convolution integral, 178 described, 71–73 of the discrete Fourier transform, 819 of the discrete-time Fourier transform, 870 of the Fourier transform, 707 of the Laplace transform, 349–51 of the *z*-transform, 501–5, 510 Time-division multiplexing (TDM), 749, 797 Time-domain analysis, 723 of continuous-time systems, 150–236 of discrete-time systems, 237–329 of the Fourier series, 598, 601 of interpolation, 785–88 state equation solution in, 933–39 two-dimensional view and, 732–33 Time-frequency duality, 702–3, 723, 753 Time invariant systems, 134 discrete-time, 262 linear. *See* Linear time-invariant systems properties of, 102–3 Time-varying systems, 134 discrete-time, 262 linear, 103 properties of, 102–3 - -Time-limited signals, 802, 805, 807 Torque, 116–18 Torsional dashpots, 116 Torsional springs, 116, 117 Total response of continuous-time systems, 195–96 of discrete-time systems, 297–98 *Traité de mécanique céleste* (Laplace), 346 Transfer functions, 522 analog filter realization with, 548–49 block diagrams and, 386–88 of continuous-time systems, 193–94, 222 of discrete-time systems, 296–97, 314, 514–15, 567–68 from the frequency response, 435 inadequacy for system description, 953 realization of, 389–99, 401, 524–25 state equations from, 916, 919–26 from state-space representations, 964–65 Translational systems, 114–16 Transpose of a matrix, 37–38 Transposed direct form II (TDFII) realization, 398, 967–69 state equations and, 920–24 *z*-transform and, 520–22, 525–26 Triangular windows, 751 Trigonometric Fourier series, 640, 652, 657–58, 667, 668 exponential, 621–37, 661 periodic signals and, 593–612, 661 sampling and, 777, 782 symmetry effect on, 607–8 Trigonometric identities, 55–56 Tukey, J. W., 824 - -Underdamped systems, 409 Uniformly convergent series, 613 Unilateral Laplace transform, 333–36, 337, 338, 345, 360, 445, 467 Unilateral *z*-transform, 489, 491, 492, 495, 554–55, 559 Uniqueness, 335 Unit delay, 517, 520, 521 Unit-gate function, 689 Unit-impulse function, 133 of discrete-time systems, 246–47, 280, 313 as a generalized function, 88–89 properties of, 86–89 Unit-impulse response of continuous-time systems, 163–68, 170, 189–93, 220–21, 222, 731 convolution with, 171 determining, 221 of discrete-time systems, 277–80, 286, 295, 313 Unit matrices, 37 Unit-step function, 84–86, 88–89 of discrete-time systems, 246–47 relational operators and, 128–30 - -### 988 Index - -Unit-triangle function, 689–90 Unrepeated roots, 198, 202, 223, 301, 314 Unstable equilibrium, 196–97 Unstable systems, 110, 263 Upper sideband (USB), 737–39, 746–48 Upsampling, 243–44 Vectors, 36–37, 641–59 basis, 648 characteristic, 910 column, 36 components of, 642–43 error, 642 MATLAB operations, 45–46 matrix multiplication by, 40 orthogonal space, 647–48 row, 36, 45, 48–50 signals as, 641–59 state, 927–30, 961 Vestigial sideband (VSB), 749 Video signals, 725, 749 Waveshaping, 615–17 Weber–Fechner law, 421 Width of the convolution integral, 172, 187 of the convolution sum, 283 Window functions, 749–55, 760–62 *z*-transform, 488–592 bilateral. *See* Bilateral z-transform difference equation solutions of, 488, 510–19, 574 direct, 488–592 discrete-time Fourier transform and, 866–67, 886–88, 898 existence of, 491–95 - -inverse. *See* inverse *z*-transform - -properties of, 501–9 stability of, 518–19 state-space analysis and, 956, 959–65 system realization and, 519–25, 567 time-reversal property, 506–7 time-shifting properties, 501–5 unilateral, 489, 491, 492, 495, 554–55, 559 *z*-domain differentiation property, 506 *z*-domain scaling property, 505 Zero matrices, 37 Zero padding, 810–11, 829–30 Zero-input response, 119, 123 of continuous-time systems, 151–63, 195–96, 203, 220–22 described, 98–100 of discrete-time systems, 270–76, 297–301, 309–11 insights into behavior of, 161–63 of the Laplace transform, 363, 368 in oscillators, 203 of the *z*-transform, 512–13 zero-state response independence from, 161 Zero-order hold (ZOH) filters, 785 Zero-state response, 119, 123 alternate interpretation, 515–18 causality and, 172–73 of continuous-time systems, 151, 161, 168–96, 221–22, 512–16 described, 98–101 of discrete-time systems, 280–98, 308–9, 311, 312, 313 of the Laplace transform, 358, 363, 366–67, 369, 370 zero-input response independence from, 161 Zeros controlling gain by, 540 filter design, 436–45 first-order, 424–27 gain suppression by, 439–40 at the origin, 422–23 - -second-order, 426–35 \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/catalog_entry.json b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/catalog_entry.json deleted file mode 100644 index 164726f996efc4465dffcc5385495e6e532f7aca..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/catalog_entry.json +++ /dev/null @@ -1,1861 +0,0 @@ -{ - "doc_id": "linear-systems-and-signals-3rd-edition-bp-lathi", - "file_name": "linear-systems-and-signals-3rd-edition-bp-lathi.pdf", - "title": "LINEAR SYSTEMS AND SIGNALS", - "author": "B. P. Lathi and R. A. Green", - "subject": "Third Edition", - "keywords": "", - "creator": "", - "producer": "Acrobat Distiller 7.0 (Windows)", - "category": "engineering", - "tags": [ - "3rd", - "and", - "lathi", - "linear", - "signals", - "systems" - ], - "publication_year": "2017", - "total_pages": 1010, - "md_url": "processed_docs/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.md", - "layout_url": "processed_docs/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.layout.json", - "pdf_url": "linear-systems-and-signals-3rd-edition-bp-lathi.pdf", - "toc": [ - { - "level": 1, - "title": "Cover", - "page": 1 - }, - { - "level": 1, - "title": "Half title", - "page": 3 - }, - { - "level": 1, - "title": "Series page", - "page": 4 - }, - { - "level": 1, - "title": "Title page", - "page": 5 - }, - { - "level": 1, - "title": "Copyright page", - "page": 6 - }, - { - "level": 1, - "title": "CONTENTS", - "page": 7 - }, - { - "level": 1, - "title": "PREFACE", - "page": 17 - }, - { - "level": 1, - "title": "B BACKGROUND", - "page": 21 - }, - { - "level": 2, - "title": "B.1 COMPLEX NUMBERS", - "page": 21 - }, - { - "level": 3, - "title": "B.1-1 A Historical Note", - "page": 21 - }, - { - "level": 3, - "title": "B.1-2 Algebra of Complex Numbers", - "page": 25 - }, - { - "level": 2, - "title": "B.2 SINUSOIDS", - "page": 36 - }, - { - "level": 3, - "title": "B.2-1 Addition of Sinusoids", - "page": 38 - }, - { - "level": 3, - "title": "B.2-2 Sinusoids in Terms of Exponentials", - "page": 40 - }, - { - "level": 2, - "title": "B.3 SKETCHING SIGNALS", - "page": 40 - }, - { - "level": 3, - "title": "B.3-1 Monotonic Exponentials", - "page": 40 - }, - { - "level": 3, - "title": "B.3-2 The Exponentially Varying Sinusoid", - "page": 42 - }, - { - "level": 2, - "title": "B.4 CRAMER’S RULE", - "page": 43 - }, - { - "level": 2, - "title": "B.5 PARTIAL FRACTION EXPANSION", - "page": 45 - }, - { - "level": 3, - "title": "B.5-1 Method of Clearing Fractions", - "page": 46 - }, - { - "level": 3, - "title": "B.5-2 The Heaviside “Cover-Up” Method", - "page": 47 - }, - { - "level": 3, - "title": "B.5-3 Repeated Factors of Q(x)", - "page": 51 - }, - { - "level": 3, - "title": "B.5-4 A Combination of Heaviside “Cover-Up” and Clearing Fractions", - "page": 52 - }, - { - "level": 3, - "title": "B.5-5 Improper F(x) with m = n", - "page": 54 - }, - { - "level": 3, - "title": "B.5-6 Modified Partial Fractions", - "page": 55 - }, - { - "level": 2, - "title": "B.6 VECTORS AND MATRICES", - "page": 56 - }, - { - "level": 3, - "title": "B.6-1 Some Definitions and Properties", - "page": 57 - }, - { - "level": 3, - "title": "B.6-2 Matrix Algebra", - "page": 58 - }, - { - "level": 2, - "title": "B.7 MATLAB: ELEMENTARY OPERATIONS", - "page": 62 - }, - { - "level": 3, - "title": "B.7-1 MATLAB Overview", - "page": 62 - }, - { - "level": 3, - "title": "B.7-2 Calculator Operations", - "page": 63 - }, - { - "level": 3, - "title": "B.7-3 Vector Operations", - "page": 65 - }, - { - "level": 3, - "title": "B.7-4 Simple Plotting", - "page": 66 - }, - { - "level": 3, - "title": "B.7-5 Element-by-Element Operations", - "page": 68 - }, - { - "level": 3, - "title": "B.7-6 Matrix Operations", - "page": 69 - }, - { - "level": 3, - "title": "B.7-7 Partial Fraction Expansions", - "page": 73 - }, - { - "level": 2, - "title": "B.8 APPENDIX: USEFUL MATHEMATICAL FORMULAS", - "page": 74 - }, - { - "level": 3, - "title": "B.8-1 Some Useful Constants", - "page": 74 - }, - { - "level": 3, - "title": "B.8-2 Complex Numbers", - "page": 74 - }, - { - "level": 3, - "title": "B.8-3 Sums", - "page": 74 - }, - { - "level": 3, - "title": "B.8-4 Taylor and Maclaurin Series", - "page": 75 - }, - { - "level": 3, - "title": "B.8-5 Power Series", - "page": 75 - }, - { - "level": 3, - "title": "B.8-6 Trigonometric Identities", - "page": 75 - }, - { - "level": 3, - "title": "B.8-7 Common Derivative Formulas", - "page": 76 - }, - { - "level": 3, - "title": "B.8-8 Indefinite Integrals", - "page": 77 - }, - { - "level": 3, - "title": "B.8-9 L’Hôpital’s Rule", - "page": 78 - }, - { - "level": 3, - "title": "B.8-10 Solution of Quadratic and Cubic Equations", - "page": 78 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 78 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 79 - }, - { - "level": 1, - "title": "1 SIGNALS AND SYSTEMS", - "page": 84 - }, - { - "level": 2, - "title": "1.1 SIZE OF A SIGNAL", - "page": 84 - }, - { - "level": 3, - "title": "1.1-1 Signal Energy", - "page": 85 - }, - { - "level": 3, - "title": "1.1-2 Signal Power", - "page": 85 - }, - { - "level": 2, - "title": "1.2 SOME USEFUL SIGNAL OPERATIONS", - "page": 91 - }, - { - "level": 3, - "title": "1.2-1 Time Shifting", - "page": 91 - }, - { - "level": 3, - "title": "1.2-2 Time Scaling", - "page": 93 - }, - { - "level": 3, - "title": "1.2-3 Time Reversal", - "page": 96 - }, - { - "level": 3, - "title": "1.2-4 Combined Operations", - "page": 97 - }, - { - "level": 2, - "title": "1.3 CLASSIFICATION OF SIGNALS", - "page": 98 - }, - { - "level": 3, - "title": "1.3-1 Continuous-Time and Discrete-Time Signals", - "page": 98 - }, - { - "level": 3, - "title": "1.3-2 Analog and Digital Signals", - "page": 98 - }, - { - "level": 3, - "title": "1.3-3 Periodic and Aperiodic Signals", - "page": 99 - }, - { - "level": 3, - "title": "1.3-4 Energy and Power Signals", - "page": 102 - }, - { - "level": 3, - "title": "1.3-5 Deterministic and Random Signals", - "page": 102 - }, - { - "level": 2, - "title": "1.4 SOME USEFUL SIGNAL MODELS", - "page": 102 - }, - { - "level": 3, - "title": "1.4-1 The Unit Step Function u(t)", - "page": 103 - }, - { - "level": 3, - "title": "1.4-2 The Unit Impulse Function δ(t)", - "page": 106 - }, - { - "level": 3, - "title": "1.4-3 The Exponential Function e^{st}", - "page": 109 - }, - { - "level": 2, - "title": "1.5 EVEN AND ODD FUNCTIONS", - "page": 112 - }, - { - "level": 3, - "title": "1.5-1 Some Properties of Even and Odd Functions", - "page": 112 - }, - { - "level": 3, - "title": "1.5-2 Even and Odd Components of a Signal", - "page": 113 - }, - { - "level": 2, - "title": "1.6 SYSTEMS", - "page": 115 - }, - { - "level": 2, - "title": "1.7 CLASSIFICATION OF SYSTEMS", - "page": 117 - }, - { - "level": 3, - "title": "1.7-1 Linear and Nonlinear Systems", - "page": 117 - }, - { - "level": 3, - "title": "1.7-2 Time-Invariant and Time-Varying Systems", - "page": 122 - }, - { - "level": 3, - "title": "1.7-3 Instantaneous and Dynamic Systems", - "page": 123 - }, - { - "level": 3, - "title": "1.7-4 Causal and Noncausal Systems", - "page": 124 - }, - { - "level": 3, - "title": "1.7-5 Continuous-Time and Discrete-Time Systems", - "page": 127 - }, - { - "level": 3, - "title": "1.7-6 Analog and Digital Systems", - "page": 129 - }, - { - "level": 3, - "title": "1.7-7 Invertible and Noninvertible Systems", - "page": 129 - }, - { - "level": 3, - "title": "1.7-8 Stable and Unstable Systems", - "page": 130 - }, - { - "level": 2, - "title": "1.8 SYSTEM MODEL: INPUT–OUTPUT DESCRIPTION", - "page": 131 - }, - { - "level": 3, - "title": "1.8-1 Electrical Systems", - "page": 131 - }, - { - "level": 3, - "title": "1.8-2 Mechanical Systems", - "page": 134 - }, - { - "level": 3, - "title": "1.8-3 Electromechanical Systems", - "page": 138 - }, - { - "level": 2, - "title": "1.9 INTERNAL AND EXTERNAL DESCRIPTIONS OF A SYSTEM", - "page": 139 - }, - { - "level": 2, - "title": "1.10 INTERNAL DESCRIPTION: THE STATE-SPACE DESCRIPTION", - "page": 141 - }, - { - "level": 2, - "title": "1.11 MATLAB: WORKING WITH FUNCTIONS", - "page": 146 - }, - { - "level": 3, - "title": "1.11-1 Anonymous Functions", - "page": 146 - }, - { - "level": 3, - "title": "1.11-2 Relational Operators and the Unit Step Function", - "page": 148 - }, - { - "level": 3, - "title": "1.11-3 Visualizing Operations on the Independent Variable", - "page": 150 - }, - { - "level": 3, - "title": "1.11-4 Numerical Integration and Estimating Signal Energy", - "page": 151 - }, - { - "level": 2, - "title": "1.12 SUMMARY", - "page": 153 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 155 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 156 - }, - { - "level": 1, - "title": "2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "page": 170 - }, - { - "level": 2, - "title": "2.1 INTRODUCTION", - "page": 170 - }, - { - "level": 2, - "title": "2.2 SYSTEM RESPONSE TO INTERNAL CONDITIONS: THE ZERO-INPUT RESPONSE", - "page": 171 - }, - { - "level": 3, - "title": "2.2-1 Some Insights into the Zero-Input Behavior of a System", - "page": 181 - }, - { - "level": 2, - "title": "2.3 THE UNIT IMPULSE RESPONSE h(t)", - "page": 183 - }, - { - "level": 2, - "title": "2.4 SYSTEM RESPONSE TO EXTERNAL INPUT: THE ZERO-STATE RESPONSE", - "page": 188 - }, - { - "level": 3, - "title": "2.4-1 The Convolution Integral", - "page": 190 - }, - { - "level": 3, - "title": "2.4-2 Graphical Understanding of Convolution Operation", - "page": 198 - }, - { - "level": 3, - "title": "2.4-3 Interconnected Systems", - "page": 210 - }, - { - "level": 3, - "title": "2.4-4 A Very Special Function for LTIC Systems: The Everlasting Exponential e^{st}", - "page": 213 - }, - { - "level": 3, - "title": "2.4-5 Total Response", - "page": 215 - }, - { - "level": 2, - "title": "2.5 SYSTEM STABILITY", - "page": 216 - }, - { - "level": 3, - "title": "2.5-1 External (BIBO) Stability", - "page": 216 - }, - { - "level": 3, - "title": "2.5-2 Internal (Asymptotic) Stability", - "page": 218 - }, - { - "level": 3, - "title": "2.5-3 Relationship Between BIBO and Asymptotic Stability", - "page": 219 - }, - { - "level": 2, - "title": "2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR", - "page": 223 - }, - { - "level": 3, - "title": "2.6-1 Dependence of System Behavior on Characteristic Modes", - "page": 223 - }, - { - "level": 3, - "title": "2.6-2 Response Time of a System: The System Time Constant", - "page": 225 - }, - { - "level": 3, - "title": "2.6-3 Time Constant and Rise Time of a System", - "page": 226 - }, - { - "level": 3, - "title": "2.6-4 Time Constant and Filtering", - "page": 227 - }, - { - "level": 3, - "title": "2.6-5 Time Constant and Pulse Dispersion (Spreading)", - "page": 229 - }, - { - "level": 3, - "title": "2.6-6 Time Constant and Rate of Information Transmission", - "page": 229 - }, - { - "level": 3, - "title": "2.6-7 The Resonance Phenomenon", - "page": 230 - }, - { - "level": 2, - "title": "2.7 MATLAB: M-FILES", - "page": 232 - }, - { - "level": 3, - "title": "2.7-1 Script M-Files", - "page": 233 - }, - { - "level": 3, - "title": "2.7-2 Function M-Files", - "page": 234 - }, - { - "level": 3, - "title": "2.7-3 For-Loops", - "page": 235 - }, - { - "level": 3, - "title": "2.7-4 Graphical Understanding of Convolution", - "page": 237 - }, - { - "level": 2, - "title": "2.8 APPENDIX: DETERMINING THE IMPULSE RESPONSE", - "page": 240 - }, - { - "level": 2, - "title": "2.9 SUMMARY", - "page": 241 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 243 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 243 - }, - { - "level": 1, - "title": "3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "page": 257 - }, - { - "level": 2, - "title": "3.1 INTRODUCTION", - "page": 257 - }, - { - "level": 3, - "title": "3.1-1 Size of a Discrete-Time Signal", - "page": 258 - }, - { - "level": 2, - "title": "3.2 USEFUL SIGNAL OPERATIONS", - "page": 260 - }, - { - "level": 2, - "title": "3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS", - "page": 265 - }, - { - "level": 3, - "title": "3.3-1 Discrete-Time Impulse Function δ[n]", - "page": 265 - }, - { - "level": 3, - "title": "3.3-2 Discrete-Time Unit Step Function u[n]", - "page": 266 - }, - { - "level": 3, - "title": "3.3-3 Discrete-Time Exponential γ^n", - "page": 267 - }, - { - "level": 3, - "title": "3.3-4 Discrete-Time Sinusoid cos(Omega n+θ)", - "page": 271 - }, - { - "level": 3, - "title": "3.3-5 Discrete-Time Complex Exponential e^{jOmega n}", - "page": 272 - }, - { - "level": 2, - "title": "3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS", - "page": 273 - }, - { - "level": 3, - "title": "3.4-1 Classification of Discrete-Time Systems", - "page": 282 - }, - { - "level": 2, - "title": "3.5 DISCRETE-TIME SYSTEM EQUATIONS", - "page": 285 - }, - { - "level": 3, - "title": "3.5-1 Recursive (Iterative) Solution of Difference Equation", - "page": 286 - }, - { - "level": 2, - "title": "3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS: THE ZERO-INPUT RESPONSE", - "page": 290 - }, - { - "level": 2, - "title": "3.7 THE UNIT IMPULSE RESPONSE h[n]", - "page": 297 - }, - { - "level": 3, - "title": "3.7-1 The Closed-Form Solution of h[n]", - "page": 298 - }, - { - "level": 2, - "title": "3.8 SYSTEM RESPONSE TO EXTERNAL INPUT: THE ZERO-STATE RESPONSE", - "page": 300 - }, - { - "level": 3, - "title": "3.8-1 Graphical Procedure for the Convolution Sum", - "page": 308 - }, - { - "level": 3, - "title": "3.8-2 Interconnected Systems", - "page": 314 - }, - { - "level": 3, - "title": "3.8-3 Total Response", - "page": 317 - }, - { - "level": 2, - "title": "3.9 SYSTEM STABILITY", - "page": 318 - }, - { - "level": 3, - "title": "3.9-1 External (BIBO) Stability", - "page": 318 - }, - { - "level": 3, - "title": "3.9-2 Internal (Asymptotic) Stability", - "page": 319 - }, - { - "level": 3, - "title": "3.9-3 Relationship Between BIBO and Asymptotic Stability", - "page": 321 - }, - { - "level": 2, - "title": "3.10 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR", - "page": 325 - }, - { - "level": 2, - "title": "3.11 MATLAB: DISCRETE-TIME SIGNALS AND SYSTEMS", - "page": 326 - }, - { - "level": 3, - "title": "3.11-1 Discrete-Time Functions and Stem Plots", - "page": 326 - }, - { - "level": 3, - "title": "3.11-2 System Responses Through Filtering", - "page": 328 - }, - { - "level": 3, - "title": "3.11-3 A Custom Filter Function", - "page": 330 - }, - { - "level": 3, - "title": "3.11-4 Discrete-Time Convolution", - "page": 331 - }, - { - "level": 2, - "title": "3.12 APPENDIX: IMPULSE RESPONSE FOR A SPECIAL CASE", - "page": 333 - }, - { - "level": 2, - "title": "3.13 SUMMARY", - "page": 333 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 334 - }, - { - "level": 1, - "title": "4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM", - "page": 350 - }, - { - "level": 2, - "title": "4.1 THE LAPLACE TRANSFORM", - "page": 350 - }, - { - "level": 3, - "title": "4.1-1 Finding the Inverse Transform", - "page": 358 - }, - { - "level": 2, - "title": "4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM", - "page": 369 - }, - { - "level": 3, - "title": "4.2-1 Time Shifting", - "page": 369 - }, - { - "level": 3, - "title": "4.2-2 Frequency Shifting", - "page": 373 - }, - { - "level": 3, - "title": "4.2-3 The Time-Differentiation Property", - "page": 374 - }, - { - "level": 3, - "title": "4.2-4 The Time-Integration Property", - "page": 376 - }, - { - "level": 3, - "title": "4.2-5 The Scaling Property", - "page": 377 - }, - { - "level": 3, - "title": "4.2-6 Time Convolution and Frequency Convolution", - "page": 377 - }, - { - "level": 2, - "title": "4.3 SOLUTION OF DIFFERENTIAL AND INTEGRO-DIFFERENTIAL EQUATIONS", - "page": 380 - }, - { - "level": 3, - "title": "4.3-1 Comments on Initial Conditions at 0^− and at 0^+", - "page": 383 - }, - { - "level": 3, - "title": "4.3-2 Zero-State Response", - "page": 386 - }, - { - "level": 3, - "title": "4.3-3 Stability", - "page": 391 - }, - { - "level": 3, - "title": "4.3-4 Inverse Systems", - "page": 393 - }, - { - "level": 2, - "title": "4.4 ANALYSIS OF ELECTRICAL NETWORKS: THE TRANSFORMED NETWORK", - "page": 393 - }, - { - "level": 3, - "title": "4.4-1 Analysis of Active Circuits", - "page": 402 - }, - { - "level": 2, - "title": "4.5 BLOCK DIAGRAMS", - "page": 406 - }, - { - "level": 2, - "title": "4.6 SYSTEM REALIZATION", - "page": 408 - }, - { - "level": 3, - "title": "4.6-1 Direct Form I Realization", - "page": 409 - }, - { - "level": 3, - "title": "4.6-2 Direct Form II Realization", - "page": 410 - }, - { - "level": 3, - "title": "4.6-3 Cascade and Parallel Realizations", - "page": 413 - }, - { - "level": 3, - "title": "4.6-4 Transposed Realization", - "page": 416 - }, - { - "level": 3, - "title": "4.6-5 Using Operational Amplifiers for System Realization", - "page": 419 - }, - { - "level": 2, - "title": "4.7 APPLICATION TO FEEDBACK AND CONTROLS", - "page": 424 - }, - { - "level": 3, - "title": "4.7-1 Analysis of a Simple Control System", - "page": 426 - }, - { - "level": 2, - "title": "4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM", - "page": 432 - }, - { - "level": 3, - "title": "4.8-1 Steady-State Response to Causal Sinusoidal Inputs", - "page": 438 - }, - { - "level": 2, - "title": "4.9 BODE PLOTS", - "page": 439 - }, - { - "level": 3, - "title": "4.9-1 Constant Ka_1a_2/b_1b_3", - "page": 442 - }, - { - "level": 3, - "title": "4.9-2 Pole (or Zero) at the Origin", - "page": 442 - }, - { - "level": 3, - "title": "4.9-3 First-Order Pole (or Zero)", - "page": 444 - }, - { - "level": 3, - "title": "4.9-4 Second-Order Pole (or Zero)", - "page": 446 - }, - { - "level": 3, - "title": "4.9-5 The Transfer Function from the Frequency Response", - "page": 455 - }, - { - "level": 2, - "title": "4.10 FILTER DESIGN BY PLACEMENT OF POLES AND ZEROS OF H(s)", - "page": 456 - }, - { - "level": 3, - "title": "4.10-1 Dependence of Frequency Response on Poles and Zeros of H(s)", - "page": 456 - }, - { - "level": 3, - "title": "4.10-2 Lowpass Filters", - "page": 459 - }, - { - "level": 3, - "title": "4.10-3 Bandpass Filters", - "page": 461 - }, - { - "level": 3, - "title": "4.10-4 Notch (Bandstop) Filters", - "page": 461 - }, - { - "level": 3, - "title": "4.10-5 Practical Filters and Their Specifications", - "page": 464 - }, - { - "level": 2, - "title": "4.11 THE BILATERAL LAPLACE TRANSFORM", - "page": 465 - }, - { - "level": 3, - "title": "4.11-1 Properties of the Bilateral Laplace Transform", - "page": 471 - }, - { - "level": 3, - "title": "4.11-2 Using the Bilateral Transform for Linear System Analysis", - "page": 472 - }, - { - "level": 2, - "title": "4.12 MATLAB: CONTINUOUS-TIME FILTERS", - "page": 475 - }, - { - "level": 3, - "title": "4.12-1 Frequency Response and Polynomial Evaluation", - "page": 476 - }, - { - "level": 3, - "title": "4.12-2 Butterworth Filters and the Find Command", - "page": 479 - }, - { - "level": 3, - "title": "4.12-3 Using Cascaded Second-Order Sections for Butterworth Filter Realization", - "page": 481 - }, - { - "level": 3, - "title": "4.12-4 Chebyshev Filters", - "page": 483 - }, - { - "level": 2, - "title": "4.13 SUMMARY", - "page": 486 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 488 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 488 - }, - { - "level": 1, - "title": "5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM", - "page": 508 - }, - { - "level": 2, - "title": "5.1 THE z-TRANSFORM", - "page": 508 - }, - { - "level": 3, - "title": "5.1-1 Inverse Transform by Partial Fraction Expansion and Tables", - "page": 515 - }, - { - "level": 3, - "title": "5.1-2 Inverse z-Transform by Power Series Expansion", - "page": 519 - }, - { - "level": 2, - "title": "5.2 SOME PROPERTIES OF THE z-TRANSFORM", - "page": 521 - }, - { - "level": 3, - "title": "5.2-1 Time-Shifting Properties", - "page": 521 - }, - { - "level": 3, - "title": "5.2-2 z-Domain Scaling Property (Multiplication by γ^n)", - "page": 525 - }, - { - "level": 3, - "title": "5.2-3 z-Domain Differentiation Property (Multiplication by n)", - "page": 526 - }, - { - "level": 3, - "title": "5.2-4 Time-Reversal Property", - "page": 526 - }, - { - "level": 3, - "title": "5.2-5 Convolution Property", - "page": 527 - }, - { - "level": 2, - "title": "5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS", - "page": 530 - }, - { - "level": 3, - "title": "5.3-1 Zero-State Response of LTID Systems: The Transfer Function", - "page": 534 - }, - { - "level": 3, - "title": "5.3-2 Stability", - "page": 538 - }, - { - "level": 3, - "title": "5.3-3 Inverse Systems", - "page": 539 - }, - { - "level": 2, - "title": "5.4 SYSTEM REALIZATION", - "page": 539 - }, - { - "level": 2, - "title": "5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS", - "page": 546 - }, - { - "level": 3, - "title": "5.5-1 The Periodic Nature of Frequency Response", - "page": 552 - }, - { - "level": 3, - "title": "5.5-2 Aliasing and Sampling Rate", - "page": 556 - }, - { - "level": 2, - "title": "5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS", - "page": 558 - }, - { - "level": 2, - "title": "5.7 DIGITAL PROCESSING OF ANALOG SIGNALS", - "page": 567 - }, - { - "level": 2, - "title": "5.8 THE BILATERAL z-TRANSFORM", - "page": 574 - }, - { - "level": 3, - "title": "5.8-1 Properties of the Bilateral z-Transform", - "page": 579 - }, - { - "level": 3, - "title": "5.8-2 Using the Bilateral z-Transform for Analysis of LTID Systems", - "page": 580 - }, - { - "level": 2, - "title": "5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS", - "page": 583 - }, - { - "level": 2, - "title": "5.10 MATLAB: DISCRETE-TIME IIR FILTERS", - "page": 585 - }, - { - "level": 3, - "title": "5.10-1 Frequency Response and Pole-Zero Plots", - "page": 586 - }, - { - "level": 3, - "title": "5.10-2 Transformation Basics", - "page": 587 - }, - { - "level": 3, - "title": "5.10-3 Transformation by First-Order Backward Difference", - "page": 588 - }, - { - "level": 3, - "title": "5.10-4 Bilinear Transformation", - "page": 589 - }, - { - "level": 3, - "title": "5.10-5 Bilinear Transformation with Prewarping", - "page": 590 - }, - { - "level": 3, - "title": "5.10-6 Example: Butterworth Filter Transformation", - "page": 591 - }, - { - "level": 3, - "title": "5.10-7 Problems Finding Polynomial Roots", - "page": 592 - }, - { - "level": 3, - "title": "5.10-8 Using Cascaded Second-Order Sections to Improve Design", - "page": 592 - }, - { - "level": 2, - "title": "5.11 SUMMARY", - "page": 594 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 595 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 595 - }, - { - "level": 1, - "title": "6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "page": 613 - }, - { - "level": 2, - "title": "6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES", - "page": 613 - }, - { - "level": 3, - "title": "6.1-1 The Fourier Spectrum", - "page": 618 - }, - { - "level": 3, - "title": "6.1-2 The Effect of Symmetry", - "page": 627 - }, - { - "level": 3, - "title": "6.1-3 Determining the Fundamental Frequency and Period", - "page": 629 - }, - { - "level": 2, - "title": "6.2 EXISTENCE AND CONVERGENCE OF THE FOURIER SERIES", - "page": 632 - }, - { - "level": 3, - "title": "6.2-1 Convergence of a Series", - "page": 633 - }, - { - "level": 3, - "title": "6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping", - "page": 635 - }, - { - "level": 2, - "title": "6.3 EXPONENTIAL FOURIER SERIES", - "page": 641 - }, - { - "level": 3, - "title": "6.3-1 Exponential Fourier Spectra", - "page": 644 - }, - { - "level": 3, - "title": "6.3-2 Parseval’s Theorem", - "page": 652 - }, - { - "level": 3, - "title": "6.3-3 Properties of the Fourier Series", - "page": 655 - }, - { - "level": 2, - "title": "6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS", - "page": 657 - }, - { - "level": 2, - "title": "6.5 GENERALIZED FOURIER SERIES:SIGNALS AS VECTORS", - "page": 661 - }, - { - "level": 3, - "title": "6.5-1 Component of a Vector", - "page": 662 - }, - { - "level": 3, - "title": "6.5-2 Signal Comparison and Component of a Signal", - "page": 663 - }, - { - "level": 3, - "title": "6.5-3 Extension to Complex Signals", - "page": 665 - }, - { - "level": 3, - "title": "6.5-4 Signal Representation by an Orthogonal Signal Set", - "page": 667 - }, - { - "level": 2, - "title": "6.6 NUMERICAL COMPUTATION OF D_n", - "page": 679 - }, - { - "level": 2, - "title": "6.7 MATLAB: FOURIER SERIES APPLICATIONS", - "page": 681 - }, - { - "level": 3, - "title": "6.7-1 Periodic Functions and the Gibbs Phenomenon", - "page": 681 - }, - { - "level": 3, - "title": "6.7-2 Optimization and Phase Spectra", - "page": 684 - }, - { - "level": 2, - "title": "6.8 SUMMARY", - "page": 687 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 688 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 689 - }, - { - "level": 1, - "title": "7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "page": 700 - }, - { - "level": 2, - "title": "7.1 APERIODIC SIGNAL REPRESENTATION BY THE FOURIER INTEGRAL", - "page": 700 - }, - { - "level": 3, - "title": "7.1-1 Physical Appreciation of the Fourier Transform", - "page": 707 - }, - { - "level": 2, - "title": "7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS", - "page": 709 - }, - { - "level": 3, - "title": "7.2-1 Connection Between the Fourier and Laplace Transforms", - "page": 720 - }, - { - "level": 2, - "title": "7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM", - "page": 721 - }, - { - "level": 2, - "title": "7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS", - "page": 741 - }, - { - "level": 3, - "title": "7.4-1 Signal Distortion During Transmission", - "page": 743 - }, - { - "level": 3, - "title": "7.4-2 Bandpass Systems and Group Delay", - "page": 746 - }, - { - "level": 2, - "title": "7.5 IDEAL AND PRACTICAL FILTERS", - "page": 750 - }, - { - "level": 2, - "title": "7.6 SIGNAL ENERGY", - "page": 753 - }, - { - "level": 2, - "title": "7.7 APPLICATION TO COMMUNICATIONS: AMPLITUDE MODULATION", - "page": 756 - }, - { - "level": 3, - "title": "7.7-1 Double-Sideband, Suppressed-Carrier (DSB-SC) Modulation", - "page": 757 - }, - { - "level": 3, - "title": "7.7-2 Amplitude Modulation (AM)", - "page": 762 - }, - { - "level": 3, - "title": "7.7-3 Single-Sideband Modulation (SSB)", - "page": 766 - }, - { - "level": 3, - "title": "7.7-4 Frequency-Division Multiplexing", - "page": 769 - }, - { - "level": 2, - "title": "7.8 DATA TRUNCATION: WINDOW FUNCTIONS", - "page": 769 - }, - { - "level": 3, - "title": "7.8-1 Using Windows in Filter Design", - "page": 775 - }, - { - "level": 2, - "title": "7.9 MATLAB: FOURIER TRANSFORM TOPICS", - "page": 775 - }, - { - "level": 3, - "title": "7.9-1 The Sinc Function and the Scaling Property", - "page": 777 - }, - { - "level": 3, - "title": "7.9-2 Parseval’s Theorem and Essential Bandwidth", - "page": 778 - }, - { - "level": 3, - "title": "7.9-3 Spectral Sampling", - "page": 779 - }, - { - "level": 3, - "title": "7.9-4 Kaiser Window Functions", - "page": 780 - }, - { - "level": 2, - "title": "7.10 SUMMARY", - "page": 782 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 783 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 784 - }, - { - "level": 1, - "title": "8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "page": 796 - }, - { - "level": 2, - "title": "8.1 THE SAMPLING THEOREM", - "page": 796 - }, - { - "level": 3, - "title": "8.1-1 Practical Sampling", - "page": 801 - }, - { - "level": 2, - "title": "8.2 SIGNAL RECONSTRUCTION", - "page": 805 - }, - { - "level": 3, - "title": "8.2-1 Practical Difficulties in Signal Reconstruction", - "page": 808 - }, - { - "level": 3, - "title": "8.2-2 Some Applications of the Sampling Theorem", - "page": 816 - }, - { - "level": 2, - "title": "8.3 ANALOG-TO-DIGITAL (A/D) CONVERSION", - "page": 819 - }, - { - "level": 2, - "title": "8.4 DUAL OF TIME SAMPLING: SPECTRAL SAMPLING", - "page": 822 - }, - { - "level": 2, - "title": "8.5 NUMERICAL COMPUTATION OF THE FOURIER TRANSFORM: THE DISCRETE FOURIER TRANSFORM", - "page": 825 - }, - { - "level": 3, - "title": "8.5-1 Some Properties of the DFT", - "page": 838 - }, - { - "level": 3, - "title": "8.5-2 Some Applications of the DFT", - "page": 840 - }, - { - "level": 2, - "title": "8.6 THE FAST FOURIER TRANSFORM (FFT)", - "page": 844 - }, - { - "level": 2, - "title": "8.7 MATLAB: THE DISCRETE FOURIER TRANSFORM", - "page": 847 - }, - { - "level": 3, - "title": "8.7-1 Computing the Discrete Fourier Transform", - "page": 847 - }, - { - "level": 3, - "title": "8.7-2 Improving the Picture with Zero Padding", - "page": 849 - }, - { - "level": 3, - "title": "8.7-3 Quantization", - "page": 851 - }, - { - "level": 2, - "title": "8.8 SUMMARY", - "page": 854 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 855 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 855 - }, - { - "level": 1, - "title": "9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "page": 865 - }, - { - "level": 2, - "title": "9.1 DISCRETE-TIME FOURIER SERIES (DTFS)", - "page": 865 - }, - { - "level": 3, - "title": "9.1-1 Periodic Signal Representation by Discrete-Time Fourier Series", - "page": 866 - }, - { - "level": 3, - "title": "9.1-2 Fourier Spectra of a Periodic Signal x[n]", - "page": 868 - }, - { - "level": 2, - "title": "9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL", - "page": 875 - }, - { - "level": 3, - "title": "9.2-1 Nature of Fourier Spectra", - "page": 878 - }, - { - "level": 3, - "title": "9.2-2 Connection Between the DTFT and the z-Transform", - "page": 886 - }, - { - "level": 2, - "title": "9.3 PROPERTIES OF THE DTFT", - "page": 887 - }, - { - "level": 2, - "title": "9.4 LTI DISCRETE-TIME SYSTEM ANALYSIS BY DTFT", - "page": 898 - }, - { - "level": 3, - "title": "9.4-1 Distortionless Transmission", - "page": 900 - }, - { - "level": 3, - "title": "9.4-2 Ideal and Practical Filters", - "page": 902 - }, - { - "level": 2, - "title": "9.5 DTFT CONNECTION WITH THE CTFT", - "page": 903 - }, - { - "level": 3, - "title": "9.5-1 Use of DFT and FFT for Numerical Computation of the DTFT", - "page": 905 - }, - { - "level": 2, - "title": "9.6 GENERALIZATION OF THE DTFT TO THE z-TRANSFORM", - "page": 906 - }, - { - "level": 2, - "title": "9.7 MATLAB: WORKING WITH THE DTFS AND THE DTFT", - "page": 909 - }, - { - "level": 3, - "title": "9.7-1 Computing the Discrete-Time Fourier Series", - "page": 909 - }, - { - "level": 3, - "title": "9.7-2 Measuring Code Performance", - "page": 911 - }, - { - "level": 3, - "title": "9.7-3 FIR Filter Design by Frequency Sampling", - "page": 912 - }, - { - "level": 2, - "title": "9.8 SUMMARY", - "page": 918 - }, - { - "level": 2, - "title": "REFERENCE", - "page": 918 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 919 - }, - { - "level": 1, - "title": "10 STATE-SPACE ANALYSIS", - "page": 928 - }, - { - "level": 2, - "title": "10.1 MATHEMATICAL PRELIMINARIES", - "page": 929 - }, - { - "level": 3, - "title": "10.1-1 Derivatives and Integrals of aMatrix", - "page": 929 - }, - { - "level": 3, - "title": "10.1-2 The Characteristic Equation of a Matrix: The Cayley–Hamilton Theorem", - "page": 930 - }, - { - "level": 3, - "title": "10.1-3 Computation of an Exponential and a Power of aMatrix", - "page": 932 - }, - { - "level": 2, - "title": "10.2 INTRODUCTION TO STATE SPACE", - "page": 933 - }, - { - "level": 2, - "title": "10.3 A SYSTEMATIC PROCEDURE TO DETERMINE STATE EQUATIONS", - "page": 936 - }, - { - "level": 3, - "title": "10.3-1 Electrical Circuits", - "page": 936 - }, - { - "level": 3, - "title": "10.3-2 State Equations from a Transfer Function", - "page": 939 - }, - { - "level": 2, - "title": "10.4 SOLUTION OF STATE EQUATIONS", - "page": 946 - }, - { - "level": 3, - "title": "10.4-1 Laplace Transform Solution of State Equations", - "page": 947 - }, - { - "level": 3, - "title": "10.4-2 Time-Domain Solution of State Equations", - "page": 953 - }, - { - "level": 2, - "title": "10.5 LINEAR TRANSFORMATION OF A STATE VECTOR", - "page": 959 - }, - { - "level": 3, - "title": "10.5-1 Diagonalization of Matrix A", - "page": 963 - }, - { - "level": 2, - "title": "10.6 CONTROLLABILITY AND OBSERVABILITY", - "page": 967 - }, - { - "level": 3, - "title": "10.6-1 Inadequacy of the Transfer Function Description of a System", - "page": 973 - }, - { - "level": 2, - "title": "10.7 STATE-SPACE ANALYSIS OF DISCRETE-TIME SYSTEMS", - "page": 973 - }, - { - "level": 3, - "title": "10.7-1 Solution in State Space", - "page": 975 - }, - { - "level": 3, - "title": "10.7-2 The z-Transform Solution", - "page": 979 - }, - { - "level": 2, - "title": "10.8 MATLAB: TOOLBOXES AND STATE-SPACE ANALYSIS", - "page": 981 - }, - { - "level": 3, - "title": "10.8-1 z-Transform Solutions to Discrete-Time, State-Space Systems", - "page": 981 - }, - { - "level": 3, - "title": "10.8-2 Transfer Functions from State-Space Representations", - "page": 984 - }, - { - "level": 3, - "title": "10.8-3 Controllability and Observability of Discrete-Time Systems", - "page": 985 - }, - { - "level": 3, - "title": "10.8-4 Matrix Exponentiation and the Matrix Exponential", - "page": 988 - }, - { - "level": 2, - "title": "10.9 SUMMARY", - "page": 989 - }, - { - "level": 2, - "title": "REFERENCES", - "page": 990 - }, - { - "level": 2, - "title": "PROBLEMS", - "page": 990 - }, - { - "level": 1, - "title": "INDEX", - "page": 995 - } - ] -} \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.layout.json b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.layout.json deleted file mode 100644 index c5416ef40062c936dea8113d9e8e9554765774a7..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.layout.json +++ /dev/null @@ -1,373289 +0,0 @@ -{ - "doc_id": "linear-systems-and-signals-3rd-edition-bp-lathi", - "file_name": "linear-systems-and-signals-3rd-edition-bp-lathi.pdf", - "title": "LINEAR SYSTEMS AND SIGNALS", - "author": "B. P. Lathi and R. A. Green", - "total_pages": 1010, - "pages": [ - { - "page_num": 1, - "width": 576.0, - "height": 720.0, - "blocks": [] - }, - { - "page_num": 2, - "width": 576.0, - "height": 720.0, - "blocks": [] - }, - { - "page_num": 3, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p3-b0", - "global_id": 1, - "bbox": [ - 127.59, - 137.97, - 442.21, - 157.16 - ], - "text": "LINEAR SYSTEMS AND SIGNALS", - "type": "text" - } - ] - }, - { - "page_num": 4, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p4-b0", - "global_id": 2, - "bbox": [ - 102.14, - 60.69, - 338.51, - 81.22 - ], - "text": "T H E O X F O R D S E R I E S I N E L E C T R I C A L\nAND COMPUTER ENGINEERING", - "type": "text" - }, - { - "block_id": "p4-b1", - "global_id": 3, - "bbox": [ - 101.84, - 89.67, - 204.96, - 98.71 - ], - "text": "Adel S. Sedra, Series Editor", - "type": "text" - }, - { - "block_id": "p4-b2", - "global_id": 4, - "bbox": [ - 101.84, - 115.57, - 465.57, - 595.87 - ], - "text": "Allen and Holberg, CMOS Analog Circuit Design, 3rd edition\nBoncelet, Probability, Statistics, and Random Signals\nBobrow, Elementary Linear Circuit Analysis, 2nd edition\nBobrow, Fundamentals of Electrical Engineering, 2nd edition\nCampbell, Fabrication Engineering at the Micro- and Nanoscale, 4th edition\nChen, Digital Signal Processing\nChen, Linear System Theory and Design, 4th edition\nChen, Signals and Systems, 3rd edition\nComer, Digital Logic and State Machine Design, 3rd edition\nComer, Microprocessor-Based System Design\nCooper and McGillem, Probabilistic Methods of Signal and System Analysis, 3rd edition\nDimitrijev, Principles of Semiconductor Device, 2nd edition\nDimitrijev, Understanding Semiconductor Devices\nFortney, Principles of Electronics: Analog & Digital\nFranco, Electric Circuits Fundamentals\nGhausi, Electronic Devices and Circuits: Discrete and Integrated\nGuru and Hiziro˘glu, Electric Machinery and Transformers, 3rd edition\nHouts, Signal Analysis in Linear Systems\nJones, Introduction to Optical Fiber Communication Systems\nKrein, Elements of Power Electronics, 2nd Edition\nKuo, Digital Control Systems, 3rd edition\nLathi and Green, Linear Systems and Signals, 3rd edition\nLathi and Ding, Modern Digital and Analog Communication Systems, 5th edition\nLathi, Signal Processing and Linear Systems\nMartin, Digital Integrated Circuit Design\nMiner, Lines and Electromagnetic Fields for Engineers\nMitra, Signals and Systems\nParhami, Computer Architecture\nParhami, Computer Arithmetic, 2nd edition\nRoberts and Sedra, SPICE, 2nd edition\nRoberts, Taenzler, and Burns, An Introduction to Mixed-Signal IC Test and Measurement, 2nd edition\nRoulston, An Introduction to the Physics of Semiconductor Devices\nSadiku, Elements of Electromagnetics, 7th edition\nSantina, Stubberud, and Hostetter, Digital Control System Design, 2nd edition\nSarma, Introduction to Electrical Engineering\nSchaumann, Xiao, and Van Valkenburg, Design of Analog Filters, 3rd edition\nSchwarz and Oldham, Electrical Engineering: An Introduction, 2nd edition\nSedra and Smith, Microelectronic Circuits, 7th edition\nStefani, Shahian, Savant, and Hostetter, Design of Feedback Control Systems, 4th edition\nTsividis, Operation and Modeling of the MOS Transistor, 3rd edition\nVan Valkenburg, Analog Filter Design\nWarner and Grung, Semiconductor Device Electronics\nWolovich, Automatic Control Systems\nYariv and Yeh, Photonics: Optical Electronics in Modern Communications, 6th edition", - "type": "text" - }, - { - "block_id": "p4-b3", - "global_id": 5, - "bbox": [ - 101.84, - 595.96, - 194.09, - 606.82 - ], - "text": "˙Zak, Systems and Control", - "type": "text" - } - ] - }, - { - "page_num": 5, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p5-b0", - "global_id": 6, - "bbox": [ - 275.71, - 115.82, - 533.64, - 176.0 - ], - "text": "LINEAR SYSTEMS\nAND SIGNALS", - "type": "text" - }, - { - "block_id": "p5-b1", - "global_id": 7, - "bbox": [ - 275.71, - 197.85, - 397.72, - 213.01 - ], - "text": "THIRD EDITION", - "type": "text" - }, - { - "block_id": "p5-b2", - "global_id": 8, - "bbox": [ - 275.71, - 291.89, - 490.14, - 309.62 - ], - "text": "B. P. Lathi and R. A. Green", - "type": "text" - }, - { - "block_id": "p5-b3", - "global_id": 9, - "bbox": [ - 277.03, - 605.34, - 404.96, - 636.73 - ], - "text": "New York\nOxford\nOXFORD UNIVERSITY PRESS\n2018", - "type": "text" - } - ] - }, - { - "page_num": 6, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p6-b0", - "global_id": 10, - "bbox": [ - 101.84, - 178.38, - 321.03, - 204.28 - ], - "text": "Oxford University Press is a department of the University of Oxford.\nIt furthers the University’s objective of excellence in research,\nscholarship, and education by publishing worldwide.", - "type": "text" - }, - { - "block_id": "p6-b1", - "global_id": 11, - "bbox": [ - 101.84, - 214.24, - 309.23, - 249.12 - ], - "text": "Oxford\nNew York\nAuckland\nCape Town\nDar es Salaam\nHong Kong\nKarachi\nKuala Lumpur\nMadrid\nMelbourne\nMexico City\nNairobi\nNew Delhi\nShanghai\nTaipei\nToronto", - "type": "text" - }, - { - "block_id": "p6-b2", - "global_id": 12, - "bbox": [ - 101.84, - 259.07, - 338.52, - 293.94 - ], - "text": "With offices in\nArgentina\nAustria\nBrazil\nChile\nCzech Republic\nFrance\nGreece\nGuatemala\nHungary\nItaly\nJapan\nPoland\nPortugal\nSingapore\nSouth Korea\nSwitzerland\nThailand\nTurkey\nUkraine\nVietnam", - "type": "text" - }, - { - "block_id": "p6-b3", - "global_id": 13, - "bbox": [ - 101.84, - 303.58, - 251.1, - 311.88 - ], - "text": "Copyright c⃝2018 by Oxford University Press", - "type": "text" - }, - { - "block_id": "p6-b4", - "global_id": 14, - "bbox": [ - 108.22, - 330.39, - 304.44, - 356.29 - ], - "text": "For titles covered by Section 112 of the US Higher Education\nOpportunity Act, please visit www.oup.com/us/he for the\nlatest information about pricing and alternate formats.", - "type": "text" - }, - { - "block_id": "p6-b5", - "global_id": 15, - "bbox": [ - 101.84, - 374.8, - 243.5, - 400.7 - ], - "text": "Published by Oxford University Press.\n198 Madison Avenue, New York, NY 10016\nhttp://www.oup.com", - "type": "text" - }, - { - "block_id": "p6-b6", - "global_id": 16, - "bbox": [ - 101.84, - 410.67, - 295.06, - 418.64 - ], - "text": "Oxford is a registered trademark of Oxford University Press.", - "type": "text" - }, - { - "block_id": "p6-b7", - "global_id": 17, - "bbox": [ - 101.84, - 428.6, - 333.19, - 463.47 - ], - "text": "All rights reserved. No part of this publication may be reproduced,\nstored in a retrieval system, or transmitted, in any form or by any means,\nelectronic, mechanical, photocopying, recording, or otherwise,\nwithout the prior permission of Oxford University Press.", - "type": "text" - }, - { - "block_id": "p6-b8", - "global_id": 18, - "bbox": [ - 101.84, - 478.41, - 373.12, - 576.05 - ], - "text": "Library of Congress Cataloging-in-Publication Data\nNames: Lathi, B. P. (Bhagwandas Pannalal), author. |\nGreen, R. A. (Roger A.), author.\nTitle: Linear systems and signals / B.P. Lathi and R.A. Green.\nDescription: Third Edition. | New York : Oxford University Press, [2018] |\nSeries: The Oxford Series in Electrical and Computer Engineering\nIdentifiers: LCCN 2017034962 | ISBN 9780190200176 (hardcover : acid-free paper)\nSubjects: LCSH: Signal processing–Mathematics. | System analysis. | Linear\ntime invariant systems. | Digital filters (Mathematics)\nClassification: LCC TK5102.5 L298 2017 | DDC 621.382/2–dc23 LC record\navailable at https://lccn.loc.gov/2017034962", - "type": "text" - }, - { - "block_id": "p6-b9", - "global_id": 19, - "bbox": [ - 101.84, - 586.01, - 189.77, - 593.98 - ], - "text": "ISBN 978–0–19–020017–6", - "type": "text" - }, - { - "block_id": "p6-b10", - "global_id": 20, - "bbox": [ - 101.84, - 608.83, - 225.83, - 616.8 - ], - "text": "Printing number: 9 8 7 6 5 4 3 2 1", - "type": "text" - }, - { - "block_id": "p6-b11", - "global_id": 21, - "bbox": [ - 101.84, - 626.77, - 285.8, - 634.74 - ], - "text": "Printed by R.R. Donnelly in the United States of America", - "type": "text" - } - ] - }, - { - "page_num": 7, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p7-b0", - "global_id": 22, - "bbox": [ - 128.12, - 100.75, - 235.11, - 121.67 - ], - "text": "CONTENTS", - "type": "text" - }, - { - "block_id": "p7-b1", - "global_id": 23, - "bbox": [ - 148.81, - 219.01, - 210.28, - 229.25 - ], - "text": "PREFACE xv", - "type": "text" - }, - { - "block_id": "p7-b2", - "global_id": 24, - "bbox": [ - 127.59, - 247.79, - 229.28, - 259.91 - ], - "text": "B BACKGROUND", - "type": "text" - }, - { - "block_id": "p7-b3", - "global_id": 25, - "bbox": [ - 127.59, - 266.24, - 234.87, - 276.2 - ], - "text": "B.1 Complex Numbers 1", - "type": "text" - }, - { - "block_id": "p7-b4", - "global_id": 26, - "bbox": [ - 127.59, - 278.19, - 321.07, - 314.06 - ], - "text": "B.1-1\nA Historical Note 1\nB.1-2\nAlgebra of Complex Numbers 5\nB.2 Sinusoids 16", - "type": "text" - }, - { - "block_id": "p7-b5", - "global_id": 27, - "bbox": [ - 127.59, - 316.05, - 346.87, - 351.92 - ], - "text": "B.2-1\nAddition of Sinusoids 18\nB.2-2\nSinusoids in Terms of Exponentials 20\nB.3 Sketching Signals 20", - "type": "text" - }, - { - "block_id": "p7-b6", - "global_id": 28, - "bbox": [ - 127.59, - 353.92, - 353.1, - 403.72 - ], - "text": "B.3-1\nMonotonic Exponentials 20\nB.3-2\nThe Exponentially Varying Sinusoid\n22\nB.4 Cramer’s Rule 23\nB.5 Partial Fraction Expansion 25", - "type": "text" - }, - { - "block_id": "p7-b7", - "global_id": 29, - "bbox": [ - 127.58, - 405.72, - 463.85, - 489.42 - ], - "text": "B.5-1\nMethod of Clearing Fractions 26\nB.5-2\nThe Heaviside “Cover-Up” Method 27\nB.5-3\nRepeated Factors of Q(x) 31\nB.5-4\nA Combination of Heaviside “Cover-Up” and Clearing Fractions 32\nB.5-5\nImproper F(x) with m = n 34\nB.5-6\nModified Partial Fractions 35\nB.6 Vectors and Matrices 36", - "type": "text" - }, - { - "block_id": "p7-b8", - "global_id": 30, - "bbox": [ - 127.58, - 491.42, - 334.86, - 527.29 - ], - "text": "B.6-1\nSome Definitions and Properties 37\nB.6-2\nMatrix Algebra 38\nB.7 MATLAB: Elementary Operations 42", - "type": "text" - }, - { - "block_id": "p7-b9", - "global_id": 31, - "bbox": [ - 127.58, - 529.28, - 334.48, - 624.94 - ], - "text": "B.7-1\nMATLAB Overview 42\nB.7-2\nCalculator Operations 43\nB.7-3\nVector Operations 45\nB.7-4\nSimple Plotting 46\nB.7-5\nElement-by-Element Operations 48\nB.7-6\nMatrix Operations 49\nB.7-7\nPartial Fraction Expansions 53\nB.8 Appendix: Useful Mathematical Formulas 54", - "type": "text" - }, - { - "block_id": "p7-b10", - "global_id": 32, - "bbox": [ - 157.46, - 626.93, - 298.84, - 636.9 - ], - "text": "B.8-1\nSome Useful Constants 54", - "type": "text" - }, - { - "block_id": "p7-b11", - "global_id": 33, - "bbox": [ - 510.5, - 656.18, - 516.12, - 666.28 - ], - "text": "v", - "type": "text" - } - ] - }, - { - "page_num": 8, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p8-b0", - "global_id": 34, - "bbox": [ - 60.0, - 62.89, - 137.48, - 71.98 - ], - "text": "vi\nContents", - "type": "text" - }, - { - "block_id": "p8-b1", - "global_id": 35, - "bbox": [ - 122.77, - 85.82, - 351.4, - 217.35 - ], - "text": "B.8-2\nComplex Numbers 54\nB.8-3\nSums 54\nB.8-4\nTaylor and Maclaurin Series 55\nB.8-5\nPower Series 55\nB.8-6\nTrigonometric Identities 55\nB.8-7\nCommon Derivative Formulas 56\nB.8-8\nIndefinite Integrals 57\nB.8-9\nL’Hôpital’s Rule 58\nB.8-10 Solution of Quadratic and Cubic Equations 58\nReferences 58\nProblems 59", - "type": "text" - }, - { - "block_id": "p8-b2", - "global_id": 36, - "bbox": [ - 101.84, - 235.77, - 256.24, - 247.9 - ], - "text": "1 SIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p8-b3", - "global_id": 37, - "bbox": [ - 101.84, - 254.21, - 201.51, - 264.18 - ], - "text": "1.1\nSize of a Signal 64", - "type": "text" - }, - { - "block_id": "p8-b4", - "global_id": 38, - "bbox": [ - 101.84, - 266.17, - 264.62, - 302.04 - ], - "text": "1.1-1\nSignal Energy 65\n1.1-2\nSignal Power 65\n1.2\nSome Useful Signal Operations 71", - "type": "text" - }, - { - "block_id": "p8-b5", - "global_id": 39, - "bbox": [ - 101.84, - 304.04, - 267.27, - 363.81 - ], - "text": "1.2-1\nTime Shifting 71\n1.2-2\nTime Scaling 73\n1.2-3\nTime Reversal 76\n1.2-4\nCombined Operations 77\n1.3\nClassification of Signals 78", - "type": "text" - }, - { - "block_id": "p8-b6", - "global_id": 40, - "bbox": [ - 101.84, - 365.82, - 357.91, - 437.56 - ], - "text": "1.3-1\nContinuous-Time and Discrete-Time Signals 78\n1.3-2\nAnalog and Digital Signals 78\n1.3-3\nPeriodic and Aperiodic Signals 79\n1.3-4\nEnergy and Power Signals 82\n1.3-5\nDeterministic and Random Signals 82\n1.4\nSome Useful Signal Models 82", - "type": "text" - }, - { - "block_id": "p8-b7", - "global_id": 41, - "bbox": [ - 101.84, - 439.14, - 305.44, - 487.34 - ], - "text": "1.4-1\nThe Unit Step Function u(t) 83\n1.4-2\nThe Unit Impulse Function δ(t) 86\n1.4-3\nThe Exponential Function est 89\n1.5\nEven and Odd Functions 92", - "type": "text" - }, - { - "block_id": "p8-b8", - "global_id": 42, - "bbox": [ - 101.84, - 489.33, - 357.06, - 539.14 - ], - "text": "1.5-1\nSome Properties of Even and Odd Functions 92\n1.5-2\nEven and Odd Components of a Signal 93\n1.6\nSystems 95\n1.7\nClassification of Systems 97", - "type": "text" - }, - { - "block_id": "p8-b9", - "global_id": 43, - "bbox": [ - 131.73, - 541.14, - 366.76, - 634.82 - ], - "text": "1.7-1\nLinear and Nonlinear Systems 97\n1.7-2\nTime-Invariant and Time-Varying Systems 102\n1.7-3\nInstantaneous and Dynamic Systems 103\n1.7-4\nCausal and Noncausal Systems 104\n1.7-5\nContinuous-Time and Discrete-Time Systems 107\n1.7-6\nAnalog and Digital Systems 109\n1.7-7\nInvertible and Noninvertible Systems 109\n1.7-8\nStable and Unstable Systems 110", - "type": "text" - } - ] - }, - { - "page_num": 9, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p9-b0", - "global_id": 44, - "bbox": [ - 437.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Contents\nvii", - "type": "text" - }, - { - "block_id": "p9-b1", - "global_id": 45, - "bbox": [ - 127.59, - 85.82, - 335.17, - 95.78 - ], - "text": "1.8\nSystem Model: Input–Output Description 111", - "type": "text" - }, - { - "block_id": "p9-b2", - "global_id": 46, - "bbox": [ - 127.59, - 97.78, - 419.95, - 261.18 - ], - "text": "1.8-1\nElectrical Systems 111\n1.8-2\nMechanical Systems 114\n1.8-3\nElectromechanical Systems 118\n1.9\nInternal and External Descriptions of a System 119\n1.10 Internal Description: The State-Space Description 121\n1.11 MATLAB: Working with Functions 126\n1.11-1 Anonymous Functions 126\n1.11-2 Relational Operators and the Unit Step Function 128\n1.11-3 Visualizing Operations on the Independent Variable 130\n1.11-4 Numerical Integration and Estimating Signal Energy 131\n1.12 Summary 133\nReferences 135\nProblems 136", - "type": "text" - }, - { - "block_id": "p9-b3", - "global_id": 47, - "bbox": [ - 127.59, - 297.6, - 485.07, - 309.73 - ], - "text": "2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p9-b4", - "global_id": 48, - "bbox": [ - 127.59, - 316.04, - 436.45, - 339.95 - ], - "text": "2.1\nIntroduction 150\n2.2\nSystem Response to Internal Conditions: The Zero-Input Response 151", - "type": "text" - }, - { - "block_id": "p9-b5", - "global_id": 49, - "bbox": [ - 127.59, - 341.95, - 433.66, - 379.81 - ], - "text": "2.2-1\nSome Insights into the Zero-Input Behavior of a System 161\n2.3\nThe Unit Impulse Response h(t) 163\n2.4\nSystem Response to External Input: The Zero-State Response 168", - "type": "text" - }, - { - "block_id": "p9-b6", - "global_id": 50, - "bbox": [ - 127.6, - 381.8, - 415.54, - 465.49 - ], - "text": "2.4-1\nThe Convolution Integral 170\n2.4-2\nGraphical Understanding of Convolution Operation 178\n2.4-3\nInterconnected Systems 190\n2.4-4\nA Very Special Function for LTIC Systems:\nThe Everlasting Exponential est 193\n2.4-5\nTotal Response 195\n2.5\nSystem Stability 196", - "type": "text" - }, - { - "block_id": "p9-b7", - "global_id": 51, - "bbox": [ - 127.6, - 467.48, - 426.42, - 515.3 - ], - "text": "2.5-1\nExternal (BIBO) Stability 196\n2.5-2\nInternal (Asymptotic) Stability 198\n2.5-3\nRelationship Between BIBO and Asymptotic Stability 199\n2.6\nIntuitive Insights into System Behavior 203", - "type": "text" - }, - { - "block_id": "p9-b8", - "global_id": 52, - "bbox": [ - 127.6, - 517.31, - 440.84, - 612.97 - ], - "text": "2.6-1\nDependence of System Behavior on Characteristic Modes 203\n2.6-2\nResponse Time of a System: The System Time Constant 205\n2.6-3\nTime Constant and Rise Time of a System 206\n2.6-4\nTime Constant and Filtering 207\n2.6-5\nTime Constant and Pulse Dispersion (Spreading) 209\n2.6-6\nTime Constant and Rate of Information Transmission 209\n2.6-7\nThe Resonance Phenomenon 210\n2.7\nMATLAB: M-Files 212", - "type": "text" - }, - { - "block_id": "p9-b9", - "global_id": 53, - "bbox": [ - 157.48, - 614.96, - 279.73, - 636.88 - ], - "text": "2.7-1\nScript M-Files 213\n2.7-2\nFunction M-Files 214", - "type": "text" - } - ] - }, - { - "page_num": 10, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p10-b0", - "global_id": 54, - "bbox": [ - 60.0, - 62.89, - 137.48, - 71.98 - ], - "text": "viii\nContents", - "type": "text" - }, - { - "block_id": "p10-b1", - "global_id": 55, - "bbox": [ - 101.85, - 85.82, - 347.46, - 161.53 - ], - "text": "2.7-3\nFor-Loops 215\n2.7-4\nGraphical Understanding of Convolution 217\n2.8\nAppendix: Determining the Impulse Response 220\n2.9\nSummary 221\nReferences 223\nProblems 223", - "type": "text" - }, - { - "block_id": "p10-b2", - "global_id": 56, - "bbox": [ - 101.84, - 188.99, - 433.91, - 201.11 - ], - "text": "3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p10-b3", - "global_id": 57, - "bbox": [ - 101.84, - 207.43, - 192.93, - 217.4 - ], - "text": "3.1\nIntroduction 237", - "type": "text" - }, - { - "block_id": "p10-b4", - "global_id": 58, - "bbox": [ - 101.84, - 219.39, - 315.44, - 257.25 - ], - "text": "3.1-1\nSize of a Discrete-Time Signal 238\n3.2\nUseful Signal Operations 240\n3.3\nSome Useful Discrete-Time Signal Models 245", - "type": "text" - }, - { - "block_id": "p10-b5", - "global_id": 59, - "bbox": [ - 101.85, - 258.82, - 349.5, - 330.96 - ], - "text": "3.3-1\nDiscrete-Time Impulse Function δ[n] 245\n3.3-2\nDiscrete-Time Unit Step Function u[n] 246\n3.3-3\nDiscrete-Time Exponential γ n 247\n3.3-4\nDiscrete-Time Sinusoid cos(n + θ) 251\n3.3-5\nDiscrete-Time Complex Exponential ejn 252\n3.4\nExamples of Discrete-Time Systems 253", - "type": "text" - }, - { - "block_id": "p10-b6", - "global_id": 60, - "bbox": [ - 101.85, - 332.97, - 345.31, - 356.87 - ], - "text": "3.4-1\nClassification of Discrete-Time Systems 262\n3.5\nDiscrete-Time System Equations 265", - "type": "text" - }, - { - "block_id": "p10-b7", - "global_id": 61, - "bbox": [ - 101.85, - 358.87, - 410.73, - 396.72 - ], - "text": "3.5-1\nRecursive (Iterative) Solution of Difference Equation 266\n3.6\nSystem Response to Internal Conditions: The Zero-Input Response 270\n3.7\nThe Unit Impulse Response h[n] 277", - "type": "text" - }, - { - "block_id": "p10-b8", - "global_id": 62, - "bbox": [ - 101.85, - 398.3, - 389.68, - 422.62 - ], - "text": "3.7-1\nThe Closed-Form Solution of h[n] 278\n3.8\nSystem Response to External Input: The Zero-State Response 280", - "type": "text" - }, - { - "block_id": "p10-b9", - "global_id": 63, - "bbox": [ - 101.85, - 424.62, - 367.95, - 472.44 - ], - "text": "3.8-1\nGraphical Procedure for the Convolution Sum 288\n3.8-2\nInterconnected Systems 294\n3.8-3\nTotal Response 297\n3.9\nSystem Stability 298", - "type": "text" - }, - { - "block_id": "p10-b10", - "global_id": 64, - "bbox": [ - 101.85, - 474.44, - 400.69, - 623.89 - ], - "text": "3.9-1\nExternal (BIBO) Stability 298\n3.9-2\nInternal (Asymptotic) Stability 299\n3.9-3\nRelationship Between BIBO and Asymptotic Stability 301\n3.10 Intuitive Insights into System Behavior 305\n3.11 MATLAB: Discrete-Time Signals and Systems 306\n3.11-1 Discrete-Time Functions and Stem Plots 306\n3.11-2 System Responses Through Filtering 308\n3.11-3 A Custom Filter Function 310\n3.11-4 Discrete-Time Convolution 311\n3.12 Appendix: Impulse Response for a Special Case 313\n3.13 Summary 313\nProblems 314", - "type": "text" - } - ] - }, - { - "page_num": 11, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p11-b0", - "global_id": 65, - "bbox": [ - 437.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Contents\nix", - "type": "text" - }, - { - "block_id": "p11-b1", - "global_id": 66, - "bbox": [ - 127.59, - 86.11, - 411.22, - 112.19 - ], - "text": "4 CONTINUOUS-TIME SYSTEM ANALYSIS USING\nTHE LAPLACE TRANSFORM", - "type": "text" - }, - { - "block_id": "p11-b2", - "global_id": 67, - "bbox": [ - 127.59, - 118.5, - 263.15, - 128.47 - ], - "text": "4.1\nThe Laplace Transform 330", - "type": "text" - }, - { - "block_id": "p11-b3", - "global_id": 68, - "bbox": [ - 127.59, - 130.46, - 338.69, - 154.37 - ], - "text": "4.1-1\nFinding the Inverse Transform 338\n4.2\nSome Properties of the Laplace Transform 349", - "type": "text" - }, - { - "block_id": "p11-b4", - "global_id": 69, - "bbox": [ - 127.59, - 156.36, - 402.6, - 240.07 - ], - "text": "4.2-1\nTime Shifting 349\n4.2-2\nFrequency Shifting 353\n4.2-3\nThe Time-Differentiation Property 354\n4.2-4\nThe Time-Integration Property 356\n4.2-5\nThe Scaling Property 357\n4.2-6\nTime Convolution and Frequency Convolution 357\n4.3\nSolution of Differential and Integro-Differential Equations 360", - "type": "text" - }, - { - "block_id": "p11-b5", - "global_id": 70, - "bbox": [ - 127.59, - 238.42, - 408.5, - 301.82 - ], - "text": "4.3-1\nComments on Initial Conditions at 0−and at 0+ 363\n4.3-2\nZero-State Response 366\n4.3-3\nStability 371\n4.3-4\nInverse Systems 373\n4.4\nAnalysis of Electrical Networks: The Transformed Network 373", - "type": "text" - }, - { - "block_id": "p11-b6", - "global_id": 71, - "bbox": [ - 127.59, - 303.82, - 318.62, - 341.67 - ], - "text": "4.4-1\nAnalysis of Active Circuits 382\n4.5\nBlock Diagrams 386\n4.6\nSystem Realization 388", - "type": "text" - }, - { - "block_id": "p11-b7", - "global_id": 72, - "bbox": [ - 127.59, - 343.67, - 422.87, - 415.41 - ], - "text": "4.6-1\nDirect Form I Realization 389\n4.6-2\nDirect Form II Realization 390\n4.6-3\nCascade and Parallel Realizations 393\n4.6-4\nTransposed Realization 396\n4.6-5\nUsing Operational Amplifiers for System Realization 399\n4.7\nApplication to Feedback and Controls 404", - "type": "text" - }, - { - "block_id": "p11-b8", - "global_id": 73, - "bbox": [ - 127.59, - 417.41, - 358.37, - 441.31 - ], - "text": "4.7-1\nAnalysis of a Simple Control System 406\n4.8\nFrequency Response of an LTIC System 412", - "type": "text" - }, - { - "block_id": "p11-b9", - "global_id": 74, - "bbox": [ - 127.59, - 443.31, - 413.12, - 467.21 - ], - "text": "4.8-1\nSteady-State Response to Causal Sinusoidal Inputs 418\n4.9\nBode Plots 419", - "type": "text" - }, - { - "block_id": "p11-b10", - "global_id": 75, - "bbox": [ - 127.59, - 468.77, - 421.15, - 626.63 - ], - "text": "4.9-1\nConstant Ka1a2/b1b3 422\n4.9-2\nPole (or Zero) at the Origin 422\n4.9-3\nFirst-Order Pole (or Zero) 424\n4.9-4\nSecond-Order Pole (or Zero) 426\n4.9-5\nThe Transfer Function from the Frequency Response 435\n4.10 Filter Design by Placement of Poles and Zeros of H(s) 436\n4.10-1 Dependence of Frequency Response on Poles\nand Zeros of H(s) 436\n4.10-2 Lowpass Filters 439\n4.10-3 Bandpass Filters 441\n4.10-4 Notch (Bandstop) Filters 441\n4.10-5 Practical Filters and Their Specifications 444\n4.11 The Bilateral Laplace Transform 445", - "type": "text" - } - ] - }, - { - "page_num": 12, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p12-b0", - "global_id": 76, - "bbox": [ - 60.0, - 62.89, - 137.47, - 71.98 - ], - "text": "x\nContents", - "type": "text" - }, - { - "block_id": "p12-b1", - "global_id": 77, - "bbox": [ - 101.85, - 85.82, - 415.8, - 221.33 - ], - "text": "4.11-1 Properties of the Bilateral Laplace Transform 451\n4.11-2 Using the Bilateral Transform for Linear System Analysis 452\n4.12 MATLAB: Continuous-Time Filters 455\n4.12-1 Frequency Response and Polynomial Evaluation 456\n4.12-2 Butterworth Filters and the Find Command 459\n4.12-3 Using Cascaded Second-Order Sections for Butterworth\nFilter Realization 461\n4.12-4 Chebyshev Filters 463\n4.13 Summary 466\nReferences 468\nProblems 468", - "type": "text" - }, - { - "block_id": "p12-b2", - "global_id": 78, - "bbox": [ - 101.84, - 241.32, - 475.43, - 253.87 - ], - "text": "5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p12-b3", - "global_id": 79, - "bbox": [ - 101.84, - 260.09, - 210.56, - 270.15 - ], - "text": "5.1\nThe z-Transform 488", - "type": "text" - }, - { - "block_id": "p12-b4", - "global_id": 80, - "bbox": [ - 101.82, - 272.15, - 423.68, - 308.01 - ], - "text": "5.1-1\nInverse Transform by Partial Fraction Expansion and Tables 495\n5.1-2\nInverse z-Transform by Power Series Expansion 499\n5.2\nSome Properties of the z-Transform 501", - "type": "text" - }, - { - "block_id": "p12-b5", - "global_id": 81, - "bbox": [ - 101.84, - 310.01, - 408.14, - 381.74 - ], - "text": "5.2-1\nTime-Shifting Properties 501\n5.2-2\nz-Domain Scaling Property (Multiplication by γ n) 505\n5.2-3\nz-Domain Differentiation Property (Multiplication by n) 506\n5.2-4\nTime-Reversal Property 506\n5.2-5\nConvolution Property 507\n5.3\nz-Transform Solution of Linear Difference Equations 510", - "type": "text" - }, - { - "block_id": "p12-b6", - "global_id": 82, - "bbox": [ - 101.83, - 383.74, - 432.59, - 445.5 - ], - "text": "5.3-1\nZero-State Response of LTID Systems: The Transfer Function 514\n5.3-2\nStability 518\n5.3-3\nInverse Systems 519\n5.4\nSystem Realization 519\n5.5\nFrequency Response of Discrete-Time Systems 526", - "type": "text" - }, - { - "block_id": "p12-b7", - "global_id": 83, - "bbox": [ - 101.83, - 447.51, - 361.16, - 511.26 - ], - "text": "5.5-1\nThe Periodic Nature of Frequency Response 532\n5.5-2\nAliasing and Sampling Rate 536\n5.6\nFrequency Response from Pole-Zero Locations 538\n5.7\nDigital Processing of Analog Signals 547\n5.8\nThe Bilateral z-Transform 554", - "type": "text" - }, - { - "block_id": "p12-b8", - "global_id": 84, - "bbox": [ - 101.81, - 513.16, - 433.4, - 634.82 - ], - "text": "5.8-1\nProperties of the Bilateral z-Transform 559\n5.8-2\nUsing the Bilateral z-Transform for Analysis of LTID Systems 560\n5.9\nConnecting the Laplace and z-Transforms 563\n5.10 MATLAB: Discrete-Time IIR Filters 565\n5.10-1 Frequency Response and Pole-Zero Plots 566\n5.10-2 Transformation Basics 567\n5.10-3 Transformation by First-Order Backward Difference 568\n5.10-4 Bilinear Transformation 569\n5.10-5 Bilinear Transformation with Prewarping 570\n5.10-6 Example: Butterworth Filter Transformation 571", - "type": "text" - } - ] - }, - { - "page_num": 13, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p13-b0", - "global_id": 85, - "bbox": [ - 437.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Contents\nxi", - "type": "text" - }, - { - "block_id": "p13-b1", - "global_id": 86, - "bbox": [ - 127.6, - 85.82, - 446.27, - 147.59 - ], - "text": "5.10-7 Problems Finding Polynomial Roots 572\n5.10-8 Using Cascaded Second-Order Sections to Improve Design 572\n5.11 Summary 574\nReferences 575\nProblems 575", - "type": "text" - }, - { - "block_id": "p13-b2", - "global_id": 87, - "bbox": [ - 127.59, - 165.52, - 497.48, - 177.64 - ], - "text": "6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p13-b3", - "global_id": 88, - "bbox": [ - 127.59, - 183.97, - 423.54, - 193.93 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series 593", - "type": "text" - }, - { - "block_id": "p13-b4", - "global_id": 89, - "bbox": [ - 127.59, - 195.92, - 419.59, - 243.75 - ], - "text": "6.1-1\nThe Fourier Spectrum 598\n6.1-2\nThe Effect of Symmetry 607\n6.1-3\nDetermining the Fundamental Frequency and Period 609\n6.2\nExistence and Convergence of the Fourier Series 612", - "type": "text" - }, - { - "block_id": "p13-b5", - "global_id": 90, - "bbox": [ - 127.59, - 245.74, - 443.77, - 281.61 - ], - "text": "6.2-1\nConvergence of a Series 613\n6.2-2\nThe Role of Amplitude and Phase Spectra in Waveshaping 615\n6.3\nExponential Fourier Series 621", - "type": "text" - }, - { - "block_id": "p13-b6", - "global_id": 91, - "bbox": [ - 127.59, - 283.61, - 353.82, - 345.37 - ], - "text": "6.3-1\nExponential Fourier Spectra 624\n6.3-2\nParseval’s Theorem 632\n6.3-3\nProperties of the Fourier Series 635\n6.4\nLTIC System Response to Periodic Inputs 637\n6.5\nGeneralized Fourier Series: Signals as Vectors 641", - "type": "text" - }, - { - "block_id": "p13-b7", - "global_id": 92, - "bbox": [ - 127.59, - 347.37, - 413.41, - 421.07 - ], - "text": "6.5-1\nComponent of a Vector 642\n6.5-2\nSignal Comparison and Component of a Signal 643\n6.5-3\nExtension to Complex Signals 645\n6.5-4\nSignal Representation by an Orthogonal Signal Set 647\n6.6\nNumerical Computation of Dn 659\n6.7\nMATLAB: Fourier Series Applications 661", - "type": "text" - }, - { - "block_id": "p13-b8", - "global_id": 93, - "bbox": [ - 127.59, - 423.06, - 397.59, - 484.83 - ], - "text": "6.7-1\nPeriodic Functions and the Gibbs Phenomenon 661\n6.7-2\nOptimization and Phase Spectra 664\n6.8\nSummary 667\nReferences 668\nProblems 669", - "type": "text" - }, - { - "block_id": "p13-b9", - "global_id": 94, - "bbox": [ - 127.59, - 502.77, - 455.27, - 528.84 - ], - "text": "7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER\nTRANSFORM", - "type": "text" - }, - { - "block_id": "p13-b10", - "global_id": 95, - "bbox": [ - 127.59, - 535.16, - 391.59, - 545.12 - ], - "text": "7.1\nAperiodic Signal Representation by the Fourier Integral 680", - "type": "text" - }, - { - "block_id": "p13-b11", - "global_id": 96, - "bbox": [ - 127.59, - 547.11, - 399.83, - 571.02 - ], - "text": "7.1-1\nPhysical Appreciation of the Fourier Transform 687\n7.2\nTransforms of Some Useful Functions 689", - "type": "text" - }, - { - "block_id": "p13-b12", - "global_id": 97, - "bbox": [ - 127.59, - 573.02, - 438.88, - 610.87 - ], - "text": "7.2-1\nConnection Between the Fourier and Laplace Transforms 700\n7.3\nSome Properties of the Fourier Transform 701\n7.4\nSignal Transmission Through LTIC Systems 721", - "type": "text" - }, - { - "block_id": "p13-b13", - "global_id": 98, - "bbox": [ - 157.47, - 612.87, - 365.16, - 634.79 - ], - "text": "7.4-1\nSignal Distortion During Transmission 723\n7.4-2\nBandpass Systems and Group Delay 726", - "type": "text" - } - ] - }, - { - "page_num": 14, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p14-b0", - "global_id": 99, - "bbox": [ - 60.0, - 62.89, - 137.47, - 71.98 - ], - "text": "xii\nContents", - "type": "text" - }, - { - "block_id": "p14-b1", - "global_id": 100, - "bbox": [ - 101.84, - 85.82, - 368.39, - 123.67 - ], - "text": "7.5\nIdeal and Practical Filters 730\n7.6\nSignal Energy 733\n7.7\nApplication to Communications: Amplitude Modulation 736", - "type": "text" - }, - { - "block_id": "p14-b2", - "global_id": 101, - "bbox": [ - 101.84, - 125.67, - 429.14, - 185.45 - ], - "text": "7.7-1\nDouble-Sideband, Suppressed-Carrier (DSB-SC) Modulation 737\n7.7-2\nAmplitude Modulation (AM) 742\n7.7-3\nSingle-Sideband Modulation (SSB) 746\n7.7-4\nFrequency-Division Multiplexing 749\n7.8\nData Truncation: Window Functions 749", - "type": "text" - }, - { - "block_id": "p14-b3", - "global_id": 102, - "bbox": [ - 101.84, - 187.44, - 313.39, - 211.35 - ], - "text": "7.8-1\nUsing Windows in Filter Design 755\n7.9\nMATLAB: Fourier Transform Topics 755", - "type": "text" - }, - { - "block_id": "p14-b4", - "global_id": 103, - "bbox": [ - 101.84, - 213.35, - 364.29, - 299.04 - ], - "text": "7.9-1\nThe Sinc Function and the Scaling Property 757\n7.9-2\nParseval’s Theorem and Essential Bandwidth 758\n7.9-3\nSpectral Sampling 759\n7.9-4\nKaiser Window Functions 760\n7.10 Summary 762\nReferences 763\nProblems 764", - "type": "text" - }, - { - "block_id": "p14-b5", - "global_id": 104, - "bbox": [ - 101.84, - 322.72, - 382.51, - 348.78 - ], - "text": "8 SAMPLING: THE BRIDGE FROM CONTINUOUS\nTO DISCRETE", - "type": "text" - }, - { - "block_id": "p14-b6", - "global_id": 105, - "bbox": [ - 101.84, - 355.11, - 238.31, - 365.08 - ], - "text": "8.1\nThe Sampling Theorem 776", - "type": "text" - }, - { - "block_id": "p14-b7", - "global_id": 106, - "bbox": [ - 101.84, - 367.07, - 260.07, - 390.98 - ], - "text": "8.1-1\nPractical Sampling 781\n8.2\nSignal Reconstruction 785", - "type": "text" - }, - { - "block_id": "p14-b8", - "global_id": 107, - "bbox": [ - 101.84, - 392.97, - 367.83, - 468.69 - ], - "text": "8.2-1\nPractical Difficulties in Signal Reconstruction 788\n8.2-2\nSome Applications of the Sampling Theorem 796\n8.3\nAnalog-to-Digital (A/D) Conversion 799\n8.4\nDual of Time Sampling: Spectral Sampling 802\n8.5\nNumerical Computation of the Fourier Transform:\nThe Discrete Fourier Transform 805", - "type": "text" - }, - { - "block_id": "p14-b9", - "global_id": 108, - "bbox": [ - 101.84, - 470.68, - 316.96, - 520.49 - ], - "text": "8.5-1\nSome Properties of the DFT 818\n8.5-2\nSome Applications of the DFT 820\n8.6\nThe Fast Fourier Transform (FFT) 824\n8.7\nMATLAB: The Discrete Fourier Transform 827", - "type": "text" - }, - { - "block_id": "p14-b10", - "global_id": 109, - "bbox": [ - 101.84, - 522.49, - 355.6, - 596.22 - ], - "text": "8.7-1\nComputing the Discrete Fourier Transform 827\n8.7-2\nImproving the Picture with Zero Padding 829\n8.7-3\nQuantization 831\n8.8\nSummary 834\nReferences 835\nProblems 835", - "type": "text" - } - ] - }, - { - "page_num": 15, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p15-b0", - "global_id": 110, - "bbox": [ - 437.66, - 62.89, - 516.14, - 71.98 - ], - "text": "Contents\nxiii", - "type": "text" - }, - { - "block_id": "p15-b1", - "global_id": 111, - "bbox": [ - 127.59, - 86.11, - 424.63, - 98.23 - ], - "text": "9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p15-b2", - "global_id": 112, - "bbox": [ - 127.59, - 104.55, - 318.59, - 114.52 - ], - "text": "9.1\nDiscrete-Time Fourier Series (DTFS) 845", - "type": "text" - }, - { - "block_id": "p15-b3", - "global_id": 113, - "bbox": [ - 127.59, - 116.51, - 464.37, - 164.34 - ], - "text": "9.1-1\nPeriodic Signal Representation by Discrete-Time Fourier Series 846\n9.1-2\nFourier Spectra of a Periodic Signal x[n] 848\n9.2\nAperiodic Signal Representation\nby Fourier Integral 855", - "type": "text" - }, - { - "block_id": "p15-b4", - "global_id": 114, - "bbox": [ - 127.59, - 166.33, - 418.55, - 216.14 - ], - "text": "9.2-1\nNature of Fourier Spectra 858\n9.2-2\nConnection Between the DTFT and the z-Transform 866\n9.3\nProperties of the DTFT 867\n9.4\nLTI Discrete-Time System Analysis by DTFT 878", - "type": "text" - }, - { - "block_id": "p15-b5", - "global_id": 115, - "bbox": [ - 127.59, - 218.14, - 321.46, - 254.01 - ], - "text": "9.4-1\nDistortionless Transmission 880\n9.4-2\nIdeal and Practical Filters 882\n9.5\nDTFT Connection with the CTFT 883", - "type": "text" - }, - { - "block_id": "p15-b6", - "global_id": 116, - "bbox": [ - 127.58, - 256.01, - 460.44, - 293.85 - ], - "text": "9.5-1\nUse of DFT and FFT for Numerical Computation of the DTFT 885\n9.6\nGeneralization of the DTFT to the z-Transform 886\n9.7\nMATLAB: Working with the DTFS and the DTFT 889", - "type": "text" - }, - { - "block_id": "p15-b7", - "global_id": 117, - "bbox": [ - 127.58, - 295.86, - 387.98, - 369.58 - ], - "text": "9.7-1\nComputing the Discrete-Time Fourier Series 889\n9.7-2\nMeasuring Code Performance 891\n9.7-3\nFIR Filter Design by Frequency Sampling 892\n9.8\nSummary 898\nReference 898\nProblems 899", - "type": "text" - }, - { - "block_id": "p15-b8", - "global_id": 118, - "bbox": [ - 127.59, - 387.28, - 283.03, - 399.4 - ], - "text": "10 STATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p15-b9", - "global_id": 119, - "bbox": [ - 127.59, - 405.73, - 453.82, - 605.0 - ], - "text": "10.1 Mathematical Preliminaries 909\n10.1-1 Derivatives and Integrals of a Matrix 909\n10.1-2 The Characteristic Equation of a Matrix:\nThe Cayley–Hamilton Theorem 910\n10.1-3 Computation of an Exponential and a Power of a Matrix 912\n10.2 Introduction to State Space 913\n10.3 A Systematic Procedure to Determine State Equations 916\n10.3-1 Electrical Circuits 916\n10.3-2 State Equations from a Transfer Function 919\n10.4 Solution of State Equations 926\n10.4-1 Laplace Transform Solution of State Equations 927\n10.4-2 Time-Domain Solution of State Equations 933\n10.5 Linear Transformation of a State Vector 939\n10.5-1 Diagonalization of Matrix A 943\n10.6 Controllability and Observability 947\n10.6-1 Inadequacy of the Transfer Function Description of a System 953", - "type": "text" - } - ] - }, - { - "page_num": 16, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p16-b0", - "global_id": 120, - "bbox": [ - 60.0, - 62.89, - 137.47, - 71.98 - ], - "text": "xiv\nContents", - "type": "text" - }, - { - "block_id": "p16-b1", - "global_id": 121, - "bbox": [ - 101.84, - 85.82, - 431.23, - 221.33 - ], - "text": "10.7 State-Space Analysis of Discrete-Time Systems 953\n10.7-1 Solution in State Space 955\n10.7-2 The z-Transform Solution 959\n10.8 MATLAB: Toolboxes and State-Space Analysis 961\n10.8-1 z-Transform Solutions to Discrete-Time, State-Space Systems 961\n10.8-2 Transfer Functions from State-Space Representations 964\n10.8-3 Controllability and Observability of Discrete-Time Systems 965\n10.8-4 Matrix Exponentiation and the Matrix Exponential 968\n10.9 Summary 969\nReferences 970\nProblems 970", - "type": "text" - }, - { - "block_id": "p16-b2", - "global_id": 122, - "bbox": [ - 123.07, - 238.25, - 176.43, - 248.49 - ], - "text": "INDEX 975", - "type": "text" - } - ] - }, - { - "page_num": 17, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p17-b0", - "global_id": 123, - "bbox": [ - 128.12, - 100.75, - 213.17, - 121.67 - ], - "text": "PREFACE", - "type": "text" - }, - { - "block_id": "p17-b1", - "global_id": 124, - "bbox": [ - 127.59, - 203.97, - 516.14, - 345.54 - ], - "text": "This book, Linear Systems and Signals, presents a comprehensive treatment of signals and\nlinear systems at an introductory level. Following our preferred style, it emphasizes a physical\nappreciation of concepts through heuristic reasoning and the use of metaphors, analogies, and\ncreative explanations. Such an approach is much different from a purely deductive technique\nthat uses mere mathematical manipulation of symbols. There is a temptation to treat engineering\nsubjects as a branch of applied mathematics. Such an approach is a perfect match to the public\nimage of engineering as a dry and dull discipline. It ignores the physical meaning behind\nvarious derivations and deprives students of intuitive grasp and the enjoyable experience of\nlogical uncovering of the subject matter. In this book, we use mathematics not so much to\nprove axiomatic theory as to support and enhance physical and intuitive understanding. Wherever\npossible, theoretical results are interpreted heuristically and are enhanced by carefully chosen\nexamples and analogies.", - "type": "text" - }, - { - "block_id": "p17-b2", - "global_id": 125, - "bbox": [ - 127.59, - 347.53, - 516.14, - 417.27 - ], - "text": "This third edition, which closely follows the organization of the second edition, has been\nrefined in many ways. Discussions are streamlined, adding or trimming material as needed.\nEquation, example, and section labeling is simplified and improved. Computer examples are fully\nupdated to reflect the most current version of MATLAB. Hundreds of added problems provide\nnew opportunities to learn and understand topics. We have taken special care to improve the text\nwithout the topic creep and bloat that commonly occurs with each new edition of a text.", - "type": "text" - }, - { - "block_id": "p17-b3", - "global_id": 126, - "bbox": [ - 127.59, - 447.88, - 348.96, - 473.39 - ], - "text": "NOTABLE FEATURES\nThe notable features of this book include the following.", - "type": "text" - }, - { - "block_id": "p17-b4", - "global_id": 127, - "bbox": [ - 144.52, - 481.36, - 516.15, - 634.79 - ], - "text": "1. Intuitive and heuristic understanding of the concepts and physical meaning of\nmathematical results are emphasized throughout. Such an approach not only leads to\ndeeper appreciation and easier comprehension of the concepts, but also makes learning\nenjoyable for students.\n2. Often, students lack an adequate background in basic material such as complex numbers,\nsinusoids, hand-sketching of functions, Cramer’s rule, partial fraction expansion, and\nmatrix algebra. We include a background chapter that addresses these basic and pervasive\ntopics in electrical engineering. Response by students has been unanimously enthusiastic.\n3. There are hundreds of worked examples in addition to drills (usually with answers)\nfor students to test their understanding. Additionally, there are over 900 end-of-chapter\nproblems of varying difficulty.\n4. Modern electrical engineering practice requires the use of computer calculation and\nsimulation, most often using the software package MATLAB. Thus, we integrate", - "type": "text" - }, - { - "block_id": "p17-b5", - "global_id": 128, - "bbox": [ - 505.36, - 656.12, - 516.12, - 666.22 - ], - "text": "xv", - "type": "text" - } - ] - }, - { - "page_num": 18, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p18-b0", - "global_id": 129, - "bbox": [ - 60.0, - 62.89, - 130.68, - 71.98 - ], - "text": "xvi\nPreface", - "type": "text" - }, - { - "block_id": "p18-b1", - "global_id": 130, - "bbox": [ - 118.78, - 85.82, - 490.39, - 203.37 - ], - "text": "MATLAB into many of the worked examples throughout the book. Additionally, each\nchapter concludes with a section devoted to learning and using MATLAB in the context\nand support of book topics. Problem sets also contain numerous computer problems.\n5. The discrete-time and continuous-time systems may be treated in sequence, or they may\nbe integrated by using a parallel approach.\n6. The summary at the end of each chapter proves helpful to students in summing up essential\ndevelopments in the chapter.\n7. There are several historical notes to enhance students’ interest in the subject. This\ninformation introduces students to the historical background that influenced the\ndevelopment of electrical engineering.", - "type": "text" - }, - { - "block_id": "p18-b2", - "global_id": 131, - "bbox": [ - 101.84, - 230.7, - 316.71, - 256.21 - ], - "text": "ORGANIZATION\nThe book may be conceived as divided into five parts:", - "type": "text" - }, - { - "block_id": "p18-b3", - "global_id": 132, - "bbox": [ - 118.78, - 264.18, - 434.54, - 321.97 - ], - "text": "1. Introduction (Chs. B and 1).\n2. Time-domain analysis of linear time-invariant (LTI) systems (Chs. 2 and 3).\n3. Frequency-domain (transform) analysis of LTI systems (Chs. 4 and 5).\n4. Signal analysis (Chs. 6, 7, 8, and 9).\n5. State-space analysis of LTI systems (Ch. 10).", - "type": "text" - }, - { - "block_id": "p18-b4", - "global_id": 133, - "bbox": [ - 101.84, - 329.93, - 490.42, - 387.72 - ], - "text": "The organization of the book permits much flexibility in teaching the continuous-time and\ndiscrete-time concepts. The natural sequence of chapters is meant to integrate continuous-time\nand discrete-time analysis. It is also possible to use a sequential approach in which all the\ncontinuous-time analysis is covered first (Chs. 1, 2, 4, 6, 7, and 8), followed by discrete-time\nanalysis (Chs. 3, 5, and 9).", - "type": "text" - }, - { - "block_id": "p18-b5", - "global_id": 134, - "bbox": [ - 101.84, - 415.04, - 490.41, - 536.19 - ], - "text": "SUGGESTIONS FOR USING THIS BOOK\nThe book can be readily tailored for a variety of courses spanning 30 to 45 lecture hours. Most of\nthe material in the first eight chapters can be covered at a brisk pace in about 45 hours. The book\ncan also be used for a 30-lecture-hour course by covering only analog material (Chs. 1, 2, 4, 6,\n7, and possibly selected topics in Ch. 8). Alternately, one can also select Chs. 1 to 5 for courses\npurely devoted to systems analysis or transform techniques. To treat continuous- and discrete-time\nsystems by using an integrated (or parallel) approach, the appropriate sequence of chapters is 1,\n2, 3, 4, 5, 6, 7, and 8. For a sequential approach, where the continuous-time analysis is followed\nby discrete-time analysis, the proper chapter sequence is 1, 2, 4, 6, 7, 8, 3, 5, and possibly 9\n(depending on the time available).", - "type": "text" - }, - { - "block_id": "p18-b6", - "global_id": 135, - "bbox": [ - 101.84, - 563.51, - 490.41, - 636.84 - ], - "text": "MATLAB\nMATLAB is a sophisticated language that serves as a powerful tool to better understand\nengineering topics, including control theory, filter design, and, of course, linear systems and\nsignals. MATLAB’s flexible programming structure promotes rapid development and analysis.\nOutstanding visualization capabilities provide unique insight into system behavior and signal\ncharacter.", - "type": "text" - } - ] - }, - { - "page_num": 19, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p19-b0", - "global_id": 136, - "bbox": [ - 444.46, - 62.89, - 516.13, - 71.98 - ], - "text": "Preface\nxvii", - "type": "text" - }, - { - "block_id": "p19-b1", - "global_id": 137, - "bbox": [ - 127.59, - 85.82, - 516.13, - 155.56 - ], - "text": "As with any language, learning MATLAB is incremental and requires practice. This\nbook provides two levels of exposure to MATLAB. First, MATLAB is integrated into many\nexamples throughout the text to reinforce concepts and perform various computations. These\nexamples utilize standard MATLAB functions as well as functions from the control system,\nsignal-processing, and symbolic math toolboxes. MATLAB has many more toolboxes available,\nbut these three are commonly available in most engineering departments.", - "type": "text" - }, - { - "block_id": "p19-b2", - "global_id": 138, - "bbox": [ - 127.59, - 157.55, - 516.16, - 263.15 - ], - "text": "A second and deeper level of exposure to MATLAB is achieved by concluding each chapter\nwith a separate MATLAB section. Taken together, these eleven sections provide a self-contained\nintroduction to the MATLAB environment that allows even novice users to quickly gain MATLAB\nproficiency and competence. These sessions provide detailed instruction on how to use MATLAB\nto solve problems in linear systems and signals. Except for the very last chapter, special care has\nbeen taken to avoid the use of toolbox functions in the MATLAB sessions. Rather, readers are\nshown the process of developing their own code. In this way, those readers without toolbox access\nare not at a disadvantage. All of this book’s MATLAB code is available for download at the OUP\ncompanion website www.oup.com/us/lathi.", - "type": "text" - }, - { - "block_id": "p19-b3", - "global_id": 139, - "bbox": [ - 127.59, - 290.44, - 516.18, - 363.78 - ], - "text": "CREDITS AND ACKNOWLEDGMENTS\nThe portraits of Gauss, Laplace, Heaviside, Fourier, and Michelson have been reprinted courtesy\nof the Smithsonian Institution Libraries. The likenesses of Cardano and Gibbs have been reprinted\ncourtesy of the Library of Congress. The engraving of Napoleon has been reprinted courtesy of\nBettmann/Corbis. The many fine cartoons throughout the text are the work of Joseph Coniglio, a\nformer student of Dr. Lathi.", - "type": "text" - }, - { - "block_id": "p19-b4", - "global_id": 140, - "bbox": [ - 127.59, - 365.77, - 516.15, - 411.6 - ], - "text": "Many individuals have helped us in the preparation of this book, as well as its earlier editions.\nWe are grateful to each and every one for helpful suggestions and comments. Book writing is an\nobsessively time-consuming activity, which causes much hardship for an author’s family. We both\nare grateful to our families for their enormous but invisible sacrifices.", - "type": "text" - }, - { - "block_id": "p19-b5", - "global_id": 141, - "bbox": [ - 469.47, - 425.45, - 516.13, - 447.36 - ], - "text": "B. P. Lathi\nR. A. Green", - "type": "text" - } - ] - }, - { - "page_num": 20, - "width": 576.0, - "height": 720.0, - "blocks": [] - }, - { - "page_num": 21, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p21-b0", - "global_id": 142, - "bbox": [ - 90.21, - 67.04, - 159.28, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p21-b1", - "global_id": 143, - "bbox": [ - 172.76, - 151.14, - 323.19, - 173.26 - ], - "text": "BACKGROUND", - "type": "text" - }, - { - "block_id": "p21-b2", - "global_id": 144, - "bbox": [ - 133.55, - 79.92, - 159.47, - 118.77 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p21-b3", - "global_id": 145, - "bbox": [ - 127.59, - 266.14, - 516.15, - 335.88 - ], - "text": "The topics discussed in this chapter are not entirely new to students taking this course. You have\nalready studied many of these topics in earlier courses or are expected to know them from your\nprevious training. Even so, this background material deserves a review because it is so pervasive\nin the area of signals and systems. Investing a little time in such a review will pay big dividends\nlater. Furthermore, this material is useful not only for this course but also for several courses that\nfollow. It will also be helpful later, as reference material in your professional career.", - "type": "text" - }, - { - "block_id": "p21-b4", - "global_id": 146, - "bbox": [ - 127.94, - 368.96, - 286.78, - 382.9 - ], - "text": "B.1 COMPLEX NUMBERS", - "type": "text" - }, - { - "block_id": "p21-b5", - "global_id": 147, - "bbox": [ - 127.59, - 388.79, - 516.15, - 518.41 - ], - "text": "Complex numbers are an extension of ordinary numbers and are an integral part of the modern\nnumber system. Complex numbers, particularly imaginary numbers, sometimes seem mysterious\nand unreal. This feeling of unreality derives from their unfamiliarity and novelty rather than their\nsupposed nonexistence! Mathematicians blundered in calling these numbers “imaginary,” for the\nterm immediately prejudices perception. Had these numbers been called by some other name, they\nwould have become demystified long ago, just as irrational numbers or negative numbers were.\nMany futile attempts have been made to ascribe some physical meaning to imaginary numbers.\nHowever, this effort is needless. In mathematics we assign symbols and operations any meaning\nwe wish as long as internal consistency is maintained. The history of mathematics is full of entities\nthat were unfamiliar and held in abhorrence until familiarity made them acceptable. This fact will\nbecome clear from the following historical note.", - "type": "text" - }, - { - "block_id": "p21-b6", - "global_id": 148, - "bbox": [ - 127.59, - 546.96, - 255.59, - 558.92 - ], - "text": "B.1-1 A Historical Note", - "type": "text" - }, - { - "block_id": "p21-b7", - "global_id": 149, - "bbox": [ - 127.59, - 565.05, - 516.16, - 598.92 - ], - "text": "Among early people the number system consisted only of natural numbers (positive integers)\nneeded to express the number of children, cattle, and quivers of arrows. These people had no need\nfor fractions. Whoever heard of two and one-half children or three and one-fourth cows!", - "type": "text" - }, - { - "block_id": "p21-b8", - "global_id": 150, - "bbox": [ - 127.59, - 600.91, - 516.13, - 634.79 - ], - "text": "However, with the advent of agriculture, people needed to measure continuously varying\nquantities, such as the length of a field and the weight of a quantity of butter. The number system,\ntherefore, was extended to include fractions. The ancient Egyptians and Babylonians knew how", - "type": "text" - }, - { - "block_id": "p21-b9", - "global_id": 151, - "bbox": [ - 511.15, - 656.12, - 516.13, - 666.22 - ], - "text": "1", - "type": "text" - } - ] - }, - { - "page_num": 22, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p22-b0", - "global_id": 152, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "2\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p22-b1", - "global_id": 153, - "bbox": [ - 101.84, - 85.72, - 490.4, - 155.56 - ], - "text": "to handle fractions, but Pythagoras discovered that some numbers (like the diagonal of a unit\nsquare) could not be expressed as a whole number or a fraction. Pythagoras, a number mystic,\nwho regarded numbers as the essence and principle of all things in the universe, was so appalled at\nhis discovery that he swore his followers to secrecy and imposed a death penalty for divulging this\nsecret [1]. These numbers, however, were included in the number system by the time of Descartes,\nand they are now known as irrational numbers.", - "type": "text" - }, - { - "block_id": "p22-b2", - "global_id": 154, - "bbox": [ - 101.84, - 157.45, - 490.41, - 299.02 - ], - "text": "Until recently, negative numbers were not a part of the number system. The concept of\nnegative numbers must have appeared absurd to early man. However, the medieval Hindus had a\nclear understanding of the significance of positive and negative numbers [2, 3]. They were also\nthe first to recognize the existence of absolute negative quantities [4]. The works of Bhaskar\n(1114–1185) on arithmetic (L¯il¯avat¯i) and algebra (B¯ijaganit) not only use the decimal system\nbut also give rules for dealing with negative quantities. Bhaskar recognized that positive numbers\nhave two square roots [5]. Much later, in Europe, the men who developed the banking system\nthat arose in Florence and Venice during the late Renaissance (fifteenth century) are credited with\nintroducing a crude form of negative numbers. The seemingly absurd subtraction of 7 from 5\nseemed reasonable when bankers began to allow their clients to draw seven gold ducats while\ntheir deposit stood at five. All that was necessary for this purpose was to write the difference, 2,\non the debit side of a ledger [6].", - "type": "text" - }, - { - "block_id": "p22-b3", - "global_id": 155, - "bbox": [ - 101.84, - 301.02, - 490.41, - 394.66 - ], - "text": "Thus, the number system was once again broadened (generalized) to include negative\nnumbers. The acceptance of negative numbers made it possible to solve equations such as x+5 = 0,\nwhich had no solution before. Yet for equations such as x2 + 1 = 0, leading to x2 = −1, the\nsolution could not be found in the real number system. It was therefore necessary to define a\ncompletely new kind of number with its square equal to −1. During the time of Descartes and\nNewton, imaginary (or complex) numbers came to be accepted as part of the number system, but\nthey were still regarded as algebraic fiction. The Swiss mathematician Leonhard Euler introduced\nthe notation i (for imaginary) around 1777 to represent", - "type": "text" - }, - { - "block_id": "p22-b4", - "global_id": 156, - "bbox": [ - 320.88, - 376.27, - 329.31, - 386.24 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p22-b5", - "global_id": 157, - "bbox": [ - 101.84, - 384.28, - 490.38, - 406.61 - ], - "text": "−1. Electrical engineers use the notation\nj instead of i to avoid confusion with the notation i often used for electrical current. Thus,", - "type": "text" - }, - { - "block_id": "p22-b6", - "global_id": 158, - "bbox": [ - 231.52, - 417.35, - 297.19, - 431.84 - ], - "text": "j2 = −1\nand", - "type": "text" - }, - { - "block_id": "p22-b7", - "global_id": 159, - "bbox": [ - 317.12, - 412.94, - 325.55, - 422.9 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p22-b8", - "global_id": 160, - "bbox": [ - 325.56, - 421.46, - 360.71, - 431.84 - ], - "text": "−1 = ±j", - "type": "text" - }, - { - "block_id": "p22-b9", - "global_id": 161, - "bbox": [ - 101.84, - 447.1, - 461.36, - 457.07 - ], - "text": "This notation allows us to determine the square root of any negative number. For example,", - "type": "text" - }, - { - "block_id": "p22-b10", - "global_id": 162, - "bbox": [ - 243.15, - 464.46, - 251.58, - 474.43 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p22-b11", - "global_id": 163, - "bbox": [ - 251.58, - 473.0, - 274.15, - 483.37 - ], - "text": "−4 =", - "type": "text" - }, - { - "block_id": "p22-b12", - "global_id": 164, - "bbox": [ - 276.19, - 464.05, - 284.62, - 474.01 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p22-b13", - "global_id": 165, - "bbox": [ - 284.62, - 464.46, - 308.95, - 483.37 - ], - "text": "4 ×\n√", - "type": "text" - }, - { - "block_id": "p22-b14", - "global_id": 166, - "bbox": [ - 308.94, - 473.0, - 349.08, - 483.37 - ], - "text": "−1 = ±2j", - "type": "text" - }, - { - "block_id": "p22-b15", - "global_id": 167, - "bbox": [ - 101.84, - 498.64, - 490.42, - 520.55 - ], - "text": "When imaginary numbers are included in the number system, the resulting numbers are called\ncomplex numbers.", - "type": "text" - }, - { - "block_id": "p22-b16", - "global_id": 168, - "bbox": [ - 102.14, - 536.94, - 277.87, - 549.06 - ], - "text": "ORIGINS OF COMPLEX NUMBERS", - "type": "text" - }, - { - "block_id": "p22-b17", - "global_id": 169, - "bbox": [ - 101.84, - 553.1, - 490.42, - 586.96 - ], - "text": "Ironically (and contrary to popular belief), it was not the solution of a quadratic equation, such\nas x2 + 1 = 0, but a cubic equation with real roots that made imaginary numbers plausible and\nacceptable to early mathematicians. They could dismiss", - "type": "text" - }, - { - "block_id": "p22-b18", - "global_id": 170, - "bbox": [ - 327.61, - 568.58, - 336.04, - 578.54 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p22-b19", - "global_id": 171, - "bbox": [ - 101.84, - 576.59, - 490.4, - 634.79 - ], - "text": "−1 as pure nonsense when it appeared\nas a solution to x2 + 1 = 0 because this equation has no real solution. But in 1545, Gerolamo\nCardano of Milan published Ars Magna (The Great Art), the most important algebraic work of the\nRenaissance. In this book, he gave a method of solving a general cubic equation in which a root\nof a negative number appeared in an intermediate step. According to his method, the solution to a", - "type": "text" - } - ] - }, - { - "page_num": 23, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p23-b0", - "global_id": 172, - "bbox": [ - 376.42, - 62.89, - 516.13, - 71.98 - ], - "text": "B.1\nComplex Numbers\n3", - "type": "text" - }, - { - "block_id": "p23-b1", - "global_id": 173, - "bbox": [ - 127.59, - 86.06, - 210.52, - 96.62 - ], - "text": "third-order equation†", - "type": "text" - }, - { - "block_id": "p23-b2", - "global_id": 174, - "bbox": [ - 291.14, - 94.08, - 352.58, - 108.58 - ], - "text": "x3 + ax + b = 0", - "type": "text" - }, - { - "block_id": "p23-b3", - "global_id": 175, - "bbox": [ - 127.59, - 116.83, - 170.91, - 126.79 - ], - "text": "is given by", - "type": "text" - }, - { - "block_id": "p23-b4", - "global_id": 176, - "bbox": [ - 230.93, - 140.98, - 245.2, - 151.25 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p23-b5", - "global_id": 177, - "bbox": [ - 247.47, - 120.48, - 255.73, - 142.9 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p23-b6", - "global_id": 178, - "bbox": [ - 255.74, - 134.31, - 269.68, - 150.94 - ], - "text": "−b", - "type": "text" - }, - { - "block_id": "p23-b7", - "global_id": 179, - "bbox": [ - 264.7, - 140.98, - 280.19, - 158.42 - ], - "text": "2 +", - "type": "text" - }, - { - "block_id": "p23-b9", - "global_id": 180, - "bbox": [ - 291.2, - 133.79, - 299.67, - 144.27 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p23-b10", - "global_id": 181, - "bbox": [ - 293.19, - 133.79, - 322.39, - 158.42 - ], - "text": "4 + a3", - "type": "text" - }, - { - "block_id": "p23-b11", - "global_id": 182, - "bbox": [ - 313.42, - 120.48, - 343.93, - 158.42 - ], - "text": "27 +\n3", - "type": "text" - }, - { - "block_id": "p23-b12", - "global_id": 183, - "bbox": [ - 343.94, - 134.31, - 357.88, - 150.94 - ], - "text": "−b", - "type": "text" - }, - { - "block_id": "p23-b13", - "global_id": 184, - "bbox": [ - 352.9, - 140.98, - 368.4, - 158.42 - ], - "text": "2 −", - "type": "text" - }, - { - "block_id": "p23-b15", - "global_id": 185, - "bbox": [ - 379.4, - 133.79, - 387.87, - 144.27 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p23-b16", - "global_id": 186, - "bbox": [ - 381.39, - 133.79, - 410.59, - 158.42 - ], - "text": "4 + a3", - "type": "text" - }, - { - "block_id": "p23-b17", - "global_id": 187, - "bbox": [ - 401.63, - 148.46, - 411.59, - 158.42 - ], - "text": "27", - "type": "text" - }, - { - "block_id": "p23-b18", - "global_id": 188, - "bbox": [ - 127.59, - 161.9, - 516.12, - 187.86 - ], - "text": "For example, to find a solution of x3 + 6x −20 = 0, we substitute a = 6,b = −20 in the foregoing\nequation to obtain", - "type": "text" - }, - { - "block_id": "p23-b19", - "global_id": 189, - "bbox": [ - 202.23, - 204.0, - 216.5, - 214.28 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p23-b20", - "global_id": 190, - "bbox": [ - 218.77, - 189.96, - 256.28, - 214.38 - ], - "text": "3\n10 +\n√", - "type": "text" - }, - { - "block_id": "p23-b21", - "global_id": 191, - "bbox": [ - 256.29, - 189.96, - 319.83, - 214.38 - ], - "text": "108 +\n3\n10 −\n√", - "type": "text" - }, - { - "block_id": "p23-b22", - "global_id": 192, - "bbox": [ - 319.85, - 195.17, - 441.49, - 214.38 - ], - "text": "108 =\n3√\n20.392 −\n3√\n0.392 = 2", - "type": "text" - }, - { - "block_id": "p23-b23", - "global_id": 193, - "bbox": [ - 127.59, - 220.88, - 516.14, - 246.83 - ], - "text": "We can readily verify that 2 is indeed a solution of x3 + 6x −20 = 0. But when Cardano tried to\nsolve the equation x3 −15x −4 = 0 by this formula, his solution was", - "type": "text" - }, - { - "block_id": "p23-b24", - "global_id": 194, - "bbox": [ - 252.79, - 262.72, - 267.06, - 273.0 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p23-b25", - "global_id": 195, - "bbox": [ - 269.33, - 248.93, - 301.86, - 273.1 - ], - "text": "3\n2 +\n√", - "type": "text" - }, - { - "block_id": "p23-b26", - "global_id": 196, - "bbox": [ - 301.87, - 262.72, - 333.9, - 273.1 - ], - "text": "−121 +", - "type": "text" - }, - { - "block_id": "p23-b27", - "global_id": 197, - "bbox": [ - 335.67, - 248.93, - 368.2, - 273.1 - ], - "text": "3\n2 −\n√", - "type": "text" - }, - { - "block_id": "p23-b28", - "global_id": 198, - "bbox": [ - 368.22, - 262.72, - 390.92, - 273.1 - ], - "text": "−121", - "type": "text" - }, - { - "block_id": "p23-b29", - "global_id": 199, - "bbox": [ - 127.59, - 283.62, - 516.13, - 317.49 - ], - "text": "What was Cardano to make of this equation in the year 1545? In those days, negative numbers\nwere themselves suspect, and a square root of a negative number was doubly preposterous! Today,\nwe know that", - "type": "text" - }, - { - "block_id": "p23-b30", - "global_id": 200, - "bbox": [ - 257.18, - 314.96, - 353.84, - 329.45 - ], - "text": "(2 ± j)3 = 2 ± j11 = 2 ±", - "type": "text" - }, - { - "block_id": "p23-b31", - "global_id": 201, - "bbox": [ - 355.38, - 310.55, - 363.81, - 320.51 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p23-b32", - "global_id": 202, - "bbox": [ - 363.83, - 319.08, - 386.53, - 329.45 - ], - "text": "−121", - "type": "text" - }, - { - "block_id": "p23-b33", - "global_id": 203, - "bbox": [ - 127.59, - 337.7, - 269.39, - 347.66 - ], - "text": "Therefore, Cardano’s formula gives", - "type": "text" - }, - { - "block_id": "p23-b34", - "global_id": 204, - "bbox": [ - 273.79, - 357.78, - 369.93, - 368.15 - ], - "text": "x = (2 + j) + (2 −j) = 4", - "type": "text" - }, - { - "block_id": "p23-b35", - "global_id": 205, - "bbox": [ - 127.59, - 374.65, - 516.13, - 400.6 - ], - "text": "We can readily verify that x = 4 is indeed a solution of x3 −15x −4 = 0. Cardano tried to\nexplain halfheartedly the presence of", - "type": "text" - }, - { - "block_id": "p23-b36", - "global_id": 206, - "bbox": [ - 282.98, - 382.21, - 291.41, - 392.18 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p23-b37", - "global_id": 207, - "bbox": [ - 127.59, - 390.22, - 516.18, - 520.15 - ], - "text": "−121 but ultimately dismissed the whole enterprise as\nbeing “as subtle as it is useless.” A generation later, however, Raphael Bombelli (1526–1573),\nafter examining Cardano’s results, proposed acceptance of imaginary numbers as a necessary\nvehicle that would transport the mathematician from the real cubic equation to its real solution.\nIn other words, although we begin and end with real numbers, we seem compelled to move into\nan unfamiliar world of imaginaries to complete our journey. To mathematicians of the day, this\nproposal seemed incredibly strange [7]. Yet they could not dismiss the idea of imaginary numbers\nso easily because this concept yielded the real solution of an equation. It took two more centuries\nfor the full importance of complex numbers to become evident in the works of Euler, Gauss, and\nCauchy. Still, Bombelli deserves credit for recognizing that such numbers have a role to play in\nalgebra [7].", - "type": "text" - }, - { - "block_id": "p23-b38", - "global_id": 208, - "bbox": [ - 127.59, - 538.24, - 426.59, - 550.46 - ], - "text": "† This equation is known as the depressed cubic equation. A general cubic equation", - "type": "text" - }, - { - "block_id": "p23-b39", - "global_id": 209, - "bbox": [ - 283.56, - 551.5, - 360.15, - 564.54 - ], - "text": "y3 + py2 + qy + r = 0", - "type": "text" - }, - { - "block_id": "p23-b40", - "global_id": 210, - "bbox": [ - 127.59, - 569.28, - 516.12, - 633.42 - ], - "text": "can always be reduced to a depressed cubic form by substituting y = x −(p/3). Therefore, any general cubic\nequation can be solved if we know the solution to the depressed cubic. The depressed cubic was independently\nsolved, first by Scipione del Ferro (1465–1526) and then by Niccolo Fontana (1499–1557). The latter is better\nknown in the history of mathematics as Tartaglia (“Stammerer”). Cardano learned the secret of the depressed\ncubic solution from Tartaglia. He then showed that by using the substitution y = x −(p/3), a general cubic is\nreduced to a depressed cubic.", - "type": "text" - } - ] - }, - { - "page_num": 24, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p24-b0", - "global_id": 211, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "4\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p24-b1", - "global_id": 212, - "bbox": [ - 101.84, - 85.72, - 490.39, - 179.47 - ], - "text": "In 1799 the German mathematician Karl Friedrich Gauss, at the ripe age of 22, proved the\nfundamental theorem of algebra, namely that every algebraic equation in one unknown has a root\nin the form of a complex number. He showed that every equation of the nth order has exactly n\nsolutions (roots), no more and no less. Gauss was also one of the first to give a coherent account\nof complex numbers and to interpret them as points in a complex plane. It is he who introduced\nthe term complex numbers and paved the way for their general and systematic use. The number\nsystem was once again broadened or generalized to include imaginary numbers. Ordinary (or real)\nnumbers became a special case of generalized (or complex) numbers.", - "type": "text" - }, - { - "block_id": "p24-b2", - "global_id": 213, - "bbox": [ - 101.84, - 181.46, - 490.39, - 203.37 - ], - "text": "The utility of complex numbers can be understood readily by an analogy with two neighboring\ncountries X and Y, as illustrated in Fig. B.1. If we want to travel from City a to City b (both in", - "type": "text" - }, - { - "block_id": "p24-b3", - "global_id": 214, - "bbox": [ - 156.77, - 446.83, - 422.04, - 456.8 - ], - "text": "Gerolamo Cardano\nKarl Friedrich Gauss", - "type": "text" - }, - { - "block_id": "p24-b4", - "global_id": 215, - "bbox": [ - 118.38, - 485.11, - 144.6, - 493.11 - ], - "text": "Country", - "type": "text" - }, - { - "block_id": "p24-b5", - "global_id": 216, - "bbox": [ - 129.04, - 494.03, - 133.93, - 502.03 - ], - "text": "X", - "type": "text" - }, - { - "block_id": "p24-b6", - "global_id": 217, - "bbox": [ - 132.81, - 594.51, - 159.03, - 602.51 - ], - "text": "Country", - "type": "text" - }, - { - "block_id": "p24-b7", - "global_id": 218, - "bbox": [ - 143.7, - 603.43, - 148.14, - 611.43 - ], - "text": "Y", - "type": "text" - }, - { - "block_id": "p24-b8", - "global_id": 219, - "bbox": [ - 149.01, - 515.86, - 153.01, - 523.86 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p24-b9", - "global_id": 220, - "bbox": [ - 211.51, - 603.99, - 215.51, - 611.99 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p24-b10", - "global_id": 221, - "bbox": [ - 207.09, - 503.66, - 213.54, - 512.13 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p24-b11", - "global_id": 222, - "bbox": [ - 213.74, - 504.43, - 217.06, - 512.66 - ], - "text": "l", - "type": "text" - }, - { - "block_id": "p24-b12", - "global_id": 223, - "bbox": [ - 217.57, - 505.2, - 221.67, - 513.5 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p24-b13", - "global_id": 224, - "bbox": [ - 221.35, - 506.43, - 227.32, - 515.15 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p24-b14", - "global_id": 225, - "bbox": [ - 226.3, - 508.85, - 232.31, - 517.19 - ], - "text": "r", - "type": "text" - }, - { - "block_id": "p24-b15", - "global_id": 226, - "bbox": [ - 230.0, - 511.7, - 238.01, - 520.49 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p24-b16", - "global_id": 227, - "bbox": [ - 234.5, - 516.26, - 242.78, - 524.31 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p24-b17", - "global_id": 228, - "bbox": [ - 238.01, - 521.08, - 245.79, - 527.58 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p24-b18", - "global_id": 229, - "bbox": [ - 240.18, - 525.53, - 248.93, - 532.03 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p24-b19", - "global_id": 230, - "bbox": [ - 242.85, - 536.65, - 250.98, - 539.73 - ], - "text": "r", - "type": "text" - }, - { - "block_id": "p24-b20", - "global_id": 231, - "bbox": [ - 242.68, - 541.39, - 250.99, - 546.07 - ], - "text": "o", - "type": "text" - }, - { - "block_id": "p24-b21", - "global_id": 232, - "bbox": [ - 241.53, - 547.24, - 250.18, - 552.81 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p24-b22", - "global_id": 233, - "bbox": [ - 240.27, - 552.99, - 248.56, - 557.58 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p24-b23", - "global_id": 234, - "bbox": [ - 238.41, - 556.76, - 247.15, - 562.95 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p24-b24", - "global_id": 235, - "bbox": [ - 162.04, - 545.22, - 190.24, - 583.75 - ], - "text": "Direct route", - "type": "text" - }, - { - "block_id": "p24-b25", - "global_id": 236, - "bbox": [ - 306.79, - 617.81, - 486.4, - 639.73 - ], - "text": "Figure B.1 Use of complex numbers can\nreduce the work.", - "type": "text" - } - ] - }, - { - "page_num": 25, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p25-b0", - "global_id": 237, - "bbox": [ - 376.42, - 62.89, - 516.13, - 71.98 - ], - "text": "B.1\nComplex Numbers\n5", - "type": "text" - }, - { - "block_id": "p25-b1", - "global_id": 238, - "bbox": [ - 127.59, - 85.72, - 516.16, - 179.47 - ], - "text": "Country X), the shortest route is through Country Y, although the journey begins and ends in\nCountry X. We may, if we desire, perform this journey by an alternate route that lies exclusively\nin X, but this alternate route is longer. In mathematics we have a similar situation with real\nnumbers (Country X) and complex numbers (Country Y). Most real-world problems start with\nreal numbers, and the final results must also be in real numbers. But the derivation of results\nis considerably simplified by using complex numbers as an intermediary. It is also possible to\nsolve any real-world problem by an alternate method, using real numbers exclusively, but such\nprocedures would increase the work needlessly.", - "type": "text" - }, - { - "block_id": "p25-b2", - "global_id": 239, - "bbox": [ - 127.59, - 205.51, - 324.38, - 217.46 - ], - "text": "B.1-2 Algebra of Complex Numbers", - "type": "text" - }, - { - "block_id": "p25-b3", - "global_id": 240, - "bbox": [ - 127.59, - 223.18, - 516.14, - 257.46 - ], - "text": "A complex number (a,b) or a + jb can be represented graphically by a point whose Cartesian\ncoordinates are (a,b) in a complex plane (Fig. B.2). Let us denote this complex number by z so\nthat", - "type": "text" - }, - { - "block_id": "p25-b4", - "global_id": 241, - "bbox": [ - 302.19, - 260.97, - 516.13, - 271.35 - ], - "text": "z = a + jb\n(B.1)", - "type": "text" - }, - { - "block_id": "p25-b5", - "global_id": 242, - "bbox": [ - 127.6, - 281.5, - 516.14, - 315.48 - ], - "text": "This representation is the Cartesian (or rectangular) form of complex number z. The numbers a\nand b (the abscissa and the ordinate) of z are the real part and the imaginary part, respectively, of\nz. They are also expressed as", - "type": "text" - }, - { - "block_id": "p25-b6", - "global_id": 243, - "bbox": [ - 260.9, - 328.3, - 382.82, - 338.68 - ], - "text": "Re z = a\nand\nIm z = b", - "type": "text" - }, - { - "block_id": "p25-b7", - "global_id": 244, - "bbox": [ - 127.6, - 351.92, - 516.11, - 373.84 - ], - "text": "Note that in this plane all real numbers lie on the horizontal axis, and all imaginary numbers lie on\nthe vertical axis.", - "type": "text" - }, - { - "block_id": "p25-b8", - "global_id": 245, - "bbox": [ - 249.98, - 451.71, - 266.65, - 459.71 - ], - "text": "Real", - "type": "text" - }, - { - "block_id": "p25-b9", - "global_id": 246, - "bbox": [ - 134.1, - 402.45, - 166.98, - 410.45 - ], - "text": "Imaginary", - "type": "text" - }, - { - "block_id": "p25-b10", - "global_id": 247, - "bbox": [ - 195.18, - 425.54, - 198.29, - 433.54 - ], - "text": "r", - "type": "text" - }, - { - "block_id": "p25-b11", - "global_id": 248, - "bbox": [ - 196.18, - 438.91, - 200.18, - 446.91 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p25-b12", - "global_id": 249, - "bbox": [ - 170.2, - 412.53, - 174.2, - 420.53 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p25-b13", - "global_id": 250, - "bbox": [ - 163.53, - 477.42, - 174.2, - 485.63 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p25-b14", - "global_id": 251, - "bbox": [ - 232.63, - 411.54, - 235.75, - 419.54 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p25-b15", - "global_id": 252, - "bbox": [ - 232.65, - 476.54, - 240.76, - 484.54 - ], - "text": "z*", - "type": "text" - }, - { - "block_id": "p25-b16", - "global_id": 253, - "bbox": [ - 230.58, - 448.04, - 234.58, - 456.04 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p25-b17", - "global_id": 254, - "bbox": [ - 286.53, - 465.24, - 505.55, - 487.15 - ], - "text": "Figure B.2 Representation of a number in the complex\nplane.", - "type": "text" - }, - { - "block_id": "p25-b18", - "global_id": 255, - "bbox": [ - 127.59, - 518.18, - 516.12, - 540.51 - ], - "text": "Complex numbers may also be expressed in terms of polar coordinates. If (r,θ) are the polar\ncoordinates of a point z = a + jb (see Fig. B.2), then", - "type": "text" - }, - { - "block_id": "p25-b19", - "global_id": 256, - "bbox": [ - 252.64, - 553.34, - 390.09, - 563.71 - ], - "text": "a = rcos θ\nand\nb = rsin θ", - "type": "text" - }, - { - "block_id": "p25-b20", - "global_id": 257, - "bbox": [ - 127.59, - 576.95, - 184.21, - 586.92 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p25-b21", - "global_id": 258, - "bbox": [ - 225.05, - 590.43, - 516.13, - 600.8 - ], - "text": "z = a + jb = rcos θ + jrsin θ = r(cos θ + jsin θ)\n(B.2)", - "type": "text" - }, - { - "block_id": "p25-b22", - "global_id": 259, - "bbox": [ - 145.52, - 610.95, - 250.55, - 621.02 - ], - "text": "Euler’s formula states that", - "type": "text" - }, - { - "block_id": "p25-b23", - "global_id": 260, - "bbox": [ - 283.19, - 620.42, - 516.13, - 634.91 - ], - "text": "ejθ = cos θ + jsin θ\n(B.3)", - "type": "text" - } - ] - }, - { - "page_num": 26, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p26-b0", - "global_id": 261, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "6\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p26-b1", - "global_id": 262, - "bbox": [ - 101.84, - 85.66, - 433.82, - 97.26 - ], - "text": "To prove Euler’s formula, we use a Maclaurin series to expand ejθ, cos θ, and sin θ:", - "type": "text" - }, - { - "block_id": "p26-b2", - "global_id": 263, - "bbox": [ - 183.95, - 107.11, - 261.59, - 125.4 - ], - "text": "ejθ = 1 + jθ + (jθ)2", - "type": "text" - }, - { - "block_id": "p26-b3", - "global_id": 264, - "bbox": [ - 248.43, - 107.11, - 294.42, - 132.47 - ], - "text": "2!\n+ (jθ)3", - "type": "text" - }, - { - "block_id": "p26-b4", - "global_id": 265, - "bbox": [ - 281.26, - 107.11, - 327.24, - 132.47 - ], - "text": "3!\n+ (jθ)4", - "type": "text" - }, - { - "block_id": "p26-b5", - "global_id": 266, - "bbox": [ - 314.08, - 107.11, - 360.08, - 132.47 - ], - "text": "4!\n+ (jθ)5", - "type": "text" - }, - { - "block_id": "p26-b6", - "global_id": 267, - "bbox": [ - 346.91, - 107.11, - 392.9, - 132.47 - ], - "text": "5!\n+ (jθ)6", - "type": "text" - }, - { - "block_id": "p26-b7", - "global_id": 268, - "bbox": [ - 379.74, - 115.02, - 418.52, - 132.47 - ], - "text": "6!\n+ · · ·", - "type": "text" - }, - { - "block_id": "p26-b8", - "global_id": 269, - "bbox": [ - 196.63, - 133.62, - 251.38, - 151.92 - ], - "text": "= 1 + jθ −θ2", - "type": "text" - }, - { - "block_id": "p26-b9", - "global_id": 270, - "bbox": [ - 243.32, - 133.62, - 276.78, - 158.99 - ], - "text": "2! −jθ3", - "type": "text" - }, - { - "block_id": "p26-b10", - "global_id": 271, - "bbox": [ - 268.72, - 133.62, - 299.41, - 158.99 - ], - "text": "3! + θ4", - "type": "text" - }, - { - "block_id": "p26-b11", - "global_id": 272, - "bbox": [ - 291.35, - 133.62, - 324.8, - 158.99 - ], - "text": "4! + jθ5", - "type": "text" - }, - { - "block_id": "p26-b12", - "global_id": 273, - "bbox": [ - 316.75, - 133.62, - 347.43, - 158.99 - ], - "text": "5! −θ6", - "type": "text" - }, - { - "block_id": "p26-b13", - "global_id": 274, - "bbox": [ - 339.37, - 141.54, - 373.06, - 158.99 - ], - "text": "6! −· · ·", - "type": "text" - }, - { - "block_id": "p26-b14", - "global_id": 275, - "bbox": [ - 173.71, - 160.14, - 232.37, - 178.43 - ], - "text": "cos θ = 1 −θ2", - "type": "text" - }, - { - "block_id": "p26-b15", - "global_id": 276, - "bbox": [ - 224.31, - 160.14, - 254.99, - 185.51 - ], - "text": "2! + θ4", - "type": "text" - }, - { - "block_id": "p26-b16", - "global_id": 277, - "bbox": [ - 246.93, - 160.14, - 277.62, - 185.51 - ], - "text": "4! −θ6", - "type": "text" - }, - { - "block_id": "p26-b17", - "global_id": 278, - "bbox": [ - 269.56, - 160.14, - 300.24, - 185.51 - ], - "text": "6! + θ8", - "type": "text" - }, - { - "block_id": "p26-b18", - "global_id": 279, - "bbox": [ - 292.18, - 168.06, - 316.1, - 185.51 - ], - "text": "8! · · ·", - "type": "text" - }, - { - "block_id": "p26-b19", - "global_id": 280, - "bbox": [ - 175.37, - 186.66, - 232.76, - 204.96 - ], - "text": "sin θ = θ −θ3", - "type": "text" - }, - { - "block_id": "p26-b20", - "global_id": 281, - "bbox": [ - 224.7, - 186.66, - 255.39, - 212.03 - ], - "text": "3! + θ5", - "type": "text" - }, - { - "block_id": "p26-b21", - "global_id": 282, - "bbox": [ - 247.33, - 186.66, - 278.01, - 212.03 - ], - "text": "5! −θ7", - "type": "text" - }, - { - "block_id": "p26-b22", - "global_id": 283, - "bbox": [ - 269.95, - 194.58, - 303.65, - 212.03 - ], - "text": "7! + · · ·", - "type": "text" - }, - { - "block_id": "p26-b23", - "global_id": 284, - "bbox": [ - 101.85, - 217.04, - 414.32, - 231.04 - ], - "text": "Clearly, it follows that ejθ = cos θ + jsin θ. Using Eq. (B.3) in Eq. (B.2) yields", - "type": "text" - }, - { - "block_id": "p26-b24", - "global_id": 285, - "bbox": [ - 281.18, - 240.53, - 490.39, - 252.63 - ], - "text": "z = rejθ\n(B.4)", - "type": "text" - }, - { - "block_id": "p26-b25", - "global_id": 286, - "bbox": [ - 101.85, - 264.16, - 336.34, - 274.22 - ], - "text": "This representation is the polar form of complex number z.", - "type": "text" - }, - { - "block_id": "p26-b26", - "global_id": 287, - "bbox": [ - 101.85, - 275.8, - 490.4, - 298.13 - ], - "text": "Summarizing, a complex number can be expressed in rectangular form a + jb or polar form\nrejθ with", - "type": "text" - }, - { - "block_id": "p26-b27", - "global_id": 288, - "bbox": [ - 215.62, - 297.91, - 257.55, - 308.29 - ], - "text": "a = rcos θ", - "type": "text" - }, - { - "block_id": "p26-b28", - "global_id": 289, - "bbox": [ - 216.45, - 298.24, - 337.01, - 320.24 - ], - "text": "b = rsin θ\nand\nr =", - "type": "text" - }, - { - "block_id": "p26-b29", - "global_id": 290, - "bbox": [ - 339.06, - 289.81, - 347.48, - 299.77 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p26-b30", - "global_id": 291, - "bbox": [ - 347.48, - 295.36, - 375.78, - 308.52 - ], - "text": "a2 + b2", - "type": "text" - }, - { - "block_id": "p26-b31", - "global_id": 292, - "bbox": [ - 322.75, - 302.49, - 371.41, - 320.88 - ], - "text": "θ = tan−1 b", - "type": "text" - }, - { - "block_id": "p26-b32", - "global_id": 293, - "bbox": [ - 367.92, - 316.65, - 371.41, - 323.62 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p26-b33", - "global_id": 294, - "bbox": [ - 372.6, - 302.49, - 490.39, - 314.36 - ], - "text": "(B.5)", - "type": "text" - }, - { - "block_id": "p26-b34", - "global_id": 295, - "bbox": [ - 101.84, - 329.03, - 490.4, - 363.0 - ], - "text": "Observe that r is the distance of the point z from the origin. For this reason, r is also called the\nmagnitude (or absolute value) of z and is denoted by |z|. Similarly, θ is called the angle of z and is\ndenoted by̸\nz. Therefore, we can also write polar form of Eq. (B.4) as", - "type": "text" - }, - { - "block_id": "p26-b35", - "global_id": 296, - "bbox": [ - 216.65, - 374.35, - 374.59, - 386.45 - ], - "text": "z = |z|ej̸\nz\nwhere |z| = r and̸\nz = θ", - "type": "text" - }, - { - "block_id": "p26-b36", - "global_id": 297, - "bbox": [ - 101.84, - 398.08, - 409.8, - 408.04 - ], - "text": "Using polar form, we see that the reciprocal of a complex number is given by", - "type": "text" - }, - { - "block_id": "p26-b37", - "global_id": 298, - "bbox": [ - 239.54, - 417.65, - 268.34, - 441.57 - ], - "text": "1\nz = 1", - "type": "text" - }, - { - "block_id": "p26-b38", - "global_id": 299, - "bbox": [ - 258.78, - 417.65, - 292.15, - 441.57 - ], - "text": "rejθ = 1", - "type": "text" - }, - { - "block_id": "p26-b39", - "global_id": 300, - "bbox": [ - 287.61, - 417.65, - 329.76, - 441.57 - ], - "text": "r e−jθ = 1", - "type": "text" - }, - { - "block_id": "p26-b40", - "global_id": 301, - "bbox": [ - 322.47, - 422.5, - 353.39, - 441.57 - ], - "text": "|z|e−j̸\nz", - "type": "text" - }, - { - "block_id": "p26-b41", - "global_id": 302, - "bbox": [ - 102.14, - 453.64, - 302.26, - 465.76 - ], - "text": "CONJUGATE OF A COMPLEX NUMBER", - "type": "text" - }, - { - "block_id": "p26-b42", - "global_id": 303, - "bbox": [ - 101.84, - 468.15, - 277.41, - 479.75 - ], - "text": "We define z∗, the conjugate of z = a + jb, as", - "type": "text" - }, - { - "block_id": "p26-b43", - "global_id": 304, - "bbox": [ - 237.62, - 488.59, - 490.39, - 503.08 - ], - "text": "z∗= a −jb = re−jθ = |z|e−j̸\nz\n(B.6)", - "type": "text" - }, - { - "block_id": "p26-b44", - "global_id": 305, - "bbox": [ - 101.84, - 513.08, - 490.39, - 548.59 - ], - "text": "The graphical representations of a number z and its conjugate z∗are depicted in Fig. B.2. Observe\nthat z∗is a mirror image of z about the horizontal axis. To find the conjugate of any number, we\nneed only replace j with −j in that number (which is the same as changing the sign of its angle).", - "type": "text" - }, - { - "block_id": "p26-b45", - "global_id": 306, - "bbox": [ - 101.84, - 550.58, - 490.41, - 572.5 - ], - "text": "The sum of a complex number and its conjugate is a real number equal to twice the real part\nof the number:", - "type": "text" - }, - { - "block_id": "p26-b46", - "global_id": 307, - "bbox": [ - 214.42, - 572.35, - 377.82, - 584.45 - ], - "text": "z + z∗= (a + jb) + (a −jb) = 2a = 2Rez", - "type": "text" - }, - { - "block_id": "p26-b47", - "global_id": 308, - "bbox": [ - 101.85, - 593.15, - 388.62, - 603.22 - ], - "text": "Thus, we see that the real part of complex number z can be computed as", - "type": "text" - }, - { - "block_id": "p26-b48", - "global_id": 309, - "bbox": [ - 269.59, - 611.29, - 320.96, - 629.86 - ], - "text": "Rez = z + z∗", - "type": "text" - }, - { - "block_id": "p26-b49", - "global_id": 310, - "bbox": [ - 307.58, - 619.9, - 490.39, - 636.94 - ], - "text": "2\n(B.7)", - "type": "text" - } - ] - }, - { - "page_num": 27, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p27-b0", - "global_id": 311, - "bbox": [ - 376.42, - 62.89, - 516.13, - 71.98 - ], - "text": "B.1\nComplex Numbers\n7", - "type": "text" - }, - { - "block_id": "p27-b1", - "global_id": 312, - "bbox": [ - 127.59, - 85.72, - 409.58, - 95.78 - ], - "text": "Similarly, the imaginary part of complex number z can be computed as", - "type": "text" - }, - { - "block_id": "p27-b2", - "global_id": 313, - "bbox": [ - 295.33, - 104.19, - 346.7, - 122.78 - ], - "text": "Imz = z −z∗", - "type": "text" - }, - { - "block_id": "p27-b3", - "global_id": 314, - "bbox": [ - 331.95, - 112.82, - 516.13, - 129.85 - ], - "text": "2j\n(B.8)", - "type": "text" - }, - { - "block_id": "p27-b4", - "global_id": 315, - "bbox": [ - 127.59, - 139.92, - 516.13, - 163.19 - ], - "text": "The product of a complex number z and its conjugate is a real number |z|2, the square of the\nmagnitude of the number:", - "type": "text" - }, - { - "block_id": "p27-b5", - "global_id": 316, - "bbox": [ - 269.78, - 162.54, - 516.13, - 177.03 - ], - "text": "zz∗= |z|ej̸\nz|z|e−j̸\nz = |z|2\n(B.9)", - "type": "text" - }, - { - "block_id": "p27-b6", - "global_id": 317, - "bbox": [ - 127.89, - 191.77, - 365.84, - 203.89 - ], - "text": "UNDERSTANDING SOME USEFUL IDENTITIES", - "type": "text" - }, - { - "block_id": "p27-b7", - "global_id": 318, - "bbox": [ - 127.59, - 204.31, - 516.14, - 253.76 - ], - "text": "In a complex plane, rejθ represents a point at a distance r from the origin and at an angle θ with\nthe horizontal axis, as shown in Fig. B.3a. For example, the number −1 is at a unit distance from\nthe origin and has an angle π or −π (more generally, π plus any integer multiple of 2π), as seen\nfrom Fig. B.3b. Therefore,", - "type": "text" - }, - { - "block_id": "p27-b8", - "global_id": 319, - "bbox": [ - 264.97, - 253.64, - 378.74, - 265.74 - ], - "text": "−1 = ej(π+2πn)\nn integer", - "type": "text" - }, - { - "block_id": "p27-b9", - "global_id": 320, - "bbox": [ - 127.59, - 274.73, - 516.13, - 296.65 - ], - "text": "The number 1, on the other hand, is also at a unit distance from the origin, but has an angle 0 (more\ngenerally, 0 plus any integer multiple of 2π). Therefore,", - "type": "text" - }, - { - "block_id": "p27-b10", - "global_id": 321, - "bbox": [ - 276.62, - 306.48, - 516.12, - 318.58 - ], - "text": "1 = ej2πn\nn integer\n(B.10)", - "type": "text" - }, - { - "block_id": "p27-b11", - "global_id": 322, - "bbox": [ - 130.35, - 328.52, - 399.62, - 340.52 - ], - "text": "The number j is at a unit distance from the origin and its angle is π", - "type": "text" - }, - { - "block_id": "p27-b12", - "global_id": 323, - "bbox": [ - 127.59, - 328.52, - 516.12, - 352.48 - ], - "text": "2 (more generally, π\n2 plus any\ninteger multiple of 2π), as seen from Fig. B.3b. Therefore,", - "type": "text" - }, - { - "block_id": "p27-b13", - "global_id": 324, - "bbox": [ - 269.12, - 360.3, - 297.5, - 374.32 - ], - "text": "j = ej( π", - "type": "text" - }, - { - "block_id": "p27-b14", - "global_id": 325, - "bbox": [ - 294.51, - 362.32, - 374.59, - 374.42 - ], - "text": "2 +2πn)\nn integer", - "type": "text" - }, - { - "block_id": "p27-b15", - "global_id": 326, - "bbox": [ - 127.59, - 386.39, - 166.52, - 396.35 - ], - "text": "Similarly,", - "type": "text" - }, - { - "block_id": "p27-b16", - "global_id": 327, - "bbox": [ - 262.51, - 394.22, - 304.11, - 408.24 - ], - "text": "−j = ej(−π", - "type": "text" - }, - { - "block_id": "p27-b17", - "global_id": 328, - "bbox": [ - 301.11, - 396.24, - 381.19, - 408.34 - ], - "text": "2 +2πn)\nn integer", - "type": "text" - }, - { - "block_id": "p27-b18", - "global_id": 329, - "bbox": [ - 127.59, - 416.92, - 499.4, - 427.29 - ], - "text": "Notice that the angle of any complex number is only known within an integer multiple of 2π.", - "type": "text" - }, - { - "block_id": "p27-b19", - "global_id": 330, - "bbox": [ - 127.59, - 427.64, - 516.13, - 463.16 - ], - "text": "This discussion shows the usefulness of the graphic picture of rejθ. This picture is also helpful\nin several other applications. For example, to determine the limit of e(α+jω)t as t →∞, we note\nthat", - "type": "text" - }, - { - "block_id": "p27-b20", - "global_id": 331, - "bbox": [ - 289.22, - 460.65, - 353.87, - 475.05 - ], - "text": "e(α+jω)t = eαtejωt", - "type": "text" - }, - { - "block_id": "p27-b21", - "global_id": 332, - "bbox": [ - 223.61, - 603.97, - 232.49, - 611.97 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p27-b22", - "global_id": 333, - "bbox": [ - 286.04, - 546.1, - 296.92, - 554.1 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p27-b23", - "global_id": 334, - "bbox": [ - 232.2, - 520.21, - 235.31, - 528.21 - ], - "text": "r", - "type": "text" - }, - { - "block_id": "p27-b24", - "global_id": 335, - "bbox": [ - 234.2, - 533.58, - 238.2, - 541.58 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p27-b25", - "global_id": 336, - "bbox": [ - 201.27, - 498.29, - 269.95, - 508.01 - ], - "text": "Im\nre ju", - "type": "text" - }, - { - "block_id": "p27-b26", - "global_id": 337, - "bbox": [ - 393.26, - 603.97, - 402.9, - 611.97 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p27-b27", - "global_id": 338, - "bbox": [ - 455.0, - 546.28, - 465.89, - 554.28 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p27-b28", - "global_id": 339, - "bbox": [ - 383.1, - 494.91, - 405.09, - 506.29 - ], - "text": "Im\nj", - "type": "text" - }, - { - "block_id": "p27-b29", - "global_id": 340, - "bbox": [ - 376.14, - 584.29, - 385.03, - 592.5 - ], - "text": "j", - "type": "text" - }, - { - "block_id": "p27-b30", - "global_id": 341, - "bbox": [ - 335.89, - 546.09, - 436.0, - 554.49 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p27-b31", - "global_id": 342, - "bbox": [ - 377.33, - 529.31, - 382.67, - 537.31 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p27-b32", - "global_id": 343, - "bbox": [ - 368.94, - 546.39, - 380.94, - 554.58 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p27-b33", - "global_id": 344, - "bbox": [ - 402.77, - 526.87, - 415.46, - 535.16 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p27-b34", - "global_id": 345, - "bbox": [ - 399.3, - 550.39, - 418.5, - 558.68 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p27-b35", - "global_id": 346, - "bbox": [ - 151.5, - 618.33, - 390.33, - 629.19 - ], - "text": "Figure B.3 Understanding some useful identities in terms of rejθ.", - "type": "text" - } - ] - }, - { - "page_num": 28, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p28-b0", - "global_id": 347, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "8\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p28-b1", - "global_id": 348, - "bbox": [ - 101.84, - 83.27, - 490.38, - 109.22 - ], - "text": "Now the magnitude of ejωt is unity regardless of the value of ω or t because ejωt = rejθ with r = 1.\nTherefore, eαt determines the behavior of e(α+jω)t as t →∞and", - "type": "text" - }, - { - "block_id": "p28-b2", - "global_id": 349, - "bbox": [ - 209.39, - 120.86, - 282.83, - 140.54 - ], - "text": "lim\nt→∞e(α+jω)t = lim", - "type": "text" - }, - { - "block_id": "p28-b3", - "global_id": 350, - "bbox": [ - 267.75, - 120.86, - 320.4, - 140.54 - ], - "text": "t→∞eαtejωt =", - "type": "text" - }, - { - "block_id": "p28-b4", - "global_id": 351, - "bbox": [ - 322.45, - 110.99, - 381.64, - 141.23 - ], - "text": "0\nα < 0\n∞\nα > 0", - "type": "text" - }, - { - "block_id": "p28-b5", - "global_id": 352, - "bbox": [ - 101.84, - 149.75, - 490.4, - 175.29 - ], - "text": "In future discussions, you will find it very useful to remember rejθ as a number at a distance r from\nthe origin and at an angle θ with the horizontal axis of the complex plane.", - "type": "text" - }, - { - "block_id": "p28-b6", - "global_id": 353, - "bbox": [ - 102.14, - 189.85, - 445.08, - 201.97 - ], - "text": "A WARNING ABOUT COMPUTING ANGLES WITH CALCULATORS", - "type": "text" - }, - { - "block_id": "p28-b7", - "global_id": 354, - "bbox": [ - 101.84, - 202.38, - 490.39, - 287.69 - ], - "text": "From the Cartesian form a + jb, we can readily compute the polar form rejθ [see Eq. (B.5)].\nCalculators provide ready conversion of rectangular into polar and vice versa. However, if a\ncalculator computes an angle of a complex number by using an inverse tangent function θ =\ntan−1(b/a), proper attention must be paid to the quadrant in which the number is located. For\ninstance, θ corresponding to the number −2 −j3 is tan−1(−3/−2). This result is not the same\nas tan−1(3/2). The former is −123.7◦, whereas the latter is 56.3◦. A calculator cannot make\nthis distinction and can give a correct answer only for angles in the first and fourth quadrants.†", - "type": "text" - }, - { - "block_id": "p28-b8", - "global_id": 355, - "bbox": [ - 101.84, - 288.04, - 490.41, - 359.42 - ], - "text": "A calculator will read tan−1(−3/−2) as tan−1(3/2), which is clearly wrong. When you are\ncomputing inverse trigonometric functions, if the angle appears in the second or third quadrant,\nthe answer of the calculator is off by 180◦. The correct answer is obtained by adding or subtracting\n180◦to the value found with the calculator (either adding or subtracting yields the correct answer).\nFor this reason, it is advisable to draw the point in the complex plane and determine the quadrant\nin which the point lies. This issue will be clarified by the following examples.", - "type": "text" - }, - { - "block_id": "p28-b9", - "global_id": 356, - "bbox": [ - 76.77, - 388.06, - 305.49, - 400.01 - ], - "text": "EXAMPLE B.1\nCartesian to Polar Form", - "type": "text" - }, - { - "block_id": "p28-b10", - "global_id": 357, - "bbox": [ - 103.16, - 413.35, - 477.01, - 423.72 - ], - "text": "Express the following numbers in polar form: (a) 2+j3, (b) −2+j1, (c) −2−j3, and (d) 1−j3.", - "type": "text" - }, - { - "block_id": "p28-b11", - "global_id": 358, - "bbox": [ - 121.09, - 446.56, - 132.71, - 456.52 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p28-b12", - "global_id": 359, - "bbox": [ - 190.11, - 459.7, - 209.5, - 469.97 - ], - "text": "|z| =", - "type": "text" - }, - { - "block_id": "p28-b14", - "global_id": 360, - "bbox": [ - 219.81, - 450.81, - 268.9, - 470.07 - ], - "text": "22 + 32 =\n√", - "type": "text" - }, - { - "block_id": "p28-b15", - "global_id": 361, - "bbox": [ - 268.91, - 451.69, - 351.58, - 470.07 - ], - "text": "13̸\nz = tan−1 3", - "type": "text" - }, - { - "block_id": "p28-b16", - "global_id": 362, - "bbox": [ - 348.09, - 451.69, - 356.79, - 472.89 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p28-b17", - "global_id": 363, - "bbox": [ - 358.84, - 457.98, - 389.56, - 470.07 - ], - "text": "= 56.3◦", - "type": "text" - }, - { - "block_id": "p28-b18", - "global_id": 364, - "bbox": [ - 103.16, - 479.04, - 477.03, - 500.96 - ], - "text": "In this case the number is in the first quadrant, and a calculator will give the correct value of\n56.3◦. Therefore (see Fig. B.4a), we can write", - "type": "text" - }, - { - "block_id": "p28-b19", - "global_id": 365, - "bbox": [ - 251.33, - 504.64, - 295.23, - 523.9 - ], - "text": "2 + j3 =\n√", - "type": "text" - }, - { - "block_id": "p28-b20", - "global_id": 366, - "bbox": [ - 295.22, - 510.07, - 327.83, - 523.9 - ], - "text": "13ej56.3◦", - "type": "text" - }, - { - "block_id": "p28-b21", - "global_id": 367, - "bbox": [ - 121.09, - 535.77, - 133.26, - 545.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p28-b22", - "global_id": 368, - "bbox": [ - 179.79, - 548.47, - 199.18, - 558.75 - ], - "text": "|z| =", - "type": "text" - }, - { - "block_id": "p28-b24", - "global_id": 369, - "bbox": [ - 209.49, - 545.59, - 263.31, - 558.85 - ], - "text": "(−2)2 + 12 =", - "type": "text" - }, - { - "block_id": "p28-b25", - "global_id": 370, - "bbox": [ - 265.36, - 539.58, - 273.79, - 549.55 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p28-b26", - "global_id": 371, - "bbox": [ - 273.79, - 540.46, - 354.2, - 558.85 - ], - "text": "5̸\nz = tan−1 1", - "type": "text" - }, - { - "block_id": "p28-b27", - "global_id": 372, - "bbox": [ - 347.99, - 554.4, - 356.91, - 561.67 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p28-b29", - "global_id": 373, - "bbox": [ - 364.17, - 547.25, - 399.88, - 558.85 - ], - "text": "= 153.4◦", - "type": "text" - }, - { - "block_id": "p28-b30", - "global_id": 374, - "bbox": [ - 103.16, - 567.82, - 477.02, - 589.73 - ], - "text": "In this case the angle is in the second quadrant (see Fig. B.4b), and therefore the answer\ngiven by the calculator, tan−1(1/−2) = −26.6◦, is off by 180◦. The correct answer is", - "type": "text" - }, - { - "block_id": "p28-b31", - "global_id": 375, - "bbox": [ - 101.84, - 621.19, - 421.34, - 633.41 - ], - "text": "† Calculators with two-argument inverse tangent functions will correctly compute angles.", - "type": "text" - } - ] - }, - { - "page_num": 29, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p29-b0", - "global_id": 376, - "bbox": [ - 376.42, - 62.89, - 516.13, - 71.98 - ], - "text": "B.1\nComplex Numbers\n9", - "type": "text" - }, - { - "block_id": "p29-b1", - "global_id": 377, - "bbox": [ - 128.9, - 84.66, - 502.77, - 120.17 - ], - "text": "(−26.6±180)◦= 153.4◦or −206.6◦. Both values are correct because they represent the same\nangle. It is a common practice to choose an angle whose numerical value is less than 180◦.\nSuch a value is called the principal value of the angle, which in this case is 153.4◦. Therefore,", - "type": "text" - }, - { - "block_id": "p29-b2", - "global_id": 378, - "bbox": [ - 274.49, - 132.74, - 315.69, - 143.12 - ], - "text": "−2 + j1 =", - "type": "text" - }, - { - "block_id": "p29-b3", - "global_id": 379, - "bbox": [ - 317.73, - 123.85, - 326.16, - 133.81 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p29-b4", - "global_id": 380, - "bbox": [ - 326.16, - 129.28, - 356.16, - 143.12 - ], - "text": "5ej153.4◦", - "type": "text" - }, - { - "block_id": "p29-b5", - "global_id": 381, - "bbox": [ - 146.84, - 154.99, - 157.9, - 164.95 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p29-b6", - "global_id": 382, - "bbox": [ - 191.55, - 167.68, - 210.94, - 177.96 - ], - "text": "|z| =", - "type": "text" - }, - { - "block_id": "p29-b8", - "global_id": 383, - "bbox": [ - 221.25, - 164.8, - 290.28, - 178.06 - ], - "text": "(−2)2 + (−3)2 =", - "type": "text" - }, - { - "block_id": "p29-b9", - "global_id": 384, - "bbox": [ - 292.33, - 158.79, - 300.75, - 168.75 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p29-b10", - "global_id": 385, - "bbox": [ - 300.76, - 159.67, - 388.86, - 178.06 - ], - "text": "13̸\nz = tan−1 −3", - "type": "text" - }, - { - "block_id": "p29-b11", - "global_id": 386, - "bbox": [ - 379.94, - 173.61, - 388.86, - 180.87 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p29-b13", - "global_id": 387, - "bbox": [ - 396.12, - 166.45, - 439.6, - 178.06 - ], - "text": "= −123.7◦", - "type": "text" - }, - { - "block_id": "p29-b14", - "global_id": 388, - "bbox": [ - 128.9, - 187.02, - 502.76, - 232.85 - ], - "text": "In this case the angle appears in the third quadrant (see Fig. B.4c), and therefore the answer\nobtained by the calculator (tan−1(−3/−2) = 56.3◦) is off by 180◦. The correct answer is\n(56.3 ± 180)◦= 236.3◦or −123.7◦. We choose the principal value −123.7◦so that (see\nFig. B.4c)", - "type": "text" - }, - { - "block_id": "p29-b15", - "global_id": 389, - "bbox": [ - 269.28, - 235.46, - 310.47, - 245.84 - ], - "text": "−2 −j3 =", - "type": "text" - }, - { - "block_id": "p29-b16", - "global_id": 390, - "bbox": [ - 312.52, - 226.57, - 320.95, - 236.53 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p29-b17", - "global_id": 391, - "bbox": [ - 320.95, - 232.0, - 361.37, - 245.84 - ], - "text": "13e−j123.7◦", - "type": "text" - }, - { - "block_id": "p29-b18", - "global_id": 392, - "bbox": [ - 146.84, - 254.72, - 159.02, - 264.68 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p29-b19", - "global_id": 393, - "bbox": [ - 201.64, - 267.41, - 221.04, - 277.69 - ], - "text": "|z| =", - "type": "text" - }, - { - "block_id": "p29-b21", - "global_id": 394, - "bbox": [ - 231.34, - 258.52, - 295.64, - 277.79 - ], - "text": "12 + (−3)2 =\n√", - "type": "text" - }, - { - "block_id": "p29-b22", - "global_id": 395, - "bbox": [ - 295.65, - 259.4, - 383.75, - 277.79 - ], - "text": "10̸\nz = tan−1 −3", - "type": "text" - }, - { - "block_id": "p29-b23", - "global_id": 396, - "bbox": [ - 377.55, - 259.4, - 388.96, - 280.6 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p29-b24", - "global_id": 397, - "bbox": [ - 391.01, - 266.18, - 429.51, - 277.79 - ], - "text": "= −71.6◦", - "type": "text" - }, - { - "block_id": "p29-b25", - "global_id": 398, - "bbox": [ - 128.9, - 286.75, - 502.74, - 308.67 - ], - "text": "In this case the angle appears in the fourth quadrant (see Fig. B.4d), and therefore the answer\ngiven by the calculator, tan−1(−3/1) = −71.6◦, is correct (see Fig. B.4d):", - "type": "text" - }, - { - "block_id": "p29-b26", - "global_id": 399, - "bbox": [ - 274.91, - 312.35, - 318.81, - 331.62 - ], - "text": "1 −j3 =\n√", - "type": "text" - }, - { - "block_id": "p29-b27", - "global_id": 400, - "bbox": [ - 318.8, - 317.79, - 355.74, - 331.62 - ], - "text": "10e−j71.6◦", - "type": "text" - }, - { - "block_id": "p29-b28", - "global_id": 401, - "bbox": [ - 207.86, - 451.25, - 218.75, - 459.25 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p29-b29", - "global_id": 402, - "bbox": [ - 121.19, - 369.39, - 130.07, - 377.39 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p29-b30", - "global_id": 403, - "bbox": [ - 175.96, - 389.04, - 197.33, - 397.33 - ], - "text": "2 j3", - "type": "text" - }, - { - "block_id": "p29-b31", - "global_id": 404, - "bbox": [ - 169.56, - 450.4, - 173.56, - 458.4 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p29-b32", - "global_id": 405, - "bbox": [ - 129.06, - 389.33, - 133.06, - 397.33 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p29-b33", - "global_id": 406, - "bbox": [ - 148.09, - 436.52, - 164.75, - 444.81 - ], - "text": "56.3", - "type": "text" - }, - { - "block_id": "p29-b34", - "global_id": 407, - "bbox": [ - 144.65, - 412.68, - 152.65, - 420.68 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p29-b35", - "global_id": 408, - "bbox": [ - 166.05, - 471.54, - 315.58, - 479.54 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p29-b36", - "global_id": 409, - "bbox": [ - 166.05, - 589.29, - 315.58, - 597.29 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p29-b37", - "global_id": 410, - "bbox": [ - 347.8, - 451.25, - 358.69, - 459.25 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p29-b38", - "global_id": 411, - "bbox": [ - 310.14, - 369.39, - 319.02, - 377.39 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p29-b39", - "global_id": 412, - "bbox": [ - 327.27, - 419.93, - 331.27, - 427.93 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p29-b40", - "global_id": 413, - "bbox": [ - 305.63, - 428.58, - 353.48, - 443.34 - ], - "text": "153.4\n5", - "type": "text" - }, - { - "block_id": "p29-b41", - "global_id": 414, - "bbox": [ - 269.53, - 450.11, - 280.2, - 458.4 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p29-b42", - "global_id": 415, - "bbox": [ - 242.68, - 419.64, - 270.72, - 427.93 - ], - "text": "2 j1", - "type": "text" - }, - { - "block_id": "p29-b43", - "global_id": 416, - "bbox": [ - 121.19, - 502.11, - 218.75, - 511.04 - ], - "text": "Re \nIm", - "type": "text" - }, - { - "block_id": "p29-b44", - "global_id": 417, - "bbox": [ - 195.56, - 517.81, - 222.89, - 526.11 - ], - "text": "123.7", - "type": "text" - }, - { - "block_id": "p29-b45", - "global_id": 418, - "bbox": [ - 142.63, - 503.71, - 153.29, - 512.01 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p29-b46", - "global_id": 419, - "bbox": [ - 117.23, - 564.57, - 197.12, - 576.86 - ], - "text": "3\n2 j3", - "type": "text" - }, - { - "block_id": "p29-b47", - "global_id": 420, - "bbox": [ - 159.45, - 529.31, - 167.45, - 537.31 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p29-b48", - "global_id": 421, - "bbox": [ - 275.77, - 502.61, - 358.69, - 511.32 - ], - "text": "Re \nIm\n1", - "type": "text" - }, - { - "block_id": "p29-b49", - "global_id": 422, - "bbox": [ - 275.3, - 564.57, - 285.97, - 572.86 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p29-b50", - "global_id": 423, - "bbox": [ - 299.02, - 517.11, - 322.35, - 525.41 - ], - "text": "71.6", - "type": "text" - }, - { - "block_id": "p29-b51", - "global_id": 424, - "bbox": [ - 293.56, - 558.38, - 301.56, - 566.38 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p29-b52", - "global_id": 425, - "bbox": [ - 378.52, - 576.63, - 497.55, - 598.55 - ], - "text": "Figure B.4 From Cartesian to\npolar form.", - "type": "text" - } - ] - }, - { - "page_num": 30, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p30-b0", - "global_id": 426, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "10\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p30-b1", - "global_id": 427, - "bbox": [ - 103.16, - 86.24, - 477.01, - 108.44 - ], - "text": "We can easily verify these results using the MATLAB abs and angle commands. To\nobtain units of degrees, we must multiply the radian result of the angle command by", - "type": "text" - }, - { - "block_id": "p30-b2", - "global_id": 428, - "bbox": [ - 103.16, - 108.39, - 477.02, - 135.23 - ], - "text": "180\nπ . Furthermore, the angle command correctly computes angles for all four quadrants of\nthe complex plane. To provide an example, let us use MATLAB to verify that −2 + j1 =\n√", - "type": "text" - }, - { - "block_id": "p30-b3", - "global_id": 429, - "bbox": [ - 111.59, - 130.68, - 211.95, - 144.02 - ], - "text": "5ej153.4◦= 2.2361ej153.4◦.", - "type": "text" - }, - { - "block_id": "p30-b4", - "global_id": 430, - "bbox": [ - 103.16, - 154.27, - 223.46, - 200.1 - ], - "text": ">>\nabs(-2+1j)\nans = 2.2361\n>>\nangle(-2+1j)*180/pi\nans = 153.4349", - "type": "text" - }, - { - "block_id": "p30-b5", - "global_id": 431, - "bbox": [ - 103.16, - 209.77, - 477.04, - 243.65 - ], - "text": "One can also use the cart2pol command to convert Cartesian to polar coordinates. Readers,\nparticularly those who are unfamiliar with MATLAB, will benefit by reading the overview in\nSec. B.7.", - "type": "text" - }, - { - "block_id": "p30-b6", - "global_id": 432, - "bbox": [ - 76.77, - 299.77, - 305.49, - 311.72 - ], - "text": "EXAMPLE B.2\nPolar to Cartesian Form", - "type": "text" - }, - { - "block_id": "p30-b7", - "global_id": 433, - "bbox": [ - 103.16, - 325.47, - 477.0, - 347.38 - ], - "text": "Represent the following numbers in the complex plane and express them in Cartesian form:\n(a) 2ejπ/3, (b) 4e−j3π/4, (c) 2ejπ/2, (d) 3e−j3π, (e) 2ej4π, and (f) 2e−j4π.", - "type": "text" - }, - { - "block_id": "p30-b8", - "global_id": 434, - "bbox": [ - 103.16, - 366.27, - 271.53, - 380.26 - ], - "text": "(a) 2ejπ/3 = 2(cos π/3 + jsin π/3) = 1 + j", - "type": "text" - }, - { - "block_id": "p30-b9", - "global_id": 435, - "bbox": [ - 271.53, - 361.51, - 279.96, - 371.47 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p30-b10", - "global_id": 436, - "bbox": [ - 103.17, - 370.3, - 363.52, - 392.21 - ], - "text": "3\n(see Fig. B.5a)\n(b) 4e−j3π/4 = 4(cos 3π/4 −jsin 3π/4) = −2", - "type": "text" - }, - { - "block_id": "p30-b11", - "global_id": 437, - "bbox": [ - 286.11, - 373.41, - 294.54, - 383.38 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p30-b12", - "global_id": 438, - "bbox": [ - 294.54, - 373.41, - 326.57, - 392.21 - ], - "text": "2 −j2\n√", - "type": "text" - }, - { - "block_id": "p30-b13", - "global_id": 439, - "bbox": [ - 103.16, - 382.25, - 410.69, - 440.04 - ], - "text": "2\n(see Fig. B.5b)\n(c) 2ejπ/2 = 2(cos π/2 + jsin π/2) = 2(0 + j1) = j2\n(see Fig. B.5c)\n(d) 3e−j3π = 3(cos 3π −jsin 3π) = 3(−1 + j0) = −3\n(see Fig. B.5d)\n(e) 2ej4π = 2(cos 4π + jsin 4π) = 2(1 + j0) = 2\n(see Fig. B.5e)\n(f) 2e−j4π = 2(cos 4π −jsin 4π) = 2(1 −j0) = 2\n(see Fig. B.5f)", - "type": "text" - }, - { - "block_id": "p30-b14", - "global_id": 440, - "bbox": [ - 103.17, - 448.0, - 477.03, - 493.83 - ], - "text": "We can readily verify these results using MATLAB. First, we use the exp function to\nrepresent a number in polar form. Next, we use the real and imag commands to determine\nthe real and imaginary components of that number. To provide an example, let us use MATLAB\nto verify the result of part (a): 2ejπ/3 = 1 + j", - "type": "text" - }, - { - "block_id": "p30-b15", - "global_id": 441, - "bbox": [ - 277.54, - 475.09, - 285.97, - 485.05 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p30-b16", - "global_id": 442, - "bbox": [ - 285.97, - 483.46, - 351.3, - 493.83 - ], - "text": "3 = 1 + j1.7321.", - "type": "text" - }, - { - "block_id": "p30-b17", - "global_id": 443, - "bbox": [ - 103.16, - 504.08, - 228.69, - 549.91 - ], - "text": ">>\nreal(2*exp(1j*pi/3))\nans = 1.0000\n>>\nimag(2*exp(1j*pi/3))\nans = 1.7321", - "type": "text" - }, - { - "block_id": "p30-b18", - "global_id": 444, - "bbox": [ - 103.16, - 559.59, - 477.03, - 569.55 - ], - "text": "Since MATLAB defaults to Cartesian form, we could have verified the entire result in one step.", - "type": "text" - }, - { - "block_id": "p30-b19", - "global_id": 445, - "bbox": [ - 103.16, - 579.8, - 239.14, - 601.72 - ], - "text": ">>\n2*exp(1j*pi/3)\nans = 1.0000 + 1.7321i", - "type": "text" - }, - { - "block_id": "p30-b20", - "global_id": 446, - "bbox": [ - 103.16, - 611.39, - 437.46, - 621.64 - ], - "text": "One can also use the pol2cart command to convert polar to Cartesian coordinates.", - "type": "text" - } - ] - }, - { - "page_num": 31, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p31-b0", - "global_id": 447, - "bbox": [ - 376.42, - 62.89, - 516.13, - 71.98 - ], - "text": "B.1\nComplex Numbers\n11", - "type": "text" - }, - { - "block_id": "p31-b1", - "global_id": 448, - "bbox": [ - 171.0, - 202.75, - 320.44, - 210.75 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p31-b2", - "global_id": 449, - "bbox": [ - 213.27, - 306.62, - 222.16, - 314.62 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p31-b3", - "global_id": 450, - "bbox": [ - 158.44, - 233.46, - 167.33, - 241.46 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p31-b4", - "global_id": 451, - "bbox": [ - 176.44, - 243.77, - 213.83, - 253.46 - ], - "text": "2e jp2 j2", - "type": "text" - }, - { - "block_id": "p31-b5", - "global_id": 452, - "bbox": [ - 171.0, - 325.41, - 179.88, - 333.41 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p31-b6", - "global_id": 453, - "bbox": [ - 355.94, - 258.96, - 364.83, - 266.96 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p31-b7", - "global_id": 454, - "bbox": [ - 313.11, - 230.46, - 322.0, - 238.46 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p31-b8", - "global_id": 455, - "bbox": [ - 310.95, - 325.41, - 320.28, - 333.41 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p31-b9", - "global_id": 456, - "bbox": [ - 253.11, - 273.76, - 355.11, - 287.46 - ], - "text": "3ej3p 3\n3p", - "type": "text" - }, - { - "block_id": "p31-b10", - "global_id": 457, - "bbox": [ - 356.94, - 382.59, - 365.83, - 390.59 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p31-b11", - "global_id": 458, - "bbox": [ - 285.11, - 356.09, - 294.0, - 364.09 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p31-b12", - "global_id": 459, - "bbox": [ - 311.3, - 432.17, - 319.93, - 440.17 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p31-b13", - "global_id": 460, - "bbox": [ - 314.86, - 365.42, - 330.86, - 373.71 - ], - "text": "4p", - "type": "text" - }, - { - "block_id": "p31-b14", - "global_id": 461, - "bbox": [ - 212.27, - 382.59, - 221.16, - 390.59 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p31-b15", - "global_id": 462, - "bbox": [ - 167.25, - 356.09, - 176.14, - 364.09 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p31-b16", - "global_id": 463, - "bbox": [ - 171.0, - 432.17, - 179.88, - 440.17 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p31-b17", - "global_id": 464, - "bbox": [ - 329.04, - 396.18, - 364.43, - 405.87 - ], - "text": "2ej4p 2", - "type": "text" - }, - { - "block_id": "p31-b18", - "global_id": 465, - "bbox": [ - 173.44, - 371.99, - 182.78, - 380.09 - ], - "text": "4p", - "type": "text" - }, - { - "block_id": "p31-b19", - "global_id": 466, - "bbox": [ - 192.44, - 396.32, - 224.89, - 405.87 - ], - "text": "2e j4p 2", - "type": "text" - }, - { - "block_id": "p31-b20", - "global_id": 467, - "bbox": [ - 213.27, - 182.0, - 222.16, - 190.0 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p31-b21", - "global_id": 468, - "bbox": [ - 126.44, - 95.84, - 135.33, - 103.84 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p31-b22", - "global_id": 469, - "bbox": [ - 176.44, - 117.87, - 196.66, - 127.56 - ], - "text": "2e jp3", - "type": "text" - }, - { - "block_id": "p31-b23", - "global_id": 470, - "bbox": [ - 169.44, - 182.84, - 173.44, - 190.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p31-b24", - "global_id": 471, - "bbox": [ - 130.29, - 120.22, - 134.29, - 128.22 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p31-b25", - "global_id": 472, - "bbox": [ - 149.44, - 141.84, - 153.44, - 149.84 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p31-b26", - "global_id": 473, - "bbox": [ - 153.59, - 160.35, - 158.92, - 168.35 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p31-b27", - "global_id": 474, - "bbox": [ - 154.25, - 169.22, - 158.25, - 177.22 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p31-b28", - "global_id": 475, - "bbox": [ - 183.4, - 283.37, - 188.74, - 291.37 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p31-b29", - "global_id": 476, - "bbox": [ - 184.07, - 292.28, - 188.07, - 300.28 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p31-b30", - "global_id": 477, - "bbox": [ - 330.51, - 156.84, - 351.51, - 165.14 - ], - "text": "2 2", - "type": "text" - }, - { - "block_id": "p31-b31", - "global_id": 478, - "bbox": [ - 313.11, - 93.84, - 364.83, - 103.34 - ], - "text": "Re\nIm", - "type": "text" - }, - { - "block_id": "p31-b33", - "global_id": 479, - "bbox": [ - 259.11, - 96.05, - 280.11, - 104.34 - ], - "text": "2 2", - "type": "text" - }, - { - "block_id": "p31-b34", - "global_id": 480, - "bbox": [ - 258.91, - 165.15, - 283.13, - 174.84 - ], - "text": "4ejp3", - "type": "text" - }, - { - "block_id": "p31-b35", - "global_id": 481, - "bbox": [ - 291.11, - 127.84, - 295.11, - 135.84 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p31-b36", - "global_id": 482, - "bbox": [ - 341.52, - 109.45, - 350.86, - 126.28 - ], - "text": "3p\n4", - "type": "text" - }, - { - "block_id": "p31-b37", - "global_id": 483, - "bbox": [ - 119.94, - 446.86, - 274.36, - 456.1 - ], - "text": "Figure B.5 From polar to Cartesian form.", - "type": "text" - }, - { - "block_id": "p31-b38", - "global_id": 484, - "bbox": [ - 127.89, - 499.65, - 339.07, - 525.73 - ], - "text": "ARITHMETICAL OPERATIONS, POWERS,\nAND ROOTS OF COMPLEX NUMBERS", - "type": "text" - }, - { - "block_id": "p31-b39", - "global_id": 485, - "bbox": [ - 127.59, - 529.76, - 516.12, - 551.67 - ], - "text": "To conveniently perform addition and subtraction, complex numbers should be expressed in\nCartesian form. Thus, if", - "type": "text" - }, - { - "block_id": "p31-b40", - "global_id": 486, - "bbox": [ - 280.49, - 551.99, - 362.21, - 566.9 - ], - "text": "z1 = 3 + j4 = 5ej53.1◦", - "type": "text" - }, - { - "block_id": "p31-b41", - "global_id": 487, - "bbox": [ - 127.59, - 576.25, - 141.97, - 586.22 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p31-b42", - "global_id": 488, - "bbox": [ - 273.79, - 589.99, - 326.94, - 601.44 - ], - "text": "z2 = 2 + j3 =", - "type": "text" - }, - { - "block_id": "p31-b43", - "global_id": 489, - "bbox": [ - 328.99, - 581.1, - 337.41, - 591.06 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p31-b44", - "global_id": 490, - "bbox": [ - 337.42, - 586.53, - 368.92, - 600.36 - ], - "text": "13ej56.3◦", - "type": "text" - }, - { - "block_id": "p31-b45", - "global_id": 491, - "bbox": [ - 127.59, - 610.79, - 144.74, - 620.76 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p31-b46", - "global_id": 492, - "bbox": [ - 248.43, - 624.53, - 395.29, - 635.99 - ], - "text": "z1 + z2 = (3 + j4) + (2 + j3) = 5 + j7", - "type": "text" - } - ] - }, - { - "page_num": 32, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p32-b0", - "global_id": 493, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "12\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p32-b1", - "global_id": 494, - "bbox": [ - 101.84, - 85.72, - 490.4, - 133.14 - ], - "text": "If z1 and z2 are given in polar form, we would need to convert them into Cartesian form for the\npurpose of adding (or subtracting). Multiplication and division, however, can be carried out in\neither Cartesian or polar form, although the latter proves to be much more convenient. This is\nbecause if z1 and z2 are expressed in polar form as", - "type": "text" - }, - { - "block_id": "p32-b2", - "global_id": 495, - "bbox": [ - 228.0, - 142.85, - 363.23, - 156.02 - ], - "text": "z1 = r1ejθ1\nand\nz2 = r2ejθ2", - "type": "text" - }, - { - "block_id": "p32-b3", - "global_id": 496, - "bbox": [ - 101.84, - 168.29, - 118.99, - 178.25 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p32-b4", - "global_id": 497, - "bbox": [ - 224.51, - 180.18, - 367.22, - 193.37 - ], - "text": "z1z2 = (r1ejθ1)(r2ejθ2) = r1r2ej(θ1+θ2)", - "type": "text" - }, - { - "block_id": "p32-b5", - "global_id": 498, - "bbox": [ - 101.84, - 202.64, - 116.22, - 212.6 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p32-b6", - "global_id": 499, - "bbox": [ - 248.06, - 213.97, - 255.42, - 238.86 - ], - "text": "z1\nz2", - "type": "text" - }, - { - "block_id": "p32-b7", - "global_id": 500, - "bbox": [ - 259.15, - 212.42, - 290.45, - 230.91 - ], - "text": "= r1ejθ1", - "type": "text" - }, - { - "block_id": "p32-b8", - "global_id": 501, - "bbox": [ - 270.17, - 213.97, - 313.07, - 238.86 - ], - "text": "r2ejθ2 = r1", - "type": "text" - }, - { - "block_id": "p32-b9", - "global_id": 502, - "bbox": [ - 305.7, - 228.02, - 313.07, - 238.86 - ], - "text": "r2", - "type": "text" - }, - { - "block_id": "p32-b10", - "global_id": 503, - "bbox": [ - 314.77, - 218.91, - 344.88, - 230.91 - ], - "text": "ej(θ1−θ2)", - "type": "text" - }, - { - "block_id": "p32-b11", - "global_id": 504, - "bbox": [ - 101.84, - 245.48, - 142.9, - 255.44 - ], - "text": "Moreover,", - "type": "text" - }, - { - "block_id": "p32-b12", - "global_id": 505, - "bbox": [ - 256.45, - 254.98, - 334.58, - 269.37 - ], - "text": "zn = (rejθ)n = rnejnθ", - "type": "text" - }, - { - "block_id": "p32-b13", - "global_id": 506, - "bbox": [ - 101.84, - 279.82, - 116.22, - 289.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p32-b14", - "global_id": 507, - "bbox": [ - 244.7, - 289.32, - 490.38, - 303.81 - ], - "text": "z1/n = (rejθ)1/n = r1/nejθ/n\n(B.11)", - "type": "text" - }, - { - "block_id": "p32-b15", - "global_id": 508, - "bbox": [ - 101.85, - 314.16, - 490.35, - 336.09 - ], - "text": "This shows that the operations of multiplication, division, powers, and roots can be carried out\nwith remarkable ease when the numbers are in polar form.", - "type": "text" - }, - { - "block_id": "p32-b16", - "global_id": 509, - "bbox": [ - 101.85, - 334.46, - 490.39, - 359.99 - ], - "text": "Strictly speaking, there are n values for z1/n (the nth root of z). To find all the n roots, we\nreexamine Eq. (B.11):", - "type": "text" - }, - { - "block_id": "p32-b17", - "global_id": 510, - "bbox": [ - 137.23, - 370.53, - 205.06, - 384.92 - ], - "text": "z1/n = [rejθ]1/n =", - "type": "text" - }, - { - "block_id": "p32-b19", - "global_id": 511, - "bbox": [ - 211.03, - 366.63, - 490.38, - 385.02 - ], - "text": "rej(θ+2πk)\n1/n = r1/nej(θ+2πk)/n\nk = 0,1,2,. . .,n −1\n(B.12)", - "type": "text" - }, - { - "block_id": "p32-b20", - "global_id": 512, - "bbox": [ - 101.84, - 394.74, - 490.38, - 420.27 - ], - "text": "The value of z1/n given in Eq. (B.11) is the principal value of z1/n, obtained by taking the nth root\nof the principal value of z, which corresponds to the case k = 0 in Eq. (B.12).", - "type": "text" - }, - { - "block_id": "p32-b21", - "global_id": 513, - "bbox": [ - 76.77, - 449.96, - 451.55, - 461.92 - ], - "text": "EXAMPLE B.3\nMultiplication and Division of Complex Numbers", - "type": "text" - }, - { - "block_id": "p32-b22", - "global_id": 514, - "bbox": [ - 103.16, - 478.17, - 421.22, - 490.04 - ], - "text": "Using both polar and Cartesian forms, determine z1z2 and z1/z2 for the numbers", - "type": "text" - }, - { - "block_id": "p32-b23", - "global_id": 515, - "bbox": [ - 173.54, - 497.66, - 363.65, - 512.57 - ], - "text": "z1 = 3 + j4 = 5ej53.1◦\nand\nz2 = 2 + j3 =", - "type": "text" - }, - { - "block_id": "p32-b24", - "global_id": 516, - "bbox": [ - 365.7, - 492.22, - 374.13, - 502.18 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p32-b25", - "global_id": 517, - "bbox": [ - 374.13, - 497.66, - 405.63, - 511.49 - ], - "text": "13ej56.3◦", - "type": "text" - }, - { - "block_id": "p32-b26", - "global_id": 518, - "bbox": [ - 103.16, - 544.28, - 237.38, - 554.25 - ], - "text": "Multiplication: Cartesian Form", - "type": "text" - }, - { - "block_id": "p32-b27", - "global_id": 519, - "bbox": [ - 177.64, - 565.87, - 402.52, - 577.32 - ], - "text": "z1z2 = (3 + j4)(2 + j3) = (6 −12) + j(8 + 9) = −6 + j17", - "type": "text" - }, - { - "block_id": "p32-b28", - "global_id": 520, - "bbox": [ - 103.17, - 588.12, - 218.93, - 598.09 - ], - "text": "Multiplication: Polar Form", - "type": "text" - }, - { - "block_id": "p32-b29", - "global_id": 521, - "bbox": [ - 203.69, - 606.24, - 266.24, - 621.16 - ], - "text": "z1z2 = (5ej53.1◦)", - "type": "text" - }, - { - "block_id": "p32-b30", - "global_id": 522, - "bbox": [ - 266.24, - 600.82, - 278.69, - 611.66 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p32-b31", - "global_id": 523, - "bbox": [ - 278.68, - 601.7, - 332.07, - 620.08 - ], - "text": "13ej56.3◦\n= 5", - "type": "text" - }, - { - "block_id": "p32-b32", - "global_id": 524, - "bbox": [ - 332.06, - 600.82, - 340.49, - 610.78 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p32-b33", - "global_id": 525, - "bbox": [ - 340.49, - 606.24, - 375.47, - 620.08 - ], - "text": "13ej109.4◦", - "type": "text" - } - ] - }, - { - "page_num": 33, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p33-b0", - "global_id": 526, - "bbox": [ - 376.42, - 62.89, - 516.13, - 71.98 - ], - "text": "B.1\nComplex Numbers\n13", - "type": "text" - }, - { - "block_id": "p33-b1", - "global_id": 527, - "bbox": [ - 128.9, - 86.28, - 300.34, - 119.06 - ], - "text": "Division: Cartesian Form\nz1\nz2", - "type": "text" - }, - { - "block_id": "p33-b2", - "global_id": 528, - "bbox": [ - 304.08, - 93.85, - 338.68, - 111.21 - ], - "text": "= 3 + j4", - "type": "text" - }, - { - "block_id": "p33-b3", - "global_id": 529, - "bbox": [ - 315.09, - 107.91, - 338.68, - 118.29 - ], - "text": "2 + j3", - "type": "text" - }, - { - "block_id": "p33-b4", - "global_id": 530, - "bbox": [ - 128.91, - 125.12, - 502.76, - 147.03 - ], - "text": "To eliminate the complex number in the denominator, we multiply both the numerator and the\ndenominator of the right-hand side by 2 −j3, the denominator’s conjugate. This yields", - "type": "text" - }, - { - "block_id": "p33-b5", - "global_id": 531, - "bbox": [ - 205.68, - 156.93, - 213.05, - 181.83 - ], - "text": "z1\nz2", - "type": "text" - }, - { - "block_id": "p33-b6", - "global_id": 532, - "bbox": [ - 216.79, - 156.62, - 289.84, - 173.57 - ], - "text": "= (3 + j4)(2 −j3)", - "type": "text" - }, - { - "block_id": "p33-b7", - "global_id": 533, - "bbox": [ - 227.8, - 156.62, - 332.8, - 181.05 - ], - "text": "(2 + j3)(2 −j3) = 18 −j1", - "type": "text" - }, - { - "block_id": "p33-b8", - "global_id": 534, - "bbox": [ - 304.13, - 156.62, - 399.99, - 181.05 - ], - "text": "22 + 32 = 18 −j1\n13\n= 18", - "type": "text" - }, - { - "block_id": "p33-b9", - "global_id": 535, - "bbox": [ - 390.03, - 157.03, - 425.98, - 181.05 - ], - "text": "13 −j 1\n13", - "type": "text" - }, - { - "block_id": "p33-b10", - "global_id": 536, - "bbox": [ - 128.91, - 190.92, - 218.55, - 200.88 - ], - "text": "Division: Polar Form", - "type": "text" - }, - { - "block_id": "p33-b11", - "global_id": 537, - "bbox": [ - 214.96, - 208.77, - 222.33, - 233.67 - ], - "text": "z1\nz2", - "type": "text" - }, - { - "block_id": "p33-b12", - "global_id": 538, - "bbox": [ - 226.07, - 205.5, - 270.3, - 225.45 - ], - "text": "=\n5ej53.1◦\n√", - "type": "text" - }, - { - "block_id": "p33-b13", - "global_id": 539, - "bbox": [ - 245.51, - 208.87, - 303.97, - 234.24 - ], - "text": "13ej56.3◦=\n5\n√", - "type": "text" - }, - { - "block_id": "p33-b14", - "global_id": 540, - "bbox": [ - 300.71, - 208.87, - 385.5, - 234.24 - ], - "text": "13\nej(53.1◦−56.3◦) =\n5\n√", - "type": "text" - }, - { - "block_id": "p33-b15", - "global_id": 541, - "bbox": [ - 382.24, - 211.98, - 416.89, - 234.24 - ], - "text": "13\ne−j3.2◦", - "type": "text" - }, - { - "block_id": "p33-b16", - "global_id": 542, - "bbox": [ - 128.9, - 243.15, - 502.77, - 265.06 - ], - "text": "It is clear from this example that multiplication and division are easier to accomplish in polar\nform than in Cartesian form.", - "type": "text" - }, - { - "block_id": "p33-b17", - "global_id": 543, - "bbox": [ - 128.9, - 267.05, - 502.76, - 290.05 - ], - "text": "These results are also easily verified using MATLAB. To provide one example, let us use\nCartesian forms in MATLAB to verify that z1z2 = −6 + j17.", - "type": "text" - }, - { - "block_id": "p33-b18", - "global_id": 544, - "bbox": [ - 128.9, - 297.23, - 275.35, - 331.1 - ], - "text": ">>\nz1 = 3+4j; z2 = 2+3j;\n>>\nz1*z2\nans = -6.0000 + 17.0000i", - "type": "text" - }, - { - "block_id": "p33-b19", - "global_id": 545, - "bbox": [ - 128.9, - 337.41, - 502.76, - 386.61 - ], - "text": "As a second example, let us use polar forms in MATLAB to verify that z1/z2 = 1.3868e−j3.2◦.\nSince MATLAB generally expects angles be represented in the natural units of radians, we\nmust use appropriate conversion factors in moving between degrees and radians (and vice\nversa).", - "type": "text" - }, - { - "block_id": "p33-b20", - "global_id": 546, - "bbox": [ - 128.91, - 394.86, - 474.11, - 452.65 - ], - "text": ">>\nz1 = 5*exp(1j*53.1*pi/180); z2 = sqrt(13)*exp(1j*56.3*pi/180);\n>>\nabs(z1/z2)\nans = 1.3868\n>>\nangle(z1/z2)*180/pi\nans = -3.2000", - "type": "text" - }, - { - "block_id": "p33-b21", - "global_id": 547, - "bbox": [ - 102.51, - 491.24, - 384.79, - 503.19 - ], - "text": "EXAMPLE B.4\nWorking with Complex Numbers", - "type": "text" - }, - { - "block_id": "p33-b22", - "global_id": 548, - "bbox": [ - 128.9, - 516.24, - 447.64, - 530.9 - ], - "text": "For z1 = 2ejπ/4 and z2 = 8ejπ/3, find the following: (a) 2z1 −z2, (b) 1/z1, (c) z1/z2", - "type": "text" - }, - { - "block_id": "p33-b23", - "global_id": 549, - "bbox": [ - 444.16, - 512.98, - 502.76, - 532.02 - ], - "text": "2, and (d)\n3√z2.", - "type": "text" - }, - { - "block_id": "p33-b24", - "global_id": 550, - "bbox": [ - 128.91, - 552.64, - 502.76, - 574.65 - ], - "text": "(a) Since subtraction cannot be performed directly in polar form, we convert z1 and z2 to\nCartesian form:", - "type": "text" - }, - { - "block_id": "p33-b25", - "global_id": 551, - "bbox": [ - 221.13, - 585.67, - 280.85, - 601.24 - ], - "text": "z1 = 2ejπ/4 = 2", - "type": "text" - }, - { - "block_id": "p33-b27", - "global_id": 552, - "bbox": [ - 288.67, - 582.8, - 311.35, - 600.16 - ], - "text": "cos π", - "type": "text" - }, - { - "block_id": "p33-b28", - "global_id": 553, - "bbox": [ - 306.37, - 582.8, - 349.3, - 607.23 - ], - "text": "4 + jsin π\n4", - "type": "text" - }, - { - "block_id": "p33-b30", - "global_id": 554, - "bbox": [ - 360.26, - 589.78, - 368.03, - 599.74 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p33-b31", - "global_id": 555, - "bbox": [ - 370.08, - 580.83, - 378.51, - 590.8 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p33-b32", - "global_id": 556, - "bbox": [ - 378.51, - 580.83, - 405.55, - 600.16 - ], - "text": "2 + j\n√", - "type": "text" - }, - { - "block_id": "p33-b33", - "global_id": 557, - "bbox": [ - 405.56, - 590.19, - 410.54, - 600.16 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p33-b34", - "global_id": 558, - "bbox": [ - 221.13, - 613.56, - 280.85, - 629.13 - ], - "text": "z2 = 8ejπ/3 = 8", - "type": "text" - }, - { - "block_id": "p33-b36", - "global_id": 559, - "bbox": [ - 288.67, - 610.69, - 311.35, - 628.05 - ], - "text": "cos π", - "type": "text" - }, - { - "block_id": "p33-b37", - "global_id": 560, - "bbox": [ - 306.37, - 610.69, - 349.3, - 635.12 - ], - "text": "3 + jsin π\n3", - "type": "text" - }, - { - "block_id": "p33-b39", - "global_id": 561, - "bbox": [ - 360.26, - 617.67, - 393.67, - 628.05 - ], - "text": "= 4 + j4", - "type": "text" - }, - { - "block_id": "p33-b40", - "global_id": 562, - "bbox": [ - 393.67, - 608.78, - 402.09, - 618.74 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p33-b41", - "global_id": 563, - "bbox": [ - 402.11, - 618.09, - 407.09, - 628.05 - ], - "text": "3", - "type": "text" - } - ] - }, - { - "page_num": 34, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p34-b0", - "global_id": 564, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "14\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p34-b1", - "global_id": 565, - "bbox": [ - 103.16, - 86.24, - 144.91, - 96.21 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p34-b2", - "global_id": 566, - "bbox": [ - 120.36, - 98.8, - 181.23, - 119.2 - ], - "text": "2z1 −z2 = 2\n√", - "type": "text" - }, - { - "block_id": "p34-b3", - "global_id": 567, - "bbox": [ - 181.23, - 98.8, - 208.28, - 118.12 - ], - "text": "2 + j\n√", - "type": "text" - }, - { - "block_id": "p34-b4", - "global_id": 568, - "bbox": [ - 208.28, - 99.73, - 217.28, - 118.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p34-b5", - "global_id": 569, - "bbox": [ - 218.82, - 107.74, - 226.59, - 117.71 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p34-b7", - "global_id": 570, - "bbox": [ - 232.16, - 98.85, - 264.18, - 118.12 - ], - "text": "4 + j4\n√", - "type": "text" - }, - { - "block_id": "p34-b8", - "global_id": 571, - "bbox": [ - 264.19, - 99.73, - 273.18, - 118.12 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p34-b9", - "global_id": 572, - "bbox": [ - 275.24, - 107.74, - 283.01, - 117.71 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p34-b11", - "global_id": 573, - "bbox": [ - 289.07, - 98.8, - 302.47, - 118.12 - ], - "text": "2\n√", - "type": "text" - }, - { - "block_id": "p34-b12", - "global_id": 574, - "bbox": [ - 302.48, - 99.73, - 327.32, - 118.12 - ], - "text": "2 −4", - "type": "text" - }, - { - "block_id": "p34-b13", - "global_id": 575, - "bbox": [ - 328.88, - 107.74, - 340.96, - 118.02 - ], - "text": "+ j", - "type": "text" - }, - { - "block_id": "p34-b15", - "global_id": 576, - "bbox": [ - 344.98, - 98.8, - 358.39, - 118.12 - ], - "text": "2\n√", - "type": "text" - }, - { - "block_id": "p34-b16", - "global_id": 577, - "bbox": [ - 358.4, - 98.85, - 387.64, - 118.12 - ], - "text": "2 −4\n√", - "type": "text" - }, - { - "block_id": "p34-b17", - "global_id": 578, - "bbox": [ - 387.65, - 99.73, - 396.65, - 118.12 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p34-b18", - "global_id": 579, - "bbox": [ - 398.7, - 107.74, - 459.79, - 118.12 - ], - "text": "= −1.17 −j4.1", - "type": "text" - }, - { - "block_id": "p34-b19", - "global_id": 580, - "bbox": [ - 121.09, - 129.99, - 133.27, - 139.95 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p34-b20", - "global_id": 581, - "bbox": [ - 246.04, - 139.59, - 253.41, - 164.38 - ], - "text": "1\nz1", - "type": "text" - }, - { - "block_id": "p34-b21", - "global_id": 582, - "bbox": [ - 257.15, - 139.59, - 334.83, - 163.61 - ], - "text": "=\n1\n2ejπ/4 = 1\n2e−jπ/4", - "type": "text" - }, - { - "block_id": "p34-b22", - "global_id": 583, - "bbox": [ - 121.09, - 169.94, - 132.15, - 179.9 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p34-b23", - "global_id": 584, - "bbox": [ - 174.57, - 181.02, - 181.94, - 205.49 - ], - "text": "z1\nz2", - "type": "text" - }, - { - "block_id": "p34-b24", - "global_id": 585, - "bbox": [ - 178.45, - 179.48, - 225.03, - 207.79 - ], - "text": "2\n=\n2ejπ/4", - "type": "text" - }, - { - "block_id": "p34-b25", - "global_id": 586, - "bbox": [ - 196.69, - 179.48, - 272.37, - 205.14 - ], - "text": "(8ejπ/3)2 = 2ejπ/4", - "type": "text" - }, - { - "block_id": "p34-b26", - "global_id": 587, - "bbox": [ - 245.51, - 181.12, - 406.3, - 205.14 - ], - "text": "64ej2π/3 = 1\n32ej(π/4−2π/3) = 1\n32e−j(5π/12)", - "type": "text" - }, - { - "block_id": "p34-b27", - "global_id": 588, - "bbox": [ - 121.09, - 210.89, - 384.69, - 224.89 - ], - "text": "(d) There are three cube roots of 8ej(π/3) = 8ej(π/3+2πk), k = 0,1,2.", - "type": "text" - }, - { - "block_id": "p34-b28", - "global_id": 589, - "bbox": [ - 138.62, - 240.64, - 191.22, - 260.2 - ], - "text": "3√z2 = z1/3\n2\n=", - "type": "text" - }, - { - "block_id": "p34-b30", - "global_id": 590, - "bbox": [ - 197.2, - 239.62, - 358.81, - 258.01 - ], - "text": "8ej(π/3+2πk)\n1/3 = 81/3\nej[(6πk+π)/3]1/3 =", - "type": "text" - }, - { - "block_id": "p34-b31", - "global_id": 591, - "bbox": [ - 360.86, - 227.2, - 368.75, - 246.14 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p34-b32", - "global_id": 592, - "bbox": [ - 360.86, - 254.1, - 368.75, - 264.06 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p34-b33", - "global_id": 593, - "bbox": [ - 373.73, - 234.34, - 435.59, - 269.86 - ], - "text": "2ejπ/9\nk = 0\n2ej7π/9\nk = 1\n2ej13π/9\nk = 2", - "type": "text" - }, - { - "block_id": "p34-b34", - "global_id": 594, - "bbox": [ - 103.17, - 280.81, - 355.67, - 291.19 - ], - "text": "The value corresponding to k = 0 is termed the principal value.", - "type": "text" - }, - { - "block_id": "p34-b35", - "global_id": 595, - "bbox": [ - 76.77, - 346.81, - 384.18, - 358.76 - ], - "text": "EXAMPLE B.5\nStandard Forms of Complex Numbers", - "type": "text" - }, - { - "block_id": "p34-b36", - "global_id": 596, - "bbox": [ - 103.16, - 375.02, - 325.51, - 385.39 - ], - "text": "Consider X(ω), a complex function of a real variable ω:", - "type": "text" - }, - { - "block_id": "p34-b37", - "global_id": 597, - "bbox": [ - 257.46, - 394.92, - 318.83, - 412.28 - ], - "text": "X(ω) = 2 + jω", - "type": "text" - }, - { - "block_id": "p34-b38", - "global_id": 598, - "bbox": [ - 291.2, - 408.97, - 321.32, - 419.35 - ], - "text": "3 + j4ω", - "type": "text" - }, - { - "block_id": "p34-b39", - "global_id": 599, - "bbox": [ - 121.09, - 441.75, - 443.31, - 467.07 - ], - "text": "(a) Express X(ω) in Cartesian form, and find its real and imaginary parts.\n(b) Express X(ω) in polar form, and find its magnitude |X(ω)| and angle̸\nX(ω).", - "type": "text" - }, - { - "block_id": "p34-b40", - "global_id": 600, - "bbox": [ - 103.16, - 495.54, - 477.01, - 529.83 - ], - "text": "(a) To obtain the real and imaginary parts of X(ω), we must eliminate imaginary terms\nin the denominator of X(ω). This is readily done by multiplying both the numerator and the\ndenominator of X(ω) by 3 −j4ω, the conjugate of the denominator 3 + j4ω so that", - "type": "text" - }, - { - "block_id": "p34-b41", - "global_id": 601, - "bbox": [ - 143.91, - 540.94, - 250.66, - 558.2 - ], - "text": "X(ω) = (2 + jω)(3 −j4ω)", - "type": "text" - }, - { - "block_id": "p34-b42", - "global_id": 602, - "bbox": [ - 177.65, - 540.0, - 331.54, - 565.37 - ], - "text": "(3 + j4ω)(3 −j4ω) = (6 + 4ω2) −j5ω", - "type": "text" - }, - { - "block_id": "p34-b43", - "global_id": 603, - "bbox": [ - 281.32, - 540.0, - 379.52, - 565.37 - ], - "text": "9 + 16ω2\n= 6 + 4ω2", - "type": "text" - }, - { - "block_id": "p34-b44", - "global_id": 604, - "bbox": [ - 346.0, - 540.94, - 434.56, - 565.37 - ], - "text": "9 + 16ω2 −j\n5ω\n9 + 16ω2", - "type": "text" - }, - { - "block_id": "p34-b45", - "global_id": 605, - "bbox": [ - 103.16, - 574.83, - 477.02, - 597.16 - ], - "text": "This is the Cartesian form of X(ω). Clearly, the real and imaginary parts Xr(ω) and Xi(ω) are\ngiven by", - "type": "text" - }, - { - "block_id": "p34-b46", - "global_id": 606, - "bbox": [ - 189.04, - 597.37, - 259.24, - 616.37 - ], - "text": "Xr(ω) = 6 + 4ω2", - "type": "text" - }, - { - "block_id": "p34-b47", - "global_id": 607, - "bbox": [ - 225.72, - 598.31, - 389.42, - 622.74 - ], - "text": "9 + 16ω2\nand\nXi(ω) =\n−5ω\n9 + 16ω2", - "type": "text" - } - ] - }, - { - "page_num": 35, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p35-b0", - "global_id": 608, - "bbox": [ - 376.42, - 62.89, - 516.13, - 71.98 - ], - "text": "B.1\nComplex Numbers\n15", - "type": "text" - }, - { - "block_id": "p35-b1", - "global_id": 609, - "bbox": [ - 146.84, - 86.29, - 159.02, - 96.25 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p35-b2", - "global_id": 610, - "bbox": [ - 157.13, - 111.45, - 218.5, - 128.81 - ], - "text": "X(ω) = 2 + jω", - "type": "text" - }, - { - "block_id": "p35-b3", - "global_id": 611, - "bbox": [ - 190.87, - 118.43, - 232.21, - 135.88 - ], - "text": "3 + j4ω =", - "type": "text" - }, - { - "block_id": "p35-b4", - "global_id": 612, - "bbox": [ - 242.13, - 103.01, - 250.56, - 112.98 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p35-b5", - "global_id": 613, - "bbox": [ - 235.46, - 108.49, - 318.81, - 128.45 - ], - "text": "4 + ω2 ejtan−1(ω/2)\n√", - "type": "text" - }, - { - "block_id": "p35-b6", - "global_id": 614, - "bbox": [ - 243.89, - 118.43, - 337.0, - 137.29 - ], - "text": "9 + 16ω2 ejtan−1(4ω/3) =", - "type": "text" - }, - { - "block_id": "p35-b8", - "global_id": 615, - "bbox": [ - 353.48, - 111.25, - 379.53, - 121.82 - ], - "text": "4 + ω2", - "type": "text" - }, - { - "block_id": "p35-b9", - "global_id": 616, - "bbox": [ - 348.5, - 114.98, - 474.04, - 135.88 - ], - "text": "9 + 16ω2 ej[tan−1(ω/2)−tan−1(4ω/3)]", - "type": "text" - }, - { - "block_id": "p35-b10", - "global_id": 617, - "bbox": [ - 128.9, - 146.79, - 341.74, - 157.16 - ], - "text": "This is the polar representation of X(ω). Observe that", - "type": "text" - }, - { - "block_id": "p35-b11", - "global_id": 618, - "bbox": [ - 173.72, - 179.68, - 209.92, - 189.96 - ], - "text": "|X(ω)| =", - "type": "text" - }, - { - "block_id": "p35-b13", - "global_id": 619, - "bbox": [ - 226.4, - 172.51, - 252.45, - 183.08 - ], - "text": "4 + ω2", - "type": "text" - }, - { - "block_id": "p35-b14", - "global_id": 620, - "bbox": [ - 221.42, - 177.96, - 372.66, - 197.13 - ], - "text": "9 + 16ω2\nand̸\nX(ω) = tan−1", - "type": "text" - }, - { - "block_id": "p35-b15", - "global_id": 621, - "bbox": [ - 374.26, - 165.7, - 388.71, - 182.66 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p35-b16", - "global_id": 622, - "bbox": [ - 383.06, - 187.17, - 388.04, - 197.13 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p35-b18", - "global_id": 623, - "bbox": [ - 398.37, - 177.96, - 428.8, - 190.06 - ], - "text": "−tan−1", - "type": "text" - }, - { - "block_id": "p35-b19", - "global_id": 624, - "bbox": [ - 430.4, - 165.7, - 449.83, - 183.08 - ], - "text": "4ω", - "type": "text" - }, - { - "block_id": "p35-b20", - "global_id": 625, - "bbox": [ - 441.69, - 187.17, - 446.67, - 197.13 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p35-b22", - "global_id": 626, - "bbox": [ - 127.89, - 238.64, - 329.95, - 250.77 - ], - "text": "LOGARITHMS OF COMPLEX NUMBERS", - "type": "text" - }, - { - "block_id": "p35-b23", - "global_id": 627, - "bbox": [ - 127.59, - 254.7, - 507.69, - 264.76 - ], - "text": "To take the natural logarithm of a complex number z, we first express z in general polar form as", - "type": "text" - }, - { - "block_id": "p35-b24", - "global_id": 628, - "bbox": [ - 240.86, - 278.09, - 402.86, - 292.58 - ], - "text": "z = rejθ = rej(θ±2πk)\nk = 0,1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p35-b25", - "global_id": 629, - "bbox": [ - 127.59, - 310.44, - 290.26, - 320.4 - ], - "text": "Taking the natural logarithm, we see that", - "type": "text" - }, - { - "block_id": "p35-b26", - "global_id": 630, - "bbox": [ - 201.64, - 337.84, - 234.0, - 348.22 - ], - "text": "lnz = ln", - "type": "text" - }, - { - "block_id": "p35-b28", - "global_id": 631, - "bbox": [ - 239.11, - 329.83, - 279.49, - 348.12 - ], - "text": "rej(θ±2πk)", - "type": "text" - }, - { - "block_id": "p35-b29", - "global_id": 632, - "bbox": [ - 281.53, - 337.84, - 442.09, - 348.22 - ], - "text": "= lnr ± j(θ + 2πk)\nk = 0,1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p35-b30", - "global_id": 633, - "bbox": [ - 127.59, - 365.66, - 516.12, - 387.99 - ], - "text": "The value of lnz for k = 0 is called the principal value of lnz and is denoted by Lnz. In this way,\nwe see that", - "type": "text" - }, - { - "block_id": "p35-b31", - "global_id": 634, - "bbox": [ - 214.4, - 403.71, - 407.0, - 415.81 - ], - "text": "ln1 = ln(1e±j2πk) = ±j2πk\nk = 0,1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p35-b32", - "global_id": 635, - "bbox": [ - 200.29, - 420.19, - 443.43, - 432.29 - ], - "text": "ln(−1) = ln[1e±jπ(2k+1)] = ±j(2k + 1)π\nk = 0,1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p35-b33", - "global_id": 636, - "bbox": [ - 216.6, - 443.31, - 247.86, - 453.69 - ], - "text": "lnj = ln", - "type": "text" - }, - { - "block_id": "p35-b35", - "global_id": 637, - "bbox": [ - 251.87, - 435.3, - 295.02, - 453.59 - ], - "text": "ejπ(1±4k)/2", - "type": "text" - }, - { - "block_id": "p35-b36", - "global_id": 638, - "bbox": [ - 297.07, - 436.33, - 350.4, - 453.59 - ], - "text": "= jπ(1 ± 4k)", - "type": "text" - }, - { - "block_id": "p35-b37", - "global_id": 639, - "bbox": [ - 328.14, - 443.31, - 435.51, - 460.76 - ], - "text": "2\nk = 0,1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p35-b38", - "global_id": 640, - "bbox": [ - 223.03, - 459.16, - 394.29, - 473.65 - ], - "text": "jj = ejlnj = e−π(1±4k)/2\nk = 0,1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p35-b39", - "global_id": 641, - "bbox": [ - 127.59, - 491.09, - 436.64, - 501.47 - ], - "text": "In all of these cases, setting k = 0 yields the principal value of the expression.", - "type": "text" - }, - { - "block_id": "p35-b40", - "global_id": 642, - "bbox": [ - 127.59, - 503.46, - 516.14, - 525.38 - ], - "text": "We can further our logarithm skills by noting that the familiar properties of logarithms hold\nfor complex arguments. Therefore, we have", - "type": "text" - }, - { - "block_id": "p35-b41", - "global_id": 643, - "bbox": [ - 268.51, - 542.83, - 374.72, - 569.22 - ], - "text": "log(z1z2) = logz1 + logz2\nlog(z1/z2) = logz1 −logz2", - "type": "text" - }, - { - "block_id": "p35-b42", - "global_id": 644, - "bbox": [ - 280.56, - 570.13, - 354.2, - 584.52 - ], - "text": "a(z1+z2) = az1 × az2", - "type": "text" - }, - { - "block_id": "p35-b43", - "global_id": 645, - "bbox": [ - 301.61, - 586.57, - 338.15, - 600.96 - ], - "text": "zc = eclnz", - "type": "text" - }, - { - "block_id": "p35-b44", - "global_id": 646, - "bbox": [ - 300.89, - 603.0, - 338.54, - 617.39 - ], - "text": "az = ezlna", - "type": "text" - } - ] - }, - { - "page_num": 36, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p36-b0", - "global_id": 647, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "16\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p36-b1", - "global_id": 648, - "bbox": [ - 102.2, - 90.78, - 202.52, - 104.73 - ], - "text": "B.2 SINUSOIDS", - "type": "text" - }, - { - "block_id": "p36-b2", - "global_id": 649, - "bbox": [ - 101.84, - 110.72, - 188.2, - 120.68 - ], - "text": "Consider the sinusoid", - "type": "text" - }, - { - "block_id": "p36-b3", - "global_id": 650, - "bbox": [ - 249.47, - 122.83, - 490.37, - 133.98 - ], - "text": "x(t) = Ccos(2πf0t + θ)\n(B.13)", - "type": "text" - }, - { - "block_id": "p36-b4", - "global_id": 651, - "bbox": [ - 101.84, - 142.56, - 156.69, - 152.52 - ], - "text": "We know that", - "type": "text" - }, - { - "block_id": "p36-b5", - "global_id": 652, - "bbox": [ - 197.73, - 154.68, - 394.51, - 165.06 - ], - "text": "cos ϕ = cos(ϕ + 2nπ)\nn = 0,±1,±2,±3,. . .", - "type": "text" - }, - { - "block_id": "p36-b6", - "global_id": 653, - "bbox": [ - 101.84, - 174.0, - 490.39, - 221.73 - ], - "text": "Therefore, cos ϕ repeats itself for every change of 2π in the angle ϕ. For the sinusoid in Eq. (B.13),\nthe angle 2πf0t+θ changes by 2π when t changes by 1/f0. Clearly, this sinusoid repeats every 1/f0\nseconds. As a result, there are f0 repetitions per second. This is the frequency of the sinusoid, and\nthe repetition interval T0 given by", - "type": "text" - }, - { - "block_id": "p36-b7", - "global_id": 654, - "bbox": [ - 280.84, - 220.79, - 309.31, - 238.82 - ], - "text": "T0 = 1", - "type": "text" - }, - { - "block_id": "p36-b8", - "global_id": 655, - "bbox": [ - 303.43, - 234.75, - 309.69, - 245.58 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p36-b9", - "global_id": 656, - "bbox": [ - 464.65, - 227.77, - 490.38, - 237.74 - ], - "text": "(B.14)", - "type": "text" - }, - { - "block_id": "p36-b10", - "global_id": 657, - "bbox": [ - 101.84, - 251.93, - 490.39, - 285.9 - ], - "text": "is the period. For the sinusoid in Eq. (B.13), C is the amplitude, f0 is the frequency (in hertz), and\nθ is the phase. Let us consider two special cases of this sinusoid when θ = 0 and θ = −π/2 as\nfollows:", - "type": "text" - }, - { - "block_id": "p36-b11", - "global_id": 658, - "bbox": [ - 235.96, - 288.05, - 356.26, - 299.2 - ], - "text": "x(t) = Ccos 2πf0t\n(θ = 0)", - "type": "text" - }, - { - "block_id": "p36-b12", - "global_id": 659, - "bbox": [ - 101.85, - 307.78, - 116.22, - 317.74 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p36-b13", - "global_id": 660, - "bbox": [ - 182.72, - 319.91, - 409.5, - 331.05 - ], - "text": "x(t) = Ccos(2πf0t −π/2) = Csin 2πf0t\n(θ = −π/2)", - "type": "text" - }, - { - "block_id": "p36-b14", - "global_id": 661, - "bbox": [ - 101.84, - 339.63, - 490.4, - 409.37 - ], - "text": "The angle or phase can be expressed in units of degrees or radians. Although the radian is the\nproper unit, in this book we shall often use the degree unit because students generally have a better\nfeel for the relative magnitudes of angles expressed in degrees rather than in radians. For example,\nwe relate better to the angle 24◦than to 0.419 radian. Remember, however, when in doubt, use the\nradian unit and, above all, be consistent. In other words, in a given problem or an expression, do\nnot mix the two units.", - "type": "text" - }, - { - "block_id": "p36-b15", - "global_id": 662, - "bbox": [ - 119.78, - 410.95, - 410.95, - 422.82 - ], - "text": "It is convenient to use the variable ω0 (radian frequency) to express 2πf0:", - "type": "text" - }, - { - "block_id": "p36-b16", - "global_id": 663, - "bbox": [ - 276.13, - 433.25, - 490.38, - 444.71 - ], - "text": "ω0 = 2πf0\n(B.15)", - "type": "text" - }, - { - "block_id": "p36-b17", - "global_id": 664, - "bbox": [ - 101.85, - 455.97, - 361.42, - 465.93 - ], - "text": "With this notation, the sinusoid in Eq. (B.13) can be expressed as", - "type": "text" - }, - { - "block_id": "p36-b18", - "global_id": 665, - "bbox": [ - 253.02, - 477.86, - 339.22, - 489.01 - ], - "text": "x(t) = Ccos(ω0t + θ)", - "type": "text" - }, - { - "block_id": "p36-b19", - "global_id": 666, - "bbox": [ - 101.85, - 500.17, - 429.53, - 512.04 - ], - "text": "in which the period T0 and frequency ω0 are given by [see Eqs. (B.14) and (B.15)]", - "type": "text" - }, - { - "block_id": "p36-b20", - "global_id": 667, - "bbox": [ - 212.07, - 520.45, - 287.01, - 545.66 - ], - "text": "T0 =\n1\nω0/2π = 2π", - "type": "text" - }, - { - "block_id": "p36-b21", - "global_id": 668, - "bbox": [ - 276.78, - 534.51, - 286.79, - 545.66 - ], - "text": "ω0", - "type": "text" - }, - { - "block_id": "p36-b22", - "global_id": 669, - "bbox": [ - 309.13, - 520.45, - 377.97, - 538.89 - ], - "text": "and\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p36-b23", - "global_id": 670, - "bbox": [ - 368.23, - 534.82, - 377.26, - 545.66 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p36-b24", - "global_id": 671, - "bbox": [ - 101.84, - 554.71, - 490.39, - 578.53 - ], - "text": "Although we shall often refer to ω0 as the frequency of the signal cos(ω0t+θ), it should be clearly\nunderstood that ω0 is the radian frequency; the hertzian frequency of this sinusoid is f0 = ω0/2π).", - "type": "text" - }, - { - "block_id": "p36-b25", - "global_id": 672, - "bbox": [ - 101.84, - 578.62, - 490.38, - 612.91 - ], - "text": "The signals Ccos ω0t and Csin ω0t are illustrated in Figs. B.6a and B.6b, respectively. A\ngeneral sinusoid Ccos(ω0t+θ) can be readily sketched by shifting the signal Ccos ω0t in Fig. B.6a\nby the appropriate amount. Consider, for example,", - "type": "text" - }, - { - "block_id": "p36-b26", - "global_id": 673, - "bbox": [ - 248.73, - 623.1, - 343.5, - 635.98 - ], - "text": "x(t) = Ccos(ω0t −60◦)", - "type": "text" - } - ] - }, - { - "page_num": 37, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p37-b0", - "global_id": 674, - "bbox": [ - 413.6, - 62.89, - 516.14, - 71.98 - ], - "text": "B.2\nSinusoids\n17", - "type": "text" - }, - { - "block_id": "p37-b1", - "global_id": 675, - "bbox": [ - 308.99, - 87.28, - 314.33, - 95.28 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p37-b2", - "global_id": 676, - "bbox": [ - 310.3, - 148.48, - 314.3, - 156.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p37-b3", - "global_id": 677, - "bbox": [ - 332.3, - 94.01, - 363.34, - 103.64 - ], - "text": "C cos v0t", - "type": "text" - }, - { - "block_id": "p37-b4", - "global_id": 678, - "bbox": [ - 160.32, - 132.84, - 174.43, - 142.66 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b5", - "global_id": 679, - "bbox": [ - 159.14, - 281.92, - 173.25, - 291.74 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b6", - "global_id": 680, - "bbox": [ - 305.8, - 226.15, - 311.13, - 234.15 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p37-b7", - "global_id": 681, - "bbox": [ - 320.8, - 281.23, - 324.8, - 289.23 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p37-b8", - "global_id": 682, - "bbox": [ - 312.37, - 334.33, - 322.02, - 342.33 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p37-b9", - "global_id": 683, - "bbox": [ - 305.8, - 355.02, - 311.13, - 363.02 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p37-b10", - "global_id": 684, - "bbox": [ - 310.3, - 418.04, - 314.3, - 426.04 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p37-b11", - "global_id": 685, - "bbox": [ - 312.76, - 468.74, - 321.64, - 476.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p37-b12", - "global_id": 686, - "bbox": [ - 257.3, - 356.4, - 288.1, - 366.03 - ], - "text": "C cos v0t", - "type": "text" - }, - { - "block_id": "p37-b13", - "global_id": 687, - "bbox": [ - 340.15, - 219.94, - 369.62, - 229.56 - ], - "text": "C sin v0t", - "type": "text" - }, - { - "block_id": "p37-b14", - "global_id": 688, - "bbox": [ - 431.87, - 151.97, - 434.1, - 159.97 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p37-b15", - "global_id": 689, - "bbox": [ - 422.13, - 283.14, - 424.36, - 291.14 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p37-b16", - "global_id": 690, - "bbox": [ - 458.87, - 418.51, - 461.1, - 426.51 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p37-b17", - "global_id": 691, - "bbox": [ - 463.73, - 150.6, - 471.18, - 160.21 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b18", - "global_id": 692, - "bbox": [ - 463.73, - 282.1, - 471.18, - 291.71 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b19", - "global_id": 693, - "bbox": [ - 312.76, - 199.53, - 321.64, - 207.53 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p37-b20", - "global_id": 694, - "bbox": [ - 333.1, - 440.7, - 340.55, - 450.3 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b21", - "global_id": 695, - "bbox": [ - 334.83, - 451.66, - 338.83, - 459.66 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p37-b22", - "global_id": 696, - "bbox": [ - 324.5, - 356.81, - 440.91, - 374.36 - ], - "text": "60\nC cos (v0t 60)", - "type": "text" - }, - { - "block_id": "p37-b23", - "global_id": 697, - "bbox": [ - 387.77, - 279.79, - 395.22, - 289.4 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b24", - "global_id": 698, - "bbox": [ - 234.08, - 279.58, - 393.5, - 298.64 - ], - "text": "2\nT0", - "type": "text" - }, - { - "block_id": "p37-b25", - "global_id": 699, - "bbox": [ - 242.33, - 290.64, - 246.33, - 298.64 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p37-b26", - "global_id": 700, - "bbox": [ - 232.57, - 123.43, - 246.68, - 133.25 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b27", - "global_id": 701, - "bbox": [ - 240.82, - 123.64, - 395.2, - 142.49 - ], - "text": "2\nT0", - "type": "text" - }, - { - "block_id": "p37-b28", - "global_id": 702, - "bbox": [ - 389.48, - 134.49, - 393.48, - 142.49 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p37-b29", - "global_id": 703, - "bbox": [ - 326.15, - 416.55, - 333.6, - 426.16 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p37-b30", - "global_id": 704, - "bbox": [ - 327.87, - 427.52, - 331.87, - 435.52 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p37-b31", - "global_id": 705, - "bbox": [ - 397.78, - 88.0, - 421.63, - 101.51 - ], - "text": "T0 1", - "type": "text" - }, - { - "block_id": "p37-b32", - "global_id": 706, - "bbox": [ - 417.01, - 96.59, - 422.24, - 106.19 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p37-b33", - "global_id": 707, - "bbox": [ - 425.76, - 87.9, - 444.44, - 99.98 - ], - "text": "2p", - "type": "text" - }, - { - "block_id": "p37-b34", - "global_id": 708, - "bbox": [ - 435.6, - 94.66, - 443.94, - 104.29 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p37-b35", - "global_id": 709, - "bbox": [ - 151.5, - 483.44, - 274.96, - 492.67 - ], - "text": "Figure B.6 Sketching a sinusoid.", - "type": "text" - }, - { - "block_id": "p37-b36", - "global_id": 710, - "bbox": [ - 127.59, - 519.23, - 516.13, - 577.42 - ], - "text": "This signal can be obtained by shifting (delaying) the signal Ccos ω0t (Fig. B.6a) to the right by a\nphase (angle) of 60◦. We know that a sinusoid undergoes a 360◦change of phase (or angle) in one\ncycle. A quarter-cycle segment corresponds to a 90◦change of angle. We therefore shift (delay)\nthe signal in Fig. B.6a by two-thirds of a quarter-cycle segment to obtain Ccos(ω0t −60◦), as\nshown in Fig. B.6c.", - "type": "text" - }, - { - "block_id": "p37-b37", - "global_id": 711, - "bbox": [ - 127.59, - 577.78, - 516.13, - 613.29 - ], - "text": "Observe that if we delay Ccos ω0t in Fig. B.6a by a quarter-cycle (angle of 90◦or π/2\nradians), we obtain the signal Csin ω0t, depicted in Fig. B.6b. This verifies the well-known\ntrigonometric identity", - "type": "text" - }, - { - "block_id": "p37-b38", - "global_id": 712, - "bbox": [ - 263.16, - 624.56, - 380.38, - 635.7 - ], - "text": "Ccos(ω0t −π/2) = Csin ω0t", - "type": "text" - } - ] - }, - { - "page_num": 38, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p38-b0", - "global_id": 713, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "18\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p38-b1", - "global_id": 714, - "bbox": [ - 101.84, - 85.4, - 455.39, - 96.55 - ], - "text": "Alternatively, if we advance Csin ω0t by a quarter-cycle, we obtain Ccos ω0t. Therefore,", - "type": "text" - }, - { - "block_id": "p38-b2", - "global_id": 715, - "bbox": [ - 237.42, - 111.74, - 354.63, - 122.88 - ], - "text": "Csin(ω0t + π/2) = Ccos ω0t", - "type": "text" - }, - { - "block_id": "p38-b3", - "global_id": 716, - "bbox": [ - 101.84, - 136.83, - 490.38, - 161.16 - ], - "text": "These observations mean that sin ω0t lags cos ω0t by 90◦(π/2 radians) and that cos ω0t leads\nsin ω0t by 90◦.", - "type": "text" - }, - { - "block_id": "p38-b4", - "global_id": 717, - "bbox": [ - 101.84, - 189.56, - 255.75, - 201.52 - ], - "text": "B.2-1 Addition of Sinusoids", - "type": "text" - }, - { - "block_id": "p38-b5", - "global_id": 718, - "bbox": [ - 101.84, - 207.64, - 490.38, - 229.56 - ], - "text": "Two sinusoids having the same frequency but different phases add to form a single sinusoid of the\nsame frequency. This fact is readily seen from the well-known trigonometric identity", - "type": "text" - }, - { - "block_id": "p38-b6", - "global_id": 719, - "bbox": [ - 196.64, - 245.51, - 395.59, - 256.66 - ], - "text": "Ccos θ cos ω0t −Csin θ sin ω0t = Ccos(ω0t + θ)", - "type": "text" - }, - { - "block_id": "p38-b7", - "global_id": 720, - "bbox": [ - 101.84, - 271.85, - 298.68, - 282.22 - ], - "text": "Setting a = Ccos θ and b = −Csin θ, we see that", - "type": "text" - }, - { - "block_id": "p38-b8", - "global_id": 721, - "bbox": [ - 219.68, - 298.17, - 490.38, - 309.32 - ], - "text": "a cos ω0t + b sin ω0t = Ccos(ω0t + θ)\n(B.16)", - "type": "text" - }, - { - "block_id": "p38-b9", - "global_id": 722, - "bbox": [ - 101.85, - 324.91, - 235.7, - 334.87 - ], - "text": "From trigonometry, we know that", - "type": "text" - }, - { - "block_id": "p38-b10", - "global_id": 723, - "bbox": [ - 207.39, - 356.47, - 224.07, - 366.75 - ], - "text": "C =", - "type": "text" - }, - { - "block_id": "p38-b12", - "global_id": 724, - "bbox": [ - 234.37, - 353.59, - 355.76, - 366.85 - ], - "text": "a2 + b2\nand\nθ = tan−1", - "type": "text" - }, - { - "block_id": "p38-b13", - "global_id": 725, - "bbox": [ - 356.26, - 342.49, - 376.92, - 359.76 - ], - "text": "−b", - "type": "text" - }, - { - "block_id": "p38-b14", - "global_id": 726, - "bbox": [ - 368.06, - 363.86, - 373.05, - 373.82 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p38-b16", - "global_id": 727, - "bbox": [ - 464.65, - 356.89, - 490.38, - 366.85 - ], - "text": "(B.17)", - "type": "text" - }, - { - "block_id": "p38-b17", - "global_id": 728, - "bbox": [ - 101.84, - 388.09, - 490.39, - 434.33 - ], - "text": "Equation (B.17) shows that C and θ are the magnitude and angle, respectively, of a complex\nnumber a −jb. In other words, a −jb = Cejθ. Hence, to find C and θ, we convert a −jb\nto polar form and the magnitude and the angle of the resulting polar number are C and θ,\nrespectively.", - "type": "text" - }, - { - "block_id": "p38-b18", - "global_id": 729, - "bbox": [ - 101.84, - 436.22, - 490.4, - 529.97 - ], - "text": "The process of adding two sinusoids with the same frequency can be clarified by using phasors\nto represent sinusoids. We represent the sinusoid Ccos(ω0t+θ) by a phasor of length C at an angle\nθ with the horizontal axis. Clearly, the sinusoid acos ω0t is represented by a horizontal phasor of\nlength a(θ = 0), while bsin ω0t = bcos(ω0t −π/2) is represented by a vertical phasor of length b\nat an angle −π/2 with the horizontal (Fig. B.7). Adding these two phasors results in a phasor of\nlength C at an angle θ, as depicted in Fig. B.7. From this figure, we verify the values of C and θ\nfound in Eq. (B.17). Proper care should be exercised in computing θ, as explained on page 8 (“A\nWarning About Computing Angles with Calculators”).", - "type": "text" - }, - { - "block_id": "p38-b19", - "global_id": 730, - "bbox": [ - 197.14, - 589.47, - 208.03, - 597.47 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p38-b20", - "global_id": 731, - "bbox": [ - 111.2, - 562.24, - 120.09, - 570.24 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p38-b21", - "global_id": 732, - "bbox": [ - 180.74, - 575.75, - 184.74, - 583.75 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p38-b22", - "global_id": 733, - "bbox": [ - 109.99, - 614.03, - 120.65, - 622.24 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p38-b23", - "global_id": 734, - "bbox": [ - 146.24, - 602.25, - 151.58, - 610.25 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p38-b24", - "global_id": 735, - "bbox": [ - 145.42, - 588.22, - 149.42, - 596.22 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p38-b25", - "global_id": 736, - "bbox": [ - 227.79, - 617.27, - 378.5, - 626.5 - ], - "text": "Figure B.7 Phasor addition of sinusoids.", - "type": "text" - } - ] - }, - { - "page_num": 39, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p39-b0", - "global_id": 737, - "bbox": [ - 413.6, - 62.89, - 516.14, - 71.98 - ], - "text": "B.2\nSinusoids\n19", - "type": "text" - }, - { - "block_id": "p39-b1", - "global_id": 738, - "bbox": [ - 102.51, - 93.91, - 324.24, - 105.87 - ], - "text": "EXAMPLE B.6\nAddition of Sinusoids", - "type": "text" - }, - { - "block_id": "p39-b2", - "global_id": 739, - "bbox": [ - 128.9, - 119.07, - 350.86, - 129.45 - ], - "text": "In the following cases, express x(t) as a single sinusoid:", - "type": "text" - }, - { - "block_id": "p39-b3", - "global_id": 740, - "bbox": [ - 146.84, - 137.0, - 228.41, - 148.15 - ], - "text": "(a) x(t) = cos ω0t −", - "type": "text" - }, - { - "block_id": "p39-b4", - "global_id": 741, - "bbox": [ - 229.96, - 128.63, - 238.38, - 138.59 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p39-b5", - "global_id": 742, - "bbox": [ - 238.39, - 137.0, - 271.6, - 148.15 - ], - "text": "3sin ω0t", - "type": "text" - }, - { - "block_id": "p39-b6", - "global_id": 743, - "bbox": [ - 146.84, - 151.95, - 277.59, - 163.1 - ], - "text": "(b) x(t) = −3cos ω0t + 4sin ω0t", - "type": "text" - }, - { - "block_id": "p39-b7", - "global_id": 744, - "bbox": [ - 146.84, - 190.8, - 276.56, - 201.18 - ], - "text": "(a) In this case, a = 1 and b = −", - "type": "text" - }, - { - "block_id": "p39-b8", - "global_id": 745, - "bbox": [ - 276.56, - 182.42, - 284.99, - 192.38 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p39-b9", - "global_id": 746, - "bbox": [ - 284.99, - 191.21, - 389.31, - 201.18 - ], - "text": "3. Using Eq. (B.17) yields", - "type": "text" - }, - { - "block_id": "p39-b10", - "global_id": 747, - "bbox": [ - 209.93, - 217.51, - 226.61, - 227.79 - ], - "text": "C =", - "type": "text" - }, - { - "block_id": "p39-b12", - "global_id": 748, - "bbox": [ - 236.91, - 208.62, - 269.2, - 227.89 - ], - "text": "12 +\n√", - "type": "text" - }, - { - "block_id": "p39-b13", - "global_id": 749, - "bbox": [ - 269.2, - 206.51, - 385.81, - 227.89 - ], - "text": "3\n2 = 2\nand\nθ = tan−1 √", - "type": "text" - }, - { - "block_id": "p39-b14", - "global_id": 750, - "bbox": [ - 382.87, - 206.51, - 395.93, - 230.7 - ], - "text": "3\n1", - "type": "text" - }, - { - "block_id": "p39-b15", - "global_id": 751, - "bbox": [ - 397.98, - 215.79, - 421.24, - 227.89 - ], - "text": "= 60◦", - "type": "text" - }, - { - "block_id": "p39-b16", - "global_id": 752, - "bbox": [ - 128.9, - 241.88, - 170.65, - 251.85 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p39-b17", - "global_id": 753, - "bbox": [ - 269.39, - 251.71, - 362.27, - 264.58 - ], - "text": "x(t) = 2cos(ω0t + 60◦)", - "type": "text" - }, - { - "block_id": "p39-b18", - "global_id": 754, - "bbox": [ - 128.9, - 272.64, - 502.75, - 318.06 - ], - "text": "We can verify this result by drawing phasors corresponding to the two sinusoids. The sinusoid\ncos ω0t is represented by a phasor of unit length at a zero angle with the horizontal. The phasor\nsin ω0t is represented by a unit phasor at an angle of −90◦with the horizontal. Therefore,\n−", - "type": "text" - }, - { - "block_id": "p39-b19", - "global_id": 755, - "bbox": [ - 136.67, - 299.72, - 145.1, - 309.68 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p39-b20", - "global_id": 756, - "bbox": [ - 145.1, - 299.72, - 333.85, - 319.25 - ], - "text": "3sin ω0t is represented by a phasor of length\n√", - "type": "text" - }, - { - "block_id": "p39-b21", - "global_id": 757, - "bbox": [ - 128.91, - 306.87, - 502.77, - 342.38 - ], - "text": "3 at 90◦with the horizontal, as depicted in\nFig. B.8a. The two phasors added yield a phasor of length 2 at 60◦with the horizontal (also\nshown in Fig. B.8a).", - "type": "text" - }, - { - "block_id": "p39-b22", - "global_id": 758, - "bbox": [ - 181.91, - 480.82, - 316.88, - 488.89 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p39-b23", - "global_id": 759, - "bbox": [ - 223.24, - 460.34, - 234.13, - 468.34 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p39-b24", - "global_id": 760, - "bbox": [ - 138.69, - 378.64, - 147.58, - 386.64 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p39-b25", - "global_id": 761, - "bbox": [ - 180.7, - 461.38, - 184.7, - 469.38 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p39-b26", - "global_id": 762, - "bbox": [ - 141.5, - 399.43, - 145.5, - 407.43 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p39-b27", - "global_id": 763, - "bbox": [ - 160.89, - 420.38, - 164.89, - 428.38 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p39-b28", - "global_id": 764, - "bbox": [ - 162.7, - 445.38, - 173.37, - 453.68 - ], - "text": "60", - "type": "text" - }, - { - "block_id": "p39-b29", - "global_id": 765, - "bbox": [ - 272.51, - 377.58, - 355.41, - 390.9 - ], - "text": "Re \nIm\n3", - "type": "text" - }, - { - "block_id": "p39-b30", - "global_id": 766, - "bbox": [ - 322.52, - 444.58, - 333.67, - 452.88 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p39-b31", - "global_id": 767, - "bbox": [ - 325.74, - 400.23, - 353.07, - 408.52 - ], - "text": "126.9", - "type": "text" - }, - { - "block_id": "p39-b32", - "global_id": 768, - "bbox": [ - 293.32, - 413.63, - 297.32, - 421.63 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p39-b33", - "global_id": 769, - "bbox": [ - 375.86, - 468.23, - 498.77, - 490.15 - ], - "text": "Figure B.8 Phasor addition of\nsinusoids.", - "type": "text" - }, - { - "block_id": "p39-b34", - "global_id": 770, - "bbox": [ - 146.84, - 506.74, - 301.26, - 517.12 - ], - "text": "Alternately, we note that a −jb = 1 + j", - "type": "text" - }, - { - "block_id": "p39-b35", - "global_id": 771, - "bbox": [ - 301.26, - 498.37, - 309.69, - 508.33 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p39-b36", - "global_id": 772, - "bbox": [ - 128.9, - 505.52, - 502.75, - 541.03 - ], - "text": "3 = 2ejπ/3. Hence, C = 2 and θ = π/3.\nObserve that a phase shift of ±π amounts to multiplication by −1. Therefore, x(t) can\nalso be expressed alternatively as", - "type": "text" - }, - { - "block_id": "p39-b37", - "global_id": 773, - "bbox": [ - 159.8, - 550.59, - 471.87, - 563.46 - ], - "text": "x(t) = −2cos(ω0t + 60◦± 180◦) = −2cos(ω0t −120◦) = −2cos(ω0t + 240◦)", - "type": "text" - }, - { - "block_id": "p39-b38", - "global_id": 774, - "bbox": [ - 128.91, - 572.75, - 363.95, - 584.34 - ], - "text": "In practice, the principal value, that is, −120◦, is preferred.", - "type": "text" - }, - { - "block_id": "p39-b39", - "global_id": 775, - "bbox": [ - 146.84, - 585.93, - 381.43, - 596.3 - ], - "text": "(b) In this case, a = −3 and b = 4. Using Eq. (B.17) yields", - "type": "text" - }, - { - "block_id": "p39-b40", - "global_id": 776, - "bbox": [ - 202.08, - 609.07, - 218.77, - 619.35 - ], - "text": "C =", - "type": "text" - }, - { - "block_id": "p39-b42", - "global_id": 777, - "bbox": [ - 229.07, - 601.06, - 378.33, - 619.45 - ], - "text": "(−3)2 + 42 = 5\nand\nθ = tan−1 −4", - "type": "text" - }, - { - "block_id": "p39-b43", - "global_id": 778, - "bbox": [ - 369.4, - 615.0, - 378.33, - 622.27 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p39-b45", - "global_id": 779, - "bbox": [ - 385.59, - 607.85, - 429.07, - 619.45 - ], - "text": "= −126.9◦", - "type": "text" - } - ] - }, - { - "page_num": 40, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p40-b0", - "global_id": 780, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "20\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p40-b1", - "global_id": 781, - "bbox": [ - 103.16, - 86.24, - 153.64, - 96.2 - ], - "text": "Observe that", - "type": "text" - }, - { - "block_id": "p40-b2", - "global_id": 782, - "bbox": [ - 228.03, - 89.77, - 264.87, - 108.16 - ], - "text": "tan−1 −4", - "type": "text" - }, - { - "block_id": "p40-b3", - "global_id": 783, - "bbox": [ - 255.95, - 103.71, - 264.87, - 110.97 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p40-b4", - "global_id": 784, - "bbox": [ - 266.07, - 89.77, - 270.09, - 99.73 - ], - "text": "̸", - "type": "text" - }, - { - "block_id": "p40-b5", - "global_id": 785, - "bbox": [ - 272.43, - 89.77, - 313.65, - 108.15 - ], - "text": "= tan−1 4", - "type": "text" - }, - { - "block_id": "p40-b6", - "global_id": 786, - "bbox": [ - 310.16, - 89.77, - 318.86, - 110.97 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p40-b7", - "global_id": 787, - "bbox": [ - 320.91, - 96.55, - 351.64, - 108.15 - ], - "text": "= 53.1◦", - "type": "text" - }, - { - "block_id": "p40-b8", - "global_id": 788, - "bbox": [ - 103.16, - 117.12, - 144.91, - 127.08 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p40-b9", - "global_id": 789, - "bbox": [ - 237.42, - 126.94, - 342.76, - 139.82 - ], - "text": "x(t) = 5cos(ω0t −126.9◦)", - "type": "text" - }, - { - "block_id": "p40-b10", - "global_id": 790, - "bbox": [ - 103.16, - 147.59, - 476.98, - 169.92 - ], - "text": "This result is readily verified in the phasor diagram in Fig. B.8b. Alternately, a−jb = −3−j4 =\n5e−j126.9◦, a fact readily confirmed using MATLAB.", - "type": "text" - }, - { - "block_id": "p40-b11", - "global_id": 791, - "bbox": [ - 103.16, - 180.17, - 265.3, - 226.0 - ], - "text": ">>\nC = abs(-3+4j)\nC = 5\n>>\ntheta = angle(-3+4j)*180/pi\ntheta = 126.8699", - "type": "text" - }, - { - "block_id": "p40-b12", - "global_id": 792, - "bbox": [ - 103.16, - 234.04, - 245.52, - 245.64 - ], - "text": "Hence, C = 5 and θ = −126.8699◦.", - "type": "text" - }, - { - "block_id": "p40-b13", - "global_id": 793, - "bbox": [ - 101.85, - 292.69, - 490.38, - 315.79 - ], - "text": "We can also perform the reverse operation, expressing Ccos(ω0t +θ) in terms of cos ω0t and\nsin ω0t by again using the trigonometric identity", - "type": "text" - }, - { - "block_id": "p40-b14", - "global_id": 794, - "bbox": [ - 197.75, - 326.1, - 394.31, - 337.25 - ], - "text": "Ccos(ω0t + θ) = Ccos θ cos ω0t −Csin θ sin ω0t", - "type": "text" - }, - { - "block_id": "p40-b15", - "global_id": 795, - "bbox": [ - 101.85, - 347.97, - 154.1, - 357.93 - ], - "text": "For example,", - "type": "text" - }, - { - "block_id": "p40-b16", - "global_id": 796, - "bbox": [ - 208.24, - 351.64, - 350.59, - 371.68 - ], - "text": "10cos(ω0t −60◦) = 5cos ω0t + 5\n√", - "type": "text" - }, - { - "block_id": "p40-b17", - "global_id": 797, - "bbox": [ - 350.6, - 360.53, - 383.81, - 371.68 - ], - "text": "3sin ω0t", - "type": "text" - }, - { - "block_id": "p40-b18", - "global_id": 798, - "bbox": [ - 101.84, - 395.43, - 327.47, - 407.38 - ], - "text": "B.2-2 Sinusoids in Terms of Exponentials", - "type": "text" - }, - { - "block_id": "p40-b19", - "global_id": 799, - "bbox": [ - 101.84, - 409.48, - 490.39, - 447.39 - ], - "text": "From Eq. (B.3), we know that ejϕ = cos ϕ + jsin ϕ and e−jϕ = cos ϕ −jsin ϕ. Adding these two\nexpressions and dividing by 2 provide an expression for cosine in terms of complex exponentials,\nwhile subtracting and scaling by 2j provide an expression for sine. That is,", - "type": "text" - }, - { - "block_id": "p40-b20", - "global_id": 800, - "bbox": [ - 181.06, - 456.87, - 220.69, - 473.81 - ], - "text": "cos ϕ = 1", - "type": "text" - }, - { - "block_id": "p40-b21", - "global_id": 801, - "bbox": [ - 215.71, - 456.87, - 362.6, - 480.89 - ], - "text": "2(ejϕ + e−jϕ)\nand\nsin ϕ = 1", - "type": "text" - }, - { - "block_id": "p40-b22", - "global_id": 802, - "bbox": [ - 356.23, - 459.32, - 490.38, - 480.89 - ], - "text": "2j(ejϕ −e−jϕ)\n(B.18)", - "type": "text" - }, - { - "block_id": "p40-b23", - "global_id": 803, - "bbox": [ - 102.2, - 504.5, - 268.93, - 518.44 - ], - "text": "B.3 SKETCHING SIGNALS", - "type": "text" - }, - { - "block_id": "p40-b24", - "global_id": 804, - "bbox": [ - 101.84, - 524.43, - 462.97, - 534.39 - ], - "text": "In this section, we discuss the sketching of a few useful signals, starting with exponentials.", - "type": "text" - }, - { - "block_id": "p40-b25", - "global_id": 805, - "bbox": [ - 101.84, - 558.91, - 268.04, - 570.87 - ], - "text": "B.3-1 Monotonic Exponentials", - "type": "text" - }, - { - "block_id": "p40-b26", - "global_id": 806, - "bbox": [ - 101.84, - 573.38, - 490.39, - 598.92 - ], - "text": "The signal e−at decays monotonically, and the signal eat grows monotonically with t (assuming\na > 0), as depicted in Fig. B.9. For the sake of simplicity, we shall consider an exponential e−at", - "type": "text" - }, - { - "block_id": "p40-b27", - "global_id": 807, - "bbox": [ - 101.84, - 600.5, - 262.24, - 610.88 - ], - "text": "starting at t = 0, as shown in Fig. B.10a.", - "type": "text" - }, - { - "block_id": "p40-b28", - "global_id": 808, - "bbox": [ - 101.84, - 609.25, - 490.41, - 634.79 - ], - "text": "The signal e−at has a unit value at t = 0. At t = 1/a, the value drops to 1/e (about 37% of its\ninitial value), as illustrated in Fig. B.10a. This time interval over which the exponential reduces by", - "type": "text" - } - ] - }, - { - "page_num": 41, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p41-b0", - "global_id": 809, - "bbox": [ - 382.55, - 62.89, - 516.14, - 71.98 - ], - "text": "B.3\nSketching Signals\n21", - "type": "text" - }, - { - "block_id": "p41-b1", - "global_id": 810, - "bbox": [ - 204.5, - 128.21, - 239.2, - 145.23 - ], - "text": "1\neat", - "type": "text" - }, - { - "block_id": "p41-b2", - "global_id": 811, - "bbox": [ - 204.3, - 162.71, - 263.18, - 171.32 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p41-b3", - "global_id": 812, - "bbox": [ - 368.88, - 128.21, - 372.88, - 136.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p41-b4", - "global_id": 813, - "bbox": [ - 391.27, - 104.87, - 399.97, - 114.32 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p41-b5", - "global_id": 814, - "bbox": [ - 367.88, - 162.71, - 409.57, - 171.32 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p41-b6", - "global_id": 815, - "bbox": [ - 211.25, - 182.74, - 366.89, - 190.74 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p41-b7", - "global_id": 816, - "bbox": [ - 151.5, - 197.44, - 287.59, - 206.68 - ], - "text": "Figure B.9 Monotonic exponentials.", - "type": "text" - }, - { - "block_id": "p41-b8", - "global_id": 817, - "bbox": [ - 216.81, - 328.96, - 384.45, - 336.96 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p41-b9", - "global_id": 818, - "bbox": [ - 322.32, - 234.69, - 326.32, - 242.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p41-b10", - "global_id": 819, - "bbox": [ - 321.68, - 306.75, - 433.9, - 316.59 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p41-b11", - "global_id": 820, - "bbox": [ - 341.5, - 244.0, - 368.27, - 253.7 - ], - "text": "e2t u(t)", - "type": "text" - }, - { - "block_id": "p41-b12", - "global_id": 821, - "bbox": [ - 363.58, - 268.5, - 377.58, - 276.5 - ], - "text": "0.37", - "type": "text" - }, - { - "block_id": "p41-b13", - "global_id": 822, - "bbox": [ - 401.08, - 285.0, - 419.08, - 293.0 - ], - "text": "0.135", - "type": "text" - }, - { - "block_id": "p41-b14", - "global_id": 823, - "bbox": [ - 360.88, - 308.74, - 406.88, - 316.74 - ], - "text": "0.5\n1", - "type": "text" - }, - { - "block_id": "p41-b15", - "global_id": 824, - "bbox": [ - 164.1, - 234.69, - 168.1, - 242.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p41-b16", - "global_id": 825, - "bbox": [ - 163.3, - 306.75, - 275.52, - 316.91 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p41-b17", - "global_id": 826, - "bbox": [ - 182.66, - 242.99, - 209.43, - 252.68 - ], - "text": "eat u(t)", - "type": "text" - }, - { - "block_id": "p41-b18", - "global_id": 827, - "bbox": [ - 246.87, - 276.6, - 283.29, - 294.88 - ], - "text": "1\ne2 0.135", - "type": "text" - }, - { - "block_id": "p41-b19", - "global_id": 828, - "bbox": [ - 208.85, - 262.94, - 238.28, - 277.6 - ], - "text": "1\ne 0.37", - "type": "text" - }, - { - "block_id": "p41-b20", - "global_id": 829, - "bbox": [ - 205.5, - 308.62, - 209.5, - 323.36 - ], - "text": "1\na", - "type": "text" - }, - { - "block_id": "p41-b21", - "global_id": 830, - "bbox": [ - 244.5, - 308.71, - 248.5, - 323.27 - ], - "text": "2\na", - "type": "text" - }, - { - "block_id": "p41-b22", - "global_id": 831, - "bbox": [ - 151.5, - 341.79, - 316.02, - 354.29 - ], - "text": "Figure B.10 Sketching (a) e−at and (b) e−2t.", - "type": "text" - }, - { - "block_id": "p41-b23", - "global_id": 832, - "bbox": [ - 127.59, - 380.55, - 516.14, - 427.97 - ], - "text": "a factor e (i.e., drops to about 37% of its value) is known as the time constant of the exponential.\nTherefore, the time constant of e−at is 1/a. Observe that the exponential is reduced to 37% of its\ninitial value over any time interval of duration 1/a. This can be shown by considering any set of\ninstants t1 and t2 separated by one time constant so that", - "type": "text" - }, - { - "block_id": "p41-b24", - "global_id": 833, - "bbox": [ - 300.05, - 442.27, - 342.47, - 460.3 - ], - "text": "t2 −t1 = 1", - "type": "text" - }, - { - "block_id": "p41-b25", - "global_id": 834, - "bbox": [ - 337.49, - 456.23, - 342.47, - 466.19 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p41-b26", - "global_id": 835, - "bbox": [ - 127.59, - 478.83, - 292.5, - 490.9 - ], - "text": "Now the ratio of e−at2 to e−at1 is given by", - "type": "text" - }, - { - "block_id": "p41-b27", - "global_id": 836, - "bbox": [ - 264.35, - 505.95, - 282.61, - 517.44 - ], - "text": "e−at2", - "type": "text" - }, - { - "block_id": "p41-b28", - "global_id": 837, - "bbox": [ - 264.35, - 507.58, - 350.07, - 531.5 - ], - "text": "e−at1 = e−a(t2−t1) = 1", - "type": "text" - }, - { - "block_id": "p41-b29", - "global_id": 838, - "bbox": [ - 345.37, - 514.15, - 380.56, - 531.5 - ], - "text": "e ≈0.37", - "type": "text" - }, - { - "block_id": "p41-b30", - "global_id": 839, - "bbox": [ - 127.59, - 545.36, - 430.96, - 555.33 - ], - "text": "We can use this fact to sketch an exponential quickly. For example, consider", - "type": "text" - }, - { - "block_id": "p41-b31", - "global_id": 840, - "bbox": [ - 300.56, - 571.0, - 342.54, - 583.0 - ], - "text": "x(t) = e−2t", - "type": "text" - }, - { - "block_id": "p41-b32", - "global_id": 841, - "bbox": [ - 127.59, - 600.5, - 516.13, - 634.79 - ], - "text": "The time constant in this case is 0.5. The value of x(t) at t = 0 is 1. At t = 0.5 (one time constant),\nit is 1/e (about 0.37). The value of x(t) continues to drop further by the factor 1/e (37%) over\nthe next half-second interval (one time constant). Thus, x(t) at t = 1 is (1/e)2. Continuing in this", - "type": "text" - } - ] - }, - { - "page_num": 42, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p42-b0", - "global_id": 842, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "22\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p42-b1", - "global_id": 843, - "bbox": [ - 101.84, - 83.78, - 490.38, - 109.31 - ], - "text": "manner, we see that x(t) = (1/e)3 at t = 1.5, and so on. A knowledge of the values of x(t) at t = 0,\n0.5, 1, and 1.5 allows us to sketch the desired signal, as shown in Fig. B.10b.†", - "type": "text" - }, - { - "block_id": "p42-b2", - "global_id": 844, - "bbox": [ - 101.84, - 109.89, - 490.38, - 133.23 - ], - "text": "For a monotonically growing exponential eat, the waveform increases by a factor e over each\ninterval of 1/a seconds.", - "type": "text" - }, - { - "block_id": "p42-b3", - "global_id": 845, - "bbox": [ - 101.84, - 158.06, - 333.13, - 170.02 - ], - "text": "B.3-2 The Exponentially Varying Sinusoid", - "type": "text" - }, - { - "block_id": "p42-b4", - "global_id": 846, - "bbox": [ - 101.84, - 176.15, - 342.99, - 186.11 - ], - "text": "We now discuss sketching an exponentially varying sinusoid", - "type": "text" - }, - { - "block_id": "p42-b5", - "global_id": 847, - "bbox": [ - 245.45, - 194.03, - 346.78, - 208.88 - ], - "text": "x(t) = Ae−at cos(ω0t + θ)", - "type": "text" - }, - { - "block_id": "p42-b6", - "global_id": 848, - "bbox": [ - 101.84, - 220.14, - 241.74, - 230.1 - ], - "text": "Let us consider a specific example:", - "type": "text" - }, - { - "block_id": "p42-b7", - "global_id": 849, - "bbox": [ - 244.49, - 238.02, - 347.75, - 252.1 - ], - "text": "x(t) = 4e−2t cos(6t −60◦)", - "type": "text" - }, - { - "block_id": "p42-b8", - "global_id": 850, - "bbox": [ - 101.85, - 260.52, - 402.21, - 274.1 - ], - "text": "We shall sketch 4e−2t and cos(6t −60◦) separately and then multiply them:", - "type": "text" - }, - { - "block_id": "p42-b9", - "global_id": 851, - "bbox": [ - 101.84, - 280.43, - 490.4, - 292.03 - ], - "text": "(a) Sketching 4e−2t. This monotonically decaying exponential has a time constant of 0.5 second", - "type": "text" - }, - { - "block_id": "p42-b10", - "global_id": 852, - "bbox": [ - 101.84, - 292.67, - 490.39, - 339.85 - ], - "text": "and an initial value of 4 at t = 0. Therefore, its values at t = 0.5, 1, 1.5, and 2 are 4/e, 4/e2,\n4/e3, and 4/e4, or about 1.47, 0.54, 0.2, and 0.07, respectively. Using these values as a guide,\nwe sketch 4e−2t, as illustrated in Fig. B.11a.\n(b) Sketching cos(6t −60◦). The procedure for sketching cos(6t −60◦) is discussed in Sec. B.2", - "type": "text" - }, - { - "block_id": "p42-b11", - "global_id": 853, - "bbox": [ - 102.4, - 341.43, - 490.38, - 387.68 - ], - "text": "(Fig. B.6c). Here, the period of the sinusoid is T0 = 2π/6 ≈1, and there is a phase delay of\n60◦, or two-thirds of a quarter-cycle, which is equivalent to a delay of about (60/360)(1) ≈1/6\nseconds (see Fig. B.11b).\n(c) Sketching 4e−2t cos(6t −60◦). We now multiply the waveforms in steps (a) and (b). This", - "type": "text" - }, - { - "block_id": "p42-b12", - "global_id": 854, - "bbox": [ - 118.44, - 388.02, - 490.4, - 435.49 - ], - "text": "multiplication amounts to forcing the sinusoid 4 cos(6t −60◦) to decrease exponentially with\na time constant of 0.5. The initial amplitude (at t = 0) is 4, decreasing to 4/e (=1.47) at\nt = 0.5, to 1.47/e(=0.54) at t = 1, and so on. This is depicted in Fig. B.11c. Note that when\ncos(6t −60◦) has a value of unity (peak amplitude),", - "type": "text" - }, - { - "block_id": "p42-b13", - "global_id": 855, - "bbox": [ - 249.75, - 443.41, - 358.44, - 457.49 - ], - "text": "4e−2t cos(6t −60◦) = 4e−2t", - "type": "text" - }, - { - "block_id": "p42-b14", - "global_id": 856, - "bbox": [ - 118.44, - 465.91, - 490.39, - 551.21 - ], - "text": "Therefore, 4e−2t cos(6t−60◦) touches 4e−2t at the instants at which the sinusoid cos(6t −60◦)\nis at its positive peaks. Clearly, 4e−2t is an envelope for positive amplitudes of 4e−2t cos(6t −\n60◦). Similar argument shows that 4e−2t cos(6t −60◦) touches −4e−2t at its negative peaks.\nTherefore, −4e−2t is an envelope for negative amplitudes of 4e−2t cos(6t −60◦). Thus, to\nsketch 4e−2t cos(6t −60◦), we first draw the envelopes 4e−2t and −4e−2t (the mirror image\nof 4e−2t about the horizontal axis), and then sketch the sinusoid cos(6t −60◦), with these\nenvelopes acting as constraints on the sinusoid’s amplitude (see Fig. B.11c).", - "type": "text" - }, - { - "block_id": "p42-b15", - "global_id": 857, - "bbox": [ - 119.78, - 555.57, - 489.75, - 569.92 - ], - "text": "In general, Ke−at cos(ω0t + θ) can be sketched in this manner, with Ke−at and −Ke−at", - "type": "text" - }, - { - "block_id": "p42-b16", - "global_id": 858, - "bbox": [ - 101.84, - 570.73, - 275.41, - 581.88 - ], - "text": "constraining the amplitude of cos(ω0t + θ).", - "type": "text" - }, - { - "block_id": "p42-b17", - "global_id": 859, - "bbox": [ - 101.84, - 599.27, - 490.39, - 633.41 - ], - "text": "† If we wish to refine the sketch further, we could consider intervals of half the time constant over which\nthe signal decays by a factor 1/√e. Thus, at t = 0.25, x(t) = 1/√e, and at t = 0.75, x(t) = 1/e√e,\nand so on.", - "type": "text" - } - ] - }, - { - "page_num": 43, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p43-b0", - "global_id": 860, - "bbox": [ - 394.49, - 62.89, - 516.13, - 71.98 - ], - "text": "B.4\nCramer’s Rule\n23", - "type": "text" - }, - { - "block_id": "p43-b1", - "global_id": 861, - "bbox": [ - 281.12, - 182.11, - 290.0, - 190.11 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p43-b2", - "global_id": 862, - "bbox": [ - 280.74, - 267.94, - 290.39, - 275.94 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p43-b3", - "global_id": 863, - "bbox": [ - 281.12, - 445.74, - 290.0, - 453.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p43-b4", - "global_id": 864, - "bbox": [ - 296.39, - 243.42, - 391.56, - 251.42 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p43-b5", - "global_id": 865, - "bbox": [ - 182.59, - 211.7, - 268.98, - 223.08 - ], - "text": "1\ncos 6t", - "type": "text" - }, - { - "block_id": "p43-b6", - "global_id": 866, - "bbox": [ - 287.25, - 202.98, - 332.79, - 211.27 - ], - "text": "cos (6t 60)", - "type": "text" - }, - { - "block_id": "p43-b7", - "global_id": 867, - "bbox": [ - 364.67, - 226.98, - 366.89, - 234.98 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p43-b8", - "global_id": 868, - "bbox": [ - 183.21, - 163.99, - 408.09, - 171.99 - ], - "text": "0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p43-b9", - "global_id": 869, - "bbox": [ - 182.09, - 93.96, - 186.09, - 101.96 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p43-b10", - "global_id": 870, - "bbox": [ - 238.09, - 125.71, - 252.09, - 133.71 - ], - "text": "1.47", - "type": "text" - }, - { - "block_id": "p43-b11", - "global_id": 871, - "bbox": [ - 291.09, - 140.71, - 356.79, - 153.71 - ], - "text": "0.54\n0.2", - "type": "text" - }, - { - "block_id": "p43-b12", - "global_id": 872, - "bbox": [ - 378.3, - 168.59, - 380.53, - 176.59 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p43-b13", - "global_id": 873, - "bbox": [ - 208.08, - 102.95, - 226.14, - 112.65 - ], - "text": "4e2t", - "type": "text" - }, - { - "block_id": "p43-b14", - "global_id": 874, - "bbox": [ - 183.16, - 293.3, - 187.16, - 301.3 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p43-b15", - "global_id": 875, - "bbox": [ - 175.26, - 428.7, - 185.93, - 437.0 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p43-b16", - "global_id": 876, - "bbox": [ - 245.39, - 366.75, - 379.32, - 381.75 - ], - "text": "1\n0.5\nt", - "type": "text" - }, - { - "block_id": "p43-b17", - "global_id": 877, - "bbox": [ - 218.03, - 404.94, - 242.75, - 414.64 - ], - "text": "4e2t", - "type": "text" - }, - { - "block_id": "p43-b18", - "global_id": 878, - "bbox": [ - 222.37, - 316.84, - 312.86, - 329.13 - ], - "text": "4e2t\n4e2t cos (6t 60)", - "type": "text" - }, - { - "block_id": "p43-b19", - "global_id": 879, - "bbox": [ - 151.5, - 460.44, - 364.25, - 469.68 - ], - "text": "Figure B.11 Sketching an exponentially varying sinusoid.", - "type": "text" - }, - { - "block_id": "p43-b20", - "global_id": 880, - "bbox": [ - 127.94, - 512.07, - 262.29, - 526.02 - ], - "text": "B.4 CRAMER’S RULE", - "type": "text" - }, - { - "block_id": "p43-b21", - "global_id": 881, - "bbox": [ - 127.59, - 532.01, - 516.11, - 554.7 - ], - "text": "Cramer’s rule offers a very convenient way to solve simultaneous linear equations. Consider a set\nof n linear simultaneous equations in n unknowns x1, x2,. . ., xn:", - "type": "text" - }, - { - "block_id": "p43-b22", - "global_id": 882, - "bbox": [ - 257.59, - 580.53, - 385.63, - 603.94 - ], - "text": "a11x1 + a12x2 + · · · + a1nxn = y1\na21x1 + a22x2 + · · · + a2nxn = y2", - "type": "text" - }, - { - "block_id": "p43-b23", - "global_id": 883, - "bbox": [ - 257.59, - 603.47, - 385.63, - 634.44 - ], - "text": "...\nan1x1 + an2x2 + · · · + annxn = yn", - "type": "text" - }, - { - "block_id": "p43-b24", - "global_id": 884, - "bbox": [ - 490.4, - 602.26, - 516.12, - 612.23 - ], - "text": "(B.19)", - "type": "text" - } - ] - }, - { - "page_num": 44, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p44-b0", - "global_id": 885, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "24\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p44-b1", - "global_id": 886, - "bbox": [ - 101.84, - 85.82, - 307.52, - 95.78 - ], - "text": "These equations can be expressed in matrix form as", - "type": "text" - }, - { - "block_id": "p44-b2", - "global_id": 887, - "bbox": [ - 204.33, - 101.87, - 211.59, - 111.83 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p44-b3", - "global_id": 888, - "bbox": [ - 204.33, - 119.34, - 211.59, - 147.7 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p44-b4", - "global_id": 889, - "bbox": [ - 216.58, - 109.48, - 296.38, - 163.08 - ], - "text": "a11\na12\n· · ·\na1n\na21\na22\n· · ·\na2n\n...\n...\n· · ·\n...\nan1\nan2\n· · ·\nann", - "type": "text" - }, - { - "block_id": "p44-b5", - "global_id": 890, - "bbox": [ - 301.87, - 101.87, - 309.13, - 111.83 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p44-b6", - "global_id": 891, - "bbox": [ - 301.87, - 119.34, - 309.13, - 147.7 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p44-b7", - "global_id": 892, - "bbox": [ - 310.24, - 101.87, - 317.5, - 111.83 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p44-b8", - "global_id": 893, - "bbox": [ - 310.24, - 119.34, - 317.5, - 147.7 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p44-b9", - "global_id": 894, - "bbox": [ - 322.49, - 109.79, - 330.39, - 132.58 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p44-b10", - "global_id": 895, - "bbox": [ - 322.49, - 132.42, - 330.39, - 163.01 - ], - "text": "...\nxn", - "type": "text" - }, - { - "block_id": "p44-b11", - "global_id": 896, - "bbox": [ - 335.88, - 101.87, - 343.14, - 111.83 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p44-b12", - "global_id": 897, - "bbox": [ - 335.88, - 119.34, - 352.96, - 147.7 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p44-b13", - "global_id": 898, - "bbox": [ - 355.01, - 101.87, - 362.27, - 111.83 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p44-b14", - "global_id": 899, - "bbox": [ - 355.01, - 119.34, - 362.27, - 147.7 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p44-b15", - "global_id": 900, - "bbox": [ - 367.25, - 109.79, - 375.16, - 132.58 - ], - "text": "y1\ny2", - "type": "text" - }, - { - "block_id": "p44-b16", - "global_id": 901, - "bbox": [ - 367.25, - 132.42, - 375.16, - 163.01 - ], - "text": "...\nyn", - "type": "text" - }, - { - "block_id": "p44-b17", - "global_id": 902, - "bbox": [ - 380.64, - 101.87, - 387.9, - 111.83 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p44-b18", - "global_id": 903, - "bbox": [ - 380.64, - 119.34, - 490.38, - 147.7 - ], - "text": "⎥⎥⎥⎦\n(B.20)", - "type": "text" - }, - { - "block_id": "p44-b19", - "global_id": 904, - "bbox": [ - 101.84, - 176.29, - 490.4, - 210.26 - ], - "text": "We denote the matrix on the left-hand side formed by the elements aij as A. The determinant of\nA is denoted by |A|. If the determinant |A| is not zero, Eq. (B.19) has a unique solution given by\nCramer’s formula", - "type": "text" - }, - { - "block_id": "p44-b20", - "global_id": 905, - "bbox": [ - 238.4, - 211.78, - 276.21, - 230.22 - ], - "text": "xk = |Dk|", - "type": "text" - }, - { - "block_id": "p44-b21", - "global_id": 906, - "bbox": [ - 261.46, - 218.76, - 490.38, - 236.13 - ], - "text": "|A|\nk = 1,2,. . .,n\n(B.21)", - "type": "text" - }, - { - "block_id": "p44-b22", - "global_id": 907, - "bbox": [ - 101.84, - 245.2, - 490.38, - 268.31 - ], - "text": "where |Dk| is obtained by replacing the kth column of |A| by the column on the right-hand side of\nEq. (B.20) (with elements y1, y2,. . ., yn).", - "type": "text" - }, - { - "block_id": "p44-b23", - "global_id": 908, - "bbox": [ - 119.78, - 269.52, - 352.07, - 279.49 - ], - "text": "We shall demonstrate the use of this rule with an example.", - "type": "text" - }, - { - "block_id": "p44-b24", - "global_id": 909, - "bbox": [ - 76.77, - 309.77, - 461.86, - 321.73 - ], - "text": "EXAMPLE B.7\nUsing Cramer’s Rule to Solve a System of Equations", - "type": "text" - }, - { - "block_id": "p44-b25", - "global_id": 910, - "bbox": [ - 103.16, - 338.39, - 464.45, - 348.36 - ], - "text": "Use Cramer’s rule to solve the following simultaneous linear equations in three unknowns:", - "type": "text" - }, - { - "block_id": "p44-b26", - "global_id": 911, - "bbox": [ - 255.7, - 359.9, - 324.49, - 371.35 - ], - "text": "2x1 + x2 + x3 = 3", - "type": "text" - }, - { - "block_id": "p44-b27", - "global_id": 912, - "bbox": [ - 255.7, - 374.84, - 324.49, - 386.3 - ], - "text": "x1 + 3x2 −x3 = 7", - "type": "text" - }, - { - "block_id": "p44-b28", - "global_id": 913, - "bbox": [ - 260.67, - 389.78, - 324.49, - 401.24 - ], - "text": "x1 + x2 + x3 = 1", - "type": "text" - }, - { - "block_id": "p44-b29", - "global_id": 914, - "bbox": [ - 103.16, - 433.04, - 308.59, - 443.0 - ], - "text": "In matrix form, these equations can be expressed as", - "type": "text" - }, - { - "block_id": "p44-b30", - "global_id": 915, - "bbox": [ - 218.86, - 446.2, - 226.12, - 456.16 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p44-b31", - "global_id": 916, - "bbox": [ - 218.86, - 464.13, - 226.12, - 474.1 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p44-b32", - "global_id": 917, - "bbox": [ - 231.1, - 454.52, - 273.74, - 488.4 - ], - "text": "2\n1\n1\n1\n3\n−1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p44-b33", - "global_id": 918, - "bbox": [ - 278.72, - 446.2, - 285.98, - 456.16 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p44-b34", - "global_id": 919, - "bbox": [ - 278.72, - 464.13, - 285.98, - 474.1 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p44-b35", - "global_id": 920, - "bbox": [ - 287.09, - 446.2, - 294.35, - 456.16 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p44-b36", - "global_id": 921, - "bbox": [ - 287.09, - 464.13, - 294.35, - 474.1 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p44-b37", - "global_id": 922, - "bbox": [ - 299.34, - 454.42, - 307.24, - 489.17 - ], - "text": "x1\nx2\nx3", - "type": "text" - }, - { - "block_id": "p44-b38", - "global_id": 923, - "bbox": [ - 312.73, - 446.2, - 319.99, - 456.16 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p44-b39", - "global_id": 924, - "bbox": [ - 312.73, - 464.13, - 329.81, - 476.13 - ], - "text": "⎦=", - "type": "text" - }, - { - "block_id": "p44-b40", - "global_id": 925, - "bbox": [ - 331.85, - 446.2, - 339.12, - 456.16 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p44-b41", - "global_id": 926, - "bbox": [ - 331.85, - 464.13, - 339.12, - 474.1 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p44-b42", - "global_id": 927, - "bbox": [ - 344.09, - 454.52, - 349.08, - 488.4 - ], - "text": "3\n7\n1", - "type": "text" - }, - { - "block_id": "p44-b43", - "global_id": 928, - "bbox": [ - 354.06, - 446.2, - 361.32, - 456.16 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p44-b44", - "global_id": 929, - "bbox": [ - 354.06, - 464.13, - 361.32, - 474.1 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p44-b45", - "global_id": 930, - "bbox": [ - 103.17, - 499.49, - 125.01, - 509.45 - ], - "text": "Here,", - "type": "text" - }, - { - "block_id": "p44-b46", - "global_id": 931, - "bbox": [ - 239.75, - 521.93, - 262.46, - 532.23 - ], - "text": "|A| =", - "type": "text" - }, - { - "block_id": "p44-b48", - "global_id": 932, - "bbox": [ - 272.72, - 510.28, - 315.36, - 544.16 - ], - "text": "2\n1\n1\n1\n3\n−1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p44-b50", - "global_id": 933, - "bbox": [ - 325.64, - 521.93, - 340.44, - 532.3 - ], - "text": "= 4", - "type": "text" - }, - { - "block_id": "p44-b51", - "global_id": 934, - "bbox": [ - 103.17, - 552.25, - 477.02, - 574.59 - ], - "text": "Since |A| = 4̸ = 0, a unique solution exists for x1, x2, and x3. This solution is provided by\nCramer’s rule [Eq. (B.21)] as follows:", - "type": "text" - }, - { - "block_id": "p44-b52", - "global_id": 935, - "bbox": [ - 224.18, - 591.18, - 254.58, - 609.2 - ], - "text": "x1 = 1", - "type": "text" - }, - { - "block_id": "p44-b53", - "global_id": 936, - "bbox": [ - 245.64, - 604.82, - 258.54, - 615.12 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p44-b55", - "global_id": 937, - "bbox": [ - 269.06, - 586.11, - 311.7, - 619.98 - ], - "text": "3\n1\n1\n7\n3\n−1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p44-b57", - "global_id": 938, - "bbox": [ - 321.96, - 591.17, - 337.96, - 608.12 - ], - "text": "= 8", - "type": "text" - }, - { - "block_id": "p44-b58", - "global_id": 939, - "bbox": [ - 332.98, - 597.75, - 356.01, - 615.2 - ], - "text": "4 = 2", - "type": "text" - } - ] - }, - { - "page_num": 45, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p45-b0", - "global_id": 940, - "bbox": [ - 347.22, - 62.89, - 516.14, - 71.98 - ], - "text": "B.5\nPartial Fraction Expansion\n25", - "type": "text" - }, - { - "block_id": "p45-b1", - "global_id": 941, - "bbox": [ - 249.92, - 92.89, - 280.32, - 110.92 - ], - "text": "x2 = 1", - "type": "text" - }, - { - "block_id": "p45-b2", - "global_id": 942, - "bbox": [ - 271.39, - 106.54, - 284.28, - 116.83 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p45-b4", - "global_id": 943, - "bbox": [ - 294.8, - 87.82, - 337.44, - 121.7 - ], - "text": "2\n3\n1\n1\n7\n−1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p45-b6", - "global_id": 944, - "bbox": [ - 347.71, - 92.89, - 363.71, - 109.83 - ], - "text": "= 4", - "type": "text" - }, - { - "block_id": "p45-b7", - "global_id": 945, - "bbox": [ - 358.73, - 99.46, - 381.75, - 116.91 - ], - "text": "4 = 1", - "type": "text" - }, - { - "block_id": "p45-b8", - "global_id": 946, - "bbox": [ - 246.03, - 136.73, - 276.44, - 154.76 - ], - "text": "x3 = 1", - "type": "text" - }, - { - "block_id": "p45-b9", - "global_id": 947, - "bbox": [ - 267.5, - 150.37, - 280.4, - 160.67 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p45-b11", - "global_id": 948, - "bbox": [ - 290.91, - 131.66, - 325.78, - 165.53 - ], - "text": "2\n1\n3\n1\n3\n7\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p45-b13", - "global_id": 949, - "bbox": [ - 336.06, - 136.31, - 359.81, - 153.26 - ], - "text": "= −8", - "type": "text" - }, - { - "block_id": "p45-b14", - "global_id": 950, - "bbox": [ - 350.95, - 143.3, - 385.63, - 160.75 - ], - "text": "4 = −2", - "type": "text" - }, - { - "block_id": "p45-b15", - "global_id": 951, - "bbox": [ - 128.9, - 173.91, - 502.76, - 207.78 - ], - "text": "MATLAB is well suited to compute Cramer’s formula, so these results are easy to verify. To\nprovide an example, let us verify that x1 = 2 using MATLAB’s det command to compute the\nneeded matrix determinants.", - "type": "text" - }, - { - "block_id": "p45-b16", - "global_id": 952, - "bbox": [ - 128.91, - 218.04, - 442.72, - 239.95 - ], - "text": ">>\nx1 = det([3 1 1;7 3 -1;1 1 1])/det([2 1 1;1 3 -1;1 1 1])\nx1 = 2.0000", - "type": "text" - }, - { - "block_id": "p45-b17", - "global_id": 953, - "bbox": [ - 127.94, - 293.49, - 361.79, - 307.44 - ], - "text": "B.5 PARTIAL FRACTION EXPANSION", - "type": "text" - }, - { - "block_id": "p45-b18", - "global_id": 954, - "bbox": [ - 127.59, - 313.43, - 516.15, - 347.3 - ], - "text": "In the analysis of linear time-invariant systems, we encounter functions that are ratios of two\npolynomials in a certain variable, say, x. Such functions are known as rational functions. A rational\nfunction F(x) can be expressed as", - "type": "text" - }, - { - "block_id": "p45-b19", - "global_id": 955, - "bbox": [ - 220.01, - 360.24, - 388.67, - 381.12 - ], - "text": "F(x) = bmxm + bm−1xm−1 + · · · + b1x + b0", - "type": "text" - }, - { - "block_id": "p45-b20", - "global_id": 956, - "bbox": [ - 258.97, - 375.04, - 381.09, - 389.07 - ], - "text": "xn + an−1xn−1 + · · · + a1x + a0", - "type": "text" - }, - { - "block_id": "p45-b21", - "global_id": 957, - "bbox": [ - 392.42, - 363.86, - 421.95, - 381.12 - ], - "text": "= P(x)", - "type": "text" - }, - { - "block_id": "p45-b22", - "global_id": 958, - "bbox": [ - 403.43, - 371.26, - 516.12, - 388.2 - ], - "text": "Q(x)\n(B.22)", - "type": "text" - }, - { - "block_id": "p45-b23", - "global_id": 959, - "bbox": [ - 127.59, - 400.83, - 516.14, - 438.32 - ], - "text": "The function F(x) is improper if m ≥n and proper if m < n.† An improper function can always\nbe separated into the sum of a polynomial in x and a proper function. Consider, for example, the\nfunction", - "type": "text" - }, - { - "block_id": "p45-b24", - "global_id": 960, - "bbox": [ - 266.15, - 441.97, - 376.36, - 462.95 - ], - "text": "F(x) = 2x3 + 9x2 + 11x + 2", - "type": "text" - }, - { - "block_id": "p45-b25", - "global_id": 961, - "bbox": [ - 314.65, - 456.77, - 359.23, - 470.03 - ], - "text": "x2 + 4x + 3", - "type": "text" - }, - { - "block_id": "p45-b26", - "global_id": 962, - "bbox": [ - 127.59, - 481.1, - 516.14, - 503.02 - ], - "text": "Because this is an improper function, we divide the numerator by the denominator until the\nremainder has a lower degree than the denominator.", - "type": "text" - }, - { - "block_id": "p45-b27", - "global_id": 963, - "bbox": [ - 247.74, - 519.59, - 332.56, - 542.42 - ], - "text": "2x + 1\nx2 + 4x + 3", - "type": "text" - }, - { - "block_id": "p45-b29", - "global_id": 964, - "bbox": [ - 307.28, - 530.42, - 386.11, - 582.26 - ], - "text": "2x3 + 9x2 + 11x + 2\n2x3 + 8x2 + 6x\nx2 + 5x + 2\nx2 + 4x + 3", - "type": "text" - }, - { - "block_id": "p45-b30", - "global_id": 965, - "bbox": [ - 359.09, - 585.84, - 379.39, - 596.21 - ], - "text": "x −1", - "type": "text" - }, - { - "block_id": "p45-b31", - "global_id": 966, - "bbox": [ - 127.59, - 621.19, - 470.72, - 633.41 - ], - "text": "† Some sources classify F(x) as strictly proper if m < n, proper if m ≤n, and improper if m > n.", - "type": "text" - } - ] - }, - { - "page_num": 46, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p46-b0", - "global_id": 967, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "26\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p46-b1", - "global_id": 968, - "bbox": [ - 101.84, - 85.46, - 245.03, - 95.84 - ], - "text": "Therefore, F(x) can be expressed as", - "type": "text" - }, - { - "block_id": "p46-b2", - "global_id": 969, - "bbox": [ - 183.34, - 106.01, - 293.54, - 126.98 - ], - "text": "F(x) = 2x3 + 9x2 + 11x + 2", - "type": "text" - }, - { - "block_id": "p46-b3", - "global_id": 970, - "bbox": [ - 231.84, - 116.6, - 351.9, - 140.61 - ], - "text": "x2 + 4x + 3\n=\n2x + 1\n \npolynomial in x", - "type": "text" - }, - { - "block_id": "p46-b4", - "global_id": 971, - "bbox": [ - 353.02, - 109.62, - 408.9, - 147.68 - ], - "text": "+\nx −1\nx2 + 4x + 3\n\n\n\nproper function", - "type": "text" - }, - { - "block_id": "p46-b5", - "global_id": 972, - "bbox": [ - 101.84, - 161.48, - 490.4, - 183.39 - ], - "text": "A proper function can be further expanded into partial fractions. The remaining discussion in this\nsection is concerned with various ways of doing this.", - "type": "text" - }, - { - "block_id": "p46-b6", - "global_id": 973, - "bbox": [ - 101.84, - 205.84, - 294.62, - 217.8 - ], - "text": "B.5-1 Method of Clearing Fractions", - "type": "text" - }, - { - "block_id": "p46-b7", - "global_id": 974, - "bbox": [ - 101.84, - 223.93, - 490.38, - 257.81 - ], - "text": "A rational function can be written as a sum of appropriate partial fractions with unknown\ncoefficients, which are determined by clearing fractions and equating the coefficients of similar\npowers on the two sides. This procedure is demonstrated by the following example.", - "type": "text" - }, - { - "block_id": "p46-b8", - "global_id": 975, - "bbox": [ - 76.77, - 283.16, - 337.38, - 295.11 - ], - "text": "EXAMPLE B.8\nMethod of Clearing Fractions", - "type": "text" - }, - { - "block_id": "p46-b9", - "global_id": 976, - "bbox": [ - 103.16, - 311.36, - 364.74, - 321.73 - ], - "text": "Expand the following rational function F(x) into partial fractions:", - "type": "text" - }, - { - "block_id": "p46-b10", - "global_id": 977, - "bbox": [ - 230.21, - 329.22, - 348.26, - 357.27 - ], - "text": "F(x) =\nx3 + 3x2 + 4x + 6\n(x + 1)(x + 2)(x + 3)2", - "type": "text" - }, - { - "block_id": "p46-b11", - "global_id": 978, - "bbox": [ - 103.16, - 382.66, - 477.02, - 405.0 - ], - "text": "This function can be expressed as a sum of partial fractions with denominators (x + 1),\n(x + 2),(x + 3), and (x + 3)2, as follows:", - "type": "text" - }, - { - "block_id": "p46-b12", - "global_id": 979, - "bbox": [ - 156.87, - 412.07, - 421.61, - 440.11 - ], - "text": "F(x) =\nx3 + 3x2 + 4x + 6\n(x + 1)(x + 2)(x + 3)2 =\nk1\nx + 1 +\nk2\nx + 2 +\nk3\nx + 3 +\nk4\n(x + 3)2", - "type": "text" - }, - { - "block_id": "p46-b13", - "global_id": 980, - "bbox": [ - 103.16, - 449.52, - 477.0, - 471.55 - ], - "text": "To determine the unknowns k1,k2,k3, and k4, we clear fractions by multiplying both sides by\n(x + 1)(x + 2)(x + 3)2 to obtain", - "type": "text" - }, - { - "block_id": "p46-b14", - "global_id": 981, - "bbox": [ - 133.05, - 478.97, - 409.05, - 494.23 - ], - "text": "x3 + 3x2 + 4x + 6 = k1(x3 + 8x2 + 21x + 18) + k2(x3 + 7x2 + 15x + 9)", - "type": "text" - }, - { - "block_id": "p46-b15", - "global_id": 982, - "bbox": [ - 215.36, - 495.41, - 385.67, - 510.67 - ], - "text": "+ k3(x3 + 6x2 + 11x + 6) + k4(x2 + 3x + 2)", - "type": "text" - }, - { - "block_id": "p46-b16", - "global_id": 983, - "bbox": [ - 203.98, - 514.52, - 384.57, - 527.41 - ], - "text": "= x3(k1 + k2 + k3) + x2(8k1 + 7k2 + 6k3 + k4)", - "type": "text" - }, - { - "block_id": "p46-b17", - "global_id": 984, - "bbox": [ - 215.36, - 530.9, - 447.13, - 542.36 - ], - "text": "+ x(21k1 + 15k2 + 11k3 + 3k4) + (18k1 + 9k2 + 6k3 + 2k4)", - "type": "text" - }, - { - "block_id": "p46-b18", - "global_id": 985, - "bbox": [ - 103.17, - 553.23, - 340.98, - 563.19 - ], - "text": "Equating coefficients of similar powers on both sides yields", - "type": "text" - }, - { - "block_id": "p46-b19", - "global_id": 986, - "bbox": [ - 285.26, - 574.73, - 349.07, - 586.19 - ], - "text": "k1 + k2 + k3 = 1", - "type": "text" - }, - { - "block_id": "p46-b20", - "global_id": 987, - "bbox": [ - 231.11, - 589.68, - 349.07, - 616.08 - ], - "text": "8k1 + 7k2 + 6k3 + k4 = 3\n21k1 + 15k2 + 11k3 + 3k4 = 4", - "type": "text" - }, - { - "block_id": "p46-b21", - "global_id": 988, - "bbox": [ - 241.07, - 619.57, - 349.07, - 631.03 - ], - "text": "18k1 + 9k2 + 6k3 + 2k4 = 6", - "type": "text" - } - ] - }, - { - "page_num": 47, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p47-b0", - "global_id": 989, - "bbox": [ - 347.22, - 62.89, - 516.14, - 71.98 - ], - "text": "B.5\nPartial Fraction Expansion\n27", - "type": "text" - }, - { - "block_id": "p47-b1", - "global_id": 990, - "bbox": [ - 128.9, - 86.23, - 337.56, - 96.2 - ], - "text": "Solution of these four simultaneous equations yields", - "type": "text" - }, - { - "block_id": "p47-b2", - "global_id": 991, - "bbox": [ - 222.27, - 107.75, - 409.39, - 119.2 - ], - "text": "k1 = 1,\nk2 = −2,\nk3 = 2,\nk4 = −3", - "type": "text" - }, - { - "block_id": "p47-b3", - "global_id": 992, - "bbox": [ - 128.91, - 130.07, - 170.65, - 140.04 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p47-b4", - "global_id": 993, - "bbox": [ - 233.34, - 139.3, - 396.62, - 163.32 - ], - "text": "F(x) =\n1\nx + 1 −\n2\nx + 2 +\n2\nx + 3 −\n3\n(x + 3)2", - "type": "text" - }, - { - "block_id": "p47-b5", - "global_id": 994, - "bbox": [ - 127.59, - 209.7, - 516.13, - 231.62 - ], - "text": "Although this method is straightforward and applicable to all situations, it is not necessarily\nthe most efficient. We now discuss other methods that can reduce numerical work considerably.", - "type": "text" - }, - { - "block_id": "p47-b6", - "global_id": 995, - "bbox": [ - 127.59, - 257.17, - 353.34, - 287.16 - ], - "text": "B.5-2 The Heaviside “Cover-Up” Method\nDISTINCT FACTORS OF Q(x)", - "type": "text" - }, - { - "block_id": "p47-b7", - "global_id": 996, - "bbox": [ - 127.59, - 290.78, - 516.13, - 313.11 - ], - "text": "We shall first consider the partial fraction expansion of F(x) = P(x)/Q(x), in which all the factors\nof Q(x) are distinct (not repeated). Consider the proper function", - "type": "text" - }, - { - "block_id": "p47-b8", - "global_id": 997, - "bbox": [ - 214.7, - 321.39, - 383.35, - 342.27 - ], - "text": "F(x) = bmxm + bm−1xm−1 + · · · + b1x + b0", - "type": "text" - }, - { - "block_id": "p47-b9", - "global_id": 998, - "bbox": [ - 253.65, - 336.19, - 375.77, - 350.22 - ], - "text": "xn + an−1xn−1 + · · · + a1x + a0", - "type": "text" - }, - { - "block_id": "p47-b10", - "global_id": 999, - "bbox": [ - 404.98, - 331.99, - 429.03, - 342.27 - ], - "text": "m < n", - "type": "text" - }, - { - "block_id": "p47-b11", - "global_id": 1000, - "bbox": [ - 235.05, - 351.97, - 357.92, - 377.18 - ], - "text": "=\nP(x)\n(x −λ1)(x −λ2)· · ·(x −λn)", - "type": "text" - }, - { - "block_id": "p47-b12", - "global_id": 1001, - "bbox": [ - 127.59, - 386.36, - 415.53, - 396.74 - ], - "text": "As seen in Ex. B.8, F(x) can be expressed as the sum of partial fractions", - "type": "text" - }, - { - "block_id": "p47-b13", - "global_id": 1002, - "bbox": [ - 243.21, - 407.22, - 298.84, - 432.12 - ], - "text": "F(x) =\nk1\nx −λ1", - "type": "text" - }, - { - "block_id": "p47-b14", - "global_id": 1003, - "bbox": [ - 302.08, - 407.22, - 336.86, - 432.12 - ], - "text": "+\nk2\nx −λ2", - "type": "text" - }, - { - "block_id": "p47-b15", - "global_id": 1004, - "bbox": [ - 340.1, - 407.22, - 398.82, - 432.05 - ], - "text": "+ · · · +\nkn\nx −λn", - "type": "text" - }, - { - "block_id": "p47-b16", - "global_id": 1005, - "bbox": [ - 490.4, - 414.31, - 516.12, - 424.27 - ], - "text": "(B.23)", - "type": "text" - }, - { - "block_id": "p47-b17", - "global_id": 1006, - "bbox": [ - 127.59, - 441.08, - 516.13, - 463.41 - ], - "text": "To determine the coefficient k1, we multiply both sides of Eq. (B.23) by x−λ1 and then let x = λ1.\nThis yields", - "type": "text" - }, - { - "block_id": "p47-b18", - "global_id": 1007, - "bbox": [ - 175.7, - 474.41, - 317.56, - 494.98 - ], - "text": "(x −λ1)F(x)|x=λ1 = k1 + k2(x −λ1)", - "type": "text" - }, - { - "block_id": "p47-b19", - "global_id": 1008, - "bbox": [ - 281.18, - 474.41, - 371.43, - 499.62 - ], - "text": "(x −λ2) + k3(x −λ1)", - "type": "text" - }, - { - "block_id": "p47-b20", - "global_id": 1009, - "bbox": [ - 335.03, - 474.41, - 448.43, - 499.62 - ], - "text": "(x −λ3) + · · · +kn(x −λ1)", - "type": "text" - }, - { - "block_id": "p47-b21", - "global_id": 1010, - "bbox": [ - 412.04, - 488.47, - 444.23, - 499.55 - ], - "text": "(x −λn)", - "type": "text" - }, - { - "block_id": "p47-b23", - "global_id": 1011, - "bbox": [ - 452.86, - 493.59, - 468.22, - 502.12 - ], - "text": "x=λ1", - "type": "text" - }, - { - "block_id": "p47-b24", - "global_id": 1012, - "bbox": [ - 127.59, - 511.56, - 385.44, - 523.12 - ], - "text": "On the right-hand side, all the terms except k1 vanish. Therefore,", - "type": "text" - }, - { - "block_id": "p47-b25", - "global_id": 1013, - "bbox": [ - 276.27, - 533.85, - 366.44, - 547.0 - ], - "text": "k1 = (x −λ1)F(x)|x=λ1", - "type": "text" - }, - { - "block_id": "p47-b26", - "global_id": 1014, - "bbox": [ - 127.59, - 557.08, - 237.65, - 567.05 - ], - "text": "Similarly, we can show that", - "type": "text" - }, - { - "block_id": "p47-b27", - "global_id": 1015, - "bbox": [ - 238.63, - 579.27, - 516.13, - 591.91 - ], - "text": "kr = (x −λr)F(x)|x=λr\nr = 1,2,. . .,n\n(B.24)", - "type": "text" - }, - { - "block_id": "p47-b28", - "global_id": 1016, - "bbox": [ - 127.6, - 602.4, - 370.7, - 612.47 - ], - "text": "This procedure also goes under the name method of residues.", - "type": "text" - } - ] - }, - { - "page_num": 48, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p48-b0", - "global_id": 1017, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "28\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p48-b1", - "global_id": 1018, - "bbox": [ - 76.77, - 93.92, - 346.09, - 105.87 - ], - "text": "EXAMPLE B.9\nHeaviside “Cover-Up” Method", - "type": "text" - }, - { - "block_id": "p48-b2", - "global_id": 1019, - "bbox": [ - 103.16, - 122.12, - 364.74, - 132.49 - ], - "text": "Expand the following rational function F(x) into partial fractions:", - "type": "text" - }, - { - "block_id": "p48-b3", - "global_id": 1020, - "bbox": [ - 181.36, - 139.98, - 397.63, - 168.04 - ], - "text": "F(x) =\n2x2 + 9x −11\n(x + 1)(x −2)(x + 3) =\nk1\nx + 1 +\nk2\nx −2 +\nk3\nx + 3", - "type": "text" - }, - { - "block_id": "p48-b4", - "global_id": 1021, - "bbox": [ - 103.16, - 199.4, - 477.02, - 269.56 - ], - "text": "To determine k1, we let x = −1 in (x + 1)F(x). Note that (x + 1)F(x) is obtained from F(x)\nby omitting the term (x + 1) from its denominator. Therefore, to compute k1 corresponding\nto the factor (x + 1), we cover up the term (x + 1) in the denominator of F(x) and then\nsubstitute x = −1 in the remaining expression. [Mentally conceal the term (x + 1) in F(x)\nwith a finger and then let x = −1 in the remaining expression.] The steps in covering up the\nfunction", - "type": "text" - }, - { - "block_id": "p48-b5", - "global_id": 1022, - "bbox": [ - 232.2, - 265.02, - 346.76, - 293.07 - ], - "text": "F(x) =\n2x2 + 9x −11\n(x + 1)(x −2)(x + 3)", - "type": "text" - }, - { - "block_id": "p48-b6", - "global_id": 1023, - "bbox": [ - 103.17, - 299.6, - 160.75, - 309.56 - ], - "text": "are as follows.", - "type": "text" - }, - { - "block_id": "p48-b7", - "global_id": 1024, - "bbox": [ - 103.17, - 319.11, - 328.93, - 329.49 - ], - "text": "Step 1. Cover up (conceal) the factor (x + 1) from F(x):", - "type": "text" - }, - { - "block_id": "p48-b8", - "global_id": 1025, - "bbox": [ - 248.49, - 336.98, - 331.68, - 365.43 - ], - "text": "2x2 + 9x −11\n(x + 1)(x −2)(x + 3)", - "type": "text" - }, - { - "block_id": "p48-b9", - "global_id": 1026, - "bbox": [ - 103.17, - 376.52, - 372.41, - 387.67 - ], - "text": "Step 2. Substitute x = −1 in the remaining expression to obtain k1:", - "type": "text" - }, - { - "block_id": "p48-b10", - "global_id": 1027, - "bbox": [ - 218.3, - 396.42, - 343.83, - 420.85 - ], - "text": "k1 =\n2 −9 −11\n(−1 −2)(−1 + 3) = −18", - "type": "text" - }, - { - "block_id": "p48-b11", - "global_id": 1028, - "bbox": [ - 328.6, - 403.4, - 361.88, - 420.85 - ], - "text": "−6 = 3", - "type": "text" - }, - { - "block_id": "p48-b12", - "global_id": 1029, - "bbox": [ - 103.16, - 430.01, - 477.0, - 452.35 - ], - "text": "Similarly, to compute k2, we cover up the factor (x −2) in F(x) and let x = 2 in the remaining\nfunction, as follows:", - "type": "text" - }, - { - "block_id": "p48-b13", - "global_id": 1030, - "bbox": [ - 173.34, - 459.11, - 278.01, - 487.57 - ], - "text": "k2 =\n2x2 + 9x −11\n(x + 1)(x −2)(x + 3)", - "type": "text" - }, - { - "block_id": "p48-b15", - "global_id": 1031, - "bbox": [ - 282.45, - 481.92, - 294.49, - 489.18 - ], - "text": "x=2", - "type": "text" - }, - { - "block_id": "p48-b16", - "global_id": 1032, - "bbox": [ - 297.04, - 462.73, - 388.8, - 487.17 - ], - "text": "=\n8 + 18 −11\n(2 + 1)(2 + 3) = 15", - "type": "text" - }, - { - "block_id": "p48-b17", - "global_id": 1033, - "bbox": [ - 378.84, - 469.72, - 406.85, - 487.17 - ], - "text": "15 = 1", - "type": "text" - }, - { - "block_id": "p48-b18", - "global_id": 1034, - "bbox": [ - 103.17, - 498.94, - 117.54, - 508.9 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p48-b19", - "global_id": 1035, - "bbox": [ - 155.08, - 504.38, - 259.73, - 532.83 - ], - "text": "k3 =\n2x2 + 9x −11\n(x + 1)(x −2)(x + 3)", - "type": "text" - }, - { - "block_id": "p48-b21", - "global_id": 1036, - "bbox": [ - 264.18, - 527.18, - 281.66, - 534.44 - ], - "text": "x=−3", - "type": "text" - }, - { - "block_id": "p48-b22", - "global_id": 1037, - "bbox": [ - 284.22, - 507.99, - 399.29, - 532.42 - ], - "text": "=\n18 −27 −11\n(−3 + 1)(−3 −2) = −20", - "type": "text" - }, - { - "block_id": "p48-b23", - "global_id": 1038, - "bbox": [ - 385.45, - 514.98, - 425.11, - 532.42 - ], - "text": "10 = −2", - "type": "text" - }, - { - "block_id": "p48-b24", - "global_id": 1039, - "bbox": [ - 103.17, - 541.21, - 144.91, - 551.18 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p48-b25", - "global_id": 1040, - "bbox": [ - 181.36, - 547.98, - 397.63, - 576.03 - ], - "text": "F(x) =\n2x2 + 9x −11\n(x + 1)(x −2)(x + 3) =\n3\nx + 1 +\n1\nx −2 −\n2\nx + 3", - "type": "text" - } - ] - }, - { - "page_num": 49, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p49-b0", - "global_id": 1041, - "bbox": [ - 347.22, - 62.89, - 516.14, - 71.98 - ], - "text": "B.5\nPartial Fraction Expansion\n29", - "type": "text" - }, - { - "block_id": "p49-b1", - "global_id": 1042, - "bbox": [ - 127.89, - 85.39, - 275.34, - 98.32 - ], - "text": "COMPLEX FACTORS OF Q(x)", - "type": "text" - }, - { - "block_id": "p49-b2", - "global_id": 1043, - "bbox": [ - 127.59, - 101.93, - 516.1, - 124.26 - ], - "text": "The procedure just given works regardless of whether the factors of Q(x) are real or complex.\nConsider, for example,", - "type": "text" - }, - { - "block_id": "p49-b3", - "global_id": 1044, - "bbox": [ - 195.85, - 130.42, - 446.63, - 158.47 - ], - "text": "F(x) =\n4x2 + 2x + 18\n(x + 1)(x2 + 4x + 13) =\n4x2 + 2x + 18\n(x + 1)(x + 2 −j3)(x + 2 + j3)", - "type": "text" - }, - { - "block_id": "p49-b4", - "global_id": 1045, - "bbox": [ - 216.21, - 161.07, - 516.12, - 185.19 - ], - "text": "=\nk1\nx + 1 +\nk2\nx + 2 −j3 +\nk3\nx + 2 + j3\n(B.25)", - "type": "text" - }, - { - "block_id": "p49-b5", - "global_id": 1046, - "bbox": [ - 127.59, - 193.68, - 151.92, - 203.64 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p49-b6", - "global_id": 1047, - "bbox": [ - 241.57, - 211.8, - 259.79, - 223.25 - ], - "text": "k1 =", - "type": "text" - }, - { - "block_id": "p49-b8", - "global_id": 1048, - "bbox": [ - 287.31, - 201.2, - 341.84, - 215.19 - ], - "text": "4x2 + 2x + 18", - "type": "text" - }, - { - "block_id": "p49-b9", - "global_id": 1049, - "bbox": [ - 272.21, - 217.86, - 359.92, - 231.12 - ], - "text": "(x + 1) (x2 + 4x + 13)", - "type": "text" - }, - { - "block_id": "p49-b11", - "global_id": 1050, - "bbox": [ - 367.33, - 228.2, - 384.8, - 235.47 - ], - "text": "x=−1", - "type": "text" - }, - { - "block_id": "p49-b12", - "global_id": 1051, - "bbox": [ - 387.36, - 211.8, - 402.16, - 222.17 - ], - "text": "= 2", - "type": "text" - }, - { - "block_id": "p49-b13", - "global_id": 1052, - "bbox": [ - 127.59, - 240.89, - 166.52, - 250.86 - ], - "text": "Similarly,", - "type": "text" - }, - { - "block_id": "p49-b14", - "global_id": 1053, - "bbox": [ - 180.6, - 269.7, - 198.81, - 281.16 - ], - "text": "k2 =", - "type": "text" - }, - { - "block_id": "p49-b16", - "global_id": 1054, - "bbox": [ - 244.19, - 259.1, - 298.72, - 273.1 - ], - "text": "4x2 + 2x + 18", - "type": "text" - }, - { - "block_id": "p49-b17", - "global_id": 1055, - "bbox": [ - 208.25, - 278.65, - 334.67, - 289.03 - ], - "text": "(x + 1) (x + 2 −j3) (x + 2 + j3)", - "type": "text" - }, - { - "block_id": "p49-b19", - "global_id": 1056, - "bbox": [ - 342.06, - 286.41, - 370.39, - 293.68 - ], - "text": "x=−2+j3", - "type": "text" - }, - { - "block_id": "p49-b20", - "global_id": 1057, - "bbox": [ - 372.95, - 269.7, - 416.19, - 280.08 - ], - "text": "= 1 + j2 =", - "type": "text" - }, - { - "block_id": "p49-b21", - "global_id": 1058, - "bbox": [ - 418.24, - 260.81, - 426.67, - 270.77 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p49-b22", - "global_id": 1059, - "bbox": [ - 426.67, - 266.24, - 456.67, - 280.08 - ], - "text": "5ej63.43◦", - "type": "text" - }, - { - "block_id": "p49-b23", - "global_id": 1060, - "bbox": [ - 180.59, - 306.86, - 198.81, - 318.32 - ], - "text": "k3 =", - "type": "text" - }, - { - "block_id": "p49-b25", - "global_id": 1061, - "bbox": [ - 244.19, - 296.26, - 298.72, - 310.26 - ], - "text": "4x2 + 2x + 18", - "type": "text" - }, - { - "block_id": "p49-b26", - "global_id": 1062, - "bbox": [ - 208.25, - 315.81, - 331.68, - 326.19 - ], - "text": "(x + 1)(x + 2 −j3) (x + 2 + j3)", - "type": "text" - }, - { - "block_id": "p49-b28", - "global_id": 1063, - "bbox": [ - 342.06, - 323.58, - 370.39, - 330.85 - ], - "text": "x=−2−j3", - "type": "text" - }, - { - "block_id": "p49-b29", - "global_id": 1064, - "bbox": [ - 372.95, - 306.86, - 416.19, - 317.24 - ], - "text": "= 1 −j2 =", - "type": "text" - }, - { - "block_id": "p49-b30", - "global_id": 1065, - "bbox": [ - 418.24, - 297.97, - 426.67, - 307.94 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p49-b31", - "global_id": 1066, - "bbox": [ - 426.67, - 303.41, - 462.11, - 317.24 - ], - "text": "5e−j63.43◦", - "type": "text" - }, - { - "block_id": "p49-b32", - "global_id": 1067, - "bbox": [ - 127.59, - 339.44, - 169.34, - 349.41 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p49-b33", - "global_id": 1068, - "bbox": [ - 239.99, - 351.19, - 302.19, - 375.21 - ], - "text": "F(x) =\n2\nx + 1 +", - "type": "text" - }, - { - "block_id": "p49-b34", - "global_id": 1069, - "bbox": [ - 304.94, - 342.4, - 313.37, - 352.36 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p49-b35", - "global_id": 1070, - "bbox": [ - 313.36, - 347.82, - 343.37, - 361.15 - ], - "text": "5ej63.43◦", - "type": "text" - }, - { - "block_id": "p49-b36", - "global_id": 1071, - "bbox": [ - 305.2, - 357.76, - 354.9, - 375.21 - ], - "text": "x + 2 −j3 +", - "type": "text" - }, - { - "block_id": "p49-b37", - "global_id": 1072, - "bbox": [ - 357.64, - 342.4, - 366.07, - 352.36 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p49-b38", - "global_id": 1073, - "bbox": [ - 366.08, - 347.82, - 401.51, - 361.15 - ], - "text": "5e−j63.43◦", - "type": "text" - }, - { - "block_id": "p49-b39", - "global_id": 1074, - "bbox": [ - 360.62, - 364.83, - 399.53, - 375.21 - ], - "text": "x + 2 + j3", - "type": "text" - }, - { - "block_id": "p49-b40", - "global_id": 1075, - "bbox": [ - 127.59, - 381.27, - 516.16, - 415.24 - ], - "text": "The coefficients k2 and k3 corresponding to the complex-conjugate factors are also conjugates of\neach other. This is generally true when the coefficients of a rational function are real. In such a\ncase, we need to compute only one of the coefficients.", - "type": "text" - }, - { - "block_id": "p49-b41", - "global_id": 1076, - "bbox": [ - 127.59, - 429.65, - 516.13, - 467.71 - ], - "text": "QUADRATIC FACTORS\nOften we are required to combine the two terms arising from complex-conjugate factors into one\nquadratic factor. For example, F(x) in Eq. (B.25) can be expressed as", - "type": "text" - }, - { - "block_id": "p49-b42", - "global_id": 1077, - "bbox": [ - 214.51, - 473.87, - 428.0, - 501.92 - ], - "text": "F(x) =\n4x2 + 2x + 18\n(x + 1)(x2 + 4x + 13) =\nk1\nx + 1 +\nc1x + c2\nx2 + 4x + 13", - "type": "text" - }, - { - "block_id": "p49-b43", - "global_id": 1078, - "bbox": [ - 127.59, - 510.0, - 411.45, - 521.56 - ], - "text": "The coefficient k1 is found by the Heaviside method to be 2. Therefore,", - "type": "text" - }, - { - "block_id": "p49-b44", - "global_id": 1079, - "bbox": [ - 230.79, - 526.23, - 516.12, - 554.27 - ], - "text": "4x2 + 2x + 18\n(x + 1)(x2 + 4x + 13) =\n2\nx + 1 +\nc1x + c2\nx2 + 4x + 13\n(B.26)", - "type": "text" - }, - { - "block_id": "p49-b45", - "global_id": 1080, - "bbox": [ - 127.59, - 562.36, - 516.12, - 596.33 - ], - "text": "The values of c1 and c2 are determined by clearing fractions and equating the coefficients of similar\npowers of x on both sides of the resulting equation. Clearing fractions on both sides of Eq. (B.26)\nyields", - "type": "text" - }, - { - "block_id": "p49-b46", - "global_id": 1081, - "bbox": [ - 211.66, - 602.42, - 418.25, - 617.7 - ], - "text": "4x2 + 2x + 18 = 2(x2 + 4x + 13) + (c1x + c2)(x + 1)", - "type": "text" - }, - { - "block_id": "p49-b47", - "global_id": 1082, - "bbox": [ - 268.27, - 618.87, - 432.05, - 634.44 - ], - "text": "= (2 + c1)x2 + (8 + c1 + c2)x + (26 + c2)", - "type": "text" - } - ] - }, - { - "page_num": 50, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p50-b0", - "global_id": 1083, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "30\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p50-b1", - "global_id": 1084, - "bbox": [ - 101.84, - 85.4, - 345.82, - 96.86 - ], - "text": "Equating terms of similar powers yields c1 = 2, c2 = −8, and", - "type": "text" - }, - { - "block_id": "p50-b2", - "global_id": 1085, - "bbox": [ - 205.05, - 106.58, - 387.17, - 134.62 - ], - "text": "4x2 + 2x + 18\n(x + 1)(x2 + 4x + 13) =\n2\nx + 1 +\n2x −8\nx2 + 4x + 13", - "type": "text" - }, - { - "block_id": "p50-b3", - "global_id": 1086, - "bbox": [ - 102.14, - 148.16, - 165.64, - 160.28 - ], - "text": "SHORTCUTS", - "type": "text" - }, - { - "block_id": "p50-b4", - "global_id": 1087, - "bbox": [ - 101.84, - 164.21, - 490.39, - 198.18 - ], - "text": "The values of c1 and c2 in Eq. (B.26) can also be determined by using shortcuts. After computing\nk1 = 2 by the Heaviside method as before, we let x = 0 on both sides of Eq. (B.26) to eliminate c1.\nThis gives us", - "type": "text" - }, - { - "block_id": "p50-b5", - "global_id": 1088, - "bbox": [ - 227.42, - 203.01, - 366.01, - 227.13 - ], - "text": "18\n13 = 2 + c2\n13\n⇒\nc2 = −8", - "type": "text" - }, - { - "block_id": "p50-b6", - "global_id": 1089, - "bbox": [ - 101.84, - 234.92, - 490.38, - 257.25 - ], - "text": "To determine c1, we multiply both sides of Eq. (B.26) by x and then let x →∞. Remember that\nwhen x →∞, only the terms of the highest power are significant. Therefore,", - "type": "text" - }, - { - "block_id": "p50-b7", - "global_id": 1090, - "bbox": [ - 235.76, - 272.1, - 356.47, - 283.55 - ], - "text": "4 = 2 + c1\n⇒\nc1 = 2", - "type": "text" - }, - { - "block_id": "p50-b8", - "global_id": 1091, - "bbox": [ - 101.84, - 297.32, - 490.41, - 343.56 - ], - "text": "In the procedure discussed here, we let x = 0 to determine c2 and then multiply both sides by\nx and let x →∞to determine c1. However, nothing is sacred about these values (x = 0 or x = ∞).\nWe use them because they reduce the number of computations involved. We could just as well use\nother convenient values for x, such as x = 1. Consider the case", - "type": "text" - }, - { - "block_id": "p50-b9", - "global_id": 1092, - "bbox": [ - 213.24, - 353.64, - 297.63, - 374.62 - ], - "text": "F(x) = 2x2 + 4x + 5", - "type": "text" - }, - { - "block_id": "p50-b10", - "global_id": 1093, - "bbox": [ - 244.61, - 357.57, - 319.78, - 381.69 - ], - "text": "x(x2 + 2x + 5) = k", - "type": "text" - }, - { - "block_id": "p50-b11", - "global_id": 1094, - "bbox": [ - 315.42, - 357.26, - 377.78, - 381.69 - ], - "text": "x +\nc1x + c2\nx2 + 2x + 5", - "type": "text" - }, - { - "block_id": "p50-b12", - "global_id": 1095, - "bbox": [ - 101.85, - 394.09, - 389.13, - 404.47 - ], - "text": "We find k = 1 by the Heaviside method in the usual manner. As a result,", - "type": "text" - }, - { - "block_id": "p50-b13", - "global_id": 1096, - "bbox": [ - 229.33, - 415.26, - 305.05, - 443.31 - ], - "text": "2x2 + 4x + 5\nx(x2 + 2x + 5) = 1", - "type": "text" - }, - { - "block_id": "p50-b14", - "global_id": 1097, - "bbox": [ - 300.33, - 418.88, - 490.38, - 443.31 - ], - "text": "x +\nc1x + c2\nx2 + 2x + 5\n(B.27)", - "type": "text" - }, - { - "block_id": "p50-b15", - "global_id": 1098, - "bbox": [ - 101.85, - 455.72, - 490.38, - 478.05 - ], - "text": "If we try letting x = 0 to determine c1 and c2, we obtain ∞on both sides. So let us choose x = 1.\nThis yields", - "type": "text" - }, - { - "block_id": "p50-b16", - "global_id": 1099, - "bbox": [ - 215.89, - 482.56, - 377.54, - 507.0 - ], - "text": "11\n8 = 1 + c1 + c2\n8\nor\nc1 + c2 = 3", - "type": "text" - }, - { - "block_id": "p50-b17", - "global_id": 1100, - "bbox": [ - 101.84, - 514.78, - 490.4, - 549.07 - ], - "text": "We can now choose some other value for x, such as x = 2, to obtain one more relationship to\nuse in determining c1 and c2. In this case, however, a simple method is to multiply both sides of\nEq. (B.27) by x and then let x →∞. This yields", - "type": "text" - }, - { - "block_id": "p50-b18", - "global_id": 1101, - "bbox": [ - 235.76, - 563.91, - 356.47, - 575.37 - ], - "text": "2 = 1 + c1\n⇒\nc1 = 1", - "type": "text" - }, - { - "block_id": "p50-b19", - "global_id": 1102, - "bbox": [ - 101.85, - 589.14, - 305.8, - 600.6 - ], - "text": "Since c1 + c2 = 3, we see that c2 = 2 and therefore,", - "type": "text" - }, - { - "block_id": "p50-b20", - "global_id": 1103, - "bbox": [ - 248.42, - 612.09, - 284.76, - 629.04 - ], - "text": "F(x) = 1", - "type": "text" - }, - { - "block_id": "p50-b21", - "global_id": 1104, - "bbox": [ - 280.04, - 611.68, - 342.61, - 636.11 - ], - "text": "x +\nx + 2\nx2 + 2x + 5", - "type": "text" - } - ] - }, - { - "page_num": 51, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p51-b0", - "global_id": 1105, - "bbox": [ - 347.22, - 62.89, - 516.14, - 71.98 - ], - "text": "B.5\nPartial Fraction Expansion\n31", - "type": "text" - }, - { - "block_id": "p51-b1", - "global_id": 1106, - "bbox": [ - 127.59, - 85.66, - 293.47, - 98.48 - ], - "text": "B.5-3 Repeated Factors of Q(x)", - "type": "text" - }, - { - "block_id": "p51-b2", - "global_id": 1107, - "bbox": [ - 127.59, - 104.19, - 419.61, - 114.57 - ], - "text": "If a function F(x) has a repeated factor in its denominator, it has the form", - "type": "text" - }, - { - "block_id": "p51-b3", - "global_id": 1108, - "bbox": [ - 234.28, - 128.69, - 408.25, - 153.9 - ], - "text": "F(x) =\nP(x)\n(x −λ)r(x −α1)(x −α2)· · ·(x −αj)", - "type": "text" - }, - { - "block_id": "p51-b4", - "global_id": 1109, - "bbox": [ - 127.59, - 168.35, - 287.53, - 178.32 - ], - "text": "Its partial fraction expansion is given by", - "type": "text" - }, - { - "block_id": "p51-b5", - "global_id": 1110, - "bbox": [ - 230.2, - 190.28, - 412.33, - 214.3 - ], - "text": "F(x) =\na0\n(x −λ)r +\na1\n(x −λ)r−1 + · · · +\nar−1\n(x −λ)", - "type": "text" - }, - { - "block_id": "p51-b6", - "global_id": 1111, - "bbox": [ - 261.93, - 216.73, - 296.94, - 241.62 - ], - "text": "+\nk1\nx −α1", - "type": "text" - }, - { - "block_id": "p51-b7", - "global_id": 1112, - "bbox": [ - 300.19, - 216.73, - 335.2, - 241.62 - ], - "text": "+\nk2\nx −α2", - "type": "text" - }, - { - "block_id": "p51-b8", - "global_id": 1113, - "bbox": [ - 338.45, - 216.69, - 395.85, - 241.55 - ], - "text": "+ · · · +\nkj\nx −αj", - "type": "text" - }, - { - "block_id": "p51-b9", - "global_id": 1114, - "bbox": [ - 490.4, - 223.81, - 516.12, - 233.78 - ], - "text": "(B.28)", - "type": "text" - }, - { - "block_id": "p51-b10", - "global_id": 1115, - "bbox": [ - 127.59, - 255.66, - 516.16, - 290.73 - ], - "text": "The coefficients k1,k2,. . .,kj corresponding to the unrepeated factors in this equation are\ndetermined by the Heaviside method, as before [Eq. (B.24)]. To find the coefficients a0,a1,\na2,. . .,ar−1, we multiply both sides of Eq. (B.28) by (x −λ)r. This gives us", - "type": "text" - }, - { - "block_id": "p51-b11", - "global_id": 1116, - "bbox": [ - 190.17, - 301.91, - 453.05, - 317.48 - ], - "text": "(x −λ)rF(x) = a0 + a1(x −λ) + a2(x −λ)2 + · · · + ar−1(x −λ)r−1", - "type": "text" - }, - { - "block_id": "p51-b12", - "global_id": 1117, - "bbox": [ - 253.48, - 327.43, - 270.7, - 338.58 - ], - "text": "+ k1", - "type": "text" - }, - { - "block_id": "p51-b13", - "global_id": 1118, - "bbox": [ - 272.4, - 319.44, - 303.32, - 330.72 - ], - "text": "(x −λ)r", - "type": "text" - }, - { - "block_id": "p51-b14", - "global_id": 1119, - "bbox": [ - 275.69, - 334.5, - 300.18, - 345.65 - ], - "text": "x −α1", - "type": "text" - }, - { - "block_id": "p51-b15", - "global_id": 1120, - "bbox": [ - 306.72, - 327.43, - 323.94, - 338.58 - ], - "text": "+ k2", - "type": "text" - }, - { - "block_id": "p51-b16", - "global_id": 1121, - "bbox": [ - 325.65, - 319.44, - 356.57, - 330.72 - ], - "text": "(x −λ)r", - "type": "text" - }, - { - "block_id": "p51-b17", - "global_id": 1122, - "bbox": [ - 328.93, - 334.5, - 353.42, - 345.65 - ], - "text": "x −α2", - "type": "text" - }, - { - "block_id": "p51-b18", - "global_id": 1123, - "bbox": [ - 359.96, - 327.43, - 401.12, - 338.51 - ], - "text": "+ · · · + kn", - "type": "text" - }, - { - "block_id": "p51-b19", - "global_id": 1124, - "bbox": [ - 402.83, - 319.44, - 433.75, - 330.72 - ], - "text": "(x −λ)r", - "type": "text" - }, - { - "block_id": "p51-b20", - "global_id": 1125, - "bbox": [ - 406.12, - 334.5, - 430.61, - 345.58 - ], - "text": "x −αn", - "type": "text" - }, - { - "block_id": "p51-b21", - "global_id": 1126, - "bbox": [ - 490.4, - 327.84, - 516.12, - 337.81 - ], - "text": "(B.29)", - "type": "text" - }, - { - "block_id": "p51-b22", - "global_id": 1127, - "bbox": [ - 127.59, - 358.36, - 339.17, - 368.74 - ], - "text": "If we let x = λ on both sides of Eq. (B.29), we obtain", - "type": "text" - }, - { - "block_id": "p51-b23", - "global_id": 1128, - "bbox": [ - 279.23, - 383.3, - 365.18, - 396.9 - ], - "text": "(x −λ)rF(x)|x=λ = a0", - "type": "text" - }, - { - "block_id": "p51-b24", - "global_id": 1129, - "bbox": [ - 127.59, - 408.06, - 516.13, - 457.51 - ], - "text": "Therefore, a0 is obtained by concealing the factor (x−λ)r in F(x) and letting x = λ in the remaining\nexpression (the Heaviside “cover-up” method). If we take the derivative (with respect to x) of both\nsides of Eq. (B.29), the right-hand side is a1+ terms containing a factor (x−λ) in their numerators.\nLetting x = λ on both sides of this equation, we obtain", - "type": "text" - }, - { - "block_id": "p51-b25", - "global_id": 1130, - "bbox": [ - 269.85, - 472.55, - 279.25, - 496.57 - ], - "text": "d\ndx", - "type": "text" - }, - { - "block_id": "p51-b27", - "global_id": 1131, - "bbox": [ - 285.51, - 477.72, - 335.39, - 489.5 - ], - "text": "(x −λ)rF(x)", - "type": "text" - }, - { - "block_id": "p51-b29", - "global_id": 1132, - "bbox": [ - 342.56, - 491.42, - 354.92, - 498.62 - ], - "text": "x=λ", - "type": "text" - }, - { - "block_id": "p51-b30", - "global_id": 1133, - "bbox": [ - 357.48, - 479.22, - 375.76, - 490.37 - ], - "text": "= a1", - "type": "text" - }, - { - "block_id": "p51-b31", - "global_id": 1134, - "bbox": [ - 127.59, - 508.26, - 516.11, - 533.79 - ], - "text": "Thus, a1 is obtained by concealing the factor (x−λ)r in F(x), taking the derivative of the remaining\nexpression, and then letting x = λ. Continuing in this manner, we find", - "type": "text" - }, - { - "block_id": "p51-b32", - "global_id": 1135, - "bbox": [ - 262.5, - 549.74, - 288.24, - 567.77 - ], - "text": "aj = 1", - "type": "text" - }, - { - "block_id": "p51-b33", - "global_id": 1136, - "bbox": [ - 282.98, - 563.38, - 288.51, - 573.66 - ], - "text": "j!", - "type": "text" - }, - { - "block_id": "p51-b34", - "global_id": 1137, - "bbox": [ - 295.29, - 548.32, - 302.48, - 559.6 - ], - "text": "dj", - "type": "text" - }, - { - "block_id": "p51-b35", - "global_id": 1138, - "bbox": [ - 293.2, - 563.11, - 304.56, - 573.66 - ], - "text": "dxj", - "type": "text" - }, - { - "block_id": "p51-b37", - "global_id": 1139, - "bbox": [ - 311.3, - 554.81, - 361.19, - 566.58 - ], - "text": "(x −λ)rF(x)", - "type": "text" - }, - { - "block_id": "p51-b39", - "global_id": 1140, - "bbox": [ - 368.35, - 568.51, - 380.72, - 575.7 - ], - "text": "x=λ", - "type": "text" - }, - { - "block_id": "p51-b40", - "global_id": 1141, - "bbox": [ - 490.4, - 556.72, - 516.12, - 566.69 - ], - "text": "(B.30)", - "type": "text" - }, - { - "block_id": "p51-b41", - "global_id": 1142, - "bbox": [ - 127.59, - 585.34, - 516.14, - 634.79 - ], - "text": "Observe that (x −λ)rF(x) is obtained from F(x) by omitting the factor (x −λ)r from its\ndenominator. Therefore, the coefficient aj is obtained by concealing the factor (x −λ)r in\nF(x), taking the jth derivative of the remaining expression, and then letting x = λ (while\ndividing by j!).", - "type": "text" - } - ] - }, - { - "page_num": 52, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p52-b0", - "global_id": 1143, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "32\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p52-b1", - "global_id": 1144, - "bbox": [ - 76.77, - 93.92, - 450.22, - 105.87 - ], - "text": "EXAMPLE B.10\nPartial Fraction Expansion with Repeated Factors", - "type": "text" - }, - { - "block_id": "p52-b2", - "global_id": 1145, - "bbox": [ - 103.16, - 122.12, - 246.27, - 132.49 - ], - "text": "Expand F(x) into partial fractions if", - "type": "text" - }, - { - "block_id": "p52-b3", - "global_id": 1146, - "bbox": [ - 229.4, - 139.98, - 349.56, - 160.96 - ], - "text": "F(x) = 4x3 + 16x2 + 23x + 13", - "type": "text" - }, - { - "block_id": "p52-b4", - "global_id": 1147, - "bbox": [ - 275.45, - 157.46, - 334.9, - 168.04 - ], - "text": "(x + 1)3(x + 2)", - "type": "text" - }, - { - "block_id": "p52-b5", - "global_id": 1148, - "bbox": [ - 103.16, - 193.84, - 198.61, - 203.8 - ], - "text": "The partial fractions are", - "type": "text" - }, - { - "block_id": "p52-b6", - "global_id": 1149, - "bbox": [ - 201.88, - 213.49, - 377.09, - 237.61 - ], - "text": "F(x) =\na0\n(x + 1)3 +\na1\n(x + 1)2 +\na2\nx + 1 +\nk\nx + 2", - "type": "text" - }, - { - "block_id": "p52-b7", - "global_id": 1150, - "bbox": [ - 103.16, - 246.78, - 477.02, - 269.11 - ], - "text": "The coefficient k is obtained by concealing the factor (x + 2) in F(x) and then substituting\nx = −2 in the remaining expression:", - "type": "text" - }, - { - "block_id": "p52-b8", - "global_id": 1151, - "bbox": [ - 216.63, - 278.68, - 324.26, - 299.66 - ], - "text": "k = 4x3 + 16x2 + 23x + 13", - "type": "text" - }, - { - "block_id": "p52-b9", - "global_id": 1152, - "bbox": [ - 250.15, - 296.56, - 309.61, - 307.14 - ], - "text": "(x + 1)3(x + 2)", - "type": "text" - }, - { - "block_id": "p52-b11", - "global_id": 1153, - "bbox": [ - 328.72, - 304.47, - 346.2, - 311.74 - ], - "text": "x=−2", - "type": "text" - }, - { - "block_id": "p52-b12", - "global_id": 1154, - "bbox": [ - 348.75, - 289.28, - 363.55, - 299.66 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p52-b13", - "global_id": 1155, - "bbox": [ - 103.17, - 318.79, - 474.69, - 333.14 - ], - "text": "To find a0, we conceal the factor (x + 1)3 in F(x) and let x = −1 in the remaining expression:", - "type": "text" - }, - { - "block_id": "p52-b14", - "global_id": 1156, - "bbox": [ - 214.44, - 341.93, - 326.45, - 363.99 - ], - "text": "a0 = 4x3 + 16x2 + 23x + 13", - "type": "text" - }, - { - "block_id": "p52-b15", - "global_id": 1157, - "bbox": [ - 252.34, - 359.09, - 311.79, - 370.39 - ], - "text": "(x + 1)3(x + 2)", - "type": "text" - }, - { - "block_id": "p52-b17", - "global_id": 1158, - "bbox": [ - 330.91, - 367.73, - 348.38, - 375.0 - ], - "text": "x=−1", - "type": "text" - }, - { - "block_id": "p52-b18", - "global_id": 1159, - "bbox": [ - 350.93, - 352.54, - 365.73, - 362.91 - ], - "text": "= 2", - "type": "text" - }, - { - "block_id": "p52-b19", - "global_id": 1160, - "bbox": [ - 103.17, - 382.04, - 477.0, - 407.57 - ], - "text": "To find a1, we conceal the factor (x + 1)3 in F(x), take the derivative of the remaining\nexpression, and then let x = −1:", - "type": "text" - }, - { - "block_id": "p52-b20", - "global_id": 1161, - "bbox": [ - 201.79, - 421.08, - 230.89, - 439.21 - ], - "text": "a1 = d", - "type": "text" - }, - { - "block_id": "p52-b21", - "global_id": 1162, - "bbox": [ - 223.82, - 435.14, - 233.22, - 445.1 - ], - "text": "dx", - "type": "text" - }, - { - "block_id": "p52-b23", - "global_id": 1163, - "bbox": [ - 244.13, - 417.15, - 332.92, - 445.6 - ], - "text": "4x3 + 16x2 + 23x + 13\n(x + 1)3(x + 2)", - "type": "text" - }, - { - "block_id": "p52-b25", - "global_id": 1164, - "bbox": [ - 343.56, - 442.94, - 361.04, - 450.2 - ], - "text": "x=−1", - "type": "text" - }, - { - "block_id": "p52-b26", - "global_id": 1165, - "bbox": [ - 363.59, - 427.75, - 378.39, - 438.13 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p52-b27", - "global_id": 1166, - "bbox": [ - 103.16, - 459.29, - 142.09, - 469.25 - ], - "text": "Similarly,", - "type": "text" - }, - { - "block_id": "p52-b28", - "global_id": 1167, - "bbox": [ - 194.73, - 472.89, - 223.12, - 490.92 - ], - "text": "a2 = 1", - "type": "text" - }, - { - "block_id": "p52-b29", - "global_id": 1168, - "bbox": [ - 216.76, - 471.54, - 237.72, - 496.91 - ], - "text": "2!\nd2", - "type": "text" - }, - { - "block_id": "p52-b30", - "global_id": 1169, - "bbox": [ - 226.89, - 486.34, - 239.8, - 496.81 - ], - "text": "dx2", - "type": "text" - }, - { - "block_id": "p52-b32", - "global_id": 1170, - "bbox": [ - 251.19, - 468.86, - 339.98, - 497.32 - ], - "text": "4x3 + 16x2 + 23x + 13\n(x + 1)3(x + 2)", - "type": "text" - }, - { - "block_id": "p52-b34", - "global_id": 1171, - "bbox": [ - 350.62, - 494.65, - 368.09, - 501.92 - ], - "text": "x=−1", - "type": "text" - }, - { - "block_id": "p52-b35", - "global_id": 1172, - "bbox": [ - 370.65, - 479.46, - 385.45, - 489.84 - ], - "text": "= 3", - "type": "text" - }, - { - "block_id": "p52-b36", - "global_id": 1173, - "bbox": [ - 103.17, - 508.01, - 144.91, - 517.97 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p52-b37", - "global_id": 1174, - "bbox": [ - 201.89, - 517.23, - 377.09, - 541.25 - ], - "text": "F(x) =\n2\n(x + 1)3 +\n1\n(x + 1)2 +\n3\nx + 1 +\n1\nx + 2", - "type": "text" - }, - { - "block_id": "p52-b38", - "global_id": 1175, - "bbox": [ - 101.84, - 582.83, - 485.67, - 594.78 - ], - "text": "B.5-4 A Combination of Heaviside “Cover-Up” and Clearing Fractions", - "type": "text" - }, - { - "block_id": "p52-b39", - "global_id": 1176, - "bbox": [ - 101.84, - 600.91, - 490.42, - 634.79 - ], - "text": "For multiple roots, especially of higher order, the Heaviside expansion method, which requires\nrepeated differentiation, can become cumbersome. For a function that contains several repeated\nand unrepeated roots, a hybrid of the two procedures proves to be the best. The simpler coefficients", - "type": "text" - } - ] - }, - { - "page_num": 53, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p53-b0", - "global_id": 1177, - "bbox": [ - 347.22, - 62.89, - 516.14, - 71.98 - ], - "text": "B.5\nPartial Fraction Expansion\n33", - "type": "text" - }, - { - "block_id": "p53-b1", - "global_id": 1178, - "bbox": [ - 127.59, - 85.82, - 516.14, - 119.69 - ], - "text": "are determined by the Heaviside method, and the remaining coefficients are found by clearing\nfractions or shortcuts, thus incorporating the best of the two methods. We demonstrate this\nprocedure by solving Ex. B.10 once again by this method.", - "type": "text" - }, - { - "block_id": "p53-b2", - "global_id": 1179, - "bbox": [ - 127.59, - 121.58, - 516.12, - 144.68 - ], - "text": "In Ex. B.10, coefficients k and a0 are relatively simple to determine by the Heaviside\nexpansion method. These values were found to be k1 = 1 and a0 = 2. Therefore,", - "type": "text" - }, - { - "block_id": "p53-b3", - "global_id": 1180, - "bbox": [ - 198.41, - 151.02, - 445.31, - 179.07 - ], - "text": "4x3 + 16x2 + 23x + 13\n(x + 1)3(x + 2)\n=\n2\n(x + 1)3 +\na1\n(x + 1)2 +\na2\nx + 1 +\n1\nx + 2", - "type": "text" - }, - { - "block_id": "p53-b4", - "global_id": 1181, - "bbox": [ - 127.59, - 188.76, - 510.86, - 200.06 - ], - "text": "We now multiply both sides of this equation by (x + 1)3(x + 2) to clear the fractions. This yields", - "type": "text" - }, - { - "block_id": "p53-b5", - "global_id": 1182, - "bbox": [ - 129.92, - 207.42, - 459.99, - 222.69 - ], - "text": "4x3 + 16x2 + 23x + 13 = 2(x + 2) + a1(x + 1)(x + 2) + a2(x + 1)2(x + 2) + (x + 1)3", - "type": "text" - }, - { - "block_id": "p53-b6", - "global_id": 1183, - "bbox": [ - 220.78, - 223.86, - 513.79, - 239.43 - ], - "text": "= (1 + a2)x3 + (a1 + 4a2 + 3)x2 + (5 + 3a1 + 5a2)x + (4 + 2a1 + 2a2 + 1)", - "type": "text" - }, - { - "block_id": "p53-b7", - "global_id": 1184, - "bbox": [ - 127.59, - 250.14, - 453.68, - 260.2 - ], - "text": "Equating coefficients of the third and second powers of x on both sides, we obtain", - "type": "text" - }, - { - "block_id": "p53-b8", - "global_id": 1185, - "bbox": [ - 251.97, - 271.24, - 323.44, - 294.65 - ], - "text": "1 + a2 = 4\na1 + 4a2 + 3 = 16", - "type": "text" - }, - { - "block_id": "p53-b10", - "global_id": 1186, - "bbox": [ - 340.67, - 271.24, - 392.94, - 294.65 - ], - "text": "⇒\na1 = 1\na2 = 3", - "type": "text" - }, - { - "block_id": "p53-b11", - "global_id": 1187, - "bbox": [ - 127.59, - 304.78, - 516.13, - 338.75 - ], - "text": "We may stop here if we wish because the two desired coefficients, a1 and a2, are now determined.\nHowever, equating the coefficients of the two remaining powers of x yields a convenient check on\nthe answer. Equating the coefficients of the x1 and x0 terms, we obtain", - "type": "text" - }, - { - "block_id": "p53-b12", - "global_id": 1188, - "bbox": [ - 275.71, - 350.23, - 368.01, - 376.62 - ], - "text": "23 = 5 + 3a1 + 5a2\n13 = 4 + 2a1 + 2a2 + 1", - "type": "text" - }, - { - "block_id": "p53-b13", - "global_id": 1189, - "bbox": [ - 127.59, - 387.02, - 516.1, - 409.35 - ], - "text": "These equations are satisfied by the values a1 = 1 and a2 = 3, found earlier, providing an additional\ncheck for our answers. Therefore,", - "type": "text" - }, - { - "block_id": "p53-b14", - "global_id": 1190, - "bbox": [ - 233.66, - 418.5, - 408.87, - 442.52 - ], - "text": "F(x) =\n2\n(x + 1)3 +\n1\n(x + 1)2 +\n3\nx + 1 +\n1\nx + 2", - "type": "text" - }, - { - "block_id": "p53-b15", - "global_id": 1191, - "bbox": [ - 127.59, - 451.97, - 269.8, - 461.93 - ], - "text": "which agrees with the earlier result.", - "type": "text" - }, - { - "block_id": "p53-b16", - "global_id": 1192, - "bbox": [ - 127.89, - 476.65, - 468.83, - 488.77 - ], - "text": "A COMBINATION OF HEAVISIDE “COVER-UP” AND SHORTCUTS", - "type": "text" - }, - { - "block_id": "p53-b17", - "global_id": 1193, - "bbox": [ - 127.59, - 492.39, - 516.12, - 514.72 - ], - "text": "In Ex. B.10, after determining the coefficients a0 = 2 and k = 1 by the Heaviside method as before,\nwe have", - "type": "text" - }, - { - "block_id": "p53-b18", - "global_id": 1194, - "bbox": [ - 198.41, - 510.18, - 445.31, - 538.24 - ], - "text": "4x3 + 16x2 + 23x + 13\n(x + 1)3(x + 2)\n=\n2\n(x + 1)3 +\na1\n(x + 1)2 +\na2\nx + 1 +\n1\nx + 2", - "type": "text" - }, - { - "block_id": "p53-b19", - "global_id": 1195, - "bbox": [ - 127.59, - 544.63, - 516.13, - 567.42 - ], - "text": "There are only two unknown coefficients, a1 and a2. If we multiply both sides of this equation by\nx and then let x →∞, we can eliminate a1. This yields", - "type": "text" - }, - { - "block_id": "p53-b20", - "global_id": 1196, - "bbox": [ - 269.91, - 578.12, - 373.81, - 589.58 - ], - "text": "4 = a2 + 1\n\r⇒\na2 = 3", - "type": "text" - }, - { - "block_id": "p53-b21", - "global_id": 1197, - "bbox": [ - 127.59, - 600.39, - 169.34, - 610.35 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p53-b22", - "global_id": 1198, - "bbox": [ - 198.41, - 607.16, - 445.31, - 635.21 - ], - "text": "4x3 + 16x2 + 23x + 13\n(x + 1)3(x + 2)\n=\n2\n(x + 1)3 +\na1\n(x + 1)2 +\n3\nx + 1 +\n1\nx + 2", - "type": "text" - } - ] - }, - { - "page_num": 54, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p54-b0", - "global_id": 1199, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "34\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p54-b1", - "global_id": 1200, - "bbox": [ - 101.84, - 85.72, - 490.36, - 107.74 - ], - "text": "There is now only one unknown a1, which can be readily found by setting x equal to any convenient\nvalue, say, x = 0. This yields", - "type": "text" - }, - { - "block_id": "p54-b2", - "global_id": 1201, - "bbox": [ - 224.62, - 118.42, - 368.81, - 142.44 - ], - "text": "13\n2 = 2 + a1 + 3 + 1\n2\n\r⇒\na1 = 1", - "type": "text" - }, - { - "block_id": "p54-b3", - "global_id": 1202, - "bbox": [ - 101.84, - 150.98, - 250.67, - 160.94 - ], - "text": "which agrees with our earlier answer.", - "type": "text" - }, - { - "block_id": "p54-b4", - "global_id": 1203, - "bbox": [ - 101.85, - 162.84, - 490.38, - 184.85 - ], - "text": "There are other possible shortcuts. For example, we can compute a0 (coefficient of the highest\npower of the repeated root), subtract this term from both sides, and then repeat the procedure.", - "type": "text" - }, - { - "block_id": "p54-b5", - "global_id": 1204, - "bbox": [ - 101.84, - 209.49, - 273.36, - 222.31 - ], - "text": "B.5-5 Improper F(x) with m = n", - "type": "text" - }, - { - "block_id": "p54-b6", - "global_id": 1205, - "bbox": [ - 101.84, - 228.44, - 490.39, - 274.27 - ], - "text": "A general method of handling an improper function is indicated in the beginning of this section.\nHowever, for the special case of when the numerator and denominator polynomials of F(x) have\nthe same degree (m = n), the procedure is the same as that for a proper function. We can show that\nfor", - "type": "text" - }, - { - "block_id": "p54-b7", - "global_id": 1206, - "bbox": [ - 207.55, - 280.45, - 370.01, - 301.33 - ], - "text": "F(x) = bnxn + bn−1xn−1 + · · · + b1x + b0", - "type": "text" - }, - { - "block_id": "p54-b8", - "global_id": 1207, - "bbox": [ - 243.4, - 295.25, - 365.53, - 309.27 - ], - "text": "xn + an−1xn−1 + · · · + a1x + a0", - "type": "text" - }, - { - "block_id": "p54-b9", - "global_id": 1208, - "bbox": [ - 227.91, - 311.14, - 283.02, - 336.03 - ], - "text": "= bn +\nk1\nx −λ1", - "type": "text" - }, - { - "block_id": "p54-b10", - "global_id": 1209, - "bbox": [ - 286.26, - 311.14, - 321.03, - 336.03 - ], - "text": "+\nk2\nx −λ2", - "type": "text" - }, - { - "block_id": "p54-b11", - "global_id": 1210, - "bbox": [ - 324.28, - 311.14, - 382.99, - 335.96 - ], - "text": "+ · · · +\nkn\nx −λn", - "type": "text" - }, - { - "block_id": "p54-b12", - "global_id": 1211, - "bbox": [ - 101.84, - 345.02, - 389.15, - 356.89 - ], - "text": "the coefficients k1,k2,. . .,kn are computed as if F(x) were proper. Thus,", - "type": "text" - }, - { - "block_id": "p54-b13", - "global_id": 1212, - "bbox": [ - 251.41, - 367.57, - 339.69, - 380.21 - ], - "text": "kr = (x −λr)F(x)|x=λr", - "type": "text" - }, - { - "block_id": "p54-b14", - "global_id": 1213, - "bbox": [ - 101.84, - 390.76, - 490.42, - 436.6 - ], - "text": "For quadratic or repeated factors, the appropriate procedures discussed in Secs. B.5-2 or B.5-3\nshould be used as if F(x) were proper. In other words, when m = n, the only difference between\nthe proper and improper case is the appearance of an extra constant bn in the latter. Otherwise, the\nprocedure remains the same. The proof is left as an exercise for the reader.", - "type": "text" - }, - { - "block_id": "p54-b15", - "global_id": 1214, - "bbox": [ - 76.77, - 465.96, - 444.91, - 491.87 - ], - "text": "EXAMPLE B.11\nPartial Fraction Expansion of Improper Rational\nFunction", - "type": "text" - }, - { - "block_id": "p54-b16", - "global_id": 1215, - "bbox": [ - 103.16, - 505.19, - 246.27, - 515.57 - ], - "text": "Expand F(x) into partial fractions if", - "type": "text" - }, - { - "block_id": "p54-b17", - "global_id": 1216, - "bbox": [ - 211.66, - 523.06, - 297.57, - 544.04 - ], - "text": "F(x) = 3x2 + 9x −20", - "type": "text" - }, - { - "block_id": "p54-b18", - "global_id": 1217, - "bbox": [ - 250.5, - 523.06, - 366.84, - 551.11 - ], - "text": "x2 + x −6\n= 3x2 + 9x −20", - "type": "text" - }, - { - "block_id": "p54-b19", - "global_id": 1218, - "bbox": [ - 311.85, - 540.74, - 367.31, - 551.11 - ], - "text": "(x −2)(x + 3)", - "type": "text" - }, - { - "block_id": "p54-b20", - "global_id": 1219, - "bbox": [ - 103.16, - 576.5, - 284.47, - 587.96 - ], - "text": "Here, m = n = 2 with bn = b2 = 3. Therefore,", - "type": "text" - }, - { - "block_id": "p54-b21", - "global_id": 1220, - "bbox": [ - 204.08, - 593.78, - 290.45, - 614.76 - ], - "text": "F(x) = 3x2 + 9x −20", - "type": "text" - }, - { - "block_id": "p54-b22", - "global_id": 1221, - "bbox": [ - 235.46, - 597.71, - 374.9, - 621.83 - ], - "text": "(x −2)(x + 3) = 3 +\nk1\nx −2 +\nk2\nx + 3", - "type": "text" - } - ] - }, - { - "page_num": 55, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p55-b0", - "global_id": 1222, - "bbox": [ - 347.22, - 62.89, - 516.14, - 71.98 - ], - "text": "B.5\nPartial Fraction Expansion\n35", - "type": "text" - }, - { - "block_id": "p55-b1", - "global_id": 1223, - "bbox": [ - 128.9, - 86.23, - 163.49, - 96.2 - ], - "text": "in which", - "type": "text" - }, - { - "block_id": "p55-b2", - "global_id": 1224, - "bbox": [ - 214.8, - 93.76, - 292.46, - 115.81 - ], - "text": "k1 = 3x2 + 9x −20", - "type": "text" - }, - { - "block_id": "p55-b3", - "global_id": 1225, - "bbox": [ - 237.46, - 111.84, - 292.93, - 122.22 - ], - "text": "(x −2)(x + 3)", - "type": "text" - }, - { - "block_id": "p55-b5", - "global_id": 1226, - "bbox": [ - 297.38, - 119.55, - 309.41, - 126.81 - ], - "text": "x=2", - "type": "text" - }, - { - "block_id": "p55-b6", - "global_id": 1227, - "bbox": [ - 311.96, - 97.37, - 374.57, - 114.73 - ], - "text": "= 12 + 18 −20", - "type": "text" - }, - { - "block_id": "p55-b7", - "global_id": 1228, - "bbox": [ - 334.66, - 97.78, - 398.82, - 121.8 - ], - "text": "(2 + 3)\n= 10", - "type": "text" - }, - { - "block_id": "p55-b8", - "global_id": 1229, - "bbox": [ - 391.35, - 104.35, - 416.87, - 121.8 - ], - "text": "5 = 2", - "type": "text" - }, - { - "block_id": "p55-b9", - "global_id": 1230, - "bbox": [ - 128.91, - 132.41, - 143.28, - 142.38 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p55-b10", - "global_id": 1231, - "bbox": [ - 208.19, - 139.93, - 285.86, - 161.99 - ], - "text": "k2 = 3x2 + 9x −20", - "type": "text" - }, - { - "block_id": "p55-b11", - "global_id": 1232, - "bbox": [ - 230.86, - 158.01, - 286.31, - 168.39 - ], - "text": "(x −2)(x + 3)", - "type": "text" - }, - { - "block_id": "p55-b13", - "global_id": 1233, - "bbox": [ - 290.77, - 165.72, - 308.24, - 172.98 - ], - "text": "x=−3", - "type": "text" - }, - { - "block_id": "p55-b14", - "global_id": 1234, - "bbox": [ - 310.79, - 143.55, - 373.4, - 160.91 - ], - "text": "= 27 −27 −20", - "type": "text" - }, - { - "block_id": "p55-b15", - "global_id": 1235, - "bbox": [ - 329.6, - 143.55, - 405.42, - 167.97 - ], - "text": "(−3 −2)\n= −20", - "type": "text" - }, - { - "block_id": "p55-b16", - "global_id": 1236, - "bbox": [ - 390.18, - 150.53, - 423.47, - 167.97 - ], - "text": "−5 = 4", - "type": "text" - }, - { - "block_id": "p55-b17", - "global_id": 1237, - "bbox": [ - 128.91, - 179.08, - 170.65, - 189.04 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p55-b18", - "global_id": 1238, - "bbox": [ - 229.82, - 185.85, - 316.19, - 206.83 - ], - "text": "F(x) = 3x2 + 9x −20", - "type": "text" - }, - { - "block_id": "p55-b19", - "global_id": 1239, - "bbox": [ - 261.2, - 189.88, - 400.65, - 213.9 - ], - "text": "(x −2)(x + 3) = 3 +\n2\nx −2 +\n4\nx + 3", - "type": "text" - }, - { - "block_id": "p55-b20", - "global_id": 1240, - "bbox": [ - 127.59, - 265.42, - 302.08, - 277.38 - ], - "text": "B.5-6 Modified Partial Fractions", - "type": "text" - }, - { - "block_id": "p55-b21", - "global_id": 1241, - "bbox": [ - 127.59, - 282.09, - 515.48, - 294.17 - ], - "text": "In finding the inverse z-transform (Ch. 5), we require partial fractions of the form kx/(x −λi)r", - "type": "text" - }, - { - "block_id": "p55-b22", - "global_id": 1242, - "bbox": [ - 127.59, - 294.04, - 516.12, - 317.39 - ], - "text": "rather than k/(x −λi)r. This can be achieved by expanding F(x)/x into partial fractions. Consider,\nfor example,", - "type": "text" - }, - { - "block_id": "p55-b23", - "global_id": 1243, - "bbox": [ - 275.8, - 322.59, - 366.69, - 343.57 - ], - "text": "F(x) = 5x2 + 20x + 18", - "type": "text" - }, - { - "block_id": "p55-b24", - "global_id": 1244, - "bbox": [ - 307.21, - 340.06, - 366.16, - 350.64 - ], - "text": "(x + 2)(x + 3)2", - "type": "text" - }, - { - "block_id": "p55-b25", - "global_id": 1245, - "bbox": [ - 127.59, - 362.19, - 251.07, - 372.25 - ], - "text": "Dividing both sides by x yields", - "type": "text" - }, - { - "block_id": "p55-b26", - "global_id": 1246, - "bbox": [ - 273.62, - 381.07, - 291.92, - 391.35 - ], - "text": "F(x)", - "type": "text" - }, - { - "block_id": "p55-b27", - "global_id": 1247, - "bbox": [ - 280.55, - 377.45, - 367.89, - 405.4 - ], - "text": "x\n= 5x2 + 20x + 18", - "type": "text" - }, - { - "block_id": "p55-b28", - "global_id": 1248, - "bbox": [ - 306.19, - 394.93, - 369.58, - 405.5 - ], - "text": "x(x + 2)(x + 3)2", - "type": "text" - }, - { - "block_id": "p55-b29", - "global_id": 1249, - "bbox": [ - 127.59, - 417.15, - 401.54, - 427.11 - ], - "text": "Expansion of the right-hand side into partial fractions as usual yields", - "type": "text" - }, - { - "block_id": "p55-b30", - "global_id": 1250, - "bbox": [ - 202.23, - 443.34, - 220.54, - 453.62 - ], - "text": "F(x)", - "type": "text" - }, - { - "block_id": "p55-b31", - "global_id": 1251, - "bbox": [ - 209.16, - 439.72, - 296.5, - 467.67 - ], - "text": "x\n= 5x2 + 20x + 18", - "type": "text" - }, - { - "block_id": "p55-b32", - "global_id": 1252, - "bbox": [ - 234.8, - 443.65, - 321.44, - 467.78 - ], - "text": "x(x + 2)(x + 3)2 = a1", - "type": "text" - }, - { - "block_id": "p55-b33", - "global_id": 1253, - "bbox": [ - 315.23, - 443.65, - 440.98, - 467.78 - ], - "text": "x +\na2\nx + 2 +\na3\n(x + 3) +\na4\n(x + 3)2", - "type": "text" - }, - { - "block_id": "p55-b34", - "global_id": 1254, - "bbox": [ - 127.59, - 482.0, - 503.6, - 493.45 - ], - "text": "Using the procedure discussed earlier, we find a1 = 1, a2 = 1, a3 = −2, and a4 = 1. Therefore,", - "type": "text" - }, - { - "block_id": "p55-b35", - "global_id": 1255, - "bbox": [ - 247.03, - 507.07, - 265.34, - 517.35 - ], - "text": "F(x)", - "type": "text" - }, - { - "block_id": "p55-b36", - "global_id": 1256, - "bbox": [ - 253.95, - 507.48, - 284.58, - 531.4 - ], - "text": "x\n= 1", - "type": "text" - }, - { - "block_id": "p55-b37", - "global_id": 1257, - "bbox": [ - 279.86, - 507.48, - 396.18, - 531.5 - ], - "text": "x +\n1\nx + 2 −\n2\nx + 3 +\n1\n(x + 3)2", - "type": "text" - }, - { - "block_id": "p55-b38", - "global_id": 1258, - "bbox": [ - 127.59, - 546.04, - 284.0, - 556.1 - ], - "text": "Now multiplying both sides by x yields", - "type": "text" - }, - { - "block_id": "p55-b39", - "global_id": 1259, - "bbox": [ - 248.22, - 571.07, - 393.8, - 595.18 - ], - "text": "F(x) = 1 +\nx\nx + 2 −\n2x\nx + 3 +\nx\n(x + 3)2", - "type": "text" - }, - { - "block_id": "p55-b40", - "global_id": 1260, - "bbox": [ - 127.59, - 608.46, - 445.75, - 620.54 - ], - "text": "This expresses F(x) as the sum of partial fractions having the form kx/(x −λi)r.", - "type": "text" - } - ] - }, - { - "page_num": 56, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p56-b0", - "global_id": 1261, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "36\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p56-b1", - "global_id": 1262, - "bbox": [ - 102.2, - 92.21, - 296.02, - 106.15 - ], - "text": "B.6 VECTORS AND MATRICES", - "type": "text" - }, - { - "block_id": "p56-b2", - "global_id": 1263, - "bbox": [ - 101.84, - 112.04, - 490.39, - 146.02 - ], - "text": "An entity specified by n numbers in a certain order (ordered n-tuple) is an n-dimensional vector.\nThus, an ordered n-tuple (x1, x2, . . ., xn) represents an n-dimensional vector x. A vector may be\nrepresented as a row (row vector):", - "type": "text" - }, - { - "block_id": "p56-b3", - "global_id": 1264, - "bbox": [ - 245.3, - 158.39, - 346.93, - 169.85 - ], - "text": "x = [ x1\nx2\n· · ·\nxn ]", - "type": "text" - }, - { - "block_id": "p56-b4", - "global_id": 1265, - "bbox": [ - 101.84, - 181.66, - 228.83, - 191.72 - ], - "text": "or as a column (column vector):", - "type": "text" - }, - { - "block_id": "p56-b5", - "global_id": 1266, - "bbox": [ - 271.24, - 215.21, - 286.04, - 225.5 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p56-b6", - "global_id": 1267, - "bbox": [ - 288.09, - 186.27, - 295.35, - 196.24 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p56-b7", - "global_id": 1268, - "bbox": [ - 288.09, - 203.75, - 295.35, - 232.1 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p56-b8", - "global_id": 1269, - "bbox": [ - 300.34, - 194.19, - 308.24, - 216.99 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p56-b9", - "global_id": 1270, - "bbox": [ - 300.34, - 216.82, - 308.24, - 247.41 - ], - "text": "...\nxn", - "type": "text" - }, - { - "block_id": "p56-b10", - "global_id": 1271, - "bbox": [ - 313.73, - 186.27, - 320.99, - 196.24 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p56-b11", - "global_id": 1272, - "bbox": [ - 313.73, - 203.75, - 320.99, - 232.1 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p56-b12", - "global_id": 1273, - "bbox": [ - 101.85, - 256.01, - 490.37, - 277.94 - ], - "text": "Simultaneous linear equations can be viewed as the transformation of one vector into another.\nConsider, for example, the m simultaneous linear equations", - "type": "text" - }, - { - "block_id": "p56-b13", - "global_id": 1274, - "bbox": [ - 231.85, - 289.97, - 359.89, - 313.39 - ], - "text": "y1 = a11x1 + a12x2 + · · · + a1nxn\ny2 = a21x1 + a22x2 + · · · + a2nxn", - "type": "text" - }, - { - "block_id": "p56-b14", - "global_id": 1275, - "bbox": [ - 228.75, - 312.91, - 362.98, - 343.88 - ], - "text": "...\nym = am1x1 + am2x2 + · · · + amnxn", - "type": "text" - }, - { - "block_id": "p56-b15", - "global_id": 1276, - "bbox": [ - 464.65, - 311.72, - 490.38, - 321.68 - ], - "text": "(B.31)", - "type": "text" - }, - { - "block_id": "p56-b16", - "global_id": 1277, - "bbox": [ - 101.85, - 355.01, - 273.21, - 365.06 - ], - "text": "If we define two column vectors x and y as", - "type": "text" - }, - { - "block_id": "p56-b17", - "global_id": 1278, - "bbox": [ - 217.93, - 398.25, - 232.73, - 408.54 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p56-b18", - "global_id": 1279, - "bbox": [ - 234.77, - 369.32, - 242.04, - 379.29 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p56-b19", - "global_id": 1280, - "bbox": [ - 234.77, - 386.79, - 242.04, - 415.15 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p56-b20", - "global_id": 1281, - "bbox": [ - 247.02, - 377.24, - 254.93, - 400.03 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p56-b21", - "global_id": 1282, - "bbox": [ - 247.02, - 399.87, - 254.93, - 430.46 - ], - "text": "...\nxn", - "type": "text" - }, - { - "block_id": "p56-b22", - "global_id": 1283, - "bbox": [ - 260.41, - 369.32, - 267.67, - 379.29 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p56-b23", - "global_id": 1284, - "bbox": [ - 260.41, - 386.79, - 337.82, - 415.15 - ], - "text": "⎥⎥⎥⎦\nand\ny =", - "type": "text" - }, - { - "block_id": "p56-b24", - "global_id": 1285, - "bbox": [ - 339.87, - 369.32, - 347.13, - 379.29 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p56-b25", - "global_id": 1286, - "bbox": [ - 339.87, - 386.79, - 347.13, - 415.15 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p56-b26", - "global_id": 1287, - "bbox": [ - 352.88, - 377.24, - 360.79, - 400.03 - ], - "text": "y1\ny2", - "type": "text" - }, - { - "block_id": "p56-b27", - "global_id": 1288, - "bbox": [ - 352.11, - 399.87, - 361.56, - 430.46 - ], - "text": "...\nym", - "type": "text" - }, - { - "block_id": "p56-b28", - "global_id": 1289, - "bbox": [ - 367.05, - 369.32, - 374.31, - 379.29 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p56-b29", - "global_id": 1290, - "bbox": [ - 367.05, - 386.79, - 374.31, - 415.15 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p56-b30", - "global_id": 1291, - "bbox": [ - 101.84, - 441.97, - 490.39, - 487.88 - ], - "text": "then Eq. (B.31) may be viewed as the relationship or the function that transforms vector x into\nvector y. Such a transformation is called a linear transformation of vectors. To perform a linear\ntransformation, we need to define the array of coefficients aij appearing in Eq. (B.31). This array\nis called a matrix and is denoted by A for convenience:", - "type": "text" - }, - { - "block_id": "p56-b31", - "global_id": 1292, - "bbox": [ - 231.87, - 521.24, - 248.87, - 531.54 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p56-b32", - "global_id": 1293, - "bbox": [ - 250.92, - 492.32, - 258.18, - 502.28 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p56-b33", - "global_id": 1294, - "bbox": [ - 250.92, - 509.79, - 258.18, - 538.15 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p56-b34", - "global_id": 1295, - "bbox": [ - 263.17, - 499.92, - 347.62, - 553.52 - ], - "text": "a11\na12\n· · ·\na1n\na21\na22\n· · ·\na2n\n...\n...\n· · ·\n...\nam1\nam2\n· · ·\namn", - "type": "text" - }, - { - "block_id": "p56-b35", - "global_id": 1296, - "bbox": [ - 353.11, - 492.32, - 360.37, - 502.28 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p56-b36", - "global_id": 1297, - "bbox": [ - 353.11, - 509.79, - 360.37, - 538.15 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p56-b37", - "global_id": 1298, - "bbox": [ - 101.85, - 564.63, - 490.38, - 586.96 - ], - "text": "A matrix with m rows and n columns is called a matrix of order (m,n) or an (m × n) matrix. For\nthe special case of m = n, the matrix is called a square matrix of order n.", - "type": "text" - }, - { - "block_id": "p56-b38", - "global_id": 1299, - "bbox": [ - 101.85, - 588.96, - 490.41, - 634.79 - ], - "text": "It should be stressed at this point that a matrix is not a number such as a determinant, but an\narray of numbers arranged in a particular order. It is convenient to abbreviate the representation\nof matrix A with the form (aij)m×n, implying a matrix of order m × n with aij as its ijth element.\nIn practice, when the order m × n is understood or need not be specified, the notation can be", - "type": "text" - } - ] - }, - { - "page_num": 57, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p57-b0", - "global_id": 1300, - "bbox": [ - 369.42, - 62.89, - 516.14, - 71.98 - ], - "text": "B.6\nVectors and Matrices\n37", - "type": "text" - }, - { - "block_id": "p57-b1", - "global_id": 1301, - "bbox": [ - 127.59, - 85.46, - 516.13, - 109.29 - ], - "text": "abbreviated to (aij). Note that the first index i of aij indicates the row and the second index j\nindicates the column of the element aij in matrix A.", - "type": "text" - }, - { - "block_id": "p57-b2", - "global_id": 1302, - "bbox": [ - 145.52, - 109.79, - 380.58, - 119.75 - ], - "text": "Equation (B.31) may now be expressed in a matrix form as", - "type": "text" - }, - { - "block_id": "p57-b3", - "global_id": 1303, - "bbox": [ - 187.84, - 122.2, - 195.1, - 132.17 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p57-b4", - "global_id": 1304, - "bbox": [ - 187.84, - 139.67, - 195.1, - 168.03 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p57-b5", - "global_id": 1305, - "bbox": [ - 200.85, - 130.12, - 208.76, - 152.91 - ], - "text": "y1\ny2", - "type": "text" - }, - { - "block_id": "p57-b6", - "global_id": 1306, - "bbox": [ - 200.08, - 152.75, - 209.53, - 183.34 - ], - "text": "...\nym", - "type": "text" - }, - { - "block_id": "p57-b7", - "global_id": 1307, - "bbox": [ - 215.02, - 122.2, - 222.28, - 132.17 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p57-b8", - "global_id": 1308, - "bbox": [ - 215.02, - 139.67, - 232.1, - 168.03 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p57-b9", - "global_id": 1309, - "bbox": [ - 234.15, - 122.2, - 241.42, - 132.16 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p57-b10", - "global_id": 1310, - "bbox": [ - 234.15, - 139.67, - 241.42, - 168.03 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p57-b11", - "global_id": 1311, - "bbox": [ - 246.39, - 129.81, - 330.85, - 183.41 - ], - "text": "a11\na12\n· · ·\na1n\na21\na22\n· · ·\na2n\n...\n...\n· · ·\n...\nam1\nam2\n· · ·\namn", - "type": "text" - }, - { - "block_id": "p57-b12", - "global_id": 1312, - "bbox": [ - 336.34, - 122.2, - 343.6, - 132.17 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p57-b13", - "global_id": 1313, - "bbox": [ - 336.34, - 139.67, - 343.6, - 168.03 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p57-b14", - "global_id": 1314, - "bbox": [ - 344.71, - 122.2, - 351.97, - 132.16 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p57-b15", - "global_id": 1315, - "bbox": [ - 344.71, - 139.67, - 351.97, - 168.03 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p57-b16", - "global_id": 1316, - "bbox": [ - 356.95, - 130.12, - 364.85, - 152.91 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p57-b17", - "global_id": 1317, - "bbox": [ - 356.95, - 152.75, - 364.85, - 183.34 - ], - "text": "...\nxn", - "type": "text" - }, - { - "block_id": "p57-b18", - "global_id": 1318, - "bbox": [ - 370.34, - 122.2, - 377.6, - 132.17 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p57-b19", - "global_id": 1319, - "bbox": [ - 370.34, - 139.67, - 516.12, - 168.03 - ], - "text": "⎥⎥⎥⎦\nor\ny = Ax\n(B.32)", - "type": "text" - }, - { - "block_id": "p57-b20", - "global_id": 1320, - "bbox": [ - 127.59, - 192.99, - 516.14, - 215.0 - ], - "text": "At this point, we have not defined the multiplication of a matrix by a vector. The quantity Ax is\nnot meaningful until such an operation has been defined.", - "type": "text" - }, - { - "block_id": "p57-b21", - "global_id": 1321, - "bbox": [ - 127.59, - 239.22, - 337.6, - 251.18 - ], - "text": "B.6-1 Some Definitions and Properties", - "type": "text" - }, - { - "block_id": "p57-b22", - "global_id": 1322, - "bbox": [ - 127.59, - 257.21, - 516.12, - 279.23 - ], - "text": "A square matrix whose elements are zero everywhere except on the main diagonal is a diagonal\nmatrix. An example of a diagonal matrix is", - "type": "text" - }, - { - "block_id": "p57-b23", - "global_id": 1323, - "bbox": [ - 292.18, - 281.38, - 299.45, - 291.34 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p57-b24", - "global_id": 1324, - "bbox": [ - 292.18, - 299.31, - 299.45, - 309.28 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p57-b25", - "global_id": 1325, - "bbox": [ - 304.42, - 289.7, - 339.3, - 323.58 - ], - "text": "2\n0\n0\n0\n1\n0\n0\n0\n5", - "type": "text" - }, - { - "block_id": "p57-b26", - "global_id": 1326, - "bbox": [ - 344.28, - 281.38, - 351.54, - 291.34 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p57-b27", - "global_id": 1327, - "bbox": [ - 344.28, - 299.31, - 351.54, - 309.28 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p57-b28", - "global_id": 1328, - "bbox": [ - 127.59, - 333.79, - 516.13, - 355.81 - ], - "text": "A diagonal matrix with unity for all its diagonal elements is called an identity matrix or a unit\nmatrix, denoted by I. This is a square matrix:", - "type": "text" - }, - { - "block_id": "p57-b29", - "global_id": 1329, - "bbox": [ - 265.32, - 393.17, - 279.02, - 403.46 - ], - "text": "I =", - "type": "text" - }, - { - "block_id": "p57-b30", - "global_id": 1330, - "bbox": [ - 281.06, - 358.25, - 288.32, - 368.22 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p57-b31", - "global_id": 1331, - "bbox": [ - 281.06, - 375.73, - 288.32, - 416.04 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p57-b32", - "global_id": 1332, - "bbox": [ - 293.31, - 365.87, - 366.15, - 430.65 - ], - "text": "1\n0\n0\n· · ·\n0\n0\n1\n0\n· · ·\n0\n0\n0\n1\n· · ·\n0\n...\n...\n...\n· · ·\n...\n0\n0\n0\n· · ·\n1", - "type": "text" - }, - { - "block_id": "p57-b33", - "global_id": 1333, - "bbox": [ - 371.13, - 358.25, - 378.39, - 368.22 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p57-b34", - "global_id": 1334, - "bbox": [ - 371.13, - 375.73, - 378.39, - 416.04 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p57-b35", - "global_id": 1335, - "bbox": [ - 127.59, - 440.68, - 516.13, - 474.97 - ], - "text": "The order of the unit matrix is sometimes indicated by a subscript. Thus, In represents the n×n unit\nmatrix (or identity matrix). However, we shall omit the subscript since order is easily understood\nby context.", - "type": "text" - }, - { - "block_id": "p57-b36", - "global_id": 1336, - "bbox": [ - 127.59, - 476.86, - 516.13, - 522.78 - ], - "text": "A matrix having all its elements zero is a zero matrix.\nA square matrix A is a symmetric matrix if aij = aji (symmetry about the main diagonal).\nTwo matrices of the same order are said to be equal if they are equal element by element.\nThus, if", - "type": "text" - }, - { - "block_id": "p57-b37", - "global_id": 1337, - "bbox": [ - 244.68, - 524.37, - 398.53, - 535.45 - ], - "text": "A = (aij)m×n\nand\nB = (bij)m×n", - "type": "text" - }, - { - "block_id": "p57-b38", - "global_id": 1338, - "bbox": [ - 127.59, - 542.77, - 291.25, - 554.64 - ], - "text": "then A = B only if aij = bij for all i and j.", - "type": "text" - }, - { - "block_id": "p57-b39", - "global_id": 1339, - "bbox": [ - 127.59, - 554.73, - 516.14, - 589.01 - ], - "text": "If the rows and columns of an m × n matrix A are interchanged so that the elements in the ith\nrow now become the elements of the ith column (for i = 1,2,. . .,m), the resulting matrix is called\nthe transpose of A and is denoted by AT. It is evident that AT is an n × m matrix. For example, if", - "type": "text" - }, - { - "block_id": "p57-b40", - "global_id": 1340, - "bbox": [ - 219.44, - 611.12, - 236.45, - 621.41 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p57-b41", - "global_id": 1341, - "bbox": [ - 238.5, - 591.16, - 245.76, - 601.12 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p57-b42", - "global_id": 1342, - "bbox": [ - 238.5, - 609.09, - 245.76, - 619.06 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p57-b43", - "global_id": 1343, - "bbox": [ - 250.74, - 599.48, - 270.67, - 633.36 - ], - "text": "2\n1\n3\n2\n1\n3", - "type": "text" - }, - { - "block_id": "p57-b44", - "global_id": 1344, - "bbox": [ - 275.64, - 591.16, - 282.91, - 601.12 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p57-b45", - "global_id": 1345, - "bbox": [ - 275.64, - 606.9, - 366.55, - 621.5 - ], - "text": "⎦,\nthen\nAT =", - "type": "text" - }, - { - "block_id": "p57-b46", - "global_id": 1346, - "bbox": [ - 368.6, - 597.13, - 413.87, - 627.38 - ], - "text": "2\n3\n1\n1\n2\n3", - "type": "text" - }, - { - "block_id": "p57-b47", - "global_id": 1347, - "bbox": [ - 418.85, - 597.13, - 424.28, - 607.1 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 58, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p58-b0", - "global_id": 1348, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "38\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p58-b1", - "global_id": 1349, - "bbox": [ - 101.84, - 83.27, - 490.39, - 109.82 - ], - "text": "Using the abbreviated notation, if A = (aij)m×n, then AT = (aji)n×m. Intuitively, further notice that\n(AT)T = A.", - "type": "text" - }, - { - "block_id": "p58-b2", - "global_id": 1350, - "bbox": [ - 101.84, - 134.56, - 217.58, - 146.52 - ], - "text": "B.6-2 Matrix Algebra", - "type": "text" - }, - { - "block_id": "p58-b3", - "global_id": 1351, - "bbox": [ - 101.84, - 152.65, - 490.36, - 186.53 - ], - "text": "We shall now define matrix operations, such as addition, subtraction, multiplication, and division\nof matrices. The definitions should be formulated so that they are useful in the manipulation of\nmatrices.", - "type": "text" - }, - { - "block_id": "p58-b4", - "global_id": 1352, - "bbox": [ - 101.84, - 201.26, - 334.13, - 227.37 - ], - "text": "ADDITION OF MATRICES\nFor two matrices A and B, both of the same order (m × n),", - "type": "text" - }, - { - "block_id": "p58-b5", - "global_id": 1353, - "bbox": [ - 139.1, - 259.41, - 156.12, - 269.7 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p58-b6", - "global_id": 1354, - "bbox": [ - 158.17, - 230.48, - 165.43, - 240.44 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p58-b7", - "global_id": 1355, - "bbox": [ - 158.17, - 247.95, - 165.43, - 276.31 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p58-b8", - "global_id": 1356, - "bbox": [ - 170.41, - 238.09, - 255.97, - 291.68 - ], - "text": "a11\na12\n· · ·\na1n\na21\na22\n· · ·\na2n\n...\n...\n· · ·\n...\nam1\nam2\n· · ·\namn", - "type": "text" - }, - { - "block_id": "p58-b9", - "global_id": 1357, - "bbox": [ - 261.46, - 230.48, - 268.72, - 240.44 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p58-b10", - "global_id": 1358, - "bbox": [ - 261.46, - 247.95, - 340.53, - 276.31 - ], - "text": "⎥⎥⎥⎦\nand\nB =", - "type": "text" - }, - { - "block_id": "p58-b11", - "global_id": 1359, - "bbox": [ - 342.58, - 230.48, - 349.84, - 240.44 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p58-b12", - "global_id": 1360, - "bbox": [ - 342.58, - 247.95, - 349.84, - 276.31 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p58-b13", - "global_id": 1361, - "bbox": [ - 354.82, - 238.09, - 440.38, - 291.68 - ], - "text": "b11\nb12\n· · ·\nb1n\nb21\nb22\n· · ·\nb2n\n...\n...\n· · ·\n...\nbm1\nbm2\n· · ·\nbmn", - "type": "text" - }, - { - "block_id": "p58-b14", - "global_id": 1362, - "bbox": [ - 445.87, - 230.48, - 453.13, - 240.44 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p58-b15", - "global_id": 1363, - "bbox": [ - 445.87, - 247.95, - 453.13, - 276.31 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p58-b16", - "global_id": 1364, - "bbox": [ - 101.85, - 301.98, - 212.4, - 312.35 - ], - "text": "we define the sum A + B as", - "type": "text" - }, - { - "block_id": "p58-b17", - "global_id": 1365, - "bbox": [ - 174.1, - 343.44, - 208.62, - 353.73 - ], - "text": "A + B =", - "type": "text" - }, - { - "block_id": "p58-b18", - "global_id": 1366, - "bbox": [ - 210.67, - 314.5, - 217.94, - 324.47 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p58-b19", - "global_id": 1367, - "bbox": [ - 210.67, - 331.98, - 217.94, - 360.33 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p58-b20", - "global_id": 1368, - "bbox": [ - 224.46, - 322.11, - 404.34, - 362.99 - ], - "text": "(a11 + b11)\n(a12 + b12)\n· · ·\n(a1n + b1n)\n(a21 + b21)\n(a22 + b22)\n· · ·\n(a2n + b2n)\n...", - "type": "text" - }, - { - "block_id": "p58-b21", - "global_id": 1369, - "bbox": [ - 222.91, - 345.05, - 405.89, - 376.02 - ], - "text": "...\n· · ·\n...\n(am1 + bm1)\n(am2 + bm2)\n· · ·\n(amn + bmn)", - "type": "text" - }, - { - "block_id": "p58-b22", - "global_id": 1370, - "bbox": [ - 410.88, - 314.51, - 418.14, - 324.47 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p58-b23", - "global_id": 1371, - "bbox": [ - 410.88, - 331.98, - 418.14, - 360.33 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p58-b24", - "global_id": 1372, - "bbox": [ - 101.85, - 384.07, - 110.14, - 394.04 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p58-b25", - "global_id": 1373, - "bbox": [ - 252.08, - 395.61, - 339.66, - 407.07 - ], - "text": "A + B = (aij + bij)m×n", - "type": "text" - }, - { - "block_id": "p58-b26", - "global_id": 1374, - "bbox": [ - 101.84, - 414.95, - 383.53, - 424.91 - ], - "text": "Note that two matrices can be added only if they are of the same order.", - "type": "text" - }, - { - "block_id": "p58-b27", - "global_id": 1375, - "bbox": [ - 101.84, - 439.64, - 344.32, - 465.75 - ], - "text": "MULTIPLICATION OF A MATRIX BY A SCALAR\nWe multiply a matrix A by a scalar c as follows:", - "type": "text" - }, - { - "block_id": "p58-b28", - "global_id": 1376, - "bbox": [ - 146.74, - 498.17, - 174.65, - 508.46 - ], - "text": "cA = c", - "type": "text" - }, - { - "block_id": "p58-b29", - "global_id": 1377, - "bbox": [ - 175.75, - 469.23, - 183.01, - 479.2 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p58-b30", - "global_id": 1378, - "bbox": [ - 175.75, - 486.71, - 183.01, - 515.06 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p58-b31", - "global_id": 1379, - "bbox": [ - 188.77, - 476.84, - 272.79, - 517.72 - ], - "text": "a11\na12\n· · ·\na1n\na21\na22\n· · ·\na2n\n...", - "type": "text" - }, - { - "block_id": "p58-b32", - "global_id": 1380, - "bbox": [ - 188.0, - 499.78, - 273.56, - 530.44 - ], - "text": "...\n· · ·\n...\nam1\nam2\n· · ·\namn", - "type": "text" - }, - { - "block_id": "p58-b33", - "global_id": 1381, - "bbox": [ - 279.05, - 469.23, - 286.32, - 479.2 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p58-b34", - "global_id": 1382, - "bbox": [ - 279.05, - 486.71, - 296.13, - 515.06 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p58-b35", - "global_id": 1383, - "bbox": [ - 298.18, - 469.23, - 305.44, - 479.2 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p58-b36", - "global_id": 1384, - "bbox": [ - 298.18, - 486.71, - 305.44, - 515.06 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p58-b37", - "global_id": 1385, - "bbox": [ - 311.2, - 476.84, - 408.48, - 517.72 - ], - "text": "ca11\nca12\n· · ·\nca1n\nca21\nca22\n· · ·\nca2n\n...", - "type": "text" - }, - { - "block_id": "p58-b38", - "global_id": 1386, - "bbox": [ - 310.43, - 499.78, - 409.25, - 530.44 - ], - "text": "...\n· · ·\n...\ncam1\ncam2\n· · ·\ncamn", - "type": "text" - }, - { - "block_id": "p58-b39", - "global_id": 1387, - "bbox": [ - 414.75, - 469.23, - 422.01, - 479.2 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p58-b40", - "global_id": 1388, - "bbox": [ - 414.75, - 486.71, - 445.49, - 515.06 - ], - "text": "⎥⎥⎥⎦= Ac", - "type": "text" - }, - { - "block_id": "p58-b41", - "global_id": 1389, - "bbox": [ - 101.85, - 540.74, - 407.13, - 551.11 - ], - "text": "Thus, we also observe that the scalar c and the matrix A commute: cA = Ac.", - "type": "text" - }, - { - "block_id": "p58-b42", - "global_id": 1390, - "bbox": [ - 102.14, - 565.84, - 242.19, - 577.96 - ], - "text": "MATRIX MULTIPLICATION", - "type": "text" - }, - { - "block_id": "p58-b43", - "global_id": 1391, - "bbox": [ - 101.84, - 581.99, - 189.64, - 591.96 - ], - "text": "We define the product", - "type": "text" - }, - { - "block_id": "p58-b44", - "global_id": 1392, - "bbox": [ - 279.67, - 593.53, - 312.56, - 603.83 - ], - "text": "AB = C", - "type": "text" - }, - { - "block_id": "p58-b45", - "global_id": 1393, - "bbox": [ - 101.84, - 612.76, - 490.39, - 634.79 - ], - "text": "in which cij, the element of C in the ith row and jth column, is found by adding the products of\nthe elements of A in the ith row multiplied by the corresponding elements of B in the jth column.", - "type": "text" - } - ] - }, - { - "page_num": 59, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p59-b0", - "global_id": 1394, - "bbox": [ - 369.42, - 62.89, - 516.14, - 71.98 - ], - "text": "B.6\nVectors and Matrices\n39", - "type": "text" - }, - { - "block_id": "p59-b1", - "global_id": 1395, - "bbox": [ - 127.59, - 85.82, - 150.0, - 95.78 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p59-b2", - "global_id": 1396, - "bbox": [ - 231.87, - 108.14, - 373.46, - 119.6 - ], - "text": "cij = ai1b1j + ai2b2j + · · · + ainbnj =", - "type": "text" - }, - { - "block_id": "p59-b3", - "global_id": 1397, - "bbox": [ - 375.51, - 98.19, - 389.6, - 108.64 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p59-b4", - "global_id": 1398, - "bbox": [ - 376.48, - 122.53, - 388.61, - 129.8 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p59-b5", - "global_id": 1399, - "bbox": [ - 390.71, - 108.46, - 516.12, - 119.22 - ], - "text": "aikbkj\n(B.33)", - "type": "text" - }, - { - "block_id": "p59-b6", - "global_id": 1400, - "bbox": [ - 127.59, - 138.88, - 266.39, - 148.84 - ], - "text": "This result is expressed as follows:", - "type": "text" - }, - { - "block_id": "p59-b7", - "global_id": 1401, - "bbox": [ - 188.21, - 161.0, - 195.47, - 170.97 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p59-b8", - "global_id": 1402, - "bbox": [ - 188.21, - 178.48, - 195.47, - 224.76 - ], - "text": "⎢⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p59-b9", - "global_id": 1403, - "bbox": [ - 200.46, - 204.78, - 268.98, - 216.55 - ], - "text": "ai1 ai2\n· · ·\nain", - "type": "text" - }, - { - "block_id": "p59-b10", - "global_id": 1404, - "bbox": [ - 274.46, - 161.0, - 281.73, - 170.96 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p59-b11", - "global_id": 1405, - "bbox": [ - 274.46, - 178.48, - 281.73, - 224.76 - ], - "text": "⎥⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p59-b12", - "global_id": 1406, - "bbox": [ - 188.21, - 236.28, - 281.74, - 258.25 - ], - "text": "A(m×n)", - "type": "text" - }, - { - "block_id": "p59-b13", - "global_id": 1407, - "bbox": [ - 282.83, - 155.03, - 290.1, - 164.99 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p59-b14", - "global_id": 1408, - "bbox": [ - 282.83, - 172.5, - 290.1, - 230.74 - ], - "text": "⎢⎢⎢⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p59-b15", - "global_id": 1409, - "bbox": [ - 318.11, - 162.64, - 328.51, - 185.43 - ], - "text": "b1j\nb2j", - "type": "text" - }, - { - "block_id": "p59-b16", - "global_id": 1410, - "bbox": [ - 295.08, - 185.27, - 352.05, - 246.36 - ], - "text": "...\n· · ·\nbij\n· · ·\n...\nbnj", - "type": "text" - }, - { - "block_id": "p59-b17", - "global_id": 1411, - "bbox": [ - 357.03, - 155.03, - 364.29, - 164.99 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p59-b18", - "global_id": 1412, - "bbox": [ - 357.03, - 172.5, - 364.29, - 230.74 - ], - "text": "⎥⎥⎥⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p59-b19", - "global_id": 1413, - "bbox": [ - 282.83, - 242.86, - 364.3, - 264.83 - ], - "text": "B(n×p)", - "type": "text" - }, - { - "block_id": "p59-b20", - "global_id": 1414, - "bbox": [ - 366.35, - 198.9, - 374.12, - 208.86 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p59-b21", - "global_id": 1415, - "bbox": [ - 376.16, - 161.0, - 383.42, - 170.96 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p59-b22", - "global_id": 1416, - "bbox": [ - 376.16, - 178.48, - 383.42, - 224.76 - ], - "text": "⎢⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p59-b23", - "global_id": 1417, - "bbox": [ - 388.4, - 204.77, - 443.26, - 215.86 - ], - "text": "· · ·\ncij\n· · ·", - "type": "text" - }, - { - "block_id": "p59-b24", - "global_id": 1418, - "bbox": [ - 448.25, - 161.0, - 455.51, - 170.96 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p59-b25", - "global_id": 1419, - "bbox": [ - 448.25, - 178.48, - 455.51, - 224.76 - ], - "text": "⎥⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p59-b26", - "global_id": 1420, - "bbox": [ - 376.16, - 236.28, - 455.52, - 258.25 - ], - "text": "C(m×p)", - "type": "text" - }, - { - "block_id": "p59-b27", - "global_id": 1421, - "bbox": [ - 127.59, - 277.66, - 516.14, - 359.43 - ], - "text": "Note carefully that if this procedure is to work, the number of columns of A must be equal to the\nnumber of rows of B. In other words, AB, the product of matrices A and B, is defined only if\nthe number of columns of A is equal to the number of rows of B. If this condition is not satisfied,\nthe product AB is not defined and is meaningless. When the number of columns of A is equal\nto the number of rows of B, matrix A is said to be conformable to matrix B for the product AB.\nObserve that if A is an m × n matrix and B is an n × p matrix, A and B are conformable for the\nproduct, and C is an m × p matrix.", - "type": "text" - }, - { - "block_id": "p59-b28", - "global_id": 1422, - "bbox": [ - 145.52, - 361.43, - 456.43, - 371.4 - ], - "text": "We demonstrate the use of the rule in Eq. (B.33) with the following examples.", - "type": "text" - }, - { - "block_id": "p59-b29", - "global_id": 1423, - "bbox": [ - 218.2, - 376.97, - 225.46, - 386.93 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p59-b30", - "global_id": 1424, - "bbox": [ - 218.2, - 394.9, - 225.46, - 404.87 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p59-b31", - "global_id": 1425, - "bbox": [ - 230.45, - 385.29, - 250.37, - 419.17 - ], - "text": "2\n3\n1\n1\n3\n1", - "type": "text" - }, - { - "block_id": "p59-b32", - "global_id": 1426, - "bbox": [ - 255.35, - 376.97, - 262.62, - 386.93 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p59-b33", - "global_id": 1427, - "bbox": [ - 255.35, - 394.9, - 262.62, - 404.87 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p59-b34", - "global_id": 1428, - "bbox": [ - 263.72, - 382.94, - 323.94, - 413.19 - ], - "text": "1\n3\n1\n2\n2\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p59-b35", - "global_id": 1429, - "bbox": [ - 328.93, - 382.94, - 334.36, - 392.91 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p59-b36", - "global_id": 1430, - "bbox": [ - 336.41, - 396.94, - 344.18, - 406.9 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p59-b37", - "global_id": 1431, - "bbox": [ - 346.23, - 376.97, - 353.49, - 386.93 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p59-b38", - "global_id": 1432, - "bbox": [ - 346.23, - 394.9, - 353.49, - 404.87 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p59-b39", - "global_id": 1433, - "bbox": [ - 358.47, - 385.29, - 413.26, - 419.17 - ], - "text": "8\n9\n5\n7\n3\n4\n2\n3\n5\n10\n4\n7", - "type": "text" - }, - { - "block_id": "p59-b40", - "global_id": 1434, - "bbox": [ - 418.24, - 376.97, - 425.51, - 386.93 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p59-b41", - "global_id": 1435, - "bbox": [ - 418.24, - 394.9, - 425.51, - 404.87 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p59-b42", - "global_id": 1436, - "bbox": [ - 272.41, - 449.61, - 323.87, - 460.08 - ], - "text": "[ 2\n1\n3 ]", - "type": "text" - }, - { - "block_id": "p59-b43", - "global_id": 1437, - "bbox": [ - 324.98, - 429.74, - 332.25, - 439.71 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p59-b44", - "global_id": 1438, - "bbox": [ - 324.98, - 447.68, - 332.25, - 457.64 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p59-b45", - "global_id": 1439, - "bbox": [ - 337.23, - 438.07, - 342.21, - 471.94 - ], - "text": "2\n1\n1", - "type": "text" - }, - { - "block_id": "p59-b46", - "global_id": 1440, - "bbox": [ - 347.19, - 429.74, - 354.45, - 439.71 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p59-b47", - "global_id": 1441, - "bbox": [ - 347.19, - 447.68, - 371.3, - 460.08 - ], - "text": "⎦= 8", - "type": "text" - }, - { - "block_id": "p59-b48", - "global_id": 1442, - "bbox": [ - 127.59, - 482.7, - 516.16, - 504.61 - ], - "text": "In both cases, the two matrices are conformable. However, if we interchange the order of the first\nmatrices as follows:", - "type": "text" - }, - { - "block_id": "p59-b49", - "global_id": 1443, - "bbox": [ - 263.78, - 505.34, - 324.0, - 535.58 - ], - "text": "1\n3\n1\n2\n2\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p59-b50", - "global_id": 1444, - "bbox": [ - 328.98, - 499.37, - 342.79, - 515.31 - ], - "text": "!⎡", - "type": "text" - }, - { - "block_id": "p59-b51", - "global_id": 1445, - "bbox": [ - 335.52, - 517.3, - 342.79, - 527.26 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p59-b52", - "global_id": 1446, - "bbox": [ - 347.76, - 507.68, - 367.69, - 541.56 - ], - "text": "2\n3\n1\n1\n3\n1", - "type": "text" - }, - { - "block_id": "p59-b53", - "global_id": 1447, - "bbox": [ - 372.67, - 499.37, - 379.93, - 509.33 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p59-b54", - "global_id": 1448, - "bbox": [ - 372.67, - 517.29, - 379.93, - 527.26 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p59-b55", - "global_id": 1449, - "bbox": [ - 127.59, - 552.32, - 461.83, - 562.28 - ], - "text": "the matrices are no longer conformable for the product. It is evident that, in general,", - "type": "text" - }, - { - "block_id": "p59-b56", - "global_id": 1450, - "bbox": [ - 301.94, - 576.2, - 341.78, - 586.49 - ], - "text": "AB̸ = BA", - "type": "text" - }, - { - "block_id": "p59-b57", - "global_id": 1451, - "bbox": [ - 127.59, - 600.83, - 516.12, - 634.79 - ], - "text": "Indeed, AB may exist and BA may not exist, or vice versa, as in our examples. We shall see later\nthat for some special matrices, AB = BA. When this is true, matrices A and B are said to commute.\nWe re-emphasize that in general, matrices do not commute.", - "type": "text" - } - ] - }, - { - "page_num": 60, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p60-b0", - "global_id": 1452, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "40\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p60-b1", - "global_id": 1453, - "bbox": [ - 101.84, - 85.85, - 490.39, - 107.87 - ], - "text": "In the matrix product AB, matrix A is said to be postmultiplied by B or matrix B is said to be\npremultiplied by A. We may also verify the following relationships:", - "type": "text" - }, - { - "block_id": "p60-b2", - "global_id": 1454, - "bbox": [ - 250.97, - 118.67, - 341.26, - 128.96 - ], - "text": "(A + B)C = AC + BC", - "type": "text" - }, - { - "block_id": "p60-b3", - "global_id": 1455, - "bbox": [ - 250.97, - 133.61, - 341.26, - 143.91 - ], - "text": "C(A + B) = CA + CB", - "type": "text" - }, - { - "block_id": "p60-b4", - "global_id": 1456, - "bbox": [ - 101.84, - 155.12, - 490.38, - 177.12 - ], - "text": "We can verify that any matrix A premultiplied or postmultiplied by the identity matrix I remains\nunchanged:", - "type": "text" - }, - { - "block_id": "p60-b5", - "global_id": 1457, - "bbox": [ - 269.58, - 178.7, - 322.65, - 188.99 - ], - "text": "AI = IA = A", - "type": "text" - }, - { - "block_id": "p60-b6", - "global_id": 1458, - "bbox": [ - 101.84, - 197.58, - 490.4, - 219.59 - ], - "text": "Of course, we must make sure that the order of I is such that the matrices are conformable for the\ncorresponding product.", - "type": "text" - }, - { - "block_id": "p60-b7", - "global_id": 1459, - "bbox": [ - 119.78, - 221.58, - 393.31, - 231.54 - ], - "text": "We give here, without proof, another important property of matrices:", - "type": "text" - }, - { - "block_id": "p60-b8", - "global_id": 1460, - "bbox": [ - 267.79, - 242.34, - 324.43, - 252.64 - ], - "text": "|AB| = |A||B|", - "type": "text" - }, - { - "block_id": "p60-b9", - "global_id": 1461, - "bbox": [ - 101.84, - 263.52, - 351.46, - 273.9 - ], - "text": "where |A| and |B| represent determinants of matrices A and B.", - "type": "text" - }, - { - "block_id": "p60-b10", - "global_id": 1462, - "bbox": [ - 101.84, - 288.45, - 490.4, - 374.34 - ], - "text": "MULTIPLICATION OF A MATRIX BY A VECTOR\nConsider Eq. (B.32), which represents Eq. (B.31). The right-hand side of Eq. (B.32) is a product of\nthe m×n matrix A and a vector x. If, for the time being, we treat the vector x as if it were an n×1\nmatrix, then the product Ax, according to the matrix multiplication rule, yields the right-hand side\nof Eq. (B.31). Thus, we may multiply a matrix by a vector by treating the vector as if it were an\nn × 1 matrix. Note that the constraint of conformability still applies. Thus, in this case, xA is not\ndefined and is meaningless.", - "type": "text" - }, - { - "block_id": "p60-b11", - "global_id": 1463, - "bbox": [ - 102.14, - 388.88, - 208.16, - 401.0 - ], - "text": "MATRIX INVERSION", - "type": "text" - }, - { - "block_id": "p60-b12", - "global_id": 1464, - "bbox": [ - 101.84, - 405.04, - 490.38, - 428.42 - ], - "text": "To define the inverse of a matrix, let us consider the set of equations represented by Eq. (B.32)\nwhen m = n:\n⎡", - "type": "text" - }, - { - "block_id": "p60-b13", - "global_id": 1465, - "bbox": [ - 204.33, - 435.93, - 211.59, - 464.28 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p60-b14", - "global_id": 1466, - "bbox": [ - 216.58, - 426.37, - 224.48, - 449.16 - ], - "text": "y1\ny2", - "type": "text" - }, - { - "block_id": "p60-b15", - "global_id": 1467, - "bbox": [ - 216.58, - 449.0, - 224.48, - 479.59 - ], - "text": "...\nyn", - "type": "text" - }, - { - "block_id": "p60-b16", - "global_id": 1468, - "bbox": [ - 229.97, - 418.46, - 237.23, - 428.42 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p60-b17", - "global_id": 1469, - "bbox": [ - 229.97, - 435.93, - 247.04, - 464.28 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p60-b18", - "global_id": 1470, - "bbox": [ - 249.09, - 418.45, - 256.36, - 428.42 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p60-b19", - "global_id": 1471, - "bbox": [ - 249.09, - 435.92, - 256.36, - 464.28 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p60-b20", - "global_id": 1472, - "bbox": [ - 261.34, - 426.06, - 341.15, - 479.66 - ], - "text": "a11\na12\n· · ·\na1n\na21\na22\n· · ·\na2n\n...\n...\n· · ·\n...\nan1\nan2\n· · ·\nann", - "type": "text" - }, - { - "block_id": "p60-b21", - "global_id": 1473, - "bbox": [ - 346.64, - 418.46, - 353.9, - 428.42 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p60-b22", - "global_id": 1474, - "bbox": [ - 346.64, - 435.93, - 353.9, - 464.28 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p60-b23", - "global_id": 1475, - "bbox": [ - 355.0, - 418.45, - 362.27, - 428.42 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p60-b24", - "global_id": 1476, - "bbox": [ - 355.0, - 435.92, - 362.27, - 464.28 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p60-b25", - "global_id": 1477, - "bbox": [ - 367.25, - 426.37, - 375.16, - 449.16 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p60-b26", - "global_id": 1478, - "bbox": [ - 367.25, - 449.0, - 375.16, - 479.59 - ], - "text": "...\nxn", - "type": "text" - }, - { - "block_id": "p60-b27", - "global_id": 1479, - "bbox": [ - 380.64, - 418.46, - 387.9, - 428.42 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p60-b28", - "global_id": 1480, - "bbox": [ - 380.64, - 435.93, - 490.38, - 464.28 - ], - "text": "⎥⎥⎥⎦\n(B.34)", - "type": "text" - }, - { - "block_id": "p60-b29", - "global_id": 1481, - "bbox": [ - 101.84, - 486.48, - 490.38, - 508.82 - ], - "text": "We can solve this set of equations for x1, x2, . . . , xn in terms of y1, y2, . . . , yn by using Cramer’s\nrule [see Eq. (B.21)]. This yields", - "type": "text" - }, - { - "block_id": "p60-b30", - "global_id": 1482, - "bbox": [ - 188.88, - 531.23, - 196.15, - 541.19 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p60-b31", - "global_id": 1483, - "bbox": [ - 188.88, - 548.7, - 196.15, - 577.05 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p60-b32", - "global_id": 1484, - "bbox": [ - 201.12, - 539.14, - 209.03, - 561.93 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p60-b33", - "global_id": 1485, - "bbox": [ - 201.12, - 561.77, - 209.03, - 592.36 - ], - "text": "...\nxn", - "type": "text" - }, - { - "block_id": "p60-b34", - "global_id": 1486, - "bbox": [ - 214.51, - 531.23, - 221.78, - 541.19 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p60-b35", - "global_id": 1487, - "bbox": [ - 214.51, - 548.7, - 231.59, - 577.05 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p60-b36", - "global_id": 1488, - "bbox": [ - 233.64, - 513.29, - 240.9, - 523.25 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p60-b37", - "global_id": 1489, - "bbox": [ - 233.64, - 530.77, - 240.9, - 594.98 - ], - "text": "⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p60-b38", - "global_id": 1490, - "bbox": [ - 247.09, - 517.72, - 267.45, - 528.87 - ], - "text": "|D11|", - "type": "text" - }, - { - "block_id": "p60-b39", - "global_id": 1491, - "bbox": [ - 250.82, - 531.78, - 263.71, - 542.08 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p60-b40", - "global_id": 1492, - "bbox": [ - 279.8, - 517.73, - 300.16, - 528.87 - ], - "text": "|D21|", - "type": "text" - }, - { - "block_id": "p60-b41", - "global_id": 1493, - "bbox": [ - 283.54, - 517.73, - 355.91, - 542.08 - ], - "text": "|A|\n· · ·\n|Dn1|", - "type": "text" - }, - { - "block_id": "p60-b42", - "global_id": 1494, - "bbox": [ - 339.28, - 531.78, - 352.18, - 542.08 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p60-b43", - "global_id": 1495, - "bbox": [ - 247.09, - 547.16, - 267.45, - 558.31 - ], - "text": "|D12|", - "type": "text" - }, - { - "block_id": "p60-b44", - "global_id": 1496, - "bbox": [ - 250.82, - 561.22, - 263.71, - 571.52 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p60-b45", - "global_id": 1497, - "bbox": [ - 279.8, - 547.16, - 300.16, - 558.31 - ], - "text": "|D22|", - "type": "text" - }, - { - "block_id": "p60-b46", - "global_id": 1498, - "bbox": [ - 283.54, - 547.16, - 355.91, - 571.52 - ], - "text": "|A|\n· · ·\n|Dn2|", - "type": "text" - }, - { - "block_id": "p60-b47", - "global_id": 1499, - "bbox": [ - 256.02, - 561.22, - 352.18, - 588.45 - ], - "text": "|A|\n...", - "type": "text" - }, - { - "block_id": "p60-b48", - "global_id": 1500, - "bbox": [ - 247.09, - 570.51, - 346.98, - 599.72 - ], - "text": "...\n· · ·\n...\n|D1n|", - "type": "text" - }, - { - "block_id": "p60-b49", - "global_id": 1501, - "bbox": [ - 250.82, - 602.62, - 263.71, - 612.92 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p60-b50", - "global_id": 1502, - "bbox": [ - 279.8, - 588.57, - 300.16, - 599.72 - ], - "text": "|D2n|", - "type": "text" - }, - { - "block_id": "p60-b51", - "global_id": 1503, - "bbox": [ - 283.54, - 588.57, - 355.91, - 612.92 - ], - "text": "|A|\n· · ·\n|Dnn|", - "type": "text" - }, - { - "block_id": "p60-b52", - "global_id": 1504, - "bbox": [ - 339.28, - 602.62, - 352.18, - 612.92 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p60-b53", - "global_id": 1505, - "bbox": [ - 362.09, - 513.29, - 369.35, - 523.25 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p60-b54", - "global_id": 1506, - "bbox": [ - 362.09, - 530.77, - 369.35, - 594.98 - ], - "text": "⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p60-b55", - "global_id": 1507, - "bbox": [ - 370.46, - 531.22, - 377.72, - 541.19 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p60-b56", - "global_id": 1508, - "bbox": [ - 370.46, - 548.7, - 377.72, - 577.05 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p60-b57", - "global_id": 1509, - "bbox": [ - 382.7, - 539.14, - 390.6, - 561.93 - ], - "text": "y1\ny2", - "type": "text" - }, - { - "block_id": "p60-b58", - "global_id": 1510, - "bbox": [ - 382.7, - 561.77, - 390.6, - 592.36 - ], - "text": "...\nyn", - "type": "text" - }, - { - "block_id": "p60-b59", - "global_id": 1511, - "bbox": [ - 396.09, - 531.23, - 403.35, - 541.19 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p60-b60", - "global_id": 1512, - "bbox": [ - 396.09, - 548.7, - 490.38, - 577.05 - ], - "text": "⎥⎥⎥⎦\n(B.35)", - "type": "text" - }, - { - "block_id": "p60-b61", - "global_id": 1513, - "bbox": [ - 101.84, - 621.65, - 490.39, - 645.48 - ], - "text": "in which |A| is the determinant of the matrix A and |Dij| is the cofactor of element aij in\nthe matrix A. The cofactor of element aij is given by (−1)i+j times the determinant of the", - "type": "text" - } - ] - }, - { - "page_num": 61, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p61-b0", - "global_id": 1514, - "bbox": [ - 369.42, - 62.89, - 516.14, - 71.98 - ], - "text": "B.6\nVectors and Matrices\n41", - "type": "text" - }, - { - "block_id": "p61-b1", - "global_id": 1515, - "bbox": [ - 127.59, - 85.54, - 516.13, - 107.87 - ], - "text": "(n −1) × (n −1) matrix that is obtained when the ith row and the jth column in matrix A are\ndeleted.", - "type": "text" - }, - { - "block_id": "p61-b2", - "global_id": 1516, - "bbox": [ - 145.52, - 109.85, - 359.01, - 119.82 - ], - "text": "We can express Eq. (B.34) in compact matrix form as", - "type": "text" - }, - { - "block_id": "p61-b3", - "global_id": 1517, - "bbox": [ - 307.35, - 131.18, - 516.12, - 141.56 - ], - "text": "y = Ax\n(B.36)", - "type": "text" - }, - { - "block_id": "p61-b4", - "global_id": 1518, - "bbox": [ - 127.59, - 151.09, - 412.16, - 163.3 - ], - "text": "We now define A−1, the inverse of a square matrix A, with the property", - "type": "text" - }, - { - "block_id": "p61-b5", - "global_id": 1519, - "bbox": [ - 266.8, - 172.82, - 376.91, - 185.03 - ], - "text": "A−1A = I\n(unit matrix)", - "type": "text" - }, - { - "block_id": "p61-b6", - "global_id": 1520, - "bbox": [ - 127.59, - 194.56, - 385.5, - 206.77 - ], - "text": "Then, premultiplying both sides of Eq. (B.36) by A−1, we obtain", - "type": "text" - }, - { - "block_id": "p61-b7", - "global_id": 1521, - "bbox": [ - 271.94, - 216.29, - 371.78, - 228.42 - ], - "text": "A−1y = A−1Ax = Ix = x", - "type": "text" - }, - { - "block_id": "p61-b8", - "global_id": 1522, - "bbox": [ - 127.59, - 240.27, - 135.89, - 250.24 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p61-b9", - "global_id": 1523, - "bbox": [ - 302.63, - 249.98, - 516.12, - 262.2 - ], - "text": "x = A−1y\n(B.37)", - "type": "text" - }, - { - "block_id": "p61-b10", - "global_id": 1524, - "bbox": [ - 127.59, - 271.07, - 347.88, - 281.03 - ], - "text": "A comparison of Eq. (B.37) with Eq. (B.35) shows that", - "type": "text" - }, - { - "block_id": "p61-b11", - "global_id": 1525, - "bbox": [ - 235.15, - 306.71, - 273.78, - 323.66 - ], - "text": "A−1 = 1", - "type": "text" - }, - { - "block_id": "p61-b12", - "global_id": 1526, - "bbox": [ - 264.83, - 320.36, - 277.73, - 330.65 - ], - "text": "|A|", - "type": "text" - }, - { - "block_id": "p61-b13", - "global_id": 1527, - "bbox": [ - 280.03, - 284.36, - 287.29, - 294.32 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p61-b14", - "global_id": 1528, - "bbox": [ - 280.03, - 301.82, - 287.29, - 330.18 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p61-b15", - "global_id": 1529, - "bbox": [ - 292.27, - 291.96, - 396.32, - 345.55 - ], - "text": "|D11|\n|D21|\n· · ·\n|Dn1|\n|D12|\n|D22|\n· · ·\n|Dn2|\n...\n...\n· · ·\n...\n|D1n|\n|D2n|\n· · ·\n|Dnn|", - "type": "text" - }, - { - "block_id": "p61-b16", - "global_id": 1530, - "bbox": [ - 401.3, - 284.36, - 408.56, - 294.32 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p61-b17", - "global_id": 1531, - "bbox": [ - 401.3, - 301.82, - 408.56, - 330.18 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p61-b18", - "global_id": 1532, - "bbox": [ - 127.59, - 355.97, - 516.15, - 437.66 - ], - "text": "One of the conditions necessary for a unique solution of Eq. (B.34) is that the number of\nequations must equal the number of unknowns. This implies that the matrix A must be a square\nmatrix. In addition, we observe from the solution as given in Eq. (B.35) that if the solution is\nto exist, |A|̸ = 0.† Therefore, the inverse exists only for a square matrix and only under the\ncondition that the determinant of the matrix be nonzero. A matrix whose determinant is nonzero\nis a nonsingular matrix. Thus, an inverse exists only for a nonsingular, square matrix. Since\nA−1A = I = AA−1, we further note that the matrices A and A−1 commute.‡", - "type": "text" - }, - { - "block_id": "p61-b19", - "global_id": 1533, - "bbox": [ - 145.52, - 439.66, - 463.77, - 449.62 - ], - "text": "The operation of matrix division can be accomplished through matrix inversion.", - "type": "text" - }, - { - "block_id": "p61-b20", - "global_id": 1534, - "bbox": [ - 102.51, - 478.52, - 397.63, - 490.47 - ], - "text": "EXAMPLE B.12\nComputing the Inverse of a Matrix", - "type": "text" - }, - { - "block_id": "p61-b21", - "global_id": 1535, - "bbox": [ - 128.9, - 502.92, - 199.57, - 517.11 - ], - "text": "Let us find A−1 if", - "type": "text" - }, - { - "block_id": "p61-b22", - "global_id": 1536, - "bbox": [ - 276.62, - 528.25, - 293.64, - 538.54 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p61-b23", - "global_id": 1537, - "bbox": [ - 295.68, - 508.29, - 302.95, - 518.25 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p61-b24", - "global_id": 1538, - "bbox": [ - 295.68, - 526.21, - 302.95, - 536.18 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p61-b25", - "global_id": 1539, - "bbox": [ - 307.93, - 516.61, - 342.79, - 550.48 - ], - "text": "2\n1\n1\n1\n2\n3\n3\n2\n1", - "type": "text" - }, - { - "block_id": "p61-b26", - "global_id": 1540, - "bbox": [ - 347.78, - 508.29, - 355.04, - 518.25 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p61-b27", - "global_id": 1541, - "bbox": [ - 347.78, - 526.21, - 355.04, - 536.18 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p61-b28", - "global_id": 1542, - "bbox": [ - 127.59, - 597.82, - 516.14, - 643.38 - ], - "text": "† These two conditions imply that the number of equations is equal to the number of unknowns and that all\nthe equations are independent.\n‡ To prove AA−1 = I, notice first that we define A−1A = I. Thus, IA = AI = A(A−1A) = (AA−1)A.\nSubtracting (AA−1)A, we see that IA −(AA−1)A = 0 or (I −AA−1)A = 0. This requires AA−1 = I.", - "type": "text" - } - ] - }, - { - "page_num": 62, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p62-b0", - "global_id": 1543, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "42\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p62-b1", - "global_id": 1544, - "bbox": [ - 103.16, - 85.96, - 125.0, - 95.93 - ], - "text": "Here,", - "type": "text" - }, - { - "block_id": "p62-b2", - "global_id": 1545, - "bbox": [ - 200.2, - 96.35, - 379.97, - 131.41 - ], - "text": "|D11| = −4,\n|D12| = 8,\n|D13| = −4\n|D21| = 1,\n|D22| = −1,\n|D23| = −1\n|D31| = 1,\n|D32| = −5,\n|D33| = 3", - "type": "text" - }, - { - "block_id": "p62-b3", - "global_id": 1546, - "bbox": [ - 103.17, - 138.73, - 204.28, - 149.11 - ], - "text": "and |A| = −4. Therefore,", - "type": "text" - }, - { - "block_id": "p62-b4", - "global_id": 1547, - "bbox": [ - 226.39, - 155.48, - 268.81, - 172.34 - ], - "text": "A−1 = −1", - "type": "text" - }, - { - "block_id": "p62-b5", - "global_id": 1548, - "bbox": [ - 263.83, - 169.53, - 268.81, - 179.49 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p62-b6", - "global_id": 1549, - "bbox": [ - 271.12, - 142.08, - 278.38, - 152.04 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p62-b7", - "global_id": 1550, - "bbox": [ - 271.12, - 160.02, - 278.38, - 169.98 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p62-b8", - "global_id": 1551, - "bbox": [ - 283.36, - 149.99, - 341.54, - 184.28 - ], - "text": "−4\n1\n1\n8\n−1\n−5\n−4\n−1\n3", - "type": "text" - }, - { - "block_id": "p62-b9", - "global_id": 1552, - "bbox": [ - 346.53, - 142.08, - 353.79, - 152.04 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p62-b10", - "global_id": 1553, - "bbox": [ - 346.53, - 160.02, - 353.79, - 169.98 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p62-b11", - "global_id": 1554, - "bbox": [ - 102.2, - 242.25, - 377.12, - 256.19 - ], - "text": "B.7 MATLAB: ELEMENTARY OPERATIONS", - "type": "text" - }, - { - "block_id": "p62-b12", - "global_id": 1555, - "bbox": [ - 101.84, - 262.03, - 243.74, - 273.98 - ], - "text": "B.7-1 MATLAB Overview", - "type": "text" - }, - { - "block_id": "p62-b13", - "global_id": 1556, - "bbox": [ - 101.84, - 276.5, - 490.38, - 325.94 - ], - "text": "Although MATLAB´l (a registered trademark of The MathWorks, Inc.) is easy to use, it can\nbe intimidating to new users. Over the years, MATLAB has evolved into a sophisticated\ncomputational package with thousands of functions and thousands of pages of documentation.\nThis section provides a brief introduction to the software environment.", - "type": "text" - }, - { - "block_id": "p62-b14", - "global_id": 1557, - "bbox": [ - 101.84, - 327.93, - 490.39, - 361.81 - ], - "text": "When MATLAB is first launched, its command window appears. When MATLAB is ready\nto accept an instruction or input, a command prompt (>>) is displayed in the command window.\nNearly all MATLAB activity is initiated at the command prompt.", - "type": "text" - }, - { - "block_id": "p62-b15", - "global_id": 1558, - "bbox": [ - 101.84, - 363.8, - 490.41, - 433.54 - ], - "text": "Entering instructions at the command prompt generally results in the creation of an object or\nobjects. Many classes of objects are possible, including functions and strings, but usually objects\nare just data. Objects are placed in what is called the MATLAB workspace. If not visible, the\nworkspace can be viewed in a separate window by typing workspace at the command prompt.\nThe workspace provides important information about each object, including the object’s name,\nsize, and class.", - "type": "text" - }, - { - "block_id": "p62-b16", - "global_id": 1559, - "bbox": [ - 101.84, - 435.53, - 490.4, - 469.69 - ], - "text": "Another way to view the workspace is the whos command. When whos is typed at the\ncommand prompt, a summary of the workspace is printed in the command window. The who\ncommand is a short version of whos that reports only the names of workspace objects.", - "type": "text" - }, - { - "block_id": "p62-b17", - "global_id": 1560, - "bbox": [ - 101.84, - 471.4, - 490.41, - 529.18 - ], - "text": "Several functions exist to remove unnecessary data and help free system resources. To remove\nspecific variables from the workspace, the clear command is typed, followed by the names of the\nvariables to be removed. Just typing clear removes all objects from the workspace. Additionally,\nthe clc command clears the command window, and the clf command clears the current figure\nwindow.", - "type": "text" - }, - { - "block_id": "p62-b18", - "global_id": 1561, - "bbox": [ - 101.84, - 531.17, - 490.41, - 588.96 - ], - "text": "Often, important data and objects created in one session need to be saved for future use. The\nsave command, followed by the desired filename, saves the entire workspace to a file, which\nhas the .mat extension. It is also possible to selectively save objects by typing save followed by\nthe filename and then the names of the objects to be saved. The load command followed by the\nfilename is used to load the data and objects contained in a MATLAB data file (.mat file).", - "type": "text" - }, - { - "block_id": "p62-b19", - "global_id": 1562, - "bbox": [ - 101.84, - 590.95, - 490.4, - 624.83 - ], - "text": "Although MATLAB does not automatically save workspace data from one session to the next,\nlines entered at the command prompt are recorded in the command history. Previous command\nlines can be viewed, copied, and executed directly from the command history window. From the", - "type": "text" - } - ] - }, - { - "page_num": 63, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p63-b0", - "global_id": 1563, - "bbox": [ - 317.22, - 62.89, - 516.13, - 71.98 - ], - "text": "B.7\nMATLAB: Elementary Operations\n43", - "type": "text" - }, - { - "block_id": "p63-b1", - "global_id": 1564, - "bbox": [ - 127.59, - 85.82, - 516.15, - 131.64 - ], - "text": "command window, pressing the up or down arrow key scrolls through previous commands and\nredisplays them at the command prompt. Typing the first few characters and then pressing the\narrow keys scrolls through the previous commands that start with the same characters. The arrow\nkeys allow command sequences to be repeated without retyping.", - "type": "text" - }, - { - "block_id": "p63-b2", - "global_id": 1565, - "bbox": [ - 127.59, - 133.64, - 516.15, - 251.2 - ], - "text": "Perhaps the most important and useful command for new users is help. To learn more about\na function, simply type help followed by the function name. Helpful text is then displayed in\nthe command window. The obvious shortcoming of help is that the function name must first\nbe known. This is especially limiting for MATLAB beginners. Fortunately, help screens often\nconclude by referencing related or similar functions. These references are an excellent way to\nlearn new MATLAB commands. Typing help help, for example, displays detailed information\non the help command itself and also provides reference to relevant functions, such as the lookfor\ncommand. The lookfor command helps locate MATLAB functions based on a keyword search.\nSimply type lookfor followed by a single keyword, and MATLAB searches for functions that\ncontain that keyword.", - "type": "text" - }, - { - "block_id": "p63-b3", - "global_id": 1566, - "bbox": [ - 127.59, - 253.19, - 516.15, - 322.93 - ], - "text": "MATLAB also has comprehensive HTML-based help. The HTML help is accessed by using\nMATLAB’s integrated help browser, which also functions as a standard web browser. The HTML\nhelp facility includes a function and topic index as well as full text-searching capabilities. Since\nHTML documents can contain graphics and special characters, HTML help can provide more\ninformation than the command-line help. After a little practice, it is easy to find information in\nMATLAB.", - "type": "text" - }, - { - "block_id": "p63-b4", - "global_id": 1567, - "bbox": [ - 127.59, - 324.92, - 516.15, - 383.0 - ], - "text": "When MATLAB graphics are created, the print command can save figures in a common file\nformat such as postscript, encapsulated postscript, JPEG, or TIFF. The format of displayed data,\nsuch as the number of digits displayed, is selected by using the format command. MATLAB help\nprovides the necessary details for both these functions. When a MATLAB session is complete, the\nexit command terminates MATLAB.", - "type": "text" - }, - { - "block_id": "p63-b5", - "global_id": 1568, - "bbox": [ - 127.59, - 406.67, - 280.5, - 418.62 - ], - "text": "B.7-2 Calculator Operations", - "type": "text" - }, - { - "block_id": "p63-b6", - "global_id": 1569, - "bbox": [ - 127.59, - 424.76, - 516.16, - 470.87 - ], - "text": "MATLAB can function as a simple calculator, working as easily with complex numbers as\nwith real numbers. Scalar addition, subtraction, multiplication, division, and exponentiation are\naccomplished using the traditional operator symbols +, -, *, /, and ^. Since MATLAB predefines\ni = j =", - "type": "text" - }, - { - "block_id": "p63-b7", - "global_id": 1570, - "bbox": [ - 163.55, - 452.19, - 171.98, - 462.15 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p63-b8", - "global_id": 1571, - "bbox": [ - 171.98, - 460.21, - 506.9, - 470.58 - ], - "text": "−1, a complex constant is readily created using Cartesian coordinates. For example,", - "type": "text" - }, - { - "block_id": "p63-b9", - "global_id": 1572, - "bbox": [ - 127.59, - 478.04, - 258.34, - 499.97 - ], - "text": ">>\nz = -3-4j\nz = -3.0000 - 4.0000i", - "type": "text" - }, - { - "block_id": "p63-b10", - "global_id": 1573, - "bbox": [ - 127.59, - 506.44, - 345.35, - 516.82 - ], - "text": "assigns the complex constant −3 −j4 to the variable z.", - "type": "text" - }, - { - "block_id": "p63-b11", - "global_id": 1574, - "bbox": [ - 127.59, - 518.7, - 516.13, - 540.73 - ], - "text": "The real and imaginary components of z are extracted by using the real and imag operators.\nIn MATLAB, the input to a function is placed parenthetically following the function name.", - "type": "text" - }, - { - "block_id": "p63-b12", - "global_id": 1575, - "bbox": [ - 127.59, - 548.19, - 331.57, - 558.15 - ], - "text": ">>\nz_real = real(z); z_imag = imag(z);", - "type": "text" - }, - { - "block_id": "p63-b13", - "global_id": 1576, - "bbox": [ - 127.59, - 565.05, - 516.16, - 611.16 - ], - "text": "When a command is terminated with a semicolon, the statement is evaluated but the results are\nnot displayed to the screen. This feature is useful when one is computing intermediate results, and\nit allows multiple instructions on a single line. Although not displayed, the results z_real = -3\nand z_imag = -4 are calculated and available for additional operations such as computing |z|.", - "type": "text" - }, - { - "block_id": "p63-b14", - "global_id": 1577, - "bbox": [ - 127.59, - 612.86, - 516.13, - 634.79 - ], - "text": "There are many ways to compute the modulus, or magnitude, of a complex quantity.\nTrigonometry confirms that z = −3 −j4, which corresponds to a 3-4-5 triangle, has modulus", - "type": "text" - } - ] - }, - { - "page_num": 64, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p64-b0", - "global_id": 1578, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "44\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p64-b1", - "global_id": 1579, - "bbox": [ - 101.84, - 88.4, - 175.59, - 98.78 - ], - "text": "|z| = |−3 −j4| =", - "type": "text" - }, - { - "block_id": "p64-b3", - "global_id": 1580, - "bbox": [ - 101.84, - 85.52, - 490.39, - 110.73 - ], - "text": "(−3)2 + (−4)2 = 5. The MATLAB sqrt command provides one way to\ncompute the required square root.", - "type": "text" - }, - { - "block_id": "p64-b4", - "global_id": 1581, - "bbox": [ - 101.84, - 119.16, - 295.37, - 141.08 - ], - "text": ">>\nz_mag = sqrt(z_real^2 + z_imag^2)\nz_mag = 5", - "type": "text" - }, - { - "block_id": "p64-b5", - "global_id": 1582, - "bbox": [ - 101.84, - 148.92, - 490.4, - 170.84 - ], - "text": "In MATLAB, most commands, including sqrt, accept inputs in a variety of forms, including\nconstants, variables, functions, expressions, and combinations thereof.", - "type": "text" - }, - { - "block_id": "p64-b6", - "global_id": 1583, - "bbox": [ - 101.85, - 164.75, - 490.36, - 195.04 - ], - "text": "The same result is also obtained by computing |z| = √zz∗. In this case, complex conjugation\nis performed by using the conj command.", - "type": "text" - }, - { - "block_id": "p64-b7", - "global_id": 1584, - "bbox": [ - 101.85, - 203.18, - 243.06, - 225.1 - ], - "text": ">>\nz_mag = sqrt(z*conj(z))\nz_mag = 5", - "type": "text" - }, - { - "block_id": "p64-b8", - "global_id": 1585, - "bbox": [ - 101.85, - 232.95, - 448.95, - 243.2 - ], - "text": "More simply, MATLAB computes absolute values directly by using the abs command.", - "type": "text" - }, - { - "block_id": "p64-b9", - "global_id": 1586, - "bbox": [ - 101.84, - 251.34, - 195.99, - 273.25 - ], - "text": ">>\nz_mag = abs(z)\nz_mag = 5", - "type": "text" - }, - { - "block_id": "p64-b10", - "global_id": 1587, - "bbox": [ - 101.84, - 283.1, - 490.38, - 305.01 - ], - "text": "In addition to magnitude, polar notation requires phase information. The angle command\nprovides the angle of a complex number.", - "type": "text" - }, - { - "block_id": "p64-b11", - "global_id": 1588, - "bbox": [ - 101.84, - 315.43, - 206.45, - 337.35 - ], - "text": ">>\nz_rad = angle(z)\nz_rad = -2.2143", - "type": "text" - }, - { - "block_id": "p64-b12", - "global_id": 1589, - "bbox": [ - 101.84, - 347.18, - 490.38, - 369.11 - ], - "text": "MATLAB expects and returns angles in a radian measure. Angles expressed in degrees require an\nappropriate conversion factor.", - "type": "text" - }, - { - "block_id": "p64-b13", - "global_id": 1590, - "bbox": [ - 101.84, - 379.52, - 243.06, - 401.44 - ], - "text": ">>\nz_deg = angle(z)*180/pi\nz_deg = -126.8699", - "type": "text" - }, - { - "block_id": "p64-b14", - "global_id": 1591, - "bbox": [ - 101.84, - 410.87, - 302.29, - 421.53 - ], - "text": "Notice, MATLAB predefines the variable pi = π.", - "type": "text" - }, - { - "block_id": "p64-b15", - "global_id": 1592, - "bbox": [ - 119.78, - 423.14, - 487.94, - 433.49 - ], - "text": "It is also possible to obtain the angle of z using a two-argument arc-tangent function, atan2.", - "type": "text" - }, - { - "block_id": "p64-b16", - "global_id": 1593, - "bbox": [ - 101.84, - 443.62, - 269.21, - 465.53 - ], - "text": ">>\nz_rad = atan2(z_imag,z_real)\nz_rad = -2.2143", - "type": "text" - }, - { - "block_id": "p64-b17", - "global_id": 1594, - "bbox": [ - 101.84, - 475.38, - 490.39, - 557.36 - ], - "text": "Unlike a single-argument arctangent function, the two-argument arctangent function ensures that\nthe angle reflects the proper quadrant. MATLAB supports a full complement of trigonometric\nfunctions: standard trigonometric functions cos, sin, tan; reciprocal trigonometric functions sec,\ncsc, cot; inverse trigonometric functions acos, asin, atan, asec, acsc, acot; and hyperbolic\nvariations cosh, sinh, tanh, sech, csch, coth, acosh, asinh, atanh, asech, acsch, and\nacoth. Of course, MATLAB comfortably supports complex arguments for any trigonometric\nfunction. As with the angle command, MATLAB trigonometric functions utilize units of radians.", - "type": "text" - }, - { - "block_id": "p64-b18", - "global_id": 1595, - "bbox": [ - 101.84, - 559.06, - 490.39, - 605.18 - ], - "text": "The concept of trigonometric functions with complex-valued arguments is rather intriguing.\nThe results can contradict what is often taught in introductory mathematics courses. For example,\na common claim is that |cos(x)| ≤1. While this is true for real x, it is not necessarily true for\ncomplex x. This is readily verified by example using MATLAB and the cos function.", - "type": "text" - }, - { - "block_id": "p64-b19", - "global_id": 1596, - "bbox": [ - 101.84, - 615.31, - 185.52, - 637.23 - ], - "text": ">>\ncos(1j)\nans = 1.5431", - "type": "text" - } - ] - }, - { - "page_num": 65, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p65-b0", - "global_id": 1597, - "bbox": [ - 317.22, - 62.89, - 516.13, - 71.98 - ], - "text": "B.7\nMATLAB: Elementary Operations\n45", - "type": "text" - }, - { - "block_id": "p65-b1", - "global_id": 1598, - "bbox": [ - 127.59, - 85.82, - 317.65, - 95.78 - ], - "text": "Problem B.1-19 investigates these ideas further.", - "type": "text" - }, - { - "block_id": "p65-b2", - "global_id": 1599, - "bbox": [ - 127.59, - 97.78, - 516.13, - 143.6 - ], - "text": "Similarly, the claim that it is impossible to take the logarithm of a negative number is false. For\nexample, the principal value of ln(−1) is jπ, a fact easily verified by means of Euler’s equation. In\nMATLAB, base-10 and base-e logarithms are computed by using the log10 and log commands,\nrespectively.", - "type": "text" - }, - { - "block_id": "p65-b3", - "global_id": 1600, - "bbox": [ - 127.59, - 156.61, - 237.41, - 178.53 - ], - "text": ">>\nlog(-1)\nans = 0 + 3.1416i", - "type": "text" - }, - { - "block_id": "p65-b4", - "global_id": 1601, - "bbox": [ - 127.59, - 200.55, - 258.59, - 212.51 - ], - "text": "B.7-3 Vector Operations", - "type": "text" - }, - { - "block_id": "p65-b5", - "global_id": 1602, - "bbox": [ - 127.59, - 218.64, - 516.14, - 288.66 - ], - "text": "The power of MATLAB becomes apparent when vector arguments replace scalar arguments.\nRather than computing one value at a time, a single expression computes many values. Typically,\nvectors are classified as row vectors or column vectors. For now, we consider the creation of row\nvectors with evenly spaced, real elements. To create such a vector, the notation a:b:c is used,\nwhere a is the initial value, b designates the step size, and c is the termination value. For example,\n0:2:11 creates the length-6 vector of even-valued integers ranging from 0 to 10.", - "type": "text" - }, - { - "block_id": "p65-b6", - "global_id": 1603, - "bbox": [ - 127.59, - 301.38, - 310.62, - 323.3 - ], - "text": ">>\nk = 0:2:11\nk =\n0\n2\n4\n6\n8\n10", - "type": "text" - }, - { - "block_id": "p65-b7", - "global_id": 1604, - "bbox": [ - 127.59, - 335.74, - 516.14, - 357.65 - ], - "text": "In this case, the termination value does not appear as an element of the vector. Negative and\nnoninteger step sizes are also permissible.", - "type": "text" - }, - { - "block_id": "p65-b8", - "global_id": 1605, - "bbox": [ - 127.59, - 370.67, - 362.93, - 392.58 - ], - "text": ">>\nk = 11:-10/3:0\nk = 11.0000\n7.6667\n4.3333\n1.0000", - "type": "text" - }, - { - "block_id": "p65-b9", - "global_id": 1606, - "bbox": [ - 127.59, - 405.02, - 340.18, - 414.98 - ], - "text": "If a step size is not specified, a value of 1 is assumed.", - "type": "text" - }, - { - "block_id": "p65-b10", - "global_id": 1607, - "bbox": [ - 127.59, - 428.0, - 467.49, - 449.91 - ], - "text": ">>\nk = 0:11\nk =\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11", - "type": "text" - }, - { - "block_id": "p65-b11", - "global_id": 1608, - "bbox": [ - 127.59, - 462.35, - 419.94, - 472.31 - ], - "text": "Vector notation provides the basis for solving a wide variety of problems.", - "type": "text" - }, - { - "block_id": "p65-b12", - "global_id": 1609, - "bbox": [ - 127.59, - 470.27, - 516.13, - 508.46 - ], - "text": "For example, consider finding the three cube roots of minus one, w3 = −1 = ej(π+2πk) for\ninteger k. Taking the cube root of each side yields w = ej(π/3+2πk/3). To find the three unique\nsolutions, use any three consecutive integer values of k and MATLAB’s exp function.", - "type": "text" - }, - { - "block_id": "p65-b13", - "global_id": 1610, - "bbox": [ - 127.59, - 521.19, - 451.86, - 543.11 - ], - "text": ">>\nk = 0:2; w = exp(1j*(pi/3 + 2*pi*k/3))\nw = 0.5000 + 0.8660i\n-1.0000 + 0.0000i\n0.5000 - 0.8660i", - "type": "text" - }, - { - "block_id": "p65-b14", - "global_id": 1611, - "bbox": [ - 127.59, - 555.13, - 341.77, - 565.51 - ], - "text": "The solutions, particularly w = −1, are easy to verify.", - "type": "text" - }, - { - "block_id": "p65-b15", - "global_id": 1612, - "bbox": [ - 145.52, - 563.46, - 384.51, - 577.46 - ], - "text": "Finding the 100 unique roots of w100 = −1 is just as simple.", - "type": "text" - }, - { - "block_id": "p65-b16", - "global_id": 1613, - "bbox": [ - 127.59, - 590.47, - 378.63, - 600.44 - ], - "text": ">>\nk = 0:99; w = exp(1j*(pi/100 + 2*pi*k/100));", - "type": "text" - }, - { - "block_id": "p65-b17", - "global_id": 1614, - "bbox": [ - 127.59, - 612.86, - 516.12, - 634.79 - ], - "text": "A semicolon concludes the final instruction to suppress the inconvenient display of all\n100 solutions. To view a particular solution, the user must use an index to specify desired elements.", - "type": "text" - } - ] - }, - { - "page_num": 66, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p66-b0", - "global_id": 1615, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "46\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p66-b1", - "global_id": 1616, - "bbox": [ - 101.84, - 85.82, - 490.42, - 107.74 - ], - "text": "MATLAB indices are integers that increase from a starting value of 1. For example, the fifth\nelement of w is extracted using an index of 5.†", - "type": "text" - }, - { - "block_id": "p66-b2", - "global_id": 1617, - "bbox": [ - 101.84, - 117.48, - 237.82, - 139.4 - ], - "text": ">>\nw(5)\nans = 0.9603 + 0.2790i", - "type": "text" - }, - { - "block_id": "p66-b3", - "global_id": 1618, - "bbox": [ - 101.84, - 148.15, - 490.39, - 182.44 - ], - "text": "Notice that this solution corresponds to k = 4. The independent variable of a function, in this case\nk, rarely serves as the index. Since k is also a vector, it can likewise be indexed. In this way, we\ncan verify that the fifth value of k is indeed 4.", - "type": "text" - }, - { - "block_id": "p66-b4", - "global_id": 1619, - "bbox": [ - 101.84, - 192.18, - 159.37, - 214.1 - ], - "text": ">>\nk(5)\nans = 4", - "type": "text" - }, - { - "block_id": "p66-b5", - "global_id": 1620, - "bbox": [ - 101.84, - 223.26, - 490.4, - 245.19 - ], - "text": "It is also possible to use a vector index to access multiple values. For example, index vector 98:100\nidentifies the last three solutions corresponding to k = [97,98,99].", - "type": "text" - }, - { - "block_id": "p66-b6", - "global_id": 1621, - "bbox": [ - 101.84, - 254.93, - 436.57, - 276.84 - ], - "text": ">>\nw(98:100)\nans = 0.9877 - 0.1564i\n0.9956 - 0.0941i\n0.9995 - 0.0314i", - "type": "text" - }, - { - "block_id": "p66-b7", - "global_id": 1622, - "bbox": [ - 101.84, - 286.01, - 490.42, - 331.84 - ], - "text": "Vector representations provide the foundation to rapidly create and explore various signals.\nConsider the simple 10 Hz sinusoid described by f(t) = sin(2π10t + π/6). Two cycles of this\nsinusoid are included in the interval 0 ≤t < 0.2. A vector t is used to uniformly represent 500 points\nover this interval.", - "type": "text" - }, - { - "block_id": "p66-b8", - "global_id": 1623, - "bbox": [ - 101.84, - 341.59, - 258.75, - 351.55 - ], - "text": ">>\nt = 0:0.2/500:0.2-0.2/500;", - "type": "text" - }, - { - "block_id": "p66-b9", - "global_id": 1624, - "bbox": [ - 101.84, - 360.3, - 302.3, - 370.68 - ], - "text": "Next, the function f(t) is evaluated at these points.", - "type": "text" - }, - { - "block_id": "p66-b10", - "global_id": 1625, - "bbox": [ - 101.84, - 380.42, - 243.06, - 390.39 - ], - "text": ">>\nf = sin(2*pi*10*t+pi/6)", - "type": "text" - }, - { - "block_id": "p66-b11", - "global_id": 1626, - "bbox": [ - 101.84, - 399.14, - 490.38, - 421.46 - ], - "text": "The value of f(t) at t = 0 is the first element of the vector and is thus obtained by using an index\nof 1.", - "type": "text" - }, - { - "block_id": "p66-b12", - "global_id": 1627, - "bbox": [ - 101.84, - 431.21, - 185.52, - 453.13 - ], - "text": ">>\nf(1)\nans = 0.5000", - "type": "text" - }, - { - "block_id": "p66-b13", - "global_id": 1628, - "bbox": [ - 101.84, - 458.68, - 490.42, - 508.12 - ], - "text": "Unfortunately, MATLAB’s indexing syntax conflicts with standard equation notation.‡ That is,\nthe MATLAB indexing command f(1) is not the same as the standard notation f(1) = f(t)|t=1.\nCare must be taken to avoid confusion; remember that the index parameter rarely reflects the\nindependent variable of a function.", - "type": "text" - }, - { - "block_id": "p66-b14", - "global_id": 1629, - "bbox": [ - 101.84, - 532.37, - 220.21, - 544.33 - ], - "text": "B.7-4 Simple Plotting", - "type": "text" - }, - { - "block_id": "p66-b15", - "global_id": 1630, - "bbox": [ - 101.84, - 550.04, - 490.4, - 572.37 - ], - "text": "MATLAB’s plot command provides a convenient way to visualize data, such as graphing f(t)\nagainst the independent variable t.", - "type": "text" - }, - { - "block_id": "p66-b16", - "global_id": 1631, - "bbox": [ - 101.84, - 582.11, - 175.07, - 592.08 - ], - "text": ">>\nplot(t,f);", - "type": "text" - }, - { - "block_id": "p66-b17", - "global_id": 1632, - "bbox": [ - 101.84, - 609.93, - 463.06, - 633.41 - ], - "text": "† Some other programming languages, such as C, begin indexing at 0. Careful attention is warranted.\n‡ MATLAB anonymous functions, considered in Sec. 1.11, are an important and useful exception.", - "type": "text" - } - ] - }, - { - "page_num": 67, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p67-b0", - "global_id": 1633, - "bbox": [ - 317.22, - 62.89, - 516.13, - 71.98 - ], - "text": "B.7\nMATLAB: Elementary Operations\n47", - "type": "text" - }, - { - "block_id": "p67-b1", - "global_id": 1634, - "bbox": [ - 172.7, - 168.66, - 404.16, - 189.76 - ], - "text": "0\n0.05\n0.1\n0.15\n0.2\nt", - "type": "text" - }, - { - "block_id": "p67-b2", - "global_id": 1635, - "bbox": [ - 162.48, - 159.44, - 170.48, - 167.44 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p67-b3", - "global_id": 1636, - "bbox": [ - 166.48, - 123.44, - 170.48, - 131.44 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p67-b4", - "global_id": 1637, - "bbox": [ - 166.48, - 87.44, - 170.48, - 95.44 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p67-b5", - "global_id": 1638, - "bbox": [ - 152.26, - 120.55, - 161.06, - 131.79 - ], - "text": "f(t)", - "type": "text" - }, - { - "block_id": "p67-b6", - "global_id": 1639, - "bbox": [ - 151.5, - 196.06, - 293.97, - 205.67 - ], - "text": "Figure B.12 f(t) = sin(2π10t + π/6).", - "type": "text" - }, - { - "block_id": "p67-b7", - "global_id": 1640, - "bbox": [ - 127.59, - 226.7, - 516.14, - 248.62 - ], - "text": "Axis labels are added using the xlabel and ylabel commands, where the desired string must be\nenclosed by single quotation marks. The result is shown in Fig. B.12.", - "type": "text" - }, - { - "block_id": "p67-b8", - "global_id": 1641, - "bbox": [ - 127.59, - 259.39, - 289.73, - 269.35 - ], - "text": ">>\nxlabel(’t’); ylabel(’f(t)’)", - "type": "text" - }, - { - "block_id": "p67-b9", - "global_id": 1642, - "bbox": [ - 127.59, - 279.55, - 387.8, - 289.8 - ], - "text": "The title command is used to add a title above the current axis.", - "type": "text" - }, - { - "block_id": "p67-b10", - "global_id": 1643, - "bbox": [ - 127.59, - 291.5, - 516.14, - 349.29 - ], - "text": "By default, MATLAB connects data points with solid lines. Plotting discrete points, such as\nthe 100 unique roots of w100 = −1, is accommodated by supplying the plot command with an\nadditional string argument. For example, the string ’o’ tells MATLAB to mark each data point\nwith a circle rather than connecting points with lines. A full description of the supported plot\noptions is available from MATLAB’s help facilities.", - "type": "text" - }, - { - "block_id": "p67-b11", - "global_id": 1644, - "bbox": [ - 127.59, - 360.06, - 378.65, - 381.98 - ], - "text": ">>\nplot(real(w),imag(w),’o’);\n>>\nxlabel(’Re(w)’); ylabel(’Im(w)’); axis equal", - "type": "text" - }, - { - "block_id": "p67-b12", - "global_id": 1645, - "bbox": [ - 127.59, - 392.17, - 516.13, - 437.99 - ], - "text": "The axis equal command ensures that the scale used for the horizontal axis is equal to the\nscale used for the vertical axis. Without axis equal, the plot would appear elliptical rather than\ncircular. Figure B.13 illustrates that the 100 unique roots of w100 = −1 lie equally spaced on the\nunit circle, a fact not easily discerned from the raw numerical data.", - "type": "text" - }, - { - "block_id": "p67-b13", - "global_id": 1646, - "bbox": [ - 127.59, - 439.99, - 516.14, - 473.86 - ], - "text": "MATLAB also includes many specialized plotting functions. For example, MATLAB\ncommands semilogx, semilogy, and loglog operate like the plot command but use base-10\nlogarithmic scales for the horizontal axis, vertical axis, and the horizontal and vertical axes,", - "type": "text" - }, - { - "block_id": "p67-b14", - "global_id": 1647, - "bbox": [ - 167.83, - 604.04, - 278.47, - 625.23 - ], - "text": "–1\n–0.5\n0\n0.5\n1\nRe(w)", - "type": "text" - }, - { - "block_id": "p67-b15", - "global_id": 1648, - "bbox": [ - 141.33, - 569.24, - 155.33, - 577.24 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p67-b16", - "global_id": 1649, - "bbox": [ - 151.33, - 543.65, - 155.33, - 551.65 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p67-b17", - "global_id": 1650, - "bbox": [ - 144.57, - 518.06, - 155.24, - 526.06 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p67-b18", - "global_id": 1651, - "bbox": [ - 130.35, - 535.39, - 139.15, - 557.38 - ], - "text": "Im(w)", - "type": "text" - }, - { - "block_id": "p67-b19", - "global_id": 1652, - "bbox": [ - 302.54, - 613.64, - 453.21, - 626.5 - ], - "text": "Figure B.13 Unique roots of w100 = −1.", - "type": "text" - } - ] - }, - { - "page_num": 68, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p68-b0", - "global_id": 1653, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "48\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p68-b1", - "global_id": 1654, - "bbox": [ - 101.84, - 85.82, - 490.42, - 143.6 - ], - "text": "respectively. Monochrome and color images can be displayed by using the image command,\nand contour plots are easily created with the contour command. Furthermore, a variety of\nthree-dimensional plotting routines are available, such as plot3, contour3, mesh, and surf.\nInformation about these instructions, including examples and related functions, is available from\nMATLAB help.", - "type": "text" - }, - { - "block_id": "p68-b2", - "global_id": 1655, - "bbox": [ - 101.84, - 168.51, - 310.54, - 180.47 - ], - "text": "B.7-5 Element-by-Element Operations", - "type": "text" - }, - { - "block_id": "p68-b3", - "global_id": 1656, - "bbox": [ - 101.84, - 186.18, - 490.4, - 208.51 - ], - "text": "Suppose a new function h(t) is desired that forces an exponential envelope on the sinusoid f(t),\nh(t) = f(t)g(t), where g(t) = e−10t. First, row vector g(t) is created.", - "type": "text" - }, - { - "block_id": "p68-b4", - "global_id": 1657, - "bbox": [ - 101.84, - 218.92, - 201.22, - 228.88 - ], - "text": ">>\ng = exp(-10*t);", - "type": "text" - }, - { - "block_id": "p68-b5", - "global_id": 1658, - "bbox": [ - 101.84, - 238.29, - 490.41, - 344.3 - ], - "text": "Given MATLAB’s vector representation of g(t) and f(t), computing h(t) requires some form of\nvector multiplication. There are three standard ways to multiply vectors: inner product, outer\nproduct, and element-by-element product. As a matrix-oriented language, MATLAB defines the\nstandard multiplication operator * according to the rules of matrix algebra: the multiplicand must\nbe conformable to the multiplier. A 1 × N row vector times an N × 1 column vector results in\nthe scalar-valued inner product. An N × 1 column vector times a 1 × M row vector results in the\nouter product, which is an N × M matrix. Matrix algebra prohibits multiplication of two row\nvectors or multiplication of two column vectors. Thus, the * operator is not used to perform\nelement-by-element multiplication.†", - "type": "text" - }, - { - "block_id": "p68-b6", - "global_id": 1659, - "bbox": [ - 101.84, - 346.3, - 490.41, - 439.95 - ], - "text": "Element-by-element operations require vectors to have the same dimensions. An error\noccurs if element-by-element operations are attempted between row and column vectors. In\nsuch cases, one vector must first be transposed to ensure both vector operands have the same\ndimensions. In MATLAB, most element-by-element operations are preceded by a period. For\nexample, element-by-element multiplication, division, and exponentiation are accomplished using\n.*, ./, and .^, respectively. Vector addition and subtraction are intrinsically element-by-element\noperations and require no period. Intuitively, we know h(t) should be the same size as both g(t)\nand f(t). Thus, h(t) is computed using element-by-element multiplication.", - "type": "text" - }, - { - "block_id": "p68-b7", - "global_id": 1660, - "bbox": [ - 101.84, - 450.35, - 169.83, - 460.32 - ], - "text": ">>\nh = f.*g;", - "type": "text" - }, - { - "block_id": "p68-b8", - "global_id": 1661, - "bbox": [ - 101.84, - 470.14, - 490.4, - 515.97 - ], - "text": "The plot command accommodates multiple curves and also allows modification of line\nproperties. This facilitates side-by-side comparison of different functions, such as h(t) and f(t).\nLine characteristics are specified by using options that follow each vector pair and are enclosed in\nsingle quotes.", - "type": "text" - }, - { - "block_id": "p68-b9", - "global_id": 1662, - "bbox": [ - 101.84, - 526.38, - 295.36, - 560.25 - ], - "text": ">>\nplot(t,f,’-k’,t,h,’:k’);\n>>\nxlabel(’t’); ylabel(’Amplitude’);\n>>\nlegend(’f(t)’,’h(t)’);", - "type": "text" - }, - { - "block_id": "p68-b10", - "global_id": 1663, - "bbox": [ - 101.84, - 569.66, - 490.39, - 591.99 - ], - "text": "Here, ’-k’ instructs MATLAB to plot f(t) using a solid black line, while ’:k’ instructs MATLAB\nto use a dotted black line to plot h(t). A legend and axis labels complete the plot, as shown in", - "type": "text" - }, - { - "block_id": "p68-b11", - "global_id": 1664, - "bbox": [ - 101.84, - 610.24, - 490.37, - 633.41 - ], - "text": "† While grossly inefficient, element-by-element multiplication can be accomplished by extracting the main\ndiagonal from the outer product of two N-length vectors.", - "type": "text" - } - ] - }, - { - "page_num": 69, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p69-b0", - "global_id": 1665, - "bbox": [ - 317.22, - 62.89, - 516.13, - 71.98 - ], - "text": "B.7\nMATLAB: Elementary Operations\n49", - "type": "text" - }, - { - "block_id": "p69-b1", - "global_id": 1666, - "bbox": [ - 179.46, - 202.67, - 410.63, - 223.76 - ], - "text": "0\n0.05\n0.1\n0.15\n0.2\nt", - "type": "text" - }, - { - "block_id": "p69-b2", - "global_id": 1667, - "bbox": [ - 169.24, - 193.45, - 177.24, - 201.45 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p69-b3", - "global_id": 1668, - "bbox": [ - 163.24, - 167.02, - 177.24, - 175.02 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p69-b4", - "global_id": 1669, - "bbox": [ - 173.24, - 140.57, - 177.24, - 148.57 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p69-b5", - "global_id": 1670, - "bbox": [ - 166.48, - 114.13, - 177.17, - 122.13 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p69-b6", - "global_id": 1671, - "bbox": [ - 173.24, - 87.7, - 177.24, - 95.7 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p69-b7", - "global_id": 1672, - "bbox": [ - 152.26, - 124.48, - 161.06, - 162.13 - ], - "text": "Amplitude", - "type": "text" - }, - { - "block_id": "p69-b8", - "global_id": 1673, - "bbox": [ - 381.58, - 104.54, - 390.77, - 111.74 - ], - "text": "f(t)", - "type": "text" - }, - { - "block_id": "p69-b9", - "global_id": 1674, - "bbox": [ - 381.58, - 115.57, - 391.97, - 122.77 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p69-b10", - "global_id": 1675, - "bbox": [ - 151.5, - 230.06, - 341.1, - 239.67 - ], - "text": "Figure B.14 Graphical comparison of f(t) and h(t).", - "type": "text" - }, - { - "block_id": "p69-b11", - "global_id": 1676, - "bbox": [ - 127.59, - 262.28, - 516.16, - 284.19 - ], - "text": "Fig. B.14. It is also possible, although more cumbersome, to use pull down menus to modify line\nproperties and to add labels and legends directly in the figure window.", - "type": "text" - }, - { - "block_id": "p69-b12", - "global_id": 1677, - "bbox": [ - 127.59, - 311.04, - 260.58, - 323.0 - ], - "text": "B.7-6 Matrix Operations", - "type": "text" - }, - { - "block_id": "p69-b13", - "global_id": 1678, - "bbox": [ - 127.59, - 329.13, - 516.14, - 351.05 - ], - "text": "Many applications require more than row vectors with evenly spaced elements; row vectors,\ncolumn vectors, and matrices with arbitrary elements are typically needed.", - "type": "text" - }, - { - "block_id": "p69-b14", - "global_id": 1679, - "bbox": [ - 127.59, - 353.04, - 516.15, - 410.82 - ], - "text": "MATLAB provides several functions to generate common, useful matrices. Given integers\nm, n, and vector x, the function eye(m) creates the m×m identity matrix; the function ones(m,n)\ncreates the m × n matrix of all ones; the function zeros(m,n) creates the m × n matrix of all\nzeros; and the function diag(x) uses vector x to create a diagonal matrix. The creation of general\nmatrices and vectors, however, requires each individual element to be specified.", - "type": "text" - }, - { - "block_id": "p69-b15", - "global_id": 1680, - "bbox": [ - 127.59, - 412.82, - 516.11, - 434.74 - ], - "text": "Vectors and matrices can be input spreadsheet style by using MATLAB’s array editor. This\ngraphical approach is rather cumbersome and is not often used. A more direct method is preferable.", - "type": "text" - }, - { - "block_id": "p69-b16", - "global_id": 1681, - "bbox": [ - 145.52, - 436.64, - 271.84, - 446.69 - ], - "text": "Consider a simple row vector r,", - "type": "text" - }, - { - "block_id": "p69-b17", - "global_id": 1682, - "bbox": [ - 287.98, - 460.23, - 355.73, - 470.7 - ], - "text": "r = [ 1\n0\n0 ]", - "type": "text" - }, - { - "block_id": "p69-b18", - "global_id": 1683, - "bbox": [ - 127.59, - 484.76, - 516.13, - 506.67 - ], - "text": "The MATLAB notation a:b:c cannot create this row vector. Rather, square brackets are used to\ncreate r.", - "type": "text" - }, - { - "block_id": "p69-b19", - "global_id": 1684, - "bbox": [ - 127.59, - 519.02, - 226.95, - 540.94 - ], - "text": ">>\nr = [1 0 0]\nr = 1\n0\n0", - "type": "text" - }, - { - "block_id": "p69-b20", - "global_id": 1685, - "bbox": [ - 127.59, - 552.71, - 516.13, - 574.62 - ], - "text": "Square brackets enclose elements of the vector, and spaces or commas are used to separate row\nelements.", - "type": "text" - }, - { - "block_id": "p69-b21", - "global_id": 1686, - "bbox": [ - 145.52, - 576.21, - 282.17, - 586.59 - ], - "text": "Next, consider the 3 × 2 matrix A,", - "type": "text" - }, - { - "block_id": "p69-b22", - "global_id": 1687, - "bbox": [ - 290.12, - 611.12, - 307.13, - 621.41 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p69-b23", - "global_id": 1688, - "bbox": [ - 309.18, - 591.16, - 316.45, - 601.12 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p69-b24", - "global_id": 1689, - "bbox": [ - 309.18, - 609.1, - 316.45, - 619.06 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p69-b25", - "global_id": 1690, - "bbox": [ - 321.42, - 599.48, - 341.35, - 633.36 - ], - "text": "2\n3\n4\n5\n0\n6", - "type": "text" - }, - { - "block_id": "p69-b26", - "global_id": 1691, - "bbox": [ - 346.33, - 591.16, - 353.6, - 601.12 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p69-b27", - "global_id": 1692, - "bbox": [ - 346.33, - 609.1, - 353.6, - 619.06 - ], - "text": "⎦", - "type": "text" - } - ] - }, - { - "page_num": 70, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p70-b0", - "global_id": 1693, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "50\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p70-b1", - "global_id": 1694, - "bbox": [ - 101.84, - 85.87, - 490.41, - 107.87 - ], - "text": "Matrix A can be viewed as a three-high stack of two-element row vectors. With a semicolon to\nseparate rows, square brackets are used to create the matrix.", - "type": "text" - }, - { - "block_id": "p70-b2", - "global_id": 1695, - "bbox": [ - 101.84, - 117.7, - 211.67, - 163.53 - ], - "text": ">>\nA = [2 3;4 5;0 6]\nA = 2\n3\n4\n5\n0\n6", - "type": "text" - }, - { - "block_id": "p70-b3", - "global_id": 1696, - "bbox": [ - 101.84, - 172.79, - 397.16, - 182.75 - ], - "text": "Each row vector needs to have the same length to create a sensible matrix.", - "type": "text" - }, - { - "block_id": "p70-b4", - "global_id": 1697, - "bbox": [ - 101.84, - 184.74, - 490.4, - 218.62 - ], - "text": "In addition to enclosing string arguments, a single quote performs the complex conjugate\ntranspose operation. In this way, row vectors become column vectors and vice versa. For example,\na column vector c is easily created by transposing row vector r.", - "type": "text" - }, - { - "block_id": "p70-b5", - "global_id": 1698, - "bbox": [ - 101.84, - 228.46, - 154.14, - 250.37 - ], - "text": ">>\nc = r’\nc = 1", - "type": "text" - }, - { - "block_id": "p70-b6", - "global_id": 1699, - "bbox": [ - 143.67, - 252.37, - 148.9, - 274.29 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p70-b7", - "global_id": 1700, - "bbox": [ - 101.84, - 283.46, - 490.37, - 305.75 - ], - "text": "Since vector r is real, the complex-conjugate transpose is just the transpose. Had r been complex,\nthe simple transpose could have been accomplished by either r.’ or (conj(r))’.", - "type": "text" - }, - { - "block_id": "p70-b8", - "global_id": 1701, - "bbox": [ - 101.84, - 307.46, - 490.41, - 365.24 - ], - "text": "More formally, square brackets are referred to as a concatenation operator. A concatenation\ncombines or connects smaller pieces into a larger whole. Concatenations can involve simple\nnumbers, such as the six-element concatenation used to create the 3×2 matrix A. It is also possible\nto concatenate larger objects, such as vectors and matrices. For example, vector c and matrix A\ncan be concatenated to form a 3 × 3 matrix B.", - "type": "text" - }, - { - "block_id": "p70-b9", - "global_id": 1702, - "bbox": [ - 101.84, - 375.08, - 201.2, - 420.9 - ], - "text": ">>\nB = [c A]\nB = 1\n2\n3\n0\n4\n5\n0\n0\n6", - "type": "text" - }, - { - "block_id": "p70-b10", - "global_id": 1703, - "bbox": [ - 101.84, - 429.75, - 490.38, - 452.08 - ], - "text": "Errors will occur if the component dimensions do not sensibly match; a 2×2 matrix would not be\nconcatenated with a 3 × 3 matrix, for example.", - "type": "text" - }, - { - "block_id": "p70-b11", - "global_id": 1704, - "bbox": [ - 101.84, - 454.08, - 490.41, - 475.99 - ], - "text": "Elements of a matrix are indexed much like vectors, except two indices are typically used to\nspecify row and column.† Element (1, 2) of matrix B, for example, is 2.", - "type": "text" - }, - { - "block_id": "p70-b12", - "global_id": 1705, - "bbox": [ - 101.84, - 485.83, - 159.37, - 507.74 - ], - "text": ">>\nB(1,2)\nans = 2", - "type": "text" - }, - { - "block_id": "p70-b13", - "global_id": 1706, - "bbox": [ - 101.84, - 517.01, - 490.4, - 538.93 - ], - "text": "Indices can likewise be vectors. For example, vector indices allow us to extract the elements\ncommon to the first two rows and last two columns of matrix B.", - "type": "text" - }, - { - "block_id": "p70-b14", - "global_id": 1707, - "bbox": [ - 101.84, - 548.76, - 185.52, - 582.63 - ], - "text": ">>\nB(1:2,2:3)\nans = 2\n3\n4\n5", - "type": "text" - }, - { - "block_id": "p70-b15", - "global_id": 1708, - "bbox": [ - 101.84, - 598.28, - 490.38, - 643.63 - ], - "text": "† Matrix elements can also be accessed by means of a single index, which enumerates along columns.\nFormally, the element from row m and column n of an M × N matrix may be obtained with a single index\n(n −1)M + m. For example, element (1, 2) of matrix B is accessed by using the index (2 −1)3 + 1 = 4. That\nis, B(4) yields 2.", - "type": "text" - } - ] - }, - { - "page_num": 71, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p71-b0", - "global_id": 1709, - "bbox": [ - 317.22, - 62.89, - 516.13, - 71.98 - ], - "text": "B.7\nMATLAB: Elementary Operations\n51", - "type": "text" - }, - { - "block_id": "p71-b1", - "global_id": 1710, - "bbox": [ - 127.59, - 85.82, - 516.13, - 119.69 - ], - "text": "One indexing technique is particularly useful and deserves special attention. A colon can be\nused to specify all elements along a specified dimension. For example, B(2,:) selects all column\nelements along the second row of B.", - "type": "text" - }, - { - "block_id": "p71-b2", - "global_id": 1711, - "bbox": [ - 127.59, - 132.38, - 237.41, - 154.3 - ], - "text": ">>\nB(2,:)\nans = 0\n4\n5", - "type": "text" - }, - { - "block_id": "p71-b3", - "global_id": 1712, - "bbox": [ - 127.59, - 166.41, - 516.15, - 200.29 - ], - "text": "Now that we understand basic vector and matrix creation, we turn our attention to using these\ntools on real problems. Consider solving a set of three linear simultaneous equations in three\nunknowns.", - "type": "text" - }, - { - "block_id": "p71-b4", - "global_id": 1713, - "bbox": [ - 296.29, - 214.26, - 370.07, - 225.72 - ], - "text": "x1 −2x2 + 3x3 = 1", - "type": "text" - }, - { - "block_id": "p71-b5", - "global_id": 1714, - "bbox": [ - 271.67, - 231.66, - 279.44, - 241.62 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p71-b6", - "global_id": 1715, - "bbox": [ - 279.43, - 222.78, - 287.86, - 232.74 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p71-b7", - "global_id": 1716, - "bbox": [ - 287.86, - 222.78, - 339.83, - 243.12 - ], - "text": "3x1 + x2 −\n√", - "type": "text" - }, - { - "block_id": "p71-b8", - "global_id": 1717, - "bbox": [ - 339.83, - 231.66, - 371.06, - 243.12 - ], - "text": "5x3 = π", - "type": "text" - }, - { - "block_id": "p71-b9", - "global_id": 1718, - "bbox": [ - 287.87, - 240.17, - 320.55, - 260.38 - ], - "text": "3x1 −\n√", - "type": "text" - }, - { - "block_id": "p71-b10", - "global_id": 1719, - "bbox": [ - 320.55, - 248.92, - 369.51, - 260.38 - ], - "text": "7x2 + x3 = e", - "type": "text" - }, - { - "block_id": "p71-b11", - "global_id": 1720, - "bbox": [ - 127.59, - 273.29, - 457.93, - 283.66 - ], - "text": "This system of equations is represented in matrix form according to Ax = y, where", - "type": "text" - }, - { - "block_id": "p71-b12", - "global_id": 1721, - "bbox": [ - 179.99, - 310.13, - 197.01, - 320.43 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p71-b13", - "global_id": 1722, - "bbox": [ - 199.05, - 290.17, - 206.32, - 300.13 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p71-b14", - "global_id": 1723, - "bbox": [ - 199.05, - 308.1, - 206.32, - 318.06 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p71-b15", - "global_id": 1724, - "bbox": [ - 211.3, - 297.21, - 286.66, - 320.07 - ], - "text": "1\n−2\n3\n−", - "type": "text" - }, - { - "block_id": "p71-b16", - "global_id": 1725, - "bbox": [ - 219.07, - 301.72, - 227.5, - 311.69 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p71-b17", - "global_id": 1726, - "bbox": [ - 227.49, - 310.1, - 281.35, - 320.48 - ], - "text": "3\n1\n−", - "type": "text" - }, - { - "block_id": "p71-b18", - "global_id": 1727, - "bbox": [ - 281.35, - 301.73, - 289.78, - 311.69 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p71-b19", - "global_id": 1728, - "bbox": [ - 219.39, - 310.52, - 294.76, - 333.24 - ], - "text": "5\n3\n−", - "type": "text" - }, - { - "block_id": "p71-b20", - "global_id": 1729, - "bbox": [ - 250.21, - 314.62, - 258.64, - 324.59 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p71-b21", - "global_id": 1730, - "bbox": [ - 258.64, - 323.28, - 286.66, - 333.24 - ], - "text": "7\n1", - "type": "text" - }, - { - "block_id": "p71-b22", - "global_id": 1731, - "bbox": [ - 299.74, - 290.17, - 307.01, - 300.13 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p71-b23", - "global_id": 1732, - "bbox": [ - 299.74, - 308.1, - 346.44, - 320.51 - ], - "text": "⎦,\nx =", - "type": "text" - }, - { - "block_id": "p71-b24", - "global_id": 1733, - "bbox": [ - 348.49, - 290.17, - 355.75, - 300.13 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p71-b25", - "global_id": 1734, - "bbox": [ - 348.49, - 308.1, - 355.75, - 318.06 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p71-b26", - "global_id": 1735, - "bbox": [ - 360.73, - 298.39, - 368.63, - 333.14 - ], - "text": "x1\nx2\nx3", - "type": "text" - }, - { - "block_id": "p71-b27", - "global_id": 1736, - "bbox": [ - 374.12, - 290.17, - 381.38, - 300.13 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p71-b28", - "global_id": 1737, - "bbox": [ - 374.12, - 308.1, - 440.18, - 320.51 - ], - "text": "⎦, and\ny =", - "type": "text" - }, - { - "block_id": "p71-b29", - "global_id": 1738, - "bbox": [ - 442.22, - 290.17, - 449.49, - 300.13 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p71-b30", - "global_id": 1739, - "bbox": [ - 442.22, - 308.1, - 449.49, - 318.06 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p71-b31", - "global_id": 1740, - "bbox": [ - 449.49, - 298.49, - 455.47, - 319.99 - ], - "text": "1\nπ", - "type": "text" - }, - { - "block_id": "p71-b32", - "global_id": 1741, - "bbox": [ - 450.76, - 322.31, - 455.19, - 332.27 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p71-b33", - "global_id": 1742, - "bbox": [ - 456.46, - 290.17, - 463.72, - 300.13 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p71-b34", - "global_id": 1743, - "bbox": [ - 456.46, - 308.1, - 463.72, - 318.06 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p71-b35", - "global_id": 1744, - "bbox": [ - 127.59, - 346.76, - 516.13, - 381.05 - ], - "text": "Although Cramer’s rule can be used to solve Ax = y, it is more convenient to solve by multiplying\nboth sides by the matrix inverse of A. That is, x = A−1Ax = A−1y. Solving for x by hand or by\ncalculator would be tedious at best, so MATLAB is used. We first create A and y.", - "type": "text" - }, - { - "block_id": "p71-b36", - "global_id": 1745, - "bbox": [ - 127.59, - 393.74, - 488.42, - 403.7 - ], - "text": ">>\nA = [1 -2 3;-sqrt(3) 1 -sqrt(5);3 -sqrt(7) 1]; y = [1;pi;exp(1)];", - "type": "text" - }, - { - "block_id": "p71-b37", - "global_id": 1746, - "bbox": [ - 127.59, - 415.82, - 382.77, - 426.07 - ], - "text": "The vector solution is found by using MATLAB’s inv function.", - "type": "text" - }, - { - "block_id": "p71-b38", - "global_id": 1747, - "bbox": [ - 127.59, - 438.46, - 211.27, - 460.39 - ], - "text": ">>\nx = inv(A)*y\nx = -1.9999", - "type": "text" - }, - { - "block_id": "p71-b39", - "global_id": 1748, - "bbox": [ - 169.42, - 462.38, - 206.04, - 484.29 - ], - "text": "-3.8998\n-1.5999", - "type": "text" - }, - { - "block_id": "p71-b40", - "global_id": 1749, - "bbox": [ - 127.59, - 495.99, - 516.14, - 530.28 - ], - "text": "It is also possible to use MATLAB’s left divide operator x = A\\y to find the same solution.\nThe left divide is generally more computationally efficient than matrix inverses. As with matrix\nmultiplication, left division requires that the two arguments be conformable.", - "type": "text" - }, - { - "block_id": "p71-b41", - "global_id": 1750, - "bbox": [ - 127.59, - 532.17, - 516.13, - 554.48 - ], - "text": "Of course, Cramer’s rule can be used to compute individual solutions, such as x1, by using\nvector indexing, concatenation, and MATLAB’s det command to compute determinants.", - "type": "text" - }, - { - "block_id": "p71-b42", - "global_id": 1751, - "bbox": [ - 127.59, - 566.88, - 300.19, - 588.8 - ], - "text": ">>\nx1 = det([y,A(:,2:3)])/det(A)\nx1 = -1.9999", - "type": "text" - }, - { - "block_id": "p71-b43", - "global_id": 1752, - "bbox": [ - 127.59, - 600.91, - 516.15, - 635.27 - ], - "text": "Another nice application of matrices is the simultaneous creation of a family of curves.\nConsider hα(t) = e−αt sin(2π10t + π/6) over 0 ≤t ≤0.2. Figure B.14 shows hα(t) for α = 0\nand α = 10. Let’s investigate the family of curves hα(t) for α = [0,1,...,10].", - "type": "text" - } - ] - }, - { - "page_num": 72, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p72-b0", - "global_id": 1753, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "52\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p72-b1", - "global_id": 1754, - "bbox": [ - 101.85, - 85.46, - 490.39, - 119.75 - ], - "text": "An inefficient way to solve this problem is create hα(t) for each α of interest. This requires 11\nindividual cases. Instead, a matrix approach allows all 11 curves to be computed simultaneously.\nFirst, a vector is created that contains the desired values of α.", - "type": "text" - }, - { - "block_id": "p72-b2", - "global_id": 1755, - "bbox": [ - 101.85, - 130.76, - 201.22, - 140.73 - ], - "text": ">>\nalpha = (0:10);", - "type": "text" - }, - { - "block_id": "p72-b3", - "global_id": 1756, - "bbox": [ - 101.85, - 150.74, - 461.31, - 161.12 - ], - "text": "By using a sampling interval of one millisecond, t = 0.001, a time vector is also created.", - "type": "text" - }, - { - "block_id": "p72-b4", - "global_id": 1757, - "bbox": [ - 101.85, - 172.14, - 222.14, - 182.1 - ], - "text": ">>\nt = (0:0.001:0.2)’;", - "type": "text" - }, - { - "block_id": "p72-b5", - "global_id": 1758, - "bbox": [ - 101.85, - 192.53, - 490.37, - 226.69 - ], - "text": "The result is a length-201 column vector. By replicating the time vector for each of the 11 curves\nrequired, a time matrix T is created. This replication can be accomplished by using an outer product\nbetween t and a 1 × 11 vector of ones.†", - "type": "text" - }, - { - "block_id": "p72-b6", - "global_id": 1759, - "bbox": [ - 101.84, - 237.41, - 211.68, - 247.37 - ], - "text": ">>\nT = t*ones(1,11);", - "type": "text" - }, - { - "block_id": "p72-b7", - "global_id": 1760, - "bbox": [ - 101.84, - 257.39, - 490.37, - 280.02 - ], - "text": "The result is a 201 × 11 matrix that has identical columns. Right multiplying T by a diagonal\nmatrix created from α, columns of T can be individually scaled and the final result is computed.", - "type": "text" - }, - { - "block_id": "p72-b8", - "global_id": 1761, - "bbox": [ - 101.84, - 290.74, - 358.13, - 300.7 - ], - "text": ">>\nH = exp(-T*diag(alpha)).*sin(2*pi*10*T+pi/6);", - "type": "text" - }, - { - "block_id": "p72-b9", - "global_id": 1762, - "bbox": [ - 101.84, - 310.72, - 490.41, - 345.01 - ], - "text": "Here, H is a 201 × 11 matrix, where each column corresponds to a different value of α. That is,\nH = [h0,h1,...,h10], where hα are column vectors. As shown in Fig. B.15, the 11 desired curves\nare simultaneously displayed by using MATLAB’s plot command, which allows matrix arguments.", - "type": "text" - }, - { - "block_id": "p72-b10", - "global_id": 1763, - "bbox": [ - 101.84, - 356.01, - 326.75, - 365.98 - ], - "text": ">>\nplot(t,H); xlabel(’t’); ylabel(’h(t)’);", - "type": "text" - }, - { - "block_id": "p72-b11", - "global_id": 1764, - "bbox": [ - 101.84, - 376.41, - 490.36, - 398.33 - ], - "text": "This example illustrates an important technique called vectorization, which increases execution\nefficiency for interpretive languages such as MATLAB. Algorithm vectorization uses matrix and", - "type": "text" - }, - { - "block_id": "p72-b12", - "global_id": 1765, - "bbox": [ - 153.72, - 548.79, - 384.89, - 569.88 - ], - "text": "0\n0.05\n0.1\n0.15\n0.2\nt", - "type": "text" - }, - { - "block_id": "p72-b13", - "global_id": 1766, - "bbox": [ - 143.5, - 539.57, - 151.5, - 547.58 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p72-b14", - "global_id": 1767, - "bbox": [ - 137.5, - 513.14, - 151.5, - 521.14 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p72-b15", - "global_id": 1768, - "bbox": [ - 147.5, - 486.69, - 151.5, - 494.69 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p72-b16", - "global_id": 1769, - "bbox": [ - 140.74, - 460.26, - 151.43, - 468.26 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p72-b17", - "global_id": 1770, - "bbox": [ - 147.5, - 433.82, - 151.5, - 441.82 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p72-b18", - "global_id": 1771, - "bbox": [ - 126.52, - 482.5, - 135.32, - 495.2 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p72-b19", - "global_id": 1772, - "bbox": [ - 125.76, - 576.19, - 272.94, - 586.0 - ], - "text": "Figure B.15 hα(t) for α = [0,1,...,10].", - "type": "text" - }, - { - "block_id": "p72-b20", - "global_id": 1773, - "bbox": [ - 101.84, - 610.24, - 490.39, - 633.67 - ], - "text": "† The repmat command provides a more flexible method to replicate or tile objects. Equivalently, T =\nrepmat(t,1,11).", - "type": "text" - } - ] - }, - { - "page_num": 73, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p73-b0", - "global_id": 1774, - "bbox": [ - 317.22, - 62.89, - 516.13, - 71.98 - ], - "text": "B.7\nMATLAB: Elementary Operations\n53", - "type": "text" - }, - { - "block_id": "p73-b1", - "global_id": 1775, - "bbox": [ - 127.59, - 85.82, - 516.16, - 107.74 - ], - "text": "vector operations to avoid manual repetition and loop structures. It takes practice and effort to\nbecome proficient at vectorization, but the worthwhile result is efficient, compact code.†", - "type": "text" - }, - { - "block_id": "p73-b2", - "global_id": 1776, - "bbox": [ - 127.59, - 138.05, - 308.73, - 150.0 - ], - "text": "B.7-7 Partial Fraction Expansions", - "type": "text" - }, - { - "block_id": "p73-b3", - "global_id": 1777, - "bbox": [ - 127.59, - 156.13, - 516.15, - 190.01 - ], - "text": "There are a wide variety of techniques and shortcuts to compute the partial fraction expansion\nof rational function F(x) = B(x)/A(x), but few are more simple than the MATLAB residue\ncommand. The basic form of this command is", - "type": "text" - }, - { - "block_id": "p73-b4", - "global_id": 1778, - "bbox": [ - 127.59, - 205.81, - 263.58, - 215.78 - ], - "text": ">>\n[R,P,K] = residue(B,A)", - "type": "text" - }, - { - "block_id": "p73-b5", - "global_id": 1779, - "bbox": [ - 127.59, - 231.0, - 516.14, - 312.7 - ], - "text": "The two input vectors B and A specify the polynomial coefficients of the numerator and\ndenominator, respectively. These vectors are ordered in descending powers of the independent\nvariable. Three vectors are output. The vector R contains the coefficients of each partial fraction,\nand vector P contains the corresponding roots of each partial fraction. For a root repeated r times,\nthe r partial fractions are ordered in ascending powers. When the rational function is not proper,\nthe vector K contains the direct terms, which are ordered in descending powers of the independent\nvariable.", - "type": "text" - }, - { - "block_id": "p73-b6", - "global_id": 1780, - "bbox": [ - 127.59, - 314.69, - 516.13, - 336.61 - ], - "text": "To demonstrate the power of the residue command, consider finding the partial fraction\nexpansion of", - "type": "text" - }, - { - "block_id": "p73-b7", - "global_id": 1781, - "bbox": [ - 214.32, - 342.47, - 296.6, - 364.66 - ], - "text": "F(x) =\nx5 + π", - "type": "text" - }, - { - "block_id": "p73-b8", - "global_id": 1782, - "bbox": [ - 249.72, - 362.71, - 263.5, - 372.99 - ], - "text": "x +", - "type": "text" - }, - { - "block_id": "p73-b9", - "global_id": 1783, - "bbox": [ - 265.04, - 354.29, - 273.47, - 364.25 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p73-b10", - "global_id": 1784, - "bbox": [ - 273.47, - 354.7, - 286.48, - 373.09 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p73-b11", - "global_id": 1785, - "bbox": [ - 286.48, - 362.71, - 300.25, - 372.99 - ], - "text": "x −", - "type": "text" - }, - { - "block_id": "p73-b12", - "global_id": 1786, - "bbox": [ - 301.8, - 354.29, - 310.23, - 364.25 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p73-b13", - "global_id": 1787, - "bbox": [ - 310.23, - 342.47, - 394.98, - 373.09 - ], - "text": "2\n3 =\nx5 + π", - "type": "text" - }, - { - "block_id": "p73-b14", - "global_id": 1788, - "bbox": [ - 337.47, - 358.61, - 355.23, - 371.77 - ], - "text": "x4 −", - "type": "text" - }, - { - "block_id": "p73-b15", - "global_id": 1789, - "bbox": [ - 356.78, - 353.13, - 365.21, - 363.09 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p73-b16", - "global_id": 1790, - "bbox": [ - 365.21, - 353.13, - 397.92, - 371.87 - ], - "text": "8x3 +\n√", - "type": "text" - }, - { - "block_id": "p73-b17", - "global_id": 1791, - "bbox": [ - 397.93, - 361.49, - 428.19, - 371.87 - ], - "text": "32x −4", - "type": "text" - }, - { - "block_id": "p73-b18", - "global_id": 1792, - "bbox": [ - 127.59, - 386.57, - 516.13, - 408.9 - ], - "text": "By hand, the partial fraction expansion of F(x) is difficult to compute. MATLAB, however, makes\nshort work of the expansion.", - "type": "text" - }, - { - "block_id": "p73-b19", - "global_id": 1793, - "bbox": [ - 127.59, - 424.71, - 530.28, - 470.54 - ], - "text": ">>\n[R,P,K] = residue([1 0 0 0 0 pi],[1 -sqrt(8) 0 sqrt(32) -4]); R.’, P.’, K\nR = 7.8888\n5.9713\n3.1107\n0.1112\nP = 1.4142\n1.4142\n1.4142\n-1.4142\nK = 1.0000\n2.8284", - "type": "text" - }, - { - "block_id": "p73-b20", - "global_id": 1794, - "bbox": [ - 127.59, - 485.35, - 388.41, - 495.73 - ], - "text": "Written in standard form, the partial fraction expansion of F(x) is", - "type": "text" - }, - { - "block_id": "p73-b21", - "global_id": 1795, - "bbox": [ - 190.01, - 511.23, - 303.02, - 528.18 - ], - "text": "F(x) = x + 2.8284 + 7.8888", - "type": "text" - }, - { - "block_id": "p73-b22", - "global_id": 1796, - "bbox": [ - 274.97, - 526.27, - 288.74, - 536.54 - ], - "text": "x −", - "type": "text" - }, - { - "block_id": "p73-b23", - "global_id": 1797, - "bbox": [ - 290.29, - 517.84, - 298.71, - 527.8 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p73-b24", - "global_id": 1798, - "bbox": [ - 298.72, - 511.23, - 350.72, - 536.64 - ], - "text": "2\n+\n5.9713", - "type": "text" - }, - { - "block_id": "p73-b25", - "global_id": 1799, - "bbox": [ - 316.96, - 526.27, - 334.45, - 536.54 - ], - "text": "(x −", - "type": "text" - }, - { - "block_id": "p73-b26", - "global_id": 1800, - "bbox": [ - 336.0, - 517.84, - 344.43, - 527.8 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p73-b27", - "global_id": 1801, - "bbox": [ - 344.43, - 511.23, - 404.13, - 536.64 - ], - "text": "2)2 +\n3.1107", - "type": "text" - }, - { - "block_id": "p73-b28", - "global_id": 1802, - "bbox": [ - 370.38, - 526.27, - 387.87, - 536.54 - ], - "text": "(x −", - "type": "text" - }, - { - "block_id": "p73-b29", - "global_id": 1803, - "bbox": [ - 389.41, - 517.84, - 397.84, - 527.8 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p73-b30", - "global_id": 1804, - "bbox": [ - 397.85, - 511.23, - 451.84, - 536.64 - ], - "text": "2)3 + 0.1112", - "type": "text" - }, - { - "block_id": "p73-b31", - "global_id": 1805, - "bbox": [ - 423.78, - 526.27, - 437.56, - 536.54 - ], - "text": "x +", - "type": "text" - }, - { - "block_id": "p73-b32", - "global_id": 1806, - "bbox": [ - 439.1, - 517.84, - 447.53, - 527.8 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p73-b33", - "global_id": 1807, - "bbox": [ - 447.54, - 526.68, - 452.52, - 536.64 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p73-b34", - "global_id": 1808, - "bbox": [ - 127.59, - 551.72, - 516.13, - 597.84 - ], - "text": "The signal–processing toolbox function residuez is similar to the residue command\nand offers more convenient expansion of certain rational functions, such as those commonly\nencountered in the study of discrete-time systems. Additional information about the residue and\nresiduez commands is available from MATLAB’s help facilities.", - "type": "text" - }, - { - "block_id": "p73-b35", - "global_id": 1809, - "bbox": [ - 127.59, - 621.19, - 425.98, - 633.41 - ], - "text": "† The benefits of vectorization are less pronounced in recent versions of MATLAB.", - "type": "text" - } - ] - }, - { - "page_num": 74, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p74-b0", - "global_id": 1810, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "54\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p74-b1", - "global_id": 1811, - "bbox": [ - 102.2, - 92.21, - 441.45, - 106.15 - ], - "text": "B.8 APPENDIX: USEFUL MATHEMATICAL FORMULAS", - "type": "text" - }, - { - "block_id": "p74-b2", - "global_id": 1812, - "bbox": [ - 101.84, - 112.14, - 384.54, - 122.1 - ], - "text": "We conclude this chapter with a selection of useful mathematical facts.", - "type": "text" - }, - { - "block_id": "p74-b3", - "global_id": 1813, - "bbox": [ - 101.84, - 157.02, - 261.73, - 168.97 - ], - "text": "B.8-1 Some Useful Constants", - "type": "text" - }, - { - "block_id": "p74-b4", - "global_id": 1814, - "bbox": [ - 101.84, - 176.87, - 177.93, - 202.19 - ], - "text": "π ≈3.1415926535\ne ≈2.7182818284", - "type": "text" - }, - { - "block_id": "p74-b5", - "global_id": 1815, - "bbox": [ - 103.04, - 206.58, - 178.33, - 230.5 - ], - "text": "1\ne ≈0.3678794411", - "type": "text" - }, - { - "block_id": "p74-b6", - "global_id": 1816, - "bbox": [ - 102.95, - 232.7, - 173.47, - 245.58 - ], - "text": "log10 2 ≈0.30103", - "type": "text" - }, - { - "block_id": "p74-b7", - "global_id": 1817, - "bbox": [ - 102.95, - 247.64, - 173.47, - 260.52 - ], - "text": "log10 3 ≈0.47712", - "type": "text" - }, - { - "block_id": "p74-b8", - "global_id": 1818, - "bbox": [ - 101.84, - 292.93, - 237.51, - 304.88 - ], - "text": "B.8-2 Complex Numbers", - "type": "text" - }, - { - "block_id": "p74-b9", - "global_id": 1819, - "bbox": [ - 101.84, - 308.66, - 147.85, - 323.05 - ], - "text": "e±jπ/2 = ±j", - "type": "text" - }, - { - "block_id": "p74-b10", - "global_id": 1820, - "bbox": [ - 101.84, - 330.68, - 132.33, - 345.08 - ], - "text": "e±jnπ =", - "type": "text" - }, - { - "block_id": "p74-b11", - "global_id": 1821, - "bbox": [ - 134.38, - 320.81, - 187.73, - 351.05 - ], - "text": "1\nneven\n−1\nnodd", - "type": "text" - }, - { - "block_id": "p74-b12", - "global_id": 1822, - "bbox": [ - 101.84, - 352.89, - 183.62, - 367.39 - ], - "text": "e±jθ = cos θ ± jsin θ", - "type": "text" - }, - { - "block_id": "p74-b13", - "global_id": 1823, - "bbox": [ - 101.84, - 372.23, - 185.3, - 384.23 - ], - "text": "a + jb = rejθ\nr =", - "type": "text" - }, - { - "block_id": "p74-b14", - "global_id": 1824, - "bbox": [ - 187.34, - 365.52, - 195.77, - 375.48 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p74-b15", - "global_id": 1825, - "bbox": [ - 195.77, - 365.94, - 278.2, - 384.33 - ], - "text": "a2 + b2, θ = tan−1 b", - "type": "text" - }, - { - "block_id": "p74-b16", - "global_id": 1826, - "bbox": [ - 274.72, - 380.1, - 278.2, - 387.07 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p74-b18", - "global_id": 1827, - "bbox": [ - 101.84, - 386.21, - 159.44, - 400.6 - ], - "text": "(rejθ)k = rkejkθ", - "type": "text" - }, - { - "block_id": "p74-b19", - "global_id": 1828, - "bbox": [ - 101.84, - 405.07, - 216.96, - 417.95 - ], - "text": "(r1ejθ1)(r2ejθ2) = r1r2ej(θ1+θ2)", - "type": "text" - }, - { - "block_id": "p74-b20", - "global_id": 1829, - "bbox": [ - 101.84, - 452.08, - 164.78, - 464.04 - ], - "text": "B.8-3 Sums", - "type": "text" - }, - { - "block_id": "p74-b21", - "global_id": 1830, - "bbox": [ - 102.95, - 467.89, - 117.05, - 478.34 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p74-b22", - "global_id": 1831, - "bbox": [ - 103.16, - 492.14, - 116.84, - 499.33 - ], - "text": "k=m", - "type": "text" - }, - { - "block_id": "p74-b23", - "global_id": 1832, - "bbox": [ - 118.16, - 467.24, - 176.05, - 488.12 - ], - "text": "rk = rn+1 −rm", - "type": "text" - }, - { - "block_id": "p74-b24", - "global_id": 1833, - "bbox": [ - 147.82, - 477.85, - 218.93, - 495.3 - ], - "text": "r −1\nr̸ = 1", - "type": "text" - }, - { - "block_id": "p74-b25", - "global_id": 1834, - "bbox": [ - 102.95, - 501.7, - 117.05, - 512.15 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p74-b26", - "global_id": 1835, - "bbox": [ - 103.93, - 526.04, - 116.07, - 533.3 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p74-b27", - "global_id": 1836, - "bbox": [ - 118.16, - 504.67, - 169.04, - 521.93 - ], - "text": "k = n(n + 1)", - "type": "text" - }, - { - "block_id": "p74-b28", - "global_id": 1837, - "bbox": [ - 149.94, - 519.14, - 154.92, - 529.1 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p74-b29", - "global_id": 1838, - "bbox": [ - 102.95, - 535.61, - 117.05, - 546.06 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p74-b30", - "global_id": 1839, - "bbox": [ - 103.93, - 559.95, - 116.07, - 567.21 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p74-b31", - "global_id": 1840, - "bbox": [ - 118.16, - 538.58, - 206.25, - 555.84 - ], - "text": "k2 = n(n + 1)(2n + 1)", - "type": "text" - }, - { - "block_id": "p74-b32", - "global_id": 1841, - "bbox": [ - 170.55, - 553.05, - 175.53, - 563.01 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p74-b33", - "global_id": 1842, - "bbox": [ - 102.95, - 569.51, - 117.05, - 579.97 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p74-b34", - "global_id": 1843, - "bbox": [ - 103.93, - 593.85, - 116.07, - 601.12 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p74-b35", - "global_id": 1844, - "bbox": [ - 118.16, - 571.25, - 233.28, - 589.74 - ], - "text": "krk = r + [n(r −1) −1]rn+1", - "type": "text" - }, - { - "block_id": "p74-b36", - "global_id": 1845, - "bbox": [ - 173.58, - 579.47, - 276.17, - 596.92 - ], - "text": "(r −1)2\nr̸ = 1", - "type": "text" - }, - { - "block_id": "p74-b37", - "global_id": 1846, - "bbox": [ - 102.95, - 603.42, - 117.05, - 613.87 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p74-b38", - "global_id": 1847, - "bbox": [ - 103.93, - 627.77, - 116.07, - 635.03 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p74-b39", - "global_id": 1848, - "bbox": [ - 118.16, - 602.78, - 333.83, - 623.65 - ], - "text": "k2 rk = r[(1 + r)(1 −rn) −2n(1 −r)rn −n2(1 −r)2rn]", - "type": "text" - }, - { - "block_id": "p74-b40", - "global_id": 1849, - "bbox": [ - 225.59, - 613.38, - 376.2, - 630.83 - ], - "text": "(1 −r)3\nr̸ = 1", - "type": "text" - } - ] - }, - { - "page_num": 75, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p75-b0", - "global_id": 1850, - "bbox": [ - 285.06, - 62.89, - 516.12, - 71.98 - ], - "text": "B.8\nAppendix: Useful Mathematical Formulas\n55", - "type": "text" - }, - { - "block_id": "p75-b1", - "global_id": 1851, - "bbox": [ - 127.59, - 86.52, - 313.72, - 98.48 - ], - "text": "B.8-4 Taylor and Maclaurin Series", - "type": "text" - }, - { - "block_id": "p75-b2", - "global_id": 1852, - "bbox": [ - 127.59, - 109.62, - 212.0, - 126.88 - ], - "text": "f(x) = f(a) + (x −a)", - "type": "text" - }, - { - "block_id": "p75-b3", - "global_id": 1853, - "bbox": [ - 194.26, - 108.68, - 273.12, - 134.06 - ], - "text": "1!\n˙f(a) + (x −a)2", - "type": "text" - }, - { - "block_id": "p75-b4", - "global_id": 1854, - "bbox": [ - 253.9, - 114.55, - 325.22, - 134.06 - ], - "text": "2!\n¨f(a) + · · · =", - "type": "text" - }, - { - "block_id": "p75-b5", - "global_id": 1855, - "bbox": [ - 327.27, - 106.43, - 341.36, - 117.11 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p75-b6", - "global_id": 1856, - "bbox": [ - 328.24, - 131.0, - 340.37, - 138.27 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p75-b7", - "global_id": 1857, - "bbox": [ - 343.67, - 108.61, - 374.49, - 119.9 - ], - "text": "(x −a)k", - "type": "text" - }, - { - "block_id": "p75-b8", - "global_id": 1858, - "bbox": [ - 355.72, - 114.88, - 401.86, - 133.95 - ], - "text": "k!\nf (k)(a)", - "type": "text" - }, - { - "block_id": "p75-b9", - "global_id": 1859, - "bbox": [ - 127.59, - 144.04, - 190.34, - 161.08 - ], - "text": "f(x) = f(0) + x", - "type": "text" - }, - { - "block_id": "p75-b10", - "global_id": 1860, - "bbox": [ - 184.27, - 142.79, - 229.84, - 168.16 - ], - "text": "1!\n˙f(0) + x2", - "type": "text" - }, - { - "block_id": "p75-b11", - "global_id": 1861, - "bbox": [ - 222.26, - 148.65, - 281.94, - 168.16 - ], - "text": "2!\n¨f(0) + · · · =", - "type": "text" - }, - { - "block_id": "p75-b12", - "global_id": 1862, - "bbox": [ - 283.99, - 140.53, - 298.08, - 151.21 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p75-b13", - "global_id": 1863, - "bbox": [ - 284.97, - 165.1, - 297.1, - 172.37 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p75-b14", - "global_id": 1864, - "bbox": [ - 300.38, - 142.72, - 307.93, - 154.0 - ], - "text": "xk", - "type": "text" - }, - { - "block_id": "p75-b15", - "global_id": 1865, - "bbox": [ - 300.79, - 148.98, - 335.29, - 168.06 - ], - "text": "k! f (k)(0)", - "type": "text" - }, - { - "block_id": "p75-b16", - "global_id": 1866, - "bbox": [ - 127.59, - 205.8, - 230.69, - 217.75 - ], - "text": "B.8-5 Power Series", - "type": "text" - }, - { - "block_id": "p75-b17", - "global_id": 1867, - "bbox": [ - 127.59, - 223.46, - 187.8, - 241.76 - ], - "text": "ex = 1 + x + x2", - "type": "text" - }, - { - "block_id": "p75-b18", - "global_id": 1868, - "bbox": [ - 180.22, - 223.46, - 209.5, - 248.83 - ], - "text": "2! + x3", - "type": "text" - }, - { - "block_id": "p75-b19", - "global_id": 1869, - "bbox": [ - 201.91, - 223.39, - 255.14, - 248.83 - ], - "text": "3! + · · · + xn", - "type": "text" - }, - { - "block_id": "p75-b20", - "global_id": 1870, - "bbox": [ - 247.55, - 231.38, - 280.78, - 248.73 - ], - "text": "n! + · · ·", - "type": "text" - }, - { - "block_id": "p75-b21", - "global_id": 1871, - "bbox": [ - 128.7, - 249.99, - 183.31, - 268.29 - ], - "text": "sin x = x −x3", - "type": "text" - }, - { - "block_id": "p75-b22", - "global_id": 1872, - "bbox": [ - 175.73, - 249.99, - 205.01, - 275.36 - ], - "text": "3! + x5", - "type": "text" - }, - { - "block_id": "p75-b23", - "global_id": 1873, - "bbox": [ - 197.42, - 249.99, - 226.7, - 275.36 - ], - "text": "5! −x7", - "type": "text" - }, - { - "block_id": "p75-b24", - "global_id": 1874, - "bbox": [ - 219.12, - 257.91, - 252.35, - 275.36 - ], - "text": "7! + · · ·", - "type": "text" - }, - { - "block_id": "p75-b25", - "global_id": 1875, - "bbox": [ - 128.7, - 276.51, - 185.49, - 294.8 - ], - "text": "cos x = 1 −x2", - "type": "text" - }, - { - "block_id": "p75-b26", - "global_id": 1876, - "bbox": [ - 177.91, - 276.51, - 207.19, - 301.88 - ], - "text": "2! + x4", - "type": "text" - }, - { - "block_id": "p75-b27", - "global_id": 1877, - "bbox": [ - 199.61, - 276.51, - 228.89, - 301.88 - ], - "text": "4! −x6", - "type": "text" - }, - { - "block_id": "p75-b28", - "global_id": 1878, - "bbox": [ - 221.31, - 276.51, - 250.59, - 301.88 - ], - "text": "6! + x8", - "type": "text" - }, - { - "block_id": "p75-b29", - "global_id": 1879, - "bbox": [ - 243.01, - 284.43, - 276.23, - 301.88 - ], - "text": "8! −· · ·", - "type": "text" - }, - { - "block_id": "p75-b30", - "global_id": 1880, - "bbox": [ - 128.7, - 303.03, - 183.86, - 321.33 - ], - "text": "tan x = x + x3", - "type": "text" - }, - { - "block_id": "p75-b31", - "global_id": 1881, - "bbox": [ - 177.65, - 303.03, - 210.53, - 328.39 - ], - "text": "3 + 2x5", - "type": "text" - }, - { - "block_id": "p75-b32", - "global_id": 1882, - "bbox": [ - 199.35, - 303.03, - 242.19, - 328.39 - ], - "text": "15 + 17x7", - "type": "text" - }, - { - "block_id": "p75-b33", - "global_id": 1883, - "bbox": [ - 226.04, - 306.83, - 327.8, - 328.39 - ], - "text": "315 + · · ·\nx2 < π2/4", - "type": "text" - }, - { - "block_id": "p75-b34", - "global_id": 1884, - "bbox": [ - 128.7, - 329.53, - 188.83, - 347.83 - ], - "text": "tanh x = x −x3", - "type": "text" - }, - { - "block_id": "p75-b35", - "global_id": 1885, - "bbox": [ - 182.64, - 329.53, - 215.52, - 354.91 - ], - "text": "3 + 2x5", - "type": "text" - }, - { - "block_id": "p75-b36", - "global_id": 1886, - "bbox": [ - 204.34, - 329.53, - 247.17, - 354.91 - ], - "text": "15 −17x7", - "type": "text" - }, - { - "block_id": "p75-b37", - "global_id": 1887, - "bbox": [ - 231.01, - 333.34, - 332.79, - 354.91 - ], - "text": "315 + · · ·\nx2 < π2/4", - "type": "text" - }, - { - "block_id": "p75-b38", - "global_id": 1888, - "bbox": [ - 127.59, - 356.08, - 241.76, - 373.44 - ], - "text": "(1 + x)n = 1 + nx + n(n −1)", - "type": "text" - }, - { - "block_id": "p75-b39", - "global_id": 1889, - "bbox": [ - 221.28, - 356.08, - 324.95, - 380.51 - ], - "text": "2!\nx2 + n(n −1)(n −2)", - "type": "text" - }, - { - "block_id": "p75-b40", - "global_id": 1890, - "bbox": [ - 290.36, - 358.95, - 367.88, - 380.51 - ], - "text": "3!\nx3 + · · · +", - "type": "text" - }, - { - "block_id": "p75-b41", - "global_id": 1891, - "bbox": [ - 369.42, - 349.08, - 381.14, - 366.35 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p75-b42", - "global_id": 1892, - "bbox": [ - 376.35, - 370.45, - 380.78, - 380.41 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p75-b44", - "global_id": 1893, - "bbox": [ - 387.86, - 358.95, - 438.76, - 373.34 - ], - "text": "xk + · · · + xn", - "type": "text" - }, - { - "block_id": "p75-b45", - "global_id": 1894, - "bbox": [ - 127.59, - 380.76, - 293.04, - 423.67 - ], - "text": "(1 + x)n ≈1 + nx\n|x| ≪1\n1\n1 −x = 1 + x + x2 + x3 + · · ·\n|x| < 1", - "type": "text" - }, - { - "block_id": "p75-b46", - "global_id": 1895, - "bbox": [ - 127.59, - 456.29, - 292.45, - 468.24 - ], - "text": "B.8-6 Trigonometric Identities", - "type": "text" - }, - { - "block_id": "p75-b47", - "global_id": 1896, - "bbox": [ - 127.59, - 473.23, - 207.83, - 487.72 - ], - "text": "e±jx = cos x ± jsin x", - "type": "text" - }, - { - "block_id": "p75-b48", - "global_id": 1897, - "bbox": [ - 128.7, - 492.34, - 165.19, - 504.06 - ], - "text": "cos x = 1", - "type": "text" - }, - { - "block_id": "p75-b49", - "global_id": 1898, - "bbox": [ - 161.7, - 490.07, - 209.29, - 506.88 - ], - "text": "2[ejx + e−jx]", - "type": "text" - }, - { - "block_id": "p75-b50", - "global_id": 1899, - "bbox": [ - 128.7, - 508.69, - 164.51, - 520.41 - ], - "text": "sin x = 1", - "type": "text" - }, - { - "block_id": "p75-b51", - "global_id": 1900, - "bbox": [ - 160.06, - 506.41, - 209.57, - 523.22 - ], - "text": "2j[ejx −e−jx]", - "type": "text" - }, - { - "block_id": "p75-b52", - "global_id": 1901, - "bbox": [ - 128.7, - 525.71, - 167.5, - 537.72 - ], - "text": "cos(x ± π", - "type": "text" - }, - { - "block_id": "p75-b53", - "global_id": 1902, - "bbox": [ - 164.01, - 527.35, - 212.12, - 540.85 - ], - "text": "2 ) = ∓sin x", - "type": "text" - }, - { - "block_id": "p75-b54", - "global_id": 1903, - "bbox": [ - 128.7, - 540.66, - 165.85, - 552.67 - ], - "text": "sin(x ± π", - "type": "text" - }, - { - "block_id": "p75-b55", - "global_id": 1904, - "bbox": [ - 127.59, - 542.3, - 212.12, - 567.61 - ], - "text": "2 ) = ±cos x\n2sin xcos x = sin 2x", - "type": "text" - }, - { - "block_id": "p75-b56", - "global_id": 1905, - "bbox": [ - 128.7, - 569.87, - 200.41, - 584.05 - ], - "text": "sin2 x + cos2 x = 1", - "type": "text" - }, - { - "block_id": "p75-b57", - "global_id": 1906, - "bbox": [ - 128.7, - 586.3, - 220.32, - 600.48 - ], - "text": "cos2 x −sin2 x = cos 2x", - "type": "text" - }, - { - "block_id": "p75-b58", - "global_id": 1907, - "bbox": [ - 128.7, - 602.74, - 168.07, - 616.92 - ], - "text": "cos2 x = 1", - "type": "text" - }, - { - "block_id": "p75-b59", - "global_id": 1908, - "bbox": [ - 164.58, - 606.55, - 217.47, - 619.74 - ], - "text": "2(1 + cos 2x)", - "type": "text" - }, - { - "block_id": "p75-b60", - "global_id": 1909, - "bbox": [ - 128.7, - 619.18, - 166.41, - 633.36 - ], - "text": "sin2 x = 1", - "type": "text" - }, - { - "block_id": "p75-b61", - "global_id": 1910, - "bbox": [ - 162.93, - 622.98, - 215.81, - 636.17 - ], - "text": "2(1 −cos 2x)", - "type": "text" - } - ] - }, - { - "page_num": 76, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p76-b0", - "global_id": 1911, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "56\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p76-b1", - "global_id": 1912, - "bbox": [ - 102.95, - 83.68, - 142.33, - 97.86 - ], - "text": "cos3 x = 1", - "type": "text" - }, - { - "block_id": "p76-b2", - "global_id": 1913, - "bbox": [ - 138.84, - 87.48, - 212.78, - 100.67 - ], - "text": "4(3cos x + cos 3x)", - "type": "text" - }, - { - "block_id": "p76-b3", - "global_id": 1914, - "bbox": [ - 102.95, - 100.12, - 140.67, - 114.29 - ], - "text": "sin3 x = 1", - "type": "text" - }, - { - "block_id": "p76-b4", - "global_id": 1915, - "bbox": [ - 137.19, - 103.91, - 207.82, - 117.11 - ], - "text": "4(3sin x −sin 3x)", - "type": "text" - }, - { - "block_id": "p76-b5", - "global_id": 1916, - "bbox": [ - 102.95, - 118.86, - 245.9, - 144.18 - ], - "text": "sin(x ± y) = sin xcos y ± cos xsin y\ncos(x ± y) = cos xcos y ∓sin xsin y", - "type": "text" - }, - { - "block_id": "p76-b6", - "global_id": 1917, - "bbox": [ - 102.95, - 147.18, - 208.04, - 164.54 - ], - "text": "tan(x ± y) = tan x ± tan y", - "type": "text" - }, - { - "block_id": "p76-b7", - "global_id": 1918, - "bbox": [ - 102.95, - 161.24, - 211.09, - 186.41 - ], - "text": "1 ∓tan xtan y\nsin xsin y = 1", - "type": "text" - }, - { - "block_id": "p76-b8", - "global_id": 1919, - "bbox": [ - 153.68, - 176.03, - 258.99, - 189.22 - ], - "text": "2[cos(x −y) −cos(x + y)]", - "type": "text" - }, - { - "block_id": "p76-b9", - "global_id": 1920, - "bbox": [ - 102.95, - 191.03, - 160.47, - 202.75 - ], - "text": "cos xcos y = 1", - "type": "text" - }, - { - "block_id": "p76-b10", - "global_id": 1921, - "bbox": [ - 156.99, - 192.38, - 262.3, - 205.57 - ], - "text": "2[cos(x −y) + cos(x + y)]", - "type": "text" - }, - { - "block_id": "p76-b11", - "global_id": 1922, - "bbox": [ - 102.95, - 207.38, - 158.82, - 219.1 - ], - "text": "sin xcos y = 1", - "type": "text" - }, - { - "block_id": "p76-b12", - "global_id": 1923, - "bbox": [ - 155.33, - 208.72, - 257.34, - 221.91 - ], - "text": "2[sin(x −y) + sin(x + y)]", - "type": "text" - }, - { - "block_id": "p76-b13", - "global_id": 1924, - "bbox": [ - 101.85, - 225.67, - 262.1, - 236.04 - ], - "text": "acos x + bsin x = Ccos(x + θ)\nC =", - "type": "text" - }, - { - "block_id": "p76-b14", - "global_id": 1925, - "bbox": [ - 264.15, - 217.23, - 272.58, - 227.2 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p76-b15", - "global_id": 1926, - "bbox": [ - 272.57, - 217.66, - 360.44, - 236.04 - ], - "text": "a2 + b2, θ = tan−1 −b", - "type": "text" - }, - { - "block_id": "p76-b16", - "global_id": 1927, - "bbox": [ - 354.25, - 231.81, - 357.74, - 238.79 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p76-b18", - "global_id": 1928, - "bbox": [ - 101.84, - 266.72, - 299.6, - 278.67 - ], - "text": "B.8-7 Common Derivative Formulas", - "type": "text" - }, - { - "block_id": "p76-b19", - "global_id": 1929, - "bbox": [ - 103.04, - 279.5, - 150.71, - 303.52 - ], - "text": "d\ndxf(u) = d", - "type": "text" - }, - { - "block_id": "p76-b20", - "global_id": 1930, - "bbox": [ - 143.37, - 279.5, - 182.33, - 303.52 - ], - "text": "duf(u)du", - "type": "text" - }, - { - "block_id": "p76-b21", - "global_id": 1931, - "bbox": [ - 103.04, - 293.56, - 182.03, - 328.33 - ], - "text": "dx\nd\ndx(uv) = udv", - "type": "text" - }, - { - "block_id": "p76-b22", - "global_id": 1932, - "bbox": [ - 148.54, - 304.31, - 185.62, - 328.33 - ], - "text": "dx + vdu", - "type": "text" - }, - { - "block_id": "p76-b23", - "global_id": 1933, - "bbox": [ - 103.04, - 318.37, - 185.32, - 355.89 - ], - "text": "dx\nd\ndx", - "type": "text" - }, - { - "block_id": "p76-b24", - "global_id": 1934, - "bbox": [ - 113.67, - 324.56, - 126.57, - 341.84 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p76-b25", - "global_id": 1935, - "bbox": [ - 121.87, - 345.93, - 126.29, - 355.89 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p76-b27", - "global_id": 1936, - "bbox": [ - 136.54, - 329.41, - 160.14, - 348.82 - ], - "text": "= v du", - "type": "text" - }, - { - "block_id": "p76-b28", - "global_id": 1937, - "bbox": [ - 153.36, - 329.41, - 184.97, - 344.33 - ], - "text": "dx −u dv", - "type": "text" - }, - { - "block_id": "p76-b29", - "global_id": 1938, - "bbox": [ - 162.67, - 336.98, - 184.97, - 355.89 - ], - "text": "dx\nv2", - "type": "text" - }, - { - "block_id": "p76-b30", - "global_id": 1939, - "bbox": [ - 103.04, - 357.66, - 115.95, - 368.94 - ], - "text": "dxn", - "type": "text" - }, - { - "block_id": "p76-b31", - "global_id": 1940, - "bbox": [ - 105.03, - 363.93, - 151.35, - 383.0 - ], - "text": "dx = nxn−1", - "type": "text" - }, - { - "block_id": "p76-b32", - "global_id": 1941, - "bbox": [ - 103.04, - 383.93, - 157.44, - 408.05 - ], - "text": "d\ndx ln(ax) = 1", - "type": "text" - }, - { - "block_id": "p76-b33", - "global_id": 1942, - "bbox": [ - 103.04, - 397.99, - 175.7, - 433.0 - ], - "text": "x\nd\ndx log(ax) = loge", - "type": "text" - }, - { - "block_id": "p76-b34", - "global_id": 1943, - "bbox": [ - 103.04, - 422.94, - 168.76, - 457.7 - ], - "text": "x\nd\ndxebx = bebx", - "type": "text" - }, - { - "block_id": "p76-b35", - "global_id": 1944, - "bbox": [ - 103.04, - 458.48, - 175.43, - 482.5 - ], - "text": "d\ndxabx = b(lna)abx", - "type": "text" - }, - { - "block_id": "p76-b36", - "global_id": 1945, - "bbox": [ - 103.04, - 483.28, - 180.9, - 507.4 - ], - "text": "d\ndx sin ax = acos ax", - "type": "text" - }, - { - "block_id": "p76-b37", - "global_id": 1946, - "bbox": [ - 103.04, - 508.08, - 188.67, - 532.21 - ], - "text": "d\ndx cos ax = −asin ax", - "type": "text" - }, - { - "block_id": "p76-b38", - "global_id": 1947, - "bbox": [ - 103.04, - 532.89, - 189.09, - 581.7 - ], - "text": "d\ndx tan ax =\na\ncos2 ax\nd\ndx(sin−1 ax) =\na\n√", - "type": "text" - }, - { - "block_id": "p76-b39", - "global_id": 1948, - "bbox": [ - 103.04, - 572.64, - 206.94, - 608.86 - ], - "text": "1 −a2x2\nd\ndx(cos−1 ax) =\n−a\n√", - "type": "text" - }, - { - "block_id": "p76-b40", - "global_id": 1949, - "bbox": [ - 103.04, - 599.8, - 208.59, - 636.11 - ], - "text": "1 −a2x2\nd\ndx(tan−1 ax) =\na\n1 + a2x2", - "type": "text" - } - ] - }, - { - "page_num": 77, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p77-b0", - "global_id": 1950, - "bbox": [ - 285.06, - 62.89, - 516.12, - 71.98 - ], - "text": "B.8\nAppendix: Useful Mathematical Formulas\n57", - "type": "text" - }, - { - "block_id": "p77-b1", - "global_id": 1951, - "bbox": [ - 127.59, - 86.52, - 265.21, - 98.48 - ], - "text": "B.8-8 Indefinite Integrals", - "type": "text" - }, - { - "block_id": "p77-b2", - "global_id": 1952, - "bbox": [ - 128.7, - 104.48, - 133.96, - 114.44 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b3", - "global_id": 1953, - "bbox": [ - 139.63, - 118.05, - 185.7, - 128.32 - ], - "text": "udv = uv −", - "type": "text" - }, - { - "block_id": "p77-b4", - "global_id": 1954, - "bbox": [ - 187.24, - 104.48, - 192.5, - 114.44 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b5", - "global_id": 1955, - "bbox": [ - 198.19, - 118.36, - 213.67, - 128.32 - ], - "text": "vdu", - "type": "text" - }, - { - "block_id": "p77-b6", - "global_id": 1956, - "bbox": [ - 128.7, - 133.59, - 133.96, - 143.55 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b7", - "global_id": 1957, - "bbox": [ - 139.62, - 147.15, - 237.3, - 157.43 - ], - "text": "f(x)˙g(x)dx = f(x)g(x) −", - "type": "text" - }, - { - "block_id": "p77-b8", - "global_id": 1958, - "bbox": [ - 238.84, - 133.59, - 244.1, - 143.55 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b9", - "global_id": 1959, - "bbox": [ - 249.79, - 145.09, - 293.25, - 157.43 - ], - "text": "˙f(x)g(x)dx", - "type": "text" - }, - { - "block_id": "p77-b10", - "global_id": 1960, - "bbox": [ - 128.7, - 162.91, - 133.96, - 172.87 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b11", - "global_id": 1961, - "bbox": [ - 139.63, - 169.9, - 199.26, - 186.85 - ], - "text": "sin axdx = −1", - "type": "text" - }, - { - "block_id": "p77-b12", - "global_id": 1962, - "bbox": [ - 194.27, - 176.79, - 226.45, - 193.92 - ], - "text": "a cos ax", - "type": "text" - }, - { - "block_id": "p77-b13", - "global_id": 1963, - "bbox": [ - 247.52, - 162.91, - 252.78, - 172.87 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b14", - "global_id": 1964, - "bbox": [ - 258.45, - 169.9, - 311.96, - 186.85 - ], - "text": "cos axdx = 1", - "type": "text" - }, - { - "block_id": "p77-b15", - "global_id": 1965, - "bbox": [ - 306.98, - 176.79, - 337.5, - 193.92 - ], - "text": "a sin ax", - "type": "text" - }, - { - "block_id": "p77-b16", - "global_id": 1966, - "bbox": [ - 128.7, - 192.23, - 133.96, - 202.19 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b17", - "global_id": 1967, - "bbox": [ - 139.63, - 199.12, - 194.07, - 216.17 - ], - "text": "sin2 axdx = x", - "type": "text" - }, - { - "block_id": "p77-b18", - "global_id": 1968, - "bbox": [ - 189.39, - 199.12, - 235.84, - 223.25 - ], - "text": "2 −sin 2ax\n4a", - "type": "text" - }, - { - "block_id": "p77-b19", - "global_id": 1969, - "bbox": [ - 258.11, - 192.23, - 263.37, - 202.19 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b20", - "global_id": 1970, - "bbox": [ - 269.04, - 199.12, - 325.14, - 216.17 - ], - "text": "cos2 axdx = x", - "type": "text" - }, - { - "block_id": "p77-b21", - "global_id": 1971, - "bbox": [ - 128.7, - 199.12, - 366.91, - 231.52 - ], - "text": "2 + sin 2ax\n4a\n#", - "type": "text" - }, - { - "block_id": "p77-b22", - "global_id": 1972, - "bbox": [ - 139.63, - 228.54, - 199.04, - 245.48 - ], - "text": "xsin axdx = 1", - "type": "text" - }, - { - "block_id": "p77-b23", - "global_id": 1973, - "bbox": [ - 192.07, - 235.11, - 279.26, - 252.46 - ], - "text": "a2 (sin ax −axcos ax)", - "type": "text" - }, - { - "block_id": "p77-b24", - "global_id": 1974, - "bbox": [ - 128.7, - 250.88, - 133.96, - 260.84 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b25", - "global_id": 1975, - "bbox": [ - 139.62, - 257.86, - 200.69, - 274.81 - ], - "text": "xcos axdx = 1", - "type": "text" - }, - { - "block_id": "p77-b26", - "global_id": 1976, - "bbox": [ - 193.72, - 264.43, - 280.92, - 281.78 - ], - "text": "a2 (cos ax + axsin ax)", - "type": "text" - }, - { - "block_id": "p77-b27", - "global_id": 1977, - "bbox": [ - 128.7, - 280.19, - 133.96, - 290.15 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b28", - "global_id": 1978, - "bbox": [ - 139.63, - 287.18, - 203.03, - 304.13 - ], - "text": "x2 sin axdx = 1", - "type": "text" - }, - { - "block_id": "p77-b29", - "global_id": 1979, - "bbox": [ - 196.05, - 290.05, - 348.63, - 311.1 - ], - "text": "a3 (2axsin ax + 2cos ax −a2x2 cos ax)", - "type": "text" - }, - { - "block_id": "p77-b30", - "global_id": 1980, - "bbox": [ - 128.7, - 309.51, - 133.96, - 319.47 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b31", - "global_id": 1981, - "bbox": [ - 139.63, - 316.5, - 204.67, - 333.45 - ], - "text": "x2 cos axdx = 1", - "type": "text" - }, - { - "block_id": "p77-b32", - "global_id": 1982, - "bbox": [ - 197.7, - 319.37, - 348.63, - 340.42 - ], - "text": "a3 (2axcos ax −2sin ax + a2x2 sin ax)", - "type": "text" - }, - { - "block_id": "p77-b33", - "global_id": 1983, - "bbox": [ - 128.7, - 338.9, - 133.96, - 348.86 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b34", - "global_id": 1984, - "bbox": [ - 139.63, - 345.47, - 256.3, - 362.83 - ], - "text": "sin axsin bxdx = sin(a −b)x", - "type": "text" - }, - { - "block_id": "p77-b35", - "global_id": 1985, - "bbox": [ - 216.99, - 345.47, - 315.01, - 369.9 - ], - "text": "2(a −b)\n−sin(a + b)x", - "type": "text" - }, - { - "block_id": "p77-b36", - "global_id": 1986, - "bbox": [ - 275.71, - 348.34, - 365.76, - 369.9 - ], - "text": "2(a + b)\na2̸ = b2", - "type": "text" - }, - { - "block_id": "p77-b37", - "global_id": 1987, - "bbox": [ - 128.7, - 369.12, - 133.96, - 379.08 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b38", - "global_id": 1988, - "bbox": [ - 139.63, - 382.68, - 219.12, - 393.06 - ], - "text": "sin axcos bxdx = −", - "type": "text" - }, - { - "block_id": "p77-b39", - "global_id": 1989, - "bbox": [ - 219.11, - 368.7, - 272.8, - 386.07 - ], - "text": "cos(a −b)x", - "type": "text" - }, - { - "block_id": "p77-b40", - "global_id": 1990, - "bbox": [ - 232.67, - 375.7, - 333.17, - 400.13 - ], - "text": "2(a −b)\n+ cos(a + b)x", - "type": "text" - }, - { - "block_id": "p77-b41", - "global_id": 1991, - "bbox": [ - 293.04, - 389.75, - 326.27, - 400.13 - ], - "text": "2(a + b)", - "type": "text" - }, - { - "block_id": "p77-b42", - "global_id": 1992, - "bbox": [ - 334.4, - 368.69, - 339.83, - 378.66 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p77-b43", - "global_id": 1993, - "bbox": [ - 360.87, - 378.57, - 390.47, - 392.96 - ], - "text": "a2̸ = b2", - "type": "text" - }, - { - "block_id": "p77-b44", - "global_id": 1994, - "bbox": [ - 128.7, - 399.38, - 133.96, - 409.35 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b45", - "global_id": 1995, - "bbox": [ - 139.63, - 405.96, - 259.6, - 423.32 - ], - "text": "cos axcos bxdx = sin(a −b)x", - "type": "text" - }, - { - "block_id": "p77-b46", - "global_id": 1996, - "bbox": [ - 220.29, - 405.96, - 318.32, - 430.4 - ], - "text": "2(a −b)\n+ sin(a + b)x", - "type": "text" - }, - { - "block_id": "p77-b47", - "global_id": 1997, - "bbox": [ - 279.01, - 408.83, - 369.08, - 430.4 - ], - "text": "2(a + b)\na2̸ = b2", - "type": "text" - }, - { - "block_id": "p77-b48", - "global_id": 1998, - "bbox": [ - 128.7, - 428.93, - 133.96, - 438.89 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b49", - "global_id": 1999, - "bbox": [ - 139.63, - 435.92, - 179.73, - 452.87 - ], - "text": "eax dx = 1", - "type": "text" - }, - { - "block_id": "p77-b50", - "global_id": 2000, - "bbox": [ - 174.75, - 440.99, - 191.93, - 459.84 - ], - "text": "aeax", - "type": "text" - }, - { - "block_id": "p77-b51", - "global_id": 2001, - "bbox": [ - 128.7, - 458.15, - 133.96, - 468.11 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b52", - "global_id": 2002, - "bbox": [ - 139.63, - 463.72, - 190.18, - 481.98 - ], - "text": "xeax dx = eax", - "type": "text" - }, - { - "block_id": "p77-b53", - "global_id": 2003, - "bbox": [ - 180.46, - 471.71, - 224.6, - 489.06 - ], - "text": "a2 (ax −1)", - "type": "text" - }, - { - "block_id": "p77-b54", - "global_id": 2004, - "bbox": [ - 128.7, - 487.36, - 133.96, - 497.32 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b55", - "global_id": 2005, - "bbox": [ - 139.63, - 492.93, - 194.19, - 511.2 - ], - "text": "x2eax dx = eax", - "type": "text" - }, - { - "block_id": "p77-b56", - "global_id": 2006, - "bbox": [ - 184.47, - 496.81, - 261.86, - 518.27 - ], - "text": "a3 (a2x2 −2ax + 2)", - "type": "text" - }, - { - "block_id": "p77-b57", - "global_id": 2007, - "bbox": [ - 128.7, - 516.57, - 133.96, - 526.54 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b58", - "global_id": 2008, - "bbox": [ - 139.63, - 522.15, - 218.77, - 540.51 - ], - "text": "eax sin bxdx =\neax", - "type": "text" - }, - { - "block_id": "p77-b59", - "global_id": 2009, - "bbox": [ - 199.13, - 530.14, - 307.81, - 547.49 - ], - "text": "a2 + b2 (asin bx −bcos bx)", - "type": "text" - }, - { - "block_id": "p77-b60", - "global_id": 2010, - "bbox": [ - 128.7, - 545.79, - 133.96, - 555.75 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p77-b61", - "global_id": 2011, - "bbox": [ - 139.63, - 551.36, - 220.43, - 569.73 - ], - "text": "eax cos bxdx =\neax", - "type": "text" - }, - { - "block_id": "p77-b62", - "global_id": 2012, - "bbox": [ - 200.79, - 559.35, - 309.46, - 576.7 - ], - "text": "a2 + b2 (acos bx + bsin bx)", - "type": "text" - }, - { - "block_id": "p77-b63", - "global_id": 2013, - "bbox": [ - 128.7, - 575.11, - 198.88, - 606.01 - ], - "text": "#\n1\nx2 + a2 dx = 1", - "type": "text" - }, - { - "block_id": "p77-b64", - "global_id": 2014, - "bbox": [ - 193.9, - 581.99, - 229.77, - 606.11 - ], - "text": "a tan−1 x", - "type": "text" - }, - { - "block_id": "p77-b65", - "global_id": 2015, - "bbox": [ - 128.7, - 596.05, - 230.06, - 635.34 - ], - "text": "a\n#\nx\nx2 + a2 dx = 1", - "type": "text" - }, - { - "block_id": "p77-b66", - "global_id": 2016, - "bbox": [ - 193.9, - 613.87, - 244.64, - 635.44 - ], - "text": "2 ln(x2 + a2)", - "type": "text" - } - ] - }, - { - "page_num": 78, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p78-b0", - "global_id": 2017, - "bbox": [ - 60.0, - 62.89, - 228.26, - 71.98 - ], - "text": "58\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p78-b1", - "global_id": 2018, - "bbox": [ - 101.84, - 86.52, - 222.64, - 98.48 - ], - "text": "B.8-9 L’Hôpital’s Rule", - "type": "text" - }, - { - "block_id": "p78-b2", - "global_id": 2019, - "bbox": [ - 101.84, - 104.19, - 378.69, - 114.57 - ], - "text": "If lim f(x)/g(x) results in the indeterministic form 0/0 or ∞/∞, then", - "type": "text" - }, - { - "block_id": "p78-b3", - "global_id": 2020, - "bbox": [ - 101.85, - 126.0, - 133.93, - 143.36 - ], - "text": "lim f(x)", - "type": "text" - }, - { - "block_id": "p78-b4", - "global_id": 2021, - "bbox": [ - 117.44, - 132.98, - 160.66, - 150.33 - ], - "text": "g(x) = lim", - "type": "text" - }, - { - "block_id": "p78-b5", - "global_id": 2022, - "bbox": [ - 162.96, - 123.94, - 179.82, - 150.33 - ], - "text": "˙f(x)\n˙g(x)", - "type": "text" - }, - { - "block_id": "p78-b6", - "global_id": 2023, - "bbox": [ - 101.84, - 173.02, - 376.26, - 184.97 - ], - "text": "B.8-10 Solution of Quadratic and Cubic Equations", - "type": "text" - }, - { - "block_id": "p78-b7", - "global_id": 2024, - "bbox": [ - 101.84, - 191.0, - 305.75, - 201.06 - ], - "text": "Any quadratic equation can be reduced to the form", - "type": "text" - }, - { - "block_id": "p78-b8", - "global_id": 2025, - "bbox": [ - 263.19, - 208.59, - 329.05, - 223.09 - ], - "text": "ax2 + bx + c = 0", - "type": "text" - }, - { - "block_id": "p78-b9", - "global_id": 2026, - "bbox": [ - 101.84, - 235.15, - 275.75, - 245.12 - ], - "text": "The solution of this equation is provided by", - "type": "text" - }, - { - "block_id": "p78-b10", - "global_id": 2027, - "bbox": [ - 253.63, - 257.33, - 293.2, - 274.59 - ], - "text": "x = −b ±", - "type": "text" - }, - { - "block_id": "p78-b11", - "global_id": 2028, - "bbox": [ - 294.75, - 248.9, - 303.18, - 258.86 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p78-b12", - "global_id": 2029, - "bbox": [ - 299.3, - 254.45, - 337.4, - 281.76 - ], - "text": "b2 −4ac\n2a", - "type": "text" - }, - { - "block_id": "p78-b13", - "global_id": 2030, - "bbox": [ - 119.78, - 289.67, - 219.65, - 299.74 - ], - "text": "A general cubic equation", - "type": "text" - }, - { - "block_id": "p78-b14", - "global_id": 2031, - "bbox": [ - 253.74, - 297.35, - 338.5, - 311.85 - ], - "text": "y3 + py2 + qy + r = 0", - "type": "text" - }, - { - "block_id": "p78-b15", - "global_id": 2032, - "bbox": [ - 101.84, - 320.82, - 277.72, - 330.88 - ], - "text": "may be reduced to the depressed cubic form", - "type": "text" - }, - { - "block_id": "p78-b16", - "global_id": 2033, - "bbox": [ - 265.39, - 338.42, - 326.84, - 352.91 - ], - "text": "x3 + ax + b = 0", - "type": "text" - }, - { - "block_id": "p78-b17", - "global_id": 2034, - "bbox": [ - 101.84, - 364.97, - 160.81, - 374.93 - ], - "text": "by substituting", - "type": "text" - }, - { - "block_id": "p78-b18", - "global_id": 2035, - "bbox": [ - 276.62, - 372.55, - 314.41, - 389.5 - ], - "text": "y = x −p", - "type": "text" - }, - { - "block_id": "p78-b19", - "global_id": 2036, - "bbox": [ - 309.43, - 386.71, - 314.41, - 396.68 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p78-b20", - "global_id": 2037, - "bbox": [ - 101.85, - 401.67, - 145.84, - 411.63 - ], - "text": "This yields", - "type": "text" - }, - { - "block_id": "p78-b21", - "global_id": 2038, - "bbox": [ - 207.0, - 412.03, - 228.53, - 423.65 - ], - "text": "a = 1", - "type": "text" - }, - { - "block_id": "p78-b22", - "global_id": 2039, - "bbox": [ - 225.05, - 412.03, - 310.16, - 426.56 - ], - "text": "3(3q −p2)\nb = 1", - "type": "text" - }, - { - "block_id": "p78-b23", - "global_id": 2040, - "bbox": [ - 304.93, - 409.75, - 385.2, - 426.56 - ], - "text": "27(2p3 −9pq + 27r)", - "type": "text" - }, - { - "block_id": "p78-b24", - "global_id": 2041, - "bbox": [ - 101.84, - 432.82, - 133.41, - 442.78 - ], - "text": "Now let", - "type": "text" - }, - { - "block_id": "p78-b25", - "global_id": 2042, - "bbox": [ - 190.86, - 455.07, - 206.77, - 465.34 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p78-b26", - "global_id": 2043, - "bbox": [ - 209.04, - 434.58, - 217.3, - 456.99 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p78-b27", - "global_id": 2044, - "bbox": [ - 217.31, - 448.4, - 231.25, - 465.03 - ], - "text": "−b", - "type": "text" - }, - { - "block_id": "p78-b28", - "global_id": 2045, - "bbox": [ - 226.27, - 455.07, - 241.76, - 472.52 - ], - "text": "2 +", - "type": "text" - }, - { - "block_id": "p78-b30", - "global_id": 2046, - "bbox": [ - 252.77, - 447.89, - 261.24, - 458.36 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p78-b31", - "global_id": 2047, - "bbox": [ - 254.76, - 447.89, - 283.95, - 472.52 - ], - "text": "4 + a3", - "type": "text" - }, - { - "block_id": "p78-b32", - "global_id": 2048, - "bbox": [ - 274.99, - 455.07, - 321.99, - 472.52 - ], - "text": "27\nB =", - "type": "text" - }, - { - "block_id": "p78-b33", - "global_id": 2049, - "bbox": [ - 324.25, - 434.58, - 332.51, - 456.99 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p78-b34", - "global_id": 2050, - "bbox": [ - 332.51, - 448.4, - 346.46, - 465.03 - ], - "text": "−b", - "type": "text" - }, - { - "block_id": "p78-b35", - "global_id": 2051, - "bbox": [ - 341.48, - 455.07, - 356.98, - 472.52 - ], - "text": "2 −", - "type": "text" - }, - { - "block_id": "p78-b37", - "global_id": 2052, - "bbox": [ - 367.98, - 447.89, - 376.45, - 458.36 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p78-b38", - "global_id": 2053, - "bbox": [ - 369.97, - 447.89, - 399.17, - 472.52 - ], - "text": "4 + a3", - "type": "text" - }, - { - "block_id": "p78-b39", - "global_id": 2054, - "bbox": [ - 390.21, - 462.56, - 400.17, - 472.52 - ], - "text": "27", - "type": "text" - }, - { - "block_id": "p78-b40", - "global_id": 2055, - "bbox": [ - 101.85, - 479.28, - 252.37, - 489.24 - ], - "text": "The solution of the depressed cubic is", - "type": "text" - }, - { - "block_id": "p78-b41", - "global_id": 2056, - "bbox": [ - 145.9, - 498.73, - 257.11, - 516.09 - ], - "text": "x = A + B,\nx = −A + B", - "type": "text" - }, - { - "block_id": "p78-b42", - "global_id": 2057, - "bbox": [ - 243.11, - 498.73, - 293.41, - 523.15 - ], - "text": "2\n+ A −B", - "type": "text" - }, - { - "block_id": "p78-b43", - "global_id": 2058, - "bbox": [ - 279.41, - 513.19, - 284.4, - 523.15 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p78-b44", - "global_id": 2059, - "bbox": [ - 294.61, - 497.18, - 303.04, - 507.14 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p78-b45", - "global_id": 2060, - "bbox": [ - 303.05, - 498.73, - 387.64, - 516.09 - ], - "text": "−3,\nx = −A + B", - "type": "text" - }, - { - "block_id": "p78-b46", - "global_id": 2061, - "bbox": [ - 373.64, - 498.73, - 423.95, - 523.15 - ], - "text": "2\n−A −B", - "type": "text" - }, - { - "block_id": "p78-b47", - "global_id": 2062, - "bbox": [ - 409.95, - 513.19, - 414.93, - 523.15 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p78-b48", - "global_id": 2063, - "bbox": [ - 425.15, - 497.18, - 433.58, - 507.14 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p78-b49", - "global_id": 2064, - "bbox": [ - 433.58, - 505.71, - 446.33, - 516.09 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p78-b50", - "global_id": 2065, - "bbox": [ - 101.85, - 531.05, - 116.22, - 541.01 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p78-b51", - "global_id": 2066, - "bbox": [ - 276.62, - 536.07, - 314.41, - 553.02 - ], - "text": "y = x −p", - "type": "text" - }, - { - "block_id": "p78-b52", - "global_id": 2067, - "bbox": [ - 309.43, - 550.23, - 314.41, - 560.2 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p78-b53", - "global_id": 2068, - "bbox": [ - 102.11, - 579.78, - 189.29, - 590.74 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p78-b54", - "global_id": 2069, - "bbox": [ - 101.84, - 597.81, - 371.2, - 606.86 - ], - "text": "1.\nAsimov, Isaac. Asimov on Numbers. Bell Publishing, New York, 1982.", - "type": "text" - }, - { - "block_id": "p78-b55", - "global_id": 2070, - "bbox": [ - 101.84, - 611.76, - 411.43, - 620.81 - ], - "text": "2.\nCalinger, R., ed. Classics of Mathematics. Moore Publishing, Oak Park, IL, 1982.", - "type": "text" - }, - { - "block_id": "p78-b56", - "global_id": 2071, - "bbox": [ - 101.84, - 625.71, - 394.27, - 634.76 - ], - "text": "3.\nHogben, Lancelot. Mathematics in the Making. Doubleday, New York, 1960.", - "type": "text" - } - ] - }, - { - "page_num": 79, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p79-b0", - "global_id": 2072, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n59", - "type": "text" - }, - { - "block_id": "p79-b1", - "global_id": 2073, - "bbox": [ - 127.59, - 85.81, - 419.0, - 94.86 - ], - "text": "4.\nCajori, Florian. A History of Mathematics, 4th ed. Chelsea, New York, 1985.", - "type": "text" - }, - { - "block_id": "p79-b2", - "global_id": 2074, - "bbox": [ - 127.59, - 99.76, - 450.46, - 108.81 - ], - "text": "5.\nEncyclopaedia Britannica. Micropaedia IV, 15th ed., vol. 11, p. 1043. Chicago, 1982.", - "type": "text" - }, - { - "block_id": "p79-b3", - "global_id": 2075, - "bbox": [ - 127.59, - 113.7, - 416.04, - 122.75 - ], - "text": "6.\nSingh, Jagjit. Great Ideas of Modern Mathematics. Dover, New York, 1959.", - "type": "text" - }, - { - "block_id": "p79-b4", - "global_id": 2076, - "bbox": [ - 127.59, - 127.65, - 392.76, - 136.7 - ], - "text": "7.\nDunham, William. Journey Through Genius. Wiley, New York, 1990.", - "type": "text" - }, - { - "block_id": "p79-b5", - "global_id": 2077, - "bbox": [ - 106.67, - 167.35, - 217.26, - 184.29 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p79-b6", - "global_id": 2078, - "bbox": [ - 87.98, - 194.36, - 288.99, - 236.58 - ], - "text": "B.1-1\nGiven a complex number w = x + jy, the com-\nplex conjugate of w is defined in rectangular\ncoordinates as w∗= x−jy. Use this fact to derive\ncomplex conjugation in polar form.", - "type": "text" - }, - { - "block_id": "p79-b7", - "global_id": 2079, - "bbox": [ - 87.98, - 242.56, - 288.96, - 262.56 - ], - "text": "B.1-2\nExpress\nthe\nfollowing\nnumbers\nin\npolar\nform:", - "type": "text" - }, - { - "block_id": "p79-b8", - "global_id": 2080, - "bbox": [ - 118.12, - 264.18, - 174.0, - 285.44 - ], - "text": "(a) wa = 1 + j\n(b) wb = 1 + ej", - "type": "text" - }, - { - "block_id": "p79-b9", - "global_id": 2081, - "bbox": [ - 118.12, - 286.1, - 212.33, - 307.37 - ], - "text": "(c) wc = −4 + j3\n(d) wd = (1 + j)(−4 + j3)", - "type": "text" - }, - { - "block_id": "p79-b10", - "global_id": 2082, - "bbox": [ - 118.63, - 304.76, - 205.95, - 318.32 - ], - "text": "(e) we = ejπ/4 + 2e−jπ/4", - "type": "text" - }, - { - "block_id": "p79-b11", - "global_id": 2083, - "bbox": [ - 119.62, - 316.78, - 164.16, - 329.28 - ], - "text": "(f) wf = 1+j", - "type": "text" - }, - { - "block_id": "p79-b12", - "global_id": 2084, - "bbox": [ - 118.12, - 324.36, - 216.54, - 352.57 - ], - "text": "2j\n(g) wg = (1 + j)/(−4 + j3)\n(h) wh = 1−j", - "type": "text" - }, - { - "block_id": "p79-b13", - "global_id": 2085, - "bbox": [ - 87.98, - 347.45, - 288.98, - 377.59 - ], - "text": "sin(j)\nB.1-3\nExpress the following numbers in Cartesian\n(rectangular) form:", - "type": "text" - }, - { - "block_id": "p79-b14", - "global_id": 2086, - "bbox": [ - 118.63, - 378.15, - 171.65, - 389.52 - ], - "text": "(a) wa = j + ej", - "type": "text" - }, - { - "block_id": "p79-b15", - "global_id": 2087, - "bbox": [ - 118.12, - 388.91, - 174.71, - 400.48 - ], - "text": "(b) wb = 3ejπ/4", - "type": "text" - }, - { - "block_id": "p79-b16", - "global_id": 2088, - "bbox": [ - 118.63, - 400.07, - 168.07, - 411.43 - ], - "text": "(c) wc = 1/ej", - "type": "text" - }, - { - "block_id": "p79-b17", - "global_id": 2089, - "bbox": [ - 118.12, - 412.09, - 212.33, - 422.4 - ], - "text": "(d) wd = (1 + j)(−4 + j3)", - "type": "text" - }, - { - "block_id": "p79-b18", - "global_id": 2090, - "bbox": [ - 118.63, - 419.79, - 205.95, - 433.36 - ], - "text": "(e) we = ejπ/4 + 2e−jπ/4", - "type": "text" - }, - { - "block_id": "p79-b19", - "global_id": 2091, - "bbox": [ - 118.12, - 430.75, - 173.42, - 455.27 - ], - "text": "(f) wf = ej + 1\n(g) wg = 1/2j", - "type": "text" - }, - { - "block_id": "p79-b20", - "global_id": 2092, - "bbox": [ - 118.12, - 452.74, - 273.08, - 468.28 - ], - "text": "(h) wh = jjj (j raised to the j raised to the j)", - "type": "text" - }, - { - "block_id": "p79-b21", - "global_id": 2093, - "bbox": [ - 87.98, - 473.29, - 288.98, - 504.25 - ], - "text": "B.1-4\nShowing all work and simplifying your answer,\ndetermine the real part of the following num-\nbers:", - "type": "text" - }, - { - "block_id": "p79-b22", - "global_id": 2094, - "bbox": [ - 118.63, - 504.5, - 158.03, - 516.18 - ], - "text": "(a) wa = 1", - "type": "text" - }, - { - "block_id": "p79-b23", - "global_id": 2095, - "bbox": [ - 118.12, - 504.6, - 207.83, - 527.14 - ], - "text": "j (j −5e2−3j)\n(b) wb = (1 + j)ln(1 + j)", - "type": "text" - }, - { - "block_id": "p79-b24", - "global_id": 2096, - "bbox": [ - 87.98, - 532.15, - 288.98, - 563.11 - ], - "text": "B.1-5\nShowing all work and simplifying your answer,\ndetermine the imaginary part of the following\nnumbers:", - "type": "text" - }, - { - "block_id": "p79-b25", - "global_id": 2097, - "bbox": [ - 118.63, - 563.46, - 179.35, - 575.04 - ], - "text": "(a) wa = −jejπ/4", - "type": "text" - }, - { - "block_id": "p79-b26", - "global_id": 2098, - "bbox": [ - 118.12, - 574.43, - 192.5, - 586.0 - ], - "text": "(b) wb = 1 −2je2−4j", - "type": "text" - }, - { - "block_id": "p79-b27", - "global_id": 2099, - "bbox": [ - 118.63, - 586.64, - 173.73, - 596.95 - ], - "text": "(c) wc = tan(j)", - "type": "text" - }, - { - "block_id": "p79-b28", - "global_id": 2100, - "bbox": [ - 87.98, - 601.96, - 230.89, - 611.01 - ], - "text": "B.1-6\nFor complex constant w, prove:", - "type": "text" - }, - { - "block_id": "p79-b29", - "global_id": 2101, - "bbox": [ - 118.12, - 611.37, - 210.37, - 632.93 - ], - "text": "(a) Re(w) = (w + w∗)/2\n(b) Im(w) = (w −w∗)/2j", - "type": "text" - }, - { - "block_id": "p79-b30", - "global_id": 2102, - "bbox": [ - 315.14, - 194.36, - 448.82, - 203.7 - ], - "text": "B.1-7\nGiven w = x −jy, determine:", - "type": "text" - }, - { - "block_id": "p79-b31", - "global_id": 2103, - "bbox": [ - 345.28, - 204.26, - 386.16, - 225.61 - ], - "text": "(a) Re(ew)\n(b) Im(ew)", - "type": "text" - }, - { - "block_id": "p79-b32", - "global_id": 2104, - "bbox": [ - 315.13, - 233.39, - 516.13, - 253.4 - ], - "text": "B.1-8\nFor arbitrary complex constants w1 and w2,\nprove or disprove the following:", - "type": "text" - }, - { - "block_id": "p79-b33", - "global_id": 2105, - "bbox": [ - 345.28, - 255.02, - 433.61, - 276.06 - ], - "text": "(a) Re(jw1) = −Im(w1)\n(b) Im(jw1) = Re(w1)", - "type": "text" - }, - { - "block_id": "p79-b34", - "global_id": 2106, - "bbox": [ - 345.28, - 276.94, - 479.78, - 298.21 - ], - "text": "(c) Re(w1) + Re(w2) = Re(w1 + w2)\n(d) Im(w1) + Im(w2) = Im(w1 + w2)", - "type": "text" - }, - { - "block_id": "p79-b35", - "global_id": 2107, - "bbox": [ - 345.78, - 298.86, - 460.2, - 308.94 - ], - "text": "(e) Re(w1)Re(w2) = Re(w1w2)", - "type": "text" - }, - { - "block_id": "p79-b36", - "global_id": 2108, - "bbox": [ - 346.77, - 309.81, - 468.64, - 319.89 - ], - "text": "(f) Im(w1)/Im(w2) = Im(w1/w2)", - "type": "text" - }, - { - "block_id": "p79-b37", - "global_id": 2109, - "bbox": [ - 315.13, - 325.38, - 472.03, - 336.95 - ], - "text": "B.1-9\nGiven w1 = 3 + j4 and w2 = 2ejπ/4.", - "type": "text" - }, - { - "block_id": "p79-b38", - "global_id": 2110, - "bbox": [ - 345.28, - 337.88, - 507.38, - 359.24 - ], - "text": "(a) Express w1 in standard polar form.\n(b) Express w2 in standard rectangular form.", - "type": "text" - }, - { - "block_id": "p79-b39", - "global_id": 2111, - "bbox": [ - 345.28, - 356.63, - 516.12, - 381.15 - ], - "text": "(c) Determine |w1|2 and |w2|2.\n(d) Express w1 + w2 in standard rectangular", - "type": "text" - }, - { - "block_id": "p79-b40", - "global_id": 2112, - "bbox": [ - 345.78, - 381.81, - 504.96, - 403.08 - ], - "text": "form.\n(e) Express w1 −w2 in standard polar form.", - "type": "text" - }, - { - "block_id": "p79-b41", - "global_id": 2113, - "bbox": [ - 345.28, - 403.63, - 516.12, - 424.99 - ], - "text": "(f) Express w1w2 in standard rectangular form.\n(g) Express w1/w2 in standard polar form.", - "type": "text" - }, - { - "block_id": "p79-b42", - "global_id": 2114, - "bbox": [ - 310.65, - 428.26, - 516.12, - 452.4 - ], - "text": "B.1-10\nRepeat Prob. B.1-9 using w1 = (3 + j4)2 and\nw2 = 2.5je−j40π.", - "type": "text" - }, - { - "block_id": "p79-b43", - "global_id": 2115, - "bbox": [ - 310.65, - 456.04, - 516.12, - 480.19 - ], - "text": "B.1-11\nRepeat Prob. B.1-9 using w1 = j + eπ/4 and\nw2 = cos(j).", - "type": "text" - }, - { - "block_id": "p79-b44", - "global_id": 2116, - "bbox": [ - 310.65, - 487.01, - 437.03, - 496.05 - ], - "text": "B.1-12\nUsing the complex plane:", - "type": "text" - }, - { - "block_id": "p79-b45", - "global_id": 2117, - "bbox": [ - 345.78, - 498.03, - 516.12, - 507.0 - ], - "text": "(a) Evaluate and locate the distinct solutions to", - "type": "text" - }, - { - "block_id": "p79-b46", - "global_id": 2118, - "bbox": [ - 345.28, - 505.02, - 516.12, - 528.92 - ], - "text": "(w)4 = −1.\n(b) Evaluate and locate the distinct solutions to", - "type": "text" - }, - { - "block_id": "p79-b47", - "global_id": 2119, - "bbox": [ - 360.71, - 527.29, - 442.04, - 539.88 - ], - "text": "(w −(1 + j2))5 = (32/", - "type": "text" - }, - { - "block_id": "p79-b48", - "global_id": 2120, - "bbox": [ - 442.04, - 522.96, - 449.62, - 531.93 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p79-b49", - "global_id": 2121, - "bbox": [ - 345.28, - 530.54, - 515.51, - 561.79 - ], - "text": "2)(1 + j).\n(c) Sketch the solution to |w −2j| = 3.\n(d) Graph w(t) = (1 + t)ejt for (−10 ≤t ≤10).", - "type": "text" - }, - { - "block_id": "p79-b50", - "global_id": 2122, - "bbox": [ - 310.65, - 565.68, - 516.13, - 611.5 - ], - "text": "B.1-13\nThe distinct solutions to (w −w1)n = w2 lie\non a circle in the complex plane, as shown in\nFig. PB.1-13. One solution is located on the\nreal axis at", - "type": "text" - }, - { - "block_id": "p79-b51", - "global_id": 2123, - "bbox": [ - 386.54, - 594.62, - 394.13, - 603.59 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p79-b52", - "global_id": 2124, - "bbox": [ - 345.28, - 602.16, - 516.12, - 622.45 - ], - "text": "3 + 1 = 2.732, and one solution is\nlocated on the imaginary axis at", - "type": "text" - }, - { - "block_id": "p79-b53", - "global_id": 2125, - "bbox": [ - 458.93, - 605.58, - 466.51, - 614.55 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p79-b54", - "global_id": 2126, - "bbox": [ - 345.28, - 613.11, - 516.12, - 634.15 - ], - "text": "3−1 = 0.732.\nDetermine w1, w2, and n.", - "type": "text" - } - ] - }, - { - "page_num": 80, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p80-b0", - "global_id": 2127, - "bbox": [ - 60.0, - 60.36, - 228.26, - 69.45 - ], - "text": "60\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p80-b1", - "global_id": 2128, - "bbox": [ - 129.48, - 216.74, - 133.48, - 224.74 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p80-b2", - "global_id": 2129, - "bbox": [ - 96.36, - 196.82, - 100.36, - 204.82 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p80-b3", - "global_id": 2130, - "bbox": [ - 92.38, - 232.66, - 150.0, - 241.62 - ], - "text": "Figure PB.1-13", - "type": "text" - }, - { - "block_id": "p80-b4", - "global_id": 2131, - "bbox": [ - 57.76, - 265.96, - 263.24, - 296.91 - ], - "text": "B.1-14\nFind the distinct solutions to each of the follow-\ning. Use MATLAB to graph each solution set in\nthe complex plane.", - "type": "text" - }, - { - "block_id": "p80-b5", - "global_id": 2132, - "bbox": [ - 92.89, - 295.28, - 141.26, - 307.88 - ], - "text": "(a) w3 = −8", - "type": "text" - }, - { - "block_id": "p80-b6", - "global_id": 2133, - "bbox": [ - 92.38, - 303.97, - 153.65, - 318.83 - ], - "text": "27\n(b) (w + 1)8 = 1", - "type": "text" - }, - { - "block_id": "p80-b7", - "global_id": 2134, - "bbox": [ - 92.39, - 317.19, - 181.37, - 340.75 - ], - "text": "(c) w2 + j = 0\n(d) 16(w −1)4 + 81 = 0", - "type": "text" - }, - { - "block_id": "p80-b8", - "global_id": 2135, - "bbox": [ - 92.89, - 339.12, - 163.14, - 351.71 - ], - "text": "(e) (w + 2j)3 = −8", - "type": "text" - }, - { - "block_id": "p80-b9", - "global_id": 2136, - "bbox": [ - 92.38, - 350.07, - 173.28, - 373.63 - ], - "text": "(f) (j −w)1.5 = 2 + j2\n(g) (w −1)2.5 = j4", - "type": "text" - }, - { - "block_id": "p80-b10", - "global_id": 2137, - "bbox": [ - 161.0, - 356.7, - 168.59, - 365.67 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p80-b11", - "global_id": 2138, - "bbox": [ - 168.59, - 364.66, - 173.07, - 373.63 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p80-b12", - "global_id": 2139, - "bbox": [ - 57.76, - 378.67, - 111.92, - 388.01 - ], - "text": "B.1-15\nIf j =", - "type": "text" - }, - { - "block_id": "p80-b13", - "global_id": 2140, - "bbox": [ - 113.77, - 371.46, - 121.35, - 380.43 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p80-b14", - "global_id": 2141, - "bbox": [ - 121.36, - 372.08, - 179.26, - 388.01 - ], - "text": "−1, what is √j?", - "type": "text" - }, - { - "block_id": "p80-b15", - "global_id": 2142, - "bbox": [ - 57.76, - 393.06, - 263.24, - 413.36 - ], - "text": "B.1-16\nFind all the values of ln(−e), expressing your\nanswer in Cartesian form.", - "type": "text" - }, - { - "block_id": "p80-b16", - "global_id": 2143, - "bbox": [ - 57.76, - 418.41, - 263.24, - 449.67 - ], - "text": "B.1-17\nDetermine all values of log10(−1), expressing\nyour answer in Cartesian form. Notice that the\nlogarithm has base 10, not e.", - "type": "text" - }, - { - "block_id": "p80-b17", - "global_id": 2144, - "bbox": [ - 57.76, - 455.01, - 263.22, - 475.01 - ], - "text": "B.1-18\nExpress the following in standard rectangular\ncoordinates:", - "type": "text" - }, - { - "block_id": "p80-b18", - "global_id": 2145, - "bbox": [ - 92.38, - 476.64, - 173.66, - 497.9 - ], - "text": "(a) wa = ln(1/(1 + j))\n(b) wb = cos(1 + j)", - "type": "text" - }, - { - "block_id": "p80-b19", - "global_id": 2146, - "bbox": [ - 92.89, - 497.49, - 153.11, - 508.86 - ], - "text": "(c) wc = (1 −j)j", - "type": "text" - }, - { - "block_id": "p80-b20", - "global_id": 2147, - "bbox": [ - 57.76, - 513.22, - 263.24, - 555.16 - ], - "text": "B.1-19\nBy constraining w to be purely imaginary, show\nthat the equation cos(w) = 2 can be represented\nas a standard quadratic equation. Solve this\nequation for w.", - "type": "text" - }, - { - "block_id": "p80-b21", - "global_id": 2148, - "bbox": [ - 57.76, - 560.5, - 263.24, - 592.78 - ], - "text": "B.1-20\nCertain integrals, although expressed in rela-\ntively simple form, are quite difficult to solve.\nFor example,", - "type": "text" - }, - { - "block_id": "p80-b22", - "global_id": 2149, - "bbox": [ - 144.45, - 576.21, - 148.56, - 585.18 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p80-b23", - "global_id": 2150, - "bbox": [ - 92.38, - 580.85, - 263.24, - 636.61 - ], - "text": "e−x2dx cannot be evaluated in\nterms of elementary functions; most calculators\nthat perform integration cannot handle this\nindefinite integral. Fortunately, you are smarter\nthan most calculators.", - "type": "text" - }, - { - "block_id": "p80-b24", - "global_id": 2151, - "bbox": [ - 320.03, - 83.52, - 490.37, - 98.09 - ], - "text": "(a) Express e−x2 using a Taylor series expan-", - "type": "text" - }, - { - "block_id": "p80-b25", - "global_id": 2152, - "bbox": [ - 319.53, - 100.09, - 490.38, - 120.01 - ], - "text": "sion.\n(b) Using your series expansion for e−x2, deter-", - "type": "text" - }, - { - "block_id": "p80-b26", - "global_id": 2153, - "bbox": [ - 334.96, - 123.32, - 352.89, - 132.28 - ], - "text": "mine", - "type": "text" - }, - { - "block_id": "p80-b27", - "global_id": 2154, - "bbox": [ - 355.14, - 115.73, - 359.25, - 124.69 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p80-b28", - "global_id": 2155, - "bbox": [ - 320.03, - 117.62, - 490.36, - 143.25 - ], - "text": "e−x2 dx.\n(c) Using a suitably truncated series, evaluate", - "type": "text" - }, - { - "block_id": "p80-b29", - "global_id": 2156, - "bbox": [ - 334.96, - 146.56, - 404.87, - 155.52 - ], - "text": "the definite integral", - "type": "text" - }, - { - "block_id": "p80-b30", - "global_id": 2157, - "bbox": [ - 407.12, - 138.96, - 416.2, - 150.2 - ], - "text": "$ 1", - "type": "text" - }, - { - "block_id": "p80-b31", - "global_id": 2158, - "bbox": [ - 411.23, - 143.59, - 444.09, - 158.49 - ], - "text": "0 e−x2dx.", - "type": "text" - }, - { - "block_id": "p80-b32", - "global_id": 2159, - "bbox": [ - 284.91, - 162.99, - 405.08, - 172.03 - ], - "text": "B.1-21\nRepeat Prob. B.1-20 for", - "type": "text" - }, - { - "block_id": "p80-b33", - "global_id": 2160, - "bbox": [ - 407.33, - 155.46, - 411.44, - 164.43 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p80-b34", - "global_id": 2161, - "bbox": [ - 414.17, - 157.36, - 441.56, - 172.03 - ], - "text": "e−x3 dx.", - "type": "text" - }, - { - "block_id": "p80-b35", - "global_id": 2162, - "bbox": [ - 284.91, - 176.99, - 405.08, - 186.03 - ], - "text": "B.1-22\nRepeat Prob. B.1-20 for", - "type": "text" - }, - { - "block_id": "p80-b36", - "global_id": 2163, - "bbox": [ - 407.33, - 169.48, - 411.44, - 178.44 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p80-b37", - "global_id": 2164, - "bbox": [ - 414.17, - 173.72, - 446.59, - 186.03 - ], - "text": "cosx2 dx.", - "type": "text" - }, - { - "block_id": "p80-b38", - "global_id": 2165, - "bbox": [ - 284.91, - 191.0, - 490.39, - 211.0 - ], - "text": "B.1-23\nFor each function, determine a suitable series\nexpansion.", - "type": "text" - }, - { - "block_id": "p80-b39", - "global_id": 2166, - "bbox": [ - 320.03, - 211.36, - 399.2, - 222.7 - ], - "text": "(a) fa(x) = (2 −x2)−1", - "type": "text" - }, - { - "block_id": "p80-b40", - "global_id": 2167, - "bbox": [ - 319.53, - 222.52, - 383.35, - 233.65 - ], - "text": "(b) fb(x) = (0.5)x", - "type": "text" - }, - { - "block_id": "p80-b41", - "global_id": 2168, - "bbox": [ - 284.91, - 236.59, - 474.27, - 246.92 - ], - "text": "B.1-24\nConsider the function f(x) = 1+x+x2+x3.", - "type": "text" - }, - { - "block_id": "p80-b42", - "global_id": 2169, - "bbox": [ - 320.03, - 248.54, - 490.37, - 257.88 - ], - "text": "(a) Express f(x) using a Taylor series with", - "type": "text" - }, - { - "block_id": "p80-b43", - "global_id": 2170, - "bbox": [ - 319.53, - 259.5, - 490.39, - 290.76 - ], - "text": "expansion point of a = 1. Explicitly write\nout every term. [Hint: See Sec. B.8-4.]\n(b) Describe a good reason why you might", - "type": "text" - }, - { - "block_id": "p80-b44", - "global_id": 2171, - "bbox": [ - 334.96, - 292.75, - 490.37, - 312.67 - ], - "text": "want to express a function that is already\na simple polynomial using such a series.", - "type": "text" - }, - { - "block_id": "p80-b45", - "global_id": 2172, - "bbox": [ - 284.91, - 317.65, - 490.38, - 348.6 - ], - "text": "B.1-25\nDetermine the Maclaurin series expansion\nof\neach\nof\nthe\nfollowing.\n[Hint:\nSee\nSec. B.8-4.]", - "type": "text" - }, - { - "block_id": "p80-b46", - "global_id": 2173, - "bbox": [ - 320.03, - 349.16, - 369.56, - 360.3 - ], - "text": "(a) fa(x) = 2x", - "type": "text" - }, - { - "block_id": "p80-b47", - "global_id": 2174, - "bbox": [ - 319.53, - 361.18, - 360.73, - 371.25 - ], - "text": "(b) fb(x) =", - "type": "text" - }, - { - "block_id": "p80-b48", - "global_id": 2175, - "bbox": [ - 362.57, - 353.97, - 370.62, - 366.3 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p80-b49", - "global_id": 2176, - "bbox": [ - 367.38, - 353.97, - 378.3, - 373.1 - ], - "text": "3\nx", - "type": "text" - }, - { - "block_id": "p80-b50", - "global_id": 2177, - "bbox": [ - 289.39, - 375.46, - 490.39, - 406.44 - ], - "text": "B.2-1\nDetermine the fundamental period T0, fre-\nquency f0, and radian frequency ω0 for the\nfollowing sinusoids:", - "type": "text" - }, - { - "block_id": "p80-b51", - "global_id": 2178, - "bbox": [ - 319.53, - 408.06, - 381.28, - 428.36 - ], - "text": "(a) cos(5πt + 3)\n(b) 7sin", - "type": "text" - }, - { - "block_id": "p80-b52", - "global_id": 2179, - "bbox": [ - 351.91, - 411.81, - 370.81, - 424.14 - ], - "text": "2t−π", - "type": "text" - }, - { - "block_id": "p80-b53", - "global_id": 2180, - "bbox": [ - 362.47, - 411.81, - 376.26, - 430.94 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p80-b54", - "global_id": 2181, - "bbox": [ - 289.39, - 433.32, - 490.38, - 475.24 - ], - "text": "B.2-2\nDetermine an expression for a sinusoid that\noscillates 15 times per second, that has a value\nof -1 at t = 0, and whose peak amplitude is 3.\nUse MATLAB to plot the signal over 0 ≤t ≤1.", - "type": "text" - }, - { - "block_id": "p80-b55", - "global_id": 2182, - "bbox": [ - 289.39, - 479.9, - 490.39, - 500.2 - ], - "text": "B.2-3\nLet x1(t) = 2cos(3t + 1) and x2(t) = −3cos\n(3t −2).", - "type": "text" - }, - { - "block_id": "p80-b56", - "global_id": 2183, - "bbox": [ - 320.03, - 501.83, - 490.39, - 512.51 - ], - "text": "(a) Determine a1 and b1 so that x1(t) =", - "type": "text" - }, - { - "block_id": "p80-b57", - "global_id": 2184, - "bbox": [ - 319.53, - 512.78, - 490.39, - 534.42 - ], - "text": "a1 cos(3t) + b1 sin(3t).\n(b) Determine a2 and b2 so that x2(t) =", - "type": "text" - }, - { - "block_id": "p80-b58", - "global_id": 2185, - "bbox": [ - 320.03, - 534.7, - 490.39, - 555.74 - ], - "text": "a2 cos(3t) + b2 sin(3t).\n(c) Determine C and θ so that x1(t) + x2(t) =", - "type": "text" - }, - { - "block_id": "p80-b59", - "global_id": 2186, - "bbox": [ - 334.96, - 556.62, - 384.79, - 565.96 - ], - "text": "Ccos(3t + θ).", - "type": "text" - }, - { - "block_id": "p80-b60", - "global_id": 2187, - "bbox": [ - 289.4, - 570.92, - 490.39, - 634.76 - ], - "text": "B.2-4\nIn addition to the traditional sine and cosine\nfunctions,\nthere\nare\nthe\nhyperbolic\nsine\nand\ncosine\nfunctions,\nwhich\nare\ndefined\nby sinh(w) = (ew −e−w)/2 and cosh(w) =\n(ew + e−w)/2. In general, the argument is a\ncomplex constant w = x + jy.", - "type": "text" - } - ] - }, - { - "page_num": 81, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p81-b0", - "global_id": 2188, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n61", - "type": "text" - }, - { - "block_id": "p81-b1", - "global_id": 2189, - "bbox": [ - 118.63, - 85.58, - 288.98, - 94.92 - ], - "text": "(a) Show that cosh(w) = cosh(x)cos(y) +", - "type": "text" - }, - { - "block_id": "p81-b2", - "global_id": 2190, - "bbox": [ - 118.13, - 96.54, - 288.99, - 116.83 - ], - "text": "jsinh(x)sin(y).\n(b) Determine a similar expression for sinh(w)", - "type": "text" - }, - { - "block_id": "p81-b3", - "global_id": 2191, - "bbox": [ - 133.56, - 118.83, - 288.99, - 149.71 - ], - "text": "in rectangular form that only uses functions\nof real arguments, such as sin(x), cosh(y),\nand so on.", - "type": "text" - }, - { - "block_id": "p81-b4", - "global_id": 2192, - "bbox": [ - 87.98, - 154.62, - 288.98, - 174.62 - ], - "text": "B.2-5\nUse Euler’s identity to solve or prove the\nfollowing:", - "type": "text" - }, - { - "block_id": "p81-b5", - "global_id": 2193, - "bbox": [ - 118.63, - 176.24, - 288.98, - 185.58 - ], - "text": "(a) Find real, positive constants c and φ for all", - "type": "text" - }, - { - "block_id": "p81-b6", - "global_id": 2194, - "bbox": [ - 118.13, - 187.2, - 288.98, - 229.42 - ], - "text": "real t such that 2.5cos(3t) −1.5sin(3t +\nπ/3) = ccos(3t + φ). Sketch the resulting\nsinusoid.\n(b) Prove that cos(θ ± φ) = cos(θ)cos(φ) ∓", - "type": "text" - }, - { - "block_id": "p81-b7", - "global_id": 2195, - "bbox": [ - 118.63, - 231.04, - 288.97, - 251.33 - ], - "text": "sin(θ)sin(φ).\n(c) Given real constants a, b, and α, complex", - "type": "text" - }, - { - "block_id": "p81-b8", - "global_id": 2196, - "bbox": [ - 133.56, - 253.24, - 233.58, - 262.29 - ], - "text": "constant w, and the fact that", - "type": "text" - }, - { - "block_id": "p81-b9", - "global_id": 2197, - "bbox": [ - 164.1, - 266.23, - 176.17, - 277.39 - ], - "text": "# b", - "type": "text" - }, - { - "block_id": "p81-b10", - "global_id": 2198, - "bbox": [ - 168.83, - 288.66, - 172.07, - 295.14 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p81-b11", - "global_id": 2199, - "bbox": [ - 177.67, - 272.51, - 215.96, - 287.77 - ], - "text": "ewx dx = 1", - "type": "text" - }, - { - "block_id": "p81-b12", - "global_id": 2200, - "bbox": [ - 210.73, - 274.72, - 258.45, - 294.04 - ], - "text": "w(ewb −ewa)", - "type": "text" - }, - { - "block_id": "p81-b13", - "global_id": 2201, - "bbox": [ - 133.56, - 303.18, - 205.68, - 312.15 - ], - "text": "evaluate the integral", - "type": "text" - }, - { - "block_id": "p81-b14", - "global_id": 2202, - "bbox": [ - 179.58, - 316.73, - 191.66, - 327.91 - ], - "text": "# b", - "type": "text" - }, - { - "block_id": "p81-b15", - "global_id": 2203, - "bbox": [ - 184.31, - 339.18, - 187.55, - 345.66 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p81-b16", - "global_id": 2204, - "bbox": [ - 193.15, - 325.6, - 242.95, - 338.28 - ], - "text": "ewx sin(αx)dx", - "type": "text" - }, - { - "block_id": "p81-b17", - "global_id": 2205, - "bbox": [ - 87.98, - 353.74, - 288.99, - 450.46 - ], - "text": "B.2-6\nA particularly boring stretch of interstate high-\nway has a posted speed limit of 70 mph.\nA highway engineer wants to install “rumble\nbars” (raised ridges on the side of the road) so\nthat cars traveling the speed limit will produce\nquarter-second bursts of 1 kHz sound every\nsecond, a strategy that is particularly effective\nat startling sleepy drivers awake. Provide design\nspecifications for the engineer.", - "type": "text" - }, - { - "block_id": "p81-b18", - "global_id": 2206, - "bbox": [ - 87.98, - 455.36, - 288.98, - 475.36 - ], - "text": "B.3-1\nBy hand, accurately sketch the following signals\nover (0 ≤t ≤1):", - "type": "text" - }, - { - "block_id": "p81-b19", - "global_id": 2207, - "bbox": [ - 118.63, - 475.71, - 171.77, - 487.06 - ], - "text": "(a) xa(t) = e−t", - "type": "text" - }, - { - "block_id": "p81-b20", - "global_id": 2208, - "bbox": [ - 118.12, - 487.94, - 196.36, - 498.02 - ], - "text": "(b) xb(t) = sin(2π5t)", - "type": "text" - }, - { - "block_id": "p81-b21", - "global_id": 2209, - "bbox": [ - 118.63, - 496.02, - 208.44, - 508.97 - ], - "text": "(c) xc(t) = e−t sin(2π5t)", - "type": "text" - }, - { - "block_id": "p81-b22", - "global_id": 2210, - "bbox": [ - 87.98, - 513.14, - 288.98, - 609.85 - ], - "text": "B.3-2\nIn 1950, the human population was approxi-\nmately 2.5 billion people. Assuming a doubling\ntime of 40 years, formulate an exponential\nmodel for human population in the form p(t) =\naebt, where t is measured in years. Sketch p(t)\nover the interval 1950 ≤t ≤2100. According\nto this model, in what year can we expect the\npopulation to reach the estimated 15 billion\ncarrying capacity of the earth?", - "type": "text" - }, - { - "block_id": "p81-b23", - "global_id": 2211, - "bbox": [ - 87.98, - 614.76, - 288.97, - 634.77 - ], - "text": "B.3-3\nDetermine an expression for an exponen-\ntially decaying sinusoid that oscillates three", - "type": "text" - }, - { - "block_id": "p81-b24", - "global_id": 2212, - "bbox": [ - 345.28, - 85.9, - 516.13, - 116.78 - ], - "text": "times per second and whose amplitude enve-\nlope decreases by 50% every 2 seconds. Use\nMATLAB to plot the signal over −2 ≤t ≤2.", - "type": "text" - }, - { - "block_id": "p81-b25", - "global_id": 2213, - "bbox": [ - 315.13, - 121.71, - 516.11, - 141.7 - ], - "text": "B.3-4\nBy hand, sketch the following against indepen-\ndent variable t:", - "type": "text" - }, - { - "block_id": "p81-b26", - "global_id": 2214, - "bbox": [ - 345.78, - 143.32, - 398.06, - 153.4 - ], - "text": "(a) xa(t) = Re", - "type": "text" - }, - { - "block_id": "p81-b28", - "global_id": 2215, - "bbox": [ - 401.66, - 136.11, - 443.89, - 152.66 - ], - "text": "2e(−1+j2π)t", - "type": "text" - }, - { - "block_id": "p81-b29", - "global_id": 2216, - "bbox": [ - 345.28, - 154.27, - 398.41, - 164.35 - ], - "text": "(b) xb(t) = Im", - "type": "text" - }, - { - "block_id": "p81-b31", - "global_id": 2217, - "bbox": [ - 402.02, - 147.07, - 448.99, - 163.61 - ], - "text": "3 −e(1−j2π)t", - "type": "text" - }, - { - "block_id": "p81-b32", - "global_id": 2218, - "bbox": [ - 345.78, - 165.24, - 412.32, - 175.32 - ], - "text": "(c) xc(t) = 3 −Im", - "type": "text" - }, - { - "block_id": "p81-b34", - "global_id": 2219, - "bbox": [ - 415.93, - 158.03, - 448.63, - 174.49 - ], - "text": "e(1−j2π)t", - "type": "text" - }, - { - "block_id": "p81-b35", - "global_id": 2220, - "bbox": [ - 315.13, - 179.49, - 503.7, - 188.53 - ], - "text": "B.4-1\nConsider the following system of equations:", - "type": "text" - }, - { - "block_id": "p81-b36", - "global_id": 2221, - "bbox": [ - 369.21, - 192.85, - 411.99, - 220.21 - ], - "text": "−1\n2\n3\n−4", - "type": "text" - }, - { - "block_id": "p81-b37", - "global_id": 2222, - "bbox": [ - 416.97, - 192.85, - 439.94, - 209.98 - ], - "text": "! x1", - "type": "text" - }, - { - "block_id": "p81-b38", - "global_id": 2223, - "bbox": [ - 432.72, - 211.15, - 439.94, - 220.95 - ], - "text": "x2", - "type": "text" - }, - { - "block_id": "p81-b39", - "global_id": 2224, - "bbox": [ - 445.42, - 192.85, - 450.31, - 201.82 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p81-b40", - "global_id": 2225, - "bbox": [ - 452.15, - 205.44, - 459.14, - 214.41 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p81-b41", - "global_id": 2226, - "bbox": [ - 460.99, - 192.85, - 482.33, - 220.21 - ], - "text": "3\n−1", - "type": "text" - }, - { - "block_id": "p81-b42", - "global_id": 2227, - "bbox": [ - 487.31, - 192.85, - 492.2, - 201.82 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p81-b43", - "global_id": 2228, - "bbox": [ - 345.28, - 231.72, - 516.13, - 273.57 - ], - "text": "Expressing all answers in rational form (ratio of\nintegers), use Cramer’s rule to determine x1 and\nx2. Perform all calculations by hand, including\nmatrix determinants.", - "type": "text" - }, - { - "block_id": "p81-b44", - "global_id": 2229, - "bbox": [ - 315.13, - 278.49, - 503.7, - 287.53 - ], - "text": "B.4-2\nConsider the following system of equations:", - "type": "text" - }, - { - "block_id": "p81-b45", - "global_id": 2230, - "bbox": [ - 367.52, - 291.94, - 374.06, - 300.91 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p81-b46", - "global_id": 2231, - "bbox": [ - 367.52, - 308.08, - 374.06, - 317.04 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p81-b47", - "global_id": 2232, - "bbox": [ - 379.04, - 299.27, - 412.42, - 330.16 - ], - "text": "1\n2\n0\n0\n3\n4\n5\n0\n6", - "type": "text" - }, - { - "block_id": "p81-b48", - "global_id": 2233, - "bbox": [ - 417.39, - 291.94, - 423.93, - 300.91 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p81-b49", - "global_id": 2234, - "bbox": [ - 417.39, - 308.08, - 423.93, - 317.04 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p81-b50", - "global_id": 2235, - "bbox": [ - 424.93, - 291.94, - 431.46, - 300.91 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p81-b51", - "global_id": 2236, - "bbox": [ - 424.93, - 308.08, - 431.46, - 317.04 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p81-b52", - "global_id": 2237, - "bbox": [ - 436.45, - 299.18, - 443.66, - 330.9 - ], - "text": "x1\nx2\nx3", - "type": "text" - }, - { - "block_id": "p81-b53", - "global_id": 2238, - "bbox": [ - 449.14, - 291.94, - 455.68, - 300.91 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p81-b54", - "global_id": 2239, - "bbox": [ - 449.14, - 308.08, - 464.52, - 318.88 - ], - "text": "⎦=", - "type": "text" - }, - { - "block_id": "p81-b55", - "global_id": 2240, - "bbox": [ - 466.36, - 291.94, - 472.9, - 300.91 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p81-b56", - "global_id": 2241, - "bbox": [ - 466.36, - 308.08, - 472.9, - 317.04 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p81-b57", - "global_id": 2242, - "bbox": [ - 477.88, - 299.27, - 482.37, - 330.16 - ], - "text": "7\n8\n9", - "type": "text" - }, - { - "block_id": "p81-b58", - "global_id": 2243, - "bbox": [ - 487.34, - 291.94, - 493.88, - 300.91 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p81-b59", - "global_id": 2244, - "bbox": [ - 487.34, - 308.08, - 493.88, - 317.04 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p81-b60", - "global_id": 2245, - "bbox": [ - 345.28, - 341.68, - 516.13, - 383.52 - ], - "text": "Expressing all answers in rational form (ratio\nof integers), use Cramer’s rule to determine x1,\nx2, and x3. Perform all calculations by hand,\nincluding matrix determinants.", - "type": "text" - }, - { - "block_id": "p81-b61", - "global_id": 2246, - "bbox": [ - 315.13, - 388.45, - 503.45, - 397.49 - ], - "text": "B.4-3\nConsider the following system of equations.", - "type": "text" - }, - { - "block_id": "p81-b62", - "global_id": 2247, - "bbox": [ - 402.75, - 409.2, - 460.63, - 419.51 - ], - "text": "x1 + x2 + x3 = 1", - "type": "text" - }, - { - "block_id": "p81-b63", - "global_id": 2248, - "bbox": [ - 393.79, - 423.15, - 460.63, - 433.46 - ], - "text": "x1 + 2x2 + 3x3 = 3", - "type": "text" - }, - { - "block_id": "p81-b64", - "global_id": 2249, - "bbox": [ - 420.25, - 437.1, - 467.62, - 447.41 - ], - "text": "x1 −x2 = −3", - "type": "text" - }, - { - "block_id": "p81-b65", - "global_id": 2250, - "bbox": [ - 345.28, - 458.44, - 516.13, - 489.67 - ], - "text": "Use Cramer’s rule to determine x1, x2, and x3.\nMatrix determinants can be computed by using\nMATLAB’s det command.", - "type": "text" - }, - { - "block_id": "p81-b66", - "global_id": 2251, - "bbox": [ - 315.13, - 494.32, - 516.13, - 514.33 - ], - "text": "B.5-1\nDetermine the constants a0, a1, and a2 of the\npartial fraction expansion", - "type": "text" - }, - { - "block_id": "p81-b67", - "global_id": 2252, - "bbox": [ - 363.61, - 522.25, - 419.14, - 543.95 - ], - "text": "F(s) =\ns\n(s + 1)3", - "type": "text" - }, - { - "block_id": "p81-b68", - "global_id": 2253, - "bbox": [ - 381.41, - 544.38, - 496.61, - 566.09 - ], - "text": "=\na0\n(s + 1)3 +\na1\n(s + 1)2 +\na2\n(s + 1)", - "type": "text" - }, - { - "block_id": "p81-b69", - "global_id": 2254, - "bbox": [ - 315.13, - 575.93, - 516.11, - 595.93 - ], - "text": "B.5-2\nCompute by hand the partial fraction expan-\nsions of the following rational functions:", - "type": "text" - }, - { - "block_id": "p81-b70", - "global_id": 2255, - "bbox": [ - 345.28, - 596.28, - 516.12, - 635.86 - ], - "text": "(a) Ha(s) =\ns2+5s+6\ns3+s2+s+1, which has denominator\npoles at s = ±j and s = −1\n(b) Hb(s) =\n1\nH1(s) = s3+s2+s+1", - "type": "text" - }, - { - "block_id": "p81-b71", - "global_id": 2256, - "bbox": [ - 425.09, - 627.96, - 449.94, - 635.55 - ], - "text": "s2+5s+6", - "type": "text" - } - ] - }, - { - "page_num": 82, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p82-b0", - "global_id": 2257, - "bbox": [ - 60.0, - 60.36, - 228.26, - 69.45 - ], - "text": "62\nCHAPTER B\nBACKGROUND", - "type": "text" - }, - { - "block_id": "p82-b1", - "global_id": 2258, - "bbox": [ - 92.38, - 86.08, - 177.47, - 113.33 - ], - "text": "(c) Hc(s) =\n1\n(s+1)2(s2+1)\n(d) Hd(s) = s2+5s+6", - "type": "text" - }, - { - "block_id": "p82-b2", - "global_id": 2259, - "bbox": [ - 62.24, - 108.39, - 263.22, - 137.87 - ], - "text": "3s2+2s+1\nB.5-3\nCompute by hand the partial fraction expan-\nsions of the following rational functions:", - "type": "text" - }, - { - "block_id": "p82-b3", - "global_id": 2260, - "bbox": [ - 92.89, - 137.5, - 171.25, - 149.58 - ], - "text": "(a) Fa(x) = (x−1)(x−2)", - "type": "text" - }, - { - "block_id": "p82-b4", - "global_id": 2261, - "bbox": [ - 145.63, - 144.62, - 164.37, - 152.21 - ], - "text": "(x−3)2", - "type": "text" - }, - { - "block_id": "p82-b5", - "global_id": 2262, - "bbox": [ - 92.38, - 155.34, - 178.09, - 170.57 - ], - "text": "(b) Fb(x) =\n(x−1)2\n(3x−1)(2x−1)", - "type": "text" - }, - { - "block_id": "p82-b6", - "global_id": 2263, - "bbox": [ - 92.89, - 173.7, - 169.23, - 187.1 - ], - "text": "(c) Fc(x) =\n(x−1)2", - "type": "text" - }, - { - "block_id": "p82-b7", - "global_id": 2264, - "bbox": [ - 139.24, - 182.14, - 180.98, - 189.73 - ], - "text": "(3x−1)2(2x−1)", - "type": "text" - }, - { - "block_id": "p82-b8", - "global_id": 2265, - "bbox": [ - 92.38, - 192.86, - 166.82, - 205.9 - ], - "text": "(d) Fd(x) = x2−5x+6", - "type": "text" - }, - { - "block_id": "p82-b9", - "global_id": 2266, - "bbox": [ - 139.61, - 200.94, - 168.44, - 208.53 - ], - "text": "2x2+8x+6", - "type": "text" - }, - { - "block_id": "p82-b10", - "global_id": 2267, - "bbox": [ - 92.89, - 211.07, - 171.31, - 224.1 - ], - "text": "(e) Fe(x) = 2x2−3x−11", - "type": "text" - }, - { - "block_id": "p82-b11", - "global_id": 2268, - "bbox": [ - 144.1, - 219.15, - 166.45, - 226.74 - ], - "text": "x2−x−2", - "type": "text" - }, - { - "block_id": "p82-b12", - "global_id": 2269, - "bbox": [ - 93.88, - 229.27, - 162.17, - 242.31 - ], - "text": "(f) Ff(x) =\n3+2x2", - "type": "text" - }, - { - "block_id": "p82-b13", - "global_id": 2270, - "bbox": [ - 138.52, - 237.36, - 168.66, - 244.95 - ], - "text": "−3+2x+x2", - "type": "text" - }, - { - "block_id": "p82-b14", - "global_id": 2271, - "bbox": [ - 92.38, - 247.48, - 179.62, - 260.52 - ], - "text": "(g) Fg(x) = x3+2x2+3x+4", - "type": "text" - }, - { - "block_id": "p82-b15", - "global_id": 2272, - "bbox": [ - 152.41, - 255.56, - 166.82, - 263.16 - ], - "text": "x2+1", - "type": "text" - }, - { - "block_id": "p82-b16", - "global_id": 2273, - "bbox": [ - 92.38, - 265.69, - 167.94, - 278.73 - ], - "text": "(h) Fh(x) = 1+2x+3x2", - "type": "text" - }, - { - "block_id": "p82-b17", - "global_id": 2274, - "bbox": [ - 141.22, - 273.77, - 166.82, - 281.36 - ], - "text": "x2+5x+6", - "type": "text" - }, - { - "block_id": "p82-b18", - "global_id": 2275, - "bbox": [ - 94.37, - 283.89, - 181.43, - 296.93 - ], - "text": "(i) Fi(x) = 3x3−x2+14x+4", - "type": "text" - }, - { - "block_id": "p82-b19", - "global_id": 2276, - "bbox": [ - 152.59, - 291.98, - 167.01, - 299.57 - ], - "text": "x2+4", - "type": "text" - }, - { - "block_id": "p82-b20", - "global_id": 2277, - "bbox": [ - 94.37, - 301.87, - 171.26, - 315.14 - ], - "text": "(j) Fj(x) = 2x−1−1+2x", - "type": "text" - }, - { - "block_id": "p82-b21", - "global_id": 2278, - "bbox": [ - 139.79, - 309.96, - 169.16, - 317.78 - ], - "text": "x−5+6x−1", - "type": "text" - }, - { - "block_id": "p82-b22", - "global_id": 2279, - "bbox": [ - 92.38, - 320.31, - 181.21, - 333.35 - ], - "text": "(k) Fk(x) = 3 −5x2−9x+23", - "type": "text" - }, - { - "block_id": "p82-b23", - "global_id": 2280, - "bbox": [ - 62.24, - 328.39, - 262.73, - 368.84 - ], - "text": "x2+x−2\nB.6-1\nA system of equations in terms of unknowns x1\nand x2 and arbitrary constants a, b, c, d, e, and f\nis given by", - "type": "text" - }, - { - "block_id": "p82-b24", - "global_id": 2281, - "bbox": [ - 153.39, - 384.14, - 202.23, - 394.45 - ], - "text": "ax1 + bx2 = c", - "type": "text" - }, - { - "block_id": "p82-b25", - "global_id": 2282, - "bbox": [ - 154.07, - 398.08, - 200.74, - 408.39 - ], - "text": "dx1 + ex2 = f", - "type": "text" - }, - { - "block_id": "p82-b26", - "global_id": 2283, - "bbox": [ - 92.89, - 434.06, - 263.23, - 443.03 - ], - "text": "(a) Represent this system of equations in", - "type": "text" - }, - { - "block_id": "p82-b27", - "global_id": 2284, - "bbox": [ - 92.39, - 445.01, - 263.23, - 464.94 - ], - "text": "matrix form.\n(b) Identify specific constants a, b, c, d, e, and", - "type": "text" - }, - { - "block_id": "p82-b28", - "global_id": 2285, - "bbox": [ - 92.89, - 466.57, - 263.23, - 497.82 - ], - "text": "f such that x1 = 3 and x2 = −2. Are the\nconstants you selected unique?\n(c) Identify nonzero constants a, b, c, d, e, and", - "type": "text" - }, - { - "block_id": "p82-b29", - "global_id": 2286, - "bbox": [ - 92.38, - 499.72, - 263.23, - 519.73 - ], - "text": "f such that no solutions x1 and x2 exist.\n(d) Identify nonzero constants a, b, c, d, e, and", - "type": "text" - }, - { - "block_id": "p82-b30", - "global_id": 2287, - "bbox": [ - 107.82, - 521.64, - 263.23, - 543.0 - ], - "text": "f such that an infinite number of solutions\nx1 and x2 exist.", - "type": "text" - }, - { - "block_id": "p82-b31", - "global_id": 2288, - "bbox": [ - 62.24, - 546.94, - 263.22, - 566.93 - ], - "text": "B.6-2\nUsing a matrix approach, solve the following\nsystem of equations:", - "type": "text" - }, - { - "block_id": "p82-b32", - "global_id": 2289, - "bbox": [ - 136.62, - 582.23, - 212.01, - 592.54 - ], - "text": "x1 + x2 + x3 + x4 = 4", - "type": "text" - }, - { - "block_id": "p82-b33", - "global_id": 2290, - "bbox": [ - 136.62, - 596.17, - 212.01, - 606.48 - ], - "text": "x1 + x2 + x3 −x4 = 2", - "type": "text" - }, - { - "block_id": "p82-b34", - "global_id": 2291, - "bbox": [ - 136.62, - 610.12, - 212.01, - 620.43 - ], - "text": "x1 + x2 −x3 −x4 = 0", - "type": "text" - }, - { - "block_id": "p82-b35", - "global_id": 2292, - "bbox": [ - 136.62, - 624.07, - 219.0, - 634.38 - ], - "text": "x1 −x2 −x3 −x4 = −2", - "type": "text" - }, - { - "block_id": "p82-b36", - "global_id": 2293, - "bbox": [ - 289.4, - 85.94, - 490.38, - 105.94 - ], - "text": "B.6-3\nUsing a matrix approach, solve the following\nsystem of equations:", - "type": "text" - }, - { - "block_id": "p82-b37", - "global_id": 2294, - "bbox": [ - 368.74, - 118.58, - 444.12, - 128.89 - ], - "text": "x1 + x2 + x3 + x4 = 1", - "type": "text" - }, - { - "block_id": "p82-b38", - "global_id": 2295, - "bbox": [ - 377.27, - 132.53, - 444.12, - 142.84 - ], - "text": "x1 −2x2 + 3x3 = 2", - "type": "text" - }, - { - "block_id": "p82-b39", - "global_id": 2296, - "bbox": [ - 381.75, - 146.48, - 444.12, - 156.79 - ], - "text": "x1 −x3 + 7x4 = 3", - "type": "text" - }, - { - "block_id": "p82-b40", - "global_id": 2297, - "bbox": [ - 365.8, - 160.42, - 444.12, - 170.73 - ], - "text": "−2x2 + 3x3 −4x4 = 4", - "type": "text" - }, - { - "block_id": "p82-b41", - "global_id": 2298, - "bbox": [ - 289.39, - 182.41, - 490.39, - 224.63 - ], - "text": "B.6-4\nA signal f(t) = acos(3t) + bsin(3t) reaches a\npeak amplitude of 5 at t = 1.8799 and has a\nzero crossing at t = 0.3091. Use a matrix-based\napproach to determine the constants a and b.", - "type": "text" - }, - { - "block_id": "p82-b42", - "global_id": 2299, - "bbox": [ - 289.4, - 229.64, - 343.74, - 238.68 - ], - "text": "B.6-5\nDefine", - "type": "text" - }, - { - "block_id": "p82-b43", - "global_id": 2300, - "bbox": [ - 343.15, - 254.67, - 356.47, - 263.94 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p82-b44", - "global_id": 2301, - "bbox": [ - 358.31, - 242.08, - 394.1, - 269.44 - ], - "text": "1\n3\n−2\n4", - "type": "text" - }, - { - "block_id": "p82-b45", - "global_id": 2302, - "bbox": [ - 399.08, - 242.08, - 403.97, - 251.05 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p82-b46", - "global_id": 2303, - "bbox": [ - 404.96, - 254.67, - 430.48, - 264.01 - ], - "text": ",\ny =", - "type": "text" - }, - { - "block_id": "p82-b47", - "global_id": 2304, - "bbox": [ - 432.33, - 242.08, - 453.68, - 258.48 - ], - "text": "−5", - "type": "text" - }, - { - "block_id": "p82-b48", - "global_id": 2305, - "bbox": [ - 445.69, - 260.47, - 450.17, - 269.44 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p82-b49", - "global_id": 2306, - "bbox": [ - 458.65, - 242.08, - 463.54, - 251.05 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p82-b50", - "global_id": 2307, - "bbox": [ - 464.54, - 255.04, - 466.78, - 264.01 - ], - "text": ",", - "type": "text" - }, - { - "block_id": "p82-b51", - "global_id": 2308, - "bbox": [ - 363.85, - 290.2, - 398.58, - 299.54 - ], - "text": "and\nz =", - "type": "text" - }, - { - "block_id": "p82-b52", - "global_id": 2309, - "bbox": [ - 400.42, - 277.61, - 436.21, - 304.97 - ], - "text": "0\n1\n−1\n0", - "type": "text" - }, - { - "block_id": "p82-b53", - "global_id": 2310, - "bbox": [ - 441.19, - 277.61, - 446.08, - 286.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p82-b54", - "global_id": 2311, - "bbox": [ - 319.53, - 314.43, - 438.59, - 323.39 - ], - "text": "By hand, calculate the following:", - "type": "text" - }, - { - "block_id": "p82-b55", - "global_id": 2312, - "bbox": [ - 319.53, - 323.95, - 365.57, - 346.28 - ], - "text": "(a) fa = yTy\n(b) fb = yyT", - "type": "text" - }, - { - "block_id": "p82-b56", - "global_id": 2313, - "bbox": [ - 319.53, - 346.93, - 365.93, - 368.2 - ], - "text": "(c) fc = xy\n(d) fd = xTy", - "type": "text" - }, - { - "block_id": "p82-b57", - "global_id": 2314, - "bbox": [ - 320.03, - 367.79, - 365.57, - 379.16 - ], - "text": "(e) fe = yTx", - "type": "text" - }, - { - "block_id": "p82-b58", - "global_id": 2315, - "bbox": [ - 319.53, - 379.81, - 375.21, - 412.04 - ], - "text": "(f) ff = xz\n(g) fg = zxz\n(h) fh = xT −z", - "type": "text" - }, - { - "block_id": "p82-b59", - "global_id": 2316, - "bbox": [ - 289.39, - 416.08, - 490.38, - 436.08 - ], - "text": "B.7-1\nUse MATLAB to produce the plots requested in\nProb. B.3-4.", - "type": "text" - }, - { - "block_id": "p82-b60", - "global_id": 2317, - "bbox": [ - 289.39, - 440.79, - 490.39, - 483.01 - ], - "text": "B.7-2\nUse MATLAB to plot the function x(t) =\ntsin(2πt) over 0 ≤t ≤10 using 501 equally\nspaced points. What is the maximum value of\nx(t) over this range of t?", - "type": "text" - }, - { - "block_id": "p82-b61", - "global_id": 2318, - "bbox": [ - 289.39, - 487.73, - 490.39, - 508.03 - ], - "text": "B.7-3\nUse MATLAB to plot x(t) = cos(t)sin(20t)\nover a suitable range of t.", - "type": "text" - }, - { - "block_id": "p82-b62", - "global_id": 2319, - "bbox": [ - 289.39, - 506.02, - 447.06, - 522.08 - ], - "text": "B.7-4\nUse MATLAB to plot x(t) = %10", - "type": "text" - }, - { - "block_id": "p82-b63", - "global_id": 2320, - "bbox": [ - 319.53, - 512.74, - 490.4, - 544.24 - ], - "text": "k=1 cos(2πkt)\nover a suitable range of t. The MATLAB\ncommand sum may prove useful.", - "type": "text" - }, - { - "block_id": "p82-b64", - "global_id": 2321, - "bbox": [ - 289.4, - 549.01, - 490.4, - 634.77 - ], - "text": "B.7-5\nWhen a bell is struck with a mallet, it pro-\nduces a ringing sound. Write an equation that\napproximates the sound produced by a small,\nlight bell. Carefully identify your assumptions.\nHow does your equation change if the bell is\nlarge and heavy? You can assess the quality\nof your models by using the MATLAB sound\ncommand to listen to your “bell.”", - "type": "text" - } - ] - }, - { - "page_num": 83, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p83-b0", - "global_id": 2322, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n63", - "type": "text" - }, - { - "block_id": "p83-b1", - "global_id": 2323, - "bbox": [ - 87.98, - 85.94, - 288.99, - 182.65 - ], - "text": "B.7-6\nYou are working on a digital quadrature\namplitude modulation (QAM) communication\nreceiver. The QAM receiver requires a pair of\nquadrature signals: cosn and sinn. These\ncan be simultaneously generated by following\na simple procedure: (1) choose a point w on the\nunit circle, (2) multiply w by itself and store the\nresult, (3) multiply w by the last result and store,\nand (4) repeat step 3.", - "type": "text" - }, - { - "block_id": "p83-b2", - "global_id": 2324, - "bbox": [ - 118.63, - 184.65, - 288.99, - 193.61 - ], - "text": "(a) Show that this method can generate the", - "type": "text" - }, - { - "block_id": "p83-b3", - "global_id": 2325, - "bbox": [ - 118.13, - 195.6, - 288.98, - 215.53 - ], - "text": "desired pair of quadrature sinusoids.\n(b) Determine a suitable value of w so that", - "type": "text" - }, - { - "block_id": "p83-b4", - "global_id": 2326, - "bbox": [ - 118.63, - 217.15, - 288.99, - 270.32 - ], - "text": "good-quality, periodic, 2π × 100,000 rad/s\nsignals can be generated. How much time is\navailable for the processing unit to compute\neach sample?\n(c) Simulate this procedure by using MATLAB", - "type": "text" - }, - { - "block_id": "p83-b5", - "global_id": 2327, - "bbox": [ - 118.13, - 272.32, - 288.99, - 292.25 - ], - "text": "and report your results.\n(d) Identify as many assumptions and limi-", - "type": "text" - }, - { - "block_id": "p83-b6", - "global_id": 2328, - "bbox": [ - 133.56, - 294.23, - 288.98, - 325.13 - ], - "text": "tations to this technique as possible. For\nexample, can your system operate correctly\nfor an indefinite period of time?", - "type": "text" - }, - { - "block_id": "p83-b7", - "global_id": 2329, - "bbox": [ - 87.98, - 330.03, - 259.65, - 339.32 - ], - "text": "B.7-7\nUsing MATLAB’s residue command,", - "type": "text" - }, - { - "block_id": "p83-b8", - "global_id": 2330, - "bbox": [ - 118.12, - 341.06, - 254.37, - 360.99 - ], - "text": "(a) Verify the results of Prob. B.5-2a.\n(b) Verify the results of Prob. B.5-2b.", - "type": "text" - }, - { - "block_id": "p83-b9", - "global_id": 2331, - "bbox": [ - 118.12, - 362.98, - 254.73, - 382.91 - ], - "text": "(c) Verify the results of Prob. B.5-2c.\n(d) Verify the results of Prob. B.5-2d.", - "type": "text" - }, - { - "block_id": "p83-b10", - "global_id": 2332, - "bbox": [ - 87.98, - 387.82, - 259.65, - 397.11 - ], - "text": "B.7-8\nUsing MATLAB’s residue command,", - "type": "text" - }, - { - "block_id": "p83-b11", - "global_id": 2333, - "bbox": [ - 118.12, - 398.84, - 254.37, - 418.77 - ], - "text": "(a) Verify the results of Prob. B.5-3a.\n(b) Verify the results of Prob. B.5-3b.", - "type": "text" - }, - { - "block_id": "p83-b12", - "global_id": 2334, - "bbox": [ - 118.12, - 420.77, - 254.73, - 440.69 - ], - "text": "(c) Verify the results of Prob. B.5-3c.\n(d) Verify the results of Prob. B.5-3d.", - "type": "text" - }, - { - "block_id": "p83-b13", - "global_id": 2335, - "bbox": [ - 118.63, - 442.68, - 254.23, - 451.65 - ], - "text": "(e) Verify the results of Prob. B.5-3e.", - "type": "text" - }, - { - "block_id": "p83-b14", - "global_id": 2336, - "bbox": [ - 118.12, - 453.65, - 254.73, - 484.53 - ], - "text": "(f) Verify the results of Prob. B.5-3f.\n(g) Verify the results of Prob. B.5-3g.\n(h) Verify the results of Prob. B.5-3h.", - "type": "text" - }, - { - "block_id": "p83-b15", - "global_id": 2337, - "bbox": [ - 118.12, - 486.52, - 254.73, - 517.41 - ], - "text": "(i) Verify the results of Prob. B.5-3i.\n(j) Verify the results of Prob. B.5-3j.\n(k) Verify the results of Prob. B.5-3k.", - "type": "text" - }, - { - "block_id": "p83-b16", - "global_id": 2338, - "bbox": [ - 87.98, - 522.32, - 288.98, - 542.31 - ], - "text": "B.7-9\nDetermine the original length-3 vectors a and b\nneed to produce the MATLAB output:", - "type": "text" - }, - { - "block_id": "p83-b17", - "global_id": 2339, - "bbox": [ - 118.12, - 551.53, - 240.52, - 582.41 - ], - "text": ">>\n[r,p,k] = residue(b,a)\nr =\n0 + 2.0000i\n0 - 2.0000i", - "type": "text" - }, - { - "block_id": "p83-b18", - "global_id": 2340, - "bbox": [ - 364.09, - 85.53, - 439.41, - 116.41 - ], - "text": "p =\n3\n-3\nk =\n0 + 1.0000i", - "type": "text" - }, - { - "block_id": "p83-b19", - "global_id": 2341, - "bbox": [ - 310.65, - 126.38, - 516.15, - 157.64 - ], - "text": "B.7-10\nLet N = [n7,n6,n5,...,n2,n1] represent the\nseven digits of your phone number. Construct\na rational function according to", - "type": "text" - }, - { - "block_id": "p83-b20", - "global_id": 2342, - "bbox": [ - 370.63, - 166.25, - 489.08, - 185.81 - ], - "text": "HN(s) = n7s2 + n6s + n5 + n4s−1", - "type": "text" - }, - { - "block_id": "p83-b21", - "global_id": 2343, - "bbox": [ - 419.5, - 179.57, - 473.95, - 192.24 - ], - "text": "n3s2 + n2s + n1", - "type": "text" - }, - { - "block_id": "p83-b22", - "global_id": 2344, - "bbox": [ - 345.28, - 201.65, - 516.14, - 222.26 - ], - "text": "Use MATLAB’s residue command to com-\npute the partial fraction expansion of HN(s).", - "type": "text" - }, - { - "block_id": "p83-b23", - "global_id": 2345, - "bbox": [ - 310.65, - 226.55, - 516.13, - 279.43 - ], - "text": "B.7-11\nWhen\nplotted\nin\nthe\ncomplex\nplane\nfor\n−π\n≤\nω\n≤\nπ,\nthe\nfunction\nf(ω)\n=\ncos(ω) + j0.1sin(2ω) results in a so-called\nLissajous figure that resembles a two-bladed\npropeller.", - "type": "text" - }, - { - "block_id": "p83-b24", - "global_id": 2346, - "bbox": [ - 345.78, - 281.42, - 516.12, - 290.64 - ], - "text": "(a) In MATLAB, create two row vectors fr and", - "type": "text" - }, - { - "block_id": "p83-b25", - "global_id": 2347, - "bbox": [ - 345.27, - 292.38, - 516.13, - 356.14 - ], - "text": "fi corresponding to the real and imaginary\nportions of f(ω), respectively, over a suit-\nable number N samples of ω. Plot the real\nportion against the imaginary portion and\nverify the figure resembles a propeller.\n(b) Let complex constant w = x + jy be repre-", - "type": "text" - }, - { - "block_id": "p83-b26", - "global_id": 2348, - "bbox": [ - 360.71, - 358.13, - 437.01, - 367.1 - ], - "text": "sented in vector form", - "type": "text" - }, - { - "block_id": "p83-b27", - "global_id": 2349, - "bbox": [ - 417.97, - 382.65, - 433.28, - 391.92 - ], - "text": "w =", - "type": "text" - }, - { - "block_id": "p83-b28", - "global_id": 2350, - "bbox": [ - 435.12, - 370.06, - 448.98, - 386.38 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p83-b29", - "global_id": 2351, - "bbox": [ - 445.0, - 388.36, - 448.98, - 397.33 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p83-b30", - "global_id": 2352, - "bbox": [ - 453.98, - 370.06, - 458.87, - 379.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p83-b31", - "global_id": 2353, - "bbox": [ - 360.71, - 409.2, - 500.77, - 418.53 - ], - "text": "Consider the 2 × 2 rotational matrix R:", - "type": "text" - }, - { - "block_id": "p83-b32", - "global_id": 2354, - "bbox": [ - 394.45, - 434.75, - 409.76, - 444.01 - ], - "text": "R =", - "type": "text" - }, - { - "block_id": "p83-b33", - "global_id": 2355, - "bbox": [ - 411.6, - 422.16, - 471.62, - 449.51 - ], - "text": "cosθ\n−sinθ\nsinθ\ncosθ", - "type": "text" - }, - { - "block_id": "p83-b34", - "global_id": 2356, - "bbox": [ - 477.5, - 422.16, - 482.39, - 431.12 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p83-b35", - "global_id": 2357, - "bbox": [ - 345.78, - 461.29, - 516.13, - 481.84 - ], - "text": "Show that Rw rotates vector w by θ radians.\n(c) Create a rotational matrix R corresponding", - "type": "text" - }, - { - "block_id": "p83-b36", - "global_id": 2358, - "bbox": [ - 345.28, - 481.94, - 516.13, - 536.63 - ], - "text": "to 10◦and multiply it by the 2×N matrix f\n= [fr;fi];. Plot the result to verify that\nthe “propeller” has indeed rotated counter-\nclockwise.\n(d) Given the matrix R determined in part (c),", - "type": "text" - }, - { - "block_id": "p83-b37", - "global_id": 2359, - "bbox": [ - 345.78, - 538.37, - 516.14, - 569.26 - ], - "text": "what is the effect of performing RRf? How\nabout RRRf? Generalize the result.\n(e) Investigate the behavior of multiplying f(ω)", - "type": "text" - }, - { - "block_id": "p83-b38", - "global_id": 2360, - "bbox": [ - 360.71, - 569.61, - 429.26, - 580.22 - ], - "text": "by the function ejθ.", - "type": "text" - } - ] - }, - { - "page_num": 84, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p84-b0", - "global_id": 2361, - "bbox": [ - 64.47, - 67.04, - 133.54, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p84-b1", - "global_id": 2362, - "bbox": [ - 147.02, - 151.14, - 391.37, - 173.26 - ], - "text": "SIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p84-b2", - "global_id": 2363, - "bbox": [ - 114.3, - 79.92, - 133.73, - 118.77 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p84-b3", - "global_id": 2364, - "bbox": [ - 101.84, - 266.14, - 490.42, - 323.93 - ], - "text": "In this chapter we shall discuss basic aspects of signals and systems. We shall also introduce\nfundamental concepts and qualitative explanations of the hows and whys of systems theory, thus\nbuilding a solid foundation for understanding the quantitative analysis in the remainder of the\nbook. For simplicity, the focus of this chapter is on continuous-time signals and systems. Chapter 3\npresents the same ideas for discrete-time signals and systems.", - "type": "text" - }, - { - "block_id": "p84-b4", - "global_id": 2365, - "bbox": [ - 102.14, - 340.8, - 148.75, - 352.93 - ], - "text": "SIGNALS", - "type": "text" - }, - { - "block_id": "p84-b5", - "global_id": 2366, - "bbox": [ - 101.84, - 356.86, - 490.41, - 438.65 - ], - "text": "A signal is a set of data or information. Examples include a telephone or a television signal,\nmonthly sales of a corporation, or daily closing prices of a stock market (e.g., the Dow Jones\naverages). In all these examples, the signals are functions of the independent variable time. This\nis not always the case, however. When an electrical charge is distributed over a body, for instance,\nthe signal is the charge density, a function of space rather than time. In this book we deal almost\nexclusively with signals that are functions of time. The discussion, however, applies equally well\nto other independent variables.", - "type": "text" - }, - { - "block_id": "p84-b6", - "global_id": 2367, - "bbox": [ - 101.84, - 455.53, - 490.43, - 565.34 - ], - "text": "SYSTEMS\nSignals may be processed further by systems, which may modify them or extract additional infor-\nmation from them. For example, an anti-aircraft gun operator may want to know the future location\nof a hostile moving target that is being tracked by his radar. Knowing the radar signal, he knows the\npast location and velocity of the target. By properly processing the radar signal (the input), he can\napproximately estimate the future location of the target. Thus, a system is an entity that processes\na set of signals (inputs) to yield another set of signals (outputs). A system may be made up of\nphysical components, as in electrical, mechanical, or hydraulic systems (hardware realization), or\nit may be an algorithm that computes an output from an input signal (software realization).", - "type": "text" - }, - { - "block_id": "p84-b7", - "global_id": 2368, - "bbox": [ - 102.2, - 592.93, - 244.25, - 606.88 - ], - "text": "1.1 SIZE OF A SIGNAL", - "type": "text" - }, - { - "block_id": "p84-b8", - "global_id": 2369, - "bbox": [ - 101.84, - 612.86, - 490.4, - 634.79 - ], - "text": "The size of any entity is a number that indicates the largeness or strength of that entity. Generally\nspeaking, the signal amplitude varies with time. How can a signal that exists over a certain time", - "type": "text" - }, - { - "block_id": "p84-b9", - "global_id": 2370, - "bbox": [ - 60.0, - 656.12, - 69.96, - 666.22 - ], - "text": "64", - "type": "text" - } - ] - }, - { - "page_num": 85, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p85-b0", - "global_id": 2371, - "bbox": [ - 393.32, - 62.89, - 516.14, - 71.98 - ], - "text": "1.1\nSize of a Signal\n65", - "type": "text" - }, - { - "block_id": "p85-b1", - "global_id": 2372, - "bbox": [ - 127.59, - 85.82, - 516.15, - 155.56 - ], - "text": "interval with varying amplitude be measured by one number that will indicate the signal size or\nsignal strength? Such a measure must consider not only the signal amplitude, but also its duration.\nFor instance, if we are to devise a single number V as a measure of the size of a human being,\nwe must consider not only his or her width (girth), but also the height. If we make a simplifying\nassumption that the shape of a person is a cylinder of variable radius r (which varies with the height\nh), then one possible measure of the size of a person of height H is the person’s volume V, given by", - "type": "text" - }, - { - "block_id": "p85-b2", - "global_id": 2373, - "bbox": [ - 284.28, - 172.53, - 308.98, - 182.81 - ], - "text": "V = π", - "type": "text" - }, - { - "block_id": "p85-b3", - "global_id": 2374, - "bbox": [ - 311.09, - 158.98, - 325.95, - 171.25 - ], - "text": "# H", - "type": "text" - }, - { - "block_id": "p85-b4", - "global_id": 2375, - "bbox": [ - 316.35, - 171.1, - 359.43, - 191.11 - ], - "text": "0\nr2(h)dh", - "type": "text" - }, - { - "block_id": "p85-b5", - "global_id": 2376, - "bbox": [ - 127.59, - 208.43, - 234.69, - 220.39 - ], - "text": "1.1-1 Signal Energy", - "type": "text" - }, - { - "block_id": "p85-b6", - "global_id": 2377, - "bbox": [ - 127.59, - 226.11, - 516.15, - 296.96 - ], - "text": "Arguing in this manner, we may consider the area under a signal x(t) as a possible measure of its\nsize, because it takes account not only of the amplitude but also of the duration. However, this will\nbe a defective measure because even for a large signal x(t), its positive and negative areas could\ncancel each other, indicating a signal of small size. This difficulty can be corrected by defining\nthe signal size as the area under |x(t)|2, which is always positive. We call this measure the signal\nenergy Ex, defined as", - "type": "text" - }, - { - "block_id": "p85-b7", - "global_id": 2378, - "bbox": [ - 284.59, - 303.04, - 304.1, - 314.5 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p85-b8", - "global_id": 2379, - "bbox": [ - 306.15, - 289.49, - 323.1, - 301.54 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p85-b9", - "global_id": 2380, - "bbox": [ - 311.42, - 314.36, - 323.97, - 321.33 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p85-b10", - "global_id": 2381, - "bbox": [ - 325.58, - 299.24, - 516.13, - 313.42 - ], - "text": "|x(t)|2 dt\n(1.1)", - "type": "text" - }, - { - "block_id": "p85-b11", - "global_id": 2382, - "bbox": [ - 127.59, - 328.46, - 379.44, - 339.92 - ], - "text": "This definition simplifies for a real-valued signal x(t) to Ex =", - "type": "text" - }, - { - "block_id": "p85-b12", - "global_id": 2383, - "bbox": [ - 382.07, - 320.45, - 395.69, - 332.5 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p85-b13", - "global_id": 2384, - "bbox": [ - 127.59, - 327.53, - 516.15, - 374.71 - ], - "text": "−∞x2(t)dt. There are also other\npossible measures of signal size, such as the area under |x(t)|. The energy measure, however, is\nnot only more tractable mathematically but is also more meaningful (as shown later) in the sense\nthat it is indicative of the energy that can be extracted from the signal.", - "type": "text" - }, - { - "block_id": "p85-b14", - "global_id": 2385, - "bbox": [ - 127.59, - 392.84, - 230.7, - 404.8 - ], - "text": "1.1-2 Signal Power", - "type": "text" - }, - { - "block_id": "p85-b15", - "global_id": 2386, - "bbox": [ - 127.59, - 410.93, - 516.14, - 444.81 - ], - "text": "Signal energy must be finite for it to be a meaningful measure of signal size. A necessary condition\nfor the energy to be finite is that the signal amplitude →0 as |t| →∞(Fig. 1.1a). Otherwise the\nintegral in Eq. (1.1) will not converge.", - "type": "text" - }, - { - "block_id": "p85-b16", - "global_id": 2387, - "bbox": [ - 127.59, - 446.38, - 516.14, - 494.12 - ], - "text": "When the amplitude of x(t) does not →0 as |t| →∞(Fig. 1.1b), the signal energy is infinite.\nA more meaningful measure of the signal size in such a case would be the time average of the\nenergy, if it exists. This measure is called the power of the signal. For a signal x(t), we define its\npower Px as", - "type": "text" - }, - { - "block_id": "p85-b17", - "global_id": 2388, - "bbox": [ - 266.95, - 501.02, - 304.77, - 512.48 - ], - "text": "Px = lim", - "type": "text" - }, - { - "block_id": "p85-b18", - "global_id": 2389, - "bbox": [ - 288.51, - 509.83, - 307.72, - 517.02 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p85-b19", - "global_id": 2390, - "bbox": [ - 310.01, - 494.45, - 315.66, - 518.37 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p85-b20", - "global_id": 2391, - "bbox": [ - 319.73, - 487.46, - 340.74, - 499.8 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p85-b21", - "global_id": 2392, - "bbox": [ - 324.99, - 512.33, - 341.61, - 519.6 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p85-b22", - "global_id": 2393, - "bbox": [ - 343.21, - 497.21, - 516.13, - 511.4 - ], - "text": "|x(t)|2 dt\n(1.2)", - "type": "text" - }, - { - "block_id": "p85-b23", - "global_id": 2394, - "bbox": [ - 127.59, - 527.43, - 422.68, - 540.23 - ], - "text": "This definition simplifies for a real-valued signal x(t) to Px = limT→∞1", - "type": "text" - }, - { - "block_id": "p85-b24", - "global_id": 2395, - "bbox": [ - 418.73, - 534.92, - 422.61, - 541.89 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p85-b25", - "global_id": 2396, - "bbox": [ - 426.55, - 520.74, - 444.24, - 533.09 - ], - "text": "$ T/2", - "type": "text" - }, - { - "block_id": "p85-b26", - "global_id": 2397, - "bbox": [ - 127.59, - 527.83, - 516.12, - 575.01 - ], - "text": "−T/2 x2(t)dt. Observe\nthat the signal power Px is the time average (mean) of the signal magnitude squared, that is, the\nmean-square value of |x(t)|. Indeed, the square root of Px is the familiar rms (root-mean-square)\nvalue of x(t).", - "type": "text" - }, - { - "block_id": "p85-b27", - "global_id": 2398, - "bbox": [ - 127.59, - 577.0, - 516.15, - 634.79 - ], - "text": "Generally, the mean of an entity averaged over a large time interval approaching infinity exists\nif the entity either is periodic or has a statistical regularity. If such a condition is not satisfied, the\naverage may not exist. For instance, a ramp signal x(t) = t increases indefinitely as |t| →∞, and\nneither the energy nor the power exists for this signal. However, the unit step function, which is\nnot periodic nor has statistical regularity, does have a finite power.", - "type": "text" - } - ] - }, - { - "page_num": 86, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p86-b0", - "global_id": 2399, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "66\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p86-b1", - "global_id": 2400, - "bbox": [ - 289.11, - 249.74, - 298.76, - 257.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p86-b2", - "global_id": 2401, - "bbox": [ - 306.28, - 187.94, - 317.7, - 196.03 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p86-b3", - "global_id": 2402, - "bbox": [ - 455.1, - 214.82, - 457.32, - 222.82 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p86-b4", - "global_id": 2403, - "bbox": [ - 307.25, - 92.12, - 318.67, - 100.2 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p86-b5", - "global_id": 2404, - "bbox": [ - 289.11, - 159.61, - 297.99, - 167.61 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p86-b6", - "global_id": 2405, - "bbox": [ - 437.19, - 144.35, - 439.41, - 152.35 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p86-b7", - "global_id": 2406, - "bbox": [ - 125.76, - 264.36, - 477.9, - 273.68 - ], - "text": "Figure 1.1 Examples of signals: (a) a signal with finite energy and (b) a signal with finite power.", - "type": "text" - }, - { - "block_id": "p86-b8", - "global_id": 2407, - "bbox": [ - 101.84, - 304.87, - 490.39, - 330.4 - ], - "text": "When x(t) is periodic, |x(t)|2 is also periodic. Hence, the power of x(t) can be computed from\nEq. (1.2) by averaging |x(t)|2 over one period.", - "type": "text" - }, - { - "block_id": "p86-b9", - "global_id": 2408, - "bbox": [ - 101.84, - 353.0, - 490.42, - 542.37 - ], - "text": "Comments. The signal energy as defined in Eq. (1.1) does not indicate the actual energy (in the\nconventional sense) of the signal because the signal energy depends not only on the signal, but also\non the load. It can, however, be interpreted as the energy dissipated in a normalized load of a 1 ohm\nresistor if a voltage x(t) were to be applied across the 1 ohm resistor [or if a current x(t) were to be\npassed through the 1 ohm resistor]. The measure of “energy” is therefore indicative of the energy\ncapability of the signal, not the actual energy. For this reason the concepts of conservation of\nenergy should not be applied to this “signal energy.” Parallel observation applies to “signal power”\ndefined in Eq. (1.2). These measures are but convenient indicators of the signal size, which prove\nuseful in many applications. For instance, if we approximate a signal x(t) by another signal g(t),\nthe error in the approximation is e(t) = x(t) −g(t). The energy (or power) of e(t) is a convenient\nindicator of the goodness of the approximation. It provides us with a quantitative measure of\ndetermining the closeness of the approximation. In communication systems, during transmission\nover a channel, message signals are corrupted by unwanted signals (noise). The quality of the\nreceived signal is judged by the relative sizes of the desired signal and the unwanted signal (noise).\nIn this case the ratio of the message signal and noise signal powers (signal-to-noise power ratio) is\na good indication of the received signal quality.", - "type": "text" - }, - { - "block_id": "p86-b10", - "global_id": 2409, - "bbox": [ - 101.84, - 564.97, - 490.41, - 634.79 - ], - "text": "Units of Energy and Power. Equation (1.1) is not correct dimensionally. This is because here\nwe are using the term energy not in its conventional sense, but to indicate the signal size. The\nsame observation applies to Eq. (1.2) for power. The units of energy and power, as defined here,\ndepend on the nature of the signal x(t). If x(t) is a voltage signal, its energy Ex has units of volts\nsquared-seconds (V2 s), and its power Px has units of volts squared. If x(t) is a current signal, these\nunits will be amperes squared-seconds (A2 s) and amperes squared, respectively.", - "type": "text" - } - ] - }, - { - "page_num": 87, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p87-b0", - "global_id": 2410, - "bbox": [ - 393.32, - 62.89, - 516.14, - 71.98 - ], - "text": "1.1\nSize of a Signal\n67", - "type": "text" - }, - { - "block_id": "p87-b1", - "global_id": 2411, - "bbox": [ - 102.51, - 93.92, - 409.24, - 105.87 - ], - "text": "EXAMPLE 1.1\nClassifying Energy and Power Signals", - "type": "text" - }, - { - "block_id": "p87-b2", - "global_id": 2412, - "bbox": [ - 128.9, - 122.53, - 360.25, - 132.49 - ], - "text": "Determine the suitable measures of the signals in Fig. 1.2.", - "type": "text" - }, - { - "block_id": "p87-b3", - "global_id": 2413, - "bbox": [ - 303.45, - 339.27, - 313.1, - 347.27 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p87-b4", - "global_id": 2414, - "bbox": [ - 463.58, - 319.23, - 465.8, - 327.23 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p87-b5", - "global_id": 2415, - "bbox": [ - 148.57, - 305.02, - 461.97, - 313.48 - ], - "text": "0\n1\n3\n2\n1\n2\n3\n4\n4", - "type": "text" - }, - { - "block_id": "p87-b6", - "global_id": 2416, - "bbox": [ - 310.34, - 322.16, - 321.01, - 330.46 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p87-b7", - "global_id": 2417, - "bbox": [ - 310.47, - 234.35, - 412.66, - 242.86 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p87-b8", - "global_id": 2418, - "bbox": [ - 303.45, - 251.47, - 312.33, - 259.47 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p87-b9", - "global_id": 2419, - "bbox": [ - 283.81, - 233.62, - 385.97, - 242.3 - ], - "text": "2\n4\n1", - "type": "text" - }, - { - "block_id": "p87-b10", - "global_id": 2420, - "bbox": [ - 318.49, - 166.74, - 329.6, - 174.82 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p87-b11", - "global_id": 2421, - "bbox": [ - 293.78, - 277.1, - 304.88, - 285.18 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p87-b12", - "global_id": 2422, - "bbox": [ - 347.2, - 197.45, - 367.42, - 207.14 - ], - "text": "2et2", - "type": "text" - }, - { - "block_id": "p87-b13", - "global_id": 2423, - "bbox": [ - 313.67, - 277.27, - 317.67, - 285.27 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p87-b14", - "global_id": 2424, - "bbox": [ - 301.51, - 169.89, - 305.51, - 177.89 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p87-b15", - "global_id": 2425, - "bbox": [ - 127.51, - 353.97, - 240.09, - 363.21 - ], - "text": "Figure 1.2 Signals for Ex. 1.1", - "type": "text" - }, - { - "block_id": "p87-b16", - "global_id": 2426, - "bbox": [ - 128.9, - 411.39, - 502.76, - 435.22 - ], - "text": "In Fig. 1.2a, the signal amplitude →0 as |t| →∞. Therefore the suitable measure for this\nsignal is its energy Ex given by", - "type": "text" - }, - { - "block_id": "p87-b17", - "global_id": 2427, - "bbox": [ - 199.86, - 452.08, - 219.39, - 463.53 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p87-b18", - "global_id": 2428, - "bbox": [ - 221.43, - 438.51, - 238.37, - 450.58 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p87-b19", - "global_id": 2429, - "bbox": [ - 226.69, - 463.39, - 239.25, - 470.37 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p87-b20", - "global_id": 2430, - "bbox": [ - 240.86, - 448.28, - 284.23, - 462.35 - ], - "text": "|x(t)|2 dt =", - "type": "text" - }, - { - "block_id": "p87-b21", - "global_id": 2431, - "bbox": [ - 286.28, - 438.51, - 299.58, - 450.87 - ], - "text": "# 0", - "type": "text" - }, - { - "block_id": "p87-b22", - "global_id": 2432, - "bbox": [ - 291.53, - 463.39, - 300.46, - 470.66 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p87-b23", - "global_id": 2433, - "bbox": [ - 300.96, - 448.28, - 335.72, - 462.45 - ], - "text": "(2)2 dt +", - "type": "text" - }, - { - "block_id": "p87-b24", - "global_id": 2434, - "bbox": [ - 337.26, - 438.51, - 354.21, - 450.58 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p87-b25", - "global_id": 2435, - "bbox": [ - 342.53, - 448.28, - 431.8, - 470.66 - ], - "text": "0\n4e−t dt = 4 + 4 = 8", - "type": "text" - }, - { - "block_id": "p87-b26", - "global_id": 2436, - "bbox": [ - 128.9, - 479.33, - 502.75, - 537.52 - ], - "text": "In Fig. 1.2b, the signal magnitude does not →0 as |t| →∞. However, it is periodic, and\ntherefore its power exists. We can use Eq. (1.2) to determine its power. We can simplify the\nprocedure for periodic signals by observing that a periodic signal repeats regularly each period\n(2 seconds in this case). Therefore, averaging |x(t)|2 over an infinitely large interval is identical\nto averaging this quantity over one period (2 seconds in this case). Thus", - "type": "text" - }, - { - "block_id": "p87-b27", - "global_id": 2437, - "bbox": [ - 241.59, - 554.53, - 267.85, - 567.34 - ], - "text": "Px = 1", - "type": "text" - }, - { - "block_id": "p87-b28", - "global_id": 2438, - "bbox": [ - 264.36, - 562.1, - 267.85, - 569.07 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p87-b29", - "global_id": 2439, - "bbox": [ - 271.26, - 542.32, - 284.57, - 554.66 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p87-b30", - "global_id": 2440, - "bbox": [ - 276.52, - 567.19, - 285.45, - 574.45 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p87-b31", - "global_id": 2441, - "bbox": [ - 287.05, - 552.07, - 337.15, - 566.16 - ], - "text": "|x(t)|2 dt = 1", - "type": "text" - }, - { - "block_id": "p87-b32", - "global_id": 2442, - "bbox": [ - 333.66, - 562.1, - 337.15, - 569.07 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p87-b33", - "global_id": 2443, - "bbox": [ - 340.57, - 542.32, - 353.87, - 554.66 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p87-b34", - "global_id": 2444, - "bbox": [ - 345.82, - 567.19, - 354.75, - 574.45 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p87-b35", - "global_id": 2445, - "bbox": [ - 356.35, - 552.07, - 388.87, - 566.16 - ], - "text": "t2 dt = 1", - "type": "text" - }, - { - "block_id": "p87-b36", - "global_id": 2446, - "bbox": [ - 385.39, - 562.1, - 388.87, - 569.07 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p87-b37", - "global_id": 2447, - "bbox": [ - 128.9, - 583.54, - 502.75, - 605.46 - ], - "text": "Recall that the signal power is the square of its rms value. Therefore, the rms value of this\nsignal is 1/", - "type": "text" - }, - { - "block_id": "p87-b38", - "global_id": 2448, - "bbox": [ - 173.99, - 586.71, - 182.42, - 596.67 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p87-b39", - "global_id": 2449, - "bbox": [ - 182.43, - 595.5, - 189.9, - 605.46 - ], - "text": "3.", - "type": "text" - } - ] - }, - { - "page_num": 88, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p88-b0", - "global_id": 2450, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "68\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p88-b1", - "global_id": 2451, - "bbox": [ - 76.77, - 93.91, - 371.52, - 105.87 - ], - "text": "EXAMPLE 1.2\nDetermining Power and RMS Value", - "type": "text" - }, - { - "block_id": "p88-b2", - "global_id": 2452, - "bbox": [ - 103.16, - 122.54, - 270.61, - 132.5 - ], - "text": "Determine the power and the rms value of", - "type": "text" - }, - { - "block_id": "p88-b3", - "global_id": 2453, - "bbox": [ - 121.09, - 140.05, - 226.1, - 151.2 - ], - "text": "(a) x(t) = C cos(ω0t + θ)", - "type": "text" - }, - { - "block_id": "p88-b4", - "global_id": 2454, - "bbox": [ - 121.09, - 155.0, - 364.27, - 180.23 - ], - "text": "(b) x(t) = C1 cos(ω1t + θ1) + C2 cos(ω2t + θ2)\nω1̸ = ω2\n(c) x(t) = Dejω0t", - "type": "text" - }, - { - "block_id": "p88-b5", - "global_id": 2455, - "bbox": [ - 103.16, - 208.79, - 477.02, - 255.03 - ], - "text": "(a) This is a periodic signal with period T0 = 2π/ω0. The suitable measure of this signal\nis its power. Because it is a periodic signal, we may compute its power by averaging its energy\nover one period T0 = 2π/ω0. However, for the sake of demonstration, we shall use Eq. (1.2) to\nsolve this problem by averaging over an infinitely large time interval.", - "type": "text" - }, - { - "block_id": "p88-b6", - "global_id": 2456, - "bbox": [ - 130.31, - 273.4, - 168.13, - 284.85 - ], - "text": "Px = lim", - "type": "text" - }, - { - "block_id": "p88-b7", - "global_id": 2457, - "bbox": [ - 151.88, - 282.2, - 171.09, - 289.4 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p88-b8", - "global_id": 2458, - "bbox": [ - 173.38, - 266.82, - 179.03, - 290.75 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p88-b9", - "global_id": 2459, - "bbox": [ - 183.1, - 259.83, - 204.1, - 272.18 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b10", - "global_id": 2460, - "bbox": [ - 188.36, - 284.71, - 204.98, - 291.98 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b11", - "global_id": 2461, - "bbox": [ - 206.58, - 269.28, - 312.31, - 284.54 - ], - "text": "C2 cos2 (ω0t + θ)dt = lim", - "type": "text" - }, - { - "block_id": "p88-b12", - "global_id": 2462, - "bbox": [ - 296.06, - 282.2, - 315.27, - 289.4 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p88-b13", - "global_id": 2463, - "bbox": [ - 317.79, - 265.47, - 328.13, - 276.69 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p88-b14", - "global_id": 2464, - "bbox": [ - 317.56, - 280.78, - 328.08, - 290.85 - ], - "text": "2T", - "type": "text" - }, - { - "block_id": "p88-b15", - "global_id": 2465, - "bbox": [ - 332.26, - 259.83, - 353.27, - 272.18 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b16", - "global_id": 2466, - "bbox": [ - 337.53, - 284.71, - 354.14, - 291.98 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b17", - "global_id": 2467, - "bbox": [ - 355.75, - 273.4, - 449.68, - 284.54 - ], - "text": "[1 + cos(2ω0t + 2θ)] dt", - "type": "text" - }, - { - "block_id": "p88-b18", - "global_id": 2468, - "bbox": [ - 142.06, - 303.33, - 168.13, - 313.71 - ], - "text": "= lim", - "type": "text" - }, - { - "block_id": "p88-b19", - "global_id": 2469, - "bbox": [ - 151.88, - 312.14, - 171.09, - 319.33 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p88-b20", - "global_id": 2470, - "bbox": [ - 173.6, - 295.41, - 183.94, - 306.62 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p88-b21", - "global_id": 2471, - "bbox": [ - 173.38, - 310.72, - 183.9, - 320.78 - ], - "text": "2T", - "type": "text" - }, - { - "block_id": "p88-b22", - "global_id": 2472, - "bbox": [ - 188.08, - 289.78, - 209.09, - 302.12 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b23", - "global_id": 2473, - "bbox": [ - 193.35, - 314.65, - 209.96, - 321.91 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b24", - "global_id": 2474, - "bbox": [ - 211.56, - 303.33, - 246.62, - 313.71 - ], - "text": "dt + lim", - "type": "text" - }, - { - "block_id": "p88-b25", - "global_id": 2475, - "bbox": [ - 230.36, - 312.14, - 249.57, - 319.33 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p88-b26", - "global_id": 2476, - "bbox": [ - 252.09, - 295.41, - 262.43, - 306.62 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p88-b27", - "global_id": 2477, - "bbox": [ - 251.86, - 310.72, - 262.38, - 320.78 - ], - "text": "2T", - "type": "text" - }, - { - "block_id": "p88-b28", - "global_id": 2478, - "bbox": [ - 265.45, - 289.78, - 286.46, - 302.12 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b29", - "global_id": 2479, - "bbox": [ - 270.72, - 314.65, - 287.34, - 321.91 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b30", - "global_id": 2480, - "bbox": [ - 288.94, - 303.33, - 359.28, - 314.48 - ], - "text": "cos(2ω0t + 2θ)dt", - "type": "text" - }, - { - "block_id": "p88-b31", - "global_id": 2481, - "bbox": [ - 103.16, - 332.06, - 477.02, - 391.19 - ], - "text": "The first term on the right-hand side is equal to C2/2. The second term, however, is zero\nbecause the integral appearing in this term represents the area under a sinusoid over a very\nlarge time interval T with T →∞. This area is at most equal to the area of half the cycle\nbecause of cancellations of the positive and negative areas of a sinusoid. The second term is\nthis area multiplied by C2/2T with T →∞. Clearly this term is zero, and", - "type": "text" - }, - { - "block_id": "p88-b32", - "global_id": 2482, - "bbox": [ - 272.69, - 401.36, - 305.79, - 420.73 - ], - "text": "Px = C2", - "type": "text" - }, - { - "block_id": "p88-b33", - "global_id": 2483, - "bbox": [ - 298.38, - 416.77, - 303.36, - 426.73 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p88-b34", - "global_id": 2484, - "bbox": [ - 103.17, - 434.74, - 477.02, - 459.08 - ], - "text": "This shows that a sinusoid of amplitude C has a power C2/2 regardless of the value of its\nfrequency ω0 (ω0̸ = 0) and phase θ. The rms value is C/", - "type": "text" - }, - { - "block_id": "p88-b35", - "global_id": 2485, - "bbox": [ - 325.64, - 439.2, - 334.07, - 449.16 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p88-b36", - "global_id": 2486, - "bbox": [ - 103.17, - 448.04, - 477.02, - 469.96 - ], - "text": "2. If the signal frequency is zero (dc\nor a constant signal of amplitude C), the reader can show that the power is C2.", - "type": "text" - }, - { - "block_id": "p88-b37", - "global_id": 2487, - "bbox": [ - 103.17, - 471.87, - 477.03, - 517.78 - ], - "text": "(b) In Ch. 6, we shall show that a sum of two sinusoids may or may not be periodic,\ndepending on whether the ratio ω1/ω2 is a rational number. Therefore, the period of this signal\nis not known. Hence, its power will be determined by averaging its energy over T seconds with\nT →∞. Thus,", - "type": "text" - }, - { - "block_id": "p88-b38", - "global_id": 2488, - "bbox": [ - 134.25, - 535.42, - 173.17, - 546.88 - ], - "text": "Px = lim", - "type": "text" - }, - { - "block_id": "p88-b39", - "global_id": 2489, - "bbox": [ - 156.92, - 544.23, - 176.13, - 551.42 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p88-b40", - "global_id": 2490, - "bbox": [ - 178.42, - 528.85, - 184.07, - 552.77 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p88-b41", - "global_id": 2491, - "bbox": [ - 188.14, - 521.86, - 209.14, - 534.2 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b42", - "global_id": 2492, - "bbox": [ - 193.4, - 546.74, - 210.02, - 554.0 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b43", - "global_id": 2493, - "bbox": [ - 210.52, - 531.62, - 375.58, - 547.29 - ], - "text": "[C1 cos(ω1t + θ1) + C2 cos(ω2t + θ2)]2 dt", - "type": "text" - }, - { - "block_id": "p88-b44", - "global_id": 2494, - "bbox": [ - 146.0, - 565.35, - 173.17, - 575.73 - ], - "text": "= lim", - "type": "text" - }, - { - "block_id": "p88-b45", - "global_id": 2495, - "bbox": [ - 156.92, - 574.16, - 176.13, - 581.36 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p88-b46", - "global_id": 2496, - "bbox": [ - 178.42, - 558.78, - 184.07, - 582.71 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p88-b47", - "global_id": 2497, - "bbox": [ - 188.14, - 551.8, - 209.14, - 564.15 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b48", - "global_id": 2498, - "bbox": [ - 193.4, - 576.67, - 210.02, - 583.94 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b49", - "global_id": 2499, - "bbox": [ - 211.62, - 565.67, - 221.75, - 576.5 - ], - "text": "C1", - "type": "text" - }, - { - "block_id": "p88-b50", - "global_id": 2500, - "bbox": [ - 222.25, - 561.24, - 326.07, - 581.36 - ], - "text": "2 cos2 (ω1t + θ1)dt + lim\nT→∞", - "type": "text" - }, - { - "block_id": "p88-b51", - "global_id": 2501, - "bbox": [ - 328.36, - 558.78, - 334.01, - 582.71 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p88-b52", - "global_id": 2502, - "bbox": [ - 338.08, - 551.8, - 359.08, - 564.15 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b53", - "global_id": 2503, - "bbox": [ - 343.34, - 576.67, - 359.96, - 583.94 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b54", - "global_id": 2504, - "bbox": [ - 361.56, - 565.67, - 371.69, - 576.5 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p88-b55", - "global_id": 2505, - "bbox": [ - 372.19, - 561.24, - 445.75, - 576.5 - ], - "text": "2 cos2 (ω2t + θ2)dt", - "type": "text" - }, - { - "block_id": "p88-b56", - "global_id": 2506, - "bbox": [ - 157.36, - 595.3, - 184.04, - 605.67 - ], - "text": "+ lim", - "type": "text" - }, - { - "block_id": "p88-b57", - "global_id": 2507, - "bbox": [ - 167.79, - 604.11, - 187.0, - 611.3 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p88-b58", - "global_id": 2508, - "bbox": [ - 189.29, - 588.63, - 215.04, - 612.64 - ], - "text": "2C1C2\nT", - "type": "text" - }, - { - "block_id": "p88-b59", - "global_id": 2509, - "bbox": [ - 218.95, - 581.74, - 239.95, - 594.08 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p88-b60", - "global_id": 2510, - "bbox": [ - 224.21, - 606.61, - 240.82, - 613.87 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p88-b61", - "global_id": 2511, - "bbox": [ - 242.43, - 595.3, - 361.43, - 606.45 - ], - "text": "cos(ω1t + θ1)cos(ω2t + θ2)dt", - "type": "text" - } - ] - }, - { - "page_num": 89, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p89-b0", - "global_id": 2512, - "bbox": [ - 393.32, - 62.89, - 516.14, - 71.98 - ], - "text": "1.1\nSize of a Signal\n69", - "type": "text" - }, - { - "block_id": "p89-b1", - "global_id": 2513, - "bbox": [ - 128.9, - 86.24, - 502.76, - 132.06 - ], - "text": "The first and second integrals on the right-hand side are the powers of the two sinusoids, which\nare C12/2 and C22/2, as found in part (a). The third term, the product of two sinusoids, can be\nexpressed as a sum of two sinusoids cos[(ω1+ω2)t+(θ1+θ2)] and cos[(ω1−ω2)t+(θ1−θ2)],\nrespectively. Now, arguing as in part (a), we see that the third term is zero. Hence, we have†", - "type": "text" - }, - { - "block_id": "p89-b2", - "global_id": 2514, - "bbox": [ - 282.61, - 142.23, - 319.49, - 161.94 - ], - "text": "Px = C12", - "type": "text" - }, - { - "block_id": "p89-b3", - "global_id": 2515, - "bbox": [ - 310.19, - 142.23, - 347.37, - 167.94 - ], - "text": "2 + C22", - "type": "text" - }, - { - "block_id": "p89-b4", - "global_id": 2516, - "bbox": [ - 338.06, - 157.98, - 343.05, - 167.94 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p89-b5", - "global_id": 2517, - "bbox": [ - 128.91, - 179.11, - 208.35, - 189.07 - ], - "text": "and the rms value is", - "type": "text" - }, - { - "block_id": "p89-b7", - "global_id": 2518, - "bbox": [ - 128.91, - 174.74, - 502.78, - 212.98 - ], - "text": "(C12 + C22)/2.\nWe can readily extend this result to a sum of any number of sinusoids with distinct\nfrequencies. Thus, if", - "type": "text" - }, - { - "block_id": "p89-b8", - "global_id": 2519, - "bbox": [ - 261.75, - 222.69, - 286.41, - 232.96 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p89-b9", - "global_id": 2520, - "bbox": [ - 288.46, - 212.51, - 302.55, - 223.18 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p89-b10", - "global_id": 2521, - "bbox": [ - 289.29, - 237.07, - 301.7, - 244.34 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p89-b11", - "global_id": 2522, - "bbox": [ - 303.66, - 222.69, - 369.91, - 234.56 - ], - "text": "Cn cos(ωnt + θn)", - "type": "text" - }, - { - "block_id": "p89-b12", - "global_id": 2523, - "bbox": [ - 128.91, - 250.63, - 463.53, - 262.09 - ], - "text": "assuming that none of the two sinusoids have identical frequencies and ωn̸ = 0, then", - "type": "text" - }, - { - "block_id": "p89-b13", - "global_id": 2524, - "bbox": [ - 286.65, - 279.33, - 312.9, - 292.13 - ], - "text": "Px = 1", - "type": "text" - }, - { - "block_id": "p89-b14", - "global_id": 2525, - "bbox": [ - 309.41, - 286.89, - 312.9, - 293.86 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p89-b15", - "global_id": 2526, - "bbox": [ - 315.2, - 270.5, - 329.3, - 281.18 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p89-b16", - "global_id": 2527, - "bbox": [ - 316.04, - 295.06, - 328.45, - 302.32 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p89-b17", - "global_id": 2528, - "bbox": [ - 330.41, - 280.99, - 340.53, - 291.75 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p89-b18", - "global_id": 2529, - "bbox": [ - 341.03, - 279.24, - 344.52, - 286.21 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p89-b19", - "global_id": 2530, - "bbox": [ - 128.9, - 311.67, - 240.02, - 322.05 - ], - "text": "If x(t) also has a dc term, as", - "type": "text" - }, - { - "block_id": "p89-b20", - "global_id": 2531, - "bbox": [ - 251.01, - 341.29, - 297.66, - 352.74 - ], - "text": "x(t) = C0 +", - "type": "text" - }, - { - "block_id": "p89-b21", - "global_id": 2532, - "bbox": [ - 299.2, - 331.12, - 313.3, - 341.78 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p89-b22", - "global_id": 2533, - "bbox": [ - 300.05, - 355.68, - 312.45, - 362.94 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p89-b23", - "global_id": 2534, - "bbox": [ - 314.41, - 341.29, - 380.66, - 353.16 - ], - "text": "Cn cos(ωnt + θn)", - "type": "text" - }, - { - "block_id": "p89-b24", - "global_id": 2535, - "bbox": [ - 128.91, - 372.63, - 146.05, - 382.6 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p89-b25", - "global_id": 2536, - "bbox": [ - 275.79, - 388.82, - 307.7, - 401.7 - ], - "text": "Px = C2", - "type": "text" - }, - { - "block_id": "p89-b26", - "global_id": 2537, - "bbox": [ - 304.0, - 388.91, - 323.76, - 403.44 - ], - "text": "0 + 1\n2", - "type": "text" - }, - { - "block_id": "p89-b27", - "global_id": 2538, - "bbox": [ - 326.06, - 380.08, - 340.16, - 390.74 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p89-b28", - "global_id": 2539, - "bbox": [ - 326.9, - 404.64, - 339.31, - 411.9 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p89-b29", - "global_id": 2540, - "bbox": [ - 341.27, - 390.56, - 351.4, - 401.33 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p89-b30", - "global_id": 2541, - "bbox": [ - 351.9, - 388.82, - 502.76, - 400.62 - ], - "text": "2\n(1.3)", - "type": "text" - }, - { - "block_id": "p89-b31", - "global_id": 2542, - "bbox": [ - 146.84, - 418.65, - 469.6, - 428.7 - ], - "text": "(c) In this case the signal is complex, and we use Eq. (1.2) to compute the power.", - "type": "text" - }, - { - "block_id": "p89-b32", - "global_id": 2543, - "bbox": [ - 256.26, - 447.06, - 294.07, - 458.51 - ], - "text": "Px = lim", - "type": "text" - }, - { - "block_id": "p89-b33", - "global_id": 2544, - "bbox": [ - 277.82, - 455.86, - 297.03, - 463.06 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p89-b34", - "global_id": 2545, - "bbox": [ - 299.32, - 440.49, - 304.97, - 464.41 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p89-b35", - "global_id": 2546, - "bbox": [ - 309.04, - 433.49, - 330.04, - 445.84 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p89-b36", - "global_id": 2547, - "bbox": [ - 314.3, - 458.37, - 330.91, - 465.63 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p89-b37", - "global_id": 2548, - "bbox": [ - 332.52, - 443.25, - 375.23, - 457.33 - ], - "text": "|Dejω0t|2 dt", - "type": "text" - }, - { - "block_id": "p89-b38", - "global_id": 2549, - "bbox": [ - 128.9, - 473.09, - 327.04, - 487.09 - ], - "text": "Recall that |ejω0t| = 1 so that |Dejω0t|2 = |D|2, and", - "type": "text" - }, - { - "block_id": "p89-b39", - "global_id": 2550, - "bbox": [ - 296.61, - 497.19, - 502.76, - 510.08 - ], - "text": "Px = |D|2\n(1.4)", - "type": "text" - }, - { - "block_id": "p89-b40", - "global_id": 2551, - "bbox": [ - 128.91, - 520.55, - 212.67, - 530.93 - ], - "text": "The rms value is |D|.", - "type": "text" - }, - { - "block_id": "p89-b41", - "global_id": 2552, - "bbox": [ - 127.6, - 567.31, - 516.14, - 590.07 - ], - "text": "Comment. In part (b) of Ex. 1.2, we have shown that the power of the sum of two sinusoids is\nequal to the sum of the powers of the sinusoids. It may appear that the power of x1(t) + x2(t)", - "type": "text" - }, - { - "block_id": "p89-b42", - "global_id": 2553, - "bbox": [ - 127.59, - 610.24, - 516.13, - 634.75 - ], - "text": "† This is true only if ω1̸ = ω2. If ω1 = ω2, the integrand of the third term contains a constant cos(θ1 −θ2),\nand the third term →2C1C2 cos(θ1 −θ2) as T →∞.", - "type": "text" - } - ] - }, - { - "page_num": 90, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p90-b0", - "global_id": 2554, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "70\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p90-b1", - "global_id": 2555, - "bbox": [ - 101.84, - 85.4, - 490.42, - 107.74 - ], - "text": "is Px1 + Px2. Unfortunately, this conclusion is not true in general. It is true only under a certain\ncondition (orthogonality), discussed later (Sec. 6.5-3).", - "type": "text" - }, - { - "block_id": "p90-b2", - "global_id": 2556, - "bbox": [ - 107.82, - 139.74, - 417.19, - 151.69 - ], - "text": "DRILL 1.1\nComputing Energy, Power, and RMS Value", - "type": "text" - }, - { - "block_id": "p90-b3", - "global_id": 2557, - "bbox": [ - 107.82, - 160.4, - 484.4, - 206.64 - ], - "text": "Show that the energies of the signals in Figs. 1.3a, 1.3b, 1.3c, and 1.3d are 4, 1, 4/3, and 4/3,\nrespectively. Observe that doubling a signal quadruples the energy, and time-shifting a signal\nhas no effect on the energy. Show also that the power of the signal in Fig. 1.3e is 0.4323. What\nis the rms value of signal in Fig. 1.3e?", - "type": "text" - }, - { - "block_id": "p90-b4", - "global_id": 2558, - "bbox": [ - 285.86, - 332.54, - 289.86, - 340.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p90-b5", - "global_id": 2559, - "bbox": [ - 286.36, - 389.29, - 452.96, - 397.54 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p90-b6", - "global_id": 2560, - "bbox": [ - 298.85, - 341.04, - 309.07, - 350.65 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p90-b7", - "global_id": 2561, - "bbox": [ - 129.07, - 389.24, - 433.94, - 397.54 - ], - "text": "1\n2\n3\n4\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p90-b8", - "global_id": 2562, - "bbox": [ - 293.63, - 405.42, - 302.51, - 413.42 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p90-b9", - "global_id": 2563, - "bbox": [ - 127.97, - 293.15, - 191.13, - 301.15 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p90-b10", - "global_id": 2564, - "bbox": [ - 127.97, - 243.11, - 448.17, - 260.28 - ], - "text": "2\n x1(t)\nx4(t)", - "type": "text" - }, - { - "block_id": "p90-b11", - "global_id": 2565, - "bbox": [ - 257.2, - 333.8, - 271.31, - 343.4 - ], - "text": "x5(t)", - "type": "text" - }, - { - "block_id": "p90-b12", - "global_id": 2566, - "bbox": [ - 238.53, - 243.11, - 344.57, - 254.72 - ], - "text": "x3(t)\nx2(t)", - "type": "text" - }, - { - "block_id": "p90-b13", - "global_id": 2567, - "bbox": [ - 153.07, - 292.81, - 155.3, - 300.81 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p90-b14", - "global_id": 2568, - "bbox": [ - 157.54, - 310.0, - 166.42, - 318.0 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p90-b15", - "global_id": 2569, - "bbox": [ - 216.46, - 293.15, - 279.63, - 301.15 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p90-b16", - "global_id": 2570, - "bbox": [ - 216.46, - 269.04, - 220.46, - 277.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p90-b17", - "global_id": 2571, - "bbox": [ - 241.57, - 292.81, - 243.8, - 300.81 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p90-b18", - "global_id": 2572, - "bbox": [ - 245.45, - 310.0, - 255.1, - 318.0 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p90-b19", - "global_id": 2573, - "bbox": [ - 306.96, - 293.15, - 370.13, - 301.15 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p90-b20", - "global_id": 2574, - "bbox": [ - 306.96, - 252.28, - 310.96, - 260.28 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p90-b21", - "global_id": 2575, - "bbox": [ - 332.07, - 292.81, - 334.3, - 300.81 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p90-b22", - "global_id": 2576, - "bbox": [ - 336.54, - 310.0, - 345.42, - 318.0 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p90-b23", - "global_id": 2577, - "bbox": [ - 460.63, - 293.15, - 464.63, - 301.15 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p90-b24", - "global_id": 2578, - "bbox": [ - 460.46, - 252.28, - 464.46, - 260.28 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p90-b25", - "global_id": 2579, - "bbox": [ - 399.46, - 292.81, - 424.3, - 301.15 - ], - "text": "t\n1", - "type": "text" - }, - { - "block_id": "p90-b26", - "global_id": 2580, - "bbox": [ - 427.32, - 310.0, - 436.65, - 318.0 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p90-b27", - "global_id": 2581, - "bbox": [ - 125.76, - 420.11, - 243.07, - 429.35 - ], - "text": "Figure 1.3 Signals for Drill 1.1", - "type": "text" - }, - { - "block_id": "p90-b28", - "global_id": 2582, - "bbox": [ - 107.82, - 481.99, - 357.52, - 493.95 - ], - "text": "DRILL 1.2\nComputing Power over a Period", - "type": "text" - }, - { - "block_id": "p90-b29", - "global_id": 2583, - "bbox": [ - 107.82, - 502.65, - 484.41, - 538.43 - ], - "text": "Redo Ex. 1.1a to find the power of a sinusoid C cos(ω0t + θ) by averaging the signal energy\nover one period T0 = 2π/ω0 (rather than averaging over the infinitely large interval). Show also\nthat the power of a dc signal x(t) = C0 is C2", - "type": "text" - }, - { - "block_id": "p90-b30", - "global_id": 2584, - "bbox": [ - 278.55, - 526.87, - 380.03, - 539.13 - ], - "text": "0, and its rms value is C0.", - "type": "text" - }, - { - "block_id": "p90-b31", - "global_id": 2585, - "bbox": [ - 107.82, - 584.82, - 463.02, - 596.77 - ], - "text": "DRILL 1.3\nPower of a Sum of Two Equal-Frequency Sinusoids", - "type": "text" - }, - { - "block_id": "p90-b32", - "global_id": 2586, - "bbox": [ - 107.82, - 601.53, - 484.41, - 629.3 - ], - "text": "Show that if ω1 = ω2, the power of x(t) = C1 cos(ω1t +θ1)+C2 cos(ω2t +θ2) is [C12 +C22 +\n2C1C2 cos(θ1 −θ2)]/2, which is not equal to the Ex. 1.2b result of (C12 + C22)/2.", - "type": "text" - } - ] - }, - { - "page_num": 91, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p91-b0", - "global_id": 2587, - "bbox": [ - 329.69, - 62.89, - 516.13, - 71.98 - ], - "text": "1.2\nSome Useful Signal Operations\n71", - "type": "text" - }, - { - "block_id": "p91-b1", - "global_id": 2588, - "bbox": [ - 127.94, - 94.37, - 387.83, - 108.31 - ], - "text": "1.2 SOME USEFUL SIGNAL OPERATIONS", - "type": "text" - }, - { - "block_id": "p91-b2", - "global_id": 2589, - "bbox": [ - 127.59, - 114.3, - 516.14, - 160.13 - ], - "text": "We discuss here three useful signal operations: shifting, scaling, and inversion. Since the\nindependent variable in our signal description is time, these operations are discussed as time\nshifting, time scaling, and time reversal (inversion). However, this discussion is valid for functions\nhaving independent variables other than time (e.g., frequency or distance).", - "type": "text" - }, - { - "block_id": "p91-b3", - "global_id": 2590, - "bbox": [ - 127.59, - 184.88, - 234.03, - 196.84 - ], - "text": "1.2-1 Time Shifting", - "type": "text" - }, - { - "block_id": "p91-b4", - "global_id": 2591, - "bbox": [ - 127.59, - 202.55, - 516.14, - 236.84 - ], - "text": "Consider a signal x(t) (Fig. 1.4a) and the same signal delayed by T seconds (Fig. 1.4b), which we\nshall denote by φ(t). Whatever happens in x(t) (Fig. 1.4a) at some instant t also happens in φ(t)\n(Fig. 1.4b) T seconds later at the instant t + T. Therefore", - "type": "text" - }, - { - "block_id": "p91-b5", - "global_id": 2592, - "bbox": [ - 233.84, - 248.38, - 409.87, - 258.76 - ], - "text": "φ(t + T) = x(t)\nand\nφ(t) = x(t −T)", - "type": "text" - }, - { - "block_id": "p91-b6", - "global_id": 2593, - "bbox": [ - 127.59, - 270.3, - 516.14, - 328.5 - ], - "text": "Therefore, to time-shift a signal by T, we replace t with t −T. Thus x(t −T) represents\nx(t) time-shifted by T seconds. If T is positive, the shift is to the right (delay), as in\nFig. 1.4b. If T is negative, the shift is to the left (advance), as in Fig. 1.4c. Clearly, x(t −2)\nis x(t) delayed (right-shifted) by 2 seconds, and x(t + 2) is x(t) advanced (left-shifted) by\n2 seconds.", - "type": "text" - }, - { - "block_id": "p91-b7", - "global_id": 2594, - "bbox": [ - 224.53, - 369.49, - 236.12, - 377.58 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p91-b8", - "global_id": 2595, - "bbox": [ - 214.94, - 427.84, - 223.82, - 435.84 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p91-b9", - "global_id": 2596, - "bbox": [ - 213.27, - 414.81, - 301.16, - 424.55 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p91-b10", - "global_id": 2597, - "bbox": [ - 224.53, - 440.91, - 275.58, - 449.2 - ], - "text": "f(t) x(t T)", - "type": "text" - }, - { - "block_id": "p91-b11", - "global_id": 2598, - "bbox": [ - 224.53, - 516.23, - 252.03, - 524.53 - ], - "text": "x(t T)", - "type": "text" - }, - { - "block_id": "p91-b12", - "global_id": 2599, - "bbox": [ - 214.56, - 501.58, - 224.21, - 509.58 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p91-b13", - "global_id": 2600, - "bbox": [ - 213.27, - 488.31, - 301.16, - 498.05 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p91-b14", - "global_id": 2601, - "bbox": [ - 214.94, - 578.18, - 223.82, - 586.18 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p91-b15", - "global_id": 2602, - "bbox": [ - 213.27, - 561.81, - 301.16, - 571.55 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p91-b16", - "global_id": 2603, - "bbox": [ - 188.16, - 497.37, - 192.6, - 505.37 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p91-b17", - "global_id": 2604, - "bbox": [ - 168.76, - 571.71, - 457.24, - 587.47 - ], - "text": "T\nFigure 1.4 Time-shifting a signal.", - "type": "text" - } - ] - }, - { - "page_num": 92, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p92-b0", - "global_id": 2605, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "72\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p92-b1", - "global_id": 2606, - "bbox": [ - 76.77, - 93.92, - 251.01, - 105.87 - ], - "text": "EXAMPLE 1.3\nTime Shifting", - "type": "text" - }, - { - "block_id": "p92-b2", - "global_id": 2607, - "bbox": [ - 103.16, - 118.92, - 477.03, - 156.41 - ], - "text": "An exponential function x(t) = e−2t shown in Fig. 1.5a is delayed by 1 second. Sketch and\nmathematically describe the delayed function. Repeat the problem with x(t) advanced by 1\nsecond.", - "type": "text" - }, - { - "block_id": "p92-b3", - "global_id": 2608, - "bbox": [ - 212.91, - 202.67, - 216.91, - 210.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p92-b4", - "global_id": 2609, - "bbox": [ - 213.91, - 246.0, - 324.09, - 255.37 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p92-b5", - "global_id": 2610, - "bbox": [ - 254.56, - 265.7, - 263.44, - 273.7 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p92-b6", - "global_id": 2611, - "bbox": [ - 294.41, - 247.37, - 298.41, - 255.37 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p92-b7", - "global_id": 2612, - "bbox": [ - 233.9, - 206.67, - 247.12, - 216.29 - ], - "text": "e2t", - "type": "text" - }, - { - "block_id": "p92-b8", - "global_id": 2613, - "bbox": [ - 226.12, - 290.54, - 230.12, - 298.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p92-b9", - "global_id": 2614, - "bbox": [ - 213.91, - 333.34, - 375.59, - 343.21 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p92-b10", - "global_id": 2615, - "bbox": [ - 254.17, - 352.56, - 263.82, - 360.56 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p92-b11", - "global_id": 2616, - "bbox": [ - 294.41, - 335.21, - 298.41, - 343.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p92-b12", - "global_id": 2617, - "bbox": [ - 320.48, - 299.28, - 346.17, - 308.89 - ], - "text": "e2(t1)", - "type": "text" - }, - { - "block_id": "p92-b13", - "global_id": 2618, - "bbox": [ - 164.28, - 378.26, - 212.91, - 394.56 - ], - "text": "e2(t1)\n1", - "type": "text" - }, - { - "block_id": "p92-b14", - "global_id": 2619, - "bbox": [ - 213.91, - 419.82, - 324.09, - 429.14 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p92-b15", - "global_id": 2620, - "bbox": [ - 254.56, - 441.13, - 263.44, - 449.13 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p92-b16", - "global_id": 2621, - "bbox": [ - 138.15, - 422.3, - 148.82, - 430.59 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p92-b17", - "global_id": 2622, - "bbox": [ - 201.49, - 192.71, - 212.59, - 200.79 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p92-b18", - "global_id": 2623, - "bbox": [ - 186.34, - 290.29, - 212.59, - 298.59 - ], - "text": "x(t 1)", - "type": "text" - }, - { - "block_id": "p92-b19", - "global_id": 2624, - "bbox": [ - 226.09, - 377.18, - 252.34, - 385.48 - ], - "text": "x(t 1)", - "type": "text" - }, - { - "block_id": "p92-b20", - "global_id": 2625, - "bbox": [ - 101.77, - 455.46, - 468.32, - 465.07 - ], - "text": "Figure 1.5 (a) Signal x(t). (b) Signal x(t) delayed by 1 second. (c) Signal x(t) advanced by 1 second.", - "type": "text" - }, - { - "block_id": "p92-b21", - "global_id": 2626, - "bbox": [ - 103.16, - 513.24, - 315.26, - 523.62 - ], - "text": "The function x(t) can be described mathematically as", - "type": "text" - }, - { - "block_id": "p92-b22", - "global_id": 2627, - "bbox": [ - 245.25, - 540.81, - 269.9, - 551.08 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p92-b23", - "global_id": 2628, - "bbox": [ - 271.95, - 526.81, - 477.01, - 557.06 - ], - "text": "e−2t\nt ≥0\n0\nt < 0\n(1.5)", - "type": "text" - }, - { - "block_id": "p92-b24", - "global_id": 2629, - "bbox": [ - 103.16, - 568.07, - 477.02, - 590.4 - ], - "text": "Let xd(t) represent the function x(t) delayed (right-shifted) by 1 second, as illustrated in\nFig. 1.5b. This function is x(t −1); its mathematical description can be obtained from x(t)", - "type": "text" - } - ] - }, - { - "page_num": 93, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p93-b0", - "global_id": 2630, - "bbox": [ - 329.69, - 62.89, - 516.13, - 71.98 - ], - "text": "1.2\nSome Useful Signal Operations\n73", - "type": "text" - }, - { - "block_id": "p93-b1", - "global_id": 2631, - "bbox": [ - 128.9, - 85.83, - 300.62, - 96.2 - ], - "text": "by replacing t with t −1 in Eq. (1.5). Thus,", - "type": "text" - }, - { - "block_id": "p93-b2", - "global_id": 2632, - "bbox": [ - 208.64, - 113.4, - 279.99, - 124.48 - ], - "text": "xd(t) = x(t −1) =", - "type": "text" - }, - { - "block_id": "p93-b3", - "global_id": 2633, - "bbox": [ - 282.04, - 99.42, - 421.83, - 129.68 - ], - "text": "e−2(t−1)\nt −1 ≥0\nor\nt ≥1\n0\nt −1 < 0\nor\nt < 1", - "type": "text" - }, - { - "block_id": "p93-b4", - "global_id": 2634, - "bbox": [ - 128.9, - 140.69, - 502.76, - 174.98 - ], - "text": "Let xa(t) represent the function x(t) advanced (left-shifted) by 1 second, as depicted in\nFig. 1.5c. This function is x(t + 1); its mathematical description can be obtained from x(t)\nby replacing t with t + 1 in Eq. (1.5). Thus,", - "type": "text" - }, - { - "block_id": "p93-b5", - "global_id": 2635, - "bbox": [ - 204.85, - 192.18, - 276.01, - 203.26 - ], - "text": "xa(t) = x(t + 1) =", - "type": "text" - }, - { - "block_id": "p93-b6", - "global_id": 2636, - "bbox": [ - 278.06, - 178.2, - 425.62, - 208.46 - ], - "text": "e−2(t+1)\nt + 1 ≥0\nor\nt ≥−1\n0\nt + 1 < 0\nor\nt < −1", - "type": "text" - }, - { - "block_id": "p93-b7", - "global_id": 2637, - "bbox": [ - 133.57, - 269.69, - 456.09, - 281.65 - ], - "text": "DRILL 1.4\nWorking with Time Delay and Time Advance", - "type": "text" - }, - { - "block_id": "p93-b8", - "global_id": 2638, - "bbox": [ - 133.57, - 290.35, - 510.16, - 349.25 - ], - "text": "Write a mathematical description of the signal x3(t) in Fig. 1.3c. Next, delay this signal by\n2 seconds. Sketch the delayed signal. Show that this delayed signal xd(t) can be described\nmathematically as xd(t) = 2(t −2) for 2 ≤t ≤3, and equal to 0 otherwise. Now repeat the\nprocedure with the signal advanced (left-shifted) by 1 second. Show that this advanced signal\nxa(t) can be described as xa(t) = 2(t + 1) for −1 ≤t ≤0, and 0 otherwise.", - "type": "text" - }, - { - "block_id": "p93-b9", - "global_id": 2639, - "bbox": [ - 127.59, - 383.9, - 229.37, - 395.85 - ], - "text": "1.2-2 Time Scaling", - "type": "text" - }, - { - "block_id": "p93-b10", - "global_id": 2640, - "bbox": [ - 127.59, - 401.88, - 516.12, - 435.85 - ], - "text": "The compression or expansion of a signal in time is known as time scaling. Consider the signal\nx(t) of Fig. 1.6a. The signal φ(t) in Fig. 1.6b is x(t) compressed in time by a factor of 2. Therefore,\nwhatever happens in x(t) at some instant t also happens to φ(t) at the instant t/2 so that", - "type": "text" - }, - { - "block_id": "p93-b11", - "global_id": 2641, - "bbox": [ - 244.8, - 451.49, - 250.55, - 461.45 - ], - "text": "φ", - "type": "text" - }, - { - "block_id": "p93-b12", - "global_id": 2642, - "bbox": [ - 251.04, - 440.49, - 261.46, - 454.78 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p93-b13", - "global_id": 2643, - "bbox": [ - 257.67, - 458.97, - 262.65, - 468.94 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p93-b15", - "global_id": 2644, - "bbox": [ - 271.32, - 451.49, - 398.92, - 461.86 - ], - "text": "= x(t)\nand\nφ(t) = x(2t)", - "type": "text" - }, - { - "block_id": "p93-b16", - "global_id": 2645, - "bbox": [ - 127.59, - 476.96, - 516.13, - 523.2 - ], - "text": "Observe that because x(t) = 0 at t = T1 and T2, we must have φ(t) = 0 at t = T1/2 and T2/2, as\nshown in Fig. 1.6b. If x(t) were recorded on a tape and played back at twice the normal recording\nspeed, we would obtain x(2t). In general, if x(t) is compressed in time by a factor a (a > 1), the\nresulting signal φ(t) is given by", - "type": "text" - }, - { - "block_id": "p93-b17", - "global_id": 2646, - "bbox": [ - 297.51, - 525.66, - 346.21, - 535.94 - ], - "text": "φ(t) = x(at)", - "type": "text" - }, - { - "block_id": "p93-b18", - "global_id": 2647, - "bbox": [ - 127.6, - 545.18, - 516.13, - 567.51 - ], - "text": "Using a similar argument, we can show that x(t) expanded (slowed down) in time by a factor\na (a > 1) is given by", - "type": "text" - }, - { - "block_id": "p93-b19", - "global_id": 2648, - "bbox": [ - 294.78, - 575.63, - 328.09, - 585.91 - ], - "text": "φ(t) = x", - "type": "text" - }, - { - "block_id": "p93-b20", - "global_id": 2649, - "bbox": [ - 328.12, - 561.64, - 339.83, - 578.92 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p93-b21", - "global_id": 2650, - "bbox": [ - 336.04, - 583.01, - 341.02, - 592.97 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p93-b23", - "global_id": 2651, - "bbox": [ - 127.59, - 600.5, - 516.14, - 634.79 - ], - "text": "Figure 1.6c shows x(t/2), which is x(t) expanded in time by a factor of 2. Observe that in a\ntime-scaling operation, the origin t = 0 is the anchor point, which remains unchanged under the\nscaling operation because at t = 0, x(t) = x(at) = x(0).", - "type": "text" - } - ] - }, - { - "page_num": 94, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p94-b0", - "global_id": 2652, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "74\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p94-b1", - "global_id": 2653, - "bbox": [ - 215.06, - 89.26, - 226.64, - 97.34 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p94-b2", - "global_id": 2654, - "bbox": [ - 104.61, - 132.08, - 319.69, - 144.93 - ], - "text": "(a)\nt", - "type": "text" - }, - { - "block_id": "p94-b3", - "global_id": 2655, - "bbox": [ - 317.47, - 210.41, - 319.69, - 218.41 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p94-b4", - "global_id": 2656, - "bbox": [ - 317.47, - 283.9, - 319.69, - 291.9 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p94-b5", - "global_id": 2657, - "bbox": [ - 213.41, - 138.34, - 217.41, - 146.34 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p94-b6", - "global_id": 2658, - "bbox": [ - 233.44, - 176.64, - 272.1, - 184.93 - ], - "text": "f(t) x(2t)", - "type": "text" - }, - { - "block_id": "p94-b7", - "global_id": 2659, - "bbox": [ - 103.84, - 205.58, - 113.49, - 213.58 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p94-b8", - "global_id": 2660, - "bbox": [ - 104.61, - 279.08, - 217.41, - 293.34 - ], - "text": "(c)\n0", - "type": "text" - }, - { - "block_id": "p94-b9", - "global_id": 2661, - "bbox": [ - 189.01, - 137.66, - 253.46, - 147.26 - ], - "text": "T1\nT2", - "type": "text" - }, - { - "block_id": "p94-b10", - "global_id": 2662, - "bbox": [ - 166.86, - 285.65, - 292.3, - 295.26 - ], - "text": "2T1\n2T2", - "type": "text" - }, - { - "block_id": "p94-b11", - "global_id": 2663, - "bbox": [ - 198.3, - 211.06, - 205.75, - 220.67 - ], - "text": "T1", - "type": "text" - }, - { - "block_id": "p94-b12", - "global_id": 2664, - "bbox": [ - 200.02, - 211.06, - 231.86, - 230.05 - ], - "text": "2\nT2", - "type": "text" - }, - { - "block_id": "p94-b13", - "global_id": 2665, - "bbox": [ - 226.14, - 222.05, - 230.14, - 230.05 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p94-b14", - "global_id": 2666, - "bbox": [ - 235.04, - 244.86, - 265.67, - 256.88 - ], - "text": "f(t) x(", - "type": "text" - }, - { - "block_id": "p94-b15", - "global_id": 2667, - "bbox": [ - 271.4, - 244.86, - 274.23, - 256.88 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p94-b16", - "global_id": 2668, - "bbox": [ - 266.35, - 241.23, - 270.35, - 258.34 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p94-b17", - "global_id": 2669, - "bbox": [ - 346.8, - 287.78, - 470.77, - 297.02 - ], - "text": "Figure 1.6 Time scaling a signal.", - "type": "text" - }, - { - "block_id": "p94-b18", - "global_id": 2670, - "bbox": [ - 101.84, - 331.4, - 490.39, - 353.74 - ], - "text": "In summary, to time-scale a signal by a factor a, we replace t with at. If a > 1, the scaling\nresults in compression, and if a < 1, the scaling results in expansion.", - "type": "text" - }, - { - "block_id": "p94-b19", - "global_id": 2671, - "bbox": [ - 76.77, - 388.52, - 373.52, - 400.48 - ], - "text": "EXAMPLE 1.4\nContinuous Time-Scaling Operation", - "type": "text" - }, - { - "block_id": "p94-b20", - "global_id": 2672, - "bbox": [ - 103.16, - 416.73, - 477.02, - 451.02 - ], - "text": "Figure 1.7a shows a signal x(t). Sketch and describe mathematically this signal\ntime-compressed by factor 3. Repeat the problem for the same signal time-expanded by\nfactor 2.", - "type": "text" - }, - { - "block_id": "p94-b21", - "global_id": 2673, - "bbox": [ - 103.16, - 473.51, - 241.94, - 483.89 - ], - "text": "The signal x(t) can be described as", - "type": "text" - }, - { - "block_id": "p94-b22", - "global_id": 2674, - "bbox": [ - 223.78, - 507.05, - 248.44, - 517.33 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p94-b23", - "global_id": 2675, - "bbox": [ - 250.49, - 486.63, - 258.38, - 505.56 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p94-b24", - "global_id": 2676, - "bbox": [ - 250.49, - 513.53, - 258.38, - 523.5 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p94-b25", - "global_id": 2677, - "bbox": [ - 258.37, - 495.0, - 355.21, - 529.28 - ], - "text": "2\n−1.5 ≤t < 0\n2e−t/2\n0 ≤t < 3\n0\notherwise", - "type": "text" - }, - { - "block_id": "p94-b26", - "global_id": 2678, - "bbox": [ - 457.92, - 507.47, - 477.01, - 517.43 - ], - "text": "(1.6)", - "type": "text" - }, - { - "block_id": "p94-b27", - "global_id": 2679, - "bbox": [ - 103.16, - 540.3, - 477.02, - 574.59 - ], - "text": "Figure 1.7b shows xc(t), which is x(t) time-compressed by factor 3; consequently, it can be\ndescribed mathematically as x(3t), which is obtained by replacing t with 3t in the right-hand\nside of Eq. (1.6). Thus,", - "type": "text" - }, - { - "block_id": "p94-b28", - "global_id": 2680, - "bbox": [ - 160.45, - 597.75, - 220.35, - 608.83 - ], - "text": "xc(t) = x(3t) =", - "type": "text" - }, - { - "block_id": "p94-b29", - "global_id": 2681, - "bbox": [ - 222.4, - 577.33, - 230.29, - 596.25 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p94-b30", - "global_id": 2682, - "bbox": [ - 222.4, - 604.23, - 230.29, - 614.19 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p94-b31", - "global_id": 2683, - "bbox": [ - 230.29, - 585.7, - 418.54, - 619.99 - ], - "text": "2\n−1.5 ≤3t < 0\nor\n−0.5 ≤t < 0\n2e−3t/2\n0 ≤3t < 3\nor\n0 ≤t < 1\n0\notherwise", - "type": "text" - } - ] - }, - { - "page_num": 95, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p95-b0", - "global_id": 2684, - "bbox": [ - 329.69, - 62.89, - 516.13, - 71.98 - ], - "text": "1.2\nSome Useful Signal Operations\n75", - "type": "text" - }, - { - "block_id": "p95-b1", - "global_id": 2685, - "bbox": [ - 342.06, - 176.22, - 344.29, - 184.22 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p95-b2", - "global_id": 2686, - "bbox": [ - 342.06, - 284.93, - 344.29, - 292.93 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p95-b3", - "global_id": 2687, - "bbox": [ - 293.06, - 395.29, - 295.29, - 403.29 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p95-b4", - "global_id": 2688, - "bbox": [ - 242.1, - 176.3, - 246.1, - 184.3 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p95-b5", - "global_id": 2689, - "bbox": [ - 235.08, - 194.99, - 243.96, - 202.99 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p95-b6", - "global_id": 2690, - "bbox": [ - 202.93, - 176.01, - 219.6, - 184.3 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p95-b7", - "global_id": 2691, - "bbox": [ - 246.37, - 111.68, - 257.95, - 119.76 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p95-b8", - "global_id": 2692, - "bbox": [ - 210.1, - 219.48, - 223.87, - 229.03 - ], - "text": "xc(t)", - "type": "text" - }, - { - "block_id": "p95-b9", - "global_id": 2693, - "bbox": [ - 249.63, - 331.28, - 263.4, - 340.83 - ], - "text": "xe(t)", - "type": "text" - }, - { - "block_id": "p95-b10", - "global_id": 2694, - "bbox": [ - 276.0, - 139.04, - 296.22, - 148.73 - ], - "text": "2et2", - "type": "text" - }, - { - "block_id": "p95-b11", - "global_id": 2695, - "bbox": [ - 251.51, - 244.74, - 274.73, - 254.43 - ], - "text": "2e3t2", - "type": "text" - }, - { - "block_id": "p95-b12", - "global_id": 2696, - "bbox": [ - 290.06, - 351.87, - 310.27, - 361.56 - ], - "text": "2et4", - "type": "text" - }, - { - "block_id": "p95-b13", - "global_id": 2697, - "bbox": [ - 294.02, - 176.3, - 298.02, - 184.3 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p95-b14", - "global_id": 2698, - "bbox": [ - 242.1, - 285.01, - 246.1, - 293.01 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p95-b15", - "global_id": 2699, - "bbox": [ - 234.7, - 303.87, - 244.35, - 311.87 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p95-b16", - "global_id": 2700, - "bbox": [ - 244.04, - 222.31, - 248.04, - 230.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p95-b17", - "global_id": 2701, - "bbox": [ - 257.23, - 285.01, - 261.23, - 293.01 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p95-b18", - "global_id": 2702, - "bbox": [ - 242.1, - 395.37, - 246.1, - 403.37 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p95-b19", - "global_id": 2703, - "bbox": [ - 235.08, - 414.1, - 243.96, - 422.1 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p95-b20", - "global_id": 2704, - "bbox": [ - 177.81, - 395.08, - 188.47, - 403.37 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p95-b21", - "global_id": 2705, - "bbox": [ - 217.75, - 284.72, - 234.42, - 293.01 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p95-b22", - "global_id": 2706, - "bbox": [ - 351.83, - 395.37, - 355.83, - 403.37 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p95-b23", - "global_id": 2707, - "bbox": [ - 233.14, - 332.75, - 237.14, - 340.75 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p95-b24", - "global_id": 2708, - "bbox": [ - 233.14, - 113.46, - 237.14, - 121.46 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p95-b25", - "global_id": 2709, - "bbox": [ - 380.44, - 389.08, - 506.34, - 423.36 - ], - "text": "Figure 1.7 (a) Signal x(t), (b)\nsignal x(3t), and (c) signal\nx(t/2).", - "type": "text" - }, - { - "block_id": "p95-b26", - "global_id": 2710, - "bbox": [ - 128.91, - 465.01, - 502.77, - 487.35 - ], - "text": "Observe that the instants t = −1.5 and 3 in x(t) correspond to the instants t = −0.5, and 1 in\nthe compressed signal x(3t).", - "type": "text" - }, - { - "block_id": "p95-b27", - "global_id": 2711, - "bbox": [ - 128.9, - 488.92, - 502.75, - 511.25 - ], - "text": "Figure 1.7c shows xe(t), which is x(t) time-expanded by factor 2; consequently, it can be\ndescribed mathematically as x(t/2), which is obtained by replacing t with t/2 in x(t). Thus,", - "type": "text" - }, - { - "block_id": "p95-b28", - "global_id": 2712, - "bbox": [ - 189.96, - 541.83, - 224.65, - 552.91 - ], - "text": "xe(t) = x", - "type": "text" - }, - { - "block_id": "p95-b29", - "global_id": 2713, - "bbox": [ - 225.78, - 530.83, - 236.2, - 545.13 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p95-b30", - "global_id": 2714, - "bbox": [ - 232.41, - 549.32, - 237.39, - 559.28 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p95-b32", - "global_id": 2715, - "bbox": [ - 246.07, - 541.83, - 253.84, - 551.79 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p95-b33", - "global_id": 2716, - "bbox": [ - 255.89, - 515.43, - 263.78, - 540.34 - ], - "text": "⎧\n⎪⎪⎨", - "type": "text" - }, - { - "block_id": "p95-b34", - "global_id": 2717, - "bbox": [ - 255.89, - 548.31, - 263.78, - 564.25 - ], - "text": "⎪⎪⎩", - "type": "text" - }, - { - "block_id": "p95-b35", - "global_id": 2718, - "bbox": [ - 263.78, - 519.63, - 345.79, - 536.67 - ], - "text": "2\n−1.5 ≤t", - "type": "text" - }, - { - "block_id": "p95-b36", - "global_id": 2719, - "bbox": [ - 342.01, - 526.3, - 440.51, - 543.75 - ], - "text": "2 < 0\nor\n−3 ≤t < 0", - "type": "text" - }, - { - "block_id": "p95-b37", - "global_id": 2720, - "bbox": [ - 263.78, - 539.0, - 414.4, - 563.13 - ], - "text": "2e−t/4\n0 ≤t\n2 < 3\nor\n0 ≤t < 6", - "type": "text" - }, - { - "block_id": "p95-b38", - "global_id": 2721, - "bbox": [ - 263.78, - 561.52, - 347.71, - 571.49 - ], - "text": "0\notherwise", - "type": "text" - }, - { - "block_id": "p95-b39", - "global_id": 2722, - "bbox": [ - 128.91, - 582.5, - 502.76, - 604.83 - ], - "text": "Observe that the instants t = −1.5 and 3 in x(t) correspond to the instants t = −3 and 6 in the\nexpanded signal x(t/2).", - "type": "text" - } - ] - }, - { - "page_num": 96, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p96-b0", - "global_id": 2723, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "76\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p96-b1", - "global_id": 2724, - "bbox": [ - 107.82, - 97.81, - 410.62, - 109.76 - ], - "text": "DRILL 1.5\nCompression and Expansion of Sinusoids", - "type": "text" - }, - { - "block_id": "p96-b2", - "global_id": 2725, - "bbox": [ - 107.82, - 118.47, - 484.43, - 188.62 - ], - "text": "Show that the time compression by an integer factor n (n > 1) of a sinusoid results in a\nsinusoid of the same amplitude and phase, but with the frequency increased n-fold. Similarly,\nthe time expansion by an integer factor n (n > 1) of a sinusoid results in a sinusoid of the same\namplitude and phase, but with the frequency reduced by a factor n. Verify your conclusion by\nsketching a sinusoid sin 2t and the same sinusoid compressed by a factor 3 and expanded by a\nfactor 2.", - "type": "text" - }, - { - "block_id": "p96-b3", - "global_id": 2726, - "bbox": [ - 101.84, - 236.56, - 210.28, - 248.51 - ], - "text": "1.2-3 Time Reversal", - "type": "text" - }, - { - "block_id": "p96-b4", - "global_id": 2727, - "bbox": [ - 101.84, - 254.23, - 490.39, - 312.44 - ], - "text": "Consider the signal x(t) in Fig. 1.8a. We can view x(t) as a rigid wire frame hinged at the vertical\naxis. To time-reverse x(t), we rotate this frame 180◦about the vertical axis. This time reversal [the\nreflection of x(t) about the vertical axis] gives us the signal φ(t) (Fig. 1.8b). Observe that whatever\nhappens in Fig. 1.8a at some instant t also happens in Fig. 1.8b at the instant −t, and vice versa.\nTherefore,", - "type": "text" - }, - { - "block_id": "p96-b5", - "global_id": 2728, - "bbox": [ - 270.37, - 327.13, - 321.86, - 337.41 - ], - "text": "φ(t) = x(−t)", - "type": "text" - }, - { - "block_id": "p96-b6", - "global_id": 2729, - "bbox": [ - 101.84, - 354.82, - 490.39, - 401.06 - ], - "text": "Thus, to time-reverse a signal we replace t with −t, and the time reversal of signal x(t) results in a\nsignal x(−t). We must remember that the reversal is performed about the vertical axis, which acts\nas an anchor or a hinge. Recall also that the reversal of x(t) about the horizontal axis results in\n−x(t).", - "type": "text" - }, - { - "block_id": "p96-b7", - "global_id": 2730, - "bbox": [ - 188.04, - 473.39, - 266.53, - 482.07 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p96-b8", - "global_id": 2731, - "bbox": [ - 180.14, - 437.61, - 184.14, - 445.61 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p96-b9", - "global_id": 2732, - "bbox": [ - 250.54, - 475.51, - 254.54, - 483.51 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p96-b10", - "global_id": 2733, - "bbox": [ - 158.91, - 461.87, - 169.57, - 470.17 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p96-b11", - "global_id": 2734, - "bbox": [ - 114.91, - 581.21, - 125.57, - 589.5 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p96-b12", - "global_id": 2735, - "bbox": [ - 192.01, - 511.45, - 200.89, - 519.45 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p96-b13", - "global_id": 2736, - "bbox": [ - 199.14, - 435.86, - 210.73, - 443.94 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p96-b14", - "global_id": 2737, - "bbox": [ - 189.33, - 484.86, - 199.99, - 493.16 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p96-b15", - "global_id": 2738, - "bbox": [ - 172.39, - 590.78, - 183.06, - 599.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p96-b16", - "global_id": 2739, - "bbox": [ - 180.54, - 579.09, - 261.34, - 587.83 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p96-b17", - "global_id": 2740, - "bbox": [ - 189.14, - 543.37, - 193.14, - 551.37 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p96-b18", - "global_id": 2741, - "bbox": [ - 192.01, - 617.22, - 201.66, - 625.22 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p96-b19", - "global_id": 2742, - "bbox": [ - 203.91, - 538.85, - 246.19, - 547.15 - ], - "text": "f(t) x(t)", - "type": "text" - }, - { - "block_id": "p96-b20", - "global_id": 2743, - "bbox": [ - 206.64, - 568.03, - 210.64, - 576.03 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p96-b21", - "global_id": 2744, - "bbox": [ - 299.8, - 617.27, - 436.1, - 626.5 - ], - "text": "Figure 1.8 Time reversal of a signal.", - "type": "text" - } - ] - }, - { - "page_num": 97, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p97-b0", - "global_id": 2745, - "bbox": [ - 329.69, - 62.89, - 516.13, - 71.98 - ], - "text": "1.2\nSome Useful Signal Operations\n77", - "type": "text" - }, - { - "block_id": "p97-b1", - "global_id": 2746, - "bbox": [ - 102.51, - 93.92, - 340.18, - 105.87 - ], - "text": "EXAMPLE 1.5\nTime Reversal of a Signal", - "type": "text" - }, - { - "block_id": "p97-b2", - "global_id": 2747, - "bbox": [ - 128.9, - 122.12, - 461.68, - 132.49 - ], - "text": "For the signal x(t) illustrated in Fig. 1.9a, sketch x(−t), which is time-reversed x(t).", - "type": "text" - }, - { - "block_id": "p97-b3", - "global_id": 2748, - "bbox": [ - 167.5, - 197.11, - 316.53, - 207.85 - ], - "text": "t\n5", - "type": "text" - }, - { - "block_id": "p97-b4", - "global_id": 2749, - "bbox": [ - 223.36, - 215.23, - 232.24, - 223.23 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p97-b6", - "global_id": 2750, - "bbox": [ - 145.47, - 199.55, - 222.13, - 207.85 - ], - "text": "7\n3\n1", - "type": "text" - }, - { - "block_id": "p97-b7", - "global_id": 2751, - "bbox": [ - 314.26, - 263.5, - 316.48, - 271.5 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p97-b8", - "global_id": 2752, - "bbox": [ - 222.98, - 281.12, - 232.63, - 289.12 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p97-b9", - "global_id": 2753, - "bbox": [ - 237.0, - 265.22, - 306.8, - 273.22 - ], - "text": "1\n3\n5\n7", - "type": "text" - }, - { - "block_id": "p97-b10", - "global_id": 2754, - "bbox": [ - 233.57, - 166.32, - 245.15, - 174.4 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p97-b11", - "global_id": 2755, - "bbox": [ - 206.67, - 236.17, - 224.92, - 244.47 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p97-b12", - "global_id": 2756, - "bbox": [ - 183.67, - 173.1, - 195.37, - 182.71 - ], - "text": "et2", - "type": "text" - }, - { - "block_id": "p97-b13", - "global_id": 2757, - "bbox": [ - 256.95, - 237.28, - 273.17, - 246.89 - ], - "text": "et2", - "type": "text" - }, - { - "block_id": "p97-b14", - "global_id": 2758, - "bbox": [ - 344.86, - 281.16, - 482.72, - 290.4 - ], - "text": "Figure 1.9 Example of time reversal.", - "type": "text" - }, - { - "block_id": "p97-b15", - "global_id": 2759, - "bbox": [ - 128.9, - 328.09, - 502.76, - 363.6 - ], - "text": "The instants −1 and −5 in x(t) are mapped into instants 1 and 5 in x(−t). Because x(t) = et/2,\nwe have x(−t) = e−t/2. The signal x(−t) is depicted in Fig. 1.9b. We can describe x(t) and\nx(−t) as", - "type": "text" - }, - { - "block_id": "p97-b16", - "global_id": 2760, - "bbox": [ - 256.0, - 370.4, - 280.66, - 380.68 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p97-b17", - "global_id": 2761, - "bbox": [ - 282.71, - 356.41, - 374.46, - 386.65 - ], - "text": "et/2\n−1 ≥t > −5\n0\notherwise", - "type": "text" - }, - { - "block_id": "p97-b18", - "global_id": 2762, - "bbox": [ - 128.9, - 394.68, - 443.99, - 405.06 - ], - "text": "and its time-reversed version x(−t) is obtained by replacing t with −t in x(t) as", - "type": "text" - }, - { - "block_id": "p97-b19", - "global_id": 2763, - "bbox": [ - 213.2, - 422.24, - 245.64, - 432.52 - ], - "text": "x(−t) =", - "type": "text" - }, - { - "block_id": "p97-b20", - "global_id": 2764, - "bbox": [ - 247.67, - 408.25, - 417.26, - 438.49 - ], - "text": "e−t/2\n−1 ≥−t > −5\nor\n1 ≤t < 5\n0\notherwise", - "type": "text" - }, - { - "block_id": "p97-b21", - "global_id": 2765, - "bbox": [ - 127.59, - 487.18, - 279.83, - 499.14 - ], - "text": "1.2-4 Combined Operations", - "type": "text" - }, - { - "block_id": "p97-b22", - "global_id": 2766, - "bbox": [ - 127.59, - 505.27, - 516.14, - 539.15 - ], - "text": "Certain complex operations require simultaneous use of more than one of the operations just\ndescribed. The most general operation involving all the three operations is x(at −b), which is\nrealized in two possible sequences of operation:", - "type": "text" - }, - { - "block_id": "p97-b23", - "global_id": 2767, - "bbox": [ - 144.52, - 546.7, - 516.14, - 604.9 - ], - "text": "1. Time-shift x(t) by b to obtain x(t−b). Now time-scale the shifted signal x(t−b) by a [i.e.,\nreplace t with at] to obtain x(at −b).\n2. Time-scale x(t) by a to obtain x(at). Now time-shift x(at) by b/a [i.e., replace t with\nt −(b/a)] to obtain x[a(t −b/a)] = x(at −b). In either case, if a is negative, time scaling\ninvolves time reversal.", - "type": "text" - }, - { - "block_id": "p97-b24", - "global_id": 2768, - "bbox": [ - 127.6, - 612.45, - 516.13, - 634.79 - ], - "text": "For example, the signal x(2t−6) can be obtained in two ways. We can delay x(t) by 6 to obtain\nx(t −6), and then time-compress this signal by factor 2 (replace t with 2t) to obtain x(2t −6).", - "type": "text" - } - ] - }, - { - "page_num": 98, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p98-b0", - "global_id": 2769, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "78\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p98-b1", - "global_id": 2770, - "bbox": [ - 101.84, - 85.46, - 490.39, - 107.79 - ], - "text": "Alternately, we can first time-compress x(t) by factor 2 to obtain x(2t), then delay this signal by 3\n(replace t with t −3) to obtain x(2t −6).", - "type": "text" - }, - { - "block_id": "p98-b2", - "global_id": 2771, - "bbox": [ - 102.2, - 143.26, - 318.15, - 157.21 - ], - "text": "1.3 CLASSIFICATION OF SIGNALS", - "type": "text" - }, - { - "block_id": "p98-b3", - "global_id": 2772, - "bbox": [ - 101.84, - 163.2, - 490.41, - 256.85 - ], - "text": "Classification helps us better understand and utilize the items around us. Cars, for example, are\nclassified as sports, offroad, family, and so forth. Knowing you have a sports car is useful in\ndeciding whether to drive on a highway or on a dirt road. Knowing you want to drive up a\nmountain, you would probably choose an offroad vehicle over a family sedan. Similarly, there\nare several classes of signals. Some signal classes are more suitable for certain applications\nthan others. Further, different signal classes often require different mathematical tools. Here we\nshall consider only the following classes of signals, which are suitable for the scope of this\nbook:", - "type": "text" - }, - { - "block_id": "p98-b4", - "global_id": 2773, - "bbox": [ - 118.78, - 264.82, - 301.16, - 322.6 - ], - "text": "1. Continuous-time and discrete-time signals\n2. Analog and digital signals\n3. Periodic and aperiodic signals\n4. Energy and power signals\n5. Deterministic and probabilistic signals", - "type": "text" - }, - { - "block_id": "p98-b5", - "global_id": 2774, - "bbox": [ - 101.84, - 346.57, - 375.61, - 358.52 - ], - "text": "1.3-1 Continuous-Time and Discrete-Time Signals", - "type": "text" - }, - { - "block_id": "p98-b6", - "global_id": 2775, - "bbox": [ - 101.84, - 364.55, - 490.39, - 422.43 - ], - "text": "A signal that is specified for a continuum of values of time t (Fig. 1.10a) is a continuous-time\nsignal, and a signal that is specified only at discrete values of t (Fig. 1.10b) is a discrete-time\nsignal. Telephone and video camera outputs are continuous-time signals, whereas the quarterly\ngross national product (GNP), monthly sales of a corporation, and stock market daily averages are\ndiscrete-time signals.", - "type": "text" - }, - { - "block_id": "p98-b7", - "global_id": 2776, - "bbox": [ - 101.84, - 453.38, - 280.66, - 465.33 - ], - "text": "1.3-2 Analog and Digital Signals", - "type": "text" - }, - { - "block_id": "p98-b8", - "global_id": 2777, - "bbox": [ - 101.84, - 471.47, - 490.4, - 636.84 - ], - "text": "The concept of continuous time is often confused with that of analog. The two are not the same.\nThe same is true of the concepts of discrete time and digital. A signal whose amplitude can\ntake on any value in a continuous range is an analog signal. This means that an analog signal\namplitude can take on an infinite number of values. A digital signal, on the other hand, is one\nwhose amplitude can take on only a finite number of values. Signals associated with a digital\ncomputer are digital because they take on only two values (binary signals). A digital signal whose\namplitudes can take on M values is an M-ary signal of which binary (M = 2) is a special case. The\nterms continuous time and discrete time qualify the nature of a signal along the time (horizontal)\naxis. The terms analog and digital, on the other hand, qualify the nature of the signal amplitude\n(vertical axis). Figure 1.11 shows examples of signals of various types. It is clear that analog\nis not necessarily continuous-time and digital need not be discrete-time. Figure 1.11c shows\nan example of an analog discrete-time signal. An analog signal can be converted into a digital\nsignal [analog-to-digital (A/D) conversion] through quantization (rounding off ), as explained\nin Sec. 8.3.", - "type": "text" - } - ] - }, - { - "page_num": 99, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p99-b0", - "global_id": 2778, - "bbox": [ - 359.14, - 62.89, - 516.14, - 71.98 - ], - "text": "1.3\nClassification of Signals\n79", - "type": "text" - }, - { - "block_id": "p99-b1", - "global_id": 2779, - "bbox": [ - 466.68, - 168.44, - 468.91, - 176.44 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p99-b2", - "global_id": 2780, - "bbox": [ - 160.07, - 87.33, - 171.65, - 95.41 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p99-b3", - "global_id": 2781, - "bbox": [ - 159.72, - 185.34, - 163.72, - 193.34 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p99-b4", - "global_id": 2782, - "bbox": [ - 316.53, - 204.19, - 325.41, - 212.19 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p99-b5", - "global_id": 2783, - "bbox": [ - 316.53, - 430.75, - 326.18, - 438.75 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p99-b6", - "global_id": 2784, - "bbox": [ - 137.94, - 313.59, - 141.94, - 321.59 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p99-b7", - "global_id": 2785, - "bbox": [ - 137.94, - 287.59, - 141.94, - 295.59 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p99-b8", - "global_id": 2786, - "bbox": [ - 133.94, - 262.89, - 141.94, - 270.89 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p99-b9", - "global_id": 2787, - "bbox": [ - 137.94, - 338.59, - 141.94, - 346.59 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p99-b10", - "global_id": 2788, - "bbox": [ - 131.28, - 364.4, - 141.94, - 372.69 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p99-b11", - "global_id": 2789, - "bbox": [ - 131.28, - 390.05, - 141.94, - 398.34 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p99-b12", - "global_id": 2790, - "bbox": [ - 152.62, - 409.81, - 512.73, - 417.89 - ], - "text": "1981\n’82\n’83\n’84\n’85\n’86\n’87\n’88\n’89\n’90\n’91\n’92\n’93\n’94", - "type": "text" - }, - { - "block_id": "p99-b13", - "global_id": 2791, - "bbox": [ - 145.04, - 234.96, - 353.67, - 255.09 - ], - "text": "Quarterly GNP : The return of recession\nIn percent change; seasonally adjusted annual rates", - "type": "text" - }, - { - "block_id": "p99-b14", - "global_id": 2792, - "bbox": [ - 357.04, - 255.08, - 505.0, - 263.08 - ], - "text": "Source : Commerce Department, news reports", - "type": "text" - }, - { - "block_id": "p99-b15", - "global_id": 2793, - "bbox": [ - 194.18, - 381.3, - 275.5, - 398.9 - ], - "text": "Two consecutive drops\nduring 1981-82 recession", - "type": "text" - }, - { - "block_id": "p99-b16", - "global_id": 2794, - "bbox": [ - 304.65, - 368.46, - 387.51, - 376.55 - ], - "text": "Three consecutive drops :", - "type": "text" - }, - { - "block_id": "p99-b17", - "global_id": 2795, - "bbox": [ - 310.98, - 376.46, - 377.19, - 384.46 - ], - "text": "Return of recession", - "type": "text" - }, - { - "block_id": "p99-b18", - "global_id": 2796, - "bbox": [ - 127.59, - 445.36, - 357.47, - 454.68 - ], - "text": "Figure 1.10 (a) Continuous-time and (b) discrete-time signals.", - "type": "text" - }, - { - "block_id": "p99-b19", - "global_id": 2797, - "bbox": [ - 127.59, - 480.98, - 327.64, - 492.94 - ], - "text": "1.3-3 Periodic and Aperiodic Signals", - "type": "text" - }, - { - "block_id": "p99-b20", - "global_id": 2798, - "bbox": [ - 127.59, - 498.65, - 390.96, - 509.8 - ], - "text": "A signal x(t) is said to be periodic if for some positive constant T0", - "type": "text" - }, - { - "block_id": "p99-b21", - "global_id": 2799, - "bbox": [ - 266.17, - 525.66, - 516.13, - 536.81 - ], - "text": "x(t) = x(t + T0)\nfor all t\n(1.7)", - "type": "text" - }, - { - "block_id": "p99-b22", - "global_id": 2800, - "bbox": [ - 127.59, - 552.99, - 516.12, - 598.92 - ], - "text": "The smallest value of T0 that satisfies the periodicity condition of Eq. (1.7) is the fundamental\nperiod of x(t). The signals in Figs. 1.2b and 1.3e are periodic signals with periods 2 and 1,\nrespectively. A signal is aperiodic if it is not periodic. Signals in Figs. 1.2a, 1.3a, 1.3b, 1.3c,\nand 1.3d are all aperiodic.", - "type": "text" - }, - { - "block_id": "p99-b23", - "global_id": 2801, - "bbox": [ - 127.59, - 600.5, - 516.13, - 636.28 - ], - "text": "By definition, a periodic signal x(t) remains unchanged when time-shifted by one period. For\nthis reason, a periodic signal must start at t = −∞: if it started at some finite instant, say, t = 0,\nthe time-shifted signal x(t + T0) would start at t = −T0 and x(t + T0) would not be the same as", - "type": "text" - } - ] - }, - { - "page_num": 100, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p100-b0", - "global_id": 2802, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "80\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p100-b1", - "global_id": 2803, - "bbox": [ - 195.28, - 187.49, - 204.16, - 195.49 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p100-b2", - "global_id": 2804, - "bbox": [ - 194.39, - 309.74, - 203.27, - 317.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p100-b3", - "global_id": 2805, - "bbox": [ - 357.35, - 187.49, - 367.0, - 195.49 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p100-b4", - "global_id": 2806, - "bbox": [ - 434.47, - 154.37, - 436.69, - 162.37 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p100-b5", - "global_id": 2807, - "bbox": [ - 252.01, - 173.36, - 254.23, - 181.36 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p100-b6", - "global_id": 2808, - "bbox": [ - 252.01, - 295.69, - 254.23, - 303.69 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p100-b7", - "global_id": 2809, - "bbox": [ - 434.47, - 274.69, - 436.69, - 282.69 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p100-b8", - "global_id": 2810, - "bbox": [ - 362.17, - 309.74, - 371.5, - 317.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p100-b9", - "global_id": 2811, - "bbox": [ - 145.01, - 94.04, - 319.08, - 102.12 - ], - "text": "x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p100-b10", - "global_id": 2812, - "bbox": [ - 144.33, - 214.31, - 331.21, - 222.39 - ], - "text": "x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p100-b11", - "global_id": 2813, - "bbox": [ - 125.76, - 324.24, - 490.38, - 346.24 - ], - "text": "Figure 1.11 Examples of signals: (a) analog, continuous time; (b) digital, continuous time;\n(c) analog, discrete time; and (d) digital, discrete time.", - "type": "text" - }, - { - "block_id": "p100-b12", - "global_id": 2814, - "bbox": [ - 469.64, - 417.02, - 471.86, - 425.02 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p100-b13", - "global_id": 2815, - "bbox": [ - 309.78, - 443.97, - 317.23, - 453.58 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p100-b14", - "global_id": 2816, - "bbox": [ - 300.5, - 371.88, - 311.61, - 379.97 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p100-b15", - "global_id": 2817, - "bbox": [ - 125.76, - 463.29, - 285.65, - 473.09 - ], - "text": "Figure 1.12 A periodic signal of period T0.", - "type": "text" - }, - { - "block_id": "p100-b16", - "global_id": 2818, - "bbox": [ - 101.84, - 492.9, - 490.39, - 515.13 - ], - "text": "x(t). Therefore, a periodic signal, by definition, must start at t = −∞and continue forever, as\nillustrated in Fig. 1.12.", - "type": "text" - }, - { - "block_id": "p100-b17", - "global_id": 2819, - "bbox": [ - 101.84, - 516.81, - 490.42, - 634.79 - ], - "text": "Another important property of a periodic signal x(t) is that x(t) can be generated by periodic\nextension of any segment of x(t) of duration T0 (the period). As a result, we can generate x(t) from\nany segment of x(t) having a duration of one period by placing this segment and the reproduction\nthereof end to end ad infinitum on either side. Figure 1.13 shows a periodic signal x(t) of period\nT0 = 6. The shaded portion of Fig. 1.13a shows a segment of x(t) starting at t = −1 and having\na duration of one period (6 seconds). This segment, when repeated forever in either direction,\nresults in the periodic signal x(t). Figure 1.13b shows another shaded segment of x(t) of duration\nT0 starting at t = 0. Again, we see that this segment, when repeated forever on either side, results\nin x(t). The reader can verify that this construction is possible with any segment of x(t) starting at\nany instant as long as the segment duration is one period.", - "type": "text" - } - ] - }, - { - "page_num": 101, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p101-b0", - "global_id": 2820, - "bbox": [ - 359.14, - 62.89, - 516.14, - 71.98 - ], - "text": "1.3\nClassification of Signals\n81", - "type": "text" - }, - { - "block_id": "p101-b1", - "global_id": 2821, - "bbox": [ - 300.56, - 148.83, - 309.44, - 156.83 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p101-b2", - "global_id": 2822, - "bbox": [ - 283.5, - 87.76, - 294.6, - 95.84 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p101-b3", - "global_id": 2823, - "bbox": [ - 283.5, - 169.46, - 294.6, - 177.54 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p101-b4", - "global_id": 2824, - "bbox": [ - 191.17, - 126.38, - 269.16, - 134.67 - ], - "text": "1\n7", - "type": "text" - }, - { - "block_id": "p101-b5", - "global_id": 2825, - "bbox": [ - 206.17, - 205.25, - 216.83, - 213.55 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p101-b6", - "global_id": 2826, - "bbox": [ - 281.5, - 126.67, - 405.21, - 134.67 - ], - "text": "0\n2\n5\n11", - "type": "text" - }, - { - "block_id": "p101-b7", - "global_id": 2827, - "bbox": [ - 300.17, - 226.74, - 309.82, - 234.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p101-b8", - "global_id": 2828, - "bbox": [ - 281.5, - 205.55, - 418.5, - 213.55 - ], - "text": "0\n6\n12", - "type": "text" - }, - { - "block_id": "p101-b9", - "global_id": 2829, - "bbox": [ - 442.13, - 129.99, - 444.35, - 137.99 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p101-b10", - "global_id": 2830, - "bbox": [ - 442.13, - 209.86, - 444.35, - 217.86 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p101-b11", - "global_id": 2831, - "bbox": [ - 151.5, - 241.19, - 516.16, - 263.11 - ], - "text": "Figure 1.13 Generation of a periodic signal by periodic extension of its segment of\none-period duration.", - "type": "text" - }, - { - "block_id": "p101-b12", - "global_id": 2832, - "bbox": [ - 127.59, - 286.07, - 516.13, - 309.89 - ], - "text": "An additional useful property of a periodic signal x(t) of period T0 is that the area under x(t)\nover any interval of duration T0 is the same; that is, for any real numbers a and b,", - "type": "text" - }, - { - "block_id": "p101-b13", - "global_id": 2833, - "bbox": [ - 264.34, - 316.0, - 289.94, - 329.61 - ], - "text": "# a+T0", - "type": "text" - }, - { - "block_id": "p101-b14", - "global_id": 2834, - "bbox": [ - 269.6, - 341.09, - 273.08, - 348.07 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p101-b15", - "global_id": 2835, - "bbox": [ - 292.06, - 329.56, - 325.75, - 339.84 - ], - "text": "x(t)dt =", - "type": "text" - }, - { - "block_id": "p101-b16", - "global_id": 2836, - "bbox": [ - 327.8, - 316.0, - 353.4, - 329.61 - ], - "text": "# b+T0", - "type": "text" - }, - { - "block_id": "p101-b17", - "global_id": 2837, - "bbox": [ - 333.05, - 341.09, - 336.54, - 348.07 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p101-b18", - "global_id": 2838, - "bbox": [ - 355.51, - 329.56, - 379.2, - 339.84 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p101-b19", - "global_id": 2839, - "bbox": [ - 127.59, - 359.54, - 516.13, - 395.0 - ], - "text": "This result follows from the fact that a periodic signal takes the same values at the intervals of T0.\nHence, the values over any segment of duration T0 are repeated in any other interval of the same\nduration. For convenience, the area under x(t) over any interval of duration T0 will be denoted by", - "type": "text" - }, - { - "block_id": "p101-b20", - "global_id": 2840, - "bbox": [ - 302.81, - 399.16, - 308.08, - 409.12 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p101-b21", - "global_id": 2841, - "bbox": [ - 308.07, - 424.25, - 314.94, - 432.56 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p101-b22", - "global_id": 2842, - "bbox": [ - 317.04, - 412.71, - 340.73, - 422.99 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p101-b23", - "global_id": 2843, - "bbox": [ - 127.59, - 443.86, - 516.15, - 490.1 - ], - "text": "It is helpful to label signals that start at t = −∞and continue forever as everlasting signals.\nThus, an everlasting signal exists over the entire interval −∞< t < ∞. The signals in Figs. 1.1b\nand 1.2b are examples of everlasting signals. Clearly, a periodic signal, by definition, is an\neverlasting signal.", - "type": "text" - }, - { - "block_id": "p101-b24", - "global_id": 2844, - "bbox": [ - 127.59, - 491.68, - 516.12, - 514.02 - ], - "text": "A signal that does not start before t = 0 is a causal signal. In other words, x(t) is a causal\nsignal if", - "type": "text" - }, - { - "block_id": "p101-b25", - "global_id": 2845, - "bbox": [ - 286.16, - 519.96, - 357.57, - 530.33 - ], - "text": "x(t) = 0\nt < 0", - "type": "text" - }, - { - "block_id": "p101-b26", - "global_id": 2846, - "bbox": [ - 127.6, - 541.8, - 516.15, - 600.0 - ], - "text": "The signals in Figs. 1.3a–1.3c are causal signals. A signal that starts before t = 0 is a noncausal\nsignal. All the signals in Figs. 1.1 and 1.2 are noncausal. Observe that an everlasting signal is\nalways noncausal but a noncausal signal is not necessarily everlasting. The everlasting signal in\nFig. 1.2b is noncausal; however, the noncausal signal in Fig. 1.2a is not everlasting. A signal that\nis zero for all t ≥0 is called an anti-causal signal.", - "type": "text" - }, - { - "block_id": "p101-b27", - "global_id": 2847, - "bbox": [ - 127.6, - 612.78, - 516.16, - 634.79 - ], - "text": "Comment. A true everlasting signal cannot be generated in practice for obvious reasons. Why\nshould we bother to postulate such a signal? In later chapters we shall see that certain signals", - "type": "text" - } - ] - }, - { - "page_num": 102, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p102-b0", - "global_id": 2848, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "82\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p102-b1", - "global_id": 2849, - "bbox": [ - 101.84, - 85.72, - 490.38, - 107.74 - ], - "text": "(e.g., an impulse and an everlasting sinusoid) that cannot be generated in practice do serve a very\nuseful purpose in the study of signals and systems.", - "type": "text" - }, - { - "block_id": "p102-b2", - "global_id": 2850, - "bbox": [ - 101.84, - 138.33, - 275.36, - 150.29 - ], - "text": "1.3-4 Energy and Power Signals", - "type": "text" - }, - { - "block_id": "p102-b3", - "global_id": 2851, - "bbox": [ - 101.84, - 156.32, - 490.4, - 238.11 - ], - "text": "A signal with finite energy is an energy signal, and a signal with finite and nonzero power is\na power signal. The signals in Figs. 1.2a and 1.2b are examples of energy and power signals,\nrespectively. Observe that power is the time average of energy. Since the averaging is over an\ninfinitely large interval, a signal with finite energy has zero power, and a signal with finite power\nhas infinite energy. Therefore, a signal cannot be both an energy signal and a power signal. If it is\none, it cannot be the other. On the other hand, there are signals that are neither energy nor power\nsignals. The ramp signal is one such case.", - "type": "text" - }, - { - "block_id": "p102-b4", - "global_id": 2852, - "bbox": [ - 101.84, - 253.85, - 490.39, - 311.71 - ], - "text": "Comments. All practical signals have finite energies and are therefore energy signals. A power\nsignal must necessarily have infinite duration; otherwise, its power, which is its energy averaged\nover an infinitely large interval, will not approach a (nonzero) limit. Clearly, it is impossible to\ngenerate a true power signal in practice because such a signal has infinite duration and infinite\nenergy.", - "type": "text" - }, - { - "block_id": "p102-b5", - "global_id": 2853, - "bbox": [ - 101.85, - 310.09, - 490.38, - 335.62 - ], - "text": "Also, because of periodic repetition, periodic signals for which the area under |x(t)|2 over one\nperiod is finite are power signals; however, not all power signals are periodic.", - "type": "text" - }, - { - "block_id": "p102-b6", - "global_id": 2854, - "bbox": [ - 107.82, - 371.43, - 325.31, - 383.39 - ], - "text": "DRILL 1.6\nNeither Energy nor Power", - "type": "text" - }, - { - "block_id": "p102-b7", - "global_id": 2855, - "bbox": [ - 107.82, - 388.89, - 484.41, - 426.37 - ], - "text": "Show that an everlasting exponential e−at is neither an energy nor a power signal for any real\nvalue of a. However, if a is imaginary, it is a power signal with power Px = 1 regardless of the\nvalue of a.", - "type": "text" - }, - { - "block_id": "p102-b8", - "global_id": 2856, - "bbox": [ - 101.84, - 469.98, - 323.8, - 481.94 - ], - "text": "1.3-5 Deterministic and Random Signals", - "type": "text" - }, - { - "block_id": "p102-b9", - "global_id": 2857, - "bbox": [ - 101.84, - 488.07, - 490.39, - 545.85 - ], - "text": "A signal whose physical description is known completely, in either a mathematical form or a\ngraphical form, is a deterministic signal. A signal whose values cannot be predicted precisely but\nare known only in terms of probabilistic description, such as mean value or mean-squared value, is\na random signal. In this book we shall exclusively deal with deterministic signals. Random signals\nare beyond the scope of this study.", - "type": "text" - }, - { - "block_id": "p102-b10", - "global_id": 2858, - "bbox": [ - 102.2, - 580.98, - 334.13, - 594.93 - ], - "text": "1.4 SOME USEFUL SIGNAL MODELS", - "type": "text" - }, - { - "block_id": "p102-b11", - "global_id": 2859, - "bbox": [ - 101.84, - 600.91, - 490.41, - 634.79 - ], - "text": "In the area of signals and systems, the step, the impulse, and the exponential functions play very\nimportant roles. Not only do they serve as a basis for representing other signals, but their use can\nsimplify many aspects of the signals and systems.", - "type": "text" - } - ] - }, - { - "page_num": 103, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p103-b0", - "global_id": 2860, - "bbox": [ - 344.33, - 62.89, - 516.14, - 71.98 - ], - "text": "1.4\nSome Useful Signal Models\n83", - "type": "text" - }, - { - "block_id": "p103-b1", - "global_id": 2861, - "bbox": [ - 127.59, - 85.66, - 307.78, - 98.48 - ], - "text": "1.4-1 The Unit Step Function u(t)", - "type": "text" - }, - { - "block_id": "p103-b2", - "global_id": 2862, - "bbox": [ - 127.59, - 104.19, - 516.15, - 138.48 - ], - "text": "In much of our discussion, the signals begin at t = 0 (causal signals). Such signals can be\nconveniently described in terms of unit step function u(t) shown in Fig. 1.14a. This function is\ndefined by", - "type": "text" - }, - { - "block_id": "p103-b3", - "global_id": 2863, - "bbox": [ - 277.23, - 147.96, - 302.42, - 158.24 - ], - "text": "u(t) =", - "type": "text" - }, - { - "block_id": "p103-b4", - "global_id": 2864, - "bbox": [ - 304.47, - 133.98, - 516.13, - 164.21 - ], - "text": "1\nt ≥0\n0\nt < 0\n(1.8)", - "type": "text" - }, - { - "block_id": "p103-b5", - "global_id": 2865, - "bbox": [ - 127.59, - 173.69, - 516.14, - 207.98 - ], - "text": "If we want a signal to start at t = 0 (so that it has a value of zero for t < 0), we need only\nmultiply the signal by u(t). For instance, the signal e−at represents an everlasting exponential that\nstarts at t = −∞. The causal form of this exponential (Fig. 1.14b) can be described as e−atu(t).", - "type": "text" - }, - { - "block_id": "p103-b6", - "global_id": 2866, - "bbox": [ - 127.59, - 209.97, - 516.15, - 279.71 - ], - "text": "The unit step function also proves very useful in specifying a function with different\nmathematical descriptions over different intervals. Examples of such functions appear in Fig. 1.7.\nThese functions have different mathematical descriptions over different segments of time, as\nseen from Eqs. (1.5) and (1.6). Such a description often proves clumsy and inconvenient in\nmathematical treatment. We can use the unit step function to describe such functions by a single\nexpression that is valid for all t.", - "type": "text" - }, - { - "block_id": "p103-b7", - "global_id": 2867, - "bbox": [ - 127.59, - 281.7, - 516.16, - 327.53 - ], - "text": "Consider, for example, the rectangular pulse depicted in Fig. 1.15a. We can express such a\npulse in terms of familiar step functions by observing that the pulse x(t) can be expressed as the\nsum of the two delayed unit step functions, as shown in Fig. 1.15b. The unit step function u(t)\ndelayed by T seconds is u(t −T). From Fig. 1.15b, it is clear that", - "type": "text" - }, - { - "block_id": "p103-b8", - "global_id": 2868, - "bbox": [ - 271.86, - 340.58, - 371.86, - 350.96 - ], - "text": "x(t) = u(t −2) −u(t −4)", - "type": "text" - }, - { - "block_id": "p103-b9", - "global_id": 2869, - "bbox": [ - 170.5, - 387.24, - 174.5, - 395.24 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p103-b10", - "global_id": 2870, - "bbox": [ - 170.5, - 453.64, - 408.9, - 462.24 - ], - "text": "t\nt\n0", - "type": "text" - }, - { - "block_id": "p103-b11", - "global_id": 2871, - "bbox": [ - 210.81, - 473.61, - 219.69, - 481.61 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p103-b12", - "global_id": 2872, - "bbox": [ - 332.0, - 388.64, - 336.0, - 396.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p103-b13", - "global_id": 2873, - "bbox": [ - 349.62, - 409.53, - 374.39, - 419.22 - ], - "text": "eatu(t)", - "type": "text" - }, - { - "block_id": "p103-b14", - "global_id": 2874, - "bbox": [ - 332.0, - 453.64, - 336.0, - 461.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p103-b15", - "global_id": 2875, - "bbox": [ - 355.92, - 473.61, - 365.57, - 481.61 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p103-b16", - "global_id": 2876, - "bbox": [ - 217.55, - 385.08, - 229.59, - 393.16 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p103-b17", - "global_id": 2877, - "bbox": [ - 151.5, - 487.42, - 389.11, - 498.3 - ], - "text": "Figure 1.14 (a) Unit step function u(t). (b) Exponential e−atu(t).", - "type": "text" - }, - { - "block_id": "p103-b18", - "global_id": 2878, - "bbox": [ - 218.01, - 613.23, - 226.89, - 621.23 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p103-b19", - "global_id": 2879, - "bbox": [ - 165.1, - 568.55, - 169.1, - 576.55 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p103-b20", - "global_id": 2880, - "bbox": [ - 165.1, - 528.09, - 169.1, - 536.09 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p103-b21", - "global_id": 2881, - "bbox": [ - 202.8, - 568.55, - 239.8, - 576.55 - ], - "text": "2\n4", - "type": "text" - }, - { - "block_id": "p103-b22", - "global_id": 2882, - "bbox": [ - 377.52, - 613.23, - 387.17, - 621.23 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p103-b23", - "global_id": 2883, - "bbox": [ - 325.5, - 568.55, - 329.5, - 576.55 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p103-b24", - "global_id": 2884, - "bbox": [ - 324.5, - 528.09, - 328.5, - 536.09 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p103-b25", - "global_id": 2885, - "bbox": [ - 362.8, - 568.55, - 366.8, - 576.55 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p103-b26", - "global_id": 2886, - "bbox": [ - 395.8, - 555.55, - 399.8, - 563.55 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p103-b27", - "global_id": 2887, - "bbox": [ - 317.83, - 595.84, - 328.5, - 604.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p103-b28", - "global_id": 2888, - "bbox": [ - 273.7, - 566.31, - 435.82, - 574.31 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p103-b29", - "global_id": 2889, - "bbox": [ - 151.5, - 627.93, - 400.76, - 637.17 - ], - "text": "Figure 1.15 Representation of a rectangular pulse by step functions.", - "type": "text" - } - ] - }, - { - "page_num": 104, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p104-b0", - "global_id": 2890, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "84\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p104-b1", - "global_id": 2891, - "bbox": [ - 76.77, - 93.92, - 447.54, - 105.87 - ], - "text": "EXAMPLE 1.6\nDescribing a Triangle Function with the Unit Step", - "type": "text" - }, - { - "block_id": "p104-b2", - "global_id": 2892, - "bbox": [ - 103.16, - 122.53, - 347.76, - 132.49 - ], - "text": "Use the unit step function to describe the signal in Fig. 1.16a.", - "type": "text" - }, - { - "block_id": "p104-b3", - "global_id": 2893, - "bbox": [ - 194.3, - 166.07, - 198.3, - 174.07 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b4", - "global_id": 2894, - "bbox": [ - 275.02, - 203.19, - 277.24, - 211.19 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p104-b5", - "global_id": 2895, - "bbox": [ - 338.34, - 289.01, - 340.56, - 297.01 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p104-b6", - "global_id": 2896, - "bbox": [ - 338.34, - 380.53, - 340.56, - 388.53 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p104-b7", - "global_id": 2897, - "bbox": [ - 226.38, - 221.32, - 235.26, - 229.32 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p104-b8", - "global_id": 2898, - "bbox": [ - 173.17, - 175.42, - 184.75, - 183.5 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p104-b9", - "global_id": 2899, - "bbox": [ - 230.67, - 204.93, - 256.67, - 212.93 - ], - "text": "2\n3", - "type": "text" - }, - { - "block_id": "p104-b10", - "global_id": 2900, - "bbox": [ - 194.3, - 166.07, - 198.3, - 174.07 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b11", - "global_id": 2901, - "bbox": [ - 166.12, - 246.74, - 168.34, - 254.74 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p104-b12", - "global_id": 2902, - "bbox": [ - 167.47, - 290.63, - 171.47, - 298.63 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b13", - "global_id": 2903, - "bbox": [ - 117.57, - 252.28, - 121.57, - 260.28 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b14", - "global_id": 2904, - "bbox": [ - 171.01, - 339.51, - 203.89, - 347.81 - ], - "text": "2(t 3)", - "type": "text" - }, - { - "block_id": "p104-b15", - "global_id": 2905, - "bbox": [ - 167.47, - 382.22, - 193.47, - 390.22 - ], - "text": "2\n3", - "type": "text" - }, - { - "block_id": "p104-b16", - "global_id": 2906, - "bbox": [ - 117.57, - 343.77, - 121.57, - 351.77 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b17", - "global_id": 2907, - "bbox": [ - 260.47, - 242.17, - 274.58, - 251.77 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p104-b18", - "global_id": 2908, - "bbox": [ - 260.47, - 333.62, - 274.58, - 343.23 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p104-b19", - "global_id": 2909, - "bbox": [ - 294.17, - 290.63, - 298.17, - 298.63 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b20", - "global_id": 2910, - "bbox": [ - 244.17, - 252.28, - 248.17, - 260.28 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b21", - "global_id": 2911, - "bbox": [ - 294.36, - 382.12, - 320.47, - 390.22 - ], - "text": "2\n3", - "type": "text" - }, - { - "block_id": "p104-b22", - "global_id": 2912, - "bbox": [ - 244.17, - 343.77, - 248.17, - 351.77 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p104-b23", - "global_id": 2913, - "bbox": [ - 225.99, - 312.82, - 235.64, - 320.82 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p104-b24", - "global_id": 2914, - "bbox": [ - 226.38, - 404.27, - 235.26, - 412.27 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p104-b25", - "global_id": 2915, - "bbox": [ - 117.57, - 266.58, - 121.57, - 274.58 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p104-b26", - "global_id": 2916, - "bbox": [ - 117.57, - 358.27, - 121.57, - 366.27 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p104-b27", - "global_id": 2917, - "bbox": [ - 191.37, - 204.93, - 195.37, - 212.93 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p104-b28", - "global_id": 2918, - "bbox": [ - 127.57, - 289.63, - 257.57, - 298.63 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p104-b29", - "global_id": 2919, - "bbox": [ - 127.57, - 381.12, - 257.57, - 389.12 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p104-b30", - "global_id": 2920, - "bbox": [ - 101.77, - 418.97, - 346.43, - 428.21 - ], - "text": "Figure 1.16 Representation of a signal defined interval by interval.", - "type": "text" - }, - { - "block_id": "p104-b31", - "global_id": 2921, - "bbox": [ - 103.16, - 476.8, - 477.01, - 522.63 - ], - "text": "The signal illustrated in Fig. 1.16a can be conveniently handled by breaking it up into the two\ncomponents x1(t) and x2(t), depicted in Figs. 1.16b and 1.16c, respectively. Here, x1(t) can be\nobtained by multiplying the ramp t by the gate pulse u(t) −u(t −2), as shown in Fig. 1.16b.\nTherefore,", - "type": "text" - }, - { - "block_id": "p104-b32", - "global_id": 2922, - "bbox": [ - 241.25, - 524.21, - 338.93, - 535.36 - ], - "text": "x1(t) = t[u(t) −u(t −2)]", - "type": "text" - }, - { - "block_id": "p104-b33", - "global_id": 2923, - "bbox": [ - 103.16, - 543.14, - 477.01, - 589.38 - ], - "text": "The signal x2(t) can be obtained by multiplying another ramp by the gate pulse illustrated in\nFig. 1.16c. This ramp has a slope −2; hence it can be described by −2t + c. Now, because the\nramp has a zero value at t = 3, the constant c = 6, and the ramp can be described by −2(t −3).\nAlso, the gate pulse in Fig. 1.16c is u(t −2) −u(t −3). Therefore,", - "type": "text" - }, - { - "block_id": "p104-b34", - "global_id": 2924, - "bbox": [ - 215.3, - 600.92, - 364.88, - 612.07 - ], - "text": "x2(t) = −2(t −3)[u(t −2) −u(t −3)]", - "type": "text" - } - ] - }, - { - "page_num": 105, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p105-b0", - "global_id": 2925, - "bbox": [ - 344.33, - 62.89, - 516.14, - 71.98 - ], - "text": "1.4\nSome Useful Signal Models\n85", - "type": "text" - }, - { - "block_id": "p105-b1", - "global_id": 2926, - "bbox": [ - 128.9, - 86.24, - 143.28, - 96.2 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p105-b2", - "global_id": 2927, - "bbox": [ - 206.85, - 107.74, - 282.0, - 118.89 - ], - "text": "x(t) = x1(t) + x2(t)", - "type": "text" - }, - { - "block_id": "p105-b3", - "global_id": 2928, - "bbox": [ - 223.74, - 122.65, - 424.8, - 133.07 - ], - "text": "= t[u(t) −u(t −2)] −2(t −3)[u(t −2) −u(t −3)]", - "type": "text" - }, - { - "block_id": "p105-b4", - "global_id": 2929, - "bbox": [ - 223.74, - 137.63, - 398.28, - 148.01 - ], - "text": "= tu(t) −3(t −2)u(t −2) + 2(t −3)u(t −3)", - "type": "text" - }, - { - "block_id": "p105-b5", - "global_id": 2930, - "bbox": [ - 102.51, - 212.51, - 481.89, - 224.46 - ], - "text": "EXAMPLE 1.7\nDescribing a Piecewise Function with the Unit Step", - "type": "text" - }, - { - "block_id": "p105-b6", - "global_id": 2931, - "bbox": [ - 128.9, - 241.03, - 400.42, - 251.1 - ], - "text": "Describe the signal in Fig. 1.7a by a single expression valid for all t.", - "type": "text" - }, - { - "block_id": "p105-b7", - "global_id": 2932, - "bbox": [ - 128.9, - 273.6, - 502.74, - 295.92 - ], - "text": "Over the interval from −1.5 to 0, the signal can be described by a constant 2, and over the\ninterval from 0 to 3, it can be described by 2e−t/2. Therefore,", - "type": "text" - }, - { - "block_id": "p105-b8", - "global_id": 2933, - "bbox": [ - 210.68, - 307.46, - 313.92, - 331.58 - ], - "text": "x(t) = 2[u(t + 1.5) −u(t)]\n\n\n\nconstant part", - "type": "text" - }, - { - "block_id": "p105-b9", - "global_id": 2934, - "bbox": [ - 316.13, - 305.74, - 414.37, - 331.58 - ], - "text": "+ 2e−t/2[u(t) −u(t −3)]\n\n\n\nexponential part", - "type": "text" - }, - { - "block_id": "p105-b10", - "global_id": 2935, - "bbox": [ - 227.56, - 335.69, - 420.98, - 347.8 - ], - "text": "= 2u(t + 1.5) −2(1 −e−t/2)u(t) −2e−t/2u(t −3)", - "type": "text" - }, - { - "block_id": "p105-b11", - "global_id": 2936, - "bbox": [ - 128.91, - 359.75, - 470.63, - 369.72 - ], - "text": "Compare this expression with the expression for the same function found in Eq. (1.6).", - "type": "text" - }, - { - "block_id": "p105-b12", - "global_id": 2937, - "bbox": [ - 133.57, - 436.68, - 406.47, - 448.64 - ], - "text": "DRILL 1.7\nUsing Reflected Unit Step Functions", - "type": "text" - }, - { - "block_id": "p105-b13", - "global_id": 2938, - "bbox": [ - 133.57, - 456.12, - 510.16, - 479.67 - ], - "text": "Show that the signals depicted in Figs. 1.17a and 1.17b can be described as u(−t) and e−atu(−t),\nrespectively.", - "type": "text" - }, - { - "block_id": "p105-b14", - "global_id": 2939, - "bbox": [ - 290.5, - 583.04, - 294.5, - 591.04 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p105-b15", - "global_id": 2940, - "bbox": [ - 267.38, - 519.62, - 409.09, - 530.42 - ], - "text": "eat u(t)\nu(t)", - "type": "text" - }, - { - "block_id": "p105-b16", - "global_id": 2941, - "bbox": [ - 300.17, - 531.58, - 304.17, - 539.58 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p105-b17", - "global_id": 2942, - "bbox": [ - 230.28, - 598.29, - 239.16, - 606.29 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p105-b18", - "global_id": 2943, - "bbox": [ - 429.61, - 583.04, - 433.61, - 591.04 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p105-b19", - "global_id": 2944, - "bbox": [ - 439.13, - 558.61, - 443.13, - 566.61 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p105-b20", - "global_id": 2945, - "bbox": [ - 229.83, - 582.91, - 351.11, - 590.91 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p105-b21", - "global_id": 2946, - "bbox": [ - 395.01, - 598.29, - 404.66, - 606.29 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p105-b22", - "global_id": 2947, - "bbox": [ - 157.47, - 612.98, - 281.52, - 622.22 - ], - "text": "Figure 1.17 Signals for Drill 1.7.", - "type": "text" - } - ] - }, - { - "page_num": 106, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p106-b0", - "global_id": 2948, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "86\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p106-b1", - "global_id": 2949, - "bbox": [ - 107.82, - 97.81, - 464.02, - 109.76 - ], - "text": "DRILL 1.8\nDescribing a Piecewise Function with the Unit Step", - "type": "text" - }, - { - "block_id": "p106-b2", - "global_id": 2950, - "bbox": [ - 107.82, - 118.88, - 345.03, - 128.85 - ], - "text": "Show that the signal shown in Fig. 1.18 can be described as", - "type": "text" - }, - { - "block_id": "p106-b3", - "global_id": 2951, - "bbox": [ - 198.84, - 140.39, - 393.39, - 150.77 - ], - "text": "x(t) = (t −1)u(t −1) −(t −2)u(t −2) −u(t −4)", - "type": "text" - }, - { - "block_id": "p106-b4", - "global_id": 2952, - "bbox": [ - 122.29, - 185.58, - 133.39, - 193.66 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p106-b5", - "global_id": 2953, - "bbox": [ - 157.19, - 238.49, - 251.89, - 248.63 - ], - "text": "t\n1\n2", - "type": "text" - }, - { - "block_id": "p106-b6", - "global_id": 2954, - "bbox": [ - 129.39, - 201.77, - 133.39, - 209.77 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p106-b7", - "global_id": 2955, - "bbox": [ - 222.99, - 240.63, - 410.89, - 256.4 - ], - "text": "4\nFigure 1.18 Signal for Drill 1.8.", - "type": "text" - }, - { - "block_id": "p106-b8", - "global_id": 2956, - "bbox": [ - 101.84, - 301.35, - 301.94, - 314.16 - ], - "text": "1.4-2 The Unit Impulse Function δ(t)", - "type": "text" - }, - { - "block_id": "p106-b9", - "global_id": 2957, - "bbox": [ - 101.84, - 319.87, - 490.37, - 342.21 - ], - "text": "The unit impulse function δ(t) is one of the most important functions in the study of signals and\nsystems. This function was first defined in two parts by P. A. M. Dirac as", - "type": "text" - }, - { - "block_id": "p106-b10", - "global_id": 2958, - "bbox": [ - 207.13, - 359.51, - 303.55, - 369.88 - ], - "text": "δ(t) = 0\nt̸ = 0\nand", - "type": "text" - }, - { - "block_id": "p106-b11", - "global_id": 2959, - "bbox": [ - 324.58, - 345.94, - 341.53, - 358.0 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p106-b12", - "global_id": 2960, - "bbox": [ - 329.85, - 370.82, - 342.41, - 377.79 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p106-b13", - "global_id": 2961, - "bbox": [ - 344.02, - 359.51, - 490.39, - 369.88 - ], - "text": "δ(t)dt = 1\n(1.9)", - "type": "text" - }, - { - "block_id": "p106-b14", - "global_id": 2962, - "bbox": [ - 101.84, - 387.59, - 490.43, - 469.29 - ], - "text": "We can visualize an impulse as a tall, narrow, rectangular pulse of unit area, as illustrated\nin Fig. 1.19b. The width of this rectangular pulse is a very small value ϵ →0. Consequently, its\nheight is a very large value 1/ϵ →∞. The unit impulse therefore can be regarded as a rectangular\npulse with a width that has become infinitesimally small, a height that has become infinitely\nlarge, and an overall area that has been maintained at unity. Thus δ(t) = 0 everywhere except at\nt = 0, where it is undefined. For this reason, a unit impulse is represented by the spearlike symbol\nin Fig. 1.19a.", - "type": "text" - }, - { - "block_id": "p106-b15", - "global_id": 2963, - "bbox": [ - 101.84, - 471.29, - 490.41, - 517.11 - ], - "text": "Other pulses, such as the exponential, triangular, or Gaussian types, may also be used in\nimpulse approximation. The important feature of the unit impulse function is not its shape but the\nfact that its effective duration (pulse width) approaches zero while its area remains at unity. For\nexample, the exponential pulse αe−αtu(t) in Fig. 1.20a becomes taller and narrower as α increases.", - "type": "text" - }, - { - "block_id": "p106-b16", - "global_id": 2964, - "bbox": [ - 151.52, - 594.3, - 319.81, - 605.9 - ], - "text": "t\nt\n0", - "type": "text" - }, - { - "block_id": "p106-b17", - "global_id": 2965, - "bbox": [ - 149.05, - 617.36, - 157.93, - 625.36 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p106-b18", - "global_id": 2966, - "bbox": [ - 285.03, - 545.82, - 289.03, - 560.75 - ], - "text": "1\ne", - "type": "text" - }, - { - "block_id": "p106-b19", - "global_id": 2967, - "bbox": [ - 309.18, - 563.49, - 312.73, - 571.49 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p106-b20", - "global_id": 2968, - "bbox": [ - 154.64, - 548.48, - 166.19, - 556.59 - ], - "text": "d(t)", - "type": "text" - }, - { - "block_id": "p106-b21", - "global_id": 2969, - "bbox": [ - 274.46, - 617.36, - 284.11, - 625.36 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p106-b22", - "global_id": 2970, - "bbox": [ - 322.64, - 563.53, - 326.64, - 571.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p106-b23", - "global_id": 2971, - "bbox": [ - 260.25, - 594.93, - 282.73, - 611.93 - ], - "text": "2\ne\n2\ne", - "type": "text" - }, - { - "block_id": "p106-b24", - "global_id": 2972, - "bbox": [ - 347.8, - 604.7, - 486.41, - 626.62 - ], - "text": "Figure 1.19 A unit impulse and its\napproximation.", - "type": "text" - } - ] - }, - { - "page_num": 107, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p107-b0", - "global_id": 2973, - "bbox": [ - 344.33, - 62.89, - 516.14, - 71.98 - ], - "text": "1.4\nSome Useful Signal Models\n87", - "type": "text" - }, - { - "block_id": "p107-b1", - "global_id": 2974, - "bbox": [ - 151.5, - 184.44, - 376.17, - 193.68 - ], - "text": "Figure 1.20 Other possible approximations to a unit impulse.", - "type": "text" - }, - { - "block_id": "p107-b2", - "global_id": 2975, - "bbox": [ - 127.59, - 213.57, - 516.11, - 235.91 - ], - "text": "In the limit as α →∞, the pulse height →∞, and its width or duration →0. Yet, the area under\nthe pulse is unity regardless of the value of α because", - "type": "text" - }, - { - "block_id": "p107-b3", - "global_id": 2976, - "bbox": [ - 288.26, - 238.82, - 305.2, - 250.87 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p107-b4", - "global_id": 2977, - "bbox": [ - 293.52, - 248.57, - 355.47, - 270.95 - ], - "text": "0\nαe−αt dt = 1", - "type": "text" - }, - { - "block_id": "p107-b5", - "global_id": 2978, - "bbox": [ - 127.59, - 279.14, - 516.11, - 301.05 - ], - "text": "The pulses in Figs. 1.20b and 1.20c behave in a similar fashion. Clearly, the exact impulse function\ncannot be generated in practice; it can only be approached.", - "type": "text" - }, - { - "block_id": "p107-b6", - "global_id": 2979, - "bbox": [ - 127.59, - 302.64, - 516.13, - 324.96 - ], - "text": "From Eq. (1.9), it follows that the function kδ(t) = 0 for all t̸ = 0, and its area is k. Thus, kδ(t)\nis an impulse function whose area is k (in contrast to the unit impulse function, whose area is 1).", - "type": "text" - }, - { - "block_id": "p107-b7", - "global_id": 2980, - "bbox": [ - 127.89, - 339.6, - 396.68, - 351.72 - ], - "text": "MULTIPLICATION OF A FUNCTION BY AN IMPULSE", - "type": "text" - }, - { - "block_id": "p107-b8", - "global_id": 2981, - "bbox": [ - 127.59, - 355.34, - 516.13, - 389.63 - ], - "text": "Let us now consider what happens when we multiply the unit impulse δ(t) by a function φ(t) that\nis known to be continuous at t = 0. Since the impulse has nonzero value only at t = 0, and the\nvalue of φ(t) at t = 0 is φ(0), we obtain", - "type": "text" - }, - { - "block_id": "p107-b9", - "global_id": 2982, - "bbox": [ - 282.68, - 400.77, - 361.05, - 411.15 - ], - "text": "φ(t)δ(t) = φ(0)δ(t)", - "type": "text" - }, - { - "block_id": "p107-b10", - "global_id": 2983, - "bbox": [ - 127.59, - 422.3, - 516.15, - 492.45 - ], - "text": "Thus, multiplication of a continuous-time function φ(t) with an unit impulse located at t = 0 results\nin an impulse, which is located at t = 0 and has strength φ(0) [the value of φ(t) at the location of\nthe impulse]. Use of exactly the same argument leads to the generalization of this result, stating\nthat provided φ(t) is continuous at t = T,φ(t) multiplied by an impulse δ(t −T) (impulse located\nat t = T) results in an impulse located at t = T and having strength φ(T) [the value of φ(t) at the\nlocation of the impulse].", - "type": "text" - }, - { - "block_id": "p107-b11", - "global_id": 2984, - "bbox": [ - 264.84, - 494.03, - 516.13, - 504.41 - ], - "text": "φ(t)δ(t −T) = φ(T)δ(t −T)\n(1.10)", - "type": "text" - }, - { - "block_id": "p107-b12", - "global_id": 2985, - "bbox": [ - 127.89, - 519.04, - 427.99, - 531.16 - ], - "text": "SAMPLING PROPERTY OF THE UNIT IMPULSE FUNCTION", - "type": "text" - }, - { - "block_id": "p107-b13", - "global_id": 2986, - "bbox": [ - 127.59, - 535.19, - 249.39, - 545.16 - ], - "text": "From Eq. (1.10) it follows that", - "type": "text" - }, - { - "block_id": "p107-b14", - "global_id": 2987, - "bbox": [ - 228.29, - 548.07, - 245.23, - 560.12 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p107-b15", - "global_id": 2988, - "bbox": [ - 233.55, - 572.95, - 246.11, - 579.92 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p107-b16", - "global_id": 2989, - "bbox": [ - 247.72, - 561.63, - 338.41, - 571.9 - ], - "text": "φ(t)δ(t −T)dt = φ(T)", - "type": "text" - }, - { - "block_id": "p107-b17", - "global_id": 2990, - "bbox": [ - 339.52, - 548.06, - 356.46, - 560.12 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p107-b18", - "global_id": 2991, - "bbox": [ - 344.77, - 572.95, - 357.33, - 579.92 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p107-b19", - "global_id": 2992, - "bbox": [ - 358.94, - 561.63, - 516.12, - 572.0 - ], - "text": "δ(t)dt = φ(T)\n(1.11)", - "type": "text" - }, - { - "block_id": "p107-b20", - "global_id": 2993, - "bbox": [ - 127.59, - 588.55, - 516.13, - 634.79 - ], - "text": "provided φ(t) is continuous at t = T. This result means that the area under the product of a function\nwith an impulse δ(t −T) is equal to the value of that function at the instant at which the unit\nimpulse is located. This property is very important and useful and is known as the sampling or\nsifting property of the unit impulse.", - "type": "text" - } - ] - }, - { - "page_num": 108, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p108-b0", - "global_id": 2994, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "88\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p108-b1", - "global_id": 2995, - "bbox": [ - 102.14, - 86.19, - 349.93, - 98.32 - ], - "text": "UNIT IMPULSE AS A GENERALIZED FUNCTION", - "type": "text" - }, - { - "block_id": "p108-b2", - "global_id": 2996, - "bbox": [ - 101.84, - 102.35, - 490.41, - 195.99 - ], - "text": "The definition of the unit impulse function given in Eq. (1.9) is not mathematically rigorous,\nwhich leads to serious difficulties. First, the impulse function does not define a unique function:\nfor example, it can be shown that δ(t)+ ˙δ(t) also satisfies Eq. (1.9) [1]. Moreover, δ(t) is not even\na true function in the ordinary sense. An ordinary function is specified by its values for all time t.\nThe impulse function is zero everywhere except at t = 0, and at this, the only interesting part of\nits range, it is undefined. These difficulties are resolved by defining the impulse as a generalized\nfunction rather than an ordinary function. A generalized function is defined by its effect on other\nfunctions instead of by its value at every instant of time.", - "type": "text" - }, - { - "block_id": "p108-b3", - "global_id": 2997, - "bbox": [ - 101.84, - 197.99, - 490.42, - 291.54 - ], - "text": "In this approach the impulse function is defined by the sampling property [Eq. (1.11)]. We\nsay nothing about what the impulse function is or what it looks like. Instead, the impulse function\nis defined in terms of its effect on a test function φ(t). We define a unit impulse as a function for\nwhich the area under its product with a function φ(t) is equal to the value of the function φ(t) at\nthe instant at which the impulse is located. It is assumed that φ(t) is continuous at the location\nof the impulse. Recall that the sampling property [Eq. (1.11)] is the consequence of the classical\n(Dirac) definition of the unit impulse in Eq. (1.9). In contrast, the sampling property [Eq. (1.11)]\ndefines the impulse function in the generalized function approach.", - "type": "text" - }, - { - "block_id": "p108-b4", - "global_id": 2998, - "bbox": [ - 101.84, - 293.63, - 490.39, - 351.42 - ], - "text": "We now present an interesting application of the generalized function definition of an impulse.\nBecause the unit step function u(t) is discontinuous at t = 0, its derivative du/dt does not exist at\nt = 0 in the ordinary sense. We now show that this derivative does exist in the generalized sense,\nand it is, in fact, δ(t). As a proof, let us evaluate the integral of (du/dt)φ(t), using integration by\nparts:", - "type": "text" - }, - { - "block_id": "p108-b5", - "global_id": 2999, - "bbox": [ - 196.06, - 357.56, - 213.01, - 369.62 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p108-b6", - "global_id": 3000, - "bbox": [ - 201.33, - 382.44, - 213.89, - 389.42 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p108-b7", - "global_id": 3001, - "bbox": [ - 216.69, - 364.14, - 237.03, - 374.42 - ], - "text": "du(t)", - "type": "text" - }, - { - "block_id": "p108-b8", - "global_id": 3002, - "bbox": [ - 222.89, - 371.13, - 308.56, - 388.47 - ], - "text": "dt φ(t)dt = u(t)φ(t)", - "type": "text" - }, - { - "block_id": "p108-b10", - "global_id": 3003, - "bbox": [ - 311.79, - 361.76, - 318.91, - 368.74 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p108-b11", - "global_id": 3004, - "bbox": [ - 311.79, - 383.33, - 324.34, - 390.3 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p108-b12", - "global_id": 3005, - "bbox": [ - 326.39, - 371.12, - 334.16, - 381.09 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p108-b13", - "global_id": 3006, - "bbox": [ - 335.71, - 357.56, - 352.66, - 369.62 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p108-b14", - "global_id": 3007, - "bbox": [ - 340.97, - 382.44, - 353.53, - 389.42 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p108-b15", - "global_id": 3008, - "bbox": [ - 355.14, - 369.13, - 395.99, - 381.4 - ], - "text": "u(t) ˙φ(t)dt", - "type": "text" - }, - { - "block_id": "p108-b16", - "global_id": 3009, - "bbox": [ - 266.34, - 399.63, - 325.57, - 410.0 - ], - "text": "= φ(∞) −0 −", - "type": "text" - }, - { - "block_id": "p108-b17", - "global_id": 3010, - "bbox": [ - 327.11, - 386.06, - 344.07, - 398.12 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p108-b18", - "global_id": 3011, - "bbox": [ - 332.39, - 397.63, - 371.17, - 418.2 - ], - "text": "0\n˙φ(t)dt", - "type": "text" - }, - { - "block_id": "p108-b19", - "global_id": 3012, - "bbox": [ - 266.34, - 419.21, - 339.47, - 431.25 - ], - "text": "= φ(∞) −φ(t)|∞", - "type": "text" - }, - { - "block_id": "p108-b20", - "global_id": 3013, - "bbox": [ - 332.35, - 420.98, - 370.9, - 433.62 - ], - "text": "0 = φ(0)", - "type": "text" - }, - { - "block_id": "p108-b21", - "global_id": 3014, - "bbox": [ - 101.84, - 444.85, - 490.4, - 467.18 - ], - "text": "This result shows that du/dt satisfies the sampling property of δ(t). Therefore it is an impulse δ(t)\nin the generalized sense—that is,", - "type": "text" - }, - { - "block_id": "p108-b22", - "global_id": 3015, - "bbox": [ - 272.41, - 469.75, - 292.75, - 480.02 - ], - "text": "du(t)", - "type": "text" - }, - { - "block_id": "p108-b23", - "global_id": 3016, - "bbox": [ - 278.61, - 476.73, - 490.38, - 494.08 - ], - "text": "dt\n= δ(t)\n(1.12)", - "type": "text" - }, - { - "block_id": "p108-b24", - "global_id": 3017, - "bbox": [ - 101.84, - 501.05, - 269.73, - 520.04 - ], - "text": "Consequently,\n# t", - "type": "text" - }, - { - "block_id": "p108-b25", - "global_id": 3018, - "bbox": [ - 263.23, - 532.64, - 275.79, - 539.61 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p108-b26", - "global_id": 3019, - "bbox": [ - 277.4, - 521.33, - 334.26, - 531.6 - ], - "text": "δ(τ)dτ = u(t)", - "type": "text" - }, - { - "block_id": "p108-b27", - "global_id": 3020, - "bbox": [ - 101.85, - 547.96, - 490.42, - 581.83 - ], - "text": "These results can also be obtained graphically from Fig. 1.19b. We observe that the area from\n−∞to t under the limiting form of δ(t) in Fig. 1.19b is zero if t < −ϵ/2 and unity if t ≥ϵ/2 with\nϵ →0. Consequently,", - "type": "text" - }, - { - "block_id": "p108-b28", - "global_id": 3021, - "bbox": [ - 237.14, - 587.56, - 248.9, - 599.84 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p108-b29", - "global_id": 3022, - "bbox": [ - 242.4, - 612.44, - 254.95, - 619.41 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p108-b30", - "global_id": 3023, - "bbox": [ - 256.56, - 601.12, - 296.01, - 611.4 - ], - "text": "δ(τ)dτ =", - "type": "text" - }, - { - "block_id": "p108-b31", - "global_id": 3024, - "bbox": [ - 298.06, - 587.13, - 348.92, - 617.37 - ], - "text": "0\nt < 0\n1\nt ≥0", - "type": "text" - }, - { - "block_id": "p108-b32", - "global_id": 3025, - "bbox": [ - 288.24, - 622.98, - 313.42, - 633.26 - ], - "text": "= u(t)", - "type": "text" - } - ] - }, - { - "page_num": 109, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p109-b0", - "global_id": 3026, - "bbox": [ - 344.33, - 62.89, - 516.14, - 71.98 - ], - "text": "1.4\nSome Useful Signal Models\n89", - "type": "text" - }, - { - "block_id": "p109-b1", - "global_id": 3027, - "bbox": [ - 127.59, - 85.82, - 516.14, - 155.56 - ], - "text": "This result shows that the unit step function can be obtained by integrating the unit impulse\nfunction. Similarly the unit ramp function x(t) = tu(t) can be obtained by integrating the unit\nstep function. We may continue with unit parabolic function t2/2 obtained by integrating the unit\nramp, and so on. On the other side, we have derivatives of impulse function, which can be defined\nas generalized functions (see Prob. 1.4-12). All these functions, derived from the unit impulse\nfunction (successive derivatives and integrals), are called singularity functions.†", - "type": "text" - }, - { - "block_id": "p109-b2", - "global_id": 3028, - "bbox": [ - 133.57, - 189.29, - 503.78, - 201.25 - ], - "text": "DRILL 1.9\nSimplifying Expressions Containing the Unit Impulse", - "type": "text" - }, - { - "block_id": "p109-b3", - "global_id": 3029, - "bbox": [ - 133.57, - 210.37, - 173.44, - 220.33 - ], - "text": "Show that", - "type": "text" - }, - { - "block_id": "p109-b4", - "global_id": 3030, - "bbox": [ - 151.5, - 224.27, - 245.58, - 238.27 - ], - "text": "(a) (t3 + 3)δ(t) = 3δ(t)", - "type": "text" - }, - { - "block_id": "p109-b5", - "global_id": 3031, - "bbox": [ - 151.5, - 245.4, - 163.67, - 255.36 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p109-b6", - "global_id": 3032, - "bbox": [ - 168.65, - 234.07, - 173.32, - 244.03 - ], - "text": "'", - "type": "text" - }, - { - "block_id": "p109-b7", - "global_id": 3033, - "bbox": [ - 173.32, - 245.48, - 184.96, - 255.44 - ], - "text": "sin", - "type": "text" - }, - { - "block_id": "p109-b9", - "global_id": 3034, - "bbox": [ - 192.6, - 238.08, - 217.58, - 255.34 - ], - "text": "t2 −π", - "type": "text" - }, - { - "block_id": "p109-b10", - "global_id": 3035, - "bbox": [ - 212.6, - 252.55, - 217.58, - 262.51 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p109-b11", - "global_id": 3036, - "bbox": [ - 219.76, - 234.07, - 229.87, - 244.03 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p109-b12", - "global_id": 3037, - "bbox": [ - 230.98, - 245.06, - 281.02, - 255.34 - ], - "text": "δ(t) = −δ(t)", - "type": "text" - }, - { - "block_id": "p109-b13", - "global_id": 3038, - "bbox": [ - 152.06, - 263.27, - 226.29, - 274.78 - ], - "text": "(c) e−2tδ(t) = δ(t)", - "type": "text" - }, - { - "block_id": "p109-b14", - "global_id": 3039, - "bbox": [ - 151.5, - 274.95, - 196.4, - 295.85 - ], - "text": "(d) ω2 + 1", - "type": "text" - }, - { - "block_id": "p109-b15", - "global_id": 3040, - "bbox": [ - 169.85, - 278.99, - 251.59, - 303.01 - ], - "text": "ω2 + 9 δ(ω −1) = 1", - "type": "text" - }, - { - "block_id": "p109-b16", - "global_id": 3041, - "bbox": [ - 246.61, - 285.56, - 287.6, - 303.01 - ], - "text": "5δ(ω −1)", - "type": "text" - }, - { - "block_id": "p109-b17", - "global_id": 3042, - "bbox": [ - 133.57, - 362.17, - 493.12, - 374.13 - ], - "text": "DRILL 1.10\nSimplifying Integrals Containing the Unit Impulse", - "type": "text" - }, - { - "block_id": "p109-b18", - "global_id": 3043, - "bbox": [ - 133.57, - 383.25, - 173.44, - 393.21 - ], - "text": "Show that", - "type": "text" - }, - { - "block_id": "p109-b19", - "global_id": 3044, - "bbox": [ - 151.5, - 404.36, - 163.11, - 414.32 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p109-b20", - "global_id": 3045, - "bbox": [ - 168.09, - 390.47, - 185.04, - 402.53 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p109-b21", - "global_id": 3046, - "bbox": [ - 173.36, - 415.35, - 185.92, - 422.32 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p109-b22", - "global_id": 3047, - "bbox": [ - 187.53, - 400.23, - 247.69, - 414.4 - ], - "text": "δ(t)e−jωt dt = 1", - "type": "text" - }, - { - "block_id": "p109-b23", - "global_id": 3048, - "bbox": [ - 151.5, - 431.97, - 163.67, - 441.93 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p109-b24", - "global_id": 3049, - "bbox": [ - 168.65, - 418.09, - 185.6, - 430.14 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p109-b25", - "global_id": 3050, - "bbox": [ - 173.92, - 442.96, - 186.47, - 449.93 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p109-b26", - "global_id": 3051, - "bbox": [ - 188.08, - 431.64, - 233.5, - 442.02 - ], - "text": "δ(t −2)cos", - "type": "text" - }, - { - "block_id": "p109-b27", - "global_id": 3052, - "bbox": [ - 235.74, - 420.64, - 252.11, - 434.93 - ], - "text": "πt", - "type": "text" - }, - { - "block_id": "p109-b28", - "global_id": 3053, - "bbox": [ - 244.83, - 439.13, - 249.81, - 449.09 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p109-b30", - "global_id": 3054, - "bbox": [ - 260.01, - 431.64, - 284.79, - 442.02 - ], - "text": "dt = 0", - "type": "text" - }, - { - "block_id": "p109-b31", - "global_id": 3055, - "bbox": [ - 152.06, - 459.59, - 163.12, - 469.56 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p109-b32", - "global_id": 3056, - "bbox": [ - 168.1, - 445.7, - 185.04, - 457.75 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p109-b33", - "global_id": 3057, - "bbox": [ - 173.36, - 470.58, - 185.92, - 477.55 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p109-b34", - "global_id": 3058, - "bbox": [ - 187.53, - 457.54, - 299.73, - 469.64 - ], - "text": "e−2(x−t)δ(2 −t)dt = e−2(x−2)", - "type": "text" - }, - { - "block_id": "p109-b35", - "global_id": 3059, - "bbox": [ - 127.59, - 513.61, - 313.7, - 528.37 - ], - "text": "1.4-3 The Exponential Function est", - "type": "text" - }, - { - "block_id": "p109-b36", - "global_id": 3060, - "bbox": [ - 127.59, - 533.08, - 516.13, - 556.42 - ], - "text": "Another important function in the area of signals and systems is the exponential signal est, where\ns is complex in general, given by", - "type": "text" - }, - { - "block_id": "p109-b37", - "global_id": 3061, - "bbox": [ - 300.5, - 560.55, - 343.02, - 570.82 - ], - "text": "s = σ + jω", - "type": "text" - }, - { - "block_id": "p109-b38", - "global_id": 3062, - "bbox": [ - 127.59, - 599.27, - 516.11, - 633.41 - ], - "text": "† Singularity functions were defined by late Prof. S. J. Mason as follows. A singularity is a point at which a\nfunction does not possess a derivative. Each of the singularity functions (or if not the function itself, then the\nfunction differentiated a finite number of times) has a singular point at the origin and is zero elsewhere [2].", - "type": "text" - } - ] - }, - { - "page_num": 110, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p110-b0", - "global_id": 3063, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "90\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p110-b1", - "global_id": 3064, - "bbox": [ - 101.84, - 85.82, - 143.59, - 95.78 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p110-b2", - "global_id": 3065, - "bbox": [ - 205.15, - 94.13, - 490.38, - 108.62 - ], - "text": "est = e(σ+jω)t = eσtejωt = eσt(cos ωt + jsin ωt)\n(1.13)", - "type": "text" - }, - { - "block_id": "p110-b3", - "global_id": 3066, - "bbox": [ - 101.84, - 116.54, - 275.09, - 128.14 - ], - "text": "Since s∗= σ −jω (the conjugate of s), then", - "type": "text" - }, - { - "block_id": "p110-b4", - "global_id": 3067, - "bbox": [ - 200.63, - 136.16, - 391.61, - 150.65 - ], - "text": "es∗t = e(σ−jω)t = eσte−jωt = eσt(cos ωt −jsin ωt)", - "type": "text" - }, - { - "block_id": "p110-b5", - "global_id": 3068, - "bbox": [ - 101.84, - 163.2, - 116.22, - 173.16 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p110-b6", - "global_id": 3069, - "bbox": [ - 247.64, - 171.92, - 302.08, - 185.99 - ], - "text": "eσt cos ωt = 1", - "type": "text" - }, - { - "block_id": "p110-b7", - "global_id": 3070, - "bbox": [ - 298.59, - 172.01, - 490.38, - 188.81 - ], - "text": "2(est + es∗t)\n(1.14)", - "type": "text" - }, - { - "block_id": "p110-b8", - "global_id": 3071, - "bbox": [ - 101.84, - 191.93, - 490.39, - 253.34 - ], - "text": "A comparison of Eq. (1.13) with Euler’s formula shows that est is a generalization of the function\nejωt, where the frequency variable jω is generalized to a complex variable s = σ + jω. For this\nreason, we designate the variable s as the complex frequency. In fact, function est encompasses a\nlarge class of functions. The following functions are either special cases of or can be expressed in\nterms of est:", - "type": "text" - }, - { - "block_id": "p110-b9", - "global_id": 3072, - "bbox": [ - 118.78, - 259.89, - 380.22, - 307.13 - ], - "text": "1. A constant k = ke0t\n(s = 0)\n2. A monotonic exponential eσt\n(ω = 0, s = σ)\n3. A sinusoid cos ωt\n(σ = 0, s = ±jω)\n4. An exponentially varying sinusoid eσt cos ωt\n(s = σ ± jω)", - "type": "text" - }, - { - "block_id": "p110-b10", - "global_id": 3073, - "bbox": [ - 101.84, - 315.11, - 273.12, - 325.07 - ], - "text": "These functions are illustrated in Fig. 1.21.", - "type": "text" - }, - { - "block_id": "p110-b11", - "global_id": 3074, - "bbox": [ - 101.84, - 326.96, - 490.39, - 360.94 - ], - "text": "The complex frequency s can be conveniently represented on a complex frequency plane (s\nplane), as depicted in Fig. 1.22. The horizontal axis is the real axis (σ axis), and the vertical\naxis is the imaginary axis (ω axis). The absolute value of the imaginary part of s is |ω| (the", - "type": "text" - }, - { - "block_id": "p110-b12", - "global_id": 3075, - "bbox": [ - 189.82, - 481.6, - 359.12, - 489.6 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p110-b13", - "global_id": 3076, - "bbox": [ - 189.82, - 592.31, - 198.7, - 600.31 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p110-b14", - "global_id": 3077, - "bbox": [ - 243.42, - 464.29, - 245.64, - 472.29 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p110-b15", - "global_id": 3078, - "bbox": [ - 193.26, - 394.17, - 202.46, - 403.64 - ], - "text": "est", - "type": "text" - }, - { - "block_id": "p110-b16", - "global_id": 3079, - "bbox": [ - 234.76, - 410.94, - 254.31, - 419.24 - ], - "text": "s 0", - "type": "text" - }, - { - "block_id": "p110-b17", - "global_id": 3080, - "bbox": [ - 228.26, - 450.94, - 249.81, - 459.24 - ], - "text": "s 0", - "type": "text" - }, - { - "block_id": "p110-b18", - "global_id": 3081, - "bbox": [ - 418.45, - 420.79, - 420.68, - 428.79 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p110-b19", - "global_id": 3082, - "bbox": [ - 181.76, - 518.62, - 201.31, - 526.92 - ], - "text": "s 0", - "type": "text" - }, - { - "block_id": "p110-b20", - "global_id": 3083, - "bbox": [ - 349.63, - 592.31, - 358.96, - 600.31 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p110-b21", - "global_id": 3084, - "bbox": [ - 400.45, - 548.43, - 402.68, - 556.43 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p110-b22", - "global_id": 3085, - "bbox": [ - 353.29, - 514.66, - 372.85, - 522.96 - ], - "text": "s 0", - "type": "text" - }, - { - "block_id": "p110-b23", - "global_id": 3086, - "bbox": [ - 215.26, - 434.94, - 250.81, - 443.24 - ], - "text": "s v 0", - "type": "text" - }, - { - "block_id": "p110-b24", - "global_id": 3087, - "bbox": [ - 366.79, - 409.94, - 386.34, - 418.24 - ], - "text": "s 0", - "type": "text" - }, - { - "block_id": "p110-b25", - "global_id": 3088, - "bbox": [ - 240.42, - 530.43, - 242.64, - 538.43 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p110-b26", - "global_id": 3089, - "bbox": [ - 125.76, - 606.63, - 320.12, - 616.24 - ], - "text": "Figure 1.21 Sinusoids of complex frequency σ + jω.", - "type": "text" - } - ] - }, - { - "page_num": 111, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p111-b0", - "global_id": 3090, - "bbox": [ - 344.33, - 62.89, - 516.14, - 71.98 - ], - "text": "1.4\nSome Useful Signal Models\n91", - "type": "text" - }, - { - "block_id": "p111-b1", - "global_id": 3091, - "bbox": [ - 325.77, - 142.68, - 333.77, - 246.67 - ], - "text": "Exponentially increasing signals", - "type": "text" - }, - { - "block_id": "p111-b2", - "global_id": 3092, - "bbox": [ - 190.86, - 98.34, - 365.43, - 106.34 - ], - "text": "Left half-plane\nRight half-plane", - "type": "text" - }, - { - "block_id": "p111-b3", - "global_id": 3093, - "bbox": [ - 352.25, - 185.15, - 381.81, - 193.15 - ], - "text": "Real axis", - "type": "text" - }, - { - "block_id": "p111-b4", - "global_id": 3094, - "bbox": [ - 273.76, - 104.51, - 281.86, - 161.84 - ], - "text": "Imaginary axis jv", - "type": "text" - }, - { - "block_id": "p111-b5", - "global_id": 3095, - "bbox": [ - 367.9, - 202.13, - 372.79, - 210.13 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p111-b6", - "global_id": 3096, - "bbox": [ - 201.27, - 142.01, - 209.27, - 247.33 - ], - "text": "Exponentially decreasing signals", - "type": "text" - }, - { - "block_id": "p111-b7", - "global_id": 3097, - "bbox": [ - 151.5, - 310.44, - 296.16, - 319.68 - ], - "text": "Figure 1.22 Complex frequency plane.", - "type": "text" - }, - { - "block_id": "p111-b8", - "global_id": 3098, - "bbox": [ - 127.59, - 347.44, - 516.16, - 526.2 - ], - "text": "radian frequency), which indicates the frequency of oscillation of est; the real part σ (the neper\nfrequency) gives information about the rate of increase or decrease of the amplitude of est. For\nsignals whose complex frequencies lie on the real axis (σ axis, where ω = 0), the frequency\nof oscillation is zero. Consequently these signals are monotonically increasing or decreasing\nexponentials (Fig. 1.21a). For signals whose frequencies lie on the imaginary axis (ω axis, where\nσ = 0), eσt = 1. Therefore, these signals are conventional sinusoids with constant amplitude\n(Fig. 1.21b). The case s = 0 (σ = ω = 0) corresponds to a constant (dc) signal because e0t = 1.\nFor the signals illustrated in Figs. 1.21c and 1.21d, both σ and ω are nonzero; the frequency s is\ncomplex and does not lie on either axis. The signal in Fig. 1.21c decays exponentially. Therefore,\nσ is negative, and s lies to the left of the imaginary axis. In contrast, the signal in Fig. 1.21d\ngrows exponentially. Therefore, σ is positive, and s lies to the right of the imaginary axis. Thus\nthe s plane (Fig. 1.21) can be separated into two parts: the left half-plane (LHP) corresponding to\nexponentially decaying signals and the right half-plane (RHP) corresponding to exponentially\ngrowing signals. The imaginary axis separates the two regions and corresponds to signals of\nconstant amplitude.", - "type": "text" - }, - { - "block_id": "p111-b9", - "global_id": 3099, - "bbox": [ - 127.59, - 524.57, - 516.14, - 621.84 - ], - "text": "An exponentially growing sinusoid e2t cos 5t, for example, can be expressed as a linear\ncombination of exponentials e(2+j5)t and e(2−j5)t with complex frequencies 2 + j5 and 2 −j5,\nrespectively, which lie in the RHP. An exponentially decaying sinusoid e−2t cos 5t can be expressed\nas a linear combination of exponentials e(−2+j5)t and e(−2−j5)t with complex frequencies −2 + j5\nand −2 −j5, respectively, which lie in the LHP. A constant-amplitude sinusoid cos 5t can be\nexpressed as a linear combination of exponentials ej5t and e−j5t with complex frequencies ±j5,\nwhich lie on the imaginary axis. Observe that the monotonic exponentials e±2t are also generalized\nsinusoids with complex frequencies ±2.", - "type": "text" - } - ] - }, - { - "page_num": 112, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p112-b0", - "global_id": 3100, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "92\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p112-b1", - "global_id": 3101, - "bbox": [ - 102.2, - 92.21, - 311.95, - 106.15 - ], - "text": "1.5 EVEN AND ODD FUNCTIONS", - "type": "text" - }, - { - "block_id": "p112-b2", - "global_id": 3102, - "bbox": [ - 101.84, - 111.73, - 490.39, - 146.02 - ], - "text": "A function xe(t) is said to be an even function of t if it is symmetrical about the vertical axis. A\nfunction xo(t) is said to be an odd function of t if it is antisymmetrical about the vertical axis.\nMathematically expressed, these symmetry conditions require", - "type": "text" - }, - { - "block_id": "p112-b3", - "global_id": 3103, - "bbox": [ - 208.29, - 164.59, - 490.38, - 175.67 - ], - "text": "xe(t) = xe(−t)\nand\nxo(t) = −xo(−t)\n(1.15)", - "type": "text" - }, - { - "block_id": "p112-b4", - "global_id": 3104, - "bbox": [ - 101.84, - 193.56, - 490.37, - 227.84 - ], - "text": "An even function has the same value at the instants t and −t for all values of t. On the other hand,\nthe value of an odd function at the instant t is the negative of its value at the instant −t. An example\neven signal and an example odd signal are shown in Figs. 1.23a and 1.23b, respectively.", - "type": "text" - }, - { - "block_id": "p112-b5", - "global_id": 3105, - "bbox": [ - 101.84, - 259.63, - 373.98, - 271.58 - ], - "text": "1.5-1 Some Properties of Even and Odd Functions", - "type": "text" - }, - { - "block_id": "p112-b6", - "global_id": 3106, - "bbox": [ - 101.84, - 277.71, - 317.74, - 287.68 - ], - "text": "Even and odd functions have the following properties:", - "type": "text" - }, - { - "block_id": "p112-b7", - "global_id": 3107, - "bbox": [ - 207.04, - 306.25, - 385.19, - 316.63 - ], - "text": "even function × odd function = odd function", - "type": "text" - }, - { - "block_id": "p112-b8", - "global_id": 3108, - "bbox": [ - 203.57, - 321.2, - 388.66, - 346.52 - ], - "text": "odd function × odd function = even function\neven function × even function = even function", - "type": "text" - }, - { - "block_id": "p112-b9", - "global_id": 3109, - "bbox": [ - 101.84, - 365.52, - 490.38, - 375.48 - ], - "text": "The proofs are trivial and follow directly from the definition of odd and even functions [Eq. (1.15)].", - "type": "text" - }, - { - "block_id": "p112-b10", - "global_id": 3110, - "bbox": [ - 176.25, - 479.44, - 185.13, - 487.44 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p112-b11", - "global_id": 3111, - "bbox": [ - 140.91, - 417.83, - 154.68, - 427.38 - ], - "text": "xe(t)", - "type": "text" - }, - { - "block_id": "p112-b12", - "global_id": 3112, - "bbox": [ - 149.93, - 462.81, - 242.37, - 473.8 - ], - "text": "a\nt\na\n0", - "type": "text" - }, - { - "block_id": "p112-b13", - "global_id": 3113, - "bbox": [ - 184.03, - 549.41, - 188.03, - 557.41 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p112-b14", - "global_id": 3114, - "bbox": [ - 175.86, - 597.29, - 185.51, - 605.29 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p112-b15", - "global_id": 3115, - "bbox": [ - 149.71, - 537.51, - 160.38, - 545.72 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p112-b16", - "global_id": 3116, - "bbox": [ - 203.73, - 546.61, - 242.36, - 557.21 - ], - "text": "t\na", - "type": "text" - }, - { - "block_id": "p112-b17", - "global_id": 3117, - "bbox": [ - 154.43, - 504.41, - 168.53, - 513.96 - ], - "text": "xo(t)", - "type": "text" - }, - { - "block_id": "p112-b18", - "global_id": 3118, - "bbox": [ - 270.8, - 597.25, - 451.16, - 606.58 - ], - "text": "Figure 1.23 Functions of t: (a) even and (b) odd.", - "type": "text" - } - ] - }, - { - "page_num": 113, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p113-b0", - "global_id": 3119, - "bbox": [ - 354.11, - 62.89, - 516.14, - 71.98 - ], - "text": "1.5\nEven and Odd Functions\n93", - "type": "text" - }, - { - "block_id": "p113-b1", - "global_id": 3120, - "bbox": [ - 127.89, - 86.19, - 158.64, - 98.32 - ], - "text": "AREA", - "type": "text" - }, - { - "block_id": "p113-b2", - "global_id": 3121, - "bbox": [ - 127.59, - 102.35, - 516.16, - 124.26 - ], - "text": "Because of the symmetries of even and odd functions about the vertical axis, it follows from\nEq. (1.15) [or Fig. 1.23] that", - "type": "text" - }, - { - "block_id": "p113-b3", - "global_id": 3122, - "bbox": [ - 212.19, - 138.63, - 225.5, - 150.9 - ], - "text": "# a", - "type": "text" - }, - { - "block_id": "p113-b4", - "global_id": 3123, - "bbox": [ - 217.45, - 163.5, - 226.37, - 170.69 - ], - "text": "−a", - "type": "text" - }, - { - "block_id": "p113-b5", - "global_id": 3124, - "bbox": [ - 227.98, - 152.19, - 272.26, - 163.27 - ], - "text": "xe(t)dt = 2", - "type": "text" - }, - { - "block_id": "p113-b6", - "global_id": 3125, - "bbox": [ - 273.36, - 138.63, - 286.68, - 150.9 - ], - "text": "# a", - "type": "text" - }, - { - "block_id": "p113-b7", - "global_id": 3126, - "bbox": [ - 278.63, - 152.19, - 350.02, - 170.76 - ], - "text": "0\nxe(t)dt\nand", - "type": "text" - }, - { - "block_id": "p113-b8", - "global_id": 3127, - "bbox": [ - 371.07, - 138.63, - 384.38, - 150.9 - ], - "text": "# a", - "type": "text" - }, - { - "block_id": "p113-b9", - "global_id": 3128, - "bbox": [ - 376.33, - 163.5, - 385.25, - 170.69 - ], - "text": "−a", - "type": "text" - }, - { - "block_id": "p113-b10", - "global_id": 3129, - "bbox": [ - 386.86, - 152.19, - 516.13, - 163.27 - ], - "text": "xo(t)dt = 0\n(1.16)", - "type": "text" - }, - { - "block_id": "p113-b11", - "global_id": 3130, - "bbox": [ - 127.59, - 191.11, - 516.13, - 224.98 - ], - "text": "These results are valid under the assumption that there is no impulse (or its derivatives) at the\norigin. The proof of these statements is obvious from the plots of even and odd functions. Formal\nproofs, left as an exercise for the reader, can be accomplished by using the definitions in Eq. (1.15).", - "type": "text" - }, - { - "block_id": "p113-b12", - "global_id": 3131, - "bbox": [ - 127.59, - 226.97, - 516.11, - 248.89 - ], - "text": "Because of their properties, study of odd and even functions proves useful in many\napplications, as will become evident in later chapters.", - "type": "text" - }, - { - "block_id": "p113-b13", - "global_id": 3132, - "bbox": [ - 127.59, - 284.9, - 370.5, - 296.85 - ], - "text": "1.5-2 Even and Odd Components of a Signal", - "type": "text" - }, - { - "block_id": "p113-b14", - "global_id": 3133, - "bbox": [ - 127.59, - 302.57, - 452.13, - 312.95 - ], - "text": "Every signal x(t) can be expressed as a sum of even and odd components because", - "type": "text" - }, - { - "block_id": "p113-b15", - "global_id": 3134, - "bbox": [ - 242.69, - 334.39, - 274.08, - 346.02 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p113-b16", - "global_id": 3135, - "bbox": [ - 269.4, - 335.75, - 330.22, - 361.67 - ], - "text": "2[x(t) + x(−t)]\n\n\n\neven", - "type": "text" - }, - { - "block_id": "p113-b17", - "global_id": 3136, - "bbox": [ - 331.33, - 334.39, - 344.88, - 345.71 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p113-b18", - "global_id": 3137, - "bbox": [ - 340.2, - 335.75, - 401.02, - 361.67 - ], - "text": "2[x(t) −x(−t)]\n\n\n\nodd", - "type": "text" - }, - { - "block_id": "p113-b19", - "global_id": 3138, - "bbox": [ - 492.07, - 336.16, - 516.13, - 346.12 - ], - "text": "(1.17)", - "type": "text" - }, - { - "block_id": "p113-b20", - "global_id": 3139, - "bbox": [ - 127.59, - 382.82, - 516.14, - 428.66 - ], - "text": "From the definitions in Eq. (1.15), we can clearly see that the first component on the right-hand\nside is an even function, while the second component is odd. This is apparent from the fact that\nreplacing t by −t in the first component yields the same function. The same maneuver in the\nsecond component yields the negative of that component.", - "type": "text" - }, - { - "block_id": "p113-b21", - "global_id": 3140, - "bbox": [ - 102.51, - 463.28, - 479.63, - 475.23 - ], - "text": "EXAMPLE 1.8\nFinding the Even and Odd Components of a Signal", - "type": "text" - }, - { - "block_id": "p113-b22", - "global_id": 3141, - "bbox": [ - 128.9, - 490.26, - 386.17, - 501.86 - ], - "text": "Find and sketch the even and odd components of x(t) = e−atu(t).", - "type": "text" - }, - { - "block_id": "p113-b23", - "global_id": 3142, - "bbox": [ - 128.9, - 524.36, - 502.75, - 547.39 - ], - "text": "Based on Eq. (1.17), we can express x(t) as a sum of the even component xe(t) and the odd\ncomponent xo(t) as", - "type": "text" - }, - { - "block_id": "p113-b24", - "global_id": 3143, - "bbox": [ - 278.46, - 548.27, - 353.22, - 559.35 - ], - "text": "x(t) = xe(t) + xo(t)", - "type": "text" - }, - { - "block_id": "p113-b25", - "global_id": 3144, - "bbox": [ - 128.91, - 567.62, - 153.24, - 577.58 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p113-b26", - "global_id": 3145, - "bbox": [ - 169.99, - 587.78, - 204.94, - 600.2 - ], - "text": "xe(t) = 1", - "type": "text" - }, - { - "block_id": "p113-b27", - "global_id": 3146, - "bbox": [ - 201.45, - 587.78, - 378.09, - 602.31 - ], - "text": "2[e−atu(t) + eatu(−t)]\nand\nxo(t) = 1", - "type": "text" - }, - { - "block_id": "p113-b28", - "global_id": 3147, - "bbox": [ - 374.6, - 587.89, - 461.68, - 602.31 - ], - "text": "2[e−atu(t) −eatu(−t)]", - "type": "text" - }, - { - "block_id": "p113-b29", - "global_id": 3148, - "bbox": [ - 128.91, - 609.81, - 456.4, - 621.41 - ], - "text": "The function e−atu(t) and its even and odd components are illustrated in Fig. 1.24.", - "type": "text" - } - ] - }, - { - "page_num": 114, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p114-b0", - "global_id": 3149, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "94\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p114-b2", - "global_id": 3150, - "bbox": [ - 246.33, - 190.36, - 255.21, - 198.36 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p114-b3", - "global_id": 3151, - "bbox": [ - 242.87, - 106.78, - 246.87, - 114.78 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p114-b4", - "global_id": 3152, - "bbox": [ - 245.95, - 277.33, - 255.59, - 285.33 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p114-b5", - "global_id": 3153, - "bbox": [ - 246.33, - 387.11, - 255.21, - 395.11 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p114-b6", - "global_id": 3154, - "bbox": [ - 379.12, - 346.1, - 381.35, - 354.1 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p114-b7", - "global_id": 3155, - "bbox": [ - 379.12, - 262.08, - 381.35, - 270.08 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p114-b8", - "global_id": 3156, - "bbox": [ - 379.12, - 175.18, - 381.35, - 183.18 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p114-b9", - "global_id": 3157, - "bbox": [ - 258.08, - 301.09, - 272.18, - 310.64 - ], - "text": "xo(t)", - "type": "text" - }, - { - "block_id": "p114-b10", - "global_id": 3158, - "bbox": [ - 258.08, - 218.22, - 271.84, - 227.76 - ], - "text": "xe(t)", - "type": "text" - }, - { - "block_id": "p114-b11", - "global_id": 3159, - "bbox": [ - 258.08, - 102.33, - 269.18, - 110.41 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p114-b12", - "global_id": 3160, - "bbox": [ - 286.05, - 138.13, - 299.26, - 147.75 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p114-b13", - "global_id": 3161, - "bbox": [ - 252.77, - 347.2, - 256.77, - 355.2 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p114-b14", - "global_id": 3162, - "bbox": [ - 252.77, - 175.98, - 256.77, - 183.98 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p114-b15", - "global_id": 3163, - "bbox": [ - 252.77, - 263.37, - 256.77, - 271.37 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p114-b16", - "global_id": 3164, - "bbox": [ - 191.92, - 234.16, - 206.64, - 250.94 - ], - "text": "1\n2 eat", - "type": "text" - }, - { - "block_id": "p114-b17", - "global_id": 3165, - "bbox": [ - 197.39, - 357.04, - 212.11, - 373.82 - ], - "text": "1\n2 eat", - "type": "text" - }, - { - "block_id": "p114-b18", - "global_id": 3166, - "bbox": [ - 296.1, - 232.77, - 315.81, - 249.55 - ], - "text": "1\n2 eat", - "type": "text" - }, - { - "block_id": "p114-b19", - "global_id": 3167, - "bbox": [ - 240.35, - 307.66, - 315.81, - 331.72 - ], - "text": "1\n2 eat\n1\n2", - "type": "text" - }, - { - "block_id": "p114-b20", - "global_id": 3168, - "bbox": [ - 238.42, - 218.29, - 242.42, - 235.07 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p114-b21", - "global_id": 3169, - "bbox": [ - 101.77, - 401.81, - 316.85, - 411.04 - ], - "text": "Figure 1.24 Finding even and odd components of a signal.", - "type": "text" - }, - { - "block_id": "p114-b22", - "global_id": 3170, - "bbox": [ - 76.77, - 485.17, - 467.85, - 511.08 - ], - "text": "EXAMPLE 1.9\nFinding the Even and Odd Components of a Complex\nSignal", - "type": "text" - }, - { - "block_id": "p114-b23", - "global_id": 3171, - "bbox": [ - 103.16, - 526.32, - 266.65, - 537.71 - ], - "text": "Find the even and odd components of ejt.", - "type": "text" - }, - { - "block_id": "p114-b24", - "global_id": 3172, - "bbox": [ - 103.16, - 560.62, - 169.85, - 570.58 - ], - "text": "From Eq. (1.17),", - "type": "text" - }, - { - "block_id": "p114-b25", - "global_id": 3173, - "bbox": [ - 255.67, - 568.05, - 324.52, - 583.24 - ], - "text": "ejt = xe(t) + xo(t)", - "type": "text" - }, - { - "block_id": "p114-b26", - "global_id": 3174, - "bbox": [ - 103.17, - 591.5, - 127.5, - 601.46 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p114-b27", - "global_id": 3175, - "bbox": [ - 154.42, - 601.7, - 189.37, - 614.13 - ], - "text": "xe(t) = 1", - "type": "text" - }, - { - "block_id": "p114-b28", - "global_id": 3176, - "bbox": [ - 185.88, - 599.43, - 351.24, - 616.24 - ], - "text": "2[ejt + e−jt] = cos t\nand\nxo(t) = 1", - "type": "text" - }, - { - "block_id": "p114-b29", - "global_id": 3177, - "bbox": [ - 347.76, - 599.43, - 425.59, - 616.24 - ], - "text": "2[ejt −e−jt] = jsin t", - "type": "text" - } - ] - }, - { - "page_num": 115, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p115-b0", - "global_id": 3178, - "bbox": [ - 420.69, - 62.89, - 516.14, - 71.98 - ], - "text": "1.6\nSystems\n95", - "type": "text" - }, - { - "block_id": "p115-b1", - "global_id": 3179, - "bbox": [ - 127.89, - 86.19, - 352.92, - 98.32 - ], - "text": "A MODIFICATION FOR COMPLEX SIGNALS", - "type": "text" - }, - { - "block_id": "p115-b2", - "global_id": 3180, - "bbox": [ - 127.59, - 102.35, - 516.16, - 196.69 - ], - "text": "While a complex signal can be decomposed into even and odd components, it is more common\nto decompose complex signals using conjugate symmetries. A complex signal x(t) is said to\nbe conjugate-symmetric if x(t) = x∗(−t). A conjugate-symmetric signal is even in the real part\nand odd in the imaginary part. Thus, a real conjugate-symmetric signal is an even signal. A\nsignal is conjugate-antisymmetric if x(t) = −x∗(−t). A conjugate-antisymmetric signal is odd\nin the real part and even in the imaginary part. A real conjugate-antisymmetric signal is an\nodd signal. Any signal x(t) can be decomposed into a conjugate-symmetric portion xcs(t) plus\na conjugate-antisymmetric portion xca(t). That is,", - "type": "text" - }, - { - "block_id": "p115-b3", - "global_id": 3181, - "bbox": [ - 281.57, - 208.36, - 362.15, - 219.44 - ], - "text": "x(t) = xcs(t) + xca(t)", - "type": "text" - }, - { - "block_id": "p115-b4", - "global_id": 3182, - "bbox": [ - 127.59, - 231.52, - 151.92, - 241.48 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p115-b5", - "global_id": 3183, - "bbox": [ - 206.54, - 239.08, - 293.16, - 258.37 - ], - "text": "xcs(t) = x(t) + x∗(−t)", - "type": "text" - }, - { - "block_id": "p115-b6", - "global_id": 3184, - "bbox": [ - 264.45, - 239.08, - 435.98, - 264.74 - ], - "text": "2\nand\nxca(t) = x(t) −x∗(−t)", - "type": "text" - }, - { - "block_id": "p115-b7", - "global_id": 3185, - "bbox": [ - 407.27, - 254.78, - 412.25, - 264.74 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p115-b8", - "global_id": 3186, - "bbox": [ - 127.59, - 270.36, - 516.16, - 304.23 - ], - "text": "The proof is similar to the one for decomposing a signal into even and odd components. As we\nshall see in later chapters, conjugate symmetries commonly occur in real-world signals and their\ntransforms.", - "type": "text" - }, - { - "block_id": "p115-b9", - "global_id": 3187, - "bbox": [ - 127.94, - 334.33, - 212.13, - 348.28 - ], - "text": "1.6 SYSTEMS", - "type": "text" - }, - { - "block_id": "p115-b10", - "global_id": 3188, - "bbox": [ - 127.59, - 354.27, - 516.15, - 400.1 - ], - "text": "As mentioned in Sec. 1.1, systems are used to process signals to allow modification or extraction of\nadditional information from the signals. A system may consist of physical components (hardware\nrealization) or of an algorithm that computes the output signal from the input signal (software\nrealization).", - "type": "text" - }, - { - "block_id": "p115-b11", - "global_id": 3189, - "bbox": [ - 127.59, - 402.09, - 516.19, - 483.79 - ], - "text": "Roughly speaking, a physical system consists of interconnected components, which are\ncharacterized by their terminal (input–output) relationships. In addition, a system is governed\nby laws of interconnection. For example, in electrical systems, the terminal relationships are\nthe familiar voltage-current relationships for the resistors, capacitors, inductors, transformers,\ntransistors, and so on, as well as the laws of interconnection (i.e., Kirchhoff’s laws). We use these\nlaws to derive mathematical equations relating the outputs to the inputs. These equations then\nrepresent a mathematical model of the system.", - "type": "text" - }, - { - "block_id": "p115-b12", - "global_id": 3190, - "bbox": [ - 127.59, - 485.78, - 516.15, - 520.42 - ], - "text": "A system can be conveniently illustrated by a “black box” with one set of accessible terminals\nwhere the input variables x1(t), x2(t), . . .,xj(t) are applied and another set of accessible terminals\nwhere the output variables y1(t), y2(t),. . .,yk(t) are observed (Fig. 1.25).", - "type": "text" - }, - { - "block_id": "p115-b13", - "global_id": 3191, - "bbox": [ - 127.59, - 521.64, - 516.13, - 543.56 - ], - "text": "The study of systems consists of three major areas: mathematical modeling, analysis, and\ndesign. Although we shall be dealing with mathematical modeling, our main concern is with", - "type": "text" - }, - { - "block_id": "p115-b14", - "global_id": 3192, - "bbox": [ - 238.4, - 565.15, - 252.51, - 574.76 - ], - "text": "y1(t)", - "type": "text" - }, - { - "block_id": "p115-b15", - "global_id": 3193, - "bbox": [ - 243.55, - 587.43, - 246.35, - 611.44 - ], - "text": "•\n•\n•", - "type": "text" - }, - { - "block_id": "p115-b16", - "global_id": 3194, - "bbox": [ - 221.4, - 587.44, - 224.2, - 611.44 - ], - "text": "•\n•\n•", - "type": "text" - }, - { - "block_id": "p115-b17", - "global_id": 3195, - "bbox": [ - 135.45, - 587.44, - 138.25, - 611.44 - ], - "text": "•\n•\n•", - "type": "text" - }, - { - "block_id": "p115-b18", - "global_id": 3196, - "bbox": [ - 153.9, - 587.44, - 156.7, - 611.44 - ], - "text": "•\n•\n•", - "type": "text" - }, - { - "block_id": "p115-b19", - "global_id": 3197, - "bbox": [ - 237.9, - 577.15, - 252.01, - 586.76 - ], - "text": "y2(t)", - "type": "text" - }, - { - "block_id": "p115-b20", - "global_id": 3198, - "bbox": [ - 237.9, - 611.15, - 251.67, - 620.7 - ], - "text": "yk(t)", - "type": "text" - }, - { - "block_id": "p115-b21", - "global_id": 3199, - "bbox": [ - 129.8, - 565.15, - 143.9, - 586.76 - ], - "text": "x1(t)\nx2(t)", - "type": "text" - }, - { - "block_id": "p115-b22", - "global_id": 3200, - "bbox": [ - 130.53, - 611.15, - 411.73, - 626.5 - ], - "text": "xj(t)\nFigure 1.25 Representation of a system.", - "type": "text" - } - ] - }, - { - "page_num": 116, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p116-b0", - "global_id": 3201, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "96\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p116-b1", - "global_id": 3202, - "bbox": [ - 101.84, - 85.82, - 490.39, - 143.6 - ], - "text": "analysis and design. The major portion of this book is devoted to the analysis problem—how to\ndetermine the system outputs for the given inputs and a given mathematical model of the system\n(or rules governing the system). To a lesser extent, we will also consider the problem of design\nor synthesis—how to construct a system that will produce a desired set of outputs for the given\ninputs.", - "type": "text" - }, - { - "block_id": "p116-b2", - "global_id": 3203, - "bbox": [ - 101.84, - 158.39, - 490.39, - 196.45 - ], - "text": "DATA NEEDED TO COMPUTE SYSTEM RESPONSE\nTo understand what data we need to compute a system response, consider a simple RC circuit with\na current source x(t) as its input (Fig. 1.26).", - "type": "text" - }, - { - "block_id": "p116-b3", - "global_id": 3204, - "bbox": [ - 119.78, - 198.04, - 257.95, - 208.41 - ], - "text": "The output voltage y(t) is given by", - "type": "text" - }, - { - "block_id": "p116-b4", - "global_id": 3205, - "bbox": [ - 237.36, - 219.28, - 302.94, - 236.22 - ], - "text": "y(t) = Rx(t) + 1", - "type": "text" - }, - { - "block_id": "p116-b5", - "global_id": 3206, - "bbox": [ - 297.02, - 233.23, - 303.66, - 243.2 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p116-b6", - "global_id": 3207, - "bbox": [ - 306.19, - 212.28, - 317.95, - 224.56 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p116-b7", - "global_id": 3208, - "bbox": [ - 311.45, - 237.17, - 324.0, - 244.14 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p116-b8", - "global_id": 3209, - "bbox": [ - 325.61, - 225.85, - 490.38, - 236.22 - ], - "text": "x(τ)dτ\n(1.18)", - "type": "text" - }, - { - "block_id": "p116-b9", - "global_id": 3210, - "bbox": [ - 101.85, - 253.2, - 490.39, - 287.49 - ], - "text": "The limits of the integral on the right-hand side are from −∞to t because this integral represents\nthe capacitor charge due to the current x(t) flowing in the capacitor, and this charge is the result\nof the current flowing in the capacitor from −∞. Now, Eq. (1.18) can be expressed as", - "type": "text" - }, - { - "block_id": "p116-b10", - "global_id": 3211, - "bbox": [ - 205.37, - 299.38, - 270.95, - 316.32 - ], - "text": "y(t) = Rx(t) + 1", - "type": "text" - }, - { - "block_id": "p116-b11", - "global_id": 3212, - "bbox": [ - 265.02, - 313.33, - 271.67, - 323.3 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p116-b12", - "global_id": 3213, - "bbox": [ - 274.19, - 292.38, - 287.5, - 304.73 - ], - "text": "# 0", - "type": "text" - }, - { - "block_id": "p116-b13", - "global_id": 3214, - "bbox": [ - 279.45, - 317.27, - 292.01, - 324.24 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p116-b14", - "global_id": 3215, - "bbox": [ - 293.62, - 299.38, - 340.87, - 316.32 - ], - "text": "x(τ)dτ + 1", - "type": "text" - }, - { - "block_id": "p116-b15", - "global_id": 3216, - "bbox": [ - 334.94, - 313.33, - 341.59, - 323.3 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p116-b16", - "global_id": 3217, - "bbox": [ - 344.11, - 292.38, - 355.87, - 304.66 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p116-b17", - "global_id": 3218, - "bbox": [ - 349.37, - 305.95, - 385.66, - 324.53 - ], - "text": "0\nx(τ)dτ", - "type": "text" - }, - { - "block_id": "p116-b18", - "global_id": 3219, - "bbox": [ - 101.85, - 333.36, - 458.16, - 344.44 - ], - "text": "The middle term on the right-hand side is vC(0), the capacitor voltage at t = 0. Therefore,", - "type": "text" - }, - { - "block_id": "p116-b19", - "global_id": 3220, - "bbox": [ - 204.09, - 354.6, - 302.68, - 372.25 - ], - "text": "y(t) = vC(0) + Rx(t) + 1", - "type": "text" - }, - { - "block_id": "p116-b20", - "global_id": 3221, - "bbox": [ - 296.76, - 368.56, - 303.4, - 378.52 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p116-b21", - "global_id": 3222, - "bbox": [ - 305.93, - 347.61, - 317.69, - 359.88 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p116-b22", - "global_id": 3223, - "bbox": [ - 311.18, - 361.17, - 388.15, - 379.75 - ], - "text": "0\nx(τ)dτ\nt ≥0", - "type": "text" - }, - { - "block_id": "p116-b23", - "global_id": 3224, - "bbox": [ - 101.85, - 388.42, - 273.94, - 398.39 - ], - "text": "This equation can be readily generalized as", - "type": "text" - }, - { - "block_id": "p116-b24", - "global_id": 3225, - "bbox": [ - 202.32, - 409.26, - 302.68, - 426.97 - ], - "text": "y(t) = vC(t0) + Rx(t) + 1", - "type": "text" - }, - { - "block_id": "p116-b25", - "global_id": 3226, - "bbox": [ - 296.76, - 423.21, - 303.4, - 433.17 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p116-b26", - "global_id": 3227, - "bbox": [ - 305.93, - 402.27, - 317.69, - 414.54 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p116-b27", - "global_id": 3228, - "bbox": [ - 311.18, - 427.36, - 316.11, - 435.66 - ], - "text": "t0", - "type": "text" - }, - { - "block_id": "p116-b28", - "global_id": 3229, - "bbox": [ - 319.42, - 415.82, - 490.38, - 426.97 - ], - "text": "x(τ)dτ\nt ≥t0\n(1.19)", - "type": "text" - }, - { - "block_id": "p116-b29", - "global_id": 3230, - "bbox": [ - 101.84, - 444.22, - 490.4, - 503.19 - ], - "text": "From Eq. (1.18), the output voltage y(t) at an instant t can be computed if we know the input\ncurrent flowing in the capacitor throughout its entire past (−∞to t). Alternatively, if we know the\ninput current x(t) from some moment t0 onward, then, using Eq. (1.19), we can still calculate y(t)\nfor t ≥t0 from a knowledge of the input current, provided we know vC(t0), the initial capacitor\nvoltage (voltage at t0). Thus vC(t0) contains all the relevant information about the circuit’s entire", - "type": "text" - }, - { - "block_id": "p116-b30", - "global_id": 3231, - "bbox": [ - 104.06, - 584.09, - 287.72, - 593.82 - ], - "text": "x(t)\nvc(t)", - "type": "text" - }, - { - "block_id": "p116-b31", - "global_id": 3232, - "bbox": [ - 316.29, - 568.64, - 327.71, - 576.72 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p116-b32", - "global_id": 3233, - "bbox": [ - 215.63, - 539.05, - 220.52, - 547.05 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p116-b33", - "global_id": 3234, - "bbox": [ - 242.88, - 585.3, - 248.22, - 593.3 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p116-b37", - "global_id": 3235, - "bbox": [ - 321.45, - 604.7, - 471.06, - 626.62 - ], - "text": "Figure 1.26 Example of a simple\nelectrical system.", - "type": "text" - } - ] - }, - { - "page_num": 117, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p117-b0", - "global_id": 3236, - "bbox": [ - 355.12, - 62.89, - 516.13, - 71.98 - ], - "text": "1.7\nClassification of Systems\n97", - "type": "text" - }, - { - "block_id": "p117-b1", - "global_id": 3237, - "bbox": [ - 127.59, - 85.46, - 516.13, - 120.53 - ], - "text": "past (−∞to t0) that we need to compute y(t) for t ≥t0. Therefore, the response of a system at t ≥t0\ncan be determined from its input(s) during the interval t0 to t and from certain initial conditions at\nt = t0.", - "type": "text" - }, - { - "block_id": "p117-b2", - "global_id": 3238, - "bbox": [ - 127.59, - 121.74, - 516.14, - 167.57 - ], - "text": "In the preceding example, we needed only one initial condition. However, in more complex\nsystems, several initial conditions may be necessary. We know, for example, that in passive RLC\nnetworks, the initial values of all inductor currents and all capacitor voltages† are needed to\ndetermine the outputs at any instant t ≥0 if the inputs are given over the interval [0,t].", - "type": "text" - }, - { - "block_id": "p117-b3", - "global_id": 3239, - "bbox": [ - 127.94, - 195.99, - 345.13, - 209.93 - ], - "text": "1.7 CLASSIFICATION OF SYSTEMS", - "type": "text" - }, - { - "block_id": "p117-b4", - "global_id": 3240, - "bbox": [ - 127.59, - 215.92, - 376.57, - 225.88 - ], - "text": "Systems may be classified broadly in the following categories:", - "type": "text" - }, - { - "block_id": "p117-b5", - "global_id": 3241, - "bbox": [ - 144.52, - 233.85, - 420.71, - 327.5 - ], - "text": "1. Linear and nonlinear systems\n2. Constant-parameter and time-varying-parameter systems\n3. Instantaneous (memoryless) and dynamic (with memory) systems\n4. Causal and noncausal systems\n5. Continuous-time and discrete-time systems\n6. Analog and digital systems\n7. Invertible and noninvertible systems\n8. Stable and unstable systems", - "type": "text" - }, - { - "block_id": "p117-b6", - "global_id": 3242, - "bbox": [ - 127.59, - 335.47, - 516.13, - 357.39 - ], - "text": "Other classifications, such as deterministic and probabilistic systems, are beyond the scope of this\ntext and are not considered.", - "type": "text" - }, - { - "block_id": "p117-b7", - "global_id": 3243, - "bbox": [ - 127.59, - 381.27, - 323.67, - 411.26 - ], - "text": "1.7-1 Linear and Nonlinear Systems\nTHE CONCEPT OF LINEARITY", - "type": "text" - }, - { - "block_id": "p117-b8", - "global_id": 3244, - "bbox": [ - 127.59, - 415.2, - 516.14, - 498.07 - ], - "text": "A system whose output is proportional to its input is an example of a linear system. But linearity\nimplies more than this; it also implies the additivity property: that is, if several inputs are acting\non a system, then the total effect on the system due to all these inputs can be determined by\nconsidering one input at a time while assuming all the other inputs to be zero. The total effect is\nthen the sum of all the component effects. This property may be expressed as follows: for a linear\nsystem, if an input x1 acting alone has an effect y1, and if another input x2, also acting alone, has\nan effect y2, then, with both inputs acting on the system, the total effect will be y1 + y2. Thus, if", - "type": "text" - }, - { - "block_id": "p117-b9", - "global_id": 3245, - "bbox": [ - 256.77, - 506.79, - 386.45, - 518.25 - ], - "text": "x1 −→y1\nand\nx2 −→y2", - "type": "text" - }, - { - "block_id": "p117-b10", - "global_id": 3246, - "bbox": [ - 127.59, - 527.28, - 209.48, - 538.84 - ], - "text": "then for all x1 and x2", - "type": "text" - }, - { - "block_id": "p117-b11", - "global_id": 3247, - "bbox": [ - 283.6, - 538.92, - 516.13, - 550.38 - ], - "text": "x1 + x2 −→y1 + y2\n(1.20)", - "type": "text" - }, - { - "block_id": "p117-b12", - "global_id": 3248, - "bbox": [ - 127.59, - 557.3, - 516.13, - 591.26 - ], - "text": "In addition, a linear system must satisfy the homogeneity or scaling property, which states that\nfor arbitrary real or imaginary number k, if an input is increased k-fold, the effect also increases\nk-fold. Thus, if", - "type": "text" - }, - { - "block_id": "p117-b13", - "global_id": 3249, - "bbox": [ - 306.85, - 592.85, - 336.87, - 603.13 - ], - "text": "x −→y", - "type": "text" - }, - { - "block_id": "p117-b14", - "global_id": 3250, - "bbox": [ - 127.59, - 621.19, - 429.98, - 633.41 - ], - "text": "† Strictly speaking, this means independent inductor currents and capacitor voltages.", - "type": "text" - } - ] - }, - { - "page_num": 118, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p118-b0", - "global_id": 3251, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "98\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p118-b1", - "global_id": 3252, - "bbox": [ - 101.84, - 85.72, - 223.58, - 95.78 - ], - "text": "then for all real or imaginary k", - "type": "text" - }, - { - "block_id": "p118-b2", - "global_id": 3253, - "bbox": [ - 276.73, - 98.61, - 490.38, - 108.98 - ], - "text": "kx −→ky\n(1.21)", - "type": "text" - }, - { - "block_id": "p118-b3", - "global_id": 3254, - "bbox": [ - 101.84, - 115.16, - 490.38, - 140.69 - ], - "text": "Thus, linearity implies two properties: homogeneity (scaling) and additivity.† Both these properties\ncan be combined into one property (superposition), which is expressed as follows: If", - "type": "text" - }, - { - "block_id": "p118-b4", - "global_id": 3255, - "bbox": [ - 231.03, - 153.06, - 360.71, - 164.52 - ], - "text": "x1 −→y1\nand\nx2 −→y2", - "type": "text" - }, - { - "block_id": "p118-b5", - "global_id": 3256, - "bbox": [ - 101.84, - 176.12, - 321.15, - 187.68 - ], - "text": "then for all inputs x1 and x2 and all constants k1 and k2,", - "type": "text" - }, - { - "block_id": "p118-b6", - "global_id": 3257, - "bbox": [ - 241.04, - 198.55, - 490.38, - 210.01 - ], - "text": "k1x1 + k2x2 −→k1y1 + k2y2\n(1.22)", - "type": "text" - }, - { - "block_id": "p118-b7", - "global_id": 3258, - "bbox": [ - 101.85, - 221.71, - 490.42, - 291.45 - ], - "text": "There is another useful way to view the linearity condition described in Eq. (1.22): the response of\na linear system is unchanged whether the operations of summing and scaling precede the system\n(sum and scale act on inputs) or follow the system (sum and scale act on outputs). Thus, linearity\nimplies commutability between a system and the operations of summing and scaling. It may appear\nthat additivity implies homogeneity. Unfortunately, homogeneity does not always follow from\nadditivity. Drill 1.11 demonstrates such a case.", - "type": "text" - }, - { - "block_id": "p118-b8", - "global_id": 3259, - "bbox": [ - 107.82, - 324.75, - 368.47, - 336.71 - ], - "text": "DRILL 1.11\nAdditivity but Not Homogeneity", - "type": "text" - }, - { - "block_id": "p118-b9", - "global_id": 3260, - "bbox": [ - 107.82, - 345.42, - 484.42, - 379.7 - ], - "text": "Show that a system with the input x(t) and the output y(t) related by y(t) = Re{x(t)} satisfies the\nadditivity property but violates the homogeneity property. Hence, such a system is not linear.\n[Hint: Show that Eq. (1.21) is not satisfied when k is complex.]", - "type": "text" - }, - { - "block_id": "p118-b10", - "global_id": 3261, - "bbox": [ - 102.14, - 410.95, - 274.93, - 423.07 - ], - "text": "RESPONSE OF A LINEAR SYSTEM", - "type": "text" - }, - { - "block_id": "p118-b11", - "global_id": 3262, - "bbox": [ - 101.84, - 427.0, - 490.39, - 449.02 - ], - "text": "For the sake of simplicity, we discuss only single-input, single-output (SISO) systems. But the\ndiscussion can be readily extended to multiple-input, multiple-output (MIMO) systems.", - "type": "text" - }, - { - "block_id": "p118-b12", - "global_id": 3263, - "bbox": [ - 101.84, - 450.6, - 490.39, - 544.66 - ], - "text": "A system’s output for t ≥0 is the result of two independent causes: the initial conditions of\nthe system (or the system state) at t = 0 and the input x(t) for t ≥0. If a system is to be linear, the\noutput must be the sum of the two components resulting from these two causes: first, the zero-input\nresponse (ZIR) that results only from the initial conditions at t = 0 with the input x(t) = 0 for t ≥0,\nand then the zero-state response (ZSR) that results only from the input x(t) for t ≥0 when the initial\nconditions (at t = 0) are assumed to be zero. When all the appropriate initial conditions are zero,\nthe system is said to be in zero state. The system output is zero when the input is zero only if the\nsystem is in zero state.", - "type": "text" - }, - { - "block_id": "p118-b13", - "global_id": 3264, - "bbox": [ - 101.85, - 546.65, - 490.4, - 568.58 - ], - "text": "In summary, a linear system response can be expressed as the sum of the zero-input and\nzero-state responses:", - "type": "text" - }, - { - "block_id": "p118-b14", - "global_id": 3265, - "bbox": [ - 180.8, - 581.36, - 411.44, - 591.32 - ], - "text": "total response = zero-input response + zero-state response", - "type": "text" - }, - { - "block_id": "p118-b15", - "global_id": 3266, - "bbox": [ - 101.84, - 610.24, - 490.38, - 633.41 - ], - "text": "† A linear system must also satisfy the additional condition of smoothness, where small changes in the\nsystem’s inputs must result in small changes in its outputs [3].", - "type": "text" - } - ] - }, - { - "page_num": 119, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p119-b0", - "global_id": 3267, - "bbox": [ - 355.12, - 62.89, - 516.13, - 71.98 - ], - "text": "1.7\nClassification of Systems\n99", - "type": "text" - }, - { - "block_id": "p119-b1", - "global_id": 3268, - "bbox": [ - 127.59, - 85.82, - 516.13, - 120.77 - ], - "text": "This property of linear systems, which permits the separation of an output into components\nresulting from the initial conditions and from the input, is called the decomposition property. For\nthe RC circuit of Fig. 1.26, the response y(t) was found to be [see Eq. (1.19) with t0 = 0]", - "type": "text" - }, - { - "block_id": "p119-b2", - "global_id": 3269, - "bbox": [ - 250.01, - 135.03, - 298.82, - 150.73 - ], - "text": "y(t) = vC(0)", - "type": "text" - }, - { - "block_id": "p119-b3", - "global_id": 3270, - "bbox": [ - 282.93, - 153.17, - 292.56, - 159.15 - ], - "text": "ZIR", - "type": "text" - }, - { - "block_id": "p119-b4", - "global_id": 3271, - "bbox": [ - 299.92, - 128.46, - 347.71, - 145.4 - ], - "text": "+Rx(t) + 1", - "type": "text" - }, - { - "block_id": "p119-b5", - "global_id": 3272, - "bbox": [ - 341.79, - 142.42, - 348.44, - 152.38 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p119-b6", - "global_id": 3273, - "bbox": [ - 350.95, - 121.47, - 362.72, - 133.75 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p119-b7", - "global_id": 3274, - "bbox": [ - 308.8, - 135.03, - 393.71, - 166.41 - ], - "text": "0\nx(τ)dτ\n\n\n\nZSR", - "type": "text" - }, - { - "block_id": "p119-b8", - "global_id": 3275, - "bbox": [ - 492.07, - 135.44, - 516.13, - 145.4 - ], - "text": "(1.23)", - "type": "text" - }, - { - "block_id": "p119-b9", - "global_id": 3276, - "bbox": [ - 127.59, - 173.97, - 516.14, - 220.21 - ], - "text": "From Eq. (1.23), it is clear that if the input x(t) = 0 for t ≥0, the output y(t) = vC(0). Hence vC(0)\nis the zero-input response of the response y(t). Similarly, if the system state (the voltage vC in\nthis case) is zero at t = 0, the output is given by the second component on the right-hand side of\nEq. (1.23). Clearly this is the zero-state response of the response y(t).", - "type": "text" - }, - { - "block_id": "p119-b10", - "global_id": 3277, - "bbox": [ - 127.59, - 222.21, - 516.16, - 303.9 - ], - "text": "In addition to the decomposition property, linearity implies that both the zero-input and\nzero-state components must obey the principle of superposition with respect to each of their\nrespective causes. For example, if we increase the initial condition k-fold, the zero-input response\nmust also increase k-fold. Similarly, if we increase the input k-fold, the zero-state response must\nalso increase k-fold. These facts can be readily verified from Eq. (1.23) for the RC circuit in\nFig. 1.26. For instance, if we double the initial condition vC(0), the zero-input response doubles;\nif we double the input x(t), the zero-state response doubles.", - "type": "text" - }, - { - "block_id": "p119-b11", - "global_id": 3278, - "bbox": [ - 102.51, - 331.9, - 492.46, - 357.8 - ], - "text": "EXAMPLE 1.10\nLinearity of Constant-Coefficient Linear Differential\nEquations", - "type": "text" - }, - { - "block_id": "p119-b12", - "global_id": 3279, - "bbox": [ - 128.9, - 374.05, - 318.19, - 384.01 - ], - "text": "Show that the system described by the equation", - "type": "text" - }, - { - "block_id": "p119-b13", - "global_id": 3280, - "bbox": [ - 277.26, - 392.76, - 297.04, - 403.03 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p119-b14", - "global_id": 3281, - "bbox": [ - 283.19, - 399.74, - 502.75, - 417.09 - ], - "text": "dt\n+ 3y(t) = x(t)\n(1.24)", - "type": "text" - }, - { - "block_id": "p119-b15", - "global_id": 3282, - "bbox": [ - 128.91, - 424.26, - 162.66, - 434.22 - ], - "text": "is linear.", - "type": "text" - }, - { - "block_id": "p119-b16", - "global_id": 3283, - "bbox": [ - 128.9, - 456.72, - 484.41, - 467.87 - ], - "text": "Let the system response to the inputs x1(t) and x2(t) be y1(t) and y2(t), respectively. Then", - "type": "text" - }, - { - "block_id": "p119-b17", - "global_id": 3284, - "bbox": [ - 198.46, - 475.84, - 222.22, - 486.98 - ], - "text": "dy1(t)", - "type": "text" - }, - { - "block_id": "p119-b18", - "global_id": 3285, - "bbox": [ - 206.38, - 475.84, - 367.93, - 500.17 - ], - "text": "dt\n+ 3y1(t) = x1(t)\nand\ndy2(t)", - "type": "text" - }, - { - "block_id": "p119-b19", - "global_id": 3286, - "bbox": [ - 352.07, - 482.82, - 434.42, - 500.17 - ], - "text": "dt\n+ 3y2(t) = x2(t)", - "type": "text" - }, - { - "block_id": "p119-b20", - "global_id": 3287, - "bbox": [ - 128.91, - 507.23, - 434.15, - 518.07 - ], - "text": "Multiplying the first equation by k1, the second by k2, and adding them yield", - "type": "text" - }, - { - "block_id": "p119-b21", - "global_id": 3288, - "bbox": [ - 193.48, - 526.15, - 439.39, - 550.17 - ], - "text": "d\ndt[k1y1(t) + k2y2(t)] + 3[k1y1(t) + k2y2(t)] = k1x1(t) + k2x2(t)", - "type": "text" - }, - { - "block_id": "p119-b22", - "global_id": 3289, - "bbox": [ - 128.9, - 557.34, - 357.2, - 567.3 - ], - "text": "But this equation is the system equation [Eq. (1.24)] with", - "type": "text" - }, - { - "block_id": "p119-b23", - "global_id": 3290, - "bbox": [ - 196.76, - 578.0, - 434.91, - 589.15 - ], - "text": "x(t) = k1x1(t) + k2x2(t)\nand\ny(t) = k1y1(t) + k2y2(t)", - "type": "text" - }, - { - "block_id": "p119-b24", - "global_id": 3291, - "bbox": [ - 128.91, - 599.08, - 502.79, - 621.41 - ], - "text": "Therefore, when the input is k1x1(t) + k2x2(t), the system response is k1y1(t) + k2y2(t).\nConsequently, the system is linear. Using this argument, we can readily generalize the result to", - "type": "text" - } - ] - }, - { - "page_num": 120, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p120-b0", - "global_id": 3292, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "100\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p120-b1", - "global_id": 3293, - "bbox": [ - 103.16, - 86.24, - 369.13, - 96.21 - ], - "text": "show that a system described by a differential equation of the form", - "type": "text" - }, - { - "block_id": "p120-b2", - "global_id": 3294, - "bbox": [ - 113.22, - 114.6, - 121.68, - 125.44 - ], - "text": "a0", - "type": "text" - }, - { - "block_id": "p120-b3", - "global_id": 3295, - "bbox": [ - 123.38, - 106.3, - 149.0, - 117.58 - ], - "text": "dNy(t)", - "type": "text" - }, - { - "block_id": "p120-b4", - "global_id": 3296, - "bbox": [ - 129.44, - 114.29, - 169.53, - 131.64 - ], - "text": "dtN\n+ a1", - "type": "text" - }, - { - "block_id": "p120-b5", - "global_id": 3297, - "bbox": [ - 171.23, - 106.09, - 205.78, - 117.58 - ], - "text": "dN−1y(t)", - "type": "text" - }, - { - "block_id": "p120-b6", - "global_id": 3298, - "bbox": [ - 177.3, - 114.29, - 300.3, - 131.64 - ], - "text": "dtN−1\n+ · · · + aNy(t) = bN−M", - "type": "text" - }, - { - "block_id": "p120-b7", - "global_id": 3299, - "bbox": [ - 302.28, - 106.3, - 328.95, - 117.58 - ], - "text": "dMx(t)", - "type": "text" - }, - { - "block_id": "p120-b8", - "global_id": 3300, - "bbox": [ - 308.36, - 114.29, - 383.91, - 131.64 - ], - "text": "dtM\n+ · · · + bN−1", - "type": "text" - }, - { - "block_id": "p120-b9", - "global_id": 3301, - "bbox": [ - 385.63, - 107.3, - 405.44, - 117.58 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p120-b10", - "global_id": 3302, - "bbox": [ - 391.57, - 114.29, - 477.01, - 131.64 - ], - "text": "dt\n+ bNx(t)\n(1.25)", - "type": "text" - }, - { - "block_id": "p120-b11", - "global_id": 3303, - "bbox": [ - 103.16, - 139.54, - 477.02, - 173.52 - ], - "text": "is a linear system. The coefficients ai and bi in this equation can be constants or functions of\ntime. Although here we proved only zero-state linearity, it can be shown that such systems are\nalso zero-input linear and have the decomposition property.", - "type": "text" - }, - { - "block_id": "p120-b12", - "global_id": 3304, - "bbox": [ - 107.82, - 247.03, - 410.73, - 272.93 - ], - "text": "DRILL 1.12\nLinearity of a Differential Equation with\nTime-Varying Parameters", - "type": "text" - }, - { - "block_id": "p120-b13", - "global_id": 3305, - "bbox": [ - 107.82, - 282.05, - 375.17, - 292.01 - ], - "text": "Show that the system described by the following equation is linear:", - "type": "text" - }, - { - "block_id": "p120-b14", - "global_id": 3306, - "bbox": [ - 240.96, - 301.59, - 260.74, - 311.87 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p120-b15", - "global_id": 3307, - "bbox": [ - 246.89, - 307.14, - 352.46, - 325.92 - ], - "text": "dt\n+ t2y(t) = (2t + 3)x(t)", - "type": "text" - }, - { - "block_id": "p120-b16", - "global_id": 3308, - "bbox": [ - 107.82, - 391.11, - 377.53, - 403.06 - ], - "text": "DRILL 1.13\nA Nonlinear Differential Equation", - "type": "text" - }, - { - "block_id": "p120-b17", - "global_id": 3309, - "bbox": [ - 107.82, - 412.18, - 390.11, - 422.14 - ], - "text": "Show that the system described by the following equation is nonlinear:", - "type": "text" - }, - { - "block_id": "p120-b18", - "global_id": 3310, - "bbox": [ - 248.95, - 431.72, - 284.73, - 448.98 - ], - "text": "y(t)dy(t)", - "type": "text" - }, - { - "block_id": "p120-b19", - "global_id": 3311, - "bbox": [ - 270.87, - 438.7, - 343.28, - 456.05 - ], - "text": "dt\n+ 3y(t) = x(t)", - "type": "text" - }, - { - "block_id": "p120-b20", - "global_id": 3312, - "bbox": [ - 101.84, - 489.12, - 490.41, - 634.79 - ], - "text": "MORE COMMENTS ON LINEAR SYSTEMS\nAlmost all systems observed in practice become nonlinear when large enough signals are applied\nto them. However, it is possible to approximate most of the nonlinear systems by linear systems for\nsmall-signal analysis. The analysis of nonlinear systems is generally difficult. Nonlinearities can\narise in so many ways that describing them with a common mathematical form is impossible. Not\nonly is each system a category in itself, but even for a given system, changes in initial conditions\nor input amplitudes may change the nature of the problem. On the other hand, the superposition\nproperty of linear systems is a powerful unifying principle that allows for a general solution.\nThe superposition property (linearity) greatly simplifies the analysis of linear systems. Because\nof the decomposition property, we can evaluate separately the two components of the output.\nThe zero-input response can be computed by assuming the input to be zero, and the zero-state\nresponse can be computed by assuming zero initial conditions. Moreover, if we express an", - "type": "text" - } - ] - }, - { - "page_num": 121, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p121-b0", - "global_id": 3313, - "bbox": [ - 355.12, - 62.89, - 516.13, - 71.98 - ], - "text": "1.7\nClassification of Systems\n101", - "type": "text" - }, - { - "block_id": "p121-b1", - "global_id": 3314, - "bbox": [ - 127.59, - 85.46, - 287.42, - 95.84 - ], - "text": "input x(t) as a sum of simpler functions,", - "type": "text" - }, - { - "block_id": "p121-b2", - "global_id": 3315, - "bbox": [ - 242.49, - 111.23, - 401.24, - 122.38 - ], - "text": "x(t) = a1x1(t) + a2x2(t) + · · · + amxm(t)", - "type": "text" - }, - { - "block_id": "p121-b3", - "global_id": 3316, - "bbox": [ - 127.59, - 137.01, - 348.87, - 147.38 - ], - "text": "then, by virtue of linearity, the response y(t) is given by", - "type": "text" - }, - { - "block_id": "p121-b4", - "global_id": 3317, - "bbox": [ - 242.51, - 162.77, - 401.22, - 173.92 - ], - "text": "y(t) = a1y1(t) + a2y2(t) + · · · + amym(t)", - "type": "text" - }, - { - "block_id": "p121-b5", - "global_id": 3318, - "bbox": [ - 127.59, - 188.55, - 516.16, - 222.84 - ], - "text": "where yk(t) is the zero-state response to an input xk(t). This apparently trivial observation has\nprofound implications. As we shall see repeatedly in later chapters, it proves extremely useful and\nopens new avenues for analyzing linear systems.", - "type": "text" - }, - { - "block_id": "p121-b6", - "global_id": 3319, - "bbox": [ - 127.59, - 224.42, - 516.16, - 366.3 - ], - "text": "For example, consider an arbitrary input x(t) such as the one shown in Fig. 1.27a. We can\napproximate x(t) with a sum of rectangular pulses of width t and of varying heights. The\napproximation improves as t →0, when the rectangular pulses become impulses spaced t\nseconds apart (with t →0).† Thus, an arbitrary input can be replaced by a weighted sum of\nimpulses spaced t (t →0) seconds apart. Therefore, if we know the system response to a\nunit impulse, we can immediately determine the system response to an arbitrary input x(t) by\nadding the system response to each impulse component of x(t). A similar situation is depicted\nin Fig. 1.27b, where x(t) is approximated by a sum of step functions of varying magnitude and\nspaced t seconds apart. The approximation improves as t becomes smaller. Therefore, if we\nknow the system response to a unit step input, we can compute the system response to any arbitrary\ninput x(t) with relative ease. Time-domain analysis of linear systems (discussed in Ch. 2) uses this\napproach.", - "type": "text" - }, - { - "block_id": "p121-b7", - "global_id": 3320, - "bbox": [ - 127.59, - 368.29, - 516.15, - 426.08 - ], - "text": "Chapters 4, 5, 6, and 7 employ the same approach but instead use sinusoids or exponentials\nas the basic signal components. We show that any arbitrary input signal can be expressed as a\nweighted sum of sinusoids (or exponentials) having various frequencies. Thus a knowledge of\nthe system response to a sinusoid enables us to determine the system response to an arbitrary\ninput x(t).", - "type": "text" - }, - { - "block_id": "p121-b8", - "global_id": 3321, - "bbox": [ - 151.71, - 451.07, - 162.82, - 459.15 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p121-b9", - "global_id": 3322, - "bbox": [ - 224.19, - 558.77, - 233.07, - 566.77 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p121-b10", - "global_id": 3323, - "bbox": [ - 181.6, - 534.65, - 275.05, - 553.16 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p121-b11", - "global_id": 3324, - "bbox": [ - 309.96, - 451.07, - 321.07, - 459.15 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p121-b12", - "global_id": 3325, - "bbox": [ - 382.16, - 558.77, - 391.81, - 566.77 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p121-b13", - "global_id": 3326, - "bbox": [ - 339.96, - 544.95, - 348.32, - 553.16 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p121-b14", - "global_id": 3327, - "bbox": [ - 431.17, - 534.65, - 433.4, - 542.65 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p121-b15", - "global_id": 3328, - "bbox": [ - 151.5, - 573.47, - 428.93, - 582.71 - ], - "text": "Figure 1.27 Signal representation in terms of impulse and step components.", - "type": "text" - }, - { - "block_id": "p121-b16", - "global_id": 3329, - "bbox": [ - 127.59, - 610.24, - 516.12, - 633.41 - ], - "text": "† Here, the discussion of a rectangular pulse approaching an impulse at t →0 is somewhat imprecise. It is\nexplained in Sec. 2.4 with more rigor.", - "type": "text" - } - ] - }, - { - "page_num": 122, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p122-b0", - "global_id": 3330, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "102\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p122-b1", - "global_id": 3331, - "bbox": [ - 101.84, - 86.52, - 364.36, - 98.48 - ], - "text": "1.7-2 Time-Invariant and Time-Varying Systems", - "type": "text" - }, - { - "block_id": "p122-b2", - "global_id": 3332, - "bbox": [ - 101.84, - 104.51, - 490.4, - 234.13 - ], - "text": "Systems whose parameters do not change with time are time-invariant (also constant-parameter)\nsystems. For such a system, if the input is delayed by T seconds, the output is the same as before\nbut delayed by T (assuming initial conditions are also delayed by T). This property is expressed\ngraphically in Fig. 1.28. We can also illustrate this property, as shown in Fig. 1.29. We can delay\nthe output y(t) of a system S by applying the output y(t) to a T second delay (Fig. 1.29a). If the\nsystem is time invariant, then the delayed output y(t−T) can also be obtained by first delaying the\ninput x(t) before applying it to the system, as shown in Fig. 1.29b. In other words, the system S and\nthe time delay commute if the system S is time invariant. This would not be true for time-varying\nsystems. Consider, for instance, a time-varying system specified by y(t) = e−tx(t). The output for\nsuch a system in Fig. 1.29a is e−(t−T)x(t −T). In contrast, the output for the system in Fig. 1.29b\nis e−tx(t −T).", - "type": "text" - }, - { - "block_id": "p122-b3", - "global_id": 3333, - "bbox": [ - 267.24, - 351.93, - 276.12, - 359.93 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p122-b4", - "global_id": 3334, - "bbox": [ - 266.86, - 472.78, - 276.51, - 480.78 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p122-b5", - "global_id": 3335, - "bbox": [ - 141.88, - 265.13, - 153.31, - 273.21 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p122-b6", - "global_id": 3336, - "bbox": [ - 150.59, - 334.69, - 237.87, - 344.43 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p122-b7", - "global_id": 3337, - "bbox": [ - 284.9, - 265.13, - 296.32, - 273.21 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p122-b8", - "global_id": 3338, - "bbox": [ - 293.77, - 334.69, - 381.05, - 344.43 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p122-b9", - "global_id": 3339, - "bbox": [ - 125.97, - 379.37, - 153.31, - 387.66 - ], - "text": "x(t T)", - "type": "text" - }, - { - "block_id": "p122-b10", - "global_id": 3340, - "bbox": [ - 150.59, - 454.83, - 237.87, - 464.42 - ], - "text": "t\n0\nT", - "type": "text" - }, - { - "block_id": "p122-b11", - "global_id": 3341, - "bbox": [ - 268.98, - 379.37, - 296.32, - 387.66 - ], - "text": "y(t T)", - "type": "text" - }, - { - "block_id": "p122-b12", - "global_id": 3342, - "bbox": [ - 293.77, - 454.68, - 398.98, - 464.42 - ], - "text": "t\n0\nT", - "type": "text" - }, - { - "block_id": "p122-b13", - "global_id": 3343, - "bbox": [ - 125.76, - 487.47, - 268.46, - 496.71 - ], - "text": "Figure 1.28 Time-invariance property.", - "type": "text" - }, - { - "block_id": "p122-b14", - "global_id": 3344, - "bbox": [ - 111.41, - 530.49, - 311.74, - 553.48 - ], - "text": "S\nx(t)\n \ny(t)\ny(t T)\nDelay\nT seconds", - "type": "text" - }, - { - "block_id": "p122-b15", - "global_id": 3345, - "bbox": [ - 114.62, - 585.69, - 251.39, - 606.85 - ], - "text": "S\nx(t)\nDelay\nT seconds", - "type": "text" - }, - { - "block_id": "p122-b16", - "global_id": 3346, - "bbox": [ - 190.31, - 585.47, - 311.74, - 593.77 - ], - "text": "y(t T)\nx(t T)", - "type": "text" - }, - { - "block_id": "p122-b17", - "global_id": 3347, - "bbox": [ - 205.51, - 563.98, - 214.39, - 571.98 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p122-b18", - "global_id": 3348, - "bbox": [ - 205.13, - 617.36, - 214.78, - 625.36 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p122-b19", - "global_id": 3349, - "bbox": [ - 326.8, - 604.7, - 454.89, - 626.62 - ], - "text": "Figure 1.29 Illustration of time-\ninvariance property.", - "type": "text" - } - ] - }, - { - "page_num": 123, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p123-b0", - "global_id": 3350, - "bbox": [ - 355.12, - 62.89, - 516.13, - 71.98 - ], - "text": "1.7\nClassification of Systems\n103", - "type": "text" - }, - { - "block_id": "p123-b1", - "global_id": 3351, - "bbox": [ - 127.59, - 85.82, - 516.14, - 155.56 - ], - "text": "It is possible to verify that the system in Fig. 1.26 is a time-invariant system. Networks\ncomposed of RLC elements and other commonly used active elements such as transistors\nare time-invariant systems. A system with an input–output relationship described by a linear\ndifferential equation of the form given in Ex. 1.10 [Eq. (1.25)] is a linear time-invariant (LTI)\nsystem when the coefficients ai and bi of such equation are constants. If these coefficients are\nfunctions of time, then the system is a linear time-varying system.", - "type": "text" - }, - { - "block_id": "p123-b2", - "global_id": 3352, - "bbox": [ - 127.59, - 157.55, - 516.15, - 203.37 - ], - "text": "The system described in Drill 1.12 is linear time varying. Another familiar example of a\ntime-varying system is the carbon microphone, in which the resistance R is a function of the\nmechanical pressure generated by sound waves on the carbon granules of the microphone. The\noutput current from the microphone is thus modulated by the sound waves, as desired.", - "type": "text" - }, - { - "block_id": "p123-b3", - "global_id": 3353, - "bbox": [ - 102.51, - 231.0, - 395.62, - 242.96 - ], - "text": "EXAMPLE 1.11\nAssessing System Time Invariance", - "type": "text" - }, - { - "block_id": "p123-b4", - "global_id": 3354, - "bbox": [ - 128.9, - 257.79, - 483.11, - 269.58 - ], - "text": "Determine the time invariance of the following systems: (a) y(t) = x(t)u(t) and (b) y(t) = d", - "type": "text" - }, - { - "block_id": "p123-b5", - "global_id": 3355, - "bbox": [ - 478.69, - 259.21, - 502.76, - 272.33 - ], - "text": "dtx(t).", - "type": "text" - }, - { - "block_id": "p123-b6", - "global_id": 3356, - "bbox": [ - 128.9, - 292.08, - 502.76, - 362.24 - ], - "text": "(a) In this case, the output equals the input for t ≥0 and is otherwise zero. Clearly,\nthe input is being modified by a time-dependent function, so the system is likely time\nvariant. We can prove that the system is not time invariant through a counterexample. Letting\nx1(t) = δ(t+1), we see that y1(t) = 0. However, x2(t) = x1(t−2) = δ(t−1) produces an output\nof y2(t) = δ(t −1), which does equal y1(t −2) = 0 as time-invariance would require. Thus,\ny(t) = x(t)u(t) is a time variant system.", - "type": "text" - }, - { - "block_id": "p123-b7", - "global_id": 3357, - "bbox": [ - 128.91, - 363.81, - 502.76, - 398.1 - ], - "text": "(b) Although it appears that x(t) is being modified by a time-dependent function, this is\nnot the case. The output of this system is simply the slope of the input. If the input is delayed,\nso too is the output. Applying input x(t) to the system produces output y(t) = d", - "type": "text" - }, - { - "block_id": "p123-b8", - "global_id": 3358, - "bbox": [ - 128.91, - 387.73, - 502.74, - 414.14 - ], - "text": "dtx(t); delaying\nthis output by T produces y(t −T) =\nd\nd(t−T)x(t −T) = d", - "type": "text" - }, - { - "block_id": "p123-b9", - "global_id": 3359, - "bbox": [ - 128.91, - 401.02, - 502.76, - 435.3 - ], - "text": "dtx(t −T). This is just the output of\nthe system to a delayed input x(t −T). Since the T-delayed output of the system to input x(t)\nequals the output of the system to the T-delayed input x(t −T), the system is time invariant.", - "type": "text" - }, - { - "block_id": "p123-b10", - "global_id": 3360, - "bbox": [ - 133.57, - 480.36, - 341.42, - 492.31 - ], - "text": "DRILL 1.14\nA Time-Variant System", - "type": "text" - }, - { - "block_id": "p123-b11", - "global_id": 3361, - "bbox": [ - 133.57, - 501.43, - 501.97, - 511.39 - ], - "text": "Show that a system described by the following equation is a time-varying-parameter system:", - "type": "text" - }, - { - "block_id": "p123-b12", - "global_id": 3362, - "bbox": [ - 281.07, - 522.94, - 362.64, - 533.32 - ], - "text": "y(t) = (sin t)x(t −2)", - "type": "text" - }, - { - "block_id": "p123-b13", - "global_id": 3363, - "bbox": [ - 133.57, - 545.17, - 423.86, - 555.23 - ], - "text": "[Hint: Show that the system fails to satisfy the time-invariance property.]", - "type": "text" - }, - { - "block_id": "p123-b14", - "global_id": 3364, - "bbox": [ - 127.59, - 594.78, - 359.53, - 606.73 - ], - "text": "1.7-3 Instantaneous and Dynamic Systems", - "type": "text" - }, - { - "block_id": "p123-b15", - "global_id": 3365, - "bbox": [ - 127.59, - 612.76, - 516.12, - 634.79 - ], - "text": "As observed earlier, a system’s output at any instant t generally depends on the entire past input.\nHowever, in a special class of systems, the output at any instant t depends only on its input at that", - "type": "text" - } - ] - }, - { - "page_num": 124, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p124-b0", - "global_id": 3366, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "104\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p124-b1", - "global_id": 3367, - "bbox": [ - 101.84, - 85.72, - 490.4, - 215.34 - ], - "text": "instant. In resistive networks, for example, any output of the network at some instant t depends\nonly on the input at the instant t. In these systems, past history is irrelevant in determining the\nresponse. Such systems are said to be instantaneous or memoryless systems. More precisely, a\nsystem is said to be instantaneous (or memoryless) if its output at any instant t depends, at most,\non the strength of its input(s) at the same instant t, and not on any past or future values of the\ninput(s). Otherwise, the system is said to be dynamic (or a system with memory). A system whose\nresponse at t is completely determined by the input signals over the past T seconds [interval from\n(t−T) to t] is a finite-memory system with a memory of T seconds. Networks containing inductive\nand capacitive elements generally have infinite memory because the response of such networks\nat any instant t is determined by their inputs over the entire past (−∞,t). This is true for the RC\ncircuit of Fig. 1.26.", - "type": "text" - }, - { - "block_id": "p124-b2", - "global_id": 3368, - "bbox": [ - 76.77, - 248.11, - 327.73, - 260.07 - ], - "text": "EXAMPLE 1.12\nAssessing System Memory", - "type": "text" - }, - { - "block_id": "p124-b3", - "global_id": 3369, - "bbox": [ - 103.16, - 276.32, - 477.02, - 298.55 - ], - "text": "Determine whether the following systems are memoryless: (a) y(t −1) = 2x(t −1), (b)\ny(t) = d", - "type": "text" - }, - { - "block_id": "p124-b4", - "global_id": 3370, - "bbox": [ - 131.03, - 288.28, - 258.23, - 301.4 - ], - "text": "dtx(t), and (c) y(t) = (t −1)x(t).", - "type": "text" - }, - { - "block_id": "p124-b5", - "global_id": 3371, - "bbox": [ - 103.16, - 321.15, - 477.0, - 355.44 - ], - "text": "(a) In this case, the output at time t −1 is just twice the input at the same time t −1. Since\nthe output at a particular time depends only on the strength of the input at the same time, the\nsystem is memoryless.", - "type": "text" - }, - { - "block_id": "p124-b6", - "global_id": 3372, - "bbox": [ - 103.16, - 357.02, - 477.03, - 403.26 - ], - "text": "(b) Although it appears that the output y(t) at time t depends on the input x(t) at the same\ntime t, we know that the slope (derivative) of x(t) cannot be determined solely from a single\npoint. There must be some memory, even if infinitesimally small, involved. This is confirmed\nby using the fundamental theorem of calculus to express the system as", - "type": "text" - }, - { - "block_id": "p124-b7", - "global_id": 3373, - "bbox": [ - 238.36, - 419.83, - 279.47, - 430.21 - ], - "text": "y(t) = lim", - "type": "text" - }, - { - "block_id": "p124-b8", - "global_id": 3374, - "bbox": [ - 265.04, - 428.74, - 280.61, - 436.01 - ], - "text": "T→0", - "type": "text" - }, - { - "block_id": "p124-b9", - "global_id": 3375, - "bbox": [ - 282.9, - 412.84, - 340.61, - 423.12 - ], - "text": "x(t) −x(t −T)", - "type": "text" - }, - { - "block_id": "p124-b10", - "global_id": 3376, - "bbox": [ - 308.61, - 427.22, - 314.15, - 437.18 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p124-b11", - "global_id": 3377, - "bbox": [ - 103.17, - 445.69, - 477.01, - 467.6 - ], - "text": "Since the output at a particular time depends on more than just the input at the same time, the\nsystem is not memoryless.", - "type": "text" - }, - { - "block_id": "p124-b12", - "global_id": 3378, - "bbox": [ - 103.17, - 469.18, - 477.03, - 503.47 - ], - "text": "(c) The output y(t) at time t is just the input x(t) at the same time t multiplied by the\n(time-dependent) coefficient t −1. Since the output at a particular time depends only on the\nstrength of the input at the same time, the system is memoryless.", - "type": "text" - }, - { - "block_id": "p124-b13", - "global_id": 3379, - "bbox": [ - 101.84, - 558.91, - 301.89, - 570.87 - ], - "text": "1.7-4 Causal and Noncausal Systems", - "type": "text" - }, - { - "block_id": "p124-b14", - "global_id": 3380, - "bbox": [ - 101.84, - 576.9, - 490.4, - 634.79 - ], - "text": "A causal (also known as a physical or nonanticipative) system is one for which the output at any\ninstant t0 depends only on the value of the input x(t) for t ≤t0. In other words, the value of the\noutput at the present instant depends only on the past and present values of the input x(t), not on\nits future values. To put it simply, in a causal system the output cannot start before the input is\napplied. If the response starts before the input, it means that the system knows the input in the", - "type": "text" - } - ] - }, - { - "page_num": 125, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p125-b0", - "global_id": 3381, - "bbox": [ - 355.12, - 62.89, - 516.13, - 71.98 - ], - "text": "1.7\nClassification of Systems\n105", - "type": "text" - }, - { - "block_id": "p125-b2", - "global_id": 3382, - "bbox": [ - 240.62, - 253.99, - 346.1, - 262.19 - ], - "text": "t\n0\n1", - "type": "text" - }, - { - "block_id": "p125-b3", - "global_id": 3383, - "bbox": [ - 232.82, - 211.39, - 236.82, - 219.39 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p125-b4", - "global_id": 3384, - "bbox": [ - 285.98, - 267.74, - 294.86, - 275.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p125-b5", - "global_id": 3385, - "bbox": [ - 273.69, - 254.19, - 333.02, - 262.19 - ], - "text": "2\n3\n4\n5", - "type": "text" - }, - { - "block_id": "p125-b6", - "global_id": 3386, - "bbox": [ - 242.23, - 189.36, - 253.34, - 197.44 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p125-b7", - "global_id": 3387, - "bbox": [ - 345.01, - 87.69, - 356.12, - 95.77 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p125-b8", - "global_id": 3388, - "bbox": [ - 343.45, - 153.35, - 413.02, - 162.05 - ], - "text": "t\n0\n1", - "type": "text" - }, - { - "block_id": "p125-b9", - "global_id": 3389, - "bbox": [ - 335.65, - 109.69, - 339.65, - 117.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p125-b10", - "global_id": 3390, - "bbox": [ - 352.52, - 167.66, - 362.17, - 175.66 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p125-b11", - "global_id": 3391, - "bbox": [ - 298.12, - 153.35, - 400.03, - 161.74 - ], - "text": "2\n3\n2\n1", - "type": "text" - }, - { - "block_id": "p125-b12", - "global_id": 3392, - "bbox": [ - 204.96, - 167.66, - 213.84, - 175.66 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p125-b13", - "global_id": 3393, - "bbox": [ - 189.7, - 153.35, - 210.1, - 161.35 - ], - "text": "0\n1", - "type": "text" - }, - { - "block_id": "p125-b14", - "global_id": 3394, - "bbox": [ - 181.9, - 110.55, - 185.9, - 118.55 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p125-b15", - "global_id": 3395, - "bbox": [ - 248.48, - 153.15, - 250.71, - 161.15 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p125-b16", - "global_id": 3396, - "bbox": [ - 191.27, - 87.69, - 202.38, - 95.77 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p125-b17", - "global_id": 3397, - "bbox": [ - 151.5, - 282.44, - 475.55, - 291.68 - ], - "text": "Figure 1.30 Input–output of a noncausal system and the causal output achieved by delay.", - "type": "text" - }, - { - "block_id": "p125-b18", - "global_id": 3398, - "bbox": [ - 127.59, - 311.51, - 516.11, - 333.42 - ], - "text": "future and acts on this knowledge before the input is applied. A system that violates the condition\nof causality is called a noncausal (or anticipative) system.", - "type": "text" - }, - { - "block_id": "p125-b19", - "global_id": 3399, - "bbox": [ - 127.59, - 331.8, - 516.14, - 393.2 - ], - "text": "Any practical system that operates in real time† must necessarily be causal. We do not yet\nknow how to build a system that can respond to future inputs (inputs not yet applied). A noncausal\nsystem is a prophetic system that knows the future input and acts on it in the present. Thus, if we\napply an input starting at t = 0 to a noncausal system, the output would begin even before t = 0.\nFor example, consider the system specified by", - "type": "text" - }, - { - "block_id": "p125-b20", - "global_id": 3400, - "bbox": [ - 272.4, - 403.39, - 516.13, - 413.77 - ], - "text": "y(t) = x(t −2) + x(t + 2)\n(1.26)", - "type": "text" - }, - { - "block_id": "p125-b21", - "global_id": 3401, - "bbox": [ - 127.6, - 423.95, - 516.15, - 494.1 - ], - "text": "For the input x(t) illustrated in Fig. 1.30a, the output y(t), as computed from Eq. (1.26) (shown in\nFig. 1.30b), starts even before the input is applied. Equation (1.26) shows that y(t), the output at t,\nis given by the sum of the input values 2 seconds before and 2 seconds after t (at t −2 and t + 2,\nrespectively). But if we are operating the system in real time at t, we do not know what the value\nof the input will be 2 seconds later. Thus it is impossible to implement this system in real time.\nFor this reason, noncausal systems are unrealizable in real time.", - "type": "text" - }, - { - "block_id": "p125-b22", - "global_id": 3402, - "bbox": [ - 102.51, - 522.42, - 358.73, - 534.38 - ], - "text": "EXAMPLE 1.13\nAssessing System Causality", - "type": "text" - }, - { - "block_id": "p125-b23", - "global_id": 3403, - "bbox": [ - 128.9, - 550.63, - 502.76, - 572.96 - ], - "text": "Determine whether the following systems are causal: (a) y(t) = x(−t), (b) y(t) = x(t + 1), and\n(c) y(t + 1) = x(t).", - "type": "text" - }, - { - "block_id": "p125-b24", - "global_id": 3404, - "bbox": [ - 127.59, - 620.2, - 516.15, - 643.38 - ], - "text": "† In real-time operations, the response to an input is essentially simultaneous (contemporaneous) with the\ninput itself.", - "type": "text" - } - ] - }, - { - "page_num": 126, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p126-b0", - "global_id": 3405, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "106\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p126-b1", - "global_id": 3406, - "bbox": [ - 103.16, - 86.29, - 477.02, - 132.2 - ], - "text": "(a) Here, the output is a reflection of the input. We can easily use a counterexample to\ndisprove the causality of this system. The input x(t) = δ(t −1), which is nonzero at t = 1,\nproduces an output y(t) = δ(t + 1), which is nonzero at t = −1, a time 2 seconds earlier than\nthe input! Clearly the system is not causal.", - "type": "text" - }, - { - "block_id": "p126-b2", - "global_id": 3407, - "bbox": [ - 103.17, - 133.77, - 477.02, - 156.11 - ], - "text": "(b) In this case, the output at time t depends on the input at future time of t + 1. Clearly\nthe system is not causal.", - "type": "text" - }, - { - "block_id": "p126-b3", - "global_id": 3408, - "bbox": [ - 103.17, - 157.69, - 477.01, - 180.01 - ], - "text": "(c) In this case, the output at time t + 1 depends on the input one second in the past, at\ntime t. Since the output does not depend on future values of the input, the system is causal.", - "type": "text" - }, - { - "block_id": "p126-b4", - "global_id": 3409, - "bbox": [ - 101.84, - 219.82, - 490.42, - 365.49 - ], - "text": "WHY STUDY NONCAUSAL SYSTEMS?\nThe foregoing discussion may suggest that noncausal systems have no practical purpose. This\nis not the case; they are valuable in the study of systems for several reasons. First, noncausal\nsystems are realizable when the independent variable is other than “time” (e.g., space). Consider,\nfor example, an electric charge of density q(x) placed along the x axis for x ≥0. This charge\ndensity produces an electric field E(x) that is present at every point on the x axis from x = −∞to\n∞. In this case the input [i.e., the charge density q(x)] starts at x = 0, but its output [the electric\nfield E(x)] begins before x = 0. Clearly, this space-charge system is noncausal. This discussion\nshows that only temporal systems (systems with time as independent variable) must be causal to\nbe realizable. The terms “before” and “after” have a special connection to causality only when the\nindependent variable is time. This connection is lost for variables other than time. Nontemporal\nsystems, such as those occurring in optics, can be noncausal and still realizable.", - "type": "text" - }, - { - "block_id": "p126-b5", - "global_id": 3410, - "bbox": [ - 101.85, - 367.48, - 490.4, - 485.04 - ], - "text": "Moreover, even for temporal systems, such as those used for signal processing, the study of\nnoncausal systems is important. In such systems we may have all input data prerecorded. This\noften happens with speech, geophysical, and meteorological signals, and with space probes. In\nsuch cases, the input’s future values are available to us. For example, suppose we had a set of\ninput signal records available for the system described by Eq. (1.26). We can then compute y(t)\nsince, for any t, we need only refer to the records to find the input’s value 2 seconds before and\n2 seconds after t. Thus, noncausal systems can be realized, although not in real time. We may\ntherefore be able to realize a noncausal system, provided we are willing to accept a time delay\nin the output. Consider a system whose output ˆy(t) is the same as y(t) in Eq. (1.26) delayed by\n2 seconds (Fig. 1.30c), so that", - "type": "text" - }, - { - "block_id": "p126-b6", - "global_id": 3411, - "bbox": [ - 233.33, - 501.71, - 358.89, - 512.09 - ], - "text": "ˆy(t) = y(t −2) = x(t −4) + x(t)", - "type": "text" - }, - { - "block_id": "p126-b7", - "global_id": 3412, - "bbox": [ - 101.85, - 528.77, - 490.39, - 598.92 - ], - "text": "Here the value of the output ˆy at any instant t is the sum of the values of the input x at t and at\nthe instant 4 seconds earlier [at (t −4)]. In this case, the output at any instant t does not depend\non future values of the input, and the system is causal. The output of this system, which is ˆy(t),\nis identical to that in Eq. (1.26) or Fig. 1.30b except for a delay of 2 seconds. Thus, a noncausal\nsystem may be realized or satisfactorily approximated in real time by using a causal system with\na delay.", - "type": "text" - }, - { - "block_id": "p126-b8", - "global_id": 3413, - "bbox": [ - 101.85, - 600.91, - 490.42, - 634.79 - ], - "text": "A third reason for studying noncausal systems is that they provide an upper bound on the\nperformance of causal systems. For example, if we wish to design a filter for separating a signal\nfrom noise, then the optimum filter is invariably a noncausal system. Although unrealizable, this", - "type": "text" - } - ] - }, - { - "page_num": 127, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p127-b0", - "global_id": 3414, - "bbox": [ - 355.12, - 62.89, - 516.13, - 71.98 - ], - "text": "1.7\nClassification of Systems\n107", - "type": "text" - }, - { - "block_id": "p127-b1", - "global_id": 3415, - "bbox": [ - 221.98, - 225.99, - 421.74, - 235.95 - ], - "text": "Noncausal systems are realizable with time delay!", - "type": "text" - }, - { - "block_id": "p127-b2", - "global_id": 3416, - "bbox": [ - 127.59, - 258.12, - 516.16, - 280.03 - ], - "text": "noncausal system’s performance acts as the upper limit on what can be achieved and gives us a\nstandard for evaluating the performance of causal filters.", - "type": "text" - }, - { - "block_id": "p127-b3", - "global_id": 3417, - "bbox": [ - 127.59, - 282.03, - 516.17, - 363.73 - ], - "text": "At first glance, noncausal systems may seem to be inscrutable. Actually, there is nothing\nmysterious about these systems and their approximate realization through physical systems with\ndelay. If we want to know what will happen one year from now, we have two choices: go to a\nprophet (an unrealizable person) who can give the answers instantly, or go to a wise man and\nallow him a delay of one year to give us the answer! If the wise man is truly wise, he may even be\nable, by studying trends, to shrewdly guess the future very closely with a delay of less than a year.\nSuch is the case with noncausal systems—nothing more and nothing less.", - "type": "text" - }, - { - "block_id": "p127-b4", - "global_id": 3418, - "bbox": [ - 133.57, - 400.24, - 327.45, - 412.19 - ], - "text": "DRILL 1.15\nA Noncausal System", - "type": "text" - }, - { - "block_id": "p127-b5", - "global_id": 3419, - "bbox": [ - 133.57, - 421.31, - 410.32, - 431.28 - ], - "text": "Show that a system described by the following equation is noncausal:", - "type": "text" - }, - { - "block_id": "p127-b6", - "global_id": 3420, - "bbox": [ - 282.68, - 449.63, - 307.31, - 459.9 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p127-b7", - "global_id": 3421, - "bbox": [ - 309.36, - 436.07, - 330.16, - 448.42 - ], - "text": "# t+5", - "type": "text" - }, - { - "block_id": "p127-b8", - "global_id": 3422, - "bbox": [ - 314.61, - 460.95, - 325.6, - 468.21 - ], - "text": "t−5", - "type": "text" - }, - { - "block_id": "p127-b9", - "global_id": 3423, - "bbox": [ - 331.78, - 449.63, - 359.83, - 459.9 - ], - "text": "x(τ)dτ", - "type": "text" - }, - { - "block_id": "p127-b10", - "global_id": 3424, - "bbox": [ - 133.57, - 477.29, - 510.15, - 487.26 - ], - "text": "Show that this system can be realized physically if we accept a delay of 5 seconds in the output.", - "type": "text" - }, - { - "block_id": "p127-b11", - "global_id": 3425, - "bbox": [ - 127.59, - 535.01, - 405.99, - 546.96 - ], - "text": "1.7-5 Continuous-Time and Discrete-Time Systems", - "type": "text" - }, - { - "block_id": "p127-b12", - "global_id": 3426, - "bbox": [ - 127.59, - 553.0, - 516.16, - 634.79 - ], - "text": "Signals defined or specified over a continuous range of time are continuous-time signals, denoted\nby symbols x(t), y(t), and so on. Systems whose inputs and outputs are continuous-time signals\nare continuous-time systems. On the other hand, signals defined only at discrete instants of time\nt0, t1, t2,. . .,tn,. . . are discrete-time signals, denoted by the symbols x(tn), y(tn), and so on, where\nn is some integer. Systems whose inputs and outputs are discrete-time signals are discrete-time\nsystems. A digital computer is a familiar example of this type of system. In practice, discrete-time\nsignals can arise from sampling continuous-time signals. For example, when the sampling is", - "type": "text" - } - ] - }, - { - "page_num": 128, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p128-b0", - "global_id": 3427, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "108\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p128-b1", - "global_id": 3428, - "bbox": [ - 101.84, - 85.4, - 383.79, - 96.55 - ], - "text": "uniform, the discrete instants t0, t1, t2, . . . are uniformly spaced so that", - "type": "text" - }, - { - "block_id": "p128-b2", - "global_id": 3429, - "bbox": [ - 245.11, - 108.86, - 346.95, - 120.32 - ], - "text": "tk+1 −tk = T\nfor all k", - "type": "text" - }, - { - "block_id": "p128-b3", - "global_id": 3430, - "bbox": [ - 101.84, - 132.73, - 490.4, - 214.44 - ], - "text": "In such case, the discrete-time signals represented by the samples of continuous-time signals\nx(t),y(t), and so on can be expressed as x(nT), y(nT), and so on; for convenience, we further\nsimplify this notation to x[n], y[n], . . ., where it is understood that x[n] = x(nT) and that n is some\ninteger. A typical discrete-time signal is shown in Fig. 1.31. A discrete-time signal may also be\nviewed as a sequence of numbers . . ., x[−1], x[0], x[1], x[2], . . .. Thus, a discrete-time system\nmay be seen as processing a sequence of numbers x[n] and yielding as an output another sequence\nof numbers y[n].", - "type": "text" - }, - { - "block_id": "p128-b4", - "global_id": 3431, - "bbox": [ - 101.85, - 216.42, - 490.41, - 345.93 - ], - "text": "Discrete-time signals arise naturally in situations that are inherently discrete time, such as\npopulation studies, amortization problems, national income models, and radar tracking. They may\nalso arise as a result of sampling continuous-time signals in sampled data systems, digital filtering,\nand the like. Digital filtering is a particularly interesting application in which continuous-time\nsignals are processed by using discrete-time systems, as shown in Fig. 1.32. A continuous-time\nsignal x(t) is first sampled to convert it into a discrete-time signal x[n], which then is processed\nby the discrete-time system to yield a discrete-time output y[n]. A continuous-time signal y(t) is\nfinally constructed from y[n]. In this manner, we can process a continuous-time signal with an\nappropriate discrete-time system such as a digital computer. Because discrete-time systems have\nseveral significant advantages over continuous-time systems, there is an accelerating trend toward\nprocessing continuous-time signals with discrete-time systems.", - "type": "text" - }, - { - "block_id": "p128-b5", - "global_id": 3432, - "bbox": [ - 138.43, - 365.95, - 184.09, - 374.4 - ], - "text": "x[n] or x(nT)", - "type": "text" - }, - { - "block_id": "p128-b6", - "global_id": 3433, - "bbox": [ - 130.51, - 424.69, - 287.35, - 433.25 - ], - "text": "n\n1\n5\n10\n2", - "type": "text" - }, - { - "block_id": "p128-b7", - "global_id": 3434, - "bbox": [ - 103.84, - 410.52, - 280.34, - 418.52 - ], - "text": "•••\n•••", - "type": "text" - }, - { - "block_id": "p128-b8", - "global_id": 3435, - "bbox": [ - 128.29, - 444.75, - 287.12, - 453.32 - ], - "text": "t\nT\n5T\n10T\n2T", - "type": "text" - }, - { - "block_id": "p128-b9", - "global_id": 3436, - "bbox": [ - 103.84, - 430.02, - 280.34, - 438.02 - ], - "text": "•••\n•••", - "type": "text" - }, - { - "block_id": "p128-b10", - "global_id": 3437, - "bbox": [ - 313.8, - 445.37, - 445.31, - 454.6 - ], - "text": "Figure 1.31 A discrete-time signal.", - "type": "text" - }, - { - "block_id": "p128-b11", - "global_id": 3438, - "bbox": [ - 172.53, - 565.84, - 440.64, - 576.55 - ], - "text": "x[n]\nx(t)\ny[n]\ny(t)\nContinuous to", - "type": "text" - }, - { - "block_id": "p128-b12", - "global_id": 3439, - "bbox": [ - 210.82, - 577.55, - 237.7, - 585.55 - ], - "text": "discrete,", - "type": "text" - }, - { - "block_id": "p128-b13", - "global_id": 3440, - "bbox": [ - 217.59, - 586.55, - 230.92, - 594.55 - ], - "text": "C/D", - "type": "text" - }, - { - "block_id": "p128-b14", - "global_id": 3441, - "bbox": [ - 281.48, - 573.05, - 325.03, - 581.05 - ], - "text": "Discrete-time", - "type": "text" - }, - { - "block_id": "p128-b15", - "global_id": 3442, - "bbox": [ - 292.14, - 582.05, - 314.37, - 590.05 - ], - "text": "system", - "type": "text" - }, - { - "block_id": "p128-b16", - "global_id": 3443, - "bbox": [ - 365.7, - 568.8, - 402.81, - 585.8 - ], - "text": "Discrete to \ncontinuous,", - "type": "text" - }, - { - "block_id": "p128-b17", - "global_id": 3444, - "bbox": [ - 377.59, - 586.8, - 390.92, - 594.8 - ], - "text": "D/C", - "type": "text" - }, - { - "block_id": "p128-b18", - "global_id": 3445, - "bbox": [ - 125.76, - 619.96, - 395.5, - 629.19 - ], - "text": "Figure 1.32 Processing continuous-time signals by discrete-time systems.", - "type": "text" - } - ] - }, - { - "page_num": 129, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p129-b0", - "global_id": 3446, - "bbox": [ - 355.12, - 62.89, - 516.13, - 71.98 - ], - "text": "1.7\nClassification of Systems\n109", - "type": "text" - }, - { - "block_id": "p129-b1", - "global_id": 3447, - "bbox": [ - 127.59, - 86.52, - 311.04, - 98.48 - ], - "text": "1.7-6 Analog and Digital Systems", - "type": "text" - }, - { - "block_id": "p129-b2", - "global_id": 3448, - "bbox": [ - 127.59, - 104.61, - 516.17, - 150.44 - ], - "text": "Analog and digital signals are discussed in Sec. 1.3-2. A system whose input and output signals are\nanalog is an analog system; a system whose input and output signals are digital is a digital system.\nA digital computer is an example of a digital (binary) system. Observe that a digital computer is a\ndigital as well as a discrete-time system.", - "type": "text" - }, - { - "block_id": "p129-b3", - "global_id": 3449, - "bbox": [ - 127.59, - 170.48, - 364.82, - 182.44 - ], - "text": "1.7-7 Invertible and Noninvertible Systems", - "type": "text" - }, - { - "block_id": "p129-b4", - "global_id": 3450, - "bbox": [ - 127.59, - 188.15, - 516.15, - 341.99 - ], - "text": "A system S performs certain operation(s) on input signal(s). If we can obtain the input x(t) back\nfrom the corresponding output y(t) by some operation, the system S is said to be invertible. When\nseveral different inputs result in the same output (as in a rectifier), it is impossible to obtain the\ninput from the output, and the system is noninvertible. Therefore, for an invertible system, it is\nessential that every input have a unique output so that there is a one-to-one mapping between an\ninput and the corresponding output. The system that achieves the inverse operation [of obtaining\nx(t) from y(t)] is the inverse system for S. For instance, if S is an ideal integrator, then its inverse\nsystem is an ideal differentiator. Consider a system S connected in tandem with its inverse Si, as\nshown in Fig. 1.33. The input x(t) to this tandem system results in signal y(t) at the output of S,\nand the signal y(t), which now acts as an input to Si, yields back the signal x(t) at the output of Si.\nThus, Si undoes the operation of S on x(t), yielding back x(t). A system whose output is equal to\nthe input (for all possible inputs) is an identity system. Cascading a system with its inverse system,\nas shown in Fig. 1.33, results in an identity system.", - "type": "text" - }, - { - "block_id": "p129-b5", - "global_id": 3451, - "bbox": [ - 127.59, - 343.57, - 516.12, - 365.9 - ], - "text": "In contrast, a rectifier, specified by an equation y(t) = |x(t)|, is noninvertible because the\nrectification operation cannot be undone.", - "type": "text" - }, - { - "block_id": "p129-b6", - "global_id": 3452, - "bbox": [ - 127.59, - 367.89, - 516.17, - 425.67 - ], - "text": "Inverse systems are very important in signal processing. In many applications, the signals are\ndistorted during the processing, and it is necessary to undo the distortion. For instance, in transmis-\nsion of data over a communication channel, the signals are distorted owing to non-ideal frequency\nresponse and finite bandwidth of a channel. It is necessary to restore the signal as closely as possi-\nble to its original shape. Such equalization is also used in audio systems and photographic systems.", - "type": "text" - }, - { - "block_id": "p129-b7", - "global_id": 3453, - "bbox": [ - 134.6, - 447.96, - 512.15, - 473.03 - ], - "text": "Si\nS\nx(t)\ny(t)\nx(t)\nFigure 1.33 A cascade of a system with its\ninverse results in an identity system.", - "type": "text" - }, - { - "block_id": "p129-b8", - "global_id": 3454, - "bbox": [ - 102.51, - 514.09, - 374.01, - 526.05 - ], - "text": "EXAMPLE 1.14\nAssessing System Invertibility", - "type": "text" - }, - { - "block_id": "p129-b9", - "global_id": 3455, - "bbox": [ - 128.9, - 542.29, - 502.75, - 564.55 - ], - "text": "Determine whether the following systems are invertible: (a) y(t) = x(−t), (b) y(t) = tx(t), and\n(c) y(t) = d", - "type": "text" - }, - { - "block_id": "p129-b10", - "global_id": 3456, - "bbox": [ - 170.32, - 554.25, - 194.4, - 567.37 - ], - "text": "dtx(t).", - "type": "text" - }, - { - "block_id": "p129-b11", - "global_id": 3457, - "bbox": [ - 128.9, - 587.46, - 502.76, - 633.37 - ], - "text": "(a) Here, the output is a reflection of the input, which does not cause any loss to the input.\nThe input can, in fact, be exactly recovered by simply reflecting the output [x(t) = y(−t)],\nwhich is to say that a reflecting system is its own inverse. Thus, y(t) = x(−t) is an invertible\nsystem.", - "type": "text" - } - ] - }, - { - "page_num": 130, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p130-b0", - "global_id": 3458, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "110\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p130-b1", - "global_id": 3459, - "bbox": [ - 121.09, - 86.47, - 458.52, - 98.19 - ], - "text": "(b) In this case, one might be tempted to recover the input from the output as x(t) = 1", - "type": "text" - }, - { - "block_id": "p130-b2", - "global_id": 3460, - "bbox": [ - 103.17, - 87.81, - 477.02, - 122.1 - ], - "text": "t y(t).\nThis approach works almost everywhere, except at t = 0 where the input value x(0) cannot be\nrecovered. Due to this single lost point, the system y(t) = tx(t) is not invertible.", - "type": "text" - }, - { - "block_id": "p130-b3", - "global_id": 3461, - "bbox": [ - 103.16, - 123.68, - 477.01, - 169.92 - ], - "text": "(c) Differentiation eliminates any dc component. For example, the inputs x1(t) = 1 and\nx2(t) = 2 both produce the same output y(t) = 0. Given only y(t) = 0, it is impossible to know\nif the original input was x1(t) = 1, x2(t) = 2, or something else entirely. Since unique inputs do\nproduce unique outputs, we know that y(t) = d", - "type": "text" - }, - { - "block_id": "p130-b4", - "global_id": 3462, - "bbox": [ - 285.46, - 159.54, - 416.09, - 172.66 - ], - "text": "dtx(t) is not an invertible system.", - "type": "text" - }, - { - "block_id": "p130-b5", - "global_id": 3463, - "bbox": [ - 101.84, - 211.24, - 290.6, - 223.2 - ], - "text": "1.7-8 Stable and Unstable Systems", - "type": "text" - }, - { - "block_id": "p130-b6", - "global_id": 3464, - "bbox": [ - 101.84, - 229.23, - 490.39, - 311.02 - ], - "text": "Systems can also be classified as stable or unstable systems. Stability can be internal or external.\nIf every bounded input applied at the input terminal results in a bounded output, the system is\nsaid to be stable externally. External stability can be ascertained by measurements at the external\nterminals (input and output) of the system. This type of stability is also known as the stability in\nthe BIBO (bounded-input/bounded-output) sense. The concept of internal stability is postponed\nto Ch. 2 because it requires some understanding of internal system behavior, introduced in that\nchapter.", - "type": "text" - }, - { - "block_id": "p130-b7", - "global_id": 3465, - "bbox": [ - 76.77, - 335.55, - 361.89, - 347.51 - ], - "text": "EXAMPLE 1.15\nAssessing System BIBO Stability", - "type": "text" - }, - { - "block_id": "p130-b8", - "global_id": 3466, - "bbox": [ - 103.16, - 362.82, - 477.01, - 386.09 - ], - "text": "Determine whether the following systems are BIBO-stable: (a) y(t) = x2(t), (b) y(t) = tx(t),\nand (c) y(t) = d", - "type": "text" - }, - { - "block_id": "p130-b9", - "global_id": 3467, - "bbox": [ - 161.46, - 375.71, - 185.53, - 388.83 - ], - "text": "dtx(t).", - "type": "text" - }, - { - "block_id": "p130-b10", - "global_id": 3468, - "bbox": [ - 103.16, - 408.92, - 477.03, - 432.0 - ], - "text": "(a) This system squares an input to produce the output. If the input is bounded, which is\nto say that |x(t)| ≤Mx < ∞for all t, then we see that", - "type": "text" - }, - { - "block_id": "p130-b11", - "global_id": 3469, - "bbox": [ - 220.29, - 438.35, - 337.35, - 452.74 - ], - "text": "|y(t)| = |x2(t)| = |x(t)|2 ≤M2", - "type": "text" - }, - { - "block_id": "p130-b12", - "global_id": 3470, - "bbox": [ - 333.47, - 442.47, - 359.89, - 455.1 - ], - "text": "x < ∞", - "type": "text" - }, - { - "block_id": "p130-b13", - "global_id": 3471, - "bbox": [ - 103.17, - 464.8, - 477.03, - 486.71 - ], - "text": "Since the output amplitude is guaranteed to be bounded for any bounded-amplitude input, the\nsystem y(t) = x2(t) is BIBO-stable.", - "type": "text" - }, - { - "block_id": "p130-b14", - "global_id": 3472, - "bbox": [ - 103.16, - 488.29, - 477.02, - 522.58 - ], - "text": "(b) We can prove that y(t) = tx(t) is not BIBO-stable with a simple example. The\nbounded-amplitude input x(t) = u(t) produces the output y(t) = tu(t) whose amplitude grows\nto infinity as t →∞. Thus, y(t) = tx(t) is a BIBO-unstable system.", - "type": "text" - }, - { - "block_id": "p130-b15", - "global_id": 3473, - "bbox": [ - 103.16, - 522.75, - 477.02, - 558.44 - ], - "text": "(c) We can prove that y(t) =\nd\ndtx(t) is not BIBO-stable with an example. The\nbounded-amplitude input x(t) = u(t) produces the output y(t) = δ(t) whose amplitude is infinite\nat t = 0. Thus, y(t) = d", - "type": "text" - }, - { - "block_id": "p130-b16", - "global_id": 3474, - "bbox": [ - 190.4, - 548.06, - 323.51, - 561.18 - ], - "text": "dtx(t) is a BIBO-unstable system.", - "type": "text" - }, - { - "block_id": "p130-b17", - "global_id": 3475, - "bbox": [ - 107.82, - 604.22, - 393.68, - 616.18 - ], - "text": "DRILL 1.16\nA Noninvertible BIBO-Stable System", - "type": "text" - }, - { - "block_id": "p130-b18", - "global_id": 3476, - "bbox": [ - 107.82, - 623.95, - 471.93, - 635.26 - ], - "text": "Show that a system described by the equation y(t) = x2(t) is noninvertible but BIBO-stable.", - "type": "text" - } - ] - }, - { - "page_num": 131, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p131-b0", - "global_id": 3477, - "bbox": [ - 288.03, - 62.89, - 516.12, - 71.98 - ], - "text": "1.8\nSystem Model: Input–Output Description\n111", - "type": "text" - }, - { - "block_id": "p131-b1", - "global_id": 3478, - "bbox": [ - 127.94, - 94.37, - 461.89, - 108.31 - ], - "text": "1.8 SYSTEM MODEL: INPUT–OUTPUT DESCRIPTION", - "type": "text" - }, - { - "block_id": "p131-b2", - "global_id": 3479, - "bbox": [ - 127.59, - 114.3, - 516.15, - 195.99 - ], - "text": "A system description in terms of the measurements at the input and output terminals is called\nthe input–output description. As mentioned earlier, systems theory encompasses a variety of\nsystems, such as electrical, mechanical, hydraulic, acoustic, electromechanical, and chemical, as\nwell as social, political, economic, and biological. The first step in analyzing any system is the\nconstruction of a system model, which is a mathematical expression or a rule that satisfactorily\napproximates the dynamical behavior of the system. In this chapter we shall consider only\ncontinuous-time systems. Modeling of discrete-time systems is discussed in Ch. 3.", - "type": "text" - }, - { - "block_id": "p131-b3", - "global_id": 3480, - "bbox": [ - 127.59, - 230.45, - 256.58, - 242.4 - ], - "text": "1.8-1 Electrical Systems", - "type": "text" - }, - { - "block_id": "p131-b4", - "global_id": 3481, - "bbox": [ - 127.59, - 248.54, - 516.18, - 342.19 - ], - "text": "To construct a system model, we must study the relationships between different variables in\nthe system. In electrical systems, for example, we must determine a satisfactory model for the\nvoltage-current relationship of each element, such as Ohm’s law for a resistor. In addition, we\nmust determine the various constraints on voltages and currents when several electrical elements\nare interconnected. These are the laws of interconnection—the well-known Kirchhoff laws for\nvoltage and current (KVL and KCL). From all these equations, we eliminate unwanted variables\nto obtain equation(s) relating the desired output variable(s) to the input(s). The following examples\ndemonstrate the procedure of deriving input–output relationships for some LTI electrical systems.", - "type": "text" - }, - { - "block_id": "p131-b5", - "global_id": 3482, - "bbox": [ - 102.51, - 376.03, - 463.33, - 387.99 - ], - "text": "EXAMPLE 1.16\nInput–Output Equation of a Series RLC Circuit", - "type": "text" - }, - { - "block_id": "p131-b6", - "global_id": 3483, - "bbox": [ - 128.9, - 404.55, - 502.75, - 426.57 - ], - "text": "For the series RLC circuit of Fig. 1.34, find the input–output equation relating the input voltage\nx(t) to the output current (loop current) y(t).", - "type": "text" - }, - { - "block_id": "p131-b7", - "global_id": 3484, - "bbox": [ - 240.43, - 461.68, - 268.21, - 469.97 - ], - "text": "R 3", - "type": "text" - }, - { - "block_id": "p131-b10", - "global_id": 3485, - "bbox": [ - 191.39, - 461.68, - 218.28, - 469.97 - ], - "text": "L 1 H", - "type": "text" - }, - { - "block_id": "p131-b11", - "global_id": 3486, - "bbox": [ - 138.69, - 506.3, - 329.66, - 518.68 - ], - "text": "x(t)\nvC(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p131-b12", - "global_id": 3487, - "bbox": [ - 285.55, - 516.3, - 311.6, - 529.8 - ], - "text": "C \nF\n2\n1", - "type": "text" - }, - { - "block_id": "p131-b13", - "global_id": 3488, - "bbox": [ - 340.43, - 541.39, - 462.72, - 550.63 - ], - "text": "Figure 1.34 Circuit for Ex. 1.16.", - "type": "text" - }, - { - "block_id": "p131-b14", - "global_id": 3489, - "bbox": [ - 128.9, - 589.96, - 375.1, - 599.92 - ], - "text": "Application of Kirchhoff’s voltage law around the loop yields", - "type": "text" - }, - { - "block_id": "p131-b15", - "global_id": 3490, - "bbox": [ - 262.18, - 611.46, - 369.49, - 622.53 - ], - "text": "vL(t) + vR(t) + vC(t) = x(t)", - "type": "text" - } - ] - }, - { - "page_num": 132, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p132-b0", - "global_id": 3491, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "112\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p132-b1", - "global_id": 3492, - "bbox": [ - 103.16, - 86.24, - 477.0, - 108.16 - ], - "text": "By using the voltage-current laws of each element (inductor, resistor, and capacitor), we can\nexpress this equation as", - "type": "text" - }, - { - "block_id": "p132-b2", - "global_id": 3493, - "bbox": [ - 218.71, - 108.55, - 238.5, - 118.82 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p132-b3", - "global_id": 3494, - "bbox": [ - 224.64, - 115.53, - 286.18, - 132.87 - ], - "text": "dt\n+ 3y(t) + 2", - "type": "text" - }, - { - "block_id": "p132-b4", - "global_id": 3495, - "bbox": [ - 287.29, - 101.97, - 299.07, - 114.24 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p132-b5", - "global_id": 3496, - "bbox": [ - 292.57, - 126.84, - 305.13, - 133.81 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p132-b6", - "global_id": 3497, - "bbox": [ - 306.74, - 115.53, - 477.01, - 125.91 - ], - "text": "y(τ)dτ = x(t)\n(1.27)", - "type": "text" - }, - { - "block_id": "p132-b7", - "global_id": 3498, - "bbox": [ - 103.17, - 140.2, - 313.2, - 150.16 - ], - "text": "Differentiating both sides of this equation, we obtain", - "type": "text" - }, - { - "block_id": "p132-b8", - "global_id": 3499, - "bbox": [ - 226.69, - 160.33, - 250.73, - 171.54 - ], - "text": "d2y(t)", - "type": "text" - }, - { - "block_id": "p132-b9", - "global_id": 3500, - "bbox": [ - 232.76, - 161.27, - 288.74, - 185.6 - ], - "text": "dt2\n+ 3dy(t)", - "type": "text" - }, - { - "block_id": "p132-b10", - "global_id": 3501, - "bbox": [ - 274.9, - 161.27, - 353.48, - 185.6 - ], - "text": "dt\n+ 2y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p132-b11", - "global_id": 3502, - "bbox": [ - 339.61, - 168.66, - 477.01, - 185.6 - ], - "text": "dt\n(1.28)", - "type": "text" - }, - { - "block_id": "p132-b12", - "global_id": 3503, - "bbox": [ - 103.16, - 193.24, - 477.01, - 215.58 - ], - "text": "This differential equation is the input–output relationship between the output y(t) and the\ninput x(t).", - "type": "text" - }, - { - "block_id": "p132-b13", - "global_id": 3504, - "bbox": [ - 101.84, - 253.15, - 490.38, - 275.48 - ], - "text": "It proves convenient to use a compact notation D for the differential operator d/dt. This\nnotation can be repeatedly applied. Thus,", - "type": "text" - }, - { - "block_id": "p132-b14", - "global_id": 3505, - "bbox": [ - 160.99, - 289.65, - 180.77, - 299.92 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p132-b15", - "global_id": 3506, - "bbox": [ - 166.92, - 288.72, - 264.59, - 313.98 - ], - "text": "dt\n≡Dy(t),\nd2y(t)", - "type": "text" - }, - { - "block_id": "p132-b16", - "global_id": 3507, - "bbox": [ - 246.62, - 288.65, - 391.81, - 313.98 - ], - "text": "dt2\n≡D2y(t),\n. . .,\ndNy(t)", - "type": "text" - }, - { - "block_id": "p132-b17", - "global_id": 3508, - "bbox": [ - 372.25, - 295.13, - 432.44, - 313.98 - ], - "text": "dtN\n≡DNy(t)", - "type": "text" - }, - { - "block_id": "p132-b18", - "global_id": 3509, - "bbox": [ - 101.85, - 325.05, - 299.12, - 335.02 - ], - "text": "With this notation, Eq. (1.28) can be expressed as", - "type": "text" - }, - { - "block_id": "p132-b19", - "global_id": 3510, - "bbox": [ - 243.02, - 345.51, - 490.38, - 360.0 - ], - "text": "(D2 + 3D + 2)y(t) = Dx(t)\n(1.29)", - "type": "text" - }, - { - "block_id": "p132-b20", - "global_id": 3511, - "bbox": [ - 101.85, - 374.61, - 490.39, - 396.94 - ], - "text": "The differential operator is the inverse of the integral operator, so we can use the operator 1/D\nto represent integration.†", - "type": "text" - }, - { - "block_id": "p132-b21", - "global_id": 3512, - "bbox": [ - 253.67, - 395.34, - 265.42, - 407.62 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p132-b22", - "global_id": 3513, - "bbox": [ - 258.92, - 420.23, - 271.48, - 427.2 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p132-b23", - "global_id": 3514, - "bbox": [ - 273.09, - 402.34, - 321.47, - 419.18 - ], - "text": "y(τ)dτ ≡1", - "type": "text" - }, - { - "block_id": "p132-b24", - "global_id": 3515, - "bbox": [ - 315.38, - 408.91, - 338.57, - 426.26 - ], - "text": "Dy(t)", - "type": "text" - }, - { - "block_id": "p132-b25", - "global_id": 3516, - "bbox": [ - 101.84, - 447.79, - 490.38, - 470.96 - ], - "text": "† Use of operator 1/D for integration generates some subtle mathematical difficulties because the operators\nD and 1/D do not commute. For instance, we know that D(1/D) = 1 because", - "type": "text" - }, - { - "block_id": "p132-b26", - "global_id": 3517, - "bbox": [ - 253.06, - 482.17, - 260.04, - 503.79 - ], - "text": "d\ndt", - "type": "text" - }, - { - "block_id": "p132-b27", - "global_id": 3518, - "bbox": [ - 262.4, - 475.58, - 277.92, - 487.14 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p132-b28", - "global_id": 3519, - "bbox": [ - 272.02, - 498.2, - 283.68, - 504.68 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p132-b29", - "global_id": 3520, - "bbox": [ - 285.17, - 488.17, - 310.4, - 497.42 - ], - "text": "y(τ)dτ", - "type": "text" - }, - { - "block_id": "p132-b30", - "global_id": 3521, - "bbox": [ - 311.48, - 475.58, - 316.37, - 484.55 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p132-b31", - "global_id": 3522, - "bbox": [ - 318.2, - 488.17, - 340.37, - 497.42 - ], - "text": "= y(t)", - "type": "text" - }, - { - "block_id": "p132-b32", - "global_id": 3523, - "bbox": [ - 101.84, - 514.48, - 490.4, - 633.42 - ], - "text": "However, (1/D)D is not necessarily unity. Use of Cramer’s rule in solving simultaneous integro-differential\nequations will always result in cancellation of operators 1/D and D. This procedure may yield erroneous\nresults when the factor D occurs in the numerator as well as in the denominator. This happens, for instance,\nin circuits with all-inductor loops or all-capacitor cut sets. To eliminate this problem, avoid the integral\noperation in system equations so that the resulting equations are differential rather than integro-differential.\nIn electrical circuits, this can be done by using charge (instead of current) variables in loops containing\ncapacitors and choosing current variables for loops without capacitors. In the literature this problem of\ncommutativity of D and 1/D is largely ignored. As mentioned earlier, such a procedure gives erroneous results\nonly in special systems, such as the circuits with all-inductor loops or all-capacitor cut sets. Fortunately such\nsystems constitute a very small fraction of the systems we deal with. For further discussion of this topic and\na correct method of handling problems involving integrals, see [4].", - "type": "text" - } - ] - }, - { - "page_num": 133, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p133-b0", - "global_id": 3524, - "bbox": [ - 288.03, - 62.89, - 516.12, - 71.98 - ], - "text": "1.8\nSystem Model: Input–Output Description\n113", - "type": "text" - }, - { - "block_id": "p133-b1", - "global_id": 3525, - "bbox": [ - 127.59, - 85.82, - 307.47, - 95.78 - ], - "text": "Consequently, Eq. (1.27) can be expressed as", - "type": "text" - }, - { - "block_id": "p133-b3", - "global_id": 3526, - "bbox": [ - 279.36, - 106.17, - 320.56, - 123.12 - ], - "text": "D + 3 + 2", - "type": "text" - }, - { - "block_id": "p133-b4", - "global_id": 3527, - "bbox": [ - 314.47, - 120.13, - 321.66, - 130.09 - ], - "text": "D", - "type": "text" - }, - { - "block_id": "p133-b6", - "global_id": 3528, - "bbox": [ - 329.58, - 112.74, - 371.09, - 123.02 - ], - "text": "y(t) = x(t)", - "type": "text" - }, - { - "block_id": "p133-b7", - "global_id": 3529, - "bbox": [ - 127.59, - 140.04, - 404.74, - 150.1 - ], - "text": "Multiplying both sides by D to differentiate the expression, we obtain", - "type": "text" - }, - { - "block_id": "p133-b8", - "global_id": 3530, - "bbox": [ - 268.76, - 157.31, - 374.96, - 171.8 - ], - "text": "(D2 + 3D + 2)y(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p133-b9", - "global_id": 3531, - "bbox": [ - 127.59, - 183.53, - 253.21, - 193.49 - ], - "text": "which is identical to Eq. (1.29).", - "type": "text" - }, - { - "block_id": "p133-b10", - "global_id": 3532, - "bbox": [ - 127.59, - 194.14, - 516.14, - 253.27 - ], - "text": "Recall that Eq. (1.29) is not an algebraic equation, and D2+3D+2 is not an algebraic term that\nmultiplies y(t); it is an operator that operates on y(t). It means that we must perform the following\noperations on y(t): take the second derivative of y(t) and add to it 3 times the first derivative of\ny(t) and 2 times y(t). Clearly, a polynomial in D multiplied by y(t) represents a certain differential\noperation on y(t).", - "type": "text" - }, - { - "block_id": "p133-b11", - "global_id": 3533, - "bbox": [ - 102.51, - 282.15, - 456.02, - 294.1 - ], - "text": "EXAMPLE 1.17\nInput–Output Equation of a Series RC Circuit", - "type": "text" - }, - { - "block_id": "p133-b12", - "global_id": 3534, - "bbox": [ - 128.9, - 310.67, - 502.75, - 332.69 - ], - "text": "Using operator notation, find the equation relating input to output for the series RC circuit of\nFig. 1.35 if the input is the voltage x(t) and output is", - "type": "text" - }, - { - "block_id": "p133-b13", - "global_id": 3535, - "bbox": [ - 146.84, - 340.25, - 264.09, - 365.56 - ], - "text": "(a) the loop current i(t)\n(b) the capacitor voltage y(t)", - "type": "text" - }, - { - "block_id": "p133-b14", - "global_id": 3536, - "bbox": [ - 182.01, - 382.58, - 213.78, - 390.88 - ], - "text": "R 15", - "type": "text" - }, - { - "block_id": "p133-b17", - "global_id": 3537, - "bbox": [ - 131.12, - 429.78, - 293.13, - 450.39 - ], - "text": "x(t)\ni(t)\ny(t)\nC \nF\n5\n1", - "type": "text" - }, - { - "block_id": "p133-b18", - "global_id": 3538, - "bbox": [ - 303.86, - 467.29, - 423.91, - 476.53 - ], - "text": "Figure 1.35 Circuit for Ex. 1.17", - "type": "text" - }, - { - "block_id": "p133-b19", - "global_id": 3539, - "bbox": [ - 146.84, - 499.83, - 299.3, - 509.88 - ], - "text": "(a) The loop equation for the circuit is", - "type": "text" - }, - { - "block_id": "p133-b20", - "global_id": 3540, - "bbox": [ - 258.19, - 520.63, - 296.52, - 537.58 - ], - "text": "Ri(t) + 1", - "type": "text" - }, - { - "block_id": "p133-b21", - "global_id": 3541, - "bbox": [ - 290.61, - 534.59, - 297.25, - 544.55 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p133-b22", - "global_id": 3542, - "bbox": [ - 299.77, - 513.65, - 311.54, - 525.93 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p133-b23", - "global_id": 3543, - "bbox": [ - 305.03, - 538.52, - 317.59, - 545.5 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p133-b24", - "global_id": 3544, - "bbox": [ - 319.19, - 527.2, - 373.47, - 537.48 - ], - "text": "i(τ)dτ = x(t)", - "type": "text" - }, - { - "block_id": "p133-b25", - "global_id": 3545, - "bbox": [ - 128.9, - 552.66, - 137.2, - 562.62 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p133-b26", - "global_id": 3546, - "bbox": [ - 258.4, - 554.37, - 311.33, - 578.31 - ], - "text": "15i(t) + 5\n# t", - "type": "text" - }, - { - "block_id": "p133-b27", - "global_id": 3547, - "bbox": [ - 304.83, - 579.24, - 317.38, - 586.21 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p133-b28", - "global_id": 3548, - "bbox": [ - 318.99, - 567.93, - 373.27, - 578.21 - ], - "text": "i(τ)dτ = x(t)", - "type": "text" - }, - { - "block_id": "p133-b29", - "global_id": 3549, - "bbox": [ - 128.9, - 592.6, - 356.06, - 602.56 - ], - "text": "With operator notation, this equation can be expressed as", - "type": "text" - }, - { - "block_id": "p133-b30", - "global_id": 3550, - "bbox": [ - 273.57, - 612.5, - 502.75, - 636.42 - ], - "text": "15i(t) + 5\nDi(t) = x(t)\n(1.30)", - "type": "text" - } - ] - }, - { - "page_num": 134, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p134-b0", - "global_id": 3551, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "114\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p134-b1", - "global_id": 3552, - "bbox": [ - 121.09, - 86.27, - 477.02, - 96.33 - ], - "text": "(b) Multiplying both sides of Eq. (1.30) by D (i.e., differentiating the equation), we obtain", - "type": "text" - }, - { - "block_id": "p134-b2", - "global_id": 3553, - "bbox": [ - 245.8, - 107.87, - 334.38, - 118.25 - ], - "text": "(15D + 5)i(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p134-b3", - "global_id": 3554, - "bbox": [ - 103.16, - 127.58, - 226.91, - 140.16 - ], - "text": "Using the fact that i(t) = C dy(t)", - "type": "text" - }, - { - "block_id": "p134-b4", - "global_id": 3555, - "bbox": [ - 217.21, - 128.45, - 244.66, - 143.29 - ], - "text": "dt = 1", - "type": "text" - }, - { - "block_id": "p134-b5", - "global_id": 3556, - "bbox": [ - 241.17, - 129.79, - 374.69, - 142.98 - ], - "text": "5Dy(t), simple substitution yields", - "type": "text" - }, - { - "block_id": "p134-b6", - "global_id": 3557, - "bbox": [ - 251.61, - 151.71, - 477.01, - 162.09 - ], - "text": "(3D + 1)y(t) = x(t)\n(1.31)", - "type": "text" - }, - { - "block_id": "p134-b7", - "global_id": 3558, - "bbox": [ - 107.82, - 214.84, - 473.66, - 240.74 - ], - "text": "DRILL 1.17\nInput–Output Equation of a Series RLC Circuit with\nInductor Voltage as Output", - "type": "text" - }, - { - "block_id": "p134-b8", - "global_id": 3559, - "bbox": [ - 107.82, - 249.45, - 484.4, - 272.48 - ], - "text": "If the inductor voltage vL(t) is taken as the output, show that the RLC circuit in Fig. 1.34 has an\ninput–output equation of (D2 + 3D + 2)vL(t) = D2x(t).", - "type": "text" - }, - { - "block_id": "p134-b9", - "global_id": 3560, - "bbox": [ - 107.82, - 318.26, - 466.35, - 344.15 - ], - "text": "DRILL 1.18\nInput–Output Equation of a Series RC Circuit with\nCapacitor Voltage as Output", - "type": "text" - }, - { - "block_id": "p134-b10", - "global_id": 3561, - "bbox": [ - 107.82, - 352.86, - 484.4, - 375.9 - ], - "text": "If the capacitor voltage vC(t) is taken as the output, show that the RLC circuit in Fig. 1.34 has\nan input–output equation of (D2 + 3D + 2)vC(t) = 2x(t).", - "type": "text" - }, - { - "block_id": "p134-b11", - "global_id": 3562, - "bbox": [ - 101.84, - 413.52, - 243.45, - 425.47 - ], - "text": "1.8-2 Mechanical Systems", - "type": "text" - }, - { - "block_id": "p134-b12", - "global_id": 3563, - "bbox": [ - 101.84, - 431.6, - 490.39, - 465.47 - ], - "text": "Planar motion can be resolved into translational (rectilinear) motion and rotational (torsional)\nmotion. Translational motion will be considered first. We shall restrict ourselves to motions in\none dimension.", - "type": "text" - }, - { - "block_id": "p134-b13", - "global_id": 3564, - "bbox": [ - 101.84, - 482.02, - 490.39, - 520.09 - ], - "text": "TRANSLATIONAL SYSTEMS\nThe basic elements used in modeling translational systems are ideal masses, linear springs, and\ndashpots providing viscous damping. The laws of various mechanical elements are now discussed.", - "type": "text" - }, - { - "block_id": "p134-b14", - "global_id": 3565, - "bbox": [ - 101.84, - 521.67, - 490.39, - 544.0 - ], - "text": "For a mass M (Fig. 1.36a), a force x(t) causes a motion y(t) and acceleration ¨y(t). From\nNewton’s law of motion,", - "type": "text" - }, - { - "block_id": "p134-b15", - "global_id": 3566, - "bbox": [ - 224.44, - 548.92, - 320.44, - 567.12 - ], - "text": "x(t) = M¨y(t) = M d2y(t)", - "type": "text" - }, - { - "block_id": "p134-b16", - "global_id": 3567, - "bbox": [ - 302.47, - 555.41, - 367.79, - 574.19 - ], - "text": "dt2\n= MD2y(t)", - "type": "text" - }, - { - "block_id": "p134-b17", - "global_id": 3568, - "bbox": [ - 101.85, - 582.47, - 490.4, - 604.81 - ], - "text": "The force x(t) required to stretch (or compress) a linear spring (Fig. 1.36b) by an amount y(t)\nis given by", - "type": "text" - }, - { - "block_id": "p134-b18", - "global_id": 3569, - "bbox": [ - 272.24, - 611.82, - 320.0, - 622.1 - ], - "text": "x(t) = Ky(t)", - "type": "text" - }, - { - "block_id": "p134-b19", - "global_id": 3570, - "bbox": [ - 101.85, - 634.69, - 249.43, - 644.75 - ], - "text": "where K is the stiffness of the spring.", - "type": "text" - } - ] - }, - { - "page_num": 135, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p135-b0", - "global_id": 3571, - "bbox": [ - 288.03, - 62.89, - 516.12, - 71.98 - ], - "text": "1.8\nSystem Model: Input–Output Description\n115", - "type": "text" - }, - { - "block_id": "p135-b1", - "global_id": 3572, - "bbox": [ - 198.8, - 122.52, - 205.46, - 130.52 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p135-b2", - "global_id": 3573, - "bbox": [ - 196.4, - 159.74, - 426.98, - 167.74 - ], - "text": "(a)\n(b)\n(c)", - "type": "text" - }, - { - "block_id": "p135-b3", - "global_id": 3574, - "bbox": [ - 292.54, - 103.31, - 297.88, - 111.31 - ], - "text": "K", - "type": "text" - }, - { - "block_id": "p135-b4", - "global_id": 3575, - "bbox": [ - 418.8, - 133.85, - 423.69, - 141.85 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p135-b5", - "global_id": 3576, - "bbox": [ - 162.55, - 114.16, - 478.31, - 127.77 - ], - "text": "x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p135-b6", - "global_id": 3577, - "bbox": [ - 344.14, - 121.52, - 355.24, - 129.6 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p135-b7", - "global_id": 3578, - "bbox": [ - 348.24, - 93.29, - 459.61, - 111.29 - ], - "text": "y(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p135-b8", - "global_id": 3579, - "bbox": [ - 227.6, - 87.2, - 238.7, - 95.28 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p135-b9", - "global_id": 3580, - "bbox": [ - 151.5, - 174.44, - 387.85, - 183.68 - ], - "text": "Figure 1.36 Some elements in translational mechanical systems.", - "type": "text" - }, - { - "block_id": "p135-b10", - "global_id": 3581, - "bbox": [ - 127.59, - 196.83, - 516.14, - 230.8 - ], - "text": "For a linear dashpot (Fig. 1.36c), which operates by virtue of viscous friction, the force\nmoving the dashpot is proportional to the relative velocity ˙y(t) of one surface with respect to the\nother. Thus", - "type": "text" - }, - { - "block_id": "p135-b11", - "global_id": 3582, - "bbox": [ - 258.02, - 235.41, - 344.55, - 252.67 - ], - "text": "x(t) = B˙y(t) = Bdy(t)", - "type": "text" - }, - { - "block_id": "p135-b12", - "global_id": 3583, - "bbox": [ - 330.7, - 242.4, - 385.69, - 259.75 - ], - "text": "dt\n= BDy(t)", - "type": "text" - }, - { - "block_id": "p135-b13", - "global_id": 3584, - "bbox": [ - 127.59, - 269.35, - 416.35, - 279.41 - ], - "text": "where B is the damping coefficient of the dashpot or the viscous friction.", - "type": "text" - }, - { - "block_id": "p135-b14", - "global_id": 3585, - "bbox": [ - 102.51, - 310.76, - 504.53, - 336.67 - ], - "text": "EXAMPLE 1.18\nInput–Output Equation for a Translational Mechanical\nSystem", - "type": "text" - }, - { - "block_id": "p135-b15", - "global_id": 3586, - "bbox": [ - 128.9, - 353.33, - 502.77, - 387.2 - ], - "text": "Find the input–output relationship for the translational mechanical system shown in Fig. 1.37a\nor its equivalent in Fig. 1.37b. The input is the force x(t), and the output is the mass position\ny(t).", - "type": "text" - }, - { - "block_id": "p135-b16", - "global_id": 3587, - "bbox": [ - 204.93, - 453.44, - 211.59, - 461.44 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p135-b17", - "global_id": 3588, - "bbox": [ - 351.27, - 434.85, - 356.61, - 442.85 - ], - "text": "K", - "type": "text" - }, - { - "block_id": "p135-b18", - "global_id": 3589, - "bbox": [ - 258.52, - 460.06, - 461.04, - 472.84 - ], - "text": "B\nFrictionless", - "type": "text" - }, - { - "block_id": "p135-b19", - "global_id": 3590, - "bbox": [ - 157.4, - 423.26, - 162.74, - 431.26 - ], - "text": "K", - "type": "text" - }, - { - "block_id": "p135-b20", - "global_id": 3591, - "bbox": [ - 158.66, - 472.07, - 163.54, - 480.07 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p135-b21", - "global_id": 3592, - "bbox": [ - 177.64, - 505.31, - 390.18, - 513.31 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p135-b22", - "global_id": 3593, - "bbox": [ - 400.27, - 453.44, - 406.93, - 461.44 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p135-b23", - "global_id": 3594, - "bbox": [ - 228.27, - 567.92, - 295.37, - 582.79 - ], - "text": "M\nBy.(t)", - "type": "text" - }, - { - "block_id": "p135-b24", - "global_id": 3595, - "bbox": [ - 227.9, - 556.14, - 244.82, - 564.22 - ], - "text": "Ky(t)", - "type": "text" - }, - { - "block_id": "p135-b25", - "global_id": 3596, - "bbox": [ - 256.26, - 437.18, - 267.36, - 445.26 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p135-b26", - "global_id": 3597, - "bbox": [ - 243.85, - 407.28, - 254.95, - 415.36 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p135-b27", - "global_id": 3598, - "bbox": [ - 338.97, - 565.59, - 350.07, - 573.67 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p135-b28", - "global_id": 3599, - "bbox": [ - 326.56, - 525.88, - 337.66, - 533.96 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p135-b29", - "global_id": 3600, - "bbox": [ - 451.26, - 436.88, - 462.37, - 444.96 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p135-b30", - "global_id": 3601, - "bbox": [ - 438.86, - 407.28, - 449.96, - 415.36 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p135-b31", - "global_id": 3602, - "bbox": [ - 287.6, - 597.87, - 296.48, - 605.87 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p135-b32", - "global_id": 3603, - "bbox": [ - 119.94, - 612.57, - 284.05, - 621.81 - ], - "text": "Figure 1.37 Mechanical system for Ex. 1.18", - "type": "text" - } - ] - }, - { - "page_num": 136, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p136-b0", - "global_id": 3604, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "116\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p136-b1", - "global_id": 3605, - "bbox": [ - 103.16, - 86.24, - 477.03, - 155.98 - ], - "text": "In mechanical systems it is helpful to draw a free-body diagram of each junction, which is\na point at which two or more elements are connected. In Fig. 1.37, the point representing the\nmass is a junction. The displacement of the mass is denoted by y(t). The spring is also stretched\nby the amount y(t), and therefore it exerts a force −Ky(t) on the mass. The dashpot exerts a\nforce −B˙y(t) on the mass, as shown in the free-body diagram (Fig. 1.37c). By Newton’s second\nlaw, the net force must be M¨y(t). Therefore,", - "type": "text" - }, - { - "block_id": "p136-b2", - "global_id": 3606, - "bbox": [ - 229.26, - 167.52, - 350.91, - 177.79 - ], - "text": "M¨y(t) = −B˙y(t) −Ky(t) + x(t)", - "type": "text" - }, - { - "block_id": "p136-b3", - "global_id": 3607, - "bbox": [ - 103.17, - 189.85, - 111.46, - 199.82 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p136-b4", - "global_id": 3608, - "bbox": [ - 234.78, - 197.28, - 345.4, - 211.67 - ], - "text": "(MD2 + BD + K)y(t) = x(t)", - "type": "text" - }, - { - "block_id": "p136-b5", - "global_id": 3609, - "bbox": [ - 102.14, - 253.03, - 223.28, - 265.16 - ], - "text": "ROTATIONAL SYSTEMS", - "type": "text" - }, - { - "block_id": "p136-b6", - "global_id": 3610, - "bbox": [ - 101.84, - 269.19, - 490.41, - 386.75 - ], - "text": "In rotational systems, the motion of a body may be defined as its motion about a certain axis.\nThe variables used to describe rotational motion are torque (in place of force), angular position\n(in place of linear position), angular velocity (in place of linear velocity), and angular acceleration\n(in place of linear acceleration). The system elements are rotational mass or moment of inertia (in\nplace of mass) and torsional springs and torsional dashpots (in place of linear springs and\ndashpots). The terminal equations for these elements are analogous to the corresponding equations\nfor translational elements. If J is the moment of inertia (or rotational mass) of a rotating body about\na certain axis, then the external torque required for this motion is equal to J (rotational mass) times\nthe angular acceleration. If θ(t) is the angular position of the body, ¨θ(t) is its angular acceleration,\nand", - "type": "text" - }, - { - "block_id": "p136-b7", - "global_id": 3611, - "bbox": [ - 223.44, - 385.15, - 324.38, - 403.44 - ], - "text": "torque = J ¨θ(t) = J d2θ(t)", - "type": "text" - }, - { - "block_id": "p136-b8", - "global_id": 3612, - "bbox": [ - 305.91, - 391.63, - 368.8, - 410.42 - ], - "text": "dt2\n= JD2θ(t)", - "type": "text" - }, - { - "block_id": "p136-b9", - "global_id": 3613, - "bbox": [ - 101.84, - 415.18, - 490.37, - 437.52 - ], - "text": "Similarly, if K is the stiffness of a torsional spring (per unit angular twist), and θ is the angular\ndisplacement of one terminal of the spring with respect to the other, then", - "type": "text" - }, - { - "block_id": "p136-b10", - "global_id": 3614, - "bbox": [ - 265.98, - 449.23, - 326.26, - 459.6 - ], - "text": "torque = Kθ(t)", - "type": "text" - }, - { - "block_id": "p136-b11", - "global_id": 3615, - "bbox": [ - 101.85, - 471.63, - 479.86, - 481.69 - ], - "text": "Finally, the torque due to viscous damping of a torsional dashpot with damping coefficient B is", - "type": "text" - }, - { - "block_id": "p136-b12", - "global_id": 3616, - "bbox": [ - 246.08, - 491.4, - 346.16, - 503.78 - ], - "text": "torque = B ˙θ(t) = BDθ(t)", - "type": "text" - }, - { - "block_id": "p136-b13", - "global_id": 3617, - "bbox": [ - 76.77, - 535.01, - 436.96, - 546.96 - ], - "text": "EXAMPLE 1.19\nInput–Output Equation for Aircraft Roll Angle", - "type": "text" - }, - { - "block_id": "p136-b14", - "global_id": 3618, - "bbox": [ - 103.16, - 563.63, - 477.04, - 621.41 - ], - "text": "The attitude of an aircraft can be controlled by three sets of surfaces (shown shaded in\nFig. 1.38): elevators, rudder, and ailerons. By manipulating these surfaces, one can set the\naircraft on a desired flight path. The roll angle ϕ(t) can be controlled by deflecting in the\nopposite direction the two aileron surfaces as shown in Fig. 1.38. Assuming only rolling\nmotion, find the equation relating the roll angle ϕ(t) to the input (deflection) θ(t).", - "type": "text" - } - ] - }, - { - "page_num": 137, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p137-b0", - "global_id": 3619, - "bbox": [ - 288.03, - 62.89, - 516.12, - 71.98 - ], - "text": "1.8\nSystem Model: Input–Output Description\n117", - "type": "text" - }, - { - "block_id": "p137-b1", - "global_id": 3620, - "bbox": [ - 364.99, - 229.96, - 368.99, - 237.96 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p137-b2", - "global_id": 3621, - "bbox": [ - 232.98, - 176.34, - 236.98, - 184.34 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p137-b3", - "global_id": 3622, - "bbox": [ - 127.51, - 190.23, - 153.82, - 198.23 - ], - "text": "Elevator", - "type": "text" - }, - { - "block_id": "p137-b4", - "global_id": 3623, - "bbox": [ - 240.7, - 245.11, - 267.01, - 253.11 - ], - "text": "Elevator", - "type": "text" - }, - { - "block_id": "p137-b5", - "global_id": 3624, - "bbox": [ - 157.69, - 216.31, - 181.24, - 224.31 - ], - "text": "Rudder", - "type": "text" - }, - { - "block_id": "p137-b6", - "global_id": 3625, - "bbox": [ - 277.5, - 137.1, - 301.94, - 145.1 - ], - "text": "Aileron", - "type": "text" - }, - { - "block_id": "p137-b7", - "global_id": 3626, - "bbox": [ - 315.66, - 234.59, - 340.1, - 242.59 - ], - "text": "Aileron", - "type": "text" - }, - { - "block_id": "p137-b8", - "global_id": 3627, - "bbox": [ - 419.39, - 110.27, - 471.87, - 124.6 - ], - "text": "x\nw", - "type": "text" - }, - { - "block_id": "p137-b9", - "global_id": 3628, - "bbox": [ - 127.51, - 259.81, - 288.38, - 269.05 - ], - "text": "Figure 1.38 Attitude control of an airplane.", - "type": "text" - }, - { - "block_id": "p137-b10", - "global_id": 3629, - "bbox": [ - 128.9, - 302.69, - 502.76, - 348.52 - ], - "text": "The aileron surfaces generate a torque about the roll axis proportional to the aileron deflection\nangle θ(t). Let this torque be cθ(t), where c is the constant of proportionality. Air friction\ndissipates the torque B ˙ϕ(t). The torque available for rolling motion is then cθ(t) −B ˙ϕ(t). If J\nis the moment of inertia of the plane about the x axis (roll axis), then", - "type": "text" - }, - { - "block_id": "p137-b11", - "global_id": 3630, - "bbox": [ - 246.41, - 360.06, - 385.26, - 370.44 - ], - "text": "net torque = J ¨ϕ(t) = cθ(t) −B ˙ϕ(t)", - "type": "text" - }, - { - "block_id": "p137-b12", - "global_id": 3631, - "bbox": [ - 128.91, - 382.39, - 143.28, - 392.35 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p137-b13", - "global_id": 3632, - "bbox": [ - 190.29, - 390.51, - 222.09, - 408.7 - ], - "text": "J d2ϕ(t)", - "type": "text" - }, - { - "block_id": "p137-b14", - "global_id": 3633, - "bbox": [ - 203.27, - 391.44, - 263.16, - 415.77 - ], - "text": "dt2\n+ Bdϕ(t)", - "type": "text" - }, - { - "block_id": "p137-b15", - "global_id": 3634, - "bbox": [ - 248.33, - 394.31, - 441.38, - 415.77 - ], - "text": "dt\n= cθ(t)\nor\n(JD2 + BD)ϕ(t) = cθ(t)", - "type": "text" - }, - { - "block_id": "p137-b16", - "global_id": 3635, - "bbox": [ - 128.91, - 420.44, - 502.77, - 430.81 - ], - "text": "This is the desired equation relating the output (roll angle ϕ(t)) to the input (aileron angle θ(t)).", - "type": "text" - }, - { - "block_id": "p137-b17", - "global_id": 3636, - "bbox": [ - 128.91, - 432.39, - 502.76, - 454.72 - ], - "text": "The roll velocity ω(t) is ˙ϕ(t). If the desired output is the roll velocity ω(t) rather than the\nroll angle ϕ(t), then the input–output equation would be", - "type": "text" - }, - { - "block_id": "p137-b18", - "global_id": 3637, - "bbox": [ - 200.46, - 464.3, - 229.46, - 481.56 - ], - "text": "J dω(t)", - "type": "text" - }, - { - "block_id": "p137-b19", - "global_id": 3638, - "bbox": [ - 214.04, - 471.28, - 431.21, - 488.64 - ], - "text": "dt\n+ Bω(t) = cθ(t)\nor\n(JD + B)ω(t) = cθ(t)", - "type": "text" - }, - { - "block_id": "p137-b20", - "global_id": 3639, - "bbox": [ - 133.57, - 558.92, - 493.14, - 584.82 - ], - "text": "DRILL 1.19\nInput–Output Equation of a Rotational Mechanical\nSystem", - "type": "text" - }, - { - "block_id": "p137-b21", - "global_id": 3640, - "bbox": [ - 133.57, - 593.53, - 510.17, - 627.81 - ], - "text": "Torque T (t) is applied to the rotational mechanical system shown in Fig. 1.39a. The torsional\nspring stiffness is K; the rotational mass (the cylinder’s moment of inertia about the shaft) is\nJ; the viscous damping coefficient between the cylinder and the ground is B. Find the equation", - "type": "text" - } - ] - }, - { - "page_num": 138, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p138-b0", - "global_id": 3641, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "118\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p138-b1", - "global_id": 3642, - "bbox": [ - 107.82, - 91.44, - 484.42, - 113.77 - ], - "text": "relating the output angle θ(t) to the input torque T (t). [Hint: A free-body diagram is shown in\nFig. 1.39b.]", - "type": "text" - }, - { - "block_id": "p138-b2", - "global_id": 3643, - "bbox": [ - 108.09, - 127.3, - 162.68, - 138.26 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p138-b3", - "global_id": 3644, - "bbox": [ - 107.82, - 141.83, - 138.92, - 160.03 - ], - "text": "J d2θ(t)", - "type": "text" - }, - { - "block_id": "p138-b4", - "global_id": 3645, - "bbox": [ - 120.45, - 142.77, - 179.26, - 167.1 - ], - "text": "dt2\n+ Bdθ(t)", - "type": "text" - }, - { - "block_id": "p138-b5", - "global_id": 3646, - "bbox": [ - 107.82, - 149.75, - 244.5, - 175.07 - ], - "text": "dt\n+ Kθ(t) = T (t)\nor", - "type": "text" - }, - { - "block_id": "p138-b6", - "global_id": 3647, - "bbox": [ - 107.82, - 176.02, - 219.04, - 189.92 - ], - "text": "(JD2 + BD + K)θ(t) = T (t)", - "type": "text" - }, - { - "block_id": "p138-b7", - "global_id": 3648, - "bbox": [ - 242.43, - 260.47, - 245.98, - 268.47 - ], - "text": "J", - "type": "text" - }, - { - "block_id": "p138-b8", - "global_id": 3649, - "bbox": [ - 181.39, - 245.57, - 186.73, - 253.57 - ], - "text": "K", - "type": "text" - }, - { - "block_id": "p138-b9", - "global_id": 3650, - "bbox": [ - 292.53, - 279.43, - 297.42, - 287.43 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p138-b10", - "global_id": 3651, - "bbox": [ - 215.02, - 316.53, - 399.96, - 324.53 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p138-b11", - "global_id": 3652, - "bbox": [ - 344.18, - 279.74, - 353.07, - 287.76 - ], - "text": "Bu", - "type": "text" - }, - { - "block_id": "p138-b12", - "global_id": 3653, - "bbox": [ - 350.67, - 273.26, - 352.67, - 281.26 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p138-b13", - "global_id": 3654, - "bbox": [ - 353.15, - 279.76, - 360.7, - 287.84 - ], - "text": "(t)", - "type": "text" - }, - { - "block_id": "p138-b14", - "global_id": 3655, - "bbox": [ - 190.77, - 237.73, - 202.32, - 245.83 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p138-b15", - "global_id": 3656, - "bbox": [ - 315.18, - 279.74, - 332.55, - 287.84 - ], - "text": "Ku(t)", - "type": "text" - }, - { - "block_id": "p138-b16", - "global_id": 3657, - "bbox": [ - 245.33, - 259.67, - 408.33, - 269.17 - ], - "text": "J\nJ", - "type": "text" - }, - { - "block_id": "p138-b17", - "global_id": 3658, - "bbox": [ - 139.3, - 334.07, - 305.92, - 343.31 - ], - "text": "Figure 1.39 Rotational system for Drill 1.19.", - "type": "text" - }, - { - "block_id": "p138-b18", - "global_id": 3659, - "bbox": [ - 101.84, - 385.0, - 279.97, - 396.95 - ], - "text": "1.8-3 Electromechanical Systems", - "type": "text" - }, - { - "block_id": "p138-b19", - "global_id": 3660, - "bbox": [ - 101.84, - 403.08, - 490.4, - 448.91 - ], - "text": "A wide variety of electromechanical systems is used to convert electrical signals into mechanical\nmotion (mechanical energy) and vice versa. Here we consider a rather simple example of an\narmature-controlled dc motor driven by a current source x(t), as shown in Fig. 1.40a. The torque\nT (t) generated in the motor is proportional to the armature current x(t). Therefore,", - "type": "text" - }, - { - "block_id": "p138-b20", - "global_id": 3661, - "bbox": [ - 267.82, - 459.69, - 324.41, - 470.77 - ], - "text": "T (t) = KTx(t)", - "type": "text" - }, - { - "block_id": "p138-b21", - "global_id": 3662, - "bbox": [ - 101.84, - 481.16, - 490.39, - 527.09 - ], - "text": "where KT is a constant of the motor. This torque drives a mechanical load whose free-body diagram\nis shown in Fig. 1.40b. The viscous damping (with coefficient B) dissipates a torque B ˙θ(t). If J is\nthe moment of inertia of the load (including the rotor of the motor), then the net torque T (t)−B ˙θ(t)\nmust be equal to J ¨θ(t):", - "type": "text" - }, - { - "block_id": "p138-b22", - "global_id": 3663, - "bbox": [ - 254.33, - 526.67, - 337.91, - 538.94 - ], - "text": "J ¨θ(t) = T (t) −B ˙θ(t)", - "type": "text" - }, - { - "block_id": "p138-b23", - "global_id": 3664, - "bbox": [ - 101.84, - 547.62, - 124.25, - 557.59 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p138-b24", - "global_id": 3665, - "bbox": [ - 230.41, - 555.05, - 361.82, - 570.24 - ], - "text": "(JD2 + BD)θ(t) = T (t) = KTx(t)", - "type": "text" - }, - { - "block_id": "p138-b25", - "global_id": 3666, - "bbox": [ - 101.85, - 578.12, - 292.33, - 588.09 - ], - "text": "which in conventional form can be expressed as", - "type": "text" - }, - { - "block_id": "p138-b26", - "global_id": 3667, - "bbox": [ - 240.11, - 597.49, - 271.2, - 615.69 - ], - "text": "J d2θ(t)", - "type": "text" - }, - { - "block_id": "p138-b27", - "global_id": 3668, - "bbox": [ - 252.75, - 598.43, - 311.55, - 622.76 - ], - "text": "dt2\n+ Bdθ(t)", - "type": "text" - }, - { - "block_id": "p138-b28", - "global_id": 3669, - "bbox": [ - 297.09, - 605.41, - 490.38, - 622.76 - ], - "text": "dt\n= KT x(t)\n(1.32)", - "type": "text" - } - ] - }, - { - "page_num": 139, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p139-b0", - "global_id": 3670, - "bbox": [ - 269.0, - 62.89, - 516.12, - 71.98 - ], - "text": "1.9\nInternal and External Descriptions of a System\n119", - "type": "text" - }, - { - "block_id": "p139-b1", - "global_id": 3671, - "bbox": [ - 158.77, - 87.45, - 210.05, - 97.0 - ], - "text": "if\nx(t)", - "type": "text" - }, - { - "block_id": "p139-b2", - "global_id": 3672, - "bbox": [ - 355.72, - 144.61, - 364.61, - 152.63 - ], - "text": "Bu", - "type": "text" - }, - { - "block_id": "p139-b3", - "global_id": 3673, - "bbox": [ - 362.53, - 138.13, - 364.53, - 146.13 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p139-b4", - "global_id": 3674, - "bbox": [ - 364.53, - 144.63, - 372.08, - 152.71 - ], - "text": "(t)", - "type": "text" - }, - { - "block_id": "p139-b5", - "global_id": 3675, - "bbox": [ - 268.57, - 124.35, - 469.91, - 140.72 - ], - "text": "B\nJ\nB\nJ", - "type": "text" - }, - { - "block_id": "p139-b6", - "global_id": 3676, - "bbox": [ - 242.63, - 166.34, - 423.75, - 174.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p139-b7", - "global_id": 3677, - "bbox": [ - 225.96, - 106.12, - 237.51, - 114.22 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p139-b8", - "global_id": 3678, - "bbox": [ - 151.5, - 181.44, - 311.38, - 190.68 - ], - "text": "Figure 1.40 Armature-controlled dc motor.", - "type": "text" - }, - { - "block_id": "p139-b9", - "global_id": 3679, - "bbox": [ - 127.94, - 228.1, - 324.54, - 257.98 - ], - "text": "1.9 INTERNAL AND EXTERNAL\nDESCRIPTIONS OF A SYSTEM", - "type": "text" - }, - { - "block_id": "p139-b10", - "global_id": 3680, - "bbox": [ - 127.59, - 263.87, - 516.17, - 453.27 - ], - "text": "The input–output relationship of a system is an external description of that system. We have\nfound an external description (not the internal description) of systems in all the examples\ndiscussed so far. This may puzzle the reader because in each of these cases, we derived the\ninput–output relationship by analyzing the internal structure of that system. Why is this not an\ninternal description? What makes a description internal? Although it is true that we did find the\ninput–output description by internal analysis of the system, we did so strictly for convenience. We\ncould have obtained the input–output description by making observations at the external (input\nand output) terminals, for example, by measuring the output for certain inputs, such as an impulse\nor a sinusoid. A description that can be obtained from measurements at the external terminals\n(even when the rest of the system is sealed inside an inaccessible black box) is an external\ndescription. Clearly, the input–output description is an external description. What, then, is an\ninternal description? An internal description is capable of providing complete information about\nall possible signals in the system. An external description may not give such complete information.\nAn external description can always be found from an internal description, but the converse is not\nnecessarily true. We shall now give an example to clarify the distinction between an external and\nan internal description.", - "type": "text" - }, - { - "block_id": "p139-b11", - "global_id": 3681, - "bbox": [ - 127.59, - 454.84, - 516.12, - 501.09 - ], - "text": "Let the circuit in Fig. 1.41a with the input x(t) and the output y(t) be enclosed inside a “black\nbox” with only the input and the output terminals accessible. To determine its external description,\nlet us apply a known voltage x(t) at the input terminals and measure the resulting output voltage\ny(t).", - "type": "text" - }, - { - "block_id": "p139-b12", - "global_id": 3682, - "bbox": [ - 127.59, - 502.98, - 516.15, - 572.81 - ], - "text": "Let us also assume that there is some initial charge Q0 present on the capacitor. The output\nvoltage will generally depend on both, the input x(t) and the initial charge Q0. To compute the\noutput resulting because of the charge Q0, assume the input x(t) = 0 (short across the input). In\nthis case, the currents in the two 2 resistors in the upper and the lower branches at the output\nterminals are equal and opposite because of the balanced nature of the circuit. Clearly, the capacitor\ncharge results in zero voltage at the output.†", - "type": "text" - }, - { - "block_id": "p139-b13", - "global_id": 3683, - "bbox": [ - 127.59, - 599.27, - 516.13, - 633.41 - ], - "text": "† The output voltage y(t) resulting because of the capacitor charge [assuming x(t) = 0] is the zero-input\nresponse, which, as argued above, is zero. The output component due to the input x(t) (assuming zero initial\ncapacitor charge) is the zero-state response. Complete analysis of this problem is given later in Ex. 1.21.", - "type": "text" - } - ] - }, - { - "page_num": 140, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p140-b0", - "global_id": 3684, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "120\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p140-b1", - "global_id": 3685, - "bbox": [ - 206.98, - 269.74, - 215.86, - 277.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p140-b2", - "global_id": 3686, - "bbox": [ - 125.97, - 156.54, - 137.08, - 164.62 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p140-b3", - "global_id": 3687, - "bbox": [ - 287.96, - 174.79, - 461.23, - 182.88 - ], - "text": "y(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p140-b4", - "global_id": 3688, - "bbox": [ - 184.08, - 142.95, - 196.3, - 151.25 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p140-b5", - "global_id": 3689, - "bbox": [ - 184.08, - 210.92, - 196.3, - 219.21 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p140-b6", - "global_id": 3690, - "bbox": [ - 264.43, - 141.9, - 276.66, - 150.19 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p140-b7", - "global_id": 3691, - "bbox": [ - 264.43, - 210.14, - 276.66, - 218.44 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p140-b8", - "global_id": 3692, - "bbox": [ - 192.58, - 99.59, - 194.81, - 107.59 - ], - "text": "i", - "type": "text" - }, - { - "block_id": "p140-b13", - "global_id": 3693, - "bbox": [ - 383.41, - 269.74, - 393.06, - 277.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p140-b14", - "global_id": 3694, - "bbox": [ - 370.36, - 94.95, - 382.58, - 103.25 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p140-b15", - "global_id": 3695, - "bbox": [ - 417.68, - 141.2, - 429.9, - 149.5 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p140-b16", - "global_id": 3696, - "bbox": [ - 417.68, - 209.84, - 429.9, - 218.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p140-b17", - "global_id": 3697, - "bbox": [ - 168.58, - 94.95, - 180.8, - 103.25 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p140-b18", - "global_id": 3698, - "bbox": [ - 197.75, - 175.75, - 263.37, - 183.75 - ], - "text": "a\nb", - "type": "text" - }, - { - "block_id": "p140-b19", - "global_id": 3699, - "bbox": [ - 206.37, - 243.88, - 254.69, - 251.88 - ], - "text": "d\nd", - "type": "text" - }, - { - "block_id": "p140-b20", - "global_id": 3700, - "bbox": [ - 202.47, - 101.24, - 258.75, - 109.24 - ], - "text": "c\nc", - "type": "text" - }, - { - "block_id": "p140-b21", - "global_id": 3701, - "bbox": [ - 412.07, - 117.63, - 414.3, - 125.63 - ], - "text": "i", - "type": "text" - }, - { - "block_id": "p140-b22", - "global_id": 3702, - "bbox": [ - 134.06, - 102.19, - 137.61, - 110.19 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p140-b23", - "global_id": 3703, - "bbox": [ - 321.73, - 109.63, - 325.28, - 117.63 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p140-b24", - "global_id": 3704, - "bbox": [ - 341.01, - 100.87, - 343.23, - 108.87 - ], - "text": "i", - "type": "text" - }, - { - "block_id": "p140-b25", - "global_id": 3705, - "bbox": [ - 316.81, - 156.54, - 327.92, - 164.62 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p140-b29", - "global_id": 3706, - "bbox": [ - 210.83, - 117.34, - 267.08, - 134.31 - ], - "text": "i\n2\ni\n2", - "type": "text" - }, - { - "block_id": "p140-b30", - "global_id": 3707, - "bbox": [ - 125.76, - 284.44, - 396.56, - 293.67 - ], - "text": "Figure 1.41 A system that cannot be described by external measurements.", - "type": "text" - }, - { - "block_id": "p140-b31", - "global_id": 3708, - "bbox": [ - 101.84, - 313.29, - 490.4, - 383.44 - ], - "text": "Now, to compute the output y(t) resulting from the input voltage x(t), we assume zero initial\ncapacitor charge (short across the capacitor terminals). The current i(t) (Fig. 1.41a), in this case,\ndivides equally between the two parallel branches because the circuit is balanced. Thus, the voltage\nacross the capacitor continues to remain zero. Therefore, for the purpose of computing the current\ni(t), the capacitor may be removed or replaced by a short. The resulting circuit is equivalent to that\nshown in Fig. 1.41b, which shows that the input x(t) sees a load of 5, and", - "type": "text" - }, - { - "block_id": "p140-b32", - "global_id": 3709, - "bbox": [ - 273.25, - 392.55, - 302.95, - 404.18 - ], - "text": "i(t) = 1", - "type": "text" - }, - { - "block_id": "p140-b33", - "global_id": 3710, - "bbox": [ - 299.47, - 393.9, - 318.98, - 407.1 - ], - "text": "5x(t)", - "type": "text" - }, - { - "block_id": "p140-b34", - "global_id": 3711, - "bbox": [ - 101.85, - 414.74, - 208.07, - 425.12 - ], - "text": "Also, because y(t) = 2i(t),", - "type": "text" - }, - { - "block_id": "p140-b35", - "global_id": 3712, - "bbox": [ - 272.42, - 425.35, - 303.78, - 436.97 - ], - "text": "y(t) = 2", - "type": "text" - }, - { - "block_id": "p140-b36", - "global_id": 3713, - "bbox": [ - 300.29, - 426.69, - 319.81, - 439.88 - ], - "text": "5x(t)", - "type": "text" - }, - { - "block_id": "p140-b37", - "global_id": 3714, - "bbox": [ - 101.84, - 445.49, - 490.42, - 527.18 - ], - "text": "This is the total response. Clearly, for the external description, the capacitor does not exist.\nNo external measurement or external observation can detect the presence of the capacitor.\nFurthermore, if the circuit is enclosed inside a “black box” so that only the external terminals are\naccessible, it is impossible to determine the currents (or voltages) inside the circuit from external\nmeasurements or observations. An internal description, however, can provide every possible signal\ninside the system. In Ex. 1.21, we shall find the internal description of this system and show that\nit is capable of determining every possible signal in the system.", - "type": "text" - }, - { - "block_id": "p140-b38", - "global_id": 3715, - "bbox": [ - 101.84, - 529.18, - 490.39, - 563.06 - ], - "text": "For most systems, the external and internal descriptions are equivalent, but there are a few\nexceptions, as in the present case, where the external description gives an inadequate picture of\nthe system. This happens when the system is uncontrollable and/or unobservable.", - "type": "text" - }, - { - "block_id": "p140-b39", - "global_id": 3716, - "bbox": [ - 101.84, - 565.05, - 490.4, - 635.56 - ], - "text": "Figure 1.42 shows structural representations of simple uncontrollable and unobservable\nsystems. In Fig. 1.42a, we note that part of the system (subsystem S2) inside the box cannot\nbe controlled by the input x(t). In Fig. 1.42b, some of the system outputs (those in subsystem\nS2) cannot be observed from the output terminals. If we try to describe either of these systems\nby applying an external input x(t) and then measuring the output y(t), the measurement will not\ncharacterize the complete system but only the part of the system (here S1) that is both controllable", - "type": "text" - } - ] - }, - { - "page_num": 141, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p141-b0", - "global_id": 3717, - "bbox": [ - 253.53, - 62.89, - 516.12, - 71.98 - ], - "text": "1.10\nInternal Description: The State-Space Description\n121", - "type": "text" - }, - { - "block_id": "p141-b1", - "global_id": 3718, - "bbox": [ - 230.56, - 173.74, - 239.44, - 181.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p141-b2", - "global_id": 3719, - "bbox": [ - 154.95, - 95.1, - 320.89, - 113.59 - ], - "text": "y(t)\n\nx(t)", - "type": "text" - }, - { - "block_id": "p141-b3", - "global_id": 3720, - "bbox": [ - 242.93, - 105.73, - 249.93, - 115.34 - ], - "text": "S1", - "type": "text" - }, - { - "block_id": "p141-b4", - "global_id": 3721, - "bbox": [ - 195.8, - 141.93, - 202.8, - 151.54 - ], - "text": "S2", - "type": "text" - }, - { - "block_id": "p141-b5", - "global_id": 3722, - "bbox": [ - 405.44, - 173.74, - 415.09, - 181.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p141-b6", - "global_id": 3723, - "bbox": [ - 339.5, - 118.37, - 350.61, - 126.45 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p141-b7", - "global_id": 3724, - "bbox": [ - 469.39, - 97.77, - 480.5, - 105.85 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p141-b8", - "global_id": 3725, - "bbox": [ - 396.46, - 110.21, - 403.46, - 119.81 - ], - "text": "S1", - "type": "text" - }, - { - "block_id": "p141-b9", - "global_id": 3726, - "bbox": [ - 440.96, - 138.65, - 447.96, - 148.26 - ], - "text": "S2", - "type": "text" - }, - { - "block_id": "p141-b10", - "global_id": 3727, - "bbox": [ - 151.5, - 188.44, - 399.56, - 197.68 - ], - "text": "Figure 1.42 Structures of uncontrollable and unobservable systems.", - "type": "text" - }, - { - "block_id": "p141-b11", - "global_id": 3728, - "bbox": [ - 127.59, - 210.9, - 516.16, - 256.73 - ], - "text": "and observable (linked to both the input and output). Such systems are undesirable in practice and\nshould be avoided in any system design. The system in Fig. 1.41a can be shown to be neither\ncontrollable nor observable. It can be represented structurally as a combination of the systems in\nFigs. 1.42a and 1.42b.", - "type": "text" - }, - { - "block_id": "p141-b12", - "global_id": 3729, - "bbox": [ - 127.94, - 281.72, - 447.11, - 311.61 - ], - "text": "1.10 INTERNAL DESCRIPTION: THE STATE-SPACE\nDESCRIPTION", - "type": "text" - }, - { - "block_id": "p141-b13", - "global_id": 3730, - "bbox": [ - 127.59, - 317.49, - 516.14, - 399.29 - ], - "text": "We shall now introduce the state-space description of a linear system, which is an internal\ndescription of a system. In this approach, we identify certain key variables, called the state\nvariables, of the system. These variables have the property that every possible signal in the system\ncan be expressed as a linear combination of these state variables. For example, we can show\nthat every possible signal in a passive RLC circuit can be expressed as a linear combination of\nindependent capacitor voltages and inductor currents, which, therefore, are state variables for the\ncircuit.", - "type": "text" - }, - { - "block_id": "p141-b14", - "global_id": 3731, - "bbox": [ - 127.59, - 401.28, - 516.15, - 459.06 - ], - "text": "To illustrate this point, consider the network in Fig. 1.43. We identify two state variables: the\ncapacitor voltage q1 and the inductor current q2. If the values of q1, q2, and the input x(t) are known\nat some instant t, we can demonstrate that every possible signal (current or voltage) in the circuit\ncan be determined at t. For example, if q1 = 10, q2 = 1, and the input x = 20 at some instant, the\nremaining voltages and currents at that instant will be", - "type": "text" - }, - { - "block_id": "p141-b15", - "global_id": 3732, - "bbox": [ - 260.48, - 472.28, - 393.27, - 483.74 - ], - "text": "i1 = (x −q1)/1 = 20 −10 = 10A", - "type": "text" - }, - { - "block_id": "p141-b16", - "global_id": 3733, - "bbox": [ - 275.92, - 487.23, - 393.27, - 498.69 - ], - "text": "v1 = x −q1 = 20 −10 = 10V", - "type": "text" - }, - { - "block_id": "p141-b17", - "global_id": 3734, - "bbox": [ - 333.91, - 502.17, - 393.27, - 513.63 - ], - "text": "v2 = q1 = 10V", - "type": "text" - }, - { - "block_id": "p141-b18", - "global_id": 3735, - "bbox": [ - 250.44, - 517.12, - 393.28, - 543.52 - ], - "text": "i2 = q1/2 = 5A\niC = i1 −i2 −q2 = 10 −5 −1 = 4A", - "type": "text" - }, - { - "block_id": "p141-b19", - "global_id": 3736, - "bbox": [ - 281.52, - 547.01, - 516.13, - 588.35 - ], - "text": "i3 = q2 = 1A\nv3 = 5q2 = 5V\nvL = q1 −v3 = 10 −5 = 5V\n(1.33)", - "type": "text" - }, - { - "block_id": "p141-b20", - "global_id": 3737, - "bbox": [ - 127.59, - 600.81, - 516.15, - 646.74 - ], - "text": "Thus all signals in this circuit are determined. Clearly, state variables consist of the key variables\nin a system; a knowledge of the state variables allows one to determine every possible output of\nthe system. Note that the state-variable description is an internal description of a system because\nit is capable of describing all possible signals in the system.", - "type": "text" - } - ] - }, - { - "page_num": 142, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p142-b0", - "global_id": 3738, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "122\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p142-b1", - "global_id": 3739, - "bbox": [ - 125.97, - 139.77, - 137.08, - 147.85 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p142-b2", - "global_id": 3740, - "bbox": [ - 183.87, - 89.19, - 196.1, - 97.48 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p142-b3", - "global_id": 3741, - "bbox": [ - 292.28, - 145.69, - 366.51, - 153.98 - ], - "text": "2 \t\n5", - "type": "text" - }, - { - "block_id": "p142-b4", - "global_id": 3742, - "bbox": [ - 319.28, - 89.48, - 331.06, - 97.48 - ], - "text": "2 H", - "type": "text" - }, - { - "block_id": "p142-b5", - "global_id": 3743, - "bbox": [ - 235.78, - 145.98, - 246.23, - 153.98 - ], - "text": "1 F", - "type": "text" - }, - { - "block_id": "p142-b6", - "global_id": 3744, - "bbox": [ - 183.38, - 111.2, - 330.32, - 124.31 - ], - "text": "v1(t)\nvL(t)", - "type": "text" - }, - { - "block_id": "p142-b7", - "global_id": 3745, - "bbox": [ - 202.88, - 137.12, - 398.49, - 147.81 - ], - "text": "q1(t)\nv3(t)", - "type": "text" - }, - { - "block_id": "p142-b8", - "global_id": 3746, - "bbox": [ - 148.38, - 92.7, - 293.93, - 102.31 - ], - "text": "i1(t)\nq2(t)", - "type": "text" - }, - { - "block_id": "p142-b9", - "global_id": 3747, - "bbox": [ - 172.13, - 107.7, - 394.16, - 118.72 - ], - "text": "iC(t)\ni2(t)\ni3(t)", - "type": "text" - }, - { - "block_id": "p142-b18", - "global_id": 3748, - "bbox": [ - 260.64, - 138.7, - 274.75, - 148.31 - ], - "text": "v2(t)", - "type": "text" - }, - { - "block_id": "p142-b19", - "global_id": 3749, - "bbox": [ - 125.76, - 199.44, - 352.32, - 208.68 - ], - "text": "Figure 1.43 Choosing suitable initial conditions in a network.", - "type": "text" - }, - { - "block_id": "p142-b20", - "global_id": 3750, - "bbox": [ - 76.77, - 250.08, - 377.49, - 262.03 - ], - "text": "EXAMPLE 1.20\nState-Space Description of a System", - "type": "text" - }, - { - "block_id": "p142-b21", - "global_id": 3751, - "bbox": [ - 103.16, - 278.7, - 476.98, - 314.07 - ], - "text": "This example illustrates how state equations may be natural and easier to determine than other\ndescriptions, such as loop or node equations. Consider again the network in Fig. 1.43 with q1\nand q2 as the state variables and write the state equations.", - "type": "text" - }, - { - "block_id": "p142-b22", - "global_id": 3752, - "bbox": [ - 103.16, - 335.07, - 477.02, - 357.4 - ], - "text": "This can be done by simple inspection of Fig. 1.43. Since ˙q1 is the current through the\ncapacitor,", - "type": "text" - }, - { - "block_id": "p142-b23", - "global_id": 3753, - "bbox": [ - 237.19, - 368.94, - 321.68, - 380.4 - ], - "text": "˙q1 = iC = i1 −i2 −q2", - "type": "text" - }, - { - "block_id": "p142-b24", - "global_id": 3754, - "bbox": [ - 248.21, - 383.89, - 342.48, - 410.29 - ], - "text": "= (x −q1) −0.5q1 −q2\n= −1.5q1 −q2 + x", - "type": "text" - }, - { - "block_id": "p142-b25", - "global_id": 3755, - "bbox": [ - 103.16, - 420.75, - 312.83, - 431.89 - ], - "text": "Also 2 ˙q2, the voltage across the inductor, is given by", - "type": "text" - }, - { - "block_id": "p142-b26", - "global_id": 3756, - "bbox": [ - 260.29, - 442.67, - 319.38, - 469.07 - ], - "text": "2˙q2 = q1 −v3\n= q1 −5q2", - "type": "text" - }, - { - "block_id": "p142-b27", - "global_id": 3757, - "bbox": [ - 103.16, - 479.94, - 111.46, - 489.9 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p142-b28", - "global_id": 3758, - "bbox": [ - 251.71, - 491.49, - 327.95, - 502.94 - ], - "text": "˙q2 = 0.5q1 −2.5q2", - "type": "text" - }, - { - "block_id": "p142-b29", - "global_id": 3759, - "bbox": [ - 103.16, - 510.83, - 216.32, - 520.79 - ], - "text": "Thus, the state equations are", - "type": "text" - }, - { - "block_id": "p142-b30", - "global_id": 3760, - "bbox": [ - 246.95, - 532.33, - 333.2, - 543.78 - ], - "text": "˙q1 = −1.5q1 −q2 + x", - "type": "text" - }, - { - "block_id": "p142-b31", - "global_id": 3761, - "bbox": [ - 256.48, - 547.28, - 477.01, - 558.73 - ], - "text": "˙q2 = 0.5q1 −2.5q2\n(1.34)", - "type": "text" - }, - { - "block_id": "p142-b32", - "global_id": 3762, - "bbox": [ - 103.17, - 569.61, - 477.02, - 627.39 - ], - "text": "This is a set of two simultaneous first-order differential equations. This set of equations\ncomprises the state equations. Once these equations have been solved for q1 and q2, everything\nelse in the circuit can be determined by using Eq. (1.33), which are known as the output\nequations. Thus, in this approach, we have two sets of equations, the state equations and the\noutput equations. Once we have solved the state equations, all possible outputs can be obtained", - "type": "text" - } - ] - }, - { - "page_num": 143, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p143-b0", - "global_id": 3763, - "bbox": [ - 253.53, - 62.89, - 516.12, - 71.98 - ], - "text": "1.10\nInternal Description: The State-Space Description\n123", - "type": "text" - }, - { - "block_id": "p143-b1", - "global_id": 3764, - "bbox": [ - 128.9, - 86.14, - 502.75, - 120.11 - ], - "text": "from the output equations. In the input–output description, an Nth-order system is described\nby an Nth-order equation. In the state-variable approach, the same system is described by N\nsimultaneous first-order state equations.†", - "type": "text" - }, - { - "block_id": "p143-b2", - "global_id": 3765, - "bbox": [ - 102.51, - 158.06, - 390.6, - 170.02 - ], - "text": "EXAMPLE 1.21\nControllability and Observability", - "type": "text" - }, - { - "block_id": "p143-b3", - "global_id": 3766, - "bbox": [ - 128.9, - 186.68, - 502.75, - 208.6 - ], - "text": "Investigate the nature of state equations and the issue of controllability and observability for\nthe circuit in Fig. 1.41a.", - "type": "text" - }, - { - "block_id": "p143-b4", - "global_id": 3767, - "bbox": [ - 128.9, - 231.51, - 502.77, - 313.21 - ], - "text": "This circuit has only one capacitor and no inductors. Hence, there is only one state variable,\nthe capacitor voltage q(t). Since C = 1 F, the capacitor current is ˙q. There are two sources in\nthis circuit: the input x(t) and the capacitor voltage q(t). The response due to x(t), assuming\nq(t) = 0, is the zero-state response, which can be found from Fig. 1.44a, where we have shorted\nthe capacitor [q(t) = 0]. The response due to q(t) assuming x(t) = 0, is the zero-input response,\nwhich can be found from Fig. 1.44b, where we have shorted x(t) to ensure x(t) = 0. It is now\ntrivial to find both the components.", - "type": "text" - }, - { - "block_id": "p143-b5", - "global_id": 3768, - "bbox": [ - 128.91, - 314.78, - 502.77, - 349.07 - ], - "text": "Figure 1.44a shows zero-state currents in every branch. It is clear that the input x(t) sees\nan effective resistance of 5 , and, hence, the current through x(t) is x/5 A, which divides in\nthe two parallel branches, resulting in the current x/10 through each branch.", - "type": "text" - }, - { - "block_id": "p143-b6", - "global_id": 3769, - "bbox": [ - 128.91, - 351.06, - 502.78, - 420.8 - ], - "text": "Examining the circuit in Fig. 1.44b for the zero-input response, we note that the capacitor\nvoltage is q and the current is ˙q. We also observe that the capacitor sees two loops in parallel,\neach with resistance 4 and current ˙q/2. Interestingly, the 3 branch is effectively shorted\nbecause the circuit is balanced, and thus the voltage across the terminals cd is zero. The\ntotal current in any branch is the sum of the currents in that branch in Figs. 1.44a and 1.44b\n(principle of superposition).", - "type": "text" - }, - { - "block_id": "p143-b7", - "global_id": 3770, - "bbox": [ - 241.71, - 432.32, - 371.17, - 442.28 - ], - "text": "Branch\nCurrent\nVoltage", - "type": "text" - }, - { - "block_id": "p143-b8", - "global_id": 3771, - "bbox": [ - 241.71, - 435.95, - 361.01, - 467.39 - ], - "text": "ca\nx\n10 + ˙q\n2\n2\n x", - "type": "text" - }, - { - "block_id": "p143-b9", - "global_id": 3772, - "bbox": [ - 353.83, - 442.96, - 382.03, - 467.39 - ], - "text": "10 + ˙q\n2", - "type": "text" - }, - { - "block_id": "p143-b11", - "global_id": 3773, - "bbox": [ - 241.71, - 463.95, - 361.01, - 495.38 - ], - "text": "cb\nx\n10 −˙q\n2\n2\n x", - "type": "text" - }, - { - "block_id": "p143-b12", - "global_id": 3774, - "bbox": [ - 353.83, - 470.95, - 382.03, - 495.38 - ], - "text": "10 −˙q\n2", - "type": "text" - }, - { - "block_id": "p143-b14", - "global_id": 3775, - "bbox": [ - 241.71, - 491.95, - 361.01, - 523.38 - ], - "text": "ad\nx\n10 −˙q\n2\n2\n x", - "type": "text" - }, - { - "block_id": "p143-b15", - "global_id": 3776, - "bbox": [ - 353.83, - 498.95, - 382.03, - 523.38 - ], - "text": "10 −˙q\n2", - "type": "text" - }, - { - "block_id": "p143-b17", - "global_id": 3777, - "bbox": [ - 241.71, - 519.94, - 361.01, - 551.37 - ], - "text": "bd\nx\n10 + ˙q\n2\n2\n x", - "type": "text" - }, - { - "block_id": "p143-b18", - "global_id": 3778, - "bbox": [ - 353.83, - 526.94, - 382.03, - 551.37 - ], - "text": "10 + ˙q\n2", - "type": "text" - }, - { - "block_id": "p143-b20", - "global_id": 3779, - "bbox": [ - 241.71, - 547.94, - 358.52, - 579.37 - ], - "text": "ec\nx\n5\n3\nx", - "type": "text" - }, - { - "block_id": "p143-b21", - "global_id": 3780, - "bbox": [ - 353.83, - 569.41, - 358.81, - 579.37 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p143-b23", - "global_id": 3781, - "bbox": [ - 241.71, - 576.07, - 345.35, - 600.19 - ], - "text": "ed\nx\n5\nx", - "type": "text" - }, - { - "block_id": "p143-b24", - "global_id": 3782, - "bbox": [ - 478.69, - 509.63, - 502.75, - 519.59 - ], - "text": "(1.35)", - "type": "text" - }, - { - "block_id": "p143-b25", - "global_id": 3783, - "bbox": [ - 127.59, - 630.16, - 516.09, - 653.34 - ], - "text": "† This assumes the system to be controllable and observable. If it is not, the input–output description equation\nwill be of an order lower than the corresponding number of state equations.", - "type": "text" - } - ] - }, - { - "page_num": 144, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p144-b0", - "global_id": 3784, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "124\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p144-b1", - "global_id": 3785, - "bbox": [ - 185.55, - 268.16, - 194.43, - 276.16 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p144-b2", - "global_id": 3786, - "bbox": [ - 94.42, - 154.86, - 105.52, - 162.94 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p144-b3", - "global_id": 3787, - "bbox": [ - 259.35, - 172.77, - 270.45, - 180.85 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p144-b4", - "global_id": 3788, - "bbox": [ - 155.97, - 140.82, - 168.19, - 149.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b5", - "global_id": 3789, - "bbox": [ - 155.97, - 208.79, - 168.19, - 217.08 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b6", - "global_id": 3790, - "bbox": [ - 235.91, - 139.76, - 248.14, - 148.06 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b7", - "global_id": 3791, - "bbox": [ - 235.91, - 208.01, - 248.14, - 216.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b12", - "global_id": 3792, - "bbox": [ - 137.03, - 92.93, - 149.26, - 101.23 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p144-b13", - "global_id": 3793, - "bbox": [ - 168.84, - 173.98, - 234.9, - 182.96 - ], - "text": "a\nb", - "type": "text" - }, - { - "block_id": "p144-b14", - "global_id": 3794, - "bbox": [ - 177.9, - 241.85, - 226.15, - 249.85 - ], - "text": "d\nd", - "type": "text" - }, - { - "block_id": "p144-b15", - "global_id": 3795, - "bbox": [ - 105.4, - 100.35, - 286.18, - 108.35 - ], - "text": "c\nc\ne\ne", - "type": "text" - }, - { - "block_id": "p144-b16", - "global_id": 3796, - "bbox": [ - 436.2, - 172.77, - 447.3, - 180.85 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p144-b17", - "global_id": 3797, - "bbox": [ - 332.92, - 140.82, - 345.15, - 149.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b18", - "global_id": 3798, - "bbox": [ - 332.92, - 208.79, - 345.15, - 217.08 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b19", - "global_id": 3799, - "bbox": [ - 412.81, - 139.76, - 425.03, - 148.06 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b20", - "global_id": 3800, - "bbox": [ - 412.81, - 208.01, - 425.03, - 216.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p144-b21", - "global_id": 3801, - "bbox": [ - 315.05, - 92.93, - 327.28, - 101.23 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p144-b22", - "global_id": 3802, - "bbox": [ - 345.05, - 173.73, - 411.75, - 182.96 - ], - "text": "a\nb", - "type": "text" - }, - { - "block_id": "p144-b23", - "global_id": 3803, - "bbox": [ - 354.75, - 241.85, - 403.15, - 249.85 - ], - "text": "d\nd", - "type": "text" - }, - { - "block_id": "p144-b24", - "global_id": 3804, - "bbox": [ - 355.0, - 100.35, - 410.62, - 108.35 - ], - "text": "c\nc", - "type": "text" - }, - { - "block_id": "p144-b25", - "global_id": 3805, - "bbox": [ - 360.22, - 268.16, - 369.87, - 276.16 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p144-b29", - "global_id": 3806, - "bbox": [ - 338.83, - 157.77, - 419.74, - 167.13 - ], - "text": "q/2\nq/2", - "type": "text" - }, - { - "block_id": "p144-b30", - "global_id": 3807, - "bbox": [ - 338.83, - 190.72, - 419.74, - 198.88 - ], - "text": "q/2\nq/2", - "type": "text" - }, - { - "block_id": "p144-b31", - "global_id": 3808, - "bbox": [ - 372.77, - 144.8, - 382.99, - 152.88 - ], - "text": "q/2", - "type": "text" - }, - { - "block_id": "p144-b32", - "global_id": 3809, - "bbox": [ - 372.77, - 209.3, - 382.99, - 217.38 - ], - "text": "q/2", - "type": "text" - }, - { - "block_id": "p144-b33", - "global_id": 3810, - "bbox": [ - 374.04, - 184.09, - 378.04, - 192.09 - ], - "text": "q", - "type": "text" - }, - { - "block_id": "p144-b34", - "global_id": 3811, - "bbox": [ - 358.27, - 167.21, - 362.27, - 175.21 - ], - "text": "q", - "type": "text" - }, - { - "block_id": "p144-b35", - "global_id": 3812, - "bbox": [ - 180.44, - 116.03, - 188.44, - 132.99 - ], - "text": "x\n10", - "type": "text" - }, - { - "block_id": "p144-b36", - "global_id": 3813, - "bbox": [ - 180.44, - 184.28, - 241.77, - 201.24 - ], - "text": "x\n10\nx\n10", - "type": "text" - }, - { - "block_id": "p144-b37", - "global_id": 3814, - "bbox": [ - 233.77, - 116.03, - 241.77, - 132.99 - ], - "text": "x\n10", - "type": "text" - }, - { - "block_id": "p144-b38", - "global_id": 3815, - "bbox": [ - 164.02, - 89.0, - 168.02, - 105.96 - ], - "text": "x\n5", - "type": "text" - }, - { - "block_id": "p144-b39", - "global_id": 3816, - "bbox": [ - 94.2, - 282.86, - 372.14, - 292.09 - ], - "text": "Figure 1.44 Analysis of a system that is neither controllable nor observable.", - "type": "text" - }, - { - "block_id": "p144-b40", - "global_id": 3817, - "bbox": [ - 103.16, - 304.41, - 477.01, - 326.74 - ], - "text": "To find the state equation, we note that the current in branch ca is (x/10)+ ˙q/2 and the current\nin branch cb is (x/10) −˙q/2. Hence, the equation around the loop acba is", - "type": "text" - }, - { - "block_id": "p144-b41", - "global_id": 3818, - "bbox": [ - 211.11, - 343.92, - 232.94, - 354.3 - ], - "text": "q = 2", - "type": "text" - }, - { - "block_id": "p144-b43", - "global_id": 3819, - "bbox": [ - 238.37, - 337.25, - 254.51, - 353.88 - ], - "text": "−x", - "type": "text" - }, - { - "block_id": "p144-b44", - "global_id": 3820, - "bbox": [ - 247.34, - 336.94, - 275.54, - 361.37 - ], - "text": "10 −˙q\n2", - "type": "text" - }, - { - "block_id": "p144-b45", - "global_id": 3821, - "bbox": [ - 276.74, - 329.94, - 282.17, - 339.9 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p144-b46", - "global_id": 3822, - "bbox": [ - 283.71, - 343.92, - 298.01, - 354.3 - ], - "text": "+ 2", - "type": "text" - }, - { - "block_id": "p144-b47", - "global_id": 3823, - "bbox": [ - 298.01, - 329.94, - 311.82, - 347.21 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p144-b48", - "global_id": 3824, - "bbox": [ - 304.64, - 336.94, - 332.85, - 361.37 - ], - "text": "10 −˙q\n2", - "type": "text" - }, - { - "block_id": "p144-b49", - "global_id": 3825, - "bbox": [ - 334.04, - 329.94, - 339.47, - 339.9 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p144-b50", - "global_id": 3826, - "bbox": [ - 341.52, - 343.92, - 369.06, - 354.3 - ], - "text": "= −2˙q", - "type": "text" - }, - { - "block_id": "p144-b51", - "global_id": 3827, - "bbox": [ - 103.16, - 369.33, - 111.46, - 379.29 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p144-b52", - "global_id": 3828, - "bbox": [ - 269.06, - 380.87, - 477.01, - 391.24 - ], - "text": "˙q = −0.5q\n(1.36)", - "type": "text" - }, - { - "block_id": "p144-b53", - "global_id": 3829, - "bbox": [ - 103.16, - 400.22, - 235.97, - 410.18 - ], - "text": "This is the desired state equation.", - "type": "text" - }, - { - "block_id": "p144-b54", - "global_id": 3830, - "bbox": [ - 103.16, - 411.76, - 477.04, - 458.0 - ], - "text": "Substitution of ˙q = −0.5q in Eq. (1.35) shows that every possible current and voltage in\nthe circuit can be expressed in terms of the state variable q and the input x, as desired. Hence,\nthe set of Eq. (1.35) is the output equation for this circuit. Once we have solved the state\nequation [Eq. (1.36)] for q, we can determine every possible output in the circuit.", - "type": "text" - }, - { - "block_id": "p144-b55", - "global_id": 3831, - "bbox": [ - 121.09, - 459.57, - 227.66, - 469.95 - ], - "text": "The output y(t) is given by", - "type": "text" - }, - { - "block_id": "p144-b56", - "global_id": 3832, - "bbox": [ - 207.84, - 487.14, - 239.5, - 497.52 - ], - "text": "y(t) = 2", - "type": "text" - }, - { - "block_id": "p144-b57", - "global_id": 3833, - "bbox": [ - 239.5, - 473.15, - 253.3, - 490.43 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p144-b58", - "global_id": 3834, - "bbox": [ - 246.12, - 480.16, - 274.33, - 504.59 - ], - "text": "10 −˙q\n2", - "type": "text" - }, - { - "block_id": "p144-b59", - "global_id": 3835, - "bbox": [ - 275.52, - 473.15, - 280.95, - 483.12 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p144-b60", - "global_id": 3836, - "bbox": [ - 282.51, - 487.14, - 296.8, - 497.52 - ], - "text": "+ 2", - "type": "text" - }, - { - "block_id": "p144-b61", - "global_id": 3837, - "bbox": [ - 296.8, - 473.15, - 310.61, - 490.43 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p144-b62", - "global_id": 3838, - "bbox": [ - 303.43, - 480.16, - 331.63, - 504.59 - ], - "text": "10 + ˙q\n2", - "type": "text" - }, - { - "block_id": "p144-b63", - "global_id": 3839, - "bbox": [ - 332.84, - 473.15, - 338.26, - 483.12 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p144-b64", - "global_id": 3840, - "bbox": [ - 340.31, - 480.57, - 356.3, - 497.52 - ], - "text": "= 2", - "type": "text" - }, - { - "block_id": "p144-b65", - "global_id": 3841, - "bbox": [ - 351.32, - 487.14, - 477.01, - 504.59 - ], - "text": "5x(t)\n(1.37)", - "type": "text" - }, - { - "block_id": "p144-b66", - "global_id": 3842, - "bbox": [ - 103.16, - 514.76, - 477.05, - 620.37 - ], - "text": "A little examination of the state and the output equations indicates the nature of this\nsystem. Equation (1.36) shows that the state q(t) is independent of the input x(t); hence the\nsystem state q cannot be controlled by the input. Moreover, Eq. (1.37) shows that the output\ny(t) does not depend on the state q(t). Thus, the system state cannot be observed from the\noutput terminals. Hence, the system is neither controllable nor observable. Such is not the case\nof other systems examined earlier. Consider, for example, the circuit in Fig. 1.43. The state\nequations [Eq. (1.34)] show that the states are influenced by the input directly or indirectly.\nHence, the system is controllable. Moreover, as Eq. (1.33) shows, every possible output is\nexpressed in terms of the state variables and the input. Hence, the states are also observable.", - "type": "text" - } - ] - }, - { - "page_num": 145, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p145-b0", - "global_id": 3843, - "bbox": [ - 253.53, - 62.89, - 516.12, - 71.98 - ], - "text": "1.10\nInternal Description: The State-Space Description\n125", - "type": "text" - }, - { - "block_id": "p145-b1", - "global_id": 3844, - "bbox": [ - 127.59, - 85.82, - 516.16, - 107.74 - ], - "text": "State-space techniques are useful not just because of their ability to provide internal system\ndescription, but for several other reasons, including the following.", - "type": "text" - }, - { - "block_id": "p145-b2", - "global_id": 3845, - "bbox": [ - 144.52, - 115.7, - 516.18, - 281.09 - ], - "text": "1. State equations of a system provide a mathematical model of great generality that can\ndescribe not just linear systems, but also nonlinear systems; not just time-invariant systems,\nbut also time-varying parameter systems; not just SISO (single-input/single-output)\nsystems,\nbut\nalso\nmultiple-input/multiple-output\n(MIMO)\nsystems.\nIndeed,\nstate\nequations are ideally suited for the analysis, synthesis, and optimization of MIMO\nsystems.\n2. Compact matrix notation and the powerful techniques of linear algebra greatly facilitate\ncomplex manipulations. Without such features, many important results of the modern\nsystem theory would have been difficult to obtain. State equations can yield a great deal of\ninformation about a system even when they are not solved explicitly.\n3. State equations lend themselves readily to digital computer simulation of complex systems\nof high order, with or without nonlinearities, and with multiple inputs and outputs.\n4. For second-order systems (N = 2), a graphical method called phase-plane analysis can be\nused on state equations, whether they are linear or nonlinear.", - "type": "text" - }, - { - "block_id": "p145-b3", - "global_id": 3846, - "bbox": [ - 127.59, - 289.05, - 516.15, - 346.84 - ], - "text": "The real benefits of the state-space approach, however, are realized for highly complex\nsystems of large order. Much of the book is devoted to introduction of the basic concepts of linear\nsystems analysis, which must necessarily begin with simpler systems without using the state-space\napproach. Chapter 10 deals with the state-space analysis of linear, time-invariant, continuous-time,\nand discrete-time systems.", - "type": "text" - }, - { - "block_id": "p145-b4", - "global_id": 3847, - "bbox": [ - 133.57, - 381.8, - 432.03, - 393.76 - ], - "text": "DRILL 1.20\nState Equations for a Series RLC Circuit", - "type": "text" - }, - { - "block_id": "p145-b5", - "global_id": 3848, - "bbox": [ - 133.57, - 402.78, - 510.15, - 437.53 - ], - "text": "Write the state equations for the series RLC circuit shown in Fig. 1.45, using the inductor current\nq1(t) and the capacitor voltage q2(t) as state variables. Express every voltage and current in this\ncircuit as a linear combination of q1, q2, and x.", - "type": "text" - }, - { - "block_id": "p145-b6", - "global_id": 3849, - "bbox": [ - 133.84, - 450.27, - 196.06, - 461.23 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p145-b7", - "global_id": 3850, - "bbox": [ - 133.57, - 468.22, - 270.13, - 479.67 - ], - "text": "q1 = −3q1 −q2 + x and q2 = 2q1.", - "type": "text" - }, - { - "block_id": "p145-b8", - "global_id": 3851, - "bbox": [ - 146.62, - 560.12, - 301.12, - 569.85 - ], - "text": "x(t)\nq2(t)", - "type": "text" - }, - { - "block_id": "p145-b9", - "global_id": 3852, - "bbox": [ - 213.6, - 540.37, - 228.15, - 549.98 - ], - "text": "q1(t)", - "type": "text" - }, - { - "block_id": "p145-b10", - "global_id": 3853, - "bbox": [ - 219.01, - 515.74, - 230.78, - 523.74 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p145-b13", - "global_id": 3854, - "bbox": [ - 214.89, - 608.96, - 227.11, - 617.25 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p145-b14", - "global_id": 3855, - "bbox": [ - 249.77, - 558.65, - 260.1, - 572.14 - ], - "text": "F\n2\n1", - "type": "text" - }, - { - "block_id": "p145-b15", - "global_id": 3856, - "bbox": [ - 314.09, - 610.29, - 441.11, - 619.53 - ], - "text": "Figure 1.45 Circuit for Drill 1.20.", - "type": "text" - } - ] - }, - { - "page_num": 146, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p146-b0", - "global_id": 3857, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "126\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p146-b1", - "global_id": 3858, - "bbox": [ - 102.2, - 94.37, - 394.53, - 108.31 - ], - "text": "1.11 MATLAB: WORKING WITH FUNCTIONS", - "type": "text" - }, - { - "block_id": "p146-b2", - "global_id": 3859, - "bbox": [ - 101.84, - 114.3, - 490.39, - 148.18 - ], - "text": "Working with functions is fundamental to signals and systems applications. MATLAB provides\nseveral methods of defining and evaluating functions. An understanding and proficient use of these\nmethods are therefore necessary and beneficial.", - "type": "text" - }, - { - "block_id": "p146-b3", - "global_id": 3860, - "bbox": [ - 101.84, - 167.64, - 262.08, - 179.59 - ], - "text": "1.11-1 Anonymous Functions", - "type": "text" - }, - { - "block_id": "p146-b4", - "global_id": 3861, - "bbox": [ - 101.84, - 185.72, - 490.37, - 231.55 - ], - "text": "Many simple functions are most conveniently represented by using MATLAB anonymous\nfunctions. An anonymous function provides a symbolic representation of a function defined in\nterms of MATLAB operators, functions, or other anonymous functions. For example, consider\ndefining the exponentially damped sinusoid f(t) = e−t cos(2πt).", - "type": "text" - }, - { - "block_id": "p146-b5", - "global_id": 3862, - "bbox": [ - 101.84, - 244.49, - 279.67, - 254.45 - ], - "text": ">>\nf = @(t) exp(-t).*cos(2*pi*t);", - "type": "text" - }, - { - "block_id": "p146-b6", - "global_id": 3863, - "bbox": [ - 101.84, - 266.81, - 490.4, - 324.59 - ], - "text": "In this context, the @ symbol identifies the expression as an anonymous function, which is assigned\na name of f. Parentheses following the @ symbol are used to identify the function’s independent\nvariables (input arguments), which in this case is the single time variable t. Input arguments, such\nas t, are local to the anonymous function and are not related to any workspace variables with the\nsame names.", - "type": "text" - }, - { - "block_id": "p146-b7", - "global_id": 3864, - "bbox": [ - 101.85, - 326.16, - 490.36, - 348.5 - ], - "text": "Once defined, f(t) can be evaluated simply by passing the input values of interest. For\nexample,", - "type": "text" - }, - { - "block_id": "p146-b8", - "global_id": 3865, - "bbox": [ - 101.85, - 361.43, - 180.3, - 383.35 - ], - "text": ">>\nt = 0; f(t)\nans = 1", - "type": "text" - }, - { - "block_id": "p146-b9", - "global_id": 3866, - "bbox": [ - 101.85, - 395.29, - 490.37, - 417.62 - ], - "text": "evaluates f(t) at t = 0, confirming the expected result of unity. The same result is obtained by\npassing t = 0 directly.", - "type": "text" - }, - { - "block_id": "p146-b10", - "global_id": 3867, - "bbox": [ - 101.85, - 430.55, - 159.37, - 452.47 - ], - "text": ">>\nf(0)\nans = 1", - "type": "text" - }, - { - "block_id": "p146-b11", - "global_id": 3868, - "bbox": [ - 101.85, - 464.83, - 490.39, - 522.61 - ], - "text": "Vector inputs allow the evaluation of multiple values simultaneously. Consider the task\nof plotting f(t) over the interval (−2 ≤t ≤2). Gross function behavior is clear: f(t) should\noscillate four times with a decaying envelope. Since accurate hand sketches are cumbersome,\nMATLAB-generated plots are an attractive alternative. As the following example illustrates, care\nmust be taken to ensure reliable results.", - "type": "text" - }, - { - "block_id": "p146-b12", - "global_id": 3869, - "bbox": [ - 101.85, - 524.19, - 490.37, - 546.52 - ], - "text": "Suppose vector t is chosen to include only the integers contained in (−2 ≤t ≤2), namely,\n[−2,−1,0,1,2].", - "type": "text" - }, - { - "block_id": "p146-b13", - "global_id": 3870, - "bbox": [ - 101.85, - 559.45, - 180.3, - 569.42 - ], - "text": ">>\nt = (-2:2);", - "type": "text" - }, - { - "block_id": "p146-b14", - "global_id": 3871, - "bbox": [ - 101.85, - 581.77, - 315.79, - 591.73 - ], - "text": "This vector input is evaluated to form a vector output.", - "type": "text" - }, - { - "block_id": "p146-b15", - "global_id": 3872, - "bbox": [ - 101.85, - 604.66, - 394.72, - 626.58 - ], - "text": ">>\nf(t)\nans = 7.3891\n2.7183\n1.0000\n0.3679\n0.1353", - "type": "text" - } - ] - }, - { - "page_num": 147, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p147-b0", - "global_id": 3873, - "bbox": [ - 309.8, - 62.89, - 516.12, - 71.98 - ], - "text": "1.11\nMATLAB: Working with Functions\n127", - "type": "text" - }, - { - "block_id": "p147-b1", - "global_id": 3874, - "bbox": [ - 127.59, - 85.82, - 394.26, - 96.07 - ], - "text": "The plot command graphs the result, which is shown in Fig. 1.46.", - "type": "text" - }, - { - "block_id": "p147-b2", - "global_id": 3875, - "bbox": [ - 127.59, - 106.58, - 326.34, - 128.49 - ], - "text": ">>\nplot(t,f(t));\n>>\nxlabel(’t’); ylabel(’f(t)’); grid;", - "type": "text" - }, - { - "block_id": "p147-b3", - "global_id": 3876, - "bbox": [ - 127.59, - 138.71, - 516.14, - 172.58 - ], - "text": "Grid lines, added by using the grid command, aid feature identification. Unfortunately, the\nplot does not illustrate the expected oscillatory behavior. More points are required to adequately\nrepresent f(t).", - "type": "text" - }, - { - "block_id": "p147-b4", - "global_id": 3877, - "bbox": [ - 127.59, - 170.96, - 516.16, - 220.69 - ], - "text": "The question, then, is how many points is enough?† If too few points are chosen, information\nis lost. If too many points are chosen, memory and time are wasted. A balance is needed. For\noscillatory functions, plotting 20 to 200 points per oscillation is normally adequate. For the present\ncase, t is chosen to give 100 points per oscillation.", - "type": "text" - }, - { - "block_id": "p147-b5", - "global_id": 3878, - "bbox": [ - 127.59, - 231.2, - 232.2, - 241.17 - ], - "text": ">>\nt = (-2:0.01:2);", - "type": "text" - }, - { - "block_id": "p147-b6", - "global_id": 3879, - "bbox": [ - 127.59, - 251.38, - 303.57, - 261.34 - ], - "text": "Again, the function is evaluated and plotted.", - "type": "text" - }, - { - "block_id": "p147-b7", - "global_id": 3880, - "bbox": [ - 166.79, - 387.87, - 397.96, - 395.87 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p147-b8", - "global_id": 3881, - "bbox": [ - 283.34, - 400.9, - 285.79, - 409.7 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p147-b9", - "global_id": 3882, - "bbox": [ - 164.24, - 378.64, - 168.24, - 386.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p147-b10", - "global_id": 3883, - "bbox": [ - 164.24, - 356.71, - 168.24, - 364.71 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p147-b11", - "global_id": 3884, - "bbox": [ - 164.24, - 334.77, - 168.24, - 342.77 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p147-b12", - "global_id": 3885, - "bbox": [ - 164.24, - 312.83, - 168.24, - 320.83 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p147-b13", - "global_id": 3886, - "bbox": [ - 164.24, - 290.9, - 168.24, - 298.9 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p147-b14", - "global_id": 3887, - "bbox": [ - 152.26, - 331.88, - 161.06, - 343.12 - ], - "text": "f(t)", - "type": "text" - }, - { - "block_id": "p147-b15", - "global_id": 3888, - "bbox": [ - 151.5, - 413.87, - 335.13, - 426.36 - ], - "text": "Figure 1.46 f(t) = e−t cos(2πt) for t = (-2:2).", - "type": "text" - }, - { - "block_id": "p147-b16", - "global_id": 3889, - "bbox": [ - 172.13, - 558.47, - 402.46, - 566.47 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p147-b17", - "global_id": 3890, - "bbox": [ - 287.84, - 571.5, - 290.28, - 580.3 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p147-b18", - "global_id": 3891, - "bbox": [ - 166.48, - 548.68, - 174.48, - 556.68 - ], - "text": "–5", - "type": "text" - }, - { - "block_id": "p147-b19", - "global_id": 3892, - "bbox": [ - 168.73, - 519.99, - 172.73, - 527.99 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p147-b20", - "global_id": 3893, - "bbox": [ - 168.73, - 490.75, - 172.73, - 498.75 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p147-b21", - "global_id": 3894, - "bbox": [ - 164.23, - 461.5, - 172.23, - 469.5 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p147-b22", - "global_id": 3895, - "bbox": [ - 152.26, - 502.48, - 161.06, - 513.72 - ], - "text": "f(t)", - "type": "text" - }, - { - "block_id": "p147-b23", - "global_id": 3896, - "bbox": [ - 151.5, - 584.48, - 358.67, - 596.97 - ], - "text": "Figure 1.47 f(t) = e−t cos(2πt) for t = (-2:0.01:2).", - "type": "text" - }, - { - "block_id": "p147-b24", - "global_id": 3897, - "bbox": [ - 127.59, - 621.19, - 445.65, - 633.41 - ], - "text": "† Sampling theory, presented later, formally addresses important aspects of this question.", - "type": "text" - } - ] - }, - { - "page_num": 148, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p148-b0", - "global_id": 3898, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "128\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p148-b1", - "global_id": 3899, - "bbox": [ - 101.84, - 86.24, - 300.6, - 108.15 - ], - "text": ">>\nplot(t,f(t));\n>>\nxlabel(’t’); ylabel(’f(t)’); grid;", - "type": "text" - }, - { - "block_id": "p148-b2", - "global_id": 3900, - "bbox": [ - 101.84, - 118.55, - 350.31, - 128.93 - ], - "text": "The result, shown in Fig. 1.47, is an accurate depiction of f(t).", - "type": "text" - }, - { - "block_id": "p148-b3", - "global_id": 3901, - "bbox": [ - 101.84, - 154.8, - 402.5, - 166.76 - ], - "text": "1.11-2 Relational Operators and the Unit Step Function", - "type": "text" - }, - { - "block_id": "p148-b4", - "global_id": 3902, - "bbox": [ - 101.84, - 172.48, - 490.4, - 206.77 - ], - "text": "The unit step function u(t) arises naturally in many practical situations. For example, a unit step can\nmodel the act of turning on a system. With the help of relational operators, anonymous functions\ncan represent the unit step function.", - "type": "text" - }, - { - "block_id": "p148-b5", - "global_id": 3903, - "bbox": [ - 101.84, - 208.76, - 490.38, - 254.87 - ], - "text": "In MATLAB, a relational operator compares two items. If the comparison is true, a logical true\n(1) is returned. If the comparison is false, a logical false (0) is returned. Sometimes called indicator\nfunctions, relational operators indicates whether a condition is true. Six relational operators are\navailable: <, >, <=, >=, ==, and ~=.", - "type": "text" - }, - { - "block_id": "p148-b6", - "global_id": 3904, - "bbox": [ - 119.78, - 256.58, - 408.39, - 266.83 - ], - "text": "The unit step function is readily defined using the >= relational operator.", - "type": "text" - }, - { - "block_id": "p148-b7", - "global_id": 3905, - "bbox": [ - 101.84, - 277.92, - 232.6, - 287.88 - ], - "text": ">>\nu = @(t) 1.0.*(t>=0);", - "type": "text" - }, - { - "block_id": "p148-b8", - "global_id": 3906, - "bbox": [ - 101.84, - 298.69, - 490.4, - 320.9 - ], - "text": "Any function with a jump discontinuity, such as the unit step, is difficult to plot. Consider plotting\nu(t) by using t = (-2:2).", - "type": "text" - }, - { - "block_id": "p148-b9", - "global_id": 3907, - "bbox": [ - 101.84, - 331.99, - 269.21, - 353.91 - ], - "text": ">>\nt = (-2:2); plot(t,u(t));\n>>\nxlabel(’t’); ylabel(’u(t)’);", - "type": "text" - }, - { - "block_id": "p148-b10", - "global_id": 3908, - "bbox": [ - 101.84, - 364.71, - 490.43, - 422.5 - ], - "text": "Two significant problems are apparent in the resulting plot, shown in Fig. 1.48. First,\nMATLAB automatically scales plot axes to tightly bound the data. In this case, this normally\ndesirable feature obscures most of the plot. Second, MATLAB connects plot data with lines,\nmaking a true jump discontinuity difficult to achieve. The coarse resolution of vector t emphasizes\nthe effect by showing an erroneous sloping line between t = −1 and t = 0.", - "type": "text" - }, - { - "block_id": "p148-b11", - "global_id": 3909, - "bbox": [ - 101.85, - 424.49, - 490.4, - 446.69 - ], - "text": "The first problem is corrected by vertically enlarging the bounding box with the axis\ncommand. The second problem is reduced, but not eliminated, by adding points to vector t.", - "type": "text" - }, - { - "block_id": "p148-b12", - "global_id": 3910, - "bbox": [ - 148.35, - 581.49, - 378.95, - 589.49 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p148-b13", - "global_id": 3911, - "bbox": [ - 264.33, - 594.52, - 266.78, - 603.32 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p148-b14", - "global_id": 3912, - "bbox": [ - 145.24, - 572.26, - 149.24, - 580.26 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p148-b15", - "global_id": 3913, - "bbox": [ - 138.49, - 528.38, - 149.17, - 536.39 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p148-b16", - "global_id": 3914, - "bbox": [ - 145.24, - 484.52, - 149.24, - 492.52 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p148-b17", - "global_id": 3915, - "bbox": [ - 126.52, - 525.15, - 135.32, - 537.86 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p148-b18", - "global_id": 3916, - "bbox": [ - 125.76, - 609.62, - 253.9, - 619.23 - ], - "text": "Figure 1.48 u(t) for t = (-2:2).", - "type": "text" - } - ] - }, - { - "page_num": 149, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p149-b0", - "global_id": 3917, - "bbox": [ - 309.8, - 62.89, - 516.12, - 71.98 - ], - "text": "1.11\nMATLAB: Working with Functions\n129", - "type": "text" - }, - { - "block_id": "p149-b1", - "global_id": 3918, - "bbox": [ - 174.09, - 181.91, - 404.69, - 189.91 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p149-b2", - "global_id": 3919, - "bbox": [ - 290.07, - 194.95, - 292.52, - 203.75 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p149-b3", - "global_id": 3920, - "bbox": [ - 170.98, - 165.37, - 174.98, - 173.37 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p149-b4", - "global_id": 3921, - "bbox": [ - 164.23, - 128.82, - 174.91, - 136.82 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p149-b5", - "global_id": 3922, - "bbox": [ - 170.98, - 92.25, - 174.98, - 100.25 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p149-b6", - "global_id": 3923, - "bbox": [ - 152.26, - 125.58, - 161.06, - 138.29 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p149-b7", - "global_id": 3924, - "bbox": [ - 151.5, - 210.07, - 385.91, - 219.68 - ], - "text": "Figure 1.49 u(t) for t = (-2:0.01:2) with axis modification.", - "type": "text" - }, - { - "block_id": "p149-b8", - "global_id": 3925, - "bbox": [ - 127.59, - 245.74, - 310.65, - 279.61 - ], - "text": ">>\nt = (-2:0.01:2);\nplot(t,u(t));\n>>\nxlabel(’t’); ylabel(’u(t)’);\n>>\naxis([-2 2 -0.1 1.1]);", - "type": "text" - }, - { - "block_id": "p149-b9", - "global_id": 3926, - "bbox": [ - 127.59, - 294.33, - 516.13, - 316.34 - ], - "text": "The four-element vector argument of axis specifies x axis minimum, x axis maximum, y axis\nminimum, and y axis maximum, respectively. The improved results are shown in Fig. 1.49.", - "type": "text" - }, - { - "block_id": "p149-b10", - "global_id": 3927, - "bbox": [ - 127.59, - 318.34, - 516.14, - 364.17 - ], - "text": "Relational operators can be combined using logical AND, logical OR, and logical negation: &,\n|, and ~, respectively. For example, (t>0)&(t<1) and ~((t<=0)|(t>=1)) both test if 0 < t < 1.\nTo demonstrate, consider defining and plotting the unit pulse p(t) = u(t) −u(t −1), as shown in\nFig. 1.50:", - "type": "text" - }, - { - "block_id": "p149-b11", - "global_id": 3928, - "bbox": [ - 127.6, - 379.55, - 368.19, - 425.38 - ], - "text": ">>\np = @(t) 1.0.*((t>=0)&(t<1));\n>>\nt = (-1:0.01:2); plot(t,p(t));\n>>\nxlabel(’t’); ylabel(’p(t) = u(t)-u(t-1)’);\n>>\naxis([-1 2 -.1 1.1]);", - "type": "text" - }, - { - "block_id": "p149-b12", - "global_id": 3929, - "bbox": [ - 127.59, - 440.19, - 516.15, - 474.36 - ], - "text": "Since anonymous functions can be constructed using other anonymous functions, we could\nhave used our previously defined unit step anonymous function to define p(t) as p = @(t)\nu(t)-u(t-1);.", - "type": "text" - }, - { - "block_id": "p149-b13", - "global_id": 3930, - "bbox": [ - 173.81, - 591.43, - 405.83, - 599.43 - ], - "text": "–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p149-b14", - "global_id": 3931, - "bbox": [ - 291.2, - 604.48, - 293.65, - 613.28 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p149-b15", - "global_id": 3932, - "bbox": [ - 170.98, - 574.89, - 174.98, - 582.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p149-b16", - "global_id": 3933, - "bbox": [ - 164.23, - 538.34, - 174.91, - 546.34 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p149-b17", - "global_id": 3934, - "bbox": [ - 170.98, - 501.78, - 174.98, - 509.78 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p149-b18", - "global_id": 3935, - "bbox": [ - 152.26, - 514.44, - 161.06, - 572.19 - ], - "text": "p(t) = u(t)-u(t-1)", - "type": "text" - }, - { - "block_id": "p149-b19", - "global_id": 3936, - "bbox": [ - 151.5, - 619.58, - 346.37, - 629.19 - ], - "text": "Figure 1.50 p(t) = u(t) −u(t −1) over (−1 ≤t ≤2).", - "type": "text" - } - ] - }, - { - "page_num": 150, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p150-b0", - "global_id": 3937, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "130\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p150-b1", - "global_id": 3938, - "bbox": [ - 101.84, - 85.82, - 490.4, - 167.8 - ], - "text": "For scalar operands, MATLAB also supports two short-circuit logical constructs. A\nshort-circuit logical AND is performed by using &&, and a short-circuit logical OR is performed by\nusing ||. Short-circuit logical operators are often more efficient than traditional logical operators\nbecause they test the second portion of the expression only when necessary. That is, when scalar\nexpression A is found false in (A&&B), scalar expression B is not evaluated, since a false result\nis already guaranteed. Similarly, scalar expression B is not evaluated when scalar expression A is\nfound true in (A||B), since a true result is already guaranteed.", - "type": "text" - }, - { - "block_id": "p150-b2", - "global_id": 3939, - "bbox": [ - 101.84, - 192.73, - 424.75, - 204.68 - ], - "text": "1.11-3 Visualizing Operations on the Independent Variable", - "type": "text" - }, - { - "block_id": "p150-b3", - "global_id": 3940, - "bbox": [ - 101.84, - 210.82, - 490.39, - 232.73 - ], - "text": "Two operations on a function’s independent variable are commonly encountered: shifting and\nscaling. Anonymous functions are well suited to investigate both operations.", - "type": "text" - }, - { - "block_id": "p150-b4", - "global_id": 3941, - "bbox": [ - 101.84, - 231.11, - 490.39, - 268.6 - ], - "text": "Consider g(t) = f(t)u(t) = e−t cos(2πt)u(t), a causal version of f(t). MATLAB easily\nmultiplies anonymous functions. Thus, we create g(t) by multiplying our anonymous functions\nfor f(t) and u(t).†", - "type": "text" - }, - { - "block_id": "p150-b5", - "global_id": 3942, - "bbox": [ - 101.84, - 279.32, - 227.37, - 289.28 - ], - "text": ">>\ng = @(t) f(t).*u(t);", - "type": "text" - }, - { - "block_id": "p150-b6", - "global_id": 3943, - "bbox": [ - 101.84, - 299.01, - 490.39, - 357.2 - ], - "text": "A combined shifting and scaling operation is represented by g(at + b), where a and b are\narbitrary real constants. As an example, consider plotting g(2t +1) over (−2 ≤t ≤2). With a = 2,\nthe function is compressed by a factor of 2, resulting in twice the oscillations per unit t. Adding\nthe condition b > 0 shifts the waveform to the left. Given anonymous function g, an accurate plot\nis nearly trivial to obtain.", - "type": "text" - }, - { - "block_id": "p150-b7", - "global_id": 3944, - "bbox": [ - 101.85, - 367.92, - 410.44, - 389.84 - ], - "text": ">>\nt = (-2:0.01:2);\n>>\nplot(t,g(2*t+1)); xlabel(’t’); ylabel(’g(2t+1)’); grid;", - "type": "text" - }, - { - "block_id": "p150-b8", - "global_id": 3945, - "bbox": [ - 101.85, - 399.98, - 490.42, - 433.86 - ], - "text": "Figure 1.51 confirms the expected waveform compression and left shift. As a final check, realize\nthat function g(·) turns on when the input argument is zero. Therefore, g(2t + 1) should turn on\nwhen 2t + 1 = 0 or at t = −0.5, a fact again confirmed by Fig. 1.51.", - "type": "text" - }, - { - "block_id": "p150-b9", - "global_id": 3946, - "bbox": [ - 151.17, - 548.34, - 381.22, - 556.34 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p150-b10", - "global_id": 3947, - "bbox": [ - 266.6, - 561.37, - 269.04, - 570.17 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p150-b11", - "global_id": 3948, - "bbox": [ - 145.24, - 539.11, - 153.24, - 547.11 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p150-b12", - "global_id": 3949, - "bbox": [ - 138.5, - 517.18, - 151.84, - 525.18 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p150-b13", - "global_id": 3950, - "bbox": [ - 147.49, - 495.24, - 151.49, - 503.24 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p150-b14", - "global_id": 3951, - "bbox": [ - 140.74, - 473.3, - 151.42, - 481.3 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p150-b15", - "global_id": 3952, - "bbox": [ - 147.49, - 451.37, - 151.49, - 459.37 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p150-b16", - "global_id": 3953, - "bbox": [ - 126.52, - 486.12, - 135.32, - 512.59 - ], - "text": "g(2t+1)", - "type": "text" - }, - { - "block_id": "p150-b17", - "global_id": 3954, - "bbox": [ - 125.76, - 576.48, - 276.27, - 586.09 - ], - "text": "Figure 1.51 g(2t + 1) over (−2 ≤t ≤2).", - "type": "text" - }, - { - "block_id": "p150-b18", - "global_id": 3955, - "bbox": [ - 101.84, - 610.24, - 490.39, - 633.67 - ], - "text": "† Although we define g in terms of f and u, the function g will not change if we later change either f or u\nunless we subsequently redefine g as well.", - "type": "text" - } - ] - }, - { - "page_num": 151, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p151-b0", - "global_id": 3956, - "bbox": [ - 309.8, - 62.89, - 516.12, - 71.98 - ], - "text": "1.11\nMATLAB: Working with Functions\n131", - "type": "text" - }, - { - "block_id": "p151-b1", - "global_id": 3957, - "bbox": [ - 176.63, - 184.93, - 406.96, - 192.93 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p151-b2", - "global_id": 3958, - "bbox": [ - 292.34, - 197.96, - 294.78, - 206.76 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p151-b3", - "global_id": 3959, - "bbox": [ - 170.98, - 175.7, - 178.98, - 183.7 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p151-b4", - "global_id": 3960, - "bbox": [ - 164.24, - 153.76, - 178.24, - 161.76 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p151-b5", - "global_id": 3961, - "bbox": [ - 174.24, - 131.83, - 178.24, - 139.83 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p151-b6", - "global_id": 3962, - "bbox": [ - 166.48, - 109.89, - 177.16, - 117.89 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p151-b7", - "global_id": 3963, - "bbox": [ - 173.23, - 87.96, - 177.23, - 95.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p151-b8", - "global_id": 3964, - "bbox": [ - 152.26, - 121.59, - 161.06, - 148.06 - ], - "text": "g(–t+1)", - "type": "text" - }, - { - "block_id": "p151-b9", - "global_id": 3965, - "bbox": [ - 151.5, - 213.06, - 304.53, - 222.67 - ], - "text": "Figure 1.52 g(−t + 1) over (−2 ≤t ≤2).", - "type": "text" - }, - { - "block_id": "p151-b10", - "global_id": 3966, - "bbox": [ - 176.06, - 355.39, - 406.96, - 363.39 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p151-b11", - "global_id": 3967, - "bbox": [ - 292.34, - 368.42, - 294.78, - 377.22 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p151-b12", - "global_id": 3968, - "bbox": [ - 170.98, - 346.16, - 178.98, - 354.16 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p151-b13", - "global_id": 3969, - "bbox": [ - 164.98, - 328.61, - 178.98, - 336.61 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p151-b14", - "global_id": 3970, - "bbox": [ - 173.23, - 311.06, - 177.23, - 319.06 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p151-b15", - "global_id": 3971, - "bbox": [ - 166.48, - 293.51, - 177.16, - 301.51 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p151-b16", - "global_id": 3972, - "bbox": [ - 173.23, - 275.97, - 177.23, - 283.97 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p151-b17", - "global_id": 3973, - "bbox": [ - 166.48, - 258.42, - 177.16, - 266.42 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p151-b18", - "global_id": 3974, - "bbox": [ - 152.26, - 299.05, - 161.06, - 311.76 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p151-b19", - "global_id": 3975, - "bbox": [ - 151.5, - 383.53, - 371.39, - 393.14 - ], - "text": "Figure 1.53 h(t) = g(2t + 1) + g(−t + 1) over (−2 ≤t ≤2).", - "type": "text" - }, - { - "block_id": "p151-b20", - "global_id": 3976, - "bbox": [ - 127.59, - 426.39, - 516.13, - 448.73 - ], - "text": "Next, consider plotting g(−t + 1) over (−2 ≤t ≤2). Since a < 0, the waveform will be\nreflected. Adding the condition b > 0 shifts the final waveform to the right.", - "type": "text" - }, - { - "block_id": "p151-b21", - "global_id": 3977, - "bbox": [ - 127.59, - 460.13, - 430.95, - 470.09 - ], - "text": ">>\nplot(t,g(-t+1)); xlabel(’t’); ylabel(’g(-t+1)’); grid;", - "type": "text" - }, - { - "block_id": "p151-b22", - "global_id": 3978, - "bbox": [ - 127.59, - 480.91, - 362.33, - 490.87 - ], - "text": "Figure 1.52 confirms both the reflection and the right shift.", - "type": "text" - }, - { - "block_id": "p151-b23", - "global_id": 3979, - "bbox": [ - 127.59, - 492.87, - 516.14, - 526.73 - ], - "text": "Up to this point, Figs. 1.51 and 1.52 could be reasonably sketched by hand. Consider plotting\nthe more complicated function h(t) = g(2t + 1) + g(−t + 1) over (−2 ≤t ≤2) (Fig. 1.53); an\naccurate hand sketch would be quite difficult. With MATLAB, the work is much less burdensome.", - "type": "text" - }, - { - "block_id": "p151-b24", - "global_id": 3980, - "bbox": [ - 127.59, - 538.14, - 462.33, - 548.1 - ], - "text": ">>\nplot(t,g(2*t+1)+g(-t+1)); xlabel(’t’); ylabel(’h(t)’); grid;", - "type": "text" - }, - { - "block_id": "p151-b25", - "global_id": 3981, - "bbox": [ - 127.59, - 582.83, - 451.17, - 594.78 - ], - "text": "1.11-4 Numerical Integration and Estimating Signal Energy", - "type": "text" - }, - { - "block_id": "p151-b26", - "global_id": 3982, - "bbox": [ - 127.59, - 600.91, - 516.13, - 634.79 - ], - "text": "Interesting signals often have nontrivial mathematical representations. Computing signal energy,\nwhich involves integrating the square of these expressions, can be a daunting task. Fortunately,\nmany difficult integrals can be accurately estimated by means of numerical integration techniques.", - "type": "text" - } - ] - }, - { - "page_num": 152, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p152-b0", - "global_id": 3983, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "132\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p152-b1", - "global_id": 3984, - "bbox": [ - 101.84, - 85.82, - 490.41, - 107.74 - ], - "text": "Even if the integration appears simple, numerical integration provides a good way to verify\nanalytical results.", - "type": "text" - }, - { - "block_id": "p152-b2", - "global_id": 3985, - "bbox": [ - 101.85, - 108.09, - 490.38, - 134.01 - ], - "text": "To start, consider the simple signal x(t) = e−t(u(t)−u(t −1)). The energy of x(t) is expressed\nas Ex =", - "type": "text" - }, - { - "block_id": "p152-b3", - "global_id": 3986, - "bbox": [ - 137.6, - 114.53, - 151.22, - 126.58 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p152-b4", - "global_id": 3987, - "bbox": [ - 142.16, - 119.25, - 200.72, - 135.31 - ], - "text": "−∞|x(t)|2 dt =", - "type": "text" - }, - { - "block_id": "p152-b5", - "global_id": 3988, - "bbox": [ - 203.78, - 114.53, - 213.76, - 126.87 - ], - "text": "$ 1", - "type": "text" - }, - { - "block_id": "p152-b6", - "global_id": 3989, - "bbox": [ - 101.84, - 119.25, - 490.4, - 180.75 - ], - "text": "0 e−2t dt. Integrating yields Ex = 0.5(1 −e−2) ≈0.4323. The energy\nintegral can also be evaluated numerically. Figure 1.27 helps illustrate the simple method of\nrectangular approximation: evaluate the integrand at points uniformly separated by t, multiply\neach by t to compute rectangle areas, and then sum over all rectangles. First, we create function\nx(t).", - "type": "text" - }, - { - "block_id": "p152-b7", - "global_id": 3990, - "bbox": [ - 101.84, - 193.29, - 295.36, - 203.25 - ], - "text": ">>\nx = @(t) exp(-t).*((t>=0)&(t<1));", - "type": "text" - }, - { - "block_id": "p152-b8", - "global_id": 3991, - "bbox": [ - 101.84, - 214.8, - 297.31, - 225.18 - ], - "text": "With t = 0.01, a suitable time vector is created.", - "type": "text" - }, - { - "block_id": "p152-b9", - "global_id": 3992, - "bbox": [ - 101.84, - 237.72, - 201.22, - 247.68 - ], - "text": ">>\nt = (0:0.01:1);", - "type": "text" - }, - { - "block_id": "p152-b10", - "global_id": 3993, - "bbox": [ - 101.84, - 259.65, - 327.06, - 269.9 - ], - "text": "The final result is computed by using the sum command.", - "type": "text" - }, - { - "block_id": "p152-b11", - "global_id": 3994, - "bbox": [ - 101.84, - 282.15, - 258.75, - 304.06 - ], - "text": ">>\nE_x = sum(x(t).*x(t)*0.01)\nE_x = 0.4367", - "type": "text" - }, - { - "block_id": "p152-b12", - "global_id": 3995, - "bbox": [ - 101.84, - 315.62, - 490.39, - 338.24 - ], - "text": "The result is not perfect, but at 1% relative error it is close. By reducing t, the approximation is\nimproved. For example, t = 0.001 yields E_x = 0.4328, or 0.1% relative error.", - "type": "text" - }, - { - "block_id": "p152-b13", - "global_id": 3996, - "bbox": [ - 101.84, - 339.94, - 490.41, - 397.73 - ], - "text": "Although simple to visualize, rectangular approximation is not the best numerical integration\ntechnique. The MATLAB function quad implements a better numerical integration technique\ncalled recursive adaptive Simpson quadrature.† To operate, quad requires a function describing\nthe integrand, the lower limit of integration, and the upper limit of integration. Notice that no t\nneeds to be specified.", - "type": "text" - }, - { - "block_id": "p152-b14", - "global_id": 3997, - "bbox": [ - 119.78, - 399.62, - 379.74, - 410.38 - ], - "text": "To use quad to estimate Ex, the integrand must first be described.", - "type": "text" - }, - { - "block_id": "p152-b15", - "global_id": 3998, - "bbox": [ - 101.84, - 422.22, - 269.21, - 432.18 - ], - "text": ">>\nx_squared = @(t) x(t).*x(t);", - "type": "text" - }, - { - "block_id": "p152-b16", - "global_id": 3999, - "bbox": [ - 101.84, - 444.05, - 244.14, - 455.61 - ], - "text": "Estimating Ex immediately follows.", - "type": "text" - }, - { - "block_id": "p152-b17", - "global_id": 4000, - "bbox": [ - 101.84, - 466.65, - 253.52, - 488.57 - ], - "text": ">>\nE_x = quad(x_squared,0,1)\nE_x = 0.4323", - "type": "text" - }, - { - "block_id": "p152-b18", - "global_id": 4001, - "bbox": [ - 101.84, - 500.12, - 275.2, - 510.5 - ], - "text": "In this case, the relative error is −0.0026%.", - "type": "text" - }, - { - "block_id": "p152-b19", - "global_id": 4002, - "bbox": [ - 101.84, - 512.49, - 490.38, - 535.49 - ], - "text": "The same techniques can be used to estimate the energy of more complex signals. Consider\ng(t), defined previously. Energy is expressed as Eg =", - "type": "text" - }, - { - "block_id": "p152-b20", - "global_id": 4003, - "bbox": [ - 312.03, - 516.01, - 325.65, - 528.06 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p152-b21", - "global_id": 4004, - "bbox": [ - 101.85, - 520.42, - 490.39, - 546.36 - ], - "text": "0 e−2t cos2 (2πt)dt. A closed-form solution\nexists, but it takes some effort. MATLAB provides an answer more quickly.", - "type": "text" - }, - { - "block_id": "p152-b22", - "global_id": 4005, - "bbox": [ - 101.85, - 558.9, - 269.21, - 568.86 - ], - "text": ">>\ng_squared = @(t) g(t).*g(t);", - "type": "text" - }, - { - "block_id": "p152-b23", - "global_id": 4006, - "bbox": [ - 101.84, - 599.27, - 490.37, - 633.41 - ], - "text": "† A comprehensive treatment of numerical integration is outside the scope of this text. Details of this particular\nmethod are not important for the current discussion; it is sufficient to say that it is better than the rectangular\napproximation.", - "type": "text" - } - ] - }, - { - "page_num": 153, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p153-b0", - "global_id": 4007, - "bbox": [ - 409.68, - 62.89, - 516.14, - 71.98 - ], - "text": "1.12\nSummary\n133", - "type": "text" - }, - { - "block_id": "p153-b1", - "global_id": 4008, - "bbox": [ - 127.59, - 85.82, - 516.14, - 119.69 - ], - "text": "Although the upper limit of integration is infinity, the exponentially decaying envelope ensures\ng(t) is effectively zero well before t = 100. Thus, an upper limit of t = 100 is used along with\nt = 0.001.", - "type": "text" - }, - { - "block_id": "p153-b2", - "global_id": 4009, - "bbox": [ - 127.59, - 129.0, - 300.19, - 162.87 - ], - "text": ">>\nt = (0:0.001:100);\n>>\nE_g = sum(g_squared(t)*0.001)\nE_g = 0.2567", - "type": "text" - }, - { - "block_id": "p153-b3", - "global_id": 4010, - "bbox": [ - 127.59, - 171.6, - 395.34, - 181.85 - ], - "text": "A slightly better approximation is obtained with the quad function.", - "type": "text" - }, - { - "block_id": "p153-b4", - "global_id": 4011, - "bbox": [ - 127.59, - 190.86, - 289.73, - 212.79 - ], - "text": ">>\nE_g = quad(g_squared,0,100)\nE_g = 0.2562", - "type": "text" - }, - { - "block_id": "p153-b5", - "global_id": 4012, - "bbox": [ - 133.57, - 237.57, - 440.59, - 249.53 - ], - "text": "DRILL 1.21\nComputing Signal Energy with MATLAB", - "type": "text" - }, - { - "block_id": "p153-b6", - "global_id": 4013, - "bbox": [ - 133.57, - 258.24, - 510.16, - 281.64 - ], - "text": "Use MATLAB to confirm that the energy of signal h(t), defined previously as h(t) = g(2t +\n1) + g(−t + 1), is Eh = 0.3768.", - "type": "text" - }, - { - "block_id": "p153-b7", - "global_id": 4014, - "bbox": [ - 127.94, - 311.99, - 227.89, - 325.93 - ], - "text": "1.12 SUMMARY", - "type": "text" - }, - { - "block_id": "p153-b8", - "global_id": 4015, - "bbox": [ - 127.59, - 331.82, - 516.14, - 377.75 - ], - "text": "A signal is a set of data or information. A system processes input signals to modify them or extract\nadditional information from them to produce output signals (response). A system may be made up\nof physical components (hardware realization), or it may be an algorithm that computes an output\nsignal from an input signal (software realization).", - "type": "text" - }, - { - "block_id": "p153-b9", - "global_id": 4016, - "bbox": [ - 127.59, - 379.74, - 516.15, - 449.48 - ], - "text": "A convenient measure of the size of a signal is its energy, if it is finite. If the signal energy is\ninfinite, the appropriate measure is its power, if it exists. The signal power is the time average of\nits energy (averaged over the entire time interval from −∞to ∞). For periodic signals, the time\naveraging need be performed over only one period in view of the periodic repetition of the signal.\nSignal power is also equal to the mean squared value of the signal (averaged over the entire time\ninterval from t = −∞to ∞).", - "type": "text" - }, - { - "block_id": "p153-b10", - "global_id": 4017, - "bbox": [ - 145.52, - 451.48, - 308.9, - 461.44 - ], - "text": "Signals can be classified in several ways.", - "type": "text" - }, - { - "block_id": "p153-b11", - "global_id": 4018, - "bbox": [ - 144.52, - 469.31, - 516.15, - 634.79 - ], - "text": "1. A continuous-time signal is specified for a continuum of values of the independent variable\n(such as time t). A discrete-time signal is specified only at a finite or a countable set of\ntime instants.\n2. An analog signal is a signal whose amplitude can take on any value over a continuum. On\nthe other hand, a signal whose amplitudes can take on only a finite number of values is a\ndigital signal. The terms discrete-time and continuous-time qualify the nature of a signal\nalong the time axis (horizontal axis). The terms analog and digital, on the other hand,\nqualify the nature of the signal amplitude (vertical axis).\n3. A periodic signal x(t) is defined by the fact that x(t) = x(t +T0) for some T0. The smallest\npositive value of T0 for which this relationship is satisfied is called the fundamental period.\nA periodic signal remains unchanged when shifted by an integer multiple of its period.\nA periodic signal x(t) can be generated by a periodic extension of any contiguous segment\nof x(t) of duration T0. Finally, a periodic signal, by definition, must exist over the entire\ntime interval −∞< t < ∞. A signal is aperiodic if it is not periodic.", - "type": "text" - } - ] - }, - { - "page_num": 154, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p154-b0", - "global_id": 4019, - "bbox": [ - 60.0, - 62.89, - 266.81, - 71.98 - ], - "text": "134\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p154-b1", - "global_id": 4020, - "bbox": [ - 118.78, - 85.4, - 490.39, - 203.37 - ], - "text": "4. An everlasting signal starts at t = −∞and continues forever to t = ∞. Hence, periodic\nsignals are everlasting signals. A causal signal is a signal that is zero for t < 0.\n5. A signal with finite energy is an energy signal. Similarly a signal with a finite and nonzero\npower (mean-square value) is a power signal. A signal can be either an energy signal or\na power signal, but not both. However, there are signals that are neither energy nor power\nsignals.\n6. A signal whose physical description is known completely in a mathematical or graphical\nform is a deterministic signal. A random signal is known only in terms of its probabilistic\ndescription such as mean value or mean-square value, rather than by its mathematical or\ngraphical form.", - "type": "text" - }, - { - "block_id": "p154-b2", - "global_id": 4021, - "bbox": [ - 101.85, - 210.93, - 490.39, - 257.18 - ], - "text": "A signal x(t) delayed by T seconds (right-shifted) can be expressed as x(t −T); on the other\nhand, x(t) advanced by T (left-shifted) is x(t + T). A signal x(t) time-compressed by a factor\na(a > 1) is expressed as x(at); on the other hand, the same signal time-expanded by factor a(a > 1)\nis x(t/a). The signal x(t) when time-reversed can be expressed as x(−t).", - "type": "text" - }, - { - "block_id": "p154-b3", - "global_id": 4022, - "bbox": [ - 101.84, - 258.75, - 490.39, - 281.09 - ], - "text": "The unit step function u(t) is very useful in representing causal signals and signals with\ndifferent mathematical descriptions over different intervals.", - "type": "text" - }, - { - "block_id": "p154-b4", - "global_id": 4023, - "bbox": [ - 101.84, - 282.66, - 490.41, - 352.82 - ], - "text": "In the classical (Dirac) definition, the unit impulse function δ(t) is characterized by unit area\nand is concentrated at a single instant t = 0. The impulse function has a sampling (or sifting)\nproperty, which states that the area under the product of a function with a unit impulse is equal to\nthe value of that function at the instant at which the impulse is located (assuming the function to\nbe continuous at the impulse location). In the modern approach, the impulse function is viewed as\na generalized function and is defined by the sampling property.", - "type": "text" - }, - { - "block_id": "p154-b5", - "global_id": 4024, - "bbox": [ - 101.84, - 353.39, - 490.41, - 376.73 - ], - "text": "The exponential function est, where s is complex, encompasses a large class of signals that\nincludes a constant, a monotonic exponential, a sinusoid, and an exponentially varying sinusoid.", - "type": "text" - }, - { - "block_id": "p154-b6", - "global_id": 4025, - "bbox": [ - 101.85, - 378.31, - 490.4, - 472.37 - ], - "text": "A real signal that is symmetrical about the vertical axis (t = 0) is an even function of time,\nand a real signal that is antisymmetrical about the vertical axis is an odd function of time. The\nproduct of an even function and an odd function is an odd function. However, the product of an\neven function and an even function or an odd function and an odd function is an even function.\nThe area under an odd function from t = −a to a is always zero regardless of the value of a. On\nthe other hand, the area under an even function from t = −a to a is two times the area under the\nsame function from t = 0 to a (or from t = −a to 0). Every signal can be expressed as a sum of\nodd and even functions of time.", - "type": "text" - }, - { - "block_id": "p154-b7", - "global_id": 4026, - "bbox": [ - 101.85, - 474.36, - 490.37, - 508.23 - ], - "text": "A system processes input signals to produce output signals (response). The input is the cause,\nand the output is its effect. In general, the output is affected by two causes: the internal conditions\nof the system (such as the initial conditions) and the external input.", - "type": "text" - }, - { - "block_id": "p154-b8", - "global_id": 4027, - "bbox": [ - 119.78, - 510.23, - 287.03, - 520.19 - ], - "text": "Systems can be classified in several ways.", - "type": "text" - }, - { - "block_id": "p154-b9", - "global_id": 4028, - "bbox": [ - 118.78, - 528.16, - 490.41, - 621.81 - ], - "text": "1. Linear systems are characterized by the linearity property, which implies superposition; if\nseveral causes (such as various inputs and initial conditions) are acting on a linear system,\nthe total output (response) is the sum of the responses from each cause, assuming that all\nthe remaining causes are absent. A system is nonlinear if superposition does not hold.\n2. In time-invariant systems, system parameters do not change with time. The parameters of\ntime-varying-parameter systems change with time.\n3. For memoryless (or instantaneous) systems, the system response at any instant t depends\nonly on the value of the input at t. For systems with memory (also known as dynamic", - "type": "text" - } - ] - }, - { - "page_num": 155, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p155-b0", - "global_id": 4029, - "bbox": [ - 409.68, - 62.89, - 516.14, - 71.98 - ], - "text": "1.12\nSummary\n135", - "type": "text" - }, - { - "block_id": "p155-b1", - "global_id": 4030, - "bbox": [ - 144.52, - 85.72, - 516.17, - 370.75 - ], - "text": "systems), the system response at any instant t depends not only on the present value of the\ninput, but also on the past values of the input (values before t).\n4. In contrast, if a system response at t also depends on the future values of the input (values of\ninput beyond t), the system is noncausal. In causal systems, the response does not depend\non the future values of the input. Because of the dependence of the response on the future\nvalues of input, the effect (response) of noncausal systems occurs before the cause. When\nthe independent variable is time (temporal systems), the noncausal systems are prophetic\nsystems, and therefore, unrealizable, although close approximation is possible with some\ntime delay in the response. Noncausal systems with independent variables other than time\n(e.g., space) are realizable.\n5. Systems whose inputs and outputs are continuous-time signals are continuous-time\nsystems; systems whose inputs and outputs are discrete-time signals are discrete-time\nsystems. If a continuous-time signal is sampled, the resulting signal is a discrete-time\nsignal. We can process a continuous-time signal by processing the samples of the signal\nwith a discrete-time system.\n6. Systems whose inputs and outputs are analog signals are analog systems; those whose\ninputs and outputs are digital signals are digital systems.\n7. If we can obtain the input x(t) back from the output y(t) of a system S by some operation,\nthe system S is said to be invertible. Otherwise the system is noninvertible.\n8. A system is stable if bounded input produces bounded output. This defines external\nstability because it can be ascertained from measurements at the external terminals\nof the system. External stability is also known as the stability in the BIBO\n(bounded-input/bounded-output) sense. Internal stability, discussed later in Ch. 2, is\nmeasured in terms of the internal behavior of the system.", - "type": "text" - }, - { - "block_id": "p155-b2", - "global_id": 4031, - "bbox": [ - 127.59, - 378.72, - 516.15, - 460.42 - ], - "text": "The system model derived from a knowledge of the internal structure of the system is its\ninternal description. In contrast, an external description is a representation of a system as seen\nfrom its input and output terminals; it can be obtained by applying a known input and measuring\nthe resulting output. In the majority of practical systems, an external description of a system so\nobtained is equivalent to its internal description. At times, however, the external description fails to\ndescribe the system adequately. Such is the case with the so-called uncontrollable or unobservable\nsystems.", - "type": "text" - }, - { - "block_id": "p155-b3", - "global_id": 4032, - "bbox": [ - 127.59, - 462.41, - 516.14, - 508.23 - ], - "text": "A system may also be described in terms of certain set of key variables called state variables.\nIn this description, an Nth-order system can be characterized by a set of N simultaneous first-order\ndifferential equations in N state variables. State equations of a system represent an internal\ndescription of that system.", - "type": "text" - }, - { - "block_id": "p155-b4", - "global_id": 4033, - "bbox": [ - 127.86, - 565.84, - 215.04, - 576.8 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p155-b5", - "global_id": 4034, - "bbox": [ - 127.59, - 583.86, - 459.14, - 592.91 - ], - "text": "1.\nPapoulis, A., The Fourier Integral and Its Applications. McGraw-Hill, New York, 1962.", - "type": "text" - }, - { - "block_id": "p155-b6", - "global_id": 4035, - "bbox": [ - 127.59, - 597.81, - 430.9, - 606.86 - ], - "text": "2.\nMason, S. J., Electronic Circuits, Signals, and Systems. Wiley, New York, 1960.", - "type": "text" - }, - { - "block_id": "p155-b7", - "global_id": 4036, - "bbox": [ - 127.59, - 611.76, - 400.09, - 620.81 - ], - "text": "3.\nKailath, T., Linear Systems. Prentice-Hall, Englewood Cliffs, NJ, 1980.", - "type": "text" - }, - { - "block_id": "p155-b8", - "global_id": 4037, - "bbox": [ - 127.59, - 625.71, - 449.64, - 634.76 - ], - "text": "4.\nLathi, B. P., Signals and Systems. Berkeley-Cambridge Press, Carmichael, CA, 1987.", - "type": "text" - } - ] - }, - { - "page_num": 156, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p156-b0", - "global_id": 4038, - "bbox": [ - 60.0, - 60.36, - 266.81, - 69.45 - ], - "text": "136\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p156-b1", - "global_id": 4039, - "bbox": [ - 80.93, - 91.05, - 191.52, - 107.98 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p156-b2", - "global_id": 4040, - "bbox": [ - 62.08, - 118.84, - 263.24, - 171.72 - ], - "text": "1.1-1\nFind the energies of the signals illustrated in\nFig. P1.1-1. Comment on the effect on energy\nof sign change, time shifting, or doubling of the\nsignal. What is the effect on the energy if the\nsignal is multiplied by k?", - "type": "text" - }, - { - "block_id": "p156-b3", - "global_id": 4041, - "bbox": [ - 62.08, - 179.48, - 263.22, - 188.52 - ], - "text": "1.1-2\nRepeat Prob. 1.1-1 for the signals in Fig. P1.1-2.", - "type": "text" - }, - { - "block_id": "p156-b4", - "global_id": 4042, - "bbox": [ - 62.08, - 196.29, - 263.24, - 205.33 - ], - "text": "1.1-3\n(a) Find the energies of the pair of signals", - "type": "text" - }, - { - "block_id": "p156-b5", - "global_id": 4043, - "bbox": [ - 90.72, - 206.94, - 263.24, - 260.12 - ], - "text": "x(t) and y(t) depicted in Figs. P1.1-3a\nand P1.1-3b. Sketch and find the energies of\nsignals x(t) + y(t) and x(t) −y(t). Can you\nmake any observation from these results?\n(b) Repeat part (a) for the signal pair illustrated", - "type": "text" - }, - { - "block_id": "p156-b6", - "global_id": 4044, - "bbox": [ - 106.16, - 262.12, - 263.25, - 282.04 - ], - "text": "in Fig. P1.1-3c. Is your observation in\npart (a) still valid?", - "type": "text" - }, - { - "block_id": "p156-b7", - "global_id": 4045, - "bbox": [ - 289.23, - 118.55, - 490.39, - 149.8 - ], - "text": "1.1-4\nFind the power of the periodic signal x(t) shown\nin Fig. P1.1-4. Find also the powers and the rms\nvalues of:", - "type": "text" - }, - { - "block_id": "p156-b8", - "global_id": 4046, - "bbox": [ - 317.86, - 151.43, - 353.65, - 171.72 - ], - "text": "(a) −x(t)\n(b) 2x(t)", - "type": "text" - }, - { - "block_id": "p156-b9", - "global_id": 4047, - "bbox": [ - 317.86, - 173.34, - 355.48, - 193.64 - ], - "text": "(c) cx(t)\nComment.", - "type": "text" - }, - { - "block_id": "p156-b10", - "global_id": 4048, - "bbox": [ - 289.23, - 198.85, - 490.4, - 284.6 - ], - "text": "1.1-5\nBy original design, a system outputs a 10-volt\npulse that is 3 seconds in duration. It is desired\nto upgrade the square-pulse output with a\n“soft-start” pulse that steps up to 10 volts in\n1-volt increments spaced every 20 milliseconds.\nDetermine the signal duration T so that the\n“soft-start” pulse has the same signal energy as\nthe original square pulse.", - "type": "text" - }, - { - "block_id": "p156-b11", - "global_id": 4049, - "bbox": [ - 392.31, - 414.09, - 401.64, - 422.09 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p156-b12", - "global_id": 4050, - "bbox": [ - 350.94, - 369.64, - 354.94, - 377.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p156-b13", - "global_id": 4051, - "bbox": [ - 344.28, - 399.86, - 354.94, - 408.16 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p156-b14", - "global_id": 4052, - "bbox": [ - 350.94, - 328.29, - 354.94, - 336.29 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p156-b15", - "global_id": 4053, - "bbox": [ - 153.36, - 359.78, - 432.21, - 378.97 - ], - "text": "t\n(a)", - "type": "text" - }, - { - "block_id": "p156-b16", - "global_id": 4054, - "bbox": [ - 111.5, - 329.48, - 274.41, - 337.48 - ], - "text": "0\n2\n2", - "type": "text" - }, - { - "block_id": "p156-b17", - "global_id": 4055, - "bbox": [ - 390.88, - 360.57, - 419.37, - 378.21 - ], - "text": "2\n3", - "type": "text" - }, - { - "block_id": "p156-b18", - "global_id": 4056, - "bbox": [ - 176.28, - 320.12, - 299.11, - 337.48 - ], - "text": "3\n3", - "type": "text" - }, - { - "block_id": "p156-b19", - "global_id": 4057, - "bbox": [ - 198.86, - 395.55, - 262.89, - 414.2 - ], - "text": "3\n5\n6", - "type": "text" - }, - { - "block_id": "p156-b20", - "global_id": 4058, - "bbox": [ - 104.83, - 342.06, - 115.5, - 350.36 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b21", - "global_id": 4059, - "bbox": [ - 111.5, - 307.12, - 115.5, - 315.12 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b22", - "global_id": 4060, - "bbox": [ - 190.55, - 330.75, - 192.78, - 338.75 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p156-b23", - "global_id": 4061, - "bbox": [ - 271.91, - 359.78, - 281.56, - 367.78 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p156-b24", - "global_id": 4062, - "bbox": [ - 230.46, - 329.48, - 234.46, - 337.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p156-b25", - "global_id": 4063, - "bbox": [ - 226.46, - 342.36, - 234.46, - 350.36 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p156-b26", - "global_id": 4064, - "bbox": [ - 230.46, - 307.12, - 234.46, - 315.12 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b27", - "global_id": 4065, - "bbox": [ - 309.48, - 330.6, - 311.71, - 338.6 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p156-b28", - "global_id": 4066, - "bbox": [ - 134.75, - 404.76, - 138.75, - 412.76 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p156-b29", - "global_id": 4067, - "bbox": [ - 128.08, - 417.34, - 138.75, - 425.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b30", - "global_id": 4068, - "bbox": [ - 134.75, - 382.4, - 138.75, - 390.4 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b31", - "global_id": 4069, - "bbox": [ - 291.02, - 406.12, - 293.24, - 414.12 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p156-b32", - "global_id": 4070, - "bbox": [ - 215.04, - 435.37, - 223.92, - 443.37 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p156-b33", - "global_id": 4071, - "bbox": [ - 104.83, - 449.54, - 156.48, - 458.51 - ], - "text": "Figure P1.1-1", - "type": "text" - }, - { - "block_id": "p156-b34", - "global_id": 4072, - "bbox": [ - 347.81, - 616.67, - 410.98, - 624.67 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p156-b35", - "global_id": 4073, - "bbox": [ - 347.81, - 543.07, - 351.81, - 551.07 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p156-b36", - "global_id": 4074, - "bbox": [ - 372.92, - 616.23, - 375.14, - 624.23 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p156-b37", - "global_id": 4075, - "bbox": [ - 374.46, - 533.08, - 388.57, - 542.69 - ], - "text": "x4(t)", - "type": "text" - }, - { - "block_id": "p156-b38", - "global_id": 4076, - "bbox": [ - 377.72, - 632.11, - 386.6, - 640.11 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p156-b39", - "global_id": 4077, - "bbox": [ - 111.36, - 532.86, - 174.52, - 540.86 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p156-b40", - "global_id": 4078, - "bbox": [ - 111.36, - 484.38, - 139.56, - 500.24 - ], - "text": "1\nx(t)", - "type": "text" - }, - { - "block_id": "p156-b41", - "global_id": 4079, - "bbox": [ - 136.47, - 530.47, - 271.9, - 540.86 - ], - "text": "t\n–1\n0", - "type": "text" - }, - { - "block_id": "p156-b42", - "global_id": 4080, - "bbox": [ - 266.96, - 491.74, - 270.96, - 499.74 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b43", - "global_id": 4081, - "bbox": [ - 289.24, - 530.47, - 291.46, - 538.47 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p156-b44", - "global_id": 4082, - "bbox": [ - 241.19, - 484.42, - 255.29, - 494.03 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p156-b45", - "global_id": 4083, - "bbox": [ - 141.36, - 549.3, - 258.3, - 557.3 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p156-b46", - "global_id": 4084, - "bbox": [ - 202.65, - 615.47, - 265.81, - 623.47 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p156-b47", - "global_id": 4085, - "bbox": [ - 202.65, - 574.6, - 206.65, - 582.6 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b48", - "global_id": 4086, - "bbox": [ - 286.26, - 614.97, - 319.81, - 623.03 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p156-b49", - "global_id": 4087, - "bbox": [ - 229.96, - 573.44, - 244.06, - 583.05 - ], - "text": "x3(t)", - "type": "text" - }, - { - "block_id": "p156-b50", - "global_id": 4088, - "bbox": [ - 259.34, - 632.11, - 268.67, - 640.11 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p156-b51", - "global_id": 4089, - "bbox": [ - 104.83, - 616.21, - 115.5, - 624.51 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b52", - "global_id": 4090, - "bbox": [ - 111.5, - 590.64, - 115.5, - 598.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p156-b53", - "global_id": 4091, - "bbox": [ - 164.58, - 576.05, - 166.8, - 584.05 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p156-b54", - "global_id": 4092, - "bbox": [ - 170.66, - 590.64, - 174.66, - 598.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b55", - "global_id": 4093, - "bbox": [ - 127.58, - 573.44, - 141.68, - 583.05 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p156-b56", - "global_id": 4094, - "bbox": [ - 170.66, - 590.64, - 174.66, - 598.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p156-b57", - "global_id": 4095, - "bbox": [ - 140.82, - 632.13, - 149.7, - 640.13 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p156-b58", - "global_id": 4096, - "bbox": [ - 104.83, - 646.3, - 156.48, - 655.26 - ], - "text": "Figure P1.1-2", - "type": "text" - } - ] - }, - { - "page_num": 157, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p157-b0", - "global_id": 4097, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n137", - "type": "text" - }, - { - "block_id": "p157-b1", - "global_id": 4098, - "bbox": [ - 282.52, - 147.94, - 291.4, - 155.94 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p157-b2", - "global_id": 4099, - "bbox": [ - 282.14, - 236.33, - 291.78, - 244.33 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p157-b3", - "global_id": 4100, - "bbox": [ - 282.52, - 313.74, - 291.4, - 321.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p157-b4", - "global_id": 4101, - "bbox": [ - 174.24, - 131.33, - 236.89, - 140.34 - ], - "text": "2\n0", - "type": "text" - }, - { - "block_id": "p157-b5", - "global_id": 4102, - "bbox": [ - 174.24, - 109.21, - 178.24, - 117.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p157-b6", - "global_id": 4103, - "bbox": [ - 190.79, - 89.9, - 201.89, - 97.98 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p157-b7", - "global_id": 4104, - "bbox": [ - 260.59, - 131.73, - 262.81, - 139.73 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p157-b8", - "global_id": 4105, - "bbox": [ - 174.24, - 204.81, - 238.49, - 213.07 - ], - "text": "0\n2p", - "type": "text" - }, - { - "block_id": "p157-b9", - "global_id": 4106, - "bbox": [ - 174.24, - 182.09, - 229.57, - 193.15 - ], - "text": "1\nsin t", - "type": "text" - }, - { - "block_id": "p157-b10", - "global_id": 4107, - "bbox": [ - 255.58, - 206.01, - 257.81, - 214.01 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p157-b11", - "global_id": 4108, - "bbox": [ - 188.32, - 170.73, - 201.89, - 180.07 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p157-b12", - "global_id": 4109, - "bbox": [ - 174.24, - 292.19, - 238.49, - 301.75 - ], - "text": "0\np", - "type": "text" - }, - { - "block_id": "p157-b13", - "global_id": 4110, - "bbox": [ - 174.24, - 267.24, - 233.57, - 278.31 - ], - "text": "1\nsin t", - "type": "text" - }, - { - "block_id": "p157-b14", - "global_id": 4111, - "bbox": [ - 255.58, - 294.14, - 257.81, - 302.14 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p157-b15", - "global_id": 4112, - "bbox": [ - 190.79, - 257.15, - 201.89, - 265.23 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p157-b16", - "global_id": 4113, - "bbox": [ - 302.9, - 294.94, - 366.85, - 303.07 - ], - "text": "p\n0", - "type": "text" - }, - { - "block_id": "p157-b17", - "global_id": 4114, - "bbox": [ - 302.9, - 272.95, - 306.9, - 280.95 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p157-b18", - "global_id": 4115, - "bbox": [ - 384.73, - 295.45, - 386.95, - 303.45 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p157-b19", - "global_id": 4116, - "bbox": [ - 321.4, - 264.43, - 332.5, - 272.51 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p157-b20", - "global_id": 4117, - "bbox": [ - 302.9, - 207.32, - 368.48, - 215.45 - ], - "text": "2p\n0", - "type": "text" - }, - { - "block_id": "p157-b21", - "global_id": 4118, - "bbox": [ - 302.9, - 185.33, - 306.9, - 193.33 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p157-b22", - "global_id": 4119, - "bbox": [ - 384.73, - 207.83, - 386.95, - 215.83 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p157-b23", - "global_id": 4120, - "bbox": [ - 321.4, - 173.93, - 332.5, - 182.01 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p157-b24", - "global_id": 4121, - "bbox": [ - 302.9, - 112.55, - 365.5, - 131.33 - ], - "text": "2\n0", - "type": "text" - }, - { - "block_id": "p157-b25", - "global_id": 4122, - "bbox": [ - 302.9, - 101.21, - 306.9, - 109.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p157-b26", - "global_id": 4123, - "bbox": [ - 334.71, - 123.7, - 386.93, - 132.01 - ], - "text": "t\n1", - "type": "text" - }, - { - "block_id": "p157-b27", - "global_id": 4124, - "bbox": [ - 321.4, - 89.9, - 332.5, - 97.98 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p157-b28", - "global_id": 4125, - "bbox": [ - 296.24, - 133.22, - 306.9, - 141.51 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p157-b29", - "global_id": 4126, - "bbox": [ - 334.71, - 124.01, - 338.71, - 132.01 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p157-b30", - "global_id": 4127, - "bbox": [ - 419.36, - 313.71, - 471.01, - 322.68 - ], - "text": "Figure P1.1-3", - "type": "text" - }, - { - "block_id": "p157-b31", - "global_id": 4128, - "bbox": [ - 330.97, - 381.31, - 469.76, - 391.04 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p157-b32", - "global_id": 4129, - "bbox": [ - 281.38, - 345.77, - 285.38, - 353.77 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p157-b33", - "global_id": 4130, - "bbox": [ - 375.17, - 383.44, - 379.17, - 391.44 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p157-b34", - "global_id": 4131, - "bbox": [ - 274.71, - 405.98, - 285.38, - 414.27 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p157-b35", - "global_id": 4132, - "bbox": [ - 303.36, - 338.14, - 314.47, - 346.23 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p157-b36", - "global_id": 4133, - "bbox": [ - 303.03, - 357.38, - 308.89, - 366.78 - ], - "text": "t3", - "type": "text" - }, - { - "block_id": "p157-b37", - "global_id": 4134, - "bbox": [ - 330.97, - 383.04, - 422.97, - 391.04 - ], - "text": "2\n6", - "type": "text" - }, - { - "block_id": "p157-b38", - "global_id": 4135, - "bbox": [ - 151.6, - 382.62, - 272.89, - 406.28 - ], - "text": "t3\n4\n6\n2", - "type": "text" - }, - { - "block_id": "p157-b39", - "global_id": 4136, - "bbox": [ - 130.58, - 420.54, - 182.22, - 429.51 - ], - "text": "Figure P1.1-4", - "type": "text" - }, - { - "block_id": "p157-b40", - "global_id": 4137, - "bbox": [ - 87.82, - 441.36, - 288.96, - 461.35 - ], - "text": "1.1-6\nDetermine the power and the rms value for each\nof the following signals:", - "type": "text" - }, - { - "block_id": "p157-b41", - "global_id": 4138, - "bbox": [ - 116.46, - 462.98, - 276.43, - 483.28 - ], - "text": "(a) 5 + 10cos(100t + π/3)\n(b) 10cos(100t + π/3)+16sin(150t + π/5)", - "type": "text" - }, - { - "block_id": "p157-b42", - "global_id": 4139, - "bbox": [ - 116.46, - 484.89, - 206.81, - 505.19 - ], - "text": "(c) (10 + 2sin3t)cos10t\n(d) 10cos5tcos10t", - "type": "text" - }, - { - "block_id": "p157-b43", - "global_id": 4140, - "bbox": [ - 116.97, - 507.1, - 185.86, - 516.15 - ], - "text": "(e) 10sin5tcos10t", - "type": "text" - }, - { - "block_id": "p157-b44", - "global_id": 4141, - "bbox": [ - 117.96, - 514.89, - 170.1, - 527.85 - ], - "text": "(f) ejαt cosω0t", - "type": "text" - }, - { - "block_id": "p157-b45", - "global_id": 4142, - "bbox": [ - 87.82, - 532.56, - 288.99, - 574.48 - ], - "text": "1.1-7\nFigure P1.1-7 shows a periodic 50% duty cycle\ndc-offset sawtooth wave x(t) with peak ampli-\ntude A. Determine the energy and power of\nx(t).", - "type": "text" - }, - { - "block_id": "p157-b46", - "global_id": 4143, - "bbox": [ - 87.82, - 579.94, - 288.99, - 632.82 - ], - "text": "1.1-8\nTwo periodic signals that differ only by a\n90-degree phase shift are considered to be\nquadrature signals. For example, cos(2πt) and\nsin(2πt) are quadrature signals. Another pair of\nquadrature signals is x(t) = sgn[cos(2πt)] and", - "type": "text" - }, - { - "block_id": "p157-b47", - "global_id": 4144, - "bbox": [ - 343.61, - 441.0, - 516.13, - 461.29 - ], - "text": "y(t) = sgn[sin(2πt)], where sgn is the sign (or\nsignum) function.", - "type": "text" - }, - { - "block_id": "p157-b48", - "global_id": 4145, - "bbox": [ - 344.12, - 462.91, - 516.12, - 473.59 - ], - "text": "(a) Plot x(t) and determine its power Px and", - "type": "text" - }, - { - "block_id": "p157-b49", - "global_id": 4146, - "bbox": [ - 343.61, - 474.16, - 516.12, - 495.51 - ], - "text": "energy Ex.\n(b) Plot y(t) and determine its power Py and", - "type": "text" - }, - { - "block_id": "p157-b50", - "global_id": 4147, - "bbox": [ - 344.12, - 496.07, - 516.14, - 516.09 - ], - "text": "energy Ey.\n(c) Consider the complex function f(t) = x(t)+", - "type": "text" - }, - { - "block_id": "p157-b51", - "global_id": 4148, - "bbox": [ - 343.61, - 517.7, - 516.13, - 548.97 - ], - "text": "jy(t). Determine the power and energy of\nf(t).\n(d) When real functions x(t) and y(t) are com-", - "type": "text" - }, - { - "block_id": "p157-b52", - "global_id": 4149, - "bbox": [ - 359.05, - 550.58, - 516.14, - 581.84 - ], - "text": "bined as f(t) = x(t) + jy(t), is it generally\ntrue that Ef = Ex + Ey and Pf = Px + Py?\nProve your answer.", - "type": "text" - }, - { - "block_id": "p157-b53", - "global_id": 4150, - "bbox": [ - 314.97, - 590.9, - 516.14, - 633.56 - ], - "text": "1.1-9\nThere are many useful properties related to\nsignal energy. Prove each of the following\nstatements. In each case, let energy signal x1(t)\nhave energy E[x1(t)], let energy signal x2(t) have", - "type": "text" - } - ] - }, - { - "page_num": 158, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p158-b0", - "global_id": 4151, - "bbox": [ - 60.0, - 60.36, - 266.81, - 69.45 - ], - "text": "138\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p158-b1", - "global_id": 4152, - "bbox": [ - 239.78, - 97.51, - 244.67, - 105.51 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p158-b2", - "global_id": 4153, - "bbox": [ - 178.4, - 139.5, - 376.44, - 149.83 - ], - "text": "O\n–T\nT\nt", - "type": "text" - }, - { - "block_id": "p158-b3", - "global_id": 4154, - "bbox": [ - 104.83, - 157.91, - 156.48, - 166.88 - ], - "text": "Figure P1.1-7", - "type": "text" - }, - { - "block_id": "p158-b4", - "global_id": 4155, - "bbox": [ - 90.72, - 178.02, - 263.24, - 198.32 - ], - "text": "energy E[x2(t)], and let T be a nonzero, finite,\nreal-valued constant.", - "type": "text" - }, - { - "block_id": "p158-b5", - "global_id": 4156, - "bbox": [ - 91.22, - 198.94, - 263.24, - 210.01 - ], - "text": "(a) Prove E[Tx1(t)] = T2E[x1(t)]. That is,", - "type": "text" - }, - { - "block_id": "p158-b6", - "global_id": 4157, - "bbox": [ - 90.72, - 211.18, - 263.24, - 242.89 - ], - "text": "amplitude scaling a signal by constant T\nscales the signal energy by T2.\n(b) Prove E[x1(t)] = E[x1(t −T)]. That is,", - "type": "text" - }, - { - "block_id": "p158-b7", - "global_id": 4158, - "bbox": [ - 91.22, - 244.14, - 263.24, - 264.8 - ], - "text": "shifting a signal does not affect its energy.\n(c) If (x1(t)̸ = 0) ⇒(x2(t) = 0) and (x2(t)̸ =", - "type": "text" - }, - { - "block_id": "p158-b8", - "global_id": 4159, - "bbox": [ - 90.72, - 265.69, - 263.24, - 330.56 - ], - "text": "0) ⇒(x1(t) = 0), then prove E[x1(t) +\nx2(t)] = E[x1(t)] + E[x2(t)]. That is, the\nenergy of the sum of two nonoverlapping\nsignals is the sum of the two individual\nenergies.\n(d) Prove E[x1(Tt)] = (1/|T|)E[x1(t)]. That is,", - "type": "text" - }, - { - "block_id": "p158-b9", - "global_id": 4160, - "bbox": [ - 106.15, - 331.72, - 263.23, - 351.74 - ], - "text": "time-scaling a signal by T reciprocally\nscales the signal energy by 1/|T|.", - "type": "text" - }, - { - "block_id": "p158-b10", - "global_id": 4161, - "bbox": [ - 57.59, - 356.79, - 263.24, - 399.01 - ], - "text": "1.1-10\nConsider the signal x(t) shown in Fig. P1.1-10.\nOutside the interval shown, x(t) is zero. Deter-\nmine the signal energy E[x(t)]. [Hint: Use the\nresults of Prob. 1.1-9.]", - "type": "text" - }, - { - "block_id": "p158-b11", - "global_id": 4162, - "bbox": [ - 124.35, - 539.38, - 256.3, - 547.38 - ], - "text": "–2\n–1\n0\n1\n2\n3", - "type": "text" - }, - { - "block_id": "p158-b12", - "global_id": 4163, - "bbox": [ - 111.76, - 514.83, - 115.76, - 522.83 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p158-b13", - "global_id": 4164, - "bbox": [ - 101.12, - 534.47, - 115.76, - 542.47 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p158-b14", - "global_id": 4165, - "bbox": [ - 105.76, - 495.25, - 115.76, - 503.25 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p158-b15", - "global_id": 4166, - "bbox": [ - 111.76, - 475.67, - 115.76, - 483.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p158-b16", - "global_id": 4167, - "bbox": [ - 105.76, - 456.09, - 115.76, - 464.09 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p158-b17", - "global_id": 4168, - "bbox": [ - 105.76, - 416.93, - 115.76, - 424.93 - ], - "text": "2.5", - "type": "text" - }, - { - "block_id": "p158-b18", - "global_id": 4169, - "bbox": [ - 111.76, - 436.51, - 115.76, - 444.51 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p158-b19", - "global_id": 4170, - "bbox": [ - 191.15, - 547.8, - 193.37, - 555.8 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p158-b20", - "global_id": 4171, - "bbox": [ - 90.08, - 473.37, - 98.16, - 484.79 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p158-b21", - "global_id": 4172, - "bbox": [ - 90.72, - 561.8, - 146.83, - 570.76 - ], - "text": "Figure P1.1-10", - "type": "text" - }, - { - "block_id": "p158-b22", - "global_id": 4173, - "bbox": [ - 57.59, - 582.22, - 219.27, - 591.26 - ], - "text": "1.1-11\n(a) Show that the power of a signal", - "type": "text" - }, - { - "block_id": "p158-b23", - "global_id": 4174, - "bbox": [ - 153.12, - 614.45, - 175.3, - 623.7 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p158-b24", - "global_id": 4175, - "bbox": [ - 177.16, - 605.34, - 189.85, - 614.9 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p158-b25", - "global_id": 4176, - "bbox": [ - 177.15, - 627.28, - 189.85, - 633.96 - ], - "text": "k=m", - "type": "text" - }, - { - "block_id": "p158-b26", - "global_id": 4177, - "bbox": [ - 190.85, - 612.74, - 215.66, - 624.47 - ], - "text": "Dkejωkt", - "type": "text" - }, - { - "block_id": "p158-b27", - "global_id": 4178, - "bbox": [ - 333.31, - 178.34, - 339.29, - 187.3 - ], - "text": "is", - "type": "text" - }, - { - "block_id": "p158-b28", - "global_id": 4179, - "bbox": [ - 385.8, - 197.06, - 403.52, - 207.37 - ], - "text": "Px =", - "type": "text" - }, - { - "block_id": "p158-b29", - "global_id": 4180, - "bbox": [ - 405.37, - 187.95, - 418.06, - 197.51 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p158-b30", - "global_id": 4181, - "bbox": [ - 405.36, - 209.9, - 418.06, - 216.58 - ], - "text": "k=m", - "type": "text" - }, - { - "block_id": "p158-b31", - "global_id": 4182, - "bbox": [ - 419.07, - 195.62, - 437.39, - 207.07 - ], - "text": "|Dk|2", - "type": "text" - }, - { - "block_id": "p158-b32", - "global_id": 4183, - "bbox": [ - 317.86, - 225.33, - 490.39, - 256.22 - ], - "text": "assuming all frequencies to be distinct, that\nis, ωi̸ = ωk for all i̸ = k.\n(b) Use the result in part (a) to determine the", - "type": "text" - }, - { - "block_id": "p158-b33", - "global_id": 4184, - "bbox": [ - 333.31, - 258.21, - 487.36, - 267.18 - ], - "text": "power of each of the signals in Prob. 1.1-6.", - "type": "text" - }, - { - "block_id": "p158-b34", - "global_id": 4185, - "bbox": [ - 284.74, - 271.98, - 490.39, - 379.95 - ], - "text": "1.1-12\nA binary signal x(t) = 0 for t < 0. For positive\ntime, x(t) toggles between one and zero as\nfollows: one for 1 second, zero for 1 second,\none for 1 second, zero for 2 seconds, one for 1\nsecond, zero for 3 seconds, and so forth. That\nis, the “on” time is always 1 second, but the\n“off” time successively increases by 1 second\nbetween each toggle. A portion of x(t) is shown\nin Fig. P1.1-12. Determine the energy and power\nof x(t).", - "type": "text" - }, - { - "block_id": "p158-b35", - "global_id": 4186, - "bbox": [ - 289.23, - 384.75, - 490.38, - 405.05 - ], - "text": "1.2-1\nFor the signal x(t) depicted in Fig. P1.2-1, sketch\nthe signals", - "type": "text" - }, - { - "block_id": "p158-b36", - "global_id": 4187, - "bbox": [ - 317.87, - 406.67, - 360.92, - 426.97 - ], - "text": "(a) x(−t)\n(b) x(t + 6)", - "type": "text" - }, - { - "block_id": "p158-b37", - "global_id": 4188, - "bbox": [ - 317.87, - 428.59, - 355.36, - 448.88 - ], - "text": "(c) x(3t)\n(d) x(t/2)", - "type": "text" - }, - { - "block_id": "p158-b38", - "global_id": 4189, - "bbox": [ - 289.23, - 453.68, - 490.39, - 473.98 - ], - "text": "1.2-2\nFor the signal x(t) illustrated in Fig. P1.2-2,\nsketch", - "type": "text" - }, - { - "block_id": "p158-b39", - "global_id": 4190, - "bbox": [ - 317.87, - 475.61, - 362.08, - 495.9 - ], - "text": "(a) x(t −4)\n(b) x(t/1.5)", - "type": "text" - }, - { - "block_id": "p158-b40", - "global_id": 4191, - "bbox": [ - 317.86, - 497.52, - 365.4, - 517.82 - ], - "text": "(c) x(−t)\n(d) x(2t −4)", - "type": "text" - }, - { - "block_id": "p158-b41", - "global_id": 4192, - "bbox": [ - 318.37, - 519.43, - 360.92, - 528.77 - ], - "text": "(e) x(2 −t)", - "type": "text" - }, - { - "block_id": "p158-b42", - "global_id": 4193, - "bbox": [ - 289.23, - 533.57, - 490.39, - 575.79 - ], - "text": "1.2-3\nIn Fig. P1.2-3, express signals x1(t), x2(t), x3(t),\nx4(t), and x5(t) in terms of signal x(t) and\nits time-shifted, time-scaled, or time-reversed\nversions.", - "type": "text" - }, - { - "block_id": "p158-b43", - "global_id": 4194, - "bbox": [ - 289.22, - 580.59, - 490.39, - 634.44 - ], - "text": "1.2-4\nFor an energy signal x(t) with energy Ex, show\nthat the energy of any one of the signals −x(t),\nx(−t), and x(t −T) is Ex. Show also that the\nenergy of x(at) as well as x(at −b) is Ex/a, but\nthe energy of ax(t) is a2Ex. This shows that time", - "type": "text" - } - ] - }, - { - "page_num": 159, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p159-b0", - "global_id": 4195, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n139", - "type": "text" - }, - { - "block_id": "p159-b1", - "global_id": 4196, - "bbox": [ - 180.46, - 91.5, - 184.46, - 99.5 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p159-b2", - "global_id": 4197, - "bbox": [ - 379.69, - 145.66, - 464.49, - 154.67 - ], - "text": "t\nFigure P1.1-12", - "type": "text" - }, - { - "block_id": "p159-b3", - "global_id": 4198, - "bbox": [ - 195.19, - 166.68, - 206.29, - 174.76 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p159-b4", - "global_id": 4199, - "bbox": [ - 185.52, - 184.38, - 387.01, - 204.86 - ], - "text": "t\n0\n6", - "type": "text" - }, - { - "block_id": "p159-b5", - "global_id": 4200, - "bbox": [ - 254.02, - 198.17, - 262.02, - 206.17 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p159-b6", - "global_id": 4201, - "bbox": [ - 178.86, - 212.23, - 189.52, - 220.52 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p159-b7", - "global_id": 4202, - "bbox": [ - 179.52, - 179.52, - 189.52, - 187.52 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p159-b8", - "global_id": 4203, - "bbox": [ - 270.22, - 198.17, - 327.92, - 206.17 - ], - "text": "15\n24", - "type": "text" - }, - { - "block_id": "p159-b9", - "global_id": 4204, - "bbox": [ - 418.37, - 213.1, - 470.01, - 222.07 - ], - "text": "Figure P1.2-1", - "type": "text" - }, - { - "block_id": "p159-b10", - "global_id": 4205, - "bbox": [ - 191.28, - 277.02, - 358.29, - 285.31 - ], - "text": "0\nt\n4", - "type": "text" - }, - { - "block_id": "p159-b11", - "global_id": 4206, - "bbox": [ - 286.32, - 232.78, - 297.43, - 240.86 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p159-b12", - "global_id": 4207, - "bbox": [ - 324.81, - 277.31, - 328.81, - 285.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p159-b13", - "global_id": 4208, - "bbox": [ - 274.36, - 242.81, - 278.36, - 250.81 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p159-b14", - "global_id": 4209, - "bbox": [ - 274.75, - 256.31, - 278.75, - 264.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p159-b15", - "global_id": 4210, - "bbox": [ - 389.97, - 290.5, - 441.62, - 299.47 - ], - "text": "Figure P1.2-2", - "type": "text" - }, - { - "block_id": "p159-b16", - "global_id": 4211, - "bbox": [ - 224.16, - 353.88, - 274.13, - 362.17 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p159-b17", - "global_id": 4212, - "bbox": [ - 253.62, - 327.8, - 257.62, - 335.8 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p159-b18", - "global_id": 4213, - "bbox": [ - 295.22, - 353.97, - 311.44, - 362.17 - ], - "text": "t\n1", - "type": "text" - }, - { - "block_id": "p159-b19", - "global_id": 4214, - "bbox": [ - 269.84, - 318.84, - 283.95, - 328.45 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p159-b20", - "global_id": 4215, - "bbox": [ - 344.43, - 353.88, - 393.17, - 362.17 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p159-b21", - "global_id": 4216, - "bbox": [ - 388.34, - 318.74, - 402.45, - 328.34 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p159-b22", - "global_id": 4217, - "bbox": [ - 413.27, - 353.97, - 429.63, - 362.17 - ], - "text": "t\n1", - "type": "text" - }, - { - "block_id": "p159-b23", - "global_id": 4218, - "bbox": [ - 151.06, - 318.78, - 381.17, - 337.55 - ], - "text": "11\nx(t)", - "type": "text" - }, - { - "block_id": "p159-b24", - "global_id": 4219, - "bbox": [ - 149.48, - 354.17, - 178.93, - 362.17 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p159-b25", - "global_id": 4220, - "bbox": [ - 138.25, - 328.31, - 142.25, - 336.31 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p159-b26", - "global_id": 4221, - "bbox": [ - 189.68, - 354.16, - 191.91, - 362.16 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p159-b27", - "global_id": 4222, - "bbox": [ - 302.76, - 455.88, - 316.86, - 465.49 - ], - "text": "x5(t)", - "type": "text" - }, - { - "block_id": "p159-b28", - "global_id": 4223, - "bbox": [ - 340.26, - 490.78, - 373.24, - 498.98 - ], - "text": "t\n1.5", - "type": "text" - }, - { - "block_id": "p159-b29", - "global_id": 4224, - "bbox": [ - 290.67, - 465.43, - 295.67, - 474.43 - ], - "text": "11", - "type": "text" - }, - { - "block_id": "p159-b30", - "global_id": 4225, - "bbox": [ - 242.13, - 490.68, - 319.18, - 498.98 - ], - "text": "1.5\n0.5\n0.5", - "type": "text" - }, - { - "block_id": "p159-b31", - "global_id": 4226, - "bbox": [ - 379.83, - 380.14, - 393.93, - 389.75 - ], - "text": "x4(t)", - "type": "text" - }, - { - "block_id": "p159-b32", - "global_id": 4227, - "bbox": [ - 300.19, - 399.12, - 445.48, - 410.64 - ], - "text": "t\n2\n2", - "type": "text" - }, - { - "block_id": "p159-b33", - "global_id": 4228, - "bbox": [ - 367.95, - 392.5, - 371.95, - 400.5 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p159-b34", - "global_id": 4229, - "bbox": [ - 353.29, - 409.66, - 384.28, - 424.59 - ], - "text": "0\n13", - "type": "text" - }, - { - "block_id": "p159-b35", - "global_id": 4230, - "bbox": [ - 210.33, - 389.58, - 224.43, - 399.19 - ], - "text": "x3(t)", - "type": "text" - }, - { - "block_id": "p159-b36", - "global_id": 4231, - "bbox": [ - 130.61, - 431.21, - 275.74, - 439.5 - ], - "text": "0\nt\n2\n2", - "type": "text" - }, - { - "block_id": "p159-b37", - "global_id": 4232, - "bbox": [ - 198.22, - 404.98, - 202.22, - 412.98 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p159-b38", - "global_id": 4233, - "bbox": [ - 198.22, - 384.09, - 202.22, - 392.09 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p159-b39", - "global_id": 4234, - "bbox": [ - 130.58, - 508.36, - 182.22, - 517.33 - ], - "text": "Figure P1.2-3", - "type": "text" - }, - { - "block_id": "p159-b40", - "global_id": 4235, - "bbox": [ - 116.46, - 529.51, - 288.98, - 593.27 - ], - "text": "inversion and time shifting do not affect signal\nenergy. On the other hand, time compression\nof a signal (a > 1) reduces the energy, and\ntime expansion of a signal (a < 1) increases the\nenergy. What is the effect on signal energy if the\nsignal is multiplied by a constant a?", - "type": "text" - }, - { - "block_id": "p159-b41", - "global_id": 4236, - "bbox": [ - 87.82, - 601.57, - 288.99, - 621.86 - ], - "text": "1.2-5\nDefine 2x(−3t + 1) = t[u(−t −1) −u(−t + 1)],\nwhere u(t) is the unit step function.", - "type": "text" - }, - { - "block_id": "p159-b42", - "global_id": 4237, - "bbox": [ - 116.97, - 623.48, - 285.44, - 632.82 - ], - "text": "(a) Plot 2x(−3t + 1) over a suitable range of t.", - "type": "text" - }, - { - "block_id": "p159-b43", - "global_id": 4238, - "bbox": [ - 343.61, - 529.19, - 482.37, - 538.53 - ], - "text": "(b) Plot x(t) over a suitable range of t.", - "type": "text" - }, - { - "block_id": "p159-b44", - "global_id": 4239, - "bbox": [ - 314.98, - 542.2, - 516.13, - 563.75 - ], - "text": "1.2-6\nConsider the signal x(t) = 2−tu(t), where u(t) is\nthe unit step function.", - "type": "text" - }, - { - "block_id": "p159-b45", - "global_id": 4240, - "bbox": [ - 343.61, - 565.38, - 516.13, - 585.67 - ], - "text": "(a) Accurately sketch x(t) over (−1 ≤t ≤1).\n(b) Accurately sketch y(t) = 0.5x(1 −2t) over", - "type": "text" - }, - { - "block_id": "p159-b46", - "global_id": 4241, - "bbox": [ - 359.05, - 587.29, - 407.49, - 596.63 - ], - "text": "(−1 ≤t ≤1).", - "type": "text" - }, - { - "block_id": "p159-b47", - "global_id": 4242, - "bbox": [ - 314.98, - 601.57, - 500.38, - 610.91 - ], - "text": "1.2-7\nDefine signals y(t) and z(t) as in Fig. P1.2-7.", - "type": "text" - }, - { - "block_id": "p159-b48", - "global_id": 4243, - "bbox": [ - 344.12, - 612.8, - 516.13, - 621.86 - ], - "text": "(a) Determine constants a, b, and c to produce", - "type": "text" - }, - { - "block_id": "p159-b49", - "global_id": 4244, - "bbox": [ - 359.05, - 623.48, - 472.45, - 632.82 - ], - "text": "z(t) = ax(bt + c) in Fig. P1.2-7.", - "type": "text" - } - ] - }, - { - "page_num": 160, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p160-b0", - "global_id": 4245, - "bbox": [ - 60.0, - 60.36, - 266.81, - 69.45 - ], - "text": "140\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p160-b1", - "global_id": 4246, - "bbox": [ - 157.69, - 93.62, - 189.43, - 103.21 - ], - "text": "y(t) = −1", - "type": "text" - }, - { - "block_id": "p160-b2", - "global_id": 4247, - "bbox": [ - 186.43, - 93.18, - 400.9, - 106.04 - ], - "text": "2 x(−3t + 2)\nz(t) = ax(bt + c)", - "type": "text" - }, - { - "block_id": "p160-b3", - "global_id": 4248, - "bbox": [ - 110.79, - 148.19, - 439.51, - 159.92 - ], - "text": "t\nt\n0\n0\n−4\n−4\n4\n4\n8\n8", - "type": "text" - }, - { - "block_id": "p160-b4", - "global_id": 4249, - "bbox": [ - 148.4, - 107.14, - 342.64, - 115.11 - ], - "text": "3\n3", - "type": "text" - }, - { - "block_id": "p160-b5", - "global_id": 4250, - "bbox": [ - 104.83, - 164.91, - 156.48, - 173.88 - ], - "text": "Figure P1.2-7", - "type": "text" - }, - { - "block_id": "p160-b6", - "global_id": 4251, - "bbox": [ - 90.72, - 185.02, - 263.24, - 194.36 - ], - "text": "(b) Determine and sketch a signal v(t) such that", - "type": "text" - }, - { - "block_id": "p160-b7", - "global_id": 4252, - "bbox": [ - 106.15, - 195.97, - 127.82, - 205.22 - ], - "text": "z(t) =", - "type": "text" - }, - { - "block_id": "p160-b8", - "global_id": 4253, - "bbox": [ - 129.66, - 188.75, - 137.31, - 199.93 - ], - "text": "$ t", - "type": "text" - }, - { - "block_id": "p160-b9", - "global_id": 4254, - "bbox": [ - 133.77, - 195.97, - 175.48, - 207.51 - ], - "text": "−∞v(τ)dτ.", - "type": "text" - }, - { - "block_id": "p160-b10", - "global_id": 4255, - "bbox": [ - 62.08, - 210.69, - 263.24, - 252.61 - ], - "text": "1.3-1\nThink of a real-world signal that is a personally\nrelevant and interesting. Describe the signal and\nthen classify it according to the six following\ncharacteristics:", - "type": "text" - }, - { - "block_id": "p160-b11", - "global_id": 4256, - "bbox": [ - 90.72, - 254.59, - 223.21, - 274.52 - ], - "text": "(a) continuous-time or discrete-time\n(b) analog or digital", - "type": "text" - }, - { - "block_id": "p160-b12", - "global_id": 4257, - "bbox": [ - 90.72, - 276.52, - 180.84, - 296.44 - ], - "text": "(c) periodic or aperiodic\n(d) energy or power", - "type": "text" - }, - { - "block_id": "p160-b13", - "global_id": 4258, - "bbox": [ - 91.22, - 298.43, - 176.37, - 307.4 - ], - "text": "(e) causal or noncausal", - "type": "text" - }, - { - "block_id": "p160-b14", - "global_id": 4259, - "bbox": [ - 90.72, - 309.4, - 263.24, - 362.19 - ], - "text": "(f) deterministic or random\nIf possible, think of a second real-world signal\nthat has the opposite six characteristics of your\nfirst signal. If such a second signal is not\npossible, carefully explain why that is the case.", - "type": "text" - }, - { - "block_id": "p160-b15", - "global_id": 4260, - "bbox": [ - 62.08, - 360.55, - 194.59, - 376.61 - ], - "text": "1.3-2\nDefine signal y(t) = %∞", - "type": "text" - }, - { - "block_id": "p160-b16", - "global_id": 4261, - "bbox": [ - 90.72, - 367.27, - 263.24, - 387.57 - ], - "text": "k=−∞x(0.5t −10k),\nwhere", - "type": "text" - }, - { - "block_id": "p160-b17", - "global_id": 4262, - "bbox": [ - 135.38, - 399.51, - 157.57, - 408.76 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p160-b18", - "global_id": 4263, - "bbox": [ - 159.41, - 386.93, - 212.4, - 414.28 - ], - "text": "e−2t\nt ≥1\n0\nt < 1", - "type": "text" - }, - { - "block_id": "p160-b19", - "global_id": 4264, - "bbox": [ - 91.22, - 433.35, - 263.23, - 442.41 - ], - "text": "(a) Determine the constant a such that the", - "type": "text" - }, - { - "block_id": "p160-b20", - "global_id": 4265, - "bbox": [ - 90.72, - 444.02, - 263.23, - 464.32 - ], - "text": "signal x(−2t + a) is borderline anticausal.\n(b) Is the signal y(t) periodic? If so, determine", - "type": "text" - }, - { - "block_id": "p160-b21", - "global_id": 4266, - "bbox": [ - 106.15, - 465.94, - 263.24, - 486.24 - ], - "text": "the period Ty. If not, explain why y(t) is not\nperiodic.", - "type": "text" - }, - { - "block_id": "p160-b22", - "global_id": 4267, - "bbox": [ - 62.08, - 491.61, - 263.24, - 522.57 - ], - "text": "1.3-3\nDetermine whether each of the following state-\nments is true or false. If the statement is false,\ndemonstrate this by proof or example.", - "type": "text" - }, - { - "block_id": "p160-b23", - "global_id": 4268, - "bbox": [ - 91.22, - 524.57, - 263.22, - 533.53 - ], - "text": "(a) Every continuous-time signal is an analog", - "type": "text" - }, - { - "block_id": "p160-b24", - "global_id": 4269, - "bbox": [ - 90.72, - 535.52, - 263.22, - 555.45 - ], - "text": "signal.\n(b) Every discrete-time signal is a digital signal.", - "type": "text" - }, - { - "block_id": "p160-b25", - "global_id": 4270, - "bbox": [ - 91.22, - 557.44, - 263.24, - 566.41 - ], - "text": "(c) If a signal is not an energy signal, then it", - "type": "text" - }, - { - "block_id": "p160-b26", - "global_id": 4271, - "bbox": [ - 90.72, - 568.4, - 262.69, - 588.33 - ], - "text": "must be a power signal and vice versa.\n(d) An energy signal must be of finite duration.", - "type": "text" - }, - { - "block_id": "p160-b27", - "global_id": 4272, - "bbox": [ - 91.22, - 590.32, - 224.44, - 599.29 - ], - "text": "(e) A power signal cannot be causal.", - "type": "text" - }, - { - "block_id": "p160-b28", - "global_id": 4273, - "bbox": [ - 92.21, - 601.28, - 245.09, - 610.24 - ], - "text": "(f) A periodic signal cannot be anticausal.", - "type": "text" - }, - { - "block_id": "p160-b29", - "global_id": 4274, - "bbox": [ - 62.08, - 615.62, - 263.24, - 635.62 - ], - "text": "1.3-4\nDetermine whether each of the following state-\nments is true or false. If the statement is", - "type": "text" - }, - { - "block_id": "p160-b30", - "global_id": 4275, - "bbox": [ - 317.86, - 185.34, - 490.37, - 205.26 - ], - "text": "false, demonstrate by proof or example why the\nstatement is false.", - "type": "text" - }, - { - "block_id": "p160-b31", - "global_id": 4276, - "bbox": [ - 318.36, - 207.26, - 490.38, - 216.22 - ], - "text": "(a) Every bounded periodic signal is a power", - "type": "text" - }, - { - "block_id": "p160-b32", - "global_id": 4277, - "bbox": [ - 317.86, - 218.22, - 490.38, - 238.14 - ], - "text": "signal.\n(b) Every bounded power signal is a periodic", - "type": "text" - }, - { - "block_id": "p160-b33", - "global_id": 4278, - "bbox": [ - 318.36, - 240.13, - 490.38, - 260.06 - ], - "text": "signal.\n(c) If an energy signal x(t) has energy E, then", - "type": "text" - }, - { - "block_id": "p160-b34", - "global_id": 4279, - "bbox": [ - 317.86, - 261.68, - 490.38, - 292.94 - ], - "text": "the energy of x(at) is E/a. Assume a is real\nand positive.\n(d) If a power signal x(t) has power P, then the", - "type": "text" - }, - { - "block_id": "p160-b35", - "global_id": 4280, - "bbox": [ - 333.31, - 294.55, - 490.38, - 314.86 - ], - "text": "power of x(at) is P/a. Assume a is real and\npositive.", - "type": "text" - }, - { - "block_id": "p160-b36", - "global_id": 4281, - "bbox": [ - 289.23, - 319.94, - 490.38, - 340.98 - ], - "text": "1.3-5\nGiven x1(t) = cos(t), x2(t) = sin(πt), and\nx3(t) = x1(t) + x2(t).", - "type": "text" - }, - { - "block_id": "p160-b37", - "global_id": 4282, - "bbox": [ - 318.37, - 342.14, - 490.38, - 352.54 - ], - "text": "(a) Determine the fundamental periods T1 and", - "type": "text" - }, - { - "block_id": "p160-b38", - "global_id": 4283, - "bbox": [ - 317.86, - 352.81, - 490.39, - 373.86 - ], - "text": "T2 of signals x1(t) and x2(t).\n(b) Show that x3(t) is not periodic, which", - "type": "text" - }, - { - "block_id": "p160-b39", - "global_id": 4284, - "bbox": [ - 318.37, - 374.74, - 490.39, - 408.68 - ], - "text": "requires T3 = k1T1 = k2T2 for some integers\nk1 and k2.\n(c) Determine the powers Px1, Px2, and Px3 of", - "type": "text" - }, - { - "block_id": "p160-b40", - "global_id": 4285, - "bbox": [ - 333.31, - 407.62, - 438.04, - 417.69 - ], - "text": "signals x1(t), x2(t), and x3(t).", - "type": "text" - }, - { - "block_id": "p160-b41", - "global_id": 4286, - "bbox": [ - 289.22, - 422.04, - 490.4, - 453.3 - ], - "text": "1.3-6\nFor any constant ω, is the function f(t) = sin(ωt)\na periodic function of the independent variable\nt? Justify your answer.", - "type": "text" - }, - { - "block_id": "p160-b42", - "global_id": 4287, - "bbox": [ - 289.22, - 458.69, - 479.34, - 467.73 - ], - "text": "1.3-7\nThe signal shown in Fig. P1.3-7 is defined as", - "type": "text" - }, - { - "block_id": "p160-b43", - "global_id": 4288, - "bbox": [ - 325.95, - 497.29, - 348.14, - 506.54 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p160-b44", - "global_id": 4289, - "bbox": [ - 349.98, - 476.22, - 357.09, - 495.94 - ], - "text": "⎧\n⎪⎨", - "type": "text" - }, - { - "block_id": "p160-b45", - "global_id": 4290, - "bbox": [ - 349.98, - 503.12, - 357.09, - 514.77 - ], - "text": "⎪⎩", - "type": "text" - }, - { - "block_id": "p160-b46", - "global_id": 4291, - "bbox": [ - 357.09, - 483.79, - 475.07, - 493.13 - ], - "text": "t\n0 ≤t < 1", - "type": "text" - }, - { - "block_id": "p160-b47", - "global_id": 4292, - "bbox": [ - 357.09, - 497.74, - 475.07, - 507.08 - ], - "text": "0.5 + 0.5cos(2πt)\n1 ≤t < 2", - "type": "text" - }, - { - "block_id": "p160-b48", - "global_id": 4293, - "bbox": [ - 357.09, - 511.69, - 475.07, - 521.03 - ], - "text": "3 −t\n2 ≤t < 3", - "type": "text" - }, - { - "block_id": "p160-b49", - "global_id": 4294, - "bbox": [ - 317.87, - 526.0, - 476.12, - 545.27 - ], - "text": "0\notherwise\nThe energy of x(t) is E ≈1.0417.", - "type": "text" - }, - { - "block_id": "p160-b50", - "global_id": 4295, - "bbox": [ - 317.86, - 546.89, - 482.01, - 567.92 - ], - "text": "(a) What is the energy of y1(t) = (1/3)x(2t)?\n(b) A periodic signal y2(t) is defined as", - "type": "text" - }, - { - "block_id": "p160-b51", - "global_id": 4296, - "bbox": [ - 350.79, - 591.27, - 376.69, - 601.35 - ], - "text": "y2(t) =", - "type": "text" - }, - { - "block_id": "p160-b52", - "global_id": 4297, - "bbox": [ - 378.53, - 575.99, - 384.72, - 584.95 - ], - "text": ")", - "type": "text" - }, - { - "block_id": "p160-b53", - "global_id": 4298, - "bbox": [ - 384.71, - 584.24, - 466.72, - 593.58 - ], - "text": "x(t)\n0 ≤t < 4", - "type": "text" - }, - { - "block_id": "p160-b54", - "global_id": 4299, - "bbox": [ - 384.72, - 598.19, - 454.01, - 608.27 - ], - "text": "y2(t + 4)\n∀t", - "type": "text" - }, - { - "block_id": "p160-b55", - "global_id": 4300, - "bbox": [ - 333.31, - 624.43, - 431.57, - 634.5 - ], - "text": "What is the power of y2(t)?", - "type": "text" - } - ] - }, - { - "page_num": 161, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p161-b0", - "global_id": 4301, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n141", - "type": "text" - }, - { - "block_id": "p161-b1", - "global_id": 4302, - "bbox": [ - 116.96, - 85.58, - 282.25, - 95.66 - ], - "text": "(c) What is the power of y3(t) = (1/3)y2(2t)?", - "type": "text" - }, - { - "block_id": "p161-b2", - "global_id": 4303, - "bbox": [ - 141.54, - 229.47, - 294.54, - 237.47 - ], - "text": "0\n0.5\n1\n1.5\n2\n2.5\n3", - "type": "text" - }, - { - "block_id": "p161-b3", - "global_id": 4304, - "bbox": [ - 137.53, - 214.74, - 141.53, - 222.74 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p161-b4", - "global_id": 4305, - "bbox": [ - 193.27, - 238.99, - 242.82, - 246.99 - ], - "text": "Time (seconds)", - "type": "text" - }, - { - "block_id": "p161-b5", - "global_id": 4306, - "bbox": [ - 120.11, - 163.23, - 128.19, - 175.14 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p161-b6", - "global_id": 4307, - "bbox": [ - 131.53, - 195.16, - 141.53, - 203.16 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p161-b7", - "global_id": 4308, - "bbox": [ - 131.53, - 175.58, - 141.53, - 183.58 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p161-b8", - "global_id": 4309, - "bbox": [ - 131.53, - 156.0, - 141.53, - 164.0 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p161-b9", - "global_id": 4310, - "bbox": [ - 131.53, - 136.42, - 141.53, - 144.42 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p161-b10", - "global_id": 4311, - "bbox": [ - 137.53, - 116.84, - 141.53, - 124.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p161-b11", - "global_id": 4312, - "bbox": [ - 116.46, - 254.91, - 168.11, - 263.87 - ], - "text": "Figure P1.3-7", - "type": "text" - }, - { - "block_id": "p161-b12", - "global_id": 4313, - "bbox": [ - 87.82, - 276.84, - 288.99, - 310.98 - ], - "text": "1.3-8\nLet y1(t) = y2(t) = t2 over 0 ≤t ≤1. Notice, this\nstatement does not require y1(t) = y2(t) for all\nt.", - "type": "text" - }, - { - "block_id": "p161-b13", - "global_id": 4314, - "bbox": [ - 116.96, - 312.59, - 288.97, - 322.67 - ], - "text": "(a) Define y1(t) as an even, periodic signal with", - "type": "text" - }, - { - "block_id": "p161-b14", - "global_id": 4315, - "bbox": [ - 116.46, - 323.56, - 288.98, - 355.55 - ], - "text": "period T1 = 2. Sketch y1(t) and determine\nits power.\n(b) Design an odd, periodic signal y2(t) with", - "type": "text" - }, - { - "block_id": "p161-b15", - "global_id": 4316, - "bbox": [ - 116.96, - 356.43, - 288.99, - 431.53 - ], - "text": "period T2 = 3 and power equal to unity.\nFully describe y2(t) and sketch the signal\nover at least one full period. [Hint: There\nare an infinite number of possible solutions\nto this problem—you need to find only one\nof them!]\n(c) We can create a complex-valued function", - "type": "text" - }, - { - "block_id": "p161-b16", - "global_id": 4317, - "bbox": [ - 116.46, - 433.15, - 289.0, - 487.06 - ], - "text": "y3(t) = y1(t) + jy2(t). Determine whether\nthis signal is periodic. If yes, determine the\nperiod T3. If no, justify why the signal is not\nperiodic.\n(d) Determine the power of y3(t) defined in", - "type": "text" - }, - { - "block_id": "p161-b17", - "global_id": 4318, - "bbox": [ - 131.89, - 488.31, - 288.96, - 508.24 - ], - "text": "part (c). The power of a complex-valued\nfunction z(t) is", - "type": "text" - }, - { - "block_id": "p161-b18", - "global_id": 4319, - "bbox": [ - 155.18, - 527.73, - 186.23, - 537.07 - ], - "text": "P = lim", - "type": "text" - }, - { - "block_id": "p161-b19", - "global_id": 4320, - "bbox": [ - 171.33, - 535.64, - 189.16, - 542.31 - ], - "text": "T→∞", - "type": "text" - }, - { - "block_id": "p161-b20", - "global_id": 4321, - "bbox": [ - 191.36, - 521.82, - 196.44, - 543.34 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p161-b21", - "global_id": 4322, - "bbox": [ - 199.22, - 515.53, - 218.44, - 526.77 - ], - "text": "# T/2", - "type": "text" - }, - { - "block_id": "p161-b22", - "global_id": 4323, - "bbox": [ - 203.96, - 537.77, - 219.38, - 544.51 - ], - "text": "−T/2", - "type": "text" - }, - { - "block_id": "p161-b23", - "global_id": 4324, - "bbox": [ - 220.88, - 526.01, - 264.64, - 536.98 - ], - "text": "z(τ)z∗(τ)dτ", - "type": "text" - }, - { - "block_id": "p161-b24", - "global_id": 4325, - "bbox": [ - 87.82, - 556.76, - 220.5, - 565.8 - ], - "text": "1.4-1\nSketch the following signals:", - "type": "text" - }, - { - "block_id": "p161-b25", - "global_id": 4326, - "bbox": [ - 116.46, - 567.42, - 197.87, - 587.71 - ], - "text": "(a) u(t −5) −u(t −7)\n(b) u(t −5) + u(t −7)", - "type": "text" - }, - { - "block_id": "p161-b26", - "global_id": 4327, - "bbox": [ - 116.46, - 588.34, - 227.45, - 609.64 - ], - "text": "(c) t2[u(t −1) −u(t −2)]\n(d) (t −4)[u(t −2) −u(t −4)]", - "type": "text" - }, - { - "block_id": "p161-b27", - "global_id": 4328, - "bbox": [ - 87.83, - 614.76, - 289.0, - 634.76 - ], - "text": "1.4-2\nExpress each of the signals in Fig. P1.4-2 by a\nsingle expression valid for all t.", - "type": "text" - }, - { - "block_id": "p161-b28", - "global_id": 4329, - "bbox": [ - 426.7, - 142.5, - 435.58, - 150.5 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p161-b29", - "global_id": 4330, - "bbox": [ - 449.79, - 127.05, - 480.42, - 136.35 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p161-b30", - "global_id": 4331, - "bbox": [ - 399.49, - 90.97, - 403.49, - 98.97 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p161-b31", - "global_id": 4332, - "bbox": [ - 376.2, - 128.06, - 386.86, - 136.35 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p161-b32", - "global_id": 4333, - "bbox": [ - 414.86, - 87.36, - 428.97, - 96.97 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p161-b33", - "global_id": 4334, - "bbox": [ - 426.31, - 250.75, - 435.96, - 258.75 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p161-b34", - "global_id": 4335, - "bbox": [ - 417.14, - 210.56, - 495.85, - 218.76 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p161-b35", - "global_id": 4336, - "bbox": [ - 361.84, - 174.89, - 365.84, - 182.89 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p161-b36", - "global_id": 4337, - "bbox": [ - 461.34, - 210.76, - 465.34, - 218.76 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p161-b37", - "global_id": 4338, - "bbox": [ - 355.17, - 235.09, - 365.84, - 243.39 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p161-b38", - "global_id": 4339, - "bbox": [ - 384.49, - 165.26, - 398.59, - 174.87 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p161-b39", - "global_id": 4340, - "bbox": [ - 384.04, - 187.05, - 389.9, - 196.44 - ], - "text": "t2", - "type": "text" - }, - { - "block_id": "p161-b40", - "global_id": 4341, - "bbox": [ - 343.61, - 266.66, - 395.25, - 275.63 - ], - "text": "Figure P1.4-2", - "type": "text" - }, - { - "block_id": "p161-b41", - "global_id": 4342, - "bbox": [ - 314.97, - 291.26, - 516.13, - 311.57 - ], - "text": "1.4-3\nLetting w(t) = t[u(t) −u(t −1)], define the peri-\nodic signal x(t) as", - "type": "text" - }, - { - "block_id": "p161-b42", - "global_id": 4343, - "bbox": [ - 351.68, - 332.45, - 373.86, - 341.7 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p161-b43", - "global_id": 4344, - "bbox": [ - 379.21, - 323.14, - 391.89, - 332.89 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p161-b44", - "global_id": 4345, - "bbox": [ - 375.7, - 345.29, - 395.39, - 351.96 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p161-b45", - "global_id": 4346, - "bbox": [ - 396.4, - 332.45, - 508.07, - 341.79 - ], - "text": "w(2t + 2k) −0.5w(2t + 2k −1)", - "type": "text" - }, - { - "block_id": "p161-b46", - "global_id": 4347, - "bbox": [ - 344.12, - 368.74, - 516.13, - 378.07 - ], - "text": "(a) Sketch w(t) and x(t). What is the fundamen-", - "type": "text" - }, - { - "block_id": "p161-b47", - "global_id": 4348, - "bbox": [ - 343.61, - 379.69, - 457.08, - 399.99 - ], - "text": "tal period T0 of signal x(t)?\n(b) Sketch y(t) = d", - "type": "text" - }, - { - "block_id": "p161-b48", - "global_id": 4349, - "bbox": [ - 344.12, - 390.65, - 516.13, - 443.83 - ], - "text": "dtx(1 −0.5t).\n(c) Determine\nthe\nenergy\nEz\nand\npower\nPz\nof\nthe\nsignal\nz(t)\n=\nx(0.5 −\n1.5t)[u(t) −u(t −1)]. Sketching z(t) should\nhelp.", - "type": "text" - }, - { - "block_id": "p161-b49", - "global_id": 4350, - "bbox": [ - 314.97, - 448.63, - 516.15, - 479.89 - ], - "text": "1.4-4\nDefine signal x(t) = u(t−1)−u(t−2.5)−2δ(t−\n4) + δ(t −6).\n(a) Sketch y(t) =", - "type": "text" - }, - { - "block_id": "p161-b50", - "global_id": 4351, - "bbox": [ - 409.61, - 463.33, - 417.25, - 474.5 - ], - "text": "$ t", - "type": "text" - }, - { - "block_id": "p161-b51", - "global_id": 4352, - "bbox": [ - 343.61, - 470.55, - 516.12, - 490.85 - ], - "text": "−∞x(τ)dτ.\n(b) Describe a simple change that can be made", - "type": "text" - }, - { - "block_id": "p161-b52", - "global_id": 4353, - "bbox": [ - 359.05, - 492.47, - 516.13, - 512.77 - ], - "text": "to the right-most delta function in x(t) so\nthat y(t) =", - "type": "text" - }, - { - "block_id": "p161-b53", - "global_id": 4354, - "bbox": [ - 398.74, - 496.2, - 406.39, - 507.38 - ], - "text": "$ t", - "type": "text" - }, - { - "block_id": "p161-b54", - "global_id": 4355, - "bbox": [ - 344.12, - 503.43, - 505.64, - 525.11 - ], - "text": "−∞x(τ)dτ has finite energy.\n(c) Sketch z(t) =", - "type": "text" - }, - { - "block_id": "p161-b55", - "global_id": 4356, - "bbox": [ - 409.11, - 508.55, - 421.57, - 519.52 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p161-b56", - "global_id": 4357, - "bbox": [ - 343.61, - 515.77, - 516.13, - 536.07 - ], - "text": "t\nx(τ)dτ.\n(d) Determine real constants A and B so that", - "type": "text" - }, - { - "block_id": "p161-b57", - "global_id": 4358, - "bbox": [ - 359.05, - 537.69, - 393.89, - 546.94 - ], - "text": "w(t) = x", - "type": "text" - }, - { - "block_id": "p161-b58", - "global_id": 4359, - "bbox": [ - 394.91, - 530.48, - 410.64, - 542.74 - ], - "text": "t−A", - "type": "text" - }, - { - "block_id": "p161-b59", - "global_id": 4360, - "bbox": [ - 403.2, - 543.07, - 407.16, - 549.54 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p161-b61", - "global_id": 4361, - "bbox": [ - 359.05, - 538.07, - 516.13, - 557.98 - ], - "text": "has a region of support\n[−2,2].", - "type": "text" - }, - { - "block_id": "p161-b62", - "global_id": 4362, - "bbox": [ - 314.97, - 563.09, - 471.52, - 572.13 - ], - "text": "1.4-5\nSimplify the following expressions:", - "type": "text" - }, - { - "block_id": "p161-b63", - "global_id": 4363, - "bbox": [ - 344.11, - 579.11, - 354.07, - 588.07 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p161-b64", - "global_id": 4364, - "bbox": [ - 359.04, - 566.14, - 384.01, - 581.79 - ], - "text": "sin t", - "type": "text" - }, - { - "block_id": "p161-b65", - "global_id": 4365, - "bbox": [ - 366.29, - 582.51, - 386.95, - 594.43 - ], - "text": "t2 + 2", - "type": "text" - }, - { - "block_id": "p161-b67", - "global_id": 4366, - "bbox": [ - 394.2, - 578.73, - 407.89, - 587.98 - ], - "text": "δ(t)", - "type": "text" - }, - { - "block_id": "p161-b68", - "global_id": 4367, - "bbox": [ - 343.61, - 605.61, - 354.07, - 614.58 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p161-b69", - "global_id": 4368, - "bbox": [ - 359.05, - 592.65, - 389.73, - 608.29 - ], - "text": "jω + 2", - "type": "text" - }, - { - "block_id": "p161-b70", - "global_id": 4369, - "bbox": [ - 366.29, - 609.01, - 390.35, - 620.94 - ], - "text": "ω2 + 9", - "type": "text" - }, - { - "block_id": "p161-b72", - "global_id": 4370, - "bbox": [ - 397.59, - 605.24, - 414.68, - 614.2 - ], - "text": "δ(ω)", - "type": "text" - }, - { - "block_id": "p161-b73", - "global_id": 4371, - "bbox": [ - 344.11, - 622.87, - 440.41, - 635.1 - ], - "text": "(c) [e−t cos(3t −60◦)]δ(t)", - "type": "text" - } - ] - }, - { - "page_num": 162, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p162-b0", - "global_id": 4372, - "bbox": [ - 60.0, - 60.36, - 266.81, - 69.45 - ], - "text": "142\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p162-b1", - "global_id": 4373, - "bbox": [ - 90.72, - 94.27, - 101.18, - 103.24 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p162-b2", - "global_id": 4374, - "bbox": [ - 106.15, - 81.32, - 123.87, - 96.37 - ], - "text": "sin", - "type": "text" - }, - { - "block_id": "p162-b3", - "global_id": 4375, - "bbox": [ - 125.86, - 79.83, - 134.48, - 91.89 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p162-b4", - "global_id": 4376, - "bbox": [ - 131.24, - 79.83, - 163.46, - 99.19 - ], - "text": "2 (t −2)", - "type": "text" - }, - { - "block_id": "p162-b5", - "global_id": 4377, - "bbox": [ - 128.1, - 97.68, - 148.75, - 109.61 - ], - "text": "t2 + 4", - "type": "text" - }, - { - "block_id": "p162-b7", - "global_id": 4378, - "bbox": [ - 170.7, - 93.9, - 198.65, - 103.24 - ], - "text": "δ(1 −t)", - "type": "text" - }, - { - "block_id": "p162-b8", - "global_id": 4379, - "bbox": [ - 91.22, - 120.78, - 101.17, - 129.75 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p162-b9", - "global_id": 4380, - "bbox": [ - 106.15, - 107.82, - 136.22, - 136.11 - ], - "text": "1\njω + 2", - "type": "text" - }, - { - "block_id": "p162-b11", - "global_id": 4381, - "bbox": [ - 143.46, - 120.41, - 174.8, - 129.75 - ], - "text": "δ(ω + 3)", - "type": "text" - }, - { - "block_id": "p162-b12", - "global_id": 4382, - "bbox": [ - 92.21, - 147.28, - 101.18, - 156.24 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p162-b13", - "global_id": 4383, - "bbox": [ - 106.16, - 134.32, - 135.86, - 149.96 - ], - "text": "sin kω", - "type": "text" - }, - { - "block_id": "p162-b14", - "global_id": 4384, - "bbox": [ - 121.7, - 153.28, - 127.57, - 162.24 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p162-b16", - "global_id": 4385, - "bbox": [ - 143.29, - 146.9, - 160.38, - 155.87 - ], - "text": "δ(ω)", - "type": "text" - }, - { - "block_id": "p162-b17", - "global_id": 4386, - "bbox": [ - 90.72, - 162.81, - 263.24, - 182.83 - ], - "text": "[Hint:\nUse\nEq.\n(1.10).\nFor\npart\n(f)\nuse\nL’Hôpital’s rule.]", - "type": "text" - }, - { - "block_id": "p162-b18", - "global_id": 4387, - "bbox": [ - 62.08, - 188.13, - 207.44, - 197.17 - ], - "text": "1.4-6\nEvaluate the following integrals:", - "type": "text" - }, - { - "block_id": "p162-b19", - "global_id": 4388, - "bbox": [ - 91.22, - 203.88, - 101.17, - 212.84 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p162-b20", - "global_id": 4389, - "bbox": [ - 106.15, - 191.3, - 121.6, - 202.27 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b21", - "global_id": 4390, - "bbox": [ - 110.89, - 213.54, - 122.55, - 220.02 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b22", - "global_id": 4391, - "bbox": [ - 124.05, - 203.5, - 177.74, - 212.75 - ], - "text": "δ(τ)x(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p162-b23", - "global_id": 4392, - "bbox": [ - 90.72, - 229.17, - 101.18, - 238.13 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p162-b24", - "global_id": 4393, - "bbox": [ - 106.16, - 216.6, - 121.6, - 227.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b25", - "global_id": 4394, - "bbox": [ - 110.89, - 238.83, - 122.55, - 245.31 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b26", - "global_id": 4395, - "bbox": [ - 124.05, - 228.79, - 177.74, - 238.04 - ], - "text": "x(τ)δ(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p162-b27", - "global_id": 4396, - "bbox": [ - 91.23, - 254.47, - 101.18, - 263.43 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p162-b28", - "global_id": 4397, - "bbox": [ - 106.16, - 241.89, - 121.6, - 252.86 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b29", - "global_id": 4398, - "bbox": [ - 110.89, - 264.13, - 122.55, - 270.61 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b30", - "global_id": 4399, - "bbox": [ - 125.04, - 250.67, - 164.31, - 263.34 - ], - "text": "δ(t)e−jωt dt", - "type": "text" - }, - { - "block_id": "p162-b31", - "global_id": 4400, - "bbox": [ - 90.72, - 279.76, - 101.18, - 288.72 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p162-b32", - "global_id": 4401, - "bbox": [ - 106.15, - 267.18, - 121.6, - 278.15 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b33", - "global_id": 4402, - "bbox": [ - 110.89, - 289.42, - 122.55, - 295.9 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b34", - "global_id": 4403, - "bbox": [ - 124.05, - 279.38, - 186.84, - 288.72 - ], - "text": "δ(2t −3)sin πtdt", - "type": "text" - }, - { - "block_id": "p162-b35", - "global_id": 4404, - "bbox": [ - 91.23, - 305.05, - 101.18, - 314.02 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p162-b36", - "global_id": 4405, - "bbox": [ - 106.16, - 292.48, - 121.6, - 303.45 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b37", - "global_id": 4406, - "bbox": [ - 110.89, - 314.72, - 122.55, - 321.2 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b38", - "global_id": 4407, - "bbox": [ - 124.05, - 301.26, - 171.41, - 314.02 - ], - "text": "δ(t + 3)e−t dt", - "type": "text" - }, - { - "block_id": "p162-b39", - "global_id": 4408, - "bbox": [ - 92.21, - 330.34, - 101.17, - 339.31 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p162-b40", - "global_id": 4409, - "bbox": [ - 106.15, - 317.77, - 121.6, - 328.74 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b41", - "global_id": 4410, - "bbox": [ - 110.89, - 340.01, - 122.55, - 346.49 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b42", - "global_id": 4411, - "bbox": [ - 123.05, - 326.27, - 186.33, - 339.31 - ], - "text": "(t3 + 4)δ(1 −t)dt", - "type": "text" - }, - { - "block_id": "p162-b43", - "global_id": 4412, - "bbox": [ - 90.72, - 355.64, - 101.18, - 364.61 - ], - "text": "(g)", - "type": "text" - }, - { - "block_id": "p162-b44", - "global_id": 4413, - "bbox": [ - 106.15, - 343.07, - 121.6, - 354.04 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b45", - "global_id": 4414, - "bbox": [ - 110.89, - 365.3, - 122.55, - 371.78 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b46", - "global_id": 4415, - "bbox": [ - 124.05, - 355.27, - 187.6, - 364.61 - ], - "text": "x(2 −t)δ(3 −t)dt", - "type": "text" - }, - { - "block_id": "p162-b47", - "global_id": 4416, - "bbox": [ - 90.72, - 380.93, - 101.18, - 389.9 - ], - "text": "(h)", - "type": "text" - }, - { - "block_id": "p162-b48", - "global_id": 4417, - "bbox": [ - 106.15, - 368.36, - 121.6, - 379.33 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b49", - "global_id": 4418, - "bbox": [ - 110.89, - 390.6, - 122.55, - 397.08 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b50", - "global_id": 4419, - "bbox": [ - 124.05, - 377.24, - 157.49, - 389.9 - ], - "text": "e(x−1) cos", - "type": "text" - }, - { - "block_id": "p162-b51", - "global_id": 4420, - "bbox": [ - 158.48, - 373.35, - 167.1, - 385.41 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p162-b52", - "global_id": 4421, - "bbox": [ - 163.86, - 373.35, - 197.44, - 392.71 - ], - "text": "2 (x −5)", - "type": "text" - }, - { - "block_id": "p162-b53", - "global_id": 4422, - "bbox": [ - 197.43, - 380.56, - 236.2, - 389.9 - ], - "text": "δ(x −3)dx", - "type": "text" - }, - { - "block_id": "p162-b54", - "global_id": 4423, - "bbox": [ - 62.08, - 400.04, - 263.23, - 420.05 - ], - "text": "1.4-7\nFor real and positive constant a, evaluate the\nfollowing integral:", - "type": "text" - }, - { - "block_id": "p162-b55", - "global_id": 4424, - "bbox": [ - 154.88, - 428.18, - 170.34, - 439.15 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b56", - "global_id": 4425, - "bbox": [ - 159.62, - 450.42, - 171.28, - 456.9 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b57", - "global_id": 4426, - "bbox": [ - 172.77, - 440.38, - 198.92, - 449.63 - ], - "text": "δ(at)dt", - "type": "text" - }, - { - "block_id": "p162-b58", - "global_id": 4427, - "bbox": [ - 62.08, - 470.19, - 263.23, - 479.53 - ], - "text": "1.4-8\n(a) Find and sketch dx/dt for the signal x(t)", - "type": "text" - }, - { - "block_id": "p162-b59", - "global_id": 4428, - "bbox": [ - 90.72, - 481.52, - 263.25, - 502.19 - ], - "text": "shown in Fig. P1.2-2.\n(b) Find and sketch d2x/dt2 for the signal x1(t)", - "type": "text" - }, - { - "block_id": "p162-b60", - "global_id": 4429, - "bbox": [ - 106.16, - 503.44, - 194.32, - 512.41 - ], - "text": "depicted in Fig. P1.4-2a.", - "type": "text" - }, - { - "block_id": "p162-b61", - "global_id": 4430, - "bbox": [ - 62.08, - 517.71, - 149.18, - 526.76 - ], - "text": "1.4-9\nFind and sketch", - "type": "text" - }, - { - "block_id": "p162-b62", - "global_id": 4431, - "bbox": [ - 152.31, - 510.2, - 159.96, - 521.37 - ], - "text": "$ t", - "type": "text" - }, - { - "block_id": "p162-b63", - "global_id": 4432, - "bbox": [ - 90.72, - 517.42, - 263.24, - 537.72 - ], - "text": "−∞x(t)dt for the signals x(t)\nillustrated in Fig. P1.4-9.", - "type": "text" - }, - { - "block_id": "p162-b64", - "global_id": 4433, - "bbox": [ - 57.59, - 543.03, - 263.25, - 573.99 - ], - "text": "1.4-10\nUsing the generalized function definition of\nimpulse [Eq. (1.11) with T = 0], show that δ(t)\nis an even function of t.", - "type": "text" - }, - { - "block_id": "p162-b65", - "global_id": 4434, - "bbox": [ - 57.59, - 579.3, - 263.24, - 599.3 - ], - "text": "1.4-11\nUsing the generalized function definition of\nimpulse [Eq. (1.11) with T = 0], show that", - "type": "text" - }, - { - "block_id": "p162-b66", - "global_id": 4435, - "bbox": [ - 149.72, - 613.38, - 186.81, - 628.64 - ], - "text": "δ(at) = 1", - "type": "text" - }, - { - "block_id": "p162-b67", - "global_id": 4436, - "bbox": [ - 179.76, - 619.3, - 204.25, - 634.91 - ], - "text": "|a|δ(t)", - "type": "text" - }, - { - "block_id": "p162-b68", - "global_id": 4437, - "bbox": [ - 328.86, - 91.84, - 332.86, - 99.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p162-b69", - "global_id": 4438, - "bbox": [ - 321.91, - 137.3, - 332.58, - 145.59 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p162-b70", - "global_id": 4439, - "bbox": [ - 328.58, - 118.76, - 463.5, - 126.97 - ], - "text": "2\n3\n1\nt\n0", - "type": "text" - }, - { - "block_id": "p162-b71", - "global_id": 4440, - "bbox": [ - 449.86, - 90.04, - 453.86, - 98.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p162-b72", - "global_id": 4441, - "bbox": [ - 396.68, - 157.45, - 405.56, - 165.45 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p162-b73", - "global_id": 4442, - "bbox": [ - 334.22, - 184.13, - 338.22, - 192.13 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p162-b74", - "global_id": 4443, - "bbox": [ - 327.27, - 229.59, - 337.94, - 237.88 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p162-b75", - "global_id": 4444, - "bbox": [ - 333.94, - 211.18, - 468.86, - 220.25 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p162-b76", - "global_id": 4445, - "bbox": [ - 396.3, - 249.74, - 405.95, - 257.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p162-b77", - "global_id": 4446, - "bbox": [ - 373.94, - 201.45, - 450.14, - 209.45 - ], - "text": "3\n2\n1", - "type": "text" - }, - { - "block_id": "p162-b78", - "global_id": 4447, - "bbox": [ - 379.4, - 223.89, - 462.37, - 232.19 - ], - "text": "1\n1\n1", - "type": "text" - }, - { - "block_id": "p162-b79", - "global_id": 4448, - "bbox": [ - 317.86, - 265.66, - 369.51, - 274.63 - ], - "text": "Figure P1.4-9", - "type": "text" - }, - { - "block_id": "p162-b80", - "global_id": 4449, - "bbox": [ - 284.74, - 283.01, - 353.75, - 292.05 - ], - "text": "1.4-12\nShow that", - "type": "text" - }, - { - "block_id": "p162-b81", - "global_id": 4450, - "bbox": [ - 359.44, - 293.15, - 374.89, - 304.12 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p162-b82", - "global_id": 4451, - "bbox": [ - 364.18, - 315.4, - 375.84, - 321.87 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p162-b83", - "global_id": 4452, - "bbox": [ - 377.33, - 303.57, - 448.82, - 314.7 - ], - "text": "˙δ(t)φ(t)dt = −˙φ(0)", - "type": "text" - }, - { - "block_id": "p162-b84", - "global_id": 4453, - "bbox": [ - 317.86, - 329.73, - 490.4, - 373.74 - ], - "text": "where φ(t) and ˙φ(t) are continuous at t = 0, and\nφ(t) →0 as t →±∞. This integral defines ˙δ(t)\nas a generalized function. [Hint: Use integration\nby parts.]", - "type": "text" - }, - { - "block_id": "p162-b85", - "global_id": 4454, - "bbox": [ - 284.74, - 375.46, - 490.39, - 431.53 - ], - "text": "1.4-13\nA sinusoid eσt cos ωt can be expressed as a sum\nof exponentials est and e−st [Eq. (1.14)] with\ncomplex frequencies s = σ +jω and s = σ −jω.\nLocate in the complex plane the frequencies of\nthe following sinusoids:", - "type": "text" - }, - { - "block_id": "p162-b86", - "global_id": 4455, - "bbox": [ - 317.86, - 433.42, - 368.91, - 453.44 - ], - "text": "(a) cos3t\n(b) e−3t cos3t", - "type": "text" - }, - { - "block_id": "p162-b87", - "global_id": 4456, - "bbox": [ - 317.86, - 452.18, - 363.85, - 475.36 - ], - "text": "(c) e2t cos3t\n(d) e−2t", - "type": "text" - }, - { - "block_id": "p162-b88", - "global_id": 4457, - "bbox": [ - 318.37, - 475.92, - 342.32, - 486.32 - ], - "text": "(e) e2t", - "type": "text" - }, - { - "block_id": "p162-b89", - "global_id": 4458, - "bbox": [ - 319.36, - 488.31, - 337.79, - 497.28 - ], - "text": "(f) 5", - "type": "text" - }, - { - "block_id": "p162-b90", - "global_id": 4459, - "bbox": [ - 289.22, - 502.18, - 490.37, - 522.18 - ], - "text": "1.5-1\nFind and sketch the odd and the even compo-\nnents of the following:", - "type": "text" - }, - { - "block_id": "p162-b91", - "global_id": 4460, - "bbox": [ - 317.86, - 523.81, - 349.63, - 544.1 - ], - "text": "(a) u(t)\n(b) tu(t)", - "type": "text" - }, - { - "block_id": "p162-b92", - "global_id": 4461, - "bbox": [ - 317.86, - 545.72, - 358.35, - 566.76 - ], - "text": "(c) sinω0t\n(d) cosω0t", - "type": "text" - }, - { - "block_id": "p162-b93", - "global_id": 4462, - "bbox": [ - 318.37, - 567.64, - 379.82, - 577.71 - ], - "text": "(e) cos(ω0t + θ)", - "type": "text" - }, - { - "block_id": "p162-b94", - "global_id": 4463, - "bbox": [ - 317.86, - 578.6, - 372.18, - 599.64 - ], - "text": "(f) sinω0tu(t)\n(g) cosω0tu(t)", - "type": "text" - }, - { - "block_id": "p162-b95", - "global_id": 4464, - "bbox": [ - 289.22, - 603.5, - 473.29, - 612.84 - ], - "text": "1.5-2\nDefine x(t) = 2u(t+1)−u(t−2)−u(t−3).", - "type": "text" - }, - { - "block_id": "p162-b96", - "global_id": 4465, - "bbox": [ - 318.37, - 614.46, - 490.38, - 624.48 - ], - "text": "(a) Letting xo(t) designate the odd portion of", - "type": "text" - }, - { - "block_id": "p162-b97", - "global_id": 4466, - "bbox": [ - 333.31, - 625.42, - 453.34, - 635.44 - ], - "text": "x(t), accurately sketch xo(1 −2t).", - "type": "text" - } - ] - }, - { - "page_num": 163, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p163-b0", - "global_id": 4467, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n143", - "type": "text" - }, - { - "block_id": "p163-b1", - "global_id": 4468, - "bbox": [ - 116.46, - 85.58, - 288.99, - 95.59 - ], - "text": "(b) Letting xe(t) designate the even portion of", - "type": "text" - }, - { - "block_id": "p163-b2", - "global_id": 4469, - "bbox": [ - 131.9, - 96.54, - 255.78, - 106.55 - ], - "text": "x(t), accurately sketch xe(2 + t/3).", - "type": "text" - }, - { - "block_id": "p163-b3", - "global_id": 4470, - "bbox": [ - 87.82, - 110.78, - 289.0, - 119.82 - ], - "text": "1.5-3\n(a) Determine even and odd components of the", - "type": "text" - }, - { - "block_id": "p163-b4", - "global_id": 4471, - "bbox": [ - 116.46, - 120.18, - 288.98, - 141.74 - ], - "text": "signal x(t) = e−2tu(t).\n(b) Show that the energy of x(t) is the sum of", - "type": "text" - }, - { - "block_id": "p163-b5", - "global_id": 4472, - "bbox": [ - 116.96, - 143.73, - 288.98, - 174.62 - ], - "text": "energies of its odd and even components\nfound in part (a).\n(c) Generalize the result in part (b) for any finite", - "type": "text" - }, - { - "block_id": "p163-b6", - "global_id": 4473, - "bbox": [ - 131.9, - 176.61, - 182.04, - 185.58 - ], - "text": "energy signal.", - "type": "text" - }, - { - "block_id": "p163-b7", - "global_id": 4474, - "bbox": [ - 87.82, - 190.18, - 288.98, - 200.2 - ], - "text": "1.5-4\n(a) If xe(t) and xo(t) are even and the odd", - "type": "text" - }, - { - "block_id": "p163-b8", - "global_id": 4475, - "bbox": [ - 131.89, - 201.15, - 288.98, - 224.68 - ], - "text": "components of a real signal x(t), then show\nthat\n# ∞", - "type": "text" - }, - { - "block_id": "p163-b9", - "global_id": 4476, - "bbox": [ - 177.7, - 235.95, - 189.37, - 242.43 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p163-b10", - "global_id": 4477, - "bbox": [ - 190.86, - 225.91, - 247.91, - 235.93 - ], - "text": "xe(t)xo(t)dt = 0", - "type": "text" - }, - { - "block_id": "p163-b11", - "global_id": 4478, - "bbox": [ - 116.46, - 248.7, - 167.79, - 257.66 - ], - "text": "(b) Show that", - "type": "text" - }, - { - "block_id": "p163-b12", - "global_id": 4479, - "bbox": [ - 164.05, - 260.92, - 179.5, - 271.9 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p163-b13", - "global_id": 4480, - "bbox": [ - 168.79, - 283.17, - 180.45, - 289.64 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p163-b14", - "global_id": 4481, - "bbox": [ - 181.94, - 273.13, - 212.27, - 282.38 - ], - "text": "x(t)dt =", - "type": "text" - }, - { - "block_id": "p163-b15", - "global_id": 4482, - "bbox": [ - 214.11, - 260.92, - 229.56, - 271.9 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p163-b16", - "global_id": 4483, - "bbox": [ - 218.84, - 283.17, - 230.5, - 289.64 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p163-b17", - "global_id": 4484, - "bbox": [ - 232.0, - 273.13, - 256.68, - 283.14 - ], - "text": "xe(t)dt", - "type": "text" - }, - { - "block_id": "p163-b18", - "global_id": 4485, - "bbox": [ - 87.82, - 298.4, - 288.99, - 341.29 - ], - "text": "1.5-5\nAn aperiodic signal is defined as x(t) =\nsin(πt)u(t), where u(t) is the continuous-time\nstep function. Is the odd portion of this signal,\nxo(t), periodic? Justify your answer.", - "type": "text" - }, - { - "block_id": "p163-b19", - "global_id": 4486, - "bbox": [ - 87.82, - 345.23, - 288.99, - 388.12 - ], - "text": "1.5-6\nAn aperiodic signal is defined as x(t) =\ncos(πt)u(t), where u(t) is the continuous-time\nstep function. Is the even portion of this signal,\nxe(t), periodic? Justify your answer.", - "type": "text" - }, - { - "block_id": "p163-b20", - "global_id": 4487, - "bbox": [ - 87.82, - 392.05, - 280.53, - 401.39 - ], - "text": "1.5-7\nConsider the signal x(t) shown in Fig. P1.5-7.", - "type": "text" - }, - { - "block_id": "p163-b21", - "global_id": 4488, - "bbox": [ - 179.31, - 430.96, - 183.31, - 438.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p163-b22", - "global_id": 4489, - "bbox": [ - 132.67, - 472.41, - 278.28, - 483.43 - ], - "text": "0\n–1\n1\nt", - "type": "text" - }, - { - "block_id": "p163-b23", - "global_id": 4490, - "bbox": [ - 90.4, - 491.17, - 142.04, - 500.14 - ], - "text": "Figure P1.5-7", - "type": "text" - }, - { - "block_id": "p163-b24", - "global_id": 4491, - "bbox": [ - 116.96, - 512.84, - 288.98, - 522.18 - ], - "text": "(a) Determine and carefully sketch v(t) =", - "type": "text" - }, - { - "block_id": "p163-b25", - "global_id": 4492, - "bbox": [ - 116.46, - 523.81, - 276.55, - 544.1 - ], - "text": "3x(−(1/2)(t + 1)).\n(b) Determine the energy and power of v(t).", - "type": "text" - }, - { - "block_id": "p163-b26", - "global_id": 4493, - "bbox": [ - 116.97, - 546.09, - 288.97, - 555.06 - ], - "text": "(c) Determine and carefully sketch the even", - "type": "text" - }, - { - "block_id": "p163-b27", - "global_id": 4494, - "bbox": [ - 116.46, - 556.68, - 288.99, - 576.97 - ], - "text": "portion of v(t), ve(t).\n(d) Let a = 2 and b = 3; sketch v(at + b),", - "type": "text" - }, - { - "block_id": "p163-b28", - "global_id": 4495, - "bbox": [ - 116.97, - 578.6, - 288.99, - 598.9 - ], - "text": "v(at) + b, av(t + b), and av(t) + b.\n(e) Let a = −3 and b = −2; sketch v(at + b),", - "type": "text" - }, - { - "block_id": "p163-b29", - "global_id": 4496, - "bbox": [ - 131.9, - 600.51, - 254.52, - 609.85 - ], - "text": "v(at) + b, av(t + b), and av(t) + b.", - "type": "text" - }, - { - "block_id": "p163-b30", - "global_id": 4497, - "bbox": [ - 87.83, - 614.46, - 288.99, - 634.77 - ], - "text": "1.5-8\nConsider the signal y(t) = (1/5)x(−2t −3)\nshown in Fig. P1.5-8.", - "type": "text" - }, - { - "block_id": "p163-b31", - "global_id": 4498, - "bbox": [ - 383.0, - 102.98, - 387.0, - 110.98 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p163-b32", - "global_id": 4499, - "bbox": [ - 322.28, - 144.36, - 508.02, - 155.14 - ], - "text": "0\n–1\n1\nt\n2", - "type": "text" - }, - { - "block_id": "p163-b33", - "global_id": 4500, - "bbox": [ - 317.55, - 162.89, - 369.19, - 171.86 - ], - "text": "Figure P1.5-8", - "type": "text" - }, - { - "block_id": "p163-b34", - "global_id": 4501, - "bbox": [ - 344.12, - 184.95, - 516.13, - 194.96 - ], - "text": "(a) Does y(t) have an odd portion, yo(t)? If", - "type": "text" - }, - { - "block_id": "p163-b35", - "global_id": 4502, - "bbox": [ - 343.61, - 195.91, - 516.14, - 238.12 - ], - "text": "so, determine and carefully sketch yo(t).\nOtherwise, explain why no odd portion\nexists.\n(b) Determine and carefully sketch the original", - "type": "text" - }, - { - "block_id": "p163-b36", - "global_id": 4503, - "bbox": [ - 359.05, - 239.74, - 398.3, - 249.08 - ], - "text": "signal x(t).", - "type": "text" - }, - { - "block_id": "p163-b37", - "global_id": 4504, - "bbox": [ - 314.97, - 253.7, - 516.13, - 274.0 - ], - "text": "1.5-9\nConsider the signal −(1/2)x(−3t + 2) shown in\nFig. P1.5-9.", - "type": "text" - }, - { - "block_id": "p163-b38", - "global_id": 4505, - "bbox": [ - 418.82, - 298.97, - 422.82, - 306.97 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p163-b39", - "global_id": 4506, - "bbox": [ - 359.75, - 340.23, - 508.32, - 351.58 - ], - "text": "0\n–1\n1\nt", - "type": "text" - }, - { - "block_id": "p163-b40", - "global_id": 4507, - "bbox": [ - 317.55, - 359.24, - 369.19, - 368.21 - ], - "text": "Figure P1.5-9", - "type": "text" - }, - { - "block_id": "p163-b41", - "global_id": 4508, - "bbox": [ - 344.12, - 375.69, - 516.12, - 384.65 - ], - "text": "(a) Determine and carefully sketch the original", - "type": "text" - }, - { - "block_id": "p163-b42", - "global_id": 4509, - "bbox": [ - 343.61, - 386.28, - 516.12, - 406.57 - ], - "text": "signal x(t).\n(b) Determine and carefully sketch the even", - "type": "text" - }, - { - "block_id": "p163-b43", - "global_id": 4510, - "bbox": [ - 344.12, - 408.19, - 516.13, - 428.49 - ], - "text": "portion of the original signal x(t).\n(c) Determine and carefully sketch the odd", - "type": "text" - }, - { - "block_id": "p163-b44", - "global_id": 4511, - "bbox": [ - 359.05, - 430.11, - 479.49, - 439.45 - ], - "text": "portion of the original signal x(t).", - "type": "text" - }, - { - "block_id": "p163-b45", - "global_id": 4512, - "bbox": [ - 310.48, - 444.37, - 516.14, - 497.24 - ], - "text": "1.5-10\nThe conjugate symmetric (or Hermitian) portion\nof a signal is defined as wcs(t) = (w(t) +\nw∗(−t))/2. Show that the real portion of wcs(t)\nis even and that the imaginary portion of wcs(t)\nis odd.", - "type": "text" - }, - { - "block_id": "p163-b46", - "global_id": 4513, - "bbox": [ - 310.48, - 502.17, - 521.36, - 555.72 - ], - "text": "1.5-11\nThe conjugate antisymmetric (or skew-Hermitian)\nportion of a signal is defined as wca(t) =\n(w(t) −w∗(−t))/2. Show that the real portion\nof wca(t) is odd and that the imaginary portion\nof wca(t) is even.", - "type": "text" - }, - { - "block_id": "p163-b47", - "global_id": 4514, - "bbox": [ - 310.48, - 558.41, - 426.88, - 569.01 - ], - "text": "1.5-12\nDefine w(t) = ej(t+π/4).", - "type": "text" - }, - { - "block_id": "p163-b48", - "global_id": 4515, - "bbox": [ - 344.11, - 571.0, - 516.12, - 579.96 - ], - "text": "(a) Referring to the definition in Prob. 1.5-10,", - "type": "text" - }, - { - "block_id": "p163-b49", - "global_id": 4516, - "bbox": [ - 343.61, - 581.59, - 516.14, - 612.84 - ], - "text": "determine wcs(t). Express your simplified\nanswer in standard rectangular form.\n(b) Referring to the definition in Prob. 1.5-11,", - "type": "text" - }, - { - "block_id": "p163-b50", - "global_id": 4517, - "bbox": [ - 359.05, - 614.46, - 516.14, - 634.76 - ], - "text": "determine wca(t). Express your simplified\nanswer in standard polar form.", - "type": "text" - } - ] - }, - { - "page_num": 164, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p164-b0", - "global_id": 4518, - "bbox": [ - 60.0, - 60.36, - 266.81, - 69.45 - ], - "text": "144\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p164-b1", - "global_id": 4519, - "bbox": [ - 57.59, - 85.64, - 263.24, - 127.86 - ], - "text": "1.5-13\nFigure P1.5-13 plots a complex signal w(t) in the\ncomplex plane over the time range (0 ≤t ≤1).\nThe time t = 0 corresponds with the origin, while\nthe time t = 1 corresponds with the point (2, 1).", - "type": "text" - }, - { - "block_id": "p164-b2", - "global_id": 4520, - "bbox": [ - 111.19, - 171.76, - 115.19, - 179.76 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p164-b3", - "global_id": 4521, - "bbox": [ - 100.83, - 221.72, - 224.05, - 232.93 - ], - "text": "0\n2\nRe\n1", - "type": "text" - }, - { - "block_id": "p164-b4", - "global_id": 4522, - "bbox": [ - 97.91, - 193.43, - 105.91, - 202.32 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p164-b5", - "global_id": 4523, - "bbox": [ - 90.72, - 240.86, - 146.83, - 249.82 - ], - "text": "Figure P1.5-13", - "type": "text" - }, - { - "block_id": "p164-b6", - "global_id": 4524, - "bbox": [ - 91.22, - 291.67, - 263.24, - 301.01 - ], - "text": "(a) In the complex plane, plot w(t) over (−1 ≤", - "type": "text" - }, - { - "block_id": "p164-b7", - "global_id": 4525, - "bbox": [ - 90.72, - 302.63, - 263.24, - 322.93 - ], - "text": "t ≤1) if w(t) is an even signal.\n(b) In the complex plane, plot w(t) over (−1 ≤", - "type": "text" - }, - { - "block_id": "p164-b8", - "global_id": 4526, - "bbox": [ - 91.22, - 324.55, - 263.24, - 344.85 - ], - "text": "t ≤1) if w(t) is an odd signal.\n(c) In the complex plane, plot w(t) over (−1 ≤", - "type": "text" - }, - { - "block_id": "p164-b9", - "global_id": 4527, - "bbox": [ - 90.72, - 346.47, - 263.24, - 377.73 - ], - "text": "t ≤1) if w(t) is a conjugate symmetric\nsignal. [Hint: See Prob. 1.5-10.]\n(d) In the complex plane, plot w(t) over (−1 ≤", - "type": "text" - }, - { - "block_id": "p164-b10", - "global_id": 4528, - "bbox": [ - 91.22, - 379.35, - 263.25, - 410.6 - ], - "text": "t ≤1) if w(t) is a conjugate antisymmetric\nsignal. [Hint: See Prob. 1.5-11.]\n(e) In the complex plane, plot as much of w(3t)", - "type": "text" - }, - { - "block_id": "p164-b11", - "global_id": 4529, - "bbox": [ - 106.16, - 412.6, - 147.5, - 421.57 - ], - "text": "as possible.", - "type": "text" - }, - { - "block_id": "p164-b12", - "global_id": 4530, - "bbox": [ - 57.59, - 428.17, - 263.23, - 471.37 - ], - "text": "1.5-14\nDefine complex signal x(t) = t2(1 + j) over\ninterval (1 ≤t ≤2). The remaining portion is\ndefined such that x(t) is a minimum-energy,\nskew-Hermitian signal.", - "type": "text" - }, - { - "block_id": "p164-b13", - "global_id": 4531, - "bbox": [ - 90.72, - 472.99, - 263.23, - 493.29 - ], - "text": "(a) Fully describe x(t) for all t.\n(b) Sketch y(t) = Re{x(t)} versus the indepen-", - "type": "text" - }, - { - "block_id": "p164-b14", - "global_id": 4532, - "bbox": [ - 91.23, - 495.19, - 263.23, - 515.21 - ], - "text": "dent variable t.\n(c) Sketch z(t) = Re{jx(−2t + 1)} versus the", - "type": "text" - }, - { - "block_id": "p164-b15", - "global_id": 4533, - "bbox": [ - 90.72, - 517.12, - 263.23, - 559.05 - ], - "text": "independent variable t.\n(d) Determine the energy and power of x(t).\n[Hint: See Prob. 1.5-11 for a definition of\nskew-Hermitian signals.]", - "type": "text" - }, - { - "block_id": "p164-b16", - "global_id": 4534, - "bbox": [ - 62.08, - 566.95, - 263.23, - 597.9 - ], - "text": "1.6-1\nWrite the input–output relationship for an\nideal integrator. Determine the zero-input and\nzero-state components of the response.", - "type": "text" - }, - { - "block_id": "p164-b17", - "global_id": 4535, - "bbox": [ - 62.08, - 605.5, - 263.23, - 658.67 - ], - "text": "1.6-2\nA force x(t) acts on a ball of mass M\n(Fig. P1.6-2). Show that the velocity v(t) of the\nball at any instant t > 0 can be determined if we\nknow the force x(t) over the interval from 0 to t\nand the ball’s initial velocity v(0).", - "type": "text" - }, - { - "block_id": "p164-b18", - "global_id": 4536, - "bbox": [ - 384.97, - 150.31, - 396.13, - 157.98 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p164-b19", - "global_id": 4537, - "bbox": [ - 335.29, - 96.85, - 346.82, - 104.53 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p164-b20", - "global_id": 4538, - "bbox": [ - 451.9, - 150.19, - 455.28, - 157.79 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p164-b21", - "global_id": 4539, - "bbox": [ - 387.13, - 95.38, - 393.46, - 102.98 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p164-b22", - "global_id": 4540, - "bbox": [ - 317.86, - 165.91, - 369.51, - 174.88 - ], - "text": "Figure P1.6-2", - "type": "text" - }, - { - "block_id": "p164-b23", - "global_id": 4541, - "bbox": [ - 289.22, - 182.13, - 490.38, - 202.13 - ], - "text": "1.6-3\nFrom your personal experience, provide an\nexample of:", - "type": "text" - }, - { - "block_id": "p164-b24", - "global_id": 4542, - "bbox": [ - 317.86, - 204.12, - 490.4, - 224.05 - ], - "text": "(a) a single-input, single-output (SISO) system\n(b) a multiple-input, single-output (MISO) sys-", - "type": "text" - }, - { - "block_id": "p164-b25", - "global_id": 4543, - "bbox": [ - 318.37, - 226.04, - 490.4, - 245.97 - ], - "text": "tem\n(c) a single-input, multiple-output (SIMO) sys-", - "type": "text" - }, - { - "block_id": "p164-b26", - "global_id": 4544, - "bbox": [ - 317.86, - 247.96, - 490.4, - 267.88 - ], - "text": "tem\n(d) a multiple-input, multiple-output (MIMO)", - "type": "text" - }, - { - "block_id": "p164-b27", - "global_id": 4545, - "bbox": [ - 333.31, - 269.88, - 358.21, - 278.84 - ], - "text": "system", - "type": "text" - }, - { - "block_id": "p164-b28", - "global_id": 4546, - "bbox": [ - 289.22, - 283.96, - 490.39, - 325.88 - ], - "text": "1.7-1\nFor the systems described by the following\nequations, with the input x(t) and output y(t),\ndetermine which of the systems are linear and\nwhich are nonlinear.", - "type": "text" - }, - { - "block_id": "p164-b29", - "global_id": 4547, - "bbox": [ - 318.36, - 333.75, - 352.3, - 349.37 - ], - "text": "(a) dy(t)", - "type": "text" - }, - { - "block_id": "p164-b30", - "global_id": 4548, - "bbox": [ - 339.83, - 338.58, - 408.86, - 355.64 - ], - "text": "dt\n+ 2y(t) = x2(t)", - "type": "text" - }, - { - "block_id": "p164-b31", - "global_id": 4549, - "bbox": [ - 317.86, - 356.64, - 352.3, - 372.27 - ], - "text": "(b) dy(t)", - "type": "text" - }, - { - "block_id": "p164-b32", - "global_id": 4550, - "bbox": [ - 339.83, - 361.48, - 414.01, - 378.54 - ], - "text": "dt\n+ 3ty(t) = t2x(t)", - "type": "text" - }, - { - "block_id": "p164-b33", - "global_id": 4551, - "bbox": [ - 318.37, - 379.54, - 389.41, - 388.88 - ], - "text": "(c) 3y(t) + 2 = x(t)", - "type": "text" - }, - { - "block_id": "p164-b34", - "global_id": 4552, - "bbox": [ - 317.86, - 391.52, - 352.3, - 407.15 - ], - "text": "(d) dy(t)", - "type": "text" - }, - { - "block_id": "p164-b35", - "global_id": 4553, - "bbox": [ - 339.83, - 396.37, - 404.38, - 413.43 - ], - "text": "dt\n+ y2(t) = x(t)", - "type": "text" - }, - { - "block_id": "p164-b36", - "global_id": 4554, - "bbox": [ - 318.37, - 423.56, - 328.32, - 432.52 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p164-b37", - "global_id": 4555, - "bbox": [ - 333.31, - 410.6, - 358.36, - 426.14 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p164-b38", - "global_id": 4556, - "bbox": [ - 345.89, - 429.83, - 352.86, - 438.8 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p164-b39", - "global_id": 4557, - "bbox": [ - 359.55, - 410.59, - 368.84, - 421.42 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p164-b40", - "global_id": 4558, - "bbox": [ - 370.74, - 423.19, - 420.96, - 432.52 - ], - "text": "+ 2y(t) = x(t)", - "type": "text" - }, - { - "block_id": "p164-b41", - "global_id": 4559, - "bbox": [ - 319.37, - 441.84, - 352.3, - 457.47 - ], - "text": "(f) dy(t)", - "type": "text" - }, - { - "block_id": "p164-b42", - "global_id": 4560, - "bbox": [ - 339.83, - 441.84, - 428.12, - 463.75 - ], - "text": "dt\n+ (sin t)y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p164-b43", - "global_id": 4561, - "bbox": [ - 415.63, - 448.13, - 456.93, - 463.75 - ], - "text": "dt\n+ 2x(t)", - "type": "text" - }, - { - "block_id": "p164-b44", - "global_id": 4562, - "bbox": [ - 317.86, - 464.75, - 352.3, - 480.38 - ], - "text": "(g) dy(t)", - "type": "text" - }, - { - "block_id": "p164-b45", - "global_id": 4563, - "bbox": [ - 339.83, - 464.75, - 424.16, - 486.65 - ], - "text": "dt\n+ 2y(t) = x(t)dx(t)", - "type": "text" - }, - { - "block_id": "p164-b46", - "global_id": 4564, - "bbox": [ - 411.67, - 477.68, - 418.64, - 486.65 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p164-b47", - "global_id": 4565, - "bbox": [ - 317.86, - 494.66, - 355.46, - 503.99 - ], - "text": "(h) y(t) =", - "type": "text" - }, - { - "block_id": "p164-b48", - "global_id": 4566, - "bbox": [ - 357.31, - 482.45, - 367.95, - 493.62 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p164-b49", - "global_id": 4567, - "bbox": [ - 362.04, - 504.69, - 373.7, - 511.17 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p164-b50", - "global_id": 4568, - "bbox": [ - 375.2, - 494.65, - 400.47, - 503.9 - ], - "text": "x(τ)dτ", - "type": "text" - }, - { - "block_id": "p164-b51", - "global_id": 4569, - "bbox": [ - 289.22, - 513.95, - 490.4, - 566.82 - ], - "text": "1.7-2\nFor the systems described by the following\nequations, with the input x(t) and output y(t),\nexplain with reasons which of the systems are\ntime-invariant parameter systems and which are\ntime-varying-parameter systems.", - "type": "text" - }, - { - "block_id": "p164-b52", - "global_id": 4570, - "bbox": [ - 317.86, - 568.45, - 384.92, - 588.75 - ], - "text": "(a) y(t) = x(t −2)\n(b) y(t) = x(−t)", - "type": "text" - }, - { - "block_id": "p164-b53", - "global_id": 4571, - "bbox": [ - 317.87, - 590.36, - 388.58, - 610.66 - ], - "text": "(c) y(t) = x(at)\n(d) y(t) = tx(t −2)", - "type": "text" - }, - { - "block_id": "p164-b54", - "global_id": 4572, - "bbox": [ - 318.37, - 618.27, - 355.47, - 627.61 - ], - "text": "(e) y(t) =", - "type": "text" - }, - { - "block_id": "p164-b55", - "global_id": 4573, - "bbox": [ - 357.31, - 606.07, - 369.38, - 617.3 - ], - "text": "# 5", - "type": "text" - }, - { - "block_id": "p164-b56", - "global_id": 4574, - "bbox": [ - 362.04, - 628.3, - 370.33, - 635.04 - ], - "text": "−5", - "type": "text" - }, - { - "block_id": "p164-b57", - "global_id": 4575, - "bbox": [ - 371.83, - 618.27, - 397.09, - 627.52 - ], - "text": "x(τ)dτ", - "type": "text" - }, - { - "block_id": "p164-b58", - "global_id": 4576, - "bbox": [ - 319.37, - 642.76, - 355.47, - 652.1 - ], - "text": "(f) y(t) =", - "type": "text" - }, - { - "block_id": "p164-b59", - "global_id": 4577, - "bbox": [ - 357.31, - 630.17, - 382.39, - 645.72 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p164-b60", - "global_id": 4578, - "bbox": [ - 369.9, - 649.4, - 376.87, - 658.37 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p164-b61", - "global_id": 4579, - "bbox": [ - 383.58, - 630.17, - 392.87, - 640.99 - ], - "text": "2", - "type": "text" - } - ] - }, - { - "page_num": 165, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p165-b0", - "global_id": 4580, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n145", - "type": "text" - }, - { - "block_id": "p165-b1", - "global_id": 4581, - "bbox": [ - 87.82, - 85.64, - 289.0, - 160.73 - ], - "text": "1.7-3\nTwo inputs, temperature T(t) and wind speed\nV(t), produce an output, wind chill W(t),\naccording to W(t) = 35.74 + 0.6215T(t) −\n35.75{V(t)}0.16 + 0.4275T(t){V(t)}0.16.\nThe\nindependent variable here is time, t. Answer\nthe following questions yes or no, and provide\nmathematical justification for each answer.", - "type": "text" - }, - { - "block_id": "p165-b2", - "global_id": 4582, - "bbox": [ - 116.46, - 162.72, - 232.27, - 182.65 - ], - "text": "(a) Is this system BIBO-stable?\n(b) Is the system memoryless?", - "type": "text" - }, - { - "block_id": "p165-b3", - "global_id": 4583, - "bbox": [ - 116.46, - 184.65, - 288.97, - 204.57 - ], - "text": "(c) Is the system causal?\n(d) For simplicity, let the wind speed be con-", - "type": "text" - }, - { - "block_id": "p165-b4", - "global_id": 4584, - "bbox": [ - 116.96, - 206.19, - 289.0, - 248.4 - ], - "text": "stant, V(t) = kV. Thus, W(t) = k1 + k2T(t)\nfor some constants k1 and k2. Is this simpli-\nfied system linear?\n(e) For simplicity, let the temperature be con-", - "type": "text" - }, - { - "block_id": "p165-b5", - "global_id": 4585, - "bbox": [ - 131.89, - 250.03, - 288.98, - 281.28 - ], - "text": "stant, T(t) = kT. Thus, W(t) = k3 +\nk4 {V(t)}0.16 for some constants k3 and k4.\nIs this simplified system linear?", - "type": "text" - }, - { - "block_id": "p165-b6", - "global_id": 4586, - "bbox": [ - 87.82, - 286.39, - 288.98, - 317.65 - ], - "text": "1.7-4\nInput voltage x(t) applied to an inverting op-amp\nfollower circuit produces output y(t) according\nto", - "type": "text" - }, - { - "block_id": "p165-b7", - "global_id": 4587, - "bbox": [ - 132.98, - 343.04, - 171.15, - 353.12 - ], - "text": "y(t + tp) =", - "type": "text" - }, - { - "block_id": "p165-b8", - "global_id": 4588, - "bbox": [ - 172.99, - 324.66, - 180.09, - 341.69 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p165-b9", - "global_id": 4589, - "bbox": [ - 172.99, - 348.87, - 180.09, - 357.84 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p165-b10", - "global_id": 4590, - "bbox": [ - 185.08, - 332.03, - 265.79, - 363.29 - ], - "text": "−Vref\nx(t) > Vref\nVref\nx(t) < −Vref\n−x(t)\notherwise", - "type": "text" - }, - { - "block_id": "p165-b11", - "global_id": 4591, - "bbox": [ - 116.46, - 379.59, - 288.98, - 432.48 - ], - "text": "where op-amp reference voltage Vref and prop-\nagation delay tp are both positive constants.\nAnswer the following questions yes or no,\nand provide mathematical justification for each\nanswer.", - "type": "text" - }, - { - "block_id": "p165-b12", - "global_id": 4592, - "bbox": [ - 116.46, - 434.48, - 232.27, - 454.41 - ], - "text": "(a) Is this system BIBO-stable?\n(b) Is the system causal?", - "type": "text" - }, - { - "block_id": "p165-b13", - "global_id": 4593, - "bbox": [ - 116.46, - 456.39, - 218.81, - 476.32 - ], - "text": "(c) Is the system invertible?\n(d) Is the system linear?", - "type": "text" - }, - { - "block_id": "p165-b14", - "global_id": 4594, - "bbox": [ - 116.96, - 478.32, - 228.28, - 487.28 - ], - "text": "(e) Is the system memoryless?", - "type": "text" - }, - { - "block_id": "p165-b15", - "global_id": 4595, - "bbox": [ - 117.95, - 489.27, - 234.41, - 498.24 - ], - "text": "(f) Is the system time invariant?", - "type": "text" - }, - { - "block_id": "p165-b16", - "global_id": 4596, - "bbox": [ - 87.82, - 503.35, - 288.98, - 523.64 - ], - "text": "1.7-5\nRepeat Prob. 1.7-4 for a system with input x(t)\nthat produces output y(t) according to", - "type": "text" - }, - { - "block_id": "p165-b17", - "global_id": 4597, - "bbox": [ - 132.87, - 545.41, - 169.3, - 554.75 - ], - "text": "y(t + 1) =", - "type": "text" - }, - { - "block_id": "p165-b18", - "global_id": 4598, - "bbox": [ - 171.15, - 532.83, - 266.4, - 560.18 - ], - "text": "−2x(t)\nwhen x(t) ≥0\n0\notherwise", - "type": "text" - }, - { - "block_id": "p165-b19", - "global_id": 4599, - "bbox": [ - 87.82, - 576.32, - 288.99, - 596.62 - ], - "text": "1.7-6\nRepeat Prob. 1.7-4 for a system with input x(t)\nthat produces output y(t) according to", - "type": "text" - }, - { - "block_id": "p165-b20", - "global_id": 4600, - "bbox": [ - 123.22, - 618.52, - 159.65, - 627.86 - ], - "text": "y(t −1) =", - "type": "text" - }, - { - "block_id": "p165-b22", - "global_id": 4601, - "bbox": [ - 172.01, - 611.69, - 245.53, - 622.46 - ], - "text": "x(t −1)\nwhen d", - "type": "text" - }, - { - "block_id": "p165-b23", - "global_id": 4602, - "bbox": [ - 172.01, - 613.11, - 276.05, - 633.41 - ], - "text": "dtx(t) ≥0\nx(t −2)\notherwise", - "type": "text" - }, - { - "block_id": "p165-b24", - "global_id": 4603, - "bbox": [ - 314.98, - 85.94, - 516.13, - 116.89 - ], - "text": "1.7-7\nRepeat Prob. 1.7-4 for a system that multiplies a\ngiven input by a ramp function, r(t) = tu(t). That\nis, y(t) = x(t)r(t).", - "type": "text" - }, - { - "block_id": "p165-b25", - "global_id": 4604, - "bbox": [ - 314.97, - 121.78, - 516.14, - 142.08 - ], - "text": "1.7-8\nRepeat Prob. 1.7-4 for a system with input x(t)\nthat produces output y(t) according to", - "type": "text" - }, - { - "block_id": "p165-b26", - "global_id": 4605, - "bbox": [ - 399.3, - 154.6, - 430.18, - 169.86 - ], - "text": "y(t) = d", - "type": "text" - }, - { - "block_id": "p165-b27", - "global_id": 4606, - "bbox": [ - 424.49, - 160.61, - 460.45, - 176.22 - ], - "text": "dt x(t −1)", - "type": "text" - }, - { - "block_id": "p165-b28", - "global_id": 4607, - "bbox": [ - 314.97, - 186.97, - 516.14, - 207.27 - ], - "text": "1.7-9\nRepeat Prob. 1.7-4 for a system with input x(t)\nthat produces output y(t) according to", - "type": "text" - }, - { - "block_id": "p165-b29", - "global_id": 4608, - "bbox": [ - 374.89, - 226.75, - 397.05, - 236.0 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p165-b30", - "global_id": 4609, - "bbox": [ - 398.89, - 214.16, - 478.68, - 241.52 - ], - "text": "x(t)\nif\nx(t) > 0\n0\nif\nx(t) ≤0", - "type": "text" - }, - { - "block_id": "p165-b31", - "global_id": 4610, - "bbox": [ - 310.48, - 255.66, - 479.0, - 264.7 - ], - "text": "1.7-10\nA continuous-time system is given by", - "type": "text" - }, - { - "block_id": "p165-b32", - "global_id": 4611, - "bbox": [ - 353.35, - 283.7, - 388.56, - 293.03 - ], - "text": "y(t) = 0.5", - "type": "text" - }, - { - "block_id": "p165-b33", - "global_id": 4612, - "bbox": [ - 389.56, - 271.49, - 405.01, - 282.46 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p165-b34", - "global_id": 4613, - "bbox": [ - 394.29, - 293.73, - 405.95, - 300.21 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p165-b35", - "global_id": 4614, - "bbox": [ - 407.45, - 283.69, - 505.33, - 292.94 - ], - "text": "x(τ)[δ(t −τ) −δ(t + τ)]dτ", - "type": "text" - }, - { - "block_id": "p165-b36", - "global_id": 4615, - "bbox": [ - 343.61, - 312.11, - 516.13, - 332.41 - ], - "text": "Recall that δ(t) designates the Dirac delta func-\ntion.", - "type": "text" - }, - { - "block_id": "p165-b37", - "global_id": 4616, - "bbox": [ - 343.61, - 334.4, - 516.11, - 354.32 - ], - "text": "(a) Explain what this system does.\n(b) Is the system BIBO-stable? Justify your", - "type": "text" - }, - { - "block_id": "p165-b38", - "global_id": 4617, - "bbox": [ - 343.61, - 356.32, - 516.12, - 387.2 - ], - "text": "answer.\n(c) Is the system linear? Justify your answer.\n(d) Is the system memoryless? Justify your", - "type": "text" - }, - { - "block_id": "p165-b39", - "global_id": 4618, - "bbox": [ - 344.12, - 389.2, - 508.69, - 409.13 - ], - "text": "answer.\n(e) Is the system causal? Justify your answer.", - "type": "text" - }, - { - "block_id": "p165-b40", - "global_id": 4619, - "bbox": [ - 345.11, - 411.11, - 516.13, - 420.08 - ], - "text": "(f) Is the system time invariant? Justify your", - "type": "text" - }, - { - "block_id": "p165-b41", - "global_id": 4620, - "bbox": [ - 359.05, - 422.08, - 386.18, - 431.04 - ], - "text": "answer.", - "type": "text" - }, - { - "block_id": "p165-b42", - "global_id": 4621, - "bbox": [ - 310.48, - 435.92, - 516.14, - 478.14 - ], - "text": "1.7-11\nFor a certain LTI system with the input x(t),\nthe output y(t) and the two initial conditions\nq1(0) and q2(0), the following observations were\nmade:", - "type": "text" - }, - { - "block_id": "p165-b43", - "global_id": 4622, - "bbox": [ - 343.61, - 486.24, - 497.52, - 528.46 - ], - "text": "x(t)\nq1(0)\nq2(0)\ny(t)\n0\n1\n−1\ne−tu(t)\n0\n2\n1\ne−t(3t + 2)u(t)\nu(t)\n−1\n−1\n2u(t)", - "type": "text" - }, - { - "block_id": "p165-b44", - "global_id": 4623, - "bbox": [ - 343.61, - 538.25, - 516.14, - 635.26 - ], - "text": "Determine y(t) when both the initial conditions\nare zero and the input x(t) is as shown in\nFig. P1.7-11. [Hint: There are three causes: the\ninput and each of the two initial conditions.\nBecause of the linearity property, if a cause is\nincreased by a factor k, the response to that cause\nalso increases by the same factor k. Moreover, if\ncauses are added, the corresponding responses\nadd.]", - "type": "text" - } - ] - }, - { - "page_num": 166, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p166-b0", - "global_id": 4624, - "bbox": [ - 60.0, - 60.36, - 266.81, - 69.45 - ], - "text": "146\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p166-b1", - "global_id": 4625, - "bbox": [ - 152.67, - 94.57, - 156.67, - 102.57 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p166-b2", - "global_id": 4626, - "bbox": [ - 95.55, - 126.45, - 222.48, - 134.74 - ], - "text": "5\n5", - "type": "text" - }, - { - "block_id": "p166-b3", - "global_id": 4627, - "bbox": [ - 95.38, - 87.24, - 106.49, - 95.32 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p166-b4", - "global_id": 4628, - "bbox": [ - 243.64, - 126.61, - 245.86, - 134.61 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p166-b5", - "global_id": 4629, - "bbox": [ - 90.72, - 142.66, - 146.83, - 151.62 - ], - "text": "Figure P1.7-11", - "type": "text" - }, - { - "block_id": "p166-b6", - "global_id": 4630, - "bbox": [ - 57.59, - 166.52, - 263.23, - 186.51 - ], - "text": "1.7-12\nA system is specified by its input–output rela-\ntionship as", - "type": "text" - }, - { - "block_id": "p166-b7", - "global_id": 4631, - "bbox": [ - 149.19, - 188.19, - 203.41, - 211.09 - ], - "text": "y(t) =\nx2(t)\ndx(t)/dt", - "type": "text" - }, - { - "block_id": "p166-b8", - "global_id": 4632, - "bbox": [ - 90.72, - 219.04, - 263.22, - 238.96 - ], - "text": "Show that the system satisfies the homogeneity\nproperty but not the additivity property.", - "type": "text" - }, - { - "block_id": "p166-b9", - "global_id": 4633, - "bbox": [ - 57.59, - 243.96, - 263.24, - 285.88 - ], - "text": "1.7-13\nShow that the circuit in Fig. P1.7-13 is zero-state\nlinear but not zero-input linear. Assume all\ndiodes to have identical (matched) characteris-\ntics. The output is the current y(t).", - "type": "text" - }, - { - "block_id": "p166-b10", - "global_id": 4634, - "bbox": [ - 76.82, - 366.66, - 180.31, - 383.31 - ], - "text": "vc\nx(t)", - "type": "text" - }, - { - "block_id": "p166-b11", - "global_id": 4635, - "bbox": [ - 135.16, - 330.69, - 272.48, - 340.3 - ], - "text": "D1\nD2\nD3", - "type": "text" - }, - { - "block_id": "p166-b12", - "global_id": 4636, - "bbox": [ - 213.18, - 392.16, - 222.07, - 400.24 - ], - "text": "2R", - "type": "text" - }, - { - "block_id": "p166-b16", - "global_id": 4637, - "bbox": [ - 113.79, - 306.27, - 124.89, - 314.35 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p166-b17", - "global_id": 4638, - "bbox": [ - 135.76, - 392.16, - 144.65, - 400.24 - ], - "text": "2R", - "type": "text" - }, - { - "block_id": "p166-b18", - "global_id": 4639, - "bbox": [ - 173.28, - 350.67, - 178.61, - 358.67 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p166-b19", - "global_id": 4640, - "bbox": [ - 262.06, - 392.16, - 266.94, - 400.16 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p166-b20", - "global_id": 4641, - "bbox": [ - 76.61, - 423.84, - 132.73, - 432.81 - ], - "text": "Figure P1.7-13", - "type": "text" - }, - { - "block_id": "p166-b21", - "global_id": 4642, - "bbox": [ - 57.59, - 446.99, - 263.24, - 510.84 - ], - "text": "1.7-14\nThe\ninductor\nL\nand\nthe\ncapacitor\nC\nin\nFig. P1.7-14 are nonlinear, which makes the\ncircuit nonlinear. The remaining three elements\nare linear. Show that the output y(t) of this\nnonlinear circuit satisfies the linearity conditions\nwith respect to the input x(t) and the initial", - "type": "text" - }, - { - "block_id": "p166-b22", - "global_id": 4643, - "bbox": [ - 317.87, - 85.9, - 490.38, - 105.83 - ], - "text": "conditions (all the initial inductor currents and\ncapacitor voltages).", - "type": "text" - }, - { - "block_id": "p166-b23", - "global_id": 4644, - "bbox": [ - 284.75, - 123.5, - 490.39, - 165.42 - ], - "text": "1.7-15\nFor the systems described by the following\nequations, with the input x(t) and output y(t),\ndetermine which are causal and which are\nnoncausal.", - "type": "text" - }, - { - "block_id": "p166-b24", - "global_id": 4645, - "bbox": [ - 317.87, - 167.04, - 384.92, - 187.34 - ], - "text": "(a) y(t) = x(t −2)\n(b) y(t) = x(−t)", - "type": "text" - }, - { - "block_id": "p166-b25", - "global_id": 4646, - "bbox": [ - 317.87, - 188.96, - 412.72, - 209.26 - ], - "text": "(c) y(t) = x(at)\na > 1\n(d) y(t) = x(at)\na < 1", - "type": "text" - }, - { - "block_id": "p166-b26", - "global_id": 4647, - "bbox": [ - 284.75, - 226.94, - 490.4, - 290.77 - ], - "text": "1.7-16\nFor the systems described by the following\nequations, with the input x(t) and output y(t),\ndetermine which are invertible and which are\nnoninvertible. For the invertible systems, find\nthe input–output relationship of the inverse\nsystem.", - "type": "text" - }, - { - "block_id": "p166-b27", - "global_id": 4648, - "bbox": [ - 318.37, - 297.53, - 355.46, - 306.87 - ], - "text": "(a) y(t) =", - "type": "text" - }, - { - "block_id": "p166-b28", - "global_id": 4649, - "bbox": [ - 357.31, - 285.32, - 367.95, - 296.5 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p166-b29", - "global_id": 4650, - "bbox": [ - 362.04, - 307.56, - 373.7, - 314.03 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p166-b30", - "global_id": 4651, - "bbox": [ - 375.2, - 297.53, - 400.47, - 306.78 - ], - "text": "x(τ)dτ", - "type": "text" - }, - { - "block_id": "p166-b31", - "global_id": 4652, - "bbox": [ - 317.86, - 315.2, - 444.87, - 325.59 - ], - "text": "(b) y(t) = xn(t), x(t) real, n integer", - "type": "text" - }, - { - "block_id": "p166-b32", - "global_id": 4653, - "bbox": [ - 318.37, - 328.34, - 376.33, - 343.97 - ], - "text": "(c) y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p166-b33", - "global_id": 4654, - "bbox": [ - 317.86, - 341.27, - 389.41, - 360.58 - ], - "text": "dt\n(d) y(t) = x(3t −6)", - "type": "text" - }, - { - "block_id": "p166-b34", - "global_id": 4655, - "bbox": [ - 318.37, - 365.19, - 389.58, - 374.53 - ], - "text": "(e) y(t) = cos[x(t)]", - "type": "text" - }, - { - "block_id": "p166-b35", - "global_id": 4656, - "bbox": [ - 319.37, - 374.89, - 412.26, - 385.49 - ], - "text": "(f) y(t) = ex(t),\nx(t) real", - "type": "text" - }, - { - "block_id": "p166-b36", - "global_id": 4657, - "bbox": [ - 284.74, - 402.86, - 490.38, - 434.87 - ], - "text": "1.7-17\nFigure P1.7-17 displays an input x1(t) to a linear\ntime-invariant (LTI) system H, the correspond-\ning output y1(t), and a second input x2(t).", - "type": "text" - }, - { - "block_id": "p166-b37", - "global_id": 4658, - "bbox": [ - 318.37, - 435.74, - 490.39, - 445.82 - ], - "text": "(a) Bill suggests that x2(t) = 2x1(3t)−x1(t−1).", - "type": "text" - }, - { - "block_id": "p166-b38", - "global_id": 4659, - "bbox": [ - 317.86, - 447.08, - 490.39, - 478.7 - ], - "text": "Is Bill correct? If yes, prove it. If not, correct\nhis error.\n(b) Bill wants to know the output y2(t) in", - "type": "text" - }, - { - "block_id": "p166-b39", - "global_id": 4660, - "bbox": [ - 333.31, - 479.58, - 490.39, - 511.57 - ], - "text": "response to the input x2(t). Provide him with\nan expression for y2(t) in terms of y1(t). Use\nMATLAB to plot y2(t).", - "type": "text" - }, - { - "block_id": "p166-b40", - "global_id": 4661, - "bbox": [ - 260.22, - 588.95, - 390.75, - 599.14 - ], - "text": "y(t)\n2 \t\n1 F", - "type": "text" - }, - { - "block_id": "p166-b43", - "global_id": 4662, - "bbox": [ - 305.85, - 544.57, - 323.63, - 552.57 - ], - "text": "0.1 H", - "type": "text" - }, - { - "block_id": "p166-b44", - "global_id": 4663, - "bbox": [ - 219.85, - 580.46, - 224.3, - 588.46 - ], - "text": "L", - "type": "text" - }, - { - "block_id": "p166-b45", - "global_id": 4664, - "bbox": [ - 219.85, - 530.56, - 225.19, - 538.56 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p166-b46", - "global_id": 4665, - "bbox": [ - 145.9, - 588.53, - 157.0, - 596.61 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p166-b47", - "global_id": 4666, - "bbox": [ - 405.63, - 626.61, - 461.74, - 635.58 - ], - "text": "Figure P1.7-14", - "type": "text" - } - ] - }, - { - "page_num": 167, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p167-b0", - "global_id": 4667, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n147", - "type": "text" - }, - { - "block_id": "p167-b1", - "global_id": 4668, - "bbox": [ - 134.64, - 89.88, - 138.64, - 97.88 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p167-b2", - "global_id": 4669, - "bbox": [ - 134.62, - 135.07, - 206.6, - 145.74 - ], - "text": "0\n2\nt\n1\n3\n4", - "type": "text" - }, - { - "block_id": "p167-b3", - "global_id": 4670, - "bbox": [ - 135.25, - 109.43, - 139.25, - 117.43 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p167-b4", - "global_id": 4671, - "bbox": [ - 250.24, - 89.88, - 254.24, - 97.88 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p167-b5", - "global_id": 4672, - "bbox": [ - 250.24, - 135.07, - 322.32, - 145.74 - ], - "text": "0\n2\nt\n1\n3\n4", - "type": "text" - }, - { - "block_id": "p167-b6", - "global_id": 4673, - "bbox": [ - 250.88, - 109.41, - 254.88, - 117.41 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p167-b7", - "global_id": 4674, - "bbox": [ - 187.11, - 87.69, - 314.69, - 97.3 - ], - "text": "x1(t)\ny1(t)", - "type": "text" - }, - { - "block_id": "p167-b8", - "global_id": 4675, - "bbox": [ - 214.07, - 101.74, - 219.85, - 109.74 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p167-b9", - "global_id": 4676, - "bbox": [ - 374.41, - 89.9, - 378.41, - 97.9 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p167-b10", - "global_id": 4677, - "bbox": [ - 374.36, - 135.07, - 448.28, - 145.74 - ], - "text": "0\n2\nt\n1\n3\n4", - "type": "text" - }, - { - "block_id": "p167-b11", - "global_id": 4678, - "bbox": [ - 375.02, - 109.64, - 379.02, - 117.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p167-b12", - "global_id": 4679, - "bbox": [ - 432.4, - 87.67, - 446.51, - 97.28 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p167-b13", - "global_id": 4680, - "bbox": [ - 130.58, - 151.91, - 186.69, - 160.88 - ], - "text": "Figure P1.7-17", - "type": "text" - }, - { - "block_id": "p167-b14", - "global_id": 4681, - "bbox": [ - 83.34, - 172.36, - 288.99, - 214.29 - ], - "text": "1.7-18\nA linear time-invariant system H acts on input\nx(t) = u(t −0.5) −u(t −1.5) to produce out-\nput y(t) = H {x(t)} = 0.5u(t) + 0.5u(t −1)\n−u(t −2).", - "type": "text" - }, - { - "block_id": "p167-b15", - "global_id": 4682, - "bbox": [ - 116.97, - 216.29, - 288.98, - 225.26 - ], - "text": "(a) Is it possible that the system is causal?", - "type": "text" - }, - { - "block_id": "p167-b16", - "global_id": 4683, - "bbox": [ - 116.46, - 227.24, - 289.0, - 269.09 - ], - "text": "Explain your answer. If not causal, deter-\nmine the shift necessary to make the system\ncausal.\n(b) Is it possible that the system is memoryless?", - "type": "text" - }, - { - "block_id": "p167-b17", - "global_id": 4684, - "bbox": [ - 116.97, - 271.08, - 288.98, - 291.01 - ], - "text": "Explain your answer.\n(c) Suppose the output y(t) is applied to an", - "type": "text" - }, - { - "block_id": "p167-b18", - "global_id": 4685, - "bbox": [ - 131.9, - 292.91, - 264.54, - 301.97 - ], - "text": "identical system H to produce output", - "type": "text" - }, - { - "block_id": "p167-b19", - "global_id": 4686, - "bbox": [ - 158.25, - 316.48, - 262.64, - 325.82 - ], - "text": "z(t) = H {y(t)} = H {H {x(t)}}", - "type": "text" - }, - { - "block_id": "p167-b20", - "global_id": 4687, - "bbox": [ - 131.9, - 340.52, - 288.99, - 371.78 - ], - "text": "If possible, determine and sketch z(t). If\nnot possible, explain why z(t) cannot be\ndetermined using the information given.", - "type": "text" - }, - { - "block_id": "p167-b21", - "global_id": 4688, - "bbox": [ - 87.82, - 376.99, - 288.99, - 408.69 - ], - "text": "1.8-1\nFor the circuit depicted in Fig. P1.8-1, find the\ndifferential equations relating outputs y1(t) and\ny2(t) to the input x(t).", - "type": "text" - }, - { - "block_id": "p167-b22", - "global_id": 4689, - "bbox": [ - 120.66, - 449.81, - 249.88, - 459.86 - ], - "text": "y2(t)\nx(t)\ny1(t)", - "type": "text" - }, - { - "block_id": "p167-b23", - "global_id": 4690, - "bbox": [ - 171.42, - 412.04, - 183.65, - 420.33 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p167-b24", - "global_id": 4691, - "bbox": [ - 216.78, - 450.33, - 228.55, - 458.33 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p167-b27", - "global_id": 4692, - "bbox": [ - 116.46, - 492.82, - 168.11, - 501.79 - ], - "text": "Figure P1.8-1", - "type": "text" - }, - { - "block_id": "p167-b28", - "global_id": 4693, - "bbox": [ - 87.82, - 522.05, - 288.99, - 553.75 - ], - "text": "1.8-2\nFor the circuit depicted in Fig. P1.8-2, find the\ndifferential equations relating outputs y1(t) and\ny2(t) to the input x(t).", - "type": "text" - }, - { - "block_id": "p167-b29", - "global_id": 4694, - "bbox": [ - 245.25, - 594.81, - 251.03, - 602.81 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p167-b30", - "global_id": 4695, - "bbox": [ - 212.24, - 575.37, - 226.35, - 584.98 - ], - "text": "y1(t)", - "type": "text" - }, - { - "block_id": "p167-b31", - "global_id": 4696, - "bbox": [ - 89.82, - 592.92, - 297.74, - 604.82 - ], - "text": "x(t)\n1 F\ny2(t)", - "type": "text" - }, - { - "block_id": "p167-b34", - "global_id": 4697, - "bbox": [ - 138.4, - 592.03, - 242.38, - 605.53 - ], - "text": "2\n1\n2\n1", - "type": "text" - }, - { - "block_id": "p167-b35", - "global_id": 4698, - "bbox": [ - 86.57, - 636.83, - 138.22, - 645.8 - ], - "text": "Figure P1.8-2", - "type": "text" - }, - { - "block_id": "p167-b36", - "global_id": 4699, - "bbox": [ - 314.97, - 172.37, - 516.14, - 247.17 - ], - "text": "1.8-3\nA simplified (one-dimensional) model of an\nautomobile suspension system is shown in\nFig. P1.8-3. In this case, the input is not a force\nbut a displacement x(t) (the road contour). Find\nthe differential equation relating the output y(t)\n(auto body displacement) to the input x(t) (the\nroad contour).", - "type": "text" - }, - { - "block_id": "p167-b37", - "global_id": 4700, - "bbox": [ - 314.97, - 261.05, - 516.14, - 346.81 - ], - "text": "1.8-4\nA\nfield-controlled\ndc\nmotor\nis\nshown\nin\nFig. P1.8-4. Its armature current ia is maintained\nconstant. The torque generated by this motor\nis proportional to the field current if (torque=\nKf if ). Find the differential equation relating the\noutput position θ to the input voltage x(t). The\nmotor and load together have a moment of\ninertia J.", - "type": "text" - }, - { - "block_id": "p167-b38", - "global_id": 4701, - "bbox": [ - 314.97, - 360.68, - 516.14, - 479.33 - ], - "text": "1.8-5\nWater flows into a tank at a rate of qi units/s\nand flows out through the outflow valve at a\nrate of q0 units/s (Fig. P1.8-5). Determine the\nequation relating the outflow q0 to the input qi.\nThe outflow rate is proportional to the head h.\nThus q0 = Rh, where R is the valve resistance.\nDetermine also the differential equation relating\nthe head h to the input qi. [Hint: The net inflow\nof water in time Δt is (qi −q0)Δt. This inflow\nis also AΔh, where A is the cross section of the\ntank.]", - "type": "text" - }, - { - "block_id": "p167-b39", - "global_id": 4702, - "bbox": [ - 314.97, - 493.21, - 516.14, - 524.9 - ], - "text": "1.8-6\nConsider the circuit shown in Fig. P1.8-6, with\ninput voltage x(t) and output currents y1(t),\ny2(t), and y3(t).", - "type": "text" - }, - { - "block_id": "p167-b40", - "global_id": 4703, - "bbox": [ - 344.11, - 526.16, - 516.13, - 535.12 - ], - "text": "(a) What is the order of this system? Explain", - "type": "text" - }, - { - "block_id": "p167-b41", - "global_id": 4704, - "bbox": [ - 343.61, - 537.11, - 516.11, - 557.04 - ], - "text": "your answer.\n(b) Determine the matrix representation for this", - "type": "text" - }, - { - "block_id": "p167-b42", - "global_id": 4705, - "bbox": [ - 344.11, - 559.04, - 516.14, - 578.96 - ], - "text": "system.\n(c) Use Cramer’s rule to determine the output", - "type": "text" - }, - { - "block_id": "p167-b43", - "global_id": 4706, - "bbox": [ - 359.05, - 580.58, - 516.13, - 600.88 - ], - "text": "current y3(t) for the input voltage x(t) =\n[2 −|cos(t)|]u(t −1).", - "type": "text" - }, - { - "block_id": "p167-b44", - "global_id": 4707, - "bbox": [ - 310.48, - 614.75, - 516.13, - 647.06 - ], - "text": "1.10-1\nWrite state equations for the parallel RLC circuit\nin Fig. P1.8-2. Use the capacitor voltage q1 and\nthe inductor current q2 as your state variables.", - "type": "text" - } - ] - }, - { - "page_num": 168, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p168-b0", - "global_id": 4708, - "bbox": [ - 60.0, - 60.36, - 266.81, - 69.45 - ], - "text": "148\nCHAPTER 1\nSIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p168-b1", - "global_id": 4709, - "bbox": [ - 236.45, - 102.25, - 243.11, - 110.25 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p168-b2", - "global_id": 4710, - "bbox": [ - 128.65, - 196.76, - 189.74, - 204.84 - ], - "text": "x (Road elevation)", - "type": "text" - }, - { - "block_id": "p168-b3", - "global_id": 4711, - "bbox": [ - 290.65, - 108.66, - 366.86, - 116.74 - ], - "text": "y (Auto body elevation)", - "type": "text" - }, - { - "block_id": "p168-b4", - "global_id": 4712, - "bbox": [ - 205.95, - 140.76, - 275.89, - 148.76 - ], - "text": "K\nB", - "type": "text" - }, - { - "block_id": "p168-b5", - "global_id": 4713, - "bbox": [ - 104.83, - 246.91, - 156.48, - 255.88 - ], - "text": "Figure P1.8-3", - "type": "text" - }, - { - "block_id": "p168-b6", - "global_id": 4714, - "bbox": [ - 145.9, - 318.15, - 404.1, - 329.99 - ], - "text": "x(t)\nB", - "type": "text" - }, - { - "block_id": "p168-b7", - "global_id": 4715, - "bbox": [ - 303.96, - 294.16, - 307.96, - 302.16 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p168-b8", - "global_id": 4716, - "bbox": [ - 208.81, - 313.78, - 355.06, - 326.23 - ], - "text": "J\nif", - "type": "text" - }, - { - "block_id": "p168-b9", - "global_id": 4717, - "bbox": [ - 201.81, - 271.18, - 323.86, - 287.22 - ], - "text": "ia constant\nRf", - "type": "text" - }, - { - "block_id": "p168-b10", - "global_id": 4718, - "bbox": [ - 250.45, - 314.89, - 257.05, - 324.44 - ], - "text": "Lf", - "type": "text" - }, - { - "block_id": "p168-b11", - "global_id": 4719, - "bbox": [ - 418.63, - 344.59, - 470.27, - 353.56 - ], - "text": "Figure P1.8-4", - "type": "text" - }, - { - "block_id": "p168-b12", - "global_id": 4720, - "bbox": [ - 270.0, - 408.88, - 274.0, - 416.88 - ], - "text": "h", - "type": "text" - }, - { - "block_id": "p168-b13", - "global_id": 4721, - "bbox": [ - 208.45, - 437.6, - 417.04, - 448.93 - ], - "text": "qi\nqo\nFigure P1.8-5", - "type": "text" - }, - { - "block_id": "p168-b14", - "global_id": 4722, - "bbox": [ - 173.1, - 500.86, - 177.62, - 508.86 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p168-b15", - "global_id": 4723, - "bbox": [ - 145.9, - 507.75, - 177.36, - 524.56 - ], - "text": "–\nx(t)", - "type": "text" - }, - { - "block_id": "p168-b16", - "global_id": 4724, - "bbox": [ - 214.54, - 526.09, - 295.37, - 535.69 - ], - "text": "y2(t)\ny1(t)", - "type": "text" - }, - { - "block_id": "p168-b17", - "global_id": 4725, - "bbox": [ - 203.63, - 464.95, - 369.59, - 474.56 - ], - "text": "R1 = 1\nR5 = 5\nR3 = 3", - "type": "text" - }, - { - "block_id": "p168-b18", - "global_id": 4726, - "bbox": [ - 352.23, - 526.09, - 366.34, - 535.69 - ], - "text": "y3(t)", - "type": "text" - }, - { - "block_id": "p168-b19", - "global_id": 4727, - "bbox": [ - 223.21, - 499.24, - 386.89, - 508.85 - ], - "text": "R2 = 2\nR4 = 4\nR6 = 6", - "type": "text" - }, - { - "block_id": "p168-b20", - "global_id": 4728, - "bbox": [ - 415.63, - 544.64, - 467.28, - 553.61 - ], - "text": "Figure P1.8-6", - "type": "text" - }, - { - "block_id": "p168-b21", - "global_id": 4729, - "bbox": [ - 90.72, - 576.45, - 263.23, - 607.33 - ], - "text": "Show that every possible current or voltage in\nthe circuit can be expressed in terms of q1, q2\nand the input x(t).", - "type": "text" - }, - { - "block_id": "p168-b22", - "global_id": 4730, - "bbox": [ - 57.59, - 612.82, - 263.24, - 632.82 - ], - "text": "1.10-2\nWrite state equations for the third-order cir-\ncuit shown in Fig. P1.10-2, using the inductor", - "type": "text" - }, - { - "block_id": "p168-b23", - "global_id": 4731, - "bbox": [ - 317.86, - 576.36, - 490.39, - 629.26 - ], - "text": "currents q1, q2 and the capacitor voltage q3 as\nstate variables. Show that every possible voltage\nor current in this circuit can be expressed as a\nlinear combination of q1, q2, q3, and the input\nx(t). Also, at some instant t, it was found that", - "type": "text" - } - ] - }, - { - "page_num": 169, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p169-b0", - "global_id": 4732, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n149", - "type": "text" - }, - { - "block_id": "p169-b1", - "global_id": 4733, - "bbox": [ - 340.5, - 131.9, - 352.72, - 140.19 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p169-b4", - "global_id": 4734, - "bbox": [ - 226.03, - 94.09, - 237.81, - 102.09 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p169-b5", - "global_id": 4735, - "bbox": [ - 176.62, - 133.61, - 188.21, - 141.69 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p169-b6", - "global_id": 4736, - "bbox": [ - 206.1, - 94.95, - 280.3, - 104.56 - ], - "text": "q1\nq2", - "type": "text" - }, - { - "block_id": "p169-b7", - "global_id": 4737, - "bbox": [ - 248.93, - 131.23, - 255.93, - 140.84 - ], - "text": "q3", - "type": "text" - }, - { - "block_id": "p169-b8", - "global_id": 4738, - "bbox": [ - 271.27, - 139.35, - 281.59, - 152.85 - ], - "text": "F\n2\n1", - "type": "text" - }, - { - "block_id": "p169-b9", - "global_id": 4739, - "bbox": [ - 294.36, - 87.99, - 306.02, - 101.49 - ], - "text": "H\n3\n1", - "type": "text" - }, - { - "block_id": "p169-b10", - "global_id": 4740, - "bbox": [ - 365.36, - 157.71, - 421.48, - 166.68 - ], - "text": "Figure P1.10-2", - "type": "text" - }, - { - "block_id": "p169-b11", - "global_id": 4741, - "bbox": [ - 116.46, - 178.76, - 288.98, - 210.02 - ], - "text": "q1 = 5, q2 = 1, q3 = 2, and x = 10. Determine\nthe voltage across and the current through every\nelement in this circuit.", - "type": "text" - }, - { - "block_id": "p169-b12", - "global_id": 4742, - "bbox": [ - 83.34, - 222.24, - 288.98, - 264.15 - ], - "text": "1.11-1\nProvide MATLAB code and output that plots\nthe odd portion xo(t) of the function x(t) =\n2−t cos(2πt)u(t−π) over a suitable-length inter-\nval using a suitable number of points.", - "type": "text" - }, - { - "block_id": "p169-b13", - "global_id": 4743, - "bbox": [ - 83.34, - 276.36, - 288.98, - 318.28 - ], - "text": "1.11-2\nProvide MATLAB code and output that plots\nthe even portion xe(t) of the function x(t) =\n2−t/2 cos(4πt)u(t −0.5) over a suitable t using\nt = 0.002 second between points.", - "type": "text" - }, - { - "block_id": "p169-b14", - "global_id": 4744, - "bbox": [ - 83.34, - 328.93, - 288.98, - 350.5 - ], - "text": "1.11-3\nDefine\nx(t) = et(1+j2π)u(−t)\nand\ny(t) =\nRe", - "type": "text" - }, - { - "block_id": "p169-b15", - "global_id": 4745, - "bbox": [ - 127.42, - 333.95, - 161.1, - 350.5 - ], - "text": "*\n2x\n −5−t", - "type": "text" - }, - { - "block_id": "p169-b16", - "global_id": 4746, - "bbox": [ - 151.97, - 333.95, - 170.28, - 353.07 - ], - "text": "2\n+", - "type": "text" - }, - { - "block_id": "p169-b17", - "global_id": 4747, - "bbox": [ - 116.96, - 341.53, - 288.99, - 361.45 - ], - "text": ".\n(a) Use MATLAB to plot Re{x(t)} versus", - "type": "text" - }, - { - "block_id": "p169-b18", - "global_id": 4748, - "bbox": [ - 131.9, - 362.98, - 288.98, - 372.41 - ], - "text": "Im{x(at)} for a = 0.5, 1, and 2 and −10 ≤", - "type": "text" - }, - { - "block_id": "p169-b19", - "global_id": 4749, - "bbox": [ - 343.61, - 178.76, - 516.13, - 210.02 - ], - "text": "t ≤10. How important is the scale factor a\non the shape of the resulting figure?\n(b) Use MATLAB to plot y(t) over −10 ≤t ≤", - "type": "text" - }, - { - "block_id": "p169-b20", - "global_id": 4750, - "bbox": [ - 344.11, - 211.92, - 516.14, - 253.86 - ], - "text": "10. Analytically determine the time t0 where\ny(t) has a jump discontinuity. Verify your\ncalculation of t0 using the plot of y(t).\n(c) Use MATLAB and numerical integration to", - "type": "text" - }, - { - "block_id": "p169-b21", - "global_id": 4751, - "bbox": [ - 343.61, - 255.48, - 516.14, - 275.77 - ], - "text": "compute the energy Ex of signal x(t).\n(d) Use MATLAB and numerical integration to", - "type": "text" - }, - { - "block_id": "p169-b22", - "global_id": 4752, - "bbox": [ - 359.05, - 277.39, - 491.88, - 288.07 - ], - "text": "compute the energy Ey of signal y(t).", - "type": "text" - }, - { - "block_id": "p169-b23", - "global_id": 4753, - "bbox": [ - 310.48, - 290.5, - 461.28, - 301.26 - ], - "text": "1.11-4\nConsider the signal x(t) = u( t", - "type": "text" - }, - { - "block_id": "p169-b24", - "global_id": 4754, - "bbox": [ - 343.61, - 291.92, - 516.13, - 317.16 - ], - "text": "2 + 1) −u(t −\n1) −δ( t\n2). Define y(t) =\n$ t−3", - "type": "text" - }, - { - "block_id": "p169-b25", - "global_id": 4755, - "bbox": [ - 343.61, - 305.23, - 516.13, - 321.32 - ], - "text": "−∞x(τ)dτ and z(t) =\n$ ∞", - "type": "text" - }, - { - "block_id": "p169-b26", - "global_id": 4756, - "bbox": [ - 343.61, - 317.58, - 494.9, - 348.83 - ], - "text": "t\nx(τ)dτ.\n(a) Using MATLAB, accurately plot y(t).\n(b) Using MATLAB, accurately plot z(t).", - "type": "text" - }, - { - "block_id": "p169-b27", - "global_id": 4757, - "bbox": [ - 344.12, - 350.46, - 516.13, - 359.8 - ], - "text": "(c) Using MATLAB, accurately plot w(t) =", - "type": "text" - }, - { - "block_id": "p169-b28", - "global_id": 4758, - "bbox": [ - 360.24, - 359.99, - 365.28, - 373.27 - ], - "text": "d\ndt", - "type": "text" - }, - { - "block_id": "p169-b30", - "global_id": 4759, - "bbox": [ - 371.12, - 361.41, - 407.08, - 370.66 - ], - "text": "y(t) + z(t)", - "type": "text" - }, - { - "block_id": "p169-b32", - "global_id": 4760, - "bbox": [ - 410.59, - 361.78, - 412.83, - 370.75 - ], - "text": ".", - "type": "text" - } - ] - }, - { - "page_num": 170, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p170-b0", - "global_id": 4761, - "bbox": [ - 64.47, - 67.04, - 133.54, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p170-b1", - "global_id": 4762, - "bbox": [ - 147.02, - 125.73, - 484.58, - 173.26 - ], - "text": "TIME-DOMAIN ANALYSIS\nOF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p170-b2", - "global_id": 4763, - "bbox": [ - 114.3, - 79.92, - 133.73, - 118.77 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p170-b3", - "global_id": 4764, - "bbox": [ - 101.84, - 266.14, - 490.4, - 300.02 - ], - "text": "In this book we consider two methods of analysis of linear time-invariant (LTI) systems:\nthe time-domain method and the frequency-domain method. In this chapter we discuss the\ntime-domain analysis of linear, time-invariant, continuous-time (LTIC) systems.", - "type": "text" - }, - { - "block_id": "p170-b4", - "global_id": 4765, - "bbox": [ - 102.2, - 329.04, - 230.8, - 342.99 - ], - "text": "2.1 INTRODUCTION", - "type": "text" - }, - { - "block_id": "p170-b5", - "global_id": 4766, - "bbox": [ - 101.84, - 348.88, - 490.39, - 382.85 - ], - "text": "For the purpose of analysis, we shall consider linear differential systems. This is the class of LTIC\nsystems introduced in Ch. 1, for which the input x(t) and the output y(t) are related by linear\ndifferential equations of the form", - "type": "text" - }, - { - "block_id": "p170-b6", - "global_id": 4767, - "bbox": [ - 164.52, - 392.44, - 190.15, - 403.72 - ], - "text": "dNy(t)", - "type": "text" - }, - { - "block_id": "p170-b7", - "global_id": 4768, - "bbox": [ - 170.59, - 400.43, - 210.67, - 417.77 - ], - "text": "dtN\n+ a1", - "type": "text" - }, - { - "block_id": "p170-b8", - "global_id": 4769, - "bbox": [ - 212.37, - 392.22, - 246.92, - 403.72 - ], - "text": "dN−1y(t)", - "type": "text" - }, - { - "block_id": "p170-b9", - "global_id": 4770, - "bbox": [ - 218.44, - 400.43, - 301.89, - 417.77 - ], - "text": "dtN−1\n+ · · · + aN−1", - "type": "text" - }, - { - "block_id": "p170-b10", - "global_id": 4771, - "bbox": [ - 303.6, - 393.44, - 323.38, - 403.72 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p170-b11", - "global_id": 4772, - "bbox": [ - 309.53, - 400.43, - 360.81, - 417.77 - ], - "text": "dt\n+ aNy(t)", - "type": "text" - }, - { - "block_id": "p170-b12", - "global_id": 4773, - "bbox": [ - 175.34, - 426.95, - 206.45, - 438.03 - ], - "text": "= bN−M", - "type": "text" - }, - { - "block_id": "p170-b13", - "global_id": 4774, - "bbox": [ - 208.43, - 418.96, - 235.1, - 430.24 - ], - "text": "dMx(t)", - "type": "text" - }, - { - "block_id": "p170-b14", - "global_id": 4775, - "bbox": [ - 214.51, - 426.95, - 277.65, - 444.3 - ], - "text": "dtM\n+ bN−M+1", - "type": "text" - }, - { - "block_id": "p170-b15", - "global_id": 4776, - "bbox": [ - 279.37, - 418.74, - 314.96, - 430.24 - ], - "text": "dM−1x(t)", - "type": "text" - }, - { - "block_id": "p170-b16", - "global_id": 4777, - "bbox": [ - 285.44, - 426.95, - 369.93, - 444.3 - ], - "text": "dtM−1\n+ · · · + bN−1", - "type": "text" - }, - { - "block_id": "p170-b17", - "global_id": 4778, - "bbox": [ - 371.64, - 419.97, - 391.44, - 430.24 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p170-b18", - "global_id": 4779, - "bbox": [ - 377.59, - 426.95, - 490.39, - 444.3 - ], - "text": "dt\n+ bNx(t)\n(2.1)", - "type": "text" - }, - { - "block_id": "p170-b19", - "global_id": 4780, - "bbox": [ - 101.84, - 451.38, - 490.39, - 473.71 - ], - "text": "where all the coefficients ai and bi are constants. Using operator notation D to represent d/dt, we\ncan express this equation as", - "type": "text" - }, - { - "block_id": "p170-b20", - "global_id": 4781, - "bbox": [ - 181.85, - 480.62, - 341.27, - 495.89 - ], - "text": "(DN + a1DN−1 + · · · + aN−1D + aN)y(t)", - "type": "text" - }, - { - "block_id": "p170-b21", - "global_id": 4782, - "bbox": [ - 193.86, - 497.06, - 410.38, - 512.32 - ], - "text": "= (bN−MDM + bN−M+1DM−1 + · · · + bN−1D + bN)x(t)", - "type": "text" - }, - { - "block_id": "p170-b22", - "global_id": 4783, - "bbox": [ - 101.85, - 522.99, - 110.14, - 532.95 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p170-b23", - "global_id": 4784, - "bbox": [ - 254.1, - 534.54, - 490.39, - 544.91 - ], - "text": "Q(D)y(t) = P(D)x(t)\n(2.2)", - "type": "text" - }, - { - "block_id": "p170-b24", - "global_id": 4785, - "bbox": [ - 101.85, - 553.2, - 271.65, - 563.58 - ], - "text": "where the polynomials Q(D) and P(D) are", - "type": "text" - }, - { - "block_id": "p170-b25", - "global_id": 4786, - "bbox": [ - 187.61, - 570.5, - 357.54, - 585.76 - ], - "text": "Q(D) = DN + a1DN−1 + · · · + aN−1D + aN", - "type": "text" - }, - { - "block_id": "p170-b26", - "global_id": 4787, - "bbox": [ - 188.71, - 586.94, - 403.69, - 602.2 - ], - "text": "P(D) = bN−MDM + bN−M+1DM−1 + · · · + bN−1D + bN", - "type": "text" - }, - { - "block_id": "p170-b27", - "global_id": 4788, - "bbox": [ - 101.84, - 612.76, - 490.38, - 634.79 - ], - "text": "Theoretically the powers M and N in the foregoing equations can take on any value. However,\npractical considerations make M > N undesirable for two reasons. In Sec. 4.3-3, we shall show that", - "type": "text" - }, - { - "block_id": "p170-b28", - "global_id": 4789, - "bbox": [ - 60.0, - 656.12, - 74.94, - 666.22 - ], - "text": "150", - "type": "text" - } - ] - }, - { - "page_num": 171, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p171-b0", - "global_id": 4790, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "2.2\nSystem Response to Internal Conditions: The Zero-Input Response\n151", - "type": "text" - }, - { - "block_id": "p171-b1", - "global_id": 4791, - "bbox": [ - 127.59, - 85.4, - 516.16, - 215.34 - ], - "text": "an LTIC system specified by Eq. (2.1) acts as an (M −N)th-order differentiator. A differentiator\nrepresents an unstable system because a bounded input like the step input results in an unbounded\noutput, δ(t). Second, noise is enhanced by a differentiator. Noise is a wideband signal containing\ncomponents of all frequencies from 0 to a very high frequency approaching ∞.† Hence, noise\ncontains a significant amount of rapidly varying components. We know that the derivative of any\nrapidly varying signal is high. Therefore, any system specified by Eq. (2.1) in which M > N will\nmagnify the high-frequency components of noise through differentiation. It is entirely possible for\nnoise to be magnified so much that it swamps the desired system output even if the noise signal at\nthe system’s input is tolerably small. Hence, practical systems generally use M ≤N. For the rest\nof this text we assume implicitly that M ≤N. For the sake of generality, we shall assume M = N\nin Eq. (2.1).", - "type": "text" - }, - { - "block_id": "p171-b2", - "global_id": 4792, - "bbox": [ - 127.59, - 217.32, - 516.16, - 251.2 - ], - "text": "In Ch. 1, we demonstrated that a system described by Eq. (2.2) is linear. Therefore, its\nresponse can be expressed as the sum of two components: the zero-input response and the\nzero-state response (decomposition property).‡ Therefore,", - "type": "text" - }, - { - "block_id": "p171-b3", - "global_id": 4793, - "bbox": [ - 206.54, - 264.28, - 437.18, - 274.24 - ], - "text": "total response = zero-input response + zero-state response", - "type": "text" - }, - { - "block_id": "p171-b4", - "global_id": 4794, - "bbox": [ - 127.59, - 286.9, - 516.14, - 345.1 - ], - "text": "The zero-input response is the system output when the input x(t) = 0, and thus it is the result of\ninternal system conditions (such as energy storages, initial conditions) alone. It is independent of\nthe external input x(t). In contrast, the zero-state response is the system output to the external input\nx(t) when the system is in zero state, meaning the absence of all internal energy storages: that is,\nall initial conditions are zero.", - "type": "text" - }, - { - "block_id": "p171-b5", - "global_id": 4795, - "bbox": [ - 127.94, - 375.5, - 462.93, - 405.39 - ], - "text": "2.2 SYSTEM RESPONSE TO INTERNAL CONDITIONS:\nTHE ZERO-INPUT RESPONSE", - "type": "text" - }, - { - "block_id": "p171-b6", - "global_id": 4796, - "bbox": [ - 127.59, - 410.97, - 476.63, - 422.12 - ], - "text": "The zero-input response y0(t) is the solution of Eq. (2.2) when the input x(t) = 0 so that", - "type": "text" - }, - { - "block_id": "p171-b7", - "global_id": 4797, - "bbox": [ - 293.13, - 434.01, - 350.59, - 445.16 - ], - "text": "Q(D)y0(t) = 0", - "type": "text" - }, - { - "block_id": "p171-b8", - "global_id": 4798, - "bbox": [ - 127.59, - 463.47, - 516.14, - 530.79 - ], - "text": "† Noise is any undesirable signal, natural or manufactured, that interferes with the desired signals in the\nsystem. Some of the sources of noise are the electromagnetic radiation from stars, the random motion of\nelectrons in system components, interference from nearby radio and television stations, transients produced\nby automobile ignition systems, and fluorescent lighting.\n‡ We can verify readily that the system described by Eq. (2.2) has the decomposition property. If y0(t) is the\nzero-input response, then, by definition,", - "type": "text" - }, - { - "block_id": "p171-b9", - "global_id": 4799, - "bbox": [ - 295.93, - 532.97, - 347.79, - 543.05 - ], - "text": "Q(D)y0(t) = 0", - "type": "text" - }, - { - "block_id": "p171-b10", - "global_id": 4800, - "bbox": [ - 127.59, - 549.59, - 339.02, - 558.93 - ], - "text": "If y(t) is the zero-state response, then y(t) is the solution of", - "type": "text" - }, - { - "block_id": "p171-b11", - "global_id": 4801, - "bbox": [ - 284.04, - 568.21, - 359.68, - 577.46 - ], - "text": "Q(D)y(t) = P(D)x(t)", - "type": "text" - }, - { - "block_id": "p171-b12", - "global_id": 4802, - "bbox": [ - 127.59, - 587.21, - 423.75, - 596.17 - ], - "text": "subject to zero initial conditions (zero-state). Adding these two equations, we have", - "type": "text" - }, - { - "block_id": "p171-b13", - "global_id": 4803, - "bbox": [ - 267.63, - 605.45, - 376.09, - 615.53 - ], - "text": "Q(D)[y0(t) + y(t)] = P(D)x(t)", - "type": "text" - }, - { - "block_id": "p171-b14", - "global_id": 4804, - "bbox": [ - 127.6, - 624.07, - 325.05, - 634.15 - ], - "text": "Clearly, y0(t) + y(t) is the general solution of Eq. (2.2).", - "type": "text" - } - ] - }, - { - "page_num": 172, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p172-b0", - "global_id": 4805, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "152\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p172-b1", - "global_id": 4806, - "bbox": [ - 101.84, - 83.6, - 110.14, - 93.56 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p172-b2", - "global_id": 4807, - "bbox": [ - 205.99, - 99.14, - 490.39, - 114.41 - ], - "text": "(DN + a1DN−1 + · · · + aN−1D + aN)y0(t) = 0\n(2.3)", - "type": "text" - }, - { - "block_id": "p172-b3", - "global_id": 4808, - "bbox": [ - 101.84, - 128.01, - 490.4, - 197.74 - ], - "text": "A solution to this equation can be obtained systematically [1]. However, we will take a shortcut\nby using heuristic reasoning. Equation (2.3) shows that a linear combination of y0(t) and its N\nsuccessive derivatives is zero, not at some values of t, but for all t. Such a result is possible if\nand only if y0(t) and all its N successive derivatives are of the same form. Otherwise their sum\ncan never add to zero for all values of t. We know that only an exponential function eλt has this\nproperty. So let us assume that", - "type": "text" - }, - { - "block_id": "p172-b4", - "global_id": 4809, - "bbox": [ - 273.18, - 205.71, - 318.42, - 218.58 - ], - "text": "y0(t) = ceλt", - "type": "text" - }, - { - "block_id": "p172-b5", - "global_id": 4810, - "bbox": [ - 101.84, - 232.18, - 223.32, - 242.15 - ], - "text": "is a solution to Eq. (2.3). Then", - "type": "text" - }, - { - "block_id": "p172-b6", - "global_id": 4811, - "bbox": [ - 244.93, - 257.14, - 307.74, - 275.27 - ], - "text": "Dy0(t) = dy0(t)", - "type": "text" - }, - { - "block_id": "p172-b7", - "global_id": 4812, - "bbox": [ - 291.89, - 262.39, - 340.84, - 281.46 - ], - "text": "dt\n= cλeλt", - "type": "text" - }, - { - "block_id": "p172-b8", - "global_id": 4813, - "bbox": [ - 240.94, - 282.73, - 311.99, - 301.8 - ], - "text": "D2y0(t) = d2y0(t)", - "type": "text" - }, - { - "block_id": "p172-b9", - "global_id": 4814, - "bbox": [ - 292.02, - 288.92, - 349.09, - 308.0 - ], - "text": "dt2\n= cλ2eλt", - "type": "text" - }, - { - "block_id": "p172-b10", - "global_id": 4815, - "bbox": [ - 276.58, - 309.22, - 279.07, - 327.16 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p172-b11", - "global_id": 4816, - "bbox": [ - 239.35, - 331.61, - 313.58, - 350.75 - ], - "text": "DNy0(t) = dNy0(t)", - "type": "text" - }, - { - "block_id": "p172-b12", - "global_id": 4817, - "bbox": [ - 292.02, - 337.88, - 352.25, - 356.95 - ], - "text": "dtN\n= cλNeλt", - "type": "text" - }, - { - "block_id": "p172-b13", - "global_id": 4818, - "bbox": [ - 101.84, - 370.36, - 293.34, - 380.32 - ], - "text": "Substituting these results in Eq. (2.3), we obtain", - "type": "text" - }, - { - "block_id": "p172-b14", - "global_id": 4819, - "bbox": [ - 210.39, - 393.16, - 381.85, - 408.42 - ], - "text": "c(λN + a1λN−1 + · · · + aN−1λ + aN)eλt = 0", - "type": "text" - }, - { - "block_id": "p172-b15", - "global_id": 4820, - "bbox": [ - 101.85, - 425.01, - 265.0, - 434.97 - ], - "text": "For a nontrivial solution of this equation,", - "type": "text" - }, - { - "block_id": "p172-b16", - "global_id": 4821, - "bbox": [ - 221.72, - 447.8, - 490.39, - 463.37 - ], - "text": "λN + a1λN−1 + · · · + aN−1λ + aN = 0\n(2.4)", - "type": "text" - }, - { - "block_id": "p172-b17", - "global_id": 4822, - "bbox": [ - 101.84, - 476.04, - 490.38, - 513.53 - ], - "text": "This result means that ceλt is indeed a solution of Eq. (2.3), provided λ satisfies Eq. (2.4). Note\nthat the polynomial in Eq. (2.4) is identical to the polynomial Q(D) in Eq. (2.3), with λ replacing\nD. Therefore, Eq. (2.4) can be expressed as", - "type": "text" - }, - { - "block_id": "p172-b18", - "global_id": 4823, - "bbox": [ - 277.66, - 530.48, - 314.58, - 540.85 - ], - "text": "Q(λ) = 0", - "type": "text" - }, - { - "block_id": "p172-b19", - "global_id": 4824, - "bbox": [ - 101.84, - 557.8, - 287.0, - 568.18 - ], - "text": "Expressing Q(λ) in factorized form, we obtain", - "type": "text" - }, - { - "block_id": "p172-b20", - "global_id": 4825, - "bbox": [ - 213.51, - 585.13, - 490.39, - 596.28 - ], - "text": "Q(λ) = (λ −λ1)(λ −λ2)· · ·(λ −λN) = 0\n(2.5)", - "type": "text" - }, - { - "block_id": "p172-b21", - "global_id": 4826, - "bbox": [ - 101.84, - 612.45, - 490.4, - 636.28 - ], - "text": "Clearly, λ has N solutions: λ1, λ2, . . ., λN, assuming that all λi are distinct. Consequently, Eq. (2.3)\nhas N possible solutions: c1eλ1t, c2eλ2t, . . ., cNeλNt, with c1, c2,. . .,cN as arbitrary constants. We", - "type": "text" - } - ] - }, - { - "page_num": 173, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p173-b0", - "global_id": 4827, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "2.2\nSystem Response to Internal Conditions: The Zero-Input Response\n153", - "type": "text" - }, - { - "block_id": "p173-b1", - "global_id": 4828, - "bbox": [ - 127.59, - 83.03, - 479.41, - 96.62 - ], - "text": "can readily show that a general solution is given by the sum of these N solutions† so that", - "type": "text" - }, - { - "block_id": "p173-b2", - "global_id": 4829, - "bbox": [ - 248.18, - 103.42, - 516.13, - 118.68 - ], - "text": "y0(t) = c1eλ1t + c2eλ2t + · · · + cNeλNt\n(2.6)", - "type": "text" - }, - { - "block_id": "p173-b3", - "global_id": 4830, - "bbox": [ - 127.59, - 128.83, - 516.12, - 151.16 - ], - "text": "where c1, c2, . . ., cN are arbitrary constants determined by N constraints (the auxiliary conditions)\non the solution.", - "type": "text" - }, - { - "block_id": "p173-b4", - "global_id": 4831, - "bbox": [ - 127.59, - 152.74, - 516.14, - 187.02 - ], - "text": "Observe that the polynomial Q(λ), which is characteristic of the system, has nothing to do\nwith the input. For this reason the polynomial Q(λ) is called the characteristic polynomial of the\nsystem. The equation", - "type": "text" - }, - { - "block_id": "p173-b5", - "global_id": 4832, - "bbox": [ - 303.4, - 188.61, - 340.32, - 198.98 - ], - "text": "Q(λ) = 0", - "type": "text" - }, - { - "block_id": "p173-b6", - "global_id": 4833, - "bbox": [ - 127.59, - 207.22, - 516.14, - 289.32 - ], - "text": "is called the characteristic equation of the system. Equation (2.5) clearly indicates that λ1, λ2,\n. . ., λN are the roots of the characteristic equation; consequently, they are called the characteristic\nroots of the system. The terms characteristic values, eigenvalues, and natural frequencies are also\nused for characteristic roots.‡ The exponentials eλit (i = 1,2,. . .,n) in the zero-input response are\nthe characteristic modes (also known as natural modes or simply as modes) of the system. There\nis a characteristic mode for each characteristic root of the system, and the zero-input response is a\nlinear combination of the characteristic modes of the system.", - "type": "text" - }, - { - "block_id": "p173-b7", - "global_id": 4834, - "bbox": [ - 127.6, - 291.32, - 516.17, - 349.1 - ], - "text": "An LTIC system’s characteristic modes comprise its single most important attribute.\nCharacteristic modes not only determine the zero-input response but also play an important role\nin determining the zero-state response. In other words, the entire behavior of a system is dictated\nprimarily by its characteristic modes. In the rest of this chapter we shall see the pervasive presence\nof characteristic modes in every aspect of system behavior.", - "type": "text" - }, - { - "block_id": "p173-b8", - "global_id": 4835, - "bbox": [ - 127.59, - 363.68, - 516.13, - 413.7 - ], - "text": "REPEATED ROOTS\nThe solution of Eq. (2.3) as given in Eq. (2.6) assumes that the N characteristic roots λ1, λ2, . . .,\nλN are distinct. If there are repeated roots (same root occurring more than once), the form of the\nsolution is modified slightly. By direct substitution we can show that the solution of the equation", - "type": "text" - }, - { - "block_id": "p173-b9", - "global_id": 4836, - "bbox": [ - 286.58, - 423.19, - 357.14, - 435.77 - ], - "text": "(D −λ)2y0(t) = 0", - "type": "text" - }, - { - "block_id": "p173-b10", - "global_id": 4837, - "bbox": [ - 127.59, - 446.33, - 170.92, - 456.29 - ], - "text": "is given by", - "type": "text" - }, - { - "block_id": "p173-b11", - "global_id": 4838, - "bbox": [ - 282.1, - 456.14, - 361.0, - 469.32 - ], - "text": "y0(t) = (c1 + c2t)eλt", - "type": "text" - }, - { - "block_id": "p173-b12", - "global_id": 4839, - "bbox": [ - 127.59, - 486.21, - 467.32, - 499.17 - ], - "text": "† To prove this assertion, assume that y1(t), y2(t), . . ., yN(t) are all solutions of Eq. (2.3). Then", - "type": "text" - }, - { - "block_id": "p173-b13", - "global_id": 4840, - "bbox": [ - 296.67, - 504.78, - 348.53, - 514.86 - ], - "text": "Q(D)y1(t) = 0", - "type": "text" - }, - { - "block_id": "p173-b14", - "global_id": 4841, - "bbox": [ - 296.67, - 518.73, - 348.53, - 528.81 - ], - "text": "Q(D)y2(t) = 0", - "type": "text" - }, - { - "block_id": "p173-b15", - "global_id": 4842, - "bbox": [ - 339.03, - 533.26, - 341.27, - 550.19 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p173-b16", - "global_id": 4843, - "bbox": [ - 295.2, - 554.8, - 348.53, - 564.81 - ], - "text": "Q(D)yN(t) = 0", - "type": "text" - }, - { - "block_id": "p173-b17", - "global_id": 4844, - "bbox": [ - 127.59, - 570.49, - 455.04, - 580.57 - ], - "text": "Multiplying these equations by c1, c2, . . ., cN, respectively, and adding them together yield", - "type": "text" - }, - { - "block_id": "p173-b18", - "global_id": 4845, - "bbox": [ - 243.02, - 586.18, - 400.7, - 596.25 - ], - "text": "Q(D)[c1y1(t) + c2y2(t) + · · · + cNyn(t)] = 0", - "type": "text" - }, - { - "block_id": "p173-b19", - "global_id": 4846, - "bbox": [ - 127.59, - 601.85, - 516.13, - 633.41 - ], - "text": "This result shows that c1y1(t) + c2y2(t) + · · · + cNyn(t) is also a solution of the homogeneous equation\n[Eq. (2.3)].\n‡ Eigenvalue is German for “characteristic value.”", - "type": "text" - } - ] - }, - { - "page_num": 174, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p174-b0", - "global_id": 4847, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "154\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p174-b1", - "global_id": 4848, - "bbox": [ - 101.84, - 83.64, - 490.38, - 109.17 - ], - "text": "In this case the root λ repeats twice. Observe that the characteristic modes in this case are eλt and\nteλt. Continuing this pattern, we can show that for the differential equation", - "type": "text" - }, - { - "block_id": "p174-b2", - "global_id": 4849, - "bbox": [ - 261.14, - 123.92, - 331.1, - 136.57 - ], - "text": "(D −λ)ry0(t) = 0", - "type": "text" - }, - { - "block_id": "p174-b3", - "global_id": 4850, - "bbox": [ - 101.85, - 150.82, - 412.86, - 162.42 - ], - "text": "the characteristic modes are eλt, teλt, t2eλt, . . ., tr−1eλt, and that the solution is", - "type": "text" - }, - { - "block_id": "p174-b4", - "global_id": 4851, - "bbox": [ - 227.43, - 176.96, - 364.18, - 190.13 - ], - "text": "y0(t) = (c1 + c2t + · · · + crtr−1)eλt", - "type": "text" - }, - { - "block_id": "p174-b5", - "global_id": 4852, - "bbox": [ - 101.84, - 205.71, - 348.02, - 215.68 - ], - "text": "Consequently, for a system with the characteristic polynomial", - "type": "text" - }, - { - "block_id": "p174-b6", - "global_id": 4853, - "bbox": [ - 216.09, - 230.43, - 376.15, - 243.08 - ], - "text": "Q(λ) = (λ −λ1)r(λ −λr+1)· · ·(λ −λN)", - "type": "text" - }, - { - "block_id": "p174-b7", - "global_id": 4854, - "bbox": [ - 101.84, - 255.36, - 446.99, - 268.94 - ], - "text": "the characteristic modes are eλ1t, teλ1t, . . ., tr−1eλ1t, eλr+1t, . . ., eλNt and the solution is", - "type": "text" - }, - { - "block_id": "p174-b8", - "global_id": 4855, - "bbox": [ - 170.96, - 281.07, - 420.65, - 296.64 - ], - "text": "y0(t) = (c1 + c2t + · · · + crtr−1)eλ1t + cr+1eλr+1t + · · · + cNeλNt", - "type": "text" - }, - { - "block_id": "p174-b9", - "global_id": 4856, - "bbox": [ - 101.84, - 312.65, - 490.38, - 374.63 - ], - "text": "COMPLEX ROOTS\nThe procedure for handling complex roots is the same as that for real roots. For complex roots,\nthe usual procedure leads to complex characteristic modes and the complex form of solution.\nHowever, it is possible to avoid the complex form altogether by selecting a real-form of solution,\nas described next.", - "type": "text" - }, - { - "block_id": "p174-b10", - "global_id": 4857, - "bbox": [ - 101.84, - 376.62, - 490.38, - 422.45 - ], - "text": "For a real system, complex roots must occur in pairs of conjugates if the coefficients of the\ncharacteristic polynomial Q(λ) are to be real. Therefore, if α + jβ is a characteristic root, α −jβ\nmust also be a characteristic root. The zero-input response corresponding to this pair of complex\nconjugate roots is", - "type": "text" - }, - { - "block_id": "p174-b11", - "global_id": 4858, - "bbox": [ - 238.7, - 426.98, - 490.39, - 442.24 - ], - "text": "y0(t) = c1e(α+jβ)t + c2e(α−jβ)t\n(2.7)", - "type": "text" - }, - { - "block_id": "p174-b12", - "global_id": 4859, - "bbox": [ - 101.84, - 454.73, - 490.39, - 477.06 - ], - "text": "For a real system, the response y0(t) must also be real. This is possible only if c1 and c2 are\nconjugates. Let", - "type": "text" - }, - { - "block_id": "p174-b13", - "global_id": 4860, - "bbox": [ - 228.01, - 481.59, - 254.18, - 499.72 - ], - "text": "c1 = c", - "type": "text" - }, - { - "block_id": "p174-b14", - "global_id": 4861, - "bbox": [ - 249.48, - 481.59, - 346.68, - 505.71 - ], - "text": "2ejθ\nand\nc2 = c", - "type": "text" - }, - { - "block_id": "p174-b15", - "global_id": 4862, - "bbox": [ - 341.98, - 486.54, - 363.02, - 505.71 - ], - "text": "2e−jθ", - "type": "text" - }, - { - "block_id": "p174-b16", - "global_id": 4863, - "bbox": [ - 101.84, - 515.21, - 145.84, - 525.18 - ], - "text": "This yields", - "type": "text" - }, - { - "block_id": "p174-b17", - "global_id": 4864, - "bbox": [ - 226.39, - 537.31, - 262.94, - 555.13 - ], - "text": "y0(t) = c", - "type": "text" - }, - { - "block_id": "p174-b18", - "global_id": 4865, - "bbox": [ - 258.24, - 537.31, - 365.21, - 561.43 - ], - "text": "2ejθe(α+jβ)t + c\n2e−jθe(α−jβ)t", - "type": "text" - }, - { - "block_id": "p174-b19", - "global_id": 4866, - "bbox": [ - 247.22, - 559.73, - 262.94, - 576.67 - ], - "text": "= c", - "type": "text" - }, - { - "block_id": "p174-b20", - "global_id": 4867, - "bbox": [ - 258.24, - 558.39, - 355.33, - 583.85 - ], - "text": "2eαt\t\nej(βt+θ) + e−j(βt+θ)", - "type": "text" - }, - { - "block_id": "p174-b21", - "global_id": 4868, - "bbox": [ - 247.22, - 582.13, - 490.39, - 596.21 - ], - "text": "= ceαt cos(βt + θ)\n(2.8)", - "type": "text" - }, - { - "block_id": "p174-b22", - "global_id": 4869, - "bbox": [ - 101.84, - 612.45, - 490.39, - 634.79 - ], - "text": "Therefore, the zero-input response corresponding to complex conjugate roots α ± jβ can be\nexpressed in a complex form [Eq. (2.7)] or a real form [Eq. (2.8)].", - "type": "text" - } - ] - }, - { - "page_num": 175, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p175-b0", - "global_id": 4870, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "2.2\nSystem Response to Internal Conditions: The Zero-Input Response\n155", - "type": "text" - }, - { - "block_id": "p175-b1", - "global_id": 4871, - "bbox": [ - 102.51, - 93.91, - 382.35, - 105.87 - ], - "text": "EXAMPLE 2.1\nFinding the Zero-Input Response", - "type": "text" - }, - { - "block_id": "p175-b2", - "global_id": 4872, - "bbox": [ - 128.9, - 121.14, - 462.18, - 132.29 - ], - "text": "Find y0(t), the zero-input response of the response for an LTIC system described by", - "type": "text" - }, - { - "block_id": "p175-b3", - "global_id": 4873, - "bbox": [ - 146.84, - 135.46, - 502.76, - 150.23 - ], - "text": "(a) the simple-root system (D2 + 3D + 2)y(t) = Dx(t) with initial conditions y0(0) = 0", - "type": "text" - }, - { - "block_id": "p175-b4", - "global_id": 4874, - "bbox": [ - 146.84, - 151.03, - 502.74, - 176.35 - ], - "text": "and ˙y0(0) = −5.\n(b) the repeated-root system (D2 + 6D + 9)y(t) = (3D + 5)x(t) with initial conditions", - "type": "text" - }, - { - "block_id": "p175-b5", - "global_id": 4875, - "bbox": [ - 147.4, - 177.93, - 502.74, - 203.25 - ], - "text": "y0(0) = 3 and ˙y0(0) = −7.\n(c) the complex-root system (D2 + 4D + 40)y(t) = (D + 2)x(t) with initial conditions", - "type": "text" - }, - { - "block_id": "p175-b6", - "global_id": 4876, - "bbox": [ - 163.44, - 204.83, - 278.06, - 215.98 - ], - "text": "y0(0) = 2 and ˙y0(0) = 16.78.", - "type": "text" - }, - { - "block_id": "p175-b7", - "global_id": 4877, - "bbox": [ - 128.9, - 230.11, - 502.76, - 291.92 - ], - "text": "(a) Note that y0(t), being the zero-input response (x(t) = 0), is the solution of (D2 +3D+\n2)y0(t) = 0. The characteristic polynomial of the system is λ2 + 3λ + 2. The characteristic\nequation of the system is therefore λ2 + 3λ + 2 = (λ + 1)(λ + 2) = 0. The characteristic roots\nof the system are λ1 = −1 and λ2 = −2, and the characteristic modes of the system are e−t and\ne−2t. Consequently, the zero-input response is", - "type": "text" - }, - { - "block_id": "p175-b8", - "global_id": 4878, - "bbox": [ - 272.49, - 296.35, - 358.53, - 311.62 - ], - "text": "y0(t) = c1e−t + c2e−2t", - "type": "text" - }, - { - "block_id": "p175-b9", - "global_id": 4879, - "bbox": [ - 128.9, - 319.82, - 293.69, - 329.78 - ], - "text": "Differentiating this expression, we obtain", - "type": "text" - }, - { - "block_id": "p175-b10", - "global_id": 4880, - "bbox": [ - 266.12, - 334.22, - 364.92, - 349.48 - ], - "text": "˙y0(t) = −c1e−t −2c2e−2t", - "type": "text" - }, - { - "block_id": "p175-b11", - "global_id": 4881, - "bbox": [ - 128.9, - 357.26, - 502.75, - 380.36 - ], - "text": "To determine the constants c1 and c2, we set t = 0 in the equations for y0(t) and ˙y0(t) and\nsubstitute the initial conditions y0(0) = 0 and ˙y0(0) = −5, yielding", - "type": "text" - }, - { - "block_id": "p175-b12", - "global_id": 4882, - "bbox": [ - 283.31, - 388.15, - 347.85, - 414.54 - ], - "text": "0 = c1 + c2\n−5 = −c1 −2c2", - "type": "text" - }, - { - "block_id": "p175-b13", - "global_id": 4883, - "bbox": [ - 128.9, - 422.33, - 445.8, - 433.88 - ], - "text": "Solving these two simultaneous equations in two unknowns for c1 and c2 yields", - "type": "text" - }, - { - "block_id": "p175-b14", - "global_id": 4884, - "bbox": [ - 259.58, - 440.95, - 372.1, - 452.4 - ], - "text": "c1 = −5\nand\nc2 = 5", - "type": "text" - }, - { - "block_id": "p175-b15", - "global_id": 4885, - "bbox": [ - 128.91, - 460.29, - 170.65, - 470.25 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p175-b16", - "global_id": 4886, - "bbox": [ - 272.03, - 467.71, - 502.76, - 482.98 - ], - "text": "y0(t) = −5e−t + 5e−2t\n(2.9)", - "type": "text" - }, - { - "block_id": "p175-b17", - "global_id": 4887, - "bbox": [ - 128.91, - 488.56, - 502.76, - 512.11 - ], - "text": "This is the zero-input response of y(t). Because y0(t) is present at t = 0−, we are justified in\nassuming that it exists for t ≥0.†", - "type": "text" - }, - { - "block_id": "p175-b18", - "global_id": 4888, - "bbox": [ - 128.9, - 510.08, - 502.78, - 559.94 - ], - "text": "(b) The characteristic polynomial is λ2 + 6λ + 9 = (λ + 3)2, and its characteristic roots\nare λ1 = −3, λ2 = −3 (repeated roots). Consequently, the characteristic modes of the system\nare e−3t and te−3t. The zero-input response, being a linear combination of the characteristic\nmodes, is given by", - "type": "text" - }, - { - "block_id": "p175-b19", - "global_id": 4889, - "bbox": [ - 273.52, - 559.79, - 357.52, - 572.97 - ], - "text": "y0(t) = (c1 + c2t)e−3t", - "type": "text" - }, - { - "block_id": "p175-b20", - "global_id": 4890, - "bbox": [ - 128.9, - 590.26, - 502.26, - 602.83 - ], - "text": "†y0(t) may be present even before t = 0−. However, we can be sure of its presence only from t = 0−", - "type": "text" - }, - { - "block_id": "p175-b21", - "global_id": 4891, - "bbox": [ - 128.9, - 604.08, - 157.94, - 613.05 - ], - "text": "onward.", - "type": "text" - } - ] - }, - { - "page_num": 176, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p176-b0", - "global_id": 4892, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "156\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p176-b1", - "global_id": 4893, - "bbox": [ - 103.16, - 85.88, - 477.01, - 97.75 - ], - "text": "We can find the arbitrary constants c1 and c2 from the initial conditions y0(0) = 3 and", - "type": "text" - }, - { - "block_id": "p176-b2", - "global_id": 4894, - "bbox": [ - 103.17, - 97.84, - 477.01, - 120.17 - ], - "text": "˙y0(0) = −7 following the procedure in part (a). The reader can show that c1 = 3 and c2 = 2.\nHence,", - "type": "text" - }, - { - "block_id": "p176-b3", - "global_id": 4895, - "bbox": [ - 231.47, - 120.03, - 348.72, - 132.9 - ], - "text": "y0(t) = (3 + 2t)e−3t\nt ≥0", - "type": "text" - }, - { - "block_id": "p176-b4", - "global_id": 4896, - "bbox": [ - 103.16, - 137.07, - 477.02, - 188.01 - ], - "text": "(c) The characteristic polynomial is λ2 + 4λ + 40 = (λ + 2 −j6)(λ + 2 + j6). The\ncharacteristic roots are −2 ± j6.† The solution can be written either in the complex form\n[Eq. (2.7)] or in the real form [Eq. (2.8)]. The complex form is y0(t) = c1eλ1t + c2eλ2t, where\nλ1 = −2+j6 and λ2 = −2−j6. Since α = −2 and β = 6, the real-form solution is [see Eq. (2.8)]", - "type": "text" - }, - { - "block_id": "p176-b5", - "global_id": 4897, - "bbox": [ - 241.04, - 191.78, - 339.13, - 206.62 - ], - "text": "y0(t) = ce−2t cos(6t + θ)", - "type": "text" - }, - { - "block_id": "p176-b6", - "global_id": 4898, - "bbox": [ - 103.17, - 214.82, - 267.95, - 224.78 - ], - "text": "Differentiating this expression, we obtain", - "type": "text" - }, - { - "block_id": "p176-b7", - "global_id": 4899, - "bbox": [ - 193.85, - 229.63, - 386.33, - 244.48 - ], - "text": "˙y0(t) = −2ce−2t cos(6t + θ) −6ce−2t sin(6t + θ)", - "type": "text" - }, - { - "block_id": "p176-b8", - "global_id": 4900, - "bbox": [ - 103.17, - 252.26, - 477.01, - 275.37 - ], - "text": "To determine the constants c and θ, we set t = 0 in the equations for y0(t) and ˙y0(t) and\nsubstitute the initial conditions y0(0) = 2 and ˙y0(0) = 16.78, yielding", - "type": "text" - }, - { - "block_id": "p176-b9", - "global_id": 4901, - "bbox": [ - 234.18, - 283.15, - 345.01, - 308.46 - ], - "text": "2 = ccosθ\n16.78 = −2ccosθ −6csinθ", - "type": "text" - }, - { - "block_id": "p176-b10", - "global_id": 4902, - "bbox": [ - 103.17, - 317.02, - 451.63, - 327.4 - ], - "text": "Solution of these two simultaneous equations in two unknowns ccosθ and csinθ yields", - "type": "text" - }, - { - "block_id": "p176-b11", - "global_id": 4903, - "bbox": [ - 209.05, - 335.95, - 371.12, - 346.32 - ], - "text": "ccosθ = 2\nand\ncsinθ = −3.463", - "type": "text" - }, - { - "block_id": "p176-b12", - "global_id": 4904, - "bbox": [ - 103.17, - 355.29, - 308.67, - 365.25 - ], - "text": "Squaring and then adding these two equations yield", - "type": "text" - }, - { - "block_id": "p176-b13", - "global_id": 4905, - "bbox": [ - 213.67, - 369.69, - 366.51, - 384.18 - ], - "text": "c2 = (2)2 + (−3.464)2 = 16 \r⇒c = 4", - "type": "text" - }, - { - "block_id": "p176-b14", - "global_id": 4906, - "bbox": [ - 103.16, - 392.74, - 311.89, - 403.11 - ], - "text": "Next, dividing csinθ = −3.463 by ccosθ = 2 yields", - "type": "text" - }, - { - "block_id": "p176-b15", - "global_id": 4907, - "bbox": [ - 258.54, - 409.65, - 320.43, - 427.01 - ], - "text": "tanθ = −3.463", - "type": "text" - }, - { - "block_id": "p176-b16", - "global_id": 4908, - "bbox": [ - 103.16, - 424.12, - 307.85, - 448.83 - ], - "text": "2\nand", - "type": "text" - }, - { - "block_id": "p176-b17", - "global_id": 4909, - "bbox": [ - 232.6, - 452.28, - 270.95, - 464.38 - ], - "text": "θ = tan−1", - "type": "text" - }, - { - "block_id": "p176-b18", - "global_id": 4910, - "bbox": [ - 272.55, - 440.02, - 310.64, - 457.39 - ], - "text": "−3.463", - "type": "text" - }, - { - "block_id": "p176-b19", - "global_id": 4911, - "bbox": [ - 293.07, - 461.49, - 298.05, - 471.45 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p176-b21", - "global_id": 4912, - "bbox": [ - 320.62, - 447.02, - 345.38, - 463.96 - ], - "text": "= −π", - "type": "text" - }, - { - "block_id": "p176-b22", - "global_id": 4913, - "bbox": [ - 340.4, - 461.49, - 345.39, - 471.45 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p176-b23", - "global_id": 4914, - "bbox": [ - 103.16, - 478.63, - 144.91, - 488.6 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p176-b24", - "global_id": 4915, - "bbox": [ - 236.5, - 488.41, - 302.43, - 503.26 - ], - "text": "y0(t) = 4e−2t cos", - "type": "text" - }, - { - "block_id": "p176-b26", - "global_id": 4916, - "bbox": [ - 310.08, - 485.13, - 336.06, - 509.56 - ], - "text": "6t −π\n3", - "type": "text" - }, - { - "block_id": "p176-b28", - "global_id": 4917, - "bbox": [ - 103.16, - 514.1, - 286.38, - 525.25 - ], - "text": "For the plot of y0(t), refer again to Fig. B.11c.", - "type": "text" - }, - { - "block_id": "p176-b29", - "global_id": 4918, - "bbox": [ - 101.84, - 576.69, - 490.39, - 610.83 - ], - "text": "† The complex conjugate roots of a second-order polynomial can be determined by using the formula in\nSec. B.8-10 or by expressing the polynomial as a sum of two squares. The latter can be accomplished by\ncompleting the square with the first two terms, as follows:", - "type": "text" - }, - { - "block_id": "p176-b30", - "global_id": 4919, - "bbox": [ - 153.3, - 616.55, - 438.94, - 629.6 - ], - "text": "λ2 + 4λ + 40 = (λ2 + 4λ + 4) + 36 = (λ + 2)2 + (6)2 = (λ + 2 −j6)(λ + 2 + j6)", - "type": "text" - } - ] - }, - { - "page_num": 177, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p177-b0", - "global_id": 4920, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "2.2\nSystem Response to Internal Conditions: The Zero-Input Response\n157", - "type": "text" - }, - { - "block_id": "p177-b1", - "global_id": 4921, - "bbox": [ - 102.51, - 93.92, - 431.07, - 105.87 - ], - "text": "EXAMPLE 2.2\nUsing MATLAB to Find Polynomial Roots", - "type": "text" - }, - { - "block_id": "p177-b2", - "global_id": 4922, - "bbox": [ - 128.9, - 118.51, - 502.76, - 144.45 - ], - "text": "Find the roots λ1 and λ2 of the polynomial λ2 + 4λ + k for three values of k: (a) k = 3, (b)\nk = 4, and (c) k = 40.", - "type": "text" - }, - { - "block_id": "p177-b3", - "global_id": 4923, - "bbox": [ - 128.9, - 167.29, - 140.52, - 177.25 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p177-b4", - "global_id": 4924, - "bbox": [ - 128.9, - 187.58, - 254.44, - 209.5 - ], - "text": ">>\nr = roots([1 4 3]).’\nr = -3\n-1", - "type": "text" - }, - { - "block_id": "p177-b5", - "global_id": 4925, - "bbox": [ - 128.9, - 218.76, - 399.52, - 241.01 - ], - "text": "For k = 3, the polynomial roots are therefore λ1 = −3 and λ2 = −1.\n(b)", - "type": "text" - }, - { - "block_id": "p177-b6", - "global_id": 4926, - "bbox": [ - 128.91, - 251.34, - 254.44, - 273.26 - ], - "text": ">>\nr = roots([1 4 4]).’\nr = -2\n-2", - "type": "text" - }, - { - "block_id": "p177-b7", - "global_id": 4927, - "bbox": [ - 128.9, - 282.52, - 367.41, - 304.77 - ], - "text": "For k = 4, the polynomial roots are therefore λ1 = λ2 = −2.\n(c)", - "type": "text" - }, - { - "block_id": "p177-b8", - "global_id": 4928, - "bbox": [ - 128.9, - 315.1, - 296.27, - 337.02 - ], - "text": ">>\nr = roots([1 4 40]).’\nr = -2.00+6.00i\n-2.00-6.00i", - "type": "text" - }, - { - "block_id": "p177-b9", - "global_id": 4929, - "bbox": [ - 128.9, - 346.28, - 441.74, - 357.74 - ], - "text": "For k = 40, the polynomial roots are therefore λ1 = −2 + j6 and λ2 = −2 −j6.", - "type": "text" - }, - { - "block_id": "p177-b10", - "global_id": 4930, - "bbox": [ - 102.51, - 410.42, - 469.27, - 422.37 - ], - "text": "EXAMPLE 2.3\nUsing MATLAB to Find the Zero-Input Response", - "type": "text" - }, - { - "block_id": "p177-b11", - "global_id": 4931, - "bbox": [ - 128.9, - 439.04, - 379.34, - 449.0 - ], - "text": "Consider an LTIC system specified by the differential equation", - "type": "text" - }, - { - "block_id": "p177-b12", - "global_id": 4932, - "bbox": [ - 248.79, - 456.43, - 382.86, - 470.92 - ], - "text": "(D2 + 4D + k)y(t) = (3D + 5)x(t)", - "type": "text" - }, - { - "block_id": "p177-b13", - "global_id": 4933, - "bbox": [ - 128.9, - 482.46, - 502.77, - 504.79 - ], - "text": "Using initial conditions y0(0) = 3 and ˙y0(0) = −7, apply MATLAB’s dsolve command to\ndetermine the zero-input response when: (a) k = 3, (b) k = 4, and (c) k = 40.", - "type": "text" - }, - { - "block_id": "p177-b14", - "global_id": 4934, - "bbox": [ - 128.9, - 527.63, - 140.52, - 537.59 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p177-b15", - "global_id": 4935, - "bbox": [ - 128.9, - 547.92, - 432.26, - 569.84 - ], - "text": ">>\ny_0 = dsolve(’D2y+4*Dy+3*y=0’,’y(0)=3’,’Dy(0)=-7’,’t’)\ny_0 = 1/exp(t) + 2/exp(3*t)", - "type": "text" - }, - { - "block_id": "p177-b16", - "global_id": 4936, - "bbox": [ - 128.9, - 575.48, - 391.88, - 601.34 - ], - "text": "For k = 3, the zero-input response is therefore y0(t) = e−t + 2e−3t.\n(b)", - "type": "text" - }, - { - "block_id": "p177-b17", - "global_id": 4937, - "bbox": [ - 128.91, - 611.68, - 432.26, - 633.6 - ], - "text": ">>\ny_0 = dsolve(’D2y+4*Dy+4*y=0’,’y(0)=3’,’Dy(0)=-7’,’t’)\ny_0 = 3/exp(2*t) - t/exp(2*t)", - "type": "text" - } - ] - }, - { - "page_num": 178, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p178-b0", - "global_id": 4938, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "158\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p178-b1", - "global_id": 4939, - "bbox": [ - 103.16, - 83.79, - 372.39, - 109.66 - ], - "text": "For k = 4, the zero-input response is therefore y0(t) = 3e−2t −te−2t.\n(c)", - "type": "text" - }, - { - "block_id": "p178-b2", - "global_id": 4940, - "bbox": [ - 103.17, - 119.99, - 411.75, - 141.91 - ], - "text": ">>\ny_0 = dsolve(’D2y+4*Dy+40*y=0’,’y(0)=3’,’Dy(0)=-7’,’t’)\ny_0 = (3*cos(6*t))/exp(2*t) - sin(6*t)/(6*exp(2*t))", - "type": "text" - }, - { - "block_id": "p178-b3", - "global_id": 4941, - "bbox": [ - 103.17, - 147.96, - 391.74, - 162.32 - ], - "text": "For k = 40, the zero-input response is therefore y0(t) = 3e−2t cos(6t) −1", - "type": "text" - }, - { - "block_id": "p178-b4", - "global_id": 4942, - "bbox": [ - 388.25, - 147.96, - 440.55, - 164.36 - ], - "text": "6e−2t sin(6t).", - "type": "text" - }, - { - "block_id": "p178-b5", - "global_id": 4943, - "bbox": [ - 107.82, - 220.76, - 451.81, - 246.67 - ], - "text": "DRILL 2.1\nFinding the Zero-Input Response of a First-Order\nSystem", - "type": "text" - }, - { - "block_id": "p178-b6", - "global_id": 4944, - "bbox": [ - 107.82, - 255.37, - 484.42, - 277.7 - ], - "text": "Find the zero-input response of an LTIC system described by (D + 5)y(t) = x(t) if the initial\ncondition is y(0) = 5.", - "type": "text" - }, - { - "block_id": "p178-b7", - "global_id": 4945, - "bbox": [ - 108.09, - 291.23, - 162.68, - 302.19 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p178-b8", - "global_id": 4946, - "bbox": [ - 107.82, - 307.95, - 198.83, - 320.32 - ], - "text": "y0(t) = 5e−5t\nt ≥0", - "type": "text" - }, - { - "block_id": "p178-b9", - "global_id": 4947, - "bbox": [ - 107.82, - 370.8, - 467.09, - 396.71 - ], - "text": "DRILL 2.2\nFinding the Zero-Input Response of a Second-Order\nSystem", - "type": "text" - }, - { - "block_id": "p178-b10", - "global_id": 4948, - "bbox": [ - 107.82, - 405.41, - 259.66, - 416.56 - ], - "text": "Letting y0(0) = 1 and ˙y0(0) = 4, solve", - "type": "text" - }, - { - "block_id": "p178-b11", - "global_id": 4949, - "bbox": [ - 257.47, - 423.21, - 334.76, - 438.48 - ], - "text": "(D2 + 2D)y0(t) = 0", - "type": "text" - }, - { - "block_id": "p178-b12", - "global_id": 4950, - "bbox": [ - 108.09, - 451.23, - 162.68, - 462.19 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p178-b13", - "global_id": 4951, - "bbox": [ - 107.82, - 467.94, - 214.68, - 480.32 - ], - "text": "y0(t) = 3 −2e−2t\nt ≥0", - "type": "text" - }, - { - "block_id": "p178-b14", - "global_id": 4952, - "bbox": [ - 102.14, - 510.28, - 469.0, - 525.15 - ], - "text": "PRACTICAL INITIAL CONDITIONS AND THE MEANING OF 0−AND 0+", - "type": "text" - }, - { - "block_id": "p178-b15", - "global_id": 4953, - "bbox": [ - 101.84, - 528.77, - 490.4, - 563.06 - ], - "text": "In Ex. 2.1 the initial conditions y0(0) and ˙y0(0) were supplied. In practical problems, we must\nderive such conditions from the physical situation. For instance, in an RLC circuit, we may be\ngiven the conditions (initial capacitor voltages, initial inductor currents, etc.).", - "type": "text" - }, - { - "block_id": "p178-b16", - "global_id": 4954, - "bbox": [ - 101.84, - 564.63, - 490.38, - 586.96 - ], - "text": "From this information, we need to derive y0(0), ˙y0(0), . . . for the desired variable as\ndemonstrated in the next example.", - "type": "text" - }, - { - "block_id": "p178-b17", - "global_id": 4955, - "bbox": [ - 101.84, - 588.55, - 490.4, - 634.79 - ], - "text": "In much of our discussion, the input is assumed to start at t = 0, unless otherwise mentioned.\nHence, t = 0 is the reference point. The conditions immediately before t = 0 (just before the input\nis applied) are the conditions at t = 0−, and those immediately after t = 0 (just after the input is\napplied) are the conditions at t = 0+ (compare this with the historical time frames BCE and CE). In", - "type": "text" - } - ] - }, - { - "page_num": 179, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p179-b0", - "global_id": 4956, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "2.2\nSystem Response to Internal Conditions: The Zero-Input Response\n159", - "type": "text" - }, - { - "block_id": "p179-b1", - "global_id": 4957, - "bbox": [ - 127.59, - 85.07, - 516.14, - 108.63 - ], - "text": "practice, we are likely to know the initial conditions at t = 0−rather than at t = 0+. The two sets\nof conditions are generally different, although in some cases they may be identical.", - "type": "text" - }, - { - "block_id": "p179-b2", - "global_id": 4958, - "bbox": [ - 127.59, - 110.21, - 516.14, - 252.09 - ], - "text": "The total response y(t) consists of two components: the zero-input response y0(t) [response\ndue to the initial conditions alone with x(t) = 0] and the zero-state response resulting from the\ninput alone with all initial conditions zero. At t = 0−, the total response y(t) consists solely of\nthe zero-input response y0(t) because the input has not started yet. Hence the initial conditions on\ny(t) are identical to those of y0(t). Thus, y(0−) = y0(0−), ˙y(0−) = ˙y0(0−), and so on. Moreover,\ny0(t) is the response due to initial conditions alone and does not depend on the input x(t). Hence,\napplication of the input at t = 0 does not affect y0(t). This means the initial conditions on y0(t)\nat t = 0−and 0+ are identical; that is, y0(0−), ˙y0(0−), . . . are identical to y0(0+), ˙y0(0+), . . .,\nrespectively. It is clear that for y0(t), there is no distinction between the initial conditions at t = 0−,\n0, and 0+. They are all the same. But this is not the case with the total response y(t), which consists\nof both the zero-input and zero-state responses. Thus, in general, y(0−)̸ = y(0+), ˙y(0−)̸ = ˙y(0+),\nand so on.", - "type": "text" - }, - { - "block_id": "p179-b3", - "global_id": 4959, - "bbox": [ - 102.51, - 321.86, - 393.61, - 333.82 - ], - "text": "EXAMPLE 2.4\nConsideration of Initial Conditions", - "type": "text" - }, - { - "block_id": "p179-b4", - "global_id": 4960, - "bbox": [ - 128.9, - 345.92, - 502.76, - 382.13 - ], - "text": "A voltage x(t) = 10e−3tu(t) is applied at the input of the RLC circuit illustrated in Fig. 2.2a.\nFind the loop current y(t) for t ≥0 if the initial inductor current is zero [y(0−) = 0] and the\ninitial capacitor voltage is 5 volts [vC(0−) = 5].", - "type": "text" - }, - { - "block_id": "p179-b5", - "global_id": 4961, - "bbox": [ - 128.9, - 403.93, - 445.84, - 414.31 - ], - "text": "The differential (loop) equation relating y(t) to x(t) was derived in Eq. (1.29) as", - "type": "text" - }, - { - "block_id": "p179-b6", - "global_id": 4962, - "bbox": [ - 262.73, - 421.73, - 368.94, - 436.22 - ], - "text": "(D2 + 3D + 2)y(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p179-b7", - "global_id": 4963, - "bbox": [ - 128.9, - 447.77, - 502.77, - 589.66 - ], - "text": "The zero-state component of y(t) resulting from the input x(t), assuming that all initial\nconditions are zero, that is, y(0−) = vC(0−) = 0, will be obtained later in Ex. 2.9. In this\nexample we shall find the zero-input reponse y0(t). For this purpose, we need two initial\nconditions, y0(0) and ˙y0(0). These conditions can be derived from the given initial conditions,\ny(0−) = 0 and vC(0−) = 5, as follows. Recall that y0(t) is the loop current when the input\nterminals are shorted so that the input x(t) = 0 (zero-input), as depicted in Fig. 2.2b. We now\ncompute y0(0) and ˙y0(0), the values of the loop current and its derivative at t = 0, from the\ninitial values of the inductor current and the capacitor voltage. Remember that the inductor\ncurrent cannot change instantaneously in the absence of an impulsive voltage. Similarly,\nthe capacitor voltage cannot change instantaneously in the absence of an impulsive current.\nTherefore, when the input terminals are shorted at t = 0, the inductor current is still zero and\nthe capacitor voltage is still 5 volts. Thus,", - "type": "text" - }, - { - "block_id": "p179-b8", - "global_id": 4964, - "bbox": [ - 297.0, - 601.19, - 334.67, - 612.34 - ], - "text": "y0(0) = 0", - "type": "text" - } - ] - }, - { - "page_num": 180, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p180-b0", - "global_id": 4965, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "160\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p180-b1", - "global_id": 4966, - "bbox": [ - 327.39, - 185.15, - 337.04, - 193.15 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p180-b2", - "global_id": 4967, - "bbox": [ - 306.44, - 93.29, - 318.21, - 101.29 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p180-b3", - "global_id": 4968, - "bbox": [ - 324.96, - 138.15, - 339.07, - 147.76 - ], - "text": "y0(t)", - "type": "text" - }, - { - "block_id": "p180-b4", - "global_id": 4969, - "bbox": [ - 355.7, - 93.0, - 367.92, - 101.29 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p180-b5", - "global_id": 4970, - "bbox": [ - 370.77, - 146.86, - 380.97, - 160.05 - ], - "text": "F\n1\n2", - "type": "text" - }, - { - "block_id": "p180-b6", - "global_id": 4971, - "bbox": [ - 399.25, - 138.17, - 414.71, - 147.71 - ], - "text": "vC(t)", - "type": "text" - }, - { - "block_id": "p180-b9", - "global_id": 4972, - "bbox": [ - 174.83, - 185.15, - 183.71, - 193.15 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p180-b10", - "global_id": 4973, - "bbox": [ - 153.49, - 92.62, - 214.98, - 100.91 - ], - "text": "3 \t\n1 H", - "type": "text" - }, - { - "block_id": "p180-b11", - "global_id": 4974, - "bbox": [ - 217.83, - 146.74, - 228.02, - 159.93 - ], - "text": "F\n1\n2", - "type": "text" - }, - { - "block_id": "p180-b12", - "global_id": 4975, - "bbox": [ - 94.42, - 137.39, - 262.02, - 148.82 - ], - "text": "vC(t)\ny(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p180-b15", - "global_id": 4976, - "bbox": [ - 94.2, - 199.84, - 211.0, - 209.08 - ], - "text": "Figure 2.1 Circuits for Ex. 2.4.", - "type": "text" - }, - { - "block_id": "p180-b16", - "global_id": 4977, - "bbox": [ - 103.16, - 236.33, - 476.99, - 259.43 - ], - "text": "To determine ˙y0(0), we use the loop equation for the circuit in Fig. 2.2b. Because the voltage\nacross the inductor is L(dy0/dt) or ˙y0(t), this equation can be written as follows:", - "type": "text" - }, - { - "block_id": "p180-b17", - "global_id": 4978, - "bbox": [ - 239.46, - 273.2, - 340.72, - 284.35 - ], - "text": "˙y0(t) + 3y0(t) + vC(t) = 0", - "type": "text" - }, - { - "block_id": "p180-b18", - "global_id": 4979, - "bbox": [ - 103.17, - 298.1, - 197.67, - 308.48 - ], - "text": "Setting t = 0, we obtain", - "type": "text" - }, - { - "block_id": "p180-b19", - "global_id": 4980, - "bbox": [ - 236.41, - 310.06, - 343.77, - 321.21 - ], - "text": "˙y0(0) + 3y0(0) + vC(0) = 0", - "type": "text" - }, - { - "block_id": "p180-b20", - "global_id": 4981, - "bbox": [ - 103.17, - 328.99, - 277.67, - 340.14 - ], - "text": "But y0(0) = 0 and vC(0) = 5. Consequently,", - "type": "text" - }, - { - "block_id": "p180-b21", - "global_id": 4982, - "bbox": [ - 267.37, - 353.89, - 312.81, - 365.04 - ], - "text": "˙y0(0) = −5", - "type": "text" - }, - { - "block_id": "p180-b22", - "global_id": 4983, - "bbox": [ - 103.17, - 379.22, - 275.26, - 389.18 - ], - "text": "Therefore, the desired initial conditions are", - "type": "text" - }, - { - "block_id": "p180-b23", - "global_id": 4984, - "bbox": [ - 221.42, - 403.71, - 358.76, - 414.85 - ], - "text": "y0(0) = 0\nand\n˙y0(0) = −5", - "type": "text" - }, - { - "block_id": "p180-b24", - "global_id": 4985, - "bbox": [ - 103.16, - 428.61, - 477.02, - 463.67 - ], - "text": "Thus, the problem reduces to finding y0(t), the zero-input component of y(t) of the system\nspecified by the equation (D2 + 3D + 2)y(t) = Dx(t), when the initial conditions are y0(0) = 0\nand ˙y0(0) = −5. We have already solved this problem in Ex. 2.1a, where we found", - "type": "text" - }, - { - "block_id": "p180-b25", - "global_id": 4986, - "bbox": [ - 226.56, - 473.31, - 353.62, - 488.58 - ], - "text": "y0(t) = −5e−t + 5e−2t\nt ≥0", - "type": "text" - }, - { - "block_id": "p180-b26", - "global_id": 4987, - "bbox": [ - 103.17, - 502.34, - 331.04, - 512.71 - ], - "text": "This is the zero-input component of the loop current y(t).", - "type": "text" - }, - { - "block_id": "p180-b27", - "global_id": 4988, - "bbox": [ - 103.17, - 511.08, - 477.01, - 572.49 - ], - "text": "It is interesting to find the initial conditions at t = 0−and 0+ for the total response y(t). Let\nus compare y(0−) and ˙y(0−) with y(0+) and ˙y(0+). The two pairs can be compared by writing\nthe loop equation for the circuit in Fig. 2.2a at t = 0−and t = 0+. The only difference between\nthe two situations is that at t = 0−, the input x(t) = 0, whereas at t = 0+, the input x(t) = 10\n[because x(t) = 10e−3t]. Hence, the two loop equations are", - "type": "text" - }, - { - "block_id": "p180-b28", - "global_id": 4989, - "bbox": [ - 229.0, - 585.3, - 346.2, - 598.1 - ], - "text": "˙y(0−) + 3y(0−) + vC(0−) = 0", - "type": "text" - }, - { - "block_id": "p180-b29", - "global_id": 4990, - "bbox": [ - 229.0, - 600.24, - 351.17, - 613.05 - ], - "text": "˙y(0+) + 3y(0+) + vC(0+) = 10", - "type": "text" - } - ] - }, - { - "page_num": 181, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p181-b0", - "global_id": 4991, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "2.2\nSystem Response to Internal Conditions: The Zero-Input Response\n161", - "type": "text" - }, - { - "block_id": "p181-b1", - "global_id": 4992, - "bbox": [ - 128.9, - 85.49, - 502.76, - 132.96 - ], - "text": "The loop current y(0+) = y(0−) = 0 because it cannot change instantaneously in the absence\nof impulsive voltage. The same is true of the capacitor voltage. Hence, vC(0+) = vC(0−) = 5.\nSubstituting these values in the foregoing equations, we obtain ˙y(0−) = −5 and ˙y(0+) = 5.\nThus,", - "type": "text" - }, - { - "block_id": "p181-b2", - "global_id": 4993, - "bbox": [ - 209.91, - 132.82, - 502.75, - 144.92 - ], - "text": "y(0−) = 0, ˙y(0−) = −5\nand\ny(0+) = 0, ˙y(0+) = 5\n(2.10)", - "type": "text" - }, - { - "block_id": "p181-b3", - "global_id": 4994, - "bbox": [ - 133.57, - 190.02, - 415.11, - 201.97 - ], - "text": "DRILL 2.3\nZero-Input Response of an RC Circuit", - "type": "text" - }, - { - "block_id": "p181-b4", - "global_id": 4995, - "bbox": [ - 133.57, - 210.68, - 510.15, - 244.97 - ], - "text": "In the circuit in Fig. 2.2a, the inductance L = 0 and the initial capacitor voltage vC(0) = 30\nvolts. Show that the zero-input component of the loop current is given by y0(t) = −10e−2t/3 for\nt ≥0.", - "type": "text" - }, - { - "block_id": "p181-b5", - "global_id": 4996, - "bbox": [ - 127.59, - 273.96, - 516.15, - 359.75 - ], - "text": "INDEPENDENCE OF THE ZERO-INPUT AND ZERO-STATE RESPONSES\nIn Ex. 2.4 we computed the zero-input component without using the input x(t). The zero-state\nresponse can be computed from the knowledge of the input x(t) alone; the initial conditions\nare assumed to be zero (system in zero state). The two components of the system response (the\nzero-input and zero-state responses) are independent of each other. The two worlds of zero-input\nresponse and zero-state response coexist side by side, neither one knowing or caring what the\nother is doing. For each component, the other is totally irrelevant.", - "type": "text" - }, - { - "block_id": "p181-b6", - "global_id": 4997, - "bbox": [ - 127.59, - 374.57, - 516.15, - 450.5 - ], - "text": "ROLE OF AUXILIARY CONDITIONS IN SOLUTION OF\nDIFFERENTIAL EQUATIONS\nThe solution of a differential equation requires additional pieces of information (the auxiliary\nconditions). Why? We now show heuristically why a differential equation does not, in general,\nhave a unique solution unless some additional constraints (or conditions) on the solution are\nknown.", - "type": "text" - }, - { - "block_id": "p181-b7", - "global_id": 4998, - "bbox": [ - 127.59, - 452.07, - 516.15, - 558.09 - ], - "text": "Differentiation operation is not invertible unless one piece of information about y(t) is given.\nTo get back y(t) from dy/dt, we must know one piece of information, such as y(0). Thus,\ndifferentiation is an irreversible (noninvertible) operation during which certain information is\nlost. To invert this operation, one piece of information about y(t) must be provided to restore\nthe original y(t). Using a similar argument, we can show that, given d2y/dt2, we can determine\ny(t) uniquely only if two additional pieces of information (constraints) about y(t) are given.\nIn general, to determine y(t) uniquely from its Nth derivative, we need N additional pieces of\ninformation (constraints) about y(t). These constraints are also called auxiliary conditions. When\nthese conditions are given at t = 0, they are called initial conditions.", - "type": "text" - }, - { - "block_id": "p181-b8", - "global_id": 4999, - "bbox": [ - 127.59, - 582.83, - 462.16, - 594.78 - ], - "text": "2.2-1 Some Insights into the Zero-Input Behavior of a System", - "type": "text" - }, - { - "block_id": "p181-b9", - "global_id": 5000, - "bbox": [ - 127.59, - 600.91, - 516.16, - 634.79 - ], - "text": "By definition, the zero-input response is the system response to its internal conditions, assuming\nthat its input is zero. Understanding this phenomenon provides interesting insight into system\nbehavior. If a system is disturbed momentarily from its rest position and if the disturbance is then", - "type": "text" - } - ] - }, - { - "page_num": 182, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p182-b0", - "global_id": 5001, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "162\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p182-b1", - "global_id": 5002, - "bbox": [ - 101.84, - 85.82, - 490.42, - 191.32 - ], - "text": "removed, the system will not come back to rest instantaneously. In general, it will come back to\nrest over a period of time and only through a special type of motion that is characteristic of the\nsystem.† For example, if we press on an automobile fender momentarily and then release it at\nt = 0, there is no external force on the automobile for t > 0.‡ The auto body will eventually come\nback to its rest (equilibrium) position, but not through any arbitrary motion. It must do so by using\nonly a form of response that is sustainable by the system on its own without any external source,\nsince the input is zero. Only characteristic modes satisfy this condition. The system uses a proper\ncombination of characteristic modes to come back to the rest position while satisfying appropriate\nboundary (or initial) conditions.", - "type": "text" - }, - { - "block_id": "p182-b2", - "global_id": 5003, - "bbox": [ - 101.84, - 193.41, - 490.43, - 299.02 - ], - "text": "If the shock absorbers of the automobile are in good condition (high damping coefficient), the\ncharacteristic modes will be monotonically decaying exponentials, and the auto body will come to\nrest rapidly without oscillation. In contrast, for poor shock absorbers (low damping coefficients),\nthe characteristic modes will be exponentially decaying sinusoids, and the body will come to rest\nthrough oscillatory motion. When a series RC circuit with an initial charge on the capacitor is\nshorted, the capacitor will start to discharge exponentially through the resistor. This response of\nthe RC circuit is caused entirely by its internal conditions and is sustained by this system without\nthe aid of any external input. The exponential current waveform is therefore the characteristic\nmode of the RC circuit.", - "type": "text" - }, - { - "block_id": "p182-b3", - "global_id": 5004, - "bbox": [ - 101.85, - 300.92, - 490.38, - 334.88 - ], - "text": "Mathematically we know that any combination of characteristic modes can be sustained by\nthe system alone without requiring an external input. This fact can be readily verified for the series\nRL circuit shown in Fig. 2.2. The loop equation for this system is", - "type": "text" - }, - { - "block_id": "p182-b4", - "global_id": 5005, - "bbox": [ - 260.13, - 344.43, - 332.11, - 354.81 - ], - "text": "(D + 2)y(t) = x(t)", - "type": "text" - }, - { - "block_id": "p182-b5", - "global_id": 5006, - "bbox": [ - 101.84, - 363.13, - 490.39, - 398.65 - ], - "text": "It has a single characteristic root λ = −2, and the characteristic mode is e−2t. We now verify that a\nloop current y(t) = ce−2t can be sustained through this circuit without any input voltage. The input\nvoltage x(t) required to drive a loop current y(t) = ce−2t is given by", - "type": "text" - }, - { - "block_id": "p182-b6", - "global_id": 5007, - "bbox": [ - 239.71, - 406.23, - 292.95, - 423.49 - ], - "text": "x(t) = Ldy(t)", - "type": "text" - }, - { - "block_id": "p182-b7", - "global_id": 5008, - "bbox": [ - 279.1, - 413.22, - 325.91, - 430.57 - ], - "text": "dt\n+ Ry(t)", - "type": "text" - }, - { - "block_id": "p182-b8", - "global_id": 5009, - "bbox": [ - 256.59, - 431.35, - 273.92, - 448.3 - ], - "text": "= d", - "type": "text" - }, - { - "block_id": "p182-b9", - "global_id": 5010, - "bbox": [ - 267.6, - 436.3, - 340.03, - 455.37 - ], - "text": "dt(ce−2t) + 2ce−2t", - "type": "text" - }, - { - "block_id": "p182-b10", - "global_id": 5011, - "bbox": [ - 256.59, - 453.95, - 352.53, - 468.44 - ], - "text": "= −2ce−2t + 2ce−2t = 0", - "type": "text" - }, - { - "block_id": "p182-b11", - "global_id": 5012, - "bbox": [ - 104.06, - 529.53, - 268.88, - 542.53 - ], - "text": "y0(t)\n2 \t\nx(t)", - "type": "text" - }, - { - "block_id": "p182-b12", - "global_id": 5013, - "bbox": [ - 184.25, - 487.4, - 196.02, - 495.4 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p182-b13", - "global_id": 5014, - "bbox": [ - 279.8, - 563.76, - 430.78, - 573.0 - ], - "text": "Figure 2.2 Modes always get a free ride.", - "type": "text" - }, - { - "block_id": "p182-b14", - "global_id": 5015, - "bbox": [ - 101.84, - 599.36, - 490.38, - 633.8 - ], - "text": "† This assumes that the system will eventually come back to its original rest (or equilibrium) position.\n‡ We ignore the force of gravity, which merely causes a constant displacement of the auto body without\naffecting the other motion.", - "type": "text" - } - ] - }, - { - "page_num": 183, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p183-b0", - "global_id": 5016, - "bbox": [ - 327.46, - 62.29, - 516.13, - 71.98 - ], - "text": "2.3\nThe Unit Impulse Response h(t)\n163", - "type": "text" - }, - { - "block_id": "p183-b1", - "global_id": 5017, - "bbox": [ - 127.59, - 83.78, - 516.12, - 109.31 - ], - "text": "Clearly, the loop current y(t) = ce−2t is sustained by the RL circuit on its own, without the necessity\nof an external input.", - "type": "text" - }, - { - "block_id": "p183-b2", - "global_id": 5018, - "bbox": [ - 127.89, - 123.79, - 304.54, - 135.91 - ], - "text": "THE RESONANCE PHENOMENON", - "type": "text" - }, - { - "block_id": "p183-b3", - "global_id": 5019, - "bbox": [ - 127.59, - 139.94, - 516.16, - 257.5 - ], - "text": "We have seen that any signal consisting of a system’s characteristic mode is sustained by the\nsystem on its own; the system offers no obstacle to such signals. Imagine what would happen if\nwe were to drive the system with an external input that is one of its characteristic modes. This\nwould be like pouring gasoline on a fire in a dry forest or hiring a child to eat ice cream. A child\nwould gladly do the job without pay. Think what would happen if he were paid by the amount\nof ice cream he ate! He would work overtime. He would work day and night, until he became\nsick. The same thing happens with a system driven by an input of the form of characteristic mode.\nThe system response grows without limit, until it burns out.† We call this behavior the resonance\nphenomenon. An intelligent discussion of this important phenomenon requires an understanding\nof the zero-state response; for this reason we postpone this topic until Sec. 2.6-7.", - "type": "text" - }, - { - "block_id": "p183-b4", - "global_id": 5020, - "bbox": [ - 127.94, - 285.25, - 376.53, - 300.2 - ], - "text": "2.3 THE UNIT IMPULSE RESPONSE h(t)", - "type": "text" - }, - { - "block_id": "p183-b5", - "global_id": 5021, - "bbox": [ - 127.59, - 305.78, - 516.15, - 399.84 - ], - "text": "In Ch. 1 we explained how a system response to an input x(t) may be found by breaking this\ninput into narrow rectangular pulses, as illustrated earlier in Fig. 1.27a, and then summing the\nsystem response to all the components. The rectangular pulses become impulses in the limit as\ntheir widths approach zero. Therefore, the system response is the sum of its responses to various\nimpulse components. This discussion shows that if we know the system response to an impulse\ninput, we can determine the system response to an arbitrary input x(t). We now discuss a method\nof determining h(t), the unit impulse response of an LTIC system described by the Nth-order\ndifferential equation [Eq. (2.1)]", - "type": "text" - }, - { - "block_id": "p183-b6", - "global_id": 5022, - "bbox": [ - 190.27, - 408.89, - 215.89, - 420.17 - ], - "text": "dNy(t)", - "type": "text" - }, - { - "block_id": "p183-b7", - "global_id": 5023, - "bbox": [ - 196.33, - 416.88, - 236.42, - 434.23 - ], - "text": "dtN\n+ a1", - "type": "text" - }, - { - "block_id": "p183-b8", - "global_id": 5024, - "bbox": [ - 238.12, - 408.67, - 272.66, - 420.17 - ], - "text": "dN−1y(t)", - "type": "text" - }, - { - "block_id": "p183-b9", - "global_id": 5025, - "bbox": [ - 244.18, - 416.88, - 327.63, - 434.23 - ], - "text": "dtN−1\n+ · · · + aN−1", - "type": "text" - }, - { - "block_id": "p183-b10", - "global_id": 5026, - "bbox": [ - 329.35, - 409.9, - 349.13, - 420.17 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p183-b11", - "global_id": 5027, - "bbox": [ - 335.27, - 416.88, - 386.55, - 434.23 - ], - "text": "dt\n+ aNy(t)", - "type": "text" - }, - { - "block_id": "p183-b12", - "global_id": 5028, - "bbox": [ - 201.09, - 443.41, - 232.19, - 454.49 - ], - "text": "= bN−M", - "type": "text" - }, - { - "block_id": "p183-b13", - "global_id": 5029, - "bbox": [ - 234.17, - 435.42, - 260.85, - 446.7 - ], - "text": "dMx(t)", - "type": "text" - }, - { - "block_id": "p183-b14", - "global_id": 5030, - "bbox": [ - 240.26, - 443.41, - 303.39, - 460.76 - ], - "text": "dtM\n+ bN−M+1", - "type": "text" - }, - { - "block_id": "p183-b15", - "global_id": 5031, - "bbox": [ - 305.11, - 435.2, - 340.7, - 446.7 - ], - "text": "dM−1x(t)", - "type": "text" - }, - { - "block_id": "p183-b16", - "global_id": 5032, - "bbox": [ - 311.18, - 443.41, - 395.67, - 460.76 - ], - "text": "dtM−1\n+ · · · + bN−1", - "type": "text" - }, - { - "block_id": "p183-b17", - "global_id": 5033, - "bbox": [ - 397.39, - 436.43, - 417.19, - 446.7 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p183-b18", - "global_id": 5034, - "bbox": [ - 403.33, - 443.41, - 454.64, - 460.76 - ], - "text": "dt\n+ bNx(t)", - "type": "text" - }, - { - "block_id": "p183-b19", - "global_id": 5035, - "bbox": [ - 127.59, - 467.3, - 516.14, - 489.63 - ], - "text": "Recall that noise considerations restrict practical systems to M ≤N. Under this constraint, the\nmost general case is M = N. Therefore, Eq. (2.1) can be expressed as", - "type": "text" - }, - { - "block_id": "p183-b20", - "global_id": 5036, - "bbox": [ - 139.98, - 496.01, - 516.13, - 511.27 - ], - "text": "(DN + a1DN−1 + · · · + aN−1D + aN)y(t) = (b0DN + b1DN−1 + · · · + bN−1D + bN)x(t)\n(2.11)", - "type": "text" - }, - { - "block_id": "p183-b21", - "global_id": 5037, - "bbox": [ - 127.59, - 520.99, - 516.16, - 603.1 - ], - "text": "Before deriving the general expression for the unit impulse response h(t), it is illuminating\nto understand qualitatively the nature of h(t). The impulse response h(t) is the system response\nto an impulse input δ(t) applied at t = 0 with all the initial conditions zero at t = 0−. An impulse\ninput δ(t) is like lightning, which strikes instantaneously and then vanishes. But in its wake, in\nthat single moment, objects that have been struck are rearranged. Similarly, an impulse input\nδ(t) appears momentarily at t = 0, and then it is gone forever. But in that moment it generates\nenergy storages; that is, it creates nonzero initial conditions instantaneously within the system at", - "type": "text" - }, - { - "block_id": "p183-b22", - "global_id": 5038, - "bbox": [ - 127.59, - 621.19, - 501.41, - 633.41 - ], - "text": "† In practice, the system in resonance is more likely to go in saturation because of high amplitude levels.", - "type": "text" - } - ] - }, - { - "page_num": 184, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p184-b0", - "global_id": 5039, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "164\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p184-b1", - "global_id": 5040, - "bbox": [ - 101.84, - 85.07, - 490.4, - 132.53 - ], - "text": "t = 0+. Although the impulse input δ(t) vanishes for t > 0 so that the system has no input after the\nimpulse has been applied, the system will still have a response generated by these newly created\ninitial conditions. The impulse response h(t), therefore, must consist of the system’s characteristic\nmodes for t ≥0+. As a result,", - "type": "text" - }, - { - "block_id": "p184-b2", - "global_id": 5041, - "bbox": [ - 208.62, - 142.89, - 383.12, - 154.99 - ], - "text": "h(t) = characteristic mode terms\nt ≥0+", - "type": "text" - }, - { - "block_id": "p184-b3", - "global_id": 5042, - "bbox": [ - 101.84, - 167.06, - 490.38, - 189.4 - ], - "text": "This response is valid for t > 0. But what happens at t = 0? At a single moment t = 0, there can at\nmost be an impulse,† so the form of the complete response h(t) is", - "type": "text" - }, - { - "block_id": "p184-b4", - "global_id": 5043, - "bbox": [ - 193.52, - 201.47, - 490.38, - 212.61 - ], - "text": "h(t) = A0δ(t) + characteristic mode terms\nt ≥0\n(2.12)", - "type": "text" - }, - { - "block_id": "p184-b5", - "global_id": 5044, - "bbox": [ - 101.85, - 223.92, - 482.56, - 234.3 - ], - "text": "because h(t) is the unit impulse response. Setting x(t) = δ(t) and y(t) = h(t) in Eq. (2.11) yields", - "type": "text" - }, - { - "block_id": "p184-b6", - "global_id": 5045, - "bbox": [ - 120.83, - 242.25, - 461.46, - 257.52 - ], - "text": "(DN + a1DN−1 + · · · + aN−1D + aN)h(t) = (b0DN + b1DN−1 + · · · + bN−1D + bN)δ(t)", - "type": "text" - }, - { - "block_id": "p184-b7", - "global_id": 5046, - "bbox": [ - 101.84, - 268.82, - 490.4, - 316.56 - ], - "text": "In this equation we substitute h(t) from Eq. (2.12) and compare the coefficients of similar\nimpulsive terms on both sides. The highest order of the derivative of impulse on both sides is\nN, with its coefficient value as A0 on the left-hand side and b0 on the right-hand side. The two\nvalues must be matched. Therefore, A0 = b0 and", - "type": "text" - }, - { - "block_id": "p184-b8", - "global_id": 5047, - "bbox": [ - 224.17, - 327.13, - 490.39, - 338.28 - ], - "text": "h(t) = b0δ(t) + characteristic modes\n(2.13)", - "type": "text" - }, - { - "block_id": "p184-b9", - "global_id": 5048, - "bbox": [ - 101.84, - 349.59, - 490.4, - 383.87 - ], - "text": "In Eq. (2.11), if M < N, b0 = 0. Hence, the impulse term b0δ(t) exists only if M = N. The unknown\ncoefficients of the N characteristic modes in h(t) in Eq. (2.13) can be determined by using the\ntechnique of impulse matching, as explained in the following example.", - "type": "text" - }, - { - "block_id": "p184-b10", - "global_id": 5049, - "bbox": [ - 76.77, - 419.11, - 398.75, - 431.07 - ], - "text": "EXAMPLE 2.5\nImpulse Response via Impulse Matching", - "type": "text" - }, - { - "block_id": "p184-b11", - "global_id": 5050, - "bbox": [ - 103.16, - 447.31, - 327.25, - 457.69 - ], - "text": "Find the impulse response h(t) for a system specified by", - "type": "text" - }, - { - "block_id": "p184-b12", - "global_id": 5051, - "bbox": [ - 225.35, - 465.12, - 477.01, - 479.61 - ], - "text": "(D2 + 5D + 6)y(t) = (D + 1)x(t)\n(2.14)", - "type": "text" - }, - { - "block_id": "p184-b13", - "global_id": 5052, - "bbox": [ - 103.16, - 512.07, - 477.02, - 534.41 - ], - "text": "In this case, b0 = 0. Hence, h(t) consists of only the characteristic modes. The characteristic\npolynomial is λ2 + 5λ + 6 = (λ + 2)(λ + 3). The roots are −2 and −3. Hence, the impulse", - "type": "text" - }, - { - "block_id": "p184-b14", - "global_id": 5053, - "bbox": [ - 101.84, - 566.4, - 490.39, - 633.41 - ], - "text": "† It might be possible for the derivatives of δ(t) to appear at the origin. However, if M ≤N, it is impossible for\nh(t) to have any derivatives of δ(t). This conclusion follows from Eq. (2.11) with x(t) = δ(t) and y(t) = h(t).\nThe coefficients of the impulse and all its derivatives must be matched on both sides of this equation. If h(t)\ncontains δ(1)(t), the first derivative of δ(t), the left-hand side of Eq. (2.11) will contain a term δ(N+1)(t). But\nthe highest-order derivative term on the right-hand side is δ(N)(t). Therefore, the two sides cannot match.\nSimilar arguments can be made against the presence of the impulse’s higher-order derivatives in h(t).", - "type": "text" - } - ] - }, - { - "page_num": 185, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p185-b0", - "global_id": 5054, - "bbox": [ - 327.46, - 62.29, - 516.13, - 71.98 - ], - "text": "2.3\nThe Unit Impulse Response h(t)\n165", - "type": "text" - }, - { - "block_id": "p185-b1", - "global_id": 5055, - "bbox": [ - 128.9, - 85.88, - 190.76, - 96.26 - ], - "text": "response h(t) is", - "type": "text" - }, - { - "block_id": "p185-b2", - "global_id": 5056, - "bbox": [ - 259.4, - 93.72, - 502.75, - 109.61 - ], - "text": "h(t) = (c1 e−2t + c2 e−3t)u(t)\n(2.15)", - "type": "text" - }, - { - "block_id": "p185-b3", - "global_id": 5057, - "bbox": [ - 128.91, - 116.77, - 360.34, - 127.15 - ], - "text": "Letting x(t) = δ(t) and y(t) = h(t) in Eq. (2.14), we obtain", - "type": "text" - }, - { - "block_id": "p185-b4", - "global_id": 5058, - "bbox": [ - 250.44, - 136.63, - 502.75, - 149.06 - ], - "text": "¨h(t) + 5˙h(t) + 6h(t) = ˙δ(t) + δ(t)\n(2.16)", - "type": "text" - }, - { - "block_id": "p185-b5", - "global_id": 5059, - "bbox": [ - 128.9, - 158.55, - 502.75, - 195.66 - ], - "text": "Recall that initial conditions h(0−) and ˙h(0−) are both zero. But the application of an impulse\nat t = 0 creates new initial conditions at t = 0+. Let h(0+) = K1 and ˙h(0+) = K2. These\njump discontinuities in h(t) and ˙h(t) at t = 0 result in impulse terms ˙h(0) = K1δ(t) and", - "type": "text" - }, - { - "block_id": "p185-b6", - "global_id": 5060, - "bbox": [ - 128.9, - 194.41, - 502.74, - 218.8 - ], - "text": "¨h(0) = K1 ˙δ(t) + K2δ(t) on the left-hand side. Matching the coefficients of impulse terms on\nboth sides of Eq. (2.16) yields", - "type": "text" - }, - { - "block_id": "p185-b7", - "global_id": 5061, - "bbox": [ - 199.97, - 230.34, - 431.7, - 241.8 - ], - "text": "5K1 + K2 = 1,\nK1 = 1\n\r⇒\nK1 = 1,K2 = −4", - "type": "text" - }, - { - "block_id": "p185-b8", - "global_id": 5062, - "bbox": [ - 128.91, - 250.21, - 502.75, - 287.63 - ], - "text": "We now use these values h(0+) = K1 = 1 and ˙h(0+) = K2 = −4 in Eq. (2.15) to find c1 and\nc2. Setting t = 0+ in Eq. (2.15), we obtain c1 + c2 = 1. Also setting t = 0+ in ˙h(t), we obtain\n−2c1 −3c1 = −4. These two simultaneous equations yield c1 = −1 and c2 = 2. Therefore,", - "type": "text" - }, - { - "block_id": "p185-b9", - "global_id": 5063, - "bbox": [ - 263.1, - 293.97, - 368.57, - 308.46 - ], - "text": "h(t) = (−e−2t + 2e−3t)u(t)", - "type": "text" - }, - { - "block_id": "p185-b10", - "global_id": 5064, - "bbox": [ - 127.59, - 345.92, - 516.13, - 367.84 - ], - "text": "Although the method used in this example is relatively simple, we can simplify it still further\nby using a modified version of impulse matching.", - "type": "text" - }, - { - "block_id": "p185-b11", - "global_id": 5065, - "bbox": [ - 127.59, - 383.99, - 516.17, - 445.97 - ], - "text": "SIMPLIFIED IMPULSE MATCHING METHOD\nThe alternate technique we present now allows us to reduce the procedure to a simple routine to\ndetermine h(t). To avoid the needless distraction, the proof for this procedure is placed in Sec. 2.8.\nThere, we show that for an LTIC system specified by Eq. (2.11), the unit impulse response h(t) is\ngiven by", - "type": "text" - }, - { - "block_id": "p185-b12", - "global_id": 5066, - "bbox": [ - 259.97, - 451.83, - 516.13, - 462.98 - ], - "text": "h(t) = b0δ(t) + [P(D)yn(t)]u(t)\n(2.17)", - "type": "text" - }, - { - "block_id": "p185-b13", - "global_id": 5067, - "bbox": [ - 127.59, - 473.6, - 516.15, - 495.94 - ], - "text": "where yn(t) is a linear combination of the characteristic modes of the system subject to the\nfollowing initial conditions:", - "type": "text" - }, - { - "block_id": "p185-b14", - "global_id": 5068, - "bbox": [ - 179.86, - 508.6, - 326.47, - 521.41 - ], - "text": "yn(0) = ˙yn(0) = ¨yn(0) = · · · = y(N−2)", - "type": "text" - }, - { - "block_id": "p185-b15", - "global_id": 5069, - "bbox": [ - 307.28, - 508.6, - 434.09, - 522.59 - ], - "text": "n\n(0) = 0\nand\ny(N−1)", - "type": "text" - }, - { - "block_id": "p185-b16", - "global_id": 5070, - "bbox": [ - 414.9, - 510.33, - 516.13, - 522.59 - ], - "text": "n\n(0) = 1\n(2.18)", - "type": "text" - }, - { - "block_id": "p185-b17", - "global_id": 5071, - "bbox": [ - 127.59, - 533.87, - 166.81, - 545.46 - ], - "text": "where y(k)", - "type": "text" - }, - { - "block_id": "p185-b18", - "global_id": 5072, - "bbox": [ - 127.59, - 535.09, - 516.12, - 557.42 - ], - "text": "n (0) is the value of the kth derivative of yn(t) at t = 0. We can express this set of conditions\nfor various values of N (the system order) as follows:", - "type": "text" - }, - { - "block_id": "p185-b19", - "global_id": 5073, - "bbox": [ - 250.57, - 571.82, - 393.15, - 612.77 - ], - "text": "N = 1 : yn(0) = 1\nN = 2 : yn(0) = 0, ˙yn(0) = 1\nN = 3 : yn(0) = ˙yn(0) = 0, ¨yn(0) = 1", - "type": "text" - }, - { - "block_id": "p185-b20", - "global_id": 5074, - "bbox": [ - 127.6, - 626.88, - 168.27, - 636.84 - ], - "text": "and so on.", - "type": "text" - } - ] - }, - { - "page_num": 186, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p186-b0", - "global_id": 5075, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "166\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p186-b1", - "global_id": 5076, - "bbox": [ - 101.84, - 85.46, - 490.38, - 108.87 - ], - "text": "As stated earlier, if the order of P(D) is less than the order of Q(D), that is, if M < N, then\nb0 = 0, and the impulse term b0δ(t) in h(t) is zero.", - "type": "text" - }, - { - "block_id": "p186-b2", - "global_id": 5077, - "bbox": [ - 76.77, - 135.8, - 459.48, - 147.75 - ], - "text": "EXAMPLE 2.6\nImpulse Response via Simplified Impulse Matching", - "type": "text" - }, - { - "block_id": "p186-b3", - "global_id": 5078, - "bbox": [ - 103.16, - 163.88, - 420.45, - 174.26 - ], - "text": "Determine the unit impulse response h(t) for a system specified by the equation", - "type": "text" - }, - { - "block_id": "p186-b5", - "global_id": 5079, - "bbox": [ - 240.15, - 181.46, - 290.21, - 195.95 - ], - "text": "D2 + 3D + 2", - "type": "text" - }, - { - "block_id": "p186-b7", - "global_id": 5080, - "bbox": [ - 295.35, - 185.57, - 477.01, - 195.95 - ], - "text": "y(t) = Dx(t)\n(2.19)", - "type": "text" - }, - { - "block_id": "p186-b8", - "global_id": 5081, - "bbox": [ - 103.16, - 228.18, - 402.33, - 238.55 - ], - "text": "This is a second-order system (N = 2) having the characteristic polynomial", - "type": "text" - }, - { - "block_id": "p186-b10", - "global_id": 5082, - "bbox": [ - 232.13, - 245.75, - 278.7, - 260.24 - ], - "text": "λ2 + 3λ + 2", - "type": "text" - }, - { - "block_id": "p186-b12", - "global_id": 5083, - "bbox": [ - 284.78, - 249.87, - 352.05, - 260.24 - ], - "text": "= (λ + 1)(λ + 2)", - "type": "text" - }, - { - "block_id": "p186-b13", - "global_id": 5084, - "bbox": [ - 103.16, - 271.56, - 397.9, - 281.93 - ], - "text": "The characteristic roots of this system are λ = −1 and λ = −2. Therefore,", - "type": "text" - }, - { - "block_id": "p186-b14", - "global_id": 5085, - "bbox": [ - 246.75, - 289.12, - 477.01, - 304.39 - ], - "text": "yn(t) = c1e−t + c2e−2t\n(2.20)", - "type": "text" - }, - { - "block_id": "p186-b15", - "global_id": 5086, - "bbox": [ - 103.17, - 315.34, - 252.89, - 325.3 - ], - "text": "Differentiation of this equation yields", - "type": "text" - }, - { - "block_id": "p186-b16", - "global_id": 5087, - "bbox": [ - 240.37, - 332.5, - 477.01, - 347.77 - ], - "text": "˙yn(t) = −c1e−t −2c2e−2t\n(2.21)", - "type": "text" - }, - { - "block_id": "p186-b17", - "global_id": 5088, - "bbox": [ - 103.17, - 358.72, - 267.49, - 368.68 - ], - "text": "The initial conditions are [see Eq. (2.18)]", - "type": "text" - }, - { - "block_id": "p186-b18", - "global_id": 5089, - "bbox": [ - 225.3, - 379.99, - 354.88, - 391.07 - ], - "text": "˙yn(0) = 1\nand\nyn(0) = 0", - "type": "text" - }, - { - "block_id": "p186-b19", - "global_id": 5090, - "bbox": [ - 103.17, - 401.68, - 477.03, - 424.01 - ], - "text": "Setting t = 0 in Eqs. (2.20) and (2.21), and substituting the initial conditions just given, we\nobtain", - "type": "text" - }, - { - "block_id": "p186-b20", - "global_id": 5091, - "bbox": [ - 261.45, - 435.32, - 318.22, - 461.72 - ], - "text": "0 = c1 + c2\n1 = −c1 −2c2", - "type": "text" - }, - { - "block_id": "p186-b21", - "global_id": 5092, - "bbox": [ - 103.16, - 472.36, - 310.06, - 482.33 - ], - "text": "Solution of these two simultaneous equations yields", - "type": "text" - }, - { - "block_id": "p186-b22", - "global_id": 5093, - "bbox": [ - 233.83, - 493.64, - 346.35, - 505.1 - ], - "text": "c1 = 1\nand\nc2 = −1", - "type": "text" - }, - { - "block_id": "p186-b23", - "global_id": 5094, - "bbox": [ - 103.16, - 515.73, - 144.91, - 525.7 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p186-b24", - "global_id": 5095, - "bbox": [ - 255.16, - 523.17, - 324.38, - 538.36 - ], - "text": "yn(t) = e−t −e−2t", - "type": "text" - }, - { - "block_id": "p186-b25", - "global_id": 5096, - "bbox": [ - 103.16, - 546.09, - 312.4, - 556.47 - ], - "text": "Moreover, according to Eq. (2.19), P(D) = D so that", - "type": "text" - }, - { - "block_id": "p186-b26", - "global_id": 5097, - "bbox": [ - 204.17, - 563.67, - 375.36, - 578.86 - ], - "text": "P(D)yn(t) = Dyn(t) = ˙yn(t) = −e−t + 2e−2t", - "type": "text" - }, - { - "block_id": "p186-b27", - "global_id": 5098, - "bbox": [ - 103.16, - 589.47, - 413.35, - 600.93 - ], - "text": "Also in this case, b0 = 0 [the second-order term is absent in P(D)]. Therefore,", - "type": "text" - }, - { - "block_id": "p186-b28", - "global_id": 5099, - "bbox": [ - 202.42, - 607.04, - 377.76, - 622.23 - ], - "text": "h(t) = [P(D)yn(t)]u(t) = (−e−t + 2e−2t)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 187, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p187-b0", - "global_id": 5100, - "bbox": [ - 327.46, - 62.29, - 516.13, - 71.98 - ], - "text": "2.3\nThe Unit Impulse Response h(t)\n167", - "type": "text" - }, - { - "block_id": "p187-b1", - "global_id": 5101, - "bbox": [ - 127.59, - 85.54, - 516.14, - 119.82 - ], - "text": "Comment. In the above discussion, we have assumed M ≤N, as specified by Eq. (2.11).\nSection 2.8 shows that the expression for h(t) applicable to all possible values of M and N is\ngiven by", - "type": "text" - }, - { - "block_id": "p187-b2", - "global_id": 5102, - "bbox": [ - 277.5, - 122.51, - 366.23, - 133.59 - ], - "text": "h(t) = P(D)[yn(t)u(t)]", - "type": "text" - }, - { - "block_id": "p187-b3", - "global_id": 5103, - "bbox": [ - 127.59, - 142.17, - 516.15, - 164.5 - ], - "text": "where yn(t) is a linear combination of the characteristic modes of the system subject to initial\nconditions [Eq. (2.18)]. This expression reduces to Eq. (2.17) when M ≤N.", - "type": "text" - }, - { - "block_id": "p187-b4", - "global_id": 5104, - "bbox": [ - 127.59, - 166.08, - 516.14, - 212.32 - ], - "text": "Determination of the impulse response h(t) using the procedures in this section is relatively\nsimple. However, in Ch. 4 we shall discuss another, even simpler method using the Laplace\ntransform. As the next example demonstrates, it is also possible to find h(t) using functions from\nMATLAB’s symbolic math toolbox.", - "type": "text" - }, - { - "block_id": "p187-b5", - "global_id": 5105, - "bbox": [ - 102.51, - 241.69, - 454.65, - 253.64 - ], - "text": "EXAMPLE 2.7\nUsing MATLAB to Find the Impulse Response", - "type": "text" - }, - { - "block_id": "p187-b6", - "global_id": 5106, - "bbox": [ - 128.9, - 269.89, - 502.76, - 280.27 - ], - "text": "Determine the impulse response h(t) for an LTIC system specified by the differential equation", - "type": "text" - }, - { - "block_id": "p187-b7", - "global_id": 5107, - "bbox": [ - 262.73, - 287.69, - 368.94, - 302.19 - ], - "text": "(D2 + 3D + 2)y(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p187-b8", - "global_id": 5108, - "bbox": [ - 128.9, - 334.64, - 502.76, - 369.63 - ], - "text": "This is a second-order system with b0 = 0. First we find the zero-input component for initial\nconditions y(0−) = 0, and ˙y(0−) = 1. Since P(D) = D, the zero-input response is differentiated\nand the impulse response immediately follows as h(t) = 0δ(t) + [Dyn(t)]u(t).", - "type": "text" - }, - { - "block_id": "p187-b9", - "global_id": 5109, - "bbox": [ - 128.9, - 379.19, - 505.49, - 401.1 - ], - "text": ">>\ny_n = dsolve(’D2y+3*Dy+2*y=0’,’y(0)=0’,’Dy(0)=1’,’t’); h = diff(y_n)\nh = 2/exp(2*t) - 1/exp(t)", - "type": "text" - }, - { - "block_id": "p187-b10", - "global_id": 5110, - "bbox": [ - 128.9, - 406.75, - 269.87, - 420.74 - ], - "text": "Therefore, h(t) = (2e−2t −e−t)u(t).", - "type": "text" - }, - { - "block_id": "p187-b11", - "global_id": 5111, - "bbox": [ - 133.57, - 472.29, - 375.6, - 484.25 - ], - "text": "DRILL 2.4\nFinding the Impulse Response", - "type": "text" - }, - { - "block_id": "p187-b12", - "global_id": 5112, - "bbox": [ - 133.57, - 493.37, - 510.15, - 515.29 - ], - "text": "Determine the unit impulse response of LTIC systems described by the following\nequations:", - "type": "text" - }, - { - "block_id": "p187-b13", - "global_id": 5113, - "bbox": [ - 151.5, - 522.84, - 275.52, - 533.22 - ], - "text": "(a) (D + 2)y(t) = (3D + 5)x(t)", - "type": "text" - }, - { - "block_id": "p187-b14", - "global_id": 5114, - "bbox": [ - 151.5, - 537.79, - 278.29, - 548.17 - ], - "text": "(b) D(D + 2)y(t) = (D + 4)x(t)", - "type": "text" - }, - { - "block_id": "p187-b15", - "global_id": 5115, - "bbox": [ - 152.06, - 549.11, - 274.3, - 563.11 - ], - "text": "(c) (D2 + 2D + 1)y(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p187-b16", - "global_id": 5116, - "bbox": [ - 133.84, - 576.63, - 196.06, - 587.59 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p187-b17", - "global_id": 5117, - "bbox": [ - 151.5, - 593.35, - 230.43, - 604.95 - ], - "text": "(a) 3δ(t) −e−2tu(t)", - "type": "text" - }, - { - "block_id": "p187-b18", - "global_id": 5118, - "bbox": [ - 151.5, - 608.29, - 223.21, - 619.9 - ], - "text": "(b) (2 −e−2t)u(t)", - "type": "text" - }, - { - "block_id": "p187-b19", - "global_id": 5119, - "bbox": [ - 152.06, - 623.23, - 222.12, - 634.84 - ], - "text": "(c) (1 −t)e−tu(t)", - "type": "text" - } - ] - }, - { - "page_num": 188, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p188-b0", - "global_id": 5120, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "168\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p188-b1", - "global_id": 5121, - "bbox": [ - 102.14, - 86.19, - 323.83, - 98.32 - ], - "text": "SYSTEM RESPONSE TO DELAYED IMPULSE", - "type": "text" - }, - { - "block_id": "p188-b2", - "global_id": 5122, - "bbox": [ - 101.84, - 101.93, - 490.4, - 160.13 - ], - "text": "If h(t) is the response of an LTIC system to the input δ(t), then h(t−T) is the response of this same\nsystem to the input δ(t −T). This conclusion follows from the time-invariance property of LTIC\nsystems. Thus, by knowing the unit impulse response h(t), we can determine the system response\nto a delayed impulse δ(t −T). Next, we put this result to good use in finding an LTIC system’s\nzero-state response.", - "type": "text" - }, - { - "block_id": "p188-b3", - "global_id": 5123, - "bbox": [ - 102.19, - 194.0, - 392.16, - 223.88 - ], - "text": "2.4 SYSTEM RESPONSE TO EXTERNAL INPUT:\nTHE ZERO-STATE RESPONSE", - "type": "text" - }, - { - "block_id": "p188-b4", - "global_id": 5124, - "bbox": [ - 101.84, - 229.87, - 490.4, - 287.66 - ], - "text": "This section is devoted to the determination of the zero-state response of an LTIC system. This\nis the system response y(t) to an input x(t) when the system is in the zero state, that is, when all\ninitial conditions are zero. We shall assume that the systems discussed in this section are in the\nzero state unless mentioned otherwise. Under these conditions, the zero-state response will be the\ntotal response of the system.", - "type": "text" - }, - { - "block_id": "p188-b5", - "global_id": 5125, - "bbox": [ - 101.84, - 289.65, - 490.42, - 347.43 - ], - "text": "We shall use the superposition property for finding the system response to an arbitrary input\nx(t). Let us define a basic pulse p(t) of unit height and width τ, starting at t = 0 as illustrated in\nFig. 2.3a. Figure 2.3b shows an input x(t) as a sum of narrow rectangular pulses. The pulse starting\nat t = nτ in Fig. 2.3b has a height x(nτ) and can be expressed as x(nτ)p(t−nτ). Now, x(t)\nis the sum of all such pulses. Hence,", - "type": "text" - }, - { - "block_id": "p188-b6", - "global_id": 5126, - "bbox": [ - 150.85, - 369.2, - 194.51, - 379.58 - ], - "text": "x(t) = lim", - "type": "text" - }, - { - "block_id": "p188-b7", - "global_id": 5127, - "bbox": [ - 177.55, - 378.11, - 198.18, - 385.38 - ], - "text": "τ→0", - "type": "text" - }, - { - "block_id": "p188-b8", - "global_id": 5128, - "bbox": [ - 199.27, - 359.74, - 213.37, - 369.7 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p188-b9", - "global_id": 5129, - "bbox": [ - 204.39, - 383.05, - 207.41, - 390.02 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p188-b10", - "global_id": 5130, - "bbox": [ - 214.48, - 369.2, - 318.42, - 379.58 - ], - "text": "x(nτ)p(t −nτ) = lim", - "type": "text" - }, - { - "block_id": "p188-b11", - "global_id": 5131, - "bbox": [ - 301.46, - 378.11, - 322.09, - 385.38 - ], - "text": "τ→0", - "type": "text" - }, - { - "block_id": "p188-b12", - "global_id": 5132, - "bbox": [ - 323.18, - 359.74, - 337.28, - 369.7 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p188-b13", - "global_id": 5133, - "bbox": [ - 328.3, - 383.05, - 331.32, - 390.02 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p188-b14", - "global_id": 5134, - "bbox": [ - 338.39, - 355.21, - 375.39, - 372.5 - ], - "text": "x(nτ)", - "type": "text" - }, - { - "block_id": "p188-b15", - "global_id": 5135, - "bbox": [ - 353.45, - 376.27, - 365.77, - 386.24 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p188-b16", - "global_id": 5136, - "bbox": [ - 376.6, - 355.21, - 382.03, - 365.18 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p188-b17", - "global_id": 5137, - "bbox": [ - 383.13, - 369.2, - 440.17, - 379.47 - ], - "text": "p(t −nτ)τ", - "type": "text" - }, - { - "block_id": "p188-b18", - "global_id": 5138, - "bbox": [ - 101.85, - 409.34, - 414.3, - 419.71 - ], - "text": "The term [x(nτ)/τ]p(t −nτ) represents a pulse p(t −nτ) with height", - "type": "text" - }, - { - "block_id": "p188-b20", - "global_id": 5139, - "bbox": [ - 421.27, - 409.34, - 468.65, - 419.61 - ], - "text": "x(nτ)/τ", - "type": "text" - }, - { - "block_id": "p188-b22", - "global_id": 5140, - "bbox": [ - 101.85, - 409.75, - 490.39, - 443.63 - ], - "text": ". As\nτ →0, the height of this strip →∞, but its area remains x(nτ). Hence, this strip approaches\nan impulse x(nτ)δ(t −nτ) as τ →0 (Fig. 2.3e). Therefore,", - "type": "text" - }, - { - "block_id": "p188-b23", - "global_id": 5141, - "bbox": [ - 219.51, - 461.41, - 263.18, - 471.79 - ], - "text": "x(t) = lim", - "type": "text" - }, - { - "block_id": "p188-b24", - "global_id": 5142, - "bbox": [ - 246.21, - 470.32, - 266.84, - 477.58 - ], - "text": "τ→0", - "type": "text" - }, - { - "block_id": "p188-b25", - "global_id": 5143, - "bbox": [ - 267.93, - 451.94, - 282.03, - 461.91 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p188-b26", - "global_id": 5144, - "bbox": [ - 273.05, - 475.26, - 276.07, - 482.23 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p188-b27", - "global_id": 5145, - "bbox": [ - 283.14, - 461.41, - 490.38, - 471.79 - ], - "text": "x(nτ)δ(t −nτ)τ\n(2.22)", - "type": "text" - }, - { - "block_id": "p188-b28", - "global_id": 5146, - "bbox": [ - 101.85, - 496.43, - 490.39, - 518.77 - ], - "text": "To find the response for this input x(t), we consider the input and the corresponding output pairs,\nas shown in Figs. 2.3c–2.3f and also shown by directed arrow notation as follows:", - "type": "text" - }, - { - "block_id": "p188-b29", - "global_id": 5147, - "bbox": [ - 265.81, - 534.9, - 332.34, - 545.27 - ], - "text": "input \r⇒output", - "type": "text" - }, - { - "block_id": "p188-b30", - "global_id": 5148, - "bbox": [ - 271.09, - 549.84, - 322.26, - 560.11 - ], - "text": "δ(t) \r⇒h(t)", - "type": "text" - }, - { - "block_id": "p188-b31", - "global_id": 5149, - "bbox": [ - 191.18, - 564.79, - 402.16, - 590.0 - ], - "text": "δ(t −nτ) \r⇒h(t −nτ)\n[x(nτ)τ]δ(t −nτ) \r⇒[x(nτ)τ]h(t −nτ)", - "type": "text" - }, - { - "block_id": "p188-b32", - "global_id": 5150, - "bbox": [ - 159.78, - 598.18, - 180.4, - 613.93 - ], - "text": "lim\nτ→0", - "type": "text" - }, - { - "block_id": "p188-b33", - "global_id": 5151, - "bbox": [ - 181.5, - 588.29, - 195.6, - 598.26 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p188-b34", - "global_id": 5152, - "bbox": [ - 186.62, - 611.61, - 189.64, - 618.58 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p188-b35", - "global_id": 5153, - "bbox": [ - 196.7, - 597.76, - 285.09, - 608.04 - ], - "text": "x(nτ)δ(t −nτ)τ", - "type": "text" - }, - { - "block_id": "p188-b36", - "global_id": 5154, - "bbox": [ - 159.78, - 614.39, - 286.3, - 633.61 - ], - "text": "x(t)\n[see Eq. (2.22)]", - "type": "text" - }, - { - "block_id": "p188-b37", - "global_id": 5155, - "bbox": [ - 288.34, - 597.76, - 323.85, - 608.14 - ], - "text": "⇒lim", - "type": "text" - }, - { - "block_id": "p188-b38", - "global_id": 5156, - "bbox": [ - 306.89, - 606.67, - 327.51, - 613.93 - ], - "text": "τ→0", - "type": "text" - }, - { - "block_id": "p188-b39", - "global_id": 5157, - "bbox": [ - 328.61, - 588.29, - 342.71, - 598.26 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p188-b40", - "global_id": 5158, - "bbox": [ - 333.73, - 611.61, - 336.75, - 618.58 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p188-b41", - "global_id": 5159, - "bbox": [ - 343.82, - 597.76, - 431.25, - 608.04 - ], - "text": "x(nτ)h(t −nτ)τ", - "type": "text" - }, - { - "block_id": "p188-b42", - "global_id": 5160, - "bbox": [ - 306.89, - 614.39, - 432.46, - 633.54 - ], - "text": "y(t)", - "type": "text" - } - ] - }, - { - "page_num": 189, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p189-b0", - "global_id": 5161, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n169", - "type": "text" - }, - { - "block_id": "p189-b1", - "global_id": 5162, - "bbox": [ - 292.11, - 297.03, - 300.99, - 305.03 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p189-b2", - "global_id": 5163, - "bbox": [ - 294.3, - 383.61, - 303.63, - 391.61 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p189-b3", - "global_id": 5164, - "bbox": [ - 295.3, - 485.43, - 304.18, - 493.43 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p189-b4", - "global_id": 5165, - "bbox": [ - 251.8, - 595.44, - 365.42, - 607.05 - ], - "text": "t\nnt\n0", - "type": "text" - }, - { - "block_id": "p189-b5", - "global_id": 5166, - "bbox": [ - 228.72, - 539.42, - 239.82, - 547.5 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p189-b6", - "global_id": 5167, - "bbox": [ - 293.74, - 613.74, - 302.69, - 621.74 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p189-b7", - "global_id": 5168, - "bbox": [ - 373.78, - 205.87, - 383.43, - 213.87 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p189-b8", - "global_id": 5169, - "bbox": [ - 181.64, - 182.81, - 328.6, - 200.43 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p189-b9", - "global_id": 5170, - "bbox": [ - 361.08, - 192.17, - 386.86, - 200.38 - ], - "text": "t nt", - "type": "text" - }, - { - "block_id": "p189-b10", - "global_id": 5171, - "bbox": [ - 325.61, - 107.31, - 347.38, - 115.6 - ], - "text": "x(nt)", - "type": "text" - }, - { - "block_id": "p189-b11", - "global_id": 5172, - "bbox": [ - 336.03, - 95.46, - 347.13, - 103.54 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p189-b12", - "global_id": 5173, - "bbox": [ - 250.37, - 182.93, - 252.59, - 190.93 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p189-b13", - "global_id": 5174, - "bbox": [ - 179.65, - 281.81, - 183.65, - 289.81 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p189-b14", - "global_id": 5175, - "bbox": [ - 181.47, - 235.14, - 193.02, - 243.24 - ], - "text": "d(t)", - "type": "text" - }, - { - "block_id": "p189-b15", - "global_id": 5176, - "bbox": [ - 250.37, - 280.49, - 252.59, - 288.49 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p189-b16", - "global_id": 5177, - "bbox": [ - 179.69, - 366.73, - 183.69, - 374.73 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p189-b17", - "global_id": 5178, - "bbox": [ - 205.7, - 319.96, - 240.8, - 328.25 - ], - "text": "d(t nt)", - "type": "text" - }, - { - "block_id": "p189-b18", - "global_id": 5179, - "bbox": [ - 193.65, - 366.44, - 252.59, - 374.65 - ], - "text": "nt\nt", - "type": "text" - }, - { - "block_id": "p189-b19", - "global_id": 5180, - "bbox": [ - 169.71, - 182.37, - 424.95, - 192.06 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p189-b20", - "global_id": 5181, - "bbox": [ - 169.75, - 122.03, - 195.12, - 138.1 - ], - "text": "p(t)\n1", - "type": "text" - }, - { - "block_id": "p189-b21", - "global_id": 5182, - "bbox": [ - 205.95, - 205.87, - 214.83, - 213.87 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p189-b22", - "global_id": 5183, - "bbox": [ - 179.8, - 468.63, - 252.43, - 476.71 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p189-b23", - "global_id": 5184, - "bbox": [ - 202.65, - 407.18, - 282.41, - 415.47 - ], - "text": "[x(nt)t]d(t nt)", - "type": "text" - }, - { - "block_id": "p189-b24", - "global_id": 5185, - "bbox": [ - 193.68, - 468.42, - 206.56, - 476.63 - ], - "text": "nt", - "type": "text" - }, - { - "block_id": "p189-b25", - "global_id": 5186, - "bbox": [ - 353.08, - 281.41, - 357.08, - 289.41 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p189-b26", - "global_id": 5187, - "bbox": [ - 335.55, - 226.62, - 347.11, - 234.7 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p189-b27", - "global_id": 5188, - "bbox": [ - 422.72, - 280.15, - 424.95, - 288.15 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p189-b28", - "global_id": 5189, - "bbox": [ - 353.06, - 366.36, - 379.23, - 374.64 - ], - "text": "nt\n0", - "type": "text" - }, - { - "block_id": "p189-b29", - "global_id": 5190, - "bbox": [ - 409.78, - 329.94, - 444.89, - 338.24 - ], - "text": "h(t nt)", - "type": "text" - }, - { - "block_id": "p189-b30", - "global_id": 5191, - "bbox": [ - 422.72, - 366.57, - 424.95, - 374.57 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p189-b31", - "global_id": 5192, - "bbox": [ - 422.56, - 468.35, - 424.79, - 476.35 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p189-b32", - "global_id": 5193, - "bbox": [ - 408.25, - 412.32, - 474.01, - 420.62 - ], - "text": "x(nt)h(t nt)t", - "type": "text" - }, - { - "block_id": "p189-b33", - "global_id": 5194, - "bbox": [ - 352.92, - 468.14, - 379.23, - 476.43 - ], - "text": "nt\n0", - "type": "text" - }, - { - "block_id": "p189-b34", - "global_id": 5195, - "bbox": [ - 330.68, - 405.46, - 347.12, - 413.76 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p189-b35", - "global_id": 5196, - "bbox": [ - 151.5, - 628.06, - 390.5, - 637.67 - ], - "text": "Figure 2.3 Finding the system response to an arbitrary input x(t).", - "type": "text" - } - ] - }, - { - "page_num": 190, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p190-b0", - "global_id": 5197, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "170\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p190-b1", - "global_id": 5198, - "bbox": [ - 101.84, - 86.06, - 146.58, - 96.62 - ], - "text": "Therefore,†", - "type": "text" - }, - { - "block_id": "p190-b2", - "global_id": 5199, - "bbox": [ - 220.0, - 107.48, - 263.63, - 117.86 - ], - "text": "y(t) = lim", - "type": "text" - }, - { - "block_id": "p190-b3", - "global_id": 5200, - "bbox": [ - 246.67, - 116.39, - 267.3, - 123.65 - ], - "text": "τ→0", - "type": "text" - }, - { - "block_id": "p190-b4", - "global_id": 5201, - "bbox": [ - 268.39, - 98.01, - 282.49, - 107.97 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p190-b5", - "global_id": 5202, - "bbox": [ - 273.51, - 121.33, - 276.53, - 128.3 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p190-b6", - "global_id": 5203, - "bbox": [ - 283.6, - 107.48, - 371.03, - 117.75 - ], - "text": "x(nτ)h(t −nτ)τ", - "type": "text" - }, - { - "block_id": "p190-b7", - "global_id": 5204, - "bbox": [ - 236.85, - 138.22, - 244.63, - 148.19 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p190-b8", - "global_id": 5205, - "bbox": [ - 246.67, - 124.66, - 263.61, - 136.72 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p190-b9", - "global_id": 5206, - "bbox": [ - 251.93, - 149.53, - 264.48, - 156.51 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p190-b10", - "global_id": 5207, - "bbox": [ - 266.09, - 138.22, - 490.38, - 148.6 - ], - "text": "x(τ)h(t −τ)dτ\n(2.23)", - "type": "text" - }, - { - "block_id": "p190-b11", - "global_id": 5208, - "bbox": [ - 101.85, - 164.84, - 490.41, - 222.94 - ], - "text": "This is the result we seek. We have obtained the system response y(t) to an arbitrary input x(t)\nin terms of the unit impulse response h(t). Knowing h(t), we can determine the response y(t) to\nany input. Observe once again the all-pervasive nature of the system’s characteristic modes. The\nsystem response to any input is determined by the impulse response, which, in turn, is made up of\ncharacteristic modes of the system.", - "type": "text" - }, - { - "block_id": "p190-b12", - "global_id": 5209, - "bbox": [ - 101.85, - 225.04, - 490.39, - 258.9 - ], - "text": "It is important to keep in mind the assumptions used in deriving Eq. (2.23). We assumed a\nlinear time-invariant (LTI) system. Linearity allowed us to use the principle of superposition, and\ntime invariance made it possible to express the system’s response to δ(t −n△τ) as h(t −n△τ).", - "type": "text" - }, - { - "block_id": "p190-b13", - "global_id": 5210, - "bbox": [ - 101.84, - 283.31, - 271.72, - 295.27 - ], - "text": "2.4-1 The Convolution Integral", - "type": "text" - }, - { - "block_id": "p190-b14", - "global_id": 5211, - "bbox": [ - 101.84, - 300.99, - 490.42, - 348.0 - ], - "text": "The zero-state response y(t) obtained in Eq. (2.23) is given by an integral that occurs frequently in\nthe physical sciences, engineering, and mathematics. For this reason this integral is given a special\nname: the convolution integral. The convolution integral of two functions x1(t) and x2(t) is denoted\nsymbolically by x1(t) ∗x2(t) and is defined as", - "type": "text" - }, - { - "block_id": "p190-b15", - "global_id": 5212, - "bbox": [ - 223.34, - 363.41, - 279.03, - 374.56 - ], - "text": "x1(t) ∗x2(t) ≡", - "type": "text" - }, - { - "block_id": "p190-b16", - "global_id": 5213, - "bbox": [ - 281.08, - 349.85, - 298.03, - 361.9 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p190-b17", - "global_id": 5214, - "bbox": [ - 286.34, - 374.73, - 298.9, - 381.7 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p190-b18", - "global_id": 5215, - "bbox": [ - 300.5, - 363.41, - 490.38, - 374.56 - ], - "text": "x1(τ)x2(t −τ)dτ\n(2.24)", - "type": "text" - }, - { - "block_id": "p190-b19", - "global_id": 5216, - "bbox": [ - 101.85, - 390.39, - 345.91, - 400.36 - ], - "text": "Some important properties of the convolution integral follow.", - "type": "text" - }, - { - "block_id": "p190-b20", - "global_id": 5217, - "bbox": [ - 101.84, - 414.92, - 490.39, - 464.94 - ], - "text": "THE COMMUTATIVE PROPERTY\nConvolution operation is commutative; that is, x1(t) ∗x2(t) = x2(t) ∗x1(t). This property can be\nproved by a change of variable. In Eq. (2.24), if we let z = t −τ so that τ = t −z and dτ = −dz,\nwe obtain", - "type": "text" - }, - { - "block_id": "p190-b21", - "global_id": 5218, - "bbox": [ - 219.21, - 479.86, - 284.72, - 491.01 - ], - "text": "x1(t) ∗x2(t) = −", - "type": "text" - }, - { - "block_id": "p190-b22", - "global_id": 5219, - "bbox": [ - 285.82, - 466.3, - 308.2, - 478.36 - ], - "text": "# −∞", - "type": "text" - }, - { - "block_id": "p190-b23", - "global_id": 5220, - "bbox": [ - 291.08, - 491.18, - 298.2, - 498.16 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p190-b24", - "global_id": 5221, - "bbox": [ - 309.81, - 479.86, - 373.02, - 491.01 - ], - "text": "x2(z)x1(t −z)dz", - "type": "text" - }, - { - "block_id": "p190-b25", - "global_id": 5222, - "bbox": [ - 267.13, - 506.9, - 274.9, - 516.86 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p190-b26", - "global_id": 5223, - "bbox": [ - 276.95, - 493.34, - 293.89, - 505.39 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p190-b27", - "global_id": 5224, - "bbox": [ - 282.2, - 518.22, - 294.76, - 525.19 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p190-b28", - "global_id": 5225, - "bbox": [ - 296.37, - 506.9, - 359.58, - 518.05 - ], - "text": "x2(z)x1(t −z)dz", - "type": "text" - }, - { - "block_id": "p190-b29", - "global_id": 5226, - "bbox": [ - 267.13, - 528.76, - 490.38, - 539.91 - ], - "text": "= x2(t) ∗x1(t)\n(2.25)", - "type": "text" - }, - { - "block_id": "p190-b30", - "global_id": 5227, - "bbox": [ - 101.84, - 558.65, - 490.4, - 592.79 - ], - "text": "† In deriving this result we have assumed a time-invariant system. If the system is time-varying, then the\nsystem response to the input δ(t−nΔτ) cannot be expressed as h(t−nΔτ) but instead has the form h(t,nΔτ).\nUse of this form modifies Eq. (2.23) to", - "type": "text" - }, - { - "block_id": "p190-b31", - "global_id": 5228, - "bbox": [ - 250.99, - 603.73, - 273.15, - 612.97 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p190-b32", - "global_id": 5229, - "bbox": [ - 275.0, - 591.52, - 290.44, - 602.49 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p190-b33", - "global_id": 5230, - "bbox": [ - 279.73, - 613.76, - 291.39, - 620.24 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p190-b34", - "global_id": 5231, - "bbox": [ - 292.89, - 603.73, - 340.19, - 613.06 - ], - "text": "x(τ)h(t,τ)dτ", - "type": "text" - }, - { - "block_id": "p190-b35", - "global_id": 5232, - "bbox": [ - 101.84, - 624.08, - 399.2, - 633.41 - ], - "text": "where h(t,τ) is the system response at instant t to a unit impulse input located at τ.", - "type": "text" - } - ] - }, - { - "page_num": 191, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p191-b0", - "global_id": 5233, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n171", - "type": "text" - }, - { - "block_id": "p191-b1", - "global_id": 5234, - "bbox": [ - 127.89, - 86.19, - 287.14, - 98.32 - ], - "text": "THE DISTRIBUTIVE PROPERTY", - "type": "text" - }, - { - "block_id": "p191-b2", - "global_id": 5235, - "bbox": [ - 127.59, - 102.35, - 279.92, - 112.31 - ], - "text": "According to the distributive property,", - "type": "text" - }, - { - "block_id": "p191-b3", - "global_id": 5236, - "bbox": [ - 223.54, - 123.53, - 516.13, - 134.67 - ], - "text": "x1(t) ∗[x2(t) + x3(t)] = x1(t) ∗x2(t) + x1(t) ∗x3(t)\n(2.26)", - "type": "text" - }, - { - "block_id": "p191-b4", - "global_id": 5237, - "bbox": [ - 127.59, - 148.55, - 283.9, - 174.66 - ], - "text": "THE ASSOCIATIVE PROPERTY\nAccording to the associative property,", - "type": "text" - }, - { - "block_id": "p191-b5", - "global_id": 5238, - "bbox": [ - 236.34, - 185.88, - 516.13, - 197.03 - ], - "text": "x1(t) ∗[x2(t) ∗x3(t)] = [x1(t) ∗x2(t)] ∗x3(t)\n(2.27)", - "type": "text" - }, - { - "block_id": "p191-b6", - "global_id": 5239, - "bbox": [ - 127.59, - 207.88, - 516.12, - 229.81 - ], - "text": "The proofs of Eqs. (2.26) and (2.27) follow directly from the definition of the convolution integral.\nThey are left as an exercise for the reader.", - "type": "text" - }, - { - "block_id": "p191-b7", - "global_id": 5240, - "bbox": [ - 127.59, - 244.46, - 243.54, - 270.57 - ], - "text": "THE SHIFT PROPERTY\nIf", - "type": "text" - }, - { - "block_id": "p191-b8", - "global_id": 5241, - "bbox": [ - 285.59, - 272.14, - 358.13, - 283.29 - ], - "text": "x1(t) ∗x2(t) = c(t)", - "type": "text" - }, - { - "block_id": "p191-b9", - "global_id": 5242, - "bbox": [ - 127.59, - 291.32, - 144.74, - 301.28 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p191-b10", - "global_id": 5243, - "bbox": [ - 230.96, - 302.87, - 412.75, - 314.02 - ], - "text": "x1(t) ∗x2(t −T) = x1(t −T) ∗x2(t) = c(t −T)", - "type": "text" - }, - { - "block_id": "p191-b11", - "global_id": 5244, - "bbox": [ - 127.59, - 322.05, - 237.33, - 332.01 - ], - "text": "More generally, we see that", - "type": "text" - }, - { - "block_id": "p191-b12", - "global_id": 5245, - "bbox": [ - 244.8, - 343.22, - 516.13, - 354.68 - ], - "text": "x1(t −T1) ∗x2(t −T2) = c(t −T1 −T2)\n(2.28)", - "type": "text" - }, - { - "block_id": "p191-b13", - "global_id": 5246, - "bbox": [ - 127.59, - 365.15, - 210.42, - 375.19 - ], - "text": "Proof. We are given", - "type": "text" - }, - { - "block_id": "p191-b14", - "global_id": 5247, - "bbox": [ - 235.75, - 382.1, - 291.44, - 393.24 - ], - "text": "x1(t) ∗x2(t) =", - "type": "text" - }, - { - "block_id": "p191-b15", - "global_id": 5248, - "bbox": [ - 293.49, - 368.53, - 310.43, - 380.59 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p191-b16", - "global_id": 5249, - "bbox": [ - 298.75, - 393.41, - 311.3, - 400.38 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p191-b17", - "global_id": 5250, - "bbox": [ - 312.91, - 382.1, - 407.97, - 393.24 - ], - "text": "x1(τ)x2(t −τ)dτ = c(t)", - "type": "text" - }, - { - "block_id": "p191-b18", - "global_id": 5251, - "bbox": [ - 127.59, - 406.6, - 169.34, - 416.56 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p191-b19", - "global_id": 5252, - "bbox": [ - 231.91, - 432.39, - 304.77, - 443.54 - ], - "text": "x1(t) ∗x2(t −T) =", - "type": "text" - }, - { - "block_id": "p191-b20", - "global_id": 5253, - "bbox": [ - 306.82, - 418.82, - 323.77, - 430.88 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p191-b21", - "global_id": 5254, - "bbox": [ - 312.08, - 443.7, - 324.64, - 450.67 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p191-b22", - "global_id": 5255, - "bbox": [ - 326.25, - 432.39, - 410.61, - 443.54 - ], - "text": "x1(τ)x2(t −T −τ)dτ", - "type": "text" - }, - { - "block_id": "p191-b23", - "global_id": 5256, - "bbox": [ - 297.0, - 454.24, - 338.8, - 464.52 - ], - "text": "= c(t −T)", - "type": "text" - }, - { - "block_id": "p191-b24", - "global_id": 5257, - "bbox": [ - 127.6, - 476.24, - 392.1, - 486.21 - ], - "text": "The equally simple proof of Eq. (2.28) follows a similar approach.", - "type": "text" - }, - { - "block_id": "p191-b25", - "global_id": 5258, - "bbox": [ - 127.89, - 500.86, - 311.66, - 512.98 - ], - "text": "CONVOLUTION WITH AN IMPULSE", - "type": "text" - }, - { - "block_id": "p191-b26", - "global_id": 5259, - "bbox": [ - 127.59, - 516.6, - 516.13, - 538.93 - ], - "text": "Convolution of a function x(t) with a unit impulse results in the function x(t) itself. By definition\nof convolution,", - "type": "text" - }, - { - "block_id": "p191-b27", - "global_id": 5260, - "bbox": [ - 256.63, - 545.11, - 304.77, - 555.39 - ], - "text": "x(t) ∗δ(t) =", - "type": "text" - }, - { - "block_id": "p191-b28", - "global_id": 5261, - "bbox": [ - 306.82, - 531.56, - 323.77, - 543.61 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p191-b29", - "global_id": 5262, - "bbox": [ - 312.08, - 556.43, - 324.64, - 563.41 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p191-b30", - "global_id": 5263, - "bbox": [ - 326.25, - 545.11, - 385.89, - 555.39 - ], - "text": "x(τ)δ(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p191-b31", - "global_id": 5264, - "bbox": [ - 127.59, - 569.27, - 516.15, - 591.61 - ], - "text": "Because δ(t −τ) is an impulse located at τ = t, according to the sampling property of the impulse\n[Eq. (1.11)], the integral here is just the value of x(τ) at τ = t, that is, x(t). Therefore,", - "type": "text" - }, - { - "block_id": "p191-b32", - "global_id": 5265, - "bbox": [ - 289.35, - 602.82, - 354.38, - 613.1 - ], - "text": "x(t) ∗δ(t) = x(t)", - "type": "text" - }, - { - "block_id": "p191-b33", - "global_id": 5266, - "bbox": [ - 127.59, - 624.83, - 332.94, - 634.79 - ], - "text": "Actually this result was derived earlier [Eq. (2.22)].", - "type": "text" - } - ] - }, - { - "page_num": 192, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p192-b0", - "global_id": 5267, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "172\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p192-b1", - "global_id": 5268, - "bbox": [ - 102.14, - 86.19, - 225.57, - 98.32 - ], - "text": "THE WIDTH PROPERTY", - "type": "text" - }, - { - "block_id": "p192-b2", - "global_id": 5269, - "bbox": [ - 101.84, - 101.93, - 490.39, - 136.22 - ], - "text": "If the durations (widths) of x1(t) and x2(t) are finite, given by T1 and T2, respectively, then the\nduration (width) of x1(t) ∗x2(t) is T1 + T2 (Fig. 2.4). The proof of this property follows readily\nfrom the graphical considerations discussed later in Sec. 2.4-2.", - "type": "text" - }, - { - "block_id": "p192-b3", - "global_id": 5270, - "bbox": [ - 159.08, - 214.44, - 166.53, - 224.05 - ], - "text": "T1", - "type": "text" - }, - { - "block_id": "p192-b4", - "global_id": 5271, - "bbox": [ - 235.9, - 196.8, - 240.9, - 206.8 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p192-b5", - "global_id": 5272, - "bbox": [ - 183.03, - 162.76, - 269.64, - 172.37 - ], - "text": "x1(t)\nx2(t)", - "type": "text" - }, - { - "block_id": "p192-b7", - "global_id": 5273, - "bbox": [ - 212.69, - 226.55, - 466.66, - 234.55 - ], - "text": "t\nt\nt", - "type": "text" - }, - { - "block_id": "p192-b8", - "global_id": 5274, - "bbox": [ - 391.51, - 162.71, - 427.72, - 172.32 - ], - "text": "x1(t) * x2(t)", - "type": "text" - }, - { - "block_id": "p192-b9", - "global_id": 5275, - "bbox": [ - 285.19, - 215.23, - 430.35, - 225.05 - ], - "text": "T2\nT1 T2", - "type": "text" - }, - { - "block_id": "p192-b10", - "global_id": 5276, - "bbox": [ - 125.76, - 241.33, - 282.51, - 250.56 - ], - "text": "Figure 2.4 Width property of convolution.", - "type": "text" - }, - { - "block_id": "p192-b11", - "global_id": 5277, - "bbox": [ - 102.14, - 288.22, - 320.77, - 300.34 - ], - "text": "ZERO-STATE RESPONSE AND CAUSALITY", - "type": "text" - }, - { - "block_id": "p192-b12", - "global_id": 5278, - "bbox": [ - 101.84, - 303.96, - 305.53, - 314.33 - ], - "text": "The (zero-state) response y(t) of an LTIC system is", - "type": "text" - }, - { - "block_id": "p192-b13", - "global_id": 5279, - "bbox": [ - 217.39, - 338.77, - 292.37, - 349.05 - ], - "text": "y(t) = x(t) ∗h(t) =", - "type": "text" - }, - { - "block_id": "p192-b14", - "global_id": 5280, - "bbox": [ - 294.42, - 325.21, - 311.36, - 337.26 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p192-b15", - "global_id": 5281, - "bbox": [ - 299.67, - 350.08, - 312.23, - 357.05 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p192-b16", - "global_id": 5282, - "bbox": [ - 313.84, - 338.77, - 490.38, - 349.15 - ], - "text": "x(τ)h(t −τ)dτ\n(2.29)", - "type": "text" - }, - { - "block_id": "p192-b17", - "global_id": 5283, - "bbox": [ - 101.85, - 374.0, - 490.43, - 419.83 - ], - "text": "In deriving Eq. (2.29), we assumed the system to be linear and time-invariant. There were no other\nrestrictions either on the system or on the input signal x(t). Since, in practice, most systems are\ncausal, their response cannot begin before the input. Furthermore, most inputs are also causal,\nwhich means they start at t = 0.", - "type": "text" - }, - { - "block_id": "p192-b18", - "global_id": 5284, - "bbox": [ - 101.85, - 421.82, - 490.37, - 467.65 - ], - "text": "Causality restriction on both signals and systems further simplifies the limits of integration\nin Eq. (2.29). By definition, the response of a causal system cannot begin before its input begins.\nConsequently, the causal system’s response to a unit impulse δ(t) (which is located at t = 0) cannot\nbegin before t = 0. Therefore, a causal system’s unit impulse response h(t) is a causal signal.", - "type": "text" - }, - { - "block_id": "p192-b19", - "global_id": 5285, - "bbox": [ - 101.84, - 469.23, - 490.4, - 539.38 - ], - "text": "It is important to remember that the integration in Eq. (2.29) is performed with respect to τ\n(not t). If the input x(t) is causal, x(τ) = 0 for τ < 0. Therefore, x(τ) = 0 for τ < 0, as illustrated\nin Fig. 2.5a. Similarly, if h(t) is causal, h(t −τ) = 0 for t −τ < 0; that is, for τ > t, as depicted in\nFig. 2.5a. Therefore, the product x(τ)h(t −τ) = 0 everywhere except over the nonshaded interval\n0 ≤τ ≤t shown in Fig. 2.5a (assuming t ≥0). Observe that if t is negative, x(τ)h(t −τ) = 0 for\nall τ, as shown in Fig. 2.5b. Therefore, Eq. (2.29) reduces to", - "type": "text" - }, - { - "block_id": "p192-b20", - "global_id": 5286, - "bbox": [ - 196.16, - 564.48, - 271.14, - 574.76 - ], - "text": "y(t) = x(t) ∗h(t) =", - "type": "text" - }, - { - "block_id": "p192-b21", - "global_id": 5287, - "bbox": [ - 273.19, - 550.5, - 287.77, - 563.0 - ], - "text": "$ t", - "type": "text" - }, - { - "block_id": "p192-b22", - "global_id": 5288, - "bbox": [ - 279.34, - 558.75, - 490.38, - 581.09 - ], - "text": "0−x(τ)h(t −τ)dτ\nt ≥0\n0\nt < 0\n(2.30)", - "type": "text" - }, - { - "block_id": "p192-b23", - "global_id": 5289, - "bbox": [ - 101.85, - 599.27, - 490.39, - 634.79 - ], - "text": "The lower limit of integration in Eq. (2.30) is taken as 0−to avoid the difficulty in integration that\ncan arise if x(t) contains an impulse at the origin. This result shows that if x(t) and h(t) are both\ncausal, the response y(t) is also causal.", - "type": "text" - } - ] - }, - { - "page_num": 193, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p193-b0", - "global_id": 5290, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n173", - "type": "text" - }, - { - "block_id": "p193-b1", - "global_id": 5291, - "bbox": [ - 278.62, - 103.94, - 323.05, - 112.24 - ], - "text": "h(t t) 0", - "type": "text" - }, - { - "block_id": "p193-b2", - "global_id": 5292, - "bbox": [ - 229.07, - 119.56, - 356.8, - 130.67 - ], - "text": "t\n0\nt 0", - "type": "text" - }, - { - "block_id": "p193-b3", - "global_id": 5293, - "bbox": [ - 221.54, - 141.86, - 230.42, - 149.86 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p193-b4", - "global_id": 5294, - "bbox": [ - 153.23, - 103.94, - 180.33, - 112.24 - ], - "text": "x(t) 0", - "type": "text" - }, - { - "block_id": "p193-b5", - "global_id": 5295, - "bbox": [ - 272.23, - 122.63, - 274.46, - 130.63 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p193-b6", - "global_id": 5296, - "bbox": [ - 229.01, - 194.96, - 312.22, - 203.06 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p193-b7", - "global_id": 5297, - "bbox": [ - 339.91, - 174.94, - 356.8, - 183.24 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p193-b8", - "global_id": 5298, - "bbox": [ - 221.16, - 215.74, - 230.81, - 223.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p193-b9", - "global_id": 5299, - "bbox": [ - 201.27, - 194.99, - 203.49, - 202.99 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p193-b10", - "global_id": 5300, - "bbox": [ - 153.24, - 178.21, - 299.0, - 186.51 - ], - "text": "h(t t) 0 \nx(t)", - "type": "text" - }, - { - "block_id": "p193-b11", - "global_id": 5301, - "bbox": [ - 367.54, - 203.08, - 464.6, - 224.99 - ], - "text": "Figure 2.5 Limits of the\nconvolution integral.", - "type": "text" - }, - { - "block_id": "p193-b12", - "global_id": 5302, - "bbox": [ - 127.59, - 247.25, - 516.15, - 269.17 - ], - "text": "Because of the convolution’s commutative property [Eq. (2.25)], we can also express\nEq. (2.30) as [assuming causal x(t) and h(t)]", - "type": "text" - }, - { - "block_id": "p193-b13", - "global_id": 5303, - "bbox": [ - 245.86, - 290.67, - 270.49, - 300.95 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p193-b14", - "global_id": 5304, - "bbox": [ - 272.54, - 270.24, - 280.43, - 289.18 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p193-b15", - "global_id": 5305, - "bbox": [ - 272.54, - 297.15, - 280.43, - 307.11 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p193-b16", - "global_id": 5306, - "bbox": [ - 280.42, - 271.44, - 292.19, - 283.72 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p193-b17", - "global_id": 5307, - "bbox": [ - 285.69, - 285.0, - 396.4, - 303.58 - ], - "text": "0−h(τ)x(t −τ)dτ\nt ≥0", - "type": "text" - }, - { - "block_id": "p193-b18", - "global_id": 5308, - "bbox": [ - 280.42, - 302.36, - 396.66, - 312.74 - ], - "text": "0\nt < 0", - "type": "text" - }, - { - "block_id": "p193-b19", - "global_id": 5309, - "bbox": [ - 127.59, - 321.86, - 516.13, - 345.42 - ], - "text": "Hereafter, the lower limit of 0−will be implied even when we write it as 0. As in Eq. (2.30), this\nresult assumes that both the input and the system are causal.", - "type": "text" - }, - { - "block_id": "p193-b20", - "global_id": 5310, - "bbox": [ - 102.51, - 373.58, - 398.27, - 385.54 - ], - "text": "EXAMPLE 2.8\nComputing the Zero-State Response", - "type": "text" - }, - { - "block_id": "p193-b21", - "global_id": 5311, - "bbox": [ - 128.9, - 400.28, - 502.77, - 423.83 - ], - "text": "For an LTIC system with the unit impulse response h(t) = e−2tu(t), determine the response y(t)\nfor the input", - "type": "text" - }, - { - "block_id": "p193-b22", - "global_id": 5312, - "bbox": [ - 288.59, - 423.69, - 343.08, - 435.69 - ], - "text": "x(t) = e−tu(t)", - "type": "text" - }, - { - "block_id": "p193-b23", - "global_id": 5313, - "bbox": [ - 128.9, - 464.97, - 443.78, - 475.35 - ], - "text": "Here both x(t) and h(t) are causal (Fig. 2.6). Hence, from Eq. (2.30), we obtain", - "type": "text" - }, - { - "block_id": "p193-b24", - "global_id": 5314, - "bbox": [ - 245.52, - 492.11, - 270.14, - 502.38 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p193-b25", - "global_id": 5315, - "bbox": [ - 272.19, - 478.54, - 283.95, - 490.82 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p193-b26", - "global_id": 5316, - "bbox": [ - 277.45, - 492.11, - 386.15, - 510.69 - ], - "text": "0\nx(τ)h(t −τ)dτ\nt ≥0", - "type": "text" - }, - { - "block_id": "p193-b27", - "global_id": 5317, - "bbox": [ - 128.91, - 518.63, - 299.44, - 530.23 - ], - "text": "Because x(t) = e−tu(t) and h(t) = e−2tu(t),", - "type": "text" - }, - { - "block_id": "p193-b28", - "global_id": 5318, - "bbox": [ - 205.13, - 539.47, - 426.53, - 551.57 - ], - "text": "x(τ) = e−τu(τ)\nand\nh(t −τ) = e−2(t−τ)u(t −τ)", - "type": "text" - }, - { - "block_id": "p193-b29", - "global_id": 5319, - "bbox": [ - 128.9, - 562.54, - 502.76, - 596.83 - ], - "text": "Remember that the integration is performed with respect to τ (not t), and the region of\nintegration is 0 ≤τ ≤t. Hence, τ ≥0 and t −τ ≥0. Therefore, u(τ) = 1 and u(t −τ) = 1;\nconsequently,", - "type": "text" - }, - { - "block_id": "p193-b30", - "global_id": 5320, - "bbox": [ - 247.78, - 604.19, - 272.4, - 614.47 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p193-b31", - "global_id": 5321, - "bbox": [ - 274.44, - 590.64, - 286.21, - 602.91 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p193-b32", - "global_id": 5322, - "bbox": [ - 279.71, - 600.39, - 383.9, - 622.78 - ], - "text": "0\ne−τe−2(t−τ) dτ\nt ≥0", - "type": "text" - } - ] - }, - { - "page_num": 194, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p194-b0", - "global_id": 5323, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "174\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p194-b1", - "global_id": 5324, - "bbox": [ - 245.72, - 273.7, - 258.94, - 283.32 - ], - "text": "e2t", - "type": "text" - }, - { - "block_id": "p194-b2", - "global_id": 5325, - "bbox": [ - 113.1, - 89.81, - 137.61, - 109.23 - ], - "text": "1\nx(t)", - "type": "text" - }, - { - "block_id": "p194-b3", - "global_id": 5326, - "bbox": [ - 136.4, - 115.63, - 146.61, - 125.24 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p194-b4", - "global_id": 5327, - "bbox": [ - 111.7, - 165.88, - 206.06, - 174.49 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p194-b5", - "global_id": 5328, - "bbox": [ - 153.51, - 185.83, - 162.39, - 193.83 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p194-b6", - "global_id": 5329, - "bbox": [ - 261.97, - 89.83, - 286.66, - 109.23 - ], - "text": "1\nh(t)", - "type": "text" - }, - { - "block_id": "p194-b7", - "global_id": 5330, - "bbox": [ - 278.89, - 115.63, - 292.11, - 125.24 - ], - "text": "e2t", - "type": "text" - }, - { - "block_id": "p194-b8", - "global_id": 5331, - "bbox": [ - 260.2, - 165.88, - 355.12, - 174.49 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p194-b9", - "global_id": 5332, - "bbox": [ - 302.13, - 185.83, - 311.77, - 193.83 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p194-b10", - "global_id": 5333, - "bbox": [ - 229.62, - 330.16, - 238.5, - 338.16 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p194-b11", - "global_id": 5334, - "bbox": [ - 219.37, - 225.93, - 253.47, - 235.55 - ], - "text": "et e2t", - "type": "text" - }, - { - "block_id": "p194-b12", - "global_id": 5335, - "bbox": [ - 164.48, - 206.58, - 188.37, - 220.55 - ], - "text": "1\ny(t)", - "type": "text" - }, - { - "block_id": "p194-b13", - "global_id": 5336, - "bbox": [ - 177.71, - 309.66, - 188.37, - 317.95 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p194-b14", - "global_id": 5337, - "bbox": [ - 182.81, - 268.21, - 287.13, - 276.32 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p194-b15", - "global_id": 5338, - "bbox": [ - 260.01, - 234.7, - 270.23, - 244.31 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p194-b16", - "global_id": 5339, - "bbox": [ - 94.2, - 344.48, - 242.89, - 354.09 - ], - "text": "Figure 2.6 Convolution of x(t) and h(t).", - "type": "text" - }, - { - "block_id": "p194-b17", - "global_id": 5340, - "bbox": [ - 103.17, - 365.19, - 477.0, - 390.72 - ], - "text": "Because this integration is with respect to τ, we can pull e−2t outside the integral, giving\nus", - "type": "text" - }, - { - "block_id": "p194-b18", - "global_id": 5341, - "bbox": [ - 176.88, - 394.31, - 218.83, - 406.31 - ], - "text": "y(t) = e−2t", - "type": "text" - }, - { - "block_id": "p194-b19", - "global_id": 5342, - "bbox": [ - 220.58, - 382.48, - 232.34, - 394.75 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p194-b20", - "global_id": 5343, - "bbox": [ - 225.83, - 391.92, - 403.3, - 414.61 - ], - "text": "0\neτ dτ = e−2t(et −1) = e−t −e−2t\nt ≥0", - "type": "text" - }, - { - "block_id": "p194-b21", - "global_id": 5344, - "bbox": [ - 103.17, - 419.85, - 335.93, - 430.22 - ], - "text": "Moreover, y(t) = 0 when t < 0 [see Eq. (2.30)]. Therefore,", - "type": "text" - }, - { - "block_id": "p194-b22", - "global_id": 5345, - "bbox": [ - 245.75, - 437.65, - 334.43, - 452.04 - ], - "text": "y(t) = (e−t −e−2t)u(t)", - "type": "text" - }, - { - "block_id": "p194-b23", - "global_id": 5346, - "bbox": [ - 103.16, - 464.1, - 251.75, - 474.06 - ], - "text": "The response is depicted in Fig. 2.6c.", - "type": "text" - }, - { - "block_id": "p194-b24", - "global_id": 5347, - "bbox": [ - 107.82, - 528.45, - 380.39, - 540.41 - ], - "text": "DRILL 2.5\nComputing the Zero-State Response", - "type": "text" - }, - { - "block_id": "p194-b25", - "global_id": 5348, - "bbox": [ - 107.82, - 547.89, - 484.41, - 571.45 - ], - "text": "For an LTIC system with the impulse response h(t) = 6e−tu(t), determine the system response\nto the input: (a) 2u(t) and (b) 3e−3tu(t).", - "type": "text" - }, - { - "block_id": "p194-b26", - "global_id": 5349, - "bbox": [ - 108.09, - 584.97, - 170.31, - 595.93 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p194-b27", - "global_id": 5350, - "bbox": [ - 125.76, - 601.69, - 209.9, - 628.23 - ], - "text": "(a) 12(1 −e−t)u(t)\n(b) 9(e−t −e−3t)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 195, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p195-b0", - "global_id": 5351, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n175", - "type": "text" - }, - { - "block_id": "p195-b1", - "global_id": 5352, - "bbox": [ - 133.57, - 97.81, - 410.11, - 109.77 - ], - "text": "DRILL 2.6\nZero-State Response with Resonance", - "type": "text" - }, - { - "block_id": "p195-b2", - "global_id": 5353, - "bbox": [ - 133.57, - 117.25, - 308.7, - 128.85 - ], - "text": "Repeat Drill 2.5 for the input x(t) = e−tu(t).", - "type": "text" - }, - { - "block_id": "p195-b3", - "global_id": 5354, - "bbox": [ - 133.84, - 142.38, - 188.43, - 153.33 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p195-b4", - "global_id": 5355, - "bbox": [ - 133.57, - 159.09, - 169.1, - 170.69 - ], - "text": "6te−tu(t)", - "type": "text" - }, - { - "block_id": "p195-b5", - "global_id": 5356, - "bbox": [ - 127.59, - 206.22, - 516.16, - 281.22 - ], - "text": "THE CONVOLUTION TABLE\nThe task of convolution is considerably simplified by a ready-made convolution table (Table 2.1).\nThis table, which lists several pairs of signals and their convolution, can conveniently determine\ny(t), a system response to an input x(t), without performing the tedious job of integration. For\ninstance, we could have readily found the convolution in Ex. 2.8 by using pair 4 (with λ1 = −1\nand λ2 = −2) to be (e−t −e−2t)u(t). The following example demonstrates the utility of this table.", - "type": "text" - }, - { - "block_id": "p195-b6", - "global_id": 5357, - "bbox": [ - 102.51, - 305.72, - 324.9, - 317.67 - ], - "text": "EXAMPLE 2.9\nConvolution by Tables", - "type": "text" - }, - { - "block_id": "p195-b7", - "global_id": 5358, - "bbox": [ - 128.9, - 333.92, - 502.75, - 356.26 - ], - "text": "Use Table 2.1 to compute the loop current y(t) of the RLC circuit in Ex. 2.4 for the input\nx(t) = 10e−3tu(t) when all the initial conditions are zero.", - "type": "text" - }, - { - "block_id": "p195-b8", - "global_id": 5359, - "bbox": [ - 128.9, - 379.17, - 380.7, - 389.13 - ], - "text": "The loop equation for this circuit [see Ex. 1.16 or Eq. (1.29)] is", - "type": "text" - }, - { - "block_id": "p195-b9", - "global_id": 5360, - "bbox": [ - 262.73, - 396.56, - 368.94, - 411.05 - ], - "text": "(D2 + 3D + 2)y(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p195-b10", - "global_id": 5361, - "bbox": [ - 128.91, - 422.59, - 397.74, - 432.96 - ], - "text": "The impulse response h(t) for this system, as obtained in Ex. 2.6, is", - "type": "text" - }, - { - "block_id": "p195-b11", - "global_id": 5362, - "bbox": [ - 268.73, - 440.4, - 362.94, - 454.89 - ], - "text": "h(t) = (2e−2t −e−t)u(t)", - "type": "text" - }, - { - "block_id": "p195-b12", - "global_id": 5363, - "bbox": [ - 128.91, - 465.2, - 344.76, - 476.8 - ], - "text": "The input is x(t) = 10e−3tu(t), and the response y(t) is", - "type": "text" - }, - { - "block_id": "p195-b13", - "global_id": 5364, - "bbox": [ - 219.46, - 484.23, - 412.21, - 498.72 - ], - "text": "y(t) = x(t) ∗h(t) = 10e−3tu(t) ∗[2e−2t −e−t]u(t)", - "type": "text" - }, - { - "block_id": "p195-b14", - "global_id": 5365, - "bbox": [ - 128.9, - 510.68, - 419.07, - 520.64 - ], - "text": "Using the distributive property of the convolution [Eq. (2.26)], we obtain", - "type": "text" - }, - { - "block_id": "p195-b15", - "global_id": 5366, - "bbox": [ - 211.37, - 530.46, - 412.01, - 542.56 - ], - "text": "y(t) = 10e−3tu(t) ∗2e−2tu(t) −10e−3tu(t) ∗e−tu(t)", - "type": "text" - }, - { - "block_id": "p195-b16", - "global_id": 5367, - "bbox": [ - 228.22, - 546.89, - 420.3, - 558.99 - ], - "text": "= 20[e−3tu(t) ∗e−2tu(t)] −10[e−3tu(t) ∗e−tu(t)]", - "type": "text" - }, - { - "block_id": "p195-b17", - "global_id": 5368, - "bbox": [ - 128.91, - 570.94, - 290.56, - 580.91 - ], - "text": "Now the use of pair 4 in Table 2.1 yields", - "type": "text" - }, - { - "block_id": "p195-b18", - "global_id": 5369, - "bbox": [ - 187.92, - 590.85, - 443.75, - 614.86 - ], - "text": "y(t) =\n20\n−3 −(−2)[e−3t −e−2t]u(t) −\n10\n−3 −(−1)[e−3t −e−t]u(t)", - "type": "text" - }, - { - "block_id": "p195-b19", - "global_id": 5370, - "bbox": [ - 204.77, - 614.96, - 375.67, - 629.45 - ], - "text": "= −20(e−3t −e−2t)u(t) + 5(e−3t −e−t)u(t)", - "type": "text" - }, - { - "block_id": "p195-b20", - "global_id": 5371, - "bbox": [ - 204.77, - 631.4, - 336.05, - 645.89 - ], - "text": "= (−5e−t + 20e−2t −15e−3t)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 196, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p196-b0", - "global_id": 5372, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "176\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p196-b1", - "global_id": 5373, - "bbox": [ - 101.84, - 86.07, - 257.0, - 95.3 - ], - "text": "TABLE 2.1\nSelect Convolution Integrals", - "type": "text" - }, - { - "block_id": "p196-b2", - "global_id": 5374, - "bbox": [ - 101.84, - 106.02, - 435.07, - 116.04 - ], - "text": "No.\nx1(t)\nx2(t)\nx1(t) ∗x2(t) = x2(t) ∗x1(t)", - "type": "text" - }, - { - "block_id": "p196-b3", - "global_id": 5375, - "bbox": [ - 110.57, - 124.35, - 367.98, - 133.69 - ], - "text": "1\nx(t)\nδ(t −T)\nx(t −T)", - "type": "text" - }, - { - "block_id": "p196-b4", - "global_id": 5376, - "bbox": [ - 110.57, - 141.88, - 363.95, - 158.76 - ], - "text": "2\neλtu(t)\nu(t)\n1 −eλt", - "type": "text" - }, - { - "block_id": "p196-b5", - "global_id": 5377, - "bbox": [ - 346.51, - 149.42, - 380.59, - 164.75 - ], - "text": "−λ\nu(t)", - "type": "text" - }, - { - "block_id": "p196-b6", - "global_id": 5378, - "bbox": [ - 110.56, - 168.35, - 355.49, - 177.69 - ], - "text": "3\nu(t)\nu(t)\ntu(t)", - "type": "text" - }, - { - "block_id": "p196-b7", - "global_id": 5379, - "bbox": [ - 110.56, - 183.88, - 375.88, - 202.76 - ], - "text": "4\neλ1tu(t)\neλ2tu(t)\neλ1t −eλ2t", - "type": "text" - }, - { - "block_id": "p196-b8", - "global_id": 5380, - "bbox": [ - 344.9, - 199.79, - 371.46, - 210.1 - ], - "text": "λ1 −λ2", - "type": "text" - }, - { - "block_id": "p196-b9", - "global_id": 5381, - "bbox": [ - 378.68, - 193.42, - 438.18, - 203.73 - ], - "text": "u(t)\nλ1̸ = λ2", - "type": "text" - }, - { - "block_id": "p196-b10", - "global_id": 5382, - "bbox": [ - 110.57, - 211.09, - 365.42, - 221.69 - ], - "text": "5\neλtu(t)\neλtu(t)\nteλtu(t)", - "type": "text" - }, - { - "block_id": "p196-b11", - "global_id": 5383, - "bbox": [ - 110.56, - 230.09, - 376.2, - 251.71 - ], - "text": "6\nteλtu(t)\neλtu(t)\n1\n2t2eλtu(t)", - "type": "text" - }, - { - "block_id": "p196-b12", - "global_id": 5384, - "bbox": [ - 110.56, - 256.33, - 359.69, - 273.22 - ], - "text": "7\ntNu(t)\neλtu(t)\nN!eλt", - "type": "text" - }, - { - "block_id": "p196-b13", - "global_id": 5385, - "bbox": [ - 341.13, - 263.88, - 384.71, - 279.21 - ], - "text": "λN+1 u(t) −", - "type": "text" - }, - { - "block_id": "p196-b14", - "global_id": 5386, - "bbox": [ - 386.1, - 254.77, - 398.79, - 264.33 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p196-b15", - "global_id": 5387, - "bbox": [ - 386.8, - 276.8, - 398.07, - 283.55 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p196-b16", - "global_id": 5388, - "bbox": [ - 411.17, - 256.33, - 436.47, - 266.84 - ], - "text": "N!tN−k", - "type": "text" - }, - { - "block_id": "p196-b17", - "global_id": 5389, - "bbox": [ - 400.98, - 263.88, - 463.28, - 279.49 - ], - "text": "λk+1(N −k)! u(t)", - "type": "text" - }, - { - "block_id": "p196-b18", - "global_id": 5390, - "bbox": [ - 110.57, - 290.31, - 430.8, - 312.3 - ], - "text": "8\ntMu(t)\ntNu(t)\nM!N!\n(M + N + 1)! tM+N+1u(t)", - "type": "text" - }, - { - "block_id": "p196-b19", - "global_id": 5391, - "bbox": [ - 110.56, - 321.1, - 435.09, - 339.98 - ], - "text": "9\nteλ1tu(t)\neλ2tu(t)\neλ2t −eλ1t + (λ1 −λ2)teλ1t", - "type": "text" - }, - { - "block_id": "p196-b20", - "global_id": 5392, - "bbox": [ - 369.29, - 330.64, - 451.72, - 347.31 - ], - "text": "(λ1 −λ2)2\nu(t)", - "type": "text" - }, - { - "block_id": "p196-b21", - "global_id": 5393, - "bbox": [ - 106.07, - 357.07, - 440.74, - 379.06 - ], - "text": "10\ntMeλtu(t)\ntNeλtu(t)\nM!N!\n(N + M + 1)! tM+N+1eλtu(t)", - "type": "text" - }, - { - "block_id": "p196-b22", - "global_id": 5394, - "bbox": [ - 106.07, - 398.93, - 300.35, - 409.54 - ], - "text": "11\ntMeλ1tu(t)\ntNeλ2tu(t)", - "type": "text" - }, - { - "block_id": "p196-b23", - "global_id": 5395, - "bbox": [ - 339.17, - 391.08, - 351.85, - 400.64 - ], - "text": "M\n\"", - "type": "text" - }, - { - "block_id": "p196-b24", - "global_id": 5396, - "bbox": [ - 339.87, - 413.13, - 351.13, - 419.87 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p196-b25", - "global_id": 5397, - "bbox": [ - 354.16, - 392.65, - 446.13, - 403.25 - ], - "text": "(−1)kM!(N + k)!tM−keλ1t", - "type": "text" - }, - { - "block_id": "p196-b26", - "global_id": 5398, - "bbox": [ - 354.04, - 400.2, - 462.88, - 416.87 - ], - "text": "k!(M −k)!(λ1 −λ2)N+k+1 u(t)", - "type": "text" - }, - { - "block_id": "p196-b27", - "global_id": 5399, - "bbox": [ - 153.86, - 437.05, - 355.12, - 447.35 - ], - "text": "λ1̸ = λ2\n+", - "type": "text" - }, - { - "block_id": "p196-b28", - "global_id": 5400, - "bbox": [ - 356.12, - 427.93, - 368.81, - 437.49 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p196-b29", - "global_id": 5401, - "bbox": [ - 356.82, - 449.97, - 368.09, - 456.72 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p196-b30", - "global_id": 5402, - "bbox": [ - 371.4, - 429.5, - 462.43, - 440.09 - ], - "text": "(−1)kN!(M + k)!tN−keλ2t", - "type": "text" - }, - { - "block_id": "p196-b31", - "global_id": 5403, - "bbox": [ - 371.0, - 437.05, - 479.47, - 453.72 - ], - "text": "k!(N −k)!(λ2 −λ1)M+k+1 u(t)", - "type": "text" - }, - { - "block_id": "p196-b32", - "global_id": 5404, - "bbox": [ - 106.07, - 461.54, - 474.38, - 480.42 - ], - "text": "12\ne−αt cos(βt + θ)u(t)\neλtu(t)\ncos(θ −φ)eλt −e−αt cos(βt + θ −φ)", - "type": "text" - }, - { - "block_id": "p196-b33", - "global_id": 5405, - "bbox": [ - 386.18, - 471.08, - 490.39, - 488.01 - ], - "text": "(α + λ)2 + β2\nu(t)", - "type": "text" - }, - { - "block_id": "p196-b34", - "global_id": 5406, - "bbox": [ - 339.17, - 497.71, - 424.75, - 508.31 - ], - "text": "φ = tan−1[−β/(α + λ)]", - "type": "text" - }, - { - "block_id": "p196-b35", - "global_id": 5407, - "bbox": [ - 106.07, - 525.47, - 411.14, - 542.35 - ], - "text": "13\neλ1tu(t)\neλ2tu(−t)\neλ1tu(t) + eλ2tu(−t)", - "type": "text" - }, - { - "block_id": "p196-b36", - "global_id": 5408, - "bbox": [ - 362.21, - 539.38, - 388.78, - 549.68 - ], - "text": "λ2 −λ1", - "type": "text" - }, - { - "block_id": "p196-b37", - "global_id": 5409, - "bbox": [ - 421.3, - 533.01, - 470.68, - 543.32 - ], - "text": "Reλ2 > Reλ1", - "type": "text" - }, - { - "block_id": "p196-b38", - "global_id": 5410, - "bbox": [ - 106.07, - 557.51, - 375.88, - 576.39 - ], - "text": "14\neλ1tu(−t)\neλ2tu(−t)\neλ1t −eλ2t", - "type": "text" - }, - { - "block_id": "p196-b39", - "global_id": 5411, - "bbox": [ - 344.9, - 573.42, - 371.46, - 583.73 - ], - "text": "λ2 −λ1", - "type": "text" - }, - { - "block_id": "p196-b40", - "global_id": 5412, - "bbox": [ - 378.68, - 567.05, - 399.51, - 576.3 - ], - "text": "u(−t)", - "type": "text" - } - ] - }, - { - "page_num": 197, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p197-b0", - "global_id": 5413, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n177", - "type": "text" - }, - { - "block_id": "p197-b1", - "global_id": 5414, - "bbox": [ - 133.57, - 97.81, - 332.77, - 109.76 - ], - "text": "DRILL 2.7\nConvolution by Tables", - "type": "text" - }, - { - "block_id": "p197-b2", - "global_id": 5415, - "bbox": [ - 133.57, - 117.24, - 324.84, - 128.85 - ], - "text": "Use Table 2.1 to show e−2tu(t) ∗(1 −e−t)u(t) =", - "type": "text" - }, - { - "block_id": "p197-b3", - "global_id": 5416, - "bbox": [ - 326.89, - 110.46, - 335.59, - 124.1 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p197-b4", - "global_id": 5417, - "bbox": [ - 332.1, - 110.46, - 415.71, - 131.97 - ], - "text": "2 −e−t + 1\n2e−2t\nu(t).", - "type": "text" - }, - { - "block_id": "p197-b5", - "global_id": 5418, - "bbox": [ - 133.57, - 182.78, - 442.34, - 194.73 - ], - "text": "DRILL 2.8\nZero-State Response by Convolution Table", - "type": "text" - }, - { - "block_id": "p197-b6", - "global_id": 5419, - "bbox": [ - 133.57, - 203.85, - 302.59, - 213.82 - ], - "text": "Rework Drills 2.5 and 2.6 using Table 2.1.", - "type": "text" - }, - { - "block_id": "p197-b7", - "global_id": 5420, - "bbox": [ - 133.57, - 266.36, - 490.5, - 278.32 - ], - "text": "DRILL 2.9\nAnother Zero-State Response by Convolution Table", - "type": "text" - }, - { - "block_id": "p197-b8", - "global_id": 5421, - "bbox": [ - 133.57, - 285.8, - 510.15, - 309.35 - ], - "text": "For an LTIC system with the unit impulse response h(t) = e−2tu(t), determine the zero-state\nresponse y(t) if the input x(t) = sin 3tu(t). [Hint: Use pair 12 from Table 2.1.]", - "type": "text" - }, - { - "block_id": "p197-b9", - "global_id": 5422, - "bbox": [ - 133.84, - 322.88, - 188.43, - 333.84 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p197-b10", - "global_id": 5423, - "bbox": [ - 134.76, - 332.45, - 186.43, - 354.02 - ], - "text": "1\n13[3e−2t +\n√", - "type": "text" - }, - { - "block_id": "p197-b11", - "global_id": 5424, - "bbox": [ - 186.45, - 332.45, - 354.35, - 354.02 - ], - "text": "13cos(3t −146.32◦)]u(t) or 1\n13[3e−2t −\n√", - "type": "text" - }, - { - "block_id": "p197-b12", - "global_id": 5425, - "bbox": [ - 354.35, - 339.6, - 451.13, - 351.2 - ], - "text": "13cos(3t + 33.68◦)]u(t)", - "type": "text" - }, - { - "block_id": "p197-b13", - "global_id": 5426, - "bbox": [ - 127.89, - 386.84, - 299.01, - 398.96 - ], - "text": "RESPONSE TO COMPLEX INPUTS", - "type": "text" - }, - { - "block_id": "p197-b14", - "global_id": 5427, - "bbox": [ - 127.59, - 402.99, - 516.13, - 436.87 - ], - "text": "The LTIC system response discussed so far applies to general input signals, real or complex.\nHowever, if the system is real, that is, if h(t) is real, then we shall show that the real part of the\ninput generates the real part of the output, and a similar conclusion applies to the imaginary part.", - "type": "text" - }, - { - "block_id": "p197-b15", - "global_id": 5428, - "bbox": [ - 127.59, - 438.44, - 516.12, - 460.77 - ], - "text": "If the input is x(t) = xr(t) + jxi(t), where xr(t) and xi(t) are the real and imaginary parts of\nx(t), then for real h(t)", - "type": "text" - }, - { - "block_id": "p197-b16", - "global_id": 5429, - "bbox": [ - 184.28, - 479.03, - 459.45, - 490.1 - ], - "text": "y(t) = h(t) ∗[xr(t) + jxi(t)] = h(t) ∗xr(t) + jh(t) ∗xi(t) = yr(t) + jyi(t)", - "type": "text" - }, - { - "block_id": "p197-b17", - "global_id": 5430, - "bbox": [ - 127.59, - 507.65, - 516.13, - 541.94 - ], - "text": "where yr(t) and yi(t) are the real and the imaginary parts of y(t). Using the right-directed-arrow\nnotation to indicate a pair of the input and the corresponding output, the foregoing result can be\nexpressed as follows. If", - "type": "text" - }, - { - "block_id": "p197-b18", - "global_id": 5431, - "bbox": [ - 227.91, - 560.2, - 415.82, - 571.28 - ], - "text": "x(t) = xr(t) + jxi(t)\n\r⇒\ny(t) = yr(t) + jyi(t)", - "type": "text" - }, - { - "block_id": "p197-b19", - "global_id": 5432, - "bbox": [ - 127.59, - 589.24, - 144.74, - 599.2 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p197-b20", - "global_id": 5433, - "bbox": [ - 222.89, - 610.84, - 516.13, - 621.92 - ], - "text": "xr(t)\n\r⇒\nyr(t)\nand\nxi(t)\n\r⇒\nyi(t)\n(2.31)", - "type": "text" - } - ] - }, - { - "page_num": 198, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p198-b0", - "global_id": 5434, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "178\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p198-b1", - "global_id": 5435, - "bbox": [ - 102.14, - 86.19, - 199.36, - 98.32 - ], - "text": "MULTIPLE INPUTS", - "type": "text" - }, - { - "block_id": "p198-b2", - "global_id": 5436, - "bbox": [ - 101.84, - 102.35, - 490.4, - 136.22 - ], - "text": "Multiple inputs to LTI systems can be treated by applying the superposition principle. Each input\nis considered separately, with all other inputs assumed to be zero. The sum of all these individual\nsystem responses constitutes the total system output when all the inputs are applied simultaneously.", - "type": "text" - }, - { - "block_id": "p198-b3", - "global_id": 5437, - "bbox": [ - 101.84, - 161.73, - 417.16, - 173.69 - ], - "text": "2.4-2 Graphical Understanding of Convolution Operation", - "type": "text" - }, - { - "block_id": "p198-b4", - "global_id": 5438, - "bbox": [ - 101.84, - 179.82, - 490.41, - 261.51 - ], - "text": "The convolution operation can be grasped readily through a graphical interpretation of the\nconvolution integral. Such an understanding is helpful in evaluating the convolution integral of\nmore complex signals. In addition, graphical convolution allows us to grasp visually or mentally\nthe convolution integral’s result, which can be of great help in sampling, filtering, and many other\nproblems. Finally, many signals have no exact mathematical description, so they can be described\nonly graphically. If two such signals are to be convolved, we have no choice but to perform their\nconvolution graphically.", - "type": "text" - }, - { - "block_id": "p198-b5", - "global_id": 5439, - "bbox": [ - 101.84, - 263.1, - 490.39, - 285.43 - ], - "text": "We shall now explain the convolution operation by convolving the signals x(t) and g(t),\nillustrated in Figs. 2.7a and 2.7b, respectively. If c(t) is the convolution of x(t) with g(t), then", - "type": "text" - }, - { - "block_id": "p198-b6", - "global_id": 5440, - "bbox": [ - 242.57, - 303.05, - 267.19, - 313.33 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p198-b7", - "global_id": 5441, - "bbox": [ - 269.23, - 289.5, - 286.18, - 301.55 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p198-b8", - "global_id": 5442, - "bbox": [ - 274.5, - 314.37, - 287.06, - 321.34 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p198-b9", - "global_id": 5443, - "bbox": [ - 288.67, - 303.05, - 348.47, - 313.32 - ], - "text": "x(τ)g(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p198-b10", - "global_id": 5444, - "bbox": [ - 101.85, - 331.07, - 490.4, - 377.31 - ], - "text": "One of the crucial points to remember here is that this integration is performed with respect to τ\nso that t is just a parameter (like a constant). This consideration is especially important when we\nsketch the graphical representations of the functions x(τ) and g(t−τ). Both these functions should\nbe sketched as functions of τ, not of t.", - "type": "text" - }, - { - "block_id": "p198-b11", - "global_id": 5445, - "bbox": [ - 101.85, - 378.88, - 490.39, - 401.21 - ], - "text": "The function x(τ) is identical to x(t), with τ replacing t (Fig. 2.7c). Therefore, x(t) and x(τ)\nwill have the same graphical representations. Similar remarks apply to g(t) and g(τ) (Fig. 2.7d).", - "type": "text" - }, - { - "block_id": "p198-b12", - "global_id": 5446, - "bbox": [ - 101.85, - 402.8, - 490.39, - 437.09 - ], - "text": "To appreciate what g(t −τ) looks like, let us start with the function g(τ) (Fig. 2.7d). Time\nreversal of this function (reflection about the vertical axis τ = 0) yields g(−τ) (Fig. 2.7e). Let us\ndenote this function by φ(τ):", - "type": "text" - }, - { - "block_id": "p198-b13", - "global_id": 5447, - "bbox": [ - 267.55, - 439.8, - 324.69, - 450.08 - ], - "text": "φ(τ) = g(−τ)", - "type": "text" - }, - { - "block_id": "p198-b14", - "global_id": 5448, - "bbox": [ - 101.85, - 459.5, - 308.78, - 469.87 - ], - "text": "Now φ(τ) shifted by t seconds is φ(τ −t), given by", - "type": "text" - }, - { - "block_id": "p198-b15", - "global_id": 5449, - "bbox": [ - 228.6, - 482.18, - 363.63, - 492.45 - ], - "text": "φ(τ −t) = g[−(τ −t)] = g(t −τ)", - "type": "text" - }, - { - "block_id": "p198-b16", - "global_id": 5450, - "bbox": [ - 101.85, - 504.86, - 490.4, - 539.15 - ], - "text": "Therefore, we first time-reverse g(τ) to obtain g(−τ) and then time-shift g(−τ) by t to obtain\ng(t −τ). For positive t, the shift is to the right (Fig. 2.7f); for negative t, the shift is to the left\n(Figs. 2.7g, 2.7h).", - "type": "text" - }, - { - "block_id": "p198-b17", - "global_id": 5451, - "bbox": [ - 101.84, - 540.72, - 490.4, - 634.79 - ], - "text": "The preceding discussion gives us a graphical interpretation of the functions x(τ) and g(t−τ).\nThe convolution c(t) is the area under the product of these two functions. Thus, to compute c(t)\nat some positive instant t = t1, we first obtain g(−τ) by inverting g(τ) about the vertical axis.\nNext, we right-shift or delay g(−τ) by t1 to obtain g(t1 −τ) (Fig. 2.7f), and then we multiply this\nfunction by x(τ), giving us the product x(τ)g(t1 −τ) (shaded portion in Fig. 2.7f). The area A1\nunder this product is c(t1), the value of c(t) at t = t1. We can therefore plot c(t1) = A1 on a curve\ndescribing c(t), as shown in Fig. 2.7i. The area under the product x(τ)g(−τ) in Fig. 2.7e is c(0),\nthe value of the convolution for t = 0 (at the origin).", - "type": "text" - } - ] - }, - { - "page_num": 199, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p199-b0", - "global_id": 5452, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n179", - "type": "text" - }, - { - "block_id": "p199-b1", - "global_id": 5453, - "bbox": [ - 228.65, - 87.41, - 239.76, - 95.49 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p199-b2", - "global_id": 5454, - "bbox": [ - 237.01, - 155.79, - 245.89, - 163.79 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p199-b3", - "global_id": 5455, - "bbox": [ - 361.01, - 87.41, - 372.56, - 95.49 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p199-b4", - "global_id": 5456, - "bbox": [ - 287.1, - 227.16, - 470.65, - 235.16 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p199-b5", - "global_id": 5457, - "bbox": [ - 224.42, - 137.19, - 228.42, - 145.19 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p199-b6", - "global_id": 5458, - "bbox": [ - 405.7, - 157.29, - 415.34, - 165.29 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p199-b7", - "global_id": 5459, - "bbox": [ - 385.02, - 137.19, - 389.02, - 145.19 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p199-b8", - "global_id": 5460, - "bbox": [ - 237.01, - 245.36, - 415.18, - 253.36 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p199-b9", - "global_id": 5461, - "bbox": [ - 175.46, - 294.11, - 184.34, - 302.11 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p199-b10", - "global_id": 5462, - "bbox": [ - 436.1, - 313.94, - 439.65, - 321.94 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p199-b11", - "global_id": 5463, - "bbox": [ - 288.68, - 138.3, - 470.39, - 146.3 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p199-b12", - "global_id": 5464, - "bbox": [ - 205.38, - 176.66, - 373.89, - 189.78 - ], - "text": "g(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p199-b13", - "global_id": 5465, - "bbox": [ - 224.42, - 226.46, - 388.92, - 234.46 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p199-b14", - "global_id": 5466, - "bbox": [ - 324.52, - 315.96, - 328.52, - 323.96 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p199-b15", - "global_id": 5467, - "bbox": [ - 278.64, - 280.9, - 298.18, - 289.19 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p199-b16", - "global_id": 5468, - "bbox": [ - 387.48, - 274.34, - 399.92, - 282.44 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p199-b17", - "global_id": 5469, - "bbox": [ - 175.23, - 369.26, - 184.34, - 377.26 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p199-b18", - "global_id": 5470, - "bbox": [ - 175.01, - 443.66, - 184.34, - 451.66 - ], - "text": "(g)", - "type": "text" - }, - { - "block_id": "p199-b19", - "global_id": 5471, - "bbox": [ - 176.79, - 594.86, - 184.34, - 602.86 - ], - "text": "(i)", - "type": "text" - }, - { - "block_id": "p199-b20", - "global_id": 5472, - "bbox": [ - 324.52, - 618.16, - 328.52, - 626.16 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p199-b21", - "global_id": 5473, - "bbox": [ - 312.15, - 568.17, - 323.25, - 576.25 - ], - "text": "c(t)", - "type": "text" - }, - { - "block_id": "p199-b22", - "global_id": 5474, - "bbox": [ - 301.91, - 617.96, - 380.63, - 627.57 - ], - "text": "t1\nt2", - "type": "text" - }, - { - "block_id": "p199-b23", - "global_id": 5475, - "bbox": [ - 306.72, - 598.48, - 385.82, - 612.49 - ], - "text": "A2\nA1", - "type": "text" - }, - { - "block_id": "p199-b24", - "global_id": 5476, - "bbox": [ - 197.19, - 138.49, - 350.35, - 146.78 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p199-b25", - "global_id": 5477, - "bbox": [ - 197.19, - 226.75, - 350.35, - 235.55 - ], - "text": "2\n1", - "type": "text" - }, - { - "block_id": "p199-b26", - "global_id": 5478, - "bbox": [ - 224.42, - 95.23, - 388.92, - 115.19 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p199-b27", - "global_id": 5479, - "bbox": [ - 224.42, - 184.51, - 388.92, - 204.46 - ], - "text": "2\n1", - "type": "text" - }, - { - "block_id": "p199-b28", - "global_id": 5480, - "bbox": [ - 306.69, - 315.76, - 317.35, - 324.05 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p199-b29", - "global_id": 5481, - "bbox": [ - 333.92, - 283.46, - 337.92, - 291.46 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p199-b30", - "global_id": 5482, - "bbox": [ - 364.02, - 315.46, - 368.02, - 323.46 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p199-b31", - "global_id": 5483, - "bbox": [ - 324.52, - 391.26, - 328.52, - 399.26 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p199-b32", - "global_id": 5484, - "bbox": [ - 254.74, - 359.94, - 287.52, - 378.76 - ], - "text": "g(t t)\nt t1 0", - "type": "text" - }, - { - "block_id": "p199-b33", - "global_id": 5485, - "bbox": [ - 235.56, - 435.74, - 268.34, - 454.56 - ], - "text": "g(t t)\nt t2 0", - "type": "text" - }, - { - "block_id": "p199-b34", - "global_id": 5486, - "bbox": [ - 203.7, - 503.52, - 243.14, - 522.34 - ], - "text": "g(t t)\nt t3 3", - "type": "text" - }, - { - "block_id": "p199-b35", - "global_id": 5487, - "bbox": [ - 394.1, - 390.67, - 413.98, - 400.49 - ], - "text": "2 t1", - "type": "text" - }, - { - "block_id": "p199-b36", - "global_id": 5488, - "bbox": [ - 439.82, - 357.63, - 452.26, - 365.74 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p199-b37", - "global_id": 5489, - "bbox": [ - 439.82, - 432.21, - 452.26, - 440.31 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p199-b38", - "global_id": 5490, - "bbox": [ - 439.82, - 506.96, - 452.26, - 515.06 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p199-b39", - "global_id": 5491, - "bbox": [ - 333.92, - 358.76, - 337.92, - 366.76 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p199-b40", - "global_id": 5492, - "bbox": [ - 323.12, - 464.02, - 368.02, - 472.02 - ], - "text": "0\n2", - "type": "text" - }, - { - "block_id": "p199-b41", - "global_id": 5493, - "bbox": [ - 175.01, - 519.66, - 184.34, - 527.66 - ], - "text": "(h)", - "type": "text" - }, - { - "block_id": "p199-b42", - "global_id": 5494, - "bbox": [ - 306.69, - 540.19, - 327.12, - 548.49 - ], - "text": "0\n1", - "type": "text" - }, - { - "block_id": "p199-b43", - "global_id": 5495, - "bbox": [ - 381.15, - 330.87, - 386.38, - 340.48 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p199-b44", - "global_id": 5496, - "bbox": [ - 336.08, - 463.73, - 355.96, - 473.55 - ], - "text": "2 t2", - "type": "text" - }, - { - "block_id": "p199-b45", - "global_id": 5497, - "bbox": [ - 364.02, - 390.96, - 368.02, - 398.96 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p199-b46", - "global_id": 5498, - "bbox": [ - 357.22, - 409.95, - 362.44, - 419.56 - ], - "text": "t2", - "type": "text" - }, - { - "block_id": "p199-b47", - "global_id": 5499, - "bbox": [ - 338.13, - 495.87, - 343.36, - 505.48 - ], - "text": "t3", - "type": "text" - }, - { - "block_id": "p199-b48", - "global_id": 5500, - "bbox": [ - 244.37, - 617.96, - 249.6, - 627.57 - ], - "text": "t3", - "type": "text" - }, - { - "block_id": "p199-b49", - "global_id": 5501, - "bbox": [ - 270.08, - 540.19, - 289.96, - 550.01 - ], - "text": "2 t3", - "type": "text" - }, - { - "block_id": "p199-b50", - "global_id": 5502, - "bbox": [ - 268.69, - 617.87, - 279.35, - 626.16 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p199-b51", - "global_id": 5503, - "bbox": [ - 384.12, - 372.08, - 392.01, - 381.69 - ], - "text": "A1", - "type": "text" - }, - { - "block_id": "p199-b52", - "global_id": 5504, - "bbox": [ - 332.02, - 444.98, - 339.91, - 454.59 - ], - "text": "A2", - "type": "text" - }, - { - "block_id": "p199-b53", - "global_id": 5505, - "bbox": [ - 436.1, - 388.94, - 439.65, - 396.94 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p199-b54", - "global_id": 5506, - "bbox": [ - 436.1, - 462.94, - 439.65, - 470.94 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p199-b55", - "global_id": 5507, - "bbox": [ - 436.1, - 538.94, - 439.65, - 546.94 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p199-b56", - "global_id": 5508, - "bbox": [ - 436.1, - 616.94, - 439.65, - 624.94 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p199-b57", - "global_id": 5509, - "bbox": [ - 151.5, - 636.44, - 382.18, - 645.67 - ], - "text": "Figure 2.7 Graphical explanation of the convolution operation.", - "type": "text" - } - ] - }, - { - "page_num": 200, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p200-b0", - "global_id": 5510, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "180\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p200-b1", - "global_id": 5511, - "bbox": [ - 101.84, - 85.46, - 490.39, - 167.57 - ], - "text": "A similar procedure is followed in computing the value of c(t) at t = t2, where t2 is negative\n(Fig. 2.7g). In this case, the function g(−τ) is shifted by a negative amount (that is, left-shifted) to\nobtain g(t2−τ). Multiplication of this function with x(τ) yields the product x(τ)g(t2−τ). The area\nunder this product is c(t2) = A2, giving us another point on the curve c(t) at t = t2 (Fig. 2.7i). This\nprocedure can be repeated for all values of t, from −∞to ∞. The result will be a curve describing\nc(t) for all time t. Note that when t ≤−3,x(τ) and g(t−τ) do not overlap (see Fig. 2.7h); therefore,\nc(t) = 0 for t ≤−3.", - "type": "text" - }, - { - "block_id": "p200-b2", - "global_id": 5512, - "bbox": [ - 102.14, - 184.34, - 336.25, - 196.46 - ], - "text": "SUMMARY OF THE GRAPHICAL PROCEDURE", - "type": "text" - }, - { - "block_id": "p200-b3", - "global_id": 5513, - "bbox": [ - 101.84, - 200.49, - 387.9, - 210.46 - ], - "text": "The procedure for graphical convolution can be summarized as follows:", - "type": "text" - }, - { - "block_id": "p200-b4", - "global_id": 5514, - "bbox": [ - 118.78, - 218.01, - 490.41, - 324.03 - ], - "text": "1. Keep the function x(τ) fixed.\n2. Visualize the function g(τ) as a rigid wire frame, and rotate (or invert) this frame about the\nvertical axis (τ = 0) to obtain g(−τ).\n3. Shift the inverted frame along the τ axis by t0 seconds. The shifted frame now represents\ng(t0 −τ).\n4. The area under the product of x(τ) and g(t0 −τ) (the shifted frame) is c(t0), the value of\nthe convolution at t = t0.\n5. Repeat this procedure, shifting the frame by different values (positive and negative) to\nobtain c(t) for all values of t.", - "type": "text" - }, - { - "block_id": "p200-b5", - "global_id": 5515, - "bbox": [ - 101.84, - 332.0, - 490.4, - 401.73 - ], - "text": "The graphical procedure discussed here appears very complicated and discouraging at first\nreading. Indeed, some people claim that convolution has driven many electrical engineering\nundergraduates to contemplate theology either for salvation or as an alternative career (IEEE\nSpectrum, March 1991, p. 60). Actually, the bark of convolution is worse than its bite. In graphical\nconvolution, we need to determine the area under the product x(τ)g(t −τ) for all values of t from\n−∞to ∞. However, a mathematical description of x(τ)g(t −τ) is generally valid over a range", - "type": "text" - }, - { - "block_id": "p200-b6", - "global_id": 5516, - "bbox": [ - 209.99, - 626.88, - 382.24, - 636.84 - ], - "text": "Convolution: Its bark is worse than its bite!", - "type": "text" - } - ] - }, - { - "page_num": 201, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p201-b0", - "global_id": 5517, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n181", - "type": "text" - }, - { - "block_id": "p201-b1", - "global_id": 5518, - "bbox": [ - 127.59, - 85.72, - 516.13, - 107.74 - ], - "text": "of t. Therefore, repeating the procedure for every value of t amounts to repeating it only a few\ntimes for different ranges of t.", - "type": "text" - }, - { - "block_id": "p201-b2", - "global_id": 5519, - "bbox": [ - 127.59, - 109.73, - 516.14, - 179.47 - ], - "text": "We can also use the commutative property of convolution to our advantage by computing\nx(t) ∗g(t) or g(t) ∗x(t), whichever is simpler. As a rule of thumb, convolution computations are\nsimplified if we choose to invert (time-reverse) the simpler of the two functions. For example, if\nthe mathematical description of g(t) is simpler than that of x(t), then x(t) ∗g(t) will be easier to\ncompute than g(t) ∗x(t). In contrast, if the mathematical description of x(t) is simpler, the reverse\nwill be true.", - "type": "text" - }, - { - "block_id": "p201-b3", - "global_id": 5520, - "bbox": [ - 127.59, - 181.46, - 516.13, - 203.37 - ], - "text": "We shall demonstrate graphical convolution with the following examples. Let us start by using\nthis graphical method to rework Ex. 2.8.", - "type": "text" - }, - { - "block_id": "p201-b4", - "global_id": 5521, - "bbox": [ - 103.92, - 233.29, - 472.34, - 245.24 - ], - "text": "EXAMPLE 2.10\nGraphical Convolution of Two Causal Functions", - "type": "text" - }, - { - "block_id": "p201-b5", - "global_id": 5522, - "bbox": [ - 128.9, - 261.54, - 437.37, - 273.15 - ], - "text": "Determine graphically y(t) = x(t) ∗h(t) for x(t) = e−tu(t) and h(t) = e−2tu(t).", - "type": "text" - }, - { - "block_id": "p201-b6", - "global_id": 5523, - "bbox": [ - 128.9, - 295.65, - 502.76, - 353.84 - ], - "text": "In Figs. 2.8a and 2.8b we have x(t) and h(t), respectively; and Fig. 2.8c shows x(τ) and h(−τ)\nas functions of τ. The function h(t −τ) is now obtained by shifting h(−τ) by t. If t is positive,\nthe shift is to the right (delay); if t is negative, the shift is to the left (advance). Figure 2.8d\nshows that for negative t, h(t −τ) [obtained by left-shifting h(−τ)] does not overlap x(τ), and\nthe product x(τ)h(t −τ) = 0, so that", - "type": "text" - }, - { - "block_id": "p201-b7", - "global_id": 5524, - "bbox": [ - 280.15, - 365.1, - 351.53, - 375.48 - ], - "text": "y(t) = 0\nt < 0", - "type": "text" - }, - { - "block_id": "p201-b8", - "global_id": 5525, - "bbox": [ - 128.91, - 386.73, - 502.77, - 409.07 - ], - "text": "Figure 2.8e shows the situation for t ≥0. Here x(τ) and h(t −τ) do overlap, but the product is\nnonzero only over the interval 0 ≤τ ≤t (shaded interval). Therefore,", - "type": "text" - }, - { - "block_id": "p201-b9", - "global_id": 5526, - "bbox": [ - 245.52, - 426.1, - 270.15, - 436.38 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p201-b10", - "global_id": 5527, - "bbox": [ - 272.19, - 412.55, - 283.95, - 424.83 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p201-b11", - "global_id": 5528, - "bbox": [ - 277.45, - 426.1, - 386.15, - 444.69 - ], - "text": "0\nx(τ)h(t −τ)dτ\nt ≥0", - "type": "text" - }, - { - "block_id": "p201-b12", - "global_id": 5529, - "bbox": [ - 128.91, - 452.62, - 502.77, - 486.91 - ], - "text": "All we need to do now is substitute correct expressions for x(τ) and h(t −τ) in this integral.\nFrom Figs. 2.8a and 2.8b, it is clear that the segments of x(t) and g(t) to be used in this\nconvolution (Fig. 2.8e) are described by", - "type": "text" - }, - { - "block_id": "p201-b13", - "global_id": 5530, - "bbox": [ - 247.58, - 496.44, - 383.45, - 508.53 - ], - "text": "x(t) = e−t\nand\nh(t) = e−2t", - "type": "text" - }, - { - "block_id": "p201-b14", - "global_id": 5531, - "bbox": [ - 128.9, - 520.21, - 170.65, - 530.17 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p201-b15", - "global_id": 5532, - "bbox": [ - 229.96, - 530.02, - 401.19, - 542.12 - ], - "text": "x(τ) = e−τ\nand\nh(t −τ) = e−2(t−τ)", - "type": "text" - }, - { - "block_id": "p201-b16", - "global_id": 5533, - "bbox": [ - 128.9, - 550.95, - 185.51, - 560.91 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p201-b17", - "global_id": 5534, - "bbox": [ - 180.76, - 577.95, - 205.38, - 588.23 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p201-b18", - "global_id": 5535, - "bbox": [ - 207.43, - 564.4, - 219.2, - 576.67 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p201-b19", - "global_id": 5536, - "bbox": [ - 212.69, - 574.15, - 304.56, - 596.53 - ], - "text": "0\ne−τe−2(t−τ) dτ = e−2t", - "type": "text" - }, - { - "block_id": "p201-b20", - "global_id": 5537, - "bbox": [ - 306.3, - 564.4, - 318.07, - 576.67 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p201-b21", - "global_id": 5538, - "bbox": [ - 311.57, - 573.84, - 450.9, - 596.53 - ], - "text": "0\neτ dτ = e−t −e−2t\nt ≥0", - "type": "text" - }, - { - "block_id": "p201-b22", - "global_id": 5539, - "bbox": [ - 128.91, - 604.47, - 269.3, - 614.84 - ], - "text": "Moreover, y(t) = 0 for t < 0 so that", - "type": "text" - }, - { - "block_id": "p201-b23", - "global_id": 5540, - "bbox": [ - 271.5, - 621.99, - 360.17, - 636.38 - ], - "text": "y(t) = (e−t −e−2t)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 202, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p202-b0", - "global_id": 5541, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "182\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p202-b1", - "global_id": 5542, - "bbox": [ - 230.06, - 218.94, - 234.06, - 226.94 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p202-b2", - "global_id": 5543, - "bbox": [ - 252.0, - 233.57, - 264.43, - 241.67 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p202-b3", - "global_id": 5544, - "bbox": [ - 319.78, - 282.95, - 323.33, - 290.95 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p202-b4", - "global_id": 5545, - "bbox": [ - 319.78, - 399.68, - 323.33, - 407.68 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p202-b5", - "global_id": 5546, - "bbox": [ - 319.78, - 517.37, - 323.33, - 525.37 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p202-b6", - "global_id": 5547, - "bbox": [ - 232.26, - 304.1, - 241.14, - 312.1 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p202-b7", - "global_id": 5548, - "bbox": [ - 208.4, - 232.78, - 227.95, - 241.07 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p202-b8", - "global_id": 5549, - "bbox": [ - 181.29, - 350.39, - 207.06, - 358.68 - ], - "text": "h(t t)", - "type": "text" - }, - { - "block_id": "p202-b9", - "global_id": 5550, - "bbox": [ - 182.1, - 485.52, - 207.87, - 493.82 - ], - "text": "h(t t)", - "type": "text" - }, - { - "block_id": "p202-b10", - "global_id": 5551, - "bbox": [ - 230.1, - 336.55, - 234.1, - 344.55 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p202-b11", - "global_id": 5552, - "bbox": [ - 252.17, - 351.08, - 264.61, - 359.18 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p202-b12", - "global_id": 5553, - "bbox": [ - 232.04, - 421.26, - 241.36, - 429.26 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p202-b13", - "global_id": 5554, - "bbox": [ - 239.6, - 283.07, - 243.6, - 291.07 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p202-b14", - "global_id": 5555, - "bbox": [ - 214.7, - 399.83, - 243.48, - 407.95 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p202-b15", - "global_id": 5556, - "bbox": [ - 230.08, - 453.48, - 234.08, - 461.48 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p202-b16", - "global_id": 5557, - "bbox": [ - 232.26, - 538.23, - 241.14, - 546.23 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p202-b17", - "global_id": 5558, - "bbox": [ - 239.47, - 517.48, - 260.92, - 525.55 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p202-b18", - "global_id": 5559, - "bbox": [ - 317.85, - 334.79, - 334.74, - 343.09 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p202-b19", - "global_id": 5560, - "bbox": [ - 317.85, - 453.72, - 334.74, - 462.02 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p202-b20", - "global_id": 5561, - "bbox": [ - 323.98, - 486.07, - 336.42, - 494.17 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p202-b21", - "global_id": 5562, - "bbox": [ - 112.92, - 93.48, - 139.07, - 109.95 - ], - "text": "1\nx(t)", - "type": "text" - }, - { - "block_id": "p202-b22", - "global_id": 5563, - "bbox": [ - 134.68, - 114.82, - 285.92, - 126.32 - ], - "text": "et\ne2t", - "type": "text" - }, - { - "block_id": "p202-b23", - "global_id": 5564, - "bbox": [ - 203.83, - 165.91, - 206.05, - 173.91 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p202-b24", - "global_id": 5565, - "bbox": [ - 153.51, - 186.13, - 162.39, - 194.13 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p202-b25", - "global_id": 5566, - "bbox": [ - 256.58, - 93.48, - 282.68, - 109.95 - ], - "text": "1\nh(t)", - "type": "text" - }, - { - "block_id": "p202-b26", - "global_id": 5567, - "bbox": [ - 347.29, - 165.64, - 349.51, - 173.64 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p202-b27", - "global_id": 5568, - "bbox": [ - 296.63, - 186.13, - 306.27, - 194.13 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p202-b28", - "global_id": 5569, - "bbox": [ - 122.68, - 165.73, - 270.06, - 173.81 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p202-b29", - "global_id": 5570, - "bbox": [ - 222.56, - 561.47, - 233.67, - 569.55 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p202-b30", - "global_id": 5571, - "bbox": [ - 239.61, - 592.96, - 322.98, - 601.18 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p202-b31", - "global_id": 5572, - "bbox": [ - 270.47, - 614.15, - 279.43, - 622.15 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p202-b32", - "global_id": 5573, - "bbox": [ - 94.2, - 628.47, - 242.89, - 638.08 - ], - "text": "Figure 2.8 Convolution of x(t) and h(t).", - "type": "text" - } - ] - }, - { - "page_num": 203, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p203-b0", - "global_id": 5574, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n183", - "type": "text" - }, - { - "block_id": "p203-b1", - "global_id": 5575, - "bbox": [ - 103.92, - 95.83, - 452.83, - 121.74 - ], - "text": "EXAMPLE 2.11\nGraphical Convolution: Causal Function and\nTwo-Sided Function", - "type": "text" - }, - { - "block_id": "p203-b2", - "global_id": 5576, - "bbox": [ - 128.9, - 136.48, - 400.65, - 146.86 - ], - "text": "Find c(t) = x(t) ∗g(t) for the signals depicted in Figs. 2.9a and 2.9b.", - "type": "text" - }, - { - "block_id": "p203-b3", - "global_id": 5577, - "bbox": [ - 128.88, - 169.36, - 502.72, - 191.69 - ], - "text": "Since x(t) is simpler than g(t), it is easier to evaluate g(t) ∗x(t) than x(t) ∗g(t). However, we\nshall intentionally take the more difficult route and evaluate x(t) ∗g(t).", - "type": "text" - }, - { - "block_id": "p203-b4", - "global_id": 5578, - "bbox": [ - 128.87, - 193.27, - 502.74, - 215.61 - ], - "text": "From x(t) and g(t) (Figs. 2.9a and 2.9b, respectively), observe that g(t) is composed of\ntwo segments. As a result, it can be described as", - "type": "text" - }, - { - "block_id": "p203-b5", - "global_id": 5579, - "bbox": [ - 255.57, - 239.76, - 280.75, - 250.04 - ], - "text": "g(t) =", - "type": "text" - }, - { - "block_id": "p203-b6", - "global_id": 5580, - "bbox": [ - 282.81, - 225.78, - 374.88, - 256.01 - ], - "text": "2e−t\nsegment A\n−2e2t\nsegment B", - "type": "text" - }, - { - "block_id": "p203-b7", - "global_id": 5581, - "bbox": [ - 128.91, - 274.35, - 170.65, - 284.31 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p203-b8", - "global_id": 5582, - "bbox": [ - 240.17, - 290.85, - 281.73, - 301.13 - ], - "text": "g(t −τ) =", - "type": "text" - }, - { - "block_id": "p203-b9", - "global_id": 5583, - "bbox": [ - 283.77, - 276.87, - 390.31, - 307.15 - ], - "text": "2e−(t−τ)\nsegment A\n−2e2(t−τ)\nsegment B", - "type": "text" - }, - { - "block_id": "p203-b10", - "global_id": 5584, - "bbox": [ - 128.9, - 315.17, - 502.74, - 337.5 - ], - "text": "The segment of x(t) that is used in convolution is x(t) = 1 so that x(τ) = 1. Figure 2.9c shows\nx(τ) and g(−τ).", - "type": "text" - }, - { - "block_id": "p203-b11", - "global_id": 5585, - "bbox": [ - 128.92, - 339.09, - 502.78, - 385.34 - ], - "text": "To compute c(t) for t ≥0, we right-shift g(−τ) to obtain g(t−τ), as illustrated in Fig. 2.9d.\nClearly, g(t −τ) overlaps with x(τ) over the shaded interval, that is, over the range τ ≥0;\nsegment A overlaps with x(τ) over the interval (0,t), while segment B overlaps with x(τ) over\n(t,∞). Remembering that x(τ) = 1, we have", - "type": "text" - }, - { - "block_id": "p203-b12", - "global_id": 5586, - "bbox": [ - 221.99, - 403.2, - 246.61, - 413.48 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p203-b13", - "global_id": 5587, - "bbox": [ - 248.66, - 389.65, - 265.59, - 401.69 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p203-b14", - "global_id": 5588, - "bbox": [ - 253.91, - 403.19, - 326.99, - 421.77 - ], - "text": "0\nx(τ)g(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p203-b15", - "global_id": 5589, - "bbox": [ - 238.83, - 430.76, - 246.6, - 440.73 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p203-b16", - "global_id": 5590, - "bbox": [ - 248.65, - 417.2, - 260.41, - 429.48 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p203-b17", - "global_id": 5591, - "bbox": [ - 253.91, - 417.2, - 333.72, - 449.34 - ], - "text": "0\n2e−(t−τ) dτ +\n ∞", - "type": "text" - }, - { - "block_id": "p203-b18", - "global_id": 5592, - "bbox": [ - 322.04, - 442.3, - 323.98, - 449.27 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p203-b19", - "global_id": 5593, - "bbox": [ - 335.33, - 426.96, - 383.72, - 441.14 - ], - "text": "−2e2(t−τ) dτ", - "type": "text" - }, - { - "block_id": "p203-b20", - "global_id": 5594, - "bbox": [ - 238.82, - 450.4, - 409.69, - 462.51 - ], - "text": "= 2(1 −e−t) −1 = 1 −2e−t\nt ≥0", - "type": "text" - }, - { - "block_id": "p203-b21", - "global_id": 5595, - "bbox": [ - 128.91, - 481.02, - 502.75, - 503.35 - ], - "text": "Figure 2.9e shows the situation for t < 0. Here the overlap is over the shaded interval, that\nis, over the range τ ≥0, where only the segment B of g(t) is involved. Therefore,", - "type": "text" - }, - { - "block_id": "p203-b22", - "global_id": 5596, - "bbox": [ - 177.97, - 527.2, - 202.59, - 537.47 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p203-b23", - "global_id": 5597, - "bbox": [ - 204.64, - 513.64, - 221.59, - 525.68 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p203-b24", - "global_id": 5598, - "bbox": [ - 209.91, - 527.19, - 294.02, - 545.76 - ], - "text": "0\nx(τ)g(t −τ)dτ =", - "type": "text" - }, - { - "block_id": "p203-b25", - "global_id": 5599, - "bbox": [ - 296.06, - 513.63, - 313.0, - 525.68 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p203-b26", - "global_id": 5600, - "bbox": [ - 301.32, - 523.38, - 453.69, - 545.76 - ], - "text": "0\n−2e2(t−τ) dτ = −e2t\nt ≤0", - "type": "text" - }, - { - "block_id": "p203-b27", - "global_id": 5601, - "bbox": [ - 128.9, - 561.32, - 170.65, - 571.28 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p203-b28", - "global_id": 5602, - "bbox": [ - 262.46, - 577.78, - 287.08, - 588.06 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p203-b29", - "global_id": 5603, - "bbox": [ - 289.13, - 563.79, - 368.0, - 594.03 - ], - "text": "1 −2e−t\nt ≥0\n−e2t\nt ≤0", - "type": "text" - }, - { - "block_id": "p203-b30", - "global_id": 5604, - "bbox": [ - 128.9, - 602.06, - 255.81, - 612.43 - ], - "text": "Figure 2.9f shows a plot of c(t).", - "type": "text" - } - ] - }, - { - "page_num": 204, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p204-b0", - "global_id": 5605, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "184\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p204-b1", - "global_id": 5606, - "bbox": [ - 148.85, - 94.6, - 350.53, - 107.84 - ], - "text": "2\nx(t)", - "type": "text" - }, - { - "block_id": "p204-b2", - "global_id": 5607, - "bbox": [ - 179.3, - 178.66, - 188.18, - 186.66 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p204-b3", - "global_id": 5608, - "bbox": [ - 136.7, - 119.9, - 140.7, - 127.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p204-b4", - "global_id": 5609, - "bbox": [ - 320.45, - 93.76, - 332.0, - 101.84 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p204-b5", - "global_id": 5610, - "bbox": [ - 395.68, - 178.66, - 405.33, - 186.66 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p204-b6", - "global_id": 5611, - "bbox": [ - 274.89, - 217.33, - 287.32, - 225.43 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p204-b7", - "global_id": 5612, - "bbox": [ - 299.88, - 345.37, - 312.31, - 353.47 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p204-b8", - "global_id": 5613, - "bbox": [ - 179.51, - 217.64, - 199.05, - 225.93 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p204-b9", - "global_id": 5614, - "bbox": [ - 166.01, - 357.63, - 191.78, - 365.92 - ], - "text": "g(t t)", - "type": "text" - }, - { - "block_id": "p204-b10", - "global_id": 5615, - "bbox": [ - 151.91, - 464.41, - 177.68, - 472.7 - ], - "text": "g(t t)", - "type": "text" - }, - { - "block_id": "p204-b11", - "global_id": 5616, - "bbox": [ - 148.7, - 151.45, - 360.45, - 159.62 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p204-b12", - "global_id": 5617, - "bbox": [ - 327.85, - 330.05, - 344.74, - 338.34 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p204-b13", - "global_id": 5618, - "bbox": [ - 292.2, - 619.15, - 300.99, - 627.15 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p204-b14", - "global_id": 5619, - "bbox": [ - 371.78, - 108.32, - 386.0, - 118.01 - ], - "text": "2et", - "type": "text" - }, - { - "block_id": "p204-b15", - "global_id": 5620, - "bbox": [ - 314.78, - 169.84, - 333.99, - 179.38 - ], - "text": "2e2t", - "type": "text" - }, - { - "block_id": "p204-b16", - "global_id": 5621, - "bbox": [ - 355.79, - 199.1, - 366.45, - 207.4 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p204-b17", - "global_id": 5622, - "bbox": [ - 407.2, - 116.65, - 435.2, - 124.65 - ], - "text": "Segment", - "type": "text" - }, - { - "block_id": "p204-b18", - "global_id": 5623, - "bbox": [ - 266.11, - 165.65, - 294.11, - 173.65 - ], - "text": "Segment", - "type": "text" - }, - { - "block_id": "p204-b19", - "global_id": 5624, - "bbox": [ - 292.2, - 275.05, - 301.08, - 283.05 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p204-b20", - "global_id": 5625, - "bbox": [ - 206.46, - 261.88, - 210.46, - 269.88 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p204-b21", - "global_id": 5626, - "bbox": [ - 291.8, - 402.05, - 301.13, - 410.05 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p204-b22", - "global_id": 5627, - "bbox": [ - 206.48, - 388.81, - 239.28, - 397.96 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p204-b23", - "global_id": 5628, - "bbox": [ - 206.7, - 358.4, - 210.7, - 366.4 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p204-b24", - "global_id": 5629, - "bbox": [ - 292.2, - 536.05, - 301.08, - 544.05 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p204-b25", - "global_id": 5630, - "bbox": [ - 179.86, - 518.41, - 210.98, - 526.48 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p204-b26", - "global_id": 5631, - "bbox": [ - 206.98, - 486.9, - 210.98, - 494.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p204-b27", - "global_id": 5632, - "bbox": [ - 327.85, - 464.04, - 344.74, - 472.34 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p204-b28", - "global_id": 5633, - "bbox": [ - 315.11, - 561.13, - 338.65, - 573.9 - ], - "text": "c(t)\n1", - "type": "text" - }, - { - "block_id": "p204-b29", - "global_id": 5634, - "bbox": [ - 308.44, - 617.11, - 319.11, - 625.4 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p204-b30", - "global_id": 5635, - "bbox": [ - 149.56, - 227.81, - 155.34, - 235.81 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p204-b31", - "global_id": 5636, - "bbox": [ - 145.72, - 363.91, - 151.5, - 371.91 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p204-b32", - "global_id": 5637, - "bbox": [ - 116.72, - 486.41, - 122.5, - 494.41 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p204-b33", - "global_id": 5638, - "bbox": [ - 439.31, - 116.77, - 445.08, - 124.77 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p204-b34", - "global_id": 5639, - "bbox": [ - 297.82, - 165.86, - 303.16, - 173.86 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p204-b35", - "global_id": 5640, - "bbox": [ - 249.12, - 288.07, - 254.46, - 296.07 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p204-b36", - "global_id": 5641, - "bbox": [ - 275.34, - 420.75, - 280.67, - 428.75 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p204-b37", - "global_id": 5642, - "bbox": [ - 232.21, - 537.44, - 237.55, - 545.44 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p204-b38", - "global_id": 5643, - "bbox": [ - 291.78, - 472.71, - 304.22, - 480.82 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p204-b39", - "global_id": 5644, - "bbox": [ - 229.07, - 152.4, - 231.46, - 160.69 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p204-b40", - "global_id": 5645, - "bbox": [ - 414.57, - 596.54, - 416.96, - 604.82 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p204-b41", - "global_id": 5646, - "bbox": [ - 304.39, - 259.93, - 307.94, - 267.93 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p204-b42", - "global_id": 5647, - "bbox": [ - 424.57, - 152.4, - 426.96, - 160.69 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p204-b43", - "global_id": 5648, - "bbox": [ - 319.39, - 389.93, - 322.94, - 397.93 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p204-b44", - "global_id": 5649, - "bbox": [ - 304.39, - 516.43, - 307.94, - 524.43 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p204-b45", - "global_id": 5650, - "bbox": [ - 94.2, - 633.47, - 242.89, - 643.08 - ], - "text": "Figure 2.9 Convolution of x(t) and g(t).", - "type": "text" - } - ] - }, - { - "page_num": 205, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p205-b0", - "global_id": 5651, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n185", - "type": "text" - }, - { - "block_id": "p205-b1", - "global_id": 5652, - "bbox": [ - 103.92, - 95.83, - 464.05, - 121.73 - ], - "text": "EXAMPLE 2.12\nGraphical Convolution of Two Finite-Duration\nFunctions", - "type": "text" - }, - { - "block_id": "p205-b2", - "global_id": 5653, - "bbox": [ - 128.9, - 136.48, - 437.41, - 146.85 - ], - "text": "Find x(t) ∗g(t) for the functions x(t) and g(t) shown in Figs. 2.10a and 2.10b.", - "type": "text" - }, - { - "block_id": "p205-b3", - "global_id": 5654, - "bbox": [ - 128.9, - 169.35, - 502.77, - 191.69 - ], - "text": "Here, x(t) has a simpler mathematical description than that of g(t), so it is preferable to\ntime-reverse x(t). Hence, we shall determine g(t) ∗x(t) rather than x(t) ∗g(t). Thus,", - "type": "text" - }, - { - "block_id": "p205-b4", - "global_id": 5655, - "bbox": [ - 237.11, - 208.43, - 312.09, - 218.71 - ], - "text": "c(t) = g(t) ∗x(t) =", - "type": "text" - }, - { - "block_id": "p205-b5", - "global_id": 5656, - "bbox": [ - 314.13, - 194.87, - 331.07, - 206.92 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p205-b6", - "global_id": 5657, - "bbox": [ - 319.39, - 219.75, - 331.95, - 226.72 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p205-b7", - "global_id": 5658, - "bbox": [ - 333.56, - 208.43, - 393.36, - 218.71 - ], - "text": "g(τ)x(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p205-b8", - "global_id": 5659, - "bbox": [ - 128.91, - 235.74, - 502.77, - 258.07 - ], - "text": "First, we determine the expressions for the segments of x(t) and g(t) used in finding c(t).\nAccording to Figs. 2.10a and 2.10b, these segments can be expressed as", - "type": "text" - }, - { - "block_id": "p205-b9", - "global_id": 5660, - "bbox": [ - 254.85, - 268.27, - 372.67, - 279.99 - ], - "text": "x(t) = 1\nand\ng(t) = 1", - "type": "text" - }, - { - "block_id": "p205-b10", - "global_id": 5661, - "bbox": [ - 369.19, - 269.93, - 376.64, - 282.81 - ], - "text": "3t", - "type": "text" - }, - { - "block_id": "p205-b11", - "global_id": 5662, - "bbox": [ - 128.9, - 291.94, - 155.19, - 301.9 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p205-b12", - "global_id": 5663, - "bbox": [ - 244.1, - 302.14, - 380.86, - 313.87 - ], - "text": "x(t −τ) = 1\nand\ng(τ) = 1", - "type": "text" - }, - { - "block_id": "p205-b13", - "global_id": 5664, - "bbox": [ - 377.38, - 303.49, - 386.38, - 316.67 - ], - "text": "3τ", - "type": "text" - }, - { - "block_id": "p205-b14", - "global_id": 5665, - "bbox": [ - 128.91, - 322.42, - 502.77, - 368.66 - ], - "text": "Figure 2.10c shows g(τ) and x(−τ), whereas Fig. 2.10d shows g(τ) and x(t −τ), which is\nx(−τ) shifted by t. Because the edges of x(−τ) are at τ = −1 and 1, the edges of x(t −τ) are\nat −1 + t and 1 + t. The two functions overlap over the interval (0,1 + t) (shaded interval) so\nthat", - "type": "text" - }, - { - "block_id": "p205-b15", - "global_id": 5666, - "bbox": [ - 152.58, - 384.96, - 177.21, - 395.23 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p205-b16", - "global_id": 5667, - "bbox": [ - 179.25, - 371.39, - 199.93, - 383.74 - ], - "text": "# 1+t", - "type": "text" - }, - { - "block_id": "p205-b17", - "global_id": 5668, - "bbox": [ - 184.51, - 384.96, - 272.5, - 403.54 - ], - "text": "0\ng(τ)x(t −τ)dτ =", - "type": "text" - }, - { - "block_id": "p205-b18", - "global_id": 5669, - "bbox": [ - 274.53, - 371.39, - 295.22, - 383.74 - ], - "text": "# 1+t", - "type": "text" - }, - { - "block_id": "p205-b19", - "global_id": 5670, - "bbox": [ - 279.8, - 396.57, - 283.28, - 403.54 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p205-b20", - "global_id": 5671, - "bbox": [ - 298.15, - 383.61, - 502.75, - 398.15 - ], - "text": "1\n3τ dτ = 1\n6(t + 1)2\n−1 ≤t ≤1\n(2.32)", - "type": "text" - }, - { - "block_id": "p205-b21", - "global_id": 5672, - "bbox": [ - 128.9, - 411.7, - 502.76, - 457.94 - ], - "text": "This situation, depicted in Fig. 2.10d, is valid only for −1 ≤t ≤1. For t ≥1 but ≤2, the\nsituation is as illustrated in Fig. 2.10e. The two functions overlap only over the range −1 + t\nto 1 + t (shaded interval). Note that the expressions for g(τ) and x(t −τ) do not change; only\nthe range of integration changes. Therefore,", - "type": "text" - }, - { - "block_id": "p205-b22", - "global_id": 5673, - "bbox": [ - 230.89, - 476.29, - 255.51, - 486.57 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p205-b23", - "global_id": 5674, - "bbox": [ - 257.56, - 462.74, - 278.24, - 475.08 - ], - "text": "# 1+t", - "type": "text" - }, - { - "block_id": "p205-b24", - "global_id": 5675, - "bbox": [ - 262.81, - 487.61, - 279.11, - 494.87 - ], - "text": "−1+t", - "type": "text" - }, - { - "block_id": "p205-b25", - "global_id": 5676, - "bbox": [ - 282.05, - 474.95, - 502.75, - 489.48 - ], - "text": "1\n3τ dτ = 2\n3t\n1 ≤t ≤2\n(2.33)", - "type": "text" - }, - { - "block_id": "p205-b26", - "global_id": 5677, - "bbox": [ - 128.91, - 503.54, - 502.76, - 561.74 - ], - "text": "Also note that the expressions in Eqs. (2.32) and (2.33) both apply at t = 1, the transition point\nbetween their respective ranges. We can readily verify that both expressions yield a value of\n2/3 at t = 1 so that c(1) = 2/3. The continuity of c(t) at transition points indicates a high\nprobability of a correct answer. Continuity of c(t) at transition points is assured as long as x(t)\nand g(t) contain no impulse functions.", - "type": "text" - }, - { - "block_id": "p205-b27", - "global_id": 5678, - "bbox": [ - 128.91, - 563.32, - 502.76, - 585.65 - ], - "text": "For t ≥2 but ≤4, the situation is as shown in Fig. 2.10f. The functions g(τ) and x(t −τ)\noverlap over the interval from −1 + t to 3 (shaded interval) so that", - "type": "text" - }, - { - "block_id": "p205-b28", - "global_id": 5679, - "bbox": [ - 203.97, - 604.01, - 228.6, - 614.29 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p205-b29", - "global_id": 5680, - "bbox": [ - 230.65, - 590.45, - 243.95, - 602.79 - ], - "text": "# 3", - "type": "text" - }, - { - "block_id": "p205-b30", - "global_id": 5681, - "bbox": [ - 235.9, - 615.32, - 252.19, - 622.58 - ], - "text": "−1+t", - "type": "text" - }, - { - "block_id": "p205-b31", - "global_id": 5682, - "bbox": [ - 255.13, - 600.39, - 502.75, - 617.19 - ], - "text": "1\n3τ dτ = −1\n6(t2 −2t −8)\n2 ≤t ≤4\n(2.34)", - "type": "text" - } - ] - }, - { - "page_num": 206, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p206-b0", - "global_id": 5683, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "186\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p206-b1", - "global_id": 5684, - "bbox": [ - 286.37, - 279.62, - 311.69, - 287.92 - ], - "text": "x(t t)", - "type": "text" - }, - { - "block_id": "p206-b2", - "global_id": 5685, - "bbox": [ - 308.52, - 258.6, - 416.33, - 269.81 - ], - "text": "1 t 2\n1", - "type": "text" - }, - { - "block_id": "p206-b3", - "global_id": 5686, - "bbox": [ - 311.98, - 328.67, - 320.86, - 336.67 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p206-b4", - "global_id": 5687, - "bbox": [ - 398.12, - 277.5, - 411.0, - 285.6 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p206-b5", - "global_id": 5688, - "bbox": [ - 318.21, - 316.15, - 415.23, - 325.18 - ], - "text": "t\n3\n1 t\n1 t", - "type": "text" - }, - { - "block_id": "p206-b6", - "global_id": 5689, - "bbox": [ - 279.37, - 166.63, - 418.39, - 179.62 - ], - "text": "x(t t)\n1 t 1\ng(t)", - "type": "text" - }, - { - "block_id": "p206-b7", - "global_id": 5690, - "bbox": [ - 399.51, - 226.54, - 403.06, - 234.54 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p206-b8", - "global_id": 5691, - "bbox": [ - 318.62, - 170.86, - 322.62, - 178.86 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p206-b9", - "global_id": 5692, - "bbox": [ - 311.76, - 238.7, - 321.09, - 246.7 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p206-b10", - "global_id": 5693, - "bbox": [ - 287.64, - 225.22, - 383.82, - 233.52 - ], - "text": "0\n3\n1 t\n1 t", - "type": "text" - }, - { - "block_id": "p206-b11", - "global_id": 5694, - "bbox": [ - 146.48, - 238.7, - 155.36, - 246.7 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p206-b12", - "global_id": 5695, - "bbox": [ - 103.88, - 170.38, - 194.65, - 178.67 - ], - "text": "x(t)\n1\ng(t)", - "type": "text" - }, - { - "block_id": "p206-b13", - "global_id": 5696, - "bbox": [ - 122.19, - 225.22, - 206.94, - 234.58 - ], - "text": "1\n1\nt\n0", - "type": "text" - }, - { - "block_id": "p206-b14", - "global_id": 5697, - "bbox": [ - 329.37, - 345.61, - 428.83, - 355.58 - ], - "text": "x(t t)\n2 t 4", - "type": "text" - }, - { - "block_id": "p206-b15", - "global_id": 5698, - "bbox": [ - 312.33, - 405.88, - 419.64, - 425.49 - ], - "text": "t\n(f)", - "type": "text" - }, - { - "block_id": "p206-b16", - "global_id": 5699, - "bbox": [ - 320.62, - 405.01, - 324.62, - 413.01 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p206-b17", - "global_id": 5700, - "bbox": [ - 308.62, - 349.7, - 312.62, - 357.7 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p206-b18", - "global_id": 5701, - "bbox": [ - 320.25, - 366.85, - 333.13, - 374.96 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p206-b19", - "global_id": 5702, - "bbox": [ - 345.15, - 404.71, - 408.72, - 413.01 - ], - "text": "3\n1 t\n1 t", - "type": "text" - }, - { - "block_id": "p206-b20", - "global_id": 5703, - "bbox": [ - 198.59, - 440.31, - 249.62, - 448.6 - ], - "text": "x(t t)\nt 4", - "type": "text" - }, - { - "block_id": "p206-b21", - "global_id": 5704, - "bbox": [ - 142.33, - 459.9, - 155.21, - 468.0 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p206-b22", - "global_id": 5705, - "bbox": [ - 123.62, - 495.61, - 249.97, - 503.9 - ], - "text": "1 t\n1 t\n0", - "type": "text" - }, - { - "block_id": "p206-b23", - "global_id": 5706, - "bbox": [ - 111.62, - 440.61, - 115.62, - 448.61 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p206-b24", - "global_id": 5707, - "bbox": [ - 177.36, - 507.52, - 186.69, - 515.52 - ], - "text": "(g)", - "type": "text" - }, - { - "block_id": "p206-b25", - "global_id": 5708, - "bbox": [ - 298.86, - 440.31, - 436.44, - 448.6 - ], - "text": "x(t t)\nt 1", - "type": "text" - }, - { - "block_id": "p206-b26", - "global_id": 5709, - "bbox": [ - 379.02, - 459.9, - 391.9, - 468.0 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p206-b27", - "global_id": 5710, - "bbox": [ - 276.75, - 495.61, - 360.03, - 515.52 - ], - "text": "1 t\n1 t\n(h)", - "type": "text" - }, - { - "block_id": "p206-b28", - "global_id": 5711, - "bbox": [ - 359.12, - 495.9, - 363.12, - 503.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p206-b29", - "global_id": 5712, - "bbox": [ - 347.12, - 440.71, - 351.12, - 448.71 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p206-b30", - "global_id": 5713, - "bbox": [ - 414.82, - 497.53, - 418.37, - 505.53 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p206-b31", - "global_id": 5714, - "bbox": [ - 271.65, - 616.2, - 279.2, - 624.2 - ], - "text": "(i)", - "type": "text" - }, - { - "block_id": "p206-b32", - "global_id": 5715, - "bbox": [ - 220.18, - 526.23, - 231.28, - 534.31 - ], - "text": "c(t)", - "type": "text" - }, - { - "block_id": "p206-b33", - "global_id": 5716, - "bbox": [ - 183.92, - 595.66, - 378.72, - 612.45 - ], - "text": "1\n0\n1\n2\n3\n4\nt", - "type": "text" - }, - { - "block_id": "p206-b34", - "global_id": 5717, - "bbox": [ - 172.95, - 568.8, - 198.16, - 578.27 - ], - "text": "(t 1)2", - "type": "text" - }, - { - "block_id": "p206-b35", - "global_id": 5718, - "bbox": [ - 242.66, - 541.74, - 250.53, - 555.27 - ], - "text": "t\n2\n3", - "type": "text" - }, - { - "block_id": "p206-b36", - "global_id": 5719, - "bbox": [ - 167.47, - 560.51, - 209.59, - 581.48 - ], - "text": "2\n3\n1\n6", - "type": "text" - }, - { - "block_id": "p206-b37", - "global_id": 5720, - "bbox": [ - 206.59, - 533.35, - 388.51, - 546.97 - ], - "text": "(t2 2t 8)\n1\n6\n4\n3", - "type": "text" - }, - { - "block_id": "p206-b38", - "global_id": 5721, - "bbox": [ - 160.39, - 78.08, - 171.49, - 86.16 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p206-b39", - "global_id": 5722, - "bbox": [ - 146.48, - 147.81, - 155.36, - 155.81 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p206-b40", - "global_id": 5723, - "bbox": [ - 142.62, - 77.18, - 307.01, - 88.23 - ], - "text": "1\ng(t)", - "type": "text" - }, - { - "block_id": "p206-b41", - "global_id": 5724, - "bbox": [ - 122.69, - 135.0, - 384.12, - 143.29 - ], - "text": "1\n1\n3", - "type": "text" - }, - { - "block_id": "p206-b42", - "global_id": 5725, - "bbox": [ - 318.62, - 87.18, - 322.62, - 95.18 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p206-b43", - "global_id": 5726, - "bbox": [ - 154.62, - 135.29, - 158.62, - 143.29 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p206-b44", - "global_id": 5727, - "bbox": [ - 311.6, - 147.81, - 321.25, - 155.81 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p206-b45", - "global_id": 5728, - "bbox": [ - 204.78, - 135.29, - 402.51, - 144.15 - ], - "text": "0\nt\nt", - "type": "text" - }, - { - "block_id": "p206-b46", - "global_id": 5729, - "bbox": [ - 103.66, - 630.53, - 256.84, - 640.14 - ], - "text": "Figure 2.10 Convolution of x(t) and g(t).", - "type": "text" - } - ] - }, - { - "page_num": 207, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p207-b0", - "global_id": 5730, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n187", - "type": "text" - }, - { - "block_id": "p207-b1", - "global_id": 5731, - "bbox": [ - 128.9, - 85.83, - 502.76, - 108.16 - ], - "text": "Both Eqs. (2.33) and (2.34) apply at the transition point t = 2. We can readily verify that\nc(2) = 4/3 when either of these expressions is used.", - "type": "text" - }, - { - "block_id": "p207-b2", - "global_id": 5732, - "bbox": [ - 128.91, - 109.73, - 502.76, - 132.07 - ], - "text": "For t ≥4, x(t −τ) has been shifted so far to the right that it no longer overlaps with g(τ)\nas depicted in Fig. 2.10g. Consequently,", - "type": "text" - }, - { - "block_id": "p207-b3", - "global_id": 5733, - "bbox": [ - 270.31, - 143.61, - 361.35, - 153.99 - ], - "text": "c(t) = 0\nt ≥4", - "type": "text" - }, - { - "block_id": "p207-b4", - "global_id": 5734, - "bbox": [ - 128.91, - 165.52, - 502.77, - 199.81 - ], - "text": "We now turn our attention to negative values of t. We have already determined c(t) up to\nt = −1. For t < −1, there is no overlap between the two functions, as illustrated in Fig. 2.10h,\nso that", - "type": "text" - }, - { - "block_id": "p207-b5", - "global_id": 5735, - "bbox": [ - 266.43, - 201.39, - 365.24, - 211.77 - ], - "text": "c(t) = 0\nt ≤−1", - "type": "text" - }, - { - "block_id": "p207-b6", - "global_id": 5736, - "bbox": [ - 146.85, - 220.74, - 285.19, - 230.7 - ], - "text": "Combining our results, we see that", - "type": "text" - }, - { - "block_id": "p207-b7", - "global_id": 5737, - "bbox": [ - 233.84, - 278.7, - 258.46, - 288.98 - ], - "text": "c(t) =", - "type": "text" - }, - { - "block_id": "p207-b8", - "global_id": 5738, - "bbox": [ - 260.51, - 234.37, - 268.4, - 277.2 - ], - "text": "⎧\n⎪⎪⎪⎪⎪⎪⎪⎪⎨", - "type": "text" - }, - { - "block_id": "p207-b9", - "global_id": 5739, - "bbox": [ - 260.51, - 285.17, - 268.4, - 319.04 - ], - "text": "⎪⎪⎪⎪⎪⎪⎪⎪⎩", - "type": "text" - }, - { - "block_id": "p207-b10", - "global_id": 5740, - "bbox": [ - 269.59, - 240.64, - 396.64, - 264.66 - ], - "text": "1\n6(t + 1)2\n−1 ≤t < 1", - "type": "text" - }, - { - "block_id": "p207-b11", - "global_id": 5741, - "bbox": [ - 269.59, - 269.94, - 388.86, - 293.96 - ], - "text": "2\n3t\n1 ≤t < 2", - "type": "text" - }, - { - "block_id": "p207-b12", - "global_id": 5742, - "bbox": [ - 268.4, - 293.28, - 282.34, - 309.81 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p207-b13", - "global_id": 5743, - "bbox": [ - 277.36, - 295.74, - 388.87, - 317.3 - ], - "text": "6(t2 −2t −8)\n2 ≤t < 4", - "type": "text" - }, - { - "block_id": "p207-b14", - "global_id": 5744, - "bbox": [ - 268.4, - 315.81, - 391.22, - 325.77 - ], - "text": "0\notherwise", - "type": "text" - }, - { - "block_id": "p207-b15", - "global_id": 5745, - "bbox": [ - 128.91, - 336.78, - 333.11, - 347.15 - ], - "text": "Figure 2.10i plots c(t) according to this expression.", - "type": "text" - }, - { - "block_id": "p207-b16", - "global_id": 5746, - "bbox": [ - 127.59, - 391.48, - 516.13, - 489.33 - ], - "text": "THE WIDTH OF CONVOLVED FUNCTIONS\nThe widths (durations) of x(t), g(t), and c(t) in Ex. 2.12 (Fig. 2.10) are 2, 3, and 5, respectively.\nNote that the width of c(t) in this case is the sum of the widths of x(t) and g(t). This observation\nis not a coincidence. Using the concept of graphical convolution, we can readily see that if x(t)\nand g(t) have the finite widths of T1 and T2 respectively, then the width of c(t) is equal to T1 +T2.\nThe reason is that the time it takes for a signal of width (duration) T1 to completely pass another\nsignal of width (duration) T2 so that they become non-overlapping is T1+T2. When the two signals\nbecome non-overlapping, the convolution goes to zero.", - "type": "text" - }, - { - "block_id": "p207-b17", - "global_id": 5747, - "bbox": [ - 133.57, - 528.75, - 399.2, - 540.7 - ], - "text": "DRILL 2.10\nInterchanging Convolution Order", - "type": "text" - }, - { - "block_id": "p207-b18", - "global_id": 5748, - "bbox": [ - 133.57, - 549.41, - 300.15, - 559.79 - ], - "text": "Rework Ex. 2.11 by evaluating g(t) ∗x(t).", - "type": "text" - }, - { - "block_id": "p207-b19", - "global_id": 5749, - "bbox": [ - 133.57, - 611.72, - 496.04, - 623.67 - ], - "text": "DRILL 2.11\nShowing Commutability Using Two Causal Signals", - "type": "text" - }, - { - "block_id": "p207-b20", - "global_id": 5750, - "bbox": [ - 133.57, - 632.38, - 457.98, - 642.76 - ], - "text": "Use graphical convolution to show that x(t) ∗g(t) = g(t) ∗x(t) = c(t) in Fig. 2.11.", - "type": "text" - } - ] - }, - { - "page_num": 208, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p208-b0", - "global_id": 5751, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "188\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p208-b1", - "global_id": 5752, - "bbox": [ - 220.15, - 114.69, - 225.15, - 124.69 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p208-b2", - "global_id": 5753, - "bbox": [ - 153.5, - 88.62, - 263.27, - 101.31 - ], - "text": "x(t)\ng(t)", - "type": "text" - }, - { - "block_id": "p208-b3", - "global_id": 5754, - "bbox": [ - 165.7, - 110.13, - 326.68, - 124.39 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p208-b4", - "global_id": 5755, - "bbox": [ - 239.58, - 89.93, - 403.6, - 104.51 - ], - "text": "1 et\n1\nc(t)", - "type": "text" - }, - { - "block_id": "p208-b5", - "global_id": 5756, - "bbox": [ - 239.58, - 140.37, - 243.58, - 148.37 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p208-b6", - "global_id": 5757, - "bbox": [ - 139.69, - 96.51, - 143.69, - 104.51 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p208-b7", - "global_id": 5758, - "bbox": [ - 140.85, - 140.37, - 144.85, - 148.37 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p208-b8", - "global_id": 5759, - "bbox": [ - 341.23, - 96.51, - 345.23, - 104.51 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p208-b9", - "global_id": 5760, - "bbox": [ - 201.56, - 140.37, - 403.5, - 148.67 - ], - "text": "0\nt\nt\nt", - "type": "text" - }, - { - "block_id": "p208-b10", - "global_id": 5761, - "bbox": [ - 131.73, - 155.06, - 336.71, - 164.67 - ], - "text": "Figure 2.11 Convolution of causal signals x(t) and g(t).", - "type": "text" - }, - { - "block_id": "p208-b11", - "global_id": 5762, - "bbox": [ - 107.82, - 239.99, - 471.04, - 265.89 - ], - "text": "DRILL 2.12\nShowing Commutability Using a Causal Signal and\nan Anticausal Signal", - "type": "text" - }, - { - "block_id": "p208-b12", - "global_id": 5763, - "bbox": [ - 107.82, - 275.01, - 295.98, - 284.98 - ], - "text": "Repeat Drill 2.11 for the functions in Fig. 2.12.", - "type": "text" - }, - { - "block_id": "p208-b13", - "global_id": 5764, - "bbox": [ - 223.09, - 344.95, - 228.09, - 354.95 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p208-b14", - "global_id": 5765, - "bbox": [ - 158.68, - 319.08, - 280.02, - 332.29 - ], - "text": "x(t)\ng(t)", - "type": "text" - }, - { - "block_id": "p208-b16", - "global_id": 5766, - "bbox": [ - 295.68, - 319.34, - 372.41, - 335.04 - ], - "text": "1\nc(t)", - "type": "text" - }, - { - "block_id": "p208-b17", - "global_id": 5767, - "bbox": [ - 286.4, - 369.52, - 290.4, - 377.52 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p208-b18", - "global_id": 5768, - "bbox": [ - 146.52, - 327.45, - 150.52, - 335.45 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p208-b19", - "global_id": 5769, - "bbox": [ - 146.52, - 369.75, - 150.52, - 377.75 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p208-b20", - "global_id": 5770, - "bbox": [ - 379.54, - 327.04, - 383.54, - 335.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p208-b21", - "global_id": 5771, - "bbox": [ - 368.26, - 369.42, - 424.74, - 377.52 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p208-b22", - "global_id": 5772, - "bbox": [ - 170.93, - 337.48, - 181.15, - 347.09 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p208-b23", - "global_id": 5773, - "bbox": [ - 205.15, - 369.38, - 307.94, - 377.68 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p208-b24", - "global_id": 5774, - "bbox": [ - 391.84, - 337.16, - 402.06, - 346.77 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p208-b25", - "global_id": 5775, - "bbox": [ - 139.3, - 384.08, - 355.23, - 393.69 - ], - "text": "Figure 2.12 Convolution of causal x(t) and anticausal g(t).", - "type": "text" - }, - { - "block_id": "p208-b26", - "global_id": 5776, - "bbox": [ - 107.82, - 469.01, - 447.13, - 480.97 - ], - "text": "DRILL 2.13\nShowing Commutability Using Shifted Signals", - "type": "text" - }, - { - "block_id": "p208-b27", - "global_id": 5777, - "bbox": [ - 107.82, - 490.09, - 295.98, - 500.05 - ], - "text": "Repeat Drill 2.11 for the functions in Fig. 2.13.", - "type": "text" - }, - { - "block_id": "p208-b28", - "global_id": 5778, - "bbox": [ - 222.2, - 589.21, - 224.42, - 597.21 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p208-b29", - "global_id": 5779, - "bbox": [ - 241.64, - 564.36, - 246.64, - 574.36 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p208-b30", - "global_id": 5780, - "bbox": [ - 160.69, - 533.82, - 315.33, - 545.9 - ], - "text": "x(t)\ng(t)", - "type": "text" - }, - { - "block_id": "p208-b32", - "global_id": 5781, - "bbox": [ - 151.31, - 539.19, - 390.88, - 553.47 - ], - "text": "1\nc(t)", - "type": "text" - }, - { - "block_id": "p208-b33", - "global_id": 5782, - "bbox": [ - 151.9, - 587.52, - 155.9, - 595.52 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p208-b34", - "global_id": 5783, - "bbox": [ - 293.86, - 539.41, - 297.86, - 547.41 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p208-b35", - "global_id": 5784, - "bbox": [ - 173.33, - 587.23, - 392.35, - 595.52 - ], - "text": "0\n0\nT\nT", - "type": "text" - }, - { - "block_id": "p208-b36", - "global_id": 5785, - "bbox": [ - 423.17, - 555.58, - 425.4, - 563.58 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p208-b37", - "global_id": 5786, - "bbox": [ - 334.7, - 589.26, - 445.09, - 597.76 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p208-b38", - "global_id": 5787, - "bbox": [ - 139.3, - 604.15, - 346.28, - 613.76 - ], - "text": "Figure 2.13 Convolution of shifted signals x(t) and g(t).", - "type": "text" - } - ] - }, - { - "page_num": 209, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p209-b0", - "global_id": 5788, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n189", - "type": "text" - }, - { - "block_id": "p209-b1", - "global_id": 5789, - "bbox": [ - 127.89, - 86.19, - 420.72, - 98.32 - ], - "text": "THE PHANTOM OF THE SIGNALS AND SYSTEMS OPERA", - "type": "text" - }, - { - "block_id": "p209-b2", - "global_id": 5790, - "bbox": [ - 127.59, - 102.35, - 516.17, - 243.82 - ], - "text": "In the study of signals and systems we often come across some signals such as an impulse, which\ncannot be generated in practice and have never been sighted by anyone.† One wonders why we\neven consider such idealized signals. The answer should be clear from our discussion so far in\nthis chapter. Even if the impulse function has no physical existence, we can compute the system\nresponse h(t) to this phantom input according to the procedure in Sec. 2.3, and knowing h(t),\nwe can compute the system response to any arbitrary input. The concept of impulse response,\ntherefore, provides an effective intermediary for computing system response to an arbitrary input.\nIn addition, the impulse response h(t) itself provides a great deal of information and insight about\nthe system behavior. In Sec. 2.6 we show that the knowledge of impulse response provides much\nvaluable information, such as the response time, pulse dispersion, and filtering properties of the\nsystem. Many other useful insights about the system behavior can be obtained by inspection\nof h(t).", - "type": "text" - }, - { - "block_id": "p209-b3", - "global_id": 5791, - "bbox": [ - 127.6, - 245.71, - 516.15, - 327.5 - ], - "text": "Similarly, in frequency-domain analysis (discussed in later chapters), we use an everlasting\nexponential (or sinusoid) to determine system response. An everlasting exponential (or sinusoid),\ntoo, is a phantom, which nobody has ever seen and which has no physical existence. But it provides\nanother effective intermediary for computing the system response to an arbitrary input. Moreover,\nthe system response to everlasting exponential (or sinusoid) provides valuable information and\ninsight regarding the system’s behavior. Clearly, idealized impulses and everlasting sinusoids are\nfriendly and helpful spirits.", - "type": "text" - }, - { - "block_id": "p209-b4", - "global_id": 5792, - "bbox": [ - 127.6, - 329.49, - 516.16, - 363.37 - ], - "text": "Interestingly, the unit impulse and the everlasting exponential (or sinusoid) are the dual of\neach other in the time-frequency duality, to be studied in Ch. 7. Actually, the time-domain and the\nfrequency-domain methods of analysis are the dual of each other.", - "type": "text" - }, - { - "block_id": "p209-b5", - "global_id": 5793, - "bbox": [ - 127.59, - 380.97, - 516.17, - 492.76 - ], - "text": "WHY CONVOLUTION? AN INTUITIVE EXPLANATION OF\nSYSTEM RESPONSE\nOn the surface, it appears rather strange that the response of linear systems (those gentlest of the\ngentle systems) should be given by such a tortuous operation of convolution, where one signal is\nfixed and the other is inverted and shifted. To understand this odd behavior, consider a hypothetical\nimpulse response h(t) that decays linearly with time (Fig. 2.14a). This response is strongest at t = 0,\nthe moment the impulse is applied, and it decays linearly at future instants so that one second later\n(at t = 1 and beyond), it ceases to exist. This means that the closer the impulse input is to an instant\nt, the stronger is its response at t.", - "type": "text" - }, - { - "block_id": "p209-b6", - "global_id": 5794, - "bbox": [ - 127.59, - 494.34, - 516.16, - 564.49 - ], - "text": "Now consider the input x(t) shown in Fig. 2.14b. To compute the system response, we break\nthe input into rectangular pulses and approximate these pulses with impulses. Generally, the\nresponse of a causal system at some instant t will be determined by all the impulse components of\nthe input before t. Each of these impulse components will have different weight in determining the\nresponse at the instant t, depending on its proximity to t. As seen earlier, the closer the impulse is to\nt, the stronger is its influence at t. The impulse at t has the greatest weight (unity) in determining", - "type": "text" - }, - { - "block_id": "p209-b7", - "global_id": 5795, - "bbox": [ - 127.59, - 588.31, - 516.15, - 633.41 - ], - "text": "† The late Prof. S. J. Mason, the inventor of signal flow graph techniques, used to tell a story of a student\nfrustrated with the impulse function. The student said, “The unit impulse is a thing that is so small you can’t\nsee it, except at one place (the origin), where it is so big you can’t see it. In other words, you can’t see it at\nall; at least I can’t!” [2].", - "type": "text" - } - ] - }, - { - "page_num": 210, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p210-b0", - "global_id": 5796, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "190\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p210-b1", - "global_id": 5797, - "bbox": [ - 214.18, - 157.37, - 216.4, - 165.37 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p210-b2", - "global_id": 5798, - "bbox": [ - 314.96, - 91.16, - 327.4, - 99.26 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p210-b3", - "global_id": 5799, - "bbox": [ - 134.21, - 158.15, - 445.75, - 167.07 - ], - "text": "0\n1\nt\nt 1\nt", - "type": "text" - }, - { - "block_id": "p210-b4", - "global_id": 5800, - "bbox": [ - 345.51, - 178.07, - 373.72, - 186.07 - ], - "text": "1 second", - "type": "text" - }, - { - "block_id": "p210-b5", - "global_id": 5801, - "bbox": [ - 174.92, - 191.74, - 360.69, - 199.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p210-b6", - "global_id": 5802, - "bbox": [ - 125.76, - 89.53, - 380.68, - 100.16 - ], - "text": "h(t)\nh(t t)", - "type": "text" - }, - { - "block_id": "p210-b7", - "global_id": 5803, - "bbox": [ - 125.76, - 206.44, - 306.27, - 215.68 - ], - "text": "Figure 2.14 Intuitive explanation of convolution.", - "type": "text" - }, - { - "block_id": "p210-b8", - "global_id": 5804, - "bbox": [ - 101.84, - 239.78, - 490.4, - 309.92 - ], - "text": "the response at t. The weight decreases linearly for all impulses before t until the instant t −1.\nThe input before t −1 has no influence (zero weight). Thus, to determine the system response\nat t, we must assign a linearly decreasing weight to impulses occurring before t, as shown in\nFig. 2.14b. This weighting function is precisely the function h(t −τ). The system response at t is\nthen determined not by the input x(τ) but by the weighted input x(τ)h(t −τ), and the summation\nof all these weighted inputs is the convolution integral.", - "type": "text" - }, - { - "block_id": "p210-b9", - "global_id": 5805, - "bbox": [ - 101.84, - 338.69, - 262.04, - 350.65 - ], - "text": "2.4-3 Interconnected Systems", - "type": "text" - }, - { - "block_id": "p210-b10", - "global_id": 5806, - "bbox": [ - 101.84, - 356.77, - 490.42, - 414.56 - ], - "text": "A larger, more complex system can often be viewed as the interconnection of several smaller\nsubsystems, each of which is easier to characterize. Knowing the characterizations of these\nsubsystems, it becomes simpler to analyze such large systems. We shall consider here two basic\ninterconnections, cascade and parallel. Figure 2.15a shows S1 and S2, two LTIC subsystems\nconnected in parallel, and Fig. 2.15b shows the same two systems connected in cascade.", - "type": "text" - }, - { - "block_id": "p210-b11", - "global_id": 5807, - "bbox": [ - 101.84, - 416.13, - 490.4, - 486.29 - ], - "text": "In Fig. 2.15a, the device depicted by the symbol inside a circle represents an adder, which\nadds signals at its inputs. Also the junction from which two (or more) branches radiate out is called\nthe pickoff node. Every branch that radiates out from the pickoff node carries the same signal (the\nsignal at the junction). In Fig. 2.15a, for instance, the junction at which the input is applied is a\npickoff node from which two branches radiate out, each of which carries the input signal at the\nnode.", - "type": "text" - }, - { - "block_id": "p210-b12", - "global_id": 5808, - "bbox": [ - 101.84, - 487.86, - 490.39, - 534.11 - ], - "text": "Let the impulse response of S1 and S2 be h1(t) and h2(t), respectively. Further assume that\ninterconnecting these systems, as shown in Fig. 2.15, does not load them. This means that the\nimpulse response of either of these systems remains unchanged whether observed when these\nsystems are unconnected or when they are interconnected.", - "type": "text" - }, - { - "block_id": "p210-b13", - "global_id": 5809, - "bbox": [ - 101.85, - 535.69, - 490.39, - 582.63 - ], - "text": "To find hp(t), the impulse response of the parallel system Sp in Fig. 2.15a, we apply an impulse\nat the input of Sp. This results in the signal δ(t) at the inputs of S1 and S2, leading to their outputs\nh1(t) and h2(t), respectively. These signals are added by the adder to yield h1(t) + h2(t) as the\noutput of Sp:", - "type": "text" - }, - { - "block_id": "p210-b14", - "global_id": 5810, - "bbox": [ - 255.73, - 589.52, - 336.51, - 600.67 - ], - "text": "hp(t) = h1(t) + h2(t)", - "type": "text" - }, - { - "block_id": "p210-b15", - "global_id": 5811, - "bbox": [ - 101.84, - 612.45, - 490.39, - 636.28 - ], - "text": "To find hc(t), the impulse response of the cascade system Sc in Fig. 2.15b, we apply the input δ(t)\nat the input of Sc, which is also the input to S1. Hence, the output of S1 is h1(t), which now acts", - "type": "text" - } - ] - }, - { - "page_num": 211, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p211-b0", - "global_id": 5812, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n191", - "type": "text" - }, - { - "block_id": "p211-b1", - "global_id": 5813, - "bbox": [ - 196.66, - 91.77, - 304.19, - 101.38 - ], - "text": "d(t)\nh1(t)", - "type": "text" - }, - { - "block_id": "p211-b2", - "global_id": 5814, - "bbox": [ - 196.66, - 126.29, - 208.21, - 134.39 - ], - "text": "d(t)", - "type": "text" - }, - { - "block_id": "p211-b3", - "global_id": 5815, - "bbox": [ - 154.8, - 108.9, - 411.67, - 118.72 - ], - "text": "d(t)\nhp(t) h1(t) h2(t)", - "type": "text" - }, - { - "block_id": "p211-b4", - "global_id": 5816, - "bbox": [ - 289.63, - 143.1, - 304.19, - 152.71 - ], - "text": "h2(t)", - "type": "text" - }, - { - "block_id": "p211-b5", - "global_id": 5817, - "bbox": [ - 154.48, - 197.55, - 406.37, - 209.71 - ], - "text": "d(t)\nh(t) h1(t) * h2(t)\nh1(t)", - "type": "text" - }, - { - "block_id": "p211-b6", - "global_id": 5818, - "bbox": [ - 154.48, - 255.05, - 405.87, - 267.21 - ], - "text": "d(t)\nh(t) h2(t) * h1(t)\nh2(t)", - "type": "text" - }, - { - "block_id": "p211-b7", - "global_id": 5819, - "bbox": [ - 154.48, - 305.99, - 264.77, - 316.21 - ], - "text": "x(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p211-b8", - "global_id": 5820, - "bbox": [ - 154.48, - 358.1, - 165.58, - 366.18 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p211-b9", - "global_id": 5821, - "bbox": [ - 154.48, - 411.95, - 165.58, - 420.04 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p211-b10", - "global_id": 5822, - "bbox": [ - 212.48, - 409.95, - 414.58, - 428.63 - ], - "text": "y(t)\ny(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p211-b11", - "global_id": 5823, - "bbox": [ - 247.27, - 101.19, - 254.27, - 110.8 - ], - "text": "S1", - "type": "text" - }, - { - "block_id": "p211-b12", - "global_id": 5824, - "bbox": [ - 247.27, - 133.59, - 254.27, - 143.19 - ], - "text": "S2", - "type": "text" - }, - { - "block_id": "p211-b14", - "global_id": 5825, - "bbox": [ - 336.44, - 87.19, - 343.44, - 96.74 - ], - "text": "Sp", - "type": "text" - }, - { - "block_id": "p211-b15", - "global_id": 5826, - "bbox": [ - 204.64, - 205.98, - 313.6, - 216.32 - ], - "text": "S1\nS2", - "type": "text" - }, - { - "block_id": "p211-b16", - "global_id": 5827, - "bbox": [ - 344.02, - 183.72, - 350.68, - 193.26 - ], - "text": "Sc", - "type": "text" - }, - { - "block_id": "p211-b17", - "global_id": 5828, - "bbox": [ - 308.64, - 263.47, - 315.64, - 273.08 - ], - "text": "S1", - "type": "text" - }, - { - "block_id": "p211-b18", - "global_id": 5829, - "bbox": [ - 206.14, - 315.42, - 210.14, - 323.42 - ], - "text": "S", - "type": "text" - }, - { - "block_id": "p211-b19", - "global_id": 5830, - "bbox": [ - 310.14, - 365.39, - 314.14, - 373.39 - ], - "text": "S", - "type": "text" - }, - { - "block_id": "p211-b20", - "global_id": 5831, - "bbox": [ - 276.03, - 414.85, - 282.25, - 431.83 - ], - "text": "d\ndt", - "type": "text" - }, - { - "block_id": "p211-b21", - "global_id": 5832, - "bbox": [ - 295.86, - 409.95, - 306.96, - 418.03 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p211-b22", - "global_id": 5833, - "bbox": [ - 189.12, - 418.58, - 339.97, - 427.43 - ], - "text": "h(t)\nh(t)", - "type": "text" - }, - { - "block_id": "p211-b23", - "global_id": 5834, - "bbox": [ - 256.09, - 357.74, - 276.07, - 365.84 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p211-b25", - "global_id": 5835, - "bbox": [ - 310.7, - 311.9, - 312.37, - 317.9 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p211-b27", - "global_id": 5836, - "bbox": [ - 206.7, - 355.2, - 362.28, - 367.88 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p211-b28", - "global_id": 5837, - "bbox": [ - 353.84, - 353.39, - 355.5, - 359.39 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p211-b30", - "global_id": 5838, - "bbox": [ - 353.84, - 304.25, - 355.5, - 310.25 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p211-b32", - "global_id": 5839, - "bbox": [ - 246.88, - 354.15, - 248.55, - 360.15 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p211-b34", - "global_id": 5840, - "bbox": [ - 379.7, - 415.78, - 381.37, - 421.78 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p211-b35", - "global_id": 5841, - "bbox": [ - 363.02, - 307.85, - 383.0, - 315.95 - ], - "text": "y(t)dt", - "type": "text" - }, - { - "block_id": "p211-b36", - "global_id": 5842, - "bbox": [ - 363.02, - 356.94, - 383.0, - 365.04 - ], - "text": "y(t)dt", - "type": "text" - }, - { - "block_id": "p211-b37", - "global_id": 5843, - "bbox": [ - 204.64, - 263.47, - 211.64, - 273.08 - ], - "text": "S2", - "type": "text" - }, - { - "block_id": "p211-b38", - "global_id": 5844, - "bbox": [ - 280.67, - 234.29, - 290.31, - 242.29 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p211-b39", - "global_id": 5845, - "bbox": [ - 281.05, - 285.17, - 289.93, - 293.17 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p211-b40", - "global_id": 5846, - "bbox": [ - 281.05, - 163.01, - 289.93, - 171.01 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p211-b41", - "global_id": 5847, - "bbox": [ - 281.01, - 446.74, - 289.96, - 454.74 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p211-b42", - "global_id": 5848, - "bbox": [ - 281.05, - 385.96, - 289.93, - 393.96 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p211-b43", - "global_id": 5849, - "bbox": [ - 280.83, - 336.06, - 290.15, - 344.06 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p211-b44", - "global_id": 5850, - "bbox": [ - 151.5, - 461.44, - 287.74, - 470.68 - ], - "text": "Figure 2.15 Interconnected systems.", - "type": "text" - }, - { - "block_id": "p211-b45", - "global_id": 5851, - "bbox": [ - 127.59, - 490.68, - 437.8, - 502.55 - ], - "text": "as the input to S2. The response of S2 to input h1(t) is h1(t) ∗h2(t). Therefore,", - "type": "text" - }, - { - "block_id": "p211-b46", - "global_id": 5852, - "bbox": [ - 282.96, - 512.32, - 360.77, - 523.47 - ], - "text": "hc(t) = h1(t) ∗h2(t)", - "type": "text" - }, - { - "block_id": "p211-b47", - "global_id": 5853, - "bbox": [ - 127.59, - 533.94, - 516.16, - 592.15 - ], - "text": "Because of the commutative property of convolution, it follows that interchanging the systems S1\nand S2, as shown in Fig. 2.15c, results in the same impulse response h1(t) ∗h2(t). This means that\nwhen several LTIC systems are cascaded, the order of systems does not affect the impulse response\nof the composite system. In other words, linear operations, performed in cascade, commute. The\norder in which they are performed is not important, at least theoretically.†", - "type": "text" - }, - { - "block_id": "p211-b48", - "global_id": 5854, - "bbox": [ - 127.59, - 610.24, - 516.12, - 633.41 - ], - "text": "† Change of order, however, could affect performance because of physical limitations and sensitivities to\nchanges in the subsystems involved.", - "type": "text" - } - ] - }, - { - "page_num": 212, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p212-b0", - "global_id": 5855, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "192\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p212-b1", - "global_id": 5856, - "bbox": [ - 101.84, - 85.82, - 490.4, - 179.47 - ], - "text": "We shall give here another interesting application of the commutative property of LTIC\nsystems. Figure 2.15d shows a cascade of two LTIC systems: a system S with impulse response\nh(t), followed by an ideal integrator. Figure 2.15e shows a cascade of the same two systems in\nreverse order; an ideal integrator followed by S. In Fig. 2.15d, if the input x(t) to S yields the\noutput y(t), then the output of the system of Fig. 2.15d is the integral of y(t). In Fig. 2.15e, the\noutput of the integrator is the integral of x(t). The output in Fig. 2.15e is identical to the output in\nFig. 2.15d. Hence, it follows that if an LTIC system response to input x(t) is y(t), then the response\nof the same system to the integral of x(t) is the integral of y(t). In other words,", - "type": "text" - }, - { - "block_id": "p212-b2", - "global_id": 5857, - "bbox": [ - 187.4, - 196.38, - 283.28, - 206.76 - ], - "text": "if x(t) \r⇒y(t)\nthen", - "type": "text" - }, - { - "block_id": "p212-b3", - "global_id": 5858, - "bbox": [ - 286.89, - 182.82, - 298.65, - 195.1 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p212-b4", - "global_id": 5859, - "bbox": [ - 292.15, - 207.7, - 304.71, - 214.67 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p212-b5", - "global_id": 5860, - "bbox": [ - 306.32, - 196.38, - 354.13, - 206.66 - ], - "text": "x(τ)dτ \r⇒", - "type": "text" - }, - { - "block_id": "p212-b6", - "global_id": 5861, - "bbox": [ - 356.17, - 182.82, - 367.94, - 195.1 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p212-b7", - "global_id": 5862, - "bbox": [ - 361.44, - 207.7, - 374.0, - 214.67 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p212-b8", - "global_id": 5863, - "bbox": [ - 375.61, - 196.38, - 403.64, - 206.66 - ], - "text": "y(τ)dτ", - "type": "text" - }, - { - "block_id": "p212-b9", - "global_id": 5864, - "bbox": [ - 101.85, - 223.62, - 490.37, - 245.55 - ], - "text": "Replacing the ideal integrator with an ideal differentiator in Figs. 2.15d and 2.15e, and following\na similar argument, we conclude that", - "type": "text" - }, - { - "block_id": "p212-b10", - "global_id": 5865, - "bbox": [ - 214.44, - 254.71, - 333.83, - 272.07 - ], - "text": "if x(t) \r⇒y(t)\nthen dx(t)", - "type": "text" - }, - { - "block_id": "p212-b11", - "global_id": 5866, - "bbox": [ - 319.96, - 254.71, - 376.6, - 279.04 - ], - "text": "dt\n\r⇒dy(t)", - "type": "text" - }, - { - "block_id": "p212-b12", - "global_id": 5867, - "bbox": [ - 362.74, - 269.08, - 370.49, - 279.04 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p212-b13", - "global_id": 5868, - "bbox": [ - 101.85, - 286.27, - 490.39, - 308.6 - ], - "text": "If we let x(t) = δ(t) and y(t) = h(t) in Fig. 2.15e, we find that g(t), the unit step response of an\nLTIC system with impulse h(t), is given by", - "type": "text" - }, - { - "block_id": "p212-b14", - "global_id": 5869, - "bbox": [ - 257.9, - 325.52, - 283.08, - 335.8 - ], - "text": "g(t) =", - "type": "text" - }, - { - "block_id": "p212-b15", - "global_id": 5870, - "bbox": [ - 285.12, - 311.96, - 296.89, - 324.23 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p212-b16", - "global_id": 5871, - "bbox": [ - 290.39, - 336.83, - 302.94, - 343.81 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p212-b17", - "global_id": 5872, - "bbox": [ - 304.55, - 325.52, - 490.38, - 335.9 - ], - "text": "h(τ)dτ\n(2.35)", - "type": "text" - }, - { - "block_id": "p212-b18", - "global_id": 5873, - "bbox": [ - 101.85, - 352.21, - 490.4, - 388.49 - ], - "text": "We can also show that the system response to ˙δ(t) is dh(t)/dt. These results can be extended to\nother singularity functions. For example, the unit ramp response of an LTIC system is the integral\nof its unit step response, and so on.", - "type": "text" - }, - { - "block_id": "p212-b19", - "global_id": 5874, - "bbox": [ - 101.84, - 403.12, - 490.39, - 465.09 - ], - "text": "INVERSE SYSTEMS\nIn Fig. 2.15b, if S1 and S2 are inverse systems with impulse response h(t) and hi(t), respectively,\nthen the impulse response of the cascade of these systems is h(t) ∗hi(t). But, the cascade of a\nsystem with its inverse is an identity system, whose output is the same as the input. In other words,\nthe unit impulse response of the cascade of inverse systems is also an unit impulse δ(t). Hence,", - "type": "text" - }, - { - "block_id": "p212-b20", - "global_id": 5875, - "bbox": [ - 261.86, - 476.22, - 490.38, - 487.3 - ], - "text": "h(t) ∗hi(t) = δ(t)\n(2.36)", - "type": "text" - }, - { - "block_id": "p212-b21", - "global_id": 5876, - "bbox": [ - 101.85, - 498.13, - 490.41, - 579.83 - ], - "text": "We shall give an interesting application of the commutative property. As seen from Eq. (2.36),\na cascade of inverse systems is an identity system. Moreover, in a cascade of several LTIC\nsubsystems, changing the order of the subsystems in any manner does not affect the impulse\nresponse of the cascade system. Using these facts, we observe that the two systems, shown in\nFig. 2.15f, are equivalent. We can compute the response of the cascade system on the right-hand\nside, by computing the response of the system inside the dotted box to the input ˙x(t). The impulse\nresponse of the dotted box is g(t), the integral of h(t), as given in Eq. (2.35). Hence, it follows that", - "type": "text" - }, - { - "block_id": "p212-b22", - "global_id": 5877, - "bbox": [ - 238.37, - 590.95, - 490.38, - 601.33 - ], - "text": "y(t) = x(t) ∗h(t) = ˙x(t) ∗g(t)\n(2.37)", - "type": "text" - }, - { - "block_id": "p212-b23", - "global_id": 5878, - "bbox": [ - 101.85, - 612.45, - 490.41, - 634.79 - ], - "text": "Recall that g(t) is the unit step response of the system. Hence, an LTIC response can also be\nobtained as a convolution of ˙x(t) (the derivative of the input) with the unit step response of the", - "type": "text" - } - ] - }, - { - "page_num": 213, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p213-b0", - "global_id": 5879, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n193", - "type": "text" - }, - { - "block_id": "p213-b1", - "global_id": 5880, - "bbox": [ - 127.59, - 85.82, - 516.13, - 119.69 - ], - "text": "system. This result can be readily extended to higher derivatives of the input. An LTIC system\nresponse is the convolution of the nth derivative of the input with the nth integral of the impulse\nresponse.", - "type": "text" - }, - { - "block_id": "p213-b2", - "global_id": 5881, - "bbox": [ - 127.59, - 147.07, - 390.55, - 172.98 - ], - "text": "2.4-4 A Very Special Function for LTIC Systems:\nThe Everlasting Exponential est", - "type": "text" - }, - { - "block_id": "p213-b3", - "global_id": 5882, - "bbox": [ - 127.59, - 179.11, - 516.16, - 272.75 - ], - "text": "There is a very special connection of LTIC systems with the everlasting exponential function\nest, where s is a complex variable, in general. We now show that the LTIC system’s (zero-state)\nresponse to everlasting exponential input est is also the same everlasting exponential (within a\nmultiplicative constant). Moreover, no other function can make the same claim. Such an input\nfor which the system response is also of the same form is called the characteristic function (also\neigenfunction) of the system. Because a sinusoid is a form of exponential (s = ±jω), everlasting\nsinusoid is also a characteristic function of an LTIC system. Note that we are talking here of an\neverlasting exponential (or sinusoid), which starts at t = −∞.", - "type": "text" - }, - { - "block_id": "p213-b4", - "global_id": 5883, - "bbox": [ - 127.59, - 274.33, - 516.12, - 296.67 - ], - "text": "If h(t) is the system’s unit impulse response, then system response y(t) to an everlasting\nexponential est is given by", - "type": "text" - }, - { - "block_id": "p213-b5", - "global_id": 5884, - "bbox": [ - 204.8, - 312.04, - 274.64, - 326.43 - ], - "text": "y(t) = h(t) ∗est =", - "type": "text" - }, - { - "block_id": "p213-b6", - "global_id": 5885, - "bbox": [ - 276.69, - 302.6, - 293.64, - 314.65 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p213-b7", - "global_id": 5886, - "bbox": [ - 281.95, - 327.47, - 294.51, - 334.45 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p213-b8", - "global_id": 5887, - "bbox": [ - 296.12, - 312.35, - 371.04, - 326.43 - ], - "text": "h(τ)es(t−τ) dτ = est", - "type": "text" - }, - { - "block_id": "p213-b9", - "global_id": 5888, - "bbox": [ - 372.77, - 302.6, - 389.72, - 314.65 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p213-b10", - "global_id": 5889, - "bbox": [ - 378.04, - 327.47, - 390.59, - 334.45 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p213-b11", - "global_id": 5890, - "bbox": [ - 392.2, - 312.35, - 437.72, - 326.43 - ], - "text": "h(τ)e−sτ dτ", - "type": "text" - }, - { - "block_id": "p213-b12", - "global_id": 5891, - "bbox": [ - 127.59, - 346.35, - 516.12, - 368.36 - ], - "text": "The integral on the right-most side is a function of a complex variable s and a constant with respect\nto t. Let us denote this term by H(s), which is also complex, in general. Thus,", - "type": "text" - }, - { - "block_id": "p213-b13", - "global_id": 5892, - "bbox": [ - 294.2, - 381.03, - 516.13, - 392.9 - ], - "text": "y(t) = H(s)est\n(2.38)", - "type": "text" - }, - { - "block_id": "p213-b14", - "global_id": 5893, - "bbox": [ - 127.59, - 407.49, - 151.92, - 417.46 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p213-b15", - "global_id": 5894, - "bbox": [ - 273.38, - 426.24, - 302.15, - 436.52 - ], - "text": "H(s) =", - "type": "text" - }, - { - "block_id": "p213-b16", - "global_id": 5895, - "bbox": [ - 304.2, - 412.68, - 321.14, - 424.73 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p213-b17", - "global_id": 5896, - "bbox": [ - 309.46, - 437.55, - 322.01, - 444.52 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p213-b18", - "global_id": 5897, - "bbox": [ - 323.62, - 422.43, - 516.13, - 436.62 - ], - "text": "h(τ)e−sτ dτ\n(2.39)", - "type": "text" - }, - { - "block_id": "p213-b19", - "global_id": 5898, - "bbox": [ - 127.59, - 454.93, - 447.88, - 465.31 - ], - "text": "Equation (2.38) is valid only for the values of s for which H(s) exists, that is, if", - "type": "text" - }, - { - "block_id": "p213-b20", - "global_id": 5899, - "bbox": [ - 450.68, - 446.91, - 464.3, - 458.96 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p213-b21", - "global_id": 5900, - "bbox": [ - 127.6, - 451.63, - 516.13, - 489.21 - ], - "text": "−∞h(τ)e−sτ dτ\nexists (or converges). The region in the s plane for which this integral converges is called the region\nof convergence for H(s). Further elaboration of the region of convergence is presented in Ch. 4.", - "type": "text" - }, - { - "block_id": "p213-b22", - "global_id": 5901, - "bbox": [ - 127.6, - 490.8, - 516.13, - 513.13 - ], - "text": "For a given s, note that H(s) is a constant. Thus, the input and the output are the same (within\na multiplicative constant) for the everlasting exponential signal.", - "type": "text" - }, - { - "block_id": "p213-b23", - "global_id": 5902, - "bbox": [ - 127.6, - 514.7, - 516.15, - 537.04 - ], - "text": "H(s), which is called the transfer function of the system, is a function of complex variable s.\nAn alternate definition of the transfer function H(s) of an LTIC system, as seen from Eq. (2.38), is", - "type": "text" - }, - { - "block_id": "p213-b24", - "global_id": 5903, - "bbox": [ - 230.44, - 550.28, - 314.2, - 567.23 - ], - "text": "H(s) = output signal", - "type": "text" - }, - { - "block_id": "p213-b25", - "global_id": 5904, - "bbox": [ - 264.94, - 564.34, - 311.7, - 574.3 - ], - "text": "input signal", - "type": "text" - }, - { - "block_id": "p213-b27", - "global_id": 5905, - "bbox": [ - 318.64, - 557.26, - 516.13, - 576.31 - ], - "text": "input=everlasting exponential est\n(2.40)", - "type": "text" - }, - { - "block_id": "p213-b28", - "global_id": 5906, - "bbox": [ - 127.59, - 588.96, - 516.12, - 610.88 - ], - "text": "The transfer function is defined for, and is meaningful to, LTIC systems only. It does not exist for\nnonlinear or time-varying systems, in general.", - "type": "text" - }, - { - "block_id": "p213-b29", - "global_id": 5907, - "bbox": [ - 127.59, - 612.86, - 516.15, - 634.79 - ], - "text": "We repeat again that this discussion is about the everlasting exponential, which starts at\nt = −∞, not the causal exponential estu(t), which starts at t = 0.", - "type": "text" - } - ] - }, - { - "page_num": 214, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p214-b0", - "global_id": 5908, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "194\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p214-b1", - "global_id": 5909, - "bbox": [ - 119.78, - 85.82, - 390.45, - 95.78 - ], - "text": "For a system specified by Eq. (2.2), the transfer function is given by", - "type": "text" - }, - { - "block_id": "p214-b2", - "global_id": 5910, - "bbox": [ - 270.26, - 104.39, - 320.22, - 121.65 - ], - "text": "H(s) = P(s)", - "type": "text" - }, - { - "block_id": "p214-b3", - "global_id": 5911, - "bbox": [ - 302.27, - 111.79, - 490.38, - 128.73 - ], - "text": "Q(s)\n(2.41)", - "type": "text" - }, - { - "block_id": "p214-b4", - "global_id": 5912, - "bbox": [ - 101.84, - 136.51, - 490.39, - 159.85 - ], - "text": "This follows readily by considering an everlasting input x(t) = est. According to Eq. (2.38), the\noutput is y(t) = H(s)est. Substitution of this x(t) and y(t) in Eq. (2.2) yields", - "type": "text" - }, - { - "block_id": "p214-b5", - "global_id": 5913, - "bbox": [ - 246.43, - 168.91, - 345.17, - 180.69 - ], - "text": "H(s)[Q(D)est] = P(D)est", - "type": "text" - }, - { - "block_id": "p214-b6", - "global_id": 5914, - "bbox": [ - 101.84, - 191.77, - 142.9, - 201.74 - ], - "text": "Moreover,", - "type": "text" - }, - { - "block_id": "p214-b7", - "global_id": 5915, - "bbox": [ - 255.29, - 200.13, - 306.3, - 218.39 - ], - "text": "Drest = drest", - "type": "text" - }, - { - "block_id": "p214-b8", - "global_id": 5916, - "bbox": [ - 292.12, - 206.61, - 336.32, - 225.47 - ], - "text": "dtr = srest", - "type": "text" - }, - { - "block_id": "p214-b9", - "global_id": 5917, - "bbox": [ - 101.84, - 229.72, - 129.76, - 239.68 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p214-b10", - "global_id": 5918, - "bbox": [ - 198.52, - 237.14, - 393.09, - 251.63 - ], - "text": "P(D)est = P(s)est\nand\nQ(D)est = Q(s)est", - "type": "text" - }, - { - "block_id": "p214-b11", - "global_id": 5919, - "bbox": [ - 101.84, - 260.11, - 158.45, - 270.07 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p214-b12", - "global_id": 5920, - "bbox": [ - 270.26, - 269.7, - 320.22, - 286.96 - ], - "text": "H(s) = P(s)", - "type": "text" - }, - { - "block_id": "p214-b13", - "global_id": 5921, - "bbox": [ - 302.27, - 283.75, - 320.78, - 294.03 - ], - "text": "Q(s)", - "type": "text" - }, - { - "block_id": "p214-b14", - "global_id": 5922, - "bbox": [ - 107.82, - 332.61, - 481.8, - 344.57 - ], - "text": "DRILL 2.14\nIdeal Integrator and Differentiator Transfer Functions", - "type": "text" - }, - { - "block_id": "p214-b15", - "global_id": 5923, - "bbox": [ - 107.82, - 353.27, - 484.43, - 399.51 - ], - "text": "Show that the transfer function of an ideal integrator is H(s) = 1/s and that of an ideal\ndifferentiator is H(s) = s. Find the answer in two ways: using Eq. (2.39) and using Eq. (2.41).\n[Hint: Find h(t) for the ideal integrator and differentiator. You also may need to use the result\nin Prob. 1.4-12.]", - "type": "text" - }, - { - "block_id": "p214-b16", - "global_id": 5924, - "bbox": [ - 102.14, - 429.3, - 348.01, - 441.42 - ], - "text": "A FUNDAMENTAL PROPERTY OF LTI SYSTEMS", - "type": "text" - }, - { - "block_id": "p214-b17", - "global_id": 5925, - "bbox": [ - 101.84, - 445.46, - 490.42, - 479.33 - ], - "text": "We can show that Eq. (2.38) is a fundamental property of LTI systems and it follows directly as a\nconsequence of linearity and time invariance. To show this let us assume that the response of an\nLTI system to an everlasting exponential est is y(s,t). If we define", - "type": "text" - }, - { - "block_id": "p214-b18", - "global_id": 5926, - "bbox": [ - 265.1, - 487.94, - 325.95, - 505.3 - ], - "text": "H(s,t) = y(s,t)", - "type": "text" - }, - { - "block_id": "p214-b19", - "global_id": 5927, - "bbox": [ - 309.95, - 501.72, - 319.02, - 512.27 - ], - "text": "est", - "type": "text" - }, - { - "block_id": "p214-b20", - "global_id": 5928, - "bbox": [ - 101.84, - 519.29, - 118.99, - 529.25 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p214-b21", - "global_id": 5929, - "bbox": [ - 260.89, - 529.34, - 330.72, - 541.22 - ], - "text": "y(s,t) = H(s,t)est", - "type": "text" - }, - { - "block_id": "p214-b22", - "global_id": 5930, - "bbox": [ - 101.84, - 546.08, - 490.39, - 571.61 - ], - "text": "Because of the time-invariance property, the system response to input es(t−T) is H(s,t −T)es(t−T),\nthat is,", - "type": "text" - }, - { - "block_id": "p214-b23", - "global_id": 5931, - "bbox": [ - 236.19, - 571.46, - 490.38, - 583.57 - ], - "text": "y(s,t −T) = H(s,t −T)es(t−T)\n(2.42)", - "type": "text" - }, - { - "block_id": "p214-b24", - "global_id": 5932, - "bbox": [ - 101.85, - 588.43, - 490.37, - 613.96 - ], - "text": "The delayed input es(t−T) represents the input est multiplied by a constant e−sT. Hence, according\nto the linearity property, the system response to es(t−T) must be y(s,t)e−sT. Hence,", - "type": "text" - }, - { - "block_id": "p214-b25", - "global_id": 5933, - "bbox": [ - 218.41, - 620.42, - 373.33, - 634.91 - ], - "text": "y(s,t −T) = y(s,t)e−sT = H(s,t)es(t−T)", - "type": "text" - } - ] - }, - { - "page_num": 215, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p215-b0", - "global_id": 5934, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "2.4\nSystem Response to External Input: The Zero-State Response\n195", - "type": "text" - }, - { - "block_id": "p215-b1", - "global_id": 5935, - "bbox": [ - 127.59, - 85.82, - 336.0, - 95.78 - ], - "text": "Comparison of this result with Eq. (2.42) shows that", - "type": "text" - }, - { - "block_id": "p215-b2", - "global_id": 5936, - "bbox": [ - 256.14, - 114.66, - 386.82, - 125.04 - ], - "text": "H(s,t) = H(s,t −T)\nfor allT", - "type": "text" - }, - { - "block_id": "p215-b3", - "global_id": 5937, - "bbox": [ - 127.6, - 143.92, - 451.51, - 154.3 - ], - "text": "This means H(s,t) is independent of t, and we can express H(s,t) = H(s). Hence,", - "type": "text" - }, - { - "block_id": "p215-b4", - "global_id": 5938, - "bbox": [ - 289.91, - 171.67, - 353.18, - 183.56 - ], - "text": "y(s,t) = H(s)est", - "type": "text" - }, - { - "block_id": "p215-b5", - "global_id": 5939, - "bbox": [ - 127.59, - 215.64, - 240.66, - 227.6 - ], - "text": "2.4-5 Total Response", - "type": "text" - }, - { - "block_id": "p215-b6", - "global_id": 5940, - "bbox": [ - 127.59, - 233.73, - 516.13, - 255.65 - ], - "text": "Assuming distinct roots, the total response of a linear system can be expressed as the sum of its\nzero-input response (ZIR) and its zero-state response (ZSR):", - "type": "text" - }, - { - "block_id": "p215-b7", - "global_id": 5941, - "bbox": [ - 245.47, - 284.04, - 310.34, - 294.42 - ], - "text": "total response =", - "type": "text" - }, - { - "block_id": "p215-b8", - "global_id": 5942, - "bbox": [ - 312.39, - 274.09, - 326.49, - 284.54 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p215-b9", - "global_id": 5943, - "bbox": [ - 313.37, - 298.44, - 325.51, - 305.7 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p215-b10", - "global_id": 5944, - "bbox": [ - 327.6, - 282.32, - 349.17, - 295.12 - ], - "text": "ckeλkt", - "type": "text" - }, - { - "block_id": "p215-b11", - "global_id": 5945, - "bbox": [ - 312.39, - 301.24, - 349.8, - 319.62 - ], - "text": "ZIR", - "type": "text" - }, - { - "block_id": "p215-b12", - "global_id": 5946, - "bbox": [ - 350.89, - 284.04, - 398.25, - 308.16 - ], - "text": "+x(t) ∗h(t)\n\n\n\nZSR", - "type": "text" - }, - { - "block_id": "p215-b13", - "global_id": 5947, - "bbox": [ - 127.59, - 336.81, - 444.6, - 346.77 - ], - "text": "For repeated roots, the zero-input component should be appropriately modified.", - "type": "text" - }, - { - "block_id": "p215-b14", - "global_id": 5948, - "bbox": [ - 127.59, - 347.13, - 516.1, - 382.64 - ], - "text": "For the series RLC circuit in Ex. 2.4 with the input x(t) = 10e−3tu(t) and the initial conditions\ny(0−) = 0,vC(0−) = 5, we determined the zero-input response in Ex. 2.1a [Eq. (2.9)]. We found\nthe zero-state response in Ex. 2.9. From the results in Exs. 2.1a and 2.9, we obtain", - "type": "text" - }, - { - "block_id": "p215-b15", - "global_id": 5949, - "bbox": [ - 181.21, - 397.4, - 305.88, - 425.64 - ], - "text": "total current = (−5e−t + 5e−2t)\n\n\n\nzero-input current", - "type": "text" - }, - { - "block_id": "p215-b16", - "global_id": 5950, - "bbox": [ - 306.97, - 397.4, - 421.96, - 425.64 - ], - "text": "+(−5e−t + 20e−2t −15e−3t)\n\n\n\nzero-state current", - "type": "text" - }, - { - "block_id": "p215-b17", - "global_id": 5951, - "bbox": [ - 442.98, - 401.52, - 516.13, - 411.89 - ], - "text": "t ≥0\n(2.43)", - "type": "text" - }, - { - "block_id": "p215-b18", - "global_id": 5952, - "bbox": [ - 127.59, - 444.12, - 390.2, - 454.09 - ], - "text": "Figure 2.16a shows the zero-input, zero-state, and total responses.", - "type": "text" - }, - { - "block_id": "p215-b19", - "global_id": 5953, - "bbox": [ - 292.56, - 547.52, - 294.79, - 555.52 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p215-b20", - "global_id": 5954, - "bbox": [ - 175.49, - 482.77, - 207.73, - 490.77 - ], - "text": "Zero state", - "type": "text" - }, - { - "block_id": "p215-b21", - "global_id": 5955, - "bbox": [ - 189.48, - 509.22, - 206.37, - 517.22 - ], - "text": "Total", - "type": "text" - }, - { - "block_id": "p215-b22", - "global_id": 5956, - "bbox": [ - 181.1, - 576.56, - 215.13, - 584.56 - ], - "text": "Zero input", - "type": "text" - }, - { - "block_id": "p215-b23", - "global_id": 5957, - "bbox": [ - 214.54, - 605.26, - 223.42, - 613.26 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p215-b24", - "global_id": 5958, - "bbox": [ - 147.37, - 477.99, - 158.47, - 486.07 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p215-b25", - "global_id": 5959, - "bbox": [ - 135.43, - 520.23, - 139.43, - 528.23 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p215-b26", - "global_id": 5960, - "bbox": [ - 393.0, - 565.26, - 415.22, - 573.26 - ], - "text": "Forced", - "type": "text" - }, - { - "block_id": "p215-b27", - "global_id": 5961, - "bbox": [ - 384.51, - 512.6, - 401.39, - 520.6 - ], - "text": "Total", - "type": "text" - }, - { - "block_id": "p215-b28", - "global_id": 5962, - "bbox": [ - 388.47, - 485.9, - 412.46, - 493.9 - ], - "text": "Natural", - "type": "text" - }, - { - "block_id": "p215-b29", - "global_id": 5963, - "bbox": [ - 482.54, - 547.51, - 484.77, - 555.51 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p215-b30", - "global_id": 5964, - "bbox": [ - 409.75, - 605.26, - 419.4, - 613.26 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p215-b31", - "global_id": 5965, - "bbox": [ - 337.97, - 477.99, - 349.08, - 486.07 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p215-b32", - "global_id": 5966, - "bbox": [ - 337.7, - 532.58, - 341.7, - 540.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p215-b33", - "global_id": 5967, - "bbox": [ - 127.59, - 619.96, - 303.2, - 629.19 - ], - "text": "Figure 2.16 Total response and its components.", - "type": "text" - } - ] - }, - { - "page_num": 216, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p216-b0", - "global_id": 5968, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "196\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p216-b1", - "global_id": 5969, - "bbox": [ - 102.14, - 86.19, - 288.72, - 98.32 - ], - "text": "NATURAL AND FORCED RESPONSE", - "type": "text" - }, - { - "block_id": "p216-b2", - "global_id": 5970, - "bbox": [ - 101.84, - 98.73, - 490.4, - 195.99 - ], - "text": "For the RLC circuit in Ex. 2.4, the characteristic modes were found to be e−t and e−2t. As we\nexpected, the zero-input response is composed exclusively of characteristic modes. Note, however,\nthat even the zero-state response [Eq. (2.43)] contains characteristic mode terms. This observation\nis generally true of LTIC systems. We can now lump together all the characteristic mode terms\nin the total response, giving us a component known as the natural response yn(t). The remainder,\nconsisting entirely of noncharacteristic mode terms, is known as the forced response yφ(t). The\ntotal response of the RLC circuit in Ex. 2.4 can be expressed in terms of natural and forced\ncomponents by regrouping the terms in Eq. (2.43) as", - "type": "text" - }, - { - "block_id": "p216-b3", - "global_id": 5971, - "bbox": [ - 174.78, - 204.78, - 309.4, - 233.85 - ], - "text": "total current = (−10e−t + 25e−2t)\n\n\n\nnatural response yn(t)", - "type": "text" - }, - { - "block_id": "p216-b4", - "global_id": 5972, - "bbox": [ - 313.66, - 207.17, - 376.89, - 234.51 - ], - "text": "+\n(−15e−3t)\n\n\n\nforced response yφ(t)", - "type": "text" - }, - { - "block_id": "p216-b5", - "global_id": 5973, - "bbox": [ - 397.92, - 208.89, - 490.38, - 219.27 - ], - "text": "t ≥0\n(2.44)", - "type": "text" - }, - { - "block_id": "p216-b6", - "global_id": 5974, - "bbox": [ - 101.85, - 247.11, - 338.44, - 257.08 - ], - "text": "Figure 2.16b shows the natural, forced, and total responses.", - "type": "text" - }, - { - "block_id": "p216-b7", - "global_id": 5975, - "bbox": [ - 101.85, - 259.07, - 490.42, - 388.59 - ], - "text": "The classical solution to a differential equation includes the natural (also called the\nhomogeneous or complementary) solution and the forced (also known as the particular) solution;\ntraditional courses on differential equations provide simplified procedures to determine these\ncomponents. Unfortunately, the classical solution lacks the engineering intuition and utility\nafforded by the zero-input and zero-state responses. The classical approach cannot separate\nthe responses arising from internal conditions and external input. While the natural and forced\nsolutions can be obtained from the zero-input and zero-state responses, the converse is not true.\nFurther, the classical method is unable to express the system response to an input x(t) as an explicit\nfunction of x(t). In fact, the classical method is restricted to a certain class of inputs and cannot\nhandle arbitrary inputs, as can the method to determine the zero-state response. For these (and\nother) reasons, we do not further detail the classical solution of differential equations.", - "type": "text" - }, - { - "block_id": "p216-b8", - "global_id": 5976, - "bbox": [ - 102.2, - 419.22, - 250.01, - 433.17 - ], - "text": "2.5 SYSTEM STABILITY", - "type": "text" - }, - { - "block_id": "p216-b9", - "global_id": 5977, - "bbox": [ - 101.84, - 439.16, - 490.38, - 473.03 - ], - "text": "Stability is an important system property. Two types of system stability are generally considered:\nexternal (BIBO) stability and internal (asymptotic) stability. Let us consider both stability types in\nturn.", - "type": "text" - }, - { - "block_id": "p216-b10", - "global_id": 5978, - "bbox": [ - 101.84, - 499.14, - 269.0, - 511.1 - ], - "text": "2.5-1 External (BIBO) Stability", - "type": "text" - }, - { - "block_id": "p216-b11", - "global_id": 5979, - "bbox": [ - 101.84, - 517.22, - 490.43, - 634.79 - ], - "text": "To understand the intuitive basis for the BIBO (bounded-input/bounded-output) stability of a\nsystem introduced in Sec. 1.7, let us examine the stability concept as applied to a right circular\ncone. Such a cone can be made to stand forever on its circular base, on its apex, or on its side. For\nthis reason, these three states of the cone are said to be equilibrium states. Qualitatively, however,\nthe three states show very different behavior. If the cone, standing on its circular base, were to\nbe disturbed slightly and then left to itself, it would eventually return to its original equilibrium\nposition. In such a case, the cone is said to be in stable equilibrium. In contrast, if the cone stands\non its apex, then the slightest disturbance will cause the cone to move farther and farther away\nfrom its equilibrium state. The cone in this case is said to be in an unstable equilibrium. The cone\nlying on its side, if disturbed, will neither go back to the original state nor continue to move farther", - "type": "text" - } - ] - }, - { - "page_num": 217, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p217-b0", - "global_id": 5980, - "bbox": [ - 389.44, - 62.89, - 516.14, - 71.98 - ], - "text": "2.5\nSystem Stability\n197", - "type": "text" - }, - { - "block_id": "p217-b1", - "global_id": 5981, - "bbox": [ - 127.59, - 85.72, - 516.14, - 155.56 - ], - "text": "away from the original state. Thus it is said to be in a neutral equilibrium. Clearly, when a system\nis in stable equilibrium, application of a small disturbance (input) produces a small response.\nIn contrast, when the system is in unstable equilibrium, even a minuscule disturbance (input)\nproduces an unbounded response. The BIBO-stability definition can be understood in the light of\nthis concept. If every bounded input produces bounded output, the system is (BIBO) stable.† In\ncontrast, if even one bounded input results in unbounded response, the system is (BIBO) unstable.", - "type": "text" - }, - { - "block_id": "p217-b2", - "global_id": 5982, - "bbox": [ - 145.52, - 157.55, - 227.48, - 167.51 - ], - "text": "For an LTIC system,", - "type": "text" - }, - { - "block_id": "p217-b3", - "global_id": 5983, - "bbox": [ - 243.14, - 184.91, - 318.11, - 195.19 - ], - "text": "y(t) = h(t) ∗x(t) =", - "type": "text" - }, - { - "block_id": "p217-b4", - "global_id": 5984, - "bbox": [ - 320.16, - 171.35, - 337.11, - 183.41 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p217-b5", - "global_id": 5985, - "bbox": [ - 325.42, - 196.22, - 337.98, - 203.2 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p217-b6", - "global_id": 5986, - "bbox": [ - 339.59, - 184.91, - 399.38, - 195.19 - ], - "text": "h(τ)x(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p217-b7", - "global_id": 5987, - "bbox": [ - 127.59, - 213.11, - 169.34, - 223.07 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p217-b8", - "global_id": 5988, - "bbox": [ - 259.89, - 230.06, - 289.96, - 240.33 - ], - "text": "|y(t)| ≤", - "type": "text" - }, - { - "block_id": "p217-b9", - "global_id": 5989, - "bbox": [ - 292.01, - 216.5, - 308.94, - 228.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p217-b10", - "global_id": 5990, - "bbox": [ - 297.26, - 241.38, - 309.82, - 248.35 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p217-b11", - "global_id": 5991, - "bbox": [ - 311.43, - 230.06, - 382.63, - 240.33 - ], - "text": "|h(τ)||x(t −τ)|dτ", - "type": "text" - }, - { - "block_id": "p217-b12", - "global_id": 5992, - "bbox": [ - 127.6, - 254.92, - 365.94, - 266.37 - ], - "text": "Moreover, if x(t) is bounded, then |x(t −τ)| < K1 < ∞, and", - "type": "text" - }, - { - "block_id": "p217-b13", - "global_id": 5993, - "bbox": [ - 272.48, - 282.43, - 314.72, - 293.58 - ], - "text": "|y(t)| ≤K1", - "type": "text" - }, - { - "block_id": "p217-b14", - "global_id": 5994, - "bbox": [ - 316.33, - 268.87, - 333.27, - 280.92 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p217-b15", - "global_id": 5995, - "bbox": [ - 321.59, - 293.74, - 334.14, - 300.71 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p217-b16", - "global_id": 5996, - "bbox": [ - 335.75, - 282.43, - 370.05, - 292.71 - ], - "text": "|h(τ)|dτ", - "type": "text" - }, - { - "block_id": "p217-b17", - "global_id": 5997, - "bbox": [ - 127.6, - 310.63, - 300.33, - 326.79 - ], - "text": "Hence for BIBO stability,\n# ∞", - "type": "text" - }, - { - "block_id": "p217-b18", - "global_id": 5998, - "bbox": [ - 288.65, - 339.61, - 301.2, - 346.59 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p217-b19", - "global_id": 5999, - "bbox": [ - 302.81, - 328.29, - 516.13, - 338.67 - ], - "text": "|h(τ)|dτ < ∞\n(2.45)", - "type": "text" - }, - { - "block_id": "p217-b20", - "global_id": 6000, - "bbox": [ - 127.59, - 353.5, - 516.15, - 423.24 - ], - "text": "This is a sufficient condition for BIBO stability. We can show that this is also a necessary condition\n(see Prob. 2.5-7). Therefore, for an LTIC system, if its impulse response h(t) is absolutely\nintegrable, the system is (BIBO) stable. Otherwise it is (BIBO) unstable. In addition, we shall\nshow in Ch. 4 that a necessary (but not sufficient) condition for an LTIC system described by\nEq. (2.1) to be BIBO-stable is M ≤N. If M > N, the system is unstable. This is one of the reasons\nto avoid systems with M > N.", - "type": "text" - }, - { - "block_id": "p217-b21", - "global_id": 6001, - "bbox": [ - 127.6, - 425.23, - 516.14, - 471.06 - ], - "text": "Because the BIBO stability of a system can be ascertained by measurements at the external\nterminals (input and output), this is an external stability criterion. It is no coincidence that the\nBIBO criterion in Eq. (2.45) is in terms of the impulse response, which is an external description\nof the system.", - "type": "text" - }, - { - "block_id": "p217-b22", - "global_id": 6002, - "bbox": [ - 127.6, - 473.05, - 516.15, - 530.84 - ], - "text": "As observed in Sec. 1.9, the internal behavior of a system is not always ascertainable from\nthe external terminals. Therefore, external (BIBO) stability may not be a correct indication of\ninternal stability. Indeed, some systems that appear stable by the BIBO criterion may be internally\nunstable. This is like a room on fire inside a house: no trace of fire is visible from outside, but the\nentire house will be burned to ashes.", - "type": "text" - }, - { - "block_id": "p217-b23", - "global_id": 6003, - "bbox": [ - 127.6, - 532.83, - 516.14, - 602.57 - ], - "text": "The BIBO stability is meaningful only for systems in which the internal and the external\ndescription are equivalent (controllable and observable systems). Fortunately, most practical\nsystems fall into this category, and whenever we apply this criterion, we implicitly assume that\nthe system, in fact, belongs to this category. Internal stability is all-inclusive, and external stability\ncan always be determined from internal stability. For this reason, we now investigate the internal\nstability criterion.", - "type": "text" - }, - { - "block_id": "p217-b24", - "global_id": 6004, - "bbox": [ - 127.59, - 621.19, - 283.23, - 633.41 - ], - "text": "† The system is assumed to be in zero state.", - "type": "text" - } - ] - }, - { - "page_num": 218, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p218-b0", - "global_id": 6005, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "198\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p218-b1", - "global_id": 6006, - "bbox": [ - 101.84, - 86.52, - 300.19, - 98.48 - ], - "text": "2.5-2 Internal (Asymptotic) Stability", - "type": "text" - }, - { - "block_id": "p218-b2", - "global_id": 6007, - "bbox": [ - 101.84, - 104.61, - 490.39, - 138.48 - ], - "text": "Because of the great variety of possible system behaviors, there are several definitions of internal\nstability in the literature. Here we shall consider a definition that is suitable for causal, linear,\ntime-invariant (LTI) systems.", - "type": "text" - }, - { - "block_id": "p218-b3", - "global_id": 6008, - "bbox": [ - 101.84, - 140.47, - 490.4, - 293.9 - ], - "text": "If, in the absence of an external input, a system remains in a particular state (or condition)\nindefinitely, then that state is said to be an equilibrium state of the system. For an LTI system,\nzero state, in which all initial conditions are zero, is an equilibrium state. Now suppose an LTI\nsystem is in zero state and we change this state by creating small nonzero initial conditions (small\ndisturbance). These initial conditions will generate signals consisting of characteristic modes in\nthe system. By analogy with the cone, if the system is stable, it should eventually return to zero\nstate. In other words, when left to itself, every mode in a stable system arising as a result of nonzero\ninitial conditions should approach 0 as t →∞. However, if even one of the modes grows with time,\nthe system will never return to zero state, and the system would be identified as unstable. In the\nborderline case, some modes neither decay to zero nor grow indefinitely, while all the remaining\nmodes decay to zero. This case is like the neutral equilibrium in the cone. Such a system is said to\nbe marginally stable. Internal stability is also called asymptotic stability or stability in the sense of\nLyapunov.", - "type": "text" - }, - { - "block_id": "p218-b4", - "global_id": 6009, - "bbox": [ - 101.84, - 295.9, - 490.39, - 354.75 - ], - "text": "For a system characterized by Eq. (2.1), we can restate the internal stability criterion in terms\nof the location of the N characteristic roots λ1, λ2, . . ., λN of the system in a complex plane. The\ncharacteristic modes are of the form eλkt or treλkt. The locations of various roots in the complex\nplane and the corresponding modes are shown in Fig. 2.17. These modes →0 as t →∞if Re\nλk < 0. In contrast, the modes →∞as t →∞if Reλk > 0.†", - "type": "text" - }, - { - "block_id": "p218-b5", - "global_id": 6010, - "bbox": [ - 101.84, - 355.66, - 490.42, - 449.32 - ], - "text": "From Fig. 2.17, we see that a system is (asymptotically) stable if all its characteristic roots lie\nin the LHP, that is, if Reλk < 0 for all k. If even a single characteristic root lies in the RHP, the\nsystem is (asymptotically) unstable. Modes due to roots on the imaginary axis (λ = ±jω0) are of\nthe form e±jω0t. Hence, if some roots are on the imaginary axis, and all the remaining roots are\nin the LHP, the system is marginally stable (assuming that the roots on the imaginary axis are not\nrepeated). If the imaginary axis roots are repeated, the characteristic modes are of the form tre±jωkt,\nwhich do grow with time indefinitely. Hence, the system is unstable. Figure 2.18 shows stability\nregions in the complex plane.", - "type": "text" - }, - { - "block_id": "p218-b6", - "global_id": 6011, - "bbox": [ - 119.78, - 451.31, - 179.02, - 461.27 - ], - "text": "To summarize:", - "type": "text" - }, - { - "block_id": "p218-b7", - "global_id": 6012, - "bbox": [ - 118.78, - 469.24, - 490.42, - 538.98 - ], - "text": "1. An LTIC system is asymptotically stable if, and only if, all the characteristic roots are in\nthe LHP. The roots may be simple (unrepeated) or repeated.\n2. An LTIC system is unstable if, and only if, one or both of the following conditions exist:\n(i) at least one root is in the RHP; (ii) there are repeated roots on the imaginary axis.\n3. An LTIC system is marginally stable if, and only if, there are no roots in the RHP, and\nthere are some unrepeated roots on the imaginary axis.", - "type": "text" - }, - { - "block_id": "p218-b8", - "global_id": 6013, - "bbox": [ - 101.84, - 561.76, - 460.95, - 573.99 - ], - "text": "† This may be seen from the fact that if α and β are the real and the imaginary parts of a root λ, then", - "type": "text" - }, - { - "block_id": "p218-b9", - "global_id": 6014, - "bbox": [ - 198.04, - 590.11, - 249.1, - 607.86 - ], - "text": "lim\nt→∞eλt = lim", - "type": "text" - }, - { - "block_id": "p218-b10", - "global_id": 6015, - "bbox": [ - 235.3, - 590.11, - 302.56, - 607.86 - ], - "text": "t→∞e(α+jβ)t = lim", - "type": "text" - }, - { - "block_id": "p218-b11", - "global_id": 6016, - "bbox": [ - 288.76, - 590.11, - 337.89, - 607.86 - ], - "text": "t→∞eαt ejβt =", - "type": "text" - }, - { - "block_id": "p218-b12", - "global_id": 6017, - "bbox": [ - 339.73, - 581.22, - 393.0, - 608.57 - ], - "text": "0\nα < 0\n∞\nα > 0", - "type": "text" - }, - { - "block_id": "p218-b13", - "global_id": 6018, - "bbox": [ - 101.84, - 622.81, - 314.62, - 633.41 - ], - "text": "This conclusion is also valid for the terms of the form treλt.", - "type": "text" - } - ] - }, - { - "page_num": 219, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p219-b0", - "global_id": 6019, - "bbox": [ - 389.44, - 62.89, - 516.14, - 71.98 - ], - "text": "2.5\nSystem Stability\n199", - "type": "text" - }, - { - "block_id": "p219-b1", - "global_id": 6020, - "bbox": [ - 204.95, - 188.72, - 404.46, - 202.51 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p219-b2", - "global_id": 6021, - "bbox": [ - 204.95, - 285.46, - 404.14, - 293.73 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p219-b3", - "global_id": 6022, - "bbox": [ - 204.95, - 382.32, - 404.43, - 390.36 - ], - "text": "(e)\n(f)", - "type": "text" - }, - { - "block_id": "p219-b4", - "global_id": 6023, - "bbox": [ - 204.73, - 479.19, - 404.46, - 487.19 - ], - "text": "(g)\n(h)", - "type": "text" - }, - { - "block_id": "p219-b5", - "global_id": 6024, - "bbox": [ - 134.43, - 87.79, - 193.75, - 95.79 - ], - "text": "Characteristic root", - "type": "text" - }, - { - "block_id": "p219-b6", - "global_id": 6025, - "bbox": [ - 151.21, - 97.39, - 176.98, - 105.39 - ], - "text": "location", - "type": "text" - }, - { - "block_id": "p219-b7", - "global_id": 6026, - "bbox": [ - 322.04, - 87.79, - 381.36, - 95.79 - ], - "text": "Characteristic root", - "type": "text" - }, - { - "block_id": "p219-b8", - "global_id": 6027, - "bbox": [ - 219.29, - 97.29, - 469.8, - 105.39 - ], - "text": "location\nZero-input response\nZero-input response", - "type": "text" - }, - { - "block_id": "p219-b9", - "global_id": 6028, - "bbox": [ - 271.66, - 166.68, - 274.04, - 174.97 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p219-b10", - "global_id": 6029, - "bbox": [ - 270.36, - 351.58, - 272.74, - 359.86 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p219-b11", - "global_id": 6030, - "bbox": [ - 281.32, - 440.73, - 283.7, - 449.01 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p219-b12", - "global_id": 6031, - "bbox": [ - 272.88, - 264.69, - 466.38, - 273.11 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p219-b13", - "global_id": 6032, - "bbox": [ - 464.24, - 166.47, - 466.62, - 174.76 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p219-b14", - "global_id": 6033, - "bbox": [ - 477.49, - 344.98, - 479.88, - 353.26 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p219-b15", - "global_id": 6034, - "bbox": [ - 473.94, - 440.98, - 476.33, - 449.27 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p219-b16", - "global_id": 6035, - "bbox": [ - 187.05, - 166.69, - 416.63, - 175.01 - ], - "text": "0\n0\n0\n0", - "type": "text" - }, - { - "block_id": "p219-b17", - "global_id": 6036, - "bbox": [ - 161.64, - 264.94, - 416.31, - 273.18 - ], - "text": "0\n0\n0\n0", - "type": "text" - }, - { - "block_id": "p219-b18", - "global_id": 6037, - "bbox": [ - 160.61, - 345.08, - 410.92, - 353.1 - ], - "text": "0\n0\n0\n0", - "type": "text" - }, - { - "block_id": "p219-b19", - "global_id": 6038, - "bbox": [ - 158.58, - 441.38, - 410.82, - 449.57 - ], - "text": "0\n0\n0\n0", - "type": "text" - }, - { - "block_id": "p219-b20", - "global_id": 6039, - "bbox": [ - 127.59, - 499.44, - 449.34, - 508.68 - ], - "text": "Figure 2.17 Location of characteristic roots and the corresponding characteristic modes.", - "type": "text" - }, - { - "block_id": "p219-b21", - "global_id": 6040, - "bbox": [ - 127.59, - 535.01, - 451.82, - 546.96 - ], - "text": "2.5-3 Relationship Between BIBO and Asymptotic Stability", - "type": "text" - }, - { - "block_id": "p219-b22", - "global_id": 6041, - "bbox": [ - 127.59, - 553.1, - 516.14, - 598.92 - ], - "text": "External stability is determined by applying an external input with zero initial conditions, while\ninternal stability is determined by applying the nonzero initial conditions and no external input.\nThis is why these stabilities are also called the zero-state stability and the zero-input stability,\nrespectively.", - "type": "text" - }, - { - "block_id": "p219-b23", - "global_id": 6042, - "bbox": [ - 127.59, - 600.5, - 516.15, - 636.28 - ], - "text": "Recall that h(t), the impulse response of an LTIC system, is a linear combination of the system\ncharacteristic modes. For an LTIC system, specified by Eq. (2.1), we can readily show that when\na characteristic root λk is in the LHP, the corresponding mode eλkt is absolutely integrable. In", - "type": "text" - } - ] - }, - { - "page_num": 220, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p220-b0", - "global_id": 6043, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "200\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p220-b1", - "global_id": 6044, - "bbox": [ - 301.34, - 172.82, - 316.01, - 180.82 - ], - "text": "Real", - "type": "text" - }, - { - "block_id": "p220-b2", - "global_id": 6045, - "bbox": [ - 156.34, - 193.73, - 269.89, - 201.73 - ], - "text": "Unstable\nStable", - "type": "text" - }, - { - "block_id": "p220-b3", - "global_id": 6046, - "bbox": [ - 151.34, - 135.53, - 270.67, - 143.82 - ], - "text": "Re l 0\nRe l 0", - "type": "text" - }, - { - "block_id": "p220-b4", - "global_id": 6047, - "bbox": [ - 137.84, - 238.73, - 194.06, - 255.73 - ], - "text": "Marginally stable\nif simple roots", - "type": "text" - }, - { - "block_id": "p220-b5", - "global_id": 6048, - "bbox": [ - 212.04, - 171.63, - 216.04, - 179.63 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p220-b6", - "global_id": 6049, - "bbox": [ - 171.96, - 98.08, - 204.84, - 106.08 - ], - "text": "Imaginary", - "type": "text" - }, - { - "block_id": "p220-b7", - "global_id": 6050, - "bbox": [ - 223.84, - 238.73, - 268.51, - 255.73 - ], - "text": "Unstable if\nmultiple roots", - "type": "text" - }, - { - "block_id": "p220-b8", - "global_id": 6051, - "bbox": [ - 335.8, - 235.07, - 486.39, - 257.0 - ], - "text": "Figure 2.18 Characteristic roots loca-\ntion and system stability.", - "type": "text" - }, - { - "block_id": "p220-b9", - "global_id": 6052, - "bbox": [ - 101.85, - 283.38, - 490.42, - 380.65 - ], - "text": "contrast, if λk is in the RHP or on the imaginary axis, eλkt is not absolutely integrable.† This\nmeans that an asymptotically stable system is BIBO-stable. Moreover, a marginally stable or\nasymptotically unstable system is BIBO-unstable. The converse is not necessarily true; that is,\nBIBO stability does not necessarily inform us about the internal stability of the system. For\ninstance, if a system is uncontrollable and/or unobservable, some modes of the system are invisible\nand/or uncontrollable from the external terminals [3]. Hence, the stability picture portrayed by\nthe external description is of questionable value. BIBO (external) stability cannot assure internal\n(asymptotic) stability, as the following example shows.", - "type": "text" - }, - { - "block_id": "p220-b10", - "global_id": 6053, - "bbox": [ - 76.77, - 412.38, - 465.5, - 424.33 - ], - "text": "EXAMPLE 2.13\nA BIBO-Stable but Asymptotically Unstable System", - "type": "text" - }, - { - "block_id": "p220-b11", - "global_id": 6054, - "bbox": [ - 103.16, - 440.59, - 477.02, - 463.69 - ], - "text": "An LTID system consists of two subsystems S1 and S2 in cascade (Fig. 2.19). The impulse\nresponse of these systems are h1(t) and h2(t), respectively, given by", - "type": "text" - }, - { - "block_id": "p220-b12", - "global_id": 6055, - "bbox": [ - 190.61, - 472.73, - 389.57, - 485.61 - ], - "text": "h1(t) = δ(t) −2e−tu(t)\nand\nh2(t) = etu(t)", - "type": "text" - }, - { - "block_id": "p220-b13", - "global_id": 6056, - "bbox": [ - 103.17, - 496.79, - 395.14, - 506.76 - ], - "text": "Comment on the BIBO and asymptotic stability of the composite system.", - "type": "text" - }, - { - "block_id": "p220-b14", - "global_id": 6057, - "bbox": [ - 101.84, - 559.1, - 461.24, - 571.69 - ], - "text": "† Consider a mode of the form eλt, where λ = α + jβ. Hence, eλt = eαtejβt and |eλt| = eαt. Therefore,", - "type": "text" - }, - { - "block_id": "p220-b15", - "global_id": 6058, - "bbox": [ - 201.16, - 580.39, - 216.61, - 591.35 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p220-b16", - "global_id": 6059, - "bbox": [ - 205.89, - 602.62, - 217.55, - 609.1 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p220-b17", - "global_id": 6060, - "bbox": [ - 219.05, - 589.17, - 272.43, - 601.84 - ], - "text": "|eλτ u(τ)|dτ =", - "type": "text" - }, - { - "block_id": "p220-b18", - "global_id": 6061, - "bbox": [ - 274.27, - 580.39, - 289.72, - 591.35 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p220-b19", - "global_id": 6062, - "bbox": [ - 279.0, - 589.17, - 322.75, - 609.37 - ], - "text": "0\neατ dτ =", - "type": "text" - }, - { - "block_id": "p220-b20", - "global_id": 6063, - "bbox": [ - 324.59, - 580.0, - 389.88, - 607.36 - ], - "text": "−1/α\nα < 0\n∞\nα ≥0", - "type": "text" - }, - { - "block_id": "p220-b21", - "global_id": 6064, - "bbox": [ - 101.84, - 622.81, - 364.9, - 633.41 - ], - "text": "This conclusion is also valid when the integrand is of the form |tkeλtu(t)|.", - "type": "text" - } - ] - }, - { - "page_num": 221, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p221-b0", - "global_id": 6065, - "bbox": [ - 389.44, - 62.89, - 516.14, - 71.98 - ], - "text": "2.5\nSystem Stability\n201", - "type": "text" - }, - { - "block_id": "p221-b1", - "global_id": 6066, - "bbox": [ - 135.85, - 96.85, - 345.34, - 114.63 - ], - "text": "S1\nS2\nx(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p221-b2", - "global_id": 6067, - "bbox": [ - 367.86, - 105.49, - 498.77, - 127.4 - ], - "text": "Figure 2.19 Composite\nsystem\nfor Ex. 2.13.", - "type": "text" - }, - { - "block_id": "p221-b3", - "global_id": 6068, - "bbox": [ - 146.84, - 144.12, - 371.01, - 154.5 - ], - "text": "The composite system impulse response h(t) is given by", - "type": "text" - }, - { - "block_id": "p221-b4", - "global_id": 6069, - "bbox": [ - 194.82, - 164.32, - 435.87, - 177.19 - ], - "text": "h(t) = h1(t) ∗h2(t) = h2(t) ∗h1(t) = etu(t) ∗[δ(t) −2e−tu(t)]", - "type": "text" - }, - { - "block_id": "p221-b5", - "global_id": 6070, - "bbox": [ - 329.94, - 186.76, - 377.95, - 198.65 - ], - "text": "= etu(t) −2", - "type": "text" - }, - { - "block_id": "p221-b6", - "global_id": 6071, - "bbox": [ - 377.94, - 174.28, - 414.23, - 191.57 - ], - "text": "et −e−t", - "type": "text" - }, - { - "block_id": "p221-b7", - "global_id": 6072, - "bbox": [ - 397.23, - 195.76, - 402.21, - 205.73 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p221-b8", - "global_id": 6073, - "bbox": [ - 416.06, - 174.28, - 421.49, - 184.24 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p221-b9", - "global_id": 6074, - "bbox": [ - 421.49, - 188.28, - 436.85, - 198.55 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p221-b10", - "global_id": 6075, - "bbox": [ - 329.94, - 208.37, - 367.54, - 220.37 - ], - "text": "= e−tu(t)", - "type": "text" - }, - { - "block_id": "p221-b11", - "global_id": 6076, - "bbox": [ - 128.91, - 232.42, - 502.75, - 278.25 - ], - "text": "If the composite cascade system were to be enclosed in a black box with only the input and\nthe output terminals accessible, any measurement from these external terminals would show\nthat the impulse response of the system is e−tu(t), without any hint of the dangerously unstable\nsystem the system is harboring within.", - "type": "text" - }, - { - "block_id": "p221-b12", - "global_id": 6077, - "bbox": [ - 128.9, - 278.61, - 502.77, - 373.89 - ], - "text": "The composite system is BIBO-stable because its impulse response, e−tu(t), is absolutely\nintegrable. Observe, however, the subsystem S2 has a characteristic root 1, which lies in the\nRHP. Hence, S2 is asymptotically unstable. Eventually, S2 will burn out (or saturate) because of\nthe unbounded characteristic response generated by intended or unintended initial conditions,\nno matter how small. We shall show in Ex. 10.12 that this composite system is observable,\nbut not controllable. If the positions of S1 and S2 were interchanged (S2 followed by S1), the\nsystem is still BIBO-stable, but asymptotically unstable. In this case, the analysis in Ex. 10.12\nshows that the composite system is controllable, but not observable.", - "type": "text" - }, - { - "block_id": "p221-b13", - "global_id": 6078, - "bbox": [ - 128.91, - 375.88, - 502.77, - 397.81 - ], - "text": "This example shows that BIBO stability does not always imply asymptotic stability.\nHowever, asymptotic stability always implies BIBO stability.", - "type": "text" - }, - { - "block_id": "p221-b14", - "global_id": 6079, - "bbox": [ - 127.59, - 432.33, - 516.13, - 478.16 - ], - "text": "Fortunately, uncontrollable and/or unobservable systems are not commonly observed in\npractice. Henceforth, in determining system stability, we shall assume that unless otherwise\nmentioned, the internal and the external descriptions of a system are equivalent, implying that\nthe system is controllable and observable.", - "type": "text" - }, - { - "block_id": "p221-b15", - "global_id": 6080, - "bbox": [ - 102.51, - 508.23, - 452.39, - 520.19 - ], - "text": "EXAMPLE 2.14\nInvestigating Asymptotic and BIBO Stability", - "type": "text" - }, - { - "block_id": "p221-b16", - "global_id": 6081, - "bbox": [ - 128.9, - 536.85, - 502.73, - 558.76 - ], - "text": "Investigate the asymptotic and the BIBO stability of LTIC system described by the following\nequations, assuming that the equations are internal system descriptions:", - "type": "text" - }, - { - "block_id": "p221-b17", - "global_id": 6082, - "bbox": [ - 146.84, - 562.71, - 323.94, - 591.64 - ], - "text": "(a) (D + 1)(D2 + 4D + 8)y(t) = (D −3)x(t)\n(b) (D −1)(D2 + 4D + 8)y(t) = (D + 2)x(t)", - "type": "text" - }, - { - "block_id": "p221-b18", - "global_id": 6083, - "bbox": [ - 146.84, - 592.59, - 331.91, - 621.53 - ], - "text": "(c) (D + 2)(D2 + 4)y(t) = (D2 + D + 1)x(t)\n(d) (D + 1)(D2 + 4)2y(t) = (D2 + 2D + 8)x(t)", - "type": "text" - } - ] - }, - { - "page_num": 222, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p222-b0", - "global_id": 6084, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "202\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p222-b1", - "global_id": 6085, - "bbox": [ - 103.16, - 94.4, - 308.47, - 104.37 - ], - "text": "The characteristic polynomials of these systems are", - "type": "text" - }, - { - "block_id": "p222-b2", - "global_id": 6086, - "bbox": [ - 121.09, - 108.3, - 355.72, - 122.29 - ], - "text": "(a) (λ + 1)(λ2 + 4λ + 8) = (λ + 1)(λ + 2 −j2)(λ + 2 + j2)", - "type": "text" - }, - { - "block_id": "p222-b3", - "global_id": 6087, - "bbox": [ - 121.09, - 123.25, - 356.28, - 137.24 - ], - "text": "(b) (λ −1)(λ2 + 4λ + 8) = (λ −1)(λ + 2 −j2)(λ + 2 + j2)", - "type": "text" - }, - { - "block_id": "p222-b4", - "global_id": 6088, - "bbox": [ - 121.09, - 138.19, - 314.77, - 167.12 - ], - "text": "(c) (λ + 2)(λ2 + 4) = (λ + 2)(λ −j2)(λ + j2)\n(d) (λ + 1)(λ2 + 4)2 = (λ + 2)(λ −j2)2(λ + j2)2", - "type": "text" - }, - { - "block_id": "p222-b5", - "global_id": 6089, - "bbox": [ - 103.16, - 175.1, - 389.45, - 185.06 - ], - "text": "Consequently, the characteristic roots of the systems are (see Fig. 2.20):", - "type": "text" - }, - { - "block_id": "p222-b6", - "global_id": 6090, - "bbox": [ - 121.09, - 192.61, - 186.79, - 202.99 - ], - "text": "(a) −1, −2 ± j2", - "type": "text" - }, - { - "block_id": "p222-b7", - "global_id": 6091, - "bbox": [ - 121.09, - 207.56, - 179.58, - 217.94 - ], - "text": "(b) 1, −2 ± j2", - "type": "text" - }, - { - "block_id": "p222-b8", - "global_id": 6092, - "bbox": [ - 121.65, - 222.5, - 170.94, - 232.88 - ], - "text": "(c) −2, ±j2", - "type": "text" - }, - { - "block_id": "p222-b9", - "global_id": 6093, - "bbox": [ - 121.09, - 237.45, - 192.0, - 247.82 - ], - "text": "(d) −1, ±j2, ±j2", - "type": "text" - }, - { - "block_id": "p222-b10", - "global_id": 6094, - "bbox": [ - 103.16, - 255.79, - 477.03, - 325.53 - ], - "text": "System (a) is asymptotically stable (all roots in LHP), system (b) is unstable (one root\nin RHP), system (c) is marginally stable (unrepeated roots on imaginary axis) and no roots\nin RHP, and system (d) is unstable (repeated roots on the imaginary axis). BIBO stability\nis readily determined from the asymptotic stability. System (a) is BIBO-stable, system (b)\nis BIBO-unstable, system (c) is BIBO-unstable, and system (d) is BIBO-unstable. We have\nassumed that these systems are controllable and observable.", - "type": "text" - }, - { - "block_id": "p222-b11", - "global_id": 6095, - "bbox": [ - 138.42, - 423.2, - 440.71, - 431.2 - ], - "text": "(d)\n(a)", - "type": "text" - }, - { - "block_id": "p222-b12", - "global_id": 6096, - "bbox": [ - 152.77, - 382.75, - 156.77, - 390.75 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p222-b13", - "global_id": 6097, - "bbox": [ - 335.31, - 423.2, - 344.19, - 431.2 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p222-b14", - "global_id": 6098, - "bbox": [ - 349.99, - 382.75, - 450.49, - 390.75 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p222-b15", - "global_id": 6099, - "bbox": [ - 236.39, - 423.2, - 246.04, - 431.2 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p222-b16", - "global_id": 6100, - "bbox": [ - 249.25, - 382.75, - 253.25, - 390.75 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p222-b17", - "global_id": 6101, - "bbox": [ - 103.16, - 437.89, - 355.2, - 447.13 - ], - "text": "Figure 2.20 Characteristic root locations for the systems of Ex. 2.14.", - "type": "text" - }, - { - "block_id": "p222-b18", - "global_id": 6102, - "bbox": [ - 107.82, - 507.23, - 422.55, - 519.19 - ], - "text": "DRILL 2.15\nAssessing Stability by Characteristic Roots", - "type": "text" - }, - { - "block_id": "p222-b19", - "global_id": 6103, - "bbox": [ - 107.82, - 528.31, - 484.42, - 550.22 - ], - "text": "For each case, plot the characteristic roots and determine asymptotic and BIBO stabilities.\nAssume the equations reflect internal descriptions.", - "type": "text" - }, - { - "block_id": "p222-b20", - "global_id": 6104, - "bbox": [ - 125.76, - 557.78, - 256.54, - 583.1 - ], - "text": "(a) D(D + 2)y(t) = 3x(t)\n(b) D2(D + 3)y(t) = (D + 5)x(t)", - "type": "text" - }, - { - "block_id": "p222-b21", - "global_id": 6105, - "bbox": [ - 126.32, - 587.66, - 280.26, - 598.04 - ], - "text": "(c) (D + 1)(D + 2)y(t) = (2D + 3)x(t)", - "type": "text" - }, - { - "block_id": "p222-b22", - "global_id": 6106, - "bbox": [ - 125.76, - 598.99, - 310.83, - 612.99 - ], - "text": "(d) (D2 + 1)(D2 + 9)y(t) = (D2 + 2D + 4)x(t)", - "type": "text" - }, - { - "block_id": "p222-b23", - "global_id": 6107, - "bbox": [ - 126.32, - 613.93, - 302.3, - 627.93 - ], - "text": "(e) (D + 1)(D2 −4D + 9)y(t) = (D + 7)x(t)", - "type": "text" - } - ] - }, - { - "page_num": 223, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p223-b0", - "global_id": 6108, - "bbox": [ - 298.64, - 62.89, - 516.15, - 71.98 - ], - "text": "2.6\nIntuitive Insights into System Behavior\n203", - "type": "text" - }, - { - "block_id": "p223-b1", - "global_id": 6109, - "bbox": [ - 133.84, - 92.38, - 196.06, - 103.34 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p223-b2", - "global_id": 6110, - "bbox": [ - 151.5, - 110.66, - 318.24, - 135.64 - ], - "text": "(a) Marginally stable, but BIBO-unstable\n(b) Unstable in both senses", - "type": "text" - }, - { - "block_id": "p223-b3", - "global_id": 6111, - "bbox": [ - 151.5, - 140.55, - 318.8, - 165.53 - ], - "text": "(c) Stable in both senses\n(d) Marginally stable, but BIBO-unstable", - "type": "text" - }, - { - "block_id": "p223-b4", - "global_id": 6112, - "bbox": [ - 152.06, - 170.44, - 264.4, - 180.48 - ], - "text": "(e) Unstable in both senses.", - "type": "text" - }, - { - "block_id": "p223-b5", - "global_id": 6113, - "bbox": [ - 127.59, - 211.23, - 516.16, - 309.07 - ], - "text": "IMPLICATIONS OF STABILITY\nAll practical signal-processing systems must be asymptotically stable. Unstable systems are\nuseless from the viewpoint of signal processing because any set of intended or unintended initial\nconditions leads to an unbounded response that either destroys the system or (more likely) leads it\nto some saturation conditions that change the nature of the system. Even if the discernible initial\nconditions are zero, stray voltages or thermal noise signals generated within the system will act as\ninitial conditions. Because of exponential growth of a mode or modes in unstable systems, a stray\nsignal, no matter how small, will eventually cause an unbounded output.", - "type": "text" - }, - { - "block_id": "p223-b6", - "global_id": 6114, - "bbox": [ - 127.59, - 311.06, - 516.17, - 381.57 - ], - "text": "Marginally stable systems, though BIBO unstable, do have one important application in the\noscillator, which is a system that generates a signal on its own without the application of an external\ninput. Consequently, the oscillator output is a zero-input response. If such a response is to be a\nsinusoid of frequency ω0, the system should be marginally stable with characteristic roots at ±jω0.\nThus, to design an oscillator of frequency ω0, we should pick a system with the characteristic\npolynomial (λ −jω0)(λ + jω0) = λ2 + ω02. A system described by the differential equation", - "type": "text" - }, - { - "block_id": "p223-b8", - "global_id": 6115, - "bbox": [ - 282.83, - 388.89, - 314.89, - 404.15 - ], - "text": "D2 + ω0", - "type": "text" - }, - { - "block_id": "p223-b9", - "global_id": 6116, - "bbox": [ - 315.4, - 384.99, - 364.9, - 403.28 - ], - "text": "2\ny(t) = x(t)", - "type": "text" - }, - { - "block_id": "p223-b10", - "global_id": 6117, - "bbox": [ - 127.59, - 416.0, - 499.87, - 425.96 - ], - "text": "will do the job. However, practical oscillators are invariably realized using nonlinear systems.", - "type": "text" - }, - { - "block_id": "p223-b11", - "global_id": 6118, - "bbox": [ - 127.94, - 455.91, - 453.22, - 469.85 - ], - "text": "2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR", - "type": "text" - }, - { - "block_id": "p223-b12", - "global_id": 6119, - "bbox": [ - 127.59, - 475.84, - 516.12, - 521.67 - ], - "text": "This section attempts to provide an understanding of what determines system behavior. Because\nof its intuitive nature, the discussion is more or less qualitative. We shall now show that the most\nimportant attributes of a system are its characteristic roots or characteristic modes because they\ndetermine not only the zero-input response but also the entire behavior of the system.", - "type": "text" - }, - { - "block_id": "p223-b13", - "global_id": 6120, - "bbox": [ - 127.59, - 547.08, - 470.76, - 559.03 - ], - "text": "2.6-1 Dependence of System Behavior on Characteristic Modes", - "type": "text" - }, - { - "block_id": "p223-b14", - "global_id": 6121, - "bbox": [ - 127.59, - 565.16, - 516.15, - 634.81 - ], - "text": "Recall that the zero-input response of a system consists of the system’s characteristic modes. For a\nstable system, these characteristic modes decay exponentially and eventually vanish. This behavior\nmay give the impression that these modes do not substantially affect system behavior in general\nand system response in particular. This impression is totally wrong! We shall now see that the\nsystem’s characteristic modes leave their imprint on every aspect of the system behavior. We may\ncompare the system’s characteristic modes (or roots) to a seed that eventually dissolves in the", - "type": "text" - } - ] - }, - { - "page_num": 224, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p224-b0", - "global_id": 6122, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "204\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p224-b1", - "global_id": 6123, - "bbox": [ - 101.84, - 85.57, - 490.37, - 107.49 - ], - "text": "ground; however, the plant that springs from it is totally determined by the seed. The imprint of\nthe seed exists on every cell of the plant.", - "type": "text" - }, - { - "block_id": "p224-b2", - "global_id": 6124, - "bbox": [ - 101.84, - 109.58, - 490.43, - 215.19 - ], - "text": "To understand this interesting phenomenon, recall that the characteristic modes of a system\nare very special to that system because it can sustain these signals without the application of an\nexternal input. In other words, the system offers a free ride and ready access to these signals. Now\nimagine what would happen if we actually drove the system with an input having the form of a\ncharacteristic mode! We would expect the system to respond strongly (this is, in fact, the resonance\nphenomenon discussed later in this section). If the input is not exactly a characteristic mode but is\nclose to such a mode, we would still expect the system response to be strong. However, if the input\nis very different from any of the characteristic modes, we would expect the system to respond\npoorly. We shall now show that these intuitive deductions are indeed true.", - "type": "text" - }, - { - "block_id": "p224-b3", - "global_id": 6125, - "bbox": [ - 211.16, - 422.25, - 381.05, - 432.22 - ], - "text": "Intuition can cut the math jungle instantly!", - "type": "text" - }, - { - "block_id": "p224-b4", - "global_id": 6126, - "bbox": [ - 101.84, - 455.14, - 490.41, - 512.92 - ], - "text": "We have devised a measure of similarity of signals later (see in Ch. 6). Here we shall take\na simpler approach. Let us restrict the system’s inputs to exponentials of the form eζt, where ζ\nis generally a complex number. The similarity of two exponential signals eζt and eλt will then be\nmeasured by the closeness of ζ and λ. If the difference ζ −λ is small, the signals are similar; if\nζ −λ is large, the signals are dissimilar.", - "type": "text" - }, - { - "block_id": "p224-b5", - "global_id": 6127, - "bbox": [ - 101.84, - 511.3, - 490.38, - 548.79 - ], - "text": "Now consider a first-order system with a single characteristic mode eλt and the input eζt. The\nimpulse response of this system is then given by Aeλt, where the exact value of A is not important\nfor this qualitative discussion. The system response y(t) is given by", - "type": "text" - }, - { - "block_id": "p224-b6", - "global_id": 6128, - "bbox": [ - 224.21, - 561.61, - 368.03, - 573.6 - ], - "text": "y(t) = h(t) ∗x(t) = Aeλtu(t) ∗eζtu(t)", - "type": "text" - }, - { - "block_id": "p224-b7", - "global_id": 6129, - "bbox": [ - 101.85, - 588.65, - 299.41, - 598.62 - ], - "text": "From the convolution table (Table 2.1), we obtain", - "type": "text" - }, - { - "block_id": "p224-b8", - "global_id": 6130, - "bbox": [ - 243.36, - 610.94, - 490.38, - 634.64 - ], - "text": "y(t) =\nA\nζ −λ[eζt −eλt]u(t)\n(2.46)", - "type": "text" - } - ] - }, - { - "page_num": 225, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p225-b0", - "global_id": 6131, - "bbox": [ - 298.64, - 62.89, - 516.15, - 71.98 - ], - "text": "2.6\nIntuitive Insights into System Behavior\n205", - "type": "text" - }, - { - "block_id": "p225-b1", - "global_id": 6132, - "bbox": [ - 127.59, - 83.64, - 516.16, - 133.08 - ], - "text": "Clearly, if the input eζt is similar to eλt, ζ −λ is small and the system response is large. The closer\nthe input x(t) to the characteristic mode, the stronger the system response. In contrast, if the input\nis very different from the natural mode, ζ −λ is large and the system responds poorly. This is\nprecisely what we set out to prove.", - "type": "text" - }, - { - "block_id": "p225-b2", - "global_id": 6133, - "bbox": [ - 127.59, - 135.07, - 516.15, - 204.81 - ], - "text": "We have proved the foregoing assertion for a single-mode (first-order) system. It can be\ngeneralized to an Nth-order system, which has N characteristic modes. The impulse response h(t)\nof such a system is a linear combination of its N modes. Therefore, if x(t) is similar to any one\nof the modes, the corresponding response will be high; if it is similar to none of the modes, the\nresponse will be small. Clearly, the characteristic modes are very influential in determining system\nresponse to a given input.", - "type": "text" - }, - { - "block_id": "p225-b3", - "global_id": 6134, - "bbox": [ - 127.59, - 206.8, - 516.15, - 252.63 - ], - "text": "It would be tempting to conclude on the basis of Eq. (2.46) that if the input is identical to the\ncharacteristic mode, so that ζ = λ, then the response goes to infinity. Remember, however, that if\nζ = λ, the numerator on the right-hand side of Eq. (2.46) also goes to zero. We shall study this\ninteresting behavior (resonance phenomenon) later in this section.", - "type": "text" - }, - { - "block_id": "p225-b4", - "global_id": 6135, - "bbox": [ - 127.59, - 254.22, - 516.13, - 276.55 - ], - "text": "We now show that mere inspection of the impulse response h(t) (which is composed of\ncharacteristic modes) reveals a great deal about the system behavior.", - "type": "text" - }, - { - "block_id": "p225-b5", - "global_id": 6136, - "bbox": [ - 127.59, - 295.01, - 460.52, - 306.96 - ], - "text": "2.6-2 Response Time of a System: The System Time Constant", - "type": "text" - }, - { - "block_id": "p225-b6", - "global_id": 6137, - "bbox": [ - 127.59, - 313.09, - 516.15, - 358.92 - ], - "text": "Like human beings, systems have a certain response time. In other words, when an input (stimulus)\nis applied to a system, a certain amount of time elapses before the system fully responds to that\ninput. This time lag or response time is called the system time constant. As we shall see, a system’s\ntime constant is equal to the width of its impulse response h(t).", - "type": "text" - }, - { - "block_id": "p225-b7", - "global_id": 6138, - "bbox": [ - 127.59, - 360.51, - 516.14, - 466.42 - ], - "text": "An input δ(t) to a system is instantaneous (zero duration), but its response h(t) has a duration\nTh. Therefore, the system requires a time Th to respond fully to this input, and we are justified\nin viewing Th as the system’s response time or time constant. We arrive at the same conclusion\nvia another argument. The output is a convolution of the input with h(t). If an input is a pulse of\nwidth Tx, then the output pulse width is Tx + Th according to the width property of convolution.\nThis conclusion shows that the system requires Th seconds to respond fully to any input. The\nsystem time constant indicates how fast the system is. A system with a smaller time constant is a\nfaster system that responds quickly to an input. A system with a relatively large time constant is a\nsluggish system that cannot respond well to rapidly varying signals.", - "type": "text" - }, - { - "block_id": "p225-b8", - "global_id": 6139, - "bbox": [ - 127.59, - 468.1, - 516.14, - 514.34 - ], - "text": "Strictly speaking, the duration of the impulse response h(t) is ∞because the characteristic\nmodes approach zero asymptotically as t →∞. However, beyond some value of t, h(t) becomes\nnegligible. It is therefore necessary to use some suitable measure of the impulse response’s\neffective width.", - "type": "text" - }, - { - "block_id": "p225-b9", - "global_id": 6140, - "bbox": [ - 127.59, - 516.33, - 516.14, - 575.61 - ], - "text": "There is no single satisfactory definition of effective signal duration (or width) applicable\nto every situation. For the situation depicted in Fig. 2.21, a reasonable definition of the duration\nh(t) would be Th, the width of the rectangular pulse ˆh(t). This rectangular pulse ˆh(t) has an area\nidentical to that of h(t) and a height identical to that of h(t) at some suitable instant t = t0. In\nFig. 2.21, t0 is chosen as the instant at which h(t) is maximum. According to this definition,†", - "type": "text" - }, - { - "block_id": "p225-b10", - "global_id": 6141, - "bbox": [ - 279.67, - 590.36, - 318.18, - 601.5 - ], - "text": "Thh(t0) =", - "type": "text" - }, - { - "block_id": "p225-b11", - "global_id": 6142, - "bbox": [ - 320.23, - 576.8, - 337.17, - 588.86 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p225-b12", - "global_id": 6143, - "bbox": [ - 325.49, - 601.67, - 338.04, - 608.65 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p225-b13", - "global_id": 6144, - "bbox": [ - 339.65, - 590.36, - 363.87, - 600.63 - ], - "text": "h(t)dt", - "type": "text" - }, - { - "block_id": "p225-b14", - "global_id": 6145, - "bbox": [ - 127.59, - 626.18, - 516.14, - 649.35 - ], - "text": "† This definition is satisfactory when h(t) is a single, mostly positive (or mostly negative) pulse. Such systems\nare lowpass systems. This definition should not be applied indiscriminately to all systems.", - "type": "text" - } - ] - }, - { - "page_num": 226, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p226-b0", - "global_id": 6146, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "206\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p226-b1", - "global_id": 6147, - "bbox": [ - 245.92, - 160.79, - 248.15, - 168.79 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p226-b2", - "global_id": 6148, - "bbox": [ - 192.99, - 99.06, - 204.55, - 107.14 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p226-b3", - "global_id": 6149, - "bbox": [ - 218.66, - 130.09, - 230.21, - 138.17 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p226-b4", - "global_id": 6150, - "bbox": [ - 123.16, - 109.61, - 137.72, - 119.22 - ], - "text": "h(t0)", - "type": "text" - }, - { - "block_id": "p226-b5", - "global_id": 6151, - "bbox": [ - 133.8, - 160.02, - 477.18, - 175.02 - ], - "text": "t0\nTh\n0\nFigure 2.21 Effective duration of an impulse response.", - "type": "text" - }, - { - "block_id": "p226-b6", - "global_id": 6152, - "bbox": [ - 101.84, - 198.6, - 110.14, - 208.57 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p226-b7", - "global_id": 6153, - "bbox": [ - 262.66, - 219.85, - 282.01, - 231.31 - ], - "text": "Th =", - "type": "text" - }, - { - "block_id": "p226-b8", - "global_id": 6154, - "bbox": [ - 285.25, - 203.76, - 298.86, - 215.82 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p226-b9", - "global_id": 6155, - "bbox": [ - 289.81, - 211.79, - 328.19, - 224.54 - ], - "text": "−∞h(t)dt", - "type": "text" - }, - { - "block_id": "p226-b10", - "global_id": 6156, - "bbox": [ - 297.23, - 220.27, - 490.38, - 238.08 - ], - "text": "h(t0)\n(2.47)", - "type": "text" - }, - { - "block_id": "p226-b11", - "global_id": 6157, - "bbox": [ - 101.84, - 246.43, - 240.21, - 256.39 - ], - "text": "Now if a system has a single mode", - "type": "text" - }, - { - "block_id": "p226-b12", - "global_id": 6158, - "bbox": [ - 266.38, - 260.15, - 325.85, - 272.15 - ], - "text": "h(t) = Aeλtu(t)", - "type": "text" - }, - { - "block_id": "p226-b13", - "global_id": 6159, - "bbox": [ - 101.85, - 283.41, - 490.37, - 305.74 - ], - "text": "with λ negative and real, then h(t) is maximum at t = 0 with value h(0) = A. Therefore, according\nto Eq. (2.47),", - "type": "text" - }, - { - "block_id": "p226-b14", - "global_id": 6160, - "bbox": [ - 244.66, - 309.98, - 272.78, - 328.01 - ], - "text": "Th = 1", - "type": "text" - }, - { - "block_id": "p226-b15", - "global_id": 6161, - "bbox": [ - 267.24, - 323.94, - 273.32, - 333.9 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p226-b16", - "global_id": 6162, - "bbox": [ - 275.62, - 302.99, - 292.57, - 315.04 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p226-b17", - "global_id": 6163, - "bbox": [ - 280.89, - 309.98, - 346.14, - 335.13 - ], - "text": "0\nAeλt dt = −1", - "type": "text" - }, - { - "block_id": "p226-b18", - "global_id": 6164, - "bbox": [ - 340.93, - 323.62, - 346.38, - 333.59 - ], - "text": "λ", - "type": "text" - }, - { - "block_id": "p226-b19", - "global_id": 6165, - "bbox": [ - 101.84, - 343.31, - 490.41, - 389.14 - ], - "text": "Thus, the time constant in this case is simply the (negative of the) reciprocal of the system’s\ncharacteristic root. For the multimode case, h(t) is a weighted sum of the system’s characteristic\nmodes, and Th is a weighted average of the time constants associated with the N modes of the\nsystem.", - "type": "text" - }, - { - "block_id": "p226-b20", - "global_id": 6166, - "bbox": [ - 101.84, - 416.51, - 359.7, - 428.46 - ], - "text": "2.6-3 Time Constant and Rise Time of a System", - "type": "text" - }, - { - "block_id": "p226-b21", - "global_id": 6167, - "bbox": [ - 101.84, - 434.58, - 490.41, - 540.19 - ], - "text": "Rise time of a system, defined as the time required for the unit step response to rise from 10% to\n90% of its steady-state value, is an indication of the speed of response.† The system time constant\nmay also be viewed from a perspective of rise time. The unit step response y(t) of a system is the\nconvolution of u(t) with h(t). Let the impulse response h(t) be a rectangular pulse of width Th,\nas shown in Fig. 2.22. This assumption simplifies the discussion, yet gives satisfactory results for\nqualitative discussion. The result of this convolution is illustrated in Fig. 2.22. Note that the output\ndoes not rise from zero to a final value instantaneously as the input rises; instead, the output takes\nTh seconds to accomplish this. Hence, the rise time Tr of the system is equal to the system time\nconstant", - "type": "text" - }, - { - "block_id": "p226-b22", - "global_id": 6168, - "bbox": [ - 280.96, - 545.68, - 310.76, - 557.14 - ], - "text": "Tr = Th", - "type": "text" - }, - { - "block_id": "p226-b23", - "global_id": 6169, - "bbox": [ - 101.84, - 567.62, - 490.37, - 591.04 - ], - "text": "This result and Fig. 2.22 show clearly that a system generally does not respond to an input\ninstantaneously. Instead, it takes time Th for the system to respond fully.", - "type": "text" - }, - { - "block_id": "p226-b24", - "global_id": 6170, - "bbox": [ - 101.84, - 610.24, - 490.37, - 633.41 - ], - "text": "† Because of varying definitions of rise time, the reader may find different results in the literature. The\nqualitative and intuitive nature of this discussion should always be kept in mind.", - "type": "text" - } - ] - }, - { - "page_num": 227, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p227-b0", - "global_id": 6171, - "bbox": [ - 298.64, - 62.89, - 516.15, - 71.98 - ], - "text": "2.6\nIntuitive Insights into System Behavior\n207", - "type": "text" - }, - { - "block_id": "p227-b1", - "global_id": 6172, - "bbox": [ - 254.71, - 114.23, - 259.71, - 124.23 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p227-b2", - "global_id": 6173, - "bbox": [ - 172.0, - 88.07, - 296.94, - 96.15 - ], - "text": "x(t)\nh(t)", - "type": "text" - }, - { - "block_id": "p227-b3", - "global_id": 6174, - "bbox": [ - 161.88, - 105.95, - 376.39, - 122.75 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p227-b4", - "global_id": 6175, - "bbox": [ - 401.91, - 88.07, - 413.01, - 96.15 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p227-b5", - "global_id": 6176, - "bbox": [ - 163.09, - 141.89, - 463.72, - 152.22 - ], - "text": "0\nTh\nt\n0\nt\n0\nTh\nt", - "type": "text" - }, - { - "block_id": "p227-b6", - "global_id": 6177, - "bbox": [ - 151.5, - 159.44, - 281.02, - 168.67 - ], - "text": "Figure 2.22 Rise time of a system.", - "type": "text" - }, - { - "block_id": "p227-b7", - "global_id": 6178, - "bbox": [ - 127.59, - 195.31, - 311.71, - 207.27 - ], - "text": "2.6-4 Time Constant and Filtering", - "type": "text" - }, - { - "block_id": "p227-b8", - "global_id": 6179, - "bbox": [ - 127.59, - 213.39, - 516.17, - 271.18 - ], - "text": "A larger time constant implies a sluggish system because the system takes longer to respond fully\nto an input. Such a system cannot respond effectively to rapid variations in the input. In contrast,\na smaller time constant indicates that a system is capable of responding to rapid variations in\nthe input. Thus, there is a direct connection between a system’s time constant and its filtering\nproperties.", - "type": "text" - }, - { - "block_id": "p227-b9", - "global_id": 6180, - "bbox": [ - 127.59, - 273.17, - 516.17, - 319.0 - ], - "text": "A high-frequency sinusoid varies rapidly with time. A system with a large time constant will\nnot be able to respond well to this input. Therefore, such a system will suppress rapidly varying\n(high-frequency) sinusoids and other high-frequency signals, thereby acting as a lowpass filter (a\nfilter allowing the transmission of low-frequency signals only). We shall now show that a system", - "type": "text" - }, - { - "block_id": "p227-b10", - "global_id": 6181, - "bbox": [ - 216.85, - 430.57, - 242.62, - 438.87 - ], - "text": "h(t t)", - "type": "text" - }, - { - "block_id": "p227-b11", - "global_id": 6182, - "bbox": [ - 274.92, - 507.65, - 284.56, - 515.65 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p227-b12", - "global_id": 6183, - "bbox": [ - 254.85, - 444.96, - 267.76, - 453.06 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p227-b13", - "global_id": 6184, - "bbox": [ - 167.35, - 476.78, - 171.35, - 484.78 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p227-b14", - "global_id": 6185, - "bbox": [ - 247.96, - 493.66, - 251.51, - 501.66 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p227-b15", - "global_id": 6186, - "bbox": [ - 317.19, - 435.15, - 328.29, - 443.23 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p227-b16", - "global_id": 6187, - "bbox": [ - 389.31, - 493.62, - 391.54, - 501.62 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p227-b17", - "global_id": 6188, - "bbox": [ - 275.14, - 409.61, - 284.02, - 417.61 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p227-b18", - "global_id": 6189, - "bbox": [ - 260.96, - 340.89, - 272.52, - 348.97 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p227-b19", - "global_id": 6190, - "bbox": [ - 245.74, - 389.93, - 313.02, - 400.41 - ], - "text": "t\nTh\n0", - "type": "text" - }, - { - "block_id": "p227-b20", - "global_id": 6191, - "bbox": [ - 275.14, - 605.26, - 284.02, - 613.26 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p227-b21", - "global_id": 6192, - "bbox": [ - 216.85, - 528.18, - 242.62, - 536.47 - ], - "text": "h(t t)", - "type": "text" - }, - { - "block_id": "p227-b22", - "global_id": 6193, - "bbox": [ - 254.85, - 544.04, - 267.76, - 552.14 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p227-b23", - "global_id": 6194, - "bbox": [ - 167.3, - 574.59, - 251.51, - 583.24 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p227-b24", - "global_id": 6195, - "bbox": [ - 301.62, - 476.78, - 305.62, - 484.78 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p227-b25", - "global_id": 6196, - "bbox": [ - 301.57, - 574.59, - 391.54, - 583.73 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p227-b26", - "global_id": 6197, - "bbox": [ - 317.08, - 535.2, - 328.66, - 543.28 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p227-b27", - "global_id": 6198, - "bbox": [ - 151.5, - 619.96, - 300.69, - 629.19 - ], - "text": "Figure 2.23 Time constant and filtering.", - "type": "text" - } - ] - }, - { - "page_num": 228, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p228-b0", - "global_id": 6199, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "208\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p228-b1", - "global_id": 6200, - "bbox": [ - 101.84, - 85.4, - 490.41, - 121.19 - ], - "text": "with a time constant Th acts as a lowpass filter having a cutoff frequency of fc = 1/Th hertz, so\nthat sinusoids with frequencies below fc Hz are transmitted reasonably well, while those with\nfrequencies above fc Hz are suppressed.", - "type": "text" - }, - { - "block_id": "p228-b2", - "global_id": 6201, - "bbox": [ - 101.84, - 121.27, - 490.4, - 263.15 - ], - "text": "To demonstrate this fact, let us determine the system response to a sinusoidal input x(t) by\nconvolving this input with the effective impulse response h(t) in Fig. 2.23a. From Figs. 2.23b\nand 2.23c we see the process of convolution of h(t) with the sinusoidal inputs of two different\nfrequencies. The sinusoid in Fig. 2.23b has a relatively high frequency, while the frequency of\nthe sinusoid in Fig. 2.23c is low. Recall that the convolution of x(t) and h(t) is equal to the area\nunder the product x(τ)h(t −τ). This area is shown shaded in Figs. 2.23b and 2.23c for the two\ncases. For the high-frequency sinusoid, it is clear from Fig. 2.23b that the area under x(τ)h(t −τ)\nis very small because its positive and negative areas nearly cancel each other out. In this case the\noutput y(t) remains periodic but has a rather small amplitude. This happens when the period of\nthe sinusoid is much smaller than the system time constant Th. In contrast, for the low-frequency\nsinusoid, the period of the sinusoid is larger than Th, rendering the partial cancellation of area under\nx(τ)h(t −τ) less effective. Consequently, the output y(t) is much larger, as depicted in Fig. 2.23c.", - "type": "text" - }, - { - "block_id": "p228-b3", - "global_id": 6202, - "bbox": [ - 101.85, - 265.15, - 490.39, - 311.68 - ], - "text": "Between these two possible extremes in system behavior, a transition point occurs when\nthe period of the sinusoid is equal to the system time constant Th. The frequency at which this\ntransition occurs is known as the cutoff frequency fc of the system. Because Th is the period of\ncutoff frequency fc,", - "type": "text" - }, - { - "block_id": "p228-b4", - "global_id": 6203, - "bbox": [ - 281.04, - 321.36, - 307.72, - 339.39 - ], - "text": "fc = 1", - "type": "text" - }, - { - "block_id": "p228-b5", - "global_id": 6204, - "bbox": [ - 300.47, - 335.32, - 309.5, - 346.09 - ], - "text": "Th", - "type": "text" - }, - { - "block_id": "p228-b6", - "global_id": 6205, - "bbox": [ - 101.84, - 358.6, - 490.41, - 440.39 - ], - "text": "The frequency fc is also known as the bandwidth of the system because the system transmits\nor passes sinusoidal components with frequencies below fc while attenuating components with\nfrequencies above fc. Of course, the transition in system behavior is gradual. There is no dramatic\nchange in system behavior at fc = 1/Th. Moreover, these results are based on an idealized\n(rectangular pulse) impulse response; in practice these results will vary somewhat, depending on\nthe exact shape of h(t). Remember that the “feel” of general system behavior is more important\nthan exact system response for this qualitative discussion.", - "type": "text" - }, - { - "block_id": "p228-b7", - "global_id": 6206, - "bbox": [ - 119.78, - 442.38, - 372.61, - 452.34 - ], - "text": "Since the system time constant is equal to its rise time, we have", - "type": "text" - }, - { - "block_id": "p228-b8", - "global_id": 6207, - "bbox": [ - 242.51, - 469.22, - 270.16, - 487.25 - ], - "text": "Tr = 1", - "type": "text" - }, - { - "block_id": "p228-b9", - "global_id": 6208, - "bbox": [ - 264.48, - 483.18, - 270.35, - 493.95 - ], - "text": "fc", - "type": "text" - }, - { - "block_id": "p228-b10", - "global_id": 6209, - "bbox": [ - 291.97, - 469.22, - 346.57, - 487.25 - ], - "text": "or\nfc = 1", - "type": "text" - }, - { - "block_id": "p228-b11", - "global_id": 6210, - "bbox": [ - 339.61, - 483.18, - 347.86, - 493.95 - ], - "text": "Tr", - "type": "text" - }, - { - "block_id": "p228-b12", - "global_id": 6211, - "bbox": [ - 466.32, - 476.21, - 490.38, - 486.17 - ], - "text": "(2.48)", - "type": "text" - }, - { - "block_id": "p228-b13", - "global_id": 6212, - "bbox": [ - 101.85, - 510.0, - 490.39, - 543.88 - ], - "text": "Thus, a system’s bandwidth is inversely proportional to its rise time. Although Eq. (2.48) was\nderived for an idealized (rectangular) impulse response, its implications are valid for lowpass\nLTIC systems, in general. For a general case, we can show that [1]", - "type": "text" - }, - { - "block_id": "p228-b14", - "global_id": 6213, - "bbox": [ - 281.35, - 560.51, - 307.36, - 578.64 - ], - "text": "fc = k", - "type": "text" - }, - { - "block_id": "p228-b15", - "global_id": 6214, - "bbox": [ - 300.78, - 574.57, - 309.03, - 585.33 - ], - "text": "Tr", - "type": "text" - }, - { - "block_id": "p228-b16", - "global_id": 6215, - "bbox": [ - 101.84, - 600.5, - 490.37, - 634.79 - ], - "text": "where the exact value of k depends on the nature of h(t). An experienced engineer often can\nestimate quickly the bandwidth of an unknown system by simply observing the system response\nto a step input on an oscilloscope.", - "type": "text" - } - ] - }, - { - "page_num": 229, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p229-b0", - "global_id": 6216, - "bbox": [ - 298.64, - 62.89, - 516.15, - 71.98 - ], - "text": "2.6\nIntuitive Insights into System Behavior\n209", - "type": "text" - }, - { - "block_id": "p229-b1", - "global_id": 6217, - "bbox": [ - 127.59, - 86.52, - 425.27, - 98.48 - ], - "text": "2.6-5 Time Constant and Pulse Dispersion (Spreading)", - "type": "text" - }, - { - "block_id": "p229-b2", - "global_id": 6218, - "bbox": [ - 127.59, - 104.61, - 516.15, - 162.39 - ], - "text": "In general, the transmission of a pulse through a system causes pulse dispersion (or spreading).\nTherefore, the output pulse is generally wider than the input pulse. This system behavior can have\nserious consequences in communication systems in which information is transmitted by pulse\namplitudes. Dispersion (or spreading) causes interference or overlap with neighboring pulses,\nthereby distorting pulse amplitudes and introducing errors in the received information.", - "type": "text" - }, - { - "block_id": "p229-b3", - "global_id": 6219, - "bbox": [ - 127.59, - 163.97, - 516.11, - 186.3 - ], - "text": "Earlier we saw that if an input x(t) is a pulse of width Tx, then Ty, the width of the output\ny(t), is", - "type": "text" - }, - { - "block_id": "p229-b4", - "global_id": 6220, - "bbox": [ - 296.59, - 192.39, - 346.63, - 203.85 - ], - "text": "Ty = Tx + Th", - "type": "text" - }, - { - "block_id": "p229-b5", - "global_id": 6221, - "bbox": [ - 127.59, - 214.75, - 516.14, - 248.63 - ], - "text": "This result shows that an input pulse spreads out (disperses) as it passes through a system. Since\nTh is also the system’s time constant or rise time, the amount of spread in the pulse is equal to the\ntime constant (or rise time) of the system.", - "type": "text" - }, - { - "block_id": "p229-b6", - "global_id": 6222, - "bbox": [ - 127.59, - 276.39, - 448.85, - 288.35 - ], - "text": "2.6-6 Time Constant and Rate of Information Transmission", - "type": "text" - }, - { - "block_id": "p229-b7", - "global_id": 6223, - "bbox": [ - 127.59, - 294.47, - 516.16, - 352.26 - ], - "text": "In pulse communications systems, which convey information through pulse amplitudes, the\nrate of information transmission is proportional to the rate of pulse transmission. We shall\ndemonstrate that to avoid the destruction of information caused by dispersion of pulses during their\ntransmission through the channel (transmission medium), the rate of information transmission\nshould not exceed the bandwidth of the communications channel.", - "type": "text" - }, - { - "block_id": "p229-b8", - "global_id": 6224, - "bbox": [ - 127.59, - 354.15, - 516.14, - 423.99 - ], - "text": "Since an input pulse spreads out by Th seconds, the consecutive pulses should be spaced Th\nseconds apart to avoid interference between pulses. Thus, the rate of pulse transmission should\nnot exceed 1/Th pulses/second. But 1/Th = fc, the channel’s bandwidth, so that we can transmit\npulses through a communications channel at a rate of fc pulses per second and still avoid significant\ninterference between the pulses. The rate of information transmission is therefore proportional to\nthe channel’s bandwidth (or to the reciprocal of its time constant).†", - "type": "text" - }, - { - "block_id": "p229-b9", - "global_id": 6225, - "bbox": [ - 127.59, - 425.98, - 516.14, - 483.77 - ], - "text": "The discussion of Secs. 2.6-2, 2.6-3, 2.6-4, 2.6-5, and 2.6-6) shows that the system time\nconstant determines much of a system’s behavior—its filtering characteristics, rise time, pulse\ndispersion, and so on. In turn, the time constant is determined by the system’s characteristic roots.\nClearly the characteristic roots and their relative amounts in the impulse response h(t) determine\nthe behavior of a system.", - "type": "text" - }, - { - "block_id": "p229-b10", - "global_id": 6226, - "bbox": [ - 102.51, - 514.26, - 472.67, - 526.22 - ], - "text": "EXAMPLE 2.15\nIntuitive Insights into Lowpass System Behavior", - "type": "text" - }, - { - "block_id": "p229-b11", - "global_id": 6227, - "bbox": [ - 128.9, - 542.78, - 502.76, - 564.8 - ], - "text": "Find the time constant Th, rise time Tr, and cutoff frequency fc for a lowpass system that\nhas impulse response h(t) = te−tu(t). Determine the maximum rate that pulses of 1 second", - "type": "text" - }, - { - "block_id": "p229-b12", - "global_id": 6228, - "bbox": [ - 127.59, - 599.27, - 516.13, - 633.41 - ], - "text": "† Theoretically, a channel of bandwidth fc can transmit correctly up to 2fc pulse amplitudes per second [4].\nOur derivation here, being very simple and qualitative, yields only half the theoretical limit. In practice it is\nnot easy to attain the upper theoretical limit.", - "type": "text" - } - ] - }, - { - "page_num": 230, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p230-b0", - "global_id": 6229, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "210\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p230-b1", - "global_id": 6230, - "bbox": [ - 103.16, - 86.24, - 477.01, - 108.16 - ], - "text": "duration can be transmitted through the system so that interference is essentially avoided\nbetween adjacent pulses at the system output.", - "type": "text" - }, - { - "block_id": "p230-b2", - "global_id": 6231, - "bbox": [ - 103.16, - 129.43, - 477.02, - 164.95 - ], - "text": "The system impulse response h(t) = te−tu(t), which looks similar to the impulse response of\nFig. 2.21, has a peak value of e−1 = 0.3679 at a time t0 = 1. According to Eq. (2.47) and using\nintegration by parts, the system time constant is therefore", - "type": "text" - }, - { - "block_id": "p230-b3", - "global_id": 6232, - "bbox": [ - 128.26, - 183.83, - 147.6, - 195.29 - ], - "text": "Th =", - "type": "text" - }, - { - "block_id": "p230-b4", - "global_id": 6233, - "bbox": [ - 150.85, - 168.25, - 164.46, - 180.31 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p230-b5", - "global_id": 6234, - "bbox": [ - 155.41, - 172.97, - 211.27, - 201.18 - ], - "text": "0 te−t dt\ne−1\n= e1", - "type": "text" - }, - { - "block_id": "p230-b7", - "global_id": 6235, - "bbox": [ - 220.79, - 175.36, - 254.12, - 194.11 - ], - "text": "−te−t∞", - "type": "text" - }, - { - "block_id": "p230-b8", - "global_id": 6236, - "bbox": [ - 247.0, - 170.27, - 282.43, - 197.63 - ], - "text": "0 +\n# ∞", - "type": "text" - }, - { - "block_id": "p230-b9", - "global_id": 6237, - "bbox": [ - 270.74, - 180.03, - 305.31, - 202.41 - ], - "text": "0\ne−t dt", - "type": "text" - }, - { - "block_id": "p230-b11", - "global_id": 6238, - "bbox": [ - 314.27, - 175.82, - 337.62, - 194.11 - ], - "text": "= e1", - "type": "text" - }, - { - "block_id": "p230-b12", - "global_id": 6239, - "bbox": [ - 337.62, - 175.36, - 379.97, - 197.32 - ], - "text": "0 −e−t∞\n0", - "type": "text" - }, - { - "block_id": "p230-b13", - "global_id": 6240, - "bbox": [ - 382.02, - 182.4, - 451.91, - 194.21 - ], - "text": "= e1(1) = 2.7183", - "type": "text" - }, - { - "block_id": "p230-b14", - "global_id": 6241, - "bbox": [ - 103.17, - 211.45, - 125.57, - 221.41 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p230-b15", - "global_id": 6242, - "bbox": [ - 142.11, - 230.64, - 381.23, - 248.67 - ], - "text": "Th = 2.7183 s,\nTr = Th = 2.7183 s,\nand\nfc = 1", - "type": "text" - }, - { - "block_id": "p230-b16", - "global_id": 6243, - "bbox": [ - 373.98, - 244.59, - 383.01, - 255.35 - ], - "text": "Th", - "type": "text" - }, - { - "block_id": "p230-b17", - "global_id": 6244, - "bbox": [ - 386.75, - 237.21, - 438.07, - 247.59 - ], - "text": "= 0.3679 Hz", - "type": "text" - }, - { - "block_id": "p230-b18", - "global_id": 6245, - "bbox": [ - 103.17, - 264.01, - 477.04, - 285.93 - ], - "text": "Due to its lowpass nature, this system will spread an input pulse of 1 second to an output with\nwidth", - "type": "text" - }, - { - "block_id": "p230-b19", - "global_id": 6246, - "bbox": [ - 214.44, - 287.51, - 365.74, - 298.97 - ], - "text": "Ty = Tx + Th = 1 + 2.7183 = 3.7183 s", - "type": "text" - }, - { - "block_id": "p230-b20", - "global_id": 6247, - "bbox": [ - 103.17, - 306.85, - 477.02, - 328.77 - ], - "text": "To avoid interference between pulses at the output, the pulse transmission rate should be no\nmore than the reciprocal of the output pulse width. That is,", - "type": "text" - }, - { - "block_id": "p230-b21", - "global_id": 6248, - "bbox": [ - 171.33, - 338.97, - 408.84, - 353.81 - ], - "text": "maximum pulse transmission rate =\n1\n3.7183 = 0.2689 pulse/s", - "type": "text" - }, - { - "block_id": "p230-b22", - "global_id": 6249, - "bbox": [ - 103.16, - 362.23, - 477.0, - 384.57 - ], - "text": "By narrowing the input pulses, the pulse transmission rate could increase up to fc = 0.3679\npulse/s.", - "type": "text" - }, - { - "block_id": "p230-b23", - "global_id": 6250, - "bbox": [ - 101.84, - 428.57, - 290.32, - 440.52 - ], - "text": "2.6-7 The Resonance Phenomenon", - "type": "text" - }, - { - "block_id": "p230-b24", - "global_id": 6251, - "bbox": [ - 101.84, - 446.64, - 490.41, - 492.47 - ], - "text": "Finally, we come to the fascinating phenomenon of resonance. As we have already mentioned\nseveral times, this phenomenon is observed when the input signal is identical or is very close to a\ncharacteristic mode of the system. For the sake of simplicity and clarity, we consider a first-order\nsystem having only a single mode, eλt. Let the impulse response of this system be†", - "type": "text" - }, - { - "block_id": "p230-b25", - "global_id": 6252, - "bbox": [ - 274.06, - 502.77, - 317.54, - 514.78 - ], - "text": "h(t) = Aeλt", - "type": "text" - }, - { - "block_id": "p230-b26", - "global_id": 6253, - "bbox": [ - 101.84, - 527.32, - 178.21, - 537.28 - ], - "text": "and let the input be", - "type": "text" - }, - { - "block_id": "p230-b27", - "global_id": 6254, - "bbox": [ - 270.33, - 537.86, - 321.28, - 549.86 - ], - "text": "x(t) = e(λ−ϵ)t", - "type": "text" - }, - { - "block_id": "p230-b28", - "global_id": 6255, - "bbox": [ - 101.84, - 559.0, - 267.61, - 569.38 - ], - "text": "The system response y(t) is then given by", - "type": "text" - }, - { - "block_id": "p230-b29", - "global_id": 6256, - "bbox": [ - 257.76, - 577.29, - 333.86, - 591.68 - ], - "text": "y(t) = Aeλt ∗e(λ−ϵ)t", - "type": "text" - }, - { - "block_id": "p230-b30", - "global_id": 6257, - "bbox": [ - 101.84, - 610.24, - 490.39, - 633.41 - ], - "text": "† For convenience, we omit multiplying x(t) and h(t) by u(t). Throughout this discussion, we assume that\nthey are causal.", - "type": "text" - } - ] - }, - { - "page_num": 231, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p231-b0", - "global_id": 6258, - "bbox": [ - 298.64, - 62.89, - 516.15, - 71.98 - ], - "text": "2.6\nIntuitive Insights into System Behavior\n211", - "type": "text" - }, - { - "block_id": "p231-b1", - "global_id": 6259, - "bbox": [ - 127.59, - 85.82, - 276.68, - 95.78 - ], - "text": "From the convolution table we obtain", - "type": "text" - }, - { - "block_id": "p231-b2", - "global_id": 6260, - "bbox": [ - 238.94, - 104.65, - 272.89, - 121.6 - ], - "text": "y(t) = A", - "type": "text" - }, - { - "block_id": "p231-b3", - "global_id": 6261, - "bbox": [ - 267.4, - 118.4, - 271.75, - 128.36 - ], - "text": "ϵ", - "type": "text" - }, - { - "block_id": "p231-b5", - "global_id": 6262, - "bbox": [ - 278.01, - 103.31, - 328.48, - 121.6 - ], - "text": "eλt −e(λ−ϵ)t", - "type": "text" - }, - { - "block_id": "p231-b6", - "global_id": 6263, - "bbox": [ - 330.52, - 109.6, - 356.6, - 121.6 - ], - "text": "= Aeλt", - "type": "text" - }, - { - "block_id": "p231-b7", - "global_id": 6264, - "bbox": [ - 357.24, - 97.34, - 396.23, - 114.71 - ], - "text": "1 −e−ϵt", - "type": "text" - }, - { - "block_id": "p231-b8", - "global_id": 6265, - "bbox": [ - 378.56, - 118.4, - 382.91, - 128.36 - ], - "text": "ϵ", - "type": "text" - }, - { - "block_id": "p231-b10", - "global_id": 6266, - "bbox": [ - 492.07, - 111.74, - 516.13, - 121.7 - ], - "text": "(2.49)", - "type": "text" - }, - { - "block_id": "p231-b11", - "global_id": 6267, - "bbox": [ - 127.59, - 138.74, - 516.08, - 161.07 - ], - "text": "Now, as ϵ →0, both the numerator and the denominator of the term in the parentheses approach\nzero. Applying L’Hôpital’s rule to this term yields", - "type": "text" - }, - { - "block_id": "p231-b12", - "global_id": 6268, - "bbox": [ - 290.3, - 171.09, - 352.79, - 189.61 - ], - "text": "lim\nϵ→0 y(t) = Ateλt", - "type": "text" - }, - { - "block_id": "p231-b13", - "global_id": 6269, - "bbox": [ - 127.59, - 198.48, - 516.17, - 244.72 - ], - "text": "Clearly, the response does not go to infinity as ϵ →0, but it acquires a factor t, which approaches\n∞as t →∞. If λ has a negative real part (so that it lies in the LHP), eλt decays faster than t and\ny(t) →0 as t →∞. The resonance phenomenon in this case is present, but its manifestation is\naborted by the signal’s own exponential decay.", - "type": "text" - }, - { - "block_id": "p231-b14", - "global_id": 6270, - "bbox": [ - 127.59, - 246.62, - 516.15, - 316.45 - ], - "text": "This discussion shows that resonance is a cumulative phenomenon, not instantaneous. It builds\nup linearly with t.† When the mode decays exponentially, the signal decays too fast for resonance\nto counteract the decay; as a result, the signal vanishes before resonance has a chance to build\nit up. However, if the mode were to decay at a rate less than 1/t, we should see the resonance\nphenomenon clearly. This specific condition would be possible if Re λ ≥0. For instance, when Re\nλ = 0 so that λ lies on the imaginary axis of the complex plane (λ = jω), the output becomes", - "type": "text" - }, - { - "block_id": "p231-b15", - "global_id": 6271, - "bbox": [ - 297.28, - 326.47, - 345.81, - 338.47 - ], - "text": "y(t) = Atejωt", - "type": "text" - }, - { - "block_id": "p231-b16", - "global_id": 6272, - "bbox": [ - 127.59, - 350.63, - 337.6, - 360.69 - ], - "text": "Here, the response does go to infinity linearly with t.", - "type": "text" - }, - { - "block_id": "p231-b17", - "global_id": 6273, - "bbox": [ - 127.59, - 361.05, - 516.16, - 420.47 - ], - "text": "For a real system, if λ = jω is a root, λ∗= −jω must also be a root; the impulse response\nis of the form Aejωt + Ae−jωt = 2Acos ωt. The response of this system to input Acosωt is\n2Acosωt ∗cosωt. The reader can show that this convolution contains a term of the form Atcos ωt.\nThe resonance phenomenon is clearly visible. The system response to its characteristic mode\nincreases linearly with time, eventually reaching ∞, as indicated in Fig. 2.24.", - "type": "text" - }, - { - "block_id": "p231-b18", - "global_id": 6274, - "bbox": [ - 127.59, - 422.05, - 516.16, - 456.34 - ], - "text": "Recall that when λ = jω, the system is marginally stable. As we have indicated, the full effect\nof resonance cannot be seen for an asymptotically stable system; only in a marginally stable system\ndoes the resonance phenomenon boost the system’s response to infinity when the system’s input", - "type": "text" - }, - { - "block_id": "p231-b19", - "global_id": 6275, - "bbox": [ - 256.42, - 531.69, - 258.81, - 539.97 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p231-b20", - "global_id": 6276, - "bbox": [ - 129.77, - 475.21, - 140.88, - 483.29 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p231-b21", - "global_id": 6277, - "bbox": [ - 284.54, - 574.43, - 483.53, - 583.66 - ], - "text": "Figure 2.24 Buildup of system response in resonance.", - "type": "text" - }, - { - "block_id": "p231-b22", - "global_id": 6278, - "bbox": [ - 127.59, - 609.86, - 516.13, - 633.41 - ], - "text": "† If the characteristic root in question repeats r times, resonance effect increases as tr−1. However, tr−1eλt →0\nas t →∞for any value of r, provided Re λ < 0 (λ in the LHP).", - "type": "text" - } - ] - }, - { - "page_num": 232, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p232-b0", - "global_id": 6279, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "212\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p232-b1", - "global_id": 6280, - "bbox": [ - 101.84, - 85.82, - 490.39, - 167.51 - ], - "text": "is a characteristic mode. But even in an asymptotically stable system, we see a manifestation of\nresonance if its characteristic roots are close to the imaginary axis so that Re λ is a small, negative\nvalue. We can show that when the characteristic roots of a system are σ ± jω0, then the system\nresponse to the input ejω0t or the sinusoid cosω0t is very large for small σ.† The system response\ndrops off rapidly as the input signal frequency moves away from ω0. This frequency-selective\nbehavior can be studied more profitably after an understanding of frequency-domain analysis has\nbeen acquired. For this reason we postpone full discussion of this subject until Ch. 4.", - "type": "text" - }, - { - "block_id": "p232-b2", - "global_id": 6281, - "bbox": [ - 101.84, - 182.4, - 490.42, - 363.93 - ], - "text": "IMPORTANCE OF THE RESONANCE PHENOMENON\nThe resonance phenomenon is very important because it allows us to design frequency-selective\nsystems by choosing their characteristic roots properly. Lowpass, bandpass, highpass, and\nbandstop filters are all examples of frequency-selective networks. In mechanical systems, the\ninadvertent presence of resonance can cause signals of such tremendous magnitude that the system\nmay fall apart. A musical note (periodic vibrations) of proper frequency can shatter glass if the\nfrequency is matched to the characteristic root of the glass, which acts as a mechanical system.\nSimilarly, a company of soldiers marching in step across a bridge amounts to applying a periodic\nforce to the bridge. If the frequency of this input force happens to be nearer to a characteristic\nroot of the bridge, the bridge may respond (vibrate) violently and collapse, even though it would\nhave been strong enough to carry many soldiers marching out of step. A case in point is the\nTacoma Narrows Bridge failure of 1940. This bridge was opened to traffic in July 1940. Within\nfour months of opening (on November 7, 1940), it collapsed in a mild gale, not because of the\nwind’s brute force but because the frequencies of wind-generated vortices, which matched the\nnatural frequencies (characteristic roots) of the bridge, caused resonance.", - "type": "text" - }, - { - "block_id": "p232-b3", - "global_id": 6282, - "bbox": [ - 101.84, - 365.92, - 490.41, - 435.67 - ], - "text": "Because of the great damage that may occur, mechanical resonance is generally to be avoided,\nespecially in structures or vibrating mechanisms. If an engine with periodic force (such as piston\nmotion) is mounted on a platform, the platform with its mass and springs should be designed so\nthat their characteristic roots are not close to the engine’s frequency of vibration. Proper design\nof this platform can not only avoid resonance, but also attenuate vibrations if the system roots are\nplaced far away from the frequency of vibration.", - "type": "text" - }, - { - "block_id": "p232-b4", - "global_id": 6283, - "bbox": [ - 102.2, - 465.25, - 255.88, - 479.2 - ], - "text": "2.7 MATLAB: M-FILES", - "type": "text" - }, - { - "block_id": "p232-b5", - "global_id": 6284, - "bbox": [ - 101.84, - 485.18, - 490.4, - 519.06 - ], - "text": "M-files are stored sequences of MATLAB commands and help simplify complicated tasks. There\nare two types of M-file: script and function. Both types are simple text files and require a .m\nfilename extension.", - "type": "text" - }, - { - "block_id": "p232-b6", - "global_id": 6285, - "bbox": [ - 101.84, - 521.05, - 490.42, - 566.87 - ], - "text": "Although M-files can be created by using any text editor, MATLAB’s built-in editor is the\npreferable choice because of its special features. As with any program, comments improve the\nreadability of an M-file. Comments begin with the % character and continue through the end of the\nline.", - "type": "text" - }, - { - "block_id": "p232-b7", - "global_id": 6286, - "bbox": [ - 101.84, - 568.86, - 490.36, - 603.03 - ], - "text": "An M-file is executed by simply typing the filename (without the .m extension). To execute,\nM-files need to be located in the current directory or any other directory in the MATLAB path.\nNew directories are easily added to the MATLAB path by using the addpath command.", - "type": "text" - }, - { - "block_id": "p232-b8", - "global_id": 6287, - "bbox": [ - 101.84, - 621.19, - 344.41, - 634.75 - ], - "text": "† This follows directly from Eq. (2.49) with λ = σ + jω0 and ϵ = σ.", - "type": "text" - } - ] - }, - { - "page_num": 233, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p233-b0", - "global_id": 6288, - "bbox": [ - 380.65, - 62.89, - 516.14, - 71.98 - ], - "text": "2.7\nMATLAB: M-Files\n213", - "type": "text" - }, - { - "block_id": "p233-b1", - "global_id": 6289, - "bbox": [ - 127.59, - 86.52, - 235.31, - 98.48 - ], - "text": "2.7-1 Script M-Files", - "type": "text" - }, - { - "block_id": "p233-b2", - "global_id": 6290, - "bbox": [ - 127.59, - 104.61, - 516.15, - 138.48 - ], - "text": "Script files, the simplest type of M-file, consist of a series of MATLAB commands. Script files\nrecord and automate a series of steps, and they are easy to modify. To demonstrate the utility of a\nscript file, consider the operational amplifier circuit shown in Fig. 2.25.", - "type": "text" - }, - { - "block_id": "p233-b3", - "global_id": 6291, - "bbox": [ - 127.59, - 140.47, - 516.13, - 187.79 - ], - "text": "The system’s characteristic modes define the circuit’s behavior and provide insight regarding\nsystem behavior. Using ideal, infinite gain difference amplifier characteristics, we first derive the\ndifferential equation that relates output y(t) to input x(t). Kirchhoff’s current law (KCL) at the\nnode shared by R1 and R3 provides", - "type": "text" - }, - { - "block_id": "p233-b4", - "global_id": 6292, - "bbox": [ - 226.2, - 197.23, - 266.7, - 207.5 - ], - "text": "x(t) −v(t)", - "type": "text" - }, - { - "block_id": "p233-b5", - "global_id": 6293, - "bbox": [ - 241.42, - 211.6, - 250.99, - 222.43 - ], - "text": "R3", - "type": "text" - }, - { - "block_id": "p233-b6", - "global_id": 6294, - "bbox": [ - 269.45, - 197.23, - 320.44, - 214.49 - ], - "text": "+ y(t) −v(t)", - "type": "text" - }, - { - "block_id": "p233-b7", - "global_id": 6295, - "bbox": [ - 295.17, - 211.6, - 304.74, - 222.43 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p233-b8", - "global_id": 6296, - "bbox": [ - 323.2, - 197.23, - 364.36, - 214.59 - ], - "text": "+ 0 −v(t)", - "type": "text" - }, - { - "block_id": "p233-b9", - "global_id": 6297, - "bbox": [ - 344.01, - 211.6, - 353.58, - 222.43 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p233-b10", - "global_id": 6298, - "bbox": [ - 367.11, - 204.21, - 418.71, - 215.36 - ], - "text": "−C2˙v(t) = 0", - "type": "text" - }, - { - "block_id": "p233-b11", - "global_id": 6299, - "bbox": [ - 127.6, - 232.35, - 313.41, - 242.32 - ], - "text": "KCL at the inverting input of the op amp gives", - "type": "text" - }, - { - "block_id": "p233-b12", - "global_id": 6300, - "bbox": [ - 287.88, - 253.25, - 302.69, - 263.53 - ], - "text": "v(t)", - "type": "text" - }, - { - "block_id": "p233-b13", - "global_id": 6301, - "bbox": [ - 290.25, - 267.61, - 299.82, - 278.45 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p233-b14", - "global_id": 6302, - "bbox": [ - 305.43, - 260.24, - 357.04, - 271.38 - ], - "text": "+ C1˙y(t) = 0", - "type": "text" - }, - { - "block_id": "p233-b15", - "global_id": 6303, - "bbox": [ - 127.59, - 288.31, - 337.9, - 298.27 - ], - "text": "Combining and simplifying the KCL equations yield", - "type": "text" - }, - { - "block_id": "p233-b16", - "global_id": 6304, - "bbox": [ - 184.41, - 310.23, - 219.09, - 327.18 - ], - "text": "¨y(t) + 1", - "type": "text" - }, - { - "block_id": "p233-b17", - "global_id": 6305, - "bbox": [ - 211.28, - 324.18, - 221.41, - 335.01 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p233-b18", - "global_id": 6306, - "bbox": [ - 224.22, - 302.81, - 239.1, - 320.19 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p233-b19", - "global_id": 6307, - "bbox": [ - 231.57, - 324.18, - 241.14, - 335.01 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p233-b20", - "global_id": 6308, - "bbox": [ - 244.39, - 310.23, - 262.43, - 327.18 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p233-b21", - "global_id": 6309, - "bbox": [ - 254.9, - 324.18, - 264.47, - 335.01 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p233-b22", - "global_id": 6310, - "bbox": [ - 267.72, - 310.23, - 285.76, - 327.18 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p233-b23", - "global_id": 6311, - "bbox": [ - 278.23, - 324.18, - 287.81, - 335.01 - ], - "text": "R3", - "type": "text" - }, - { - "block_id": "p233-b25", - "global_id": 6312, - "bbox": [ - 296.77, - 310.23, - 364.54, - 335.01 - ], - "text": "˙y(t) +\n1\nR1R2C1C2", - "type": "text" - }, - { - "block_id": "p233-b26", - "global_id": 6313, - "bbox": [ - 366.24, - 310.23, - 442.78, - 335.01 - ], - "text": "y(t) = −\n1\nR1R3C1C2", - "type": "text" - }, - { - "block_id": "p233-b27", - "global_id": 6314, - "bbox": [ - 444.47, - 316.8, - 459.31, - 327.08 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p233-b28", - "global_id": 6315, - "bbox": [ - 127.59, - 345.76, - 516.13, - 367.69 - ], - "text": "which is the desired constant coefficient differential equation. Thus, the characteristic equation is\ngiven by", - "type": "text" - }, - { - "block_id": "p233-b29", - "global_id": 6316, - "bbox": [ - 189.59, - 370.35, - 218.88, - 387.3 - ], - "text": "λ2 + 1", - "type": "text" - }, - { - "block_id": "p233-b30", - "global_id": 6317, - "bbox": [ - 211.09, - 384.31, - 221.21, - 395.14 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p233-b31", - "global_id": 6318, - "bbox": [ - 222.91, - 362.93, - 237.78, - 380.31 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p233-b32", - "global_id": 6319, - "bbox": [ - 230.27, - 384.31, - 239.84, - 395.14 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p233-b33", - "global_id": 6320, - "bbox": [ - 243.08, - 370.35, - 261.12, - 387.3 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p233-b34", - "global_id": 6321, - "bbox": [ - 253.59, - 384.31, - 263.16, - 395.14 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p233-b35", - "global_id": 6322, - "bbox": [ - 266.41, - 370.35, - 284.46, - 387.3 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p233-b36", - "global_id": 6323, - "bbox": [ - 276.93, - 384.31, - 286.5, - 395.14 - ], - "text": "R3", - "type": "text" - }, - { - "block_id": "p233-b38", - "global_id": 6324, - "bbox": [ - 294.35, - 370.35, - 352.77, - 395.14 - ], - "text": "λ +\n1\nR1R2C1C2", - "type": "text" - }, - { - "block_id": "p233-b39", - "global_id": 6325, - "bbox": [ - 356.52, - 372.81, - 516.13, - 388.07 - ], - "text": "= (a0λ2 + a1λ + a2) = 0\n(2.50)", - "type": "text" - }, - { - "block_id": "p233-b40", - "global_id": 6326, - "bbox": [ - 127.59, - 400.72, - 501.52, - 415.79 - ], - "text": "The roots λ1 and λ2 of Eq. (2.50) establish the nature of the characteristic modes eλ1t and eλ2t.", - "type": "text" - }, - { - "block_id": "p233-b41", - "global_id": 6327, - "bbox": [ - 127.59, - 415.88, - 516.14, - 474.08 - ], - "text": "As a first case, assign nominal component values of R1 = R2 = R3 = 10 k and C1 = C2 = 1\nµF. A series of MATLAB commands allows convenient computation of the roots λ = [λ1;λ2].\nAlthough λ can be determined using the quadratic equation, MATLAB’s roots command is more\nconvenient. The roots command requires an input vector that contains the polynomial coefficients\nin descending order. Even if a coefficient is zero, it must still be included in the vector.", - "type": "text" - }, - { - "block_id": "p233-b42", - "global_id": 6328, - "bbox": [ - 263.02, - 542.15, - 270.91, - 551.76 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p233-b43", - "global_id": 6329, - "bbox": [ - 313.35, - 553.71, - 317.35, - 561.71 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p233-b44", - "global_id": 6330, - "bbox": [ - 312.43, - 581.26, - 316.94, - 589.26 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p233-b45", - "global_id": 6331, - "bbox": [ - 159.55, - 567.4, - 170.65, - 575.48 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p233-b46", - "global_id": 6332, - "bbox": [ - 163.1, - 575.67, - 167.1, - 583.67 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p233-b47", - "global_id": 6333, - "bbox": [ - 162.84, - 558.3, - 167.35, - 566.3 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p233-b48", - "global_id": 6334, - "bbox": [ - 263.02, - 491.65, - 270.91, - 501.26 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p233-b49", - "global_id": 6335, - "bbox": [ - 195.02, - 542.15, - 202.91, - 551.76 - ], - "text": "R3", - "type": "text" - }, - { - "block_id": "p233-b50", - "global_id": 6336, - "bbox": [ - 212.7, - 576.59, - 221.03, - 586.19 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p233-b51", - "global_id": 6337, - "bbox": [ - 280.7, - 526.08, - 289.03, - 535.69 - ], - "text": "C1", - "type": "text" - }, - { - "block_id": "p233-b52", - "global_id": 6338, - "bbox": [ - 377.14, - 581.18, - 388.25, - 589.26 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p233-b53", - "global_id": 6339, - "bbox": [ - 380.7, - 589.44, - 384.7, - 597.44 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p233-b54", - "global_id": 6340, - "bbox": [ - 245.67, - 572.07, - 384.95, - 586.96 - ], - "text": "+\nv(t)", - "type": "text" - }, - { - "block_id": "p233-b55", - "global_id": 6341, - "bbox": [ - 249.89, - 587.14, - 253.89, - 595.14 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p233-b56", - "global_id": 6342, - "bbox": [ - 248.96, - 569.78, - 253.48, - 577.78 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p233-b57", - "global_id": 6343, - "bbox": [ - 151.5, - 619.96, - 300.49, - 629.19 - ], - "text": "Figure 2.25 Operation-amplifier circuit.", - "type": "text" - } - ] - }, - { - "page_num": 234, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p234-b0", - "global_id": 6344, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "214\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p234-b1", - "global_id": 6345, - "bbox": [ - 101.84, - 86.24, - 447.05, - 108.15 - ], - "text": "% CH2MP1.m : Chapter 2, MATLAB Program 1\n% Script M-file determines characteristic roots of op-amp circuit.", - "type": "text" - }, - { - "block_id": "p234-b2", - "global_id": 6346, - "bbox": [ - 101.84, - 122.1, - 426.13, - 191.85 - ], - "text": "% Set component values:\nR = [1e4, 1e4, 1e4]; C = [1e-6, 1e-6];\n% Determine coefficients for characteristic equation:\nA = [1, (1/R(1)+1/R(2)+1/R(3))/C(2), 1/(R(1)*R(2)*C(1)*C(2))];\n% Determine characteristic roots:\nlambda = roots(A);", - "type": "text" - }, - { - "block_id": "p234-b3", - "global_id": 6347, - "bbox": [ - 101.84, - 205.98, - 490.4, - 239.86 - ], - "text": "A script file is created by placing these commands in a text file, which in this case is named\nCH2MP1.m. While comment lines improve program clarity, their removal does not affect program\nfunctionality. The program is executed by typing", - "type": "text" - }, - { - "block_id": "p234-b4", - "global_id": 6348, - "bbox": [ - 101.84, - 254.58, - 154.14, - 264.55 - ], - "text": ">>\nCH2MP1", - "type": "text" - }, - { - "block_id": "p234-b5", - "global_id": 6349, - "bbox": [ - 101.84, - 278.69, - 490.41, - 300.61 - ], - "text": "After execution, all the resulting variables are available in the workspace. For example, to\nview the characteristic roots, type", - "type": "text" - }, - { - "block_id": "p234-b6", - "global_id": 6350, - "bbox": [ - 101.84, - 315.33, - 216.9, - 337.24 - ], - "text": ">>\nlambda\nlambda = -261.8034", - "type": "text" - }, - { - "block_id": "p234-b7", - "global_id": 6351, - "bbox": [ - 175.05, - 339.24, - 216.89, - 349.2 - ], - "text": "-38.1966", - "type": "text" - }, - { - "block_id": "p234-b8", - "global_id": 6352, - "bbox": [ - 101.84, - 359.73, - 458.69, - 373.3 - ], - "text": "Thus, the characteristic modes are simple decaying exponentials: e−261.8034t and e−38.1966t.", - "type": "text" - }, - { - "block_id": "p234-b9", - "global_id": 6353, - "bbox": [ - 101.85, - 375.3, - 490.4, - 409.46 - ], - "text": "Script files permit simple or incremental changes, thereby saving significant effort. Consider\nwhat happens when capacitor C1 is changed from 1.0 µF to 1.0 nF. Changing CH2MP1.m so that C\n= [1e-9, 1e-6] allows computation of the new characteristic roots:", - "type": "text" - }, - { - "block_id": "p234-b10", - "global_id": 6354, - "bbox": [ - 101.85, - 423.89, - 222.14, - 457.77 - ], - "text": ">>\nCH2MP1\n>>\nlambda\nlambda = 1.0e+003 *", - "type": "text" - }, - { - "block_id": "p234-b11", - "global_id": 6355, - "bbox": [ - 169.82, - 459.76, - 258.74, - 481.67 - ], - "text": "-0.1500 + 3.1587i\n-0.1500 - 3.1587i", - "type": "text" - }, - { - "block_id": "p234-b12", - "global_id": 6356, - "bbox": [ - 101.85, - 495.82, - 490.42, - 541.65 - ], - "text": "Perhaps surprisingly, the characteristic modes are now complex exponentials capable of supporting\noscillations. The imaginary portion of λ dictates an oscillation rate of 3158.7 rad/s or about 503\nHz. The real portion dictates the rate of decay. The time expected to reduce the amplitude to 25%\nis approximately t = ln0.25/Re(λ) ≈0.01 second.", - "type": "text" - }, - { - "block_id": "p234-b13", - "global_id": 6357, - "bbox": [ - 101.84, - 570.87, - 225.52, - 582.83 - ], - "text": "2.7-2 Function M-Files", - "type": "text" - }, - { - "block_id": "p234-b14", - "global_id": 6358, - "bbox": [ - 101.84, - 588.96, - 490.41, - 634.79 - ], - "text": "It is inconvenient to modify and save a script file each time a change of parameters is desired.\nFunction M-files provide a sensible alternative. Unlike script M-files, function M-files can accept\ninput arguments as well as return outputs. Functions truly extend the MATLAB language in ways\nthat script files cannot.", - "type": "text" - } - ] - }, - { - "page_num": 235, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p235-b0", - "global_id": 6359, - "bbox": [ - 380.65, - 62.89, - 516.14, - 71.98 - ], - "text": "2.7\nMATLAB: M-Files\n215", - "type": "text" - }, - { - "block_id": "p235-b1", - "global_id": 6360, - "bbox": [ - 127.59, - 85.82, - 516.12, - 107.74 - ], - "text": "Syntactically, a function M-file is identical to a script M-file except for the first line. The\ngeneral form of the first line is", - "type": "text" - }, - { - "block_id": "p235-b2", - "global_id": 6361, - "bbox": [ - 154.53, - 120.14, - 489.2, - 130.81 - ], - "text": "function [output1, ..., outputN] = filename(input1, ..., inputM)", - "type": "text" - }, - { - "block_id": "p235-b3", - "global_id": 6362, - "bbox": [ - 127.59, - 143.34, - 516.16, - 177.22 - ], - "text": "For example, consider modification of CH2MP1.m to make function CH2MP2.m. Component\nvalues are passed to the function as two separate inputs: a length-3 vector of resistor values and a\nlength-2 vector of capacitor values. The characteristic roots are returned as a 2×1 complex vector.", - "type": "text" - }, - { - "block_id": "p235-b4", - "global_id": 6363, - "bbox": [ - 127.59, - 188.33, - 457.1, - 258.08 - ], - "text": "function [lambda] = CH2MP2(R,C)\n% CH2MP2.m : Chapter 2, MATLAB Program 2\n% Function M-file finds characteristic roots of op-amp circuit.\n% INPUTS:\nR = length-3 vector of resistances\n%\nC = length-2 vector of capacitances\n% OUTPUTS:\nlambda = characteristic roots", - "type": "text" - }, - { - "block_id": "p235-b5", - "global_id": 6364, - "bbox": [ - 127.59, - 272.02, - 451.87, - 317.85 - ], - "text": "% Determine coefficients for characteristic equation:\nA = [1, (1/R(1)+1/R(2)+1/R(3))/C(2), 1/(R(1)*R(2)*C(1)*C(2))];\n% Determine characteristic roots:\nlambda = roots(A);", - "type": "text" - }, - { - "block_id": "p235-b6", - "global_id": 6365, - "bbox": [ - 127.59, - 328.39, - 516.13, - 362.26 - ], - "text": "As with script M-files, function M-files execute by typing the name at the command prompt.\nHowever, inputs must also be included. For example, CH2MP2 easily confirms the oscillatory modes\nof the preceding example.", - "type": "text" - }, - { - "block_id": "p235-b7", - "global_id": 6366, - "bbox": [ - 127.59, - 373.38, - 383.87, - 395.3 - ], - "text": ">>\nlambda = CH2MP2([1e4, 1e4, 1e4],[1e-9, 1e-6])\nlambda = 1.0e+003 *", - "type": "text" - }, - { - "block_id": "p235-b8", - "global_id": 6367, - "bbox": [ - 195.57, - 397.29, - 284.48, - 419.21 - ], - "text": "-0.1500 + 3.1587i\n-0.1500 - 3.1587i", - "type": "text" - }, - { - "block_id": "p235-b9", - "global_id": 6368, - "bbox": [ - 127.59, - 429.75, - 516.16, - 535.65 - ], - "text": "Although scripts and functions have similarities, they also have distinct differences that are\nworth pointing out. Scripts operate on workspace data; either functions must be supplied data\nthrough inputs or they must create their own data. Unless passed as an output, variables and data\ncreated by functions remain local to the function; variables or data generated by scripts are global\nand are added to the workspace. To emphasize this point, consider polynomial coefficient vector\nA, which is created and used in both CH2MP1.m and CH2MP2.m. Following execution of function\nCH2MP2, the variable A is not added to the workspace. Following execution of script CH2MP1,\nhowever, A is available in the workspace. Recall, the workspace is easily viewed by typing either\nwho or whos.", - "type": "text" - }, - { - "block_id": "p235-b10", - "global_id": 6369, - "bbox": [ - 127.59, - 560.97, - 212.91, - 572.93 - ], - "text": "2.7-3 For-Loops", - "type": "text" - }, - { - "block_id": "p235-b11", - "global_id": 6370, - "bbox": [ - 127.59, - 579.06, - 516.14, - 636.84 - ], - "text": "Real resistors and capacitors never exactly equal their nominal values. Suppose that the circuit\ncomponents are measured as R1 = 10.322 k, R2 = 9.952 k, R3 = 10.115 k, C1 = 1.120 nF, and\nC2 = 1.320 µF. These values are consistent with the 10 and 25% tolerance resistor and capacitor\nvalues commonly and readily available. CH2MP2.m uses these component values to calculate the\nnew values of λ.", - "type": "text" - } - ] - }, - { - "page_num": 236, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p236-b0", - "global_id": 6371, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "216\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p236-b1", - "global_id": 6372, - "bbox": [ - 101.84, - 86.24, - 405.21, - 108.15 - ], - "text": ">>\nlambda = CH2MP2([10322,9592,10115],[1.12e-9, 1.32e-6])\nlambda = 1.0e+003 *", - "type": "text" - }, - { - "block_id": "p236-b2", - "global_id": 6373, - "bbox": [ - 169.82, - 110.15, - 258.73, - 132.07 - ], - "text": "-0.1136 + 2.6113i\n-0.1136 - 2.6113i", - "type": "text" - }, - { - "block_id": "p236-b3", - "global_id": 6374, - "bbox": [ - 101.84, - 141.95, - 490.4, - 187.78 - ], - "text": "Now the natural modes oscillate at 2611.3 rad/s or about 416 Hz. Decay to 25% amplitude is\nexpected in t = ln0.25/(−113.6) ≈0.012 second. These values, which differ significantly from\nthe nominal values of 503 Hz and t ≈0.01 second, warrant a more formal investigation of the\neffect of component variations on the locations of the characteristic roots.", - "type": "text" - }, - { - "block_id": "p236-b4", - "global_id": 6375, - "bbox": [ - 101.84, - 189.78, - 490.39, - 247.55 - ], - "text": "It is sensible to look at three values for each component: the nominal value, a low value, and\na high value. Low and high values are based on component tolerances. For example, a 10% 1 k\nresistor could have an expected low value of 1000(1 −0.1) = 900 and an expected high value\nof 1000(1+0.1) = 1100 . For the five passive components in the design, 35 = 243 permutations\nare possible.", - "type": "text" - }, - { - "block_id": "p236-b5", - "global_id": 6376, - "bbox": [ - 101.84, - 249.54, - 490.39, - 283.71 - ], - "text": "Using either CH2MP1.m or CH2MP2.m to solve each of the 243 cases would be very tedious and\nboring. For-loops help automate repetitive tasks such as this. In MATLAB, the general structure\nof a for statement is", - "type": "text" - }, - { - "block_id": "p236-b6", - "global_id": 6377, - "bbox": [ - 147.09, - 295.17, - 445.16, - 305.83 - ], - "text": "for variable = expression, statement, ..., statement, end", - "type": "text" - }, - { - "block_id": "p236-b7", - "global_id": 6378, - "bbox": [ - 101.84, - 317.71, - 469.75, - 327.67 - ], - "text": "Five nested for-loops, one for each passive component, are required for the present example.", - "type": "text" - }, - { - "block_id": "p236-b8", - "global_id": 6379, - "bbox": [ - 101.84, - 338.13, - 483.66, - 360.05 - ], - "text": "% CH2MP3.m : Chapter 2, MATLAB Program 3\n% Script M-file determines characteristic roots over a range of component", - "type": "text" - }, - { - "block_id": "p236-b9", - "global_id": 6380, - "bbox": [ - 112.3, - 362.05, - 148.91, - 372.01 - ], - "text": "values.", - "type": "text" - }, - { - "block_id": "p236-b10", - "global_id": 6381, - "bbox": [ - 101.84, - 385.95, - 352.91, - 443.74 - ], - "text": "% Pre-allocate memory for all computed roots:\nlambda = zeros(2,243);\n% Initialize index to identify each permutation:\np=0;\nfor R1 = 1e4*[0.9,1.0,1.1],", - "type": "text" - }, - { - "block_id": "p236-b11", - "global_id": 6382, - "bbox": [ - 122.76, - 445.73, - 263.98, - 455.69 - ], - "text": "for R2 = 1e4*[0.9,1.0,1.1],", - "type": "text" - }, - { - "block_id": "p236-b12", - "global_id": 6383, - "bbox": [ - 143.68, - 457.68, - 284.89, - 467.64 - ], - "text": "for R3 = 1e4*[0.9,1.0,1.1],", - "type": "text" - }, - { - "block_id": "p236-b13", - "global_id": 6384, - "bbox": [ - 164.59, - 469.64, - 321.5, - 479.6 - ], - "text": "for C1 = 1e-9*[0.75,1.0,1.25],", - "type": "text" - }, - { - "block_id": "p236-b14", - "global_id": 6385, - "bbox": [ - 185.51, - 481.59, - 342.41, - 491.56 - ], - "text": "for C2 = 1e-6*[0.75,1.0,1.25],", - "type": "text" - }, - { - "block_id": "p236-b15", - "global_id": 6386, - "bbox": [ - 101.84, - 493.55, - 420.87, - 575.25 - ], - "text": "p = p+1;\nlambda(:,p) = CH2MP2([R1 R2 R3],[C1 C2]);\nend\nend\nend\nend\nend", - "type": "text" - }, - { - "block_id": "p236-b16", - "global_id": 6387, - "bbox": [ - 101.84, - 589.19, - 337.21, - 599.15 - ], - "text": "plot(real(lambda(:)),imag(lambda(:)),’kx’,...", - "type": "text" - }, - { - "block_id": "p236-b17", - "global_id": 6388, - "bbox": [ - 101.84, - 601.15, - 499.36, - 646.98 - ], - "text": "real(lambda(:,1)),imag(lambda(:,1)),’kv’,...\nreal(lambda(:,end)),imag(lambda(:,end)),’k^’)\nxlabel(’Real’),ylabel(’Imaginary’)\nlegend(’Char. Roots’,’Min. Val. Roots’,’Max. Val. Roots’,’Location’,’West’);", - "type": "text" - } - ] - }, - { - "page_num": 237, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p237-b0", - "global_id": 6389, - "bbox": [ - 380.65, - 62.89, - 516.14, - 71.98 - ], - "text": "2.7\nMATLAB: M-Files\n217", - "type": "text" - }, - { - "block_id": "p237-b1", - "global_id": 6390, - "bbox": [ - 157.06, - 268.68, - 507.56, - 289.76 - ], - "text": "–240\n–220\n–200\n–180\n–160\n–140\n–120\n–100\nReal", - "type": "text" - }, - { - "block_id": "p237-b2", - "global_id": 6391, - "bbox": [ - 140.34, - 259.45, - 160.34, - 267.45 - ], - "text": "–5000", - "type": "text" - }, - { - "block_id": "p237-b3", - "global_id": 6392, - "bbox": [ - 156.83, - 173.57, - 160.83, - 181.57 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p237-b4", - "global_id": 6393, - "bbox": [ - 143.33, - 87.7, - 159.33, - 95.7 - ], - "text": "5000", - "type": "text" - }, - { - "block_id": "p237-b5", - "global_id": 6394, - "bbox": [ - 128.36, - 160.0, - 137.16, - 196.16 - ], - "text": "Imaginary", - "type": "text" - }, - { - "block_id": "p237-b6", - "global_id": 6395, - "bbox": [ - 203.18, - 162.93, - 237.59, - 170.13 - ], - "text": "Char. Roots", - "type": "text" - }, - { - "block_id": "p237-b7", - "global_id": 6396, - "bbox": [ - 203.18, - 173.81, - 249.59, - 181.01 - ], - "text": "Min. Val. Roots", - "type": "text" - }, - { - "block_id": "p237-b8", - "global_id": 6397, - "bbox": [ - 203.18, - 184.68, - 250.78, - 191.88 - ], - "text": "Max. Val. Roots", - "type": "text" - }, - { - "block_id": "p237-b9", - "global_id": 6398, - "bbox": [ - 127.59, - 296.44, - 390.85, - 305.68 - ], - "text": "Figure 2.26 Effect of component values on characteristic root locations.", - "type": "text" - }, - { - "block_id": "p237-b10", - "global_id": 6399, - "bbox": [ - 127.59, - 321.23, - 516.16, - 379.43 - ], - "text": "The command lambda = zeros(2,243) preallocates a 2×243 array to store the computed\nroots. When necessary, MATLAB performs dynamic memory allocation, so this command is not\nstrictly necessary. However, preallocation significantly improves script execution speed. Notice\nalso that it would be nearly useless to call script CH2MP1 from within the nested loop; script file\nparameters cannot be changed during execution.", - "type": "text" - }, - { - "block_id": "p237-b11", - "global_id": 6400, - "bbox": [ - 127.59, - 381.42, - 516.16, - 522.89 - ], - "text": "The plot instruction is quite long. Long commands can be broken across several lines by\nterminating intermediate lines with three dots (...). The three dots tell MATLAB to continue\nthe present command to the next line. Black x’s locate roots of each permutation. The command\nlambda(:) vectorizes the 2 × 243 matrix lambda into a 486 × 1 vector. This is necessary in\nthis case to ensure that a proper legend is generated. Because of loop order, permutation p = 1\ncorresponds to the case of all components at the smallest values and permutation p = 243\ncorresponds to the case of all components at the largest values. This information is used to\nseparately highlight the minimum and maximum cases using down-triangles (▽) and up-triangles\n(△), respectively. In addition to terminating each for loop, end is used to indicate the final index\nalong a particular dimension, which eliminates the need to remember the particular size of a\nvariable. An overloaded function, such as end, serves multiple uses and is typically interpreted\nbased on context.", - "type": "text" - }, - { - "block_id": "p237-b12", - "global_id": 6401, - "bbox": [ - 127.59, - 524.88, - 516.14, - 558.76 - ], - "text": "The graphical results provided by CH2MP3 are shown in Fig. 2.26. Between extremes, root\noscillations vary from 365 to 745 Hz and decay times to 25% amplitude vary from 6.2 to 12.7 ms.\nClearly, this circuit’s behavior is quite sensitive to ordinary component variations.", - "type": "text" - }, - { - "block_id": "p237-b13", - "global_id": 6402, - "bbox": [ - 127.59, - 582.83, - 384.13, - 594.78 - ], - "text": "2.7-4 Graphical Understanding of Convolution", - "type": "text" - }, - { - "block_id": "p237-b14", - "global_id": 6403, - "bbox": [ - 127.59, - 600.5, - 516.14, - 635.08 - ], - "text": "MATLAB graphics effectively illustrate the convolution process. Consider the case of y(t) = x(t)∗\nh(t), where x(t) = 1.5sin(πt)(u(t) −u(t −1)) and h(t) = 1.5(u(t) −u(t −1.5)) −u(t −2) + u(t −\n2.5). Program CH2MP4 steps through the convolution over the time interval (−0.25 ≤t ≤3.75).", - "type": "text" - } - ] - }, - { - "page_num": 238, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p238-b0", - "global_id": 6404, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "218\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p238-b1", - "global_id": 6405, - "bbox": [ - 101.84, - 86.3, - 376.92, - 104.23 - ], - "text": "% CH2MP4.m : Chapter 2, MATLAB Program 4\n% Script M-file graphically demonstrates the convolution process.", - "type": "text" - }, - { - "block_id": "p238-b2", - "global_id": 6406, - "bbox": [ - 101.84, - 116.19, - 351.53, - 193.9 - ], - "text": "figure(1) % Create figure window and make visible on screen\nu = @(t) 1.0*(t>=0);\nx = @(t) 1.5*sin(pi*t).*(u(t)-u(t-1));\nh = @(t) 1.5*(u(t)-u(t-1.5))-u(t-2)+u(t-2.5);\ndtau = 0.005; tau = -1:dtau:4;\nti = 0; tvec = -.25:.1:3.75;\ny = NaN*zeros(1,length(tvec)); % Pre-allocate memory\nfor t = tvec,", - "type": "text" - }, - { - "block_id": "p238-b3", - "global_id": 6407, - "bbox": [ - 118.78, - 195.89, - 431.93, - 253.67 - ], - "text": "ti = ti+1; % Time index\nxh = x(t-tau).*h(tau); lxh = length(xh);\ny(ti) = sum(xh.*dtau); % Trapezoidal approximation of convolution integral\nsubplot(2,1,1),plot(tau,h(tau),’k-’,tau,x(t-tau),’k--’,t,0,’ok’);\naxis([tau(1) tau(end) -2.0 2.5]);\npatch([tau(1:end-1);tau(1:end-1);tau(2:end);tau(2:end)],...", - "type": "text" - }, - { - "block_id": "p238-b4", - "global_id": 6408, - "bbox": [ - 101.84, - 255.67, - 482.71, - 343.33 - ], - "text": "[zeros(1,lxh-1);xh(1:end-1);xh(2:end);zeros(1,lxh-1)],...\n[.8 .8 .8],’edgecolor’,’none’);\nxlabel(’\\tau’); title(’h(\\tau) [solid], x(t-\\tau) [dashed], h(\\tau)x(t-\\tau) [gray]’);\nc = get(gca,’children’); set(gca,’children’,[c(2);c(3);c(4);c(1)]);\nsubplot(2,1,2),plot(tvec,y,’k’,tvec(ti),y(ti),’ok’);\nxlabel(’t’); ylabel(’y(t) = \\int h(\\tau)x(t-\\tau) d\\tau’);\naxis([tau(1) tau(end) -1.0 2.0]); grid;\ndrawnow;\nend", - "type": "text" - }, - { - "block_id": "p238-b5", - "global_id": 6409, - "bbox": [ - 101.84, - 353.88, - 490.4, - 424.04 - ], - "text": "At each step, the program plots h(τ), x(t −τ), and shades the area h(τ)x(t −τ) gray. This gray\narea, which reflects the integral of h(τ)x(t −τ), is also the desired result, y(t). Figures 2.27, 2.28,\nand 2.29 display the convolution process at times t of 0.75, 2.25, and 2.85 seconds, respectively.\nThese figures help illustrate how the regions of integration change with time. Figure 2.27 has\nlimits of integration from 0 to (t = 0.75). Figure 2.28 has two regions of integration, with limits\n(t −1 = 1.25) to 1.5 and 2.0 to (t = 2.25). The last plot, Fig. 2.29, has limits from 2.0 to 2.5.", - "type": "text" - }, - { - "block_id": "p238-b6", - "global_id": 6410, - "bbox": [ - 101.84, - 426.03, - 490.41, - 591.7 - ], - "text": "Several comments regarding CH2MP4 are in order. The command figure(1) opens the first\nfigure window and, more important, makes sure it is visible. Anonymous functions are used to\nrepresent the functions u(t), x(t), and h(t). NaN, standing for not-a-number, usually results from\noperations such as 0/0 or ∞−∞. MATLAB refuses to plot NaN values, so preallocating y(t)\nwith NaNs ensures that MATLAB displays only values of y(t) that have been computed. As its\nname suggests, length returns the length of the input vector. The subplot(a,b,c) command\npartitions the current figure window into an a-by-b matrix of axes and selects axes c for use.\nSubplots facilitate graphical comparison by allowing multiple axes in a single figure window. The\npatch command is used to create the gray-shaded area for h(τ)x(t −τ). In CH2MP4, the get and\nset commands are used to reorder plot objects so that the gray area does not obscure other lines.\nDetails of the patch, get, and set commands, as used in CH2MP4, are somewhat advanced and\nare not pursued here.† MATLAB also prints most Greek letters if the Greek name is preceded\nby a backslash (\\) character. For example, \\tau in the xlabel command produces the symbol\nτ in the plot’s axis label. Similarly, an integral sign is produced by \\int. Finally, the drawnow", - "type": "text" - }, - { - "block_id": "p238-b7", - "global_id": 6411, - "bbox": [ - 101.84, - 610.24, - 490.39, - 633.67 - ], - "text": "† Interested students should consult the MATLAB help facilities for further information. Actually, the get\nand set commands are extremely powerful and can help modify plots in almost any conceivable way.", - "type": "text" - } - ] - }, - { - "page_num": 239, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p239-b0", - "global_id": 6412, - "bbox": [ - 380.65, - 62.89, - 516.14, - 71.98 - ], - "text": "2.8\nMATLAB: M-Files\n219", - "type": "text" - }, - { - "block_id": "p239-b1", - "global_id": 6413, - "bbox": [ - 162.47, - 160.59, - 511.2, - 177.82 - ], - "text": "–1\n–0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n3.5\n4\n–2", - "type": "text" - }, - { - "block_id": "p239-b2", - "global_id": 6414, - "bbox": [ - 166.47, - 131.92, - 170.47, - 139.92 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p239-b3", - "global_id": 6415, - "bbox": [ - 166.47, - 103.26, - 170.47, - 111.26 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p239-b4", - "global_id": 6416, - "bbox": [ - 267.16, - 88.11, - 417.98, - 96.33 - ], - "text": "h(τ) [solid], x(t–τ) [dashed], h(τ)×(t–τ) [gray]", - "type": "text" - }, - { - "block_id": "p239-b5", - "global_id": 6417, - "bbox": [ - 169.3, - 272.56, - 511.2, - 293.75 - ], - "text": "–1\n–0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n3.5\n4\nt", - "type": "text" - }, - { - "block_id": "p239-b6", - "global_id": 6418, - "bbox": [ - 339.32, - 181.28, - 342.83, - 189.28 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p239-b7", - "global_id": 6419, - "bbox": [ - 162.47, - 263.34, - 170.47, - 271.34 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p239-b8", - "global_id": 6420, - "bbox": [ - 166.47, - 241.59, - 170.47, - 249.59 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p239-b9", - "global_id": 6421, - "bbox": [ - 166.47, - 219.84, - 170.47, - 227.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p239-b10", - "global_id": 6422, - "bbox": [ - 166.47, - 198.1, - 170.47, - 206.1 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p239-b11", - "global_id": 6423, - "bbox": [ - 152.42, - 199.58, - 160.72, - 267.32 - ], - "text": "y(t) = ∫ h(τ)×(t–τ) dτ", - "type": "text" - }, - { - "block_id": "p239-b12", - "global_id": 6424, - "bbox": [ - 151.5, - 300.07, - 367.51, - 309.68 - ], - "text": "Figure 2.27 Graphical convolution at step t = 0.75 second.", - "type": "text" - }, - { - "block_id": "p239-b13", - "global_id": 6425, - "bbox": [ - 162.47, - 412.17, - 170.47, - 420.17 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p239-b14", - "global_id": 6426, - "bbox": [ - 166.47, - 383.5, - 170.47, - 391.5 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p239-b15", - "global_id": 6427, - "bbox": [ - 166.47, - 354.84, - 170.47, - 362.84 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p239-b16", - "global_id": 6428, - "bbox": [ - 169.62, - 524.14, - 511.2, - 545.33 - ], - "text": "–1\n–0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n3.5\n4\nt", - "type": "text" - }, - { - "block_id": "p239-b17", - "global_id": 6429, - "bbox": [ - 169.58, - 421.85, - 511.2, - 442.72 - ], - "text": "–1\n–0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n3.5\n4\nτ", - "type": "text" - }, - { - "block_id": "p239-b18", - "global_id": 6430, - "bbox": [ - 162.47, - 514.91, - 170.47, - 522.91 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p239-b19", - "global_id": 6431, - "bbox": [ - 166.47, - 493.16, - 170.47, - 501.16 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p239-b20", - "global_id": 6432, - "bbox": [ - 166.47, - 471.42, - 170.47, - 479.42 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p239-b21", - "global_id": 6433, - "bbox": [ - 166.47, - 449.67, - 170.47, - 457.67 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p239-b22", - "global_id": 6434, - "bbox": [ - 267.22, - 339.68, - 418.04, - 347.91 - ], - "text": "h(τ) [solid], x(t–τ) [dashed], h(τ)×(t–τ) [gray]", - "type": "text" - }, - { - "block_id": "p239-b23", - "global_id": 6435, - "bbox": [ - 152.42, - 450.27, - 160.72, - 518.01 - ], - "text": "y(t) = ∫ h(τ)×(t–τ) dτ", - "type": "text" - }, - { - "block_id": "p239-b24", - "global_id": 6436, - "bbox": [ - 151.5, - 551.63, - 371.01, - 561.24 - ], - "text": "Figure 2.28 Graphical convolution at step t = 2.25 seconds.", - "type": "text" - }, - { - "block_id": "p239-b25", - "global_id": 6437, - "bbox": [ - 127.59, - 588.96, - 516.13, - 634.79 - ], - "text": "command forces MATLAB to update the graphics window for each loop iteration. Although slow,\nthis creates an animation-like effect. Replacing drawnow with the pause command allows users\nto manually step through the convolution process. The pause command still forces the graphics\nwindow to update, but the program will not continue until a key is pressed.", - "type": "text" - } - ] - }, - { - "page_num": 240, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p240-b0", - "global_id": 6438, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "220\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p240-b1", - "global_id": 6439, - "bbox": [ - 136.73, - 160.03, - 485.46, - 177.82 - ], - "text": "–1\n–0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n3.5\n4\n–2", - "type": "text" - }, - { - "block_id": "p240-b2", - "global_id": 6440, - "bbox": [ - 140.73, - 131.92, - 144.73, - 139.92 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p240-b3", - "global_id": 6441, - "bbox": [ - 140.73, - 103.26, - 144.73, - 111.26 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p240-b4", - "global_id": 6442, - "bbox": [ - 144.13, - 272.56, - 485.46, - 293.75 - ], - "text": "–1\n–0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n3.5\n4\nt", - "type": "text" - }, - { - "block_id": "p240-b5", - "global_id": 6443, - "bbox": [ - 314.43, - 180.15, - 317.94, - 188.15 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p240-b6", - "global_id": 6444, - "bbox": [ - 136.73, - 263.34, - 144.73, - 271.34 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p240-b7", - "global_id": 6445, - "bbox": [ - 140.73, - 241.59, - 144.73, - 249.59 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p240-b8", - "global_id": 6446, - "bbox": [ - 140.73, - 219.84, - 144.73, - 227.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p240-b9", - "global_id": 6447, - "bbox": [ - 140.73, - 198.1, - 144.73, - 206.1 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p240-b10", - "global_id": 6448, - "bbox": [ - 242.7, - 87.26, - 393.52, - 95.48 - ], - "text": "h(τ) [solid], x(t–τ) [dashed], h(τ)×(t–τ) [gray]", - "type": "text" - }, - { - "block_id": "p240-b11", - "global_id": 6449, - "bbox": [ - 126.68, - 199.48, - 134.98, - 267.22 - ], - "text": "y(t) = ∫ h(τ)×(t–τ) dτ", - "type": "text" - }, - { - "block_id": "p240-b12", - "global_id": 6450, - "bbox": [ - 125.76, - 300.07, - 345.25, - 309.68 - ], - "text": "Figure 2.29 Graphical convolution at step t = 2.85 seconds.", - "type": "text" - }, - { - "block_id": "p240-b13", - "global_id": 6451, - "bbox": [ - 102.2, - 337.46, - 459.22, - 351.4 - ], - "text": "2.8 APPENDIX: DETERMINING THE IMPULSE RESPONSE", - "type": "text" - }, - { - "block_id": "p240-b14", - "global_id": 6452, - "bbox": [ - 101.84, - 357.29, - 490.39, - 379.31 - ], - "text": "In Eq. (2.13), we showed that for an LTIC system S specified by Eq. (2.11), the unit impulse\nresponse h(t) can be expressed as", - "type": "text" - }, - { - "block_id": "p240-b15", - "global_id": 6453, - "bbox": [ - 224.17, - 391.73, - 490.39, - 402.87 - ], - "text": "h(t) = b0δ(t) + characteristic modes\n(2.51)", - "type": "text" - }, - { - "block_id": "p240-b16", - "global_id": 6454, - "bbox": [ - 101.84, - 414.84, - 490.38, - 436.85 - ], - "text": "To determine the characteristic mode terms in Eq. (2.51), let us consider a system S0 whose input\nx(t) and the corresponding output w(t) are related by", - "type": "text" - }, - { - "block_id": "p240-b17", - "global_id": 6455, - "bbox": [ - 263.34, - 449.28, - 490.38, - 459.66 - ], - "text": "Q(D)w(t) = x(t)\n(2.52)", - "type": "text" - }, - { - "block_id": "p240-b18", - "global_id": 6456, - "bbox": [ - 101.84, - 472.08, - 490.39, - 543.72 - ], - "text": "Observe that both the systems S and S0 have the same characteristic polynomial; namely, Q(λ),\nand, consequently, the same characteristic modes. Moreover, S0 is the same as S with P(D) = 1, that\nis, b0 = 0. Therefore, according to Eq. (2.51), the impulse response of S0 consists of characteristic\nmode terms only without an impulse at t = 0. Let us denote this impulse response of S0 by yn(t).\nObserve that yn(t) consists of characteristic modes of S and therefore may be viewed as a zero-input\nresponse of S. Now yn(t) is the response of S0 to input δ(t). Therefore, according to Eq. (2.52),", - "type": "text" - }, - { - "block_id": "p240-b19", - "global_id": 6457, - "bbox": [ - 262.28, - 554.64, - 329.97, - 565.72 - ], - "text": "Q(D)yn(t) = δ(t)", - "type": "text" - }, - { - "block_id": "p240-b20", - "global_id": 6458, - "bbox": [ - 101.84, - 577.86, - 110.14, - 587.82 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p240-b21", - "global_id": 6459, - "bbox": [ - 203.6, - 586.6, - 388.64, - 601.86 - ], - "text": "(DN + a1DN−1 + · · · + aN1D + aN)yn(t) = δ(t)", - "type": "text" - }, - { - "block_id": "p240-b22", - "global_id": 6460, - "bbox": [ - 101.84, - 610.94, - 110.14, - 620.9 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p240-b23", - "global_id": 6461, - "bbox": [ - 183.38, - 622.08, - 198.07, - 634.08 - ], - "text": "y(N)", - "type": "text" - }, - { - "block_id": "p240-b24", - "global_id": 6462, - "bbox": [ - 187.8, - 622.08, - 252.41, - 636.44 - ], - "text": "n (t) + a1y(N−1)", - "type": "text" - }, - { - "block_id": "p240-b25", - "global_id": 6463, - "bbox": [ - 233.22, - 622.08, - 330.69, - 636.07 - ], - "text": "n\n(t) + · · · + aN−1y(1)", - "type": "text" - }, - { - "block_id": "p240-b26", - "global_id": 6464, - "bbox": [ - 322.0, - 623.8, - 408.86, - 636.44 - ], - "text": "n (t) + aNyn(t) = δ(t)", - "type": "text" - } - ] - }, - { - "page_num": 241, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p241-b0", - "global_id": 6465, - "bbox": [ - 414.16, - 62.89, - 516.14, - 71.98 - ], - "text": "2.9\nSummary\n221", - "type": "text" - }, - { - "block_id": "p241-b1", - "global_id": 6466, - "bbox": [ - 127.59, - 85.8, - 167.69, - 97.41 - ], - "text": "where y(k)", - "type": "text" - }, - { - "block_id": "p241-b2", - "global_id": 6467, - "bbox": [ - 127.59, - 87.03, - 516.14, - 109.36 - ], - "text": "n (t) represents the kth derivative of yn(t). The right-hand side contains a single impulse\nterm, δ(t). This is possible only if y(N−1)", - "type": "text" - }, - { - "block_id": "p241-b3", - "global_id": 6468, - "bbox": [ - 127.59, - 98.98, - 516.13, - 121.21 - ], - "text": "n\n(t) has a unit jump discontinuity at t = 0, so that\ny(N)", - "type": "text" - }, - { - "block_id": "p241-b4", - "global_id": 6469, - "bbox": [ - 127.59, - 110.93, - 516.12, - 133.97 - ], - "text": "n (t) = δ(t). Moreover, the lower-order terms cannot have any jump discontinuity because this\nwould mean the presence of the derivatives of δ(t). Therefore yn(0) = y(1)", - "type": "text" - }, - { - "block_id": "p241-b5", - "global_id": 6470, - "bbox": [ - 406.65, - 121.67, - 487.13, - 135.53 - ], - "text": "n (0) = · · · = y(N−2)", - "type": "text" - }, - { - "block_id": "p241-b6", - "global_id": 6471, - "bbox": [ - 127.59, - 122.9, - 516.14, - 145.93 - ], - "text": "n\n(0) = 0\n(no discontinuity at t = 0), and the N initial conditions on yn(t) are", - "type": "text" - }, - { - "block_id": "p241-b7", - "global_id": 6472, - "bbox": [ - 193.6, - 154.37, - 239.4, - 167.18 - ], - "text": "yn(0) = y(1)", - "type": "text" - }, - { - "block_id": "p241-b8", - "global_id": 6473, - "bbox": [ - 230.72, - 154.37, - 312.72, - 168.74 - ], - "text": "n (0) = · · · = y(N−2)", - "type": "text" - }, - { - "block_id": "p241-b9", - "global_id": 6474, - "bbox": [ - 293.54, - 154.37, - 420.35, - 168.36 - ], - "text": "n\n(0) = 0\nand\ny(N−1)", - "type": "text" - }, - { - "block_id": "p241-b10", - "global_id": 6475, - "bbox": [ - 401.16, - 156.11, - 516.13, - 168.36 - ], - "text": "n\n(0) = 1\n(2.53)", - "type": "text" - }, - { - "block_id": "p241-b11", - "global_id": 6476, - "bbox": [ - 127.59, - 177.35, - 516.13, - 199.68 - ], - "text": "This discussion means that yn(t) is the zero-input response of the system S subject to initial\nconditions [Eq. (2.53)].", - "type": "text" - }, - { - "block_id": "p241-b12", - "global_id": 6477, - "bbox": [ - 127.59, - 201.27, - 516.12, - 223.6 - ], - "text": "We now show that for the same input x(t) to both systems, S and S0, their respective outputs\ny(t) and w(t) are related by", - "type": "text" - }, - { - "block_id": "p241-b13", - "global_id": 6478, - "bbox": [ - 289.66, - 225.17, - 516.13, - 235.55 - ], - "text": "y(t) = P(D)w(t)\n(2.54)", - "type": "text" - }, - { - "block_id": "p241-b14", - "global_id": 6479, - "bbox": [ - 127.59, - 243.78, - 434.42, - 254.15 - ], - "text": "To prove this result, we operate on both sides of Eq. (2.52) by P(D) to obtain", - "type": "text" - }, - { - "block_id": "p241-b15", - "global_id": 6480, - "bbox": [ - 268.38, - 265.02, - 375.34, - 275.3 - ], - "text": "Q(D)P(D)w(t) = P(D)x(t)", - "type": "text" - }, - { - "block_id": "p241-b16", - "global_id": 6481, - "bbox": [ - 127.6, - 286.7, - 429.2, - 296.66 - ], - "text": "Comparison of this equation with Eq. (2.2) leads immediately to Eq. (2.54).", - "type": "text" - }, - { - "block_id": "p241-b17", - "global_id": 6482, - "bbox": [ - 127.59, - 298.23, - 516.15, - 356.43 - ], - "text": "Now if the input x(t) = δ(t), the output of S0 is yn(t), and the output of S, according to\nEq. (2.54), is P(D)yn(t). This output is h(t), the unit impulse response of S. Note, however, that\nbecause it is an impulse response of a causal system S0, the function yn(t) is causal. To incorporate\nthis fact we must represent this function as yn(t)u(t). Now it follows that h(t), the unit impulse\nresponse of the system S, is given by", - "type": "text" - }, - { - "block_id": "p241-b18", - "global_id": 6483, - "bbox": [ - 277.5, - 367.31, - 516.13, - 378.39 - ], - "text": "h(t) = P(D)[yn(t)u(t)]\n(2.55)", - "type": "text" - }, - { - "block_id": "p241-b19", - "global_id": 6484, - "bbox": [ - 127.59, - 388.56, - 516.15, - 410.9 - ], - "text": "where yn(t) is a linear combination of the characteristic modes of the system subject to initial\nconditions (2.53).", - "type": "text" - }, - { - "block_id": "p241-b20", - "global_id": 6485, - "bbox": [ - 127.59, - 412.47, - 516.14, - 495.35 - ], - "text": "The right-hand side of Eq. (2.55) is a linear combination of the derivatives of yn(t)u(t).\nEvaluating these derivatives is clumsy and inconvenient because of the presence of u(t). The\nderivatives will generate an impulse and its derivatives at the origin. Fortunately when M ≤N\n[Eq. (2.11)], we can avoid this difficulty by using the observation in Eq. (2.51), which asserts that\nat t = 0 (the origin), h(t) = b0δ(t). Therefore, we need not bother to find h(t) at the origin. This\nsimplification means that instead of deriving P(D)[yn(t)u(t)], we can derive P(D)yn(t) and add to\nit the term b0δ(t) so that", - "type": "text" - }, - { - "block_id": "p241-b21", - "global_id": 6486, - "bbox": [ - 251.24, - 505.46, - 392.48, - 531.55 - ], - "text": "h(t) = b0δ(t) + P(D)yn(t)\nt ≥0\n= b0δ(t) + [P(D)yn(t)]u(t)", - "type": "text" - }, - { - "block_id": "p241-b22", - "global_id": 6487, - "bbox": [ - 127.59, - 541.66, - 516.13, - 563.99 - ], - "text": "This expression is valid when M ≤N [the form given in Eq. (2.11)]. When M > N, Eq. (2.55)\nshould be used.", - "type": "text" - }, - { - "block_id": "p241-b23", - "global_id": 6488, - "bbox": [ - 127.94, - 592.93, - 220.22, - 606.88 - ], - "text": "2.9 SUMMARY", - "type": "text" - }, - { - "block_id": "p241-b24", - "global_id": 6489, - "bbox": [ - 127.59, - 612.86, - 516.17, - 634.79 - ], - "text": "This chapter discusses time-domain analysis of LTIC systems. The total response of a linear system\nis a sum of the zero-input response and zero-state response. The zero-input response is the system", - "type": "text" - } - ] - }, - { - "page_num": 242, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p242-b0", - "global_id": 6490, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "222\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p242-b1", - "global_id": 6491, - "bbox": [ - 101.84, - 85.82, - 490.4, - 131.64 - ], - "text": "response generated only by the internal conditions (initial conditions) of the system, assuming that\nthe external input is zero; hence the adjective “zero-input.” The zero-state response is the system\nresponse generated by the external input, assuming that all initial conditions are zero, that is, when\nthe system is in zero state.", - "type": "text" - }, - { - "block_id": "p242-b2", - "global_id": 6492, - "bbox": [ - 101.84, - 133.64, - 490.41, - 191.42 - ], - "text": "Every system can sustain certain forms of response on its own with no external input (zero\ninput). These forms are intrinsic characteristics of the system; that is, they do not depend on any\nexternal input. For this reason they are called characteristic modes of the system. Needless to say,\nthe zero-input response is made up of characteristic modes chosen in a combination required to\nsatisfy the initial conditions of the system. For an Nth-order system, there are N distinct modes.", - "type": "text" - }, - { - "block_id": "p242-b3", - "global_id": 6493, - "bbox": [ - 101.84, - 193.41, - 490.41, - 263.15 - ], - "text": "The unit impulse function is an idealized mathematical model of a signal that cannot be\ngenerated in practice.† Nevertheless, introduction of such a signal as an intermediary is very\nhelpful in analysis of signals and systems. The unit impulse response of a system is a combination\nof the characteristic modes of the system‡ because the impulse δ(t) = 0 for t > 0. Therefore, the\nsystem response for t > 0 must necessarily be a zero-input response, which, as seen earlier, is a\ncombination of characteristic modes.", - "type": "text" - }, - { - "block_id": "p242-b4", - "global_id": 6494, - "bbox": [ - 101.84, - 265.15, - 490.41, - 382.71 - ], - "text": "The zero-state response (response due to external input) of a linear system can be obtained by\nbreaking the input into simpler components and then adding the responses to all the components.\nIn this chapter we represent an arbitrary input x(t) as a sum of narrow rectangular pulses [staircase\napproximation of x(t)]. In the limit as the pulse width →0, the rectangular pulse components\napproach impulses. Knowing the impulse response of the system, we can find the system response\nto all the impulse components and add them to yield the system response to the input x(t). The sum\nof the responses to the impulse components is in the form of an integral, known as the convolution\nintegral. The system response is obtained as the convolution of the input x(t) with the system’s\nimpulse response h(t). Therefore, the knowledge of the system’s impulse response allows us to\ndetermine the system response to any arbitrary input.", - "type": "text" - }, - { - "block_id": "p242-b5", - "global_id": 6495, - "bbox": [ - 101.85, - 381.08, - 490.39, - 430.52 - ], - "text": "LTIC systems have a very special relationship to the everlasting exponential signal est because\nthe response of an LTIC system to such an input signal is the same signal within a multiplicative\nconstant. The response of an LTIC system to the everlasting exponential input est is H(s)est, where\nH(s) is the transfer function of the system.", - "type": "text" - }, - { - "block_id": "p242-b6", - "global_id": 6496, - "bbox": [ - 101.85, - 432.52, - 490.43, - 490.3 - ], - "text": "If every bounded input results in a bounded output, the system is stable in the\nbounded-input/bounded-output (BIBO) sense. An LTIC system is BIBO-stable if and only if its\nimpulse response is absolutely integrable. Otherwise, it is BIBO-unstable. BIBO stability is a\nstability seen from external terminals of the system. Hence, it is also called external stability or\nzero-state stability.", - "type": "text" - }, - { - "block_id": "p242-b7", - "global_id": 6497, - "bbox": [ - 101.85, - 492.29, - 490.41, - 573.99 - ], - "text": "In contrast, internal stability (or the zero-input stability) examines the system stability from\ninside. When some initial conditions are applied to a system in zero state, then, if the system\neventually returns to zero state, the system is said to be stable in the asymptotic or Lyapunov\nsense. If the system’s response increases without bound, it is unstable. If the system does not\ngo to zero state and the response does not increase indefinitely, the system is marginally stable.\nThe internal stability criterion, in terms of the location of a system’s characteristic roots, can be\nsummarized as follows:", - "type": "text" - }, - { - "block_id": "p242-b8", - "global_id": 6498, - "bbox": [ - 101.84, - 598.98, - 490.37, - 633.41 - ], - "text": "† However, it can be closely approximated by a narrow pulse of unit area and having a width that is much\nsmaller than the time constant of an LTIC system in which it is used.\n‡ There is the possibility of an impulse in addition to the characteristic modes.", - "type": "text" - } - ] - }, - { - "page_num": 243, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p243-b0", - "global_id": 6499, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n223", - "type": "text" - }, - { - "block_id": "p243-b1", - "global_id": 6500, - "bbox": [ - 144.52, - 85.82, - 516.17, - 155.56 - ], - "text": "1. An LTIC system is asymptotically stable if, and only if, all the characteristic roots are in\nthe LHP. The roots may be repeated or unrepeated.\n2. An LTIC system is unstable if, and only if, either one or both of the following conditions\nexist: (i) at least one root is in the RHP; (ii) there are repeated roots on the imaginary axis.\n3. An LTIC system is marginally stable if, and only if, there are no roots in the RHP, and\nthere are some unrepeated roots on the imaginary axis.", - "type": "text" - }, - { - "block_id": "p243-b2", - "global_id": 6501, - "bbox": [ - 127.59, - 163.53, - 516.12, - 221.31 - ], - "text": "It is possible for a system to be externally (BIBO) stable but internally unstable. When\na system is controllable and observable, its external and internal descriptions are equivalent.\nHence, external (BIBO) and internal (asymptotic) stabilities are equivalent and provide the same\ninformation. Such a BIBO-stable system is also asymptotically stable, and vice versa. Similarly, a\nBIBO-unstable system is either marginally stable or asymptotically unstable system.", - "type": "text" - }, - { - "block_id": "p243-b3", - "global_id": 6502, - "bbox": [ - 127.59, - 223.3, - 516.15, - 316.95 - ], - "text": "The characteristic behavior of a system is extremely important because it determines not only\nthe system response to internal conditions (zero-input behavior), but also the system response\nto external inputs (zero-state behavior) and the system stability. The system response to external\ninputs is determined by the impulse response, which itself is made up of characteristic modes. The\nwidth of the impulse response is called the time constant of the system, which indicates how fast\nthe system can respond to an input. The time constant plays an important role in determining such\ndiverse system behaviors as the response time and filtering properties of the system, dispersion of\npulses, and the rate of pulse transmission through the system.", - "type": "text" - }, - { - "block_id": "p243-b4", - "global_id": 6503, - "bbox": [ - 127.86, - 341.22, - 215.04, - 352.18 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p243-b5", - "global_id": 6504, - "bbox": [ - 127.59, - 359.24, - 449.64, - 368.3 - ], - "text": "1.\nLathi, B. P., Signals and Systems. Berkeley-Cambridge Press, Carmichael, CA, 1987.", - "type": "text" - }, - { - "block_id": "p243-b6", - "global_id": 6505, - "bbox": [ - 127.59, - 373.19, - 430.9, - 382.25 - ], - "text": "2.\nMason, S. J., Electronic Circuits, Signals, and Systems. Wiley, New York, 1960.", - "type": "text" - }, - { - "block_id": "p243-b7", - "global_id": 6506, - "bbox": [ - 127.59, - 387.14, - 396.6, - 396.2 - ], - "text": "3.\nKailath, T., Linear System. Prentice-Hall, Englewood Cliffs, NJ, 1980.", - "type": "text" - }, - { - "block_id": "p243-b8", - "global_id": 6507, - "bbox": [ - 127.59, - 401.08, - 516.12, - 421.1 - ], - "text": "4.\nLathi, B. P., Modern Digital and Analog Communication Systems, 3rd ed. Oxford University Press, New\nYork, 1998.", - "type": "text" - }, - { - "block_id": "p243-b9", - "global_id": 6508, - "bbox": [ - 106.67, - 447.17, - 217.26, - 464.1 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p243-b10", - "global_id": 6509, - "bbox": [ - 87.82, - 474.34, - 288.99, - 517.3 - ], - "text": "2.2-1\nDetermine the constants c1, c2, λ1, and λ2 for\neach of the following second-order systems,\nwhich have zero-input responses of the form\nyzir(t) = c1eλ1t + c2eλ2t.", - "type": "text" - }, - { - "block_id": "p243-b11", - "global_id": 6510, - "bbox": [ - 116.96, - 518.18, - 288.98, - 527.52 - ], - "text": "(a) ¨y(t) + 2˙y(t) + 5y(t) = ¨x(t) −5x(t) with", - "type": "text" - }, - { - "block_id": "p243-b12", - "global_id": 6511, - "bbox": [ - 116.46, - 529.14, - 288.98, - 549.43 - ], - "text": "yzir(0) = 2 and ˙yzir(0) = 0.\n(b) ¨y(t) + 2˙y(t) + 5y(t) = ¨x(t) −5x(t) with", - "type": "text" - }, - { - "block_id": "p243-b13", - "global_id": 6512, - "bbox": [ - 116.96, - 551.06, - 233.86, - 572.93 - ], - "text": "yzir(0) = 4 and ˙yzir(0) = −1.\n(c)\nd2", - "type": "text" - }, - { - "block_id": "p243-b14", - "global_id": 6513, - "bbox": [ - 133.09, - 562.16, - 175.66, - 576.24 - ], - "text": "dt2 y(t) + 2 d", - "type": "text" - }, - { - "block_id": "p243-b15", - "global_id": 6514, - "bbox": [ - 116.46, - 563.59, - 288.98, - 594.85 - ], - "text": "dty(t) = x(t) with yzir(0) = 1 and\n˙yzir(0) = 2.\n(d) (D2 +2D+10){y(t)} = (D5 −D){x(t)} with", - "type": "text" - }, - { - "block_id": "p243-b16", - "global_id": 6515, - "bbox": [ - 116.97, - 596.47, - 204.95, - 616.76 - ], - "text": "yzir(0) = ˙yzir(0) = 1.\n(e) (D2 + 7", - "type": "text" - }, - { - "block_id": "p243-b17", - "global_id": 6516, - "bbox": [ - 131.9, - 606.07, - 288.98, - 628.47 - ], - "text": "2D + 3\n2){y(t)} = (D + 2){x(t)} with\nyzir(0) = 3 and ¨yzir(0) = −8. [Caution: The", - "type": "text" - }, - { - "block_id": "p243-b18", - "global_id": 6517, - "bbox": [ - 345.11, - 474.6, - 516.12, - 506.67 - ], - "text": "second IC is given in terms of the second\nderivative, not the first derivative].\n(f) 13y(t) + 4 d", - "type": "text" - }, - { - "block_id": "p243-b19", - "global_id": 6518, - "bbox": [ - 397.28, - 494.37, - 435.44, - 509.19 - ], - "text": "dty(t) + d2", - "type": "text" - }, - { - "block_id": "p243-b20", - "global_id": 6519, - "bbox": [ - 428.42, - 495.9, - 500.54, - 509.98 - ], - "text": "dt2 y(t) = 2x(t) −4 d", - "type": "text" - }, - { - "block_id": "p243-b21", - "global_id": 6520, - "bbox": [ - 359.05, - 497.33, - 516.14, - 539.54 - ], - "text": "dtx(t)\nwith yzir(0) = 3 and ¨yzir(0) = −15. [Caution:\nThe second IC is given in terms of the\nsecond derivative, not the first derivative].", - "type": "text" - }, - { - "block_id": "p243-b22", - "global_id": 6521, - "bbox": [ - 314.97, - 544.76, - 516.13, - 575.72 - ], - "text": "2.2-2\nConsider a linear time-invariant system with\ninput x(t) and output y(t) that is described by the\ndifferential equation", - "type": "text" - }, - { - "block_id": "p243-b23", - "global_id": 6522, - "bbox": [ - 358.69, - 586.67, - 501.06, - 599.72 - ], - "text": "(D + 1)(D2 −1){y(t)} = (D5 −1){x(t)}", - "type": "text" - }, - { - "block_id": "p243-b24", - "global_id": 6523, - "bbox": [ - 343.61, - 614.37, - 516.14, - 634.67 - ], - "text": "Furthermore,\nassume\ny(0) = ˙y(0) = ¨y(0)\n= 1.", - "type": "text" - } - ] - }, - { - "page_num": 244, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p244-b0", - "global_id": 6524, - "bbox": [ - 60.0, - 60.36, - 419.1, - 69.45 - ], - "text": "224\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p244-b1", - "global_id": 6525, - "bbox": [ - 90.72, - 85.9, - 263.24, - 105.82 - ], - "text": "(a) What is the order of this system?\n(b) What are the characteristic roots of this", - "type": "text" - }, - { - "block_id": "p244-b2", - "global_id": 6526, - "bbox": [ - 91.22, - 107.81, - 263.25, - 128.48 - ], - "text": "system?\n(c) Determine the zero-input response yzir(t).", - "type": "text" - }, - { - "block_id": "p244-b3", - "global_id": 6527, - "bbox": [ - 106.15, - 129.74, - 185.59, - 138.7 - ], - "text": "Simplify your answer.", - "type": "text" - }, - { - "block_id": "p244-b4", - "global_id": 6528, - "bbox": [ - 62.08, - 143.3, - 263.23, - 174.57 - ], - "text": "2.2-3\nA real LTIC system with input x(t) and\noutput y(t) is described by the following\nconstant-coefficient linear differential equation:", - "type": "text" - }, - { - "block_id": "p244-b5", - "global_id": 6529, - "bbox": [ - 112.41, - 181.28, - 241.55, - 194.32 - ], - "text": "(D3 + 9D){y(t)} = (2D3 + 1){x(t)}.", - "type": "text" - }, - { - "block_id": "p244-b6", - "global_id": 6530, - "bbox": [ - 91.22, - 205.12, - 263.23, - 214.09 - ], - "text": "(a) What is the characteristic equation of this", - "type": "text" - }, - { - "block_id": "p244-b7", - "global_id": 6531, - "bbox": [ - 90.72, - 216.08, - 263.23, - 236.0 - ], - "text": "system?\n(b) What are the characteristic modes of this", - "type": "text" - }, - { - "block_id": "p244-b8", - "global_id": 6532, - "bbox": [ - 91.22, - 238.0, - 263.25, - 301.76 - ], - "text": "system?\n(c) Assuming\nyzir(0) = 4,\n˙yzir(0) = −18,\nand ¨yzir(0) = 0, determine this system’s\nzero-input response yzir(t). Simplify yzir(t)\nto include only real terms (i.e., no j’s should\nappear in your answer).", - "type": "text" - }, - { - "block_id": "p244-b9", - "global_id": 6533, - "bbox": [ - 62.08, - 306.66, - 250.33, - 315.71 - ], - "text": "2.2-4\nAn LTIC system is specified by the equation", - "type": "text" - }, - { - "block_id": "p244-b10", - "global_id": 6534, - "bbox": [ - 118.64, - 322.42, - 235.33, - 335.47 - ], - "text": "(D2 + 5D + 6)y(t) = (D + 1)x(t)", - "type": "text" - }, - { - "block_id": "p244-b11", - "global_id": 6535, - "bbox": [ - 90.72, - 357.22, - 263.24, - 399.8 - ], - "text": "(a) Find\nthe\ncharacteristic\npolynomial,\ncharacteristic equation, characteristic roots,\nand characteristic modes of this system.\n(b) Find y0(t), the zero-input component of", - "type": "text" - }, - { - "block_id": "p244-b12", - "global_id": 6536, - "bbox": [ - 106.15, - 400.69, - 263.24, - 421.73 - ], - "text": "the response y(t) for t ≥0, if the initial\nconditions are y0(0−) = 2 and ˙y0(0−) = −1.", - "type": "text" - }, - { - "block_id": "p244-b13", - "global_id": 6537, - "bbox": [ - 62.08, - 425.89, - 170.29, - 434.93 - ], - "text": "2.2-5\nRepeat Prob. 2.2-4 for", - "type": "text" - }, - { - "block_id": "p244-b14", - "global_id": 6538, - "bbox": [ - 129.11, - 441.65, - 224.84, - 454.69 - ], - "text": "(D2 + 4D + 4)y(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p244-b15", - "global_id": 6539, - "bbox": [ - 90.72, - 463.85, - 198.83, - 475.2 - ], - "text": "and y0(0−) = 3, ˙y0(0−) = −4.", - "type": "text" - }, - { - "block_id": "p244-b16", - "global_id": 6540, - "bbox": [ - 62.08, - 479.37, - 170.29, - 488.41 - ], - "text": "2.2-6\nRepeat Prob. 2.2-4 for", - "type": "text" - }, - { - "block_id": "p244-b17", - "global_id": 6541, - "bbox": [ - 127.64, - 498.82, - 226.33, - 508.16 - ], - "text": "D(D + 1)y(t) = (D + 2)x(t)", - "type": "text" - }, - { - "block_id": "p244-b18", - "global_id": 6542, - "bbox": [ - 90.73, - 517.32, - 182.86, - 528.66 - ], - "text": "and y0(0−) = ˙y0(0−) = 1.", - "type": "text" - }, - { - "block_id": "p244-b19", - "global_id": 6543, - "bbox": [ - 62.08, - 532.83, - 170.29, - 541.87 - ], - "text": "2.2-7\nRepeat Prob. 2.2-4 for", - "type": "text" - }, - { - "block_id": "p244-b20", - "global_id": 6544, - "bbox": [ - 126.77, - 548.59, - 227.2, - 561.64 - ], - "text": "(D2 + 9)y(t) = (3D + 2)x(t)", - "type": "text" - }, - { - "block_id": "p244-b21", - "global_id": 6545, - "bbox": [ - 90.72, - 570.8, - 191.84, - 582.13 - ], - "text": "and y0(0−) = 0, ˙y0(0−) = 6.", - "type": "text" - }, - { - "block_id": "p244-b22", - "global_id": 6546, - "bbox": [ - 62.08, - 586.3, - 170.29, - 595.34 - ], - "text": "2.2-8\nRepeat Prob. 2.2-4 for", - "type": "text" - }, - { - "block_id": "p244-b23", - "global_id": 6547, - "bbox": [ - 114.15, - 602.06, - 239.81, - 615.11 - ], - "text": "(D2 + 4D + 13)y(t) = 4(D + 2)x(t)", - "type": "text" - }, - { - "block_id": "p244-b24", - "global_id": 6548, - "bbox": [ - 90.72, - 624.27, - 210.51, - 635.61 - ], - "text": "with y0(0−) = 5, ˙y0(0−) = 15.98.", - "type": "text" - }, - { - "block_id": "p244-b25", - "global_id": 6549, - "bbox": [ - 289.23, - 85.94, - 397.44, - 94.98 - ], - "text": "2.2-9\nRepeat Prob. 2.2-4 for", - "type": "text" - }, - { - "block_id": "p244-b26", - "global_id": 6550, - "bbox": [ - 351.05, - 104.64, - 457.21, - 117.69 - ], - "text": "D2(D + 1)y(t) = (D2 + 2)x(t)", - "type": "text" - }, - { - "block_id": "p244-b27", - "global_id": 6551, - "bbox": [ - 317.86, - 129.79, - 488.24, - 141.13 - ], - "text": "with y0(0−) = 4, ˙y0(0−) = 3, and ¨y0(0−) = −1.", - "type": "text" - }, - { - "block_id": "p244-b28", - "global_id": 6552, - "bbox": [ - 284.74, - 145.47, - 397.44, - 154.51 - ], - "text": "2.2-10\nRepeat Prob. 2.2-4 for", - "type": "text" - }, - { - "block_id": "p244-b29", - "global_id": 6553, - "bbox": [ - 342.55, - 164.18, - 465.71, - 177.22 - ], - "text": "(D + 1)(D2 + 5D + 6)y(t) = Dx(t)", - "type": "text" - }, - { - "block_id": "p244-b30", - "global_id": 6554, - "bbox": [ - 317.86, - 189.33, - 488.24, - 200.67 - ], - "text": "with y0(0−) = 2, ˙y0(0−) = −1, and ¨y0(0−) = 5.", - "type": "text" - }, - { - "block_id": "p244-b31", - "global_id": 6555, - "bbox": [ - 284.74, - 205.02, - 490.38, - 236.71 - ], - "text": "2.2-11\nA system is described by a constant-coefficient\nlinear differential equation and has zero-input\nresponse given by y0(t) = 2e−t + 3.", - "type": "text" - }, - { - "block_id": "p244-b32", - "global_id": 6556, - "bbox": [ - 318.37, - 237.97, - 490.39, - 246.94 - ], - "text": "(a) Is it possible for the system’s characteristic", - "type": "text" - }, - { - "block_id": "p244-b33", - "global_id": 6557, - "bbox": [ - 317.86, - 248.56, - 490.39, - 279.81 - ], - "text": "equation to be λ + 1 = 0? Justify your\nanswer.\n(b) Is it possible for the system’s characteristic", - "type": "text" - }, - { - "block_id": "p244-b34", - "global_id": 6558, - "bbox": [ - 333.31, - 281.81, - 383.68, - 290.78 - ], - "text": "equation to be", - "type": "text" - }, - { - "block_id": "p244-b35", - "global_id": 6559, - "bbox": [ - 385.71, - 273.9, - 393.3, - 282.86 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p244-b36", - "global_id": 6560, - "bbox": [ - 318.37, - 278.18, - 490.39, - 312.69 - ], - "text": "3(λ2 +λ) = 0? Justify your\nanswer.\n(c) Is it possible for the system’s characteristic", - "type": "text" - }, - { - "block_id": "p244-b37", - "global_id": 6561, - "bbox": [ - 333.31, - 311.05, - 490.38, - 334.61 - ], - "text": "equation to be λ(λ + 1)2 = 0? Justify your\nanswer.", - "type": "text" - }, - { - "block_id": "p244-b38", - "global_id": 6562, - "bbox": [ - 284.74, - 339.69, - 490.4, - 371.39 - ], - "text": "2.2-12\nConsider the circuit of Fig. P2.2-12. Using\noperator notation, this system can be described\nas (D + a1){y(t)} = (b0D + b1){x(t)}.", - "type": "text" - }, - { - "block_id": "p244-b39", - "global_id": 6563, - "bbox": [ - 318.37, - 372.55, - 490.39, - 382.95 - ], - "text": "(a) Determine the constants a1, b0, and b1 in", - "type": "text" - }, - { - "block_id": "p244-b40", - "global_id": 6564, - "bbox": [ - 317.86, - 383.51, - 490.39, - 415.46 - ], - "text": "terms of the system components R, Rf ,\nand C.\n(b) Assume that R = 300 k, Rf = 1.2 M, and", - "type": "text" - }, - { - "block_id": "p244-b41", - "global_id": 6565, - "bbox": [ - 333.31, - 416.1, - 490.39, - 437.14 - ], - "text": "C = 5 µF. What is the zero-input response\ny0(t) of this system, assuming vC(0) = 1 V?", - "type": "text" - }, - { - "block_id": "p244-b42", - "global_id": 6566, - "bbox": [ - 311.93, - 492.96, - 323.8, - 510.31 - ], - "text": "+\nx(t)", - "type": "text" - }, - { - "block_id": "p244-b43", - "global_id": 6567, - "bbox": [ - 314.76, - 511.09, - 320.97, - 519.06 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p244-b44", - "global_id": 6568, - "bbox": [ - 342.87, - 477.99, - 388.38, - 489.06 - ], - "text": "R\nC", - "type": "text" - }, - { - "block_id": "p244-b45", - "global_id": 6569, - "bbox": [ - 372.11, - 498.91, - 399.53, - 507.87 - ], - "text": "+vc(t)−", - "type": "text" - }, - { - "block_id": "p244-b46", - "global_id": 6570, - "bbox": [ - 422.12, - 488.73, - 426.78, - 494.71 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p244-b47", - "global_id": 6571, - "bbox": [ - 422.12, - 506.73, - 426.78, - 512.71 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p244-b48", - "global_id": 6572, - "bbox": [ - 436.76, - 457.07, - 443.29, - 465.78 - ], - "text": "Rf", - "type": "text" - }, - { - "block_id": "p244-b49", - "global_id": 6573, - "bbox": [ - 481.14, - 501.83, - 492.98, - 519.18 - ], - "text": "+\ny(t)", - "type": "text" - }, - { - "block_id": "p244-b50", - "global_id": 6574, - "bbox": [ - 483.96, - 520.22, - 490.18, - 528.19 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p244-b51", - "global_id": 6575, - "bbox": [ - 311.89, - 540.58, - 368.01, - 549.55 - ], - "text": "Figure P2.2-12", - "type": "text" - }, - { - "block_id": "p244-b52", - "global_id": 6576, - "bbox": [ - 289.22, - 571.26, - 490.39, - 635.1 - ], - "text": "2.3-1\nDetermine the characteristic equation, charac-\nteristic modes, and impulse response h(t) for\neach of the following real LTIC systems. Since\nthe systems are real, express each h(t) using\nonly real terms (i.e., no j’s should appear in your\nanswers).", - "type": "text" - } - ] - }, - { - "page_num": 245, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p245-b0", - "global_id": 6577, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n225", - "type": "text" - }, - { - "block_id": "p245-b1", - "global_id": 6578, - "bbox": [ - 116.44, - 83.82, - 253.31, - 107.39 - ], - "text": "(a) (D2 + 1){y(t)} = 2D{x(t)}\n(b) (D3 + D){y(t)} = (2D3 + 1){x(t)}", - "type": "text" - }, - { - "block_id": "p245-b2", - "global_id": 6579, - "bbox": [ - 116.96, - 107.62, - 140.12, - 119.91 - ], - "text": "(c)\nd2", - "type": "text" - }, - { - "block_id": "p245-b3", - "global_id": 6580, - "bbox": [ - 133.09, - 109.15, - 175.58, - 123.23 - ], - "text": "dt2 y(t) + 2 d", - "type": "text" - }, - { - "block_id": "p245-b4", - "global_id": 6581, - "bbox": [ - 171.47, - 110.57, - 247.25, - 122.43 - ], - "text": "dty(t) + 5y(t) = 8x(t)", - "type": "text" - }, - { - "block_id": "p245-b5", - "global_id": 6582, - "bbox": [ - 87.83, - 124.84, - 289.0, - 144.85 - ], - "text": "2.3-2\nFind the unit impulse response of a system\nspecified by the equation", - "type": "text" - }, - { - "block_id": "p245-b6", - "global_id": 6583, - "bbox": [ - 144.39, - 152.86, - 261.04, - 165.91 - ], - "text": "(D2 + 4D + 3)y(t) = (D + 5)x(t)", - "type": "text" - }, - { - "block_id": "p245-b7", - "global_id": 6584, - "bbox": [ - 87.79, - 177.94, - 196.0, - 186.98 - ], - "text": "2.3-3\nRepeat Prob. 2.3-2 for", - "type": "text" - }, - { - "block_id": "p245-b8", - "global_id": 6585, - "bbox": [ - 129.87, - 195.0, - 275.53, - 208.05 - ], - "text": "(D2 + 5D + 6)y(t) = (D2 + 7D + 11)x(t)", - "type": "text" - }, - { - "block_id": "p245-b9", - "global_id": 6586, - "bbox": [ - 87.81, - 220.08, - 288.93, - 240.07 - ], - "text": "2.3-4\nRepeat Prob. 2.3-2 for the first-order allpass\nfilter specified by the equation", - "type": "text" - }, - { - "block_id": "p245-b10", - "global_id": 6587, - "bbox": [ - 153.1, - 251.82, - 252.3, - 261.15 - ], - "text": "(D + 1)y(t) = −(D −1)x(t)", - "type": "text" - }, - { - "block_id": "p245-b11", - "global_id": 6588, - "bbox": [ - 87.79, - 273.19, - 288.95, - 293.18 - ], - "text": "2.3-5\nFind the unit impulse response of an LTIC\nsystem specified by the equation", - "type": "text" - }, - { - "block_id": "p245-b12", - "global_id": 6589, - "bbox": [ - 142.11, - 301.2, - 263.28, - 314.24 - ], - "text": "(D2 + 6D + 9)y(t) = (2D + 9)x(t)", - "type": "text" - }, - { - "block_id": "p245-b13", - "global_id": 6590, - "bbox": [ - 87.8, - 326.27, - 288.98, - 368.19 - ], - "text": "2.3-6\nDetermine and plot the unit impulse response\nh(t) of the op-amp circuit of Fig. P2.2-12,\nassuming that R = 300 k, Rf = 1.2 M, and\nC = 5 µF.", - "type": "text" - }, - { - "block_id": "p245-b14", - "global_id": 6591, - "bbox": [ - 87.82, - 372.82, - 288.99, - 404.08 - ], - "text": "2.3-7\nA causal LTIC system with input x(t) and output\ny(t) is described by the constant coefficient\nintegral equation", - "type": "text" - }, - { - "block_id": "p245-b15", - "global_id": 6592, - "bbox": [ - 143.14, - 419.99, - 164.86, - 429.24 - ], - "text": "y(t) +", - "type": "text" - }, - { - "block_id": "p245-b17", - "global_id": 6593, - "bbox": [ - 176.1, - 407.79, - 222.4, - 429.33 - ], - "text": "3y(t)dt +", - "type": "text" - }, - { - "block_id": "p245-b18", - "global_id": 6594, - "bbox": [ - 153.12, - 419.99, - 262.29, - 440.66 - ], - "text": "2y(t)dt =", - "type": "text" - }, - { - "block_id": "p245-b19", - "global_id": 6595, - "bbox": [ - 168.81, - 443.9, - 198.67, - 453.14 - ], - "text": "x(t)dt −", - "type": "text" - }, - { - "block_id": "p245-b21", - "global_id": 6596, - "bbox": [ - 221.62, - 443.9, - 245.35, - 453.23 - ], - "text": "x(t)dt.", - "type": "text" - }, - { - "block_id": "p245-b22", - "global_id": 6597, - "bbox": [ - 116.96, - 476.3, - 288.95, - 485.26 - ], - "text": "(a) Express this system as a constant coefficient", - "type": "text" - }, - { - "block_id": "p245-b23", - "global_id": 6598, - "bbox": [ - 116.46, - 487.26, - 288.98, - 518.15 - ], - "text": "linear\ndifferential\nequation\nin\nstandard\noperator form.\n(b) Determine the characteristic modes of this", - "type": "text" - }, - { - "block_id": "p245-b24", - "global_id": 6599, - "bbox": [ - 131.89, - 520.15, - 159.04, - 529.12 - ], - "text": "system.", - "type": "text" - }, - { - "block_id": "p245-b25", - "global_id": 6600, - "bbox": [ - 344.11, - 85.6, - 516.11, - 94.94 - ], - "text": "(c) Determine the impulse response h(t) of this", - "type": "text" - }, - { - "block_id": "p245-b26", - "global_id": 6601, - "bbox": [ - 359.03, - 96.94, - 386.18, - 105.9 - ], - "text": "system.", - "type": "text" - }, - { - "block_id": "p245-b27", - "global_id": 6602, - "bbox": [ - 314.96, - 120.15, - 516.13, - 163.1 - ], - "text": "2.4-1\nLet f(t) = h1(t)∗h2(t), where h1(t) and h2(t) are\nshown in Fig. P2.4-1. In the following, use the\ngraphical convolution procedure where you flip\nand shift h2(t).", - "type": "text" - }, - { - "block_id": "p245-b28", - "global_id": 6603, - "bbox": [ - 344.11, - 163.99, - 516.14, - 174.06 - ], - "text": "(a) Plot h1(τ) and h2(t −τ) as functions of τ.", - "type": "text" - }, - { - "block_id": "p245-b29", - "global_id": 6604, - "bbox": [ - 343.62, - 175.32, - 516.14, - 206.21 - ], - "text": "Clearly label the plots, including necessary\nfunction parameterizations.\n(b) Determine the (piecewise) regions of f(t)", - "type": "text" - }, - { - "block_id": "p245-b30", - "global_id": 6605, - "bbox": [ - 344.12, - 208.2, - 516.14, - 250.06 - ], - "text": "and set up the corresponding integrals\nthat describe f(t) in those regions. Do not\nevaluate the integrals, only set them up!\n(c) Determine f(1), which is f(t) evaluated at", - "type": "text" - }, - { - "block_id": "p245-b31", - "global_id": 6606, - "bbox": [ - 359.05, - 251.69, - 499.65, - 261.02 - ], - "text": "t = 1. Provide a number, not a formula.", - "type": "text" - }, - { - "block_id": "p245-b32", - "global_id": 6607, - "bbox": [ - 314.98, - 275.3, - 516.15, - 298.98 - ], - "text": "2.4-2\nConsider\nsignals\nh(t)\n=\nu(t + 3) −\n2u(t + 1) + u(t −1)\nand\nx(t) = cos(t)", - "type": "text" - }, - { - "block_id": "p245-b33", - "global_id": 6608, - "bbox": [ - 347.15, - 297.23, - 437.69, - 306.57 - ], - "text": "u(t −π/2) −u(t −3π/2)", - "type": "text" - }, - { - "block_id": "p245-b35", - "global_id": 6609, - "bbox": [ - 343.61, - 297.19, - 516.13, - 317.49 - ], - "text": ". Let y(t) = x(t) ∗\nh(t).", - "type": "text" - }, - { - "block_id": "p245-b36", - "global_id": 6610, - "bbox": [ - 344.11, - 319.1, - 516.12, - 329.78 - ], - "text": "(a) Determine the last time tlast that y(t) is", - "type": "text" - }, - { - "block_id": "p245-b37", - "global_id": 6611, - "bbox": [ - 343.61, - 330.44, - 516.14, - 362.66 - ], - "text": "nonzero. That is, find the smallest value\ntlast such that y(t) = 0 for all t > tlast.\n(b) Determine the approximate time tmax where", - "type": "text" - }, - { - "block_id": "p245-b38", - "global_id": 6612, - "bbox": [ - 359.05, - 362.94, - 427.65, - 372.28 - ], - "text": "y(t) is a maximum.", - "type": "text" - }, - { - "block_id": "p245-b39", - "global_id": 6613, - "bbox": [ - 314.97, - 386.55, - 516.13, - 439.73 - ], - "text": "2.4-3\nConsider\nsignals\nh(t)\n=\n−u(t + 2) +\n3u(t −1) −2u(t −\n5\n2)\nand\nx(t)\n=\nsin(t)[u(t + 2π) −u(t + π)].\nDetermine\nthe\napproximate time tmin where y(t) = x(t)∗h(t) is a\nminimum. Note, the minimum value of y(t)̸ = 0!", - "type": "text" - }, - { - "block_id": "p245-b40", - "global_id": 6614, - "bbox": [ - 314.97, - 454.01, - 516.13, - 529.1 - ], - "text": "2.4-4\nAn LTIC system has impulse response h(t) =\n3u(t −2). For input x(t) shown in Fig. P2.4-4,\nuse the graphical convolution procedure to\ndetermine\nyzsr(t) = h(t) ∗x(t).\nAccurately\nsketch yzsr(t). When solving for yzsr(t), flip\nand shift x(t) and explicitly show all integration\nsteps—even if apparently trivial!", - "type": "text" - }, - { - "block_id": "p245-b41", - "global_id": 6615, - "bbox": [ - 175.78, - 552.34, - 491.07, - 561.11 - ], - "text": "h1(t)\nh2(t)", - "type": "text" - }, - { - "block_id": "p245-b42", - "global_id": 6616, - "bbox": [ - 515.73, - 601.86, - 517.94, - 609.83 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p245-b43", - "global_id": 6617, - "bbox": [ - 293.81, - 575.02, - 296.02, - 582.99 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p245-b44", - "global_id": 6618, - "bbox": [ - 415.81, - 605.49, - 451.19, - 613.49 - ], - "text": "2\n1", - "type": "text" - }, - { - "block_id": "p245-b45", - "global_id": 6619, - "bbox": [ - 141.15, - 578.66, - 151.35, - 586.66 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p245-b46", - "global_id": 6620, - "bbox": [ - 496.32, - 605.52, - 500.31, - 613.49 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p245-b47", - "global_id": 6621, - "bbox": [ - 248.89, - 578.69, - 252.87, - 586.66 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p245-b48", - "global_id": 6622, - "bbox": [ - 197.22, - 563.27, - 201.21, - 571.24 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p245-b49", - "global_id": 6623, - "bbox": [ - 158.07, - 593.98, - 168.28, - 601.98 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p245-b50", - "global_id": 6624, - "bbox": [ - 475.82, - 580.9, - 479.8, - 588.88 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p245-b51", - "global_id": 6625, - "bbox": [ - 475.82, - 564.15, - 479.8, - 572.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p245-b52", - "global_id": 6626, - "bbox": [ - 127.59, - 626.82, - 179.22, - 635.79 - ], - "text": "Figure P2.4-1", - "type": "text" - } - ] - }, - { - "page_num": 246, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p246-b0", - "global_id": 6627, - "bbox": [ - 60.0, - 60.36, - 419.11, - 69.45 - ], - "text": "226\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p246-b1", - "global_id": 6628, - "bbox": [ - 145.08, - 94.41, - 156.95, - 102.38 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p246-b2", - "global_id": 6629, - "bbox": [ - 96.02, - 154.61, - 257.82, - 166.25 - ], - "text": "t\n2\n1\n1", - "type": "text" - }, - { - "block_id": "p246-b3", - "global_id": 6630, - "bbox": [ - 182.81, - 142.71, - 186.8, - 150.68 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p246-b4", - "global_id": 6631, - "bbox": [ - 203.84, - 158.28, - 207.83, - 166.25 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p246-b5", - "global_id": 6632, - "bbox": [ - 224.72, - 142.71, - 228.7, - 150.68 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p246-b6", - "global_id": 6633, - "bbox": [ - 245.62, - 158.43, - 249.61, - 166.4 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p246-b7", - "global_id": 6634, - "bbox": [ - 145.94, - 115.26, - 149.93, - 123.23 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p246-b8", - "global_id": 6635, - "bbox": [ - 147.14, - 130.82, - 150.13, - 142.85 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p246-b9", - "global_id": 6636, - "bbox": [ - 128.2, - 165.91, - 138.6, - 175.51 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p246-b10", - "global_id": 6637, - "bbox": [ - 135.6, - 171.95, - 138.59, - 177.93 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p246-b11", - "global_id": 6638, - "bbox": [ - 90.72, - 188.66, - 142.35, - 197.62 - ], - "text": "Figure P2.4-4", - "type": "text" - }, - { - "block_id": "p246-b12", - "global_id": 6639, - "bbox": [ - 62.08, - 222.76, - 263.24, - 264.68 - ], - "text": "2.4-5\nSuppose an LTIC system has impulse response\nh(t) and input x(t) = u(t). Figure P2.4-5 shows\nx(t) and h(t + 1), respectively. Be careful!\nFigure P2.4-5 shows h(t + 1), not h(t).", - "type": "text" - }, - { - "block_id": "p246-b13", - "global_id": 6640, - "bbox": [ - 99.75, - 294.75, - 244.91, - 303.06 - ], - "text": "x(t)\nh(t + 1)", - "type": "text" - }, - { - "block_id": "p246-b14", - "global_id": 6641, - "bbox": [ - 227.05, - 309.25, - 245.09, - 318.6 - ], - "text": "1 −t2", - "type": "text" - }, - { - "block_id": "p246-b15", - "global_id": 6642, - "bbox": [ - 73.61, - 339.13, - 271.52, - 350.48 - ], - "text": "t\nt\n0\n0\n−1\n−1\n1\n1\n2\n2", - "type": "text" - }, - { - "block_id": "p246-b16", - "global_id": 6643, - "bbox": [ - 89.13, - 304.86, - 214.22, - 317.43 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p246-b17", - "global_id": 6644, - "bbox": [ - 58.83, - 362.69, - 110.47, - 371.66 - ], - "text": "Figure P2.4-5", - "type": "text" - }, - { - "block_id": "p246-b18", - "global_id": 6645, - "bbox": [ - 90.71, - 396.49, - 263.26, - 427.75 - ], - "text": "(a) Is\nsystem\nh(t)\ncausal?\nMathematically\njustify your answer.\n(b) Use the graphical convolution procedure to", - "type": "text" - }, - { - "block_id": "p246-b19", - "global_id": 6646, - "bbox": [ - 106.14, - 429.37, - 263.24, - 471.59 - ], - "text": "determine yzsr(t) = x(t) ∗h(t). Accurately\nsketch yzsr(t). When solving for yzsr(t),\nflip and shift x(t) and explicitly show all\nintegration steps.", - "type": "text" - }, - { - "block_id": "p246-b20", - "global_id": 6647, - "bbox": [ - 289.22, - 85.95, - 490.4, - 116.91 - ], - "text": "2.4-6\nRepeat\nProb.\n2.4-5\nusing\nthe\nsignals\nof\nFig. P2.4-6, rather than those of Fig. P2.4-5. Be\ncareful! Figure P2.4-6 shows h(t −1), not h(t).", - "type": "text" - }, - { - "block_id": "p246-b21", - "global_id": 6648, - "bbox": [ - 325.21, - 146.96, - 462.36, - 155.26 - ], - "text": "x(t)\nh(t −1)", - "type": "text" - }, - { - "block_id": "p246-b22", - "global_id": 6649, - "bbox": [ - 460.84, - 180.46, - 466.19, - 189.72 - ], - "text": "t2", - "type": "text" - }, - { - "block_id": "p246-b23", - "global_id": 6650, - "bbox": [ - 300.34, - 191.61, - 485.4, - 203.03 - ], - "text": "t\nt\n0\n0\n−1\n−1\n1\n1\n2\n2", - "type": "text" - }, - { - "block_id": "p246-b24", - "global_id": 6651, - "bbox": [ - 314.9, - 157.5, - 431.92, - 169.95 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p246-b25", - "global_id": 6652, - "bbox": [ - 285.98, - 212.72, - 337.62, - 221.69 - ], - "text": "Figure P2.4-6", - "type": "text" - }, - { - "block_id": "p246-b26", - "global_id": 6653, - "bbox": [ - 289.22, - 247.27, - 490.39, - 311.11 - ], - "text": "2.4-7\nSuppose an LTIC system has impulse response\nh(t) and input x(t), both shown in Fig. P2.4-7.\nUse the graphical convolution procedure to\ndetermine yzsr(t) = x(t)∗h(t). Accurately sketch\nyzsr(t). When solving for yzsr(t), flip and shift\nx(t) and explicitly show all integration steps.", - "type": "text" - }, - { - "block_id": "p246-b27", - "global_id": 6654, - "bbox": [ - 289.22, - 316.29, - 490.38, - 359.24 - ], - "text": "2.4-8\nAn LTIC system has impulse response h(t), as\nshown in Fig. P2.4-8. Let t have units of seconds.\nLet the input be x(t) = u(−t −2) and designate\nthe output as yzsr(t) = x(t) ∗h(t).", - "type": "text" - }, - { - "block_id": "p246-b28", - "global_id": 6655, - "bbox": [ - 318.36, - 360.5, - 490.41, - 369.47 - ], - "text": "(a) Use the graphical convolution procedure", - "type": "text" - }, - { - "block_id": "p246-b29", - "global_id": 6656, - "bbox": [ - 317.85, - 371.08, - 490.4, - 402.34 - ], - "text": "where h(t) is flipped and shifted to deter-\nmine yzsr(t). Accurately plot your result.\n(b) Use the graphical convolution procedure", - "type": "text" - }, - { - "block_id": "p246-b30", - "global_id": 6657, - "bbox": [ - 333.29, - 403.96, - 490.34, - 424.99 - ], - "text": "where x(t) is flipped and shifted to deter-\nmine yzsr(t). Accurately plot your result.", - "type": "text" - }, - { - "block_id": "p246-b31", - "global_id": 6658, - "bbox": [ - 289.21, - 429.43, - 490.39, - 471.66 - ], - "text": "2.4-9\nIf c(t) = x(t) ∗g(t), then show that Ac = AxAg,\nwhere Ax,Ag, and Ac are the areas under x(t),\ng(t), and c(t), respectively. Verify this area\nproperty of convolution in Exs. 2.10 and 2.12.", - "type": "text" - }, - { - "block_id": "p246-b32", - "global_id": 6659, - "bbox": [ - 152.37, - 498.76, - 355.48, - 506.98 - ], - "text": "x(t)\nh(t)", - "type": "text" - }, - { - "block_id": "p246-b33", - "global_id": 6660, - "bbox": [ - 106.48, - 551.94, - 436.52, - 563.65 - ], - "text": "t\nt\n0\n0\n−2\n−2\n−1\n−1\n1\n1\n2\n3\n4\n4", - "type": "text" - }, - { - "block_id": "p246-b34", - "global_id": 6661, - "bbox": [ - 206.1, - 540.19, - 381.04, - 548.2 - ], - "text": "3\n2", - "type": "text" - }, - { - "block_id": "p246-b35", - "global_id": 6662, - "bbox": [ - 210.41, - 586.25, - 220.61, - 594.55 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p246-b36", - "global_id": 6663, - "bbox": [ - 314.79, - 508.74, - 318.77, - 516.71 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p246-b37", - "global_id": 6664, - "bbox": [ - 381.41, - 605.93, - 391.61, - 614.23 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p246-b38", - "global_id": 6665, - "bbox": [ - 101.84, - 626.82, - 153.48, - 635.79 - ], - "text": "Figure P2.4-7", - "type": "text" - } - ] - }, - { - "page_num": 247, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p247-b0", - "global_id": 6666, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n227", - "type": "text" - }, - { - "block_id": "p247-b1", - "global_id": 6667, - "bbox": [ - 83.34, - 85.64, - 288.99, - 138.82 - ], - "text": "2.4-10\nIf x(t) ∗g(t) = c(t), then show that x(at) ∗\ng(at) = |1/a|c(at). This time-scaling property\nof convolution states that if both x(t) and g(t)\nare time-scaled by a, their convolution is also\ntime-scaled by a (and multiplied by |1/a|).", - "type": "text" - }, - { - "block_id": "p247-b2", - "global_id": 6668, - "bbox": [ - 173.93, - 165.83, - 186.21, - 174.05 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p247-b3", - "global_id": 6669, - "bbox": [ - 124.87, - 212.45, - 286.67, - 224.06 - ], - "text": "t\n0\n−2\n−1\n1\n2\n3\n4\n5", - "type": "text" - }, - { - "block_id": "p247-b4", - "global_id": 6670, - "bbox": [ - 163.1, - 181.08, - 167.08, - 189.05 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p247-b5", - "global_id": 6671, - "bbox": [ - 116.46, - 236.83, - 168.11, - 245.8 - ], - "text": "Figure P2.4-8", - "type": "text" - }, - { - "block_id": "p247-b6", - "global_id": 6672, - "bbox": [ - 83.34, - 266.97, - 288.98, - 319.84 - ], - "text": "2.4-11\nShow that the convolution of an odd and an even\nfunction is an odd function and the convolution\nof two odd or two even functions is an even\nfunction. [Hint: Use the time-scaling property\nof convolution in Prob. 2.4-10.]", - "type": "text" - }, - { - "block_id": "p247-b7", - "global_id": 6673, - "bbox": [ - 83.34, - 324.87, - 288.99, - 399.67 - ], - "text": "2.4-12\nSuppose an LTIC system has impulse response\nh(t) = (1 −t)[u(t) −u(t −1)] and input x(t) =\nu(−t −1)+u(t −1). Use the graphical convolu-\ntion procedure to determine yzsr(t) = x(t) ∗h(t).\nAccurately sketch yzsr(t). When solving for\nyzsr(t), flip and shift h(t), explicitly show all\nintegration steps, and simplify your answer.", - "type": "text" - }, - { - "block_id": "p247-b8", - "global_id": 6674, - "bbox": [ - 83.34, - 403.14, - 288.99, - 413.73 - ], - "text": "2.4-13\nUsing direct integration, find e−atu(t) ∗e−btu(t).", - "type": "text" - }, - { - "block_id": "p247-b9", - "global_id": 6675, - "bbox": [ - 83.34, - 418.47, - 288.99, - 438.76 - ], - "text": "2.4-14\nUsing\ndirect\nintegration,\nfind\nu(t) ∗u(t),\ne−atu(t) ∗e−atu(t), and tu(t) ∗u(t).", - "type": "text" - }, - { - "block_id": "p247-b10", - "global_id": 6676, - "bbox": [ - 83.34, - 443.5, - 288.98, - 463.8 - ], - "text": "2.4-15\nUsing direct integration, find sin tu(t) ∗u(t) and\ncos tu(t) ∗u(t).", - "type": "text" - }, - { - "block_id": "p247-b11", - "global_id": 6677, - "bbox": [ - 83.34, - 468.83, - 287.99, - 477.87 - ], - "text": "2.4-16\nThe unit impulse response of an LTIC system is", - "type": "text" - }, - { - "block_id": "p247-b12", - "global_id": 6678, - "bbox": [ - 177.83, - 488.98, - 227.62, - 499.94 - ], - "text": "h(t) = e−tu(t)", - "type": "text" - }, - { - "block_id": "p247-b13", - "global_id": 6679, - "bbox": [ - 116.46, - 512.85, - 288.99, - 533.15 - ], - "text": "Find this system’s (zero-state) response y(t) if\nthe input x(t) is:", - "type": "text" - }, - { - "block_id": "p247-b14", - "global_id": 6680, - "bbox": [ - 116.46, - 534.78, - 157.17, - 555.07 - ], - "text": "(a) u(t)\n(b) e−tu(t)", - "type": "text" - }, - { - "block_id": "p247-b15", - "global_id": 6681, - "bbox": [ - 116.46, - 555.43, - 288.98, - 598.91 - ], - "text": "(c) e−2tu(t)\n(d) sin3tu(t)\nUse the convolution table (Table 2.1) to find\nyour answers.", - "type": "text" - }, - { - "block_id": "p247-b16", - "global_id": 6682, - "bbox": [ - 83.34, - 603.94, - 200.51, - 612.98 - ], - "text": "2.4-17\nRepeat Prob. 2.4-16 for", - "type": "text" - }, - { - "block_id": "p247-b17", - "global_id": 6683, - "bbox": [ - 158.75, - 622.09, - 246.7, - 635.14 - ], - "text": "h(t) = [2e−3t −e−2t]u(t)", - "type": "text" - }, - { - "block_id": "p247-b18", - "global_id": 6684, - "bbox": [ - 343.61, - 85.58, - 424.47, - 94.92 - ], - "text": "and if the input x(t) is:", - "type": "text" - }, - { - "block_id": "p247-b19", - "global_id": 6685, - "bbox": [ - 343.61, - 96.54, - 384.32, - 116.83 - ], - "text": "(a) u(t)\n(b) e−tu(t)", - "type": "text" - }, - { - "block_id": "p247-b20", - "global_id": 6686, - "bbox": [ - 344.12, - 117.19, - 387.56, - 127.79 - ], - "text": "(c) e−2tu(t)", - "type": "text" - }, - { - "block_id": "p247-b21", - "global_id": 6687, - "bbox": [ - 310.48, - 132.7, - 427.66, - 141.74 - ], - "text": "2.4-18\nRepeat Prob. 2.4-16 for", - "type": "text" - }, - { - "block_id": "p247-b22", - "global_id": 6688, - "bbox": [ - 389.31, - 151.26, - 470.43, - 162.31 - ], - "text": "h(t) = (1 −2t)e−2tu(t)", - "type": "text" - }, - { - "block_id": "p247-b23", - "global_id": 6689, - "bbox": [ - 343.61, - 173.54, - 419.58, - 182.87 - ], - "text": "and input x(t) = u(t).", - "type": "text" - }, - { - "block_id": "p247-b24", - "global_id": 6690, - "bbox": [ - 310.48, - 187.78, - 427.66, - 196.82 - ], - "text": "2.4-19\nRepeat Prob. 2.4-16 for", - "type": "text" - }, - { - "block_id": "p247-b25", - "global_id": 6691, - "bbox": [ - 390.08, - 204.72, - 469.66, - 217.4 - ], - "text": "h(t) = 4e−2t cos3tu(t)", - "type": "text" - }, - { - "block_id": "p247-b26", - "global_id": 6692, - "bbox": [ - 343.61, - 228.62, - 477.28, - 237.96 - ], - "text": "and each of the following inputs x(t):", - "type": "text" - }, - { - "block_id": "p247-b27", - "global_id": 6693, - "bbox": [ - 343.61, - 239.59, - 384.32, - 259.88 - ], - "text": "(a) u(t)\n(b) e−tu(t)", - "type": "text" - }, - { - "block_id": "p247-b28", - "global_id": 6694, - "bbox": [ - 310.48, - 264.79, - 427.66, - 273.83 - ], - "text": "2.4-20\nRepeat Prob. 2.4-16 for", - "type": "text" - }, - { - "block_id": "p247-b29", - "global_id": 6695, - "bbox": [ - 404.98, - 283.35, - 454.77, - 294.31 - ], - "text": "h(t) = e−tu(t)", - "type": "text" - }, - { - "block_id": "p247-b30", - "global_id": 6696, - "bbox": [ - 343.61, - 305.63, - 477.28, - 314.97 - ], - "text": "and each of the following inputs x(t):", - "type": "text" - }, - { - "block_id": "p247-b31", - "global_id": 6697, - "bbox": [ - 343.61, - 315.32, - 400.68, - 336.88 - ], - "text": "(a) e−2tu(t)\n(b) e−2(t−3)u(t)", - "type": "text" - }, - { - "block_id": "p247-b32", - "global_id": 6698, - "bbox": [ - 343.61, - 337.24, - 518.63, - 358.81 - ], - "text": "(c) e−2tu(t −3)\n(d) The gate pulse depicted in Fig. P2.4-20—and", - "type": "text" - }, - { - "block_id": "p247-b33", - "global_id": 6699, - "bbox": [ - 359.05, - 360.42, - 445.11, - 369.76 - ], - "text": "provide a sketch of y(t).", - "type": "text" - }, - { - "block_id": "p247-b34", - "global_id": 6700, - "bbox": [ - 373.92, - 389.88, - 385.02, - 397.96 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p247-b35", - "global_id": 6701, - "bbox": [ - 368.26, - 437.02, - 461.95, - 445.44 - ], - "text": "0\nt\n1", - "type": "text" - }, - { - "block_id": "p247-b36", - "global_id": 6702, - "bbox": [ - 363.6, - 399.91, - 367.6, - 407.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p247-b37", - "global_id": 6703, - "bbox": [ - 343.61, - 453.35, - 399.72, - 462.32 - ], - "text": "Figure P2.4-20", - "type": "text" - }, - { - "block_id": "p247-b38", - "global_id": 6704, - "bbox": [ - 310.48, - 482.07, - 516.12, - 502.08 - ], - "text": "2.4-21\nA first-order allpass filter impulse response is\ngiven by", - "type": "text" - }, - { - "block_id": "p247-b39", - "global_id": 6705, - "bbox": [ - 387.51, - 511.6, - 472.24, - 522.64 - ], - "text": "h(t) = −δ(t) + 2e−tu(t)", - "type": "text" - }, - { - "block_id": "p247-b40", - "global_id": 6706, - "bbox": [ - 344.11, - 535.25, - 516.14, - 544.21 - ], - "text": "(a) Find the zero-state response of this filter for", - "type": "text" - }, - { - "block_id": "p247-b41", - "global_id": 6707, - "bbox": [ - 343.61, - 544.77, - 516.12, - 566.13 - ], - "text": "the input etu(−t).\n(b) Sketch the input and the corresponding", - "type": "text" - }, - { - "block_id": "p247-b42", - "global_id": 6708, - "bbox": [ - 359.05, - 568.12, - 429.75, - 577.08 - ], - "text": "zero-state response.", - "type": "text" - }, - { - "block_id": "p247-b43", - "global_id": 6709, - "bbox": [ - 310.48, - 581.69, - 516.13, - 612.95 - ], - "text": "2.4-22\nFigure P2.4-22 shows the input x(t) and the\nimpulse response h(t) for an LTIC system. Let\nthe output be y(t).", - "type": "text" - }, - { - "block_id": "p247-b44", - "global_id": 6710, - "bbox": [ - 344.12, - 614.57, - 516.13, - 634.87 - ], - "text": "(a) By\ninspection\nof\nx(t)\nand\nh(t),\nfind\ny(−1),y(0),y(1),y(2),y(3),y(4),y(5),\nand", - "type": "text" - } - ] - }, - { - "page_num": 248, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p248-b0", - "global_id": 6711, - "bbox": [ - 60.0, - 60.36, - 419.1, - 69.45 - ], - "text": "228\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p248-b1", - "global_id": 6712, - "bbox": [ - 124.32, - 87.3, - 302.82, - 99.33 - ], - "text": "h(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p248-b2", - "global_id": 6713, - "bbox": [ - 112.0, - 132.54, - 351.98, - 140.7 - ], - "text": "t\n0\nt", - "type": "text" - }, - { - "block_id": "p248-b3", - "global_id": 6714, - "bbox": [ - 111.23, - 97.3, - 115.23, - 105.3 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p248-b4", - "global_id": 6715, - "bbox": [ - 145.13, - 132.39, - 333.6, - 140.7 - ], - "text": "3\n2\n3\n3", - "type": "text" - }, - { - "block_id": "p248-b5", - "global_id": 6716, - "bbox": [ - 281.75, - 92.38, - 285.75, - 100.38 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p248-b6", - "global_id": 6717, - "bbox": [ - 281.17, - 132.68, - 285.17, - 140.68 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p248-b7", - "global_id": 6718, - "bbox": [ - 101.84, - 154.65, - 157.97, - 163.62 - ], - "text": "Figure P2.4-22", - "type": "text" - }, - { - "block_id": "p248-b8", - "global_id": 6719, - "bbox": [ - 90.72, - 177.77, - 263.24, - 230.94 - ], - "text": "y(6). Thus, by merely examining x(t) and\nh(t), you are required to see what the result\nof convolution yields at t = −1, 0, 1, 2, 3, 4,\n5, and 6.\n(b) Find the system response to the input x(t).", - "type": "text" - }, - { - "block_id": "p248-b9", - "global_id": 6720, - "bbox": [ - 57.59, - 238.48, - 263.26, - 324.23 - ], - "text": "2.4-23\nThe zero-state response of an LTIC system to\nan input x(t) = 2e−2tu(t) is y(t) = [4e−2t +\n6e−3t]u(t). Find the impulse response of the sys-\ntem. [Hint: We have not yet developed a method\nof finding h(t) from the knowledge of the input\nand the corresponding output. Knowing the form\nof x(t) and y(t), you will have to make the best\nguess of the general form of h(t).]", - "type": "text" - }, - { - "block_id": "p248-b10", - "global_id": 6721, - "bbox": [ - 57.59, - 328.22, - 263.25, - 351.77 - ], - "text": "2.4-24\nSketch the functions x(t) = 1/(t2 + 1) and u(t).\nNow find x(t) ∗u(t) and sketch the result.", - "type": "text" - }, - { - "block_id": "p248-b11", - "global_id": 6722, - "bbox": [ - 57.59, - 359.02, - 263.24, - 379.32 - ], - "text": "2.4-25\nFigure P2.4-25 shows x(t) and g(t). Find and\nsketch c(t) = x(t) ∗g(t).", - "type": "text" - }, - { - "block_id": "p248-b12", - "global_id": 6723, - "bbox": [ - 57.6, - 386.56, - 263.24, - 406.86 - ], - "text": "2.4-26\nFind and sketch c(t) = x(t) ∗g(t) for the\nfunctions depicted in Fig. P2.4-26.", - "type": "text" - }, - { - "block_id": "p248-b13", - "global_id": 6724, - "bbox": [ - 57.6, - 414.1, - 263.23, - 434.4 - ], - "text": "2.4-27\nFind and sketch c(t) = x1(t) ∗x2(t) for the pairs\nof functions illustrated in Fig. P2.4-27.", - "type": "text" - }, - { - "block_id": "p248-b14", - "global_id": 6725, - "bbox": [ - 57.59, - 441.64, - 263.24, - 461.94 - ], - "text": "2.4-28\nUse Eq. (2.37) to find the convolution of x(t)\nand w(t), shown in Fig. P2.4-28.", - "type": "text" - }, - { - "block_id": "p248-b15", - "global_id": 6726, - "bbox": [ - 284.74, - 177.83, - 490.39, - 220.04 - ], - "text": "2.4-29\nDetermine H(s), the transfer function of an ideal\ntime delay of T seconds. Find your answer\nby two methods: using Eq. (2.39) and using\nEq. (2.40).", - "type": "text" - }, - { - "block_id": "p248-b16", - "global_id": 6727, - "bbox": [ - 284.74, - 229.24, - 490.39, - 249.53 - ], - "text": "2.4-30\nDetermine\ny(t) = x(t) ∗h(t)\nfor\nthe\nsignals\ndepicted in Fig. P2.4-30.", - "type": "text" - }, - { - "block_id": "p248-b17", - "global_id": 6728, - "bbox": [ - 284.74, - 259.02, - 490.4, - 322.86 - ], - "text": "2.4-31\nTwo linear time-invariant systems, each with\nimpulse response h(t), are connected in cascade.\nRefer to Fig. P2.4-31. Given input x(t) = u(t),\ndetermine y(1). That is, determine the step\nresponse at time t = 1 for the cascaded system\nshown.", - "type": "text" - }, - { - "block_id": "p248-b18", - "global_id": 6729, - "bbox": [ - 284.74, - 332.35, - 490.4, - 352.35 - ], - "text": "2.4-32\nConsider\nthe\nelectric\ncircuit\nshown\nin\nFig. P2.4-32.", - "type": "text" - }, - { - "block_id": "p248-b19", - "global_id": 6730, - "bbox": [ - 318.37, - 354.34, - 490.38, - 363.31 - ], - "text": "(a) Determine the differential equation that", - "type": "text" - }, - { - "block_id": "p248-b20", - "global_id": 6731, - "bbox": [ - 333.31, - 364.93, - 490.39, - 385.89 - ], - "text": "relates the input x(t) to output y(t). Recall\nthat iC(t) = C dvC(t)", - "type": "text" - }, - { - "block_id": "p248-b21", - "global_id": 6732, - "bbox": [ - 389.83, - 373.75, - 469.27, - 387.74 - ], - "text": "dt\nand vL(t) = L diL(t)", - "type": "text" - }, - { - "block_id": "p248-b22", - "global_id": 6733, - "bbox": [ - 317.86, - 376.25, - 490.38, - 396.18 - ], - "text": "dt .\n(b) Find the characteristic equation for this", - "type": "text" - }, - { - "block_id": "p248-b23", - "global_id": 6734, - "bbox": [ - 318.37, - 398.18, - 490.39, - 429.06 - ], - "text": "circuit, and express the root(s) of the\ncharacteristic equation in terms of L and C.\n(c) Determine the zero-input response given", - "type": "text" - }, - { - "block_id": "p248-b24", - "global_id": 6735, - "bbox": [ - 333.31, - 431.06, - 490.39, - 462.68 - ], - "text": "an initial capacitor voltage of one volt and\nan initial inductor current of zero amps.\nThat is, find y0(t) given vC(0) = 1 V and", - "type": "text" - }, - { - "block_id": "p248-b25", - "global_id": 6736, - "bbox": [ - 110.78, - 480.46, - 121.88, - 488.54 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p248-b26", - "global_id": 6737, - "bbox": [ - 201.37, - 517.64, - 203.6, - 525.64 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p248-b27", - "global_id": 6738, - "bbox": [ - 117.62, - 482.61, - 257.33, - 500.99 - ], - "text": "1\n1\nsin t", - "type": "text" - }, - { - "block_id": "p248-b28", - "global_id": 6739, - "bbox": [ - 148.44, - 515.47, - 153.78, - 523.47 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p248-b29", - "global_id": 6740, - "bbox": [ - 181.32, - 504.95, - 190.66, - 513.06 - ], - "text": "2p", - "type": "text" - }, - { - "block_id": "p248-b30", - "global_id": 6741, - "bbox": [ - 117.62, - 517.36, - 342.16, - 525.51 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p248-b31", - "global_id": 6742, - "bbox": [ - 266.36, - 480.55, - 277.92, - 488.63 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p248-b32", - "global_id": 6743, - "bbox": [ - 253.22, - 517.43, - 257.22, - 525.43 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p248-b33", - "global_id": 6744, - "bbox": [ - 101.84, - 542.18, - 157.97, - 551.14 - ], - "text": "Figure P2.4-25", - "type": "text" - }, - { - "block_id": "p248-b34", - "global_id": 6745, - "bbox": [ - 110.79, - 567.78, - 274.04, - 575.86 - ], - "text": "g(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p248-b35", - "global_id": 6746, - "bbox": [ - 118.44, - 566.79, - 256.6, - 584.64 - ], - "text": "1\n1\nsin t", - "type": "text" - }, - { - "block_id": "p248-b36", - "global_id": 6747, - "bbox": [ - 147.28, - 598.84, - 152.61, - 606.84 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p248-b37", - "global_id": 6748, - "bbox": [ - 185.81, - 588.77, - 195.15, - 596.87 - ], - "text": "2p", - "type": "text" - }, - { - "block_id": "p248-b38", - "global_id": 6749, - "bbox": [ - 200.5, - 601.0, - 342.18, - 610.72 - ], - "text": "2p\nt\nt", - "type": "text" - }, - { - "block_id": "p248-b39", - "global_id": 6750, - "bbox": [ - 101.84, - 625.88, - 157.97, - 634.84 - ], - "text": "Figure P2.4-26", - "type": "text" - } - ] - }, - { - "page_num": 249, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p249-b0", - "global_id": 6751, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n229", - "type": "text" - }, - { - "block_id": "p249-b1", - "global_id": 6752, - "bbox": [ - 207.92, - 158.22, - 216.8, - 166.22 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p249-b2", - "global_id": 6753, - "bbox": [ - 170.52, - 179.21, - 277.52, - 188.82 - ], - "text": "x1(t)\nx2(t)", - "type": "text" - }, - { - "block_id": "p249-b3", - "global_id": 6754, - "bbox": [ - 191.01, - 260.13, - 256.62, - 269.74 - ], - "text": "x2(t)\nx1(t)", - "type": "text" - }, - { - "block_id": "p249-b4", - "global_id": 6755, - "bbox": [ - 131.08, - 219.79, - 234.65, - 228.09 - ], - "text": "2\n2", - "type": "text" - }, - { - "block_id": "p249-b5", - 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"type": "text" - }, - { - "block_id": "p249-b40", - "global_id": 6791, - "bbox": [ - 344.64, - 362.88, - 461.62, - 374.5 - ], - "text": "e2t\net", - "type": "text" - }, - { - "block_id": "p249-b41", - "global_id": 6792, - "bbox": [ - 333.32, - 259.95, - 448.42, - 270.12 - ], - "text": "x2(t)\nx1(t)", - "type": "text" - }, - { - "block_id": "p249-b42", - "global_id": 6793, - "bbox": [ - 383.4, - 302.09, - 477.73, - 310.09 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p249-b43", - "global_id": 6794, - "bbox": [ - 424.32, - 267.76, - 428.32, - 275.76 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p249-b44", - "global_id": 6795, - "bbox": [ - 323.67, - 302.76, - 469.11, - 310.76 - ], - "text": "0\n3\n0", - "type": "text" - }, - { - "block_id": "p249-b45", - "global_id": 6796, - "bbox": [ - 322.91, - 267.76, - 326.91, - 275.76 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p249-b46", - "global_id": 6797, - "bbox": [ - 362.41, - 282.42, - 372.63, - 292.03 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p249-b47", - "global_id": 6798, - "bbox": [ - 336.97, - 97.07, - 488.62, - 106.67 - ], - "text": "x1(t)\nx2(t)", - "type": "text" - }, - { - "block_id": "p249-b48", - "global_id": 6799, - "bbox": [ - 323.67, - 137.52, - 477.73, - 146.45 - ], - "text": "5 3\nt\nt\n0", - "type": "text" - }, - { - "block_id": "p249-b49", - "global_id": 6800, - "bbox": [ - 320.93, - 115.17, - 325.82, - 123.17 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p249-b50", - "global_id": 6801, - "bbox": [ - 355.71, - 138.06, - 375.21, - 146.06 - ], - "text": "5\n3", - "type": "text" - }, - { - "block_id": "p249-b51", - "global_id": 6802, - "bbox": [ - 461.51, - 102.27, - 466.4, - 110.27 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p249-b52", - "global_id": 6803, - "bbox": [ - 463.61, - 138.56, - 467.61, - 146.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p249-b53", - "global_id": 6804, - "bbox": [ - 336.32, - 179.21, - 488.62, - 188.82 - ], - "text": "x1(t)\nx2(t)", - "type": "text" - }, - { - "block_id": "p249-b54", - "global_id": 6805, - "bbox": [ - 383.4, - 219.79, - 477.73, - 228.14 - ], - "text": "3\nt\nt", - "type": "text" - }, - { - "block_id": "p249-b55", - "global_id": 6806, - "bbox": [ - 464.11, - 180.91, - 468.11, - 188.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p249-b56", - "global_id": 6807, - "bbox": [ - 323.67, - 220.91, - 467.61, - 228.91 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p249-b57", - "global_id": 6808, - "bbox": [ - 323.41, - 185.91, - 327.41, - 193.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p249-b58", - "global_id": 6809, - "bbox": [ - 362.41, - 199.14, - 372.63, - 208.76 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p249-b59", - "global_id": 6810, - "bbox": [ - 143.66, - 274.69, - 164.03, - 292.88 - ], - "text": "1\nt2 1", - "type": "text" - }, - { - "block_id": "p249-b60", - "global_id": 6811, - "bbox": [ - 127.59, - 434.65, - 183.71, - 443.62 - ], - "text": "Figure P2.4-27", - "type": "text" - }, - { - "block_id": "p249-b61", - "global_id": 6812, - "bbox": [ - 177.95, - 534.12, - 320.5, - 542.12 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p249-b62", - "global_id": 6813, - "bbox": [ - 307.55, - 459.08, - 320.43, - 467.16 - ], - "text": "w(t)", - "type": "text" - }, - { - "block_id": "p249-b63", - "global_id": 6814, - "bbox": [ - 276.41, - 473.81, - 280.41, - 481.81 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p249-b64", - "global_id": 6815, - "bbox": [ - 277.4, - 500.97, - 281.4, - 508.97 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p249-b65", - "global_id": 6816, - "bbox": [ - 270.73, - 516.59, - 281.4, - 524.88 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p249-b66", - "global_id": 6817, - "bbox": [ - 306.04, - 490.2, - 360.59, - 509.33 - ], - "text": "2\n1\nt", - "type": "text" - }, - { - "block_id": "p249-b67", - "global_id": 6818, - "bbox": [ - 150.78, - 459.12, - 161.88, - 467.2 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p249-b68", - "global_id": 6819, - "bbox": [ - 137.78, - 501.61, - 141.78, - 509.61 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p249-b69", - "global_id": 6820, - "bbox": [ - 136.51, - 473.81, - 140.51, - 481.81 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p249-b70", - "global_id": 6821, - "bbox": [ - 199.68, - 501.32, - 221.59, - 509.61 - ], - "text": "2\nt", - "type": "text" - }, - { - "block_id": "p249-b71", - "global_id": 6822, - "bbox": [ - 127.59, - 551.03, - 183.71, - 560.0 - ], - "text": "Figure P2.4-28", - "type": "text" - }, - { - "block_id": "p249-b72", - "global_id": 6823, - "bbox": [ - 116.46, - 573.97, - 288.98, - 605.97 - ], - "text": "iL(0) = 0 A. [Hint: The coefficient(s) in\ny0(t) are independent of L and C.]\n(d) Plot y0(t) for t ≥0. Does the zero-input", - "type": "text" - }, - { - "block_id": "p249-b73", - "global_id": 6824, - "bbox": [ - 131.89, - 607.22, - 288.97, - 627.15 - ], - "text": "response, which is caused solely by initial\nconditions, ever “die out”?", - "type": "text" - }, - { - "block_id": "p249-b74", - "global_id": 6825, - "bbox": [ - 344.11, - 573.97, - 516.13, - 583.31 - ], - "text": "(e) Determine the total response y(t) to the", - "type": "text" - }, - { - "block_id": "p249-b75", - "global_id": 6826, - "bbox": [ - 359.04, - 583.67, - 516.13, - 627.15 - ], - "text": "input\nx(t) = e−tu(t).\nAssume\nan\ninitial\ninductor current of iL(0−) = 0 A, an initial\ncapacitor voltage of vC(0−) = 1 V, L = 1 H,\nand C = 1 F.", - "type": "text" - } - ] - }, - { - "page_num": 250, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p250-b0", - "global_id": 6827, - "bbox": [ - 60.0, - 60.36, - 419.1, - 69.45 - ], - "text": "230\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p250-b1", - "global_id": 6828, - "bbox": [ - 341.48, - 87.82, - 353.03, - 95.9 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p250-b2", - "global_id": 6829, - "bbox": [ - 296.5, - 141.2, - 389.71, - 151.78 - ], - "text": "t\n1\n–1", - "type": "text" - }, - { - "block_id": "p250-b3", - "global_id": 6830, - "bbox": [ - 325.61, - 101.95, - 329.61, - 109.95 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p250-b4", - "global_id": 6831, - "bbox": [ - 129.26, - 141.2, - 247.07, - 151.78 - ], - "text": "t\n1\n–1", - "type": "text" - }, - { - "block_id": "p250-b5", - "global_id": 6832, - "bbox": [ - 158.3, - 101.95, - 162.3, - 109.95 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p250-b6", - "global_id": 6833, - "bbox": [ - 226.43, - 143.78, - 230.43, - 151.78 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p250-b7", - "global_id": 6834, - "bbox": [ - 174.18, - 87.82, - 185.28, - 95.9 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p250-b8", - "global_id": 6835, - "bbox": [ - 101.84, - 165.65, - 157.97, - 174.62 - ], - "text": "Figure P2.4-30", - "type": "text" - }, - { - "block_id": "p250-b9", - "global_id": 6836, - "bbox": [ - 125.75, - 246.2, - 238.32, - 255.97 - ], - "text": "t\n1\n–1", - "type": "text" - }, - { - "block_id": "p250-b10", - "global_id": 6837, - "bbox": [ - 154.57, - 204.32, - 158.57, - 212.32 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p250-b11", - "global_id": 6838, - "bbox": [ - 220.11, - 247.97, - 224.11, - 255.97 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p250-b12", - "global_id": 6839, - "bbox": [ - 169.87, - 192.37, - 181.42, - 200.45 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p250-b13", - "global_id": 6840, - "bbox": [ - 268.59, - 221.1, - 383.7, - 229.18 - ], - "text": "x(t)\ny(t)\nh(t)\nh(t)", - "type": "text" - }, - { - "block_id": "p250-b14", - "global_id": 6841, - "bbox": [ - 101.84, - 269.77, - 157.97, - 278.74 - ], - "text": "Figure P2.4-31", - "type": "text" - }, - { - "block_id": "p250-b15", - "global_id": 6842, - "bbox": [ - 128.12, - 307.36, - 223.55, - 317.48 - ], - "text": "+\n+", - "type": "text" - }, - { - "block_id": "p250-b16", - "global_id": 6843, - "bbox": [ - 102.06, - 314.98, - 226.4, - 333.18 - ], - "text": "–\n–\nx(t)\ny(t)\nC", - "type": "text" - }, - { - "block_id": "p250-b17", - "global_id": 6844, - "bbox": [ - 164.26, - 302.23, - 168.71, - 310.23 - ], - "text": "L", - "type": "text" - }, - { - "block_id": "p250-b18", - "global_id": 6845, - "bbox": [ - 101.84, - 349.88, - 157.97, - 358.85 - ], - "text": "Figure P2.4-32", - "type": "text" - }, - { - "block_id": "p250-b19", - "global_id": 6846, - "bbox": [ - 172.47, - 445.08, - 362.25, - 453.08 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p250-b20", - "global_id": 6847, - "bbox": [ - 166.66, - 380.75, - 173.66, - 390.36 - ], - "text": "h1", - "type": "text" - }, - { - "block_id": "p250-b21", - "global_id": 6848, - "bbox": [ - 166.66, - 423.95, - 173.66, - 433.56 - ], - "text": "h2", - "type": "text" - }, - { - "block_id": "p250-b22", - "global_id": 6849, - "bbox": [ - 102.06, - 402.35, - 426.7, - 412.94 - ], - "text": "x(t)\nyp(t)\nx(t)\nh1\nh2\nys(t)", - "type": "text" - }, - { - "block_id": "p250-b23", - "global_id": 6850, - "bbox": [ - 101.84, - 462.0, - 157.97, - 470.96 - ], - "text": "Figure P2.4-33", - "type": "text" - }, - { - "block_id": "p250-b24", - "global_id": 6851, - "bbox": [ - 57.59, - 485.67, - 263.26, - 517.37 - ], - "text": "2.4-33\nTwo LTIC systems have impulse response\nfunctions given by h1(t) = (1−t)[u(t)−u(t−1)]\nand h2(t) = t[u(t + 2) −u(t −2)].", - "type": "text" - }, - { - "block_id": "p250-b25", - "global_id": 6852, - "bbox": [ - 91.23, - 518.25, - 263.23, - 528.33 - ], - "text": "(a) Carefully sketch the functions h1(t) and", - "type": "text" - }, - { - "block_id": "p250-b26", - "global_id": 6853, - "bbox": [ - 90.72, - 529.21, - 263.21, - 549.51 - ], - "text": "h2(t).\n(b) Assume that the two systems are connected", - "type": "text" - }, - { - "block_id": "p250-b27", - "global_id": 6854, - "bbox": [ - 91.22, - 551.5, - 263.25, - 593.35 - ], - "text": "in parallel, as shown in Fig. P2.4-33a.\nCarefully\nplot\nthe\nequivalent\nimpulse\nresponse function, hp(t).\n(c) Assume that the two systems are connected", - "type": "text" - }, - { - "block_id": "p250-b28", - "global_id": 6855, - "bbox": [ - 106.15, - 595.34, - 263.25, - 626.9 - ], - "text": "in cascade, as shown in Fig. P2.4-33b.\nCarefully\nplot\nthe\nequivalent\nimpulse\nresponse function, hs(t).", - "type": "text" - }, - { - "block_id": "p250-b29", - "global_id": 6856, - "bbox": [ - 284.74, - 485.67, - 471.47, - 494.71 - ], - "text": "2.4-34\nConsider the circuit shown in Fig. P2.4-34.", - "type": "text" - }, - { - "block_id": "p250-b30", - "global_id": 6857, - "bbox": [ - 317.87, - 496.34, - 490.39, - 538.55 - ], - "text": "(a) Find\nthe\noutput\ny(t)\ngiven\nan\ninitial\ncapacitor voltage of y(0) = 2 volts and an\ninput x(t) = u(t).\n(b) Given an input x(t) = u(t −1), determine", - "type": "text" - }, - { - "block_id": "p250-b31", - "global_id": 6858, - "bbox": [ - 333.31, - 540.17, - 490.39, - 560.47 - ], - "text": "the initial capacitor voltage y(0) so that the\noutput y(t) is 0.5 volt at t = 2 seconds.", - "type": "text" - }, - { - "block_id": "p250-b32", - "global_id": 6859, - "bbox": [ - 348.37, - 586.9, - 445.26, - 598.27 - ], - "text": "+\n+", - "type": "text" - }, - { - "block_id": "p250-b33", - "global_id": 6860, - "bbox": [ - 322.06, - 594.44, - 448.55, - 613.97 - ], - "text": "–\n–\nx(t)\ny(t)\nC", - "type": "text" - }, - { - "block_id": "p250-b34", - "global_id": 6861, - "bbox": [ - 383.53, - 582.97, - 388.42, - 590.97 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p250-b35", - "global_id": 6862, - "bbox": [ - 317.86, - 629.59, - 373.98, - 638.56 - ], - "text": "Figure P2.4-34", - "type": "text" - } - ] - }, - { - "page_num": 251, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p251-b0", - "global_id": 6863, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n231", - "type": "text" - }, - { - "block_id": "p251-b1", - "global_id": 6864, - "bbox": [ - 83.34, - 85.64, - 288.98, - 116.9 - ], - "text": "2.4-35\nAn analog signal is given by x(t) = t[u(t) −\nu(t −1)], as shown in Fig. P2.4-35. Determine\nand plot y(t) = x(t) ∗x(2t).", - "type": "text" - }, - { - "block_id": "p251-b2", - "global_id": 6865, - "bbox": [ - 150.56, - 196.05, - 262.11, - 206.36 - ], - "text": "t\n1\n–1", - "type": "text" - }, - { - "block_id": "p251-b3", - "global_id": 6866, - "bbox": [ - 184.53, - 154.63, - 188.53, - 162.63 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p251-b4", - "global_id": 6867, - "bbox": [ - 116.46, - 220.34, - 172.58, - 229.3 - ], - "text": "Figure P2.4-35", - "type": "text" - }, - { - "block_id": "p251-b5", - "global_id": 6868, - "bbox": [ - 83.34, - 252.89, - 288.98, - 272.88 - ], - "text": "2.4-36\nConsider\nthe\nelectric\ncircuit\nshown\nin\nFig. P2.4-36.", - "type": "text" - }, - { - "block_id": "p251-b6", - "global_id": 6869, - "bbox": [ - 116.96, - 274.88, - 288.98, - 283.85 - ], - "text": "(a) Determine the differential equation that", - "type": "text" - }, - { - "block_id": "p251-b7", - "global_id": 6870, - "bbox": [ - 131.89, - 285.47, - 288.98, - 305.76 - ], - "text": "relates the input current x(t) to output\ncurrent y(t). Recall that", - "type": "text" - }, - { - "block_id": "p251-b8", - "global_id": 6871, - "bbox": [ - 182.46, - 319.04, - 237.23, - 335.34 - ], - "text": "vL(t) = LdiL(t)", - "type": "text" - }, - { - "block_id": "p251-b9", - "global_id": 6872, - "bbox": [ - 223.44, - 331.97, - 230.42, - 340.94 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p251-b10", - "global_id": 6873, - "bbox": [ - 116.46, - 352.9, - 288.98, - 361.87 - ], - "text": "(b) Find the characteristic equation for this", - "type": "text" - }, - { - "block_id": "p251-b11", - "global_id": 6874, - "bbox": [ - 116.96, - 363.86, - 288.99, - 394.74 - ], - "text": "circuit, and express the root(s) of the char-\nacteristic equation in terms of L1, L2, and R.\n(c) Determine the zero-input response given", - "type": "text" - }, - { - "block_id": "p251-b12", - "global_id": 6875, - "bbox": [ - 131.9, - 396.74, - 288.98, - 427.62 - ], - "text": "initial inductor currents of one ampere each.\nThat is, find y0(t) given iL1(0) = iL2(0) =\n1 A.", - "type": "text" - }, - { - "block_id": "p251-b13", - "global_id": 6876, - "bbox": [ - 96.75, - 475.01, - 282.39, - 484.61 - ], - "text": "x(t)\ny(t)\nL1\nL2\nR", - "type": "text" - }, - { - "block_id": "p251-b14", - "global_id": 6877, - "bbox": [ - 92.55, - 518.2, - 148.67, - 527.17 - ], - "text": "Figure P2.4-36", - "type": "text" - }, - { - "block_id": "p251-b15", - "global_id": 6878, - "bbox": [ - 310.48, - 85.94, - 516.13, - 127.86 - ], - "text": "2.4-37\nAn LTI system has step response given by\ng(t) = e−tu(t) −e−2tu(t). Determine the output\nof this system y(t) given an input x(t) =\nδ(t −π) −cos(", - "type": "text" - }, - { - "block_id": "p251-b16", - "global_id": 6879, - "bbox": [ - 398.44, - 110.98, - 406.03, - 119.94 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p251-b17", - "global_id": 6880, - "bbox": [ - 406.02, - 118.52, - 429.92, - 127.86 - ], - "text": "3)u(t).", - "type": "text" - }, - { - "block_id": "p251-b18", - "global_id": 6881, - "bbox": [ - 310.48, - 132.67, - 516.14, - 196.81 - ], - "text": "2.4-38\nThe periodic signal x(t) shown in Fig. P2.4-38\nis input to a system with impulse response\nfunction h(t) = t[u(t) −u(t −1.5)], also shown\nin Fig. P2.4-38. Use convolution to determine\nthe output y(t) of this system. Plot y(t) over\n(−3 ≤t ≤3).", - "type": "text" - }, - { - "block_id": "p251-b19", - "global_id": 6882, - "bbox": [ - 310.48, - 201.92, - 516.14, - 221.93 - ], - "text": "2.4-39\nConsider\nthe\nelectric\ncircuit\nshown\nin\nFig. P2.4-39.", - "type": "text" - }, - { - "block_id": "p251-b20", - "global_id": 6883, - "bbox": [ - 344.12, - 223.91, - 516.12, - 232.88 - ], - "text": "(a) Determine the differential equation relating", - "type": "text" - }, - { - "block_id": "p251-b21", - "global_id": 6884, - "bbox": [ - 343.61, - 234.5, - 516.13, - 254.8 - ], - "text": "input x(t) to output y(t).\n(b) Determine the output y(t) in response", - "type": "text" - }, - { - "block_id": "p251-b22", - "global_id": 6885, - "bbox": [ - 359.05, - 255.16, - 516.13, - 300.95 - ], - "text": "to the input x(t) = 4te−3t/2u(t). Assume\ncomponent values of R = 1 , C1 = 1 F, and\nC2 = 2 F, and initial capacitor voltages of\nVC1 = 2 V and VC2 = 1 V.", - "type": "text" - }, - { - "block_id": "p251-b23", - "global_id": 6886, - "bbox": [ - 375.35, - 342.55, - 497.49, - 353.92 - ], - "text": "+\n+", - "type": "text" - }, - { - "block_id": "p251-b24", - "global_id": 6887, - "bbox": [ - 347.81, - 350.09, - 500.79, - 369.62 - ], - "text": "–\n–\nx(t)\ny(t)\nC2", - "type": "text" - }, - { - "block_id": "p251-b25", - "global_id": 6888, - "bbox": [ - 408.5, - 339.89, - 447.44, - 349.5 - ], - "text": "R\nC1", - "type": "text" - }, - { - "block_id": "p251-b26", - "global_id": 6889, - "bbox": [ - 343.61, - 385.24, - 399.72, - 394.21 - ], - "text": "Figure P2.4-39", - "type": "text" - }, - { - "block_id": "p251-b27", - "global_id": 6890, - "bbox": [ - 310.48, - 415.92, - 516.13, - 447.18 - ], - "text": "2.4-40\nAn LTIC system has impulse response h(t) =\n3e−|t|.\n(a) Is the system causal? Mathematically justify", - "type": "text" - }, - { - "block_id": "p251-b28", - "global_id": 6891, - "bbox": [ - 343.61, - 449.17, - 516.13, - 469.1 - ], - "text": "your answer.\n(b) Determine the zero-state response of this", - "type": "text" - }, - { - "block_id": "p251-b29", - "global_id": 6892, - "bbox": [ - 359.05, - 470.72, - 490.38, - 480.06 - ], - "text": "system if the input is x(t) = u(2 −t).", - "type": "text" - }, - { - "block_id": "p251-b30", - "global_id": 6893, - "bbox": [ - 310.48, - 485.17, - 516.14, - 527.09 - ], - "text": "2.4-41\nA cardiovascular researcher is attempting to\nmodel the human heart. He has recorded\nventricular\npressure,\nwhich\nhe\nbelieves\ncorresponds to the heart’s impulse response", - "type": "text" - }, - { - "block_id": "p251-b31", - "global_id": 6894, - "bbox": [ - 225.79, - 548.31, - 236.9, - 556.39 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p251-b32", - "global_id": 6895, - "bbox": [ - 180.73, - 603.85, - 312.02, - 613.31 - ], - "text": "t\n1\n–1", - "type": "text" - }, - { - "block_id": "p251-b33", - "global_id": 6896, - "bbox": [ - 214.64, - 560.39, - 218.64, - 568.39 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p251-b34", - "global_id": 6897, - "bbox": [ - 144.79, - 603.85, - 296.79, - 611.85 - ], - "text": "2\n–2", - "type": "text" - }, - { - "block_id": "p251-b35", - "global_id": 6898, - "bbox": [ - 433.79, - 548.31, - 445.35, - 556.31 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p251-b36", - "global_id": 6899, - "bbox": [ - 389.25, - 602.31, - 502.02, - 611.85 - ], - "text": "t\n1\n–1", - "type": "text" - }, - { - "block_id": "p251-b37", - "global_id": 6900, - "bbox": [ - 422.65, - 560.39, - 426.65, - 568.39 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p251-b38", - "global_id": 6901, - "bbox": [ - 127.59, - 626.82, - 183.71, - 635.79 - ], - "text": "Figure P2.4-38", - "type": "text" - } - ] - }, - { - "page_num": 252, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p252-b0", - "global_id": 6902, - "bbox": [ - 60.0, - 60.36, - 419.1, - 69.45 - ], - "text": "232\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p252-b1", - "global_id": 6903, - "bbox": [ - 90.72, - 85.58, - 263.24, - 149.71 - ], - "text": "function\nh(t),\nas\nshown\nin\nFig.\nP2.4-41.\nComment\non\nthe\nfunction\nh(t)\nshown\nin\nFig. P2.4-41. Can you establish any system\nproperties, such as causality or stability? Do\nthe data suggest any reason to suspect that the\nmeasurement is not a true impulse response?", - "type": "text" - }, - { - "block_id": "p252-b2", - "global_id": 6904, - "bbox": [ - 112.98, - 289.31, - 258.99, - 311.9 - ], - "text": "0\n0\n0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45\nt (seconds)", - "type": "text" - }, - { - "block_id": "p252-b3", - "global_id": 6905, - "bbox": [ - 94.36, - 228.34, - 102.44, - 240.53 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p252-b4", - "global_id": 6906, - "bbox": [ - 261.77, - 294.25, - 271.77, - 302.25 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p252-b5", - "global_id": 6907, - "bbox": [ - 104.98, - 172.79, - 116.98, - 180.79 - ], - "text": "140", - "type": "text" - }, - { - "block_id": "p252-b6", - "global_id": 6908, - "bbox": [ - 104.98, - 189.44, - 116.98, - 197.44 - ], - "text": "120", - "type": "text" - }, - { - "block_id": "p252-b7", - "global_id": 6909, - "bbox": [ - 104.98, - 206.08, - 116.98, - 214.08 - ], - "text": "100", - "type": "text" - }, - { - "block_id": "p252-b8", - "global_id": 6910, - "bbox": [ - 108.98, - 222.73, - 116.98, - 230.73 - ], - "text": "80", - "type": "text" - }, - { - "block_id": "p252-b9", - "global_id": 6911, - "bbox": [ - 108.98, - 239.37, - 116.98, - 247.37 - ], - "text": "60", - "type": "text" - }, - { - "block_id": "p252-b10", - "global_id": 6912, - "bbox": [ - 108.98, - 256.02, - 116.98, - 264.02 - ], - "text": "40", - "type": "text" - }, - { - "block_id": "p252-b11", - "global_id": 6913, - "bbox": [ - 108.98, - 272.66, - 116.98, - 280.66 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p252-b12", - "global_id": 6914, - "bbox": [ - 90.72, - 319.82, - 146.83, - 328.79 - ], - "text": "Figure P2.4-41", - "type": "text" - }, - { - "block_id": "p252-b13", - "global_id": 6915, - "bbox": [ - 57.59, - 345.1, - 263.24, - 360.01 - ], - "text": "2.4-42\nConsider\nan\nintegrator\nsystem,\ny(t)\n=\n$ t", - "type": "text" - }, - { - "block_id": "p252-b14", - "global_id": 6916, - "bbox": [ - 91.22, - 356.06, - 263.24, - 377.1 - ], - "text": "−∞x(τ)dτ.\n(a) What is the unit impulse response hi(t) of", - "type": "text" - }, - { - "block_id": "p252-b15", - "global_id": 6917, - "bbox": [ - 90.72, - 378.35, - 263.21, - 398.28 - ], - "text": "this system?\n(b) If two such integrators are put in parallel,", - "type": "text" - }, - { - "block_id": "p252-b16", - "global_id": 6918, - "bbox": [ - 91.22, - 400.28, - 263.25, - 431.16 - ], - "text": "what is the resulting impulse response\nhp(t)?\n(c) If two such integrators are put in series, what", - "type": "text" - }, - { - "block_id": "p252-b17", - "global_id": 6919, - "bbox": [ - 106.16, - 432.78, - 246.28, - 442.86 - ], - "text": "is the resulting impulse response hs(t)?", - "type": "text" - }, - { - "block_id": "p252-b18", - "global_id": 6920, - "bbox": [ - 57.59, - 447.05, - 263.23, - 468.03 - ], - "text": "2.4-43\nThe autocorrelation of a function x(t) is given\nby rxx(t) =", - "type": "text" - }, - { - "block_id": "p252-b19", - "global_id": 6921, - "bbox": [ - 134.5, - 450.8, - 146.96, - 461.76 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p252-b20", - "global_id": 6922, - "bbox": [ - 90.72, - 458.02, - 263.23, - 489.27 - ], - "text": "−∞x(τ)x(τ −t)dτ. This equation\nis computed in a manner nearly identical to\nconvolution.", - "type": "text" - }, - { - "block_id": "p252-b21", - "global_id": 6923, - "bbox": [ - 90.72, - 490.89, - 263.24, - 511.87 - ], - "text": "(a) Show rxx(t) = x(t) ∗x(−t).\n(b) Determine and plot rxx(t) for the signal x(t)", - "type": "text" - }, - { - "block_id": "p252-b22", - "global_id": 6924, - "bbox": [ - 106.15, - 512.81, - 263.24, - 533.78 - ], - "text": "depicted in Fig. P2.4-43. [Hint: rxx(t) =\nrxx(−t).]", - "type": "text" - }, - { - "block_id": "p252-b23", - "global_id": 6925, - "bbox": [ - 202.74, - 602.37, - 263.97, - 611.87 - ], - "text": "t\n1", - "type": "text" - }, - { - "block_id": "p252-b24", - "global_id": 6926, - "bbox": [ - 173.46, - 547.06, - 184.57, - 555.14 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p252-b25", - "global_id": 6927, - "bbox": [ - 126.67, - 603.87, - 134.67, - 611.87 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p252-b26", - "global_id": 6928, - "bbox": [ - 160.64, - 560.17, - 164.64, - 568.17 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p252-b27", - "global_id": 6929, - "bbox": [ - 238.74, - 603.87, - 242.74, - 611.87 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p252-b28", - "global_id": 6930, - "bbox": [ - 90.72, - 625.88, - 146.83, - 634.84 - ], - "text": "Figure P2.4-43", - "type": "text" - }, - { - "block_id": "p252-b29", - "global_id": 6931, - "bbox": [ - 284.74, - 85.94, - 490.39, - 116.89 - ], - "text": "2.4-44\nConsider the circuit shown in Fig. P2.4-44. This\ncircuit functions as an integrator. Assume ideal\nop-amp behavior and recall that", - "type": "text" - }, - { - "block_id": "p252-b30", - "global_id": 6932, - "bbox": [ - 373.98, - 127.15, - 433.09, - 143.45 - ], - "text": "iC(t) = C dVC(t)", - "type": "text" - }, - { - "block_id": "p252-b31", - "global_id": 6933, - "bbox": [ - 417.38, - 140.08, - 424.36, - 149.05 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p252-b32", - "global_id": 6934, - "bbox": [ - 364.34, - 207.66, - 369.23, - 215.66 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p252-b33", - "global_id": 6935, - "bbox": [ - 434.49, - 168.45, - 439.83, - 176.45 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p252-b34", - "global_id": 6936, - "bbox": [ - 407.76, - 196.49, - 411.76, - 204.49 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p252-b35", - "global_id": 6937, - "bbox": [ - 407.76, - 225.48, - 412.27, - 233.48 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p252-b36", - "global_id": 6938, - "bbox": [ - 328.71, - 210.84, - 339.82, - 218.92 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p252-b37", - "global_id": 6939, - "bbox": [ - 332.26, - 220.46, - 336.26, - 228.46 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p252-b38", - "global_id": 6940, - "bbox": [ - 332.01, - 201.38, - 336.52, - 209.38 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p252-b39", - "global_id": 6941, - "bbox": [ - 472.22, - 226.14, - 483.33, - 234.22 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p252-b40", - "global_id": 6942, - "bbox": [ - 475.78, - 235.76, - 479.78, - 243.76 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p252-b41", - "global_id": 6943, - "bbox": [ - 475.52, - 216.68, - 480.03, - 224.68 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p252-b42", - "global_id": 6944, - "bbox": [ - 317.86, - 267.62, - 373.98, - 276.59 - ], - "text": "Figure P2.4-44", - "type": "text" - }, - { - "block_id": "p252-b43", - "global_id": 6945, - "bbox": [ - 318.37, - 297.12, - 490.38, - 306.08 - ], - "text": "(a) Determine the differential equation that", - "type": "text" - }, - { - "block_id": "p252-b44", - "global_id": 6946, - "bbox": [ - 317.86, - 307.71, - 490.39, - 360.88 - ], - "text": "relates the input x(t) to the output y(t).\n(b) This\ncircuit\ndoes\nnot\nbehave\nwell\nat\ndc. Demonstrate this by computing the\nzero-state response y(t) for a unit step input\nx(t) = u(t).", - "type": "text" - }, - { - "block_id": "p252-b45", - "global_id": 6947, - "bbox": [ - 284.74, - 365.84, - 490.4, - 429.68 - ], - "text": "2.4-45\nDerive the result in Eq. (2.37) in another way. As\nmentioned in Ch. 1 (Fig. 1.27b), it is possible to\nexpress an input in terms of its step components,\nas shown in Fig. P2.4-45. Find the system\nresponse as a sum of the responses to the step\ncomponents of the input.", - "type": "text" - }, - { - "block_id": "p252-b46", - "global_id": 6948, - "bbox": [ - 362.92, - 547.77, - 390.02, - 556.07 - ], - "text": "t nt", - "type": "text" - }, - { - "block_id": "p252-b47", - "global_id": 6949, - "bbox": [ - 322.07, - 456.54, - 374.06, - 469.3 - ], - "text": "dt t\nx(t)", - "type": "text" - }, - { - "block_id": "p252-b48", - "global_id": 6950, - "bbox": [ - 442.58, - 543.6, - 444.8, - 551.6 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p252-b49", - "global_id": 6951, - "bbox": [ - 354.46, - 452.67, - 362.01, - 460.67 - ], - "text": "dx", - "type": "text" - }, - { - "block_id": "p252-b50", - "global_id": 6952, - "bbox": [ - 317.86, - 563.98, - 373.98, - 572.95 - ], - "text": "Figure P2.4-45", - "type": "text" - }, - { - "block_id": "p252-b51", - "global_id": 6953, - "bbox": [ - 284.74, - 593.41, - 490.38, - 614.15 - ], - "text": "2.4-46\nShow that an LTIC system response to an\neverlasting sinusoid cosω0t is given by", - "type": "text" - }, - { - "block_id": "p252-b52", - "global_id": 6954, - "bbox": [ - 339.79, - 625.53, - 468.46, - 635.61 - ], - "text": "y(t) = |H(jω0)| cos[ω0t +̸ H(jω0)]", - "type": "text" - } - ] - }, - { - "page_num": 253, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p253-b0", - "global_id": 6955, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n233", - "type": "text" - }, - { - "block_id": "p253-b1", - "global_id": 6956, - "bbox": [ - 116.46, - 85.9, - 138.36, - 94.86 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p253-b2", - "global_id": 6957, - "bbox": [ - 157.59, - 99.34, - 188.54, - 108.59 - ], - "text": "H(jω) =", - "type": "text" - }, - { - "block_id": "p253-b3", - "global_id": 6958, - "bbox": [ - 190.38, - 87.13, - 205.84, - 98.11 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p253-b4", - "global_id": 6959, - "bbox": [ - 195.12, - 109.38, - 206.78, - 115.85 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p253-b5", - "global_id": 6960, - "bbox": [ - 208.28, - 95.91, - 247.69, - 108.59 - ], - "text": "h(t)e−jωt dt", - "type": "text" - }, - { - "block_id": "p253-b6", - "global_id": 6961, - "bbox": [ - 116.46, - 121.83, - 288.99, - 141.75 - ], - "text": "assuming the integral on the right-hand side\nexists.", - "type": "text" - }, - { - "block_id": "p253-b7", - "global_id": 6962, - "bbox": [ - 83.34, - 146.65, - 288.98, - 188.58 - ], - "text": "2.4-47\nA line charge is located along the x axis with a\ncharge density Q(x) coulombs per meter. Show\nthat the electric field E(x) produced by this line\ncharge at a point x is given by", - "type": "text" - }, - { - "block_id": "p253-b8", - "global_id": 6963, - "bbox": [ - 169.29, - 199.1, - 236.16, - 208.34 - ], - "text": "E(x) = Q(x) ∗h(x)", - "type": "text" - }, - { - "block_id": "p253-b9", - "global_id": 6964, - "bbox": [ - 116.46, - 217.95, - 288.98, - 272.12 - ], - "text": "where h(x) = 1/4πϵx2. [Hint: The charge\nover an interval τ located at τ = nτ is\nQ(nτ)τ. Also by Coulomb’s law, the electric\nfield E(r) at a distance r from a charge q\ncoulombs is given by E(r) = q/4πϵr2.]", - "type": "text" - }, - { - "block_id": "p253-b10", - "global_id": 6965, - "bbox": [ - 83.34, - 277.03, - 288.98, - 329.91 - ], - "text": "2.4-48\nA system is called complex if a real-valued\ninput can produce a complex-valued output.\nSuppose a linear time-invariant complex system\nhas impulse response h(t) = j[u(−t + 2) −\nu(−t)].", - "type": "text" - }, - { - "block_id": "p253-b11", - "global_id": 6966, - "bbox": [ - 116.46, - 331.89, - 288.99, - 351.82 - ], - "text": "(a) Is this system causal? Explain.\n(b) Use convolution to determine the zero-state", - "type": "text" - }, - { - "block_id": "p253-b12", - "global_id": 6967, - "bbox": [ - 116.96, - 353.44, - 288.98, - 395.66 - ], - "text": "response y1(t) of this system in response to\nthe unit-duration pulse x1(t) = u(t) −u(t −\n1).\n(c) Using the result from part (a), determine", - "type": "text" - }, - { - "block_id": "p253-b13", - "global_id": 6968, - "bbox": [ - 131.89, - 397.27, - 288.99, - 418.32 - ], - "text": "the zero-state response y2(t) in response to\nx2(t) = 2u(t −1) −u(t −2) −u(t −3).", - "type": "text" - }, - { - "block_id": "p253-b14", - "global_id": 6969, - "bbox": [ - 87.83, - 422.49, - 288.99, - 486.32 - ], - "text": "2.5-1\nExplain,\nwith\nreasons,\nwhether\nthe\nLTIC\nsystems described by the following equations\nare (i) stable or unstable in the BIBO sense; (ii)\nasymptotically stable, unstable, or marginally\nstable. Assume that the systems are controllable\nand observable.", - "type": "text" - }, - { - "block_id": "p253-b15", - "global_id": 6970, - "bbox": [ - 116.46, - 484.68, - 255.07, - 508.23 - ], - "text": "(a) (D2 + 8D + 12)y(t) = (D −1)x(t)\n(b) D(D2 + 3D + 2)y(t) = (D + 5)x(t)", - "type": "text" - }, - { - "block_id": "p253-b16", - "global_id": 6971, - "bbox": [ - 116.46, - 506.6, - 280.5, - 530.16 - ], - "text": "(c) D2(D2 + 2)y(t) = x(t)\n(d) (D + 1)(D2 −6D + 5)y(t) = (3D + 1)x(t)", - "type": "text" - }, - { - "block_id": "p253-b17", - "global_id": 6972, - "bbox": [ - 87.83, - 535.06, - 248.6, - 544.1 - ], - "text": "2.5-2\nRepeat Prob. 2.5-1 for the following:", - "type": "text" - }, - { - "block_id": "p253-b18", - "global_id": 6973, - "bbox": [ - 116.46, - 542.46, - 259.77, - 566.02 - ], - "text": "(a) (D + 1)(D2 + 2D + 5)2y(t) = x(t)\n(b) (D + 1)(D2 + 9)y(t) = (2D + 9)x(t)", - "type": "text" - }, - { - "block_id": "p253-b19", - "global_id": 6974, - "bbox": [ - 116.46, - 564.39, - 273.71, - 587.94 - ], - "text": "(c) (D + 1)(D2 + 9)2y(t) = (2D + 9)x(t)\n(d) (D2 + 1)(D2 + 4)(D2 + 9)y(t) = 3Dx(t)", - "type": "text" - }, - { - "block_id": "p253-b20", - "global_id": 6975, - "bbox": [ - 87.82, - 592.85, - 288.98, - 612.84 - ], - "text": "2.5-3\nConsider an LTIC system with unit impulse\nresponse\nh(t)\n=\net \t 2", - "type": "text" - }, - { - "block_id": "p253-b21", - "global_id": 6976, - "bbox": [ - 210.61, - 596.29, - 288.99, - 616.03 - ], - "text": "3 cos( 3\n2t) + 1\n3 sin(πt)", - "type": "text" - }, - { - "block_id": "p253-b22", - "global_id": 6977, - "bbox": [ - 116.46, - 614.46, - 288.99, - 634.76 - ], - "text": "u(123 −t).\nIs\nthis\nsystem\nBIBO-stable?\nMathematically justify your answer.", - "type": "text" - }, - { - "block_id": "p253-b23", - "global_id": 6978, - "bbox": [ - 314.97, - 85.94, - 516.13, - 105.94 - ], - "text": "2.5-4\nConsider an LTIC system with unit impulse\nresponse h(t) = 1", - "type": "text" - }, - { - "block_id": "p253-b24", - "global_id": 6979, - "bbox": [ - 344.12, - 96.6, - 516.13, - 116.9 - ], - "text": "t u(t −T).\n(a) Determine, if possible, the value(s) of T for", - "type": "text" - }, - { - "block_id": "p253-b25", - "global_id": 6980, - "bbox": [ - 343.61, - 118.89, - 516.13, - 138.82 - ], - "text": "which this system is causal.\n(b) Determine, if possible, the value(s) of T for", - "type": "text" - }, - { - "block_id": "p253-b26", - "global_id": 6981, - "bbox": [ - 343.61, - 140.81, - 481.35, - 160.74 - ], - "text": "which this system is BIBO-stable.\nJustify all answers mathematically.", - "type": "text" - }, - { - "block_id": "p253-b27", - "global_id": 6982, - "bbox": [ - 314.97, - 166.18, - 516.14, - 208.1 - ], - "text": "2.5-5\nYou are given the choice of a system that is\nguaranteed internally stable or a system that\nis guaranteed externally stable. Which do you\nchoose? Why?", - "type": "text" - }, - { - "block_id": "p253-b28", - "global_id": 6983, - "bbox": [ - 314.97, - 213.55, - 516.13, - 233.55 - ], - "text": "2.5-6\nFor a certain LTIC system, the impulse response\nh(t) = u(t).", - "type": "text" - }, - { - "block_id": "p253-b29", - "global_id": 6984, - "bbox": [ - 344.12, - 235.54, - 516.14, - 244.51 - ], - "text": "(a) Determine the characteristic root(s) of this", - "type": "text" - }, - { - "block_id": "p253-b30", - "global_id": 6985, - "bbox": [ - 343.61, - 246.5, - 516.11, - 266.42 - ], - "text": "system.\n(b) Is this system asymptotically or marginally", - "type": "text" - }, - { - "block_id": "p253-b31", - "global_id": 6986, - "bbox": [ - 343.61, - 268.42, - 481.55, - 299.3 - ], - "text": "stable, or is it unstable?\n(c) Is this system BIBO-stable?\n(d) What can this system be used for?", - "type": "text" - }, - { - "block_id": "p253-b32", - "global_id": 6987, - "bbox": [ - 314.98, - 304.75, - 516.14, - 457.6 - ], - "text": "2.5-7\nIn Sec. 2.5 we demonstrated that for an LTIC\nsystem, the condition of Eq. (2.45) is sufficient\nfor BIBO stability. Show that this is also a\nnecessary condition for BIBO stability in such\nsystems. In other words, show that if Eq. (2.45)\nis not satisfied, then there exists a bounded\ninput that produces an unbounded output.\n[Hint: Assume that a system exists for which\nh(t) violates Eq. (2.45) and yet produces an\noutput that is bounded for every bounded input.\nEstablish the contradiction in this statement by\nconsidering an input x(t) defined by x(t1−τ) = 1\nwhen h(τ) ≥0 and x(t1 −τ) = −1 when\nh(τ) < 0, where t1 is some fixed instant.]", - "type": "text" - }, - { - "block_id": "p253-b33", - "global_id": 6988, - "bbox": [ - 314.97, - 461.71, - 516.14, - 492.66 - ], - "text": "2.5-8\nAn analog LTIC system with impulse response\nfunction h(t) = u(t + 2) −u(t −2) is presented\nwith an input x(t) = t(u(t) −u(t −2)).", - "type": "text" - }, - { - "block_id": "p253-b34", - "global_id": 6989, - "bbox": [ - 344.12, - 494.66, - 516.13, - 503.63 - ], - "text": "(a) Determine and plot the system output", - "type": "text" - }, - { - "block_id": "p253-b35", - "global_id": 6990, - "bbox": [ - 343.62, - 505.24, - 516.11, - 525.54 - ], - "text": "y(t) = x(t) ∗h(t).\n(b) Is this system stable? Is this system causal?", - "type": "text" - }, - { - "block_id": "p253-b36", - "global_id": 6991, - "bbox": [ - 359.05, - 527.54, - 435.01, - 536.5 - ], - "text": "Justify your answers.", - "type": "text" - }, - { - "block_id": "p253-b37", - "global_id": 6992, - "bbox": [ - 314.98, - 541.94, - 516.14, - 583.86 - ], - "text": "2.5-9\nA system has an impulse response function\nshaped like a rectangular pulse, h(t) = u(t) −\nu(t −1). Is the system stable? Is the system\ncausal?", - "type": "text" - }, - { - "block_id": "p253-b38", - "global_id": 6993, - "bbox": [ - 310.49, - 589.31, - 516.14, - 609.31 - ], - "text": "2.5-10\nA continuous-time LTI system has impulse\nresponse function h(t) = %∞", - "type": "text" - }, - { - "block_id": "p253-b39", - "global_id": 6994, - "bbox": [ - 343.61, - 598.91, - 505.43, - 631.23 - ], - "text": "i=0(0.5)iδ(t−i).\n(a) Is the system causal? Prove your answer.\n(b) Is the system stable? Prove your answer.", - "type": "text" - } - ] - }, - { - "page_num": 254, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p254-b0", - "global_id": 6995, - "bbox": [ - 60.0, - 60.36, - 419.11, - 69.45 - ], - "text": "234\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p254-b1", - "global_id": 6996, - "bbox": [ - 62.08, - 85.94, - 263.24, - 127.87 - ], - "text": "2.6-1\nData at a rate of 1 million pulses per second are\nto be transmitted over a certain communications\nchannel. The unit step response g(t) for this\nchannel is shown in Fig. P2.6-1.", - "type": "text" - }, - { - "block_id": "p254-b2", - "global_id": 6997, - "bbox": [ - 91.22, - 129.87, - 263.22, - 138.84 - ], - "text": "(a) Can this channel transmit data at the", - "type": "text" - }, - { - "block_id": "p254-b3", - "global_id": 6998, - "bbox": [ - 90.71, - 140.83, - 263.24, - 193.65 - ], - "text": "required rate? Explain your answer.\n(b) Can\nan\naudio\nsignal\nconsisting\nof\ncomponents with frequencies up to 15\nkHz be transmitted over this channel with\nreasonable fidelity?", - "type": "text" - }, - { - "block_id": "p254-b4", - "global_id": 6999, - "bbox": [ - 90.72, - 275.48, - 142.35, - 284.45 - ], - "text": "Figure P2.6-1", - "type": "text" - }, - { - "block_id": "p254-b5", - "global_id": 7000, - "bbox": [ - 62.08, - 297.17, - 263.24, - 350.34 - ], - "text": "2.6-2\nDetermine a frequency ω that will cause the\ninput x(t) = cos(ωt) to produce a strong\nresponse when applied to the system described\nby (D2 + 2D + 13/4){y(t)} = x(t). Carefully\nexplain your choice.", - "type": "text" - }, - { - "block_id": "p254-b6", - "global_id": 7001, - "bbox": [ - 62.07, - 355.26, - 263.24, - 430.06 - ], - "text": "2.6-3\nFigure P2.6-3 shows the impulse response\nh(t) of a lowpass LTIC system. Determine\nthe peak amplitude A and time constant Th\nso that rectangular impulse response ˆh(t) is\nan appropriate approximation of h(t). The two\ngraphs of Fig. P2.6-3 are not necessarily drawn\nto the same scale.", - "type": "text" - }, - { - "block_id": "p254-b7", - "global_id": 7002, - "bbox": [ - 96.88, - 444.22, - 108.39, - 452.19 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p254-b8", - "global_id": 7003, - "bbox": [ - 122.16, - 444.43, - 228.01, - 465.15 - ], - "text": "4t 2t2\nˆh(t)", - "type": "text" - }, - { - "block_id": "p254-b9", - "global_id": 7004, - "bbox": [ - 70.18, - 494.16, - 268.1, - 507.46 - ], - "text": "t\nt\n0\n0\n1\n1\n2\nTh", - "type": "text" - }, - { - "block_id": "p254-b10", - "global_id": 7005, - "bbox": [ - 85.7, - 471.64, - 89.68, - 479.61 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p254-b11", - "global_id": 7006, - "bbox": [ - 87.09, - 453.25, - 210.77, - 470.36 - ], - "text": "2\nA", - "type": "text" - }, - { - "block_id": "p254-b12", - "global_id": 7007, - "bbox": [ - 57.84, - 516.07, - 109.47, - 525.03 - ], - "text": "Figure P2.6-3", - "type": "text" - }, - { - "block_id": "p254-b13", - "global_id": 7008, - "bbox": [ - 62.08, - 538.04, - 263.24, - 569.02 - ], - "text": "2.6-4\nA\ncertain\ncommunication\nchannel\nhas\na\nbandwidth of 10 kHz. A pulse of 0.5 ms duration\nis transmitted over this channel.", - "type": "text" - }, - { - "block_id": "p254-b14", - "global_id": 7009, - "bbox": [ - 91.22, - 571.01, - 263.23, - 579.97 - ], - "text": "(a) Determine the width (duration) of the", - "type": "text" - }, - { - "block_id": "p254-b15", - "global_id": 7010, - "bbox": [ - 90.72, - 581.97, - 263.23, - 601.9 - ], - "text": "received pulse.\n(b) Find the maximum rate at which these", - "type": "text" - }, - { - "block_id": "p254-b16", - "global_id": 7011, - "bbox": [ - 106.15, - 603.9, - 263.25, - 634.79 - ], - "text": "pulses can be transmitted over this channel\nwithout interference between the successive\npulses.", - "type": "text" - }, - { - "block_id": "p254-b17", - "global_id": 7012, - "bbox": [ - 289.22, - 85.97, - 490.37, - 105.97 - ], - "text": "2.6-5\nA first-order LTIC system has a characteristic\nroot λ = −104.", - "type": "text" - }, - { - "block_id": "p254-b18", - "global_id": 7013, - "bbox": [ - 318.37, - 107.84, - 490.38, - 117.57 - ], - "text": "(a) Determine Tr, the rise time of its unit step", - "type": "text" - }, - { - "block_id": "p254-b19", - "global_id": 7014, - "bbox": [ - 317.86, - 118.89, - 478.73, - 138.83 - ], - "text": "input response.\n(b) Determine the bandwidth of this system.", - "type": "text" - }, - { - "block_id": "p254-b20", - "global_id": 7015, - "bbox": [ - 318.36, - 140.82, - 490.38, - 149.78 - ], - "text": "(c) Determine the rate at which the information", - "type": "text" - }, - { - "block_id": "p254-b21", - "global_id": 7016, - "bbox": [ - 333.3, - 151.78, - 490.37, - 171.71 - ], - "text": "pulses can be transmitted through this\nsystem.", - "type": "text" - }, - { - "block_id": "p254-b22", - "global_id": 7017, - "bbox": [ - 289.22, - 176.62, - 490.38, - 196.62 - ], - "text": "2.6-6\nA lowpass system with a 6 MHz cutoff\nfrequency needs to transmit data pulse that are", - "type": "text" - }, - { - "block_id": "p254-b23", - "global_id": 7018, - "bbox": [ - 317.86, - 196.86, - 490.38, - 219.85 - ], - "text": "500\n6\nns wide. Determine a suitable transmission\nrate Frate (pulses/s) for this system.", - "type": "text" - }, - { - "block_id": "p254-b24", - "global_id": 7019, - "bbox": [ - 289.22, - 223.13, - 490.38, - 265.35 - ], - "text": "2.6-7\nSketch an impulse response h(t) of a non-causal\nLP system that has an approximate cutoff\nfrequency of 5 kHz. Since many solutions are\npossible, be sure to properly justify your answer.", - "type": "text" - }, - { - "block_id": "p254-b25", - "global_id": 7020, - "bbox": [ - 289.21, - 270.27, - 490.38, - 334.09 - ], - "text": "2.6-8\nTwo LTIC transmission channels are available:\nthe first has impulse response h1(t) = u(t) −\nu(t −1) and the second has impulse response\nh2(t) = δ(t) + 0.5δ(t −1) + 0.25δ(t −2).\nExplain which channel is better suited for the\ntransmission of high-speed digital data (pulses).", - "type": "text" - }, - { - "block_id": "p254-b26", - "global_id": 7021, - "bbox": [ - 289.21, - 339.0, - 490.37, - 369.97 - ], - "text": "2.6-9\nConsider a linear time-invariant system with\nimpulse response h(t) shown in Fig. P2.6-9.\nOutside the interval shown, h(t) = 0.", - "type": "text" - }, - { - "block_id": "p254-b27", - "global_id": 7022, - "bbox": [ - 330.73, - 497.47, - 357.73, - 514.29 - ], - "text": "0\n–0.2", - "type": "text" - }, - { - "block_id": "p254-b28", - "global_id": 7023, - "bbox": [ - 367.06, - 506.29, - 488.38, - 522.71 - ], - "text": "0.5\n1\n1.5\n2\n2.5\n3\n3.5\n4\nt", - "type": "text" - }, - { - "block_id": "p254-b29", - "global_id": 7024, - "bbox": [ - 321.51, - 440.98, - 329.59, - 452.53 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p254-b30", - "global_id": 7025, - "bbox": [ - 335.53, - 389.24, - 345.53, - 397.24 - ], - "text": "1.2", - "type": "text" - }, - { - "block_id": "p254-b31", - "global_id": 7026, - "bbox": [ - 341.53, - 404.7, - 345.53, - 412.7 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p254-b32", - "global_id": 7027, - "bbox": [ - 335.53, - 420.17, - 345.53, - 428.17 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p254-b33", - "global_id": 7028, - "bbox": [ - 335.53, - 435.63, - 345.53, - 443.63 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p254-b34", - "global_id": 7029, - "bbox": [ - 335.53, - 451.1, - 345.53, - 459.1 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p254-b35", - "global_id": 7030, - "bbox": [ - 335.53, - 466.56, - 345.53, - 474.56 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p254-b36", - "global_id": 7031, - "bbox": [ - 341.53, - 482.02, - 345.53, - 490.02 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p254-b37", - "global_id": 7032, - "bbox": [ - 317.86, - 530.71, - 369.5, - 539.68 - ], - "text": "Figure P2.6-9", - "type": "text" - }, - { - "block_id": "p254-b38", - "global_id": 7033, - "bbox": [ - 318.36, - 548.99, - 490.39, - 559.39 - ], - "text": "(a) What is the rise time Tr of this system?", - "type": "text" - }, - { - "block_id": "p254-b39", - "global_id": 7034, - "bbox": [ - 317.87, - 560.05, - 490.4, - 601.9 - ], - "text": "Remember, rise time is the time between the\napplication of a unit step and the moment at\nwhich the system has “fully” responded.\n(b) Suppose h(t) represents the response of a", - "type": "text" - }, - { - "block_id": "p254-b40", - "global_id": 7035, - "bbox": [ - 333.32, - 603.89, - 490.41, - 634.79 - ], - "text": "communication channel. What conditions\nmight cause the channel to have such an\nimpulse response? What is the maximum", - "type": "text" - } - ] - }, - { - "page_num": 255, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p255-b0", - "global_id": 7036, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n235", - "type": "text" - }, - { - "block_id": "p255-b1", - "global_id": 7037, - "bbox": [ - 116.96, - 85.9, - 288.98, - 127.74 - ], - "text": "average number of pulses per unit time\nthat can be transmitted without causing\ninterference? Justify your answer.\n(c) Determine the system output y(t) = x(t) ∗", - "type": "text" - }, - { - "block_id": "p255-b2", - "global_id": 7038, - "bbox": [ - 131.9, - 129.36, - 288.97, - 149.66 - ], - "text": "h(t) for x(t) = [u(t −2) −u(t)]. Accurately\nsketch y(t) over (0 ≤t ≤10).", - "type": "text" - }, - { - "block_id": "p255-b3", - "global_id": 7039, - "bbox": [ - 83.34, - 156.03, - 288.98, - 176.03 - ], - "text": "2.6-10\nA lowpass LTIC system has impulse response\nh(t) = −te−tu(t).", - "type": "text" - }, - { - "block_id": "p255-b4", - "global_id": 7040, - "bbox": [ - 116.46, - 177.66, - 288.97, - 197.95 - ], - "text": "(a) Accurately sketch h(t).\n(b) Describe a rectangular impulse response", - "type": "text" - }, - { - "block_id": "p255-b5", - "global_id": 7041, - "bbox": [ - 131.89, - 197.72, - 288.99, - 230.83 - ], - "text": "ˆh(t) as an appropriate approximation of h(t).\nWhat is the approximate cutoff frequency of\nthis system?", - "type": "text" - }, - { - "block_id": "p255-b6", - "global_id": 7042, - "bbox": [ - 83.34, - 237.21, - 288.98, - 257.21 - ], - "text": "2.6-11\nA lowpass LTIC system has impulse response\nh(t), as shown in Fig. P2.4-8.", - "type": "text" - }, - { - "block_id": "p255-b7", - "global_id": 7043, - "bbox": [ - 116.97, - 259.2, - 288.97, - 268.17 - ], - "text": "(a) As discussed in Sec. 2.6-2, determine a", - "type": "text" - }, - { - "block_id": "p255-b8", - "global_id": 7044, - "bbox": [ - 116.46, - 267.94, - 288.98, - 291.42 - ], - "text": "rectangular approximation ˆh(t) to h(t).\n(b) Using ˆh(t), what is the time constant Th of", - "type": "text" - }, - { - "block_id": "p255-b9", - "global_id": 7045, - "bbox": [ - 116.96, - 292.08, - 288.99, - 312.01 - ], - "text": "this lowpass system?\n(c) Using ˆh(t), what is the approximate radian", - "type": "text" - }, - { - "block_id": "p255-b10", - "global_id": 7046, - "bbox": [ - 116.46, - 313.62, - 288.99, - 334.89 - ], - "text": "cutoff frequency ωc of this lowpass system?\n(d) Assuming a frequency ω0 ≪ωc, what", - "type": "text" - }, - { - "block_id": "p255-b11", - "global_id": 7047, - "bbox": [ - 131.89, - 335.54, - 288.98, - 356.58 - ], - "text": "is the system response y(t) to the input\nx(t) = sin(ω0t + π/3)?", - "type": "text" - }, - { - "block_id": "p255-b12", - "global_id": 7048, - "bbox": [ - 83.34, - 362.21, - 288.98, - 437.01 - ], - "text": "2.6-12\nA first CT lowpass system with time constant\nT1 = 4 µs is put in series with a second CT\nlowpass system with time constant T2 = 2 µs.\nMake an educated sketch of the overall impulse\nresponse function hseries(t). What is the time\nconstant Tseries of the overall series-connected\nsystem?", - "type": "text" - }, - { - "block_id": "p255-b13", - "global_id": 7049, - "bbox": [ - 87.82, - 443.09, - 288.98, - 474.35 - ], - "text": "2.7-1\nAn LTIC system with input x(t) and output\ny(t) is described by the following constant\ncoefficient linear differential equation:", - "type": "text" - }, - { - "block_id": "p255-b14", - "global_id": 7050, - "bbox": [ - 143.26, - 486.98, - 262.19, - 500.02 - ], - "text": "(D4 −16){y(t)} = (D −2){x(t)}.", - "type": "text" - }, - { - "block_id": "p255-b15", - "global_id": 7051, - "bbox": [ - 116.97, - 516.73, - 288.97, - 525.69 - ], - "text": "(a) What are the 4 characteristic roots of this", - "type": "text" - }, - { - "block_id": "p255-b16", - "global_id": 7052, - "bbox": [ - 116.46, - 527.31, - 288.98, - 569.52 - ], - "text": "system (λ1, λ2, λ3, and λ4)? Determine the\nroots by hand and then verify your answers\nusing MATLAB’s roots command.\n(b) From Eq. (2.17), computing h(t) requires a", - "type": "text" - }, - { - "block_id": "p255-b17", - "global_id": 7053, - "bbox": [ - 131.89, - 564.42, - 196.94, - 581.16 - ], - "text": "signal ˜yn(t) = %4", - "type": "text" - }, - { - "block_id": "p255-b18", - "global_id": 7054, - "bbox": [ - 131.89, - 569.88, - 288.99, - 635.95 - ], - "text": "k=1 ckeλkt. First, determine\na matrix representation of the system of\nequations needed to solve for the four\ncoefficients ck. Second, write MATLAB\ncode that computes the length-4 column\nvector of coefficients ck.", - "type": "text" - }, - { - "block_id": "p255-b19", - "global_id": 7055, - "bbox": [ - 314.97, - 86.08, - 424.88, - 96.79 - ], - "text": "2.7-2\nDefine x(t) = 2u(t + 2", - "type": "text" - }, - { - "block_id": "p255-b20", - "global_id": 7056, - "bbox": [ - 343.61, - 87.45, - 516.12, - 108.49 - ], - "text": "3) −2u(t). Further, define\nthe periodic signal h1(t) as", - "type": "text" - }, - { - "block_id": "p255-b21", - "global_id": 7057, - "bbox": [ - 370.13, - 124.65, - 396.53, - 134.72 - ], - "text": "h1(t) =", - "type": "text" - }, - { - "block_id": "p255-b22", - "global_id": 7058, - "bbox": [ - 398.38, - 112.06, - 483.44, - 140.15 - ], - "text": "t\n0 ≤t < 1\nh1(t + 1)\n∀t", - "type": "text" - }, - { - "block_id": "p255-b23", - "global_id": 7059, - "bbox": [ - 343.61, - 150.61, - 516.13, - 171.65 - ], - "text": "Lastly, define the aperiodic signal h2(t) in terms\nof h1(t) as", - "type": "text" - }, - { - "block_id": "p255-b24", - "global_id": 7060, - "bbox": [ - 371.0, - 182.62, - 488.74, - 192.7 - ], - "text": "h2(t) = h1(t)[u(t −1) −u(t −2)]", - "type": "text" - }, - { - "block_id": "p255-b25", - "global_id": 7061, - "bbox": [ - 344.12, - 214.64, - 516.14, - 224.72 - ], - "text": "(a) Use MATLAB to plot x(t), h1(t), and h2(t)", - "type": "text" - }, - { - "block_id": "p255-b26", - "global_id": 7062, - "bbox": [ - 343.61, - 225.6, - 516.12, - 245.9 - ], - "text": "over the interval −2.5 ≤t ≤3.5.\n(b) Using the graphical convolution procedure,", - "type": "text" - }, - { - "block_id": "p255-b27", - "global_id": 7063, - "bbox": [ - 344.11, - 247.51, - 516.12, - 267.82 - ], - "text": "compute y2(t) = x(t) ∗h2(t).\n(c) Compute by hand and then MATLAB", - "type": "text" - }, - { - "block_id": "p255-b28", - "global_id": 7064, - "bbox": [ - 359.05, - 269.44, - 516.14, - 300.69 - ], - "text": "plot y1(t) = x(t) ∗h1(t). Modify program\nCH2MP4.m in Sec. 2.7-4 to validate your\nanalytical result.", - "type": "text" - }, - { - "block_id": "p255-b29", - "global_id": 7065, - "bbox": [ - 314.97, - 305.61, - 516.13, - 325.61 - ], - "text": "2.7-3\nConsider the circuit shown in Fig. P2.7-3.\nAssume ideal op-amp behavior and recall that", - "type": "text" - }, - { - "block_id": "p255-b30", - "global_id": 7066, - "bbox": [ - 399.72, - 335.47, - 458.83, - 351.77 - ], - "text": "iC(t) = C dVC(t)", - "type": "text" - }, - { - "block_id": "p255-b31", - "global_id": 7067, - "bbox": [ - 443.12, - 348.4, - 450.1, - 357.37 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p255-b32", - "global_id": 7068, - "bbox": [ - 343.61, - 365.71, - 516.13, - 418.6 - ], - "text": "Without a feedback resistor Rf , the circuit\nfunctions as an integrator and is unstable,\nparticularly at dc. A feedback resistor Rf\ncorrects this problem and results in a stable\ncircuit that functions as a “lossy” integrator.", - "type": "text" - }, - { - "block_id": "p255-b33", - "global_id": 7069, - "bbox": [ - 342.63, - 501.98, - 353.73, - 510.06 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p255-b34", - "global_id": 7070, - "bbox": [ - 497.24, - 513.16, - 508.34, - 521.24 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p255-b35", - "global_id": 7071, - "bbox": [ - 376.44, - 496.97, - 387.56, - 507.05 - ], - "text": "Rin", - "type": "text" - }, - { - "block_id": "p255-b36", - "global_id": 7072, - "bbox": [ - 456.89, - 470.67, - 462.23, - 478.67 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p255-b37", - "global_id": 7073, - "bbox": [ - 428.16, - 486.41, - 432.16, - 494.41 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p255-b38", - "global_id": 7074, - "bbox": [ - 346.18, - 511.01, - 432.41, - 521.24 - ], - "text": "+\n–", - "type": "text" - }, - { - "block_id": "p255-b39", - "global_id": 7075, - "bbox": [ - 345.92, - 493.12, - 350.44, - 501.12 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p255-b40", - "global_id": 7076, - "bbox": [ - 500.79, - 522.19, - 504.79, - 530.19 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p255-b41", - "global_id": 7077, - "bbox": [ - 500.53, - 504.3, - 505.05, - 512.3 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p255-b42", - "global_id": 7078, - "bbox": [ - 430.1, - 450.15, - 437.21, - 460.15 - ], - "text": "Rf", - "type": "text" - }, - { - "block_id": "p255-b43", - "global_id": 7079, - "bbox": [ - 330.66, - 552.51, - 382.3, - 561.47 - ], - "text": "Figure P2.7-3", - "type": "text" - }, - { - "block_id": "p255-b44", - "global_id": 7080, - "bbox": [ - 344.11, - 571.64, - 516.13, - 580.61 - ], - "text": "(a) Determine the differential equation that", - "type": "text" - }, - { - "block_id": "p255-b45", - "global_id": 7081, - "bbox": [ - 343.61, - 582.23, - 516.13, - 624.45 - ], - "text": "relates the input x(t) to the output y(t).\nWhat is the corresponding characteristic\nequation?\n(b) To demonstrate that this “lossy” integrator is", - "type": "text" - }, - { - "block_id": "p255-b46", - "global_id": 7082, - "bbox": [ - 359.05, - 626.44, - 516.14, - 635.4 - ], - "text": "well behaved at dc, determine the zero-state", - "type": "text" - } - ] - }, - { - "page_num": 256, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p256-b0", - "global_id": 7083, - "bbox": [ - 60.0, - 62.89, - 419.1, - 71.98 - ], - "text": "236\nCHAPTER 2\nTIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p256-b1", - "global_id": 7084, - "bbox": [ - 391.68, - 153.2, - 396.19, - 161.2 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p256-b2", - "global_id": 7085, - "bbox": [ - 391.93, - 168.67, - 395.93, - 176.67 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p256-b3", - "global_id": 7086, - "bbox": [ - 110.2, - 154.22, - 121.31, - 162.3 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p256-b4", - "global_id": 7087, - "bbox": [ - 388.38, - 161.66, - 399.49, - 169.74 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p256-b5", - "global_id": 7088, - "bbox": [ - 189.93, - 87.82, - 351.27, - 106.28 - ], - "text": "C2\nC1", - "type": "text" - }, - { - "block_id": "p256-b6", - "global_id": 7089, - "bbox": [ - 113.5, - 145.45, - 118.01, - 153.45 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p256-b7", - "global_id": 7090, - "bbox": [ - 113.75, - 158.72, - 341.74, - 171.15 - ], - "text": "–\n+", - "type": "text" - }, - { - "block_id": "p256-b8", - "global_id": 7091, - "bbox": [ - 184.23, - 141.02, - 341.48, - 157.87 - ], - "text": "–\n+", - "type": "text" - }, - { - "block_id": "p256-b9", - "global_id": 7092, - "bbox": [ - 184.48, - 132.17, - 188.48, - 140.17 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p256-b10", - "global_id": 7093, - "bbox": [ - 143.57, - 118.81, - 151.46, - 128.41 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p256-b11", - "global_id": 7094, - "bbox": [ - 282.66, - 167.49, - 290.55, - 177.1 - ], - "text": "R3", - "type": "text" - }, - { - "block_id": "p256-b12", - "global_id": 7095, - "bbox": [ - 282.66, - 125.78, - 290.55, - 135.39 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p256-b13", - "global_id": 7096, - "bbox": [ - 101.84, - 218.65, - 153.49, - 227.62 - ], - "text": "Figure P2.7-4", - "type": "text" - }, - { - "block_id": "p256-b14", - "global_id": 7097, - "bbox": [ - 91.23, - 242.25, - 263.24, - 273.5 - ], - "text": "response y(t) given a unit step input x(t) =\nu(t).\n(c) Investigate the effect of 10% resistor and", - "type": "text" - }, - { - "block_id": "p256-b15", - "global_id": 7098, - "bbox": [ - 106.16, - 275.5, - 263.25, - 295.42 - ], - "text": "25% capacitor tolerances on the system’s\ncharacteristic root(s).", - "type": "text" - }, - { - "block_id": "p256-b16", - "global_id": 7099, - "bbox": [ - 62.08, - 300.33, - 263.24, - 332.25 - ], - "text": "2.7-4\nConsider\nthe\nelectric\ncircuit\nshown\nin\nFig. P2.7-4. Let C1 = C2 = 10 µF, R1 = R2 = 100\nk, and R3 = 50 k.", - "type": "text" - }, - { - "block_id": "p256-b17", - "global_id": 7100, - "bbox": [ - 91.22, - 333.28, - 263.22, - 342.25 - ], - "text": "(a) Determine the corresponding differential", - "type": "text" - }, - { - "block_id": "p256-b18", - "global_id": 7101, - "bbox": [ - 90.72, - 344.24, - 263.24, - 375.86 - ], - "text": "equation describing this circuit. Is the circuit\nBIBO-stable?\n(b) Determine the zero-input response y0(t) if", - "type": "text" - }, - { - "block_id": "p256-b19", - "global_id": 7102, - "bbox": [ - 106.15, - 377.12, - 263.24, - 397.04 - ], - "text": "the output of each op amp initially reads one\nvolt.", - "type": "text" - }, - { - "block_id": "p256-b20", - "global_id": 7103, - "bbox": [ - 318.37, - 242.25, - 490.38, - 251.59 - ], - "text": "(c) Determine the zero-state response y(t) to a", - "type": "text" - }, - { - "block_id": "p256-b21", - "global_id": 7104, - "bbox": [ - 317.86, - 253.2, - 490.37, - 273.5 - ], - "text": "step input x(t) = u(t).\n(d) Investigate the effect of 10% resistor and", - "type": "text" - }, - { - "block_id": "p256-b22", - "global_id": 7105, - "bbox": [ - 333.31, - 275.5, - 490.37, - 295.42 - ], - "text": "25% capacitor tolerances on the system’s\ncharacteristic roots.", - "type": "text" - }, - { - "block_id": "p256-b23", - "global_id": 7106, - "bbox": [ - 289.23, - 300.03, - 490.44, - 375.12 - ], - "text": "2.7-5\nInput x(t) = 3[u(t)−u(t−1)]+2[u(t−2)−u(t−\n3)] is applied to a lowpass LTIC system with\nimpulse response h(t) = (4 −t)[u(t) −u(t −2)]\nto produce output y(t) = x(t) ∗h(t). Modify\nprogram CH2MP4.m in Sec. 2.7-4 to perform\nthe graphical convolution procedure to produce\na plot of y(t).", - "type": "text" - } - ] - }, - { - "page_num": 257, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p257-b0", - "global_id": 7107, - "bbox": [ - 90.21, - 67.04, - 159.28, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p257-b1", - "global_id": 7108, - "bbox": [ - 172.76, - 125.73, - 471.24, - 173.26 - ], - "text": "TIME-DOMAIN ANALYSIS\nOF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p257-b2", - "global_id": 7109, - "bbox": [ - 140.04, - 79.92, - 159.47, - 118.77 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p257-b3", - "global_id": 7110, - "bbox": [ - 127.59, - 266.14, - 516.14, - 311.97 - ], - "text": "In this chapter we introduce the basic concepts of discrete-time signals and systems. Furthermore,\nwe explore the time-domain analysis of linear, time-invariant, discrete-time (LTID) systems. We\nshow how to compute the zero-input response, determine the unit impulse response, and use\nconvolution to evaluate the zero-state response.", - "type": "text" - }, - { - "block_id": "p257-b4", - "global_id": 7111, - "bbox": [ - 127.94, - 335.69, - 256.54, - 349.64 - ], - "text": "3.1 INTRODUCTION", - "type": "text" - }, - { - "block_id": "p257-b5", - "global_id": 7112, - "bbox": [ - 127.59, - 355.53, - 516.15, - 437.32 - ], - "text": "A discrete-time signal is basically a sequence of numbers. Such signals arise naturally in\ninherently discrete-time situations such as population studies, amortization problems, national\nincome models, and radar tracking. They may also arise as a result of sampling continuous-time\nsignals in sampled data systems and digital filtering. Such signals can be denoted by x[n], y[n], and\nso on, where the variable n takes integer values, and x[n] denotes the nth number in the sequence\nlabeled x. In this notation, the discrete-time variable n is enclosed in square brackets instead of\nparentheses, which we have reserved for enclosing continuous-time variables, such as t.", - "type": "text" - }, - { - "block_id": "p257-b6", - "global_id": 7113, - "bbox": [ - 127.6, - 439.21, - 516.14, - 485.14 - ], - "text": "Systems whose inputs and outputs are discrete-time signals are called discrete-time systems. A\ndigital computer is a familiar example of this type of system. A discrete-time signal is a sequence\nof numbers, and a discrete-time system processes a sequence of numbers x[n] to yield another\nsequence y[n] as the output.†", - "type": "text" - }, - { - "block_id": "p257-b7", - "global_id": 7114, - "bbox": [ - 127.59, - 486.72, - 516.14, - 580.78 - ], - "text": "A discrete-time signal, when obtained by uniform sampling of a continuous-time signal x(t),\ncan also be expressed as x(nT), where T is the sampling interval and n, the discrete variable taking\non integer values. Thus, x(nT) denotes the value of the signal x(t) at t = nT. The signal x(nT) is\na sequence of numbers (sample values), and hence, by definition, is a discrete-time signal. Such\na signal can also be denoted by the customary discrete-time notation x[n], where x[n] = x(nT). A\ntypical discrete-time signal is depicted in Fig. 3.1, which shows both forms of notation. By way\nof an example, a continuous-time exponential x(t) = e−t, when sampled every T = 0.1 seconds,\nresults in a discrete-time signal x(nT) given by", - "type": "text" - }, - { - "block_id": "p257-b8", - "global_id": 7115, - "bbox": [ - 277.99, - 588.62, - 365.22, - 603.01 - ], - "text": "x(nT) = e−nT = e−0.1n", - "type": "text" - }, - { - "block_id": "p257-b9", - "global_id": 7116, - "bbox": [ - 127.59, - 621.19, - 352.7, - 633.41 - ], - "text": "† There may be more than one input and more than one output.", - "type": "text" - }, - { - "block_id": "p257-b10", - "global_id": 7117, - "bbox": [ - 501.19, - 656.12, - 516.13, - 666.22 - ], - "text": "237", - "type": "text" - } - ] - }, - { - "page_num": 258, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p258-b0", - "global_id": 7118, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "238\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p258-b1", - "global_id": 7119, - "bbox": [ - 136.81, - 89.32, - 185.65, - 97.4 - ], - "text": "x[n]\nor x(nT)", - "type": "text" - }, - { - "block_id": "p258-b2", - "global_id": 7120, - "bbox": [ - 130.51, - 145.98, - 286.63, - 155.64 - ], - "text": "n\n1\n5\n10\n2", - "type": "text" - }, - { - "block_id": "p258-b3", - "global_id": 7121, - "bbox": [ - 103.84, - 135.26, - 279.64, - 141.26 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p258-b4", - "global_id": 7122, - "bbox": [ - 128.29, - 166.03, - 286.21, - 175.74 - ], - "text": "t\nT\n5T\n10T\n2T", - "type": "text" - }, - { - "block_id": "p258-b5", - "global_id": 7123, - "bbox": [ - 103.84, - 154.76, - 279.64, - 160.76 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p258-b6", - "global_id": 7124, - "bbox": [ - 313.8, - 167.79, - 440.81, - 177.03 - ], - "text": "Figure 3.1 A discrete-time signal.", - "type": "text" - }, - { - "block_id": "p258-b7", - "global_id": 7125, - "bbox": [ - 169.53, - 291.42, - 440.64, - 299.64 - ], - "text": "x[n]\nx(t)\ny[n]\ny(t)\nContinuous to", - "type": "text" - }, - { - "block_id": "p258-b8", - "global_id": 7126, - "bbox": [ - 210.82, - 300.42, - 237.7, - 308.42 - ], - "text": "discrete,", - "type": "text" - }, - { - "block_id": "p258-b9", - "global_id": 7127, - "bbox": [ - 217.11, - 309.42, - 231.4, - 317.42 - ], - "text": "C/D", - "type": "text" - }, - { - "block_id": "p258-b10", - "global_id": 7128, - "bbox": [ - 281.48, - 295.92, - 325.03, - 303.92 - ], - "text": "Discrete-time", - "type": "text" - }, - { - "block_id": "p258-b11", - "global_id": 7129, - "bbox": [ - 292.14, - 304.92, - 314.37, - 312.92 - ], - "text": "system", - "type": "text" - }, - { - "block_id": "p258-b12", - "global_id": 7130, - "bbox": [ - 365.7, - 291.42, - 402.81, - 308.42 - ], - "text": "Discrete to \ncontinuous,", - "type": "text" - }, - { - "block_id": "p258-b13", - "global_id": 7131, - "bbox": [ - 377.59, - 309.42, - 390.92, - 317.42 - ], - "text": "D/C", - "type": "text" - }, - { - "block_id": "p258-b14", - "global_id": 7132, - "bbox": [ - 125.76, - 342.83, - 431.36, - 352.07 - ], - "text": "Figure 3.2 Processing a continuous-time signal by means of a discrete-time system.", - "type": "text" - }, - { - "block_id": "p258-b15", - "global_id": 7133, - "bbox": [ - 101.84, - 377.15, - 490.4, - 411.43 - ], - "text": "Clearly, this signal is a function of n and may be expressed as x[n]. Such representation is more\nconvenient and will be followed throughout this book, even for signals resulting from sampling\ncontinuous-time signals.", - "type": "text" - }, - { - "block_id": "p258-b16", - "global_id": 7134, - "bbox": [ - 101.85, - 413.42, - 490.41, - 519.02 - ], - "text": "Digital filters can process continuous-time signals by discrete-time systems, using appropriate\ninterfaces at the input and the output, as illustrated in Fig. 3.2. A continuous-time signal x(t) is first\nsampled to convert it into a discrete-time signal x[n], which is then processed by a discrete-time\nsystem to yield the output y[n]. A continuous-time signal y(t) is finally constructed from y[n]. We\nshall use the notations C/D and D/C for conversion from continuous to discrete time and from\ndiscrete to continuous time. By using the interfaces in this manner, we can use an appropriate\ndiscrete-time system to process a continuous-time signal. As we shall see later in our discussion,\ndiscrete-time systems have several advantages over continuous-time systems. For this reason, there\nis an accelerating trend toward processing continuous-time signals with discrete-time systems.", - "type": "text" - }, - { - "block_id": "p258-b17", - "global_id": 7135, - "bbox": [ - 101.84, - 548.77, - 296.92, - 560.72 - ], - "text": "3.1-1 Size of a Discrete-Time Signal", - "type": "text" - }, - { - "block_id": "p258-b18", - "global_id": 7136, - "bbox": [ - 101.84, - 566.85, - 490.39, - 589.47 - ], - "text": "Arguing along the lines similar to those used for continuous-time signals, the size of a discrete-time\nsignal x[n] will be measured by its energy Ex, defined by", - "type": "text" - }, - { - "block_id": "p258-b19", - "global_id": 7137, - "bbox": [ - 261.16, - 613.42, - 280.68, - 624.88 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p258-b20", - "global_id": 7138, - "bbox": [ - 286.42, - 603.25, - 300.52, - 613.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p258-b21", - "global_id": 7139, - "bbox": [ - 282.73, - 627.27, - 304.2, - 634.46 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p258-b22", - "global_id": 7140, - "bbox": [ - 305.33, - 611.99, - 490.39, - 623.8 - ], - "text": "|x[n]|2\n(3.1)", - "type": "text" - } - ] - }, - { - "page_num": 259, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p259-b0", - "global_id": 7141, - "bbox": [ - 403.16, - 62.89, - 516.13, - 71.98 - ], - "text": "3.1\nIntroduction\n239", - "type": "text" - }, - { - "block_id": "p259-b1", - "global_id": 7142, - "bbox": [ - 127.59, - 85.4, - 516.14, - 131.64 - ], - "text": "This definition is valid for real or complex x[n]. For this measure to be meaningful, the energy of a\nsignal must be finite. A necessary condition for the energy to be finite is that the signal amplitude\nmust →0 as |n| →∞. Otherwise the sum in Eq. (3.1) will not converge. If Ex is finite, the signal\nis called an energy signal.", - "type": "text" - }, - { - "block_id": "p259-b2", - "global_id": 7143, - "bbox": [ - 127.6, - 133.23, - 516.13, - 168.21 - ], - "text": "In some cases, for instance, when the amplitude of x[n] does not →0 as |n| →∞, then the\nsignal energy is infinite, and a more meaningful measure of the signal in such a case would be the\ntime average of the energy (if it exists), which is the signal power Px, defined by", - "type": "text" - }, - { - "block_id": "p259-b3", - "global_id": 7144, - "bbox": [ - 264.33, - 197.87, - 302.47, - 209.33 - ], - "text": "Px = lim", - "type": "text" - }, - { - "block_id": "p259-b4", - "global_id": 7145, - "bbox": [ - 285.9, - 206.68, - 305.76, - 213.88 - ], - "text": "N→∞", - "type": "text" - }, - { - "block_id": "p259-b5", - "global_id": 7146, - "bbox": [ - 308.05, - 191.3, - 336.13, - 215.33 - ], - "text": "1\n2N + 1", - "type": "text" - }, - { - "block_id": "p259-b6", - "global_id": 7147, - "bbox": [ - 338.44, - 187.92, - 352.53, - 198.37 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p259-b7", - "global_id": 7148, - "bbox": [ - 340.23, - 212.16, - 350.31, - 219.36 - ], - "text": "−N", - "type": "text" - }, - { - "block_id": "p259-b8", - "global_id": 7149, - "bbox": [ - 353.64, - 196.44, - 378.89, - 208.15 - ], - "text": "|x[n]|2", - "type": "text" - }, - { - "block_id": "p259-b9", - "global_id": 7150, - "bbox": [ - 127.59, - 238.5, - 516.13, - 308.66 - ], - "text": "In this equation, the sum is divided by 2N + 1 because there are 2N + 1 samples in the interval\nfrom −N to N. For periodic signals, the time averaging need be performed over only one period\nin view of the periodic repetition of the signal. If Px is finite and nonzero, the signal is called a\npower signal. As in the continuous-time case, a discrete-time signal can either be an energy signal\nor a power signal, but cannot be both at the same time. Some signals are neither energy nor power\nsignals.", - "type": "text" - }, - { - "block_id": "p259-b10", - "global_id": 7151, - "bbox": [ - 102.51, - 342.3, - 386.68, - 354.25 - ], - "text": "EXAMPLE 3.1\nComputing DT Energy and Power", - "type": "text" - }, - { - "block_id": "p259-b11", - "global_id": 7152, - "bbox": [ - 128.9, - 370.51, - 502.77, - 392.84 - ], - "text": "Find the energy of the signal x[n] = n(u[n] −u[n −6]), shown in Fig. 3.3a and the power for\nthe periodic signal y[n] in Fig. 3.3b.", - "type": "text" - }, - { - "block_id": "p259-b12", - "global_id": 7153, - "bbox": [ - 128.9, - 415.75, - 184.05, - 425.71 - ], - "text": "By definition,", - "type": "text" - }, - { - "block_id": "p259-b13", - "global_id": 7154, - "bbox": [ - 282.05, - 436.91, - 301.57, - 448.37 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p259-b14", - "global_id": 7155, - "bbox": [ - 303.62, - 427.03, - 317.72, - 437.41 - ], - "text": "5\n\"", - "type": "text" - }, - { - "block_id": "p259-b15", - "global_id": 7156, - "bbox": [ - 304.46, - 451.3, - 316.86, - 458.56 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p259-b16", - "global_id": 7157, - "bbox": [ - 318.82, - 432.8, - 349.62, - 447.29 - ], - "text": "n2 = 55", - "type": "text" - }, - { - "block_id": "p259-b17", - "global_id": 7158, - "bbox": [ - 146.84, - 464.86, - 421.89, - 476.73 - ], - "text": "A periodic signal x[n] with period N0 is characterized by the fact that", - "type": "text" - }, - { - "block_id": "p259-b18", - "global_id": 7159, - "bbox": [ - 283.08, - 486.77, - 348.59, - 497.92 - ], - "text": "x[n] = x[n + N0]", - "type": "text" - }, - { - "block_id": "p259-b19", - "global_id": 7160, - "bbox": [ - 128.9, - 509.01, - 502.78, - 578.84 - ], - "text": "The smallest value of N0 for which the preceding equation holds is the fundamental period.\nSuch a signal is called N0 periodic. Figure 3.3b shows an example of a periodic signal y[n] of\nperiod N0 = 6 because each period contains 6 samples. Note that if the first sample is taken\nat n = 0, the last sample is at n = N0 −1 = 5, not at n = N0 = 6. Because the signal y[n] is\nperiodic, its power Py can be found by averaging its energy over one period. Averaging the\nenergy over one period, we obtain", - "type": "text" - }, - { - "block_id": "p259-b20", - "global_id": 7161, - "bbox": [ - 276.62, - 593.43, - 304.35, - 611.46 - ], - "text": "Py = 1", - "type": "text" - }, - { - "block_id": "p259-b21", - "global_id": 7162, - "bbox": [ - 299.36, - 607.49, - 304.35, - 617.45 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p259-b22", - "global_id": 7163, - "bbox": [ - 306.65, - 590.12, - 320.75, - 600.5 - ], - "text": "5\n\"", - "type": "text" - }, - { - "block_id": "p259-b23", - "global_id": 7164, - "bbox": [ - 307.49, - 614.4, - 319.9, - 621.66 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p259-b24", - "global_id": 7165, - "bbox": [ - 321.86, - 593.43, - 353.85, - 610.38 - ], - "text": "n2 = 55", - "type": "text" - }, - { - "block_id": "p259-b25", - "global_id": 7166, - "bbox": [ - 346.37, - 607.49, - 351.35, - 617.45 - ], - "text": "6", - "type": "text" - } - ] - }, - { - "page_num": 260, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p260-b0", - "global_id": 7167, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "240\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p260-b1", - "global_id": 7168, - "bbox": [ - 191.04, - 90.99, - 204.4, - 99.07 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p260-b2", - "global_id": 7169, - "bbox": [ - 190.09, - 129.34, - 216.09, - 137.34 - ], - "text": "0\n3", - "type": "text" - }, - { - "block_id": "p260-b3", - "global_id": 7170, - "bbox": [ - 179.09, - 100.99, - 183.09, - 108.99 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p260-b4", - "global_id": 7171, - "bbox": [ - 208.14, - 151.78, - 217.02, - 159.78 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p260-b5", - "global_id": 7172, - "bbox": [ - 230.77, - 129.34, - 234.77, - 137.34 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p260-b6", - "global_id": 7173, - "bbox": [ - 94.2, - 195.5, - 387.14, - 201.94 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p260-b7", - "global_id": 7174, - "bbox": [ - 170.6, - 172.84, - 183.48, - 180.92 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p260-b8", - "global_id": 7175, - "bbox": [ - 366.27, - 208.9, - 370.27, - 216.9 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p260-b9", - "global_id": 7176, - "bbox": [ - 265.71, - 129.05, - 269.71, - 137.05 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p260-b10", - "global_id": 7177, - "bbox": [ - 190.09, - 210.69, - 194.09, - 218.69 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p260-b11", - "global_id": 7178, - "bbox": [ - 198.09, - 182.34, - 202.09, - 190.34 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p260-b12", - "global_id": 7179, - "bbox": [ - 207.75, - 233.15, - 217.4, - 241.15 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p260-b13", - "global_id": 7180, - "bbox": [ - 126.59, - 210.4, - 348.44, - 218.69 - ], - "text": "6\n12\n18\n6", - "type": "text" - }, - { - "block_id": "p260-b14", - "global_id": 7181, - "bbox": [ - 94.2, - 247.78, - 326.41, - 257.09 - ], - "text": "Figure 3.3 (a) Energy and (b) power computations for a signal.", - "type": "text" - }, - { - "block_id": "p260-b15", - "global_id": 7182, - "bbox": [ - 107.82, - 317.49, - 470.29, - 329.45 - ], - "text": "DRILL 3.1\nDT Signal Classification: Energy, Power, and Neither", - "type": "text" - }, - { - "block_id": "p260-b16", - "global_id": 7183, - "bbox": [ - 107.82, - 337.15, - 484.42, - 372.44 - ], - "text": "Show that the signal x[n] = anu[n] is an energy signal of energy Ex = 1/(1 −|a|2) if |a| < 1,\nthat it is a power signal of power Px = 0.5 if |a| = 1, and that it is neither an energy signal nor\na power signal if |a| > 1.", - "type": "text" - }, - { - "block_id": "p260-b17", - "global_id": 7184, - "bbox": [ - 102.2, - 413.24, - 320.03, - 427.18 - ], - "text": "3.2 USEFUL SIGNAL OPERATIONS", - "type": "text" - }, - { - "block_id": "p260-b18", - "global_id": 7185, - "bbox": [ - 101.84, - 433.07, - 490.38, - 455.09 - ], - "text": "Signal operations for shifting, and scaling, as discussed for continuous-time signals also apply,\nwith some modifications, to discrete-time signals.", - "type": "text" - }, - { - "block_id": "p260-b19", - "global_id": 7186, - "bbox": [ - 102.14, - 470.61, - 153.39, - 482.73 - ], - "text": "SHIFTING", - "type": "text" - }, - { - "block_id": "p260-b20", - "global_id": 7187, - "bbox": [ - 101.84, - 486.35, - 490.4, - 520.63 - ], - "text": "Consider a signal x[n] (Fig. 3.4a) and the same signal delayed (right-shifted) by 5 units (Fig. 3.4b),\nwhich we shall denote by xs[n].† Using the argument employed for a similar operation in\ncontinuous-time signals (Sec. 1.2), we obtain", - "type": "text" - }, - { - "block_id": "p260-b21", - "global_id": 7188, - "bbox": [ - 264.6, - 533.76, - 327.62, - 544.84 - ], - "text": "xs[n] = x[n −5]", - "type": "text" - }, - { - "block_id": "p260-b22", - "global_id": 7189, - "bbox": [ - 101.84, - 557.27, - 490.39, - 603.51 - ], - "text": "Therefore, to shift a sequence by M units (M integer), we replace n with n −M. Thus x[n −M]\nrepresents x[n] shifted by M units. If M is positive, the shift is to the right (delay). If M is negative,\nthe shift is to the left (advance). Accordingly, x[n −5] is x[n] delayed (right-shifted) by 5 units,\nand x[n + 5] is x[n] advanced (left-shifted) by 5 units.", - "type": "text" - }, - { - "block_id": "p260-b23", - "global_id": 7190, - "bbox": [ - 101.84, - 623.19, - 490.39, - 657.33 - ], - "text": "† The terms “delay” and “advance” are meaningful only when the independent variable is time. For other\nindependent variables, such as frequency or distance, it is more appropriate to refer to the “right shift” and\n“left shift” of a sequence.", - "type": "text" - } - ] - }, - { - "page_num": 261, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p261-b0", - "global_id": 7191, - "bbox": [ - 353.74, - 62.89, - 516.13, - 71.98 - ], - "text": "3.2\nUseful Signal Operations\n241", - "type": "text" - }, - { - "block_id": "p261-b1", - "global_id": 7192, - "bbox": [ - 281.54, - 279.86, - 438.1, - 287.86 - ], - "text": "12\n8\n10\n15\n0", - "type": "text" - }, - { - "block_id": "p261-b2", - "global_id": 7193, - "bbox": [ - 293.38, - 203.26, - 346.8, - 213.03 - ], - "text": "xs[n] x[n 5]", - "type": "text" - }, - { - "block_id": "p261-b3", - "global_id": 7194, - "bbox": [ - 293.38, - 320.56, - 338.8, - 330.32 - ], - "text": "xr[n] x[n]", - "type": "text" - }, - { - "block_id": "p261-b4", - "global_id": 7195, - "bbox": [ - 293.38, - 436.11, - 367.94, - 444.41 - ], - "text": "x[k n] for k 5", - "type": "text" - }, - { - "block_id": "p261-b5", - "global_id": 7196, - "bbox": [ - 276.25, - 201.05, - 280.25, - 209.05 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p261-b6", - "global_id": 7197, - "bbox": [ - 328.94, - 294.15, - 338.59, - 302.15 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p261-b7", - "global_id": 7198, - "bbox": [ - 394.86, - 221.69, - 421.67, - 231.39 - ], - "text": "(0.9)n5", - "type": "text" - }, - { - "block_id": "p261-b8", - "global_id": 7199, - "bbox": [ - 489.64, - 395.86, - 493.64, - 403.86 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p261-b9", - "global_id": 7200, - "bbox": [ - 489.64, - 509.29, - 493.64, - 517.29 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p261-b10", - "global_id": 7201, - "bbox": [ - 489.64, - 280.67, - 493.64, - 288.67 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p261-b11", - "global_id": 7202, - "bbox": [ - 175.66, - 393.83, - 190.32, - 402.12 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p261-b12", - "global_id": 7203, - "bbox": [ - 174.76, - 509.01, - 189.43, - 517.3 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p261-b13", - "global_id": 7204, - "bbox": [ - 217.83, - 393.83, - 258.63, - 402.12 - ], - "text": "3\n6", - "type": "text" - }, - { - "block_id": "p261-b14", - "global_id": 7205, - "bbox": [ - 227.72, - 509.01, - 258.47, - 517.3 - ], - "text": "3\n5", - "type": "text" - }, - { - "block_id": "p261-b15", - "global_id": 7206, - "bbox": [ - 281.55, - 394.12, - 285.55, - 402.12 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p261-b16", - "global_id": 7207, - "bbox": [ - 276.51, - 315.48, - 280.51, - 323.48 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p261-b17", - "global_id": 7208, - "bbox": [ - 341.85, - 394.12, - 438.1, - 402.12 - ], - "text": "6\n15", - "type": "text" - }, - { - "block_id": "p261-b18", - "global_id": 7209, - "bbox": [ - 329.33, - 409.89, - 338.21, - 417.89 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p261-b19", - "global_id": 7210, - "bbox": [ - 329.33, - 180.37, - 338.21, - 188.37 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p261-b20", - "global_id": 7211, - "bbox": [ - 312.09, - 166.02, - 438.1, - 174.02 - ], - "text": "8\n6\n3\n10\n15", - "type": "text" - }, - { - "block_id": "p261-b21", - "global_id": 7212, - "bbox": [ - 268.68, - 87.49, - 291.25, - 97.95 - ], - "text": "x[n] 1", - "type": "text" - }, - { - "block_id": "p261-b22", - "global_id": 7213, - "bbox": [ - 281.98, - 166.02, - 285.98, - 174.02 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p261-b23", - "global_id": 7214, - "bbox": [ - 344.64, - 110.38, - 362.97, - 119.91 - ], - "text": "(0.9)n", - "type": "text" - }, - { - "block_id": "p261-b24", - "global_id": 7215, - "bbox": [ - 489.64, - 166.89, - 493.64, - 174.89 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p261-b25", - "global_id": 7216, - "bbox": [ - 281.55, - 509.3, - 285.55, - 517.3 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p261-b26", - "global_id": 7217, - "bbox": [ - 277.07, - 431.56, - 281.07, - 439.56 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p261-b27", - "global_id": 7218, - "bbox": [ - 341.85, - 509.3, - 438.1, - 517.3 - ], - "text": "6\n15", - "type": "text" - }, - { - "block_id": "p261-b28", - "global_id": 7219, - "bbox": [ - 329.1, - 522.74, - 338.43, - 530.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p261-b29", - "global_id": 7220, - "bbox": [ - 238.37, - 459.65, - 264.83, - 469.35 - ], - "text": "(0.9)kn", - "type": "text" - }, - { - "block_id": "p261-b30", - "global_id": 7221, - "bbox": [ - 197.51, - 340.11, - 220.84, - 349.81 - ], - "text": "(0.9)n", - "type": "text" - }, - { - "block_id": "p261-b31", - "global_id": 7222, - "bbox": [ - 301.35, - 509.3, - 305.35, - 517.3 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p261-b32", - "global_id": 7223, - "bbox": [ - 151.5, - 537.44, - 331.45, - 546.67 - ], - "text": "Figure 3.4 Shifting and time reversal of a signal.", - "type": "text" - }, - { - "block_id": "p261-b33", - "global_id": 7224, - "bbox": [ - 133.57, - 584.82, - 320.46, - 596.77 - ], - "text": "DRILL 3.2\nLeft-Shift Operation", - "type": "text" - }, - { - "block_id": "p261-b34", - "global_id": 7225, - "bbox": [ - 133.57, - 602.28, - 510.15, - 627.81 - ], - "text": "Show that x[n] in Fig. 3.4a left-shifted by 3 units can be expressed as 0.729(0.9)n for 0 ≤n ≤7,\nand zero otherwise. Sketch the shifted signal.", - "type": "text" - } - ] - }, - { - "page_num": 262, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p262-b0", - "global_id": 7226, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "242\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p262-b1", - "global_id": 7227, - "bbox": [ - 107.82, - 105.17, - 303.35, - 117.13 - ], - "text": "DRILL 3.3\nRight-Shift Operation", - "type": "text" - }, - { - "block_id": "p262-b2", - "global_id": 7228, - "bbox": [ - 107.82, - 125.83, - 484.41, - 148.16 - ], - "text": "Show that x[−k −n] can be obtained from x[n] by first right-shifting x[n] by k units and then\ntime-reversing this shifted signal.", - "type": "text" - }, - { - "block_id": "p262-b3", - "global_id": 7229, - "bbox": [ - 102.14, - 179.77, - 188.36, - 191.89 - ], - "text": "TIME REVERSAL", - "type": "text" - }, - { - "block_id": "p262-b4", - "global_id": 7230, - "bbox": [ - 101.84, - 195.51, - 490.38, - 229.79 - ], - "text": "To time-reverse x[n] in Fig. 3.4a, we rotate x[n] about the vertical axis to obtain the time-reversed\nsignal xr[n] shown in Fig. 3.4c. Using the argument employed for a similar operation in\ncontinuous-time signals (Sec. 1.2), we obtain", - "type": "text" - }, - { - "block_id": "p262-b5", - "global_id": 7231, - "bbox": [ - 268.56, - 242.1, - 323.67, - 253.18 - ], - "text": "xr[n] = x[−n]", - "type": "text" - }, - { - "block_id": "p262-b6", - "global_id": 7232, - "bbox": [ - 101.85, - 264.79, - 490.39, - 299.07 - ], - "text": "Therefore, to time-reverse a signal, we replace n with −n so that x[−n] is the time-reversed x[n].\nFor example, if x[n] = (0.9)n for 3 ≤n ≤10, then xr[n] = (0.9)−n for 3 ≤−n ≤10; that is,\n−3 ≥n ≥−10, as shown in Fig. 3.4c.", - "type": "text" - }, - { - "block_id": "p262-b7", - "global_id": 7233, - "bbox": [ - 101.84, - 300.66, - 490.39, - 334.94 - ], - "text": "The origin n = 0 is the anchor point, which remains unchanged under time-reversal operation\nbecause at n = 0, x[n] = x[−n] = x[0]. Note that while the reversal of x[n] about the vertical axis\nis x[−n], the reversal of x[n] about the horizontal axis is −x[n].", - "type": "text" - }, - { - "block_id": "p262-b8", - "global_id": 7234, - "bbox": [ - 76.77, - 363.72, - 324.72, - 375.67 - ], - "text": "EXAMPLE 3.2\nTime Reversal and Shifting", - "type": "text" - }, - { - "block_id": "p262-b9", - "global_id": 7235, - "bbox": [ - 103.16, - 391.92, - 472.48, - 402.3 - ], - "text": "In the convolution operation, discussed later, we need to find the function x[k −n] from x[n].", - "type": "text" - }, - { - "block_id": "p262-b10", - "global_id": 7236, - "bbox": [ - 103.16, - 424.8, - 477.02, - 483.0 - ], - "text": "This can be done in two steps: (i) time-reverse the signal x[n] to obtain x[−n]; (ii) now,\nright-shift x[−n] by k. Recall that right-shifting is accomplished by replacing n with n −k.\nHence, right-shifting x[−n] by k units is x[−(n −k)] = x[k −n]. Figure 3.4d shows x[5 −n],\nobtained this way. We first time-reverse x[n] to obtain x[−n] in Fig. 3.4c. Next, we shift x[−n]\nby k = 5 to obtain x[k −n] = x[5 −n], as shown in Fig. 3.4d.", - "type": "text" - }, - { - "block_id": "p262-b11", - "global_id": 7237, - "bbox": [ - 103.17, - 484.99, - 477.05, - 530.82 - ], - "text": "In this particular example, the order of the two operations employed is interchangeable.\nWe can first left-shift x[k] to obtain x[n + 5]. Next, we time-reverse x[n + 5] to obtain\nx[−n + 5] = x[5 −n]. The reader is encouraged to verify that this procedure yields the same\nresult, as in Fig. 3.4d.", - "type": "text" - }, - { - "block_id": "p262-b12", - "global_id": 7238, - "bbox": [ - 107.82, - 584.82, - 260.88, - 596.77 - ], - "text": "DRILL 3.4\nTime Reversal", - "type": "text" - }, - { - "block_id": "p262-b13", - "global_id": 7239, - "bbox": [ - 107.82, - 602.28, - 484.38, - 628.51 - ], - "text": "Sketch the signal x[n] = e−0.5n for −3 ≤n ≤2, and zero otherwise. Sketch the corresponding\ntime-reversed signal and show that it can be expressed as xr[n] = e0.5n for −2 ≤n ≤3.", - "type": "text" - } - ] - }, - { - "page_num": 263, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p263-b0", - "global_id": 7240, - "bbox": [ - 353.74, - 62.89, - 516.13, - 71.98 - ], - "text": "3.2\nUseful Signal Operations\n243", - "type": "text" - }, - { - "block_id": "p263-b1", - "global_id": 7241, - "bbox": [ - 127.89, - 86.19, - 392.55, - 112.27 - ], - "text": "SAMPLING RATE ALTERATION: DOWNSAMPLING,\nUPSAMPLING, AND INTERPOLATION", - "type": "text" - }, - { - "block_id": "p263-b2", - "global_id": 7242, - "bbox": [ - 127.59, - 116.3, - 516.12, - 150.87 - ], - "text": "Alteration of the sampling rate is somewhat similar to time-scaling in continuous-time signals.\nConsider a signal x[n] compressed by factor M. Compressing a signal x[n] by factor M yields\nxd[n] given by", - "type": "text" - }, - { - "block_id": "p263-b3", - "global_id": 7243, - "bbox": [ - 293.63, - 159.15, - 350.08, - 170.23 - ], - "text": "xd[n] = x[Mn]", - "type": "text" - }, - { - "block_id": "p263-b4", - "global_id": 7244, - "bbox": [ - 127.59, - 183.43, - 516.15, - 265.12 - ], - "text": "Because of the restriction that discrete-time signals are defined only for integer values of the\nargument, we must restrict M to integer values. The values of x[Mn] at n = 0,1,2,3,. . . are x[0],\nx[M], x[2M], x[3M], . . . . This means x[Mn] selects every Mth sample of x[n] and deletes all the\nsamples in between. It reduces the number of samples by factor M. If x[n] is obtained by sampling\na continuous-time signal, this operation implies reducing the sampling rate by factor M. For this\nreason, this operation is commonly called downsampling. Figure 3.5a shows a signal x[n] and\nFig. 3.5b shows the signal x[2n], which is obtained by deleting odd-numbered samples of x[n].†", - "type": "text" - }, - { - "block_id": "p263-b5", - "global_id": 7245, - "bbox": [ - 127.59, - 267.11, - 516.14, - 312.94 - ], - "text": "In the continuous-time case, time compression merely speeds up the signal without loss of any\ndata. In contrast, downsampling x[n] generally causes loss of data. Under certain conditions—for\nexample, if x[n] is the result of oversampling some continuous-time signal—then xd[n] may still\nretain the complete information about x[n].", - "type": "text" - }, - { - "block_id": "p263-b6", - "global_id": 7246, - "bbox": [ - 127.6, - 314.52, - 516.11, - 337.55 - ], - "text": "An interpolated signal is generated in two steps; first, we expand x[n] by an integer factor L\nto obtain the expanded signal xe[n], as", - "type": "text" - }, - { - "block_id": "p263-b7", - "global_id": 7247, - "bbox": [ - 238.51, - 358.97, - 267.96, - 370.04 - ], - "text": "xe[n] =", - "type": "text" - }, - { - "block_id": "p263-b8", - "global_id": 7248, - "bbox": [ - 270.01, - 344.98, - 516.13, - 375.23 - ], - "text": "x[n/L]\nn = 0, ±L ± 2L,. . .,\n0\notherwise\n(3.2)", - "type": "text" - }, - { - "block_id": "p263-b9", - "global_id": 7249, - "bbox": [ - 127.59, - 391.11, - 516.14, - 450.01 - ], - "text": "To understand this expression, consider a simple case of expanding x[n] by a factor 2 (L = 2).\nWhen n is odd, n/2 is noninteger, and xe[n] = 0. That is, xe[1] = xe[3] = xe[5],. . . are all zero,\nas depicted in Fig. 3.5c. Moreover, n/2 is integer for even n, and the values of xe[n] = x[n/2] for\nn = 0,2,4,6,. . ., are x[0], x[1], x[2], x[3], . . . , as shown in Fig. 3.5c. In general, for n = 0,1,2,. . .,\nxe[n] is given by the sequence", - "type": "text" - }, - { - "block_id": "p263-b10", - "global_id": 7250, - "bbox": [ - 207.43, - 465.79, - 275.73, - 490.39 - ], - "text": "x[0],0,0,. . .,0,0\n\n\n\nL−1 zeros", - "type": "text" - }, - { - "block_id": "p263-b11", - "global_id": 7251, - "bbox": [ - 275.71, - 465.79, - 347.62, - 490.39 - ], - "text": ",x[1],0,0,. . .,0,0\n\n\n\nL−1 zeros", - "type": "text" - }, - { - "block_id": "p263-b12", - "global_id": 7252, - "bbox": [ - 347.61, - 465.79, - 419.51, - 490.39 - ], - "text": ",x[2],0,0,. . .,0,0\n\n\n\nL−1 zeros", - "type": "text" - }, - { - "block_id": "p263-b13", - "global_id": 7253, - "bbox": [ - 419.5, - 465.79, - 436.3, - 476.16 - ], - "text": ",. . .", - "type": "text" - }, - { - "block_id": "p263-b14", - "global_id": 7254, - "bbox": [ - 127.59, - 505.1, - 516.13, - 539.39 - ], - "text": "Thus, the sampling rate of xe[n] is L times that of x[n]. Hence, this operation is commonly called\nupsampling. The upsampled signal xe[n] contains all the data of x[n], although in an expanded\nform.", - "type": "text" - }, - { - "block_id": "p263-b15", - "global_id": 7255, - "bbox": [ - 127.59, - 541.38, - 516.14, - 587.21 - ], - "text": "In the expanded signal in Fig. 3.5c, the missing (zero-valued) odd-numbered samples can\nbe reconstructed from the non-zero-valued samples by using some suitable interpolation formula.\nFigure 3.5d shows such an interpolated signal xi[n], where the missing samples are constructed\nby using an interpolating filter. The optimum interpolating filter is usually an ideal lowpass", - "type": "text" - }, - { - "block_id": "p263-b16", - "global_id": 7256, - "bbox": [ - 127.59, - 610.24, - 516.15, - 634.09 - ], - "text": "† Odd-numbered samples of x[n] can be retained (and even-numbered samples deleted) by using the\ntransformation xd[n] = x[2n + 1].", - "type": "text" - } - ] - }, - { - "page_num": 264, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p264-b0", - "global_id": 7257, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "244\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p264-b1", - "global_id": 7258, - "bbox": [ - 162.79, - 378.48, - 448.68, - 394.67 - ], - "text": "2\n4\n6\n8\nn", - "type": "text" - }, - { - "block_id": "p264-b2", - "global_id": 7259, - "bbox": [ - 232.98, - 274.18, - 236.98, - 282.18 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p264-b3", - "global_id": 7260, - "bbox": [ - 221.16, - 378.48, - 229.16, - 386.48 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p264-b4", - "global_id": 7261, - "bbox": [ - 162.79, - 268.98, - 229.16, - 276.98 - ], - "text": "2\n4\n6\n8\n10", - "type": "text" - }, - { - "block_id": "p264-b5", - "global_id": 7262, - "bbox": [ - 236.07, - 378.48, - 304.21, - 386.48 - ], - "text": "12\n14\n16\n18\n20", - "type": "text" - }, - { - "block_id": "p264-b6", - "global_id": 7263, - "bbox": [ - 162.79, - 154.83, - 304.21, - 162.83 - ], - "text": "2\n4\n6\n8\n10\n12\n14\n16\n18\n20", - "type": "text" - }, - { - "block_id": "p264-b7", - "global_id": 7264, - "bbox": [ - 311.56, - 378.48, - 454.56, - 386.48 - ], - "text": "24\n28\n22\n26\n30\n32\n34\n36\n38\n40", - "type": "text" - }, - { - "block_id": "p264-b8", - "global_id": 7265, - "bbox": [ - 293.09, - 509.74, - 302.42, - 517.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p264-b9", - "global_id": 7266, - "bbox": [ - 131.21, - 311.96, - 146.76, - 321.51 - ], - "text": "xe[n]", - "type": "text" - }, - { - "block_id": "p264-b10", - "global_id": 7267, - "bbox": [ - 130.88, - 204.36, - 146.76, - 213.91 - ], - "text": "xd[n]", - "type": "text" - }, - { - "block_id": "p264-b11", - "global_id": 7268, - "bbox": [ - 133.89, - 88.23, - 146.77, - 96.31 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p264-b12", - "global_id": 7269, - "bbox": [ - 162.8, - 487.93, - 454.56, - 504.12 - ], - "text": "n\n2\n4\n6\n8\n10\n12\n14\n16\n18\n20\n24\n28\n22\n26\n30\n32\n34\n36\n38\n40", - "type": "text" - }, - { - "block_id": "p264-b13", - "global_id": 7270, - "bbox": [ - 132.25, - 422.54, - 344.48, - 440.18 - ], - "text": "Interpolation\nxi[n]", - "type": "text" - }, - { - "block_id": "p264-b14", - "global_id": 7271, - "bbox": [ - 293.32, - 400.17, - 302.2, - 408.17 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p264-b15", - "global_id": 7272, - "bbox": [ - 348.22, - 330.71, - 387.33, - 338.71 - ], - "text": "Upsampling", - "type": "text" - }, - { - "block_id": "p264-b16", - "global_id": 7273, - "bbox": [ - 292.93, - 288.69, - 302.58, - 296.69 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p264-b17", - "global_id": 7274, - "bbox": [ - 206.71, - 224.23, - 255.6, - 232.23 - ], - "text": "Downsampling", - "type": "text" - }, - { - "block_id": "p264-b18", - "global_id": 7275, - "bbox": [ - 190.05, - 203.13, - 235.47, - 212.89 - ], - "text": "xd[n] x[2n]", - "type": "text" - }, - { - "block_id": "p264-b19", - "global_id": 7276, - "bbox": [ - 293.32, - 182.04, - 302.2, - 190.04 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p264-b20", - "global_id": 7277, - "bbox": [ - 287.97, - 162.56, - 291.97, - 170.56 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p264-b21", - "global_id": 7278, - "bbox": [ - 319.2, - 307.88, - 330.58, - 322.75 - ], - "text": "]\n[n", - "type": "text" - }, - { - "block_id": "p264-b22", - "global_id": 7279, - "bbox": [ - 289.05, - 311.67, - 326.89, - 324.44 - ], - "text": "2\nxe[n] x", - "type": "text" - }, - { - "block_id": "p264-b23", - "global_id": 7280, - "bbox": [ - 125.76, - 524.44, - 475.34, - 533.68 - ], - "text": "Figure 3.5 Compression (downsampling) and expansion (upsampling, interpolation) of a signal.", - "type": "text" - }, - { - "block_id": "p264-b24", - "global_id": 7281, - "bbox": [ - 101.84, - 554.79, - 490.41, - 612.57 - ], - "text": "filter, which is realizable only approximately. In practice, we may use an interpolation that is\nnonoptimum but realizable. The process of filtering to interpolate the zero-valued samples is called\ninterpolation. Since the interpolated data are computed from the existing data, interpolation does\nnot result in gain of information. While further discussion of interpolation is beyond our scope,\nDrill 3.5 and Prob. 3.11-10 introduce the idea of linear interpolation.", - "type": "text" - } - ] - }, - { - "page_num": 265, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p265-b0", - "global_id": 7282, - "bbox": [ - 287.05, - 62.89, - 516.14, - 71.98 - ], - "text": "3.3\nSome Useful Discrete-Time Signal Models\n245", - "type": "text" - }, - { - "block_id": "p265-b1", - "global_id": 7283, - "bbox": [ - 133.57, - 97.81, - 364.65, - 109.77 - ], - "text": "DRILL 3.5\nExpansion and Interpolation", - "type": "text" - }, - { - "block_id": "p265-b2", - "global_id": 7284, - "bbox": [ - 133.57, - 118.47, - 510.17, - 153.46 - ], - "text": "A signal x[n] is expanded by factor 2 to obtain signal x[n/2]. The odd-numbered samples (n\nodd) in this signal have zero value. Show that the linearly interpolated odd-numbered samples\nare given by xi[n] = (1/2){x[n −1] + x[n + 1]}.", - "type": "text" - }, - { - "block_id": "p265-b3", - "global_id": 7285, - "bbox": [ - 127.94, - 202.55, - 468.57, - 216.49 - ], - "text": "3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS", - "type": "text" - }, - { - "block_id": "p265-b4", - "global_id": 7286, - "bbox": [ - 127.59, - 222.48, - 516.16, - 244.4 - ], - "text": "We now discuss some important discrete-time signal models that are encountered frequently in the\nstudy of discrete-time signals and systems.", - "type": "text" - }, - { - "block_id": "p265-b5", - "global_id": 7287, - "bbox": [ - 127.59, - 273.89, - 358.36, - 286.7 - ], - "text": "3.3-1 Discrete-Time Impulse Function δ[n]", - "type": "text" - }, - { - "block_id": "p265-b6", - "global_id": 7288, - "bbox": [ - 127.59, - 292.42, - 516.11, - 314.75 - ], - "text": "The discrete-time counterpart of the continuous-time impulse function δ(t) is δ[n], a Kronecker\ndelta function, defined by", - "type": "text" - }, - { - "block_id": "p265-b7", - "global_id": 7289, - "bbox": [ - 280.58, - 330.35, - 306.71, - 340.63 - ], - "text": "δ[n] =", - "type": "text" - }, - { - "block_id": "p265-b8", - "global_id": 7290, - "bbox": [ - 308.76, - 316.37, - 361.94, - 346.62 - ], - "text": "1\nn = 0\n0\nn̸ = 0", - "type": "text" - }, - { - "block_id": "p265-b9", - "global_id": 7291, - "bbox": [ - 127.59, - 360.59, - 516.16, - 406.41 - ], - "text": "This function, also called the unit impulse sequence, is shown in Fig. 3.6a. The shifted impulse\nsequence δ[n −m] is depicted in Fig. 3.6b. Unlike its continuous-time counterpart δ(t) (the Dirac\ndelta), the Kronecker delta is a very simple function, requiring no special esoteric knowledge of\ndistribution theory.", - "type": "text" - }, - { - "block_id": "p265-b10", - "global_id": 7292, - "bbox": [ - 219.38, - 433.23, - 232.71, - 441.34 - ], - "text": "d[n]", - "type": "text" - }, - { - "block_id": "p265-b11", - "global_id": 7293, - "bbox": [ - 211.94, - 503.53, - 220.82, - 511.53 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p265-b12", - "global_id": 7294, - "bbox": [ - 207.81, - 449.95, - 211.81, - 457.95 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p265-b13", - "global_id": 7295, - "bbox": [ - 211.56, - 593.83, - 221.21, - 601.83 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p265-b14", - "global_id": 7296, - "bbox": [ - 294.22, - 489.84, - 298.22, - 497.84 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p265-b15", - "global_id": 7297, - "bbox": [ - 219.38, - 524.28, - 249.15, - 532.58 - ], - "text": "d[n m]", - "type": "text" - }, - { - "block_id": "p265-b16", - "global_id": 7298, - "bbox": [ - 209.81, - 540.58, - 213.81, - 548.58 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p265-b17", - "global_id": 7299, - "bbox": [ - 294.22, - 580.09, - 298.22, - 588.09 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p265-b18", - "global_id": 7300, - "bbox": [ - 259.68, - 561.79, - 268.98, - 567.79 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p265-b19", - "global_id": 7301, - "bbox": [ - 277.38, - 580.28, - 283.16, - 588.28 - ], - "text": "m", - "type": "text" - }, - { - "block_id": "p265-b20", - "global_id": 7302, - "bbox": [ - 326.55, - 569.21, - 498.86, - 603.09 - ], - "text": "Figure 3.6 Discrete-time impulse function:\n(a) unit impulse sequence and (b) shifted\nimpulse sequence.", - "type": "text" - } - ] - }, - { - "page_num": 266, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p266-b0", - "global_id": 7303, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "246\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p266-b1", - "global_id": 7304, - "bbox": [ - 101.84, - 85.41, - 340.44, - 98.22 - ], - "text": "3.3-2 Discrete-Time Unit Step Function u[n]", - "type": "text" - }, - { - "block_id": "p266-b2", - "global_id": 7305, - "bbox": [ - 101.84, - 103.93, - 454.85, - 114.31 - ], - "text": "The discrete-time counterpart of the unit step function u(t) is u[n] (Fig. 3.7a), defined by", - "type": "text" - }, - { - "block_id": "p266-b3", - "global_id": 7306, - "bbox": [ - 248.38, - 133.96, - 274.8, - 144.23 - ], - "text": "u[n] =", - "type": "text" - }, - { - "block_id": "p266-b4", - "global_id": 7307, - "bbox": [ - 276.85, - 119.97, - 343.85, - 150.21 - ], - "text": "1\nfor n ≥0\n0\nfor n < 0", - "type": "text" - }, - { - "block_id": "p266-b5", - "global_id": 7308, - "bbox": [ - 101.85, - 163.63, - 490.39, - 185.96 - ], - "text": "If we want a signal to start at n = 0 (so that it has a zero value for all n < 0), we need only\nmultiply the signal by u[n].", - "type": "text" - }, - { - "block_id": "p266-b6", - "global_id": 7309, - "bbox": [ - 157.35, - 208.95, - 170.68, - 217.03 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p266-b7", - "global_id": 7310, - "bbox": [ - 156.4, - 246.65, - 182.4, - 254.65 - ], - "text": "0\n3", - "type": "text" - }, - { - "block_id": "p266-b8", - "global_id": 7311, - "bbox": [ - 145.58, - 219.35, - 149.58, - 227.35 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p266-b9", - "global_id": 7312, - "bbox": [ - 173.0, - 269.82, - 181.88, - 277.82 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p266-b10", - "global_id": 7313, - "bbox": [ - 197.1, - 246.13, - 236.25, - 254.65 - ], - "text": "5\nn", - "type": "text" - }, - { - "block_id": "p266-b11", - "global_id": 7314, - "bbox": [ - 172.62, - 375.24, - 182.26, - 383.24 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p266-b12", - "global_id": 7315, - "bbox": [ - 157.22, - 291.08, - 170.1, - 299.16 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p266-b13", - "global_id": 7316, - "bbox": [ - 156.93, - 350.84, - 290.33, - 359.7 - ], - "text": "n\n0", - "type": "text" - }, - { - "block_id": "p266-b14", - "global_id": 7317, - "bbox": [ - 145.63, - 305.13, - 149.63, - 313.13 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p266-b15", - "global_id": 7318, - "bbox": [ - 198.55, - 351.7, - 256.53, - 359.7 - ], - "text": "5\n10", - "type": "text" - }, - { - "block_id": "p266-b16", - "global_id": 7319, - "bbox": [ - 145.63, - 326.06, - 149.63, - 334.06 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p266-b17", - "global_id": 7320, - "bbox": [ - 230.89, - 230.98, - 240.19, - 236.98 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p266-b18", - "global_id": 7321, - "bbox": [ - 317.8, - 362.5, - 486.41, - 384.5 - ], - "text": "Figure 3.7 (a) A discrete-time unit step\nfunction u[n] and (b) its application.", - "type": "text" - }, - { - "block_id": "p266-b19", - "global_id": 7322, - "bbox": [ - 76.77, - 443.29, - 461.47, - 469.19 - ], - "text": "EXAMPLE 3.3\nDescribing Signals with Unit Step and Unit Impulse\nFunctions", - "type": "text" - }, - { - "block_id": "p266-b20", - "global_id": 7323, - "bbox": [ - 103.16, - 482.51, - 424.08, - 492.89 - ], - "text": "Describe the signal x[n] shown in Fig. 3.7b by a single expression valid for all n.", - "type": "text" - }, - { - "block_id": "p266-b21", - "global_id": 7324, - "bbox": [ - 103.16, - 515.39, - 477.01, - 537.73 - ], - "text": "There are many different ways of viewing x[n]. Although each way of viewing yields a different\nexpression, they are all equivalent. We shall consider here just one possible expression.", - "type": "text" - }, - { - "block_id": "p266-b22", - "global_id": 7325, - "bbox": [ - 103.16, - 539.3, - 477.01, - 574.36 - ], - "text": "The signal x[n] can be broken into three components: (1) a ramp component x1[n] from\nn = 0 to 4, (2) a scaled step component x2[n] from n = 5 to 10, and (3) an impulse component\nx3[n] represented by the negative spike at n = 8. Let us consider each one separately.", - "type": "text" - }, - { - "block_id": "p266-b23", - "global_id": 7326, - "bbox": [ - 103.17, - 575.17, - 477.02, - 621.41 - ], - "text": "We express x1[n] = n(u[n]−u[n−5]) to account for the signal from n = 0 to 4. Assuming\nthat the spike at n = 8 does not exist, we can express x2[n] = 4(u[n −5] −u[n −11]) to account\nfor the signal from n = 5 to 10. Once these two components have been added, the only part\nthat is unaccounted for is a spike of amplitude −2 at n = 8, which can be represented by", - "type": "text" - } - ] - }, - { - "page_num": 267, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p267-b0", - "global_id": 7327, - "bbox": [ - 287.05, - 62.89, - 516.14, - 71.98 - ], - "text": "3.3\nSome Useful Discrete-Time Signal Models\n247", - "type": "text" - }, - { - "block_id": "p267-b1", - "global_id": 7328, - "bbox": [ - 128.9, - 85.83, - 238.6, - 96.98 - ], - "text": "x3[n] = −2δ[n −8]. Hence,", - "type": "text" - }, - { - "block_id": "p267-b2", - "global_id": 7329, - "bbox": [ - 163.39, - 107.74, - 273.14, - 118.89 - ], - "text": "x[n] = x1[n] + x2[n] + x3[n]", - "type": "text" - }, - { - "block_id": "p267-b3", - "global_id": 7330, - "bbox": [ - 181.51, - 122.68, - 468.27, - 133.06 - ], - "text": "= n(u[n] −u[n −5]) + 4(u[n −5] −u[n −11]) −2δ[n −8]\nfor all n", - "type": "text" - }, - { - "block_id": "p267-b4", - "global_id": 7331, - "bbox": [ - 128.91, - 144.92, - 502.77, - 190.85 - ], - "text": "We stress again that the expression is valid for all values of n. The reader can find several other\nequivalent expressions for x[n]. For example, one may consider a scaled step function from\nn = 0 to 10, subtract a ramp over the range n = 0 to 3, and subtract the spike. You can also play\nwith breaking n into different ranges for your expression.", - "type": "text" - }, - { - "block_id": "p267-b5", - "global_id": 7332, - "bbox": [ - 127.59, - 239.05, - 319.67, - 253.82 - ], - "text": "3.3-3 Discrete-Time Exponential γ n", - "type": "text" - }, - { - "block_id": "p267-b6", - "global_id": 7333, - "bbox": [ - 127.59, - 256.33, - 425.57, - 269.91 - ], - "text": "A continuous-time exponential eλt can be expressed in an alternate form as", - "type": "text" - }, - { - "block_id": "p267-b7", - "global_id": 7334, - "bbox": [ - 254.04, - 282.5, - 389.68, - 296.99 - ], - "text": "eλt = γ t\n(γ = eλ or λ = lnγ )", - "type": "text" - }, - { - "block_id": "p267-b8", - "global_id": 7335, - "bbox": [ - 127.59, - 310.09, - 516.13, - 336.03 - ], - "text": "For example, e−0.3t = (0.7408)t because e−0.3 = 0.7408. Conversely, 4t = e1.386t because e1.386 = 4,\nthat is, ln4 = 1.386. In the study of continuous-time signals and systems, we prefer the form eλt", - "type": "text" - }, - { - "block_id": "p267-b9", - "global_id": 7336, - "bbox": [ - 127.59, - 334.4, - 516.13, - 371.9 - ], - "text": "rather than γ t. In contrast, the exponential form γ n is preferable in the study of discrete-time\nsignals and systems, as will become apparent later. The discrete-time exponential γ n can also be\nexpressed by using a natural base, as", - "type": "text" - }, - { - "block_id": "p267-b10", - "global_id": 7337, - "bbox": [ - 252.62, - 384.49, - 391.1, - 398.98 - ], - "text": "eλn = γ n\n(γ = eλ or λ = lnγ )", - "type": "text" - }, - { - "block_id": "p267-b11", - "global_id": 7338, - "bbox": [ - 127.59, - 414.68, - 515.63, - 426.06 - ], - "text": "Because of unfamiliarity with exponentials with bases other than e, exponentials of the form γ n", - "type": "text" - }, - { - "block_id": "p267-b12", - "global_id": 7339, - "bbox": [ - 127.59, - 428.05, - 516.14, - 438.01 - ], - "text": "may seem inconvenient and confusing at first. The reader is urged to plot some exponentials to", - "type": "text" - }, - { - "block_id": "p267-b13", - "global_id": 7340, - "bbox": [ - 127.59, - 439.0, - 364.11, - 453.0 - ], - "text": "acquire a sense of these functions. Also observe that γ −n =", - "type": "text" - }, - { - "block_id": "p267-b15", - "global_id": 7341, - "bbox": [ - 372.78, - 441.27, - 376.83, - 455.52 - ], - "text": "1\nγ", - "type": "text" - }, - { - "block_id": "p267-b16", - "global_id": 7342, - "bbox": [ - 378.59, - 431.62, - 387.51, - 443.44 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p267-b17", - "global_id": 7343, - "bbox": [ - 388.01, - 443.03, - 390.5, - 453.0 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p267-b18", - "global_id": 7344, - "bbox": [ - 133.57, - 492.78, - 414.46, - 504.74 - ], - "text": "DRILL 3.6\nEquivalent Forms of DT Exponentials", - "type": "text" - }, - { - "block_id": "p267-b19", - "global_id": 7345, - "bbox": [ - 151.5, - 515.81, - 510.15, - 529.79 - ], - "text": "(a) Show that (i) (0.25)−n = 4n, (ii) 4−n = (0.25)n, (iii) e2t = (7.389)t, (iv) e−2t =", - "type": "text" - }, - { - "block_id": "p267-b20", - "global_id": 7346, - "bbox": [ - 151.5, - 527.76, - 507.46, - 556.69 - ], - "text": "(0.1353)t = (7.389)−t, (v) e3n = (20.086)n, and (vi) e−1.5n = (0.2231)n = (4.4817)−n.\n(b) Show that (i) 2n = e0.693n, (ii) (0.5)n = e−0.693n, and (iii) (0.8)−n = e0.2231n.", - "type": "text" - }, - { - "block_id": "p267-b21", - "global_id": 7347, - "bbox": [ - 127.59, - 585.34, - 516.14, - 634.79 - ], - "text": "Nature of γ n. The signal eλn grows exponentially with n if Reλ > 0 (λ in the RHP), and decays\nexponentially if Reλ < 0 (λ in the LHP). It is constant or oscillates with constant amplitude if\nReλ = 0 (λ on the imaginary axis). Clearly, the location of λ in the complex plane indicates\nwhether the signal eλn will grow exponentially, decay exponentially, or oscillate with constant", - "type": "text" - } - ] - }, - { - "page_num": 268, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p268-b0", - "global_id": 7348, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "248\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p268-b1", - "global_id": 7349, - "bbox": [ - 343.05, - 170.02, - 455.74, - 178.32 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p268-b2", - "global_id": 7350, - "bbox": [ - 222.5, - 92.91, - 238.06, - 100.91 - ], - "text": "RHP", - "type": "text" - }, - { - "block_id": "p268-b3", - "global_id": 7351, - "bbox": [ - 122.07, - 221.25, - 146.3, - 229.54 - ], - "text": "l Plane", - "type": "text" - }, - { - "block_id": "p268-b4", - "global_id": 7352, - "bbox": [ - 186.65, - 254.74, - 407.12, - 262.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p268-b5", - "global_id": 7353, - "bbox": [ - 426.49, - 94.82, - 471.38, - 102.82 - ], - "text": "Exponentially", - "type": "text" - }, - { - "block_id": "p268-b6", - "global_id": 7354, - "bbox": [ - 432.49, - 103.82, - 465.37, - 111.82 - ], - "text": "increasing", - "type": "text" - }, - { - "block_id": "p268-b7", - "global_id": 7355, - "bbox": [ - 462.28, - 170.07, - 471.17, - 178.07 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p268-b8", - "global_id": 7356, - "bbox": [ - 137.63, - 124.96, - 145.63, - 206.06 - ], - "text": "Exponentially decreasing", - "type": "text" - }, - { - "block_id": "p268-b9", - "global_id": 7357, - "bbox": [ - 230.02, - 124.64, - 238.02, - 204.41 - ], - "text": "Exponentially increasing", - "type": "text" - }, - { - "block_id": "p268-b10", - "global_id": 7358, - "bbox": [ - 259.05, - 170.6, - 267.93, - 178.6 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p268-b11", - "global_id": 7359, - "bbox": [ - 147.68, - 92.91, - 398.26, - 107.2 - ], - "text": "Im\nLHP\nIm", - "type": "text" - }, - { - "block_id": "p268-b12", - "global_id": 7360, - "bbox": [ - 380.72, - 159.56, - 425.6, - 167.56 - ], - "text": "Exponentially", - "type": "text" - }, - { - "block_id": "p268-b13", - "global_id": 7361, - "bbox": [ - 386.06, - 168.56, - 420.26, - 176.56 - ], - "text": "decreasing", - "type": "text" - }, - { - "block_id": "p268-b14", - "global_id": 7362, - "bbox": [ - 441.71, - 221.25, - 465.94, - 229.54 - ], - "text": "g Plane", - "type": "text" - }, - { - "block_id": "p268-b15", - "global_id": 7363, - "bbox": [ - 101.84, - 269.07, - 307.42, - 278.68 - ], - "text": "Figure 3.8 The λ plane, the γ plane, and their mapping.", - "type": "text" - }, - { - "block_id": "p268-b16", - "global_id": 7364, - "bbox": [ - 101.84, - 304.88, - 490.37, - 339.17 - ], - "text": "amplitude (Fig. 3.8a). A constant signal (λ = 0) is also an oscillation with zero frequency. We now\nfind a similar criterion for determining the nature of γ n from the location of γ in the complex\nplane.", - "type": "text" - }, - { - "block_id": "p268-b17", - "global_id": 7365, - "bbox": [ - 101.84, - 339.53, - 490.4, - 422.87 - ], - "text": "Figure 3.8a shows a complex plane (λ plane). Consider a signal ejn. In this case, λ = j lies\non the imaginary axis (Fig. 3.8a), and therefore is a constant-amplitude oscillating signal. This\nsignal ejn can be expressed as γ n, where γ = ej. Because the magnitude of ej is unity, |γ | = 1.\nHence, when λ lies on the imaginary axis, the corresponding γ lies on a circle of unit radius,\ncentered at the origin (the unit circle illustrated in Fig. 3.8b). Therefore, a signal γ n oscillates with\nconstant amplitude if γ lies on the unit circle. Thus, the imaginary axis in the λ plane maps into\nthe unit circle in the γ plane.", - "type": "text" - }, - { - "block_id": "p268-b18", - "global_id": 7366, - "bbox": [ - 101.84, - 423.21, - 490.39, - 458.73 - ], - "text": "Next consider the signal eλn, where λ lies in the left half-plane in Fig. 3.8a. This means\nλ = a + jb, where a is negative (a < 0). In this case, the signal decays exponentially. This signal\ncan be expressed as γ n, where", - "type": "text" - }, - { - "block_id": "p268-b19", - "global_id": 7367, - "bbox": [ - 251.08, - 465.37, - 340.65, - 479.76 - ], - "text": "γ = eλ = ea+jb = ea ejb", - "type": "text" - }, - { - "block_id": "p268-b20", - "global_id": 7368, - "bbox": [ - 101.84, - 494.95, - 116.22, - 504.91 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p268-b21", - "global_id": 7369, - "bbox": [ - 215.07, - 514.17, - 377.17, - 526.05 - ], - "text": "|γ | = |ea||ejb| = ea\nbecause |ejb| = 1", - "type": "text" - }, - { - "block_id": "p268-b22", - "global_id": 7370, - "bbox": [ - 101.84, - 537.1, - 490.39, - 598.92 - ], - "text": "Also, a is negative (a < 0). Hence, |γ | = ea < 1. This result means that the corresponding γ lies\ninside the unit circle. Therefore, a signal γ n decays exponentially if γ lies within the unit circle\n(Fig. 3.8b). If, in the preceding case we select a to be positive (λ in the right half-plane), then\n|γ | > 1, and γ lies outside the unit circle. Therefore, a signal γ n grows exponentially if γ lies\noutside the unit circle (Fig. 3.8b).", - "type": "text" - }, - { - "block_id": "p268-b23", - "global_id": 7371, - "bbox": [ - 101.84, - 600.5, - 490.38, - 634.79 - ], - "text": "To summarize, the imaginary axis in the λ plane maps into the unit circle in the γ plane. The\nleft half-plane in the λ plane maps into the inside of the unit circle and the right half of the λ plane\nmaps into the outside of the unit circle in the γ plane, as depicted in Fig. 3.8.", - "type": "text" - } - ] - }, - { - "page_num": 269, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p269-b0", - "global_id": 7372, - "bbox": [ - 287.05, - 62.89, - 516.14, - 71.98 - ], - "text": "3.3\nSome Useful Discrete-Time Signal Models\n249", - "type": "text" - }, - { - "block_id": "p269-b1", - "global_id": 7373, - "bbox": [ - 217.81, - 220.81, - 226.69, - 228.81 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p269-b2", - "global_id": 7374, - "bbox": [ - 160.38, - 91.97, - 164.38, - 99.97 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p269-b3", - "global_id": 7375, - "bbox": [ - 160.95, - 188.1, - 258.59, - 196.1 - ], - "text": "2\n4\n6\n1\n3\n5\n0", - "type": "text" - }, - { - "block_id": "p269-b4", - "global_id": 7376, - "bbox": [ - 196.5, - 103.96, - 215.3, - 113.49 - ], - "text": "(0.8)n", - "type": "text" - }, - { - "block_id": "p269-b5", - "global_id": 7377, - "bbox": [ - 271.63, - 187.65, - 275.63, - 195.65 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p269-b6", - "global_id": 7378, - "bbox": [ - 393.86, - 220.81, - 403.5, - 228.81 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p269-b7", - "global_id": 7379, - "bbox": [ - 464.55, - 180.62, - 468.55, - 188.62 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p269-b8", - "global_id": 7380, - "bbox": [ - 337.57, - 167.72, - 472.05, - 175.72 - ], - "text": "2\n4\n6\n7\n8\n1\n3\n5\n0", - "type": "text" - }, - { - "block_id": "p269-b9", - "global_id": 7381, - "bbox": [ - 337.57, - 91.97, - 398.47, - 107.43 - ], - "text": "(0.8)n\n1", - "type": "text" - }, - { - "block_id": "p269-b10", - "global_id": 7382, - "bbox": [ - 330.9, - 232.43, - 341.57, - 240.72 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p269-b11", - "global_id": 7383, - "bbox": [ - 195.87, - 356.0, - 275.96, - 364.41 - ], - "text": "n\n2\n4\n6", - "type": "text" - }, - { - "block_id": "p269-b12", - "global_id": 7384, - "bbox": [ - 160.74, - 260.5, - 164.74, - 268.5 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p269-b13", - "global_id": 7385, - "bbox": [ - 161.12, - 356.41, - 243.68, - 364.41 - ], - "text": "1\n3\n5\n0", - "type": "text" - }, - { - "block_id": "p269-b14", - "global_id": 7386, - "bbox": [ - 217.81, - 376.75, - 226.69, - 384.75 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p269-b15", - "global_id": 7387, - "bbox": [ - 337.53, - 356.0, - 452.35, - 364.05 - ], - "text": "n\n2\n4\n6\n1\n3\n5\n0", - "type": "text" - }, - { - "block_id": "p269-b16", - "global_id": 7388, - "bbox": [ - 196.64, - 271.97, - 391.81, - 281.5 - ], - "text": "(1.1)n\n(0.5)n", - "type": "text" - }, - { - "block_id": "p269-b17", - "global_id": 7389, - "bbox": [ - 394.02, - 376.75, - 403.34, - 384.75 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p269-b18", - "global_id": 7390, - "bbox": [ - 151.5, - 390.0, - 307.48, - 400.68 - ], - "text": "Figure 3.9 Discrete-time exponentials γ n.", - "type": "text" - }, - { - "block_id": "p269-b19", - "global_id": 7391, - "bbox": [ - 127.59, - 434.35, - 516.14, - 507.7 - ], - "text": "Plots of (0.8)n and (−0.8)n appear in Figs. 3.9a and 3.9b, respectively. Plots of (0.5)n and\n(1.1)n appear in Figs. 3.9c and 3.9d, respectively. These plots verify our earlier conclusions about\nthe location of γ and the nature of signal growth. Observe that a signal (−|γ |)n alternates sign\nsuccessively (is positive for even values of n and negative for odd values of n, as depicted in\nFig. 3.9b). Also, the exponential (0.5)n decays faster than (0.8)n because 0.5 is closer to the origin\nthan 0.8. The exponential (0.5)n can also be expressed as 2−n because (0.5)−1 = 2.", - "type": "text" - }, - { - "block_id": "p269-b20", - "global_id": 7392, - "bbox": [ - 133.57, - 548.95, - 358.66, - 560.91 - ], - "text": "DRILL 3.7\nSketching DT Exponentials", - "type": "text" - }, - { - "block_id": "p269-b21", - "global_id": 7393, - "bbox": [ - 133.57, - 568.39, - 510.16, - 627.81 - ], - "text": "Sketch the following signals: (a) (1)n, (b) (−1)n, (c) (0.5)n, (d) (−0.5)n, (e) (0.5)−n, (f) 2−n,\nand (g) (−2)n. Express these exponentials as γ n, and plot γ in the complex plane for each case.\nVerify that γ n decays exponentially with n if γ lies inside the unit circle and that γ n grows\nwith n if γ is outside the unit circle. If γ is on the unit circle, γ n is constant or oscillates with a\nconstant amplitude.", - "type": "text" - } - ] - }, - { - "page_num": 270, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p270-b0", - "global_id": 7394, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "250\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p270-b1", - "global_id": 7395, - "bbox": [ - 101.84, - 86.0, - 490.36, - 107.93 - ], - "text": "Accurately hand-sketching DT signals can be tedious and difficult. As the next example\nshows, MATLAB is particularly well suited to plot DT signals, including exponentials.", - "type": "text" - }, - { - "block_id": "p270-b2", - "global_id": 7396, - "bbox": [ - 76.77, - 155.0, - 397.63, - 166.95 - ], - "text": "EXAMPLE 3.4\nPlotting DT Exponentials with MATLAB", - "type": "text" - }, - { - "block_id": "p270-b3", - "global_id": 7397, - "bbox": [ - 103.16, - 182.2, - 477.01, - 206.24 - ], - "text": "Use MATLAB to plot the following discrete-time signals over (0 ≤n ≤8): (a) xa[n] = (0.8)n,\n(b) xb[n] = (−0.8)n, (c) xc[n] = (0.5)n, and (d) xd[n] = (1.1)n.", - "type": "text" - }, - { - "block_id": "p270-b4", - "global_id": 7398, - "bbox": [ - 103.16, - 228.45, - 477.02, - 262.32 - ], - "text": "To begin, we use anonymous functions to represent each of the four signals. Next, we plot\nthese functions over the desired range of n. The results, shown in Fig. 3.10, match the earlier\nFig. 3.9 plots of the same signals.", - "type": "text" - }, - { - "block_id": "p270-b5", - "global_id": 7399, - "bbox": [ - 103.16, - 282.54, - 469.29, - 352.28 - ], - "text": ">>\nn = (0:8); x_a = @(n) (0.8).^n; x_b = @(n) (-0.8).^(n);\n>>\nx_c = @(n) (0.5).^n; x_d = @(n) (1.1).^n;\n>>\nsubplot(2,2,1); stem(n,x_a(n),’k’); ylabel(’x_a[n]’); xlabel(’n’);\n>>\nsubplot(2,2,2); stem(n,x_b(n),’k’); ylabel(’x_b[n]’); xlabel(’n’);\n>>\nsubplot(2,2,3); stem(n,x_c(n),’k’); ylabel(’x_c[n]’); xlabel(’n’);\n>>\nsubplot(2,2,4); stem(n,x_d(n),’k’); ylabel(’x_d[n]’); xlabel(’n’);", - "type": "text" - }, - { - "block_id": "p270-b6", - "global_id": 7400, - "bbox": [ - 132.68, - 449.36, - 136.68, - 457.36 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p270-b7", - "global_id": 7401, - "bbox": [ - 125.93, - 417.49, - 136.61, - 425.49 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p270-b8", - "global_id": 7402, - "bbox": [ - 132.68, - 385.62, - 136.68, - 393.62 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p270-b9", - "global_id": 7403, - "bbox": [ - 111.43, - 412.85, - 120.95, - 428.84 - ], - "text": "xa[n]", - "type": "text" - }, - { - "block_id": "p270-b10", - "global_id": 7404, - "bbox": [ - 320.93, - 449.36, - 327.6, - 457.36 - ], - "text": "-1", - "type": "text" - }, - { - "block_id": "p270-b11", - "global_id": 7405, - "bbox": [ - 323.18, - 417.49, - 327.18, - 425.49 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p270-b12", - "global_id": 7406, - "bbox": [ - 323.18, - 385.61, - 327.18, - 393.61 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p270-b13", - "global_id": 7407, - "bbox": [ - 306.43, - 412.52, - 315.95, - 428.84 - ], - "text": "xb[n]", - "type": "text" - }, - { - "block_id": "p270-b14", - "global_id": 7408, - "bbox": [ - 138.89, - 561.33, - 477.68, - 578.84 - ], - "text": "0\n2\n4\n6\n8\n0\n2\n4\n6\n8\nn", - "type": "text" - }, - { - "block_id": "p270-b15", - "global_id": 7409, - "bbox": [ - 132.68, - 552.11, - 136.68, - 560.11 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p270-b16", - "global_id": 7410, - "bbox": [ - 125.93, - 519.86, - 136.61, - 527.86 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p270-b17", - "global_id": 7411, - "bbox": [ - 132.68, - 487.62, - 136.68, - 495.62 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p270-b18", - "global_id": 7412, - "bbox": [ - 111.43, - 515.6, - 120.95, - 531.59 - ], - "text": "xc[n]", - "type": "text" - }, - { - "block_id": "p270-b19", - "global_id": 7413, - "bbox": [ - 401.4, - 570.04, - 405.8, - 578.84 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p270-b20", - "global_id": 7414, - "bbox": [ - 139.18, - 455.1, - 477.96, - 472.6 - ], - "text": "0\n2\n4\n6\n8\n0\n2\n4\n6\n8\nn\nn", - "type": "text" - }, - { - "block_id": "p270-b21", - "global_id": 7415, - "bbox": [ - 323.18, - 552.11, - 327.18, - 560.11 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p270-b22", - "global_id": 7416, - "bbox": [ - 323.18, - 530.61, - 327.18, - 538.61 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p270-b23", - "global_id": 7417, - "bbox": [ - 323.18, - 509.12, - 327.18, - 517.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p270-b24", - "global_id": 7418, - "bbox": [ - 323.18, - 487.62, - 327.18, - 495.62 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p270-b25", - "global_id": 7419, - "bbox": [ - 308.68, - 515.27, - 318.2, - 531.59 - ], - "text": "xd[n]", - "type": "text" - }, - { - "block_id": "p270-b26", - "global_id": 7420, - "bbox": [ - 110.73, - 585.52, - 235.26, - 594.76 - ], - "text": "Figure 3.10 DT plots for Ex. 3.4.", - "type": "text" - } - ] - }, - { - "page_num": 271, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p271-b0", - "global_id": 7421, - "bbox": [ - 287.05, - 62.89, - 516.14, - 71.98 - ], - "text": "3.3\nSome Useful Discrete-Time Signal Models\n251", - "type": "text" - }, - { - "block_id": "p271-b1", - "global_id": 7422, - "bbox": [ - 127.59, - 85.66, - 351.81, - 98.48 - ], - "text": "3.3-4 Discrete-Time Sinusoid cos(n + θ)", - "type": "text" - }, - { - "block_id": "p271-b2", - "global_id": 7423, - "bbox": [ - 127.59, - 104.19, - 516.13, - 138.48 - ], - "text": "A general discrete-time sinusoid can be expressed as Ccos(n+θ), where C is the amplitude, and\nθ is the phase in radians. Also, n is an angle in radians. Hence, the dimensions of the frequency\n are radians per sample. This sinusoid may also be expressed as", - "type": "text" - }, - { - "block_id": "p271-b3", - "global_id": 7424, - "bbox": [ - 250.97, - 158.91, - 392.73, - 169.29 - ], - "text": "Ccos(n + θ) = Ccos(2πFn + θ)", - "type": "text" - }, - { - "block_id": "p271-b4", - "global_id": 7425, - "bbox": [ - 127.59, - 189.71, - 516.13, - 225.5 - ], - "text": "where F = /2π. Therefore, the dimensions of the discrete-time frequency F are (radians/2π)\nper sample, which is equal to cycles per sample. This means if N0 is the period (samples/cycle) of\nthe sinusoid, then the frequency of the sinusoid F = 1/N0 (samples/cycle).", - "type": "text" - }, - { - "block_id": "p271-b5", - "global_id": 7426, - "bbox": [ - 145.52, - 223.94, - 351.88, - 235.96 - ], - "text": "Figure 3.11 shows a discrete-time sinusoid cos( π", - "type": "text" - }, - { - "block_id": "p271-b6", - "global_id": 7427, - "bbox": [ - 127.59, - 223.94, - 516.14, - 259.87 - ], - "text": "12n + π\n4 ). For this case, the frequency is\n = π/12 radians/sample. Alternately, the frequency is F = 1/24 cycles/sample. In other words,\nthere are 24 samples in one cycle of the sinusoid.", - "type": "text" - }, - { - "block_id": "p271-b7", - "global_id": 7428, - "bbox": [ - 145.52, - 261.45, - 255.88, - 271.83 - ], - "text": "Because cos(−x) = cos(x),", - "type": "text" - }, - { - "block_id": "p271-b8", - "global_id": 7429, - "bbox": [ - 261.24, - 292.25, - 382.47, - 302.63 - ], - "text": "cos(−n + θ) = cos(n −θ)", - "type": "text" - }, - { - "block_id": "p271-b9", - "global_id": 7430, - "bbox": [ - 127.59, - 323.06, - 516.13, - 345.39 - ], - "text": "This shows that both cos(n+θ) and cos(−n+θ) have the same frequency (). Therefore, the\nfrequency of cos(n + θ) is ||.", - "type": "text" - }, - { - "block_id": "p271-b10", - "global_id": 7431, - "bbox": [ - 339.51, - 435.61, - 455.53, - 448.3 - ], - "text": "3\n15\n27\nn", - "type": "text" - }, - { - "block_id": "p271-b11", - "global_id": 7432, - "bbox": [ - 330.37, - 386.66, - 395.04, - 403.88 - ], - "text": "1\ncos(\n)\np\n12\np", - "type": "text" - }, - { - "block_id": "p271-b12", - "global_id": 7433, - "bbox": [ - 370.11, - 391.03, - 390.35, - 403.88 - ], - "text": "4\nn", - "type": "text" - }, - { - "block_id": "p271-b13", - "global_id": 7434, - "bbox": [ - 167.46, - 435.31, - 327.73, - 443.61 - ], - "text": "33\n21\n9\n0", - "type": "text" - }, - { - "block_id": "p271-b14", - "global_id": 7435, - "bbox": [ - 151.5, - 482.83, - 313.82, - 494.07 - ], - "text": "Figure 3.11 A discrete-time sinusoid cos( π", - "type": "text" - }, - { - "block_id": "p271-b15", - "global_id": 7436, - "bbox": [ - 308.96, - 482.83, - 343.41, - 496.61 - ], - "text": "12n + π\n4 ).", - "type": "text" - }, - { - "block_id": "p271-b16", - "global_id": 7437, - "bbox": [ - 127.89, - 533.72, - 386.4, - 559.78 - ], - "text": "SAMPLED CONTINUOUS-TIME SINUSOID YIELDS\nA DISCRETE-TIME SINUSOID", - "type": "text" - }, - { - "block_id": "p271-b17", - "global_id": 7438, - "bbox": [ - 127.59, - 563.4, - 516.12, - 585.73 - ], - "text": "A continuous-time sinusoid cosωt sampled every T seconds yields a discrete-time sequence whose\nnth element (at t = nT) is cosωnT. Thus, the sampled signal x[n] is given by", - "type": "text" - }, - { - "block_id": "p271-b18", - "global_id": 7439, - "bbox": [ - 232.5, - 606.16, - 410.46, - 616.54 - ], - "text": "x[n] = cosωnT = cosn\nwhere = ωT", - "type": "text" - } - ] - }, - { - "page_num": 272, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p272-b0", - "global_id": 7440, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "252\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p272-b1", - "global_id": 7441, - "bbox": [ - 101.84, - 85.4, - 490.39, - 107.74 - ], - "text": "Thus, a continuous-time sinusoid cosωt sampled every T seconds yields a discrete-time sinusoid\ncosn, where = ωT.†", - "type": "text" - }, - { - "block_id": "p272-b2", - "global_id": 7442, - "bbox": [ - 101.84, - 137.49, - 352.88, - 152.25 - ], - "text": "3.3-5 Discrete-Time Complex Exponential ejn", - "type": "text" - }, - { - "block_id": "p272-b3", - "global_id": 7443, - "bbox": [ - 101.84, - 154.76, - 430.54, - 168.35 - ], - "text": "Using Euler’s formula, we can express an exponential ejn in terms of sinusoids as", - "type": "text" - }, - { - "block_id": "p272-b4", - "global_id": 7444, - "bbox": [ - 163.87, - 183.57, - 428.36, - 198.06 - ], - "text": "ejn = (cosn + jsinn)\nand\ne−jn = (cosn −jsinn)", - "type": "text" - }, - { - "block_id": "p272-b5", - "global_id": 7445, - "bbox": [ - 101.85, - 214.11, - 490.37, - 239.74 - ], - "text": "These equations show that the frequency of both ejn and e−jn is (radians/sample). Therefore,\nthe frequency of ejn is ||.", - "type": "text" - }, - { - "block_id": "p272-b6", - "global_id": 7446, - "bbox": [ - 119.78, - 241.32, - 259.64, - 251.69 - ], - "text": "Observe that for r = 1 and θ = n,", - "type": "text" - }, - { - "block_id": "p272-b7", - "global_id": 7447, - "bbox": [ - 275.23, - 266.92, - 315.8, - 281.32 - ], - "text": "ejn = rejθ", - "type": "text" - }, - { - "block_id": "p272-b8", - "global_id": 7448, - "bbox": [ - 101.84, - 297.56, - 490.39, - 323.09 - ], - "text": "This equation shows that the magnitude and angle of ejn are 1 and n, respectively. In the\ncomplex plane, ejn is a point on a unit circle at an angle n.", - "type": "text" - }, - { - "block_id": "p272-b9", - "global_id": 7449, - "bbox": [ - 76.77, - 373.92, - 384.71, - 385.88 - ], - "text": "EXAMPLE 3.5\nPlotting a DT Sinusoid with MATLAB", - "type": "text" - }, - { - "block_id": "p272-b10", - "global_id": 7450, - "bbox": [ - 103.16, - 402.13, - 339.48, - 412.51 - ], - "text": "Using MATLAB, plot the discrete-time sinusoid x[n] = cos", - "type": "text" - }, - { - "block_id": "p272-b11", - "global_id": 7451, - "bbox": [ - 340.58, - 394.12, - 351.02, - 407.47 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p272-b12", - "global_id": 7452, - "bbox": [ - 345.8, - 394.12, - 381.1, - 415.32 - ], - "text": "12n + π\n4", - "type": "text" - }, - { - "block_id": "p272-b13", - "global_id": 7453, - "bbox": [ - 381.1, - 402.55, - 383.59, - 412.51 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p272-b14", - "global_id": 7454, - "bbox": [ - 103.16, - 435.42, - 476.99, - 469.29 - ], - "text": "We represent the desired sinusoid using an anonymous function. Next, we plot this function\nover the desired range of n. The result, shown in Fig. 3.12, matches the plot of the same signal\nshown in Fig. 3.11.", - "type": "text" - }, - { - "block_id": "p272-b15", - "global_id": 7455, - "bbox": [ - 103.16, - 479.54, - 390.83, - 501.47 - ], - "text": ">>\nn = (-30:30); x = @(n) cos(n*pi/12+pi/4);\n>>\nclf; stem(n,x(n),’k’); ylabel(’x[n]’); xlabel(’n’);", - "type": "text" - }, - { - "block_id": "p272-b16", - "global_id": 7456, - "bbox": [ - 101.84, - 540.5, - 490.4, - 618.48 - ], - "text": "† Superficially, it may appear that a discrete-time sinusoid is a continuous-time sinusoid’s cousin in a\nstriped suit. However, some of the properties of discrete-time sinusoids are very different from those of\ncontinuous-time sinusoids. For instance, not every discrete-time sinusoid is periodic. A sinusoid cosn is\nperiodic only if is a rational multiple of 2π. Also, discrete-time sinusoids are bandlimited to = π.\nAny sinusoid with ≥π can always be expressed as a sinusoid of some frequency ≤π. These peculiar\nproperties are the direct consequence of the fact that the period of a discrete-time sinusoid must be an integer.\nThese topics are discussed in Chs. 5 and 9.", - "type": "text" - } - ] - }, - { - "page_num": 273, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p273-b0", - "global_id": 7457, - "bbox": [ - 312.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.4\nExamples of Discrete-Time Systems\n253", - "type": "text" - }, - { - "block_id": "p273-b1", - "global_id": 7458, - "bbox": [ - 136.07, - 164.01, - 482.16, - 185.2 - ], - "text": "–30\n–20\n–10\n0\n10\n20\n30\nn", - "type": "text" - }, - { - "block_id": "p273-b2", - "global_id": 7459, - "bbox": [ - 130.92, - 154.78, - 138.92, - 162.78 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p273-b3", - "global_id": 7460, - "bbox": [ - 134.92, - 118.78, - 138.92, - 126.78 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p273-b4", - "global_id": 7461, - "bbox": [ - 134.92, - 82.77, - 138.92, - 90.77 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p273-b5", - "global_id": 7462, - "bbox": [ - 120.7, - 114.34, - 129.5, - 129.0 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p273-b6", - "global_id": 7463, - "bbox": [ - 119.94, - 191.88, - 260.42, - 201.12 - ], - "text": "Figure 3.12 Sinusoid plot for Ex. 3.5.", - "type": "text" - }, - { - "block_id": "p273-b7", - "global_id": 7464, - "bbox": [ - 127.94, - 252.54, - 413.27, - 266.49 - ], - "text": "3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p273-b8", - "global_id": 7465, - "bbox": [ - 127.59, - 272.48, - 516.13, - 318.3 - ], - "text": "We shall give here four examples of discrete-time systems. In the first two examples, the signals\nare inherently of the discrete-time variety. In the third and fourth examples, a continuous-time\nsignal is processed by a discrete-time system, as illustrated in Fig. 3.2, by discretizing the signal\nthrough sampling.", - "type": "text" - }, - { - "block_id": "p273-b9", - "global_id": 7466, - "bbox": [ - 103.92, - 342.31, - 293.44, - 354.27 - ], - "text": "EXAMPLE 3.6\nSavings Account", - "type": "text" - }, - { - "block_id": "p273-b10", - "global_id": 7467, - "bbox": [ - 128.9, - 372.24, - 502.76, - 418.17 - ], - "text": "A person makes a deposit (the input) in a bank regularly at an interval of T (say, 1 month). The\nbank pays a certain interest on the account balance during the period T and mails out a periodic\nstatement of the account balance (the output) to the depositor. Find the equation relating the\noutput y[n] (the balance) to the input x[n] (the deposit).", - "type": "text" - }, - { - "block_id": "p273-b11", - "global_id": 7468, - "bbox": [ - 128.9, - 441.09, - 350.52, - 451.05 - ], - "text": "In this case, the signals are inherently discrete time. Let", - "type": "text" - }, - { - "block_id": "p273-b12", - "global_id": 7469, - "bbox": [ - 212.47, - 462.6, - 414.69, - 487.91 - ], - "text": "x[n] = deposit made at the nth discrete instant\ny[n] = account balance at the nth instant computed", - "type": "text" - }, - { - "block_id": "p273-b13", - "global_id": 7470, - "bbox": [ - 228.53, - 492.48, - 419.2, - 502.85 - ], - "text": "immediately after receipt of the nth deposit x[n]", - "type": "text" - }, - { - "block_id": "p273-b14", - "global_id": 7471, - "bbox": [ - 224.43, - 507.42, - 361.87, - 517.8 - ], - "text": "r = interest per dollar per period T", - "type": "text" - }, - { - "block_id": "p273-b15", - "global_id": 7472, - "bbox": [ - 128.91, - 529.34, - 502.76, - 551.68 - ], - "text": "The balance y[n] is the sum of (i) the previous balance y[n −1], (ii) the interest on y[n −1]\nduring the period T, and (iii) the deposit x[n]", - "type": "text" - }, - { - "block_id": "p273-b16", - "global_id": 7473, - "bbox": [ - 249.15, - 563.22, - 382.48, - 573.59 - ], - "text": "y[n] = y[n −1] + ry[n −1] + x[n]", - "type": "text" - }, - { - "block_id": "p273-b17", - "global_id": 7474, - "bbox": [ - 267.24, - 578.16, - 363.24, - 588.53 - ], - "text": "= (1 + r)y[n −1] + x[n]", - "type": "text" - }, - { - "block_id": "p273-b18", - "global_id": 7475, - "bbox": [ - 128.91, - 600.49, - 137.2, - 610.46 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p273-b19", - "global_id": 7476, - "bbox": [ - 241.62, - 612.03, - 502.76, - 622.41 - ], - "text": "y[n] −ay[n −1] = x[n]\na = 1 + r\n(3.3)", - "type": "text" - }, - { - "block_id": "p273-b20", - "global_id": 7477, - "bbox": [ - 128.91, - 630.96, - 501.51, - 641.34 - ], - "text": "In this example the deposit x[n] is the input (cause) and the balance y[n] is the output (effect).", - "type": "text" - } - ] - }, - { - "page_num": 274, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p274-b0", - "global_id": 7478, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "254\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p274-b1", - "global_id": 7479, - "bbox": [ - 103.16, - 86.24, - 477.04, - 144.03 - ], - "text": "A withdrawal from the account is a negative deposit. Therefore, this formulation can\nhandle deposits as well as withdrawals. It also applies to a loan payment problem with the\ninitial value y[0] = −M, where M is the amount of the loan. A loan is an initial deposit with a\nnegative value. Alternately, we may treat a loan of M dollars taken at n = 0 as an input of −M\nat n = 0 (see Prob. 3.8-23).", - "type": "text" - }, - { - "block_id": "p274-b2", - "global_id": 7480, - "bbox": [ - 103.16, - 145.92, - 477.02, - 167.93 - ], - "text": "We can express Eq. (3.3) in an alternate form. The choice of index n in Eq. (3.3) is\ncompletely arbitrary, so we can substitute n + 1 for n to obtain", - "type": "text" - }, - { - "block_id": "p274-b3", - "global_id": 7481, - "bbox": [ - 236.3, - 179.43, - 477.01, - 189.81 - ], - "text": "y[n + 1] −ay[n] = x[n + 1]\n(3.4)", - "type": "text" - }, - { - "block_id": "p274-b4", - "global_id": 7482, - "bbox": [ - 103.16, - 201.3, - 477.02, - 235.58 - ], - "text": "We also could have obtained Eq. (3.4) directly by realizing that y[n+1], the balance at instant\n(n + 1), is the sum of y[n] plus ry[n] (the interest on y[n]) plus the deposit (input) x[n + 1] at\ninstant (n + 1).", - "type": "text" - }, - { - "block_id": "p274-b5", - "global_id": 7483, - "bbox": [ - 103.17, - 237.58, - 477.03, - 295.36 - ], - "text": "The difference equation in Eq. (3.3) uses delays, whereas the form in Eq. (3.4) uses\nadvances. Thus, Eq. (3.3) is said to be in delay form and Eq. (3.4) is said to be in advance\nform. The delay form is more natural because operation of delay is causal, hence realizable.\nIn contrast, advance operation, being noncausal, is unrealizable. We use the advance form\nprimarily for its mathematical convenience over the delay form.†", - "type": "text" - }, - { - "block_id": "p274-b6", - "global_id": 7484, - "bbox": [ - 103.16, - 297.36, - 477.02, - 367.09 - ], - "text": "We shall now represent this system in a block diagram form, which is basically a road map\nto a hardware (or software) realization of the system. For this purpose, the causal (realizable)\ndelay form in Eq. (3.3) will be used. There are three basic operations in this equation: addition,\nscalar multiplication, and delay. Figure 3.13 shows their schematic representation. In addition,\nwe also have a pickoff node (Fig. 3.13d), which is used to provide multiple copies of a signal\nat its input.", - "type": "text" - }, - { - "block_id": "p274-b7", - "global_id": 7485, - "bbox": [ - 113.0, - 386.71, - 218.14, - 403.02 - ], - "text": "x[n]\ny[n] x[n] w[n]", - "type": "text" - }, - { - "block_id": "p274-b8", - "global_id": 7486, - "bbox": [ - 108.75, - 425.52, - 122.68, - 433.19 - ], - "text": "w[n]", - "type": "text" - }, - { - "block_id": "p274-b9", - "global_id": 7487, - "bbox": [ - 160.33, - 440.87, - 168.76, - 448.47 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p274-b10", - "global_id": 7488, - "bbox": [ - 274.81, - 387.79, - 278.61, - 395.39 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p274-b11", - "global_id": 7489, - "bbox": [ - 240.4, - 394.92, - 340.99, - 402.8 - ], - "text": "y[n] ax[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p274-b12", - "global_id": 7490, - "bbox": [ - 286.11, - 440.87, - 295.28, - 448.47 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p274-b13", - "global_id": 7491, - "bbox": [ - 163.55, - 461.54, - 302.01, - 469.42 - ], - "text": "y[n] x[n – 1]\nx[n]", - "type": "text" - }, - { - "block_id": "p274-b14", - "global_id": 7492, - "bbox": [ - 205.76, - 470.07, - 211.24, - 477.67 - ], - "text": "D", - "type": "text" - }, - { - "block_id": "p274-b15", - "global_id": 7493, - "bbox": [ - 219.99, - 491.98, - 228.43, - 499.58 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p274-b16", - "global_id": 7494, - "bbox": [ - 194.97, - 511.97, - 252.97, - 519.64 - ], - "text": "x[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p274-b17", - "global_id": 7495, - "bbox": [ - 217.62, - 544.83, - 229.85, - 552.5 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p274-b18", - "global_id": 7496, - "bbox": [ - 219.3, - 558.79, - 228.16, - 566.39 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p274-b19", - "global_id": 7497, - "bbox": [ - 103.16, - 575.79, - 369.11, - 585.03 - ], - "text": "Figure 3.13 Schematic representations of basic operations on sequences.", - "type": "text" - }, - { - "block_id": "p274-b20", - "global_id": 7498, - "bbox": [ - 103.16, - 606.81, - 477.04, - 641.31 - ], - "text": "† Use of the advance form results in discrete-time system equations that are identical in form to those for\ncontinuous-time systems. This will become apparent later. In transform analysis, advance form leads to\nthe more convenient variable z instead of the clumsy z−1 that arises from delay form.", - "type": "text" - } - ] - }, - { - "page_num": 275, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p275-b0", - "global_id": 7499, - "bbox": [ - 312.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.4\nExamples of Discrete-Time Systems\n255", - "type": "text" - }, - { - "block_id": "p275-b1", - "global_id": 7500, - "bbox": [ - 135.07, - 88.08, - 322.15, - 96.17 - ], - "text": "y[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p275-b2", - "global_id": 7501, - "bbox": [ - 193.07, - 135.87, - 220.61, - 144.16 - ], - "text": "y[n 1]", - "type": "text" - }, - { - "block_id": "p275-b3", - "global_id": 7502, - "bbox": [ - 132.0, - 118.54, - 183.56, - 130.41 - ], - "text": "ay[n 1]\na", - "type": "text" - }, - { - "block_id": "p275-b4", - "global_id": 7503, - "bbox": [ - 165.27, - 95.21, - 307.16, - 112.23 - ], - "text": "N", - "type": "text" - }, - { - "block_id": "p275-b5", - "global_id": 7504, - "bbox": [ - 234.42, - 129.87, - 240.19, - 137.87 - ], - "text": "D", - "type": "text" - }, - { - "block_id": "p275-b6", - "global_id": 7505, - "bbox": [ - 128.9, - 153.54, - 331.2, - 162.78 - ], - "text": "Figure 3.14 Realization of the savings account system.", - "type": "text" - }, - { - "block_id": "p275-b7", - "global_id": 7506, - "bbox": [ - 128.9, - 182.68, - 502.77, - 264.37 - ], - "text": "Figure 3.14 shows in block diagram form a system represented by Eq. (3.3). To understand\nthis realization, it is helpful to rewrite Eq. (3.3) as y[n] = ay[n −1] + x[n] (a = 1 + r). Now,\nassume that the output y[n] is available at the pickoff node N. Unit delay of y[n] results in\ny[n −1], which is multiplied by a scalar of value a to yield ay[n −1]. Next, we generate y[n]\nby adding the input x[n] and ay[n −1].† Observe that node N is a pickoff node, from which\ntwo copies of the output signal flow out: one as the feedback signal and the other as the output\nsignal.", - "type": "text" - }, - { - "block_id": "p275-b8", - "global_id": 7507, - "bbox": [ - 102.51, - 303.75, - 279.38, - 315.7 - ], - "text": "EXAMPLE 3.7\nSales Estimate", - "type": "text" - }, - { - "block_id": "p275-b9", - "global_id": 7508, - "bbox": [ - 128.9, - 328.68, - 502.77, - 386.87 - ], - "text": "During semester n, x[n] students enroll in a course requiring a certain textbook while the\npublisher sells y[n] new copies of the same book. On the average, one-quarter of students\nwith books in salable condition resell the texts at the end of the semester, and the book life is\nthree semesters. Write the equation relating y[n], the new books sold by the publisher, to x[n],\nthe number of students enrolled in the nth semester, assuming that every student buys a book.", - "type": "text" - }, - { - "block_id": "p275-b10", - "global_id": 7509, - "bbox": [ - 128.9, - 409.37, - 502.8, - 503.44 - ], - "text": "In the nth semester, the total books x[n] sold to students must be equal to y[n] (new books\nfrom the publisher) plus the used books from students enrolled in the preceding two semesters\n(because the book life is only three semesters). There are y[n −1] new books sold in semester\n(n −1), and one-quarter of these books, that is, (1/4)y[n −1], will be resold in the nth\nsemester. Also, y[n −2] new books are sold in semester n −2, and one-quarter of these,\nthat is, (1/4)y[n −2], will be resold in semester (n −1). Again, a quarter of these, that is,\n(1/16)y[n −2], will be resold in the nth semester. Therefore, x[n] must be equal to the sum of\ny[n], (1/4)y[n −1], and (1/16)y[n −2].", - "type": "text" - }, - { - "block_id": "p275-b11", - "global_id": 7510, - "bbox": [ - 243.47, - 512.91, - 275.06, - 524.53 - ], - "text": "y[n] + 1", - "type": "text" - }, - { - "block_id": "p275-b12", - "global_id": 7511, - "bbox": [ - 271.57, - 512.91, - 502.76, - 527.45 - ], - "text": "4y[n −1] + 1\n16y[n −2] = x[n]\n(3.5)", - "type": "text" - }, - { - "block_id": "p275-b13", - "global_id": 7512, - "bbox": [ - 128.91, - 535.87, - 502.79, - 557.79 - ], - "text": "Equation (3.5) can also be expressed in an alternative form by realizing that this equation is\nvalid for any value of n. Therefore, replacing n by n + 2, we obtain", - "type": "text" - }, - { - "block_id": "p275-b14", - "global_id": 7513, - "bbox": [ - 235.54, - 567.27, - 282.98, - 578.99 - ], - "text": "y[n + 2] + 1", - "type": "text" - }, - { - "block_id": "p275-b15", - "global_id": 7514, - "bbox": [ - 279.49, - 567.27, - 502.76, - 581.81 - ], - "text": "4y[n + 1] + 1\n16y[n] = x[n + 2]\n(3.6)", - "type": "text" - }, - { - "block_id": "p275-b16", - "global_id": 7515, - "bbox": [ - 128.91, - 590.22, - 287.35, - 600.19 - ], - "text": "This is the alternative form of Eq. (3.5).", - "type": "text" - }, - { - "block_id": "p275-b17", - "global_id": 7516, - "bbox": [ - 127.59, - 631.66, - 516.13, - 654.83 - ], - "text": "† A unit delay represents 1 unit of time delay. In this example, 1 unit of delay in the output corresponds to\nperiod T for the actual output.", - "type": "text" - } - ] - }, - { - "page_num": 276, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p276-b0", - "global_id": 7517, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "256\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p276-b1", - "global_id": 7518, - "bbox": [ - 103.16, - 86.24, - 477.02, - 108.15 - ], - "text": "To facilitate a realization of a system with this input–output equation, we rewrite\nthe delay-form Eq. (3.5) as y[n] = −1", - "type": "text" - }, - { - "block_id": "p276-b2", - "global_id": 7519, - "bbox": [ - 103.17, - 96.44, - 477.02, - 120.12 - ], - "text": "4y[n −1] −1\n16y[n −2] + x[n]. Figure 3.15 shows a\ncorresponding hardware realization using two unit delays in cascade.†", - "type": "text" - }, - { - "block_id": "p276-b3", - "global_id": 7520, - "bbox": [ - 114.21, - 160.72, - 225.89, - 178.65 - ], - "text": "D\nx[n]", - "type": "text" - }, - { - "block_id": "p276-b5", - "global_id": 7521, - "bbox": [ - 228.37, - 139.9, - 241.25, - 147.98 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p276-b6", - "global_id": 7522, - "bbox": [ - 180.71, - 178.05, - 193.59, - 186.13 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p276-b7", - "global_id": 7523, - "bbox": [ - 299.16, - 170.4, - 304.94, - 178.4 - ], - "text": "D", - "type": "text" - }, - { - "block_id": "p276-b9", - "global_id": 7524, - "bbox": [ - 250.18, - 160.5, - 350.72, - 168.8 - ], - "text": "y[n 1]\ny[n 2]", - "type": "text" - }, - { - "block_id": "p276-b10", - "global_id": 7525, - "bbox": [ - 219.15, - 200.29, - 290.48, - 213.77 - ], - "text": "1\n4\n1\n16", - "type": "text" - }, - { - "block_id": "p276-b12", - "global_id": 7526, - "bbox": [ - 103.16, - 228.33, - 381.82, - 237.57 - ], - "text": "Figure 3.15 Realization of the system representing sales estimate in Ex. 3.7.", - "type": "text" - }, - { - "block_id": "p276-b13", - "global_id": 7527, - "bbox": [ - 76.77, - 310.24, - 292.6, - 322.2 - ], - "text": "EXAMPLE 3.8\nDigital Differentiator", - "type": "text" - }, - { - "block_id": "p276-b14", - "global_id": 7528, - "bbox": [ - 103.16, - 338.85, - 477.0, - 360.77 - ], - "text": "Design a discrete-time system, like the one in Fig. 3.2, to differentiate continuous-time signals.\nThis differentiator is used in an audio system having an input signal bandwidth below 20 kHz.", - "type": "text" - }, - { - "block_id": "p276-b15", - "global_id": 7529, - "bbox": [ - 103.16, - 383.27, - 477.02, - 429.51 - ], - "text": "In this case, the output y(t) is required to be the derivative of the input x(t). The discrete-time\nprocessor (system) G processes the samples of x(t) to produce the discrete-time output y[n].\nLet x[n] and y[n] represent the samples T seconds apart of the signals x(t) and y(t), respectively,\nthat is,", - "type": "text" - }, - { - "block_id": "p276-b16", - "global_id": 7530, - "bbox": [ - 211.9, - 431.1, - 477.01, - 441.47 - ], - "text": "x[n] = x(nT)\nand\ny[n] = y(nT)\n(3.7)", - "type": "text" - }, - { - "block_id": "p276-b17", - "global_id": 7531, - "bbox": [ - 103.17, - 450.02, - 477.02, - 472.35 - ], - "text": "The signals x[n] and y[n] are the input and the output for the discrete-time system G. Now, we\nrequire that", - "type": "text" - }, - { - "block_id": "p276-b18", - "global_id": 7532, - "bbox": [ - 265.66, - 471.98, - 313.33, - 489.23 - ], - "text": "y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p276-b19", - "global_id": 7533, - "bbox": [ - 103.17, - 486.35, - 307.21, - 511.29 - ], - "text": "dt\nTherefore, at t = nT (see Fig. 3.16a),", - "type": "text" - }, - { - "block_id": "p276-b20", - "global_id": 7534, - "bbox": [ - 188.97, - 521.48, - 246.18, - 538.74 - ], - "text": "y(nT) = dx(t)", - "type": "text" - }, - { - "block_id": "p276-b21", - "global_id": 7535, - "bbox": [ - 232.31, - 535.85, - 240.06, - 545.82 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p276-b23", - "global_id": 7536, - "bbox": [ - 250.62, - 540.68, - 265.48, - 547.87 - ], - "text": "t=nT", - "type": "text" - }, - { - "block_id": "p276-b24", - "global_id": 7537, - "bbox": [ - 268.58, - 528.47, - 292.82, - 538.84 - ], - "text": "= lim", - "type": "text" - }, - { - "block_id": "p276-b25", - "global_id": 7538, - "bbox": [ - 278.39, - 537.38, - 293.97, - 544.65 - ], - "text": "T→0", - "type": "text" - }, - { - "block_id": "p276-b26", - "global_id": 7539, - "bbox": [ - 296.26, - 521.9, - 391.16, - 545.92 - ], - "text": "1\nT [x(nT) −x[(n −1)T]]", - "type": "text" - }, - { - "block_id": "p276-b27", - "global_id": 7540, - "bbox": [ - 101.84, - 613.22, - 490.38, - 647.36 - ], - "text": "† The comments in the preceding footnote apply here also. Although 1 unit of delay in this example is one\nsemester, we need not use this value in the hardware realization. Any value other than one semester results in\na time-scaled output.", - "type": "text" - } - ] - }, - { - "page_num": 277, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p277-b0", - "global_id": 7541, - "bbox": [ - 312.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.4\nExamples of Discrete-Time Systems\n257", - "type": "text" - }, - { - "block_id": "p277-b1", - "global_id": 7542, - "bbox": [ - 419.16, - 335.88, - 430.26, - 343.96 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p277-b2", - "global_id": 7543, - "bbox": [ - 249.19, - 92.66, - 260.3, - 100.74 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p277-b3", - "global_id": 7544, - "bbox": [ - 281.62, - 181.84, - 353.44, - 190.57 - ], - "text": "t\n(n 1)T nT", - "type": "text" - }, - { - "block_id": "p277-b4", - "global_id": 7545, - "bbox": [ - 179.17, - 236.81, - 329.18, - 253.54 - ], - "text": "x[n]\nC/D", - "type": "text" - }, - { - "block_id": "p277-b5", - "global_id": 7546, - "bbox": [ - 274.35, - 282.21, - 280.13, - 290.21 - ], - "text": "D", - "type": "text" - }, - { - "block_id": "p277-b6", - "global_id": 7547, - "bbox": [ - 298.1, - 308.43, - 307.75, - 316.43 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p277-b7", - "global_id": 7548, - "bbox": [ - 152.84, - 236.81, - 451.13, - 253.31 - ], - "text": "D/C\nx(t)\ny(t)\ny[n]", - "type": "text" - }, - { - "block_id": "p277-b9", - "global_id": 7549, - "bbox": [ - 270.68, - 218.31, - 276.45, - 226.31 - ], - "text": "G", - "type": "text" - }, - { - "block_id": "p277-b10", - "global_id": 7550, - "bbox": [ - 273.73, - 236.81, - 286.61, - 244.89 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p277-b11", - "global_id": 7551, - "bbox": [ - 329.66, - 272.88, - 357.21, - 281.18 - ], - "text": "x[n 1]", - "type": "text" - }, - { - "block_id": "p277-b12", - "global_id": 7552, - "bbox": [ - 130.14, - 333.57, - 141.25, - 341.65 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p277-b13", - "global_id": 7553, - "bbox": [ - 160.02, - 345.94, - 175.79, - 354.02 - ], - "text": "tu(t)", - "type": "text" - }, - { - "block_id": "p277-b14", - "global_id": 7554, - "bbox": [ - 222.69, - 333.57, - 235.56, - 341.65 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p277-b15", - "global_id": 7555, - "bbox": [ - 275.32, - 397.04, - 277.54, - 405.04 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p277-b16", - "global_id": 7556, - "bbox": [ - 168.33, - 380.91, - 170.55, - 388.91 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p277-b17", - "global_id": 7557, - "bbox": [ - 384.62, - 395.67, - 386.84, - 403.67 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p277-b18", - "global_id": 7558, - "bbox": [ - 237.24, - 380.02, - 453.04, - 388.86 - ], - "text": "t\n5\n10", - "type": "text" - }, - { - "block_id": "p277-b19", - "global_id": 7559, - "bbox": [ - 235.02, - 398.18, - 265.46, - 406.26 - ], - "text": "5T\n10T", - "type": "text" - }, - { - "block_id": "p277-b20", - "global_id": 7560, - "bbox": [ - 330.64, - 336.06, - 343.52, - 344.15 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p277-b21", - "global_id": 7561, - "bbox": [ - 418.04, - 381.18, - 422.49, - 389.18 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p277-b22", - "global_id": 7562, - "bbox": [ - 149.5, - 419.14, - 438.76, - 427.14 - ], - "text": "(c)\n(d)\n(e)\n(f)", - "type": "text" - }, - { - "block_id": "p277-b23", - "global_id": 7563, - "bbox": [ - 224.12, - 108.64, - 242.57, - 116.73 - ], - "text": "x[nT]", - "type": "text" - }, - { - "block_id": "p277-b24", - "global_id": 7564, - "bbox": [ - 204.29, - 126.43, - 242.57, - 134.73 - ], - "text": "x[(n 1)T]", - "type": "text" - }, - { - "block_id": "p277-b25", - "global_id": 7565, - "bbox": [ - 120.69, - 379.98, - 363.54, - 390.71 - ], - "text": "n\nn\n0", - "type": "text" - }, - { - "block_id": "p277-b26", - "global_id": 7566, - "bbox": [ - 213.19, - 398.26, - 217.19, - 406.26 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p277-b27", - "global_id": 7567, - "bbox": [ - 213.11, - 380.02, - 217.11, - 388.02 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p277-b28", - "global_id": 7568, - "bbox": [ - 328.72, - 396.48, - 373.46, - 404.56 - ], - "text": "10T\nT", - "type": "text" - }, - { - "block_id": "p277-b29", - "global_id": 7569, - "bbox": [ - 298.48, - 197.48, - 307.36, - 205.48 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p277-b30", - "global_id": 7570, - "bbox": [ - 354.33, - 229.61, - 358.78, - 245.97 - ], - "text": "1\nT", - "type": "text" - }, - { - "block_id": "p277-b31", - "global_id": 7571, - "bbox": [ - 264.65, - 357.87, - 381.18, - 372.23 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p277-b32", - "global_id": 7572, - "bbox": [ - 119.94, - 433.84, - 312.46, - 443.08 - ], - "text": "Figure 3.16 Digital differentiator and its realization.", - "type": "text" - }, - { - "block_id": "p277-b33", - "global_id": 7573, - "bbox": [ - 146.84, - 470.75, - 455.88, - 480.71 - ], - "text": "By using the notation in Eq. (3.7), the foregoing equation can be expressed as", - "type": "text" - }, - { - "block_id": "p277-b34", - "global_id": 7574, - "bbox": [ - 256.07, - 497.22, - 298.42, - 507.6 - ], - "text": "y[n] = lim", - "type": "text" - }, - { - "block_id": "p277-b35", - "global_id": 7575, - "bbox": [ - 283.98, - 506.13, - 299.56, - 513.39 - ], - "text": "T→0", - "type": "text" - }, - { - "block_id": "p277-b36", - "global_id": 7576, - "bbox": [ - 301.86, - 490.65, - 375.56, - 514.57 - ], - "text": "1\nT {x[n] −x[n −1]}", - "type": "text" - }, - { - "block_id": "p277-b37", - "global_id": 7577, - "bbox": [ - 128.91, - 522.97, - 502.76, - 556.94 - ], - "text": "This is the input–output relationship for G required to achieve our objective. In practice, the\nsampling interval T cannot be zero. Assuming T to be sufficiently small, the equation just\ngiven can be expressed as", - "type": "text" - }, - { - "block_id": "p277-b38", - "global_id": 7578, - "bbox": [ - 264.42, - 556.92, - 299.16, - 573.86 - ], - "text": "y[n] = 1", - "type": "text" - }, - { - "block_id": "p277-b39", - "global_id": 7579, - "bbox": [ - 293.51, - 563.49, - 502.76, - 580.84 - ], - "text": "T {x[n] −x[n −1]}\n(3.8)", - "type": "text" - }, - { - "block_id": "p277-b40", - "global_id": 7580, - "bbox": [ - 128.91, - 585.63, - 502.76, - 619.6 - ], - "text": "The approximation improves as T approaches 0. A discrete-time processor G to realize\nEq. (3.8) is shown inside the shaded box in Fig. 3.16b. The system in Fig. 3.16b acts as\na differentiator. This example shows how a continuous-time signal can be processed by a", - "type": "text" - } - ] - }, - { - "page_num": 278, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p278-b0", - "global_id": 7581, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "258\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p278-b1", - "global_id": 7582, - "bbox": [ - 103.16, - 86.14, - 477.04, - 108.16 - ], - "text": "discrete-time system. The considerations for determining the sampling interval T are discussed\nin Chs. 5 and 8, where it is shown that to process frequencies below 20 kHz, the proper choice is", - "type": "text" - }, - { - "block_id": "p278-b2", - "global_id": 7583, - "bbox": [ - 198.41, - 118.09, - 381.77, - 142.11 - ], - "text": "T ≤\n1\n2 × highest frequency =\n1\n40,000 = 25 µs", - "type": "text" - }, - { - "block_id": "p278-b3", - "global_id": 7584, - "bbox": [ - 103.17, - 152.05, - 477.03, - 197.88 - ], - "text": "To see how well this method works, let us consider the differentiator in Fig. 3.16b with a\nramp input x(t) = t, depicted in Fig. 3.16c. If the system were to act as a differentiator, then\nthe output y(t) of the system should be the unit step function u(t). Let us investigate how the\nsystem performs this particular operation and how well the system achieves the objective.", - "type": "text" - }, - { - "block_id": "p278-b4", - "global_id": 7585, - "bbox": [ - 103.17, - 199.46, - 477.01, - 221.79 - ], - "text": "The samples of the input x(t) = t at the interval of T seconds act as the input to the\ndiscrete-time system G. These samples, denoted by a compact notation x[n], are, therefore,", - "type": "text" - }, - { - "block_id": "p278-b5", - "global_id": 7586, - "bbox": [ - 222.81, - 233.34, - 357.36, - 258.65 - ], - "text": "x[n] = x(t)|t=nT = t|t=nT\nt ≥0\n= nT\nn ≥0", - "type": "text" - }, - { - "block_id": "p278-b6", - "global_id": 7587, - "bbox": [ - 103.17, - 270.19, - 477.03, - 400.12 - ], - "text": "Figure 3.16d shows the sampled signal x[n]. This signal acts as an input to the discrete-time\nsystem G. Figure 3.16b shows that the operation of G consists of subtracting a sample from the\npreceding (delayed) sample and then multiplying the difference with 1/T. From Fig. 3.16d, it\nis clear that the difference between the successive samples is a constant nT −(n−1)T = T for\nall samples, except for the sample at n = 0 (because there is no preceding sample at n = 0).\nThe output of G is 1/T times the difference T, which is unity for all values of n, except n = 0,\nwhere it is zero. Therefore, the output y[n] of G consists of samples of unit values for n ≥1, as\nillustrated in Fig. 3.16e. The D/C (discrete-time to continuous-time) converter converts these\nsamples into a continuous-time signal y(t), as shown in Fig. 3.16f. Ideally, the output should\nhave been y(t) = u(t). This deviation from the ideal is due to our use of a nonzero sampling\ninterval T. As T approaches zero, the output y(t) approaches the desired output u(t).", - "type": "text" - }, - { - "block_id": "p278-b7", - "global_id": 7588, - "bbox": [ - 103.16, - 402.01, - 477.03, - 471.85 - ], - "text": "The digital differentiator in Eq. (3.8) is an example of what is known as the backward\ndifference system. The reason for calling it so is obvious from Fig. 3.16a. To compute the\nderivative of y(t), we are using the difference between the present sample value and the\npreceding (backward) sample value. If we use the difference between the next (forward) sample\nat t = (n + 1)T and the present sample at t = nT, we obtain a forward difference form of\ndifferentiator as", - "type": "text" - }, - { - "block_id": "p278-b8", - "global_id": 7589, - "bbox": [ - 238.67, - 469.78, - 273.42, - 486.71 - ], - "text": "y[n] = 1", - "type": "text" - }, - { - "block_id": "p278-b9", - "global_id": 7590, - "bbox": [ - 267.77, - 476.34, - 477.01, - 493.69 - ], - "text": "T {x[n + 1] −x[n]}\n(3.9)", - "type": "text" - }, - { - "block_id": "p278-b10", - "global_id": 7591, - "bbox": [ - 76.77, - 534.82, - 270.93, - 546.78 - ], - "text": "EXAMPLE 3.9\nDigital Integrator", - "type": "text" - }, - { - "block_id": "p278-b11", - "global_id": 7592, - "bbox": [ - 103.16, - 563.45, - 442.84, - 573.41 - ], - "text": "Design a digital integrator along the same lines as the digital differentiator in Ex. 3.8.", - "type": "text" - }, - { - "block_id": "p278-b12", - "global_id": 7593, - "bbox": [ - 103.16, - 595.91, - 360.37, - 606.29 - ], - "text": "For an integrator, the input x(t) and the output y(t) are related by", - "type": "text" - }, - { - "block_id": "p278-b13", - "global_id": 7594, - "bbox": [ - 252.41, - 623.61, - 277.03, - 633.89 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p278-b14", - "global_id": 7595, - "bbox": [ - 279.08, - 610.06, - 290.85, - 622.33 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p278-b15", - "global_id": 7596, - "bbox": [ - 284.34, - 634.93, - 296.9, - 641.9 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p278-b16", - "global_id": 7597, - "bbox": [ - 298.51, - 623.61, - 326.57, - 633.89 - ], - "text": "x(τ)dτ", - "type": "text" - } - ] - }, - { - "page_num": 279, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p279-b0", - "global_id": 7598, - "bbox": [ - 312.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.4\nExamples of Discrete-Time Systems\n259", - "type": "text" - }, - { - "block_id": "p279-b1", - "global_id": 7599, - "bbox": [ - 128.9, - 85.83, - 275.89, - 96.21 - ], - "text": "Therefore, at t = nT (see Fig. 3.16a),", - "type": "text" - }, - { - "block_id": "p279-b2", - "global_id": 7600, - "bbox": [ - 264.37, - 118.66, - 313.81, - 129.04 - ], - "text": "y(nT) = lim", - "type": "text" - }, - { - "block_id": "p279-b3", - "global_id": 7601, - "bbox": [ - 299.38, - 127.58, - 314.96, - 134.84 - ], - "text": "T→0", - "type": "text" - }, - { - "block_id": "p279-b4", - "global_id": 7602, - "bbox": [ - 319.61, - 108.71, - 333.71, - 119.16 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p279-b5", - "global_id": 7603, - "bbox": [ - 316.05, - 132.95, - 337.26, - 140.14 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p279-b6", - "global_id": 7604, - "bbox": [ - 338.37, - 118.66, - 366.51, - 128.94 - ], - "text": "x(kT)T", - "type": "text" - }, - { - "block_id": "p279-b7", - "global_id": 7605, - "bbox": [ - 128.9, - 153.04, - 502.75, - 175.37 - ], - "text": "Using the usual notation x(kT) = x[k],y(nT) = y[n], and so on, this equation can be expressed\nas", - "type": "text" - }, - { - "block_id": "p279-b8", - "global_id": 7606, - "bbox": [ - 270.84, - 182.82, - 313.17, - 193.2 - ], - "text": "y[n] = lim", - "type": "text" - }, - { - "block_id": "p279-b9", - "global_id": 7607, - "bbox": [ - 298.74, - 183.14, - 320.95, - 199.0 - ], - "text": "T→0T", - "type": "text" - }, - { - "block_id": "p279-b10", - "global_id": 7608, - "bbox": [ - 326.39, - 172.87, - 340.49, - 183.32 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p279-b11", - "global_id": 7609, - "bbox": [ - 322.83, - 197.11, - 344.03, - 204.3 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p279-b12", - "global_id": 7610, - "bbox": [ - 345.15, - 182.82, - 360.82, - 193.1 - ], - "text": "x[k]", - "type": "text" - }, - { - "block_id": "p279-b13", - "global_id": 7611, - "bbox": [ - 128.91, - 211.16, - 427.69, - 221.54 - ], - "text": "Assuming that T is small enough to justify the assumption T →0, we have", - "type": "text" - }, - { - "block_id": "p279-b14", - "global_id": 7612, - "bbox": [ - 279.17, - 243.99, - 312.62, - 254.27 - ], - "text": "y[n] = T", - "type": "text" - }, - { - "block_id": "p279-b15", - "global_id": 7613, - "bbox": [ - 318.05, - 234.04, - 332.15, - 244.5 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p279-b16", - "global_id": 7614, - "bbox": [ - 314.49, - 258.28, - 335.7, - 265.48 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p279-b17", - "global_id": 7615, - "bbox": [ - 336.81, - 244.0, - 502.75, - 254.37 - ], - "text": "x[k]\n(3.10)", - "type": "text" - }, - { - "block_id": "p279-b18", - "global_id": 7616, - "bbox": [ - 128.91, - 278.63, - 502.75, - 300.65 - ], - "text": "This equation represents an example of accumulator system. This digital integrator equation\ncan be expressed in an alternate form. From Eq. (3.10), it follows that", - "type": "text" - }, - { - "block_id": "p279-b19", - "global_id": 7617, - "bbox": [ - 269.69, - 315.18, - 502.75, - 325.55 - ], - "text": "y[n] −y[n −1] = Tx[n]\n(3.11)", - "type": "text" - }, - { - "block_id": "p279-b20", - "global_id": 7618, - "bbox": [ - 128.91, - 340.49, - 502.78, - 398.28 - ], - "text": "This is an alternate description for the digital integrator. Equations (3.10) and (3.11) are\nequivalent; the one can be derived from the other. Observe that the form of Eq. (3.11) is\nsimilar to that of Eq. (3.3). Hence, the block diagram representation of a digital integrator\nin the form of Eq. (3.11) is identical to that in Fig. 3.14 with a = 1 and the input multiplied\nby T.", - "type": "text" - }, - { - "block_id": "p279-b21", - "global_id": 7619, - "bbox": [ - 127.59, - 451.26, - 516.17, - 634.79 - ], - "text": "RECURSIVE AND NONRECURSIVE FORMS OF\nDIFFERENCE EQUATION\nIf Eq. (3.11) expresses Eq. (3.10) in another form, what is the difference between these two forms?\nWhich form is preferable? To answer these questions, let us examine how the output is computed\nby each of these forms. In Eq. (3.10), the output y[n] at any instant n is computed by adding all\nthe past input values till n. This can mean a large number of additions. In contrast, Eq. (3.11) can\nbe expressed as y[n] = y[n−1]+Tx[n]. Hence, computation of y[n] involves addition of only two\nvalues: the preceding output value y[n −1] and the present input value x[n]. The computations\nare done recursively by using the preceding output values. For example, if the input starts at\nn = 0, we first compute y[0]. Then we use the computed value y[0] to compute y[1]. Knowing\ny[1], we compute y[2], and so on. The computations are recursive. This is why the form of Eq.\n(3.11) is called recursive form and the form of Eq. (3.10) is called nonrecursive form. Clearly,\n“recursive” and “nonrecursive” describe two different ways of presenting the same information.\nEquations (3.3), (3.5), and (3.11) are examples of recursive form, and Eqs. (3.8) and (3.10) are\nexamples of nonrecursive form.", - "type": "text" - } - ] - }, - { - "page_num": 280, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p280-b0", - "global_id": 7620, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "260\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p280-b1", - "global_id": 7621, - "bbox": [ - 102.14, - 86.19, - 301.49, - 112.27 - ], - "text": "KINSHIP OF DIFFERENCE EQUATIONS\nTO DIFFERENTIAL EQUATIONS", - "type": "text" - }, - { - "block_id": "p280-b2", - "global_id": 7622, - "bbox": [ - 101.84, - 116.3, - 490.4, - 138.21 - ], - "text": "We now show that a digitized version of a differential equation results in a difference equation.\nLet us consider a simple first-order differential equation", - "type": "text" - }, - { - "block_id": "p280-b3", - "global_id": 7623, - "bbox": [ - 257.82, - 150.98, - 277.6, - 161.25 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p280-b4", - "global_id": 7624, - "bbox": [ - 263.75, - 157.96, - 490.38, - 175.31 - ], - "text": "dt\n+ cy(t) = x(t)\n(3.12)", - "type": "text" - }, - { - "block_id": "p280-b5", - "global_id": 7625, - "bbox": [ - 101.84, - 186.14, - 490.39, - 220.43 - ], - "text": "Consider uniform samples of x(t) at intervals of T seconds. As usual, we use the notation x[n] to\ndenote x(nT), the nth sample of x(t). Similarly, y[n] denotes y[nT], the nth sample of y(t). From\nthe basic definition of a derivative, we can express Eq. (3.12) at t = nT as", - "type": "text" - }, - { - "block_id": "p280-b6", - "global_id": 7626, - "bbox": [ - 227.0, - 240.53, - 242.57, - 256.29 - ], - "text": "lim\nT→0", - "type": "text" - }, - { - "block_id": "p280-b7", - "global_id": 7627, - "bbox": [ - 245.97, - 233.13, - 304.75, - 243.5 - ], - "text": "y[n] −y[n −1]", - "type": "text" - }, - { - "block_id": "p280-b8", - "global_id": 7628, - "bbox": [ - 272.22, - 240.11, - 365.23, - 257.46 - ], - "text": "T\n+ cy[n] = x[n]", - "type": "text" - }, - { - "block_id": "p280-b9", - "global_id": 7629, - "bbox": [ - 101.84, - 269.04, - 466.95, - 279.11 - ], - "text": "Clearing the fractions and rearranging the terms yield (assuming nonzero, but very small T)", - "type": "text" - }, - { - "block_id": "p280-b10", - "global_id": 7630, - "bbox": [ - 246.55, - 293.83, - 490.38, - 304.21 - ], - "text": "y[n] + αy[n −1] = βx[n]\n(3.13)", - "type": "text" - }, - { - "block_id": "p280-b11", - "global_id": 7631, - "bbox": [ - 101.85, - 319.35, - 126.18, - 329.31 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p280-b12", - "global_id": 7632, - "bbox": [ - 221.96, - 331.58, - 368.29, - 356.01 - ], - "text": "α =\n−1\n1 + cT\nand\nβ =\nT\n1 + cT", - "type": "text" - }, - { - "block_id": "p280-b13", - "global_id": 7633, - "bbox": [ - 101.85, - 364.82, - 302.33, - 374.78 - ], - "text": "We can also express Eq. (3.13) in advance form as", - "type": "text" - }, - { - "block_id": "p280-b14", - "global_id": 7634, - "bbox": [ - 238.62, - 389.5, - 353.59, - 399.88 - ], - "text": "y[n + 1] + αy[n] = βx[n + 1]", - "type": "text" - }, - { - "block_id": "p280-b15", - "global_id": 7635, - "bbox": [ - 101.85, - 415.01, - 490.4, - 508.66 - ], - "text": "It is clear that a differential equation can be approximated by a difference equation of the\nsame order. In this way, we can approximate an nth-order differential equation by a difference\nequation of nth order. Indeed, a digital computer solves differential equations by using an\nequivalent difference equation, which can be solved by means of simple operations of addition,\nmultiplication, and shifting. Recall that a computer can perform only these simple operations. It\nmust necessarily approximate complex operation like differentiation and integration in terms of\nsuch simple operations. The approximation can be made as close to the exact answer as possible\nby choosing sufficiently small value for T.", - "type": "text" - }, - { - "block_id": "p280-b16", - "global_id": 7636, - "bbox": [ - 101.85, - 510.65, - 490.41, - 556.48 - ], - "text": "At this stage, we have not developed tools required to choose a suitable value of the sampling\ninterval T. This subject is discussed in Ch. 5 and also in Ch. 8. In Sec. 5.7, we shall discuss a\nsystematic procedure (impulse invariance method) for finding a discrete-time system with which\nto realize an Nth-order LTIC system.", - "type": "text" - }, - { - "block_id": "p280-b17", - "global_id": 7637, - "bbox": [ - 102.14, - 572.8, - 299.3, - 584.93 - ], - "text": "ORDER OF A DIFFERENCE EQUATION", - "type": "text" - }, - { - "block_id": "p280-b18", - "global_id": 7638, - "bbox": [ - 101.84, - 588.96, - 490.42, - 634.79 - ], - "text": "Equations (3.3), (3.5), (3.9), (3.11), and (3.13) are examples of difference equations. The\nhighest-order difference of the output signal or the input signal, whichever is higher, represents\nthe order of the difference equation. Hence, Eqs. (3.3), (3.9), (3.11), and (3.13) are first-order\ndifference equations, whereas Eq. (3.5) is of the second order.", - "type": "text" - } - ] - }, - { - "page_num": 281, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p281-b0", - "global_id": 7639, - "bbox": [ - 312.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.4\nExamples of Discrete-Time Systems\n261", - "type": "text" - }, - { - "block_id": "p281-b1", - "global_id": 7640, - "bbox": [ - 133.57, - 97.81, - 346.71, - 109.77 - ], - "text": "DRILL 3.8\nDigital Integrator Design", - "type": "text" - }, - { - "block_id": "p281-b2", - "global_id": 7641, - "bbox": [ - 133.57, - 118.47, - 510.16, - 152.76 - ], - "text": "Design a digital integrator in Ex. 3.9 using the fact that for an integrator, the output y(t) and\nthe input x(t) are related by dy(t)/dt = x(t). Approximation (similar to that in Ex. 3.8) of this\nequation at t = nT yields the recursive form in Eq. (3.11).", - "type": "text" - }, - { - "block_id": "p281-b3", - "global_id": 7642, - "bbox": [ - 127.59, - 184.81, - 516.16, - 320.51 - ], - "text": "ANALOG, DIGITAL, CONTINUOUS-TIME,\nAND DISCRETE-TIME SYSTEMS\nThe basic difference between continuous-time systems and analog systems, as also between\ndiscrete-time and digital systems, is fully explained in Secs. 1.7-5 and 1.7-6.† Historically,\ndiscrete-time systems have been realized with digital computers, where continuous-time signals\nare processed through digitized samples rather than unquantized samples. Therefore, the terms\ndigital filters and discrete-time systems are used synonymously in the literature. This distinction is\nirrelevant in the analysis of discrete-time systems. For this reason, we follow this loose convention\nin this book, where the term digital filter implies a discrete-time system, and analog filter means\ncontinuous-time system. Moreover, the terms C/D (continuous-to-discrete-time ) and D/C will\noccasionally be used interchangeably with terms A/D (analog-to-digital) and D/A, respectively.", - "type": "text" - }, - { - "block_id": "p281-b4", - "global_id": 7643, - "bbox": [ - 127.89, - 336.46, - 383.23, - 348.58 - ], - "text": "ADVANTAGES OF DIGITAL SIGNAL PROCESSING", - "type": "text" - }, - { - "block_id": "p281-b5", - "global_id": 7644, - "bbox": [ - 144.52, - 352.61, - 516.16, - 589.72 - ], - "text": "1. Digital systems operation can tolerate considerable variation in signal values, and hence\nare less sensitive to changes in the component parameter values due to temperature\nvariation, aging, and other factors. This results in greater degree of precision and stability.\nSince digital systems are binary circuits, their accuracy can be increased by using more\ncomplex circuitry to increase word length, subject to cost limitations.\n2. Digital systems do not require any factory adjustment and can be easily duplicated in\nvolume without having to worry about precise component values. They can be fully\nintegrated, and even highly complex systems can be placed on a single chip by using VLSI\n(very-large-scale integrated) circuits.\n3. Digital filters are more flexible. Their characteristics can be easily altered simply\nby changing the program. Digital hardware implementation permits the use of\nmicroprocessors, miniprocessors, digital switching, and large-scale integrated circuits.\n4. A greater variety of filters can be realized by digital systems.\n5. Digital signals can be stored easily and inexpensively on various media (e.g., magnetic,\noptical, and solid state) without deterioration of signal quality. It is also possible (and\nincreasingly popular) to search and select information from distant electronic storehouses,\nsuch as the cloud.\n6. Digital signals can be coded to yield extremely low error rates and high fidelity, as well\nas privacy. Also, more sophisticated signal-processing algorithms can be used to process\ndigital signals.", - "type": "text" - }, - { - "block_id": "p281-b6", - "global_id": 7645, - "bbox": [ - 127.59, - 610.24, - 516.11, - 633.41 - ], - "text": "† The terms discrete-time and continuous-time qualify the nature of a signal along the time axis (horizontal\naxis). The terms analog and digital, in contrast, qualify the nature of the signal amplitude (vertical axis).", - "type": "text" - } - ] - }, - { - "page_num": 282, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p282-b0", - "global_id": 7646, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "262\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p282-b1", - "global_id": 7647, - "bbox": [ - 118.78, - 85.82, - 490.4, - 167.51 - ], - "text": "7. Digital filters can be easily time-shared and therefore can serve a number of inputs\nsimultaneously. Moreover, it is easier and more efficient to multiplex several digital signals\non the same channel.\n8. Reproduction with digital messages is extremely reliable without deterioration. Analog\nmessages such as photocopies and films, for example, lose quality at each successive stage\nof reproduction and have to be transported physically from one distant place to another,\noften at relatively high cost.", - "type": "text" - }, - { - "block_id": "p282-b2", - "global_id": 7648, - "bbox": [ - 101.84, - 175.48, - 490.4, - 221.31 - ], - "text": "One must weigh these advantages against such disadvantages as increased system complexity due\nto use of A/D and D/A interfaces, limited range of frequencies available in practice (affordable\nrates are gigahertz or less), and use of more power than is needed for the passive analog circuits.\nDigital systems use power-consuming active devices.", - "type": "text" - }, - { - "block_id": "p282-b3", - "global_id": 7649, - "bbox": [ - 101.84, - 251.8, - 348.38, - 263.75 - ], - "text": "3.4-1 Classification of Discrete-Time Systems", - "type": "text" - }, - { - "block_id": "p282-b4", - "global_id": 7650, - "bbox": [ - 101.84, - 269.88, - 490.37, - 303.75 - ], - "text": "Before examining the nature of discrete-time system equations, let us consider the concepts of\nlinearity, time invariance (or shift invariance), and causality, which apply to discrete-time systems\nalso.", - "type": "text" - }, - { - "block_id": "p282-b5", - "global_id": 7651, - "bbox": [ - 101.84, - 321.35, - 490.4, - 371.38 - ], - "text": "LINEARITY AND TIME INVARIANCE\nFor discrete-time systems, the definition of linearity is identical to that for continuous-time\nsystems, as given in Eq. (1.22). We can show that the systems in Exs. 3.6, 3.7, 3.8, and 3.9 are all\nlinear.", - "type": "text" - }, - { - "block_id": "p282-b6", - "global_id": 7652, - "bbox": [ - 101.84, - 373.27, - 490.4, - 467.02 - ], - "text": "Time invariance (or shift invariance) for discrete-time systems is also defined in a way similar\nto that for continuous-time systems. Systems whose parameters do not change with time (with n)\nare time-invariant or shift-invariant (also constant-parameter) systems. For such a system, if the\ninput is delayed by k units or samples, the output is the same as before but delayed by k samples\n(assuming the initial conditions also are delayed by k). The systems in Exs. 3.6, 3.7, 3.8, and 3.9\nare time-invariant because the coefficients in the system equations are constants (independent of\nn). If these coefficients were functions of n (time), then the systems would be linear time-varying\nsystems. Consider, for example, a system described by", - "type": "text" - }, - { - "block_id": "p282-b7", - "global_id": 7653, - "bbox": [ - 267.2, - 482.57, - 325.02, - 494.57 - ], - "text": "y[n] = e−nx[n]", - "type": "text" - }, - { - "block_id": "p282-b8", - "global_id": 7654, - "bbox": [ - 101.85, - 511.94, - 490.37, - 535.05 - ], - "text": "For this system, let a signal x1[n] yield the output y1[n], and another input x2[n] yield the output\ny2[n]. Then", - "type": "text" - }, - { - "block_id": "p282-b9", - "global_id": 7655, - "bbox": [ - 203.24, - 542.72, - 389.0, - 555.6 - ], - "text": "y1[n] = e−nx1[n]\nand\ny2[n] = e−nx2[n]", - "type": "text" - }, - { - "block_id": "p282-b10", - "global_id": 7656, - "bbox": [ - 101.85, - 569.11, - 233.08, - 580.26 - ], - "text": "If we let x2[n] = x1[n −N0], then", - "type": "text" - }, - { - "block_id": "p282-b11", - "global_id": 7657, - "bbox": [ - 202.77, - 595.03, - 389.46, - 607.9 - ], - "text": "y2[n] = e−nx2[n] = e−nx1[n −N0]̸ = y1[n −N0]", - "type": "text" - }, - { - "block_id": "p282-b12", - "global_id": 7658, - "bbox": [ - 101.85, - 624.83, - 294.64, - 634.79 - ], - "text": "Clearly, this is a time-varying parameter system.", - "type": "text" - } - ] - }, - { - "page_num": 283, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p283-b0", - "global_id": 7659, - "bbox": [ - 312.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.4\nExamples of Discrete-Time Systems\n263", - "type": "text" - }, - { - "block_id": "p283-b1", - "global_id": 7660, - "bbox": [ - 127.89, - 86.19, - 324.35, - 98.32 - ], - "text": "CAUSAL AND NONCAUSAL SYSTEMS", - "type": "text" - }, - { - "block_id": "p283-b2", - "global_id": 7661, - "bbox": [ - 127.59, - 102.25, - 516.14, - 148.18 - ], - "text": "A causal (also known as a physical or nonanticipative) system is one for which the output at any\ninstant n = k depends only on the value of the input x[n] for n ≤k. In other words, the value of the\noutput at the present instant depends only on the past and present values of the input x[n], not on\nits future values. As we shall see, the systems in Exs. 3.6, 3.7, 3.8, and 3.9 are all causal.", - "type": "text" - }, - { - "block_id": "p283-b3", - "global_id": 7662, - "bbox": [ - 127.59, - 168.23, - 516.15, - 254.12 - ], - "text": "INVERTIBLE AND NONINVERTIBLE SYSTEMS\nA discrete-time system S is invertible if an inverse system Si exists such that the cascade of S and\nSi results in an identity system. An identity system is defined as one whose output is identical to\nthe input. In other words, for an invertible system, the input can be uniquely determined from the\ncorresponding output. For every input there is a unique output. When a signal is processed through\nsuch a system, its input can be reconstructed from the corresponding output. There is no loss of\ninformation when a signal is processed through an invertible system.", - "type": "text" - }, - { - "block_id": "p283-b4", - "global_id": 7663, - "bbox": [ - 127.59, - 256.11, - 516.13, - 325.85 - ], - "text": "A cascade of a unit delay with a unit advance results in an identity system because the output\nof such a cascaded system is identical to the input. Clearly, the inverse of an ideal unit delay\nis ideal unit advance, which is a noncausal (and unrealizable) system. In contrast, a compressor\ny[n] = x[Mn] is not invertible because this operation loses all but every Mth sample of the input,\nand, generally, the input cannot be reconstructed. Similarly, operations, such as y[n] = cosx[n] or\ny[n] = |x[n]|, are not invertible.", - "type": "text" - }, - { - "block_id": "p283-b5", - "global_id": 7664, - "bbox": [ - 133.57, - 363.93, - 275.63, - 375.88 - ], - "text": "DRILL 3.9\nInvertibility", - "type": "text" - }, - { - "block_id": "p283-b6", - "global_id": 7665, - "bbox": [ - 133.57, - 384.59, - 510.16, - 406.92 - ], - "text": "Show that a system specified by equation y[n] = ax[n] + b is invertible but that the system\ny[n] = |x[n]|2 is noninvertible.", - "type": "text" - }, - { - "block_id": "p283-b7", - "global_id": 7666, - "bbox": [ - 127.89, - 442.95, - 305.51, - 455.08 - ], - "text": "STABLE AND UNSTABLE SYSTEMS", - "type": "text" - }, - { - "block_id": "p283-b8", - "global_id": 7667, - "bbox": [ - 127.59, - 459.01, - 516.17, - 528.85 - ], - "text": "The concept of stability is similar to that in continuous-time systems. Stability can be internal\nor external. If every bounded input applied at the input terminal results in a bounded output, the\nsystem is said to be stable externally. External stability can be ascertained by measurements at\nthe external terminals of the system. This type of stability is also known as the stability in the\nBIBO (bounded-input/bounded-output) sense. Both internal and external stability are discussed in\ngreater detail in Sec. 3.9.", - "type": "text" - }, - { - "block_id": "p283-b9", - "global_id": 7668, - "bbox": [ - 127.89, - 548.89, - 414.47, - 561.01 - ], - "text": "MEMORYLESS SYSTEMS AND SYSTEMS WITH MEMORY", - "type": "text" - }, - { - "block_id": "p283-b10", - "global_id": 7669, - "bbox": [ - 127.59, - 565.05, - 516.17, - 634.79 - ], - "text": "The concepts of memoryless (or instantaneous) systems and those with memory (or dynamic) are\nidentical to the corresponding concepts of the continuous-time case. A system is memoryless if\nits response at any instant n depends at most on the input at the same instant n. The output at any\ninstant of a system with memory generally depends on the past, present, and future values of the\ninput. For example, y[n] = sinx[n] is an example of instantaneous system, and y[n]−y[n−1] = x[n]\nis an example of a dynamic system or a system with memory.", - "type": "text" - } - ] - }, - { - "page_num": 284, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p284-b0", - "global_id": 7670, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "264\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p284-b1", - "global_id": 7671, - "bbox": [ - 76.77, - 93.64, - 375.85, - 105.6 - ], - "text": "EXAMPLE 3.10\nInvestigating DT System Properties", - "type": "text" - }, - { - "block_id": "p284-b2", - "global_id": 7672, - "bbox": [ - 103.16, - 121.85, - 477.02, - 144.18 - ], - "text": "Consider a DT system described as y[n + 1] = x[n + 1]x[n]. Determine whether the system is\n(a) linear, (b) time-invariant, (c) causal, (d) invertible, (e) BIBO-stable, and (f) memoryless.", - "type": "text" - }, - { - "block_id": "p284-b3", - "global_id": 7673, - "bbox": [ - 103.16, - 167.1, - 476.97, - 189.02 - ], - "text": "Let us delay the input–output equation by one to obtain the equivalent but more convenient\nrepresentation of y[n] = x[n]x[n −1].", - "type": "text" - }, - { - "block_id": "p284-b4", - "global_id": 7674, - "bbox": [ - 103.17, - 190.93, - 477.01, - 212.92 - ], - "text": "(a) Linearity requires both homogeneity and additivity. Let us first investigate\nhomogeneity. Assuming x[n] \r⇒y[n], we see that", - "type": "text" - }, - { - "block_id": "p284-b5", - "global_id": 7675, - "bbox": [ - 197.83, - 223.03, - 382.34, - 234.85 - ], - "text": "ax[n] \r⇒(ax[n])(ax[n −1]) = a2y[n]̸ = ay[n]", - "type": "text" - }, - { - "block_id": "p284-b6", - "global_id": 7676, - "bbox": [ - 103.17, - 246.8, - 342.17, - 256.76 - ], - "text": "Thus, the system does not satisfy the homogeneity property.", - "type": "text" - }, - { - "block_id": "p284-b7", - "global_id": 7677, - "bbox": [ - 103.17, - 258.34, - 477.01, - 281.45 - ], - "text": "The system also does not satisfy the additivity property. Assuming x1[n] \r⇒y1[n] and\nx2[n] \r⇒y2[n], we see that input x[n] = x1[n] + x2[n] produces output y[n] as", - "type": "text" - }, - { - "block_id": "p284-b8", - "global_id": 7678, - "bbox": [ - 148.03, - 292.21, - 324.33, - 303.36 - ], - "text": "y[n] = (x1[n] + x2[n])(x1[n −1] + x2[n −1])", - "type": "text" - }, - { - "block_id": "p284-b9", - "global_id": 7679, - "bbox": [ - 166.12, - 307.15, - 432.14, - 318.3 - ], - "text": "= x1[n]x1[n −1] + x2[n]x2[n −1] + x1[n]x2[n −1] + x2[n]x1[n −1]", - "type": "text" - }, - { - "block_id": "p284-b10", - "global_id": 7680, - "bbox": [ - 166.12, - 322.1, - 360.39, - 348.19 - ], - "text": "= y1[n] + y2[n] + x1[n]x2[n −1] + x2[n]x1[n −1]̸\n= y1[n] + y2[n]", - "type": "text" - }, - { - "block_id": "p284-b11", - "global_id": 7681, - "bbox": [ - 103.17, - 359.38, - 237.12, - 369.34 - ], - "text": "Clearly, additivity is not satisfied.", - "type": "text" - }, - { - "block_id": "p284-b12", - "global_id": 7682, - "bbox": [ - 103.17, - 371.33, - 477.04, - 393.25 - ], - "text": "Since the system does not satisfy both the homogeneity and additivity properties, we\nconclude that the system is not linear.", - "type": "text" - }, - { - "block_id": "p284-b13", - "global_id": 7683, - "bbox": [ - 103.16, - 395.16, - 477.02, - 429.11 - ], - "text": "(b) To be time-invariant, a shift in any input should cause a corresponding shift in\nrespective output. Assume that x[n] \r⇒y[n]. Applying a delay version of this input to the\nsystem yields", - "type": "text" - }, - { - "block_id": "p284-b14", - "global_id": 7684, - "bbox": [ - 142.11, - 440.66, - 438.06, - 451.04 - ], - "text": "x[n −N] \r⇒x[n −N]x[n −1 −N] = x[(n −N)]x[(n −N) −1] = y[n −N]", - "type": "text" - }, - { - "block_id": "p284-b15", - "global_id": 7685, - "bbox": [ - 103.17, - 462.99, - 477.01, - 484.9 - ], - "text": "Since shifting an input causes a corresponding shift in the output, we conclude that the system\nis time-invariant.", - "type": "text" - }, - { - "block_id": "p284-b16", - "global_id": 7686, - "bbox": [ - 103.17, - 486.8, - 477.02, - 520.77 - ], - "text": "(c) To be causal, an output value cannot depend on any future input values. The output y\nat time n depends on the input x at present and past times n and n−1. Since the current output\ndoes not depend on future input values, the system is causal.", - "type": "text" - }, - { - "block_id": "p284-b17", - "global_id": 7687, - "bbox": [ - 103.17, - 522.68, - 477.03, - 568.59 - ], - "text": "(d) For a system to be invertible, every input must generate a unique output, which allows\nexact recovery of the input from the output. Consider two inputs to this system: x1[n] = 1 and\nx2[n] = −1. Both inputs generate the same output: y1[n] = y2[n] = 1. Since unique inputs do\nnot always generate unique outputs, we conclude that the system is not invertible.", - "type": "text" - }, - { - "block_id": "p284-b18", - "global_id": 7688, - "bbox": [ - 103.17, - 570.5, - 477.0, - 604.46 - ], - "text": "(e) To be BIBO-stable, any bounded input must generate a bounded output. A bounded\ninput satisfies |x[n]| ≤Mx < ∞for all n. Given this condition, the system output magnitude\nbehaves as", - "type": "text" - }, - { - "block_id": "p284-b19", - "global_id": 7689, - "bbox": [ - 187.1, - 604.6, - 370.54, - 616.41 - ], - "text": "|y[n]| = |x[n]x[n −1]| = |x[n]||x[n −1]| ≤M2", - "type": "text" - }, - { - "block_id": "p284-b20", - "global_id": 7690, - "bbox": [ - 366.66, - 606.03, - 393.08, - 618.68 - ], - "text": "x < ∞", - "type": "text" - } - ] - }, - { - "page_num": 285, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p285-b0", - "global_id": 7691, - "bbox": [ - 325.13, - 62.89, - 516.14, - 71.98 - ], - "text": "3.5\nDiscrete-Time System Equations\n265", - "type": "text" - }, - { - "block_id": "p285-b1", - "global_id": 7692, - "bbox": [ - 128.9, - 86.24, - 502.79, - 108.16 - ], - "text": "Since any bounded input is guaranteed to produce a bounded output, it follows that the system\nis BIBO-stable.", - "type": "text" - }, - { - "block_id": "p285-b2", - "global_id": 7693, - "bbox": [ - 128.9, - 110.06, - 502.76, - 144.02 - ], - "text": "(f) To be memoryless, a system’s output can only depend on the strength of the current\ninput. Since the output y at time n depends on the input x not only at present time n but also on\npast time n −1, we see that the system is not memoryless.", - "type": "text" - }, - { - "block_id": "p285-b3", - "global_id": 7694, - "bbox": [ - 127.94, - 189.07, - 394.41, - 203.02 - ], - "text": "3.5 DISCRETE-TIME SYSTEM EQUATIONS", - "type": "text" - }, - { - "block_id": "p285-b4", - "global_id": 7695, - "bbox": [ - 127.59, - 209.01, - 516.15, - 230.92 - ], - "text": "In this section we discuss time-domain analysis of LTID (linear, time-invariant, discrete-time\nsystems). With minor differences, the procedure is parallel to that for continuous-time systems.", - "type": "text" - }, - { - "block_id": "p285-b5", - "global_id": 7696, - "bbox": [ - 127.89, - 246.07, - 261.27, - 258.19 - ], - "text": "DIFFERENCE EQUATIONS", - "type": "text" - }, - { - "block_id": "p285-b6", - "global_id": 7697, - "bbox": [ - 127.59, - 262.22, - 516.17, - 379.78 - ], - "text": "Equations (3.3), (3.5), (3.8), and (3.13) are examples of difference equations. Equations (3.3),\n(3.8), and (3.13) are first-order difference equations, and Eq. (3.5) is a second-order difference\nequation. All these equations are linear, with constant (not time-varying) coefficients.† Before\ngiving a general form of an Nth-order linear difference equation, we recall that a difference\nequation can be written in two forms: the first form uses delay terms such as y[n −1], y[n −2],\nx[n −1], x[n −2], and so on; and the alternate form uses advance terms such as y[n + 1], y[n + 2],\nand so on. Although the delay form is more natural, we shall often prefer the advance form, not\njust for the general notational convenience, but also for resulting notational uniformity with the\noperator form for differential equations. This facilitates the commonality of the solutions and\nconcepts for continuous-time and discrete-time systems.", - "type": "text" - }, - { - "block_id": "p285-b7", - "global_id": 7698, - "bbox": [ - 145.53, - 381.77, - 447.14, - 391.74 - ], - "text": "We start here with a general difference equation, written in advance form as", - "type": "text" - }, - { - "block_id": "p285-b8", - "global_id": 7699, - "bbox": [ - 162.48, - 404.1, - 399.93, - 415.25 - ], - "text": "y[n + N] + a1y[n + N −1] + · · · + aN−1y[n + 1] + aNy[n] =", - "type": "text" - }, - { - "block_id": "p285-b9", - "global_id": 7700, - "bbox": [ - 182.4, - 419.05, - 516.13, - 430.2 - ], - "text": "bN−Mx[n + M] + bN−M+1x[n + M −1] + · · · + bN−1x[n + 1] + bNx[n]\n(3.14)", - "type": "text" - }, - { - "block_id": "p285-b10", - "global_id": 7701, - "bbox": [ - 127.59, - 441.79, - 516.11, - 477.57 - ], - "text": "This is a linear difference equation whose order is max(N,M). We have assumed the coefficient\nof y[n + N] to be unity (a0 = 1) without loss of generality. If a0̸ = 1, we can divide the equation\nthroughout by a0 to normalize the equation to have a0 = 1.", - "type": "text" - }, - { - "block_id": "p285-b11", - "global_id": 7702, - "bbox": [ - 127.89, - 491.22, - 257.2, - 503.34 - ], - "text": "CAUSALITY CONDITION", - "type": "text" - }, - { - "block_id": "p285-b12", - "global_id": 7703, - "bbox": [ - 127.59, - 507.38, - 516.15, - 553.2 - ], - "text": "For a causal system, the output cannot depend on future input values. This means that when the\nsystem equation is in the advance form of Eq. (3.14), causality requires M ≤N. If M were to be\ngreater than N, then y[n+N], the output at n+N would depend on x[n+M], which is the input at\nthe later instant n + M. For a general causal case, M = N, and Eq. (3.14) can be expressed as", - "type": "text" - }, - { - "block_id": "p285-b13", - "global_id": 7704, - "bbox": [ - 192.99, - 565.57, - 430.44, - 576.72 - ], - "text": "y[n + N] + a1y[n + N −1] + · · · + aN−1y[n + 1] + aNy[n] =", - "type": "text" - }, - { - "block_id": "p285-b14", - "global_id": 7705, - "bbox": [ - 212.91, - 580.51, - 516.13, - 592.28 - ], - "text": "b0x[n + N] + b1 x[n + N −1] + · · · + bN−1x[n + 1] + bNx[n]\n(3.15)", - "type": "text" - }, - { - "block_id": "p285-b15", - "global_id": 7706, - "bbox": [ - 127.59, - 611.23, - 516.12, - 645.37 - ], - "text": "† Equations such as (3.3), (3.5), (3.8), and (3.13) are considered to be linear according to the classical\ndefinition of linearity. Some authors label such equations as incrementally linear. We prefer the classical\ndefinition. It is just a matter of individual choice and makes no difference in the final results.", - "type": "text" - } - ] - }, - { - "page_num": 286, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p286-b0", - "global_id": 7707, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "266\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p286-b1", - "global_id": 7708, - "bbox": [ - 101.84, - 85.72, - 490.39, - 131.64 - ], - "text": "where some of the coefficients on either side can be zero. In this Nth-order equation, a0, the\ncoefficient of y[n+N], is normalized to unity. Equation (3.15) is valid for all values of n. Therefore,\nit is still valid if we replace n by n −N throughout the equation [see Eqs. (3.3) and (3.4)]. Such\nreplacement yields a delay-form alternative:", - "type": "text" - }, - { - "block_id": "p286-b2", - "global_id": 7709, - "bbox": [ - 167.8, - 144.28, - 405.25, - 155.43 - ], - "text": "y[n] + a1y[n −1] + · · · + aN−1y[n −N + 1] + aNy[n −N] =", - "type": "text" - }, - { - "block_id": "p286-b3", - "global_id": 7710, - "bbox": [ - 187.72, - 159.23, - 490.38, - 170.38 - ], - "text": "b0x[n] + b1x[n −1] + · · · + bN−1x[n −N + 1] + bNx[n −N]\n(3.16)", - "type": "text" - }, - { - "block_id": "p286-b4", - "global_id": 7711, - "bbox": [ - 101.84, - 195.45, - 421.21, - 207.4 - ], - "text": "3.5-1 Recursive (Iterative) Solution of Difference Equation", - "type": "text" - }, - { - "block_id": "p286-b5", - "global_id": 7712, - "bbox": [ - 101.84, - 213.53, - 245.01, - 223.5 - ], - "text": "Equation (3.16) can be expressed as", - "type": "text" - }, - { - "block_id": "p286-b6", - "global_id": 7713, - "bbox": [ - 190.68, - 236.13, - 401.53, - 247.28 - ], - "text": "y[n] = −a1y[n −1] −a2y[n −2] −· · · −aNy[n −N]", - "type": "text" - }, - { - "block_id": "p286-b7", - "global_id": 7714, - "bbox": [ - 220.14, - 251.07, - 490.38, - 262.22 - ], - "text": "+ b0x[n] + b1x[n −1] + · · · + bNx[n −N]\n(3.17)", - "type": "text" - }, - { - "block_id": "p286-b8", - "global_id": 7715, - "bbox": [ - 101.84, - 274.09, - 490.43, - 475.75 - ], - "text": "In Eq. (3.17), y[n] is computed from 2N + 1 pieces of information; the preceding N values of\nthe output: y[n −1], y[n −2], . . . , y[n −N], and the preceding N values of the input: x[n −1],\nx[n −2], . . . , x[n −N], and the present value of the input x[n]. Initially, to compute y[0], the\nN initial conditions y[−1], y[−2], . . . , y[−N] serve as the preceding N output values. Hence,\nknowing the N initial conditions and the input, we can determine recursively the entire output\ny[0], y[1], y[2], y[3], . . . , one value at a time. For instance, to find y[0] we set n = 0 in Eq. (3.17).\nThe left-hand side is y[0], and the right-hand side is expressed in terms of N initial conditions\ny[−1], y[−2], . . . , y[−N] and the input x[0] if x[n] is causal (because of causality, other input\nterms x[−n] = 0). Similarly, knowing y[0] and the input, we can compute y[1] by setting n = 1\nin Eq. (3.17). Knowing y[0] and y[1], we find y[2], and so on. Thus, we can use this recursive\nprocedure to find the complete response y[0], y[1], y[2], . . .. For this reason, this equation is\nclassed as a recursive form. This method basically reflects the manner in which a computer would\nsolve a recursive difference equation, given the input and initial conditions. Equation (3.17) [or\nEq. (3.16)] is nonrecursive if all the N −1 coefficients ai = 0 (i = 1,2,. . .,N −1). In this case,\nit can be seen that y[n] is computed only from the input values and without using any previous\noutputs. Generally speaking, the recursive procedure applies only to equations in the recursive\nform. The recursive (iterative) procedure is demonstrated by the following examples.", - "type": "text" - }, - { - "block_id": "p286-b9", - "global_id": 7716, - "bbox": [ - 76.77, - 505.29, - 477.56, - 517.24 - ], - "text": "EXAMPLE 3.11\nIterative Solution to a First-Order Difference Equation", - "type": "text" - }, - { - "block_id": "p286-b10", - "global_id": 7717, - "bbox": [ - 103.16, - 533.91, - 168.2, - 543.87 - ], - "text": "Solve iteratively", - "type": "text" - }, - { - "block_id": "p286-b11", - "global_id": 7718, - "bbox": [ - 240.49, - 545.45, - 339.68, - 555.82 - ], - "text": "y[n] −0.5y[n −1] = x[n]", - "type": "text" - }, - { - "block_id": "p286-b12", - "global_id": 7719, - "bbox": [ - 103.17, - 563.44, - 477.01, - 586.71 - ], - "text": "with initial condition y[−1] = 16 and causal input x[n] = n2u[n]. This equation can be\nexpressed as", - "type": "text" - }, - { - "block_id": "p286-b13", - "global_id": 7720, - "bbox": [ - 240.5, - 588.29, - 477.01, - 598.66 - ], - "text": "y[n] = 0.5y[n −1] + x[n]\n(3.18)", - "type": "text" - } - ] - }, - { - "page_num": 287, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p287-b0", - "global_id": 7721, - "bbox": [ - 325.13, - 62.89, - 516.14, - 71.98 - ], - "text": "3.5\nDiscrete-Time System Equations\n267", - "type": "text" - }, - { - "block_id": "p287-b1", - "global_id": 7722, - "bbox": [ - 128.9, - 85.83, - 284.37, - 96.2 - ], - "text": "If we set n = 0 in Eq. (3.18), we obtain", - "type": "text" - }, - { - "block_id": "p287-b2", - "global_id": 7723, - "bbox": [ - 270.27, - 107.74, - 361.36, - 118.12 - ], - "text": "y[0] = 0.5y[−1] + x[0]", - "type": "text" - }, - { - "block_id": "p287-b3", - "global_id": 7724, - "bbox": [ - 288.37, - 122.69, - 360.73, - 133.07 - ], - "text": "= 0.5(16) + 0 = 8", - "type": "text" - }, - { - "block_id": "p287-b4", - "global_id": 7725, - "bbox": [ - 128.91, - 144.6, - 502.76, - 166.93 - ], - "text": "Now, setting n = 1 in Eq. (3.18) and using the value y[0] = 8 (computed in the first step) and\nx[1] = (1)2 = 1, we obtain", - "type": "text" - }, - { - "block_id": "p287-b5", - "global_id": 7726, - "bbox": [ - 267.39, - 164.41, - 364.28, - 178.89 - ], - "text": "y[1] = 0.5(8) + (1)2 = 5", - "type": "text" - }, - { - "block_id": "p287-b6", - "global_id": 7727, - "bbox": [ - 128.9, - 187.45, - 502.76, - 209.77 - ], - "text": "Next, setting n = 2 in Eq. (3.18) and using the value y[1] = 5 (computed in the previous step)\nand x[2] = (2)2, we obtain", - "type": "text" - }, - { - "block_id": "p287-b7", - "global_id": 7728, - "bbox": [ - 263.65, - 207.24, - 368.01, - 221.73 - ], - "text": "y[2] = 0.5(5) + (2)2 = 6.5", - "type": "text" - }, - { - "block_id": "p287-b8", - "global_id": 7729, - "bbox": [ - 128.91, - 230.7, - 305.69, - 240.66 - ], - "text": "Continuing in this way iteratively, we obtain", - "type": "text" - }, - { - "block_id": "p287-b9", - "global_id": 7730, - "bbox": [ - 247.46, - 248.09, - 369.25, - 262.58 - ], - "text": "y[3] = 0.5(6.5) + (3)2 = 12.25", - "type": "text" - }, - { - "block_id": "p287-b10", - "global_id": 7731, - "bbox": [ - 247.46, - 264.52, - 384.18, - 279.01 - ], - "text": "y[4] = 0.5(12.25) + (4)2 = 22.125", - "type": "text" - }, - { - "block_id": "p287-b11", - "global_id": 7732, - "bbox": [ - 269.17, - 283.61, - 271.66, - 301.54 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p287-b12", - "global_id": 7733, - "bbox": [ - 128.91, - 313.08, - 287.18, - 323.45 - ], - "text": "The output y[n] is depicted in Fig. 3.17.", - "type": "text" - }, - { - "block_id": "p287-b13", - "global_id": 7734, - "bbox": [ - 140.32, - 341.43, - 153.2, - 349.51 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p287-b14", - "global_id": 7735, - "bbox": [ - 149.2, - 375.84, - 153.2, - 383.84 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p287-b15", - "global_id": 7736, - "bbox": [ - 180.04, - 387.08, - 214.79, - 405.68 - ], - "text": "5\n6.5", - "type": "text" - }, - { - "block_id": "p287-b16", - "global_id": 7737, - "bbox": [ - 227.45, - 345.34, - 245.45, - 353.34 - ], - "text": "12.25", - "type": "text" - }, - { - "block_id": "p287-b17", - "global_id": 7738, - "bbox": [ - 159.87, - 431.63, - 506.35, - 453.56 - ], - "text": "0\n1\n2\n3\n4\n5 n\nFigure 3.17 Iterative\nsolution\nof\na\ndifference equation.", - "type": "text" - }, - { - "block_id": "p287-b18", - "global_id": 7739, - "bbox": [ - 127.59, - 490.23, - 516.13, - 536.06 - ], - "text": "We now present one more example of iterative solution—this time for a second-order\nequation. The iterative method can be applied to a difference equation in delay form or advance\nform. In Ex. 3.11 we considered the former. Let us now apply the iterative method to the advance\nform.", - "type": "text" - }, - { - "block_id": "p287-b19", - "global_id": 7740, - "bbox": [ - 102.51, - 559.04, - 465.78, - 584.95 - ], - "text": "EXAMPLE 3.12\nIterative Solution to a Second-Order Difference\nEquation", - "type": "text" - }, - { - "block_id": "p287-b20", - "global_id": 7741, - "bbox": [ - 128.9, - 601.6, - 193.94, - 611.57 - ], - "text": "Solve iteratively", - "type": "text" - }, - { - "block_id": "p287-b21", - "global_id": 7742, - "bbox": [ - 210.55, - 623.11, - 421.07, - 633.49 - ], - "text": "y[n + 2] −y[n + 1] + 0.24y[n] = x[n + 2] −2x[n + 1]", - "type": "text" - } - ] - }, - { - "page_num": 288, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p288-b0", - "global_id": 7743, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "268\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p288-b1", - "global_id": 7744, - "bbox": [ - 103.16, - 85.83, - 477.02, - 108.16 - ], - "text": "with initial conditions y[−1] = 2, y[−2] = 1 and a causal input x[n] = nu[n]. The system\nequation can be expressed as", - "type": "text" - }, - { - "block_id": "p288-b2", - "global_id": 7745, - "bbox": [ - 184.82, - 119.7, - 477.02, - 130.07 - ], - "text": "y[n + 2] = y[n + 1] −0.24y[n] + x[n + 2] −2x[n + 1]\n(3.19)", - "type": "text" - }, - { - "block_id": "p288-b3", - "global_id": 7746, - "bbox": [ - 103.16, - 162.54, - 477.02, - 184.87 - ], - "text": "Setting n = −2 in Eq. (3.19) and then substituting y[−1] = 2, y[−2] = 1, x[0] = x[−1] = 0, we\nobtain", - "type": "text" - }, - { - "block_id": "p288-b4", - "global_id": 7747, - "bbox": [ - 222.79, - 186.45, - 357.4, - 196.83 - ], - "text": "y[0] = 2 −0.24(1) + 0 −0 = 1.76", - "type": "text" - }, - { - "block_id": "p288-b5", - "global_id": 7748, - "bbox": [ - 103.17, - 205.38, - 477.02, - 227.71 - ], - "text": "Setting n = −1 in Eq. (3.19) and then substituting y[0] = 1.76, y[−1] = 2, x[1] = 1, x[0] = 0,\nwe obtain", - "type": "text" - }, - { - "block_id": "p288-b6", - "global_id": 7749, - "bbox": [ - 216.56, - 229.29, - 363.63, - 239.67 - ], - "text": "y[1] = 1.76 −0.24(2) + 1 −0 = 2.28", - "type": "text" - }, - { - "block_id": "p288-b7", - "global_id": 7750, - "bbox": [ - 103.17, - 248.22, - 477.02, - 270.54 - ], - "text": "Setting n = 0 in Eq. (3.19) and then substituting y[0] = 1.76, y[1] = 2.28, x[2] = 2, and x[1] = 1\nyield", - "type": "text" - }, - { - "block_id": "p288-b8", - "global_id": 7751, - "bbox": [ - 199.14, - 272.13, - 381.03, - 282.51 - ], - "text": "y[2] = 2.28 −0.24(1.76) + 2 −2(1) = 1.8576", - "type": "text" - }, - { - "block_id": "p288-b9", - "global_id": 7752, - "bbox": [ - 103.17, - 291.47, - 143.84, - 301.43 - ], - "text": "and so on.", - "type": "text" - }, - { - "block_id": "p288-b10", - "global_id": 7753, - "bbox": [ - 121.1, - 303.42, - 433.2, - 313.38 - ], - "text": "With MATLAB, we can readily verify and extend these recursive calculations.", - "type": "text" - }, - { - "block_id": "p288-b11", - "global_id": 7754, - "bbox": [ - 103.16, - 323.09, - 462.91, - 390.84 - ], - "text": ">>\nn = -2:5; y = [1,2,zeros(1,length(n)-2)]; x = [0,0,n(3:end)];\n>>\nfor k = 1:length(n)-2,\n>>\ny(k+2) = y(k+1)-0.24*y(k)+x(k+2)-2*x(k+1);\n>>\nend\n>>\nn,y\nn = -2\n-1\n0\n1\n2\n3\n4\n5\ny =\n1.0000\n2.0000\n1.7600\n2.2800\n1.8576\n0.3104\n-2.1354\n-5.2099", - "type": "text" - }, - { - "block_id": "p288-b12", - "global_id": 7755, - "bbox": [ - 101.84, - 419.3, - 490.42, - 501.09 - ], - "text": "Note carefully the recursive nature of the computations. From the N initial conditions (and\nthe input), we obtained y[0] first. Then, using this value of y[0] and the preceding N −1 initial\nconditions (along with the input), we find y[1]. Next, using y[0], y[1] along with the past N −2\ninitial conditions and input, we obtained y[2], and so on. This method is general and can be applied\nto a recursive difference equation of any order. It is interesting that the hardware realization of\nEq. (3.18) depicted in Fig. 3.14 (with a = 0.5) generates the solution precisely in this (iterative)\nfashion.", - "type": "text" - }, - { - "block_id": "p288-b13", - "global_id": 7756, - "bbox": [ - 107.82, - 524.17, - 421.36, - 536.13 - ], - "text": "DRILL 3.10\nIterative Solution to a Difference Equation", - "type": "text" - }, - { - "block_id": "p288-b14", - "global_id": 7757, - "bbox": [ - 107.82, - 544.82, - 357.71, - 555.2 - ], - "text": "Using the iterative method, find the first three terms of y[n] for", - "type": "text" - }, - { - "block_id": "p288-b15", - "global_id": 7758, - "bbox": [ - 250.26, - 566.75, - 341.98, - 577.12 - ], - "text": "y[n + 1] −2y[n] = x[n]", - "type": "text" - }, - { - "block_id": "p288-b16", - "global_id": 7759, - "bbox": [ - 107.83, - 588.66, - 405.25, - 599.04 - ], - "text": "The initial condition is y[−1] = 10 and the input x[n] = 2 starting at n = 0.", - "type": "text" - }, - { - "block_id": "p288-b17", - "global_id": 7760, - "bbox": [ - 108.09, - 612.56, - 162.68, - 623.52 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p288-b18", - "global_id": 7761, - "bbox": [ - 107.82, - 630.5, - 248.27, - 640.88 - ], - "text": "y[0] = 20, y[1] = 42, and y[2] = 86", - "type": "text" - } - ] - }, - { - "page_num": 289, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p289-b0", - "global_id": 7762, - "bbox": [ - 325.13, - 62.89, - 516.14, - 71.98 - ], - "text": "3.5\nDiscrete-Time System Equations\n269", - "type": "text" - }, - { - "block_id": "p289-b1", - "global_id": 7763, - "bbox": [ - 127.59, - 85.82, - 516.15, - 155.56 - ], - "text": "We shall see in the future that the solution of a difference equation obtained in this direct\n(iterative) way is useful in many situations. Despite the many uses of this method, a closed-form\nsolution of a difference equation is far more useful in the study of system behavior and its\ndependence on the input and various system parameters. For this reason we shall develop a\nsystematic procedure to analyze discrete-time systems along lines similar to those used for\ncontinuous-time systems.", - "type": "text" - }, - { - "block_id": "p289-b2", - "global_id": 7764, - "bbox": [ - 127.59, - 170.34, - 516.14, - 232.32 - ], - "text": "OPERATOR NOTATION\nIn difference equations, it is convenient to use operator notation similar to that used in differential\nequations for the sake of compactness. In continuous-time systems, we used the operator D to\ndenote the operation of differentiation. For discrete-time systems, we shall use the operator E to\ndenote the operation for advancing a sequence by one time unit. Thus,", - "type": "text" - }, - { - "block_id": "p289-b3", - "global_id": 7765, - "bbox": [ - 290.65, - 243.96, - 356.58, - 254.33 - ], - "text": "Ex[n] ≡x[n + 1]", - "type": "text" - }, - { - "block_id": "p289-b4", - "global_id": 7766, - "bbox": [ - 286.44, - 258.96, - 356.58, - 270.77 - ], - "text": "E2x[n] ≡x[n + 2]", - "type": "text" - }, - { - "block_id": "p289-b5", - "global_id": 7767, - "bbox": [ - 318.49, - 275.37, - 320.98, - 293.29 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p289-b6", - "global_id": 7768, - "bbox": [ - 284.86, - 297.75, - 358.84, - 309.53 - ], - "text": "ENx[n] ≡x[n + N]", - "type": "text" - }, - { - "block_id": "p289-b7", - "global_id": 7769, - "bbox": [ - 127.6, - 321.69, - 516.16, - 343.6 - ], - "text": "Let us use this advance operator notation to represent several systems investigated earlier. The\nfirst-order difference equation of a savings account is [see Eq. (3.4)]", - "type": "text" - }, - { - "block_id": "p289-b8", - "global_id": 7770, - "bbox": [ - 268.08, - 355.24, - 375.63, - 365.62 - ], - "text": "y[n + 1] −ay[n] = x[n + 1]", - "type": "text" - }, - { - "block_id": "p289-b9", - "global_id": 7771, - "bbox": [ - 127.6, - 377.68, - 366.5, - 387.64 - ], - "text": "Using the operator notation, we can express this equation as", - "type": "text" - }, - { - "block_id": "p289-b10", - "global_id": 7772, - "bbox": [ - 213.93, - 399.28, - 429.78, - 409.65 - ], - "text": "Ey[n] −ay[n] = Ex[n]\nor\n(E −a)y[n] = Ex[n]", - "type": "text" - }, - { - "block_id": "p289-b11", - "global_id": 7773, - "bbox": [ - 127.6, - 421.71, - 417.18, - 431.67 - ], - "text": "Similarly, the second-order book sales estimate described by Eq. (3.6) as", - "type": "text" - }, - { - "block_id": "p289-b12", - "global_id": 7774, - "bbox": [ - 241.57, - 441.96, - 289.01, - 453.69 - ], - "text": "y[n + 2] + 1", - "type": "text" - }, - { - "block_id": "p289-b13", - "global_id": 7775, - "bbox": [ - 285.52, - 441.96, - 402.14, - 456.51 - ], - "text": "4y[n + 1] + 1\n16y[n] = x[n + 2]", - "type": "text" - }, - { - "block_id": "p289-b14", - "global_id": 7776, - "bbox": [ - 127.6, - 465.74, - 286.8, - 475.7 - ], - "text": "can be expressed in operator notation as", - "type": "text" - }, - { - "block_id": "p289-b16", - "global_id": 7777, - "bbox": [ - 267.92, - 483.23, - 293.77, - 497.62 - ], - "text": "E2 + 1", - "type": "text" - }, - { - "block_id": "p289-b17", - "global_id": 7778, - "bbox": [ - 290.29, - 479.33, - 325.54, - 500.53 - ], - "text": "4E + 1\n16", - "type": "text" - }, - { - "block_id": "p289-b18", - "global_id": 7779, - "bbox": [ - 325.54, - 486.4, - 379.81, - 497.62 - ], - "text": "y[n] = E2x[n]", - "type": "text" - }, - { - "block_id": "p289-b19", - "global_id": 7780, - "bbox": [ - 127.59, - 509.68, - 487.73, - 519.74 - ], - "text": "The general Nth-order advance-form difference equation of Eq. (3.15) can be expressed as", - "type": "text" - }, - { - "block_id": "p289-b20", - "global_id": 7781, - "bbox": [ - 153.4, - 527.27, - 490.31, - 542.53 - ], - "text": "(EN + a1EN−1 + · · · + aN−1E + aN)y[n] = (b0EN + b1EN−1 + · · · + bN−1E + bN)x[n]", - "type": "text" - }, - { - "block_id": "p289-b21", - "global_id": 7782, - "bbox": [ - 127.6, - 553.81, - 135.89, - 563.77 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p289-b22", - "global_id": 7783, - "bbox": [ - 280.28, - 565.5, - 516.13, - 575.87 - ], - "text": "Q[E]y[n] = P[E]x[n]\n(3.20)", - "type": "text" - }, - { - "block_id": "p289-b23", - "global_id": 7784, - "bbox": [ - 127.6, - 584.52, - 356.88, - 594.9 - ], - "text": "where Q[E] and P[E] are Nth-order polynomial operators", - "type": "text" - }, - { - "block_id": "p289-b24", - "global_id": 7785, - "bbox": [ - 234.1, - 602.42, - 399.73, - 617.7 - ], - "text": "Q[E] = EN + a1EN−1 + · · · + aN−1E + aN", - "type": "text" - }, - { - "block_id": "p289-b25", - "global_id": 7786, - "bbox": [ - 235.2, - 618.87, - 408.7, - 634.13 - ], - "text": "P[E] = b0EN + b1EN−1 + · · · + bN−1E + bN", - "type": "text" - } - ] - }, - { - "page_num": 290, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p290-b0", - "global_id": 7787, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "270\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p290-b1", - "global_id": 7788, - "bbox": [ - 102.14, - 86.19, - 357.85, - 98.32 - ], - "text": "RESPONSE OF LINEAR DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p290-b2", - "global_id": 7789, - "bbox": [ - 101.84, - 102.35, - 490.41, - 160.13 - ], - "text": "Following the procedure used for continuous-time systems, we can show that Eq. (3.20) is a\nlinear equation (with constant coefficients). A system described by such an equation is a linear,\ntime-invariant, discrete-time (LTID) system. We can verify, as in the case of LTIC systems (see the\nfootnote on page 151), that the general solution of Eq. (3.20) consists of zero-input and zero-state\ncomponents.", - "type": "text" - }, - { - "block_id": "p290-b3", - "global_id": 7790, - "bbox": [ - 102.19, - 182.64, - 437.19, - 212.53 - ], - "text": "3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS:\nTHE ZERO-INPUT RESPONSE", - "type": "text" - }, - { - "block_id": "p290-b4", - "global_id": 7791, - "bbox": [ - 101.84, - 218.11, - 419.89, - 229.26 - ], - "text": "The zero-input response y0[n] is the solution of Eq. (3.20) with x[n] = 0; that is,", - "type": "text" - }, - { - "block_id": "p290-b5", - "global_id": 7792, - "bbox": [ - 267.61, - 236.03, - 324.63, - 247.18 - ], - "text": "Q[E]y0[n] = 0", - "type": "text" - }, - { - "block_id": "p290-b6", - "global_id": 7793, - "bbox": [ - 101.85, - 254.39, - 110.14, - 264.35 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p290-b7", - "global_id": 7794, - "bbox": [ - 206.69, - 263.61, - 490.38, - 278.87 - ], - "text": "(EN + a1EN−1 + · · · + aN−1E + aN)y0[n] = 0\n(3.21)", - "type": "text" - }, - { - "block_id": "p290-b8", - "global_id": 7795, - "bbox": [ - 101.84, - 288.27, - 490.41, - 346.05 - ], - "text": "Although we can solve this equation systematically, even a cursory examination points to the\nsolution. This equation states that a linear combination of y0[n] and advanced y0[n] is zero, not for\nsome values of n, but for all n. Such a situation is possible if and only if y0[n] and advanced y0[n]\nhave the same form. Only an exponential function γ n has this property, as the following equation\nindicates:", - "type": "text" - }, - { - "block_id": "p290-b9", - "global_id": 7796, - "bbox": [ - 250.2, - 345.32, - 341.54, - 359.71 - ], - "text": "Ek{γ n} = γ n+k = γ kγ n", - "type": "text" - }, - { - "block_id": "p290-b10", - "global_id": 7797, - "bbox": [ - 101.84, - 366.36, - 490.39, - 391.9 - ], - "text": "This expression shows that γ n advanced by k units is a constant (γ k) times γ n. Therefore, the\nsolution of Eq. (3.21) must be of the form†", - "type": "text" - }, - { - "block_id": "p290-b11", - "global_id": 7798, - "bbox": [ - 272.67, - 397.95, - 490.38, - 410.61 - ], - "text": "y0[n] = cγ n\n(3.22)", - "type": "text" - }, - { - "block_id": "p290-b12", - "global_id": 7799, - "bbox": [ - 101.85, - 416.16, - 490.39, - 439.72 - ], - "text": "To determine c and γ , we substitute this solution in Eq. (3.21). Since Eky0[n] = y0[n+k] = cγ n+k,\nthis produces", - "type": "text" - }, - { - "block_id": "p290-b13", - "global_id": 7800, - "bbox": [ - 208.22, - 438.98, - 384.01, - 454.25 - ], - "text": "c(γ N + a1γ N−1 + · · · + aN−1γ + aN)γ n = 0", - "type": "text" - }, - { - "block_id": "p290-b14", - "global_id": 7801, - "bbox": [ - 101.84, - 463.64, - 265.0, - 473.61 - ], - "text": "For a nontrivial solution of this equation,", - "type": "text" - }, - { - "block_id": "p290-b15", - "global_id": 7802, - "bbox": [ - 220.79, - 477.05, - 490.38, - 492.62 - ], - "text": "γ N + a1γ n−1 + · · · + aN−1γ + aN = 0\n(3.23)", - "type": "text" - }, - { - "block_id": "p290-b16", - "global_id": 7803, - "bbox": [ - 101.85, - 499.51, - 110.14, - 509.47 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p290-b17", - "global_id": 7804, - "bbox": [ - 277.48, - 512.86, - 314.75, - 523.23 - ], - "text": "Q[γ ] = 0", - "type": "text" - }, - { - "block_id": "p290-b18", - "global_id": 7805, - "bbox": [ - 101.84, - 529.78, - 490.37, - 555.32 - ], - "text": "Our solution cγ n [Eq. (3.22)] is correct, provided γ satisfies Eq. (3.23). Now, Q[γ ] is an Nth-order\npolynomial and can be expressed in the factored form (assuming all distinct roots):", - "type": "text" - }, - { - "block_id": "p290-b19", - "global_id": 7806, - "bbox": [ - 228.44, - 562.87, - 363.8, - 574.02 - ], - "text": "(γ −γ1)(γ −γ2)· · ·(γ −γN) = 0", - "type": "text" - }, - { - "block_id": "p290-b20", - "global_id": 7807, - "bbox": [ - 101.85, - 579.66, - 487.4, - 592.68 - ], - "text": "Clearly, γ has N solutions γ1, γ2, . . . , γN and, therefore, Eq. (3.21) also has N solutions c1γ n", - "type": "text" - }, - { - "block_id": "p290-b21", - "global_id": 7808, - "bbox": [ - 101.85, - 581.22, - 490.39, - 603.91 - ], - "text": "1 ,\nc2γ n", - "type": "text" - }, - { - "block_id": "p290-b22", - "global_id": 7809, - "bbox": [ - 115.25, - 591.62, - 490.37, - 606.05 - ], - "text": "2 , . . . , cnγ n\nN. In such a case, we have shown that the general solution is a linear combination", - "type": "text" - }, - { - "block_id": "p290-b23", - "global_id": 7810, - "bbox": [ - 101.84, - 623.19, - 490.36, - 646.36 - ], - "text": "† A signal of the form nmγ n also satisfies this requirement under certain conditions (repeated roots), discussed\nlater.", - "type": "text" - } - ] - }, - { - "page_num": 291, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p291-b0", - "global_id": 7811, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "3.6\nSystem Response to Internal Conditions: The Zero-Input Response\n271", - "type": "text" - }, - { - "block_id": "p291-b1", - "global_id": 7812, - "bbox": [ - 127.59, - 85.72, - 349.54, - 95.78 - ], - "text": "of the N solutions (see the footnote on page 153). Thus,", - "type": "text" - }, - { - "block_id": "p291-b2", - "global_id": 7813, - "bbox": [ - 254.59, - 101.84, - 304.98, - 114.49 - ], - "text": "y0[n] = c1γ n", - "type": "text" - }, - { - "block_id": "p291-b3", - "global_id": 7814, - "bbox": [ - 299.9, - 101.84, - 388.63, - 115.98 - ], - "text": "1 + c2γ n\n2 + · · · + cnγ n\nN", - "type": "text" - }, - { - "block_id": "p291-b4", - "global_id": 7815, - "bbox": [ - 127.59, - 121.27, - 516.14, - 191.42 - ], - "text": "where γ1, γ2, . . . , γn are the roots of Eq. (3.23) and c1, c2, . . . , cn are arbitrary constants\ndetermined from N auxiliary conditions, generally given in the form of initial conditions. The\npolynomial Q[γ ] is called the characteristic polynomial of the system, and Q[γ ] = 0 [Eq. (3.23)] is\nthe characteristic equation of the system. Moreover, γ1, γ2, . . . , γN, the roots of the characteristic\nequation, are called characteristic roots or characteristic values (also eigenvalues) of the system.\nThe exponentials γ n", - "type": "text" - }, - { - "block_id": "p291-b5", - "global_id": 7816, - "bbox": [ - 127.59, - 181.04, - 516.13, - 215.33 - ], - "text": "i (i = 1,2,. . .,N) are the characteristic modes or natural modes of the system.\nA characteristic mode corresponds to each characteristic root of the system, and the zero-input\nresponse is a linear combination of the characteristic modes of the system.", - "type": "text" - }, - { - "block_id": "p291-b6", - "global_id": 7817, - "bbox": [ - 102.51, - 243.21, - 495.57, - 269.12 - ], - "text": "EXAMPLE 3.13\nZero-Input Response of a Second-Order System with\nReal Roots", - "type": "text" - }, - { - "block_id": "p291-b7", - "global_id": 7818, - "bbox": [ - 128.9, - 281.92, - 346.0, - 291.88 - ], - "text": "The LTID system described by the difference equation", - "type": "text" - }, - { - "block_id": "p291-b8", - "global_id": 7819, - "bbox": [ - 225.73, - 299.44, - 405.92, - 309.82 - ], - "text": "y[n + 2] −0.6y[n + 1] −0.16y[n] = 5x[n + 2]", - "type": "text" - }, - { - "block_id": "p291-b9", - "global_id": 7820, - "bbox": [ - 128.9, - 316.15, - 502.77, - 351.66 - ], - "text": "has input x[n] = 4−nu[n] and initial conditions y[−1] = 0 and y[−2] = 25/4. Determine\nthe zero-input response y0[n]. The zero-state response of this system is considered later, in\nEx. 3.21.", - "type": "text" - }, - { - "block_id": "p291-b10", - "global_id": 7821, - "bbox": [ - 128.9, - 374.57, - 301.58, - 384.54 - ], - "text": "The system equation in operator notation is", - "type": "text" - }, - { - "block_id": "p291-b11", - "global_id": 7822, - "bbox": [ - 248.36, - 387.97, - 383.3, - 402.47 - ], - "text": "(E2 −0.6E −0.16)y[n] = 5E2x[n]", - "type": "text" - }, - { - "block_id": "p291-b12", - "global_id": 7823, - "bbox": [ - 128.91, - 410.44, - 257.03, - 420.4 - ], - "text": "The characteristic polynomial is", - "type": "text" - }, - { - "block_id": "p291-b13", - "global_id": 7824, - "bbox": [ - 238.15, - 423.85, - 393.51, - 438.34 - ], - "text": "γ 2 −0.6γ −0.16 = (γ + 0.2)(γ −0.8)", - "type": "text" - }, - { - "block_id": "p291-b14", - "global_id": 7825, - "bbox": [ - 128.91, - 446.3, - 245.95, - 456.27 - ], - "text": "The characteristic equation is", - "type": "text" - }, - { - "block_id": "p291-b15", - "global_id": 7826, - "bbox": [ - 270.06, - 457.85, - 361.62, - 468.23 - ], - "text": "(γ + 0.2)(γ −0.8) = 0", - "type": "text" - }, - { - "block_id": "p291-b16", - "global_id": 7827, - "bbox": [ - 128.91, - 475.84, - 443.71, - 487.3 - ], - "text": "The characteristic roots are γ1 = −0.2 and γ2 = 0.8. The zero-input response is", - "type": "text" - }, - { - "block_id": "p291-b17", - "global_id": 7828, - "bbox": [ - 258.29, - 489.66, - 502.75, - 504.93 - ], - "text": "y0[n] = c1(−0.2)n + c2(0.8)n\n(3.24)", - "type": "text" - }, - { - "block_id": "p291-b18", - "global_id": 7829, - "bbox": [ - 128.9, - 511.71, - 502.75, - 534.81 - ], - "text": "To determine arbitrary constants c1 and c2, we set n = −1 and −2 in Eq. (3.24), then substitute\ny0[−1] = 0 and y0[−2] = 25/4 to obtain†", - "type": "text" - }, - { - "block_id": "p291-b19", - "global_id": 7830, - "bbox": [ - 243.6, - 540.12, - 310.15, - 567.32 - ], - "text": "0 = −5c1 + 5\n4c2\n25\n4 = 25c1 + 25\n16c2", - "type": "text" - }, - { - "block_id": "p291-b21", - "global_id": 7831, - "bbox": [ - 332.86, - 540.1, - 389.26, - 557.51 - ], - "text": "⇒\nc1 = 1", - "type": "text" - }, - { - "block_id": "p291-b22", - "global_id": 7832, - "bbox": [ - 364.31, - 547.67, - 389.26, - 565.28 - ], - "text": "5\nc2 = 4", - "type": "text" - }, - { - "block_id": "p291-b23", - "global_id": 7833, - "bbox": [ - 385.77, - 560.04, - 389.26, - 567.02 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p291-b24", - "global_id": 7834, - "bbox": [ - 127.59, - 613.72, - 516.13, - 648.6 - ], - "text": "† The initial conditions y[−1] and y[−2] are the conditions given on the total response. But because the input\ndoes not start until n = 0, the zero-state response is zero for n < 0. Hence, at n = −1 and −2 the total response\nconsists of the zero-input component only so that y[−1] = y0[−1] and y[−2] = y0[−2].", - "type": "text" - } - ] - }, - { - "page_num": 292, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p292-b0", - "global_id": 7835, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "272\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p292-b1", - "global_id": 7836, - "bbox": [ - 103.16, - 86.24, - 144.91, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p292-b2", - "global_id": 7837, - "bbox": [ - 214.33, - 96.43, - 250.9, - 108.93 - ], - "text": "y0[n] = 1", - "type": "text" - }, - { - "block_id": "p292-b3", - "global_id": 7838, - "bbox": [ - 247.42, - 94.17, - 365.85, - 110.97 - ], - "text": "5(−0.2)n + 4\n5(0.8)n\nn ≥0", - "type": "text" - }, - { - "block_id": "p292-b4", - "global_id": 7839, - "bbox": [ - 103.16, - 117.13, - 477.01, - 139.04 - ], - "text": "The reader can verify this solution by computing the first few terms using the iterative method\n(see Exs. 3.11 and 3.12).", - "type": "text" - }, - { - "block_id": "p292-b5", - "global_id": 7840, - "bbox": [ - 107.82, - 210.87, - 430.56, - 222.82 - ], - "text": "DRILL 3.11\nZero-Input Response of First-Order Systems", - "type": "text" - }, - { - "block_id": "p292-b6", - "global_id": 7841, - "bbox": [ - 107.82, - 231.94, - 484.38, - 253.87 - ], - "text": "Find and sketch the zero-input response for the systems described by the following\nequations:", - "type": "text" - }, - { - "block_id": "p292-b7", - "global_id": 7842, - "bbox": [ - 125.76, - 261.42, - 262.36, - 271.8 - ], - "text": "(a) y[n + 1] −0.8y[n] = 3x[n + 1]", - "type": "text" - }, - { - "block_id": "p292-b8", - "global_id": 7843, - "bbox": [ - 125.76, - 276.37, - 262.92, - 286.74 - ], - "text": "(b) y[n + 1] + 0.8y[n] = 3x[n + 1]", - "type": "text" - }, - { - "block_id": "p292-b9", - "global_id": 7844, - "bbox": [ - 107.82, - 294.3, - 484.42, - 316.63 - ], - "text": "In each case the initial condition is y[−1] = 10. Verify the solutions by computing the first three\nterms using the iterative method.", - "type": "text" - }, - { - "block_id": "p292-b10", - "global_id": 7845, - "bbox": [ - 108.09, - 330.15, - 170.31, - 341.11 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p292-b11", - "global_id": 7846, - "bbox": [ - 125.76, - 347.09, - 170.7, - 358.47 - ], - "text": "(a) 8(0.8)n", - "type": "text" - }, - { - "block_id": "p292-b12", - "global_id": 7847, - "bbox": [ - 125.76, - 362.03, - 186.79, - 373.41 - ], - "text": "(b) −8(−0.8)n", - "type": "text" - }, - { - "block_id": "p292-b13", - "global_id": 7848, - "bbox": [ - 107.82, - 438.8, - 477.69, - 464.7 - ], - "text": "DRILL 3.12\nZero-Input Response of a Second-Order System with\nReal Roots", - "type": "text" - }, - { - "block_id": "p292-b14", - "global_id": 7849, - "bbox": [ - 107.82, - 473.82, - 373.97, - 483.78 - ], - "text": "Find the zero-input response of a system described by the equation", - "type": "text" - }, - { - "block_id": "p292-b15", - "global_id": 7850, - "bbox": [ - 195.03, - 493.33, - 397.17, - 503.71 - ], - "text": "y[n] + 0.3y[n −1] −0.1y[n −2] = x[n] + 2x[n −1]", - "type": "text" - }, - { - "block_id": "p292-b16", - "global_id": 7851, - "bbox": [ - 107.82, - 515.25, - 484.42, - 537.59 - ], - "text": "The initial conditions are y0[−1] = 1 and y0[−2] = 33. Verify the solution by computing the\nfirst three terms iteratively.", - "type": "text" - }, - { - "block_id": "p292-b17", - "global_id": 7852, - "bbox": [ - 108.09, - 551.1, - 162.68, - 562.06 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p292-b18", - "global_id": 7853, - "bbox": [ - 107.82, - 565.43, - 210.56, - 580.2 - ], - "text": "y0[n] = (0.2)n + 2(−0.5)n", - "type": "text" - }, - { - "block_id": "p292-b19", - "global_id": 7854, - "bbox": [ - 101.84, - 612.86, - 490.4, - 634.79 - ], - "text": "Section 3.5-1 introduced the method of recursion to solve difference equations. As the next\nexample illustrates, the zero-input response can likewise be found through recursion. Since it does", - "type": "text" - } - ] - }, - { - "page_num": 293, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p293-b0", - "global_id": 7855, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "3.6\nSystem Response to Internal Conditions: The Zero-Input Response\n273", - "type": "text" - }, - { - "block_id": "p293-b1", - "global_id": 7856, - "bbox": [ - 127.59, - 85.82, - 516.14, - 107.74 - ], - "text": "not provide a closed-form solution, recursion is generally not the preferred method of solving\ndifference equations.", - "type": "text" - }, - { - "block_id": "p293-b2", - "global_id": 7857, - "bbox": [ - 102.51, - 139.02, - 434.5, - 150.98 - ], - "text": "EXAMPLE 3.14\nIterative Solution to Zero-Input Response", - "type": "text" - }, - { - "block_id": "p293-b3", - "global_id": 7858, - "bbox": [ - 128.9, - 167.23, - 502.76, - 201.51 - ], - "text": "Using the initial conditions y[−1] = 2 and y[−2] = 1, use MATLAB to iteratively compute\nand then plot the zero-input response for the system described by (E2 −1.56E + 0.81)y[n] =\n(E + 3)x[n].", - "type": "text" - }, - { - "block_id": "p293-b4", - "global_id": 7859, - "bbox": [ - 128.9, - 224.17, - 450.53, - 272.0 - ], - "text": ">>\nn = (-2:20)’; y = [1;2;zeros(length(n)-2,1)];\n>>\nfor k = 1:length(n)-2,\n>>\ny(k+2) = 1.56*y(k+1)-0.81*y(k);\n>>\nend;\n>>\nclf; stem(n,y,’k’); xlabel(’n’); ylabel(’y[n]’); axis([-2 20 -1.5 2.5]);", - "type": "text" - }, - { - "block_id": "p293-b5", - "global_id": 7860, - "bbox": [ - 167.52, - 383.47, - 376.46, - 404.66 - ], - "text": "0\n5\n10\n15\n20\nn", - "type": "text" - }, - { - "block_id": "p293-b6", - "global_id": 7861, - "bbox": [ - 136.98, - 365.25, - 144.98, - 373.25 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p293-b7", - "global_id": 7862, - "bbox": [ - 140.98, - 347.25, - 144.98, - 355.25 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p293-b8", - "global_id": 7863, - "bbox": [ - 140.98, - 329.25, - 144.98, - 337.25 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p293-b9", - "global_id": 7864, - "bbox": [ - 140.98, - 311.25, - 144.98, - 319.25 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p293-b10", - "global_id": 7865, - "bbox": [ - 126.76, - 333.81, - 135.56, - 348.47 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p293-b11", - "global_id": 7866, - "bbox": [ - 126.0, - 411.34, - 295.34, - 420.58 - ], - "text": "Figure 3.18 Zero-input response for Ex. 3.14.", - "type": "text" - }, - { - "block_id": "p293-b12", - "global_id": 7867, - "bbox": [ - 127.59, - 465.7, - 516.14, - 503.76 - ], - "text": "REPEATED ROOTS\nSo far we have assumed the system to have N distinct characteristic roots γ1, γ2, . . . , γN with\ncorresponding characteristic modes γ n", - "type": "text" - }, - { - "block_id": "p293-b13", - "global_id": 7868, - "bbox": [ - 127.59, - 492.25, - 516.14, - 539.63 - ], - "text": "1 , γ n\n2 , . . . , γ n\nN. If two or more roots coincide (repeated roots),\nthe form of characteristic modes is modified. Direct substitution shows that if a root γ repeats r\ntimes (root of multiplicity r), the corresponding characteristic modes for this root are γ n, nγ n,\nn2γ n, . . . , nr−1γ n. Thus, if the characteristic equation of a system is", - "type": "text" - }, - { - "block_id": "p293-b14", - "global_id": 7869, - "bbox": [ - 219.52, - 554.25, - 424.21, - 566.9 - ], - "text": "Q[γ ] = (γ −γ1)r(γ −γr+1)(γ −γr+2)· · ·(γ −γN)", - "type": "text" - }, - { - "block_id": "p293-b15", - "global_id": 7870, - "bbox": [ - 127.59, - 582.67, - 304.96, - 592.63 - ], - "text": "then the zero-input response of the system is", - "type": "text" - }, - { - "block_id": "p293-b16", - "global_id": 7871, - "bbox": [ - 164.72, - 604.65, - 334.93, - 620.22 - ], - "text": "y0[n] = (c1 + c2n + c3n2 + · · · + crnr−1)γ n", - "type": "text" - }, - { - "block_id": "p293-b17", - "global_id": 7872, - "bbox": [ - 329.85, - 607.26, - 417.99, - 621.41 - ], - "text": "1 + cr+1γ n\nr+1 + cr+2γ n", - "type": "text" - }, - { - "block_id": "p293-b18", - "global_id": 7873, - "bbox": [ - 412.91, - 607.26, - 478.51, - 621.41 - ], - "text": "r+2 + · · · + cnγ n", - "type": "text" - }, - { - "block_id": "p293-b19", - "global_id": 7874, - "bbox": [ - 473.43, - 614.05, - 478.08, - 621.03 - ], - "text": "N", - "type": "text" - } - ] - }, - { - "page_num": 294, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p294-b0", - "global_id": 7875, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "274\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p294-b1", - "global_id": 7876, - "bbox": [ - 76.77, - 93.92, - 469.82, - 119.82 - ], - "text": "EXAMPLE 3.15\nZero-Input Response of a Second-Order System with\nRepeated Roots", - "type": "text" - }, - { - "block_id": "p294-b2", - "global_id": 7877, - "bbox": [ - 103.16, - 136.48, - 361.59, - 146.45 - ], - "text": "Consider a second-order difference equation with repeated roots:", - "type": "text" - }, - { - "block_id": "p294-b3", - "global_id": 7878, - "bbox": [ - 217.79, - 153.87, - 362.37, - 168.36 - ], - "text": "(E2 + 6E + 9)y[n] = (2E2 + 6E)x[n]", - "type": "text" - }, - { - "block_id": "p294-b4", - "global_id": 7879, - "bbox": [ - 103.17, - 179.91, - 477.01, - 203.01 - ], - "text": "Determine the zero-input response y0[n] if the initial conditions are y0[−1] = −1/3 and\ny0[−2] = −2/9.", - "type": "text" - }, - { - "block_id": "p294-b5", - "global_id": 7880, - "bbox": [ - 103.16, - 221.12, - 477.01, - 259.03 - ], - "text": "The characteristic polynomial is γ 2 +6γ +9 = (γ +3)2, and we have a repeated characteristic\nroot at γ = −3. The characteristic modes are (−3)n and n(−3)n. Hence, the zero-input response\nis", - "type": "text" - }, - { - "block_id": "p294-b6", - "global_id": 7881, - "bbox": [ - 242.01, - 259.1, - 337.66, - 272.06 - ], - "text": "y0[n] = (c1 + c2n)(−3)n", - "type": "text" - }, - { - "block_id": "p294-b7", - "global_id": 7882, - "bbox": [ - 103.16, - 279.84, - 477.01, - 301.86 - ], - "text": "Although we can determine the constants c1 and c2 from the initial conditions following a\nprocedure similar to Ex. 3.13, we instead use MATLAB to perform the needed calculations.", - "type": "text" - }, - { - "block_id": "p294-b8", - "global_id": 7883, - "bbox": [ - 103.17, - 312.11, - 479.75, - 334.03 - ], - "text": ">>\nc = inv([(-3)^(-1) -1*(-3)^(-1);(-3)^(-2) -2*(-3)^(-2)])*[-1/3;-2/9]\nc = 4", - "type": "text" - }, - { - "block_id": "p294-b9", - "global_id": 7884, - "bbox": [ - 145.0, - 336.02, - 150.23, - 345.99 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p294-b10", - "global_id": 7885, - "bbox": [ - 103.16, - 355.66, - 230.16, - 365.62 - ], - "text": "Thus, the zero-input response is", - "type": "text" - }, - { - "block_id": "p294-b11", - "global_id": 7886, - "bbox": [ - 224.69, - 375.66, - 355.49, - 388.31 - ], - "text": "y0[n] = (4 + 3n)(−3)n\nn ≥0", - "type": "text" - }, - { - "block_id": "p294-b12", - "global_id": 7887, - "bbox": [ - 101.84, - 425.31, - 490.4, - 487.29 - ], - "text": "COMPLEX ROOTS\nAs in the case of continuous-time systems, the complex roots of a discrete-time system will occur\nin pairs of conjugates if the system equation coefficients are real. Complex roots can be treated\nexactly as we would treat real roots. However, just as in the case of continuous-time systems, we\ncan also use the real form of solution as an alternative.", - "type": "text" - }, - { - "block_id": "p294-b13", - "global_id": 7888, - "bbox": [ - 101.85, - 487.64, - 490.37, - 511.2 - ], - "text": "First we express the complex conjugate roots γ and γ ∗in polar form. If |γ | is the magnitude\nand β is the angle of γ , then", - "type": "text" - }, - { - "block_id": "p294-b14", - "global_id": 7889, - "bbox": [ - 222.18, - 522.32, - 369.07, - 534.42 - ], - "text": "γ = |γ |ejβ\nand\nγ ∗= |γ |e−jβ", - "type": "text" - }, - { - "block_id": "p294-b15", - "global_id": 7890, - "bbox": [ - 101.84, - 547.69, - 243.94, - 557.65 - ], - "text": "The zero-input response is given by", - "type": "text" - }, - { - "block_id": "p294-b16", - "global_id": 7891, - "bbox": [ - 196.42, - 566.38, - 395.3, - 581.64 - ], - "text": "y0[n] = c1γ n + c2(γ ∗)n = c1|γ |nejβn + c2|γ |ne−jβn", - "type": "text" - }, - { - "block_id": "p294-b17", - "global_id": 7892, - "bbox": [ - 101.84, - 593.71, - 447.0, - 605.58 - ], - "text": "For a real system, c1 and c2 must be conjugates so that y0[n] is a real function of n. Let", - "type": "text" - }, - { - "block_id": "p294-b18", - "global_id": 7893, - "bbox": [ - 228.01, - 612.82, - 254.18, - 630.95 - ], - "text": "c1 = c", - "type": "text" - }, - { - "block_id": "p294-b19", - "global_id": 7894, - "bbox": [ - 249.48, - 612.82, - 346.68, - 636.94 - ], - "text": "2ejθ\nand\nc2 = c", - "type": "text" - }, - { - "block_id": "p294-b20", - "global_id": 7895, - "bbox": [ - 341.98, - 617.77, - 363.02, - 636.94 - ], - "text": "2e−jθ", - "type": "text" - } - ] - }, - { - "page_num": 295, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p295-b0", - "global_id": 7896, - "bbox": [ - 189.55, - 62.89, - 516.13, - 71.98 - ], - "text": "3.6\nSystem Response to Internal Conditions: The Zero-Input Response\n275", - "type": "text" - }, - { - "block_id": "p295-b1", - "global_id": 7897, - "bbox": [ - 127.59, - 85.82, - 148.05, - 95.78 - ], - "text": "Then", - "type": "text" - }, - { - "block_id": "p295-b2", - "global_id": 7898, - "bbox": [ - 210.76, - 97.67, - 248.55, - 115.49 - ], - "text": "y0[n] = c", - "type": "text" - }, - { - "block_id": "p295-b3", - "global_id": 7899, - "bbox": [ - 243.85, - 96.33, - 349.92, - 121.79 - ], - "text": "2|γ |n \t\nej(βn+θ) + e−j(βn+θ)", - "type": "text" - }, - { - "block_id": "p295-b4", - "global_id": 7900, - "bbox": [ - 351.97, - 100.64, - 516.13, - 114.71 - ], - "text": "= c|γ |n cos(βn + θ)\n(3.25)", - "type": "text" - }, - { - "block_id": "p295-b5", - "global_id": 7901, - "bbox": [ - 127.6, - 130.82, - 516.13, - 153.15 - ], - "text": "where c and θ are arbitrary constants determined from the auxiliary conditions. This is the solution\nin real form, which avoids dealing with complex numbers.", - "type": "text" - }, - { - "block_id": "p295-b6", - "global_id": 7902, - "bbox": [ - 102.51, - 184.47, - 495.57, - 210.37 - ], - "text": "EXAMPLE 3.16\nZero-Input Response of a Second-Order System with\nComplex Roots", - "type": "text" - }, - { - "block_id": "p295-b7", - "global_id": 7903, - "bbox": [ - 128.9, - 227.04, - 429.76, - 237.0 - ], - "text": "Consider a second-order difference equation with complex-conjugate roots:", - "type": "text" - }, - { - "block_id": "p295-b8", - "global_id": 7904, - "bbox": [ - 238.71, - 244.42, - 392.94, - 258.91 - ], - "text": "(E2 −1.56E + 0.81)y[n] = (E + 3)x[n]", - "type": "text" - }, - { - "block_id": "p295-b9", - "global_id": 7905, - "bbox": [ - 128.91, - 270.46, - 502.75, - 281.61 - ], - "text": "Determine the zero-input response y0[n] if the initial conditions are y0[−1] = 2 and y0[−2] = 1.", - "type": "text" - }, - { - "block_id": "p295-b10", - "global_id": 7906, - "bbox": [ - 128.9, - 299.72, - 502.76, - 337.62 - ], - "text": "The characteristic polynomial is (γ 2 −1.56γ + 0.81) = (γ −0.78 −j0.45)(γ −0.78 + j0.45).\nThe characteristic roots are 0.78 ± j0.45; that is, 0.9e±j(π/6). We could immediately write the\nsolution as", - "type": "text" - }, - { - "block_id": "p295-b11", - "global_id": 7907, - "bbox": [ - 239.38, - 335.09, - 391.77, - 350.35 - ], - "text": "y0[n] = c(0.9)nejπn/6 + c∗(0.9)ne−jπn/6", - "type": "text" - }, - { - "block_id": "p295-b12", - "global_id": 7908, - "bbox": [ - 128.9, - 358.13, - 502.76, - 380.46 - ], - "text": "Setting n = −1 and −2 and using the initial conditions y0[−1] = 2 and y0[−2] = 1, we find\nc = 1.1550 −j0.2025 = 1.1726e−j0.1735 and c∗= 1.1550 + j0.2025 = 1.1726ej0.1735.", - "type": "text" - }, - { - "block_id": "p295-b13", - "global_id": 7909, - "bbox": [ - 128.9, - 394.16, - 450.48, - 422.04 - ], - "text": ">>\ngamma = roots([1 -1.56 0.81]);\n>>\nc = inv([gamma(1)^(-1) gamma(2)^(-1);gamma(1)^(-2) gamma(2)^(-2)])*[2;1]\nc = 1.1550 - 0.2025i", - "type": "text" - }, - { - "block_id": "p295-b14", - "global_id": 7910, - "bbox": [ - 162.77, - 424.03, - 230.47, - 432.0 - ], - "text": "1.1550 + 0.2025i", - "type": "text" - }, - { - "block_id": "p295-b15", - "global_id": 7911, - "bbox": [ - 128.9, - 446.21, - 502.76, - 480.09 - ], - "text": "Alternately, we could also find the unknown coefficient by using the real form of the\nsolution, as given in Eq. (3.25). In the present case, the roots are 0.9e±j(π/6). Hence, |γ | = 0.9\nand β = π/6, and the zero-input response, according to Eq. (3.25), is given by", - "type": "text" - }, - { - "block_id": "p295-b16", - "global_id": 7912, - "bbox": [ - 255.98, - 493.57, - 330.54, - 508.42 - ], - "text": "y0[n] = c(0.9)n cos", - "type": "text" - }, - { - "block_id": "p295-b17", - "global_id": 7913, - "bbox": [ - 331.65, - 483.28, - 345.55, - 500.25 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p295-b18", - "global_id": 7914, - "bbox": [ - 340.57, - 483.28, - 375.69, - 514.72 - ], - "text": "6 n + θ", - "type": "text" - }, - { - "block_id": "p295-b19", - "global_id": 7915, - "bbox": [ - 128.91, - 524.48, - 502.77, - 547.59 - ], - "text": "To determine the constants c and θ, we set n = −1 and −2 in this equation and substitute the\ninitial conditions y0[−1] = 2 and y0[−2] = 1 to obtain", - "type": "text" - }, - { - "block_id": "p295-b20", - "global_id": 7916, - "bbox": [ - 202.35, - 552.88, - 256.26, - 584.32 - ], - "text": "2 = c\n0.9 cos", - "type": "text" - }, - { - "block_id": "p295-b21", - "global_id": 7917, - "bbox": [ - 256.26, - 559.88, - 271.2, - 576.83 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p295-b22", - "global_id": 7918, - "bbox": [ - 266.23, - 552.88, - 296.37, - 584.32 - ], - "text": "6 + θ", - "type": "text" - }, - { - "block_id": "p295-b23", - "global_id": 7919, - "bbox": [ - 298.42, - 560.2, - 317.87, - 577.14 - ], - "text": "= c", - "type": "text" - }, - { - "block_id": "p295-b24", - "global_id": 7920, - "bbox": [ - 309.44, - 574.35, - 321.89, - 584.32 - ], - "text": "0.9", - "type": "text" - }, - { - "block_id": "p295-b25", - "global_id": 7921, - "bbox": [ - 324.19, - 549.89, - 340.0, - 561.47 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p295-b26", - "global_id": 7922, - "bbox": [ - 335.79, - 560.3, - 405.74, - 584.32 - ], - "text": "3\n2 cos θ + 1\n2 sin θ", - "type": "text" - }, - { - "block_id": "p295-b28", - "global_id": 7923, - "bbox": [ - 202.35, - 594.07, - 259.84, - 618.19 - ], - "text": "1 =\nc\n(0.9)2 cos", - "type": "text" - }, - { - "block_id": "p295-b30", - "global_id": 7924, - "bbox": [ - 267.68, - 593.76, - 282.63, - 610.71 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p295-b31", - "global_id": 7925, - "bbox": [ - 277.65, - 586.75, - 307.79, - 618.19 - ], - "text": "3 + θ", - "type": "text" - }, - { - "block_id": "p295-b32", - "global_id": 7926, - "bbox": [ - 309.84, - 594.07, - 338.28, - 618.19 - ], - "text": "=\nc\n0.81", - "type": "text" - }, - { - "block_id": "p295-b34", - "global_id": 7927, - "bbox": [ - 347.97, - 594.17, - 385.45, - 618.19 - ], - "text": "1\n2 cos θ +", - "type": "text" - }, - { - "block_id": "p295-b35", - "global_id": 7928, - "bbox": [ - 388.19, - 585.38, - 396.62, - 595.34 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p295-b36", - "global_id": 7929, - "bbox": [ - 392.41, - 594.17, - 422.13, - 618.19 - ], - "text": "3\n2 sin θ", - "type": "text" - } - ] - }, - { - "page_num": 296, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p296-b0", - "global_id": 7930, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "276\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p296-b1", - "global_id": 7931, - "bbox": [ - 103.16, - 84.02, - 111.46, - 93.99 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p296-b2", - "global_id": 7932, - "bbox": [ - 239.92, - 98.59, - 248.35, - 108.56 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p296-b3", - "global_id": 7933, - "bbox": [ - 240.4, - 107.38, - 350.46, - 131.4 - ], - "text": "3\n1.8 ccos θ + 1\n1.8 csin θ = 2", - "type": "text" - }, - { - "block_id": "p296-b4", - "global_id": 7934, - "bbox": [ - 230.92, - 134.84, - 286.38, - 158.86 - ], - "text": "1\n1.62 ccos θ +", - "type": "text" - }, - { - "block_id": "p296-b5", - "global_id": 7935, - "bbox": [ - 291.14, - 126.04, - 299.57, - 136.01 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p296-b6", - "global_id": 7936, - "bbox": [ - 289.12, - 134.84, - 350.46, - 158.86 - ], - "text": "3\n1.62 csin θ = 1", - "type": "text" - }, - { - "block_id": "p296-b7", - "global_id": 7937, - "bbox": [ - 103.17, - 169.32, - 477.02, - 191.64 - ], - "text": "These are two simultaneous equations in two unknowns ccos θ and csin θ. Solution of these\nequations yields", - "type": "text" - }, - { - "block_id": "p296-b8", - "global_id": 7938, - "bbox": [ - 255.86, - 206.17, - 316.54, - 216.55 - ], - "text": "ccos θ = 2.308", - "type": "text" - }, - { - "block_id": "p296-b9", - "global_id": 7939, - "bbox": [ - 257.51, - 221.12, - 324.31, - 231.5 - ], - "text": "csin θ = −0.397", - "type": "text" - }, - { - "block_id": "p296-b10", - "global_id": 7940, - "bbox": [ - 103.17, - 246.02, - 233.21, - 256.4 - ], - "text": "Dividing csin θ by ccos θ yields", - "type": "text" - }, - { - "block_id": "p296-b11", - "global_id": 7941, - "bbox": [ - 218.66, - 268.91, - 281.67, - 286.27 - ], - "text": "tan θ = −0.397", - "type": "text" - }, - { - "block_id": "p296-b12", - "global_id": 7942, - "bbox": [ - 233.05, - 268.91, - 361.52, - 306.3 - ], - "text": "2.308 = −0.172\n1\nθ = tan−1(−0.172) = −0.17 rad", - "type": "text" - }, - { - "block_id": "p296-b13", - "global_id": 7943, - "bbox": [ - 103.16, - 320.82, - 376.46, - 331.2 - ], - "text": "Substituting θ = −0.17 radian in ccos θ = 2.308 yields c = 2.34 and", - "type": "text" - }, - { - "block_id": "p296-b14", - "global_id": 7944, - "bbox": [ - 201.37, - 347.68, - 288.95, - 362.53 - ], - "text": "y0[n] = 2.34(0.9)n cos", - "type": "text" - }, - { - "block_id": "p296-b15", - "global_id": 7945, - "bbox": [ - 291.17, - 337.39, - 305.07, - 354.36 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p296-b16", - "global_id": 7946, - "bbox": [ - 300.09, - 337.39, - 347.26, - 368.83 - ], - "text": "6 n −0.17", - "type": "text" - }, - { - "block_id": "p296-b17", - "global_id": 7947, - "bbox": [ - 357.23, - 351.38, - 378.8, - 361.75 - ], - "text": "n ≥0", - "type": "text" - }, - { - "block_id": "p296-b18", - "global_id": 7948, - "bbox": [ - 103.17, - 381.57, - 477.02, - 415.86 - ], - "text": "Observe that here we have used radian units for both β and θ. We also could have used the\ndegree unit, although this practice is not recommended. The important consideration is to be\nconsistent and to use the same units for both β and θ.", - "type": "text" - }, - { - "block_id": "p296-b19", - "global_id": 7949, - "bbox": [ - 107.82, - 480.84, - 477.69, - 506.74 - ], - "text": "DRILL 3.13\nZero-Input Response of a Second-Order System with\nComplex Roots", - "type": "text" - }, - { - "block_id": "p296-b20", - "global_id": 7950, - "bbox": [ - 107.82, - 515.86, - 373.97, - 525.83 - ], - "text": "Find the zero-input response of a system described by the equation", - "type": "text" - }, - { - "block_id": "p296-b21", - "global_id": 7951, - "bbox": [ - 247.76, - 537.36, - 344.46, - 547.74 - ], - "text": "y[n] + 4y[n −2] = 2x[n]", - "type": "text" - }, - { - "block_id": "p296-b22", - "global_id": 7952, - "bbox": [ - 107.82, - 559.29, - 278.18, - 570.44 - ], - "text": "The initial conditions are y0[−1] = −1/(2", - "type": "text" - }, - { - "block_id": "p296-b23", - "global_id": 7953, - "bbox": [ - 278.18, - 550.86, - 286.6, - 560.82 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p296-b24", - "global_id": 7954, - "bbox": [ - 286.62, - 550.86, - 382.52, - 570.44 - ], - "text": "2) and y0[−2] = 1/(4\n√", - "type": "text" - }, - { - "block_id": "p296-b25", - "global_id": 7955, - "bbox": [ - 107.82, - 559.29, - 484.41, - 581.62 - ], - "text": "2). Verify the solution by\ncomputing the first three terms iteratively.", - "type": "text" - }, - { - "block_id": "p296-b26", - "global_id": 7956, - "bbox": [ - 108.09, - 595.14, - 162.68, - 606.1 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p296-b27", - "global_id": 7957, - "bbox": [ - 107.82, - 609.88, - 170.49, - 624.23 - ], - "text": "y0[n] = (2)n cos", - "type": "text" - }, - { - "block_id": "p296-b28", - "global_id": 7958, - "bbox": [ - 171.6, - 605.07, - 181.0, - 618.42 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p296-b29", - "global_id": 7959, - "bbox": [ - 177.51, - 605.07, - 213.52, - 626.9 - ], - "text": "2 n −3π\n4", - "type": "text" - } - ] - }, - { - "page_num": 297, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p297-b0", - "global_id": 7960, - "bbox": [ - 326.35, - 62.29, - 516.13, - 71.98 - ], - "text": "3.7\nThe Unit Impulse Response h[n]\n277", - "type": "text" - }, - { - "block_id": "p297-b1", - "global_id": 7961, - "bbox": [ - 127.94, - 93.36, - 379.58, - 108.31 - ], - "text": "3.7 THE UNIT IMPULSE RESPONSE h[n]", - "type": "text" - }, - { - "block_id": "p297-b2", - "global_id": 7962, - "bbox": [ - 127.59, - 114.2, - 347.34, - 124.26 - ], - "text": "Consider an nth-order system specified by the equation", - "type": "text" - }, - { - "block_id": "p297-b3", - "global_id": 7963, - "bbox": [ - 153.41, - 131.93, - 490.31, - 147.19 - ], - "text": "(EN + a1EN−1 + · · · + aN−1E + aN)y[n] = (b0EN + b1EN−1 + · · · + bN−1E + bN)x[n]", - "type": "text" - }, - { - "block_id": "p297-b4", - "global_id": 7964, - "bbox": [ - 127.6, - 158.63, - 135.89, - 168.59 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p297-b5", - "global_id": 7965, - "bbox": [ - 280.28, - 170.53, - 363.43, - 180.81 - ], - "text": "Q[E]y[n] = P[E]x[n]", - "type": "text" - }, - { - "block_id": "p297-b6", - "global_id": 7966, - "bbox": [ - 127.6, - 189.7, - 516.14, - 212.03 - ], - "text": "The unit impulse response h[n] is the solution of this equation for the input δ[n] with all the initial\nconditions zero; that is,", - "type": "text" - }, - { - "block_id": "p297-b7", - "global_id": 7967, - "bbox": [ - 279.9, - 213.98, - 516.13, - 224.35 - ], - "text": "Q[E]h[n] = P[E]δ[n]\n(3.26)", - "type": "text" - }, - { - "block_id": "p297-b8", - "global_id": 7968, - "bbox": [ - 127.6, - 233.57, - 235.8, - 243.53 - ], - "text": "subject to initial conditions", - "type": "text" - }, - { - "block_id": "p297-b9", - "global_id": 7969, - "bbox": [ - 251.42, - 255.31, - 392.32, - 265.69 - ], - "text": "h[−1] = h[−2] = · · · = h[−N] = 0", - "type": "text" - }, - { - "block_id": "p297-b10", - "global_id": 7970, - "bbox": [ - 127.6, - 277.47, - 516.15, - 299.81 - ], - "text": "Equation (3.26) can be solved to determine h[n] iteratively or in a closed form. The following\nexample demonstrates the iterative solution.", - "type": "text" - }, - { - "block_id": "p297-b11", - "global_id": 7971, - "bbox": [ - 102.51, - 328.92, - 473.98, - 340.87 - ], - "text": "EXAMPLE 3.17\nIterative Determination of the Impulse Response", - "type": "text" - }, - { - "block_id": "p297-b12", - "global_id": 7972, - "bbox": [ - 128.9, - 357.13, - 502.76, - 379.45 - ], - "text": "Iteratively compute the first two values of the impulse response h[n] of a system described by\nthe equation", - "type": "text" - }, - { - "block_id": "p297-b13", - "global_id": 7973, - "bbox": [ - 233.65, - 381.04, - 398.01, - 391.42 - ], - "text": "y[n] −0.6y[n −1] −0.16y[n −2] = 5x[n]", - "type": "text" - }, - { - "block_id": "p297-b14", - "global_id": 7974, - "bbox": [ - 128.9, - 420.89, - 502.75, - 443.22 - ], - "text": "To determine the unit impulse response, we let the input x[n] = δ[n] and the output y[n] = h[n]\nin the system’s difference equation to obtain", - "type": "text" - }, - { - "block_id": "p297-b15", - "global_id": 7975, - "bbox": [ - 232.7, - 454.76, - 398.97, - 465.13 - ], - "text": "h[n] −0.6h[n −1] −0.16h[n −2] = 5δ[n]", - "type": "text" - }, - { - "block_id": "p297-b16", - "global_id": 7976, - "bbox": [ - 128.91, - 476.68, - 347.77, - 487.06 - ], - "text": "subject to zero initial state; that is, h[−1] = h[−2] = 0.", - "type": "text" - }, - { - "block_id": "p297-b17", - "global_id": 7977, - "bbox": [ - 146.85, - 488.63, - 289.61, - 499.01 - ], - "text": "Setting n = 0 in this equation yields", - "type": "text" - }, - { - "block_id": "p297-b18", - "global_id": 7978, - "bbox": [ - 219.75, - 510.55, - 411.93, - 520.92 - ], - "text": "h[0] −0.6(0) −0.16(0) = 5(1)\n\r⇒\nh[0] = 5", - "type": "text" - }, - { - "block_id": "p297-b19", - "global_id": 7979, - "bbox": [ - 146.85, - 532.47, - 404.96, - 542.85 - ], - "text": "Setting n = 1 in the same equation and using h[0] = 5, we obtain", - "type": "text" - }, - { - "block_id": "p297-b20", - "global_id": 7980, - "bbox": [ - 219.75, - 554.39, - 411.92, - 564.76 - ], - "text": "h[1] −0.6(5) −0.16(0) = 5(0)\n\r⇒\nh[1] = 3", - "type": "text" - }, - { - "block_id": "p297-b21", - "global_id": 7981, - "bbox": [ - 127.59, - 600.5, - 516.14, - 634.79 - ], - "text": "Continuing this way, we can determine any number of terms of h[n]. Unfortunately, such a\nsolution does not yield a closed-form expression for h[n]. Nevertheless, determining a few values\nof h[n] can be useful in determining the closed-form solution, as the following development shows.", - "type": "text" - } - ] - }, - { - "page_num": 298, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p298-b0", - "global_id": 7982, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "278\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p298-b1", - "global_id": 7983, - "bbox": [ - 101.84, - 85.66, - 315.91, - 98.48 - ], - "text": "3.7-1 The Closed-Form Solution of h[n]", - "type": "text" - }, - { - "block_id": "p298-b2", - "global_id": 7984, - "bbox": [ - 101.84, - 104.19, - 490.38, - 150.44 - ], - "text": "Recall that h[n] is the system response to input δ[n], which is zero for n > 0. We know that when\nthe input is zero, only the characteristic modes can be sustained by the system. Therefore, h[n]\nmust be made up of characteristic modes for n > 0. At n = 0, it may have some nonzero value A0\nso that a general form of h[n] can be expressed as†", - "type": "text" - }, - { - "block_id": "p298-b3", - "global_id": 7985, - "bbox": [ - 245.15, - 162.5, - 490.38, - 173.64 - ], - "text": "h[n] = A0δ[n] + yc[n]u[n]\n(3.27)", - "type": "text" - }, - { - "block_id": "p298-b4", - "global_id": 7986, - "bbox": [ - 101.84, - 184.93, - 490.4, - 219.99 - ], - "text": "where yc[n] is a linear combination of the characteristic modes. We now substitute Eq. (3.27)\nin Eq. (3.26) to obtain Q[E](A0δ[n] + yc[n]u[n]) = P[E]δ[n]. Because yc[n] is made up of\ncharacteristic modes, Q[E]yc[n]u[n] = 0, and we obtain A0Q[E]δ[n] = P[E]δ[n], that is,", - "type": "text" - }, - { - "block_id": "p298-b5", - "global_id": 7987, - "bbox": [ - 145.28, - 231.27, - 446.96, - 242.73 - ], - "text": "A0 (δ[n + N] + a1δ[n + N −1] + · · · + aNδ[n]) = b0δ[n + N] + · · · + bNδ[n]", - "type": "text" - }, - { - "block_id": "p298-b6", - "global_id": 7988, - "bbox": [ - 101.85, - 253.71, - 490.39, - 264.09 - ], - "text": "Setting n = 0 in this equation and using the fact that δ[m] = 0 for all m̸ = 0, and δ[0] = 1, we obtain", - "type": "text" - }, - { - "block_id": "p298-b7", - "global_id": 7989, - "bbox": [ - 236.9, - 274.29, - 353.22, - 292.42 - ], - "text": "A0aN = bN\n\r⇒\nA0 = bN", - "type": "text" - }, - { - "block_id": "p298-b8", - "global_id": 7990, - "bbox": [ - 343.59, - 288.35, - 353.22, - 299.12 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p298-b9", - "global_id": 7991, - "bbox": [ - 466.32, - 281.38, - 490.38, - 291.34 - ], - "text": "(3.28)", - "type": "text" - }, - { - "block_id": "p298-b10", - "global_id": 7992, - "bbox": [ - 101.85, - 308.54, - 132.75, - 319.1 - ], - "text": "Hence,‡", - "type": "text" - }, - { - "block_id": "p298-b11", - "global_id": 7993, - "bbox": [ - 243.16, - 318.88, - 282.45, - 335.82 - ], - "text": "h[n] = bN", - "type": "text" - }, - { - "block_id": "p298-b12", - "global_id": 7994, - "bbox": [ - 272.82, - 332.94, - 282.45, - 343.7 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p298-b13", - "global_id": 7995, - "bbox": [ - 285.67, - 325.55, - 490.38, - 336.63 - ], - "text": "δ[n] + yc[n]u[n]\n(3.29)", - "type": "text" - }, - { - "block_id": "p298-b14", - "global_id": 7996, - "bbox": [ - 101.84, - 349.47, - 490.39, - 407.67 - ], - "text": "The N unknown coefficients in yc[n] (on the right-hand side) can be determined from a knowledge\nof N values of h[n]. Fortunately, it is a straightforward task to determine values of h[n] iteratively,\nas demonstrated in Ex. 3.17. We compute N values h[0], h[1], h[2], . . . , h[N −1] iteratively. Now,\nsetting n = 0, 1, 2, . . . , N −1 in Eq. (3.29), we can determine the N unknowns in yc[n]. This point\nwill become clear in the following example.", - "type": "text" - }, - { - "block_id": "p298-b15", - "global_id": 7997, - "bbox": [ - 76.77, - 436.92, - 472.84, - 448.88 - ], - "text": "EXAMPLE 3.18\nClosed-Form Determination of the Impulse Response", - "type": "text" - }, - { - "block_id": "p298-b16", - "global_id": 7998, - "bbox": [ - 103.16, - 465.13, - 467.92, - 475.51 - ], - "text": "Determine the unit impulse response h[n] for a system in Ex. 3.17 specified by the equation", - "type": "text" - }, - { - "block_id": "p298-b17", - "global_id": 7999, - "bbox": [ - 207.91, - 487.05, - 372.27, - 497.42 - ], - "text": "y[n] −0.6y[n −1] −0.16y[n −2] = 5x[n]", - "type": "text" - }, - { - "block_id": "p298-b18", - "global_id": 8000, - "bbox": [ - 101.84, - 555.14, - 490.37, - 578.32 - ], - "text": "† We assume that the term yc[n] consists of characteristic modes for n > 0 only. To reflect this behavior,\nthe characteristic terms should be expressed in the form γ n", - "type": "text" - }, - { - "block_id": "p298-b19", - "global_id": 8001, - "bbox": [ - 101.84, - 568.98, - 490.39, - 589.95 - ], - "text": "j u[n −1]. But because u[n −1] = u[n] −δ[n],\ncjγ n", - "type": "text" - }, - { - "block_id": "p298-b20", - "global_id": 8002, - "bbox": [ - 112.63, - 578.76, - 173.42, - 591.85 - ], - "text": "j u[n −1] = cjγ n", - "type": "text" - }, - { - "block_id": "p298-b21", - "global_id": 8003, - "bbox": [ - 168.75, - 578.76, - 421.08, - 591.85 - ], - "text": "j u[n] −cjδ[n], and yc[n] can be expressed in terms of exponentials γ n", - "type": "text" - }, - { - "block_id": "p298-b22", - "global_id": 8004, - "bbox": [ - 101.84, - 579.94, - 490.39, - 633.41 - ], - "text": "j u[n] (which start at\nn = 0), plus an impulse at n = 0.\n‡ If aN = 0, then A0 cannot be determined by Eq. (3.28). In such a case, we show in Sec. 3.12 that h[n] is\nof the form A0δ[n] + A1δ[n −1] + yc[n]u[n]. We have here N + 2 unknowns, which can be determined from\nN + 2 values h[0],h[1],. . .,h[N + 1] found iteratively.", - "type": "text" - } - ] - }, - { - "page_num": 299, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p299-b0", - "global_id": 8005, - "bbox": [ - 326.35, - 62.29, - 516.13, - 71.98 - ], - "text": "3.7\nThe Unit Impulse Response h[n]\n279", - "type": "text" - }, - { - "block_id": "p299-b1", - "global_id": 8006, - "bbox": [ - 128.9, - 86.24, - 345.7, - 96.2 - ], - "text": "This equation can be expressed in the advance form as", - "type": "text" - }, - { - "block_id": "p299-b2", - "global_id": 8007, - "bbox": [ - 225.73, - 103.76, - 405.92, - 114.14 - ], - "text": "y[n + 2] −0.6y[n + 1] −0.16y[n] = 5x[n + 2]", - "type": "text" - }, - { - "block_id": "p299-b3", - "global_id": 8008, - "bbox": [ - 128.91, - 122.1, - 250.66, - 132.06 - ], - "text": "or in advance operator form as", - "type": "text" - }, - { - "block_id": "p299-b4", - "global_id": 8009, - "bbox": [ - 248.36, - 135.51, - 383.3, - 150.0 - ], - "text": "(E2 −0.6E −0.16)y[n] = 5E2x[n]", - "type": "text" - }, - { - "block_id": "p299-b5", - "global_id": 8010, - "bbox": [ - 128.91, - 157.97, - 257.03, - 167.93 - ], - "text": "The characteristic polynomial is", - "type": "text" - }, - { - "block_id": "p299-b6", - "global_id": 8011, - "bbox": [ - 238.15, - 171.38, - 393.51, - 185.87 - ], - "text": "γ 2 −0.6γ −0.16 = (γ + 0.2)(γ −0.8)", - "type": "text" - }, - { - "block_id": "p299-b7", - "global_id": 8012, - "bbox": [ - 128.91, - 190.22, - 367.26, - 203.8 - ], - "text": "The characteristic modes are (−0.2)n and (0.8)n. Therefore,", - "type": "text" - }, - { - "block_id": "p299-b8", - "global_id": 8013, - "bbox": [ - 258.49, - 207.24, - 372.68, - 222.5 - ], - "text": "yc[n] = c1(−0.2)n + c2(0.8)n", - "type": "text" - }, - { - "block_id": "p299-b9", - "global_id": 8014, - "bbox": [ - 128.9, - 229.29, - 502.74, - 251.62 - ], - "text": "Inspecting the system difference equation, we see that aN = −0.16 and bN = 0. Therefore,\naccording to Eq. (3.29),", - "type": "text" - }, - { - "block_id": "p299-b10", - "global_id": 8015, - "bbox": [ - 248.39, - 249.08, - 383.28, - 264.34 - ], - "text": "h[n] = [c1(−0.2)n + c2(0.8)n]u[n]", - "type": "text" - }, - { - "block_id": "p299-b11", - "global_id": 8016, - "bbox": [ - 128.91, - 272.13, - 502.76, - 306.41 - ], - "text": "To determine c1 and c2, we need to find two values of h[n] iteratively. From Ex. 3.17, we know\nthat h[0] = 5 and h[1] = 3. Setting n = 0 and 1 in our expression for h[n] and using the fact\nthat h[0] = 5 and h[1] = 3, we obtain", - "type": "text" - }, - { - "block_id": "p299-b12", - "global_id": 8017, - "bbox": [ - 238.37, - 312.94, - 315.06, - 336.35 - ], - "text": "5 = c1 + c2\n3 = −0.2c1 + 0.8c2", - "type": "text" - }, - { - "block_id": "p299-b14", - "global_id": 8018, - "bbox": [ - 337.79, - 312.94, - 394.49, - 336.35 - ], - "text": "⇒\nc1 = 1\nc2 = 4", - "type": "text" - }, - { - "block_id": "p299-b15", - "global_id": 8019, - "bbox": [ - 128.91, - 342.65, - 170.65, - 352.61 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p299-b16", - "global_id": 8020, - "bbox": [ - 253.14, - 354.19, - 279.56, - 364.47 - ], - "text": "h[n] =", - "type": "text" - }, - { - "block_id": "p299-b18", - "global_id": 8021, - "bbox": [ - 285.53, - 346.18, - 360.82, - 364.57 - ], - "text": "(−0.2)n + 4(0.8)n", - "type": "text" - }, - { - "block_id": "p299-b19", - "global_id": 8022, - "bbox": [ - 361.92, - 354.19, - 378.52, - 364.47 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p299-b20", - "global_id": 8023, - "bbox": [ - 133.57, - 432.82, - 506.46, - 444.77 - ], - "text": "DRILL 3.14\nClosed-Form Determination of the Impulse Response", - "type": "text" - }, - { - "block_id": "p299-b21", - "global_id": 8024, - "bbox": [ - 133.57, - 453.47, - 510.15, - 475.81 - ], - "text": "Find h[n], the unit impulse response of the LTID systems specified by the following\nequations:", - "type": "text" - }, - { - "block_id": "p299-b22", - "global_id": 8025, - "bbox": [ - 151.5, - 483.36, - 254.83, - 493.74 - ], - "text": "(a) y[n + 1] −y[n] = x[n]", - "type": "text" - }, - { - "block_id": "p299-b23", - "global_id": 8026, - "bbox": [ - 151.5, - 498.31, - 381.66, - 508.68 - ], - "text": "(b) y[n] −5y[n −1] + 6y[n −2] = 8x[n −1] −19x[n −2]", - "type": "text" - }, - { - "block_id": "p299-b24", - "global_id": 8027, - "bbox": [ - 152.06, - 513.25, - 376.13, - 523.62 - ], - "text": "(c) y[n + 2] −4y[n + 1] + 4y[n] = 2x[n + 2] −2x[n + 1]", - "type": "text" - }, - { - "block_id": "p299-b25", - "global_id": 8028, - "bbox": [ - 151.51, - 528.2, - 265.35, - 538.57 - ], - "text": "(d) y[n] = 2x[n] −2x[n −1]", - "type": "text" - }, - { - "block_id": "p299-b26", - "global_id": 8029, - "bbox": [ - 133.84, - 555.08, - 196.06, - 566.04 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p299-b27", - "global_id": 8030, - "bbox": [ - 151.5, - 573.03, - 229.0, - 598.26 - ], - "text": "(a) h[n] = u[n −1]\n(b) h[n] = −19", - "type": "text" - }, - { - "block_id": "p299-b28", - "global_id": 8031, - "bbox": [ - 207.83, - 579.96, - 250.03, - 601.47 - ], - "text": "6 δ[n] +\n\t 3", - "type": "text" - }, - { - "block_id": "p299-b29", - "global_id": 8032, - "bbox": [ - 246.55, - 579.96, - 321.29, - 601.16 - ], - "text": "2(2)n + 5\n3(3)n\nu[n]", - "type": "text" - }, - { - "block_id": "p299-b30", - "global_id": 8033, - "bbox": [ - 151.5, - 601.91, - 266.37, - 628.23 - ], - "text": "(c) h[n] = (2 + n)2nu[n]\n(d) h[n] = 2δ[n] −2δ[n −1]", - "type": "text" - } - ] - }, - { - "page_num": 300, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p300-b0", - "global_id": 8034, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "280\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p300-b1", - "global_id": 8035, - "bbox": [ - 76.77, - 93.91, - 459.17, - 105.87 - ], - "text": "EXAMPLE 3.19\nFiltering Perspective of the Unit Impulse Response", - "type": "text" - }, - { - "block_id": "p300-b2", - "global_id": 8036, - "bbox": [ - 103.16, - 122.54, - 324.49, - 132.79 - ], - "text": "Use the MATLAB filter command to solve Ex. 3.18.", - "type": "text" - }, - { - "block_id": "p300-b3", - "global_id": 8037, - "bbox": [ - 103.16, - 155.41, - 477.04, - 201.24 - ], - "text": "There are several ways to find the impulse response using MATLAB. In this method, we first\nspecify the unit impulse function, which will serve as our input. Vectors a and b are created to\nspecify the system. The filter command is then used to determine the impulse response. In\nfact, this method can be used to determine the zero-state response for any input.", - "type": "text" - }, - { - "block_id": "p300-b4", - "global_id": 8038, - "bbox": [ - 103.16, - 211.49, - 375.14, - 257.32 - ], - "text": ">>\nn = (0:19); delta = @(n) 1.0.*(n==0);\n>>\na = [1 -0.6 -0.16]; b = [5 0 0];\n>>\nh = filter(b,a,delta(n));\n>>\nclf; stem(n,h,’k’); xlabel(’n’); ylabel(’h[n]’);", - "type": "text" - }, - { - "block_id": "p300-b5", - "global_id": 8039, - "bbox": [ - 120.74, - 366.71, - 349.99, - 387.9 - ], - "text": "0\n5\n10\n15\n20\nn", - "type": "text" - }, - { - "block_id": "p300-b6", - "global_id": 8040, - "bbox": [ - 114.51, - 357.49, - 118.51, - 365.49 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p300-b7", - "global_id": 8041, - "bbox": [ - 114.51, - 333.48, - 118.51, - 341.48 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p300-b8", - "global_id": 8042, - "bbox": [ - 114.51, - 309.49, - 118.51, - 317.49 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p300-b9", - "global_id": 8043, - "bbox": [ - 114.51, - 285.48, - 118.51, - 293.48 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p300-b10", - "global_id": 8044, - "bbox": [ - 102.53, - 317.43, - 111.33, - 332.09 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p300-b11", - "global_id": 8045, - "bbox": [ - 101.77, - 394.58, - 259.41, - 403.82 - ], - "text": "Figure 3.19 Impulse response for Ex. 3.19", - "type": "text" - }, - { - "block_id": "p300-b12", - "global_id": 8046, - "bbox": [ - 101.84, - 452.16, - 490.39, - 474.5 - ], - "text": "Comment. Although it is relatively simple to determine the impulse response h[n] by using the\nprocedure in this section, in Ch. 5 we shall discuss the much simpler method of the z-transform.", - "type": "text" - }, - { - "block_id": "p300-b13", - "global_id": 8047, - "bbox": [ - 102.19, - 517.21, - 392.16, - 547.1 - ], - "text": "3.8 SYSTEM RESPONSE TO EXTERNAL INPUT:\nTHE ZERO-STATE RESPONSE", - "type": "text" - }, - { - "block_id": "p300-b14", - "global_id": 8048, - "bbox": [ - 101.84, - 552.68, - 490.4, - 634.79 - ], - "text": "The zero-state response y[n] is the system response to an input x[n] when the system is in the\nzero state. In this section we shall assume that systems are in the zero state unless mentioned\notherwise, so that the zero-state response will be the total response of the system. Here we follow\nthe procedure parallel to that used in the continuous-time case by expressing an arbitrary input\nx[n] as a sum of impulse components. A signal x[n] in Fig. 3.20a can be expressed as a sum\nof impulse components, such as those depicted in Figs. 3.20b–3.20f. The component of x[n] at\nn = m is x[m]δ[n −m], and x[n] is the sum of all these components summed from m = −∞to ∞.", - "type": "text" - } - ] - }, - { - "page_num": 301, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p301-b0", - "global_id": 8049, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n281", - "type": "text" - }, - { - "block_id": "p301-b1", - "global_id": 8050, - "bbox": [ - 193.45, - 95.28, - 206.33, - 103.36 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p301-b2", - "global_id": 8051, - "bbox": [ - 129.58, - 145.91, - 264.21, - 159.55 - ], - "text": "(a)\nn", - "type": "text" - }, - { - "block_id": "p301-b3", - "global_id": 8052, - "bbox": [ - 260.21, - 230.18, - 264.21, - 238.18 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p301-b4", - "global_id": 8053, - "bbox": [ - 233.06, - 151.96, - 252.16, - 159.96 - ], - "text": "3\n4", - "type": "text" - }, - { - "block_id": "p301-b5", - "global_id": 8054, - "bbox": [ - 129.58, - 492.68, - 221.97, - 506.73 - ], - "text": "(f)\n2", - "type": "text" - }, - { - "block_id": "p301-b6", - "global_id": 8055, - "bbox": [ - 129.58, - 225.2, - 161.57, - 239.51 - ], - "text": "(b)\n2", - "type": "text" - }, - { - "block_id": "p301-b7", - "global_id": 8056, - "bbox": [ - 129.58, - 294.93, - 176.67, - 308.78 - ], - "text": "(c)\n1", - "type": "text" - }, - { - "block_id": "p301-b8", - "global_id": 8057, - "bbox": [ - 129.58, - 360.79, - 138.91, - 368.79 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p301-b9", - "global_id": 8058, - "bbox": [ - 129.58, - 426.71, - 206.57, - 440.77 - ], - "text": "(e)\n1", - "type": "text" - }, - { - "block_id": "p301-b10", - "global_id": 8059, - "bbox": [ - 192.25, - 456.42, - 233.77, - 464.72 - ], - "text": "x[2]d[n 2]", - "type": "text" - }, - { - "block_id": "p301-b11", - "global_id": 8060, - "bbox": [ - 192.25, - 389.46, - 233.77, - 397.76 - ], - "text": "x[1]d[n 1]", - "type": "text" - }, - { - "block_id": "p301-b12", - "global_id": 8061, - "bbox": [ - 194.58, - 324.73, - 221.26, - 332.83 - ], - "text": "x[0]d[n]", - "type": "text" - }, - { - "block_id": "p301-b13", - "global_id": 8062, - "bbox": [ - 192.91, - 259.68, - 241.08, - 267.97 - ], - "text": "x[1]d[n 1]", - "type": "text" - }, - { - "block_id": "p301-b14", - "global_id": 8063, - "bbox": [ - 192.91, - 189.95, - 240.93, - 198.25 - ], - "text": "x[2]d[n 2]", - "type": "text" - }, - { - "block_id": "p301-b15", - "global_id": 8064, - "bbox": [ - 150.91, - 150.92, - 221.97, - 159.96 - ], - "text": "2\n1\n2 1", - "type": "text" - }, - { - "block_id": "p301-b16", - "global_id": 8065, - "bbox": [ - 260.21, - 300.57, - 264.21, - 308.57 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p301-b17", - "global_id": 8066, - "bbox": [ - 260.21, - 366.39, - 264.21, - 374.39 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p301-b18", - "global_id": 8067, - "bbox": [ - 260.21, - 432.42, - 264.21, - 440.42 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p301-b19", - "global_id": 8068, - "bbox": [ - 260.21, - 498.3, - 264.21, - 506.3 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p301-b20", - "global_id": 8069, - "bbox": [ - 291.55, - 485.66, - 512.14, - 508.0 - ], - "text": "Figure 3.20 Representation of an arbitrary signal x[n]\nin terms of impulse components.", - "type": "text" - }, - { - "block_id": "p301-b21", - "global_id": 8070, - "bbox": [ - 127.59, - 541.78, - 169.34, - 551.75 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p301-b22", - "global_id": 8071, - "bbox": [ - 220.65, - 573.98, - 423.08, - 584.35 - ], - "text": "x[n] = x[0]δ[n] + x[1]δ[n −1] + x[2]δ[n −2] + · · ·", - "type": "text" - }, - { - "block_id": "p301-b23", - "global_id": 8072, - "bbox": [ - 250.14, - 588.92, - 406.26, - 599.3 - ], - "text": "+ x[−1]δ[n + 1] + x[−2]δ[n + 2] + · · ·", - "type": "text" - }, - { - "block_id": "p301-b24", - "global_id": 8073, - "bbox": [ - 238.78, - 613.42, - 246.55, - 623.38 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p301-b25", - "global_id": 8074, - "bbox": [ - 254.16, - 603.25, - 268.26, - 613.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p301-b26", - "global_id": 8075, - "bbox": [ - 249.69, - 627.27, - 272.71, - 634.46 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p301-b27", - "global_id": 8076, - "bbox": [ - 273.83, - 613.42, - 516.13, - 623.8 - ], - "text": "x[m]δ[n −m]\n(3.30)", - "type": "text" - } - ] - }, - { - "page_num": 302, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p302-b0", - "global_id": 8077, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "282\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p302-b1", - "global_id": 8078, - "bbox": [ - 101.84, - 85.4, - 490.38, - 119.69 - ], - "text": "For a linear system, if we know the system response to impulse δ[n], we can obtain the system\nresponse to any arbitrary input by summing the system response to various impulse components.\nLet h[n] be the system response to impulse input δ[n]. We shall use the notation", - "type": "text" - }, - { - "block_id": "p302-b2", - "global_id": 8079, - "bbox": [ - 269.77, - 131.72, - 322.47, - 141.99 - ], - "text": "x[n] \r⇒y[n]", - "type": "text" - }, - { - "block_id": "p302-b3", - "global_id": 8080, - "bbox": [ - 101.85, - 154.54, - 401.23, - 164.5 - ], - "text": "to indicate the input and the corresponding response of the system. Thus, if", - "type": "text" - }, - { - "block_id": "p302-b4", - "global_id": 8081, - "bbox": [ - 269.37, - 176.53, - 322.87, - 186.8 - ], - "text": "δ[n] \r⇒h[n]", - "type": "text" - }, - { - "block_id": "p302-b5", - "global_id": 8082, - "bbox": [ - 101.85, - 199.35, - 227.34, - 209.31 - ], - "text": "then because of time invariance", - "type": "text" - }, - { - "block_id": "p302-b6", - "global_id": 8083, - "bbox": [ - 251.31, - 211.62, - 340.92, - 221.89 - ], - "text": "δ[n −m] \r⇒h[n −m]", - "type": "text" - }, - { - "block_id": "p302-b7", - "global_id": 8084, - "bbox": [ - 101.85, - 231.45, - 196.74, - 241.42 - ], - "text": "and because of linearity", - "type": "text" - }, - { - "block_id": "p302-b8", - "global_id": 8085, - "bbox": [ - 233.03, - 243.72, - 359.2, - 254.0 - ], - "text": "x[m]δ[n −m] \r⇒x[m]h[n −m]", - "type": "text" - }, - { - "block_id": "p302-b9", - "global_id": 8086, - "bbox": [ - 101.85, - 263.56, - 220.76, - 273.52 - ], - "text": "and again because of linearity", - "type": "text" - }, - { - "block_id": "p302-b10", - "global_id": 8087, - "bbox": [ - 202.28, - 283.5, - 216.38, - 294.17 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p302-b11", - "global_id": 8088, - "bbox": [ - 197.81, - 307.52, - 220.83, - 314.71 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p302-b12", - "global_id": 8089, - "bbox": [ - 221.96, - 293.67, - 274.59, - 303.95 - ], - "text": "x[m]δ[n −m]", - "type": "text" - }, - { - "block_id": "p302-b13", - "global_id": 8090, - "bbox": [ - 197.81, - 310.82, - 274.6, - 329.81 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p302-b14", - "global_id": 8091, - "bbox": [ - 287.72, - 293.67, - 304.22, - 303.63 - ], - "text": "⇒", - "type": "text" - }, - { - "block_id": "p302-b15", - "global_id": 8092, - "bbox": [ - 321.8, - 283.5, - 335.9, - 294.17 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p302-b16", - "global_id": 8093, - "bbox": [ - 317.34, - 307.52, - 340.36, - 314.71 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p302-b17", - "global_id": 8094, - "bbox": [ - 341.48, - 293.67, - 394.41, - 303.95 - ], - "text": "x[m]h[n −m]", - "type": "text" - }, - { - "block_id": "p302-b18", - "global_id": 8095, - "bbox": [ - 317.34, - 310.82, - 394.44, - 329.81 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p302-b19", - "global_id": 8096, - "bbox": [ - 101.84, - 340.99, - 490.39, - 363.32 - ], - "text": "The left-hand side is x[n] [see Eq. (3.30)], and the right-hand side is the system response y[n] to\ninput x[n]. Therefore,†", - "type": "text" - }, - { - "block_id": "p302-b20", - "global_id": 8097, - "bbox": [ - 243.62, - 373.76, - 269.48, - 384.04 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p302-b21", - "global_id": 8098, - "bbox": [ - 275.99, - 363.59, - 290.09, - 374.26 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p302-b22", - "global_id": 8099, - "bbox": [ - 271.53, - 387.61, - 294.55, - 394.8 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p302-b23", - "global_id": 8100, - "bbox": [ - 295.67, - 373.76, - 490.38, - 384.14 - ], - "text": "x[m]h[n −m]\n(3.31)", - "type": "text" - }, - { - "block_id": "p302-b24", - "global_id": 8101, - "bbox": [ - 101.85, - 402.15, - 490.4, - 424.48 - ], - "text": "The summation on the right-hand side is known as the convolution sum of x[n] and h[n], and is\nrepresented symbolically by x[n] ∗h[n]", - "type": "text" - }, - { - "block_id": "p302-b25", - "global_id": 8102, - "bbox": [ - 231.17, - 444.63, - 281.94, - 454.9 - ], - "text": "x[n] ∗h[n] =", - "type": "text" - }, - { - "block_id": "p302-b26", - "global_id": 8103, - "bbox": [ - 288.46, - 434.45, - 302.56, - 445.13 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p302-b27", - "global_id": 8104, - "bbox": [ - 283.99, - 458.48, - 307.01, - 465.67 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p302-b28", - "global_id": 8105, - "bbox": [ - 308.13, - 444.63, - 361.06, - 454.9 - ], - "text": "x[m]h[n −m]", - "type": "text" - }, - { - "block_id": "p302-b29", - "global_id": 8106, - "bbox": [ - 102.14, - 478.54, - 320.85, - 490.66 - ], - "text": "PROPERTIES OF THE CONVOLUTION SUM", - "type": "text" - }, - { - "block_id": "p302-b30", - "global_id": 8107, - "bbox": [ - 101.84, - 494.69, - 490.4, - 540.52 - ], - "text": "The structure of the convolution sum is similar to that of the convolution integral. Moreover,\nthe properties of the convolution sum are similar to those of the convolution integral. We shall\nenumerate these properties here without proof. The proofs are similar to those for the convolution\nintegral and may be derived by the reader.", - "type": "text" - }, - { - "block_id": "p302-b31", - "global_id": 8108, - "bbox": [ - 101.84, - 559.1, - 490.38, - 593.24 - ], - "text": "† In deriving this result, we have assumed a time-invariant system. The system response to input δ[n −m]\nfor a time-varying system cannot be expressed as h[n −m]; instead, it has the form h[n, m]. Using this form,\nEq. (3.31) is modified as follows:", - "type": "text" - }, - { - "block_id": "p302-b32", - "global_id": 8109, - "bbox": [ - 251.32, - 602.4, - 274.59, - 611.65 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p302-b33", - "global_id": 8110, - "bbox": [ - 280.78, - 593.09, - 293.47, - 602.86 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p302-b34", - "global_id": 8111, - "bbox": [ - 276.43, - 614.71, - 297.82, - 621.39 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p302-b35", - "global_id": 8112, - "bbox": [ - 298.82, - 602.4, - 340.92, - 611.74 - ], - "text": "x[m]h[n, m]", - "type": "text" - } - ] - }, - { - "page_num": 303, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p303-b0", - "global_id": 8113, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n283", - "type": "text" - }, - { - "block_id": "p303-b1", - "global_id": 8114, - "bbox": [ - 127.59, - 85.87, - 246.85, - 95.83 - ], - "text": "The Commutative Property.", - "type": "text" - }, - { - "block_id": "p303-b2", - "global_id": 8115, - "bbox": [ - 267.59, - 104.02, - 376.13, - 115.17 - ], - "text": "x1[n] ∗x2[n] = x2[n] ∗x1[n]", - "type": "text" - }, - { - "block_id": "p303-b3", - "global_id": 8116, - "bbox": [ - 127.59, - 128.64, - 239.47, - 138.6 - ], - "text": "The Distributive Property.", - "type": "text" - }, - { - "block_id": "p303-b4", - "global_id": 8117, - "bbox": [ - 218.82, - 154.58, - 424.91, - 165.72 - ], - "text": "x1[n] ∗(x2[n] + x3[n]) = x1[n] ∗x2[n] + x1[n] ∗x3[n]", - "type": "text" - }, - { - "block_id": "p303-b5", - "global_id": 8118, - "bbox": [ - 127.59, - 181.18, - 236.35, - 191.14 - ], - "text": "The Associative Property.", - "type": "text" - }, - { - "block_id": "p303-b6", - "global_id": 8119, - "bbox": [ - 231.84, - 207.13, - 411.88, - 218.28 - ], - "text": "x1[n] ∗(x2[n] ∗x3[n]) = (x1[n] ∗x2[n]) ∗x3[n]", - "type": "text" - }, - { - "block_id": "p303-b7", - "global_id": 8120, - "bbox": [ - 127.59, - 233.73, - 234.37, - 243.77 - ], - "text": "The Shifting Property. If", - "type": "text" - }, - { - "block_id": "p303-b8", - "global_id": 8121, - "bbox": [ - 283.74, - 251.88, - 359.98, - 263.03 - ], - "text": "x1[n] ∗x2[n] = c[n]", - "type": "text" - }, - { - "block_id": "p303-b9", - "global_id": 8122, - "bbox": [ - 127.6, - 275.58, - 144.74, - 285.54 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p303-b10", - "global_id": 8123, - "bbox": [ - 249.82, - 293.66, - 516.13, - 304.81 - ], - "text": "x1[n −m] ∗x2[n −p] = c[n −m −p]\n(3.32)", - "type": "text" - }, - { - "block_id": "p303-b11", - "global_id": 8124, - "bbox": [ - 127.6, - 318.27, - 272.4, - 328.23 - ], - "text": "The Convolution with an Impulse.", - "type": "text" - }, - { - "block_id": "p303-b12", - "global_id": 8125, - "bbox": [ - 287.56, - 344.22, - 356.16, - 354.49 - ], - "text": "x[n] ∗δ[n] = x[n]", - "type": "text" - }, - { - "block_id": "p303-b13", - "global_id": 8126, - "bbox": [ - 127.59, - 370.49, - 516.14, - 440.64 - ], - "text": "The Width Property. If x1[n] and x2[n] have finite widths of W1 and W2, respectively, then the\nwidth of x1[n] ∗x2[n] is W1 + W2. The width of a signal is 1 less than the number of its elements\n(length). Thus the signal in Fig. 3.22h has six elements (length of 6) but a width of only 5.\nAlternately, the property may be stated in terms of lengths as follows: if x1[n] and x2[n] have\nfinite lengths of L1 and L2 elements, respectively, then the length of x1[n] ∗x2[n] is L1 + L2 −1\nelements.", - "type": "text" - }, - { - "block_id": "p303-b14", - "global_id": 8127, - "bbox": [ - 127.59, - 457.55, - 516.15, - 555.39 - ], - "text": "CAUSALITY AND ZERO-STATE RESPONSE\nIn deriving Eq. (3.31), we assumed the system to be linear and time-invariant. There were no other\nrestrictions on either the input signal or the system. In our applications, almost all the input signals\nare causal, and a majority of the systems are also causal. These restrictions further simplify the\nlimits of the sum in Eq. (3.31). If the input x[n] is causal, x[m] = 0 for m < 0. Similarly, if the\nsystem is causal (i.e., if h[n] is causal), then h[x] = 0 for negative x so that h[n −m] = 0 when\nm > n. Therefore, if x[n] and h[n] are both causal, the product x[m]h[n−m] = 0 for m < 0 and for\nm > n, and it is nonzero only for the range 0 ≤m ≤n. Therefore, Eq. (3.31) in this case reduces to", - "type": "text" - }, - { - "block_id": "p303-b15", - "global_id": 8128, - "bbox": [ - 273.84, - 579.22, - 299.7, - 589.5 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p303-b16", - "global_id": 8129, - "bbox": [ - 301.74, - 569.27, - 315.84, - 579.72 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p303-b17", - "global_id": 8130, - "bbox": [ - 301.81, - 593.61, - 315.76, - 600.87 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p303-b18", - "global_id": 8131, - "bbox": [ - 316.94, - 579.22, - 516.13, - 589.6 - ], - "text": "x[m]h[n −m]\n(3.33)", - "type": "text" - }, - { - "block_id": "p303-b19", - "global_id": 8132, - "bbox": [ - 127.6, - 614.93, - 516.09, - 636.84 - ], - "text": "We shall evaluate the convolution sum first by an analytical method and later with graphical\naid.", - "type": "text" - } - ] - }, - { - "page_num": 304, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p304-b0", - "global_id": 8133, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "284\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p304-b1", - "global_id": 8134, - "bbox": [ - 76.77, - 93.91, - 347.96, - 105.87 - ], - "text": "EXAMPLE 3.20\nConvolution of Causal Signals", - "type": "text" - }, - { - "block_id": "p304-b2", - "global_id": 8135, - "bbox": [ - 103.16, - 122.12, - 230.68, - 132.5 - ], - "text": "Determine c[n] = x[n] ∗g[n] for", - "type": "text" - }, - { - "block_id": "p304-b3", - "global_id": 8136, - "bbox": [ - 194.31, - 142.54, - 385.87, - 154.41 - ], - "text": "x[n] = (0.8)nu[n]\nand\ng[n] = (0.3)nu[n]", - "type": "text" - }, - { - "block_id": "p304-b4", - "global_id": 8137, - "bbox": [ - 103.16, - 187.29, - 137.13, - 197.25 - ], - "text": "We have", - "type": "text" - }, - { - "block_id": "p304-b5", - "global_id": 8138, - "bbox": [ - 237.6, - 204.91, - 263.45, - 215.18 - ], - "text": "c[n] =", - "type": "text" - }, - { - "block_id": "p304-b6", - "global_id": 8139, - "bbox": [ - 269.97, - 194.73, - 284.07, - 205.4 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p304-b7", - "global_id": 8140, - "bbox": [ - 265.5, - 218.75, - 288.52, - 225.95 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p304-b8", - "global_id": 8141, - "bbox": [ - 289.64, - 204.91, - 342.57, - 215.18 - ], - "text": "x[m]g[n −m]", - "type": "text" - }, - { - "block_id": "p304-b9", - "global_id": 8142, - "bbox": [ - 103.17, - 233.21, - 139.96, - 243.17 - ], - "text": "Note that", - "type": "text" - }, - { - "block_id": "p304-b10", - "global_id": 8143, - "bbox": [ - 168.02, - 243.04, - 412.15, - 255.13 - ], - "text": "x[m] = (0.8)mu[m]\nand\ng[n −m] = (0.3)n−mu[n −m]", - "type": "text" - }, - { - "block_id": "p304-b11", - "global_id": 8144, - "bbox": [ - 103.17, - 263.68, - 330.31, - 274.06 - ], - "text": "Both x[n] and g[n] are causal. Therefore [see Eq. (3.33)],", - "type": "text" - }, - { - "block_id": "p304-b12", - "global_id": 8145, - "bbox": [ - 171.92, - 296.52, - 197.77, - 306.8 - ], - "text": "c[n] =", - "type": "text" - }, - { - "block_id": "p304-b13", - "global_id": 8146, - "bbox": [ - 199.82, - 286.57, - 213.91, - 297.02 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p304-b14", - "global_id": 8147, - "bbox": [ - 199.89, - 310.91, - 213.84, - 318.18 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p304-b15", - "global_id": 8148, - "bbox": [ - 215.03, - 296.53, - 277.79, - 306.8 - ], - "text": "x[m]g[n −m] =", - "type": "text" - }, - { - "block_id": "p304-b16", - "global_id": 8149, - "bbox": [ - 279.83, - 286.57, - 293.93, - 297.02 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p304-b17", - "global_id": 8150, - "bbox": [ - 279.9, - 310.91, - 293.85, - 318.18 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p304-b18", - "global_id": 8151, - "bbox": [ - 293.93, - 294.79, - 408.25, - 306.9 - ], - "text": "(0.8)mu[m](0.3)n−mu[n −m]", - "type": "text" - }, - { - "block_id": "p304-b19", - "global_id": 8152, - "bbox": [ - 103.17, - 330.45, - 477.02, - 364.73 - ], - "text": "In this summation, m lies between 0 and n (0 ≤m ≤n). Therefore, if n ≥0, then both m and\nn−m ≥0 so that u[m] = u[n−m] = 1. If n < 0, m is negative because m lies between 0 and n,\nand u[m] = 0. Therefore,", - "type": "text" - }, - { - "block_id": "p304-b20", - "global_id": 8153, - "bbox": [ - 208.72, - 384.3, - 234.57, - 394.58 - ], - "text": "c[n] =", - "type": "text" - }, - { - "block_id": "p304-b21", - "global_id": 8154, - "bbox": [ - 236.62, - 370.32, - 261.66, - 383.03 - ], - "text": "%n", - "type": "text" - }, - { - "block_id": "p304-b22", - "global_id": 8155, - "bbox": [ - 258.17, - 374.61, - 365.29, - 400.56 - ], - "text": "m=0(0.8)m (0.3)n−m\nn ≥0\n0\nn < 0", - "type": "text" - }, - { - "block_id": "p304-b23", - "global_id": 8156, - "bbox": [ - 103.16, - 412.7, - 111.46, - 422.66 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p304-b24", - "global_id": 8157, - "bbox": [ - 230.28, - 428.61, - 281.56, - 440.49 - ], - "text": "c[n] = (0.3)n", - "type": "text" - }, - { - "block_id": "p304-b25", - "global_id": 8158, - "bbox": [ - 283.16, - 420.16, - 297.26, - 430.61 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p304-b26", - "global_id": 8159, - "bbox": [ - 283.24, - 444.5, - 297.19, - 451.76 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p304-b27", - "global_id": 8160, - "bbox": [ - 298.37, - 416.13, - 318.74, - 433.5 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p304-b28", - "global_id": 8161, - "bbox": [ - 306.29, - 437.6, - 318.74, - 447.56 - ], - "text": "0.3", - "type": "text" - }, - { - "block_id": "p304-b29", - "global_id": 8162, - "bbox": [ - 319.94, - 416.13, - 331.7, - 427.94 - ], - "text": "m", - "type": "text" - }, - { - "block_id": "p304-b30", - "global_id": 8163, - "bbox": [ - 333.3, - 430.11, - 349.9, - 440.39 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p304-b31", - "global_id": 8164, - "bbox": [ - 103.17, - 458.11, - 451.96, - 468.49 - ], - "text": "This is a geometric progression with common ratio (0.8/0.3). From Sec. B.8-3 we have", - "type": "text" - }, - { - "block_id": "p304-b32", - "global_id": 8165, - "bbox": [ - 216.47, - 478.96, - 345.39, - 499.94 - ], - "text": "c[n] = (0.3)n (0.8)n+1 −(0.3)n+1", - "type": "text" - }, - { - "block_id": "p304-b33", - "global_id": 8166, - "bbox": [ - 274.14, - 489.57, - 363.7, - 507.02 - ], - "text": "(0.3)n(0.8 −0.3) u[n]", - "type": "text" - }, - { - "block_id": "p304-b34", - "global_id": 8167, - "bbox": [ - 234.57, - 507.11, - 349.06, - 521.6 - ], - "text": "= 2[(0.8)n+1 −(0.3)n+1]u[n]", - "type": "text" - }, - { - "block_id": "p304-b35", - "global_id": 8168, - "bbox": [ - 107.82, - 592.21, - 355.83, - 604.16 - ], - "text": "DRILL 3.15\nConvolution of Causal Signals", - "type": "text" - }, - { - "block_id": "p304-b36", - "global_id": 8169, - "bbox": [ - 107.82, - 611.65, - 306.77, - 623.25 - ], - "text": "Show that (0.8)nu[n] ∗u[n] = 5[1 −(0.8)n+1]u[n].", - "type": "text" - } - ] - }, - { - "page_num": 305, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p305-b0", - "global_id": 8170, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n285", - "type": "text" - }, - { - "block_id": "p305-b1", - "global_id": 8171, - "bbox": [ - 127.89, - 86.19, - 319.62, - 98.32 - ], - "text": "CONVOLUTION SUM FROM A TABLE", - "type": "text" - }, - { - "block_id": "p305-b2", - "global_id": 8172, - "bbox": [ - 127.59, - 102.35, - 516.15, - 136.22 - ], - "text": "Just as in the continuous-time case, we have prepared a table (Table 3.1) from which convolution\nsums may be determined directly for a variety of signal pairs. For example, the convolution in\nEx. 3.20 can be read directly from this table (pair 4) as", - "type": "text" - }, - { - "block_id": "p305-b3", - "global_id": 8173, - "bbox": [ - 165.31, - 149.6, - 343.55, - 170.58 - ], - "text": "(0.8)nu[n] ∗(0.3)nu[n] = (0.8)n+1 −(0.3)n+1", - "type": "text" - }, - { - "block_id": "p305-b4", - "global_id": 8174, - "bbox": [ - 287.95, - 156.09, - 478.4, - 177.65 - ], - "text": "0.8 −0.3\nu[n] = 2[(0.8)n+1 −(0.3)n+1]u[n]", - "type": "text" - }, - { - "block_id": "p305-b5", - "global_id": 8175, - "bbox": [ - 145.52, - 192.15, - 462.73, - 202.11 - ], - "text": "We shall demonstrate the use of the convolution table in the following example.", - "type": "text" - }, - { - "block_id": "p305-b6", - "global_id": 8176, - "bbox": [ - 127.59, - 248.03, - 271.44, - 257.27 - ], - "text": "TABLE 3.1\nSelect Convolution Sums", - "type": "text" - }, - { - "block_id": "p305-b7", - "global_id": 8177, - "bbox": [ - 127.59, - 267.98, - 416.57, - 278.01 - ], - "text": "No.\nx1[n]\nx2[n]\nx1[n] ∗x2[n] = x2[n] ∗x1[n]", - "type": "text" - }, - { - "block_id": "p305-b8", - "global_id": 8178, - "bbox": [ - 136.3, - 286.31, - 341.18, - 295.65 - ], - "text": "1\nδ[n −k]\nx[n]\nx[n −k]", - "type": "text" - }, - { - "block_id": "p305-b9", - "global_id": 8179, - "bbox": [ - 136.3, - 314.45, - 274.87, - 324.85 - ], - "text": "2\nγ nu[n]\nu[n]", - "type": "text" - }, - { - "block_id": "p305-b10", - "global_id": 8180, - "bbox": [ - 312.8, - 302.92, - 350.61, - 318.57 - ], - "text": "1 −γ n+1", - "type": "text" - }, - { - "block_id": "p305-b11", - "global_id": 8181, - "bbox": [ - 324.9, - 321.87, - 343.67, - 331.21 - ], - "text": "1 −γ", - "type": "text" - }, - { - "block_id": "p305-b12", - "global_id": 8182, - "bbox": [ - 352.31, - 302.92, - 357.2, - 311.89 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p305-b13", - "global_id": 8183, - "bbox": [ - 358.2, - 315.51, - 373.14, - 324.76 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p305-b14", - "global_id": 8184, - "bbox": [ - 136.31, - 338.42, - 353.19, - 347.76 - ], - "text": "3\nu[n]\nu[n]\n(n + 1)u[n]", - "type": "text" - }, - { - "block_id": "p305-b15", - "global_id": 8185, - "bbox": [ - 136.31, - 368.18, - 170.79, - 378.7 - ], - "text": "4\nγ n", - "type": "text" - }, - { - "block_id": "p305-b16", - "global_id": 8186, - "bbox": [ - 166.12, - 368.18, - 269.1, - 381.37 - ], - "text": "1 u[n]\nγ n", - "type": "text" - }, - { - "block_id": "p305-b17", - "global_id": 8187, - "bbox": [ - 264.43, - 369.36, - 284.54, - 381.37 - ], - "text": "2 u[n]", - "type": "text" - }, - { - "block_id": "p305-b19", - "global_id": 8188, - "bbox": [ - 319.57, - 361.07, - 337.03, - 371.88 - ], - "text": "γ n+1", - "type": "text" - }, - { - "block_id": "p305-b20", - "global_id": 8189, - "bbox": [ - 324.07, - 361.07, - 364.78, - 374.41 - ], - "text": "1\n−γ n+1", - "type": "text" - }, - { - "block_id": "p305-b21", - "global_id": 8190, - "bbox": [ - 329.29, - 367.94, - 355.06, - 386.03 - ], - "text": "2\nγ1 −γ2", - "type": "text" - }, - { - "block_id": "p305-b23", - "global_id": 8191, - "bbox": [ - 373.03, - 369.36, - 432.83, - 379.66 - ], - "text": "u[n]\nγ1̸ = γ2", - "type": "text" - }, - { - "block_id": "p305-b24", - "global_id": 8192, - "bbox": [ - 136.3, - 390.76, - 343.92, - 406.39 - ], - "text": "5\nu[n]\nnu[n]\nn(n + 1)", - "type": "text" - }, - { - "block_id": "p305-b25", - "global_id": 8193, - "bbox": [ - 326.73, - 397.05, - 360.06, - 412.75 - ], - "text": "2\nu[n]", - "type": "text" - }, - { - "block_id": "p305-b26", - "global_id": 8194, - "bbox": [ - 136.3, - 424.23, - 279.35, - 434.63 - ], - "text": "6\nγ nu[n]\nnu[n]", - "type": "text" - }, - { - "block_id": "p305-b27", - "global_id": 8195, - "bbox": [ - 312.8, - 412.7, - 396.61, - 428.34 - ], - "text": "γ (γ n −1) + n(1 −γ )", - "type": "text" - }, - { - "block_id": "p305-b28", - "global_id": 8196, - "bbox": [ - 342.44, - 431.32, - 372.57, - 440.99 - ], - "text": "(1 −γ )2", - "type": "text" - }, - { - "block_id": "p305-b29", - "global_id": 8197, - "bbox": [ - 397.81, - 412.7, - 402.7, - 421.67 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p305-b30", - "global_id": 8198, - "bbox": [ - 403.69, - 425.29, - 418.62, - 434.54 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p305-b31", - "global_id": 8199, - "bbox": [ - 136.3, - 447.2, - 388.74, - 460.49 - ], - "text": "7\nnu[n]\nnu[n]\n1\n6n(n −1)(n + 1)u[n]", - "type": "text" - }, - { - "block_id": "p305-b32", - "global_id": 8200, - "bbox": [ - 136.31, - 470.42, - 362.86, - 480.82 - ], - "text": "8\nγ nu[n]\nγ nu[n]\n(n + 1)γ nu[n]", - "type": "text" - }, - { - "block_id": "p305-b33", - "global_id": 8201, - "bbox": [ - 136.31, - 498.54, - 175.27, - 509.06 - ], - "text": "9\nnγ n", - "type": "text" - }, - { - "block_id": "p305-b34", - "global_id": 8202, - "bbox": [ - 170.6, - 498.54, - 269.1, - 511.73 - ], - "text": "1 u[n]\nγ n", - "type": "text" - }, - { - "block_id": "p305-b35", - "global_id": 8203, - "bbox": [ - 264.43, - 493.44, - 350.19, - 516.39 - ], - "text": "2 u[n]\nγ1γ2\n(γ1 −γ2)2", - "type": "text" - }, - { - "block_id": "p305-b37", - "global_id": 8204, - "bbox": [ - 357.77, - 498.22, - 366.94, - 508.69 - ], - "text": "γ n", - "type": "text" - }, - { - "block_id": "p305-b38", - "global_id": 8205, - "bbox": [ - 362.27, - 493.44, - 423.63, - 516.16 - ], - "text": "2 −γ n\n1 + γ1 −γ2\nγ2", - "type": "text" - }, - { - "block_id": "p305-b39", - "global_id": 8206, - "bbox": [ - 425.32, - 498.22, - 438.97, - 508.97 - ], - "text": "nγ n", - "type": "text" - }, - { - "block_id": "p305-b40", - "global_id": 8207, - "bbox": [ - 434.3, - 504.42, - 437.54, - 510.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p305-b41", - "global_id": 8208, - "bbox": [ - 439.48, - 487.13, - 444.37, - 496.1 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p305-b42", - "global_id": 8209, - "bbox": [ - 445.37, - 499.72, - 505.16, - 510.03 - ], - "text": "u[n]\nγ1̸ = γ2", - "type": "text" - }, - { - "block_id": "p305-b43", - "global_id": 8210, - "bbox": [ - 131.82, - 521.44, - 516.14, - 542.97 - ], - "text": "10\n|γ1|n cos(βn + θ)u[n]\n|γ2|nu[n]\n1\nR[|γ1|n+1 cos[β(n+1)+θ−φ] −|γ2|n+1 cos(θ−φ)]u[n]", - "type": "text" - }, - { - "block_id": "p305-b44", - "global_id": 8211, - "bbox": [ - 330.74, - 552.17, - 345.05, - 561.42 - ], - "text": "R =", - "type": "text" - }, - { - "block_id": "p305-b46", - "global_id": 8212, - "bbox": [ - 350.43, - 548.92, - 454.42, - 562.25 - ], - "text": "|γ1|2 + |γ2|2 −2|γ1||γ2|cosβ", - "type": "text" - }, - { - "block_id": "p305-b47", - "global_id": 8213, - "bbox": [ - 455.05, - 544.96, - 468.1, - 555.75 - ], - "text": "1/2", - "type": "text" - }, - { - "block_id": "p305-b48", - "global_id": 8214, - "bbox": [ - 330.74, - 579.16, - 366.29, - 589.75 - ], - "text": "φ = tan−1", - "type": "text" - }, - { - "block_id": "p305-b49", - "global_id": 8215, - "bbox": [ - 367.79, - 567.83, - 436.69, - 596.87 - ], - "text": "(|γ1|sinβ)\n(|γ1|cosβ −|γ2|)", - "type": "text" - }, - { - "block_id": "p305-b50", - "global_id": 8216, - "bbox": [ - 437.9, - 567.83, - 442.78, - 576.8 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p305-b51", - "global_id": 8217, - "bbox": [ - 131.82, - 604.72, - 170.79, - 615.23 - ], - "text": "11\nγ n", - "type": "text" - }, - { - "block_id": "p305-b52", - "global_id": 8218, - "bbox": [ - 166.12, - 604.72, - 269.1, - 617.9 - ], - "text": "1 u[−(n + 1)]\nγ n", - "type": "text" - }, - { - "block_id": "p305-b53", - "global_id": 8219, - "bbox": [ - 264.43, - 599.6, - 339.76, - 622.57 - ], - "text": "2 u[n]\nγ2\nγ1 −γ2", - "type": "text" - }, - { - "block_id": "p305-b54", - "global_id": 8220, - "bbox": [ - 341.46, - 604.38, - 350.63, - 614.86 - ], - "text": "γ n", - "type": "text" - }, - { - "block_id": "p305-b55", - "global_id": 8221, - "bbox": [ - 345.96, - 599.6, - 402.8, - 622.57 - ], - "text": "2 u[n] +\nγ1\nγ1 −γ2", - "type": "text" - }, - { - "block_id": "p305-b56", - "global_id": 8222, - "bbox": [ - 404.5, - 604.38, - 413.66, - 614.86 - ], - "text": "γ n", - "type": "text" - }, - { - "block_id": "p305-b57", - "global_id": 8223, - "bbox": [ - 409.0, - 605.89, - 512.41, - 617.57 - ], - "text": "1 u[−(n + 1)]\n|γ1| > |γ2|", - "type": "text" - } - ] - }, - { - "page_num": 306, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p306-b0", - "global_id": 8224, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "286\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p306-b1", - "global_id": 8225, - "bbox": [ - 76.77, - 93.91, - 305.13, - 105.87 - ], - "text": "EXAMPLE 3.21\nConvolution by Tables", - "type": "text" - }, - { - "block_id": "p306-b2", - "global_id": 8226, - "bbox": [ - 103.16, - 122.12, - 477.02, - 144.45 - ], - "text": "Using Table 3.1, find the (zero-state) response y[n] of an LTID system described by the\nequation", - "type": "text" - }, - { - "block_id": "p306-b3", - "global_id": 8227, - "bbox": [ - 199.98, - 146.04, - 380.18, - 156.41 - ], - "text": "y[n + 2] −0.6y[n + 1] −0.16y[n] = 5x[n + 2]", - "type": "text" - }, - { - "block_id": "p306-b4", - "global_id": 8228, - "bbox": [ - 103.17, - 163.74, - 210.81, - 175.34 - ], - "text": "if the input x[n] = 4−nu[n].", - "type": "text" - }, - { - "block_id": "p306-b5", - "global_id": 8229, - "bbox": [ - 103.16, - 196.61, - 477.02, - 220.17 - ], - "text": "The input can be expressed as x[n] = 4−nu[n] = (1/4)nu[n] = (0.25)nu[n]. The unit impulse\nresponse of this system, obtained in Ex. 3.18, is", - "type": "text" - }, - { - "block_id": "p306-b6", - "global_id": 8230, - "bbox": [ - 228.56, - 227.59, - 351.62, - 242.08 - ], - "text": "h[n] = [(−0.2)n + 4(0.8)n]u[n]", - "type": "text" - }, - { - "block_id": "p306-b7", - "global_id": 8231, - "bbox": [ - 103.17, - 254.04, - 144.91, - 264.01 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p306-b8", - "global_id": 8232, - "bbox": [ - 170.12, - 275.55, - 238.97, - 285.82 - ], - "text": "y[n] = x[n] ∗h[n]", - "type": "text" - }, - { - "block_id": "p306-b9", - "global_id": 8233, - "bbox": [ - 188.21, - 290.08, - 250.21, - 301.96 - ], - "text": "= (0.25)nu[n] ∗", - "type": "text" - }, - { - "block_id": "p306-b11", - "global_id": 8234, - "bbox": [ - 255.69, - 290.08, - 360.25, - 301.96 - ], - "text": "(−0.2)nu[n] + 4(0.8)nu[n]", - "type": "text" - }, - { - "block_id": "p306-b13", - "global_id": 8235, - "bbox": [ - 188.21, - 305.02, - 410.06, - 316.91 - ], - "text": "= (0.25)nu[n] ∗(−0.2)nu[n] + (0.25)nu[n] ∗4(0.8)nu[n]", - "type": "text" - }, - { - "block_id": "p306-b14", - "global_id": 8236, - "bbox": [ - 103.17, - 328.86, - 362.67, - 338.82 - ], - "text": "We use pair 4 (Table 3.1) to find the foregoing convolution sums.", - "type": "text" - }, - { - "block_id": "p306-b15", - "global_id": 8237, - "bbox": [ - 149.29, - 356.92, - 175.14, - 367.19 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p306-b16", - "global_id": 8238, - "bbox": [ - 177.19, - 342.93, - 272.51, - 360.31 - ], - "text": "(0.25)n+1 −(−0.2)n+1", - "type": "text" - }, - { - "block_id": "p306-b17", - "global_id": 8239, - "bbox": [ - 200.45, - 346.31, - 372.21, - 374.37 - ], - "text": "0.25 −(−0.2)\n+ 4(0.25)n+1 −(0.8)n+1", - "type": "text" - }, - { - "block_id": "p306-b18", - "global_id": 8240, - "bbox": [ - 311.62, - 363.99, - 352.36, - 374.37 - ], - "text": "0.25 −0.8", - "type": "text" - }, - { - "block_id": "p306-b19", - "global_id": 8241, - "bbox": [ - 373.92, - 342.93, - 379.35, - 352.89 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p306-b20", - "global_id": 8242, - "bbox": [ - 379.35, - 356.91, - 395.94, - 367.19 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p306-b21", - "global_id": 8243, - "bbox": [ - 167.37, - 375.11, - 430.9, - 389.6 - ], - "text": "= (2.22[(0.25)n+1 −(−0.2)n+1] −7.27[(0.25)n+1 −(0.8)n+1])u[n]", - "type": "text" - }, - { - "block_id": "p306-b22", - "global_id": 8244, - "bbox": [ - 167.37, - 391.55, - 393.38, - 406.04 - ], - "text": "= [−5.05(0.25)n+1 −2.22(−0.2)n+1 + 7.27(0.8)n+1]u[n]", - "type": "text" - }, - { - "block_id": "p306-b23", - "global_id": 8245, - "bbox": [ - 121.09, - 417.99, - 188.88, - 427.95 - ], - "text": "Recognizing that", - "type": "text" - }, - { - "block_id": "p306-b24", - "global_id": 8246, - "bbox": [ - 262.1, - 425.42, - 317.58, - 439.5 - ], - "text": "γ n+1 = γ (γ )n", - "type": "text" - }, - { - "block_id": "p306-b25", - "global_id": 8247, - "bbox": [ - 103.16, - 448.46, - 192.63, - 458.84 - ], - "text": "we can express y[n] as", - "type": "text" - }, - { - "block_id": "p306-b26", - "global_id": 8248, - "bbox": [ - 178.95, - 466.27, - 401.24, - 480.75 - ], - "text": "y[n] = [−1.26(0.25)n + 0.444(−0.2)n + 5.81(0.8)n]u[n]", - "type": "text" - }, - { - "block_id": "p306-b27", - "global_id": 8249, - "bbox": [ - 197.03, - 481.21, - 394.22, - 495.7 - ], - "text": "= [−1.26(4)−n + 0.444(−0.2)n + 5.81(0.8)n]u[n]", - "type": "text" - }, - { - "block_id": "p306-b28", - "global_id": 8250, - "bbox": [ - 107.82, - 548.5, - 313.0, - 560.46 - ], - "text": "DRILL 3.16\nConvolution by Tables", - "type": "text" - }, - { - "block_id": "p306-b29", - "global_id": 8251, - "bbox": [ - 107.82, - 569.57, - 213.59, - 579.53 - ], - "text": "Use Table 3.1 to show that", - "type": "text" - }, - { - "block_id": "p306-b30", - "global_id": 8252, - "bbox": [ - 125.76, - 585.87, - 308.89, - 612.41 - ], - "text": "(a) (0.8)n+1u[n] ∗u[n] = 4[1 −0.8(0.8)n]u[n]\n(b) n3−nu[n] ∗(0.2)nu[n] = 15", - "type": "text" - }, - { - "block_id": "p306-b31", - "global_id": 8253, - "bbox": [ - 243.57, - 594.02, - 253.92, - 615.22 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p306-b32", - "global_id": 8254, - "bbox": [ - 253.92, - 598.42, - 287.11, - 612.41 - ], - "text": "(0.2)n −", - "type": "text" - }, - { - "block_id": "p306-b34", - "global_id": 8255, - "bbox": [ - 292.67, - 594.02, - 323.4, - 615.22 - ], - "text": "1 −2\n3n", - "type": "text" - }, - { - "block_id": "p306-b35", - "global_id": 8256, - "bbox": [ - 323.4, - 594.02, - 358.32, - 612.41 - ], - "text": "3−n\nu[n]", - "type": "text" - }, - { - "block_id": "p306-b36", - "global_id": 8257, - "bbox": [ - 126.32, - 622.73, - 237.17, - 637.27 - ], - "text": "(c) e−nu[n] ∗2−nu[n] =\n2\n2−e", - "type": "text" - }, - { - "block_id": "p306-b38", - "global_id": 8258, - "bbox": [ - 243.8, - 620.46, - 273.01, - 634.35 - ], - "text": "e−n −e", - "type": "text" - }, - { - "block_id": "p306-b39", - "global_id": 8259, - "bbox": [ - 269.71, - 610.09, - 294.23, - 637.27 - ], - "text": "22−n\n!", - "type": "text" - }, - { - "block_id": "p306-b40", - "global_id": 8260, - "bbox": [ - 294.24, - 624.07, - 310.83, - 634.35 - ], - "text": "u[n]", - "type": "text" - } - ] - }, - { - "page_num": 307, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p307-b0", - "global_id": 8261, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n287", - "type": "text" - }, - { - "block_id": "p307-b1", - "global_id": 8262, - "bbox": [ - 102.51, - 93.91, - 468.66, - 105.87 - ], - "text": "EXAMPLE 3.22\nFiltering Perspective of the Zero-State Response", - "type": "text" - }, - { - "block_id": "p307-b2", - "global_id": 8263, - "bbox": [ - 128.9, - 122.54, - 502.75, - 144.45 - ], - "text": "Use the MATLAB filter command to compute and sketch the zero-state response for the\nsystem described by (E2 + 0.5E −1)y[n] = (2E2 + 6E)x[n] and the input x[n] = 4−nu[n].", - "type": "text" - }, - { - "block_id": "p307-b3", - "global_id": 8264, - "bbox": [ - 128.9, - 167.37, - 502.75, - 189.28 - ], - "text": "We solve this problem using the same approach as Ex. 3.19. Although the input is bounded\nand quickly decays to zero, the system itself is unstable and an unbounded output results.", - "type": "text" - }, - { - "block_id": "p307-b4", - "global_id": 8265, - "bbox": [ - 128.9, - 198.99, - 459.0, - 226.89 - ], - "text": ">>\nn = (0:11); x = @(n) 4.^(-n).*(n>=0);\n>>\na = [1 0.5 -1]; b = [2 6 0]; y = filter(b,a,x(n));\n>>\nclf; stem(n,y,’k’); xlabel(’n’); ylabel(’y[n]’); axis([-0.5 11.5 -20 25]);", - "type": "text" - }, - { - "block_id": "p307-b5", - "global_id": 8266, - "bbox": [ - 162.16, - 338.36, - 354.16, - 359.55 - ], - "text": "0\n2\n4\n6\n8\n10\nn", - "type": "text" - }, - { - "block_id": "p307-b6", - "global_id": 8267, - "bbox": [ - 138.73, - 329.14, - 149.4, - 337.14 - ], - "text": "-20", - "type": "text" - }, - { - "block_id": "p307-b7", - "global_id": 8268, - "bbox": [ - 145.48, - 297.14, - 149.48, - 305.14 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p307-b8", - "global_id": 8269, - "bbox": [ - 140.98, - 265.14, - 148.98, - 273.14 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p307-b9", - "global_id": 8270, - "bbox": [ - 126.76, - 288.71, - 135.56, - 303.37 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p307-b10", - "global_id": 8271, - "bbox": [ - 126.0, - 366.23, - 291.1, - 375.47 - ], - "text": "Figure 3.21 Zero-state response for Ex. 3.22", - "type": "text" - }, - { - "block_id": "p307-b11", - "global_id": 8272, - "bbox": [ - 127.59, - 417.73, - 516.14, - 479.71 - ], - "text": "RESPONSE TO COMPLEX INPUTS\nAs in the case of real continuous-time systems, we can show that for an LTID system with real\nh[n], if the input and the output are expressed in terms of their real and imaginary parts, then the\nreal part of the input generates the real part of the response and the imaginary part of the input\ngenerates the imaginary part. Thus, if", - "type": "text" - }, - { - "block_id": "p307-b12", - "global_id": 8273, - "bbox": [ - 215.29, - 490.13, - 428.42, - 501.21 - ], - "text": "x[n] = xr[n] + jxi[n]\nand\ny[n] = yr[n] + jyi[n]", - "type": "text" - }, - { - "block_id": "p307-b13", - "global_id": 8274, - "bbox": [ - 127.59, - 511.34, - 449.57, - 521.31 - ], - "text": "using the right-directed arrow to indicate the input–output pair, we can show that", - "type": "text" - }, - { - "block_id": "p307-b14", - "global_id": 8275, - "bbox": [ - 236.26, - 531.73, - 516.13, - 542.81 - ], - "text": "xr[n] \r⇒yr[n]\nand\nxi[n] \r⇒yi[n]\n(3.34)", - "type": "text" - }, - { - "block_id": "p307-b15", - "global_id": 8276, - "bbox": [ - 127.6, - 552.93, - 409.08, - 562.9 - ], - "text": "The proof is similar to that used to derive Eq. (2.31) for LTIC systems.", - "type": "text" - }, - { - "block_id": "p307-b16", - "global_id": 8277, - "bbox": [ - 127.89, - 577.35, - 225.1, - 589.47 - ], - "text": "MULTIPLE INPUTS", - "type": "text" - }, - { - "block_id": "p307-b17", - "global_id": 8278, - "bbox": [ - 127.59, - 593.51, - 516.14, - 627.37 - ], - "text": "Multiple inputs to LTI systems can be treated by applying the superposition principle. Each input\nis considered separately, with all other inputs assumed to be zero. The sum of all these individual\nsystem responses constitutes the total system output when all the inputs are applied simultaneously.", - "type": "text" - } - ] - }, - { - "page_num": 308, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p308-b0", - "global_id": 8279, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "288\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p308-b1", - "global_id": 8280, - "bbox": [ - 107.82, - 97.81, - 345.18, - 109.77 - ], - "text": "DRILL 3.17\nResponse to Multiple Inputs", - "type": "text" - }, - { - "block_id": "p308-b2", - "global_id": 8281, - "bbox": [ - 107.82, - 118.47, - 484.4, - 152.76 - ], - "text": "Show that the system described by y[n] −0.6y[n −1] −0.16y[n −2] = 5x[n] responds to input\nx[n] = δ[n] + 4−nu[n] with output y[n] = [−1.26(4)−n + 1.444(−0.2)n + 9.81(0.8)n]u[n].\n[Hint: Use the results of Exs. 3.18 and 3.21.]", - "type": "text" - }, - { - "block_id": "p308-b3", - "global_id": 8282, - "bbox": [ - 101.84, - 191.06, - 384.27, - 203.02 - ], - "text": "3.8-1 Graphical Procedure for the Convolution Sum", - "type": "text" - }, - { - "block_id": "p308-b4", - "global_id": 8283, - "bbox": [ - 101.84, - 209.15, - 490.36, - 231.07 - ], - "text": "The steps in evaluating the convolution sum are parallel to those followed in evaluating the\nconvolution integral. The convolution sum of causal signals x[n] and g[n] is given by", - "type": "text" - }, - { - "block_id": "p308-b5", - "global_id": 8284, - "bbox": [ - 248.1, - 248.54, - 273.95, - 258.82 - ], - "text": "c[n] =", - "type": "text" - }, - { - "block_id": "p308-b6", - "global_id": 8285, - "bbox": [ - 275.99, - 238.59, - 290.09, - 249.05 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p308-b7", - "global_id": 8286, - "bbox": [ - 276.07, - 262.94, - 290.02, - 270.21 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p308-b8", - "global_id": 8287, - "bbox": [ - 291.2, - 248.54, - 344.13, - 258.82 - ], - "text": "x[m]g[n −m]", - "type": "text" - }, - { - "block_id": "p308-b9", - "global_id": 8288, - "bbox": [ - 101.84, - 277.49, - 490.4, - 311.78 - ], - "text": "We first plot x[m] and g[n −m] as functions of m (not n), because the summation is over m.\nFunctions x[m] and g[m] are the same as x[n] and g[n], plotted, respectively, as functions of m (see\nFig. 3.22). The convolution operation can be performed as follows:", - "type": "text" - }, - { - "block_id": "p308-b10", - "global_id": 8289, - "bbox": [ - 118.78, - 319.33, - 490.4, - 401.44 - ], - "text": "1. Invert g[m] about the vertical axis (m = 0) to obtain g[−m] (Fig. 3.22d). Figure 3.22e\nshows both x[m] and g[−m].\n2. Shift g[−m] by n units to obtain g[n −m]. For n > 0, the shift is to the right (delay); for\nn < 0, the shift is to the left (advance). Figure 3.22f shows g[n −m] for n > 0; for n < 0,\nsee Fig. 3.22g.\n3. Next we multiply x[m] and g[n−m] and add all the products to obtain c[n]. The procedure\nis repeated for each value of n over the range −∞to ∞.", - "type": "text" - }, - { - "block_id": "p308-b11", - "global_id": 8290, - "bbox": [ - 101.85, - 409.41, - 490.39, - 443.29 - ], - "text": "We shall demonstrate by an example the graphical procedure for finding the convolution sum.\nAlthough both the functions in this example are causal, this procedure is applicable to the general\ncase.", - "type": "text" - }, - { - "block_id": "p308-b12", - "global_id": 8291, - "bbox": [ - 76.77, - 469.22, - 432.97, - 481.18 - ], - "text": "EXAMPLE 3.23\nGraphical Procedure for the Convolution Sum", - "type": "text" - }, - { - "block_id": "p308-b13", - "global_id": 8292, - "bbox": [ - 103.16, - 496.47, - 476.99, - 506.85 - ], - "text": "Find c[n] = x[n]∗g[n], where x[n] and g[n] are depicted in Figs. 3.22a and 3.22b, respectively.", - "type": "text" - }, - { - "block_id": "p308-b14", - "global_id": 8293, - "bbox": [ - 103.16, - 529.76, - 155.06, - 539.72 - ], - "text": "We are given", - "type": "text" - }, - { - "block_id": "p308-b15", - "global_id": 8294, - "bbox": [ - 210.9, - 539.8, - 368.78, - 551.67 - ], - "text": "x[n] = (0.8)n\nand\ng[n] = (0.3)n", - "type": "text" - }, - { - "block_id": "p308-b16", - "global_id": 8295, - "bbox": [ - 103.16, - 559.68, - 144.91, - 569.65 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p308-b17", - "global_id": 8296, - "bbox": [ - 194.75, - 569.5, - 384.92, - 581.6 - ], - "text": "x[m] = (0.8)m\nand\ng[n −m] = (0.3)n−m", - "type": "text" - } - ] - }, - { - "page_num": 309, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p309-b0", - "global_id": 8297, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n289", - "type": "text" - }, - { - "block_id": "p309-b1", - "global_id": 8298, - "bbox": [ - 191.77, - 100.02, - 195.77, - 108.02 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p309-b2", - "global_id": 8299, - "bbox": [ - 252.81, - 179.82, - 261.69, - 187.82 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p309-b3", - "global_id": 8300, - "bbox": [ - 202.89, - 166.11, - 263.83, - 174.11 - ], - "text": "0\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p309-b4", - "global_id": 8301, - "bbox": [ - 202.89, - 269.4, - 263.83, - 277.4 - ], - "text": "0\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p309-b5", - "global_id": 8302, - "bbox": [ - 203.44, - 378.43, - 264.86, - 386.43 - ], - "text": "0\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p309-b6", - "global_id": 8303, - "bbox": [ - 227.43, - 100.19, - 246.23, - 109.72 - ], - "text": "(0.8)n", - "type": "text" - }, - { - "block_id": "p309-b7", - "global_id": 8304, - "bbox": [ - 304.78, - 165.0, - 454.84, - 173.0 - ], - "text": "n\nn", - "type": "text" - }, - { - "block_id": "p309-b8", - "global_id": 8305, - "bbox": [ - 471.25, - 603.66, - 475.25, - 611.66 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p309-b9", - "global_id": 8306, - "bbox": [ - 303.01, - 268.43, - 454.84, - 276.43 - ], - "text": "m\nm", - "type": "text" - }, - { - "block_id": "p309-b10", - "global_id": 8307, - "bbox": [ - 282.54, - 377.42, - 288.32, - 385.42 - ], - "text": "m", - "type": "text" - }, - { - "block_id": "p309-b11", - "global_id": 8308, - "bbox": [ - 282.54, - 486.68, - 288.32, - 494.68 - ], - "text": "m", - "type": "text" - }, - { - "block_id": "p309-b12", - "global_id": 8309, - "bbox": [ - 282.54, - 593.92, - 288.32, - 601.92 - ], - "text": "m", - "type": "text" - }, - { - "block_id": "p309-b13", - "global_id": 8310, - "bbox": [ - 193.88, - 85.73, - 206.76, - 93.82 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p309-b14", - "global_id": 8311, - "bbox": [ - 345.17, - 537.07, - 349.17, - 545.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p309-b15", - "global_id": 8312, - "bbox": [ - 346.15, - 594.97, - 470.29, - 602.97 - ], - "text": "0\n1\n2\n3\n4\n5", - "type": "text" - }, - { - "block_id": "p309-b16", - "global_id": 8313, - "bbox": [ - 411.03, - 617.17, - 420.36, - 625.17 - ], - "text": "(h)", - "type": "text" - }, - { - "block_id": "p309-b17", - "global_id": 8314, - "bbox": [ - 192.21, - 522.78, - 370.61, - 536.84 - ], - "text": "c[n]\n1", - "type": "text" - }, - { - "block_id": "p309-b18", - "global_id": 8315, - "bbox": [ - 202.76, - 617.17, - 212.08, - 625.17 - ], - "text": "(g)", - "type": "text" - }, - { - "block_id": "p309-b19", - "global_id": 8316, - "bbox": [ - 183.18, - 594.82, - 186.74, - 602.82 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p309-b20", - "global_id": 8317, - "bbox": [ - 146.23, - 539.14, - 241.52, - 550.35 - ], - "text": "g[n m]\nx[m]", - "type": "text" - }, - { - "block_id": "p309-b21", - "global_id": 8318, - "bbox": [ - 203.55, - 595.02, - 207.55, - 603.02 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p309-b22", - "global_id": 8319, - "bbox": [ - 181.38, - 200.52, - 196.04, - 208.61 - ], - "text": "x[m]", - "type": "text" - }, - { - "block_id": "p309-b23", - "global_id": 8320, - "bbox": [ - 253.2, - 283.21, - 262.08, - 291.21 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p309-b24", - "global_id": 8321, - "bbox": [ - 227.43, - 209.34, - 247.09, - 218.88 - ], - "text": "(0.8)m", - "type": "text" - }, - { - "block_id": "p309-b25", - "global_id": 8322, - "bbox": [ - 356.67, - 99.98, - 360.67, - 107.98 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p309-b26", - "global_id": 8323, - "bbox": [ - 403.91, - 179.82, - 413.56, - 187.82 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p309-b27", - "global_id": 8324, - "bbox": [ - 357.74, - 166.05, - 426.68, - 174.05 - ], - "text": "0\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p309-b28", - "global_id": 8325, - "bbox": [ - 358.17, - 87.4, - 404.62, - 109.72 - ], - "text": "(0.3)n\ng[n]", - "type": "text" - }, - { - "block_id": "p309-b29", - "global_id": 8326, - "bbox": [ - 437.31, - 201.43, - 459.07, - 209.73 - ], - "text": "g[m]", - "type": "text" - }, - { - "block_id": "p309-b30", - "global_id": 8327, - "bbox": [ - 404.07, - 283.21, - 413.4, - 291.21 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p309-b31", - "global_id": 8328, - "bbox": [ - 365.67, - 269.09, - 437.93, - 277.38 - ], - "text": "0\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p309-b32", - "global_id": 8329, - "bbox": [ - 395.18, - 224.92, - 419.83, - 234.61 - ], - "text": "(0.3)m", - "type": "text" - }, - { - "block_id": "p309-b33", - "global_id": 8330, - "bbox": [ - 104.0, - 372.55, - 112.88, - 380.55 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p309-b34", - "global_id": 8331, - "bbox": [ - 194.24, - 304.13, - 198.24, - 312.13 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p309-b35", - "global_id": 8332, - "bbox": [ - 155.02, - 316.97, - 176.79, - 325.27 - ], - "text": "g[m]", - "type": "text" - }, - { - "block_id": "p309-b36", - "global_id": 8333, - "bbox": [ - 133.98, - 378.13, - 191.02, - 386.43 - ], - "text": "1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p309-b37", - "global_id": 8334, - "bbox": [ - 240.52, - 313.82, - 255.18, - 321.9 - ], - "text": "x[m]", - "type": "text" - }, - { - "block_id": "p309-b38", - "global_id": 8335, - "bbox": [ - 192.2, - 422.64, - 196.2, - 430.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p309-b39", - "global_id": 8336, - "bbox": [ - 167.53, - 481.96, - 264.05, - 495.95 - ], - "text": "(f)\nk", - "type": "text" - }, - { - "block_id": "p309-b40", - "global_id": 8337, - "bbox": [ - 286.22, - 447.8, - 300.88, - 455.88 - ], - "text": "x[m]", - "type": "text" - }, - { - "block_id": "p309-b41", - "global_id": 8338, - "bbox": [ - 212.98, - 416.29, - 242.75, - 424.58 - ], - "text": "g[n m]", - "type": "text" - }, - { - "block_id": "p309-b42", - "global_id": 8339, - "bbox": [ - 203.34, - 488.02, - 207.34, - 496.02 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p309-b43", - "global_id": 8340, - "bbox": [ - 282.2, - 420.55, - 300.86, - 428.84 - ], - "text": "n 0", - "type": "text" - }, - { - "block_id": "p309-b44", - "global_id": 8341, - "bbox": [ - 262.95, - 531.3, - 281.61, - 539.59 - ], - "text": "n 0", - "type": "text" - }, - { - "block_id": "p309-b45", - "global_id": 8342, - "bbox": [ - 311.74, - 156.96, - 321.04, - 162.96 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p309-b46", - "global_id": 8343, - "bbox": [ - 311.74, - 260.29, - 321.04, - 266.29 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p309-b47", - "global_id": 8344, - "bbox": [ - 104.0, - 631.49, - 322.48, - 641.1 - ], - "text": "Figure 3.22 Graphical procedure to convolve x[n] and g[n].", - "type": "text" - } - ] - }, - { - "page_num": 310, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p310-b0", - "global_id": 8345, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "290\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p310-b1", - "global_id": 8346, - "bbox": [ - 103.16, - 85.83, - 477.01, - 108.16 - ], - "text": "Figure 3.22f shows the general situation for n ≥0. The two functions x[m] and g[n−m] overlap\nover the interval 0 ≤m ≤n. Therefore,", - "type": "text" - }, - { - "block_id": "p310-b2", - "global_id": 8347, - "bbox": [ - 169.57, - 130.89, - 195.43, - 141.17 - ], - "text": "c[n] =", - "type": "text" - }, - { - "block_id": "p310-b3", - "global_id": 8348, - "bbox": [ - 197.48, - 120.95, - 211.57, - 131.4 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p310-b4", - "global_id": 8349, - "bbox": [ - 197.55, - 145.29, - 211.5, - 152.55 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p310-b5", - "global_id": 8350, - "bbox": [ - 212.69, - 130.89, - 265.61, - 141.17 - ], - "text": "x[m]g[n −m]", - "type": "text" - }, - { - "block_id": "p310-b6", - "global_id": 8351, - "bbox": [ - 187.66, - 164.8, - 195.43, - 174.77 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p310-b7", - "global_id": 8352, - "bbox": [ - 197.48, - 154.85, - 211.57, - 165.3 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p310-b8", - "global_id": 8353, - "bbox": [ - 197.55, - 179.19, - 211.5, - 186.46 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p310-b9", - "global_id": 8354, - "bbox": [ - 211.58, - 163.08, - 270.83, - 175.18 - ], - "text": "(0.8)m(0.3)n−m", - "type": "text" - }, - { - "block_id": "p310-b10", - "global_id": 8355, - "bbox": [ - 187.66, - 197.2, - 220.84, - 209.09 - ], - "text": "= (0.3)n", - "type": "text" - }, - { - "block_id": "p310-b11", - "global_id": 8356, - "bbox": [ - 222.45, - 188.76, - 236.55, - 199.21 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p310-b12", - "global_id": 8357, - "bbox": [ - 222.52, - 213.1, - 236.48, - 220.37 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p310-b13", - "global_id": 8358, - "bbox": [ - 237.66, - 184.72, - 258.03, - 202.11 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p310-b14", - "global_id": 8359, - "bbox": [ - 245.58, - 206.2, - 258.03, - 216.17 - ], - "text": "0.3", - "type": "text" - }, - { - "block_id": "p310-b15", - "global_id": 8360, - "bbox": [ - 259.22, - 184.72, - 270.99, - 196.55 - ], - "text": "m", - "type": "text" - }, - { - "block_id": "p310-b16", - "global_id": 8361, - "bbox": [ - 187.66, - 220.64, - 410.59, - 235.13 - ], - "text": "= 2[(0.8)n+1 −(0.3)n+1]\nn ≥0\n(see Sec. B.8-3)", - "type": "text" - }, - { - "block_id": "p310-b17", - "global_id": 8362, - "bbox": [ - 103.16, - 250.66, - 456.44, - 261.03 - ], - "text": "For n < 0, there is no overlap between x[m] and g[n −m], as shown in Fig. 3.22g, so that", - "type": "text" - }, - { - "block_id": "p310-b18", - "global_id": 8363, - "bbox": [ - 252.77, - 276.56, - 327.41, - 286.93 - ], - "text": "c[n] = 0\nn < 0", - "type": "text" - }, - { - "block_id": "p310-b19", - "global_id": 8364, - "bbox": [ - 103.16, - 302.87, - 224.64, - 312.83 - ], - "text": "Combining pieces, we see that", - "type": "text" - }, - { - "block_id": "p310-b20", - "global_id": 8365, - "bbox": [ - 223.8, - 324.25, - 356.39, - 338.74 - ], - "text": "c[n] = 2[(0.8)n+1 −(0.3)n+1]u[n]", - "type": "text" - }, - { - "block_id": "p310-b21", - "global_id": 8366, - "bbox": [ - 103.17, - 354.68, - 317.3, - 364.64 - ], - "text": "which agrees with the result found earlier in Ex. 3.20.", - "type": "text" - }, - { - "block_id": "p310-b22", - "global_id": 8367, - "bbox": [ - 107.82, - 434.37, - 440.85, - 446.32 - ], - "text": "DRILL 3.18\nGraphical Procedure for the Convolution Sum", - "type": "text" - }, - { - "block_id": "p310-b23", - "global_id": 8368, - "bbox": [ - 107.82, - 454.02, - 327.77, - 465.4 - ], - "text": "Find (0.8)nu[n] ∗u[n] graphically and sketch the result.", - "type": "text" - }, - { - "block_id": "p310-b24", - "global_id": 8369, - "bbox": [ - 108.09, - 478.93, - 162.68, - 489.89 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p310-b25", - "global_id": 8370, - "bbox": [ - 107.82, - 495.64, - 185.48, - 507.24 - ], - "text": "5(1 −(0.8)n+1)u[n]", - "type": "text" - }, - { - "block_id": "p310-b26", - "global_id": 8371, - "bbox": [ - 101.84, - 546.9, - 490.41, - 634.79 - ], - "text": "AN ALTERNATIVE FORM OF GRAPHICAL PROCEDURE:\nTHE SLIDING-TAPE METHOD\nThis algorithm is convenient when the sequences x[n] and g[n] are short or when they are available\nonly in graphical form. The algorithm is basically the same as the graphical procedure in Fig. 3.22.\nThe only difference is that instead of presenting the data as graphical plots, we display it as a\nsequence of numbers on tapes. Otherwise the procedure is the same, as will become clear in the\nfollowing example.", - "type": "text" - } - ] - }, - { - "page_num": 311, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p311-b0", - "global_id": 8372, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n291", - "type": "text" - }, - { - "block_id": "p311-b1", - "global_id": 8373, - "bbox": [ - 102.51, - 93.92, - 462.72, - 105.87 - ], - "text": "EXAMPLE 3.24\nSliding-Tape Method for the Convolution Sum", - "type": "text" - }, - { - "block_id": "p311-b2", - "global_id": 8374, - "bbox": [ - 128.9, - 122.12, - 502.76, - 144.45 - ], - "text": "Use the sliding-tape method to convolve the two sequences x[n] and g[n] depicted in\nFigs. 3.23a and 3.23b, respectively.", - "type": "text" - }, - { - "block_id": "p311-b3", - "global_id": 8375, - "bbox": [ - 128.9, - 166.96, - 502.77, - 249.06 - ], - "text": "In this procedure we write the sequences x[n]andg[n] in the slots of two tapes: x tape and g\ntape (Fig. 3.23c). Now leave the x tape stationary (to correspond to x[m]). The g[−m] tape is\nobtained by inverting the g[m] tape about the origin (m = 0) so that the slots corresponding to\nx[0] and g[0] remain aligned (Fig. 3.23d). We now shift the inverted tape by n slots, multiply\nvalues on two tapes in adjacent slots, and add all the products to find c[n]. Figures 3.23d–3.23i\nshow the cases for n = 0–5. Figures 3.23j, 3.23k, and 3.23l show the cases for n = −1,−2, and\n−3, respectively.", - "type": "text" - }, - { - "block_id": "p311-b4", - "global_id": 8376, - "bbox": [ - 146.84, - 250.64, - 334.93, - 261.01 - ], - "text": "For the case of n = 0, for example (Fig. 3.23d),", - "type": "text" - }, - { - "block_id": "p311-b5", - "global_id": 8377, - "bbox": [ - 228.47, - 281.53, - 403.2, - 291.9 - ], - "text": "c[0] = (−2 × 1) + (−1 × 1) + (0 × 1) = −3", - "type": "text" - }, - { - "block_id": "p311-b6", - "global_id": 8378, - "bbox": [ - 128.91, - 312.41, - 218.65, - 322.78 - ], - "text": "For n = 1 (Fig. 3.23e),", - "type": "text" - }, - { - "block_id": "p311-b7", - "global_id": 8379, - "bbox": [ - 208.89, - 343.29, - 422.78, - 353.67 - ], - "text": "c[1] = (−2 × 1) + (−1 × 1) + (0 × 1) + (1 × 1) = −2", - "type": "text" - }, - { - "block_id": "p311-b8", - "global_id": 8380, - "bbox": [ - 128.91, - 374.59, - 167.84, - 384.56 - ], - "text": "Similarly,", - "type": "text" - }, - { - "block_id": "p311-b9", - "global_id": 8381, - "bbox": [ - 154.01, - 405.06, - 399.31, - 415.44 - ], - "text": "c[2] = (−2 × 1) + (−1 × 1) + (0 × 1) + (1 × 1) + (2 × 1) = 0", - "type": "text" - }, - { - "block_id": "p311-b10", - "global_id": 8382, - "bbox": [ - 154.01, - 420.01, - 477.67, - 460.26 - ], - "text": "c[3] = (−2 × 1) + (−1 × 1) + (0 × 1) + (1 × 1) + (2 × 1) + (3 × 1) = 3\nc[4] = (−2 × 1) + (−1 × 1) + (0 × 1) + (1 × 1) + (2 × 1) + (3 × 1) + (4 × 1) = 7\nc[5] = (−2 × 1) + (−1 × 1) + (0 × 1) + (1 × 1) + (2 × 1) + (3 × 1) + (4 × 1) = 7", - "type": "text" - }, - { - "block_id": "p311-b11", - "global_id": 8383, - "bbox": [ - 128.91, - 480.78, - 298.23, - 491.15 - ], - "text": "Figure 3.23i shows that c[n] = 7 for n ≥4.", - "type": "text" - }, - { - "block_id": "p311-b12", - "global_id": 8384, - "bbox": [ - 128.9, - 492.73, - 502.76, - 527.02 - ], - "text": "Similarly, we compute c[n] for negative n by sliding the tape backward, one slot at a time,\nas shown in the plots corresponding to n = −1, −2, and −3, respectively (Figs. 3.23j, 3.23k,\nand 3.23l).", - "type": "text" - }, - { - "block_id": "p311-b13", - "global_id": 8385, - "bbox": [ - 244.18, - 547.53, - 387.49, - 572.85 - ], - "text": "c[−1] = (−2 × 1) + (−1 × 1) = −3\nc[−2] = (−2 × 1) = −2", - "type": "text" - }, - { - "block_id": "p311-b14", - "global_id": 8386, - "bbox": [ - 244.18, - 577.42, - 284.84, - 587.79 - ], - "text": "c[−3] = 0", - "type": "text" - }, - { - "block_id": "p311-b15", - "global_id": 8387, - "bbox": [ - 128.91, - 608.3, - 446.03, - 618.67 - ], - "text": "Figure 3.23l shows that c[n] = 0 for n ≤3. Figure 3.23m shows the plot of c[n].", - "type": "text" - } - ] - }, - { - "page_num": 312, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p312-b0", - "global_id": 8388, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "292\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p312-b1", - "global_id": 8389, - "bbox": [ - 175.97, - 83.18, - 188.2, - 90.86 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p312-b2", - "global_id": 8390, - "bbox": [ - 207.97, - 158.19, - 216.41, - 165.79 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p312-b3", - "global_id": 8391, - "bbox": [ - 184.4, - 105.41, - 356.73, - 113.68 - ], - "text": "4\ng[n]", - "type": "text" - }, - { - "block_id": "p312-b4", - "global_id": 8392, - "bbox": [ - 381.28, - 158.19, - 390.14, - 165.79 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p312-b5", - "global_id": 8393, - "bbox": [ - 164.79, - 132.97, - 356.25, - 143.54 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p312-b6", - "global_id": 8394, - "bbox": [ - 196.99, - 146.88, - 406.37, - 155.57 - ], - "text": "1 2 3 4 5\n1\n3\n5\n6\n7", - "type": "text" - }, - { - "block_id": "p312-b7", - "global_id": 8395, - "bbox": [ - 188.57, - 192.96, - 197.0, - 200.56 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p312-b8", - "global_id": 8396, - "bbox": [ - 244.59, - 182.94, - 262.53, - 190.62 - ], - "text": "x tape", - "type": "text" - }, - { - "block_id": "p312-b9", - "global_id": 8397, - "bbox": [ - 252.03, - 197.95, - 270.4, - 205.63 - ], - "text": "g tape", - "type": "text" - }, - { - "block_id": "p312-b10", - "global_id": 8398, - "bbox": [ - 208.31, - 236.73, - 376.96, - 244.41 - ], - "text": "Rotate the g tape about the vertical axis as shown in (d)", - "type": "text" - }, - { - "block_id": "p312-b11", - "global_id": 8399, - "bbox": [ - 188.14, - 288.45, - 197.0, - 296.05 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p312-b12", - "global_id": 8400, - "bbox": [ - 266.92, - 147.8, - 415.81, - 155.69 - ], - "text": "n\nn", - "type": "text" - }, - { - "block_id": "p312-b13", - "global_id": 8401, - "bbox": [ - 265.77, - 185.08, - 338.27, - 192.96 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b14", - "global_id": 8402, - "bbox": [ - 290.77, - 200.56, - 360.12, - 208.16 - ], - "text": "1\n1\n1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b15", - "global_id": 8403, - "bbox": [ - 265.77, - 292.98, - 377.02, - 301.5 - ], - "text": "c[0] 3\n0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b16", - "global_id": 8404, - "bbox": [ - 312.59, - 309.01, - 330.32, - 316.89 - ], - "text": "n 0", - "type": "text" - }, - { - "block_id": "p312-b17", - "global_id": 8405, - "bbox": [ - 240.73, - 294.23, - 261.41, - 302.11 - ], - "text": "g[m]", - "type": "text" - }, - { - "block_id": "p312-b18", - "global_id": 8406, - "bbox": [ - 301.23, - 279.81, - 315.15, - 287.49 - ], - "text": "x[m]", - "type": "text" - }, - { - "block_id": "p312-b19", - "global_id": 8407, - "bbox": [ - 282.93, - 264.17, - 302.35, - 272.05 - ], - "text": "m 0", - "type": "text" - }, - { - "block_id": "p312-b20", - "global_id": 8408, - "bbox": [ - 225.24, - 309.1, - 294.59, - 316.7 - ], - "text": "1\n1\n1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b21", - "global_id": 8409, - "bbox": [ - 367.54, - 335.89, - 386.95, - 343.77 - ], - "text": "m 0", - "type": "text" - }, - { - "block_id": "p312-b22", - "global_id": 8410, - "bbox": [ - 105.06, - 354.33, - 113.5, - 361.93 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p312-b23", - "global_id": 8411, - "bbox": [ - 105.06, - 403.47, - 113.26, - 411.07 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p312-b24", - "global_id": 8412, - "bbox": [ - 105.06, - 453.23, - 114.08, - 460.83 - ], - "text": "(g)", - "type": "text" - }, - { - "block_id": "p312-b25", - "global_id": 8413, - "bbox": [ - 105.06, - 503.2, - 114.23, - 510.8 - ], - "text": "(h)", - "type": "text" - }, - { - "block_id": "p312-b26", - "global_id": 8414, - "bbox": [ - 213.21, - 374.11, - 230.94, - 381.99 - ], - "text": "n 1", - "type": "text" - }, - { - "block_id": "p312-b27", - "global_id": 8415, - "bbox": [ - 248.48, - 557.48, - 274.65, - 565.36 - ], - "text": "c[5] 7", - "type": "text" - }, - { - "block_id": "p312-b28", - "global_id": 8416, - "bbox": [ - 248.48, - 507.84, - 274.65, - 515.72 - ], - "text": "c[4] 7", - "type": "text" - }, - { - "block_id": "p312-b29", - "global_id": 8417, - "bbox": [ - 248.48, - 458.29, - 274.65, - 466.17 - ], - "text": "c[3] 3", - "type": "text" - }, - { - "block_id": "p312-b30", - "global_id": 8418, - "bbox": [ - 248.48, - 408.49, - 274.65, - 416.37 - ], - "text": "c[2] 0", - "type": "text" - }, - { - "block_id": "p312-b31", - "global_id": 8419, - "bbox": [ - 248.48, - 359.4, - 280.98, - 367.29 - ], - "text": "c[1] 2", - "type": "text" - }, - { - "block_id": "p312-b32", - "global_id": 8420, - "bbox": [ - 105.06, - 552.66, - 112.24, - 560.26 - ], - "text": "(i)", - "type": "text" - }, - { - "block_id": "p312-b33", - "global_id": 8421, - "bbox": [ - 174.32, - 335.89, - 193.73, - 343.77 - ], - "text": "m 0", - "type": "text" - }, - { - "block_id": "p312-b34", - "global_id": 8422, - "bbox": [ - 328.97, - 533.73, - 341.21, - 541.41 - ], - "text": "c[n]", - "type": "text" - }, - { - "block_id": "p312-b35", - "global_id": 8423, - "bbox": [ - 337.41, - 559.35, - 341.21, - 566.95 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p312-b36", - "global_id": 8424, - "bbox": [ - 432.44, - 576.33, - 436.24, - 583.93 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p312-b37", - "global_id": 8425, - "bbox": [ - 337.41, - 543.82, - 341.21, - 551.42 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p312-b38", - "global_id": 8426, - "bbox": [ - 359.39, - 576.6, - 390.65, - 584.2 - ], - "text": "5\n2", - "type": "text" - }, - { - "block_id": "p312-b39", - "global_id": 8427, - "bbox": [ - 295.43, - 354.41, - 473.28, - 364.42 - ], - "text": "(j)\nc[1] 3", - "type": "text" - }, - { - "block_id": "p312-b40", - "global_id": 8428, - "bbox": [ - 434.45, - 406.61, - 473.28, - 414.49 - ], - "text": "c[2] 2", - "type": "text" - }, - { - "block_id": "p312-b41", - "global_id": 8429, - "bbox": [ - 434.45, - 456.04, - 466.95, - 463.92 - ], - "text": "c[3] 0", - "type": "text" - }, - { - "block_id": "p312-b42", - "global_id": 8430, - "bbox": [ - 295.43, - 403.81, - 304.29, - 411.41 - ], - "text": "(k)", - "type": "text" - }, - { - "block_id": "p312-b43", - "global_id": 8431, - "bbox": [ - 295.43, - 453.62, - 302.61, - 461.22 - ], - "text": "(l)", - "type": "text" - }, - { - "block_id": "p312-b44", - "global_id": 8432, - "bbox": [ - 295.43, - 533.81, - 306.41, - 541.41 - ], - "text": "(m)", - "type": "text" - }, - { - "block_id": "p312-b45", - "global_id": 8433, - "bbox": [ - 400.59, - 371.96, - 422.61, - 379.84 - ], - "text": "n = 1", - "type": "text" - }, - { - "block_id": "p312-b46", - "global_id": 8434, - "bbox": [ - 412.79, - 421.37, - 436.85, - 429.26 - ], - "text": "n 2", - "type": "text" - }, - { - "block_id": "p312-b47", - "global_id": 8435, - "bbox": [ - 423.24, - 471.72, - 447.3, - 479.61 - ], - "text": "n 3", - "type": "text" - }, - { - "block_id": "p312-b48", - "global_id": 8436, - "bbox": [ - 308.49, - 576.32, - 318.63, - 584.2 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p312-b49", - "global_id": 8437, - "bbox": [ - 157.12, - 359.41, - 229.62, - 367.29 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b50", - "global_id": 8438, - "bbox": [ - 127.51, - 374.88, - 196.86, - 382.48 - ], - "text": "1\n1\n1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b51", - "global_id": 8439, - "bbox": [ - 224.29, - 423.19, - 242.02, - 431.07 - ], - "text": "n 2", - "type": "text" - }, - { - "block_id": "p312-b52", - "global_id": 8440, - "bbox": [ - 157.24, - 408.49, - 229.73, - 416.37 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b53", - "global_id": 8441, - "bbox": [ - 138.54, - 423.97, - 207.88, - 431.57 - ], - "text": "1\n1\n1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b54", - "global_id": 8442, - "bbox": [ - 235.23, - 472.99, - 252.96, - 480.87 - ], - "text": "n 3", - "type": "text" - }, - { - "block_id": "p312-b55", - "global_id": 8443, - "bbox": [ - 157.24, - 458.29, - 229.73, - 466.17 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b56", - "global_id": 8444, - "bbox": [ - 149.48, - 473.77, - 218.83, - 481.37 - ], - "text": "1\n1\n1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b57", - "global_id": 8445, - "bbox": [ - 246.15, - 522.55, - 263.88, - 530.43 - ], - "text": "n 4", - "type": "text" - }, - { - "block_id": "p312-b58", - "global_id": 8446, - "bbox": [ - 157.24, - 507.84, - 229.73, - 515.72 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b59", - "global_id": 8447, - "bbox": [ - 160.38, - 523.32, - 229.72, - 530.92 - ], - "text": "1\n1\n1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b60", - "global_id": 8448, - "bbox": [ - 346.54, - 359.17, - 419.04, - 367.05 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b61", - "global_id": 8449, - "bbox": [ - 316.92, - 374.65, - 364.42, - 382.25 - ], - "text": "1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b62", - "global_id": 8450, - "bbox": [ - 346.54, - 458.29, - 419.04, - 466.17 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b63", - "global_id": 8451, - "bbox": [ - 327.86, - 473.93, - 342.59, - 481.53 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p312-b64", - "global_id": 8452, - "bbox": [ - 346.54, - 408.57, - 419.04, - 416.45 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b65", - "global_id": 8453, - "bbox": [ - 327.85, - 424.05, - 353.5, - 431.65 - ], - "text": "1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b66", - "global_id": 8454, - "bbox": [ - 256.92, - 572.19, - 274.65, - 580.07 - ], - "text": "n 5", - "type": "text" - }, - { - "block_id": "p312-b67", - "global_id": 8455, - "bbox": [ - 157.24, - 557.48, - 229.73, - 565.36 - ], - "text": "0\n1\n2\n3\n4\n21", - "type": "text" - }, - { - "block_id": "p312-b68", - "global_id": 8456, - "bbox": [ - 149.45, - 572.96, - 240.66, - 580.56 - ], - "text": "1\n1\n1\n1\n1\n1\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p312-b69", - "global_id": 8457, - "bbox": [ - 294.58, - 124.83, - 299.33, - 134.33 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p312-b70", - "global_id": 8458, - "bbox": [ - 103.16, - 601.88, - 342.69, - 611.11 - ], - "text": "Figure 3.23 Sliding-tape algorithm for discrete-time convolution.", - "type": "text" - } - ] - }, - { - "page_num": 313, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p313-b0", - "global_id": 8459, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n293", - "type": "text" - }, - { - "block_id": "p313-b1", - "global_id": 8460, - "bbox": [ - 133.57, - 97.81, - 470.6, - 109.76 - ], - "text": "DRILL 3.19\nSliding-Tape Method for the Convolution Sum", - "type": "text" - }, - { - "block_id": "p313-b2", - "global_id": 8461, - "bbox": [ - 133.57, - 118.47, - 510.15, - 140.81 - ], - "text": "Use the graphical procedure of Ex. 3.24 (sliding-tape technique) to show that x[n]∗g[n] = c[n]\nin Fig. 3.24. Verify the width property of convolution.", - "type": "text" - }, - { - "block_id": "p313-b3", - "global_id": 8462, - "bbox": [ - 182.7, - 207.0, - 190.69, - 214.2 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p313-b4", - "global_id": 8463, - "bbox": [ - 164.53, - 156.87, - 176.12, - 164.14 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p313-b5", - "global_id": 8464, - "bbox": [ - 154.93, - 181.7, - 158.53, - 188.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p313-b6", - "global_id": 8465, - "bbox": [ - 218.18, - 193.79, - 221.78, - 200.99 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p313-b7", - "global_id": 8466, - "bbox": [ - 154.93, - 165.14, - 158.53, - 180.66 - ], - "text": "2\n3", - "type": "text" - }, - { - "block_id": "p313-b8", - "global_id": 8467, - "bbox": [ - 155.22, - 194.31, - 312.5, - 201.61 - ], - "text": "1 2 3 4 5 6\n0\n1 2 3 4 5", - "type": "text" - }, - { - "block_id": "p313-b9", - "global_id": 8468, - "bbox": [ - 256.94, - 156.87, - 268.94, - 164.14 - ], - "text": "g[n]", - "type": "text" - }, - { - "block_id": "p313-b10", - "global_id": 8469, - "bbox": [ - 289.65, - 207.0, - 298.34, - 214.2 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p313-b11", - "global_id": 8470, - "bbox": [ - 264.45, - 181.7, - 268.05, - 188.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p313-b12", - "global_id": 8471, - "bbox": [ - 200.45, - 306.75, - 289.59, - 313.95 - ], - "text": "1 2 3 4 5 6 7 8 9 1011", - "type": "text" - }, - { - "block_id": "p313-b13", - "global_id": 8472, - "bbox": [ - 172.96, - 240.29, - 184.55, - 247.57 - ], - "text": "c[n]", - "type": "text" - }, - { - "block_id": "p313-b14", - "global_id": 8473, - "bbox": [ - 241.89, - 321.24, - 249.88, - 328.44 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p313-b15", - "global_id": 8474, - "bbox": [ - 188.1, - 294.08, - 191.7, - 301.28 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p313-b16", - "global_id": 8475, - "bbox": [ - 303.18, - 306.4, - 306.78, - 313.6 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p313-b17", - "global_id": 8476, - "bbox": [ - 321.31, - 194.0, - 324.91, - 201.2 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p313-b18", - "global_id": 8477, - "bbox": [ - 188.1, - 277.94, - 191.7, - 285.14 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p313-b19", - "global_id": 8478, - "bbox": [ - 188.8, - 306.75, - 192.4, - 313.95 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p313-b20", - "global_id": 8479, - "bbox": [ - 188.1, - 254.59, - 191.7, - 261.79 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p313-b21", - "global_id": 8480, - "bbox": [ - 188.1, - 231.23, - 191.7, - 247.35 - ], - "text": "9\n8", - "type": "text" - }, - { - "block_id": "p313-b22", - "global_id": 8481, - "bbox": [ - 350.19, - 320.5, - 478.72, - 329.74 - ], - "text": "Figure 3.24 Signals for Drill 3.19.", - "type": "text" - }, - { - "block_id": "p313-b23", - "global_id": 8482, - "bbox": [ - 102.51, - 376.2, - 484.2, - 402.11 - ], - "text": "EXAMPLE 3.25\nConvolution of Two Finite-Duration Signals Using\nMATLAB", - "type": "text" - }, - { - "block_id": "p313-b24", - "global_id": 8483, - "bbox": [ - 128.9, - 415.43, - 502.76, - 437.76 - ], - "text": "For the signals x[n] and g[n] depicted in Fig. 3.24, use MATLAB to compute and plot\nc[n] = x[n] ∗g[n].", - "type": "text" - }, - { - "block_id": "p313-b25", - "global_id": 8484, - "bbox": [ - 128.9, - 460.42, - 454.77, - 498.28 - ], - "text": ">>\nx = [0 1 2 3 2 1]; g = [1 1 1 1 1 1];\n>>\nn = (0:1:length(x)+length(g)-2);\n>>\nc = conv(x,g);\n>>\nclf; stem(n,c,’k’); xlabel(’n’); ylabel(’c[n]’);\naxis([-0.5 10.5 0 10]);", - "type": "text" - }, - { - "block_id": "p313-b26", - "global_id": 8485, - "bbox": [ - 153.56, - 593.05, - 362.5, - 614.23 - ], - "text": "0\n2\n4\n6\n8\n10\nn", - "type": "text" - }, - { - "block_id": "p313-b27", - "global_id": 8486, - "bbox": [ - 137.18, - 583.82, - 141.18, - 591.82 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p313-b28", - "global_id": 8487, - "bbox": [ - 137.18, - 547.82, - 141.18, - 555.82 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p313-b29", - "global_id": 8488, - "bbox": [ - 132.68, - 511.81, - 140.68, - 519.81 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p313-b30", - "global_id": 8489, - "bbox": [ - 120.7, - 543.87, - 129.5, - 558.04 - ], - "text": "c[n]", - "type": "text" - }, - { - "block_id": "p313-b31", - "global_id": 8490, - "bbox": [ - 119.94, - 620.92, - 283.79, - 630.16 - ], - "text": "Figure 3.25 Convolution result for Ex. 3.25.", - "type": "text" - } - ] - }, - { - "page_num": 314, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p314-b0", - "global_id": 8491, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "294\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p314-b1", - "global_id": 8492, - "bbox": [ - 101.84, - 86.52, - 262.04, - 98.48 - ], - "text": "3.8-2 Interconnected Systems", - "type": "text" - }, - { - "block_id": "p314-b2", - "global_id": 8493, - "bbox": [ - 101.84, - 104.61, - 490.41, - 198.26 - ], - "text": "As with continuous-time case, we can determine the impulse response of systems connected in\nparallel (Fig. 3.26a) and cascade (Figs. 3.26b, 3.26c). We can use arguments identical to those\nused for the continuous-time systems in Sec. 2.4-3 to show that if two LTID systems S1 and S2\nwith impulse responses h1[n] and h2[n], respectively, are connected in parallel, the composite\nparallel system impulse response is h1[n] + h2[n]. Similarly, if these systems are connected\nin cascade, the impulse response of the composite system is h1[n] ∗h2[n]. Moreover, because\nh1[n] ∗h2[n] = h2[n] ∗h1[n], linear systems commute. Their orders can be interchanged without\naffecting the composite system behavior.", - "type": "text" - }, - { - "block_id": "p314-b3", - "global_id": 8494, - "bbox": [ - 263.89, - 222.24, - 280.22, - 231.85 - ], - "text": "h1[n]", - "type": "text" - }, - { - "block_id": "p314-b4", - "global_id": 8495, - "bbox": [ - 225.87, - 293.07, - 234.75, - 301.07 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p314-b5", - "global_id": 8496, - "bbox": [ - 225.48, - 363.81, - 235.13, - 371.81 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p314-b6", - "global_id": 8497, - "bbox": [ - 225.87, - 412.16, - 234.75, - 420.16 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p314-b7", - "global_id": 8498, - "bbox": [ - 225.64, - 475.19, - 234.97, - 483.19 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p314-b8", - "global_id": 8499, - "bbox": [ - 225.87, - 535.67, - 234.75, - 543.67 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p314-b9", - "global_id": 8500, - "bbox": [ - 137.23, - 239.9, - 150.56, - 248.0 - ], - "text": "d[n]", - "type": "text" - }, - { - "block_id": "p314-b10", - "global_id": 8501, - "bbox": [ - 169.41, - 223.38, - 182.74, - 231.48 - ], - "text": "d[n]", - "type": "text" - }, - { - "block_id": "p314-b11", - "global_id": 8502, - "bbox": [ - 169.41, - 256.59, - 182.74, - 264.7 - ], - "text": "d[n]", - "type": "text" - }, - { - "block_id": "p314-b12", - "global_id": 8503, - "bbox": [ - 138.59, - 330.34, - 151.92, - 338.44 - ], - "text": "d[n]", - "type": "text" - }, - { - "block_id": "p314-b13", - "global_id": 8504, - "bbox": [ - 136.59, - 384.06, - 149.92, - 392.16 - ], - "text": "d[n]", - "type": "text" - }, - { - "block_id": "p314-b14", - "global_id": 8505, - "bbox": [ - 319.99, - 239.48, - 390.71, - 249.3 - ], - "text": "hp[n] h1[n] h2[n]", - "type": "text" - }, - { - "block_id": "p314-b15", - "global_id": 8506, - "bbox": [ - 324.66, - 330.87, - 365.32, - 340.48 - ], - "text": "h1[n] * h2[n]", - "type": "text" - }, - { - "block_id": "p314-b16", - "global_id": 8507, - "bbox": [ - 323.98, - 385.07, - 364.63, - 394.68 - ], - "text": "h2[n] * h1[n]", - "type": "text" - }, - { - "block_id": "p314-b17", - "global_id": 8508, - "bbox": [ - 263.89, - 271.28, - 280.22, - 280.89 - ], - "text": "h2[n]", - "type": "text" - }, - { - "block_id": "p314-b18", - "global_id": 8509, - "bbox": [ - 229.06, - 384.08, - 245.39, - 393.69 - ], - "text": "h2[n]", - "type": "text" - }, - { - "block_id": "p314-b19", - "global_id": 8510, - "bbox": [ - 229.06, - 330.36, - 245.39, - 339.97 - ], - "text": "h1[n]", - "type": "text" - }, - { - "block_id": "p314-b20", - "global_id": 8511, - "bbox": [ - 136.81, - 443.04, - 149.69, - 451.12 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p314-b21", - "global_id": 8512, - "bbox": [ - 136.81, - 505.2, - 149.69, - 513.28 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p314-b22", - "global_id": 8513, - "bbox": [ - 230.79, - 443.04, - 243.67, - 451.12 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p314-b23", - "global_id": 8514, - "bbox": [ - 223.07, - 230.91, - 230.07, - 240.52 - ], - "text": "S1", - "type": "text" - }, - { - "block_id": "p314-b24", - "global_id": 8515, - "bbox": [ - 223.07, - 264.3, - 230.07, - 273.91 - ], - "text": "S2", - "type": "text" - }, - { - "block_id": "p314-b26", - "global_id": 8516, - "bbox": [ - 310.7, - 217.66, - 317.7, - 227.2 - ], - "text": "Sp", - "type": "text" - }, - { - "block_id": "p314-b27", - "global_id": 8517, - "bbox": [ - 182.07, - 336.33, - 292.57, - 345.94 - ], - "text": "S1\nS2", - "type": "text" - }, - { - "block_id": "p314-b28", - "global_id": 8518, - "bbox": [ - 321.45, - 314.07, - 328.11, - 323.62 - ], - "text": "Sc", - "type": "text" - }, - { - "block_id": "p314-b29", - "global_id": 8519, - "bbox": [ - 182.07, - 390.59, - 293.07, - 400.2 - ], - "text": "S2\nS1", - "type": "text" - }, - { - "block_id": "p314-b30", - "global_id": 8520, - "bbox": [ - 183.57, - 450.55, - 187.57, - 458.55 - ], - "text": "S", - "type": "text" - }, - { - "block_id": "p314-b31", - "global_id": 8521, - "bbox": [ - 287.57, - 512.71, - 291.57, - 520.71 - ], - "text": "S", - "type": "text" - }, - { - "block_id": "p314-b32", - "global_id": 8522, - "bbox": [ - 325.62, - 497.55, - 354.18, - 512.75 - ], - "text": "y[k]\n\nk", - "type": "text" - }, - { - "block_id": "p314-b33", - "global_id": 8523, - "bbox": [ - 331.7, - 492.32, - 334.7, - 498.32 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p314-b34", - "global_id": 8524, - "bbox": [ - 327.6, - 435.39, - 356.14, - 450.59 - ], - "text": "y[k]\n\nk", - "type": "text" - }, - { - "block_id": "p314-b35", - "global_id": 8525, - "bbox": [ - 333.68, - 430.16, - 336.68, - 436.16 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p314-b36", - "global_id": 8526, - "bbox": [ - 224.53, - 497.55, - 253.16, - 512.75 - ], - "text": "x[k]\n\nk", - "type": "text" - }, - { - "block_id": "p314-b37", - "global_id": 8527, - "bbox": [ - 230.61, - 492.32, - 233.61, - 498.32 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p314-b38", - "global_id": 8528, - "bbox": [ - 175.05, - 512.42, - 194.71, - 527.62 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p314-b39", - "global_id": 8529, - "bbox": [ - 181.12, - 507.18, - 184.12, - 513.18 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p314-b40", - "global_id": 8530, - "bbox": [ - 279.68, - 451.32, - 299.34, - 466.51 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p314-b41", - "global_id": 8531, - "bbox": [ - 285.76, - 446.08, - 288.76, - 452.08 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p314-b42", - "global_id": 8532, - "bbox": [ - 125.76, - 550.36, - 261.98, - 559.6 - ], - "text": "Figure 3.26 Interconnected systems.", - "type": "text" - }, - { - "block_id": "p314-b43", - "global_id": 8533, - "bbox": [ - 102.14, - 584.76, - 198.26, - 596.88 - ], - "text": "INVERSE SYSTEMS", - "type": "text" - }, - { - "block_id": "p314-b44", - "global_id": 8534, - "bbox": [ - 101.84, - 600.5, - 490.39, - 634.79 - ], - "text": "If the two systems in cascade are the inverse of each other, with impulse responses h[n] and hi[n],\nrespectively, then the impulse response of the cascade of these systems is h[n] ∗hi[n]. But, the\ncascade of a system with its inverse is an identity system, whose output is the same as the input.", - "type": "text" - } - ] - }, - { - "page_num": 315, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p315-b0", - "global_id": 8535, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n295", - "type": "text" - }, - { - "block_id": "p315-b1", - "global_id": 8536, - "bbox": [ - 127.59, - 85.4, - 434.88, - 95.78 - ], - "text": "Hence, the unit impulse response of an identity system is δ[n]. Consequently,", - "type": "text" - }, - { - "block_id": "p315-b2", - "global_id": 8537, - "bbox": [ - 285.81, - 109.51, - 357.9, - 120.59 - ], - "text": "h[n] ∗hi[n] = δ[n]", - "type": "text" - }, - { - "block_id": "p315-b3", - "global_id": 8538, - "bbox": [ - 127.59, - 134.04, - 516.13, - 155.95 - ], - "text": "As an example, we show that an accumulator system and a backward difference system are the\ninverse of each other. An accumulator system is specified by†", - "type": "text" - }, - { - "block_id": "p315-b4", - "global_id": 8539, - "bbox": [ - 288.91, - 177.62, - 314.76, - 187.89 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p315-b5", - "global_id": 8540, - "bbox": [ - 320.37, - 167.66, - 334.47, - 178.11 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p315-b6", - "global_id": 8541, - "bbox": [ - 316.81, - 191.9, - 338.01, - 199.1 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p315-b7", - "global_id": 8542, - "bbox": [ - 339.14, - 177.62, - 516.12, - 187.99 - ], - "text": "x[k]\n(3.35)", - "type": "text" - }, - { - "block_id": "p315-b8", - "global_id": 8543, - "bbox": [ - 127.59, - 211.55, - 316.81, - 221.51 - ], - "text": "The backward difference system is specified by", - "type": "text" - }, - { - "block_id": "p315-b9", - "global_id": 8544, - "bbox": [ - 278.48, - 235.24, - 516.13, - 245.62 - ], - "text": "y[n] = x[n] −x[n −1]\n(3.36)", - "type": "text" - }, - { - "block_id": "p315-b10", - "global_id": 8545, - "bbox": [ - 127.6, - 259.35, - 434.22, - 270.5 - ], - "text": "From Eq. (3.35), we find hacc[n], the impulse response of the accumulator, as", - "type": "text" - }, - { - "block_id": "p315-b11", - "global_id": 8546, - "bbox": [ - 269.39, - 291.39, - 305.59, - 302.54 - ], - "text": "hacc[n] =", - "type": "text" - }, - { - "block_id": "p315-b12", - "global_id": 8547, - "bbox": [ - 311.19, - 281.44, - 325.29, - 291.89 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p315-b13", - "global_id": 8548, - "bbox": [ - 307.64, - 305.68, - 328.84, - 312.88 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p315-b14", - "global_id": 8549, - "bbox": [ - 329.96, - 291.39, - 374.33, - 301.67 - ], - "text": "δ[k] = u[n]", - "type": "text" - }, - { - "block_id": "p315-b15", - "global_id": 8550, - "bbox": [ - 127.59, - 324.91, - 516.15, - 347.24 - ], - "text": "Similarly, from Eq. (3.36), hbdf[n], the impulse response of the backward difference system is\ngiven by", - "type": "text" - }, - { - "block_id": "p315-b16", - "global_id": 8551, - "bbox": [ - 273.07, - 352.11, - 370.63, - 363.26 - ], - "text": "hbdf[n] = δ[n] −δ[n −1]", - "type": "text" - }, - { - "block_id": "p315-b17", - "global_id": 8552, - "bbox": [ - 127.6, - 373.64, - 200.5, - 383.6 - ], - "text": "We can verify that", - "type": "text" - }, - { - "block_id": "p315-b18", - "global_id": 8553, - "bbox": [ - 201.25, - 397.33, - 442.48, - 408.79 - ], - "text": "hacc ∗hbdf = u[n] ∗{δ[n] −δ[n −1]} = u[n] −u[n −1] = δ[n]", - "type": "text" - }, - { - "block_id": "p315-b19", - "global_id": 8554, - "bbox": [ - 127.6, - 421.86, - 516.11, - 455.72 - ], - "text": "Roughly speaking, a discrete-time accumulator is analogous to a continuous-time integrator, and\na backward difference system is analogous to a differentiator. We have already encountered\nexamples of these systems in Exs. 3.8 and 3.9 (digital differentiator and integrator).", - "type": "text" - }, - { - "block_id": "p315-b20", - "global_id": 8555, - "bbox": [ - 127.89, - 461.79, - 265.61, - 483.68 - ], - "text": "SYSTEM RESPONSE TO %n", - "type": "text" - }, - { - "block_id": "p315-b21", - "global_id": 8556, - "bbox": [ - 261.13, - 470.75, - 309.05, - 485.83 - ], - "text": "k=−∞x[k]", - "type": "text" - }, - { - "block_id": "p315-b22", - "global_id": 8557, - "bbox": [ - 127.59, - 487.3, - 516.15, - 557.44 - ], - "text": "Figure 3.26d shows a cascade of two LTID systems: a system S with impulse response h[n],\nfollowed by an accumulator. Figure 3.26e shows a cascade of the same two systems in reverse\norder: an accumulator followed by S. In Fig. 3.26d, if the input x[n] to S results in the output\ny[n], then the output of the system in Fig. 3.26d is the %y[k]. In Fig. 3.26e, the output of the\naccumulator is the sum %x[k]. Because the output of the system in Fig. 3.26e is identical to that\nof system Fig. 3.26d, it follows that", - "type": "text" - }, - { - "block_id": "p315-b23", - "global_id": 8558, - "bbox": [ - 221.91, - 579.11, - 322.76, - 589.48 - ], - "text": "if x[n] \r⇒y[n],\nthen", - "type": "text" - }, - { - "block_id": "p315-b24", - "global_id": 8559, - "bbox": [ - 328.81, - 569.15, - 342.9, - 579.61 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p315-b25", - "global_id": 8560, - "bbox": [ - 325.25, - 593.39, - 346.46, - 600.59 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p315-b26", - "global_id": 8561, - "bbox": [ - 347.57, - 579.11, - 381.8, - 589.38 - ], - "text": "x[k] \r⇒", - "type": "text" - }, - { - "block_id": "p315-b27", - "global_id": 8562, - "bbox": [ - 387.4, - 569.15, - 401.5, - 579.61 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p315-b28", - "global_id": 8563, - "bbox": [ - 383.85, - 593.39, - 405.06, - 600.59 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p315-b29", - "global_id": 8564, - "bbox": [ - 406.17, - 579.11, - 421.82, - 589.38 - ], - "text": "y[k]", - "type": "text" - }, - { - "block_id": "p315-b30", - "global_id": 8565, - "bbox": [ - 127.59, - 621.19, - 456.52, - 633.41 - ], - "text": "† Equations (3.35) and (3.36) are identical to Eqs. (3.10) and (3.8), respectively, with T = 1.", - "type": "text" - } - ] - }, - { - "page_num": 316, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p316-b0", - "global_id": 8566, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "296\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p316-b1", - "global_id": 8567, - "bbox": [ - 101.84, - 85.4, - 490.4, - 107.74 - ], - "text": "If we let x[n] = δ[n] and y[n] = h[n], we find that g[n], the unit step response of an LTID system\nwith impulse response h[n], is given by", - "type": "text" - }, - { - "block_id": "p316-b2", - "global_id": 8568, - "bbox": [ - 262.62, - 128.2, - 289.04, - 138.47 - ], - "text": "g[n] =", - "type": "text" - }, - { - "block_id": "p316-b3", - "global_id": 8569, - "bbox": [ - 294.64, - 118.24, - 308.74, - 128.69 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p316-b4", - "global_id": 8570, - "bbox": [ - 291.09, - 142.49, - 312.29, - 149.68 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p316-b5", - "global_id": 8571, - "bbox": [ - 313.41, - 128.2, - 490.38, - 138.57 - ], - "text": "h[k]\n(3.37)", - "type": "text" - }, - { - "block_id": "p316-b6", - "global_id": 8572, - "bbox": [ - 101.85, - 160.93, - 310.98, - 170.89 - ], - "text": "The reader can readily prove the inverse relationship", - "type": "text" - }, - { - "block_id": "p316-b7", - "global_id": 8573, - "bbox": [ - 251.93, - 183.42, - 340.28, - 193.8 - ], - "text": "h[n] = g[n] −g[n −1]", - "type": "text" - }, - { - "block_id": "p316-b8", - "global_id": 8574, - "bbox": [ - 102.14, - 209.02, - 468.18, - 235.1 - ], - "text": "A VERY SPECIAL FUNCTION FOR LTID SYSTEMS: THE EVERLASTING\nEXPONENTIAL zn", - "type": "text" - }, - { - "block_id": "p316-b9", - "global_id": 8575, - "bbox": [ - 101.84, - 239.12, - 490.4, - 296.91 - ], - "text": "In Sec. 2.4-4, we showed that there exists one signal for which the response of an LTIC system\nis the same as the input within a multiplicative constant. The response of an LTIC system to an\neverlasting exponential input est is H(s)est, where H(s) is the system transfer function. We now\nshow that for an LTID system, the same role is played by an everlasting exponential zn. The system\nresponse y[n] in this case is given by", - "type": "text" - }, - { - "block_id": "p316-b10", - "global_id": 8576, - "bbox": [ - 248.78, - 307.94, - 308.94, - 319.71 - ], - "text": "y[n] = h[n] ∗zn", - "type": "text" - }, - { - "block_id": "p316-b11", - "global_id": 8577, - "bbox": [ - 266.87, - 333.94, - 274.64, - 343.9 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p316-b12", - "global_id": 8578, - "bbox": [ - 281.16, - 323.76, - 295.26, - 334.43 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p316-b13", - "global_id": 8579, - "bbox": [ - 276.69, - 347.78, - 299.71, - 354.98 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p316-b14", - "global_id": 8580, - "bbox": [ - 300.83, - 332.21, - 337.47, - 344.21 - ], - "text": "h[m]zn−m", - "type": "text" - }, - { - "block_id": "p316-b15", - "global_id": 8581, - "bbox": [ - 266.87, - 366.49, - 284.05, - 378.27 - ], - "text": "= zn", - "type": "text" - }, - { - "block_id": "p316-b16", - "global_id": 8582, - "bbox": [ - 290.12, - 357.82, - 304.22, - 368.49 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p316-b17", - "global_id": 8583, - "bbox": [ - 285.65, - 381.84, - 308.67, - 389.03 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p316-b18", - "global_id": 8584, - "bbox": [ - 309.8, - 366.27, - 342.96, - 378.27 - ], - "text": "h[m]z−m", - "type": "text" - }, - { - "block_id": "p316-b19", - "global_id": 8585, - "bbox": [ - 101.84, - 399.86, - 490.39, - 422.2 - ], - "text": "For causal h[n], the limits on the sum on the right-hand side would range from 0 to ∞. In any case,\nthis sum is a function of z. Assuming that this sum converges, let us denote it by H[z]. Thus,", - "type": "text" - }, - { - "block_id": "p316-b20", - "global_id": 8586, - "bbox": [ - 269.16, - 433.23, - 490.38, - 445.1 - ], - "text": "y[n] = H[z]zn\n(3.38)", - "type": "text" - }, - { - "block_id": "p316-b21", - "global_id": 8587, - "bbox": [ - 101.85, - 458.05, - 126.18, - 468.01 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p316-b22", - "global_id": 8588, - "bbox": [ - 252.21, - 477.14, - 280.18, - 487.42 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p316-b23", - "global_id": 8589, - "bbox": [ - 286.7, - 466.97, - 300.79, - 477.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p316-b24", - "global_id": 8590, - "bbox": [ - 282.23, - 490.99, - 305.25, - 498.19 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p316-b25", - "global_id": 8591, - "bbox": [ - 306.37, - 475.41, - 490.38, - 487.52 - ], - "text": "h[m]z−m\n(3.39)", - "type": "text" - }, - { - "block_id": "p316-b26", - "global_id": 8592, - "bbox": [ - 101.85, - 506.34, - 490.38, - 540.32 - ], - "text": "Equation (3.38) is valid only for values of z for which the sum on the right-hand side of Eq. (3.39)\nexists (converges). Note that H[z] is a constant for a given z. Thus, the input and the output are the\nsame (within a multiplicative constant) for the everlasting exponential input zn.", - "type": "text" - }, - { - "block_id": "p316-b27", - "global_id": 8593, - "bbox": [ - 101.85, - 541.89, - 490.4, - 564.23 - ], - "text": "H[z], which is called the transfer function of the system, is a function of the complex variable\nz. An alternate definition of the transfer function H[z] of an LTID system from Eq. (3.38) is", - "type": "text" - }, - { - "block_id": "p316-b28", - "global_id": 8594, - "bbox": [ - 205.24, - 575.83, - 289.41, - 592.78 - ], - "text": "H[z] = output signal", - "type": "text" - }, - { - "block_id": "p316-b29", - "global_id": 8595, - "bbox": [ - 240.14, - 589.89, - 286.91, - 599.85 - ], - "text": "input signal", - "type": "text" - }, - { - "block_id": "p316-b31", - "global_id": 8596, - "bbox": [ - 293.84, - 582.81, - 490.38, - 601.86 - ], - "text": "input=everlasting exponential zn\n(3.40)", - "type": "text" - }, - { - "block_id": "p316-b32", - "global_id": 8597, - "bbox": [ - 101.85, - 612.86, - 490.38, - 634.79 - ], - "text": "The transfer function is defined for, and is meaningful to, LTID systems only. It does not exist for\nnonlinear or time-varying systems in general.", - "type": "text" - } - ] - }, - { - "page_num": 317, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p317-b0", - "global_id": 8598, - "bbox": [ - 212.35, - 62.89, - 516.13, - 71.98 - ], - "text": "3.8\nSystem Response to External Input: The Zero-State Response\n297", - "type": "text" - }, - { - "block_id": "p317-b1", - "global_id": 8599, - "bbox": [ - 127.59, - 85.82, - 516.15, - 107.74 - ], - "text": "We repeat again that in this discussion we are talking of the everlasting exponential, which\nstarts at n = −∞, not the causal exponential znu[n], which starts at n = 0.", - "type": "text" - }, - { - "block_id": "p317-b2", - "global_id": 8600, - "bbox": [ - 145.52, - 109.73, - 421.18, - 119.69 - ], - "text": "For a system specified by Eq. (3.20), the transfer function is given by", - "type": "text" - }, - { - "block_id": "p317-b3", - "global_id": 8601, - "bbox": [ - 296.8, - 127.58, - 345.18, - 144.84 - ], - "text": "H[z] = P[z]", - "type": "text" - }, - { - "block_id": "p317-b4", - "global_id": 8602, - "bbox": [ - 328.01, - 134.97, - 516.13, - 151.91 - ], - "text": "Q[z]\n(3.41)", - "type": "text" - }, - { - "block_id": "p317-b5", - "global_id": 8603, - "bbox": [ - 127.59, - 158.62, - 516.13, - 181.95 - ], - "text": "This follows readily by considering an everlasting input x[n] = zn. According to Eq. (3.40), the\noutput is y[n] = H[z]zn. Substitution of this x[n] and y[n] in Eq. (3.20) yields", - "type": "text" - }, - { - "block_id": "p317-b6", - "global_id": 8604, - "bbox": [ - 275.16, - 190.49, - 368.06, - 202.27 - ], - "text": "H[z]{Q[E]zn} = P[E]zn", - "type": "text" - }, - { - "block_id": "p317-b7", - "global_id": 8605, - "bbox": [ - 127.59, - 212.82, - 168.65, - 222.79 - ], - "text": "Moreover,", - "type": "text" - }, - { - "block_id": "p317-b8", - "global_id": 8606, - "bbox": [ - 285.07, - 220.25, - 358.16, - 234.64 - ], - "text": "Ekzn = zn+k = zkzn", - "type": "text" - }, - { - "block_id": "p317-b9", - "global_id": 8607, - "bbox": [ - 127.59, - 242.96, - 155.51, - 252.92 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p317-b10", - "global_id": 8608, - "bbox": [ - 230.41, - 250.38, - 412.82, - 264.87 - ], - "text": "P[E]zn = P[z]zn\nand\nQ[E]zn = Q[z]zn", - "type": "text" - }, - { - "block_id": "p317-b11", - "global_id": 8609, - "bbox": [ - 127.59, - 273.09, - 184.2, - 283.05 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p317-b12", - "global_id": 8610, - "bbox": [ - 296.8, - 282.47, - 345.18, - 299.73 - ], - "text": "H[z] = P[z]", - "type": "text" - }, - { - "block_id": "p317-b13", - "global_id": 8611, - "bbox": [ - 328.01, - 296.52, - 345.72, - 306.8 - ], - "text": "Q[z]", - "type": "text" - }, - { - "block_id": "p317-b14", - "global_id": 8612, - "bbox": [ - 133.57, - 344.85, - 374.96, - 356.81 - ], - "text": "DRILL 3.20\nDT System Transfer Function", - "type": "text" - }, - { - "block_id": "p317-b15", - "global_id": 8613, - "bbox": [ - 133.57, - 365.93, - 510.15, - 399.8 - ], - "text": "Show that the transfer function of the digital differentiator in Ex. 3.8 (big shaded block in\nFig. 3.16b) is given by H[z] = (z −1)/Tz, and the transfer function of an unit delay, specified\nby y[n] = x[n −1], is given by 1/z.", - "type": "text" - }, - { - "block_id": "p317-b16", - "global_id": 8614, - "bbox": [ - 127.59, - 435.17, - 240.66, - 447.13 - ], - "text": "3.8-3 Total Response", - "type": "text" - }, - { - "block_id": "p317-b17", - "global_id": 8615, - "bbox": [ - 127.59, - 453.26, - 516.1, - 475.17 - ], - "text": "The total response of an LTID system can be expressed as a sum of the zero-input and zero-state\nresponses:", - "type": "text" - }, - { - "block_id": "p317-b18", - "global_id": 8616, - "bbox": [ - 247.85, - 486.27, - 312.73, - 496.65 - ], - "text": "total response =", - "type": "text" - }, - { - "block_id": "p317-b19", - "global_id": 8617, - "bbox": [ - 314.77, - 476.32, - 328.87, - 486.77 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p317-b20", - "global_id": 8618, - "bbox": [ - 316.39, - 500.66, - 327.25, - 507.92 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p317-b21", - "global_id": 8619, - "bbox": [ - 329.98, - 484.77, - 346.93, - 497.35 - ], - "text": "cjγ n", - "type": "text" - }, - { - "block_id": "p317-b22", - "global_id": 8620, - "bbox": [ - 341.85, - 491.56, - 343.79, - 498.54 - ], - "text": "j", - "type": "text" - }, - { - "block_id": "p317-b24", - "global_id": 8621, - "bbox": [ - 326.29, - 517.21, - 335.92, - 523.19 - ], - "text": "ZIR", - "type": "text" - }, - { - "block_id": "p317-b25", - "global_id": 8622, - "bbox": [ - 348.52, - 486.27, - 398.37, - 510.21 - ], - "text": "+x[n] ∗h[n]\n\n\n\nZSR", - "type": "text" - }, - { - "block_id": "p317-b26", - "global_id": 8623, - "bbox": [ - 127.59, - 529.3, - 516.15, - 622.95 - ], - "text": "In this expression, the zero-input response should be appropriately modified for the case of\nrepeated roots. We have developed procedures to determine these two components. From the\nsystem equation, we find the characteristic roots and characteristic modes. The zero-input response\nis a linear combination of the characteristic modes. From the system equation, we also determine\nh[n], the impulse response, as discussed in Sec. 3.7. Knowing h[n] and the input x[n], we find the\nzero-state response as the convolution of x[n] and h[n]. The arbitrary constants c1,c2,. . .,cn in the\nzero-input response are determined from the n initial conditions. For the system described by the\nequation", - "type": "text" - }, - { - "block_id": "p317-b27", - "global_id": 8624, - "bbox": [ - 231.75, - 624.53, - 411.95, - 634.91 - ], - "text": "y[n + 2] −0.6y[n + 1] −0.16y[n] = 5x[n + 2]", - "type": "text" - } - ] - }, - { - "page_num": 318, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p318-b0", - "global_id": 8625, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "298\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p318-b1", - "global_id": 8626, - "bbox": [ - 101.84, - 85.07, - 490.42, - 120.58 - ], - "text": "with initial conditions y[−1] = 0,y[−2] = 25/4 and input x[n] = (4)−nu[n], we have determined\nthe two components of the response in Exs. 3.13 and 3.21, respectively. From the results in these\nexamples, the total response for n ≥0 is", - "type": "text" - }, - { - "block_id": "p318-b2", - "global_id": 8627, - "bbox": [ - 121.8, - 129.87, - 279.5, - 144.36 - ], - "text": "total response = 0.2(−0.2)n + 0.8(0.8)n", - "type": "text" - }, - { - "block_id": "p318-b3", - "global_id": 8628, - "bbox": [ - 188.72, - 139.73, - 280.02, - 158.1 - ], - "text": "ZIR", - "type": "text" - }, - { - "block_id": "p318-b4", - "global_id": 8629, - "bbox": [ - 281.11, - 129.87, - 445.85, - 144.36 - ], - "text": "+0.444(−0.2)n + 5.81(0.8)n −1.26(4)−n", - "type": "text" - }, - { - "block_id": "p318-b5", - "global_id": 8630, - "bbox": [ - 289.99, - 139.73, - 446.37, - 158.1 - ], - "text": "ZSR", - "type": "text" - }, - { - "block_id": "p318-b6", - "global_id": 8631, - "bbox": [ - 466.32, - 134.4, - 490.38, - 144.36 - ], - "text": "(3.42)", - "type": "text" - }, - { - "block_id": "p318-b7", - "global_id": 8632, - "bbox": [ - 102.14, - 171.31, - 288.72, - 183.44 - ], - "text": "NATURAL AND FORCED RESPONSE", - "type": "text" - }, - { - "block_id": "p318-b8", - "global_id": 8633, - "bbox": [ - 101.84, - 183.85, - 490.4, - 257.2 - ], - "text": "The characteristic modes of this system are (−0.2)n and (0.8)n. The zero-input response is made\nup of characteristic modes exclusively, as expected, but the characteristic modes also appear in\nthe zero-state response. When all the characteristic mode terms in the total response are lumped\ntogether, the resulting component is the natural response. The remaining part of the total response\nthat is made up of noncharacteristic modes is the forced response. For the present case, Eq. (3.42)\nyields", - "type": "text" - }, - { - "block_id": "p318-b9", - "global_id": 8634, - "bbox": [ - 159.72, - 257.46, - 332.36, - 271.95 - ], - "text": "total response = 0.644(−0.2)n + 6.61(0.8)n", - "type": "text" - }, - { - "block_id": "p318-b10", - "global_id": 8635, - "bbox": [ - 226.65, - 267.32, - 332.88, - 285.7 - ], - "text": "natural response", - "type": "text" - }, - { - "block_id": "p318-b11", - "global_id": 8636, - "bbox": [ - 333.99, - 259.85, - 389.39, - 271.95 - ], - "text": "+−1.26(4)−n", - "type": "text" - }, - { - "block_id": "p318-b12", - "global_id": 8637, - "bbox": [ - 342.86, - 267.32, - 389.92, - 285.7 - ], - "text": "forced response", - "type": "text" - }, - { - "block_id": "p318-b13", - "global_id": 8638, - "bbox": [ - 410.94, - 261.58, - 432.52, - 271.95 - ], - "text": "n ≥0", - "type": "text" - }, - { - "block_id": "p318-b14", - "global_id": 8639, - "bbox": [ - 101.85, - 295.71, - 490.4, - 413.27 - ], - "text": "Just like differential equations, the classical solution to difference equations includes the\nnatural and forced responses, a decomposition that lacks the engineering intuition and utility\nafforded by the zero-input and zero-state responses. The classical approach cannot separate\nthe responses arising from internal conditions and external input. While the natural and forced\nsolutions can be obtained from the zero-input and zero-state responses, the converse is not true.\nFurther, the classical method is unable to express the system response to an input x[n] as an explicit\nfunction of x[n]. In fact, the classical method is restricted to a certain class of inputs and cannot\nhandle arbitrary inputs as can the method to determine the zero-state response. For these (and\nother) reasons, we do not further detail the classical approach and its direct calculation of the\nforced and natural responses.", - "type": "text" - }, - { - "block_id": "p318-b15", - "global_id": 8640, - "bbox": [ - 102.2, - 444.42, - 250.01, - 458.37 - ], - "text": "3.9 SYSTEM STABILITY", - "type": "text" - }, - { - "block_id": "p318-b16", - "global_id": 8641, - "bbox": [ - 101.84, - 464.36, - 490.4, - 510.18 - ], - "text": "The concepts and criteria for the BIBO (external) stability and internal (asymptotic) stability\nfor discrete-time systems are identical to those corresponding to continuous-time systems. The\ncomments in Sec. 2.5 for LTIC systems concerning the distinction between external and internal\nstability are also valid for LTID systems. Let us begin with external (BIBO) stability.", - "type": "text" - }, - { - "block_id": "p318-b17", - "global_id": 8642, - "bbox": [ - 101.84, - 536.8, - 269.0, - 548.75 - ], - "text": "3.9-1 External (BIBO) Stability", - "type": "text" - }, - { - "block_id": "p318-b18", - "global_id": 8643, - "bbox": [ - 101.84, - 554.89, - 144.73, - 564.85 - ], - "text": "Recall that", - "type": "text" - }, - { - "block_id": "p318-b19", - "global_id": 8644, - "bbox": [ - 217.21, - 575.29, - 295.89, - 585.57 - ], - "text": "y[n] = h[n] ∗x[n] =", - "type": "text" - }, - { - "block_id": "p318-b20", - "global_id": 8645, - "bbox": [ - 302.41, - 565.12, - 316.51, - 575.78 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p318-b21", - "global_id": 8646, - "bbox": [ - 297.94, - 589.14, - 320.96, - 596.33 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p318-b22", - "global_id": 8647, - "bbox": [ - 322.08, - 575.29, - 375.01, - 585.57 - ], - "text": "h[m]x[n −m]", - "type": "text" - }, - { - "block_id": "p318-b23", - "global_id": 8648, - "bbox": [ - 101.84, - 605.47, - 116.22, - 615.43 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p318-b24", - "global_id": 8649, - "bbox": [ - 186.94, - 626.37, - 218.5, - 636.65 - ], - "text": "|y[n]| =", - "type": "text" - }, - { - "block_id": "p318-b26", - "global_id": 8650, - "bbox": [ - 228.25, - 616.2, - 242.35, - 626.87 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p318-b27", - "global_id": 8651, - "bbox": [ - 223.78, - 640.22, - 246.8, - 647.42 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p318-b28", - "global_id": 8652, - "bbox": [ - 247.92, - 626.37, - 300.85, - 636.65 - ], - "text": "h[m]x[n −m]", - "type": "text" - }, - { - "block_id": "p318-b29", - "global_id": 8653, - "bbox": [ - 300.85, - 608.94, - 313.67, - 642.82 - ], - "text": "≤", - "type": "text" - }, - { - "block_id": "p318-b30", - "global_id": 8654, - "bbox": [ - 320.17, - 616.2, - 334.27, - 626.87 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p318-b31", - "global_id": 8655, - "bbox": [ - 315.71, - 640.22, - 338.73, - 647.42 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p318-b32", - "global_id": 8656, - "bbox": [ - 339.85, - 626.37, - 405.29, - 636.65 - ], - "text": "|h[m]||x[n −m]|", - "type": "text" - } - ] - }, - { - "page_num": 319, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p319-b0", - "global_id": 8657, - "bbox": [ - 389.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.9\nSystem Stability\n299", - "type": "text" - }, - { - "block_id": "p319-b1", - "global_id": 8658, - "bbox": [ - 127.59, - 85.4, - 328.18, - 96.86 - ], - "text": "If x[n] is bounded, then |x[n −m]| < K1 < ∞, and", - "type": "text" - }, - { - "block_id": "p319-b2", - "global_id": 8659, - "bbox": [ - 274.99, - 121.38, - 318.47, - 132.53 - ], - "text": "|y[n]| ≤K1", - "type": "text" - }, - { - "block_id": "p319-b3", - "global_id": 8660, - "bbox": [ - 324.55, - 111.21, - 338.65, - 121.88 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p319-b4", - "global_id": 8661, - "bbox": [ - 320.08, - 135.23, - 343.1, - 142.43 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p319-b5", - "global_id": 8662, - "bbox": [ - 344.22, - 121.38, - 368.74, - 131.66 - ], - "text": "|h[m]|", - "type": "text" - }, - { - "block_id": "p319-b6", - "global_id": 8663, - "bbox": [ - 127.6, - 158.88, - 491.48, - 168.84 - ], - "text": "Clearly the output is bounded if the summation on the right-hand side is bounded; that is, if", - "type": "text" - }, - { - "block_id": "p319-b7", - "global_id": 8664, - "bbox": [ - 280.85, - 184.52, - 294.94, - 195.2 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p319-b8", - "global_id": 8665, - "bbox": [ - 277.15, - 208.55, - 298.62, - 215.74 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p319-b9", - "global_id": 8666, - "bbox": [ - 299.74, - 194.7, - 516.13, - 206.15 - ], - "text": "|h[n]| < K2 < ∞\n(3.43)", - "type": "text" - }, - { - "block_id": "p319-b10", - "global_id": 8667, - "bbox": [ - 127.59, - 232.2, - 516.13, - 266.07 - ], - "text": "This is a sufficient condition for BIBO stability. We can show that this is also a necessary\ncondition (see Prob. 3.9-1). Therefore, if the impulse response h[n] of an LTID system is absolutely\nsummable, the system is (BIBO) stable. Otherwise it is unstable.", - "type": "text" - }, - { - "block_id": "p319-b11", - "global_id": 8668, - "bbox": [ - 127.6, - 268.07, - 516.12, - 289.98 - ], - "text": "All the comments about the nature of external and internal stability in Ch. 2 apply to\ndiscrete-time case. We shall not elaborate them further.", - "type": "text" - }, - { - "block_id": "p319-b12", - "global_id": 8669, - "bbox": [ - 127.59, - 320.93, - 325.94, - 332.89 - ], - "text": "3.9-2 Internal (Asymptotic) Stability", - "type": "text" - }, - { - "block_id": "p319-b13", - "global_id": 8670, - "bbox": [ - 127.59, - 339.01, - 516.12, - 372.89 - ], - "text": "For LTID systems, as in the case of LTIC systems, internal stability, called asymptotical stability\nor stability in the sense of Lyapunov (also the zero-input stability), is defined in terms of the\nzero-input response of a system.", - "type": "text" - }, - { - "block_id": "p319-b14", - "global_id": 8671, - "bbox": [ - 127.59, - 374.89, - 516.15, - 408.75 - ], - "text": "For an LTID system specified by a difference equation in the form of Eq. (3.15) [or\nEq. (3.20)], the zero-input response consists of the characteristic modes of the system. The mode\ncorresponding to a characteristic root γ is γ n. To be more general, let γ be complex so that", - "type": "text" - }, - { - "block_id": "p319-b15", - "global_id": 8672, - "bbox": [ - 246.98, - 422.38, - 396.24, - 436.87 - ], - "text": "γ = |γ |ejβ\nand\nγ n = |γ |nejβn", - "type": "text" - }, - { - "block_id": "p319-b16", - "global_id": 8673, - "bbox": [ - 127.59, - 451.4, - 516.14, - 476.94 - ], - "text": "Since the magnitude of ejβn is always unity regardless of the value of n, the magnitude of γ n is\n|γ |n. Therefore,", - "type": "text" - }, - { - "block_id": "p319-b17", - "global_id": 8674, - "bbox": [ - 250.77, - 491.06, - 409.83, - 519.99 - ], - "text": "if |γ | < 1,\nthen γ n →0 as n →∞\nif |γ | > 1,\nthen γ n →∞as n →∞", - "type": "text" - }, - { - "block_id": "p319-b18", - "global_id": 8675, - "bbox": [ - 233.89, - 520.95, - 397.71, - 534.94 - ], - "text": "and if |γ | = 1,\nthen |γ |n = 1 for all n", - "type": "text" - }, - { - "block_id": "p319-b19", - "global_id": 8676, - "bbox": [ - 127.59, - 553.1, - 516.13, - 575.01 - ], - "text": "The characteristic modes corresponding to characteristic roots at various locations in the complex\nplane appear in Fig. 3.27.", - "type": "text" - }, - { - "block_id": "p319-b20", - "global_id": 8677, - "bbox": [ - 127.59, - 577.0, - 516.15, - 634.79 - ], - "text": "These results can be grasped more effectively in terms of the location of characteristic roots\nin the complex plane. Figure 3.28 shows a circle of unit radius, centered at the origin in a\ncomplex plane. Our discussion shows that if all characteristic roots of the system lie inside the\nunit circle, |γi| < 1 for all i and the system is asymptotically stable. On the other hand, even if one\ncharacteristic root lies outside the unit circle, the system is unstable. If none of the characteristic", - "type": "text" - } - ] - }, - { - "page_num": 320, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p320-b0", - "global_id": 8678, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "300\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p320-b1", - "global_id": 8679, - "bbox": [ - 251.58, - 319.13, - 255.58, - 327.13 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p320-b2", - "global_id": 8680, - "bbox": [ - 268.28, - 458.56, - 440.69, - 466.56 - ], - "text": "n\nn", - "type": "text" - }, - { - "block_id": "p320-b3", - "global_id": 8681, - "bbox": [ - 442.69, - 319.13, - 446.69, - 327.13 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p320-b4", - "global_id": 8682, - "bbox": [ - 187.58, - 458.56, - 191.58, - 466.56 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p320-b5", - "global_id": 8683, - "bbox": [ - 187.28, - 581.47, - 440.69, - 589.47 - ], - "text": "n\nn\nn", - "type": "text" - }, - { - "block_id": "p320-b6", - "global_id": 8684, - "bbox": [ - 293.8, - 211.96, - 302.68, - 219.96 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p320-b7", - "global_id": 8685, - "bbox": [ - 293.42, - 355.96, - 303.07, - 363.96 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p320-b8", - "global_id": 8686, - "bbox": [ - 293.58, - 597.26, - 302.91, - 605.26 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p320-b9", - "global_id": 8687, - "bbox": [ - 293.8, - 479.96, - 302.68, - 487.96 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p320-b10", - "global_id": 8688, - "bbox": [ - 192.12, - 179.16, - 459.95, - 187.16 - ], - "text": "n\nn\nn\nn", - "type": "text" - }, - { - "block_id": "p320-b11", - "global_id": 8689, - "bbox": [ - 239.33, - 103.4, - 287.87, - 111.4 - ], - "text": "Complex plane", - "type": "text" - }, - { - "block_id": "p320-b12", - "global_id": 8690, - "bbox": [ - 192.42, - 565.03, - 284.8, - 571.03 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p320-b13", - "global_id": 8691, - "bbox": [ - 191.16, - 444.33, - 281.56, - 450.33 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p320-b14", - "global_id": 8692, - "bbox": [ - 125.76, - 619.43, - 440.25, - 628.67 - ], - "text": "Figure 3.27 Characteristic roots locations and the corresponding characteristic modes.", - "type": "text" - } - ] - }, - { - "page_num": 321, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p321-b0", - "global_id": 8693, - "bbox": [ - 389.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.9\nSystem Stability\n301", - "type": "text" - }, - { - "block_id": "p321-b1", - "global_id": 8694, - "bbox": [ - 127.59, - 85.82, - 516.14, - 143.6 - ], - "text": "roots lie outside the unit circle, but some simple (unrepeated) roots lie on the circle itself, the\nsystem is marginally stable. If two or more characteristic roots coincide on the unit circle (repeated\nroots), the system is unstable. The reason is that for repeated roots, the zero-input response is of the\nform nr−1γ n, and if |γ | = 1, then |nr−1γ n| = nr−1 →∞as n →∞.† Note, however, that repeated\nroots inside the unit circle do not cause instability.", - "type": "text" - }, - { - "block_id": "p321-b2", - "global_id": 8695, - "bbox": [ - 254.81, - 245.41, - 259.7, - 253.41 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p321-b3", - "global_id": 8696, - "bbox": [ - 247.26, - 232.57, - 276.29, - 240.98 - ], - "text": "g\ng", - "type": "text" - }, - { - "block_id": "p321-b4", - "global_id": 8697, - "bbox": [ - 246.14, - 273.07, - 266.14, - 281.07 - ], - "text": "Stable", - "type": "text" - }, - { - "block_id": "p321-b5", - "global_id": 8698, - "bbox": [ - 141.07, - 198.66, - 197.28, - 206.66 - ], - "text": "Marginally stable", - "type": "text" - }, - { - "block_id": "p321-b6", - "global_id": 8699, - "bbox": [ - 272.44, - 183.57, - 300.88, - 191.57 - ], - "text": "Unstable", - "type": "text" - }, - { - "block_id": "p321-b7", - "global_id": 8700, - "bbox": [ - 180.12, - 256.95, - 336.83, - 265.66 - ], - "text": "Re\n1\n1", - "type": "text" - }, - { - "block_id": "p321-b8", - "global_id": 8701, - "bbox": [ - 227.66, - 186.37, - 236.55, - 194.37 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p321-b9", - "global_id": 8702, - "bbox": [ - 357.54, - 311.05, - 512.13, - 332.96 - ], - "text": "Figure 3.28 Characteristic root loca-\ntions and system stability.", - "type": "text" - }, - { - "block_id": "p321-b10", - "global_id": 8703, - "bbox": [ - 145.52, - 358.39, - 204.75, - 368.35 - ], - "text": "To summarize:", - "type": "text" - }, - { - "block_id": "p321-b11", - "global_id": 8704, - "bbox": [ - 144.52, - 376.32, - 516.16, - 458.01 - ], - "text": "1. An LTID system is asymptotically stable if, and only if, all the characteristic roots are\ninside the unit circle. The roots may be simple or repeated.\n2. An LTID system is unstable if, and only if, either one or both of the following conditions\nexist: (i) at least one root is outside the unit circle; (ii) there are repeated roots on the unit\ncircle.\n3. An LTID system is marginally stable if and only if there are no roots outside the unit circle\nand there are some unrepeated roots on the unit circle.", - "type": "text" - }, - { - "block_id": "p321-b12", - "global_id": 8705, - "bbox": [ - 127.59, - 487.07, - 451.82, - 499.02 - ], - "text": "3.9-3 Relationship Between BIBO and Asymptotic Stability", - "type": "text" - }, - { - "block_id": "p321-b13", - "global_id": 8706, - "bbox": [ - 127.59, - 505.15, - 516.1, - 527.78 - ], - "text": "For LTID systems, the relation between the two types of stability is similar to those in LTIC\nsystems. For a system specified by Eq. (3.15), we can readily show that if a characteristic root γk", - "type": "text" - }, - { - "block_id": "p321-b14", - "global_id": 8707, - "bbox": [ - 127.59, - 555.44, - 516.14, - 611.5 - ], - "text": "† If the development of discrete-time systems is parallel to that of continuous-time systems, we wonder why\nthe parallel breaks down here. Why, for instance, are LHP and RHP not the regions demarcating stability and\ninstability? The reason lies in the form of the characteristic modes. In continuous-time systems, we chose the\nform of characteristic mode as eλit. In discrete-time systems, for computational convenience, we choose the\nform to be γ n", - "type": "text" - }, - { - "block_id": "p321-b15", - "global_id": 8708, - "bbox": [ - 127.59, - 599.27, - 516.12, - 633.41 - ], - "text": "i . Had we chosen this form to be eλin where γi = eλi, then the LHP and RHP (for the location of\nλi) again would demarcate stability and instability. The reason is that if γ = eλ, |γ | = 1 implies |eλ| = 1, and\ntherefore λ = jω. This shows that the unit circle in γ plane maps into the imaginary axis in the λ plane.", - "type": "text" - } - ] - }, - { - "page_num": 322, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p322-b0", - "global_id": 8709, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "302\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p322-b1", - "global_id": 8710, - "bbox": [ - 101.84, - 84.3, - 306.92, - 95.81 - ], - "text": "is inside the unit circle, the corresponding mode γ n", - "type": "text" - }, - { - "block_id": "p322-b2", - "global_id": 8711, - "bbox": [ - 101.85, - 85.43, - 490.39, - 107.77 - ], - "text": "k is absolutely summable. In contrast, if γk lies\noutside the unit circle, or on the unit circle, γ n", - "type": "text" - }, - { - "block_id": "p322-b3", - "global_id": 8712, - "bbox": [ - 280.9, - 97.21, - 403.19, - 110.57 - ], - "text": "k is not absolutely summable.†", - "type": "text" - }, - { - "block_id": "p322-b4", - "global_id": 8713, - "bbox": [ - 101.84, - 109.76, - 490.39, - 155.58 - ], - "text": "This means that an asymptotically stable system is BIBO-stable. Moreover, a marginally\nstable or asymptotically unstable system is BIBO-unstable. The converse is not necessarily true.\nThe stability picture portrayed by the external description is of questionable value. BIBO (external)\nstability cannot ensure internal (asymptotic) stability, as the following example shows.", - "type": "text" - }, - { - "block_id": "p322-b5", - "global_id": 8714, - "bbox": [ - 76.77, - 195.48, - 465.5, - 207.44 - ], - "text": "EXAMPLE 3.26\nA BIBO-Stable but Asymptotically Unstable System", - "type": "text" - }, - { - "block_id": "p322-b6", - "global_id": 8715, - "bbox": [ - 103.16, - 223.69, - 477.02, - 246.79 - ], - "text": "An LTID systems consists of two subsystems S1 and S2 in cascade (Fig. 3.29). The impulse\nresponse of these systems are h1[n] and h2[n], respectively, given by", - "type": "text" - }, - { - "block_id": "p322-b7", - "global_id": 8716, - "bbox": [ - 178.94, - 256.06, - 401.23, - 268.72 - ], - "text": "h1[n] = 4δ[n] −3(0.5)nu[n]\nand\nh2[n] = 2nu[n]", - "type": "text" - }, - { - "block_id": "p322-b8", - "global_id": 8717, - "bbox": [ - 103.17, - 279.9, - 386.52, - 289.86 - ], - "text": "Investigate the BIBO and asymptotic stability of the composite system.", - "type": "text" - }, - { - "block_id": "p322-b9", - "global_id": 8718, - "bbox": [ - 118.96, - 340.38, - 480.6, - 368.69 - ], - "text": "S1\nx[n]\ny[n]\nS2\nFigure 3.29 Composite\nsystem\nfor Ex. 3.26.", - "type": "text" - }, - { - "block_id": "p322-b10", - "global_id": 8719, - "bbox": [ - 121.09, - 387.4, - 346.5, - 397.78 - ], - "text": "The composite system impulse response h[n] is given by", - "type": "text" - }, - { - "block_id": "p322-b11", - "global_id": 8720, - "bbox": [ - 140.81, - 407.81, - 410.8, - 420.47 - ], - "text": "h[n] = h1[n] ∗h2[n] = h2[n] ∗h1[n] = 2nu[n] ∗(4δ[n] −3(0.5)nu[n])", - "type": "text" - }, - { - "block_id": "p322-b12", - "global_id": 8721, - "bbox": [ - 282.1, - 430.74, - 345.74, - 442.62 - ], - "text": "= 4(2)nu[n] −3", - "type": "text" - }, - { - "block_id": "p322-b13", - "global_id": 8722, - "bbox": [ - 346.84, - 418.25, - 414.52, - 435.63 - ], - "text": "2n+1 −(0.5)n+1", - "type": "text" - }, - { - "block_id": "p322-b14", - "global_id": 8723, - "bbox": [ - 370.1, - 439.31, - 398.4, - 449.69 - ], - "text": "2 −0.5", - "type": "text" - }, - { - "block_id": "p322-b15", - "global_id": 8724, - "bbox": [ - 416.24, - 418.25, - 421.67, - 428.22 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p322-b16", - "global_id": 8725, - "bbox": [ - 422.77, - 432.24, - 439.36, - 442.51 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p322-b17", - "global_id": 8726, - "bbox": [ - 282.1, - 452.55, - 332.38, - 464.44 - ], - "text": "= (0.5)nu[n]", - "type": "text" - }, - { - "block_id": "p322-b18", - "global_id": 8727, - "bbox": [ - 103.16, - 476.39, - 477.04, - 522.22 - ], - "text": "If the composite cascade system were to be enclosed in a black box with only the input and\nthe output terminals accessible, any measurement from these external terminals would show\nthat the impulse response of the system is (0.5)nu[n], without any hint of the unstable system\nsheltered inside the composite system.", - "type": "text" - }, - { - "block_id": "p322-b19", - "global_id": 8728, - "bbox": [ - 101.84, - 555.57, - 316.92, - 567.79 - ], - "text": "† This conclusion follows from the fact that (see Sec. B.8-3)", - "type": "text" - }, - { - "block_id": "p322-b20", - "global_id": 8729, - "bbox": [ - 210.12, - 575.87, - 222.81, - 585.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p322-b21", - "global_id": 8730, - "bbox": [ - 206.49, - 597.5, - 226.44, - 604.18 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p322-b22", - "global_id": 8731, - "bbox": [ - 227.44, - 577.57, - 239.51, - 594.16 - ], - "text": "γ n", - "type": "text" - }, - { - "block_id": "p322-b23", - "global_id": 8732, - "bbox": [ - 234.85, - 589.8, - 237.72, - 596.28 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p322-b24", - "global_id": 8733, - "bbox": [ - 240.02, - 577.57, - 266.71, - 594.44 - ], - "text": "u[n] =", - "type": "text" - }, - { - "block_id": "p322-b25", - "global_id": 8734, - "bbox": [ - 268.55, - 575.87, - 281.24, - 585.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p322-b26", - "global_id": 8735, - "bbox": [ - 269.14, - 598.12, - 280.66, - 604.87 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p322-b27", - "global_id": 8736, - "bbox": [ - 282.24, - 579.28, - 385.75, - 601.57 - ], - "text": "|γk|n =\n1\n1 −|γk|\n|γk| < 1", - "type": "text" - }, - { - "block_id": "p322-b28", - "global_id": 8737, - "bbox": [ - 101.84, - 613.11, - 490.4, - 633.41 - ], - "text": "Moreover, if |γ | ≥1, the sum diverges and goes to ∞. These conclusions are valid also for the modes of the\nform nrγ n", - "type": "text" - }, - { - "block_id": "p322-b29", - "global_id": 8738, - "bbox": [ - 133.67, - 624.45, - 141.08, - 636.08 - ], - "text": "k .", - "type": "text" - } - ] - }, - { - "page_num": 323, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p323-b0", - "global_id": 8739, - "bbox": [ - 389.44, - 62.89, - 516.14, - 71.98 - ], - "text": "3.9\nSystem Stability\n303", - "type": "text" - }, - { - "block_id": "p323-b1", - "global_id": 8740, - "bbox": [ - 128.9, - 84.88, - 502.76, - 144.09 - ], - "text": "The composite system is BIBO-stable because its impulse response (0.5)nu[n] is\nabsolutely summable. However, the system S2 is asymptotically unstable because its\ncharacteristic root, 2, lies outside the unit circle. This system will eventually burn out\n(or saturate) because of the unbounded characteristic response generated by intended or\nunintended initial conditions, no matter how small.", - "type": "text" - }, - { - "block_id": "p323-b2", - "global_id": 8741, - "bbox": [ - 128.9, - 146.08, - 502.78, - 203.86 - ], - "text": "The system is asymptotically unstable, though BIBO-stable. This example shows\nthat BIBO stability does not necessarily ensure asymptotic stability when a system is\nuncontrollable, unobservable, or both. The internal and the external descriptions of a system\nare equivalent only when the system is controllable and observable. In such a case, BIBO\nstability means the system is asymptotically stable, and vice versa.", - "type": "text" - }, - { - "block_id": "p323-b3", - "global_id": 8742, - "bbox": [ - 127.59, - 242.3, - 516.13, - 288.13 - ], - "text": "Fortunately, uncontrollable or unobservable systems are not common in practice. Henceforth,\nin determining system stability, we shall assume that unless otherwise mentioned, the internal and\nthe external descriptions of the system are equivalent, implying that the system is controllable and\nobservable.", - "type": "text" - }, - { - "block_id": "p323-b4", - "global_id": 8743, - "bbox": [ - 102.51, - 317.06, - 452.39, - 329.01 - ], - "text": "EXAMPLE 3.27\nInvestigating Asymptotic and BIBO Stability", - "type": "text" - }, - { - "block_id": "p323-b5", - "global_id": 8744, - "bbox": [ - 128.9, - 345.08, - 502.76, - 366.99 - ], - "text": "Determine the internal and external stability of systems specified by the following equations.\nIn each case plot the characteristic roots in the complex plane.", - "type": "text" - }, - { - "block_id": "p323-b6", - "global_id": 8745, - "bbox": [ - 146.84, - 374.56, - 379.49, - 399.87 - ], - "text": "(a) y[n + 2] + 2.5y[n + 1] + y[n] = x[n + 1] −2x[n]\n(b) y[n] −y[n −1] + 0.21y[n −2] = 2x[n −1] + 3x[n −2]", - "type": "text" - }, - { - "block_id": "p323-b7", - "global_id": 8746, - "bbox": [ - 147.4, - 403.09, - 258.63, - 414.82 - ], - "text": "(c) y[n + 3] + 2y[n + 2] + 3", - "type": "text" - }, - { - "block_id": "p323-b8", - "global_id": 8747, - "bbox": [ - 146.84, - 403.09, - 368.27, - 429.76 - ], - "text": "2y[n + 1] + 1\n2y[n] = x[n + 1]\n(d) (E2 −E + 1)2y[n] = (3E + 1)x[n]", - "type": "text" - }, - { - "block_id": "p323-b9", - "global_id": 8748, - "bbox": [ - 146.84, - 458.57, - 289.07, - 468.61 - ], - "text": "(a) The characteristic polynomial is", - "type": "text" - }, - { - "block_id": "p323-b10", - "global_id": 8749, - "bbox": [ - 248.11, - 474.84, - 383.55, - 489.33 - ], - "text": "γ 2 + 2.5γ + 1 = (γ + 0.5)(γ + 2)", - "type": "text" - }, - { - "block_id": "p323-b11", - "global_id": 8750, - "bbox": [ - 128.91, - 499.67, - 502.78, - 522.01 - ], - "text": "The characteristic roots are −0.5 and −2. Because | −2| > 1 (−2 lies outside the unit circle),\nthe system is BIBO-unstable and also asymptotically unstable (Fig. 3.30a).", - "type": "text" - }, - { - "block_id": "p323-b12", - "global_id": 8751, - "bbox": [ - 146.84, - 523.92, - 289.63, - 533.96 - ], - "text": "(b) The characteristic polynomial is", - "type": "text" - }, - { - "block_id": "p323-b13", - "global_id": 8752, - "bbox": [ - 244.38, - 540.19, - 387.29, - 554.68 - ], - "text": "γ 2 −γ + 0.21 = (γ −0.3)(γ −0.7)", - "type": "text" - }, - { - "block_id": "p323-b14", - "global_id": 8753, - "bbox": [ - 128.91, - 565.44, - 502.77, - 587.36 - ], - "text": "The characteristic roots are 0.3 and 0.7, both of which lie inside the unit circle. The system is\nBIBO-stable and asymptotically stable (Fig. 3.30b).", - "type": "text" - }, - { - "block_id": "p323-b15", - "global_id": 8754, - "bbox": [ - 146.84, - 589.27, - 288.51, - 599.31 - ], - "text": "(c) The characteristic polynomial is", - "type": "text" - }, - { - "block_id": "p323-b16", - "global_id": 8755, - "bbox": [ - 145.68, - 605.54, - 198.24, - 620.03 - ], - "text": "γ 3 + 2γ 2 + 3", - "type": "text" - }, - { - "block_id": "p323-b17", - "global_id": 8756, - "bbox": [ - 194.76, - 601.64, - 268.53, - 623.15 - ], - "text": "2γ + 1\n2 = (γ + 1)", - "type": "text" - }, - { - "block_id": "p323-b18", - "global_id": 8757, - "bbox": [ - 268.55, - 606.04, - 312.14, - 619.62 - ], - "text": "γ 2 + γ + 1", - "type": "text" - }, - { - "block_id": "p323-b19", - "global_id": 8758, - "bbox": [ - 308.65, - 601.64, - 317.36, - 622.84 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p323-b20", - "global_id": 8759, - "bbox": [ - 319.4, - 609.65, - 485.96, - 620.03 - ], - "text": "= (γ + 1)(γ + 0.5 −j0.5)(γ + 0.5 + j0.5)", - "type": "text" - } - ] - }, - { - "page_num": 324, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p324-b0", - "global_id": 8760, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "304\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p324-b1", - "global_id": 8761, - "bbox": [ - 103.16, - 85.83, - 477.03, - 120.12 - ], - "text": "The characteristic roots are −1, −0.5 ± j0.5 (Fig. 3.30c). One of the characteristic roots\nis on the unit circle and the remaining two roots are inside the unit circle. The system is\nBIBO-unstable but marginally stable.", - "type": "text" - }, - { - "block_id": "p324-b2", - "global_id": 8762, - "bbox": [ - 121.09, - 122.02, - 263.88, - 132.07 - ], - "text": "(d) The characteristic polynomial is", - "type": "text" - }, - { - "block_id": "p324-b3", - "global_id": 8763, - "bbox": [ - 192.89, - 152.17, - 258.03, - 166.66 - ], - "text": "(γ 2 −γ + 1)2 =", - "type": "text" - }, - { - "block_id": "p324-b5", - "global_id": 8764, - "bbox": [ - 265.51, - 154.94, - 287.65, - 166.25 - ], - "text": "γ −1", - "type": "text" - }, - { - "block_id": "p324-b6", - "global_id": 8765, - "bbox": [ - 284.16, - 148.8, - 309.58, - 169.79 - ], - "text": "2 −j\n√", - "type": "text" - }, - { - "block_id": "p324-b7", - "global_id": 8766, - "bbox": [ - 306.64, - 145.29, - 329.11, - 169.48 - ], - "text": "3\n2\n2", - "type": "text" - }, - { - "block_id": "p324-b8", - "global_id": 8767, - "bbox": [ - 329.11, - 154.94, - 351.26, - 166.25 - ], - "text": "γ −1", - "type": "text" - }, - { - "block_id": "p324-b9", - "global_id": 8768, - "bbox": [ - 347.78, - 148.8, - 373.19, - 169.79 - ], - "text": "2 + j\n√", - "type": "text" - }, - { - "block_id": "p324-b10", - "global_id": 8769, - "bbox": [ - 370.24, - 145.29, - 386.79, - 169.48 - ], - "text": "3\n2\n2", - "type": "text" - }, - { - "block_id": "p324-b11", - "global_id": 8770, - "bbox": [ - 103.16, - 196.14, - 251.03, - 206.52 - ], - "text": "The characteristic roots are (1/2)±j(", - "type": "text" - }, - { - "block_id": "p324-b12", - "global_id": 8771, - "bbox": [ - 251.03, - 187.77, - 259.46, - 197.73 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p324-b13", - "global_id": 8772, - "bbox": [ - 103.17, - 192.93, - 477.01, - 218.48 - ], - "text": "3/2) = 1e±j(π/3) repeated twice, and they lie on the unit\ncircle (Fig. 3.30d). The system is BIBO-unstable and asymptotically unstable.", - "type": "text" - }, - { - "block_id": "p324-b14", - "global_id": 8773, - "bbox": [ - 182.12, - 397.12, - 191.0, - 405.12 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p324-b15", - "global_id": 8774, - "bbox": [ - 216.76, - 257.45, - 265.42, - 265.45 - ], - "text": "Complex plane", - "type": "text" - }, - { - "block_id": "p324-b16", - "global_id": 8775, - "bbox": [ - 182.12, - 564.25, - 191.0, - 572.25 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p324-b17", - "global_id": 8776, - "bbox": [ - 372.9, - 397.12, - 382.22, - 405.12 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p324-b18", - "global_id": 8777, - "bbox": [ - 225.96, - 319.45, - 229.96, - 327.45 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p324-b19", - "global_id": 8778, - "bbox": [ - 155.93, - 486.71, - 172.6, - 495.01 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p324-b20", - "global_id": 8779, - "bbox": [ - 107.23, - 319.78, - 173.66, - 328.07 - ], - "text": "0.5\n2", - "type": "text" - }, - { - "block_id": "p324-b21", - "global_id": 8780, - "bbox": [ - 189.0, - 499.33, - 205.66, - 507.63 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p324-b22", - "global_id": 8781, - "bbox": [ - 136.09, - 486.71, - 146.76, - 495.01 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p324-b23", - "global_id": 8782, - "bbox": [ - 382.92, - 320.08, - 407.61, - 328.08 - ], - "text": "0.3 0.7", - "type": "text" - }, - { - "block_id": "p324-b24", - "global_id": 8783, - "bbox": [ - 189.0, - 467.64, - 402.72, - 482.0 - ], - "text": "p3\n3p4", - "type": "text" - }, - { - "block_id": "p324-b25", - "global_id": 8784, - "bbox": [ - 372.9, - 564.25, - 382.22, - 572.25 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p324-b26", - "global_id": 8785, - "bbox": [ - 327.78, - 485.43, - 421.36, - 494.04 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p324-b27", - "global_id": 8786, - "bbox": [ - 327.78, - 319.78, - 338.44, - 328.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p324-b28", - "global_id": 8787, - "bbox": [ - 103.66, - 578.95, - 352.21, - 588.19 - ], - "text": "Figure 3.30 Characteristic root locations for the system of Ex. 3.27.", - "type": "text" - } - ] - }, - { - "page_num": 325, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p325-b0", - "global_id": 8788, - "bbox": [ - 294.16, - 62.89, - 516.15, - 71.98 - ], - "text": "3.10\nIntuitive Insights into System Behavior\n305", - "type": "text" - }, - { - "block_id": "p325-b1", - "global_id": 8789, - "bbox": [ - 133.57, - 97.81, - 448.29, - 109.77 - ], - "text": "DRILL 3.21\nAssessing Stability by Characteristic Roots", - "type": "text" - }, - { - "block_id": "p325-b2", - "global_id": 8790, - "bbox": [ - 133.57, - 118.89, - 510.17, - 140.8 - ], - "text": "Using the complex plane, locate the characteristic roots of the following systems, and use the\ncharacteristic root locations to determine external and internal stability of each system.", - "type": "text" - }, - { - "block_id": "p325-b3", - "global_id": 8791, - "bbox": [ - 151.5, - 144.74, - 339.03, - 173.68 - ], - "text": "(a) (E + 1)(E2 + 6E + 25)y[n] = 3Ex[n]\n(b) (E −1)2(E + 0.5)y[n] = (E2 + 2E + 3)x[n]", - "type": "text" - }, - { - "block_id": "p325-b4", - "global_id": 8792, - "bbox": [ - 133.84, - 193.18, - 196.06, - 204.14 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p325-b5", - "global_id": 8793, - "bbox": [ - 133.57, - 211.54, - 344.97, - 221.5 - ], - "text": "Both systems are BIBO-and asymptotically unstable.", - "type": "text" - }, - { - "block_id": "p325-b6", - "global_id": 8794, - "bbox": [ - 127.94, - 264.58, - 460.89, - 278.53 - ], - "text": "3.10 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR", - "type": "text" - }, - { - "block_id": "p325-b7", - "global_id": 8795, - "bbox": [ - 127.59, - 284.52, - 516.15, - 318.39 - ], - "text": "The intuitive insights into the behavior of continuous-time systems and their qualitative proofs,\ndiscussed in Sec. 2.6, also apply to discrete-time systems. For this reason, we shall merely mention\nhere without discussion some of the insights presented in Sec. 2.6.", - "type": "text" - }, - { - "block_id": "p325-b8", - "global_id": 8796, - "bbox": [ - 127.59, - 320.38, - 516.16, - 473.81 - ], - "text": "The system’s entire (zero-input and zero-state) behavior is strongly influenced by the\ncharacteristic roots (or modes) of the system. The system responds strongly to input signals similar\nto its characteristic modes and poorly to inputs very different from its characteristic modes. In\nfact, when the input is a characteristic mode of the system, the response goes to infinity, provided\nthe mode is a nondecaying signal. This is the resonance phenomenon. The width of an impulse\nresponse h[n] indicates the response time (time required to respond fully to an input) of the system.\nIt is the time constant of the system.† Discrete-time pulses are generally dispersed when passed\nthrough a discrete-time system. The amount of dispersion (or spreading out) is equal to the system\ntime constant (or width of h[n]). The system time constant also determines the rate at which\nthe system can transmit information. A smaller time constant corresponds to a higher rate of\ninformation transmission, and vice versa. We keep in mind that concepts such as time constant\nand pulse dispersion only coarsely illustrate system behavior. Let us illustrate these ideas with an\nexample.", - "type": "text" - }, - { - "block_id": "p325-b9", - "global_id": 8797, - "bbox": [ - 102.51, - 506.02, - 493.58, - 517.98 - ], - "text": "EXAMPLE 3.28\nIntuitive Insights into Lowpass DT System Behavior", - "type": "text" - }, - { - "block_id": "p325-b10", - "global_id": 8798, - "bbox": [ - 128.9, - 534.64, - 502.74, - 556.56 - ], - "text": "Determine the time constant, rise time, pulse dispersion, and filter characteristics of a lowpass\nDT system with impulse response h[n] = 2(0.6)nu[n].", - "type": "text" - }, - { - "block_id": "p325-b11", - "global_id": 8799, - "bbox": [ - 127.59, - 610.24, - 516.13, - 633.41 - ], - "text": "† This part of the discussion applies to systems with impulse response h[n] that is a mostly positive (or mostly\nnegative) pulse.", - "type": "text" - } - ] - }, - { - "page_num": 326, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p326-b0", - "global_id": 8800, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "306\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p326-b1", - "global_id": 8801, - "bbox": [ - 103.16, - 85.83, - 477.01, - 132.07 - ], - "text": "Since h[n] resembles a single, mostly positive pulse, we know that the DT system is lowpass.\nSimilar to the CT case shown in Sec. 2.6, we can determine the time constant Th as the width\nof a rectangle that approximates h[n]. This rectangle possesses the same peak height and total\nsum (area), as does h[n]. The peak of h[n] is 2, and the total sum (area) is", - "type": "text" - }, - { - "block_id": "p326-b2", - "global_id": 8802, - "bbox": [ - 235.87, - 141.56, - 249.97, - 152.23 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p326-b3", - "global_id": 8803, - "bbox": [ - 236.71, - 166.12, - 249.11, - 173.39 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p326-b4", - "global_id": 8804, - "bbox": [ - 251.07, - 144.75, - 344.32, - 169.19 - ], - "text": "2(0.6)n = 2 1 −0\n1 −0.6 = 5", - "type": "text" - }, - { - "block_id": "p326-b5", - "global_id": 8805, - "bbox": [ - 103.16, - 182.98, - 476.52, - 205.0 - ], - "text": "Since the width of a DT signal is 1 less than its length, we see that the time constant Th\n(rectangle width) is", - "type": "text" - }, - { - "block_id": "p326-b6", - "global_id": 8806, - "bbox": [ - 174.04, - 214.93, - 290.93, - 232.96 - ], - "text": "Th = rectangle width = area", - "type": "text" - }, - { - "block_id": "p326-b7", - "global_id": 8807, - "bbox": [ - 270.19, - 214.93, - 330.18, - 238.95 - ], - "text": "height −1 = 5", - "type": "text" - }, - { - "block_id": "p326-b8", - "global_id": 8808, - "bbox": [ - 325.2, - 221.5, - 406.14, - 238.95 - ], - "text": "2 −1 = 1.5 samples", - "type": "text" - }, - { - "block_id": "p326-b9", - "global_id": 8809, - "bbox": [ - 103.16, - 248.89, - 464.71, - 258.85 - ], - "text": "Since time constant, rise time, pulse dispersion are all given by the same value, we see that", - "type": "text" - }, - { - "block_id": "p326-b10", - "global_id": 8810, - "bbox": [ - 161.84, - 270.4, - 418.34, - 281.86 - ], - "text": "time constant = rise time = pulse dispersion = Th = 1.5 samples", - "type": "text" - }, - { - "block_id": "p326-b11", - "global_id": 8811, - "bbox": [ - 103.17, - 292.73, - 477.04, - 314.64 - ], - "text": "The approximate cutoff frequency of our DT system can be determined as the frequency of a\nDT sinusoid whose period equals the length of the rectangle approximation to h[n]. That is,", - "type": "text" - }, - { - "block_id": "p326-b12", - "global_id": 8812, - "bbox": [ - 198.68, - 324.59, - 322.08, - 349.69 - ], - "text": "cutoff frequency =\n1\nTh + 1 = 2", - "type": "text" - }, - { - "block_id": "p326-b13", - "global_id": 8813, - "bbox": [ - 317.09, - 331.57, - 381.48, - 348.61 - ], - "text": "5 cycles/sample", - "type": "text" - }, - { - "block_id": "p326-b14", - "global_id": 8814, - "bbox": [ - 103.16, - 357.55, - 400.08, - 367.92 - ], - "text": "Equivalently, we can express the cutoff frequency as 4π/5 radians/sample.", - "type": "text" - }, - { - "block_id": "p326-b15", - "global_id": 8815, - "bbox": [ - 103.17, - 369.82, - 477.03, - 415.75 - ], - "text": "Notice that Th is not an integer and thus lacks a clear physical meaning for our DT system.\nHow, for example, can it take 1.5 samples for our DT system to fully respond to an input? We\ncan put our minds at ease by remembering the approximate nature of Th, which is meant to\nprovide only a rough understanding of system behavior.", - "type": "text" - }, - { - "block_id": "p326-b16", - "global_id": 8816, - "bbox": [ - 102.2, - 472.43, - 471.04, - 486.38 - ], - "text": "3.11 MATLAB: DISCRETE-TIME SIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p326-b17", - "global_id": 8817, - "bbox": [ - 101.84, - 492.37, - 490.4, - 526.24 - ], - "text": "MATLAB is naturally and ideally suited to discrete-time signals and systems. Many special\nfunctions are available for discrete-time data operations, including the stem, filter, and conv\ncommands. In this section, we investigate and apply these and other commands.", - "type": "text" - }, - { - "block_id": "p326-b18", - "global_id": 8818, - "bbox": [ - 101.84, - 556.92, - 356.68, - 568.88 - ], - "text": "3.11-1 Discrete-Time Functions and Stem Plots", - "type": "text" - }, - { - "block_id": "p326-b19", - "global_id": 8819, - "bbox": [ - 101.84, - 571.39, - 490.41, - 608.88 - ], - "text": "Consider the discrete-time function f[n] = e−n/5 cos(πn/5)u[n]. In MATLAB, there are many\nways to represent f[n] including M-files or, for particular n, explicit command line evaluation. In\nthis example, however, we use an anonymous function.", - "type": "text" - }, - { - "block_id": "p326-b20", - "global_id": 8820, - "bbox": [ - 101.85, - 625.05, - 331.98, - 635.01 - ], - "text": ">>\nf = @(n) exp(-n/5).*cos(pi*n/5).*(n>=0);", - "type": "text" - } - ] - }, - { - "page_num": 327, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p327-b0", - "global_id": 8821, - "bbox": [ - 268.15, - 62.89, - 516.13, - 71.98 - ], - "text": "3.11\nMATLAB: Discrete-Time Signals and Systems\n307", - "type": "text" - }, - { - "block_id": "p327-b1", - "global_id": 8822, - "bbox": [ - 127.59, - 85.72, - 516.15, - 143.6 - ], - "text": "A true discrete-time function is undefined (or zero) for noninteger n. Although anonymous\nfunction f is intended as a discrete-time function, its present construction does not restrict n to\nbe integer, and it can therefore be misused. For example, MATLAB dutifully returns 0.8606 to\nf(0.5) when a NaN (not-a-number) or zero is more appropriate. The user is responsible for\nappropriate function use.", - "type": "text" - }, - { - "block_id": "p327-b2", - "global_id": 8823, - "bbox": [ - 127.59, - 145.18, - 516.14, - 167.51 - ], - "text": "Next, consider plotting the discrete-time function f[n] over (−10 ≤n ≤10). The stem\ncommand simplifies this task.", - "type": "text" - }, - { - "block_id": "p327-b3", - "global_id": 8824, - "bbox": [ - 127.59, - 188.95, - 294.96, - 222.82 - ], - "text": ">>\nn = (-10:10)’;\n>>\nstem(n,f(n),’k’);\n>>\nxlabel(’n’); ylabel(’f[n]’);", - "type": "text" - }, - { - "block_id": "p327-b4", - "global_id": 8825, - "bbox": [ - 127.59, - 243.69, - 516.15, - 277.55 - ], - "text": "Here, stem operates much like the plot command: dependent variable f(n) is plotted against\nindependent variable n with black lines. The stem command emphasizes the discrete-time nature\nof the data, as Fig. 3.31 illustrates.", - "type": "text" - }, - { - "block_id": "p327-b5", - "global_id": 8826, - "bbox": [ - 127.6, - 279.55, - 516.15, - 349.28 - ], - "text": "For discrete-time functions, the operations of shifting, inversion, and scaling can have\nsurprising results. Compare f[−2n] with f[−2n + 1]. Contrary to the continuous case, the second\nis not a shifted version of the first. We can use separate subplots, each over (−10 ≤n ≤10),\nto help illustrate this fact. Notice that unlike the plot command, the stem command cannot\nsimultaneously plot multiple functions on a single axis; overlapping stem lines would make such\nplots difficult to read anyway.", - "type": "text" - }, - { - "block_id": "p327-b6", - "global_id": 8827, - "bbox": [ - 127.6, - 370.72, - 519.87, - 392.64 - ], - "text": ">>\nsubplot(2,1,1); stem(n,f(-2*n),’k’); ylabel(’f[-2n]’);\n>>\nsubplot(2,1,2); stem(n,f(-2*n+1),’k’); ylabel(’f[-2n+1]’); xlabel(’n’);", - "type": "text" - }, - { - "block_id": "p327-b7", - "global_id": 8828, - "bbox": [ - 127.6, - 413.09, - 516.14, - 435.41 - ], - "text": "The results are shown in Fig. 3.32. Interestingly, the original function f[n] can be recovered by\ninterleaving samples of f[−2n] and f[−2n + 1] and then time-reflecting the result.", - "type": "text" - }, - { - "block_id": "p327-b8", - "global_id": 8829, - "bbox": [ - 127.6, - 437.41, - 516.16, - 483.24 - ], - "text": "Care must always be taken to ensure that MATLAB performs the desired computations. Our\nanonymous function f is a case in point: although it correctly downsamples, it does not properly\nupsample (see Prob. 3.11-2). MATLAB does what it is told, but it is not always told how to do\neverything correctly!", - "type": "text" - }, - { - "block_id": "p327-b9", - "global_id": 8830, - "bbox": [ - 176.07, - 592.09, - 408.7, - 613.28 - ], - "text": "–10\n–5\n0\n5\n10\nn", - "type": "text" - }, - { - "block_id": "p327-b10", - "global_id": 8831, - "bbox": [ - 163.23, - 582.86, - 177.23, - 590.86 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p327-b11", - "global_id": 8832, - "bbox": [ - 173.23, - 558.86, - 177.23, - 566.86 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p327-b12", - "global_id": 8833, - "bbox": [ - 166.48, - 534.86, - 177.16, - 542.86 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p327-b13", - "global_id": 8834, - "bbox": [ - 173.23, - 510.86, - 177.23, - 518.86 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p327-b14", - "global_id": 8835, - "bbox": [ - 152.26, - 542.77, - 161.06, - 555.96 - ], - "text": "f[n]", - "type": "text" - }, - { - "block_id": "p327-b15", - "global_id": 8836, - "bbox": [ - 151.5, - 619.58, - 294.49, - 629.19 - ], - "text": "Figure 3.31 f[n] over (−10 ≤n ≤10).", - "type": "text" - } - ] - }, - { - "page_num": 328, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p328-b0", - "global_id": 8837, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "308\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p328-b1", - "global_id": 8838, - "bbox": [ - 137.49, - 156.59, - 382.96, - 173.81 - ], - "text": "–10\n–5\n10\n–0.5", - "type": "text" - }, - { - "block_id": "p328-b2", - "global_id": 8839, - "bbox": [ - 147.49, - 133.58, - 151.49, - 141.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p328-b3", - "global_id": 8840, - "bbox": [ - 140.74, - 110.59, - 151.42, - 118.59 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p328-b4", - "global_id": 8841, - "bbox": [ - 147.49, - 87.59, - 151.49, - 95.59 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p328-b5", - "global_id": 8842, - "bbox": [ - 126.52, - 113.32, - 135.32, - 135.31 - ], - "text": "f[–2n]", - "type": "text" - }, - { - "block_id": "p328-b6", - "global_id": 8843, - "bbox": [ - 147.78, - 268.57, - 213.91, - 276.57 - ], - "text": "–10\n–5", - "type": "text" - }, - { - "block_id": "p328-b7", - "global_id": 8844, - "bbox": [ - 265.46, - 165.81, - 325.85, - 173.81 - ], - "text": "0\n5", - "type": "text" - }, - { - "block_id": "p328-b8", - "global_id": 8845, - "bbox": [ - 265.45, - 268.57, - 382.96, - 289.76 - ], - "text": "0\n5\n10\nn", - "type": "text" - }, - { - "block_id": "p328-b9", - "global_id": 8846, - "bbox": [ - 137.49, - 259.34, - 151.49, - 267.34 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p328-b10", - "global_id": 8847, - "bbox": [ - 147.49, - 236.08, - 151.49, - 244.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p328-b11", - "global_id": 8848, - "bbox": [ - 140.74, - 212.83, - 151.42, - 220.83 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p328-b12", - "global_id": 8849, - "bbox": [ - 147.49, - 189.59, - 151.49, - 197.59 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p328-b13", - "global_id": 8850, - "bbox": [ - 126.52, - 211.59, - 135.32, - 242.94 - ], - "text": "f[–2n+1]", - "type": "text" - }, - { - "block_id": "p328-b14", - "global_id": 8851, - "bbox": [ - 125.76, - 296.07, - 337.65, - 305.68 - ], - "text": "Figure 3.32 f[−2n] and f[−2n + 1] over (−10 ≤n ≤10).", - "type": "text" - }, - { - "block_id": "p328-b15", - "global_id": 8852, - "bbox": [ - 101.84, - 327.62, - 339.78, - 339.58 - ], - "text": "3.11-2 System Responses Through Filtering", - "type": "text" - }, - { - "block_id": "p328-b16", - "global_id": 8853, - "bbox": [ - 101.84, - 345.71, - 490.38, - 367.62 - ], - "text": "MATLAB’s filter command provides an efficient way to evaluate the system response of a\nconstant coefficient linear difference equation represented in delay form as", - "type": "text" - }, - { - "block_id": "p328-b17", - "global_id": 8854, - "bbox": [ - 234.77, - 379.47, - 248.86, - 389.92 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p328-b18", - "global_id": 8855, - "bbox": [ - 235.75, - 403.81, - 247.88, - 411.07 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p328-b19", - "global_id": 8856, - "bbox": [ - 249.98, - 389.42, - 299.99, - 400.5 - ], - "text": "aky[n −k] =", - "type": "text" - }, - { - "block_id": "p328-b20", - "global_id": 8857, - "bbox": [ - 302.03, - 379.47, - 316.13, - 389.92 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p328-b21", - "global_id": 8858, - "bbox": [ - 303.01, - 403.81, - 315.15, - 411.07 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p328-b22", - "global_id": 8859, - "bbox": [ - 317.24, - 389.42, - 490.38, - 400.5 - ], - "text": "bkx[n −k]\n(3.44)", - "type": "text" - }, - { - "block_id": "p328-b23", - "global_id": 8860, - "bbox": [ - 101.84, - 421.09, - 490.39, - 467.33 - ], - "text": "In the simplest form, filter requires three input arguments: a length-(N + 1) vector of\nfeedforward coefficients [b0,b1,...,bN], a length-(N + 1) vector of feedback coefficients\n[a0,a1,...,aN], and an input vector.† Since no initial conditions are specified, the output\ncorresponds to the system’s zero-state response.", - "type": "text" - }, - { - "block_id": "p328-b24", - "global_id": 8861, - "bbox": [ - 101.85, - 468.92, - 490.39, - 503.2 - ], - "text": "To serve as an example, consider a system described by y[n]−y[n−1]+y[n−2] = x[n]. When\nx[n] = δ[n], the zero-state response is equal to the impulse response h[n], which we compute over\n(0 ≤n ≤30).", - "type": "text" - }, - { - "block_id": "p328-b25", - "global_id": 8862, - "bbox": [ - 101.85, - 514.19, - 363.37, - 571.97 - ], - "text": ">>\nb = [1 0 0]; a = [1 -1 1];\n>>\nn = (0:30)’; delta = @(n) 1.0.*(n==0);\n>>\nh = filter(b,a,delta(n));\n>>\nclf; stem(n,h,’k’); axis([-.5 30.5 -1.1 1.1]);\n>>\nxlabel(’n’); ylabel(’h[n]’);", - "type": "text" - }, - { - "block_id": "p328-b26", - "global_id": 8863, - "bbox": [ - 101.84, - 599.27, - 490.37, - 634.75 - ], - "text": "† It is important to pay close attention to the inevitable notational differences found throughout engineering\ndocuments. In MATLAB help documents, coefficient subscripts begin at 1 rather than 0 to better conform\nwith MATLAB indexing conventions. That is, MATLAB labels a0 as a(1), b0 as b(1), and so forth.", - "type": "text" - } - ] - }, - { - "page_num": 329, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p329-b0", - "global_id": 8864, - "bbox": [ - 268.15, - 62.89, - 516.13, - 71.98 - ], - "text": "3.11\nMATLAB: Discrete-Time Signals and Systems\n309", - "type": "text" - }, - { - "block_id": "p329-b1", - "global_id": 8865, - "bbox": [ - 176.32, - 165.57, - 398.36, - 186.76 - ], - "text": "0\n5\n10\n15\n20\n25\n30\nn", - "type": "text" - }, - { - "block_id": "p329-b2", - "global_id": 8866, - "bbox": [ - 162.48, - 153.08, - 170.48, - 161.08 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p329-b3", - "global_id": 8867, - "bbox": [ - 166.48, - 120.34, - 170.48, - 128.34 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p329-b4", - "global_id": 8868, - "bbox": [ - 166.48, - 87.63, - 170.48, - 95.63 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p329-b5", - "global_id": 8869, - "bbox": [ - 152.26, - 116.3, - 161.06, - 130.96 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p329-b6", - "global_id": 8870, - "bbox": [ - 151.5, - 193.06, - 350.23, - 202.68 - ], - "text": "Figure 3.33 h[n] for y[n] −y[n −1] + y[n −2] = x[n].", - "type": "text" - }, - { - "block_id": "p329-b7", - "global_id": 8871, - "bbox": [ - 177.21, - 312.91, - 406.45, - 334.1 - ], - "text": "0\n5\n10\n15\n20\n25\n30\nn", - "type": "text" - }, - { - "block_id": "p329-b8", - "global_id": 8872, - "bbox": [ - 162.99, - 303.68, - 174.99, - 311.68 - ], - "text": "–20", - "type": "text" - }, - { - "block_id": "p329-b9", - "global_id": 8873, - "bbox": [ - 170.99, - 267.68, - 174.99, - 275.68 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p329-b10", - "global_id": 8874, - "bbox": [ - 166.49, - 231.68, - 174.49, - 239.68 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p329-b11", - "global_id": 8875, - "bbox": [ - 152.26, - 263.25, - 161.06, - 277.91 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p329-b12", - "global_id": 8876, - "bbox": [ - 151.5, - 340.41, - 423.17, - 350.02 - ], - "text": "Figure 3.34 Resonant zero-state response y[n] for x[n] = cos(2πn/6)u[n].", - "type": "text" - }, - { - "block_id": "p329-b13", - "global_id": 8877, - "bbox": [ - 127.59, - 377.09, - 516.13, - 399.42 - ], - "text": "As shown in Fig. 3.33, h[n] appears to be (N0 = 6)-periodic for n ≥0. Since periodic signals\nare not absolutely summable, %∞", - "type": "text" - }, - { - "block_id": "p329-b14", - "global_id": 8878, - "bbox": [ - 127.59, - 389.04, - 516.14, - 423.33 - ], - "text": "n=−∞|h[n]| is not finite and the system is not BIBO-stable.\nFurthermore, the sinusoidal input x[n] = cos(2πn/6)u[n], which is (N0 = 6)-periodic for n ≥0,\nshould generate a resonant zero-state response.", - "type": "text" - }, - { - "block_id": "p329-b15", - "global_id": 8879, - "bbox": [ - 127.59, - 440.57, - 373.42, - 474.44 - ], - "text": ">>\nx = @(n) cos(2*pi*n/6).*(n>=0);\n>>\ny = filter(b,a,x(n));\n>>\nstem(n,y,’k’); xlabel(’n’); ylabel(’y[n]’);", - "type": "text" - }, - { - "block_id": "p329-b16", - "global_id": 8880, - "bbox": [ - 127.59, - 491.09, - 516.13, - 536.92 - ], - "text": "The response’s linear envelope, shown in Fig. 3.34, confirms a resonant response. The\ncharacteristic equation of the system is γ 2 −γ + 1, which has roots γ = e±jπ/3. Since the input\nx[n] = cos(2πn/6)u[n] = (1/2)(ejπn/3 + e−jπn/3)u[n] coincides with the characteristic roots, a\nresonant response is guaranteed.", - "type": "text" - }, - { - "block_id": "p329-b17", - "global_id": 8881, - "bbox": [ - 127.59, - 538.91, - 516.11, - 572.79 - ], - "text": "By adding initial conditions, the filter command can also compute a system’s zero-input\nresponse and total response. Continuing the preceding example, consider finding the zero-input\nresponse for y[−1] = 1 and y[−2] = 2 over (0 ≤n ≤30).", - "type": "text" - }, - { - "block_id": "p329-b18", - "global_id": 8882, - "bbox": [ - 127.6, - 590.02, - 410.03, - 635.85 - ], - "text": ">>\nz_i = filtic(b,a,[1 2]);\n>>\ny_0 = filter(b,a,zeros(size(n)),z_i);\n>>\nstem(n,y_0,’k’); xlabel(’n’); ylabel(’y_{0} [n]’);\n>>\naxis([-0.5 30.5 -2.1 2.1]);", - "type": "text" - } - ] - }, - { - "page_num": 330, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p330-b0", - "global_id": 8883, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "310\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p330-b1", - "global_id": 8884, - "bbox": [ - 153.04, - 167.57, - 375.07, - 188.76 - ], - "text": "0\n5\n10\n15\n20\n25\n30\nn", - "type": "text" - }, - { - "block_id": "p330-b2", - "global_id": 8885, - "bbox": [ - 139.21, - 156.63, - 147.21, - 164.63 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p330-b3", - "global_id": 8886, - "bbox": [ - 143.21, - 122.34, - 147.21, - 130.34 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p330-b4", - "global_id": 8887, - "bbox": [ - 143.21, - 88.07, - 147.21, - 96.07 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p330-b5", - "global_id": 8888, - "bbox": [ - 126.45, - 117.25, - 135.98, - 135.58 - ], - "text": "y0 [n]", - "type": "text" - }, - { - "block_id": "p330-b6", - "global_id": 8889, - "bbox": [ - 125.76, - 195.06, - 375.99, - 205.14 - ], - "text": "Figure 3.35 Zero-input response y0[n] for y[−1] = 1 and y[−2] = 2.", - "type": "text" - }, - { - "block_id": "p330-b7", - "global_id": 8890, - "bbox": [ - 101.84, - 225.82, - 490.41, - 319.46 - ], - "text": "There are many physical ways to implement a particular equation. MATLAB implements\nEq. (3.44) by using the popular direct form II transposed structure.† Consequently, initial\nconditions must be compatible with this implementation structure. The signal-processing toolbox\nfunction filtic converts the traditional y[−1], y[−2], ..., y[−N] initial conditions for use with\nthe filter command. An input of zero is created with the zeros command. The dimensions of\nthis zero input are made to match the vector n by using the size command. Finally, _{ } forces\nsubscript text in the graphics window, and ^{ } forces superscript text. The results are shown in\nFig. 3.35.", - "type": "text" - }, - { - "block_id": "p330-b8", - "global_id": 8891, - "bbox": [ - 101.85, - 321.05, - 490.39, - 343.67 - ], - "text": "Given y[−1] = 1 and y[−2] = 2 and an input x[n] = cos(2πn/6)u[n], the total response is\neasy to obtain with the filter command.", - "type": "text" - }, - { - "block_id": "p330-b9", - "global_id": 8892, - "bbox": [ - 101.85, - 354.26, - 284.91, - 364.22 - ], - "text": ">>\ny_total = filter(b,a,x(n),z_i);", - "type": "text" - }, - { - "block_id": "p330-b10", - "global_id": 8893, - "bbox": [ - 101.85, - 374.54, - 490.36, - 396.45 - ], - "text": "Summing the zero-state and zero-input response gives the same result. Computing the total\nabsolute error provides a check.", - "type": "text" - }, - { - "block_id": "p330-b11", - "global_id": 8894, - "bbox": [ - 101.85, - 407.34, - 263.99, - 429.26 - ], - "text": ">>\nsum(abs(y_total-(y + y_0)))\nans = 1.8430e-014", - "type": "text" - }, - { - "block_id": "p330-b12", - "global_id": 8895, - "bbox": [ - 101.85, - 439.56, - 373.99, - 449.52 - ], - "text": "Within computer round-off, both methods return the same sequence.", - "type": "text" - }, - { - "block_id": "p330-b13", - "global_id": 8896, - "bbox": [ - 101.84, - 474.92, - 276.68, - 486.87 - ], - "text": "3.11-3 A Custom Filter Function", - "type": "text" - }, - { - "block_id": "p330-b14", - "global_id": 8897, - "bbox": [ - 101.84, - 493.0, - 490.39, - 539.91 - ], - "text": "The filtic command is available only if the signal-processing toolbox is installed. To\naccommodate installations without the signal-processing toolbox and to help develop your\nMATLAB skills, consider writing a function similar in syntax to filter that directly uses the\nICs y[−1], y[−2], ..., y[−N]. Normalizing a0 = 1 and solving Eq. (3.44) for y[n] yield", - "type": "text" - }, - { - "block_id": "p330-b15", - "global_id": 8898, - "bbox": [ - 221.31, - 560.52, - 247.17, - 570.8 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p330-b16", - "global_id": 8899, - "bbox": [ - 249.22, - 550.57, - 263.32, - 561.02 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p330-b17", - "global_id": 8900, - "bbox": [ - 250.2, - 574.92, - 262.34, - 582.18 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p330-b18", - "global_id": 8901, - "bbox": [ - 264.42, - 560.52, - 313.95, - 571.6 - ], - "text": "bkx[n −k] −", - "type": "text" - }, - { - "block_id": "p330-b19", - "global_id": 8902, - "bbox": [ - 315.52, - 550.57, - 329.62, - 561.02 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p330-b20", - "global_id": 8903, - "bbox": [ - 316.49, - 574.92, - 328.63, - 582.18 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p330-b21", - "global_id": 8904, - "bbox": [ - 330.72, - 560.52, - 370.91, - 571.6 - ], - "text": "aky[n −k]", - "type": "text" - }, - { - "block_id": "p330-b22", - "global_id": 8905, - "bbox": [ - 101.85, - 592.51, - 389.42, - 602.47 - ], - "text": "This recursive form provides a good basis for our custom filter function.", - "type": "text" - }, - { - "block_id": "p330-b23", - "global_id": 8906, - "bbox": [ - 101.84, - 621.19, - 408.15, - 633.41 - ], - "text": "† Implementation structures, such as direct form II transposed, are discussed in Ch. 4.", - "type": "text" - } - ] - }, - { - "page_num": 331, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p331-b0", - "global_id": 8907, - "bbox": [ - 268.15, - 62.89, - 516.13, - 71.98 - ], - "text": "3.11\nMATLAB: Discrete-Time Signals and Systems\n311", - "type": "text" - }, - { - "block_id": "p331-b1", - "global_id": 8908, - "bbox": [ - 127.59, - 86.24, - 467.54, - 179.88 - ], - "text": "function [y] = CH3MP1(b,a,x,yi);\n% CH3MP1.m : Chapter 3, MATLAB Program 1\n% Function M-file filters data x to create y\n% INPUTS:\nb = vector of feedforward coefficients\n%\na = vector of feedback coefficients\n%\nx = input data vector\n%\nyi = vector of initial conditions [y[-1], y[-2], ...]\n% OUTPUTS:\ny = vector of filtered output data", - "type": "text" - }, - { - "block_id": "p331-b2", - "global_id": 8909, - "bbox": [ - 127.59, - 193.84, - 509.4, - 251.62 - ], - "text": "yi = flipud(yi(:)); % Properly format IC’s.\ny = [yi;zeros(length(x),1)]; % Preinitialize y, beginning with IC’s.\nx = [zeros(length(yi),1);x(:)]; % Append x with zeros to match size of y.\nb = b/a(1);a = a/a(1); % Normalize coefficients.\nfor n = length(yi)+1:length(y),", - "type": "text" - }, - { - "block_id": "p331-b3", - "global_id": 8910, - "bbox": [ - 148.51, - 253.61, - 268.8, - 263.58 - ], - "text": "for nb = 0:length(b)-1,", - "type": "text" - }, - { - "block_id": "p331-b4", - "global_id": 8911, - "bbox": [ - 148.51, - 265.57, - 436.17, - 299.44 - ], - "text": "y(n) = y(n) + b(nb+1)*x(n-nb); % Feedforward terms.\nend\nfor na = 1:length(a)-1,", - "type": "text" - }, - { - "block_id": "p331-b5", - "global_id": 8912, - "bbox": [ - 127.59, - 301.43, - 436.18, - 347.26 - ], - "text": "y(n) = y(n) - a(na+1)*y(n-na); % Feedback terms.\nend\nend\ny = y(length(yi)+1:end); % Strip off IC’s for final output.", - "type": "text" - }, - { - "block_id": "p331-b6", - "global_id": 8913, - "bbox": [ - 127.59, - 361.89, - 516.14, - 431.63 - ], - "text": "Most instructions in CH3MP1 have been discussed; now we turn to the flipud instruction. The\nflip up-down command flipud reverses the order of elements in a column vector. Although not\nused here, the flip left-right command fliplr reverses the order of elements in a row vector. Note\nthat typing help filename displays the first contiguous set of comment lines in an M-file. Thus,\nit is good programming practice to document M-files, as in CH3MP1, with an initial block of clear\ncomment lines.", - "type": "text" - }, - { - "block_id": "p331-b7", - "global_id": 8914, - "bbox": [ - 127.59, - 433.62, - 516.14, - 456.31 - ], - "text": "As an exercise, the reader should verify that CH3MP1 correctly computes the impulse response\nh[n], the zero-state response y[n], the zero-input response y0[n], and the total response y[n]+y0[n].", - "type": "text" - }, - { - "block_id": "p331-b8", - "global_id": 8915, - "bbox": [ - 127.59, - 485.25, - 313.05, - 497.2 - ], - "text": "3.11-4 Discrete-Time Convolution", - "type": "text" - }, - { - "block_id": "p331-b9", - "global_id": 8916, - "bbox": [ - 127.59, - 503.33, - 516.14, - 549.16 - ], - "text": "Convolution of two finite-duration discrete-time signals is accomplished by using the conv\ncommand. For example, the discrete-time convolution of two length-4 rectangular pulses, g[n] =\n(u[n]−u[n−4])∗(u[n]−u[n−4]), is a length-(4+4−1 = 7) triangle. Representing u[n]−u[n−4]\nby the vector [1,1,1,1], the convolution is computed by", - "type": "text" - }, - { - "block_id": "p331-b10", - "global_id": 8917, - "bbox": [ - 127.6, - 564.36, - 373.38, - 586.29 - ], - "text": ">>\nconv([1 1 1 1],[1 1 1 1])\nans = 1\n2\n3\n4\n3\n2\n1", - "type": "text" - }, - { - "block_id": "p331-b11", - "global_id": 8918, - "bbox": [ - 127.6, - 600.5, - 516.15, - 635.08 - ], - "text": "Notice that (u[n+4]−u[n])∗(u[n]−u[n−4]) is also computed by conv([1 1 1 1],[1 1\n1 1]) and obviously yields the same result. The difference between these two cases is the regions\nof support: (0 ≤n ≤6) for the first and (−4 ≤n ≤2) for the second. Although the conv command", - "type": "text" - } - ] - }, - { - "page_num": 332, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p332-b0", - "global_id": 8919, - "bbox": [ - 60.0, - 62.89, - 400.04, - 71.98 - ], - "text": "312\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p332-b1", - "global_id": 8920, - "bbox": [ - 101.84, - 85.4, - 489.89, - 108.82 - ], - "text": "does not compute the region of support, it is relatively easy to obtain. If vector w begins at n = nw\nand vector v begins at n = nv, then conv(w,v) begins at n = nw + nv.", - "type": "text" - }, - { - "block_id": "p332-b2", - "global_id": 8921, - "bbox": [ - 101.84, - 109.73, - 490.42, - 179.47 - ], - "text": "In general, the conv command cannot properly convolve infinite-duration signals. This is not\ntoo surprising, since computers themselves cannot store an infinite-duration signal. For special\ncases, however, conv can correctly compute a portion of such convolution problems. Consider\nthe common case of convolving two causal signals. By passing the first N samples of each, conv\nreturns a length-(2N −1) sequence. The first N samples of this sequence are valid; the remaining\nN −1 samples are not.", - "type": "text" - }, - { - "block_id": "p332-b3", - "global_id": 8922, - "bbox": [ - 101.84, - 181.05, - 490.4, - 215.34 - ], - "text": "To illustrate this point, reconsider the zero-state response y[n] over (0 ≤n ≤30) for system\ny[n]−y[n−1]+y[n−2] = x[n] given input x[n] = cos(2πn/6)u[n]. The results obtained by using\na filtering approach are shown in Fig. 3.34.", - "type": "text" - }, - { - "block_id": "p332-b4", - "global_id": 8923, - "bbox": [ - 101.85, - 216.91, - 490.39, - 239.25 - ], - "text": "The response can also be computed using convolution according to y[n] = h[n] ∗x[n]. The\nimpulse response of this system is†", - "type": "text" - }, - { - "block_id": "p332-b5", - "global_id": 8924, - "bbox": [ - 210.38, - 260.07, - 236.79, - 270.35 - ], - "text": "h[n] =", - "type": "text" - }, - { - "block_id": "p332-b7", - "global_id": 8925, - "bbox": [ - 245.0, - 253.5, - 309.7, - 270.45 - ], - "text": "cos(πn/3) + 1\n√", - "type": "text" - }, - { - "block_id": "p332-b8", - "global_id": 8926, - "bbox": [ - 308.93, - 260.07, - 357.99, - 278.87 - ], - "text": "3\nsin(πn/3)", - "type": "text" - }, - { - "block_id": "p332-b10", - "global_id": 8927, - "bbox": [ - 365.27, - 260.07, - 381.86, - 270.35 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p332-b11", - "global_id": 8928, - "bbox": [ - 101.85, - 291.01, - 490.38, - 313.34 - ], - "text": "Both h[n] and x[n] are causal and have infinite duration, so conv can be used to obtain a portion\nof the convolution.", - "type": "text" - }, - { - "block_id": "p332-b12", - "global_id": 8929, - "bbox": [ - 101.85, - 327.24, - 494.12, - 361.11 - ], - "text": ">>\nu = @(n) 1.0.*(n>=0); h = @(n) (cos(pi*n/3)+sin(pi*n/3)/sqrt(3)).*u(n);\n>>\ny = conv(h(n),x(n));\n>>\nstem([0:60],y,’k’); xlabel(’n’); ylabel(’y[n]’);", - "type": "text" - }, - { - "block_id": "p332-b13", - "global_id": 8930, - "bbox": [ - 101.85, - 374.43, - 490.4, - 408.3 - ], - "text": "The conv output is fully displayed in Fig. 3.36. As expected, the results are correct over\n(0 ≤n ≤30). The remaining values are clearly incorrect; the output envelope should continue to\ngrow, not decay. Normally, these incorrect values are not displayed.", - "type": "text" - }, - { - "block_id": "p332-b14", - "global_id": 8931, - "bbox": [ - 101.85, - 422.2, - 379.06, - 432.16 - ], - "text": ">>\nstem(n,y(1:31),’k’); xlabel(’n’); ylabel(’y[n]’);", - "type": "text" - }, - { - "block_id": "p332-b15", - "global_id": 8932, - "bbox": [ - 101.85, - 445.48, - 269.54, - 455.44 - ], - "text": "The resulting plot is identical to Fig. 3.34.", - "type": "text" - }, - { - "block_id": "p332-b16", - "global_id": 8933, - "bbox": [ - 128.18, - 556.76, - 468.88, - 577.95 - ], - "text": "0\n10\n20\n30\n40\n50\n60\nn", - "type": "text" - }, - { - "block_id": "p332-b17", - "global_id": 8934, - "bbox": [ - 113.16, - 547.54, - 125.16, - 555.54 - ], - "text": "–20", - "type": "text" - }, - { - "block_id": "p332-b18", - "global_id": 8935, - "bbox": [ - 121.16, - 511.54, - 125.16, - 519.54 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p332-b19", - "global_id": 8936, - "bbox": [ - 116.84, - 475.53, - 124.84, - 483.53 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p332-b20", - "global_id": 8937, - "bbox": [ - 102.61, - 507.25, - 111.41, - 521.91 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p332-b21", - "global_id": 8938, - "bbox": [ - 101.84, - 584.26, - 344.04, - 593.87 - ], - "text": "Figure 3.36 y[n] for x[n] = cos(2πn/6)u[n] computed with conv.", - "type": "text" - }, - { - "block_id": "p332-b22", - "global_id": 8939, - "bbox": [ - 101.84, - 621.19, - 339.52, - 633.41 - ], - "text": "† Techniques to analytically determine h[n] are presented in Ch. 5.", - "type": "text" - } - ] - }, - { - "page_num": 333, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p333-b0", - "global_id": 8940, - "bbox": [ - 409.68, - 62.89, - 516.14, - 71.98 - ], - "text": "3.13\nSummary\n313", - "type": "text" - }, - { - "block_id": "p333-b1", - "global_id": 8941, - "bbox": [ - 127.94, - 92.21, - 363.96, - 122.09 - ], - "text": "3.12 APPENDIX: IMPULSE RESPONSE\nFOR A SPECIAL CASE", - "type": "text" - }, - { - "block_id": "p333-b2", - "global_id": 8942, - "bbox": [ - 127.59, - 127.67, - 516.13, - 151.08 - ], - "text": "When aN = 0, A0 = bN/aN becomes indeterminate, and the procedure needs to be modified slightly.\nWhen aN = 0, Q[E] can be expressed as E ˆQ[E], and Eq. (3.26) can be expressed as", - "type": "text" - }, - { - "block_id": "p333-b3", - "global_id": 8943, - "bbox": [ - 203.23, - 159.65, - 440.47, - 172.09 - ], - "text": "E ˆQ[E]h[n] = P[E]δ[n] = P[E]{Eδ[n −1]} = EP[E]δ[n −1]", - "type": "text" - }, - { - "block_id": "p333-b4", - "global_id": 8944, - "bbox": [ - 127.6, - 184.2, - 155.51, - 194.17 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p333-b5", - "global_id": 8945, - "bbox": [ - 271.97, - 193.94, - 371.75, - 206.38 - ], - "text": "ˆQ[E]h[n] = P[E]δ[n −1]", - "type": "text" - }, - { - "block_id": "p333-b6", - "global_id": 8946, - "bbox": [ - 127.6, - 215.09, - 516.14, - 249.38 - ], - "text": "In this case the input vanishes not for n ≥1, but for n ≥2. Therefore, the response consists not\nonly of the zero-input term and an impulse A0δ[n] (at n = 0), but also of an impulse A1δ[n−1] (at\nn = 1). Therefore,", - "type": "text" - }, - { - "block_id": "p333-b7", - "global_id": 8947, - "bbox": [ - 244.34, - 251.21, - 399.37, - 262.36 - ], - "text": "h[n] = A0δ[n] + A1δ[n −1] + yc[n]u[n]", - "type": "text" - }, - { - "block_id": "p333-b8", - "global_id": 8948, - "bbox": [ - 127.59, - 270.31, - 516.14, - 328.5 - ], - "text": "We can determine the unknowns A0, A1, and the N −1 coefficients in yc[n] from the N + 1\nnumber of initial values h[0], h[1], . . . , h[N], determined as usual from the iterative solution\nof the equation Q[E]h[n] = P[E]δ[n].† Similarly, if aN = aN−1 = 0, we need to use the form\nh[n] = A0δ[n]+A1δ[n−1]+A2δ[n−2]+yc[n]u[n]. The N +1 unknown constants are determined\nfrom the N + 1 values h[0], h[1], . . . , h[N], determined iteratively, and so on.", - "type": "text" - }, - { - "block_id": "p333-b9", - "global_id": 8949, - "bbox": [ - 127.94, - 357.95, - 227.89, - 371.9 - ], - "text": "3.13 SUMMARY", - "type": "text" - }, - { - "block_id": "p333-b10", - "global_id": 8950, - "bbox": [ - 127.59, - 377.88, - 516.15, - 447.62 - ], - "text": "This chapter discusses time-domain analysis of LTID (linear, time-invariant, discrete-time)\nsystems. The analysis is parallel to that of LTIC systems, with some minor differences.\nDiscrete-time systems are described by difference equations. For an Nth-order system, N auxiliary\nconditions must be specified for a unique solution. Characteristic modes are discrete-time\nexponentials of the form γ n corresponding to an unrepeated root γ , and the modes are of the\nform niγ n corresponding to a repeated root γ .", - "type": "text" - }, - { - "block_id": "p333-b11", - "global_id": 8951, - "bbox": [ - 127.59, - 449.2, - 516.13, - 483.49 - ], - "text": "The unit impulse function δ[n] is a sequence of a single number of unit value at n = 0. The\nunit impulse response h[n] of a discrete-time system is a linear combination of its characteristic\nmodes.‡", - "type": "text" - }, - { - "block_id": "p333-b12", - "global_id": 8952, - "bbox": [ - 127.59, - 485.48, - 516.18, - 567.17 - ], - "text": "The zero-state response (response due to external input) of a linear system is obtained by\nbreaking the input into impulse components and then adding the system responses to all the\nimpulse components. The sum of the system responses to the impulse components is in the form of\na sum, known as the convolution sum, whose structure and properties are similar to the convolution\nintegral. The system response is obtained as the convolution sum of the input x[n] with the system’s\nimpulse response h[n]. Therefore, the knowledge of the system’s impulse response allows us to\ndetermine the system response to any arbitrary input.", - "type": "text" - }, - { - "block_id": "p333-b13", - "global_id": 8953, - "bbox": [ - 127.59, - 565.55, - 516.14, - 591.08 - ], - "text": "LTID systems have a very special relationship to the everlasting exponential signal zn because\nthe response of an LTID system to such an input signal is the same signal within a multiplicative", - "type": "text" - }, - { - "block_id": "p333-b14", - "global_id": 8954, - "bbox": [ - 127.59, - 609.56, - 454.89, - 633.41 - ], - "text": "† ˆQ[γ ] is now an (N −1)-order polynomial. Hence there are only N −1 unknowns in yc[n].\n‡ There is a possibility of an impulse δ[n] in addition to characteristic modes.", - "type": "text" - } - ] - }, - { - "page_num": 334, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p334-b0", - "global_id": 8955, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "314\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p334-b1", - "global_id": 8956, - "bbox": [ - 101.84, - 82.2, - 490.38, - 107.74 - ], - "text": "constant. The response of an LTID system to the everlasting exponential input zn is H[z]zn, where\nH[z] is the transfer function of the system.", - "type": "text" - }, - { - "block_id": "p334-b2", - "global_id": 8957, - "bbox": [ - 101.85, - 109.73, - 490.4, - 143.6 - ], - "text": "The external stability criterion, the bounded-input/bounded-output (BIBO) stability criterion,\nstates that a system is stable if and only if every bounded input produces a bounded output.\nOtherwise the system is unstable.", - "type": "text" - }, - { - "block_id": "p334-b3", - "global_id": 8958, - "bbox": [ - 101.85, - 145.59, - 490.38, - 167.51 - ], - "text": "The internal stability criterion can be stated in terms of the location of characteristic roots of\nthe system as follows:", - "type": "text" - }, - { - "block_id": "p334-b4", - "global_id": 8959, - "bbox": [ - 118.79, - 175.48, - 490.42, - 257.18 - ], - "text": "1. An LTID system is asymptotically stable if and only if all the characteristic roots are inside\nthe unit circle. The roots may be repeated or unrepeated.\n2. An LTID system is unstable if and only if either one or both of the following conditions\nexist: (i) at least one root is outside the unit circle; (ii) there are repeated roots on the unit\ncircle.\n3. An LTID system is marginally stable if and only if there are no roots outside the unit circle\nand some unrepeated roots on the unit circle.", - "type": "text" - }, - { - "block_id": "p334-b5", - "global_id": 8960, - "bbox": [ - 119.79, - 265.15, - 490.37, - 275.11 - ], - "text": "An asymptotically stable system is always BIBO-stable. The converse is not necessarily true.", - "type": "text" - }, - { - "block_id": "p334-b6", - "global_id": 8961, - "bbox": [ - 80.93, - 295.77, - 191.52, - 312.71 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p334-b7", - "global_id": 8962, - "bbox": [ - 62.08, - 323.07, - 263.24, - 343.07 - ], - "text": "3.1-1\nFind the energy of the signals depicted in\nFig. P3.1-1.", - "type": "text" - }, - { - "block_id": "p334-b8", - "global_id": 8963, - "bbox": [ - 62.08, - 347.98, - 263.23, - 367.99 - ], - "text": "3.1-2\nFind the power of the signals illustrated in\nFig. P3.1-2.", - "type": "text" - }, - { - "block_id": "p334-b9", - "global_id": 8964, - "bbox": [ - 62.08, - 369.71, - 263.24, - 404.85 - ], - "text": "3.1-3\nShow that the power of a signal Dej(2π/N0)n is\n|D|2. Hence, show that the power of a signal\nx[n] = %N0−1", - "type": "text" - }, - { - "block_id": "p334-b10", - "global_id": 8965, - "bbox": [ - 124.33, - 388.88, - 240.35, - 407.33 - ], - "text": "r=0 Drejr(2π/N0)n is Px = %N0−1", - "type": "text" - }, - { - "block_id": "p334-b11", - "global_id": 8966, - "bbox": [ - 90.72, - 394.6, - 263.24, - 415.9 - ], - "text": "r=0 |Dr|2.\nUse the fact that", - "type": "text" - }, - { - "block_id": "p334-b12", - "global_id": 8967, - "bbox": [ - 103.67, - 417.57, - 119.52, - 427.9 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p334-b13", - "global_id": 8968, - "bbox": [ - 105.96, - 440.38, - 117.22, - 447.12 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p334-b14", - "global_id": 8969, - "bbox": [ - 120.52, - 425.09, - 174.21, - 436.7 - ], - "text": "ej(r−m)2πk/N0 =", - "type": "text" - }, - { - "block_id": "p334-b15", - "global_id": 8970, - "bbox": [ - 176.06, - 414.86, - 244.11, - 442.22 - ], - "text": "N0\nr = m\n0\notherwise", - "type": "text" - }, - { - "block_id": "p334-b16", - "global_id": 8971, - "bbox": [ - 62.08, - 453.25, - 263.25, - 462.29 - ], - "text": "3.1-4\n(a) Determine even and odd components of the", - "type": "text" - }, - { - "block_id": "p334-b17", - "global_id": 8972, - "bbox": [ - 90.72, - 462.85, - 263.23, - 484.2 - ], - "text": "signal x[n] = (0.8)nu[n].\n(b) Show that the energy of x[n] is the sum of", - "type": "text" - }, - { - "block_id": "p334-b18", - "global_id": 8973, - "bbox": [ - 91.22, - 486.2, - 263.24, - 517.08 - ], - "text": "energies of its odd and even components\nfound in part (a).\n(c) Generalize the result in part (b) for any finite", - "type": "text" - }, - { - "block_id": "p334-b19", - "global_id": 8974, - "bbox": [ - 106.16, - 519.08, - 156.3, - 528.04 - ], - "text": "energy signal.", - "type": "text" - }, - { - "block_id": "p334-b20", - "global_id": 8975, - "bbox": [ - 62.08, - 532.65, - 263.24, - 542.67 - ], - "text": "3.1-5\n(a) If xe[n] and xo[n] are the even and the odd", - "type": "text" - }, - { - "block_id": "p334-b21", - "global_id": 8976, - "bbox": [ - 90.72, - 543.61, - 263.24, - 587.16 - ], - "text": "components of causal energy signal x[n],\nthen determine Exe and Ex0, and show that\nExe + Ex0 = Ex.\n(b) Show that the cross-energy of xe and xo is", - "type": "text" - }, - { - "block_id": "p334-b22", - "global_id": 8977, - "bbox": [ - 106.15, - 587.82, - 149.99, - 596.78 - ], - "text": "zero, that is,", - "type": "text" - }, - { - "block_id": "p334-b23", - "global_id": 8978, - "bbox": [ - 152.28, - 604.26, - 164.97, - 614.02 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p334-b24", - "global_id": 8979, - "bbox": [ - 148.65, - 625.88, - 168.6, - 632.56 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p334-b25", - "global_id": 8980, - "bbox": [ - 169.6, - 613.57, - 220.74, - 623.58 - ], - "text": "xe[n]xo[n] = 0", - "type": "text" - }, - { - "block_id": "p334-b26", - "global_id": 8981, - "bbox": [ - 289.22, - 323.07, - 342.07, - 332.11 - ], - "text": "3.1-6\nDefine", - "type": "text" - }, - { - "block_id": "p334-b27", - "global_id": 8982, - "bbox": [ - 360.11, - 347.1, - 383.4, - 356.35 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p334-b28", - "global_id": 8983, - "bbox": [ - 385.25, - 334.52, - 403.82, - 347.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p334-b29", - "global_id": 8984, - "bbox": [ - 399.46, - 334.72, - 441.97, - 362.23 - ], - "text": "3\nn\nn ≥0\nAn\nn < 0", - "type": "text" - }, - { - "block_id": "p334-b30", - "global_id": 8985, - "bbox": [ - 318.37, - 382.72, - 490.39, - 393.12 - ], - "text": "(a) Determine the energy Ex and power Px of", - "type": "text" - }, - { - "block_id": "p334-b31", - "global_id": 8986, - "bbox": [ - 333.31, - 392.04, - 378.32, - 402.74 - ], - "text": "x[n] if A = 1", - "type": "text" - }, - { - "block_id": "p334-b32", - "global_id": 8987, - "bbox": [ - 317.86, - 393.78, - 490.39, - 415.04 - ], - "text": "2.\n(b) Determine the energy Ex and power Px of", - "type": "text" - }, - { - "block_id": "p334-b33", - "global_id": 8988, - "bbox": [ - 318.37, - 415.32, - 490.39, - 436.96 - ], - "text": "x[n] if A = 1.\n(c) Determine the energy Ex and power Px of", - "type": "text" - }, - { - "block_id": "p334-b34", - "global_id": 8989, - "bbox": [ - 333.31, - 437.24, - 380.61, - 446.57 - ], - "text": "x[n] if A = 2.", - "type": "text" - }, - { - "block_id": "p334-b35", - "global_id": 8990, - "bbox": [ - 289.22, - 452.11, - 490.39, - 472.13 - ], - "text": "3.1-7\nDetermine the energy Ex and power Px of the\ncomplex DT signal x[n] = Re", - "type": "text" - }, - { - "block_id": "p334-b36", - "global_id": 8991, - "bbox": [ - 424.81, - 455.58, - 464.99, - 472.13 - ], - "text": "*\n3(ejπ/4)n+", - "type": "text" - }, - { - "block_id": "p334-b37", - "global_id": 8992, - "bbox": [ - 289.22, - 477.38, - 490.39, - 497.68 - ], - "text": "3.2-1\nIf the energy of a signal x[n] is Ex, then find the\nenergy of the following:", - "type": "text" - }, - { - "block_id": "p334-b38", - "global_id": 8993, - "bbox": [ - 317.86, - 499.3, - 364.02, - 519.6 - ], - "text": "(a) x[−n]\n(b) x[n −m]", - "type": "text" - }, - { - "block_id": "p334-b39", - "global_id": 8994, - "bbox": [ - 317.86, - 521.22, - 451.46, - 541.52 - ], - "text": "(c) x[m −n]\n(d) Kx[n] (m integer and K constant)", - "type": "text" - }, - { - "block_id": "p334-b40", - "global_id": 8995, - "bbox": [ - 289.22, - 546.77, - 490.39, - 578.03 - ], - "text": "3.2-2\nIf the power of a periodic signal x[n] is Px, find\nand comment on the powers and the rms values\nof the following:", - "type": "text" - }, - { - "block_id": "p334-b41", - "global_id": 8996, - "bbox": [ - 317.86, - 579.64, - 354.76, - 599.94 - ], - "text": "(a) −x[n]\n(b) x[−n]", - "type": "text" - }, - { - "block_id": "p334-b42", - "global_id": 8997, - "bbox": [ - 317.86, - 601.57, - 405.71, - 621.86 - ], - "text": "(c) x[n −m] (m integer)\n(d) cx[n]", - "type": "text" - }, - { - "block_id": "p334-b43", - "global_id": 8998, - "bbox": [ - 318.37, - 623.48, - 405.71, - 632.82 - ], - "text": "(e) x[m −n] (m integer)", - "type": "text" - } - ] - }, - { - "page_num": 335, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p335-b0", - "global_id": 8999, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n315", - "type": "text" - }, - { - "block_id": "p335-b1", - "global_id": 9000, - "bbox": [ - 213.97, - 247.53, - 217.97, - 255.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p335-b2", - "global_id": 9001, - "bbox": [ - 203.04, - 205.18, - 207.04, - 213.18 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p335-b3", - "global_id": 9002, - "bbox": [ - 235.67, - 247.53, - 239.67, - 255.53 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p335-b4", - "global_id": 9003, - "bbox": [ - 176.06, - 229.68, - 207.04, - 242.84 - ], - "text": "3\n3", - "type": "text" - }, - { - "block_id": "p335-b5", - "global_id": 9004, - "bbox": [ - 214.61, - 89.61, - 334.85, - 97.69 - ], - "text": "x[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p335-b6", - "global_id": 9005, - "bbox": [ - 214.61, - 189.33, - 349.55, - 197.41 - ], - "text": "x[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p335-b7", - "global_id": 9006, - "bbox": [ - 213.97, - 147.65, - 239.97, - 155.65 - ], - "text": "0\n3", - "type": "text" - }, - { - "block_id": "p335-b8", - "global_id": 9007, - "bbox": [ - 204.97, - 102.65, - 208.97, - 110.65 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p335-b9", - "global_id": 9008, - "bbox": [ - 321.47, - 147.65, - 347.47, - 155.65 - ], - "text": "0\n3", - "type": "text" - }, - { - "block_id": "p335-b10", - "global_id": 9009, - "bbox": [ - 311.97, - 105.3, - 315.97, - 113.3 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p335-b11", - "global_id": 9010, - "bbox": [ - 369.47, - 147.65, - 373.47, - 155.65 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p335-b12", - "global_id": 9011, - "bbox": [ - 374.03, - 247.67, - 378.03, - 255.67 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p335-b13", - "global_id": 9012, - "bbox": [ - 258.03, - 147.57, - 396.03, - 155.57 - ], - "text": "n\nn", - "type": "text" - }, - { - "block_id": "p335-b14", - "global_id": 9013, - "bbox": [ - 258.03, - 247.53, - 340.17, - 255.67 - ], - "text": "n\n0", - "type": "text" - }, - { - "block_id": "p335-b15", - "global_id": 9014, - "bbox": [ - 325.17, - 216.68, - 329.17, - 224.68 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p335-b16", - "global_id": 9015, - "bbox": [ - 307.73, - 247.23, - 353.87, - 255.53 - ], - "text": "2\n2", - "type": "text" - }, - { - "block_id": "p335-b17", - "global_id": 9016, - "bbox": [ - 220.86, - 169.0, - 229.74, - 177.0 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p335-b18", - "global_id": 9017, - "bbox": [ - 220.86, - 290.73, - 229.74, - 298.73 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p335-b19", - "global_id": 9018, - "bbox": [ - 355.35, - 169.0, - 364.99, - 177.0 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p335-b20", - "global_id": 9019, - "bbox": [ - 355.51, - 290.71, - 479.01, - 299.67 - ], - "text": "(d)\nFigure P3.1-1", - "type": "text" - }, - { - "block_id": "p335-b21", - "global_id": 9020, - "bbox": [ - 415.55, - 479.4, - 423.55, - 487.4 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p335-b22", - "global_id": 9021, - "bbox": [ - 451.35, - 465.61, - 460.65, - 471.61 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p335-b23", - "global_id": 9022, - "bbox": [ - 300.35, - 437.05, - 304.35, - 445.05 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p335-b24", - "global_id": 9023, - "bbox": [ - 246.45, - 479.1, - 447.35, - 487.4 - ], - "text": "3\n0\n6\n15\n6", - "type": "text" - }, - { - "block_id": "p335-b25", - "global_id": 9024, - "bbox": [ - 293.68, - 509.24, - 304.35, - 517.53 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p335-b26", - "global_id": 9025, - "bbox": [ - 218.82, - 479.1, - 229.49, - 487.39 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p335-b27", - "global_id": 9026, - "bbox": [ - 161.34, - 465.61, - 202.42, - 474.89 - ], - "text": "12\n• • •", - "type": "text" - }, - { - "block_id": "p335-b28", - "global_id": 9027, - "bbox": [ - 300.41, - 564.92, - 342.82, - 580.89 - ], - "text": "1\nan", - "type": "text" - }, - { - "block_id": "p335-b29", - "global_id": 9028, - "bbox": [ - 177.28, - 602.83, - 433.72, - 614.13 - ], - "text": "2N0\n2N0\nN0\nN0\n0", - "type": "text" - }, - { - "block_id": "p335-b30", - "global_id": 9029, - "bbox": [ - 305.01, - 400.75, - 313.89, - 408.75 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p335-b31", - "global_id": 9030, - "bbox": [ - 303.37, - 529.92, - 313.02, - 537.92 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p335-b32", - "global_id": 9031, - "bbox": [ - 308.91, - 621.66, - 317.79, - 629.66 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p335-b33", - "global_id": 9032, - "bbox": [ - 311.67, - 550.56, - 324.55, - 558.64 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p335-b34", - "global_id": 9033, - "bbox": [ - 311.67, - 421.4, - 324.55, - 429.48 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p335-b35", - "global_id": 9034, - "bbox": [ - 311.67, - 321.41, - 324.55, - 329.49 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p335-b36", - "global_id": 9035, - "bbox": [ - 412.17, - 379.47, - 420.17, - 387.47 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p335-b37", - "global_id": 9036, - "bbox": [ - 436.96, - 365.68, - 446.26, - 371.68 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p335-b38", - "global_id": 9037, - "bbox": [ - 219.16, - 379.17, - 391.45, - 387.47 - ], - "text": "9\n9\n6\n3\n0\n3\n6", - "type": "text" - }, - { - "block_id": "p335-b39", - "global_id": 9038, - "bbox": [ - 301.96, - 334.42, - 305.96, - 342.42 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p335-b40", - "global_id": 9039, - "bbox": [ - 469.43, - 601.6, - 473.43, - 609.6 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p335-b41", - "global_id": 9040, - "bbox": [ - 451.89, - 480.23, - 455.89, - 488.23 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p335-b42", - "global_id": 9041, - "bbox": [ - 431.32, - 379.76, - 435.32, - 387.76 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p335-b43", - "global_id": 9042, - "bbox": [ - 194.46, - 364.8, - 203.76, - 370.8 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p335-b44", - "global_id": 9043, - "bbox": [ - 130.58, - 635.83, - 182.22, - 644.8 - ], - "text": "Figure P3.1-2", - "type": "text" - } - ] - }, - { - "page_num": 336, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p336-b0", - "global_id": 9044, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "316\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p336-b1", - "global_id": 9045, - "bbox": [ - 62.08, - 85.64, - 263.24, - 132.72 - ], - "text": "3.2-3\nLetting ↓identify n = 0, define the nonzero\nvalues of signal x[n] as [−1,2,−3,4,−5,4,−3,\n↓\n2,−1].\n(a) Using vector form, represent signal y[n] =", - "type": "text" - }, - { - "block_id": "p336-b2", - "global_id": 9046, - "bbox": [ - 90.72, - 134.34, - 263.24, - 165.6 - ], - "text": "x[−3n + 2]. Be sure to identify the n = 0\nelement.\n(b) Using vector form, represent signal z[n] =", - "type": "text" - }, - { - "block_id": "p336-b3", - "global_id": 9047, - "bbox": [ - 106.15, - 167.22, - 263.24, - 187.52 - ], - "text": "x[n/2 −3]. Be sure to identify the n = 0\nelement.", - "type": "text" - }, - { - "block_id": "p336-b4", - "global_id": 9048, - "bbox": [ - 62.08, - 193.36, - 263.24, - 202.69 - ], - "text": "3.2-4\nLetting\n↓\nidentify\nn\n=\n0,\ndefine\nthe", - "type": "text" - }, - { - "block_id": "p336-b5", - "global_id": 9049, - "bbox": [ - 90.72, - 209.18, - 255.52, - 218.52 - ], - "text": "nonzero\nvalues\nof\nsignal\nx[n]\nas\n[−1,", - "type": "text" - }, - { - "block_id": "p336-b6", - "global_id": 9050, - "bbox": [ - 90.72, - 203.6, - 263.24, - 229.48 - ], - "text": "↓\n2,\n−3,4,−5,4,−3,2,−1].", - "type": "text" - }, - { - "block_id": "p336-b7", - "global_id": 9051, - "bbox": [ - 91.22, - 231.37, - 263.24, - 241.77 - ], - "text": "(a) Determine the energy Ex and power Px of", - "type": "text" - }, - { - "block_id": "p336-b8", - "global_id": 9052, - "bbox": [ - 90.72, - 242.05, - 263.24, - 262.35 - ], - "text": "the signal x[n].\n(b) Using vector form, represent signal y[n] =", - "type": "text" - }, - { - "block_id": "p336-b9", - "global_id": 9053, - "bbox": [ - 91.23, - 263.97, - 263.24, - 295.23 - ], - "text": "x[2(n + 2)]. Be sure to identify the n = 0\nelement.\n(c) Using vector form, represent signal z[n] =", - "type": "text" - }, - { - "block_id": "p336-b10", - "global_id": 9054, - "bbox": [ - 106.16, - 295.22, - 132.86, - 306.09 - ], - "text": "x[−n−6", - "type": "text" - }, - { - "block_id": "p336-b11", - "global_id": 9055, - "bbox": [ - 106.16, - 296.84, - 263.24, - 317.14 - ], - "text": "3 ]. Be sure to identify the n = 0\nelement.", - "type": "text" - }, - { - "block_id": "p336-b12", - "global_id": 9056, - "bbox": [ - 62.08, - 322.98, - 263.24, - 343.27 - ], - "text": "3.2-5\nLetting ↓identify the n = 0 value, describe a\n4-periodic signal w[n] using vector notation as", - "type": "text" - }, - { - "block_id": "p336-b13", - "global_id": 9057, - "bbox": [ - 90.72, - 349.76, - 140.59, - 359.1 - ], - "text": "[· · · ,1,2,3,4,", - "type": "text" - }, - { - "block_id": "p336-b14", - "global_id": 9058, - "bbox": [ - 91.22, - 344.18, - 263.24, - 371.4 - ], - "text": "↓\n1,2,3,4,1,2,3,4,· · ·].\n(a) Determine the energy Ex and power Px of", - "type": "text" - }, - { - "block_id": "p336-b15", - "global_id": 9059, - "bbox": [ - 90.72, - 371.68, - 263.24, - 393.32 - ], - "text": "signal x[n] = w[2n].\n(b) Determine the energy Ey and power Py of", - "type": "text" - }, - { - "block_id": "p336-b16", - "global_id": 9060, - "bbox": [ - 106.15, - 392.17, - 182.6, - 402.93 - ], - "text": "signal y[n] = w[2 −n", - "type": "text" - }, - { - "block_id": "p336-b17", - "global_id": 9061, - "bbox": [ - 179.36, - 393.59, - 189.02, - 405.52 - ], - "text": "3].", - "type": "text" - }, - { - "block_id": "p336-b18", - "global_id": 9062, - "bbox": [ - 62.08, - 408.77, - 263.24, - 472.82 - ], - "text": "3.2-6\nLet DT signal x[n] have values [1, 2, 3, 4, 5, 6]\nfor 0 ≤n ≤5 and let DT signal y[n] have\nvalues [5, 0, 0, 3, 0, 0, 1] for 0 ≤n ≤6. Both\nsignals are zero outside the ranges given. Fur-\nther, define a 6-periodic replication of y[n] as\n˜y[n] = %∞", - "type": "text" - }, - { - "block_id": "p336-b19", - "global_id": 9063, - "bbox": [ - 91.22, - 463.57, - 263.24, - 485.2 - ], - "text": "k=−∞y[n −6k] .\n(a) Determine the energy Ex and power Px of", - "type": "text" - }, - { - "block_id": "p336-b20", - "global_id": 9064, - "bbox": [ - 90.72, - 485.48, - 263.22, - 505.79 - ], - "text": "signal x[n].\n(b) Determine the smallest-magnitude integers", - "type": "text" - }, - { - "block_id": "p336-b21", - "global_id": 9065, - "bbox": [ - 106.15, - 505.39, - 238.07, - 518.08 - ], - "text": "N1, N2, and N3 such that y[n] = x[ N1n", - "type": "text" - }, - { - "block_id": "p336-b22", - "global_id": 9066, - "bbox": [ - 91.22, - 507.4, - 263.24, - 529.09 - ], - "text": "N2 +N3].\n(c) Determine the energy E˜y and power P˜y of", - "type": "text" - }, - { - "block_id": "p336-b23", - "global_id": 9067, - "bbox": [ - 106.15, - 529.32, - 122.83, - 538.66 - ], - "text": "˜y[n].", - "type": "text" - }, - { - "block_id": "p336-b24", - "global_id": 9068, - "bbox": [ - 62.08, - 544.79, - 263.22, - 564.79 - ], - "text": "3.2-7\nFor the signal shown in Fig. P3.1-1b, sketch the\nfollowing signals:", - "type": "text" - }, - { - "block_id": "p336-b25", - "global_id": 9069, - "bbox": [ - 90.72, - 566.41, - 134.88, - 586.7 - ], - "text": "(a) x[−n]\n(b) x[n + 6]", - "type": "text" - }, - { - "block_id": "p336-b26", - "global_id": 9070, - "bbox": [ - 90.72, - 588.33, - 134.88, - 608.63 - ], - "text": "(c) x[n −6]\n(d) x[3n]", - "type": "text" - }, - { - "block_id": "p336-b27", - "global_id": 9071, - "bbox": [ - 91.22, - 612.48, - 110.14, - 621.54 - ], - "text": "(e) x", - "type": "text" - }, - { - "block_id": "p336-b28", - "global_id": 9072, - "bbox": [ - 110.16, - 602.31, - 120.04, - 615.16 - ], - "text": "'n", - "type": "text" - }, - { - "block_id": "p336-b29", - "global_id": 9073, - "bbox": [ - 115.56, - 618.94, - 120.04, - 627.91 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p336-b30", - "global_id": 9074, - "bbox": [ - 121.24, - 602.3, - 125.44, - 611.27 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p336-b31", - "global_id": 9075, - "bbox": [ - 92.21, - 625.76, - 134.88, - 635.1 - ], - "text": "(f) x[3 −n]", - "type": "text" - }, - { - "block_id": "p336-b32", - "global_id": 9076, - "bbox": [ - 289.22, - 85.94, - 490.39, - 105.94 - ], - "text": "3.2-8\nRepeat Prob. 3.2-7 for the signal depicted in\nFig. P3.1-1c.", - "type": "text" - }, - { - "block_id": "p336-b33", - "global_id": 9077, - "bbox": [ - 289.22, - 110.54, - 490.39, - 179.05 - ], - "text": "3.2-9\nLetting ↓identify n = 0, consider a DT\nsignal x[n] whose nonzero values are given\nas\nx[n]\n=\n[1, −3, 2, 2, 3, −2, −1, 1, 2, −3,\n↓\n3, 3, −2, 1, −3, 2, 3, −1].\nAccurately\nsketch\ny[n] = x[−1 −2n] and z[n] = x[−2 + n/3] over\n−5 ≤n ≤4.", - "type": "text" - }, - { - "block_id": "p336-b34", - "global_id": 9078, - "bbox": [ - 287.15, - 183.96, - 344.49, - 193.0 - ], - "text": "3.2-10\nDefine", - "type": "text" - }, - { - "block_id": "p336-b35", - "global_id": 9079, - "bbox": [ - 360.11, - 201.31, - 383.4, - 210.56 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p336-b36", - "global_id": 9080, - "bbox": [ - 385.25, - 188.73, - 403.82, - 201.25 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p336-b37", - "global_id": 9081, - "bbox": [ - 400.58, - 188.92, - 441.97, - 216.43 - ], - "text": "2\nn\nn ≥0\n0\nn < 0", - "type": "text" - }, - { - "block_id": "p336-b38", - "global_id": 9082, - "bbox": [ - 317.86, - 230.39, - 490.38, - 250.32 - ], - "text": "Determine and locate the two largest non-zero\nvalues of:", - "type": "text" - }, - { - "block_id": "p336-b39", - "global_id": 9083, - "bbox": [ - 317.86, - 251.93, - 385.31, - 272.97 - ], - "text": "(a) ya[n] = x[2n]\n(b) yb[n] = x[n/3]", - "type": "text" - }, - { - "block_id": "p336-b40", - "global_id": 9084, - "bbox": [ - 317.86, - 273.86, - 402.35, - 294.89 - ], - "text": "(c) yc[n] = x[3n + 1]\n(d) yd[n] = x[−2n + 5]", - "type": "text" - }, - { - "block_id": "p336-b41", - "global_id": 9085, - "bbox": [ - 318.37, - 295.77, - 412.9, - 305.85 - ], - "text": "(e) ye[n] = x[−(n + 8)/2]", - "type": "text" - }, - { - "block_id": "p336-b42", - "global_id": 9086, - "bbox": [ - 289.22, - 310.02, - 490.38, - 330.01 - ], - "text": "3.3-1\nSketch, and find the power of, the following\nsignals:", - "type": "text" - }, - { - "block_id": "p336-b43", - "global_id": 9087, - "bbox": [ - 318.37, - 330.58, - 347.72, - 340.98 - ], - "text": "(a) (1)n", - "type": "text" - }, - { - "block_id": "p336-b44", - "global_id": 9088, - "bbox": [ - 317.86, - 341.54, - 354.71, - 351.94 - ], - "text": "(b) (−1)n", - "type": "text" - }, - { - "block_id": "p336-b45", - "global_id": 9089, - "bbox": [ - 317.86, - 353.55, - 370.14, - 373.85 - ], - "text": "(c) u[n]\n(d) (−1)nu[n]", - "type": "text" - }, - { - "block_id": "p336-b46", - "global_id": 9090, - "bbox": [ - 318.37, - 377.8, - 345.25, - 386.77 - ], - "text": "(e) cos", - "type": "text" - }, - { - "block_id": "p336-b47", - "global_id": 9091, - "bbox": [ - 346.25, - 367.53, - 357.03, - 380.11 - ], - "text": "'π", - "type": "text" - }, - { - "block_id": "p336-b48", - "global_id": 9092, - "bbox": [ - 352.55, - 371.14, - 379.97, - 393.13 - ], - "text": "3 n + π\n6", - "type": "text" - }, - { - "block_id": "p336-b49", - "global_id": 9093, - "bbox": [ - 382.06, - 367.53, - 386.26, - 376.5 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p336-b50", - "global_id": 9094, - "bbox": [ - 289.22, - 394.39, - 353.75, - 403.43 - ], - "text": "3.3-2\nShow that", - "type": "text" - }, - { - "block_id": "p336-b51", - "global_id": 9095, - "bbox": [ - 317.86, - 405.05, - 451.29, - 425.35 - ], - "text": "(a) δ[n] + δ[n −1] = u[n] −u[n −2]\n(b) 2n−1 sin", - "type": "text" - }, - { - "block_id": "p336-b52", - "global_id": 9096, - "bbox": [ - 361.27, - 408.8, - 373.85, - 421.06 - ], - "text": "πn", - "type": "text" - }, - { - "block_id": "p336-b53", - "global_id": 9097, - "bbox": [ - 368.34, - 408.8, - 378.66, - 427.93 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p336-b54", - "global_id": 9098, - "bbox": [ - 378.66, - 414.65, - 406.72, - 425.26 - ], - "text": "u[n]= 1", - "type": "text" - }, - { - "block_id": "p336-b55", - "global_id": 9099, - "bbox": [ - 403.48, - 408.8, - 441.17, - 428.53 - ], - "text": "2 2n sin\n πn", - "type": "text" - }, - { - "block_id": "p336-b56", - "global_id": 9100, - "bbox": [ - 435.66, - 408.8, - 445.97, - 427.93 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p336-b57", - "global_id": 9101, - "bbox": [ - 317.86, - 416.01, - 473.18, - 449.46 - ], - "text": "u[n−1]\n(c) n(n −1)γ nu[n] = n(n −1)γ nu[n −2]\n(d) (u[n] + (−1)nu[n])sin", - "type": "text" - }, - { - "block_id": "p336-b58", - "global_id": 9102, - "bbox": [ - 414.01, - 430.22, - 430.86, - 443.08 - ], - "text": "πn", - "type": "text" - }, - { - "block_id": "p336-b59", - "global_id": 9103, - "bbox": [ - 423.23, - 446.86, - 427.71, - 455.83 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p336-b61", - "global_id": 9104, - "bbox": [ - 438.78, - 440.12, - 482.73, - 449.46 - ], - "text": "= 0 for all n", - "type": "text" - }, - { - "block_id": "p336-b62", - "global_id": 9105, - "bbox": [ - 318.37, - 456.53, - 422.02, - 467.14 - ], - "text": "(e) (u[n]+(−1)n+1u[n])cos", - "type": "text" - }, - { - "block_id": "p336-b63", - "global_id": 9106, - "bbox": [ - 423.02, - 447.9, - 439.87, - 460.76 - ], - "text": "πn", - "type": "text" - }, - { - "block_id": "p336-b64", - "global_id": 9107, - "bbox": [ - 432.23, - 464.54, - 436.72, - 473.5 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p336-b66", - "global_id": 9108, - "bbox": [ - 447.53, - 457.8, - 490.38, - 467.14 - ], - "text": "= 0 for all n", - "type": "text" - }, - { - "block_id": "p336-b67", - "global_id": 9109, - "bbox": [ - 289.22, - 474.67, - 421.9, - 483.71 - ], - "text": "3.3-3\nSketch the following signals:", - "type": "text" - }, - { - "block_id": "p336-b68", - "global_id": 9110, - "bbox": [ - 317.86, - 485.33, - 401.5, - 505.62 - ], - "text": "(a) u[n −2] −u[n −6]\n(b) n{u[n] −u[n −7]}", - "type": "text" - }, - { - "block_id": "p336-b69", - "global_id": 9111, - "bbox": [ - 317.86, - 507.25, - 440.57, - 527.55 - ], - "text": "(c) (n −2){u[n −2] −u[n −6]}\n(d) (−n + 8){u[n −6] −u[n −9]}", - "type": "text" - }, - { - "block_id": "p336-b70", - "global_id": 9112, - "bbox": [ - 318.37, - 529.16, - 490.45, - 538.5 - ], - "text": "(e) (n−2){u[n−2]−u[n−6]}+(−n+8){u[n−", - "type": "text" - }, - { - "block_id": "p336-b71", - "global_id": 9113, - "bbox": [ - 333.31, - 540.12, - 383.08, - 549.46 - ], - "text": "6] −u[n −9]}", - "type": "text" - }, - { - "block_id": "p336-b72", - "global_id": 9114, - "bbox": [ - 289.22, - 554.37, - 490.37, - 574.37 - ], - "text": "3.3-4\nDescribe each of the signals in Fig. P3.1-1 by a\nsingle expression valid for all n.", - "type": "text" - }, - { - "block_id": "p336-b73", - "global_id": 9115, - "bbox": [ - 289.22, - 576.1, - 490.38, - 599.28 - ], - "text": "3.3-5\nWhy are DT signals of the form zn so important\nto the study of LTID systems?", - "type": "text" - }, - { - "block_id": "p336-b74", - "global_id": 9116, - "bbox": [ - 289.22, - 604.19, - 490.39, - 635.14 - ], - "text": "3.3-6\nExplain the similarities and differences between\nthe Kronecker delta function δ[n] and the Dirac\ndelta function δ(t).", - "type": "text" - } - ] - }, - { - "page_num": 337, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p337-b0", - "global_id": 9117, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n317", - "type": "text" - }, - { - "block_id": "p337-b1", - "global_id": 9118, - "bbox": [ - 87.82, - 85.68, - 288.98, - 107.25 - ], - "text": "3.3-7\nThe following signals are in the form eλn.\nExpress them in the form γ n:", - "type": "text" - }, - { - "block_id": "p337-b2", - "global_id": 9119, - "bbox": [ - 116.96, - 107.6, - 152.26, - 118.2 - ], - "text": "(a) e−0.5n", - "type": "text" - }, - { - "block_id": "p337-b3", - "global_id": 9120, - "bbox": [ - 116.46, - 118.76, - 147.21, - 129.16 - ], - "text": "(b) e0.5n", - "type": "text" - }, - { - "block_id": "p337-b4", - "global_id": 9121, - "bbox": [ - 116.96, - 129.52, - 150.5, - 140.12 - ], - "text": "(c) e−jπn", - "type": "text" - }, - { - "block_id": "p337-b5", - "global_id": 9122, - "bbox": [ - 116.46, - 140.48, - 145.45, - 151.08 - ], - "text": "(d) ejπn", - "type": "text" - }, - { - "block_id": "p337-b6", - "global_id": 9123, - "bbox": [ - 116.46, - 152.7, - 288.99, - 227.79 - ], - "text": "In each case show the locations of λ and γ in\nthe complex plane. Verify that an exponential is\ngrowing if γ lies outside the unit circle (or if λ\nlies in the RHP), is decaying if γ lies within the\nunit circle (or if λ lies in the LHP), and has a\nconstant amplitude if γ lies on the unit circle (or\nif λ lies on the imaginary axis).", - "type": "text" - }, - { - "block_id": "p337-b7", - "global_id": 9124, - "bbox": [ - 87.82, - 238.22, - 288.99, - 258.22 - ], - "text": "3.3-8\nExpress the following signals, which are in the\nform eλn, in the form γ n:", - "type": "text" - }, - { - "block_id": "p337-b8", - "global_id": 9125, - "bbox": [ - 116.96, - 258.57, - 163.61, - 269.17 - ], - "text": "(a) e−(1+jπ)n", - "type": "text" - }, - { - "block_id": "p337-b9", - "global_id": 9126, - "bbox": [ - 116.46, - 269.53, - 163.61, - 280.14 - ], - "text": "(b) e−(1−jπ)n", - "type": "text" - }, - { - "block_id": "p337-b10", - "global_id": 9127, - "bbox": [ - 116.96, - 280.49, - 158.56, - 291.09 - ], - "text": "(c) e(1+jπ)n", - "type": "text" - }, - { - "block_id": "p337-b11", - "global_id": 9128, - "bbox": [ - 116.46, - 291.45, - 158.56, - 302.05 - ], - "text": "(d) e(1−jπ)n", - "type": "text" - }, - { - "block_id": "p337-b12", - "global_id": 9129, - "bbox": [ - 116.96, - 302.41, - 173.89, - 313.01 - ], - "text": "(e) e−[1+j(π/3)]n", - "type": "text" - }, - { - "block_id": "p337-b13", - "global_id": 9130, - "bbox": [ - 117.95, - 313.37, - 168.84, - 323.97 - ], - "text": "(f) e[1−j(π/3)]n", - "type": "text" - }, - { - "block_id": "p337-b14", - "global_id": 9131, - "bbox": [ - 87.82, - 334.39, - 288.99, - 387.27 - ], - "text": "3.3-9\nThe concepts of even and odd functions for\ndiscrete-time signals are identical to those of the\ncontinuous-time signals discussed in Sec. 1.5.\nUsing these concepts, find and sketch the odd\nand the even components of the following:", - "type": "text" - }, - { - "block_id": "p337-b15", - "global_id": 9132, - "bbox": [ - 116.46, - 388.9, - 151.31, - 409.19 - ], - "text": "(a) u[n]\n(b) nu[n]", - "type": "text" - }, - { - "block_id": "p337-b16", - "global_id": 9133, - "bbox": [ - 116.96, - 413.14, - 142.36, - 422.1 - ], - "text": "(c) sin", - "type": "text" - }, - { - "block_id": "p337-b17", - "global_id": 9134, - "bbox": [ - 142.36, - 402.87, - 159.21, - 415.73 - ], - "text": "πn", - "type": "text" - }, - { - "block_id": "p337-b18", - "global_id": 9135, - "bbox": [ - 151.58, - 419.5, - 156.06, - 428.47 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p337-b20", - "global_id": 9136, - "bbox": [ - 116.46, - 430.81, - 143.84, - 439.78 - ], - "text": "(d) cos", - "type": "text" - }, - { - "block_id": "p337-b21", - "global_id": 9137, - "bbox": [ - 143.85, - 420.54, - 160.7, - 433.4 - ], - "text": "πn", - "type": "text" - }, - { - "block_id": "p337-b22", - "global_id": 9138, - "bbox": [ - 153.07, - 437.18, - 157.56, - 446.14 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p337-b24", - "global_id": 9139, - "bbox": [ - 87.82, - 452.53, - 288.98, - 483.79 - ], - "text": "3.4-1\nA cash register output y[n] represents the total\ncost of n items rung up by a cashier. The input\nx[n] is the cost of the nth item.", - "type": "text" - }, - { - "block_id": "p337-b25", - "global_id": 9140, - "bbox": [ - 116.97, - 485.4, - 288.98, - 494.74 - ], - "text": "(a) Write the difference equation relating y[n]", - "type": "text" - }, - { - "block_id": "p337-b26", - "global_id": 9141, - "bbox": [ - 116.46, - 496.37, - 288.97, - 516.67 - ], - "text": "to x[n].\n(b) Realize this system using a time-delay ele-", - "type": "text" - }, - { - "block_id": "p337-b27", - "global_id": 9142, - "bbox": [ - 131.9, - 518.65, - 152.06, - 527.62 - ], - "text": "ment.", - "type": "text" - }, - { - "block_id": "p337-b28", - "global_id": 9143, - "bbox": [ - 87.82, - 537.75, - 288.98, - 634.77 - ], - "text": "3.4-2\nLet p[n] be the population of a certain country\nat the beginning of the nth year. The birth and\ndeath rates of the population during any year\nare 3.3 and 1.3%, respectively. If i[n] is the\ntotal number of immigrants entering the country\nduring the nth year, write the difference equation\nrelating p[n + 1], p[n], and i[n]. Assume that\nthe immigrants enter the country throughout the\nyear at a uniform rate.", - "type": "text" - }, - { - "block_id": "p337-b29", - "global_id": 9144, - "bbox": [ - 314.97, - 85.94, - 516.13, - 193.62 - ], - "text": "3.4-3\nA moving average is used to detect a trend of\na rapidly fluctuating variable, such as the stock\nmarket average. A variable may fluctuate (up\nand down) daily, masking its long-term (secular)\ntrend. We can discern the long-term trend by\nsmoothing or averaging the past N values of the\nvariable. For the stock market average, we may\nconsider a 5-day moving average y[n] to be the\nmean of the past 5 days’ market closing values\nx[n],x[n −1],. . .,x[n −4].", - "type": "text" - }, - { - "block_id": "p337-b30", - "global_id": 9145, - "bbox": [ - 344.11, - 195.23, - 516.13, - 204.56 - ], - "text": "(a) Write the difference equation relating y[n]", - "type": "text" - }, - { - "block_id": "p337-b31", - "global_id": 9146, - "bbox": [ - 343.61, - 206.19, - 516.13, - 226.49 - ], - "text": "to the input x[n].\n(b) Use time-delay elements to realize the 5-day", - "type": "text" - }, - { - "block_id": "p337-b32", - "global_id": 9147, - "bbox": [ - 359.05, - 228.48, - 438.1, - 237.44 - ], - "text": "moving-average filter.", - "type": "text" - }, - { - "block_id": "p337-b33", - "global_id": 9148, - "bbox": [ - 314.97, - 242.5, - 510.64, - 251.54 - ], - "text": "3.4-4\nThe digital integrator in Ex. 3.9 is specified by", - "type": "text" - }, - { - "block_id": "p337-b34", - "global_id": 9149, - "bbox": [ - 388.35, - 264.56, - 471.39, - 273.9 - ], - "text": "y[n] −y[n −1] = Tx[n]", - "type": "text" - }, - { - "block_id": "p337-b35", - "global_id": 9150, - "bbox": [ - 343.61, - 286.93, - 516.12, - 318.18 - ], - "text": "If an input u[n] is applied to such an integrator,\nshow that the output is (n + 1)Tu[n], which\napproaches the desired ramp nTu[n] as T →0.", - "type": "text" - }, - { - "block_id": "p337-b36", - "global_id": 9151, - "bbox": [ - 314.97, - 323.24, - 516.12, - 343.24 - ], - "text": "3.4-5\nApproximate the following second-order differ-\nential equation with a difference equation.", - "type": "text" - }, - { - "block_id": "p337-b37", - "global_id": 9152, - "bbox": [ - 372.19, - 354.91, - 393.99, - 365.15 - ], - "text": "d2y(t)", - "type": "text" - }, - { - "block_id": "p337-b38", - "global_id": 9153, - "bbox": [ - 377.66, - 362.18, - 412.68, - 377.8 - ], - "text": "dt2\n+ a1", - "type": "text" - }, - { - "block_id": "p337-b39", - "global_id": 9154, - "bbox": [ - 414.38, - 355.9, - 432.19, - 365.15 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p337-b40", - "global_id": 9155, - "bbox": [ - 419.71, - 362.18, - 488.74, - 377.8 - ], - "text": "dt\n+ a0y(t) = x(t)", - "type": "text" - }, - { - "block_id": "p337-b41", - "global_id": 9156, - "bbox": [ - 314.97, - 387.29, - 516.13, - 407.59 - ], - "text": "3.4-6\nLetting ↓identify n = 0, define the nonzero\nvalues\nof\nsignal\ng[n]\nin\nvector\nform\nas", - "type": "text" - }, - { - "block_id": "p337-b42", - "global_id": 9157, - "bbox": [ - 343.61, - 414.07, - 399.66, - 423.41 - ], - "text": "[1,2,3,4,5,4,3,", - "type": "text" - }, - { - "block_id": "p337-b43", - "global_id": 9158, - "bbox": [ - 343.61, - 408.49, - 516.14, - 445.32 - ], - "text": "↓\n2,1]. The impulse response of\nan LTID system is defined in terms of g[n] as\nh[n] = g[−2n −1].", - "type": "text" - }, - { - "block_id": "p337-b44", - "global_id": 9159, - "bbox": [ - 344.12, - 446.95, - 516.12, - 456.29 - ], - "text": "(a) Express the nonzero values of h[n] in vector", - "type": "text" - }, - { - "block_id": "p337-b45", - "global_id": 9160, - "bbox": [ - 343.61, - 457.91, - 516.13, - 478.2 - ], - "text": "form, taking care to identify the n = 0 point.\n(b) Write a constant-coefficient linear differ-", - "type": "text" - }, - { - "block_id": "p337-b46", - "global_id": 9161, - "bbox": [ - 344.12, - 479.83, - 516.13, - 511.08 - ], - "text": "ence equation (input x[n] and output y[n])\nthat has impulse response h[n].\n(c) Show that the system is both linear and", - "type": "text" - }, - { - "block_id": "p337-b47", - "global_id": 9162, - "bbox": [ - 343.61, - 513.08, - 516.12, - 533.0 - ], - "text": "time-invariant.\n(d) Determine, if possible, whether the system", - "type": "text" - }, - { - "block_id": "p337-b48", - "global_id": 9163, - "bbox": [ - 344.12, - 534.99, - 516.12, - 554.92 - ], - "text": "is BIBO-stable.\n(e) Determine, if possible, whether the system", - "type": "text" - }, - { - "block_id": "p337-b49", - "global_id": 9164, - "bbox": [ - 345.11, - 556.92, - 516.12, - 576.84 - ], - "text": "is memoryless.\n(f) Determine, if possible, whether the system", - "type": "text" - }, - { - "block_id": "p337-b50", - "global_id": 9165, - "bbox": [ - 359.05, - 578.83, - 391.91, - 587.8 - ], - "text": "is causal.", - "type": "text" - }, - { - "block_id": "p337-b51", - "global_id": 9166, - "bbox": [ - 314.97, - 592.85, - 516.13, - 612.85 - ], - "text": "3.4-7\nAn LTID system has an impulse response func-\ntion h[n] = u[−(5 −n)/3].", - "type": "text" - }, - { - "block_id": "p337-b52", - "global_id": 9167, - "bbox": [ - 344.12, - 614.84, - 516.13, - 623.81 - ], - "text": "(a) Using an accurate sketch or vector represen-", - "type": "text" - }, - { - "block_id": "p337-b53", - "global_id": 9168, - "bbox": [ - 359.05, - 625.43, - 467.87, - 634.77 - ], - "text": "tation, graphically depict h[n].", - "type": "text" - } - ] - }, - { - "page_num": 338, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p338-b0", - "global_id": 9169, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "318\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p338-b1", - "global_id": 9170, - "bbox": [ - 90.72, - 85.9, - 263.23, - 94.86 - ], - "text": "(b) Determine, if possible, whether the system", - "type": "text" - }, - { - "block_id": "p338-b2", - "global_id": 9171, - "bbox": [ - 91.22, - 96.86, - 263.23, - 116.78 - ], - "text": "is BIBO-stable.\n(c) Determine, if possible, whether the system", - "type": "text" - }, - { - "block_id": "p338-b3", - "global_id": 9172, - "bbox": [ - 90.72, - 118.77, - 263.23, - 138.7 - ], - "text": "is memoryless.\n(d) Determine, if possible, whether the system", - "type": "text" - }, - { - "block_id": "p338-b4", - "global_id": 9173, - "bbox": [ - 106.15, - 140.69, - 139.01, - 149.66 - ], - "text": "is causal.", - "type": "text" - }, - { - "block_id": "p338-b5", - "global_id": 9174, - "bbox": [ - 62.08, - 154.75, - 263.23, - 196.68 - ], - "text": "3.4-8\nThe voltage at the nth node of a resistive ladder\nin Fig. P3.4-8 is v[n], (n = 0,1,2,. . .,N). Show\nthat v[n] satisfies the second-order difference\nequation", - "type": "text" - }, - { - "block_id": "p338-b6", - "global_id": 9175, - "bbox": [ - 98.06, - 208.74, - 254.69, - 223.99 - ], - "text": "v[n + 2] −Av[n + 1] + v[n] = 0\nA = 2 + 1", - "type": "text" - }, - { - "block_id": "p338-b7", - "global_id": 9176, - "bbox": [ - 250.21, - 221.3, - 254.69, - 230.27 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p338-b8", - "global_id": 9177, - "bbox": [ - 90.72, - 240.48, - 263.24, - 260.5 - ], - "text": "[Hint: Consider the node equation at the nth\nnode with voltage v[n].]", - "type": "text" - }, - { - "block_id": "p338-b9", - "global_id": 9178, - "bbox": [ - 62.08, - 265.61, - 263.24, - 318.49 - ], - "text": "3.4-9\nDetermine whether each of the following state-\nments is true or false. If the statement is false,\ndemonstrate by proof or example why the state-\nment is false. If the statement is true, explain\nwhy.", - "type": "text" - }, - { - "block_id": "p338-b10", - "global_id": 9179, - "bbox": [ - 91.22, - 320.47, - 263.22, - 329.44 - ], - "text": "(a) A discrete-time signal with finite power", - "type": "text" - }, - { - "block_id": "p338-b11", - "global_id": 9180, - "bbox": [ - 106.15, - 331.44, - 203.86, - 340.4 - ], - "text": "cannot be an energy signal.", - "type": "text" - }, - { - "block_id": "p338-b12", - "global_id": 9181, - "bbox": [ - 317.86, - 85.9, - 490.37, - 94.86 - ], - "text": "(b) A discrete-time signal with infinite energy", - "type": "text" - }, - { - "block_id": "p338-b13", - "global_id": 9182, - "bbox": [ - 318.37, - 96.86, - 490.39, - 116.78 - ], - "text": "must be a power signal.\n(c) The system described by y[n] = (n+1)x[n]", - "type": "text" - }, - { - "block_id": "p338-b14", - "global_id": 9183, - "bbox": [ - 317.86, - 118.78, - 490.39, - 138.7 - ], - "text": "is causal.\n(d) The system described by y[n −1] = x[n] is", - "type": "text" - }, - { - "block_id": "p338-b15", - "global_id": 9184, - "bbox": [ - 318.37, - 140.69, - 490.39, - 160.62 - ], - "text": "causal.\n(e) If an energy signal x[n] has energy E, then", - "type": "text" - }, - { - "block_id": "p338-b16", - "global_id": 9185, - "bbox": [ - 333.31, - 162.24, - 433.86, - 171.58 - ], - "text": "the energy of x[an] is E/|a|.", - "type": "text" - }, - { - "block_id": "p338-b17", - "global_id": 9186, - "bbox": [ - 284.75, - 176.56, - 490.39, - 229.43 - ], - "text": "3.4-10\nA linear time-invariant system produces output\ny1[n] in response to input x1[n], as shown in\nFig. P3.4-10. Determine and sketch the output\ny2[n] that results when input x2[n] is applied to\nthe same system.", - "type": "text" - }, - { - "block_id": "p338-b18", - "global_id": 9187, - "bbox": [ - 284.74, - 234.42, - 407.51, - 243.46 - ], - "text": "3.4-11\nA system is described by", - "type": "text" - }, - { - "block_id": "p338-b19", - "global_id": 9188, - "bbox": [ - 334.03, - 261.76, - 363.58, - 272.37 - ], - "text": "y[n] = 1", - "type": "text" - }, - { - "block_id": "p338-b20", - "global_id": 9189, - "bbox": [ - 360.34, - 268.57, - 363.58, - 275.04 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p338-b21", - "global_id": 9190, - "bbox": [ - 369.28, - 253.81, - 381.97, - 263.57 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p338-b22", - "global_id": 9191, - "bbox": [ - 365.77, - 275.96, - 385.46, - 282.64 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p338-b23", - "global_id": 9192, - "bbox": [ - 386.47, - 263.12, - 474.22, - 272.37 - ], - "text": "x[k](δ[n −k] + δ[n + k])", - "type": "text" - }, - { - "block_id": "p338-b24", - "global_id": 9193, - "bbox": [ - 317.86, - 298.56, - 490.36, - 318.49 - ], - "text": "(a) Explain what this system does.\n(b) Is the system BIBO-stable? Justify your", - "type": "text" - }, - { - "block_id": "p338-b25", - "global_id": 9194, - "bbox": [ - 318.36, - 320.47, - 480.95, - 340.4 - ], - "text": "answer.\n(c) Is the system linear? Justify your answer.", - "type": "text" - }, - { - "block_id": "p338-b26", - "global_id": 9195, - "bbox": [ - 104.83, - 396.03, - 109.72, - 404.03 - ], - "text": "V", - "type": "text" - }, - { - "block_id": "p338-b27", - "global_id": 9196, - "bbox": [ - 136.53, - 358.49, - 215.58, - 366.49 - ], - "text": "R\nR\nR", - "type": "text" - }, - { - "block_id": "p338-b28", - "global_id": 9197, - "bbox": [ - 248.04, - 380.97, - 457.44, - 388.97 - ], - "text": "R\nR\nR\nR\nR", - "type": "text" - }, - { - "block_id": "p338-b29", - "global_id": 9198, - "bbox": [ - 258.57, - 359.39, - 480.2, - 368.27 - ], - "text": "v[n 1]\nv[n]\nv[n 1]\nv[N 1]\nv[N]", - "type": "text" - }, - { - "block_id": "p338-b30", - "global_id": 9199, - "bbox": [ - 165.75, - 395.36, - 448.54, - 403.36 - ], - "text": "aR\naR\naR\naR\naR\naR\naR", - "type": "text" - }, - { - "block_id": "p338-b31", - "global_id": 9200, - "bbox": [ - 104.83, - 431.73, - 156.48, - 440.69 - ], - "text": "Figure P3.4-8", - "type": "text" - }, - { - "block_id": "p338-b32", - "global_id": 9201, - "bbox": [ - 157.39, - 606.43, - 237.64, - 626.02 - ], - "text": "–2\n0\n2\n4\n–2", - "type": "text" - }, - { - "block_id": "p338-b33", - "global_id": 9202, - "bbox": [ - 157.39, - 593.19, - 165.39, - 601.19 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p338-b34", - "global_id": 9203, - "bbox": [ - 161.39, - 579.92, - 165.39, - 587.92 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p338-b35", - "global_id": 9204, - "bbox": [ - 161.39, - 566.64, - 165.39, - 574.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p338-b36", - "global_id": 9205, - "bbox": [ - 161.39, - 553.4, - 165.39, - 561.4 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p338-b37", - "global_id": 9206, - "bbox": [ - 157.39, - 511.62, - 165.39, - 519.62 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p338-b38", - "global_id": 9207, - "bbox": [ - 157.39, - 498.37, - 165.39, - 506.37 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p338-b39", - "global_id": 9208, - "bbox": [ - 161.39, - 485.11, - 165.39, - 493.11 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p338-b40", - "global_id": 9209, - "bbox": [ - 161.39, - 471.85, - 165.39, - 479.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p338-b41", - "global_id": 9210, - "bbox": [ - 161.39, - 458.6, - 165.39, - 466.6 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p338-b42", - "global_id": 9211, - "bbox": [ - 207.39, - 625.62, - 211.39, - 633.62 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p338-b43", - "global_id": 9212, - "bbox": [ - 287.68, - 618.02, - 358.76, - 633.62 - ], - "text": "–2\n0\n2\n4\nn", - "type": "text" - }, - { - "block_id": "p338-b44", - "global_id": 9213, - "bbox": [ - 166.56, - 523.2, - 237.64, - 538.8 - ], - "text": "–2\n0\n2\n4\nn", - "type": "text" - }, - { - "block_id": "p338-b45", - "global_id": 9214, - "bbox": [ - 287.68, - 523.2, - 358.77, - 538.8 - ], - "text": "–2\n0\n2\n4\nn", - "type": "text" - }, - { - "block_id": "p338-b46", - "global_id": 9215, - "bbox": [ - 146.3, - 480.57, - 155.9, - 591.26 - ], - "text": "x1[n]\nx2[n]", - "type": "text" - }, - { - "block_id": "p338-b47", - "global_id": 9216, - "bbox": [ - 278.48, - 606.43, - 286.48, - 614.43 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p338-b48", - "global_id": 9217, - "bbox": [ - 278.48, - 593.19, - 286.48, - 601.19 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p338-b49", - "global_id": 9218, - "bbox": [ - 282.48, - 579.92, - 286.48, - 587.92 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p338-b50", - "global_id": 9219, - "bbox": [ - 282.48, - 566.64, - 286.48, - 574.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p338-b51", - "global_id": 9220, - "bbox": [ - 282.48, - 553.4, - 286.48, - 561.4 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p338-b52", - "global_id": 9221, - "bbox": [ - 278.48, - 511.62, - 286.48, - 519.62 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p338-b53", - "global_id": 9222, - "bbox": [ - 278.48, - 498.37, - 286.48, - 506.37 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p338-b54", - "global_id": 9223, - "bbox": [ - 282.48, - 485.11, - 286.48, - 493.11 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p338-b55", - "global_id": 9224, - "bbox": [ - 282.48, - 471.85, - 286.48, - 479.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p338-b56", - "global_id": 9225, - "bbox": [ - 282.48, - 458.6, - 286.48, - 466.6 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p338-b57", - "global_id": 9226, - "bbox": [ - 267.39, - 480.57, - 276.99, - 591.26 - ], - "text": "y1[n]\ny2[n]", - "type": "text" - }, - { - "block_id": "p338-b58", - "global_id": 9227, - "bbox": [ - 387.63, - 625.67, - 443.74, - 634.64 - ], - "text": "Figure P3.4-10", - "type": "text" - } - ] - }, - { - "page_num": 339, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p339-b0", - "global_id": 9228, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n319", - "type": "text" - }, - { - "block_id": "p339-b1", - "global_id": 9229, - "bbox": [ - 116.46, - 85.9, - 288.96, - 94.86 - ], - "text": "(d) Is the system memoryless? Justify your", - "type": "text" - }, - { - "block_id": "p339-b2", - "global_id": 9230, - "bbox": [ - 116.96, - 96.86, - 281.56, - 116.78 - ], - "text": "answer.\n(e) Is the system causal? Justify your answer.", - "type": "text" - }, - { - "block_id": "p339-b3", - "global_id": 9231, - "bbox": [ - 117.95, - 118.77, - 288.96, - 127.74 - ], - "text": "(f) Is the system time-invariant? Justify your", - "type": "text" - }, - { - "block_id": "p339-b4", - "global_id": 9232, - "bbox": [ - 131.89, - 129.74, - 159.03, - 138.7 - ], - "text": "answer.", - "type": "text" - }, - { - "block_id": "p339-b5", - "global_id": 9233, - "bbox": [ - 83.34, - 143.69, - 240.36, - 152.73 - ], - "text": "3.4-12\nA discrete-time system is given by", - "type": "text" - }, - { - "block_id": "p339-b6", - "global_id": 9234, - "bbox": [ - 167.47, - 163.18, - 236.78, - 185.17 - ], - "text": "y[n + 1] =\nx[n]\nx[n + 1]", - "type": "text" - }, - { - "block_id": "p339-b7", - "global_id": 9235, - "bbox": [ - 116.96, - 202.62, - 288.96, - 211.58 - ], - "text": "(a) Is the system BIBO-stable? Justify your", - "type": "text" - }, - { - "block_id": "p339-b8", - "global_id": 9236, - "bbox": [ - 116.46, - 213.58, - 288.97, - 233.5 - ], - "text": "answer.\n(b) Is the system memoryless? Justify your", - "type": "text" - }, - { - "block_id": "p339-b9", - "global_id": 9237, - "bbox": [ - 116.96, - 235.5, - 281.56, - 255.42 - ], - "text": "answer.\n(c) Is the system causal? Justify your answer.", - "type": "text" - }, - { - "block_id": "p339-b10", - "global_id": 9238, - "bbox": [ - 83.34, - 260.12, - 288.98, - 302.33 - ], - "text": "3.4-13\nExplain why the continuous-time system y(t) =\nx(2t) is always invertible and yet the correspond-\ning discrete-time system y[n] = x[2n] is not\ninvertible.", - "type": "text" - }, - { - "block_id": "p339-b11", - "global_id": 9239, - "bbox": [ - 83.34, - 307.33, - 288.96, - 327.34 - ], - "text": "3.4-14\nConsider the input–output relationships of two\nsimilar discrete-time systems:", - "type": "text" - }, - { - "block_id": "p339-b12", - "global_id": 9240, - "bbox": [ - 156.24, - 342.1, - 195.56, - 352.18 - ], - "text": "y1[n] = sin", - "type": "text" - }, - { - "block_id": "p339-b13", - "global_id": 9241, - "bbox": [ - 196.55, - 332.21, - 208.01, - 344.78 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p339-b14", - "global_id": 9242, - "bbox": [ - 203.54, - 332.21, - 233.74, - 357.81 - ], - "text": "2 n + 1", - "type": "text" - }, - { - "block_id": "p339-b15", - "global_id": 9243, - "bbox": [ - 234.74, - 342.11, - 249.2, - 351.35 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p339-b16", - "global_id": 9244, - "bbox": [ - 116.46, - 366.8, - 129.4, - 375.77 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p339-b17", - "global_id": 9245, - "bbox": [ - 152.9, - 378.74, - 192.21, - 388.82 - ], - "text": "y2[n] = sin", - "type": "text" - }, - { - "block_id": "p339-b18", - "global_id": 9246, - "bbox": [ - 193.21, - 368.84, - 204.67, - 381.43 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p339-b19", - "global_id": 9247, - "bbox": [ - 200.19, - 368.84, - 237.1, - 394.45 - ], - "text": "2 (n + 1)", - "type": "text" - }, - { - "block_id": "p339-b20", - "global_id": 9248, - "bbox": [ - 238.09, - 378.74, - 252.55, - 387.99 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p339-b21", - "global_id": 9249, - "bbox": [ - 116.46, - 400.08, - 288.99, - 421.12 - ], - "text": "Explain why x[n] can be recovered from y1[n]\nyet x[n] cannot be recovered from y2[n].", - "type": "text" - }, - { - "block_id": "p339-b22", - "global_id": 9250, - "bbox": [ - 83.34, - 425.38, - 288.98, - 456.34 - ], - "text": "3.4-15\nConsider a system that multiplies a given input\nby a ramp function, r[n]. That is, y[n] =\nx[n]r[n].", - "type": "text" - }, - { - "block_id": "p339-b23", - "global_id": 9251, - "bbox": [ - 116.96, - 458.33, - 288.96, - 467.3 - ], - "text": "(a) Is the system BIBO-stable? Justify your", - "type": "text" - }, - { - "block_id": "p339-b24", - "global_id": 9252, - "bbox": [ - 116.46, - 469.29, - 279.57, - 489.22 - ], - "text": "answer.\n(b) Is the system linear? Justify your answer.", - "type": "text" - }, - { - "block_id": "p339-b25", - "global_id": 9253, - "bbox": [ - 116.96, - 491.21, - 288.97, - 500.18 - ], - "text": "(c) Is the system memoryless? Justify your", - "type": "text" - }, - { - "block_id": "p339-b26", - "global_id": 9254, - "bbox": [ - 116.46, - 502.17, - 281.56, - 522.09 - ], - "text": "answer.\n(d) Is the system causal? Justify your answer.", - "type": "text" - }, - { - "block_id": "p339-b27", - "global_id": 9255, - "bbox": [ - 116.96, - 524.09, - 288.97, - 533.06 - ], - "text": "(e) Is the system time-invariant? Justify your", - "type": "text" - }, - { - "block_id": "p339-b28", - "global_id": 9256, - "bbox": [ - 131.89, - 535.04, - 159.03, - 544.01 - ], - "text": "answer.", - "type": "text" - }, - { - "block_id": "p339-b29", - "global_id": 9257, - "bbox": [ - 83.34, - 549.01, - 288.99, - 634.77 - ], - "text": "3.4-16\nA jet-powered car is filmed using a camera\noperating at 60 frames per second. Let variable\nn designate the film frame, where n = 0 corre-\nsponds to engine ignition (film before ignition is\ndiscarded). By analyzing each frame of the film,\nit is possible to determine the car position x[n],\nmeasured in meters, from the original starting\nposition x[0] = 0.", - "type": "text" - }, - { - "block_id": "p339-b30", - "global_id": 9258, - "bbox": [ - 343.61, - 85.9, - 516.12, - 105.83 - ], - "text": "From physics, we know that velocity is the time\nderivative of position:", - "type": "text" - }, - { - "block_id": "p339-b31", - "global_id": 9259, - "bbox": [ - 406.43, - 114.22, - 437.32, - 129.47 - ], - "text": "v(t) = d", - "type": "text" - }, - { - "block_id": "p339-b32", - "global_id": 9260, - "bbox": [ - 431.63, - 120.22, - 453.32, - 135.83 - ], - "text": "dt x(t)", - "type": "text" - }, - { - "block_id": "p339-b33", - "global_id": 9261, - "bbox": [ - 343.61, - 142.7, - 516.14, - 162.63 - ], - "text": "Furthermore, we know that acceleration is the\ntime derivative of velocity:", - "type": "text" - }, - { - "block_id": "p339-b34", - "global_id": 9262, - "bbox": [ - 406.19, - 171.03, - 437.58, - 186.28 - ], - "text": "a(t) = d", - "type": "text" - }, - { - "block_id": "p339-b35", - "global_id": 9263, - "bbox": [ - 431.89, - 177.03, - 453.55, - 192.64 - ], - "text": "dt v(t)", - "type": "text" - }, - { - "block_id": "p339-b36", - "global_id": 9264, - "bbox": [ - 343.61, - 199.51, - 516.13, - 230.39 - ], - "text": "We can estimate the car velocity from the\nfilm data by using a simple difference equation\nv[n] = k(x[n] −x[n −1]).", - "type": "text" - }, - { - "block_id": "p339-b37", - "global_id": 9265, - "bbox": [ - 344.11, - 232.3, - 516.13, - 241.36 - ], - "text": "(a) Determine the appropriate constant k to", - "type": "text" - }, - { - "block_id": "p339-b38", - "global_id": 9266, - "bbox": [ - 343.61, - 242.98, - 516.12, - 263.27 - ], - "text": "ensure v[n] has units of meters per second.\n(b) Determine a standard-form constant coef-", - "type": "text" - }, - { - "block_id": "p339-b39", - "global_id": 9267, - "bbox": [ - 359.05, - 265.27, - 516.13, - 329.03 - ], - "text": "ficient difference equation that outputs an\nestimate of acceleration, a[n], using an input\nof position, x[n]. Identify the advantages\nand shortcomings of estimating acceleration\na(t) with a[n]. What is the impulse response\nh[n] for this system?", - "type": "text" - }, - { - "block_id": "p339-b40", - "global_id": 9268, - "bbox": [ - 314.97, - 333.94, - 516.13, - 375.86 - ], - "text": "3.5-1\nAn LTID system is described by a constant\ncoefficient linear difference equation 2y[n] +\n2y[n −1] = x[n −1].\n(a) Express this system in standard advance", - "type": "text" - }, - { - "block_id": "p339-b41", - "global_id": 9269, - "bbox": [ - 343.61, - 377.85, - 516.12, - 397.77 - ], - "text": "operator form.\n(b) Using recursion, determine the first 5 values", - "type": "text" - }, - { - "block_id": "p339-b42", - "global_id": 9270, - "bbox": [ - 344.12, - 399.39, - 516.12, - 419.7 - ], - "text": "of the system impulse response h[n].\n(c) Using recursion, determine the first 5 values", - "type": "text" - }, - { - "block_id": "p339-b43", - "global_id": 9271, - "bbox": [ - 343.61, - 421.68, - 516.14, - 452.57 - ], - "text": "of the system zero-state response to input\nx[n] = 2u[n].\n(d) Using recursion, determine for (0 ≤n ≤4)", - "type": "text" - }, - { - "block_id": "p339-b44", - "global_id": 9272, - "bbox": [ - 359.05, - 454.18, - 516.13, - 463.52 - ], - "text": "the system zero-input response if y[−1] = 1.", - "type": "text" - }, - { - "block_id": "p339-b45", - "global_id": 9273, - "bbox": [ - 314.97, - 468.43, - 491.1, - 477.47 - ], - "text": "3.5-2\nSolve recursively (first three terms only):", - "type": "text" - }, - { - "block_id": "p339-b46", - "global_id": 9274, - "bbox": [ - 343.61, - 479.09, - 516.13, - 499.39 - ], - "text": "(a) y[n + 1] −0.5y[n] = 0, with y[−1] = 10\n(b) y[n + 1] + 2y[n] = x[n + 1], with x[n] =", - "type": "text" - }, - { - "block_id": "p339-b47", - "global_id": 9275, - "bbox": [ - 359.05, - 499.74, - 440.78, - 510.35 - ], - "text": "e−nu[n] and y[−1] = 0", - "type": "text" - }, - { - "block_id": "p339-b48", - "global_id": 9276, - "bbox": [ - 314.97, - 515.26, - 516.15, - 535.25 - ], - "text": "3.5-3\nSolve the following equation recursively (first\nthree terms only):", - "type": "text" - }, - { - "block_id": "p339-b49", - "global_id": 9277, - "bbox": [ - 363.13, - 545.42, - 496.6, - 554.75 - ], - "text": "y[n] −0.6y[n −1] −0.16y[n −2] = 0", - "type": "text" - }, - { - "block_id": "p339-b50", - "global_id": 9278, - "bbox": [ - 343.61, - 565.28, - 359.55, - 574.25 - ], - "text": "with", - "type": "text" - }, - { - "block_id": "p339-b51", - "global_id": 9279, - "bbox": [ - 384.3, - 575.87, - 475.44, - 585.21 - ], - "text": "y[−1] = −25, y[−2] = 0.", - "type": "text" - }, - { - "block_id": "p339-b52", - "global_id": 9280, - "bbox": [ - 314.97, - 593.39, - 516.13, - 635.31 - ], - "text": "3.5-4\nSolve recursively the second-order difference\nEq. (3.6) for sales estimate (first three terms\nonly), assuming y[−1] = y[−2] = 0 and x[n] =\n100u[n].", - "type": "text" - } - ] - }, - { - "page_num": 340, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p340-b0", - "global_id": 9281, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "320\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p340-b1", - "global_id": 9282, - "bbox": [ - 62.08, - 85.94, - 263.25, - 105.94 - ], - "text": "3.5-5\nSolve the following equation recursively (first\nthree terms only):", - "type": "text" - }, - { - "block_id": "p340-b2", - "global_id": 9283, - "bbox": [ - 122.26, - 120.15, - 231.48, - 129.49 - ], - "text": "y[n + 2] + 3y[n + 1] + 2y[n] =", - "type": "text" - }, - { - "block_id": "p340-b3", - "global_id": 9284, - "bbox": [ - 131.24, - 134.1, - 231.68, - 143.44 - ], - "text": "x[n + 2] + 3x[n + 1] + 3x[n]", - "type": "text" - }, - { - "block_id": "p340-b4", - "global_id": 9285, - "bbox": [ - 90.72, - 156.58, - 261.23, - 166.98 - ], - "text": "with x[n] = (3)nu[n], y[−1] = 3, and y[−2] = 2", - "type": "text" - }, - { - "block_id": "p340-b5", - "global_id": 9286, - "bbox": [ - 62.08, - 172.15, - 170.29, - 181.19 - ], - "text": "3.5-6\nRepeat Prob. 3.5-5 for", - "type": "text" - }, - { - "block_id": "p340-b6", - "global_id": 9287, - "bbox": [ - 94.97, - 195.4, - 258.99, - 204.73 - ], - "text": "y[n] + 2y[n −1] + y[n −2] = 2x[n] −x[n −1]", - "type": "text" - }, - { - "block_id": "p340-b7", - "global_id": 9288, - "bbox": [ - 90.72, - 217.68, - 263.24, - 228.29 - ], - "text": "withx[n] = (3)−nu[n],y[−1] = 2,andy[−2] = 3.", - "type": "text" - }, - { - "block_id": "p340-b8", - "global_id": 9289, - "bbox": [ - 62.08, - 233.16, - 263.23, - 275.38 - ], - "text": "3.6-1\nGiven y0[−1] = 3 and y0[−2] = −1, determine\nthe closed-form expression of the zero-input\nresponse y0[n] of an LTID system described by\nthe equation y[n]+ 1", - "type": "text" - }, - { - "block_id": "p340-b9", - "global_id": 9290, - "bbox": [ - 90.72, - 264.67, - 263.24, - 287.28 - ], - "text": "6y[n−1]−1\n6y[n−2] = 1\n3x[n]\n+ 2", - "type": "text" - }, - { - "block_id": "p340-b10", - "global_id": 9291, - "bbox": [ - 98.91, - 278.31, - 134.31, - 290.23 - ], - "text": "3x[n −2].", - "type": "text" - }, - { - "block_id": "p340-b11", - "global_id": 9292, - "bbox": [ - 62.08, - 292.82, - 111.01, - 301.86 - ], - "text": "3.6-2\nSolve", - "type": "text" - }, - { - "block_id": "p340-b12", - "global_id": 9293, - "bbox": [ - 119.21, - 316.07, - 234.75, - 325.41 - ], - "text": "y[n + 2] + 3y[n + 1] + 2y[n] = 0", - "type": "text" - }, - { - "block_id": "p340-b13", - "global_id": 9294, - "bbox": [ - 90.72, - 339.61, - 191.29, - 348.95 - ], - "text": "if y[−1] = 0 and y[−2] = 1.", - "type": "text" - }, - { - "block_id": "p340-b14", - "global_id": 9295, - "bbox": [ - 62.08, - 354.12, - 111.01, - 363.16 - ], - "text": "3.6-3\nSolve", - "type": "text" - }, - { - "block_id": "p340-b15", - "global_id": 9296, - "bbox": [ - 121.45, - 377.38, - 232.51, - 386.71 - ], - "text": "y[n + 2] + 2y[n + 1] + y[n] = 0", - "type": "text" - }, - { - "block_id": "p340-b16", - "global_id": 9297, - "bbox": [ - 90.72, - 400.92, - 191.29, - 410.26 - ], - "text": "if y[−1] = 1 and y[−2] = 1.", - "type": "text" - }, - { - "block_id": "p340-b17", - "global_id": 9298, - "bbox": [ - 62.08, - 415.43, - 111.01, - 424.47 - ], - "text": "3.6-4\nSolve", - "type": "text" - }, - { - "block_id": "p340-b18", - "global_id": 9299, - "bbox": [ - 119.21, - 438.67, - 234.75, - 448.01 - ], - "text": "y[n + 2] −2y[n + 1] + 2y[n] = 0", - "type": "text" - }, - { - "block_id": "p340-b19", - "global_id": 9300, - "bbox": [ - 90.72, - 462.23, - 191.29, - 471.57 - ], - "text": "if y[−1] = 1 and y[−2] = 0.", - "type": "text" - }, - { - "block_id": "p340-b20", - "global_id": 9301, - "bbox": [ - 62.08, - 476.72, - 263.23, - 496.73 - ], - "text": "3.6-5\nFor the general Nth-order difference Eq. (3.16),\nletting", - "type": "text" - }, - { - "block_id": "p340-b21", - "global_id": 9302, - "bbox": [ - 130.28, - 510.94, - 223.67, - 521.25 - ], - "text": "a1 = a2 = · · · = aN−1 = 0", - "type": "text" - }, - { - "block_id": "p340-b22", - "global_id": 9303, - "bbox": [ - 90.72, - 534.77, - 263.24, - 554.79 - ], - "text": "results in a general causal Nth-order LTI nonre-\ncursive difference equation", - "type": "text" - }, - { - "block_id": "p340-b23", - "global_id": 9304, - "bbox": [ - 93.83, - 568.99, - 260.14, - 579.07 - ], - "text": "y[n] = b0x[n] + b1x[n −1] + · · · + bNx[n −N]", - "type": "text" - }, - { - "block_id": "p340-b24", - "global_id": 9305, - "bbox": [ - 90.72, - 592.92, - 263.25, - 634.76 - ], - "text": "Show that the characteristic roots for this system\nare zero—hence, that the zero-input response is\nzero. Consequently, the total response consists\nof the zero-state component only.", - "type": "text" - }, - { - "block_id": "p340-b25", - "global_id": 9306, - "bbox": [ - 289.22, - 85.94, - 490.39, - 116.89 - ], - "text": "3.6-6\nLeonardo Pisano Fibonacci, a famous thirteenth-\ncentury mathematician, generated the sequence\nof integers", - "type": "text" - }, - { - "block_id": "p340-b26", - "global_id": 9307, - "bbox": [ - 350.29, - 128.23, - 457.97, - 137.57 - ], - "text": "{0,1,1,2,3,5,8,13,21,34,...}", - "type": "text" - }, - { - "block_id": "p340-b27", - "global_id": 9308, - "bbox": [ - 317.86, - 149.27, - 490.39, - 191.11 - ], - "text": "while addressing, oddly enough, a problem\ninvolving rabbit reproduction. An element of the\nFibonacci sequence is the sum of the previous\ntwo.", - "type": "text" - }, - { - "block_id": "p340-b28", - "global_id": 9309, - "bbox": [ - 317.86, - 193.11, - 490.39, - 267.83 - ], - "text": "(a) Find\nthe\nconstant-coefficient\ndifference\nequation whose zero-input response f[n]\nwith auxiliary conditions f[1] = 0 and\nf[2] = 1 is a Fibonacci sequence. Given f[n]\nis the system output, what is the system\ninput?\n(b) What are the characteristic roots of this", - "type": "text" - }, - { - "block_id": "p340-b29", - "global_id": 9310, - "bbox": [ - 318.37, - 269.82, - 490.37, - 289.74 - ], - "text": "system? Is the system stable?\n(c) Designating 0 and 1 as the first and second", - "type": "text" - }, - { - "block_id": "p340-b30", - "global_id": 9311, - "bbox": [ - 333.31, - 291.74, - 490.4, - 322.62 - ], - "text": "Fibonacci numbers, determine the fiftieth\nFibonacci number. Determine the one thou-\nsandth Fibonacci number.", - "type": "text" - }, - { - "block_id": "p340-b31", - "global_id": 9312, - "bbox": [ - 289.22, - 327.23, - 490.39, - 402.33 - ], - "text": "3.6-7\nFind v[n], the voltage at the nth node of the\nresistive ladder depicted in Fig. P3.4-8, if V =\n100 volts and a = 2. [Hint 1: Consider the\nnode equation at the nth node with voltage v[n].\nHint 2: See Prob. 3.4-8 for the equation for\nv[n]. The auxiliary conditions are v[0] = 100 and\nv[N] = 0.]", - "type": "text" - }, - { - "block_id": "p340-b32", - "global_id": 9313, - "bbox": [ - 289.22, - 406.94, - 490.39, - 427.22 - ], - "text": "3.6-8\nConsider the discrete-time system y[n] + y[n −\n1] + 0.25y[n −2] =\n√", - "type": "text" - }, - { - "block_id": "p340-b33", - "global_id": 9314, - "bbox": [ - 317.86, - 417.88, - 490.39, - 449.15 - ], - "text": "3x[n −8]. Find the zero\ninput response, y0[n], if y0[−1] = 1 and y0[1] =\n1.", - "type": "text" - }, - { - "block_id": "p340-b34", - "global_id": 9315, - "bbox": [ - 289.22, - 453.75, - 490.4, - 506.93 - ], - "text": "3.6-9\nProvide\na\nstandard-form\npolynomial\nQ(X)\nsuch that Q(E){y[n]} = x[n] corresponds to\na marginally stable third-order LTID system\nand Q(D){y(t)} = x(t) corresponds to a stable\nthird-order LTIC system.", - "type": "text" - }, - { - "block_id": "p340-b35", - "global_id": 9316, - "bbox": [ - 289.23, - 511.54, - 490.39, - 531.83 - ], - "text": "3.7-1\nFind the unit impulse response h[n] of systems\nspecified by the following equations:", - "type": "text" - }, - { - "block_id": "p340-b36", - "global_id": 9317, - "bbox": [ - 317.87, - 533.46, - 415.84, - 553.76 - ], - "text": "(a) y[n + 1] + 2y[n] = x[n]\n(b) y[n] + 2y[n −1] = x[n]", - "type": "text" - }, - { - "block_id": "p340-b37", - "global_id": 9318, - "bbox": [ - 289.23, - 558.36, - 490.4, - 600.58 - ], - "text": "3.7-2\nDetermine the unit impulse response h[n] of the\nfollowing systems. In each case, use recursion\nto verify the n = 3 value of the closed-form\nexpression of h[n].", - "type": "text" - }, - { - "block_id": "p340-b38", - "global_id": 9319, - "bbox": [ - 317.86, - 598.94, - 465.54, - 622.5 - ], - "text": "(a) (E2 + 1){y[n]} = (E + 0.5){x[n]}\n(b) y[n] −y[n −1] + 0.25y[n −2] = x[n]", - "type": "text" - }, - { - "block_id": "p340-b39", - "global_id": 9320, - "bbox": [ - 318.37, - 622.75, - 361.95, - 633.45 - ], - "text": "(c) y[n] −1", - "type": "text" - }, - { - "block_id": "p340-b40", - "global_id": 9321, - "bbox": [ - 358.71, - 622.75, - 481.0, - 636.03 - ], - "text": "6y[n −1] −1\n6y[n −2] = 1\n3x[n −2]", - "type": "text" - } - ] - }, - { - "page_num": 341, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p341-b0", - "global_id": 9322, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n321", - "type": "text" - }, - { - "block_id": "p341-b1", - "global_id": 9323, - "bbox": [ - 116.46, - 86.08, - 160.55, - 96.79 - ], - "text": "(d) y[n] + 1", - "type": "text" - }, - { - "block_id": "p341-b2", - "global_id": 9324, - "bbox": [ - 116.96, - 86.08, - 265.34, - 109.06 - ], - "text": "6y[n −1] −1\n6y[n −2] = 1\n3x[n]\n(e) y[n] + 1", - "type": "text" - }, - { - "block_id": "p341-b3", - "global_id": 9325, - "bbox": [ - 117.96, - 99.73, - 215.58, - 121.28 - ], - "text": "4y[n −2] = x[n]\n(f) (E2 −4", - "type": "text" - }, - { - "block_id": "p341-b4", - "global_id": 9326, - "bbox": [ - 116.46, - 108.68, - 246.65, - 133.64 - ], - "text": "9){y[n]} = (E2 + 1){x[n]}\n(g) (E2 −1", - "type": "text" - }, - { - "block_id": "p341-b5", - "global_id": 9327, - "bbox": [ - 116.46, - 122.93, - 253.48, - 145.84 - ], - "text": "4)(E + 1\n2){y[n]} = E3{x[n]}\n(h) (E −1", - "type": "text" - }, - { - "block_id": "p341-b6", - "global_id": 9328, - "bbox": [ - 151.9, - 135.51, - 209.62, - 148.42 - ], - "text": "2)2{y[n]} = x[n]", - "type": "text" - }, - { - "block_id": "p341-b7", - "global_id": 9329, - "bbox": [ - 87.82, - 150.54, - 288.98, - 170.84 - ], - "text": "3.7-3\nConsider a DT system with input x[n] and output\ny[n] described by the difference equation", - "type": "text" - }, - { - "block_id": "p341-b8", - "global_id": 9330, - "bbox": [ - 130.57, - 183.3, - 274.87, - 192.64 - ], - "text": "4y[n + 1] + y[n −1] = 8x[n + 1] + 8x[n]", - "type": "text" - }, - { - "block_id": "p341-b9", - "global_id": 9331, - "bbox": [ - 116.46, - 216.44, - 288.97, - 236.37 - ], - "text": "(a) What is the order of this system?\n(b) Determine the characteristic mode(s) of the", - "type": "text" - }, - { - "block_id": "p341-b10", - "global_id": 9332, - "bbox": [ - 116.97, - 238.36, - 288.99, - 258.29 - ], - "text": "system.\n(c) Determine a closed-form expression for the", - "type": "text" - }, - { - "block_id": "p341-b11", - "global_id": 9333, - "bbox": [ - 131.9, - 259.91, - 246.47, - 269.25 - ], - "text": "system’s impulse response h[n].", - "type": "text" - }, - { - "block_id": "p341-b12", - "global_id": 9334, - "bbox": [ - 87.83, - 274.24, - 288.97, - 294.24 - ], - "text": "3.7-4\nRepeat Prob. 3.7-3 for a system described by the\ndifference equation", - "type": "text" - }, - { - "block_id": "p341-b13", - "global_id": 9335, - "bbox": [ - 116.59, - 305.16, - 163.0, - 320.41 - ], - "text": "y[n + 3] −3", - "type": "text" - }, - { - "block_id": "p341-b14", - "global_id": 9336, - "bbox": [ - 156.27, - 305.16, - 288.86, - 326.78 - ], - "text": "10y[n + 2] −1\n10y[n + 1] = 2x[n + 1]", - "type": "text" - }, - { - "block_id": "p341-b15", - "global_id": 9337, - "bbox": [ - 87.82, - 335.9, - 196.03, - 344.94 - ], - "text": "3.7-5\nRepeat Prob. 3.7-1 for", - "type": "text" - }, - { - "block_id": "p341-b16", - "global_id": 9338, - "bbox": [ - 155.03, - 353.7, - 250.41, - 366.74 - ], - "text": "(E2 −6E + 9)y[n] = Ex[n]", - "type": "text" - }, - { - "block_id": "p341-b17", - "global_id": 9339, - "bbox": [ - 87.82, - 379.5, - 196.03, - 388.54 - ], - "text": "3.7-6\nRepeat Prob. 3.7-1 for", - "type": "text" - }, - { - "block_id": "p341-b18", - "global_id": 9340, - "bbox": [ - 116.97, - 401.01, - 288.47, - 410.35 - ], - "text": "y[n] −6y[n−1] + 25y[n−2] = 2x[n] −4x[n−1]", - "type": "text" - }, - { - "block_id": "p341-b19", - "global_id": 9341, - "bbox": [ - 87.82, - 423.1, - 288.97, - 443.12 - ], - "text": "3.7-7\n(a) For\nthe\ngeneral\nNth-order\ndifference\nEq. (3.16), letting", - "type": "text" - }, - { - "block_id": "p341-b20", - "global_id": 9342, - "bbox": [ - 154.3, - 455.59, - 266.58, - 465.89 - ], - "text": "a0 = a1 = a2 = · · · = aN−1 = 0", - "type": "text" - }, - { - "block_id": "p341-b21", - "global_id": 9343, - "bbox": [ - 131.9, - 477.67, - 288.98, - 497.69 - ], - "text": "results in a general causal Nth-order LTI\nnonrecursive difference equation", - "type": "text" - }, - { - "block_id": "p341-b22", - "global_id": 9344, - "bbox": [ - 174.28, - 518.75, - 197.55, - 528.0 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p341-b23", - "global_id": 9345, - "bbox": [ - 199.4, - 509.64, - 212.08, - 519.2 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p341-b24", - "global_id": 9346, - "bbox": [ - 200.69, - 531.68, - 210.78, - 538.42 - ], - "text": "i=0", - "type": "text" - }, - { - "block_id": "p341-b25", - "global_id": 9347, - "bbox": [ - 213.08, - 518.75, - 246.6, - 528.76 - ], - "text": "bix[n −i]", - "type": "text" - }, - { - "block_id": "p341-b26", - "global_id": 9348, - "bbox": [ - 131.89, - 548.71, - 288.99, - 634.76 - ], - "text": "Find the impulse response h[n] for this\nsystem. [Hint: The characteristic equation\nfor this case is γ n = 0. Hence, all the\ncharacteristic roots are zero. In this case,\nyc[n] = 0, and the approach in Sec. 3.7 does\nnot work. Use a direct method to find h[n]\nby realizing that h[n] is the response to unit\nimpulse input.]", - "type": "text" - }, - { - "block_id": "p341-b27", - "global_id": 9349, - "bbox": [ - 343.61, - 85.9, - 516.13, - 94.86 - ], - "text": "(b) Find the impulse response of a non-", - "type": "text" - }, - { - "block_id": "p341-b28", - "global_id": 9350, - "bbox": [ - 359.05, - 96.86, - 516.13, - 116.78 - ], - "text": "recursive LTID system described by the\nequation", - "type": "text" - }, - { - "block_id": "p341-b29", - "global_id": 9351, - "bbox": [ - 372.56, - 131.59, - 502.6, - 140.92 - ], - "text": "y[n] = 3x[n] −5x[n −1] −2x[n −3]", - "type": "text" - }, - { - "block_id": "p341-b30", - "global_id": 9352, - "bbox": [ - 359.05, - 156.11, - 516.14, - 252.74 - ], - "text": "Observe that the impulse response has only\na finite (N) number of nonzero elements.\nFor this reason, such systems are called\nfinite-impulse response (FIR) systems. For\na general recursive case [Eq. (3.20)], the\nimpulse response has an infinite number\nof nonzero elements, and such systems\nare called infinite-impulse response (IIR)\nsystems.", - "type": "text" - }, - { - "block_id": "p341-b31", - "global_id": 9353, - "bbox": [ - 314.98, - 257.68, - 451.59, - 267.02 - ], - "text": "3.8-1\nThe\nconvolution\ny[n]\n=", - "type": "text" - }, - { - "block_id": "p341-b32", - "global_id": 9354, - "bbox": [ - 459.91, - 250.47, - 469.57, - 262.79 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p341-b33", - "global_id": 9355, - "bbox": [ - 464.71, - 250.47, - 505.21, - 269.6 - ], - "text": "2n u[n + 5]", - "type": "text" - }, - { - "block_id": "p341-b34", - "global_id": 9356, - "bbox": [ - 343.61, - 257.68, - 516.13, - 278.02 - ], - "text": "∗\n(3nu[−n −2]) can be represented as", - "type": "text" - }, - { - "block_id": "p341-b35", - "global_id": 9357, - "bbox": [ - 378.95, - 297.96, - 402.22, - 307.21 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p341-b36", - "global_id": 9358, - "bbox": [ - 404.06, - 285.37, - 474.07, - 313.47 - ], - "text": "C1(γ1)n\nn < N\nC2(γ2)n\nn ≥N", - "type": "text" - }, - { - "block_id": "p341-b37", - "global_id": 9359, - "bbox": [ - 343.61, - 327.34, - 516.12, - 348.0 - ], - "text": "Using the graphical convolution procedure,\ndetermine constants C1, C2, γ1, γ2, and N.", - "type": "text" - }, - { - "block_id": "p341-b38", - "global_id": 9360, - "bbox": [ - 314.97, - 352.49, - 516.13, - 372.49 - ], - "text": "3.8-2\nUse the graphical convolution procedure to\ndetermine the following:", - "type": "text" - }, - { - "block_id": "p341-b39", - "global_id": 9361, - "bbox": [ - 344.11, - 374.11, - 516.13, - 384.19 - ], - "text": "(a) ya[n] = u[n] ∗(u[n −5] −u[n −9] +", - "type": "text" - }, - { - "block_id": "p341-b40", - "global_id": 9362, - "bbox": [ - 343.61, - 383.81, - 426.34, - 406.11 - ], - "text": "(0.5)(n−8)u[n −9])\n(b) yb[n] = ( 1", - "type": "text" - }, - { - "block_id": "p341-b41", - "global_id": 9363, - "bbox": [ - 392.44, - 392.77, - 451.31, - 407.95 - ], - "text": "2)|n| ∗u[−n + 5]", - "type": "text" - }, - { - "block_id": "p341-b42", - "global_id": 9364, - "bbox": [ - 314.97, - 407.05, - 516.13, - 441.56 - ], - "text": "3.8-3\nLet x[n] = (0.5)n (u[n + 4] −u[n −4]) be input\ninto an LTID system with an impulse response\ngiven by", - "type": "text" - }, - { - "block_id": "p341-b43", - "global_id": 9365, - "bbox": [ - 351.48, - 461.54, - 375.25, - 470.79 - ], - "text": "h[n] =", - "type": "text" - }, - { - "block_id": "p341-b44", - "global_id": 9366, - "bbox": [ - 377.09, - 448.95, - 502.09, - 476.31 - ], - "text": "2\n[(n mod 6) < 4] and [n ≥0]\n0\notherwise", - "type": "text" - }, - { - "block_id": "p341-b45", - "global_id": 9367, - "bbox": [ - 343.61, - 490.61, - 516.13, - 544.51 - ], - "text": "Recall, (n mod p) is the remainder of the divi-\nsion n/p. The system is described according to\nthe difference equation y[n]−y[n−6] = 2x[n]+\n2x[n −1] + 2x[n −2] + 2x[n −3].\n(a) Determine the six characteristic roots (γ1", - "type": "text" - }, - { - "block_id": "p341-b46", - "global_id": 9368, - "bbox": [ - 343.61, - 545.4, - 516.13, - 565.7 - ], - "text": "through γ6) of the system.\n(b) Determine the value of y[10], the zero-state", - "type": "text" - }, - { - "block_id": "p341-b47", - "global_id": 9369, - "bbox": [ - 359.04, - 567.31, - 516.14, - 609.53 - ], - "text": "output of system h[n] in response to x[n] at\ntime n = 10. Express your result in decimal\nform to at least three decimal places (e.g.,\ny[10] = 3.142).", - "type": "text" - }, - { - "block_id": "p341-b48", - "global_id": 9370, - "bbox": [ - 314.97, - 614.46, - 516.13, - 634.77 - ], - "text": "3.8-4\nAn LTID system has impulse response h[n] =\n(0.5)(n+3) (u[n] −u[n + 6]). A 6-periodic DT", - "type": "text" - } - ] - }, - { - "page_num": 342, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p342-b0", - "global_id": 9371, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "322\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p342-b1", - "global_id": 9372, - "bbox": [ - 90.72, - 85.52, - 190.75, - 94.86 - ], - "text": "input signal x[n] is given by", - "type": "text" - }, - { - "block_id": "p342-b2", - "global_id": 9373, - "bbox": [ - 92.77, - 122.34, - 116.07, - 131.59 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p342-b3", - "global_id": 9374, - "bbox": [ - 117.92, - 98.58, - 125.02, - 120.99 - ], - "text": "⎧\n⎪⎪⎨", - "type": "text" - }, - { - "block_id": "p342-b4", - "global_id": 9375, - "bbox": [ - 117.92, - 128.16, - 125.02, - 142.51 - ], - "text": "⎪⎪⎩", - "type": "text" - }, - { - "block_id": "p342-b5", - "global_id": 9376, - "bbox": [ - 130.0, - 105.85, - 251.77, - 148.07 - ], - "text": "1\nn = 0, ±3, ±6, ±9, ±12, . . .\n2\nn = 1, 1 ± 6, 1 ± 12, . . .\n3\nn = 2, 2 ± 6, 2 ± 12, . . .\n0\notherwise", - "type": "text" - }, - { - "block_id": "p342-b6", - "global_id": 9377, - "bbox": [ - 258.94, - 122.72, - 261.18, - 131.68 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p342-b7", - "global_id": 9378, - "bbox": [ - 91.22, - 163.69, - 263.25, - 173.03 - ], - "text": "(a) Is system h[n] causal? Mathematically jus-", - "type": "text" - }, - { - "block_id": "p342-b8", - "global_id": 9379, - "bbox": [ - 90.72, - 175.02, - 263.23, - 194.94 - ], - "text": "tify your answer.\n(b) Determine the value of y[12], the zero-state", - "type": "text" - }, - { - "block_id": "p342-b9", - "global_id": 9380, - "bbox": [ - 106.16, - 196.57, - 263.24, - 238.78 - ], - "text": "output of system h[n] in response to x[n] at\ntime n = 12. Express your result in decimal\nform to at least three decimal places (e.g.,\ny[12] = 1.234).", - "type": "text" - }, - { - "block_id": "p342-b10", - "global_id": 9381, - "bbox": [ - 62.08, - 243.39, - 263.24, - 263.69 - ], - "text": "3.8-5\nFind the (zero-state) response y[n] of an LTID\nsystem whose unit impulse response is", - "type": "text" - }, - { - "block_id": "p342-b11", - "global_id": 9382, - "bbox": [ - 138.62, - 273.53, - 215.34, - 284.37 - ], - "text": "h[n] = (−2)nu[n −1]", - "type": "text" - }, - { - "block_id": "p342-b12", - "global_id": 9383, - "bbox": [ - 90.72, - 294.44, - 263.24, - 326.96 - ], - "text": "and the input is x[n] = e−nu[n + 1]. Find your\nanswer by computing the convolution sum and\nalso by using Table 3.1.", - "type": "text" - }, - { - "block_id": "p342-b13", - "global_id": 9384, - "bbox": [ - 62.08, - 331.58, - 263.24, - 351.87 - ], - "text": "3.8-6\nFind the (zero-state) response y[n] of an LTID\nsystem if the input is x[n] = 3n−1u[n + 2], and", - "type": "text" - }, - { - "block_id": "p342-b14", - "global_id": 9385, - "bbox": [ - 109.31, - 361.84, - 139.36, - 372.46 - ], - "text": "h[n] = 1", - "type": "text" - }, - { - "block_id": "p342-b15", - "global_id": 9386, - "bbox": [ - 136.12, - 361.94, - 244.64, - 375.12 - ], - "text": "2[δ[n −2] −(−2)n+1]u[n −3]", - "type": "text" - }, - { - "block_id": "p342-b16", - "global_id": 9387, - "bbox": [ - 62.08, - 383.88, - 263.24, - 404.18 - ], - "text": "3.8-7\nFind the (zero-state) response y[n] of an LTID\nsystem if the input x[n] = (3)n+2u[n + 1], and", - "type": "text" - }, - { - "block_id": "p342-b17", - "global_id": 9388, - "bbox": [ - 112.75, - 411.82, - 241.21, - 424.87 - ], - "text": "h[n] = [(2)n−2 + 3(−5)n+2]u[n −1]", - "type": "text" - }, - { - "block_id": "p342-b18", - "global_id": 9389, - "bbox": [ - 62.08, - 436.2, - 263.24, - 456.5 - ], - "text": "3.8-8\nFind the (zero-state) response y[n] of an LTID\nsystem if the input x[n] = (3)−n+2u[n + 3], and", - "type": "text" - }, - { - "block_id": "p342-b19", - "global_id": 9390, - "bbox": [ - 123.01, - 466.13, - 230.95, - 477.17 - ], - "text": "h[n] = 3(n −2)(2)n−3u[n −4]", - "type": "text" - }, - { - "block_id": "p342-b20", - "global_id": 9391, - "bbox": [ - 62.08, - 488.52, - 263.24, - 508.82 - ], - "text": "3.8-9\nFind the (zero-state) response y[n] of an LTID\nsystem if its input x[n] = (2)nu[n −1], and", - "type": "text" - }, - { - "block_id": "p342-b21", - "global_id": 9392, - "bbox": [ - 119.32, - 519.11, - 172.79, - 531.79 - ], - "text": "h[n] = (3)n cos", - "type": "text" - }, - { - "block_id": "p342-b22", - "global_id": 9393, - "bbox": [ - 174.79, - 512.55, - 186.25, - 525.12 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p342-b23", - "global_id": 9394, - "bbox": [ - 181.77, - 512.55, - 218.7, - 538.15 - ], - "text": "3 n −0.5", - "type": "text" - }, - { - "block_id": "p342-b24", - "global_id": 9395, - "bbox": [ - 219.7, - 522.45, - 234.64, - 531.69 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p342-b25", - "global_id": 9396, - "bbox": [ - 90.72, - 546.09, - 230.95, - 555.06 - ], - "text": "Find your answer using only Table 3.1.", - "type": "text" - }, - { - "block_id": "p342-b26", - "global_id": 9397, - "bbox": [ - 57.59, - 559.97, - 263.23, - 579.96 - ], - "text": "3.8-10\nConsider\nan\nLTID\nsystem\n(“system\n1”)\ndescribed by (E −1", - "type": "text" - }, - { - "block_id": "p342-b27", - "global_id": 9398, - "bbox": [ - 91.22, - 570.53, - 263.24, - 591.66 - ], - "text": "2){y[n]} = x[n].\n(a) Determine the impulse response h1[n] for", - "type": "text" - }, - { - "block_id": "p342-b28", - "global_id": 9399, - "bbox": [ - 90.72, - 592.92, - 263.24, - 612.84 - ], - "text": "system 1. Simplify your answer.\n(b) Determine the step response s[n] for system", - "type": "text" - }, - { - "block_id": "p342-b29", - "global_id": 9400, - "bbox": [ - 106.15, - 614.84, - 263.24, - 634.76 - ], - "text": "1 (the step response is the output in response\nto a unit step input). Simplify your answer.", - "type": "text" - }, - { - "block_id": "p342-b30", - "global_id": 9401, - "bbox": [ - 318.37, - 85.52, - 490.39, - 95.6 - ], - "text": "(c) Determine the impulse response hcascade[n]", - "type": "text" - }, - { - "block_id": "p342-b31", - "global_id": 9402, - "bbox": [ - 333.31, - 96.86, - 490.38, - 127.74 - ], - "text": "of system 1 cascaded with an LTID system\nwith impulse response h2[n] = −3u[n−13].\nSimplify your answer.", - "type": "text" - }, - { - "block_id": "p342-b32", - "global_id": 9403, - "bbox": [ - 284.74, - 132.75, - 490.39, - 163.71 - ], - "text": "3.8-11\nDerive the results in entries 1, 2, and 3 in\nTable 3.1. [Hint: You may need to use the\ninformation in Sec. B.8-3.]", - "type": "text" - }, - { - "block_id": "p342-b33", - "global_id": 9404, - "bbox": [ - 284.74, - 168.71, - 490.39, - 188.72 - ], - "text": "3.8-12\nDerive the results in entries 4, 5, and 6 in\nTable 3.1.", - "type": "text" - }, - { - "block_id": "p342-b34", - "global_id": 9405, - "bbox": [ - 284.74, - 193.73, - 490.38, - 224.68 - ], - "text": "3.8-13\nDerive the results in entries 7 and 8 in Table 3.1.\n[Hint: You may need to use the information in\nSec. B.8-3.]", - "type": "text" - }, - { - "block_id": "p342-b35", - "global_id": 9406, - "bbox": [ - 284.74, - 229.7, - 490.39, - 260.66 - ], - "text": "3.8-14\nDerive the results in entries 9 and 11 in\nTable 3.1. [Hint: You may need to use the\ninformation in Sec. B.8-3.]", - "type": "text" - }, - { - "block_id": "p342-b36", - "global_id": 9407, - "bbox": [ - 284.74, - 265.66, - 490.39, - 285.67 - ], - "text": "3.8-15\nFind the total response of a system specified by\nthe equation", - "type": "text" - }, - { - "block_id": "p342-b37", - "global_id": 9408, - "bbox": [ - 355.72, - 298.27, - 452.53, - 307.61 - ], - "text": "y[n + 1] + 2y[n] = x[n + 1]", - "type": "text" - }, - { - "block_id": "p342-b38", - "global_id": 9409, - "bbox": [ - 317.86, - 318.94, - 475.3, - 329.55 - ], - "text": "if y[−1] = 10, and the input x[n] = e−nu[n].", - "type": "text" - }, - { - "block_id": "p342-b39", - "global_id": 9410, - "bbox": [ - 284.74, - 334.56, - 490.39, - 365.51 - ], - "text": "3.8-16\nFind an LTID system (zero-state) response if\nits impulse response h[n] = (0.5)nu[n], and the\ninput x[n] is", - "type": "text" - }, - { - "block_id": "p342-b40", - "global_id": 9411, - "bbox": [ - 317.86, - 366.07, - 365.47, - 387.43 - ], - "text": "(a) 2nu[n]\n(b) 2n−3u[n]", - "type": "text" - }, - { - "block_id": "p342-b41", - "global_id": 9412, - "bbox": [ - 317.86, - 388.0, - 490.39, - 420.31 - ], - "text": "(c) 2nu[n −2]\n[Hint: You may need to use the convolution shift\nproperty of Eq. (3.32).]", - "type": "text" - }, - { - "block_id": "p342-b42", - "global_id": 9413, - "bbox": [ - 284.74, - 425.32, - 442.3, - 434.36 - ], - "text": "3.8-17\nFor a system specified by equation", - "type": "text" - }, - { - "block_id": "p342-b43", - "global_id": 9414, - "bbox": [ - 362.85, - 446.96, - 445.41, - 456.3 - ], - "text": "y[n] = x[n] −2x[n −1]", - "type": "text" - }, - { - "block_id": "p342-b44", - "global_id": 9415, - "bbox": [ - 317.86, - 468.91, - 490.41, - 522.09 - ], - "text": "Find the system response to input x[n] = u[n].\nWhat is the order of the system? What type of\nsystem (recursive or nonrecursive) is this? Is the\nknowledge of initial condition(s) necessary to\nfind the system response? Explain.", - "type": "text" - }, - { - "block_id": "p342-b45", - "global_id": 9416, - "bbox": [ - 284.74, - 527.09, - 490.38, - 536.13 - ], - "text": "3.8-18\n(a) A discrete-time LTI system is shown in", - "type": "text" - }, - { - "block_id": "p342-b46", - "global_id": 9417, - "bbox": [ - 317.86, - 538.12, - 490.39, - 634.76 - ], - "text": "Fig. P3.8-18. Express the overall impulse\nresponse of the system, h[n], in terms of\nh1[n], h2[n], h3[n], h4[n], and h5[n].\n(b) Two\nLTID\nsystems\nin\ncascade\nhave\nimpulse\nresponse\nh1[n]\nand\nh2[n],\nrespectively.\nShow\nthat\nif\nh1[n]\n=\n(0.9)nu[n] −0.5(0.9)n−1u[n −1]\nand\nh2[n] = (0.5)nu[n] −0.9(0.5)n−1u[n −1],\nthe cascade system is an identity system.", - "type": "text" - } - ] - }, - { - "page_num": 343, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p343-b0", - "global_id": 9418, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n323", - "type": "text" - }, - { - "block_id": "p343-b1", - "global_id": 9419, - "bbox": [ - 92.52, - 128.96, - 296.33, - 137.2 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p343-b2", - "global_id": 9420, - "bbox": [ - 143.75, - 94.79, - 161.01, - 111.94 - ], - "text": "h1[n]", - "type": "text" - }, - { - "block_id": "p343-b4", - "global_id": 9421, - "bbox": [ - 220.31, - 94.81, - 237.46, - 112.06 - ], - "text": "h2[n]", - "type": "text" - }, - { - "block_id": "p343-b5", - "global_id": 9422, - "bbox": [ - 142.59, - 155.69, - 159.74, - 172.95 - ], - "text": "h4[n]", - "type": "text" - }, - { - "block_id": "p343-b6", - "global_id": 9423, - "bbox": [ - 182.61, - 129.9, - 198.93, - 139.51 - ], - "text": "h5[n]", - "type": "text" - }, - { - "block_id": "p343-b7", - "global_id": 9424, - "bbox": [ - 221.75, - 155.09, - 239.01, - 172.24 - ], - "text": "h3[n]", - "type": "text" - }, - { - "block_id": "p343-b8", - "global_id": 9425, - "bbox": [ - 89.56, - 189.66, - 145.68, - 198.62 - ], - "text": "Figure P3.8-18", - "type": "text" - }, - { - "block_id": "p343-b9", - "global_id": 9426, - "bbox": [ - 83.34, - 218.47, - 288.98, - 227.51 - ], - "text": "3.8-19\n(a) Show that for a causal system, Eq. (3.37)", - "type": "text" - }, - { - "block_id": "p343-b10", - "global_id": 9427, - "bbox": [ - 131.9, - 229.51, - 218.9, - 238.48 - ], - "text": "can also be expressed as", - "type": "text" - }, - { - "block_id": "p343-b11", - "global_id": 9428, - "bbox": [ - 176.36, - 263.08, - 200.13, - 272.33 - ], - "text": "g[n] =", - "type": "text" - }, - { - "block_id": "p343-b12", - "global_id": 9429, - "bbox": [ - 201.98, - 253.97, - 214.67, - 263.53 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p343-b13", - "global_id": 9430, - "bbox": [ - 202.69, - 276.01, - 213.95, - 282.76 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p343-b14", - "global_id": 9431, - "bbox": [ - 215.67, - 263.08, - 244.52, - 272.33 - ], - "text": "h[n −k]", - "type": "text" - }, - { - "block_id": "p343-b15", - "global_id": 9432, - "bbox": [ - 116.46, - 298.43, - 288.98, - 307.4 - ], - "text": "(b) How would the expressions in part (a)", - "type": "text" - }, - { - "block_id": "p343-b16", - "global_id": 9433, - "bbox": [ - 131.89, - 309.39, - 255.91, - 318.36 - ], - "text": "change if the system is not causal?", - "type": "text" - }, - { - "block_id": "p343-b17", - "global_id": 9434, - "bbox": [ - 83.34, - 323.56, - 288.98, - 365.77 - ], - "text": "3.8-20\nAn LTID system with input x[n] and output y[n]\nhas impulse response h[n] = 2(u[n + 2] −u[n −\n3]).\n(a) Write a constant-coefficient linear differ-", - "type": "text" - }, - { - "block_id": "p343-b18", - "global_id": 9435, - "bbox": [ - 116.46, - 367.77, - 288.98, - 409.61 - ], - "text": "ence equation that has the given impulse\nresponse. [Hint: First express h[n] in terms\nof delta functions δ[n].]\n(b) Using graphical convolution, determine the", - "type": "text" - }, - { - "block_id": "p343-b19", - "global_id": 9436, - "bbox": [ - 131.89, - 411.6, - 288.98, - 442.49 - ], - "text": "zero-state output of this system in response\nto the anticausal input x[n] = 2nu[−n]. A\nsimplified closed-form solution is required.", - "type": "text" - }, - { - "block_id": "p343-b20", - "global_id": 9437, - "bbox": [ - 83.34, - 447.98, - 288.98, - 457.02 - ], - "text": "3.8-21\nConsider three LTID systems: system 1 has im-", - "type": "text" - }, - { - "block_id": "p343-b21", - "global_id": 9438, - "bbox": [ - 116.46, - 463.51, - 217.09, - 473.58 - ], - "text": "pulse response h1[n] = [", - "type": "text" - }, - { - "block_id": "p343-b22", - "global_id": 9439, - "bbox": [ - 217.09, - 457.93, - 288.98, - 473.65 - ], - "text": "↓\n2, −3, 4], system 2", - "type": "text" - }, - { - "block_id": "p343-b23", - "global_id": 9440, - "bbox": [ - 116.46, - 479.37, - 257.83, - 489.45 - ], - "text": "has\nimpulse\nresponse\nh2[n] = [", - "type": "text" - }, - { - "block_id": "p343-b24", - "global_id": 9441, - "bbox": [ - 116.46, - 473.75, - 288.99, - 510.64 - ], - "text": "↓\n0, 0, −6,\n−9, 3], and system 3 is an identity system\n(output equals input).", - "type": "text" - }, - { - "block_id": "p343-b25", - "global_id": 9442, - "bbox": [ - 116.96, - 512.25, - 288.99, - 521.59 - ], - "text": "(a) Determine the overall impulse response h[n]", - "type": "text" - }, - { - "block_id": "p343-b26", - "global_id": 9443, - "bbox": [ - 116.46, - 523.59, - 288.98, - 554.47 - ], - "text": "if system 1 is connected in cascade with a\nparallel connection of systems 2 and 3.\n(b) For input x[n] = u[−n], determine the", - "type": "text" - }, - { - "block_id": "p343-b27", - "global_id": 9444, - "bbox": [ - 131.9, - 556.09, - 270.92, - 566.17 - ], - "text": "zero-state response yzsr[n] of system 2.", - "type": "text" - }, - { - "block_id": "p343-b28", - "global_id": 9445, - "bbox": [ - 83.34, - 570.92, - 288.98, - 634.77 - ], - "text": "3.8-22\nIn the savings account problem described in\nEx. 3.6, a person deposits $500 at the beginning\nof every month, starting at n = 0 with the\nexception at n = 4, when instead of depositing\n$500, she withdraws $1000. Find y[n] if the\ninterest rate is 1% per month (r = 0.01).", - "type": "text" - }, - { - "block_id": "p343-b29", - "global_id": 9446, - "bbox": [ - 310.48, - 85.92, - 516.14, - 116.9 - ], - "text": "3.8-23\nTo pay off a loan of M dollars in N number\nof payments using a fixed monthly payment of\nP dollars, show that", - "type": "text" - }, - { - "block_id": "p343-b30", - "global_id": 9447, - "bbox": [ - 396.01, - 138.23, - 461.65, - 159.94 - ], - "text": "P =\nrM\n1 −(1 + r)−N", - "type": "text" - }, - { - "block_id": "p343-b31", - "global_id": 9448, - "bbox": [ - 343.61, - 181.67, - 516.13, - 333.2 - ], - "text": "where r is the interest rate per dollar per month.\n[Hint: This problem can be modeled by Eq. (3.3)\nwith the payments of P dollars starting at n = 1.\nThe problem can be approached in two ways.\nFirst, consider the loan as the initial condition\ny0[0] = −M, and the input x[n] = Pu[n −1].\nThe loan balance is the sum of the zero-input\ncomponent (due to the initial condition) and\nthe zero-state component h[n] ∗x[n]. Second,\nconsider the loan as an input −M at n = 0 along\nwith the input due to payments. The loan balance\nis now exclusively a zero-state component h[n]∗\nx[n]. Because the loan is paid off in N payments,\nset y[N] = 0.]", - "type": "text" - }, - { - "block_id": "p343-b32", - "global_id": 9449, - "bbox": [ - 310.48, - 339.33, - 516.14, - 479.88 - ], - "text": "3.8-24\nA person receives an automobile loan of $10,000\nfrom a bank at the interest rate of 1.5% per\nmonth. His monthly payment is $500, with the\nfirst payment due one month after he receives\nthe loan. Compute the number of payments\nrequired to pay off the loan. Note that the last\npayment may not be exactly $500. [Hint: Follow\nthe procedure in Prob. 3.8-23 to determine the\nbalance y[n]. To determine N, the number of\npayments, set y[N] = 0. In general, N will not\nbe an integer. The number of payments K is\nthe largest integer ≤N. The residual payment is\n|y[K]|.]", - "type": "text" - }, - { - "block_id": "p343-b33", - "global_id": 9450, - "bbox": [ - 310.48, - 485.7, - 516.14, - 516.96 - ], - "text": "3.8-25\nLetting ↓identify the n = 0 values, use the\nsliding-tape method to determine the follow-\ning:", - "type": "text" - }, - { - "block_id": "p343-b34", - "global_id": 9451, - "bbox": [ - 344.12, - 523.48, - 380.07, - 533.79 - ], - "text": "(a) ya = [", - "type": "text" - }, - { - "block_id": "p343-b35", - "global_id": 9452, - "bbox": [ - 380.07, - 517.86, - 476.51, - 533.63 - ], - "text": "↓\n2,3,−2,−3] ∗[−10,\n↓\n0,−5]", - "type": "text" - }, - { - "block_id": "p343-b36", - "global_id": 9453, - "bbox": [ - 343.61, - 540.01, - 387.15, - 550.31 - ], - "text": "(b) yb = [2,", - "type": "text" - }, - { - "block_id": "p343-b37", - "global_id": 9454, - "bbox": [ - 388.15, - 533.62, - 471.65, - 549.34 - ], - "text": "↓\n−1,3,−2] ∗[−1,−4,1,", - "type": "text" - }, - { - "block_id": "p343-b38", - "global_id": 9455, - "bbox": [ - 472.65, - 533.62, - 487.11, - 549.34 - ], - "text": "↓\n−2]", - "type": "text" - }, - { - "block_id": "p343-b39", - "global_id": 9456, - "bbox": [ - 344.12, - 555.77, - 380.07, - 566.08 - ], - "text": "(c) yc = [", - "type": "text" - }, - { - "block_id": "p343-b40", - "global_id": 9457, - "bbox": [ - 380.07, - 550.14, - 481.93, - 565.92 - ], - "text": "↓\n0,0,3,2,1,2,3] ∗[2,3,−2,\n↓\n1]", - "type": "text" - }, - { - "block_id": "p343-b41", - "global_id": 9458, - "bbox": [ - 343.61, - 572.3, - 402.6, - 582.61 - ], - "text": "(d) yd = [5,0,0,", - "type": "text" - }, - { - "block_id": "p343-b42", - "global_id": 9459, - "bbox": [ - 403.6, - 565.91, - 457.67, - 581.64 - ], - "text": "↓\n−2,8] ∗[−1,1,", - "type": "text" - }, - { - "block_id": "p343-b43", - "global_id": 9460, - "bbox": [ - 458.67, - 566.68, - 496.28, - 582.4 - ], - "text": "↓\n3,3,−2,3]", - "type": "text" - }, - { - "block_id": "p343-b44", - "global_id": 9461, - "bbox": [ - 344.11, - 588.83, - 390.14, - 599.14 - ], - "text": "(e) ye = ([1,", - "type": "text" - }, - { - "block_id": "p343-b45", - "global_id": 9462, - "bbox": [ - 391.14, - 582.44, - 416.04, - 598.17 - ], - "text": "↓\n−1] ∗[", - "type": "text" - }, - { - "block_id": "p343-b46", - "global_id": 9463, - "bbox": [ - 416.04, - 582.44, - 513.52, - 598.98 - ], - "text": "↓\n1,−1]) ∗([\n↓\n1,−1] ∗[1,\n↓\n−1])", - "type": "text" - }, - { - "block_id": "p343-b47", - "global_id": 9464, - "bbox": [ - 345.11, - 605.35, - 389.42, - 615.66 - ], - "text": "(f) yf = ([2,", - "type": "text" - }, - { - "block_id": "p343-b48", - "global_id": 9465, - "bbox": [ - 390.42, - 598.97, - 415.32, - 614.69 - ], - "text": "↓\n−1] ∗[", - "type": "text" - }, - { - "block_id": "p343-b49", - "global_id": 9466, - "bbox": [ - 343.61, - 598.97, - 516.12, - 636.61 - ], - "text": "↓\n1,−2]) ∗([\n↓\n1,−2] ∗[2,\n↓\n−1])\nOutside the values shown, assume all signals are\nzero.", - "type": "text" - } - ] - }, - { - "page_num": 344, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p344-b0", - "global_id": 9467, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "324\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p344-b1", - "global_id": 9468, - "bbox": [ - 57.59, - 85.64, - 263.24, - 105.94 - ], - "text": "3.8-26\nLet ↓identify the n = 0 and consider DT signals\nx[n] and h[n] whose nonzero values are given as", - "type": "text" - }, - { - "block_id": "p344-b2", - "global_id": 9469, - "bbox": [ - 90.72, - 112.43, - 144.85, - 121.76 - ], - "text": "x[n] = [1, 2, 3,", - "type": "text" - }, - { - "block_id": "p344-b3", - "global_id": 9470, - "bbox": [ - 90.72, - 106.85, - 263.24, - 143.68 - ], - "text": "↓\n4, 5] and h[n] = [−2, −1, 1,\n↓\n2].\nUse DT convolution (any method) to determine\ny[n] = (2x[n −30]) ∗", - "type": "text" - }, - { - "block_id": "p344-b5", - "global_id": 9471, - "bbox": [ - 178.5, - 132.98, - 189.92, - 143.31 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p344-b6", - "global_id": 9472, - "bbox": [ - 186.69, - 127.13, - 228.42, - 146.26 - ], - "text": "2h[n −10]", - "type": "text" - }, - { - "block_id": "p344-b7", - "global_id": 9473, - "bbox": [ - 90.72, - 134.71, - 263.25, - 176.56 - ], - "text": ". Express\nyour result in vector notation, making sure to\nindicate the time index of the leftmost (nonzero)\nelement.", - "type": "text" - }, - { - "block_id": "p344-b8", - "global_id": 9474, - "bbox": [ - 57.59, - 189.31, - 244.87, - 198.35 - ], - "text": "3.8-27\nUsing the sliding-tape algorithm, show that", - "type": "text" - }, - { - "block_id": "p344-b9", - "global_id": 9475, - "bbox": [ - 90.72, - 199.96, - 263.28, - 220.26 - ], - "text": "(a) u[n] ∗u[n] = (n + 1)u[n]\n(b) (u[n] −u[n −m])∗u[n] = (n+1)u[n]−(n−", - "type": "text" - }, - { - "block_id": "p344-b10", - "global_id": 9476, - "bbox": [ - 106.16, - 221.89, - 161.43, - 231.22 - ], - "text": "m + 1)u[n −m]", - "type": "text" - }, - { - "block_id": "p344-b11", - "global_id": 9477, - "bbox": [ - 57.59, - 243.67, - 263.23, - 263.97 - ], - "text": "3.8-28\nUsing the sliding-tape algorithm, find x[n]∗g[n]\nfor the signals shown in Fig. P3.8-28.", - "type": "text" - }, - { - "block_id": "p344-b12", - "global_id": 9478, - "bbox": [ - 284.75, - 85.94, - 490.4, - 105.94 - ], - "text": "3.8-29\nRepeat Prob. 3.8-28 for the signals shown in\nFig. P3.8-29.", - "type": "text" - }, - { - "block_id": "p344-b13", - "global_id": 9479, - "bbox": [ - 284.75, - 113.45, - 490.4, - 133.46 - ], - "text": "3.8-30\nRepeat Prob. 3.8-28 for the signals shown in\nFig. P3.8-30.", - "type": "text" - }, - { - "block_id": "p344-b14", - "global_id": 9480, - "bbox": [ - 284.75, - 140.67, - 490.39, - 150.01 - ], - "text": "3.8-31\nLetting ↓identify n = 0, define the nonzero", - "type": "text" - }, - { - "block_id": "p344-b15", - "global_id": 9481, - "bbox": [ - 317.87, - 156.5, - 438.67, - 165.84 - ], - "text": "values of signal x[n] as [1,2,", - "type": "text" - }, - { - "block_id": "p344-b16", - "global_id": 9482, - "bbox": [ - 317.86, - 150.91, - 490.39, - 176.79 - ], - "text": "↓\n2]. Similarly,\ndefine the non-zero values of signal y[n] as", - "type": "text" - }, - { - "block_id": "p344-b17", - "global_id": 9483, - "bbox": [ - 317.86, - 183.27, - 362.94, - 192.61 - ], - "text": "[3,4,6,6,11,", - "type": "text" - }, - { - "block_id": "p344-b18", - "global_id": 9484, - "bbox": [ - 317.86, - 177.7, - 490.37, - 214.54 - ], - "text": "↓\n2,−2]. Using the sliding-tape algo-\nrithm as the basis for your work, determine the\nsignal h[n] so that y[n] = x[n] ∗h[n].", - "type": "text" - }, - { - "block_id": "p344-b19", - "global_id": 9485, - "bbox": [ - 284.74, - 222.05, - 490.39, - 263.97 - ], - "text": "3.8-32\nThe convolution sum in Eq. (3.33) can be\nexpressed in a matrix form as y = Hx, where y is\na column vector containing y[0],y[1],. . .y[n]; x\nis a column vector containing x[0],x[1],. . .x[n];", - "type": "text" - }, - { - "block_id": "p344-b20", - "global_id": 9486, - "bbox": [ - 168.02, - 284.69, - 180.9, - 292.77 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p344-b21", - "global_id": 9487, - "bbox": [ - 239.94, - 353.04, - 243.94, - 361.04 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p344-b22", - "global_id": 9488, - "bbox": [ - 176.23, - 300.32, - 180.23, - 308.32 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p344-b23", - "global_id": 9489, - "bbox": [ - 228.98, - 353.49, - 232.98, - 361.49 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p344-b24", - "global_id": 9490, - "bbox": [ - 338.86, - 311.99, - 352.19, - 320.07 - ], - "text": "g[n]", - "type": "text" - }, - { - "block_id": "p344-b25", - "global_id": 9491, - "bbox": [ - 357.89, - 339.34, - 361.89, - 347.34 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p344-b26", - "global_id": 9492, - "bbox": [ - 307.72, - 352.48, - 367.44, - 362.38 - ], - "text": "n\n4", - "type": "text" - }, - { - "block_id": "p344-b27", - "global_id": 9493, - "bbox": [ - 271.23, - 330.88, - 276.23, - 340.88 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p344-b28", - "global_id": 9494, - "bbox": [ - 404.63, - 354.35, - 460.74, - 363.32 - ], - "text": "Figure P3.8-28", - "type": "text" - }, - { - "block_id": "p344-b29", - "global_id": 9495, - "bbox": [ - 215.51, - 382.73, - 228.39, - 390.81 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p344-b30", - "global_id": 9496, - "bbox": [ - 272.21, - 421.86, - 276.21, - 429.86 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p344-b31", - "global_id": 9497, - "bbox": [ - 205.78, - 391.93, - 209.78, - 399.93 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p344-b32", - "global_id": 9498, - "bbox": [ - 104.83, - 421.95, - 313.27, - 430.54 - ], - "text": "5\n10\n5\n10", - "type": "text" - }, - { - "block_id": "p344-b33", - "global_id": 9499, - "bbox": [ - 409.65, - 382.73, - 422.98, - 390.81 - ], - "text": "g[n]", - "type": "text" - }, - { - "block_id": "p344-b34", - "global_id": 9500, - "bbox": [ - 466.61, - 421.86, - 470.61, - 429.86 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p344-b35", - "global_id": 9501, - "bbox": [ - 398.88, - 397.86, - 402.88, - 405.86 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p344-b36", - "global_id": 9502, - "bbox": [ - 350.39, - 421.64, - 456.55, - 429.93 - ], - "text": "5\n5", - "type": "text" - }, - { - "block_id": "p344-b37", - "global_id": 9503, - "bbox": [ - 324.1, - 404.2, - 329.1, - 414.2 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p344-b38", - "global_id": 9504, - "bbox": [ - 104.83, - 436.71, - 160.95, - 445.67 - ], - "text": "Figure P3.8-29", - "type": "text" - }, - { - "block_id": "p344-b39", - "global_id": 9505, - "bbox": [ - 205.3, - 458.78, - 218.18, - 466.86 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p344-b40", - "global_id": 9506, - "bbox": [ - 264.17, - 527.87, - 273.05, - 535.87 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p344-b41", - "global_id": 9507, - "bbox": [ - 241.37, - 514.25, - 245.37, - 522.25 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p344-b42", - "global_id": 9508, - "bbox": [ - 195.82, - 473.78, - 199.82, - 481.78 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p344-b43", - "global_id": 9509, - "bbox": [ - 168.89, - 514.03, - 232.02, - 522.33 - ], - "text": "3\n3", - "type": "text" - }, - { - "block_id": "p344-b44", - "global_id": 9510, - "bbox": [ - 333.04, - 458.78, - 346.37, - 466.86 - ], - "text": "g[n]", - "type": "text" - }, - { - "block_id": "p344-b45", - "global_id": 9511, - "bbox": [ - 369.29, - 514.25, - 373.29, - 522.25 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p344-b46", - "global_id": 9512, - "bbox": [ - 323.45, - 473.78, - 327.45, - 481.78 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p344-b47", - "global_id": 9513, - "bbox": [ - 296.81, - 514.0, - 359.94, - 522.29 - ], - "text": "3\n3", - "type": "text" - }, - { - "block_id": "p344-b48", - "global_id": 9514, - "bbox": [ - 118.42, - 548.48, - 131.3, - 556.56 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p344-b49", - "global_id": 9515, - "bbox": [ - 263.78, - 614.19, - 273.43, - 622.19 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p344-b50", - "global_id": 9516, - "bbox": [ - 241.37, - 600.82, - 245.37, - 608.82 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p344-b51", - "global_id": 9517, - "bbox": [ - 108.62, - 566.91, - 112.62, - 574.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p344-b52", - "global_id": 9518, - "bbox": [ - 107.32, - 600.9, - 228.02, - 608.9 - ], - "text": "12\n6\n0", - "type": "text" - }, - { - "block_id": "p344-b53", - "global_id": 9519, - "bbox": [ - 412.46, - 548.48, - 425.79, - 556.56 - ], - "text": "g[n]", - "type": "text" - }, - { - "block_id": "p344-b54", - "global_id": 9520, - "bbox": [ - 421.41, - 600.61, - 425.41, - 608.61 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p344-b55", - "global_id": 9521, - "bbox": [ - 412.46, - 566.91, - 416.46, - 574.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p344-b56", - "global_id": 9522, - "bbox": [ - 302.43, - 600.4, - 416.96, - 608.7 - ], - "text": "12\n6\n0", - "type": "text" - }, - { - "block_id": "p344-b57", - "global_id": 9523, - "bbox": [ - 266.1, - 496.03, - 271.1, - 506.03 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p344-b58", - "global_id": 9524, - "bbox": [ - 266.1, - 582.56, - 271.1, - 592.56 - ], - "text": "*", - "type": "text" - }, - { - "block_id": "p344-b59", - "global_id": 9525, - "bbox": [ - 104.83, - 628.36, - 160.95, - 637.33 - ], - "text": "Figure P3.8-30", - "type": "text" - } - ] - }, - { - "page_num": 345, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p345-b0", - "global_id": 9526, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n325", - "type": "text" - }, - { - "block_id": "p345-b1", - "global_id": 9527, - "bbox": [ - 116.46, - 85.6, - 277.11, - 94.93 - ], - "text": "and H is a lower triangular matrix defined as", - "type": "text" - }, - { - "block_id": "p345-b2", - "global_id": 9528, - "bbox": [ - 120.93, - 124.99, - 136.93, - 133.96 - ], - "text": "H =", - "type": "text" - }, - { - "block_id": "p345-b3", - "global_id": 9529, - "bbox": [ - 138.78, - 98.94, - 145.32, - 107.91 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p345-b4", - "global_id": 9530, - "bbox": [ - 138.78, - 114.67, - 145.32, - 140.19 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p345-b5", - "global_id": 9531, - "bbox": [ - 150.3, - 104.91, - 267.76, - 143.35 - ], - "text": "h[0]\n0\n0\n. . .\n0\nh[1]\nh[0]\n0\n. . .\n0\n...", - "type": "text" - }, - { - "block_id": "p345-b6", - "global_id": 9532, - "bbox": [ - 188.68, - 126.4, - 190.92, - 143.35 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p345-b7", - "global_id": 9533, - "bbox": [ - 150.3, - 126.41, - 272.99, - 154.31 - ], - "text": "...\n· · ·\n...\nh[n]\nh[n −1]\n. . .\n. . .\nh[0]", - "type": "text" - }, - { - "block_id": "p345-b8", - "global_id": 9534, - "bbox": [ - 277.98, - 98.95, - 284.51, - 107.91 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p345-b9", - "global_id": 9535, - "bbox": [ - 277.98, - 114.68, - 284.51, - 140.19 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p345-b10", - "global_id": 9536, - "bbox": [ - 116.46, - 164.06, - 288.99, - 293.95 - ], - "text": "Knowing h[n] and the output y[n], we can\ndetermine the input x[n] according to x = H−1y.\nThis operation is the reverse of convolution and\nis known as deconvolution. Moreover, knowing\nx[n] and y[n], we can determine h[n]. This can\nbe done by expressing the foregoing matrix\nequation as n + 1 simultaneous equations in\nterms of n + 1 unknowns h[0], h[1], . . . ,\nh[n]. These equations can readily be solved\niteratively. Thus, we can synthesize a system that\nyields a certain output y[n] for a given input\nx[n].", - "type": "text" - }, - { - "block_id": "p345-b11", - "global_id": 9537, - "bbox": [ - 116.96, - 295.57, - 288.98, - 304.91 - ], - "text": "(a) Design a system (i.e., determine h[n]) that", - "type": "text" - }, - { - "block_id": "p345-b12", - "global_id": 9538, - "bbox": [ - 116.46, - 306.9, - 288.98, - 348.74 - ], - "text": "will yield the output sequence (8, 12, 14, 15,\n15.5, 15.75, . . .) for the input sequence (1,\n1, 1, 1, 1, 1, . . .).\n(b) For a system with the impulse response", - "type": "text" - }, - { - "block_id": "p345-b13", - "global_id": 9539, - "bbox": [ - 131.89, - 350.36, - 288.97, - 381.62 - ], - "text": "sequence (1, 2, 4, . . .), the output sequence\nwas (1, 7/3, 43/9, . . .). Determine the input\nsequence.", - "type": "text" - }, - { - "block_id": "p345-b14", - "global_id": 9540, - "bbox": [ - 83.34, - 386.53, - 288.97, - 406.53 - ], - "text": "3.8-33\nA second-order LTID system has zero-input\nresponse", - "type": "text" - }, - { - "block_id": "p345-b15", - "global_id": 9541, - "bbox": [ - 146.46, - 416.84, - 173.48, - 426.92 - ], - "text": "y0[n] =", - "type": "text" - }, - { - "block_id": "p345-b17", - "global_id": 9542, - "bbox": [ - 178.85, - 409.63, - 243.78, - 428.76 - ], - "text": "3,2 1\n3,2 1\n9,2 1\n27,...", - "type": "text" - }, - { - "block_id": "p345-b18", - "global_id": 9543, - "bbox": [ - 166.48, - 439.48, - 173.48, - 448.45 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p345-b19", - "global_id": 9544, - "bbox": [ - 175.32, - 430.16, - 188.01, - 439.93 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p345-b20", - "global_id": 9545, - "bbox": [ - 176.02, - 452.41, - 187.29, - 459.16 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p345-b21", - "global_id": 9546, - "bbox": [ - 189.0, - 429.58, - 193.9, - 438.55 - ], - "text": ",", - "type": "text" - }, - { - "block_id": "p345-b22", - "global_id": 9547, - "bbox": [ - 193.9, - 432.28, - 216.21, - 448.82 - ], - "text": "2 +\n 1", - "type": "text" - }, - { - "block_id": "p345-b23", - "global_id": 9548, - "bbox": [ - 212.98, - 429.58, - 229.4, - 451.41 - ], - "text": "3\nk-", - "type": "text" - }, - { - "block_id": "p345-b24", - "global_id": 9549, - "bbox": [ - 230.39, - 439.48, - 258.97, - 448.73 - ], - "text": "δ[n −k]", - "type": "text" - }, - { - "block_id": "p345-b25", - "global_id": 9550, - "bbox": [ - 116.96, - 472.53, - 288.98, - 481.49 - ], - "text": "(a) Determine the characteristic equation of this", - "type": "text" - }, - { - "block_id": "p345-b26", - "global_id": 9551, - "bbox": [ - 116.46, - 479.86, - 288.98, - 503.41 - ], - "text": "system, a0γ 2 + a1γ + a2 = 0.\n(b) Find a bounded, causal input with infinite", - "type": "text" - }, - { - "block_id": "p345-b27", - "global_id": 9552, - "bbox": [ - 131.9, - 505.4, - 288.99, - 525.32 - ], - "text": "duration that would cause a strong response\nfrom this system. Justify your choice.", - "type": "text" - }, - { - "block_id": "p345-b28", - "global_id": 9553, - "bbox": [ - 344.12, - 85.9, - 516.14, - 94.86 - ], - "text": "(c) Find a bounded, causal input with infinite", - "type": "text" - }, - { - "block_id": "p345-b29", - "global_id": 9554, - "bbox": [ - 359.05, - 96.86, - 516.13, - 116.78 - ], - "text": "duration that would cause a weak response\nfrom this system. Justify your choice.", - "type": "text" - }, - { - "block_id": "p345-b30", - "global_id": 9555, - "bbox": [ - 310.48, - 123.37, - 516.13, - 166.03 - ], - "text": "3.8-34\nAn LTID filter has an impulse response function\ngiven by h1[n] = δ[n + 2] −δ[n −2]. A second\nLTID system has an impulse response function\ngiven by h2[n] = n(u[n + 4] −u[n −4]).", - "type": "text" - }, - { - "block_id": "p345-b31", - "global_id": 9556, - "bbox": [ - 344.12, - 166.91, - 516.12, - 176.99 - ], - "text": "(a) Carefully sketch the functions h1[n] and", - "type": "text" - }, - { - "block_id": "p345-b32", - "global_id": 9557, - "bbox": [ - 343.61, - 177.87, - 516.1, - 198.17 - ], - "text": "h2[n] over (−10 ≤n ≤10).\n(b) Assume that the two systems are connected", - "type": "text" - }, - { - "block_id": "p345-b33", - "global_id": 9558, - "bbox": [ - 344.12, - 200.16, - 516.13, - 252.97 - ], - "text": "in parallel, as shown in Fig. P3.8-34a.\nDetermine the impulse response hp[n] for\nthe parallel system in terms of h1[n] and\nh2[n]. Sketch hp[n] over (−10 ≤n ≤10).\n(c) Assume that the two systems are connected", - "type": "text" - }, - { - "block_id": "p345-b34", - "global_id": 9559, - "bbox": [ - 359.05, - 254.95, - 516.13, - 297.53 - ], - "text": "in cascade, as shown in Fig. P3.8-34b.\nDetermine the impulse response hs[n] for\nthe cascade system in terms of h1[n] and\nh2[n]. Sketch hs[n] over (−10 ≤n ≤10).", - "type": "text" - }, - { - "block_id": "p345-b35", - "global_id": 9560, - "bbox": [ - 310.48, - 303.39, - 516.13, - 334.35 - ], - "text": "3.8-35\nThis problem investigates an interesting applica-\ntion of discrete-time convolution: the expansion\nof certain polynomial expressions.", - "type": "text" - }, - { - "block_id": "p345-b36", - "global_id": 9561, - "bbox": [ - 344.12, - 334.98, - 516.14, - 345.31 - ], - "text": "(a) By hand, expand (z3+z2+z+1)2. Compare", - "type": "text" - }, - { - "block_id": "p345-b37", - "global_id": 9562, - "bbox": [ - 343.61, - 346.93, - 516.12, - 367.23 - ], - "text": "the coefficients to [1,1,1,1] ∗[1,1,1,1].\n(b) Formulate a relationship between discrete-", - "type": "text" - }, - { - "block_id": "p345-b38", - "global_id": 9563, - "bbox": [ - 344.11, - 369.22, - 516.13, - 411.06 - ], - "text": "time convolution and the expansion of\nconstant-coefficient\npolynomial\nexpres-\nsions.\n(c) Use convolution to expand (z−4 −2z−3 +", - "type": "text" - }, - { - "block_id": "p345-b39", - "global_id": 9564, - "bbox": [ - 343.61, - 411.42, - 516.13, - 432.98 - ], - "text": "3z−2)4.\n(d) Use convolution to expand (z5 +2z4 +3z2 +", - "type": "text" - }, - { - "block_id": "p345-b40", - "global_id": 9565, - "bbox": [ - 359.05, - 431.34, - 437.1, - 443.94 - ], - "text": "5)2(z−4 −5z−2 + 13).", - "type": "text" - }, - { - "block_id": "p345-b41", - "global_id": 9566, - "bbox": [ - 310.48, - 450.53, - 516.13, - 525.33 - ], - "text": "3.8-36\nJoe likes coffee, and he drinks his coffee accord-\ning to a very particular routine. He begins by\nadding two teaspoons of sugar to his mug, which\nhe then fills to the brim with hot coffee. He\ndrinks 2/3 of the mug’s contents, adds another\ntwo teaspoons of sugar, and tops the mug off\nwith steaming hot coffee. This refill procedure", - "type": "text" - }, - { - "block_id": "p345-b42", - "global_id": 9567, - "bbox": [ - 197.5, - 548.84, - 204.5, - 558.45 - ], - "text": "h1", - "type": "text" - }, - { - "block_id": "p345-b43", - "global_id": 9568, - "bbox": [ - 197.5, - 592.04, - 204.5, - 601.65 - ], - "text": "h2", - "type": "text" - }, - { - "block_id": "p345-b44", - "global_id": 9569, - "bbox": [ - 130.8, - 571.48, - 284.38, - 581.03 - ], - "text": "x[n]\nyp[n]", - "type": "text" - }, - { - "block_id": "p345-b45", - "global_id": 9570, - "bbox": [ - 203.15, - 615.14, - 394.92, - 623.14 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p345-b46", - "global_id": 9571, - "bbox": [ - 319.2, - 570.44, - 461.31, - 581.03 - ], - "text": "x[n]\nh1\nh2\nys[n]", - "type": "text" - }, - { - "block_id": "p345-b47", - "global_id": 9572, - "bbox": [ - 130.58, - 629.31, - 186.69, - 638.27 - ], - "text": "Figure P3.8-34", - "type": "text" - } - ] - }, - { - "page_num": 346, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p346-b0", - "global_id": 9573, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "326\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p346-b1", - "global_id": 9574, - "bbox": [ - 90.72, - 85.9, - 263.25, - 204.45 - ], - "text": "continues, sometimes for many, many cups of\ncoffee. Joe has noted that his coffee tends to taste\nsweeter with the number of refills.\nLet independent variable n designate the coffee\nrefill number. In this way, n = 0 indicates the\nfirst cup of coffee, n = 1 is the first refill, and\nso forth. Let x[n] represent the sugar (measured\nin teaspoons) added into the system (a coffee\nmug) on refill n. Let y[n] designate the amount\nof sugar (again, teaspoons) contained in the mug\non refill n.", - "type": "text" - }, - { - "block_id": "p346-b2", - "global_id": 9575, - "bbox": [ - 91.22, - 206.45, - 263.23, - 215.41 - ], - "text": "(a) The sugar (teaspoons) in Joe’s coffee can be", - "type": "text" - }, - { - "block_id": "p346-b3", - "global_id": 9576, - "bbox": [ - 90.72, - 217.41, - 263.24, - 281.16 - ], - "text": "represented using a standard second-order\nconstant\ncoefficient\ndifference\nequation\ny[n] + a1y[n −1] + a2y[n −2] = b0x[n] +\nb1x[n −1] + b2x[n −2]. Determine the\nconstants a1, a2, b0, b1, and b2.\n(b) Determine x[n], the driving function to this", - "type": "text" - }, - { - "block_id": "p346-b4", - "global_id": 9577, - "bbox": [ - 91.22, - 283.16, - 263.24, - 303.08 - ], - "text": "system.\n(c) Solve the difference equation for y[n].", - "type": "text" - }, - { - "block_id": "p346-b5", - "global_id": 9578, - "bbox": [ - 90.72, - 305.07, - 263.25, - 357.88 - ], - "text": "This requires finding the total solution. Joe\nalways starts with a clean mug from the\ndishwasher, so y[−1] (the sugar content\nbefore the first cup) is zero.\n(d) Determine the steady-state value of y[n].", - "type": "text" - }, - { - "block_id": "p346-b6", - "global_id": 9579, - "bbox": [ - 106.15, - 359.5, - 263.24, - 401.71 - ], - "text": "That is, what is y[n] as n →∞? If possible,\nsuggest a way of modifying x[n] so that\nthe sugar content of Joe’s coffee remains a\nconstant for all nonnegative n.", - "type": "text" - }, - { - "block_id": "p346-b7", - "global_id": 9580, - "bbox": [ - 57.59, - 406.62, - 263.24, - 459.5 - ], - "text": "3.8-37\nA system is called complex if a real-valued input\ncan produce a complex-valued output. Con-\nsider a causal complex system described by a\nfirst-order constant coefficient linear difference\nequation:", - "type": "text" - }, - { - "block_id": "p346-b8", - "global_id": 9581, - "bbox": [ - 127.84, - 471.04, - 226.13, - 480.38 - ], - "text": "(jE + 0.5)y[n] = (−5E)x[n]", - "type": "text" - }, - { - "block_id": "p346-b9", - "global_id": 9582, - "bbox": [ - 91.23, - 491.3, - 263.24, - 500.26 - ], - "text": "(a) Determine the impulse response function", - "type": "text" - }, - { - "block_id": "p346-b10", - "global_id": 9583, - "bbox": [ - 90.72, - 501.89, - 263.24, - 522.19 - ], - "text": "h[n] for this system.\n(b) Given input x[n] = u[n −5] and initial con-", - "type": "text" - }, - { - "block_id": "p346-b11", - "global_id": 9584, - "bbox": [ - 106.15, - 523.81, - 263.24, - 544.1 - ], - "text": "dition y0[−1] = j, determine the system’s\ntotal output y[n] for n ≥0.", - "type": "text" - }, - { - "block_id": "p346-b12", - "global_id": 9585, - "bbox": [ - 57.59, - 549.01, - 263.24, - 579.96 - ], - "text": "3.8-38\nA\ndiscrete-time\nLTI\nsystem\nhas\nimpulse\nresponse\nfunction\nh[n]\n=\nn(u[n −2] −\nu[n + 2]).", - "type": "text" - }, - { - "block_id": "p346-b13", - "global_id": 9586, - "bbox": [ - 91.23, - 581.59, - 263.23, - 590.93 - ], - "text": "(a) Carefully sketch the function h[n] over", - "type": "text" - }, - { - "block_id": "p346-b14", - "global_id": 9587, - "bbox": [ - 90.72, - 592.55, - 263.24, - 612.84 - ], - "text": "(−5 ≤n ≤5).\n(b) Determine the difference equation represen-", - "type": "text" - }, - { - "block_id": "p346-b15", - "global_id": 9588, - "bbox": [ - 106.16, - 614.47, - 263.23, - 634.77 - ], - "text": "tation of this system, using y[n] to designate\nthe output and x[n] to designate the input.", - "type": "text" - }, - { - "block_id": "p346-b16", - "global_id": 9589, - "bbox": [ - 284.74, - 85.64, - 490.39, - 127.86 - ], - "text": "3.8-39\nConsider three discrete-time signals: x[n], y[n],\nand z[n]. Denoting convolution as ∗, iden-\ntify the expression(s) that is(are) equivalent to\nx[n](y[n] ∗z[n]):", - "type": "text" - }, - { - "block_id": "p346-b17", - "global_id": 9590, - "bbox": [ - 317.86, - 129.48, - 411.47, - 149.77 - ], - "text": "(a) (x[n] ∗y[n])z[n]\n(b) (x[n]y[n]) ∗(x[n]z[n])", - "type": "text" - }, - { - "block_id": "p346-b18", - "global_id": 9591, - "bbox": [ - 317.86, - 151.4, - 397.02, - 182.65 - ], - "text": "(c) (x[n]y[n]) ∗z[n]\n(d) none of the above\nJustify your answer!", - "type": "text" - }, - { - "block_id": "p346-b19", - "global_id": 9592, - "bbox": [ - 284.74, - 187.75, - 490.39, - 208.05 - ], - "text": "3.8-40\nA causal system with input x[n] and output y[n]\nis described by", - "type": "text" - }, - { - "block_id": "p346-b20", - "global_id": 9593, - "bbox": [ - 362.85, - 224.52, - 445.4, - 233.86 - ], - "text": "y[n] −ny[n −1] = x[n]", - "type": "text" - }, - { - "block_id": "p346-b21", - "global_id": 9594, - "bbox": [ - 318.37, - 249.72, - 490.37, - 258.68 - ], - "text": "(a) By recursion, determine the first six nonzero", - "type": "text" - }, - { - "block_id": "p346-b22", - "global_id": 9595, - "bbox": [ - 317.86, - 260.3, - 490.4, - 303.25 - ], - "text": "values of h[n], the response to x[n] = δ[n].\nDo you think this system is BIBO-stable?\nWhy?\n(b) Compute yR[4] recursively from yR[n] −", - "type": "text" - }, - { - "block_id": "p346-b23", - "global_id": 9596, - "bbox": [ - 318.37, - 304.13, - 490.39, - 358.05 - ], - "text": "nyR[n−1] = x[n], assuming all initial condi-\ntions are zero and x[n] = u[n]. The subscript\nR is only used to emphasize a recursive\nsolution.\n(c) Define yC[n] = x[n]∗h[n]. Using x[n] = u[n]", - "type": "text" - }, - { - "block_id": "p346-b24", - "global_id": 9597, - "bbox": [ - 317.86, - 358.93, - 490.39, - 401.14 - ], - "text": "and h[n] from part (a), compute yC[4]. The\nsubscript C is only used to emphasize a\nconvolution solution.\n(d) In this chapter, both recursion and convo-", - "type": "text" - }, - { - "block_id": "p346-b25", - "global_id": 9598, - "bbox": [ - 333.31, - 403.14, - 490.4, - 477.86 - ], - "text": "lution are presented as potential methods\nto compute the zero-state response (ZSR)\nof a discrete-time system. Comparing parts\n(b) and (c), we see that yR[4]̸ = yC[4].\nWhy are the two results not the same?\nWhich method, if any, yields the correct\nZSR value?", - "type": "text" - }, - { - "block_id": "p346-b26", - "global_id": 9599, - "bbox": [ - 289.22, - 483.25, - 490.4, - 634.76 - ], - "text": "3.9-1\nIn Sec. 3.9-1 we showed that for BIBO stability\nin an LTID system, it is sufficient for its impulse\nresponse h[n] to satisfy Eq. (3.43). Show that\nthis is also a necessary condition for the system\nto be BIBO-stable. In other words, show that if\nEq. (3.43) is not satisfied, there exists a bounded\ninput that produces unbounded output. [Hint:\nAssume that a system exists for which h[n]\nviolates Eq. (3.43), yet its output is bounded for\nevery bounded input. Establish the contradiction\nin this statement by considering an input x[n]\ndefined by x[n1 −m] = 1 when h[m] > 0 and\nx[n1−m] = −1 when h[m] < 0, where n1 is some\nfixed integer.]", - "type": "text" - } - ] - }, - { - "page_num": 347, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p347-b0", - "global_id": 9600, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n327", - "type": "text" - }, - { - "block_id": "p347-b1", - "global_id": 9601, - "bbox": [ - 87.82, - 85.94, - 288.98, - 138.82 - ], - "text": "3.9-2\nEach of the following equations specifies an\nLTID system. Determine whether each of these\nsystems is BIBO-stable or -unstable. Deter-\nmine also whether each is asymptotically stable,\nunstable, or marginally stable.", - "type": "text" - }, - { - "block_id": "p347-b2", - "global_id": 9602, - "bbox": [ - 116.96, - 140.43, - 288.98, - 149.77 - ], - "text": "(a) y[n + 2] + 0.6y[n + 1] −0.16y[n] = x[n +", - "type": "text" - }, - { - "block_id": "p347-b3", - "global_id": 9603, - "bbox": [ - 116.46, - 151.4, - 288.98, - 171.69 - ], - "text": "1] −2x[n]\n(b) y[n] + 3y[n −1] + 2y[n −2] = x[n −1] +", - "type": "text" - }, - { - "block_id": "p347-b4", - "global_id": 9604, - "bbox": [ - 116.96, - 173.31, - 165.88, - 193.61 - ], - "text": "2x[n −2]\n(c) (E −1)2", - "type": "text" - }, - { - "block_id": "p347-b5", - "global_id": 9605, - "bbox": [ - 165.89, - 182.91, - 185.78, - 193.52 - ], - "text": "E + 1", - "type": "text" - }, - { - "block_id": "p347-b6", - "global_id": 9606, - "bbox": [ - 182.55, - 177.06, - 190.6, - 196.19 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p347-b7", - "global_id": 9607, - "bbox": [ - 116.46, - 184.27, - 268.62, - 204.56 - ], - "text": "y[n] = x[n]\n(d) y[n] + 2y[n −1] + 0.96y[n −2] = x[n]", - "type": "text" - }, - { - "block_id": "p347-b8", - "global_id": 9608, - "bbox": [ - 116.96, - 206.19, - 288.97, - 215.53 - ], - "text": "(e) y[n]+y[n−1]−2y[n−2] = x[n]+2x[n−1]", - "type": "text" - }, - { - "block_id": "p347-b9", - "global_id": 9609, - "bbox": [ - 117.96, - 213.89, - 232.22, - 226.49 - ], - "text": "(f) (E2 −1)(E2 + 1)y[n] = x[n]", - "type": "text" - }, - { - "block_id": "p347-b10", - "global_id": 9610, - "bbox": [ - 87.82, - 231.69, - 288.99, - 306.49 - ], - "text": "3.9-3\nConsider two LTIC systems in cascade, as\nillustrated in Fig. 3.29. The impulse response of\nthe system S1 is h1[n] = 2n u[n] and the impulse\nresponse of the system S2 is h2[n] = δ[n] −\n2δ[n−1]. Is the cascaded system asymptotically\nstable or unstable? Determine the BIBO stability\nof the composite system.", - "type": "text" - }, - { - "block_id": "p347-b11", - "global_id": 9611, - "bbox": [ - 87.82, - 311.7, - 289.0, - 397.45 - ], - "text": "3.9-4\nFigure P3.9-4 locates the characteristic roots of\nten causal, LTID systems, labeled A through J.\nEach system has only two roots and is described\nusing operator notation as Q(E)y[n] = P(E)x[n].\nAll plots are drawn to scale, with the unit\ncircle shown for reference. For each of the\nfollowing parts, identify all the answers that are\ncorrect.", - "type": "text" - }, - { - "block_id": "p347-b12", - "global_id": 9612, - "bbox": [ - 116.46, - 399.44, - 288.98, - 419.37 - ], - "text": "(a) Identify all systems that are unstable.\n(b) Assuming all systems have P(E) = E2,", - "type": "text" - }, - { - "block_id": "p347-b13", - "global_id": 9613, - "bbox": [ - 116.96, - 421.35, - 288.99, - 463.2 - ], - "text": "identify all systems that are real. Recall that\na real system always generates a real-valued\nresponse to a real-valued input.\n(c) Identify all systems that support oscillatory", - "type": "text" - }, - { - "block_id": "p347-b14", - "global_id": 9614, - "bbox": [ - 131.89, - 465.19, - 184.7, - 474.16 - ], - "text": "natural modes.", - "type": "text" - }, - { - "block_id": "p347-b15", - "global_id": 9615, - "bbox": [ - 343.61, - 85.9, - 516.13, - 94.86 - ], - "text": "(d) Identify all systems that have at least one", - "type": "text" - }, - { - "block_id": "p347-b16", - "global_id": 9616, - "bbox": [ - 344.11, - 95.22, - 516.13, - 116.78 - ], - "text": "mode whose envelop decays at a rate of 2−n.\n(e) Identify all systems that have only one", - "type": "text" - }, - { - "block_id": "p347-b17", - "global_id": 9617, - "bbox": [ - 359.05, - 118.77, - 381.2, - 127.74 - ], - "text": "mode.", - "type": "text" - }, - { - "block_id": "p347-b18", - "global_id": 9618, - "bbox": [ - 314.97, - 132.65, - 516.13, - 152.64 - ], - "text": "3.9-5\nA\ndiscrete-time\nLTI\nsystem\nhas\nimpulse\nresponse given by", - "type": "text" - }, - { - "block_id": "p347-b19", - "global_id": 9619, - "bbox": [ - 381.44, - 165.8, - 430.12, - 175.04 - ], - "text": "h[n] = δ[n] +", - "type": "text" - }, - { - "block_id": "p347-b20", - "global_id": 9620, - "bbox": [ - 431.51, - 158.59, - 439.56, - 170.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p347-b21", - "global_id": 9621, - "bbox": [ - 436.32, - 158.59, - 478.3, - 177.72 - ], - "text": "3\nn u[n −1]", - "type": "text" - }, - { - "block_id": "p347-b22", - "global_id": 9622, - "bbox": [ - 344.12, - 187.66, - 516.1, - 196.63 - ], - "text": "(a) Is the system stable? Is the system causal?", - "type": "text" - }, - { - "block_id": "p347-b23", - "global_id": 9623, - "bbox": [ - 343.61, - 198.62, - 508.18, - 218.54 - ], - "text": "Justify your answers.\n(b) Plot the signal x[n] = u[n −3] −u[n + 3].", - "type": "text" - }, - { - "block_id": "p347-b24", - "global_id": 9624, - "bbox": [ - 344.12, - 220.54, - 516.14, - 229.5 - ], - "text": "(c) Determine the system’s zero-state response", - "type": "text" - }, - { - "block_id": "p347-b25", - "global_id": 9625, - "bbox": [ - 359.05, - 231.13, - 516.13, - 251.42 - ], - "text": "y[n] to the input x[n] = u[n −3] −u[n + 3].\nPlot y[n] over (−10 ≤n ≤10).", - "type": "text" - }, - { - "block_id": "p347-b26", - "global_id": 9626, - "bbox": [ - 314.97, - 256.33, - 516.12, - 276.33 - ], - "text": "3.9-6\nAn LTID system has an impulse response given\nby", - "type": "text" - }, - { - "block_id": "p347-b27", - "global_id": 9627, - "bbox": [ - 406.92, - 282.13, - 430.69, - 291.38 - ], - "text": "h[n] =", - "type": "text" - }, - { - "block_id": "p347-b28", - "global_id": 9628, - "bbox": [ - 432.53, - 274.92, - 440.57, - 287.25 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p347-b29", - "global_id": 9629, - "bbox": [ - 437.34, - 274.92, - 452.33, - 294.05 - ], - "text": "2\n|n|", - "type": "text" - }, - { - "block_id": "p347-b30", - "global_id": 9630, - "bbox": [ - 343.61, - 303.0, - 508.69, - 322.92 - ], - "text": "(a) Is the system causal? Justify your answer.\n(b) Compute %∞", - "type": "text" - }, - { - "block_id": "p347-b31", - "global_id": 9631, - "bbox": [ - 344.12, - 313.58, - 516.13, - 344.84 - ], - "text": "n=−∞|h[n]|. Is this system\nBIBO-stable?\n(c) Compute the energy and power of input", - "type": "text" - }, - { - "block_id": "p347-b32", - "global_id": 9632, - "bbox": [ - 343.61, - 346.46, - 516.13, - 366.76 - ], - "text": "signal x[n] = 3u[n −5].\n(d) Using input x[n] = 3u[n −5], determine the", - "type": "text" - }, - { - "block_id": "p347-b33", - "global_id": 9633, - "bbox": [ - 359.05, - 368.75, - 516.14, - 389.3 - ], - "text": "zero-state response of this system at time\nn = 10. That is, determine yzsr[10].", - "type": "text" - }, - { - "block_id": "p347-b34", - "global_id": 9634, - "bbox": [ - 310.48, - 393.59, - 516.13, - 424.54 - ], - "text": "3.10-1\nDetermine a constant coefficient linear dif-\nference equation that describes a system for\nwhich the input x[n] = 2( 1", - "type": "text" - }, - { - "block_id": "p347-b35", - "global_id": 9635, - "bbox": [ - 343.61, - 414.14, - 516.13, - 435.5 - ], - "text": "3)nu[−n −4] causes\nresonance.", - "type": "text" - }, - { - "block_id": "p347-b36", - "global_id": 9636, - "bbox": [ - 310.48, - 440.11, - 516.13, - 482.32 - ], - "text": "3.10-2\nIf one exists, determine a real input x[n] that\nwill cause resonance in the causal LTID system\ndescribed by (E2 +1){y[n]} = (E+0.5){x[n]}. If\nno such input exists, explain why not.", - "type": "text" - }, - { - "block_id": "p347-b37", - "global_id": 9637, - "bbox": [ - 132.66, - 506.57, - 420.09, - 515.57 - ], - "text": "A\nB\nC\nD\nE", - "type": "text" - }, - { - "block_id": "p347-b38", - "global_id": 9638, - "bbox": [ - 132.66, - 574.74, - 418.5, - 583.74 - ], - "text": "F\nG\nH\nI\nJ", - "type": "text" - }, - { - "block_id": "p347-b39", - "global_id": 9639, - "bbox": [ - 130.58, - 629.31, - 182.22, - 638.27 - ], - "text": "Figure P3.9-4", - "type": "text" - } - ] - }, - { - "page_num": 348, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p348-b0", - "global_id": 9640, - "bbox": [ - 60.0, - 60.36, - 400.04, - 69.45 - ], - "text": "328\nCHAPTER 3\nTIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p348-b1", - "global_id": 9641, - "bbox": [ - 57.59, - 85.94, - 263.24, - 160.73 - ], - "text": "3.10-3\nConsider two lowpass LTID systems, one\nwith infinite-duration impulse response h1[n] =\n−(0.5)nu[n] and the other with finite-duration\nimpulse response h2[n] = 2(u[n] −u[n −4]).\nWhich system (1, 2, both, or neither) would\nmore efficiently transmit a binary communica-\ntion signal? Carefully justify your result.", - "type": "text" - }, - { - "block_id": "p348-b2", - "global_id": 9642, - "bbox": [ - 57.59, - 165.64, - 263.24, - 185.64 - ], - "text": "3.11-1\nWrite a MATLAB program that recursively\ncomputes and then plots the solution to y[n] −", - "type": "text" - }, - { - "block_id": "p348-b3", - "global_id": 9643, - "bbox": [ - 90.72, - 185.89, - 263.24, - 218.51 - ], - "text": "1\n3y[n −1] + 1\n2y[n −2] = x[n] for (0 ≤n ≤100)\ngiven x[n] = δ[n] + u[n −50] and y[−2] =\ny[−1] = 2.", - "type": "text" - }, - { - "block_id": "p348-b4", - "global_id": 9644, - "bbox": [ - 57.59, - 223.13, - 263.24, - 254.38 - ], - "text": "3.11-2\nConsider the discrete-time function f[n] =\ne−n/5 cos(πn/5)u[n]. Section 3.11 uses anony-\nmous functions in describing DT signals.", - "type": "text" - }, - { - "block_id": "p348-b5", - "global_id": 9645, - "bbox": [ - 75.77, - 264.59, - 264.09, - 273.56 - ], - "text": "f = @(n) exp(-n/5).*cos(pi*n/5).*(n>=0);", - "type": "text" - }, - { - "block_id": "p348-b6", - "global_id": 9646, - "bbox": [ - 90.72, - 283.28, - 263.24, - 369.2 - ], - "text": "While this anonymous function operates cor-\nrectly for a downsampling operation such as\nf[2n], it does not operate correctly for an\nupsampling operation, such as f[n/2]. Mod-\nify the anonymous function f so that it also\ncorrectly accommodates upsampling operations.\nTest your code by computing and plotting\nf(n/2) over (−10 ≤n ≤10).", - "type": "text" - }, - { - "block_id": "p348-b7", - "global_id": 9647, - "bbox": [ - 57.59, - 373.86, - 263.24, - 393.86 - ], - "text": "3.11-3\nWrite MATLAB code to compute and plot the\nDT convolutions of Prob. 3.8-25.", - "type": "text" - }, - { - "block_id": "p348-b8", - "global_id": 9648, - "bbox": [ - 57.59, - 398.76, - 263.25, - 528.36 - ], - "text": "3.11-4\nAn indecisive student contemplates whether\nhe should stay home or take his final exam,\nwhich is being held 2 miles away. Starting at\nhome, the student travels half the distance to\nthe exam location before changing his mind.\nThe student turns around and travels half the\ndistance between his current location and his\nhome before changing his mind again. This\nprocess of changing direction and traveling\nhalf the remaining distance continues until the\nstudent either reaches a destination or dies from\nexhaustion.", - "type": "text" - }, - { - "block_id": "p348-b9", - "global_id": 9649, - "bbox": [ - 91.22, - 530.35, - 263.23, - 539.32 - ], - "text": "(a) Determine a suitable difference equation", - "type": "text" - }, - { - "block_id": "p348-b10", - "global_id": 9650, - "bbox": [ - 90.72, - 541.31, - 263.24, - 561.24 - ], - "text": "description of this system.\n(b) Use MATLAB to simulate the difference", - "type": "text" - }, - { - "block_id": "p348-b11", - "global_id": 9651, - "bbox": [ - 91.22, - 563.23, - 263.24, - 616.04 - ], - "text": "equation in part (a). Where does the student\nend up as n →∞? How does your answer\nchange if the student goes two-thirds the\nway each time, rather than halfway?\n(c) Determine a closed-form solution to the", - "type": "text" - }, - { - "block_id": "p348-b12", - "global_id": 9652, - "bbox": [ - 106.16, - 618.03, - 263.25, - 637.95 - ], - "text": "equation in part (a). Use this solution to\nverify the results in part (b).", - "type": "text" - }, - { - "block_id": "p348-b13", - "global_id": 9653, - "bbox": [ - 284.74, - 85.65, - 490.38, - 105.95 - ], - "text": "3.11-5\nThe cross-correlation function between x[n] and\ny[n] is given as", - "type": "text" - }, - { - "block_id": "p348-b14", - "global_id": 9654, - "bbox": [ - 356.99, - 125.05, - 385.66, - 135.06 - ], - "text": "rxy[k] =", - "type": "text" - }, - { - "block_id": "p348-b15", - "global_id": 9655, - "bbox": [ - 391.14, - 115.74, - 403.83, - 125.5 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p348-b16", - "global_id": 9656, - "bbox": [ - 387.5, - 137.36, - 407.45, - 144.04 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p348-b17", - "global_id": 9657, - "bbox": [ - 408.46, - 125.05, - 451.27, - 134.3 - ], - "text": "x[n]y[n −k]", - "type": "text" - }, - { - "block_id": "p348-b18", - "global_id": 9658, - "bbox": [ - 317.86, - 154.18, - 490.39, - 196.4 - ], - "text": "Notice that rxy[k] is quite similar to the con-\nvolution sum. The independent variable k cor-\nresponds to the relative shift between the two\ninputs.", - "type": "text" - }, - { - "block_id": "p348-b19", - "global_id": 9659, - "bbox": [ - 318.37, - 198.02, - 490.39, - 208.04 - ], - "text": "(a) Express rxy[k] in terms of convolution. Is", - "type": "text" - }, - { - "block_id": "p348-b20", - "global_id": 9660, - "bbox": [ - 317.86, - 208.99, - 490.39, - 229.28 - ], - "text": "rxy[k] = ryx[k]?\n(b) Cross-correlation is said to indicate similar-", - "type": "text" - }, - { - "block_id": "p348-b21", - "global_id": 9661, - "bbox": [ - 318.37, - 231.27, - 490.37, - 262.16 - ], - "text": "ity between two signals. Do you agree? Why\nor why not?\n(c) If x[n] and y[n] are both finite duration,", - "type": "text" - }, - { - "block_id": "p348-b22", - "global_id": 9662, - "bbox": [ - 317.86, - 264.15, - 490.39, - 382.71 - ], - "text": "MATLAB’s conv command is well suited to\ncompute rxy[k]. Write a MATLAB function\nthat computes the cross-correlation function\nusing the conv command. Four vectors are\npassed to the function (x, y, nx, and ny)\ncorresponding to the inputs x[n], y[n], and\ntheir respective time vectors. Notice that x\nand y are not necessarily the same length.\nTwo outputs should be created (rxy and k)\ncorresponding to rxy[k] and its shift vector.\n(d) Test your code from part (c) using x[n] =", - "type": "text" - }, - { - "block_id": "p348-b23", - "global_id": 9663, - "bbox": [ - 333.3, - 384.32, - 490.42, - 448.46 - ], - "text": "u[n −5] −u[n −10] over (0 ≤n = nx ≤20)\nand y[n] = u[−n−15]−u[−n−10]+δ[n−\n2] over (−20 ≤n = ny ≤10). Plot the result\nrxy as a function of the shift vector k. What\nshift k gives the largest magnitude of rxy[k]?\nDoes this make sense?", - "type": "text" - }, - { - "block_id": "p348-b24", - "global_id": 9664, - "bbox": [ - 284.74, - 453.36, - 490.39, - 484.32 - ], - "text": "3.11-6\nSuppose a vector x exists in the MATLAB\nworkspace, corresponding to a finite-duration\nDT signal x[n]", - "type": "text" - }, - { - "block_id": "p348-b25", - "global_id": 9665, - "bbox": [ - 318.36, - 486.32, - 490.38, - 495.28 - ], - "text": "(a) Write a MATLAB function that, when", - "type": "text" - }, - { - "block_id": "p348-b26", - "global_id": 9666, - "bbox": [ - 317.86, - 497.28, - 490.39, - 528.16 - ], - "text": "passed vector x, computes and returns Ex,\nthe energy of x[n].\n(b) Write a MATLAB function that, when", - "type": "text" - }, - { - "block_id": "p348-b27", - "global_id": 9667, - "bbox": [ - 333.3, - 530.16, - 490.39, - 572.0 - ], - "text": "passed vector x, computes and returns Px,\nthe power of x[n]. Assume that x[n] is\nperiodic and that vector x contains data for\nan integer number of periods of x[n].", - "type": "text" - }, - { - "block_id": "p348-b28", - "global_id": 9668, - "bbox": [ - 284.74, - 576.6, - 490.38, - 596.9 - ], - "text": "3.11-7\nA causal N-point max filter assigns y[n] to the\nmaximum of {x[n],...,x[n −(N −1)]}.", - "type": "text" - }, - { - "block_id": "p348-b29", - "global_id": 9669, - "bbox": [ - 318.36, - 598.9, - 490.39, - 607.87 - ], - "text": "(a) Write a MATLAB function that performs", - "type": "text" - }, - { - "block_id": "p348-b30", - "global_id": 9670, - "bbox": [ - 333.3, - 609.76, - 490.38, - 641.0 - ], - "text": "N-point max filtering on a length-M input\nvector x. The two function inputs are vector\nx and scalar N. To create the length-M output", - "type": "text" - } - ] - }, - { - "page_num": 349, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p349-b0", - "global_id": 9671, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n329", - "type": "text" - }, - { - "block_id": "p349-b1", - "global_id": 9672, - "bbox": [ - 116.46, - 85.9, - 288.99, - 127.74 - ], - "text": "vector y, initially pad the input vector with\nN −1 zeros. The MATLAB command max\nmay be helpful.\n(b) Test your filter and MATLAB code by", - "type": "text" - }, - { - "block_id": "p349-b2", - "global_id": 9673, - "bbox": [ - 131.9, - 129.36, - 288.98, - 171.58 - ], - "text": "filtering a length-45 input defined as x[n] =\ncos(πn/5) + δ[n −30] −δ[n −35]. Sepa-\nrately plot the results for N = 4, N = 8, and\nN = 12. Comment on the filter behavior.", - "type": "text" - }, - { - "block_id": "p349-b3", - "global_id": 9674, - "bbox": [ - 83.34, - 176.18, - 288.98, - 196.48 - ], - "text": "3.11-8\nA causal N-point min filter assigns y[n] to the\nminimum of {x[n],...,x[n −(N −1)]}.", - "type": "text" - }, - { - "block_id": "p349-b4", - "global_id": 9675, - "bbox": [ - 116.96, - 198.48, - 288.99, - 207.45 - ], - "text": "(a) Write a MATLAB function that performs", - "type": "text" - }, - { - "block_id": "p349-b5", - "global_id": 9676, - "bbox": [ - 116.46, - 209.34, - 288.99, - 284.16 - ], - "text": "N-point min filtering on a length-M input\nvector x. The two function inputs are vector\nx and scalar N. To create the length-M output\nvector y, initially pad the input vector with\nN −1 zeros. The MATLAB command min\nmay be helpful.\n(b) Test your filter and MATLAB code by", - "type": "text" - }, - { - "block_id": "p349-b6", - "global_id": 9677, - "bbox": [ - 131.89, - 285.77, - 288.98, - 327.99 - ], - "text": "filtering a length-45 input defined as x[n] =\ncos(πn/5) + δ[n −30] −δ[n −35]. Sepa-\nrately plot the results for N = 4,N = 8, and\nN = 12. Comment on the filter behavior.", - "type": "text" - }, - { - "block_id": "p349-b7", - "global_id": 9678, - "bbox": [ - 83.34, - 332.6, - 288.98, - 396.73 - ], - "text": "3.11-9\nA causal N-point median filter assigns y[n] to the\nmedian of {x[n],...,x[n−(N −1)]}. The median\nis found by sorting sequence {x[n],...,x[n −\n(N −1)]} and choosing the middle value (odd\nN) or the average of the two middle values (even\nN).", - "type": "text" - }, - { - "block_id": "p349-b8", - "global_id": 9679, - "bbox": [ - 116.96, - 398.72, - 288.99, - 407.69 - ], - "text": "(a) Write a MATLAB function that performs", - "type": "text" - }, - { - "block_id": "p349-b9", - "global_id": 9680, - "bbox": [ - 116.46, - 409.59, - 288.98, - 484.4 - ], - "text": "N-point median filtering on a length-M\ninput vector x. The two function inputs are\nvector x and scalar N. To create the length-M\noutput vector y, initially pad the input vector\nwith N −1 zeros. The MATLAB command\nsort or median may be helpful.\n(b) Test your filter and MATLAB code by", - "type": "text" - }, - { - "block_id": "p349-b10", - "global_id": 9681, - "bbox": [ - 131.89, - 486.02, - 288.98, - 528.24 - ], - "text": "filtering a length-45 input defined as x[n] =\ncos(πn/5) + δ[n −30] −δ[n −35]. Sepa-\nrately plot the results for N = 4, N = 8, and\nN = 12. Comment on the filter behavior.", - "type": "text" - }, - { - "block_id": "p349-b11", - "global_id": 9682, - "bbox": [ - 78.86, - 532.84, - 288.99, - 586.02 - ], - "text": "3.11-10\nRecall that y[n] = x[n/N] represents an upsam-\nple by N operation. An interpolation filter\nreplaces the inserted zeros with more realistic\nvalues. A linear interpolation filter has impulse\nresponse", - "type": "text" - }, - { - "block_id": "p349-b12", - "global_id": 9683, - "bbox": [ - 138.2, - 608.29, - 161.97, - 617.54 - ], - "text": "h[n] =", - "type": "text" - }, - { - "block_id": "p349-b13", - "global_id": 9684, - "bbox": [ - 172.76, - 598.97, - 185.76, - 608.73 - ], - "text": "N−1\n\"", - "type": "text" - }, - { - "block_id": "p349-b14", - "global_id": 9685, - "bbox": [ - 163.81, - 621.25, - 194.71, - 628.0 - ], - "text": "k=−(N−1)", - "type": "text" - }, - { - "block_id": "p349-b16", - "global_id": 9686, - "bbox": [ - 201.77, - 595.28, - 218.95, - 620.39 - ], - "text": "1 −", - "type": "text" - }, - { - "block_id": "p349-b17", - "global_id": 9687, - "bbox": [ - 220.15, - 602.29, - 226.13, - 623.9 - ], - "text": "k\nN", - "type": "text" - }, - { - "block_id": "p349-b20", - "global_id": 9688, - "bbox": [ - 237.82, - 608.29, - 267.25, - 617.54 - ], - "text": "δ(n −k)", - "type": "text" - }, - { - "block_id": "p349-b21", - "global_id": 9689, - "bbox": [ - 344.11, - 90.74, - 516.13, - 99.71 - ], - "text": "(a) Determine a constant coefficient difference", - "type": "text" - }, - { - "block_id": "p349-b22", - "global_id": 9690, - "bbox": [ - 343.61, - 101.33, - 516.12, - 121.63 - ], - "text": "equation that has impulse response h[n].\n(b) The impulse response h[n] is noncausal.", - "type": "text" - }, - { - "block_id": "p349-b23", - "global_id": 9691, - "bbox": [ - 344.12, - 123.62, - 516.15, - 165.47 - ], - "text": "What is the smallest time shift necessary to\nmake the filter causal? What is the effect of\nthis shift on the behavior of the filter?\n(c) Write a MATLAB function that will com-", - "type": "text" - }, - { - "block_id": "p349-b24", - "global_id": 9692, - "bbox": [ - 343.61, - 167.46, - 516.14, - 231.22 - ], - "text": "pute the parameters necessary to implement\nan interpolation filter using MATLAB’s\nfilter command. That is, your function\nshould output filter vectors b and a given an\ninput scalar N.\n(d) Test your filter and MATLAB code. To do", - "type": "text" - }, - { - "block_id": "p349-b25", - "global_id": 9693, - "bbox": [ - 359.05, - 232.85, - 516.13, - 286.01 - ], - "text": "this, create x[n] = cos(n) for (0 ≤n ≤9).\nUpsample x[n] by N = 10 to create a new\nsignal xup[n]. Design the corresponding N =\n10 linear interpolation filter, filter xup[n] to\nproduce y[n], and plot the results.", - "type": "text" - }, - { - "block_id": "p349-b26", - "global_id": 9694, - "bbox": [ - 306.0, - 290.91, - 516.13, - 321.88 - ], - "text": "3.11-11\nA causal N-point moving-average filter has\nimpulse\nresponse\nh[n] = (u[n] −u[n −\nN])/N.", - "type": "text" - }, - { - "block_id": "p349-b27", - "global_id": 9695, - "bbox": [ - 344.11, - 323.88, - 516.12, - 332.84 - ], - "text": "(a) Determine a constant-coefficient difference", - "type": "text" - }, - { - "block_id": "p349-b28", - "global_id": 9696, - "bbox": [ - 343.61, - 334.47, - 516.12, - 354.76 - ], - "text": "equation that has impulse response h[n].\n(b) Write a MATLAB function that will com-", - "type": "text" - }, - { - "block_id": "p349-b29", - "global_id": 9697, - "bbox": [ - 344.11, - 356.75, - 516.14, - 420.51 - ], - "text": "pute the parameters necessary to implement\nan N-point moving-average filter using\nMATLAB’s filter command. That is,\nyour function should output filter vectors b\nand a given a scalar input N.\n(c) Test your filter and MATLAB code by", - "type": "text" - }, - { - "block_id": "p349-b30", - "global_id": 9698, - "bbox": [ - 343.61, - 422.14, - 516.13, - 475.32 - ], - "text": "filtering a length-45 input defined as x[n] =\ncos(πn/5) + δ[n −30] −δ[n −35]. Sepa-\nrately plot the results for N = 4, N = 8, and\nN = 12. Comment on the filter behavior.\n(d) Problem 3.11-10 introduces linear interpo-", - "type": "text" - }, - { - "block_id": "p349-b31", - "global_id": 9699, - "bbox": [ - 359.04, - 477.3, - 516.14, - 628.73 - ], - "text": "lation filters, for use following an upsam-\nple by N operation. Within a scale fac-\ntor, show that a cascade of two N-point\nmoving-average filters is equivalent to the\nlinear interpolation filter. What is the scale\nfactor difference? Test this idea with MAT-\nLAB. Create x[n] = cos(n) for (0 ≤n ≤\n9). Upsample x[n] by N = 10 to create\na new signal xup[n]. Design an N = 10\nmoving-average filter. Filter xup[n] twice\nand scale to produce y[n]. Plot the results.\nDoes the output from the cascaded pair of\nmoving-average filters linearly interpolate\nthe upsampled data?", - "type": "text" - } - ] - }, - { - "page_num": 350, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p350-b0", - "global_id": 9700, - "bbox": [ - 64.47, - 67.04, - 133.54, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p350-b1", - "global_id": 9701, - "bbox": [ - 147.02, - 100.33, - 440.38, - 173.26 - ], - "text": "CONTINUOUS-TIME SYSTEM\nANALYSIS USING THE\nLAPLACE TRANSFORM", - "type": "text" - }, - { - "block_id": "p350-b2", - "global_id": 9702, - "bbox": [ - 114.3, - 79.92, - 133.73, - 118.77 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p350-b3", - "global_id": 9703, - "bbox": [ - 101.84, - 266.14, - 490.39, - 359.79 - ], - "text": "Because of the linearity (superposition) property of linear time-invariant systems, we can find the\nresponse of these systems by breaking the input x(t) into several components and then summing the\nsystem response to all the components of x(t). We have already used this procedure in time-domain\nanalysis, in which the input x(t) is broken into impulsive components. In the frequency-domain\nanalysis developed in this chapter, we break up the input x(t) into exponentials of the form est,\nwhere the parameter s is the complex frequency of the signal est, as explained in Sec. 1.4-3. This\nmethod offers an insight into the system behavior complementary to that seen in the time-domain\nanalysis. In fact, the time-domain and the frequency-domain methods are duals of each other.", - "type": "text" - }, - { - "block_id": "p350-b4", - "global_id": 9704, - "bbox": [ - 101.85, - 361.37, - 490.37, - 383.7 - ], - "text": "The tool that makes it possible to represent arbitrary input x(t) in terms of exponential\ncomponents is the Laplace transform, which is discussed in the following section.", - "type": "text" - }, - { - "block_id": "p350-b5", - "global_id": 9705, - "bbox": [ - 102.2, - 412.16, - 303.35, - 426.1 - ], - "text": "4.1 THE LAPLACE TRANSFORM", - "type": "text" - }, - { - "block_id": "p350-b6", - "global_id": 9706, - "bbox": [ - 101.84, - 431.68, - 329.79, - 442.05 - ], - "text": "For a signal x(t), its Laplace transform X(s) is defined by", - "type": "text" - }, - { - "block_id": "p350-b7", - "global_id": 9707, - "bbox": [ - 252.06, - 457.25, - 279.71, - 467.53 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p350-b8", - "global_id": 9708, - "bbox": [ - 281.75, - 443.7, - 298.69, - 455.76 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p350-b9", - "global_id": 9709, - "bbox": [ - 287.01, - 468.57, - 299.57, - 475.55 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p350-b10", - "global_id": 9710, - "bbox": [ - 301.18, - 453.46, - 490.39, - 467.63 - ], - "text": "x(t)e−st dt\n(4.1)", - "type": "text" - }, - { - "block_id": "p350-b11", - "global_id": 9711, - "bbox": [ - 101.84, - 482.91, - 448.52, - 493.29 - ], - "text": "The signal x(t) is said to be the inverse Laplace transform of X(s). It can be shown that", - "type": "text" - }, - { - "block_id": "p350-b12", - "global_id": 9712, - "bbox": [ - 240.4, - 503.32, - 283.01, - 527.34 - ], - "text": "x(t) =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p350-b13", - "global_id": 9713, - "bbox": [ - 285.32, - 496.32, - 312.73, - 508.61 - ], - "text": "# c+j∞", - "type": "text" - }, - { - "block_id": "p350-b14", - "global_id": 9714, - "bbox": [ - 290.58, - 521.21, - 308.17, - 528.4 - ], - "text": "c−j∞", - "type": "text" - }, - { - "block_id": "p350-b15", - "global_id": 9715, - "bbox": [ - 314.34, - 506.09, - 490.39, - 520.26 - ], - "text": "X(s)est ds\n(4.2)", - "type": "text" - }, - { - "block_id": "p350-b16", - "global_id": 9716, - "bbox": [ - 101.84, - 536.64, - 490.36, - 558.66 - ], - "text": "where c is a constant chosen to ensure the convergence of the integral in Eq. (4.1), as explained\nlater. See also [1].", - "type": "text" - }, - { - "block_id": "p350-b17", - "global_id": 9717, - "bbox": [ - 101.85, - 560.24, - 490.39, - 582.57 - ], - "text": "This pair of equations is known as the bilateral Laplace transform pair, where X(s) is the\ndirect Laplace transform of x(t) and x(t) is the inverse Laplace transform of X(s). Symbolically,", - "type": "text" - }, - { - "block_id": "p350-b18", - "global_id": 9718, - "bbox": [ - 206.24, - 590.73, - 386.0, - 602.83 - ], - "text": "X(s) = L[x(t)]\nand\nx(t) = L−1[X(s)]", - "type": "text" - }, - { - "block_id": "p350-b19", - "global_id": 9719, - "bbox": [ - 101.85, - 613.13, - 138.64, - 623.1 - ], - "text": "Note that", - "type": "text" - }, - { - "block_id": "p350-b20", - "global_id": 9720, - "bbox": [ - 187.29, - 622.95, - 404.95, - 635.05 - ], - "text": "L−1{L[x(t)]} = x(t)\nand\nL{L−1[X(s)]} = X(s)", - "type": "text" - }, - { - "block_id": "p350-b21", - "global_id": 9721, - "bbox": [ - 60.01, - 656.12, - 74.94, - 666.22 - ], - "text": "330", - "type": "text" - } - ] - }, - { - "page_num": 351, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p351-b0", - "global_id": 9722, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n331", - "type": "text" - }, - { - "block_id": "p351-b1", - "global_id": 9723, - "bbox": [ - 127.59, - 85.82, - 516.13, - 107.74 - ], - "text": "It is also common practice to use a bidirectional arrow to indicate a Laplace transform pair, as\nfollows:", - "type": "text" - }, - { - "block_id": "p351-b2", - "global_id": 9724, - "bbox": [ - 293.92, - 109.67, - 349.81, - 119.94 - ], - "text": "x(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p351-b3", - "global_id": 9725, - "bbox": [ - 127.59, - 129.24, - 516.15, - 175.07 - ], - "text": "The Laplace transform, defined in this way, can handle signals existing over the entire time\ninterval from −∞to ∞(causal and noncausal signals). For this reason it is called the bilateral (or\ntwo-sided) Laplace transform. Later we shall consider a special case—the unilateral or one-sided\nLaplace transform—which can handle only causal signals.", - "type": "text" - }, - { - "block_id": "p351-b4", - "global_id": 9726, - "bbox": [ - 127.89, - 189.92, - 351.96, - 202.04 - ], - "text": "LINEARITY OF THE LAPLACE TRANSFORM", - "type": "text" - }, - { - "block_id": "p351-b5", - "global_id": 9727, - "bbox": [ - 127.59, - 206.07, - 516.14, - 227.99 - ], - "text": "We now prove that the Laplace transform is a linear operator by showing that the principle of\nsuperposition holds, implying that if", - "type": "text" - }, - { - "block_id": "p351-b6", - "global_id": 9728, - "bbox": [ - 231.35, - 239.76, - 412.37, - 250.91 - ], - "text": "x1(t) ⇐⇒X1(s)\nand\nx2(t) ⇐⇒X2(s)", - "type": "text" - }, - { - "block_id": "p351-b7", - "global_id": 9729, - "bbox": [ - 127.59, - 262.32, - 144.74, - 272.29 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p351-b8", - "global_id": 9730, - "bbox": [ - 241.28, - 274.22, - 402.44, - 285.37 - ], - "text": "a1x1(t) + a2x2(t) ⇐⇒a1X1(s) + a2X2(s)", - "type": "text" - }, - { - "block_id": "p351-b9", - "global_id": 9731, - "bbox": [ - 127.59, - 293.8, - 265.48, - 303.76 - ], - "text": "The proof is simple. By definition,", - "type": "text" - }, - { - "block_id": "p351-b10", - "global_id": 9732, - "bbox": [ - 197.53, - 320.82, - 288.35, - 332.0 - ], - "text": "L[a1x1(t) + a2x2(t)] =", - "type": "text" - }, - { - "block_id": "p351-b11", - "global_id": 9733, - "bbox": [ - 290.39, - 307.3, - 307.33, - 319.35 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p351-b12", - "global_id": 9734, - "bbox": [ - 295.65, - 332.17, - 308.21, - 339.14 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p351-b13", - "global_id": 9735, - "bbox": [ - 308.71, - 317.05, - 405.72, - 332.0 - ], - "text": "[a1x1(t) + a2x2(t)]e−st dt", - "type": "text" - }, - { - "block_id": "p351-b14", - "global_id": 9736, - "bbox": [ - 280.58, - 348.46, - 298.86, - 359.61 - ], - "text": "= a1", - "type": "text" - }, - { - "block_id": "p351-b15", - "global_id": 9737, - "bbox": [ - 300.46, - 334.91, - 317.41, - 346.96 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p351-b16", - "global_id": 9738, - "bbox": [ - 305.73, - 359.78, - 318.29, - 366.76 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p351-b17", - "global_id": 9739, - "bbox": [ - 319.9, - 344.66, - 382.19, - 359.61 - ], - "text": "x1(t)e−st dt + a2", - "type": "text" - }, - { - "block_id": "p351-b18", - "global_id": 9740, - "bbox": [ - 383.8, - 334.91, - 400.74, - 346.96 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p351-b19", - "global_id": 9741, - "bbox": [ - 389.06, - 359.78, - 401.62, - 366.76 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p351-b20", - "global_id": 9742, - "bbox": [ - 403.23, - 344.66, - 446.01, - 359.61 - ], - "text": "x2(t)e−st dt", - "type": "text" - }, - { - "block_id": "p351-b21", - "global_id": 9743, - "bbox": [ - 280.58, - 370.32, - 516.13, - 381.47 - ], - "text": "= a1X1(s) + a2X2(s)\n(4.3)", - "type": "text" - }, - { - "block_id": "p351-b22", - "global_id": 9744, - "bbox": [ - 127.59, - 392.89, - 308.34, - 402.85 - ], - "text": "This result can be extended to any finite sum.", - "type": "text" - }, - { - "block_id": "p351-b23", - "global_id": 9745, - "bbox": [ - 127.89, - 411.72, - 339.6, - 423.85 - ], - "text": "THE REGION OF CONVERGENCE (ROC)", - "type": "text" - }, - { - "block_id": "p351-b24", - "global_id": 9746, - "bbox": [ - 127.59, - 427.78, - 516.14, - 461.75 - ], - "text": "The region of convergence (ROC), also called the region of existence, for the Laplace transform,\nX(s), is the set of values of s (the region in the complex plane) for which the integral in Eq. (4.1)\nconverges. This concept will become clear in the following example.", - "type": "text" - }, - { - "block_id": "p351-b25", - "global_id": 9747, - "bbox": [ - 102.51, - 484.88, - 487.94, - 496.84 - ], - "text": "EXAMPLE 4.1\nLaplace Transform and ROC of a Causal Exponential", - "type": "text" - }, - { - "block_id": "p351-b26", - "global_id": 9748, - "bbox": [ - 128.9, - 511.86, - 421.18, - 523.47 - ], - "text": "For a signal x(t) = e−atu(t), find the Laplace transform X(s) and its ROC.", - "type": "text" - }, - { - "block_id": "p351-b27", - "global_id": 9749, - "bbox": [ - 128.9, - 546.38, - 184.05, - 556.34 - ], - "text": "By definition,", - "type": "text" - }, - { - "block_id": "p351-b28", - "global_id": 9750, - "bbox": [ - 263.54, - 563.24, - 291.2, - 573.52 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p351-b29", - "global_id": 9751, - "bbox": [ - 293.25, - 549.69, - 310.19, - 561.74 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p351-b30", - "global_id": 9752, - "bbox": [ - 298.51, - 574.56, - 311.07, - 581.53 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p351-b31", - "global_id": 9753, - "bbox": [ - 312.68, - 559.44, - 367.94, - 573.52 - ], - "text": "e−atu(t)e−st dt", - "type": "text" - }, - { - "block_id": "p351-b32", - "global_id": 9754, - "bbox": [ - 128.91, - 587.57, - 323.41, - 597.94 - ], - "text": "Because u(t) = 0 for t < 0 and u(t) = 1 for t ≥0,", - "type": "text" - }, - { - "block_id": "p351-b33", - "global_id": 9755, - "bbox": [ - 196.33, - 615.27, - 223.98, - 625.54 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p351-b34", - "global_id": 9756, - "bbox": [ - 226.03, - 601.7, - 242.96, - 613.76 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p351-b35", - "global_id": 9757, - "bbox": [ - 231.28, - 611.46, - 294.48, - 633.84 - ], - "text": "0\ne−ate−st dt =", - "type": "text" - }, - { - "block_id": "p351-b36", - "global_id": 9758, - "bbox": [ - 296.52, - 601.71, - 313.47, - 613.76 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p351-b37", - "global_id": 9759, - "bbox": [ - 301.79, - 608.7, - 423.85, - 633.84 - ], - "text": "0\ne−(s+a)t dt = −\n1\ns + ae−(s+a)t", - "type": "text" - }, - { - "block_id": "p351-b39", - "global_id": 9760, - "bbox": [ - 427.72, - 605.91, - 434.84, - 612.88 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p351-b40", - "global_id": 9761, - "bbox": [ - 427.72, - 615.68, - 502.76, - 634.73 - ], - "text": "0\n(4.4)", - "type": "text" - } - ] - }, - { - "page_num": 352, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p352-b0", - "global_id": 9762, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "332\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p352-b1", - "global_id": 9763, - "bbox": [ - 103.16, - 84.25, - 477.03, - 109.78 - ], - "text": "Note that s is complex and as t →∞, the term e−(s+a)t does not necessarily vanish. Here we\nrecall that for a complex number z = α + jβ,", - "type": "text" - }, - { - "block_id": "p352-b2", - "global_id": 9764, - "bbox": [ - 236.08, - 113.23, - 343.47, - 127.61 - ], - "text": "e−zt = e−(α+jβ)t = e−αte−jβt", - "type": "text" - }, - { - "block_id": "p352-b3", - "global_id": 9765, - "bbox": [ - 103.16, - 131.65, - 477.01, - 157.6 - ], - "text": "Now |e−jβt| = 1 regardless of the value of βt. Therefore, as t →∞, e−zt →0 only if α > 0,\nand e−zt →∞if α < 0. Thus,", - "type": "text" - }, - { - "block_id": "p352-b4", - "global_id": 9766, - "bbox": [ - 231.75, - 165.97, - 274.65, - 185.65 - ], - "text": "lim\nt→∞e−zt =", - "type": "text" - }, - { - "block_id": "p352-b5", - "global_id": 9767, - "bbox": [ - 276.7, - 156.09, - 477.01, - 186.34 - ], - "text": "0\nRe z > 0\n∞\nRe z < 0\n(4.5)", - "type": "text" - }, - { - "block_id": "p352-b6", - "global_id": 9768, - "bbox": [ - 103.16, - 193.72, - 134.33, - 203.68 - ], - "text": "Clearly,", - "type": "text" - }, - { - "block_id": "p352-b7", - "global_id": 9769, - "bbox": [ - 214.29, - 206.79, - 271.32, - 226.47 - ], - "text": "lim\nt→∞e−(s+a)t =", - "type": "text" - }, - { - "block_id": "p352-b8", - "global_id": 9770, - "bbox": [ - 273.37, - 196.92, - 364.7, - 227.16 - ], - "text": "0\nRe(s + a) > 0\n∞\nRe(s + a) < 0", - "type": "text" - }, - { - "block_id": "p352-b9", - "global_id": 9771, - "bbox": [ - 103.17, - 235.54, - 245.11, - 245.5 - ], - "text": "Use of this result in Eq. (4.4) yields", - "type": "text" - }, - { - "block_id": "p352-b10", - "global_id": 9772, - "bbox": [ - 226.68, - 251.46, - 353.5, - 275.38 - ], - "text": "X(s) =\n1\ns + a\nRe(s + a) > 0", - "type": "text" - }, - { - "block_id": "p352-b11", - "global_id": 9773, - "bbox": [ - 103.16, - 277.88, - 111.46, - 287.84 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p352-b12", - "global_id": 9774, - "bbox": [ - 220.79, - 285.76, - 477.01, - 309.68 - ], - "text": "e−atu(t) ⇐⇒\n1\ns + a\nRe s > −a\n(4.6)", - "type": "text" - }, - { - "block_id": "p352-b13", - "global_id": 9775, - "bbox": [ - 103.16, - 315.04, - 477.03, - 361.28 - ], - "text": "The ROC of X(s) is Re s > −a, as shown in the shaded area in Fig. 4.1a. This fact means that\nthe integral defining X(s) in Eq. (4.4) exists only for the values of s in the shaded region in\nFig. 4.1a. For other values of s, the integral in Eq. (4.4) does not converge. For this reason, the\nshaded region is called the ROC (or the region of existence) for X(s).", - "type": "text" - }, - { - "block_id": "p352-b14", - "global_id": 9776, - "bbox": [ - 438.23, - 443.23, - 452.89, - 451.23 - ], - "text": "Real", - "type": "text" - }, - { - "block_id": "p352-b15", - "global_id": 9777, - "bbox": [ - 438.23, - 538.8, - 452.89, - 546.8 - ], - "text": "Real", - "type": "text" - }, - { - "block_id": "p352-b16", - "global_id": 9778, - "bbox": [ - 375.55, - 395.01, - 384.44, - 403.01 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p352-b17", - "global_id": 9779, - "bbox": [ - 390.84, - 431.83, - 417.39, - 439.9 - ], - "text": "0\nc", - "type": "text" - }, - { - "block_id": "p352-b18", - "global_id": 9780, - "bbox": [ - 399.63, - 482.41, - 422.05, - 490.62 - ], - "text": "c j", - "type": "text" - }, - { - "block_id": "p352-b19", - "global_id": 9781, - "bbox": [ - 412.81, - 394.72, - 434.9, - 402.93 - ], - "text": "c j", - "type": "text" - }, - { - "block_id": "p352-b20", - "global_id": 9782, - "bbox": [ - 229.51, - 433.13, - 231.73, - 441.13 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p352-b21", - "global_id": 9783, - "bbox": [ - 214.5, - 523.81, - 216.73, - 531.81 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p352-b22", - "global_id": 9784, - "bbox": [ - 159.34, - 388.02, - 163.34, - 396.02 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p352-b23", - "global_id": 9785, - "bbox": [ - 170.73, - 522.77, - 181.39, - 541.18 - ], - "text": "0\n1", - "type": "text" - }, - { - "block_id": "p352-b24", - "global_id": 9786, - "bbox": [ - 391.97, - 518.07, - 400.86, - 526.07 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p352-b25", - "global_id": 9787, - "bbox": [ - 356.34, - 536.77, - 394.04, - 546.07 - ], - "text": "0\na", - "type": "text" - }, - { - "block_id": "p352-b26", - "global_id": 9788, - "bbox": [ - 351.73, - 441.52, - 362.4, - 449.73 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p352-b27", - "global_id": 9789, - "bbox": [ - 272.66, - 481.25, - 281.54, - 489.25 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p352-b28", - "global_id": 9790, - "bbox": [ - 272.27, - 577.16, - 281.92, - 585.16 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p352-b29", - "global_id": 9791, - "bbox": [ - 128.04, - 560.33, - 165.18, - 570.03 - ], - "text": "eatu(t)", - "type": "text" - }, - { - "block_id": "p352-b30", - "global_id": 9792, - "bbox": [ - 183.17, - 408.33, - 208.42, - 418.03 - ], - "text": "eatu(t)", - "type": "text" - }, - { - "block_id": "p352-b31", - "global_id": 9793, - "bbox": [ - 154.9, - 369.15, - 439.24, - 377.23 - ], - "text": "Signal x(t)\nRegion of convergence", - "type": "text" - }, - { - "block_id": "p352-b32", - "global_id": 9794, - "bbox": [ - 103.16, - 592.16, - 477.02, - 615.71 - ], - "text": "Figure 4.1 Signals (a) e−atu(t) and (b) −e−atu(−t) have the same Laplace transform but\ndifferent regions of convergence.", - "type": "text" - } - ] - }, - { - "page_num": 353, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p353-b0", - "global_id": 9795, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n333", - "type": "text" - }, - { - "block_id": "p353-b1", - "global_id": 9796, - "bbox": [ - 127.89, - 86.19, - 451.07, - 98.32 - ], - "text": "REGION OF CONVERGENCE FOR FINITE-DURATION SIGNALS", - "type": "text" - }, - { - "block_id": "p353-b2", - "global_id": 9797, - "bbox": [ - 127.59, - 101.93, - 516.14, - 197.07 - ], - "text": "A finite-duration signal xf (t) is a signal that is nonzero only for t1 ≤t ≤t2, where both t1 and t2\nare finite numbers and t2 > t1. For a finite-duration, absolutely integrable signal, the ROC is the\nentire s plane. This is clear from the fact that if xf (t) is absolutely integrable and a finite-duration\nsignal, then x(t)e−σt is also absolutely integrable for any value of σ because the integration is over\nthe finite range of t only. Hence, the Laplace transform of such a signal converges for every value\nof s. This means that the ROC of a general signal x(t) remains unaffected by the addition of any\nabsolutely integrable, finite-duration signal xf (t) to x(t). In other words, if R represents the ROC\nof a signal x(t), then the ROC of a signal x(t) + xf (t) is also R.", - "type": "text" - }, - { - "block_id": "p353-b3", - "global_id": 9798, - "bbox": [ - 127.59, - 211.62, - 516.17, - 393.15 - ], - "text": "ROLE OF THE REGION OF CONVERGENCE\nThe ROC is required for evaluating the inverse Laplace transform x(t) from X(s), as defined by\nEq. (4.2). The operation of finding the inverse transform requires an integration in the complex\nplane, which needs some explanation. The path of integration is along c+jω, with ω varying from\n−∞to ∞.† Moreover, the path of integration must lie in the ROC (or existence) for X(s). For\nthe signal e−atu(t), this is possible if c > −a. One possible path of integration is shown (dotted)\nin Fig. 4.1a. Thus, to obtain x(t) from X(s), the integration in Eq. (4.2) is performed along this\npath. When we integrate [1/(s + a)]est along this path, the result is e−atu(t). Such integration in\nthe complex plane requires a background in the theory of functions of complex variables. We can\navoid this integration by compiling a table of Laplace transforms (Table 4.1), where the Laplace\ntransform pairs are tabulated for a variety of signals. To find the inverse Laplace transform of,\nsay, 1/(s + a), instead of using the complex integral of Eq. (4.2), we look up the table and find\nthe inverse Laplace transform to be e−atu(t) (assuming that the ROC is Re s > −a). Although\nthe table given here is rather short, it comprises the functions of most practical interest. A more\ncomprehensive table appears in Doetsch [2].", - "type": "text" - }, - { - "block_id": "p353-b4", - "global_id": 9799, - "bbox": [ - 127.59, - 408.76, - 516.09, - 446.83 - ], - "text": "THE UNILATERAL LAPLACE TRANSFORM\nTo understand the need for defining unilateral transform, let us find the Laplace transform of signal\nx(t) illustrated in Fig. 4.1b:", - "type": "text" - }, - { - "block_id": "p353-b5", - "global_id": 9800, - "bbox": [ - 285.1, - 449.34, - 358.61, - 461.34 - ], - "text": "x(t) = −e−atu(−t)", - "type": "text" - }, - { - "block_id": "p353-b6", - "global_id": 9801, - "bbox": [ - 127.59, - 472.17, - 281.44, - 482.13 - ], - "text": "The Laplace transform of this signal is", - "type": "text" - }, - { - "block_id": "p353-b7", - "global_id": 9802, - "bbox": [ - 261.8, - 500.77, - 289.45, - 511.05 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p353-b8", - "global_id": 9803, - "bbox": [ - 291.5, - 487.21, - 308.45, - 499.26 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p353-b9", - "global_id": 9804, - "bbox": [ - 296.77, - 512.08, - 309.32, - 519.05 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p353-b10", - "global_id": 9805, - "bbox": [ - 310.92, - 496.96, - 381.73, - 511.05 - ], - "text": "−e−atu(−t)e−st dt", - "type": "text" - }, - { - "block_id": "p353-b11", - "global_id": 9806, - "bbox": [ - 127.59, - 529.84, - 337.88, - 540.22 - ], - "text": "Because u(−t) = 1 for t < 0 and u(−t) = 0 for t > 0,", - "type": "text" - }, - { - "block_id": "p353-b12", - "global_id": 9807, - "bbox": [ - 193.76, - 560.81, - 221.42, - 571.08 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p353-b13", - "global_id": 9808, - "bbox": [ - 223.46, - 547.25, - 236.77, - 559.59 - ], - "text": "# 0", - "type": "text" - }, - { - "block_id": "p353-b14", - "global_id": 9809, - "bbox": [ - 228.72, - 572.13, - 241.27, - 579.1 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p353-b15", - "global_id": 9810, - "bbox": [ - 242.88, - 557.01, - 310.38, - 571.08 - ], - "text": "−e−ate−st dt = −", - "type": "text" - }, - { - "block_id": "p353-b16", - "global_id": 9811, - "bbox": [ - 311.48, - 547.25, - 324.8, - 559.59 - ], - "text": "# 0", - "type": "text" - }, - { - "block_id": "p353-b17", - "global_id": 9812, - "bbox": [ - 316.75, - 572.13, - 329.3, - 579.1 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p353-b18", - "global_id": 9813, - "bbox": [ - 330.91, - 554.24, - 433.02, - 578.16 - ], - "text": "e−(s+a)t dt =\n1\ns + ae−(s+a)t", - "type": "text" - }, - { - "block_id": "p353-b20", - "global_id": 9814, - "bbox": [ - 436.9, - 551.74, - 440.39, - 558.71 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p353-b21", - "global_id": 9815, - "bbox": [ - 436.9, - 573.01, - 449.46, - 579.98 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p353-b22", - "global_id": 9816, - "bbox": [ - 127.59, - 599.27, - 516.15, - 633.41 - ], - "text": "† The discussion about the path of convergence is rather complicated, requiring the concepts of contour\nintegration and understanding of the theory of complex variables. For this reason, the discussion here is\nsomewhat simplified.", - "type": "text" - } - ] - }, - { - "page_num": 354, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p354-b0", - "global_id": 9817, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "334\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p354-b1", - "global_id": 9818, - "bbox": [ - 101.84, - 86.07, - 311.48, - 95.3 - ], - "text": "TABLE 4.1\nSelect (Unilateral) Laplace Transform Pairs", - "type": "text" - }, - { - "block_id": "p354-b2", - "global_id": 9819, - "bbox": [ - 101.84, - 106.02, - 390.12, - 115.28 - ], - "text": "No.\nx(t)\nX(s)", - "type": "text" - }, - { - "block_id": "p354-b3", - "global_id": 9820, - "bbox": [ - 106.33, - 124.35, - 378.14, - 133.69 - ], - "text": "1\nδ(t)\n1", - "type": "text" - }, - { - "block_id": "p354-b4", - "global_id": 9821, - "bbox": [ - 106.33, - 146.07, - 379.33, - 167.6 - ], - "text": "2\nu(t)\n1\ns", - "type": "text" - }, - { - "block_id": "p354-b5", - "global_id": 9822, - "bbox": [ - 106.33, - 173.7, - 381.58, - 195.23 - ], - "text": "3\ntu(t)\n1\ns2", - "type": "text" - }, - { - "block_id": "p354-b6", - "global_id": 9823, - "bbox": [ - 106.33, - 201.07, - 389.87, - 222.97 - ], - "text": "4\ntnu(t)\nn!\nsn+1", - "type": "text" - }, - { - "block_id": "p354-b7", - "global_id": 9824, - "bbox": [ - 106.33, - 229.08, - 393.03, - 250.61 - ], - "text": "5\neλtu(t)\n1\ns −λ", - "type": "text" - }, - { - "block_id": "p354-b8", - "global_id": 9825, - "bbox": [ - 106.33, - 256.71, - 402.96, - 278.24 - ], - "text": "6\nteλtu(t)\n1\n(s −λ)2", - "type": "text" - }, - { - "block_id": "p354-b9", - "global_id": 9826, - "bbox": [ - 106.33, - 284.07, - 411.25, - 305.98 - ], - "text": "7\ntneλtu(t)\nn!\n(s −λ)n+1", - "type": "text" - }, - { - "block_id": "p354-b10", - "global_id": 9827, - "bbox": [ - 106.33, - 309.83, - 399.58, - 331.44 - ], - "text": "8a\ncos btu(t)\ns\ns2 + b2", - "type": "text" - }, - { - "block_id": "p354-b11", - "global_id": 9828, - "bbox": [ - 106.33, - 337.33, - 399.58, - 358.94 - ], - "text": "8b\nsin btu(t)\nb\ns2 + b2", - "type": "text" - }, - { - "block_id": "p354-b12", - "global_id": 9829, - "bbox": [ - 106.33, - 363.8, - 420.53, - 385.69 - ], - "text": "9a\ne−at cos btu(t)\ns + a\n(s + a)2 + b2", - "type": "text" - }, - { - "block_id": "p354-b13", - "global_id": 9830, - "bbox": [ - 106.33, - 391.58, - 420.53, - 413.19 - ], - "text": "9b\ne−at sin btu(t)\nb\n(s + a)2 + b2", - "type": "text" - }, - { - "block_id": "p354-b14", - "global_id": 9831, - "bbox": [ - 101.84, - 418.98, - 489.19, - 434.61 - ], - "text": "10a\nre−at cos(bt + θ)u(t)\n(rcos θ)s + (arcos θ −brsin θ)", - "type": "text" - }, - { - "block_id": "p354-b15", - "global_id": 9832, - "bbox": [ - 395.95, - 429.04, - 468.1, - 440.98 - ], - "text": "s2 + 2as + (a2 + b2)", - "type": "text" - }, - { - "block_id": "p354-b16", - "global_id": 9833, - "bbox": [ - 101.84, - 446.82, - 403.03, - 463.71 - ], - "text": "10b\nre−at cos(bt + θ)u(t)\n0.5rejθ", - "type": "text" - }, - { - "block_id": "p354-b17", - "global_id": 9834, - "bbox": [ - 374.85, - 446.82, - 452.24, - 469.98 - ], - "text": "s + a −jb + 0.5re−jθ", - "type": "text" - }, - { - "block_id": "p354-b18", - "global_id": 9835, - "bbox": [ - 421.54, - 460.73, - 456.06, - 469.98 - ], - "text": "s + a + jb", - "type": "text" - }, - { - "block_id": "p354-b19", - "global_id": 9836, - "bbox": [ - 101.84, - 475.58, - 418.08, - 497.58 - ], - "text": "10c\nre−at cos(bt + θ)u(t)\nAs + B\ns2 + 2as + c", - "type": "text" - }, - { - "block_id": "p354-b20", - "global_id": 9837, - "bbox": [ - 190.53, - 514.86, - 203.06, - 524.11 - ], - "text": "r =", - "type": "text" - }, - { - "block_id": "p354-b22", - "global_id": 9838, - "bbox": [ - 213.53, - 505.32, - 275.44, - 517.91 - ], - "text": "A2c + B2 −2ABa", - "type": "text" - }, - { - "block_id": "p354-b23", - "global_id": 9839, - "bbox": [ - 233.5, - 520.89, - 254.98, - 530.47 - ], - "text": "c −a2", - "type": "text" - }, - { - "block_id": "p354-b24", - "global_id": 9840, - "bbox": [ - 190.53, - 541.84, - 225.29, - 552.44 - ], - "text": "θ = tan−1", - "type": "text" - }, - { - "block_id": "p354-b25", - "global_id": 9841, - "bbox": [ - 226.79, - 530.51, - 264.17, - 546.07 - ], - "text": "Aa −B", - "type": "text" - }, - { - "block_id": "p354-b26", - "global_id": 9842, - "bbox": [ - 234.05, - 551.09, - 239.52, - 560.05 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p354-b27", - "global_id": 9843, - "bbox": [ - 239.53, - 543.15, - 247.11, - 552.11 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p354-b28", - "global_id": 9844, - "bbox": [ - 247.1, - 550.48, - 268.58, - 560.05 - ], - "text": "c −a2", - "type": "text" - }, - { - "block_id": "p354-b30", - "global_id": 9845, - "bbox": [ - 190.53, - 566.84, - 203.85, - 576.09 - ], - "text": "b =", - "type": "text" - }, - { - "block_id": "p354-b31", - "global_id": 9846, - "bbox": [ - 205.69, - 559.18, - 213.28, - 568.15 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p354-b32", - "global_id": 9847, - "bbox": [ - 213.28, - 566.51, - 234.76, - 576.09 - ], - "text": "c −a2", - "type": "text" - }, - { - "block_id": "p354-b33", - "global_id": 9848, - "bbox": [ - 101.84, - 593.83, - 195.63, - 604.42 - ], - "text": "10d\ne−at", - "type": "text" - }, - { - "block_id": "p354-b35", - "global_id": 9849, - "bbox": [ - 202.13, - 588.8, - 265.9, - 604.42 - ], - "text": "Acos bt + B −Aa", - "type": "text" - }, - { - "block_id": "p354-b36", - "global_id": 9850, - "bbox": [ - 251.04, - 595.37, - 287.51, - 610.71 - ], - "text": "b\nsin bt", - "type": "text" - }, - { - "block_id": "p354-b37", - "global_id": 9851, - "bbox": [ - 287.68, - 582.5, - 292.56, - 591.47 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p354-b38", - "global_id": 9852, - "bbox": [ - 293.56, - 588.8, - 418.08, - 610.8 - ], - "text": "u(t)\nAs + B\ns2 + 2as + c", - "type": "text" - }, - { - "block_id": "p354-b39", - "global_id": 9853, - "bbox": [ - 190.53, - 618.84, - 203.85, - 628.08 - ], - "text": "b =", - "type": "text" - }, - { - "block_id": "p354-b40", - "global_id": 9854, - "bbox": [ - 205.7, - 611.18, - 213.28, - 620.14 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p354-b41", - "global_id": 9855, - "bbox": [ - 213.28, - 618.51, - 234.76, - 628.09 - ], - "text": "c −a2", - "type": "text" - } - ] - }, - { - "page_num": 355, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p355-b0", - "global_id": 9856, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n335", - "type": "text" - }, - { - "block_id": "p355-b1", - "global_id": 9857, - "bbox": [ - 127.59, - 85.82, - 229.73, - 95.78 - ], - "text": "Equation (4.5) shows that", - "type": "text" - }, - { - "block_id": "p355-b2", - "global_id": 9858, - "bbox": [ - 248.58, - 104.51, - 395.14, - 124.18 - ], - "text": "lim\nt→−∞e−(s+a)t = 0\nRe (s + a) < 0", - "type": "text" - }, - { - "block_id": "p355-b3", - "global_id": 9859, - "bbox": [ - 127.59, - 135.47, - 155.51, - 145.43 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p355-b4", - "global_id": 9860, - "bbox": [ - 265.19, - 146.64, - 378.53, - 170.56 - ], - "text": "X(s) =\n1\ns + a\nRe s < −a", - "type": "text" - }, - { - "block_id": "p355-b5", - "global_id": 9861, - "bbox": [ - 127.59, - 176.83, - 516.15, - 296.03 - ], - "text": "The signal −e−atu(−t) and its ROC (Re s < −a) are depicted in Fig. 4.1b. Note that the Laplace\ntransforms for the signals e−atu(t) and −e−atu(−t) are identical except for their regions of\nconvergence. Therefore, for a given X(s), there may be more than one inverse transform, depending\non the ROC. In other words, unless the ROC is specified, there is no one-to-one correspondence\nbetween X(s) and x(t). This fact increases the complexity in using the Laplace transform. The\ncomplexity is the result of trying to handle causal as well as noncausal signals. If we restrict all\nour signals to the causal type, such an ambiguity does not arise. There is only one inverse transform\nof X(s) = 1/(s + a), namely, e−atu(t). To find x(t) from X(s), we need not even specify the ROC.\nIn summary, if all signals are restricted to the causal type, then, for a given X(s), there is only one\ninverse transform x(t).†", - "type": "text" - }, - { - "block_id": "p355-b6", - "global_id": 9862, - "bbox": [ - 127.59, - 298.03, - 516.14, - 343.85 - ], - "text": "The unilateral Laplace transform is a special case of the bilateral Laplace transform in which\nall signals are restricted to being causal; consequently, the limits of integration for the integral in\nEq. (4.1) can be taken from 0 to ∞. Therefore, the unilateral Laplace transform X(s) of a signal\nx(t) is defined as", - "type": "text" - }, - { - "block_id": "p355-b7", - "global_id": 9863, - "bbox": [ - 278.24, - 352.28, - 305.89, - 362.56 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p355-b8", - "global_id": 9864, - "bbox": [ - 307.94, - 338.73, - 324.87, - 350.78 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p355-b9", - "global_id": 9865, - "bbox": [ - 313.19, - 348.48, - 516.13, - 370.86 - ], - "text": "0−x(t)e−st dt\n(4.7)", - "type": "text" - }, - { - "block_id": "p355-b10", - "global_id": 9866, - "bbox": [ - 127.59, - 375.02, - 516.14, - 460.33 - ], - "text": "We choose 0−(rather than 0+ used in some texts) as the lower limit of integration. This convention\nnot only ensures inclusion of an impulse function at t = 0, but also allows us to use initial\nconditions at 0−(rather than at 0+) in the solution of differential equations via the Laplace\ntransform. In practice, we are likely to know the initial conditions before the input is applied\n(at 0−), not after the input is applied (at 0+). Indeed, the very meaning of the term “initial\nconditions” implies conditions at t = 0−(conditions before the input is applied). Detailed analysis\nof desirability of using t = 0−appears in Sec. 4.3.", - "type": "text" - }, - { - "block_id": "p355-b11", - "global_id": 9867, - "bbox": [ - 127.59, - 462.32, - 516.14, - 532.06 - ], - "text": "The unilateral Laplace transform simplifies the system analysis problem considerably because\nof its uniqueness property, which says that for a given X(s), there is a unique inverse transform.\nBut there is a price for this simplification: we cannot analyze noncausal systems or use noncausal\ninputs. However, in most practical problems, this restriction is of little consequence. For this\nreason, we shall first consider the unilateral Laplace transform and its application to system\nanalysis. (The bilateral Laplace transform is discussed later, in Sec. 4.11.)", - "type": "text" - }, - { - "block_id": "p355-b12", - "global_id": 9868, - "bbox": [ - 127.59, - 534.05, - 516.14, - 579.89 - ], - "text": "Basically there is no difference between the unilateral and the bilateral Laplace transform. The\nunilateral transform is the bilateral transform that deals with a subclass of signals starting at t = 0\n(causal signals). Therefore, the expression [Eq. (4.2)] for the inverse Laplace transform remains\nunchanged. In practice, the term Laplace transform means the unilateral Laplace transform.", - "type": "text" - }, - { - "block_id": "p355-b13", - "global_id": 9869, - "bbox": [ - 127.59, - 599.27, - 515.64, - 611.5 - ], - "text": "† Actually, X(s) specifies x(t) within a null function n(t), which has the property that the area under |n(t)|2", - "type": "text" - }, - { - "block_id": "p355-b14", - "global_id": 9870, - "bbox": [ - 127.59, - 613.11, - 516.12, - 633.41 - ], - "text": "is zero over any finite interval 0 to t (t > 0) (Lerch’s theorem). For example, if two functions are identical\neverywhere except at finite number of points, they differ by a null function.", - "type": "text" - } - ] - }, - { - "page_num": 356, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p356-b0", - "global_id": 9871, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "336\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p356-b1", - "global_id": 9872, - "bbox": [ - 102.14, - 86.19, - 326.81, - 98.32 - ], - "text": "EXISTENCE OF THE LAPLACE TRANSFORM", - "type": "text" - }, - { - "block_id": "p356-b2", - "global_id": 9873, - "bbox": [ - 101.84, - 101.93, - 490.38, - 124.26 - ], - "text": "The variable s in the Laplace transform is complex in general, and it can be expressed as s = σ +jω.\nBy definition,", - "type": "text" - }, - { - "block_id": "p356-b3", - "global_id": 9874, - "bbox": [ - 204.51, - 133.0, - 232.17, - 143.28 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p356-b4", - "global_id": 9875, - "bbox": [ - 234.22, - 119.45, - 251.16, - 131.5 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p356-b5", - "global_id": 9876, - "bbox": [ - 239.48, - 119.45, - 320.58, - 151.59 - ], - "text": "0−x(t)e−st dt =\n# ∞", - "type": "text" - }, - { - "block_id": "p356-b6", - "global_id": 9877, - "bbox": [ - 308.9, - 129.2, - 387.54, - 151.59 - ], - "text": "0−[x(t)e−σt]e−jωt dt", - "type": "text" - }, - { - "block_id": "p356-b7", - "global_id": 9878, - "bbox": [ - 101.85, - 158.24, - 428.49, - 169.84 - ], - "text": "Because |ejωt| = 1, the integral on the right-hand side of this equation converges if", - "type": "text" - }, - { - "block_id": "p356-b8", - "global_id": 9879, - "bbox": [ - 251.57, - 174.38, - 268.52, - 186.43 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p356-b9", - "global_id": 9880, - "bbox": [ - 256.83, - 198.03, - 264.98, - 206.51 - ], - "text": "0−", - "type": "text" - }, - { - "block_id": "p356-b10", - "global_id": 9881, - "bbox": [ - 270.12, - 179.46, - 490.39, - 198.31 - ], - "text": "x(t)e−σt dt < ∞\n(4.8)", - "type": "text" - }, - { - "block_id": "p356-b11", - "global_id": 9882, - "bbox": [ - 101.84, - 216.31, - 490.41, - 251.68 - ], - "text": "Hence the existence of the Laplace transform is guaranteed if the integral in Eq. (4.8) is finite for\nsome value of σ. Any signal that grows no faster than an exponential signal Meσ0t for some M and\nσ0 satisfies the condition of Eq. (4.8). Thus, if for some M and σ0,", - "type": "text" - }, - { - "block_id": "p356-b12", - "global_id": 9883, - "bbox": [ - 268.8, - 261.23, - 490.39, - 273.33 - ], - "text": "|x(t)| ≤Meσ0t\n(4.9)", - "type": "text" - }, - { - "block_id": "p356-b13", - "global_id": 9884, - "bbox": [ - 101.84, - 283.94, - 490.39, - 357.3 - ], - "text": "we can choose σ > σ0 to satisfy Eq. (4.8).† The signal et2, in contrast, grows at a rate faster\nthan eσ0t, and consequently is not Laplace–transformable.‡ Fortunately such signals (which are not\nLaplace–transformable) are of little consequence from either a practical or a theoretical viewpoint.\nIf σ0 is the smallest value of σ for which the integral in Eq. (4.8) is finite, σ0 is called the abscissa\nof convergence and the ROC of X(s) is Re s > σ0. The abscissa of convergence for e−atu(t) is −a\n(the ROC is Re s > −a).", - "type": "text" - }, - { - "block_id": "p356-b14", - "global_id": 9885, - "bbox": [ - 76.77, - 386.53, - 477.12, - 398.49 - ], - "text": "EXAMPLE 4.2\nBilateral Laplace Transform of Common Causal Signals", - "type": "text" - }, - { - "block_id": "p356-b15", - "global_id": 9886, - "bbox": [ - 103.16, - 414.73, - 454.48, - 425.88 - ], - "text": "Determine the Laplace transform of the following: (a) δ(t), (b) u(t), and (c) cos ω0tu(t).", - "type": "text" - }, - { - "block_id": "p356-b16", - "global_id": 9887, - "bbox": [ - 121.09, - 447.94, - 132.71, - 457.91 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p356-b17", - "global_id": 9888, - "bbox": [ - 240.84, - 464.47, - 279.37, - 474.74 - ], - "text": "L[δ(t)] =", - "type": "text" - }, - { - "block_id": "p356-b18", - "global_id": 9889, - "bbox": [ - 281.42, - 450.91, - 298.36, - 462.96 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p356-b19", - "global_id": 9890, - "bbox": [ - 286.67, - 460.67, - 339.16, - 483.05 - ], - "text": "0−δ(t)e−st dt", - "type": "text" - }, - { - "block_id": "p356-b20", - "global_id": 9891, - "bbox": [ - 103.17, - 488.22, - 355.98, - 498.6 - ], - "text": "Using the sampling property [Eq. (1.11) with T = 0], we obtain", - "type": "text" - }, - { - "block_id": "p356-b21", - "global_id": 9892, - "bbox": [ - 242.13, - 510.13, - 338.05, - 520.51 - ], - "text": "L[δ(t)] = 1\nfor all s", - "type": "text" - }, - { - "block_id": "p356-b22", - "global_id": 9893, - "bbox": [ - 103.17, - 532.46, - 129.74, - 542.43 - ], - "text": "that is,", - "type": "text" - }, - { - "block_id": "p356-b23", - "global_id": 9894, - "bbox": [ - 243.21, - 544.01, - 336.97, - 554.39 - ], - "text": "δ(t) ⇐⇒1\nfor all s", - "type": "text" - }, - { - "block_id": "p356-b24", - "global_id": 9895, - "bbox": [ - 101.84, - 586.65, - 490.39, - 633.41 - ], - "text": "† The condition of Eq. (4.9) is sufficient but not necessary for the existence of the Laplace transform. For\nexample, x(t) = 1/√t is infinite at t = 0, and Eq. (4.9) cannot be satisfied; but the transform of 1/√t exists\nand is given by √π/s.\n‡ However, if we consider a truncated (finite-duration) signal et2, the Laplace transform exists.", - "type": "text" - } - ] - }, - { - "page_num": 357, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p357-b0", - "global_id": 9896, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n337", - "type": "text" - }, - { - "block_id": "p357-b1", - "global_id": 9897, - "bbox": [ - 146.84, - 85.95, - 470.79, - 96.33 - ], - "text": "(b) To find the Laplace transform of u(t), recall that u(t) = 1 for t ≥0. Therefore,", - "type": "text" - }, - { - "block_id": "p357-b2", - "global_id": 9898, - "bbox": [ - 210.72, - 114.04, - 250.51, - 124.42 - ], - "text": "L[u(t)] =", - "type": "text" - }, - { - "block_id": "p357-b3", - "global_id": 9899, - "bbox": [ - 252.56, - 100.52, - 269.49, - 112.58 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p357-b4", - "global_id": 9900, - "bbox": [ - 257.81, - 100.52, - 339.45, - 132.66 - ], - "text": "0−u(t)e−st dt =\n# ∞", - "type": "text" - }, - { - "block_id": "p357-b5", - "global_id": 9901, - "bbox": [ - 327.77, - 107.52, - 407.94, - 132.66 - ], - "text": "0−e−st dt = −1\ns e−st", - "type": "text" - }, - { - "block_id": "p357-b7", - "global_id": 9902, - "bbox": [ - 411.81, - 104.72, - 418.93, - 111.7 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p357-b8", - "global_id": 9903, - "bbox": [ - 411.81, - 125.07, - 419.96, - 133.55 - ], - "text": "0−", - "type": "text" - }, - { - "block_id": "p357-b9", - "global_id": 9904, - "bbox": [ - 242.74, - 135.15, - 258.73, - 152.09 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p357-b10", - "global_id": 9905, - "bbox": [ - 254.3, - 141.72, - 313.69, - 159.07 - ], - "text": "s\nRe s > 0", - "type": "text" - }, - { - "block_id": "p357-b11", - "global_id": 9906, - "bbox": [ - 128.9, - 166.66, - 411.08, - 177.03 - ], - "text": "We also could have obtained this result from Eq. (4.6) by letting a = 0.", - "type": "text" - }, - { - "block_id": "p357-b12", - "global_id": 9907, - "bbox": [ - 146.84, - 177.26, - 258.05, - 189.76 - ], - "text": "(c) Because cos ω0tu(t) = 1", - "type": "text" - }, - { - "block_id": "p357-b13", - "global_id": 9908, - "bbox": [ - 254.56, - 174.99, - 389.91, - 191.8 - ], - "text": "2[ejω0t + e−jω0t]u(t), we know that", - "type": "text" - }, - { - "block_id": "p357-b14", - "global_id": 9909, - "bbox": [ - 229.14, - 200.54, - 305.72, - 213.03 - ], - "text": "L[cos ω0tu(t)] = 1", - "type": "text" - }, - { - "block_id": "p357-b15", - "global_id": 9910, - "bbox": [ - 302.24, - 200.65, - 402.53, - 215.38 - ], - "text": "2 L[ejω0tu(t) + e−jω0tu(t)]", - "type": "text" - }, - { - "block_id": "p357-b16", - "global_id": 9911, - "bbox": [ - 128.91, - 224.21, - 248.2, - 234.17 - ], - "text": "From Eq. (4.6), it follows that", - "type": "text" - }, - { - "block_id": "p357-b17", - "global_id": 9912, - "bbox": [ - 181.2, - 244.79, - 258.18, - 262.51 - ], - "text": "L[cos ω0tu(t)] = 1", - "type": "text" - }, - { - "block_id": "p357-b18", - "global_id": 9913, - "bbox": [ - 253.19, - 258.84, - 258.18, - 268.8 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p357-b19", - "global_id": 9914, - "bbox": [ - 259.37, - 237.37, - 293.52, - 269.58 - ], - "text": "1\ns −jω0", - "type": "text" - }, - { - "block_id": "p357-b20", - "global_id": 9915, - "bbox": [ - 296.76, - 244.79, - 334.8, - 269.58 - ], - "text": "+\n1\ns + jω0", - "type": "text" - }, - { - "block_id": "p357-b21", - "global_id": 9916, - "bbox": [ - 336.5, - 237.37, - 341.93, - 247.33 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p357-b22", - "global_id": 9917, - "bbox": [ - 361.85, - 251.36, - 450.47, - 261.74 - ], - "text": "Re(s ± jω) = Res > 0", - "type": "text" - }, - { - "block_id": "p357-b23", - "global_id": 9918, - "bbox": [ - 242.18, - 269.5, - 502.75, - 294.39 - ], - "text": "=\ns\ns2 + ω02\nRe s > 0\n(4.10)", - "type": "text" - }, - { - "block_id": "p357-b24", - "global_id": 9919, - "bbox": [ - 127.59, - 322.33, - 516.15, - 380.52 - ], - "text": "For the unilateral Laplace transform, there is a unique inverse transform of X(s); consequently,\nthere is no need to specify the ROC explicitly. For this reason, we shall generally ignore any\nmention of the ROC for unilateral transforms. Recall, also, that in the unilateral Laplace transform\nit is understood that every signal x(t) is zero for t < 0, and it is appropriate to indicate this fact by\nmultiplying the signal by u(t).", - "type": "text" - }, - { - "block_id": "p357-b25", - "global_id": 9920, - "bbox": [ - 133.57, - 397.7, - 459.27, - 409.66 - ], - "text": "DRILL 4.1\nBilateral Laplace Transform of Gate Functions", - "type": "text" - }, - { - "block_id": "p357-b26", - "global_id": 9921, - "bbox": [ - 133.57, - 418.36, - 510.16, - 440.7 - ], - "text": "By direct integration, find the Laplace transform X(s) and the region of convergence of X(s) for\nthe gate functions shown in Fig. 4.2.", - "type": "text" - }, - { - "block_id": "p357-b27", - "global_id": 9922, - "bbox": [ - 173.07, - 509.89, - 216.98, - 517.89 - ], - "text": "0\n2", - "type": "text" - }, - { - "block_id": "p357-b28", - "global_id": 9923, - "bbox": [ - 173.48, - 466.49, - 177.48, - 474.49 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p357-b29", - "global_id": 9924, - "bbox": [ - 218.58, - 523.37, - 227.46, - 531.37 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p357-b30", - "global_id": 9925, - "bbox": [ - 270.74, - 509.89, - 422.07, - 517.89 - ], - "text": "t\nt\n0", - "type": "text" - }, - { - "block_id": "p357-b31", - "global_id": 9926, - "bbox": [ - 367.3, - 523.37, - 376.95, - 531.37 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p357-b32", - "global_id": 9927, - "bbox": [ - 396.47, - 509.89, - 400.47, - 517.89 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p357-b33", - "global_id": 9928, - "bbox": [ - 322.17, - 466.34, - 326.17, - 474.34 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p357-b34", - "global_id": 9929, - "bbox": [ - 361.97, - 509.89, - 365.97, - 517.89 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p357-b35", - "global_id": 9930, - "bbox": [ - 183.61, - 458.93, - 343.6, - 467.21 - ], - "text": "x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p357-b36", - "global_id": 9931, - "bbox": [ - 157.47, - 538.06, - 303.16, - 547.3 - ], - "text": "Figure 4.2 Gate functions for Drill 4.1.", - "type": "text" - }, - { - "block_id": "p357-b37", - "global_id": 9932, - "bbox": [ - 133.84, - 566.58, - 196.06, - 577.53 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p357-b38", - "global_id": 9933, - "bbox": [ - 151.5, - 580.88, - 174.27, - 597.82 - ], - "text": "(a) 1", - "type": "text" - }, - { - "block_id": "p357-b39", - "global_id": 9934, - "bbox": [ - 169.84, - 585.73, - 249.34, - 604.8 - ], - "text": "s (1 −e−2s) for all s", - "type": "text" - }, - { - "block_id": "p357-b40", - "global_id": 9935, - "bbox": [ - 151.5, - 605.83, - 174.83, - 622.77 - ], - "text": "(b) 1", - "type": "text" - }, - { - "block_id": "p357-b41", - "global_id": 9936, - "bbox": [ - 170.4, - 608.7, - 265.36, - 629.75 - ], - "text": "s (1 −e−2s)e−2s for all s", - "type": "text" - } - ] - }, - { - "page_num": 358, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p358-b0", - "global_id": 9937, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "338\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p358-b1", - "global_id": 9938, - "bbox": [ - 101.84, - 86.52, - 298.63, - 98.48 - ], - "text": "4.1-1 Finding the Inverse Transform", - "type": "text" - }, - { - "block_id": "p358-b2", - "global_id": 9939, - "bbox": [ - 101.84, - 104.61, - 490.4, - 174.35 - ], - "text": "Finding the inverse Laplace transform by using Eq. (4.2) requires integration in the complex plane,\na subject beyond the scope of this book (but see, e.g., [3]). For our purpose, we can find the inverse\ntransforms from Table 4.1. All we need is to express X(s) as a sum of simpler functions of the forms\nlisted in the table. Most of the transforms X(s) of practical interest are rational functions, that is,\nratios of polynomials in s. Such functions can be expressed as a sum of simpler functions by using\npartial fraction expansion (see Sec. B.5).", - "type": "text" - }, - { - "block_id": "p358-b3", - "global_id": 9940, - "bbox": [ - 101.85, - 175.92, - 490.4, - 210.21 - ], - "text": "Values of s for which X(s) = 0 are called the zeros of X(s); the values of s for which X(s) →∞\nare called the poles of X(s). If X(s) is a rational function of the form P(s)/Q(s), the roots of P(s)\nare the zeros and the roots of Q(s) are the poles of X(s).", - "type": "text" - }, - { - "block_id": "p358-b4", - "global_id": 9941, - "bbox": [ - 78.18, - 242.74, - 379.27, - 254.7 - ], - "text": "EXAMPLE 4.3\nInverse Unilateral Laplace Transform", - "type": "text" - }, - { - "block_id": "p358-b5", - "global_id": 9942, - "bbox": [ - 103.16, - 272.78, - 296.84, - 282.74 - ], - "text": "Find the inverse unilateral Laplace transforms of", - "type": "text" - }, - { - "block_id": "p358-b6", - "global_id": 9943, - "bbox": [ - 121.09, - 288.28, - 177.34, - 312.72 - ], - "text": "(a)\n7s −6\ns2 −s −6", - "type": "text" - }, - { - "block_id": "p358-b7", - "global_id": 9944, - "bbox": [ - 121.09, - 311.9, - 182.87, - 339.95 - ], - "text": "(b)\n2s2 + 5\ns2 + 3s + 2", - "type": "text" - }, - { - "block_id": "p358-b8", - "global_id": 9945, - "bbox": [ - 121.65, - 341.23, - 203.58, - 365.66 - ], - "text": "(c)\n6(s + 34)\ns(s2 + 10s + 34)", - "type": "text" - }, - { - "block_id": "p358-b9", - "global_id": 9946, - "bbox": [ - 121.09, - 367.79, - 197.24, - 392.22 - ], - "text": "(d)\n8s + 10\n(s + 1)(s + 2)3", - "type": "text" - }, - { - "block_id": "p358-b10", - "global_id": 9947, - "bbox": [ - 103.16, - 414.04, - 477.03, - 483.78 - ], - "text": "In no case is the inverse transform of these functions directly available in Table 4.1. Rather,\nwe need to expand these functions into partial fractions, as discussed in Sec. B.5-1. Today,\nit is very easy to find partial fractions via software such as MATLAB. However, just as the\navailability of a calculator does not obviate the need for learning the mechanics of arithmetical\noperations (addition, multiplication, etc.), the widespread availability of computers does not\neliminate the need to learn the mechanics of partial fraction expansion.", - "type": "text" - }, - { - "block_id": "p358-b11", - "global_id": 9948, - "bbox": [ - 121.09, - 485.69, - 132.71, - 495.65 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p358-b12", - "global_id": 9949, - "bbox": [ - 213.4, - 494.87, - 365.58, - 519.3 - ], - "text": "X(s) =\n7s −6\n(s + 2)(s −3) =\nk1\ns + 2 +\nk2\ns −3", - "type": "text" - }, - { - "block_id": "p358-b13", - "global_id": 9950, - "bbox": [ - 103.16, - 525.48, - 477.02, - 559.77 - ], - "text": "To determine k1, corresponding to the term (s + 2), we cover up (conceal) the term (s + 2) in\nX(s) and substitute s = −2 (the value of s that makes s + 2 = 0) in the remaining expression\n(see Sec. B.5-2):", - "type": "text" - }, - { - "block_id": "p358-b14", - "global_id": 9951, - "bbox": [ - 208.85, - 559.63, - 284.63, - 584.47 - ], - "text": "k1 =\n7s −6\n(s + 2)(s −3)", - "type": "text" - }, - { - "block_id": "p358-b16", - "global_id": 9952, - "bbox": [ - 289.07, - 578.82, - 306.14, - 586.08 - ], - "text": "s=−2", - "type": "text" - }, - { - "block_id": "p358-b17", - "global_id": 9953, - "bbox": [ - 308.69, - 559.63, - 353.27, - 576.58 - ], - "text": "= −14 −6", - "type": "text" - }, - { - "block_id": "p358-b18", - "global_id": 9954, - "bbox": [ - 322.2, - 566.62, - 371.33, - 584.06 - ], - "text": "−2 −3 = 4", - "type": "text" - }, - { - "block_id": "p358-b19", - "global_id": 9955, - "bbox": [ - 103.16, - 592.49, - 477.02, - 614.83 - ], - "text": "Similarly, to determine k2 corresponding to the term (s −3), we cover up the term (s −3) in\nX(s) and substitute s = 3 in the remaining expression", - "type": "text" - }, - { - "block_id": "p358-b20", - "global_id": 9956, - "bbox": [ - 215.45, - 625.03, - 291.23, - 649.87 - ], - "text": "k2 =\n7s −6\n(s + 2)(s −3)", - "type": "text" - }, - { - "block_id": "p358-b22", - "global_id": 9957, - "bbox": [ - 295.67, - 644.22, - 307.31, - 651.48 - ], - "text": "s=3", - "type": "text" - }, - { - "block_id": "p358-b23", - "global_id": 9958, - "bbox": [ - 309.86, - 625.03, - 346.67, - 642.39 - ], - "text": "= 21 −6", - "type": "text" - }, - { - "block_id": "p358-b24", - "global_id": 9959, - "bbox": [ - 323.37, - 632.01, - 364.73, - 649.46 - ], - "text": "3 + 2 = 3", - "type": "text" - } - ] - }, - { - "page_num": 359, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p359-b0", - "global_id": 9960, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n339", - "type": "text" - }, - { - "block_id": "p359-b1", - "global_id": 9961, - "bbox": [ - 128.9, - 86.24, - 170.65, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p359-b2", - "global_id": 9962, - "bbox": [ - 239.14, - 95.04, - 502.75, - 119.48 - ], - "text": "X(s) =\n7s −6\n(s + 2)(s −3) =\n4\ns + 2 +\n3\ns −3\n(4.11)", - "type": "text" - }, - { - "block_id": "p359-b3", - "global_id": 9963, - "bbox": [ - 129.2, - 130.87, - 262.85, - 142.99 - ], - "text": "CHECKING THE ANSWER", - "type": "text" - }, - { - "block_id": "p359-b4", - "global_id": 9964, - "bbox": [ - 128.9, - 147.03, - 502.76, - 192.85 - ], - "text": "It is easy to make a mistake in partial fraction computations. Fortunately it is simple to check\nthe answer by recognizing that X(s) and its partial fractions must be equal for every value of s\nif the partial fractions are correct. Let us verify this assertion in Eq. (4.11) for some convenient\nvalue, say, s = 0. Substitution of s = 0 in Eq. (4.11) yields†", - "type": "text" - }, - { - "block_id": "p359-b5", - "global_id": 9965, - "bbox": [ - 288.57, - 202.4, - 343.1, - 212.78 - ], - "text": "1 = 2 −1 = 1", - "type": "text" - }, - { - "block_id": "p359-b6", - "global_id": 9966, - "bbox": [ - 128.91, - 222.74, - 502.77, - 244.66 - ], - "text": "We can now be sure of our answer with a high margin of confidence. Using pair 5 of Table 4.1\nin Eq. (4.11), we obtain", - "type": "text" - }, - { - "block_id": "p359-b7", - "global_id": 9967, - "bbox": [ - 218.56, - 258.13, - 261.06, - 270.13 - ], - "text": "x(t) = L−1", - "type": "text" - }, - { - "block_id": "p359-b8", - "global_id": 9968, - "bbox": [ - 262.67, - 245.86, - 323.3, - 277.3 - ], - "text": "4\ns + 2 +\n3\ns −3", - "type": "text" - }, - { - "block_id": "p359-b10", - "global_id": 9969, - "bbox": [ - 333.28, - 255.74, - 413.11, - 270.23 - ], - "text": "= (4e−2t + 3e3t)u(t)", - "type": "text" - }, - { - "block_id": "p359-b11", - "global_id": 9970, - "bbox": [ - 146.84, - 285.53, - 159.02, - 295.49 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p359-b12", - "global_id": 9971, - "bbox": [ - 243.78, - 292.67, - 386.68, - 320.73 - ], - "text": "X(s) =\n2s2 + 5\ns2 + 3s + 2 =\n2s2 + 5\n(s + 1)(s + 2)", - "type": "text" - }, - { - "block_id": "p359-b13", - "global_id": 9972, - "bbox": [ - 128.91, - 325.91, - 502.76, - 372.15 - ], - "text": "Observe that X(s) is an improper function with M = N. In such a case, we can express\nX(s) as a sum of the coefficient of the highest power in the numerator plus partial fractions\ncorresponding to the poles of X(s) (see Sec. B.5-5). In the present case, the coefficient of the\nhighest power in the numerator is 2. Therefore,", - "type": "text" - }, - { - "block_id": "p359-b14", - "global_id": 9973, - "bbox": [ - 265.51, - 379.85, - 364.96, - 403.98 - ], - "text": "X(s) = 2 +\nk1\ns + 1 +\nk2\ns + 2", - "type": "text" - }, - { - "block_id": "p359-b15", - "global_id": 9974, - "bbox": [ - 128.9, - 410.59, - 153.24, - 420.55 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p359-b16", - "global_id": 9975, - "bbox": [ - 237.08, - 416.02, - 312.86, - 444.48 - ], - "text": "k1 =\n2s2 + 5\n(s + 1)(s + 2)", - "type": "text" - }, - { - "block_id": "p359-b18", - "global_id": 9976, - "bbox": [ - 317.3, - 438.83, - 334.37, - 446.09 - ], - "text": "s=−1", - "type": "text" - }, - { - "block_id": "p359-b19", - "global_id": 9977, - "bbox": [ - 336.93, - 419.64, - 372.65, - 437.0 - ], - "text": "= 2 + 5", - "type": "text" - }, - { - "block_id": "p359-b20", - "global_id": 9978, - "bbox": [ - 347.94, - 426.62, - 394.59, - 444.07 - ], - "text": "−1 + 2 = 7", - "type": "text" - }, - { - "block_id": "p359-b21", - "global_id": 9979, - "bbox": [ - 128.91, - 451.87, - 143.28, - 461.84 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p359-b22", - "global_id": 9980, - "bbox": [ - 230.7, - 457.31, - 306.48, - 485.77 - ], - "text": "k2 =\n2s2 + 5\n(s + 1)(s + 2)", - "type": "text" - }, - { - "block_id": "p359-b24", - "global_id": 9981, - "bbox": [ - 310.93, - 480.11, - 328.0, - 487.38 - ], - "text": "s=−2", - "type": "text" - }, - { - "block_id": "p359-b25", - "global_id": 9982, - "bbox": [ - 330.55, - 460.93, - 366.28, - 478.29 - ], - "text": "= 8 + 5", - "type": "text" - }, - { - "block_id": "p359-b26", - "global_id": 9983, - "bbox": [ - 341.57, - 467.91, - 400.95, - 485.36 - ], - "text": "−2 + 1 = −13", - "type": "text" - }, - { - "block_id": "p359-b27", - "global_id": 9984, - "bbox": [ - 128.91, - 493.17, - 170.65, - 503.13 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p359-b28", - "global_id": 9985, - "bbox": [ - 265.51, - 502.38, - 360.08, - 526.4 - ], - "text": "X(s) = 2 +\n7\ns + 1 −13", - "type": "text" - }, - { - "block_id": "p359-b29", - "global_id": 9986, - "bbox": [ - 128.9, - 516.03, - 364.96, - 541.0 - ], - "text": "s + 2\nFrom Table 4.1, pairs 1 and 5, we obtain", - "type": "text" - }, - { - "block_id": "p359-b30", - "global_id": 9987, - "bbox": [ - 248.48, - 546.44, - 383.19, - 560.93 - ], - "text": "x(t) = 2δ(t) + (7e−t −13e−2t)u(t)", - "type": "text" - }, - { - "block_id": "p359-b31", - "global_id": 9988, - "bbox": [ - 128.9, - 580.21, - 502.76, - 625.68 - ], - "text": "† Because X(s) = ∞at its poles, we should avoid the pole values (−2 and 3 in the present case) for\nchecking. The answers may check even if partial fractions are wrong. This situation can occur when two\nor more errors cancel their effects. But the chances of this problem arising for randomly selected values\nof s are extremely small.", - "type": "text" - } - ] - }, - { - "page_num": 360, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p360-b0", - "global_id": 9989, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "340\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p360-b1", - "global_id": 9990, - "bbox": [ - 121.09, - 86.29, - 132.15, - 96.25 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p360-b2", - "global_id": 9991, - "bbox": [ - 186.85, - 105.49, - 392.12, - 129.93 - ], - "text": "X(s) =\n6(s + 34)\ns(s2 + 10s + 34) =\n6(s + 34)\ns(s + 5 −j3)(s + 5 + j3)", - "type": "text" - }, - { - "block_id": "p360-b3", - "global_id": 9992, - "bbox": [ - 206.73, - 132.83, - 225.64, - 149.78 - ], - "text": "= k1", - "type": "text" - }, - { - "block_id": "p360-b4", - "global_id": 9993, - "bbox": [ - 220.0, - 131.22, - 314.05, - 156.95 - ], - "text": "s +\nk2\ns + 5 −j3 +\nk∗", - "type": "text" - }, - { - "block_id": "p360-b5", - "global_id": 9994, - "bbox": [ - 291.02, - 138.09, - 329.36, - 156.95 - ], - "text": "2\ns + 5 + j3", - "type": "text" - }, - { - "block_id": "p360-b6", - "global_id": 9995, - "bbox": [ - 121.09, - 165.22, - 265.7, - 178.37 - ], - "text": "Note that the coefficients (k2 and k∗", - "type": "text" - }, - { - "block_id": "p360-b7", - "global_id": 9996, - "bbox": [ - 103.17, - 166.91, - 477.04, - 188.83 - ], - "text": "2) of the conjugate terms must also be conjugate (see\nSec. B.5). Now", - "type": "text" - }, - { - "block_id": "p360-b8", - "global_id": 9997, - "bbox": [ - 170.21, - 198.66, - 256.36, - 223.5 - ], - "text": "k1 =\n6(s + 34)\ns(s2 + 10s + 34)", - "type": "text" - }, - { - "block_id": "p360-b10", - "global_id": 9998, - "bbox": [ - 260.83, - 217.85, - 272.46, - 225.11 - ], - "text": "s=0", - "type": "text" - }, - { - "block_id": "p360-b11", - "global_id": 9999, - "bbox": [ - 275.01, - 198.66, - 311.87, - 216.02 - ], - "text": "= 6 × 34", - "type": "text" - }, - { - "block_id": "p360-b12", - "global_id": 10000, - "bbox": [ - 293.96, - 205.65, - 329.92, - 223.09 - ], - "text": "34\n= 6", - "type": "text" - }, - { - "block_id": "p360-b13", - "global_id": 10001, - "bbox": [ - 170.21, - 228.15, - 288.2, - 253.0 - ], - "text": "k2 =\n6(s + 34)\ns(s + 5 −j3)(s + 5 + j3)", - "type": "text" - }, - { - "block_id": "p360-b15", - "global_id": 10002, - "bbox": [ - 292.66, - 247.35, - 320.58, - 254.61 - ], - "text": "s=−5+j3", - "type": "text" - }, - { - "block_id": "p360-b16", - "global_id": 10003, - "bbox": [ - 323.15, - 228.15, - 364.12, - 245.51 - ], - "text": "= 29 + j3", - "type": "text" - }, - { - "block_id": "p360-b17", - "global_id": 10004, - "bbox": [ - 334.16, - 235.14, - 409.96, - 252.59 - ], - "text": "−3 −j5 = −3 + j4", - "type": "text" - }, - { - "block_id": "p360-b18", - "global_id": 10005, - "bbox": [ - 103.17, - 264.54, - 144.91, - 274.5 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p360-b19", - "global_id": 10006, - "bbox": [ - 264.11, - 274.36, - 272.33, - 286.35 - ], - "text": "k∗", - "type": "text" - }, - { - "block_id": "p360-b20", - "global_id": 10007, - "bbox": [ - 268.53, - 276.08, - 316.06, - 288.72 - ], - "text": "2 = −3 −j4", - "type": "text" - }, - { - "block_id": "p360-b21", - "global_id": 10008, - "bbox": [ - 103.16, - 293.73, - 333.58, - 306.88 - ], - "text": "To use pair 10b of Table 4.1, we need to express k2 and k∗", - "type": "text" - }, - { - "block_id": "p360-b22", - "global_id": 10009, - "bbox": [ - 329.78, - 295.42, - 391.64, - 308.3 - ], - "text": "2 in polar form.", - "type": "text" - }, - { - "block_id": "p360-b23", - "global_id": 10010, - "bbox": [ - 191.74, - 319.63, - 232.94, - 330.01 - ], - "text": "−3 + j4 =", - "type": "text" - }, - { - "block_id": "p360-b25", - "global_id": 10011, - "bbox": [ - 247.26, - 311.62, - 387.93, - 330.01 - ], - "text": "32 + 42\nejtan−1(4/−3) = 5ejtan−1(4/−3)", - "type": "text" - }, - { - "block_id": "p360-b26", - "global_id": 10012, - "bbox": [ - 103.16, - 340.33, - 477.01, - 363.88 - ], - "text": "Observe that tan−1(4/−3)̸ = tan−1(−4/3). This fact is evident in Fig. 4.3. For further\ndiscussion of this topic, see Ex. B.1.", - "type": "text" - }, - { - "block_id": "p360-b27", - "global_id": 10013, - "bbox": [ - 179.82, - 405.76, - 186.52, - 413.85 - ], - "text": "j4", - "type": "text" - }, - { - "block_id": "p360-b28", - "global_id": 10014, - "bbox": [ - 144.5, - 446.46, - 155.17, - 454.75 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p360-b29", - "global_id": 10015, - "bbox": [ - 204.97, - 479.83, - 226.5, - 488.13 - ], - "text": "3 j4", - "type": "text" - }, - { - "block_id": "p360-b30", - "global_id": 10016, - "bbox": [ - 117.41, - 405.55, - 145.44, - 413.85 - ], - "text": "3 j4", - "type": "text" - }, - { - "block_id": "p360-b31", - "global_id": 10017, - "bbox": [ - 158.12, - 428.66, - 210.6, - 437.42 - ], - "text": "5\n126.9", - "type": "text" - }, - { - "block_id": "p360-b32", - "global_id": 10018, - "bbox": [ - 193.07, - 449.91, - 216.4, - 458.2 - ], - "text": "53.1", - "type": "text" - }, - { - "block_id": "p360-b33", - "global_id": 10019, - "bbox": [ - 103.16, - 503.43, - 298.2, - 514.29 - ], - "text": "Figure 4.3 Visualizing tan−1(−4/3)̸ = tan−1(4/−3).", - "type": "text" - }, - { - "block_id": "p360-b34", - "global_id": 10020, - "bbox": [ - 121.09, - 544.15, - 243.25, - 554.12 - ], - "text": "From Fig. 4.3, we observe that", - "type": "text" - }, - { - "block_id": "p360-b35", - "global_id": 10021, - "bbox": [ - 242.83, - 562.2, - 336.34, - 577.12 - ], - "text": "k2 = −3 + j4 = 5ej126.9◦", - "type": "text" - }, - { - "block_id": "p360-b36", - "global_id": 10022, - "bbox": [ - 103.16, - 587.99, - 112.03, - 597.95 - ], - "text": "so", - "type": "text" - }, - { - "block_id": "p360-b37", - "global_id": 10023, - "bbox": [ - 261.56, - 597.81, - 269.79, - 609.81 - ], - "text": "k∗", - "type": "text" - }, - { - "block_id": "p360-b38", - "global_id": 10024, - "bbox": [ - 265.98, - 596.08, - 317.59, - 612.18 - ], - "text": "2 = 5e−j126.9◦", - "type": "text" - } - ] - }, - { - "page_num": 361, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p361-b0", - "global_id": 10025, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n341", - "type": "text" - }, - { - "block_id": "p361-b1", - "global_id": 10026, - "bbox": [ - 128.9, - 86.23, - 170.65, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p361-b2", - "global_id": 10027, - "bbox": [ - 245.69, - 97.46, - 281.57, - 114.4 - ], - "text": "X(s) = 6", - "type": "text" - }, - { - "block_id": "p361-b3", - "global_id": 10028, - "bbox": [ - 277.13, - 94.09, - 328.49, - 121.38 - ], - "text": "s + 5ej126.9◦", - "type": "text" - }, - { - "block_id": "p361-b4", - "global_id": 10029, - "bbox": [ - 294.83, - 94.09, - 382.81, - 121.48 - ], - "text": "s + 5 −j3 + 5e−j126.9◦", - "type": "text" - }, - { - "block_id": "p361-b5", - "global_id": 10030, - "bbox": [ - 346.43, - 111.1, - 384.77, - 121.48 - ], - "text": "s + 5 + j3", - "type": "text" - }, - { - "block_id": "p361-b6", - "global_id": 10031, - "bbox": [ - 128.9, - 127.77, - 304.1, - 137.73 - ], - "text": "From Table 4.1 (pairs 2 and 10b), we obtain", - "type": "text" - }, - { - "block_id": "p361-b7", - "global_id": 10032, - "bbox": [ - 236.55, - 144.5, - 395.11, - 158.58 - ], - "text": "x(t) = [6 + 10e−5t cos(3t + 126.9◦)]u(t)", - "type": "text" - }, - { - "block_id": "p361-b8", - "global_id": 10033, - "bbox": [ - 129.2, - 173.04, - 416.19, - 185.16 - ], - "text": "ALTERNATIVE METHOD USING QUADRATIC FACTORS", - "type": "text" - }, - { - "block_id": "p361-b9", - "global_id": 10034, - "bbox": [ - 128.9, - 189.19, - 502.76, - 235.02 - ], - "text": "The foregoing procedure involves considerable manipulation of complex numbers. Pair 10c\n(Table 4.1) indicates that the inverse transform of quadratic terms (with complex conjugate\npoles) can be found directly without having to find first-order partial fractions. We discussed\nsuch a procedure in Sec. B.5-2. For this purpose, we shall express X(s) as", - "type": "text" - }, - { - "block_id": "p361-b10", - "global_id": 10035, - "bbox": [ - 222.77, - 243.53, - 340.53, - 267.96 - ], - "text": "X(s) =\n6(s + 34)\ns(s2 + 10s + 34) = k1", - "type": "text" - }, - { - "block_id": "p361-b11", - "global_id": 10036, - "bbox": [ - 334.9, - 243.53, - 407.68, - 267.96 - ], - "text": "s +\nAs + B\ns2 + 10s + 34", - "type": "text" - }, - { - "block_id": "p361-b12", - "global_id": 10037, - "bbox": [ - 128.9, - 275.99, - 488.44, - 287.45 - ], - "text": "We have already determined that k1 = 6 by the (Heaviside) “cover-up” method. Therefore,", - "type": "text" - }, - { - "block_id": "p361-b13", - "global_id": 10038, - "bbox": [ - 240.53, - 294.87, - 324.48, - 319.31 - ], - "text": "6(s + 34)\ns(s2 + 10s + 34) = 6", - "type": "text" - }, - { - "block_id": "p361-b14", - "global_id": 10039, - "bbox": [ - 320.05, - 294.87, - 391.11, - 319.31 - ], - "text": "s +\nAs + B\ns2 + 10s + 34", - "type": "text" - }, - { - "block_id": "p361-b15", - "global_id": 10040, - "bbox": [ - 128.9, - 325.29, - 425.24, - 339.29 - ], - "text": "Clearing the fractions by multiplying both sides by s(s2 + 10s + 34) yields", - "type": "text" - }, - { - "block_id": "p361-b16", - "global_id": 10041, - "bbox": [ - 235.26, - 345.64, - 396.36, - 360.13 - ], - "text": "6(s + 34) = (6 + A)s2 + (60 + B)s + 204", - "type": "text" - }, - { - "block_id": "p361-b17", - "global_id": 10042, - "bbox": [ - 128.9, - 367.39, - 375.59, - 380.98 - ], - "text": "Now, equating the coefficients of s2 and s on both sides yields", - "type": "text" - }, - { - "block_id": "p361-b18", - "global_id": 10043, - "bbox": [ - 255.52, - 391.44, - 376.14, - 401.82 - ], - "text": "A = −6\nand\nB = −54", - "type": "text" - }, - { - "block_id": "p361-b19", - "global_id": 10044, - "bbox": [ - 128.9, - 412.7, - 143.28, - 422.66 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p361-b20", - "global_id": 10045, - "bbox": [ - 263.97, - 420.58, - 299.85, - 437.52 - ], - "text": "X(s) = 6", - "type": "text" - }, - { - "block_id": "p361-b21", - "global_id": 10046, - "bbox": [ - 295.41, - 420.16, - 366.48, - 444.6 - ], - "text": "s +\n−6s −54\ns2 + 10s + 34", - "type": "text" - }, - { - "block_id": "p361-b22", - "global_id": 10047, - "bbox": [ - 128.9, - 449.68, - 502.78, - 471.6 - ], - "text": "We now use pairs 2 and 10c to find the inverse Laplace transform. The parameters for pair 10c\nare A = −6, B = −54, a = 5, c = 34, b =", - "type": "text" - }, - { - "block_id": "p361-b23", - "global_id": 10048, - "bbox": [ - 294.81, - 452.79, - 303.24, - 462.75 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p361-b24", - "global_id": 10049, - "bbox": [ - 303.23, - 458.34, - 363.7, - 471.6 - ], - "text": "c −a2 = 3, and", - "type": "text" - }, - { - "block_id": "p361-b25", - "global_id": 10050, - "bbox": [ - 187.73, - 492.32, - 201.66, - 502.6 - ], - "text": "r =", - "type": "text" - }, - { - "block_id": "p361-b27", - "global_id": 10051, - "bbox": [ - 213.15, - 482.46, - 281.59, - 495.72 - ], - "text": "A2c + B2 −2ABa", - "type": "text" - }, - { - "block_id": "p361-b28", - "global_id": 10052, - "bbox": [ - 235.25, - 485.34, - 399.09, - 509.67 - ], - "text": "c −a2\n= 10\nθ = tan−1\nAa −B", - "type": "text" - }, - { - "block_id": "p361-b29", - "global_id": 10053, - "bbox": [ - 365.69, - 501.12, - 371.78, - 511.09 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p361-b30", - "global_id": 10054, - "bbox": [ - 371.78, - 492.38, - 380.21, - 502.34 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p361-b31", - "global_id": 10055, - "bbox": [ - 380.21, - 490.6, - 443.42, - 511.09 - ], - "text": "c −a2 = 126.9◦", - "type": "text" - }, - { - "block_id": "p361-b32", - "global_id": 10056, - "bbox": [ - 128.9, - 519.96, - 170.65, - 529.93 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p361-b33", - "global_id": 10057, - "bbox": [ - 236.55, - 527.8, - 395.11, - 541.88 - ], - "text": "x(t) = [6 + 10e−5t cos(3t + 126.9◦)]u(t)", - "type": "text" - }, - { - "block_id": "p361-b34", - "global_id": 10058, - "bbox": [ - 128.9, - 550.31, - 271.12, - 560.27 - ], - "text": "which agrees with the earlier result.", - "type": "text" - }, - { - "block_id": "p361-b35", - "global_id": 10059, - "bbox": [ - 128.9, - 574.74, - 491.6, - 600.85 - ], - "text": "SHORTCUTS\nThe partial fractions with quadratic terms also can be obtained by using shortcuts. We have", - "type": "text" - }, - { - "block_id": "p361-b36", - "global_id": 10060, - "bbox": [ - 224.48, - 609.36, - 339.33, - 633.79 - ], - "text": "X(s) =\n6(s + 34)\ns(s2 + 10s + 34) = 6", - "type": "text" - }, - { - "block_id": "p361-b37", - "global_id": 10061, - "bbox": [ - 334.9, - 609.36, - 405.96, - 633.79 - ], - "text": "s +\nAs + B\ns2 + 10s + 34", - "type": "text" - } - ] - }, - { - "page_num": 362, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p362-b0", - "global_id": 10062, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "342\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p362-b1", - "global_id": 10063, - "bbox": [ - 103.16, - 86.13, - 477.0, - 120.11 - ], - "text": "We can determine A by eliminating B on the right-hand side. This step can be accomplished\nby multiplying both sides of the equation for X(s) by s and then letting s →∞. This procedure\nyields", - "type": "text" - }, - { - "block_id": "p362-b2", - "global_id": 10064, - "bbox": [ - 235.08, - 121.7, - 345.1, - 132.07 - ], - "text": "0 = 6 + A\n\r⇒\nA = −6", - "type": "text" - }, - { - "block_id": "p362-b3", - "global_id": 10065, - "bbox": [ - 103.16, - 141.04, - 144.91, - 151.0 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p362-b4", - "global_id": 10066, - "bbox": [ - 214.78, - 149.9, - 298.73, - 174.34 - ], - "text": "6(s + 34)\ns(s2 + 10s + 34) = 6", - "type": "text" - }, - { - "block_id": "p362-b5", - "global_id": 10067, - "bbox": [ - 294.3, - 149.9, - 365.37, - 174.34 - ], - "text": "s +\n−6s + B\ns2 + 10s + 34", - "type": "text" - }, - { - "block_id": "p362-b6", - "global_id": 10068, - "bbox": [ - 103.16, - 180.45, - 440.81, - 190.82 - ], - "text": "To find B, we let s take on any convenient value, say, s = 1, in this equation to obtain", - "type": "text" - }, - { - "block_id": "p362-b7", - "global_id": 10069, - "bbox": [ - 218.49, - 200.35, - 362.88, - 224.78 - ], - "text": "210\n45 = 6 + B −6\n45\n\r⇒\nB = −54", - "type": "text" - }, - { - "block_id": "p362-b8", - "global_id": 10070, - "bbox": [ - 103.16, - 232.66, - 298.76, - 242.62 - ], - "text": "a result that agrees with the answer found earlier.", - "type": "text" - }, - { - "block_id": "p362-b9", - "global_id": 10071, - "bbox": [ - 121.09, - 244.53, - 133.26, - 254.5 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p362-b10", - "global_id": 10072, - "bbox": [ - 167.0, - 253.72, - 411.98, - 278.15 - ], - "text": "X(s) =\n8s + 10\n(s + 1)(s + 2)3 =\nk1\ns + 1 +\na0\n(s + 2)3 +\na1\n(s + 2)2 +\na2\ns + 2", - "type": "text" - }, - { - "block_id": "p362-b11", - "global_id": 10073, - "bbox": [ - 103.16, - 284.68, - 127.5, - 294.64 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p362-b12", - "global_id": 10074, - "bbox": [ - 203.32, - 302.78, - 282.58, - 327.62 - ], - "text": "k1 =\n8s + 10\n(s + 1)(s + 2)3", - "type": "text" - }, - { - "block_id": "p362-b14", - "global_id": 10075, - "bbox": [ - 287.52, - 321.97, - 304.59, - 329.23 - ], - "text": "s=−1", - "type": "text" - }, - { - "block_id": "p362-b15", - "global_id": 10076, - "bbox": [ - 307.15, - 309.77, - 321.95, - 320.14 - ], - "text": "= 2", - "type": "text" - }, - { - "block_id": "p362-b16", - "global_id": 10077, - "bbox": [ - 202.76, - 332.28, - 282.58, - 357.13 - ], - "text": "a0 =\n8s + 10\n(s + 1)(s + 2)3", - "type": "text" - }, - { - "block_id": "p362-b18", - "global_id": 10078, - "bbox": [ - 287.52, - 351.47, - 304.59, - 358.73 - ], - "text": "s=−2", - "type": "text" - }, - { - "block_id": "p362-b19", - "global_id": 10079, - "bbox": [ - 307.15, - 339.26, - 321.95, - 349.64 - ], - "text": "= 6", - "type": "text" - }, - { - "block_id": "p362-b20", - "global_id": 10080, - "bbox": [ - 202.76, - 371.75, - 221.55, - 383.21 - ], - "text": "a1 =", - "type": "text" - }, - { - "block_id": "p362-b21", - "global_id": 10081, - "bbox": [ - 223.59, - 354.78, - 230.47, - 364.74 - ], - "text": ")", - "type": "text" - }, - { - "block_id": "p362-b22", - "global_id": 10082, - "bbox": [ - 231.66, - 365.08, - 240.51, - 389.1 - ], - "text": "d\nds", - "type": "text" - }, - { - "block_id": "p362-b24", - "global_id": 10083, - "bbox": [ - 250.21, - 364.77, - 308.0, - 389.62 - ], - "text": "8s + 10\n(s + 1)(s + 2)3", - "type": "text" - }, - { - "block_id": "p362-b25", - "global_id": 10084, - "bbox": [ - 309.71, - 354.78, - 322.76, - 364.74 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p362-b26", - "global_id": 10085, - "bbox": [ - 322.76, - 386.95, - 339.83, - 394.21 - ], - "text": "s=−2", - "type": "text" - }, - { - "block_id": "p362-b27", - "global_id": 10086, - "bbox": [ - 342.38, - 371.75, - 364.95, - 382.13 - ], - "text": "= −2", - "type": "text" - }, - { - "block_id": "p362-b28", - "global_id": 10087, - "bbox": [ - 202.76, - 399.98, - 229.77, - 418.01 - ], - "text": "a2 = 1", - "type": "text" - }, - { - "block_id": "p362-b29", - "global_id": 10088, - "bbox": [ - 224.79, - 414.04, - 229.77, - 424.0 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p362-b30", - "global_id": 10089, - "bbox": [ - 232.07, - 389.58, - 238.95, - 399.54 - ], - "text": ")", - "type": "text" - }, - { - "block_id": "p362-b31", - "global_id": 10090, - "bbox": [ - 241.94, - 398.63, - 250.68, - 409.85 - ], - "text": "d2", - "type": "text" - }, - { - "block_id": "p362-b32", - "global_id": 10091, - "bbox": [ - 240.14, - 413.43, - 252.49, - 423.9 - ], - "text": "ds2", - "type": "text" - }, - { - "block_id": "p362-b34", - "global_id": 10092, - "bbox": [ - 262.66, - 399.57, - 320.46, - 424.41 - ], - "text": "8s + 10\n(s + 1)(s + 2)3", - "type": "text" - }, - { - "block_id": "p362-b35", - "global_id": 10093, - "bbox": [ - 322.16, - 389.58, - 335.23, - 399.54 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p362-b36", - "global_id": 10094, - "bbox": [ - 335.23, - 421.74, - 352.3, - 429.01 - ], - "text": "s=−2", - "type": "text" - }, - { - "block_id": "p362-b37", - "global_id": 10095, - "bbox": [ - 354.85, - 406.55, - 377.42, - 416.93 - ], - "text": "= −2", - "type": "text" - }, - { - "block_id": "p362-b38", - "global_id": 10096, - "bbox": [ - 103.17, - 438.09, - 144.91, - 448.05 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p362-b39", - "global_id": 10097, - "bbox": [ - 203.28, - 447.31, - 375.7, - 471.34 - ], - "text": "X(s) =\n2\ns + 1 +\n6\n(s + 2)3 −\n2\n(s + 2)2 −\n2\ns + 2", - "type": "text" - }, - { - "block_id": "p362-b40", - "global_id": 10098, - "bbox": [ - 103.16, - 477.86, - 117.54, - 487.82 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p362-b41", - "global_id": 10099, - "bbox": [ - 216.65, - 485.29, - 363.53, - 499.78 - ], - "text": "x(t) = [2e−t + (3t2 −2t −2)e−2t]u(t)", - "type": "text" - }, - { - "block_id": "p362-b42", - "global_id": 10100, - "bbox": [ - 103.16, - 514.51, - 477.0, - 590.43 - ], - "text": "ALTERNATIVE METHOD: A HYBRID OF HEAVISIDE\nAND CLEARING FRACTIONS\nIn this method, the simpler coefficients k1 and a0 are determined by the Heaviside\n“cover-up” procedure, as discussed earlier. To determine the remaining coefficients, we use the\nclearing-fraction method. Using the values k1 = 2 and a0 = 6 obtained earlier by the Heaviside\n“cover-up” method, we have", - "type": "text" - }, - { - "block_id": "p362-b43", - "global_id": 10101, - "bbox": [ - 183.05, - 599.96, - 397.13, - 624.39 - ], - "text": "8s + 10\n(s + 1)(s + 2)3 =\n2\ns + 1 +\n6\n(s + 2)3 +\na1\n(s + 2)2 +\na2\ns + 2", - "type": "text" - } - ] - }, - { - "page_num": 363, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p363-b0", - "global_id": 10102, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n343", - "type": "text" - }, - { - "block_id": "p363-b1", - "global_id": 10103, - "bbox": [ - 128.9, - 86.47, - 502.75, - 109.74 - ], - "text": "We now clear fractions by multiplying both sides of the equation by (s + 1)(s + 2)3. This\nprocedure yields†", - "type": "text" - }, - { - "block_id": "p363-b2", - "global_id": 10104, - "bbox": [ - 154.77, - 115.99, - 427.25, - 131.26 - ], - "text": "8s + 10 = 2(s + 2)3 + 6(s + 1) + a1(s + 1)(s + 2) + a2(s + 1)(s + 2)2", - "type": "text" - }, - { - "block_id": "p363-b3", - "global_id": 10105, - "bbox": [ - 186.51, - 132.43, - 476.9, - 148.0 - ], - "text": "= (2 + a2)s3 + (12 + a1 + 5a2)s2 + (30 + 3a1 + 8a2)s + (22 + 2a1 + 4a2)", - "type": "text" - }, - { - "block_id": "p363-b4", - "global_id": 10106, - "bbox": [ - 128.91, - 154.09, - 360.83, - 167.67 - ], - "text": "Equating coefficients of s3 and s2 on both sides, we obtain", - "type": "text" - }, - { - "block_id": "p363-b5", - "global_id": 10107, - "bbox": [ - 224.71, - 178.04, - 347.91, - 189.49 - ], - "text": "0 = (2 + a2)\n\r⇒\na2 = −2", - "type": "text" - }, - { - "block_id": "p363-b6", - "global_id": 10108, - "bbox": [ - 224.71, - 192.99, - 406.95, - 204.44 - ], - "text": "0 = 12 + a1 + 5a2 = 2 + a1\n\r⇒\na1 = −2", - "type": "text" - }, - { - "block_id": "p363-b7", - "global_id": 10109, - "bbox": [ - 128.9, - 214.04, - 502.76, - 248.02 - ], - "text": "We can stop here if we wish, since the two desired coefficients a1 and a2 have already been\nfound. However, equating the coefficients of s1 and s0 serves as a check on our answers. This\nstep yields", - "type": "text" - }, - { - "block_id": "p363-b8", - "global_id": 10110, - "bbox": [ - 275.12, - 258.39, - 356.03, - 284.8 - ], - "text": "8 = 30 + 3a1 + 8a2\n10 = 22 + 2a1 + 4a2", - "type": "text" - }, - { - "block_id": "p363-b9", - "global_id": 10111, - "bbox": [ - 128.9, - 294.08, - 502.77, - 316.41 - ], - "text": "Substitution of a1 = a2 = −2, obtained earlier, satisfies these equations. This step confirms the\ncorrectness of our answers.", - "type": "text" - }, - { - "block_id": "p363-b10", - "global_id": 10112, - "bbox": [ - 128.9, - 330.86, - 502.78, - 428.72 - ], - "text": "ANOTHER ALTERNATIVE: A HYBRID OF HEAVISIDE\nAND SHORTCUTS\nIn this method, the simpler coefficients k1 and a0 are determined by the Heaviside “cover-up”\nprocedure, as discussed earlier. The usual shortcuts are then used to determine the remaining\ncoefficients. Using the values k1 = 2 and a0 = 6, determined earlier by the Heaviside method,\nwe have\n8s + 10\n(s + 1)(s + 2)3 =\n2\ns + 1 +\n6\n(s + 2)3 +\na1\n(s + 2)2 +\na2\ns + 2", - "type": "text" - }, - { - "block_id": "p363-b11", - "global_id": 10113, - "bbox": [ - 128.9, - 434.25, - 502.76, - 457.35 - ], - "text": "There are two unknowns, a1 and a2. If we multiply both sides by s and then let s →∞, we\neliminate a1. This procedure yields", - "type": "text" - }, - { - "block_id": "p363-b12", - "global_id": 10114, - "bbox": [ - 260.0, - 466.95, - 371.67, - 478.4 - ], - "text": "0 = 2 + a2\n\r⇒\na2 = −2", - "type": "text" - }, - { - "block_id": "p363-b13", - "global_id": 10115, - "bbox": [ - 128.9, - 488.11, - 170.65, - 498.08 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p363-b14", - "global_id": 10116, - "bbox": [ - 208.79, - 496.92, - 422.88, - 521.35 - ], - "text": "8s + 10\n(s + 1)(s + 2)3 =\n2\ns + 1 +\n6\n(s + 2)3 +\na1\n(s + 2)2 −\n2\ns + 2", - "type": "text" - }, - { - "block_id": "p363-b15", - "global_id": 10117, - "bbox": [ - 128.9, - 527.19, - 502.77, - 549.21 - ], - "text": "There is now only one unknown, a1. This value can be determined readily by setting s equal to\nany convenient value, say, s = 0. This step yields", - "type": "text" - }, - { - "block_id": "p363-b16", - "global_id": 10118, - "bbox": [ - 239.26, - 557.88, - 393.6, - 582.0 - ], - "text": "10\n8 = 2 + 3\n4 + a1\n4 −1\n\r⇒\na1 = −2", - "type": "text" - }, - { - "block_id": "p363-b17", - "global_id": 10119, - "bbox": [ - 128.9, - 586.9, - 502.75, - 632.0 - ], - "text": "† We could have cleared fractions without finding k1 and a0. This alternative, however, proves more\nlaborious because it increases the number of unknowns to 4. By predetermining k1 and a0, we reduce\nthe unknowns to 2. Moreover, this method provides a convenient check on the solution. This hybrid\nprocedure achieves the best of both methods.", - "type": "text" - } - ] - }, - { - "page_num": 364, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p364-b0", - "global_id": 10120, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "344\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p364-b1", - "global_id": 10121, - "bbox": [ - 76.77, - 93.91, - 403.63, - 105.87 - ], - "text": "EXAMPLE 4.4\nInverse Laplace Transform with MATLAB", - "type": "text" - }, - { - "block_id": "p364-b2", - "global_id": 10122, - "bbox": [ - 103.16, - 122.46, - 477.01, - 144.38 - ], - "text": "Using the MATLAB residue command, determine the inverse Laplace transform of each of\nthe following functions:", - "type": "text" - }, - { - "block_id": "p364-b3", - "global_id": 10123, - "bbox": [ - 121.09, - 147.88, - 215.56, - 175.93 - ], - "text": "(a) Xa(s) =\n2s2 + 5\ns2 + 3s + 2", - "type": "text" - }, - { - "block_id": "p364-b4", - "global_id": 10124, - "bbox": [ - 121.09, - 175.12, - 230.49, - 203.17 - ], - "text": "(b) Xb(s) =\n2s2 + 7s + 4\n(s + 1)(s + 2)2", - "type": "text" - }, - { - "block_id": "p364-b5", - "global_id": 10125, - "bbox": [ - 121.65, - 203.26, - 244.78, - 231.3 - ], - "text": "(c) Xc(s) =\n8s2 + 21s + 19\n(s + 2)(s2 + s + 7)", - "type": "text" - }, - { - "block_id": "p364-b6", - "global_id": 10126, - "bbox": [ - 103.16, - 253.13, - 477.02, - 275.04 - ], - "text": "In each case, we use the MATLAB residue command to perform the necessary partial fraction\nexpansions. The inverse Laplace transform follows using Table 4.1.", - "type": "text" - }, - { - "block_id": "p364-b7", - "global_id": 10127, - "bbox": [ - 121.09, - 278.95, - 132.71, - 288.91 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p364-b8", - "global_id": 10128, - "bbox": [ - 103.16, - 299.16, - 275.75, - 333.04 - ], - "text": ">>\nnum = [2 0 5]; den = [1 3 2];\n>>\n[r, p, k] = residue(num,den)\nr = -13", - "type": "text" - }, - { - "block_id": "p364-b9", - "global_id": 10129, - "bbox": [ - 124.08, - 335.03, - 160.69, - 380.86 - ], - "text": "7\np =\n-2\n-1\nk =\n2", - "type": "text" - }, - { - "block_id": "p364-b10", - "global_id": 10130, - "bbox": [ - 103.16, - 386.43, - 462.06, - 401.11 - ], - "text": "Therefore, Xa(s) = −13/(s + 2) + 7/(s + 1) + 2 and xa(t) = (−13e−2t + 7e−t)u(t) + 2δ(t).", - "type": "text" - }, - { - "block_id": "p364-b11", - "global_id": 10131, - "bbox": [ - 121.09, - 408.3, - 133.27, - 418.27 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p364-b12", - "global_id": 10132, - "bbox": [ - 103.17, - 428.52, - 401.29, - 498.26 - ], - "text": ">>\nnum = [2 7 4]; den = [conv([1 1],conv([1 2],[1 2]))];\n>>\n[r, p, k] = residue(num,den)\nr =\n3\n2\n-1\np = -2", - "type": "text" - }, - { - "block_id": "p364-b13", - "global_id": 10133, - "bbox": [ - 124.08, - 500.25, - 155.46, - 534.13 - ], - "text": "-2\n-1\nk = []", - "type": "text" - }, - { - "block_id": "p364-b14", - "global_id": 10134, - "bbox": [ - 103.17, - 539.7, - 470.89, - 554.39 - ], - "text": "Therefore, Xb(s) = 3/(s + 2) + 2/(s + 2)2 −1/(s + 1) and xb(t) = (3e−2t + 2te−2t −e−t)u(t).", - "type": "text" - }, - { - "block_id": "p364-b15", - "global_id": 10135, - "bbox": [ - 103.17, - 567.55, - 477.0, - 589.56 - ], - "text": "(c) In this case, a few calculations are needed beyond the results of the residue command\nso that pair 10b of Table 4.1 can be utilized.", - "type": "text" - }, - { - "block_id": "p364-b16", - "global_id": 10136, - "bbox": [ - 103.17, - 599.73, - 359.44, - 621.64 - ], - "text": ">>\nnum = [8 21 19]; den = [conv([1 2],[1 1 7])];\n>>\n[r, p, k]= residue(num,den)", - "type": "text" - } - ] - }, - { - "page_num": 365, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p365-b0", - "global_id": 10137, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n345", - "type": "text" - }, - { - "block_id": "p365-b1", - "global_id": 10138, - "bbox": [ - 149.82, - 85.83, - 254.42, - 131.66 - ], - "text": "r =\n3.5000-0.48113i\n3.5000+0.48113i\n1.0000\np = -0.5000+2.5981i", - "type": "text" - }, - { - "block_id": "p365-b2", - "global_id": 10139, - "bbox": [ - 128.9, - 133.65, - 296.28, - 191.44 - ], - "text": "-0.5000-2.5981i\n-2.0000\nk = []\n>>\nang = angle(r), mag = abs(r)\nang = -0.13661", - "type": "text" - }, - { - "block_id": "p365-b3", - "global_id": 10140, - "bbox": [ - 149.82, - 193.43, - 223.04, - 251.22 - ], - "text": "0.13661\n0\nmag =\n3.5329\n3.5329\n1.0000", - "type": "text" - }, - { - "block_id": "p365-b4", - "global_id": 10141, - "bbox": [ - 128.9, - 260.89, - 151.31, - 270.85 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p365-b5", - "global_id": 10142, - "bbox": [ - 206.85, - 270.05, - 338.64, - 295.71 - ], - "text": "Xc(s) =\n1\ns + 2 + 3.5329e−j0.13661", - "type": "text" - }, - { - "block_id": "p365-b6", - "global_id": 10143, - "bbox": [ - 273.89, - 270.27, - 417.42, - 295.71 - ], - "text": "s + 0.5 −j2.5981 +\n3.5329ej0.13661", - "type": "text" - }, - { - "block_id": "p365-b7", - "global_id": 10144, - "bbox": [ - 355.39, - 285.33, - 423.6, - 295.71 - ], - "text": "s + 0.5 + j2.5981", - "type": "text" - }, - { - "block_id": "p365-b8", - "global_id": 10145, - "bbox": [ - 128.9, - 302.53, - 143.28, - 312.49 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p365-b9", - "global_id": 10146, - "bbox": [ - 205.58, - 309.97, - 426.07, - 325.16 - ], - "text": "xc(t) = [e−2t + 1.7665e−0.5t cos(2.5981t −0.1366)]u(t).", - "type": "text" - }, - { - "block_id": "p365-b10", - "global_id": 10147, - "bbox": [ - 102.51, - 361.08, - 474.0, - 386.99 - ], - "text": "EXAMPLE 4.5\nSymbolic Laplace and Inverse Laplace Transforms\nwith MATLAB", - "type": "text" - }, - { - "block_id": "p365-b11", - "global_id": 10148, - "bbox": [ - 128.9, - 400.73, - 400.43, - 410.69 - ], - "text": "Using MATLAB’s symbolic math toolbox, determine the following:", - "type": "text" - }, - { - "block_id": "p365-b12", - "global_id": 10149, - "bbox": [ - 146.84, - 418.24, - 428.81, - 429.32 - ], - "text": "(a) the direct unilateral Laplace transform of xa(t) = sin(at) + cos(bt)", - "type": "text" - }, - { - "block_id": "p365-b13", - "global_id": 10150, - "bbox": [ - 146.84, - 429.57, - 421.43, - 444.27 - ], - "text": "(b) the inverse unilateral Laplace transform of Xb(s) = as2/(s2 + b2)", - "type": "text" - }, - { - "block_id": "p365-b14", - "global_id": 10151, - "bbox": [ - 128.91, - 466.4, - 502.75, - 489.1 - ], - "text": "(a) Here, we use the sym command to symbolically define our variables and expression for\nxa(t), and then we use the laplace command to compute the (unilateral) Laplace transform.", - "type": "text" - }, - { - "block_id": "p365-b15", - "global_id": 10152, - "bbox": [ - 128.9, - 498.65, - 338.1, - 532.52 - ], - "text": ">>\nsyms a b t; x_a = sin(a*t)+cos(b*t);\n>>\nX_a = laplace(x_a);\nX_a = a/(a^2 + s^2) + s/(b^2 + s^2)", - "type": "text" - }, - { - "block_id": "p365-b16", - "global_id": 10153, - "bbox": [ - 128.9, - 540.36, - 502.76, - 564.11 - ], - "text": "Therefore, Xa(s) =\na\ns2+a2 +\ns\ns2+b2 . It is also easy to use MATLAB to determine Xa(s) in standard\nrational form.", - "type": "text" - }, - { - "block_id": "p365-b17", - "global_id": 10154, - "bbox": [ - 128.91, - 574.37, - 510.71, - 596.28 - ], - "text": ">>\nX_a = collect(X_a)\nX_a = (a^2*s + a*b^2 + a*s^2 + s^3)/(s^4 + (a^2 + b^2)*s^2 + a^2*b^2)", - "type": "text" - }, - { - "block_id": "p365-b18", - "global_id": 10155, - "bbox": [ - 128.91, - 602.43, - 310.12, - 616.62 - ], - "text": "Thus, we also see that Xa(s) =\ns3+as2+a2s+ab2", - "type": "text" - }, - { - "block_id": "p365-b19", - "global_id": 10156, - "bbox": [ - 128.9, - 611.25, - 502.78, - 641.57 - ], - "text": "s4+(a2+b2)s2+a2b2\n(b) A similar approach is taken for the inverse Laplace transform, except that the\nilaplace command is used rather than the laplace command.", - "type": "text" - } - ] - }, - { - "page_num": 366, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p366-b0", - "global_id": 10157, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "346\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p366-b1", - "global_id": 10158, - "bbox": [ - 103.16, - 86.66, - 312.35, - 120.53 - ], - "text": ">>\nsyms a b s; X_b = (a*s^2)/(s^2+b^2);\n>>\nx_b = ilaplace(X_b)\nx_b = a*dirac(t) - a*b*sin(b*t)", - "type": "text" - }, - { - "block_id": "p366-b2", - "global_id": 10159, - "bbox": [ - 103.16, - 129.79, - 266.15, - 140.87 - ], - "text": "Therefore, xb(t) = aδ(t) −absin(bt)u(t).", - "type": "text" - }, - { - "block_id": "p366-b3", - "global_id": 10160, - "bbox": [ - 107.82, - 191.95, - 284.14, - 203.9 - ], - "text": "DRILL 4.2\nLaplace Transform", - "type": "text" - }, - { - "block_id": "p366-b4", - "global_id": 10161, - "bbox": [ - 107.82, - 208.99, - 484.41, - 234.94 - ], - "text": "Show that the Laplace transform of 10e−3t cos (4t + 53.13◦) is (6s −14)/(s2 + 6s + 25). Use\nTable 4.1.", - "type": "text" - }, - { - "block_id": "p366-b5", - "global_id": 10162, - "bbox": [ - 107.82, - 278.68, - 327.63, - 290.64 - ], - "text": "DRILL 4.3\nInverse Laplace Transform", - "type": "text" - }, - { - "block_id": "p366-b6", - "global_id": 10163, - "bbox": [ - 107.82, - 299.76, - 315.92, - 309.72 - ], - "text": "Find the inverse Laplace transform of the following:", - "type": "text" - }, - { - "block_id": "p366-b7", - "global_id": 10164, - "bbox": [ - 125.76, - 315.26, - 186.97, - 339.69 - ], - "text": "(a)\ns + 17\ns2 + 4s −5", - "type": "text" - }, - { - "block_id": "p366-b8", - "global_id": 10165, - "bbox": [ - 125.76, - 340.92, - 222.12, - 365.35 - ], - "text": "(b)\n3s −5\n(s + 1)(s2 + 2s + 5)", - "type": "text" - }, - { - "block_id": "p366-b9", - "global_id": 10166, - "bbox": [ - 126.32, - 367.48, - 201.34, - 391.91 - ], - "text": "(c)\n16s + 43\n(s −2)(s + 3)2", - "type": "text" - }, - { - "block_id": "p366-b10", - "global_id": 10167, - "bbox": [ - 108.09, - 409.98, - 170.31, - 420.94 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p366-b11", - "global_id": 10168, - "bbox": [ - 125.76, - 424.31, - 208.88, - 453.16 - ], - "text": "(a) (3et −2e−5t)u(t)\n(b)", - "type": "text" - }, - { - "block_id": "p366-b13", - "global_id": 10169, - "bbox": [ - 146.84, - 439.25, - 187.57, - 453.24 - ], - "text": "−2e−t + 5", - "type": "text" - }, - { - "block_id": "p366-b14", - "global_id": 10170, - "bbox": [ - 184.08, - 434.86, - 273.24, - 456.06 - ], - "text": "2e−t cos(2t −36.87◦)", - "type": "text" - }, - { - "block_id": "p366-b15", - "global_id": 10171, - "bbox": [ - 126.32, - 442.87, - 288.6, - 468.18 - ], - "text": "u(t)\n(c) [3e2t + (t −3)e−3t]u(t)", - "type": "text" - }, - { - "block_id": "p366-b16", - "global_id": 10172, - "bbox": [ - 102.14, - 499.09, - 429.01, - 525.15 - ], - "text": "A HISTORICAL NOTE: MARQUIS PIERRE-SIMON DE LAPLACE\n(1749–1827)", - "type": "text" - }, - { - "block_id": "p366-b17", - "global_id": 10173, - "bbox": [ - 101.84, - 529.18, - 490.4, - 563.06 - ], - "text": "The Laplace transform is named after the great French mathematician and astronomer Laplace,\nwho first presented the transform and its applications to differential equations in a paper published\nin 1779.", - "type": "text" - }, - { - "block_id": "p366-b18", - "global_id": 10174, - "bbox": [ - 101.84, - 565.05, - 490.4, - 634.79 - ], - "text": "Laplace developed the foundations of potential theory and made important contributions to\nspecial functions, probability theory, astronomy, and celestial mechanics. In his Exposition du\nsystème du monde (1796), Laplace formulated a nebular hypothesis of cosmic origin and tried to\nexplain the universe as a pure mechanism. In his Traité de mécanique céleste (celestial mechanics),\nwhich completed the work of Newton, Laplace used mathematics and physics to subject the solar\nsystem and all heavenly bodies to the laws of motion and the principle of gravitation. Newton had", - "type": "text" - } - ] - }, - { - "page_num": 367, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p367-b0", - "global_id": 10175, - "bbox": [ - 361.84, - 62.89, - 516.13, - 71.98 - ], - "text": "4.1\nThe Laplace Transform\n347", - "type": "text" - }, - { - "block_id": "p367-b1", - "global_id": 10176, - "bbox": [ - 229.06, - 310.94, - 414.64, - 320.9 - ], - "text": "Pierre-Simon de Laplace and Oliver Heaviside", - "type": "text" - }, - { - "block_id": "p367-b2", - "global_id": 10177, - "bbox": [ - 127.59, - 341.86, - 516.15, - 423.55 - ], - "text": "been unable to explain the irregularities of some heavenly bodies; in desperation, he concluded\nthat God himself must intervene now and then to prevent such catastrophes as Jupiter eventually\nfalling into the sun (and the moon into the earth), as predicted by Newton’s calculations. Laplace\nproposed to show that these irregularities would correct themselves periodically and that a little\npatience—in Jupiter’s case, 929 years—would see everything returning automatically to order;\nthus there was no reason why the solar and the stellar systems could not continue to operate by the\nlaws of Newton and Laplace to the end of time [4].", - "type": "text" - }, - { - "block_id": "p367-b3", - "global_id": 10178, - "bbox": [ - 127.59, - 425.44, - 516.15, - 495.28 - ], - "text": "Laplace presented a copy of Mécanique céleste to Napoleon, who, after reading the book,\ntook Laplace to task for not including God in his scheme: “You have written this huge book on the\nsystem of the world without once mentioning the author of the universe.” “Sire,” Laplace retorted,\n“I had no need of that hypothesis.” Napoleon was not amused, and when he reported this reply to\nanother great mathematician-astronomer, Louis de Lagrange, the latter remarked, “Ah, but that is\na fine hypothesis. It explains so many things” [5].", - "type": "text" - }, - { - "block_id": "p367-b4", - "global_id": 10179, - "bbox": [ - 127.59, - 497.28, - 516.11, - 543.11 - ], - "text": "Napoleon, following his policy of honoring and promoting scientists, made Laplace the\nminister of the interior. To Napoleon’s dismay, however, the new appointee attempted to bring\n“the spirit of infinitesimals” into administration, and so Laplace was transferred hastily to the\nSenate.", - "type": "text" - }, - { - "block_id": "p367-b5", - "global_id": 10180, - "bbox": [ - 127.59, - 560.85, - 516.16, - 634.79 - ], - "text": "OLIVER HEAVISIDE (1850–1925)\nAlthough Laplace published his transform method to solve differential equations in 1779, the\nmethod did not catch on until a century later. It was rediscovered independently in a rather\nawkward form by an eccentric British engineer, Oliver Heaviside (1850–1925), one of the tragic\nfigures in the history of science and engineering. Despite his prolific contributions to electrical\nengineering, he was severely criticized during his lifetime and was neglected later to the point that", - "type": "text" - } - ] - }, - { - "page_num": 368, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p368-b0", - "global_id": 10181, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "348\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p368-b1", - "global_id": 10182, - "bbox": [ - 101.84, - 85.82, - 490.42, - 155.56 - ], - "text": "hardly a textbook today mentions his name or credits him with contributions. Nevertheless, his\nstudies had a major impact on many aspects of modern electrical engineering. It was Heaviside\nwho made transatlantic communication possible by inventing cable loading, but few mention him\nas a pioneer or an innovator in telephony. It was Heaviside who suggested the use of inductive\ncable loading, but the credit is given to M. Pupin, who was not even responsible for building the\nfirst loading coil.† In addition, Heaviside was [6]:", - "type": "text" - }, - { - "block_id": "p368-b2", - "global_id": 10183, - "bbox": [ - 119.78, - 169.51, - 490.39, - 346.84 - ], - "text": "• The first to find a solution to the distortionless transmission line.\n• The innovator of lowpass filters.\n• The first to write Maxwell’s equations in modern form.\n• The codiscoverer of rate energy transfer by an electromagnetic field.\n• An early champion of the now-common phasor analysis.\n• An important contributor to the development of vector analysis. In fact, he essentially\ncreated the subject independently of Gibbs [7].\n• An originator of the use of operational mathematics used to solve linear integro-differential\nequations, which eventually led to rediscovery of the ignored Laplace transform.\n• The first to theorize (along with Kennelly of Harvard) that a conducting layer (the\nKennelly–Heaviside layer) of atmosphere exists, which allows radio waves to follow earth’s\ncurvature instead of traveling off into space in a straight line.\n• The first to posit that an electrical charge would increase in mass as its velocity increases,\nan anticipation of an aspect of Einstein’s special theory of relativity [8]. He also forecast\nthe possibility of superconductivity.", - "type": "text" - }, - { - "block_id": "p368-b3", - "global_id": 10184, - "bbox": [ - 101.84, - 360.79, - 490.43, - 418.57 - ], - "text": "Heaviside was a self-made, self-educated man. Although his formal education ended with\nelementary school, he eventually became a pragmatically successful mathematical physicist. He\nbegan his career as a telegrapher, but increasing deafness forced him to retire at the age of 24.\nHe then devoted himself to the study of electricity. His creative work was disdained by many\nprofessional mathematicians because of his lack of formal education and his unorthodox methods.", - "type": "text" - }, - { - "block_id": "p368-b4", - "global_id": 10185, - "bbox": [ - 101.84, - 420.56, - 490.4, - 562.04 - ], - "text": "Heaviside had the misfortune to be criticized both by mathematicians, who faulted him for\nlack of rigor, and by men of practice, who faulted him for using too much mathematics and\nthereby confusing students. Many mathematicians, trying to find solutions to the distortionless\ntransmission line, failed because no rigorous tools were available at the time. Heaviside\nsucceeded because he used mathematics not with rigor, but with insight and intuition. Using\nhis much maligned operational method, Heaviside successfully attacked problems that the rigid\nmathematicians could not solve, problems such as the flow-of-heat in a body of spatially varying\nconductivity. Heaviside brilliantly used this method in 1895 to demonstrate a fatal flaw in Lord\nKelvin’s determination of the geological age of the earth by secular cooling; he used the same\nflow-of-heat theory as for his cable analysis. Yet the mathematicians of the Royal Society remained\nunmoved and were not the least impressed by the fact that Heaviside had found the answer to\nproblems no one else could solve. Many mathematicians who examined his work dismissed it", - "type": "text" - }, - { - "block_id": "p368-b5", - "global_id": 10186, - "bbox": [ - 101.84, - 588.31, - 490.38, - 633.41 - ], - "text": "† Heaviside developed the theory for cable loading, George Campbell built the first loading coil, and the\ntelephone circuits using Campbell’s coils were in operation before Pupin published his paper. In the legal\nfight over the patent, however, Pupin won the battle: he was a shrewd self-promoter, and Campbell had poor\nlegal support.", - "type": "text" - } - ] - }, - { - "page_num": 369, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p369-b0", - "global_id": 10187, - "bbox": [ - 287.44, - 62.89, - 516.14, - 71.98 - ], - "text": "4.2\nSome Properties of the Laplace Transform\n349", - "type": "text" - }, - { - "block_id": "p369-b1", - "global_id": 10188, - "bbox": [ - 127.59, - 85.82, - 516.15, - 107.74 - ], - "text": "with contempt, asserting that his methods were either complete nonsense or a rehash of known\nideas [6].", - "type": "text" - }, - { - "block_id": "p369-b2", - "global_id": 10189, - "bbox": [ - 127.59, - 109.73, - 516.14, - 167.51 - ], - "text": "Sir William Preece, the chief engineer of the British Post Office, a savage critic of Heaviside,\nridiculed Heaviside’s work as too theoretical and, therefore, leading to faulty conclusions.\nHeaviside’s work on transmission lines and loading was dismissed by the British Post Office\nand might have remained hidden, had not Lord Kelvin himself publicly expressed admiration\nfor it [6].", - "type": "text" - }, - { - "block_id": "p369-b3", - "global_id": 10190, - "bbox": [ - 127.59, - 169.51, - 516.18, - 227.29 - ], - "text": "Heaviside’s operational calculus may be formally inaccurate, but in fact it anticipated the\noperational methods developed in more recent years [9]. Although his method was not fully\nunderstood, it provided correct results. When Heaviside was attacked for the vague meaning of\nhis operational calculus, his pragmatic reply was, “Shall I refuse my dinner because I do not fully\nunderstand the process of digestion?”", - "type": "text" - }, - { - "block_id": "p369-b4", - "global_id": 10191, - "bbox": [ - 127.59, - 229.29, - 516.15, - 275.11 - ], - "text": "Heaviside lived as a bachelor hermit, often in near-squalid conditions, and died largely\nunnoticed, in poverty. His life demonstrates the persistent arrogance and snobbishness of the\nintellectual establishment, which does not respect creativity unless it is presented in the strict\nlanguage of the establishment.", - "type": "text" - }, - { - "block_id": "p369-b5", - "global_id": 10192, - "bbox": [ - 127.94, - 308.24, - 472.38, - 322.19 - ], - "text": "4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM", - "type": "text" - }, - { - "block_id": "p369-b6", - "global_id": 10193, - "bbox": [ - 127.59, - 328.18, - 516.16, - 385.96 - ], - "text": "Properties of the Laplace transform are useful not only in the derivation of the Laplace transform\nof functions but also in the solutions of linear integro-differential equations. A glance at Eqs. (4.2)\nand (4.1) shows that there is a certain measure of symmetry in going from x(t) to X(s), and vice\nversa. This symmetry or duality is also carried over to the properties of the Laplace transform.\nThis fact will be evident in the following development.", - "type": "text" - }, - { - "block_id": "p369-b7", - "global_id": 10194, - "bbox": [ - 127.59, - 387.95, - 516.17, - 409.87 - ], - "text": "We are already familiar with two properties: linearity [Eq. (4.3)] and the uniqueness property\nof the Laplace transform discussed earlier.", - "type": "text" - }, - { - "block_id": "p369-b8", - "global_id": 10195, - "bbox": [ - 127.59, - 438.47, - 234.03, - 450.43 - ], - "text": "4.2-1 Time Shifting", - "type": "text" - }, - { - "block_id": "p369-b9", - "global_id": 10196, - "bbox": [ - 127.59, - 456.55, - 283.95, - 466.52 - ], - "text": "The time-shifting property states that if", - "type": "text" - }, - { - "block_id": "p369-b10", - "global_id": 10197, - "bbox": [ - 293.92, - 476.07, - 349.81, - 486.34 - ], - "text": "x(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p369-b11", - "global_id": 10198, - "bbox": [ - 127.59, - 495.99, - 184.68, - 507.45 - ], - "text": "then for t0 ≥0", - "type": "text" - }, - { - "block_id": "p369-b12", - "global_id": 10199, - "bbox": [ - 275.86, - 512.0, - 516.13, - 524.87 - ], - "text": "x(t −t0) ⇐⇒X(s)e−st0\n(4.12)", - "type": "text" - }, - { - "block_id": "p369-b13", - "global_id": 10200, - "bbox": [ - 127.59, - 536.5, - 516.14, - 570.79 - ], - "text": "Observe that x(t) starts at t = 0, and, therefore, x(t −t0) starts at t = t0. This fact is implicit, but is\nnot explicitly indicated in Eq. (4.12). This often leads to inadvertent errors. To avoid such a pitfall,\nwe should restate the property as follows. If", - "type": "text" - }, - { - "block_id": "p369-b14", - "global_id": 10201, - "bbox": [ - 286.24, - 580.34, - 357.49, - 590.62 - ], - "text": "x(t)u(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p369-b15", - "global_id": 10202, - "bbox": [ - 127.6, - 600.68, - 144.74, - 610.64 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p369-b16", - "global_id": 10203, - "bbox": [ - 237.73, - 616.27, - 405.99, - 629.45 - ], - "text": "x(t −t0)u(t −t0) ⇐⇒X(s)e−st0\nt0 ≥0", - "type": "text" - } - ] - }, - { - "page_num": 370, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p370-b0", - "global_id": 10204, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "350\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p370-b1", - "global_id": 10205, - "bbox": [ - 101.84, - 85.87, - 127.78, - 95.83 - ], - "text": "Proof.", - "type": "text" - }, - { - "block_id": "p370-b2", - "global_id": 10206, - "bbox": [ - 196.06, - 119.74, - 285.94, - 130.92 - ], - "text": "L[x(t −t0)u(t −t0)] =", - "type": "text" - }, - { - "block_id": "p370-b3", - "global_id": 10207, - "bbox": [ - 287.99, - 106.21, - 304.94, - 118.27 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p370-b4", - "global_id": 10208, - "bbox": [ - 293.26, - 115.97, - 395.98, - 138.36 - ], - "text": "0\nx(t −t0)u(t −t0)e−st dt", - "type": "text" - }, - { - "block_id": "p370-b5", - "global_id": 10209, - "bbox": [ - 101.85, - 156.19, - 214.51, - 167.64 - ], - "text": "Setting t −t0 = τ, we obtain", - "type": "text" - }, - { - "block_id": "p370-b6", - "global_id": 10210, - "bbox": [ - 200.79, - 181.4, - 290.67, - 192.58 - ], - "text": "L[x(t −t0)u(t −t0)] =", - "type": "text" - }, - { - "block_id": "p370-b7", - "global_id": 10211, - "bbox": [ - 292.72, - 167.88, - 309.66, - 179.93 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p370-b8", - "global_id": 10212, - "bbox": [ - 297.97, - 192.75, - 308.33, - 201.28 - ], - "text": "−t0", - "type": "text" - }, - { - "block_id": "p370-b9", - "global_id": 10213, - "bbox": [ - 311.27, - 177.63, - 390.24, - 191.71 - ], - "text": "x(τ)u(τ)e−s(τ+t0) dτ", - "type": "text" - }, - { - "block_id": "p370-b10", - "global_id": 10214, - "bbox": [ - 101.85, - 207.73, - 490.39, - 230.07 - ], - "text": "Because u(τ) = 0 for τ < 0 and u(τ) = 1 for τ ≥0, the limits of integration can be taken from 0\nto ∞. Thus,", - "type": "text" - }, - { - "block_id": "p370-b11", - "global_id": 10215, - "bbox": [ - 207.98, - 244.18, - 297.86, - 255.36 - ], - "text": "L[x(t −t0)u(t −t0)] =", - "type": "text" - }, - { - "block_id": "p370-b12", - "global_id": 10216, - "bbox": [ - 299.91, - 230.66, - 316.85, - 242.72 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p370-b13", - "global_id": 10217, - "bbox": [ - 305.16, - 240.42, - 379.52, - 262.8 - ], - "text": "0\nx(τ)e−s(τ+t0) dτ", - "type": "text" - }, - { - "block_id": "p370-b14", - "global_id": 10218, - "bbox": [ - 290.09, - 269.6, - 317.39, - 281.6 - ], - "text": "= e−st0", - "type": "text" - }, - { - "block_id": "p370-b15", - "global_id": 10219, - "bbox": [ - 319.51, - 257.77, - 336.46, - 269.83 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p370-b16", - "global_id": 10220, - "bbox": [ - 324.77, - 267.52, - 383.05, - 289.91 - ], - "text": "0\nx(τ)e−sτ dτ", - "type": "text" - }, - { - "block_id": "p370-b17", - "global_id": 10221, - "bbox": [ - 290.09, - 290.95, - 335.23, - 302.95 - ], - "text": "= X(s)e−st0", - "type": "text" - }, - { - "block_id": "p370-b18", - "global_id": 10222, - "bbox": [ - 101.84, - 312.6, - 490.39, - 336.33 - ], - "text": "Note that x(t −t0)u(t −t0) is the signal x(t)u(t) delayed by t0 seconds. The time-shifting property\nstates that delaying a signal by t0 seconds amounts to multiplying its transform e−st0.", - "type": "text" - }, - { - "block_id": "p370-b19", - "global_id": 10223, - "bbox": [ - 101.84, - 336.82, - 490.39, - 359.61 - ], - "text": "This property of the unilateral Laplace transform holds only for positive t0 because if t0 were\nnegative, the signal x(t −t0)u(t −t0) may not be causal.", - "type": "text" - }, - { - "block_id": "p370-b20", - "global_id": 10224, - "bbox": [ - 101.84, - 360.42, - 490.4, - 394.7 - ], - "text": "We can readily verify this property in Drill 4.1. If the signal in Fig. 4.2a is x(t)u(t), then\nthe signal in Fig. 4.2b is x(t −2)u(t −2). The Laplace transform for the pulse in Fig. 4.2a is\n(1/s)(1−e−2s). Therefore, the Laplace transform for the pulse in Fig. 4.2b is (1/s)(1−e−2s)e−2s.", - "type": "text" - }, - { - "block_id": "p370-b21", - "global_id": 10225, - "bbox": [ - 101.85, - 396.7, - 490.4, - 430.57 - ], - "text": "The time-shifting property proves very convenient in finding the Laplace transform\nof functions with different descriptions over different intervals, as the following example\ndemonstrates.", - "type": "text" - }, - { - "block_id": "p370-b22", - "global_id": 10226, - "bbox": [ - 76.77, - 463.84, - 450.27, - 475.8 - ], - "text": "EXAMPLE 4.6\nLaplace Transform and the Time-Shifting Property", - "type": "text" - }, - { - "block_id": "p370-b23", - "global_id": 10227, - "bbox": [ - 103.16, - 492.05, - 326.07, - 502.43 - ], - "text": "Find the Laplace transform of x(t) depicted in Fig. 4.4a.", - "type": "text" - }, - { - "block_id": "p370-b24", - "global_id": 10228, - "bbox": [ - 103.16, - 525.34, - 477.03, - 583.12 - ], - "text": "Describing mathematically a function such as the one in Fig. 4.4a is discussed in Sec. 1.4. The\nfunction x(t) in Fig. 4.4a can be described as a sum of two components shown in Fig. 4.4b. The\nequation for the first component is t−1 over 1 ≤t ≤2 so that this component can be described\nby (t −1)[u(t −1)−u(t −2)]. The second component can be described by u(t −2)−u(t −4).\nTherefore,", - "type": "text" - }, - { - "block_id": "p370-b25", - "global_id": 10229, - "bbox": [ - 171.78, - 594.66, - 395.45, - 605.04 - ], - "text": "x(t) = (t −1)[u(t −1) −u(t −2)] + [u(t −2) −u(t −4)]", - "type": "text" - }, - { - "block_id": "p370-b26", - "global_id": 10230, - "bbox": [ - 188.67, - 609.61, - 477.01, - 619.98 - ], - "text": "= (t −1)u(t −1) −(t −1)u(t −2) + u(t −2) −u(t −4)\n(4.13)", - "type": "text" - } - ] - }, - { - "page_num": 371, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p371-b0", - "global_id": 10231, - "bbox": [ - 287.44, - 62.89, - 516.14, - 71.98 - ], - "text": "4.2\nSome Properties of the Laplace Transform\n351", - "type": "text" - }, - { - "block_id": "p371-b1", - "global_id": 10232, - "bbox": [ - 169.5, - 197.16, - 366.11, - 205.16 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p371-b2", - "global_id": 10233, - "bbox": [ - 123.14, - 146.94, - 212.43, - 155.41 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p371-b3", - "global_id": 10234, - "bbox": [ - 123.14, - 103.91, - 127.14, - 111.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p371-b4", - "global_id": 10235, - "bbox": [ - 145.14, - 147.41, - 201.64, - 155.41 - ], - "text": "4\n3\n2\n1", - "type": "text" - }, - { - "block_id": "p371-b5", - "global_id": 10236, - "bbox": [ - 135.94, - 87.93, - 147.05, - 96.01 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p371-b7", - "global_id": 10237, - "bbox": [ - 257.74, - 146.94, - 452.21, - 155.41 - ], - "text": "t\nt\n0", - "type": "text" - }, - { - "block_id": "p371-b8", - "global_id": 10238, - "bbox": [ - 257.74, - 103.91, - 261.74, - 111.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p371-b9", - "global_id": 10239, - "bbox": [ - 279.74, - 147.41, - 364.72, - 155.41 - ], - "text": "2\n1\n0", - "type": "text" - }, - { - "block_id": "p371-b10", - "global_id": 10240, - "bbox": [ - 360.72, - 103.91, - 364.72, - 111.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p371-b11", - "global_id": 10241, - "bbox": [ - 400.22, - 147.41, - 439.22, - 155.41 - ], - "text": "4\n2", - "type": "text" - }, - { - "block_id": "p371-b12", - "global_id": 10242, - "bbox": [ - 270.72, - 87.93, - 388.06, - 97.54 - ], - "text": "x2(t)\nx1(t)", - "type": "text" - }, - { - "block_id": "p371-b14", - "global_id": 10243, - "bbox": [ - 288.56, - 176.26, - 305.45, - 184.55 - ], - "text": "t 1", - "type": "text" - }, - { - "block_id": "p371-b15", - "global_id": 10244, - "bbox": [ - 119.94, - 211.49, - 347.49, - 221.1 - ], - "text": "Figure 4.4 Finding a piecewise representation of a signal x(t).", - "type": "text" - }, - { - "block_id": "p371-b16", - "global_id": 10245, - "bbox": [ - 128.9, - 233.41, - 502.77, - 279.65 - ], - "text": "The first term on the right-hand side is the signal tu(t) delayed by 1 second. Also, the third\nand fourth terms are the signal u(t) delayed by 2 and 4 seconds, respectively. The second term,\nhowever, cannot be interpreted as a delayed version of any entry in Table 4.1. For this reason,\nwe rearrange it as", - "type": "text" - }, - { - "block_id": "p371-b17", - "global_id": 10246, - "bbox": [ - 188.83, - 291.2, - 442.84, - 301.57 - ], - "text": "(t −1)u(t −2) = (t −2 + 1)u(t −2) = (t −2)u(t −2) + u(t −2)", - "type": "text" - }, - { - "block_id": "p371-b18", - "global_id": 10247, - "bbox": [ - 128.91, - 313.11, - 502.77, - 335.44 - ], - "text": "We have now expressed the second term in the desired form as tu(t) delayed by 2 seconds plus\nu(t) delayed by 2 seconds. With this result, Eq. (4.13) can be expressed as", - "type": "text" - }, - { - "block_id": "p371-b19", - "global_id": 10248, - "bbox": [ - 218.57, - 346.99, - 413.1, - 357.36 - ], - "text": "x(t) = (t −1)u(t −1) −(t −2)u(t −2) −u(t −4)", - "type": "text" - }, - { - "block_id": "p371-b20", - "global_id": 10249, - "bbox": [ - 128.91, - 365.7, - 389.5, - 379.28 - ], - "text": "Application of the time-shifting property to tu(t) ⇐⇒1/s2 yields", - "type": "text" - }, - { - "block_id": "p371-b21", - "global_id": 10250, - "bbox": [ - 182.99, - 389.21, - 271.26, - 406.16 - ], - "text": "(t −1)u(t −1) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p371-b22", - "global_id": 10251, - "bbox": [ - 264.84, - 389.21, - 429.49, - 413.13 - ], - "text": "s2 e−s\nand\n(t −2)u(t −2) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p371-b23", - "global_id": 10252, - "bbox": [ - 423.07, - 394.06, - 448.17, - 413.13 - ], - "text": "s2 e−2s", - "type": "text" - }, - { - "block_id": "p371-b24", - "global_id": 10253, - "bbox": [ - 128.9, - 421.14, - 147.73, - 431.1 - ], - "text": "Also", - "type": "text" - }, - { - "block_id": "p371-b25", - "global_id": 10254, - "bbox": [ - 226.55, - 429.01, - 271.32, - 445.86 - ], - "text": "u(t) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p371-b26", - "global_id": 10255, - "bbox": [ - 266.88, - 429.01, - 387.35, - 452.93 - ], - "text": "s\nand\nu(t −4) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p371-b27", - "global_id": 10256, - "bbox": [ - 382.92, - 433.86, - 404.6, - 452.93 - ], - "text": "s e−4s", - "type": "text" - }, - { - "block_id": "p371-b28", - "global_id": 10257, - "bbox": [ - 128.9, - 457.95, - 170.65, - 467.91 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p371-b29", - "global_id": 10258, - "bbox": [ - 253.08, - 467.16, - 290.4, - 484.11 - ], - "text": "X(s) = 1", - "type": "text" - }, - { - "block_id": "p371-b30", - "global_id": 10259, - "bbox": [ - 283.97, - 467.16, - 324.59, - 491.08 - ], - "text": "s2 e−s −1", - "type": "text" - }, - { - "block_id": "p371-b31", - "global_id": 10260, - "bbox": [ - 318.17, - 467.16, - 360.83, - 491.08 - ], - "text": "s2 e−2s −1", - "type": "text" - }, - { - "block_id": "p371-b32", - "global_id": 10261, - "bbox": [ - 356.4, - 472.01, - 378.08, - 491.08 - ], - "text": "s e−4s", - "type": "text" - }, - { - "block_id": "p371-b33", - "global_id": 10262, - "bbox": [ - 102.51, - 533.61, - 469.34, - 559.52 - ], - "text": "EXAMPLE 4.7\nInverse Laplace Transform and the Time-Shifting\nProperty", - "type": "text" - }, - { - "block_id": "p371-b34", - "global_id": 10263, - "bbox": [ - 128.9, - 576.17, - 278.59, - 586.14 - ], - "text": "Find the inverse Laplace transform of", - "type": "text" - }, - { - "block_id": "p371-b35", - "global_id": 10264, - "bbox": [ - 272.62, - 589.04, - 356.22, - 607.53 - ], - "text": "X(s) = s + 3 + 5e−2s", - "type": "text" - }, - { - "block_id": "p371-b36", - "global_id": 10265, - "bbox": [ - 303.52, - 604.33, - 357.83, - 614.7 - ], - "text": "(s + 1)(s + 2)", - "type": "text" - } - ] - }, - { - "page_num": 372, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p372-b0", - "global_id": 10266, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "352\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p372-b1", - "global_id": 10267, - "bbox": [ - 103.16, - 84.21, - 477.02, - 109.74 - ], - "text": "Observe the exponential term e−2s in the numerator of X(s), indicating time delay. In such a\ncase, we should separate X(s) into terms with and without a delay factor, as", - "type": "text" - }, - { - "block_id": "p372-b2", - "global_id": 10268, - "bbox": [ - 213.54, - 116.86, - 299.96, - 157.12 - ], - "text": "X(s) =\ns + 3\n(s + 1)(s + 2)\n\n\n\nX1(s)", - "type": "text" - }, - { - "block_id": "p372-b3", - "global_id": 10269, - "bbox": [ - 301.06, - 115.63, - 348.54, - 133.8 - ], - "text": "+\n5e−2s", - "type": "text" - }, - { - "block_id": "p372-b4", - "global_id": 10270, - "bbox": [ - 309.93, - 130.92, - 366.65, - 146.62 - ], - "text": "(s + 1)(s + 2)", - "type": "text" - }, - { - "block_id": "p372-b5", - "global_id": 10271, - "bbox": [ - 323.67, - 148.68, - 352.4, - 158.42 - ], - "text": "X2(s)e−2s", - "type": "text" - }, - { - "block_id": "p372-b6", - "global_id": 10272, - "bbox": [ - 103.16, - 164.29, - 127.49, - 174.25 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p372-b7", - "global_id": 10273, - "bbox": [ - 211.62, - 177.73, - 367.35, - 202.16 - ], - "text": "X1(s) =\ns + 3\n(s + 1)(s + 2) =\n2\ns + 1 −\n1\ns + 2", - "type": "text" - }, - { - "block_id": "p372-b8", - "global_id": 10274, - "bbox": [ - 211.62, - 204.7, - 367.35, - 228.72 - ], - "text": "X2(s) =\n5\n(s + 1)(s + 2) =\n5\ns + 1 −\n5\ns + 2", - "type": "text" - }, - { - "block_id": "p372-b9", - "global_id": 10275, - "bbox": [ - 103.17, - 234.25, - 144.91, - 244.22 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p372-b10", - "global_id": 10276, - "bbox": [ - 241.27, - 247.66, - 338.91, - 262.92 - ], - "text": "x1(t) = (2e−t −e−2t)u(t)", - "type": "text" - }, - { - "block_id": "p372-b11", - "global_id": 10277, - "bbox": [ - 241.27, - 264.09, - 338.91, - 279.35 - ], - "text": "x2(t) = 5(e−t −e−2t)u(t)", - "type": "text" - }, - { - "block_id": "p372-b12", - "global_id": 10278, - "bbox": [ - 103.17, - 286.55, - 158.47, - 296.52 - ], - "text": "Also, because", - "type": "text" - }, - { - "block_id": "p372-b13", - "global_id": 10279, - "bbox": [ - 240.15, - 296.37, - 339.54, - 309.24 - ], - "text": "X(s) = X1(s) + X2(s)e−2s", - "type": "text" - }, - { - "block_id": "p372-b14", - "global_id": 10280, - "bbox": [ - 103.16, - 317.43, - 154.07, - 327.39 - ], - "text": "we can write", - "type": "text" - }, - { - "block_id": "p372-b15", - "global_id": 10281, - "bbox": [ - 180.95, - 334.95, - 271.94, - 346.1 - ], - "text": "x(t) = x1(t) + x2(t −2)", - "type": "text" - }, - { - "block_id": "p372-b16", - "global_id": 10282, - "bbox": [ - 197.84, - 347.32, - 290.48, - 361.81 - ], - "text": "= (2e−t −e−2t)u(t) + 5", - "type": "text" - }, - { - "block_id": "p372-b18", - "global_id": 10283, - "bbox": [ - 295.52, - 343.42, - 366.91, - 361.71 - ], - "text": "e−(t−2) −e−2(t−2)", - "type": "text" - }, - { - "block_id": "p372-b19", - "global_id": 10284, - "bbox": [ - 368.02, - 351.43, - 399.22, - 361.81 - ], - "text": "u(t −2)", - "type": "text" - }, - { - "block_id": "p372-b20", - "global_id": 10285, - "bbox": [ - 107.82, - 426.57, - 458.14, - 438.52 - ], - "text": "DRILL 4.4\nLaplace Transform and the Time-Shifting Property", - "type": "text" - }, - { - "block_id": "p372-b21", - "global_id": 10286, - "bbox": [ - 107.82, - 447.64, - 356.04, - 457.61 - ], - "text": "Find the Laplace transform of the signal illustrated in Fig. 4.5.", - "type": "text" - }, - { - "block_id": "p372-b22", - "global_id": 10287, - "bbox": [ - 108.09, - 471.13, - 162.68, - 482.09 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p372-b23", - "global_id": 10288, - "bbox": [ - 109.02, - 485.43, - 195.31, - 509.35 - ], - "text": "1\ns2 (1 −3e−2s + 2e−3s)", - "type": "text" - }, - { - "block_id": "p372-b24", - "global_id": 10289, - "bbox": [ - 133.39, - 610.08, - 259.85, - 618.24 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p372-b25", - "global_id": 10290, - "bbox": [ - 133.39, - 546.99, - 137.39, - 554.99 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p372-b26", - "global_id": 10291, - "bbox": [ - 166.89, - 610.24, - 227.99, - 618.24 - ], - "text": "3\n2\n1", - "type": "text" - }, - { - "block_id": "p372-b27", - "global_id": 10292, - "bbox": [ - 125.94, - 527.8, - 137.04, - 535.89 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p372-b28", - "global_id": 10293, - "bbox": [ - 286.35, - 610.29, - 402.42, - 619.53 - ], - "text": "Figure 4.5 Signal for Drill 4.4.", - "type": "text" - } - ] - }, - { - "page_num": 373, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p373-b0", - "global_id": 10294, - "bbox": [ - 287.44, - 62.89, - 516.14, - 71.98 - ], - "text": "4.2\nSome Properties of the Laplace Transform\n353", - "type": "text" - }, - { - "block_id": "p373-b1", - "global_id": 10295, - "bbox": [ - 133.57, - 97.81, - 477.22, - 123.71 - ], - "text": "DRILL 4.5\nInverse Laplace Transform and the Time-Shifting\nProperty", - "type": "text" - }, - { - "block_id": "p373-b2", - "global_id": 10296, - "bbox": [ - 133.57, - 128.6, - 354.06, - 147.18 - ], - "text": "Find the inverse Laplace transform of X(s) =\n3e−2s", - "type": "text" - }, - { - "block_id": "p373-b3", - "global_id": 10297, - "bbox": [ - 316.64, - 137.22, - 374.64, - 154.26 - ], - "text": "(s −1)(s + 2).", - "type": "text" - }, - { - "block_id": "p373-b4", - "global_id": 10298, - "bbox": [ - 133.57, - 166.35, - 188.43, - 186.25 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p373-b5", - "global_id": 10299, - "bbox": [ - 137.49, - 176.29, - 198.24, - 194.58 - ], - "text": "et−2 −e−2(t−2)", - "type": "text" - }, - { - "block_id": "p373-b6", - "global_id": 10300, - "bbox": [ - 198.24, - 184.3, - 229.45, - 194.68 - ], - "text": "u(t −2)", - "type": "text" - }, - { - "block_id": "p373-b7", - "global_id": 10301, - "bbox": [ - 127.59, - 225.56, - 263.23, - 237.52 - ], - "text": "4.2-2 Frequency Shifting", - "type": "text" - }, - { - "block_id": "p373-b8", - "global_id": 10302, - "bbox": [ - 127.59, - 243.65, - 305.9, - 253.61 - ], - "text": "The frequency-shifting property states that if", - "type": "text" - }, - { - "block_id": "p373-b9", - "global_id": 10303, - "bbox": [ - 293.92, - 263.82, - 349.81, - 274.09 - ], - "text": "x(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p373-b10", - "global_id": 10304, - "bbox": [ - 127.59, - 284.81, - 144.74, - 294.77 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p373-b11", - "global_id": 10305, - "bbox": [ - 277.96, - 292.23, - 516.13, - 307.5 - ], - "text": "x(t)es0t ⇐⇒X(s −s0)\n(4.14)", - "type": "text" - }, - { - "block_id": "p373-b12", - "global_id": 10306, - "bbox": [ - 127.59, - 315.02, - 516.13, - 336.94 - ], - "text": "Observe the symmetry (or duality) between this property and the time-shifting property of\nEq. (4.12).", - "type": "text" - }, - { - "block_id": "p373-b13", - "global_id": 10307, - "bbox": [ - 127.59, - 346.15, - 153.53, - 356.11 - ], - "text": "Proof.", - "type": "text" - }, - { - "block_id": "p373-b14", - "global_id": 10308, - "bbox": [ - 192.45, - 359.56, - 243.79, - 371.34 - ], - "text": "L[x(t)es0t] =", - "type": "text" - }, - { - "block_id": "p373-b15", - "global_id": 10309, - "bbox": [ - 245.84, - 347.5, - 262.78, - 359.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p373-b16", - "global_id": 10310, - "bbox": [ - 251.1, - 347.5, - 345.39, - 379.65 - ], - "text": "0−x(t)es0te−st dt =\n# ∞", - "type": "text" - }, - { - "block_id": "p373-b17", - "global_id": 10311, - "bbox": [ - 333.71, - 357.26, - 451.28, - 379.65 - ], - "text": "0−x(t)e−(s−s0)t dt = X(s −s0)", - "type": "text" - }, - { - "block_id": "p373-b18", - "global_id": 10312, - "bbox": [ - 102.51, - 397.96, - 357.13, - 409.92 - ], - "text": "EXAMPLE 4.8\nFrequency-Shifting Property", - "type": "text" - }, - { - "block_id": "p373-b19", - "global_id": 10313, - "bbox": [ - 128.9, - 426.58, - 432.35, - 436.54 - ], - "text": "Derive pair 9a in Table 4.1 from pair 8a and the frequency-shifting property.", - "type": "text" - }, - { - "block_id": "p373-b20", - "global_id": 10314, - "bbox": [ - 128.9, - 459.46, - 165.84, - 469.42 - ], - "text": "Pair 8a is", - "type": "text" - }, - { - "block_id": "p373-b21", - "global_id": 10315, - "bbox": [ - 269.23, - 464.33, - 360.73, - 488.35 - ], - "text": "cos btu(t) ⇐⇒\ns\ns2 + b2", - "type": "text" - }, - { - "block_id": "p373-b22", - "global_id": 10316, - "bbox": [ - 128.9, - 493.66, - 425.75, - 505.11 - ], - "text": "From the frequency-shifting property [Eq. (4.14)] with s0 = −a, we obtain", - "type": "text" - }, - { - "block_id": "p373-b23", - "global_id": 10317, - "bbox": [ - 249.08, - 512.57, - 380.88, - 536.9 - ], - "text": "e−at cos btu(t) ⇐⇒\ns + a\n(s + a)2 + b2", - "type": "text" - }, - { - "block_id": "p373-b24", - "global_id": 10318, - "bbox": [ - 133.57, - 578.51, - 365.0, - 590.46 - ], - "text": "DRILL 4.6\nFrequency-Shifting Property", - "type": "text" - }, - { - "block_id": "p373-b25", - "global_id": 10319, - "bbox": [ - 133.57, - 599.58, - 428.15, - 609.55 - ], - "text": "Derive pair 6 in Table 4.1 from pair 3 and the frequency-shifting property.", - "type": "text" - } - ] - }, - { - "page_num": 374, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p374-b0", - "global_id": 10320, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "354\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p374-b1", - "global_id": 10321, - "bbox": [ - 101.84, - 85.82, - 490.41, - 107.74 - ], - "text": "We are now ready to consider the two most important properties of the Laplace transform:\ntime differentiation and time integration.", - "type": "text" - }, - { - "block_id": "p374-b2", - "global_id": 10322, - "bbox": [ - 101.84, - 131.86, - 321.31, - 143.82 - ], - "text": "4.2-3 The Time-Differentiation Property", - "type": "text" - }, - { - "block_id": "p374-b3", - "global_id": 10323, - "bbox": [ - 101.84, - 149.35, - 287.49, - 159.91 - ], - "text": "The time-differentiation property states that if†", - "type": "text" - }, - { - "block_id": "p374-b4", - "global_id": 10324, - "bbox": [ - 268.17, - 170.2, - 324.06, - 180.48 - ], - "text": "x(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p374-b5", - "global_id": 10325, - "bbox": [ - 101.85, - 191.28, - 266.72, - 209.09 - ], - "text": "then\ndx(t)", - "type": "text" - }, - { - "block_id": "p374-b6", - "global_id": 10326, - "bbox": [ - 252.86, - 204.07, - 346.52, - 223.14 - ], - "text": "dt\n⇐⇒sX(s) −x(0−)", - "type": "text" - }, - { - "block_id": "p374-b7", - "global_id": 10327, - "bbox": [ - 101.84, - 227.53, - 368.86, - 237.49 - ], - "text": "Repeating this property a second time (differentiating twice) yields", - "type": "text" - }, - { - "block_id": "p374-b8", - "global_id": 10328, - "bbox": [ - 224.01, - 246.41, - 248.08, - 257.62 - ], - "text": "d2x(t)", - "type": "text" - }, - { - "block_id": "p374-b9", - "global_id": 10329, - "bbox": [ - 230.09, - 252.6, - 369.41, - 271.67 - ], - "text": "dt2\n⇐⇒s2X(s) −sx(0−) −˙x(0−)", - "type": "text" - }, - { - "block_id": "p374-b10", - "global_id": 10330, - "bbox": [ - 101.85, - 278.42, - 224.43, - 288.39 - ], - "text": "Repeated differentiation yields", - "type": "text" - }, - { - "block_id": "p374-b11", - "global_id": 10331, - "bbox": [ - 171.3, - 295.71, - 195.37, - 307.0 - ], - "text": "dnx(t)", - "type": "text" - }, - { - "block_id": "p374-b12", - "global_id": 10332, - "bbox": [ - 177.38, - 301.98, - 422.13, - 321.06 - ], - "text": "dtn\n⇐⇒snX(s) −sn−1x(0−) −sn−2˙x(0−) −· · · −x(n−1)(0−)", - "type": "text" - }, - { - "block_id": "p374-b13", - "global_id": 10333, - "bbox": [ - 198.61, - 330.11, - 243.46, - 341.89 - ], - "text": "= snX(s) −", - "type": "text" - }, - { - "block_id": "p374-b14", - "global_id": 10334, - "bbox": [ - 245.0, - 321.66, - 259.1, - 332.11 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p374-b15", - "global_id": 10335, - "bbox": [ - 245.98, - 346.0, - 258.11, - 353.27 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p374-b16", - "global_id": 10336, - "bbox": [ - 260.2, - 329.89, - 490.38, - 341.99 - ], - "text": "sn−kx(k−1)(0−)\n(4.15)", - "type": "text" - }, - { - "block_id": "p374-b17", - "global_id": 10337, - "bbox": [ - 101.85, - 359.72, - 241.13, - 373.29 - ], - "text": "where x(r)(0−) is drx/dtr at t = 0−.", - "type": "text" - }, - { - "block_id": "p374-b18", - "global_id": 10338, - "bbox": [ - 101.84, - 382.55, - 127.78, - 392.52 - ], - "text": "Proof.", - "type": "text" - }, - { - "block_id": "p374-b19", - "global_id": 10339, - "bbox": [ - 237.2, - 397.78, - 244.06, - 407.74 - ], - "text": "L", - "type": "text" - }, - { - "block_id": "p374-b20", - "global_id": 10340, - "bbox": [ - 245.17, - 383.8, - 271.61, - 401.07 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p374-b21", - "global_id": 10341, - "bbox": [ - 257.74, - 405.17, - 265.49, - 415.13 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p374-b22", - "global_id": 10342, - "bbox": [ - 272.81, - 383.8, - 278.24, - 393.76 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p374-b23", - "global_id": 10343, - "bbox": [ - 280.29, - 397.78, - 288.06, - 407.75 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p374-b24", - "global_id": 10344, - "bbox": [ - 290.11, - 384.22, - 307.05, - 396.28 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p374-b25", - "global_id": 10345, - "bbox": [ - 295.37, - 390.8, - 329.66, - 416.37 - ], - "text": "0−\ndx(t)", - "type": "text" - }, - { - "block_id": "p374-b26", - "global_id": 10346, - "bbox": [ - 315.79, - 393.98, - 354.86, - 415.13 - ], - "text": "dt e−st dt", - "type": "text" - }, - { - "block_id": "p374-b27", - "global_id": 10347, - "bbox": [ - 101.84, - 421.79, - 223.72, - 431.75 - ], - "text": "Integrating by parts, we obtain", - "type": "text" - }, - { - "block_id": "p374-b28", - "global_id": 10348, - "bbox": [ - 211.78, - 447.69, - 218.64, - 457.65 - ], - "text": "L", - "type": "text" - }, - { - "block_id": "p374-b29", - "global_id": 10349, - "bbox": [ - 219.75, - 433.69, - 246.19, - 450.98 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p374-b30", - "global_id": 10350, - "bbox": [ - 232.33, - 455.08, - 240.07, - 465.04 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p374-b31", - "global_id": 10351, - "bbox": [ - 247.39, - 433.69, - 252.82, - 443.66 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p374-b32", - "global_id": 10352, - "bbox": [ - 254.87, - 439.22, - 305.02, - 457.96 - ], - "text": "= x(t)e−st∞", - "type": "text" - }, - { - "block_id": "p374-b33", - "global_id": 10353, - "bbox": [ - 297.9, - 434.12, - 339.84, - 461.18 - ], - "text": "0−+ s\n# ∞", - "type": "text" - }, - { - "block_id": "p374-b34", - "global_id": 10354, - "bbox": [ - 328.16, - 443.88, - 380.27, - 466.26 - ], - "text": "0−x(t)e−st dt", - "type": "text" - }, - { - "block_id": "p374-b35", - "global_id": 10355, - "bbox": [ - 101.85, - 470.92, - 490.39, - 496.86 - ], - "text": "For the Laplace integral to converge [i.e., for X(s) to exist], it is necessary that x(t)e−st →0 as\nt →∞for the values of s in the ROC for X(s). Thus,", - "type": "text" - }, - { - "block_id": "p374-b36", - "global_id": 10356, - "bbox": [ - 238.09, - 512.37, - 244.95, - 522.33 - ], - "text": "L", - "type": "text" - }, - { - "block_id": "p374-b37", - "global_id": 10357, - "bbox": [ - 246.06, - 498.39, - 272.49, - 515.66 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p374-b38", - "global_id": 10358, - "bbox": [ - 258.63, - 519.76, - 266.38, - 529.72 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p374-b39", - "global_id": 10359, - "bbox": [ - 273.7, - 498.38, - 279.13, - 508.35 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p374-b40", - "global_id": 10360, - "bbox": [ - 281.18, - 510.65, - 354.15, - 522.75 - ], - "text": "= −x(0−) + sX(s)", - "type": "text" - }, - { - "block_id": "p374-b41", - "global_id": 10361, - "bbox": [ - 101.84, - 538.74, - 327.06, - 548.7 - ], - "text": "Repeated application of this procedure yields Eq. (4.15).", - "type": "text" - }, - { - "block_id": "p374-b42", - "global_id": 10362, - "bbox": [ - 101.84, - 588.09, - 476.43, - 600.31 - ], - "text": "† The dual of the time-differentiation property is the frequency-differentiation property, which states that", - "type": "text" - }, - { - "block_id": "p374-b43", - "global_id": 10363, - "bbox": [ - 261.04, - 603.64, - 312.08, - 618.89 - ], - "text": "tx(t) ⇐⇒−d", - "type": "text" - }, - { - "block_id": "p374-b44", - "global_id": 10364, - "bbox": [ - 305.97, - 609.64, - 331.21, - 625.26 - ], - "text": "dsX(s)", - "type": "text" - } - ] - }, - { - "page_num": 375, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p375-b0", - "global_id": 10365, - "bbox": [ - 287.44, - 62.89, - 516.14, - 71.98 - ], - "text": "4.2\nSome Properties of the Laplace Transform\n355", - "type": "text" - }, - { - "block_id": "p375-b1", - "global_id": 10366, - "bbox": [ - 102.51, - 93.91, - 513.45, - 105.87 - ], - "text": "EXAMPLE 4.9\nLaplace Transform and the Time-Differentiation Property", - "type": "text" - }, - { - "block_id": "p375-b2", - "global_id": 10367, - "bbox": [ - 128.9, - 122.12, - 502.75, - 144.45 - ], - "text": "Find the Laplace transform of the signal x(t) in Fig. 4.6a by using Table 4.1 and the\ntime-differentiation and time-shifting properties of the Laplace transform.", - "type": "text" - }, - { - "block_id": "p375-b3", - "global_id": 10368, - "bbox": [ - 348.6, - 434.59, - 350.83, - 442.59 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p375-b4", - "global_id": 10369, - "bbox": [ - 348.6, - 311.42, - 350.83, - 319.42 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p375-b5", - "global_id": 10370, - "bbox": [ - 348.6, - 239.15, - 350.83, - 247.15 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p375-b6", - "global_id": 10371, - "bbox": [ - 236.9, - 433.83, - 275.73, - 441.83 - ], - "text": "3\n2", - "type": "text" - }, - { - "block_id": "p375-b7", - "global_id": 10372, - "bbox": [ - 236.9, - 311.51, - 279.65, - 319.51 - ], - "text": "3\n2", - "type": "text" - }, - { - "block_id": "p375-b8", - "global_id": 10373, - "bbox": [ - 158.66, - 239.44, - 278.65, - 247.44 - ], - "text": "0\n2\n3", - "type": "text" - }, - { - "block_id": "p375-b9", - "global_id": 10374, - "bbox": [ - 252.16, - 496.0, - 261.04, - 504.0 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p375-b10", - "global_id": 10375, - "bbox": [ - 158.4, - 433.83, - 162.4, - 441.83 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p375-b11", - "global_id": 10376, - "bbox": [ - 158.66, - 392.19, - 162.66, - 400.19 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p375-b12", - "global_id": 10377, - "bbox": [ - 251.78, - 363.02, - 261.43, - 371.02 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p375-b13", - "global_id": 10378, - "bbox": [ - 158.66, - 311.13, - 162.66, - 319.13 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p375-b14", - "global_id": 10379, - "bbox": [ - 151.73, - 482.18, - 162.4, - 490.47 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p375-b15", - "global_id": 10380, - "bbox": [ - 152.0, - 340.87, - 162.66, - 349.17 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p375-b16", - "global_id": 10381, - "bbox": [ - 158.66, - 196.33, - 162.66, - 204.33 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p375-b17", - "global_id": 10382, - "bbox": [ - 158.66, - 287.17, - 162.66, - 295.17 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p375-b18", - "global_id": 10383, - "bbox": [ - 140.37, - 214.18, - 151.48, - 222.26 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p375-b19", - "global_id": 10384, - "bbox": [ - 252.16, - 252.82, - 261.04, - 260.82 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p375-b20", - "global_id": 10385, - "bbox": [ - 142.81, - 284.53, - 150.36, - 292.53 - ], - "text": "dx", - "type": "text" - }, - { - "block_id": "p375-b21", - "global_id": 10386, - "bbox": [ - 143.48, - 293.33, - 149.7, - 301.33 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p375-b22", - "global_id": 10387, - "bbox": [ - 139.46, - 400.36, - 150.49, - 409.76 - ], - "text": "d2x", - "type": "text" - }, - { - "block_id": "p375-b23", - "global_id": 10388, - "bbox": [ - 140.12, - 410.66, - 149.83, - 420.06 - ], - "text": "dt2", - "type": "text" - }, - { - "block_id": "p375-b24", - "global_id": 10389, - "bbox": [ - 375.43, - 459.43, - 506.34, - 505.26 - ], - "text": "Figure 4.6 Finding the Laplace\ntransform of a piecewise-linear\nfunction\nby\nusing\nthe\ntime-\ndifferentiation property.", - "type": "text" - }, - { - "block_id": "p375-b25", - "global_id": 10390, - "bbox": [ - 128.91, - 535.1, - 502.74, - 569.39 - ], - "text": "Figures 4.6b and 4.6c show the first two derivatives of x(t). Recall that the derivative\nat a point of jump discontinuity is an impulse of strength equal to the amount of jump [see\nEq. (1.12)]. Therefore,", - "type": "text" - }, - { - "block_id": "p375-b26", - "global_id": 10391, - "bbox": [ - 243.36, - 569.59, - 267.42, - 580.8 - ], - "text": "d2x(t)", - "type": "text" - }, - { - "block_id": "p375-b27", - "global_id": 10392, - "bbox": [ - 249.43, - 577.51, - 389.5, - 594.86 - ], - "text": "dt2\n= δ(t) −3δ(t −2) + 2δ(t −3)", - "type": "text" - } - ] - }, - { - "page_num": 376, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p376-b0", - "global_id": 10393, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "356\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p376-b1", - "global_id": 10394, - "bbox": [ - 103.16, - 86.24, - 284.68, - 96.2 - ], - "text": "The Laplace transform of this equation yields", - "type": "text" - }, - { - "block_id": "p376-b2", - "global_id": 10395, - "bbox": [ - 198.4, - 114.29, - 205.27, - 124.26 - ], - "text": "L", - "type": "text" - }, - { - "block_id": "p376-b3", - "global_id": 10396, - "bbox": [ - 206.38, - 100.31, - 238.36, - 117.58 - ], - "text": "d2x(t)", - "type": "text" - }, - { - "block_id": "p376-b4", - "global_id": 10397, - "bbox": [ - 220.37, - 121.17, - 231.79, - 131.64 - ], - "text": "dt2", - "type": "text" - }, - { - "block_id": "p376-b6", - "global_id": 10398, - "bbox": [ - 248.33, - 114.29, - 381.77, - 124.67 - ], - "text": "= L[δ(t) −3δ(t −2) + 2δ(t −3)]", - "type": "text" - }, - { - "block_id": "p376-b7", - "global_id": 10399, - "bbox": [ - 103.17, - 141.91, - 477.01, - 163.83 - ], - "text": "Using the time-differentiation property of Eq. (4.15), the time-shifting property of Eq. (4.12),\nand the facts that x(0−) = ˙x(0−) = 0, and δ(t) ⇐⇒1, we obtain", - "type": "text" - }, - { - "block_id": "p376-b8", - "global_id": 10400, - "bbox": [ - 220.55, - 171.25, - 359.1, - 185.75 - ], - "text": "s2X(s) −0 −0 = 1 −3e−2s + 2e−3s", - "type": "text" - }, - { - "block_id": "p376-b9", - "global_id": 10401, - "bbox": [ - 103.16, - 197.7, - 144.91, - 207.67 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p376-b10", - "global_id": 10402, - "bbox": [ - 231.49, - 206.92, - 268.81, - 223.87 - ], - "text": "X(s) = 1", - "type": "text" - }, - { - "block_id": "p376-b11", - "global_id": 10403, - "bbox": [ - 262.39, - 209.38, - 348.68, - 230.84 - ], - "text": "s2 (1 −3e−2s + 2e−3s)", - "type": "text" - }, - { - "block_id": "p376-b12", - "global_id": 10404, - "bbox": [ - 103.17, - 235.85, - 281.42, - 245.82 - ], - "text": "which confirms the earlier result in Drill 4.4.", - "type": "text" - }, - { - "block_id": "p376-b13", - "global_id": 10405, - "bbox": [ - 101.84, - 287.09, - 299.62, - 299.04 - ], - "text": "4.2-4 The Time-Integration Property", - "type": "text" - }, - { - "block_id": "p376-b14", - "global_id": 10406, - "bbox": [ - 101.84, - 304.58, - 273.75, - 315.14 - ], - "text": "The time-integration property states that if†", - "type": "text" - }, - { - "block_id": "p376-b15", - "global_id": 10407, - "bbox": [ - 268.17, - 326.71, - 324.06, - 336.99 - ], - "text": "x(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p376-b16", - "global_id": 10408, - "bbox": [ - 101.85, - 349.08, - 118.99, - 359.05 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p376-b17", - "global_id": 10409, - "bbox": [ - 135.61, - 365.61, - 147.38, - 377.88 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p376-b18", - "global_id": 10410, - "bbox": [ - 140.88, - 372.19, - 258.13, - 397.75 - ], - "text": "0−x(τ)dτ ⇐⇒X(s)\ns\nand", - "type": "text" - }, - { - "block_id": "p376-b19", - "global_id": 10411, - "bbox": [ - 279.17, - 365.61, - 290.94, - 377.88 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p376-b20", - "global_id": 10412, - "bbox": [ - 284.44, - 390.49, - 296.99, - 397.46 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p376-b21", - "global_id": 10413, - "bbox": [ - 298.6, - 372.19, - 370.11, - 389.45 - ], - "text": "x(τ)dτ ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p376-b22", - "global_id": 10414, - "bbox": [ - 359.25, - 379.17, - 380.62, - 396.52 - ], - "text": "s\n+", - "type": "text" - }, - { - "block_id": "p376-b23", - "global_id": 10415, - "bbox": [ - 383.36, - 363.08, - 398.01, - 375.42 - ], - "text": "$ 0−", - "type": "text" - }, - { - "block_id": "p376-b24", - "global_id": 10416, - "bbox": [ - 387.93, - 371.11, - 430.15, - 383.86 - ], - "text": "−∞x(τ)dτ", - "type": "text" - }, - { - "block_id": "p376-b25", - "global_id": 10417, - "bbox": [ - 405.42, - 379.58, - 490.38, - 396.52 - ], - "text": "s\n(4.16)", - "type": "text" - }, - { - "block_id": "p376-b26", - "global_id": 10418, - "bbox": [ - 101.84, - 406.92, - 316.09, - 416.96 - ], - "text": "Proof. To prove the first part of Eq. (4.16), we define", - "type": "text" - }, - { - "block_id": "p376-b27", - "global_id": 10419, - "bbox": [ - 260.12, - 434.32, - 285.29, - 444.6 - ], - "text": "g(t) =", - "type": "text" - }, - { - "block_id": "p376-b28", - "global_id": 10420, - "bbox": [ - 287.34, - 420.77, - 299.11, - 433.04 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p376-b29", - "global_id": 10421, - "bbox": [ - 292.61, - 434.32, - 330.92, - 452.91 - ], - "text": "0−x(τ)dτ", - "type": "text" - }, - { - "block_id": "p376-b30", - "global_id": 10422, - "bbox": [ - 101.85, - 461.51, - 369.52, - 493.23 - ], - "text": "so that\nd\ndtg(t) = x(t)\nand\ng(0−) = 0", - "type": "text" - }, - { - "block_id": "p376-b31", - "global_id": 10423, - "bbox": [ - 101.85, - 498.27, - 131.39, - 508.23 - ], - "text": "Now, if", - "type": "text" - }, - { - "block_id": "p376-b32", - "global_id": 10424, - "bbox": [ - 267.58, - 509.87, - 324.66, - 520.14 - ], - "text": "g(t) ⇐⇒G(s)", - "type": "text" - }, - { - "block_id": "p376-b33", - "global_id": 10425, - "bbox": [ - 101.85, - 529.24, - 118.99, - 539.21 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p376-b34", - "global_id": 10426, - "bbox": [ - 207.67, - 544.43, - 244.24, - 554.7 - ], - "text": "X(s) = L", - "type": "text" - }, - { - "block_id": "p376-b35", - "global_id": 10427, - "bbox": [ - 245.34, - 530.44, - 258.29, - 547.72 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p376-b36", - "global_id": 10428, - "bbox": [ - 251.96, - 544.43, - 276.45, - 561.78 - ], - "text": "dtg(t)", - "type": "text" - }, - { - "block_id": "p376-b37", - "global_id": 10429, - "bbox": [ - 276.45, - 530.44, - 281.88, - 540.41 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p376-b38", - "global_id": 10430, - "bbox": [ - 283.93, - 542.71, - 384.58, - 554.8 - ], - "text": "= sG(s) −g(0−) = sG(s)", - "type": "text" - }, - { - "block_id": "p376-b39", - "global_id": 10431, - "bbox": [ - 101.84, - 580.24, - 451.71, - 592.45 - ], - "text": "† The dual of the time-integration property is the frequency-integration property, which states that", - "type": "text" - }, - { - "block_id": "p376-b40", - "global_id": 10432, - "bbox": [ - 258.01, - 598.53, - 271.37, - 607.78 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p376-b41", - "global_id": 10433, - "bbox": [ - 263.36, - 604.81, - 291.61, - 620.43 - ], - "text": "t\n⇐⇒", - "type": "text" - }, - { - "block_id": "p376-b42", - "global_id": 10434, - "bbox": [ - 293.45, - 592.61, - 308.9, - 603.58 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p376-b43", - "global_id": 10435, - "bbox": [ - 298.19, - 615.06, - 300.71, - 621.53 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p376-b44", - "global_id": 10436, - "bbox": [ - 310.4, - 604.81, - 335.44, - 614.06 - ], - "text": "X(z)dz", - "type": "text" - } - ] - }, - { - "page_num": 377, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p377-b0", - "global_id": 10437, - "bbox": [ - 287.44, - 62.89, - 516.14, - 71.98 - ], - "text": "4.2\nSome Properties of the Laplace Transform\n357", - "type": "text" - }, - { - "block_id": "p377-b1", - "global_id": 10438, - "bbox": [ - 127.59, - 85.82, - 169.34, - 95.78 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p377-b2", - "global_id": 10439, - "bbox": [ - 296.57, - 97.35, - 345.97, - 114.61 - ], - "text": "G(s) = X(s)", - "type": "text" - }, - { - "block_id": "p377-b3", - "global_id": 10440, - "bbox": [ - 127.59, - 111.73, - 338.98, - 142.13 - ], - "text": "s\nor\n# t", - "type": "text" - }, - { - "block_id": "p377-b4", - "global_id": 10441, - "bbox": [ - 283.01, - 136.42, - 364.78, - 161.99 - ], - "text": "0−x(τ)dτ ⇐⇒X(s)\ns", - "type": "text" - }, - { - "block_id": "p377-b5", - "global_id": 10442, - "bbox": [ - 127.59, - 169.35, - 332.18, - 179.32 - ], - "text": "To prove the second part of Eq. (4.16), observe that", - "type": "text" - }, - { - "block_id": "p377-b6", - "global_id": 10443, - "bbox": [ - 239.09, - 187.16, - 250.85, - 199.43 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p377-b7", - "global_id": 10444, - "bbox": [ - 244.35, - 212.04, - 256.91, - 219.01 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p377-b8", - "global_id": 10445, - "bbox": [ - 258.52, - 200.72, - 297.6, - 210.99 - ], - "text": "x(τ)dτ =", - "type": "text" - }, - { - "block_id": "p377-b9", - "global_id": 10446, - "bbox": [ - 299.65, - 187.16, - 317.61, - 199.5 - ], - "text": "# 0−", - "type": "text" - }, - { - "block_id": "p377-b10", - "global_id": 10447, - "bbox": [ - 304.91, - 212.04, - 317.47, - 219.01 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p377-b11", - "global_id": 10448, - "bbox": [ - 319.73, - 200.72, - 358.3, - 210.99 - ], - "text": "x(τ)dτ +", - "type": "text" - }, - { - "block_id": "p377-b12", - "global_id": 10449, - "bbox": [ - 359.85, - 187.16, - 371.61, - 199.43 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p377-b13", - "global_id": 10450, - "bbox": [ - 365.11, - 200.72, - 403.42, - 219.3 - ], - "text": "0−x(τ)dτ", - "type": "text" - }, - { - "block_id": "p377-b14", - "global_id": 10451, - "bbox": [ - 127.6, - 229.74, - 516.15, - 252.08 - ], - "text": "Note that the first term on the right-hand side is a constant for t ≥0. Taking the Laplace transform\nof the foregoing equation and using the first part of Eq. (4.16), we obtain", - "type": "text" - }, - { - "block_id": "p377-b15", - "global_id": 10452, - "bbox": [ - 245.17, - 262.45, - 256.94, - 274.73 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p377-b16", - "global_id": 10453, - "bbox": [ - 250.43, - 287.32, - 262.99, - 294.3 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p377-b17", - "global_id": 10454, - "bbox": [ - 264.6, - 276.01, - 315.03, - 286.29 - ], - "text": "x(τ)dτ ⇐⇒", - "type": "text" - }, - { - "block_id": "p377-b18", - "global_id": 10455, - "bbox": [ - 318.27, - 259.92, - 332.91, - 272.27 - ], - "text": "$ 0−", - "type": "text" - }, - { - "block_id": "p377-b19", - "global_id": 10456, - "bbox": [ - 322.83, - 267.94, - 365.05, - 280.7 - ], - "text": "−∞x(τ)dτ", - "type": "text" - }, - { - "block_id": "p377-b20", - "global_id": 10457, - "bbox": [ - 340.33, - 269.03, - 397.36, - 293.36 - ], - "text": "s\n+ X(s)", - "type": "text" - }, - { - "block_id": "p377-b21", - "global_id": 10458, - "bbox": [ - 386.5, - 283.4, - 390.37, - 293.36 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p377-b22", - "global_id": 10459, - "bbox": [ - 127.59, - 318.11, - 272.88, - 330.06 - ], - "text": "4.2-5 The Scaling Property", - "type": "text" - }, - { - "block_id": "p377-b23", - "global_id": 10460, - "bbox": [ - 127.59, - 336.2, - 260.68, - 346.16 - ], - "text": "The scaling property states that if", - "type": "text" - }, - { - "block_id": "p377-b24", - "global_id": 10461, - "bbox": [ - 293.92, - 350.41, - 349.81, - 360.68 - ], - "text": "x(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p377-b25", - "global_id": 10462, - "bbox": [ - 127.59, - 371.13, - 183.17, - 381.5 - ], - "text": "then for a > 0", - "type": "text" - }, - { - "block_id": "p377-b26", - "global_id": 10463, - "bbox": [ - 282.99, - 382.79, - 332.2, - 399.63 - ], - "text": "x(at) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p377-b27", - "global_id": 10464, - "bbox": [ - 327.21, - 389.67, - 339.47, - 406.7 - ], - "text": "aX", - "type": "text" - }, - { - "block_id": "p377-b28", - "global_id": 10465, - "bbox": [ - 339.91, - 375.36, - 352.26, - 392.65 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p377-b29", - "global_id": 10466, - "bbox": [ - 347.83, - 396.74, - 352.81, - 406.7 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p377-b31", - "global_id": 10467, - "bbox": [ - 127.59, - 415.41, - 516.14, - 449.7 - ], - "text": "The proof is given in Ch. 7. Note that a is restricted to positive values because if x(t) is causal,\nthen x(at) is anticausal (is zero for t ≥0) for negative a, and anticausal signals are not permitted\nin the (unilateral) Laplace transform.", - "type": "text" - }, - { - "block_id": "p377-b32", - "global_id": 10468, - "bbox": [ - 145.52, - 449.87, - 441.69, - 461.66 - ], - "text": "Recall that x(at) is the signal x(t) time-compressed by the factor a, and X( s", - "type": "text" - }, - { - "block_id": "p377-b33", - "global_id": 10469, - "bbox": [ - 127.59, - 451.28, - 516.14, - 509.38 - ], - "text": "a) is X(s) expanded\nalong the s scale by the same factor a (see Sec. 1.2-2). The scaling property states that time\ncompression of a signal by a factor a causes expansion of its Laplace transform in the s scale\nby the same factor. Similarly, time expansion x(t) causes compression of X(s) in the s scale by the\nsame factor.", - "type": "text" - }, - { - "block_id": "p377-b34", - "global_id": 10470, - "bbox": [ - 127.59, - 536.02, - 413.7, - 547.97 - ], - "text": "4.2-6 Time Convolution and Frequency Convolution", - "type": "text" - }, - { - "block_id": "p377-b35", - "global_id": 10471, - "bbox": [ - 127.59, - 554.09, - 281.98, - 564.06 - ], - "text": "Another pair of properties states that if", - "type": "text" - }, - { - "block_id": "p377-b36", - "global_id": 10472, - "bbox": [ - 231.35, - 577.39, - 412.37, - 588.54 - ], - "text": "x1(t) ⇐⇒X1(s)\nand\nx2(t) ⇐⇒X2(s)", - "type": "text" - }, - { - "block_id": "p377-b37", - "global_id": 10473, - "bbox": [ - 127.59, - 601.4, - 257.33, - 611.46 - ], - "text": "then (time-convolution property)", - "type": "text" - }, - { - "block_id": "p377-b38", - "global_id": 10474, - "bbox": [ - 265.94, - 624.78, - 516.13, - 635.93 - ], - "text": "x1(t) ∗x2(t) ⇐⇒X1(s)X2(s)\n(4.17)", - "type": "text" - } - ] - }, - { - "page_num": 378, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p378-b0", - "global_id": 10475, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "358\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p378-b1", - "global_id": 10476, - "bbox": [ - 101.84, - 85.72, - 250.58, - 95.78 - ], - "text": "and (frequency-convolution property)", - "type": "text" - }, - { - "block_id": "p378-b2", - "global_id": 10477, - "bbox": [ - 228.32, - 110.4, - 363.92, - 134.42 - ], - "text": "x1(t)x2(t) ⇐⇒\n1\n2πj[X1(s) ∗X2(s)]", - "type": "text" - }, - { - "block_id": "p378-b3", - "global_id": 10478, - "bbox": [ - 101.85, - 149.05, - 490.4, - 170.96 - ], - "text": "Observe the symmetry (or duality) between the two properties. Proofs of these properties are\npostponed to Ch. 7.", - "type": "text" - }, - { - "block_id": "p378-b4", - "global_id": 10479, - "bbox": [ - 101.85, - 172.55, - 490.38, - 194.88 - ], - "text": "Equation (2.39) indicates that H(s), the transfer function of an LTIC system, is the Laplace\ntransform of the system’s impulse response h(t); that is,", - "type": "text" - }, - { - "block_id": "p378-b5", - "global_id": 10480, - "bbox": [ - 267.36, - 211.22, - 324.88, - 221.5 - ], - "text": "h(t) ⇐⇒H(s)", - "type": "text" - }, - { - "block_id": "p378-b6", - "global_id": 10481, - "bbox": [ - 101.85, - 237.94, - 490.39, - 272.23 - ], - "text": "If the system is causal, h(t) is causal, and, according to Eq. (2.39), H(s) is the unilateral Laplace\ntransform of h(t). Similarly, if the system is noncausal, h(t) is noncausal, and H(s) is the bilateral\ntransform of h(t).", - "type": "text" - }, - { - "block_id": "p378-b7", - "global_id": 10482, - "bbox": [ - 101.85, - 273.81, - 490.4, - 296.15 - ], - "text": "We can apply the time-convolution property to the LTIC input–output relationship y(t) =\nx(t) ∗h(t) to obtain", - "type": "text" - }, - { - "block_id": "p378-b8", - "global_id": 10483, - "bbox": [ - 262.99, - 304.93, - 490.39, - 315.31 - ], - "text": "Y(s) = X(s)H(s)\n(4.18)", - "type": "text" - }, - { - "block_id": "p378-b9", - "global_id": 10484, - "bbox": [ - 101.84, - 328.66, - 490.39, - 350.99 - ], - "text": "The response y(t) is the zero-state response of the LTIC system to the input x(t). From Eq. (4.18),\nit follows that", - "type": "text" - }, - { - "block_id": "p378-b10", - "global_id": 10485, - "bbox": [ - 218.68, - 355.77, - 268.41, - 373.03 - ], - "text": "H(s) = Y(s)", - "type": "text" - }, - { - "block_id": "p378-b11", - "global_id": 10486, - "bbox": [ - 250.69, - 355.77, - 372.36, - 380.1 - ], - "text": "X(s) = L[zero-state response]", - "type": "text" - }, - { - "block_id": "p378-b12", - "global_id": 10487, - "bbox": [ - 310.57, - 363.17, - 490.38, - 380.21 - ], - "text": "L[input]\n(4.19)", - "type": "text" - }, - { - "block_id": "p378-b13", - "global_id": 10488, - "bbox": [ - 101.85, - 391.6, - 490.39, - 413.83 - ], - "text": "This may be considered an alternate definition of the LTIC system transfer function H(s). It is the\nratio of the transform of zero-state response to the transform of the input.", - "type": "text" - }, - { - "block_id": "p378-b14", - "global_id": 10489, - "bbox": [ - 76.77, - 457.16, - 332.05, - 469.12 - ], - "text": "EXAMPLE 4.10\nTime-Convolution Property", - "type": "text" - }, - { - "block_id": "p378-b15", - "global_id": 10490, - "bbox": [ - 103.16, - 484.37, - 477.01, - 495.75 - ], - "text": "Use the time-convolution property of the Laplace transform to determine c(t) = eatu(t)∗ebtu(t).", - "type": "text" - }, - { - "block_id": "p378-b16", - "global_id": 10491, - "bbox": [ - 103.16, - 518.66, - 227.44, - 528.63 - ], - "text": "From Eq. (4.17), it follows that", - "type": "text" - }, - { - "block_id": "p378-b17", - "global_id": 10492, - "bbox": [ - 195.64, - 539.24, - 316.26, - 563.16 - ], - "text": "C(s) =\n1\n(s −a)(s −b) =\n1\na −b", - "type": "text" - }, - { - "block_id": "p378-b18", - "global_id": 10493, - "bbox": [ - 318.58, - 531.82, - 377.92, - 563.16 - ], - "text": "1\ns −a −\n1\ns −b", - "type": "text" - }, - { - "block_id": "p378-b19", - "global_id": 10494, - "bbox": [ - 379.11, - 531.82, - 384.54, - 541.79 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p378-b20", - "global_id": 10495, - "bbox": [ - 103.16, - 573.43, - 281.36, - 583.39 - ], - "text": "The inverse transform of this equation yields", - "type": "text" - }, - { - "block_id": "p378-b21", - "global_id": 10496, - "bbox": [ - 237.84, - 593.33, - 342.35, - 617.25 - ], - "text": "c(t) =\n1\na −b(eat −ebt)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 379, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p379-b0", - "global_id": 10497, - "bbox": [ - 287.44, - 62.89, - 516.14, - 71.98 - ], - "text": "4.2\nSome Properties of the Laplace Transform\n359", - "type": "text" - }, - { - "block_id": "p379-b1", - "global_id": 10498, - "bbox": [ - 127.89, - 86.19, - 279.2, - 98.32 - ], - "text": "INITIAL AND FINAL VALUES", - "type": "text" - }, - { - "block_id": "p379-b2", - "global_id": 10499, - "bbox": [ - 127.59, - 101.93, - 516.14, - 136.22 - ], - "text": "In certain applications, it is desirable to know the values of x(t) as t →0 and t →∞[initial\nand final values of x(t)] from the knowledge of its Laplace transform X(s). Initial and final value\ntheorems provide such information.", - "type": "text" - }, - { - "block_id": "p379-b3", - "global_id": 10500, - "bbox": [ - 127.6, - 137.8, - 516.11, - 160.13 - ], - "text": "The initial value theorem states that if x(t) and its derivative dx/dt are both Laplace\ntransformable, then", - "type": "text" - }, - { - "block_id": "p379-b4", - "global_id": 10501, - "bbox": [ - 284.37, - 159.98, - 334.44, - 172.09 - ], - "text": "x(0+) = lim", - "type": "text" - }, - { - "block_id": "p379-b5", - "global_id": 10502, - "bbox": [ - 319.04, - 161.71, - 516.13, - 177.27 - ], - "text": "s→∞sX(s)\n(4.20)", - "type": "text" - }, - { - "block_id": "p379-b6", - "global_id": 10503, - "bbox": [ - 127.59, - 183.68, - 369.42, - 193.64 - ], - "text": "provided the limit on the right-hand side of Eq. (4.20) exists.", - "type": "text" - }, - { - "block_id": "p379-b7", - "global_id": 10504, - "bbox": [ - 145.52, - 195.23, - 498.07, - 205.61 - ], - "text": "The final value theorem states that if both x(t) and dx/dt are Laplace transformable, then", - "type": "text" - }, - { - "block_id": "p379-b8", - "global_id": 10505, - "bbox": [ - 281.2, - 216.42, - 339.43, - 231.98 - ], - "text": "lim\nt→∞x(t) = lim", - "type": "text" - }, - { - "block_id": "p379-b9", - "global_id": 10506, - "bbox": [ - 325.85, - 216.42, - 516.13, - 232.6 - ], - "text": "s→0sX(s)\n(4.21)", - "type": "text" - }, - { - "block_id": "p379-b10", - "global_id": 10507, - "bbox": [ - 127.59, - 241.22, - 516.12, - 263.56 - ], - "text": "provided sX(s) has no poles in the RHP or on the imaginary axis. To prove these theorems, we\nbegin by setting n = 1 in Eq. (4.15). Using the definition of the Laplace transform, we see that", - "type": "text" - }, - { - "block_id": "p379-b11", - "global_id": 10508, - "bbox": [ - 217.11, - 277.97, - 282.31, - 290.07 - ], - "text": "sX(s) −x(0−) =", - "type": "text" - }, - { - "block_id": "p379-b12", - "global_id": 10509, - "bbox": [ - 284.36, - 266.14, - 301.3, - 278.2 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p379-b13", - "global_id": 10510, - "bbox": [ - 289.62, - 272.71, - 323.91, - 298.28 - ], - "text": "0−\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b14", - "global_id": 10511, - "bbox": [ - 310.04, - 275.89, - 349.1, - 297.04 - ], - "text": "dt e−st dt", - "type": "text" - }, - { - "block_id": "p379-b15", - "global_id": 10512, - "bbox": [ - 274.54, - 309.56, - 282.31, - 319.52 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p379-b16", - "global_id": 10513, - "bbox": [ - 284.35, - 295.99, - 302.33, - 308.35 - ], - "text": "# 0+", - "type": "text" - }, - { - "block_id": "p379-b17", - "global_id": 10514, - "bbox": [ - 289.62, - 302.57, - 325.44, - 328.14 - ], - "text": "0−\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b18", - "global_id": 10515, - "bbox": [ - 311.57, - 305.76, - 360.14, - 326.91 - ], - "text": "dt e−st dt +", - "type": "text" - }, - { - "block_id": "p379-b19", - "global_id": 10516, - "bbox": [ - 361.68, - 295.99, - 378.63, - 308.06 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p379-b20", - "global_id": 10517, - "bbox": [ - 366.95, - 302.57, - 401.24, - 328.14 - ], - "text": "0+\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b21", - "global_id": 10518, - "bbox": [ - 387.37, - 305.76, - 426.44, - 326.91 - ], - "text": "dt e−st dt", - "type": "text" - }, - { - "block_id": "p379-b22", - "global_id": 10519, - "bbox": [ - 274.54, - 336.66, - 299.19, - 346.94 - ], - "text": "= x(t)", - "type": "text" - }, - { - "block_id": "p379-b23", - "global_id": 10520, - "bbox": [ - 299.19, - 328.19, - 310.57, - 344.13 - ], - "text": "0+", - "type": "text" - }, - { - "block_id": "p379-b24", - "global_id": 10521, - "bbox": [ - 302.43, - 323.1, - 339.39, - 350.15 - ], - "text": "0−+\n# ∞", - "type": "text" - }, - { - "block_id": "p379-b25", - "global_id": 10522, - "bbox": [ - 327.71, - 329.68, - 362.0, - 355.25 - ], - "text": "0+\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b26", - "global_id": 10523, - "bbox": [ - 348.13, - 332.86, - 387.2, - 354.01 - ], - "text": "dt e−st dt", - "type": "text" - }, - { - "block_id": "p379-b27", - "global_id": 10524, - "bbox": [ - 274.54, - 362.04, - 350.16, - 374.14 - ], - "text": "= x(0+) −x(0−) +", - "type": "text" - }, - { - "block_id": "p379-b28", - "global_id": 10525, - "bbox": [ - 351.7, - 350.21, - 368.65, - 362.26 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p379-b29", - "global_id": 10526, - "bbox": [ - 356.97, - 356.78, - 391.26, - 382.35 - ], - "text": "0+\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b30", - "global_id": 10527, - "bbox": [ - 377.39, - 359.96, - 416.45, - 381.11 - ], - "text": "dt e−st dt", - "type": "text" - }, - { - "block_id": "p379-b31", - "global_id": 10528, - "bbox": [ - 127.59, - 390.2, - 169.34, - 400.17 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p379-b32", - "global_id": 10529, - "bbox": [ - 255.77, - 404.62, - 321.47, - 416.73 - ], - "text": "sX(s) = x(0+) +", - "type": "text" - }, - { - "block_id": "p379-b33", - "global_id": 10530, - "bbox": [ - 323.01, - 392.79, - 339.97, - 404.84 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p379-b34", - "global_id": 10531, - "bbox": [ - 328.28, - 399.37, - 362.57, - 424.92 - ], - "text": "0+\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b35", - "global_id": 10532, - "bbox": [ - 348.7, - 402.54, - 387.76, - 423.7 - ], - "text": "dt e−st dt", - "type": "text" - }, - { - "block_id": "p379-b36", - "global_id": 10533, - "bbox": [ - 127.59, - 430.16, - 141.97, - 440.12 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p379-b37", - "global_id": 10534, - "bbox": [ - 231.18, - 452.48, - 332.43, - 469.76 - ], - "text": "lim\ns→∞sX(s) = x(0+) + lim", - "type": "text" - }, - { - "block_id": "p379-b38", - "global_id": 10535, - "bbox": [ - 317.03, - 462.57, - 334.53, - 469.76 - ], - "text": "s→∞", - "type": "text" - }, - { - "block_id": "p379-b39", - "global_id": 10536, - "bbox": [ - 336.74, - 440.64, - 353.69, - 452.7 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p379-b40", - "global_id": 10537, - "bbox": [ - 342.01, - 447.22, - 376.3, - 472.79 - ], - "text": "0+\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b41", - "global_id": 10538, - "bbox": [ - 362.44, - 450.4, - 401.5, - 471.55 - ], - "text": "dt e−st dt", - "type": "text" - }, - { - "block_id": "p379-b42", - "global_id": 10539, - "bbox": [ - 273.55, - 479.58, - 315.49, - 491.69 - ], - "text": "= x(0+) +", - "type": "text" - }, - { - "block_id": "p379-b43", - "global_id": 10540, - "bbox": [ - 317.03, - 467.75, - 333.98, - 479.8 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p379-b44", - "global_id": 10541, - "bbox": [ - 322.3, - 474.33, - 356.59, - 499.89 - ], - "text": "0+\ndx(t)", - "type": "text" - }, - { - "block_id": "p379-b45", - "global_id": 10542, - "bbox": [ - 342.72, - 488.7, - 350.47, - 498.66 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p379-b47", - "global_id": 10543, - "bbox": [ - 364.33, - 470.3, - 403.5, - 496.87 - ], - "text": "lim\ns→∞e−st", - "type": "text" - }, - { - "block_id": "p379-b48", - "global_id": 10544, - "bbox": [ - 404.6, - 481.62, - 412.35, - 491.59 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p379-b49", - "global_id": 10545, - "bbox": [ - 273.55, - 500.94, - 306.17, - 513.04 - ], - "text": "= x(0+)", - "type": "text" - }, - { - "block_id": "p379-b50", - "global_id": 10546, - "bbox": [ - 127.59, - 523.85, - 516.14, - 570.1 - ], - "text": "Comment. The initial value theorem applies only if X(s) is strictly proper (M < N), because for\nM ≥N, lims→∞sX(s) does not exist, and the theorem does not apply. In such a case, we can still\nfind the answer by using long division to express X(s) as a polynomial in s plus a strictly proper\nfraction, where M < N. For example, by using long division, we can express", - "type": "text" - }, - { - "block_id": "p379-b51", - "global_id": 10547, - "bbox": [ - 242.91, - 576.86, - 305.08, - 590.86 - ], - "text": "s3 + 3s2 + s + 1", - "type": "text" - }, - { - "block_id": "p379-b52", - "global_id": 10548, - "bbox": [ - 252.28, - 580.8, - 400.79, - 604.92 - ], - "text": "s2 + 2s + 1\n= (s + 1) −\n2s\ns2 + 2s + 1", - "type": "text" - }, - { - "block_id": "p379-b53", - "global_id": 10549, - "bbox": [ - 127.59, - 612.45, - 516.14, - 634.79 - ], - "text": "The inverse transform of the polynomial in s is in terms of δ(t), and its derivatives, which are zero\nat t = 0+. In the foregoing case, the inverse transform of s + 1 is ˙δ(t) + δ(t). Hence, the desired", - "type": "text" - } - ] - }, - { - "page_num": 380, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p380-b0", - "global_id": 10550, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "360\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p380-b1", - "global_id": 10551, - "bbox": [ - 101.84, - 85.07, - 490.38, - 108.63 - ], - "text": "x(0+) is the value of the remainder (strictly proper) fraction, for which the initial value theorem\napplies. In the present case,", - "type": "text" - }, - { - "block_id": "p380-b2", - "global_id": 10552, - "bbox": [ - 233.23, - 115.12, - 283.3, - 127.22 - ], - "text": "x(0+) = lim", - "type": "text" - }, - { - "block_id": "p380-b3", - "global_id": 10553, - "bbox": [ - 267.9, - 125.21, - 285.4, - 132.4 - ], - "text": "s→∞", - "type": "text" - }, - { - "block_id": "p380-b4", - "global_id": 10554, - "bbox": [ - 301.17, - 108.92, - 321.28, - 120.23 - ], - "text": "−2s2", - "type": "text" - }, - { - "block_id": "p380-b5", - "global_id": 10555, - "bbox": [ - 289.75, - 116.84, - 359.0, - 134.29 - ], - "text": "s2 + 2s + 1 = −2", - "type": "text" - }, - { - "block_id": "p380-b6", - "global_id": 10556, - "bbox": [ - 119.78, - 139.56, - 436.23, - 149.93 - ], - "text": "To prove the final value theorem, we let n = 1 and s →0 in Eq. (4.15) to obtain", - "type": "text" - }, - { - "block_id": "p380-b7", - "global_id": 10557, - "bbox": [ - 180.91, - 165.12, - 282.24, - 183.02 - ], - "text": "lim\ns→0[sX(s) −x(0−)] = lim", - "type": "text" - }, - { - "block_id": "p380-b8", - "global_id": 10558, - "bbox": [ - 268.66, - 175.76, - 282.52, - 183.02 - ], - "text": "s→0", - "type": "text" - }, - { - "block_id": "p380-b9", - "global_id": 10559, - "bbox": [ - 284.73, - 153.29, - 301.68, - 165.34 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p380-b10", - "global_id": 10560, - "bbox": [ - 290.0, - 159.87, - 324.29, - 185.43 - ], - "text": "0−\ndx(t)", - "type": "text" - }, - { - "block_id": "p380-b11", - "global_id": 10561, - "bbox": [ - 310.42, - 163.04, - 359.48, - 184.2 - ], - "text": "dt e−st dt =", - "type": "text" - }, - { - "block_id": "p380-b12", - "global_id": 10562, - "bbox": [ - 361.53, - 153.29, - 378.48, - 165.34 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p380-b13", - "global_id": 10563, - "bbox": [ - 366.8, - 159.87, - 401.08, - 185.43 - ], - "text": "0−\ndx(t)", - "type": "text" - }, - { - "block_id": "p380-b14", - "global_id": 10564, - "bbox": [ - 387.22, - 167.16, - 411.15, - 184.2 - ], - "text": "dt\ndt", - "type": "text" - }, - { - "block_id": "p380-b15", - "global_id": 10565, - "bbox": [ - 258.84, - 191.85, - 283.49, - 202.13 - ], - "text": "= x(t)", - "type": "text" - }, - { - "block_id": "p380-b16", - "global_id": 10566, - "bbox": [ - 283.49, - 183.38, - 293.85, - 199.32 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p380-b17", - "global_id": 10567, - "bbox": [ - 286.73, - 190.13, - 374.21, - 207.41 - ], - "text": "0−= lim\nt→∞x(t) −x(0−)", - "type": "text" - }, - { - "block_id": "p380-b18", - "global_id": 10568, - "bbox": [ - 101.85, - 217.23, - 316.86, - 227.2 - ], - "text": "a deduction that leads to the desired result, Eq. (4.21).", - "type": "text" - }, - { - "block_id": "p380-b19", - "global_id": 10569, - "bbox": [ - 101.85, - 236.8, - 490.39, - 295.0 - ], - "text": "Comment. The final value theorem applies only if the poles of X(s) are in the LHP (including\ns = 0). If X(s) has a pole in the RHP, x(t) contains an exponentially growing term and x(∞) does\nnot exist. If there is a pole on the imaginary axis, then x(t) contains an oscillating term and x(∞)\ndoes not exist. However, if there is a pole at the origin, then x(t) contains a constant term, and\nhence, x(∞) exists and is a constant.", - "type": "text" - }, - { - "block_id": "p380-b20", - "global_id": 10570, - "bbox": [ - 76.77, - 323.6, - 308.77, - 335.55 - ], - "text": "EXAMPLE 4.11\nInitial and Final Values", - "type": "text" - }, - { - "block_id": "p380-b21", - "global_id": 10571, - "bbox": [ - 103.16, - 348.88, - 440.49, - 359.25 - ], - "text": "Determine the initial and final values of y(t) if its Laplace transform Y(s) is given by", - "type": "text" - }, - { - "block_id": "p380-b22", - "global_id": 10572, - "bbox": [ - 246.78, - 368.84, - 332.19, - 393.27 - ], - "text": "Y(s) =\n10(2s + 3)\ns(s2 + 2s + 5)", - "type": "text" - }, - { - "block_id": "p380-b23", - "global_id": 10573, - "bbox": [ - 103.16, - 419.08, - 235.42, - 429.04 - ], - "text": "Equations (4.20) and (4.21) yield", - "type": "text" - }, - { - "block_id": "p380-b24", - "global_id": 10574, - "bbox": [ - 202.43, - 443.89, - 252.48, - 455.99 - ], - "text": "y(0+) = lim", - "type": "text" - }, - { - "block_id": "p380-b25", - "global_id": 10575, - "bbox": [ - 237.08, - 445.61, - 304.43, - 461.17 - ], - "text": "s→∞sY(s) = lim", - "type": "text" - }, - { - "block_id": "p380-b26", - "global_id": 10576, - "bbox": [ - 289.03, - 453.98, - 306.53, - 461.17 - ], - "text": "s→∞", - "type": "text" - }, - { - "block_id": "p380-b27", - "global_id": 10577, - "bbox": [ - 308.83, - 438.63, - 377.75, - 463.06 - ], - "text": "10(2s + 3)\n(s2 + 2s + 5) = 0", - "type": "text" - }, - { - "block_id": "p380-b28", - "global_id": 10578, - "bbox": [ - 203.18, - 472.23, - 250.66, - 482.61 - ], - "text": "y(∞) = lim", - "type": "text" - }, - { - "block_id": "p380-b29", - "global_id": 10579, - "bbox": [ - 237.08, - 472.23, - 298.98, - 488.4 - ], - "text": "s→0sY(s) = lim", - "type": "text" - }, - { - "block_id": "p380-b30", - "global_id": 10580, - "bbox": [ - 285.4, - 481.14, - 299.26, - 488.4 - ], - "text": "s→0", - "type": "text" - }, - { - "block_id": "p380-b31", - "global_id": 10581, - "bbox": [ - 301.57, - 465.25, - 370.48, - 489.68 - ], - "text": "10(2s + 3)\n(s2 + 2s + 5) = 6", - "type": "text" - }, - { - "block_id": "p380-b32", - "global_id": 10582, - "bbox": [ - 119.78, - 525.74, - 444.65, - 535.7 - ], - "text": "Table 4.2 summarizes the most important unilateral Laplace transform properties.", - "type": "text" - }, - { - "block_id": "p380-b33", - "global_id": 10583, - "bbox": [ - 102.2, - 565.04, - 346.65, - 594.93 - ], - "text": "4.3 SOLUTION OF DIFFERENTIAL AND\nINTEGRO-DIFFERENTIAL EQUATIONS", - "type": "text" - }, - { - "block_id": "p380-b34", - "global_id": 10584, - "bbox": [ - 101.84, - 600.91, - 490.4, - 634.79 - ], - "text": "The time-differentiation property of the Laplace transform has set the stage for solving linear\ndifferential (or integro-differential) equations with constant coefficients. Because dky/dtk ⇐⇒\nskY(s), the Laplace transform of a differential equation is an algebraic equation that can be readily", - "type": "text" - } - ] - }, - { - "page_num": 381, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p381-b0", - "global_id": 10585, - "bbox": [ - 223.7, - 62.89, - 516.12, - 71.98 - ], - "text": "4.3\nSolution of Differential and Integro-Differential Equations\n361", - "type": "text" - }, - { - "block_id": "p381-b1", - "global_id": 10586, - "bbox": [ - 127.59, - 86.07, - 325.66, - 95.3 - ], - "text": "TABLE 4.2\nUnilateral Laplace Transform Properties", - "type": "text" - }, - { - "block_id": "p381-b2", - "global_id": 10587, - "bbox": [ - 127.59, - 106.02, - 309.42, - 115.28 - ], - "text": "Operation\nx(t)\nX(s)", - "type": "text" - }, - { - "block_id": "p381-b3", - "global_id": 10588, - "bbox": [ - 127.59, - 124.35, - 341.53, - 146.64 - ], - "text": "Addition\nx1(t) + x2(t)\nX1(s) + X2(s)\nScalar multiplication\nkx(t)\nkX(s)", - "type": "text" - }, - { - "block_id": "p381-b4", - "global_id": 10589, - "bbox": [ - 127.59, - 148.74, - 240.83, - 164.36 - ], - "text": "Time differentiation\ndx(t)", - "type": "text" - }, - { - "block_id": "p381-b5", - "global_id": 10590, - "bbox": [ - 228.35, - 153.77, - 343.0, - 170.64 - ], - "text": "dt\nsX(s) −x(0−)", - "type": "text" - }, - { - "block_id": "p381-b6", - "global_id": 10591, - "bbox": [ - 222.99, - 170.96, - 244.81, - 181.21 - ], - "text": "d2x(t)", - "type": "text" - }, - { - "block_id": "p381-b7", - "global_id": 10592, - "bbox": [ - 228.46, - 176.98, - 380.74, - 193.85 - ], - "text": "dt2\ns2X(s) −sx(0−) −˙x(0−)", - "type": "text" - }, - { - "block_id": "p381-b8", - "global_id": 10593, - "bbox": [ - 222.99, - 194.17, - 244.81, - 204.42 - ], - "text": "d3x(t)", - "type": "text" - }, - { - "block_id": "p381-b9", - "global_id": 10594, - "bbox": [ - 228.46, - 200.2, - 418.48, - 217.07 - ], - "text": "dt3\ns3X(s) −s2x(0−) −s˙x(0−) −¨x(0−)", - "type": "text" - }, - { - "block_id": "p381-b10", - "global_id": 10595, - "bbox": [ - 222.99, - 215.84, - 244.81, - 226.15 - ], - "text": "dnx(t)", - "type": "text" - }, - { - "block_id": "p381-b11", - "global_id": 10596, - "bbox": [ - 228.46, - 222.12, - 324.64, - 238.79 - ], - "text": "dtn\nsnX(s) −", - "type": "text" - }, - { - "block_id": "p381-b12", - "global_id": 10597, - "bbox": [ - 326.97, - 216.46, - 336.34, - 225.42 - ], - "text": "n%", - "type": "text" - }, - { - "block_id": "p381-b13", - "global_id": 10598, - "bbox": [ - 326.03, - 233.42, - 337.3, - 240.16 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p381-b14", - "global_id": 10599, - "bbox": [ - 338.28, - 221.91, - 390.88, - 232.52 - ], - "text": "sn−kx(k−1)(0−)", - "type": "text" - }, - { - "block_id": "p381-b15", - "global_id": 10600, - "bbox": [ - 127.59, - 246.03, - 187.15, - 255.0 - ], - "text": "Time integration", - "type": "text" - }, - { - "block_id": "p381-b16", - "global_id": 10601, - "bbox": [ - 221.81, - 233.46, - 232.44, - 244.63 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p381-b17", - "global_id": 10602, - "bbox": [ - 221.8, - 239.74, - 315.9, - 267.98 - ], - "text": "0−x(τ)dτ\n1\ns X(s)\n# t", - "type": "text" - }, - { - "block_id": "p381-b18", - "global_id": 10603, - "bbox": [ - 226.54, - 279.05, - 238.2, - 285.52 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p381-b19", - "global_id": 10604, - "bbox": [ - 239.7, - 263.1, - 331.35, - 284.63 - ], - "text": "x(τ)dτ\n1\ns X(s) + 1", - "type": "text" - }, - { - "block_id": "p381-b20", - "global_id": 10605, - "bbox": [ - 327.36, - 275.66, - 330.84, - 284.63 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p381-b21", - "global_id": 10606, - "bbox": [ - 333.54, - 256.81, - 349.89, - 268.04 - ], - "text": "# 0−", - "type": "text" - }, - { - "block_id": "p381-b22", - "global_id": 10607, - "bbox": [ - 338.27, - 279.05, - 349.93, - 285.52 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p381-b23", - "global_id": 10608, - "bbox": [ - 351.88, - 269.02, - 373.21, - 278.26 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p381-b24", - "global_id": 10609, - "bbox": [ - 127.59, - 284.0, - 365.19, - 308.29 - ], - "text": "Time shifting\nx(t −t0)u(t −t0)\nX(s)e−st0\nt0 ≥0\nFrequency shifting\nx(t)es0t\nX(s −s0)", - "type": "text" - }, - { - "block_id": "p381-b25", - "global_id": 10610, - "bbox": [ - 127.59, - 309.66, - 314.7, - 325.28 - ], - "text": "Frequency\n−tx(t)\ndX(s)", - "type": "text" - }, - { - "block_id": "p381-b26", - "global_id": 10611, - "bbox": [ - 136.56, - 322.59, - 308.41, - 339.43 - ], - "text": "ds\ndifferentiation", - "type": "text" - }, - { - "block_id": "p381-b27", - "global_id": 10612, - "bbox": [ - 127.59, - 341.81, - 236.35, - 357.43 - ], - "text": "Frequency integration\nx(t)", - "type": "text" - }, - { - "block_id": "p381-b28", - "global_id": 10613, - "bbox": [ - 228.35, - 354.74, - 230.84, - 363.71 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p381-b29", - "global_id": 10614, - "bbox": [ - 292.96, - 335.9, - 308.41, - 346.86 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p381-b30", - "global_id": 10615, - "bbox": [ - 297.69, - 358.33, - 300.21, - 364.81 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p381-b31", - "global_id": 10616, - "bbox": [ - 309.91, - 348.09, - 334.94, - 357.34 - ], - "text": "X(z)dz", - "type": "text" - }, - { - "block_id": "p381-b32", - "global_id": 10617, - "bbox": [ - 127.59, - 363.3, - 305.31, - 384.83 - ], - "text": "Scaling\nx(at),a ≥0\n1\naX", - "type": "text" - }, - { - "block_id": "p381-b33", - "global_id": 10618, - "bbox": [ - 305.71, - 356.63, - 316.94, - 372.18 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p381-b34", - "global_id": 10619, - "bbox": [ - 312.96, - 375.87, - 317.44, - 384.83 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p381-b36", - "global_id": 10620, - "bbox": [ - 127.59, - 385.4, - 331.75, - 395.48 - ], - "text": "Time convolution\nx1(t) ∗x2(t)\nX1(s)X2(s)", - "type": "text" - }, - { - "block_id": "p381-b37", - "global_id": 10621, - "bbox": [ - 127.59, - 397.16, - 354.85, - 418.78 - ], - "text": "Frequency convolution\nx1(t)x2(t)\n1\n2πjX1(s) ∗X2(s)", - "type": "text" - }, - { - "block_id": "p381-b38", - "global_id": 10622, - "bbox": [ - 127.59, - 417.69, - 376.0, - 433.0 - ], - "text": "Initial value\nx(0+)\nlim\ns→∞sX(s)\n(n > m)", - "type": "text" - }, - { - "block_id": "p381-b39", - "global_id": 10623, - "bbox": [ - 127.59, - 433.08, - 429.08, - 447.82 - ], - "text": "Final value\nx(∞)\nlim\ns→0sX(s)\n[poles of sX(s) in LHP]", - "type": "text" - }, - { - "block_id": "p381-b40", - "global_id": 10624, - "bbox": [ - 127.59, - 456.77, - 516.13, - 491.06 - ], - "text": "solved for Y(s). Next we take the inverse Laplace transform of Y(s) to find the desired solution\ny(t). The following examples demonstrate the Laplace transform procedure for solving linear\ndifferential equations with constant coefficients.", - "type": "text" - }, - { - "block_id": "p381-b41", - "global_id": 10625, - "bbox": [ - 102.51, - 515.25, - 481.64, - 541.16 - ], - "text": "EXAMPLE 4.12\nLaplace Transform to Solve a Second-Order Linear\nDifferential Equation", - "type": "text" - }, - { - "block_id": "p381-b42", - "global_id": 10626, - "bbox": [ - 128.9, - 557.81, - 329.08, - 567.78 - ], - "text": "Solve the second-order linear differential equation", - "type": "text" - }, - { - "block_id": "p381-b43", - "global_id": 10627, - "bbox": [ - 251.09, - 575.2, - 380.56, - 589.7 - ], - "text": "(D2 + 5D + 6)y(t) = (D + 1)x(t)", - "type": "text" - }, - { - "block_id": "p381-b44", - "global_id": 10628, - "bbox": [ - 128.91, - 600.01, - 443.52, - 611.61 - ], - "text": "for the initial conditions y(0−) = 2 and ˙y(0−) = 1 and the input x(t) = e−4tu(t).", - "type": "text" - } - ] - }, - { - "page_num": 382, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p382-b0", - "global_id": 10629, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "362\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p382-b1", - "global_id": 10630, - "bbox": [ - 103.16, - 86.24, - 164.59, - 96.2 - ], - "text": "The equation is", - "type": "text" - }, - { - "block_id": "p382-b2", - "global_id": 10631, - "bbox": [ - 213.84, - 96.41, - 237.88, - 107.62 - ], - "text": "d2y(t)", - "type": "text" - }, - { - "block_id": "p382-b3", - "global_id": 10632, - "bbox": [ - 219.9, - 97.35, - 275.89, - 121.68 - ], - "text": "dt2\n+ 5dy(t)", - "type": "text" - }, - { - "block_id": "p382-b4", - "global_id": 10633, - "bbox": [ - 262.05, - 97.35, - 340.63, - 121.68 - ], - "text": "dt\n+ 6y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p382-b5", - "global_id": 10634, - "bbox": [ - 326.76, - 104.33, - 477.01, - 121.68 - ], - "text": "dt\n+ x(t)\n(4.22)", - "type": "text" - }, - { - "block_id": "p382-b6", - "global_id": 10635, - "bbox": [ - 103.16, - 126.41, - 116.44, - 136.37 - ], - "text": "Let", - "type": "text" - }, - { - "block_id": "p382-b7", - "global_id": 10636, - "bbox": [ - 262.28, - 137.95, - 317.91, - 148.23 - ], - "text": "y(t) ⇐⇒Y(s)", - "type": "text" - }, - { - "block_id": "p382-b8", - "global_id": 10637, - "bbox": [ - 103.17, - 157.29, - 190.59, - 167.26 - ], - "text": "Then from Eq. (4.15),", - "type": "text" - }, - { - "block_id": "p382-b9", - "global_id": 10638, - "bbox": [ - 216.42, - 166.88, - 236.2, - 177.16 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p382-b10", - "global_id": 10639, - "bbox": [ - 222.34, - 172.14, - 364.95, - 191.21 - ], - "text": "dt\n⇐⇒sY(s) −y(0−) = sY(s) −2", - "type": "text" - }, - { - "block_id": "p382-b11", - "global_id": 10640, - "bbox": [ - 103.17, - 196.22, - 117.54, - 206.18 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p382-b12", - "global_id": 10641, - "bbox": [ - 181.68, - 204.34, - 205.72, - 215.55 - ], - "text": "d2y(t)", - "type": "text" - }, - { - "block_id": "p382-b13", - "global_id": 10642, - "bbox": [ - 187.75, - 210.53, - 399.67, - 229.61 - ], - "text": "dt2\n⇐⇒s2Y(s) −sy(0−) −˙y(0−) = s2Y(s) −2s −1", - "type": "text" - }, - { - "block_id": "p382-b14", - "global_id": 10643, - "bbox": [ - 103.17, - 234.56, - 221.31, - 246.16 - ], - "text": "Moreover, for x(t) = e−4tu(t),", - "type": "text" - }, - { - "block_id": "p382-b15", - "global_id": 10644, - "bbox": [ - 154.71, - 255.32, - 261.84, - 279.76 - ], - "text": "X(s) =\n1\ns + 4\nand\ndx(t)", - "type": "text" - }, - { - "block_id": "p382-b16", - "global_id": 10645, - "bbox": [ - 247.98, - 255.64, - 424.27, - 279.76 - ], - "text": "dt\n⇐⇒sX(s) −x(0−) =\ns\ns + 4 −0 =\ns\ns + 4", - "type": "text" - }, - { - "block_id": "p382-b17", - "global_id": 10646, - "bbox": [ - 103.16, - 288.36, - 317.88, - 298.33 - ], - "text": "Taking the Laplace transform of Eq. (4.22), we obtain", - "type": "text" - }, - { - "block_id": "p382-b18", - "global_id": 10647, - "bbox": [ - 176.11, - 308.16, - 402.87, - 332.28 - ], - "text": "[s2Y(s) −2s −1] + 5[sY(s) −2] + 6Y(s) =\ns\ns + 4 +\n1\ns + 4", - "type": "text" - }, - { - "block_id": "p382-b19", - "global_id": 10648, - "bbox": [ - 103.17, - 340.54, - 477.0, - 362.86 - ], - "text": "Collecting all the terms of Y(s) and the remaining terms separately on the left-hand side, we\nobtain", - "type": "text" - }, - { - "block_id": "p382-b20", - "global_id": 10649, - "bbox": [ - 214.87, - 360.37, - 364.12, - 377.73 - ], - "text": "(s2 + 5s + 6)Y(s) −(2s + 11) = s + 1", - "type": "text" - }, - { - "block_id": "p382-b21", - "global_id": 10650, - "bbox": [ - 344.4, - 367.77, - 477.01, - 384.81 - ], - "text": "s + 4\n(4.23)", - "type": "text" - }, - { - "block_id": "p382-b22", - "global_id": 10651, - "bbox": [ - 103.16, - 390.42, - 144.91, - 400.38 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p382-b23", - "global_id": 10652, - "bbox": [ - 178.54, - 400.81, - 327.8, - 418.17 - ], - "text": "(s2 + 5s + 6)Y(s) = (2s + 11) + s + 1", - "type": "text" - }, - { - "block_id": "p382-b24", - "global_id": 10653, - "bbox": [ - 308.08, - 397.2, - 400.42, - 425.24 - ], - "text": "s + 4 = 2s2 + 20s + 45", - "type": "text" - }, - { - "block_id": "p382-b25", - "global_id": 10654, - "bbox": [ - 361.39, - 414.87, - 381.12, - 425.24 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p382-b26", - "global_id": 10655, - "bbox": [ - 103.16, - 430.87, - 117.54, - 440.83 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p382-b27", - "global_id": 10656, - "bbox": [ - 187.27, - 436.3, - 391.69, - 464.35 - ], - "text": "Y(s) =\n2s2 + 20s + 45\n(s2 + 5s + 6)(s + 4) =\n2s2 + 20s + 45\n(s + 2)(s + 3)(s + 4)", - "type": "text" - }, - { - "block_id": "p382-b28", - "global_id": 10657, - "bbox": [ - 103.16, - 470.87, - 333.07, - 480.84 - ], - "text": "Expanding the right-hand side into partial fractions yields", - "type": "text" - }, - { - "block_id": "p382-b29", - "global_id": 10658, - "bbox": [ - 231.31, - 490.36, - 281.65, - 507.72 - ], - "text": "Y(s) = 13/2", - "type": "text" - }, - { - "block_id": "p382-b30", - "global_id": 10659, - "bbox": [ - 261.98, - 490.36, - 345.13, - 514.79 - ], - "text": "s + 2 −\n3\ns + 3 −3/2", - "type": "text" - }, - { - "block_id": "p382-b31", - "global_id": 10660, - "bbox": [ - 327.96, - 504.42, - 347.68, - 514.79 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p382-b32", - "global_id": 10661, - "bbox": [ - 103.17, - 523.4, - 315.39, - 533.36 - ], - "text": "The inverse Laplace transform of this equation yields", - "type": "text" - }, - { - "block_id": "p382-b33", - "global_id": 10662, - "bbox": [ - 219.65, - 544.91, - 244.28, - 555.18 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p382-b34", - "global_id": 10663, - "bbox": [ - 246.33, - 536.9, - 258.51, - 550.53 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p382-b35", - "global_id": 10664, - "bbox": [ - 253.28, - 536.9, - 477.01, - 558.72 - ], - "text": "2 e−2t −3e−3t −3\n2e−4t\nu(t)\n(4.24)", - "type": "text" - }, - { - "block_id": "p382-b36", - "global_id": 10665, - "bbox": [ - 101.85, - 600.91, - 490.41, - 634.79 - ], - "text": "Example 4.12 demonstrates the ease with which the Laplace transform can solve linear\ndifferential equations with constant coefficients. The method is general and can solve a linear\ndifferential equation with constant coefficients of any order.", - "type": "text" - } - ] - }, - { - "page_num": 383, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p383-b0", - "global_id": 10666, - "bbox": [ - 223.7, - 62.89, - 516.12, - 71.98 - ], - "text": "4.3\nSolution of Differential and Integro-Differential Equations\n363", - "type": "text" - }, - { - "block_id": "p383-b1", - "global_id": 10667, - "bbox": [ - 127.89, - 86.19, - 449.59, - 98.32 - ], - "text": "ZERO-INPUT AND ZERO-STATE COMPONENTS OF RESPONSE", - "type": "text" - }, - { - "block_id": "p383-b2", - "global_id": 10668, - "bbox": [ - 127.59, - 102.35, - 516.16, - 184.04 - ], - "text": "The Laplace transform method gives the total response, which includes zero-input and zero-state\ncomponents. It is possible to separate the two components if we so desire. The initial condition\nterms in the response give rise to the zero-input response. For instance, in Ex. 4.12, the terms\nattributable to initial conditions y(0−) = 2 and ˙y(0−) = 1 in Eq. (4.23) generate the zero-input\nresponse. These initial condition terms are −(2s + 11), as seen in Eq. (4.23). The terms on the\nright-hand side are exclusively due to the input. Equation (4.23) is reproduced below with the\nproper labeling of the terms", - "type": "text" - }, - { - "block_id": "p383-b4", - "global_id": 10669, - "bbox": [ - 249.79, - 196.42, - 293.22, - 210.91 - ], - "text": "s2 + 5s + 6", - "type": "text" - }, - { - "block_id": "p383-b6", - "global_id": 10670, - "bbox": [ - 298.35, - 193.55, - 396.74, - 210.91 - ], - "text": "Y(s) −(2s + 11) = s + 1", - "type": "text" - }, - { - "block_id": "p383-b7", - "global_id": 10671, - "bbox": [ - 377.02, - 207.6, - 396.74, - 217.98 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p383-b8", - "global_id": 10672, - "bbox": [ - 127.59, - 226.58, - 236.18, - 242.98 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p383-b9", - "global_id": 10673, - "bbox": [ - 236.18, - 236.92, - 279.6, - 251.41 - ], - "text": "s2 + 5s + 6", - "type": "text" - }, - { - "block_id": "p383-b11", - "global_id": 10674, - "bbox": [ - 284.74, - 241.03, - 371.28, - 256.73 - ], - "text": "Y(s) =\n(2s + 11)", - "type": "text" - }, - { - "block_id": "p383-b12", - "global_id": 10675, - "bbox": [ - 316.34, - 259.17, - 369.14, - 265.15 - ], - "text": "initial condition terms", - "type": "text" - }, - { - "block_id": "p383-b13", - "global_id": 10676, - "bbox": [ - 372.38, - 234.05, - 406.26, - 251.31 - ], - "text": "+ s + 1", - "type": "text" - }, - { - "block_id": "p383-b14", - "global_id": 10677, - "bbox": [ - 381.25, - 248.1, - 411.56, - 262.9 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p383-b15", - "global_id": 10678, - "bbox": [ - 382.87, - 265.34, - 409.94, - 271.31 - ], - "text": "input terms", - "type": "text" - }, - { - "block_id": "p383-b16", - "global_id": 10679, - "bbox": [ - 127.59, - 279.46, - 169.34, - 289.42 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p383-b17", - "global_id": 10680, - "bbox": [ - 217.88, - 298.22, - 293.19, - 335.48 - ], - "text": "Y(s) =\n2s + 11\ns2 + 5s + 6\n\n\n\nZIR", - "type": "text" - }, - { - "block_id": "p383-b18", - "global_id": 10681, - "bbox": [ - 294.3, - 298.22, - 383.6, - 336.39 - ], - "text": "+\ns + 1\n(s + 4)(s2 + 5s + 6)\n\n\n\nZSR", - "type": "text" - }, - { - "block_id": "p383-b19", - "global_id": 10682, - "bbox": [ - 237.55, - 346.57, - 245.32, - 356.54 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p383-b20", - "global_id": 10683, - "bbox": [ - 247.37, - 332.59, - 306.7, - 364.02 - ], - "text": "7\ns + 2 −\n5\ns + 3", - "type": "text" - }, - { - "block_id": "p383-b21", - "global_id": 10684, - "bbox": [ - 307.9, - 332.59, - 313.33, - 342.55 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p383-b22", - "global_id": 10685, - "bbox": [ - 314.87, - 346.57, - 322.64, - 356.54 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p383-b23", - "global_id": 10686, - "bbox": [ - 324.19, - 332.59, - 353.23, - 349.97 - ], - "text": "−1/2", - "type": "text" - }, - { - "block_id": "p383-b24", - "global_id": 10687, - "bbox": [ - 332.16, - 339.59, - 416.67, - 364.02 - ], - "text": "s + 2 +\n2\ns + 3 −3/2", - "type": "text" - }, - { - "block_id": "p383-b25", - "global_id": 10688, - "bbox": [ - 399.49, - 353.65, - 419.21, - 364.02 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p383-b26", - "global_id": 10689, - "bbox": [ - 420.41, - 332.59, - 425.84, - 342.55 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p383-b27", - "global_id": 10690, - "bbox": [ - 127.59, - 374.18, - 332.38, - 384.15 - ], - "text": "Taking the inverse transform of this equation yields", - "type": "text" - }, - { - "block_id": "p383-b28", - "global_id": 10691, - "bbox": [ - 205.71, - 395.68, - 230.33, - 405.95 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p383-b30", - "global_id": 10692, - "bbox": [ - 232.38, - 387.67, - 309.55, - 421.54 - ], - "text": "7e−2t −5e−3t\nu(t)\n\n\n\nZIR", - "type": "text" - }, - { - "block_id": "p383-b31", - "global_id": 10693, - "bbox": [ - 310.64, - 395.68, - 318.41, - 405.64 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p383-b33", - "global_id": 10694, - "bbox": [ - 323.54, - 394.33, - 335.99, - 405.64 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p383-b34", - "global_id": 10695, - "bbox": [ - 319.52, - 387.67, - 438.01, - 421.6 - ], - "text": "2e−2t + 2e−3t −3\n2e−4t\nu(t)\n\n\n\nZSR", - "type": "text" - }, - { - "block_id": "p383-b35", - "global_id": 10696, - "bbox": [ - 127.59, - 442.54, - 416.83, - 457.3 - ], - "text": "4.3-1 Comments on Initial Conditions at 0−and at 0+", - "type": "text" - }, - { - "block_id": "p383-b36", - "global_id": 10697, - "bbox": [ - 127.59, - 461.79, - 516.15, - 569.03 - ], - "text": "The initial conditions in Ex. 4.12 are y(0−) = 2 and ˙y(0−) = 1. If we let t = 0 in the total response\nin Eq. (4.24), we find y(0) = 2 and ˙y(0) = 2, which is at odds with the given initial conditions.\nWhy? Because the initial conditions are given at t = 0−(just before the input is applied), when only\nthe zero-input response is present. The zero-state response is the result of the input x(t) applied\nat t = 0. Hence, this component does not exist at t = 0−. Consequently, the initial conditions at\nt = 0−are satisfied by the zero-input response, not by the total response. We can readily verify in\nthis example that the zero-input response does indeed satisfy the given initial conditions at t = 0−.\nIt is the total response that satisfies the initial conditions at t = 0+, which are generally different\nfrom the initial conditions at 0−.", - "type": "text" - }, - { - "block_id": "p383-b37", - "global_id": 10698, - "bbox": [ - 127.59, - 570.61, - 516.15, - 640.76 - ], - "text": "There also exists a L+ version of the Laplace transform, which uses the initial conditions at\nt = 0+ rather than at 0−(as in our present L−version). The L+ version, which was in vogue till the\nearly 1960s, is identical to the L−version except the limits of Laplace integral [Eq. (4.7)] are from\n0+ to ∞. Hence, by definition, the origin t = 0 is excluded from the domain. This version, still used\nin some math books, has some serious difficulties. For instance, the Laplace transform of δ(t) is\nzero because δ(t) = 0 for t ≥0+. Moreover, this approach is rather clumsy in the theoretical study", - "type": "text" - } - ] - }, - { - "page_num": 384, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p384-b0", - "global_id": 10699, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "364\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p384-b1", - "global_id": 10700, - "bbox": [ - 101.84, - 85.82, - 490.41, - 192.92 - ], - "text": "of linear systems because the response obtained cannot be separated into zero-input and zero-state\ncomponents. As we know, the zero-state component represents the system response as an explicit\nfunction of the input, and without knowing this component, it is not possible to assess the effect\nof the input on the system response in a general way. The L+ version can separate the response\nin terms of the natural and the forced components, which are not as interesting as the zero-input\nand the zero-state components. Note that we can always determine the natural and the forced\ncomponents from the zero-input and the zero-state components [e.g., Eq. (2.44) from Eq. (2.43)],\nbut the converse is not true. Because of these and some other problems, electrical engineers\n(wisely) started discarding the L+ version in the early 1960s.", - "type": "text" - }, - { - "block_id": "p384-b2", - "global_id": 10701, - "bbox": [ - 101.84, - 193.41, - 490.39, - 275.11 - ], - "text": "It is interesting to note the time-domain duals of these two Laplace versions. The classical\nmethod is the dual of the L+ method, and the convolution (zero-input/zero-state) method is the dual\nof the L−method. The first pair uses the initial conditions at 0+, and the second pair uses those at\nt = 0−. The first pair (the classical method and the L+ version) is awkward in the theoretical study\nof linear system analysis. It was no coincidence that the L−version was adopted immediately\nafter the introduction to the electrical engineering community of state-space analysis (which uses\nzero-input/zero-state separation of the output).", - "type": "text" - }, - { - "block_id": "p384-b3", - "global_id": 10702, - "bbox": [ - 107.82, - 311.89, - 457.8, - 337.79 - ], - "text": "DRILL 4.7\nLaplace Transform to Solve a Second-Order Linear\nDifferential Equation", - "type": "text" - }, - { - "block_id": "p384-b4", - "global_id": 10703, - "bbox": [ - 107.82, - 346.91, - 130.35, - 356.87 - ], - "text": "Solve", - "type": "text" - }, - { - "block_id": "p384-b5", - "global_id": 10704, - "bbox": [ - 217.38, - 355.03, - 241.42, - 366.24 - ], - "text": "d2y(t)", - "type": "text" - }, - { - "block_id": "p384-b6", - "global_id": 10705, - "bbox": [ - 223.44, - 355.96, - 279.43, - 380.29 - ], - "text": "dt2\n+ 4dy(t)", - "type": "text" - }, - { - "block_id": "p384-b7", - "global_id": 10706, - "bbox": [ - 265.58, - 355.96, - 349.15, - 380.29 - ], - "text": "dt\n+ 3y(t) = 2dx(t)", - "type": "text" - }, - { - "block_id": "p384-b8", - "global_id": 10707, - "bbox": [ - 335.28, - 362.95, - 376.05, - 380.29 - ], - "text": "dt\n+ x(t)", - "type": "text" - }, - { - "block_id": "p384-b9", - "global_id": 10708, - "bbox": [ - 107.82, - 384.55, - 410.1, - 396.16 - ], - "text": "for the input x(t) = u(t). The initial conditions are y(0−) = 1 and ˙y(0−) = 2.", - "type": "text" - }, - { - "block_id": "p384-b10", - "global_id": 10709, - "bbox": [ - 108.09, - 409.68, - 162.68, - 420.64 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p384-b11", - "global_id": 10710, - "bbox": [ - 107.82, - 426.28, - 139.17, - 437.9 - ], - "text": "y(t) = 1", - "type": "text" - }, - { - "block_id": "p384-b12", - "global_id": 10711, - "bbox": [ - 135.69, - 424.01, - 228.19, - 440.81 - ], - "text": "3(1 + 9e−t −7e−3t)u(t)", - "type": "text" - }, - { - "block_id": "p384-b13", - "global_id": 10712, - "bbox": [ - 76.77, - 490.18, - 429.95, - 502.13 - ], - "text": "EXAMPLE 4.13\nLaplace Transform to Solve an Electric Circuit", - "type": "text" - }, - { - "block_id": "p384-b14", - "global_id": 10713, - "bbox": [ - 103.16, - 518.39, - 477.0, - 540.72 - ], - "text": "In the circuit of Fig. 4.7a, the switch is in the closed position for a long time before t = 0, when\nit is opened instantaneously. Find the inductor current y(t) for t ≥0.", - "type": "text" - }, - { - "block_id": "p384-b15", - "global_id": 10714, - "bbox": [ - 103.16, - 563.63, - 477.01, - 621.41 - ], - "text": "When the switch is in the closed position (for a long time), the inductor current is 2 amperes\nand the capacitor voltage is 10 volts. When the switch is opened, the circuit is equivalent to\nthat depicted in Fig. 4.7b, with the initial inductor current y(0−) = 2 and the initial capacitor\nvoltage vC(0−) = 10. The input voltage is 10 volts, starting at t = 0, and, therefore, can be\nrepresented by 10u(t).", - "type": "text" - } - ] - }, - { - "page_num": 385, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p385-b0", - "global_id": 10715, - "bbox": [ - 223.7, - 62.89, - 516.12, - 71.98 - ], - "text": "4.3\nSolution of Differential and Integro-Differential Equations\n365", - "type": "text" - }, - { - "block_id": "p385-b1", - "global_id": 10716, - "bbox": [ - 410.92, - 134.21, - 422.34, - 142.29 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p385-b2", - "global_id": 10717, - "bbox": [ - 392.25, - 90.76, - 453.74, - 99.06 - ], - "text": "2 \t\n1 H", - "type": "text" - }, - { - "block_id": "p385-b3", - "global_id": 10718, - "bbox": [ - 324.63, - 139.87, - 344.18, - 147.95 - ], - "text": "10u(t)", - "type": "text" - }, - { - "block_id": "p385-b4", - "global_id": 10719, - "bbox": [ - 235.29, - 156.77, - 252.18, - 165.07 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p385-b5", - "global_id": 10720, - "bbox": [ - 128.9, - 135.91, - 500.09, - 148.53 - ], - "text": "10 V\nvC(t)", - "type": "text" - }, - { - "block_id": "p385-b6", - "global_id": 10721, - "bbox": [ - 413.36, - 183.59, - 422.69, - 191.59 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p385-b9", - "global_id": 10722, - "bbox": [ - 258.99, - 97.46, - 271.21, - 105.75 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p385-b10", - "global_id": 10723, - "bbox": [ - 234.51, - 126.21, - 246.73, - 134.5 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p385-b11", - "global_id": 10724, - "bbox": [ - 184.87, - 97.75, - 196.64, - 105.75 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p385-b12", - "global_id": 10725, - "bbox": [ - 223.96, - 183.59, - 232.84, - 191.59 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p385-b13", - "global_id": 10726, - "bbox": [ - 283.63, - 144.0, - 468.8, - 161.81 - ], - "text": "1\n5 F\n1\n5 F", - "type": "text" - }, - { - "block_id": "p385-b14", - "global_id": 10727, - "bbox": [ - 303.35, - 220.57, - 314.77, - 228.65 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p385-b15", - "global_id": 10728, - "bbox": [ - 159.92, - 116.6, - 171.34, - 124.68 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p385-b16", - "global_id": 10729, - "bbox": [ - 420.36, - 281.63, - 422.59, - 289.63 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p385-b17", - "global_id": 10730, - "bbox": [ - 288.66, - 207.29, - 292.66, - 215.29 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p385-b18", - "global_id": 10731, - "bbox": [ - 288.06, - 282.05, - 292.06, - 290.05 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p385-b19", - "global_id": 10732, - "bbox": [ - 312.42, - 311.15, - 321.3, - 319.15 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p385-b20", - "global_id": 10733, - "bbox": [ - 128.9, - 325.85, - 341.25, - 335.08 - ], - "text": "Figure 4.7 Analysis of a network with a switching action.", - "type": "text" - }, - { - "block_id": "p385-b21", - "global_id": 10734, - "bbox": [ - 146.84, - 350.33, - 330.31, - 360.29 - ], - "text": "The loop equation of the circuit in Fig. 4.7b is", - "type": "text" - }, - { - "block_id": "p385-b22", - "global_id": 10735, - "bbox": [ - 239.21, - 369.69, - 258.99, - 379.97 - ], - "text": "dy(t)", - "type": "text" - }, - { - "block_id": "p385-b23", - "global_id": 10736, - "bbox": [ - 245.14, - 376.68, - 306.69, - 394.03 - ], - "text": "dt\n+ 2y(t) + 5", - "type": "text" - }, - { - "block_id": "p385-b24", - "global_id": 10737, - "bbox": [ - 307.79, - 363.11, - 319.56, - 375.39 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p385-b25", - "global_id": 10738, - "bbox": [ - 313.06, - 387.99, - 325.62, - 394.96 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p385-b26", - "global_id": 10739, - "bbox": [ - 327.23, - 376.68, - 502.75, - 387.05 - ], - "text": "y(τ)dτ = 10u(t)\n(4.25)", - "type": "text" - }, - { - "block_id": "p385-b27", - "global_id": 10740, - "bbox": [ - 128.9, - 403.39, - 135.54, - 413.35 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p385-b28", - "global_id": 10741, - "bbox": [ - 288.02, - 414.93, - 343.65, - 425.2 - ], - "text": "y(t) ⇐⇒Y(s)", - "type": "text" - }, - { - "block_id": "p385-b29", - "global_id": 10742, - "bbox": [ - 128.91, - 433.8, - 261.94, - 451.6 - ], - "text": "then\ndy(t)", - "type": "text" - }, - { - "block_id": "p385-b30", - "global_id": 10743, - "bbox": [ - 248.09, - 446.59, - 390.7, - 465.66 - ], - "text": "dt\n⇐⇒sY(s) −y(0−) = sY(s) −2", - "type": "text" - }, - { - "block_id": "p385-b31", - "global_id": 10744, - "bbox": [ - 128.91, - 470.2, - 207.75, - 480.16 - ], - "text": "and [see Eq. (4.16)]", - "type": "text" - }, - { - "block_id": "p385-b32", - "global_id": 10745, - "bbox": [ - 239.28, - 478.79, - 251.05, - 491.07 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p385-b33", - "global_id": 10746, - "bbox": [ - 244.55, - 503.66, - 257.1, - 510.64 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p385-b34", - "global_id": 10747, - "bbox": [ - 258.71, - 485.36, - 329.96, - 502.62 - ], - "text": "y(τ)dτ ⇐⇒Y(s)", - "type": "text" - }, - { - "block_id": "p385-b35", - "global_id": 10748, - "bbox": [ - 319.22, - 492.35, - 340.48, - 509.7 - ], - "text": "s\n+", - "type": "text" - }, - { - "block_id": "p385-b36", - "global_id": 10749, - "bbox": [ - 343.23, - 476.26, - 357.87, - 488.61 - ], - "text": "$ 0−", - "type": "text" - }, - { - "block_id": "p385-b37", - "global_id": 10750, - "bbox": [ - 347.79, - 484.28, - 389.99, - 497.04 - ], - "text": "−∞y(τ)dτ", - "type": "text" - }, - { - "block_id": "p385-b38", - "global_id": 10751, - "bbox": [ - 365.27, - 499.73, - 369.15, - 509.7 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p385-b39", - "global_id": 10752, - "bbox": [ - 128.91, - 520.7, - 325.1, - 531.07 - ], - "text": "Because y(t) is the capacitor current, the integral", - "type": "text" - }, - { - "block_id": "p385-b40", - "global_id": 10753, - "bbox": [ - 327.9, - 512.67, - 342.55, - 525.01 - ], - "text": "$ 0−", - "type": "text" - }, - { - "block_id": "p385-b41", - "global_id": 10754, - "bbox": [ - 128.9, - 519.47, - 502.75, - 543.02 - ], - "text": "−∞y(τ)dτ is qC(0−), the capacitor charge\nat t = 0−, which is given by C times the capacitor voltage at t = 0−. Therefore,", - "type": "text" - }, - { - "block_id": "p385-b42", - "global_id": 10755, - "bbox": [ - 221.64, - 548.14, - 239.61, - 560.48 - ], - "text": "# 0−", - "type": "text" - }, - { - "block_id": "p385-b43", - "global_id": 10756, - "bbox": [ - 226.91, - 573.02, - 239.46, - 579.99 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p385-b44", - "global_id": 10757, - "bbox": [ - 241.72, - 559.98, - 374.59, - 572.78 - ], - "text": "y(τ)dτ = qC(0−) = CvC(0−) = 1", - "type": "text" - }, - { - "block_id": "p385-b45", - "global_id": 10758, - "bbox": [ - 371.1, - 561.7, - 410.02, - 574.89 - ], - "text": "5(10) = 2", - "type": "text" - }, - { - "block_id": "p385-b46", - "global_id": 10759, - "bbox": [ - 128.91, - 588.41, - 272.54, - 602.41 - ], - "text": "and\n# t", - "type": "text" - }, - { - "block_id": "p385-b47", - "global_id": 10760, - "bbox": [ - 266.04, - 615.0, - 278.59, - 621.98 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p385-b48", - "global_id": 10761, - "bbox": [ - 280.19, - 596.7, - 351.46, - 613.96 - ], - "text": "y(τ)dτ ⇐⇒Y(s)", - "type": "text" - }, - { - "block_id": "p385-b49", - "global_id": 10762, - "bbox": [ - 340.71, - 597.11, - 369.69, - 621.03 - ], - "text": "s\n+ 2", - "type": "text" - }, - { - "block_id": "p385-b50", - "global_id": 10763, - "bbox": [ - 365.26, - 611.07, - 369.14, - 621.03 - ], - "text": "s", - "type": "text" - } - ] - }, - { - "page_num": 386, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p386-b0", - "global_id": 10764, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "366\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p386-b1", - "global_id": 10765, - "bbox": [ - 103.16, - 86.24, - 333.38, - 96.21 - ], - "text": "Using these results, the Laplace transform of Eq. (4.25) is", - "type": "text" - }, - { - "block_id": "p386-b2", - "global_id": 10766, - "bbox": [ - 213.04, - 105.78, - 318.5, - 123.14 - ], - "text": "sY(s) −2 + 2Y(s) + 5Y(s)", - "type": "text" - }, - { - "block_id": "p386-b3", - "global_id": 10767, - "bbox": [ - 305.27, - 106.2, - 341.72, - 130.12 - ], - "text": "s\n+ 10", - "type": "text" - }, - { - "block_id": "p386-b4", - "global_id": 10768, - "bbox": [ - 334.81, - 106.2, - 365.94, - 130.12 - ], - "text": "s = 10", - "type": "text" - }, - { - "block_id": "p386-b5", - "global_id": 10769, - "bbox": [ - 103.16, - 120.15, - 362.9, - 146.2 - ], - "text": "s\nor", - "type": "text" - }, - { - "block_id": "p386-b6", - "global_id": 10770, - "bbox": [ - 253.32, - 143.66, - 290.1, - 160.61 - ], - "text": "s + 2 + 5", - "type": "text" - }, - { - "block_id": "p386-b7", - "global_id": 10771, - "bbox": [ - 285.67, - 157.61, - 289.54, - 167.57 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p386-b8", - "global_id": 10772, - "bbox": [ - 291.29, - 136.24, - 296.72, - 146.2 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p386-b9", - "global_id": 10773, - "bbox": [ - 297.83, - 150.23, - 332.29, - 160.61 - ], - "text": "Y(s) = 2", - "type": "text" - }, - { - "block_id": "p386-b10", - "global_id": 10774, - "bbox": [ - 103.17, - 174.87, - 117.54, - 184.83 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p386-b11", - "global_id": 10775, - "bbox": [ - 252.44, - 182.64, - 326.54, - 206.76 - ], - "text": "Y(s) =\n2s\ns2 + 2s + 5", - "type": "text" - }, - { - "block_id": "p386-b12", - "global_id": 10776, - "bbox": [ - 103.17, - 212.03, - 477.02, - 234.36 - ], - "text": "To find the inverse Laplace transform of Y(s), we use pair 10c (Table 4.1) with values A = 2,\nB = 0, a = 1, and c = 5. This yields", - "type": "text" - }, - { - "block_id": "p386-b13", - "global_id": 10777, - "bbox": [ - 160.04, - 250.05, - 173.96, - 260.33 - ], - "text": "r =", - "type": "text" - }, - { - "block_id": "p386-b15", - "global_id": 10778, - "bbox": [ - 185.47, - 241.16, - 213.93, - 263.55 - ], - "text": "20\n4 =\n√", - "type": "text" - }, - { - "block_id": "p386-b16", - "global_id": 10779, - "bbox": [ - 213.93, - 250.05, - 247.27, - 260.43 - ], - "text": "5,\nb =", - "type": "text" - }, - { - "block_id": "p386-b18", - "global_id": 10780, - "bbox": [ - 257.58, - 242.04, - 381.65, - 260.43 - ], - "text": "c −a2 = 2\nand\nθ = tan−1 2", - "type": "text" - }, - { - "block_id": "p386-b19", - "global_id": 10781, - "bbox": [ - 378.16, - 242.04, - 386.85, - 263.24 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p386-b20", - "global_id": 10782, - "bbox": [ - 388.91, - 248.83, - 419.64, - 260.43 - ], - "text": "= 26.6◦", - "type": "text" - }, - { - "block_id": "p386-b21", - "global_id": 10783, - "bbox": [ - 103.16, - 273.52, - 144.91, - 283.49 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p386-b22", - "global_id": 10784, - "bbox": [ - 224.58, - 285.07, - 249.2, - 295.35 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p386-b23", - "global_id": 10785, - "bbox": [ - 251.25, - 276.18, - 259.68, - 286.14 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p386-b24", - "global_id": 10786, - "bbox": [ - 259.68, - 281.37, - 355.6, - 295.45 - ], - "text": "5e−t cos(2t + 26.6◦)u(t)", - "type": "text" - }, - { - "block_id": "p386-b25", - "global_id": 10787, - "bbox": [ - 103.17, - 304.41, - 245.97, - 314.37 - ], - "text": "This response is shown in Fig. 4.7c.", - "type": "text" - }, - { - "block_id": "p386-b26", - "global_id": 10788, - "bbox": [ - 103.17, - 323.92, - 477.03, - 406.03 - ], - "text": "Comment. In our discussion so far, we have multiplied input signals by u(t), implying that\nthe signals are zero prior to t = 0. This is needlessly restrictive. These signals can have any\narbitrary value prior to t = 0. As long as the initial conditions at t = 0 are specified, we need\nonly the knowledge of the input for t ≥0 to compute the response for t ≥0. Some authors use\nthe notation 1(t) to denote a function that is equal to u(t) for t ≥0 and that has arbitrary value\nfor negative t. We have abstained from this usage to avoid needless confusion caused by the\nintroduction of a new function, which is very similar to u(t).", - "type": "text" - }, - { - "block_id": "p386-b27", - "global_id": 10789, - "bbox": [ - 101.84, - 454.85, - 244.13, - 466.81 - ], - "text": "4.3-2 Zero-State Response", - "type": "text" - }, - { - "block_id": "p386-b28", - "global_id": 10790, - "bbox": [ - 101.84, - 472.84, - 347.58, - 482.9 - ], - "text": "Consider an Nth-order LTIC system specified by the equation", - "type": "text" - }, - { - "block_id": "p386-b29", - "global_id": 10791, - "bbox": [ - 254.1, - 498.14, - 338.13, - 508.42 - ], - "text": "Q(D)y(t) = P(D)x(t)", - "type": "text" - }, - { - "block_id": "p386-b30", - "global_id": 10792, - "bbox": [ - 101.84, - 524.18, - 110.14, - 534.14 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p386-b31", - "global_id": 10793, - "bbox": [ - 114.23, - 545.27, - 490.38, - 560.53 - ], - "text": "(DN + a1DN−1 + · · · + aN−1D + aN)y(t) = (b0DN + b1DN−1 + · · · + bN−1D + bN)x(t)\n(4.26)", - "type": "text" - }, - { - "block_id": "p386-b32", - "global_id": 10794, - "bbox": [ - 101.85, - 575.42, - 490.4, - 609.28 - ], - "text": "We shall now find the general expression for the zero-state response of an LTIC system.\nZero-state response y(t), by definition, is the system response to an input when the system is\ninitially relaxed (in zero state). Therefore, y(t) satisfies Eq. (4.26) with zero initial conditions", - "type": "text" - }, - { - "block_id": "p386-b33", - "global_id": 10795, - "bbox": [ - 202.03, - 622.81, - 390.21, - 634.91 - ], - "text": "y(0−) = ˙y(0−) = ¨y(0−) = · · · = y(N−1)(0−) = 0", - "type": "text" - } - ] - }, - { - "page_num": 387, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p387-b0", - "global_id": 10796, - "bbox": [ - 223.7, - 62.89, - 516.12, - 71.98 - ], - "text": "4.3\nSolution of Differential and Integro-Differential Equations\n367", - "type": "text" - }, - { - "block_id": "p387-b1", - "global_id": 10797, - "bbox": [ - 127.59, - 85.46, - 288.93, - 95.84 - ], - "text": "Moreover, the input x(t) is causal so that", - "type": "text" - }, - { - "block_id": "p387-b2", - "global_id": 10798, - "bbox": [ - 227.71, - 104.33, - 416.01, - 116.43 - ], - "text": "x(0−) = ˙x(0−) = ¨x(0−) = · · · = x(N−1)(0−) = 0", - "type": "text" - }, - { - "block_id": "p387-b3", - "global_id": 10799, - "bbox": [ - 127.59, - 127.06, - 140.87, - 137.02 - ], - "text": "Let", - "type": "text" - }, - { - "block_id": "p387-b4", - "global_id": 10800, - "bbox": [ - 238.99, - 138.6, - 404.75, - 148.98 - ], - "text": "y(t) ⇐⇒Y(s)\nand\nx(t) ⇐⇒X(s)", - "type": "text" - }, - { - "block_id": "p387-b5", - "global_id": 10801, - "bbox": [ - 127.6, - 157.28, - 263.43, - 167.25 - ], - "text": "Because of zero initial conditions,", - "type": "text" - }, - { - "block_id": "p387-b6", - "global_id": 10802, - "bbox": [ - 264.68, - 173.61, - 312.4, - 191.88 - ], - "text": "Dry(t) = dr", - "type": "text" - }, - { - "block_id": "p387-b7", - "global_id": 10803, - "bbox": [ - 303.1, - 180.1, - 378.49, - 198.95 - ], - "text": "dtr y(t) ⇐⇒srY(s)", - "type": "text" - }, - { - "block_id": "p387-b8", - "global_id": 10804, - "bbox": [ - 264.3, - 200.04, - 312.79, - 218.3 - ], - "text": "Dkx(t) = dk", - "type": "text" - }, - { - "block_id": "p387-b9", - "global_id": 10805, - "bbox": [ - 303.1, - 206.52, - 379.42, - 225.38 - ], - "text": "dtk x(t) ⇐⇒skX(s)", - "type": "text" - }, - { - "block_id": "p387-b10", - "global_id": 10806, - "bbox": [ - 127.59, - 232.06, - 338.93, - 242.02 - ], - "text": "Therefore, the Laplace transform of Eq. (4.26) yields", - "type": "text" - }, - { - "block_id": "p387-b11", - "global_id": 10807, - "bbox": [ - 159.06, - 248.11, - 484.67, - 263.38 - ], - "text": "(sN + a1sN−1 + · · · + aN−1s + aN)Y(s) = (b0sN + b1sN−1 + · · · + bN−1s + bN)X(s)", - "type": "text" - }, - { - "block_id": "p387-b12", - "global_id": 10808, - "bbox": [ - 127.6, - 273.24, - 135.89, - 283.2 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p387-b13", - "global_id": 10809, - "bbox": [ - 203.62, - 278.67, - 369.57, - 299.55 - ], - "text": "Y(s) = b0sN + b1sN−1 + · · · + bN−1s + bN", - "type": "text" - }, - { - "block_id": "p387-b14", - "global_id": 10810, - "bbox": [ - 238.77, - 293.46, - 365.08, - 307.49 - ], - "text": "sN + a1sN−1 + · · · + aN−1s + aN", - "type": "text" - }, - { - "block_id": "p387-b15", - "global_id": 10811, - "bbox": [ - 371.68, - 282.29, - 420.53, - 299.55 - ], - "text": "X(s) = P(s)", - "type": "text" - }, - { - "block_id": "p387-b16", - "global_id": 10812, - "bbox": [ - 402.58, - 289.27, - 440.11, - 306.62 - ], - "text": "Q(s)X(s)", - "type": "text" - }, - { - "block_id": "p387-b17", - "global_id": 10813, - "bbox": [ - 127.6, - 312.56, - 405.93, - 322.94 - ], - "text": "But we have shown in Eq. (4.18) that Y(s) = H(s)X(s). Consequently,", - "type": "text" - }, - { - "block_id": "p387-b18", - "global_id": 10814, - "bbox": [ - 296.01, - 331.2, - 345.97, - 348.46 - ], - "text": "H(s) = P(s)", - "type": "text" - }, - { - "block_id": "p387-b19", - "global_id": 10815, - "bbox": [ - 328.01, - 338.6, - 516.13, - 355.53 - ], - "text": "Q(s)\n(4.27)", - "type": "text" - }, - { - "block_id": "p387-b20", - "global_id": 10816, - "bbox": [ - 127.59, - 363.83, - 516.15, - 385.76 - ], - "text": "This is the transfer function of a linear differential system specified in Eq. (4.26). The same result\nhas been derived earlier in Eq. (2.41) using an alternate (time-domain) approach.", - "type": "text" - }, - { - "block_id": "p387-b21", - "global_id": 10817, - "bbox": [ - 127.6, - 387.33, - 516.14, - 421.62 - ], - "text": "We have shown that Y(s), the Laplace transform of the zero-state response y(t), is the product\nof X(s) and H(s), where X(s) is the Laplace transform of the input x(t) and H(s) is the system\ntransfer function [relating the particular output y(t) to the input x(t)].", - "type": "text" - }, - { - "block_id": "p387-b22", - "global_id": 10818, - "bbox": [ - 127.59, - 436.02, - 516.14, - 498.0 - ], - "text": "INTUITIVE INTERPRETATION OF THE LAPLACE TRANSFORM\nSo far we have treated the Laplace transform as a machine that converts linear integro-differential\nequations into algebraic equations. There is no physical understanding of how this is accomplished\nor what it means. We now discuss a more intuitive interpretation and meaning of the Laplace\ntransform.", - "type": "text" - }, - { - "block_id": "p387-b23", - "global_id": 10819, - "bbox": [ - 127.59, - 496.38, - 516.15, - 533.86 - ], - "text": "In Ch. 2, Eq. (2.38), we showed that LTI system response to an everlasting exponential est is\nH(s)est. If we could express every signal as a linear combination of everlasting exponentials of the\nform est, we could readily obtain the system response to any input. For example, if", - "type": "text" - }, - { - "block_id": "p387-b24", - "global_id": 10820, - "bbox": [ - 284.84, - 553.6, - 309.5, - 563.88 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p387-b25", - "global_id": 10821, - "bbox": [ - 311.54, - 543.65, - 325.64, - 554.09 - ], - "text": "K\n\"", - "type": "text" - }, - { - "block_id": "p387-b26", - "global_id": 10822, - "bbox": [ - 312.52, - 567.99, - 324.66, - 575.25 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p387-b27", - "global_id": 10823, - "bbox": [ - 326.74, - 552.1, - 358.26, - 564.68 - ], - "text": "X(si)esit", - "type": "text" - }, - { - "block_id": "p387-b28", - "global_id": 10824, - "bbox": [ - 127.59, - 583.26, - 370.79, - 593.64 - ], - "text": "the response of an LTIC system to such input x(t) is given by", - "type": "text" - }, - { - "block_id": "p387-b29", - "global_id": 10825, - "bbox": [ - 274.17, - 613.38, - 298.79, - 623.65 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p387-b30", - "global_id": 10826, - "bbox": [ - 300.83, - 603.42, - 314.93, - 613.87 - ], - "text": "K\n\"", - "type": "text" - }, - { - "block_id": "p387-b31", - "global_id": 10827, - "bbox": [ - 301.82, - 627.77, - 313.95, - 635.03 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p387-b32", - "global_id": 10828, - "bbox": [ - 316.04, - 611.88, - 368.93, - 624.46 - ], - "text": "X(si)H(si)esit", - "type": "text" - } - ] - }, - { - "page_num": 388, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p388-b0", - "global_id": 10829, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "368\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p388-b1", - "global_id": 10830, - "bbox": [ - 101.84, - 85.82, - 490.39, - 119.69 - ], - "text": "Unfortunately, the class of signals that can be expressed in this form is very small. However,\nwe can express almost all signals of practical utility as a sum of everlasting exponentials over a\ncontinuum of frequencies. This is precisely what the Laplace transform in Eq. (4.2) does.", - "type": "text" - }, - { - "block_id": "p388-b2", - "global_id": 10831, - "bbox": [ - 240.4, - 130.23, - 283.01, - 154.25 - ], - "text": "x(t) =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p388-b3", - "global_id": 10832, - "bbox": [ - 285.32, - 123.25, - 312.73, - 135.52 - ], - "text": "# c+j∞", - "type": "text" - }, - { - "block_id": "p388-b4", - "global_id": 10833, - "bbox": [ - 290.58, - 148.12, - 308.17, - 155.31 - ], - "text": "c−j∞", - "type": "text" - }, - { - "block_id": "p388-b5", - "global_id": 10834, - "bbox": [ - 314.34, - 133.0, - 490.38, - 147.18 - ], - "text": "X(s)est ds\n(4.28)", - "type": "text" - }, - { - "block_id": "p388-b6", - "global_id": 10835, - "bbox": [ - 101.85, - 163.82, - 490.39, - 186.15 - ], - "text": "Invoking the linearity property of the Laplace transform, we can find the system response y(t) to\ninput x(t) in Eq. (4.28) as†", - "type": "text" - }, - { - "block_id": "p388-b7", - "global_id": 10836, - "bbox": [ - 197.41, - 197.77, - 240.0, - 221.79 - ], - "text": "y(t) =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p388-b8", - "global_id": 10837, - "bbox": [ - 242.3, - 190.77, - 271.83, - 203.06 - ], - "text": "# c′+j∞", - "type": "text" - }, - { - "block_id": "p388-b9", - "global_id": 10838, - "bbox": [ - 247.56, - 214.44, - 267.27, - 222.85 - ], - "text": "c′−j∞", - "type": "text" - }, - { - "block_id": "p388-b10", - "global_id": 10839, - "bbox": [ - 273.44, - 200.54, - 490.38, - 214.71 - ], - "text": "X(s)H(s)est ds = L−1X(s)H(s)\n(4.29)", - "type": "text" - }, - { - "block_id": "p388-b11", - "global_id": 10840, - "bbox": [ - 101.85, - 231.71, - 133.01, - 241.67 - ], - "text": "Clearly,", - "type": "text" - }, - { - "block_id": "p388-b12", - "global_id": 10841, - "bbox": [ - 262.99, - 243.24, - 329.26, - 253.52 - ], - "text": "Y(s) = X(s)H(s)", - "type": "text" - }, - { - "block_id": "p388-b13", - "global_id": 10842, - "bbox": [ - 101.85, - 262.02, - 490.4, - 307.85 - ], - "text": "We can now represent the transformed version of the system, as depicted in Fig. 4.8a. The input\nX(s) is the Laplace transform of x(t), and the output Y(s) is the Laplace transform of (the zero-input\nresponse) y(t). The system is described by the transfer function H(s). The output Y(s) is the\nproduct X(s)H(s).", - "type": "text" - }, - { - "block_id": "p388-b14", - "global_id": 10843, - "bbox": [ - 101.84, - 308.42, - 490.4, - 439.35 - ], - "text": "Recall that s is the complex frequency of est. This explains why the Laplace transform\nmethod is also called the frequency-domain method. Note that X(s),Y(s), and H(s) are the\nfrequency-domain representations of x(t),y(t), and h(t), respectively. We may view the boxes\nmarked L and L−1 in Fig. 4.8a as the interfaces that convert the time-domain entities into the\ncorresponding frequency-domain entities, and vice versa. All real-life signals begin in the time\ndomain, and the final answers must also be in the time domain. First, we convert the time-domain\ninput(s) into the frequency-domain counterparts. The problem itself is solved in the frequency\ndomain, resulting in the answer Y(s), also in the frequency domain. Finally, we convert Y(s) to\ny(t). Solving the problem is relatively simpler in the frequency domain than in the time domain.\nHenceforth, we shall omit the explicit representation of the interface boxes L and L−1, representing\nsignals and systems in the frequency domain, as shown in Fig. 4.8b.", - "type": "text" - }, - { - "block_id": "p388-b15", - "global_id": 10844, - "bbox": [ - 125.76, - 577.32, - 348.12, - 586.56 - ], - "text": "Figure 4.8 Alternate interpretation of the Laplace transform.", - "type": "text" - }, - { - "block_id": "p388-b16", - "global_id": 10845, - "bbox": [ - 101.84, - 610.24, - 490.38, - 633.41 - ], - "text": "† Recall that H(s) has its own region of validity. Hence, the limits of integration for the integral in Eq. (4.28)\nare modified in Eq. (4.29) to accommodate the region of existence (validity) of X(s) as well as H(s).", - "type": "text" - } - ] - }, - { - "page_num": 389, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p389-b0", - "global_id": 10846, - "bbox": [ - 223.7, - 62.89, - 516.12, - 71.98 - ], - "text": "4.3\nSolution of Differential and Integro-Differential Equations\n369", - "type": "text" - }, - { - "block_id": "p389-b1", - "global_id": 10847, - "bbox": [ - 102.51, - 93.92, - 484.98, - 105.87 - ], - "text": "EXAMPLE 4.14\nLaplace Transform to Find the Zero-State Response", - "type": "text" - }, - { - "block_id": "p389-b2", - "global_id": 10848, - "bbox": [ - 128.9, - 122.12, - 397.61, - 132.5 - ], - "text": "Find the response y(t) of an LTIC system described by the equation", - "type": "text" - }, - { - "block_id": "p389-b3", - "global_id": 10849, - "bbox": [ - 239.59, - 138.68, - 263.62, - 149.89 - ], - "text": "d2y(t)", - "type": "text" - }, - { - "block_id": "p389-b4", - "global_id": 10850, - "bbox": [ - 245.64, - 139.62, - 301.64, - 163.95 - ], - "text": "dt2\n+ 5dy(t)", - "type": "text" - }, - { - "block_id": "p389-b5", - "global_id": 10851, - "bbox": [ - 287.79, - 139.62, - 366.37, - 163.95 - ], - "text": "dt\n+ 6y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p389-b6", - "global_id": 10852, - "bbox": [ - 352.51, - 146.6, - 393.27, - 163.95 - ], - "text": "dt\n+ x(t)", - "type": "text" - }, - { - "block_id": "p389-b7", - "global_id": 10853, - "bbox": [ - 128.9, - 167.91, - 502.76, - 191.46 - ], - "text": "if the input x(t) = 3e−5tu(t) and all the initial conditions are zero; that is, the system is in the\nzero state.", - "type": "text" - }, - { - "block_id": "p389-b8", - "global_id": 10854, - "bbox": [ - 128.9, - 214.38, - 220.5, - 224.34 - ], - "text": "The system equation is", - "type": "text" - }, - { - "block_id": "p389-b9", - "global_id": 10855, - "bbox": [ - 249.98, - 221.8, - 307.49, - 250.79 - ], - "text": "(D2 + 5D + 6)\n\n\n\nQ(D)", - "type": "text" - }, - { - "block_id": "p389-b10", - "global_id": 10856, - "bbox": [ - 308.6, - 225.92, - 365.74, - 241.62 - ], - "text": "y(t) = (D + 1)", - "type": "text" - }, - { - "block_id": "p389-b11", - "global_id": 10857, - "bbox": [ - 343.26, - 243.6, - 357.75, - 250.79 - ], - "text": "P(D)", - "type": "text" - }, - { - "block_id": "p389-b12", - "global_id": 10858, - "bbox": [ - 366.85, - 225.92, - 381.68, - 236.19 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p389-b13", - "global_id": 10859, - "bbox": [ - 128.91, - 258.71, - 170.65, - 268.67 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p389-b14", - "global_id": 10860, - "bbox": [ - 261.13, - 267.58, - 311.09, - 284.84 - ], - "text": "H(s) = P(s)", - "type": "text" - }, - { - "block_id": "p389-b15", - "global_id": 10861, - "bbox": [ - 293.15, - 267.58, - 369.33, - 292.01 - ], - "text": "Q(s) =\ns + 1\ns2 + 5s + 6", - "type": "text" - }, - { - "block_id": "p389-b16", - "global_id": 10862, - "bbox": [ - 128.91, - 298.55, - 150.22, - 308.51 - ], - "text": "Also,", - "type": "text" - }, - { - "block_id": "p389-b17", - "global_id": 10863, - "bbox": [ - 259.11, - 307.77, - 371.36, - 331.79 - ], - "text": "X(s) = L[3e−5tu(t)] =\n3\ns + 5", - "type": "text" - }, - { - "block_id": "p389-b18", - "global_id": 10864, - "bbox": [ - 128.9, - 337.41, - 143.28, - 347.37 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p389-b19", - "global_id": 10865, - "bbox": [ - 209.18, - 350.92, - 366.51, - 375.34 - ], - "text": "Y(s) = X(s)H(s) =\n3(s + 1)\n(s + 5)(s2 + 5s + 6)", - "type": "text" - }, - { - "block_id": "p389-b20", - "global_id": 10866, - "bbox": [ - 228.84, - 377.53, - 351.83, - 401.97 - ], - "text": "=\n3(s + 1)\n(s + 5)(s + 2)(s + 3) = −2", - "type": "text" - }, - { - "block_id": "p389-b21", - "global_id": 10867, - "bbox": [ - 335.59, - 377.94, - 421.29, - 401.97 - ], - "text": "s + 5 −\n1\ns + 2 +\n3\ns + 3", - "type": "text" - }, - { - "block_id": "p389-b22", - "global_id": 10868, - "bbox": [ - 128.9, - 407.5, - 323.99, - 417.46 - ], - "text": "The inverse Laplace transform of this equation is", - "type": "text" - }, - { - "block_id": "p389-b23", - "global_id": 10869, - "bbox": [ - 247.5, - 420.9, - 384.17, - 435.39 - ], - "text": "y(t) = (−2e−5t −e−2t + 3e−3t)u(t)", - "type": "text" - }, - { - "block_id": "p389-b24", - "global_id": 10870, - "bbox": [ - 102.51, - 484.2, - 497.59, - 496.15 - ], - "text": "EXAMPLE 4.15\nLaplace Transform to Find System Transfer Functions", - "type": "text" - }, - { - "block_id": "p389-b25", - "global_id": 10871, - "bbox": [ - 128.9, - 512.82, - 265.61, - 522.78 - ], - "text": "Show that the transfer function of:", - "type": "text" - }, - { - "block_id": "p389-b26", - "global_id": 10872, - "bbox": [ - 146.84, - 529.11, - 300.46, - 540.71 - ], - "text": "(a) an ideal delay of T seconds is e−sT", - "type": "text" - }, - { - "block_id": "p389-b27", - "global_id": 10873, - "bbox": [ - 146.84, - 545.6, - 265.57, - 555.66 - ], - "text": "(b) an ideal differentiator is s", - "type": "text" - }, - { - "block_id": "p389-b28", - "global_id": 10874, - "bbox": [ - 147.4, - 560.23, - 260.94, - 570.6 - ], - "text": "(c) an ideal integrator is 1/s", - "type": "text" - }, - { - "block_id": "p389-b29", - "global_id": 10875, - "bbox": [ - 128.9, - 593.1, - 502.75, - 615.43 - ], - "text": "(a) Ideal Delay. For an ideal delay of T seconds, the input x(t) and output y(t) are related\nby", - "type": "text" - }, - { - "block_id": "p389-b30", - "global_id": 10876, - "bbox": [ - 195.99, - 615.29, - 435.67, - 627.39 - ], - "text": "y(t) = x(t −T)\nand\nY(s) = X(s)e−sT\n[see Eq. (4.12)]", - "type": "text" - } - ] - }, - { - "page_num": 390, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p390-b0", - "global_id": 10877, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "370\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p390-b1", - "global_id": 10878, - "bbox": [ - 121.09, - 86.24, - 162.84, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p390-b2", - "global_id": 10879, - "bbox": [ - 249.89, - 95.1, - 299.63, - 112.36 - ], - "text": "H(s) = Y(s)", - "type": "text" - }, - { - "block_id": "p390-b3", - "global_id": 10880, - "bbox": [ - 281.9, - 100.37, - 477.01, - 119.44 - ], - "text": "X(s) = e−sT\n(4.30)", - "type": "text" - }, - { - "block_id": "p390-b4", - "global_id": 10881, - "bbox": [ - 103.17, - 125.78, - 477.02, - 148.11 - ], - "text": "(b) Ideal Differentiator. For an ideal differentiator, the input x(t) and the output y(t) are\nrelated by", - "type": "text" - }, - { - "block_id": "p390-b5", - "global_id": 10882, - "bbox": [ - 265.65, - 147.74, - 313.32, - 165.0 - ], - "text": "y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p390-b6", - "global_id": 10883, - "bbox": [ - 103.17, - 162.11, - 307.21, - 187.04 - ], - "text": "dt\nThe Laplace transform of this equation yields", - "type": "text" - }, - { - "block_id": "p390-b7", - "global_id": 10884, - "bbox": [ - 194.04, - 197.36, - 386.14, - 208.96 - ], - "text": "Y(s) = sX(s)\n[x(0−) = 0 for a causal signal]", - "type": "text" - }, - { - "block_id": "p390-b8", - "global_id": 10885, - "bbox": [ - 103.17, - 220.91, - 117.54, - 230.88 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p390-b9", - "global_id": 10886, - "bbox": [ - 256.7, - 228.44, - 306.43, - 245.7 - ], - "text": "H(s) = Y(s)", - "type": "text" - }, - { - "block_id": "p390-b10", - "global_id": 10887, - "bbox": [ - 288.71, - 235.42, - 477.01, - 252.77 - ], - "text": "X(s) = s\n(4.31)", - "type": "text" - }, - { - "block_id": "p390-b11", - "global_id": 10888, - "bbox": [ - 121.09, - 258.65, - 457.98, - 270.25 - ], - "text": "(c) Ideal Integrator. For an ideal integrator with zero initial state, that is, y(0−) = 0,", - "type": "text" - }, - { - "block_id": "p390-b12", - "global_id": 10889, - "bbox": [ - 210.88, - 287.59, - 235.51, - 297.86 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p390-b13", - "global_id": 10890, - "bbox": [ - 237.55, - 274.02, - 249.31, - 286.3 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p390-b14", - "global_id": 10891, - "bbox": [ - 242.81, - 281.02, - 350.27, - 306.17 - ], - "text": "0\nx(τ)dτ\nand\nY(s) = 1", - "type": "text" - }, - { - "block_id": "p390-b15", - "global_id": 10892, - "bbox": [ - 345.84, - 287.59, - 369.3, - 304.94 - ], - "text": "s X(s)", - "type": "text" - }, - { - "block_id": "p390-b16", - "global_id": 10893, - "bbox": [ - 103.17, - 314.74, - 144.91, - 324.7 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p390-b17", - "global_id": 10894, - "bbox": [ - 271.0, - 323.95, - 307.99, - 340.9 - ], - "text": "H(s) = 1", - "type": "text" - }, - { - "block_id": "p390-b18", - "global_id": 10895, - "bbox": [ - 303.56, - 330.94, - 477.01, - 347.88 - ], - "text": "s\n(4.32)", - "type": "text" - }, - { - "block_id": "p390-b19", - "global_id": 10896, - "bbox": [ - 107.82, - 413.34, - 438.95, - 439.24 - ], - "text": "DRILL 4.8\nDifferential Equation and Zero-State Response\nfrom a System Transfer Function", - "type": "text" - }, - { - "block_id": "p390-b20", - "global_id": 10897, - "bbox": [ - 107.82, - 448.36, - 278.57, - 458.32 - ], - "text": "For an LTIC system with transfer function,", - "type": "text" - }, - { - "block_id": "p390-b21", - "global_id": 10898, - "bbox": [ - 257.8, - 467.85, - 333.23, - 492.28 - ], - "text": "H(s) =\ns + 5\ns2 + 4s + 3", - "type": "text" - }, - { - "block_id": "p390-b22", - "global_id": 10899, - "bbox": [ - 125.76, - 514.68, - 428.25, - 525.06 - ], - "text": "(a) Describe the differential equation relating the input x(t) and output y(t).", - "type": "text" - }, - { - "block_id": "p390-b23", - "global_id": 10900, - "bbox": [ - 125.76, - 528.4, - 484.41, - 540.01 - ], - "text": "(b) Find the system response y(t) to the input x(t) = e−2tu(t) if the system is initially in", - "type": "text" - }, - { - "block_id": "p390-b24", - "global_id": 10901, - "bbox": [ - 142.36, - 542.0, - 182.75, - 551.96 - ], - "text": "zero state.", - "type": "text" - }, - { - "block_id": "p390-b25", - "global_id": 10902, - "bbox": [ - 108.09, - 571.46, - 170.31, - 582.42 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p390-b26", - "global_id": 10903, - "bbox": [ - 125.76, - 586.0, - 167.59, - 604.21 - ], - "text": "(a) d2y(t)", - "type": "text" - }, - { - "block_id": "p390-b27", - "global_id": 10904, - "bbox": [ - 149.61, - 586.93, - 205.61, - 611.27 - ], - "text": "dt2\n+ 4dy(t)", - "type": "text" - }, - { - "block_id": "p390-b28", - "global_id": 10905, - "bbox": [ - 191.76, - 586.93, - 270.34, - 611.27 - ], - "text": "dt\n+ 3y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p390-b29", - "global_id": 10906, - "bbox": [ - 256.47, - 593.92, - 302.21, - 611.27 - ], - "text": "dt\n+ 5x(t)", - "type": "text" - }, - { - "block_id": "p390-b30", - "global_id": 10907, - "bbox": [ - 125.76, - 609.84, - 268.33, - 623.83 - ], - "text": "(b) y(t) = (2e−t −3e−2t + e−3t)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 391, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p391-b0", - "global_id": 10908, - "bbox": [ - 223.7, - 62.89, - 516.12, - 71.98 - ], - "text": "4.3\nSolution of Differential and Integro-Differential Equations\n371", - "type": "text" - }, - { - "block_id": "p391-b1", - "global_id": 10909, - "bbox": [ - 127.59, - 86.52, - 205.09, - 98.48 - ], - "text": "4.3-3 Stability", - "type": "text" - }, - { - "block_id": "p391-b2", - "global_id": 10910, - "bbox": [ - 127.59, - 104.19, - 516.14, - 234.13 - ], - "text": "Equation (4.27) shows that the denominator of H(s) is Q(s), which is apparently identical to the\ncharacteristic polynomial Q(λ) defined in Ch. 2. Does this mean that the denominator of H(s)\nis the characteristic polynomial of the system? This may or may not be the case, since if P(s)\nand Q(s) in Eq. (4.27) have any common factors, they cancel out, and the effective denominator\nof H(s) is not necessarily equal to Q(s). Recall also that the system transfer function H(s), like\nh(t), is defined in terms of measurements at the external terminals. Consequently, H(s) and h(t)\nare both external descriptions of the system. In contrast, the characteristic polynomial Q(s) is an\ninternal description. Clearly, we can determine only external stability, that is, BIBO stability, from\nH(s). If all the poles of H(s) are in LHP, all the terms in h(t) are decaying exponentials, and h(t)\nis absolutely integrable [see Eq. (2.45)].† Consequently, the system is BIBO-stable. Otherwise the\nsystem is BIBO-unstable.", - "type": "text" - }, - { - "block_id": "p391-b3", - "global_id": 10911, - "bbox": [ - 255.92, - 431.51, - 387.8, - 441.48 - ], - "text": "Beware of right half-plane poles!", - "type": "text" - }, - { - "block_id": "p391-b4", - "global_id": 10912, - "bbox": [ - 127.59, - 459.31, - 516.14, - 505.55 - ], - "text": "So far, we have assumed that H(s) is a proper function, that is, M ≤N. We now show that\nif H(s) is improper, that is, if M > N, the system is BIBO-unstable. In such a case, using long\ndivision, we obtain H(s) = R(s) + H′(s), where R(s) is an (M −N)th-order polynomial and H′(s)\nis a proper transfer function. For example,", - "type": "text" - }, - { - "block_id": "p391-b5", - "global_id": 10913, - "bbox": [ - 235.46, - 511.34, - 334.61, - 532.22 - ], - "text": "H(s) = s3 + 4s2 + 4s + 5", - "type": "text" - }, - { - "block_id": "p391-b6", - "global_id": 10914, - "bbox": [ - 279.32, - 511.34, - 407.05, - 539.4 - ], - "text": "s2 + 3s + 2\n= s + s2 + 2s + 5", - "type": "text" - }, - { - "block_id": "p391-b7", - "global_id": 10915, - "bbox": [ - 363.63, - 526.14, - 407.05, - 539.4 - ], - "text": "s2 + 3s + 2", - "type": "text" - }, - { - "block_id": "p391-b8", - "global_id": 10916, - "bbox": [ - 127.59, - 546.22, - 516.15, - 592.15 - ], - "text": "As shown in Eq. (4.31), the term s is the transfer function of an ideal differentiator. If we apply step\nfunction (bounded input) to this system, the output will contain an impulse (unbounded output).\nClearly, the system is BIBO-unstable. Moreover, such a system greatly amplifies noise because\ndifferentiation enhances higher frequencies, which generally predominate in a noise signal. These", - "type": "text" - }, - { - "block_id": "p391-b9", - "global_id": 10917, - "bbox": [ - 127.59, - 610.24, - 516.13, - 633.41 - ], - "text": "† Values of s for which H(s) is ∞are the poles of H(s). Thus, poles of H(s) are the values of s for which the\ndenominator of H(s) is zero.", - "type": "text" - } - ] - }, - { - "page_num": 392, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p392-b0", - "global_id": 10918, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "372\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p392-b1", - "global_id": 10919, - "bbox": [ - 101.84, - 85.4, - 490.4, - 107.74 - ], - "text": "are two good reasons to avoid improper systems (M > N). In our future discussion, we shall\nimplicitly assume that the systems are proper, unless stated otherwise.", - "type": "text" - }, - { - "block_id": "p392-b2", - "global_id": 10920, - "bbox": [ - 101.84, - 109.32, - 490.4, - 167.51 - ], - "text": "If P(s) and Q(s) do not have common factors, then the denominator of H(s) is identical to\nQ(s), the characteristic polynomial of the system. In this case, we can determine internal stability\nby using the criterion described in Sec. 2.5. Thus, if P(s) and Q(s) have no common factors,\nthe asymptotic stability criterion in Sec. 2.5 can be restated in terms of the poles of the transfer\nfunction of a system, as follows:", - "type": "text" - }, - { - "block_id": "p392-b3", - "global_id": 10921, - "bbox": [ - 118.78, - 175.48, - 490.39, - 257.18 - ], - "text": "1. An LTIC system is asymptotically stable if and only if all the poles of its transfer function\nH(s) are in the LHP. The poles may be simple or repeated.\n2. An LTIC system is unstable if and only if either one or both of the following conditions\nexist: (i) at least one pole of H(s) is in the RHP; (ii) there are repeated poles of H(s) on the\nimaginary axis.\n3. An LTIC system is marginally stable if and only if there are no poles of H(s) in the RHP\nand some unrepeated poles on the imaginary axis.", - "type": "text" - }, - { - "block_id": "p392-b4", - "global_id": 10922, - "bbox": [ - 101.84, - 264.74, - 415.87, - 275.11 - ], - "text": "The locations of zeros of H(s) have no role in determining the system stability.", - "type": "text" - }, - { - "block_id": "p392-b5", - "global_id": 10923, - "bbox": [ - 76.77, - 307.83, - 351.26, - 319.78 - ], - "text": "EXAMPLE 4.16\nBIBO and Asymptotic Stability", - "type": "text" - }, - { - "block_id": "p392-b6", - "global_id": 10924, - "bbox": [ - 103.16, - 336.04, - 477.01, - 370.33 - ], - "text": "Figure 4.9a shows a cascade connection of two LTIC systems S1 followed by S2. The transfer\nfunctions of these systems are H1(s) = 1/(s −1) and H2(s) = (s −1)/(s + 1), respectively.\nDetermine the BIBO and asymptotic stability of the composite (cascade) system.", - "type": "text" - }, - { - "block_id": "p392-b7", - "global_id": 10925, - "bbox": [ - 172.64, - 448.95, - 181.52, - 456.95 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p392-b8", - "global_id": 10926, - "bbox": [ - 172.26, - 506.91, - 181.91, - 514.91 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p392-b9", - "global_id": 10927, - "bbox": [ - 116.15, - 415.08, - 236.92, - 436.06 - ], - "text": "y(t)\nx(t)\ns – 1\ns 1\n1\ns 1", - "type": "text" - }, - { - "block_id": "p392-b10", - "global_id": 10928, - "bbox": [ - 168.74, - 477.24, - 186.51, - 494.09 - ], - "text": "1\ns 1", - "type": "text" - }, - { - "block_id": "p392-b11", - "global_id": 10929, - "bbox": [ - 116.15, - 472.73, - 236.92, - 480.81 - ], - "text": "y(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p392-b12", - "global_id": 10930, - "bbox": [ - 251.69, - 494.25, - 480.6, - 516.16 - ], - "text": "Figure 4.9 Distinction between BIBO and asymptotic\nstability.", - "type": "text" - }, - { - "block_id": "p392-b13", - "global_id": 10931, - "bbox": [ - 103.16, - 534.05, - 477.02, - 568.34 - ], - "text": "If the impulse responses of S1 and S2 are h1(t) and h2(t), respectively, then the impulse\nresponse of the cascade system is h(t) = h1(t)∗h2(t). Hence, H(s) = H1(s)H2(s). In the present\ncase,", - "type": "text" - }, - { - "block_id": "p392-b14", - "global_id": 10932, - "bbox": [ - 222.13, - 574.84, - 250.9, - 585.11 - ], - "text": "H(s) =", - "type": "text" - }, - { - "block_id": "p392-b15", - "global_id": 10933, - "bbox": [ - 252.94, - 560.85, - 280.59, - 592.29 - ], - "text": "1\ns −1", - "type": "text" - }, - { - "block_id": "p392-b16", - "global_id": 10934, - "bbox": [ - 281.78, - 560.85, - 316.15, - 578.23 - ], - "text": "s −1", - "type": "text" - }, - { - "block_id": "p392-b17", - "global_id": 10935, - "bbox": [ - 296.43, - 581.91, - 316.15, - 592.29 - ], - "text": "s + 1", - "type": "text" - }, - { - "block_id": "p392-b19", - "global_id": 10936, - "bbox": [ - 326.12, - 568.26, - 356.86, - 592.29 - ], - "text": "=\n1\ns + 1", - "type": "text" - }, - { - "block_id": "p392-b20", - "global_id": 10937, - "bbox": [ - 103.16, - 599.08, - 477.03, - 621.41 - ], - "text": "The pole of S1 at s = 1 cancels with the zero at s = 1 of S2. This results in a composite system\nhaving a single pole at s = −1. If the composite cascade system were to be enclosed inside a", - "type": "text" - } - ] - }, - { - "page_num": 393, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p393-b0", - "global_id": 10938, - "bbox": [ - 217.76, - 62.89, - 516.12, - 71.98 - ], - "text": "4.4\nAnalysis of Electrical Networks: The Transformed Network\n373", - "type": "text" - }, - { - "block_id": "p393-b1", - "global_id": 10939, - "bbox": [ - 128.9, - 86.24, - 502.75, - 120.11 - ], - "text": "black box with only the input and the output terminals accessible, any measurement from these\nexternal terminals would show that the transfer function of the system is 1/(s+1), without any\nhint of the fact that the system is housing an unstable system (Fig. 4.9b).", - "type": "text" - }, - { - "block_id": "p393-b2", - "global_id": 10940, - "bbox": [ - 128.91, - 120.47, - 502.75, - 144.02 - ], - "text": "The impulse response of the cascade system is h(t) = e−tu(t), which is absolutely\nintegrable. Consequently, the system is BIBO-stable.", - "type": "text" - }, - { - "block_id": "p393-b3", - "global_id": 10941, - "bbox": [ - 128.9, - 145.6, - 502.78, - 215.75 - ], - "text": "To determine the asymptotic stability, we note that S1 has one characteristic root at 1, and\nS2 also has one root at −1. Recall that the two systems are independent (one does not load the\nother), and the characteristic modes generated in each subsystem are independent of the other.\nClearly, the mode et will not be eliminated by the presence of S2. Hence, the composite system\nhas two characteristic roots, located at ±1, and the system is asymptotically unstable, though\nBIBO-stable.", - "type": "text" - }, - { - "block_id": "p393-b4", - "global_id": 10942, - "bbox": [ - 128.91, - 217.33, - 502.77, - 311.4 - ], - "text": "Interchanging the positions of S1 and S2 makes no difference in this conclusion. This\nexample shows that BIBO stability can be misleading. If a system is asymptotically unstable,\nit will destroy itself (or, more likely, lead to saturation condition) because of unchecked growth\nof the response due to intended or unintended stray initial conditions. BIBO stability is not\ngoing to save the system. Control systems are often compensated to realize certain desirable\ncharacteristics. One should never try to stabilize an unstable system by canceling its RHP\npole(s) with RHP zero(s). Such a misguided attempt will fail, not because of the practical\nimpossibility of exact cancellation but for the more fundamental reason, as just explained.", - "type": "text" - }, - { - "block_id": "p393-b5", - "global_id": 10943, - "bbox": [ - 133.57, - 364.56, - 378.91, - 376.52 - ], - "text": "DRILL 4.9\nBIBO and Asymptotic Stability", - "type": "text" - }, - { - "block_id": "p393-b6", - "global_id": 10944, - "bbox": [ - 133.57, - 385.64, - 408.92, - 395.6 - ], - "text": "Show that an ideal integrator is marginally stable but BIBO-unstable.", - "type": "text" - }, - { - "block_id": "p393-b7", - "global_id": 10945, - "bbox": [ - 127.59, - 435.39, - 246.64, - 447.34 - ], - "text": "4.3-4 Inverse Systems", - "type": "text" - }, - { - "block_id": "p393-b8", - "global_id": 10946, - "bbox": [ - 127.59, - 453.06, - 516.13, - 476.1 - ], - "text": "If H(s) is the transfer function of a system S, then Si, its inverse system has a transfer function\nHi(s) given by", - "type": "text" - }, - { - "block_id": "p393-b9", - "global_id": 10947, - "bbox": [ - 294.79, - 475.37, - 347.74, - 499.29 - ], - "text": "Hi(s) =\n1\nH(s)", - "type": "text" - }, - { - "block_id": "p393-b10", - "global_id": 10948, - "bbox": [ - 127.59, - 505.49, - 516.14, - 540.48 - ], - "text": "This follows from the fact the cascade of S with its inverse system Si is an identity system, with\nimpulse response δ(t), implying H(s)Hi(s) = 1. For example, an ideal integrator and its inverse,\nan ideal differentiator, have transfer functions 1/s and s, respectively, leading to H(s)Hi(s) = 1.", - "type": "text" - }, - { - "block_id": "p393-b11", - "global_id": 10949, - "bbox": [ - 127.94, - 565.04, - 405.29, - 594.93 - ], - "text": "4.4 ANALYSIS OF ELECTRICAL NETWORKS:\nTHE TRANSFORMED NETWORK", - "type": "text" - }, - { - "block_id": "p393-b12", - "global_id": 10950, - "bbox": [ - 127.59, - 600.91, - 516.12, - 634.79 - ], - "text": "Example 4.12 shows how electrical networks may be analyzed by writing the integro-differential\nequation(s) of the system and then solving these equations by the Laplace transform. We now\nshow that it is also possible to analyze electrical networks directly without having to write the", - "type": "text" - } - ] - }, - { - "page_num": 394, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p394-b0", - "global_id": 10951, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "374\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p394-b1", - "global_id": 10952, - "bbox": [ - 101.84, - 85.82, - 490.42, - 131.64 - ], - "text": "integro-differential equations. This procedure is considerably simpler because it permits us to treat\nan electrical network as if it were a resistive network. For this purpose, we need to represent a\nnetwork in the “frequency domain” where all the voltages and currents are represented by their\nLaplace transforms.", - "type": "text" - }, - { - "block_id": "p394-b2", - "global_id": 10953, - "bbox": [ - 101.84, - 133.23, - 490.38, - 155.56 - ], - "text": "For the sake of simplicity, let us first discuss the case with zero initial conditions. If v(t) and\ni(t) are the voltage across and the current through an inductor of L henries, then", - "type": "text" - }, - { - "block_id": "p394-b3", - "global_id": 10954, - "bbox": [ - 269.74, - 165.22, - 321.3, - 182.48 - ], - "text": "v(t) = Ldi(t)", - "type": "text" - }, - { - "block_id": "p394-b4", - "global_id": 10955, - "bbox": [ - 308.27, - 179.59, - 316.01, - 189.56 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p394-b5", - "global_id": 10956, - "bbox": [ - 101.84, - 197.64, - 389.06, - 207.6 - ], - "text": "The Laplace transform of this equation (assuming zero initial current) is", - "type": "text" - }, - { - "block_id": "p394-b6", - "global_id": 10957, - "bbox": [ - 268.83, - 219.22, - 323.41, - 229.5 - ], - "text": "V(s) = LsI(s)", - "type": "text" - }, - { - "block_id": "p394-b7", - "global_id": 10958, - "bbox": [ - 101.85, - 241.23, - 490.4, - 263.56 - ], - "text": "Similarly, for a capacitor of C farads, the voltage-current relationship is i(t) = C(dv/dt) and its\nLaplace transform, assuming zero initial capacitor voltage, yields I(s) = CsV(s); that is,", - "type": "text" - }, - { - "block_id": "p394-b8", - "global_id": 10959, - "bbox": [ - 267.09, - 273.57, - 306.06, - 290.52 - ], - "text": "V(s) = 1", - "type": "text" - }, - { - "block_id": "p394-b9", - "global_id": 10960, - "bbox": [ - 298.3, - 280.14, - 325.16, - 297.49 - ], - "text": "CsI(s)", - "type": "text" - }, - { - "block_id": "p394-b10", - "global_id": 10961, - "bbox": [ - 101.84, - 305.23, - 490.4, - 327.55 - ], - "text": "For a resistor of R ohms, the voltage-current relationship is v(t) = Ri(t), and its Laplace transform\nis", - "type": "text" - }, - { - "block_id": "p394-b11", - "global_id": 10962, - "bbox": [ - 270.49, - 329.26, - 321.74, - 339.53 - ], - "text": "V(s) = RI(s)", - "type": "text" - }, - { - "block_id": "p394-b12", - "global_id": 10963, - "bbox": [ - 101.85, - 348.68, - 490.4, - 406.47 - ], - "text": "Thus, in the “frequency domain,” the voltage-current relationships of an inductor and a capacitor\nare algebraic; these elements behave like resistors of “resistance” Ls and 1/Cs, respectively. The\ngeneralized “resistance” of an element is called its impedance and is given by the ratio V(s)/I(s)\nfor the element (under zero initial conditions). The impedances of a resistor of R ohms, an inductor\nof L henries, and a capacitance of C farads are R, Ls, and 1/Cs, respectively.", - "type": "text" - }, - { - "block_id": "p394-b13", - "global_id": 10964, - "bbox": [ - 101.84, - 408.46, - 490.4, - 443.03 - ], - "text": "Also, the interconnection constraints (Kirchhoff’s laws) remain valid for voltages and currents\nin the frequency domain. To demonstrate this point, let vj(t) (j = 1,2,. . .,k) be the voltages across\nk elements in a loop and let ij(t)(j = 1,2,. . .,m) be the j currents entering a node. Then", - "type": "text" - }, - { - "block_id": "p394-b14", - "global_id": 10965, - "bbox": [ - 219.98, - 454.28, - 234.08, - 464.73 - ], - "text": "k\n\"", - "type": "text" - }, - { - "block_id": "p394-b15", - "global_id": 10966, - "bbox": [ - 221.6, - 478.63, - 232.46, - 485.89 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p394-b16", - "global_id": 10967, - "bbox": [ - 235.18, - 464.23, - 303.58, - 475.31 - ], - "text": "vj(t) = 0\nand", - "type": "text" - }, - { - "block_id": "p394-b17", - "global_id": 10968, - "bbox": [ - 324.61, - 454.28, - 338.71, - 464.73 - ], - "text": "m\n\"", - "type": "text" - }, - { - "block_id": "p394-b18", - "global_id": 10969, - "bbox": [ - 326.23, - 478.63, - 337.09, - 485.89 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p394-b19", - "global_id": 10970, - "bbox": [ - 339.82, - 464.23, - 372.25, - 475.31 - ], - "text": "ij(t) = 0", - "type": "text" - }, - { - "block_id": "p394-b20", - "global_id": 10971, - "bbox": [ - 101.84, - 497.01, - 129.54, - 506.97 - ], - "text": "Now if", - "type": "text" - }, - { - "block_id": "p394-b21", - "global_id": 10972, - "bbox": [ - 210.92, - 508.67, - 381.32, - 519.75 - ], - "text": "vj(t) ⇐⇒Vj(s)\nand\nij(t) ⇐⇒Ij(s)", - "type": "text" - }, - { - "block_id": "p394-b22", - "global_id": 10973, - "bbox": [ - 101.85, - 528.09, - 118.99, - 538.06 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p394-b23", - "global_id": 10974, - "bbox": [ - 217.94, - 537.27, - 232.04, - 547.71 - ], - "text": "k\n\"", - "type": "text" - }, - { - "block_id": "p394-b24", - "global_id": 10975, - "bbox": [ - 219.56, - 561.61, - 230.42, - 568.87 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p394-b25", - "global_id": 10976, - "bbox": [ - 233.15, - 547.22, - 304.13, - 558.3 - ], - "text": "Vj(s) = 0\nand", - "type": "text" - }, - { - "block_id": "p394-b26", - "global_id": 10977, - "bbox": [ - 325.17, - 537.27, - 339.27, - 547.71 - ], - "text": "m\n\"", - "type": "text" - }, - { - "block_id": "p394-b27", - "global_id": 10978, - "bbox": [ - 326.79, - 561.61, - 337.65, - 568.87 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p394-b28", - "global_id": 10979, - "bbox": [ - 340.38, - 547.22, - 374.29, - 558.3 - ], - "text": "Ij(s) = 0", - "type": "text" - }, - { - "block_id": "p394-b29", - "global_id": 10980, - "bbox": [ - 101.85, - 577.0, - 490.41, - 634.79 - ], - "text": "This result shows that if we represent all the voltages and currents in an electrical network by\ntheir Laplace transforms, we can treat the network as if it consisted of the “resistances” R, Ls,\nand 1/Cs corresponding to a resistor R, an inductor L, and a capacitor C, respectively. The system\nequations (loop or node) are now algebraic. Moreover, the simplification techniques that have been\ndeveloped for resistive circuits—equivalent series and parallel impedances, voltage and current", - "type": "text" - } - ] - }, - { - "page_num": 395, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p395-b0", - "global_id": 10981, - "bbox": [ - 217.76, - 62.89, - 516.12, - 71.98 - ], - "text": "4.4\nAnalysis of Electrical Networks: The Transformed Network\n375", - "type": "text" - }, - { - "block_id": "p395-b1", - "global_id": 10982, - "bbox": [ - 127.59, - 85.82, - 516.12, - 107.74 - ], - "text": "divider rules, Thévenin and Norton theorems—can be applied to general electrical networks. The\nfollowing examples demonstrate these concepts.", - "type": "text" - }, - { - "block_id": "p395-b2", - "global_id": 10983, - "bbox": [ - 102.51, - 146.06, - 420.18, - 158.02 - ], - "text": "EXAMPLE 4.17\nTransform Analysis of a Simple Circuit", - "type": "text" - }, - { - "block_id": "p395-b3", - "global_id": 10984, - "bbox": [ - 128.9, - 174.27, - 502.77, - 184.65 - ], - "text": "Find the loop current i(t) in the circuit shown in Fig. 4.10a if all the initial conditions are zero.", - "type": "text" - }, - { - "block_id": "p395-b4", - "global_id": 10985, - "bbox": [ - 208.69, - 326.21, - 398.71, - 334.21 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p395-b5", - "global_id": 10986, - "bbox": [ - 393.25, - 280.8, - 404.36, - 288.88 - ], - "text": "I(s)", - "type": "text" - }, - { - "block_id": "p395-b6", - "global_id": 10987, - "bbox": [ - 377.36, - 233.95, - 430.4, - 243.02 - ], - "text": "3\ns", - "type": "text" - }, - { - "block_id": "p395-b7", - "global_id": 10988, - "bbox": [ - 215.62, - 278.62, - 470.58, - 293.57 - ], - "text": "2\ns\n10\ns\ni(t)", - "type": "text" - }, - { - "block_id": "p395-b8", - "global_id": 10989, - "bbox": [ - 194.73, - 233.68, - 256.21, - 242.58 - ], - "text": "3 \t\n1 H", - "type": "text" - }, - { - "block_id": "p395-b9", - "global_id": 10990, - "bbox": [ - 127.52, - 279.18, - 298.74, - 293.25 - ], - "text": "10u(t)\n1\n2 F", - "type": "text" - }, - { - "block_id": "p395-b10", - "global_id": 10991, - "bbox": [ - 127.52, - 340.82, - 337.31, - 350.14 - ], - "text": "Figure 4.10 (a) A circuit and (b) its transformed version.", - "type": "text" - }, - { - "block_id": "p395-b11", - "global_id": 10992, - "bbox": [ - 128.9, - 377.81, - 502.77, - 495.37 - ], - "text": "In the first step, we represent the circuit in the frequency domain, as illustrated in\nFig. 4.10b. All the voltages and currents are represented by their Laplace transforms. The\nvoltage 10u(t) is represented by 10/s and the (unknown) current i(t) is represented by its\nLaplace transform I(s). All the circuit elements are represented by their respective impedances.\nThe inductor of 1 henry is represented by s, the capacitor of 1/2 farad is represented by\n2/s, and the resistor of 3 ohms is represented by 3. We now consider the frequency-domain\nrepresentation of voltages and currents. The voltage across any element is I(s) times its\nimpedance. Therefore, the total voltage drop in the loop is I(s) times the total loop impedance,\nand it must be equal to V(s), (transform of) the input voltage. The total impedance in the loop\nis", - "type": "text" - }, - { - "block_id": "p395-b12", - "global_id": 10993, - "bbox": [ - 253.4, - 494.86, - 319.39, - 511.81 - ], - "text": "Z(s) = s + 3 + 2", - "type": "text" - }, - { - "block_id": "p395-b13", - "global_id": 10994, - "bbox": [ - 314.95, - 490.84, - 377.07, - 518.78 - ], - "text": "s = s2 + 3s + 2", - "type": "text" - }, - { - "block_id": "p395-b14", - "global_id": 10995, - "bbox": [ - 128.91, - 508.82, - 413.46, - 533.82 - ], - "text": "s\nThe input“voltage” is V(s) = 10/s. Therefore, the “loop current” I(s) is", - "type": "text" - }, - { - "block_id": "p395-b15", - "global_id": 10996, - "bbox": [ - 158.6, - 543.4, - 204.95, - 560.66 - ], - "text": "I(s) = V(s)", - "type": "text" - }, - { - "block_id": "p395-b16", - "global_id": 10997, - "bbox": [ - 187.19, - 543.4, - 466.99, - 567.83 - ], - "text": "Z(s) =\n10/s\n(s2 + 3s + 2)/s =\n10\ns2 + 3s + 2 =\n10\n(s + 1)(s + 2) =\n10\ns + 1 −10", - "type": "text" - }, - { - "block_id": "p395-b17", - "global_id": 10998, - "bbox": [ - 452.16, - 557.46, - 471.88, - 567.83 - ], - "text": "s + 2", - "type": "text" - }, - { - "block_id": "p395-b18", - "global_id": 10999, - "bbox": [ - 128.9, - 577.63, - 380.42, - 587.59 - ], - "text": "The inverse transform of this equation yields the desired result:", - "type": "text" - }, - { - "block_id": "p395-b19", - "global_id": 11000, - "bbox": [ - 267.34, - 595.02, - 364.33, - 609.51 - ], - "text": "i(t) = 10(e−t −e−2t)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 396, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p396-b0", - "global_id": 11001, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "376\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p396-b1", - "global_id": 11002, - "bbox": [ - 102.14, - 86.19, - 286.12, - 98.32 - ], - "text": "INITIAL CONDITION GENERATORS", - "type": "text" - }, - { - "block_id": "p396-b2", - "global_id": 11003, - "bbox": [ - 101.84, - 102.35, - 490.41, - 207.95 - ], - "text": "The discussion in which we assumed zero initial conditions can be readily extended to the case\nof nonzero initial conditions because the initial condition in a capacitor or an inductor can be\nrepresented by an equivalent source. We now show that a capacitor C with an initial voltage v(0−)\n(Fig. 4.11a) can be represented in the frequency domain by an uncharged capacitor of impedance\n1/Cs in series with a voltage source of value v(0−)/s (Fig. 4.11b) or as the same uncharged\ncapacitor in parallel with a current source of value Cv(0−) (Fig. 4.11c). Similarly, an inductor\nL with an initial current i(0−) (Fig. 4.11d) can be represented in the frequency domain by an\ninductor of impedance Ls in series with a voltage source of value Li(0−) (Fig. 4.11e) or by the\nsame inductor in parallel with a current source of value i(0−)/s (Fig. 4.11f).", - "type": "text" - }, - { - "block_id": "p396-b3", - "global_id": 11004, - "bbox": [ - 361.4, - 237.26, - 372.51, - 245.34 - ], - "text": "I(s)", - "type": "text" - }, - { - "block_id": "p396-b4", - "global_id": 11005, - "bbox": [ - 173.74, - 289.86, - 347.07, - 300.55 - ], - "text": "V(s)\nv(0)", - "type": "text" - }, - { - "block_id": "p396-b5", - "global_id": 11006, - "bbox": [ - 133.57, - 237.26, - 143.35, - 245.34 - ], - "text": "i(t)", - "type": "text" - }, - { - "block_id": "p396-b6", - "global_id": 11007, - "bbox": [ - 107.24, - 289.86, - 464.26, - 301.64 - ], - "text": "v(t)\nC\nCv(0)", - "type": "text" - }, - { - "block_id": "p396-b7", - "global_id": 11008, - "bbox": [ - 136.2, - 347.34, - 406.22, - 355.34 - ], - "text": "(a)\n(b)\n(c)", - "type": "text" - }, - { - "block_id": "p396-b8", - "global_id": 11009, - "bbox": [ - 122.34, - 485.34, - 406.3, - 493.34 - ], - "text": "(d)\n(e)\n(f)", - "type": "text" - }, - { - "block_id": "p396-b19", - "global_id": 11010, - "bbox": [ - 251.4, - 237.26, - 262.51, - 245.34 - ], - "text": "I(s)", - "type": "text" - }, - { - "block_id": "p396-b20", - "global_id": 11011, - "bbox": [ - 133.57, - 377.26, - 262.51, - 385.34 - ], - "text": "i(t)\nI(s)", - "type": "text" - }, - { - "block_id": "p396-b21", - "global_id": 11012, - "bbox": [ - 225.74, - 289.86, - 239.07, - 297.94 - ], - "text": "V(s)", - "type": "text" - }, - { - "block_id": "p396-b22", - "global_id": 11013, - "bbox": [ - 361.4, - 377.26, - 372.51, - 385.34 - ], - "text": "I(s)", - "type": "text" - }, - { - "block_id": "p396-b23", - "global_id": 11014, - "bbox": [ - 107.24, - 429.31, - 347.07, - 437.94 - ], - "text": "V(s)\nv(t)\nL\nV(s)", - "type": "text" - }, - { - "block_id": "p396-b24", - "global_id": 11015, - "bbox": [ - 291.24, - 450.45, - 313.2, - 460.14 - ], - "text": "Li(0)", - "type": "text" - }, - { - "block_id": "p396-b25", - "global_id": 11016, - "bbox": [ - 287.21, - 409.26, - 295.25, - 417.26 - ], - "text": "Ls", - "type": "text" - }, - { - "block_id": "p396-b26", - "global_id": 11017, - "bbox": [ - 397.04, - 426.87, - 405.08, - 434.87 - ], - "text": "Ls", - "type": "text" - }, - { - "block_id": "p396-b31", - "global_id": 11018, - "bbox": [ - 289.99, - 262.03, - 298.44, - 278.68 - ], - "text": "1\nCs", - "type": "text" - }, - { - "block_id": "p396-b32", - "global_id": 11019, - "bbox": [ - 292.08, - 287.7, - 375.1, - 305.77 - ], - "text": "1\nCs\nv(0)", - "type": "text" - }, - { - "block_id": "p396-b33", - "global_id": 11020, - "bbox": [ - 299.46, - 306.12, - 302.57, - 314.12 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p396-b34", - "global_id": 11021, - "bbox": [ - 441.71, - 425.7, - 458.74, - 435.39 - ], - "text": "i(0)", - "type": "text" - }, - { - "block_id": "p396-b35", - "global_id": 11022, - "bbox": [ - 448.67, - 435.22, - 451.78, - 443.22 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p396-b36", - "global_id": 11023, - "bbox": [ - 101.84, - 509.33, - 362.82, - 518.57 - ], - "text": "Figure 4.11 Initial condition generators for a capacitor and an inductor.", - "type": "text" - }, - { - "block_id": "p396-b37", - "global_id": 11024, - "bbox": [ - 119.78, - 545.92, - 453.47, - 555.89 - ], - "text": "To prove this point, consider the terminal relationship of the capacitor in Fig. 4.11a:", - "type": "text" - }, - { - "block_id": "p396-b38", - "global_id": 11025, - "bbox": [ - 269.09, - 567.31, - 321.94, - 584.57 - ], - "text": "i(t) = Cdv(t)", - "type": "text" - }, - { - "block_id": "p396-b39", - "global_id": 11026, - "bbox": [ - 308.1, - 581.68, - 315.84, - 591.64 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p396-b40", - "global_id": 11027, - "bbox": [ - 101.85, - 601.49, - 283.36, - 611.45 - ], - "text": "The Laplace transform of this equation yields", - "type": "text" - }, - { - "block_id": "p396-b41", - "global_id": 11028, - "bbox": [ - 248.02, - 623.1, - 344.21, - 635.21 - ], - "text": "I(s) = C[sV(s) −v(0−)]", - "type": "text" - } - ] - }, - { - "page_num": 397, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p397-b0", - "global_id": 11029, - "bbox": [ - 217.76, - 62.89, - 516.12, - 71.98 - ], - "text": "4.4\nAnalysis of Electrical Networks: The Transformed Network\n377", - "type": "text" - }, - { - "block_id": "p397-b1", - "global_id": 11030, - "bbox": [ - 127.59, - 85.82, - 266.19, - 95.78 - ], - "text": "This equation can be rearranged as", - "type": "text" - }, - { - "block_id": "p397-b2", - "global_id": 11031, - "bbox": [ - 274.81, - 108.06, - 313.78, - 125.01 - ], - "text": "V(s) = 1", - "type": "text" - }, - { - "block_id": "p397-b3", - "global_id": 11032, - "bbox": [ - 306.03, - 106.43, - 367.72, - 131.98 - ], - "text": "CsI(s) + v(0−)", - "type": "text" - }, - { - "block_id": "p397-b4", - "global_id": 11033, - "bbox": [ - 354.39, - 115.05, - 516.12, - 131.98 - ], - "text": "s\n(4.33)", - "type": "text" - }, - { - "block_id": "p397-b5", - "global_id": 11034, - "bbox": [ - 127.59, - 141.09, - 516.15, - 187.34 - ], - "text": "Observe that V(s) is the voltage (in the frequency domain) across the charged capacitor and I(s)/Cs\nis the voltage across the same capacitor without any charge. Therefore, the charged capacitor can\nbe represented by the uncharged capacitor in series with a voltage source of value v(0−)/s, as\ndepicted in Fig. 4.11b. Equation (4.33) can also be rearranged as", - "type": "text" - }, - { - "block_id": "p397-b6", - "global_id": 11035, - "bbox": [ - 269.36, - 198.73, - 308.34, - 215.68 - ], - "text": "V(s) = 1", - "type": "text" - }, - { - "block_id": "p397-b7", - "global_id": 11036, - "bbox": [ - 300.58, - 203.58, - 374.36, - 222.65 - ], - "text": "Cs[I(s) + Cv(0−)]", - "type": "text" - }, - { - "block_id": "p397-b8", - "global_id": 11037, - "bbox": [ - 127.59, - 231.75, - 516.14, - 266.04 - ], - "text": "This equation shows that the charged capacitor voltage V(s) is equal to the uncharged capacitor\nvoltage caused by a current I(s) + Cv(0−). This result is reflected precisely in Fig. 4.11c, where\nthe current through the uncharged capacitor is I(s) + Cv(0−).†", - "type": "text" - }, - { - "block_id": "p397-b9", - "global_id": 11038, - "bbox": [ - 145.52, - 268.03, - 361.79, - 277.99 - ], - "text": "For the inductor in Fig. 4.11d, the terminal equation is", - "type": "text" - }, - { - "block_id": "p397-b10", - "global_id": 11039, - "bbox": [ - 295.48, - 289.04, - 347.04, - 306.3 - ], - "text": "v(t) = Ldi(t)", - "type": "text" - }, - { - "block_id": "p397-b11", - "global_id": 11040, - "bbox": [ - 334.02, - 303.41, - 341.76, - 313.37 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p397-b12", - "global_id": 11041, - "bbox": [ - 127.59, - 322.82, - 141.97, - 332.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p397-b13", - "global_id": 11042, - "bbox": [ - 238.28, - 334.83, - 516.13, - 346.93 - ], - "text": "V(s) = L[sI(s) −i(0−)] = LsI(s) −Li(0−)\n(4.34)", - "type": "text" - }, - { - "block_id": "p397-b14", - "global_id": 11043, - "bbox": [ - 127.59, - 357.35, - 432.37, - 367.32 - ], - "text": "This expression is consistent with Fig. 4.11e. We can rearrange Eq. (4.34) as", - "type": "text" - }, - { - "block_id": "p397-b15", - "global_id": 11044, - "bbox": [ - 271.95, - 386.17, - 311.39, - 396.45 - ], - "text": "V(s) = Ls", - "type": "text" - }, - { - "block_id": "p397-b17", - "global_id": 11045, - "bbox": [ - 316.83, - 377.96, - 365.15, - 396.45 - ], - "text": "I(s) −i(0−)", - "type": "text" - }, - { - "block_id": "p397-b18", - "global_id": 11046, - "bbox": [ - 352.65, - 393.56, - 356.52, - 403.52 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p397-b19", - "global_id": 11047, - "bbox": [ - 366.34, - 372.19, - 371.77, - 382.15 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p397-b20", - "global_id": 11048, - "bbox": [ - 127.59, - 415.25, - 305.93, - 425.21 - ], - "text": "This expression is consistent with Fig. 4.11f.", - "type": "text" - }, - { - "block_id": "p397-b21", - "global_id": 11049, - "bbox": [ - 127.59, - 427.2, - 516.14, - 544.77 - ], - "text": "Let us rework Ex. 4.13 using these concepts. Figure 4.12a shows the circuit in Fig. 4.7b\nwith the initial conditions y(0−) = 2 and vC(0−) = 10. Figure 4.12b shows the frequency-domain\nrepresentation (transformed circuit) of the circuit in Fig. 4.12a. The resistor is represented by\nits impedance 2; the inductor with initial current of 2 amperes is represented according to the\narrangement in Fig. 4.11e with a series voltage source Ly(0−) = 2. The capacitor with initial\nvoltage of 10 volts is represented according to the arrangement in Fig. 4.11b with a series voltage\nsource v(0−)/s = 10/s. Note that the impedance of the inductor is s and that of the capacitor is\n5/s. The input of 10u(t) is represented by its Laplace transform 10/s.\nThe total voltage in the loop is (10/s) + 2 −(10/s) = 2, and the loop impedance is\n(s + 2 + (5/s)). Therefore,", - "type": "text" - }, - { - "block_id": "p397-b22", - "global_id": 11050, - "bbox": [ - 255.01, - 546.69, - 387.5, - 570.8 - ], - "text": "Y(s) =\n2\ns + 2 + 5/s =\n2s\ns2 + 2s + 5", - "type": "text" - }, - { - "block_id": "p397-b23", - "global_id": 11051, - "bbox": [ - 127.59, - 579.06, - 306.67, - 589.02 - ], - "text": "which confirms our earlier result in Ex. 4.13.", - "type": "text" - }, - { - "block_id": "p397-b24", - "global_id": 11052, - "bbox": [ - 127.59, - 604.26, - 516.13, - 649.35 - ], - "text": "† In the time domain, a charged capacitor C with initial voltage v(0−) can be represented as the same capacitor\nuncharged in series with a voltage source v(0−)u(t), or in parallel with a current source Cv(0−)δ(t). Similarly,\nan inductor L with initial current i(0−) can be represented by the same inductor with zero initial current in\nseries with a voltage source Li(0−)δ(t) or with a parallel current source i(0−)u(t).", - "type": "text" - } - ] - }, - { - "page_num": 398, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p398-b0", - "global_id": 11053, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "378\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p398-b1", - "global_id": 11054, - "bbox": [ - 205.98, - 186.74, - 389.79, - 194.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p398-b2", - "global_id": 11055, - "bbox": [ - 212.98, - 138.15, - 224.08, - 146.23 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p398-b3", - "global_id": 11056, - "bbox": [ - 192.76, - 91.24, - 204.53, - 99.24 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p398-b4", - "global_id": 11057, - "bbox": [ - 125.76, - 139.4, - 145.31, - 147.48 - ], - "text": "10u(t)", - "type": "text" - }, - { - "block_id": "p398-b7", - "global_id": 11058, - "bbox": [ - 287.08, - 138.43, - 295.08, - 146.43 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p398-b8", - "global_id": 11059, - "bbox": [ - 242.02, - 90.93, - 254.24, - 99.23 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p398-b9", - "global_id": 11060, - "bbox": [ - 168.4, - 113.09, - 201.42, - 122.78 - ], - "text": "y(0) 2", - "type": "text" - }, - { - "block_id": "p398-b10", - "global_id": 11061, - "bbox": [ - 256.52, - 135.38, - 397.59, - 149.44 - ], - "text": "1\n5 F\nY(s)", - "type": "text" - }, - { - "block_id": "p398-b11", - "global_id": 11062, - "bbox": [ - 358.24, - 87.98, - 424.9, - 100.07 - ], - "text": "2\ns\n2", - "type": "text" - }, - { - "block_id": "p398-b12", - "global_id": 11063, - "bbox": [ - 313.13, - 132.74, - 321.13, - 147.4 - ], - "text": "10\ns", - "type": "text" - }, - { - "block_id": "p398-b13", - "global_id": 11064, - "bbox": [ - 427.13, - 151.73, - 435.13, - 166.4 - ], - "text": "10\ns", - "type": "text" - }, - { - "block_id": "p398-b14", - "global_id": 11065, - "bbox": [ - 431.62, - 122.14, - 435.62, - 136.8 - ], - "text": "5\ns", - "type": "text" - }, - { - "block_id": "p398-b15", - "global_id": 11066, - "bbox": [ - 125.76, - 201.44, - 427.48, - 210.68 - ], - "text": "Figure 4.12 A circuit and its transformed version with initial-condition generators.", - "type": "text" - }, - { - "block_id": "p398-b16", - "global_id": 11067, - "bbox": [ - 76.77, - 245.69, - 443.56, - 257.65 - ], - "text": "EXAMPLE 4.18\nTransformed Analysis of a Circuit with a Switch", - "type": "text" - }, - { - "block_id": "p398-b17", - "global_id": 11068, - "bbox": [ - 103.16, - 273.89, - 477.02, - 297.0 - ], - "text": "The switch in the circuit of Fig. 4.13a is in the closed position for a long time before t = 0,\nwhen it is opened instantaneously. Find the currents y1(t) and y2(t) for t ≥0.", - "type": "text" - }, - { - "block_id": "p398-b18", - "global_id": 11069, - "bbox": [ - 363.51, - 561.4, - 372.39, - 569.4 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p398-b21", - "global_id": 11070, - "bbox": [ - 380.09, - 539.87, - 384.09, - 547.87 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p398-b22", - "global_id": 11071, - "bbox": [ - 380.39, - 455.53, - 384.39, - 463.53 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p398-b23", - "global_id": 11072, - "bbox": [ - 411.49, - 498.1, - 424.81, - 506.18 - ], - "text": "V(s)", - "type": "text" - }, - { - "block_id": "p398-b24", - "global_id": 11073, - "bbox": [ - 407.24, - 461.5, - 420.13, - 469.58 - ], - "text": "Z(s)", - "type": "text" - }, - { - "block_id": "p398-b25", - "global_id": 11074, - "bbox": [ - 182.73, - 561.83, - 192.06, - 569.83 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p398-b26", - "global_id": 11075, - "bbox": [ - 363.1, - 496.52, - 378.99, - 506.13 - ], - "text": "Y1(s)", - "type": "text" - }, - { - "block_id": "p398-b27", - "global_id": 11076, - "bbox": [ - 202.94, - 356.44, - 244.86, - 368.12 - ], - "text": "1 F\n4 V", - "type": "text" - }, - { - "block_id": "p398-b28", - "global_id": 11077, - "bbox": [ - 236.33, - 334.83, - 250.44, - 344.44 - ], - "text": "y1(t)", - "type": "text" - }, - { - "block_id": "p398-b29", - "global_id": 11078, - "bbox": [ - 376.06, - 346.73, - 411.29, - 367.05 - ], - "text": "vC(0) 16\n y2(0) 4", - "type": "text" - }, - { - "block_id": "p398-b30", - "global_id": 11079, - "bbox": [ - 319.42, - 376.26, - 333.52, - 385.87 - ], - "text": "y2(t)", - "type": "text" - }, - { - "block_id": "p398-b31", - "global_id": 11080, - "bbox": [ - 321.61, - 332.23, - 333.83, - 340.53 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p398-b32", - "global_id": 11081, - "bbox": [ - 231.34, - 382.25, - 248.23, - 390.55 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p398-b33", - "global_id": 11082, - "bbox": [ - 193.47, - 330.38, - 219.98, - 339.92 - ], - "text": "vC", - "type": "text" - }, - { - "block_id": "p398-b34", - "global_id": 11083, - "bbox": [ - 168.33, - 371.62, - 385.21, - 385.21 - ], - "text": "20 V\n\t\nH\n1\n2", - "type": "text" - }, - { - "block_id": "p398-b35", - "global_id": 11084, - "bbox": [ - 292.57, - 494.81, - 296.57, - 509.47 - ], - "text": "s\n4", - "type": "text" - }, - { - "block_id": "p398-b36", - "global_id": 11085, - "bbox": [ - 336.08, - 442.59, - 340.08, - 457.13 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p398-b37", - "global_id": 11086, - "bbox": [ - 275.38, - 419.73, - 284.26, - 427.73 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p398-b38", - "global_id": 11087, - "bbox": [ - 194.16, - 473.47, - 197.16, - 486.79 - ], - "text": "1\n5", - "type": "text" - }, - { - "block_id": "p398-b39", - "global_id": 11088, - "bbox": [ - 250.32, - 526.47, - 254.32, - 534.47 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p398-b41", - "global_id": 11089, - "bbox": [ - 152.68, - 546.85, - 165.56, - 554.93 - ], - "text": "Z(s)", - "type": "text" - }, - { - "block_id": "p398-b42", - "global_id": 11090, - "bbox": [ - 173.22, - 448.94, - 226.09, - 464.39 - ], - "text": "1\na", - "type": "text" - }, - { - "block_id": "p398-b43", - "global_id": 11091, - "bbox": [ - 173.22, - 541.89, - 177.22, - 549.89 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p398-b44", - "global_id": 11092, - "bbox": [ - 115.73, - 496.96, - 230.92, - 523.66 - ], - "text": "Y1(s)\nY2(s)\n20\ns", - "type": "text" - }, - { - "block_id": "p398-b45", - "global_id": 11093, - "bbox": [ - 249.83, - 474.44, - 253.83, - 491.05 - ], - "text": "s\n2", - "type": "text" - }, - { - "block_id": "p398-b46", - "global_id": 11094, - "bbox": [ - 124.65, - 440.84, - 158.29, - 458.14 - ], - "text": "16\ns\n1\ns", - "type": "text" - }, - { - "block_id": "p398-b47", - "global_id": 11095, - "bbox": [ - 268.37, - 372.16, - 271.37, - 385.48 - ], - "text": "1\n5", - "type": "text" - }, - { - "block_id": "p398-b48", - "global_id": 11096, - "bbox": [ - 101.77, - 576.54, - 417.47, - 585.77 - ], - "text": "Figure 4.13 Using initial condition generators and Thévenin equivalent representation.", - "type": "text" - }, - { - "block_id": "p398-b49", - "global_id": 11097, - "bbox": [ - 103.16, - 599.5, - 477.02, - 622.49 - ], - "text": "Inspection of this circuit shows that when the switch is closed and the steady-state\nconditions are reached, the capacitor voltage vC = 16 volts, and the inductor current y2 = 4", - "type": "text" - } - ] - }, - { - "page_num": 399, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p399-b0", - "global_id": 11098, - "bbox": [ - 217.76, - 62.89, - 516.12, - 71.98 - ], - "text": "4.4\nAnalysis of Electrical Networks: The Transformed Network\n379", - "type": "text" - }, - { - "block_id": "p399-b1", - "global_id": 11099, - "bbox": [ - 128.9, - 85.49, - 502.76, - 145.69 - ], - "text": "amperes. Therefore, when the switch is opened (at t = 0), the initial conditions are vC(0−) = 16\nand y2(0−) = 4. Figure 4.13b shows the transformed version of the circuit in Fig. 4.13a. We\nhave used equivalent sources to account for the initial conditions. The initial capacitor voltage\nof 16 volts is represented by a series voltage of 16/s and the initial inductor current of 4\namperes is represented by a source of value Ly2(0−) = 2.", - "type": "text" - }, - { - "block_id": "p399-b2", - "global_id": 11100, - "bbox": [ - 146.84, - 146.9, - 491.74, - 156.87 - ], - "text": "From Fig. 4.13b, the loop equations can be written directly in the frequency domain as", - "type": "text" - }, - { - "block_id": "p399-b3", - "global_id": 11101, - "bbox": [ - 259.15, - 162.46, - 280.0, - 173.61 - ], - "text": "Y1(s)", - "type": "text" - }, - { - "block_id": "p399-b4", - "global_id": 11102, - "bbox": [ - 267.63, - 162.88, - 298.23, - 186.8 - ], - "text": "s\n+ 1", - "type": "text" - }, - { - "block_id": "p399-b5", - "global_id": 11103, - "bbox": [ - 293.24, - 162.88, - 376.63, - 186.9 - ], - "text": "5[Y1(s) −Y2(s)] = 4\ns", - "type": "text" - }, - { - "block_id": "p399-b6", - "global_id": 11104, - "bbox": [ - 244.47, - 187.83, - 258.42, - 204.36 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p399-b7", - "global_id": 11105, - "bbox": [ - 253.44, - 187.73, - 375.45, - 219.96 - ], - "text": "5Y1(s) + 6\n5Y2(s) + s\n2Y2(s) = 2\n 1", - "type": "text" - }, - { - "block_id": "p399-b8", - "global_id": 11106, - "bbox": [ - 266.73, - 212.99, - 286.58, - 227.83 - ], - "text": "s + 1", - "type": "text" - }, - { - "block_id": "p399-b9", - "global_id": 11107, - "bbox": [ - 283.09, - 212.99, - 314.68, - 227.52 - ], - "text": "5\n−1", - "type": "text" - }, - { - "block_id": "p399-b10", - "global_id": 11108, - "bbox": [ - 269.64, - 220.55, - 314.69, - 236.69 - ], - "text": "5\n−1", - "type": "text" - }, - { - "block_id": "p399-b11", - "global_id": 11109, - "bbox": [ - 278.61, - 225.31, - 319.17, - 240.22 - ], - "text": "5\n6\n5 + s\n2", - "type": "text" - }, - { - "block_id": "p399-b12", - "global_id": 11110, - "bbox": [ - 320.37, - 206.45, - 353.17, - 225.51 - ], - "text": "! Y1(s)", - "type": "text" - }, - { - "block_id": "p399-b13", - "global_id": 11111, - "bbox": [ - 332.33, - 226.31, - 353.17, - 237.46 - ], - "text": "Y2(s)", - "type": "text" - }, - { - "block_id": "p399-b14", - "global_id": 11112, - "bbox": [ - 353.17, - 206.45, - 358.6, - 216.41 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p399-b15", - "global_id": 11113, - "bbox": [ - 360.65, - 220.43, - 368.42, - 230.39 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p399-b16", - "global_id": 11114, - "bbox": [ - 370.46, - 206.45, - 380.58, - 220.17 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p399-b17", - "global_id": 11115, - "bbox": [ - 376.34, - 220.69, - 381.32, - 236.91 - ], - "text": "s\n2", - "type": "text" - }, - { - "block_id": "p399-b18", - "global_id": 11116, - "bbox": [ - 381.77, - 206.45, - 387.2, - 216.41 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p399-b19", - "global_id": 11117, - "bbox": [ - 146.84, - 244.5, - 351.9, - 254.46 - ], - "text": "Application of Cramer’s rule to this equation yields", - "type": "text" - }, - { - "block_id": "p399-b20", - "global_id": 11118, - "bbox": [ - 206.3, - 260.06, - 390.18, - 284.49 - ], - "text": "Y1(s) =\n24(s + 2)\ns2 + 7s + 12 =\n24(s + 2)\n(s + 3)(s + 4) = −24", - "type": "text" - }, - { - "block_id": "p399-b21", - "global_id": 11119, - "bbox": [ - 371.45, - 260.47, - 419.28, - 284.49 - ], - "text": "s + 3 + 48", - "type": "text" - }, - { - "block_id": "p399-b22", - "global_id": 11120, - "bbox": [ - 404.44, - 274.12, - 424.16, - 284.49 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p399-b23", - "global_id": 11121, - "bbox": [ - 128.9, - 290.03, - 143.28, - 299.99 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p399-b24", - "global_id": 11122, - "bbox": [ - 253.91, - 297.46, - 377.75, - 312.72 - ], - "text": "y1(t) = (−24e−3t + 48e−4t)u(t)", - "type": "text" - }, - { - "block_id": "p399-b25", - "global_id": 11123, - "bbox": [ - 128.91, - 320.7, - 209.34, - 330.66 - ], - "text": "Similarly, we obtain", - "type": "text" - }, - { - "block_id": "p399-b26", - "global_id": 11124, - "bbox": [ - 240.6, - 330.29, - 384.99, - 354.72 - ], - "text": "Y2(s) =\n4(s + 7)\ns2 + 7s + 12 =\n16\ns + 3 −12", - "type": "text" - }, - { - "block_id": "p399-b27", - "global_id": 11125, - "bbox": [ - 370.15, - 344.35, - 389.87, - 354.72 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p399-b28", - "global_id": 11126, - "bbox": [ - 128.9, - 360.13, - 143.28, - 370.09 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p399-b29", - "global_id": 11127, - "bbox": [ - 257.8, - 367.57, - 373.87, - 382.83 - ], - "text": "y2(t) = (16e−3t −12e−4t)u(t)", - "type": "text" - }, - { - "block_id": "p399-b30", - "global_id": 11128, - "bbox": [ - 128.91, - 390.4, - 502.76, - 436.64 - ], - "text": "We also could have used Thévenin’s theorem to compute Y1(s) and Y2(s) by replacing\nthe circuit to the right of the capacitor (right of terminals ab) with its Thévenin equivalent, as\nshown in Fig. 4.13c. Figure 4.13b shows that the Thévenin impedance Z(s) and the Thévenin\nsource V(s) are", - "type": "text" - }, - { - "block_id": "p399-b31", - "global_id": 11129, - "bbox": [ - 255.25, - 460.0, - 282.4, - 470.28 - ], - "text": "Z(s) =", - "type": "text" - }, - { - "block_id": "p399-b32", - "global_id": 11130, - "bbox": [ - 286.85, - 442.17, - 291.83, - 466.19 - ], - "text": "1\n5", - "type": "text" - }, - { - "block_id": "p399-b33", - "global_id": 11131, - "bbox": [ - 294.13, - 437.74, - 305.18, - 452.03 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p399-b34", - "global_id": 11132, - "bbox": [ - 300.75, - 437.74, - 328.2, - 466.19 - ], - "text": "2 + 1", - "type": "text" - }, - { - "block_id": "p399-b35", - "global_id": 11133, - "bbox": [ - 287.39, - 453.02, - 372.15, - 490.34 - ], - "text": "1\n5 + s\n2 + 1\n=\ns + 2\n5s + 12", - "type": "text" - }, - { - "block_id": "p399-b36", - "global_id": 11134, - "bbox": [ - 254.43, - 509.08, - 282.4, - 519.36 - ], - "text": "V(s) =", - "type": "text" - }, - { - "block_id": "p399-b37", - "global_id": 11135, - "bbox": [ - 298.82, - 491.24, - 312.76, - 507.78 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p399-b38", - "global_id": 11136, - "bbox": [ - 286.85, - 502.1, - 376.03, - 539.41 - ], - "text": "5\n1\n5 + s\n2 + 1\n2 =\n−4\n5s + 12", - "type": "text" - }, - { - "block_id": "p399-b39", - "global_id": 11137, - "bbox": [ - 128.9, - 542.96, - 340.82, - 554.11 - ], - "text": "According to Fig. 4.13c, the current Y1(s) is given by", - "type": "text" - }, - { - "block_id": "p399-b40", - "global_id": 11138, - "bbox": [ - 248.75, - 577.14, - 279.4, - 588.29 - ], - "text": "Y1(s) =", - "type": "text" - }, - { - "block_id": "p399-b41", - "global_id": 11139, - "bbox": [ - 283.84, - 559.28, - 319.04, - 583.2 - ], - "text": "4\ns −V(s)", - "type": "text" - }, - { - "block_id": "p399-b42", - "global_id": 11140, - "bbox": [ - 284.24, - 583.46, - 318.63, - 607.38 - ], - "text": "1\ns + Z(s)", - "type": "text" - }, - { - "block_id": "p399-b43", - "global_id": 11141, - "bbox": [ - 322.29, - 570.16, - 381.71, - 594.59 - ], - "text": "=\n24(s + 2)\ns2 + 7s + 12", - "type": "text" - }, - { - "block_id": "p399-b44", - "global_id": 11142, - "bbox": [ - 128.9, - 611.03, - 443.11, - 622.18 - ], - "text": "which confirms the earlier result. We may determine Y2(s) in a similar manner.", - "type": "text" - } - ] - }, - { - "page_num": 400, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p400-b0", - "global_id": 11143, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "380\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p400-b1", - "global_id": 11144, - "bbox": [ - 76.77, - 93.92, - 481.81, - 105.87 - ], - "text": "EXAMPLE 4.19\nTransformed Analysis of a Coupled Inductive Network", - "type": "text" - }, - { - "block_id": "p400-b2", - "global_id": 11145, - "bbox": [ - 103.16, - 122.12, - 477.03, - 156.41 - ], - "text": "The switch in the circuit in Fig. 4.14a is at position a for a long time before t = 0, when it is\nmoved instantaneously to position b. Determine the current y1(t) and the output voltage v0(t)\nfor t ≥0.", - "type": "text" - }, - { - "block_id": "p400-b3", - "global_id": 11146, - "bbox": [ - 103.16, - 179.32, - 477.04, - 202.01 - ], - "text": "Just before switching, the values of the loop currents are 2 and 1, respectively, that is,\ny1(0−) = 2 and y2(0−) = 1.", - "type": "text" - }, - { - "block_id": "p400-b4", - "global_id": 11147, - "bbox": [ - 103.17, - 203.24, - 477.04, - 273.74 - ], - "text": "The equivalent circuits for two types of inductive coupling are illustrated in Figs. 4.14b and\n4.14c. For our situation, the circuit in Fig. 4.14c applies. Figure 4.14d shows the transformed\nversion of the circuit in Fig. 4.14a after switching. Note that the inductors L1 + M, L2 + M,\nand −M are 3, 4, and −1 henries with impedances 3s, 4s, and −s respectively. The initial\ncondition voltages in the three branches are (L1 + M)y1(0−) = 6, (L2 + M)y2(0−) = 4, and\n−M[y1(0−) −y2(0−)] = −1, respectively. The two loop equations of the circuit are", - "type": "text" - }, - { - "block_id": "p400-b5", - "global_id": 11148, - "bbox": [ - 214.13, - 288.89, - 349.0, - 306.61 - ], - "text": "(2s + 3)Y1(s) + (s −1)Y2(s) = 10", - "type": "text" - }, - { - "block_id": "p400-b6", - "global_id": 11149, - "bbox": [ - 342.08, - 295.46, - 366.04, - 312.81 - ], - "text": "s + 5", - "type": "text" - }, - { - "block_id": "p400-b7", - "global_id": 11150, - "bbox": [ - 214.13, - 315.0, - 342.82, - 326.15 - ], - "text": "(s −1)Y1(s) + (3s + 2)Y2(s) = 5", - "type": "text" - }, - { - "block_id": "p400-b8", - "global_id": 11151, - "bbox": [ - 103.17, - 343.31, - 286.88, - 374.91 - ], - "text": "or\n 2s + 3\ns −1\ns −1\n3s + 2", - "type": "text" - }, - { - "block_id": "p400-b9", - "global_id": 11152, - "bbox": [ - 286.9, - 344.67, - 319.69, - 363.73 - ], - "text": "! Y1(s)", - "type": "text" - }, - { - "block_id": "p400-b10", - "global_id": 11153, - "bbox": [ - 298.86, - 364.53, - 319.69, - 375.68 - ], - "text": "Y2(s)", - "type": "text" - }, - { - "block_id": "p400-b11", - "global_id": 11154, - "bbox": [ - 319.7, - 344.67, - 351.45, - 358.39 - ], - "text": "! 5s+10", - "type": "text" - }, - { - "block_id": "p400-b12", - "global_id": 11155, - "bbox": [ - 339.68, - 358.92, - 344.66, - 375.14 - ], - "text": "s\n5", - "type": "text" - }, - { - "block_id": "p400-b13", - "global_id": 11156, - "bbox": [ - 352.66, - 344.67, - 358.09, - 354.63 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p400-b14", - "global_id": 11157, - "bbox": [ - 103.17, - 383.16, - 215.59, - 394.3 - ], - "text": "Solving for Y1(s), we obtain", - "type": "text" - }, - { - "block_id": "p400-b15", - "global_id": 11158, - "bbox": [ - 185.13, - 407.0, - 270.6, - 428.75 - ], - "text": "Y1(s) = 2s2 + 9s + 4", - "type": "text" - }, - { - "block_id": "p400-b16", - "global_id": 11159, - "bbox": [ - 219.02, - 411.03, - 293.01, - 435.05 - ], - "text": "s(s2 + 3s + 1) = 4", - "type": "text" - }, - { - "block_id": "p400-b17", - "global_id": 11160, - "bbox": [ - 288.57, - 411.03, - 393.84, - 435.05 - ], - "text": "s −\n1\ns + 0.382 −\n1\ns + 2.618", - "type": "text" - }, - { - "block_id": "p400-b18", - "global_id": 11161, - "bbox": [ - 103.17, - 450.54, - 144.91, - 460.5 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p400-b19", - "global_id": 11162, - "bbox": [ - 221.88, - 457.97, - 358.3, - 473.23 - ], - "text": "y1(t) = (4 −e−0.382t −e−2.618t)u(t)", - "type": "text" - }, - { - "block_id": "p400-b20", - "global_id": 11163, - "bbox": [ - 103.17, - 481.43, - 142.09, - 491.39 - ], - "text": "Similarly,", - "type": "text" - }, - { - "block_id": "p400-b21", - "global_id": 11164, - "bbox": [ - 185.13, - 488.91, - 293.01, - 516.97 - ], - "text": "Y2(s) =\ns2 + 2s + 2\ns(s2 + 3s + 1) = 2", - "type": "text" - }, - { - "block_id": "p400-b22", - "global_id": 11165, - "bbox": [ - 288.57, - 492.95, - 393.84, - 516.97 - ], - "text": "s −\n1.618\ns + 0.382 +\n0.618\ns + 2.618", - "type": "text" - }, - { - "block_id": "p400-b23", - "global_id": 11166, - "bbox": [ - 103.17, - 523.49, - 117.55, - 533.46 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p400-b24", - "global_id": 11167, - "bbox": [ - 199.47, - 530.92, - 380.71, - 546.18 - ], - "text": "y2(t) = (2 −1.618e−0.382t + 0.618e−2.618t)u(t)", - "type": "text" - }, - { - "block_id": "p400-b25", - "global_id": 11168, - "bbox": [ - 103.17, - 554.37, - 225.82, - 564.33 - ], - "text": "The output voltage is therefore", - "type": "text" - }, - { - "block_id": "p400-b26", - "global_id": 11169, - "bbox": [ - 184.15, - 577.74, - 396.04, - 593.01 - ], - "text": "v0(t) = y2(t) = (2 −1.618e−0.382t + 0.618e−2.618t)u(t)", - "type": "text" - } - ] - }, - { - "page_num": 401, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p401-b0", - "global_id": 11170, - "bbox": [ - 217.76, - 62.89, - 516.12, - 71.98 - ], - "text": "4.4\nAnalysis of Electrical Networks: The Transformed Network\n381", - "type": "text" - }, - { - "block_id": "p401-b1", - "global_id": 11171, - "bbox": [ - 315.87, - 208.79, - 324.75, - 216.79 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p401-b2", - "global_id": 11172, - "bbox": [ - 262.32, - 422.58, - 385.2, - 432.23 - ], - "text": "Y1(s)\nY2(s)", - "type": "text" - }, - { - "block_id": "p401-b3", - "global_id": 11173, - "bbox": [ - 233.8, - 361.84, - 237.8, - 369.84 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p401-b4", - "global_id": 11174, - "bbox": [ - 316.69, - 496.16, - 326.02, - 504.16 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p401-b5", - "global_id": 11175, - "bbox": [ - 264.85, - 358.71, - 396.26, - 370.47 - ], - "text": "6\n3s\n4\n4s", - "type": "text" - }, - { - "block_id": "p401-b6", - "global_id": 11176, - "bbox": [ - 320.53, - 395.32, - 330.3, - 403.54 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p401-b7", - "global_id": 11177, - "bbox": [ - 311.73, - 419.28, - 423.1, - 432.12 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p401-b8", - "global_id": 11178, - "bbox": [ - 214.08, - 331.91, - 223.73, - 339.91 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p401-b9", - "global_id": 11179, - "bbox": [ - 368.68, - 266.61, - 375.34, - 274.61 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p401-b10", - "global_id": 11180, - "bbox": [ - 412.52, - 331.91, - 421.4, - 339.91 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p401-b11", - "global_id": 11181, - "bbox": [ - 211.19, - 278.78, - 411.16, - 294.12 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p401-b12", - "global_id": 11182, - "bbox": [ - 142.46, - 271.64, - 202.28, - 281.25 - ], - "text": "L1\nL2", - "type": "text" - }, - { - "block_id": "p401-b13", - "global_id": 11183, - "bbox": [ - 168.47, - 238.94, - 175.14, - 246.94 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p401-b14", - "global_id": 11184, - "bbox": [ - 462.36, - 278.56, - 475.69, - 286.78 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p401-b17", - "global_id": 11185, - "bbox": [ - 175.93, - 154.81, - 191.7, - 162.81 - ], - "text": "10 V", - "type": "text" - }, - { - "block_id": "p401-b18", - "global_id": 11186, - "bbox": [ - 196.2, - 107.98, - 238.89, - 122.16 - ], - "text": "R1\nt 0", - "type": "text" - }, - { - "block_id": "p401-b19", - "global_id": 11187, - "bbox": [ - 134.43, - 154.53, - 356.51, - 165.06 - ], - "text": "y1(t)\n5 V\ny2(t)", - "type": "text" - }, - { - "block_id": "p401-b20", - "global_id": 11188, - "bbox": [ - 297.3, - 90.98, - 303.97, - 98.98 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p401-b21", - "global_id": 11189, - "bbox": [ - 189.73, - 116.36, - 193.28, - 124.36 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p401-b22", - "global_id": 11190, - "bbox": [ - 206.4, - 134.91, - 210.4, - 142.91 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p401-b23", - "global_id": 11191, - "bbox": [ - 425.56, - 89.69, - 496.46, - 99.51 - ], - "text": "R1 2, R2 R3 1", - "type": "text" - }, - { - "block_id": "p401-b24", - "global_id": 11192, - "bbox": [ - 425.56, - 103.69, - 507.66, - 113.51 - ], - "text": "M 1, L1 2, L2 3", - "type": "text" - }, - { - "block_id": "p401-b25", - "global_id": 11193, - "bbox": [ - 376.14, - 168.22, - 384.03, - 177.83 - ], - "text": "R3", - "type": "text" - }, - { - "block_id": "p401-b26", - "global_id": 11194, - "bbox": [ - 273.9, - 129.35, - 356.35, - 138.96 - ], - "text": "L2\nL1", - "type": "text" - }, - { - "block_id": "p401-b27", - "global_id": 11195, - "bbox": [ - 294.27, - 152.66, - 425.25, - 164.84 - ], - "text": "R2\nv0(t)", - "type": "text" - }, - { - "block_id": "p401-b28", - "global_id": 11196, - "bbox": [ - 230.14, - 237.15, - 492.56, - 247.16 - ], - "text": "L1 M\nL1 M\nL2 M\nL2 M", - "type": "text" - }, - { - "block_id": "p401-b29", - "global_id": 11197, - "bbox": [ - 337.43, - 271.28, - 396.88, - 280.88 - ], - "text": "L1\nL2", - "type": "text" - }, - { - "block_id": "p401-b30", - "global_id": 11198, - "bbox": [ - 191.57, - 418.83, - 199.57, - 433.43 - ], - "text": "10\ns", - "type": "text" - }, - { - "block_id": "p401-b31", - "global_id": 11199, - "bbox": [ - 321.29, - 454.28, - 325.29, - 462.28 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p401-b32", - "global_id": 11200, - "bbox": [ - 128.9, - 510.85, - 449.21, - 520.09 - ], - "text": "Figure 4.14 Solution of a coupled inductive network by the transformed circuit method.", - "type": "text" - }, - { - "block_id": "p401-b33", - "global_id": 11201, - "bbox": [ - 133.57, - 574.9, - 510.13, - 586.86 - ], - "text": "DRILL 4.10\nTransformed Analysis of an RLC Circuit with a Switch", - "type": "text" - }, - { - "block_id": "p401-b34", - "global_id": 11202, - "bbox": [ - 133.57, - 595.56, - 510.15, - 630.55 - ], - "text": "For the RLC circuit in Fig. 4.15, the input is switched on at t = 0. The initial conditions are\ny(0−) = 2 amperes and vC(0−) = 50 volts. Find the loop current y(t) and the capacitor voltage\nvC(t) for t ≥0.", - "type": "text" - }, - { - "block_id": "p401-b35", - "global_id": 11203, - "bbox": [ - 133.84, - 643.37, - 196.06, - 654.33 - ], - "text": "ANSWERS", - "type": "text" - } - ] - }, - { - "page_num": 402, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p402-b0", - "global_id": 11204, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "382\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p402-b1", - "global_id": 11205, - "bbox": [ - 107.82, - 93.96, - 144.46, - 104.34 - ], - "text": "y(t) = 10", - "type": "text" - }, - { - "block_id": "p402-b2", - "global_id": 11206, - "bbox": [ - 144.45, - 85.54, - 152.88, - 95.5 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p402-b3", - "global_id": 11207, - "bbox": [ - 152.88, - 90.76, - 440.97, - 105.04 - ], - "text": "2e−t cos(2t + 81.8◦)u(t) and vC(t) = [24 + 31.62e−t cos(2t −34.7◦)]u(t)", - "type": "text" - }, - { - "block_id": "p402-b4", - "global_id": 11208, - "bbox": [ - 158.66, - 139.4, - 254.6, - 147.69 - ], - "text": "t 0\n2 \t\n1 H", - "type": "text" - }, - { - "block_id": "p402-b5", - "global_id": 11209, - "bbox": [ - 117.39, - 182.71, - 330.1, - 193.46 - ], - "text": "0.2 F\nvC(t)\n24 V", - "type": "text" - }, - { - "block_id": "p402-b8", - "global_id": 11210, - "bbox": [ - 211.4, - 186.17, - 222.83, - 194.26 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p402-b9", - "global_id": 11211, - "bbox": [ - 340.35, - 212.04, - 467.37, - 221.28 - ], - "text": "Figure 4.15 Circuit for Drill 4.10.", - "type": "text" - }, - { - "block_id": "p402-b10", - "global_id": 11212, - "bbox": [ - 101.84, - 273.01, - 278.62, - 284.96 - ], - "text": "4.4-1 Analysis of Active Circuits", - "type": "text" - }, - { - "block_id": "p402-b11", - "global_id": 11213, - "bbox": [ - 101.84, - 291.09, - 490.4, - 336.92 - ], - "text": "Although we have considered examples of only passive networks so far, the circuit analysis\nprocedure using the Laplace transform is also applicable to active circuits. All that is needed is\nto replace the active elements with their mathematical models (or equivalent circuits) and proceed\nas before.", - "type": "text" - }, - { - "block_id": "p402-b12", - "global_id": 11214, - "bbox": [ - 101.84, - 338.91, - 490.4, - 396.69 - ], - "text": "The operational amplifier (depicted by the triangular symbol in Fig. 4.16a) is a well-known\nelement in modern electronic circuits. The terminals with the positive and the negative signs\ncorrespond to noninverting and inverting terminals, respectively. This means that the polarity of\nthe output voltage v2 is the same as that of the input voltage at the terminal marked by the positive\nsign (noninverting). The opposite is true for the inverting terminal, marked by the negative sign.", - "type": "text" - }, - { - "block_id": "p402-b13", - "global_id": 11215, - "bbox": [ - 101.84, - 398.69, - 490.41, - 468.42 - ], - "text": "Figure 4.16b shows the model (equivalent circuit) of the operational amplifier (op amp) in\nFig. 4.16a. A typical op amp has a very large gain. The output voltage v2 = −Av1, where A is\ntypically 105 to 106. The input impedance is very high, of the order of 1012 , and the output\nimpedance is very low (50–100 ). For most applications, we are justified in assuming the gain A\nand the input impedance to be infinite and the output impedance to be zero. For this reason we see\nan ideal voltage source at the output.", - "type": "text" - }, - { - "block_id": "p402-b14", - "global_id": 11216, - "bbox": [ - 101.85, - 470.32, - 490.39, - 517.74 - ], - "text": "Consider now the operational amplifier with resistors Ra and Rb connected, as shown in\nFig. 4.16c. This configuration is known as the noninverting amplifier. Observe that the input\npolarities in this configuration are inverted in comparison to those in Fig. 4.16a. We now show\nthat the output voltage v2 and the input voltage v1 in this case are related by", - "type": "text" - }, - { - "block_id": "p402-b15", - "global_id": 11217, - "bbox": [ - 230.75, - 530.12, - 359.78, - 543.63 - ], - "text": "v2 = Kv1,\nwhere K = 1 + Rb", - "type": "text" - }, - { - "block_id": "p402-b16", - "global_id": 11218, - "bbox": [ - 352.54, - 538.32, - 359.78, - 546.11 - ], - "text": "Ra", - "type": "text" - }, - { - "block_id": "p402-b17", - "global_id": 11219, - "bbox": [ - 101.84, - 558.89, - 490.4, - 617.37 - ], - "text": "First, we recognize that because the input impedance and the gain of the operational amplifier\napproach infinity, the input current ix and the input voltage vx in Fig. 4.16c are infinitesimal and\nmay be taken as zero. The dependent source in this case is Avx instead of −Avx because of the input\npolarity inversion. The dependent source Avx (see Fig. 4.16b) at the output will generate current\nio, as illustrated in Fig. 4.16c. Now", - "type": "text" - }, - { - "block_id": "p402-b18", - "global_id": 11220, - "bbox": [ - 263.37, - 624.83, - 328.36, - 636.29 - ], - "text": "v2 = (Rb + Ra)io", - "type": "text" - } - ] - }, - { - "page_num": 403, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p403-b0", - "global_id": 11221, - "bbox": [ - 217.76, - 62.89, - 516.12, - 71.98 - ], - "text": "4.4\nAnalysis of Electrical Networks: The Transformed Network\n383", - "type": "text" - }, - { - "block_id": "p403-b2", - "global_id": 11222, - "bbox": [ - 285.97, - 119.04, - 300.07, - 128.65 - ], - "text": "v2(t)", - "type": "text" - }, - { - "block_id": "p403-b3", - "global_id": 11223, - "bbox": [ - 194.07, - 95.44, - 361.18, - 105.05 - ], - "text": "v1(t)\nv1(t)", - "type": "text" - }, - { - "block_id": "p403-b4", - "global_id": 11224, - "bbox": [ - 415.39, - 119.04, - 476.88, - 135.54 - ], - "text": "v2(t)\nAv1(t)", - "type": "text" - }, - { - "block_id": "p403-b5", - "global_id": 11225, - "bbox": [ - 404.26, - 169.7, - 413.91, - 177.7 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p403-b6", - "global_id": 11226, - "bbox": [ - 232.95, - 288.74, - 409.61, - 296.74 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p403-b14", - "global_id": 11227, - "bbox": [ - 232.95, - 169.7, - 241.83, - 177.7 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p403-b15", - "global_id": 11228, - "bbox": [ - 151.5, - 228.92, - 165.6, - 238.53 - ], - "text": "v1(t)", - "type": "text" - }, - { - "block_id": "p403-b16", - "global_id": 11229, - "bbox": [ - 211.47, - 193.05, - 215.47, - 201.05 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p403-b17", - "global_id": 11230, - "bbox": [ - 223.27, - 237.19, - 323.29, - 256.76 - ], - "text": "v2(t)\nRa\nRb", - "type": "text" - }, - { - "block_id": "p403-b18", - "global_id": 11231, - "bbox": [ - 285.77, - 216.61, - 319.57, - 227.97 - ], - "text": "io(t)", - "type": "text" - }, - { - "block_id": "p403-b21", - "global_id": 11232, - "bbox": [ - 181.31, - 190.23, - 193.75, - 199.78 - ], - "text": "ix(t)", - "type": "text" - }, - { - "block_id": "p403-b22", - "global_id": 11233, - "bbox": [ - 200.56, - 208.69, - 206.77, - 218.23 - ], - "text": "vx", - "type": "text" - }, - { - "block_id": "p403-b25", - "global_id": 11234, - "bbox": [ - 358.62, - 233.82, - 469.62, - 246.31 - ], - "text": "v1(t)\nKv1(t)\nv2(t)", - "type": "text" - }, - { - "block_id": "p403-b31", - "global_id": 11235, - "bbox": [ - 151.5, - 303.44, - 371.27, - 312.67 - ], - "text": "Figure 4.16 Operational amplifier and its equivalent circuit.", - "type": "text" - }, - { - "block_id": "p403-b32", - "global_id": 11236, - "bbox": [ - 127.59, - 341.18, - 160.52, - 351.14 - ], - "text": "and also", - "type": "text" - }, - { - "block_id": "p403-b33", - "global_id": 11237, - "bbox": [ - 279.5, - 364.72, - 363.72, - 376.17 - ], - "text": "v1 = vx + Raio = Raio", - "type": "text" - }, - { - "block_id": "p403-b34", - "global_id": 11238, - "bbox": [ - 127.59, - 392.06, - 169.34, - 402.02 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p403-b35", - "global_id": 11239, - "bbox": [ - 265.4, - 413.04, - 273.31, - 437.93 - ], - "text": "v2\nv1", - "type": "text" - }, - { - "block_id": "p403-b36", - "global_id": 11240, - "bbox": [ - 277.06, - 412.72, - 318.58, - 429.98 - ], - "text": "= Rb + Ra", - "type": "text" - }, - { - "block_id": "p403-b37", - "global_id": 11241, - "bbox": [ - 298.54, - 427.09, - 308.11, - 437.86 - ], - "text": "Ra", - "type": "text" - }, - { - "block_id": "p403-b38", - "global_id": 11242, - "bbox": [ - 322.33, - 413.04, - 358.76, - 430.08 - ], - "text": "= 1 + Rb", - "type": "text" - }, - { - "block_id": "p403-b39", - "global_id": 11243, - "bbox": [ - 349.19, - 427.09, - 358.76, - 437.86 - ], - "text": "Ra", - "type": "text" - }, - { - "block_id": "p403-b40", - "global_id": 11244, - "bbox": [ - 362.5, - 419.71, - 378.97, - 429.98 - ], - "text": "= K", - "type": "text" - }, - { - "block_id": "p403-b41", - "global_id": 11245, - "bbox": [ - 127.59, - 449.31, - 135.89, - 459.27 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p403-b42", - "global_id": 11246, - "bbox": [ - 293.81, - 472.84, - 349.91, - 483.99 - ], - "text": "v2(t) = Kv1(t)", - "type": "text" - }, - { - "block_id": "p403-b43", - "global_id": 11247, - "bbox": [ - 127.59, - 500.18, - 432.92, - 510.15 - ], - "text": "The equivalent circuit of the noninverting amplifier is depicted in Fig. 4.16d.", - "type": "text" - }, - { - "block_id": "p403-b44", - "global_id": 11248, - "bbox": [ - 102.51, - 543.15, - 443.44, - 555.1 - ], - "text": "EXAMPLE 4.20\nTransform Analysis of a Sallen–Key Circuit", - "type": "text" - }, - { - "block_id": "p403-b45", - "global_id": 11249, - "bbox": [ - 128.9, - 571.66, - 502.77, - 606.34 - ], - "text": "The circuit in Fig. 4.17a is called the Sallen–Key circuit, which is frequently used in filter\ndesign. Find the transfer function H(s) relating the output voltage vo(t) to the input voltage\nvi(t).", - "type": "text" - } - ] - }, - { - "page_num": 404, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p404-b0", - "global_id": 11250, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "384\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p404-b1", - "global_id": 11251, - "bbox": [ - 94.2, - 416.77, - 304.06, - 426.09 - ], - "text": "Figure 4.17 (a) Sallen–Key circuit and (b) its equivalent.", - "type": "text" - }, - { - "block_id": "p404-b2", - "global_id": 11252, - "bbox": [ - 121.09, - 453.76, - 213.06, - 463.72 - ], - "text": "We are required to find", - "type": "text" - }, - { - "block_id": "p404-b3", - "global_id": 11253, - "bbox": [ - 262.8, - 463.34, - 316.19, - 480.59 - ], - "text": "H(s) = Vo(s)", - "type": "text" - }, - { - "block_id": "p404-b4", - "global_id": 11254, - "bbox": [ - 295.58, - 477.39, - 315.42, - 488.47 - ], - "text": "Vi(s)", - "type": "text" - }, - { - "block_id": "p404-b5", - "global_id": 11255, - "bbox": [ - 103.17, - 494.29, - 267.27, - 504.26 - ], - "text": "assuming all initial conditions to be zero.", - "type": "text" - }, - { - "block_id": "p404-b6", - "global_id": 11256, - "bbox": [ - 103.17, - 506.26, - 477.02, - 552.08 - ], - "text": "Figure 4.17b shows the transformed version of the circuit in Fig. 4.17a. The noninverting\namplifier is replaced by its equivalent circuit. All the voltages are replaced by their Laplace\ntransforms, and all the circuit elements are shown by their impedances. All the initial\nconditions are assumed to be zero, as required for determining H(s).", - "type": "text" - }, - { - "block_id": "p404-b7", - "global_id": 11257, - "bbox": [ - 103.17, - 554.07, - 477.03, - 576.7 - ], - "text": "We shall use node analysis to derive the result. There are two unknown node voltages,\nVa(s) and Vb(s), requiring two node equations.", - "type": "text" - }, - { - "block_id": "p404-b8", - "global_id": 11258, - "bbox": [ - 103.17, - 577.57, - 477.02, - 613.35 - ], - "text": "At node a, IR1(s), the current in R1 (leaving the node a), is [Va(s) −Vi(s)]/R1. Similarly,\nIR2(s), the current in R2 (leaving the node a), is [Va(s) −Vb(s)]/R2, and IC1(s), the current in\ncapacitor C1 (leaving the node a), is [Va(s) −Vo(s)]C1s = [Va(s) −KVb(s)]C1s.", - "type": "text" - } - ] - }, - { - "page_num": 405, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p405-b0", - "global_id": 11259, - "bbox": [ - 217.76, - 62.89, - 516.12, - 71.98 - ], - "text": "4.5\nAnalysis of Electrical Networks: The Transformed Network\n385", - "type": "text" - }, - { - "block_id": "p405-b1", - "global_id": 11260, - "bbox": [ - 146.84, - 86.23, - 351.82, - 96.2 - ], - "text": "The sum of all the three currents is zero. Therefore,", - "type": "text" - }, - { - "block_id": "p405-b2", - "global_id": 11261, - "bbox": [ - 201.78, - 105.07, - 253.87, - 116.15 - ], - "text": "Va(s) −Vi(s)", - "type": "text" - }, - { - "block_id": "p405-b3", - "global_id": 11262, - "bbox": [ - 222.78, - 119.43, - 232.35, - 130.27 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p405-b4", - "global_id": 11263, - "bbox": [ - 256.61, - 105.07, - 320.76, - 122.33 - ], - "text": "+ Va(s) −Vb(s)", - "type": "text" - }, - { - "block_id": "p405-b5", - "global_id": 11264, - "bbox": [ - 288.91, - 119.43, - 298.48, - 130.27 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p405-b6", - "global_id": 11265, - "bbox": [ - 323.5, - 112.06, - 431.08, - 123.2 - ], - "text": "+ [Va(s) −KVb(s)]C1s = 0", - "type": "text" - }, - { - "block_id": "p405-b7", - "global_id": 11266, - "bbox": [ - 128.9, - 136.56, - 215.0, - 155.08 - ], - "text": "or\n 1", - "type": "text" - }, - { - "block_id": "p405-b8", - "global_id": 11267, - "bbox": [ - 207.47, - 159.08, - 217.05, - 169.91 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p405-b9", - "global_id": 11268, - "bbox": [ - 220.29, - 145.12, - 238.33, - 162.07 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p405-b10", - "global_id": 11269, - "bbox": [ - 230.81, - 159.08, - 240.38, - 169.91 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p405-b11", - "global_id": 11270, - "bbox": [ - 243.63, - 151.69, - 267.45, - 162.84 - ], - "text": "+ C1s", - "type": "text" - }, - { - "block_id": "p405-b13", - "global_id": 11271, - "bbox": [ - 275.28, - 151.69, - 305.98, - 162.77 - ], - "text": "Va(s) −", - "type": "text" - }, - { - "block_id": "p405-b14", - "global_id": 11272, - "bbox": [ - 307.53, - 137.7, - 322.98, - 155.08 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p405-b15", - "global_id": 11273, - "bbox": [ - 315.46, - 159.08, - 325.03, - 169.91 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p405-b16", - "global_id": 11274, - "bbox": [ - 328.27, - 151.69, - 358.74, - 162.84 - ], - "text": "+ KC1s", - "type": "text" - }, - { - "block_id": "p405-b18", - "global_id": 11275, - "bbox": [ - 366.58, - 145.12, - 408.54, - 162.77 - ], - "text": "Vb(s) = 1", - "type": "text" - }, - { - "block_id": "p405-b19", - "global_id": 11276, - "bbox": [ - 401.02, - 159.08, - 410.59, - 169.91 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p405-b20", - "global_id": 11277, - "bbox": [ - 412.28, - 151.69, - 432.12, - 162.77 - ], - "text": "Vi(s)", - "type": "text" - }, - { - "block_id": "p405-b21", - "global_id": 11278, - "bbox": [ - 128.91, - 176.22, - 306.46, - 186.29 - ], - "text": "Similarly, the node equation at node b yields", - "type": "text" - }, - { - "block_id": "p405-b22", - "global_id": 11279, - "bbox": [ - 257.22, - 195.86, - 310.85, - 206.94 - ], - "text": "Vb(s) −Va(s)", - "type": "text" - }, - { - "block_id": "p405-b23", - "global_id": 11280, - "bbox": [ - 279.0, - 210.24, - 288.57, - 221.07 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p405-b24", - "global_id": 11281, - "bbox": [ - 313.59, - 202.85, - 375.64, - 214.0 - ], - "text": "+ C2sVb(s) = 0", - "type": "text" - }, - { - "block_id": "p405-b25", - "global_id": 11282, - "bbox": [ - 128.91, - 227.36, - 137.2, - 237.33 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p405-b26", - "global_id": 11283, - "bbox": [ - 244.29, - 235.91, - 260.78, - 252.45 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p405-b27", - "global_id": 11284, - "bbox": [ - 253.25, - 249.87, - 262.82, - 260.71 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p405-b28", - "global_id": 11285, - "bbox": [ - 264.52, - 242.48, - 295.22, - 253.56 - ], - "text": "Va(s) +", - "type": "text" - }, - { - "block_id": "p405-b29", - "global_id": 11286, - "bbox": [ - 296.76, - 228.5, - 312.21, - 245.88 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p405-b30", - "global_id": 11287, - "bbox": [ - 304.68, - 249.87, - 314.25, - 260.71 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p405-b31", - "global_id": 11288, - "bbox": [ - 317.5, - 242.48, - 341.33, - 253.63 - ], - "text": "+ C2s", - "type": "text" - }, - { - "block_id": "p405-b33", - "global_id": 11289, - "bbox": [ - 349.16, - 242.48, - 387.39, - 253.56 - ], - "text": "Vb(s) = 0", - "type": "text" - }, - { - "block_id": "p405-b34", - "global_id": 11290, - "bbox": [ - 128.91, - 266.77, - 502.76, - 289.09 - ], - "text": "The two node equations in two unknown node voltages Va(s) and Vb(s) can be expressed in\nmatrix form as", - "type": "text" - }, - { - "block_id": "p405-b35", - "global_id": 11291, - "bbox": [ - 203.94, - 290.24, - 336.34, - 321.56 - ], - "text": "G1 + G2 + C1s\n−(G2 + KC1s)\n−G2\n(G2 + C2s)", - "type": "text" - }, - { - "block_id": "p405-b36", - "global_id": 11292, - "bbox": [ - 336.33, - 290.24, - 368.57, - 309.23 - ], - "text": "! Va(s)", - "type": "text" - }, - { - "block_id": "p405-b37", - "global_id": 11293, - "bbox": [ - 347.19, - 310.1, - 368.57, - 321.18 - ], - "text": "Vb(s)", - "type": "text" - }, - { - "block_id": "p405-b38", - "global_id": 11294, - "bbox": [ - 368.57, - 290.24, - 374.0, - 300.2 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p405-b39", - "global_id": 11295, - "bbox": [ - 376.04, - 304.22, - 383.82, - 314.19 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p405-b40", - "global_id": 11296, - "bbox": [ - 385.86, - 290.24, - 422.31, - 309.3 - ], - "text": "G1Vi(s)", - "type": "text" - }, - { - "block_id": "p405-b41", - "global_id": 11297, - "bbox": [ - 404.31, - 310.52, - 409.29, - 320.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p405-b42", - "global_id": 11298, - "bbox": [ - 422.3, - 290.24, - 427.73, - 300.2 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p405-b43", - "global_id": 11299, - "bbox": [ - 128.91, - 331.85, - 153.24, - 341.81 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p405-b44", - "global_id": 11300, - "bbox": [ - 253.2, - 339.73, - 284.98, - 357.76 - ], - "text": "G1 = 1", - "type": "text" - }, - { - "block_id": "p405-b45", - "global_id": 11301, - "bbox": [ - 277.45, - 353.69, - 287.02, - 364.52 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p405-b46", - "global_id": 11302, - "bbox": [ - 308.64, - 339.73, - 374.72, - 357.76 - ], - "text": "and\nG2 = 1", - "type": "text" - }, - { - "block_id": "p405-b47", - "global_id": 11303, - "bbox": [ - 128.9, - 353.69, - 376.76, - 379.99 - ], - "text": "R2\nApplication of Cramer’s rule yields", - "type": "text" - }, - { - "block_id": "p405-b48", - "global_id": 11304, - "bbox": [ - 197.33, - 389.58, - 218.71, - 400.66 - ], - "text": "Vb(s)", - "type": "text" - }, - { - "block_id": "p405-b49", - "global_id": 11305, - "bbox": [ - 198.11, - 389.89, - 433.84, - 415.09 - ], - "text": "Vi(s) =\nG1G2\nC1C2s2 + [G1C2 + G2C2 + G2C1(1 −K)]s + G1G2", - "type": "text" - }, - { - "block_id": "p405-b50", - "global_id": 11306, - "bbox": [ - 221.96, - 416.79, - 269.24, - 434.67 - ], - "text": "=\nω02", - "type": "text" - }, - { - "block_id": "p405-b51", - "global_id": 11307, - "bbox": [ - 232.97, - 428.9, - 291.51, - 442.93 - ], - "text": "s2 + 2αs + ω02", - "type": "text" - }, - { - "block_id": "p405-b52", - "global_id": 11308, - "bbox": [ - 128.9, - 451.5, - 153.23, - 461.46 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p405-b53", - "global_id": 11309, - "bbox": [ - 211.64, - 469.25, - 257.32, - 486.29 - ], - "text": "K = 1 + Rb", - "type": "text" - }, - { - "block_id": "p405-b54", - "global_id": 11310, - "bbox": [ - 247.75, - 483.3, - 257.32, - 494.07 - ], - "text": "Ra", - "type": "text" - }, - { - "block_id": "p405-b55", - "global_id": 11311, - "bbox": [ - 278.94, - 475.92, - 323.26, - 487.07 - ], - "text": "and\nω0", - "type": "text" - }, - { - "block_id": "p405-b56", - "global_id": 11312, - "bbox": [ - 323.76, - 469.25, - 362.67, - 494.14 - ], - "text": "2 = G1G2\nC1C2", - "type": "text" - }, - { - "block_id": "p405-b57", - "global_id": 11313, - "bbox": [ - 177.67, - 469.35, - 418.33, - 516.61 - ], - "text": "=\n1\nR1R2C1C2\n2α = G1C2 + G2C2 + G2C1(1 −K)\nC1C2", - "type": "text" - }, - { - "block_id": "p405-b58", - "global_id": 11314, - "bbox": [ - 322.68, - 491.82, - 353.9, - 516.61 - ], - "text": "=\n1\nR1C1", - "type": "text" - }, - { - "block_id": "p405-b59", - "global_id": 11315, - "bbox": [ - 357.14, - 491.82, - 387.87, - 516.61 - ], - "text": "+\n1\nR2C1", - "type": "text" - }, - { - "block_id": "p405-b60", - "global_id": 11316, - "bbox": [ - 391.11, - 491.82, - 421.82, - 516.61 - ], - "text": "+\n1\nR2C2", - "type": "text" - }, - { - "block_id": "p405-b61", - "global_id": 11317, - "bbox": [ - 423.52, - 498.39, - 453.98, - 508.77 - ], - "text": "(1 −K)", - "type": "text" - }, - { - "block_id": "p405-b62", - "global_id": 11318, - "bbox": [ - 128.91, - 524.84, - 148.02, - 534.8 - ], - "text": "Now", - "type": "text" - }, - { - "block_id": "p405-b63", - "global_id": 11319, - "bbox": [ - 285.2, - 536.38, - 346.47, - 547.46 - ], - "text": "Vo(s) = KVb(s)", - "type": "text" - }, - { - "block_id": "p405-b64", - "global_id": 11320, - "bbox": [ - 128.91, - 555.72, - 170.65, - 565.68 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p405-b65", - "global_id": 11321, - "bbox": [ - 230.48, - 566.12, - 283.87, - 583.38 - ], - "text": "H(s) = Vo(s)", - "type": "text" - }, - { - "block_id": "p405-b66", - "global_id": 11322, - "bbox": [ - 263.27, - 566.12, - 326.71, - 591.25 - ], - "text": "Vi(s) = K Vb(s)", - "type": "text" - }, - { - "block_id": "p405-b67", - "global_id": 11323, - "bbox": [ - 306.09, - 565.18, - 380.82, - 591.25 - ], - "text": "Vi(s) =\nKω02", - "type": "text" - }, - { - "block_id": "p405-b68", - "global_id": 11324, - "bbox": [ - 340.95, - 577.29, - 399.49, - 591.32 - ], - "text": "s2 + 2αs + ω02", - "type": "text" - } - ] - }, - { - "page_num": 406, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p406-b0", - "global_id": 11325, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "386\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p406-b1", - "global_id": 11326, - "bbox": [ - 102.2, - 94.37, - 249.83, - 108.31 - ], - "text": "4.5 BLOCK DIAGRAMS", - "type": "text" - }, - { - "block_id": "p406-b2", - "global_id": 11327, - "bbox": [ - 101.84, - 114.3, - 490.42, - 195.99 - ], - "text": "Large systems may consist of an enormous number of components or elements. As anyone who has\nseen the circuit diagram of a radio or a television receiver can appreciate, analyzing such systems\nall at once could be next to impossible. In such cases, it is convenient to represent a system by\nsuitably interconnected subsystems, each of which can be readily analyzed. Each subsystem can\nbe characterized in terms of its input–output relationships. A linear system can be characterized by\nits transfer function H(s). Figure 4.18a shows a block diagram of a system with a transfer function\nH(s) and its input and output X(s) and Y(s), respectively.", - "type": "text" - }, - { - "block_id": "p406-b3", - "global_id": 11328, - "bbox": [ - 101.84, - 197.99, - 490.41, - 255.77 - ], - "text": "Subsystems may be interconnected by using cascade, parallel, and feedback interconnections\n(Figs. 4.18b, 4.18c, 4.18d), the three elementary types. When transfer functions appear in cascade,\nas depicted in Fig. 4.18b, then, as shown earlier, the transfer function of the overall system is\nthe product of the two transfer functions. This result can also be proved by observing that in\nFig. 4.18b", - "type": "text" - }, - { - "block_id": "p406-b4", - "global_id": 11329, - "bbox": [ - 229.57, - 262.0, - 282.0, - 286.33 - ], - "text": "Y(s)\nX(s) = W(s)", - "type": "text" - }, - { - "block_id": "p406-b5", - "global_id": 11330, - "bbox": [ - 262.91, - 276.06, - 280.75, - 286.33 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p406-b6", - "global_id": 11331, - "bbox": [ - 285.49, - 262.0, - 363.86, - 286.33 - ], - "text": "Y(s)\nW(s) = H1(s)H2(s)", - "type": "text" - }, - { - "block_id": "p406-b7", - "global_id": 11332, - "bbox": [ - 134.2, - 366.44, - 282.3, - 374.52 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p406-b8", - "global_id": 11333, - "bbox": [ - 135.72, - 448.57, - 282.76, - 456.66 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p406-b9", - "global_id": 11334, - "bbox": [ - 322.21, - 366.44, - 435.82, - 374.74 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p406-b10", - "global_id": 11335, - "bbox": [ - 322.21, - 448.26, - 435.82, - 456.57 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p406-b11", - "global_id": 11336, - "bbox": [ - 201.31, - 366.44, - 216.41, - 374.52 - ], - "text": "W(s)", - "type": "text" - }, - { - "block_id": "p406-b14", - "global_id": 11337, - "bbox": [ - 279.6, - 400.96, - 289.24, - 408.96 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p406-b15", - "global_id": 11338, - "bbox": [ - 279.98, - 507.68, - 288.86, - 515.68 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p406-b16", - "global_id": 11339, - "bbox": [ - 279.76, - 605.26, - 289.08, - 613.26 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p406-b18", - "global_id": 11340, - "bbox": [ - 279.98, - 342.9, - 288.86, - 350.9 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p406-b19", - "global_id": 11341, - "bbox": [ - 207.89, - 316.87, - 390.31, - 325.62 - ], - "text": "X(s)\nY(s)\nH(s)", - "type": "text" - }, - { - "block_id": "p406-b20", - "global_id": 11342, - "bbox": [ - 166.26, - 375.56, - 250.97, - 385.16 - ], - "text": "H1(s)\nH2(s)", - "type": "text" - }, - { - "block_id": "p406-b21", - "global_id": 11343, - "bbox": [ - 195.86, - 432.69, - 213.08, - 442.29 - ], - "text": "H1(s)", - "type": "text" - }, - { - "block_id": "p406-b22", - "global_id": 11344, - "bbox": [ - 195.86, - 481.26, - 213.08, - 490.87 - ], - "text": "H2(s)", - "type": "text" - }, - { - "block_id": "p406-b23", - "global_id": 11345, - "bbox": [ - 360.93, - 375.56, - 396.33, - 385.16 - ], - "text": "H1(s)H2(s)", - "type": "text" - }, - { - "block_id": "p406-b24", - "global_id": 11346, - "bbox": [ - 133.95, - 531.03, - 435.82, - 547.98 - ], - "text": "Y(s)\nX(s)\nE(s)\nY(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p406-b26", - "global_id": 11347, - "bbox": [ - 156.42, - 539.78, - 215.51, - 549.78 - ], - "text": "G(s)", - "type": "text" - }, - { - "block_id": "p406-b27", - "global_id": 11348, - "bbox": [ - 200.81, - 579.84, - 215.51, - 587.92 - ], - "text": "H(s)", - "type": "text" - }, - { - "block_id": "p406-b28", - "global_id": 11349, - "bbox": [ - 356.87, - 534.52, - 399.97, - 553.88 - ], - "text": "G(s)\n1 G(s)H(s)", - "type": "text" - }, - { - "block_id": "p406-b29", - "global_id": 11350, - "bbox": [ - 356.05, - 457.45, - 401.14, - 467.27 - ], - "text": "H1(s) H2(s)", - "type": "text" - }, - { - "block_id": "p406-b30", - "global_id": 11351, - "bbox": [ - 125.76, - 619.96, - 376.33, - 629.19 - ], - "text": "Figure 4.18 Elementary connections of blocks and their equivalents.", - "type": "text" - } - ] - }, - { - "page_num": 407, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p407-b0", - "global_id": 11352, - "bbox": [ - 390.17, - 62.89, - 516.13, - 71.98 - ], - "text": "4.5\nBlock Diagrams\n387", - "type": "text" - }, - { - "block_id": "p407-b1", - "global_id": 11353, - "bbox": [ - 127.59, - 85.82, - 516.15, - 167.51 - ], - "text": "We can extend this result to any number of transfer functions in cascade. It follows from this\ndiscussion that the subsystems in cascade can be interchanged without affecting the overall transfer\nfunction. This commutation property of LTI systems follows directly from the commutative (and\nassociative) property of convolution. We have already proved this property in Sec. 2.4-3. Every\npossible ordering of the subsystems yields the same overall transfer function. However, there may\nbe practical consequences (such as sensitivity to parameter variation) affecting the behavior of\ndifferent ordering.", - "type": "text" - }, - { - "block_id": "p407-b2", - "global_id": 11354, - "bbox": [ - 127.59, - 169.09, - 516.14, - 203.37 - ], - "text": "Similarly, when two transfer functions, H1(s) and H2(s), appear in parallel, as illustrated in\nFig. 4.18c, the overall transfer function is given by H1(s) + H2(s), the sum of the two transfer\nfunctions. The proof is trivial. This result can be extended to any number of systems in parallel.", - "type": "text" - }, - { - "block_id": "p407-b3", - "global_id": 11355, - "bbox": [ - 127.59, - 205.37, - 516.13, - 239.25 - ], - "text": "When the output is fed back to the input, as shown in Fig. 4.18d, the overall transfer function\nY(s)/X(s) can be computed as follows. The inputs to the adder are X(s) and −H(s)Y(s). Therefore,\nE(s), the output of the adder, is", - "type": "text" - }, - { - "block_id": "p407-b4", - "global_id": 11356, - "bbox": [ - 274.49, - 240.82, - 369.25, - 251.1 - ], - "text": "E(s) = X(s) −H(s)Y(s)", - "type": "text" - }, - { - "block_id": "p407-b5", - "global_id": 11357, - "bbox": [ - 127.6, - 260.0, - 141.99, - 269.96 - ], - "text": "But", - "type": "text" - }, - { - "block_id": "p407-b6", - "global_id": 11358, - "bbox": [ - 261.93, - 275.19, - 327.54, - 285.47 - ], - "text": "Y(s) = G(s)E(s)", - "type": "text" - }, - { - "block_id": "p407-b7", - "global_id": 11359, - "bbox": [ - 281.6, - 290.13, - 381.83, - 300.41 - ], - "text": "= G(s)[X(s) −H(s)Y(s)]", - "type": "text" - }, - { - "block_id": "p407-b8", - "global_id": 11360, - "bbox": [ - 127.61, - 312.14, - 169.35, - 322.1 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p407-b9", - "global_id": 11361, - "bbox": [ - 259.0, - 323.68, - 384.74, - 334.05 - ], - "text": "Y(s)[1 + G(s)H(s)] = G(s)X(s)", - "type": "text" - }, - { - "block_id": "p407-b10", - "global_id": 11362, - "bbox": [ - 127.6, - 342.86, - 516.13, - 374.81 - ], - "text": "so that\nY(s)\nX(s) =\nG(s)\n1 + G(s)H(s)\n(4.35)", - "type": "text" - }, - { - "block_id": "p407-b11", - "global_id": 11363, - "bbox": [ - 127.59, - 381.18, - 516.12, - 403.09 - ], - "text": "Therefore, the feedback loop can be replaced by a single block with the transfer function shown\nin Eq. (4.35) (see Fig. 4.18d).", - "type": "text" - }, - { - "block_id": "p407-b12", - "global_id": 11364, - "bbox": [ - 127.59, - 405.08, - 516.16, - 499.51 - ], - "text": "In deriving these equations, we implicitly assume that when the output of one subsystem is\nconnected to the input of another subsystem, the latter does not load the former. For example,\nthe transfer function H1(s) in Fig. 4.18b is computed by assuming that the second subsystem\nH2(s) was not connected. This is the same as assuming that H2(s) does not load H1(s). In other\nwords, the input–output relationship of H1(s) will remain unchanged regardless of whether H2(s)\nis connected. Many modern circuits use op amps with high input impedances, so this assumption\nis justified. When such an assumption is not valid, H1(s) must be computed under operating\nconditions [i.e., with H2(s) connected].", - "type": "text" - }, - { - "block_id": "p407-b13", - "global_id": 11365, - "bbox": [ - 102.51, - 525.14, - 461.38, - 551.05 - ], - "text": "EXAMPLE 4.21\nTransfer Functions of Feedback Systems Using\nMATLAB", - "type": "text" - }, - { - "block_id": "p407-b14", - "global_id": 11366, - "bbox": [ - 128.9, - 564.37, - 502.77, - 598.66 - ], - "text": "Consider the feedback system of Fig. 4.18d with G(s) = K/(s(s + 8)) and H(s) = 1. Use\nMATLAB to determine the transfer function for each of the following cases: (a) K = 7, (b)\nK = 16, and (c) K = 80.", - "type": "text" - }, - { - "block_id": "p407-b15", - "global_id": 11367, - "bbox": [ - 128.9, - 621.58, - 426.86, - 631.83 - ], - "text": "We solve these cases using the control system toolbox function feedback.", - "type": "text" - } - ] - }, - { - "page_num": 408, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p408-b0", - "global_id": 11368, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "388\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p408-b1", - "global_id": 11369, - "bbox": [ - 103.16, - 86.29, - 114.78, - 96.25 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p408-b2", - "global_id": 11370, - "bbox": [ - 103.16, - 106.58, - 458.79, - 128.5 - ], - "text": ">>\nH = tf(1,1); K = 7; G = tf([0 0 K],[1 8 0]); TFa = feedback(G,H)\nHa =", - "type": "text" - }, - { - "block_id": "p408-b3", - "global_id": 11371, - "bbox": [ - 134.54, - 130.5, - 202.53, - 164.36 - ], - "text": "7\n-------------\ns^2 + 8 s + 7", - "type": "text" - }, - { - "block_id": "p408-b4", - "global_id": 11372, - "bbox": [ - 103.16, - 170.01, - 225.43, - 195.88 - ], - "text": "Thus, Ha(s) = 7/(s2 + 8s + 7).\n(b)", - "type": "text" - }, - { - "block_id": "p408-b5", - "global_id": 11373, - "bbox": [ - 103.16, - 206.21, - 464.04, - 228.13 - ], - "text": ">>\nH = tf(1,1); K = 16; G = tf([0 0 K],[1 8 0]); TFb = feedback(G,H)\nHb =", - "type": "text" - }, - { - "block_id": "p408-b6", - "global_id": 11374, - "bbox": [ - 134.54, - 230.12, - 207.76, - 263.99 - ], - "text": "16\n--------------\ns^2 + 8 s + 16", - "type": "text" - }, - { - "block_id": "p408-b7", - "global_id": 11375, - "bbox": [ - 103.16, - 269.64, - 235.39, - 295.5 - ], - "text": "Thus, Hb(s) = 16/(s2 + 8s + 16).\n(c)", - "type": "text" - }, - { - "block_id": "p408-b8", - "global_id": 11376, - "bbox": [ - 103.16, - 305.84, - 464.04, - 327.75 - ], - "text": ">>\nH = tf(1,1); K = 80; G = tf([0 0 K],[1 8 0]); TFc = feedback(G,H)\nHc =", - "type": "text" - }, - { - "block_id": "p408-b9", - "global_id": 11377, - "bbox": [ - 134.54, - 329.75, - 207.76, - 363.62 - ], - "text": "80\n--------------\ns^2 + 8 s + 80", - "type": "text" - }, - { - "block_id": "p408-b10", - "global_id": 11378, - "bbox": [ - 103.16, - 369.27, - 235.0, - 383.95 - ], - "text": "Thus, Hc(s) = 80/(s2 + 8s + 80).", - "type": "text" - }, - { - "block_id": "p408-b11", - "global_id": 11379, - "bbox": [ - 102.2, - 440.76, - 272.17, - 454.71 - ], - "text": "4.6 SYSTEM REALIZATION", - "type": "text" - }, - { - "block_id": "p408-b12", - "global_id": 11380, - "bbox": [ - 101.84, - 460.59, - 490.37, - 482.61 - ], - "text": "We now develop a systematic method for realization (or implementation) of an arbitrary Nth-order\ntransfer function. The most general transfer function with M = N is given by", - "type": "text" - }, - { - "block_id": "p408-b13", - "global_id": 11381, - "bbox": [ - 211.42, - 496.67, - 378.7, - 517.54 - ], - "text": "H(s) = b0sN + b1sN−1 + · · · + bN−1s + bN", - "type": "text" - }, - { - "block_id": "p408-b14", - "global_id": 11382, - "bbox": [ - 247.91, - 511.46, - 374.22, - 525.49 - ], - "text": "sN + a1sN−1 + · · · + aN−1s + aN", - "type": "text" - }, - { - "block_id": "p408-b15", - "global_id": 11383, - "bbox": [ - 466.32, - 507.68, - 490.38, - 517.64 - ], - "text": "(4.36)", - "type": "text" - }, - { - "block_id": "p408-b16", - "global_id": 11384, - "bbox": [ - 101.85, - 541.13, - 490.43, - 634.79 - ], - "text": "Since realization is basically a synthesis problem, there is no unique way of realizing a system.\nA given transfer function can be realized in many different ways. A transfer function H(s) can be\nrealized by using integrators or differentiators along with adders and multipliers. We avoid use of\ndifferentiators for practical reasons discussed in Secs. 2.1 and 4.3-3. Hence, in our implementation,\nwe shall use integrators along with scalar multipliers and adders. We are already familiar with\nrepresentation of all these elements except the integrator. The integrator can be represented by a\nbox with integral sign (time-domain representation, Fig. 4.19a) or by a box with transfer function\n1/s (frequency-domain representation, Fig. 4.19b).", - "type": "text" - } - ] - }, - { - "page_num": 409, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p409-b0", - "global_id": 11385, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n389", - "type": "text" - }, - { - "block_id": "p409-b1", - "global_id": 11386, - "bbox": [ - 157.74, - 91.04, - 327.48, - 103.38 - ], - "text": "x(t)\nX(s)\n\ny(t)", - "type": "text" - }, - { - "block_id": "p409-b2", - "global_id": 11387, - "bbox": [ - 252.4, - 89.41, - 431.91, - 104.33 - ], - "text": "tx(t)d(t)\n0 \n1s\nY(s) X(s)", - "type": "text" - }, - { - "block_id": "p409-b3", - "global_id": 11388, - "bbox": [ - 216.93, - 126.74, - 380.03, - 134.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p409-b4", - "global_id": 11389, - "bbox": [ - 197.64, - 99.07, - 362.66, - 113.93 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p409-b5", - "global_id": 11390, - "bbox": [ - 151.5, - 141.36, - 474.63, - 150.67 - ], - "text": "Figure 4.19 (a) Time-domain and (b) frequency-domain representations of an integrator.", - "type": "text" - }, - { - "block_id": "p409-b6", - "global_id": 11391, - "bbox": [ - 127.59, - 172.66, - 296.41, - 184.61 - ], - "text": "4.6-1 Direct Form I Realization", - "type": "text" - }, - { - "block_id": "p409-b7", - "global_id": 11392, - "bbox": [ - 127.59, - 190.65, - 516.14, - 212.66 - ], - "text": "Rather than realize the general Nth-order system described by Eq. (4.36), we begin with a specific\ncase of the following third-order system and then extend the results to the Nth-order case:", - "type": "text" - }, - { - "block_id": "p409-b8", - "global_id": 11393, - "bbox": [ - 216.27, - 230.49, - 335.85, - 251.37 - ], - "text": "H(s) = b0s3 + b1s2 + b2s + b3", - "type": "text" - }, - { - "block_id": "p409-b9", - "global_id": 11394, - "bbox": [ - 252.77, - 245.29, - 331.37, - 259.32 - ], - "text": "s3 + a1s2 + a2s + a3", - "type": "text" - }, - { - "block_id": "p409-b10", - "global_id": 11395, - "bbox": [ - 339.6, - 241.09, - 347.37, - 251.06 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p409-b11", - "global_id": 11396, - "bbox": [ - 350.61, - 223.14, - 380.1, - 241.26 - ], - "text": "b0 + b1", - "type": "text" - }, - { - "block_id": "p409-b12", - "global_id": 11397, - "bbox": [ - 374.18, - 223.14, - 402.33, - 247.16 - ], - "text": "s + b2", - "type": "text" - }, - { - "block_id": "p409-b13", - "global_id": 11398, - "bbox": [ - 394.42, - 223.14, - 424.56, - 247.16 - ], - "text": "s2 + b3", - "type": "text" - }, - { - "block_id": "p409-b14", - "global_id": 11399, - "bbox": [ - 416.65, - 236.68, - 424.01, - 247.16 - ], - "text": "s3", - "type": "text" - }, - { - "block_id": "p409-b15", - "global_id": 11400, - "bbox": [ - 352.6, - 244.89, - 400.34, - 268.91 - ], - "text": "1 + a1\ns + a2", - "type": "text" - }, - { - "block_id": "p409-b16", - "global_id": 11401, - "bbox": [ - 392.43, - 244.89, - 422.57, - 268.91 - ], - "text": "s2 + a3", - "type": "text" - }, - { - "block_id": "p409-b17", - "global_id": 11402, - "bbox": [ - 414.66, - 258.43, - 422.02, - 268.91 - ], - "text": "s3", - "type": "text" - }, - { - "block_id": "p409-b18", - "global_id": 11403, - "bbox": [ - 145.52, - 277.34, - 239.32, - 287.71 - ], - "text": "We can express H(s) as", - "type": "text" - }, - { - "block_id": "p409-b19", - "global_id": 11404, - "bbox": [ - 216.36, - 314.63, - 245.13, - 324.91 - ], - "text": "H(s) =", - "type": "text" - }, - { - "block_id": "p409-b21", - "global_id": 11405, - "bbox": [ - 253.9, - 307.96, - 283.4, - 326.09 - ], - "text": "b0 + b1", - "type": "text" - }, - { - "block_id": "p409-b22", - "global_id": 11406, - "bbox": [ - 277.48, - 307.96, - 305.63, - 331.98 - ], - "text": "s + b2", - "type": "text" - }, - { - "block_id": "p409-b23", - "global_id": 11407, - "bbox": [ - 297.71, - 307.96, - 327.86, - 331.98 - ], - "text": "s2 + b3", - "type": "text" - }, - { - "block_id": "p409-b24", - "global_id": 11408, - "bbox": [ - 319.94, - 321.51, - 327.31, - 331.98 - ], - "text": "s3", - "type": "text" - }, - { - "block_id": "p409-b26", - "global_id": 11409, - "bbox": [ - 247.18, - 330.08, - 336.28, - 350.55 - ], - "text": "H1(s)", - "type": "text" - }, - { - "block_id": "p409-b27", - "global_id": 11410, - "bbox": [ - 337.38, - 291.68, - 345.34, - 301.64 - ], - "text": "⎛", - "type": "text" - }, - { - "block_id": "p409-b28", - "global_id": 11411, - "bbox": [ - 337.38, - 308.06, - 384.86, - 325.56 - ], - "text": "⎜⎝\n1", - "type": "text" - }, - { - "block_id": "p409-b29", - "global_id": 11412, - "bbox": [ - 346.54, - 318.44, - 394.28, - 342.46 - ], - "text": "1 + a1\ns + a2", - "type": "text" - }, - { - "block_id": "p409-b30", - "global_id": 11413, - "bbox": [ - 386.36, - 318.44, - 416.51, - 342.46 - ], - "text": "s2 + a3", - "type": "text" - }, - { - "block_id": "p409-b31", - "global_id": 11414, - "bbox": [ - 408.59, - 331.98, - 415.96, - 342.46 - ], - "text": "s3", - "type": "text" - }, - { - "block_id": "p409-b32", - "global_id": 11415, - "bbox": [ - 419.4, - 291.68, - 427.36, - 301.64 - ], - "text": "⎞", - "type": "text" - }, - { - "block_id": "p409-b33", - "global_id": 11416, - "bbox": [ - 419.4, - 309.16, - 427.36, - 325.56 - ], - "text": "⎟⎠", - "type": "text" - }, - { - "block_id": "p409-b34", - "global_id": 11417, - "bbox": [ - 337.38, - 337.07, - 427.37, - 357.53 - ], - "text": "H2(s)", - "type": "text" - }, - { - "block_id": "p409-b35", - "global_id": 11418, - "bbox": [ - 127.59, - 367.81, - 516.14, - 414.83 - ], - "text": "We can realize H(s) as a cascade of transfer function H1(s) followed by H2(s), as depicted in\nFig. 4.20a, where the output of H1(s) is denoted by W(s). Because of the commutative property of\nLTI system transfer functions in cascade, we can also realize H(s) as a cascade of H2(s) followed\nby H1(s), as illustrated in Fig. 4.20b, where the (intermediate) output of H2(s) is denoted by V(s).", - "type": "text" - }, - { - "block_id": "p409-b36", - "global_id": 11419, - "bbox": [ - 153.04, - 438.93, - 457.77, - 456.04 - ], - "text": "X(s)\nY(s)\nW(s)\nX(s)\nY(s)\nV(s)", - "type": "text" - }, - { - "block_id": "p409-b37", - "global_id": 11420, - "bbox": [ - 217.12, - 473.67, - 393.32, - 481.67 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p409-b38", - "global_id": 11421, - "bbox": [ - 185.25, - 446.84, - 426.08, - 456.45 - ], - "text": "H1(s)\nH1(s)\nH2(s)\nH2(s)", - "type": "text" - }, - { - "block_id": "p409-b39", - "global_id": 11422, - "bbox": [ - 151.5, - 488.36, - 367.34, - 497.6 - ], - "text": "Figure 4.20 Realization of a transfer function in two steps.", - "type": "text" - }, - { - "block_id": "p409-b40", - "global_id": 11423, - "bbox": [ - 145.52, - 523.47, - 431.33, - 534.62 - ], - "text": "The output of H1(s) in Fig. 4.20a is given by W(s) = H1(s)X(s). Hence,", - "type": "text" - }, - { - "block_id": "p409-b41", - "global_id": 11424, - "bbox": [ - 251.74, - 551.81, - 281.9, - 562.08 - ], - "text": "W(s) =", - "type": "text" - }, - { - "block_id": "p409-b43", - "global_id": 11425, - "bbox": [ - 290.67, - 545.14, - 320.16, - 563.27 - ], - "text": "b0 + b1", - "type": "text" - }, - { - "block_id": "p409-b44", - "global_id": 11426, - "bbox": [ - 314.24, - 545.14, - 342.39, - 569.16 - ], - "text": "s + b2", - "type": "text" - }, - { - "block_id": "p409-b45", - "global_id": 11427, - "bbox": [ - 334.48, - 545.14, - 364.62, - 569.16 - ], - "text": "s2 + b3", - "type": "text" - }, - { - "block_id": "p409-b46", - "global_id": 11428, - "bbox": [ - 356.71, - 558.68, - 364.07, - 569.16 - ], - "text": "s3", - "type": "text" - }, - { - "block_id": "p409-b48", - "global_id": 11429, - "bbox": [ - 374.15, - 551.81, - 516.13, - 562.18 - ], - "text": "X(s)\n(4.37)", - "type": "text" - }, - { - "block_id": "p409-b49", - "global_id": 11430, - "bbox": [ - 127.59, - 579.85, - 516.15, - 602.18 - ], - "text": "Also, the output Y(s) and the input W(s) of H2(s) in Fig. 4.20a are related by Y(s) = H2(s)W(s).\nHence,", - "type": "text" - }, - { - "block_id": "p409-b50", - "global_id": 11431, - "bbox": [ - 255.14, - 606.86, - 285.3, - 617.14 - ], - "text": "W(s) =", - "type": "text" - }, - { - "block_id": "p409-b52", - "global_id": 11432, - "bbox": [ - 292.77, - 600.19, - 340.51, - 624.21 - ], - "text": "1 + a1\ns + a2", - "type": "text" - }, - { - "block_id": "p409-b53", - "global_id": 11433, - "bbox": [ - 332.6, - 600.19, - 362.73, - 624.21 - ], - "text": "s2 + a3", - "type": "text" - }, - { - "block_id": "p409-b54", - "global_id": 11434, - "bbox": [ - 354.82, - 613.74, - 362.18, - 624.21 - ], - "text": "s3", - "type": "text" - }, - { - "block_id": "p409-b56", - "global_id": 11435, - "bbox": [ - 370.97, - 606.86, - 516.13, - 617.24 - ], - "text": "Y(s)\n(4.38)", - "type": "text" - } - ] - }, - { - "page_num": 410, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p410-b0", - "global_id": 11436, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "390\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p410-b13", - "global_id": 11437, - "bbox": [ - 184.78, - 240.74, - 390.9, - 248.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p410-b14", - "global_id": 11438, - "bbox": [ - 336.29, - 129.96, - 343.29, - 139.57 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p410-b15", - "global_id": 11439, - "bbox": [ - 336.29, - 183.68, - 352.29, - 193.29 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p410-b16", - "global_id": 11440, - "bbox": [ - 336.29, - 220.8, - 344.29, - 230.35 - ], - "text": "bN", - "type": "text" - }, - { - "block_id": "p410-b17", - "global_id": 11441, - "bbox": [ - 418.18, - 128.75, - 431.84, - 138.57 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p410-b18", - "global_id": 11442, - "bbox": [ - 418.18, - 182.47, - 440.84, - 192.29 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p410-b19", - "global_id": 11443, - "bbox": [ - 410.85, - 219.59, - 425.51, - 229.35 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p410-b20", - "global_id": 11444, - "bbox": [ - 301.67, - 87.91, - 468.95, - 97.82 - ], - "text": "W(s)\nb0\nX(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p410-b21", - "global_id": 11445, - "bbox": [ - 446.55, - 207.57, - 450.55, - 222.16 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b22", - "global_id": 11446, - "bbox": [ - 446.55, - 153.32, - 450.55, - 167.91 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b23", - "global_id": 11447, - "bbox": [ - 446.55, - 114.8, - 450.55, - 129.39 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b24", - "global_id": 11448, - "bbox": [ - 318.2, - 208.57, - 322.2, - 223.16 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b25", - "global_id": 11449, - "bbox": [ - 318.2, - 153.32, - 322.2, - 167.91 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b26", - "global_id": 11450, - "bbox": [ - 318.2, - 114.07, - 322.2, - 128.66 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b27", - "global_id": 11451, - "bbox": [ - 139.57, - 87.21, - 195.39, - 96.82 - ], - "text": "W(s)\nb0", - "type": "text" - }, - { - "block_id": "p410-b28", - "global_id": 11452, - "bbox": [ - 139.57, - 129.96, - 146.57, - 139.57 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p410-b29", - "global_id": 11453, - "bbox": [ - 139.57, - 166.56, - 146.57, - 176.17 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p410-b30", - "global_id": 11454, - "bbox": [ - 139.57, - 202.6, - 146.57, - 212.21 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p410-b31", - "global_id": 11455, - "bbox": [ - 220.86, - 128.75, - 234.52, - 138.57 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p410-b32", - "global_id": 11456, - "bbox": [ - 220.86, - 165.34, - 234.52, - 175.17 - ], - "text": "a2", - "type": "text" - }, - { - "block_id": "p410-b33", - "global_id": 11457, - "bbox": [ - 220.86, - 201.39, - 234.52, - 211.21 - ], - "text": "a3", - "type": "text" - }, - { - "block_id": "p410-b34", - "global_id": 11458, - "bbox": [ - 104.84, - 87.91, - 271.87, - 95.99 - ], - "text": "X(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p410-b35", - "global_id": 11459, - "bbox": [ - 249.69, - 190.57, - 253.69, - 205.16 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b36", - "global_id": 11460, - "bbox": [ - 249.69, - 153.32, - 253.69, - 167.91 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b37", - "global_id": 11461, - "bbox": [ - 249.69, - 114.07, - 253.69, - 128.66 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b38", - "global_id": 11462, - "bbox": [ - 121.34, - 190.57, - 125.34, - 205.16 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b39", - "global_id": 11463, - "bbox": [ - 121.34, - 153.32, - 125.34, - 167.91 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b40", - "global_id": 11464, - "bbox": [ - 121.34, - 114.07, - 125.34, - 128.66 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p410-b41", - "global_id": 11465, - "bbox": [ - 101.84, - 255.35, - 432.26, - 264.68 - ], - "text": "Figure 4.21 Direct form I realization of an LTIC system: (a) third-order and (b) Nth-order.", - "type": "text" - }, - { - "block_id": "p410-b42", - "global_id": 11466, - "bbox": [ - 101.84, - 285.88, - 490.39, - 344.84 - ], - "text": "We shall first realize H1(s). Equation (4.37) shows that the output W(s) can be synthesized by\nadding the input b0X(s) to b1(X(s)/s),b2(X(s)/s2), and b3(X(s)/s3). Because the transfer function\nof an integrator is 1/s, the signals X(s)/s,X(s)/s2, and X(s)/s3 can be obtained by successive\nintegration of the input x(t). The left-half section of Fig. 4.21a shows how W(s) can be synthesized\nfrom X(s), according to Eq. (4.37). Hence, this section represents a realization of H1(s).", - "type": "text" - }, - { - "block_id": "p410-b43", - "global_id": 11467, - "bbox": [ - 101.85, - 345.65, - 490.38, - 367.98 - ], - "text": "To complete the picture, we shall realize H2(s), which is specified by Eq. (4.38). We can\nrearrange Eq. (4.38) as", - "type": "text" - }, - { - "block_id": "p410-b44", - "global_id": 11468, - "bbox": [ - 223.08, - 373.78, - 282.22, - 384.06 - ], - "text": "Y(s) = W(s) −", - "type": "text" - }, - { - "block_id": "p410-b45", - "global_id": 11469, - "bbox": [ - 283.76, - 362.78, - 298.85, - 377.94 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p410-b46", - "global_id": 11470, - "bbox": [ - 292.93, - 367.11, - 321.08, - 391.13 - ], - "text": "s + a2", - "type": "text" - }, - { - "block_id": "p410-b47", - "global_id": 11471, - "bbox": [ - 313.16, - 367.11, - 343.31, - 391.13 - ], - "text": "s2 + a3", - "type": "text" - }, - { - "block_id": "p410-b48", - "global_id": 11472, - "bbox": [ - 335.39, - 380.65, - 342.76, - 391.13 - ], - "text": "s3", - "type": "text" - }, - { - "block_id": "p410-b50", - "global_id": 11473, - "bbox": [ - 351.54, - 373.78, - 490.38, - 384.16 - ], - "text": "Y(s)\n(4.39)", - "type": "text" - }, - { - "block_id": "p410-b51", - "global_id": 11474, - "bbox": [ - 101.84, - 395.15, - 490.39, - 492.42 - ], - "text": "Hence, to obtain Y(s), we subtract a1Y(s)/s, a2Y(s)/s2, and a3Y(s)/s3 from W(s). We have already\nobtained W(s) from the first step [output of H1(s)]. To obtain signals Y(s)/s, Y(s)/s2, and Y(s)/s3,\nwe assume that we already have the desired output Y(s). Successive integration of Y(s) yields the\nneeded signals Y(s)/s, Y(s)/s2, and Y(s)/s3. We now synthesize the final output Y(s) according\nto Eq. (4.39), as seen in the right-half section of Fig. 4.21a.† The left-half section in Fig. 4.21a\nrepresents H1(s) and the right-half is H2(s). We can generalize this procedure, known as the direct\nform I (DFI) realization, for any value of N. This procedure requires 2N integrators to realize an\nNth-order transfer function, as shown in Fig. 4.21b.", - "type": "text" - }, - { - "block_id": "p410-b52", - "global_id": 11475, - "bbox": [ - 101.84, - 518.22, - 275.31, - 530.18 - ], - "text": "4.6-2 Direct Form II Realization", - "type": "text" - }, - { - "block_id": "p410-b53", - "global_id": 11476, - "bbox": [ - 101.84, - 535.89, - 490.39, - 559.0 - ], - "text": "In the direct form I, we realize H(s) by implementing H1(s) followed by H2(s), as shown in\nFig. 4.20a. We can also realize H(s), as shown in Fig. 4.20b, where H2(s) is followed by H1(s).", - "type": "text" - }, - { - "block_id": "p410-b54", - "global_id": 11477, - "bbox": [ - 101.84, - 577.36, - 490.4, - 633.41 - ], - "text": "† It may seem odd that we first assumed the existence of Y(s), integrated it successively, and then in turn\ngenerated Y(s) from W(s) and the three successive integrals of signal Y(s). This procedure poses a dilemma\nsimilar to “Which came first, the chicken or the egg?” The problem here is satisfactorily resolved by writing\nthe expression for Y(s) at the output of the right-hand adder (at the top) in Fig. 4.21a and verifying that this\nexpression is indeed the same as Eq. (4.38).", - "type": "text" - } - ] - }, - { - "page_num": 411, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p411-b0", - "global_id": 11478, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n391", - "type": "text" - }, - { - "block_id": "p411-b13", - "global_id": 11479, - "bbox": [ - 234.43, - 239.74, - 422.45, - 247.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p411-b14", - "global_id": 11480, - "bbox": [ - 221.08, - 113.34, - 225.08, - 128.38 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b15", - "global_id": 11481, - "bbox": [ - 221.08, - 152.81, - 225.08, - 167.85 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b16", - "global_id": 11482, - "bbox": [ - 221.08, - 208.3, - 225.08, - 223.34 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b17", - "global_id": 11483, - "bbox": [ - 253.01, - 113.34, - 257.01, - 128.38 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b18", - "global_id": 11484, - "bbox": [ - 415.83, - 113.57, - 419.83, - 128.61 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b19", - "global_id": 11485, - "bbox": [ - 415.83, - 152.85, - 419.83, - 167.89 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b20", - "global_id": 11486, - "bbox": [ - 415.83, - 207.19, - 419.83, - 222.23 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b21", - "global_id": 11487, - "bbox": [ - 253.01, - 152.81, - 257.01, - 167.85 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b22", - "global_id": 11488, - "bbox": [ - 253.01, - 207.3, - 257.01, - 222.34 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p411-b23", - "global_id": 11489, - "bbox": [ - 154.73, - 87.72, - 485.13, - 95.8 - ], - "text": "V(s)\nY(s)\nY(s)\nX(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p411-b24", - "global_id": 11490, - "bbox": [ - 194.81, - 129.0, - 208.48, - 138.83 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p411-b25", - "global_id": 11491, - "bbox": [ - 190.31, - 182.55, - 212.98, - 192.37 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p411-b26", - "global_id": 11492, - "bbox": [ - 194.31, - 218.86, - 208.98, - 228.62 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p411-b27", - "global_id": 11493, - "bbox": [ - 273.44, - 87.6, - 280.44, - 97.21 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p411-b28", - "global_id": 11494, - "bbox": [ - 272.94, - 218.08, - 280.94, - 227.62 - ], - "text": "bN", - "type": "text" - }, - { - "block_id": "p411-b29", - "global_id": 11495, - "bbox": [ - 268.82, - 183.76, - 284.82, - 193.37 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p411-b30", - "global_id": 11496, - "bbox": [ - 273.32, - 129.22, - 280.32, - 138.83 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p411-b31", - "global_id": 11497, - "bbox": [ - 433.25, - 87.35, - 440.25, - 96.96 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p411-b32", - "global_id": 11498, - "bbox": [ - 433.25, - 219.08, - 441.25, - 228.62 - ], - "text": "bN", - "type": "text" - }, - { - "block_id": "p411-b33", - "global_id": 11499, - "bbox": [ - 433.25, - 182.76, - 449.25, - 192.37 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p411-b34", - "global_id": 11500, - "bbox": [ - 388.42, - 129.0, - 440.25, - 138.83 - ], - "text": "b1\na1", - "type": "text" - }, - { - "block_id": "p411-b35", - "global_id": 11501, - "bbox": [ - 388.42, - 182.55, - 411.08, - 192.37 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p411-b36", - "global_id": 11502, - "bbox": [ - 388.42, - 218.86, - 403.08, - 228.62 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p411-b37", - "global_id": 11503, - "bbox": [ - 151.5, - 254.35, - 400.74, - 263.68 - ], - "text": "Figure 4.22 Direct form II realization of an Nth-order LTIC system.", - "type": "text" - }, - { - "block_id": "p411-b38", - "global_id": 11504, - "bbox": [ - 127.59, - 284.72, - 516.13, - 319.46 - ], - "text": "This procedure is known as the direct form II realization. Figure 4.22a shows direct form II\nrealization, where we have interchanged sections representing H1(s) and H2(s) in Fig. 4.21b. The\noutput of H2(s) in this case is denoted by V(s).‡", - "type": "text" - }, - { - "block_id": "p411-b39", - "global_id": 11505, - "bbox": [ - 127.59, - 320.69, - 516.13, - 390.42 - ], - "text": "An interesting observation in Fig. 4.22a is that the input signal to both the chains of integrators\nis V(s). Clearly, the outputs of integrators in the left-side chain are identical to the corresponding\noutputs of the right-side integrator chain, thus making the right-side chain redundant. We can\neliminate this chain and obtain the required signals from the left-side chain, as shown in Fig. 4.22b.\nThis implementation halves the number of integrators to N, and, thus, is more efficient in hardware\nutilization than either Figs. 4.21b or 4.22a. This is the direct form II (DFII) realization.", - "type": "text" - }, - { - "block_id": "p411-b40", - "global_id": 11506, - "bbox": [ - 127.6, - 392.01, - 516.17, - 450.2 - ], - "text": "An Nth-order differential equation with N = M has a property that its implementation requires\na minimum of N integrators. A realization is canonic if the number of integrators used in the\nrealization is equal to the order of the transfer function realized. Thus, canonic realization has no\nredundant integrators. The DFII form in Fig. 4.22b is a canonic realization, and is also called the\ndirect canonic form. Note that the DFI is noncanonic.", - "type": "text" - }, - { - "block_id": "p411-b41", - "global_id": 11507, - "bbox": [ - 127.59, - 452.19, - 516.16, - 509.98 - ], - "text": "The direct form I realization (Fig. 4.22b) implements zeros first [the left-half section\nrepresented by H1(s)] followed by realization of poles [the right-half section represented by\nH2(s)] of H(s). In contrast, canonic direct implements poles first followed by zeros. Although\nboth these realizations result in the same transfer function, they generally behave differently from\nthe viewpoint of sensitivity to parameter variations.", - "type": "text" - }, - { - "block_id": "p411-b42", - "global_id": 11508, - "bbox": [ - 127.59, - 550.94, - 436.58, - 563.16 - ], - "text": "‡ The reader can show that the equations relating X(s),V(s), and Y(s) in Fig. 4.22a are", - "type": "text" - }, - { - "block_id": "p411-b43", - "global_id": 11509, - "bbox": [ - 244.67, - 574.0, - 296.15, - 583.25 - ], - "text": "V(s) = X(s) −", - "type": "text" - }, - { - "block_id": "p411-b44", - "global_id": 11510, - "bbox": [ - 297.54, - 564.1, - 311.33, - 577.8 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p411-b45", - "global_id": 11511, - "bbox": [ - 305.97, - 568.0, - 331.72, - 589.62 - ], - "text": "s + a2", - "type": "text" - }, - { - "block_id": "p411-b46", - "global_id": 11512, - "bbox": [ - 324.49, - 568.0, - 374.74, - 589.62 - ], - "text": "s2 + · · · + aN", - "type": "text" - }, - { - "block_id": "p411-b47", - "global_id": 11513, - "bbox": [ - 366.43, - 579.98, - 374.25, - 589.62 - ], - "text": "sN", - "type": "text" - }, - { - "block_id": "p411-b49", - "global_id": 11514, - "bbox": [ - 382.71, - 574.0, - 399.07, - 583.25 - ], - "text": "V(s)", - "type": "text" - }, - { - "block_id": "p411-b50", - "global_id": 11515, - "bbox": [ - 127.59, - 594.88, - 140.53, - 603.85 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p411-b51", - "global_id": 11516, - "bbox": [ - 249.33, - 608.17, - 274.02, - 617.42 - ], - "text": "Y(s) =", - "type": "text" - }, - { - "block_id": "p411-b53", - "global_id": 11517, - "bbox": [ - 280.75, - 602.17, - 307.66, - 618.48 - ], - "text": "b0 + b1", - "type": "text" - }, - { - "block_id": "p411-b54", - "global_id": 11518, - "bbox": [ - 302.31, - 602.17, - 328.06, - 623.79 - ], - "text": "s + b2", - "type": "text" - }, - { - "block_id": "p411-b55", - "global_id": 11519, - "bbox": [ - 320.83, - 602.17, - 371.08, - 623.79 - ], - "text": "s2 + · · · + bN", - "type": "text" - }, - { - "block_id": "p411-b56", - "global_id": 11520, - "bbox": [ - 362.77, - 614.15, - 370.58, - 623.79 - ], - "text": "sN", - "type": "text" - }, - { - "block_id": "p411-b58", - "global_id": 11521, - "bbox": [ - 378.04, - 608.17, - 394.4, - 617.42 - ], - "text": "V(s)", - "type": "text" - } - ] - }, - { - "page_num": 412, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p412-b0", - "global_id": 11522, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "392\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p412-b1", - "global_id": 11523, - "bbox": [ - 76.77, - 93.91, - 365.2, - 105.86 - ], - "text": "EXAMPLE 4.22\nCanonic Direct Form Realizations", - "type": "text" - }, - { - "block_id": "p412-b2", - "global_id": 11524, - "bbox": [ - 103.16, - 119.61, - 402.85, - 129.58 - ], - "text": "Find the canonic direct form realization of the following transfer functions:", - "type": "text" - }, - { - "block_id": "p412-b3", - "global_id": 11525, - "bbox": [ - 121.09, - 135.53, - 158.62, - 159.55 - ], - "text": "(a)\n5\ns + 7", - "type": "text" - }, - { - "block_id": "p412-b4", - "global_id": 11526, - "bbox": [ - 121.09, - 158.67, - 159.17, - 182.79 - ], - "text": "(b)\ns\ns + 7", - "type": "text" - }, - { - "block_id": "p412-b5", - "global_id": 11527, - "bbox": [ - 121.65, - 184.01, - 158.62, - 201.29 - ], - "text": "(c) s + 5", - "type": "text" - }, - { - "block_id": "p412-b6", - "global_id": 11528, - "bbox": [ - 138.89, - 198.07, - 158.62, - 208.44 - ], - "text": "s + 7", - "type": "text" - }, - { - "block_id": "p412-b7", - "global_id": 11529, - "bbox": [ - 103.16, - 209.67, - 397.13, - 248.74 - ], - "text": "(d)\n4s + 28\ns2 + 6s + 5\nAll four of these transfer functions are special cases of H(s) in Eq. (4.36).", - "type": "text" - }, - { - "block_id": "p412-b8", - "global_id": 11530, - "bbox": [ - 103.16, - 271.25, - 477.01, - 293.57 - ], - "text": "(a) The transfer function 5/(s + 7) is of the first order (N = 1); therefore, we need only\none integrator for its realization. The feedback and feedforward coefficients are", - "type": "text" - }, - { - "block_id": "p412-b9", - "global_id": 11531, - "bbox": [ - 217.47, - 305.11, - 362.71, - 316.57 - ], - "text": "a1 = 7\nand\nb0 = 0,\nb1 = 5", - "type": "text" - }, - { - "block_id": "p412-b10", - "global_id": 11532, - "bbox": [ - 103.16, - 327.04, - 477.01, - 397.18 - ], - "text": "The realization is depicted in Fig. 4.23a. Because N = 1, there is a single feedback connection\nfrom the output of the integrator to the input adder with coefficient a1 = 7. For N = 1, generally,\nthere are N + 1 = 2 feedforward connections. However, in this case, b0 = 0, and there is only\none feedforward connection with coefficient b1 = 5 from the output of the integrator to the\noutput adder. Because there is only one input signal to the output adder, we can do away with\nthe adder, as shown in Fig. 4.23a.", - "type": "text" - }, - { - "block_id": "p412-b11", - "global_id": 11533, - "bbox": [ - 121.09, - 399.1, - 133.26, - 409.06 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p412-b12", - "global_id": 11534, - "bbox": [ - 263.62, - 406.18, - 315.36, - 430.3 - ], - "text": "H(s) =\ns\ns + 7", - "type": "text" - }, - { - "block_id": "p412-b13", - "global_id": 11535, - "bbox": [ - 103.16, - 435.5, - 477.0, - 457.84 - ], - "text": "In this first-order transfer function, b1 = 0. The realization is shown in Fig. 4.23b. Because\nthere is only one signal to be added at the output adder, we can discard the adder.", - "type": "text" - }, - { - "block_id": "p412-b14", - "global_id": 11536, - "bbox": [ - 121.09, - 459.75, - 132.15, - 469.71 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p412-b15", - "global_id": 11537, - "bbox": [ - 263.62, - 468.93, - 315.36, - 486.19 - ], - "text": "H(s) = s + 5", - "type": "text" - }, - { - "block_id": "p412-b16", - "global_id": 11538, - "bbox": [ - 295.64, - 482.99, - 315.36, - 493.36 - ], - "text": "s + 7", - "type": "text" - }, - { - "block_id": "p412-b17", - "global_id": 11539, - "bbox": [ - 103.16, - 498.63, - 477.02, - 532.92 - ], - "text": "The realization appears in Fig. 4.23c. Here H(s) is a first-order transfer function with a1 = 7\nand b0 = 1, b1 = 5. There is a single feedback connection (with coefficient 7) from the\nintegrator output to the input adder. There are two feedforward connections (Fig. 4.23c).†", - "type": "text" - }, - { - "block_id": "p412-b18", - "global_id": 11540, - "bbox": [ - 103.16, - 556.61, - 436.88, - 569.19 - ], - "text": "† When M = N (as in this case), H(s) can also be realized in another way by recognizing that", - "type": "text" - }, - { - "block_id": "p412-b19", - "global_id": 11541, - "bbox": [ - 259.02, - 579.23, - 319.98, - 600.85 - ], - "text": "H(s) = 1 −\n2\ns + 7", - "type": "text" - }, - { - "block_id": "p412-b20", - "global_id": 11542, - "bbox": [ - 103.16, - 609.38, - 469.29, - 618.72 - ], - "text": "We now realize H(s) as a parallel combination of two transfer functions, as indicated by this equation.", - "type": "text" - } - ] - }, - { - "page_num": 413, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p413-b0", - "global_id": 11543, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n393", - "type": "text" - }, - { - "block_id": "p413-b1", - "global_id": 11544, - "bbox": [ - 146.84, - 86.28, - 159.02, - 96.25 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p413-b2", - "global_id": 11545, - "bbox": [ - 277.51, - 95.47, - 352.95, - 119.9 - ], - "text": "H(s) =\n4s + 28\ns2 + 6s + 5", - "type": "text" - }, - { - "block_id": "p413-b3", - "global_id": 11546, - "bbox": [ - 128.9, - 125.11, - 502.75, - 147.44 - ], - "text": "This is a second-order system with b0 = 0, b1 = 4, b2 = 28, a1 = 6, and a2 = 5. Figure 4.23d\nshows a realization with two feedback connections and two feedforward connections.", - "type": "text" - }, - { - "block_id": "p413-b4", - "global_id": 11547, - "bbox": [ - 133.12, - 164.59, - 146.45, - 173.44 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p413-b5", - "global_id": 11548, - "bbox": [ - 216.74, - 209.31, - 229.63, - 217.39 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p413-b6", - "global_id": 11549, - "bbox": [ - 174.13, - 222.53, - 225.18, - 230.82 - ], - "text": "7\n5", - "type": "text" - }, - { - "block_id": "p413-b7", - "global_id": 11550, - "bbox": [ - 201.61, - 236.96, - 210.49, - 244.96 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p413-b8", - "global_id": 11551, - "bbox": [ - 133.95, - 257.66, - 278.87, - 265.75 - ], - "text": "X(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p413-b9", - "global_id": 11552, - "bbox": [ - 177.76, - 314.87, - 231.13, - 323.17 - ], - "text": "7\n5", - "type": "text" - }, - { - "block_id": "p413-b10", - "global_id": 11553, - "bbox": [ - 201.61, - 328.88, - 210.49, - 336.88 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p413-b11", - "global_id": 11554, - "bbox": [ - 303.74, - 165.36, - 399.74, - 173.44 - ], - "text": "X(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p413-b12", - "global_id": 11555, - "bbox": [ - 347.03, - 222.78, - 357.69, - 231.08 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p413-b13", - "global_id": 11556, - "bbox": [ - 371.23, - 236.96, - 380.88, - 244.96 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p413-b14", - "global_id": 11557, - "bbox": [ - 303.93, - 257.66, - 317.26, - 265.75 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p413-b15", - "global_id": 11558, - "bbox": [ - 435.12, - 302.93, - 448.01, - 311.01 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p413-b16", - "global_id": 11559, - "bbox": [ - 350.57, - 316.72, - 361.23, - 325.02 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p413-b17", - "global_id": 11560, - "bbox": [ - 347.56, - 361.72, - 403.12, - 370.02 - ], - "text": "5\n28", - "type": "text" - }, - { - "block_id": "p413-b18", - "global_id": 11561, - "bbox": [ - 393.89, - 317.02, - 397.89, - 325.02 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p413-b19", - "global_id": 11562, - "bbox": [ - 371.23, - 375.22, - 380.56, - 383.22 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p413-b23", - "global_id": 11563, - "bbox": [ - 203.0, - 189.97, - 207.0, - 204.82 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p413-b24", - "global_id": 11564, - "bbox": [ - 204.5, - 283.03, - 377.92, - 299.59 - ], - "text": "1\ns\n1\ns", - "type": "text" - }, - { - "block_id": "p413-b25", - "global_id": 11565, - "bbox": [ - 373.92, - 328.38, - 377.92, - 343.23 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p413-b26", - "global_id": 11566, - "bbox": [ - 373.92, - 190.81, - 377.92, - 205.67 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p413-b27", - "global_id": 11567, - "bbox": [ - 128.9, - 392.23, - 299.62, - 401.84 - ], - "text": "Figure 4.23 Realizations of H(s) for Ex. 4.22.", - "type": "text" - }, - { - "block_id": "p413-b28", - "global_id": 11568, - "bbox": [ - 133.57, - 453.36, - 393.51, - 465.32 - ], - "text": "DRILL 4.11\nCanonic Direct Form Realization", - "type": "text" - }, - { - "block_id": "p413-b29", - "global_id": 11569, - "bbox": [ - 133.57, - 474.44, - 281.17, - 484.4 - ], - "text": "Give the canonic direct realization of", - "type": "text" - }, - { - "block_id": "p413-b30", - "global_id": 11570, - "bbox": [ - 281.04, - 492.18, - 361.46, - 516.3 - ], - "text": "H(s) =\n2s\ns2 + 6s + 25", - "type": "text" - }, - { - "block_id": "p413-b31", - "global_id": 11571, - "bbox": [ - 127.59, - 551.02, - 341.58, - 562.97 - ], - "text": "4.6-3 Cascade and Parallel Realizations", - "type": "text" - }, - { - "block_id": "p413-b32", - "global_id": 11572, - "bbox": [ - 127.59, - 568.69, - 516.13, - 602.98 - ], - "text": "An Nth-order transfer function H(s) can be expressed as a product or a sum of N first-order transfer\nfunctions. Accordingly, we can also realize H(s) as a cascade (series) or parallel form of these N\nfirst-order transfer functions. Consider, for instance, the transfer function in part (d) of Ex. 4.22.", - "type": "text" - }, - { - "block_id": "p413-b33", - "global_id": 11573, - "bbox": [ - 283.54, - 611.68, - 358.98, - 636.11 - ], - "text": "H(s) =\n4s + 28\ns2 + 6s + 5", - "type": "text" - } - ] - }, - { - "page_num": 414, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p414-b0", - "global_id": 11574, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "394\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p414-b1", - "global_id": 11575, - "bbox": [ - 101.84, - 85.46, - 195.64, - 95.84 - ], - "text": "We can express H(s) as", - "type": "text" - }, - { - "block_id": "p414-b2", - "global_id": 11576, - "bbox": [ - 205.32, - 116.16, - 302.66, - 140.6 - ], - "text": "H(s) =\n4s + 28\n(s + 1)(s + 5) =", - "type": "text" - }, - { - "block_id": "p414-b3", - "global_id": 11577, - "bbox": [ - 304.71, - 109.16, - 342.31, - 126.54 - ], - "text": "4s + 28", - "type": "text" - }, - { - "block_id": "p414-b4", - "global_id": 11578, - "bbox": [ - 317.61, - 130.22, - 337.33, - 140.6 - ], - "text": "s + 1", - "type": "text" - }, - { - "block_id": "p414-b6", - "global_id": 11579, - "bbox": [ - 304.71, - 136.62, - 350.25, - 157.09 - ], - "text": "H1(s)", - "type": "text" - }, - { - "block_id": "p414-b7", - "global_id": 11580, - "bbox": [ - 351.35, - 109.16, - 378.99, - 140.6 - ], - "text": "1\ns + 5", - "type": "text" - }, - { - "block_id": "p414-b9", - "global_id": 11581, - "bbox": [ - 351.35, - 136.62, - 386.92, - 157.09 - ], - "text": "H2(s)", - "type": "text" - }, - { - "block_id": "p414-b10", - "global_id": 11582, - "bbox": [ - 101.84, - 176.71, - 327.06, - 187.08 - ], - "text": "We can also express H(s) as a sum of partial fractions as", - "type": "text" - }, - { - "block_id": "p414-b11", - "global_id": 11583, - "bbox": [ - 219.31, - 206.73, - 340.82, - 235.58 - ], - "text": "H(s) =\n4s + 28\n(s + 1)(s + 5) =\n6\ns + 1", - "type": "text" - }, - { - "block_id": "p414-b12", - "global_id": 11584, - "bbox": [ - 321.54, - 237.56, - 337.98, - 246.09 - ], - "text": "H3(s)", - "type": "text" - }, - { - "block_id": "p414-b13", - "global_id": 11585, - "bbox": [ - 341.93, - 207.14, - 372.93, - 235.58 - ], - "text": "−\n2\ns + 5", - "type": "text" - }, - { - "block_id": "p414-b14", - "global_id": 11586, - "bbox": [ - 353.65, - 237.56, - 370.08, - 246.09 - ], - "text": "H4(s)", - "type": "text" - }, - { - "block_id": "p414-b15", - "global_id": 11587, - "bbox": [ - 101.84, - 265.77, - 490.38, - 312.01 - ], - "text": "These equations give us the option of realizing H(s) as a cascade of H1(s) and H2(s), as shown\nin Fig. 4.24a, or a parallel of H3(s) and H4(s), as depicted in Fig. 4.24b. Each of the first-order\ntransfer functions in Fig. 4.24 can be implemented by using canonic direct realizations, discussed\nearlier.", - "type": "text" - }, - { - "block_id": "p414-b16", - "global_id": 11588, - "bbox": [ - 101.84, - 314.01, - 490.43, - 371.79 - ], - "text": "This discussion by no means exhausts all the possibilities. In the cascade form alone, there\nare different ways of grouping the factors in the numerator and the denominator of H(s), and each\ngrouping can be realized in DFI or canonic direct form. Accordingly, several cascade forms are\npossible. In Sec. 4.6-4, we shall discuss yet another form that essentially doubles the numbers of\nrealizations discussed so far.", - "type": "text" - }, - { - "block_id": "p414-b17", - "global_id": 11589, - "bbox": [ - 101.84, - 373.78, - 490.39, - 455.47 - ], - "text": "From a practical viewpoint, parallel and cascade forms are preferable because parallel and\ncertain cascade forms are numerically less sensitive than canonic direct form to small parameter\nvariations in the system. Qualitatively, this difference can be explained by the fact that in a canonic\nrealization all the coefficients interact with each other, and a change in any coefficient will be\nmagnified through its repeated influence from feedback and feedforward connections. In a parallel\nrealization, in contrast, the change in a coefficient will affect only a localized segment; the case\nwith a cascade realization is similar.", - "type": "text" - }, - { - "block_id": "p414-b18", - "global_id": 11590, - "bbox": [ - 101.84, - 457.06, - 490.4, - 503.3 - ], - "text": "In the examples of cascade and parallel realization, we have separated H(s) into first-order\nfactors. For H(s) of higher orders, we could group H(s) into factors, not all of which are necessarily\nof the first order. For example, if H(s) is a third-order transfer function, we could realize this\nfunction as a cascade (or a parallel) combination of a first-order and a second-order factor.", - "type": "text" - }, - { - "block_id": "p414-b19", - "global_id": 11591, - "bbox": [ - 187.09, - 605.26, - 356.32, - 613.26 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p414-b20", - "global_id": 11592, - "bbox": [ - 125.99, - 552.27, - 251.46, - 572.74 - ], - "text": "Y(s)\nX(s)\n4s 28\ns 1", - "type": "text" - }, - { - "block_id": "p414-b21", - "global_id": 11593, - "bbox": [ - 208.62, - 556.01, - 401.58, - 572.88 - ], - "text": "1\ns 5", - "type": "text" - }, - { - "block_id": "p414-b24", - "global_id": 11594, - "bbox": [ - 278.1, - 552.68, - 424.05, - 561.58 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p414-b25", - "global_id": 11595, - "bbox": [ - 339.76, - 534.21, - 357.54, - 550.84 - ], - "text": "6\ns 1", - "type": "text" - }, - { - "block_id": "p414-b26", - "global_id": 11596, - "bbox": [ - 338.44, - 577.89, - 356.21, - 594.35 - ], - "text": "2\ns 5", - "type": "text" - }, - { - "block_id": "p414-b27", - "global_id": 11597, - "bbox": [ - 125.76, - 619.58, - 465.51, - 629.19 - ], - "text": "Figure 4.24 Realization of (4s + 28)/[(s + 1)(s + 5)]: (a) cascade form and (b) parallel form.", - "type": "text" - } - ] - }, - { - "page_num": 415, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p415-b0", - "global_id": 11598, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n395", - "type": "text" - }, - { - "block_id": "p415-b1", - "global_id": 11599, - "bbox": [ - 127.89, - 86.19, - 380.82, - 98.32 - ], - "text": "REALIZATION OF COMPLEX CONJUGATE POLES", - "type": "text" - }, - { - "block_id": "p415-b2", - "global_id": 11600, - "bbox": [ - 127.59, - 101.93, - 516.14, - 124.26 - ], - "text": "The complex poles in H(s) should be realized as a second-order (quadratic) factor because we\ncannot implement multiplication by complex numbers. Consider, for example,", - "type": "text" - }, - { - "block_id": "p415-b3", - "global_id": 11601, - "bbox": [ - 243.78, - 133.45, - 358.78, - 157.89 - ], - "text": "H(s) =\n10s + 50\n(s + 3)(s2 + 4s + 13)", - "type": "text" - }, - { - "block_id": "p415-b4", - "global_id": 11602, - "bbox": [ - 264.78, - 160.01, - 394.48, - 184.45 - ], - "text": "=\n10s + 50\n(s + 3)(s + 2 −j3)(s + 2 + j3)", - "type": "text" - }, - { - "block_id": "p415-b5", - "global_id": 11603, - "bbox": [ - 264.78, - 186.88, - 398.73, - 211.31 - ], - "text": "=\n2\ns + 3 −\n1 + j2\ns + 2 −j3 −\n1 −j2\ns + 2 + j3", - "type": "text" - }, - { - "block_id": "p415-b6", - "global_id": 11604, - "bbox": [ - 127.59, - 220.37, - 516.16, - 266.62 - ], - "text": "We cannot realize first-order transfer functions individually with the poles −2 ± j3 because they\nrequire multiplication by complex numbers in the feedback and the feedforward paths. Therefore,\nwe need to combine the conjugate poles and realize them as a second-order transfer function.† In\nthe present example, we can create a cascade realization from H(s) expressed in product form as", - "type": "text" - }, - { - "block_id": "p415-b7", - "global_id": 11605, - "bbox": [ - 256.54, - 283.47, - 285.31, - 293.75 - ], - "text": "H(s) =", - "type": "text" - }, - { - "block_id": "p415-b8", - "global_id": 11606, - "bbox": [ - 287.35, - 269.48, - 310.12, - 286.87 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p415-b9", - "global_id": 11607, - "bbox": [ - 295.27, - 290.54, - 314.99, - 300.92 - ], - "text": "s + 3", - "type": "text" - }, - { - "block_id": "p415-b10", - "global_id": 11608, - "bbox": [ - 316.2, - 269.48, - 379.25, - 300.92 - ], - "text": "s + 5\ns2 + 4s + 13", - "type": "text" - }, - { - "block_id": "p415-b12", - "global_id": 11609, - "bbox": [ - 127.59, - 310.4, - 453.95, - 320.78 - ], - "text": "Similarly, we can create a parallel realization from H(s) expressed in sum form as", - "type": "text" - }, - { - "block_id": "p415-b13", - "global_id": 11610, - "bbox": [ - 264.56, - 329.97, - 377.95, - 354.4 - ], - "text": "H(s) =\n2\ns + 3 −\n2s −8\ns2 + 4s + 13", - "type": "text" - }, - { - "block_id": "p415-b14", - "global_id": 11611, - "bbox": [ - 127.89, - 365.29, - 315.21, - 377.41 - ], - "text": "REALIZATION OF REPEATED POLES", - "type": "text" - }, - { - "block_id": "p415-b15", - "global_id": 11612, - "bbox": [ - 127.59, - 381.45, - 516.15, - 415.31 - ], - "text": "When repeated poles occur, the procedure for canonic and cascade realization is exactly the same\nas before. For a parallel realization, however, the procedure requires special handling, as explained\nin Ex. 4.23.", - "type": "text" - }, - { - "block_id": "p415-b16", - "global_id": 11613, - "bbox": [ - 102.51, - 437.1, - 313.9, - 449.05 - ], - "text": "EXAMPLE 4.23\nParallel Realization", - "type": "text" - }, - { - "block_id": "p415-b17", - "global_id": 11614, - "bbox": [ - 128.9, - 462.8, - 273.33, - 472.76 - ], - "text": "Determine the parallel realization of", - "type": "text" - }, - { - "block_id": "p415-b18", - "global_id": 11615, - "bbox": [ - 214.35, - 480.25, - 304.72, - 501.23 - ], - "text": "H(s) = 7s2 + 37s + 51", - "type": "text" - }, - { - "block_id": "p415-b19", - "global_id": 11616, - "bbox": [ - 246.4, - 484.28, - 415.62, - 508.3 - ], - "text": "(s + 2)(s + 3)2 =\n5\ns + 2 +\n2\ns + 3 −\n3\n(s + 3)2", - "type": "text" - }, - { - "block_id": "p415-b20", - "global_id": 11617, - "bbox": [ - 128.9, - 534.11, - 502.78, - 567.97 - ], - "text": "This third-order transfer function should require no more than three integrators. But if we try to\nrealize each of the three partial fractions separately, we require four integrators because of the\none second-order term. This difficulty can be avoided by observing that the terms 1/(s+3) and", - "type": "text" - }, - { - "block_id": "p415-b21", - "global_id": 11618, - "bbox": [ - 127.59, - 609.24, - 516.12, - 643.38 - ], - "text": "† It is possible to realize complex, conjugate poles indirectly by using a cascade of two first-order transfer\nfunctions and feedback. A transfer function with poles −a ± jb can be realized by using a cascade of two\nidentical first-order transfer functions, each having a pole at −a (see Prob. 4.6-15).", - "type": "text" - } - ] - }, - { - "page_num": 416, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p416-b0", - "global_id": 11619, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "396\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p416-b1", - "global_id": 11620, - "bbox": [ - 230.5, - 95.54, - 234.5, - 103.54 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p416-b2", - "global_id": 11621, - "bbox": [ - 283.87, - 185.98, - 287.87, - 193.98 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p416-b3", - "global_id": 11622, - "bbox": [ - 257.61, - 139.97, - 350.24, - 155.04 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p416-b6", - "global_id": 11623, - "bbox": [ - 158.81, - 98.03, - 176.59, - 114.78 - ], - "text": "1\ns 2", - "type": "text" - }, - { - "block_id": "p416-b7", - "global_id": 11624, - "bbox": [ - 156.81, - 188.23, - 174.59, - 204.98 - ], - "text": "1\ns 3", - "type": "text" - }, - { - "block_id": "p416-b8", - "global_id": 11625, - "bbox": [ - 236.31, - 188.23, - 254.09, - 204.98 - ], - "text": "1\ns 3", - "type": "text" - }, - { - "block_id": "p416-b9", - "global_id": 11626, - "bbox": [ - 97.2, - 137.38, - 380.33, - 146.82 - ], - "text": "X(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p416-b10", - "global_id": 11627, - "bbox": [ - 94.2, - 213.6, - 348.72, - 226.47 - ], - "text": "Figure 4.25 Parallel realization of (7s2 + 37s + 51)/((s + 2)(s + 3)2).", - "type": "text" - }, - { - "block_id": "p416-b11", - "global_id": 11628, - "bbox": [ - 103.16, - 250.52, - 477.0, - 288.0 - ], - "text": "1/(s + 3)2 can be realized with a cascade of two subsystems, each having a transfer function\n1/(s + 3), as shown in Fig. 4.25. Each of the three first-order transfer functions in Fig. 4.25\nmay now be realized as in Fig. 4.23.", - "type": "text" - }, - { - "block_id": "p416-b12", - "global_id": 11629, - "bbox": [ - 107.82, - 356.12, - 426.53, - 368.07 - ], - "text": "DRILL 4.12\nCanonic, Cascade, and Parallel Realizations", - "type": "text" - }, - { - "block_id": "p416-b13", - "global_id": 11630, - "bbox": [ - 107.82, - 377.2, - 317.27, - 387.16 - ], - "text": "Find the canonic, cascade, and parallel realization of", - "type": "text" - }, - { - "block_id": "p416-b14", - "global_id": 11631, - "bbox": [ - 213.25, - 397.36, - 304.69, - 421.79 - ], - "text": "H(s) =\ns + 3\ns2 + 7s + 10 =", - "type": "text" - }, - { - "block_id": "p416-b15", - "global_id": 11632, - "bbox": [ - 306.74, - 390.35, - 334.38, - 407.73 - ], - "text": "s + 3", - "type": "text" - }, - { - "block_id": "p416-b16", - "global_id": 11633, - "bbox": [ - 314.66, - 411.41, - 334.38, - 421.79 - ], - "text": "s + 2", - "type": "text" - }, - { - "block_id": "p416-b17", - "global_id": 11634, - "bbox": [ - 335.58, - 390.35, - 371.06, - 421.79 - ], - "text": "1\ns + 5", - "type": "text" - }, - { - "block_id": "p416-b19", - "global_id": 11635, - "bbox": [ - 101.84, - 463.28, - 260.06, - 475.23 - ], - "text": "4.6-4 Transposed Realization", - "type": "text" - }, - { - "block_id": "p416-b20", - "global_id": 11636, - "bbox": [ - 101.84, - 481.26, - 490.39, - 515.23 - ], - "text": "Two realizations are said to be equivalent if they have the same transfer function. A simple way\nto generate an equivalent realization from a given realization is to use its transpose. To generate a\ntranspose of any realization, we change the given realization as follows:", - "type": "text" - }, - { - "block_id": "p416-b21", - "global_id": 11637, - "bbox": [ - 118.78, - 523.21, - 443.02, - 557.07 - ], - "text": "1. Reverse all the arrow directions without changing the scalar multiplier values.\n2. Replace pickoff nodes by adders and vice versa.\n3. Replace the input X(s) with the output Y(s) and vice versa.", - "type": "text" - }, - { - "block_id": "p416-b22", - "global_id": 11638, - "bbox": [ - 101.84, - 565.05, - 490.39, - 610.88 - ], - "text": "Figure 4.26a shows the transposed version of the canonic direct form realization in Fig. 4.22b\nfound according to the rules just listed. Figure 4.26b is Fig. 4.26a reoriented in the conventional\nform so that the input X(s) appears at the left and the output Y(s) appears at the right. Observe that\nthis realization is also canonic.", - "type": "text" - }, - { - "block_id": "p416-b23", - "global_id": 11639, - "bbox": [ - 101.84, - 612.86, - 490.37, - 634.79 - ], - "text": "Rather than prove the theorem on equivalence of the transposed realizations, we shall verify\nthat the transfer function of the realization in Fig. 4.26b is identical to that in Eq. (4.36).", - "type": "text" - } - ] - }, - { - "page_num": 417, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p417-b0", - "global_id": 11640, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n397", - "type": "text" - }, - { - "block_id": "p417-b8", - "global_id": 11641, - "bbox": [ - 380.98, - 296.74, - 390.95, - 304.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p417-b9", - "global_id": 11642, - "bbox": [ - 318.92, - 87.28, - 454.4, - 95.37 - ], - "text": "X(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p417-b10", - "global_id": 11643, - "bbox": [ - 397.33, - 210.63, - 419.99, - 220.45 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p417-b11", - "global_id": 11644, - "bbox": [ - 401.83, - 141.96, - 415.49, - 151.78 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p417-b12", - "global_id": 11645, - "bbox": [ - 356.51, - 266.63, - 415.99, - 276.39 - ], - "text": "aN\nbN", - "type": "text" - }, - { - "block_id": "p417-b13", - "global_id": 11646, - "bbox": [ - 352.51, - 210.85, - 368.51, - 220.45 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p417-b14", - "global_id": 11647, - "bbox": [ - 357.01, - 142.18, - 364.01, - 151.78 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p417-b15", - "global_id": 11648, - "bbox": [ - 356.6, - 87.58, - 363.6, - 97.19 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p417-b16", - "global_id": 11649, - "bbox": [ - 218.43, - 296.74, - 227.31, - 304.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p417-b17", - "global_id": 11650, - "bbox": [ - 159.46, - 88.0, - 291.69, - 96.08 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p417-b18", - "global_id": 11651, - "bbox": [ - 237.57, - 210.85, - 253.57, - 220.45 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p417-b19", - "global_id": 11652, - "bbox": [ - 242.07, - 142.18, - 249.07, - 151.78 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p417-b20", - "global_id": 11653, - "bbox": [ - 190.08, - 266.63, - 249.57, - 276.39 - ], - "text": "bN\naN", - "type": "text" - }, - { - "block_id": "p417-b21", - "global_id": 11654, - "bbox": [ - 186.08, - 210.63, - 208.74, - 220.45 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p417-b22", - "global_id": 11655, - "bbox": [ - 190.58, - 141.96, - 204.24, - 151.78 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p417-b23", - "global_id": 11656, - "bbox": [ - 242.07, - 87.41, - 249.07, - 97.01 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p417-b24", - "global_id": 11657, - "bbox": [ - 221.65, - 119.49, - 225.65, - 134.34 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p417-b25", - "global_id": 11658, - "bbox": [ - 221.65, - 173.46, - 225.65, - 188.31 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p417-b26", - "global_id": 11659, - "bbox": [ - 221.65, - 243.54, - 225.65, - 258.39 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p417-b27", - "global_id": 11660, - "bbox": [ - 384.19, - 119.49, - 388.19, - 134.34 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p417-b28", - "global_id": 11661, - "bbox": [ - 384.19, - 173.46, - 388.19, - 188.31 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p417-b29", - "global_id": 11662, - "bbox": [ - 384.19, - 243.54, - 388.19, - 258.39 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p417-b30", - "global_id": 11663, - "bbox": [ - 151.5, - 311.35, - 462.5, - 320.67 - ], - "text": "Figure 4.26 Realization of an Nth-order LTI transfer function in the transposed form.", - "type": "text" - }, - { - "block_id": "p417-b31", - "global_id": 11664, - "bbox": [ - 127.59, - 340.87, - 516.11, - 363.2 - ], - "text": "Figure 4.26b shows that Y(s) is being fed back through N paths. The fed-back signal appearing\nat the input of the top adder is", - "type": "text" - }, - { - "block_id": "p417-b32", - "global_id": 11665, - "bbox": [ - 233.2, - 363.46, - 257.35, - 381.61 - ], - "text": "−a1", - "type": "text" - }, - { - "block_id": "p417-b33", - "global_id": 11666, - "bbox": [ - 247.55, - 370.46, - 287.35, - 394.79 - ], - "text": "s\n+ −a2", - "type": "text" - }, - { - "block_id": "p417-b34", - "global_id": 11667, - "bbox": [ - 275.55, - 370.46, - 351.79, - 394.79 - ], - "text": "s2 + · · · + −aN−1", - "type": "text" - }, - { - "block_id": "p417-b35", - "global_id": 11668, - "bbox": [ - 329.49, - 370.46, - 382.96, - 394.79 - ], - "text": "sN−1 + −aN", - "type": "text" - }, - { - "block_id": "p417-b36", - "global_id": 11669, - "bbox": [ - 370.0, - 384.25, - 378.53, - 394.79 - ], - "text": "sN", - "type": "text" - }, - { - "block_id": "p417-b38", - "global_id": 11670, - "bbox": [ - 392.91, - 377.44, - 410.53, - 387.72 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p417-b39", - "global_id": 11671, - "bbox": [ - 127.59, - 404.75, - 437.72, - 415.12 - ], - "text": "The signal X(s), fed to the top adder through N + 1 forward paths, contributes", - "type": "text" - }, - { - "block_id": "p417-b41", - "global_id": 11672, - "bbox": [ - 256.56, - 422.69, - 286.04, - 440.82 - ], - "text": "b0 + b1", - "type": "text" - }, - { - "block_id": "p417-b42", - "global_id": 11673, - "bbox": [ - 280.12, - 422.69, - 342.72, - 446.71 - ], - "text": "s + · · · + bN−1", - "type": "text" - }, - { - "block_id": "p417-b43", - "global_id": 11674, - "bbox": [ - 324.3, - 422.69, - 366.12, - 446.71 - ], - "text": "sN−1 + bN", - "type": "text" - }, - { - "block_id": "p417-b44", - "global_id": 11675, - "bbox": [ - 357.04, - 436.17, - 365.57, - 446.71 - ], - "text": "sN", - "type": "text" - }, - { - "block_id": "p417-b46", - "global_id": 11676, - "bbox": [ - 376.06, - 429.36, - 393.9, - 439.64 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p417-b47", - "global_id": 11677, - "bbox": [ - 127.59, - 456.68, - 500.52, - 467.05 - ], - "text": "The output Y(s) is equal to the sum of these two signals (feed forward and feed back). Hence,", - "type": "text" - }, - { - "block_id": "p417-b48", - "global_id": 11678, - "bbox": [ - 219.02, - 481.29, - 246.44, - 491.57 - ], - "text": "Y(s) =", - "type": "text" - }, - { - "block_id": "p417-b49", - "global_id": 11679, - "bbox": [ - 248.49, - 467.31, - 272.64, - 485.46 - ], - "text": "−a1", - "type": "text" - }, - { - "block_id": "p417-b50", - "global_id": 11680, - "bbox": [ - 262.85, - 474.31, - 302.64, - 498.64 - ], - "text": "s\n+ −a2", - "type": "text" - }, - { - "block_id": "p417-b51", - "global_id": 11681, - "bbox": [ - 290.85, - 474.31, - 367.08, - 498.64 - ], - "text": "s2 + · · · + −aN−1", - "type": "text" - }, - { - "block_id": "p417-b52", - "global_id": 11682, - "bbox": [ - 344.79, - 474.31, - 398.25, - 498.64 - ], - "text": "sN−1 + −aN", - "type": "text" - }, - { - "block_id": "p417-b53", - "global_id": 11683, - "bbox": [ - 385.29, - 488.1, - 393.82, - 498.64 - ], - "text": "sN", - "type": "text" - }, - { - "block_id": "p417-b55", - "global_id": 11684, - "bbox": [ - 407.1, - 481.29, - 424.71, - 491.57 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p417-b56", - "global_id": 11685, - "bbox": [ - 250.04, - 509.18, - 257.81, - 519.15 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p417-b58", - "global_id": 11686, - "bbox": [ - 267.19, - 502.51, - 296.69, - 520.64 - ], - "text": "b0 + b1", - "type": "text" - }, - { - "block_id": "p417-b59", - "global_id": 11687, - "bbox": [ - 290.77, - 502.51, - 353.36, - 526.53 - ], - "text": "s + · · · + bN−1", - "type": "text" - }, - { - "block_id": "p417-b60", - "global_id": 11688, - "bbox": [ - 334.94, - 502.51, - 376.76, - 526.53 - ], - "text": "sN−1 + bN", - "type": "text" - }, - { - "block_id": "p417-b61", - "global_id": 11689, - "bbox": [ - 367.68, - 515.99, - 376.21, - 526.53 - ], - "text": "sN", - "type": "text" - }, - { - "block_id": "p417-b63", - "global_id": 11690, - "bbox": [ - 385.6, - 509.18, - 403.43, - 519.46 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p417-b64", - "global_id": 11691, - "bbox": [ - 145.52, - 536.91, - 508.17, - 548.3 - ], - "text": "Transporting all the Y(s) terms to the left side and multiplying throughout by sN, we obtain", - "type": "text" - }, - { - "block_id": "p417-b65", - "global_id": 11692, - "bbox": [ - 159.06, - 555.76, - 484.67, - 571.03 - ], - "text": "(sN + a1sN−1 + · · · + aN−1s + aN)Y(s) = (b0sN + b1sN−1 + · · · + bN−1s + bN)X(s)", - "type": "text" - }, - { - "block_id": "p417-b66", - "global_id": 11693, - "bbox": [ - 127.6, - 582.26, - 184.21, - 592.22 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p417-b67", - "global_id": 11694, - "bbox": [ - 221.11, - 593.42, - 270.85, - 610.68 - ], - "text": "H(s) = Y(s)", - "type": "text" - }, - { - "block_id": "p417-b68", - "global_id": 11695, - "bbox": [ - 253.12, - 589.81, - 420.49, - 617.76 - ], - "text": "X(s) = b0sN + b1sN−1 + · · · + bN−1s + bN", - "type": "text" - }, - { - "block_id": "p417-b69", - "global_id": 11696, - "bbox": [ - 289.7, - 604.6, - 416.01, - 618.63 - ], - "text": "sN + a1sN−1 + · · · + aN−1s + aN", - "type": "text" - }, - { - "block_id": "p417-b70", - "global_id": 11697, - "bbox": [ - 127.59, - 624.41, - 389.2, - 634.79 - ], - "text": "Hence, the transfer function H(s) is identical to that in Eq. (4.36).", - "type": "text" - } - ] - }, - { - "page_num": 418, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p418-b0", - "global_id": 11698, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "398\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p418-b1", - "global_id": 11699, - "bbox": [ - 101.84, - 85.82, - 490.4, - 107.74 - ], - "text": "We have essentially doubled the number of possible realizations. Every realization that was\nfound earlier has a transpose. Note that the transpose of a transpose results in the same realization.", - "type": "text" - }, - { - "block_id": "p418-b2", - "global_id": 11700, - "bbox": [ - 76.77, - 139.54, - 314.08, - 151.5 - ], - "text": "EXAMPLE 4.24\nTransposed Realizations", - "type": "text" - }, - { - "block_id": "p418-b3", - "global_id": 11701, - "bbox": [ - 103.16, - 168.16, - 476.99, - 190.08 - ], - "text": "Find the transpose canonic direct realizations for parts (a) and (d) of Ex. 4.22 (Figs. 4.23c and\n4.23d). The transfer functions are:", - "type": "text" - }, - { - "block_id": "p418-b4", - "global_id": 11702, - "bbox": [ - 121.09, - 195.24, - 158.62, - 212.52 - ], - "text": "(a) s + 5", - "type": "text" - }, - { - "block_id": "p418-b5", - "global_id": 11703, - "bbox": [ - 138.89, - 209.3, - 158.62, - 219.67 - ], - "text": "s + 7", - "type": "text" - }, - { - "block_id": "p418-b6", - "global_id": 11704, - "bbox": [ - 103.16, - 220.9, - 360.07, - 259.91 - ], - "text": "(b)\n4s + 28\ns2 + 6s + 5\nBoth these realizations are special cases of the one in Fig. 4.26b.", - "type": "text" - }, - { - "block_id": "p418-b7", - "global_id": 11705, - "bbox": [ - 103.16, - 282.41, - 477.01, - 329.73 - ], - "text": "(a) In this case, N = 1 with a1 = 7,b0 = 1,b1 = 5. The desired realization can be obtained\nby transposing Fig. 4.23c. However, we already have the general model of the transposed\nrealization in Fig. 4.26b. The desired solution is a special case of Fig. 4.26b with N = 1 and\na1 = 7,b0 = 1,b1 = 5, as shown in Fig. 4.27a.", - "type": "text" - }, - { - "block_id": "p418-b8", - "global_id": 11706, - "bbox": [ - 103.17, - 330.24, - 477.01, - 352.57 - ], - "text": "(b) In this case, N = 2 with b0 = 0, b1 = 4, b2 = 28, a1 = 6, a2 = 5. Using the model of\nFig. 4.26b, we obtain the desired realization, as shown in Fig. 4.27b.", - "type": "text" - }, - { - "block_id": "p418-b9", - "global_id": 11707, - "bbox": [ - 161.14, - 503.24, - 331.35, - 511.24 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p418-b12", - "global_id": 11708, - "bbox": [ - 138.18, - 487.83, - 196.87, - 496.13 - ], - "text": "7\n5", - "type": "text" - }, - { - "block_id": "p418-b13", - "global_id": 11709, - "bbox": [ - 100.33, - 421.2, - 232.8, - 429.29 - ], - "text": "X(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p418-b14", - "global_id": 11710, - "bbox": [ - 164.7, - 451.97, - 168.7, - 466.8 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p418-b16", - "global_id": 11711, - "bbox": [ - 301.09, - 427.54, - 359.4, - 442.65 - ], - "text": "6\n4", - "type": "text" - }, - { - "block_id": "p418-b17", - "global_id": 11712, - "bbox": [ - 299.16, - 487.83, - 359.4, - 496.13 - ], - "text": "5\n28", - "type": "text" - }, - { - "block_id": "p418-b18", - "global_id": 11713, - "bbox": [ - 263.09, - 421.2, - 276.41, - 429.29 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p418-b19", - "global_id": 11714, - "bbox": [ - 378.76, - 370.85, - 391.65, - 378.94 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p418-b20", - "global_id": 11715, - "bbox": [ - 327.45, - 451.97, - 331.45, - 466.8 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p418-b21", - "global_id": 11716, - "bbox": [ - 328.03, - 395.77, - 332.03, - 410.61 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p418-b22", - "global_id": 11717, - "bbox": [ - 94.2, - 517.4, - 468.05, - 539.73 - ], - "text": "Figure 4.27 Transposed canonic direct form realizations of (a) (s + 5)/(s + 7) and (b)\n(4s + 28)/(s2 + 6s + 5).", - "type": "text" - }, - { - "block_id": "p418-b23", - "global_id": 11718, - "bbox": [ - 107.82, - 603.93, - 321.95, - 615.88 - ], - "text": "DRILL 4.13\nTransposed Realizations", - "type": "text" - }, - { - "block_id": "p418-b24", - "global_id": 11719, - "bbox": [ - 107.82, - 624.59, - 484.4, - 634.96 - ], - "text": "Find the transposed DFI and transposed canonic direct (TDFII) realizations of H(s) in Drill 4.11", - "type": "text" - } - ] - }, - { - "page_num": 419, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p419-b0", - "global_id": 11720, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n399", - "type": "text" - }, - { - "block_id": "p419-b1", - "global_id": 11721, - "bbox": [ - 127.59, - 86.52, - 447.83, - 98.48 - ], - "text": "4.6-5 Using Operational Amplifiers for System Realization", - "type": "text" - }, - { - "block_id": "p419-b2", - "global_id": 11722, - "bbox": [ - 127.59, - 104.61, - 516.14, - 150.44 - ], - "text": "In this section, we discuss practical implementation of the realizations described in Sec. 4.6-4.\nEarlier we saw that the basic elements required for the synthesis of an LTIC system (or a given\ntransfer function) are (scalar) multipliers, integrators, and adders. All these elements can be\nrealized by operational amplifier (op-amp) circuits.", - "type": "text" - }, - { - "block_id": "p419-b3", - "global_id": 11723, - "bbox": [ - 127.59, - 167.89, - 516.13, - 229.88 - ], - "text": "OPERATIONAL AMPLIFIER CIRCUITS\nFigure 4.28 shows an op-amp circuit in the frequency domain (the transformed circuit). Because\nthe input impedance of the op amp is infinite (very high), all the current I(s) flows in the feedback\npath, as illustrated. Moreover Vx(s), the voltage at the input of the op amp, is zero (very small)\nbecause of the infinite (very large) gain of the op amp. Therefore, for all practical purposes,", - "type": "text" - }, - { - "block_id": "p419-b4", - "global_id": 11724, - "bbox": [ - 285.52, - 246.86, - 358.21, - 258.32 - ], - "text": "Y(s) = −I(s)Zf (s)", - "type": "text" - }, - { - "block_id": "p419-b5", - "global_id": 11725, - "bbox": [ - 127.59, - 274.23, - 232.55, - 285.69 - ], - "text": "Moreover, because vx ≈0,", - "type": "text" - }, - { - "block_id": "p419-b6", - "global_id": 11726, - "bbox": [ - 298.25, - 291.83, - 344.29, - 309.09 - ], - "text": "I(s) = X(s)", - "type": "text" - }, - { - "block_id": "p419-b7", - "global_id": 11727, - "bbox": [ - 326.69, - 305.88, - 344.03, - 316.16 - ], - "text": "Z(s)", - "type": "text" - }, - { - "block_id": "p419-b8", - "global_id": 11728, - "bbox": [ - 127.59, - 328.24, - 337.68, - 338.2 - ], - "text": "Substitution of the second equation in the first yields", - "type": "text" - }, - { - "block_id": "p419-b9", - "global_id": 11729, - "bbox": [ - 282.97, - 353.23, - 341.73, - 370.53 - ], - "text": "Y(s) = −Zf (s)", - "type": "text" - }, - { - "block_id": "p419-b10", - "global_id": 11730, - "bbox": [ - 322.89, - 360.25, - 360.76, - 377.6 - ], - "text": "Z(s) X(s)", - "type": "text" - }, - { - "block_id": "p419-b11", - "global_id": 11731, - "bbox": [ - 127.6, - 392.67, - 392.39, - 402.63 - ], - "text": "Therefore, the op-amp circuit in Fig. 4.28 has the transfer function", - "type": "text" - }, - { - "block_id": "p419-b12", - "global_id": 11732, - "bbox": [ - 291.22, - 417.66, - 351.31, - 434.95 - ], - "text": "H(s) = −Zf (s)", - "type": "text" - }, - { - "block_id": "p419-b13", - "global_id": 11733, - "bbox": [ - 332.48, - 431.75, - 349.82, - 442.02 - ], - "text": "Z(s)", - "type": "text" - }, - { - "block_id": "p419-b14", - "global_id": 11734, - "bbox": [ - 127.59, - 456.74, - 516.13, - 479.07 - ], - "text": "By properly choosing Z(s) and Zf (s), we can obtain a variety of transfer functions, as the following\ndevelopment shows.", - "type": "text" - }, - { - "block_id": "p419-b18", - "global_id": 11735, - "bbox": [ - 150.92, - 577.31, - 309.33, - 592.02 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p419-b19", - "global_id": 11736, - "bbox": [ - 160.95, - 549.62, - 174.32, - 557.7 - ], - "text": "Z(s)", - "type": "text" - }, - { - "block_id": "p419-b20", - "global_id": 11737, - "bbox": [ - 197.51, - 509.12, - 255.93, - 528.24 - ], - "text": "Zf (s)\nI(s)", - "type": "text" - }, - { - "block_id": "p419-b21", - "global_id": 11738, - "bbox": [ - 191.37, - 551.69, - 248.37, - 576.24 - ], - "text": "Vx(s)\nI(s)\n–\n+", - "type": "text" - }, - { - "block_id": "p419-b22", - "global_id": 11739, - "bbox": [ - 319.55, - 604.7, - 512.13, - 626.62 - ], - "text": "Figure 4.28 A\nbasic\ninverting\nconfiguration\nop-amp circuit.", - "type": "text" - } - ] - }, - { - "page_num": 420, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p420-b0", - "global_id": 11740, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "400\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p420-b1", - "global_id": 11741, - "bbox": [ - 102.14, - 86.19, - 240.15, - 98.32 - ], - "text": "THE SCALAR MULTIPLIER", - "type": "text" - }, - { - "block_id": "p420-b2", - "global_id": 11742, - "bbox": [ - 101.84, - 101.93, - 490.39, - 124.26 - ], - "text": "If we use a resistor Rf in the feedback and a resistor R at the input (Fig. 4.29a), then Zf (s) = Rf ,\nZ(s) = R, and", - "type": "text" - }, - { - "block_id": "p420-b3", - "global_id": 11743, - "bbox": [ - 270.86, - 127.02, - 318.66, - 143.99 - ], - "text": "H(s) = −Rf", - "type": "text" - }, - { - "block_id": "p420-b4", - "global_id": 11744, - "bbox": [ - 312.36, - 141.1, - 318.45, - 151.06 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p420-b5", - "global_id": 11745, - "bbox": [ - 101.85, - 157.84, - 490.4, - 204.08 - ], - "text": "The system acts as a scalar multiplier (or an amplifier) with a negative gain Rf /R. A positive\ngain can be obtained by using two such multipliers in cascade or by using a single noninverting\namplifier, as depicted in Fig. 4.16c. Figure 4.29a also shows the compact symbol used in circuit\ndiagrams for a scalar multiplier.", - "type": "text" - }, - { - "block_id": "p420-b6", - "global_id": 11746, - "bbox": [ - 101.84, - 219.97, - 490.38, - 258.03 - ], - "text": "THE INTEGRATOR\nIf we use a capacitor C in the feedback and a resistor R at the input (Fig. 4.29b), then Zf (s) = 1/Cs,\nZ(s) = R, and", - "type": "text" - }, - { - "block_id": "p420-b7", - "global_id": 11747, - "bbox": [ - 258.74, - 268.28, - 287.51, - 278.56 - ], - "text": "H(s) =", - "type": "text" - }, - { - "block_id": "p420-b9", - "global_id": 11748, - "bbox": [ - 296.29, - 261.71, - 314.22, - 278.24 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p420-b10", - "global_id": 11749, - "bbox": [ - 305.24, - 275.66, - 317.97, - 285.62 - ], - "text": "RC", - "type": "text" - }, - { - "block_id": "p420-b11", - "global_id": 11750, - "bbox": [ - 319.39, - 254.29, - 332.29, - 271.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p420-b12", - "global_id": 11751, - "bbox": [ - 327.86, - 275.66, - 331.74, - 285.62 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p420-b13", - "global_id": 11752, - "bbox": [ - 101.84, - 294.8, - 490.37, - 317.12 - ], - "text": "The system acts as an ideal integrator with a gain −1/RC. Figure 4.29b also shows the compact\nsymbol used in circuit diagrams for an integrator.", - "type": "text" - }, - { - "block_id": "p420-b14", - "global_id": 11753, - "bbox": [ - 273.03, - 605.26, - 282.68, - 613.26 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p420-b15", - "global_id": 11754, - "bbox": [ - 273.42, - 457.57, - 282.3, - 465.57 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p420-b16", - "global_id": 11755, - "bbox": [ - 238.9, - 376.24, - 244.54, - 400.79 - ], - "text": "–\n+", - "type": "text" - }, - { - "block_id": "p420-b17", - "global_id": 11756, - "bbox": [ - 147.16, - 403.32, - 305.5, - 416.09 - ], - "text": "Y(s)\nF(s)", - "type": "text" - }, - { - "block_id": "p420-b18", - "global_id": 11757, - "bbox": [ - 243.08, - 335.67, - 249.64, - 345.21 - ], - "text": "Rf", - "type": "text" - }, - { - "block_id": "p420-b19", - "global_id": 11758, - "bbox": [ - 166.41, - 363.85, - 171.3, - 371.85 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p420-b22", - "global_id": 11759, - "bbox": [ - 343.73, - 397.48, - 423.09, - 405.56 - ], - "text": "Y(s)\nF(s)", - "type": "text" - }, - { - "block_id": "p420-b23", - "global_id": 11760, - "bbox": [ - 379.03, - 405.56, - 382.58, - 413.56 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p420-b24", - "global_id": 11761, - "bbox": [ - 394.31, - 430.26, - 408.52, - 438.47 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p420-b25", - "global_id": 11762, - "bbox": [ - 408.51, - 422.96, - 421.73, - 432.72 - ], - "text": "Rf", - "type": "text" - }, - { - "block_id": "p420-b26", - "global_id": 11763, - "bbox": [ - 415.29, - 434.76, - 420.18, - 442.76 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p420-b27", - "global_id": 11764, - "bbox": [ - 148.16, - 551.55, - 382.02, - 561.43 - ], - "text": "F(s)", - "type": "text" - }, - { - "block_id": "p420-b28", - "global_id": 11765, - "bbox": [ - 165.41, - 511.68, - 170.3, - 519.68 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p420-b29", - "global_id": 11766, - "bbox": [ - 292.61, - 556.34, - 305.5, - 564.42 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p420-b31", - "global_id": 11767, - "bbox": [ - 295.72, - 570.37, - 419.48, - 582.82 - ], - "text": "k 1", - "type": "text" - }, - { - "block_id": "p420-b32", - "global_id": 11768, - "bbox": [ - 412.21, - 578.9, - 422.44, - 586.9 - ], - "text": "RC", - "type": "text" - }, - { - "block_id": "p420-b33", - "global_id": 11769, - "bbox": [ - 384.2, - 550.13, - 387.75, - 564.66 - ], - "text": "k\ns", - "type": "text" - }, - { - "block_id": "p420-b34", - "global_id": 11770, - "bbox": [ - 241.43, - 478.43, - 249.88, - 494.68 - ], - "text": "1\nCs", - "type": "text" - }, - { - "block_id": "p420-b35", - "global_id": 11771, - "bbox": [ - 345.2, - 546.18, - 424.24, - 554.26 - ], - "text": "Y(s)\nF(s)", - "type": "text" - }, - { - "block_id": "p420-b36", - "global_id": 11772, - "bbox": [ - 238.9, - 524.48, - 244.54, - 549.02 - ], - "text": "–\n+", - "type": "text" - }, - { - "block_id": "p420-b37", - "global_id": 11773, - "bbox": [ - 125.76, - 619.88, - 341.11, - 629.19 - ], - "text": "Figure 4.29 (a) Op-amp inverting amplifier. (b) Integrator.", - "type": "text" - } - ] - }, - { - "page_num": 421, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p421-b0", - "global_id": 11774, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n401", - "type": "text" - }, - { - "block_id": "p421-b1", - "global_id": 11775, - "bbox": [ - 151.5, - 254.44, - 350.74, - 263.68 - ], - "text": "Figure 4.30 Op-amp summing and amplifying circuit.", - "type": "text" - }, - { - "block_id": "p421-b2", - "global_id": 11776, - "bbox": [ - 127.59, - 277.92, - 516.15, - 341.39 - ], - "text": "THE ADDER\nConsider now the circuit in Fig. 4.30a with r inputs X1(s), X2(s), . . . , Xr(s). As usual, the input\nvoltage Vx(s) ≃0 because the op-amp gain →∞. Moreover, the current going into the op amp is\nvery small (≃0) because the input impedance →∞. Therefore, the total current in the feedback\nresistor Rf is I1(s) + I2(s) + · · · + Ir(s). Moreover, because Vx(s) = 0,", - "type": "text" - }, - { - "block_id": "p421-b3", - "global_id": 11777, - "bbox": [ - 259.43, - 346.26, - 309.38, - 364.36 - ], - "text": "Ij(s) = Xj(s)", - "type": "text" - }, - { - "block_id": "p421-b4", - "global_id": 11778, - "bbox": [ - 295.2, - 360.67, - 303.22, - 371.43 - ], - "text": "Rj", - "type": "text" - }, - { - "block_id": "p421-b5", - "global_id": 11779, - "bbox": [ - 330.5, - 353.28, - 384.07, - 363.65 - ], - "text": "j = 1,2,. . .,r", - "type": "text" - }, - { - "block_id": "p421-b6", - "global_id": 11780, - "bbox": [ - 127.59, - 377.44, - 148.9, - 387.41 - ], - "text": "Also,", - "type": "text" - }, - { - "block_id": "p421-b7", - "global_id": 11781, - "bbox": [ - 224.78, - 394.96, - 379.1, - 406.41 - ], - "text": "Y(s) = −Rf [I1(s) + I2(s) + · · · + Ir(s)]", - "type": "text" - }, - { - "block_id": "p421-b8", - "global_id": 11782, - "bbox": [ - 244.43, - 416.98, - 262.02, - 426.94 - ], - "text": "= −", - "type": "text" - }, - { - "block_id": "p421-b9", - "global_id": 11783, - "bbox": [ - 262.03, - 402.99, - 276.93, - 421.04 - ], - "text": "Rf", - "type": "text" - }, - { - "block_id": "p421-b10", - "global_id": 11784, - "bbox": [ - 268.65, - 424.37, - 278.22, - 435.2 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p421-b11", - "global_id": 11785, - "bbox": [ - 279.92, - 410.27, - 321.65, - 428.13 - ], - "text": "X1(s) + Rf", - "type": "text" - }, - { - "block_id": "p421-b12", - "global_id": 11786, - "bbox": [ - 313.36, - 424.37, - 322.93, - 435.2 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p421-b13", - "global_id": 11787, - "bbox": [ - 324.63, - 410.27, - 390.04, - 428.13 - ], - "text": "X2(s) + · · · + Rf", - "type": "text" - }, - { - "block_id": "p421-b14", - "global_id": 11788, - "bbox": [ - 382.06, - 424.37, - 390.86, - 435.13 - ], - "text": "Rr", - "type": "text" - }, - { - "block_id": "p421-b15", - "global_id": 11789, - "bbox": [ - 392.75, - 416.98, - 413.52, - 428.06 - ], - "text": "Xr(s)", - "type": "text" - }, - { - "block_id": "p421-b16", - "global_id": 11790, - "bbox": [ - 413.52, - 402.99, - 418.95, - 412.95 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p421-b17", - "global_id": 11791, - "bbox": [ - 244.43, - 438.8, - 388.07, - 449.94 - ], - "text": "= k1X1(s) + k2X2(s) + · · · + krXr(s)", - "type": "text" - }, - { - "block_id": "p421-b18", - "global_id": 11792, - "bbox": [ - 127.59, - 457.14, - 151.93, - 467.1 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p421-b19", - "global_id": 11793, - "bbox": [ - 302.64, - 464.46, - 338.36, - 482.93 - ], - "text": "ki = −Rf", - "type": "text" - }, - { - "block_id": "p421-b20", - "global_id": 11794, - "bbox": [ - 127.59, - 478.86, - 516.11, - 528.51 - ], - "text": "Ri\nClearly, the circuit in Fig. 4.30 serves an adder and an amplifier with any desired gain for each of\nthe input signals. Figure 4.30b shows the compact symbol used in circuit diagrams for an adder\nwith r inputs.", - "type": "text" - }, - { - "block_id": "p421-b21", - "global_id": 11795, - "bbox": [ - 102.51, - 546.87, - 320.55, - 558.83 - ], - "text": "EXAMPLE 4.25\nOp-Amp Realization", - "type": "text" - }, - { - "block_id": "p421-b22", - "global_id": 11796, - "bbox": [ - 128.9, - 574.51, - 437.67, - 584.47 - ], - "text": "Use op-amp circuits to realize the canonic direct form of the transfer function", - "type": "text" - }, - { - "block_id": "p421-b23", - "global_id": 11797, - "bbox": [ - 275.02, - 592.02, - 355.43, - 616.46 - ], - "text": "H(s) =\n2s + 5\ns2 + 4s + 10", - "type": "text" - } - ] - }, - { - "page_num": 422, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p422-b0", - "global_id": 11798, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "402\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p422-b1", - "global_id": 11799, - "bbox": [ - 156.9, - 230.02, - 165.78, - 238.02 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p422-b4", - "global_id": 11800, - "bbox": [ - 88.81, - 103.71, - 168.06, - 113.18 - ], - "text": "X(s)\ns2W(s)", - "type": "text" - }, - { - "block_id": "p422-b5", - "global_id": 11801, - "bbox": [ - 156.01, - 131.13, - 160.01, - 145.67 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p422-b6", - "global_id": 11802, - "bbox": [ - 156.0, - 182.88, - 160.0, - 197.42 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p422-b7", - "global_id": 11803, - "bbox": [ - 132.45, - 170.99, - 143.12, - 179.29 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p422-b8", - "global_id": 11804, - "bbox": [ - 128.68, - 156.67, - 229.89, - 164.75 - ], - "text": "sW(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p422-b9", - "global_id": 11805, - "bbox": [ - 461.81, - 139.5, - 474.7, - 147.58 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p422-b10", - "global_id": 11806, - "bbox": [ - 174.54, - 156.75, - 178.54, - 164.75 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p422-b11", - "global_id": 11807, - "bbox": [ - 126.17, - 214.8, - 181.67, - 223.1 - ], - "text": "W(s)\n5\n10", - "type": "text" - }, - { - "block_id": "p422-b12", - "global_id": 11808, - "bbox": [ - 360.54, - 230.02, - 370.19, - 238.02 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p422-b13", - "global_id": 11809, - "bbox": [ - 338.88, - 194.51, - 353.54, - 202.8 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p422-b14", - "global_id": 11810, - "bbox": [ - 322.17, - 173.03, - 332.84, - 181.32 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p422-b15", - "global_id": 11811, - "bbox": [ - 428.7, - 153.92, - 432.7, - 161.92 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p422-b16", - "global_id": 11812, - "bbox": [ - 409.98, - 118.37, - 413.98, - 126.37 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p422-b17", - "global_id": 11813, - "bbox": [ - 258.24, - 138.2, - 453.05, - 159.05 - ], - "text": "s2W(s)\nsW(s)\nW(s)\nX(s)\n1\ns", - "type": "text" - }, - { - "block_id": "p422-b18", - "global_id": 11814, - "bbox": [ - 394.21, - 144.51, - 398.21, - 159.05 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p422-b19", - "global_id": 11815, - "bbox": [ - 275.56, - 401.59, - 284.44, - 409.59 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p422-b20", - "global_id": 11816, - "bbox": [ - 392.07, - 308.83, - 402.74, - 317.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p422-b21", - "global_id": 11817, - "bbox": [ - 151.6, - 314.36, - 402.74, - 333.56 - ], - "text": "5\n1", - "type": "text" - }, - { - "block_id": "p422-b22", - "global_id": 11818, - "bbox": [ - 151.27, - 327.72, - 165.62, - 336.02 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p422-b23", - "global_id": 11819, - "bbox": [ - 151.6, - 341.18, - 161.95, - 349.48 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p422-b24", - "global_id": 11820, - "bbox": [ - 119.48, - 307.01, - 132.8, - 315.09 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p422-b25", - "global_id": 11821, - "bbox": [ - 193.52, - 318.48, - 375.06, - 327.95 - ], - "text": "s2W(s)\nsW(s)\nW(s)", - "type": "text" - }, - { - "block_id": "p422-b26", - "global_id": 11822, - "bbox": [ - 426.02, - 310.51, - 438.91, - 318.59 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p422-b27", - "global_id": 11823, - "bbox": [ - 246.61, - 327.45, - 338.73, - 335.75 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p422-b28", - "global_id": 11824, - "bbox": [ - 341.09, - 281.62, - 351.75, - 289.92 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p422-b29", - "global_id": 11825, - "bbox": [ - 196.12, - 361.94, - 206.79, - 370.24 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p422-b36", - "global_id": 11826, - "bbox": [ - 275.13, - 611.11, - 284.46, - 619.11 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p422-b37", - "global_id": 11827, - "bbox": [ - 91.38, - 516.03, - 152.2, - 525.35 - ], - "text": "X(s)\n100 k", - "type": "text" - }, - { - "block_id": "p422-b38", - "global_id": 11828, - "bbox": [ - 133.02, - 532.11, - 208.3, - 554.98 - ], - "text": "10 k\t\n10 k\t\n25 k", - "type": "text" - }, - { - "block_id": "p422-b39", - "global_id": 11829, - "bbox": [ - 169.36, - 500.85, - 193.59, - 509.15 - ], - "text": "100 k", - "type": "text" - }, - { - "block_id": "p422-b40", - "global_id": 11830, - "bbox": [ - 202.02, - 520.12, - 226.24, - 528.41 - ], - "text": "100 k", - "type": "text" - }, - { - "block_id": "p422-b41", - "global_id": 11831, - "bbox": [ - 237.87, - 562.88, - 258.1, - 571.18 - ], - "text": "10 k", - "type": "text" - }, - { - "block_id": "p422-b42", - "global_id": 11832, - "bbox": [ - 237.12, - 498.5, - 324.3, - 506.86 - ], - "text": "10 F\n10 F", - "type": "text" - }, - { - "block_id": "p422-b43", - "global_id": 11833, - "bbox": [ - 285.48, - 451.57, - 309.7, - 459.86 - ], - "text": "100 k", - "type": "text" - }, - { - "block_id": "p422-b44", - "global_id": 11834, - "bbox": [ - 323.36, - 438.56, - 347.59, - 446.86 - ], - "text": "100 k", - "type": "text" - }, - { - "block_id": "p422-b45", - "global_id": 11835, - "bbox": [ - 356.63, - 504.76, - 376.85, - 513.05 - ], - "text": "50 k", - "type": "text" - }, - { - "block_id": "p422-b46", - "global_id": 11836, - "bbox": [ - 411.66, - 492.1, - 435.89, - 500.4 - ], - "text": "100 k", - "type": "text" - }, - { - "block_id": "p422-b47", - "global_id": 11837, - "bbox": [ - 271.99, - 531.72, - 469.76, - 545.2 - ], - "text": "20 k\t\n100 k\t\nY(s)", - "type": "text" - }, - { - "block_id": "p422-b50", - "global_id": 11838, - "bbox": [ - 85.48, - 623.57, - 419.79, - 636.43 - ], - "text": "Figure 4.31 Op-amp realization of a second-order transfer function (2s + 5)/(s2 + 4s + 10).", - "type": "text" - } - ] - }, - { - "page_num": 423, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p423-b0", - "global_id": 11839, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "4.6\nSystem Realization\n403", - "type": "text" - }, - { - "block_id": "p423-b1", - "global_id": 11840, - "bbox": [ - 128.9, - 86.24, - 502.79, - 215.75 - ], - "text": "The basic canonic realization is shown in Fig. 4.31a. The same realization with horizontal\nreorientation is shown in Fig. 4.31b. Signals at various points are also indicated in the\nrealization. For convenience, we denote the output of the last integrator by W(s). Consequently,\nthe signals at the inputs of the two integrators are sW(s) and s2W(s), as shown in Figs. 4.31a\nand 4.31b. Op-amp elements (multipliers, integrators, and adders) change the polarity of the\noutput signals. To incorporate this fact, we modify the canonic realization in Fig. 4.31b to that\ndepicted in Fig. 4.31c. In Fig. 4.31b, the successive outputs of the adder and the integrators\nare s2W(s),sW(s), and W(s), respectively. Because of polarity reversals in op-amp circuits,\nthese outputs are −s2W(s),sW(s), and −W(s), respectively, in Fig. 4.31c. This polarity\nreversal requires corresponding modifications in the signs of feedback and feedforward gains.\nAccording to Fig. 4.31b,", - "type": "text" - }, - { - "block_id": "p423-b2", - "global_id": 11841, - "bbox": [ - 246.28, - 225.86, - 385.39, - 237.67 - ], - "text": "s2W(s) = X(s) −4sW(s) −10W(s)", - "type": "text" - }, - { - "block_id": "p423-b3", - "global_id": 11842, - "bbox": [ - 128.91, - 249.63, - 170.65, - 259.59 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p423-b4", - "global_id": 11843, - "bbox": [ - 238.51, - 259.73, - 393.16, - 271.54 - ], - "text": "−s2W(s) = −X(s) + 4sW(s) + 10W(s)", - "type": "text" - }, - { - "block_id": "p423-b5", - "global_id": 11844, - "bbox": [ - 128.91, - 280.51, - 502.74, - 302.43 - ], - "text": "Because the adder gains are always negative (see Fig. 4.30b), we rewrite the foregoing equation\nas", - "type": "text" - }, - { - "block_id": "p423-b6", - "global_id": 11845, - "bbox": [ - 218.29, - 302.57, - 413.38, - 314.38 - ], - "text": "−s2W(s) = −1[X(s)] −4[−sW(s)] −10[−W(s)]", - "type": "text" - }, - { - "block_id": "p423-b7", - "global_id": 11846, - "bbox": [ - 128.91, - 323.35, - 502.76, - 405.04 - ], - "text": "Figure 4.31c shows the implementation of this equation. The hardware realization appears in\nFig. 4.31d. Both integrators have a unity gain, which requires RC = 1. We have used R = 100\nk and C = 10 µF. The gain of 10 in the outer feedback path is obtained in the adder by\nchoosing the feedback resistor of the adder to be 100 k and an input resistor of 10 k.\nSimilarly, the gain of 4 in the inner feedback path is obtained by using the corresponding input\nresistor of 25 k. The gains of 2 and 5, required in the feedforward connections, are obtained\nby using a feedback resistor of 100 k and input resistors of 50 and 20 k, respectively.†", - "type": "text" - }, - { - "block_id": "p423-b8", - "global_id": 11847, - "bbox": [ - 128.9, - 407.04, - 502.77, - 452.87 - ], - "text": "The op-amp realization in Fig. 4.31 is not necessarily the one that uses the fewest op\namps. This example is given just to illustrate a systematic procedure for designing an op-amp\ncircuit of an arbitrary transfer function. There are more efficient circuits (such as Sallen–Key\nor biquad) that use fewer op amps to realize a second-order transfer function.", - "type": "text" - }, - { - "block_id": "p423-b9", - "global_id": 11848, - "bbox": [ - 133.57, - 496.23, - 428.36, - 508.18 - ], - "text": "DRILL 4.14\nTransfer Functions of Op-Amp Circuits", - "type": "text" - }, - { - "block_id": "p423-b10", - "global_id": 11849, - "bbox": [ - 133.57, - 516.89, - 510.14, - 539.99 - ], - "text": "Show that the transfer functions of the op-amp circuits in Figs. 4.32a and 4.32b are H1(s) and\nH2(s), respectively, where", - "type": "text" - }, - { - "block_id": "p423-b11", - "global_id": 11850, - "bbox": [ - 226.29, - 549.39, - 277.63, - 567.56 - ], - "text": "H1(s) = −Rf", - "type": "text" - }, - { - "block_id": "p423-b12", - "global_id": 11851, - "bbox": [ - 267.45, - 563.8, - 273.54, - 573.76 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p423-b13", - "global_id": 11852, - "bbox": [ - 280.34, - 542.41, - 307.99, - 573.76 - ], - "text": "a\ns + a", - "type": "text" - }, - { - "block_id": "p423-b15", - "global_id": 11853, - "bbox": [ - 335.84, - 549.84, - 372.0, - 575.25 - ], - "text": "a =\n1\nRf Cf", - "type": "text" - }, - { - "block_id": "p423-b16", - "global_id": 11854, - "bbox": [ - 226.29, - 578.16, - 277.87, - 595.98 - ], - "text": "H2(s) = −C", - "type": "text" - }, - { - "block_id": "p423-b17", - "global_id": 11855, - "bbox": [ - 269.61, - 592.22, - 278.19, - 602.98 - ], - "text": "Cf", - "type": "text" - }, - { - "block_id": "p423-b18", - "global_id": 11856, - "bbox": [ - 280.9, - 570.84, - 308.54, - 588.12 - ], - "text": "s + b", - "type": "text" - }, - { - "block_id": "p423-b19", - "global_id": 11857, - "bbox": [ - 288.82, - 591.9, - 308.54, - 602.18 - ], - "text": "s + a", - "type": "text" - }, - { - "block_id": "p423-b21", - "global_id": 11858, - "bbox": [ - 336.39, - 578.26, - 372.56, - 603.67 - ], - "text": "a =\n1\nRf Cf", - "type": "text" - }, - { - "block_id": "p423-b22", - "global_id": 11859, - "bbox": [ - 385.24, - 578.26, - 412.25, - 595.21 - ], - "text": "b = 1", - "type": "text" - }, - { - "block_id": "p423-b23", - "global_id": 11860, - "bbox": [ - 403.28, - 592.22, - 416.01, - 602.18 - ], - "text": "RC", - "type": "text" - }, - { - "block_id": "p423-b24", - "global_id": 11861, - "bbox": [ - 127.59, - 630.16, - 516.13, - 653.34 - ], - "text": "† It is possible to avoid the two inverting op amps (with gain −1) in Fig. 4.31d by adding signal sW(s) to the\ninput and output adders directly, using the noninverting amplifier configuration in Fig. 4.16d.", - "type": "text" - } - ] - }, - { - "page_num": 424, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p424-b0", - "global_id": 11862, - "bbox": [ - 60.0, - 62.89, - 332.51, - 71.98 - ], - "text": "404\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p424-b1", - "global_id": 11863, - "bbox": [ - 205.95, - 233.48, - 384.63, - 241.48 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p424-b2", - "global_id": 11864, - "bbox": [ - 241.01, - 120.3, - 248.01, - 129.85 - ], - "text": "Cf", - "type": "text" - }, - { - "block_id": "p424-b3", - "global_id": 11865, - "bbox": [ - 172.15, - 146.41, - 247.56, - 155.96 - ], - "text": "Rf\nR", - "type": "text" - }, - { - "block_id": "p424-b4", - "global_id": 11866, - "bbox": [ - 274.55, - 172.64, - 280.19, - 182.64 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p424-b5", - "global_id": 11867, - "bbox": [ - 274.87, - 201.03, - 279.87, - 211.03 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p424-b6", - "global_id": 11868, - "bbox": [ - 156.75, - 184.46, - 284.06, - 196.53 - ], - "text": "X(s)\nY(s)", - "type": "text" - }, - { - "block_id": "p424-b7", - "global_id": 11869, - "bbox": [ - 340.35, - 180.9, - 345.69, - 188.9 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p424-b8", - "global_id": 11870, - "bbox": [ - 341.97, - 134.82, - 346.85, - 142.82 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p424-b9", - "global_id": 11871, - "bbox": [ - 412.6, - 146.4, - 419.6, - 155.95 - ], - "text": "Cf", - "type": "text" - }, - { - "block_id": "p424-b10", - "global_id": 11872, - "bbox": [ - 418.72, - 121.92, - 425.27, - 131.46 - ], - "text": "Rf", - "type": "text" - }, - { - "block_id": "p424-b11", - "global_id": 11873, - "bbox": [ - 321.34, - 184.46, - 335.14, - 192.54 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p424-b12", - "global_id": 11874, - "bbox": [ - 448.39, - 172.64, - 454.03, - 182.64 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p424-b13", - "global_id": 11875, - "bbox": [ - 448.71, - 201.03, - 453.71, - 211.03 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p424-b14", - "global_id": 11876, - "bbox": [ - 444.53, - 188.45, - 457.9, - 196.53 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p424-b15", - "global_id": 11877, - "bbox": [ - 231.58, - 158.72, - 237.22, - 183.26 - ], - "text": "–\n+", - "type": "text" - }, - { - "block_id": "p424-b16", - "global_id": 11878, - "bbox": [ - 403.59, - 158.72, - 409.23, - 183.26 - ], - "text": "–\n+", - "type": "text" - }, - { - "block_id": "p424-b17", - "global_id": 11879, - "bbox": [ - 131.73, - 253.38, - 291.85, - 262.62 - ], - "text": "Figure 4.32 Op-amp circuits for Drill 4.14.", - "type": "text" - }, - { - "block_id": "p424-b18", - "global_id": 11880, - "bbox": [ - 102.2, - 308.8, - 418.98, - 322.75 - ], - "text": "4.7 APPLICATION TO FEEDBACK AND CONTROLS", - "type": "text" - }, - { - "block_id": "p424-b19", - "global_id": 11881, - "bbox": [ - 101.82, - 328.32, - 490.41, - 398.49 - ], - "text": "Generally, systems are designed to produce a desired output y(t) for a given input x(t). Using\nthe given performance criteria, we can design a system, as shown in Fig. 4.33a. Ideally, such an\nopen-loop system should yield the desired output. In practice, however, the system characteristics\nchange with time, as a result of aging or replacement of some components, or because of changes\nin the operating environment. Such variations cause changes in the output for the same input.\nClearly, this is undesirable in precision systems.", - "type": "text" - }, - { - "block_id": "p424-b20", - "global_id": 11882, - "bbox": [ - 181.65, - 450.55, - 190.53, - 458.55 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p424-b21", - "global_id": 11883, - "bbox": [ - 121.81, - 418.55, - 236.4, - 434.84 - ], - "text": "x(t)\ny(t)\nG(s)", - "type": "text" - }, - { - "block_id": "p424-b23", - "global_id": 11884, - "bbox": [ - 181.47, - 529.49, - 191.12, - 537.49 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p424-b24", - "global_id": 11885, - "bbox": [ - 112.91, - 472.17, - 260.2, - 480.48 - ], - "text": "y(t)\ne(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p424-b26", - "global_id": 11886, - "bbox": [ - 183.99, - 480.72, - 198.68, - 488.8 - ], - "text": "G(s)", - "type": "text" - }, - { - "block_id": "p424-b27", - "global_id": 11887, - "bbox": [ - 278.8, - 516.75, - 486.4, - 538.76 - ], - "text": "Figure 4.33 (a) Open-loop and (b) closed-loop\n(feedback) systems.", - "type": "text" - }, - { - "block_id": "p424-b28", - "global_id": 11888, - "bbox": [ - 101.85, - 565.05, - 490.42, - 634.81 - ], - "text": "A possible solution to this problem is to add a signal component to the input that is not\na predetermined function of time but will change to counteract the effects of changing system\ncharacteristics and the environment. In short, we must provide a correction at the system input\nto account for the undesired changes just mentioned. Yet since these changes are generally\nunpredictable, it is not clear how to preprogram appropriate corrections to the input. However,\nthe difference between the actual output and the desired output gives an indication of the suitable", - "type": "text" - } - ] - }, - { - "page_num": 425, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p425-b0", - "global_id": 11889, - "bbox": [ - 302.53, - 62.89, - 516.13, - 71.98 - ], - "text": "4.7\nApplication to Feedback and Controls\n405", - "type": "text" - }, - { - "block_id": "p425-b1", - "global_id": 11890, - "bbox": [ - 127.59, - 85.82, - 516.1, - 107.74 - ], - "text": "correction to be applied to the system input. It may be possible to counteract the variations by\nfeeding the output (or some function of output) back to the input.", - "type": "text" - }, - { - "block_id": "p425-b2", - "global_id": 11891, - "bbox": [ - 127.59, - 109.73, - 516.15, - 287.07 - ], - "text": "We unconsciously apply this principle in daily life. Consider an example of marketing a\ncertain product. The optimum price of the product is the value that maximizes the profit of a\nmerchant. The output in this case is the profit, and the input is the price of the item. The output\n(profit) can be controlled (within limits) by varying the input (price). The merchant may price\nthe product too high initially, in which case, he will sell too few items, reducing the profit. Using\nfeedback of the profit (output), he adjusts the price (input), to maximize his profit. If there is a\nsudden or unexpected change in the business environment, such as a strike-imposed shutdown of\na large factory in town, the demand for the item goes down, thus reducing his output (profit). He\nadjusts his input (reduces price) using the feedback of the output (profit) in a way that will optimize\nhis profit in the changed circumstances. If the town suddenly becomes more prosperous because a\nnew factory opens, he will increase the price to maximize the profit. Thus, by continuous feedback\nof the output to the input, he realizes his goal of maximum profit (optimum output) in any given\ncircumstances. We observe thousands of examples of feedback systems around us in everyday life.\nMost social, economical, educational, and political processes are, in fact, feedback processes. A\nblock diagram of such a system, called the feedback or closed-loop system, is shown in Fig. 4.33b.", - "type": "text" - }, - { - "block_id": "p425-b3", - "global_id": 11892, - "bbox": [ - 127.59, - 289.06, - 516.15, - 382.71 - ], - "text": "A feedback system can address the problems arising because of unwanted disturbances such as\nrandom-noise signals in electronic systems, a gust of wind affecting a tracking antenna, a meteorite\nhitting a spacecraft, and the rolling motion of antiaircraft gun platforms mounted on ships or\nmoving tanks. Feedback may also be used to reduce nonlinearities in a system or to control its\nrise time (or bandwidth). Feedback is used to achieve, with a given system, the desired objective\nwithin a given tolerance, despite partial ignorance of the system and the environment. A feedback\nsystem, thus, has an ability for supervision and self-correction in the face of changes in the system\nparameters and external disturbances (change in the environment).", - "type": "text" - }, - { - "block_id": "p425-b4", - "global_id": 11893, - "bbox": [ - 127.59, - 384.29, - 516.13, - 418.58 - ], - "text": "Consider the feedback amplifier in Fig. 4.34. Let the forward amplifier gain G = 10,000.\nOne-hundredth of the output is fed back to the input (H = 0.01). The gain T of the feedback\namplifier is obtained by [see Eq. (4.35)]", - "type": "text" - }, - { - "block_id": "p425-b5", - "global_id": 11894, - "bbox": [ - 256.57, - 428.17, - 387.14, - 452.29 - ], - "text": "T =\nG\n1 + GH = 10,000\n1 + 100 = 99.01", - "type": "text" - }, - { - "block_id": "p425-b6", - "global_id": 11895, - "bbox": [ - 127.59, - 460.55, - 516.14, - 482.56 - ], - "text": "Suppose that because of aging or replacement of some transistors, the gain G of the forward\namplifier changes from 10,000 to 20,000. The new gain of the feedback amplifier is given by", - "type": "text" - }, - { - "block_id": "p425-b7", - "global_id": 11896, - "bbox": [ - 259.06, - 492.15, - 384.66, - 516.28 - ], - "text": "T =\nG\n1 + GH = 20,000\n1 + 200 = 99.5", - "type": "text" - }, - { - "block_id": "p425-b8", - "global_id": 11897, - "bbox": [ - 127.59, - 524.54, - 516.15, - 558.5 - ], - "text": "Surprisingly, 100% variation in the forward gain G causes only 0.5% variation in the feedback\namplifier gain T. Such reduced sensitivity to parameter variations is a must in precision amplifiers.\nIn this example, we reduced the sensitivity of gain to parameter variations at the cost of forward", - "type": "text" - }, - { - "block_id": "p425-b10", - "global_id": 11898, - "bbox": [ - 138.65, - 577.4, - 285.7, - 585.48 - ], - "text": "y(t)\ne(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p425-b12", - "global_id": 11899, - "bbox": [ - 214.19, - 585.89, - 219.97, - 593.89 - ], - "text": "G", - "type": "text" - }, - { - "block_id": "p425-b13", - "global_id": 11900, - "bbox": [ - 211.88, - 612.55, - 503.86, - 626.5 - ], - "text": "H\nFigure 4.34 Effects of negative and positive feedback.", - "type": "text" - } - ] - }, - { - "page_num": 426, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p426-b0", - "global_id": 11901, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "406\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p426-b1", - "global_id": 11902, - "bbox": [ - 101.84, - 85.82, - 490.38, - 107.74 - ], - "text": "gain, which is reduced from 10,000 to 99. There is no dearth of forward gain (obtained by\ncascading stages). But low sensitivity is extremely precious in precision systems.", - "type": "text" - }, - { - "block_id": "p426-b2", - "global_id": 11903, - "bbox": [ - 101.84, - 109.73, - 490.39, - 143.6 - ], - "text": "Now, consider what happens when we add (instead of subtract) the signal fed back to the\ninput. Such addition means the sign on the feedback connection is + instead of −(which is same\nas changing the sign of H in Fig. 4.34). Consequently,", - "type": "text" - }, - { - "block_id": "p426-b3", - "global_id": 11904, - "bbox": [ - 270.49, - 153.76, - 320.09, - 177.89 - ], - "text": "T =\nG\n1 −GH", - "type": "text" - }, - { - "block_id": "p426-b4", - "global_id": 11905, - "bbox": [ - 101.84, - 186.91, - 326.47, - 198.5 - ], - "text": "If we let G = 10,000 as before and H = 0.9 × 10−4, then", - "type": "text" - }, - { - "block_id": "p426-b5", - "global_id": 11906, - "bbox": [ - 225.46, - 208.2, - 366.76, - 232.22 - ], - "text": "T =\n10,000\n1 −0.9(104)(10−4) = 100,000", - "type": "text" - }, - { - "block_id": "p426-b6", - "global_id": 11907, - "bbox": [ - 101.85, - 242.21, - 490.36, - 264.12 - ], - "text": "Suppose that because of aging or replacement of some transistors, the gain of the forward amplifier\nchanges to 11,000. The new gain of the feedback amplifier is", - "type": "text" - }, - { - "block_id": "p426-b7", - "global_id": 11908, - "bbox": [ - 215.0, - 274.53, - 377.22, - 298.55 - ], - "text": "T =\n11,000\n1 −0.9(11,000)(10−4) = 1,100,000", - "type": "text" - }, - { - "block_id": "p426-b8", - "global_id": 11909, - "bbox": [ - 101.84, - 308.44, - 490.39, - 354.37 - ], - "text": "Observe that in this case, a mere 10% increase in the forward gain G caused 1000% increase\nin the gain T (from 100,000 to 1,100,000). Clearly, the amplifier is very sensitive to parameter\nvariations. This behavior is exactly opposite of what was observed earlier, when the signal fed\nback was subtracted from the input.", - "type": "text" - }, - { - "block_id": "p426-b9", - "global_id": 11910, - "bbox": [ - 101.84, - 356.36, - 490.42, - 461.96 - ], - "text": "What is the difference between the two situations? Crudely speaking, the former case is called\nthe negative feedback and the latter is the positive feedback. The positive feedback increases\nsystem gain but tends to make the system more sensitive to parameter variations. It can also lead to\ninstability. In our example, if G were to be 111,111, then GH = 1, T = ∞, and the system would\nbecome unstable because the signal fed back was exactly equal to the input signal itself, since\nGH = 1. Hence, once a signal has been applied, no matter how small and how short in duration,\nit comes back to reinforce the input undiminished, which further passes to the output, and is fed\nback again and again and again. In essence, the signal perpetuates itself forever. This perpetuation,\neven when the input ceases to exist, is precisely the symptom of instability.", - "type": "text" - }, - { - "block_id": "p426-b10", - "global_id": 11911, - "bbox": [ - 101.84, - 463.95, - 490.42, - 533.69 - ], - "text": "Generally speaking, a feedback system cannot be described in black and white terms, such as\npositive or negative. Usually H is a frequency-dependent component, more accurately represented\nby H(s); hence it varies with frequency. Consequently, what was negative feedback at lower\nfrequencies can turn into positive feedback at higher frequencies and may give rise to instability.\nThis is one of the serious aspects of feedback systems, which warrants a designer’s careful\nattention.", - "type": "text" - }, - { - "block_id": "p426-b11", - "global_id": 11912, - "bbox": [ - 101.84, - 558.91, - 333.11, - 570.87 - ], - "text": "4.7-1 Analysis of a Simple Control System", - "type": "text" - }, - { - "block_id": "p426-b12", - "global_id": 11913, - "bbox": [ - 101.84, - 577.0, - 490.42, - 635.49 - ], - "text": "Figure 4.35a represents an automatic position control system, which can be used to control the\nangular position of a heavy object (e.g., a tracking antenna, an anti-aircraft gun mount, or the\nposition of a ship). The input θi is the desired angular position of the object, which can be set\nat any given value. The actual angular position θo of the object (the output) is measured by a\npotentiometer whose wiper is mounted on the output shaft. The difference between the input θi", - "type": "text" - } - ] - }, - { - "page_num": 427, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p427-b0", - "global_id": 11914, - "bbox": [ - 302.53, - 62.89, - 516.13, - 71.98 - ], - "text": "4.7\nApplication to Feedback and Controls\n407", - "type": "text" - }, - { - "block_id": "p427-b1", - "global_id": 11915, - "bbox": [ - 127.59, - 85.4, - 516.13, - 155.56 - ], - "text": "(set at the desired output position) and the output θo (actual position) is amplified; the amplified\noutput, which is proportional to θi −θo, is applied to the motor input. If θi −θo = 0 (the output\nbeing equal to the desired angle), there is no input to the motor, and the motor stops. But if θo̸ = θi,\nthere will be a nonzero input to the motor, which will turn the shaft until θo = θi. It is evident that\nby setting the input potentiometer at a desired position in this system, we can control the angular\nposition of a heavy remote object.", - "type": "text" - }, - { - "block_id": "p427-b2", - "global_id": 11916, - "bbox": [ - 127.59, - 157.45, - 516.15, - 216.83 - ], - "text": "The block diagram of this system is shown in Fig. 4.35b. The amplifier gain is K, where K is\nadjustable. Let the motor (with load) transfer function that relates the output angle θo to the motor\ninput voltage be G(s) [for a starting point, see Eq. (1.32)]. This feedback arrangement is identical\nto that in Fig. 4.18d with H(s) = 1. Hence, T(s), the (closed-loop) system transfer function relating\nthe output θo to the input θi, is", - "type": "text" - }, - { - "block_id": "p427-b3", - "global_id": 11917, - "bbox": [ - 267.85, - 226.54, - 291.14, - 237.62 - ], - "text": "o(s)", - "type": "text" - }, - { - "block_id": "p427-b4", - "global_id": 11918, - "bbox": [ - 268.62, - 226.54, - 375.86, - 251.68 - ], - "text": "i(s) = T(s) =\nKG(s)\n1 + KG(s)", - "type": "text" - }, - { - "block_id": "p427-b5", - "global_id": 11919, - "bbox": [ - 127.59, - 262.11, - 516.16, - 284.02 - ], - "text": "From this equation, we shall investigate the behavior of the automatic position control system in\nFig. 4.35a for a step and a ramp input.", - "type": "text" - }, - { - "block_id": "p427-b6", - "global_id": 11920, - "bbox": [ - 127.59, - 299.57, - 516.15, - 421.33 - ], - "text": "STEP INPUT\nIf we desire to change the angular position of the object instantaneously, we need to apply a step\ninput. We may then want to know how long the system takes to position itself at the desired\nangle, whether it reaches the desired angle, and whether it reaches the desired position smoothly\n(monotonically) or oscillates about the final position. If the system oscillates, we may want to\nknow how long it takes for the oscillations to settle down. All these questions can be readily\nanswered by finding the output θo(t) when the input θi(t) = u(t). A step input implies instantaneous\nchange in the angle. This input would be one of the most difficult to follow; if the system can\nperform well for this input, it is likely to give a good account of itself under most other expected\nsituations. This is why we test control systems for a step input.", - "type": "text" - }, - { - "block_id": "p427-b7", - "global_id": 11921, - "bbox": [ - 145.52, - 422.9, - 331.31, - 433.98 - ], - "text": "For the step input θi(t) = u(t), i(s) = 1/s and", - "type": "text" - }, - { - "block_id": "p427-b8", - "global_id": 11922, - "bbox": [ - 258.91, - 444.9, - 300.24, - 462.55 - ], - "text": "o(s) = 1", - "type": "text" - }, - { - "block_id": "p427-b9", - "global_id": 11923, - "bbox": [ - 295.8, - 444.48, - 383.61, - 468.92 - ], - "text": "s T(s) =\nKG(s)\ns[1 + KG(s)]", - "type": "text" - }, - { - "block_id": "p427-b10", - "global_id": 11924, - "bbox": [ - 127.59, - 479.7, - 516.11, - 502.03 - ], - "text": "Let the motor (with load) transfer function relating the load angle θo(t) to the motor input voltage\nbe G(s) = 1/(s(s + 8)). This yields", - "type": "text" - }, - { - "block_id": "p427-b11", - "global_id": 11925, - "bbox": [ - 235.48, - 532.92, - 268.58, - 544.0 - ], - "text": "o(s) =", - "type": "text" - }, - { - "block_id": "p427-b12", - "global_id": 11926, - "bbox": [ - 288.31, - 513.35, - 319.34, - 537.47 - ], - "text": "K\ns(s + 8)", - "type": "text" - }, - { - "block_id": "p427-b13", - "global_id": 11927, - "bbox": [ - 271.82, - 546.79, - 275.69, - 556.76 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p427-b15", - "global_id": 11928, - "bbox": [ - 281.12, - 539.81, - 329.2, - 563.93 - ], - "text": "1 +\nK\ns(s + 8)", - "type": "text" - }, - { - "block_id": "p427-b16", - "global_id": 11929, - "bbox": [ - 330.4, - 526.25, - 407.03, - 550.37 - ], - "text": "! =\nK\ns(s2 + 8s + K)", - "type": "text" - }, - { - "block_id": "p427-b17", - "global_id": 11930, - "bbox": [ - 127.59, - 575.62, - 422.58, - 585.68 - ], - "text": "Let us investigate the system behavior for three different values of gain K.", - "type": "text" - }, - { - "block_id": "p427-b18", - "global_id": 11931, - "bbox": [ - 145.52, - 587.26, - 188.23, - 597.64 - ], - "text": "For K = 7,", - "type": "text" - }, - { - "block_id": "p427-b19", - "global_id": 11932, - "bbox": [ - 196.88, - 611.19, - 379.67, - 635.21 - ], - "text": "o(s) =\n7\ns(s2 + 8s + 7) =\n7\ns(s + 1)(s + 7) = 1", - "type": "text" - }, - { - "block_id": "p427-b20", - "global_id": 11933, - "bbox": [ - 375.24, - 617.76, - 390.18, - 635.11 - ], - "text": "s −", - "type": "text" - }, - { - "block_id": "p427-b21", - "global_id": 11934, - "bbox": [ - 392.92, - 608.71, - 423.17, - 635.21 - ], - "text": "7\n6\ns + 1 +", - "type": "text" - }, - { - "block_id": "p427-b22", - "global_id": 11935, - "bbox": [ - 425.92, - 608.71, - 445.64, - 635.21 - ], - "text": "1\n6\ns + 7", - "type": "text" - } - ] - }, - { - "page_num": 428, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p428-b0", - "global_id": 11936, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "408\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p428-b3", - "global_id": 11937, - "bbox": [ - 421.09, - 468.23, - 423.31, - 476.23 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p428-b4", - "global_id": 11938, - "bbox": [ - 384.71, - 617.99, - 387.09, - 626.27 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p428-b5", - "global_id": 11939, - "bbox": [ - 291.25, - 237.77, - 300.13, - 245.77 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p428-b6", - "global_id": 11940, - "bbox": [ - 311.02, - 123.66, - 328.68, - 133.21 - ], - "text": "Em(s)", - "type": "text" - }, - { - "block_id": "p428-b7", - "global_id": 11941, - "bbox": [ - 357.29, - 87.59, - 401.73, - 105.19 - ], - "text": "Output\npotentiometer", - "type": "text" - }, - { - "block_id": "p428-b8", - "global_id": 11942, - "bbox": [ - 168.68, - 111.36, - 174.35, - 120.93 - ], - "text": "ui", - "type": "text" - }, - { - "block_id": "p428-b9", - "global_id": 11943, - "bbox": [ - 125.76, - 87.59, - 170.2, - 105.19 - ], - "text": "Input\npotentiometer", - "type": "text" - }, - { - "block_id": "p428-b10", - "global_id": 11944, - "bbox": [ - 231.82, - 119.71, - 261.15, - 137.31 - ], - "text": "dc\namplifier", - "type": "text" - }, - { - "block_id": "p428-b11", - "global_id": 11945, - "bbox": [ - 397.44, - 103.53, - 424.7, - 116.83 - ], - "text": "uo\nuo", - "type": "text" - }, - { - "block_id": "p428-b12", - "global_id": 11946, - "bbox": [ - 291.25, - 490.07, - 300.13, - 498.07 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p428-b13", - "global_id": 11947, - "bbox": [ - 280.99, - 474.49, - 284.99, - 482.49 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p428-b14", - "global_id": 11948, - "bbox": [ - 157.0, - 372.48, - 161.0, - 380.48 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p428-b15", - "global_id": 11949, - "bbox": [ - 151.0, - 352.98, - 226.93, - 371.79 - ], - "text": "1.2\nK 80", - "type": "text" - }, - { - "block_id": "p428-b16", - "global_id": 11950, - "bbox": [ - 213.29, - 394.59, - 237.29, - 402.89 - ], - "text": "K 16", - "type": "text" - }, - { - "block_id": "p428-b17", - "global_id": 11951, - "bbox": [ - 223.32, - 420.45, - 243.32, - 428.75 - ], - "text": "K 7", - "type": "text" - }, - { - "block_id": "p428-b18", - "global_id": 11952, - "bbox": [ - 186.65, - 473.31, - 403.99, - 482.86 - ], - "text": "4\ntp", - "type": "text" - }, - { - "block_id": "p428-b19", - "global_id": 11953, - "bbox": [ - 153.53, - 416.55, - 159.57, - 426.12 - ], - "text": "uo", - "type": "text" - }, - { - "block_id": "p428-b20", - "global_id": 11954, - "bbox": [ - 301.32, - 510.97, - 311.32, - 518.97 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p428-b21", - "global_id": 11955, - "bbox": [ - 227.71, - 546.12, - 252.59, - 554.12 - ], - "text": "Desired", - "type": "text" - }, - { - "block_id": "p428-b22", - "global_id": 11956, - "bbox": [ - 269.19, - 573.57, - 290.52, - 581.57 - ], - "text": "Actual", - "type": "text" - }, - { - "block_id": "p428-b23", - "global_id": 11957, - "bbox": [ - 291.02, - 631.74, - 300.35, - 639.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p428-b24", - "global_id": 11958, - "bbox": [ - 153.77, - 528.41, - 159.81, - 537.97 - ], - "text": "uo", - "type": "text" - }, - { - "block_id": "p428-b25", - "global_id": 11959, - "bbox": [ - 290.78, - 326.27, - 300.59, - 334.27 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p428-b26", - "global_id": 11960, - "bbox": [ - 144.91, - 270.1, - 150.58, - 279.67 - ], - "text": "ui", - "type": "text" - }, - { - "block_id": "p428-b27", - "global_id": 11961, - "bbox": [ - 305.53, - 278.64, - 319.74, - 286.72 - ], - "text": "G(s)", - "type": "text" - }, - { - "block_id": "p428-b28", - "global_id": 11962, - "bbox": [ - 287.97, - 258.81, - 337.3, - 266.81 - ], - "text": "Motor and load", - "type": "text" - }, - { - "block_id": "p428-b29", - "global_id": 11963, - "bbox": [ - 380.23, - 270.1, - 386.27, - 279.67 - ], - "text": "uo", - "type": "text" - }, - { - "block_id": "p428-b30", - "global_id": 11964, - "bbox": [ - 221.86, - 258.81, - 253.41, - 266.81 - ], - "text": "Amplifier", - "type": "text" - }, - { - "block_id": "p428-b31", - "global_id": 11965, - "bbox": [ - 234.97, - 278.59, - 240.3, - 286.59 - ], - "text": "K", - "type": "text" - }, - { - "block_id": "p428-b32", - "global_id": 11966, - "bbox": [ - 125.76, - 646.24, - 490.38, - 668.24 - ], - "text": "Figure 4.35 (a) An automatic position control system. (b) Its block diagram. (c) The unit\nstep response. (d) The unit ramp response.", - "type": "text" - } - ] - }, - { - "page_num": 429, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p429-b0", - "global_id": 11967, - "bbox": [ - 302.53, - 62.89, - 516.13, - 71.98 - ], - "text": "4.7\nApplication to Feedback and Controls\n409", - "type": "text" - }, - { - "block_id": "p429-b1", - "global_id": 11968, - "bbox": [ - 127.59, - 85.82, - 141.97, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p429-b2", - "global_id": 11969, - "bbox": [ - 260.89, - 97.36, - 289.46, - 108.44 - ], - "text": "θo(t) =", - "type": "text" - }, - { - "block_id": "p429-b4", - "global_id": 11970, - "bbox": [ - 295.52, - 89.35, - 382.82, - 110.55 - ], - "text": "1 −7\n6e−t + 1\n6e−7t\nu(t)", - "type": "text" - }, - { - "block_id": "p429-b5", - "global_id": 11971, - "bbox": [ - 127.59, - 115.8, - 516.11, - 137.73 - ], - "text": "This response, illustrated in Fig. 4.35c, shows that the system reaches the desired angle, but at a\nrather leisurely pace. To speed up the response let us increase the gain to, say, 80.", - "type": "text" - }, - { - "block_id": "p429-b6", - "global_id": 11972, - "bbox": [ - 145.52, - 139.3, - 193.21, - 149.68 - ], - "text": "For K = 80,", - "type": "text" - }, - { - "block_id": "p429-b7", - "global_id": 11973, - "bbox": [ - 218.38, - 157.1, - 424.11, - 181.13 - ], - "text": "o(s) =\n80\ns(s2 + 8s + 80) =\n80\ns(s + 4 −j8)(s + 4 + j8)", - "type": "text" - }, - { - "block_id": "p429-b8", - "global_id": 11974, - "bbox": [ - 243.71, - 188.37, - 259.71, - 205.32 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p429-b9", - "global_id": 11975, - "bbox": [ - 255.28, - 194.94, - 270.22, - 212.29 - ], - "text": "s +", - "type": "text" - }, - { - "block_id": "p429-b10", - "global_id": 11976, - "bbox": [ - 277.05, - 179.82, - 282.95, - 186.79 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p429-b11", - "global_id": 11977, - "bbox": [ - 279.99, - 184.35, - 307.41, - 201.12 - ], - "text": "5\n4 ej153◦", - "type": "text" - }, - { - "block_id": "p429-b12", - "global_id": 11978, - "bbox": [ - 272.97, - 194.94, - 321.83, - 212.39 - ], - "text": "s + 4 −j8 +", - "type": "text" - }, - { - "block_id": "p429-b13", - "global_id": 11979, - "bbox": [ - 325.93, - 179.82, - 331.83, - 186.79 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p429-b14", - "global_id": 11980, - "bbox": [ - 328.89, - 184.35, - 361.75, - 201.12 - ], - "text": "5\n4 e−j153◦", - "type": "text" - }, - { - "block_id": "p429-b15", - "global_id": 11981, - "bbox": [ - 324.58, - 202.01, - 362.91, - 212.39 - ], - "text": "s + 4 + j8", - "type": "text" - }, - { - "block_id": "p429-b16", - "global_id": 11982, - "bbox": [ - 127.59, - 220.42, - 141.97, - 230.38 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p429-b17", - "global_id": 11983, - "bbox": [ - 241.55, - 231.96, - 270.11, - 243.04 - ], - "text": "θo(t) =", - "type": "text" - }, - { - "block_id": "p429-b18", - "global_id": 11984, - "bbox": [ - 272.16, - 220.96, - 276.83, - 230.92 - ], - "text": "'", - "type": "text" - }, - { - "block_id": "p429-b19", - "global_id": 11985, - "bbox": [ - 276.83, - 224.47, - 299.79, - 242.33 - ], - "text": "1 +\n√", - "type": "text" - }, - { - "block_id": "p429-b20", - "global_id": 11986, - "bbox": [ - 296.83, - 220.96, - 385.71, - 245.77 - ], - "text": "5\n2 e−4t cos(8t + 153◦)\n(", - "type": "text" - }, - { - "block_id": "p429-b21", - "global_id": 11987, - "bbox": [ - 386.81, - 231.96, - 402.17, - 242.23 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p429-b22", - "global_id": 11988, - "bbox": [ - 127.59, - 252.7, - 516.14, - 395.67 - ], - "text": "This response, also depicted in Fig. 4.35c, achieves the goal of reaching the final position at a\nfaster rate than that in the earlier case (K = 7). Unfortunately the improvement is achieved at the\ncost of ringing (oscillations) with high overshoot. In the present case, the percent overshoot (PO) is\n21%. The response reaches its peak value at peak time tp = 0.393 second. The rise time, defined as\nthe time required for the response to rise from 10% to 90% of its steady-state value, indicates the\nspeed of response.† In the present case tr = 0.175 second. The steady-state value of the response\nis unity so that the steady-state error is zero. Theoretically it takes infinite time for the response to\nreach the desired value of unity. In practice, however, we may consider the response to have settled\nto the final value if it closely approaches the final value. A widely accepted measure of closeness\nis within 2% of the final value. The time required for the response to reach and stay within 2% of\nthe final value is called the settling time ts.‡ In Fig. 4.35c, we find ts ≈1 second (when K = 80). A\ngood system has a small overshoot, small tr and ts and a small steady-state error.", - "type": "text" - }, - { - "block_id": "p429-b23", - "global_id": 11989, - "bbox": [ - 127.59, - 396.17, - 516.17, - 465.9 - ], - "text": "A large overshoot, as in the present case, may be unacceptable in many applications. Let\nus try to determine K (the gain) that yields the fastest response without oscillations. Complex\ncharacteristic roots lead to oscillations; to avoid oscillations, the characteristic roots should be\nreal. In the present case, the characteristic polynomial is s2 +8s+K. For K > 16, the characteristic\nroots are complex; for K < 16, the roots are real. The fastest response without oscillations is\nobtained by choosing K = 16. We now consider this case.", - "type": "text" - }, - { - "block_id": "p429-b24", - "global_id": 11990, - "bbox": [ - 145.52, - 467.49, - 193.21, - 477.86 - ], - "text": "For K = 16,", - "type": "text" - }, - { - "block_id": "p429-b25", - "global_id": 11991, - "bbox": [ - 200.27, - 485.29, - 364.86, - 509.31 - ], - "text": "o(s) =\n16\ns(s2 + 8s + 16) =\n16\ns(s + 4)2 = 1", - "type": "text" - }, - { - "block_id": "p429-b26", - "global_id": 11992, - "bbox": [ - 360.43, - 485.29, - 441.75, - 509.31 - ], - "text": "s −\n1\ns + 4 −\n4\n(s + 4)2", - "type": "text" - }, - { - "block_id": "p429-b27", - "global_id": 11993, - "bbox": [ - 127.59, - 517.02, - 141.97, - 526.99 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p429-b28", - "global_id": 11994, - "bbox": [ - 264.07, - 526.85, - 379.66, - 539.65 - ], - "text": "θo(t) = [1 −(4t + 1)e−4t]u(t)", - "type": "text" - }, - { - "block_id": "p429-b29", - "global_id": 11995, - "bbox": [ - 127.59, - 546.6, - 516.13, - 580.89 - ], - "text": "This response also appears in Fig. 4.35c. The system with K > 16 is said to be underdamped\n(oscillatory response), whereas the system with K < 16 is said to be overdamped. For K = 16, the\nsystem is said to be critically damped.", - "type": "text" - }, - { - "block_id": "p429-b30", - "global_id": 11996, - "bbox": [ - 127.59, - 598.89, - 516.12, - 634.09 - ], - "text": "† Delay time td, defined as the time required for the response to reach 50% of its steady-state value, is another\nindication of speed. For the present case, td = 0.141 second.\n‡ Typical percentage values used are 2 to 5% for ts.", - "type": "text" - } - ] - }, - { - "page_num": 430, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p430-b0", - "global_id": 11997, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "410\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p430-b1", - "global_id": 11998, - "bbox": [ - 101.84, - 85.82, - 490.39, - 155.56 - ], - "text": "There is a trade-off between undesirable overshoot and rise time. Reducing overshoots leads\nto higher rise time (sluggish system). In practice, a small overshoot, which is still faster than\nthe critical damping, may be acceptable. Note that percent overshoot PO and peak time tp are\nmeaningless for the overdamped or critically damped cases. In addition to adjusting gain K, we\nmay need to augment the system with some type of compensator if the specifications on overshoot\nand the speed of response are too stringent.", - "type": "text" - }, - { - "block_id": "p430-b2", - "global_id": 11999, - "bbox": [ - 102.14, - 173.23, - 171.19, - 185.35 - ], - "text": "RAMP INPUT", - "type": "text" - }, - { - "block_id": "p430-b3", - "global_id": 12000, - "bbox": [ - 101.84, - 189.39, - 490.39, - 235.92 - ], - "text": "If the anti-aircraft gun in Fig. 4.35a is tracking an enemy plane moving with a uniform velocity,\nthe gun-position angle must increase linearly with t. Hence, the input in this case is a ramp; that\nis, θi(t) = tu(t). Let us find the response of the system to this input when K = 80. In this case,\ni(s) = 1/s2, and", - "type": "text" - }, - { - "block_id": "p430-b4", - "global_id": 12001, - "bbox": [ - 186.85, - 250.96, - 321.38, - 274.97 - ], - "text": "o(s) =\n80\ns2(s2 + 8s + 80) = −0.1", - "type": "text" - }, - { - "block_id": "p430-b5", - "global_id": 12002, - "bbox": [ - 313.23, - 250.96, - 341.07, - 274.87 - ], - "text": "s + 1", - "type": "text" - }, - { - "block_id": "p430-b6", - "global_id": 12003, - "bbox": [ - 334.65, - 250.54, - 399.77, - 274.87 - ], - "text": "s2 + 0.1(s −2)", - "type": "text" - }, - { - "block_id": "p430-b7", - "global_id": 12004, - "bbox": [ - 355.77, - 261.72, - 404.18, - 274.97 - ], - "text": "s2 + 8s + 80", - "type": "text" - }, - { - "block_id": "p430-b8", - "global_id": 12005, - "bbox": [ - 101.84, - 290.38, - 193.73, - 300.34 - ], - "text": "Use of Table 4.1 yields", - "type": "text" - }, - { - "block_id": "p430-b9", - "global_id": 12006, - "bbox": [ - 201.23, - 317.76, - 229.8, - 328.84 - ], - "text": "θo(t) =", - "type": "text" - }, - { - "block_id": "p430-b11", - "global_id": 12007, - "bbox": [ - 235.77, - 316.42, - 285.37, - 328.14 - ], - "text": "−0.1 + t + 1", - "type": "text" - }, - { - "block_id": "p430-b12", - "global_id": 12008, - "bbox": [ - 281.88, - 309.75, - 374.54, - 330.96 - ], - "text": "8e−8t cos(8t + 36.87◦)", - "type": "text" - }, - { - "block_id": "p430-b13", - "global_id": 12009, - "bbox": [ - 375.64, - 317.76, - 391.0, - 328.04 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p430-b14", - "global_id": 12010, - "bbox": [ - 101.85, - 345.56, - 490.4, - 391.82 - ], - "text": "This response, sketched in Fig. 4.35d, shows that there is a steady-state error er = 0.1 radian. In\nmany cases such a small steady-state error may be tolerable. If, however, a zero steady-state error\nto a ramp input is required, this system in its present form is unsatisfactory. We must add some\nform of compensator to the system.", - "type": "text" - }, - { - "block_id": "p430-b15", - "global_id": 12011, - "bbox": [ - 76.77, - 423.75, - 478.14, - 449.65 - ], - "text": "EXAMPLE 4.26\nStep and Ramp Responses of Feedback Systems Using\nMATLAB", - "type": "text" - }, - { - "block_id": "p430-b16", - "global_id": 12012, - "bbox": [ - 103.16, - 462.98, - 477.01, - 497.26 - ], - "text": "Using the feedback system of Fig. 4.18d with G(s) = K/(s(s + 8)) and H(s) = 1, determine\nthe step response for each of the following cases: (a) K = 7, (b) K = 16, and (c) K = 80.\nAdditionally, find the unit ramp response when (d) K = 80.", - "type": "text" - }, - { - "block_id": "p430-b17", - "global_id": 12013, - "bbox": [ - 103.16, - 520.18, - 477.04, - 554.34 - ], - "text": "Example 4.21 computes the transfer functions of these feedback systems in a simple way. In\nthis example, the conv command is used to demonstrate polynomial multiplication of the two\ndenominator factors of G(s). Step responses are computed by using the step command.", - "type": "text" - }, - { - "block_id": "p430-b18", - "global_id": 12014, - "bbox": [ - 121.09, - 553.97, - 142.1, - 563.93 - ], - "text": "(a–c)", - "type": "text" - }, - { - "block_id": "p430-b19", - "global_id": 12015, - "bbox": [ - 103.16, - 573.72, - 420.54, - 621.54 - ], - "text": ">>\nH = tf(1,1); K = 7; G = tf([K],conv([1 0],[1 8])); Ha = feedback(G,H);\n>>\nH = tf(1,1); K = 16; G = tf([K],conv([1 0],[1 8])); Hb = feedback(G,H);\n>>\nH = tf(1,1); K = 80; G = tf([K],conv([1 0],[1 8])); Hc = feedback(G,H);\n>>\nclf; step(Ha,’k-’,Hb,’k--’,Hc,’k-.’);\n>>\nlegend(’K = 7’,’K = 16’,’K = 80’,’Location’,’best’);", - "type": "text" - } - ] - }, - { - "page_num": 431, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p431-b0", - "global_id": 12016, - "bbox": [ - 302.53, - 62.89, - 516.13, - 71.98 - ], - "text": "4.7\nApplication to Feedback and Controls\n411", - "type": "text" - }, - { - "block_id": "p431-b1", - "global_id": 12017, - "bbox": [ - 145.66, - 209.99, - 437.01, - 231.17 - ], - "text": "6\n4\n2\n0\nTime (seconds)", - "type": "text" - }, - { - "block_id": "p431-b2", - "global_id": 12018, - "bbox": [ - 132.68, - 164.34, - 143.36, - 172.34 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p431-b3", - "global_id": 12019, - "bbox": [ - 139.43, - 127.9, - 143.43, - 135.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p431-b4", - "global_id": 12020, - "bbox": [ - 120.7, - 135.1, - 129.5, - 172.74 - ], - "text": "Amplitude", - "type": "text" - }, - { - "block_id": "p431-b5", - "global_id": 12021, - "bbox": [ - 282.9, - 90.8, - 336.92, - 99.6 - ], - "text": "Step Response", - "type": "text" - }, - { - "block_id": "p431-b6", - "global_id": 12022, - "bbox": [ - 452.78, - 164.25, - 465.64, - 171.45 - ], - "text": "K=7", - "type": "text" - }, - { - "block_id": "p431-b7", - "global_id": 12023, - "bbox": [ - 452.78, - 175.13, - 469.24, - 182.33 - ], - "text": "K=16", - "type": "text" - }, - { - "block_id": "p431-b8", - "global_id": 12024, - "bbox": [ - 452.78, - 186.0, - 469.24, - 193.2 - ], - "text": "K=80", - "type": "text" - }, - { - "block_id": "p431-b9", - "global_id": 12025, - "bbox": [ - 119.94, - 237.84, - 270.36, - 247.08 - ], - "text": "Figure 4.36 Step responses for Ex. 4.26.", - "type": "text" - }, - { - "block_id": "p431-b10", - "global_id": 12026, - "bbox": [ - 128.9, - 262.72, - 502.78, - 308.63 - ], - "text": "(d) The unit ramp response is equivalent to the integral of the unit step response. We\ncan obtain the ramp response by taking the step response of the system in cascade with an\nintegrator. To help highlight waveform detail, we compute the ramp response over the short\ntime interval of 0 ≤t ≤1.5.", - "type": "text" - }, - { - "block_id": "p431-b11", - "global_id": 12027, - "bbox": [ - 128.9, - 318.88, - 390.42, - 340.8 - ], - "text": ">>\nt = 0:.001:1.5; Hd = series(Hc,tf([1],[1 0]));\n>>\nstep(Hd,’k-’,t); title(’Unit Ramp Response’);", - "type": "text" - }, - { - "block_id": "p431-b12", - "global_id": 12028, - "bbox": [ - 162.2, - 493.19, - 504.01, - 514.36 - ], - "text": "0.5\n0\n1.5\n1\nTime (seconds)", - "type": "text" - }, - { - "block_id": "p431-b13", - "global_id": 12029, - "bbox": [ - 149.22, - 449.96, - 159.89, - 457.96 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p431-b14", - "global_id": 12030, - "bbox": [ - 155.97, - 415.96, - 159.97, - 423.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p431-b15", - "global_id": 12031, - "bbox": [ - 149.22, - 381.95, - 159.89, - 389.95 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p431-b16", - "global_id": 12032, - "bbox": [ - 137.25, - 418.29, - 146.05, - 455.94 - ], - "text": "Amplitude", - "type": "text" - }, - { - "block_id": "p431-b17", - "global_id": 12033, - "bbox": [ - 286.7, - 373.99, - 365.9, - 382.79 - ], - "text": "Unit Ramp Response", - "type": "text" - }, - { - "block_id": "p431-b18", - "global_id": 12034, - "bbox": [ - 136.48, - 520.67, - 335.44, - 530.28 - ], - "text": "Figure 4.37 Ramp response for Ex. 4.26 with K = 80.", - "type": "text" - }, - { - "block_id": "p431-b19", - "global_id": 12035, - "bbox": [ - 127.89, - 572.8, - 259.66, - 584.93 - ], - "text": "DESIGN SPECIFICATIONS", - "type": "text" - }, - { - "block_id": "p431-b20", - "global_id": 12036, - "bbox": [ - 127.59, - 588.96, - 516.15, - 634.79 - ], - "text": "Now the reader has some idea of the various specifications a control system might require.\nGenerally, a control system is designed to meet given transient specifications, steady-state error\nspecifications, and sensitivity specifications. Transient specifications include overshoot, rise time,\nand settling time of the response to step input. The steady-state error is the difference between", - "type": "text" - } - ] - }, - { - "page_num": 432, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p432-b0", - "global_id": 12037, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "412\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p432-b1", - "global_id": 12038, - "bbox": [ - 101.84, - 85.82, - 490.42, - 131.64 - ], - "text": "the desired response and the actual response to a test input in steady state. The system should\nalso satisfy a specified sensitivity specifications to some system parameter variations, or to certain\ndisturbances. Above all, the system must remain stable under operating conditions. Discussion of\ndesign procedures used to realize given specifications is beyond the scope of this book.", - "type": "text" - }, - { - "block_id": "p432-b2", - "global_id": 12039, - "bbox": [ - 102.2, - 162.89, - 414.28, - 176.84 - ], - "text": "4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM", - "type": "text" - }, - { - "block_id": "p432-b3", - "global_id": 12040, - "bbox": [ - 101.84, - 182.83, - 490.4, - 228.65 - ], - "text": "Filtering is an important area of signal processing. Filtering characteristics of a system are\nindicated by its response to sinusoids of various frequencies varying from 0 to ∞. Such\ncharacteristics are called the frequency response of the system. In this section, we shall find the\nfrequency response of LTIC systems.", - "type": "text" - }, - { - "block_id": "p432-b4", - "global_id": 12041, - "bbox": [ - 101.84, - 230.65, - 490.4, - 264.52 - ], - "text": "In Sec. 2.4-4 we showed that an LTIC system response to an everlasting exponential input\nx(t) = est is also an everlasting exponential H(s)est. As before, we use an arrow directed from the\ninput to the output to represent an input–output pair:", - "type": "text" - }, - { - "block_id": "p432-b5", - "global_id": 12042, - "bbox": [ - 266.64, - 273.92, - 490.38, - 288.41 - ], - "text": "est \r⇒H(s)est\n(4.40)", - "type": "text" - }, - { - "block_id": "p432-b6", - "global_id": 12043, - "bbox": [ - 101.85, - 301.92, - 260.76, - 312.29 - ], - "text": "Setting s = jω in this relationship yields", - "type": "text" - }, - { - "block_id": "p432-b7", - "global_id": 12044, - "bbox": [ - 259.9, - 321.69, - 490.38, - 336.18 - ], - "text": "ejωt \r⇒H(jω)ejωt\n(4.41)", - "type": "text" - }, - { - "block_id": "p432-b8", - "global_id": 12045, - "bbox": [ - 101.85, - 348.46, - 357.58, - 360.06 - ], - "text": "Noting that cosωt is the real part of ejωt, use of Eq. (2.31) yields", - "type": "text" - }, - { - "block_id": "p432-b9", - "global_id": 12046, - "bbox": [ - 245.83, - 371.85, - 490.38, - 383.95 - ], - "text": "cosωt \r⇒Re[H(jω)ejωt]\n(4.42)", - "type": "text" - }, - { - "block_id": "p432-b10", - "global_id": 12047, - "bbox": [ - 101.84, - 397.46, - 270.99, - 407.84 - ], - "text": "We can express H(jω) in the polar form as", - "type": "text" - }, - { - "block_id": "p432-b11", - "global_id": 12048, - "bbox": [ - 248.33, - 419.62, - 343.41, - 431.62 - ], - "text": "H(jω) = |H(jω)|ej̸ H(jω)", - "type": "text" - }, - { - "block_id": "p432-b12", - "global_id": 12049, - "bbox": [ - 101.84, - 445.64, - 245.32, - 455.61 - ], - "text": "With this result, Eq. (4.42) becomes", - "type": "text" - }, - { - "block_id": "p432-b13", - "global_id": 12050, - "bbox": [ - 222.22, - 469.12, - 370.02, - 479.49 - ], - "text": "cosωt \r⇒|H(jω)|cos[ωt +̸ H(jω)]", - "type": "text" - }, - { - "block_id": "p432-b14", - "global_id": 12051, - "bbox": [ - 101.84, - 493.0, - 417.21, - 503.38 - ], - "text": "In other words, the system response y(t) to a sinusoidal input cosωt is given by", - "type": "text" - }, - { - "block_id": "p432-b15", - "global_id": 12052, - "bbox": [ - 231.21, - 516.89, - 361.02, - 527.27 - ], - "text": "y(t) = |H(jω)|cos[ωt +̸ H(jω)]", - "type": "text" - }, - { - "block_id": "p432-b16", - "global_id": 12053, - "bbox": [ - 101.84, - 540.78, - 470.71, - 551.15 - ], - "text": "Using a similar argument, we can show that the system response to a sinusoid cos(ωt + θ) is", - "type": "text" - }, - { - "block_id": "p432-b17", - "global_id": 12054, - "bbox": [ - 223.08, - 564.66, - 490.38, - 575.04 - ], - "text": "y(t) = |H(jω)|cos[ωt + θ +̸ H(jω)]\n(4.43)", - "type": "text" - }, - { - "block_id": "p432-b18", - "global_id": 12055, - "bbox": [ - 101.84, - 588.96, - 490.41, - 634.79 - ], - "text": "This result is valid only for BIBO-stable systems. The frequency response is meaningless for\nBIBO-unstable systems. This follows from the fact that the frequency response in Eq. (4.41) is\nobtained by setting s = jω in Eq. (4.40). But, as shown in Sec. 2.4-4 [Eqs. (2.38) and (2.39)],\nEq. (4.40) applies only for the values of s for which H(s) exists. For BIBO-unstable systems, the", - "type": "text" - } - ] - }, - { - "page_num": 433, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p433-b0", - "global_id": 12056, - "bbox": [ - 296.76, - 62.89, - 516.14, - 71.98 - ], - "text": "4.8\nFrequency Response of an LTIC System\n413", - "type": "text" - }, - { - "block_id": "p433-b1", - "global_id": 12057, - "bbox": [ - 127.59, - 85.46, - 516.14, - 107.79 - ], - "text": "ROC for H(s) does not include the ω axis where s = jω [see Eq. (4.10)]. This means that H(s)\nwhen s = jω is meaningless for BIBO-unstable systems.†", - "type": "text" - }, - { - "block_id": "p433-b2", - "global_id": 12058, - "bbox": [ - 127.59, - 109.38, - 516.14, - 287.13 - ], - "text": "Equation (4.43) shows that for a sinusoidal input of radian frequency ω, the system response\nis also a sinusoid of the same frequency ω. The amplitude of the output sinusoid is |H(jω)| times\nthe input amplitude, and the phase of the output sinusoid is shifted by̸\nH(jω) with respect to\nthe input phase (see later Fig. 4.38 in Ex. 4.27). For instance, a certain system with |H(j10)| = 3\nand̸\nH(j10) = −30◦amplifies a sinusoid of frequency ω = 10 by a factor of 3 and delays its\nphase by 30◦. The system response to an input 5cos(10t + 50◦) is 3 × 5cos(10t + 50◦−30◦) =\n15cos(10t + 20◦).\nClearly |H(jω)| is the amplitude gain of the system, and a plot of |H(jω)| versus ω shows the\namplitude gain as a function of frequency ω. We shall call |H(jω)| the amplitude response. It also\ngoes under the name magnitude response.‡ Similarly,̸\nH(jω) is the phase response, and a plot of̸\nH(jω) versus ω shows how the system modifies or changes the phase of the input sinusoid. Plots\nof the magnitude response |H(jω)| and phase response̸\nH(jω) show at a glance how a system\nresponds to sinusoids of various frequencies. Observe that H(jω) has the information of |H(jω)|\nand̸\nH(jω) and is therefore termed the frequency response of the system. Clearly, the frequency\nresponse of a system represents its filtering characteristics.", - "type": "text" - }, - { - "block_id": "p433-b3", - "global_id": 12059, - "bbox": [ - 102.51, - 316.81, - 319.26, - 328.77 - ], - "text": "EXAMPLE 4.27\nFrequency Response", - "type": "text" - }, - { - "block_id": "p433-b4", - "global_id": 12060, - "bbox": [ - 128.9, - 345.44, - 502.73, - 367.35 - ], - "text": "Find the frequency response (amplitude and phase responses) of a system whose transfer\nfunction is", - "type": "text" - }, - { - "block_id": "p433-b5", - "global_id": 12061, - "bbox": [ - 285.63, - 364.86, - 344.84, - 382.12 - ], - "text": "H(s) = s + 0.1", - "type": "text" - }, - { - "block_id": "p433-b6", - "global_id": 12062, - "bbox": [ - 321.38, - 378.92, - 341.1, - 389.29 - ], - "text": "s + 5", - "type": "text" - }, - { - "block_id": "p433-b7", - "global_id": 12063, - "bbox": [ - 128.9, - 394.56, - 340.72, - 404.94 - ], - "text": "Also, find the system response y(t) if the input x(t) is", - "type": "text" - }, - { - "block_id": "p433-b8", - "global_id": 12064, - "bbox": [ - 146.84, - 412.8, - 186.68, - 422.86 - ], - "text": "(a) cos 2t", - "type": "text" - }, - { - "block_id": "p433-b9", - "global_id": 12065, - "bbox": [ - 146.84, - 426.21, - 223.55, - 437.81 - ], - "text": "(b) cos(10t −50◦)", - "type": "text" - }, - { - "block_id": "p433-b10", - "global_id": 12066, - "bbox": [ - 128.9, - 466.7, - 176.21, - 476.67 - ], - "text": "In this case,", - "type": "text" - }, - { - "block_id": "p433-b11", - "global_id": 12067, - "bbox": [ - 280.01, - 475.51, - 350.45, - 492.77 - ], - "text": "H(jω) = jω + 0.1", - "type": "text" - }, - { - "block_id": "p433-b12", - "global_id": 12068, - "bbox": [ - 321.38, - 489.56, - 346.72, - 499.94 - ], - "text": "jω + 5", - "type": "text" - }, - { - "block_id": "p433-b13", - "global_id": 12069, - "bbox": [ - 127.59, - 533.22, - 516.16, - 633.41 - ], - "text": "† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains\nnondecaying natural mode terms of the form cosω0t or eat cosω0t (a > 0). Hence, the response of such a\nsystem to a sinusoid cosωt will contain not just the sinusoid of frequency ω, but also nondecaying natural\nmodes, rendering the concept of frequency response meaningless.\n‡ Strictly speaking, |H(ω)| is magnitude response. There is a fine distinction between amplitude and\nmagnitude. Amplitude A can be positive and negative. In contrast, the magnitude |A| is always nonnegative.\nWe refrain from relying on this useful distinction between amplitude and magnitude in the interest of\navoiding proliferation of essentially similar entities. This is also why we shall use the “amplitude” (instead\nof “magnitude”) spectrum for |H(ω)|.", - "type": "text" - } - ] - }, - { - "page_num": 434, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p434-b0", - "global_id": 12070, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "414\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p434-b1", - "global_id": 12071, - "bbox": [ - 103.16, - 86.24, - 144.91, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p434-b2", - "global_id": 12072, - "bbox": [ - 142.39, - 114.28, - 182.47, - 124.56 - ], - "text": "|H(jω)| =", - "type": "text" - }, - { - "block_id": "p434-b3", - "global_id": 12073, - "bbox": [ - 185.71, - 98.87, - 194.14, - 108.83 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p434-b4", - "global_id": 12074, - "bbox": [ - 189.45, - 104.42, - 233.14, - 124.3 - ], - "text": "ω2 + 0.01\n√", - "type": "text" - }, - { - "block_id": "p434-b5", - "global_id": 12075, - "bbox": [ - 197.88, - 119.89, - 229.41, - 133.15 - ], - "text": "ω2 + 25", - "type": "text" - }, - { - "block_id": "p434-b6", - "global_id": 12076, - "bbox": [ - 254.27, - 112.56, - 351.75, - 124.66 - ], - "text": "and̸\nH(jω) = tan−1", - "type": "text" - }, - { - "block_id": "p434-b7", - "global_id": 12077, - "bbox": [ - 353.36, - 100.3, - 370.67, - 117.26 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p434-b8", - "global_id": 12078, - "bbox": [ - 361.28, - 121.77, - 373.74, - 131.73 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p434-b10", - "global_id": 12079, - "bbox": [ - 383.21, - 112.56, - 413.62, - 124.66 - ], - "text": "−tan−1", - "type": "text" - }, - { - "block_id": "p434-b11", - "global_id": 12080, - "bbox": [ - 415.23, - 100.3, - 429.68, - 117.26 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p434-b12", - "global_id": 12081, - "bbox": [ - 424.02, - 121.77, - 429.0, - 131.73 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p434-b14", - "global_id": 12082, - "bbox": [ - 103.16, - 141.28, - 477.03, - 175.57 - ], - "text": "Both the amplitude and the phase response are depicted in Fig. 4.38a as functions of ω. These\nplots furnish the complete information about the frequency response of the system to sinusoidal\ninputs.", - "type": "text" - }, - { - "block_id": "p434-b15", - "global_id": 12083, - "bbox": [ - 121.09, - 177.15, - 287.06, - 187.53 - ], - "text": "(a) For the input x(t) = cos 2t, ω = 2, and", - "type": "text" - }, - { - "block_id": "p434-b16", - "global_id": 12084, - "bbox": [ - 167.77, - 206.73, - 206.11, - 217.11 - ], - "text": "|H(j2)| =", - "type": "text" - }, - { - "block_id": "p434-b18", - "global_id": 12085, - "bbox": [ - 213.09, - 196.88, - 262.29, - 216.6 - ], - "text": "(2)2 + 0.01", - "type": "text" - }, - { - "block_id": "p434-b19", - "global_id": 12086, - "bbox": [ - 221.34, - 212.75, - 258.56, - 226.0 - ], - "text": "(2)2 + 25", - "type": "text" - }, - { - "block_id": "p434-b20", - "global_id": 12087, - "bbox": [ - 265.56, - 206.73, - 297.77, - 217.11 - ], - "text": "= 0.372̸", - "type": "text" - }, - { - "block_id": "p434-b21", - "global_id": 12088, - "bbox": [ - 173.46, - 234.85, - 229.25, - 246.95 - ], - "text": "H(j2) = tan−1", - "type": "text" - }, - { - "block_id": "p434-b22", - "global_id": 12089, - "bbox": [ - 230.86, - 222.59, - 247.5, - 239.97 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p434-b23", - "global_id": 12090, - "bbox": [ - 238.79, - 244.06, - 251.24, - 254.03 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p434-b25", - "global_id": 12091, - "bbox": [ - 260.7, - 234.85, - 291.12, - 246.95 - ], - "text": "−tan−1", - "type": "text" - }, - { - "block_id": "p434-b26", - "global_id": 12092, - "bbox": [ - 292.73, - 222.59, - 305.63, - 239.97 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p434-b27", - "global_id": 12093, - "bbox": [ - 300.65, - 244.06, - 305.63, - 254.03 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p434-b29", - "global_id": 12094, - "bbox": [ - 315.6, - 234.85, - 411.9, - 246.95 - ], - "text": "= 87.1◦−21.8◦= 65.3◦", - "type": "text" - }, - { - "block_id": "p434-b30", - "global_id": 12095, - "bbox": [ - 294.96, - 437.85, - 377.47, - 445.85 - ], - "text": "10\n2\n0", - "type": "text" - }, - { - "block_id": "p434-b31", - "global_id": 12096, - "bbox": [ - 274.93, - 454.17, - 283.81, - 462.17 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p434-b32", - "global_id": 12097, - "bbox": [ - 107.84, - 305.84, - 111.84, - 313.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p434-b33", - "global_id": 12098, - "bbox": [ - 175.87, - 377.57, - 193.87, - 385.57 - ], - "text": "0.894", - "type": "text" - }, - { - "block_id": "p434-b34", - "global_id": 12099, - "bbox": [ - 122.67, - 415.01, - 140.67, - 423.01 - ], - "text": "0.372", - "type": "text" - }, - { - "block_id": "p434-b35", - "global_id": 12100, - "bbox": [ - 309.48, - 380.99, - 326.14, - 389.28 - ], - "text": "65.3", - "type": "text" - }, - { - "block_id": "p434-b36", - "global_id": 12101, - "bbox": [ - 369.56, - 407.01, - 380.22, - 415.3 - ], - "text": "26", - "type": "text" - }, - { - "block_id": "p434-b37", - "global_id": 12102, - "bbox": [ - 117.68, - 288.16, - 142.14, - 296.46 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p434-b38", - "global_id": 12103, - "bbox": [ - 327.14, - 329.89, - 353.9, - 338.18 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p434-b39", - "global_id": 12104, - "bbox": [ - 107.9, - 433.55, - 444.39, - 443.73 - ], - "text": "10\n0\n2\nv\nv", - "type": "text" - }, - { - "block_id": "p434-b40", - "global_id": 12105, - "bbox": [ - 262.95, - 591.89, - 272.59, - 599.89 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p434-b41", - "global_id": 12106, - "bbox": [ - 147.13, - 478.79, - 464.82, - 487.45 - ], - "text": "y(t) 0.372 cos(2t 65.3)\nx(t) cos 2t", - "type": "text" - }, - { - "block_id": "p434-b42", - "global_id": 12107, - "bbox": [ - 201.82, - 521.73, - 393.14, - 541.59 - ], - "text": "0\n2\n6\nt", - "type": "text" - }, - { - "block_id": "p434-b43", - "global_id": 12108, - "bbox": [ - 101.77, - 606.58, - 287.06, - 615.82 - ], - "text": "Figure 4.38 Responses for the system of Ex. 4.27.", - "type": "text" - } - ] - }, - { - "page_num": 435, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p435-b0", - "global_id": 12109, - "bbox": [ - 296.76, - 62.89, - 516.14, - 71.98 - ], - "text": "4.8\nFrequency Response of an LTIC System\n415", - "type": "text" - }, - { - "block_id": "p435-b1", - "global_id": 12110, - "bbox": [ - 128.9, - 86.24, - 502.77, - 144.02 - ], - "text": "We also could have read these values directly from the frequency response plots in Fig. 4.38a\ncorresponding to ω = 2. This result means that for a sinusoidal input with frequency ω = 2, the\namplitude gain of the system is 0.372, and the phase shift is 65.3◦. In other words, the output\namplitude is 0.372 times the input amplitude, and the phase of the output is shifted with respect\nto that of the input by 65.3◦. Therefore, the system response to the input cos 2t is", - "type": "text" - }, - { - "block_id": "p435-b2", - "global_id": 12111, - "bbox": [ - 259.72, - 153.84, - 371.95, - 165.94 - ], - "text": "y(t) = 0.372cos(2t + 65.3◦)", - "type": "text" - }, - { - "block_id": "p435-b3", - "global_id": 12112, - "bbox": [ - 128.9, - 176.26, - 502.78, - 199.81 - ], - "text": "The input cos 2t and the corresponding system response 0.372cos(2t + 65.3◦) are illustrated\nin Fig. 4.38b.", - "type": "text" - }, - { - "block_id": "p435-b4", - "global_id": 12113, - "bbox": [ - 128.91, - 200.17, - 502.77, - 235.68 - ], - "text": "(b) For the input cos(10t −50◦), instead of computing the values |H(jω)| and̸\nH(jω)\nas in part (a), we shall read them directly from the frequency response plots in Fig. 4.38a\ncorresponding to ω = 10. These are", - "type": "text" - }, - { - "block_id": "p435-b5", - "global_id": 12114, - "bbox": [ - 225.2, - 245.5, - 405.96, - 257.59 - ], - "text": "|H(j10)| = 0.894\nand̸\nH(j10) = 26◦", - "type": "text" - }, - { - "block_id": "p435-b6", - "global_id": 12115, - "bbox": [ - 128.9, - 269.14, - 502.78, - 303.42 - ], - "text": "Therefore, for a sinusoidal input of frequency ω = 10, the output sinusoid amplitude is 0.894\ntimes the input amplitude, and the output sinusoid is shifted with respect to the input sinusoid\nby 26◦. Therefore, the system response y(t) to an input cos(10t −50◦) is", - "type": "text" - }, - { - "block_id": "p435-b7", - "global_id": 12116, - "bbox": [ - 201.08, - 313.24, - 430.58, - 325.35 - ], - "text": "y(t) = 0.894cos(10t −50◦+ 26◦) = 0.894cos(10t −24◦)", - "type": "text" - }, - { - "block_id": "p435-b8", - "global_id": 12117, - "bbox": [ - 128.91, - 335.66, - 502.76, - 395.08 - ], - "text": "If the input were sin(10t −50◦), the response would be 0.894sin(10t −50◦+ 26◦) =\n0.894 sin(10t −24◦).\nThe frequency response plots in Fig. 4.38a show that the system has highpass filtering\ncharacteristics; it responds well to sinusoids of higher frequencies (ω well above 5), and\nsuppresses sinusoids of lower frequencies (ω well below 5).", - "type": "text" - }, - { - "block_id": "p435-b9", - "global_id": 12118, - "bbox": [ - 129.2, - 409.81, - 401.38, - 421.93 - ], - "text": "PLOTTING FREQUENCY RESPONSE WITH MATLAB", - "type": "text" - }, - { - "block_id": "p435-b10", - "global_id": 12119, - "bbox": [ - 128.9, - 425.97, - 502.76, - 459.84 - ], - "text": "It is simple to use MATLAB to create magnitude and phase response plots. Here, we consider\ntwo methods. In the first method, we use an anonymous function to define the transfer function\nH(s) and then obtain the frequency response plots by substituting jω for s.", - "type": "text" - }, - { - "block_id": "p435-b11", - "global_id": 12120, - "bbox": [ - 128.9, - 470.09, - 458.42, - 503.96 - ], - "text": ">>\nH = @(s) (s+0.1)./(s+5); omega = 0:.01:20;\n>>\nsubplot(1,2,1); plot(omega,abs(H(1j*omega)),’k-’);\n>>\nsubplot(1,2,2); plot(omega,angle(H(1j*omega))*180/pi,’k-’);", - "type": "text" - }, - { - "block_id": "p435-b12", - "global_id": 12121, - "bbox": [ - 128.9, - 513.64, - 502.76, - 535.85 - ], - "text": "In the second method, we define vectors that contain the numerator and denominator\ncoefficients of H(s) and then use the freqs command to compute frequency response.", - "type": "text" - }, - { - "block_id": "p435-b13", - "global_id": 12122, - "bbox": [ - 128.91, - 545.8, - 479.31, - 579.68 - ], - "text": ">>\nB = [1 0.1]; A = [1 5]; H = freqs(B,A,omega); omega = 0:.01:20;\n>>\nsubplot(1,2,1); plot(omega,abs(H),’k-’);\n>>\nsubplot(1,2,2); plot(omega,angle(H)*180/pi,’k-’);", - "type": "text" - }, - { - "block_id": "p435-b14", - "global_id": 12123, - "bbox": [ - 128.91, - 589.35, - 343.07, - 599.32 - ], - "text": "Both approaches generate plots that match Fig. 4.38a.", - "type": "text" - } - ] - }, - { - "page_num": 436, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p436-b0", - "global_id": 12124, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "416\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p436-b1", - "global_id": 12125, - "bbox": [ - 76.77, - 93.92, - 455.59, - 119.82 - ], - "text": "EXAMPLE 4.28\nFrequency Responses of Delay, Differentiator, and\nIntegrator Systems", - "type": "text" - }, - { - "block_id": "p436-b2", - "global_id": 12126, - "bbox": [ - 103.16, - 136.38, - 476.25, - 158.4 - ], - "text": "Find and sketch the frequency responses (magnitude and phase) for (a) an ideal delay of T\nseconds, (b) an ideal differentiator, and (c) an ideal integrator.", - "type": "text" - }, - { - "block_id": "p436-b3", - "global_id": 12127, - "bbox": [ - 121.09, - 180.91, - 468.19, - 191.28 - ], - "text": "(a) Ideal delay of T seconds. The transfer function of an ideal delay is [see Eq. (4.30)]", - "type": "text" - }, - { - "block_id": "p436-b4", - "global_id": 12128, - "bbox": [ - 265.94, - 205.08, - 313.2, - 217.08 - ], - "text": "H(s) = e−sT", - "type": "text" - }, - { - "block_id": "p436-b5", - "global_id": 12129, - "bbox": [ - 103.16, - 233.12, - 144.91, - 243.09 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p436-b6", - "global_id": 12130, - "bbox": [ - 261.16, - 242.94, - 317.98, - 254.94 - ], - "text": "H(jω) = e−jωT", - "type": "text" - }, - { - "block_id": "p436-b7", - "global_id": 12131, - "bbox": [ - 103.16, - 264.0, - 159.77, - 273.96 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p436-b8", - "global_id": 12132, - "bbox": [ - 207.98, - 275.55, - 371.43, - 285.93 - ], - "text": "|H(jω)| = 1\nand̸\nH(jω) = −ωT", - "type": "text" - }, - { - "block_id": "p436-b9", - "global_id": 12133, - "bbox": [ - 103.16, - 294.89, - 477.04, - 388.54 - ], - "text": "These amplitude and phase responses are shown in Fig. 4.39a. The amplitude response is\nconstant (unity) for all frequencies. The phase shift increases linearly with frequency with a\nslope of −T. This result can be explained physically by recognizing that if a sinusoid cosωt\nis passed through an ideal delay of T seconds, the output is cosω(t −T). The output sinusoid\namplitude is the same as that of the input for all values of ω. Therefore, the amplitude response\n(gain) is unity for all frequencies. Moreover, the output cosω(t −T) = cos(ωt −ωT) has a\nphase shift −ωT with respect to the input cosωt. Therefore, the phase response is linearly\nproportional to the frequency ω with a slope −T.", - "type": "text" - }, - { - "block_id": "p436-b10", - "global_id": 12134, - "bbox": [ - 103.16, - 390.45, - 477.03, - 412.45 - ], - "text": "(b) An ideal differentiator. The transfer function of an ideal differentiator is [see\nEq. (4.31)]", - "type": "text" - }, - { - "block_id": "p436-b11", - "global_id": 12135, - "bbox": [ - 272.74, - 414.02, - 307.43, - 424.3 - ], - "text": "H(s) = s", - "type": "text" - }, - { - "block_id": "p436-b12", - "global_id": 12136, - "bbox": [ - 103.16, - 433.37, - 144.91, - 443.34 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p436-b13", - "global_id": 12137, - "bbox": [ - 248.75, - 443.19, - 330.92, - 455.19 - ], - "text": "H(jω) = jω = ωejπ/2", - "type": "text" - }, - { - "block_id": "p436-b14", - "global_id": 12138, - "bbox": [ - 103.16, - 464.25, - 159.77, - 474.21 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p436-b15", - "global_id": 12139, - "bbox": [ - 212.83, - 471.38, - 365.16, - 488.74 - ], - "text": "|H(jω)| = ω\nand̸\nH(jω) = π", - "type": "text" - }, - { - "block_id": "p436-b16", - "global_id": 12140, - "bbox": [ - 103.16, - 485.85, - 477.02, - 594.25 - ], - "text": "2\nThese amplitude and phase responses are depicted in Fig. 4.39b. The amplitude response\nincreases linearly with frequency, and phase response is constant (π/2) for all frequencies.\nThis result can be explained physically by recognizing that if a sinusoid cosωt is passed\nthrough an ideal differentiator, the output is −ωsin ωt = ωcos[ωt + (π/2)]. Therefore, the\noutput sinusoid amplitude is ω times the input amplitude; that is, the amplitude response (gain)\nincreases linearly with frequency ω. Moreover, the output sinusoid undergoes a phase shift\nπ/2 with respect to the input cosωt. Therefore, the phase response is constant (π/2) with\nfrequency.", - "type": "text" - } - ] - }, - { - "page_num": 437, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p437-b0", - "global_id": 12141, - "bbox": [ - 296.76, - 62.89, - 516.14, - 71.98 - ], - "text": "4.8\nFrequency Response of an LTIC System\n417", - "type": "text" - }, - { - "block_id": "p437-b1", - "global_id": 12142, - "bbox": [ - 270.24, - 246.08, - 405.94, - 257.08 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p437-b2", - "global_id": 12143, - "bbox": [ - 138.9, - 148.02, - 142.9, - 156.02 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p437-b3", - "global_id": 12144, - "bbox": [ - 148.64, - 127.6, - 304.02, - 137.94 - ], - "text": "H( jv)\nH( jv)", - "type": "text" - }, - { - "block_id": "p437-b4", - "global_id": 12145, - "bbox": [ - 419.13, - 117.84, - 443.58, - 126.14 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p437-b5", - "global_id": 12146, - "bbox": [ - 149.67, - 218.56, - 176.43, - 226.86 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p437-b6", - "global_id": 12147, - "bbox": [ - 138.86, - 175.48, - 142.86, - 183.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p437-b7", - "global_id": 12148, - "bbox": [ - 139.87, - 248.5, - 143.87, - 256.5 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p437-b8", - "global_id": 12149, - "bbox": [ - 208.82, - 174.15, - 404.16, - 185.48 - ], - "text": "0\n0\nv", - "type": "text" - }, - { - "block_id": "p437-b9", - "global_id": 12150, - "bbox": [ - 208.82, - 236.8, - 214.16, - 244.8 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p437-b10", - "global_id": 12151, - "bbox": [ - 339.77, - 176.15, - 345.1, - 184.15 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p437-b11", - "global_id": 12152, - "bbox": [ - 339.77, - 248.15, - 345.1, - 256.15 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p437-b12", - "global_id": 12153, - "bbox": [ - 470.34, - 173.15, - 475.68, - 181.15 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p437-b13", - "global_id": 12154, - "bbox": [ - 470.34, - 234.8, - 475.68, - 242.8 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p437-b14", - "global_id": 12155, - "bbox": [ - 279.26, - 206.2, - 306.03, - 214.5 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p437-b15", - "global_id": 12156, - "bbox": [ - 260.9, - 223.73, - 274.24, - 232.02 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p437-b16", - "global_id": 12157, - "bbox": [ - 168.89, - 88.38, - 471.21, - 96.38 - ], - "text": "Ideal Delay\nIdeal Differentiator\nIdeal Integrator", - "type": "text" - }, - { - "block_id": "p437-b17", - "global_id": 12158, - "bbox": [ - 181.85, - 283.16, - 451.03, - 291.16 - ], - "text": "(a)\n(b)\n(c)", - "type": "text" - }, - { - "block_id": "p437-b18", - "global_id": 12159, - "bbox": [ - 409.93, - 217.12, - 436.7, - 225.42 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p437-b19", - "global_id": 12160, - "bbox": [ - 384.24, - 259.73, - 404.24, - 268.02 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p437-b20", - "global_id": 12161, - "bbox": [ - 128.9, - 309.48, - 456.9, - 318.79 - ], - "text": "Figure 4.39 Frequency response of an ideal (a) delay, (b) differentiator, and (c) integrator.", - "type": "text" - }, - { - "block_id": "p437-b21", - "global_id": 12162, - "bbox": [ - 128.9, - 354.62, - 502.79, - 424.36 - ], - "text": "In an ideal differentiator, the amplitude response (gain) is proportional to frequency\n[|H(jω)| = ω] so that the higher-frequency components are enhanced (see Fig. 4.39b). All\npractical signals are contaminated with noise, which, by its nature, is a broadband (rapidly\nvarying) signal containing components of very high frequencies. A differentiator can increase\nthe noise disproportionately to the point of drowning out the desired signal. This is why ideal\ndifferentiators are avoided in practice.", - "type": "text" - }, - { - "block_id": "p437-b22", - "global_id": 12163, - "bbox": [ - 146.84, - 426.27, - 489.06, - 436.32 - ], - "text": "(c) An ideal integrator. The transfer function of an ideal integrator is [see Eq. (4.32)]", - "type": "text" - }, - { - "block_id": "p437-b23", - "global_id": 12164, - "bbox": [ - 296.73, - 446.25, - 333.73, - 463.2 - ], - "text": "H(s) = 1", - "type": "text" - }, - { - "block_id": "p437-b24", - "global_id": 12165, - "bbox": [ - 329.3, - 460.21, - 333.17, - 470.17 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p437-b25", - "global_id": 12166, - "bbox": [ - 128.91, - 478.17, - 170.65, - 488.14 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p437-b26", - "global_id": 12167, - "bbox": [ - 256.98, - 487.39, - 301.85, - 504.34 - ], - "text": "H(jω) = 1", - "type": "text" - }, - { - "block_id": "p437-b27", - "global_id": 12168, - "bbox": [ - 294.61, - 486.98, - 328.9, - 511.31 - ], - "text": "jω = −j", - "type": "text" - }, - { - "block_id": "p437-b28", - "global_id": 12169, - "bbox": [ - 320.27, - 487.39, - 349.02, - 511.0 - ], - "text": "ω = 1", - "type": "text" - }, - { - "block_id": "p437-b29", - "global_id": 12170, - "bbox": [ - 343.17, - 492.24, - 374.18, - 511.0 - ], - "text": "ωe−jπ/2", - "type": "text" - }, - { - "block_id": "p437-b30", - "global_id": 12171, - "bbox": [ - 128.9, - 518.24, - 185.51, - 528.21 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p437-b31", - "global_id": 12172, - "bbox": [ - 233.49, - 528.18, - 282.67, - 545.13 - ], - "text": "|H(jω)| = 1", - "type": "text" - }, - { - "block_id": "p437-b32", - "global_id": 12173, - "bbox": [ - 276.82, - 527.77, - 395.98, - 551.79 - ], - "text": "ω\nand̸\nH(jω) = −π", - "type": "text" - }, - { - "block_id": "p437-b33", - "global_id": 12174, - "bbox": [ - 128.91, - 542.24, - 502.79, - 614.88 - ], - "text": "2\nThese amplitude and phase responses are illustrated in Fig. 4.39c. The amplitude response is\ninversely proportional to frequency, and the phase shift is constant (−π/2) with frequency.\nThis result can be explained physically by recognizing that if a sinusoid cosωt is passed\nthrough an ideal integrator, the output is (1/ω)sin ωt = (1/ω)cos[ωt −(π/2)]. Therefore, the\namplitude response is inversely proportional to ω, and the phase response is constant (−π/2)", - "type": "text" - } - ] - }, - { - "page_num": 438, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p438-b0", - "global_id": 12175, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "418\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p438-b1", - "global_id": 12176, - "bbox": [ - 103.16, - 83.46, - 477.03, - 132.91 - ], - "text": "with frequency.† Because its gain is 1/ω, the ideal integrator suppresses higher-frequency\ncomponents but enhances lower-frequency components with ω < 1. Consequently, noise\nsignals (if they do not contain an appreciable amount of very-low-frequency components) are\nsuppressed (smoothed out) by an integrator.", - "type": "text" - }, - { - "block_id": "p438-b2", - "global_id": 12177, - "bbox": [ - 107.82, - 199.79, - 407.41, - 211.75 - ], - "text": "DRILL 4.15\nSinusoidal Response of an LTIC System", - "type": "text" - }, - { - "block_id": "p438-b3", - "global_id": 12178, - "bbox": [ - 107.82, - 220.87, - 305.37, - 230.83 - ], - "text": "Find the response of an LTIC system specified by", - "type": "text" - }, - { - "block_id": "p438-b4", - "global_id": 12179, - "bbox": [ - 217.38, - 241.0, - 241.42, - 252.21 - ], - "text": "d2y(t)", - "type": "text" - }, - { - "block_id": "p438-b5", - "global_id": 12180, - "bbox": [ - 223.44, - 241.93, - 279.43, - 266.27 - ], - "text": "dt2\n+ 3dy(t)", - "type": "text" - }, - { - "block_id": "p438-b6", - "global_id": 12181, - "bbox": [ - 265.58, - 241.93, - 344.17, - 266.27 - ], - "text": "dt\n+ 2y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p438-b7", - "global_id": 12182, - "bbox": [ - 330.3, - 248.92, - 376.04, - 266.27 - ], - "text": "dt\n+ 5x(t)", - "type": "text" - }, - { - "block_id": "p438-b8", - "global_id": 12183, - "bbox": [ - 107.82, - 272.69, - 272.27, - 284.3 - ], - "text": "if the input is a sinusoid 20sin(3t + 35◦).", - "type": "text" - }, - { - "block_id": "p438-b9", - "global_id": 12184, - "bbox": [ - 108.09, - 297.81, - 162.68, - 308.77 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p438-b10", - "global_id": 12185, - "bbox": [ - 107.82, - 314.53, - 195.61, - 326.14 - ], - "text": "10.23sin(3t −61.91◦)", - "type": "text" - }, - { - "block_id": "p438-b11", - "global_id": 12186, - "bbox": [ - 101.84, - 383.07, - 410.78, - 395.03 - ], - "text": "4.8-1 Steady-State Response to Causal Sinusoidal Inputs", - "type": "text" - }, - { - "block_id": "p438-b12", - "global_id": 12187, - "bbox": [ - 101.84, - 401.16, - 490.4, - 446.99 - ], - "text": "So far we have discussed the LTIC system response to everlasting sinusoidal inputs (starting at\nt = −∞). In practice, we are more interested in causal sinusoidal inputs (sinusoids starting at\nt = 0). Consider the input ejωtu(t), which starts at t = 0 rather than at t = −∞. In this case\nX(s) = 1/(s + jω). Moreover, according to Eq. (4.27), H(s) = P(s)/Q(s), where Q(s) is the", - "type": "text" - }, - { - "block_id": "p438-b13", - "global_id": 12188, - "bbox": [ - 101.84, - 478.73, - 490.41, - 633.42 - ], - "text": "† A puzzling aspect of this result is that in deriving the transfer function of the integrator in Eq. (4.32), we\nhave assumed that the input starts at t = 0. In contrast, in deriving its frequency response, we assume that the\neverlasting exponential input ejωt starts at t = −∞. There appears to be a fundamental contradiction between\nthe everlasting input, which starts at t = −∞, and the integrator, which opens its gates only at t = 0. Of\nwhat use is everlasting input, since the integrator starts integrating at t = 0? The answer is that the integrator\ngates are always open, and integration begins whenever the input starts. We restricted the input to start at\nt = 0 in deriving Eq. (4.32) because we were finding the transfer function using the unilateral transform,\nwhere the inputs begin at t = 0. So the integrator starting to integrate at t = 0 is restricted because of the\nlimitations of the unilateral transform method, not because of the limitations of the integrator itself. If we\nwere to find the integrator transfer function using Eq. (2.40), where there is no such restriction on the input,\nwe would still find the transfer function of an integrator as 1/s. Similarly, even if we were to use the bilateral\nLaplace transform, where t starts at −∞, we would find the transfer function of an integrator to be 1/s.\nThe transfer function of a system is the property of the system and does not depend on the method used to\nfind it.", - "type": "text" - } - ] - }, - { - "page_num": 439, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p439-b0", - "global_id": 12189, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n419", - "type": "text" - }, - { - "block_id": "p439-b1", - "global_id": 12190, - "bbox": [ - 127.59, - 83.03, - 447.32, - 97.39 - ], - "text": "characteristic polynomial given by Q(s) = (s −λ1)(s −λ2)· · · (s −λN).† Hence,", - "type": "text" - }, - { - "block_id": "p439-b2", - "global_id": 12191, - "bbox": [ - 209.36, - 106.63, - 433.17, - 131.84 - ], - "text": "Y(s) = X(s)H(s) =\nP(s)\n(s −λ1)(s −λ2)· · · (s −λN)(s −jω)", - "type": "text" - }, - { - "block_id": "p439-b3", - "global_id": 12192, - "bbox": [ - 127.59, - 141.23, - 516.13, - 176.27 - ], - "text": "In the partial fraction expansion of the right-hand side, let the coefficients corresponding to the N\nterms (s −λ1), (s −λ2), . . . , (s −λN) be k1, k2, . . . , kN. The coefficient corresponding to the last\nterm (s −jω) is P(s)/Q(s)|s=jω = H(jω). Hence,", - "type": "text" - }, - { - "block_id": "p439-b4", - "global_id": 12193, - "bbox": [ - 268.09, - 195.87, - 295.53, - 206.14 - ], - "text": "Y(s) =", - "type": "text" - }, - { - "block_id": "p439-b5", - "global_id": 12194, - "bbox": [ - 297.58, - 185.91, - 311.67, - 196.36 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p439-b6", - "global_id": 12195, - "bbox": [ - 299.19, - 210.26, - 310.05, - 217.52 - ], - "text": "i=1", - "type": "text" - }, - { - "block_id": "p439-b7", - "global_id": 12196, - "bbox": [ - 313.97, - 189.2, - 336.11, - 214.02 - ], - "text": "ki\ns −λi", - "type": "text" - }, - { - "block_id": "p439-b8", - "global_id": 12197, - "bbox": [ - 339.35, - 188.88, - 374.43, - 206.14 - ], - "text": "+ H(jω)", - "type": "text" - }, - { - "block_id": "p439-b9", - "global_id": 12198, - "bbox": [ - 350.03, - 202.94, - 374.07, - 213.22 - ], - "text": "s −jω", - "type": "text" - }, - { - "block_id": "p439-b10", - "global_id": 12199, - "bbox": [ - 127.59, - 227.65, - 141.97, - 237.61 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p439-b11", - "global_id": 12200, - "bbox": [ - 236.04, - 245.72, - 260.67, - 255.99 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p439-b12", - "global_id": 12201, - "bbox": [ - 268.55, - 235.77, - 282.65, - 246.21 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p439-b13", - "global_id": 12202, - "bbox": [ - 270.17, - 260.11, - 281.03, - 267.37 - ], - "text": "i=1", - "type": "text" - }, - { - "block_id": "p439-b14", - "global_id": 12203, - "bbox": [ - 283.76, - 241.92, - 320.05, - 256.8 - ], - "text": "kieλit u(t)", - "type": "text" - }, - { - "block_id": "p439-b16", - "global_id": 12204, - "bbox": [ - 262.72, - 275.07, - 325.9, - 282.47 - ], - "text": "transient component ytr(t)", - "type": "text" - }, - { - "block_id": "p439-b17", - "global_id": 12205, - "bbox": [ - 327.0, - 241.92, - 407.68, - 271.04 - ], - "text": "+\nH(jω)ejωt u(t)\n\n\n\nsteady-state component yss(t)", - "type": "text" - }, - { - "block_id": "p439-b18", - "global_id": 12206, - "bbox": [ - 127.59, - 287.7, - 516.12, - 313.24 - ], - "text": "For an asymptotically stable system, the characteristic mode terms eλit decay with time, and,\ntherefore, constitute the so-called transient component of the response. The last term H(jω)ejωt", - "type": "text" - }, - { - "block_id": "p439-b19", - "global_id": 12207, - "bbox": [ - 127.59, - 315.13, - 427.43, - 325.19 - ], - "text": "persists forever, and is the steady-state component of the response given by", - "type": "text" - }, - { - "block_id": "p439-b20", - "global_id": 12208, - "bbox": [ - 278.78, - 335.45, - 364.94, - 348.32 - ], - "text": "yss(t) = H(jω)ejωtu(t)", - "type": "text" - }, - { - "block_id": "p439-b21", - "global_id": 12209, - "bbox": [ - 127.59, - 356.32, - 516.14, - 405.77 - ], - "text": "This result also explains why an everlasting exponential input ejωt results in the total response\nH(jω)ejωt for BIBO systems. Because the input started at t = −∞, at any finite time the decaying\ntransient component has long vanished, leaving only the steady-state component. Hence, the total\nresponse appears to be H(jω)ejωt.", - "type": "text" - }, - { - "block_id": "p439-b22", - "global_id": 12210, - "bbox": [ - 127.59, - 407.34, - 516.13, - 430.37 - ], - "text": "From the argument that led to Eq. (4.43), it follows that for a causal sinusoidal input cosωt,\nthe steady-state response yss(t) is given by", - "type": "text" - }, - { - "block_id": "p439-b23", - "global_id": 12211, - "bbox": [ - 246.31, - 441.65, - 397.41, - 452.8 - ], - "text": "yss(t) = |H(jω)|cos[ωt +̸ H(jω)]u(t)", - "type": "text" - }, - { - "block_id": "p439-b24", - "global_id": 12212, - "bbox": [ - 127.59, - 464.0, - 516.14, - 486.33 - ], - "text": "In summary, |H(jω)|cos[ωt +̸ H(jω)] is the total response to everlasting sinusoid cosωt. In\ncontrast, it is the steady-state response to the same input applied at t = 0.", - "type": "text" - }, - { - "block_id": "p439-b25", - "global_id": 12213, - "bbox": [ - 127.94, - 516.05, - 235.58, - 530.0 - ], - "text": "4.9 BODE PLOTS", - "type": "text" - }, - { - "block_id": "p439-b26", - "global_id": 12214, - "bbox": [ - 127.59, - 535.57, - 516.15, - 593.77 - ], - "text": "Sketching frequency response plots (|H(jω)| and̸\nH(jω) versus ω) is considerably facilitated by\nthe use of logarithmic scales. The amplitude and phase response plots as a function of ω on a\nlogarithmic scale are known as Bode plots. By using the asymptotic behavior of the amplitude and\nthe phase responses, we can sketch these plots with remarkable ease, even for higher-order transfer\nfunctions.", - "type": "text" - }, - { - "block_id": "p439-b27", - "global_id": 12215, - "bbox": [ - 127.59, - 610.24, - 516.1, - 633.41 - ], - "text": "† For simplicity, we have assumed nonrepeating characteristic roots. The procedure is readily modified for\nrepeated roots, and the same conclusion results.", - "type": "text" - } - ] - }, - { - "page_num": 440, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p440-b0", - "global_id": 12216, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "420\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p440-b1", - "global_id": 12217, - "bbox": [ - 119.78, - 85.82, - 321.22, - 95.78 - ], - "text": "Let us consider a system with the transfer function", - "type": "text" - }, - { - "block_id": "p440-b2", - "global_id": 12218, - "bbox": [ - 232.59, - 105.26, - 490.38, - 130.47 - ], - "text": "H(s) =\nK(s + a1)(s + a2)\ns(s + b1)(s2 + b2s + b3)\n(4.44)", - "type": "text" - }, - { - "block_id": "p440-b3", - "global_id": 12219, - "bbox": [ - 101.84, - 136.67, - 490.39, - 162.61 - ], - "text": "where the second-order factor (s2 + b2s + b3) is assumed to have complex conjugate roots.† We\nshall rearrange Eq. (4.44) in the form", - "type": "text" - }, - { - "block_id": "p440-b4", - "global_id": 12220, - "bbox": [ - 210.39, - 186.59, - 266.47, - 203.53 - ], - "text": "H(s) = Ka1a2", - "type": "text" - }, - { - "block_id": "p440-b5", - "global_id": 12221, - "bbox": [ - 245.72, - 200.65, - 263.15, - 211.48 - ], - "text": "b1b3", - "type": "text" - }, - { - "block_id": "p440-b6", - "global_id": 12222, - "bbox": [ - 284.36, - 165.71, - 298.69, - 182.99 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p440-b7", - "global_id": 12223, - "bbox": [ - 292.27, - 187.08, - 300.74, - 197.92 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p440-b8", - "global_id": 12224, - "bbox": [ - 303.98, - 179.69, - 318.28, - 190.07 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p440-b9", - "global_id": 12225, - "bbox": [ - 318.28, - 165.71, - 339.35, - 182.99 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p440-b10", - "global_id": 12226, - "bbox": [ - 332.93, - 187.08, - 341.39, - 197.92 - ], - "text": "a2", - "type": "text" - }, - { - "block_id": "p440-b11", - "global_id": 12227, - "bbox": [ - 344.64, - 179.69, - 358.93, - 190.07 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p440-b13", - "global_id": 12228, - "bbox": [ - 269.36, - 208.03, - 273.24, - 218.0 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p440-b14", - "global_id": 12229, - "bbox": [ - 273.24, - 193.74, - 287.58, - 211.01 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p440-b15", - "global_id": 12230, - "bbox": [ - 281.16, - 215.11, - 289.62, - 225.94 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p440-b16", - "global_id": 12231, - "bbox": [ - 292.87, - 207.72, - 307.17, - 218.1 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p440-b17", - "global_id": 12232, - "bbox": [ - 307.16, - 193.74, - 329.74, - 211.01 - ], - "text": "s2", - "type": "text" - }, - { - "block_id": "p440-b18", - "global_id": 12233, - "bbox": [ - 321.81, - 215.11, - 330.28, - 225.94 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p440-b19", - "global_id": 12234, - "bbox": [ - 333.53, - 201.05, - 352.51, - 218.0 - ], - "text": "+ b2", - "type": "text" - }, - { - "block_id": "p440-b20", - "global_id": 12235, - "bbox": [ - 344.04, - 215.11, - 352.51, - 225.94 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p440-b21", - "global_id": 12236, - "bbox": [ - 354.2, - 207.72, - 373.93, - 218.1 - ], - "text": "s + 1", - "type": "text" - }, - { - "block_id": "p440-b23", - "global_id": 12237, - "bbox": [ - 101.84, - 235.24, - 116.22, - 245.21 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p440-b24", - "global_id": 12238, - "bbox": [ - 197.01, - 257.26, - 258.72, - 274.2 - ], - "text": "H(jω) = Ka1a2", - "type": "text" - }, - { - "block_id": "p440-b25", - "global_id": 12239, - "bbox": [ - 237.96, - 271.31, - 255.4, - 282.15 - ], - "text": "b1b3", - "type": "text" - }, - { - "block_id": "p440-b27", - "global_id": 12240, - "bbox": [ - 293.35, - 243.39, - 319.7, - 268.59 - ], - "text": "1 + jω\na1", - "type": "text" - }, - { - "block_id": "p440-b29", - "global_id": 12241, - "bbox": [ - 334.54, - 243.39, - 360.89, - 268.59 - ], - "text": "1 + jω\na2", - "type": "text" - }, - { - "block_id": "p440-b31", - "global_id": 12242, - "bbox": [ - 261.61, - 278.39, - 270.91, - 288.66 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p440-b33", - "global_id": 12243, - "bbox": [ - 277.83, - 271.41, - 304.17, - 296.61 - ], - "text": "1 + jω\nb1", - "type": "text" - }, - { - "block_id": "p440-b35", - "global_id": 12244, - "bbox": [ - 317.72, - 271.4, - 353.03, - 296.61 - ], - "text": "1 + jb2ω\nb3", - "type": "text" - }, - { - "block_id": "p440-b36", - "global_id": 12245, - "bbox": [ - 355.98, - 270.47, - 386.9, - 288.35 - ], - "text": "+ (jω)2", - "type": "text" - }, - { - "block_id": "p440-b37", - "global_id": 12246, - "bbox": [ - 372.46, - 285.78, - 380.93, - 296.61 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p440-b38", - "global_id": 12247, - "bbox": [ - 388.59, - 264.4, - 394.02, - 274.37 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p440-b39", - "global_id": 12248, - "bbox": [ - 101.84, - 302.62, - 490.38, - 324.95 - ], - "text": "This equation shows that H(jω) is a complex function of ω. The amplitude response |H(jω)| and\nthe phase response̸\nH(jω) are given by", - "type": "text" - }, - { - "block_id": "p440-b40", - "global_id": 12249, - "bbox": [ - 193.2, - 355.59, - 233.28, - 365.87 - ], - "text": "|H(jω)| =", - "type": "text" - }, - { - "block_id": "p440-b42", - "global_id": 12250, - "bbox": [ - 239.77, - 348.92, - 263.84, - 359.76 - ], - "text": "Ka1a2", - "type": "text" - }, - { - "block_id": "p440-b43", - "global_id": 12251, - "bbox": [ - 243.09, - 362.98, - 260.52, - 373.81 - ], - "text": "b1b3", - "type": "text" - }, - { - "block_id": "p440-b45", - "global_id": 12252, - "bbox": [ - 300.25, - 327.59, - 329.83, - 355.48 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p440-b46", - "global_id": 12253, - "bbox": [ - 320.8, - 349.42, - 329.26, - 360.26 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p440-b48", - "global_id": 12254, - "bbox": [ - 334.46, - 327.59, - 364.04, - 355.48 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p440-b49", - "global_id": 12255, - "bbox": [ - 355.01, - 349.42, - 363.47, - 360.26 - ], - "text": "a2", - "type": "text" - }, - { - "block_id": "p440-b51", - "global_id": 12256, - "bbox": [ - 271.08, - 370.05, - 286.28, - 380.33 - ], - "text": "|jω|", - "type": "text" - }, - { - "block_id": "p440-b52", - "global_id": 12257, - "bbox": [ - 286.28, - 355.61, - 315.85, - 383.51 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p440-b53", - "global_id": 12258, - "bbox": [ - 306.82, - 377.44, - 315.28, - 388.28 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p440-b55", - "global_id": 12259, - "bbox": [ - 320.49, - 355.61, - 359.03, - 383.51 - ], - "text": "1 + jb2ω", - "type": "text" - }, - { - "block_id": "p440-b56", - "global_id": 12260, - "bbox": [ - 346.9, - 377.44, - 355.37, - 388.28 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p440-b57", - "global_id": 12261, - "bbox": [ - 361.98, - 362.13, - 392.91, - 380.02 - ], - "text": "+ (jω)2", - "type": "text" - }, - { - "block_id": "p440-b58", - "global_id": 12262, - "bbox": [ - 378.47, - 377.44, - 386.93, - 388.28 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p440-b60", - "global_id": 12263, - "bbox": [ - 466.32, - 356.0, - 490.38, - 365.97 - ], - "text": "(4.45)", - "type": "text" - }, - { - "block_id": "p440-b61", - "global_id": 12264, - "bbox": [ - 101.85, - 397.57, - 116.22, - 407.54 - ], - "text": "and̸", - "type": "text" - }, - { - "block_id": "p440-b62", - "global_id": 12265, - "bbox": [ - 189.76, - 422.57, - 224.14, - 432.84 - ], - "text": "H(jω) ≠", - "type": "text" - }, - { - "block_id": "p440-b63", - "global_id": 12266, - "bbox": [ - 231.82, - 408.58, - 263.81, - 426.73 - ], - "text": "Ka1a2", - "type": "text" - }, - { - "block_id": "p440-b64", - "global_id": 12267, - "bbox": [ - 243.06, - 429.96, - 260.49, - 440.79 - ], - "text": "b1b3", - "type": "text" - }, - { - "block_id": "p440-b66", - "global_id": 12268, - "bbox": [ - 273.79, - 422.57, - 281.56, - 432.53 - ], - "text": "+̸", - "type": "text" - }, - { - "block_id": "p440-b68", - "global_id": 12269, - "bbox": [ - 295.47, - 415.58, - 321.81, - 440.79 - ], - "text": "1 + jω\na1", - "type": "text" - }, - { - "block_id": "p440-b70", - "global_id": 12270, - "bbox": [ - 331.48, - 422.57, - 339.25, - 432.53 - ], - "text": "+̸", - "type": "text" - }, - { - "block_id": "p440-b72", - "global_id": 12271, - "bbox": [ - 353.16, - 415.58, - 379.5, - 440.79 - ], - "text": "1 + jω\na2", - "type": "text" - }, - { - "block_id": "p440-b74", - "global_id": 12272, - "bbox": [ - 227.74, - 451.37, - 261.51, - 461.64 - ], - "text": "−̸ jω −̸", - "type": "text" - }, - { - "block_id": "p440-b76", - "global_id": 12273, - "bbox": [ - 275.42, - 444.38, - 301.76, - 469.59 - ], - "text": "1 + jω\nb1", - "type": "text" - }, - { - "block_id": "p440-b78", - "global_id": 12274, - "bbox": [ - 311.43, - 451.37, - 319.2, - 461.33 - ], - "text": "−̸", - "type": "text" - }, - { - "block_id": "p440-b80", - "global_id": 12275, - "bbox": [ - 331.81, - 444.38, - 367.12, - 469.59 - ], - "text": "1 + jb2ω\nb3", - "type": "text" - }, - { - "block_id": "p440-b81", - "global_id": 12276, - "bbox": [ - 370.06, - 443.45, - 400.99, - 461.33 - ], - "text": "+ (jω)2", - "type": "text" - }, - { - "block_id": "p440-b82", - "global_id": 12277, - "bbox": [ - 386.55, - 458.76, - 395.01, - 469.59 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p440-b83", - "global_id": 12278, - "bbox": [ - 402.69, - 437.38, - 408.12, - 447.35 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p440-b84", - "global_id": 12279, - "bbox": [ - 466.32, - 451.78, - 490.38, - 461.74 - ], - "text": "(4.46)", - "type": "text" - }, - { - "block_id": "p440-b85", - "global_id": 12280, - "bbox": [ - 101.84, - 478.89, - 490.39, - 512.76 - ], - "text": "From Eq. (4.46) we see that the phase function consists of the addition of terms of four kinds: (i)\nthe phase of a constant, (ii) the phase of jω, which is 90◦for all values of ω, (iii) the phase for the\nfirst-order term of the form 1 + jω/a, and (iv) the phase of the second-order term", - "type": "text" - }, - { - "block_id": "p440-b87", - "global_id": 12281, - "bbox": [ - 260.68, - 523.76, - 295.98, - 548.97 - ], - "text": "1 + jb2ω\nb3", - "type": "text" - }, - { - "block_id": "p440-b88", - "global_id": 12282, - "bbox": [ - 298.93, - 522.83, - 329.86, - 540.71 - ], - "text": "+ (jω)2", - "type": "text" - }, - { - "block_id": "p440-b89", - "global_id": 12283, - "bbox": [ - 315.42, - 538.14, - 323.89, - 548.97 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p440-b90", - "global_id": 12284, - "bbox": [ - 331.55, - 516.76, - 336.98, - 526.73 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p440-b91", - "global_id": 12285, - "bbox": [ - 101.84, - 557.86, - 490.41, - 592.15 - ], - "text": "We can plot these basic phase functions for ω in the range 0 to ∞and then, using these plots, we\ncan construct the phase function of any transfer function by properly adding these basic responses.\nNote that if a particular term is in the numerator, its phase is added, but if the term is in the", - "type": "text" - }, - { - "block_id": "p440-b92", - "global_id": 12286, - "bbox": [ - 101.84, - 610.24, - 490.4, - 633.41 - ], - "text": "† Coefficients a1, a2 and b1, b2, b3 used in this section are not to be confused with those used in the\nrepresentation of Nth-order LTIC system equations given earlier [Eqs. (2.1) or (4.26)].", - "type": "text" - } - ] - }, - { - "page_num": 441, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p441-b0", - "global_id": 12287, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n421", - "type": "text" - }, - { - "block_id": "p441-b1", - "global_id": 12288, - "bbox": [ - 127.59, - 85.46, - 516.14, - 131.71 - ], - "text": "denominator, its phase is subtracted. This makes it easy to plot the phase function̸\nH(jω) as a\nfunction of ω. Computation of |H(jω)|, unlike that of the phase function, however, involves the\nmultiplication and division of various terms. This is a formidable task, especially when we have\nto plot this function for the entire range of ω (0 to ∞).", - "type": "text" - }, - { - "block_id": "p441-b2", - "global_id": 12289, - "bbox": [ - 127.59, - 133.7, - 516.14, - 263.21 - ], - "text": "We know that a log operation converts multiplication and division to addition and subtraction.\nSo, instead of plotting |H(jω)|, why not plot log |H(jω)| to simplify our task? We can take\nadvantage of the fact that logarithmic units are desirable in several applications, where the\nvariables considered have a very large range of variation. This is particularly true in frequency\nresponse plots, where we may have to plot frequency response over a range from a very low\nfrequency, near 0, to a very high frequency, in the range of 1010 or higher. A plot on a linear\nscale of frequencies for such a large range will bury much of the useful information at lower\nfrequencies. Also, the amplitude response may have a very large dynamic range from a low of\n10−6 to a high of 106. A linear plot would be unsuitable for such a situation. Therefore, logarithmic\nplots not only simplify our task of plotting, but, fortunately, they are also desirable in this\nsituation.", - "type": "text" - }, - { - "block_id": "p441-b3", - "global_id": 12290, - "bbox": [ - 127.59, - 265.2, - 516.14, - 334.95 - ], - "text": "There is another important reason for using logarithmic scale. The Weber–Fechner law (first\nobserved by Weber in 1834) states that human senses (sight, touch, hearing, etc.) generally\nrespond in a logarithmic way. For instance, when we hear sound at two different power levels,\nwe judge one sound twice as loud when the ratio of the two sound powers is 10. Human senses\nrespond to equal ratios of power, not equal increments in power [10]. This is clearly a logarithmic\nresponse.†", - "type": "text" - }, - { - "block_id": "p441-b4", - "global_id": 12291, - "bbox": [ - 127.59, - 336.83, - 516.11, - 360.95 - ], - "text": "The logarithmic unit is the decibel and is equal to 20 times the logarithm of the quantity\n(log to the base 10). Therefore, 20log10 |H(jω)| is simply the log amplitude in decibels (dB).‡", - "type": "text" - }, - { - "block_id": "p441-b5", - "global_id": 12292, - "bbox": [ - 127.59, - 360.43, - 516.14, - 394.72 - ], - "text": "Thus, instead of plotting |H(jω)|, we shall plot 20log10 |H(jω)| as a function of ω. These plots\n(log amplitude and phase) are called Bode plots. For the transfer function in Eq. (4.45), the log\namplitude is", - "type": "text" - }, - { - "block_id": "p441-b6", - "global_id": 12293, - "bbox": [ - 164.79, - 401.67, - 259.98, - 429.57 - ], - "text": "20log|H(jω)| = 20log", - "type": "text" - }, - { - "block_id": "p441-b7", - "global_id": 12294, - "bbox": [ - 261.18, - 409.44, - 285.25, - 420.28 - ], - "text": "Ka1a2", - "type": "text" - }, - { - "block_id": "p441-b8", - "global_id": 12295, - "bbox": [ - 264.5, - 423.5, - 281.93, - 434.34 - ], - "text": "b1b3", - "type": "text" - }, - { - "block_id": "p441-b9", - "global_id": 12296, - "bbox": [ - 286.95, - 401.67, - 324.84, - 429.57 - ], - "text": "+ 20log", - "type": "text" - }, - { - "block_id": "p441-b10", - "global_id": 12297, - "bbox": [ - 324.85, - 401.67, - 354.44, - 429.57 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p441-b11", - "global_id": 12298, - "bbox": [ - 345.4, - 423.5, - 353.87, - 434.34 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p441-b12", - "global_id": 12299, - "bbox": [ - 355.83, - 401.67, - 393.72, - 429.57 - ], - "text": "+ 20log", - "type": "text" - }, - { - "block_id": "p441-b13", - "global_id": 12300, - "bbox": [ - 393.74, - 401.67, - 423.32, - 429.57 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p441-b14", - "global_id": 12301, - "bbox": [ - 414.29, - 423.5, - 422.75, - 434.34 - ], - "text": "a2", - "type": "text" - }, - { - "block_id": "p441-b15", - "global_id": 12302, - "bbox": [ - 424.72, - 401.67, - 478.93, - 429.57 - ], - "text": "−20log|jω|", - "type": "text" - }, - { - "block_id": "p441-b16", - "global_id": 12303, - "bbox": [ - 233.39, - 444.91, - 266.49, - 455.29 - ], - "text": "−20log", - "type": "text" - }, - { - "block_id": "p441-b17", - "global_id": 12304, - "bbox": [ - 266.51, - 430.47, - 296.09, - 458.36 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p441-b18", - "global_id": 12305, - "bbox": [ - 287.05, - 452.3, - 295.52, - 463.14 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p441-b19", - "global_id": 12306, - "bbox": [ - 297.48, - 430.47, - 335.37, - 458.36 - ], - "text": "−20log", - "type": "text" - }, - { - "block_id": "p441-b20", - "global_id": 12307, - "bbox": [ - 335.39, - 430.47, - 373.94, - 458.36 - ], - "text": "1 + jb2ω", - "type": "text" - }, - { - "block_id": "p441-b21", - "global_id": 12308, - "bbox": [ - 360.42, - 452.3, - 368.89, - 463.14 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p441-b22", - "global_id": 12309, - "bbox": [ - 376.88, - 436.99, - 407.81, - 454.88 - ], - "text": "+ (jω)2", - "type": "text" - }, - { - "block_id": "p441-b23", - "global_id": 12310, - "bbox": [ - 393.37, - 452.3, - 401.84, - 463.14 - ], - "text": "b3", - "type": "text" - }, - { - "block_id": "p441-b24", - "global_id": 12311, - "bbox": [ - 409.51, - 430.47, - 516.13, - 458.36 - ], - "text": "(4.47)", - "type": "text" - }, - { - "block_id": "p441-b25", - "global_id": 12312, - "bbox": [ - 127.59, - 476.51, - 516.14, - 511.88 - ], - "text": "The term 20log(Ka1a2/b1b3) is a constant. We observe that the log amplitude is a sum of four\nbasic terms corresponding to a constant, a pole or zero at the origin (20log|jω|), a first-order pole\nor zero (20log|1+jω/a|), and complex-conjugate poles or zeros (20log|1+jωb2/b3 +(jω)2/b3|).", - "type": "text" - }, - { - "block_id": "p441-b26", - "global_id": 12313, - "bbox": [ - 127.59, - 533.22, - 516.15, - 635.59 - ], - "text": "† Observe that the frequencies of musical notes are spaced logarithmically (not linearly). The octave is a ratio\nof 2. The frequencies of the same note in the successive octaves have a ratio of 2. On the Western musical\nscale, there are 12 distinct notes in each octave. The frequency of each note is about 6% higher than the\nfrequency of the preceding note. Thus, the successive notes are separated not by some constant frequency,\nbut by constant ratio of 1.06.\n‡ Originally, the unit bel (after the inventor of telephone, Alexander Graham Bell) was introduced to\nrepresent power ratio as log10 P2/P1 bels. A tenth of this unit is a decibel, as in 10 log10 P2/P1 decibels.\nSince the power ratio of two signals is proportional to the amplitude ratio squared, or |H(jω)|2, we have\n10 log10 P2/P1 = 10 log10 |H(jω)|2 = 20 log10 |H(jω)| dB.", - "type": "text" - } - ] - }, - { - "page_num": 442, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p442-b0", - "global_id": 12314, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "422\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p442-b1", - "global_id": 12315, - "bbox": [ - 101.84, - 85.4, - 490.36, - 107.74 - ], - "text": "We can sketch these four basic terms as functions of ω and use them to construct the log-amplitude\nplot of any desired transfer function. Let us discuss each of the terms.", - "type": "text" - }, - { - "block_id": "p442-b2", - "global_id": 12316, - "bbox": [ - 101.84, - 132.16, - 241.44, - 145.64 - ], - "text": "4.9-1 Constant Ka1a2/b1b3", - "type": "text" - }, - { - "block_id": "p442-b3", - "global_id": 12317, - "bbox": [ - 101.84, - 150.69, - 490.38, - 184.98 - ], - "text": "The log amplitude of the constant Ka1a2/b1b2 term is also a constant, 20log|Ka1a2/b1b3|. The\nphase contribution from this term is zero for positive value and π for negative value of the constant\n(complex constants can have different phases).", - "type": "text" - }, - { - "block_id": "p442-b4", - "global_id": 12318, - "bbox": [ - 101.84, - 210.25, - 279.34, - 222.21 - ], - "text": "4.9-2 Pole (or Zero) at the Origin", - "type": "text" - }, - { - "block_id": "p442-b5", - "global_id": 12319, - "bbox": [ - 101.84, - 228.13, - 425.3, - 254.24 - ], - "text": "LOG MAGNITUDE\nA pole at the origin gives rise to the term −20log|jω|, which can be expressed as", - "type": "text" - }, - { - "block_id": "p442-b6", - "global_id": 12320, - "bbox": [ - 246.55, - 266.31, - 345.49, - 276.69 - ], - "text": "−20log|jω| = −20logω", - "type": "text" - }, - { - "block_id": "p442-b7", - "global_id": 12321, - "bbox": [ - 101.85, - 288.75, - 490.39, - 311.08 - ], - "text": "This function can be plotted as a function of ω. However, we can effect further simplification by\nusing the logarithmic scale for the variable ω itself. Let us define a new variable u such that", - "type": "text" - }, - { - "block_id": "p442-b8", - "global_id": 12322, - "bbox": [ - 277.41, - 323.15, - 314.62, - 333.52 - ], - "text": "u = logω", - "type": "text" - }, - { - "block_id": "p442-b9", - "global_id": 12323, - "bbox": [ - 101.84, - 346.0, - 129.76, - 355.97 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p442-b10", - "global_id": 12324, - "bbox": [ - 259.13, - 358.34, - 333.09, - 368.71 - ], - "text": "−20logω = −20u", - "type": "text" - }, - { - "block_id": "p442-b11", - "global_id": 12325, - "bbox": [ - 101.84, - 377.78, - 490.4, - 450.45 - ], - "text": "The log-amplitude function −20u is plotted as a function of u in Fig. 4.40a. This is a straight\nline with a slope of −20. It crosses the u axis at u = 0. The ω-scale (u = logω) also appears in\nFig. 4.40a. Semilog graphs can be conveniently used for plotting, and we can directly plot ω on\nsemilog paper. A ratio of 10 is a decade, and a ratio of 2 is known as an octave. Furthermore, a\ndecade along the ω scale is equivalent to 1 unit along the u scale. We can also show that a ratio of\n2 (an octave) along the ω scale equals to 0.3010 (which is log10 2) along the u scale.†", - "type": "text" - }, - { - "block_id": "p442-b12", - "global_id": 12326, - "bbox": [ - 101.84, - 466.97, - 490.38, - 491.49 - ], - "text": "† This point can be shown as follows. Let ω1 and ω2 along the ω scale correspond to u1 and u2 along the u\nscale so that logω1 = u1 and logω2 = u2. Then", - "type": "text" - }, - { - "block_id": "p442-b13", - "global_id": 12327, - "bbox": [ - 214.12, - 498.53, - 378.11, - 509.77 - ], - "text": "u2 −u1 = log10 ω2 −log10 ω1 = log10(ω2/ω1)", - "type": "text" - }, - { - "block_id": "p442-b14", - "global_id": 12328, - "bbox": [ - 101.84, - 516.64, - 129.74, - 525.6 - ], - "text": "Thus, if", - "type": "text" - }, - { - "block_id": "p442-b15", - "global_id": 12329, - "bbox": [ - 227.54, - 527.49, - 364.68, - 537.57 - ], - "text": "(ω2/ω1) = 10\n(which is a decade)", - "type": "text" - }, - { - "block_id": "p442-b16", - "global_id": 12330, - "bbox": [ - 101.84, - 543.59, - 117.28, - 552.56 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p442-b17", - "global_id": 12331, - "bbox": [ - 255.89, - 554.43, - 336.35, - 566.04 - ], - "text": "u2 −u1 = log10 10 = 1", - "type": "text" - }, - { - "block_id": "p442-b18", - "global_id": 12332, - "bbox": [ - 101.84, - 570.54, - 122.51, - 579.5 - ], - "text": "and if", - "type": "text" - }, - { - "block_id": "p442-b19", - "global_id": 12333, - "bbox": [ - 228.44, - 581.39, - 363.78, - 591.47 - ], - "text": "(ω2/ω1) = 2\n(which is an octave)", - "type": "text" - }, - { - "block_id": "p442-b20", - "global_id": 12334, - "bbox": [ - 101.84, - 597.49, - 117.28, - 606.46 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p442-b21", - "global_id": 12335, - "bbox": [ - 248.04, - 608.34, - 344.19, - 619.95 - ], - "text": "u2 −u1 = log10 2 = 0.3010", - "type": "text" - } - ] - }, - { - "page_num": 443, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p443-b0", - "global_id": 12336, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n423", - "type": "text" - }, - { - "block_id": "p443-b1", - "global_id": 12337, - "bbox": [ - 137.25, - 320.33, - 158.58, - 328.62 - ], - "text": "150", - "type": "text" - }, - { - "block_id": "p443-b2", - "global_id": 12338, - "bbox": [ - 141.25, - 297.88, - 158.58, - 315.62 - ], - "text": "90\n50", - "type": "text" - }, - { - "block_id": "p443-b3", - "global_id": 12339, - "bbox": [ - 147.91, - 266.02, - 158.58, - 294.53 - ], - "text": "0\n50\n90", - "type": "text" - }, - { - "block_id": "p443-b4", - "global_id": 12340, - "bbox": [ - 143.91, - 252.16, - 158.58, - 260.45 - ], - "text": "150", - "type": "text" - }, - { - "block_id": "p443-b5", - "global_id": 12341, - "bbox": [ - 129.39, - 281.3, - 137.39, - 299.97 - ], - "text": "Phase", - "type": "text" - }, - { - "block_id": "p443-b6", - "global_id": 12342, - "bbox": [ - 145.33, - 333.4, - 497.08, - 351.77 - ], - "text": "0.01\n0.1\nv 1\n10\n100\n50\n5\n0.5\n0.05\n(u 0)\n(u 1)\n(u 2)\n(u 1)\n(u 2)", - "type": "text" - }, - { - "block_id": "p443-b7", - "global_id": 12343, - "bbox": [ - 143.91, - 194.23, - 158.58, - 202.52 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p443-b8", - "global_id": 12344, - "bbox": [ - 143.91, - 177.53, - 158.58, - 185.83 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p443-b9", - "global_id": 12345, - "bbox": [ - 143.91, - 159.71, - 158.58, - 168.01 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p443-b10", - "global_id": 12346, - "bbox": [ - 154.58, - 142.19, - 158.58, - 150.19 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p443-b11", - "global_id": 12347, - "bbox": [ - 150.58, - 123.83, - 158.58, - 131.83 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p443-b12", - "global_id": 12348, - "bbox": [ - 150.58, - 106.01, - 158.58, - 114.01 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p443-b13", - "global_id": 12349, - "bbox": [ - 150.58, - 87.98, - 158.58, - 95.98 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p443-b14", - "global_id": 12350, - "bbox": [ - 154.53, - 201.52, - 491.51, - 210.83 - ], - "text": "0.01\n0.1\nv 1\n10\n100\n50\n5\n0.5\n0.05", - "type": "text" - }, - { - "block_id": "p443-b15", - "global_id": 12351, - "bbox": [ - 128.23, - 129.69, - 136.53, - 177.91 - ], - "text": "20 log H (dB)", - "type": "text" - }, - { - "block_id": "p443-b16", - "global_id": 12352, - "bbox": [ - 146.2, - 212.61, - 497.51, - 220.91 - ], - "text": "(u 0)\n(u 1)\n(u 2)\n(u 1)\n(u 2)", - "type": "text" - }, - { - "block_id": "p443-b17", - "global_id": 12353, - "bbox": [ - 494.64, - 148.19, - 499.98, - 156.19 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p443-b18", - "global_id": 12354, - "bbox": [ - 494.64, - 292.25, - 499.98, - 300.25 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p443-b19", - "global_id": 12355, - "bbox": [ - 319.79, - 227.82, - 328.67, - 235.82 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p443-b20", - "global_id": 12356, - "bbox": [ - 319.57, - 358.75, - 328.89, - 366.75 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p443-b21", - "global_id": 12357, - "bbox": [ - 127.59, - 373.36, - 431.9, - 382.68 - ], - "text": "Figure 4.40 (a) Amplitude and (b) phase responses of a pole or a zero at the origin.", - "type": "text" - }, - { - "block_id": "p443-b22", - "global_id": 12358, - "bbox": [ - 127.59, - 414.58, - 516.14, - 462.91 - ], - "text": "Note that equal increments in u are equivalent to equal ratios on the ω scale. Thus, 1 unit along\nthe u scale is the same as one decade along the ω scale. This means that the amplitude plot has a\nslope of −20 dB/decade or −20(0.3010) = −6.02 dB/octave (commonly stated as −6 dB/octave).\nMoreover, the amplitude plot crosses the ω axis at ω = 1, since u = log10 ω = 0 when ω = 1.", - "type": "text" - }, - { - "block_id": "p443-b23", - "global_id": 12359, - "bbox": [ - 127.59, - 462.39, - 516.14, - 496.68 - ], - "text": "For the case of a zero at the origin, the log-amplitude term is 20 log ω. This is a straight line\npassing through ω = 1 and having a slope of 20 dB/decade (or 6 dB/octave). This plot is a mirror\nimage about the ω axis of the plot for a pole at the origin and is shown dashed in Fig. 4.40a.", - "type": "text" - }, - { - "block_id": "p443-b24", - "global_id": 12360, - "bbox": [ - 127.59, - 517.32, - 485.02, - 543.43 - ], - "text": "PHASE\nThe phase function corresponding to the pole at the origin is −̸ jω [see Eq. (4.46)]. Thus,̸", - "type": "text" - }, - { - "block_id": "p443-b25", - "global_id": 12361, - "bbox": [ - 278.22, - 565.05, - 370.62, - 577.15 - ], - "text": "H(jω) = −̸ jω = −90◦", - "type": "text" - }, - { - "block_id": "p443-b26", - "global_id": 12362, - "bbox": [ - 127.59, - 599.27, - 516.13, - 634.79 - ], - "text": "The phase is constant (−90◦) for all values of ω, as depicted in Fig. 4.40b. For a zero at the origin,\nthe phase is̸\njω = 90◦. This is a mirror image of the phase plot for a pole at the origin and is\nshown dashed in Fig. 4.40b.", - "type": "text" - } - ] - }, - { - "page_num": 444, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p444-b0", - "global_id": 12363, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "424\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p444-b1", - "global_id": 12364, - "bbox": [ - 101.84, - 86.52, - 270.69, - 98.48 - ], - "text": "4.9-3 First-Order Pole (or Zero)", - "type": "text" - }, - { - "block_id": "p444-b2", - "global_id": 12365, - "bbox": [ - 101.84, - 104.4, - 490.38, - 142.47 - ], - "text": "THE LOG MAGNITUDE\nThe log amplitude of a first-order pole at −a is −20log|1+jω/a|. Let us investigate the asymptotic\nbehavior of this function for extreme values of ω (ω ≪a and ω ≫a).", - "type": "text" - }, - { - "block_id": "p444-b3", - "global_id": 12366, - "bbox": [ - 101.85, - 150.02, - 162.72, - 160.4 - ], - "text": "(a) For ω ≪a,", - "type": "text" - }, - { - "block_id": "p444-b4", - "global_id": 12367, - "bbox": [ - 238.34, - 167.19, - 269.9, - 177.57 - ], - "text": "−20log", - "type": "text" - }, - { - "block_id": "p444-b5", - "global_id": 12368, - "bbox": [ - 269.91, - 152.75, - 299.5, - 180.65 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p444-b6", - "global_id": 12369, - "bbox": [ - 292.45, - 174.58, - 297.43, - 184.54 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p444-b7", - "global_id": 12370, - "bbox": [ - 300.88, - 152.75, - 370.5, - 180.65 - ], - "text": "≈−20log1 = 0", - "type": "text" - }, - { - "block_id": "p444-b8", - "global_id": 12371, - "bbox": [ - 101.85, - 190.61, - 414.1, - 200.99 - ], - "text": "Hence, the log-amplitude function →0 asymptotically for ω ≪a (Fig. 4.41a).", - "type": "text" - }, - { - "block_id": "p444-b9", - "global_id": 12372, - "bbox": [ - 101.85, - 208.54, - 283.74, - 218.91 - ], - "text": "(a) For the other extreme case, where ω ≫a,", - "type": "text" - }, - { - "block_id": "p444-b10", - "global_id": 12373, - "bbox": [ - 154.84, - 234.06, - 186.39, - 244.44 - ], - "text": "−20log", - "type": "text" - }, - { - "block_id": "p444-b11", - "global_id": 12374, - "bbox": [ - 186.41, - 219.62, - 215.99, - 247.51 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p444-b12", - "global_id": 12375, - "bbox": [ - 208.95, - 241.45, - 213.93, - 251.41 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p444-b13", - "global_id": 12376, - "bbox": [ - 217.39, - 219.62, - 264.04, - 247.51 - ], - "text": "≈−20log", - "type": "text" - }, - { - "block_id": "p444-b14", - "global_id": 12377, - "bbox": [ - 264.06, - 220.08, - 278.51, - 237.04 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p444-b15", - "global_id": 12378, - "bbox": [ - 272.85, - 241.45, - 277.83, - 251.41 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p444-b17", - "global_id": 12379, - "bbox": [ - 288.67, - 234.06, - 454.0, - 244.44 - ], - "text": "= −20logω + 20loga = −20u + 20loga", - "type": "text" - }, - { - "block_id": "p444-b18", - "global_id": 12380, - "bbox": [ - 101.84, - 259.65, - 490.4, - 293.94 - ], - "text": "This represents a straight line (when plotted as a function of u, the log of ω) with a slope of −20\ndB/decade (or −6 dB/octave). When ω = a, the log amplitude is zero. Hence, this line crosses the\nω axis at ω = a, as illustrated in Fig. 4.41a. Note that the asymptotes in (a) and (b) meet at ω = a.", - "type": "text" - }, - { - "block_id": "p444-b19", - "global_id": 12381, - "bbox": [ - 119.78, - 295.93, - 275.97, - 305.89 - ], - "text": "The exact log amplitude for this pole is", - "type": "text" - }, - { - "block_id": "p444-b20", - "global_id": 12382, - "bbox": [ - 172.01, - 323.52, - 203.56, - 333.9 - ], - "text": "−20log", - "type": "text" - }, - { - "block_id": "p444-b21", - "global_id": 12383, - "bbox": [ - 203.58, - 309.07, - 233.16, - 336.97 - ], - "text": "1 + jω", - "type": "text" - }, - { - "block_id": "p444-b22", - "global_id": 12384, - "bbox": [ - 226.12, - 330.9, - 231.11, - 340.86 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p444-b23", - "global_id": 12385, - "bbox": [ - 234.56, - 309.07, - 281.22, - 336.97 - ], - "text": "= −20log", - "type": "text" - }, - { - "block_id": "p444-b25", - "global_id": 12386, - "bbox": [ - 287.95, - 315.6, - 315.21, - 333.9 - ], - "text": "1 + ω2", - "type": "text" - }, - { - "block_id": "p444-b26", - "global_id": 12387, - "bbox": [ - 305.87, - 330.4, - 314.34, - 340.86 - ], - "text": "a2", - "type": "text" - }, - { - "block_id": "p444-b27", - "global_id": 12388, - "bbox": [ - 316.91, - 309.53, - 333.88, - 321.42 - ], - "text": "1/2", - "type": "text" - }, - { - "block_id": "p444-b28", - "global_id": 12389, - "bbox": [ - 336.43, - 323.52, - 377.8, - 333.9 - ], - "text": "= −10log", - "type": "text" - }, - { - "block_id": "p444-b30", - "global_id": 12390, - "bbox": [ - 384.55, - 315.6, - 411.81, - 333.9 - ], - "text": "1 + ω2", - "type": "text" - }, - { - "block_id": "p444-b31", - "global_id": 12391, - "bbox": [ - 402.47, - 330.4, - 410.93, - 340.86 - ], - "text": "a2", - "type": "text" - }, - { - "block_id": "p444-b33", - "global_id": 12392, - "bbox": [ - 101.85, - 349.52, - 490.41, - 431.22 - ], - "text": "This exact log magnitude function also appears in Fig. 4.41a. Observe that the actual and the\nasymptotic plots are very close. A maximum error of 3 dB occurs at ω = a. This frequency is\nknown as the corner frequency or break frequency. The error everywhere else is less than 3 dB.\nA plot of the error as a function of ω is shown in Fig. 4.42a. This figure shows that the error at 1\noctave above or below the corner frequency is 1 dB and the error at 2 octaves above or below the\ncorner frequency is 0.3 dB. The actual plot can be obtained by adding the error to the asymptotic\nplot.", - "type": "text" - }, - { - "block_id": "p444-b34", - "global_id": 12393, - "bbox": [ - 101.85, - 432.79, - 490.4, - 467.08 - ], - "text": "The amplitude response for a zero at −a (shown dotted in Fig. 4.41a) is identical to that of\nthe pole at −a with a sign change and therefore is the mirror image (about the 0 dB line) of the\namplitude plot for a pole at −a.", - "type": "text" - }, - { - "block_id": "p444-b35", - "global_id": 12394, - "bbox": [ - 102.14, - 481.41, - 138.0, - 493.53 - ], - "text": "PHASE", - "type": "text" - }, - { - "block_id": "p444-b36", - "global_id": 12395, - "bbox": [ - 101.84, - 497.15, - 267.67, - 507.53 - ], - "text": "The phase for the first-order pole at −a is̸", - "type": "text" - }, - { - "block_id": "p444-b37", - "global_id": 12396, - "bbox": [ - 220.97, - 523.09, - 265.17, - 533.37 - ], - "text": "H(jω) = −̸", - "type": "text" - }, - { - "block_id": "p444-b39", - "global_id": 12397, - "bbox": [ - 277.53, - 516.11, - 303.88, - 540.44 - ], - "text": "1 + jω\na", - "type": "text" - }, - { - "block_id": "p444-b41", - "global_id": 12398, - "bbox": [ - 314.04, - 521.37, - 353.83, - 533.47 - ], - "text": "= −tan−1", - "type": "text" - }, - { - "block_id": "p444-b42", - "global_id": 12399, - "bbox": [ - 354.34, - 509.1, - 368.78, - 526.07 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p444-b43", - "global_id": 12400, - "bbox": [ - 363.13, - 530.48, - 368.11, - 540.44 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p444-b45", - "global_id": 12401, - "bbox": [ - 101.84, - 548.69, - 382.2, - 559.06 - ], - "text": "Let us investigate the asymptotic behavior of this function. For ω ≪a,", - "type": "text" - }, - { - "block_id": "p444-b46", - "global_id": 12402, - "bbox": [ - 261.17, - 572.9, - 291.14, - 585.01 - ], - "text": "−tan−1", - "type": "text" - }, - { - "block_id": "p444-b47", - "global_id": 12403, - "bbox": [ - 291.64, - 560.64, - 306.09, - 577.61 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p444-b48", - "global_id": 12404, - "bbox": [ - 300.44, - 582.01, - 305.42, - 591.97 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p444-b50", - "global_id": 12405, - "bbox": [ - 316.26, - 574.63, - 331.06, - 585.01 - ], - "text": "≈0", - "type": "text" - }, - { - "block_id": "p444-b51", - "global_id": 12406, - "bbox": [ - 101.84, - 600.22, - 163.41, - 610.6 - ], - "text": "and, for ω ≫a,", - "type": "text" - }, - { - "block_id": "p444-b52", - "global_id": 12407, - "bbox": [ - 252.8, - 615.37, - 282.77, - 627.48 - ], - "text": "−tan−1", - "type": "text" - }, - { - "block_id": "p444-b53", - "global_id": 12408, - "bbox": [ - 283.28, - 603.11, - 297.73, - 620.08 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p444-b54", - "global_id": 12409, - "bbox": [ - 292.08, - 624.49, - 297.06, - 634.45 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p444-b56", - "global_id": 12410, - "bbox": [ - 307.89, - 615.37, - 338.92, - 627.48 - ], - "text": "≈−90◦", - "type": "text" - } - ] - }, - { - "page_num": 445, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p445-b0", - "global_id": 12411, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n425", - "type": "text" - }, - { - "block_id": "p445-b1", - "global_id": 12412, - "bbox": [ - 136.73, - 240.65, - 151.39, - 248.95 - ], - "text": "18", - "type": "text" - }, - { - "block_id": "p445-b2", - "global_id": 12413, - "bbox": [ - 136.73, - 216.29, - 151.39, - 224.58 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p445-b3", - "global_id": 12414, - "bbox": [ - 140.73, - 191.92, - 151.39, - 200.21 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p445-b4", - "global_id": 12415, - "bbox": [ - 147.39, - 168.28, - 151.39, - 176.28 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p445-b5", - "global_id": 12416, - "bbox": [ - 147.39, - 143.18, - 151.39, - 151.18 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p445-b6", - "global_id": 12417, - "bbox": [ - 143.39, - 118.81, - 151.39, - 126.81 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p445-b7", - "global_id": 12418, - "bbox": [ - 143.39, - 94.14, - 151.39, - 102.14 - ], - "text": "18", - "type": "text" - }, - { - "block_id": "p445-b8", - "global_id": 12419, - "bbox": [ - 123.58, - 146.08, - 131.88, - 194.93 - ], - "text": "20 log H (dB)", - "type": "text" - }, - { - "block_id": "p445-b9", - "global_id": 12420, - "bbox": [ - 337.08, - 119.73, - 350.41, - 127.73 - ], - "text": "1dB", - "type": "text" - }, - { - "block_id": "p445-b10", - "global_id": 12421, - "bbox": [ - 336.83, - 213.48, - 350.16, - 221.48 - ], - "text": "1dB", - "type": "text" - }, - { - "block_id": "p445-b11", - "global_id": 12422, - "bbox": [ - 222.62, - 143.73, - 406.03, - 159.98 - ], - "text": "Asymptotes\nAsymptote", - "type": "text" - }, - { - "block_id": "p445-b12", - "global_id": 12423, - "bbox": [ - 313.83, - 272.16, - 322.71, - 280.16 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p445-b13", - "global_id": 12424, - "bbox": [ - 311.36, - 137.92, - 324.7, - 145.92 - ], - "text": "3dB", - "type": "text" - }, - { - "block_id": "p445-b14", - "global_id": 12425, - "bbox": [ - 287.02, - 188.98, - 324.71, - 205.83 - ], - "text": "3dB\n1dB", - "type": "text" - }, - { - "block_id": "p445-b15", - "global_id": 12426, - "bbox": [ - 286.9, - 146.14, - 300.24, - 154.14 - ], - "text": "1dB", - "type": "text" - }, - { - "block_id": "p445-b16", - "global_id": 12427, - "bbox": [ - 300.91, - 234.15, - 314.24, - 240.37 - ], - "text": "s a", - "type": "text" - }, - { - "block_id": "p445-b17", - "global_id": 12428, - "bbox": [ - 306.08, - 227.29, - 309.08, - 233.29 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p445-b18", - "global_id": 12429, - "bbox": [ - 145.45, - 257.93, - 490.06, - 266.82 - ], - "text": "v a\n10a\n100a\n0.1a\n0.01a", - "type": "text" - }, - { - "block_id": "p445-b19", - "global_id": 12430, - "bbox": [ - 145.16, - 472.17, - 489.39, - 481.06 - ], - "text": "v a\n10a\n100a\n0.1a\n0.01a", - "type": "text" - }, - { - "block_id": "p445-b20", - "global_id": 12431, - "bbox": [ - 312.94, - 485.74, - 322.59, - 493.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p445-b21", - "global_id": 12432, - "bbox": [ - 337.52, - 340.45, - 372.63, - 348.45 - ], - "text": "Asymptote", - "type": "text" - }, - { - "block_id": "p445-b22", - "global_id": 12433, - "bbox": [ - 350.52, - 417.73, - 385.63, - 425.73 - ], - "text": "Asymptote", - "type": "text" - }, - { - "block_id": "p445-b23", - "global_id": 12434, - "bbox": [ - 217.14, - 356.88, - 252.25, - 364.88 - ], - "text": "Asymptote", - "type": "text" - }, - { - "block_id": "p445-b24", - "global_id": 12435, - "bbox": [ - 144.4, - 377.76, - 151.06, - 386.06 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p445-b25", - "global_id": 12436, - "bbox": [ - 119.36, - 377.84, - 127.36, - 396.5 - ], - "text": "Phase", - "type": "text" - }, - { - "block_id": "p445-b26", - "global_id": 12437, - "bbox": [ - 133.74, - 418.35, - 151.06, - 426.64 - ], - "text": "45", - "type": "text" - }, - { - "block_id": "p445-b27", - "global_id": 12438, - "bbox": [ - 133.74, - 461.7, - 151.06, - 470.0 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p445-b28", - "global_id": 12439, - "bbox": [ - 140.4, - 336.32, - 151.06, - 344.61 - ], - "text": "45", - "type": "text" - }, - { - "block_id": "p445-b29", - "global_id": 12440, - "bbox": [ - 140.4, - 293.69, - 151.06, - 301.99 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p445-b30", - "global_id": 12441, - "bbox": [ - 501.39, - 175.06, - 506.73, - 183.06 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p445-b31", - "global_id": 12442, - "bbox": [ - 501.2, - 386.98, - 506.54, - 394.98 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p445-b32", - "global_id": 12443, - "bbox": [ - 283.94, - 101.57, - 314.83, - 109.86 - ], - "text": "For s a", - "type": "text" - }, - { - "block_id": "p445-b33", - "global_id": 12444, - "bbox": [ - 286.23, - 229.98, - 297.34, - 237.98 - ], - "text": "For", - "type": "text" - }, - { - "block_id": "p445-b34", - "global_id": 12445, - "bbox": [ - 305.74, - 307.22, - 336.63, - 315.51 - ], - "text": "For s a", - "type": "text" - }, - { - "block_id": "p445-b35", - "global_id": 12446, - "bbox": [ - 319.12, - 453.07, - 332.45, - 459.29 - ], - "text": "s a", - "type": "text" - }, - { - "block_id": "p445-b36", - "global_id": 12447, - "bbox": [ - 304.45, - 446.2, - 327.29, - 456.9 - ], - "text": "1\nFor", - "type": "text" - }, - { - "block_id": "p445-b37", - "global_id": 12448, - "bbox": [ - 127.59, - 505.74, - 455.11, - 515.35 - ], - "text": "Figure 4.41 (a) Amplitude and (b) phase responses of a first-order pole or zero at s = −a.", - "type": "text" - }, - { - "block_id": "p445-b38", - "global_id": 12449, - "bbox": [ - 127.59, - 553.1, - 516.14, - 634.79 - ], - "text": "The actual plot along with the asymptotes is depicted in Fig. 4.41b. In this case, we use a three-line\nsegment asymptotic plot for greater accuracy. The asymptotes are a phase angle of 0◦for ω ≤a/10,\na phase angle of −90◦for ω ≥10a, and a straight line with a slope −45◦/decade connecting\nthese two asymptotes (from ω = a/10 to 10a) crossing the ω axis at ω = a/10. It can be seen\nfrom Fig. 4.41b that the asymptotes are very close to the curve and the maximum error is 5.7◦.\nFigure 4.42b plots the error as a function of ω; the actual plot can be obtained by adding the error\nto the asymptotic plot.", - "type": "text" - } - ] - }, - { - "page_num": 446, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p446-b0", - "global_id": 12450, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "426\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p446-b1", - "global_id": 12451, - "bbox": [ - 282.62, - 205.44, - 302.62, - 213.73 - ], - "text": "v a", - "type": "text" - }, - { - "block_id": "p446-b2", - "global_id": 12452, - "bbox": [ - 282.92, - 375.93, - 302.92, - 384.23 - ], - "text": "v a", - "type": "text" - }, - { - "block_id": "p446-b3", - "global_id": 12453, - "bbox": [ - 121.67, - 206.93, - 462.97, - 215.01 - ], - "text": "5a\n10a\n0.2a\n0.1a\n2a\n0.5a", - "type": "text" - }, - { - "block_id": "p446-b4", - "global_id": 12454, - "bbox": [ - 102.54, - 130.05, - 110.54, - 163.59 - ], - "text": "Error (dB)", - "type": "text" - }, - { - "block_id": "p446-b5", - "global_id": 12455, - "bbox": [ - 118.97, - 183.08, - 126.97, - 191.08 - ], - "text": "–3", - "type": "text" - }, - { - "block_id": "p446-b6", - "global_id": 12456, - "bbox": [ - 118.97, - 151.23, - 126.97, - 159.23 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p446-b7", - "global_id": 12457, - "bbox": [ - 118.97, - 119.38, - 126.97, - 127.38 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p446-b8", - "global_id": 12458, - "bbox": [ - 122.97, - 87.53, - 126.97, - 95.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p446-b9", - "global_id": 12459, - "bbox": [ - 112.97, - 167.15, - 126.97, - 175.15 - ], - "text": "–2.5", - "type": "text" - }, - { - "block_id": "p446-b10", - "global_id": 12460, - "bbox": [ - 112.97, - 135.3, - 126.97, - 143.3 - ], - "text": "–1.5", - "type": "text" - }, - { - "block_id": "p446-b11", - "global_id": 12461, - "bbox": [ - 112.49, - 103.45, - 126.97, - 111.45 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p446-b12", - "global_id": 12462, - "bbox": [ - 461.89, - 198.58, - 467.22, - 206.58 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p446-b13", - "global_id": 12463, - "bbox": [ - 113.56, - 367.73, - 126.89, - 376.03 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p446-b14", - "global_id": 12464, - "bbox": [ - 113.56, - 347.5, - 126.89, - 355.79 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p446-b15", - "global_id": 12465, - "bbox": [ - 113.56, - 326.27, - 126.89, - 334.57 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p446-b16", - "global_id": 12466, - "bbox": [ - 120.23, - 305.06, - 126.89, - 313.36 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p446-b17", - "global_id": 12467, - "bbox": [ - 120.23, - 283.81, - 126.89, - 292.11 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p446-b18", - "global_id": 12468, - "bbox": [ - 120.23, - 262.58, - 126.89, - 270.88 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p446-b19", - "global_id": 12469, - "bbox": [ - 120.23, - 241.35, - 126.89, - 249.65 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p446-b20", - "global_id": 12470, - "bbox": [ - 119.92, - 376.27, - 464.92, - 384.35 - ], - "text": "10a\n100a\n0.1a\n0.01a\n50a\n5a\n0.5a\n0.05a", - "type": "text" - }, - { - "block_id": "p446-b21", - "global_id": 12471, - "bbox": [ - 103.83, - 301.56, - 111.83, - 337.77 - ], - "text": "Phase error", - "type": "text" - }, - { - "block_id": "p446-b22", - "global_id": 12472, - "bbox": [ - 461.89, - 366.99, - 467.22, - 374.99 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p446-b23", - "global_id": 12473, - "bbox": [ - 288.38, - 220.34, - 297.26, - 228.34 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p446-b24", - "global_id": 12474, - "bbox": [ - 288.25, - 390.74, - 297.58, - 398.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p446-b25", - "global_id": 12475, - "bbox": [ - 101.84, - 405.07, - 390.54, - 414.68 - ], - "text": "Figure 4.42 Errors in asymptotic approximation of a first-order pole at s = −a.", - "type": "text" - }, - { - "block_id": "p446-b26", - "global_id": 12476, - "bbox": [ - 101.84, - 435.01, - 490.39, - 469.3 - ], - "text": "The phase for a zero at −a (shown dotted in Fig. 4.41b) is identical to that of the pole at −a\nwith a sign change, and therefore is the mirror image (about the 0◦line) of the phase plot for a\npole at −a.", - "type": "text" - }, - { - "block_id": "p446-b27", - "global_id": 12477, - "bbox": [ - 101.84, - 494.28, - 285.99, - 506.24 - ], - "text": "4.9-4 Second-Order Pole (or Zero)", - "type": "text" - }, - { - "block_id": "p446-b28", - "global_id": 12478, - "bbox": [ - 101.84, - 508.34, - 490.39, - 534.99 - ], - "text": "Let us consider the second-order pole in Eq. (4.44). The denominator term is s2 + b2s + b3. We\nshall introduce the often-used standard form s2 + 2ζωns + ω2", - "type": "text" - }, - { - "block_id": "p446-b29", - "global_id": 12479, - "bbox": [ - 101.85, - 520.29, - 490.39, - 546.25 - ], - "text": "n instead of s2 + b2s + b3. With this\nform, the log amplitude function for the second-order term in Eq. (4.47) becomes", - "type": "text" - }, - { - "block_id": "p446-b30", - "global_id": 12480, - "bbox": [ - 234.89, - 565.72, - 266.45, - 576.1 - ], - "text": "−20log", - "type": "text" - }, - { - "block_id": "p446-b31", - "global_id": 12481, - "bbox": [ - 267.57, - 551.28, - 309.61, - 579.17 - ], - "text": "1 + 2jζ ω", - "type": "text" - }, - { - "block_id": "p446-b32", - "global_id": 12482, - "bbox": [ - 301.18, - 572.8, - 311.2, - 583.88 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p446-b33", - "global_id": 12483, - "bbox": [ - 314.44, - 565.72, - 322.21, - 575.69 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p446-b34", - "global_id": 12484, - "bbox": [ - 323.76, - 551.74, - 341.49, - 569.02 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p446-b35", - "global_id": 12485, - "bbox": [ - 331.69, - 572.8, - 341.7, - 583.88 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p446-b36", - "global_id": 12486, - "bbox": [ - 343.39, - 551.28, - 357.34, - 579.17 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p446-b37", - "global_id": 12487, - "bbox": [ - 101.85, - 593.58, - 200.91, - 603.54 - ], - "text": "and the phase function is", - "type": "text" - }, - { - "block_id": "p446-b38", - "global_id": 12488, - "bbox": [ - 241.03, - 614.11, - 248.8, - 624.08 - ], - "text": "−̸", - "type": "text" - }, - { - "block_id": "p446-b40", - "global_id": 12489, - "bbox": [ - 261.73, - 607.13, - 302.12, - 632.27 - ], - "text": "1 + 2jζ ω\nωn", - "type": "text" - }, - { - "block_id": "p446-b41", - "global_id": 12490, - "bbox": [ - 305.36, - 614.12, - 313.13, - 624.08 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p446-b42", - "global_id": 12491, - "bbox": [ - 314.68, - 600.12, - 332.41, - 617.41 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p446-b43", - "global_id": 12492, - "bbox": [ - 322.61, - 621.19, - 332.62, - 632.27 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p446-b44", - "global_id": 12493, - "bbox": [ - 334.3, - 597.13, - 351.2, - 612.02 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p446-b45", - "global_id": 12494, - "bbox": [ - 466.32, - 614.53, - 490.38, - 624.49 - ], - "text": "(4.48)", - "type": "text" - } - ] - }, - { - "page_num": 447, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p447-b0", - "global_id": 12495, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n427", - "type": "text" - }, - { - "block_id": "p447-b1", - "global_id": 12496, - "bbox": [ - 127.89, - 86.19, - 249.2, - 98.32 - ], - "text": "THE LOG MAGNITUDE", - "type": "text" - }, - { - "block_id": "p447-b2", - "global_id": 12497, - "bbox": [ - 127.59, - 102.35, - 246.46, - 112.31 - ], - "text": "The log amplitude is given by", - "type": "text" - }, - { - "block_id": "p447-b3", - "global_id": 12498, - "bbox": [ - 220.45, - 130.56, - 318.93, - 140.94 - ], - "text": "log amplitude = −20log", - "type": "text" - }, - { - "block_id": "p447-b4", - "global_id": 12499, - "bbox": [ - 320.07, - 116.12, - 351.17, - 144.02 - ], - "text": "1 + 2jζ", - "type": "text" - }, - { - "block_id": "p447-b5", - "global_id": 12500, - "bbox": [ - 352.47, - 116.58, - 368.82, - 133.54 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p447-b6", - "global_id": 12501, - "bbox": [ - 360.4, - 137.64, - 370.42, - 148.72 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b8", - "global_id": 12502, - "bbox": [ - 380.38, - 130.56, - 388.16, - 140.53 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p447-b9", - "global_id": 12503, - "bbox": [ - 389.7, - 116.58, - 407.43, - 133.86 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p447-b10", - "global_id": 12504, - "bbox": [ - 397.62, - 137.64, - 407.63, - 148.72 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b11", - "global_id": 12505, - "bbox": [ - 409.33, - 116.12, - 516.13, - 144.02 - ], - "text": "2\n(4.49)", - "type": "text" - }, - { - "block_id": "p447-b12", - "global_id": 12506, - "bbox": [ - 127.59, - 156.79, - 286.94, - 167.87 - ], - "text": "For ω ≪ωn, the log amplitude becomes", - "type": "text" - }, - { - "block_id": "p447-b13", - "global_id": 12507, - "bbox": [ - 261.13, - 177.71, - 382.59, - 188.09 - ], - "text": "log amplitude ≈−20log1 = 0", - "type": "text" - }, - { - "block_id": "p447-b14", - "global_id": 12508, - "bbox": [ - 127.59, - 198.65, - 258.77, - 209.73 - ], - "text": "For ω ≫ωn, the log amplitude is", - "type": "text" - }, - { - "block_id": "p447-b15", - "global_id": 12509, - "bbox": [ - 205.11, - 227.28, - 303.61, - 237.65 - ], - "text": "log amplitude ≈−20log", - "type": "text" - }, - { - "block_id": "p447-b18", - "global_id": 12510, - "bbox": [ - 315.8, - 220.29, - 333.18, - 237.24 - ], - "text": "−ω", - "type": "text" - }, - { - "block_id": "p447-b19", - "global_id": 12511, - "bbox": [ - 324.77, - 234.35, - 334.78, - 245.43 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b20", - "global_id": 12512, - "bbox": [ - 336.47, - 212.83, - 394.95, - 240.73 - ], - "text": "2 = −40log", - "type": "text" - }, - { - "block_id": "p447-b21", - "global_id": 12513, - "bbox": [ - 396.07, - 213.29, - 412.41, - 230.26 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p447-b22", - "global_id": 12514, - "bbox": [ - 403.99, - 234.35, - 414.0, - 245.43 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b24", - "global_id": 12515, - "bbox": [ - 262.23, - 249.09, - 516.13, - 260.55 - ], - "text": "= −40logω −40logωn = −40u −40logωn\n(4.50)", - "type": "text" - }, - { - "block_id": "p447-b25", - "global_id": 12516, - "bbox": [ - 127.59, - 270.03, - 516.15, - 316.27 - ], - "text": "The two asymptotes are zero for ω < ωn and −40u −40logωn for ω > ωn. The second asymptote\nis a straight line with a slope of −40 dB/decade (or −12 dB/octave) when plotted against the log\nω scale. It begins at ω = ωn [see Eq. (4.50)]. The asymptotes are depicted in Fig. 4.43a. The exact\nlog amplitude is given by [see Eq. (4.49)]", - "type": "text" - }, - { - "block_id": "p447-b26", - "global_id": 12517, - "bbox": [ - 202.13, - 335.44, - 300.62, - 345.82 - ], - "text": "log amplitude = −20log", - "type": "text" - }, - { - "block_id": "p447-b28", - "global_id": 12518, - "bbox": [ - 313.34, - 321.45, - 345.52, - 345.82 - ], - "text": "1 −\n ω", - "type": "text" - }, - { - "block_id": "p447-b29", - "global_id": 12519, - "bbox": [ - 337.1, - 342.51, - 347.12, - 353.59 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b30", - "global_id": 12520, - "bbox": [ - 348.81, - 321.45, - 368.44, - 333.34 - ], - "text": "2!2", - "type": "text" - }, - { - "block_id": "p447-b31", - "global_id": 12521, - "bbox": [ - 370.48, - 334.01, - 393.84, - 345.82 - ], - "text": "+ 4ζ 2", - "type": "text" - }, - { - "block_id": "p447-b32", - "global_id": 12522, - "bbox": [ - 394.35, - 321.45, - 410.69, - 338.42 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p447-b33", - "global_id": 12523, - "bbox": [ - 402.27, - 342.51, - 412.28, - 353.59 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b34", - "global_id": 12524, - "bbox": [ - 413.98, - 321.45, - 441.09, - 333.34 - ], - "text": "21/2", - "type": "text" - }, - { - "block_id": "p447-b35", - "global_id": 12525, - "bbox": [ - 492.07, - 335.85, - 516.13, - 345.82 - ], - "text": "(4.51)", - "type": "text" - }, - { - "block_id": "p447-b36", - "global_id": 12526, - "bbox": [ - 127.59, - 362.57, - 516.15, - 420.77 - ], - "text": "The log amplitude in this case involves a parameter ζ, resulting in a different plot for each value\nof ζ. For complex-conjugate poles,† ζ < 1. Hence, we must sketch a family of curves for a number\nof values of ζ in the range 0 to 1. This is illustrated in Fig. 4.43a. The error between the actual plot\nand the asymptotes is shown in Fig. 4.44. The actual plot can be obtained by adding the error to\nthe asymptotic plot.", - "type": "text" - }, - { - "block_id": "p447-b37", - "global_id": 12527, - "bbox": [ - 127.59, - 422.76, - 516.11, - 456.63 - ], - "text": "For second-order zeros (complex-conjugate zeros), the plots are mirror images (about the 0 dB\nline) of the plots depicted in Fig. 4.43a. Note the resonance phenomenon of the complex-conjugate\npoles. This phenomenon is barely noticeable for ζ > 0.707 but becomes pronounced as ζ →0.", - "type": "text" - }, - { - "block_id": "p447-b38", - "global_id": 12528, - "bbox": [ - 127.59, - 471.12, - 410.64, - 497.23 - ], - "text": "PHASE\nThe phase function for second-order poles, as apparent in Eq. (4.48), is̸", - "type": "text" - }, - { - "block_id": "p447-b39", - "global_id": 12529, - "bbox": [ - 259.12, - 526.65, - 325.52, - 538.75 - ], - "text": "H(jω) = −tan−1", - "type": "text" - }, - { - "block_id": "p447-b40", - "global_id": 12530, - "bbox": [ - 327.13, - 499.45, - 334.39, - 509.41 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p447-b41", - "global_id": 12531, - "bbox": [ - 327.13, - 516.92, - 334.39, - 545.27 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p447-b42", - "global_id": 12532, - "bbox": [ - 340.23, - 500.82, - 367.13, - 525.19 - ], - "text": "2ζ\n ω", - "type": "text" - }, - { - "block_id": "p447-b43", - "global_id": 12533, - "bbox": [ - 358.71, - 521.89, - 368.72, - 532.97 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b45", - "global_id": 12534, - "bbox": [ - 335.59, - 530.0, - 367.79, - 554.36 - ], - "text": "1 −\n ω", - "type": "text" - }, - { - "block_id": "p447-b46", - "global_id": 12535, - "bbox": [ - 359.36, - 551.06, - 369.37, - 562.14 - ], - "text": "ωn", - "type": "text" - }, - { - "block_id": "p447-b47", - "global_id": 12536, - "bbox": [ - 371.07, - 530.0, - 381.28, - 541.89 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p447-b48", - "global_id": 12537, - "bbox": [ - 382.98, - 499.45, - 390.24, - 509.41 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p447-b49", - "global_id": 12538, - "bbox": [ - 382.98, - 516.92, - 516.13, - 545.27 - ], - "text": "⎥⎥⎥⎦\n(4.52)", - "type": "text" - }, - { - "block_id": "p447-b50", - "global_id": 12539, - "bbox": [ - 127.59, - 569.93, - 177.4, - 581.01 - ], - "text": "For ω ≪ωn,̸", - "type": "text" - }, - { - "block_id": "p447-b51", - "global_id": 12540, - "bbox": [ - 303.97, - 581.89, - 345.39, - 592.26 - ], - "text": "H(jω) ≈0", - "type": "text" - }, - { - "block_id": "p447-b52", - "global_id": 12541, - "bbox": [ - 127.59, - 610.24, - 516.14, - 633.41 - ], - "text": "† For ζ ≥1, the two poles in the second-order factor are no longer complex but real, and each of these two\nreal poles can be dealt with as a separate first-order factor.", - "type": "text" - } - ] - }, - { - "page_num": 448, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p448-b0", - "global_id": 12542, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "428\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p448-b1", - "global_id": 12543, - "bbox": [ - 123.43, - 267.77, - 138.09, - 276.06 - ], - "text": "40", - "type": "text" - }, - { - "block_id": "p448-b2", - "global_id": 12544, - "bbox": [ - 120.77, - 342.03, - 138.09, - 350.33 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p448-b3", - "global_id": 12545, - "bbox": [ - 120.77, - 372.3, - 138.09, - 380.6 - ], - "text": "60", - "type": "text" - }, - { - "block_id": "p448-b4", - "global_id": 12546, - "bbox": [ - 120.77, - 402.58, - 138.09, - 410.87 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p448-b5", - "global_id": 12547, - "bbox": [ - 116.77, - 432.85, - 138.09, - 441.14 - ], - "text": "120", - "type": "text" - }, - { - "block_id": "p448-b6", - "global_id": 12548, - "bbox": [ - 116.77, - 463.12, - 138.09, - 471.41 - ], - "text": "150", - "type": "text" - }, - { - "block_id": "p448-b7", - "global_id": 12549, - "bbox": [ - 116.77, - 491.36, - 138.09, - 499.65 - ], - "text": "180", - "type": "text" - }, - { - "block_id": "p448-b8", - "global_id": 12550, - "bbox": [ - 123.43, - 238.82, - 138.09, - 247.11 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p448-b9", - "global_id": 12551, - "bbox": [ - 123.43, - 208.59, - 138.09, - 216.88 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p448-b10", - "global_id": 12552, - "bbox": [ - 123.43, - 178.35, - 138.09, - 186.65 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p448-b11", - "global_id": 12553, - "bbox": [ - 134.09, - 148.42, - 138.09, - 156.42 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p448-b12", - "global_id": 12554, - "bbox": [ - 130.09, - 118.18, - 138.09, - 126.18 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p448-b13", - "global_id": 12555, - "bbox": [ - 130.09, - 87.95, - 138.09, - 95.95 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p448-b14", - "global_id": 12556, - "bbox": [ - 131.69, - 275.73, - 476.65, - 285.3 - ], - "text": "10vn\n5vn\n2vn\n0.1vn\n0.2vn\n0.5vn\nvn", - "type": "text" - }, - { - "block_id": "p448-b15", - "global_id": 12557, - "bbox": [ - 131.69, - 500.2, - 476.65, - 509.77 - ], - "text": "10vn\n5vn\n2vn\n0.1vn\n0.2vn\n0.5vn\nvn", - "type": "text" - }, - { - "block_id": "p448-b16", - "global_id": 12558, - "bbox": [ - 111.07, - 158.07, - 119.36, - 206.6 - ], - "text": "20 log H (dB)", - "type": "text" - }, - { - "block_id": "p448-b17", - "global_id": 12559, - "bbox": [ - 238.51, - 122.54, - 273.62, - 130.54 - ], - "text": "Asymptote", - "type": "text" - }, - { - "block_id": "p448-b18", - "global_id": 12560, - "bbox": [ - 359.57, - 164.48, - 394.68, - 172.48 - ], - "text": "Asymptote", - "type": "text" - }, - { - "block_id": "p448-b19", - "global_id": 12561, - "bbox": [ - 302.08, - 182.26, - 321.64, - 213.08 - ], - "text": "z 1\n0.707\n0.5", - "type": "text" - }, - { - "block_id": "p448-b20", - "global_id": 12562, - "bbox": [ - 323.17, - 104.4, - 347.39, - 112.7 - ], - "text": "z 0.1", - "type": "text" - }, - { - "block_id": "p448-b21", - "global_id": 12563, - "bbox": [ - 334.08, - 115.72, - 344.08, - 135.34 - ], - "text": "0.2\n0.3", - "type": "text" - }, - { - "block_id": "p448-b22", - "global_id": 12564, - "bbox": [ - 131.43, - 311.76, - 138.09, - 320.06 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p448-b23", - "global_id": 12565, - "bbox": [ - 340.8, - 325.32, - 375.91, - 333.32 - ], - "text": "Asymptote", - "type": "text" - }, - { - "block_id": "p448-b24", - "global_id": 12566, - "bbox": [ - 229.43, - 481.79, - 264.55, - 489.79 - ], - "text": "Asymptote", - "type": "text" - }, - { - "block_id": "p448-b25", - "global_id": 12567, - "bbox": [ - 192.09, - 385.96, - 210.31, - 394.25 - ], - "text": "z 1", - "type": "text" - }, - { - "block_id": "p448-b26", - "global_id": 12568, - "bbox": [ - 194.66, - 363.17, - 212.66, - 382.09 - ], - "text": "0.707\n0.5", - "type": "text" - }, - { - "block_id": "p448-b27", - "global_id": 12569, - "bbox": [ - 327.18, - 337.32, - 351.39, - 345.61 - ], - "text": "z 0.1", - "type": "text" - }, - { - "block_id": "p448-b28", - "global_id": 12570, - "bbox": [ - 345.82, - 348.63, - 355.82, - 368.25 - ], - "text": "0.2\n0.3", - "type": "text" - }, - { - "block_id": "p448-b29", - "global_id": 12571, - "bbox": [ - 102.54, - 397.29, - 110.54, - 415.95 - ], - "text": "Phase", - "type": "text" - }, - { - "block_id": "p448-b30", - "global_id": 12572, - "bbox": [ - 482.42, - 153.28, - 487.76, - 161.28 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p448-b31", - "global_id": 12573, - "bbox": [ - 482.42, - 317.26, - 487.76, - 325.26 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p448-b32", - "global_id": 12574, - "bbox": [ - 300.23, - 291.23, - 309.11, - 299.23 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p448-b33", - "global_id": 12575, - "bbox": [ - 299.69, - 515.74, - 309.66, - 523.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p448-b34", - "global_id": 12576, - "bbox": [ - 101.84, - 530.44, - 346.62, - 539.68 - ], - "text": "Figure 4.43 Amplitude and phase response of a second-order pole.", - "type": "text" - }, - { - "block_id": "p448-b35", - "global_id": 12577, - "bbox": [ - 101.84, - 562.38, - 151.65, - 573.46 - ], - "text": "For ω ≫ωn,̸", - "type": "text" - }, - { - "block_id": "p448-b36", - "global_id": 12578, - "bbox": [ - 267.36, - 576.95, - 329.99, - 589.05 - ], - "text": "H(jω) ≃−180◦", - "type": "text" - }, - { - "block_id": "p448-b37", - "global_id": 12579, - "bbox": [ - 101.84, - 599.27, - 490.4, - 636.28 - ], - "text": "Hence, the phase →−180◦as ω →∞. As in the case of amplitude, we also have a family of\nphase plots for various values of ζ, as illustrated in Fig. 4.43b. A convenient asymptote for the\nphase of complex-conjugate poles is a step function that is 0◦for ω < ωn and −180◦for ω > ωn.", - "type": "text" - } - ] - }, - { - "page_num": 449, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p449-b0", - "global_id": 12580, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n429", - "type": "text" - }, - { - "block_id": "p449-b1", - "global_id": 12581, - "bbox": [ - 146.76, - 256.19, - 157.43, - 264.49 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p449-b2", - "global_id": 12582, - "bbox": [ - 153.43, - 212.14, - 157.43, - 220.14 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p449-b3", - "global_id": 12583, - "bbox": [ - 153.43, - 167.8, - 157.43, - 175.8 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p449-b4", - "global_id": 12584, - "bbox": [ - 149.43, - 122.23, - 157.43, - 130.23 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p449-b5", - "global_id": 12585, - "bbox": [ - 151.26, - 274.24, - 489.65, - 284.0 - ], - "text": "0.1vn\n0.2vn\n0.5vn\n2vn\n5vn\n10vn", - "type": "text" - }, - { - "block_id": "p449-b6", - "global_id": 12586, - "bbox": [ - 232.08, - 244.8, - 252.74, - 262.81 - ], - "text": "z 1\n0.707", - "type": "text" - }, - { - "block_id": "p449-b7", - "global_id": 12587, - "bbox": [ - 239.82, - 231.69, - 249.82, - 239.69 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p449-b8", - "global_id": 12588, - "bbox": [ - 351.31, - 128.91, - 375.53, - 137.21 - ], - "text": "z 0.1", - "type": "text" - }, - { - "block_id": "p449-b9", - "global_id": 12589, - "bbox": [ - 361.0, - 140.0, - 371.0, - 148.0 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p449-b10", - "global_id": 12590, - "bbox": [ - 361.0, - 152.37, - 371.0, - 160.37 - ], - "text": "0.3", - "type": "text" - }, - { - "block_id": "p449-b11", - "global_id": 12591, - "bbox": [ - 146.76, - 265.5, - 157.43, - 273.8 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p449-b12", - "global_id": 12592, - "bbox": [ - 149.43, - 87.71, - 157.43, - 95.71 - ], - "text": "14", - "type": "text" - }, - { - "block_id": "p449-b13", - "global_id": 12593, - "bbox": [ - 128.29, - 163.5, - 136.29, - 197.04 - ], - "text": "Error (dB)", - "type": "text" - }, - { - "block_id": "p449-b14", - "global_id": 12594, - "bbox": [ - 308.8, - 273.1, - 333.14, - 282.86 - ], - "text": "v vn", - "type": "text" - }, - { - "block_id": "p449-b15", - "global_id": 12595, - "bbox": [ - 150.65, - 397.42, - 157.31, - 405.72 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p449-b16", - "global_id": 12596, - "bbox": [ - 139.98, - 456.57, - 157.31, - 464.86 - ], - "text": "60", - "type": "text" - }, - { - "block_id": "p449-b17", - "global_id": 12597, - "bbox": [ - 139.98, - 485.96, - 157.31, - 494.25 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p449-b18", - "global_id": 12598, - "bbox": [ - 139.98, - 426.7, - 157.31, - 435.0 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p449-b19", - "global_id": 12599, - "bbox": [ - 146.65, - 367.7, - 157.31, - 375.99 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p449-b20", - "global_id": 12600, - "bbox": [ - 146.65, - 337.81, - 157.31, - 346.1 - ], - "text": "60", - "type": "text" - }, - { - "block_id": "p449-b21", - "global_id": 12601, - "bbox": [ - 146.65, - 309.23, - 157.31, - 317.53 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p449-b22", - "global_id": 12602, - "bbox": [ - 128.29, - 382.88, - 136.29, - 419.09 - ], - "text": "Phase error", - "type": "text" - }, - { - "block_id": "p449-b23", - "global_id": 12603, - "bbox": [ - 382.98, - 317.95, - 401.52, - 326.25 - ], - "text": "z 1", - "type": "text" - }, - { - "block_id": "p449-b24", - "global_id": 12604, - "bbox": [ - 389.21, - 329.19, - 433.32, - 371.83 - ], - "text": "0.707\n0.5\n0.3\n0.2\n0.1", - "type": "text" - }, - { - "block_id": "p449-b25", - "global_id": 12605, - "bbox": [ - 498.21, - 403.2, - 503.54, - 411.2 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p449-b26", - "global_id": 12606, - "bbox": [ - 498.21, - 217.95, - 503.54, - 225.95 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p449-b27", - "global_id": 12607, - "bbox": [ - 316.52, - 289.93, - 325.4, - 297.93 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p449-b28", - "global_id": 12608, - "bbox": [ - 316.18, - 510.74, - 325.99, - 518.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p449-b29", - "global_id": 12609, - "bbox": [ - 151.26, - 495.14, - 489.75, - 505.47 - ], - "text": "0.1vn\n0.2vn\n0.5vn\n2vn\n5vn\n10vn\nv vn", - "type": "text" - }, - { - "block_id": "p449-b30", - "global_id": 12610, - "bbox": [ - 127.59, - 525.44, - 403.53, - 534.68 - ], - "text": "Figure 4.44 Errors in the asymptotic approximation of a second-order pole.", - "type": "text" - }, - { - "block_id": "p449-b31", - "global_id": 12611, - "bbox": [ - 127.59, - 564.67, - 516.12, - 587.0 - ], - "text": "Error plots for such an asymptote are shown in Fig. 4.44 for various values of ζ. The exact phase\nis the asymptotic value plus the error.", - "type": "text" - }, - { - "block_id": "p449-b32", - "global_id": 12612, - "bbox": [ - 127.59, - 588.99, - 516.12, - 610.91 - ], - "text": "For complex-conjugate zeros, the amplitude and phase plots are mirror images of those for\ncomplex conjugate-poles.", - "type": "text" - }, - { - "block_id": "p449-b33", - "global_id": 12613, - "bbox": [ - 145.52, - 612.9, - 451.34, - 622.86 - ], - "text": "We shall demonstrate the application of these techniques with two examples.", - "type": "text" - } - ] - }, - { - "page_num": 450, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p450-b0", - "global_id": 12614, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "430\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p450-b1", - "global_id": 12615, - "bbox": [ - 76.77, - 93.92, - 465.18, - 119.82 - ], - "text": "EXAMPLE 4.29\nBode Plots for Second-Order Transfer Function with\nReal Roots", - "type": "text" - }, - { - "block_id": "p450-b2", - "global_id": 12616, - "bbox": [ - 103.16, - 133.56, - 272.94, - 143.53 - ], - "text": "Sketch Bode plots for the transfer function", - "type": "text" - }, - { - "block_id": "p450-b3", - "global_id": 12617, - "bbox": [ - 243.83, - 153.1, - 330.95, - 170.46 - ], - "text": "H(s) = 20s(s + 100)", - "type": "text" - }, - { - "block_id": "p450-b4", - "global_id": 12618, - "bbox": [ - 275.85, - 167.16, - 335.13, - 177.54 - ], - "text": "(s + 2)(s + 10)", - "type": "text" - }, - { - "block_id": "p450-b5", - "global_id": 12619, - "bbox": [ - 103.16, - 203.13, - 323.71, - 229.24 - ], - "text": "MAGNITUDE PLOT\nFirst, we write the transfer function in normalized form", - "type": "text" - }, - { - "block_id": "p450-b6", - "global_id": 12620, - "bbox": [ - 161.46, - 252.29, - 229.26, - 269.65 - ], - "text": "H(s) = 20 × 100", - "type": "text" - }, - { - "block_id": "p450-b7", - "global_id": 12621, - "bbox": [ - 198.45, - 266.35, - 224.29, - 276.73 - ], - "text": "2 × 10", - "type": "text" - }, - { - "block_id": "p450-b8", - "global_id": 12622, - "bbox": [ - 245.58, - 246.03, - 249.46, - 255.99 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p450-b10", - "global_id": 12623, - "bbox": [ - 256.18, - 239.04, - 288.16, - 263.16 - ], - "text": "1 +\ns\n100", - "type": "text" - }, - { - "block_id": "p450-b13", - "global_id": 12624, - "bbox": [ - 238.4, - 266.16, - 260.43, - 290.28 - ], - "text": "1 + s\n2", - "type": "text" - }, - { - "block_id": "p450-b15", - "global_id": 12625, - "bbox": [ - 275.07, - 266.16, - 302.08, - 290.28 - ], - "text": "1 + s\n10", - "type": "text" - }, - { - "block_id": "p450-b16", - "global_id": 12626, - "bbox": [ - 303.28, - 258.84, - 338.0, - 269.65 - ], - "text": "= 100", - "type": "text" - }, - { - "block_id": "p450-b17", - "global_id": 12627, - "bbox": [ - 353.11, - 246.03, - 356.99, - 255.99 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p450-b19", - "global_id": 12628, - "bbox": [ - 363.71, - 239.04, - 395.7, - 263.16 - ], - "text": "1 +\ns\n100", - "type": "text" - }, - { - "block_id": "p450-b22", - "global_id": 12629, - "bbox": [ - 345.93, - 266.16, - 367.96, - 290.28 - ], - "text": "1 + s\n2", - "type": "text" - }, - { - "block_id": "p450-b24", - "global_id": 12630, - "bbox": [ - 382.6, - 266.16, - 409.61, - 290.28 - ], - "text": "1 + s\n10", - "type": "text" - }, - { - "block_id": "p450-b26", - "global_id": 12631, - "bbox": [ - 103.17, - 300.04, - 477.04, - 346.28 - ], - "text": "Here, the constant term is 100; that is, 40 dB (20 log100 = 40). This term can be added to the\nplot by simply relabeling the horizontal axis (from which the asymptotes begin) as the 40 dB\nline (see Fig. 4.45a). Such a step implies shifting the horizontal axis upward by 40 dB. This is\nprecisely what is desired.", - "type": "text" - }, - { - "block_id": "p450-b27", - "global_id": 12632, - "bbox": [ - 103.16, - 347.86, - 477.01, - 370.19 - ], - "text": "In addition, we have two first-order poles at −2 and −10, one zero at the origin, and one\nzero at −100.", - "type": "text" - }, - { - "block_id": "p450-b28", - "global_id": 12633, - "bbox": [ - 103.16, - 380.07, - 477.0, - 402.07 - ], - "text": "Step 1. For each of these terms, we draw an asymptotic plot as follows (shown in Fig. 4.45a\nby dashed lines):", - "type": "text" - }, - { - "block_id": "p450-b29", - "global_id": 12634, - "bbox": [ - 121.09, - 409.63, - 477.01, - 420.01 - ], - "text": "(a) For the zero at the origin, draw a straight line with a slope of 20 dB/decade passing", - "type": "text" - }, - { - "block_id": "p450-b30", - "global_id": 12635, - "bbox": [ - 137.7, - 421.58, - 197.24, - 431.96 - ], - "text": "through ω = 1.", - "type": "text" - }, - { - "block_id": "p450-b31", - "global_id": 12636, - "bbox": [ - 121.09, - 436.53, - 477.01, - 446.91 - ], - "text": "(b) For the pole at −2, draw a straight line with a slope of −20 dB/decade (for ω > 2)", - "type": "text" - }, - { - "block_id": "p450-b32", - "global_id": 12637, - "bbox": [ - 137.7, - 448.48, - 300.4, - 458.86 - ], - "text": "beginning at the corner frequency ω = 2.", - "type": "text" - }, - { - "block_id": "p450-b33", - "global_id": 12638, - "bbox": [ - 121.65, - 463.43, - 477.02, - 473.81 - ], - "text": "(c) For the pole at −10, draw a straight line with a slope of −20 dB/decade beginning at", - "type": "text" - }, - { - "block_id": "p450-b34", - "global_id": 12639, - "bbox": [ - 121.09, - 475.38, - 477.02, - 500.7 - ], - "text": "the corner frequency ω = 10.\n(d) For the zero at −100, draw a straight line with a slope of 20 dB/decade beginning at", - "type": "text" - }, - { - "block_id": "p450-b35", - "global_id": 12640, - "bbox": [ - 137.7, - 502.28, - 258.48, - 512.66 - ], - "text": "the corner frequency ω = 100.", - "type": "text" - }, - { - "block_id": "p450-b36", - "global_id": 12641, - "bbox": [ - 103.17, - 522.54, - 424.58, - 532.59 - ], - "text": "Step 2. Add all the asymptotes, as depicted in Fig. 4.45a by solid line segments.", - "type": "text" - }, - { - "block_id": "p450-b37", - "global_id": 12642, - "bbox": [ - 103.17, - 542.47, - 331.47, - 552.51 - ], - "text": "Step 3. Apply the following corrections (see Fig. 4.42a):", - "type": "text" - }, - { - "block_id": "p450-b38", - "global_id": 12643, - "bbox": [ - 121.09, - 560.06, - 477.01, - 570.44 - ], - "text": "(a) The correction at ω = 1 because of the corner frequency at ω = 2 is −1 dB. The", - "type": "text" - }, - { - "block_id": "p450-b39", - "global_id": 12644, - "bbox": [ - 137.69, - 572.02, - 477.02, - 606.31 - ], - "text": "correction at ω = 1 because of the corner frequencies at ω = 10 and ω = 100 is quite\nsmall (see Fig. 4.42a) and may be ignored. Hence, the net correction at ω = 1 is\n−1 dB.", - "type": "text" - } - ] - }, - { - "page_num": 451, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p451-b0", - "global_id": 12645, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n431", - "type": "text" - }, - { - "block_id": "p451-b1", - "global_id": 12646, - "bbox": [ - 148.42, - 261.99, - 156.42, - 269.99 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p451-b2", - "global_id": 12647, - "bbox": [ - 148.42, - 232.27, - 156.42, - 240.27 - ], - "text": "25", - "type": "text" - }, - { - "block_id": "p451-b3", - "global_id": 12648, - "bbox": [ - 148.42, - 203.34, - 156.42, - 211.34 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p451-b4", - "global_id": 12649, - "bbox": [ - 148.42, - 174.41, - 156.42, - 182.41 - ], - "text": "35", - "type": "text" - }, - { - "block_id": "p451-b5", - "global_id": 12650, - "bbox": [ - 148.42, - 144.6, - 156.42, - 152.6 - ], - "text": "40", - "type": "text" - }, - { - "block_id": "p451-b6", - "global_id": 12651, - "bbox": [ - 148.42, - 115.58, - 156.42, - 123.58 - ], - "text": "45", - "type": "text" - }, - { - "block_id": "p451-b7", - "global_id": 12652, - "bbox": [ - 148.42, - 86.64, - 156.42, - 94.64 - ], - "text": "50", - "type": "text" - }, - { - "block_id": "p451-b8", - "global_id": 12653, - "bbox": [ - 328.25, - 106.69, - 380.03, - 114.69 - ], - "text": "Asymptotic plot", - "type": "text" - }, - { - "block_id": "p451-b9", - "global_id": 12654, - "bbox": [ - 341.66, - 124.16, - 374.32, - 132.16 - ], - "text": "Exact plot", - "type": "text" - }, - { - "block_id": "p451-b10", - "global_id": 12655, - "bbox": [ - 237.95, - 154.15, - 480.13, - 170.61 - ], - "text": "1\n2\n5\n10\n20\n100\n1000\n400\nv", - "type": "text" - }, - { - "block_id": "p451-b11", - "global_id": 12656, - "bbox": [ - 126.82, - 163.81, - 135.11, - 212.51 - ], - "text": "20 log H (dB)", - "type": "text" - }, - { - "block_id": "p451-b12", - "global_id": 12657, - "bbox": [ - 138.99, - 427.73, - 156.32, - 436.02 - ], - "text": "45", - "type": "text" - }, - { - "block_id": "p451-b13", - "global_id": 12658, - "bbox": [ - 149.66, - 384.2, - 156.32, - 392.49 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p451-b14", - "global_id": 12659, - "bbox": [ - 145.66, - 340.33, - 156.32, - 348.63 - ], - "text": "45", - "type": "text" - }, - { - "block_id": "p451-b15", - "global_id": 12660, - "bbox": [ - 145.66, - 293.84, - 156.32, - 302.14 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p451-b16", - "global_id": 12661, - "bbox": [ - 138.99, - 470.74, - 156.32, - 479.04 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p451-b17", - "global_id": 12662, - "bbox": [ - 236.08, - 391.8, - 488.65, - 399.8 - ], - "text": "1\n2\n5\n10\n20\n100\n1000\n400", - "type": "text" - }, - { - "block_id": "p451-b18", - "global_id": 12663, - "bbox": [ - 437.15, - 415.97, - 442.49, - 423.97 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p451-b19", - "global_id": 12664, - "bbox": [ - 412.08, - 432.84, - 463.86, - 440.84 - ], - "text": "Asymptotic plot", - "type": "text" - }, - { - "block_id": "p451-b20", - "global_id": 12665, - "bbox": [ - 293.64, - 359.57, - 326.3, - 367.57 - ], - "text": "Exact plot", - "type": "text" - }, - { - "block_id": "p451-b21", - "global_id": 12666, - "bbox": [ - 177.22, - 391.8, - 187.22, - 399.8 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p451-b22", - "global_id": 12667, - "bbox": [ - 129.74, - 391.06, - 137.74, - 409.72 - ], - "text": "Phase", - "type": "text" - }, - { - "block_id": "p451-b23", - "global_id": 12668, - "bbox": [ - 308.13, - 272.99, - 317.01, - 280.99 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p451-b24", - "global_id": 12669, - "bbox": [ - 310.99, - 487.17, - 320.64, - 495.17 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p451-b25", - "global_id": 12670, - "bbox": [ - 126.18, - 501.79, - 416.78, - 511.1 - ], - "text": "Figure 4.45 (a) Amplitude and (b) phase responses of the second-order system.", - "type": "text" - }, - { - "block_id": "p451-b26", - "global_id": 12671, - "bbox": [ - 146.84, - 530.39, - 502.76, - 540.77 - ], - "text": "(b) The correction at ω = 2 because of the corner frequency at ω = 2 is −3 dB, and the", - "type": "text" - }, - { - "block_id": "p451-b27", - "global_id": 12672, - "bbox": [ - 147.4, - 542.35, - 502.77, - 591.57 - ], - "text": "correction because of the corner frequency at ω = 10 is −0.17 dB. The correction\nbecause of the corner frequency ω = 100 can be safely ignored. Hence the net\ncorrection at ω = 2 is −3.17 dB.\n(c) The correction at ω = 10 because of the corner frequency at ω = 10 is −3 dB, and", - "type": "text" - }, - { - "block_id": "p451-b28", - "global_id": 12673, - "bbox": [ - 163.44, - 593.15, - 502.75, - 615.49 - ], - "text": "the correction because of the corner frequency at ω = 2 is −0.17 dB. The correction\nbecause of ω = 100 can be ignored. Hence the net correction at ω = 10 is −3.17 dB.", - "type": "text" - } - ] - }, - { - "page_num": 452, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p452-b0", - "global_id": 12674, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "432\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p452-b1", - "global_id": 12675, - "bbox": [ - 121.09, - 85.96, - 477.01, - 96.33 - ], - "text": "(d) The correction at ω = 100 because of the corner frequency at ω = 100 is 3 dB, and", - "type": "text" - }, - { - "block_id": "p452-b2", - "global_id": 12676, - "bbox": [ - 137.7, - 98.32, - 422.37, - 108.28 - ], - "text": "the corrections because of the other corner frequencies may be ignored.", - "type": "text" - }, - { - "block_id": "p452-b3", - "global_id": 12677, - "bbox": [ - 121.65, - 113.19, - 477.01, - 123.23 - ], - "text": "(e) In addition to the corrections at corner frequencies, we may consider corrections at", - "type": "text" - }, - { - "block_id": "p452-b4", - "global_id": 12678, - "bbox": [ - 137.7, - 124.81, - 477.02, - 171.05 - ], - "text": "intermediate points for more accurate plots. For instance, the corrections at ω = 4\nbecause of corner frequencies at ω = 2 and 10 are −1 and about −0.65, totaling\n−1.65 dB. In the same way, the corrections at ω = 5 because of corner frequencies at\nω = 2 and 10 are −0.65 and −1, totaling −1.65 dB.", - "type": "text" - }, - { - "block_id": "p452-b5", - "global_id": 12679, - "bbox": [ - 121.09, - 179.02, - 431.98, - 188.99 - ], - "text": "With these corrections, the resulting amplitude plot is illustrated in Fig. 4.45a.", - "type": "text" - }, - { - "block_id": "p452-b6", - "global_id": 12680, - "bbox": [ - 103.16, - 203.72, - 367.75, - 229.83 - ], - "text": "PHASE PLOT\nWe draw the asymptotes corresponding to each of the four factors:", - "type": "text" - }, - { - "block_id": "p452-b7", - "global_id": 12681, - "bbox": [ - 121.09, - 236.16, - 477.01, - 262.71 - ], - "text": "(a) The zero at the origin causes a 90◦phase shift.\n(b) The pole at s = −2 has an asymptote with a zero value for −∞< ω < 0.2 and a slope", - "type": "text" - }, - { - "block_id": "p452-b8", - "global_id": 12682, - "bbox": [ - 121.65, - 263.06, - 477.01, - 301.56 - ], - "text": "of −45◦/decade beginning at ω = 0.2 and going up to ω = 20. The asymptotic value\nfor ω > 20 is −90◦.\n(c) The pole at s = −10 has an asymptote with a zero value for −∞< ω < 1 and a slope", - "type": "text" - }, - { - "block_id": "p452-b9", - "global_id": 12683, - "bbox": [ - 137.7, - 301.92, - 477.0, - 325.47 - ], - "text": "of −45◦/decade beginning at ω = 1 and going up to ω = 100. The asymptotic value\nfor ω > 100 is −90◦.", - "type": "text" - }, - { - "block_id": "p452-b10", - "global_id": 12684, - "bbox": [ - 121.09, - 330.04, - 477.02, - 340.42 - ], - "text": "(d) The zero at s = −100 has an asymptote with a zero value for −∞< ω < 10 and a", - "type": "text" - }, - { - "block_id": "p452-b11", - "global_id": 12685, - "bbox": [ - 137.7, - 340.77, - 477.02, - 388.24 - ], - "text": "slope of 45◦/decade beginning at ω = 10 and going up to ω = 1000. The asymptotic\nvalue for ω > 1000 is 90◦. All the asymptotes are added, as shown in Fig. 4.45b.\nThe appropriate corrections are applied from Fig. 4.42b, and the exact phase plot is\ndepicted in Fig. 4.45b.", - "type": "text" - }, - { - "block_id": "p452-b12", - "global_id": 12686, - "bbox": [ - 76.77, - 455.21, - 465.18, - 481.12 - ], - "text": "EXAMPLE 4.30\nBode Plots for Second-Order Transfer Function with\nComplex Poles", - "type": "text" - }, - { - "block_id": "p452-b13", - "global_id": 12687, - "bbox": [ - 103.16, - 497.79, - 415.97, - 507.75 - ], - "text": "Sketch the amplitude and phase response (Bode plots) for the transfer function", - "type": "text" - }, - { - "block_id": "p452-b14", - "global_id": 12688, - "bbox": [ - 206.47, - 526.13, - 288.7, - 543.49 - ], - "text": "H(s) = 10(s + 100)", - "type": "text" - }, - { - "block_id": "p452-b15", - "global_id": 12689, - "bbox": [ - 238.48, - 533.11, - 314.9, - 550.56 - ], - "text": "s2 + 2s + 100 = 10", - "type": "text" - }, - { - "block_id": "p452-b16", - "global_id": 12690, - "bbox": [ - 327.71, - 515.17, - 359.7, - 539.29 - ], - "text": "1 +\ns\n100", - "type": "text" - }, - { - "block_id": "p452-b17", - "global_id": 12691, - "bbox": [ - 316.1, - 539.65, - 367.28, - 565.02 - ], - "text": "1 + s\n50 + s2", - "type": "text" - }, - { - "block_id": "p452-b18", - "global_id": 12692, - "bbox": [ - 356.37, - 555.06, - 371.31, - 565.02 - ], - "text": "100", - "type": "text" - }, - { - "block_id": "p452-b19", - "global_id": 12693, - "bbox": [ - 103.46, - 581.35, - 203.77, - 593.47 - ], - "text": "MAGNITUDE PLOT", - "type": "text" - }, - { - "block_id": "p452-b20", - "global_id": 12694, - "bbox": [ - 103.16, - 597.09, - 477.02, - 631.37 - ], - "text": "Here, the constant term is 10: that is, 20 dB(20 log10 = 20). To add this term, we simply\nlabel the horizontal axis (from which the asymptotes begin) as the 20 dB line, as before (see\nFig. 4.46a).", - "type": "text" - } - ] - }, - { - "page_num": 453, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p453-b0", - "global_id": 12695, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n433", - "type": "text" - }, - { - "block_id": "p453-b1", - "global_id": 12696, - "bbox": [ - 163.52, - 190.16, - 167.52, - 198.16 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p453-b2", - "global_id": 12697, - "bbox": [ - 158.06, - 163.77, - 166.06, - 171.77 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p453-b3", - "global_id": 12698, - "bbox": [ - 158.06, - 137.37, - 166.06, - 145.37 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p453-b4", - "global_id": 12699, - "bbox": [ - 158.06, - 111.89, - 166.06, - 119.89 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p453-b5", - "global_id": 12700, - "bbox": [ - 171.71, - 142.36, - 512.36, - 151.83 - ], - "text": "1\n2\n5\n10\n20\n100\n103", - "type": "text" - }, - { - "block_id": "p453-b6", - "global_id": 12701, - "bbox": [ - 500.88, - 381.98, - 511.88, - 391.45 - ], - "text": "103", - "type": "text" - }, - { - "block_id": "p453-b7", - "global_id": 12702, - "bbox": [ - 447.3, - 143.83, - 461.63, - 163.07 - ], - "text": "400\nv", - "type": "text" - }, - { - "block_id": "p453-b8", - "global_id": 12703, - "bbox": [ - 447.3, - 395.37, - 452.63, - 403.37 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p453-b9", - "global_id": 12704, - "bbox": [ - 267.98, - 185.32, - 319.76, - 193.32 - ], - "text": "Asymptotic plot", - "type": "text" - }, - { - "block_id": "p453-b10", - "global_id": 12705, - "bbox": [ - 320.48, - 122.08, - 353.14, - 130.08 - ], - "text": "Exact plot", - "type": "text" - }, - { - "block_id": "p453-b11", - "global_id": 12706, - "bbox": [ - 137.12, - 180.09, - 145.42, - 228.94 - ], - "text": "20 log H (dB)", - "type": "text" - }, - { - "block_id": "p453-b12", - "global_id": 12707, - "bbox": [ - 330.25, - 292.23, - 339.13, - 300.23 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p453-b13", - "global_id": 12708, - "bbox": [ - 522.54, - 130.55, - 527.87, - 138.55 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p453-b14", - "global_id": 12709, - "bbox": [ - 523.38, - 370.25, - 528.71, - 378.25 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p453-b15", - "global_id": 12710, - "bbox": [ - 172.93, - 383.45, - 426.32, - 391.45 - ], - "text": "1\n2\n5\n10\n100\n1000\n500", - "type": "text" - }, - { - "block_id": "p453-b16", - "global_id": 12711, - "bbox": [ - 329.87, - 513.11, - 339.51, - 521.11 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p453-b17", - "global_id": 12712, - "bbox": [ - 307.85, - 383.45, - 465.45, - 391.45 - ], - "text": "50\n3000", - "type": "text" - }, - { - "block_id": "p453-b18", - "global_id": 12713, - "bbox": [ - 355.69, - 471.27, - 407.47, - 479.27 - ], - "text": "Asymptotic plot", - "type": "text" - }, - { - "block_id": "p453-b19", - "global_id": 12714, - "bbox": [ - 271.19, - 459.0, - 303.85, - 467.0 - ], - "text": "Exact plot", - "type": "text" - }, - { - "block_id": "p453-b20", - "global_id": 12715, - "bbox": [ - 151.35, - 268.12, - 166.01, - 276.42 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p453-b21", - "global_id": 12716, - "bbox": [ - 151.35, - 215.34, - 166.01, - 223.64 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p453-b22", - "global_id": 12717, - "bbox": [ - 151.35, - 241.73, - 166.01, - 250.02 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p453-b23", - "global_id": 12718, - "bbox": [ - 158.61, - 316.47, - 169.28, - 324.77 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p453-b24", - "global_id": 12719, - "bbox": [ - 158.61, - 343.08, - 169.28, - 351.38 - ], - "text": "50", - "type": "text" - }, - { - "block_id": "p453-b25", - "global_id": 12720, - "bbox": [ - 162.61, - 376.77, - 169.28, - 385.07 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p453-b26", - "global_id": 12721, - "bbox": [ - 151.95, - 410.42, - 169.28, - 418.72 - ], - "text": "50", - "type": "text" - }, - { - "block_id": "p453-b27", - "global_id": 12722, - "bbox": [ - 147.95, - 444.1, - 169.28, - 452.39 - ], - "text": "100", - "type": "text" - }, - { - "block_id": "p453-b28", - "global_id": 12723, - "bbox": [ - 147.95, - 478.67, - 169.28, - 486.97 - ], - "text": "150", - "type": "text" - }, - { - "block_id": "p453-b29", - "global_id": 12724, - "bbox": [ - 147.95, - 498.76, - 169.28, - 507.05 - ], - "text": "180", - "type": "text" - }, - { - "block_id": "p453-b30", - "global_id": 12725, - "bbox": [ - 138.01, - 412.24, - 146.01, - 430.9 - ], - "text": "Phase", - "type": "text" - }, - { - "block_id": "p453-b31", - "global_id": 12726, - "bbox": [ - 136.48, - 527.73, - 427.09, - 537.04 - ], - "text": "Figure 4.46 (a) Amplitude and (b) phase responses of the second-order system.", - "type": "text" - }, - { - "block_id": "p453-b32", - "global_id": 12727, - "bbox": [ - 128.9, - 564.3, - 502.74, - 586.63 - ], - "text": "In addition, we have a real zero at s = −100 and a pair of complex conjugate poles. When\nwe express the second-order factor in standard form,", - "type": "text" - }, - { - "block_id": "p453-b33", - "global_id": 12728, - "bbox": [ - 250.57, - 594.06, - 380.59, - 609.26 - ], - "text": "s2 + 2s + 100 = s2 + 2ζωns + ω2", - "type": "text" - }, - { - "block_id": "p453-b34", - "global_id": 12729, - "bbox": [ - 376.91, - 603.46, - 380.4, - 610.44 - ], - "text": "n", - "type": "text" - } - ] - }, - { - "page_num": 454, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p454-b0", - "global_id": 12730, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "434\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p454-b1", - "global_id": 12731, - "bbox": [ - 103.16, - 86.24, - 135.71, - 96.2 - ], - "text": "we have", - "type": "text" - }, - { - "block_id": "p454-b2", - "global_id": 12732, - "bbox": [ - 231.86, - 97.78, - 348.33, - 109.23 - ], - "text": "ωn = 10\nand\nζ = 0.1", - "type": "text" - }, - { - "block_id": "p454-b3", - "global_id": 12733, - "bbox": [ - 103.17, - 117.7, - 477.02, - 151.99 - ], - "text": "Step 1. Draw an asymptote of −40 dB/decade (−12 dB/octave) starting at ω = 10 for the\ncomplex conjugate poles, and draw another asymptote of 20 dB/decade starting at ω = 100\nfor the (real) zero.", - "type": "text" - }, - { - "block_id": "p454-b4", - "global_id": 12734, - "bbox": [ - 103.17, - 161.88, - 224.09, - 171.92 - ], - "text": "Step 2. Add both asymptotes.", - "type": "text" - }, - { - "block_id": "p454-b5", - "global_id": 12735, - "bbox": [ - 103.17, - 181.47, - 477.02, - 251.63 - ], - "text": "Step 3. Apply the correction at ω = 100, where the correction because of the corner frequency\nω = 100 is 3 dB. The correction because of the corner frequency ω = 10, as seen from\nFig. 4.44a for ζ = 0.1, can be safely ignored. Next, the correction at ω = 10 because of the\ncorner frequency ω = 10 is 13.90 dB (see Fig. 4.44a for ζ = 0.1). The correction because of\nthe real zero at −100 can be safely ignored at ω = 10. We may find corrections at a few more\npoints. The resulting plot is illustrated in Fig. 4.46a.", - "type": "text" - }, - { - "block_id": "p454-b6", - "global_id": 12736, - "bbox": [ - 103.16, - 266.36, - 477.02, - 340.28 - ], - "text": "PHASE PLOT\nThe asymptote for the complex conjugate poles is a step function with a jump of −180◦at\nω = 10. The asymptote for the zero at s = −100 is zero for ω ≤10 and is a straight line with a\nslope of 45◦/decade, starting at ω = 10 and going to ω = 1000. For ω ≥1000, the asymptote\nis 90◦. The two asymptotes add to give the sawtooth shown in Fig. 4.46b. We now apply the\ncorrections from Figs. 4.42b and 4.44b to obtain the exact plot.", - "type": "text" - }, - { - "block_id": "p454-b7", - "global_id": 12737, - "bbox": [ - 128.01, - 450.7, - 140.01, - 458.7 - ], - "text": "–60", - "type": "text" - }, - { - "block_id": "p454-b8", - "global_id": 12738, - "bbox": [ - 128.01, - 433.74, - 140.01, - 441.74 - ], - "text": "–40", - "type": "text" - }, - { - "block_id": "p454-b9", - "global_id": 12739, - "bbox": [ - 128.01, - 416.79, - 140.01, - 424.79 - ], - "text": "–20", - "type": "text" - }, - { - "block_id": "p454-b10", - "global_id": 12740, - "bbox": [ - 136.01, - 399.85, - 140.01, - 407.85 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p454-b11", - "global_id": 12741, - "bbox": [ - 132.01, - 382.89, - 140.01, - 390.89 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p454-b12", - "global_id": 12742, - "bbox": [ - 132.01, - 365.94, - 140.01, - 373.94 - ], - "text": "40", - "type": "text" - }, - { - "block_id": "p454-b13", - "global_id": 12743, - "bbox": [ - 116.75, - 385.99, - 125.55, - 442.45 - ], - "text": "Magnitude (dB)", - "type": "text" - }, - { - "block_id": "p454-b14", - "global_id": 12744, - "bbox": [ - 280.81, - 358.26, - 335.32, - 367.06 - ], - "text": "Bode Diagram", - "type": "text" - }, - { - "block_id": "p454-b15", - "global_id": 12745, - "bbox": [ - 137.56, - 578.88, - 484.82, - 589.11 - ], - "text": "10 0\n10 1\n10 2\n10 3\n10 4", - "type": "text" - }, - { - "block_id": "p454-b16", - "global_id": 12746, - "bbox": [ - 275.95, - 593.34, - 338.26, - 602.14 - ], - "text": "Frequency (rad/s)", - "type": "text" - }, - { - "block_id": "p454-b17", - "global_id": 12747, - "bbox": [ - 124.01, - 569.94, - 140.01, - 577.94 - ], - "text": "–180", - "type": "text" - }, - { - "block_id": "p454-b18", - "global_id": 12748, - "bbox": [ - 124.01, - 548.76, - 140.01, - 556.76 - ], - "text": "–135", - "type": "text" - }, - { - "block_id": "p454-b19", - "global_id": 12749, - "bbox": [ - 128.01, - 527.57, - 140.01, - 535.57 - ], - "text": "–90", - "type": "text" - }, - { - "block_id": "p454-b20", - "global_id": 12750, - "bbox": [ - 128.01, - 506.38, - 140.01, - 514.38 - ], - "text": "–45", - "type": "text" - }, - { - "block_id": "p454-b21", - "global_id": 12751, - "bbox": [ - 136.01, - 485.2, - 140.01, - 493.2 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p454-b22", - "global_id": 12752, - "bbox": [ - 111.5, - 514.03, - 120.3, - 555.33 - ], - "text": "Phase (deg)", - "type": "text" - }, - { - "block_id": "p454-b23", - "global_id": 12753, - "bbox": [ - 110.73, - 608.82, - 323.67, - 618.06 - ], - "text": "Figure 4.47 MATLAB-generated Bode plots for Ex. 4.30.", - "type": "text" - } - ] - }, - { - "page_num": 455, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p455-b0", - "global_id": 12754, - "bbox": [ - 411.08, - 62.89, - 516.14, - 71.98 - ], - "text": "4.9\nBode Plots\n435", - "type": "text" - }, - { - "block_id": "p455-b1", - "global_id": 12755, - "bbox": [ - 128.9, - 86.62, - 502.77, - 136.93 - ], - "text": "BODE PLOTS WITH MATLAB\nBode plots make it relatively simple to hand-draw straight-line approximations to a system’s\nmagnitude and frequency responses. To produce exact Bode plots, we turn to MATLAB and\nits bode command.", - "type": "text" - }, - { - "block_id": "p455-b2", - "global_id": 12756, - "bbox": [ - 128.9, - 146.9, - 332.89, - 156.86 - ], - "text": ">>\nbode(tf([10 1000],[1 2 100]),’k-’);", - "type": "text" - }, - { - "block_id": "p455-b3", - "global_id": 12757, - "bbox": [ - 128.9, - 166.53, - 471.27, - 176.5 - ], - "text": "The resulting MATLAB plots, shown in Fig. 4.47, match the plots shown in Fig. 4.46.", - "type": "text" - }, - { - "block_id": "p455-b4", - "global_id": 12758, - "bbox": [ - 127.59, - 213.0, - 516.12, - 258.92 - ], - "text": "Comment. These two examples demonstrate that actual frequency response plots are very close\nto asymptotic plots, which are so easy to construct. Thus, by mere inspection of H(s) and its poles\nand zeros, one can rapidly construct a mental image of the frequency response of a system. This is\nthe principal virtue of Bode plots.", - "type": "text" - }, - { - "block_id": "p455-b5", - "global_id": 12759, - "bbox": [ - 127.59, - 273.8, - 516.14, - 383.76 - ], - "text": "POLES AND ZEROS IN THE RIGHT HALF-PLANE\nIn our discussion so far, we have assumed the poles and zeros of the transfer function to be in the\nleft half-plane. What if some of the poles and/or zeros of H(s) lie in the RHP? If there is a pole in\nthe RHP, the system is unstable. Such systems are useless for any signal-processing application.\nFor this reason, we shall consider only the case of the RHP zero. The term corresponding to RHP\nzero at s = a is (s/a) −1, and the corresponding frequency response is (jω/a) −1. The amplitude\nresponse is", - "type": "text" - }, - { - "block_id": "p455-b6", - "global_id": 12760, - "bbox": [ - 276.68, - 363.33, - 285.98, - 373.61 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p455-b7", - "global_id": 12761, - "bbox": [ - 278.94, - 370.32, - 303.21, - 387.66 - ], - "text": "a −1", - "type": "text" - }, - { - "block_id": "p455-b8", - "global_id": 12762, - "bbox": [ - 303.21, - 355.86, - 316.27, - 383.76 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p455-b9", - "global_id": 12763, - "bbox": [ - 318.32, - 356.32, - 336.45, - 373.29 - ], - "text": "ω2", - "type": "text" - }, - { - "block_id": "p455-b10", - "global_id": 12764, - "bbox": [ - 327.11, - 370.31, - 353.99, - 387.66 - ], - "text": "a2 + 1", - "type": "text" - }, - { - "block_id": "p455-b11", - "global_id": 12765, - "bbox": [ - 353.99, - 356.32, - 370.98, - 367.31 - ], - "text": "1/2", - "type": "text" - }, - { - "block_id": "p455-b12", - "global_id": 12766, - "bbox": [ - 127.59, - 394.82, - 516.14, - 429.11 - ], - "text": "This shows that the amplitude response of an RHP zero at s = a is identical to that of an LHP zero\nor s = −a. Therefore, the log amplitude plots remain unchanged whether the zeros are in the LHP\nor the RHP. However, the phase corresponding to the RHP zero at s = a is̸", - "type": "text" - }, - { - "block_id": "p455-b13", - "global_id": 12767, - "bbox": [ - 191.55, - 432.6, - 208.76, - 449.88 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p455-b14", - "global_id": 12768, - "bbox": [ - 201.73, - 446.59, - 226.0, - 463.94 - ], - "text": "a −1", - "type": "text" - }, - { - "block_id": "p455-b16", - "global_id": 12769, - "bbox": [ - 234.78, - 446.59, - 259.55, - 456.55 - ], - "text": "≠\n−", - "type": "text" - }, - { - "block_id": "p455-b18", - "global_id": 12770, - "bbox": [ - 267.83, - 439.61, - 294.17, - 463.94 - ], - "text": "1 −jω\na", - "type": "text" - }, - { - "block_id": "p455-b20", - "global_id": 12771, - "bbox": [ - 304.34, - 444.87, - 353.09, - 456.97 - ], - "text": "= π + tan−1", - "type": "text" - }, - { - "block_id": "p455-b21", - "global_id": 12772, - "bbox": [ - 354.71, - 432.6, - 376.92, - 449.57 - ], - "text": "−ω", - "type": "text" - }, - { - "block_id": "p455-b22", - "global_id": 12773, - "bbox": [ - 367.38, - 453.98, - 372.36, - 463.94 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p455-b24", - "global_id": 12774, - "bbox": [ - 387.09, - 444.87, - 435.85, - 456.97 - ], - "text": "= π −tan−1", - "type": "text" - }, - { - "block_id": "p455-b25", - "global_id": 12775, - "bbox": [ - 436.35, - 432.6, - 450.79, - 449.57 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p455-b26", - "global_id": 12776, - "bbox": [ - 445.14, - 453.98, - 450.12, - 463.94 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p455-b28", - "global_id": 12777, - "bbox": [ - 127.59, - 474.44, - 425.74, - 486.05 - ], - "text": "whereas the phase corresponding to the LHP zero at s = −a is tan−1(ω/a).", - "type": "text" - }, - { - "block_id": "p455-b29", - "global_id": 12778, - "bbox": [ - 145.52, - 486.69, - 442.02, - 498.71 - ], - "text": "The complex-conjugate zeros in the RHP give rise to a term s2−2ζωns+ω2", - "type": "text" - }, - { - "block_id": "p455-b30", - "global_id": 12779, - "bbox": [ - 127.59, - 488.04, - 516.12, - 510.66 - ], - "text": "n, which is identical\nto the term s2 +2ζωns+ω2", - "type": "text" - }, - { - "block_id": "p455-b31", - "global_id": 12780, - "bbox": [ - 127.59, - 499.58, - 516.14, - 521.91 - ], - "text": "n with a sign change in ζ. Hence, from Eqs. (4.51) and (4.52), it follows\nthat the amplitudes are identical, but the phases are of opposite signs for the two terms.", - "type": "text" - }, - { - "block_id": "p455-b32", - "global_id": 12781, - "bbox": [ - 127.59, - 523.81, - 516.13, - 557.78 - ], - "text": "Systems whose poles and zeros are restricted to the LHP are classified as minimum phase\nsystems. Minimum phase systems are particularly desirable because the system and its inverse are\nboth stable.", - "type": "text" - }, - { - "block_id": "p455-b33", - "global_id": 12782, - "bbox": [ - 127.59, - 582.83, - 444.22, - 594.78 - ], - "text": "4.9-5 The Transfer Function from the Frequency Response", - "type": "text" - }, - { - "block_id": "p455-b34", - "global_id": 12783, - "bbox": [ - 127.59, - 600.91, - 516.15, - 634.79 - ], - "text": "In the preceding section we were given the transfer function of a system. From a knowledge of\nthe transfer function, we developed techniques for determining the system response to sinusoidal\ninputs. We can also reverse the procedure to determine the transfer function of a minimum phase", - "type": "text" - } - ] - }, - { - "page_num": 456, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p456-b0", - "global_id": 12784, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "436\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p456-b1", - "global_id": 12785, - "bbox": [ - 101.84, - 85.82, - 490.41, - 251.2 - ], - "text": "system from the system’s response to sinusoids. This application has significant practical utility. If\nwe are given a system in a black box with only the input and output terminals available, the transfer\nfunction has to be determined by experimental measurements at the input and output terminals.\nThe frequency response to sinusoidal inputs is one of the possibilities that is very attractive because\nthe measurements involved are so simple. One needs only to apply a sinusoidal signal at the input\nand observe the output. We find the amplitude gain |H(jω)| and the output phase shift̸\nH(jω)\n(with respect to the input sinusoid) for various values of ω over the entire range from 0 to ∞. This\ninformation yields the frequency response plots (Bode plots) when plotted against log ω. From\nthese plots we determine the appropriate asymptotes by taking advantage of the fact that the slopes\nof all asymptotes must be multiples of ±20 dB/decade if the transfer function is a rational function\n(function that is a ratio of two polynomials in s). From the asymptotes, the corner frequencies are\nobtained. Corner frequencies determine the poles and zeros of the transfer function. Because of the\nambiguity about the location of zeros since LHP and RHP zeros (zeros at s = ±a) have identical\nmagnitudes, this procedure works only for minimum phase systems.", - "type": "text" - }, - { - "block_id": "p456-b2", - "global_id": 12786, - "bbox": [ - 102.19, - 282.94, - 403.2, - 312.83 - ], - "text": "4.10 FILTER DESIGN BY PLACEMENT OF POLES\nAND ZEROS OF H(s)", - "type": "text" - }, - { - "block_id": "p456-b3", - "global_id": 12787, - "bbox": [ - 101.84, - 318.82, - 490.37, - 340.74 - ], - "text": "In this section we explore the strong dependence of frequency response on the location of poles\nand zeros of H(s). This dependence points to a simple intuitive procedure to filter design.", - "type": "text" - }, - { - "block_id": "p456-b4", - "global_id": 12788, - "bbox": [ - 101.84, - 367.96, - 385.6, - 393.86 - ], - "text": "4.10-1 Dependence of Frequency Response on Poles\nand Zeros of H(s)", - "type": "text" - }, - { - "block_id": "p456-b5", - "global_id": 12789, - "bbox": [ - 101.84, - 399.99, - 490.41, - 421.91 - ], - "text": "Frequency response of a system is basically the information about the filtering capability of the\nsystem. A system transfer function can be expressed as", - "type": "text" - }, - { - "block_id": "p456-b6", - "global_id": 12790, - "bbox": [ - 202.8, - 433.96, - 252.77, - 451.22 - ], - "text": "H(s) = P(s)", - "type": "text" - }, - { - "block_id": "p456-b7", - "global_id": 12791, - "bbox": [ - 234.81, - 440.95, - 274.84, - 458.3 - ], - "text": "Q(s) = b0", - "type": "text" - }, - { - "block_id": "p456-b8", - "global_id": 12792, - "bbox": [ - 276.53, - 433.96, - 388.24, - 459.17 - ], - "text": "(s −z1)(s −z2)· · ·(s −zN)\n(s −λ1)(s −λ2)· · ·(s −λN)", - "type": "text" - }, - { - "block_id": "p456-b9", - "global_id": 12793, - "bbox": [ - 101.84, - 470.03, - 490.39, - 492.36 - ], - "text": "where z1, z2, . . . , zN are λ1, λ2, . . . , λN are the poles of H(s). Now the value of the transfer\nfunction H(s) at some frequency s = p is", - "type": "text" - }, - { - "block_id": "p456-b10", - "global_id": 12794, - "bbox": [ - 210.02, - 511.4, - 264.29, - 522.86 - ], - "text": "H(s)|s=p = b0", - "type": "text" - }, - { - "block_id": "p456-b11", - "global_id": 12795, - "bbox": [ - 265.99, - 504.41, - 490.38, - 529.62 - ], - "text": "(p −z1)(p −z2)· · ·(p −zN)\n(p −λ1)(p −λ2)· · ·(p −λN)\n(4.53)", - "type": "text" - }, - { - "block_id": "p456-b12", - "global_id": 12796, - "bbox": [ - 101.84, - 540.72, - 490.4, - 636.28 - ], - "text": "This equation consists of factors of the form p−zi and p−λi. The factor p−zi is a complex number\nrepresented by a vector drawn from point z to the point p in the complex plane, as illustrated in\nFig. 4.48a. The length of this line segment is |p −zi|, the magnitude of p −zi. The angle of this\ndirected line segment (with the horizontal axis) is̸\n(p −zi). To compute H(s) at s = p, we draw\nline segments from all poles and zeros of H(s) to the point p, as shown in Fig. 4.48b. The vector\nconnecting a zero zi to the point p is p −zi. Let the length of this vector be ri, and let its angle\nwith the horizontal axis be φi. Then p−zi = riejφi. Similarly, the vector connecting a pole λi to the\npoint p is p −λi = diejθi, where di and θi are the length and the angle (with the horizontal axis),", - "type": "text" - } - ] - }, - { - "page_num": 457, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p457-b0", - "global_id": 12797, - "bbox": [ - 237.2, - 62.29, - 516.13, - 71.98 - ], - "text": "4.10\nFilter Design by Placement of Poles and Zeros of H(s)\n437", - "type": "text" - }, - { - "block_id": "p457-b1", - "global_id": 12798, - "bbox": [ - 249.9, - 190.54, - 260.78, - 198.54 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p457-b2", - "global_id": 12799, - "bbox": [ - 248.1, - 99.03, - 256.99, - 107.03 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p457-b3", - "global_id": 12800, - "bbox": [ - 362.74, - 231.74, - 372.54, - 239.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p457-b4", - "global_id": 12801, - "bbox": [ - 429.86, - 190.79, - 438.75, - 198.79 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p457-b5", - "global_id": 12802, - "bbox": [ - 205.54, - 231.74, - 214.42, - 239.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p457-b6", - "global_id": 12803, - "bbox": [ - 390.56, - 99.03, - 399.45, - 107.03 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p457-b7", - "global_id": 12804, - "bbox": [ - 310.66, - 127.46, - 317.66, - 137.07 - ], - "text": "d1", - "type": "text" - }, - { - "block_id": "p457-b8", - "global_id": 12805, - "bbox": [ - 209.1, - 87.6, - 355.87, - 95.9 - ], - "text": "p\np", - "type": "text" - }, - { - "block_id": "p457-b9", - "global_id": 12806, - "bbox": [ - 169.05, - 109.85, - 230.92, - 129.77 - ], - "text": "p\np zi", - "type": "text" - }, - { - "block_id": "p457-b10", - "global_id": 12807, - "bbox": [ - 164.42, - 135.7, - 169.2, - 145.24 - ], - "text": "zi", - "type": "text" - }, - { - "block_id": "p457-b11", - "global_id": 12808, - "bbox": [ - 189.54, - 155.69, - 194.32, - 165.24 - ], - "text": "zi", - "type": "text" - }, - { - "block_id": "p457-b12", - "global_id": 12809, - "bbox": [ - 366.49, - 141.38, - 404.5, - 155.06 - ], - "text": "r1\nr2", - "type": "text" - }, - { - "block_id": "p457-b13", - "global_id": 12810, - "bbox": [ - 367.88, - 188.96, - 373.99, - 198.57 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p457-b14", - "global_id": 12811, - "bbox": [ - 337.53, - 143.38, - 344.53, - 152.99 - ], - "text": "d2", - "type": "text" - }, - { - "block_id": "p457-b15", - "global_id": 12812, - "bbox": [ - 420.47, - 188.96, - 426.58, - 198.57 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p457-b16", - "global_id": 12813, - "bbox": [ - 375.44, - 174.28, - 438.93, - 184.16 - ], - "text": "f2\nf1", - "type": "text" - }, - { - "block_id": "p457-b17", - "global_id": 12814, - "bbox": [ - 315.58, - 206.2, - 322.58, - 215.83 - ], - "text": "u2", - "type": "text" - }, - { - "block_id": "p457-b18", - "global_id": 12815, - "bbox": [ - 291.07, - 151.74, - 298.52, - 161.37 - ], - "text": "l1", - "type": "text" - }, - { - "block_id": "p457-b19", - "global_id": 12816, - "bbox": [ - 291.5, - 212.05, - 298.95, - 221.67 - ], - "text": "l2", - "type": "text" - }, - { - "block_id": "p457-b20", - "global_id": 12817, - "bbox": [ - 315.58, - 147.82, - 322.58, - 157.45 - ], - "text": "u1", - "type": "text" - }, - { - "block_id": "p457-b21", - "global_id": 12818, - "bbox": [ - 151.5, - 246.07, - 456.2, - 255.68 - ], - "text": "Figure 4.48 Vector representations of (a) complex numbers and (b) factors of H(s).", - "type": "text" - }, - { - "block_id": "p457-b22", - "global_id": 12819, - "bbox": [ - 127.59, - 275.32, - 402.25, - 286.4 - ], - "text": "respectively, of the vector p −λi. Now from Eq. (4.53) it follows that", - "type": "text" - }, - { - "block_id": "p457-b23", - "global_id": 12820, - "bbox": [ - 210.8, - 302.8, - 265.06, - 314.25 - ], - "text": "H(s)|s=p = b0", - "type": "text" - }, - { - "block_id": "p457-b24", - "global_id": 12821, - "bbox": [ - 266.76, - 294.59, - 374.43, - 321.02 - ], - "text": "(r1ejφ1)(r2ejφ2)· · ·(rNejφN)\n(d1ejθ1)(d2ejθ2)· · ·(dNejθN)", - "type": "text" - }, - { - "block_id": "p457-b25", - "global_id": 12822, - "bbox": [ - 246.78, - 328.96, - 265.06, - 340.11 - ], - "text": "= b0", - "type": "text" - }, - { - "block_id": "p457-b26", - "global_id": 12823, - "bbox": [ - 266.76, - 321.98, - 309.61, - 347.49 - ], - "text": "r1r2 · · ·rN\nd1d2 · · ·dN", - "type": "text" - }, - { - "block_id": "p457-b27", - "global_id": 12824, - "bbox": [ - 312.83, - 327.24, - 432.42, - 339.24 - ], - "text": "ej[(φ1+φ2+···+φN)−(θ1+θ2+···+θN)]", - "type": "text" - }, - { - "block_id": "p457-b28", - "global_id": 12825, - "bbox": [ - 127.59, - 354.88, - 166.86, - 364.84 - ], - "text": "Therefore", - "type": "text" - }, - { - "block_id": "p457-b29", - "global_id": 12826, - "bbox": [ - 192.23, - 369.32, - 249.35, - 380.78 - ], - "text": "|H(s)|s=p = b0", - "type": "text" - }, - { - "block_id": "p457-b30", - "global_id": 12827, - "bbox": [ - 251.05, - 362.34, - 293.88, - 387.86 - ], - "text": "r1r2 · · ·rN\nd1d2 · · ·dN", - "type": "text" - }, - { - "block_id": "p457-b31", - "global_id": 12828, - "bbox": [ - 298.05, - 369.32, - 316.34, - 380.47 - ], - "text": "= b0", - "type": "text" - }, - { - "block_id": "p457-b32", - "global_id": 12829, - "bbox": [ - 318.03, - 362.65, - 516.13, - 386.78 - ], - "text": "product of distances of zeros to p\nproduct of distances of poles to p\n(4.54)", - "type": "text" - }, - { - "block_id": "p457-b33", - "global_id": 12830, - "bbox": [ - 127.59, - 393.28, - 141.97, - 403.24 - ], - "text": "and̸", - "type": "text" - }, - { - "block_id": "p457-b34", - "global_id": 12831, - "bbox": [ - 188.98, - 413.89, - 409.35, - 425.35 - ], - "text": "H(s)|s=p = (φ1 + φ2 + · · · + φN) −(θ1 + θ2 + · · · + θN)", - "type": "text" - }, - { - "block_id": "p457-b35", - "global_id": 12832, - "bbox": [ - 224.96, - 428.83, - 516.13, - 439.21 - ], - "text": "= sum of angles of zeros to p −sum of angles of poles to p\n(4.55)", - "type": "text" - }, - { - "block_id": "p457-b36", - "global_id": 12833, - "bbox": [ - 127.59, - 449.87, - 516.13, - 508.06 - ], - "text": "Here, we have assumed positive b0. If b0 is negative, there is an additional phase π. Using this\nprocedure, we can determine H(s) for any value of s. To compute the frequency response H(jω),\nwe use s = jω (a point on the imaginary axis), connect all poles and zeros to the point jω, and\ndetermine |H(jω)| and̸ H(jω) from Eqs. (4.54) and (4.55). We repeat this procedure for all values\nof ω from 0 to ∞to obtain the frequency response.", - "type": "text" - }, - { - "block_id": "p457-b37", - "global_id": 12834, - "bbox": [ - 127.59, - 522.58, - 516.16, - 584.55 - ], - "text": "GAIN ENHANCEMENT BY A POLE\nTo understand the effect of poles and zeros on the frequency response, consider a hypothetical case\nof a single pole −α + jω0, as depicted in Fig. 4.49a. To find the amplitude response |H(jω)| for a\ncertain value of ω, we connect the pole to the point jω (Fig. 4.49a). If the length of this line is d,\nthen |H(jω)| is proportional to 1/d,", - "type": "text" - }, - { - "block_id": "p457-b38", - "global_id": 12835, - "bbox": [ - 296.0, - 584.28, - 345.97, - 601.23 - ], - "text": "|H(jω)| = K", - "type": "text" - }, - { - "block_id": "p457-b39", - "global_id": 12836, - "bbox": [ - 340.3, - 591.36, - 516.12, - 608.3 - ], - "text": "d\n(4.56)", - "type": "text" - }, - { - "block_id": "p457-b40", - "global_id": 12837, - "bbox": [ - 127.59, - 612.45, - 516.12, - 635.56 - ], - "text": "where the exact value of constant K is not important at this point. As ω increases from zero,\nd decreases progressively until ω reaches the value ω0. As ω increases beyond ω0, d increases", - "type": "text" - } - ] - }, - { - "page_num": 458, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p458-b0", - "global_id": 12838, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "438\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p458-b1", - "global_id": 12839, - "bbox": [ - 160.69, - 415.74, - 170.02, - 423.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p458-b2", - "global_id": 12840, - "bbox": [ - 176.09, - 345.07, - 184.98, - 353.07 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p458-b3", - "global_id": 12841, - "bbox": [ - 176.09, - 172.9, - 184.98, - 180.9 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p458-b4", - "global_id": 12842, - "bbox": [ - 148.36, - 127.7, - 152.36, - 135.7 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p458-b5", - "global_id": 12843, - "bbox": [ - 168.36, - 356.46, - 172.36, - 364.46 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p458-b6", - "global_id": 12844, - "bbox": [ - 160.92, - 237.78, - 169.8, - 245.78 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p458-b7", - "global_id": 12845, - "bbox": [ - 151.06, - 300.66, - 154.17, - 308.66 - ], - "text": "r", - "type": "text" - }, - { - "block_id": "p458-b8", - "global_id": 12846, - "bbox": [ - 168.36, - 184.58, - 172.36, - 192.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p458-b9", - "global_id": 12847, - "bbox": [ - 225.26, - 370.44, - 229.26, - 378.44 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p458-b10", - "global_id": 12848, - "bbox": [ - 268.97, - 415.74, - 277.85, - 423.74 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p458-b11", - "global_id": 12849, - "bbox": [ - 225.36, - 197.91, - 229.36, - 205.91 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p458-b12", - "global_id": 12850, - "bbox": [ - 268.58, - 237.78, - 278.39, - 245.78 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p458-b13", - "global_id": 12851, - "bbox": [ - 357.86, - 369.64, - 361.86, - 377.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p458-b14", - "global_id": 12852, - "bbox": [ - 402.73, - 415.74, - 411.68, - 423.74 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p458-b15", - "global_id": 12853, - "bbox": [ - 357.86, - 174.48, - 361.86, - 182.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p458-b16", - "global_id": 12854, - "bbox": [ - 402.77, - 237.78, - 411.65, - 245.78 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p458-b17", - "global_id": 12855, - "bbox": [ - 168.36, - 95.72, - 177.25, - 103.72 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p458-b18", - "global_id": 12856, - "bbox": [ - 168.36, - 267.89, - 177.25, - 275.89 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p458-b19", - "global_id": 12857, - "bbox": [ - 152.63, - 199.48, - 159.3, - 207.69 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p458-b20", - "global_id": 12858, - "bbox": [ - 153.14, - 369.35, - 158.92, - 377.57 - ], - "text": "r", - "type": "text" - }, - { - "block_id": "p458-b21", - "global_id": 12859, - "bbox": [ - 169.45, - 131.71, - 177.01, - 139.73 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p458-b22", - "global_id": 12860, - "bbox": [ - 130.11, - 182.12, - 141.66, - 190.31 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p458-b23", - "global_id": 12861, - "bbox": [ - 130.11, - 354.01, - 141.66, - 362.2 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p458-b24", - "global_id": 12862, - "bbox": [ - 164.17, - 113.73, - 174.73, - 123.36 - ], - "text": "jv0", - "type": "text" - }, - { - "block_id": "p458-b25", - "global_id": 12863, - "bbox": [ - 253.01, - 197.81, - 261.34, - 207.44 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p458-b26", - "global_id": 12864, - "bbox": [ - 254.41, - 370.34, - 262.75, - 379.97 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p458-b27", - "global_id": 12865, - "bbox": [ - 169.45, - 303.0, - 177.01, - 311.02 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p458-b28", - "global_id": 12866, - "bbox": [ - 149.81, - 108.06, - 156.81, - 117.69 - ], - "text": "u1", - "type": "text" - }, - { - "block_id": "p458-b29", - "global_id": 12867, - "bbox": [ - 152.44, - 216.41, - 159.44, - 226.04 - ], - "text": "u2", - "type": "text" - }, - { - "block_id": "p458-b30", - "global_id": 12868, - "bbox": [ - 355.32, - 334.74, - 360.66, - 342.74 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p458-b31", - "global_id": 12869, - "bbox": [ - 348.66, - 197.87, - 360.66, - 206.06 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p458-b32", - "global_id": 12870, - "bbox": [ - 151.97, - 279.26, - 160.31, - 288.89 - ], - "text": "f1", - "type": "text" - }, - { - "block_id": "p458-b33", - "global_id": 12871, - "bbox": [ - 152.33, - 389.81, - 160.67, - 399.43 - ], - "text": "f2", - "type": "text" - }, - { - "block_id": "p458-b34", - "global_id": 12872, - "bbox": [ - 318.38, - 370.24, - 453.02, - 378.24 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p458-b35", - "global_id": 12873, - "bbox": [ - 317.94, - 196.36, - 323.28, - 204.36 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p458-b36", - "global_id": 12874, - "bbox": [ - 447.24, - 175.51, - 452.58, - 183.51 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p458-b37", - "global_id": 12875, - "bbox": [ - 235.0, - 110.34, - 259.3, - 118.64 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p458-b38", - "global_id": 12876, - "bbox": [ - 233.18, - 283.14, - 257.47, - 291.44 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p458-b39", - "global_id": 12877, - "bbox": [ - 367.08, - 119.11, - 394.01, - 127.4 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p458-b40", - "global_id": 12878, - "bbox": [ - 366.92, - 312.68, - 393.85, - 320.97 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p458-b41", - "global_id": 12879, - "bbox": [ - 164.17, - 285.64, - 174.73, - 295.26 - ], - "text": "jv0", - "type": "text" - }, - { - "block_id": "p458-b42", - "global_id": 12880, - "bbox": [ - 125.76, - 430.43, - 483.87, - 439.67 - ], - "text": "Figure 4.49 The role of poles and zeros in determining the frequency response of an LTIC system.", - "type": "text" - }, - { - "block_id": "p458-b43", - "global_id": 12881, - "bbox": [ - 101.84, - 468.99, - 490.39, - 599.7 - ], - "text": "progressively. Therefore, according to Eq. (4.56), the amplitude response |H(jω)| increases from\nω = 0 until ω = ω0, and it decreases continuously as ω increases beyond ω0, as illustrated in\nFig. 4.49b. Therefore, a pole at −α + jω0 results in a frequency-selective behavior that enhances\nthe gain at the frequency ω0 (resonance). Moreover, as the pole moves closer to the imaginary axis\n(as α is reduced), this enhancement (resonance) becomes more pronounced. This is because α,\nthe distance between the pole and jω0 (d corresponding to jω0), becomes smaller, which increases\nthe gain K/d. In the extreme case, when α = 0 (pole on the imaginary axis), the gain at ω0 goes\nto infinity. Repeated poles further enhance the frequency-selective effect. To summarize, we can\nenhance a gain at a frequency ω0 by placing a pole opposite the point jω0. The closer the pole is\nto jω0, the higher is the gain at ω0, and the gain variation is more rapid (more frequency selective)\nin the vicinity of frequency ω0. Note that a pole must be placed in the LHP for stability.", - "type": "text" - }, - { - "block_id": "p458-b44", - "global_id": 12882, - "bbox": [ - 101.84, - 600.91, - 490.39, - 634.79 - ], - "text": "Here we have considered the effect of a single complex pole on the system gain. For a\nreal system, a complex pole −α + jω0 must accompany its conjugate −α −jω0. We can readily\nshow that the presence of the conjugate pole does not appreciably change the frequency-selective", - "type": "text" - } - ] - }, - { - "page_num": 459, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p459-b0", - "global_id": 12883, - "bbox": [ - 237.2, - 62.29, - 516.13, - 71.98 - ], - "text": "4.10\nFilter Design by Placement of Poles and Zeros of H(s)\n439", - "type": "text" - }, - { - "block_id": "p459-b1", - "global_id": 12884, - "bbox": [ - 127.59, - 82.76, - 516.14, - 144.15 - ], - "text": "behavior in the vicinity of ω0. This is because the gain in this case is K/dd′, where d′ is the distance\nof a point jω from the conjugate pole −α −jω0. Because the conjugate pole is far from jω0, there\nis no dramatic change in the length d′ as ω varies in the vicinity of ω0. There is a gradual increase\nin the value of d′ as ω increases, which leaves the frequency-selective behavior as it was originally,\nwith only minor changes.", - "type": "text" - }, - { - "block_id": "p459-b2", - "global_id": 12885, - "bbox": [ - 127.59, - 158.53, - 516.15, - 256.37 - ], - "text": "GAIN SUPPRESSION BY A ZERO\nUsing the same argument, we observe that zeros at −α ± jω0 (Fig. 4.49d) will have exactly the\nopposite effect of suppressing the gain in the vicinity of ω0, as shown in Fig. 4.49e). A zero on\nthe imaginary axis at jω0 will totally suppress the gain (zero gain) at frequency ω0. Repeated zeros\nwill further enhance the effect. Also, a closely placed pair of a pole and a zero (dipole) tend to\ncancel out each other’s influence on the frequency response. Clearly, a proper placement of poles\nand zeros can yield a variety of frequency-selective behavior. We can use these observations to\ndesign lowpass, highpass, bandpass, and bandstop (or notch) filters.", - "type": "text" - }, - { - "block_id": "p459-b3", - "global_id": 12886, - "bbox": [ - 127.59, - 258.36, - 516.14, - 352.01 - ], - "text": "Phase response can also be computed graphically. In Fig. 4.49a, angles formed by the complex\nconjugate poles −α±jω0 at ω = 0 (the origin) are equal and opposite. As ω increases from 0 up, the\nangle θ1 (due to the pole −α +jω0), which has a negative value at ω = 0, is reduced in magnitude;\nthe angle θ2 because of the pole −α −jω0, which has a positive value at ω = 0, increases in\nmagnitude. As a result, θ1 + θ2, the sum of the two angles, increases continuously, approaching a\nvalue π as ω →∞. The resulting phase response̸\nH(jω) = −(θ1 +θ2) is illustrated in Fig. 4.49c.\nSimilar arguments apply to zeros at −α ± jω0. The resulting phase response̸\nH(jω) = (φ1 + φ2)\nis depicted in Fig. 4.49f.", - "type": "text" - }, - { - "block_id": "p459-b4", - "global_id": 12887, - "bbox": [ - 127.59, - 354.01, - 516.11, - 375.92 - ], - "text": "We now focus on simple filters, using the intuitive insights gained in this discussion. The\ndiscussion is essentially qualitative.", - "type": "text" - }, - { - "block_id": "p459-b5", - "global_id": 12888, - "bbox": [ - 127.59, - 399.95, - 249.27, - 411.9 - ], - "text": "4.10-2 Lowpass Filters", - "type": "text" - }, - { - "block_id": "p459-b6", - "global_id": 12889, - "bbox": [ - 127.59, - 417.62, - 516.15, - 451.91 - ], - "text": "A typical lowpass filter has a maximum gain at ω = 0. Because a pole enhances the gain at\nfrequencies in its vicinity, we need to place a pole (or poles) on the real axis opposite the origin\n(jω = 0), as shown in Fig. 4.50a. The transfer function of this system is", - "type": "text" - }, - { - "block_id": "p459-b7", - "global_id": 12890, - "bbox": [ - 292.82, - 457.57, - 349.2, - 482.71 - ], - "text": "H(s) =\nωc\ns + ωc", - "type": "text" - }, - { - "block_id": "p459-b8", - "global_id": 12891, - "bbox": [ - 127.59, - 489.55, - 516.13, - 513.37 - ], - "text": "We have chosen the numerator of H(s) to be ωc to normalize the dc gain H(0) to unity. If d is the\ndistance from the pole −ωc to a point jω (Fig. 4.50a), then", - "type": "text" - }, - { - "block_id": "p459-b9", - "global_id": 12892, - "bbox": [ - 294.54, - 517.54, - 347.49, - 534.8 - ], - "text": "|H(jω)| = ωc", - "type": "text" - }, - { - "block_id": "p459-b10", - "global_id": 12893, - "bbox": [ - 340.3, - 531.91, - 345.28, - 541.87 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p459-b11", - "global_id": 12894, - "bbox": [ - 127.59, - 548.06, - 516.13, - 582.35 - ], - "text": "with H(0) = 1. As ω increases, d increases and |H(jω)| decreases monotonically with ω, as\nillustrated in Fig. 4.50d with label N = 1. This is clearly a lowpass filter with gain enhanced\nin the vicinity of ω = 0.", - "type": "text" - }, - { - "block_id": "p459-b12", - "global_id": 12895, - "bbox": [ - 127.59, - 596.72, - 516.16, - 635.49 - ], - "text": "WALL OF POLES\nAn ideal lowpass filter characteristic (shaded in Fig. 4.50d) has a constant gain of unity up to\nfrequency ωc. Then the gain drops suddenly to 0 for ω > ωc. To achieve the ideal lowpass", - "type": "text" - } - ] - }, - { - "page_num": 460, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p460-b0", - "global_id": 12896, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "440\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p460-b1", - "global_id": 12897, - "bbox": [ - 181.04, - 210.77, - 189.92, - 218.77 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p460-b2", - "global_id": 12898, - "bbox": [ - 188.48, - 92.81, - 208.48, - 101.11 - ], - "text": "N 1", - "type": "text" - }, - { - "block_id": "p460-b3", - "global_id": 12899, - "bbox": [ - 188.48, - 149.17, - 217.93, - 157.23 - ], - "text": "Re \n0", - "type": "text" - }, - { - "block_id": "p460-b4", - "global_id": 12900, - "bbox": [ - 160.13, - 125.72, - 196.84, - 134.36 - ], - "text": "jv\nd", - "type": "text" - }, - { - "block_id": "p460-b5", - "global_id": 12901, - "bbox": [ - 175.31, - 277.69, - 199.13, - 285.99 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p460-b6", - "global_id": 12902, - "bbox": [ - 209.74, - 341.3, - 213.74, - 349.3 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p460-b7", - "global_id": 12903, - "bbox": [ - 302.46, - 356.74, - 311.79, - 364.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p460-b8", - "global_id": 12904, - "bbox": [ - 295.42, - 241.73, - 340.08, - 250.02 - ], - "text": "Ideal (N \r)", - "type": "text" - }, - { - "block_id": "p460-b9", - "global_id": 12905, - "bbox": [ - 331.42, - 281.98, - 351.42, - 290.27 - ], - "text": "N 1", - "type": "text" - }, - { - "block_id": "p460-b10", - "global_id": 12906, - "bbox": [ - 333.06, - 301.12, - 353.06, - 309.41 - ], - "text": "N 2", - "type": "text" - }, - { - "block_id": "p460-b11", - "global_id": 12907, - "bbox": [ - 292.42, - 316.83, - 343.56, - 335.13 - ], - "text": "N 4\n8\n10", - "type": "text" - }, - { - "block_id": "p460-b12", - "global_id": 12908, - "bbox": [ - 210.54, - 238.23, - 214.54, - 246.23 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p460-b13", - "global_id": 12909, - "bbox": [ - 277.02, - 339.25, - 400.0, - 348.82 - ], - "text": "vc\nv", - "type": "text" - }, - { - "block_id": "p460-b14", - "global_id": 12910, - "bbox": [ - 411.61, - 210.77, - 420.49, - 218.77 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p460-b15", - "global_id": 12911, - "bbox": [ - 422.18, - 114.81, - 442.18, - 123.11 - ], - "text": "N 5", - "type": "text" - }, - { - "block_id": "p460-b16", - "global_id": 12912, - "bbox": [ - 418.54, - 149.31, - 445.72, - 157.31 - ], - "text": "Re\n0", - "type": "text" - }, - { - "block_id": "p460-b17", - "global_id": 12913, - "bbox": [ - 419.03, - 92.04, - 429.25, - 101.6 - ], - "text": "jvc", - "type": "text" - }, - { - "block_id": "p460-b18", - "global_id": 12914, - "bbox": [ - 417.56, - 192.39, - 434.45, - 202.15 - ], - "text": "jvc", - "type": "text" - }, - { - "block_id": "p460-b19", - "global_id": 12915, - "bbox": [ - 295.5, - 210.77, - 305.31, - 218.77 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p460-b20", - "global_id": 12916, - "bbox": [ - 303.45, - 149.32, - 330.87, - 157.32 - ], - "text": "Re\n0", - "type": "text" - }, - { - "block_id": "p460-b21", - "global_id": 12917, - "bbox": [ - 306.54, - 87.62, - 316.76, - 97.18 - ], - "text": "jvc", - "type": "text" - }, - { - "block_id": "p460-b22", - "global_id": 12918, - "bbox": [ - 304.48, - 195.72, - 321.37, - 205.48 - ], - "text": "jvc", - "type": "text" - }, - { - "block_id": "p460-b23", - "global_id": 12919, - "bbox": [ - 125.76, - 149.06, - 140.42, - 158.63 - ], - "text": "2vc", - "type": "text" - }, - { - "block_id": "p460-b24", - "global_id": 12920, - "bbox": [ - 125.76, - 371.44, - 480.22, - 380.68 - ], - "text": "Figure 4.50 Pole-zero configuration and the amplitude response of a lowpass (Butterworth) filter.", - "type": "text" - }, - { - "block_id": "p460-b25", - "global_id": 12921, - "bbox": [ - 101.84, - 401.61, - 490.42, - 592.01 - ], - "text": "characteristic, we need enhanced gain over the entire frequency band from 0 to ωc. We know\nthat to enhance a gain at any frequency ω, we need to place a pole opposite ω. To achieve an\nenhanced gain for all frequencies over the band (0 to ωc), we need to place a pole opposite every\nfrequency in this band. In other words, we need a continuous wall of poles facing the imaginary\naxis opposite the frequency band 0 to ωc (and from 0 to −ωc for conjugate poles), as depicted in\nFig. 4.50b. At this point, the optimum shape of this wall is not obvious because our arguments\nare qualitative and intuitive. Yet, it is certain that to have enhanced gain (constant gain) at every\nfrequency over this range, we need an infinite number of poles on this wall. We can show that\nfor a maximally flat† response over the frequency range (0 to ωc), the wall is a semicircle with an\ninfinite number of poles uniformly distributed along the wall [11]. In practice, we compromise by\nusing a finite number (N) of poles with less-than-ideal characteristics. Figure 4.50c shows the pole\nconfiguration for a fifth-order (N = 5) filter. The amplitude response for various values of N is\nillustrated in Fig. 4.50d. As N →∞, the filter response approaches the ideal. This family of filters\nis known as the Butterworth filters. There are also other families. In Chebyshev filters, the wall\nshape is a semiellipse rather than a semicircle. The characteristics of a Chebyshev filter are inferior\nto those of Butterworth over the passband (0,ωc), where the characteristics show a rippling effect", - "type": "text" - }, - { - "block_id": "p460-b26", - "global_id": 12922, - "bbox": [ - 101.84, - 610.24, - 490.37, - 633.41 - ], - "text": "† Maximally flat amplitude response means the first 2N −1 derivatives of |H(jω)| with respect to ω are zero\nat ω = 0.", - "type": "text" - } - ] - }, - { - "page_num": 461, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p461-b0", - "global_id": 12923, - "bbox": [ - 237.2, - 62.29, - 516.13, - 71.98 - ], - "text": "4.10\nFilter Design by Placement of Poles and Zeros of H(s)\n441", - "type": "text" - }, - { - "block_id": "p461-b1", - "global_id": 12924, - "bbox": [ - 127.59, - 85.4, - 516.1, - 107.74 - ], - "text": "instead of the maximally flat response of Butterworth. But in the stopband (ω > ωc), Chebyshev\nbehavior is superior in the sense that Chebyshev filter gain drops faster than that of the Butterworth.", - "type": "text" - }, - { - "block_id": "p461-b2", - "global_id": 12925, - "bbox": [ - 127.59, - 132.46, - 253.92, - 144.41 - ], - "text": "4.10-3 Bandpass Filters", - "type": "text" - }, - { - "block_id": "p461-b3", - "global_id": 12926, - "bbox": [ - 127.59, - 150.53, - 516.15, - 220.28 - ], - "text": "The shaded characteristic in Fig. 4.51b shows the ideal bandpass filter gain. In the bandpass\nfilter, the gain is enhanced over the entire passband. Our earlier discussion indicates that this\ncan be realized by a wall of poles opposite the imaginary axis in front of the passband centered\nat ω0. (There is also a wall of conjugate poles opposite −ω0.) Ideally, an infinite number of\npoles is required. In practice, we compromise by using a finite number of poles and accepting\nless-than-ideal characteristics (Fig. 4.51).", - "type": "text" - }, - { - "block_id": "p461-b4", - "global_id": 12927, - "bbox": [ - 127.59, - 244.99, - 297.41, - 256.95 - ], - "text": "4.10-4 Notch (Bandstop) Filters", - "type": "text" - }, - { - "block_id": "p461-b5", - "global_id": 12928, - "bbox": [ - 127.59, - 263.08, - 516.15, - 428.45 - ], - "text": "An ideal notch filter amplitude response (shaded in Fig. 4.52b) is a complement of the amplitude\nresponse of an ideal bandpass filter. Its gain is zero over a small band centered at some frequency\nω0 and is unity over the remaining frequencies. Realization of such a characteristic requires an\ninfinite number of poles and zeros. Let us consider a practical second-order notch filter to obtain\nzero gain at a frequency ω = ω0. For this purpose, we must have zeros at ±jω0. The requirement\nof unity gain at ω = ∞requires the number of poles to be equal to the number of zeros (M = N).\nThis ensures that for very large values of ω, the product of the distances of poles from ω will\nbe equal to the product of the distances of zeros from ω. Moreover, unity gain at ω = 0 requires\na pole and the corresponding zero to be equidistant from the origin. For example, if we use two\n(complex-conjugate) zeros, we must have two poles; the distance from the origin of the poles\nand of the zeros should be the same. This requirement can be met by placing the two conjugate\npoles on the semicircle of radius ω0, as depicted in Fig. 4.52a. The poles can be anywhere on the\nsemicircle to satisfy the equidistance condition. Let the two conjugate poles be at angles ±θ with\nrespect to the negative real axis. Recall that a pole and a zero in the same vicinity tend to cancel out", - "type": "text" - }, - { - "block_id": "p461-b6", - "global_id": 12929, - "bbox": [ - 210.36, - 605.26, - 219.24, - 613.26 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p461-b7", - "global_id": 12930, - "bbox": [ - 240.95, - 525.47, - 251.83, - 533.47 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p461-b8", - "global_id": 12931, - "bbox": [ - 184.1, - 454.23, - 192.99, - 462.23 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p461-b9", - "global_id": 12932, - "bbox": [ - 217.75, - 525.53, - 221.75, - 533.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p461-b10", - "global_id": 12933, - "bbox": [ - 360.45, - 572.94, - 435.09, - 582.74 - ], - "text": "v0\nv", - "type": "text" - }, - { - "block_id": "p461-b11", - "global_id": 12934, - "bbox": [ - 394.67, - 495.16, - 410.66, - 503.16 - ], - "text": "Ideal", - "type": "text" - }, - { - "block_id": "p461-b12", - "global_id": 12935, - "bbox": [ - 356.62, - 605.26, - 366.43, - 613.26 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p461-b13", - "global_id": 12936, - "bbox": [ - 219.46, - 473.95, - 230.02, - 483.58 - ], - "text": "jv0", - "type": "text" - }, - { - "block_id": "p461-b14", - "global_id": 12937, - "bbox": [ - 217.38, - 563.26, - 234.6, - 573.08 - ], - "text": "jv0", - "type": "text" - }, - { - "block_id": "p461-b15", - "global_id": 12938, - "bbox": [ - 281.93, - 480.92, - 305.91, - 489.22 - ], - "text": "H( jv)", - "type": "text" - }, - { - "block_id": "p461-b16", - "global_id": 12939, - "bbox": [ - 151.5, - 619.88, - 484.6, - 629.19 - ], - "text": "Figure 4.51 (a) Pole-zero configuration and (b) the amplitude response of a bandpass filter.", - "type": "text" - } - ] - }, - { - "page_num": 462, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p462-b0", - "global_id": 12940, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "442\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p462-b1", - "global_id": 12941, - "bbox": [ - 392.01, - 188.81, - 408.0, - 196.81 - ], - "text": "Ideal", - "type": "text" - }, - { - "block_id": "p462-b2", - "global_id": 12942, - "bbox": [ - 184.62, - 247.68, - 193.5, - 255.68 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p462-b3", - "global_id": 12943, - "bbox": [ - 215.2, - 169.01, - 224.09, - 177.01 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p462-b4", - "global_id": 12944, - "bbox": [ - 172.11, - 96.79, - 181.0, - 104.79 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p462-b5", - "global_id": 12945, - "bbox": [ - 191.96, - 167.92, - 195.96, - 175.92 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p462-b6", - "global_id": 12946, - "bbox": [ - 199.81, - 101.81, - 222.57, - 109.9 - ], - "text": "s plane", - "type": "text" - }, - { - "block_id": "p462-b7", - "global_id": 12947, - "bbox": [ - 193.68, - 117.71, - 204.24, - 127.34 - ], - "text": "jv0", - "type": "text" - }, - { - "block_id": "p462-b8", - "global_id": 12948, - "bbox": [ - 194.06, - 207.44, - 211.28, - 217.26 - ], - "text": "jv0", - "type": "text" - }, - { - "block_id": "p462-b9", - "global_id": 12949, - "bbox": [ - 175.96, - 155.81, - 179.96, - 163.81 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p462-b10", - "global_id": 12950, - "bbox": [ - 356.43, - 247.74, - 366.08, - 255.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p462-b11", - "global_id": 12951, - "bbox": [ - 252.03, - 122.78, - 275.37, - 131.08 - ], - "text": "H(jv)", - "type": "text" - }, - { - "block_id": "p462-b12", - "global_id": 12952, - "bbox": [ - 342.84, - 216.41, - 442.24, - 226.31 - ], - "text": "v0\nv", - "type": "text" - }, - { - "block_id": "p462-b13", - "global_id": 12953, - "bbox": [ - 381.61, - 165.25, - 406.93, - 173.55 - ], - "text": "u 60", - "type": "text" - }, - { - "block_id": "p462-b14", - "global_id": 12954, - "bbox": [ - 375.01, - 147.92, - 400.33, - 156.21 - ], - "text": "u 80", - "type": "text" - }, - { - "block_id": "p462-b15", - "global_id": 12955, - "bbox": [ - 376.61, - 131.82, - 401.94, - 140.11 - ], - "text": "u 87", - "type": "text" - }, - { - "block_id": "p462-b16", - "global_id": 12956, - "bbox": [ - 283.63, - 214.3, - 287.63, - 222.3 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p462-b17", - "global_id": 12957, - "bbox": [ - 270.21, - 137.67, - 274.21, - 145.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p462-b18", - "global_id": 12958, - "bbox": [ - 125.76, - 262.37, - 486.47, - 271.68 - ], - "text": "Figure 4.52 (a) Pole-zero configuration and (b) the amplitude response of a bandstop (notch) filter.", - "type": "text" - }, - { - "block_id": "p462-b19", - "global_id": 12959, - "bbox": [ - 101.84, - 305.56, - 490.39, - 339.84 - ], - "text": "each other’s influences. Therefore, placing poles closer to zeros (selecting θ closer to π/2) results\nin a rapid recovery of the gain from value 0 to 1 as we move away from ω0 in either direction.\nFigure 4.52b shows the gain |H(jω)| for three different values of θ.", - "type": "text" - }, - { - "block_id": "p462-b20", - "global_id": 12960, - "bbox": [ - 76.77, - 375.73, - 288.52, - 387.69 - ], - "text": "EXAMPLE 4.31\nNotch Filter Design", - "type": "text" - }, - { - "block_id": "p462-b21", - "global_id": 12961, - "bbox": [ - 103.16, - 404.35, - 409.11, - 414.31 - ], - "text": "Design a second-order notch filter to suppress 60 Hz hum in a radio receiver.", - "type": "text" - }, - { - "block_id": "p462-b22", - "global_id": 12962, - "bbox": [ - 103.16, - 436.81, - 477.02, - 460.64 - ], - "text": "We use the poles and zeros in Fig. 4.52a with ω0 = 120π. The zeros are at s = ±jω0. The two\npoles are at −ω0 cos θ ± jω0 sin θ. The filter transfer function is (with ω0 = 120π)", - "type": "text" - }, - { - "block_id": "p462-b23", - "global_id": 12963, - "bbox": [ - 157.99, - 468.73, - 388.29, - 494.65 - ], - "text": "H(s) =\n(s −jω0)(s + jω0)\n(s + ω0 cos θ + jω0 sin θ)(s + ω0 cos θ −jω0 sin θ)", - "type": "text" - }, - { - "block_id": "p462-b24", - "global_id": 12964, - "bbox": [ - 178.99, - 493.61, - 248.76, - 514.27 - ], - "text": "=\ns2 + ω2", - "type": "text" - }, - { - "block_id": "p462-b25", - "global_id": 12965, - "bbox": [ - 190.0, - 502.83, - 278.59, - 523.71 - ], - "text": "0\ns2 + (2ω0 cos θ)s + ω2", - "type": "text" - }, - { - "block_id": "p462-b26", - "global_id": 12966, - "bbox": [ - 274.9, - 493.71, - 420.97, - 524.41 - ], - "text": "0\n=\ns2 + 142122.3\ns2 + (753.98cos θ)s + 142122.3", - "type": "text" - }, - { - "block_id": "p462-b27", - "global_id": 12967, - "bbox": [ - 103.16, - 532.98, - 117.54, - 542.94 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p462-b28", - "global_id": 12968, - "bbox": [ - 185.44, - 538.41, - 344.49, - 559.28 - ], - "text": "|H(jω)| =\n−ω2 + 142122.3", - "type": "text" - }, - { - "block_id": "p462-b29", - "global_id": 12969, - "bbox": [ - 237.03, - 555.03, - 393.05, - 568.27 - ], - "text": "(−ω2 + 142122.3)2 + (753.98ωcos θ)2", - "type": "text" - }, - { - "block_id": "p462-b30", - "global_id": 12970, - "bbox": [ - 103.16, - 575.17, - 477.03, - 621.41 - ], - "text": "The closer the poles are to the zeros (the closer θ is to π/2), the faster the gain recovery from 0\nto 1 on either side of ω0 = 120π. Figure 4.52b shows the amplitude response for three different\nvalues of θ. This example is a case of very simple design. To achieve zero gain over a band,\nwe need an infinite number of poles as well as an infinite number of zeros.", - "type": "text" - } - ] - }, - { - "page_num": 463, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p463-b0", - "global_id": 12971, - "bbox": [ - 237.2, - 62.29, - 516.13, - 71.98 - ], - "text": "4.10\nFilter Design by Placement of Poles and Zeros of H(s)\n443", - "type": "text" - }, - { - "block_id": "p463-b1", - "global_id": 12972, - "bbox": [ - 128.9, - 86.24, - 502.75, - 120.12 - ], - "text": "MATLAB easily computes and plots the magnitude response curves of Fig. 4.52b. To\nillustrate, let us plot the magnitude response using θ = 60◦over a frequency range of\n0 ≤f ≤150 Hz. The result, shown in Fig. 4.53, matches the θ = 60◦case of Fig. 4.52b.", - "type": "text" - }, - { - "block_id": "p463-b2", - "global_id": 12973, - "bbox": [ - 128.91, - 130.37, - 484.57, - 176.2 - ], - "text": ">>\nf = (0:.01:150); omega0 = 2*pi*60; theta = 60*pi/180;\n>>\nH = @(s) (s.^2+omega0^2)./(s.^2+2*omega0*cos(theta)*s+omega0^2);\n>>\nplot(f,abs(H(1j*2*pi*f)),’k-’);\n>>\nxlabel(’f [Hz]’); ylabel(’|H(j2\\pi f)|’);", - "type": "text" - }, - { - "block_id": "p463-b3", - "global_id": 12974, - "bbox": [ - 145.98, - 285.46, - 376.35, - 306.55 - ], - "text": "0\n50\n100\n150\nf [Hz]", - "type": "text" - }, - { - "block_id": "p463-b4", - "global_id": 12975, - "bbox": [ - 139.13, - 276.25, - 143.13, - 284.25 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p463-b5", - "global_id": 12976, - "bbox": [ - 132.38, - 240.24, - 143.06, - 248.24 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p463-b6", - "global_id": 12977, - "bbox": [ - 139.13, - 204.25, - 143.13, - 212.25 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p463-b7", - "global_id": 12978, - "bbox": [ - 120.58, - 227.24, - 128.88, - 259.47 - ], - "text": "|H(j2 π f)|", - "type": "text" - }, - { - "block_id": "p463-b8", - "global_id": 12979, - "bbox": [ - 119.94, - 311.6, - 347.72, - 322.48 - ], - "text": "Figure 4.53 Magnitude response for notch filter with θ = 60◦.", - "type": "text" - }, - { - "block_id": "p463-b9", - "global_id": 12980, - "bbox": [ - 133.57, - 384.32, - 446.36, - 396.28 - ], - "text": "DRILL 4.16\nMagnitude Response from Pole-Zero Plots", - "type": "text" - }, - { - "block_id": "p463-b10", - "global_id": 12981, - "bbox": [ - 133.57, - 405.4, - 510.18, - 439.26 - ], - "text": "Use the qualitative method of sketching the frequency response to show that the system with\nthe pole-zero configuration in Fig. 4.54a is a highpass filter and the configuration in Fig. 4.54b\nis a bandpass filter.", - "type": "text" - }, - { - "block_id": "p463-b11", - "global_id": 12982, - "bbox": [ - 188.62, - 605.88, - 340.26, - 613.88 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p463-b12", - "global_id": 12983, - "bbox": [ - 240.19, - 472.85, - 262.63, - 480.93 - ], - "text": "s plane", - "type": "text" - }, - { - "block_id": "p463-b13", - "global_id": 12984, - "bbox": [ - 141.36, - 550.09, - 208.14, - 561.03 - ], - "text": "v0\nRe", - "type": "text" - }, - { - "block_id": "p463-b14", - "global_id": 12985, - "bbox": [ - 180.49, - 494.31, - 189.38, - 502.31 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p463-b15", - "global_id": 12986, - "bbox": [ - 339.51, - 505.93, - 350.07, - 515.56 - ], - "text": "jv0", - "type": "text" - }, - { - "block_id": "p463-b16", - "global_id": 12987, - "bbox": [ - 340.51, - 550.09, - 349.4, - 558.09 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p463-b17", - "global_id": 12988, - "bbox": [ - 321.63, - 494.31, - 330.52, - 502.31 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p463-b18", - "global_id": 12989, - "bbox": [ - 369.52, - 581.27, - 506.17, - 615.14 - ], - "text": "Figure 4.54 Pole-zero configura-\ntion of (a) a highpass filter and (b)\na bandpass filter.", - "type": "text" - } - ] - }, - { - "page_num": 464, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p464-b0", - "global_id": 12990, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "444\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p464-b1", - "global_id": 12991, - "bbox": [ - 101.84, - 86.52, - 357.33, - 98.48 - ], - "text": "4.10-5 Practical Filters and Their Specifications", - "type": "text" - }, - { - "block_id": "p464-b2", - "global_id": 12992, - "bbox": [ - 101.84, - 104.61, - 490.39, - 150.44 - ], - "text": "For ideal filters, everything is black and white; the gains are either zero or unity over certain bands.\nAs we saw earlier, real life does not permit such a worldview. Things have to be gray or shades\nof gray. In practice, we can realize a variety of filter characteristics that can only approach ideal\ncharacteristics.", - "type": "text" - }, - { - "block_id": "p464-b3", - "global_id": 12993, - "bbox": [ - 101.84, - 152.43, - 490.42, - 269.99 - ], - "text": "An ideal filter has a passband (unity gain) and a stopband (zero gain) with a sudden transition\nfrom the passband to the stopband. There is no transition band. For practical (or realizable) filters,\non the other hand, the transition from the passband to the stopband (or vice versa) is gradual and\ntakes place over a finite band of frequencies. Moreover, for realizable filters, the gain cannot be\nzero over a finite band (Paley–Wiener condition). As a result, there can be no true stopband for\npractical filters. We therefore define a stopband to be a band over which the gain is below some\nsmall number Gs, as illustrated in Fig. 4.55. Similarly, we define a passband to be a band over\nwhich the gain is between 1 and some number Gp (Gp < 1), as shown in Fig. 4.55. We have\nselected the passband gain of unity for convenience. It could be any constant. Usually the gains\nare specified in terms of decibels. This is simply 20 times the log (to base 10) of the gain. Thus,", - "type": "text" - }, - { - "block_id": "p464-b4", - "global_id": 12994, - "bbox": [ - 257.67, - 278.55, - 334.57, - 293.39 - ], - "text": "ˆG(dB) = 20log10 G", - "type": "text" - }, - { - "block_id": "p464-b5", - "global_id": 12995, - "bbox": [ - 101.84, - 302.01, - 252.14, - 311.98 - ], - "text": "A gain of unity is 0 dB and a gain of", - "type": "text" - }, - { - "block_id": "p464-b6", - "global_id": 12996, - "bbox": [ - 255.11, - 293.17, - 263.53, - 303.13 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p464-b7", - "global_id": 12997, - "bbox": [ - 101.85, - 302.01, - 490.38, - 335.88 - ], - "text": "2 is 3.01 dB, usually approximated by 3 dB. Sometimes\nthe specification may be in terms of attenuation, which is the negative of the gain in dB. Thus, a\ngain of 1/", - "type": "text" - }, - { - "block_id": "p464-b8", - "global_id": 12998, - "bbox": [ - 141.89, - 317.08, - 150.31, - 327.04 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p464-b9", - "global_id": 12999, - "bbox": [ - 150.33, - 325.5, - 372.92, - 335.88 - ], - "text": "2, that is, 0.707, is −3 dB, but is an attenuation of 3 dB.", - "type": "text" - }, - { - "block_id": "p464-b10", - "global_id": 13000, - "bbox": [ - 336.3, - 447.89, - 425.31, - 457.52 - ], - "text": "vp2\nvs2\nvp1\nvs1", - "type": "text" - }, - { - "block_id": "p464-b11", - "global_id": 13001, - "bbox": [ - 336.47, - 587.5, - 420.31, - 597.13 - ], - "text": "vs2\nvp2\nvs1\nvp1", - "type": "text" - }, - { - "block_id": "p464-b12", - "global_id": 13002, - "bbox": [ - 296.53, - 527.51, - 305.3, - 537.05 - ], - "text": "Gp", - "type": "text" - }, - { - "block_id": "p464-b13", - "global_id": 13003, - "bbox": [ - 201.94, - 587.5, - 242.44, - 597.07 - ], - "text": "vp\nvs", - "type": "text" - }, - { - "block_id": "p464-b14", - "global_id": 13004, - "bbox": [ - 256.3, - 452.99, - 261.64, - 460.99 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p464-b15", - "global_id": 13005, - "bbox": [ - 125.84, - 362.56, - 315.96, - 370.86 - ], - "text": "H( jv)\nH( jv)", - "type": "text" - }, - { - "block_id": "p464-b16", - "global_id": 13006, - "bbox": [ - 131.62, - 386.39, - 140.4, - 395.94 - ], - "text": "Gp", - "type": "text" - }, - { - "block_id": "p464-b17", - "global_id": 13007, - "bbox": [ - 131.28, - 433.81, - 139.39, - 443.36 - ], - "text": "Gs", - "type": "text" - }, - { - "block_id": "p464-b18", - "global_id": 13008, - "bbox": [ - 136.88, - 448.89, - 214.61, - 458.9 - ], - "text": "vp\nvs\n0", - "type": "text" - }, - { - "block_id": "p464-b19", - "global_id": 13009, - "bbox": [ - 136.4, - 378.22, - 140.4, - 386.22 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p464-b20", - "global_id": 13010, - "bbox": [ - 302.74, - 450.9, - 439.28, - 460.99 - ], - "text": "v\n0", - "type": "text" - }, - { - "block_id": "p464-b21", - "global_id": 13011, - "bbox": [ - 301.39, - 378.22, - 305.39, - 386.22 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p464-b22", - "global_id": 13012, - "bbox": [ - 256.39, - 591.6, - 261.73, - 599.6 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p464-b23", - "global_id": 13013, - "bbox": [ - 125.76, - 498.18, - 315.96, - 506.47 - ], - "text": "H( jv)\nH( jv)", - "type": "text" - }, - { - "block_id": "p464-b24", - "global_id": 13014, - "bbox": [ - 136.88, - 589.51, - 140.88, - 597.51 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p464-b25", - "global_id": 13015, - "bbox": [ - 135.36, - 516.84, - 139.36, - 524.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p464-b26", - "global_id": 13016, - "bbox": [ - 433.86, - 591.59, - 439.19, - 599.59 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p464-b27", - "global_id": 13017, - "bbox": [ - 297.19, - 571.42, - 305.3, - 580.97 - ], - "text": "Gs", - "type": "text" - }, - { - "block_id": "p464-b28", - "global_id": 13018, - "bbox": [ - 302.74, - 589.51, - 306.74, - 597.51 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p464-b29", - "global_id": 13019, - "bbox": [ - 301.3, - 516.84, - 305.3, - 524.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p464-b30", - "global_id": 13020, - "bbox": [ - 199.19, - 466.65, - 208.07, - 474.65 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p464-b31", - "global_id": 13021, - "bbox": [ - 199.19, - 605.26, - 208.07, - 613.26 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p464-b32", - "global_id": 13022, - "bbox": [ - 365.09, - 466.65, - 374.89, - 474.65 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p464-b33", - "global_id": 13023, - "bbox": [ - 365.33, - 605.26, - 374.65, - 613.26 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p464-b34", - "global_id": 13024, - "bbox": [ - 130.58, - 525.01, - 139.36, - 534.55 - ], - "text": "Gp", - "type": "text" - }, - { - "block_id": "p464-b35", - "global_id": 13025, - "bbox": [ - 131.25, - 571.42, - 139.36, - 580.97 - ], - "text": "Gs", - "type": "text" - }, - { - "block_id": "p464-b36", - "global_id": 13026, - "bbox": [ - 296.61, - 387.39, - 305.39, - 396.94 - ], - "text": "Gp", - "type": "text" - }, - { - "block_id": "p464-b37", - "global_id": 13027, - "bbox": [ - 297.28, - 432.81, - 305.39, - 442.36 - ], - "text": "Gs", - "type": "text" - }, - { - "block_id": "p464-b38", - "global_id": 13028, - "bbox": [ - 125.76, - 619.96, - 412.89, - 629.19 - ], - "text": "Figure 4.55 Passband, stopband, and transition band in filters of various types.", - "type": "text" - } - ] - }, - { - "page_num": 465, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p465-b0", - "global_id": 13029, - "bbox": [ - 322.07, - 62.89, - 516.13, - 71.98 - ], - "text": "4.11\nThe Bilateral Laplace Transform\n445", - "type": "text" - }, - { - "block_id": "p465-b1", - "global_id": 13030, - "bbox": [ - 127.59, - 85.72, - 516.17, - 191.42 - ], - "text": "In a typical design procedure, Gp (minimum passband gain) and Gs (maximum stopband\ngain) are specified. Figure 4.55 shows the passband, the stopband, and the transition band for\ntypical lowpass, bandpass, highpass, and bandstop filters. Fortunately, the highpass, bandpass, and\nbandstop filters can be obtained from a basic lowpass filter by simple frequency transformations.\nFor example, replacing s with ωc/s in the lowpass filter transfer function results in a highpass filter.\nSimilarly, other frequency transformations yield the bandpass and bandstop filters. Hence, it is\nnecessary to develop a design procedure only for a basic lowpass filter. Then, by using appropriate\ntransformations, we can design filters of other types. The design procedures are beyond our scope\nhere and will not be discussed. The interested reader is referred to [1].", - "type": "text" - }, - { - "block_id": "p465-b2", - "global_id": 13031, - "bbox": [ - 127.94, - 227.76, - 412.4, - 241.7 - ], - "text": "4.11 THE BILATERAL LAPLACE TRANSFORM", - "type": "text" - }, - { - "block_id": "p465-b3", - "global_id": 13032, - "bbox": [ - 127.59, - 247.69, - 516.12, - 281.57 - ], - "text": "Situations involving noncausal signals and/or systems cannot be handled by the (unilateral)\nLaplace transform discussed so far. These cases can be analyzed by the bilateral (or two-sided)\nLaplace transform defined by", - "type": "text" - }, - { - "block_id": "p465-b4", - "global_id": 13033, - "bbox": [ - 277.79, - 299.04, - 305.45, - 309.32 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p465-b5", - "global_id": 13034, - "bbox": [ - 307.5, - 285.49, - 324.43, - 297.55 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p465-b6", - "global_id": 13035, - "bbox": [ - 312.75, - 310.36, - 325.31, - 317.34 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p465-b7", - "global_id": 13036, - "bbox": [ - 326.92, - 295.24, - 365.75, - 309.32 - ], - "text": "x(t)e−st dt", - "type": "text" - }, - { - "block_id": "p465-b8", - "global_id": 13037, - "bbox": [ - 127.59, - 330.42, - 385.49, - 340.79 - ], - "text": "and x(t) can be obtained from X(s) by the inverse transformation", - "type": "text" - }, - { - "block_id": "p465-b9", - "global_id": 13038, - "bbox": [ - 266.14, - 359.53, - 308.76, - 383.55 - ], - "text": "x(t) =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p465-b10", - "global_id": 13039, - "bbox": [ - 311.07, - 352.54, - 338.47, - 364.82 - ], - "text": "# c+j∞", - "type": "text" - }, - { - "block_id": "p465-b11", - "global_id": 13040, - "bbox": [ - 316.32, - 377.42, - 333.91, - 384.62 - ], - "text": "c−j∞", - "type": "text" - }, - { - "block_id": "p465-b12", - "global_id": 13041, - "bbox": [ - 340.08, - 362.3, - 377.57, - 376.38 - ], - "text": "X(s)est ds", - "type": "text" - }, - { - "block_id": "p465-b13", - "global_id": 13042, - "bbox": [ - 127.59, - 401.66, - 516.14, - 435.54 - ], - "text": "Observe that the unilateral Laplace transform discussed so far is a special case of the bilateral\nLaplace transform, where the signals are restricted to the causal type. Basically, the two transforms\nare the same. For this reason we use the same notation for the bilateral Laplace transform.", - "type": "text" - }, - { - "block_id": "p465-b14", - "global_id": 13043, - "bbox": [ - 127.59, - 435.89, - 516.14, - 507.27 - ], - "text": "Earlier we showed that the Laplace transforms of e−atu(t) and of −e−atu(−t) are identical. The\nonly difference is in their regions of convergence (ROC). The ROC for the former is Res > −a;\nthat for the latter is Res < −a, as illustrated in Fig. 4.1. Clearly, the inverse Laplace transform\nof X(s) is not unique unless the ROC is specified. If we restrict all our signals to the causal type,\nhowever, this ambiguity does not arise. The inverse transform of 1/(s+a) is e−atu(t). Thus, in the\nunilateral Laplace transform, we can ignore the ROC in determining the inverse transform of X(s).", - "type": "text" - }, - { - "block_id": "p465-b15", - "global_id": 13044, - "bbox": [ - 127.6, - 509.27, - 516.14, - 543.13 - ], - "text": "We now show that any bilateral transform can be expressed in terms of two unilateral\ntransforms. It is, therefore, possible to evaluate bilateral transforms from a table of unilateral\ntransforms.", - "type": "text" - }, - { - "block_id": "p465-b16", - "global_id": 13045, - "bbox": [ - 127.59, - 544.72, - 516.14, - 579.0 - ], - "text": "Consider the function x(t) appearing in Fig. 4.56a. We separate x(t) into two components, x1(t)\nand x2(t), representing the positive time (causal) component and the negative time (anticausal)\ncomponent of x(t), respectively (Figs. 4.56b and 4.56c):", - "type": "text" - }, - { - "block_id": "p465-b17", - "global_id": 13046, - "bbox": [ - 230.0, - 597.59, - 413.71, - 608.74 - ], - "text": "x1(t) = x(t)u(t)\nand\nx2(t) = x(t)u(−t)", - "type": "text" - } - ] - }, - { - "page_num": 466, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p466-b0", - "global_id": 13047, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "446\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p466-b1", - "global_id": 13048, - "bbox": [ - 205.74, - 278.07, - 219.85, - 287.67 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p466-b2", - "global_id": 13049, - "bbox": [ - 205.81, - 91.37, - 216.91, - 99.45 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p466-b3", - "global_id": 13050, - "bbox": [ - 205.39, - 143.63, - 209.39, - 151.63 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p466-b4", - "global_id": 13051, - "bbox": [ - 185.39, - 188.39, - 199.49, - 198.0 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p466-b5", - "global_id": 13052, - "bbox": [ - 205.34, - 234.7, - 209.34, - 242.7 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p466-b6", - "global_id": 13053, - "bbox": [ - 205.42, - 337.64, - 209.42, - 345.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p466-b7", - "global_id": 13054, - "bbox": [ - 205.43, - 440.21, - 209.43, - 448.21 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p466-b8", - "global_id": 13055, - "bbox": [ - 178.73, - 378.43, - 199.5, - 388.25 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p466-b9", - "global_id": 13056, - "bbox": [ - 198.15, - 164.0, - 207.03, - 172.0 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p466-b10", - "global_id": 13057, - "bbox": [ - 197.69, - 255.12, - 207.5, - 263.12 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p466-b11", - "global_id": 13058, - "bbox": [ - 198.15, - 357.92, - 207.03, - 365.92 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p466-b12", - "global_id": 13059, - "bbox": [ - 197.93, - 460.73, - 207.26, - 468.73 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p466-b13", - "global_id": 13060, - "bbox": [ - 281.5, - 442.68, - 283.73, - 450.68 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p466-b14", - "global_id": 13061, - "bbox": [ - 281.5, - 340.97, - 283.73, - 348.97 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p466-b15", - "global_id": 13062, - "bbox": [ - 281.5, - 236.48, - 283.73, - 244.48 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p466-b16", - "global_id": 13063, - "bbox": [ - 281.5, - 146.18, - 283.73, - 154.18 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p466-b17", - "global_id": 13064, - "bbox": [ - 312.8, - 448.07, - 486.4, - 469.99 - ], - "text": "Figure 4.56 Expressing a signal as a sum of\ncausal and anticausal components.", - "type": "text" - }, - { - "block_id": "p466-b18", - "global_id": 13065, - "bbox": [ - 119.78, - 514.13, - 319.59, - 524.51 - ], - "text": "The bilateral Laplace transform of x(t) is given by", - "type": "text" - }, - { - "block_id": "p466-b19", - "global_id": 13066, - "bbox": [ - 213.55, - 562.78, - 241.21, - 573.05 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p466-b20", - "global_id": 13067, - "bbox": [ - 243.25, - 549.22, - 260.2, - 561.28 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p466-b21", - "global_id": 13068, - "bbox": [ - 248.52, - 574.1, - 261.07, - 581.07 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p466-b22", - "global_id": 13069, - "bbox": [ - 262.68, - 558.98, - 301.51, - 573.06 - ], - "text": "x(t)e−st dt", - "type": "text" - }, - { - "block_id": "p466-b23", - "global_id": 13070, - "bbox": [ - 233.43, - 597.13, - 241.2, - 607.1 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p466-b24", - "global_id": 13071, - "bbox": [ - 243.25, - 583.58, - 261.23, - 595.92 - ], - "text": "# 0−", - "type": "text" - }, - { - "block_id": "p466-b25", - "global_id": 13072, - "bbox": [ - 248.52, - 608.45, - 261.07, - 615.43 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p466-b26", - "global_id": 13073, - "bbox": [ - 263.33, - 593.33, - 315.62, - 608.28 - ], - "text": "x2(t)e−st dt +", - "type": "text" - }, - { - "block_id": "p466-b27", - "global_id": 13074, - "bbox": [ - 317.16, - 583.58, - 334.11, - 595.63 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p466-b28", - "global_id": 13075, - "bbox": [ - 322.43, - 593.33, - 378.49, - 615.72 - ], - "text": "0−x1(t)e−st dt", - "type": "text" - }, - { - "block_id": "p466-b29", - "global_id": 13076, - "bbox": [ - 233.43, - 622.98, - 490.38, - 634.13 - ], - "text": "= X2(s) + X1(s)\n(4.57)", - "type": "text" - } - ] - }, - { - "page_num": 467, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p467-b0", - "global_id": 13077, - "bbox": [ - 322.07, - 62.89, - 516.13, - 71.98 - ], - "text": "4.11\nThe Bilateral Laplace Transform\n447", - "type": "text" - }, - { - "block_id": "p467-b1", - "global_id": 13078, - "bbox": [ - 127.59, - 85.46, - 516.11, - 108.57 - ], - "text": "where X1(s) is the Laplace transform of the causal component x1(t), and X2(s) is the Laplace\ntransform of the anticausal component x2(t). Consider X2(s), given by", - "type": "text" - }, - { - "block_id": "p467-b2", - "global_id": 13079, - "bbox": [ - 235.86, - 127.4, - 267.06, - 138.55 - ], - "text": "X2(s) =", - "type": "text" - }, - { - "block_id": "p467-b3", - "global_id": 13080, - "bbox": [ - 269.11, - 113.84, - 287.07, - 126.19 - ], - "text": "# 0−", - "type": "text" - }, - { - "block_id": "p467-b4", - "global_id": 13081, - "bbox": [ - 274.36, - 138.72, - 286.92, - 145.7 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p467-b5", - "global_id": 13082, - "bbox": [ - 289.19, - 123.61, - 341.97, - 138.55 - ], - "text": "x2(t)e−st dt =", - "type": "text" - }, - { - "block_id": "p467-b6", - "global_id": 13083, - "bbox": [ - 344.02, - 113.84, - 360.96, - 125.9 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p467-b7", - "global_id": 13084, - "bbox": [ - 349.27, - 123.61, - 407.68, - 145.99 - ], - "text": "0+ x2(−t)est dt", - "type": "text" - }, - { - "block_id": "p467-b8", - "global_id": 13085, - "bbox": [ - 127.59, - 155.05, - 169.34, - 165.02 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p467-b9", - "global_id": 13086, - "bbox": [ - 266.71, - 171.2, - 305.68, - 182.35 - ], - "text": "X2(−s) =", - "type": "text" - }, - { - "block_id": "p467-b10", - "global_id": 13087, - "bbox": [ - 307.73, - 157.64, - 324.67, - 169.7 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p467-b11", - "global_id": 13088, - "bbox": [ - 312.98, - 167.4, - 376.83, - 189.78 - ], - "text": "0+ x2(−t)e−st dt", - "type": "text" - }, - { - "block_id": "p467-b12", - "global_id": 13089, - "bbox": [ - 127.59, - 195.0, - 516.1, - 229.29 - ], - "text": "If x(t) has any impulse or its derivative(s) at the origin, they are included in x1(t). Consequently,\nx2(t) = 0 at the origin; that is, x2(0) = 0. Hence, the lower limit on the integration in the preceding\nequation can be taken as 0−instead of 0+. Therefore,", - "type": "text" - }, - { - "block_id": "p467-b13", - "global_id": 13090, - "bbox": [ - 266.71, - 246.14, - 305.68, - 257.29 - ], - "text": "X2(−s) =", - "type": "text" - }, - { - "block_id": "p467-b14", - "global_id": 13091, - "bbox": [ - 307.73, - 232.59, - 324.67, - 244.64 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p467-b15", - "global_id": 13092, - "bbox": [ - 312.98, - 242.34, - 376.83, - 264.72 - ], - "text": "0−x2(−t)e−st dt", - "type": "text" - }, - { - "block_id": "p467-b16", - "global_id": 13093, - "bbox": [ - 127.59, - 272.93, - 516.13, - 296.04 - ], - "text": "Because x2(−t) is causal (Fig. 4.56d), X2(−s) can be found from the unilateral transform table.\nChanging the sign of s in X2(−s) yields X2(s).", - "type": "text" - }, - { - "block_id": "p467-b17", - "global_id": 13094, - "bbox": [ - 127.59, - 296.84, - 516.12, - 319.18 - ], - "text": "To summarize, the bilateral transform X(s) in Eq. (4.57) can be computed from the unilateral\ntransforms in two steps:", - "type": "text" - }, - { - "block_id": "p467-b18", - "global_id": 13095, - "bbox": [ - 144.52, - 326.73, - 516.14, - 372.97 - ], - "text": "1. Split x(t) into its causal and anticausal components, x1(t) and x2(t), respectively.\n2. Since the signals x1(t) and x2(−t) are both causal, take the (unilateral) Laplace transform\nof x1(t) and add to it the (unilateral) Laplace transform of x2(−t), with s replaced by −s.\nThis procedure gives the (bilateral) Laplace transform of x(t).", - "type": "text" - }, - { - "block_id": "p467-b19", - "global_id": 13096, - "bbox": [ - 127.59, - 380.53, - 516.14, - 438.72 - ], - "text": "Since x1(t) and x2(−t) are both causal, X1(s) and X2(−s) are both unilateral Laplace\ntransforms. Let σc1 and σc2 be the abscissas of convergence of X1(s) and X2(−s), respectively.\nThis statement implies that X1(s) exists for all s with Res > σc1, and X2(−s) exists for all s with\nRes > σc2. Therefore, X2(s) exists for all s with Res < −σc2.† Therefore, X(s) = X1(s) + X2(s)\nexists for all s such that", - "type": "text" - }, - { - "block_id": "p467-b20", - "global_id": 13097, - "bbox": [ - 285.68, - 440.31, - 357.54, - 451.76 - ], - "text": "σc1 < Res < −σc2", - "type": "text" - }, - { - "block_id": "p467-b21", - "global_id": 13098, - "bbox": [ - 127.59, - 459.23, - 516.14, - 517.43 - ], - "text": "The regions of convergence of X1(s), X2(s), and X(s) are shown in Fig. 4.57. Because X(s) is\nfinite for all values of s lying in the strip of convergence (σc1 < Res < −σc2), poles of X(s) must\nlie outside this strip. The poles of X(s) arising from the causal component x1(t) lie to the left of\nthe strip (region) of convergence, and those arising from its anticausal component x2(t) lie to its\nright (see Fig. 4.57). This fact is of crucial importance in finding the inverse bilateral transform.", - "type": "text" - }, - { - "block_id": "p467-b22", - "global_id": 13099, - "bbox": [ - 127.59, - 519.0, - 516.14, - 589.16 - ], - "text": "This result can be generalized to left-sided and right-sided signals. We define a signal x(t) as\na right-sided signal if x(t) = 0 for t < T1 for some finite positive or negative number T1. A causal\nsignal is always a right-sided signal, but the converse is not necessarily true. A signal is said to\nleft-sided if it is zero for t > T2 for some finite, positive, or negative number T2. An anticausal\nsignal is always a left-sided signal, but the converse is not necessarily true. A two-sided signal is\nof infinite duration on both positive and negative sides of t and is neither right-sided nor left-sided.", - "type": "text" - }, - { - "block_id": "p467-b23", - "global_id": 13100, - "bbox": [ - 127.59, - 591.15, - 516.14, - 613.06 - ], - "text": "We can show that the conclusions for ROC for causal signals also hold for right-sided signals,\nand those for anticausal signals hold for left-sided signals. In other words, if x(t) is causal or", - "type": "text" - }, - { - "block_id": "p467-b24", - "global_id": 13101, - "bbox": [ - 127.59, - 631.15, - 469.99, - 643.38 - ], - "text": "† For instance, if x(t) exists for all t > 10, then x(−t), its time-inverted form, exists for t < −10.", - "type": "text" - } - ] - }, - { - "page_num": 468, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p468-b0", - "global_id": 13102, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "448\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p468-b1", - "global_id": 13103, - "bbox": [ - 313.8, - 233.07, - 486.4, - 255.0 - ], - "text": "Figure 4.57 Regions of convergence for\ncausal, anticausal, and combined signals.", - "type": "text" - }, - { - "block_id": "p468-b2", - "global_id": 13104, - "bbox": [ - 101.84, - 282.56, - 490.39, - 304.89 - ], - "text": "right-sided, the poles of X(s) lie to the left of the ROC, and if x(t) is anticausal or left-sided, the\npoles of X(s) lie to the right of the ROC.", - "type": "text" - }, - { - "block_id": "p468-b3", - "global_id": 13105, - "bbox": [ - 101.85, - 306.88, - 490.41, - 376.62 - ], - "text": "To prove this generalization, we observe that a right-sided signal can be expressed as\nx(t) + xf (t), where x(t) is a causal signal and xf (t) is some finite-duration signal. The ROC of\nany finite-duration signal is the entire s-plane (no finite poles). Hence, the ROC of the right-sided\nsignal x(t) + xf (t) is the region common to the ROCs of x(t) and xf (t), which is same as the ROC\nfor x(t). This proves the generalization for right-sided signals. We can use a similar argument to\ngeneralize the result for left-sided signals. Let us find the bilateral Laplace transform of", - "type": "text" - }, - { - "block_id": "p468-b4", - "global_id": 13106, - "bbox": [ - 247.61, - 391.49, - 490.38, - 403.37 - ], - "text": "x(t) = ebtu(−t) + eatu(t)\n(4.58)", - "type": "text" - }, - { - "block_id": "p468-b5", - "global_id": 13107, - "bbox": [ - 101.84, - 420.16, - 360.57, - 430.12 - ], - "text": "We already know the Laplace transform of the causal component", - "type": "text" - }, - { - "block_id": "p468-b6", - "global_id": 13108, - "bbox": [ - 234.12, - 444.89, - 490.38, - 468.81 - ], - "text": "eatu(t) ⇐⇒\n1\ns −a\nRes > a\n(4.59)", - "type": "text" - }, - { - "block_id": "p468-b7", - "global_id": 13109, - "bbox": [ - 101.85, - 482.41, - 326.08, - 494.57 - ], - "text": "For the anticausal component, x2(t) = ebtu(−t), we have", - "type": "text" - }, - { - "block_id": "p468-b8", - "global_id": 13110, - "bbox": [ - 208.31, - 508.56, - 383.93, - 532.48 - ], - "text": "x2(−t) = e−btu(t) ⇐⇒\n1\ns + b\nRes > −b", - "type": "text" - }, - { - "block_id": "p468-b9", - "global_id": 13111, - "bbox": [ - 101.84, - 546.02, - 128.13, - 555.98 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p468-b10", - "global_id": 13112, - "bbox": [ - 221.15, - 560.74, - 313.58, - 585.07 - ], - "text": "X2(s) =\n1\n−s + b = −1", - "type": "text" - }, - { - "block_id": "p468-b11", - "global_id": 13113, - "bbox": [ - 297.34, - 567.72, - 371.09, - 585.07 - ], - "text": "s −b\nRes < b", - "type": "text" - }, - { - "block_id": "p468-b12", - "global_id": 13114, - "bbox": [ - 101.84, - 595.63, - 143.59, - 605.59 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p468-b13", - "global_id": 13115, - "bbox": [ - 230.24, - 611.68, - 304.49, - 628.94 - ], - "text": "ebtu(−t) ⇐⇒−1", - "type": "text" - }, - { - "block_id": "p468-b14", - "global_id": 13116, - "bbox": [ - 288.25, - 618.66, - 490.38, - 636.01 - ], - "text": "s −b\nRes < b\n(4.60)", - "type": "text" - } - ] - }, - { - "page_num": 469, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p469-b0", - "global_id": 13117, - "bbox": [ - 322.07, - 62.89, - 516.13, - 71.98 - ], - "text": "4.11\nThe Bilateral Laplace Transform\n449", - "type": "text" - }, - { - "block_id": "p469-b1", - "global_id": 13118, - "bbox": [ - 127.59, - 85.46, - 322.0, - 95.84 - ], - "text": "and the Laplace transform of x(t) in Eq. (4.58) is", - "type": "text" - }, - { - "block_id": "p469-b2", - "global_id": 13119, - "bbox": [ - 215.56, - 106.23, - 428.17, - 130.15 - ], - "text": "X(s) = −\n1\ns −b +\n1\ns −a\nRes > a\nand\nRes < b", - "type": "text" - }, - { - "block_id": "p469-b3", - "global_id": 13120, - "bbox": [ - 235.44, - 131.33, - 516.12, - 155.66 - ], - "text": "=\na −b\n(s −b)(s −a)\na < Res < b\n(4.61)", - "type": "text" - }, - { - "block_id": "p469-b4", - "global_id": 13121, - "bbox": [ - 127.6, - 165.38, - 516.14, - 223.57 - ], - "text": "Figure 4.58 shows x(t) and the ROC of X(s) for various values of a and b. Equation (4.61)\nindicates that the ROC of X(s) does not exist if a > b, which is precisely the case in Fig. 4.58f.\nObserve that the poles of X(s) are outside (on the edges) of the ROC. The poles of X(s) because of\nthe anticausal component of x(t) lie to the right of the ROC, and those due to the causal component\nof x(t) lie to its left.", - "type": "text" - }, - { - "block_id": "p469-b5", - "global_id": 13122, - "bbox": [ - 127.6, - 225.15, - 516.14, - 247.48 - ], - "text": "When X(s) is expressed as a sum of several terms, the ROC for X(s) is the intersection of\n(region common to) the ROCs of all the terms. In general, if x(t) = %k", - "type": "text" - }, - { - "block_id": "p469-b6", - "global_id": 13123, - "bbox": [ - 127.59, - 237.11, - 516.14, - 272.1 - ], - "text": "i=1 xi(t), then the ROC for\nX(s) is the intersection of the ROCs (region common to all ROCs) for the transforms X1(s), X2(s),\n. . . , Xk(s).", - "type": "text" - }, - { - "block_id": "p469-b7", - "global_id": 13124, - "bbox": [ - 213.47, - 384.04, - 222.35, - 392.04 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p469-b8", - "global_id": 13125, - "bbox": [ - 213.19, - 494.83, - 222.52, - 502.83 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p469-b9", - "global_id": 13126, - "bbox": [ - 213.42, - 605.26, - 222.3, - 613.26 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p469-b10", - "global_id": 13127, - "bbox": [ - 424.75, - 374.52, - 434.08, - 382.52 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p469-b11", - "global_id": 13128, - "bbox": [ - 424.97, - 484.51, - 433.85, - 492.51 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p469-b12", - "global_id": 13129, - "bbox": [ - 425.42, - 595.18, - 433.41, - 603.18 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p469-b13", - "global_id": 13130, - "bbox": [ - 159.16, - 526.56, - 170.75, - 534.64 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p469-b14", - "global_id": 13131, - "bbox": [ - 167.05, - 581.7, - 200.84, - 590.79 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p469-b15", - "global_id": 13132, - "bbox": [ - 153.45, - 552.49, - 161.67, - 561.94 - ], - "text": "ebt", - "type": "text" - }, - { - "block_id": "p469-b16", - "global_id": 13133, - "bbox": [ - 159.63, - 415.75, - 170.73, - 423.83 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p469-b17", - "global_id": 13134, - "bbox": [ - 168.05, - 471.37, - 200.84, - 480.45 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p469-b18", - "global_id": 13135, - "bbox": [ - 191.6, - 443.14, - 199.82, - 452.59 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p469-b19", - "global_id": 13136, - "bbox": [ - 388.54, - 516.22, - 399.65, - 524.3 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p469-b20", - "global_id": 13137, - "bbox": [ - 378.49, - 572.73, - 382.49, - 580.73 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p469-b21", - "global_id": 13138, - "bbox": [ - 396.73, - 547.34, - 404.95, - 556.8 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p469-b22", - "global_id": 13139, - "bbox": [ - 363.46, - 524.43, - 371.68, - 533.89 - ], - "text": "ebt", - "type": "text" - }, - { - "block_id": "p469-b23", - "global_id": 13140, - "bbox": [ - 400.64, - 530.52, - 419.3, - 538.82 - ], - "text": "a b", - "type": "text" - }, - { - "block_id": "p469-b24", - "global_id": 13141, - "bbox": [ - 410.07, - 571.62, - 412.29, - 579.62 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p469-b25", - "global_id": 13142, - "bbox": [ - 388.19, - 406.4, - 399.29, - 414.48 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p469-b26", - "global_id": 13143, - "bbox": [ - 378.99, - 462.16, - 382.99, - 470.16 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p469-b27", - "global_id": 13144, - "bbox": [ - 396.73, - 435.54, - 404.95, - 444.99 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p469-b28", - "global_id": 13145, - "bbox": [ - 365.04, - 413.63, - 373.26, - 423.08 - ], - "text": "ebt", - "type": "text" - }, - { - "block_id": "p469-b29", - "global_id": 13146, - "bbox": [ - 410.07, - 461.07, - 412.29, - 469.07 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p469-b30", - "global_id": 13147, - "bbox": [ - 371.07, - 296.91, - 382.18, - 304.99 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p469-b31", - "global_id": 13148, - "bbox": [ - 378.99, - 350.68, - 382.99, - 358.68 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p469-b32", - "global_id": 13149, - "bbox": [ - 365.04, - 317.45, - 405.95, - 330.83 - ], - "text": "eat\nebt", - "type": "text" - }, - { - "block_id": "p469-b33", - "global_id": 13150, - "bbox": [ - 410.07, - 349.62, - 412.29, - 357.62 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p469-b34", - "global_id": 13151, - "bbox": [ - 298.71, - 564.92, - 303.6, - 572.92 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p469-b35", - "global_id": 13152, - "bbox": [ - 255.65, - 528.56, - 263.21, - 536.58 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p469-b36", - "global_id": 13153, - "bbox": [ - 252.53, - 570.32, - 279.08, - 581.65 - ], - "text": "0\nb", - "type": "text" - }, - { - "block_id": "p469-b37", - "global_id": 13154, - "bbox": [ - 474.13, - 535.04, - 515.01, - 552.04 - ], - "text": "No region of\nconvergence", - "type": "text" - }, - { - "block_id": "p469-b38", - "global_id": 13155, - "bbox": [ - 464.99, - 560.31, - 468.99, - 568.31 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p469-b39", - "global_id": 13156, - "bbox": [ - 467.26, - 518.36, - 474.82, - 526.38 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p469-b40", - "global_id": 13157, - "bbox": [ - 446.54, - 451.2, - 476.54, - 459.28 - ], - "text": "0\nb\na", - "type": "text" - }, - { - "block_id": "p469-b41", - "global_id": 13158, - "bbox": [ - 467.26, - 408.14, - 474.82, - 416.16 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p469-b42", - "global_id": 13159, - "bbox": [ - 446.54, - 339.36, - 503.75, - 347.92 - ], - "text": "0\na\nb", - "type": "text" - }, - { - "block_id": "p469-b43", - "global_id": 13160, - "bbox": [ - 467.26, - 297.44, - 474.82, - 305.46 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p469-b44", - "global_id": 13161, - "bbox": [ - 298.6, - 454.09, - 303.49, - 462.09 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p469-b45", - "global_id": 13162, - "bbox": [ - 255.54, - 417.96, - 263.1, - 425.98 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p469-b46", - "global_id": 13163, - "bbox": [ - 252.42, - 460.99, - 279.92, - 469.09 - ], - "text": "0\na", - "type": "text" - }, - { - "block_id": "p469-b47", - "global_id": 13164, - "bbox": [ - 159.75, - 305.81, - 170.85, - 313.89 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p469-b48", - "global_id": 13165, - "bbox": [ - 167.55, - 360.71, - 200.84, - 369.84 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p469-b49", - "global_id": 13166, - "bbox": [ - 185.99, - 328.98, - 194.21, - 338.43 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p469-b50", - "global_id": 13167, - "bbox": [ - 298.6, - 343.78, - 303.49, - 351.78 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p469-b51", - "global_id": 13168, - "bbox": [ - 255.54, - 308.06, - 263.1, - 316.08 - ], - "text": "jv", - "type": "text" - }, - { - "block_id": "p469-b52", - "global_id": 13169, - "bbox": [ - 239.42, - 349.45, - 256.42, - 358.19 - ], - "text": "0\na", - "type": "text" - }, - { - "block_id": "p469-b53", - "global_id": 13170, - "bbox": [ - 127.59, - 619.96, - 412.95, - 629.19 - ], - "text": "Figure 4.58 Various two exponential signals and their regions of convergence.", - "type": "text" - } - ] - }, - { - "page_num": 470, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p470-b0", - "global_id": 13171, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "450\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p470-b1", - "global_id": 13172, - "bbox": [ - 76.77, - 93.91, - 375.2, - 105.87 - ], - "text": "EXAMPLE 4.32\nInverse Bilateral Laplace Transform", - "type": "text" - }, - { - "block_id": "p470-b2", - "global_id": 13173, - "bbox": [ - 103.16, - 122.34, - 287.98, - 132.3 - ], - "text": "Find the inverse bilateral Laplace transform of", - "type": "text" - }, - { - "block_id": "p470-b3", - "global_id": 13174, - "bbox": [ - 246.88, - 141.44, - 332.09, - 165.87 - ], - "text": "X(s) =\n−3\n(s + 2)(s −1)", - "type": "text" - }, - { - "block_id": "p470-b4", - "global_id": 13175, - "bbox": [ - 103.16, - 174.71, - 360.85, - 185.09 - ], - "text": "if the ROC is (a) −2 < Res < 1, (b) Res > 1, and (c) Res < −2.", - "type": "text" - }, - { - "block_id": "p470-b5", - "global_id": 13176, - "bbox": [ - 121.09, - 207.92, - 132.71, - 217.88 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p470-b6", - "global_id": 13177, - "bbox": [ - 247.69, - 217.52, - 331.29, - 241.53 - ], - "text": "X(s) =\n1\ns + 2 −\n1\ns −1", - "type": "text" - }, - { - "block_id": "p470-b7", - "global_id": 13178, - "bbox": [ - 103.16, - 246.61, - 477.01, - 292.85 - ], - "text": "Now, X(s) has poles at −2 and 1. The strip of convergence is −2 < Res < 1. The pole at −2,\nbeing to the left of the strip of convergence, corresponds to a causal signal. The pole at 1, being\nto the right of the strip of convergence, corresponds to an anticausal signal. Equations (4.59)\nand (4.60) yield", - "type": "text" - }, - { - "block_id": "p470-b8", - "global_id": 13179, - "bbox": [ - 240.6, - 292.71, - 339.57, - 304.7 - ], - "text": "x(t) = e−2tu(t) + etu(−t)", - "type": "text" - }, - { - "block_id": "p470-b9", - "global_id": 13180, - "bbox": [ - 103.16, - 313.5, - 477.02, - 335.49 - ], - "text": "(b) Both poles lie to the left of the ROC, so both poles correspond to causal signals.\nTherefore,", - "type": "text" - }, - { - "block_id": "p470-b10", - "global_id": 13181, - "bbox": [ - 248.46, - 332.96, - 331.72, - 347.36 - ], - "text": "x(t) = (e−2t −et)u(t)", - "type": "text" - }, - { - "block_id": "p470-b11", - "global_id": 13182, - "bbox": [ - 228.83, - 375.96, - 241.94, - 384.04 - ], - "text": "x (t)", - "type": "text" - }, - { - "block_id": "p470-b12", - "global_id": 13183, - "bbox": [ - 247.78, - 452.57, - 251.78, - 460.57 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p470-b13", - "global_id": 13184, - "bbox": [ - 247.24, - 384.92, - 251.24, - 392.92 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p470-b14", - "global_id": 13185, - "bbox": [ - 140.68, - 451.44, - 305.31, - 460.57 - ], - "text": "4\n2\nt", - "type": "text" - }, - { - "block_id": "p470-b15", - "global_id": 13186, - "bbox": [ - 240.36, - 470.79, - 249.24, - 478.79 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p470-b16", - "global_id": 13187, - "bbox": [ - 111.5, - 491.64, - 122.6, - 499.72 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p470-b17", - "global_id": 13188, - "bbox": [ - 225.1, - 513.51, - 229.1, - 521.51 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p470-b18", - "global_id": 13189, - "bbox": [ - 210.17, - 526.1, - 212.39, - 534.1 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p470-b19", - "global_id": 13190, - "bbox": [ - 157.2, - 591.89, - 167.0, - 599.89 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p470-b20", - "global_id": 13191, - "bbox": [ - 100.07, - 526.16, - 104.07, - 534.16 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p470-b21", - "global_id": 13192, - "bbox": [ - 318.81, - 591.89, - 327.69, - 599.89 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p470-b22", - "global_id": 13193, - "bbox": [ - 381.16, - 526.32, - 385.16, - 534.32 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p470-b23", - "global_id": 13194, - "bbox": [ - 362.98, - 491.64, - 374.09, - 499.72 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p470-b24", - "global_id": 13195, - "bbox": [ - 400.54, - 526.25, - 402.77, - 534.25 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p470-b25", - "global_id": 13196, - "bbox": [ - 246.52, - 513.17, - 257.19, - 521.47 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p470-b26", - "global_id": 13197, - "bbox": [ - 94.2, - 606.21, - 350.5, - 615.82 - ], - "text": "Figure 4.59 Three possible inverse transforms of −3/((s + 2)(s −1)).", - "type": "text" - } - ] - }, - { - "page_num": 471, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p471-b0", - "global_id": 13198, - "bbox": [ - 322.07, - 62.89, - 516.13, - 71.98 - ], - "text": "4.11\nThe Bilateral Laplace Transform\n451", - "type": "text" - }, - { - "block_id": "p471-b1", - "global_id": 13199, - "bbox": [ - 128.9, - 86.29, - 502.75, - 108.28 - ], - "text": "(c) Both poles lie to the right of the region of convergence, so both poles correspond to\nanticausal signals, and", - "type": "text" - }, - { - "block_id": "p471-b2", - "global_id": 13200, - "bbox": [ - 266.43, - 105.76, - 365.23, - 120.14 - ], - "text": "x(t) = (−e−2t + et)u(−t)", - "type": "text" - }, - { - "block_id": "p471-b3", - "global_id": 13201, - "bbox": [ - 128.91, - 128.8, - 502.76, - 151.12 - ], - "text": "Figure 4.59 shows the three inverse transforms corresponding to the same X(s) but with\ndifferent regions of convergence.", - "type": "text" - }, - { - "block_id": "p471-b4", - "global_id": 13202, - "bbox": [ - 127.59, - 192.57, - 408.03, - 204.52 - ], - "text": "4.11-1 Properties of the Bilateral Laplace Transform", - "type": "text" - }, - { - "block_id": "p471-b5", - "global_id": 13203, - "bbox": [ - 127.59, - 210.65, - 516.13, - 246.02 - ], - "text": "Properties of the bilateral Laplace transform are similar to those of the unilateral transform. We\nshall merely state the properties here without proofs. Let the ROC of X(s) be a < Re s < b.\nSimilarly, let the ROC of Xi(s) be ai < Re s < bi for (i = 1,2).", - "type": "text" - }, - { - "block_id": "p471-b6", - "global_id": 13204, - "bbox": [ - 127.89, - 258.8, - 186.25, - 270.93 - ], - "text": "LINEARITY", - "type": "text" - }, - { - "block_id": "p471-b7", - "global_id": 13205, - "bbox": [ - 241.28, - 286.5, - 402.44, - 297.65 - ], - "text": "a1x1(t) + a2x2(t) ⇐⇒a1X1(s) + a2X2(s)", - "type": "text" - }, - { - "block_id": "p471-b8", - "global_id": 13206, - "bbox": [ - 127.59, - 304.52, - 516.13, - 327.62 - ], - "text": "The ROC for a1X1(s) + a2X2(s) is the region common to (intersection of) the ROCs for X1(s) and\nX2(s).", - "type": "text" - }, - { - "block_id": "p471-b9", - "global_id": 13207, - "bbox": [ - 127.89, - 341.12, - 189.69, - 353.24 - ], - "text": "TIME SHIFT", - "type": "text" - }, - { - "block_id": "p471-b10", - "global_id": 13208, - "bbox": [ - 276.59, - 367.09, - 366.09, - 379.09 - ], - "text": "x(t −T) ⇐⇒X(s)e−sT", - "type": "text" - }, - { - "block_id": "p471-b11", - "global_id": 13209, - "bbox": [ - 127.59, - 383.63, - 347.92, - 397.21 - ], - "text": "The ROC for X(s)e−sT is identical to the ROC for X(s).", - "type": "text" - }, - { - "block_id": "p471-b12", - "global_id": 13210, - "bbox": [ - 127.89, - 411.48, - 227.99, - 423.6 - ], - "text": "FREQUENCY SHIFT", - "type": "text" - }, - { - "block_id": "p471-b13", - "global_id": 13211, - "bbox": [ - 277.96, - 435.06, - 365.76, - 450.32 - ], - "text": "x(t)es0t ⇐⇒X(s −s0)", - "type": "text" - }, - { - "block_id": "p471-b14", - "global_id": 13212, - "bbox": [ - 127.59, - 457.18, - 383.29, - 468.33 - ], - "text": "The ROC for X(s −s0) is a + c < Re s < b + c, where c = Re s0.", - "type": "text" - }, - { - "block_id": "p471-b15", - "global_id": 13213, - "bbox": [ - 127.89, - 481.84, - 255.88, - 493.96 - ], - "text": "TIME DIFFERENTIATION", - "type": "text" - }, - { - "block_id": "p471-b16", - "global_id": 13214, - "bbox": [ - 289.49, - 505.41, - 309.3, - 515.69 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p471-b17", - "global_id": 13215, - "bbox": [ - 295.43, - 512.4, - 355.44, - 529.75 - ], - "text": "dt\n⇐⇒sX(s)", - "type": "text" - }, - { - "block_id": "p471-b18", - "global_id": 13216, - "bbox": [ - 127.59, - 533.49, - 516.14, - 555.82 - ], - "text": "The ROC for sX(s) contains the ROC for X(s) and may be larger than that of X(s) under certain\nconditions [e.g., if X(s) has a first-order pole at s = 0, it is canceled by the factor s in sX(s)].", - "type": "text" - }, - { - "block_id": "p471-b19", - "global_id": 13217, - "bbox": [ - 127.89, - 570.1, - 233.27, - 582.23 - ], - "text": "TIME INTEGRATION", - "type": "text" - }, - { - "block_id": "p471-b20", - "global_id": 13218, - "bbox": [ - 272.71, - 587.86, - 284.48, - 600.15 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p471-b21", - "global_id": 13219, - "bbox": [ - 277.97, - 612.74, - 290.53, - 619.72 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p471-b22", - "global_id": 13220, - "bbox": [ - 292.14, - 601.43, - 371.01, - 611.7 - ], - "text": "x(τ)dτ ⇐⇒X(s)/s", - "type": "text" - }, - { - "block_id": "p471-b23", - "global_id": 13221, - "bbox": [ - 127.6, - 624.84, - 304.43, - 635.21 - ], - "text": "The ROC for sX(s) is max (a,0) < Re s < b.", - "type": "text" - } - ] - }, - { - "page_num": 472, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p472-b0", - "global_id": 13222, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "452\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p472-b1", - "global_id": 13223, - "bbox": [ - 102.14, - 86.19, - 181.3, - 98.32 - ], - "text": "TIME SCALING", - "type": "text" - }, - { - "block_id": "p472-b2", - "global_id": 13224, - "bbox": [ - 252.39, - 115.56, - 306.46, - 132.41 - ], - "text": "x(βt) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p472-b3", - "global_id": 13225, - "bbox": [ - 297.95, - 122.44, - 317.25, - 139.17 - ], - "text": "|β|X", - "type": "text" - }, - { - "block_id": "p472-b4", - "global_id": 13226, - "bbox": [ - 317.69, - 108.14, - 330.7, - 125.42 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p472-b5", - "global_id": 13227, - "bbox": [ - 325.61, - 129.2, - 331.23, - 139.17 - ], - "text": "β", - "type": "text" - }, - { - "block_id": "p472-b7", - "global_id": 13228, - "bbox": [ - 101.85, - 153.57, - 490.41, - 187.85 - ], - "text": "The ROC for X(s/β) is βa < Re s < βb. For β > 1,x(βt) represents time compression and the\ncorresponding ROC expands by factor β. For 0 > β > 1, x(βt) represents time expansion and the\ncorresponding ROC is compressed by factor β.", - "type": "text" - }, - { - "block_id": "p472-b8", - "global_id": 13229, - "bbox": [ - 102.14, - 206.16, - 215.06, - 218.29 - ], - "text": "TIME CONVOLUTION", - "type": "text" - }, - { - "block_id": "p472-b9", - "global_id": 13230, - "bbox": [ - 240.19, - 244.58, - 352.04, - 255.73 - ], - "text": "x1(t) ∗x2(t) ⇐⇒X1(s)X2(s)", - "type": "text" - }, - { - "block_id": "p472-b10", - "global_id": 13231, - "bbox": [ - 101.85, - 270.67, - 488.2, - 281.82 - ], - "text": "The ROC for X1(s)X2(s) is the region common to (intersection of) the ROCs for X1(s) and X2(s).", - "type": "text" - }, - { - "block_id": "p472-b11", - "global_id": 13232, - "bbox": [ - 102.14, - 299.36, - 253.37, - 311.48 - ], - "text": "FREQUENCY CONVOLUTION", - "type": "text" - }, - { - "block_id": "p472-b12", - "global_id": 13233, - "bbox": [ - 204.76, - 335.77, - 281.47, - 359.79 - ], - "text": "x1(t)x2(t) ⇐⇒\n1\n2πj", - "type": "text" - }, - { - "block_id": "p472-b13", - "global_id": 13234, - "bbox": [ - 282.67, - 328.78, - 310.09, - 341.05 - ], - "text": "# c+j∞", - "type": "text" - }, - { - "block_id": "p472-b14", - "global_id": 13235, - "bbox": [ - 287.94, - 353.65, - 305.53, - 360.85 - ], - "text": "c−j∞", - "type": "text" - }, - { - "block_id": "p472-b15", - "global_id": 13236, - "bbox": [ - 311.7, - 342.34, - 387.46, - 353.49 - ], - "text": "X1(w)X2(s −w)dw", - "type": "text" - }, - { - "block_id": "p472-b16", - "global_id": 13237, - "bbox": [ - 101.84, - 374.66, - 320.01, - 386.12 - ], - "text": "The ROC for X1(s) ∗X2(s) is a1 + a2 < Re s < b1 + b2.", - "type": "text" - }, - { - "block_id": "p472-b17", - "global_id": 13238, - "bbox": [ - 102.14, - 403.35, - 188.36, - 415.48 - ], - "text": "TIME REVERSAL", - "type": "text" - }, - { - "block_id": "p472-b18", - "global_id": 13239, - "bbox": [ - 260.41, - 441.77, - 331.83, - 452.05 - ], - "text": "x(−t) ⇐⇒X(−s)", - "type": "text" - }, - { - "block_id": "p472-b19", - "global_id": 13240, - "bbox": [ - 101.85, - 467.86, - 262.46, - 478.23 - ], - "text": "The ROC for X(−s) is −b < Re s < −a.", - "type": "text" - }, - { - "block_id": "p472-b20", - "global_id": 13241, - "bbox": [ - 101.84, - 510.14, - 450.66, - 522.1 - ], - "text": "4.11-2 Using the Bilateral Transform for Linear System Analysis", - "type": "text" - }, - { - "block_id": "p472-b21", - "global_id": 13242, - "bbox": [ - 101.84, - 528.23, - 490.37, - 562.1 - ], - "text": "Since the bilateral Laplace transform can handle noncausal signals, we can analyze noncausal\nLTIC systems using the bilateral Laplace transform. We have shown that the (zero-state) output\ny(t) is given by", - "type": "text" - }, - { - "block_id": "p472-b22", - "global_id": 13243, - "bbox": [ - 252.93, - 572.69, - 339.33, - 584.69 - ], - "text": "y(t) = L−1[X(s)H(s)]", - "type": "text" - }, - { - "block_id": "p472-b23", - "global_id": 13244, - "bbox": [ - 101.85, - 600.5, - 490.4, - 634.79 - ], - "text": "This expression is valid only if X(s)H(s) exists. The ROC of X(s)H(s) is the region in which both\nX(s) and H(s) exist. In other words, the ROC of X(s)H(s) is the region common to the regions of\nconvergence of both X(s) and H(s). These ideas are clarified in the following examples.", - "type": "text" - } - ] - }, - { - "page_num": 473, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p473-b0", - "global_id": 13245, - "bbox": [ - 322.07, - 62.89, - 516.13, - 71.98 - ], - "text": "4.11\nThe Bilateral Laplace Transform\n453", - "type": "text" - }, - { - "block_id": "p473-b1", - "global_id": 13246, - "bbox": [ - 102.51, - 93.92, - 416.84, - 105.87 - ], - "text": "EXAMPLE 4.33\nCircuit Response to a Noncausal Input", - "type": "text" - }, - { - "block_id": "p473-b2", - "global_id": 13247, - "bbox": [ - 128.9, - 121.73, - 416.19, - 132.11 - ], - "text": "Find the current y(t) for the RC circuit in Fig. 4.60a if the voltage x(t) is", - "type": "text" - }, - { - "block_id": "p473-b3", - "global_id": 13248, - "bbox": [ - 269.07, - 141.35, - 362.59, - 153.13 - ], - "text": "x(t) = etu(t) + e2tu(−t)", - "type": "text" - }, - { - "block_id": "p473-b4", - "global_id": 13249, - "bbox": [ - 170.13, - 315.49, - 181.71, - 323.57 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p473-b5", - "global_id": 13250, - "bbox": [ - 182.71, - 340.23, - 186.71, - 348.23 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p473-b6", - "global_id": 13251, - "bbox": [ - 182.58, - 390.55, - 191.9, - 398.55 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p473-b7", - "global_id": 13252, - "bbox": [ - 119.94, - 371.27, - 417.02, - 379.56 - ], - "text": "2\n2\n0\n2\n2", - "type": "text" - }, - { - "block_id": "p473-b8", - "global_id": 13253, - "bbox": [ - 332.33, - 315.49, - 343.92, - 323.57 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p473-b9", - "global_id": 13254, - "bbox": [ - 214.51, - 371.56, - 376.07, - 385.04 - ], - "text": "t\nt\n0", - "type": "text" - }, - { - "block_id": "p473-b10", - "global_id": 13255, - "bbox": [ - 334.4, - 347.08, - 345.92, - 355.37 - ], - "text": "23", - "type": "text" - }, - { - "block_id": "p473-b11", - "global_id": 13256, - "bbox": [ - 344.38, - 390.55, - 353.26, - 398.55 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p473-b12", - "global_id": 13257, - "bbox": [ - 263.02, - 294.77, - 271.9, - 302.77 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p473-b13", - "global_id": 13258, - "bbox": [ - 207.11, - 254.75, - 327.34, - 265.65 - ], - "text": "x(t)\ny(t)\n1 F", - "type": "text" - }, - { - "block_id": "p473-b14", - "global_id": 13259, - "bbox": [ - 261.89, - 215.51, - 274.12, - 223.81 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p473-b15", - "global_id": 13260, - "bbox": [ - 119.94, - 405.24, - 321.9, - 414.48 - ], - "text": "Figure 4.60 Response of a circuit to a noncausal input.", - "type": "text" - }, - { - "block_id": "p473-b16", - "global_id": 13261, - "bbox": [ - 146.84, - 433.37, - 351.59, - 443.75 - ], - "text": "The transfer function H(s) of the circuit is given by", - "type": "text" - }, - { - "block_id": "p473-b17", - "global_id": 13262, - "bbox": [ - 259.07, - 450.38, - 372.59, - 474.5 - ], - "text": "H(s) =\ns\ns + 1\nRes > −1", - "type": "text" - }, - { - "block_id": "p473-b18", - "global_id": 13263, - "bbox": [ - 128.9, - 481.96, - 502.74, - 504.29 - ], - "text": "Because h(t) is a causal function, the ROC of H(s) is Res > −1. Next, the bilateral Laplace\ntransform of x(t) is given by", - "type": "text" - }, - { - "block_id": "p473-b19", - "global_id": 13264, - "bbox": [ - 204.31, - 513.03, - 427.36, - 537.47 - ], - "text": "X(s) =\n1\ns −1 −\n1\ns −2 =\n−1\n(s −1)(s −2)\n1 < Res < 2", - "type": "text" - }, - { - "block_id": "p473-b20", - "global_id": 13265, - "bbox": [ - 128.9, - 545.84, - 347.63, - 556.21 - ], - "text": "The response y(t) is the inverse transform of X(s)H(s):", - "type": "text" - }, - { - "block_id": "p473-b21", - "global_id": 13266, - "bbox": [ - 171.26, - 570.88, - 213.72, - 582.88 - ], - "text": "y(t) = L−1", - "type": "text" - }, - { - "block_id": "p473-b22", - "global_id": 13267, - "bbox": [ - 215.34, - 558.62, - 303.42, - 590.05 - ], - "text": "−s\n(s + 1)(s −1)(s −2)", - "type": "text" - }, - { - "block_id": "p473-b23", - "global_id": 13268, - "bbox": [ - 304.64, - 558.62, - 310.07, - 568.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p473-b24", - "global_id": 13269, - "bbox": [ - 312.11, - 570.88, - 337.72, - 582.57 - ], - "text": "= L−1", - "type": "text" - }, - { - "block_id": "p473-b25", - "global_id": 13270, - "bbox": [ - 339.33, - 558.62, - 350.94, - 576.0 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p473-b26", - "global_id": 13271, - "bbox": [ - 345.96, - 566.03, - 391.3, - 590.05 - ], - "text": "6\n1\ns + 1 + 1", - "type": "text" - }, - { - "block_id": "p473-b27", - "global_id": 13272, - "bbox": [ - 386.32, - 566.03, - 431.66, - 590.05 - ], - "text": "2\n1\ns −1 −2", - "type": "text" - }, - { - "block_id": "p473-b28", - "global_id": 13273, - "bbox": [ - 426.68, - 566.03, - 453.77, - 590.05 - ], - "text": "3\n1\ns −2", - "type": "text" - }, - { - "block_id": "p473-b29", - "global_id": 13274, - "bbox": [ - 454.97, - 558.62, - 460.4, - 568.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p473-b30", - "global_id": 13275, - "bbox": [ - 128.9, - 599.08, - 502.77, - 621.41 - ], - "text": "The ROC of X(s)H(s) is that ROC common to both X(s) and H(s). This is 1 < Res < 2. The\npoles s = ±1 lie to the left of the ROC and, therefore, correspond to causal signals; the pole", - "type": "text" - } - ] - }, - { - "page_num": 474, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p474-b0", - "global_id": 13276, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "454\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p474-b1", - "global_id": 13277, - "bbox": [ - 103.16, - 85.83, - 426.72, - 96.21 - ], - "text": "s = 2 lies to the right of the ROC and thus represents an anticausal signal. Hence,", - "type": "text" - }, - { - "block_id": "p474-b2", - "global_id": 13278, - "bbox": [ - 215.2, - 106.4, - 246.55, - 118.02 - ], - "text": "y(t) = 1", - "type": "text" - }, - { - "block_id": "p474-b3", - "global_id": 13279, - "bbox": [ - 243.06, - 106.4, - 364.98, - 120.94 - ], - "text": "6e−tu(t) + 1\n2etu(t) + 2\n3e2tu(−t)", - "type": "text" - }, - { - "block_id": "p474-b4", - "global_id": 13280, - "bbox": [ - 103.17, - 129.66, - 314.06, - 140.04 - ], - "text": "Figure 4.60c shows y(t). Note that in this example, if", - "type": "text" - }, - { - "block_id": "p474-b5", - "global_id": 13281, - "bbox": [ - 236.15, - 149.85, - 344.03, - 161.86 - ], - "text": "x(t) = e−4tu(t) + e−2tu(−t)", - "type": "text" - }, - { - "block_id": "p474-b6", - "global_id": 13282, - "bbox": [ - 103.16, - 173.5, - 477.04, - 195.83 - ], - "text": "then the ROC of X(s) is −4 < Res < −2. Here no region of convergence exists for X(s)H(s).\nHence, the response y(t) goes to infinity.", - "type": "text" - }, - { - "block_id": "p474-b7", - "global_id": 13283, - "bbox": [ - 76.77, - 288.6, - 360.24, - 300.56 - ], - "text": "EXAMPLE 4.34\nResponse of a Noncausal System", - "type": "text" - }, - { - "block_id": "p474-b8", - "global_id": 13284, - "bbox": [ - 103.16, - 316.81, - 384.43, - 327.19 - ], - "text": "Find the response y(t) of a noncausal system with the transfer function", - "type": "text" - }, - { - "block_id": "p474-b9", - "global_id": 13285, - "bbox": [ - 237.21, - 336.71, - 285.45, - 353.97 - ], - "text": "H(s) = −1", - "type": "text" - }, - { - "block_id": "p474-b10", - "global_id": 13286, - "bbox": [ - 269.22, - 343.7, - 342.96, - 361.15 - ], - "text": "s −1\nRes < 1", - "type": "text" - }, - { - "block_id": "p474-b11", - "global_id": 13287, - "bbox": [ - 103.16, - 369.7, - 211.51, - 381.3 - ], - "text": "to the input x(t) = e−2tu(t).", - "type": "text" - }, - { - "block_id": "p474-b12", - "global_id": 13288, - "bbox": [ - 103.16, - 404.21, - 137.13, - 414.18 - ], - "text": "We have", - "type": "text" - }, - { - "block_id": "p474-b13", - "global_id": 13289, - "bbox": [ - 233.88, - 412.09, - 346.29, - 436.11 - ], - "text": "X(s) =\n1\ns + 2\nRes > −2", - "type": "text" - }, - { - "block_id": "p474-b14", - "global_id": 13290, - "bbox": [ - 103.17, - 441.73, - 117.54, - 451.7 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p474-b15", - "global_id": 13291, - "bbox": [ - 222.68, - 449.2, - 356.3, - 473.63 - ], - "text": "Y(s) = X(s)H(s) =\n−1\n(s −1)(s + 2)", - "type": "text" - }, - { - "block_id": "p474-b16", - "global_id": 13292, - "bbox": [ - 103.16, - 479.8, - 426.12, - 490.18 - ], - "text": "The ROC of X(s)H(s) is the region −2 < Res < 1. By partial fraction expansion,", - "type": "text" - }, - { - "block_id": "p474-b17", - "global_id": 13293, - "bbox": [ - 207.74, - 499.7, - 260.81, - 516.96 - ], - "text": "Y(s) = −1/3", - "type": "text" - }, - { - "block_id": "p474-b18", - "global_id": 13294, - "bbox": [ - 239.75, - 499.7, - 291.26, - 524.14 - ], - "text": "s −1 + 1/3", - "type": "text" - }, - { - "block_id": "p474-b19", - "global_id": 13295, - "bbox": [ - 274.08, - 506.69, - 372.45, - 524.14 - ], - "text": "s + 2\n−2 < Res < 1", - "type": "text" - }, - { - "block_id": "p474-b20", - "global_id": 13296, - "bbox": [ - 103.16, - 532.74, - 117.54, - 542.7 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p474-b21", - "global_id": 13297, - "bbox": [ - 234.37, - 542.94, - 265.72, - 554.56 - ], - "text": "y(t) = 1", - "type": "text" - }, - { - "block_id": "p474-b22", - "global_id": 13298, - "bbox": [ - 262.23, - 543.06, - 345.82, - 557.48 - ], - "text": "3[etu(−t) + e−2tu(t)]", - "type": "text" - }, - { - "block_id": "p474-b23", - "global_id": 13299, - "bbox": [ - 103.17, - 563.22, - 477.04, - 621.41 - ], - "text": "Note that the pole of H(s) lies in the RHP at 1. Yet the system is not unstable. The pole(s) in\nthe RHP may indicate instability or noncausality, depending on its location with respect to the\nregion of convergence of H(s). For example, if H(s) = −1/(s−1) with Res > 1, the system is\ncausal and unstable, with h(t) = −etu(t). In contrast, if H(s) = −1/(s −1) with Res < 1, the\nsystem is noncausal and stable, with h(t) = etu(−t).", - "type": "text" - } - ] - }, - { - "page_num": 475, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p475-b0", - "global_id": 13300, - "bbox": [ - 309.38, - 62.89, - 516.12, - 71.98 - ], - "text": "4.12\nMATLAB: Continuous-Time Filters\n455", - "type": "text" - }, - { - "block_id": "p475-b1", - "global_id": 13301, - "bbox": [ - 102.51, - 93.92, - 418.86, - 105.87 - ], - "text": "EXAMPLE 4.35\nSystem Response to a Noncausal Input", - "type": "text" - }, - { - "block_id": "p475-b2", - "global_id": 13302, - "bbox": [ - 128.9, - 122.13, - 367.83, - 132.5 - ], - "text": "Find the response y(t) of a system with the transfer function", - "type": "text" - }, - { - "block_id": "p475-b3", - "global_id": 13303, - "bbox": [ - 259.07, - 142.44, - 372.59, - 166.46 - ], - "text": "H(s) =\n1\ns + 5\nRes > −5", - "type": "text" - }, - { - "block_id": "p475-b4", - "global_id": 13304, - "bbox": [ - 128.9, - 175.06, - 180.92, - 185.03 - ], - "text": "and the input", - "type": "text" - }, - { - "block_id": "p475-b5", - "global_id": 13305, - "bbox": [ - 263.63, - 184.88, - 368.03, - 196.88 - ], - "text": "x(t) = e−tu(t) + e−2tu(−t)", - "type": "text" - }, - { - "block_id": "p475-b6", - "global_id": 13306, - "bbox": [ - 128.9, - 226.45, - 502.77, - 261.52 - ], - "text": "The input x(t) is of the type depicted in Fig. 4.58f, and the region of convergence for X(s) does\nnot exist. In this case, we must determine separately the system response to each of the two\ninput components, x1(t) = e−tu(t) and x2(t) = e−2tu(−t).", - "type": "text" - }, - { - "block_id": "p475-b7", - "global_id": 13307, - "bbox": [ - 257.86, - 270.68, - 373.81, - 294.7 - ], - "text": "X1(s) =\n1\ns + 1\nRes > −1", - "type": "text" - }, - { - "block_id": "p475-b8", - "global_id": 13308, - "bbox": [ - 257.86, - 295.92, - 308.53, - 314.05 - ], - "text": "X2(s) = −1", - "type": "text" - }, - { - "block_id": "p475-b9", - "global_id": 13309, - "bbox": [ - 292.3, - 302.9, - 373.81, - 320.35 - ], - "text": "s + 2\nRes < −2", - "type": "text" - }, - { - "block_id": "p475-b10", - "global_id": 13310, - "bbox": [ - 128.9, - 328.6, - 439.58, - 339.75 - ], - "text": "If y1(t) and y2(t) are the system responses to x1(t) and x2(t), respectively, then", - "type": "text" - }, - { - "block_id": "p475-b11", - "global_id": 13311, - "bbox": [ - 207.35, - 348.51, - 327.0, - 372.94 - ], - "text": "Y1(s) =\n1\n(s + 1)(s + 5) = 1/4", - "type": "text" - }, - { - "block_id": "p475-b12", - "global_id": 13312, - "bbox": [ - 309.81, - 348.51, - 359.98, - 372.94 - ], - "text": "s + 1 −1/4", - "type": "text" - }, - { - "block_id": "p475-b13", - "global_id": 13313, - "bbox": [ - 342.81, - 355.49, - 424.32, - 372.94 - ], - "text": "s + 5\nRes > −1", - "type": "text" - }, - { - "block_id": "p475-b14", - "global_id": 13314, - "bbox": [ - 128.9, - 382.46, - 155.19, - 392.42 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p475-b15", - "global_id": 13315, - "bbox": [ - 266.57, - 392.66, - 301.91, - 405.15 - ], - "text": "y1(t) = 1", - "type": "text" - }, - { - "block_id": "p475-b16", - "global_id": 13316, - "bbox": [ - 298.42, - 390.38, - 365.1, - 407.19 - ], - "text": "4(e−t −e−5t)u(t)", - "type": "text" - }, - { - "block_id": "p475-b17", - "global_id": 13317, - "bbox": [ - 128.91, - 413.34, - 143.28, - 423.3 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p475-b18", - "global_id": 13318, - "bbox": [ - 193.7, - 420.81, - 318.57, - 445.24 - ], - "text": "Y2(s) =\n−1\n(s + 2)(s + 5) = −1/3", - "type": "text" - }, - { - "block_id": "p475-b19", - "global_id": 13319, - "bbox": [ - 297.5, - 420.81, - 349.02, - 445.24 - ], - "text": "s + 2 + 1/3", - "type": "text" - }, - { - "block_id": "p475-b20", - "global_id": 13320, - "bbox": [ - 331.84, - 427.79, - 437.97, - 445.24 - ], - "text": "s + 5\n−5 < Res < −2", - "type": "text" - }, - { - "block_id": "p475-b21", - "global_id": 13321, - "bbox": [ - 128.9, - 451.77, - 155.19, - 461.73 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p475-b22", - "global_id": 13322, - "bbox": [ - 253.65, - 461.96, - 288.99, - 474.46 - ], - "text": "y2(t) = 1", - "type": "text" - }, - { - "block_id": "p475-b23", - "global_id": 13323, - "bbox": [ - 285.5, - 462.08, - 378.01, - 476.5 - ], - "text": "3[e−2tu(−t) + e−5tu(t)]", - "type": "text" - }, - { - "block_id": "p475-b24", - "global_id": 13324, - "bbox": [ - 128.91, - 482.65, - 170.65, - 492.61 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p475-b25", - "global_id": 13325, - "bbox": [ - 204.97, - 492.84, - 296.64, - 505.34 - ], - "text": "y(t) = y1(t) + y2(t) = 1", - "type": "text" - }, - { - "block_id": "p475-b26", - "global_id": 13326, - "bbox": [ - 293.15, - 486.18, - 356.45, - 507.39 - ], - "text": "3e−2tu(−t) +\n 1", - "type": "text" - }, - { - "block_id": "p475-b27", - "global_id": 13327, - "bbox": [ - 352.96, - 486.18, - 426.7, - 507.39 - ], - "text": "4e−t + 1\n12e−5t\nu(t)", - "type": "text" - }, - { - "block_id": "p475-b28", - "global_id": 13328, - "bbox": [ - 127.94, - 569.02, - 418.72, - 582.97 - ], - "text": "4.12 MATLAB: CONTINUOUS-TIME FILTERS", - "type": "text" - }, - { - "block_id": "p475-b29", - "global_id": 13329, - "bbox": [ - 127.59, - 588.96, - 516.13, - 634.79 - ], - "text": "Continuous-time filters are essential to many if not most engineering systems, and MATLAB\nis an excellent assistant for filter design and analysis. Although a comprehensive treatment of\ncontinuous-time filter techniques is outside the scope of this book, quality filters can be designed\nand realized with minimal additional theory.", - "type": "text" - } - ] - }, - { - "page_num": 476, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p476-b0", - "global_id": 13330, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "456\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p476-b1", - "global_id": 13331, - "bbox": [ - 101.84, - 85.82, - 490.39, - 143.6 - ], - "text": "A simple yet practical example demonstrates basic filtering concepts. Telephone voice signals\nare often lowpass-filtered to eliminate frequencies above a cutoff of 3 kHz, or ωc = 3000(2π) ≈\n18,850 rad/s. Filtering maintains satisfactory speech quality and reduces signal bandwidth, thereby\nincreasing the phone company’s call capacity. How, then, do we design and realize an acceptable\n3 kHz lowpass filter?", - "type": "text" - }, - { - "block_id": "p476-b2", - "global_id": 13332, - "bbox": [ - 101.84, - 168.57, - 402.54, - 180.53 - ], - "text": "4.12-1 Frequency Response and Polynomial Evaluation", - "type": "text" - }, - { - "block_id": "p476-b3", - "global_id": 13333, - "bbox": [ - 101.84, - 186.66, - 490.4, - 221.23 - ], - "text": "Magnitude response plots help assess a filter’s performance and quality. The magnitude response\nof an ideal filter is a brick-wall function with unity passband gain and perfect stopband attenuation.\nFor a lowpass filter with cutoff frequency ωc, the ideal magnitude response is", - "type": "text" - }, - { - "block_id": "p476-b4", - "global_id": 13334, - "bbox": [ - 232.43, - 237.93, - 286.11, - 249.08 - ], - "text": "|Hideal(jω)| =", - "type": "text" - }, - { - "block_id": "p476-b5", - "global_id": 13335, - "bbox": [ - 288.16, - 223.95, - 353.13, - 254.89 - ], - "text": "1\n|ω| ≤ωc\n0\n|ω| > ωc", - "type": "text" - }, - { - "block_id": "p476-b6", - "global_id": 13336, - "bbox": [ - 101.84, - 265.77, - 490.41, - 287.69 - ], - "text": "Unfortunately, ideal filters cannot be implemented in practice. Realizable filters require\ncompromises, although good designs will closely approximate the desired brick-wall response.", - "type": "text" - }, - { - "block_id": "p476-b7", - "global_id": 13337, - "bbox": [ - 101.84, - 289.68, - 490.41, - 311.6 - ], - "text": "A realizable LTIC system often has a rational transfer function that is represented in the\ns-domain as", - "type": "text" - }, - { - "block_id": "p476-b8", - "global_id": 13338, - "bbox": [ - 216.09, - 325.31, - 265.83, - 342.57 - ], - "text": "H(s) = Y(s)", - "type": "text" - }, - { - "block_id": "p476-b9", - "global_id": 13339, - "bbox": [ - 248.1, - 325.31, - 297.59, - 349.64 - ], - "text": "X(s) = B(s)", - "type": "text" - }, - { - "block_id": "p476-b10", - "global_id": 13340, - "bbox": [ - 280.2, - 332.29, - 308.61, - 349.64 - ], - "text": "A(s) =", - "type": "text" - }, - { - "block_id": "p476-b11", - "global_id": 13341, - "bbox": [ - 312.71, - 310.52, - 323.12, - 320.48 - ], - "text": "M%", - "type": "text" - }, - { - "block_id": "p476-b12", - "global_id": 13342, - "bbox": [ - 311.85, - 329.4, - 323.99, - 336.66 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p476-b13", - "global_id": 13343, - "bbox": [ - 325.1, - 316.77, - 374.32, - 329.08 - ], - "text": "bk+N−MsM−k", - "type": "text" - }, - { - "block_id": "p476-b14", - "global_id": 13344, - "bbox": [ - 324.24, - 339.26, - 334.65, - 349.23 - ], - "text": "N%", - "type": "text" - }, - { - "block_id": "p476-b15", - "global_id": 13345, - "bbox": [ - 323.38, - 358.13, - 335.51, - 365.4 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p476-b16", - "global_id": 13346, - "bbox": [ - 336.63, - 346.25, - 362.8, - 357.81 - ], - "text": "aksN−k", - "type": "text" - }, - { - "block_id": "p476-b17", - "global_id": 13347, - "bbox": [ - 101.84, - 371.97, - 490.39, - 394.3 - ], - "text": "Frequency response H(jω) is obtained by letting s = jω, where frequency ω is in radians per\nsecond.", - "type": "text" - }, - { - "block_id": "p476-b18", - "global_id": 13348, - "bbox": [ - 101.84, - 395.88, - 490.39, - 442.12 - ], - "text": "MATLAB is ideally suited to evaluate frequency response functions. Defining a length-(N +\n1) coefficient vector A = [a0,a1,...,aN] and a length-(M + 1) coefficient vector B =\n[bN−M,bN−M+1,..., bN], program CH4MP1 computes H(jω) for each frequency in the input vector\nω.", - "type": "text" - }, - { - "block_id": "p476-b19", - "global_id": 13349, - "bbox": [ - 101.84, - 452.58, - 420.9, - 534.28 - ], - "text": "function [H] = CH4MP1(B,A,omega);\n% CH4MP1.m : Chapter 4, MATLAB Program 1\n% Function M-file computes frequency response for LTIC system\n% INPUTS:\nB = vector of feedforward coefficients\n%\nA = vector of feedback coefficients\n%\nomega = vector of frequencies [rad/s].\n% OUTPUTS:\nH =\nfrequency response", - "type": "text" - }, - { - "block_id": "p476-b20", - "global_id": 13350, - "bbox": [ - 101.84, - 548.23, - 326.75, - 558.19 - ], - "text": "H = polyval(B,j*omega)./polyval(A,j*omega);", - "type": "text" - }, - { - "block_id": "p476-b21", - "global_id": 13351, - "bbox": [ - 101.84, - 568.08, - 490.41, - 601.96 - ], - "text": "The function polyval efficiently evaluates simple polynomials and makes the program nearly\ntrivial. For example, when A is the vector of coefficients [a0,a1,. . .,aN], polyval (A,j*omega)\ncomputes", - "type": "text" - }, - { - "block_id": "p476-b22", - "global_id": 13352, - "bbox": [ - 268.6, - 603.42, - 282.69, - 613.87 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p476-b23", - "global_id": 13353, - "bbox": [ - 269.57, - 627.77, - 281.7, - 635.03 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p476-b24", - "global_id": 13354, - "bbox": [ - 283.8, - 611.66, - 323.02, - 624.46 - ], - "text": "ak(jω)N−k", - "type": "text" - } - ] - }, - { - "page_num": 477, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p477-b0", - "global_id": 13355, - "bbox": [ - 309.38, - 62.89, - 516.12, - 71.98 - ], - "text": "4.12\nMATLAB: Continuous-Time Filters\n457", - "type": "text" - }, - { - "block_id": "p477-b1", - "global_id": 13356, - "bbox": [ - 127.59, - 85.82, - 516.11, - 108.03 - ], - "text": "for each value of the frequency vector omega. It is also possible to compute frequency responses\nby using the signal-processing toolbox function freqs.", - "type": "text" - }, - { - "block_id": "p477-b2", - "global_id": 13357, - "bbox": [ - 127.89, - 132.81, - 399.81, - 145.36 - ], - "text": "DESIGN AND EVALUATION OF A SIMPLE RC FILTER", - "type": "text" - }, - { - "block_id": "p477-b3", - "global_id": 13358, - "bbox": [ - 127.59, - 149.29, - 516.13, - 183.97 - ], - "text": "One of the simplest lowpass filters is realized by using an RC circuit, as shown in Fig. 4.61.\nThis one-pole system has transfer function HRC(s) = (RCs + 1)−1 and magnitude response\n|HRC(jω)| = |(jωRC + 1)−1| = 1/", - "type": "text" - }, - { - "block_id": "p477-b5", - "global_id": 13359, - "bbox": [ - 127.59, - 172.69, - 516.14, - 207.17 - ], - "text": "1 + (RCω)2. Independent of component values R and C, this\ncircuit has many desirable characteristics, such as unity gain at ω = 0 and magnitude response that\nmonotonically decreases to zero as ω →∞.", - "type": "text" - }, - { - "block_id": "p477-b6", - "global_id": 13360, - "bbox": [ - 129.8, - 264.3, - 233.92, - 273.36 - ], - "text": "x(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p477-b7", - "global_id": 13361, - "bbox": [ - 170.66, - 236.87, - 175.55, - 244.87 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p477-b8", - "global_id": 13362, - "bbox": [ - 186.4, - 266.47, - 191.73, - 274.47 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p477-b9", - "global_id": 13363, - "bbox": [ - 226.37, - 279.64, - 230.37, - 287.64 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p477-b10", - "global_id": 13364, - "bbox": [ - 226.11, - 252.29, - 230.62, - 260.29 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p477-b11", - "global_id": 13365, - "bbox": [ - 133.35, - 279.64, - 137.35, - 287.64 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p477-b12", - "global_id": 13366, - "bbox": [ - 133.09, - 252.29, - 137.61, - 260.29 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p477-b13", - "global_id": 13367, - "bbox": [ - 244.55, - 280.55, - 340.7, - 289.88 - ], - "text": "Figure 4.61 An RC filter.", - "type": "text" - }, - { - "block_id": "p477-b14", - "global_id": 13368, - "bbox": [ - 127.59, - 327.56, - 516.12, - 350.28 - ], - "text": "Components R and C are chosen to set the desired 3 kHz cutoff frequency. For many filter\ntypes, the cutoff frequency corresponds to the half-power point, or |HRC(jωc)| = 1/", - "type": "text" - }, - { - "block_id": "p477-b15", - "global_id": 13369, - "bbox": [ - 469.13, - 330.78, - 477.56, - 340.74 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p477-b16", - "global_id": 13370, - "bbox": [ - 127.59, - 339.62, - 516.13, - 361.54 - ], - "text": "2. Assign\nC a realistic capacitance of 1 nF, then the required resistance is computed by R = 1/", - "type": "text" - }, - { - "block_id": "p477-b18", - "global_id": 13371, - "bbox": [ - 127.59, - 350.96, - 516.14, - 375.34 - ], - "text": "C2ω2c =\n1/", - "type": "text" - }, - { - "block_id": "p477-b19", - "global_id": 13372, - "bbox": [ - 145.51, - 364.48, - 222.11, - 375.34 - ], - "text": "(10−9)2(2π3000)2.", - "type": "text" - }, - { - "block_id": "p477-b20", - "global_id": 13373, - "bbox": [ - 127.59, - 395.17, - 441.37, - 417.1 - ], - "text": ">>\nomega_c = 2*pi*3000; C = 1e-9; R = 1/sqrt(C^2*omega_c^2)\nR = 5.3052e+004", - "type": "text" - }, - { - "block_id": "p477-b21", - "global_id": 13374, - "bbox": [ - 127.59, - 435.94, - 516.14, - 458.97 - ], - "text": "The root of this first-order RC filter is directly related to the cutoff frequency, λ = −1/RC =\n−18,850 = −ωc.", - "type": "text" - }, - { - "block_id": "p477-b22", - "global_id": 13375, - "bbox": [ - 127.59, - 460.16, - 516.14, - 482.18 - ], - "text": "To evaluate the RC filter performance, the magnitude response is plotted over the mostly\naudible frequency range (0 ≤f ≤20 kHz).", - "type": "text" - }, - { - "block_id": "p477-b23", - "global_id": 13376, - "bbox": [ - 127.59, - 502.02, - 509.4, - 547.85 - ], - "text": ">>\nf = linspace(0,20000,200); Hmag_RC = abs(CH4MP1([1],[R*C 1],f*2*pi));\n>>\nplot(f,abs(f*2*pi)<=omega_c,’k-’,f,Hmag_RC,’k--’);\n>>\naxis([0 20000 -0.05 1.05]); xlabel(’f [Hz]’); ylabel(’|H(j2\\pi f)|’);\n>>\nlegend(’Ideal’,’First-order RC’,’location’,’best’);", - "type": "text" - }, - { - "block_id": "p477-b24", - "global_id": 13377, - "bbox": [ - 127.59, - 567.1, - 516.11, - 589.31 - ], - "text": "The linspace(X1,X2,N) command generates an N-length vector of linearly spaced points\nbetween X1 and X2.", - "type": "text" - }, - { - "block_id": "p477-b25", - "global_id": 13378, - "bbox": [ - 127.59, - 590.91, - 516.15, - 636.84 - ], - "text": "As shown in Fig. 4.62, the first-order RC response is indeed lowpass with a half-power\ncutoff frequency equal to 3 kHz. It rather poorly approximates the desired brick-wall response:\nthe passband is not very flat, and stopband attenuation increases very slowly to less than 20 dB at\n20 kHz.", - "type": "text" - } - ] - }, - { - "page_num": 478, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p478-b0", - "global_id": 13379, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "458\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p478-b1", - "global_id": 13380, - "bbox": [ - 151.17, - 198.42, - 489.23, - 220.07 - ], - "text": "0\n0.2\n0.4\n0.6\n0.8\n1\n1.2\n1.4\n1.6\n1.8\n2\nf [Hz]\n×104", - "type": "text" - }, - { - "block_id": "p478-b2", - "global_id": 13381, - "bbox": [ - 144.95, - 184.42, - 148.95, - 192.42 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p478-b3", - "global_id": 13382, - "bbox": [ - 138.2, - 165.32, - 148.87, - 173.32 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p478-b4", - "global_id": 13383, - "bbox": [ - 138.2, - 146.23, - 148.87, - 154.24 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p478-b5", - "global_id": 13384, - "bbox": [ - 138.2, - 127.15, - 148.87, - 135.15 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p478-b6", - "global_id": 13385, - "bbox": [ - 138.2, - 108.05, - 148.87, - 116.06 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p478-b7", - "global_id": 13386, - "bbox": [ - 144.95, - 88.97, - 148.95, - 96.97 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p478-b8", - "global_id": 13387, - "bbox": [ - 126.4, - 123.53, - 134.69, - 155.92 - ], - "text": "|H(j2 π f)|", - "type": "text" - }, - { - "block_id": "p478-b9", - "global_id": 13388, - "bbox": [ - 420.79, - 96.54, - 435.18, - 103.74 - ], - "text": "Ideal", - "type": "text" - }, - { - "block_id": "p478-b10", - "global_id": 13389, - "bbox": [ - 420.79, - 107.57, - 462.99, - 114.77 - ], - "text": "First-order RC", - "type": "text" - }, - { - "block_id": "p478-b11", - "global_id": 13390, - "bbox": [ - 125.76, - 227.07, - 384.27, - 237.08 - ], - "text": "Figure 4.62 Magnitude response |HRC(j2πf)| of a first-order RC filter.", - "type": "text" - }, - { - "block_id": "p478-b12", - "global_id": 13391, - "bbox": [ - 125.97, - 308.5, - 137.08, - 316.58 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p478-b13", - "global_id": 13392, - "bbox": [ - 150.43, - 282.87, - 155.31, - 290.87 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p478-b14", - "global_id": 13393, - "bbox": [ - 129.52, - 318.05, - 133.52, - 326.05 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p478-b15", - "global_id": 13394, - "bbox": [ - 129.27, - 299.11, - 133.78, - 307.11 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p478-b16", - "global_id": 13395, - "bbox": [ - 430.73, - 308.54, - 441.84, - 316.62 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p478-b17", - "global_id": 13396, - "bbox": [ - 320.53, - 282.87, - 325.41, - 290.87 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p478-b18", - "global_id": 13397, - "bbox": [ - 434.29, - 318.09, - 438.29, - 326.09 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p478-b19", - "global_id": 13398, - "bbox": [ - 344.83, - 299.15, - 438.54, - 315.17 - ], - "text": "+\nC", - "type": "text" - }, - { - "block_id": "p478-b20", - "global_id": 13399, - "bbox": [ - 387.13, - 293.3, - 391.64, - 301.3 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p478-b21", - "global_id": 13400, - "bbox": [ - 387.13, - 277.1, - 391.13, - 285.1 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p478-b22", - "global_id": 13401, - "bbox": [ - 174.73, - 307.17, - 180.06, - 315.17 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p478-b23", - "global_id": 13402, - "bbox": [ - 217.03, - 293.3, - 221.54, - 301.3 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p478-b24", - "global_id": 13403, - "bbox": [ - 217.03, - 277.1, - 221.03, - 285.1 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p478-b25", - "global_id": 13404, - "bbox": [ - 125.76, - 336.82, - 252.03, - 346.14 - ], - "text": "Figure 4.63 A cascaded RC filter.", - "type": "text" - }, - { - "block_id": "p478-b26", - "global_id": 13405, - "bbox": [ - 102.14, - 382.66, - 402.75, - 395.21 - ], - "text": "A CASCADED RC FILTER AND POLYNOMIAL EXPANSION", - "type": "text" - }, - { - "block_id": "p478-b27", - "global_id": 13406, - "bbox": [ - 101.84, - 399.14, - 490.4, - 457.02 - ], - "text": "A first-order RC filter is destined for poor performance; one pole is simply insufficient to obtain\ngood results. A cascade of RC circuits increases the number of poles and improves the filter\nresponse. To simplify the analysis and prevent loading between stages, we employ op-amp\nfollowers to buffer the output of each stage, as shown in Fig. 4.63. A cascade of N stages results\nin an Nth-order filter with transfer function given by", - "type": "text" - }, - { - "block_id": "p478-b28", - "global_id": 13407, - "bbox": [ - 218.33, - 470.51, - 372.98, - 485.81 - ], - "text": "Hcascade(s) = [HRC(s)]N = (RCs + 1)−N", - "type": "text" - }, - { - "block_id": "p478-b29", - "global_id": 13408, - "bbox": [ - 101.84, - 502.69, - 490.38, - 525.02 - ], - "text": "Upon choosing a cascade of 10 stages and C = 1 nF, a 3 kHz cutoff frequency is obtained by\nsetting R =", - "type": "text" - }, - { - "block_id": "p478-b30", - "global_id": 13409, - "bbox": [ - 148.86, - 506.21, - 157.29, - 516.17 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p478-b31", - "global_id": 13410, - "bbox": [ - 157.29, - 506.21, - 241.75, - 525.72 - ], - "text": "21/10 −1/(Cωc) =\n√", - "type": "text" - }, - { - "block_id": "p478-b32", - "global_id": 13411, - "bbox": [ - 241.75, - 511.76, - 329.92, - 525.02 - ], - "text": "21/10 −1/(6π(10)−6).", - "type": "text" - }, - { - "block_id": "p478-b33", - "global_id": 13412, - "bbox": [ - 101.84, - 541.37, - 290.13, - 563.28 - ], - "text": ">>\nR = sqrt(2^(1/10)-1)/(C*omega_c)\nR = 1.4213e+004", - "type": "text" - }, - { - "block_id": "p478-b34", - "global_id": 13413, - "bbox": [ - 101.84, - 578.65, - 490.39, - 636.84 - ], - "text": "This cascaded filter has a 10th-order pole at λ = −1/RC and no finite zeros. To compute\nthe magnitude response, polynomial coefficient vectors A and B are needed. Setting B = [1]\nensures there are no finite zeros or, equivalently, that all zeros are at infinity. The poly command,\nwhich expands a vector of roots into a corresponding vector of polynomial coefficients, is used to\nobtain A.", - "type": "text" - } - ] - }, - { - "page_num": 479, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p479-b0", - "global_id": 13414, - "bbox": [ - 309.38, - 62.89, - 516.12, - 71.98 - ], - "text": "4.12\nMATLAB: Continuous-Time Filters\n459", - "type": "text" - }, - { - "block_id": "p479-b1", - "global_id": 13415, - "bbox": [ - 127.59, - 86.24, - 509.4, - 144.02 - ], - "text": ">>\nB = 1; A = poly(-1/(R*C)*ones(10,1));A = A/A(end);\n>>\nHmag_cascade = abs(CH4MP1(B,A,f*2*pi));\n>>\nplot(f,abs(f*2*pi)<=omega_c,’k-’,f,Hmag_cascade,’k--’);\n>>\naxis([0 20000 -0.05 1.05]); xlabel(’f [Hz]’); ylabel(’|H(j2\\pi f)|’);\n>>\nlegend(’Ideal’,’Tenth-order RC cascade’,’location’,’best’);", - "type": "text" - }, - { - "block_id": "p479-b2", - "global_id": 13416, - "bbox": [ - 127.59, - 153.69, - 516.13, - 187.56 - ], - "text": "Notice that scaling a polynomial by a constant does not change its roots. Conversely, the roots of\na polynomial specify a polynomial within a scale factor. The command A = A/A(end) properly\nscales the denominator polynomial to ensure unity gain at ω = 0.", - "type": "text" - }, - { - "block_id": "p479-b3", - "global_id": 13417, - "bbox": [ - 127.59, - 189.46, - 516.13, - 223.44 - ], - "text": "The magnitude response plot of the tenth-order RC cascade is shown in Fig. 4.64. Compared\nwith the simple RC response of Fig. 4.62, the passband remains relatively unchanged, but stopband\nattenuation is greatly improved to over 60 dB at 20 kHz.", - "type": "text" - }, - { - "block_id": "p479-b4", - "global_id": 13418, - "bbox": [ - 176.91, - 352.94, - 515.09, - 374.58 - ], - "text": "0\n0.2\n0.4\n0.6\n0.8\n1\n1.2\n1.4\n1.6\n1.8\n2\nf [Hz]\n×10 4", - "type": "text" - }, - { - "block_id": "p479-b5", - "global_id": 13419, - "bbox": [ - 170.69, - 338.93, - 174.69, - 346.94 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p479-b6", - "global_id": 13420, - "bbox": [ - 163.94, - 319.83, - 174.61, - 327.83 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p479-b7", - "global_id": 13421, - "bbox": [ - 163.94, - 300.75, - 174.61, - 308.75 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p479-b8", - "global_id": 13422, - "bbox": [ - 163.94, - 281.66, - 174.61, - 289.66 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p479-b9", - "global_id": 13423, - "bbox": [ - 163.94, - 262.57, - 174.61, - 270.57 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p479-b10", - "global_id": 13424, - "bbox": [ - 170.69, - 243.48, - 174.69, - 251.48 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p479-b11", - "global_id": 13425, - "bbox": [ - 152.14, - 278.15, - 160.43, - 309.73 - ], - "text": "|H(j2 π f)|", - "type": "text" - }, - { - "block_id": "p479-b12", - "global_id": 13426, - "bbox": [ - 401.53, - 251.05, - 415.92, - 258.25 - ], - "text": "Ideal", - "type": "text" - }, - { - "block_id": "p479-b13", - "global_id": 13427, - "bbox": [ - 401.53, - 262.08, - 471.49, - 269.28 - ], - "text": "Tenth-order RC cascade", - "type": "text" - }, - { - "block_id": "p479-b14", - "global_id": 13428, - "bbox": [ - 151.5, - 381.58, - 436.51, - 391.66 - ], - "text": "Figure 4.64 Magnitude response |Hcascade(j2πf)| of a tenth-order RC cascade.", - "type": "text" - }, - { - "block_id": "p479-b15", - "global_id": 13429, - "bbox": [ - 127.59, - 418.26, - 399.4, - 433.46 - ], - "text": "4.12-2 Butterworth Filters and the Find Command", - "type": "text" - }, - { - "block_id": "p479-b16", - "global_id": 13430, - "bbox": [ - 127.59, - 439.6, - 516.16, - 509.34 - ], - "text": "The pole location of a first-order lowpass filter is necessarily fixed by the cutoff frequency. There\nis little reason, however, to place all the poles of a 10th-order filter at one location. Better pole\nplacement will improve our filter’s magnitude response. One strategy, discussed in Sec. 4.10, is to\nplace a wall of poles opposite the passband frequencies. A semicircular wall of poles leads to the\nButterworth family of filters, and a semi-elliptical shape leads to the Chebyshev family of filters.\nButterworth filters are considered first.", - "type": "text" - }, - { - "block_id": "p479-b17", - "global_id": 13431, - "bbox": [ - 127.59, - 510.91, - 516.13, - 557.15 - ], - "text": "To begin, notice that a transfer function H(s) with real coefficients has a squared magnitude\nresponse given by |H(jω)|2 = H(jω)H∗(jω) = H(jω)H(−jω) = H(s)H(−s)|s=jω. Thus, half the\npoles of |H(jω)|2 correspond to the filter H(s) and the other half correspond to H(−s). Filters that\nare both stable and causal require H(s) to include only left-half-plane poles.", - "type": "text" - }, - { - "block_id": "p479-b18", - "global_id": 13432, - "bbox": [ - 145.52, - 559.15, - 375.45, - 569.12 - ], - "text": "The squared magnitude response of a Butterworth filter is", - "type": "text" - }, - { - "block_id": "p479-b19", - "global_id": 13433, - "bbox": [ - 262.26, - 579.04, - 379.35, - 603.77 - ], - "text": "|HBW(jω)|2 =\n1\n1 + (jω/jωc)2N", - "type": "text" - }, - { - "block_id": "p479-b20", - "global_id": 13434, - "bbox": [ - 127.59, - 612.76, - 516.13, - 634.79 - ], - "text": "This function has the same appealing characteristics as the first-order RC filter: a gain that is unity\nat ω = 0 and monotonically decreases to zero as ω →∞. By construction, the half-power gain", - "type": "text" - } - ] - }, - { - "page_num": 480, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p480-b0", - "global_id": 13435, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "460\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p480-b1", - "global_id": 13436, - "bbox": [ - 139.34, - 235.55, - 278.6, - 243.55 - ], - "text": "–2 –1.5 –1 –0.5\n0\n0.5\n1\n1.5\n2", - "type": "text" - }, - { - "block_id": "p480-b2", - "global_id": 13437, - "bbox": [ - 127.76, - 223.19, - 136.24, - 231.19 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p480-b3", - "global_id": 13438, - "bbox": [ - 122.24, - 206.83, - 136.24, - 214.83 - ], - "text": "–1.5", - "type": "text" - }, - { - "block_id": "p480-b4", - "global_id": 13439, - "bbox": [ - 128.24, - 190.47, - 136.24, - 198.47 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p480-b5", - "global_id": 13440, - "bbox": [ - 121.44, - 174.11, - 136.24, - 182.11 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p480-b6", - "global_id": 13441, - "bbox": [ - 132.24, - 157.75, - 136.24, - 165.75 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p480-b7", - "global_id": 13442, - "bbox": [ - 126.24, - 141.4, - 136.24, - 149.4 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p480-b8", - "global_id": 13443, - "bbox": [ - 132.24, - 125.03, - 136.24, - 133.03 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p480-b9", - "global_id": 13444, - "bbox": [ - 126.24, - 108.67, - 136.24, - 116.67 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p480-b10", - "global_id": 13445, - "bbox": [ - 132.24, - 92.31, - 136.24, - 100.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p480-b11", - "global_id": 13446, - "bbox": [ - 106.06, - 132.55, - 115.54, - 189.76 - ], - "text": "Imaginary (3 104)", - "type": "text" - }, - { - "block_id": "p480-b12", - "global_id": 13447, - "bbox": [ - 191.65, - 247.26, - 230.64, - 256.74 - ], - "text": "Real (3 104)", - "type": "text" - }, - { - "block_id": "p480-b13", - "global_id": 13448, - "bbox": [ - 293.8, - 232.46, - 486.39, - 259.07 - ], - "text": "Figure 4.65 Roots of |HBW(jω)|2 for N = 10 and\nωc = 3000(2π).", - "type": "text" - }, - { - "block_id": "p480-b14", - "global_id": 13449, - "bbox": [ - 101.84, - 286.82, - 490.41, - 321.1 - ], - "text": "occurs at ωc. Perhaps most importantly, however, the first 2N −1 derivatives of |HBW(jω)| with\nrespect to ω are zero at ω = 0. Put another way, the passband is constrained to be very flat for low\nfrequencies. For this reason, Butterworth filters are sometimes called maximally flat filters.", - "type": "text" - }, - { - "block_id": "p480-b15", - "global_id": 13450, - "bbox": [ - 101.84, - 323.1, - 490.39, - 381.65 - ], - "text": "As discussed in Sec. B.7, the roots of minus 1 must lie equally spaced on a circle centered\nat the origin. Thus, the 2N poles of |HBW(jω)|2 naturally lie equally spaced on a circle of radius\nωc centered at the origin. Figure 4.65 displays the 20 poles corresponding to the case N = 10 and\nωc = 3000(2π) rad/s. An Nth-order Butterworth filter that is both causal and stable uses the N\nleft-half-plane poles of |HBW(jω)|2.", - "type": "text" - }, - { - "block_id": "p480-b16", - "global_id": 13451, - "bbox": [ - 119.78, - 381.53, - 465.77, - 393.6 - ], - "text": "To design a 10th-order Butterworth filter, we first compute the 20 poles of |HBW(jω)|2:", - "type": "text" - }, - { - "block_id": "p480-b17", - "global_id": 13452, - "bbox": [ - 101.85, - 409.6, - 436.58, - 419.56 - ], - "text": ">>\nN=10; poles = roots([(1j*omega_c)^(-2*N),zeros(1,2*N-1),1]);", - "type": "text" - }, - { - "block_id": "p480-b18", - "global_id": 13453, - "bbox": [ - 101.85, - 435.75, - 490.38, - 469.63 - ], - "text": "The find command is a powerful and useful function that returns the indices of a vector’s nonzero\nelements. Combined with relational operators, the find command allows us to extract the 10\nleft-half-plane roots that correspond to the poles of our Butterworth filter.", - "type": "text" - }, - { - "block_id": "p480-b19", - "global_id": 13454, - "bbox": [ - 101.85, - 486.39, - 321.52, - 496.35 - ], - "text": ">>\nBW_poles = poles(find(real(poles)<0));", - "type": "text" - }, - { - "block_id": "p480-b20", - "global_id": 13455, - "bbox": [ - 101.85, - 512.46, - 443.54, - 522.5 - ], - "text": "To compute the magnitude response, these roots are converted to coefficient vector A.", - "type": "text" - }, - { - "block_id": "p480-b21", - "global_id": 13456, - "bbox": [ - 101.85, - 539.27, - 483.66, - 585.1 - ], - "text": ">>\nA = poly(BW_poles); A = A/A(end); Hmag_BW = abs(CH4MP1(B,A,f*2*pi));\n>>\nplot(f,abs(f*2*pi)<=omega_c,’k-’,f,Hmag_BW,’k--’);\n>>\naxis([0 20000 -0.05 1.05]); xlabel(’f [Hz]’); ylabel(’|H(j2\\pi f)|’);\n>>\nlegend(’Ideal’,’Tenth-order Butterworth’,’location’,’best’);", - "type": "text" - }, - { - "block_id": "p480-b22", - "global_id": 13457, - "bbox": [ - 101.85, - 601.28, - 490.4, - 647.11 - ], - "text": "The magnitude response plot of the Butterworth filter is shown in Fig. 4.66. The Butterworth\nresponse closely approximates the brick-wall function and provides excellent filter characteristics:\nflat passband, rapid transition to the stopband, and excellent stopband attenuation (>40 dB at\n5 kHz).", - "type": "text" - } - ] - }, - { - "page_num": 481, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p481-b0", - "global_id": 13458, - "bbox": [ - 309.38, - 62.89, - 516.13, - 71.98 - ], - "text": "4.12\nMATLAB: Continuous-Time Filters\n461", - "type": "text" - }, - { - "block_id": "p481-b1", - "global_id": 13459, - "bbox": [ - 176.9, - 198.42, - 514.99, - 220.06 - ], - "text": "0\n0.2\n0.4\n0.6\n0.8\n1\n1.2\n1.4\n1.6\n1.8\n2\nf [Hz]\n×104", - "type": "text" - }, - { - "block_id": "p481-b2", - "global_id": 13460, - "bbox": [ - 170.69, - 184.42, - 174.69, - 192.42 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p481-b3", - "global_id": 13461, - "bbox": [ - 163.93, - 165.32, - 174.6, - 173.32 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p481-b4", - "global_id": 13462, - "bbox": [ - 163.93, - 146.23, - 174.6, - 154.23 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p481-b5", - "global_id": 13463, - "bbox": [ - 163.93, - 127.14, - 174.6, - 135.14 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p481-b6", - "global_id": 13464, - "bbox": [ - 163.93, - 108.04, - 174.6, - 116.04 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p481-b7", - "global_id": 13465, - "bbox": [ - 170.69, - 88.96, - 174.69, - 96.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p481-b8", - "global_id": 13466, - "bbox": [ - 152.14, - 123.63, - 160.43, - 155.22 - ], - "text": "|H(j2 π f)|", - "type": "text" - }, - { - "block_id": "p481-b9", - "global_id": 13467, - "bbox": [ - 403.04, - 96.53, - 417.43, - 103.73 - ], - "text": "Ideal", - "type": "text" - }, - { - "block_id": "p481-b10", - "global_id": 13468, - "bbox": [ - 403.04, - 107.57, - 474.02, - 114.77 - ], - "text": "Tenth-order Butterworth", - "type": "text" - }, - { - "block_id": "p481-b11", - "global_id": 13469, - "bbox": [ - 151.5, - 227.07, - 447.3, - 237.15 - ], - "text": "Figure 4.66 Magnitude response |HBW(j2πf)| of a tenth-order Butterworth filter.", - "type": "text" - }, - { - "block_id": "p481-b12", - "global_id": 13470, - "bbox": [ - 127.59, - 263.06, - 379.44, - 288.95 - ], - "text": "4.12-3 Using Cascaded Second-Order Sections\nfor Butterworth Filter Realization", - "type": "text" - }, - { - "block_id": "p481-b13", - "global_id": 13471, - "bbox": [ - 127.59, - 294.98, - 516.14, - 317.01 - ], - "text": "For our RC filters, realization preceded design. For our Butterworth filter, however, design has\npreceded realization. For our Butterworth filter to be useful, we must be able to implement it.", - "type": "text" - }, - { - "block_id": "p481-b14", - "global_id": 13472, - "bbox": [ - 127.59, - 318.58, - 516.13, - 412.64 - ], - "text": "Since the transfer function HBW(s) is known, the differential equation is also known.\nTherefore, it is possible to try to implement the design by using op-amp integrators, summers, and\nscalar multipliers. Unfortunately, this approach will not work well. To understand why, consider\nthe denominator coefficients a0 = 1.766×10−43 and a10 = 1. The smallest coefficient is 43 orders\nof magnitude smaller than the largest coefficient! It is practically impossible to accurately realize\nsuch a broad range in scale values. To understand this, skeptics should try to find realistic resistors\nsuch that Rf /R = 1.766×10−43. Additionally, small component variations will cause large changes\nin actual pole location.", - "type": "text" - }, - { - "block_id": "p481-b15", - "global_id": 13473, - "bbox": [ - 127.6, - 414.65, - 516.14, - 472.44 - ], - "text": "A better approach is to cascade five second-order sections, where each section implements\none complex conjugate pair of poles. By pairing poles in complex conjugate pairs, each of the\nresulting second-order sections has real coefficients. With this approach, the smallest coefficients\nare only about nine orders of magnitude smaller than the largest coefficients. Furthermore, pole\nplacement is typically less sensitive to component variations for cascaded structures.", - "type": "text" - }, - { - "block_id": "p481-b16", - "global_id": 13474, - "bbox": [ - 127.6, - 474.44, - 516.17, - 496.36 - ], - "text": "The Sallen–Key circuit shown in Fig. 4.67 provides a good way to realize a pair of\ncomplex-conjugate poles.† The transfer function of this circuit is", - "type": "text" - }, - { - "block_id": "p481-b17", - "global_id": 13475, - "bbox": [ - 183.38, - 530.67, - 221.11, - 541.82 - ], - "text": "HSK(s) =", - "type": "text" - }, - { - "block_id": "p481-b18", - "global_id": 13476, - "bbox": [ - 278.1, - 511.44, - 319.0, - 536.23 - ], - "text": "1\nR1R2C1C2", - "type": "text" - }, - { - "block_id": "p481-b19", - "global_id": 13477, - "bbox": [ - 224.35, - 541.35, - 241.54, - 554.51 - ], - "text": "s2 +", - "type": "text" - }, - { - "block_id": "p481-b20", - "global_id": 13478, - "bbox": [ - 243.08, - 530.23, - 271.21, - 562.45 - ], - "text": "1\nR1C1", - "type": "text" - }, - { - "block_id": "p481-b21", - "global_id": 13479, - "bbox": [ - 274.45, - 537.66, - 305.18, - 562.45 - ], - "text": "+\n1\nR2C1", - "type": "text" - }, - { - "block_id": "p481-b23", - "global_id": 13480, - "bbox": [ - 314.7, - 537.66, - 371.54, - 562.45 - ], - "text": "s +\n1\nR1R2C1C2", - "type": "text" - }, - { - "block_id": "p481-b24", - "global_id": 13481, - "bbox": [ - 376.49, - 522.6, - 428.18, - 540.63 - ], - "text": "=\nω2", - "type": "text" - }, - { - "block_id": "p481-b25", - "global_id": 13482, - "bbox": [ - 424.5, - 529.18, - 427.98, - 536.15 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p481-b26", - "global_id": 13483, - "bbox": [ - 387.49, - 541.35, - 404.68, - 554.51 - ], - "text": "s2 +", - "type": "text" - }, - { - "block_id": "p481-b27", - "global_id": 13484, - "bbox": [ - 406.22, - 530.23, - 424.16, - 548.4 - ], - "text": "ω0", - "type": "text" - }, - { - "block_id": "p481-b28", - "global_id": 13485, - "bbox": [ - 415.81, - 551.62, - 423.0, - 561.58 - ], - "text": "Q", - "type": "text" - }, - { - "block_id": "p481-b30", - "global_id": 13486, - "bbox": [ - 433.69, - 543.24, - 458.64, - 554.51 - ], - "text": "s + ω2", - "type": "text" - }, - { - "block_id": "p481-b31", - "global_id": 13487, - "bbox": [ - 454.96, - 549.83, - 458.45, - 556.8 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p481-b32", - "global_id": 13488, - "bbox": [ - 127.59, - 576.6, - 516.15, - 598.94 - ], - "text": "Geometrically, ω0 is the distance from the origin to the poles and Q = 1/2cosψ, where ψ is\nthe angle between the negative real axis and the pole. Termed the “quality factor” of a circuit, Q", - "type": "text" - }, - { - "block_id": "p481-b33", - "global_id": 13489, - "bbox": [ - 127.59, - 622.19, - 516.13, - 646.71 - ], - "text": "† A more general version of the Sallen–Key circuit has a resistor Ra from the negative terminal to ground and\na resistor Rb between the negative terminal and the output. In Fig. 4.67, Ra = ∞and Rb = 0.", - "type": "text" - } - ] - }, - { - "page_num": 482, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p482-b0", - "global_id": 13490, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "462\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p482-b1", - "global_id": 13491, - "bbox": [ - 111.72, - 134.26, - 122.82, - 142.34 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p482-b2", - "global_id": 13492, - "bbox": [ - 115.27, - 143.29, - 119.27, - 151.29 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p482-b3", - "global_id": 13493, - "bbox": [ - 115.01, - 125.4, - 119.53, - 133.4 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p482-b4", - "global_id": 13494, - "bbox": [ - 310.86, - 143.18, - 321.96, - 151.27 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p482-b5", - "global_id": 13495, - "bbox": [ - 314.41, - 152.21, - 318.41, - 160.21 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p482-b6", - "global_id": 13496, - "bbox": [ - 314.15, - 134.32, - 318.67, - 142.32 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p482-b7", - "global_id": 13497, - "bbox": [ - 264.65, - 115.2, - 269.17, - 123.2 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p482-b8", - "global_id": 13498, - "bbox": [ - 265.17, - 130.5, - 269.17, - 138.5 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p482-b9", - "global_id": 13499, - "bbox": [ - 190.97, - 109.16, - 198.85, - 118.77 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p482-b10", - "global_id": 13500, - "bbox": [ - 251.74, - 89.61, - 260.08, - 99.22 - ], - "text": "C1", - "type": "text" - }, - { - "block_id": "p482-b11", - "global_id": 13501, - "bbox": [ - 136.83, - 108.76, - 144.72, - 118.36 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p482-b12", - "global_id": 13502, - "bbox": [ - 220.56, - 138.51, - 228.89, - 148.11 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p482-b13", - "global_id": 13503, - "bbox": [ - 340.8, - 170.78, - 474.74, - 180.02 - ], - "text": "Figure 4.67 Sallen–Key filter stage.", - "type": "text" - }, - { - "block_id": "p482-b14", - "global_id": 13504, - "bbox": [ - 101.84, - 202.2, - 490.19, - 224.53 - ], - "text": "provides a measure of the peakedness of the response. High-Q filters have poles close to the ω\naxis, which boost the magnitude response near those frequencies.", - "type": "text" - }, - { - "block_id": "p482-b15", - "global_id": 13505, - "bbox": [ - 101.84, - 226.52, - 490.42, - 284.31 - ], - "text": "Although many ways exist to determine suitable component values, a simple method is to\nassign R1 a realistic value and then let R2 = R1, C1 = 2Q/ω0R1, and C2 = 1/2Qω0R2. Butterworth\npoles are a distance ωc from the origin, so ω0 = ωc. For our 10th-order Butterworth filter, the angles\nψ are regularly spaced at 9, 27, 45, 63, and 81 degrees. MATLAB program CH4MP2 automates the\ntask of computing component values and magnitude responses for each stage.", - "type": "text" - }, - { - "block_id": "p482-b16", - "global_id": 13506, - "bbox": [ - 101.84, - 293.41, - 402.3, - 321.3 - ], - "text": "% CH4MP2.m : Chapter 4, MATLAB Program 2\n% Script M-file computes Sallen-Key component values and magnitude\n% responses for each of the five cascaded second-order filter sections.", - "type": "text" - }, - { - "block_id": "p482-b17", - "global_id": 13507, - "bbox": [ - 101.84, - 333.26, - 436.15, - 381.08 - ], - "text": "omega_0 = 3000*2*pi; % Filter cut-off frequency\npsi = [9 27 45 63 81]*pi/180; % Butterworth pole angles\nf = linspace(0,6000,200); % Frequency range for magnitude response calculations\nHmag_SK = zeros(5,200); % Pre-allocate array for magnitude responses\nfor stage = 1:5,", - "type": "text" - }, - { - "block_id": "p482-b18", - "global_id": 13508, - "bbox": [ - 118.78, - 383.07, - 355.77, - 410.97 - ], - "text": "Q = 1/(2*cos(psi(stage))); % Compute Q for current stage\n% Compute and display filter components to the screen:\ndisp([’Stage ’,num2str(stage),...", - "type": "text" - }, - { - "block_id": "p482-b19", - "global_id": 13509, - "bbox": [ - 101.84, - 412.96, - 427.69, - 500.62 - ], - "text": "’ (Q = ’,num2str(Q),...\n’):\nR1 = R2 = ’,num2str(56000),...\n’, C1 = ’,num2str(2*Q/(omega_0*56000)),...\n’, C2 = ’,num2str(1/(2*Q*omega_0*56000))]);\nB = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute filter coefficients\nHmag_SK(stage,:) = abs(CH4MP1(B,A,2*pi*f)); % Compute magnitude response\nend\nplot(f,Hmag_SK,’k’,f,prod(Hmag_SK),’k:’)\nxlabel(’f [Hz]’); ylabel(’Magnitude Response’)", - "type": "text" - }, - { - "block_id": "p482-b20", - "global_id": 13510, - "bbox": [ - 101.84, - 510.25, - 490.38, - 568.03 - ], - "text": "The disp command displays a character string to the screen. Character strings must be enclosed\nin single quotation marks. The num2str command converts numbers to character strings and\nfacilitates the formatted display of information. The prod command multiplies along the columns\nof a matrix; it computes the total magnitude response as the product of the magnitude responses\nof the five stages.", - "type": "text" - }, - { - "block_id": "p482-b21", - "global_id": 13511, - "bbox": [ - 119.78, - 570.03, - 335.75, - 579.99 - ], - "text": "Executing the program produces the following output:", - "type": "text" - }, - { - "block_id": "p482-b22", - "global_id": 13512, - "bbox": [ - 101.84, - 589.08, - 427.71, - 646.87 - ], - "text": ">>\nCH4MP2\nStage 1 (Q = 0.50623):\nR1 = R2 = 56000, C1 = 9.5916e-10, C2 = 9.3569e-10\nStage 2 (Q = 0.56116):\nR1 = R2 = 56000, C1 = 1.0632e-09, C2 = 8.441e-10\nStage 3 (Q = 0.70711):\nR1 = R2 = 56000, C1 = 1.3398e-09, C2 = 6.6988e-10\nStage 4 (Q = 1.1013):\nR1 = R2 = 56000, C1 = 2.0867e-09, C2 = 4.3009e-10\nStage 5 (Q = 3.1962):\nR1 = R2 = 56000, C1 = 6.0559e-09, C2 = 1.482e-10", - "type": "text" - } - ] - }, - { - "page_num": 483, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p483-b0", - "global_id": 13513, - "bbox": [ - 309.38, - 62.89, - 516.12, - 71.98 - ], - "text": "4.12\nMATLAB: Continuous-Time Filters\n463", - "type": "text" - }, - { - "block_id": "p483-b1", - "global_id": 13514, - "bbox": [ - 153.31, - 268.68, - 497.06, - 289.76 - ], - "text": "0\n1000\n2000\n3000\n4000\n5000\n6000\nf [Hz]", - "type": "text" - }, - { - "block_id": "p483-b2", - "global_id": 13515, - "bbox": [ - 147.08, - 259.46, - 151.08, - 267.46 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p483-b3", - "global_id": 13516, - "bbox": [ - 140.33, - 234.92, - 151.01, - 242.92 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p483-b4", - "global_id": 13517, - "bbox": [ - 147.08, - 210.38, - 151.08, - 218.38 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p483-b5", - "global_id": 13518, - "bbox": [ - 140.33, - 185.85, - 151.01, - 193.85 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p483-b6", - "global_id": 13519, - "bbox": [ - 147.08, - 161.32, - 151.08, - 169.32 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p483-b7", - "global_id": 13520, - "bbox": [ - 140.33, - 136.77, - 151.01, - 144.77 - ], - "text": "2.5", - "type": "text" - }, - { - "block_id": "p483-b8", - "global_id": 13521, - "bbox": [ - 147.08, - 112.24, - 151.08, - 120.24 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p483-b9", - "global_id": 13522, - "bbox": [ - 140.33, - 87.71, - 151.01, - 95.71 - ], - "text": "3.5", - "type": "text" - }, - { - "block_id": "p483-b10", - "global_id": 13523, - "bbox": [ - 128.36, - 145.0, - 137.16, - 219.06 - ], - "text": "Magnitude Response", - "type": "text" - }, - { - "block_id": "p483-b11", - "global_id": 13524, - "bbox": [ - 127.59, - 296.44, - 355.9, - 305.68 - ], - "text": "Figure 4.68 Magnitude responses for Sallen–Key filter stages.", - "type": "text" - }, - { - "block_id": "p483-b12", - "global_id": 13525, - "bbox": [ - 127.59, - 329.09, - 516.13, - 410.79 - ], - "text": "Since all the component values are practical, this filter is possible to implement. Figure 4.68\ndisplays the magnitude responses for all five stages (solid lines). The total response (dotted line)\nconfirms a 10th-order Butterworth response. Stage 5, which has the largest Q and implements the\npair of conjugate poles nearest the ω axis, is the most peaked response. Stage 1, which has the\nsmallest Q and implements the pair of conjugate poles furthest from the ω axis, is the least peaked\nresponse. In practice, it is best to order high-Q stages last; this reduces the risk that the high gains\nwill saturate the filter hardware.", - "type": "text" - }, - { - "block_id": "p483-b13", - "global_id": 13526, - "bbox": [ - 127.59, - 438.46, - 262.57, - 450.41 - ], - "text": "4.12-4 Chebyshev Filters", - "type": "text" - }, - { - "block_id": "p483-b14", - "global_id": 13527, - "bbox": [ - 127.59, - 456.44, - 516.15, - 502.37 - ], - "text": "Like an order-N Butterworth lowpass filter (LPF), an order-N Chebyshev LPF is an all-pole filter\nthat possesses many desirable characteristics. Compared with an equal-order Butterworth filter,\nthe Chebyshev filter achieves better stopband attenuation and reduced transition bandwidth by\nallowing an adjustable amount of ripple within the passband.", - "type": "text" - }, - { - "block_id": "p483-b15", - "global_id": 13528, - "bbox": [ - 145.52, - 504.37, - 371.41, - 514.33 - ], - "text": "The squared magnitude response of a Chebyshev filter is", - "type": "text" - }, - { - "block_id": "p483-b16", - "global_id": 13529, - "bbox": [ - 262.23, - 527.18, - 380.29, - 553.68 - ], - "text": "|HC(jω)|2 =\n1\n1 + ϵ2C2\nN(ω/ωc)", - "type": "text" - }, - { - "block_id": "p483-b17", - "global_id": 13530, - "bbox": [ - 127.59, - 564.89, - 516.14, - 587.23 - ], - "text": "where ϵ controls the passband ripple, CN(ω/ωc) is a degree-N Chebyshev polynomial, and ωc is\nthe radian cutoff frequency. Several characteristics of Chebyshev LPFs are noteworthy:", - "type": "text" - }, - { - "block_id": "p483-b18", - "global_id": 13531, - "bbox": [ - 145.52, - 600.76, - 516.12, - 635.75 - ], - "text": "• An order-N Chebyshev LPF is equi-ripple in the passband (|ω| ≤ωc), has a total of N\nmaxima and minima over (0 ≤ω ≤ωc), and is monotonic decreasing in the stopband\n(|ω| > ωc).", - "type": "text" - } - ] - }, - { - "page_num": 484, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p484-b0", - "global_id": 13532, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "464\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p484-b1", - "global_id": 13533, - "bbox": [ - 119.78, - 79.56, - 444.27, - 98.37 - ], - "text": "• In the passband, the maximum gain is 1 and the minimum gain is 1/\n√", - "type": "text" - }, - { - "block_id": "p484-b2", - "global_id": 13534, - "bbox": [ - 128.85, - 87.79, - 490.38, - 111.09 - ], - "text": "1 + ϵ2. For\nodd-valued N, |H(j0)| = 1. For even-valued N, |HC(j0)| = 1/", - "type": "text" - }, - { - "block_id": "p484-b3", - "global_id": 13535, - "bbox": [ - 372.17, - 91.51, - 380.6, - 101.48 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p484-b4", - "global_id": 13536, - "bbox": [ - 119.78, - 99.75, - 407.83, - 124.27 - ], - "text": "1 + ϵ2.\n• Ripple is controlled by setting ϵ =\n√", - "type": "text" - }, - { - "block_id": "p484-b5", - "global_id": 13537, - "bbox": [ - 119.78, - 111.02, - 490.4, - 174.86 - ], - "text": "10R/10 −1, where R is the allowable passband ripple\nexpressed in decibels. Reducing ϵ adversely affects filter performance (see Prob. 4.12-10).\n• Unlike Butterworth filters, the cutoff frequency ωc rarely specifies the 3 dB point. For ϵ̸ = 1,\n|HC(jωc)|2 = 1/(1+ϵ2)̸ = 0.5. The cutoff frequency ωc simply indicates the frequency after\nwhich |HC(jω)| < 1/", - "type": "text" - }, - { - "block_id": "p484-b6", - "global_id": 13538, - "bbox": [ - 212.17, - 155.27, - 220.6, - 165.23 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p484-b7", - "global_id": 13539, - "bbox": [ - 220.61, - 163.51, - 247.84, - 174.09 - ], - "text": "1 + ϵ2.", - "type": "text" - }, - { - "block_id": "p484-b8", - "global_id": 13540, - "bbox": [ - 119.78, - 187.61, - 308.35, - 198.69 - ], - "text": "The Chebyshev polynomial CN(x) is defined as", - "type": "text" - }, - { - "block_id": "p484-b9", - "global_id": 13541, - "bbox": [ - 204.36, - 206.89, - 387.86, - 219.69 - ], - "text": "CN(x) = cos[N cos−1(x)] = cosh[N cosh−1(x)]", - "type": "text" - }, - { - "block_id": "p484-b10", - "global_id": 13542, - "bbox": [ - 101.84, - 229.6, - 490.39, - 252.64 - ], - "text": "In this form, it is difficult to verify that CN(x) is a degree-N polynomial in x. A recursive form of\nCN(x) makes this fact more clear (see Prob. 4.12-13).", - "type": "text" - }, - { - "block_id": "p484-b11", - "global_id": 13543, - "bbox": [ - 234.97, - 262.55, - 357.25, - 273.7 - ], - "text": "CN(x) = 2xCN−1(x) −CN−2(x)", - "type": "text" - }, - { - "block_id": "p484-b12", - "global_id": 13544, - "bbox": [ - 101.84, - 283.55, - 490.38, - 318.53 - ], - "text": "With C0(x) = 1 and C1(x) = x, the recursive form shows that any CN is a linear combination\nof degree-N polynomials and is therefore a degree-N polynomial itself. For N ≥2, MATLAB\nprogram CH4MP3 generates the (N + 1) coefficients of Chebyshev polynomial CN(x).", - "type": "text" - }, - { - "block_id": "p484-b13", - "global_id": 13545, - "bbox": [ - 101.85, - 327.62, - 441.82, - 397.36 - ], - "text": "function [C_N] = CH4MP3(N);\n% CH4MP3.m : Chapter 4, MATLAB Program 3\n% Function M-file computes Chebyshev polynomial coefficients\n% using the recursion relation C_N(x) = 2xC_{N-1}(x) - C_{N-2}(x)\n% INPUTS:\nN = degree of Chebyshev polynomial\n% OUTPUTS:\nC_N =\nvector of Chebyshev polynomial coefficients", - "type": "text" - }, - { - "block_id": "p484-b14", - "global_id": 13546, - "bbox": [ - 101.85, - 411.3, - 447.04, - 433.23 - ], - "text": "C_Nm2 = 1; C_Nm1 = [1 0];\n% Initial polynomial coefficients:\nfor t = 2:N;", - "type": "text" - }, - { - "block_id": "p484-b15", - "global_id": 13547, - "bbox": [ - 101.85, - 435.22, - 504.58, - 469.09 - ], - "text": "C_N = 2*conv([1 0],C_Nm1)-[zeros(1,length(C_Nm1)-length(C_Nm2)+1),C_Nm2];\nC_Nm2 = C_Nm1; C_Nm1 = C_N;\nend", - "type": "text" - }, - { - "block_id": "p484-b16", - "global_id": 13548, - "bbox": [ - 101.85, - 474.27, - 490.39, - 500.99 - ], - "text": "As examples, consider C2(x) = 2xC1(x) −C0(x) = 2x(x) −1 = 2x2 −1 and C3(x) = 2xC2(x) −\nC1(x) = 2x(2x2 −1) −x = 4x3 −3x. CH4MP3 easily confirms these cases.", - "type": "text" - }, - { - "block_id": "p484-b17", - "global_id": 13549, - "bbox": [ - 101.84, - 510.0, - 258.73, - 555.83 - ], - "text": ">>\nCH4MP3(2)\nans =\n2\n0\n-1\n>>\nCH4MP3(3)\nans =\n4\n0\n-3\n0", - "type": "text" - }, - { - "block_id": "p484-b18", - "global_id": 13550, - "bbox": [ - 101.84, - 560.89, - 490.36, - 598.92 - ], - "text": "Since CN(ω/ωc) is a degree-N polynomial, |HC(jω)|2 is an all-pole rational function with 2N\nfinite poles. Similar to the Butterworth case, the N poles specifying a causal and stable Chebyshev\nfilter can be found by selecting the N left-half-plane roots of 1 + ϵ2C2", - "type": "text" - }, - { - "block_id": "p484-b19", - "global_id": 13551, - "bbox": [ - 101.85, - 588.55, - 490.38, - 634.79 - ], - "text": "N[s/(jωc)].\nRoot locations and dc gain are sufficient to specify a Chebyshev filter for a given N and ϵ. To\ndemonstrate, consider the design of an order-8 Chebyshev filter with cutoff frequency fc = 1 kHz\nand allowable passband ripple R = 1 dB. First, filter parameters are specified.", - "type": "text" - } - ] - }, - { - "page_num": 485, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p485-b0", - "global_id": 13552, - "bbox": [ - 309.38, - 62.89, - 516.12, - 71.98 - ], - "text": "4.12\nMATLAB: Continuous-Time Filters\n465", - "type": "text" - }, - { - "block_id": "p485-b1", - "global_id": 13553, - "bbox": [ - 127.59, - 85.41, - 326.31, - 107.33 - ], - "text": ">>\nomega_c = 2*pi*1000; R = 1; N = 8;\n>>\nepsilon = sqrt(10^(R/10)-1);", - "type": "text" - }, - { - "block_id": "p485-b2", - "global_id": 13554, - "bbox": [ - 127.59, - 118.36, - 516.12, - 140.69 - ], - "text": "The coefficients of CN[s/(jωc)] are obtained with the help of CH4MP3, and then the coefficients of\n[1 + ϵ2C2", - "type": "text" - }, - { - "block_id": "p485-b3", - "global_id": 13555, - "bbox": [ - 162.29, - 130.31, - 503.52, - 142.71 - ], - "text": "N(s/(jωc))] are computed by using convolution to perform polynomial multiplication.", - "type": "text" - }, - { - "block_id": "p485-b4", - "global_id": 13556, - "bbox": [ - 127.59, - 152.7, - 399.56, - 174.63 - ], - "text": ">>\nCN\n= CH4MP3(N).*((1/(1j*omega_c)).^[N:-1:0]);\n>>\nCP = epsilon^2*conv(CN,CN); CP(end) = CP(end)+1;", - "type": "text" - }, - { - "block_id": "p485-b5", - "global_id": 13557, - "bbox": [ - 127.59, - 186.07, - 490.42, - 196.03 - ], - "text": "Next, the polynomial roots are found, and the left-half-plane poles are retained and plotted.", - "type": "text" - }, - { - "block_id": "p485-b6", - "global_id": 13558, - "bbox": [ - 127.59, - 208.05, - 477.99, - 253.88 - ], - "text": ">>\npoles = roots(CP); i = find(real(poles)<0); C_poles = poles(i);\n>>\nplot(real(C_poles),imag(C_poles),’kx’); axis equal;\n>>\naxis(omega_c*[-1.1 1.1 -1.1 1.1]);\n>>\nxlabel(’Real’); ylabel(’Imaginary’);", - "type": "text" - }, - { - "block_id": "p485-b7", - "global_id": 13559, - "bbox": [ - 127.59, - 261.7, - 490.52, - 275.28 - ], - "text": "As shown in Fig. 4.69, the roots of a Chebyshev filter lie on an ellipse† (see Prob. 4.12-14).", - "type": "text" - }, - { - "block_id": "p485-b8", - "global_id": 13560, - "bbox": [ - 177.7, - 459.71, - 308.09, - 480.9 - ], - "text": "–5000\n0\n5000\nReal", - "type": "text" - }, - { - "block_id": "p485-b9", - "global_id": 13561, - "bbox": [ - 142.82, - 440.25, - 162.82, - 448.25 - ], - "text": "–6000", - "type": "text" - }, - { - "block_id": "p485-b10", - "global_id": 13562, - "bbox": [ - 142.82, - 417.79, - 162.82, - 425.79 - ], - "text": "–4000", - "type": "text" - }, - { - "block_id": "p485-b11", - "global_id": 13563, - "bbox": [ - 142.82, - 395.32, - 162.82, - 403.32 - ], - "text": "–2000", - "type": "text" - }, - { - "block_id": "p485-b12", - "global_id": 13564, - "bbox": [ - 158.82, - 372.86, - 162.82, - 380.86 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p485-b13", - "global_id": 13565, - "bbox": [ - 146.82, - 350.4, - 162.82, - 358.4 - ], - "text": "2000", - "type": "text" - }, - { - "block_id": "p485-b14", - "global_id": 13566, - "bbox": [ - 146.82, - 327.94, - 162.82, - 335.94 - ], - "text": "4000", - "type": "text" - }, - { - "block_id": "p485-b15", - "global_id": 13567, - "bbox": [ - 146.82, - 305.47, - 162.82, - 313.47 - ], - "text": "6000", - "type": "text" - }, - { - "block_id": "p485-b16", - "global_id": 13568, - "bbox": [ - 130.35, - 359.29, - 139.15, - 395.46 - ], - "text": "Imaginary", - "type": "text" - }, - { - "block_id": "p485-b17", - "global_id": 13569, - "bbox": [ - 333.54, - 448.26, - 512.15, - 482.14 - ], - "text": "Figure 4.69 Pole-zero plot for an order-8\nChebyshev LPF with fc = 1 kHz and R = 1\ndB.", - "type": "text" - }, - { - "block_id": "p485-b18", - "global_id": 13570, - "bbox": [ - 127.59, - 510.84, - 516.12, - 533.05 - ], - "text": "To compute the filter’s magnitude response, the poles are expanded into a polynomial, the dc\ngain is set based on the even value of N, and CH4MP1 is used.", - "type": "text" - }, - { - "block_id": "p485-b19", - "global_id": 13571, - "bbox": [ - 127.59, - 544.78, - 462.33, - 590.61 - ], - "text": ">>\nA = poly(C_poles); B = A(end)/sqrt(1+epsilon^2);\n>>\nomega = linspace(0,2*pi*2000,2001); H_C = CH4MP1(B,A,omega);\n>>\nplot(omega/2/pi,abs(H_C),’k’); axis([0 2000 0 1.1]);\n>>\nxlabel(’f [Hz]’); ylabel(’|H_C(j2\\pi f)|’);", - "type": "text" - }, - { - "block_id": "p485-b20", - "global_id": 13572, - "bbox": [ - 127.59, - 610.24, - 516.11, - 633.41 - ], - "text": "† E. A. Guillemin demonstrates a wonderful relationship between the Chebyshev ellipse and the Butterworth\ncircle in his book Synthesis of Passive Networks (Wiley, New York, 1957).", - "type": "text" - } - ] - }, - { - "page_num": 486, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p486-b0", - "global_id": 13573, - "bbox": [ - 60.0, - 62.89, - 332.52, - 71.98 - ], - "text": "466\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p486-b1", - "global_id": 13574, - "bbox": [ - 130.26, - 199.67, - 474.0, - 220.75 - ], - "text": "0\n200\n400\n600\n800\n1000\n1200\n1400\n1600\n1800\n2000\nf [Hz]", - "type": "text" - }, - { - "block_id": "p486-b2", - "global_id": 13575, - "bbox": [ - 124.03, - 190.44, - 128.03, - 198.44 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p486-b3", - "global_id": 13576, - "bbox": [ - 117.29, - 171.22, - 127.95, - 179.22 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p486-b4", - "global_id": 13577, - "bbox": [ - 117.29, - 151.99, - 127.95, - 159.99 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p486-b5", - "global_id": 13578, - "bbox": [ - 117.29, - 132.76, - 127.95, - 140.76 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p486-b6", - "global_id": 13579, - "bbox": [ - 117.29, - 113.53, - 127.95, - 121.53 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p486-b7", - "global_id": 13580, - "bbox": [ - 124.03, - 94.31, - 128.03, - 102.31 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p486-b8", - "global_id": 13581, - "bbox": [ - 102.48, - 123.03, - 112.31, - 157.58 - ], - "text": "|HC(j2π f)|", - "type": "text" - }, - { - "block_id": "p486-b9", - "global_id": 13582, - "bbox": [ - 101.84, - 227.07, - 450.93, - 237.37 - ], - "text": "Figure 4.70 Magnitude responses for an order-8 Chebyshev LPF with fc = 1 kHz and R = 1 dB.", - "type": "text" - }, - { - "block_id": "p486-b10", - "global_id": 13583, - "bbox": [ - 101.84, - 259.0, - 490.41, - 305.91 - ], - "text": "As seen in Fig. 4.70, the magnitude response exhibits correct Chebyshev filter characteristics:\npassband ripples are equal in height and never exceed R = 1 dB; there are a total of N = 8 maxima\nand minima in the passband; and the gain rapidly and monotonically decreases after the cutoff\nfrequency of fc = 1 kHz.", - "type": "text" - }, - { - "block_id": "p486-b11", - "global_id": 13584, - "bbox": [ - 101.84, - 306.83, - 490.37, - 328.74 - ], - "text": "For higher-order filters, polynomial rooting may not provide reliable results. Fortunately,\nChebyshev roots can also be determined analytically. For", - "type": "text" - }, - { - "block_id": "p486-b12", - "global_id": 13585, - "bbox": [ - 202.99, - 340.77, - 250.93, - 359.21 - ], - "text": "φk = 2k + 1", - "type": "text" - }, - { - "block_id": "p486-b13", - "global_id": 13586, - "bbox": [ - 232.11, - 341.18, - 337.85, - 365.2 - ], - "text": "2N\nπ\nand\nξ = 1", - "type": "text" - }, - { - "block_id": "p486-b14", - "global_id": 13587, - "bbox": [ - 331.74, - 346.03, - 366.82, - 365.2 - ], - "text": "N sinh−1", - "type": "text" - }, - { - "block_id": "p486-b15", - "global_id": 13588, - "bbox": [ - 368.42, - 333.77, - 381.33, - 351.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p486-b16", - "global_id": 13589, - "bbox": [ - 376.38, - 354.83, - 380.73, - 364.79 - ], - "text": "ϵ", - "type": "text" - }, - { - "block_id": "p486-b18", - "global_id": 13590, - "bbox": [ - 101.84, - 377.2, - 198.72, - 387.16 - ], - "text": "the Chebyshev poles are", - "type": "text" - }, - { - "block_id": "p486-b19", - "global_id": 13591, - "bbox": [ - 204.91, - 400.53, - 387.32, - 412.4 - ], - "text": "pk = ωc sinh(ξ)sin(φk) + jωc cosh(ξ)cos(φk)", - "type": "text" - }, - { - "block_id": "p486-b20", - "global_id": 13592, - "bbox": [ - 101.84, - 424.68, - 490.39, - 446.6 - ], - "text": "Continuing the same example, the poles are recomputed and again plotted. The result is identical\nto Fig. 4.69.", - "type": "text" - }, - { - "block_id": "p486-b21", - "global_id": 13593, - "bbox": [ - 101.84, - 458.68, - 441.82, - 516.46 - ], - "text": ">>\nk = [1:N]; xi = 1/N*asinh(1/epsilon); phi = (k*2-1)/(2*N)*pi;\n>>\nC_poles = omega_c*(-sinh(xi)*sin(phi)+1j*cosh(xi)*cos(phi));\n>>\nplot(real(C_poles),imag(C_poles),’kx’); axis equal;\n>>\naxis(omega_c*[-1.1 1.1 -1.1 1.1]);\n>>\nxlabel(’Real’); ylabel(’Imaginary’);", - "type": "text" - }, - { - "block_id": "p486-b22", - "global_id": 13594, - "bbox": [ - 101.84, - 527.96, - 490.4, - 561.83 - ], - "text": "As in the case of high-order Butterworth filters, a cascade of second-order filter sections\nfacilitates practical implementation of Chebyshev filters. Problems 4.12-5 and 4.12-8 use\nsecond-order Sallen–Key circuit stages to investigate such implementations.", - "type": "text" - }, - { - "block_id": "p486-b23", - "global_id": 13595, - "bbox": [ - 102.2, - 592.93, - 202.15, - 606.88 - ], - "text": "4.13 SUMMARY", - "type": "text" - }, - { - "block_id": "p486-b24", - "global_id": 13596, - "bbox": [ - 101.84, - 612.86, - 490.39, - 634.79 - ], - "text": "This chapter discusses analysis of LTIC (linear, time-invariant, continuous-time) systems by the\nLaplace transform, which transforms integro-differential equations of such systems into algebraic", - "type": "text" - } - ] - }, - { - "page_num": 487, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p487-b0", - "global_id": 13597, - "bbox": [ - 409.68, - 62.89, - 516.14, - 71.98 - ], - "text": "4.13\nSummary\n467", - "type": "text" - }, - { - "block_id": "p487-b1", - "global_id": 13598, - "bbox": [ - 127.59, - 85.82, - 516.15, - 119.69 - ], - "text": "equations. Therefore solving these integro-differential equations reduces to solving algebraic\nequations. The Laplace transform method cannot be used for time-varying-parameter systems or\nfor nonlinear systems in general.", - "type": "text" - }, - { - "block_id": "p487-b2", - "global_id": 13599, - "bbox": [ - 127.59, - 121.27, - 516.15, - 215.34 - ], - "text": "The transfer function H(s) of an LTIC system is the Laplace transform of its impulse response.\nIt may also be defined as a ratio of the Laplace transform of the output to the Laplace transform\nof the input when all initial conditions are zero (system in zero state). If X(s) is the Laplace\ntransform of the input x(t) and Y(s) is the Laplace transform of the corresponding output y(t)\n(when all initial conditions are zero), then Y(s) = X(s)H(s). For an LTIC system described by\nan Nth-order differential equation Q(D)y(t) = P(D)x(t), the transfer function H(s) = P(s)/Q(s).\nLike the impulse response h(t), the transfer function H(s) is also an external description of the\nsystem.", - "type": "text" - }, - { - "block_id": "p487-b3", - "global_id": 13600, - "bbox": [ - 127.6, - 217.33, - 516.16, - 275.11 - ], - "text": "Electrical circuit analysis can also be carried out by using a transformed circuit method,\nin which all signals (voltages and currents) are represented by their Laplace transforms, all\nelements by their impedances (or admittances), and initial conditions by their equivalent sources\n(initial condition generators). In this method, a network can be analyzed as if it were a resistive\ncircuit.", - "type": "text" - }, - { - "block_id": "p487-b4", - "global_id": 13601, - "bbox": [ - 127.59, - 277.1, - 516.16, - 334.88 - ], - "text": "Large systems can be depicted by suitably interconnected subsystems represented by\nblocks. Each subsystem, being a smaller system, can be readily analyzed and represented by\nits input–output relationship, such as its transfer function. Analysis of large systems can be\ncarried out with the knowledge of input–output relationships of its subsystems and the nature\nof interconnection of various subsystems.", - "type": "text" - }, - { - "block_id": "p487-b5", - "global_id": 13602, - "bbox": [ - 127.59, - 336.88, - 516.15, - 394.66 - ], - "text": "LTIC systems can be realized by scalar multipliers, adders, and integrators. A given transfer\nfunction can be synthesized in many different ways, such as canonic, cascade, and parallel.\nMoreover, every realization has a transpose, which also has the same transfer function. In practice,\nall the building blocks (scalar multipliers, adders, and integrators) can be obtained from operational\namplifiers.", - "type": "text" - }, - { - "block_id": "p487-b6", - "global_id": 13603, - "bbox": [ - 127.59, - 393.04, - 516.14, - 502.25 - ], - "text": "The system response to an everlasting exponential est is also an everlasting exponential\nH(s)est. Consequently, the system response to an everlasting exponential ejωt is H(jω)ejωt. Hence,\nH(jω) is the frequency response of the system. For a sinusoidal input of unit amplitude and having\nfrequency ω, the system response is also a sinusoid of the same frequency (ω) with amplitude\n|H(jω)|, and its phase is shifted by̸\nH(jω) with respect to the input sinusoid. For this reason\n|H(jω)| is called the amplitude response (gain) and̸\nH(jω) is called the phase response of the\nsystem. Amplitude and phase response of a system indicate the filtering characteristics of the\nsystem. The general nature of the filtering characteristics of a system can be quickly determined\nfrom a knowledge of the location of poles and zeros of the system transfer function.", - "type": "text" - }, - { - "block_id": "p487-b7", - "global_id": 13604, - "bbox": [ - 127.59, - 504.25, - 516.16, - 621.81 - ], - "text": "Most of the input signals and practical systems are causal. Consequently we are required\nmost of the time to deal with causal signals. When all signals must be causal, the Laplace\ntransform analysis is greatly simplified; the region of convergence of a signal becomes irrelevant\nto the analysis process. This special case of the Laplace transform (which is restricted to\ncausal signals) is called the unilateral Laplace transform. Much of the chapter deals with this\nvariety of Laplace transform. Section 4.11 discusses the general Laplace transform (the bilateral\nLaplace transform), which can handle causal and noncausal signals and systems. In the bilateral\ntransform, the inverse transform of X(s) is not unique but depends on the region of convergence\nof X(s). Thus, the region of convergence plays a very crucial role in the bilateral Laplace\ntransform.", - "type": "text" - } - ] - }, - { - "page_num": 488, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p488-b0", - "global_id": 13605, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "468\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p488-b1", - "global_id": 13606, - "bbox": [ - 102.11, - 86.41, - 189.29, - 97.37 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p488-b2", - "global_id": 13607, - "bbox": [ - 104.09, - 104.44, - 482.65, - 113.49 - ], - "text": "1.\nLathi, B. P. Signal Processing and Linear Systems, 1st ed. Oxford University Press, New York, 1998.", - "type": "text" - }, - { - "block_id": "p488-b3", - "global_id": 13608, - "bbox": [ - 104.08, - 118.38, - 490.42, - 138.4 - ], - "text": "2.\nDoetsch, G. Introduction to the Theory and Applications of the Laplace Transformation with a Table of\nLaplace Transformations. Springer-Verlag, New York, 1974.", - "type": "text" - }, - { - "block_id": "p488-b4", - "global_id": 13609, - "bbox": [ - 104.08, - 143.29, - 490.38, - 163.3 - ], - "text": "3.\nLePage, W. R. Complex Variables and the Laplace Transforms for Engineers. McGraw-Hill, New York,\n1961.", - "type": "text" - }, - { - "block_id": "p488-b5", - "global_id": 13610, - "bbox": [ - 104.08, - 168.19, - 490.38, - 188.21 - ], - "text": "4.\nDurant, Will, and Ariel Durant. The Age of Napoleon, Part XI in The Story of Civilization Series. Simon\n& Schuster, New York, 1975.", - "type": "text" - }, - { - "block_id": "p488-b6", - "global_id": 13611, - "bbox": [ - 104.08, - 193.1, - 368.58, - 202.16 - ], - "text": "5.\nBell, E. T. Men of Mathematics. Simon & Schuster, New York, 1937.", - "type": "text" - }, - { - "block_id": "p488-b7", - "global_id": 13612, - "bbox": [ - 104.08, - 207.05, - 490.38, - 227.06 - ], - "text": "6.\nNahin, P. J. “Oliver Heaviside: Genius and Curmudgeon.” IEEE Spectrum, vol. 20, pp. 63–69, July\n1983.", - "type": "text" - }, - { - "block_id": "p488-b8", - "global_id": 13613, - "bbox": [ - 104.08, - 231.96, - 333.15, - 241.01 - ], - "text": "7.\nBerkey, D. Calculus, 2nd ed. Saunders, Philadelphia, 1988.", - "type": "text" - }, - { - "block_id": "p488-b9", - "global_id": 13614, - "bbox": [ - 104.08, - 245.9, - 393.42, - 254.96 - ], - "text": "8.\nEncyclopaedia Britannica. Micropaedia IV, 15th ed., p. 981, Chicago, 1982.", - "type": "text" - }, - { - "block_id": "p488-b10", - "global_id": 13615, - "bbox": [ - 104.08, - 259.85, - 418.98, - 268.91 - ], - "text": "9.\nChurchill, R. V. Operational Mathematics, 2nd ed. McGraw-Hill, New York, 1958.", - "type": "text" - }, - { - "block_id": "p488-b11", - "global_id": 13616, - "bbox": [ - 99.6, - 273.8, - 401.32, - 282.86 - ], - "text": "10.\nTruxal, J. G. The Age of Electronic Messages. McGraw-Hill, New York, 1990.", - "type": "text" - }, - { - "block_id": "p488-b12", - "global_id": 13617, - "bbox": [ - 99.6, - 287.75, - 427.0, - 296.81 - ], - "text": "11.\nVan Valkenberg, M. Analog Filter Design. Oxford University Press, New York, 1982.", - "type": "text" - }, - { - "block_id": "p488-b13", - "global_id": 13618, - "bbox": [ - 80.93, - 332.53, - 191.52, - 349.46 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p488-b14", - "global_id": 13619, - "bbox": [ - 62.08, - 360.12, - 263.25, - 391.09 - ], - "text": "4.1-1\nBy direct integration [Eq. (4.1)] find the Laplace\ntransforms and the region of convergence of the\nfollowing functions:", - "type": "text" - }, - { - "block_id": "p488-b15", - "global_id": 13620, - "bbox": [ - 90.72, - 392.7, - 157.86, - 413.0 - ], - "text": "(a) u(t) −u(t −1)\n(b) te−tu(t)", - "type": "text" - }, - { - "block_id": "p488-b16", - "global_id": 13621, - "bbox": [ - 90.72, - 414.62, - 162.02, - 434.92 - ], - "text": "(c) tcos ω0tu(t)\n(d) (e2t −2e−t)u(t)", - "type": "text" - }, - { - "block_id": "p488-b17", - "global_id": 13622, - "bbox": [ - 91.22, - 436.54, - 175.39, - 446.62 - ], - "text": "(e) cos ω1t cos ω2tu(t)", - "type": "text" - }, - { - "block_id": "p488-b18", - "global_id": 13623, - "bbox": [ - 90.72, - 447.5, - 178.06, - 478.76 - ], - "text": "(f) cosh(at)u(t)\n(g) sinh(at)u(t)\n(h) e−2t cos(5t + θ)u(t)", - "type": "text" - }, - { - "block_id": "p488-b19", - "global_id": 13624, - "bbox": [ - 62.08, - 485.18, - 263.25, - 516.14 - ], - "text": "4.1-2\nBy direct integration [Eq. (4.1)] find the Laplace\ntransforms and the region of convergence of the\nfollowing functions:", - "type": "text" - }, - { - "block_id": "p488-b20", - "global_id": 13625, - "bbox": [ - 91.23, - 516.49, - 186.67, - 527.1 - ], - "text": "(a) e−2tu(t −5) + δ(t −1)", - "type": "text" - }, - { - "block_id": "p488-b21", - "global_id": 13626, - "bbox": [ - 317.87, - 360.21, - 405.26, - 370.6 - ], - "text": "(b) πe3tu(t + 5) −δ(2t)", - "type": "text" - }, - { - "block_id": "p488-b22", - "global_id": 13627, - "bbox": [ - 318.37, - 365.51, - 349.29, - 381.57 - ], - "text": "(c) %∞", - "type": "text" - }, - { - "block_id": "p488-b23", - "global_id": 13628, - "bbox": [ - 342.67, - 372.23, - 413.88, - 383.76 - ], - "text": "k=0 δ(t −kT), T > 0", - "type": "text" - }, - { - "block_id": "p488-b24", - "global_id": 13629, - "bbox": [ - 289.22, - 388.99, - 490.4, - 408.99 - ], - "text": "4.1-3\nBy direct integration find the Laplace transforms\nof the signals shown in Fig. P4.1-3.", - "type": "text" - }, - { - "block_id": "p488-b25", - "global_id": 13630, - "bbox": [ - 289.22, - 416.41, - 490.4, - 436.41 - ], - "text": "4.1-4\nFind the inverse (unilateral) Laplace transforms\nof the following functions:", - "type": "text" - }, - { - "block_id": "p488-b26", - "global_id": 13631, - "bbox": [ - 318.37, - 436.12, - 373.74, - 458.11 - ], - "text": "(a)\n2s + 5\ns2 + 5s + 6", - "type": "text" - }, - { - "block_id": "p488-b27", - "global_id": 13632, - "bbox": [ - 317.86, - 458.61, - 378.23, - 480.6 - ], - "text": "(b)\n3s + 5\ns2 + 4s + 13", - "type": "text" - }, - { - "block_id": "p488-b28", - "global_id": 13633, - "bbox": [ - 318.37, - 481.66, - 365.48, - 498.28 - ], - "text": "(c)\n(s + 1)2", - "type": "text" - }, - { - "block_id": "p488-b29", - "global_id": 13634, - "bbox": [ - 334.49, - 492.72, - 369.26, - 504.65 - ], - "text": "s2 −s −6", - "type": "text" - }, - { - "block_id": "p488-b30", - "global_id": 13635, - "bbox": [ - 317.86, - 505.52, - 366.17, - 527.14 - ], - "text": "(d)\n5\ns2(s + 2)", - "type": "text" - }, - { - "block_id": "p488-b31", - "global_id": 13636, - "bbox": [ - 117.63, - 554.28, - 266.56, - 566.83 - ], - "text": "1\n1\nsin t", - "type": "text" - }, - { - "block_id": "p488-b32", - "global_id": 13637, - "bbox": [ - 144.27, - 588.71, - 267.45, - 597.76 - ], - "text": "p\n1", - "type": "text" - }, - { - "block_id": "p488-b33", - "global_id": 13638, - "bbox": [ - 143.29, - 618.18, - 265.41, - 626.18 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p488-b34", - "global_id": 13639, - "bbox": [ - 335.09, - 562.36, - 346.16, - 570.66 - ], - "text": "1e", - "type": "text" - }, - { - "block_id": "p488-b35", - "global_id": 13640, - "bbox": [ - 369.8, - 589.76, - 373.8, - 597.76 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p488-b36", - "global_id": 13641, - "bbox": [ - 392.12, - 567.12, - 402.34, - 576.74 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p488-b37", - "global_id": 13642, - "bbox": [ - 377.42, - 618.18, - 386.3, - 626.18 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p488-b38", - "global_id": 13643, - "bbox": [ - 173.11, - 589.46, - 417.36, - 597.46 - ], - "text": "t\nt\nt", - "type": "text" - }, - { - "block_id": "p488-b39", - "global_id": 13644, - "bbox": [ - 104.83, - 632.35, - 156.48, - 641.31 - ], - "text": "Figure P4.1-3", - "type": "text" - } - ] - }, - { - "page_num": 489, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p489-b0", - "global_id": 13645, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n469", - "type": "text" - }, - { - "block_id": "p489-b1", - "global_id": 13646, - "bbox": [ - 193.28, - 95.18, - 331.89, - 103.4 - ], - "text": "x(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p489-b2", - "global_id": 13647, - "bbox": [ - 185.91, - 142.28, - 394.77, - 153.74 - ], - "text": "t\nt\n0\n0\n1\n1\n2\n2\n3\n3", - "type": "text" - }, - { - "block_id": "p489-b3", - "global_id": 13648, - "bbox": [ - 182.37, - 110.92, - 313.18, - 118.89 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p489-b4", - "global_id": 13649, - "bbox": [ - 414.37, - 150.71, - 466.02, - 159.68 - ], - "text": "Figure P4.2-6", - "type": "text" - }, - { - "block_id": "p489-b5", - "global_id": 13650, - "bbox": [ - 116.96, - 173.45, - 203.47, - 195.44 - ], - "text": "(e)\n2s + 1\n(s + 1)(s2 + 2s + 2)", - "type": "text" - }, - { - "block_id": "p489-b6", - "global_id": 13651, - "bbox": [ - 117.95, - 196.76, - 164.27, - 218.75 - ], - "text": "(f)\ns + 2\ns(s + 1)2", - "type": "text" - }, - { - "block_id": "p489-b7", - "global_id": 13652, - "bbox": [ - 116.46, - 220.44, - 185.24, - 242.06 - ], - "text": "(g)\n1\n(s + 1)(s + 2)4", - "type": "text" - }, - { - "block_id": "p489-b8", - "global_id": 13653, - "bbox": [ - 116.46, - 243.38, - 210.7, - 265.36 - ], - "text": "(h)\ns + 1\ns(s + 2)2(s2 + 4s + 5)", - "type": "text" - }, - { - "block_id": "p489-b9", - "global_id": 13654, - "bbox": [ - 118.45, - 267.25, - 173.27, - 283.86 - ], - "text": "(i)\ns3", - "type": "text" - }, - { - "block_id": "p489-b10", - "global_id": 13655, - "bbox": [ - 87.82, - 278.29, - 290.98, - 313.42 - ], - "text": "(s + 1)2(s2 + 2s + 5)\n4.2-1\nSuppose a CT signal x(t)=2[u(t −2) −u(t + 1)]\nhas a transform X(s).", - "type": "text" - }, - { - "block_id": "p489-b11", - "global_id": 13656, - "bbox": [ - 116.97, - 313.78, - 200.98, - 325.12 - ], - "text": "(a) If Ya(s) = e−5ssX", - "type": "text" - }, - { - "block_id": "p489-b13", - "global_id": 13657, - "bbox": [ - 204.98, - 313.69, - 222.68, - 324.29 - ], - "text": "s + 1", - "type": "text" - }, - { - "block_id": "p489-b14", - "global_id": 13658, - "bbox": [ - 219.45, - 307.83, - 227.5, - 326.97 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p489-b15", - "global_id": 13659, - "bbox": [ - 116.46, - 315.41, - 288.98, - 347.03 - ], - "text": ", determine and\nsketch the corresponding signal ya(t).\n(b) If Yb(s) = 2−ssX(s −2), determine and", - "type": "text" - }, - { - "block_id": "p489-b16", - "global_id": 13660, - "bbox": [ - 131.89, - 347.92, - 266.67, - 358.0 - ], - "text": "sketch the corresponding signal yb(t).", - "type": "text" - }, - { - "block_id": "p489-b17", - "global_id": 13661, - "bbox": [ - 87.82, - 362.64, - 288.98, - 404.55 - ], - "text": "4.2-2\nFind the Laplace transforms of the follow-\ning functions using only Table 4.1 and the\ntime-shifting property (if needed) of the unilat-\neral Laplace transform:", - "type": "text" - }, - { - "block_id": "p489-b18", - "global_id": 13662, - "bbox": [ - 116.46, - 406.18, - 185.38, - 426.47 - ], - "text": "(a) u(t) −u(t −1)\n(b) e−(t−τ)u(t −τ)", - "type": "text" - }, - { - "block_id": "p489-b19", - "global_id": 13663, - "bbox": [ - 116.46, - 426.83, - 171.92, - 448.39 - ], - "text": "(c) e−(t−τ)u(t)\n(d) e−tu(t −τ)", - "type": "text" - }, - { - "block_id": "p489-b20", - "global_id": 13664, - "bbox": [ - 116.96, - 448.75, - 174.41, - 459.35 - ], - "text": "(e) te−tu(t −τ)", - "type": "text" - }, - { - "block_id": "p489-b21", - "global_id": 13665, - "bbox": [ - 116.46, - 460.97, - 212.59, - 492.96 - ], - "text": "(f) sin[ω0(t −τ)]u(t −τ)\n(g) sin[ω0(t −τ)]u(t)\n(h) sin ω0tu(t −τ)", - "type": "text" - }, - { - "block_id": "p489-b22", - "global_id": 13666, - "bbox": [ - 118.45, - 493.85, - 220.16, - 514.15 - ], - "text": "(i) tsin(t)u(t)\n(j) (1 −t)cos(t −1)u(t −1)", - "type": "text" - }, - { - "block_id": "p489-b23", - "global_id": 13667, - "bbox": [ - 87.82, - 519.51, - 289.0, - 561.43 - ], - "text": "4.2-3\nUsing only Table 4.1 and the time-shifting\nproperty, determine the Laplace transform of the\nsignals in Fig. P4.1-3. [Hint: See Sec. 1.4 for dis-\ncussion of expressing such signals analytically.]", - "type": "text" - }, - { - "block_id": "p489-b24", - "global_id": 13668, - "bbox": [ - 87.83, - 566.81, - 288.97, - 586.72 - ], - "text": "4.2-4\nProve the frequency-differentiation property,\n−tx(t) ⇐⇒d", - "type": "text" - }, - { - "block_id": "p489-b25", - "global_id": 13669, - "bbox": [ - 116.46, - 577.47, - 288.99, - 597.77 - ], - "text": "dsX(s). This property holds for both\nthe unilateral and bilateral Laplace transforms.", - "type": "text" - }, - { - "block_id": "p489-b26", - "global_id": 13670, - "bbox": [ - 87.82, - 601.59, - 270.87, - 612.18 - ], - "text": "4.2-5\nConsider the signal x(t) = te−2(t−3)u(t−2).", - "type": "text" - }, - { - "block_id": "p489-b27", - "global_id": 13671, - "bbox": [ - 116.97, - 614.09, - 288.99, - 623.14 - ], - "text": "(a) Determine the unilateral Laplace transform", - "type": "text" - }, - { - "block_id": "p489-b28", - "global_id": 13672, - "bbox": [ - 131.9, - 624.67, - 195.27, - 635.08 - ], - "text": "Xu(s) = Lu {x(t)}.", - "type": "text" - }, - { - "block_id": "p489-b29", - "global_id": 13673, - "bbox": [ - 343.61, - 173.73, - 516.14, - 182.79 - ], - "text": "(b) Determine the bilateral Laplace transform", - "type": "text" - }, - { - "block_id": "p489-b30", - "global_id": 13674, - "bbox": [ - 359.05, - 184.31, - 415.35, - 193.74 - ], - "text": "X(s) = L{x(t)}.", - "type": "text" - }, - { - "block_id": "p489-b31", - "global_id": 13675, - "bbox": [ - 314.97, - 198.55, - 516.13, - 218.85 - ], - "text": "4.2-6\nConsider the signals x(t) and y(t), as shown in\nFig. P4.2-6.", - "type": "text" - }, - { - "block_id": "p489-b32", - "global_id": 13676, - "bbox": [ - 344.12, - 220.47, - 516.14, - 229.81 - ], - "text": "(a) Using the definition, compute X(s), the", - "type": "text" - }, - { - "block_id": "p489-b33", - "global_id": 13677, - "bbox": [ - 343.61, - 231.42, - 516.13, - 251.72 - ], - "text": "bilateral Laplace transform of x(t).\n(b) Using Laplace transform properties, express", - "type": "text" - }, - { - "block_id": "p489-b34", - "global_id": 13678, - "bbox": [ - 359.05, - 253.35, - 516.13, - 306.52 - ], - "text": "Y(s), the bilateral Laplace transform of y(t),\nas a function of X(s), the bilateral Laplace\ntransform of x(t). Simplify as much as\npossible without substituting your answer\nfrom part (a).", - "type": "text" - }, - { - "block_id": "p489-b35", - "global_id": 13679, - "bbox": [ - 314.97, - 311.62, - 516.14, - 331.62 - ], - "text": "4.2-7\nFind the inverse Laplace transforms of the\nfollowing functions:", - "type": "text" - }, - { - "block_id": "p489-b36", - "global_id": 13680, - "bbox": [ - 344.12, - 331.61, - 403.96, - 348.5 - ], - "text": "(a) (2s + 5)e−2s", - "type": "text" - }, - { - "block_id": "p489-b37", - "global_id": 13681, - "bbox": [ - 362.73, - 342.94, - 401.97, - 354.86 - ], - "text": "s2 + 5s + 6", - "type": "text" - }, - { - "block_id": "p489-b38", - "global_id": 13682, - "bbox": [ - 343.61, - 353.67, - 399.48, - 378.92 - ], - "text": "(b)\nse−3s + 2\ns2 + 2s + 2", - "type": "text" - }, - { - "block_id": "p489-b39", - "global_id": 13683, - "bbox": [ - 344.11, - 377.76, - 399.67, - 396.64 - ], - "text": "(c) e−(s−1) + 3", - "type": "text" - }, - { - "block_id": "p489-b40", - "global_id": 13684, - "bbox": [ - 360.34, - 391.07, - 399.58, - 403.0 - ], - "text": "s2 −2s + 5", - "type": "text" - }, - { - "block_id": "p489-b41", - "global_id": 13685, - "bbox": [ - 343.61, - 401.81, - 411.63, - 420.69 - ], - "text": "(d) e−s + e−2s + 1", - "type": "text" - }, - { - "block_id": "p489-b42", - "global_id": 13686, - "bbox": [ - 314.97, - 415.12, - 516.14, - 451.44 - ], - "text": "s2 + 3s + 2\n4.2-8\nUsing ROC σ > 0, determine the inverse\nLaplace transform of X(s) = s−1 d", - "type": "text" - }, - { - "block_id": "p489-b43", - "global_id": 13687, - "bbox": [ - 461.24, - 447.49, - 467.0, - 453.96 - ], - "text": "ds", - "type": "text" - }, - { - "block_id": "p489-b45", - "global_id": 13688, - "bbox": [ - 475.27, - 438.93, - 487.29, - 447.16 - ], - "text": "e−2s", - "type": "text" - }, - { - "block_id": "p489-b46", - "global_id": 13689, - "bbox": [ - 480.28, - 447.49, - 482.8, - 453.96 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p489-b48", - "global_id": 13690, - "bbox": [ - 493.87, - 442.48, - 496.12, - 451.44 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p489-b49", - "global_id": 13691, - "bbox": [ - 314.97, - 458.63, - 516.14, - 489.59 - ], - "text": "4.2-9\nThe Laplace transform of a causal periodic sig-\nnal can be determined from the knowledge of the\nLaplace transform of its first cycle (period).", - "type": "text" - }, - { - "block_id": "p489-b50", - "global_id": 13692, - "bbox": [ - 344.11, - 491.21, - 516.14, - 533.43 - ], - "text": "(a) If\nthe\nLaplace\ntransform\nof\nx(t)\nin\nFig.\nP4.2-9a\nis\nX(s),\nthen\nshow\nthat\nG(s),\nthe\nLaplace\ntransform\nof\ng(t)\n(Fig. P4.2-9b), is", - "type": "text" - }, - { - "block_id": "p489-b51", - "global_id": 13693, - "bbox": [ - 382.38, - 545.05, - 492.8, - 567.04 - ], - "text": "G(s) =\nX(s)\n1 −e−sT0\nRes > 0", - "type": "text" - }, - { - "block_id": "p489-b52", - "global_id": 13694, - "bbox": [ - 343.61, - 577.78, - 516.14, - 586.75 - ], - "text": "(b) Use this result to find the Laplace transform", - "type": "text" - }, - { - "block_id": "p489-b53", - "global_id": 13695, - "bbox": [ - 359.04, - 588.37, - 515.34, - 597.71 - ], - "text": "of the signal p(t) illustrated in Fig. P4.2-9c.", - "type": "text" - }, - { - "block_id": "p489-b54", - "global_id": 13696, - "bbox": [ - 310.48, - 602.51, - 516.13, - 633.77 - ], - "text": "4.2-10\nStarting only with the fact that δ(t) ⇐⇒1, build\npairs 2 through 10b in Table 4.1, using various\nproperties of the Laplace transform.", - "type": "text" - } - ] - }, - { - "page_num": 490, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p490-b0", - "global_id": 13697, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "470\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p490-b1", - "global_id": 13698, - "bbox": [ - 438.08, - 135.58, - 440.31, - 143.58 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p490-b2", - "global_id": 13699, - "bbox": [ - 372.04, - 230.76, - 374.26, - 238.76 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p490-b3", - "global_id": 13700, - "bbox": [ - 159.02, - 139.72, - 161.25, - 147.72 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p490-b4", - "global_id": 13701, - "bbox": [ - 110.22, - 87.45, - 121.32, - 95.53 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p490-b5", - "global_id": 13702, - "bbox": [ - 136.54, - 156.83, - 145.42, - 164.83 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p490-b6", - "global_id": 13703, - "bbox": [ - 209.27, - 87.45, - 220.82, - 95.53 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p490-b7", - "global_id": 13704, - "bbox": [ - 325.2, - 156.83, - 335.01, - 164.83 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p490-b8", - "global_id": 13705, - "bbox": [ - 284.13, - 135.56, - 425.8, - 145.28 - ], - "text": "T0\n2T0\n3T0", - "type": "text" - }, - { - "block_id": "p490-b9", - "global_id": 13706, - "bbox": [ - 156.22, - 177.66, - 167.77, - 185.74 - ], - "text": "p(t)", - "type": "text" - }, - { - "block_id": "p490-b10", - "global_id": 13707, - "bbox": [ - 266.14, - 251.74, - 275.02, - 259.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p490-b11", - "global_id": 13708, - "bbox": [ - 163.97, - 189.99, - 167.97, - 197.99 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p490-b12", - "global_id": 13709, - "bbox": [ - 187.09, - 230.69, - 361.16, - 238.69 - ], - "text": "2\n8\n10\n16\n18\n24", - "type": "text" - }, - { - "block_id": "p490-b13", - "global_id": 13710, - "bbox": [ - 104.83, - 265.91, - 156.48, - 274.88 - ], - "text": "Figure P4.2-9", - "type": "text" - }, - { - "block_id": "p490-b14", - "global_id": 13711, - "bbox": [ - 57.59, - 286.38, - 263.24, - 295.42 - ], - "text": "4.2-11\n(a) Find the Laplace transform of the pulses", - "type": "text" - }, - { - "block_id": "p490-b15", - "global_id": 13712, - "bbox": [ - 90.72, - 297.41, - 263.24, - 339.25 - ], - "text": "in Fig. 4.2 by using only the time-different\niation property, the time-shifting property,\nand the fact that δ(t) ⇐⇒1.\n(b) In Ex. 4.9, the Laplace transform of x(t)", - "type": "text" - }, - { - "block_id": "p490-b16", - "global_id": 13713, - "bbox": [ - 106.15, - 341.25, - 263.24, - 394.05 - ], - "text": "is found by finding the Laplace transform\nof d2x/dt2. Find the Laplace transform of\nx(t) in that example by finding the Laplace\ntransform of dx/dt and using Table 4.1, if\nnecessary.", - "type": "text" - }, - { - "block_id": "p490-b17", - "global_id": 13714, - "bbox": [ - 57.59, - 398.95, - 263.23, - 418.95 - ], - "text": "4.2-12\nDetermine the inverse unilateral Laplace trans-\nform of", - "type": "text" - }, - { - "block_id": "p490-b18", - "global_id": 13715, - "bbox": [ - 129.13, - 426.82, - 171.84, - 448.34 - ], - "text": "X(s) =\n1\nes+3", - "type": "text" - }, - { - "block_id": "p490-b19", - "global_id": 13716, - "bbox": [ - 195.57, - 425.45, - 202.3, - 435.69 - ], - "text": "s2", - "type": "text" - }, - { - "block_id": "p490-b20", - "global_id": 13717, - "bbox": [ - 174.74, - 439.09, - 223.64, - 448.43 - ], - "text": "(s + 1)(s + 2)", - "type": "text" - }, - { - "block_id": "p490-b21", - "global_id": 13718, - "bbox": [ - 57.59, - 456.24, - 263.24, - 476.24 - ], - "text": "4.2-13\nSince 13 is such a lucky number, determine the\ninverse Laplace transform of X(s) = 1/(s + 1)13", - "type": "text" - }, - { - "block_id": "p490-b22", - "global_id": 13719, - "bbox": [ - 90.72, - 477.86, - 263.24, - 498.16 - ], - "text": "given region of convergence σ > −1. [Hint:\nWhat is the nth derivative of 1/(s + a)?]", - "type": "text" - }, - { - "block_id": "p490-b23", - "global_id": 13720, - "bbox": [ - 57.59, - 503.06, - 263.23, - 523.07 - ], - "text": "4.2-14\nIt is difficult to compute the Laplace transform\nX(s) of signal", - "type": "text" - }, - { - "block_id": "p490-b24", - "global_id": 13721, - "bbox": [ - 154.61, - 531.22, - 184.32, - 546.46 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p490-b25", - "global_id": 13722, - "bbox": [ - 180.76, - 537.13, - 199.35, - 552.75 - ], - "text": "t u(t)", - "type": "text" - }, - { - "block_id": "p490-b26", - "global_id": 13723, - "bbox": [ - 90.72, - 559.15, - 263.22, - 579.08 - ], - "text": "by using direct integration. Instead, properties\nprovide a simpler method.", - "type": "text" - }, - { - "block_id": "p490-b27", - "global_id": 13724, - "bbox": [ - 91.22, - 581.07, - 263.24, - 590.03 - ], - "text": "(a) Use Laplace transform properties to express", - "type": "text" - }, - { - "block_id": "p490-b28", - "global_id": 13725, - "bbox": [ - 90.72, - 591.66, - 263.23, - 622.91 - ], - "text": "the Laplace transform of tx(t) in terms of the\nunknown quantity X(s).\n(b) Use the definition to determine the Laplace", - "type": "text" - }, - { - "block_id": "p490-b29", - "global_id": 13726, - "bbox": [ - 106.15, - 624.54, - 195.06, - 633.87 - ], - "text": "transform of y(t) = tx(t).", - "type": "text" - }, - { - "block_id": "p490-b30", - "global_id": 13727, - "bbox": [ - 318.37, - 286.02, - 490.38, - 295.36 - ], - "text": "(c) Solve for X(s) by using the two pieces", - "type": "text" - }, - { - "block_id": "p490-b31", - "global_id": 13728, - "bbox": [ - 333.31, - 297.28, - 486.69, - 306.32 - ], - "text": "from ()(a) and ()(b). Simplify your answer.", - "type": "text" - }, - { - "block_id": "p490-b32", - "global_id": 13729, - "bbox": [ - 289.23, - 311.55, - 490.37, - 331.55 - ], - "text": "4.3-1\nUse the Laplace transform to solve the following\ndifferential equations:", - "type": "text" - }, - { - "block_id": "p490-b33", - "global_id": 13730, - "bbox": [ - 318.37, - 329.92, - 490.39, - 342.51 - ], - "text": "(a) (D2 + 3D + 2)y(t) = Dx(t) if y(0−) =", - "type": "text" - }, - { - "block_id": "p490-b34", - "global_id": 13731, - "bbox": [ - 317.86, - 342.86, - 490.39, - 364.43 - ], - "text": "˙y(0−) = 0 and x(t) = u(t)\n(b) (D2 + 4D + 4)y(t) = (D + 1)x(t) if y(0−) =", - "type": "text" - }, - { - "block_id": "p490-b35", - "global_id": 13732, - "bbox": [ - 318.37, - 364.79, - 490.39, - 386.35 - ], - "text": "2, ˙y(0−) = 1 and x(t) = e−tu(t)\n(c) (D2 +6D+25)y(t) = (D+2)x(t) if y(0−) =", - "type": "text" - }, - { - "block_id": "p490-b36", - "global_id": 13733, - "bbox": [ - 333.31, - 386.7, - 433.42, - 397.31 - ], - "text": "˙y(0−) = 1 and x(t) = 25u(t)", - "type": "text" - }, - { - "block_id": "p490-b37", - "global_id": 13734, - "bbox": [ - 289.23, - 402.54, - 490.38, - 444.46 - ], - "text": "4.3-2\nSolve the differential equations in Prob. 4.3-1\nusing the Laplace transform. In each case deter-\nmine the zero-input and zero-state components\nof the solution.", - "type": "text" - }, - { - "block_id": "p490-b38", - "global_id": 13735, - "bbox": [ - 289.23, - 449.7, - 490.39, - 469.7 - ], - "text": "4.3-3\nConsider a causal LTIC system described by the\ndifferential equation", - "type": "text" - }, - { - "block_id": "p490-b39", - "global_id": 13736, - "bbox": [ - 355.61, - 484.59, - 452.66, - 493.93 - ], - "text": "2˙y(t) + 6y(t) = ˙x(t) −4x(t)", - "type": "text" - }, - { - "block_id": "p490-b40", - "global_id": 13737, - "bbox": [ - 318.37, - 509.2, - 490.4, - 518.17 - ], - "text": "(a) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p490-b41", - "global_id": 13738, - "bbox": [ - 317.86, - 518.53, - 490.39, - 540.08 - ], - "text": "mine the ZIR yzir(t) if y(0−) = −3.\n(b) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p490-b42", - "global_id": 13739, - "bbox": [ - 333.31, - 541.71, - 490.39, - 562.0 - ], - "text": "mine the ZSR yzsr(t) to the input x(t) =\neδ(t −π).", - "type": "text" - }, - { - "block_id": "p490-b43", - "global_id": 13740, - "bbox": [ - 289.23, - 567.24, - 490.39, - 587.24 - ], - "text": "4.3-4\nConsider a causal LTIC system described by the\ndifferential equation", - "type": "text" - }, - { - "block_id": "p490-b44", - "global_id": 13741, - "bbox": [ - 344.05, - 602.13, - 464.22, - 611.47 - ], - "text": "¨y(t) + 3˙y(t) + 2y(t) = 2˙x(t) −x(t)", - "type": "text" - } - ] - }, - { - "page_num": 491, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p491-b0", - "global_id": 13742, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n471", - "type": "text" - }, - { - "block_id": "p491-b1", - "global_id": 13743, - "bbox": [ - 116.96, - 85.9, - 288.99, - 94.86 - ], - "text": "(a) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p491-b2", - "global_id": 13744, - "bbox": [ - 116.46, - 95.22, - 288.99, - 127.74 - ], - "text": "mine the ZIR yzir(t) if ˙y(0−) = 2 and\ny(0−) = −3.\n(b) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p491-b3", - "global_id": 13745, - "bbox": [ - 131.9, - 129.36, - 288.98, - 149.65 - ], - "text": "mine the ZSR yzsr(t) to the input x(t) =\nu(t).", - "type": "text" - }, - { - "block_id": "p491-b4", - "global_id": 13746, - "bbox": [ - 87.82, - 159.66, - 289.0, - 201.58 - ], - "text": "4.3-5\nSolve the following simultaneous differential\nequations using the Laplace transform, assum-\ning all initial conditions to be zero and the input\nx(t) = u(t):", - "type": "text" - }, - { - "block_id": "p491-b5", - "global_id": 13747, - "bbox": [ - 116.97, - 203.2, - 231.75, - 213.28 - ], - "text": "(a) (D + 3)y1(t) −2y2(t) = x(t)", - "type": "text" - }, - { - "block_id": "p491-b6", - "global_id": 13748, - "bbox": [ - 116.46, - 214.16, - 245.82, - 235.19 - ], - "text": "−2y1(t) + (2D + 4)y2(t) = 0\n(b) (D + 2)y1(t) −(D + 1)y2(t) = 0", - "type": "text" - }, - { - "block_id": "p491-b7", - "global_id": 13749, - "bbox": [ - 116.46, - 236.08, - 288.97, - 268.07 - ], - "text": "−(D + 1)y1(t) + (2D + 1)y2(t) = x(t)\nDetermine the transfer functions relating outputs\ny1(t) and y2(t) to the input x(t).", - "type": "text" - }, - { - "block_id": "p491-b8", - "global_id": 13750, - "bbox": [ - 87.82, - 277.34, - 288.99, - 308.3 - ], - "text": "4.3-6\nConsider a causal LTIC system described by\n˙y(t) + 2y(t) = ˙x(t).\n(a) Determine the transfer function H(s) for this", - "type": "text" - }, - { - "block_id": "p491-b9", - "global_id": 13751, - "bbox": [ - 116.46, - 310.29, - 288.98, - 330.21 - ], - "text": "system.\n(b) Using your result from part (a), determine", - "type": "text" - }, - { - "block_id": "p491-b10", - "global_id": 13752, - "bbox": [ - 116.97, - 331.84, - 288.97, - 352.13 - ], - "text": "the impulse response h(t) for this system.\n(c) Using Laplace transform techniques, deter-", - "type": "text" - }, - { - "block_id": "p491-b11", - "global_id": 13753, - "bbox": [ - 131.9, - 353.75, - 288.98, - 374.05 - ], - "text": "mine the output y(t) if the input is x(t) =\ne−tu(t) and y(0−) =", - "type": "text" - }, - { - "block_id": "p491-b12", - "global_id": 13754, - "bbox": [ - 205.98, - 357.13, - 213.57, - 366.09 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p491-b13", - "global_id": 13755, - "bbox": [ - 213.57, - 365.09, - 220.29, - 374.05 - ], - "text": "2.", - "type": "text" - }, - { - "block_id": "p491-b14", - "global_id": 13756, - "bbox": [ - 87.82, - 384.05, - 288.99, - 404.06 - ], - "text": "4.3-7\nRepeat Prob. 4.3-6 for a causal LTIC system\ndescribed by 3y(t) + ˙y(t) + ˙x(t) = 0.", - "type": "text" - }, - { - "block_id": "p491-b15", - "global_id": 13757, - "bbox": [ - 87.82, - 414.06, - 288.98, - 445.02 - ], - "text": "4.3-8\nFor the circuit in Fig. P4.3-8, the switch is in the\nopen position for a long time before t = 0, when\nit is closed instantaneously.", - "type": "text" - }, - { - "block_id": "p491-b16", - "global_id": 13758, - "bbox": [ - 116.96, - 447.01, - 288.98, - 455.98 - ], - "text": "(a) Write loop equations (in time domain) for", - "type": "text" - }, - { - "block_id": "p491-b17", - "global_id": 13759, - "bbox": [ - 116.46, - 457.6, - 288.98, - 478.63 - ], - "text": "t ≥0.\n(b) Solve for y1(t) and y2(t) by taking the", - "type": "text" - }, - { - "block_id": "p491-b18", - "global_id": 13760, - "bbox": [ - 131.89, - 479.89, - 288.97, - 499.81 - ], - "text": "Laplace transform of loop equations found\nin part (a).", - "type": "text" - }, - { - "block_id": "p491-b19", - "global_id": 13761, - "bbox": [ - 314.97, - 85.94, - 516.12, - 116.89 - ], - "text": "4.3-9\nFor each of the systems described by the fol-\nlowing differential equations, find the system\ntransfer function:", - "type": "text" - }, - { - "block_id": "p491-b20", - "global_id": 13762, - "bbox": [ - 344.12, - 118.3, - 382.03, - 134.92 - ], - "text": "(a) d2y(t)", - "type": "text" - }, - { - "block_id": "p491-b21", - "global_id": 13763, - "bbox": [ - 365.7, - 119.3, - 420.66, - 141.19 - ], - "text": "dt2\n+11dy(t)", - "type": "text" - }, - { - "block_id": "p491-b22", - "global_id": 13764, - "bbox": [ - 408.2, - 119.3, - 487.62, - 141.19 - ], - "text": "dt\n+24y(t) = 5dx(t)", - "type": "text" - }, - { - "block_id": "p491-b23", - "global_id": 13765, - "bbox": [ - 475.14, - 125.58, - 516.13, - 141.19 - ], - "text": "dt\n+3x(t)", - "type": "text" - }, - { - "block_id": "p491-b24", - "global_id": 13766, - "bbox": [ - 343.61, - 142.22, - 382.03, - 158.84 - ], - "text": "(b) d3y(t)", - "type": "text" - }, - { - "block_id": "p491-b25", - "global_id": 13767, - "bbox": [ - 365.7, - 142.22, - 420.47, - 165.11 - ], - "text": "dt3\n+ 6d2y(t)", - "type": "text" - }, - { - "block_id": "p491-b26", - "global_id": 13768, - "bbox": [ - 404.14, - 143.2, - 459.42, - 165.11 - ], - "text": "dt2\n−11dy(t)", - "type": "text" - }, - { - "block_id": "p491-b27", - "global_id": 13769, - "bbox": [ - 446.94, - 149.5, - 488.2, - 165.11 - ], - "text": "dt\n+ 6y(t)", - "type": "text" - }, - { - "block_id": "p491-b28", - "global_id": 13770, - "bbox": [ - 360.89, - 163.64, - 397.22, - 180.25 - ], - "text": "= 3d2x(t)", - "type": "text" - }, - { - "block_id": "p491-b29", - "global_id": 13771, - "bbox": [ - 380.87, - 164.63, - 431.71, - 186.53 - ], - "text": "dt2\n+ 7dx(t)", - "type": "text" - }, - { - "block_id": "p491-b30", - "global_id": 13772, - "bbox": [ - 419.22, - 170.92, - 460.52, - 186.53 - ], - "text": "dt\n+ 5x(t)", - "type": "text" - }, - { - "block_id": "p491-b31", - "global_id": 13773, - "bbox": [ - 344.12, - 190.03, - 382.03, - 206.65 - ], - "text": "(c) d4y(t)", - "type": "text" - }, - { - "block_id": "p491-b32", - "global_id": 13774, - "bbox": [ - 365.7, - 191.03, - 416.49, - 212.93 - ], - "text": "dt4\n+ 4dy(t)", - "type": "text" - }, - { - "block_id": "p491-b33", - "global_id": 13775, - "bbox": [ - 404.02, - 191.03, - 451.88, - 212.93 - ], - "text": "dt\n= 3dx(t)", - "type": "text" - }, - { - "block_id": "p491-b34", - "global_id": 13776, - "bbox": [ - 439.39, - 197.31, - 480.7, - 212.93 - ], - "text": "dt\n+ 2x(t)", - "type": "text" - }, - { - "block_id": "p491-b35", - "global_id": 13777, - "bbox": [ - 343.61, - 213.95, - 382.03, - 230.56 - ], - "text": "(d) d2y(t)", - "type": "text" - }, - { - "block_id": "p491-b36", - "global_id": 13778, - "bbox": [ - 365.7, - 214.94, - 436.04, - 236.84 - ], - "text": "dt2\n−y(t) = dx(t)", - "type": "text" - }, - { - "block_id": "p491-b37", - "global_id": 13779, - "bbox": [ - 423.55, - 221.23, - 460.37, - 236.84 - ], - "text": "dt\n−x(t)", - "type": "text" - }, - { - "block_id": "p491-b38", - "global_id": 13780, - "bbox": [ - 310.48, - 238.3, - 516.14, - 291.18 - ], - "text": "4.3-10\nFor each of the systems specified by the fol-\nlowing transfer functions, find the differential\nequation relating the output y(t) to the input x(t),\nassuming that the systems are controllable and\nobservable:", - "type": "text" - }, - { - "block_id": "p491-b39", - "global_id": 13781, - "bbox": [ - 344.12, - 289.03, - 427.22, - 311.02 - ], - "text": "(a) H(s) =\ns + 5\ns2 + 3s + 8", - "type": "text" - }, - { - "block_id": "p491-b40", - "global_id": 13782, - "bbox": [ - 343.61, - 309.83, - 448.71, - 335.07 - ], - "text": "(b) H(s) =\ns2 + 3s + 5\ns3 + 8s2 + 5s + 7", - "type": "text" - }, - { - "block_id": "p491-b41", - "global_id": 13783, - "bbox": [ - 344.11, - 333.88, - 431.7, - 352.76 - ], - "text": "(c) H(s) = 5s2 + 7s + 2", - "type": "text" - }, - { - "block_id": "p491-b42", - "global_id": 13784, - "bbox": [ - 310.48, - 347.19, - 469.23, - 370.17 - ], - "text": "s2 −2s + 5\n4.3-11\nFor a system with transfer function", - "type": "text" - }, - { - "block_id": "p491-b43", - "global_id": 13785, - "bbox": [ - 395.19, - 370.87, - 463.37, - 392.86 - ], - "text": "H(s) =\n2s + 3\ns2 + 2s + 5", - "type": "text" - }, - { - "block_id": "p491-b44", - "global_id": 13786, - "bbox": [ - 344.11, - 402.69, - 516.13, - 411.65 - ], - "text": "(a) Find the (zero-state) response for inputs", - "type": "text" - }, - { - "block_id": "p491-b45", - "global_id": 13787, - "bbox": [ - 343.61, - 413.28, - 516.14, - 466.45 - ], - "text": "x1(t) = 10u(t) and x2(t) = u(t −5).\n(b) For\nthis\nsystem\nwrite\nthe\ndifferential\nequation relating the output y(t) to the\ninput x(t), assuming that the systems are\ncontrollable and observable.", - "type": "text" - }, - { - "block_id": "p491-b46", - "global_id": 13788, - "bbox": [ - 310.49, - 471.46, - 469.24, - 480.5 - ], - "text": "4.3-12\nFor a system with transfer function", - "type": "text" - }, - { - "block_id": "p491-b47", - "global_id": 13789, - "bbox": [ - 404.07, - 479.31, - 454.48, - 501.01 - ], - "text": "H(s) =\ns\ns2 + 9", - "type": "text" - }, - { - "block_id": "p491-b48", - "global_id": 13790, - "bbox": [ - 217.34, - 513.86, - 234.23, - 522.16 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p491-b49", - "global_id": 13791, - "bbox": [ - 171.42, - 586.8, - 419.83, - 600.8 - ], - "text": "v0(t)\n40 V", - "type": "text" - }, - { - "block_id": "p491-b50", - "global_id": 13792, - "bbox": [ - 288.74, - 545.67, - 348.49, - 555.28 - ], - "text": "y2(t)\ny1(t)", - "type": "text" - }, - { - "block_id": "p491-b51", - "global_id": 13793, - "bbox": [ - 355.73, - 588.6, - 366.18, - 596.6 - ], - "text": "1 F", - "type": "text" - }, - { - "block_id": "p491-b52", - "global_id": 13794, - "bbox": [ - 291.74, - 610.25, - 303.52, - 618.25 - ], - "text": "2 H", - "type": "text" - }, - { - "block_id": "p491-b53", - "global_id": 13795, - "bbox": [ - 216.47, - 560.4, - 416.11, - 583.2 - ], - "text": "5 \t\n4 \t\n1", - "type": "text" - }, - { - "block_id": "p491-b54", - "global_id": 13796, - "bbox": [ - 409.45, - 623.27, - 486.01, - 635.58 - ], - "text": "Figure P4.3-8", - "type": "text" - } - ] - }, - { - "page_num": 492, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p492-b0", - "global_id": 13797, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "472\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p492-b1", - "global_id": 13798, - "bbox": [ - 91.22, - 85.9, - 263.22, - 94.86 - ], - "text": "(a) Find the (zero-state) response if the input", - "type": "text" - }, - { - "block_id": "p492-b2", - "global_id": 13799, - "bbox": [ - 90.72, - 95.22, - 263.24, - 149.66 - ], - "text": "x(t) = (1 −e−t)u(t)\n(b) For\nthis\nsystem\nwrite\nthe\ndifferential\nequation relating the output y(t) to the\ninput x(t), assuming that the systems are\ncontrollable and observable.", - "type": "text" - }, - { - "block_id": "p492-b3", - "global_id": 13800, - "bbox": [ - 57.59, - 154.57, - 236.4, - 163.61 - ], - "text": "4.3-13\nConsider a system with transfer function", - "type": "text" - }, - { - "block_id": "p492-b4", - "global_id": 13801, - "bbox": [ - 90.72, - 164.89, - 263.24, - 207.25 - ], - "text": "H(s) =\ns + 5\ns2 + 5s + 6\nFind the (zero-state) response for the following\ninputs:", - "type": "text" - }, - { - "block_id": "p492-b5", - "global_id": 13802, - "bbox": [ - 90.72, - 207.6, - 162.41, - 229.91 - ], - "text": "(a) xa(t) = e−3tu(t)\n(b) xb(t) = e−4tu(t)", - "type": "text" - }, - { - "block_id": "p492-b6", - "global_id": 13803, - "bbox": [ - 90.72, - 229.53, - 189.43, - 251.82 - ], - "text": "(c) xc(t) = e−4(t−5)u(t −5)\n(d) xd(t) = e−4(t−5)u(t)", - "type": "text" - }, - { - "block_id": "p492-b7", - "global_id": 13804, - "bbox": [ - 90.72, - 251.44, - 263.23, - 283.96 - ], - "text": "(e) xe(t) = e−4tu(t −5)\nAssuming that the system H(s) is controllable\nand observable,", - "type": "text" - }, - { - "block_id": "p492-b8", - "global_id": 13805, - "bbox": [ - 92.22, - 285.96, - 263.22, - 294.92 - ], - "text": "(f) write the differential equation relating the", - "type": "text" - }, - { - "block_id": "p492-b9", - "global_id": 13806, - "bbox": [ - 106.16, - 296.54, - 205.57, - 305.88 - ], - "text": "output y(t) to the input x(t).", - "type": "text" - }, - { - "block_id": "p492-b10", - "global_id": 13807, - "bbox": [ - 57.59, - 310.79, - 263.24, - 352.7 - ], - "text": "4.3-14\nAn LTI system has a step response given by\ns(t) = e−tu(t) −e−2tu(t). Determine the output\nof this system y(t) given an input x(t) = δ(t −\nπ) −cos(", - "type": "text" - }, - { - "block_id": "p492-b11", - "global_id": 13808, - "bbox": [ - 126.42, - 335.82, - 134.0, - 344.79 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p492-b12", - "global_id": 13809, - "bbox": [ - 134.0, - 343.36, - 157.91, - 352.7 - ], - "text": "3)u(t).", - "type": "text" - }, - { - "block_id": "p492-b13", - "global_id": 13810, - "bbox": [ - 57.59, - 357.61, - 263.24, - 399.53 - ], - "text": "4.3-15\nFor an LTIC system with zero initial conditions\n(system initially in zero state), if an input x(t)\nproduces an output y(t), then using the Laplace\ntransform, show the following:", - "type": "text" - }, - { - "block_id": "p492-b14", - "global_id": 13811, - "bbox": [ - 90.72, - 401.15, - 257.62, - 421.45 - ], - "text": "(a) The input dx/dt produces an output dy/dt.\n(b) The input", - "type": "text" - }, - { - "block_id": "p492-b15", - "global_id": 13812, - "bbox": [ - 148.68, - 404.89, - 156.32, - 416.05 - ], - "text": "$ t", - "type": "text" - }, - { - "block_id": "p492-b16", - "global_id": 13813, - "bbox": [ - 106.16, - 412.11, - 263.23, - 428.37 - ], - "text": "0 x(τ)dτ produces an output\n$ t", - "type": "text" - }, - { - "block_id": "p492-b17", - "global_id": 13814, - "bbox": [ - 106.16, - 424.41, - 263.24, - 455.67 - ], - "text": "0 y(τ)dτ. Hence, show that the unit step\nresponse of a system is an integral of the\nimpulse response; that is,", - "type": "text" - }, - { - "block_id": "p492-b18", - "global_id": 13815, - "bbox": [ - 199.05, - 439.11, - 206.7, - 450.28 - ], - "text": "$ t", - "type": "text" - }, - { - "block_id": "p492-b19", - "global_id": 13816, - "bbox": [ - 203.16, - 446.33, - 237.38, - 458.64 - ], - "text": "0 h(τ)dτ.", - "type": "text" - }, - { - "block_id": "p492-b20", - "global_id": 13817, - "bbox": [ - 57.59, - 460.57, - 263.25, - 502.49 - ], - "text": "4.3-16\nDiscuss asymptotic and BIBO stabilities for\nthe systems described by the following trans-\nfer functions, assuming that the systems are\ncontrollable and observable:", - "type": "text" - }, - { - "block_id": "p492-b21", - "global_id": 13818, - "bbox": [ - 91.23, - 506.38, - 146.59, - 528.37 - ], - "text": "(a)\n(s + 5)\ns2 + 3s + 2", - "type": "text" - }, - { - "block_id": "p492-b22", - "global_id": 13819, - "bbox": [ - 90.72, - 532.85, - 139.02, - 554.84 - ], - "text": "(b)\ns + 5\ns2(s + 2)", - "type": "text" - }, - { - "block_id": "p492-b23", - "global_id": 13820, - "bbox": [ - 91.22, - 560.2, - 135.29, - 575.82 - ], - "text": "(c) s(s + 2)", - "type": "text" - }, - { - "block_id": "p492-b24", - "global_id": 13821, - "bbox": [ - 112.44, - 572.85, - 130.19, - 582.19 - ], - "text": "s + 5", - "type": "text" - }, - { - "block_id": "p492-b25", - "global_id": 13822, - "bbox": [ - 90.72, - 586.67, - 135.29, - 608.66 - ], - "text": "(d)\ns + 5\ns(s + 2)", - "type": "text" - }, - { - "block_id": "p492-b26", - "global_id": 13823, - "bbox": [ - 91.22, - 613.97, - 146.59, - 635.96 - ], - "text": "(e)\ns + 5\ns2 −2s + 3", - "type": "text" - }, - { - "block_id": "p492-b27", - "global_id": 13824, - "bbox": [ - 284.74, - 85.94, - 490.37, - 116.89 - ], - "text": "4.3-17\nRepeat Prob. 4.3-16 for systems described by the\nfollowing differential equations. Systems may\nbe uncontrollable and/or unobservable.", - "type": "text" - }, - { - "block_id": "p492-b28", - "global_id": 13825, - "bbox": [ - 317.87, - 115.26, - 450.0, - 138.82 - ], - "text": "(a) (D2 + 3D + 2)y(t) = (D + 3)x(t)\n(b) (D2 + 3D + 2)y(t) = (D + 1)x(t)", - "type": "text" - }, - { - "block_id": "p492-b29", - "global_id": 13826, - "bbox": [ - 317.87, - 137.18, - 450.0, - 160.73 - ], - "text": "(c) (D2 + D −2)y(t) = (D −1)x(t)\n(d) (D2 −3D + 2)y(t) = (D −1)x(t)", - "type": "text" - }, - { - "block_id": "p492-b30", - "global_id": 13827, - "bbox": [ - 289.23, - 166.26, - 490.4, - 208.18 - ], - "text": "4.4-1\nThe circuit shown in Fig. P4.4-1 has system\nfunction given by H(s) =\n1\n1+RCs. Let R = 2 and\nC = 3 and use Laplace transform techniques to\nsolve the following.", - "type": "text" - }, - { - "block_id": "p492-b31", - "global_id": 13828, - "bbox": [ - 318.37, - 209.8, - 490.37, - 219.14 - ], - "text": "(a) Find the output y(t) given an initial capaci-", - "type": "text" - }, - { - "block_id": "p492-b32", - "global_id": 13829, - "bbox": [ - 317.86, - 219.49, - 490.39, - 252.01 - ], - "text": "tor voltage of y(0−) = 3 and an input x(t) =\nu(t).\n(b) Given an input x(t) = u(t −3), determine", - "type": "text" - }, - { - "block_id": "p492-b33", - "global_id": 13830, - "bbox": [ - 333.31, - 252.37, - 490.39, - 273.94 - ], - "text": "the initial capacitor voltage y(0−) so that the\noutput y(t) is 1 volt at t = 6 seconds.", - "type": "text" - }, - { - "block_id": "p492-b34", - "global_id": 13831, - "bbox": [ - 342.64, - 331.99, - 348.86, - 339.96 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p492-b35", - "global_id": 13832, - "bbox": [ - 327.46, - 319.12, - 339.33, - 327.34 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p492-b36", - "global_id": 13833, - "bbox": [ - 342.64, - 305.84, - 379.56, - 314.12 - ], - "text": "+\nR", - "type": "text" - }, - { - "block_id": "p492-b37", - "global_id": 13834, - "bbox": [ - 390.84, - 319.94, - 396.15, - 327.91 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p492-b38", - "global_id": 13835, - "bbox": [ - 419.8, - 305.28, - 426.01, - 313.25 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p492-b39", - "global_id": 13836, - "bbox": [ - 420.33, - 319.01, - 432.16, - 327.23 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p492-b40", - "global_id": 13837, - "bbox": [ - 419.8, - 332.86, - 486.62, - 347.59 - ], - "text": "−\nFigure P4.4-1", - "type": "text" - }, - { - "block_id": "p492-b41", - "global_id": 13838, - "bbox": [ - 289.22, - 367.06, - 490.38, - 398.02 - ], - "text": "4.4-2\nConsider the circuit shown in Fig. P4.4-2.\nUse Laplace transform techniques to solve the\nfollowing.", - "type": "text" - }, - { - "block_id": "p492-b42", - "global_id": 13839, - "bbox": [ - 317.86, - 400.02, - 490.38, - 441.86 - ], - "text": "(a) Determine\nthe\nstandard-form,\nconstant-\ncoefficient differential equation description\nof this circuit.\n(b) Letting R = C = 1, determine the total", - "type": "text" - }, - { - "block_id": "p492-b43", - "global_id": 13840, - "bbox": [ - 333.31, - 442.22, - 490.39, - 464.45 - ], - "text": "response y(t) to input x(t) = 3e−tu(t) and\ninitial capacitor voltage of vC(0−) = 5.", - "type": "text" - }, - { - "block_id": "p492-b44", - "global_id": 13841, - "bbox": [ - 342.64, - 529.04, - 348.86, - 537.01 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p492-b45", - "global_id": 13842, - "bbox": [ - 327.46, - 516.17, - 339.33, - 524.39 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p492-b46", - "global_id": 13843, - "bbox": [ - 342.64, - 503.19, - 348.86, - 511.16 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p492-b47", - "global_id": 13844, - "bbox": [ - 374.69, - 495.69, - 379.56, - 503.66 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p492-b48", - "global_id": 13845, - "bbox": [ - 394.46, - 502.63, - 399.33, - 510.6 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p492-b49", - "global_id": 13846, - "bbox": [ - 390.84, - 531.39, - 396.15, - 539.36 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p492-b50", - "global_id": 13847, - "bbox": [ - 419.8, - 499.93, - 426.01, - 507.9 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p492-b51", - "global_id": 13848, - "bbox": [ - 420.33, - 516.05, - 432.16, - 524.27 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p492-b52", - "global_id": 13849, - "bbox": [ - 419.8, - 532.31, - 426.01, - 540.28 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p492-b53", - "global_id": 13850, - "bbox": [ - 434.98, - 542.87, - 486.62, - 551.84 - ], - "text": "Figure P4.4-2", - "type": "text" - }, - { - "block_id": "p492-b54", - "global_id": 13851, - "bbox": [ - 289.22, - 571.01, - 490.39, - 635.14 - ], - "text": "4.4-3\nFind the zero-state response y(t) of the network\nin Fig. P4.4-3 if the input voltage x(t) = te−tu(t).\nFind the transfer function relating the output\nY(s) to the input X(s). From the transfer func-\ntion, write the differential equation relating y(t)\nto x(t).", - "type": "text" - } - ] - }, - { - "page_num": 493, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p493-b0", - "global_id": 13852, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n473", - "type": "text" - }, - { - "block_id": "p493-b3", - "global_id": 13853, - "bbox": [ - 185.69, - 130.35, - 196.14, - 138.35 - ], - "text": "1 F", - "type": "text" - }, - { - "block_id": "p493-b4", - "global_id": 13854, - "bbox": [ - 236.56, - 88.77, - 248.34, - 96.77 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p493-b5", - "global_id": 13855, - "bbox": [ - 119.71, - 128.31, - 300.32, - 141.68 - ], - "text": "x(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p493-b6", - "global_id": 13856, - "bbox": [ - 171.76, - 87.77, - 183.98, - 96.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p493-b7", - "global_id": 13857, - "bbox": [ - 247.44, - 129.74, - 259.66, - 138.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p493-b8", - "global_id": 13858, - "bbox": [ - 87.82, - 180.66, - 288.98, - 224.11 - ], - "text": "Figure P4.4-3\n4.4-4\nThe switch in the circuit of Fig. P4.4-4 is closed\nfor a long time and then opened instantaneously\nat t = 0. Find and sketch the current y(t).", - "type": "text" - }, - { - "block_id": "p493-b9", - "global_id": 13859, - "bbox": [ - 119.49, - 275.7, - 135.27, - 283.7 - ], - "text": "10 V", - "type": "text" - }, - { - "block_id": "p493-b10", - "global_id": 13860, - "bbox": [ - 209.98, - 236.84, - 222.2, - 245.14 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p493-b11", - "global_id": 13861, - "bbox": [ - 277.56, - 268.64, - 294.44, - 276.94 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p493-b12", - "global_id": 13862, - "bbox": [ - 189.61, - 254.89, - 200.72, - 262.97 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p493-b13", - "global_id": 13863, - "bbox": [ - 163.48, - 237.62, - 175.26, - 245.62 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p493-b14", - "global_id": 13864, - "bbox": [ - 232.97, - 274.35, - 244.74, - 282.35 - ], - "text": "2 H", - "type": "text" - }, - { - "block_id": "p493-b15", - "global_id": 13865, - "bbox": [ - 87.82, - 315.34, - 288.98, - 347.83 - ], - "text": "Figure P4.4-4\n4.4-5\nFind the current y(t) for the parallel resonant\ncircuit in Fig. P4.4-5 if the input is:", - "type": "text" - }, - { - "block_id": "p493-b16", - "global_id": 13866, - "bbox": [ - 116.46, - 349.46, - 288.97, - 391.67 - ], - "text": "(a) x(t) = Acos ω0tu(t)\n(b) x(t) = Asin ω0tu(t)\nAssume all initial conditions to be zero and, in\nboth cases, ω2", - "type": "text" - }, - { - "block_id": "p493-b17", - "global_id": 13867, - "bbox": [ - 163.93, - 382.33, - 200.62, - 394.06 - ], - "text": "0 = 1/LC.", - "type": "text" - }, - { - "block_id": "p493-b18", - "global_id": 13868, - "bbox": [ - 199.15, - 449.66, - 296.65, - 457.37 - ], - "text": "C\nL", - "type": "text" - }, - { - "block_id": "p493-b19", - "global_id": 13869, - "bbox": [ - 166.27, - 405.58, - 176.59, - 413.09 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p493-b20", - "global_id": 13870, - "bbox": [ - 119.69, - 449.71, - 130.47, - 457.22 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p493-b21", - "global_id": 13871, - "bbox": [ - 116.46, - 496.9, - 168.11, - 505.86 - ], - "text": "Figure P4.4-5", - "type": "text" - }, - { - "block_id": "p493-b22", - "global_id": 13872, - "bbox": [ - 314.97, - 85.64, - 516.13, - 116.9 - ], - "text": "4.4-6\nFind the loop currents y1(t) and y2(t) for t ≥0\nin the circuit of Fig. P4.4-6a for the input x(t) in\nFig. P4.4-6b.", - "type": "text" - }, - { - "block_id": "p493-b23", - "global_id": 13873, - "bbox": [ - 314.97, - 121.84, - 516.13, - 164.42 - ], - "text": "4.4-7\nFor the network in Fig. P4.4-7, the switch is in\na closed position for a long time before t = 0,\nwhen it is opened instantaneously. Find y1(t) and\nvs(t) for t ≥0.", - "type": "text" - }, - { - "block_id": "p493-b24", - "global_id": 13874, - "bbox": [ - 309.57, - 227.68, - 513.0, - 238.2 - ], - "text": "t 0\n1 F\n10 V", - "type": "text" - }, - { - "block_id": "p493-b25", - "global_id": 13875, - "bbox": [ - 366.72, - 276.25, - 378.94, - 284.55 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p493-b26", - "global_id": 13876, - "bbox": [ - 336.44, - 183.24, - 457.72, - 195.32 - ], - "text": "2 H\n1 H\ny1(t)\ny2(t)", - "type": "text" - }, - { - "block_id": "p493-b27", - "global_id": 13877, - "bbox": [ - 382.63, - 227.62, - 396.06, - 237.16 - ], - "text": "vs(t)", - "type": "text" - }, - { - "block_id": "p493-b30", - "global_id": 13878, - "bbox": [ - 309.57, - 290.73, - 361.22, - 299.69 - ], - "text": "Figure P4.4-7", - "type": "text" - }, - { - "block_id": "p493-b31", - "global_id": 13879, - "bbox": [ - 314.97, - 312.95, - 516.13, - 344.22 - ], - "text": "4.4-8\nFind the output voltage v0(t) for t ≥0 for the\ncircuit in Fig. P4.4-8, if the input x(t) = 100u(t).\nThe system is in the zero state initially.", - "type": "text" - }, - { - "block_id": "p493-b32", - "global_id": 13880, - "bbox": [ - 314.97, - 348.86, - 516.14, - 380.78 - ], - "text": "4.4-9\nFind the output voltage y(t) for the network in\nFig. P4.4-9 for the initial conditions iL(0) = 1 A\nand vC(0) = 3 V.", - "type": "text" - }, - { - "block_id": "p493-b33", - "global_id": 13881, - "bbox": [ - 310.48, - 385.05, - 516.13, - 426.97 - ], - "text": "4.4-10\nFor the network in Fig. P4.4-10, the switch is in\nposition a for a long time and then is moved to\nposition b instantaneously at t = 0. Determine\nthe current y(t) for t > 0.", - "type": "text" - }, - { - "block_id": "p493-b34", - "global_id": 13882, - "bbox": [ - 310.48, - 431.91, - 473.12, - 440.95 - ], - "text": "4.4-11\nConsider the circuit of Fig. P4.4-11.", - "type": "text" - }, - { - "block_id": "p493-b35", - "global_id": 13883, - "bbox": [ - 344.12, - 442.95, - 516.15, - 451.91 - ], - "text": "(a) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p493-b36", - "global_id": 13884, - "bbox": [ - 343.61, - 453.9, - 516.13, - 484.79 - ], - "text": "mine the system’s standard-form transfer\nfunction H(s).\n(b) Using transform-domain techniques and let-", - "type": "text" - }, - { - "block_id": "p493-b37", - "global_id": 13885, - "bbox": [ - 359.05, - 486.4, - 516.13, - 517.66 - ], - "text": "ting R = L = 1, determine the circuit’s\nzero-state response yzsr(t) to the input x(t) =\ne−2tu(t −1).", - "type": "text" - }, - { - "block_id": "p493-b38", - "global_id": 13886, - "bbox": [ - 382.83, - 581.12, - 386.83, - 589.12 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p493-b39", - "global_id": 13887, - "bbox": [ - 382.83, - 608.97, - 386.83, - 616.97 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p493-b40", - "global_id": 13888, - "bbox": [ - 374.33, - 559.97, - 378.33, - 567.97 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p493-b41", - "global_id": 13889, - "bbox": [ - 453.42, - 607.85, - 455.64, - 615.85 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p493-b42", - "global_id": 13890, - "bbox": [ - 221.32, - 627.09, - 410.66, - 635.09 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p493-b43", - "global_id": 13891, - "bbox": [ - 388.11, - 548.08, - 399.22, - 556.17 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p493-b44", - "global_id": 13892, - "bbox": [ - 130.8, - 594.34, - 142.38, - 602.42 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p493-b45", - "global_id": 13893, - "bbox": [ - 138.0, - 562.09, - 150.23, - 570.38 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p493-b46", - "global_id": 13894, - "bbox": [ - 235.63, - 580.46, - 320.23, - 588.88 - ], - "text": "1 \t\n1", - "type": "text" - }, - { - "block_id": "p493-b47", - "global_id": 13895, - "bbox": [ - 185.07, - 534.97, - 196.85, - 542.97 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p493-b48", - "global_id": 13896, - "bbox": [ - 183.91, - 560.9, - 198.01, - 570.51 - ], - "text": "y1(t)", - "type": "text" - }, - { - "block_id": "p493-b49", - "global_id": 13897, - "bbox": [ - 252.78, - 534.97, - 264.55, - 542.97 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p493-b50", - "global_id": 13898, - "bbox": [ - 251.61, - 561.06, - 265.72, - 570.67 - ], - "text": "y2(t)", - "type": "text" - }, - { - "block_id": "p493-b51", - "global_id": 13899, - "bbox": [ - 130.58, - 641.27, - 182.22, - 650.23 - ], - "text": "Figure P4.4-6", - "type": "text" - } - ] - }, - { - "page_num": 494, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p494-b0", - "global_id": 13900, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "474\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p494-b1", - "global_id": 13901, - "bbox": [ - 209.14, - 139.55, - 223.24, - 149.16 - ], - "text": "y1(t)", - "type": "text" - }, - { - "block_id": "p494-b2", - "global_id": 13902, - "bbox": [ - 246.46, - 118.74, - 258.23, - 126.74 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p494-b3", - "global_id": 13903, - "bbox": [ - 204.52, - 89.86, - 216.74, - 98.15 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p494-b4", - "global_id": 13904, - "bbox": [ - 145.9, - 137.25, - 396.98, - 149.16 - ], - "text": "v0(t)\nx(t)\ny2(t)", - "type": "text" - }, - { - "block_id": "p494-b5", - "global_id": 13905, - "bbox": [ - 287.83, - 119.24, - 299.61, - 127.24 - ], - "text": "4 H", - "type": "text" - }, - { - "block_id": "p494-b6", - "global_id": 13906, - "bbox": [ - 362.41, - 136.5, - 374.63, - 144.8 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p494-b7", - "global_id": 13907, - "bbox": [ - 258.38, - 92.54, - 287.48, - 100.84 - ], - "text": "M 2 H", - "type": "text" - }, - { - "block_id": "p494-b9", - "global_id": 13908, - "bbox": [ - 386.39, - 167.45, - 463.27, - 181.67 - ], - "text": "Figure P4.4-8", - "type": "text" - }, - { - "block_id": "p494-b10", - "global_id": 13909, - "bbox": [ - 145.68, - 258.35, - 367.92, - 274.81 - ], - "text": "y(t)\n2u(t)\n1 F", - "type": "text" - }, - { - "block_id": "p494-b11", - "global_id": 13910, - "bbox": [ - 220.9, - 230.83, - 226.46, - 240.38 - ], - "text": "iL", - "type": "text" - }, - { - "block_id": "p494-b12", - "global_id": 13911, - "bbox": [ - 286.14, - 256.79, - 299.91, - 266.33 - ], - "text": "vc(t)", - "type": "text" - }, - { - "block_id": "p494-b17", - "global_id": 13912, - "bbox": [ - 309.0, - 203.58, - 321.26, - 217.35 - ], - "text": "1\n8", - "type": "text" - }, - { - "block_id": "p494-b18", - "global_id": 13913, - "bbox": [ - 207.75, - 253.71, - 222.49, - 267.77 - ], - "text": "1\n13 H", - "type": "text" - }, - { - "block_id": "p494-b19", - "global_id": 13914, - "bbox": [ - 321.71, - 245.74, - 333.97, - 259.5 - ], - "text": "1\n8", - "type": "text" - }, - { - "block_id": "p494-b20", - "global_id": 13915, - "bbox": [ - 382.63, - 293.84, - 434.27, - 302.81 - ], - "text": "Figure P4.4-9", - "type": "text" - }, - { - "block_id": "p494-b21", - "global_id": 13916, - "bbox": [ - 178.98, - 344.53, - 182.53, - 352.53 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p494-b22", - "global_id": 13917, - "bbox": [ - 196.48, - 353.03, - 200.48, - 361.03 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p494-b23", - "global_id": 13918, - "bbox": [ - 179.04, - 330.41, - 195.93, - 338.71 - ], - "text": "t 0", - "type": "text" - }, - { - "block_id": "p494-b24", - "global_id": 13919, - "bbox": [ - 104.83, - 371.67, - 181.93, - 390.67 - ], - "text": "100 V\n10 V", - "type": "text" - }, - { - "block_id": "p494-b25", - "global_id": 13920, - "bbox": [ - 239.15, - 332.36, - 250.57, - 340.45 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p494-b26", - "global_id": 13921, - "bbox": [ - 254.28, - 371.48, - 330.12, - 379.77 - ], - "text": "1 \t\n1", - "type": "text" - }, - { - "block_id": "p494-b27", - "global_id": 13922, - "bbox": [ - 270.57, - 327.86, - 344.79, - 336.15 - ], - "text": "2 \t\n2", - "type": "text" - }, - { - "block_id": "p494-b28", - "global_id": 13923, - "bbox": [ - 374.98, - 370.22, - 386.76, - 378.22 - ], - "text": "1 H", - "type": "text" - }, - { - "block_id": "p494-b29", - "global_id": 13924, - "bbox": [ - 216.84, - 322.67, - 227.32, - 336.73 - ], - "text": "1\n5 F", - "type": "text" - }, - { - "block_id": "p494-b30", - "global_id": 13925, - "bbox": [ - 104.83, - 414.18, - 160.95, - 423.14 - ], - "text": "Figure P4.4-10", - "type": "text" - }, - { - "block_id": "p494-b31", - "global_id": 13926, - "bbox": [ - 91.22, - 451.15, - 263.22, - 460.12 - ], - "text": "(c) Using transform-domain techniques and let-", - "type": "text" - }, - { - "block_id": "p494-b32", - "global_id": 13927, - "bbox": [ - 106.15, - 461.73, - 263.24, - 493.0 - ], - "text": "ting R = 2L = 1, determine the circuit’s\nzero-state response yzsr(t) to the input x(t) =\ne−2tu(t −1).", - "type": "text" - }, - { - "block_id": "p494-b33", - "global_id": 13928, - "bbox": [ - 115.49, - 564.21, - 121.7, - 572.18 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p494-b34", - "global_id": 13929, - "bbox": [ - 100.31, - 551.34, - 112.18, - 559.56 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p494-b35", - "global_id": 13930, - "bbox": [ - 115.49, - 538.36, - 121.7, - 546.33 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p494-b36", - "global_id": 13931, - "bbox": [ - 145.55, - 530.98, - 172.18, - 545.77 - ], - "text": "2R\nR", - "type": "text" - }, - { - "block_id": "p494-b37", - "global_id": 13932, - "bbox": [ - 170.72, - 566.6, - 175.15, - 574.57 - ], - "text": "L", - "type": "text" - }, - { - "block_id": "p494-b38", - "global_id": 13933, - "bbox": [ - 192.64, - 535.1, - 198.86, - 543.07 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p494-b39", - "global_id": 13934, - "bbox": [ - 193.18, - 551.23, - 205.01, - 559.45 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p494-b40", - "global_id": 13935, - "bbox": [ - 192.64, - 567.48, - 198.86, - 575.45 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p494-b41", - "global_id": 13936, - "bbox": [ - 207.83, - 578.04, - 263.95, - 587.01 - ], - "text": "Figure P4.4-11", - "type": "text" - }, - { - "block_id": "p494-b42", - "global_id": 13937, - "bbox": [ - 57.59, - 612.82, - 263.24, - 632.82 - ], - "text": "4.4-12\nShow that the transfer function that relates the\noutput voltage y(t) to the input voltage x(t) for", - "type": "text" - }, - { - "block_id": "p494-b43", - "global_id": 13938, - "bbox": [ - 317.86, - 451.15, - 484.89, - 460.12 - ], - "text": "the op-amp circuit in Fig. P4.4-12a is given by", - "type": "text" - }, - { - "block_id": "p494-b44", - "global_id": 13939, - "bbox": [ - 352.57, - 471.47, - 395.62, - 486.72 - ], - "text": "H(s) = Ka", - "type": "text" - }, - { - "block_id": "p494-b45", - "global_id": 13940, - "bbox": [ - 381.5, - 477.85, - 440.28, - 493.09 - ], - "text": "s + a\nwhere", - "type": "text" - }, - { - "block_id": "p494-b46", - "global_id": 13941, - "bbox": [ - 352.57, - 494.95, - 393.9, - 510.3 - ], - "text": "K = 1 + Rb", - "type": "text" - }, - { - "block_id": "p494-b47", - "global_id": 13942, - "bbox": [ - 385.18, - 507.61, - 393.9, - 517.34 - ], - "text": "Ra", - "type": "text" - }, - { - "block_id": "p494-b48", - "global_id": 13943, - "bbox": [ - 404.56, - 495.05, - 450.9, - 510.3 - ], - "text": "and\na = 1", - "type": "text" - }, - { - "block_id": "p494-b49", - "global_id": 13944, - "bbox": [ - 442.84, - 507.61, - 454.3, - 516.58 - ], - "text": "RC", - "type": "text" - }, - { - "block_id": "p494-b50", - "global_id": 13945, - "bbox": [ - 317.86, - 527.72, - 490.38, - 547.64 - ], - "text": "and that the transfer function for the circuit in\nFig. P4.4-12b is given by", - "type": "text" - }, - { - "block_id": "p494-b51", - "global_id": 13946, - "bbox": [ - 380.19, - 559.0, - 426.87, - 580.62 - ], - "text": "H(s) =\nKs\ns + a", - "type": "text" - }, - { - "block_id": "p494-b52", - "global_id": 13947, - "bbox": [ - 284.74, - 601.86, - 490.38, - 632.82 - ], - "text": "4.4-13\nFor\nthe\nsecond-order\nop-amp\ncircuit\nin\nFig. P4.4-13, show that the transfer function\nH(s) relating the output voltage y(t) to the input", - "type": "text" - } - ] - }, - { - "page_num": 495, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p495-b0", - "global_id": 13948, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n475", - "type": "text" - }, - { - "block_id": "p495-b1", - "global_id": 13949, - "bbox": [ - 210.1, - 165.17, - 405.35, - 173.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p495-b2", - "global_id": 13950, - "bbox": [ - 358.46, - 87.98, - 363.79, - 95.98 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p495-b3", - "global_id": 13951, - "bbox": [ - 374.16, - 121.22, - 379.05, - 129.22 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p495-b4", - "global_id": 13952, - "bbox": [ - 416.79, - 97.76, - 481.71, - 115.35 - ], - "text": "+\n–", - "type": "text" - }, - { - "block_id": "p495-b6", - "global_id": 13953, - "bbox": [ - 187.33, - 118.84, - 484.09, - 141.85 - ], - "text": "x(t)\ny(t)\nRa\nRb\nC", - "type": "text" - }, - { - "block_id": "p495-b7", - "global_id": 13954, - "bbox": [ - 172.92, - 88.14, - 177.81, - 96.14 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p495-b8", - "global_id": 13955, - "bbox": [ - 130.8, - 118.84, - 142.38, - 126.92 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p495-b11", - "global_id": 13956, - "bbox": [ - 230.77, - 123.54, - 297.97, - 142.5 - ], - "text": "y(t)\nRa\nRb", - "type": "text" - }, - { - "block_id": "p495-b12", - "global_id": 13957, - "bbox": [ - 231.56, - 97.78, - 236.35, - 115.24 - ], - "text": "+\n–", - "type": "text" - }, - { - "block_id": "p495-b13", - "global_id": 13958, - "bbox": [ - 130.58, - 179.91, - 186.69, - 188.87 - ], - "text": "Figure P4.4-12", - "type": "text" - }, - { - "block_id": "p495-b14", - "global_id": 13959, - "bbox": [ - 171.64, - 285.1, - 182.74, - 293.18 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p495-b15", - "global_id": 13960, - "bbox": [ - 216.88, - 245.53, - 286.7, - 253.83 - ], - "text": "1 \t\n1", - "type": "text" - }, - { - "block_id": "p495-b18", - "global_id": 13961, - "bbox": [ - 405.04, - 290.55, - 416.14, - 298.63 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p495-b19", - "global_id": 13962, - "bbox": [ - 321.13, - 210.81, - 331.61, - 224.87 - ], - "text": "1\n3 F", - "type": "text" - }, - { - "block_id": "p495-b20", - "global_id": 13963, - "bbox": [ - 343.92, - 243.33, - 356.17, - 257.09 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p495-b21", - "global_id": 13964, - "bbox": [ - 246.04, - 258.72, - 334.57, - 283.27 - ], - "text": "1\n6 F\n–\n+", - "type": "text" - }, - { - "block_id": "p495-b22", - "global_id": 13965, - "bbox": [ - 430.37, - 324.37, - 486.49, - 333.33 - ], - "text": "Figure P4.4-13", - "type": "text" - }, - { - "block_id": "p495-b23", - "global_id": 13966, - "bbox": [ - 101.74, - 398.04, - 113.6, - 415.38 - ], - "text": "+\nx(t)", - "type": "text" - }, - { - "block_id": "p495-b24", - "global_id": 13967, - "bbox": [ - 104.56, - 416.18, - 110.78, - 424.15 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p495-b25", - "global_id": 13968, - "bbox": [ - 144.42, - 385.92, - 215.68, - 399.78 - ], - "text": "2\n−", - "type": "text" - }, - { - "block_id": "p495-b26", - "global_id": 13969, - "bbox": [ - 211.02, - 411.81, - 215.68, - 417.78 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p495-b27", - "global_id": 13970, - "bbox": [ - 207.41, - 361.65, - 215.82, - 369.62 - ], - "text": "1F", - "type": "text" - }, - { - "block_id": "p495-b28", - "global_id": 13971, - "bbox": [ - 270.02, - 390.65, - 280.4, - 402.83 - ], - "text": "1\n3", - "type": "text" - }, - { - "block_id": "p495-b29", - "global_id": 13972, - "bbox": [ - 325.32, - 402.81, - 329.98, - 408.78 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p495-b30", - "global_id": 13973, - "bbox": [ - 325.32, - 420.81, - 329.98, - 426.78 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p495-b31", - "global_id": 13974, - "bbox": [ - 270.02, - 338.51, - 280.4, - 350.69 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p495-b32", - "global_id": 13975, - "bbox": [ - 338.82, - 374.76, - 349.01, - 383.06 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p495-b33", - "global_id": 13976, - "bbox": [ - 499.54, - 358.32, - 511.38, - 375.67 - ], - "text": "+\ny(t)", - "type": "text" - }, - { - "block_id": "p495-b34", - "global_id": 13977, - "bbox": [ - 502.36, - 376.7, - 508.58, - 384.67 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p495-b35", - "global_id": 13978, - "bbox": [ - 398.22, - 403.84, - 444.28, - 417.78 - ], - "text": "3\n−", - "type": "text" - }, - { - "block_id": "p495-b36", - "global_id": 13979, - "bbox": [ - 439.62, - 429.81, - 444.28, - 435.78 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p495-b37", - "global_id": 13980, - "bbox": [ - 453.12, - 383.76, - 463.31, - 392.06 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p495-b38", - "global_id": 13981, - "bbox": [ - 144.92, - 419.51, - 155.3, - 431.69 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p495-b39", - "global_id": 13982, - "bbox": [ - 107.67, - 479.96, - 163.78, - 488.93 - ], - "text": "Figure P4.4-14", - "type": "text" - }, - { - "block_id": "p495-b40", - "global_id": 13983, - "bbox": [ - 116.46, - 503.16, - 199.5, - 512.5 - ], - "text": "voltage x(t) is given by", - "type": "text" - }, - { - "block_id": "p495-b41", - "global_id": 13984, - "bbox": [ - 165.8, - 525.14, - 238.46, - 547.13 - ], - "text": "H(s) =\n−s\ns2 + 8s + 12", - "type": "text" - }, - { - "block_id": "p495-b42", - "global_id": 13985, - "bbox": [ - 83.34, - 570.83, - 272.89, - 579.87 - ], - "text": "4.4-14\nConsider the op-amp circuit of Fig. P4.4-14.", - "type": "text" - }, - { - "block_id": "p495-b43", - "global_id": 13986, - "bbox": [ - 116.96, - 581.87, - 288.96, - 590.83 - ], - "text": "(a) Determine the standard-form transfer func-", - "type": "text" - }, - { - "block_id": "p495-b44", - "global_id": 13987, - "bbox": [ - 116.46, - 592.46, - 288.97, - 612.75 - ], - "text": "tion H(s) of this system.\n(b) Determine the standard-form constant coef-", - "type": "text" - }, - { - "block_id": "p495-b45", - "global_id": 13988, - "bbox": [ - 131.89, - 614.75, - 288.98, - 634.67 - ], - "text": "ficient linear differential equation descrip-\ntion of this circuit.", - "type": "text" - }, - { - "block_id": "p495-b46", - "global_id": 13989, - "bbox": [ - 344.11, - 503.49, - 516.14, - 512.45 - ], - "text": "(c) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p495-b47", - "global_id": 13990, - "bbox": [ - 343.61, - 514.07, - 516.14, - 545.32 - ], - "text": "mine the circuit’s zero-state response yzsr(t)\nto the input x(t) = e2tu(t + 1).\n(d) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p495-b48", - "global_id": 13991, - "bbox": [ - 359.05, - 546.94, - 516.14, - 578.2 - ], - "text": "mine the circuit’s zero-input response yzir(t)\nif the t = 0−capacitor voltage (first op-amp\noutput voltage) is 3 volts.", - "type": "text" - }, - { - "block_id": "p495-b49", - "global_id": 13992, - "bbox": [ - 310.48, - 590.9, - 516.14, - 610.91 - ], - "text": "4.4-15\nWe desire the op-amp circuit of Fig. P4.4-15 to\nbehave as ˙y(t)−1.5y(t) = −3˙x(t)+0.75x(t).", - "type": "text" - }, - { - "block_id": "p495-b50", - "global_id": 13993, - "bbox": [ - 344.12, - 612.8, - 516.13, - 623.2 - ], - "text": "(a) Determine resistors R1, R2, and R3 so that", - "type": "text" - }, - { - "block_id": "p495-b51", - "global_id": 13994, - "bbox": [ - 359.05, - 623.85, - 516.12, - 632.82 - ], - "text": "the circuit’s input–output behavior follows", - "type": "text" - } - ] - }, - { - "page_num": 496, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p496-b0", - "global_id": 13995, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "476\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p496-b1", - "global_id": 13996, - "bbox": [ - 98.9, - 136.23, - 110.77, - 153.58 - ], - "text": "+\nx(t)", - "type": "text" - }, - { - "block_id": "p496-b2", - "global_id": 13997, - "bbox": [ - 101.73, - 154.37, - 107.94, - 162.34 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p496-b3", - "global_id": 13998, - "bbox": [ - 142.51, - 123.17, - 150.36, - 131.94 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p496-b4", - "global_id": 13999, - "bbox": [ - 208.19, - 132.01, - 212.85, - 137.98 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p496-b5", - "global_id": 14000, - "bbox": [ - 208.19, - 150.01, - 212.85, - 155.98 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p496-b6", - "global_id": 14001, - "bbox": [ - 197.86, - 98.01, - 219.71, - 106.31 - ], - "text": "100μF", - "type": "text" - }, - { - "block_id": "p496-b7", - "global_id": 14002, - "bbox": [ - 267.61, - 132.17, - 275.46, - 140.94 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p496-b8", - "global_id": 14003, - "bbox": [ - 322.49, - 141.01, - 327.15, - 146.98 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p496-b9", - "global_id": 14004, - "bbox": [ - 322.49, - 159.01, - 327.15, - 164.98 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p496-b10", - "global_id": 14005, - "bbox": [ - 267.61, - 79.97, - 275.46, - 88.74 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p496-b11", - "global_id": 14006, - "bbox": [ - 336.91, - 111.95, - 344.76, - 120.71 - ], - "text": "R3", - "type": "text" - }, - { - "block_id": "p496-b12", - "global_id": 14007, - "bbox": [ - 381.51, - 154.11, - 393.35, - 171.46 - ], - "text": "+\ny(t)", - "type": "text" - }, - { - "block_id": "p496-b13", - "global_id": 14008, - "bbox": [ - 384.33, - 172.5, - 390.54, - 180.47 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p496-b14", - "global_id": 14009, - "bbox": [ - 142.51, - 151.97, - 150.36, - 160.74 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p496-b15", - "global_id": 14010, - "bbox": [ - 104.83, - 206.41, - 160.95, - 215.38 - ], - "text": "Figure P4.4-15", - "type": "text" - }, - { - "block_id": "p496-b16", - "global_id": 14011, - "bbox": [ - 98.9, - 295.53, - 110.77, - 312.88 - ], - "text": "+\nx(t)", - "type": "text" - }, - { - "block_id": "p496-b17", - "global_id": 14012, - "bbox": [ - 101.73, - 313.67, - 107.94, - 321.64 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p496-b18", - "global_id": 14013, - "bbox": [ - 138.6, - 283.34, - 212.85, - 297.28 - ], - "text": "5 k\n−", - "type": "text" - }, - { - "block_id": "p496-b19", - "global_id": 14014, - "bbox": [ - 208.19, - 309.31, - 212.85, - 315.28 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p496-b20", - "global_id": 14015, - "bbox": [ - 197.86, - 257.31, - 219.71, - 265.61 - ], - "text": "100μF", - "type": "text" - }, - { - "block_id": "p496-b21", - "global_id": 14016, - "bbox": [ - 267.61, - 291.47, - 275.46, - 300.24 - ], - "text": "R2", - "type": "text" - }, - { - "block_id": "p496-b22", - "global_id": 14017, - "bbox": [ - 322.49, - 300.31, - 327.15, - 306.28 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p496-b23", - "global_id": 14018, - "bbox": [ - 322.49, - 318.31, - 327.15, - 324.28 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p496-b24", - "global_id": 14019, - "bbox": [ - 267.61, - 248.18, - 352.01, - 259.35 - ], - "text": "R1\n100μF", - "type": "text" - }, - { - "block_id": "p496-b25", - "global_id": 14020, - "bbox": [ - 378.0, - 301.34, - 427.05, - 315.28 - ], - "text": "5 k\n−", - "type": "text" - }, - { - "block_id": "p496-b26", - "global_id": 14021, - "bbox": [ - 422.39, - 327.31, - 427.05, - 333.28 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p496-b27", - "global_id": 14022, - "bbox": [ - 432.9, - 281.18, - 449.06, - 289.48 - ], - "text": "5 k", - "type": "text" - }, - { - "block_id": "p496-b28", - "global_id": 14023, - "bbox": [ - 381.51, - 245.01, - 393.35, - 262.36 - ], - "text": "+\ny(t)", - "type": "text" - }, - { - "block_id": "p496-b29", - "global_id": 14024, - "bbox": [ - 384.33, - 263.4, - 390.54, - 271.37 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p496-b30", - "global_id": 14025, - "bbox": [ - 138.6, - 321.14, - 275.46, - 337.98 - ], - "text": "5 k\nR3", - "type": "text" - }, - { - "block_id": "p496-b31", - "global_id": 14026, - "bbox": [ - 104.83, - 378.31, - 160.95, - 387.28 - ], - "text": "Figure P4.4-16", - "type": "text" - }, - { - "block_id": "p496-b32", - "global_id": 14027, - "bbox": [ - 90.72, - 409.72, - 263.25, - 440.98 - ], - "text": "the desired differential equation of ˙y(t) −\n1.5y(t) = −3˙x(t) + 0.75x(t).\n(b) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p496-b33", - "global_id": 14028, - "bbox": [ - 91.22, - 442.6, - 263.25, - 484.82 - ], - "text": "mine the circuit’s zero-input response yzir(t)\nif the t = 0 capacitor voltage (first op-amp\noutput voltage) is 2 volts.\n(c) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p496-b34", - "global_id": 14029, - "bbox": [ - 90.72, - 486.44, - 263.25, - 517.69 - ], - "text": "mine the impulse response h(t) of this\ncircuit.\n(d) Determine the circuit’s zero-state response", - "type": "text" - }, - { - "block_id": "p496-b35", - "global_id": 14030, - "bbox": [ - 106.16, - 519.31, - 227.22, - 529.39 - ], - "text": "yzsr(t) to the input x(t) = u(t −2).", - "type": "text" - }, - { - "block_id": "p496-b36", - "global_id": 14031, - "bbox": [ - 57.59, - 536.11, - 263.25, - 556.1 - ], - "text": "4.4-16\nWe desire the op-amp circuit of Fig. P4.4-16 to\nbehave as", - "type": "text" - }, - { - "block_id": "p496-b37", - "global_id": 14032, - "bbox": [ - 127.58, - 539.55, - 137.53, - 548.51 - ], - "text": "$ $", - "type": "text" - }, - { - "block_id": "p496-b38", - "global_id": 14033, - "bbox": [ - 140.27, - 545.41, - 167.56, - 556.01 - ], - "text": "y(t)+ 2", - "type": "text" - }, - { - "block_id": "p496-b39", - "global_id": 14034, - "bbox": [ - 164.32, - 539.55, - 173.86, - 558.69 - ], - "text": "5\n$", - "type": "text" - }, - { - "block_id": "p496-b40", - "global_id": 14035, - "bbox": [ - 176.6, - 545.41, - 203.89, - 556.01 - ], - "text": "y(t)+ 1", - "type": "text" - }, - { - "block_id": "p496-b41", - "global_id": 14036, - "bbox": [ - 200.65, - 539.55, - 238.89, - 558.69 - ], - "text": "5y(t) =\n$ $", - "type": "text" - }, - { - "block_id": "p496-b42", - "global_id": 14037, - "bbox": [ - 90.72, - 546.76, - 263.25, - 559.47 - ], - "text": "x(t)−\n$", - "type": "text" - }, - { - "block_id": "p496-b43", - "global_id": 14038, - "bbox": [ - 91.22, - 557.73, - 263.24, - 579.37 - ], - "text": "x(t)\n(a) Determine the resistors R1, R2, and R3 to", - "type": "text" - }, - { - "block_id": "p496-b44", - "global_id": 14039, - "bbox": [ - 90.72, - 580.01, - 263.25, - 599.94 - ], - "text": "produce the desired behavior.\n(b) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p496-b45", - "global_id": 14040, - "bbox": [ - 106.15, - 601.57, - 263.25, - 632.82 - ], - "text": "mine the circuit’s zero-input response yzir(t)\nif the t = 0 capacitor voltages (first two\nop-amp outputs) are each 1 volt.", - "type": "text" - }, - { - "block_id": "p496-b46", - "global_id": 14041, - "bbox": [ - 284.74, - 410.08, - 490.37, - 419.12 - ], - "text": "4.4-17\n(a) Using the initial and final value theorems,", - "type": "text" - }, - { - "block_id": "p496-b47", - "global_id": 14042, - "bbox": [ - 333.31, - 421.12, - 490.39, - 452.0 - ], - "text": "find the initial and final value of the\nzero-state response of a system with the\ntransfer function", - "type": "text" - }, - { - "block_id": "p496-b48", - "global_id": 14043, - "bbox": [ - 372.68, - 468.96, - 449.82, - 487.84 - ], - "text": "H(s) = 6s2 + 3s + 10", - "type": "text" - }, - { - "block_id": "p496-b49", - "global_id": 14044, - "bbox": [ - 403.85, - 482.28, - 447.58, - 494.21 - ], - "text": "2s2 + 6s + 5", - "type": "text" - }, - { - "block_id": "p496-b50", - "global_id": 14045, - "bbox": [ - 317.86, - 513.57, - 486.95, - 533.87 - ], - "text": "and input x(t) = u(t).\n(b) Repeat part (a) for the input x(t) = e−tu(t).", - "type": "text" - }, - { - "block_id": "p496-b51", - "global_id": 14046, - "bbox": [ - 318.37, - 531.88, - 486.88, - 550.76 - ], - "text": "(c) Find y(0+) and y(∞) if Y(s) = s2 + 5s + 6", - "type": "text" - }, - { - "block_id": "p496-b52", - "global_id": 14047, - "bbox": [ - 447.63, - 541.79, - 490.32, - 557.13 - ], - "text": "s2 + 3s + 2.", - "type": "text" - }, - { - "block_id": "p496-b53", - "global_id": 14048, - "bbox": [ - 317.86, - 555.29, - 490.39, - 576.38 - ], - "text": "(d) Find\ny(0+)\nand\ny(∞)\nif\nY(s)\n=\ns3 + 4s2 + 10s + 7", - "type": "text" - }, - { - "block_id": "p496-b54", - "global_id": 14049, - "bbox": [ - 347.48, - 573.7, - 403.16, - 589.03 - ], - "text": "s2 + 2s + 3\n.", - "type": "text" - }, - { - "block_id": "p496-b55", - "global_id": 14050, - "bbox": [ - 289.22, - 597.92, - 490.39, - 634.02 - ], - "text": "4.5-1\nConsider two LTIC systems. The first has trans-\nfer function H1(s) =\n2s\ns+1, and the second has\ntransfer function H2(s) =\n1\nse3(s−1) .", - "type": "text" - } - ] - }, - { - "page_num": 497, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p497-b0", - "global_id": 14051, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n477", - "type": "text" - }, - { - "block_id": "p497-b1", - "global_id": 14052, - "bbox": [ - 201.67, - 165.74, - 210.55, - 173.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p497-b2", - "global_id": 14053, - "bbox": [ - 142.47, - 87.35, - 173.24, - 97.18 - ], - "text": "R1 2", - "type": "text" - }, - { - "block_id": "p497-b3", - "global_id": 14054, - "bbox": [ - 143.54, - 127.05, - 255.28, - 136.87 - ], - "text": "R2 2 \t\nR4 1", - "type": "text" - }, - { - "block_id": "p497-b4", - "global_id": 14055, - "bbox": [ - 225.47, - 87.35, - 256.25, - 97.17 - ], - "text": "R3 1", - "type": "text" - }, - { - "block_id": "p497-b5", - "global_id": 14056, - "bbox": [ - 378.62, - 165.74, - 388.42, - 173.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p497-b6", - "global_id": 14057, - "bbox": [ - 343.55, - 89.3, - 355.78, - 97.6 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p497-b7", - "global_id": 14058, - "bbox": [ - 353.9, - 127.76, - 431.54, - 136.35 - ], - "text": "2 \t\n1", - "type": "text" - }, - { - "block_id": "p497-b8", - "global_id": 14059, - "bbox": [ - 405.66, - 89.3, - 417.88, - 97.6 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p497-b9", - "global_id": 14060, - "bbox": [ - 300.84, - 124.66, - 466.2, - 133.69 - ], - "text": "x(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p497-b12", - "global_id": 14061, - "bbox": [ - 130.58, - 179.91, - 182.22, - 188.87 - ], - "text": "Figure P4.5-2", - "type": "text" - }, - { - "block_id": "p497-b13", - "global_id": 14062, - "bbox": [ - 116.96, - 200.36, - 288.98, - 209.32 - ], - "text": "(a) Determine the overall impulse response", - "type": "text" - }, - { - "block_id": "p497-b14", - "global_id": 14063, - "bbox": [ - 116.46, - 210.95, - 288.98, - 242.2 - ], - "text": "hs(t) if the two systems are connected in\nseries.\n(b) Determine the overall impulse response", - "type": "text" - }, - { - "block_id": "p497-b15", - "global_id": 14064, - "bbox": [ - 131.89, - 243.82, - 288.98, - 264.12 - ], - "text": "hp(t) if the two systems are connected in\nparallel.", - "type": "text" - }, - { - "block_id": "p497-b16", - "global_id": 14065, - "bbox": [ - 87.82, - 278.13, - 288.99, - 331.01 - ], - "text": "4.5-2\nFigure P4.5-2a shows two resistive ladder seg-\nments. The transfer function of each segment\n(ratio of output to input voltage) is 1/2.\nFigure P4.5-2b shows these two segments con-\nnected in cascade.", - "type": "text" - }, - { - "block_id": "p497-b17", - "global_id": 14066, - "bbox": [ - 116.96, - 333.0, - 288.96, - 341.97 - ], - "text": "(a) Is the transfer function (ratio of output", - "type": "text" - }, - { - "block_id": "p497-b18", - "global_id": 14067, - "bbox": [ - 116.46, - 343.96, - 288.98, - 374.85 - ], - "text": "to input voltage) of this cascaded network\n(1/2)(1/2) = 1/4?\n(b) If your answer is affirmative, verify the", - "type": "text" - }, - { - "block_id": "p497-b19", - "global_id": 14068, - "bbox": [ - 116.96, - 376.84, - 288.99, - 419.64 - ], - "text": "answer by direct computation of the transfer\nfunction. Does this computation confirm the\nearlier value 1/4? If not, why?\n(c) Repeat the problem with R3 = R4 = 20 k.", - "type": "text" - }, - { - "block_id": "p497-b20", - "global_id": 14069, - "bbox": [ - 131.89, - 420.67, - 288.99, - 440.6 - ], - "text": "Does this result suggest the answer to the\nproblem in part (b)?", - "type": "text" - }, - { - "block_id": "p497-b21", - "global_id": 14070, - "bbox": [ - 87.82, - 454.61, - 288.99, - 529.4 - ], - "text": "4.5-3\nIn communication channels, transmitted signal\nis propagated simultaneously by several paths of\nvarying lengths. This causes the signal to reach\nthe destination with varying time delays and\nvarying gains. Such a system generally distorts\nthe received signal. For error-free communica-\ntion, it is necessary to undo this distortion as", - "type": "text" - }, - { - "block_id": "p497-b22", - "global_id": 14071, - "bbox": [ - 343.61, - 200.36, - 516.13, - 220.29 - ], - "text": "much as possible by using the system that is\ninverse of the channel model.", - "type": "text" - }, - { - "block_id": "p497-b23", - "global_id": 14072, - "bbox": [ - 453.79, - 267.48, - 457.79, - 275.48 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p497-b24", - "global_id": 14073, - "bbox": [ - 355.41, - 234.37, - 498.51, - 242.45 - ], - "text": "y(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p497-b25", - "global_id": 14074, - "bbox": [ - 402.87, - 241.31, - 474.63, - 251.31 - ], - "text": "Delay T", - "type": "text" - }, - { - "block_id": "p497-b26", - "global_id": 14075, - "bbox": [ - 395.77, - 274.64, - 437.53, - 282.93 - ], - "text": "Delay T t", - "type": "text" - }, - { - "block_id": "p497-b27", - "global_id": 14076, - "bbox": [ - 343.61, - 294.6, - 395.25, - 303.56 - ], - "text": "Figure P4.5-3", - "type": "text" - }, - { - "block_id": "p497-b28", - "global_id": 14077, - "bbox": [ - 343.61, - 319.95, - 516.14, - 504.26 - ], - "text": "For simplicity, let us assume that a signal is\npropagated by two paths whose time delays dif-\nfer by τ seconds. The channel over the intended\npath has a delay of T seconds and unity gain.\nThe signal over the unintended path has a delay\nof T + τ seconds and gain a. Such a channel\ncan be modeled, as shown in Fig. P4.5-3. Find\nthe inverse system transfer function to correct\nthe delay distortion and show that the inverse\nsystem can be realized by a feedback system.\nThe inverse system should be causal to be\nrealizable. [Hint: We want to correct only the\ndistortion caused by the relative delay τ seconds.\nFor distortionless transmission, the signal may\nbe delayed. What is important is to maintain the\nshape of x(t). Thus, a received signal of the form\ncx(t −T) is considered to be distortionless.]", - "type": "text" - }, - { - "block_id": "p497-b29", - "global_id": 14078, - "bbox": [ - 314.97, - 509.4, - 516.14, - 529.41 - ], - "text": "4.5-4\nDiscuss BIBO stability of the feedback systems\ndepicted in Fig. P4.5-4. For the system in", - "type": "text" - }, - { - "block_id": "p497-b30", - "global_id": 14079, - "bbox": [ - 208.54, - 614.19, - 405.51, - 622.19 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p497-b31", - "global_id": 14080, - "bbox": [ - 139.04, - 559.16, - 282.98, - 567.24 - ], - "text": "y(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p497-b32", - "global_id": 14081, - "bbox": [ - 212.96, - 595.18, - 216.96, - 603.18 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p497-b34", - "global_id": 14082, - "bbox": [ - 161.69, - 562.64, - 226.42, - 579.11 - ], - "text": "1\ns 1", - "type": "text" - }, - { - "block_id": "p497-b35", - "global_id": 14083, - "bbox": [ - 322.44, - 559.16, - 474.38, - 567.24 - ], - "text": "y(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p497-b37", - "global_id": 14084, - "bbox": [ - 345.1, - 562.48, - 429.43, - 579.06 - ], - "text": "K\ns(s 2)(s 4)", - "type": "text" - }, - { - "block_id": "p497-b38", - "global_id": 14085, - "bbox": [ - 130.58, - 628.36, - 182.22, - 637.33 - ], - "text": "Figure P4.5-4", - "type": "text" - } - ] - }, - { - "page_num": 498, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p498-b0", - "global_id": 14086, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "478\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p498-b1", - "global_id": 14087, - "bbox": [ - 90.72, - 85.64, - 263.24, - 105.94 - ], - "text": "Fig. P4.5-4b, consider three cases: (a) K = 10,\n(b) K = 50, and (c) K = 48.", - "type": "text" - }, - { - "block_id": "p498-b2", - "global_id": 14088, - "bbox": [ - 62.08, - 110.84, - 117.61, - 119.88 - ], - "text": "4.6-1\nRealize", - "type": "text" - }, - { - "block_id": "p498-b3", - "global_id": 14089, - "bbox": [ - 125.26, - 127.11, - 227.53, - 149.11 - ], - "text": "H(s) =\ns(s + 2)\n(s + 1)(s + 3)(s + 4)", - "type": "text" - }, - { - "block_id": "p498-b4", - "global_id": 14090, - "bbox": [ - 90.72, - 158.11, - 249.33, - 167.08 - ], - "text": "by canonic direct, series, and parallel forms.", - "type": "text" - }, - { - "block_id": "p498-b5", - "global_id": 14091, - "bbox": [ - 62.08, - 171.98, - 263.25, - 202.94 - ], - "text": "4.6-2\nRealize the transfer function in Prob. 4.6-1 by\nusing the transposed form of the realizations\nfound in Prob. 4.6-1.", - "type": "text" - }, - { - "block_id": "p498-b6", - "global_id": 14092, - "bbox": [ - 62.08, - 207.84, - 170.29, - 216.89 - ], - "text": "4.6-3\nRepeat Prob. 4.6-1 for", - "type": "text" - }, - { - "block_id": "p498-b7", - "global_id": 14093, - "bbox": [ - 91.22, - 216.65, - 205.46, - 238.64 - ], - "text": "(a) H(s) =\n3s(s + 2)\n(s + 1)(s2 + 2s + 2)", - "type": "text" - }, - { - "block_id": "p498-b8", - "global_id": 14094, - "bbox": [ - 62.08, - 237.96, - 263.22, - 293.64 - ], - "text": "(b) H(s) =\n2s −4\n(s + 2)(s2 + 4)\n4.6-4\nRealize the transfer functions in Prob. 4.6-3 by\nusing the transposed form of the realizations\nfound in Prob. 4.6-3.", - "type": "text" - }, - { - "block_id": "p498-b9", - "global_id": 14095, - "bbox": [ - 62.08, - 298.55, - 170.29, - 307.59 - ], - "text": "4.6-5\nRepeat Prob. 4.6-1 for", - "type": "text" - }, - { - "block_id": "p498-b10", - "global_id": 14096, - "bbox": [ - 131.62, - 316.6, - 221.14, - 338.58 - ], - "text": "H(s) =\n2s + 3\n5s(s + 2)2(s + 3)", - "type": "text" - }, - { - "block_id": "p498-b11", - "global_id": 14097, - "bbox": [ - 62.08, - 347.63, - 263.25, - 378.59 - ], - "text": "4.6-6\nRealize the transfer function in Prob. 4.6-5 by\nusing the transposed form of the realizations\nfound in Prob. 4.6-5.", - "type": "text" - }, - { - "block_id": "p498-b12", - "global_id": 14098, - "bbox": [ - 62.08, - 383.5, - 170.29, - 392.54 - ], - "text": "4.6-7\nRepeat Prob. 4.6-1 for", - "type": "text" - }, - { - "block_id": "p498-b13", - "global_id": 14099, - "bbox": [ - 125.25, - 401.6, - 227.53, - 423.59 - ], - "text": "H(s) =\ns(s + 1)(s + 2)\n(s + 5)(s + 6)(s + 8)", - "type": "text" - }, - { - "block_id": "p498-b14", - "global_id": 14100, - "bbox": [ - 62.08, - 432.64, - 263.25, - 463.59 - ], - "text": "4.6-8\nRealize the transfer function in Prob. 4.6-7 by\nusing the transposed form of the realizations\nfound in Prob. 4.6-7.", - "type": "text" - }, - { - "block_id": "p498-b15", - "global_id": 14101, - "bbox": [ - 62.08, - 468.5, - 170.29, - 477.54 - ], - "text": "4.6-9\nRepeat Prob. 4.6-1 for", - "type": "text" - }, - { - "block_id": "p498-b16", - "global_id": 14102, - "bbox": [ - 123.38, - 487.12, - 193.97, - 503.64 - ], - "text": "H(s) =\ns3", - "type": "text" - }, - { - "block_id": "p498-b17", - "global_id": 14103, - "bbox": [ - 152.32, - 500.43, - 229.4, - 510.1 - ], - "text": "(s + 1)2(s + 2)(s + 3)", - "type": "text" - }, - { - "block_id": "p498-b18", - "global_id": 14104, - "bbox": [ - 57.59, - 519.14, - 263.23, - 550.1 - ], - "text": "4.6-10\nRealize the transfer function in Prob. 4.6-9 by\nusing the transposed form of the realizations\nfound in Prob. 4.6-9.", - "type": "text" - }, - { - "block_id": "p498-b19", - "global_id": 14105, - "bbox": [ - 57.59, - 555.01, - 170.29, - 564.05 - ], - "text": "4.6-11\nRepeat Prob. 4.6-1 for", - "type": "text" - }, - { - "block_id": "p498-b20", - "global_id": 14106, - "bbox": [ - 124.49, - 573.62, - 193.97, - 590.15 - ], - "text": "H(s) =\ns3", - "type": "text" - }, - { - "block_id": "p498-b21", - "global_id": 14107, - "bbox": [ - 153.41, - 584.68, - 228.27, - 596.61 - ], - "text": "(s + 1)(s2 + 4s + 13)", - "type": "text" - }, - { - "block_id": "p498-b22", - "global_id": 14108, - "bbox": [ - 57.59, - 605.65, - 263.22, - 636.61 - ], - "text": "4.6-12\nRealize the transfer function in Prob. 4.6-11 by\nusing the transposed form of the realizations\nfound in Prob. 4.6-11.", - "type": "text" - }, - { - "block_id": "p498-b23", - "global_id": 14109, - "bbox": [ - 284.74, - 85.94, - 490.39, - 105.94 - ], - "text": "4.6-13\nDraw a TDFII block realization of a causal\nLTIC system with transfer function H(s) =", - "type": "text" - }, - { - "block_id": "p498-b24", - "global_id": 14110, - "bbox": [ - 317.86, - 105.36, - 490.4, - 127.86 - ], - "text": "(s−2j)(s+2j)\n(s−j)(s+j)(s+2). Give two reasons why TDFII\ntends to be a good structure.", - "type": "text" - }, - { - "block_id": "p498-b25", - "global_id": 14111, - "bbox": [ - 284.74, - 132.77, - 490.38, - 163.72 - ], - "text": "4.6-14\nConsider a causal LTIC system with transfer\nfunction H(s) =\n(s−2j)(s+2j)(s−3j)(s+3j)\n9(s+1)(s+2)(s+1−j)(s+1+j).\n(a) Realize H(s) using a single fourth-order real", - "type": "text" - }, - { - "block_id": "p498-b26", - "global_id": 14112, - "bbox": [ - 317.86, - 165.72, - 490.39, - 196.6 - ], - "text": "TDFII structure. Is this block realization\nunique? Explain.\n(b) Realize H(s) using a cascade of second-", - "type": "text" - }, - { - "block_id": "p498-b27", - "global_id": 14113, - "bbox": [ - 318.37, - 198.6, - 490.37, - 229.48 - ], - "text": "order real DFII structures. Is this block\nrealization unique? Explain.\n(c) Realize H(s) using a parallel connection", - "type": "text" - }, - { - "block_id": "p498-b28", - "global_id": 14114, - "bbox": [ - 333.31, - 231.47, - 490.4, - 251.39 - ], - "text": "of second-order real DFI structures. Is this\nblock realization unique? Explain.", - "type": "text" - }, - { - "block_id": "p498-b29", - "global_id": 14115, - "bbox": [ - 284.74, - 256.3, - 490.39, - 320.14 - ], - "text": "4.6-15\nIn this problem we show how a pair of complex\nconjugate poles may be realized by using a\ncascade of two first-order transfer functions and\nfeedback. Show that the transfer functions of the\nblock diagrams in Figs. P4.6-15a and P4.6-15b\nare:", - "type": "text" - }, - { - "block_id": "p498-b30", - "global_id": 14116, - "bbox": [ - 318.37, - 319.14, - 328.32, - 328.11 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p498-b31", - "global_id": 14117, - "bbox": [ - 359.22, - 335.94, - 436.8, - 357.47 - ], - "text": "Ha(s) =\n1\n(s + a)2 + b2", - "type": "text" - }, - { - "block_id": "p498-b32", - "global_id": 14118, - "bbox": [ - 317.86, - 360.25, - 463.27, - 395.14 - ], - "text": "=\n1\ns2 + 2as + (a2 + b2)\n(b)", - "type": "text" - }, - { - "block_id": "p498-b33", - "global_id": 14119, - "bbox": [ - 359.04, - 401.72, - 436.98, - 423.62 - ], - "text": "Hb(s) =\ns + a\n(s + a)2 + b2", - "type": "text" - }, - { - "block_id": "p498-b34", - "global_id": 14120, - "bbox": [ - 381.27, - 425.14, - 463.45, - 447.13 - ], - "text": "=\ns + a\ns2 + 2as + (a2 + b2)", - "type": "text" - }, - { - "block_id": "p498-b35", - "global_id": 14121, - "bbox": [ - 318.37, - 454.92, - 490.38, - 479.83 - ], - "text": "Hence, show that the transfer function of the\nblock diagram in Fig. P4.6-15c is\n(c)", - "type": "text" - }, - { - "block_id": "p498-b36", - "global_id": 14122, - "bbox": [ - 359.22, - 487.17, - 436.8, - 509.07 - ], - "text": "Hc(s) =\nAs + B\n(s + a)2 + b2", - "type": "text" - }, - { - "block_id": "p498-b37", - "global_id": 14123, - "bbox": [ - 381.09, - 511.34, - 463.27, - 533.33 - ], - "text": "=\nAs + B\ns2 + 2as + (a2 + b2)", - "type": "text" - }, - { - "block_id": "p498-b38", - "global_id": 14124, - "bbox": [ - 284.74, - 541.17, - 490.38, - 561.17 - ], - "text": "4.6-16\nShow op-amp realizations of the following\ntransfer functions:", - "type": "text" - }, - { - "block_id": "p498-b39", - "global_id": 14125, - "bbox": [ - 318.37, - 563.01, - 351.35, - 578.63 - ], - "text": "(a) −10", - "type": "text" - }, - { - "block_id": "p498-b40", - "global_id": 14126, - "bbox": [ - 334.5, - 575.65, - 352.26, - 584.99 - ], - "text": "s + 5", - "type": "text" - }, - { - "block_id": "p498-b41", - "global_id": 14127, - "bbox": [ - 317.86, - 588.86, - 352.26, - 610.48 - ], - "text": "(b)\n10\ns + 5", - "type": "text" - }, - { - "block_id": "p498-b42", - "global_id": 14128, - "bbox": [ - 318.37, - 613.97, - 352.26, - 629.59 - ], - "text": "(c) s + 2", - "type": "text" - }, - { - "block_id": "p498-b43", - "global_id": 14129, - "bbox": [ - 334.5, - 626.62, - 352.26, - 635.96 - ], - "text": "s + 5", - "type": "text" - } - ] - }, - { - "page_num": 499, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p499-b0", - "global_id": 14130, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n479", - "type": "text" - }, - { - "block_id": "p499-b1", - "global_id": 14131, - "bbox": [ - 202.89, - 188.27, - 211.77, - 196.27 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p499-b2", - "global_id": 14132, - "bbox": [ - 309.94, - 324.74, - 318.82, - 332.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p499-b3", - "global_id": 14133, - "bbox": [ - 399.87, - 188.27, - 409.68, - 196.27 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p499-b4", - "global_id": 14134, - "bbox": [ - 329.99, - 118.87, - 343.32, - 126.95 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p499-b5", - "global_id": 14135, - "bbox": [ - 445.01, - 88.18, - 457.9, - 96.26 - ], - "text": "Y(s)", - "type": "text" - }, - { - "block_id": "p499-b6", - "global_id": 14136, - "bbox": [ - 383.19, - 122.27, - 400.97, - 139.19 - ], - "text": "1\ns a", - "type": "text" - }, - { - "block_id": "p499-b7", - "global_id": 14137, - "bbox": [ - 439.19, - 122.27, - 456.97, - 139.19 - ], - "text": "1\ns a", - "type": "text" - }, - { - "block_id": "p499-b8", - "global_id": 14138, - "bbox": [ - 412.42, - 167.08, - 419.74, - 176.48 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p499-b10", - "global_id": 14139, - "bbox": [ - 132.56, - 118.68, - 275.56, - 126.86 - ], - "text": "Y(s)\nX(s)", - "type": "text" - }, - { - "block_id": "p499-b12", - "global_id": 14140, - "bbox": [ - 188.19, - 122.98, - 205.97, - 139.9 - ], - "text": "1\ns a", - "type": "text" - }, - { - "block_id": "p499-b13", - "global_id": 14141, - "bbox": [ - 226.69, - 122.98, - 244.47, - 139.9 - ], - "text": "1\ns a", - "type": "text" - }, - { - "block_id": "p499-b14", - "global_id": 14142, - "bbox": [ - 144.02, - 250.51, - 478.34, - 270.91 - ], - "text": "X(s)\nY(s)\n1\ns a", - "type": "text" - }, - { - "block_id": "p499-b15", - "global_id": 14143, - "bbox": [ - 295.99, - 253.99, - 400.1, - 270.91 - ], - "text": "1\ns a\nB aA", - "type": "text" - }, - { - "block_id": "p499-b17", - "global_id": 14144, - "bbox": [ - 337.94, - 216.57, - 342.82, - 224.57 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p499-b20", - "global_id": 14145, - "bbox": [ - 261.38, - 303.68, - 268.38, - 313.07 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p499-b21", - "global_id": 14146, - "bbox": [ - 212.92, - 167.08, - 220.24, - 176.48 - ], - "text": "b2", - "type": "text" - }, - { - "block_id": "p499-b22", - "global_id": 14147, - "bbox": [ - 130.58, - 338.91, - 186.69, - 347.88 - ], - "text": "Figure P4.6-15", - "type": "text" - }, - { - "block_id": "p499-b23", - "global_id": 14148, - "bbox": [ - 374.98, - 361.24, - 477.0, - 369.32 - ], - "text": "x(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p499-b24", - "global_id": 14149, - "bbox": [ - 433.32, - 399.49, - 443.32, - 407.49 - ], - "text": "0.9", - "type": "text" - }, - { - "block_id": "p499-b25", - "global_id": 14150, - "bbox": [ - 427.08, - 372.72, - 448.7, - 382.48 - ], - "text": "s vc", - "type": "text" - }, - { - "block_id": "p499-b26", - "global_id": 14151, - "bbox": [ - 135.46, - 361.71, - 441.81, - 372.06 - ], - "text": "vc\nx(t)\ny(t)\nx(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p499-b27", - "global_id": 14152, - "bbox": [ - 290.52, - 399.49, - 294.52, - 407.49 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p499-b28", - "global_id": 14153, - "bbox": [ - 162.74, - 417.88, - 442.76, - 425.88 - ], - "text": "(b)\n(c)\n(a)", - "type": "text" - }, - { - "block_id": "p499-b29", - "global_id": 14154, - "bbox": [ - 157.13, - 372.72, - 399.0, - 388.67 - ], - "text": "s vc", - "type": "text" - }, - { - "block_id": "p499-b30", - "global_id": 14155, - "bbox": [ - 164.02, - 362.49, - 171.86, - 372.06 - ], - "text": "vc", - "type": "text" - }, - { - "block_id": "p499-b31", - "global_id": 14156, - "bbox": [ - 284.19, - 372.72, - 305.8, - 382.48 - ], - "text": "s vc", - "type": "text" - }, - { - "block_id": "p499-b32", - "global_id": 14157, - "bbox": [ - 251.79, - 362.49, - 403.93, - 378.25 - ], - "text": "vc", - "type": "text" - }, - { - "block_id": "p499-b33", - "global_id": 14158, - "bbox": [ - 130.58, - 432.05, - 182.22, - 441.02 - ], - "text": "Figure P4.7-1", - "type": "text" - }, - { - "block_id": "p499-b34", - "global_id": 14159, - "bbox": [ - 83.34, - 452.87, - 288.97, - 472.86 - ], - "text": "4.6-17\nShow two different op-amp circuit realizations\nof the transfer function", - "type": "text" - }, - { - "block_id": "p499-b35", - "global_id": 14160, - "bbox": [ - 156.24, - 479.01, - 202.92, - 494.54 - ], - "text": "H(s) = s + 2", - "type": "text" - }, - { - "block_id": "p499-b36", - "global_id": 14161, - "bbox": [ - 185.17, - 479.38, - 248.01, - 501.0 - ], - "text": "s + 5 = 1 −\n3\ns + 5", - "type": "text" - }, - { - "block_id": "p499-b37", - "global_id": 14162, - "bbox": [ - 83.34, - 508.21, - 288.98, - 528.21 - ], - "text": "4.6-18\nShow an op-amp canonic direct realization of\nthe transfer function", - "type": "text" - }, - { - "block_id": "p499-b38", - "global_id": 14163, - "bbox": [ - 165.8, - 534.35, - 238.46, - 556.35 - ], - "text": "H(s) =\n3s + 7\ns2 + 4s + 10", - "type": "text" - }, - { - "block_id": "p499-b39", - "global_id": 14164, - "bbox": [ - 83.34, - 566.55, - 288.98, - 586.55 - ], - "text": "4.6-19\nShow an op-amp canonic direct realization of\nthe transfer function", - "type": "text" - }, - { - "block_id": "p499-b40", - "global_id": 14165, - "bbox": [ - 165.8, - 591.0, - 236.22, - 609.79 - ], - "text": "H(s) = s2 + 5s + 2", - "type": "text" - }, - { - "block_id": "p499-b41", - "global_id": 14166, - "bbox": [ - 194.72, - 604.31, - 238.46, - 616.24 - ], - "text": "s2 + 4s + 13", - "type": "text" - }, - { - "block_id": "p499-b42", - "global_id": 14167, - "bbox": [ - 83.34, - 623.46, - 288.96, - 643.46 - ], - "text": "4.6-20\nConsider a system described by a constant-\ncoefficient linear differential equation as d", - "type": "text" - }, - { - "block_id": "p499-b43", - "global_id": 14168, - "bbox": [ - 269.31, - 634.12, - 288.99, - 645.97 - ], - "text": "dty(t)", - "type": "text" - }, - { - "block_id": "p499-b44", - "global_id": 14169, - "bbox": [ - 345.19, - 452.86, - 415.99, - 463.62 - ], - "text": "+ 2y(t) = x(t) −3 d", - "type": "text" - }, - { - "block_id": "p499-b45", - "global_id": 14170, - "bbox": [ - 343.61, - 454.28, - 516.14, - 507.46 - ], - "text": "dtx(t). Draw an op-amp real-\nization of this system if resistors and inductors\nare available but not capacitors. Would using\ninductors rather than capacitors in this circuit\npose any problem? Explain.", - "type": "text" - }, - { - "block_id": "p499-b46", - "global_id": 14171, - "bbox": [ - 314.97, - 515.19, - 516.13, - 557.78 - ], - "text": "4.7-1\nFeedback can be used to increase (or decrease)\nthe system bandwidth. Consider the system\nin Fig. P4.7-1a with transfer function G(s) =\nωc/(s + ωc).", - "type": "text" - }, - { - "block_id": "p499-b47", - "global_id": 14172, - "bbox": [ - 344.11, - 559.1, - 516.13, - 568.06 - ], - "text": "(a) Show that the 3 dB bandwidth of this system", - "type": "text" - }, - { - "block_id": "p499-b48", - "global_id": 14173, - "bbox": [ - 343.61, - 569.68, - 516.14, - 600.94 - ], - "text": "is ωc and the dc gain is unity; that is,\n|H(j0)| = 1.\n(b) To increase the bandwidth of this system,", - "type": "text" - }, - { - "block_id": "p499-b49", - "global_id": 14174, - "bbox": [ - 359.05, - 602.56, - 516.13, - 644.77 - ], - "text": "we use negative feedback with H(s) = 9, as\ndepicted in Fig. P4.7-1b. Show that the 3 dB\nbandwidth of this system is 10ωc. What is\nthe dc gain?", - "type": "text" - } - ] - }, - { - "page_num": 500, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p500-b0", - "global_id": 14175, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "480\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p500-b1", - "global_id": 14176, - "bbox": [ - 91.22, - 85.9, - 263.23, - 94.86 - ], - "text": "(c) To decrease the bandwidth of this system,", - "type": "text" - }, - { - "block_id": "p500-b2", - "global_id": 14177, - "bbox": [ - 90.72, - 96.49, - 263.23, - 149.66 - ], - "text": "we use positive feedback with H(s) = −0.9,\nas illustrated in Fig. P4.7-1c. Show that the\n3 dB bandwidth of this system is ωc/10.\nWhat is the dc gain?\n(d) The system gain at dc times its 3 dB", - "type": "text" - }, - { - "block_id": "p500-b3", - "global_id": 14178, - "bbox": [ - 106.15, - 151.56, - 263.24, - 215.41 - ], - "text": "bandwidth is the gain-bandwidth product\nof a system. Show that this product is the\nsame for all the three systems in Fig. P4.7-1.\nThis result shows that if we increase the\nbandwidth, the gain decreases and vice\nversa.", - "type": "text" - }, - { - "block_id": "p500-b4", - "global_id": 14179, - "bbox": [ - 62.08, - 220.88, - 263.24, - 264.48 - ], - "text": "4.8-1\nSuppose an engineer builds a controllable,\nobservable LTIC system with transfer function\nH(s) =\ns2+4\n2s2+4s+4.\n(a) By direct calculation, compute the magni-", - "type": "text" - }, - { - "block_id": "p500-b5", - "global_id": 14180, - "bbox": [ - 90.72, - 266.1, - 263.24, - 319.28 - ], - "text": "tude response at frequencies ω = 0, 1, 2,\n3, 5, 10, and ∞. Use these calculations to\nroughly sketch the magnitude response over\n0 ≤ω ≤10.\n(b) To test the system, the engineer connects", - "type": "text" - }, - { - "block_id": "p500-b6", - "global_id": 14181, - "bbox": [ - 91.23, - 321.27, - 263.25, - 417.9 - ], - "text": "a signal generator to the system in hopes\nto measure the magnitude response using a\nstandard oscilloscope. What type of signal\nshould the engineer input into the system\nto make the measurements? How should\nthe engineer make the measurements? Pro-\nvide sufficient detail to fully justify your\nanswers.\n(c) Suppose the engineer accidentally con-", - "type": "text" - }, - { - "block_id": "p500-b7", - "global_id": 14182, - "bbox": [ - 106.16, - 418.25, - 259.8, - 433.36 - ], - "text": "structs the system H−1(s) =\n1\nH(s) = 2s2+4s+4", - "type": "text" - }, - { - "block_id": "p500-b8", - "global_id": 14183, - "bbox": [ - 106.15, - 421.58, - 263.24, - 452.46 - ], - "text": "s2+4\n.\nWhat impact will this mistake have on his\ntests?", - "type": "text" - }, - { - "block_id": "p500-b9", - "global_id": 14184, - "bbox": [ - 62.08, - 457.93, - 263.24, - 477.93 - ], - "text": "4.8-2\nFor an LTIC system described by the transfer\nfunction", - "type": "text" - }, - { - "block_id": "p500-b10", - "global_id": 14185, - "bbox": [ - 142.29, - 484.2, - 210.47, - 506.19 - ], - "text": "H(s) =\ns + 2\ns2 + 5s + 4", - "type": "text" - }, - { - "block_id": "p500-b11", - "global_id": 14186, - "bbox": [ - 90.72, - 517.65, - 263.23, - 537.58 - ], - "text": "find the response to the following everlasting\nsinusoidal inputs:", - "type": "text" - }, - { - "block_id": "p500-b12", - "global_id": 14187, - "bbox": [ - 90.72, - 537.94, - 163.89, - 559.5 - ], - "text": "(a) 5cos(2t + 30◦)\n(b) 10sin(2t + 45◦)", - "type": "text" - }, - { - "block_id": "p500-b13", - "global_id": 14188, - "bbox": [ - 90.72, - 559.85, - 249.11, - 581.41 - ], - "text": "(c) 10cos(3t + 40◦)\nObserve that these are everlasting sinusoids.", - "type": "text" - }, - { - "block_id": "p500-b14", - "global_id": 14189, - "bbox": [ - 62.08, - 586.88, - 263.24, - 606.88 - ], - "text": "4.8-3\nFor an LTIC system described by the transfer\nfunction", - "type": "text" - }, - { - "block_id": "p500-b15", - "global_id": 14190, - "bbox": [ - 147.83, - 613.15, - 204.45, - 635.14 - ], - "text": "H(s) =\ns + 3\n(s + 2)2", - "type": "text" - }, - { - "block_id": "p500-b16", - "global_id": 14191, - "bbox": [ - 317.86, - 85.9, - 490.39, - 105.83 - ], - "text": "find the steady-state system response to the\nfollowing inputs:", - "type": "text" - }, - { - "block_id": "p500-b17", - "global_id": 14192, - "bbox": [ - 317.86, - 107.44, - 396.39, - 127.74 - ], - "text": "(a) 10u(t)\n(b) cos(2t + 60◦)u(t)", - "type": "text" - }, - { - "block_id": "p500-b18", - "global_id": 14193, - "bbox": [ - 317.86, - 128.1, - 394.91, - 149.65 - ], - "text": "(c) sin(3t −45◦)u(t)\n(d) ej3tu(t)", - "type": "text" - }, - { - "block_id": "p500-b19", - "global_id": 14194, - "bbox": [ - 289.22, - 154.63, - 490.38, - 174.63 - ], - "text": "4.8-4\nFor an allpass filter specified by the transfer\nfunction", - "type": "text" - }, - { - "block_id": "p500-b20", - "global_id": 14195, - "bbox": [ - 371.11, - 173.53, - 435.96, - 189.06 - ], - "text": "H(s) = −(s −10)", - "type": "text" - }, - { - "block_id": "p500-b21", - "global_id": 14196, - "bbox": [ - 406.87, - 186.18, - 429.11, - 195.52 - ], - "text": "s + 10", - "type": "text" - }, - { - "block_id": "p500-b22", - "global_id": 14197, - "bbox": [ - 317.86, - 202.04, - 490.38, - 221.96 - ], - "text": "find the system response to the following (ever-\nlasting) inputs:", - "type": "text" - }, - { - "block_id": "p500-b23", - "global_id": 14198, - "bbox": [ - 318.37, - 222.32, - 345.25, - 232.92 - ], - "text": "(a) ejωt", - "type": "text" - }, - { - "block_id": "p500-b24", - "global_id": 14199, - "bbox": [ - 317.86, - 234.54, - 376.27, - 243.88 - ], - "text": "(b) cos(ωt + θ)", - "type": "text" - }, - { - "block_id": "p500-b25", - "global_id": 14200, - "bbox": [ - 317.86, - 245.78, - 352.74, - 265.8 - ], - "text": "(c) cos t\n(d) sin 2t", - "type": "text" - }, - { - "block_id": "p500-b26", - "global_id": 14201, - "bbox": [ - 318.37, - 267.7, - 358.7, - 276.75 - ], - "text": "(e) cos 10t", - "type": "text" - }, - { - "block_id": "p500-b27", - "global_id": 14202, - "bbox": [ - 317.86, - 278.66, - 432.98, - 298.68 - ], - "text": "(f) cos 100t\nComment on the filter response.", - "type": "text" - }, - { - "block_id": "p500-b28", - "global_id": 14203, - "bbox": [ - 289.23, - 303.65, - 490.39, - 334.61 - ], - "text": "4.8-5\nThe pole-zero plot of a second-order system\nH(s) is shown in Fig. P4.8-5. The dc response\nof this system is minus 1, H(j0) = −1.", - "type": "text" - }, - { - "block_id": "p500-b29", - "global_id": 14204, - "bbox": [ - 318.37, - 332.97, - 490.39, - 346.31 - ], - "text": "(a) Letting H(s) = k(s2 + b1s + b2)/(s2 +a1s+", - "type": "text" - }, - { - "block_id": "p500-b30", - "global_id": 14205, - "bbox": [ - 317.86, - 347.19, - 490.39, - 378.44 - ], - "text": "a2), determine the constants k, b1, b2, a1,\nand a2.\n(b) What is the output y(t) of this system in", - "type": "text" - }, - { - "block_id": "p500-b31", - "global_id": 14206, - "bbox": [ - 333.3, - 380.06, - 490.39, - 400.36 - ], - "text": "response to the input x(t) = 4 + cos(t/2 +\nπ/3)?", - "type": "text" - }, - { - "block_id": "p500-b32", - "global_id": 14207, - "bbox": [ - 333.67, - 535.18, - 415.29, - 550.73 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\ns", - "type": "text" - }, - { - "block_id": "p500-b33", - "global_id": 14208, - "bbox": [ - 312.53, - 473.27, - 320.53, - 478.61 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p500-b34", - "global_id": 14209, - "bbox": [ - 329.67, - 529.24, - 487.97, - 543.18 - ], - "text": "0.5\n1\n1.5\n2\n–2", - "type": "text" - }, - { - "block_id": "p500-b35", - "global_id": 14210, - "bbox": [ - 323.67, - 515.82, - 337.67, - 523.82 - ], - "text": "–1.5", - "type": "text" - }, - { - "block_id": "p500-b36", - "global_id": 14211, - "bbox": [ - 329.67, - 501.4, - 337.67, - 509.4 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p500-b37", - "global_id": 14212, - "bbox": [ - 322.87, - 486.96, - 337.67, - 494.96 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p500-b38", - "global_id": 14213, - "bbox": [ - 333.67, - 472.56, - 337.67, - 480.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p500-b39", - "global_id": 14214, - "bbox": [ - 327.67, - 458.14, - 337.67, - 466.14 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p500-b40", - "global_id": 14215, - "bbox": [ - 333.67, - 443.72, - 337.67, - 451.72 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p500-b41", - "global_id": 14216, - "bbox": [ - 327.67, - 429.29, - 337.67, - 437.29 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p500-b42", - "global_id": 14217, - "bbox": [ - 333.67, - 414.87, - 337.67, - 422.87 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p500-b43", - "global_id": 14218, - "bbox": [ - 307.9, - 558.69, - 359.55, - 567.66 - ], - "text": "Figure P4.8-5", - "type": "text" - }, - { - "block_id": "p500-b44", - "global_id": 14219, - "bbox": [ - 289.22, - 581.96, - 490.38, - 613.23 - ], - "text": "4.8-6\nConsider a CT system described by (D+1)(D+\n2){y(t)} = x(t −1). Notice that this differential\nequation is in terms of x(t −1), not x(t)!", - "type": "text" - }, - { - "block_id": "p500-b45", - "global_id": 14220, - "bbox": [ - 318.37, - 614.84, - 490.39, - 624.18 - ], - "text": "(a) Determine the output y(t) given input", - "type": "text" - }, - { - "block_id": "p500-b46", - "global_id": 14221, - "bbox": [ - 333.31, - 625.8, - 364.07, - 635.14 - ], - "text": "x(t) = 1.", - "type": "text" - } - ] - }, - { - "page_num": 501, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p501-b0", - "global_id": 14222, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n481", - "type": "text" - }, - { - "block_id": "p501-b1", - "global_id": 14223, - "bbox": [ - 116.46, - 85.58, - 288.98, - 94.92 - ], - "text": "(b) Determine the output y(t) given input x(t) =", - "type": "text" - }, - { - "block_id": "p501-b2", - "global_id": 14224, - "bbox": [ - 131.9, - 96.54, - 155.44, - 105.88 - ], - "text": "cos(t).", - "type": "text" - }, - { - "block_id": "p501-b3", - "global_id": 14225, - "bbox": [ - 87.82, - 110.95, - 288.98, - 120.29 - ], - "text": "4.8-7\nAn LTIC system has transfer function H(s) =", - "type": "text" - }, - { - "block_id": "p501-b4", - "global_id": 14226, - "bbox": [ - 116.46, - 120.48, - 288.98, - 142.2 - ], - "text": "4s\ns2+2s+37\n=\n4s\n(s+1+6j)(s+1−6j).\nDetermine\nthe\nsteady-state output in response to input x(t) =", - "type": "text" - }, - { - "block_id": "p501-b5", - "global_id": 14227, - "bbox": [ - 117.66, - 142.46, - 198.76, - 155.74 - ], - "text": "1\n3ej(6t+π/3)u(6t + π/3).", - "type": "text" - }, - { - "block_id": "p501-b6", - "global_id": 14228, - "bbox": [ - 87.82, - 158.23, - 288.99, - 200.44 - ], - "text": "4.9-1\nSuppose a real first-order lowpass system H(s)\nhas unity gain in the passband, one finite pole\nat s = −2, and one finite zero at an unspecified\nlocation.", - "type": "text" - }, - { - "block_id": "p501-b7", - "global_id": 14229, - "bbox": [ - 116.97, - 202.44, - 288.98, - 211.41 - ], - "text": "(a) Determine the location of the system zero", - "type": "text" - }, - { - "block_id": "p501-b8", - "global_id": 14230, - "bbox": [ - 116.46, - 213.4, - 288.99, - 266.2 - ], - "text": "so that the filter achieves 40 dB of stop-\nband attenuation. Sketch the corresponding\nstraight-line Bode approximation of the\nsystem magnitude response.\n(b) Determine the location of the system zero", - "type": "text" - }, - { - "block_id": "p501-b9", - "global_id": 14231, - "bbox": [ - 131.9, - 268.19, - 288.99, - 310.04 - ], - "text": "so that the filter achieves 30 dB of stop-\nband attenuation. Sketch the corresponding\nstraight-line Bode approximation of the\nsystem magnitude response.", - "type": "text" - }, - { - "block_id": "p501-b10", - "global_id": 14232, - "bbox": [ - 87.82, - 315.41, - 288.98, - 335.41 - ], - "text": "4.9-2\nRepeat Prob. 4.9-1 for a highpass rather than a\nlowpass system.", - "type": "text" - }, - { - "block_id": "p501-b11", - "global_id": 14233, - "bbox": [ - 87.82, - 340.78, - 288.98, - 371.74 - ], - "text": "4.9-3\nRepeat Prob. 4.9-1 for a second-order system\nthat has a pair of repeated poles and a pair of\nrepeated zeros.", - "type": "text" - }, - { - "block_id": "p501-b12", - "global_id": 14234, - "bbox": [ - 87.82, - 377.1, - 288.97, - 397.1 - ], - "text": "4.9-4\nSketch Bode plots for the following transfer\nfunctions:", - "type": "text" - }, - { - "block_id": "p501-b13", - "global_id": 14235, - "bbox": [ - 116.96, - 395.01, - 186.48, - 416.99 - ], - "text": "(a)\ns(s + 100)\n(s + 2)(s + 20)", - "type": "text" - }, - { - "block_id": "p501-b14", - "global_id": 14236, - "bbox": [ - 116.46, - 419.36, - 190.96, - 434.98 - ], - "text": "(b) (s + 10)(s + 20)", - "type": "text" - }, - { - "block_id": "p501-b15", - "global_id": 14237, - "bbox": [ - 141.71, - 431.68, - 182.34, - 441.35 - ], - "text": "s2(s + 100)", - "type": "text" - }, - { - "block_id": "p501-b16", - "global_id": 14238, - "bbox": [ - 116.96, - 443.72, - 203.65, - 465.71 - ], - "text": "(c)\n(s + 10)(s + 200)\n(s + 20)2(s + 1000)", - "type": "text" - }, - { - "block_id": "p501-b17", - "global_id": 14239, - "bbox": [ - 314.97, - 85.94, - 423.18, - 94.98 - ], - "text": "4.9-5\nRepeat Prob. 4.9-4 for", - "type": "text" - }, - { - "block_id": "p501-b18", - "global_id": 14240, - "bbox": [ - 344.11, - 95.25, - 400.79, - 111.86 - ], - "text": "(a)\ns2", - "type": "text" - }, - { - "block_id": "p501-b19", - "global_id": 14241, - "bbox": [ - 360.24, - 106.3, - 435.1, - 118.24 - ], - "text": "(s + 1)(s2 + 4s + 16)", - "type": "text" - }, - { - "block_id": "p501-b20", - "global_id": 14242, - "bbox": [ - 343.61, - 118.66, - 455.27, - 140.37 - ], - "text": "(b)\ns\n(s + 1)(s2 + 14.14s + 100)", - "type": "text" - }, - { - "block_id": "p501-b21", - "global_id": 14243, - "bbox": [ - 344.12, - 142.73, - 434.32, - 164.72 - ], - "text": "(c)\n(s + 10)\ns(s2 + 14.14s + 100)", - "type": "text" - }, - { - "block_id": "p501-b22", - "global_id": 14244, - "bbox": [ - 314.97, - 173.38, - 516.13, - 215.3 - ], - "text": "4.9-6\nUsing the lowest order possible, determine a\nsystem function H(s) with real-valued roots that\nmatches the frequency response in Fig. P4.9-6.\nVerify your answer with MATLAB.", - "type": "text" - }, - { - "block_id": "p501-b23", - "global_id": 14245, - "bbox": [ - 314.97, - 226.15, - 516.14, - 322.86 - ], - "text": "4.9-7\nA graduate student recently implemented an\nanalog phase lock loop (PLL) as part of his the-\nsis. His PLL consists of four basic components:\na phase/frequency detector, a charge pump, a\nloop filter, and a voltage-controlled oscillator.\nThis problem considers only the loop filter,\nwhich is shown in Fig. P4.9-7a. The loop filter\ninput is the current x(t), and the output is the\nvoltage y(t).", - "type": "text" - }, - { - "block_id": "p501-b24", - "global_id": 14246, - "bbox": [ - 344.12, - 324.86, - 516.13, - 333.83 - ], - "text": "(a) Derive the loop filter’s transfer function", - "type": "text" - }, - { - "block_id": "p501-b25", - "global_id": 14247, - "bbox": [ - 343.61, - 335.44, - 516.13, - 355.74 - ], - "text": "H(s). Express H(s) in standard form.\n(b) Figure P4.9-7b provides four possible fre-", - "type": "text" - }, - { - "block_id": "p501-b26", - "global_id": 14248, - "bbox": [ - 344.12, - 357.74, - 516.13, - 432.46 - ], - "text": "quency response plots, labeled A through D.\nEach log-log plot is drawn to the same scale,\nand line slopes are either 20 dB/decade,\n0 dB/decade, or −20 dB/decade. Clearly\nidentify which plot(s), if any, could repre-\nsent the loop filter.\n(c) Holding the other components constant,", - "type": "text" - }, - { - "block_id": "p501-b27", - "global_id": 14249, - "bbox": [ - 359.05, - 434.45, - 516.14, - 465.33 - ], - "text": "what is the general effect of increasing the\nresistance R on the magnitude response for\nlow-frequency inputs?", - "type": "text" - }, - { - "block_id": "p501-b28", - "global_id": 14250, - "bbox": [ - 143.11, - 597.94, - 310.6, - 613.52 - ], - "text": "10–1\n100\n101\n102\n103\n104\n–10", - "type": "text" - }, - { - "block_id": "p501-b29", - "global_id": 14251, - "bbox": [ - 147.11, - 587.4, - 155.11, - 595.4 - ], - "text": "–5", - "type": "text" - }, - { - "block_id": "p501-b30", - "global_id": 14252, - "bbox": [ - 147.11, - 483.57, - 155.11, - 583.86 - ], - "text": "0\n5\n10\n15\n20\n25\n30\n35\n40", - "type": "text" - }, - { - "block_id": "p501-b31", - "global_id": 14253, - "bbox": [ - 216.2, - 615.03, - 244.41, - 623.14 - ], - "text": "v (rad/s)", - "type": "text" - }, - { - "block_id": "p501-b32", - "global_id": 14254, - "bbox": [ - 131.22, - 524.89, - 139.32, - 564.38 - ], - "text": "|H(jv)| (dB)", - "type": "text" - }, - { - "block_id": "p501-b33", - "global_id": 14255, - "bbox": [ - 331.43, - 489.34, - 396.98, - 508.75 - ], - "text": "Bode approximation\nTrue |H( jv)|", - "type": "text" - }, - { - "block_id": "p501-b34", - "global_id": 14256, - "bbox": [ - 130.58, - 629.31, - 182.22, - 638.27 - ], - "text": "Figure P4.9-6", - "type": "text" - } - ] - }, - { - "page_num": 502, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p502-b0", - "global_id": 14257, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "482\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p502-b1", - "global_id": 14258, - "bbox": [ - 90.72, - 85.9, - 263.23, - 94.86 - ], - "text": "(d) Holding the other components constant,", - "type": "text" - }, - { - "block_id": "p502-b2", - "global_id": 14259, - "bbox": [ - 106.15, - 96.86, - 263.24, - 127.74 - ], - "text": "what is the general effect of increasing the\nresistance R on the magnitude response for\nhigh-frequency inputs?", - "type": "text" - }, - { - "block_id": "p502-b3", - "global_id": 14260, - "bbox": [ - 100.61, - 296.84, - 198.7, - 304.84 - ], - "text": "A\nB", - "type": "text" - }, - { - "block_id": "p502-b4", - "global_id": 14261, - "bbox": [ - 100.61, - 339.81, - 199.14, - 347.81 - ], - "text": "C\nD", - "type": "text" - }, - { - "block_id": "p502-b5", - "global_id": 14262, - "bbox": [ - 170.37, - 394.0, - 180.02, - 402.0 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p502-b6", - "global_id": 14263, - "bbox": [ - 98.62, - 157.93, - 109.72, - 166.02 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p502-b7", - "global_id": 14264, - "bbox": [ - 240.68, - 180.88, - 251.78, - 188.97 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p502-b8", - "global_id": 14265, - "bbox": [ - 170.76, - 238.26, - 179.64, - 246.26 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p502-b9", - "global_id": 14266, - "bbox": [ - 148.47, - 190.45, - 248.23, - 199.17 - ], - "text": "R\n–", - "type": "text" - }, - { - "block_id": "p502-b10", - "global_id": 14267, - "bbox": [ - 144.91, - 159.85, - 248.49, - 178.77 - ], - "text": "+\nC1", - "type": "text" - }, - { - "block_id": "p502-b11", - "global_id": 14268, - "bbox": [ - 204.15, - 179.49, - 212.49, - 189.1 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p502-b12", - "global_id": 14269, - "bbox": [ - 90.72, - 409.91, - 142.36, - 418.88 - ], - "text": "Figure P4.9-7", - "type": "text" - }, - { - "block_id": "p502-b13", - "global_id": 14270, - "bbox": [ - 57.59, - 440.71, - 263.24, - 485.12 - ], - "text": "4.10-1\nA causal LTIC system H(s) =\n2(s−4j)(s+4j)\n(s+1+2j)(s+1−2j)\nhas input x(t) = −1 + 2cos(2t) −3sin(4t +\nπ/3) + 4cos(10t). Below, perform accurate cal-\nculations at ω = 0, ±2, ±4, and ±10.", - "type": "text" - }, - { - "block_id": "p502-b14", - "global_id": 14271, - "bbox": [ - 91.23, - 487.12, - 263.24, - 496.08 - ], - "text": "(a) Using the graphical method of Sec. 4.10-1,", - "type": "text" - }, - { - "block_id": "p502-b15", - "global_id": 14272, - "bbox": [ - 90.72, - 498.08, - 263.24, - 528.96 - ], - "text": "accurately sketch the magnitude response\n|H(jω)| over −10 ≤ω ≤10.\n(b) Using the graphical method of Sec. 4.10-1,", - "type": "text" - }, - { - "block_id": "p502-b16", - "global_id": 14273, - "bbox": [ - 91.22, - 530.95, - 263.24, - 561.84 - ], - "text": "accurately\nsketch\nthe\nphase\nresponse̸\nH(jω) over −10 ≤ω ≤10.\n(c) Approximate the system output y(t) in", - "type": "text" - }, - { - "block_id": "p502-b17", - "global_id": 14274, - "bbox": [ - 106.15, - 563.45, - 198.46, - 572.79 - ], - "text": "response to the input x(t).", - "type": "text" - }, - { - "block_id": "p502-b18", - "global_id": 14275, - "bbox": [ - 57.59, - 578.74, - 263.24, - 609.7 - ], - "text": "4.10-2\nThe pole-zero plot of a second-order system\nH(s) is shown in Fig. P4.10-2. The dc response\nof this system is minus 2, H(j0) = −2.", - "type": "text" - }, - { - "block_id": "p502-b19", - "global_id": 14276, - "bbox": [ - 91.23, - 610.05, - 203.63, - 622.93 - ], - "text": "(a) Letting H(s) = k s2+b1s+b2", - "type": "text" - }, - { - "block_id": "p502-b20", - "global_id": 14277, - "bbox": [ - 106.15, - 613.96, - 263.23, - 636.1 - ], - "text": "s2+a1s+a2 , determine the\nconstants k, b1, b2, a1, and a2.", - "type": "text" - }, - { - "block_id": "p502-b21", - "global_id": 14278, - "bbox": [ - 317.86, - 85.89, - 490.38, - 94.86 - ], - "text": "(b) Using the graphical method of Sec. 4.10-1,", - "type": "text" - }, - { - "block_id": "p502-b22", - "global_id": 14279, - "bbox": [ - 318.37, - 96.86, - 490.4, - 138.7 - ], - "text": "hand-sketch\nthe\nmagnitude\nresponse\n|H(jω)| over −10 ≤ω ≤10. Verify your\nsketch with MATLAB.\n(c) Using the graphical method of Sec. 4.10-1,", - "type": "text" - }, - { - "block_id": "p502-b23", - "global_id": 14280, - "bbox": [ - 317.86, - 140.32, - 490.4, - 182.53 - ], - "text": "hand-sketch the phase response̸\nH(jω)\nover −10 ≤ω ≤10. Verify your sketch with\nMATLAB.\n(d) What is the output y(t) in response to input", - "type": "text" - }, - { - "block_id": "p502-b24", - "global_id": 14281, - "bbox": [ - 333.31, - 184.16, - 487.06, - 193.5 - ], - "text": "x(t) = −3 + cos(3t+π/3) −sin(4t−π/8)?", - "type": "text" - }, - { - "block_id": "p502-b25", - "global_id": 14282, - "bbox": [ - 397.98, - 313.95, - 406.84, - 321.92 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p502-b26", - "global_id": 14283, - "bbox": [ - 366.76, - 225.96, - 375.61, - 233.93 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p502-b27", - "global_id": 14284, - "bbox": [ - 346.26, - 316.82, - 356.46, - 325.13 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p502-b28", - "global_id": 14285, - "bbox": [ - 371.07, - 367.38, - 381.28, - 375.68 - ], - "text": "−4", - "type": "text" - }, - { - "block_id": "p502-b29", - "global_id": 14286, - "bbox": [ - 371.07, - 352.81, - 381.28, - 361.11 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p502-b30", - "global_id": 14287, - "bbox": [ - 371.01, - 265.43, - 375.0, - 273.4 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p502-b31", - "global_id": 14288, - "bbox": [ - 370.92, - 250.71, - 374.91, - 258.68 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p502-b32", - "global_id": 14289, - "bbox": [ - 317.86, - 409.12, - 373.98, - 418.09 - ], - "text": "Figure P4.10-2", - "type": "text" - }, - { - "block_id": "p502-b33", - "global_id": 14290, - "bbox": [ - 284.74, - 444.99, - 490.4, - 486.91 - ], - "text": "4.10-3\nUsing the graphical method of Sec. 4.10-1, draw\na rough sketch of the amplitude and phase\nresponses of an LTIC system described by the\ntransfer function", - "type": "text" - }, - { - "block_id": "p502-b34", - "global_id": 14291, - "bbox": [ - 347.86, - 505.99, - 420.52, - 524.78 - ], - "text": "H(s) = s2 −2s + 50", - "type": "text" - }, - { - "block_id": "p502-b35", - "global_id": 14292, - "bbox": [ - 376.79, - 519.31, - 420.52, - 531.24 - ], - "text": "s2 + 2s + 50", - "type": "text" - }, - { - "block_id": "p502-b36", - "global_id": 14293, - "bbox": [ - 366.76, - 532.79, - 459.21, - 548.05 - ], - "text": "= (s −1 −j7)(s −1 + j7)", - "type": "text" - }, - { - "block_id": "p502-b37", - "global_id": 14294, - "bbox": [ - 376.79, - 545.44, - 459.21, - 554.78 - ], - "text": "(s + 1 −j7)(s + 1 + j7)", - "type": "text" - }, - { - "block_id": "p502-b38", - "global_id": 14295, - "bbox": [ - 317.86, - 577.67, - 412.07, - 586.64 - ], - "text": "What kind of filter is this?", - "type": "text" - }, - { - "block_id": "p502-b39", - "global_id": 14296, - "bbox": [ - 284.74, - 592.85, - 490.4, - 634.77 - ], - "text": "4.10-4\nUsing the graphical method of Sec. 4.10-1, draw\na rough sketch of the amplitude and phase\nresponses of LTIC systems whose pole-zero\nplots are shown in Fig. P4.10-4.", - "type": "text" - } - ] - }, - { - "page_num": 503, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p503-b0", - "global_id": 14297, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n483", - "type": "text" - }, - { - "block_id": "p503-b1", - "global_id": 14298, - "bbox": [ - 274.72, - 221.74, - 284.53, - 229.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p503-b2", - "global_id": 14299, - "bbox": [ - 217.09, - 87.63, - 239.53, - 95.71 - ], - "text": "s plane", - "type": "text" - }, - { - "block_id": "p503-b3", - "global_id": 14300, - "bbox": [ - 286.19, - 165.89, - 297.08, - 173.89 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p503-b4", - "global_id": 14301, - "bbox": [ - 266.52, - 110.11, - 275.4, - 118.11 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p503-b5", - "global_id": 14302, - "bbox": [ - 229.68, - 170.09, - 261.26, - 178.88 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p503-b6", - "global_id": 14303, - "bbox": [ - 172.55, - 221.74, - 181.43, - 229.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p503-b7", - "global_id": 14304, - "bbox": [ - 183.56, - 165.89, - 194.45, - 173.89 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p503-b8", - "global_id": 14305, - "bbox": [ - 163.89, - 110.11, - 172.77, - 118.11 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p503-b9", - "global_id": 14306, - "bbox": [ - 127.46, - 170.09, - 161.21, - 178.88 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p503-b10", - "global_id": 14307, - "bbox": [ - 116.46, - 237.66, - 172.58, - 246.63 - ], - "text": "Figure P4.10-4", - "type": "text" - }, - { - "block_id": "p503-b11", - "global_id": 14308, - "bbox": [ - 83.34, - 262.31, - 288.98, - 295.76 - ], - "text": "4.10-5\nA causal LTIC system H(s) =\n(s−3j)(s+3j)\n3(s+2+j)(s+2−j)\nhas input x(t) = cos(t) + sin(3t + π/3) +\ncos(100t).", - "type": "text" - }, - { - "block_id": "p503-b12", - "global_id": 14309, - "bbox": [ - 116.97, - 297.75, - 288.98, - 306.71 - ], - "text": "(a) Using the graphical method of Sec. 4.10-1,", - "type": "text" - }, - { - "block_id": "p503-b13", - "global_id": 14310, - "bbox": [ - 116.46, - 308.34, - 288.98, - 361.52 - ], - "text": "sketch the magnitude response |H(jω)| over\n−10 ≤ω ≤10.\n(b) Determine\nthe\nsystem\noutput\ny(t)\nin\nresponse to the input x(t).\n(c) Suppose we create a second causal sys-", - "type": "text" - }, - { - "block_id": "p503-b14", - "global_id": 14311, - "bbox": [ - 131.89, - 363.13, - 288.99, - 405.35 - ], - "text": "tem with transfer function H2(s) = H(−s).\nSketch this system’s pole/zero plot. What is\nthe response y2(t) of system H2(s) to input\nx(t)?", - "type": "text" - }, - { - "block_id": "p503-b15", - "global_id": 14312, - "bbox": [ - 83.34, - 410.68, - 288.98, - 463.56 - ], - "text": "4.10-6\nDesign a second-order bandpass filter with cen-\nter frequency ω = 10. The gain should be zero at\nω = 0 and at ω = ∞. Select poles at −a ± j10.\nLeave your answer in terms of a. Explain the\ninfluence of a on the frequency response.", - "type": "text" - }, - { - "block_id": "p503-b16", - "global_id": 14313, - "bbox": [ - 83.34, - 468.58, - 288.99, - 565.6 - ], - "text": "4.10-7\nThe\nLTIC\nsystem\ndescribed\nby\nH(s) =\n(s −1)/(s + 1) has unity magnitude response\n|H(jω)| = 1. Positive Pat claims that the output\ny(t) of this system is equal the input x(t), since\nthe system is allpass. Cynical Cynthia doesn’t\nthink so. “This is signals and systems class,”\nshe complains. “It has to be more complicated!”\nWho is correct, Pat or Cynthia? Justify your\nanswer.", - "type": "text" - }, - { - "block_id": "p503-b17", - "global_id": 14314, - "bbox": [ - 83.34, - 570.93, - 288.99, - 634.76 - ], - "text": "4.10-8\nTwo students, Amy and Jeff, disagree about\nan analog system function given by H1(s) = s.\nSensible Jeff claims the system has a zero at\ns = 0. Rebellious Amy, however, notes that the\nsystem function can be rewritten as H1(s) =\n1/s−1 and claims that this implies a system pole", - "type": "text" - }, - { - "block_id": "p503-b18", - "global_id": 14315, - "bbox": [ - 343.61, - 85.52, - 516.13, - 106.56 - ], - "text": "at s = ∞. Who is correct? Why? What are the\npoles and zeros of the system H2(s) = 1/s?", - "type": "text" - }, - { - "block_id": "p503-b19", - "global_id": 14316, - "bbox": [ - 310.48, - 110.43, - 516.14, - 163.61 - ], - "text": "4.10-9\nA rational transfer function H(s) is often used\nto represent an analog filter. Why must H(s) be\nstrictly proper for lowpass and bandpass filters?\nWhy must H(s) be proper for highpass and\nbandstop filters?", - "type": "text" - }, - { - "block_id": "p503-b20", - "global_id": 14317, - "bbox": [ - 306.0, - 168.5, - 516.12, - 199.47 - ], - "text": "4.10-10\nFor a given filter order N, why is the stopband\nattenuation rate of an all-pole lowpass filter\nbetter than filters with finite zeros?", - "type": "text" - }, - { - "block_id": "p503-b21", - "global_id": 14318, - "bbox": [ - 306.0, - 204.38, - 516.11, - 225.11 - ], - "text": "4.10-11\nIs\nit\npossible,\nwith\nreal\ncoefficients\n([k,b1,b2,a1,a2] ∈R), for a system", - "type": "text" - }, - { - "block_id": "p503-b22", - "global_id": 14319, - "bbox": [ - 389.38, - 231.9, - 468.66, - 250.69 - ], - "text": "H(s) = k s2 + b1s + b2", - "type": "text" - }, - { - "block_id": "p503-b23", - "global_id": 14320, - "bbox": [ - 422.44, - 245.22, - 468.66, - 257.88 - ], - "text": "s2 + a1s + a2", - "type": "text" - }, - { - "block_id": "p503-b24", - "global_id": 14321, - "bbox": [ - 343.61, - 266.14, - 516.14, - 286.07 - ], - "text": "to function as a lowpass filter? Explain your\nanswer.", - "type": "text" - }, - { - "block_id": "p503-b25", - "global_id": 14322, - "bbox": [ - 306.0, - 290.98, - 516.14, - 442.48 - ], - "text": "4.10-12\nNick recently built a simple second-order But-\nterworth lowpass filter for his home stereo.\nAlthough the system performs fairly well, Nick\nis an overachiever and hopes to improve the\nsystem performance. Unfortunately, Nick is\nlazy and doesn’t want to design another filter.\nThinking “Twice the filtering gives twice the\nperformance,” he suggests filtering the audio\nsignal not once but twice with a cascade of\ntwo identical filters. His overworked, underpaid\nsignals professor is skeptical and states, “If you\nare using identical filters, it makes no difference\nwhether you filter once or twice!” Who is\ncorrect? Why?", - "type": "text" - }, - { - "block_id": "p503-b26", - "global_id": 14323, - "bbox": [ - 306.0, - 447.39, - 516.13, - 467.4 - ], - "text": "4.10-13\nAn LTIC system impulse response is given by\nh(t) = u(t) −u(t −1).", - "type": "text" - }, - { - "block_id": "p503-b27", - "global_id": 14324, - "bbox": [ - 344.12, - 469.01, - 516.13, - 478.35 - ], - "text": "(a) Determine the transfer function H(s). Using", - "type": "text" - }, - { - "block_id": "p503-b28", - "global_id": 14325, - "bbox": [ - 343.62, - 479.97, - 516.15, - 544.1 - ], - "text": "H(s), determine and plot the magnitude\nresponse |H(jω)|. Which type of filter most\naccurately describes the behavior of this\nsystem: lowpass, highpass, bandpass, or\nbandstop?\n(b) What are the poles and zeros of H(s)?", - "type": "text" - }, - { - "block_id": "p503-b29", - "global_id": 14326, - "bbox": [ - 344.12, - 546.1, - 516.15, - 566.03 - ], - "text": "Explain your answer.\n(c) Can you determine the impulse response", - "type": "text" - }, - { - "block_id": "p503-b30", - "global_id": 14327, - "bbox": [ - 359.05, - 568.02, - 516.15, - 609.86 - ], - "text": "of the inverse system? If so, provide it. If\nnot, suggest a method that could be used\nto approximate the impulse response of the\ninverse system.", - "type": "text" - }, - { - "block_id": "p503-b31", - "global_id": 14328, - "bbox": [ - 306.01, - 614.46, - 516.13, - 634.76 - ], - "text": "4.10-14\nAn ideal lowpass filter HLP(s) has magnitude\nresponse that is unity for low frequencies", - "type": "text" - } - ] - }, - { - "page_num": 504, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p504-b0", - "global_id": 14329, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "484\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p504-b1", - "global_id": 14330, - "bbox": [ - 90.72, - 85.9, - 263.25, - 160.62 - ], - "text": "and zero for high frequencies. An ideal high-\npass filter HHP(s) has an opposite magnitude\nresponse: zero for low frequencies and unity\nfor high frequencies. A student suggests a pos-\nsible lowpass-to-highpass filter transformation:\nHHP(s) = 1 −HLP(s). In general, will this\ntransformation work? Explain your answer.", - "type": "text" - }, - { - "block_id": "p504-b2", - "global_id": 14331, - "bbox": [ - 53.11, - 165.53, - 263.23, - 196.48 - ], - "text": "4.10-15\nAn LTIC system has a rational transfer function\nH(s). When appropriate, assume that all initial\nconditions are zero.", - "type": "text" - }, - { - "block_id": "p504-b3", - "global_id": 14332, - "bbox": [ - 91.22, - 198.48, - 263.23, - 207.45 - ], - "text": "(a) Is is possible for this system to output", - "type": "text" - }, - { - "block_id": "p504-b4", - "global_id": 14333, - "bbox": [ - 90.72, - 209.06, - 263.23, - 240.32 - ], - "text": "y(t) = sin(100πt)u(t) in response to an\ninput x(t) = cos(100πt)u(t)? Explain.\n(b) Is is possible for this system to output", - "type": "text" - }, - { - "block_id": "p504-b5", - "global_id": 14334, - "bbox": [ - 91.23, - 241.94, - 263.24, - 273.2 - ], - "text": "y(t) = sin(100πt)u(t) in response to an\ninput x(t) = sin(50πt)u(t)? Explain.\n(c) Is is possible for this system to output y(t) =", - "type": "text" - }, - { - "block_id": "p504-b6", - "global_id": 14335, - "bbox": [ - 90.72, - 274.82, - 263.24, - 306.07 - ], - "text": "sin(100πt) in response to an input x(t) =\ncos(100πt)? Explain.\n(d) Is is possible for this system to output y(t) =", - "type": "text" - }, - { - "block_id": "p504-b7", - "global_id": 14336, - "bbox": [ - 106.15, - 307.69, - 263.24, - 327.99 - ], - "text": "sin(100πt) in response to an input x(t) =\nsin(50πt)? Explain.", - "type": "text" - }, - { - "block_id": "p504-b8", - "global_id": 14337, - "bbox": [ - 57.59, - 332.9, - 263.24, - 352.89 - ], - "text": "4.11-1\nFind the ROC, if it exists, of the (bilateral)\nLaplace transform of the following signals:", - "type": "text" - }, - { - "block_id": "p504-b9", - "global_id": 14338, - "bbox": [ - 91.22, - 353.25, - 121.92, - 363.86 - ], - "text": "(a) etu(t)", - "type": "text" - }, - { - "block_id": "p504-b10", - "global_id": 14339, - "bbox": [ - 90.72, - 364.21, - 126.97, - 374.82 - ], - "text": "(b) e−tu(t)", - "type": "text" - }, - { - "block_id": "p504-b11", - "global_id": 14340, - "bbox": [ - 91.22, - 374.55, - 127.51, - 396.17 - ], - "text": "(c)\n1\n1 + t2", - "type": "text" - }, - { - "block_id": "p504-b12", - "global_id": 14341, - "bbox": [ - 90.72, - 399.03, - 127.4, - 420.65 - ], - "text": "(d)\n1\n1 + et", - "type": "text" - }, - { - "block_id": "p504-b13", - "global_id": 14342, - "bbox": [ - 91.22, - 419.8, - 122.71, - 431.73 - ], - "text": "(e) e−kt2", - "type": "text" - }, - { - "block_id": "p504-b14", - "global_id": 14343, - "bbox": [ - 57.59, - 436.64, - 263.24, - 489.51 - ], - "text": "4.11-2\nUsing the definition and direct integration,\nfind the (bilateral) Laplace transform and the\ncorresponding region of convergence for the\nfollowing signals. If the Laplace transform does\nnot exist, carefully explain why.", - "type": "text" - }, - { - "block_id": "p504-b15", - "global_id": 14344, - "bbox": [ - 90.72, - 489.87, - 187.99, - 512.18 - ], - "text": "(a) xa(t) = e(−1−j)tu(1 −t)\n(b) xb(t) = j(t+1)u(−t −1)", - "type": "text" - }, - { - "block_id": "p504-b16", - "global_id": 14345, - "bbox": [ - 90.72, - 511.79, - 218.62, - 534.09 - ], - "text": "(c) xc(t) = ejπ/3u(2 −t) + jδ(t −5)\n(d) xd(t) = 1 + 1 = 2", - "type": "text" - }, - { - "block_id": "p504-b17", - "global_id": 14346, - "bbox": [ - 91.22, - 533.71, - 245.46, - 545.05 - ], - "text": "(e) xe(t) = 3u(−t) + e−2t[u(t) −u(t −10)]", - "type": "text" - }, - { - "block_id": "p504-b18", - "global_id": 14347, - "bbox": [ - 92.22, - 544.67, - 228.15, - 556.01 - ], - "text": "(f) xf(t) = et−2u(1 −t) + e−2tu(t + 1)", - "type": "text" - }, - { - "block_id": "p504-b19", - "global_id": 14348, - "bbox": [ - 57.59, - 559.88, - 263.25, - 580.18 - ], - "text": "4.11-3\nDetermine the bilateral Laplace transform X(s)\nof the signal", - "type": "text" - }, - { - "block_id": "p504-b20", - "global_id": 14349, - "bbox": [ - 119.48, - 591.47, - 141.67, - 600.72 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p504-b22", - "global_id": 14350, - "bbox": [ - 147.05, - 589.96, - 174.26, - 600.72 - ], - "text": "etu(−t)", - "type": "text" - }, - { - "block_id": "p504-b24", - "global_id": 14351, - "bbox": [ - 179.19, - 591.47, - 234.49, - 600.81 - ], - "text": "∗[tcos(2t)u(t)]", - "type": "text" - }, - { - "block_id": "p504-b25", - "global_id": 14352, - "bbox": [ - 57.59, - 612.1, - 263.24, - 621.43 - ], - "text": "4.11-4\nA signal has bilateral Laplace transform X(s) =", - "type": "text" - }, - { - "block_id": "p504-b26", - "global_id": 14353, - "bbox": [ - 91.92, - 621.05, - 263.21, - 635.57 - ], - "text": "(s+1)\n(s−2)(s−3) but unknown region of convergence.", - "type": "text" - }, - { - "block_id": "p504-b27", - "global_id": 14354, - "bbox": [ - 317.86, - 85.89, - 490.38, - 105.82 - ], - "text": "What ROC results in the smallest maximum\namplitude of x(t)? Justify your answer.", - "type": "text" - }, - { - "block_id": "p504-b28", - "global_id": 14355, - "bbox": [ - 284.74, - 110.73, - 490.38, - 141.69 - ], - "text": "4.11-5\nFind the (bilateral) Laplace transform and the\ncorresponding region of convergence for the\nfollowing signals:", - "type": "text" - }, - { - "block_id": "p504-b29", - "global_id": 14356, - "bbox": [ - 318.37, - 142.05, - 347.96, - 152.65 - ], - "text": "(a) e−|t|", - "type": "text" - }, - { - "block_id": "p504-b30", - "global_id": 14357, - "bbox": [ - 317.86, - 151.39, - 365.89, - 163.6 - ], - "text": "(b) e−|t| cos t", - "type": "text" - }, - { - "block_id": "p504-b31", - "global_id": 14358, - "bbox": [ - 317.86, - 164.17, - 393.76, - 185.53 - ], - "text": "(c) etu(t) + e2tu(−t)\n(d) e−tu(t)", - "type": "text" - }, - { - "block_id": "p504-b32", - "global_id": 14359, - "bbox": [ - 318.37, - 185.89, - 354.12, - 196.48 - ], - "text": "(e) etu(−t)", - "type": "text" - }, - { - "block_id": "p504-b33", - "global_id": 14360, - "bbox": [ - 319.36, - 195.13, - 411.33, - 208.18 - ], - "text": "(f) cosω0tu(t) + et u(−t)", - "type": "text" - }, - { - "block_id": "p504-b34", - "global_id": 14361, - "bbox": [ - 284.74, - 212.35, - 490.4, - 232.36 - ], - "text": "4.11-6\nFind the inverse (bilateral) Laplace transforms\nof the following functions:", - "type": "text" - }, - { - "block_id": "p504-b35", - "global_id": 14362, - "bbox": [ - 318.37, - 232.06, - 455.57, - 254.05 - ], - "text": "(a)\n2s + 5\n(s + 2)(s + 3)\n−3 < σ < −2", - "type": "text" - }, - { - "block_id": "p504-b36", - "global_id": 14363, - "bbox": [ - 317.86, - 253.37, - 438.8, - 275.37 - ], - "text": "(b)\n2s −5\n(s −2)(s −3)\n2 < σ < 3", - "type": "text" - }, - { - "block_id": "p504-b37", - "global_id": 14364, - "bbox": [ - 318.37, - 278.67, - 430.62, - 300.67 - ], - "text": "(c)\n2s + 3\n(s + 1)(s + 2)\nσ > −1", - "type": "text" - }, - { - "block_id": "p504-b38", - "global_id": 14365, - "bbox": [ - 317.86, - 303.97, - 430.62, - 325.96 - ], - "text": "(d)\n2s + 3\n(s + 1)(s + 2)\nσ < −2", - "type": "text" - }, - { - "block_id": "p504-b39", - "global_id": 14366, - "bbox": [ - 318.37, - 327.58, - 473.02, - 352.82 - ], - "text": "(e)\n3s2 −2s −17\n(s + 1)(s + 3)(s −5)\n−1 < σ < 5", - "type": "text" - }, - { - "block_id": "p504-b40", - "global_id": 14367, - "bbox": [ - 284.74, - 355.55, - 334.31, - 364.59 - ], - "text": "4.11-7\nFind", - "type": "text" - }, - { - "block_id": "p504-b41", - "global_id": 14368, - "bbox": [ - 353.32, - 368.66, - 367.94, - 379.33 - ], - "text": "L−1", - "type": "text" - }, - { - "block_id": "p504-b42", - "global_id": 14369, - "bbox": [ - 369.44, - 357.78, - 448.88, - 386.08 - ], - "text": "2s2 −2s −6\n(s + 1)(s −1)(s + 2)", - "type": "text" - }, - { - "block_id": "p504-b43", - "global_id": 14370, - "bbox": [ - 450.06, - 357.78, - 454.94, - 366.75 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p504-b44", - "global_id": 14371, - "bbox": [ - 317.86, - 393.38, - 365.08, - 402.35 - ], - "text": "if the ROC is", - "type": "text" - }, - { - "block_id": "p504-b45", - "global_id": 14372, - "bbox": [ - 317.86, - 403.96, - 371.15, - 424.26 - ], - "text": "(a) Re s > 1\n(b) Re s < −2", - "type": "text" - }, - { - "block_id": "p504-b46", - "global_id": 14373, - "bbox": [ - 317.86, - 425.89, - 391.06, - 446.18 - ], - "text": "(c) −1 < Res < 1\n(d) −2 < Res < −1", - "type": "text" - }, - { - "block_id": "p504-b47", - "global_id": 14374, - "bbox": [ - 284.74, - 451.09, - 490.39, - 482.05 - ], - "text": "4.11-8\nFor a causal LTIC system having a transfer\nfunction H(s) = 1/(s + 1), find the output y(t)\nif the input x(t) is given by", - "type": "text" - }, - { - "block_id": "p504-b48", - "global_id": 14375, - "bbox": [ - 318.37, - 482.41, - 354.24, - 493.01 - ], - "text": "(a) e−|t|/2", - "type": "text" - }, - { - "block_id": "p504-b49", - "global_id": 14376, - "bbox": [ - 317.86, - 493.57, - 393.76, - 503.97 - ], - "text": "(b) etu(t) + e2tu(−t)", - "type": "text" - }, - { - "block_id": "p504-b50", - "global_id": 14377, - "bbox": [ - 317.86, - 504.32, - 413.19, - 525.88 - ], - "text": "(c) e−t/2u(t) + e−t/4u(−t)\n(d) e2tu(t) + etu(−t)", - "type": "text" - }, - { - "block_id": "p504-b51", - "global_id": 14378, - "bbox": [ - 318.37, - 526.24, - 413.19, - 536.84 - ], - "text": "(e) e−t/4u(t) + e−t/2u(−t)", - "type": "text" - }, - { - "block_id": "p504-b52", - "global_id": 14379, - "bbox": [ - 319.36, - 537.2, - 407.1, - 547.81 - ], - "text": "(f) e−3tu(t) + e−2tu(−t)", - "type": "text" - }, - { - "block_id": "p504-b53", - "global_id": 14380, - "bbox": [ - 284.74, - 552.42, - 490.39, - 572.71 - ], - "text": "4.11-9\nThe autocorrelation function rxx(t) of a signal\nx(t) is given by", - "type": "text" - }, - { - "block_id": "p504-b54", - "global_id": 14381, - "bbox": [ - 353.08, - 589.05, - 381.02, - 599.06 - ], - "text": "rxx(t) =", - "type": "text" - }, - { - "block_id": "p504-b55", - "global_id": 14382, - "bbox": [ - 382.86, - 576.84, - 398.31, - 587.8 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p504-b56", - "global_id": 14383, - "bbox": [ - 387.59, - 599.08, - 399.25, - 605.55 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p504-b57", - "global_id": 14384, - "bbox": [ - 400.75, - 589.05, - 454.11, - 598.3 - ], - "text": "x(τ)x(τ + t)dτ", - "type": "text" - }, - { - "block_id": "p504-b58", - "global_id": 14385, - "bbox": [ - 317.86, - 614.79, - 490.39, - 635.08 - ], - "text": "Derive an expression for Rxx(s) = L(rxx(t)) in\nterms of X(s), where X(s) = L(x(t)).", - "type": "text" - } - ] - }, - { - "page_num": 505, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p505-b0", - "global_id": 14386, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n485", - "type": "text" - }, - { - "block_id": "p505-b1", - "global_id": 14387, - "bbox": [ - 78.86, - 85.94, - 272.56, - 94.98 - ], - "text": "4.11-10\nDetermine the inverse Laplace transform of", - "type": "text" - }, - { - "block_id": "p505-b2", - "global_id": 14388, - "bbox": [ - 177.59, - 109.18, - 210.0, - 124.43 - ], - "text": "X(s) = 2", - "type": "text" - }, - { - "block_id": "p505-b3", - "global_id": 14389, - "bbox": [ - 206.01, - 109.09, - 226.16, - 130.71 - ], - "text": "s + s", - "type": "text" - }, - { - "block_id": "p505-b4", - "global_id": 14390, - "bbox": [ - 222.17, - 121.83, - 226.66, - 130.8 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p505-b5", - "global_id": 14391, - "bbox": [ - 116.46, - 142.79, - 280.36, - 152.13 - ], - "text": "given that the region of convergence is σ < 0.", - "type": "text" - }, - { - "block_id": "p505-b6", - "global_id": 14392, - "bbox": [ - 78.86, - 157.16, - 288.98, - 199.37 - ], - "text": "4.11-11\nAn absolutely integrable signal x(t) has a pole\nat s = π. It is possible that other poles may\nbe present. Recall that an absolutely integrable\nsignal satisfies", - "type": "text" - }, - { - "block_id": "p505-b7", - "global_id": 14393, - "bbox": [ - 170.55, - 207.62, - 186.01, - 218.58 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p505-b8", - "global_id": 14394, - "bbox": [ - 175.28, - 229.85, - 186.95, - 236.33 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p505-b9", - "global_id": 14395, - "bbox": [ - 188.44, - 219.72, - 234.89, - 229.07 - ], - "text": "|x(t)|dt < ∞", - "type": "text" - }, - { - "block_id": "p505-b10", - "global_id": 14396, - "bbox": [ - 116.46, - 255.86, - 250.1, - 276.16 - ], - "text": "(a) Can x(t) be left-sided? Explain.\n(b) Can x(t) be right-sided? Explain.", - "type": "text" - }, - { - "block_id": "p505-b11", - "global_id": 14397, - "bbox": [ - 116.46, - 277.77, - 271.81, - 298.07 - ], - "text": "(c) Can x(t) be two-sided? Explain.\n(d) Can x(t) be of finite duration? Explain.", - "type": "text" - }, - { - "block_id": "p505-b12", - "global_id": 14398, - "bbox": [ - 78.86, - 303.1, - 288.98, - 325.69 - ], - "text": "4.11-12\nUsing ROC σ < 0, determine the inverse\nLaplace transform of X(s) = s d", - "type": "text" - }, - { - "block_id": "p505-b13", - "global_id": 14399, - "bbox": [ - 229.1, - 321.73, - 234.86, - 328.2 - ], - "text": "ds", - "type": "text" - }, - { - "block_id": "p505-b15", - "global_id": 14400, - "bbox": [ - 243.14, - 313.16, - 255.15, - 321.4 - ], - "text": "e−2s", - "type": "text" - }, - { - "block_id": "p505-b16", - "global_id": 14401, - "bbox": [ - 248.14, - 321.73, - 250.66, - 328.2 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p505-b18", - "global_id": 14402, - "bbox": [ - 116.46, - 316.63, - 288.98, - 349.57 - ], - "text": ". [Hint:\nUse Laplace transform properties to avoid\ntedious calculus.]", - "type": "text" - }, - { - "block_id": "p505-b19", - "global_id": 14403, - "bbox": [ - 78.86, - 354.9, - 288.97, - 385.86 - ], - "text": "4.11-13\nWith the assistance of Laplace transform prop-\nerties, determine the inverse bilateral Laplace\ntransform x(t) of signal", - "type": "text" - }, - { - "block_id": "p505-b20", - "global_id": 14404, - "bbox": [ - 156.3, - 400.66, - 189.97, - 415.92 - ], - "text": "X(s) = 2", - "type": "text" - }, - { - "block_id": "p505-b21", - "global_id": 14405, - "bbox": [ - 184.23, - 400.66, - 207.89, - 422.19 - ], - "text": "es + 1", - "type": "text" - }, - { - "block_id": "p505-b22", - "global_id": 14406, - "bbox": [ - 203.9, - 413.23, - 207.39, - 422.19 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p505-b24", - "global_id": 14407, - "bbox": [ - 214.96, - 400.66, - 243.06, - 425.1 - ], - "text": "es\n4\ns\n3 + 2", - "type": "text" - }, - { - "block_id": "p505-b25", - "global_id": 14408, - "bbox": [ - 244.25, - 393.99, - 249.14, - 402.96 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p505-b26", - "global_id": 14409, - "bbox": [ - 116.46, - 437.32, - 242.98, - 446.75 - ], - "text": "where the ROC is −6 < Re{s} < 0.", - "type": "text" - }, - { - "block_id": "p505-b27", - "global_id": 14410, - "bbox": [ - 78.86, - 452.08, - 288.98, - 483.04 - ], - "text": "4.11-14\nWith the assistance of Laplace transform prop-\nerties, determine the inverse bilateral Laplace\ntransform x(t) of signal", - "type": "text" - }, - { - "block_id": "p505-b28", - "global_id": 14411, - "bbox": [ - 151.28, - 497.43, - 188.8, - 513.96 - ], - "text": "X(s) = d7", - "type": "text" - }, - { - "block_id": "p505-b29", - "global_id": 14412, - "bbox": [ - 179.21, - 510.75, - 190.42, - 520.33 - ], - "text": "ds7", - "type": "text" - }, - { - "block_id": "p505-b30", - "global_id": 14413, - "bbox": [ - 193.11, - 492.12, - 230.78, - 507.67 - ], - "text": "e−4s", - "type": "text" - }, - { - "block_id": "p505-b31", - "global_id": 14414, - "bbox": [ - 199.19, - 511.08, - 248.09, - 520.42 - ], - "text": "(s + 2)(s + 3)", - "type": "text" - }, - { - "block_id": "p505-b32", - "global_id": 14415, - "bbox": [ - 249.27, - 492.12, - 254.16, - 501.08 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p505-b33", - "global_id": 14416, - "bbox": [ - 116.46, - 534.35, - 249.98, - 543.79 - ], - "text": "where the ROC is −3 < Re{s} < −2.", - "type": "text" - }, - { - "block_id": "p505-b34", - "global_id": 14417, - "bbox": [ - 78.86, - 549.11, - 288.98, - 591.03 - ], - "text": "4.11-15\nUsing the definition, compute the bilateral\nLaplace transform, including the region of\nconvergence (ROC), of the following complex-\nvalued functions:", - "type": "text" - }, - { - "block_id": "p505-b35", - "global_id": 14418, - "bbox": [ - 116.46, - 591.59, - 210.73, - 613.69 - ], - "text": "(a) x1(t) = (j + ejt)u(t)\n(b) x2(t) = jcosh(t)u(−t)", - "type": "text" - }, - { - "block_id": "p505-b36", - "global_id": 14419, - "bbox": [ - 116.96, - 611.15, - 172.32, - 624.65 - ], - "text": "(c) x3(t) = ej( π", - "type": "text" - }, - { - "block_id": "p505-b37", - "global_id": 14420, - "bbox": [ - 116.46, - 612.97, - 252.29, - 635.61 - ], - "text": "4 )u(−t + 1) + jδ(t −5)\n(d) x4(t) = jtu(−t) + δ(t −π)", - "type": "text" - }, - { - "block_id": "p505-b38", - "global_id": 14421, - "bbox": [ - 306.0, - 85.64, - 516.12, - 105.94 - ], - "text": "4.11-16\nA bounded-amplitude signal x(t) has bilateral\nLaplace transform X(s) given by", - "type": "text" - }, - { - "block_id": "p505-b39", - "global_id": 14422, - "bbox": [ - 390.87, - 114.42, - 448.24, - 131.01 - ], - "text": "X(s) =\ns2s", - "type": "text" - }, - { - "block_id": "p505-b40", - "global_id": 14423, - "bbox": [ - 418.79, - 128.13, - 467.69, - 137.47 - ], - "text": "(s −1)(s + 1)", - "type": "text" - }, - { - "block_id": "p505-b41", - "global_id": 14424, - "bbox": [ - 344.12, - 153.89, - 516.12, - 162.86 - ], - "text": "(a) Determine the corresponding region of con-", - "type": "text" - }, - { - "block_id": "p505-b42", - "global_id": 14425, - "bbox": [ - 343.61, - 164.85, - 499.66, - 184.77 - ], - "text": "vergence.\n(b) Determine the time-domain signal x(t).", - "type": "text" - }, - { - "block_id": "p505-b43", - "global_id": 14426, - "bbox": [ - 310.48, - 189.38, - 516.13, - 215.82 - ], - "text": "4.12-1\nExpress the polynomial C20(x) in standard form.\nThat is, determine the coefficients ak of C20(x) =\n%20", - "type": "text" - }, - { - "block_id": "p505-b44", - "global_id": 14427, - "bbox": [ - 352.98, - 210.04, - 394.97, - 223.12 - ], - "text": "k=0 akx20−k.", - "type": "text" - }, - { - "block_id": "p505-b45", - "global_id": 14428, - "bbox": [ - 310.48, - 225.55, - 453.36, - 234.59 - ], - "text": "4.12-2\nConsider an LTIC system with", - "type": "text" - }, - { - "block_id": "p505-b46", - "global_id": 14429, - "bbox": [ - 343.61, - 244.5, - 516.73, - 266.12 - ], - "text": "H(s) =\n1\ns3 + 4s2 + 8s + 8 =\n1\n(s2 + 2s + 4)(s + 2)", - "type": "text" - }, - { - "block_id": "p505-b47", - "global_id": 14430, - "bbox": [ - 362.5, - 268.8, - 494.37, - 290.42 - ], - "text": "=\n1\n(s −2ej2π/3)(s −2e−j2π/3)(s + 2).", - "type": "text" - }, - { - "block_id": "p505-b48", - "global_id": 14431, - "bbox": [ - 344.11, - 306.85, - 516.12, - 315.81 - ], - "text": "(a) Write MATLAB code that accurately plots", - "type": "text" - }, - { - "block_id": "p505-b49", - "global_id": 14432, - "bbox": [ - 343.61, - 317.44, - 516.13, - 348.69 - ], - "text": "the system magnitude response |H(jω)| over\n−10 ≤ω ≤10.\n(b) Write MATLAB code that accurately plots", - "type": "text" - }, - { - "block_id": "p505-b50", - "global_id": 14433, - "bbox": [ - 344.11, - 350.31, - 516.14, - 382.91 - ], - "text": "the system phase response over −10 ≤ω ≤\n10.\n(c) Determine the max value ymax of output y(t)", - "type": "text" - }, - { - "block_id": "p505-b51", - "global_id": 14434, - "bbox": [ - 343.61, - 383.19, - 516.13, - 403.48 - ], - "text": "in response to input x(t) = 2−sin(2t+π/3).\n(d) Draw a parallel representation of this system", - "type": "text" - }, - { - "block_id": "p505-b52", - "global_id": 14435, - "bbox": [ - 359.05, - 405.48, - 516.14, - 436.36 - ], - "text": "using real DFI structures of order 2 or less.\n[Hint: Use MATLAB to perform a partial\nfraction expansion of H(s).]", - "type": "text" - }, - { - "block_id": "p505-b53", - "global_id": 14436, - "bbox": [ - 310.48, - 441.27, - 516.14, - 461.27 - ], - "text": "4.12-3\nConsider the op-amp circuit of Fig. P4.12-3.\nFurther, let RC = 1.", - "type": "text" - }, - { - "block_id": "p505-b54", - "global_id": 14437, - "bbox": [ - 344.11, - 463.26, - 516.14, - 472.23 - ], - "text": "(a) From Fig. P4.12-3, determine the (simpli-", - "type": "text" - }, - { - "block_id": "p505-b55", - "global_id": 14438, - "bbox": [ - 343.61, - 474.22, - 516.14, - 505.1 - ], - "text": "fied, standard form, rational) transfer func-\ntion H(s).\n(b) Use MATLAB to accurately plot |H(jω)|", - "type": "text" - }, - { - "block_id": "p505-b56", - "global_id": 14439, - "bbox": [ - 344.12, - 506.73, - 516.12, - 527.03 - ], - "text": "over −10 ≤ω ≤10.\n(c) Determine the output of this system in", - "type": "text" - }, - { - "block_id": "p505-b57", - "global_id": 14440, - "bbox": [ - 343.61, - 528.64, - 516.13, - 548.94 - ], - "text": "response to x(t) = cos(10t) −1.\n(d) The circuit of Fig. P4.12-3 contains two", - "type": "text" - }, - { - "block_id": "p505-b58", - "global_id": 14441, - "bbox": [ - 359.05, - 550.94, - 516.14, - 636.62 - ], - "text": "capacitors. Suppose one capacitor must be\na 25% tolerance part, while the other must\nbe a 10% tolerance part. If the goal is to\npreserve the original magnitude response,\nshould you use the 25% tolerance capacitor\nwith the first op-amp or the second op-amp?\nJustify your answer with appropriate MAT-\nLAB simulations.", - "type": "text" - } - ] - }, - { - "page_num": 506, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p506-b0", - "global_id": 14442, - "bbox": [ - 60.0, - 60.36, - 332.52, - 69.45 - ], - "text": "486\nCHAPTER 4\nCONTINUOUS-TIME SYSTEM ANALYSIS", - "type": "text" - }, - { - "block_id": "p506-b1", - "global_id": 14443, - "bbox": [ - 98.9, - 136.23, - 110.77, - 153.58 - ], - "text": "+\nx(t)", - "type": "text" - }, - { - "block_id": "p506-b2", - "global_id": 14444, - "bbox": [ - 101.73, - 154.37, - 107.94, - 162.34 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p506-b3", - "global_id": 14445, - "bbox": [ - 144.24, - 124.37, - 149.11, - 132.34 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p506-b4", - "global_id": 14446, - "bbox": [ - 208.19, - 132.01, - 212.85, - 137.98 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p506-b5", - "global_id": 14447, - "bbox": [ - 208.19, - 150.01, - 212.85, - 155.98 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p506-b6", - "global_id": 14448, - "bbox": [ - 206.04, - 99.67, - 211.35, - 107.64 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p506-b7", - "global_id": 14449, - "bbox": [ - 269.34, - 133.37, - 274.21, - 141.34 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p506-b8", - "global_id": 14450, - "bbox": [ - 322.49, - 141.01, - 327.15, - 146.98 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p506-b9", - "global_id": 14451, - "bbox": [ - 322.49, - 159.01, - 327.15, - 164.98 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p506-b10", - "global_id": 14452, - "bbox": [ - 263.49, - 79.31, - 280.08, - 87.61 - ], - "text": "R/10", - "type": "text" - }, - { - "block_id": "p506-b11", - "global_id": 14453, - "bbox": [ - 338.34, - 110.11, - 343.65, - 118.08 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p506-b12", - "global_id": 14454, - "bbox": [ - 383.64, - 142.37, - 388.51, - 150.34 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p506-b13", - "global_id": 14455, - "bbox": [ - 422.39, - 150.01, - 427.05, - 155.98 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p506-b14", - "global_id": 14456, - "bbox": [ - 422.39, - 168.01, - 427.05, - 173.98 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p506-b15", - "global_id": 14457, - "bbox": [ - 438.54, - 119.33, - 443.41, - 127.3 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p506-b16", - "global_id": 14458, - "bbox": [ - 481.41, - 163.11, - 493.25, - 180.46 - ], - "text": "+\ny(t)", - "type": "text" - }, - { - "block_id": "p506-b17", - "global_id": 14459, - "bbox": [ - 484.23, - 181.5, - 490.44, - 189.47 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p506-b18", - "global_id": 14460, - "bbox": [ - 138.39, - 154.91, - 278.08, - 172.21 - ], - "text": "R/2\nR/26", - "type": "text" - }, - { - "block_id": "p506-b19", - "global_id": 14461, - "bbox": [ - 104.83, - 219.01, - 160.95, - 227.98 - ], - "text": "Figure P4.12-3", - "type": "text" - }, - { - "block_id": "p506-b20", - "global_id": 14462, - "bbox": [ - 153.85, - 286.85, - 164.95, - 294.93 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p506-b21", - "global_id": 14463, - "bbox": [ - 157.4, - 295.8, - 161.4, - 303.8 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p506-b22", - "global_id": 14464, - "bbox": [ - 157.14, - 278.07, - 161.65, - 286.07 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p506-b23", - "global_id": 14465, - "bbox": [ - 350.15, - 295.69, - 361.26, - 303.78 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p506-b24", - "global_id": 14466, - "bbox": [ - 353.7, - 304.64, - 357.7, - 312.64 - ], - "text": "–", - "type": "text" - }, - { - "block_id": "p506-b25", - "global_id": 14467, - "bbox": [ - 304.98, - 286.69, - 357.96, - 294.91 - ], - "text": "–\n+", - "type": "text" - }, - { - "block_id": "p506-b26", - "global_id": 14468, - "bbox": [ - 232.23, - 261.45, - 309.5, - 279.96 - ], - "text": "+\nR2", - "type": "text" - }, - { - "block_id": "p506-b27", - "global_id": 14469, - "bbox": [ - 290.82, - 243.24, - 299.16, - 252.85 - ], - "text": "C1", - "type": "text" - }, - { - "block_id": "p506-b28", - "global_id": 14470, - "bbox": [ - 178.97, - 261.45, - 186.86, - 271.06 - ], - "text": "R1", - "type": "text" - }, - { - "block_id": "p506-b29", - "global_id": 14471, - "bbox": [ - 261.55, - 291.1, - 269.88, - 300.7 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p506-b30", - "global_id": 14472, - "bbox": [ - 383.62, - 322.95, - 439.75, - 331.92 - ], - "text": "Figure P4.12-4", - "type": "text" - }, - { - "block_id": "p506-b31", - "global_id": 14473, - "bbox": [ - 57.59, - 349.85, - 263.24, - 380.8 - ], - "text": "4.12-4\nDesign an order-12 Butterworth lowpass filter\nwith a cutoff frequency of ωc = 2π5000 by\ncompleting the following.", - "type": "text" - }, - { - "block_id": "p506-b32", - "global_id": 14474, - "bbox": [ - 91.22, - 382.8, - 263.24, - 391.76 - ], - "text": "(a) Locate and plot the filter’s poles and zeros", - "type": "text" - }, - { - "block_id": "p506-b33", - "global_id": 14475, - "bbox": [ - 90.72, - 393.75, - 263.24, - 435.59 - ], - "text": "in the complex plane. Plot the correspond-\ning magnitude response |HLP(jω)| to verify\nproper design.\n(b) Setting all resistor values to 100,000, deter-", - "type": "text" - }, - { - "block_id": "p506-b34", - "global_id": 14476, - "bbox": [ - 106.16, - 437.59, - 263.24, - 545.19 - ], - "text": "mine the capacitor values to implement the\nfilter using a cascade of six second-order\nSallen–Key circuit sections. The form of a\nSallen–Key stage is shown in Fig. P4.12-4.\nOn a single plot, plot the magnitude\nresponse of each section as well as the over-\nall magnitude response. Identify the poles\nthat correspond to each section’s magnitude\nresponse curve. Are the capacitor values\nrealistic?", - "type": "text" - }, - { - "block_id": "p506-b35", - "global_id": 14477, - "bbox": [ - 57.59, - 552.57, - 263.24, - 605.45 - ], - "text": "4.12-5\nRather\nthan\na\nButterworth\nfilter,\nrepeat\nProb. 4.12-4 for a Chebyshev LPF with R = 3\ndB of passband ripple. Since each Sallen–Key\nstage is constrained to have unity gain at dc, an\noverall gain error of 1/", - "type": "text" - }, - { - "block_id": "p506-b36", - "global_id": 14478, - "bbox": [ - 173.3, - 588.45, - 180.88, - 597.42 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p506-b37", - "global_id": 14479, - "bbox": [ - 180.88, - 593.88, - 253.82, - 605.44 - ], - "text": "1 + ϵ2 is acceptable.", - "type": "text" - }, - { - "block_id": "p506-b38", - "global_id": 14480, - "bbox": [ - 57.59, - 612.82, - 263.25, - 634.16 - ], - "text": "4.12-6\nAn analog lowpass filter with cutoff frequency\nωc can be transformed into a highpass filter", - "type": "text" - }, - { - "block_id": "p506-b39", - "global_id": 14481, - "bbox": [ - 317.86, - 349.43, - 490.4, - 380.68 - ], - "text": "with cutoff frequency ωc by using an RC–CR\ntransformation rule: each resistor Ri is replaced\nby a capacitor C′", - "type": "text" - }, - { - "block_id": "p506-b40", - "global_id": 14482, - "bbox": [ - 317.86, - 371.35, - 490.36, - 392.99 - ], - "text": "i = 1/Riωc and each capacitor\nCi is replaced by a resistor R′", - "type": "text" - }, - { - "block_id": "p506-b41", - "global_id": 14483, - "bbox": [ - 317.86, - 382.31, - 490.39, - 424.52 - ], - "text": "i = 1/Ciωc.\nUse this rule to design an order-8 Butterworth\nhighpass filter with ωc = 2π4000 by completing\nthe following.", - "type": "text" - }, - { - "block_id": "p506-b42", - "global_id": 14484, - "bbox": [ - 318.37, - 426.51, - 490.39, - 435.48 - ], - "text": "(a) Design an order-8 Butterworth lowpass", - "type": "text" - }, - { - "block_id": "p506-b43", - "global_id": 14485, - "bbox": [ - 317.86, - 437.1, - 490.39, - 534.11 - ], - "text": "filter with ωc = 2π4000 by using four\nsecond-order Sallen–Key circuit stages, the\nform of which is shown in Fig. P4.12-4.\nGive resistor and capacitor values for each\nstage. Choose the resistors so that the\nRC–CR transformation will result in 1 nF\ncapacitors. At this point, are the component\nvalues realistic?\n(b) Draw an RC–CR transformed Sallen–Key", - "type": "text" - }, - { - "block_id": "p506-b44", - "global_id": 14486, - "bbox": [ - 333.31, - 536.11, - 490.38, - 566.99 - ], - "text": "circuit stage. Determine the transfer func-\ntion H(s) of the transformed stage in terms\nof the variables R′", - "type": "text" - }, - { - "block_id": "p506-b45", - "global_id": 14487, - "bbox": [ - 318.37, - 556.34, - 490.39, - 577.95 - ], - "text": "1, R′\n2, C′\n1, and C′\n2.\n(c) Transform the LPF designed in part (a) by", - "type": "text" - }, - { - "block_id": "p506-b46", - "global_id": 14488, - "bbox": [ - 333.31, - 579.85, - 490.39, - 632.74 - ], - "text": "using an RC–CR transformation. Give the\nresistor and capacitor values for each stage.\nAre the component values realistic?\nUsing H(s) derived in part (b), plot the\nmagnitude response of each section as well", - "type": "text" - } - ] - }, - { - "page_num": 507, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p507-b0", - "global_id": 14489, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n487", - "type": "text" - }, - { - "block_id": "p507-b1", - "global_id": 14490, - "bbox": [ - 131.89, - 85.9, - 288.98, - 149.66 - ], - "text": "as the overall magnitude response. Does\nthe overall response look like a highpass\nButterworth filter?\nPlot the HPF system poles and zeros in the\ncomplex s plane. How do these locations\ncompare with those of the Butterworth LPF?", - "type": "text" - }, - { - "block_id": "p507-b2", - "global_id": 14491, - "bbox": [ - 83.34, - 154.27, - 288.98, - 185.52 - ], - "text": "4.12-7\nRepeat Prob. 4.12-6, using ωc = 2π1500 and an\norder-16 filter. That is, eight second-order stages\nneed to be designed.", - "type": "text" - }, - { - "block_id": "p507-b3", - "global_id": 14492, - "bbox": [ - 83.34, - 190.43, - 288.98, - 254.26 - ], - "text": "4.12-8\nRather\nthan\na\nButterworth\nfilter,\nrepeat\nProb. 4.12-6 for a Chebyshev LPF with R = 3 dB\nof passband ripple. Since each transformed\nSallen–Key stage is constrained to have unity\ngain at ω = ∞, an overall gain error of\n1/\n√", - "type": "text" - }, - { - "block_id": "p507-b4", - "global_id": 14493, - "bbox": [ - 132.74, - 242.7, - 205.68, - 254.26 - ], - "text": "1 + ϵ2 is acceptable.", - "type": "text" - }, - { - "block_id": "p507-b5", - "global_id": 14494, - "bbox": [ - 83.34, - 259.17, - 288.99, - 335.86 - ], - "text": "4.12-9\nThe MATLAB signal-processing toolbox func-\ntion butter helps design analog Butterworth\nfilters. Use MATLAB help to learn how butter\nworks. For each of the following cases, design\nthe filter, plot the filter’s poles and zeros in the\ncomplex s plane, and plot the decibel magnitude\nresponse 20log10 |H(jω)|:", - "type": "text" - }, - { - "block_id": "p507-b6", - "global_id": 14495, - "bbox": [ - 116.97, - 335.96, - 289.0, - 344.93 - ], - "text": "(a) Design a sixth-order analog lowpass filter", - "type": "text" - }, - { - "block_id": "p507-b7", - "global_id": 14496, - "bbox": [ - 116.46, - 346.54, - 289.0, - 366.84 - ], - "text": "with ωc = 2π3500.\n(b) Design a sixth-order analog highpass filter", - "type": "text" - }, - { - "block_id": "p507-b8", - "global_id": 14497, - "bbox": [ - 116.96, - 368.47, - 289.0, - 388.76 - ], - "text": "with ωc = 2π3500.\n(c) Design a sixth-order analog bandpass filter", - "type": "text" - }, - { - "block_id": "p507-b9", - "global_id": 14498, - "bbox": [ - 116.46, - 390.75, - 289.0, - 410.68 - ], - "text": "with a passband between 2 and 4 kHz.\n(d) Design a sixth-order analog bandstop filter", - "type": "text" - }, - { - "block_id": "p507-b10", - "global_id": 14499, - "bbox": [ - 131.89, - 412.67, - 268.1, - 421.64 - ], - "text": "with a stopband between 2 and 4 kHz.", - "type": "text" - }, - { - "block_id": "p507-b11", - "global_id": 14500, - "bbox": [ - 78.86, - 426.55, - 288.98, - 446.8 - ], - "text": "4.12-10\nThe MATLAB signal-processing toolbox func-\ntion cheby1 helps design analog Chebyshev", - "type": "text" - }, - { - "block_id": "p507-b12", - "global_id": 14501, - "bbox": [ - 343.61, - 85.9, - 516.14, - 160.62 - ], - "text": "type I filters. A Chebyshev type I filter has a\npassband ripple and a smooth stopband. Set-\nting the passband ripple to Rp = 3 dB, repeat\nProb. 4.12-9 using the cheby1 command. With\nall other parameters held constant, what is the\ngeneral effect of reducing Rp, the allowable\npassband ripple?", - "type": "text" - }, - { - "block_id": "p507-b13", - "global_id": 14502, - "bbox": [ - 306.0, - 165.54, - 516.14, - 262.25 - ], - "text": "4.12-11\nThe MATLAB signal-processing toolbox func-\ntion cheby2 helps design analog Chebyshev\ntype II filters. A Chebyshev type II filter has\na smooth passband and ripple in the stopband.\nSetting the stopband ripple Rs = 20 dB down,\nrepeat Prob. 4.12-9 using the cheby2 command.\nWith all other parameters held constant, what is\nthe general effect of increasing Rs, the minimum\nstopband attenuation?", - "type": "text" - }, - { - "block_id": "p507-b14", - "global_id": 14503, - "bbox": [ - 306.0, - 267.16, - 516.14, - 341.96 - ], - "text": "4.12-12\nThe MATLAB signal-processing toolbox func-\ntion ellip helps design analog elliptic filters.\nAn elliptic filter has ripple in both the passband\nand the stopband. Setting the passband ripple to\nRp = 3 dB and the stopband ripple Rs = 20 dB\ndown, repeat Prob. 4.12-9 using the ellip\ncommand.", - "type": "text" - }, - { - "block_id": "p507-b15", - "global_id": 14504, - "bbox": [ - 306.0, - 344.97, - 516.16, - 378.58 - ], - "text": "4.12-13\nUsing the definition CN(x)=cosh(N cosh−1(x)),\nprove\nthe\nrecursive\nrelation\nCN(x)\n=\n2xCN−1(x) −CN−2(x).", - "type": "text" - }, - { - "block_id": "p507-b16", - "global_id": 14505, - "bbox": [ - 306.0, - 382.76, - 516.14, - 446.59 - ], - "text": "4.12-14\nProve that the poles of a Chebyshev filter,\nwhich are located at pk = ωc sinh(ξ)sin(φk) +\njωc cosh(ξ)cos(φk), lie on an ellipse. [Hint: The\nequation of an ellipse in the x–y plane is (x/a)2+\n(y/b)2 = 1, where constants a and b define the\nmajor and minor axes of the ellipse.]", - "type": "text" - } - ] - }, - { - "page_num": 508, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p508-b0", - "global_id": 14506, - "bbox": [ - 64.47, - 67.04, - 133.54, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p508-b1", - "global_id": 14507, - "bbox": [ - 146.48, - 100.33, - 401.31, - 173.26 - ], - "text": "DISCRETE-TIME SYSTEM\nANALYSIS USING THE\nz-TRANSFORM", - "type": "text" - }, - { - "block_id": "p508-b2", - "global_id": 14508, - "bbox": [ - 114.3, - 79.92, - 133.73, - 118.77 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p508-b3", - "global_id": 14509, - "bbox": [ - 101.84, - 266.04, - 490.41, - 335.88 - ], - "text": "The counterpart of the Laplace transform for discrete-time systems is the z-transform. The\nLaplace transform converts integro-differential equations into algebraic equations. In the same\nway, the z-transforms changes difference equations into algebraic equations, thereby simplifying\nthe analysis of discrete-time systems. The z-transform method of analysis of discrete-time systems\nparallels the Laplace transform method of analysis of continuous-time systems, with some minor\ndifferences. In fact, we shall see that the z-transform is the Laplace transform in disguise.", - "type": "text" - }, - { - "block_id": "p508-b4", - "global_id": 14510, - "bbox": [ - 101.84, - 337.87, - 490.41, - 431.52 - ], - "text": "The behavior of discrete-time systems is similar to that of continuous-time systems (with\nsome differences). The frequency-domain analysis of discrete-time systems is based on the fact\n(proved in Sec. 3.8-2) that the response of a linear, time-invariant, discrete-time (LTID) system to\nan everlasting exponential zn is the same exponential (within a multiplicative constant) given by\nH[z]zn. We then express an input x[n] as a sum of (everlasting) exponentials of the form zn. The\nsystem response to x[n] is then found as a sum of the system’s responses to all these exponential\ncomponents. The tool that allows us to represent an arbitrary input x[n] as a sum of (everlasting)\nexponentials of the form zn is the z-transform.", - "type": "text" - }, - { - "block_id": "p508-b5", - "global_id": 14511, - "bbox": [ - 102.2, - 459.02, - 251.42, - 473.97 - ], - "text": "5.1 THE z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p508-b6", - "global_id": 14512, - "bbox": [ - 101.84, - 479.54, - 294.97, - 489.92 - ], - "text": "We define X[z], the direct z-transform of x[n], as", - "type": "text" - }, - { - "block_id": "p508-b7", - "global_id": 14513, - "bbox": [ - 255.68, - 507.4, - 282.54, - 517.68 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p508-b8", - "global_id": 14514, - "bbox": [ - 288.28, - 497.23, - 302.38, - 507.9 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p508-b9", - "global_id": 14515, - "bbox": [ - 284.59, - 521.25, - 306.06, - 528.44 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p508-b10", - "global_id": 14516, - "bbox": [ - 307.18, - 505.68, - 490.39, - 517.78 - ], - "text": "x[n]z−n\n(5.1)", - "type": "text" - }, - { - "block_id": "p508-b11", - "global_id": 14517, - "bbox": [ - 101.84, - 536.72, - 490.4, - 559.04 - ], - "text": "where z is a complex variable. The signal x[n], which is the inverse z-transform of X[z], can be\nobtained from X[z] by using the following inverse z-transformation:", - "type": "text" - }, - { - "block_id": "p508-b12", - "global_id": 14518, - "bbox": [ - 245.69, - 567.41, - 289.54, - 591.43 - ], - "text": "x[n] =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p508-b13", - "global_id": 14519, - "bbox": [ - 291.83, - 560.42, - 490.39, - 584.35 - ], - "text": "5\nX[z]zn−1 dz\n(5.2)", - "type": "text" - }, - { - "block_id": "p508-b14", - "global_id": 14520, - "bbox": [ - 101.84, - 600.91, - 149.66, - 610.87 - ], - "text": "The symbol", - "type": "text" - }, - { - "block_id": "p508-b15", - "global_id": 14521, - "bbox": [ - 101.85, - 592.47, - 490.41, - 634.79 - ], - "text": "6\nindicates an integration in counterclockwise direction around a closed path in the\ncomplex plane (see Fig. 5.1). We derive this z-transform pair later, in Ch. 9, as an extension of the\ndiscrete-time Fourier transform pair.", - "type": "text" - }, - { - "block_id": "p508-b16", - "global_id": 14522, - "bbox": [ - 60.0, - 656.12, - 74.94, - 666.22 - ], - "text": "488", - "type": "text" - } - ] - }, - { - "page_num": 509, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p509-b0", - "global_id": 14523, - "bbox": [ - 388.32, - 62.57, - 516.13, - 71.98 - ], - "text": "5.1\nThe z-Transform\n489", - "type": "text" - }, - { - "block_id": "p509-b1", - "global_id": 14524, - "bbox": [ - 127.59, - 85.82, - 516.15, - 119.69 - ], - "text": "As in the case of the Laplace transform, we need not worry about this integral at this point\nbecause inverse z-transforms of many signals of engineering interest can be found in a z-transform\ntable. The direct and inverse z-transforms can be expressed symbolically as", - "type": "text" - }, - { - "block_id": "p509-b2", - "global_id": 14525, - "bbox": [ - 229.67, - 129.71, - 414.06, - 141.81 - ], - "text": "X[z] = Z{x[n]}\nand\nx[n] = Z−1{X[z]}", - "type": "text" - }, - { - "block_id": "p509-b3", - "global_id": 14526, - "bbox": [ - 127.59, - 153.96, - 176.29, - 163.93 - ], - "text": "or simply as", - "type": "text" - }, - { - "block_id": "p509-b4", - "global_id": 14527, - "bbox": [ - 293.7, - 165.81, - 350.03, - 176.08 - ], - "text": "x[n] ⇐⇒X[z]", - "type": "text" - }, - { - "block_id": "p509-b5", - "global_id": 14528, - "bbox": [ - 127.6, - 185.36, - 164.38, - 195.32 - ], - "text": "Note that", - "type": "text" - }, - { - "block_id": "p509-b6", - "global_id": 14529, - "bbox": [ - 210.32, - 195.47, - 433.41, - 207.57 - ], - "text": "Z−1[Z{x[n]}] = x[n]\nand\nZ[Z−1{X[z]}] = X[z]", - "type": "text" - }, - { - "block_id": "p509-b7", - "global_id": 14530, - "bbox": [ - 127.59, - 221.98, - 389.21, - 248.52 - ], - "text": "LINEARITY OF THE z-TRANSFORM\nLike the Laplace transform, the z-transform is a linear operator. If", - "type": "text" - }, - { - "block_id": "p509-b8", - "global_id": 14531, - "bbox": [ - 230.91, - 260.26, - 412.81, - 271.41 - ], - "text": "x1[n] ⇐⇒X1[z]\nand\nx2[n] ⇐⇒X2[z]", - "type": "text" - }, - { - "block_id": "p509-b9", - "global_id": 14532, - "bbox": [ - 127.6, - 282.79, - 144.74, - 292.75 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p509-b10", - "global_id": 14533, - "bbox": [ - 240.84, - 294.63, - 402.89, - 305.78 - ], - "text": "a1x1[n] + a2x2[n] ⇐⇒a1X1[z] + a2X2[z]", - "type": "text" - }, - { - "block_id": "p509-b11", - "global_id": 14534, - "bbox": [ - 127.59, - 314.07, - 516.11, - 336.1 - ], - "text": "The proof is trivial and follows from the definition of the z-transform. This result can be extended\nto finite sums.", - "type": "text" - }, - { - "block_id": "p509-b12", - "global_id": 14535, - "bbox": [ - 127.89, - 350.5, - 303.49, - 363.05 - ], - "text": "THE UNILATERAL z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p509-b13", - "global_id": 14536, - "bbox": [ - 127.59, - 367.08, - 516.14, - 448.78 - ], - "text": "For the same reasons discussed in Ch. 4, we find it convenient to consider the unilateral\nz-transform. As seen for the Laplace case, the bilateral transform has some complications because\nof the non-uniqueness of the inverse transform. In contrast, the unilateral transform has a unique\ninverse. This fact simplifies the analysis problem considerably, but at a price: the unilateral version\ncan handle only causal signals and systems. Fortunately, most of the practical cases are causal. The\nmore general bilateral z-transform is discussed later, in Sec. 5.8. In practice, the term z-transform\ngenerally means the unilateral z-transform.", - "type": "text" - }, - { - "block_id": "p509-b14", - "global_id": 14537, - "bbox": [ - 127.59, - 450.67, - 516.15, - 496.59 - ], - "text": "In a basic sense, there is no difference between the unilateral and the bilateral z-transform.\nThe unilateral transform is the bilateral transform that deals with a subclass of signals starting at\nn = 0 (causal signals). Hence, the definition of the unilateral transform is the same as that of the\nbilateral [Eq. (5.1)], except that the limits of the sum are from 0 to ∞:", - "type": "text" - }, - { - "block_id": "p509-b15", - "global_id": 14538, - "bbox": [ - 285.12, - 516.47, - 311.98, - 526.74 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p509-b16", - "global_id": 14539, - "bbox": [ - 314.03, - 506.29, - 328.13, - 516.96 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p509-b17", - "global_id": 14540, - "bbox": [ - 314.86, - 530.85, - 327.27, - 538.12 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p509-b18", - "global_id": 14541, - "bbox": [ - 329.23, - 514.73, - 516.13, - 526.84 - ], - "text": "x[n]z−n\n(5.3)", - "type": "text" - }, - { - "block_id": "p509-b19", - "global_id": 14542, - "bbox": [ - 127.59, - 547.91, - 509.56, - 557.97 - ], - "text": "The expression for the inverse z-transform in Eq. (5.2) remains valid for the unilateral case also.", - "type": "text" - }, - { - "block_id": "p509-b20", - "global_id": 14543, - "bbox": [ - 127.89, - 572.0, - 380.35, - 584.93 - ], - "text": "THE REGION OF CONVERGENCE (ROC) OF X[z]", - "type": "text" - }, - { - "block_id": "p509-b21", - "global_id": 14544, - "bbox": [ - 127.59, - 588.55, - 516.15, - 634.79 - ], - "text": "The sum in Eq. (5.1) [or Eq. (5.3)] defining the direct z-transform X[z] may not converge (exist)\nfor all values of z. The values of z (the region in the complex plane) for which the sum in\nEq. (5.1) converges (or exists) are called the region of existence, or more commonly the region\nof convergence (ROC), for X[z]. This concept will become clear in the following example.", - "type": "text" - } - ] - }, - { - "page_num": 510, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p510-b0", - "global_id": 14545, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "490\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p510-b1", - "global_id": 14546, - "bbox": [ - 76.77, - 93.91, - 423.02, - 105.87 - ], - "text": "EXAMPLE 5.1\nBilateral z-Transform of a Causal Exponential", - "type": "text" - }, - { - "block_id": "p510-b2", - "global_id": 14547, - "bbox": [ - 103.16, - 120.67, - 385.35, - 132.05 - ], - "text": "Find the z-transform and the corresponding ROC for the signal γ nu[n].", - "type": "text" - }, - { - "block_id": "p510-b3", - "global_id": 14548, - "bbox": [ - 103.16, - 154.96, - 158.31, - 164.93 - ], - "text": "By definition,", - "type": "text" - }, - { - "block_id": "p510-b4", - "global_id": 14549, - "bbox": [ - 247.79, - 174.63, - 274.65, - 184.9 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p510-b5", - "global_id": 14550, - "bbox": [ - 276.7, - 164.46, - 290.8, - 175.13 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p510-b6", - "global_id": 14551, - "bbox": [ - 277.54, - 189.03, - 289.95, - 196.29 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p510-b7", - "global_id": 14552, - "bbox": [ - 291.9, - 172.91, - 331.89, - 184.91 - ], - "text": "γ nu[n]z−n", - "type": "text" - }, - { - "block_id": "p510-b8", - "global_id": 14553, - "bbox": [ - 103.16, - 202.13, - 214.35, - 212.51 - ], - "text": "Since u[n] = 1 for all n ≥0,", - "type": "text" - }, - { - "block_id": "p510-b9", - "global_id": 14554, - "bbox": [ - 168.52, - 230.68, - 195.37, - 240.96 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p510-b10", - "global_id": 14555, - "bbox": [ - 197.42, - 220.51, - 211.52, - 231.18 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p510-b11", - "global_id": 14556, - "bbox": [ - 198.27, - 245.07, - 210.67, - 252.33 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p510-b12", - "global_id": 14557, - "bbox": [ - 212.62, - 216.7, - 225.54, - 233.66 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p510-b13", - "global_id": 14558, - "bbox": [ - 221.9, - 238.07, - 225.78, - 248.03 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p510-b14", - "global_id": 14559, - "bbox": [ - 228.34, - 216.7, - 238.55, - 228.51 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p510-b15", - "global_id": 14560, - "bbox": [ - 241.1, - 230.68, - 265.21, - 241.06 - ], - "text": "= 1 +", - "type": "text" - }, - { - "block_id": "p510-b16", - "global_id": 14561, - "bbox": [ - 266.76, - 216.7, - 279.69, - 233.66 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p510-b17", - "global_id": 14562, - "bbox": [ - 276.05, - 238.07, - 279.92, - 248.03 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p510-b19", - "global_id": 14563, - "bbox": [ - 290.75, - 230.68, - 298.52, - 240.64 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p510-b20", - "global_id": 14564, - "bbox": [ - 300.07, - 216.7, - 313.0, - 233.66 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p510-b21", - "global_id": 14565, - "bbox": [ - 309.35, - 238.07, - 313.23, - 248.03 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p510-b22", - "global_id": 14566, - "bbox": [ - 315.78, - 216.7, - 325.99, - 228.58 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p510-b23", - "global_id": 14567, - "bbox": [ - 328.04, - 230.68, - 335.81, - 240.64 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p510-b24", - "global_id": 14568, - "bbox": [ - 337.35, - 216.7, - 350.28, - 233.66 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p510-b25", - "global_id": 14569, - "bbox": [ - 346.64, - 238.07, - 350.52, - 248.03 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p510-b26", - "global_id": 14570, - "bbox": [ - 353.08, - 216.7, - 363.29, - 228.58 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p510-b27", - "global_id": 14571, - "bbox": [ - 365.33, - 230.68, - 477.01, - 241.06 - ], - "text": "+ · · · + · · ·\n(5.4)", - "type": "text" - }, - { - "block_id": "p510-b28", - "global_id": 14572, - "bbox": [ - 103.17, - 261.14, - 424.97, - 271.1 - ], - "text": "It is helpful to remember the geometric progression and its sum [see Sec. B.8-3]:", - "type": "text" - }, - { - "block_id": "p510-b29", - "global_id": 14573, - "bbox": [ - 199.89, - 280.15, - 380.29, - 304.17 - ], - "text": "1 + x + x2 + x3 + · · · =\n1\n1 −x\nif\n|x| < 1", - "type": "text" - }, - { - "block_id": "p510-b30", - "global_id": 14574, - "bbox": [ - 103.17, - 311.88, - 281.38, - 321.84 - ], - "text": "Applying this relationship to Eq. (5.4) yields", - "type": "text" - }, - { - "block_id": "p510-b31", - "global_id": 14575, - "bbox": [ - 236.94, - 331.56, - 281.95, - 348.41 - ], - "text": "X[z] =\n1", - "type": "text" - }, - { - "block_id": "p510-b32", - "global_id": 14576, - "bbox": [ - 267.05, - 341.63, - 289.33, - 365.97 - ], - "text": "1 −γ\nz", - "type": "text" - }, - { - "block_id": "p510-b34", - "global_id": 14577, - "bbox": [ - 313.45, - 331.15, - 318.45, - 341.11 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p510-b35", - "global_id": 14578, - "bbox": [ - 314.81, - 345.52, - 318.68, - 355.48 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p510-b36", - "global_id": 14579, - "bbox": [ - 321.24, - 323.69, - 341.33, - 351.59 - ], - "text": "< 1", - "type": "text" - }, - { - "block_id": "p510-b37", - "global_id": 14580, - "bbox": [ - 256.03, - 365.21, - 477.01, - 389.23 - ], - "text": "=\nz\nz −γ\n|z| > |γ |\n(5.5)", - "type": "text" - }, - { - "block_id": "p510-b38", - "global_id": 14581, - "bbox": [ - 103.17, - 397.86, - 477.03, - 432.14 - ], - "text": "Observe that X[z] exists only for |z| > |γ |. For |z| < |γ |, the sum in Eq. (5.4) does not converge;\nit goes to infinity. Therefore, the ROC of X[z] is the shaded region outside the circle of radius\n|γ |, centered at the origin, in the z-plane, as depicted in Fig. 5.1b.", - "type": "text" - }, - { - "block_id": "p510-b39", - "global_id": 14582, - "bbox": [ - 111.73, - 476.04, - 133.61, - 485.57 - ], - "text": "g ku[n]", - "type": "text" - }, - { - "block_id": "p510-b40", - "global_id": 14583, - "bbox": [ - 425.2, - 531.29, - 434.09, - 539.29 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p510-b41", - "global_id": 14584, - "bbox": [ - 339.26, - 460.07, - 348.14, - 468.07 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p510-b42", - "global_id": 14585, - "bbox": [ - 352.37, - 518.46, - 360.85, - 526.65 - ], - "text": "g", - "type": "text" - }, - { - "block_id": "p510-b43", - "global_id": 14586, - "bbox": [ - 108.2, - 583.66, - 420.01, - 600.66 - ], - "text": "k\n0\n1\n2\n3\n4\n5\n6\nRegion of\nconvergence", - "type": "text" - }, - { - "block_id": "p510-b44", - "global_id": 14587, - "bbox": [ - 94.2, - 607.6, - 340.66, - 618.26 - ], - "text": "Figure 5.1 γ nu[n] and the region of convergence of its z-transform.", - "type": "text" - } - ] - }, - { - "page_num": 511, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p511-b0", - "global_id": 14588, - "bbox": [ - 388.32, - 62.57, - 516.13, - 71.98 - ], - "text": "5.1\nThe z-Transform\n491", - "type": "text" - }, - { - "block_id": "p511-b1", - "global_id": 14589, - "bbox": [ - 128.91, - 84.88, - 502.77, - 132.13 - ], - "text": "Later in Eq. (5.52), we show that the z-transform of another signal, −γ nu[−(n + 1)], is\nalso z/(z −γ ). However, the ROC in this case is |z| < |γ |. Clearly, the inverse z-transform of\nz/(z −γ ) is not unique. However, if we restrict the inverse transform to be causal, then the\ninverse transform is unique, namely, γ nu[n].", - "type": "text" - }, - { - "block_id": "p511-b2", - "global_id": 14590, - "bbox": [ - 127.59, - 170.69, - 516.16, - 312.58 - ], - "text": "The ROC is required for evaluating x[n] from X[z], according to Eq. (5.2). The integral in\nEq. (5.2) is a contour integral, implying integration in a counterclockwise direction along a closed\npath centered at the origin and satisfying the condition |z| > |γ |. Thus, any circular path centered at\nthe origin and with a radius greater than |γ | (Fig. 5.1b) will suffice. We can show that the integral\nin Eq. (5.2) along any such path (with a radius greater than |γ |) yields the same result, namely,\nx[n].† Such integration in the complex plane requires a background in the theory of functions of\ncomplex variables. We can avoid this integration by compiling a table of z-transforms (Table 5.1),\nwhere z-transform pairs are tabulated for a variety of signals. To find the inverse z-transform of\nsay, z/(z −γ ), instead of using the complex integration in Eq. (5.2), we consult the table and find\nthe inverse z-transform of z/(z−γ ) as γ nu[n]. Because of the uniqueness property of the unilateral\nz-transform, there is only one inverse for each X[z]. Although the table given here is rather short,\nit comprises the functions of most practical interest.", - "type": "text" - }, - { - "block_id": "p511-b3", - "global_id": 14591, - "bbox": [ - 127.6, - 314.47, - 516.16, - 372.36 - ], - "text": "The situation of the z-transform regarding the uniqueness of the inverse transform is parallel to\nthat of the Laplace transform. For the bilateral case, the inverse z-transform is not unique unless the\nROC is specified. For the unilateral case, the inverse transform is unique; the region of convergence\nneed not be specified to determine the inverse z-transform. For this reason, we shall ignore the ROC\nin the unilateral z-transform Table 5.1.", - "type": "text" - }, - { - "block_id": "p511-b4", - "global_id": 14592, - "bbox": [ - 127.89, - 389.19, - 308.21, - 401.74 - ], - "text": "EXISTENCE OF THE z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p511-b5", - "global_id": 14593, - "bbox": [ - 127.59, - 405.77, - 182.73, - 415.74 - ], - "text": "By definition,", - "type": "text" - }, - { - "block_id": "p511-b6", - "global_id": 14594, - "bbox": [ - 262.36, - 433.04, - 289.21, - 443.31 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p511-b7", - "global_id": 14595, - "bbox": [ - 291.26, - 422.86, - 305.36, - 433.53 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p511-b8", - "global_id": 14596, - "bbox": [ - 292.1, - 447.42, - 304.51, - 454.69 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p511-b9", - "global_id": 14597, - "bbox": [ - 306.46, - 428.92, - 345.65, - 443.31 - ], - "text": "x[n]z−n =", - "type": "text" - }, - { - "block_id": "p511-b10", - "global_id": 14598, - "bbox": [ - 347.7, - 422.86, - 361.79, - 433.53 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p511-b11", - "global_id": 14599, - "bbox": [ - 348.54, - 447.42, - 360.95, - 454.69 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p511-b12", - "global_id": 14600, - "bbox": [ - 364.1, - 426.05, - 380.16, - 436.33 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p511-b13", - "global_id": 14601, - "bbox": [ - 368.21, - 439.84, - 375.57, - 450.39 - ], - "text": "zn", - "type": "text" - }, - { - "block_id": "p511-b14", - "global_id": 14602, - "bbox": [ - 127.59, - 466.35, - 320.29, - 476.42 - ], - "text": "The existence of the z-transform is guaranteed if", - "type": "text" - }, - { - "block_id": "p511-b15", - "global_id": 14603, - "bbox": [ - 273.99, - 501.15, - 306.29, - 511.42 - ], - "text": "|X[z]| ≤", - "type": "text" - }, - { - "block_id": "p511-b16", - "global_id": 14604, - "bbox": [ - 308.33, - 490.97, - 322.43, - 501.65 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p511-b17", - "global_id": 14605, - "bbox": [ - 309.18, - 515.54, - 321.58, - 522.81 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p511-b18", - "global_id": 14606, - "bbox": [ - 324.73, - 494.16, - 346.5, - 504.44 - ], - "text": "|x[n]|", - "type": "text" - }, - { - "block_id": "p511-b19", - "global_id": 14607, - "bbox": [ - 328.83, - 501.15, - 369.73, - 518.5 - ], - "text": "|z|n < ∞", - "type": "text" - }, - { - "block_id": "p511-b20", - "global_id": 14608, - "bbox": [ - 127.59, - 536.03, - 458.56, - 547.55 - ], - "text": "for some |z|. Any signal x[n] that grows no faster than an exponential signal rn", - "type": "text" - }, - { - "block_id": "p511-b21", - "global_id": 14609, - "bbox": [ - 127.59, - 537.49, - 516.13, - 559.5 - ], - "text": "0, for some r0,\nsatisfies this condition. Thus, if", - "type": "text" - }, - { - "block_id": "p511-b22", - "global_id": 14610, - "bbox": [ - 268.42, - 574.61, - 309.39, - 586.38 - ], - "text": "|x[n]| ≤rn", - "type": "text" - }, - { - "block_id": "p511-b23", - "global_id": 14611, - "bbox": [ - 305.68, - 576.42, - 516.13, - 588.45 - ], - "text": "0\nfor some r0\n(5.6)", - "type": "text" - }, - { - "block_id": "p511-b24", - "global_id": 14612, - "bbox": [ - 127.59, - 610.24, - 516.09, - 633.41 - ], - "text": "† Indeed, the path need not even be circular. It can have any odd shape, as long as it encloses the pole(s) of\nX[z] and the path of integration is counterclockwise.", - "type": "text" - } - ] - }, - { - "page_num": 512, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p512-b0", - "global_id": 14613, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "492\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p512-b1", - "global_id": 14614, - "bbox": [ - 101.84, - 85.98, - 287.35, - 95.3 - ], - "text": "TABLE 5.1\nSelect (Unilateral) z-Transform Pairs", - "type": "text" - }, - { - "block_id": "p512-b2", - "global_id": 14615, - "bbox": [ - 101.84, - 106.02, - 405.95, - 115.28 - ], - "text": "No.\nx[n]\nX[z]", - "type": "text" - }, - { - "block_id": "p512-b3", - "global_id": 14616, - "bbox": [ - 106.33, - 123.08, - 401.4, - 133.69 - ], - "text": "1\nδ[n −k]\nz−k", - "type": "text" - }, - { - "block_id": "p512-b4", - "global_id": 14617, - "bbox": [ - 106.33, - 137.83, - 408.94, - 159.54 - ], - "text": "2\nu[n]\nz\nz −1", - "type": "text" - }, - { - "block_id": "p512-b5", - "global_id": 14618, - "bbox": [ - 106.33, - 162.3, - 418.88, - 184.01 - ], - "text": "3\nnu[n]\nz\n(z −1)2", - "type": "text" - }, - { - "block_id": "p512-b6", - "global_id": 14619, - "bbox": [ - 106.33, - 188.71, - 419.25, - 210.7 - ], - "text": "4\nn2u[n]\nz(z + 1)\n(z −1)3", - "type": "text" - }, - { - "block_id": "p512-b7", - "global_id": 14620, - "bbox": [ - 106.33, - 213.65, - 440.61, - 232.54 - ], - "text": "5\nn3u[n]\nz(z2 + 4z + 1)", - "type": "text" - }, - { - "block_id": "p512-b8", - "global_id": 14621, - "bbox": [ - 401.8, - 229.23, - 429.5, - 238.9 - ], - "text": "(z −1)4", - "type": "text" - }, - { - "block_id": "p512-b9", - "global_id": 14622, - "bbox": [ - 106.33, - 241.66, - 408.96, - 263.28 - ], - "text": "6\nγ nu[n]\nz\nz −γ", - "type": "text" - }, - { - "block_id": "p512-b10", - "global_id": 14623, - "bbox": [ - 106.33, - 268.39, - 408.96, - 289.91 - ], - "text": "7\nγ n−1u[n −1]\n1\nz −γ", - "type": "text" - }, - { - "block_id": "p512-b11", - "global_id": 14624, - "bbox": [ - 106.33, - 292.49, - 420.32, - 314.39 - ], - "text": "8\nnγ nu[n]\nγ z\n(z −γ )2", - "type": "text" - }, - { - "block_id": "p512-b12", - "global_id": 14625, - "bbox": [ - 106.33, - 319.18, - 426.51, - 334.8 - ], - "text": "9\nn2γ nu[n]\nγ z(z + γ )", - "type": "text" - }, - { - "block_id": "p512-b13", - "global_id": 14626, - "bbox": [ - 394.03, - 331.51, - 423.17, - 341.08 - ], - "text": "(z −γ )3", - "type": "text" - }, - { - "block_id": "p512-b14", - "global_id": 14627, - "bbox": [ - 101.84, - 345.88, - 295.74, - 361.5 - ], - "text": "10\nn(n −1)(n −2)· · ·(n −m + 1)", - "type": "text" - }, - { - "block_id": "p512-b15", - "global_id": 14628, - "bbox": [ - 230.3, - 346.16, - 430.04, - 367.78 - ], - "text": "γ mm!\nγ nu[n]\nz\n(z −γ )m+1", - "type": "text" - }, - { - "block_id": "p512-b16", - "global_id": 14629, - "bbox": [ - 101.84, - 372.57, - 478.64, - 394.55 - ], - "text": "11a\n|γ |n cos βnu[n]\nz(z −|γ |cos β)\nz2 −(2|γ |cos β)z + |γ |2", - "type": "text" - }, - { - "block_id": "p512-b17", - "global_id": 14630, - "bbox": [ - 101.84, - 399.2, - 478.64, - 421.19 - ], - "text": "11b\n|γ |n sin βnu[n]\nz|γ |sin β\nz2 −(2|γ |cos β)z + |γ |2", - "type": "text" - }, - { - "block_id": "p512-b18", - "global_id": 14631, - "bbox": [ - 101.84, - 425.89, - 489.19, - 441.52 - ], - "text": "12a\nr|γ |n cos(βn + θ)u[n]\nrz[zcos θ −|γ |cos(β −θ)]", - "type": "text" - }, - { - "block_id": "p512-b19", - "global_id": 14632, - "bbox": [ - 396.22, - 435.96, - 483.67, - 447.89 - ], - "text": "z2 −(2|γ |cos β)z + |γ |2", - "type": "text" - }, - { - "block_id": "p512-b20", - "global_id": 14633, - "bbox": [ - 101.84, - 452.74, - 425.51, - 469.62 - ], - "text": "12b\nr|γ |n cos(βn + θ)u[n]\nγ = |γ |ejβ\n(0.5rejθ)z", - "type": "text" - }, - { - "block_id": "p512-b21", - "global_id": 14634, - "bbox": [ - 398.74, - 452.74, - 477.05, - 475.9 - ], - "text": "z −γ\n+ (0.5re−jθ)z", - "type": "text" - }, - { - "block_id": "p512-b22", - "global_id": 14635, - "bbox": [ - 445.82, - 466.06, - 468.4, - 475.9 - ], - "text": "z −γ ∗", - "type": "text" - }, - { - "block_id": "p512-b23", - "global_id": 14636, - "bbox": [ - 101.84, - 480.7, - 444.72, - 502.68 - ], - "text": "12c\nr|γ |n cos(βn + θ)u[n]\nz(Az + B)\nz2 + 2az + |γ |2", - "type": "text" - }, - { - "block_id": "p512-b24", - "global_id": 14637, - "bbox": [ - 193.7, - 518.41, - 206.23, - 527.66 - ], - "text": "r =", - "type": "text" - }, - { - "block_id": "p512-b26", - "global_id": 14638, - "bbox": [ - 216.7, - 508.87, - 289.42, - 521.46 - ], - "text": "A2|γ |2 + B2 −2AaB", - "type": "text" - }, - { - "block_id": "p512-b27", - "global_id": 14639, - "bbox": [ - 236.66, - 522.19, - 268.96, - 534.03 - ], - "text": "|γ |2 −a2", - "type": "text" - }, - { - "block_id": "p512-b28", - "global_id": 14640, - "bbox": [ - 192.7, - 537.88, - 243.48, - 553.5 - ], - "text": "β = cos−1 −a", - "type": "text" - }, - { - "block_id": "p512-b29", - "global_id": 14641, - "bbox": [ - 232.21, - 550.53, - 243.27, - 559.49 - ], - "text": "|γ |", - "type": "text" - }, - { - "block_id": "p512-b30", - "global_id": 14642, - "bbox": [ - 193.7, - 564.39, - 266.62, - 580.01 - ], - "text": "θ = tan−1\nAa −B", - "type": "text" - }, - { - "block_id": "p512-b31", - "global_id": 14643, - "bbox": [ - 231.16, - 578.88, - 236.63, - 587.85 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p512-b33", - "global_id": 14644, - "bbox": [ - 244.07, - 576.01, - 276.37, - 587.85 - ], - "text": "|γ |2 −a2", - "type": "text" - } - ] - }, - { - "page_num": 513, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p513-b0", - "global_id": 14645, - "bbox": [ - 388.32, - 62.57, - 516.13, - 71.98 - ], - "text": "5.1\nThe z-Transform\n493", - "type": "text" - }, - { - "block_id": "p513-b1", - "global_id": 14646, - "bbox": [ - 127.59, - 85.82, - 144.74, - 95.78 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p513-b2", - "global_id": 14647, - "bbox": [ - 236.74, - 129.18, - 269.03, - 139.45 - ], - "text": "|X[z]| ≤", - "type": "text" - }, - { - "block_id": "p513-b3", - "global_id": 14648, - "bbox": [ - 271.08, - 119.0, - 285.18, - 129.68 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p513-b4", - "global_id": 14649, - "bbox": [ - 271.92, - 143.57, - 284.32, - 150.84 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p513-b5", - "global_id": 14650, - "bbox": [ - 286.28, - 115.19, - 302.43, - 133.34 - ], - "text": "r0", - "type": "text" - }, - { - "block_id": "p513-b6", - "global_id": 14651, - "bbox": [ - 294.2, - 136.25, - 303.78, - 146.53 - ], - "text": "|z|", - "type": "text" - }, - { - "block_id": "p513-b7", - "global_id": 14652, - "bbox": [ - 304.97, - 115.19, - 315.18, - 127.01 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p513-b8", - "global_id": 14653, - "bbox": [ - 317.73, - 122.61, - 345.15, - 139.14 - ], - "text": "=\n1", - "type": "text" - }, - { - "block_id": "p513-b9", - "global_id": 14654, - "bbox": [ - 328.74, - 132.98, - 355.37, - 157.0 - ], - "text": "1 −r0\n|z|", - "type": "text" - }, - { - "block_id": "p513-b10", - "global_id": 14655, - "bbox": [ - 377.69, - 129.18, - 406.49, - 140.33 - ], - "text": "|z| > r0", - "type": "text" - }, - { - "block_id": "p513-b11", - "global_id": 14656, - "bbox": [ - 127.59, - 180.54, - 516.13, - 204.29 - ], - "text": "Therefore, X[z] exists for |z| > r0. Almost all practical signals satisfy Eq. (5.6) and are therefore\nz-transformable. Some signal models (e.g., γ n2) grow faster than the exponential signal rn", - "type": "text" - }, - { - "block_id": "p513-b12", - "global_id": 14657, - "bbox": [ - 127.59, - 194.33, - 516.14, - 240.16 - ], - "text": "0 (for\nany r0) and do not satisfy Eq. (5.6) and therefore are not z-transformable. Fortunately, such\nsignals are of little practical or theoretical interest. Even such signals over a finite interval are\nz-transformable.", - "type": "text" - }, - { - "block_id": "p513-b13", - "global_id": 14658, - "bbox": [ - 102.51, - 276.42, - 316.61, - 288.38 - ], - "text": "EXAMPLE 5.2\nBilateral z-Transform", - "type": "text" - }, - { - "block_id": "p513-b14", - "global_id": 14659, - "bbox": [ - 128.9, - 301.71, - 502.75, - 312.09 - ], - "text": "Find the z-transforms of: (a) δ[n], (b) u[n], (c) cos βnu[n], and (d) the signal shown in Fig. 5.2.", - "type": "text" - }, - { - "block_id": "p513-b15", - "global_id": 14660, - "bbox": [ - 271.55, - 430.27, - 275.55, - 438.27 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p513-b16", - "global_id": 14661, - "bbox": [ - 165.85, - 376.1, - 169.85, - 384.1 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p513-b17", - "global_id": 14662, - "bbox": [ - 168.4, - 428.05, - 235.4, - 436.05 - ], - "text": "0\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p513-b18", - "global_id": 14663, - "bbox": [ - 177.44, - 363.35, - 190.32, - 371.43 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p513-b19", - "global_id": 14664, - "bbox": [ - 301.43, - 434.68, - 417.24, - 443.92 - ], - "text": "Figure 5.2 Signal for Ex. 5.2d.", - "type": "text" - }, - { - "block_id": "p513-b20", - "global_id": 14665, - "bbox": [ - 146.84, - 474.28, - 243.22, - 484.25 - ], - "text": "Recall that by definition", - "type": "text" - }, - { - "block_id": "p513-b21", - "global_id": 14666, - "bbox": [ - 209.16, - 503.91, - 236.02, - 514.19 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p513-b22", - "global_id": 14667, - "bbox": [ - 238.07, - 493.74, - 252.16, - 504.41 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p513-b23", - "global_id": 14668, - "bbox": [ - 238.9, - 518.3, - 251.31, - 525.56 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p513-b24", - "global_id": 14669, - "bbox": [ - 253.27, - 496.93, - 338.7, - 514.29 - ], - "text": "x[n]z−n = x[0] + x[1]", - "type": "text" - }, - { - "block_id": "p513-b25", - "global_id": 14670, - "bbox": [ - 328.74, - 496.93, - 368.03, - 521.26 - ], - "text": "z\n+ x[2]", - "type": "text" - }, - { - "block_id": "p513-b26", - "global_id": 14671, - "bbox": [ - 356.07, - 496.93, - 397.36, - 521.26 - ], - "text": "z2 + x[3]", - "type": "text" - }, - { - "block_id": "p513-b27", - "global_id": 14672, - "bbox": [ - 385.4, - 503.91, - 502.76, - 521.26 - ], - "text": "z3 + · · ·\n(5.7)", - "type": "text" - }, - { - "block_id": "p513-b28", - "global_id": 14673, - "bbox": [ - 146.84, - 534.98, - 438.82, - 545.36 - ], - "text": "(a) For x[n] = δ[n], x[0] = 1 and x[2] = x[3] = x[4] = · · · = 0. Therefore,", - "type": "text" - }, - { - "block_id": "p513-b29", - "global_id": 14674, - "bbox": [ - 268.41, - 556.89, - 363.27, - 567.27 - ], - "text": "δ[n] ⇐⇒1\nfor all z", - "type": "text" - }, - { - "block_id": "p513-b30", - "global_id": 14675, - "bbox": [ - 146.85, - 578.81, - 387.39, - 589.19 - ], - "text": "(b) For x[n] = u[n], x[0] = x[1] = x[3] = · · · = 1. Therefore,", - "type": "text" - }, - { - "block_id": "p513-b31", - "global_id": 14676, - "bbox": [ - 256.69, - 598.7, - 307.62, - 615.65 - ], - "text": "X[z] = 1 + 1", - "type": "text" - }, - { - "block_id": "p513-b32", - "global_id": 14677, - "bbox": [ - 303.18, - 598.7, - 327.3, - 622.62 - ], - "text": "z + 1", - "type": "text" - }, - { - "block_id": "p513-b33", - "global_id": 14678, - "bbox": [ - 320.87, - 598.7, - 348.42, - 622.62 - ], - "text": "z2 + 1", - "type": "text" - }, - { - "block_id": "p513-b34", - "global_id": 14679, - "bbox": [ - 342.0, - 605.27, - 374.99, - 622.62 - ], - "text": "z3 + · · ·", - "type": "text" - } - ] - }, - { - "page_num": 514, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p514-b0", - "global_id": 14680, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "494\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p514-b1", - "global_id": 14681, - "bbox": [ - 103.16, - 86.24, - 300.27, - 96.2 - ], - "text": "This geometric sum simplifies [see Sec. B.8-3] to", - "type": "text" - }, - { - "block_id": "p514-b2", - "global_id": 14682, - "bbox": [ - 236.97, - 109.81, - 281.17, - 126.65 - ], - "text": "X[z] =\n1", - "type": "text" - }, - { - "block_id": "p514-b3", - "global_id": 14683, - "bbox": [ - 267.06, - 122.69, - 289.09, - 146.61 - ], - "text": "1 −1\nz", - "type": "text" - }, - { - "block_id": "p514-b5", - "global_id": 14684, - "bbox": [ - 316.95, - 109.81, - 321.93, - 133.73 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p514-b6", - "global_id": 14685, - "bbox": [ - 323.12, - 101.93, - 343.22, - 129.82 - ], - "text": "< 1", - "type": "text" - }, - { - "block_id": "p514-b7", - "global_id": 14686, - "bbox": [ - 256.05, - 145.85, - 334.33, - 169.97 - ], - "text": "=\nz\nz −1\n|z| > 1", - "type": "text" - }, - { - "block_id": "p514-b8", - "global_id": 14687, - "bbox": [ - 103.17, - 181.58, - 144.91, - 191.54 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p514-b9", - "global_id": 14688, - "bbox": [ - 235.95, - 188.29, - 344.23, - 212.42 - ], - "text": "u[n] ⇐⇒\nz\nz −1\n|z| > 1", - "type": "text" - }, - { - "block_id": "p514-b10", - "global_id": 14689, - "bbox": [ - 121.09, - 215.5, - 419.82, - 229.5 - ], - "text": "(c) Recall that cos βn = (ejβn + e−jβn)/2. Moreover, according to Eq. (5.5),", - "type": "text" - }, - { - "block_id": "p514-b11", - "global_id": 14690, - "bbox": [ - 202.83, - 239.91, - 377.34, - 263.93 - ], - "text": "e±jβnu[n] ⇐⇒\nz\nz −e±jβ\n|z| > |e±jβ| = 1", - "type": "text" - }, - { - "block_id": "p514-b12", - "global_id": 14691, - "bbox": [ - 103.16, - 275.65, - 144.91, - 285.61 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p514-b13", - "global_id": 14692, - "bbox": [ - 165.91, - 285.55, - 201.0, - 302.5 - ], - "text": "X[z] = 1", - "type": "text" - }, - { - "block_id": "p514-b14", - "global_id": 14693, - "bbox": [ - 196.02, - 299.61, - 201.0, - 309.57 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p514-b15", - "global_id": 14694, - "bbox": [ - 203.3, - 278.13, - 279.67, - 309.47 - ], - "text": "z\nz −ejβ +\nz\nz −e−jβ", - "type": "text" - }, - { - "block_id": "p514-b16", - "global_id": 14695, - "bbox": [ - 281.88, - 278.13, - 287.31, - 288.09 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p514-b17", - "global_id": 14696, - "bbox": [ - 289.35, - 285.14, - 414.26, - 309.57 - ], - "text": "=\nz(z −cos β)\nz2 −2zcos β + 1\n|z| > 1", - "type": "text" - }, - { - "block_id": "p514-b18", - "global_id": 14697, - "bbox": [ - 103.17, - 316.47, - 477.01, - 338.79 - ], - "text": "(d) Here x[0] = x[1] = x[2] = x[3] = x[4] = 1 and x[5] = x[6] = · · · = 0. Therefore,\naccording to Eq. (5.7),", - "type": "text" - }, - { - "block_id": "p514-b19", - "global_id": 14698, - "bbox": [ - 153.73, - 353.3, - 204.66, - 370.25 - ], - "text": "X[z] = 1 + 1", - "type": "text" - }, - { - "block_id": "p514-b20", - "global_id": 14699, - "bbox": [ - 200.22, - 353.3, - 224.34, - 377.22 - ], - "text": "z + 1", - "type": "text" - }, - { - "block_id": "p514-b21", - "global_id": 14700, - "bbox": [ - 217.91, - 353.3, - 245.46, - 377.22 - ], - "text": "z2 + 1", - "type": "text" - }, - { - "block_id": "p514-b22", - "global_id": 14701, - "bbox": [ - 239.04, - 353.3, - 266.58, - 377.22 - ], - "text": "z3 + 1", - "type": "text" - }, - { - "block_id": "p514-b23", - "global_id": 14702, - "bbox": [ - 260.16, - 349.27, - 358.18, - 377.22 - ], - "text": "z4 = z4 + z3 + z2 + z + 1", - "type": "text" - }, - { - "block_id": "p514-b24", - "global_id": 14703, - "bbox": [ - 316.3, - 359.87, - 426.46, - 377.22 - ], - "text": "z4\nfor all z̸ = 0", - "type": "text" - }, - { - "block_id": "p514-b25", - "global_id": 14704, - "bbox": [ - 103.17, - 388.94, - 477.02, - 422.82 - ], - "text": "We can also express this result in a more compact form by summing the geometric progression\non the right-hand side of the foregoing equation. From the result in Sec. B.8-3 with r = 1/z,m =\n0, and n = 4, we obtain", - "type": "text" - }, - { - "block_id": "p514-b26", - "global_id": 14705, - "bbox": [ - 208.92, - 457.88, - 235.78, - 468.16 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p514-b27", - "global_id": 14706, - "bbox": [ - 239.02, - 430.34, - 251.92, - 447.72 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p514-b28", - "global_id": 14707, - "bbox": [ - 247.49, - 451.72, - 251.36, - 461.68 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p514-b29", - "global_id": 14708, - "bbox": [ - 253.11, - 430.34, - 263.32, - 442.23 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p514-b30", - "global_id": 14709, - "bbox": [ - 265.37, - 444.33, - 273.14, - 454.29 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p514-b31", - "global_id": 14710, - "bbox": [ - 274.69, - 430.34, - 287.6, - 447.72 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p514-b32", - "global_id": 14711, - "bbox": [ - 283.16, - 451.72, - 287.04, - 461.68 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p514-b33", - "global_id": 14712, - "bbox": [ - 288.79, - 430.34, - 299.0, - 442.23 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p514-b34", - "global_id": 14713, - "bbox": [ - 258.84, - 464.2, - 280.86, - 488.12 - ], - "text": "1\nz −1", - "type": "text" - }, - { - "block_id": "p514-b35", - "global_id": 14714, - "bbox": [ - 302.74, - 451.22, - 371.26, - 475.33 - ], - "text": "=\nz\nz −1(1 −z−5)", - "type": "text" - }, - { - "block_id": "p514-b36", - "global_id": 14715, - "bbox": [ - 107.82, - 564.5, - 298.73, - 576.45 - ], - "text": "DRILL 5.1\nBilateral z-Transform", - "type": "text" - }, - { - "block_id": "p514-b37", - "global_id": 14716, - "bbox": [ - 125.76, - 591.45, - 341.89, - 601.51 - ], - "text": "(a) Find the z-transform of a signal shown in Fig. 5.3.", - "type": "text" - }, - { - "block_id": "p514-b38", - "global_id": 14717, - "bbox": [ - 125.76, - 606.08, - 406.22, - 616.46 - ], - "text": "(b) Use pair 12a (Table 5.1) to find the z-transform of x[n] = 20.65(", - "type": "text" - }, - { - "block_id": "p514-b39", - "global_id": 14718, - "bbox": [ - 406.22, - 597.66, - 414.64, - 607.62 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p514-b40", - "global_id": 14719, - "bbox": [ - 142.36, - 602.88, - 484.41, - 628.41 - ], - "text": "2)n cos[(π/4)n −\n1.415]u[n].", - "type": "text" - } - ] - }, - { - "page_num": 515, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p515-b0", - "global_id": 14720, - "bbox": [ - 388.32, - 62.57, - 516.13, - 71.98 - ], - "text": "5.1\nThe z-Transform\n495", - "type": "text" - }, - { - "block_id": "p515-b1", - "global_id": 14721, - "bbox": [ - 133.84, - 92.39, - 196.06, - 103.34 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p515-b2", - "global_id": 14722, - "bbox": [ - 151.5, - 104.24, - 300.61, - 125.14 - ], - "text": "(a) X[z] = z5 + z4 + +z3 + z2 + z + 1", - "type": "text" - }, - { - "block_id": "p515-b3", - "global_id": 14723, - "bbox": [ - 245.48, - 108.17, - 420.47, - 132.29 - ], - "text": "z9\nor\nz\nz −1(z−4 −z−10)", - "type": "text" - }, - { - "block_id": "p515-b4", - "global_id": 14724, - "bbox": [ - 151.5, - 133.59, - 225.78, - 150.87 - ], - "text": "(b) z(3.2z + 17.2)", - "type": "text" - }, - { - "block_id": "p515-b5", - "global_id": 14725, - "bbox": [ - 176.1, - 144.77, - 219.52, - 158.02 - ], - "text": "z2 −2z + 2", - "type": "text" - }, - { - "block_id": "p515-b6", - "global_id": 14726, - "bbox": [ - 318.18, - 247.81, - 322.18, - 255.81 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p515-b7", - "global_id": 14727, - "bbox": [ - 157.7, - 190.99, - 161.7, - 198.99 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p515-b8", - "global_id": 14728, - "bbox": [ - 158.83, - 245.81, - 299.23, - 253.81 - ], - "text": "4\n5\n6\n7\n8\n9\n0", - "type": "text" - }, - { - "block_id": "p515-b9", - "global_id": 14729, - "bbox": [ - 174.53, - 180.11, - 187.41, - 188.19 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p515-b10", - "global_id": 14730, - "bbox": [ - 352.09, - 250.58, - 472.13, - 259.82 - ], - "text": "Figure 5.3 Signal for Drill 5.1a.", - "type": "text" - }, - { - "block_id": "p515-b11", - "global_id": 14731, - "bbox": [ - 127.59, - 313.25, - 483.09, - 325.2 - ], - "text": "5.1-1 Inverse Transform by Partial Fraction Expansion and Tables", - "type": "text" - }, - { - "block_id": "p515-b12", - "global_id": 14732, - "bbox": [ - 127.59, - 331.34, - 516.16, - 413.03 - ], - "text": "As in the Laplace transform, we shall avoid the integration in the complex plane required to find\nthe inverse z-transform [Eq. (5.2)] by using the (unilateral) transform table (Table 5.1). Many of\nthe transforms X[z] of practical interest are rational functions (ratio of polynomials in z), which\ncan be expressed as a sum of partial fractions, whose inverse transforms can be readily found in a\ntable of transform. The partial fraction method works because for every transformable x[n] defined\nfor n ≥0, there is a corresponding unique X[z] defined for |z| > r0 (where r0 is some constant),\nand vice versa.", - "type": "text" - }, - { - "block_id": "p515-b13", - "global_id": 14733, - "bbox": [ - 102.51, - 444.05, - 474.02, - 456.01 - ], - "text": "EXAMPLE 5.3\nInverse z-Transform by Partial Fraction Expansion", - "type": "text" - }, - { - "block_id": "p515-b14", - "global_id": 14734, - "bbox": [ - 128.9, - 472.58, - 255.65, - 482.64 - ], - "text": "Find the inverse z-transforms of", - "type": "text" - }, - { - "block_id": "p515-b15", - "global_id": 14735, - "bbox": [ - 146.84, - 486.86, - 218.94, - 511.3 - ], - "text": "(a)\n8z −19\n(z −2)(z −3)", - "type": "text" - }, - { - "block_id": "p515-b16", - "global_id": 14736, - "bbox": [ - 146.84, - 511.39, - 234.85, - 532.28 - ], - "text": "(b) z(2z2 −11z + 12)", - "type": "text" - }, - { - "block_id": "p515-b17", - "global_id": 14737, - "bbox": [ - 170.89, - 528.86, - 228.68, - 539.44 - ], - "text": "(z −1)(z −2)3", - "type": "text" - }, - { - "block_id": "p515-b18", - "global_id": 14738, - "bbox": [ - 147.4, - 541.63, - 247.62, - 566.06 - ], - "text": "(c)\n2z(3z + 17)\n(z −1)(z2 −6z + 25)", - "type": "text" - }, - { - "block_id": "p515-b19", - "global_id": 14739, - "bbox": [ - 146.84, - 577.5, - 333.21, - 587.88 - ], - "text": "(a) Expanding X[z] into partial fractions yields", - "type": "text" - }, - { - "block_id": "p515-b20", - "global_id": 14740, - "bbox": [ - 239.54, - 597.4, - 390.93, - 621.83 - ], - "text": "X[z] =\n8z −19\n(z −2)(z −3) =\n3\nz −2 +\n5\nz −3", - "type": "text" - } - ] - }, - { - "page_num": 516, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p516-b0", - "global_id": 14741, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "496\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p516-b1", - "global_id": 14742, - "bbox": [ - 103.16, - 86.23, - 236.01, - 96.2 - ], - "text": "From Table 5.1, pair 7, we obtain", - "type": "text" - }, - { - "block_id": "p516-b2", - "global_id": 14743, - "bbox": [ - 220.84, - 102.07, - 477.01, - 116.56 - ], - "text": "x[n] = [3(2)n−1 + 5(3)n−1]u[n −1]\n(5.8)", - "type": "text" - }, - { - "block_id": "p516-b3", - "global_id": 14744, - "bbox": [ - 103.17, - 126.55, - 477.03, - 220.61 - ], - "text": "If we expand rational X[z] into partial fractions directly, we shall always obtain an answer\nthat is multiplied by u[n −1] because of the nature of pair 7 in Table 5.1. This form is rather\nawkward as well as inconvenient. We prefer the form that contains u[n] rather than u[n−1]. A\nglance at Table 5.1 shows that the z-transform of every signal that is multiplied by u[n] has a\nfactor z in the numerator. This observation suggests that we expand X[z] into modified partial\nfractions, where each term has a factor z in the numerator. This goal can be accomplished\nby expanding X[z]/z into partial fractions and then multiplying both sides by z. We shall\ndemonstrate this procedure by reworking part (a). For this case,", - "type": "text" - }, - { - "block_id": "p516-b4", - "global_id": 14745, - "bbox": [ - 185.47, - 228.63, - 202.51, - 238.91 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p516-b5", - "global_id": 14746, - "bbox": [ - 192.05, - 228.63, - 324.02, - 253.07 - ], - "text": "z\n=\n8z −19\nz(z −2)(z −3) = (−19/6)", - "type": "text" - }, - { - "block_id": "p516-b6", - "global_id": 14747, - "bbox": [ - 304.68, - 228.63, - 359.37, - 252.97 - ], - "text": "z\n+ (3/2)", - "type": "text" - }, - { - "block_id": "p516-b7", - "global_id": 14748, - "bbox": [ - 338.48, - 228.63, - 394.71, - 253.07 - ], - "text": "z −2 + (5/3)", - "type": "text" - }, - { - "block_id": "p516-b8", - "global_id": 14749, - "bbox": [ - 373.82, - 242.69, - 393.54, - 253.07 - ], - "text": "z −3", - "type": "text" - }, - { - "block_id": "p516-b9", - "global_id": 14750, - "bbox": [ - 103.17, - 260.92, - 238.49, - 270.99 - ], - "text": "Multiplying both sides by z yields", - "type": "text" - }, - { - "block_id": "p516-b10", - "global_id": 14751, - "bbox": [ - 210.66, - 280.04, - 258.48, - 296.89 - ], - "text": "X[z] = −19", - "type": "text" - }, - { - "block_id": "p516-b11", - "global_id": 14752, - "bbox": [ - 251.02, - 280.04, - 276.73, - 304.06 - ], - "text": "6 + 3\n2", - "type": "text" - }, - { - "block_id": "p516-b12", - "global_id": 14753, - "bbox": [ - 279.03, - 272.63, - 306.68, - 304.06 - ], - "text": "z\nz −2", - "type": "text" - }, - { - "block_id": "p516-b14", - "global_id": 14754, - "bbox": [ - 316.15, - 280.04, - 331.65, - 296.99 - ], - "text": "+ 5", - "type": "text" - }, - { - "block_id": "p516-b15", - "global_id": 14755, - "bbox": [ - 326.67, - 294.1, - 331.65, - 304.06 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p516-b16", - "global_id": 14756, - "bbox": [ - 333.95, - 272.63, - 361.59, - 304.06 - ], - "text": "z\nz −3", - "type": "text" - }, - { - "block_id": "p516-b18", - "global_id": 14757, - "bbox": [ - 103.16, - 312.68, - 288.07, - 322.64 - ], - "text": "From pairs 1 and 6 in Table 5.1, it follows that", - "type": "text" - }, - { - "block_id": "p516-b19", - "global_id": 14758, - "bbox": [ - 213.48, - 331.28, - 257.35, - 342.89 - ], - "text": "x[n] = −19", - "type": "text" - }, - { - "block_id": "p516-b20", - "global_id": 14759, - "bbox": [ - 252.12, - 324.61, - 294.33, - 346.12 - ], - "text": "6 δ[n] +\n\t 3", - "type": "text" - }, - { - "block_id": "p516-b21", - "global_id": 14760, - "bbox": [ - 290.84, - 324.61, - 477.01, - 345.81 - ], - "text": "2(2)n + 5\n3(3)n\nu[n]\n(5.9)", - "type": "text" - }, - { - "block_id": "p516-b22", - "global_id": 14761, - "bbox": [ - 103.17, - 352.98, - 477.03, - 411.18 - ], - "text": "The reader can verify that this answer is equivalent to that in Eq. (5.8) by computing x[n]\nin both cases for n = 0,1,2,3,. . ., and comparing the results. The form in Eq. (5.9) is more\nconvenient than that in Eq. (5.8). For this reason, we shall always expand X[z]/z rather than\nX[z] into partial fractions and then multiply both sides by z to obtain modified partial fractions\nof X[z], which have a factor z in the numerator.", - "type": "text" - }, - { - "block_id": "p516-b23", - "global_id": 14762, - "bbox": [ - 121.1, - 413.09, - 133.27, - 423.05 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p516-b24", - "global_id": 14763, - "bbox": [ - 239.6, - 420.23, - 339.36, - 441.11 - ], - "text": "X[z] = z(2z2 −11z + 12)", - "type": "text" - }, - { - "block_id": "p516-b25", - "global_id": 14764, - "bbox": [ - 275.39, - 437.71, - 333.18, - 448.29 - ], - "text": "(z −1)(z −2)3", - "type": "text" - }, - { - "block_id": "p516-b26", - "global_id": 14765, - "bbox": [ - 103.16, - 454.03, - 117.54, - 463.99 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p516-b27", - "global_id": 14766, - "bbox": [ - 163.64, - 463.07, - 180.69, - 473.35 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p516-b28", - "global_id": 14767, - "bbox": [ - 170.22, - 459.47, - 253.3, - 487.41 - ], - "text": "z\n= 2z2 −11z + 12", - "type": "text" - }, - { - "block_id": "p516-b29", - "global_id": 14768, - "bbox": [ - 194.97, - 463.39, - 416.53, - 487.51 - ], - "text": "(z −1)(z −2)3 =\nk\nz −1 +\na0\n(z −2)3 +\na1\n(z −2)2 +\na2\n(z −2)", - "type": "text" - }, - { - "block_id": "p516-b30", - "global_id": 14769, - "bbox": [ - 103.16, - 493.26, - 127.5, - 503.22 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p516-b31", - "global_id": 14770, - "bbox": [ - 233.06, - 503.21, - 310.28, - 524.17 - ], - "text": "k = 2z2 −11z + 12", - "type": "text" - }, - { - "block_id": "p516-b32", - "global_id": 14771, - "bbox": [ - 251.95, - 521.09, - 309.75, - 531.66 - ], - "text": "(z −1)(z −2)3", - "type": "text" - }, - { - "block_id": "p516-b34", - "global_id": 14772, - "bbox": [ - 314.74, - 529.0, - 326.37, - 536.26 - ], - "text": "z=1", - "type": "text" - }, - { - "block_id": "p516-b35", - "global_id": 14773, - "bbox": [ - 328.92, - 513.8, - 351.49, - 524.17 - ], - "text": "= −3", - "type": "text" - }, - { - "block_id": "p516-b36", - "global_id": 14774, - "bbox": [ - 228.7, - 538.02, - 310.28, - 560.08 - ], - "text": "a0 = 2z2 −11z + 12", - "type": "text" - }, - { - "block_id": "p516-b37", - "global_id": 14775, - "bbox": [ - 251.95, - 555.16, - 309.75, - 566.47 - ], - "text": "(z −1)(z −2)3", - "type": "text" - }, - { - "block_id": "p516-b39", - "global_id": 14776, - "bbox": [ - 314.74, - 563.81, - 326.37, - 571.07 - ], - "text": "z=2", - "type": "text" - }, - { - "block_id": "p516-b40", - "global_id": 14777, - "bbox": [ - 328.92, - 548.62, - 351.49, - 559.0 - ], - "text": "= −2", - "type": "text" - }, - { - "block_id": "p516-b41", - "global_id": 14778, - "bbox": [ - 103.16, - 578.61, - 144.91, - 588.57 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p516-b42", - "global_id": 14779, - "bbox": [ - 163.64, - 597.4, - 180.69, - 607.67 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p516-b43", - "global_id": 14780, - "bbox": [ - 170.22, - 593.78, - 253.3, - 621.73 - ], - "text": "z\n= 2z2 −11z + 12", - "type": "text" - }, - { - "block_id": "p516-b44", - "global_id": 14781, - "bbox": [ - 194.97, - 597.4, - 283.82, - 621.83 - ], - "text": "(z −1)(z −2)3 = −3", - "type": "text" - }, - { - "block_id": "p516-b45", - "global_id": 14782, - "bbox": [ - 267.58, - 597.71, - 477.01, - 621.83 - ], - "text": "z −1 −\n2\n(z −2)3 +\na1\n(z −2)2 +\na2\n(z −2)\n(5.10)", - "type": "text" - } - ] - }, - { - "page_num": 517, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p517-b0", - "global_id": 14783, - "bbox": [ - 388.32, - 62.57, - 516.13, - 71.98 - ], - "text": "5.1\nThe z-Transform\n497", - "type": "text" - }, - { - "block_id": "p517-b1", - "global_id": 14784, - "bbox": [ - 128.9, - 86.14, - 502.75, - 108.93 - ], - "text": "We can determine a1 and a2 by clearing fractions. Or we may use a shortcut. For example, to\ndetermine a2, we multiply both sides of Eq. (5.10) by z and let z →∞. This yields", - "type": "text" - }, - { - "block_id": "p517-b2", - "global_id": 14785, - "bbox": [ - 244.14, - 119.7, - 387.52, - 131.16 - ], - "text": "0 = −3 −0 + 0 + a2\n\r⇒\na2 = 3", - "type": "text" - }, - { - "block_id": "p517-b3", - "global_id": 14786, - "bbox": [ - 128.91, - 141.93, - 502.75, - 163.95 - ], - "text": "This result leaves only one unknown, a1, which is readily determined by letting z take any\nconvenient value, say, z = 0, on both sides of Eq. (5.10). This produces", - "type": "text" - }, - { - "block_id": "p517-b4", - "global_id": 14787, - "bbox": [ - 273.08, - 173.79, - 358.6, - 197.91 - ], - "text": "12\n8 = 3 + 1\n4 + a1\n4 −3\n2", - "type": "text" - }, - { - "block_id": "p517-b5", - "global_id": 14788, - "bbox": [ - 128.9, - 205.37, - 262.35, - 216.83 - ], - "text": "which yields a1 = −1. Therefore,", - "type": "text" - }, - { - "block_id": "p517-b6", - "global_id": 14789, - "bbox": [ - 229.42, - 225.27, - 246.47, - 235.55 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p517-b7", - "global_id": 14790, - "bbox": [ - 236.0, - 225.27, - 276.95, - 249.6 - ], - "text": "z\n= −3", - "type": "text" - }, - { - "block_id": "p517-b8", - "global_id": 14791, - "bbox": [ - 260.72, - 225.68, - 402.24, - 249.7 - ], - "text": "z −1 −\n2\n(z −2)3 −\n1\n(z −2)2 +\n3\nz −2", - "type": "text" - }, - { - "block_id": "p517-b9", - "global_id": 14792, - "bbox": [ - 128.9, - 259.22, - 143.28, - 269.18 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p517-b10", - "global_id": 14793, - "bbox": [ - 218.07, - 264.09, - 412.39, - 288.21 - ], - "text": "X[z] = −3\nz\nz −1 −2\nz\n(z −2)3 −\nz\n(z −2)2 + 3\nz\nz −2", - "type": "text" - }, - { - "block_id": "p517-b11", - "global_id": 14794, - "bbox": [ - 128.9, - 294.74, - 318.5, - 304.7 - ], - "text": "Now the use of Table 5.1, pairs 6 and 10, yields", - "type": "text" - }, - { - "block_id": "p517-b12", - "global_id": 14795, - "bbox": [ - 213.82, - 321.88, - 239.71, - 332.16 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p517-b14", - "global_id": 14796, - "bbox": [ - 247.18, - 314.9, - 310.19, - 332.26 - ], - "text": "−3 −2n(n −1)", - "type": "text" - }, - { - "block_id": "p517-b15", - "global_id": 14797, - "bbox": [ - 291.11, - 315.21, - 344.87, - 339.33 - ], - "text": "8\n(2)n −n", - "type": "text" - }, - { - "block_id": "p517-b16", - "global_id": 14798, - "bbox": [ - 339.89, - 307.9, - 400.14, - 339.33 - ], - "text": "2(2)n + 3(2)n\n!", - "type": "text" - }, - { - "block_id": "p517-b17", - "global_id": 14799, - "bbox": [ - 401.25, - 321.88, - 417.85, - 332.16 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p517-b18", - "global_id": 14800, - "bbox": [ - 231.94, - 349.78, - 249.53, - 359.75 - ], - "text": "= −", - "type": "text" - }, - { - "block_id": "p517-b20", - "global_id": 14801, - "bbox": [ - 254.96, - 335.79, - 345.65, - 367.23 - ], - "text": "3 + 1\n4(n2 + n −12)2n\n!", - "type": "text" - }, - { - "block_id": "p517-b21", - "global_id": 14802, - "bbox": [ - 345.66, - 349.78, - 362.25, - 360.06 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p517-b22", - "global_id": 14803, - "bbox": [ - 146.84, - 377.45, - 225.49, - 387.41 - ], - "text": "(c) Complex Poles.", - "type": "text" - }, - { - "block_id": "p517-b23", - "global_id": 14804, - "bbox": [ - 192.19, - 396.91, - 438.24, - 421.34 - ], - "text": "X[z] =\n2z(3z + 17)\n(z −1)(z2 −6z + 25) =\n2z(3z + 17)\n(z −1)(z −3 −j4)(z −3 + j4)", - "type": "text" - }, - { - "block_id": "p517-b24", - "global_id": 14805, - "bbox": [ - 128.91, - 430.75, - 502.76, - 488.94 - ], - "text": "The poles of X[z] are 1, 3 + j4, and 3 −j4. Whenever there are complex-conjugate poles, the\nproblem can be worked out in two ways. In the first method we expand X[z] into (modified)\nfirst-order partial fractions. In the second method, rather than obtain one factor corresponding\nto each complex-conjugate pole, we obtain quadratic factors corresponding to each pair of\ncomplex-conjugate poles. This procedure is explained next.", - "type": "text" - }, - { - "block_id": "p517-b25", - "global_id": 14806, - "bbox": [ - 129.2, - 509.66, - 322.38, - 521.78 - ], - "text": "METHOD OF FIRST-ORDER FACTORS", - "type": "text" - }, - { - "block_id": "p517-b26", - "global_id": 14807, - "bbox": [ - 192.19, - 521.46, - 209.24, - 531.74 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p517-b27", - "global_id": 14808, - "bbox": [ - 198.77, - 521.46, - 439.44, - 545.89 - ], - "text": "z\n=\n2(3z + 17)\n(z −1)(z2 −6z + 25) =\n2(3z + 17)\n(z −1)(z −3 −j4)(z −3 + j4)", - "type": "text" - }, - { - "block_id": "p517-b28", - "global_id": 14809, - "bbox": [ - 128.9, - 552.3, - 437.5, - 562.68 - ], - "text": "We find the partial fraction of X[z]/z using the Heaviside “cover-up” method:", - "type": "text" - }, - { - "block_id": "p517-b29", - "global_id": 14810, - "bbox": [ - 237.67, - 573.79, - 254.71, - 584.06 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p517-b30", - "global_id": 14811, - "bbox": [ - 244.25, - 572.56, - 341.87, - 598.22 - ], - "text": "z\n=\n2\nz −1 + 1.6e−j2.246", - "type": "text" - }, - { - "block_id": "p517-b31", - "global_id": 14812, - "bbox": [ - 302.99, - 572.78, - 391.81, - 598.22 - ], - "text": "z −3 −j4 + 1.6ej2.246", - "type": "text" - }, - { - "block_id": "p517-b32", - "global_id": 14813, - "bbox": [ - 355.65, - 587.84, - 393.99, - 598.22 - ], - "text": "z −3 + j4", - "type": "text" - } - ] - }, - { - "page_num": 518, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p518-b0", - "global_id": 14814, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "498\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p518-b1", - "global_id": 14815, - "bbox": [ - 103.16, - 86.24, - 117.54, - 96.2 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p518-b2", - "global_id": 14816, - "bbox": [ - 165.33, - 98.08, - 413.65, - 122.2 - ], - "text": "X[z] = 2\nz\nz −1 + (1.6e−j2.246)\nz\nz −3 −j4 + (1.6ej2.246)\nz\nz −3 + j4", - "type": "text" - }, - { - "block_id": "p518-b3", - "global_id": 14817, - "bbox": [ - 103.17, - 128.62, - 477.02, - 174.86 - ], - "text": "The inverse transform of the first term on the right-hand side is 2u[n]. The inverse transform of\nthe remaining two terms (complex conjugate poles) can be obtained from pair 12b (Table 5.1)\nby identifying r/2 = 1.6, θ = −2.246 rad, γ = 3 + j4 = 5ej0.927, so that |γ | = 5, β = 0.927.\nTherefore,", - "type": "text" - }, - { - "block_id": "p518-b4", - "global_id": 14818, - "bbox": [ - 200.35, - 179.71, - 379.8, - 193.79 - ], - "text": "x[n] = [2 + 3.2(5)n cos(0.927n −2.246)]u[n]", - "type": "text" - }, - { - "block_id": "p518-b5", - "global_id": 14819, - "bbox": [ - 103.46, - 212.51, - 290.16, - 224.63 - ], - "text": "METHOD OF QUADRATIC FACTORS", - "type": "text" - }, - { - "block_id": "p518-b6", - "global_id": 14820, - "bbox": [ - 185.1, - 233.17, - 202.14, - 243.44 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p518-b7", - "global_id": 14821, - "bbox": [ - 191.69, - 233.17, - 395.06, - 257.6 - ], - "text": "z\n=\n2(3z + 17)\n(z −1)(z2 −6z + 25) =\n2\nz −1 +\nAz + B\nz2 −6z + 25", - "type": "text" - }, - { - "block_id": "p518-b8", - "global_id": 14822, - "bbox": [ - 103.16, - 263.71, - 323.57, - 274.09 - ], - "text": "Multiplying both sides by z and letting z →∞, we find", - "type": "text" - }, - { - "block_id": "p518-b9", - "global_id": 14823, - "bbox": [ - 245.05, - 282.64, - 335.14, - 293.01 - ], - "text": "0 = 2 + A \r⇒A = −2", - "type": "text" - }, - { - "block_id": "p518-b10", - "global_id": 14824, - "bbox": [ - 103.17, - 301.99, - 117.54, - 311.95 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p518-b11", - "global_id": 14825, - "bbox": [ - 200.75, - 316.49, - 379.41, - 340.92 - ], - "text": "2(3z + 17)\n(z −1)(z2 −6z + 25) =\n2\nz −1 +\n−2z + B\nz2 −6z + 25", - "type": "text" - }, - { - "block_id": "p518-b12", - "global_id": 14826, - "bbox": [ - 103.17, - 347.03, - 391.79, - 357.41 - ], - "text": "To find B, we let z take any convenient value, say, z = 0. This step yields", - "type": "text" - }, - { - "block_id": "p518-b13", - "global_id": 14827, - "bbox": [ - 233.04, - 363.94, - 250.77, - 374.32 - ], - "text": "−34", - "type": "text" - }, - { - "block_id": "p518-b14", - "global_id": 14828, - "bbox": [ - 236.93, - 364.26, - 348.33, - 388.38 - ], - "text": "25 = −2 + B\n25 \r⇒B = 16", - "type": "text" - }, - { - "block_id": "p518-b15", - "global_id": 14829, - "bbox": [ - 103.17, - 393.27, - 144.91, - 403.23 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p518-b16", - "global_id": 14830, - "bbox": [ - 233.74, - 409.05, - 250.78, - 419.33 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p518-b17", - "global_id": 14831, - "bbox": [ - 240.32, - 409.05, - 346.42, - 433.48 - ], - "text": "z\n=\n2\nz −1 +\n−2z + 16\nz2 −6z + 25", - "type": "text" - }, - { - "block_id": "p518-b18", - "global_id": 14832, - "bbox": [ - 103.17, - 439.12, - 117.54, - 449.08 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p518-b19", - "global_id": 14833, - "bbox": [ - 233.57, - 453.62, - 345.41, - 478.05 - ], - "text": "X[z] =\n2z\nz −1 + z(−2z + 16)", - "type": "text" - }, - { - "block_id": "p518-b20", - "global_id": 14834, - "bbox": [ - 296.83, - 464.8, - 345.22, - 478.05 - ], - "text": "z2 −6z + 25", - "type": "text" - }, - { - "block_id": "p518-b21", - "global_id": 14835, - "bbox": [ - 103.16, - 483.28, - 461.54, - 493.66 - ], - "text": "We now use pair 12c, where we identify A = −2, B = 16, |γ | = 5, and a = −3. Therefore,", - "type": "text" - }, - { - "block_id": "p518-b22", - "global_id": 14836, - "bbox": [ - 169.09, - 509.79, - 183.02, - 520.07 - ], - "text": "r =", - "type": "text" - }, - { - "block_id": "p518-b24", - "global_id": 14837, - "bbox": [ - 194.51, - 502.81, - 340.55, - 527.24 - ], - "text": "100 + 256 −192\n25 −9\n= 3.2,\nβ = cos−1", - "type": "text" - }, - { - "block_id": "p518-b25", - "global_id": 14838, - "bbox": [ - 342.16, - 495.79, - 355.06, - 513.18 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p518-b26", - "global_id": 14839, - "bbox": [ - 350.08, - 517.28, - 355.06, - 527.24 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p518-b28", - "global_id": 14840, - "bbox": [ - 365.03, - 509.79, - 411.08, - 520.17 - ], - "text": "= 0.927rad", - "type": "text" - }, - { - "block_id": "p518-b29", - "global_id": 14841, - "bbox": [ - 103.17, - 534.42, - 117.54, - 544.38 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p518-b30", - "global_id": 14842, - "bbox": [ - 230.76, - 544.93, - 288.33, - 563.31 - ], - "text": "θ = tan−1 −10", - "type": "text" - }, - { - "block_id": "p518-b31", - "global_id": 14843, - "bbox": [ - 277.67, - 558.86, - 286.59, - 566.13 - ], - "text": "−8", - "type": "text" - }, - { - "block_id": "p518-b33", - "global_id": 14844, - "bbox": [ - 295.59, - 552.94, - 349.41, - 563.31 - ], - "text": "= −2.246rad", - "type": "text" - }, - { - "block_id": "p518-b34", - "global_id": 14845, - "bbox": [ - 103.16, - 572.28, - 129.45, - 582.24 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p518-b35", - "global_id": 14846, - "bbox": [ - 200.35, - 587.09, - 379.8, - 601.17 - ], - "text": "x[n] = [2 + 3.2(5)n cos(0.927n −2.246)]u[n]", - "type": "text" - } - ] - }, - { - "page_num": 519, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p519-b0", - "global_id": 14847, - "bbox": [ - 388.32, - 62.57, - 516.13, - 71.98 - ], - "text": "5.1\nThe z-Transform\n499", - "type": "text" - }, - { - "block_id": "p519-b1", - "global_id": 14848, - "bbox": [ - 133.57, - 97.81, - 481.89, - 109.76 - ], - "text": "DRILL 5.2\nInverse z-Transform by Partial Fraction Expansion", - "type": "text" - }, - { - "block_id": "p519-b2", - "global_id": 14849, - "bbox": [ - 133.57, - 118.78, - 354.42, - 128.85 - ], - "text": "Find the inverse z-transform of the following functions:", - "type": "text" - }, - { - "block_id": "p519-b3", - "global_id": 14850, - "bbox": [ - 151.5, - 134.44, - 231.07, - 158.88 - ], - "text": "(a)\nz(2z −1)\n(z −1)(z + 0.5)", - "type": "text" - }, - { - "block_id": "p519-b4", - "global_id": 14851, - "bbox": [ - 151.5, - 161.42, - 231.62, - 185.44 - ], - "text": "(b)\n1\n(z −1)(z + 0.5)", - "type": "text" - }, - { - "block_id": "p519-b5", - "global_id": 14852, - "bbox": [ - 152.06, - 187.98, - 234.55, - 212.0 - ], - "text": "(c)\n9\n(z + 2)(z −0.5)2", - "type": "text" - }, - { - "block_id": "p519-b6", - "global_id": 14853, - "bbox": [ - 151.5, - 214.18, - 228.22, - 238.62 - ], - "text": "(d)\n5z(z −1)\nz2 −1.6z + 0.8", - "type": "text" - }, - { - "block_id": "p519-b7", - "global_id": 14854, - "bbox": [ - 133.57, - 245.69, - 157.92, - 255.75 - ], - "text": "[Hint:", - "type": "text" - }, - { - "block_id": "p519-b8", - "global_id": 14855, - "bbox": [ - 160.41, - 237.0, - 168.84, - 246.96 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p519-b9", - "global_id": 14856, - "bbox": [ - 168.83, - 237.0, - 211.24, - 255.75 - ], - "text": "0.8 = 2/\n√", - "type": "text" - }, - { - "block_id": "p519-b10", - "global_id": 14857, - "bbox": [ - 211.25, - 245.79, - 222.04, - 255.75 - ], - "text": "5.]", - "type": "text" - }, - { - "block_id": "p519-b11", - "global_id": 14858, - "bbox": [ - 133.84, - 269.28, - 196.06, - 280.24 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p519-b12", - "global_id": 14859, - "bbox": [ - 151.5, - 287.55, - 163.11, - 297.51 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p519-b13", - "global_id": 14860, - "bbox": [ - 168.09, - 279.21, - 176.7, - 292.85 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p519-b14", - "global_id": 14861, - "bbox": [ - 173.22, - 279.21, - 246.81, - 300.72 - ], - "text": "3 + 4\n3(−0.5)n\nu[n]", - "type": "text" - }, - { - "block_id": "p519-b15", - "global_id": 14862, - "bbox": [ - 151.5, - 303.58, - 207.02, - 313.96 - ], - "text": "(b) −2δ[n] +", - "type": "text" - }, - { - "block_id": "p519-b16", - "global_id": 14863, - "bbox": [ - 208.56, - 295.57, - 217.18, - 309.21 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p519-b17", - "global_id": 14864, - "bbox": [ - 213.7, - 295.57, - 287.29, - 317.08 - ], - "text": "3 + 4\n3(−0.5)n\nu[n]", - "type": "text" - }, - { - "block_id": "p519-b18", - "global_id": 14865, - "bbox": [ - 152.06, - 314.91, - 384.38, - 328.91 - ], - "text": "(c) 18δ[n] −[0.72(−2)n + 17.28(0.5)n −14.4n(0.5)n]u[n]", - "type": "text" - }, - { - "block_id": "p519-b19", - "global_id": 14866, - "bbox": [ - 151.5, - 327.29, - 179.23, - 345.09 - ], - "text": "(d)\n5\n√", - "type": "text" - }, - { - "block_id": "p519-b20", - "global_id": 14867, - "bbox": [ - 174.55, - 326.78, - 195.57, - 347.98 - ], - "text": "5\n2\n 2\n√", - "type": "text" - }, - { - "block_id": "p519-b21", - "global_id": 14868, - "bbox": [ - 195.03, - 326.78, - 307.88, - 349.16 - ], - "text": "5\nn cos(0.464n + 0.464)u[n]", - "type": "text" - }, - { - "block_id": "p519-b22", - "global_id": 14869, - "bbox": [ - 127.59, - 401.53, - 417.33, - 414.34 - ], - "text": "5.1-2 Inverse z-Transform by Power Series Expansion", - "type": "text" - }, - { - "block_id": "p519-b23", - "global_id": 14870, - "bbox": [ - 127.59, - 420.47, - 182.73, - 430.43 - ], - "text": "By definition,", - "type": "text" - }, - { - "block_id": "p519-b24", - "global_id": 14871, - "bbox": [ - 223.11, - 461.43, - 249.97, - 471.71 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p519-b25", - "global_id": 14872, - "bbox": [ - 252.02, - 451.26, - 266.12, - 461.93 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p519-b26", - "global_id": 14873, - "bbox": [ - 252.86, - 475.82, - 265.26, - 483.08 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p519-b27", - "global_id": 14874, - "bbox": [ - 267.22, - 459.71, - 296.09, - 471.71 - ], - "text": "x[n]z−n", - "type": "text" - }, - { - "block_id": "p519-b28", - "global_id": 14875, - "bbox": [ - 242.2, - 485.39, - 296.21, - 502.75 - ], - "text": "= x[0] + x[1]", - "type": "text" - }, - { - "block_id": "p519-b29", - "global_id": 14876, - "bbox": [ - 286.25, - 485.39, - 325.55, - 509.72 - ], - "text": "z\n+ x[2]", - "type": "text" - }, - { - "block_id": "p519-b30", - "global_id": 14877, - "bbox": [ - 313.59, - 485.39, - 354.88, - 509.72 - ], - "text": "z2 + x[3]", - "type": "text" - }, - { - "block_id": "p519-b31", - "global_id": 14878, - "bbox": [ - 342.92, - 492.37, - 380.02, - 509.72 - ], - "text": "z3 + · · ·", - "type": "text" - }, - { - "block_id": "p519-b32", - "global_id": 14879, - "bbox": [ - 242.2, - 509.03, - 420.6, - 523.52 - ], - "text": "= x[0]z0 + x[1]z−1 + x[2]z−2 + x[3]z−3 + · · ·", - "type": "text" - }, - { - "block_id": "p519-b33", - "global_id": 14880, - "bbox": [ - 127.59, - 545.16, - 516.15, - 592.63 - ], - "text": "This result is a power series in z−1. Therefore, if we can expand X[z] into the power series in z−1,\nthe coefficients of this power series can be identified as x[0], x[1], x[2], x[3], . . .. A rational X[z]\ncan be expanded into a power series of z−1 by dividing its numerator by the denominator. Consider,\nfor example,", - "type": "text" - }, - { - "block_id": "p519-b34", - "global_id": 14881, - "bbox": [ - 206.09, - 607.16, - 409.42, - 635.21 - ], - "text": "X[z] =\nz2(7z −2)\n(z −0.2)(z −0.5)(z −1) =\n7z3 −2z2", - "type": "text" - }, - { - "block_id": "p519-b35", - "global_id": 14882, - "bbox": [ - 346.87, - 621.95, - 436.41, - 635.21 - ], - "text": "z3 −1.7z2 + 0.8z −0.1", - "type": "text" - } - ] - }, - { - "page_num": 520, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p520-b0", - "global_id": 14883, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "500\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p520-b1", - "global_id": 14884, - "bbox": [ - 101.84, - 85.76, - 490.38, - 109.31 - ], - "text": "To obtain a series expansion in powers of z−1, we divide the numerator by the denominator as\nfollows:", - "type": "text" - }, - { - "block_id": "p520-b2", - "global_id": 14885, - "bbox": [ - 167.99, - 110.11, - 421.01, - 136.85 - ], - "text": "7 + 9.9z−1 + 11.23z−2 + 11.87z−3 + · · ·\nz3 −1.7z2 + 0.8z −0.1", - "type": "text" - }, - { - "block_id": "p520-b4", - "global_id": 14886, - "bbox": [ - 262.68, - 122.86, - 298.73, - 136.85 - ], - "text": "7z3 −2z2", - "type": "text" - }, - { - "block_id": "p520-b5", - "global_id": 14887, - "bbox": [ - 262.68, - 134.81, - 423.22, - 172.72 - ], - "text": "7z3 −11.9z2 + 5.60z −0.7\n9.9z2 −5.60z + 0.7\n9.9z2 −16.83z + 7.92 −0.99z−1", - "type": "text" - }, - { - "block_id": "p520-b6", - "global_id": 14888, - "bbox": [ - 323.05, - 173.07, - 423.71, - 184.67 - ], - "text": "11.23z −7.22 + 0.99z−1", - "type": "text" - }, - { - "block_id": "p520-b7", - "global_id": 14889, - "bbox": [ - 323.05, - 185.03, - 423.69, - 196.63 - ], - "text": "11.23z −19.09 + 8.98z−1", - "type": "text" - }, - { - "block_id": "p520-b8", - "global_id": 14890, - "bbox": [ - 360.22, - 196.98, - 423.71, - 208.58 - ], - "text": "11.87 −7.99z−1", - "type": "text" - }, - { - "block_id": "p520-b9", - "global_id": 14891, - "bbox": [ - 101.84, - 220.51, - 124.25, - 230.47 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p520-b10", - "global_id": 14892, - "bbox": [ - 146.97, - 235.28, - 445.26, - 260.65 - ], - "text": "X[z] =\nz2(7z −2)\n(z −0.2)(z −0.5)(z −1) = 7 + 9.9z−1 + 11.23z−2 + 11.87z−3 + · · ·", - "type": "text" - }, - { - "block_id": "p520-b11", - "global_id": 14893, - "bbox": [ - 101.85, - 270.72, - 143.59, - 280.68 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p520-b12", - "global_id": 14894, - "bbox": [ - 192.24, - 287.58, - 399.99, - 297.95 - ], - "text": "x[0] = 7, x[1] = 9.9, x[2] = 11.23, x[3] = 11.87,. . .", - "type": "text" - }, - { - "block_id": "p520-b13", - "global_id": 14895, - "bbox": [ - 101.85, - 310.05, - 490.4, - 344.34 - ], - "text": "Although this procedure yields x[n] directly, it does not provide a closed-form solution. For\nthis reason, it is not very useful unless we want to know only the first few terms of the sequence\nx[n].", - "type": "text" - }, - { - "block_id": "p520-b14", - "global_id": 14896, - "bbox": [ - 107.82, - 378.4, - 391.39, - 390.35 - ], - "text": "DRILL 5.3\nInverse z-Transform by Long Division", - "type": "text" - }, - { - "block_id": "p520-b15", - "global_id": 14897, - "bbox": [ - 107.82, - 397.84, - 484.4, - 421.39 - ], - "text": "Using long division to find the power series in z−1, show that the inverse z-transform of\nz/(z−0.5) is (0.5)nu[n] or (2)−nu[n].", - "type": "text" - }, - { - "block_id": "p520-b16", - "global_id": 14898, - "bbox": [ - 101.84, - 453.06, - 490.39, - 491.92 - ], - "text": "RELATIONSHIP BETWEEN h[n] AND H[z]\nFor an LTID system, if h[n] is its unit impulse response, then from Eq. (3.39), where we defined\nH[z], the system transfer function, we write", - "type": "text" - }, - { - "block_id": "p520-b17", - "global_id": 14899, - "bbox": [ - 254.86, - 515.14, - 282.83, - 525.42 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p520-b18", - "global_id": 14900, - "bbox": [ - 288.58, - 504.97, - 302.67, - 515.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p520-b19", - "global_id": 14901, - "bbox": [ - 284.88, - 528.99, - 306.35, - 536.18 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p520-b20", - "global_id": 14902, - "bbox": [ - 307.48, - 513.41, - 490.38, - 525.52 - ], - "text": "h[n]z−n\n(5.11)", - "type": "text" - }, - { - "block_id": "p520-b21", - "global_id": 14903, - "bbox": [ - 101.85, - 549.57, - 490.39, - 571.91 - ], - "text": "For causal systems, the limits on the sum are from n = 0 to ∞. This equation shows that the\ntransfer function H[z] is the z-transform of the impulse response h[n] of an LTID system; that is,", - "type": "text" - }, - { - "block_id": "p520-b22", - "global_id": 14904, - "bbox": [ - 267.14, - 586.99, - 325.11, - 597.27 - ], - "text": "h[n] ⇐⇒H[z]", - "type": "text" - }, - { - "block_id": "p520-b23", - "global_id": 14905, - "bbox": [ - 101.85, - 612.45, - 490.4, - 634.79 - ], - "text": "This important result relates the time-domain specification h[n] of a system to H[z], the\nfrequency-domain specification of a system. The result is parallel to that for LTIC systems.", - "type": "text" - } - ] - }, - { - "page_num": 521, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p521-b0", - "global_id": 14906, - "bbox": [ - 313.93, - 62.57, - 516.13, - 71.98 - ], - "text": "5.2\nSome Properties of the z-Transform\n501", - "type": "text" - }, - { - "block_id": "p521-b1", - "global_id": 14907, - "bbox": [ - 133.57, - 97.81, - 439.04, - 109.77 - ], - "text": "DRILL 5.4\nImpulse Response by Inverse z-Transform", - "type": "text" - }, - { - "block_id": "p521-b2", - "global_id": 14908, - "bbox": [ - 133.57, - 118.47, - 455.97, - 128.85 - ], - "text": "Redo Drill 3.14 by taking the inverse z-transform of H[z], as given by Eq. (3.41).", - "type": "text" - }, - { - "block_id": "p521-b3", - "global_id": 14909, - "bbox": [ - 127.94, - 167.93, - 420.45, - 182.88 - ], - "text": "5.2 SOME PROPERTIES OF THE z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p521-b4", - "global_id": 14910, - "bbox": [ - 127.59, - 188.77, - 516.14, - 222.74 - ], - "text": "The z-transform properties are useful in the derivation of z-transforms of many functions and also\nin the solution of linear difference equations with constant coefficients. Here we consider a few\nimportant properties of the z-transform.", - "type": "text" - }, - { - "block_id": "p521-b5", - "global_id": 14911, - "bbox": [ - 127.59, - 224.32, - 516.13, - 258.61 - ], - "text": "In our discussion, the variable n appearing in signals, such as x[n] and y[n], may or may not\nstand for time. However, in most applications of our interest, n is proportional to time. For this\nreason, we shall loosely refer to the variable n as time.", - "type": "text" - }, - { - "block_id": "p521-b6", - "global_id": 14912, - "bbox": [ - 127.59, - 282.7, - 293.79, - 294.66 - ], - "text": "5.2-1 Time-Shifting Properties", - "type": "text" - }, - { - "block_id": "p521-b7", - "global_id": 14913, - "bbox": [ - 127.59, - 300.38, - 516.16, - 358.57 - ], - "text": "In the following discussion of the shift property, we deal with shifted signals x[n]u[n], x[n−k]u[n−\nk], x[n−k]u[n], and x[n+k]u[n]. Unless we physically understand the meaning of such shifts, our\nunderstanding of the shift property remains mechanical rather than intuitive or heuristic. For this\nreason, using a hypothetical signal x[n], we have illustrated various shifted signals for k = 1 in\nFig. 5.4.", - "type": "text" - }, - { - "block_id": "p521-b8", - "global_id": 14914, - "bbox": [ - 127.59, - 372.98, - 245.97, - 399.09 - ], - "text": "RIGHT SHIFT (DELAY)\nIf", - "type": "text" - }, - { - "block_id": "p521-b9", - "global_id": 14915, - "bbox": [ - 285.4, - 400.67, - 358.33, - 410.95 - ], - "text": "x[n]u[n] ⇐⇒X[z]", - "type": "text" - }, - { - "block_id": "p521-b10", - "global_id": 14916, - "bbox": [ - 127.6, - 419.36, - 144.74, - 429.32 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p521-b11", - "global_id": 14917, - "bbox": [ - 265.86, - 427.23, - 359.63, - 444.18 - ], - "text": "x[n −1]u[n −1] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p521-b12", - "global_id": 14918, - "bbox": [ - 355.19, - 433.8, - 516.13, - 451.15 - ], - "text": "z X[z]\n(5.12)", - "type": "text" - }, - { - "block_id": "p521-b13", - "global_id": 14919, - "bbox": [ - 127.59, - 456.24, - 170.18, - 466.2 - ], - "text": "In general,", - "type": "text" - }, - { - "block_id": "p521-b14", - "global_id": 14920, - "bbox": [ - 261.44, - 466.17, - 361.84, - 483.02 - ], - "text": "x[n −m]u[n −m] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p521-b15", - "global_id": 14921, - "bbox": [ - 354.64, - 472.74, - 516.13, - 490.09 - ], - "text": "zm X[z]\n(5.13)", - "type": "text" - }, - { - "block_id": "p521-b16", - "global_id": 14922, - "bbox": [ - 127.59, - 494.89, - 168.65, - 504.86 - ], - "text": "Moreover,", - "type": "text" - }, - { - "block_id": "p521-b17", - "global_id": 14923, - "bbox": [ - 256.44, - 504.12, - 334.35, - 521.06 - ], - "text": "x[n −1]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p521-b18", - "global_id": 14924, - "bbox": [ - 329.91, - 510.69, - 516.13, - 528.04 - ], - "text": "z X[z] + x[−1]\n(5.14)", - "type": "text" - }, - { - "block_id": "p521-b19", - "global_id": 14925, - "bbox": [ - 127.59, - 533.12, - 301.63, - 543.08 - ], - "text": "Repeated application of this property yields", - "type": "text" - }, - { - "block_id": "p521-b20", - "global_id": 14926, - "bbox": [ - 171.43, - 552.38, - 249.35, - 569.32 - ], - "text": "x[n −2]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p521-b21", - "global_id": 14927, - "bbox": [ - 244.92, - 566.33, - 248.79, - 576.3 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p521-b22", - "global_id": 14928, - "bbox": [ - 251.65, - 544.96, - 263.25, - 562.34 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p521-b23", - "global_id": 14929, - "bbox": [ - 258.82, - 558.95, - 316.19, - 576.3 - ], - "text": "z X[z] + x[−1]", - "type": "text" - }, - { - "block_id": "p521-b24", - "global_id": 14930, - "bbox": [ - 316.2, - 544.96, - 321.63, - 554.92 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p521-b25", - "global_id": 14931, - "bbox": [ - 323.18, - 552.38, - 375.83, - 569.32 - ], - "text": "+ x[−2] = 1", - "type": "text" - }, - { - "block_id": "p521-b26", - "global_id": 14932, - "bbox": [ - 369.4, - 552.38, - 412.54, - 576.3 - ], - "text": "z2 X[z] + 1", - "type": "text" - }, - { - "block_id": "p521-b27", - "global_id": 14933, - "bbox": [ - 408.11, - 558.95, - 472.26, - 576.3 - ], - "text": "z x[−1] + x[−2]", - "type": "text" - }, - { - "block_id": "p521-b28", - "global_id": 14934, - "bbox": [ - 127.6, - 585.15, - 261.09, - 595.22 - ], - "text": "In general, for integer value of m,", - "type": "text" - }, - { - "block_id": "p521-b29", - "global_id": 14935, - "bbox": [ - 232.08, - 611.66, - 363.14, - 623.65 - ], - "text": "x[n −m]u[n] ⇐⇒z−mX[z] + z−m", - "type": "text" - }, - { - "block_id": "p521-b30", - "global_id": 14936, - "bbox": [ - 364.73, - 603.42, - 378.83, - 613.87 - ], - "text": "m\n\"", - "type": "text" - }, - { - "block_id": "p521-b31", - "global_id": 14937, - "bbox": [ - 365.58, - 627.77, - 377.99, - 635.03 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p521-b32", - "global_id": 14938, - "bbox": [ - 379.94, - 611.88, - 516.13, - 623.75 - ], - "text": "x[−n]zn\n(5.15)", - "type": "text" - } - ] - }, - { - "page_num": 522, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p522-b0", - "global_id": 14939, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "502\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p522-b1", - "global_id": 14940, - "bbox": [ - 170.43, - 164.58, - 179.31, - 172.58 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p522-b2", - "global_id": 14941, - "bbox": [ - 170.05, - 259.39, - 179.69, - 267.39 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p522-b3", - "global_id": 14942, - "bbox": [ - 170.43, - 354.68, - 179.31, - 362.68 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p522-b4", - "global_id": 14943, - "bbox": [ - 170.21, - 449.68, - 179.53, - 457.68 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p522-b5", - "global_id": 14944, - "bbox": [ - 170.43, - 544.73, - 179.31, - 552.73 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p522-b6", - "global_id": 14945, - "bbox": [ - 178.32, - 90.09, - 191.2, - 98.17 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p522-b7", - "global_id": 14946, - "bbox": [ - 235.37, - 149.51, - 239.37, - 157.51 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p522-b8", - "global_id": 14947, - "bbox": [ - 235.82, - 244.49, - 239.82, - 252.49 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p522-b9", - "global_id": 14948, - "bbox": [ - 177.37, - 148.0, - 221.36, - 156.0 - ], - "text": "0\n5", - "type": "text" - }, - { - "block_id": "p522-b10", - "global_id": 14949, - "bbox": [ - 168.37, - 103.06, - 172.37, - 111.06 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p522-b11", - "global_id": 14950, - "bbox": [ - 123.75, - 147.71, - 134.41, - 156.0 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p522-b12", - "global_id": 14951, - "bbox": [ - 178.32, - 190.02, - 205.17, - 198.1 - ], - "text": "x[n]u[n]", - "type": "text" - }, - { - "block_id": "p522-b13", - "global_id": 14952, - "bbox": [ - 177.37, - 243.11, - 220.91, - 251.11 - ], - "text": "0\n5", - "type": "text" - }, - { - "block_id": "p522-b14", - "global_id": 14953, - "bbox": [ - 168.37, - 199.68, - 172.37, - 207.68 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p522-b15", - "global_id": 14954, - "bbox": [ - 123.75, - 242.81, - 134.41, - 251.11 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p522-b16", - "global_id": 14955, - "bbox": [ - 178.32, - 284.63, - 234.5, - 292.93 - ], - "text": "x[n 1]u[n 1]", - "type": "text" - }, - { - "block_id": "p522-b17", - "global_id": 14956, - "bbox": [ - 177.37, - 337.76, - 229.77, - 345.76 - ], - "text": "0\n6", - "type": "text" - }, - { - "block_id": "p522-b18", - "global_id": 14957, - "bbox": [ - 166.62, - 295.11, - 170.62, - 303.11 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p522-b19", - "global_id": 14958, - "bbox": [ - 133.44, - 337.46, - 144.1, - 345.76 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p522-b20", - "global_id": 14959, - "bbox": [ - 178.32, - 379.79, - 219.83, - 388.08 - ], - "text": "x[n 1]u[n]", - "type": "text" - }, - { - "block_id": "p522-b21", - "global_id": 14960, - "bbox": [ - 177.37, - 433.09, - 230.29, - 441.09 - ], - "text": "0\n6", - "type": "text" - }, - { - "block_id": "p522-b22", - "global_id": 14961, - "bbox": [ - 166.62, - 389.92, - 170.62, - 397.92 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p522-b23", - "global_id": 14962, - "bbox": [ - 133.44, - 432.79, - 144.1, - 441.09 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p522-b24", - "global_id": 14963, - "bbox": [ - 178.32, - 474.91, - 219.83, - 483.21 - ], - "text": "x[n 1]u[n]", - "type": "text" - }, - { - "block_id": "p522-b25", - "global_id": 14964, - "bbox": [ - 177.37, - 528.15, - 211.79, - 536.15 - ], - "text": "0\n4", - "type": "text" - }, - { - "block_id": "p522-b26", - "global_id": 14965, - "bbox": [ - 177.63, - 485.05, - 181.63, - 493.05 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p522-b27", - "global_id": 14966, - "bbox": [ - 114.06, - 527.85, - 124.72, - 536.15 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p522-b28", - "global_id": 14967, - "bbox": [ - 235.37, - 339.31, - 239.37, - 347.31 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p522-b29", - "global_id": 14968, - "bbox": [ - 235.37, - 434.51, - 239.37, - 442.51 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p522-b30", - "global_id": 14969, - "bbox": [ - 235.37, - 529.59, - 239.37, - 537.59 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p522-b31", - "global_id": 14970, - "bbox": [ - 267.8, - 544.41, - 447.01, - 554.02 - ], - "text": "Figure 5.4 A signal x[n] and its shifted versions.", - "type": "text" - }, - { - "block_id": "p522-b32", - "global_id": 14971, - "bbox": [ - 101.84, - 600.5, - 490.39, - 634.79 - ], - "text": "A look at Eqs. (5.12) and (5.14) shows that they are identical except for the extra term x[−1]\nin Eq. (5.14). We see from Figs. 5.4c and 5.4d that x[n −1]u[n] is the same as x[n −1]u[n −1]\nplus x[−1]δ[n]. Hence, the difference between their transforms is x[−1].", - "type": "text" - } - ] - }, - { - "page_num": 523, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p523-b0", - "global_id": 14972, - "bbox": [ - 313.93, - 62.57, - 516.13, - 71.98 - ], - "text": "5.2\nSome Properties of the z-Transform\n503", - "type": "text" - }, - { - "block_id": "p523-b1", - "global_id": 14973, - "bbox": [ - 127.59, - 85.85, - 263.67, - 95.91 - ], - "text": "Proof. For the integer value of m,", - "type": "text" - }, - { - "block_id": "p523-b2", - "global_id": 14974, - "bbox": [ - 225.19, - 117.35, - 319.19, - 127.63 - ], - "text": "Z{x[n −m]u[n −m]} =", - "type": "text" - }, - { - "block_id": "p523-b3", - "global_id": 14975, - "bbox": [ - 321.24, - 107.18, - 335.33, - 117.86 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p523-b4", - "global_id": 14976, - "bbox": [ - 322.07, - 131.75, - 334.48, - 139.01 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p523-b5", - "global_id": 14977, - "bbox": [ - 336.44, - 115.63, - 418.01, - 127.63 - ], - "text": "x[n −m]u[n −m]z−n", - "type": "text" - }, - { - "block_id": "p523-b6", - "global_id": 14978, - "bbox": [ - 127.59, - 150.07, - 516.12, - 172.4 - ], - "text": "Recall that x[n −m]u[n −m] = 0 for n < m so that the limits on the summation on the right-hand\nside can be taken from n = m to ∞. Therefore,", - "type": "text" - }, - { - "block_id": "p523-b7", - "global_id": 14979, - "bbox": [ - 225.49, - 193.12, - 319.48, - 203.4 - ], - "text": "Z{x[n −m]u[n −m]} =", - "type": "text" - }, - { - "block_id": "p523-b8", - "global_id": 14980, - "bbox": [ - 321.53, - 182.95, - 335.63, - 193.63 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p523-b9", - "global_id": 14981, - "bbox": [ - 321.61, - 206.97, - 335.56, - 214.16 - ], - "text": "n=m", - "type": "text" - }, - { - "block_id": "p523-b10", - "global_id": 14982, - "bbox": [ - 336.74, - 191.4, - 383.66, - 203.4 - ], - "text": "x[n −m]z−n", - "type": "text" - }, - { - "block_id": "p523-b11", - "global_id": 14983, - "bbox": [ - 311.71, - 226.68, - 319.49, - 236.65 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p523-b12", - "global_id": 14984, - "bbox": [ - 321.53, - 216.51, - 335.63, - 227.19 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p523-b13", - "global_id": 14985, - "bbox": [ - 322.69, - 241.07, - 334.48, - 248.34 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p523-b14", - "global_id": 14986, - "bbox": [ - 336.74, - 224.96, - 379.79, - 236.96 - ], - "text": "x[r]z−(r+m)", - "type": "text" - }, - { - "block_id": "p523-b15", - "global_id": 14987, - "bbox": [ - 311.71, - 254.21, - 329.93, - 271.15 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p523-b16", - "global_id": 14988, - "bbox": [ - 322.73, - 267.58, - 331.65, - 278.13 - ], - "text": "zm", - "type": "text" - }, - { - "block_id": "p523-b17", - "global_id": 14989, - "bbox": [ - 334.44, - 250.6, - 348.54, - 261.27 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p523-b18", - "global_id": 14990, - "bbox": [ - 335.59, - 275.16, - 347.39, - 282.43 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p523-b19", - "global_id": 14991, - "bbox": [ - 349.65, - 254.21, - 397.79, - 271.15 - ], - "text": "x[r]z−r = 1", - "type": "text" - }, - { - "block_id": "p523-b20", - "global_id": 14992, - "bbox": [ - 390.59, - 260.78, - 418.24, - 278.13 - ], - "text": "zm X[z]", - "type": "text" - }, - { - "block_id": "p523-b21", - "global_id": 14993, - "bbox": [ - 127.59, - 310.41, - 242.88, - 320.37 - ], - "text": "To prove Eq. (5.15), we have", - "type": "text" - }, - { - "block_id": "p523-b22", - "global_id": 14994, - "bbox": [ - 213.9, - 341.82, - 289.82, - 352.1 - ], - "text": "Z{x[n −m]u[n]} =", - "type": "text" - }, - { - "block_id": "p523-b23", - "global_id": 14995, - "bbox": [ - 291.87, - 331.65, - 305.97, - 342.32 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p523-b24", - "global_id": 14996, - "bbox": [ - 292.72, - 356.21, - 305.12, - 363.47 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p523-b25", - "global_id": 14997, - "bbox": [ - 307.08, - 337.71, - 364.33, - 352.1 - ], - "text": "x[n −m]z−n =", - "type": "text" - }, - { - "block_id": "p523-b26", - "global_id": 14998, - "bbox": [ - 368.72, - 331.65, - 382.82, - 342.32 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p523-b27", - "global_id": 14999, - "bbox": [ - 366.38, - 355.67, - 385.16, - 362.86 - ], - "text": "r=−m", - "type": "text" - }, - { - "block_id": "p523-b28", - "global_id": 15000, - "bbox": [ - 386.27, - 340.1, - 429.32, - 352.1 - ], - "text": "x[r]z−(r+m)", - "type": "text" - }, - { - "block_id": "p523-b29", - "global_id": 15001, - "bbox": [ - 282.05, - 375.69, - 306.22, - 387.69 - ], - "text": "= z−m", - "type": "text" - }, - { - "block_id": "p523-b30", - "global_id": 15002, - "bbox": [ - 307.83, - 360.44, - 330.46, - 377.91 - ], - "text": "−1\n\"", - "type": "text" - }, - { - "block_id": "p523-b31", - "global_id": 15003, - "bbox": [ - 314.02, - 391.26, - 332.8, - 398.46 - ], - "text": "r=−m", - "type": "text" - }, - { - "block_id": "p523-b32", - "global_id": 15004, - "bbox": [ - 333.91, - 373.3, - 371.11, - 387.69 - ], - "text": "x[r]z−r +", - "type": "text" - }, - { - "block_id": "p523-b33", - "global_id": 15005, - "bbox": [ - 372.66, - 367.24, - 386.76, - 377.91 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p523-b34", - "global_id": 15006, - "bbox": [ - 373.81, - 391.8, - 385.6, - 399.07 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p523-b35", - "global_id": 15007, - "bbox": [ - 387.86, - 375.69, - 415.08, - 387.69 - ], - "text": "x[r]z−r", - "type": "text" - }, - { - "block_id": "p523-b37", - "global_id": 15008, - "bbox": [ - 282.05, - 409.59, - 306.22, - 421.59 - ], - "text": "= z−m", - "type": "text" - }, - { - "block_id": "p523-b38", - "global_id": 15009, - "bbox": [ - 307.83, - 401.36, - 321.92, - 411.8 - ], - "text": "m\n\"", - "type": "text" - }, - { - "block_id": "p523-b39", - "global_id": 15010, - "bbox": [ - 308.67, - 425.7, - 321.08, - 432.96 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p523-b40", - "global_id": 15011, - "bbox": [ - 323.04, - 407.2, - 397.5, - 421.59 - ], - "text": "x[−n]zn + z−mX[z]", - "type": "text" - }, - { - "block_id": "p523-b41", - "global_id": 15012, - "bbox": [ - 127.59, - 446.28, - 255.42, - 472.39 - ], - "text": "LEFT SHIFT (ADVANCE)\nIf", - "type": "text" - }, - { - "block_id": "p523-b42", - "global_id": 15013, - "bbox": [ - 285.4, - 476.64, - 358.33, - 486.92 - ], - "text": "x[n]u[n] ⇐⇒X[z]", - "type": "text" - }, - { - "block_id": "p523-b43", - "global_id": 15014, - "bbox": [ - 127.6, - 497.76, - 144.74, - 507.72 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p523-b44", - "global_id": 15015, - "bbox": [ - 260.13, - 511.97, - 383.6, - 522.34 - ], - "text": "x[n + 1]u[n] ⇐⇒zX[z] −zx[0]", - "type": "text" - }, - { - "block_id": "p523-b45", - "global_id": 15016, - "bbox": [ - 127.6, - 533.09, - 301.63, - 543.05 - ], - "text": "Repeated application of this property yields", - "type": "text" - }, - { - "block_id": "p523-b46", - "global_id": 15017, - "bbox": [ - 186.14, - 554.93, - 457.58, - 566.74 - ], - "text": "x[n + 2]u[n] ⇐⇒z{z(X[z] −zx[0]) −x[1]} = z2X[z] −z2x[0] −zx[1]", - "type": "text" - }, - { - "block_id": "p523-b47", - "global_id": 15018, - "bbox": [ - 127.59, - 580.37, - 247.54, - 590.44 - ], - "text": "and for the integer value of m,", - "type": "text" - }, - { - "block_id": "p523-b48", - "global_id": 15019, - "bbox": [ - 238.68, - 611.88, - 358.86, - 623.65 - ], - "text": "x[n + m]u[n] ⇐⇒zmX[z] −zm", - "type": "text" - }, - { - "block_id": "p523-b49", - "global_id": 15020, - "bbox": [ - 360.46, - 603.21, - 374.56, - 613.87 - ], - "text": "m−1\n\"", - "type": "text" - }, - { - "block_id": "p523-b50", - "global_id": 15021, - "bbox": [ - 361.3, - 627.77, - 373.71, - 635.03 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p523-b51", - "global_id": 15022, - "bbox": [ - 375.67, - 611.66, - 516.13, - 623.75 - ], - "text": "x[n]z−n\n(5.16)", - "type": "text" - } - ] - }, - { - "page_num": 524, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p524-b0", - "global_id": 15023, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "504\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p524-b1", - "global_id": 15024, - "bbox": [ - 101.84, - 85.87, - 187.92, - 95.91 - ], - "text": "Proof. By definition,", - "type": "text" - }, - { - "block_id": "p524-b2", - "global_id": 15025, - "bbox": [ - 197.92, - 116.9, - 273.86, - 127.18 - ], - "text": "Z{x[n + m]u[n]} =", - "type": "text" - }, - { - "block_id": "p524-b3", - "global_id": 15026, - "bbox": [ - 275.9, - 106.73, - 290.0, - 117.4 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p524-b4", - "global_id": 15027, - "bbox": [ - 276.74, - 131.29, - 289.15, - 138.55 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p524-b5", - "global_id": 15028, - "bbox": [ - 291.11, - 115.18, - 338.03, - 127.18 - ], - "text": "x[n + m]z−n", - "type": "text" - }, - { - "block_id": "p524-b6", - "global_id": 15029, - "bbox": [ - 266.08, - 151.0, - 273.86, - 160.97 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p524-b7", - "global_id": 15030, - "bbox": [ - 275.9, - 140.83, - 290.0, - 151.51 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p524-b8", - "global_id": 15031, - "bbox": [ - 276.27, - 164.85, - 289.62, - 172.04 - ], - "text": "r=m", - "type": "text" - }, - { - "block_id": "p524-b9", - "global_id": 15032, - "bbox": [ - 291.11, - 149.28, - 334.15, - 161.28 - ], - "text": "x[r]z−(r−m)", - "type": "text" - }, - { - "block_id": "p524-b10", - "global_id": 15033, - "bbox": [ - 266.08, - 183.06, - 284.82, - 194.84 - ], - "text": "= zm", - "type": "text" - }, - { - "block_id": "p524-b11", - "global_id": 15034, - "bbox": [ - 286.42, - 174.39, - 300.51, - 185.07 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p524-b12", - "global_id": 15035, - "bbox": [ - 286.79, - 198.41, - 300.14, - 205.61 - ], - "text": "r=m", - "type": "text" - }, - { - "block_id": "p524-b13", - "global_id": 15036, - "bbox": [ - 301.62, - 182.84, - 328.84, - 194.84 - ], - "text": "x[r]z−r", - "type": "text" - }, - { - "block_id": "p524-b14", - "global_id": 15037, - "bbox": [ - 266.08, - 218.12, - 284.82, - 229.89 - ], - "text": "= zm", - "type": "text" - }, - { - "block_id": "p524-b15", - "global_id": 15038, - "bbox": [ - 285.31, - 205.63, - 305.94, - 220.11 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p524-b16", - "global_id": 15039, - "bbox": [ - 292.99, - 234.01, - 304.79, - 241.27 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p524-b17", - "global_id": 15040, - "bbox": [ - 307.05, - 215.51, - 344.25, - 229.89 - ], - "text": "x[r]z−r −", - "type": "text" - }, - { - "block_id": "p524-b18", - "global_id": 15041, - "bbox": [ - 345.8, - 209.45, - 359.9, - 220.11 - ], - "text": "m−1\n\"", - "type": "text" - }, - { - "block_id": "p524-b19", - "global_id": 15042, - "bbox": [ - 346.95, - 234.01, - 358.74, - 241.27 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p524-b20", - "global_id": 15043, - "bbox": [ - 361.01, - 217.9, - 388.23, - 229.89 - ], - "text": "x[r]z−r", - "type": "text" - }, - { - "block_id": "p524-b21", - "global_id": 15044, - "bbox": [ - 388.88, - 205.63, - 394.31, - 215.6 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p524-b22", - "global_id": 15045, - "bbox": [ - 266.08, - 253.69, - 322.13, - 265.47 - ], - "text": "= zmX[z] −zm", - "type": "text" - }, - { - "block_id": "p524-b23", - "global_id": 15046, - "bbox": [ - 323.73, - 245.02, - 337.83, - 255.7 - ], - "text": "m−1\n\"", - "type": "text" - }, - { - "block_id": "p524-b24", - "global_id": 15047, - "bbox": [ - 324.88, - 269.59, - 336.68, - 276.86 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p524-b25", - "global_id": 15048, - "bbox": [ - 338.94, - 253.47, - 366.16, - 265.47 - ], - "text": "x[r]z−r", - "type": "text" - }, - { - "block_id": "p524-b26", - "global_id": 15049, - "bbox": [ - 76.77, - 302.6, - 412.07, - 314.55 - ], - "text": "EXAMPLE 5.4\nz-Transform Using the Right-Shift Property", - "type": "text" - }, - { - "block_id": "p524-b27", - "global_id": 15050, - "bbox": [ - 103.16, - 330.8, - 337.01, - 341.17 - ], - "text": "Find the z-transform of the signal x[n] depicted in Fig. 5.5.", - "type": "text" - }, - { - "block_id": "p524-b28", - "global_id": 15051, - "bbox": [ - 115.76, - 394.76, - 128.64, - 402.84 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p524-b29", - "global_id": 15052, - "bbox": [ - 125.15, - 472.06, - 224.01, - 480.06 - ], - "text": "0\n1\n2\n3\n4\n5", - "type": "text" - }, - { - "block_id": "p524-b30", - "global_id": 15053, - "bbox": [ - 136.36, - 407.18, - 140.36, - 415.18 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p524-b31", - "global_id": 15054, - "bbox": [ - 275.43, - 474.04, - 417.45, - 488.0 - ], - "text": "n\nFigure 5.5 Signal for Ex. 5.4.", - "type": "text" - }, - { - "block_id": "p524-b32", - "global_id": 15055, - "bbox": [ - 103.17, - 517.95, - 477.02, - 540.28 - ], - "text": "The signal x[n] can be expressed as a product of n and a gate pulse u[n] −u[n −6].\nTherefore,", - "type": "text" - }, - { - "block_id": "p524-b33", - "global_id": 15056, - "bbox": [ - 199.12, - 551.82, - 381.04, - 562.2 - ], - "text": "x[n] = n{u[n] −u[n −6]} = nu[n] −nu[n −6]", - "type": "text" - }, - { - "block_id": "p524-b34", - "global_id": 15057, - "bbox": [ - 103.17, - 573.74, - 477.03, - 596.07 - ], - "text": "We cannot find the z-transform of nu[n −6] directly by using the right-shift property\n[Eq. (5.13)]. So we rearrange it in terms of (n −6)u[n −6] as follows:", - "type": "text" - }, - { - "block_id": "p524-b35", - "global_id": 15058, - "bbox": [ - 205.39, - 607.61, - 342.29, - 617.99 - ], - "text": "x[n] = nu[n] −(n −6 + 6)u[n −6]", - "type": "text" - }, - { - "block_id": "p524-b36", - "global_id": 15059, - "bbox": [ - 223.51, - 622.56, - 374.72, - 632.94 - ], - "text": "= nu[n] −(n −6)u[n −6] −6u[n −6]", - "type": "text" - } - ] - }, - { - "page_num": 525, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p525-b0", - "global_id": 15060, - "bbox": [ - 313.93, - 62.57, - 516.13, - 71.98 - ], - "text": "5.2\nSome Properties of the z-Transform\n505", - "type": "text" - }, - { - "block_id": "p525-b1", - "global_id": 15061, - "bbox": [ - 128.9, - 86.13, - 502.77, - 108.16 - ], - "text": "We can now find the z-transform of the bracketed term by using the right-shift property\n[Eq. (5.13)]. Because u[n] ⇐⇒z/(z −1),", - "type": "text" - }, - { - "block_id": "p525-b2", - "global_id": 15062, - "bbox": [ - 247.17, - 118.1, - 310.46, - 135.04 - ], - "text": "u[n −6] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p525-b3", - "global_id": 15063, - "bbox": [ - 304.04, - 131.54, - 311.4, - 142.02 - ], - "text": "z6", - "type": "text" - }, - { - "block_id": "p525-b4", - "global_id": 15064, - "bbox": [ - 314.29, - 117.99, - 383.29, - 142.12 - ], - "text": "z\nz −1 =\n1\nz5(z −1)", - "type": "text" - }, - { - "block_id": "p525-b5", - "global_id": 15065, - "bbox": [ - 128.91, - 151.86, - 273.72, - 163.18 - ], - "text": "Also, because nu[n] ⇐⇒z/(z −1)2,", - "type": "text" - }, - { - "block_id": "p525-b6", - "global_id": 15066, - "bbox": [ - 225.34, - 172.97, - 316.89, - 189.91 - ], - "text": "(n −6)u[n −6] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p525-b7", - "global_id": 15067, - "bbox": [ - 310.47, - 186.41, - 317.84, - 196.89 - ], - "text": "z6", - "type": "text" - }, - { - "block_id": "p525-b8", - "global_id": 15068, - "bbox": [ - 320.72, - 172.87, - 404.62, - 196.99 - ], - "text": "z\n(z −1)2 =\n1\nz5(z −1)2", - "type": "text" - }, - { - "block_id": "p525-b9", - "global_id": 15069, - "bbox": [ - 128.9, - 206.5, - 170.65, - 216.46 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p525-b10", - "global_id": 15070, - "bbox": [ - 205.49, - 213.27, - 424.96, - 241.32 - ], - "text": "X[z] =\nz\n(z −1)2 −\n1\nz5(z −1)2 −\n6\nz5(z −1) = z6 −6z + 5", - "type": "text" - }, - { - "block_id": "p525-b11", - "global_id": 15071, - "bbox": [ - 383.76, - 230.75, - 422.25, - 241.32 - ], - "text": "z5(z −1)2", - "type": "text" - }, - { - "block_id": "p525-b12", - "global_id": 15072, - "bbox": [ - 133.57, - 296.72, - 445.68, - 308.68 - ], - "text": "DRILL 5.5\nz-Transform Using the Right-Shift Property", - "type": "text" - }, - { - "block_id": "p525-b13", - "global_id": 15073, - "bbox": [ - 133.57, - 317.39, - 510.15, - 339.71 - ], - "text": "Using only the fact that u[n] ⇐⇒z/(z −1) and the right-shift property [Eq. (5.13)], find the\nz-transforms of the signals in Figs. 5.2 and 5.3.", - "type": "text" - }, - { - "block_id": "p525-b14", - "global_id": 15074, - "bbox": [ - 133.84, - 353.24, - 196.06, - 364.2 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p525-b15", - "global_id": 15075, - "bbox": [ - 133.57, - 371.59, - 243.96, - 381.56 - ], - "text": "See Ex. 5.2d and Drill 5.1a.", - "type": "text" - }, - { - "block_id": "p525-b16", - "global_id": 15076, - "bbox": [ - 127.59, - 418.44, - 428.74, - 433.2 - ], - "text": "5.2-2 z-Domain Scaling Property (Multiplication by γ n)", - "type": "text" - }, - { - "block_id": "p525-b17", - "global_id": 15077, - "bbox": [ - 127.59, - 439.24, - 516.12, - 461.26 - ], - "text": "Scaling in the z-domain is equivalent to multiplying a time-domain signal by an exponential. That\nis, if", - "type": "text" - }, - { - "block_id": "p525-b18", - "global_id": 15078, - "bbox": [ - 285.4, - 463.73, - 358.34, - 474.0 - ], - "text": "x[n]u[n] ⇐⇒X[z]", - "type": "text" - }, - { - "block_id": "p525-b19", - "global_id": 15079, - "bbox": [ - 127.6, - 483.67, - 144.74, - 493.63 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p525-b20", - "global_id": 15080, - "bbox": [ - 274.89, - 498.19, - 347.44, - 509.97 - ], - "text": "γ nx[n]u[n] ⇐⇒X", - "type": "text" - }, - { - "block_id": "p525-b21", - "global_id": 15081, - "bbox": [ - 348.99, - 485.7, - 360.85, - 502.98 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p525-b22", - "global_id": 15082, - "bbox": [ - 355.61, - 506.76, - 360.61, - 516.73 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p525-b23", - "global_id": 15083, - "bbox": [ - 363.4, - 485.7, - 368.83, - 495.67 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p525-b24", - "global_id": 15084, - "bbox": [ - 492.06, - 500.1, - 516.12, - 510.07 - ], - "text": "(5.17)", - "type": "text" - }, - { - "block_id": "p525-b25", - "global_id": 15085, - "bbox": [ - 127.59, - 525.96, - 153.53, - 535.92 - ], - "text": "Proof.", - "type": "text" - }, - { - "block_id": "p525-b26", - "global_id": 15086, - "bbox": [ - 201.31, - 543.07, - 269.76, - 554.85 - ], - "text": "Z{γ nx[n]u[n]} =", - "type": "text" - }, - { - "block_id": "p525-b27", - "global_id": 15087, - "bbox": [ - 271.81, - 534.4, - 285.91, - 545.07 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p525-b28", - "global_id": 15088, - "bbox": [ - 272.65, - 558.96, - 285.05, - 566.22 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p525-b29", - "global_id": 15089, - "bbox": [ - 287.01, - 540.46, - 336.77, - 554.85 - ], - "text": "γ nx[n]z−n =", - "type": "text" - }, - { - "block_id": "p525-b30", - "global_id": 15090, - "bbox": [ - 338.82, - 534.4, - 352.92, - 545.07 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p525-b31", - "global_id": 15091, - "bbox": [ - 339.67, - 558.96, - 352.08, - 566.22 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p525-b32", - "global_id": 15092, - "bbox": [ - 354.03, - 544.57, - 370.1, - 554.85 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p525-b33", - "global_id": 15093, - "bbox": [ - 371.21, - 530.59, - 384.36, - 547.87 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p525-b34", - "global_id": 15094, - "bbox": [ - 379.12, - 551.65, - 384.13, - 561.61 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p525-b35", - "global_id": 15095, - "bbox": [ - 386.92, - 530.59, - 402.56, - 542.4 - ], - "text": "−n", - "type": "text" - }, - { - "block_id": "p525-b36", - "global_id": 15096, - "bbox": [ - 405.12, - 544.57, - 421.02, - 554.85 - ], - "text": "= X", - "type": "text" - }, - { - "block_id": "p525-b37", - "global_id": 15097, - "bbox": [ - 422.57, - 530.59, - 434.43, - 547.87 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p525-b38", - "global_id": 15098, - "bbox": [ - 429.19, - 551.65, - 434.19, - 561.61 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p525-b39", - "global_id": 15099, - "bbox": [ - 436.99, - 530.59, - 442.42, - 540.55 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p525-b40", - "global_id": 15100, - "bbox": [ - 133.57, - 596.77, - 412.46, - 608.73 - ], - "text": "DRILL 5.6\nUsing the z-Domain Scaling Property", - "type": "text" - }, - { - "block_id": "p525-b41", - "global_id": 15101, - "bbox": [ - 133.57, - 617.85, - 462.23, - 627.81 - ], - "text": "Use Eq. (5.17) to derive pairs 6 and 8 in Table 5.1 from pairs 2 and 3, respectively.", - "type": "text" - } - ] - }, - { - "page_num": 526, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p526-b0", - "global_id": 15102, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "506\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p526-b1", - "global_id": 15103, - "bbox": [ - 101.84, - 85.66, - 438.48, - 98.48 - ], - "text": "5.2-3 z-Domain Differentiation Property (Multiplication by n)", - "type": "text" - }, - { - "block_id": "p526-b2", - "global_id": 15104, - "bbox": [ - 101.84, - 104.51, - 485.7, - 114.57 - ], - "text": "Multiplying a signal by n in the time domain produces differentiation in the z-domain. That is, if", - "type": "text" - }, - { - "block_id": "p526-b3", - "global_id": 15105, - "bbox": [ - 259.66, - 126.32, - 332.58, - 136.6 - ], - "text": "x[n]u[n] ⇐⇒X[z]", - "type": "text" - }, - { - "block_id": "p526-b4", - "global_id": 15106, - "bbox": [ - 101.85, - 148.87, - 118.99, - 158.83 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p526-b5", - "global_id": 15107, - "bbox": [ - 245.72, - 156.83, - 326.21, - 173.78 - ], - "text": "nx[n]u[n] ⇐⇒−z d", - "type": "text" - }, - { - "block_id": "p526-b6", - "global_id": 15108, - "bbox": [ - 319.43, - 163.5, - 490.38, - 180.85 - ], - "text": "dzX[z]\n(5.18)", - "type": "text" - }, - { - "block_id": "p526-b7", - "global_id": 15109, - "bbox": [ - 101.85, - 187.85, - 127.78, - 197.81 - ], - "text": "Proof.", - "type": "text" - }, - { - "block_id": "p526-b8", - "global_id": 15110, - "bbox": [ - 213.0, - 227.48, - 232.62, - 244.43 - ], - "text": "−z d", - "type": "text" - }, - { - "block_id": "p526-b9", - "global_id": 15111, - "bbox": [ - 225.84, - 227.48, - 284.41, - 251.5 - ], - "text": "dzX[z] = −z d", - "type": "text" - }, - { - "block_id": "p526-b10", - "global_id": 15112, - "bbox": [ - 277.63, - 241.54, - 286.48, - 251.5 - ], - "text": "dz", - "type": "text" - }, - { - "block_id": "p526-b11", - "global_id": 15113, - "bbox": [ - 288.79, - 223.98, - 302.89, - 234.65 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p526-b12", - "global_id": 15114, - "bbox": [ - 289.63, - 248.54, - 302.04, - 255.81 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p526-b13", - "global_id": 15115, - "bbox": [ - 303.99, - 232.43, - 332.86, - 244.43 - ], - "text": "x[n]z−n", - "type": "text" - }, - { - "block_id": "p526-b14", - "global_id": 15116, - "bbox": [ - 254.97, - 268.25, - 276.43, - 278.53 - ], - "text": "= −z", - "type": "text" - }, - { - "block_id": "p526-b15", - "global_id": 15117, - "bbox": [ - 277.54, - 258.08, - 291.64, - 268.75 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p526-b16", - "global_id": 15118, - "bbox": [ - 278.38, - 282.64, - 290.79, - 289.91 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p526-b17", - "global_id": 15119, - "bbox": [ - 292.74, - 266.53, - 343.27, - 278.53 - ], - "text": "−nx[n]z−n−1", - "type": "text" - }, - { - "block_id": "p526-b18", - "global_id": 15120, - "bbox": [ - 254.97, - 302.36, - 262.74, - 312.32 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p526-b19", - "global_id": 15121, - "bbox": [ - 264.79, - 292.18, - 278.89, - 302.86 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p526-b20", - "global_id": 15122, - "bbox": [ - 265.63, - 316.74, - 278.03, - 324.01 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p526-b21", - "global_id": 15123, - "bbox": [ - 279.99, - 298.24, - 379.24, - 312.63 - ], - "text": "nx[n]z−n = Z{nx[n]u[n]}", - "type": "text" - }, - { - "block_id": "p526-b22", - "global_id": 15124, - "bbox": [ - 107.82, - 366.94, - 429.01, - 378.89 - ], - "text": "DRILL 5.7\nUsing the z-Domain Differentiation Property", - "type": "text" - }, - { - "block_id": "p526-b23", - "global_id": 15125, - "bbox": [ - 107.82, - 388.01, - 484.41, - 409.94 - ], - "text": "Use Eq. (5.18) to derive pairs 3 and 4 in Table 5.1 from pair 2. Similarly, derive pairs 8 and 9\nfrom pair 6.", - "type": "text" - }, - { - "block_id": "p526-b24", - "global_id": 15126, - "bbox": [ - 101.84, - 447.13, - 261.43, - 459.09 - ], - "text": "5.2-4 Time-Reversal Property", - "type": "text" - }, - { - "block_id": "p526-b25", - "global_id": 15127, - "bbox": [ - 101.84, - 465.21, - 108.48, - 475.17 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p526-b26", - "global_id": 15128, - "bbox": [ - 267.96, - 477.08, - 324.29, - 487.36 - ], - "text": "x[n] ⇐⇒X[z]", - "type": "text" - }, - { - "block_id": "p526-b27", - "global_id": 15129, - "bbox": [ - 101.85, - 496.04, - 121.98, - 506.6 - ], - "text": "then†", - "type": "text" - }, - { - "block_id": "p526-b28", - "global_id": 15130, - "bbox": [ - 259.24, - 508.5, - 333.0, - 518.88 - ], - "text": "x[−n] ⇐⇒X[1/z]", - "type": "text" - }, - { - "block_id": "p526-b29", - "global_id": 15131, - "bbox": [ - 101.85, - 528.97, - 127.78, - 538.94 - ], - "text": "Proof.", - "type": "text" - }, - { - "block_id": "p526-b30", - "global_id": 15132, - "bbox": [ - 240.71, - 547.01, - 289.75, - 557.29 - ], - "text": "Z{x[−n]} =", - "type": "text" - }, - { - "block_id": "p526-b31", - "global_id": 15133, - "bbox": [ - 295.49, - 536.84, - 309.59, - 547.51 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p526-b32", - "global_id": 15134, - "bbox": [ - 291.8, - 560.86, - 313.27, - 568.05 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p526-b33", - "global_id": 15135, - "bbox": [ - 314.39, - 545.28, - 351.02, - 557.29 - ], - "text": "x[−n]z−n", - "type": "text" - }, - { - "block_id": "p526-b34", - "global_id": 15136, - "bbox": [ - 101.84, - 586.67, - 374.01, - 598.9 - ], - "text": "† For complex signal x[n], the time-reversal property is modified as follows:", - "type": "text" - }, - { - "block_id": "p526-b35", - "global_id": 15137, - "bbox": [ - 257.12, - 605.11, - 335.11, - 616.15 - ], - "text": "x∗[−n] ⇐⇒X∗[1/z∗]", - "type": "text" - } - ] - }, - { - "page_num": 527, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p527-b0", - "global_id": 15138, - "bbox": [ - 313.93, - 62.57, - 516.13, - 71.98 - ], - "text": "5.2\nSome Properties of the z-Transform\n507", - "type": "text" - }, - { - "block_id": "p527-b1", - "global_id": 15139, - "bbox": [ - 127.59, - 85.72, - 326.57, - 95.78 - ], - "text": "Changing the sign of the dummy variable n yields", - "type": "text" - }, - { - "block_id": "p527-b2", - "global_id": 15140, - "bbox": [ - 261.78, - 117.65, - 310.83, - 127.92 - ], - "text": "Z{x[−n]} =", - "type": "text" - }, - { - "block_id": "p527-b3", - "global_id": 15141, - "bbox": [ - 316.58, - 107.48, - 330.67, - 118.14 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p527-b4", - "global_id": 15142, - "bbox": [ - 312.88, - 131.5, - 334.35, - 138.69 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p527-b5", - "global_id": 15143, - "bbox": [ - 335.47, - 116.15, - 358.9, - 127.92 - ], - "text": "x[n]zn", - "type": "text" - }, - { - "block_id": "p527-b6", - "global_id": 15144, - "bbox": [ - 303.06, - 151.7, - 310.83, - 161.67 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p527-b7", - "global_id": 15145, - "bbox": [ - 316.58, - 141.53, - 330.67, - 152.2 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p527-b8", - "global_id": 15146, - "bbox": [ - 312.88, - 165.55, - 334.35, - 172.75 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p527-b9", - "global_id": 15147, - "bbox": [ - 335.47, - 149.97, - 381.42, - 162.08 - ], - "text": "x[n](1/z)−n", - "type": "text" - }, - { - "block_id": "p527-b10", - "global_id": 15148, - "bbox": [ - 303.06, - 177.2, - 339.58, - 187.58 - ], - "text": "= X[1/z]", - "type": "text" - }, - { - "block_id": "p527-b11", - "global_id": 15149, - "bbox": [ - 127.6, - 201.31, - 516.14, - 223.64 - ], - "text": "The region of convergence is also inverted; that is, if the ROC of x[n] is |z| > |γ |, then the ROC of\nx[−n] is |z| < 1/|γ |.", - "type": "text" - }, - { - "block_id": "p527-b12", - "global_id": 15150, - "bbox": [ - 133.57, - 257.53, - 393.55, - 269.48 - ], - "text": "DRILL 5.8\nUsing the Time-Reversal Property", - "type": "text" - }, - { - "block_id": "p527-b13", - "global_id": 15151, - "bbox": [ - 133.57, - 278.19, - 510.16, - 300.52 - ], - "text": "Use the time-reversal property and pair 2 in Table 5.1 to show that u[−n] ⇐⇒−1/(z−1) with\nthe ROC |z| < 1.", - "type": "text" - }, - { - "block_id": "p527-b14", - "global_id": 15152, - "bbox": [ - 127.59, - 340.43, - 277.2, - 352.38 - ], - "text": "5.2-5 Convolution Property", - "type": "text" - }, - { - "block_id": "p527-b15", - "global_id": 15153, - "bbox": [ - 127.59, - 357.92, - 303.48, - 368.47 - ], - "text": "The time-convolution property states that if‡", - "type": "text" - }, - { - "block_id": "p527-b16", - "global_id": 15154, - "bbox": [ - 229.67, - 382.21, - 414.06, - 393.36 - ], - "text": "x1[n] ⇐⇒X1[z]\nand\nx2[n] ⇐⇒X2[z],", - "type": "text" - }, - { - "block_id": "p527-b17", - "global_id": 15155, - "bbox": [ - 127.59, - 406.64, - 220.14, - 416.7 - ], - "text": "then (time convolution)", - "type": "text" - }, - { - "block_id": "p527-b18", - "global_id": 15156, - "bbox": [ - 265.5, - 421.58, - 516.13, - 432.73 - ], - "text": "x1[n] ∗x2[n] ⇐⇒X1[z]X2[z]\n(5.19)", - "type": "text" - }, - { - "block_id": "p527-b19", - "global_id": 15157, - "bbox": [ - 127.59, - 444.03, - 516.14, - 466.04 - ], - "text": "Proof. This property applies to causal as well as noncausal sequences. We shall prove it for the\nmore general case of noncausal sequences, where the convolution sum ranges from −∞to ∞.", - "type": "text" - }, - { - "block_id": "p527-b20", - "global_id": 15158, - "bbox": [ - 145.52, - 468.03, - 179.49, - 477.99 - ], - "text": "We have", - "type": "text" - }, - { - "block_id": "p527-b21", - "global_id": 15159, - "bbox": [ - 223.34, - 498.3, - 306.14, - 509.45 - ], - "text": "Z{x1[n] ∗x2[n]} = Z", - "type": "text" - }, - { - "block_id": "p527-b22", - "global_id": 15160, - "bbox": [ - 308.04, - 481.33, - 332.8, - 498.8 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p527-b23", - "global_id": 15161, - "bbox": [ - 314.23, - 512.15, - 337.25, - 519.35 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p527-b24", - "global_id": 15162, - "bbox": [ - 338.37, - 498.3, - 398.69, - 509.45 - ], - "text": "x1[m]x2[n −m]", - "type": "text" - }, - { - "block_id": "p527-b26", - "global_id": 15163, - "bbox": [ - 289.11, - 532.36, - 296.88, - 542.32 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p527-b27", - "global_id": 15164, - "bbox": [ - 302.62, - 522.19, - 316.71, - 532.85 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p527-b28", - "global_id": 15165, - "bbox": [ - 298.93, - 546.21, - 320.4, - 553.4 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p527-b29", - "global_id": 15166, - "bbox": [ - 321.52, - 530.64, - 334.32, - 542.64 - ], - "text": "z−n", - "type": "text" - }, - { - "block_id": "p527-b30", - "global_id": 15167, - "bbox": [ - 340.39, - 522.19, - 354.49, - 532.85 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p527-b31", - "global_id": 15168, - "bbox": [ - 335.92, - 546.21, - 358.95, - 553.4 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p527-b32", - "global_id": 15169, - "bbox": [ - 360.06, - 532.36, - 420.38, - 543.51 - ], - "text": "x1[m]x2[n −m]", - "type": "text" - }, - { - "block_id": "p527-b33", - "global_id": 15170, - "bbox": [ - 127.59, - 574.01, - 371.61, - 586.23 - ], - "text": "‡ There is also the frequency-convolution property, which states that", - "type": "text" - }, - { - "block_id": "p527-b34", - "global_id": 15171, - "bbox": [ - 245.3, - 595.58, - 317.0, - 617.2 - ], - "text": "x1[n]x2[n] ⇐⇒\n1\n2πj", - "type": "text" - }, - { - "block_id": "p527-b35", - "global_id": 15172, - "bbox": [ - 319.19, - 589.29, - 357.41, - 611.57 - ], - "text": "5\nX1[u]X2", - "type": "text" - }, - { - "block_id": "p527-b36", - "global_id": 15173, - "bbox": [ - 358.9, - 591.59, - 368.29, - 604.46 - ], - "text": "' z", - "type": "text" - }, - { - "block_id": "p527-b37", - "global_id": 15174, - "bbox": [ - 364.3, - 608.14, - 368.79, - 617.11 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p527-b38", - "global_id": 15175, - "bbox": [ - 369.98, - 591.59, - 374.19, - 600.56 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p527-b39", - "global_id": 15176, - "bbox": [ - 375.18, - 598.08, - 398.41, - 610.74 - ], - "text": "u−1 du", - "type": "text" - } - ] - }, - { - "page_num": 528, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p528-b0", - "global_id": 15177, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "508\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p528-b1", - "global_id": 15178, - "bbox": [ - 101.84, - 85.82, - 290.45, - 95.78 - ], - "text": "Interchanging the order of summation, we have", - "type": "text" - }, - { - "block_id": "p528-b2", - "global_id": 15179, - "bbox": [ - 197.95, - 116.27, - 270.76, - 127.42 - ], - "text": "Z[x1[n] ∗x2[n]] =", - "type": "text" - }, - { - "block_id": "p528-b3", - "global_id": 15180, - "bbox": [ - 277.28, - 106.1, - 291.38, - 116.78 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p528-b4", - "global_id": 15181, - "bbox": [ - 272.81, - 130.12, - 295.83, - 137.31 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p528-b5", - "global_id": 15182, - "bbox": [ - 296.95, - 116.27, - 319.19, - 127.42 - ], - "text": "x1[m]", - "type": "text" - }, - { - "block_id": "p528-b6", - "global_id": 15183, - "bbox": [ - 323.99, - 106.1, - 338.09, - 116.78 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p528-b7", - "global_id": 15184, - "bbox": [ - 320.3, - 130.12, - 341.78, - 137.31 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p528-b8", - "global_id": 15185, - "bbox": [ - 342.89, - 114.55, - 393.77, - 127.42 - ], - "text": "x2[n −m]z−n", - "type": "text" - }, - { - "block_id": "p528-b9", - "global_id": 15186, - "bbox": [ - 262.99, - 150.33, - 270.76, - 160.29 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p528-b10", - "global_id": 15187, - "bbox": [ - 277.28, - 140.15, - 291.38, - 150.83 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p528-b11", - "global_id": 15188, - "bbox": [ - 272.81, - 164.18, - 295.83, - 171.37 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p528-b12", - "global_id": 15189, - "bbox": [ - 296.95, - 150.33, - 319.19, - 161.48 - ], - "text": "x1[m]", - "type": "text" - }, - { - "block_id": "p528-b13", - "global_id": 15190, - "bbox": [ - 323.68, - 140.15, - 337.78, - 150.83 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p528-b14", - "global_id": 15191, - "bbox": [ - 320.3, - 164.18, - 341.16, - 171.37 - ], - "text": "r=−∞", - "type": "text" - }, - { - "block_id": "p528-b15", - "global_id": 15192, - "bbox": [ - 342.28, - 148.61, - 389.29, - 161.48 - ], - "text": "x2[r]z−(r+m)", - "type": "text" - }, - { - "block_id": "p528-b16", - "global_id": 15193, - "bbox": [ - 262.99, - 184.38, - 270.76, - 194.35 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p528-b17", - "global_id": 15194, - "bbox": [ - 277.28, - 174.21, - 291.38, - 184.89 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p528-b18", - "global_id": 15195, - "bbox": [ - 272.81, - 198.23, - 295.83, - 205.43 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p528-b19", - "global_id": 15196, - "bbox": [ - 296.95, - 182.66, - 333.54, - 195.53 - ], - "text": "x1[m]z−m", - "type": "text" - }, - { - "block_id": "p528-b20", - "global_id": 15197, - "bbox": [ - 338.54, - 174.21, - 352.63, - 184.89 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p528-b21", - "global_id": 15198, - "bbox": [ - 335.14, - 198.23, - 356.01, - 205.43 - ], - "text": "r=−∞", - "type": "text" - }, - { - "block_id": "p528-b22", - "global_id": 15199, - "bbox": [ - 357.12, - 182.66, - 388.31, - 195.53 - ], - "text": "x2[r]z−r", - "type": "text" - }, - { - "block_id": "p528-b23", - "global_id": 15200, - "bbox": [ - 262.99, - 209.88, - 313.98, - 221.03 - ], - "text": "= X1[z]X2[z]", - "type": "text" - }, - { - "block_id": "p528-b24", - "global_id": 15201, - "bbox": [ - 102.14, - 235.4, - 234.64, - 247.53 - ], - "text": "LTID SYSTEM RESPONSE", - "type": "text" - }, - { - "block_id": "p528-b25", - "global_id": 15202, - "bbox": [ - 101.84, - 251.15, - 490.39, - 273.47 - ], - "text": "It is interesting to apply the time-convolution property to the LTID input–output equation y[n] =\nx[n] ∗h[n]. Since h[n] ⇐⇒H[z], it follows from Eq. (5.19) that", - "type": "text" - }, - { - "block_id": "p528-b26", - "global_id": 15203, - "bbox": [ - 264.18, - 285.84, - 490.38, - 296.22 - ], - "text": "Y[z] = X[z]H[z]\n(5.20)", - "type": "text" - }, - { - "block_id": "p528-b27", - "global_id": 15204, - "bbox": [ - 107.82, - 339.94, - 357.86, - 351.89 - ], - "text": "DRILL 5.9\nUsing the Convolution Property", - "type": "text" - }, - { - "block_id": "p528-b28", - "global_id": 15205, - "bbox": [ - 107.82, - 361.01, - 484.41, - 382.93 - ], - "text": "Use the time-convolution property and appropriate pairs in Table 5.1 to show that\nu[n] ∗u[n −1] = nu[n].", - "type": "text" - }, - { - "block_id": "p528-b29", - "global_id": 15206, - "bbox": [ - 101.84, - 413.58, - 312.29, - 439.69 - ], - "text": "INITIAL AND FINAL VALUES\nFor a causal x[n], the initial value theorem states that", - "type": "text" - }, - { - "block_id": "p528-b30", - "global_id": 15207, - "bbox": [ - 264.33, - 452.05, - 307.67, - 462.43 - ], - "text": "x[0] = lim", - "type": "text" - }, - { - "block_id": "p528-b31", - "global_id": 15208, - "bbox": [ - 292.27, - 452.05, - 327.91, - 467.62 - ], - "text": "z→∞X[z]", - "type": "text" - }, - { - "block_id": "p528-b32", - "global_id": 15209, - "bbox": [ - 101.85, - 478.72, - 290.62, - 488.68 - ], - "text": "This result follows immediately from Eq. (5.7).", - "type": "text" - }, - { - "block_id": "p528-b33", - "global_id": 15210, - "bbox": [ - 119.78, - 490.26, - 471.75, - 500.63 - ], - "text": "If (z −1)X[z] has no poles outside the unit circle, then the final value theorem states that", - "type": "text" - }, - { - "block_id": "p528-b34", - "global_id": 15211, - "bbox": [ - 241.51, - 513.0, - 306.24, - 529.01 - ], - "text": "lim\nN→∞x[N] = lim", - "type": "text" - }, - { - "block_id": "p528-b35", - "global_id": 15212, - "bbox": [ - 292.67, - 513.0, - 350.73, - 529.18 - ], - "text": "z→1(z −1)X[z]", - "type": "text" - }, - { - "block_id": "p528-b36", - "global_id": 15213, - "bbox": [ - 101.85, - 540.21, - 247.31, - 550.17 - ], - "text": "This can be shown from the fact that", - "type": "text" - }, - { - "block_id": "p528-b37", - "global_id": 15214, - "bbox": [ - 199.01, - 566.13, - 279.05, - 576.5 - ], - "text": "x[n] −x[n −1] ⇐⇒", - "type": "text" - }, - { - "block_id": "p528-b39", - "global_id": 15215, - "bbox": [ - 287.25, - 559.56, - 309.27, - 583.48 - ], - "text": "1 −1\nz", - "type": "text" - }, - { - "block_id": "p528-b41", - "global_id": 15216, - "bbox": [ - 317.74, - 559.14, - 392.03, - 576.4 - ], - "text": "X[z] = (z −1)X[z]", - "type": "text" - }, - { - "block_id": "p528-b42", - "global_id": 15217, - "bbox": [ - 367.99, - 573.52, - 371.87, - 583.48 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p528-b43", - "global_id": 15218, - "bbox": [ - 101.85, - 594.58, - 116.22, - 604.54 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p528-b44", - "global_id": 15219, - "bbox": [ - 217.58, - 606.44, - 261.77, - 616.82 - ], - "text": "(z −1)X[z]", - "type": "text" - }, - { - "block_id": "p528-b45", - "global_id": 15220, - "bbox": [ - 237.73, - 613.42, - 272.78, - 630.77 - ], - "text": "z\n=", - "type": "text" - }, - { - "block_id": "p528-b46", - "global_id": 15221, - "bbox": [ - 278.53, - 603.25, - 292.63, - 613.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p528-b47", - "global_id": 15222, - "bbox": [ - 274.83, - 627.27, - 296.31, - 634.46 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p528-b48", - "global_id": 15223, - "bbox": [ - 296.32, - 611.7, - 375.33, - 623.8 - ], - "text": "{x[n] −x[n −1]}z−n", - "type": "text" - } - ] - }, - { - "page_num": 529, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p529-b0", - "global_id": 15224, - "bbox": [ - 313.93, - 62.57, - 516.13, - 71.98 - ], - "text": "5.2\nSome Properties of the z-Transform\n509", - "type": "text" - }, - { - "block_id": "p529-b1", - "global_id": 15225, - "bbox": [ - 127.59, - 85.82, - 141.97, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p529-b2", - "global_id": 15226, - "bbox": [ - 155.57, - 123.53, - 169.44, - 139.3 - ], - "text": "lim\nz→1", - "type": "text" - }, - { - "block_id": "p529-b3", - "global_id": 15227, - "bbox": [ - 172.85, - 116.14, - 217.05, - 126.51 - ], - "text": "(z −1)X[z]", - "type": "text" - }, - { - "block_id": "p529-b4", - "global_id": 15228, - "bbox": [ - 193.01, - 123.12, - 243.68, - 140.47 - ], - "text": "z\n= lim", - "type": "text" - }, - { - "block_id": "p529-b5", - "global_id": 15229, - "bbox": [ - 230.1, - 123.12, - 313.62, - 139.3 - ], - "text": "z→1(z −1)X[z] = lim", - "type": "text" - }, - { - "block_id": "p529-b6", - "global_id": 15230, - "bbox": [ - 300.03, - 123.53, - 331.58, - 140.02 - ], - "text": "z→1 lim", - "type": "text" - }, - { - "block_id": "p529-b7", - "global_id": 15231, - "bbox": [ - 315.0, - 131.94, - 334.86, - 139.13 - ], - "text": "N→∞", - "type": "text" - }, - { - "block_id": "p529-b8", - "global_id": 15232, - "bbox": [ - 339.66, - 113.17, - 353.76, - 123.62 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p529-b9", - "global_id": 15233, - "bbox": [ - 335.96, - 136.97, - 357.44, - 144.16 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p529-b10", - "global_id": 15234, - "bbox": [ - 357.45, - 119.01, - 465.43, - 133.5 - ], - "text": "{x[n] −x[n −1]}z−n = lim", - "type": "text" - }, - { - "block_id": "p529-b11", - "global_id": 15235, - "bbox": [ - 448.85, - 123.12, - 488.13, - 139.13 - ], - "text": "N→∞x[N]", - "type": "text" - }, - { - "block_id": "p529-b12", - "global_id": 15236, - "bbox": [ - 145.52, - 162.66, - 388.2, - 172.73 - ], - "text": "All these properties of the z-transform are listed in Table 5.2.", - "type": "text" - }, - { - "block_id": "p529-b13", - "global_id": 15237, - "bbox": [ - 127.59, - 196.15, - 263.42, - 205.47 - ], - "text": "TABLE 5.2\nz-Transform Properties", - "type": "text" - }, - { - "block_id": "p529-b14", - "global_id": 15238, - "bbox": [ - 127.59, - 216.19, - 382.14, - 225.45 - ], - "text": "Operation\nx[n]\nX[z]", - "type": "text" - }, - { - "block_id": "p529-b15", - "global_id": 15239, - "bbox": [ - 127.59, - 234.52, - 413.31, - 244.6 - ], - "text": "Addition\nx1[n] + x2[n]\nX1[z] + X2[z]", - "type": "text" - }, - { - "block_id": "p529-b16", - "global_id": 15240, - "bbox": [ - 127.59, - 254.44, - 386.0, - 263.78 - ], - "text": "Scalar multiplication\nax[n]\naX[z]", - "type": "text" - }, - { - "block_id": "p529-b17", - "global_id": 15241, - "bbox": [ - 127.59, - 273.18, - 392.57, - 294.7 - ], - "text": "Right shifting\nx[n −m]u[n −m]\n1\nzm X[z]", - "type": "text" - }, - { - "block_id": "p529-b18", - "global_id": 15242, - "bbox": [ - 253.89, - 300.57, - 410.11, - 322.1 - ], - "text": "x[n −m]u[n]\n1\nzm X[z] + 1", - "type": "text" - }, - { - "block_id": "p529-b19", - "global_id": 15243, - "bbox": [ - 403.53, - 312.46, - 411.7, - 322.1 - ], - "text": "zm", - "type": "text" - }, - { - "block_id": "p529-b20", - "global_id": 15244, - "bbox": [ - 414.39, - 297.38, - 427.07, - 306.94 - ], - "text": "m\n\"", - "type": "text" - }, - { - "block_id": "p529-b21", - "global_id": 15245, - "bbox": [ - 414.97, - 319.42, - 426.5, - 326.16 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p529-b22", - "global_id": 15246, - "bbox": [ - 428.08, - 304.98, - 456.26, - 315.74 - ], - "text": "x[−n]zn", - "type": "text" - }, - { - "block_id": "p529-b23", - "global_id": 15247, - "bbox": [ - 253.89, - 325.22, - 419.63, - 346.75 - ], - "text": "x[n −1]u[n]\n1\nz X[z] + x[−1]", - "type": "text" - }, - { - "block_id": "p529-b24", - "global_id": 15248, - "bbox": [ - 253.89, - 349.87, - 406.58, - 371.4 - ], - "text": "x[n −2]u[n]\n1\nz2 X[z] + 1", - "type": "text" - }, - { - "block_id": "p529-b25", - "global_id": 15249, - "bbox": [ - 402.6, - 355.78, - 460.47, - 371.4 - ], - "text": "z x[−1] + x[−2]", - "type": "text" - }, - { - "block_id": "p529-b26", - "global_id": 15250, - "bbox": [ - 253.89, - 374.51, - 407.95, - 396.04 - ], - "text": "x[n −3]u[n]\n1\nz3 X[z] + 1", - "type": "text" - }, - { - "block_id": "p529-b27", - "global_id": 15251, - "bbox": [ - 402.1, - 374.51, - 447.44, - 396.04 - ], - "text": "z2 x[−1] + 1", - "type": "text" - }, - { - "block_id": "p529-b28", - "global_id": 15252, - "bbox": [ - 443.45, - 380.43, - 501.33, - 396.04 - ], - "text": "z x[−2] + x[−3]", - "type": "text" - }, - { - "block_id": "p529-b29", - "global_id": 15253, - "bbox": [ - 127.59, - 408.34, - 408.13, - 418.73 - ], - "text": "Left shifting\nx[n + m]u[n]\nzmX[z] −zm", - "type": "text" - }, - { - "block_id": "p529-b30", - "global_id": 15254, - "bbox": [ - 409.61, - 400.08, - 422.57, - 409.84 - ], - "text": "m−1\n\"", - "type": "text" - }, - { - "block_id": "p529-b31", - "global_id": 15255, - "bbox": [ - 410.33, - 422.33, - 421.85, - 429.07 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p529-b32", - "global_id": 15256, - "bbox": [ - 423.57, - 407.68, - 449.81, - 418.64 - ], - "text": "x[n]z−n", - "type": "text" - }, - { - "block_id": "p529-b33", - "global_id": 15257, - "bbox": [ - 253.89, - 429.31, - 412.75, - 438.65 - ], - "text": "x[n + 1]u[n]\nzX[z] −zx[0]", - "type": "text" - }, - { - "block_id": "p529-b34", - "global_id": 15258, - "bbox": [ - 253.89, - 448.25, - 447.94, - 458.58 - ], - "text": "x[n + 2]u[n]\nz2X[z] −z2x[0] −zx[1]", - "type": "text" - }, - { - "block_id": "p529-b35", - "global_id": 15259, - "bbox": [ - 253.89, - 468.18, - 479.4, - 478.51 - ], - "text": "x[n + 3]u[n]\nz3X[z] −z3x[0] −z2x[1] −zx[2]", - "type": "text" - }, - { - "block_id": "p529-b36", - "global_id": 15260, - "bbox": [ - 127.59, - 493.37, - 371.65, - 503.76 - ], - "text": "Multiplication by γ n\nγ nx[n]u[n]\nX", - "type": "text" - }, - { - "block_id": "p529-b37", - "global_id": 15261, - "bbox": [ - 373.05, - 481.84, - 383.84, - 497.39 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p529-b38", - "global_id": 15262, - "bbox": [ - 379.12, - 500.79, - 383.63, - 509.76 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p529-b39", - "global_id": 15263, - "bbox": [ - 386.25, - 481.84, - 391.14, - 490.81 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p529-b40", - "global_id": 15264, - "bbox": [ - 127.59, - 512.94, - 383.96, - 528.28 - ], - "text": "Multiplication by n\nnx[n]u[n]\n−z d", - "type": "text" - }, - { - "block_id": "p529-b41", - "global_id": 15265, - "bbox": [ - 377.86, - 518.94, - 402.36, - 534.56 - ], - "text": "dzX[z]", - "type": "text" - }, - { - "block_id": "p529-b42", - "global_id": 15266, - "bbox": [ - 127.59, - 538.86, - 390.21, - 548.2 - ], - "text": "Time reversal\nx[−n]\nX[1/z]", - "type": "text" - }, - { - "block_id": "p529-b43", - "global_id": 15267, - "bbox": [ - 127.59, - 558.79, - 403.53, - 568.87 - ], - "text": "Time convolution\nx1[n] ∗x2[n]\nX1[z]X2[z]", - "type": "text" - }, - { - "block_id": "p529-b44", - "global_id": 15268, - "bbox": [ - 127.59, - 578.71, - 398.76, - 592.77 - ], - "text": "Initial value\nx[0]\nlim\nz→∞X[z]", - "type": "text" - }, - { - "block_id": "p529-b45", - "global_id": 15269, - "bbox": [ - 127.59, - 598.64, - 510.17, - 613.38 - ], - "text": "Final value\nlim\nN→∞x[N]\nlim\nz→1(z −1)X[z]\nPoles of (z −1)X[z]", - "type": "text" - }, - { - "block_id": "p529-b46", - "global_id": 15270, - "bbox": [ - 437.18, - 614.23, - 510.15, - 623.2 - ], - "text": "inside the unit circle", - "type": "text" - } - ] - }, - { - "page_num": 530, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p530-b0", - "global_id": 15271, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "510\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p530-b1", - "global_id": 15272, - "bbox": [ - 102.2, - 91.2, - 450.75, - 122.09 - ], - "text": "5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE\nEQUATIONS", - "type": "text" - }, - { - "block_id": "p530-b2", - "global_id": 15273, - "bbox": [ - 101.84, - 128.09, - 490.41, - 185.87 - ], - "text": "The time-shifting (left-shift or right-shift) property has set the stage for solving linear difference\nequations with constant coefficients. As in the case of the Laplace transform with differential\nequations, the z-transform converts difference equations into algebraic equations that are readily\nsolved to find the solution in the z domain. Taking the inverse z-transform of the z-domain solution\nyields the desired time-domain solution. The following examples demonstrate the procedure.", - "type": "text" - }, - { - "block_id": "p530-b3", - "global_id": 15274, - "bbox": [ - 76.77, - 206.89, - 466.98, - 218.85 - ], - "text": "EXAMPLE 5.5\nz-Transform Solution of a Linear Difference Equation", - "type": "text" - }, - { - "block_id": "p530-b4", - "global_id": 15275, - "bbox": [ - 103.16, - 235.08, - 125.7, - 245.04 - ], - "text": "Solve", - "type": "text" - }, - { - "block_id": "p530-b5", - "global_id": 15276, - "bbox": [ - 193.98, - 246.61, - 386.16, - 256.99 - ], - "text": "y[n + 2] −5y[n + 1] + 6y[n] = 3x[n + 1] + 5x[n]", - "type": "text" - }, - { - "block_id": "p530-b6", - "global_id": 15277, - "bbox": [ - 103.17, - 263.87, - 459.74, - 275.48 - ], - "text": "if the initial conditions are y[−1] = 11/6, y[−2] = 37/36, and the input x[n] = (2)−nu[n].", - "type": "text" - }, - { - "block_id": "p530-b7", - "global_id": 15278, - "bbox": [ - 103.16, - 298.39, - 477.04, - 380.08 - ], - "text": "As we shall see, difference equations can be solved by using the right-shift or the left-\nshift property. Because the difference equation here is in advance form, the use of the left-shift\nproperty in Eq. (5.16) may seem appropriate for its solution. Unfortunately, this left-shift\nproperty requires a knowledge of auxiliary conditions y[0], y[1], . . . , y[N −1] rather than of\nthe initial conditions y[−1], y[−2], . . . , y[−n], which are generally given. This difficulty can\nbe overcome by expressing the difference equation in delay form (obtained by replacing n with\nn −2) and then using the right-shift property.† The resulting delay-form difference equation is", - "type": "text" - }, - { - "block_id": "p530-b8", - "global_id": 15279, - "bbox": [ - 186.05, - 390.75, - 477.01, - 401.13 - ], - "text": "y[n] −5y[n −1] + 6y[n −2] = 3x[n −1] + 5x[n −2]\n(5.21)", - "type": "text" - }, - { - "block_id": "p530-b9", - "global_id": 15280, - "bbox": [ - 103.17, - 412.1, - 477.03, - 481.94 - ], - "text": "We now use the right-shift property to take the z-transform of this equation. But before\nproceeding, we must be clear about the meaning of a term like y[n −1] here. Does it mean\ny[n −1]u[n −1] or y[n −1]u[n]? In any equation, we must have some time reference n = 0,\nand every term is referenced from this instant. Hence, y[n−k] means y[n−k]u[n]. Remember\nalso that although we are considering the situation for n ≥0, y[n] is present even before n = 0\n(in the form of initial conditions). Now", - "type": "text" - }, - { - "block_id": "p530-b10", - "global_id": 15281, - "bbox": [ - 175.5, - 492.6, - 248.17, - 502.88 - ], - "text": "y[n]u[n] ⇐⇒Y[z]", - "type": "text" - }, - { - "block_id": "p530-b11", - "global_id": 15282, - "bbox": [ - 159.65, - 507.37, - 237.53, - 524.32 - ], - "text": "y[n −1]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p530-b12", - "global_id": 15283, - "bbox": [ - 233.1, - 507.37, - 308.26, - 531.29 - ], - "text": "z Y[z] + y[−1] = 1", - "type": "text" - }, - { - "block_id": "p530-b13", - "global_id": 15284, - "bbox": [ - 303.83, - 507.37, - 348.3, - 531.29 - ], - "text": "z Y[z] + 11", - "type": "text" - }, - { - "block_id": "p530-b14", - "global_id": 15285, - "bbox": [ - 340.84, - 521.43, - 345.82, - 531.39 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p530-b15", - "global_id": 15286, - "bbox": [ - 159.65, - 533.05, - 238.97, - 549.99 - ], - "text": "y[n −2]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p530-b16", - "global_id": 15287, - "bbox": [ - 232.55, - 533.05, - 275.47, - 556.97 - ], - "text": "z2 Y[z] + 1", - "type": "text" - }, - { - "block_id": "p530-b17", - "global_id": 15288, - "bbox": [ - 271.03, - 533.05, - 354.64, - 556.97 - ], - "text": "z y[−1] + y[−2] = 1", - "type": "text" - }, - { - "block_id": "p530-b18", - "global_id": 15289, - "bbox": [ - 348.22, - 533.05, - 396.11, - 556.97 - ], - "text": "z2 Y[z] + 11", - "type": "text" - }, - { - "block_id": "p530-b19", - "global_id": 15290, - "bbox": [ - 386.7, - 533.05, - 419.34, - 557.07 - ], - "text": "6z + 37\n36", - "type": "text" - }, - { - "block_id": "p530-b20", - "global_id": 15291, - "bbox": [ - 103.17, - 564.39, - 233.79, - 574.77 - ], - "text": "Noting that for causal input x[n],", - "type": "text" - }, - { - "block_id": "p530-b21", - "global_id": 15292, - "bbox": [ - 221.57, - 585.44, - 358.61, - 595.81 - ], - "text": "x[−1] = x[−2] = · · · = x[−n] = 0", - "type": "text" - }, - { - "block_id": "p530-b22", - "global_id": 15293, - "bbox": [ - 101.84, - 626.67, - 490.38, - 649.85 - ], - "text": "† Another approach is to find y[0], y[1], y[2], . . . , y[N −1] from y[−1], y[−2], . . . , y[−n] iteratively, as in\nSec. 3.5-1, and then apply the left-shift property to the advance-form difference equation.", - "type": "text" - } - ] - }, - { - "page_num": 531, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p531-b0", - "global_id": 15294, - "bbox": [ - 246.93, - 62.57, - 516.11, - 71.98 - ], - "text": "5.3\nz-Transform Solution of Linear Difference Equations\n511", - "type": "text" - }, - { - "block_id": "p531-b1", - "global_id": 15295, - "bbox": [ - 128.9, - 86.24, - 169.33, - 96.21 - ], - "text": "We obtain", - "type": "text" - }, - { - "block_id": "p531-b2", - "global_id": 15296, - "bbox": [ - 202.93, - 91.11, - 427.54, - 115.23 - ], - "text": "x[n] = (2)−nu[n] = (2−1)nu[n] = (0.5)nu[n] ⇐⇒\nz\nz −0.5", - "type": "text" - }, - { - "block_id": "p531-b3", - "global_id": 15297, - "bbox": [ - 166.11, - 134.82, - 244.03, - 151.76 - ], - "text": "x[n −1]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p531-b4", - "global_id": 15298, - "bbox": [ - 239.6, - 134.82, - 315.01, - 158.74 - ], - "text": "z X[z] + x[−1] = 1", - "type": "text" - }, - { - "block_id": "p531-b5", - "global_id": 15299, - "bbox": [ - 310.58, - 148.77, - 314.45, - 158.74 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p531-b6", - "global_id": 15300, - "bbox": [ - 317.4, - 134.71, - 401.91, - 158.84 - ], - "text": "z\nz −0.5 + 0 =\n1\nz −0.5", - "type": "text" - }, - { - "block_id": "p531-b7", - "global_id": 15301, - "bbox": [ - 166.11, - 160.49, - 245.47, - 177.44 - ], - "text": "x[n −2]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p531-b8", - "global_id": 15302, - "bbox": [ - 239.05, - 160.49, - 282.19, - 184.41 - ], - "text": "z2 X[z] + 1", - "type": "text" - }, - { - "block_id": "p531-b9", - "global_id": 15303, - "bbox": [ - 277.76, - 160.49, - 361.42, - 184.41 - ], - "text": "z x[−1] + x[−2] = 1", - "type": "text" - }, - { - "block_id": "p531-b10", - "global_id": 15304, - "bbox": [ - 354.99, - 160.49, - 464.35, - 184.51 - ], - "text": "z2 X[z] + 0 + 0 =\n1\nz(z −0.5)", - "type": "text" - }, - { - "block_id": "p531-b11", - "global_id": 15305, - "bbox": [ - 128.91, - 198.01, - 171.5, - 207.98 - ], - "text": "In general,", - "type": "text" - }, - { - "block_id": "p531-b12", - "global_id": 15306, - "bbox": [ - 267.06, - 207.95, - 345.23, - 224.8 - ], - "text": "x[n −r]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p531-b13", - "global_id": 15307, - "bbox": [ - 339.13, - 214.52, - 364.61, - 231.87 - ], - "text": "zr X[z]", - "type": "text" - }, - { - "block_id": "p531-b14", - "global_id": 15308, - "bbox": [ - 128.91, - 237.5, - 466.91, - 247.57 - ], - "text": "Taking the z-transform of Eq. (5.21) and substituting the foregoing results, we obtain", - "type": "text" - }, - { - "block_id": "p531-b15", - "global_id": 15309, - "bbox": [ - 172.53, - 268.74, - 205.19, - 279.11 - ], - "text": "Y[z] −5", - "type": "text" - }, - { - "block_id": "p531-b16", - "global_id": 15310, - "bbox": [ - 206.3, - 254.75, - 217.91, - 272.13 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p531-b17", - "global_id": 15311, - "bbox": [ - 213.48, - 262.17, - 257.95, - 286.09 - ], - "text": "z Y[z] + 11", - "type": "text" - }, - { - "block_id": "p531-b18", - "global_id": 15312, - "bbox": [ - 250.49, - 276.22, - 255.47, - 286.19 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p531-b19", - "global_id": 15313, - "bbox": [ - 259.15, - 254.75, - 264.58, - 264.71 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p531-b20", - "global_id": 15314, - "bbox": [ - 266.13, - 268.74, - 280.42, - 279.11 - ], - "text": "+ 6", - "type": "text" - }, - { - "block_id": "p531-b21", - "global_id": 15315, - "bbox": [ - 281.53, - 254.75, - 294.58, - 272.13 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p531-b22", - "global_id": 15316, - "bbox": [ - 288.16, - 262.17, - 336.06, - 286.09 - ], - "text": "z2 Y[z] + 11", - "type": "text" - }, - { - "block_id": "p531-b23", - "global_id": 15317, - "bbox": [ - 326.65, - 262.17, - 359.28, - 286.19 - ], - "text": "6z + 37\n36", - "type": "text" - }, - { - "block_id": "p531-b24", - "global_id": 15318, - "bbox": [ - 360.48, - 254.75, - 365.91, - 264.71 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p531-b25", - "global_id": 15319, - "bbox": [ - 367.96, - 262.17, - 457.93, - 286.19 - ], - "text": "=\n3\nz −0.5 +\n5\nz(z −0.5)", - "type": "text" - }, - { - "block_id": "p531-b26", - "global_id": 15320, - "bbox": [ - 128.91, - 298.13, - 211.82, - 309.24 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p531-b27", - "global_id": 15321, - "bbox": [ - 211.82, - 306.69, - 253.53, - 330.61 - ], - "text": "1 −5\nz + 6", - "type": "text" - }, - { - "block_id": "p531-b28", - "global_id": 15322, - "bbox": [ - 247.11, - 320.13, - 254.48, - 330.61 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p531-b30", - "global_id": 15323, - "bbox": [ - 264.0, - 313.26, - 290.14, - 323.53 - ], - "text": "Y[z] −", - "type": "text" - }, - { - "block_id": "p531-b32", - "global_id": 15324, - "bbox": [ - 298.41, - 306.69, - 325.41, - 330.61 - ], - "text": "3 −11\nz", - "type": "text" - }, - { - "block_id": "p531-b34", - "global_id": 15325, - "bbox": [ - 335.39, - 306.69, - 502.75, - 330.71 - ], - "text": "=\n3\nz −0.5 +\n5\nz(z −0.5)\n(5.22)", - "type": "text" - }, - { - "block_id": "p531-b35", - "global_id": 15326, - "bbox": [ - 128.91, - 337.89, - 216.62, - 347.85 - ], - "text": "from which we obtain", - "type": "text" - }, - { - "block_id": "p531-b36", - "global_id": 15327, - "bbox": [ - 235.03, - 357.27, - 395.41, - 378.25 - ], - "text": "(z2 −5z + 6)Y[z] = z(3z2 −9.5z + 10.5)", - "type": "text" - }, - { - "block_id": "p531-b37", - "global_id": 15328, - "bbox": [ - 338.3, - 374.94, - 372.92, - 385.32 - ], - "text": "(z −0.5)", - "type": "text" - }, - { - "block_id": "p531-b38", - "global_id": 15329, - "bbox": [ - 128.9, - 398.82, - 155.19, - 408.78 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p531-b39", - "global_id": 15330, - "bbox": [ - 257.54, - 404.25, - 369.98, - 425.13 - ], - "text": "Y[z] = z(3z2 −9.5z + 10.5)", - "type": "text" - }, - { - "block_id": "p531-b40", - "global_id": 15331, - "bbox": [ - 287.42, - 419.04, - 372.91, - 432.3 - ], - "text": "(z −0.5)(z2 −5z + 6)", - "type": "text" - }, - { - "block_id": "p531-b41", - "global_id": 15332, - "bbox": [ - 128.9, - 438.83, - 143.28, - 448.79 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p531-b42", - "global_id": 15333, - "bbox": [ - 194.85, - 447.87, - 211.67, - 458.15 - ], - "text": "Y[z]", - "type": "text" - }, - { - "block_id": "p531-b43", - "global_id": 15334, - "bbox": [ - 201.32, - 444.25, - 361.15, - 472.31 - ], - "text": "z\n=\n3z2 −9.5z + 10.5\n(z −0.5)(z −2)(z −3) = (26/15)", - "type": "text" - }, - { - "block_id": "p531-b44", - "global_id": 15335, - "bbox": [ - 331.55, - 447.87, - 396.49, - 472.31 - ], - "text": "z −0.5 −(7/3)", - "type": "text" - }, - { - "block_id": "p531-b45", - "global_id": 15336, - "bbox": [ - 375.6, - 447.87, - 436.81, - 472.31 - ], - "text": "z −2 + (18/5)", - "type": "text" - }, - { - "block_id": "p531-b46", - "global_id": 15337, - "bbox": [ - 413.43, - 461.93, - 433.15, - 472.31 - ], - "text": "z −3", - "type": "text" - }, - { - "block_id": "p531-b47", - "global_id": 15338, - "bbox": [ - 128.91, - 478.83, - 170.65, - 488.79 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p531-b48", - "global_id": 15339, - "bbox": [ - 215.84, - 488.73, - 255.67, - 505.68 - ], - "text": "Y[z] = 26", - "type": "text" - }, - { - "block_id": "p531-b49", - "global_id": 15340, - "bbox": [ - 245.72, - 502.79, - 255.67, - 512.75 - ], - "text": "15", - "type": "text" - }, - { - "block_id": "p531-b50", - "global_id": 15341, - "bbox": [ - 257.98, - 481.32, - 293.1, - 512.75 - ], - "text": "z\nz −0.5", - "type": "text" - }, - { - "block_id": "p531-b52", - "global_id": 15342, - "bbox": [ - 302.57, - 488.73, - 318.06, - 505.26 - ], - "text": "−7", - "type": "text" - }, - { - "block_id": "p531-b53", - "global_id": 15343, - "bbox": [ - 313.08, - 502.79, - 318.06, - 512.75 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p531-b54", - "global_id": 15344, - "bbox": [ - 320.37, - 481.32, - 348.01, - 512.75 - ], - "text": "z\nz −2", - "type": "text" - }, - { - "block_id": "p531-b56", - "global_id": 15345, - "bbox": [ - 357.49, - 488.73, - 377.96, - 505.68 - ], - "text": "+ 18", - "type": "text" - }, - { - "block_id": "p531-b57", - "global_id": 15346, - "bbox": [ - 370.49, - 502.79, - 375.47, - 512.75 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p531-b58", - "global_id": 15347, - "bbox": [ - 380.27, - 481.32, - 407.91, - 512.75 - ], - "text": "z\nz −3", - "type": "text" - }, - { - "block_id": "p531-b60", - "global_id": 15348, - "bbox": [ - 128.9, - 519.93, - 143.28, - 529.9 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p531-b61", - "global_id": 15349, - "bbox": [ - 238.15, - 531.47, - 264.01, - 541.75 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p531-b62", - "global_id": 15350, - "bbox": [ - 266.06, - 523.46, - 278.15, - 537.1 - ], - "text": "26", - "type": "text" - }, - { - "block_id": "p531-b63", - "global_id": 15351, - "bbox": [ - 271.18, - 523.46, - 502.75, - 544.97 - ], - "text": "15(0.5)n −7\n3(2)n + 18\n5 (3)n\nu[n]\n(5.23)", - "type": "text" - }, - { - "block_id": "p531-b64", - "global_id": 15352, - "bbox": [ - 127.6, - 588.96, - 516.14, - 634.79 - ], - "text": "This example demonstrates the ease with which linear difference equations with constant\ncoefficients can be solved by the z-transform. This method is general: it can be used to solve a\nsingle difference equation or a set of simultaneous difference equations of any order as long as the\nequations are linear with constant coefficients.", - "type": "text" - } - ] - }, - { - "page_num": 532, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p532-b0", - "global_id": 15353, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "512\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p532-b1", - "global_id": 15354, - "bbox": [ - 101.84, - 85.87, - 490.42, - 143.73 - ], - "text": "Comment.\nSometimes, instead of initial conditions y[−1], y[−2], . . . , y[−n], auxiliary conditions y[0],\ny[1], . . . , y[N −1] are given to solve a difference equation. In this case, the equation can\nbe solved by expressing it in the advance form and then using the left-shift property (see\nDrill 5.11).", - "type": "text" - }, - { - "block_id": "p532-b2", - "global_id": 15355, - "bbox": [ - 107.82, - 187.55, - 480.83, - 199.5 - ], - "text": "DRILL 5.10\nz-Transform Solution of a Linear Difference Equation", - "type": "text" - }, - { - "block_id": "p532-b3", - "global_id": 15356, - "bbox": [ - 107.82, - 208.21, - 484.41, - 230.55 - ], - "text": "Solve the following equation if the initial conditions y[−1] = 2, y[−2] = 0, and the input\nx[n] = u[n]:", - "type": "text" - }, - { - "block_id": "p532-b4", - "global_id": 15357, - "bbox": [ - 201.6, - 230.78, - 249.04, - 242.5 - ], - "text": "y[n + 2] −5", - "type": "text" - }, - { - "block_id": "p532-b5", - "global_id": 15358, - "bbox": [ - 245.56, - 230.78, - 390.6, - 245.32 - ], - "text": "6y[n + 1] + 1\n6y[n] = 5x[n + 1] −x[n]", - "type": "text" - }, - { - "block_id": "p532-b6", - "global_id": 15359, - "bbox": [ - 108.09, - 263.0, - 162.68, - 273.96 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p532-b7", - "global_id": 15360, - "bbox": [ - 107.82, - 280.94, - 133.68, - 291.22 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p532-b9", - "global_id": 15361, - "bbox": [ - 139.66, - 272.93, - 179.15, - 291.32 - ], - "text": "12 −15\n 1", - "type": "text" - }, - { - "block_id": "p532-b10", - "global_id": 15362, - "bbox": [ - 175.66, - 272.93, - 207.38, - 294.13 - ], - "text": "2\nn + 14", - "type": "text" - }, - { - "block_id": "p532-b11", - "global_id": 15363, - "bbox": [ - 202.15, - 272.93, - 217.27, - 294.13 - ], - "text": "3\n 1", - "type": "text" - }, - { - "block_id": "p532-b12", - "global_id": 15364, - "bbox": [ - 213.79, - 272.93, - 230.39, - 294.13 - ], - "text": "3\nn", - "type": "text" - }, - { - "block_id": "p532-b13", - "global_id": 15365, - "bbox": [ - 231.5, - 280.94, - 248.09, - 291.22 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p532-b14", - "global_id": 15366, - "bbox": [ - 107.82, - 353.39, - 462.43, - 380.15 - ], - "text": "DRILL 5.11\nDifference Equation Solution Using y[0], y[1], . . . ,\ny[N −1]", - "type": "text" - }, - { - "block_id": "p532-b15", - "global_id": 15367, - "bbox": [ - 107.82, - 388.86, - 484.4, - 411.19 - ], - "text": "Solve the following equation if the auxiliary conditions are y[0] = 1, y[1] = 2, and the input\nx[n] = u[n]:", - "type": "text" - }, - { - "block_id": "p532-b16", - "global_id": 15368, - "bbox": [ - 194.57, - 412.76, - 397.61, - 423.14 - ], - "text": "y[n] + 3y[n −1] + 2y[n −2] = x[n −1] + 3x[n −2]", - "type": "text" - }, - { - "block_id": "p532-b17", - "global_id": 15369, - "bbox": [ - 108.09, - 443.64, - 162.68, - 454.6 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p532-b18", - "global_id": 15370, - "bbox": [ - 107.82, - 461.58, - 133.68, - 471.85 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p532-b19", - "global_id": 15371, - "bbox": [ - 135.73, - 453.57, - 144.34, - 467.21 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p532-b20", - "global_id": 15372, - "bbox": [ - 140.85, - 453.57, - 248.09, - 475.08 - ], - "text": "3 + 2(−1)n −5\n3(−2)n\nu[n]", - "type": "text" - }, - { - "block_id": "p532-b21", - "global_id": 15373, - "bbox": [ - 102.14, - 510.57, - 348.17, - 522.69 - ], - "text": "ZERO-INPUT AND ZERO-STATE COMPONENTS", - "type": "text" - }, - { - "block_id": "p532-b22", - "global_id": 15374, - "bbox": [ - 101.84, - 526.73, - 490.41, - 572.55 - ], - "text": "In Ex. 5.5 we found the total solution of the difference equation. It is relatively easy to separate the\nsolution into zero-input and zero-state components. All we have to do is to separate the response\ninto terms arising from the input and terms arising from initial conditions (IC). We can separate\nthe response in Eq. (5.22) as follows:", - "type": "text" - }, - { - "block_id": "p532-b24", - "global_id": 15375, - "bbox": [ - 192.1, - 595.82, - 233.82, - 619.74 - ], - "text": "1 −5\nz + 6", - "type": "text" - }, - { - "block_id": "p532-b25", - "global_id": 15376, - "bbox": [ - 227.39, - 609.26, - 234.76, - 619.74 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p532-b27", - "global_id": 15377, - "bbox": [ - 244.29, - 602.39, - 270.43, - 612.66 - ], - "text": "Y[z] −", - "type": "text" - }, - { - "block_id": "p532-b29", - "global_id": 15378, - "bbox": [ - 278.69, - 595.82, - 305.7, - 619.74 - ], - "text": "3 −11\nz", - "type": "text" - }, - { - "block_id": "p532-b32", - "global_id": 15379, - "bbox": [ - 282.42, - 628.26, - 303.17, - 634.24 - ], - "text": "IC terms", - "type": "text" - }, - { - "block_id": "p532-b33", - "global_id": 15380, - "bbox": [ - 315.67, - 595.82, - 406.86, - 633.57 - ], - "text": "=\n3\nz −0.5 +\n5\nz(z −0.5)\n\n\n\ninput terms", - "type": "text" - } - ] - }, - { - "page_num": 533, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p533-b0", - "global_id": 15381, - "bbox": [ - 246.93, - 62.57, - 516.11, - 71.98 - ], - "text": "5.3\nz-Transform Solution of Linear Difference Equations\n513", - "type": "text" - }, - { - "block_id": "p533-b1", - "global_id": 15382, - "bbox": [ - 127.59, - 85.82, - 238.52, - 99.13 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p533-b2", - "global_id": 15383, - "bbox": [ - 238.53, - 96.58, - 280.24, - 120.5 - ], - "text": "1 −5\nz + 6", - "type": "text" - }, - { - "block_id": "p533-b3", - "global_id": 15384, - "bbox": [ - 273.82, - 110.02, - 281.18, - 120.5 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p533-b5", - "global_id": 15385, - "bbox": [ - 290.7, - 103.15, - 317.34, - 113.43 - ], - "text": "Y[z] =", - "type": "text" - }, - { - "block_id": "p533-b7", - "global_id": 15386, - "bbox": [ - 326.11, - 96.58, - 353.12, - 120.5 - ], - "text": "3 −11\nz", - "type": "text" - }, - { - "block_id": "p533-b9", - "global_id": 15387, - "bbox": [ - 319.38, - 116.62, - 361.04, - 135.0 - ], - "text": "IC terms", - "type": "text" - }, - { - "block_id": "p533-b10", - "global_id": 15388, - "bbox": [ - 362.14, - 96.17, - 407.53, - 113.11 - ], - "text": "+ (3z + 5)", - "type": "text" - }, - { - "block_id": "p533-b11", - "global_id": 15389, - "bbox": [ - 371.02, - 110.22, - 411.91, - 125.92 - ], - "text": "z(z −0.5)", - "type": "text" - }, - { - "block_id": "p533-b12", - "global_id": 15390, - "bbox": [ - 377.94, - 128.37, - 405.01, - 134.34 - ], - "text": "input terms", - "type": "text" - }, - { - "block_id": "p533-b13", - "global_id": 15391, - "bbox": [ - 127.59, - 141.04, - 266.91, - 154.61 - ], - "text": "Multiplying both sides by z2 yields", - "type": "text" - }, - { - "block_id": "p533-b14", - "global_id": 15392, - "bbox": [ - 237.39, - 167.64, - 357.95, - 195.88 - ], - "text": "(z2 −5z + 6)Y[z] = z(3z −11)\n\n\n\nIC terms", - "type": "text" - }, - { - "block_id": "p533-b15", - "global_id": 15393, - "bbox": [ - 359.05, - 164.77, - 405.13, - 182.03 - ], - "text": "+ z(3z + 5)", - "type": "text" - }, - { - "block_id": "p533-b16", - "global_id": 15394, - "bbox": [ - 367.92, - 178.83, - 406.33, - 202.06 - ], - "text": "z −0.5\n\n\n\ninput terms", - "type": "text" - }, - { - "block_id": "p533-b17", - "global_id": 15395, - "bbox": [ - 127.59, - 213.79, - 141.97, - 223.75 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p533-b18", - "global_id": 15396, - "bbox": [ - 235.09, - 222.17, - 307.75, - 239.43 - ], - "text": "Y[z] = z(3z −11)", - "type": "text" - }, - { - "block_id": "p533-b19", - "global_id": 15397, - "bbox": [ - 263.77, - 233.35, - 310.74, - 259.46 - ], - "text": "z2 −5z + 6\n\n\n\nzero-input response", - "type": "text" - }, - { - "block_id": "p533-b20", - "global_id": 15398, - "bbox": [ - 311.86, - 222.18, - 408.63, - 260.34 - ], - "text": "+\nz(3z + 5)\n(z −0.5)(z2 −5z + 6)\n\n\n\nzero-state response", - "type": "text" - }, - { - "block_id": "p533-b21", - "global_id": 15399, - "bbox": [ - 127.59, - 269.08, - 460.4, - 279.05 - ], - "text": "We expand both terms on the right-hand side into modified partial fractions to yield", - "type": "text" - }, - { - "block_id": "p533-b22", - "global_id": 15400, - "bbox": [ - 156.43, - 296.8, - 183.06, - 307.08 - ], - "text": "Y[z] =", - "type": "text" - }, - { - "block_id": "p533-b24", - "global_id": 15401, - "bbox": [ - 190.54, - 282.82, - 224.27, - 314.25 - ], - "text": "5\n\nz\nz −2", - "type": "text" - }, - { - "block_id": "p533-b26", - "global_id": 15402, - "bbox": [ - 233.75, - 296.8, - 248.04, - 307.18 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p533-b27", - "global_id": 15403, - "bbox": [ - 249.15, - 282.82, - 276.8, - 314.25 - ], - "text": "z\nz −3", - "type": "text" - }, - { - "block_id": "p533-b28", - "global_id": 15404, - "bbox": [ - 278.0, - 282.82, - 290.15, - 292.78 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p533-b30", - "global_id": 15405, - "bbox": [ - 214.15, - 322.68, - 261.11, - 328.65 - ], - "text": "zero-input response", - "type": "text" - }, - { - "block_id": "p533-b31", - "global_id": 15406, - "bbox": [ - 291.26, - 296.8, - 299.03, - 306.77 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p533-b32", - "global_id": 15407, - "bbox": [ - 300.13, - 282.82, - 316.72, - 300.2 - ], - "text": "26", - "type": "text" - }, - { - "block_id": "p533-b33", - "global_id": 15408, - "bbox": [ - 306.77, - 304.29, - 316.72, - 314.25 - ], - "text": "15", - "type": "text" - }, - { - "block_id": "p533-b34", - "global_id": 15409, - "bbox": [ - 319.02, - 282.82, - 354.14, - 314.25 - ], - "text": "z\nz −0.5", - "type": "text" - }, - { - "block_id": "p533-b36", - "global_id": 15410, - "bbox": [ - 363.62, - 290.23, - 384.09, - 306.77 - ], - "text": "−22", - "type": "text" - }, - { - "block_id": "p533-b37", - "global_id": 15411, - "bbox": [ - 376.62, - 304.29, - 381.6, - 314.25 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p533-b38", - "global_id": 15412, - "bbox": [ - 386.4, - 282.82, - 414.04, - 314.25 - ], - "text": "z\nz −2", - "type": "text" - }, - { - "block_id": "p533-b40", - "global_id": 15413, - "bbox": [ - 423.51, - 290.23, - 443.99, - 307.18 - ], - "text": "+ 28", - "type": "text" - }, - { - "block_id": "p533-b41", - "global_id": 15414, - "bbox": [ - 436.52, - 304.29, - 441.5, - 314.25 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p533-b42", - "global_id": 15415, - "bbox": [ - 446.29, - 282.82, - 473.93, - 314.25 - ], - "text": "z\nz −3", - "type": "text" - }, - { - "block_id": "p533-b43", - "global_id": 15416, - "bbox": [ - 475.14, - 282.82, - 487.29, - 292.78 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p533-b45", - "global_id": 15417, - "bbox": [ - 370.89, - 322.68, - 416.54, - 328.65 - ], - "text": "zero-state response", - "type": "text" - }, - { - "block_id": "p533-b46", - "global_id": 15418, - "bbox": [ - 127.59, - 340.37, - 141.97, - 350.33 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p533-b47", - "global_id": 15419, - "bbox": [ - 197.41, - 358.34, - 304.08, - 386.63 - ], - "text": "y[n] = (5(2)n −2(3)n)u[n]\n\n\n\nzero-input response", - "type": "text" - }, - { - "block_id": "p533-b48", - "global_id": 15420, - "bbox": [ - 305.19, - 362.45, - 312.96, - 372.41 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p533-b49", - "global_id": 15421, - "bbox": [ - 314.06, - 354.44, - 326.25, - 368.08 - ], - "text": "26", - "type": "text" - }, - { - "block_id": "p533-b50", - "global_id": 15422, - "bbox": [ - 314.06, - 354.44, - 446.31, - 388.44 - ], - "text": "15(0.5)n −22\n3 (2)n + 28\n5 (3)n\nu[n]\n\n\n\nzero-state response", - "type": "text" - }, - { - "block_id": "p533-b51", - "global_id": 15423, - "bbox": [ - 215.51, - 394.19, - 223.28, - 404.15 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p533-b53", - "global_id": 15424, - "bbox": [ - 229.25, - 392.84, - 241.7, - 404.15 - ], - "text": "−7", - "type": "text" - }, - { - "block_id": "p533-b54", - "global_id": 15425, - "bbox": [ - 238.21, - 386.18, - 361.65, - 407.68 - ], - "text": "3(2)n + 18\n5 (3)n + 26\n15(0.5)n\nu[n]", - "type": "text" - }, - { - "block_id": "p533-b55", - "global_id": 15426, - "bbox": [ - 127.59, - 417.1, - 294.72, - 427.06 - ], - "text": "which agrees with the result in Eq. (5.23).", - "type": "text" - }, - { - "block_id": "p533-b56", - "global_id": 15427, - "bbox": [ - 133.57, - 460.23, - 480.23, - 472.18 - ], - "text": "DRILL 5.12\nSeparating Zero-Input and Zero-State Responses", - "type": "text" - }, - { - "block_id": "p533-b57", - "global_id": 15428, - "bbox": [ - 133.57, - 481.3, - 156.1, - 491.27 - ], - "text": "Solve", - "type": "text" - }, - { - "block_id": "p533-b58", - "global_id": 15429, - "bbox": [ - 227.35, - 491.5, - 274.79, - 503.23 - ], - "text": "y[n + 2] −5", - "type": "text" - }, - { - "block_id": "p533-b59", - "global_id": 15430, - "bbox": [ - 271.31, - 491.5, - 416.34, - 506.04 - ], - "text": "6y[n + 1] + 1\n6y[n] = 5x[n + 1] −x[n]", - "type": "text" - }, - { - "block_id": "p533-b60", - "global_id": 15431, - "bbox": [ - 133.57, - 511.78, - 510.16, - 534.11 - ], - "text": "if the initial conditions are y[−1] = 2, y[−2] = 0, and the input x[n] = u[n]. Separate the\nresponse into zero-input and zero-state responses.", - "type": "text" - }, - { - "block_id": "p533-b61", - "global_id": 15432, - "bbox": [ - 133.84, - 547.63, - 188.43, - 558.59 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p533-b62", - "global_id": 15433, - "bbox": [ - 202.16, - 572.55, - 228.02, - 582.82 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p533-b64", - "global_id": 15434, - "bbox": [ - 235.5, - 564.54, - 250.29, - 582.92 - ], - "text": "3\n 1", - "type": "text" - }, - { - "block_id": "p533-b65", - "global_id": 15435, - "bbox": [ - 246.8, - 564.54, - 275.04, - 585.74 - ], - "text": "2\nn −4", - "type": "text" - }, - { - "block_id": "p533-b66", - "global_id": 15436, - "bbox": [ - 271.55, - 564.54, - 286.03, - 585.74 - ], - "text": "3\n 1", - "type": "text" - }, - { - "block_id": "p533-b67", - "global_id": 15437, - "bbox": [ - 282.55, - 561.55, - 300.66, - 585.74 - ], - "text": "3\nn", - "type": "text" - }, - { - "block_id": "p533-b68", - "global_id": 15438, - "bbox": [ - 230.07, - 572.55, - 318.36, - 601.41 - ], - "text": "u[n]\n\n\n\nzero-input response", - "type": "text" - }, - { - "block_id": "p533-b69", - "global_id": 15439, - "bbox": [ - 319.47, - 572.55, - 327.24, - 582.51 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p533-b71", - "global_id": 15440, - "bbox": [ - 333.77, - 564.54, - 374.38, - 582.92 - ], - "text": "12 −18\n 1", - "type": "text" - }, - { - "block_id": "p533-b72", - "global_id": 15441, - "bbox": [ - 370.89, - 564.54, - 399.42, - 585.74 - ], - "text": "2\nn + 6", - "type": "text" - }, - { - "block_id": "p533-b73", - "global_id": 15442, - "bbox": [ - 400.52, - 564.54, - 409.23, - 578.18 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p533-b74", - "global_id": 15443, - "bbox": [ - 405.74, - 561.55, - 423.85, - 585.74 - ], - "text": "3\nn", - "type": "text" - }, - { - "block_id": "p533-b75", - "global_id": 15444, - "bbox": [ - 328.34, - 572.55, - 441.56, - 601.41 - ], - "text": "u[n]\n\n\n\nzero-state response", - "type": "text" - }, - { - "block_id": "p533-b76", - "global_id": 15445, - "bbox": [ - 220.25, - 609.84, - 228.02, - 619.8 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p533-b77", - "global_id": 15446, - "bbox": [ - 230.07, - 598.84, - 234.74, - 608.81 - ], - "text": "'", - "type": "text" - }, - { - "block_id": "p533-b78", - "global_id": 15447, - "bbox": [ - 234.74, - 601.83, - 275.34, - 620.22 - ], - "text": "12 −15\n 1", - "type": "text" - }, - { - "block_id": "p533-b79", - "global_id": 15448, - "bbox": [ - 271.86, - 601.83, - 303.58, - 623.03 - ], - "text": "2\nn + 14", - "type": "text" - }, - { - "block_id": "p533-b80", - "global_id": 15449, - "bbox": [ - 298.34, - 601.83, - 314.57, - 623.03 - ], - "text": "3\n 1", - "type": "text" - }, - { - "block_id": "p533-b81", - "global_id": 15450, - "bbox": [ - 311.09, - 598.84, - 328.45, - 623.03 - ], - "text": "3\nn(", - "type": "text" - }, - { - "block_id": "p533-b82", - "global_id": 15451, - "bbox": [ - 329.55, - 609.84, - 346.15, - 620.12 - ], - "text": "u[n]", - "type": "text" - } - ] - }, - { - "page_num": 534, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p534-b0", - "global_id": 15452, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "514\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p534-b1", - "global_id": 15453, - "bbox": [ - 101.84, - 86.52, - 465.43, - 98.48 - ], - "text": "5.3-1 Zero-State Response of LTID Systems: The Transfer Function", - "type": "text" - }, - { - "block_id": "p534-b2", - "global_id": 15454, - "bbox": [ - 101.84, - 104.51, - 390.76, - 114.57 - ], - "text": "Consider an Nth-order LTID system specified by the difference equation", - "type": "text" - }, - { - "block_id": "p534-b3", - "global_id": 15455, - "bbox": [ - 254.54, - 126.55, - 337.69, - 136.83 - ], - "text": "Q[E]y[n] = P[E]x[n]", - "type": "text" - }, - { - "block_id": "p534-b4", - "global_id": 15456, - "bbox": [ - 101.85, - 149.32, - 110.14, - 159.28 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p534-b5", - "global_id": 15457, - "bbox": [ - 196.71, - 167.15, - 354.73, - 182.41 - ], - "text": "(EN + a1EN−1 + · · · + aN−1E + aN)y[n]", - "type": "text" - }, - { - "block_id": "p534-b6", - "global_id": 15458, - "bbox": [ - 218.68, - 183.58, - 395.52, - 198.84 - ], - "text": "= (b0EN + b1EN−1 + · · · + bN−1E + bN)x[n]", - "type": "text" - }, - { - "block_id": "p534-b7", - "global_id": 15459, - "bbox": [ - 101.85, - 210.47, - 110.14, - 220.43 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p534-b8", - "global_id": 15460, - "bbox": [ - 182.31, - 232.41, - 409.93, - 243.56 - ], - "text": "y[n + N] + a1y[n + N −1] + · · · + aN−1y[n + 1] + aNy[n]", - "type": "text" - }, - { - "block_id": "p534-b9", - "global_id": 15461, - "bbox": [ - 204.27, - 247.35, - 490.38, - 258.5 - ], - "text": "= b0x[n + N] + · · · + bN−1x[n + 1] + bNx[n]\n(5.24)", - "type": "text" - }, - { - "block_id": "p534-b10", - "global_id": 15462, - "bbox": [ - 101.85, - 270.12, - 490.4, - 303.99 - ], - "text": "We now derive the general expression for the zero-state response: that is, the system response to\ninput x[n] when all the initial conditions y[−1] = y[−2] = · · · = y[−N] = 0 (zero state). The input\nx[n] is assumed to be causal so that x[−1] = x[−2] = · · · = x[−N] = 0.", - "type": "text" - }, - { - "block_id": "p534-b11", - "global_id": 15463, - "bbox": [ - 119.79, - 305.99, - 319.12, - 315.95 - ], - "text": "Equation (5.24) can be expressed in delay form as", - "type": "text" - }, - { - "block_id": "p534-b12", - "global_id": 15464, - "bbox": [ - 202.06, - 327.92, - 349.31, - 339.07 - ], - "text": "y[n] + a1y[n −1] + · · · + aNy[n −N]", - "type": "text" - }, - { - "block_id": "p534-b13", - "global_id": 15465, - "bbox": [ - 224.03, - 342.87, - 490.38, - 354.02 - ], - "text": "= b0x[n] + b1x[n −1] + · · · + bNx[n −N]\n(5.25)", - "type": "text" - }, - { - "block_id": "p534-b14", - "global_id": 15466, - "bbox": [ - 101.85, - 365.23, - 289.5, - 375.61 - ], - "text": "Because y[−r] = x[−r] = 0 for r = 1,2,. . .,N,", - "type": "text" - }, - { - "block_id": "p534-b15", - "global_id": 15467, - "bbox": [ - 204.11, - 385.86, - 286.42, - 402.7 - ], - "text": "y[n −m]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p534-b16", - "global_id": 15468, - "bbox": [ - 279.23, - 392.43, - 306.65, - 409.78 - ], - "text": "zm Y[z]", - "type": "text" - }, - { - "block_id": "p534-b17", - "global_id": 15469, - "bbox": [ - 204.09, - 411.53, - 286.42, - 428.38 - ], - "text": "x[n −m]u[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p534-b18", - "global_id": 15470, - "bbox": [ - 279.23, - 418.1, - 387.56, - 435.45 - ], - "text": "zm X[z]\nm = 1,2,. . .,N", - "type": "text" - }, - { - "block_id": "p534-b19", - "global_id": 15471, - "bbox": [ - 101.85, - 444.52, - 283.27, - 454.58 - ], - "text": "Now the z-transform of Eq. (5.25) is given by", - "type": "text" - }, - { - "block_id": "p534-b21", - "global_id": 15472, - "bbox": [ - 166.25, - 465.53, - 213.99, - 489.55 - ], - "text": "1 + a1\nz + a2", - "type": "text" - }, - { - "block_id": "p534-b22", - "global_id": 15473, - "bbox": [ - 206.07, - 465.53, - 261.31, - 489.55 - ], - "text": "z2 + · · · + aN", - "type": "text" - }, - { - "block_id": "p534-b23", - "global_id": 15474, - "bbox": [ - 252.23, - 479.01, - 260.76, - 489.55 - ], - "text": "zN", - "type": "text" - }, - { - "block_id": "p534-b25", - "global_id": 15475, - "bbox": [ - 271.27, - 472.2, - 297.9, - 482.48 - ], - "text": "Y[z] =", - "type": "text" - }, - { - "block_id": "p534-b27", - "global_id": 15476, - "bbox": [ - 306.67, - 465.53, - 336.17, - 483.66 - ], - "text": "b0 + b1", - "type": "text" - }, - { - "block_id": "p534-b28", - "global_id": 15477, - "bbox": [ - 330.26, - 465.53, - 358.39, - 489.55 - ], - "text": "z + b2", - "type": "text" - }, - { - "block_id": "p534-b29", - "global_id": 15478, - "bbox": [ - 350.48, - 465.53, - 405.73, - 489.55 - ], - "text": "z2 + · · · + bN", - "type": "text" - }, - { - "block_id": "p534-b30", - "global_id": 15479, - "bbox": [ - 396.65, - 479.01, - 405.18, - 489.55 - ], - "text": "zN", - "type": "text" - }, - { - "block_id": "p534-b32", - "global_id": 15480, - "bbox": [ - 415.67, - 472.2, - 432.72, - 482.48 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p534-b33", - "global_id": 15481, - "bbox": [ - 101.85, - 498.13, - 262.94, - 511.7 - ], - "text": "Multiplication of both sides by zN yields", - "type": "text" - }, - { - "block_id": "p534-b34", - "global_id": 15482, - "bbox": [ - 199.89, - 519.57, - 351.37, - 534.83 - ], - "text": "(zN + a1zN−1 + · · · + aN−1z + aN)Y[z]", - "type": "text" - }, - { - "block_id": "p534-b35", - "global_id": 15483, - "bbox": [ - 221.86, - 536.0, - 392.35, - 551.28 - ], - "text": "= (b0zN + b1zN−1 + · · · + bN−1z + bN)X[z]", - "type": "text" - }, - { - "block_id": "p534-b36", - "global_id": 15484, - "bbox": [ - 101.85, - 562.9, - 143.59, - 572.86 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p534-b37", - "global_id": 15485, - "bbox": [ - 196.69, - 590.66, - 223.32, - 600.94 - ], - "text": "Y[z] =", - "type": "text" - }, - { - "block_id": "p534-b38", - "global_id": 15486, - "bbox": [ - 225.37, - 576.68, - 368.56, - 594.83 - ], - "text": "b0zN + b1zN−1 + · · · + bN−1z + bN", - "type": "text" - }, - { - "block_id": "p534-b39", - "global_id": 15487, - "bbox": [ - 237.77, - 594.86, - 364.08, - 608.88 - ], - "text": "zN + a1zN−1 + · · · + aN−1z + aN", - "type": "text" - }, - { - "block_id": "p534-b41", - "global_id": 15488, - "bbox": [ - 378.51, - 590.66, - 395.56, - 600.94 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p534-b42", - "global_id": 15489, - "bbox": [ - 215.55, - 610.76, - 243.72, - 628.02 - ], - "text": "= P[z]", - "type": "text" - }, - { - "block_id": "p534-b43", - "global_id": 15490, - "bbox": [ - 226.57, - 617.74, - 262.51, - 635.09 - ], - "text": "Q[z]X[z]", - "type": "text" - } - ] - }, - { - "page_num": 535, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p535-b0", - "global_id": 15491, - "bbox": [ - 246.93, - 62.57, - 516.11, - 71.98 - ], - "text": "5.3\nz-Transform Solution of Linear Difference Equations\n515", - "type": "text" - }, - { - "block_id": "p535-b1", - "global_id": 15492, - "bbox": [ - 127.59, - 85.4, - 416.99, - 95.78 - ], - "text": "We have shown in Eq. (5.20) that Y[z] = X[z]H[z]. Hence, it follows that", - "type": "text" - }, - { - "block_id": "p535-b2", - "global_id": 15493, - "bbox": [ - 221.58, - 105.62, - 269.95, - 122.88 - ], - "text": "H[z] = P[z]", - "type": "text" - }, - { - "block_id": "p535-b3", - "global_id": 15494, - "bbox": [ - 252.79, - 102.0, - 420.03, - 129.95 - ], - "text": "Q[z] = b0zN + b1zN−1 + · · · + bN−1z + bN", - "type": "text" - }, - { - "block_id": "p535-b4", - "global_id": 15495, - "bbox": [ - 289.23, - 116.79, - 415.54, - 130.82 - ], - "text": "zN + a1zN−1 + · · · + aN−1z + aN", - "type": "text" - }, - { - "block_id": "p535-b5", - "global_id": 15496, - "bbox": [ - 492.07, - 113.01, - 516.13, - 122.98 - ], - "text": "(5.26)", - "type": "text" - }, - { - "block_id": "p535-b6", - "global_id": 15497, - "bbox": [ - 127.59, - 138.63, - 516.12, - 160.54 - ], - "text": "As in the case of LTIC systems, this result leads to an alternative definition of the LTID system\ntransfer function as the ratio of Y[z] to X[z] (assuming all initial conditions zero).", - "type": "text" - }, - { - "block_id": "p535-b7", - "global_id": 15498, - "bbox": [ - 244.65, - 168.82, - 292.8, - 186.08 - ], - "text": "H[z] ≡Y[z]", - "type": "text" - }, - { - "block_id": "p535-b8", - "global_id": 15499, - "bbox": [ - 275.86, - 168.82, - 397.87, - 193.16 - ], - "text": "X[z] = Z[zero-state response]", - "type": "text" - }, - { - "block_id": "p535-b9", - "global_id": 15500, - "bbox": [ - 334.95, - 182.88, - 370.08, - 193.26 - ], - "text": "Z[input]", - "type": "text" - }, - { - "block_id": "p535-b10", - "global_id": 15501, - "bbox": [ - 127.59, - 204.82, - 516.14, - 255.27 - ], - "text": "ALTERNATE INTERPRETATION OF THE z-TRANSFORM\nSo far we have treated the z-transform as a machine that converts linear difference equations into\nalgebraic equations. There is no physical understanding of how this is accomplished or what it\nmeans. We now discuss more intuitive interpretation and meaning of the z-transform.", - "type": "text" - }, - { - "block_id": "p535-b11", - "global_id": 15502, - "bbox": [ - 127.59, - 257.26, - 516.14, - 303.08 - ], - "text": "In Ch. 3, Eq. (3.38), we showed that the LTID system response to an everlasting exponential\nzn is H[z]zn. If we could express every discrete-time signal as a linear combination of everlasting\nexponentials of the form zn, we could readily obtain the system response to any input. For\nexample, if", - "type": "text" - }, - { - "block_id": "p535-b12", - "global_id": 15503, - "bbox": [ - 285.98, - 314.18, - 311.87, - 324.46 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p535-b13", - "global_id": 15504, - "bbox": [ - 313.92, - 304.23, - 328.02, - 314.69 - ], - "text": "K\n\"", - "type": "text" - }, - { - "block_id": "p535-b14", - "global_id": 15505, - "bbox": [ - 314.9, - 328.57, - 327.04, - 335.83 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p535-b15", - "global_id": 15506, - "bbox": [ - 329.13, - 312.68, - 357.24, - 325.26 - ], - "text": "X[zk]zn", - "type": "text" - }, - { - "block_id": "p535-b16", - "global_id": 15507, - "bbox": [ - 353.75, - 314.6, - 516.13, - 326.45 - ], - "text": "k\n(5.27)", - "type": "text" - }, - { - "block_id": "p535-b17", - "global_id": 15508, - "bbox": [ - 127.59, - 341.9, - 350.15, - 351.86 - ], - "text": "the response of an LTID system to this input is given by", - "type": "text" - }, - { - "block_id": "p535-b18", - "global_id": 15509, - "bbox": [ - 275.07, - 371.66, - 300.93, - 381.93 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p535-b19", - "global_id": 15510, - "bbox": [ - 302.98, - 361.7, - 317.07, - 372.15 - ], - "text": "K\n\"", - "type": "text" - }, - { - "block_id": "p535-b20", - "global_id": 15511, - "bbox": [ - 303.95, - 386.04, - 316.08, - 393.31 - ], - "text": "k=1", - "type": "text" - }, - { - "block_id": "p535-b21", - "global_id": 15512, - "bbox": [ - 318.18, - 370.15, - 368.16, - 382.73 - ], - "text": "X[zk]H[zk]zn", - "type": "text" - }, - { - "block_id": "p535-b22", - "global_id": 15513, - "bbox": [ - 364.67, - 376.95, - 367.77, - 383.92 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p535-b23", - "global_id": 15514, - "bbox": [ - 127.59, - 401.72, - 516.14, - 435.6 - ], - "text": "Unfortunately, a very small class of signals can be expressed in the form of Eq. (5.27). However,\nwe can express almost all signals of practical utility as a sum of everlasting exponentials over a\ncontinuum of values of z. This is precisely what the z-transform in Eq. (5.2) does.", - "type": "text" - }, - { - "block_id": "p535-b24", - "global_id": 15515, - "bbox": [ - 271.42, - 444.27, - 315.28, - 468.29 - ], - "text": "x[n] =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p535-b25", - "global_id": 15516, - "bbox": [ - 317.58, - 437.28, - 516.12, - 461.21 - ], - "text": "5\nX[z]zn−1 dz\n(5.28)", - "type": "text" - }, - { - "block_id": "p535-b26", - "global_id": 15517, - "bbox": [ - 127.59, - 476.42, - 516.13, - 498.75 - ], - "text": "Invoking the linearity property of the z-transform, we can find the system response y[n] to input\nx[n] in Eq. (5.28) as†", - "type": "text" - }, - { - "block_id": "p535-b27", - "global_id": 15518, - "bbox": [ - 226.43, - 507.41, - 270.25, - 531.43 - ], - "text": "y[n] =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p535-b28", - "global_id": 15519, - "bbox": [ - 272.56, - 500.43, - 417.32, - 524.26 - ], - "text": "5\nX[z]H[z]zn−1 dz = Z−1{X[z]H[z]}", - "type": "text" - }, - { - "block_id": "p535-b29", - "global_id": 15520, - "bbox": [ - 127.6, - 539.98, - 158.76, - 549.95 - ], - "text": "Clearly,", - "type": "text" - }, - { - "block_id": "p535-b30", - "global_id": 15521, - "bbox": [ - 289.93, - 551.52, - 353.82, - 561.8 - ], - "text": "Y[z] = X[z]H[z]", - "type": "text" - }, - { - "block_id": "p535-b31", - "global_id": 15522, - "bbox": [ - 127.6, - 570.22, - 516.15, - 592.15 - ], - "text": "This viewpoint of finding the response of LTID system is illustrated in Fig. 5.6a. Just as in\ncontinuous-time systems, we can model discrete-time systems in the transformed manner by", - "type": "text" - }, - { - "block_id": "p535-b32", - "global_id": 15523, - "bbox": [ - 127.59, - 610.24, - 516.12, - 633.41 - ], - "text": "† In computing y[n], the contour along which the integration is performed is modified to consider the ROC of\nX[z] as well as H[z]. We ignore this consideration in this intuitive discussion.", - "type": "text" - } - ] - }, - { - "page_num": 536, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p536-b0", - "global_id": 15524, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "516\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p536-b1", - "global_id": 15525, - "bbox": [ - 125.76, - 211.44, - 355.53, - 220.68 - ], - "text": "Figure 5.6 The transformed representation of an LTID system.", - "type": "text" - }, - { - "block_id": "p536-b2", - "global_id": 15526, - "bbox": [ - 101.84, - 254.77, - 490.41, - 276.79 - ], - "text": "representing all signals by their z-transforms and all system components (or elements) by their\ntransfer functions, as shown in Fig. 5.6b.", - "type": "text" - }, - { - "block_id": "p536-b3", - "global_id": 15527, - "bbox": [ - 101.85, - 278.37, - 490.37, - 300.7 - ], - "text": "The result Y[z] = H[z]X[z] greatly facilitates derivation of the system response to a given\ninput. We shall demonstrate this assertion by an example.", - "type": "text" - }, - { - "block_id": "p536-b4", - "global_id": 15528, - "bbox": [ - 76.77, - 336.54, - 448.6, - 348.49 - ], - "text": "EXAMPLE 5.6\nTransfer Function to Find the Zero-State Response", - "type": "text" - }, - { - "block_id": "p536-b5", - "global_id": 15529, - "bbox": [ - 103.16, - 364.74, - 416.3, - 375.12 - ], - "text": "Find the response y[n] of an LTID system described by the difference equation", - "type": "text" - }, - { - "block_id": "p536-b6", - "global_id": 15530, - "bbox": [ - 186.51, - 386.66, - 393.63, - 397.04 - ], - "text": "y[n + 2] + y[n + 1] + 0.16y[n] = x[n + 1] + 0.32x[n]", - "type": "text" - }, - { - "block_id": "p536-b7", - "global_id": 15531, - "bbox": [ - 103.17, - 408.99, - 111.47, - 418.96 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p536-b8", - "global_id": 15532, - "bbox": [ - 215.47, - 416.42, - 364.7, - 430.91 - ], - "text": "(E2 + E + 0.16)y[n] = (E + 0.32)x[n]", - "type": "text" - }, - { - "block_id": "p536-b9", - "global_id": 15533, - "bbox": [ - 103.17, - 438.24, - 477.01, - 461.8 - ], - "text": "for the input x[n] = (−2)−nu[n] and with all the initial conditions zero (system in the zero\nstate).", - "type": "text" - }, - { - "block_id": "p536-b10", - "global_id": 15534, - "bbox": [ - 103.16, - 484.71, - 253.78, - 494.67 - ], - "text": "From the difference equation, we find", - "type": "text" - }, - { - "block_id": "p536-b11", - "global_id": 15535, - "bbox": [ - 232.45, - 504.2, - 280.82, - 521.46 - ], - "text": "H[z] = P[z]", - "type": "text" - }, - { - "block_id": "p536-b12", - "global_id": 15536, - "bbox": [ - 263.66, - 504.19, - 346.52, - 528.63 - ], - "text": "Q[z] =\nz + 0.32\nz2 + z + 0.16", - "type": "text" - }, - { - "block_id": "p536-b13", - "global_id": 15537, - "bbox": [ - 103.16, - 538.1, - 363.21, - 549.7 - ], - "text": "For the input x[n] = (−2)−nu[n] = [(−2)−1]nu(n) = (−0.5)nu[n],", - "type": "text" - }, - { - "block_id": "p536-b14", - "global_id": 15538, - "bbox": [ - 260.84, - 557.13, - 318.13, - 581.24 - ], - "text": "X[z] =\nz\nz + 0.5", - "type": "text" - }, - { - "block_id": "p536-b15", - "global_id": 15539, - "bbox": [ - 103.16, - 589.88, - 117.54, - 599.84 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p536-b16", - "global_id": 15540, - "bbox": [ - 204.54, - 597.4, - 374.41, - 621.83 - ], - "text": "Y[z] = X[z]H[z] =\nz(z + 0.32)\n(z2 + z + 0.16)(z + 0.5)", - "type": "text" - } - ] - }, - { - "page_num": 537, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p537-b0", - "global_id": 15541, - "bbox": [ - 246.93, - 62.57, - 516.11, - 71.98 - ], - "text": "5.3\nz-Transform Solution of Linear Difference Equations\n517", - "type": "text" - }, - { - "block_id": "p537-b1", - "global_id": 15542, - "bbox": [ - 128.9, - 86.24, - 170.65, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p537-b2", - "global_id": 15543, - "bbox": [ - 194.73, - 111.04, - 211.55, - 121.32 - ], - "text": "Y[z]", - "type": "text" - }, - { - "block_id": "p537-b3", - "global_id": 15544, - "bbox": [ - 201.2, - 111.04, - 436.91, - 135.48 - ], - "text": "z\n=\n(z + 0.32)\n(z2 + z + 0.16)(z + 0.5) =\n(z + 0.32)\n(z + 0.2)(z + 0.8)(z + 0.5)", - "type": "text" - }, - { - "block_id": "p537-b4", - "global_id": 15545, - "bbox": [ - 214.79, - 137.6, - 333.92, - 162.04 - ], - "text": "=\n2/3\nz + 0.2 −\n8/3\nz + 0.8 +\n2\nz + 0.5", - "type": "text" - }, - { - "block_id": "p537-b5", - "global_id": 15546, - "bbox": [ - 128.9, - 176.64, - 155.19, - 186.61 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p537-b6", - "global_id": 15547, - "bbox": [ - 214.54, - 201.14, - 249.4, - 218.09 - ], - "text": "Y[z] = 2", - "type": "text" - }, - { - "block_id": "p537-b7", - "global_id": 15548, - "bbox": [ - 244.42, - 215.2, - 249.4, - 225.16 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p537-b8", - "global_id": 15549, - "bbox": [ - 251.7, - 193.73, - 286.81, - 225.16 - ], - "text": "z\nz + 0.2", - "type": "text" - }, - { - "block_id": "p537-b10", - "global_id": 15550, - "bbox": [ - 296.29, - 201.14, - 311.79, - 217.68 - ], - "text": "−8", - "type": "text" - }, - { - "block_id": "p537-b11", - "global_id": 15551, - "bbox": [ - 306.81, - 215.2, - 311.79, - 225.16 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p537-b12", - "global_id": 15552, - "bbox": [ - 314.09, - 193.73, - 349.2, - 225.16 - ], - "text": "z\nz + 0.8", - "type": "text" - }, - { - "block_id": "p537-b14", - "global_id": 15553, - "bbox": [ - 358.68, - 207.71, - 372.98, - 218.09 - ], - "text": "+ 2", - "type": "text" - }, - { - "block_id": "p537-b15", - "global_id": 15554, - "bbox": [ - 374.08, - 193.73, - 409.2, - 225.16 - ], - "text": "z\nz + 0.5", - "type": "text" - }, - { - "block_id": "p537-b17", - "global_id": 15555, - "bbox": [ - 128.9, - 241.31, - 143.28, - 251.27 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p537-b18", - "global_id": 15556, - "bbox": [ - 222.41, - 268.8, - 248.26, - 279.07 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p537-b19", - "global_id": 15557, - "bbox": [ - 250.31, - 260.79, - 258.91, - 274.43 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p537-b20", - "global_id": 15558, - "bbox": [ - 255.43, - 260.79, - 409.26, - 281.99 - ], - "text": "3(−0.2)n −8\n3(−0.8)n + 2(−0.5)n\nu[n]", - "type": "text" - }, - { - "block_id": "p537-b21", - "global_id": 15559, - "bbox": [ - 102.51, - 343.89, - 383.67, - 355.84 - ], - "text": "EXAMPLE 5.7\nTransfer Function of a Unit Delay", - "type": "text" - }, - { - "block_id": "p537-b22", - "global_id": 15560, - "bbox": [ - 128.9, - 372.09, - 339.49, - 382.47 - ], - "text": "Show that the transfer function of a unit delay is 1/z.", - "type": "text" - }, - { - "block_id": "p537-b23", - "global_id": 15561, - "bbox": [ - 128.9, - 404.97, - 432.11, - 415.34 - ], - "text": "If the input to the unit delay is x[n]u[n], then its output (Fig. 5.7) is given by", - "type": "text" - }, - { - "block_id": "p537-b24", - "global_id": 15562, - "bbox": [ - 269.69, - 426.88, - 361.94, - 437.26 - ], - "text": "y[n] = x[n −1]u[n −1]", - "type": "text" - }, - { - "block_id": "p537-b25", - "global_id": 15563, - "bbox": [ - 128.91, - 449.12, - 348.04, - 459.18 - ], - "text": "The z-transform of this equation yields [see Eq. (5.12)]", - "type": "text" - }, - { - "block_id": "p537-b26", - "global_id": 15564, - "bbox": [ - 265.76, - 469.12, - 300.62, - 486.06 - ], - "text": "Y[z] = 1", - "type": "text" - }, - { - "block_id": "p537-b27", - "global_id": 15565, - "bbox": [ - 296.19, - 475.69, - 365.93, - 493.04 - ], - "text": "z X[z] = H[z]X[z]", - "type": "text" - }, - { - "block_id": "p537-b28", - "global_id": 15566, - "bbox": [ - 128.91, - 501.76, - 344.5, - 511.72 - ], - "text": "It follows that the transfer function of the unit delay is", - "type": "text" - }, - { - "block_id": "p537-b29", - "global_id": 15567, - "bbox": [ - 297.14, - 521.66, - 333.34, - 538.61 - ], - "text": "H[z] = 1", - "type": "text" - }, - { - "block_id": "p537-b30", - "global_id": 15568, - "bbox": [ - 328.9, - 535.62, - 332.78, - 545.58 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p537-b31", - "global_id": 15569, - "bbox": [ - 155.85, - 582.11, - 182.54, - 590.19 - ], - "text": "x[n]u[n]", - "type": "text" - }, - { - "block_id": "p537-b32", - "global_id": 15570, - "bbox": [ - 161.53, - 599.79, - 176.86, - 607.88 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p537-b33", - "global_id": 15571, - "bbox": [ - 259.96, - 581.89, - 317.98, - 590.19 - ], - "text": "x[n 1]u[n 1]", - "type": "text" - }, - { - "block_id": "p537-b34", - "global_id": 15572, - "bbox": [ - 263.09, - 598.62, - 307.82, - 613.07 - ], - "text": "Y[z] \nX[z]\n1\nz", - "type": "text" - }, - { - "block_id": "p537-b35", - "global_id": 15573, - "bbox": [ - 231.03, - 588.18, - 506.35, - 614.4 - ], - "text": "1\nz\nFigure 5.7 Ideal unit delay and its transfer\nfunction.", - "type": "text" - } - ] - }, - { - "page_num": 538, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p538-b0", - "global_id": 15574, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "518\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p538-b1", - "global_id": 15575, - "bbox": [ - 107.82, - 97.81, - 465.76, - 123.71 - ], - "text": "DRILL 5.13\nTransfer Function to Find Zero-State Response and\nDifference Equation", - "type": "text" - }, - { - "block_id": "p538-b2", - "global_id": 15576, - "bbox": [ - 107.82, - 132.83, - 388.95, - 142.79 - ], - "text": "A discrete-time system is described by the following transfer function:", - "type": "text" - }, - { - "block_id": "p538-b3", - "global_id": 15577, - "bbox": [ - 249.01, - 152.31, - 342.01, - 176.75 - ], - "text": "H[z] =\nz −0.5\n(z + 0.5)(z −1)", - "type": "text" - }, - { - "block_id": "p538-b4", - "global_id": 15578, - "bbox": [ - 125.76, - 187.96, - 479.14, - 199.56 - ], - "text": "(a) Find the system response to input x[n] = 3−(n+1)u[n] if all initial conditions are zero.", - "type": "text" - }, - { - "block_id": "p538-b5", - "global_id": 15579, - "bbox": [ - 125.76, - 204.14, - 469.19, - 214.51 - ], - "text": "(b) Write the difference equation relating the output y[n] to input x[n] for this system.", - "type": "text" - }, - { - "block_id": "p538-b6", - "global_id": 15580, - "bbox": [ - 108.09, - 234.02, - 170.31, - 244.97 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p538-b7", - "global_id": 15581, - "bbox": [ - 125.76, - 250.61, - 174.94, - 262.25 - ], - "text": "(a) y[n] = 1", - "type": "text" - }, - { - "block_id": "p538-b8", - "global_id": 15582, - "bbox": [ - 171.45, - 243.94, - 184.75, - 265.14 - ], - "text": "3\n\t 1", - "type": "text" - }, - { - "block_id": "p538-b9", - "global_id": 15583, - "bbox": [ - 181.26, - 243.94, - 272.93, - 265.45 - ], - "text": "2 −0.8(−0.5)n + 0.3\n 1", - "type": "text" - }, - { - "block_id": "p538-b10", - "global_id": 15584, - "bbox": [ - 269.44, - 243.94, - 286.05, - 265.14 - ], - "text": "3\nn", - "type": "text" - }, - { - "block_id": "p538-b11", - "global_id": 15585, - "bbox": [ - 286.05, - 251.95, - 302.64, - 262.23 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p538-b12", - "global_id": 15586, - "bbox": [ - 125.76, - 266.9, - 352.53, - 277.28 - ], - "text": "(b) y[n + 2] −0.5y[n + 1] −0.5y[n] = x[n + 1] −0.5x[n]", - "type": "text" - }, - { - "block_id": "p538-b13", - "global_id": 15587, - "bbox": [ - 101.84, - 316.98, - 179.35, - 328.94 - ], - "text": "5.3-2 Stability", - "type": "text" - }, - { - "block_id": "p538-b14", - "global_id": 15588, - "bbox": [ - 101.84, - 334.66, - 490.41, - 464.58 - ], - "text": "Equation (5.26) shows that the denominator of H[z] is Q[z], which is apparently identical to the\ncharacteristic polynomial Q[γ ] defined in Ch. 3. Does this mean that the denominator of H[z] is\nthe characteristic polynomial of the system? This may or may not be the case: if P[z] and Q[z]\nin Eq. (5.26) have any common factors, they cancel out, and the effective denominator of H[z]\nis not necessarily equal to Q[z]. Recall also that the system transfer function H[z], like h[n], is\ndefined in terms of measurements at the external terminals. Consequently, H[z] and h[n] are both\nexternal descriptions of the system. In contrast, the characteristic polynomial Q[z] is an internal\ndescription. Clearly, we can determine only external stability, that is, BIBO stability, from H[z].\nIf all the poles of H[z] are within the unit circle, all the terms in h[n] are decaying exponentials,\nand as shown in Sec. 3.9, h[n] is absolutely summable. Consequently, the system is BIBO-stable.\nOtherwise the system is BIBO-unstable.", - "type": "text" - }, - { - "block_id": "p538-b15", - "global_id": 15589, - "bbox": [ - 101.84, - 466.16, - 490.4, - 512.41 - ], - "text": "If P[z] and Q[z] do not have common factors, then the denominator of H[z] is identical to\nQ[z].† The poles of H[z] are the characteristic roots of the system. We can now determine internal\nstability. The internal stability criterion in Sec. 3.9-2 can be restated in terms of the poles of H[z],\nas follows.", - "type": "text" - }, - { - "block_id": "p538-b16", - "global_id": 15590, - "bbox": [ - 118.79, - 520.37, - 490.39, - 578.16 - ], - "text": "1. An LTID system is asymptotically stable if and only if all the poles of its transfer function\nH[z] are within the unit circle. The poles may be repeated or simple.\n2. An LTID system is unstable if and only if either one or both of the following conditions\nexist: (i) at least one pole of H[z] is outside the unit circle; (ii) there are repeated poles of\nH[z] on the unit circle.", - "type": "text" - }, - { - "block_id": "p538-b17", - "global_id": 15591, - "bbox": [ - 101.84, - 600.27, - 490.38, - 645.37 - ], - "text": "† There is no way of determining whether any common factors in P[z] and Q[z] were canceled out. This is\nbecause in our derivation of H[z], we generally get the final result after the cancellations have been effected.\nWhen we use internal description of the system to derive Q[z], however, we find pure Q[z] unaffected by any\ncommon factor in P[z].", - "type": "text" - } - ] - }, - { - "page_num": 539, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p539-b0", - "global_id": 15592, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "5.4\nSystem Realization\n519", - "type": "text" - }, - { - "block_id": "p539-b1", - "global_id": 15593, - "bbox": [ - 144.52, - 85.4, - 516.13, - 107.74 - ], - "text": "3. An LTID system is marginally stable if and only if there are no poles of H[z] outside the\nunit circle, and there are some simple poles on the unit circle.", - "type": "text" - }, - { - "block_id": "p539-b2", - "global_id": 15594, - "bbox": [ - 133.57, - 146.87, - 436.34, - 158.82 - ], - "text": "DRILL 5.14\nTransfer Function to Determine Stability", - "type": "text" - }, - { - "block_id": "p539-b3", - "global_id": 15595, - "bbox": [ - 133.57, - 167.53, - 510.15, - 189.86 - ], - "text": "Show that an accumulator whose impulse response is h[n] = u[n] is marginally stable but\nBIBO-unstable.", - "type": "text" - }, - { - "block_id": "p539-b4", - "global_id": 15596, - "bbox": [ - 127.59, - 221.48, - 246.64, - 233.44 - ], - "text": "5.3-3 Inverse Systems", - "type": "text" - }, - { - "block_id": "p539-b5", - "global_id": 15597, - "bbox": [ - 127.59, - 239.16, - 516.14, - 262.19 - ], - "text": "If H[z] is the transfer function of a system S, then Si, its inverse system, has a transfer function\nHi[z] given by", - "type": "text" - }, - { - "block_id": "p539-b6", - "global_id": 15598, - "bbox": [ - 295.59, - 262.26, - 346.95, - 286.17 - ], - "text": "Hi[z] =\n1\nH[z]", - "type": "text" - }, - { - "block_id": "p539-b7", - "global_id": 15599, - "bbox": [ - 127.59, - 292.76, - 516.12, - 339.71 - ], - "text": "This follows from the fact the inverse system Si undoes the operation of S. Hence, if H[z] is placed\nin cascade with Hi[z], the transfer function of the composite system (identity system) is unity. For\nexample, an accumulator whose transfer function is H[z] = z/(z −1) and a backward difference\nsystem whose transfer function is Hi[z] = (z −1)/z are inverse of each other. Similarly if", - "type": "text" - }, - { - "block_id": "p539-b8", - "global_id": 15600, - "bbox": [ - 292.06, - 349.06, - 350.47, - 366.32 - ], - "text": "H[z] = z −0.4", - "type": "text" - }, - { - "block_id": "p539-b9", - "global_id": 15601, - "bbox": [ - 323.27, - 363.11, - 350.46, - 373.49 - ], - "text": "z −0.7", - "type": "text" - }, - { - "block_id": "p539-b10", - "global_id": 15602, - "bbox": [ - 127.59, - 382.64, - 275.64, - 392.61 - ], - "text": "its inverse system transfer function is", - "type": "text" - }, - { - "block_id": "p539-b11", - "global_id": 15603, - "bbox": [ - 291.06, - 402.66, - 351.45, - 420.72 - ], - "text": "Hi[z] = z −0.7", - "type": "text" - }, - { - "block_id": "p539-b12", - "global_id": 15604, - "bbox": [ - 324.26, - 416.72, - 351.45, - 427.1 - ], - "text": "z −0.4", - "type": "text" - }, - { - "block_id": "p539-b13", - "global_id": 15605, - "bbox": [ - 127.59, - 435.84, - 383.07, - 446.92 - ], - "text": "as required by the property H[z]Hi[z] = 1. Hence, it follows that", - "type": "text" - }, - { - "block_id": "p539-b14", - "global_id": 15606, - "bbox": [ - 285.81, - 458.29, - 357.9, - 469.37 - ], - "text": "h[n] ∗hi[n] = δ[n]", - "type": "text" - }, - { - "block_id": "p539-b15", - "global_id": 15607, - "bbox": [ - 133.57, - 499.78, - 303.22, - 511.73 - ], - "text": "DRILL 5.15\nInverse Systems", - "type": "text" - }, - { - "block_id": "p539-b16", - "global_id": 15608, - "bbox": [ - 133.57, - 520.85, - 510.13, - 542.77 - ], - "text": "Find the impulse responses of an accumulator and a first-order backward difference system.\nShow that the convolution of the two impulse responses yields δ[n].", - "type": "text" - }, - { - "block_id": "p539-b17", - "global_id": 15609, - "bbox": [ - 127.94, - 580.98, - 297.92, - 594.93 - ], - "text": "5.4 SYSTEM REALIZATION", - "type": "text" - }, - { - "block_id": "p539-b18", - "global_id": 15610, - "bbox": [ - 127.59, - 600.91, - 516.15, - 646.74 - ], - "text": "Because of the similarity between LTIC and LTID systems, conventions for block diagrams and\nrules of interconnection for LTID are identical to those for continuous-time (LTIC) systems. It is\nnot necessary to rederive these relationships. We shall merely restate them to refresh the reader’s\nmemory.", - "type": "text" - } - ] - }, - { - "page_num": 540, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p540-b0", - "global_id": 15611, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "520\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p540-b1", - "global_id": 15612, - "bbox": [ - 101.84, - 85.82, - 490.4, - 143.6 - ], - "text": "The block diagram representations of the basic operations, such as an adder, a scalar\nmultiplier, unit delay, and pickoff points, is shown in Fig. 3.13. In our development, the unit delay,\nwhich is represented by a box marked D in Fig. 3.13, will be represented by its transfer function\n1/z. All the signals will also be represented in terms of their z-transforms. Thus, the input and the\noutput will be labeled X[z] and Y[z], respectively.", - "type": "text" - }, - { - "block_id": "p540-b2", - "global_id": 15613, - "bbox": [ - 101.85, - 145.18, - 490.41, - 179.47 - ], - "text": "When two systems with transfer functions H1[z] and H2[z] are connected in cascade (as\nin Fig. 4.18b), the transfer function of the composite system is H1[z]H2[z]. If the same two\nsystems are connected in parallel (as in Fig. 4.18c), the transfer function of the composite", - "type": "text" - }, - { - "block_id": "p540-b3", - "global_id": 15614, - "bbox": [ - 130.89, - 210.5, - 470.48, - 218.58 - ], - "text": "X[z]\nY[z]\nX[z]\nW[z]\nY[z]", - "type": "text" - }, - { - "block_id": "p540-b4", - "global_id": 15615, - "bbox": [ - 208.69, - 362.91, - 406.71, - 370.91 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p540-b16", - "global_id": 15616, - "bbox": [ - 369.62, - 252.6, - 383.29, - 262.42 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p540-b17", - "global_id": 15617, - "bbox": [ - 369.86, - 306.3, - 392.05, - 316.12 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p540-b18", - "global_id": 15618, - "bbox": [ - 369.62, - 342.15, - 384.29, - 351.91 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p540-b19", - "global_id": 15619, - "bbox": [ - 245.61, - 252.6, - 259.28, - 262.42 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p540-b20", - "global_id": 15620, - "bbox": [ - 245.85, - 306.3, - 268.04, - 316.12 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p540-b21", - "global_id": 15621, - "bbox": [ - 245.61, - 342.59, - 260.28, - 352.35 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p540-b22", - "global_id": 15622, - "bbox": [ - 416.26, - 210.5, - 423.26, - 220.11 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p540-b23", - "global_id": 15623, - "bbox": [ - 416.26, - 342.37, - 424.26, - 351.91 - ], - "text": "bN", - "type": "text" - }, - { - "block_id": "p540-b24", - "global_id": 15624, - "bbox": [ - 416.5, - 306.51, - 432.02, - 316.12 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p540-b25", - "global_id": 15625, - "bbox": [ - 416.26, - 252.81, - 423.26, - 262.42 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p540-b26", - "global_id": 15626, - "bbox": [ - 162.92, - 210.5, - 169.92, - 220.11 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p540-b27", - "global_id": 15627, - "bbox": [ - 162.92, - 341.8, - 170.92, - 351.35 - ], - "text": "bN", - "type": "text" - }, - { - "block_id": "p540-b28", - "global_id": 15628, - "bbox": [ - 163.16, - 306.51, - 178.68, - 316.12 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p540-b29", - "global_id": 15629, - "bbox": [ - 162.92, - 252.81, - 169.92, - 262.42 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p540-b30", - "global_id": 15630, - "bbox": [ - 232.22, - 384.6, - 364.79, - 392.68 - ], - "text": "X[z]\nY[z]", - "type": "text" - }, - { - "block_id": "p540-b31", - "global_id": 15631, - "bbox": [ - 295.07, - 592.93, - 303.95, - 600.93 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p540-b36", - "global_id": 15632, - "bbox": [ - 268.76, - 383.87, - 275.76, - 393.47 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p540-b37", - "global_id": 15633, - "bbox": [ - 268.76, - 563.35, - 276.76, - 572.89 - ], - "text": "bN", - "type": "text" - }, - { - "block_id": "p540-b38", - "global_id": 15634, - "bbox": [ - 269.0, - 507.58, - 284.51, - 517.18 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p540-b39", - "global_id": 15635, - "bbox": [ - 268.76, - 438.51, - 327.77, - 448.33 - ], - "text": "b1\na1", - "type": "text" - }, - { - "block_id": "p540-b40", - "global_id": 15636, - "bbox": [ - 314.35, - 507.36, - 336.53, - 517.18 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p540-b41", - "global_id": 15637, - "bbox": [ - 314.11, - 563.13, - 328.77, - 572.89 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p540-b42", - "global_id": 15638, - "bbox": [ - 298.27, - 469.82, - 302.27, - 484.86 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b43", - "global_id": 15639, - "bbox": [ - 298.27, - 415.88, - 302.27, - 430.92 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b44", - "global_id": 15640, - "bbox": [ - 298.27, - 539.62, - 302.27, - 554.66 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b45", - "global_id": 15641, - "bbox": [ - 274.21, - 330.41, - 278.21, - 345.45 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b46", - "global_id": 15642, - "bbox": [ - 274.21, - 276.16, - 278.21, - 291.2 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b47", - "global_id": 15643, - "bbox": [ - 145.31, - 236.91, - 278.21, - 251.95 - ], - "text": "1\nz\n1\nz", - "type": "text" - }, - { - "block_id": "p540-b48", - "global_id": 15644, - "bbox": [ - 145.31, - 276.16, - 149.31, - 291.2 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b49", - "global_id": 15645, - "bbox": [ - 145.31, - 330.41, - 149.31, - 345.45 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b50", - "global_id": 15646, - "bbox": [ - 400.06, - 330.41, - 404.06, - 345.45 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b51", - "global_id": 15647, - "bbox": [ - 400.06, - 276.16, - 404.06, - 291.2 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b52", - "global_id": 15648, - "bbox": [ - 400.06, - 236.91, - 404.06, - 251.95 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p540-b53", - "global_id": 15649, - "bbox": [ - 125.76, - 607.28, - 490.38, - 629.31 - ], - "text": "Figure 5.8 Realization of an Nth-order causal LTID system transfer function by using\n(a) DFI, (b) canonic direct (DFII), and (c) the transpose form of DFII.", - "type": "text" - } - ] - }, - { - "page_num": 541, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p541-b0", - "global_id": 15650, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "5.4\nSystem Realization\n521", - "type": "text" - }, - { - "block_id": "p541-b1", - "global_id": 15651, - "bbox": [ - 127.59, - 85.4, - 516.13, - 107.74 - ], - "text": "system is H1[z] + H2[z]. For a feedback system (as in Fig. 4.18d), the transfer function is\nG[z]/(1 + G[z]H[z]).", - "type": "text" - }, - { - "block_id": "p541-b2", - "global_id": 15652, - "bbox": [ - 127.59, - 109.63, - 516.15, - 203.37 - ], - "text": "We now consider a systematic method for realization (or simulation) of an arbitrary Nth-order\nLTID transfer function. Since realization is basically a synthesis problem, there is no unique way of\nrealizing a system. A given transfer function can be realized in many different ways. We present\nhere the two forms of direct realization. Each of these forms can be executed in several other\nways, such as cascade and parallel. Furthermore, a system can be realized by the transposed\nversion of any known realization of that system. This artifice doubles the number of system\nrealizations. A transfer function H[z] can be realized by using time delays along with adders and\nmultipliers.", - "type": "text" - }, - { - "block_id": "p541-b3", - "global_id": 15653, - "bbox": [ - 127.6, - 205.27, - 516.13, - 227.29 - ], - "text": "We shall consider a realization of a general Nth-order causal LTID system, whose transfer\nfunction is given by", - "type": "text" - }, - { - "block_id": "p541-b4", - "global_id": 15654, - "bbox": [ - 237.56, - 230.54, - 404.04, - 251.42 - ], - "text": "H[z] = b0zN + b1zN−1 + · · · + bN−1z + bN", - "type": "text" - }, - { - "block_id": "p541-b5", - "global_id": 15655, - "bbox": [ - 273.26, - 245.34, - 399.56, - 259.36 - ], - "text": "zN + a1zN−1 + · · · + aN−1z + aN", - "type": "text" - }, - { - "block_id": "p541-b6", - "global_id": 15656, - "bbox": [ - 492.07, - 241.55, - 516.13, - 251.51 - ], - "text": "(5.29)", - "type": "text" - }, - { - "block_id": "p541-b7", - "global_id": 15657, - "bbox": [ - 127.59, - 269.17, - 516.17, - 350.97 - ], - "text": "This equation is identical to the transfer function of a general Nth-order proper LTIC system given\nin Eq. (4.36). The only difference is that the variable z in the former is replaced by the variable s\nin the latter. Hence, the procedure for realizing an LTID transfer function is identical to that for\nthe LTIC transfer function with the basic element 1/s (integrator) replaced by the element 1/z\n(unit delay). The reader is encouraged to follow the steps in Sec. 4.6 and rederive the results for\nthe LTID transfer function in Eq. (5.29). Here we shall merely reproduce the realizations from\nSec. 4.6 with integrators (1/s) replaced by unit delays (1/z).", - "type": "text" - }, - { - "block_id": "p541-b8", - "global_id": 15658, - "bbox": [ - 127.6, - 352.96, - 516.17, - 422.7 - ], - "text": "The direct form I (DFI) is shown in Fig. 5.8a, the canonic direct form (DFII) is shown in\nFig. 5.8b and the transpose of canonic direct is shown in Fig. 5.8c. The DFII and its transpose are\ncanonic because they require N delays, which is the minimum number needed to implement the\nNth-order LTID transfer function in Eq. (5.29). In contrast, the form DFI is a noncanonic because\nit generally requires 2N delays. The DFII realization in Fig. 5.8b is also called a canonic direct\nform.", - "type": "text" - }, - { - "block_id": "p541-b9", - "global_id": 15659, - "bbox": [ - 102.51, - 451.54, - 445.75, - 463.5 - ], - "text": "EXAMPLE 5.8\nCanonical Realizations of Transfer Functions", - "type": "text" - }, - { - "block_id": "p541-b10", - "global_id": 15660, - "bbox": [ - 128.9, - 477.25, - 502.73, - 487.21 - ], - "text": "Find the canonic direct and the transposed canonic direct realizations of the following transfer", - "type": "text" - }, - { - "block_id": "p541-b11", - "global_id": 15661, - "bbox": [ - 128.9, - 486.77, - 258.0, - 511.2 - ], - "text": "functions: (a)\n2\nz + 5, (b) 4z + 28", - "type": "text" - }, - { - "block_id": "p541-b12", - "global_id": 15662, - "bbox": [ - 233.3, - 486.77, - 384.69, - 511.2 - ], - "text": "z + 1 , (c)\nz\nz + 7, and (d)\n4z + 28\nz2 + 6z + 5.", - "type": "text" - }, - { - "block_id": "p541-b13", - "global_id": 15663, - "bbox": [ - 128.9, - 526.63, - 422.07, - 537.01 - ], - "text": "All four of these transfer functions are special cases of H[z] in Eq. (5.29).", - "type": "text" - }, - { - "block_id": "p541-b14", - "global_id": 15664, - "bbox": [ - 146.84, - 538.91, - 158.46, - 548.88 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p541-b15", - "global_id": 15665, - "bbox": [ - 289.76, - 548.51, - 340.71, - 572.53 - ], - "text": "H[z] =\n2\nz + 5", - "type": "text" - }, - { - "block_id": "p541-b16", - "global_id": 15666, - "bbox": [ - 128.9, - 577.76, - 502.76, - 600.09 - ], - "text": "For this case, the transfer function is of the first order (N = 1); therefore, we need only one\ndelay for its realization. The feedback and feedforward coefficients are", - "type": "text" - }, - { - "block_id": "p541-b17", - "global_id": 15667, - "bbox": [ - 243.77, - 611.63, - 387.9, - 623.08 - ], - "text": "a1 = 5\nand\nb0 = 0,\nb1 = 2", - "type": "text" - } - ] - }, - { - "page_num": 542, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p542-b0", - "global_id": 15668, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "522\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p542-b1", - "global_id": 15669, - "bbox": [ - 103.16, - 85.82, - 477.03, - 132.07 - ], - "text": "We use Fig. 5.8 as our model and reduce it to the case of N = 1. Figure 5.9a shows the canonic\ndirect (DFII) form, and Fig. 5.9b its transpose. The two realizations are almost the same. The\nminor difference is that in the DFII form, the gain 2 is provided at the output, and in the\ntranspose, the same gain is provided at the input.", - "type": "text" - }, - { - "block_id": "p542-b2", - "global_id": 15670, - "bbox": [ - 153.08, - 235.73, - 283.59, - 243.73 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p542-b3", - "global_id": 15671, - "bbox": [ - 106.15, - 159.88, - 119.48, - 167.96 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p542-b4", - "global_id": 15672, - "bbox": [ - 192.39, - 204.13, - 205.76, - 212.21 - ], - "text": "Y[z]", - "type": "text" - }, - { - "block_id": "p542-b5", - "global_id": 15673, - "bbox": [ - 228.1, - 214.5, - 241.43, - 222.58 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p542-b6", - "global_id": 15674, - "bbox": [ - 318.18, - 152.04, - 331.55, - 160.12 - ], - "text": "Y[z]", - "type": "text" - }, - { - "block_id": "p542-b8", - "global_id": 15675, - "bbox": [ - 146.16, - 212.74, - 307.71, - 229.15 - ], - "text": "5\n2\n\n2\n5", - "type": "text" - }, - { - "block_id": "p542-b9", - "global_id": 15676, - "bbox": [ - 175.11, - 186.23, - 179.11, - 201.17 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p542-b10", - "global_id": 15677, - "bbox": [ - 276.86, - 177.46, - 280.86, - 192.31 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p542-b11", - "global_id": 15678, - "bbox": [ - 346.12, - 199.16, - 473.04, - 244.99 - ], - "text": "Figure 5.9 Realization\nof transfer function 2/(z + 5):\n(a) canonic direct form and (b)\nits transpose.", - "type": "text" - }, - { - "block_id": "p542-b12", - "global_id": 15679, - "bbox": [ - 121.09, - 263.29, - 362.99, - 285.13 - ], - "text": "In a similar way, we realize the remaining transfer functions.\n(b)", - "type": "text" - }, - { - "block_id": "p542-b13", - "global_id": 15680, - "bbox": [ - 259.04, - 284.35, - 319.93, - 301.71 - ], - "text": "H[z] = 4z + 28", - "type": "text" - }, - { - "block_id": "p542-b14", - "global_id": 15681, - "bbox": [ - 295.23, - 298.4, - 314.96, - 308.78 - ], - "text": "z + 1", - "type": "text" - }, - { - "block_id": "p542-b15", - "global_id": 15682, - "bbox": [ - 103.16, - 314.0, - 477.02, - 336.34 - ], - "text": "In this case also, the transfer function is of the first order (N = 1); therefore, we need only one\ndelay for its realization. The feedback and feedforward coefficients are", - "type": "text" - }, - { - "block_id": "p542-b16", - "global_id": 15683, - "bbox": [ - 214.98, - 347.88, - 365.2, - 359.33 - ], - "text": "a1 = 1\nand\nb0 = 4,\nb1 = 28", - "type": "text" - }, - { - "block_id": "p542-b17", - "global_id": 15684, - "bbox": [ - 103.17, - 369.61, - 388.65, - 380.17 - ], - "text": "Figure 5.10 illustrates the canonic direct and its transpose for this case.†", - "type": "text" - }, - { - "block_id": "p542-b18", - "global_id": 15685, - "bbox": [ - 177.15, - 491.0, - 356.28, - 499.0 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p542-b19", - "global_id": 15686, - "bbox": [ - 107.38, - 399.94, - 257.01, - 408.93 - ], - "text": "X[z]\nY[z]", - "type": "text" - }, - { - "block_id": "p542-b20", - "global_id": 15687, - "bbox": [ - 153.07, - 459.19, - 163.73, - 467.49 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p542-b22", - "global_id": 15688, - "bbox": [ - 200.45, - 459.49, - 208.45, - 467.49 - ], - "text": "28", - "type": "text" - }, - { - "block_id": "p542-b23", - "global_id": 15689, - "bbox": [ - 196.45, - 399.94, - 424.0, - 409.3 - ], - "text": "4\nX[z]\nY[z]", - "type": "text" - }, - { - "block_id": "p542-b24", - "global_id": 15690, - "bbox": [ - 318.77, - 475.52, - 326.77, - 483.52 - ], - "text": "28", - "type": "text" - }, - { - "block_id": "p542-b26", - "global_id": 15691, - "bbox": [ - 369.2, - 475.22, - 379.87, - 483.52 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p542-b27", - "global_id": 15692, - "bbox": [ - 320.89, - 401.3, - 324.89, - 409.3 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p542-b29", - "global_id": 15693, - "bbox": [ - 179.63, - 427.1, - 353.53, - 450.14 - ], - "text": "1\nz\n1\nz", - "type": "text" - }, - { - "block_id": "p542-b30", - "global_id": 15694, - "bbox": [ - 103.16, - 508.02, - 433.19, - 517.63 - ], - "text": "Figure 5.10 Realization of (4z + 28)/(z + 1): (a) canonic direct form and (b) its transpose.", - "type": "text" - }, - { - "block_id": "p542-b31", - "global_id": 15695, - "bbox": [ - 103.16, - 547.41, - 477.02, - 570.94 - ], - "text": "† Transfer functions with N = M may also be expressed as a sum of a constant and a strictly proper\ntransfer function. For example,", - "type": "text" - }, - { - "block_id": "p542-b32", - "global_id": 15696, - "bbox": [ - 239.48, - 570.65, - 294.42, - 586.28 - ], - "text": "H[z] = 4z + 28", - "type": "text" - }, - { - "block_id": "p542-b33", - "global_id": 15697, - "bbox": [ - 272.18, - 571.02, - 335.11, - 592.64 - ], - "text": "z + 1\n= 4 + 24", - "type": "text" - }, - { - "block_id": "p542-b34", - "global_id": 15698, - "bbox": [ - 321.75, - 583.3, - 339.51, - 592.64 - ], - "text": "z + 1", - "type": "text" - }, - { - "block_id": "p542-b35", - "global_id": 15699, - "bbox": [ - 103.16, - 598.52, - 409.08, - 607.48 - ], - "text": "Hence, this transfer function can also be realized as two transfer functions in parallel.", - "type": "text" - } - ] - }, - { - "page_num": 543, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p543-b0", - "global_id": 15700, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "5.4\nSystem Realization\n523", - "type": "text" - }, - { - "block_id": "p543-b1", - "global_id": 15701, - "bbox": [ - 146.84, - 86.28, - 157.9, - 96.25 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p543-b2", - "global_id": 15702, - "bbox": [ - 289.76, - 93.38, - 340.71, - 117.5 - ], - "text": "H[z] =\nz\nz + 7", - "type": "text" - }, - { - "block_id": "p543-b3", - "global_id": 15703, - "bbox": [ - 128.9, - 122.72, - 502.74, - 145.05 - ], - "text": "Here N = 1 and b0 = 1,b1 = 0 and a1 = 7. Figure 5.11 shows the direct and the transposed\nrealizations. Observe that the realizations are almost alike.", - "type": "text" - }, - { - "block_id": "p543-b4", - "global_id": 15704, - "bbox": [ - 133.12, - 182.22, - 347.07, - 190.3 - ], - "text": "X[z]\nY[z]\nX[z]\nY[z]", - "type": "text" - }, - { - "block_id": "p543-b5", - "global_id": 15705, - "bbox": [ - 180.78, - 261.74, - 306.28, - 269.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p543-b7", - "global_id": 15706, - "bbox": [ - 178.8, - 240.93, - 189.47, - 249.23 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p543-b9", - "global_id": 15707, - "bbox": [ - 294.95, - 248.0, - 305.62, - 256.29 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p543-b10", - "global_id": 15708, - "bbox": [ - 205.36, - 208.97, - 281.51, - 235.01 - ], - "text": "1\nz\n1\nz", - "type": "text" - }, - { - "block_id": "p543-b11", - "global_id": 15709, - "bbox": [ - 128.9, - 278.76, - 429.01, - 288.37 - ], - "text": "Figure 5.11 Realization of z/(z + 7): (a) canonic direct form and (b) its transpose.", - "type": "text" - }, - { - "block_id": "p543-b12", - "global_id": 15710, - "bbox": [ - 146.84, - 328.11, - 159.02, - 338.08 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p543-b13", - "global_id": 15711, - "bbox": [ - 277.91, - 337.3, - 352.55, - 361.73 - ], - "text": "H[z] =\n4z + 28\nz2 + 6z + 5", - "type": "text" - }, - { - "block_id": "p543-b14", - "global_id": 15712, - "bbox": [ - 128.91, - 366.95, - 502.75, - 389.28 - ], - "text": "This is a second-order system (N = 2) with b0 = 0, b1 = 4, b2 = 28, a1 = 6, a2 = 5. Figure 5.12\nshows the canonic direct and transposed canonic direct realizations.", - "type": "text" - }, - { - "block_id": "p543-b15", - "global_id": 15713, - "bbox": [ - 133.12, - 440.61, - 146.45, - 448.69 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p543-b16", - "global_id": 15714, - "bbox": [ - 266.28, - 483.52, - 321.73, - 491.95 - ], - "text": "Y[z]\nX[z]", - "type": "text" - }, - { - "block_id": "p543-b17", - "global_id": 15715, - "bbox": [ - 426.43, - 428.61, - 439.8, - 436.7 - ], - "text": "Y[z]", - "type": "text" - }, - { - "block_id": "p543-b19", - "global_id": 15716, - "bbox": [ - 159.62, - 490.89, - 254.46, - 508.18 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p543-b20", - "global_id": 15717, - "bbox": [ - 178.62, - 543.68, - 234.18, - 551.98 - ], - "text": "28\n5", - "type": "text" - }, - { - "block_id": "p543-b21", - "global_id": 15718, - "bbox": [ - 225.05, - 490.23, - 406.24, - 508.18 - ], - "text": "4\n6", - "type": "text" - }, - { - "block_id": "p543-b22", - "global_id": 15719, - "bbox": [ - 348.34, - 552.18, - 403.9, - 560.48 - ], - "text": "28\n5", - "type": "text" - }, - { - "block_id": "p543-b23", - "global_id": 15720, - "bbox": [ - 347.0, - 499.41, - 351.0, - 507.41 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p543-b24", - "global_id": 15721, - "bbox": [ - 202.8, - 566.03, - 380.33, - 574.03 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p543-b25", - "global_id": 15722, - "bbox": [ - 373.56, - 519.11, - 377.56, - 534.15 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p543-b26", - "global_id": 15723, - "bbox": [ - 205.0, - 455.36, - 377.39, - 481.9 - ], - "text": "1\nz\n1\nz", - "type": "text" - }, - { - "block_id": "p543-b27", - "global_id": 15724, - "bbox": [ - 204.99, - 510.61, - 208.99, - 525.65 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p543-b28", - "global_id": 15725, - "bbox": [ - 128.9, - 579.8, - 480.43, - 592.66 - ], - "text": "Figure 5.12 Realization of (4z + 28)/(z2 + 6z + 5): (a) canonic direct form and (b) its transpose.", - "type": "text" - } - ] - }, - { - "page_num": 544, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p544-b0", - "global_id": 15726, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "524\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p544-b1", - "global_id": 15727, - "bbox": [ - 107.82, - 97.81, - 453.45, - 109.76 - ], - "text": "DRILL 5.16\nRealization of a Second-Order Transfer Function", - "type": "text" - }, - { - "block_id": "p544-b2", - "global_id": 15728, - "bbox": [ - 107.82, - 118.88, - 220.98, - 128.84 - ], - "text": "Realize the transfer function", - "type": "text" - }, - { - "block_id": "p544-b3", - "global_id": 15729, - "bbox": [ - 255.7, - 126.67, - 335.32, - 150.79 - ], - "text": "H[z] =\n2z\nz2 + 6z + 25", - "type": "text" - }, - { - "block_id": "p544-b4", - "global_id": 15730, - "bbox": [ - 101.84, - 184.03, - 490.41, - 257.97 - ], - "text": "REALIZATION OF FINITE IMPULSE RESPONSE (FIR) FILTERS\nSo far we have been quite general in our development of realization techniques. They can be\napplied to infinite impulse response (IIR) or FIR filters. For FIR filters, the coefficients ai = 0 for\nall i̸ = 0.† Hence, FIR filters can be readily implemented by means of the schemes developed so\nfar by eliminating all branches with ai coefficients. The condition ai = 0 implies that all the poles\nof a FIR filter are at z = 0.", - "type": "text" - }, - { - "block_id": "p544-b5", - "global_id": 15731, - "bbox": [ - 76.77, - 289.36, - 323.06, - 301.32 - ], - "text": "EXAMPLE 5.9\nRealization of an FIR Filter", - "type": "text" - }, - { - "block_id": "p544-b6", - "global_id": 15732, - "bbox": [ - 103.16, - 311.03, - 425.62, - 325.03 - ], - "text": "Realize H[z] = (z3 + 4z2 + 5z + 2)/z3 using canonic direct and transposed forms.", - "type": "text" - }, - { - "block_id": "p544-b7", - "global_id": 15733, - "bbox": [ - 103.16, - 347.53, - 196.16, - 357.9 - ], - "text": "We can express H[z] as", - "type": "text" - }, - { - "block_id": "p544-b8", - "global_id": 15734, - "bbox": [ - 240.31, - 355.43, - 338.66, - 376.31 - ], - "text": "H[z] = z3 + 4z2 + 5z + 2", - "type": "text" - }, - { - "block_id": "p544-b9", - "global_id": 15735, - "bbox": [ - 301.17, - 372.91, - 308.54, - 383.38 - ], - "text": "z3", - "type": "text" - }, - { - "block_id": "p544-b10", - "global_id": 15736, - "bbox": [ - 226.48, - 468.7, - 235.36, - 476.7 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p544-b11", - "global_id": 15737, - "bbox": [ - 226.1, - 565.01, - 235.74, - 573.01 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p544-b12", - "global_id": 15738, - "bbox": [ - 113.15, - 516.76, - 126.47, - 524.84 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p544-b13", - "global_id": 15739, - "bbox": [ - 167.91, - 489.31, - 181.44, - 497.39 - ], - "text": "Y[z]", - "type": "text" - }, - { - "block_id": "p544-b15", - "global_id": 15740, - "bbox": [ - 210.61, - 544.77, - 332.58, - 552.77 - ], - "text": "5\n2\n4", - "type": "text" - }, - { - "block_id": "p544-b16", - "global_id": 15741, - "bbox": [ - 135.56, - 438.65, - 331.99, - 455.76 - ], - "text": "X[z]\nY[z]", - "type": "text" - }, - { - "block_id": "p544-b18", - "global_id": 15742, - "bbox": [ - 201.0, - 436.32, - 284.3, - 448.07 - ], - "text": "5\n2\n4", - "type": "text" - }, - { - "block_id": "p544-b19", - "global_id": 15743, - "bbox": [ - 185.23, - 522.11, - 315.48, - 537.35 - ], - "text": "1\nz\n1\nz\n1\nz", - "type": "text" - }, - { - "block_id": "p544-b20", - "global_id": 15744, - "bbox": [ - 175.02, - 444.0, - 179.02, - 459.04 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p544-b21", - "global_id": 15745, - "bbox": [ - 220.2, - 444.07, - 224.2, - 459.11 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p544-b22", - "global_id": 15746, - "bbox": [ - 264.95, - 444.07, - 268.95, - 459.11 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p544-b23", - "global_id": 15747, - "bbox": [ - 349.12, - 552.35, - 455.34, - 574.27 - ], - "text": "Figure 5.13 Realization of\n(z3 + 4z2 + 5z + 2)/z3.", - "type": "text" - }, - { - "block_id": "p544-b24", - "global_id": 15748, - "bbox": [ - 101.84, - 621.19, - 353.04, - 634.75 - ], - "text": "† This statement is true for all i̸ = 0 because a0 is assumed to be unity.", - "type": "text" - } - ] - }, - { - "page_num": 545, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p545-b0", - "global_id": 15749, - "bbox": [ - 377.65, - 62.89, - 516.13, - 71.98 - ], - "text": "5.4\nSystem Realization\n525", - "type": "text" - }, - { - "block_id": "p545-b1", - "global_id": 15750, - "bbox": [ - 128.9, - 85.83, - 502.77, - 144.02 - ], - "text": "For H[z], b0 = 1, b1 = 4, b2 = 5, and b3 = 2. Hence, we obtain the canonic direct realization,\nshown in Fig. 5.13a. We have shown the horizontal orientation because it is easier to see\nthat this filter is basically a tapped delay line. That is why this structure is also known as\na tapped delay line or transversal filter. Figure 5.13b shows the corresponding transposed\nimplementation.", - "type": "text" - }, - { - "block_id": "p545-b2", - "global_id": 15751, - "bbox": [ - 127.59, - 182.82, - 516.13, - 234.83 - ], - "text": "CASCADE AND PARALLEL REALIZATIONS,\nCOMPLEX AND REPEATED POLES\nThe considerations and observations for cascade and parallel realizations as well as complex and\nmultiple poles are identical to those discussed for LTIC systems in Sec. 4.6-3.", - "type": "text" - }, - { - "block_id": "p545-b3", - "global_id": 15752, - "bbox": [ - 133.57, - 269.28, - 469.57, - 295.17 - ], - "text": "DRILL 5.17\nCascade and Parallel Realizations of a Transfer\nFunction", - "type": "text" - }, - { - "block_id": "p545-b4", - "global_id": 15753, - "bbox": [ - 133.57, - 304.3, - 510.12, - 326.22 - ], - "text": "Find canonic direct realizations of the following transfer function by using the cascade and\nparallel forms. The specific cascade decomposition is as follows:", - "type": "text" - }, - { - "block_id": "p545-b5", - "global_id": 15754, - "bbox": [ - 239.39, - 336.42, - 330.03, - 360.85 - ], - "text": "H[z] =\nz + 3\nz2 + 7z + 10 =", - "type": "text" - }, - { - "block_id": "p545-b6", - "global_id": 15755, - "bbox": [ - 332.08, - 329.42, - 359.73, - 346.79 - ], - "text": "z + 3", - "type": "text" - }, - { - "block_id": "p545-b7", - "global_id": 15756, - "bbox": [ - 340.0, - 350.47, - 359.73, - 360.85 - ], - "text": "z + 2", - "type": "text" - }, - { - "block_id": "p545-b8", - "global_id": 15757, - "bbox": [ - 360.93, - 329.41, - 396.4, - 360.85 - ], - "text": "1\nz + 5", - "type": "text" - }, - { - "block_id": "p545-b10", - "global_id": 15758, - "bbox": [ - 127.59, - 393.48, - 516.15, - 491.32 - ], - "text": "DO ALL REALIZATIONS LEAD TO THE SAME PERFORMANCE?\nFor a given transfer function, we have presented here several possible different realizations (DFI,\ncanonic form DFII, and its transpose). There are also cascade and parallel versions, and there\nare many possible grouping of the factors in the numerator and the denominator of H[z], leading\nto different realizations. We can also use various combinations of these forms in implementing\ndifferent subsections of a system. Moreover, the transpose of each version doubles the number.\nHowever, this discussion by no means exhausts all the possibilities. Transforming variables affords\nlimitless potential realizations of the same transfer function.", - "type": "text" - }, - { - "block_id": "p545-b11", - "global_id": 15759, - "bbox": [ - 127.59, - 493.32, - 516.16, - 634.79 - ], - "text": "Theoretically, all these realizations are equivalent; that is, they lead to the same transfer\nfunction. This, however, is true only when we implement them with infinite precision. In practice,\nfinite wordlength restriction causes each realization to behave differently in terms of sensitivity\nto parameter variation, stability, frequency response distortion error, and so on. These effects\nare serious for higher-order transfer functions, which require correspondingly higher numbers of\ndelay elements. The finite wordlength errors that plague these implementations are coefficient\nquantization, overflow errors, and round-off errors. From a practical viewpoint, parallel and\ncascade forms using low-order filters minimize the effects of finite wordlength. Parallel and certain\ncascade forms are numerically less sensitive than the canonic direct form to small parameter\nvariations in the system. In the canonic direct form structure with large N, a small change in a\nfilter coefficient due to parameter quantization results in a large change in the location of the\npoles and the zeros of the system. Qualitatively, this difference can be explained by the fact that", - "type": "text" - } - ] - }, - { - "page_num": 546, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p546-b0", - "global_id": 15760, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "526\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p546-b1", - "global_id": 15761, - "bbox": [ - 101.84, - 85.82, - 490.42, - 215.34 - ], - "text": "in a direct form (or its transpose), all the coefficients interact with each other, and a change in\nany coefficient will be magnified through its repeated influence from feedback and feedforward\nconnections. In a parallel realization, in contrast, a change in a coefficient will affect only a\nlocalized segment; the case of a cascade realization is similar. For this reason, the most popular\ntechnique for minimizing finite wordlength effects is to design filters by using cascade or parallel\nforms employing low-order filters. In practice, high-order filters are realized by using multiple\nsecond-order sections in cascade, because second-order filters not only are easier to design but\nare less susceptible to coefficient quantization and round-off errors, and their implementations\nallow easier data word scaling to reduce the potential overflow effects of data word-size growth.\nA cascaded system using second-order building blocks usually requires fewer multiplications for\na given filter frequency response [1].", - "type": "text" - }, - { - "block_id": "p546-b2", - "global_id": 15762, - "bbox": [ - 101.85, - 216.91, - 490.4, - 299.02 - ], - "text": "There are several ways to pair the poles and zeros of an Nth-order H[z] into a cascade of\nsecond-order sections, and several ways to order the resulting sections. Quantizing error will be\ndifferent for each combination. Although several papers published provide guidelines in predicting\nand minimizing finite wordlength errors, it is advisable to resort to computer simulation of the filter\ndesign. This way, one can vary filter hardware characteristic, such as coefficient wordlengths,\naccumulator register sizes, sequencing of cascaded sections, and input signal sets. Such an\napproach is both reliable and economical [1].", - "type": "text" - }, - { - "block_id": "p546-b3", - "global_id": 15763, - "bbox": [ - 102.2, - 327.36, - 469.11, - 341.31 - ], - "text": "5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p546-b4", - "global_id": 15764, - "bbox": [ - 101.84, - 347.29, - 490.39, - 369.21 - ], - "text": "For (asymptotically or BIBO-stable) continuous-time systems, we showed that the system\nresponse to an input ejωt is H(jω)ejωt and that the response to an input cos ωt is |H(jω)|cos[ωt +̸", - "type": "text" - }, - { - "block_id": "p546-b5", - "global_id": 15765, - "bbox": [ - 101.84, - 370.79, - 490.38, - 405.07 - ], - "text": "H(jω)]. Similar results hold for discrete-time systems. We now show that for an (asymptotically\nor BIBO-stable) LTID system, the system response to an input ejn is H[ej]ejn and the response\nto an input cos n is |H[ej]|cos(n +̸ H[ej]).", - "type": "text" - }, - { - "block_id": "p546-b6", - "global_id": 15766, - "bbox": [ - 101.84, - 407.07, - 490.4, - 452.9 - ], - "text": "The proof is similar to the one used for continuous-time systems. In Sec. 3.8-2, we showed\nthat an LTID system response to an (everlasting) exponential zn is also an (everlasting) exponential\nH[z]zn. This result is valid only for values of z for which H[z], as defined in Eq. (5.11), exists\n(converges). As usual, we represent this input–output relationship by a directed arrow notation as", - "type": "text" - }, - { - "block_id": "p546-b7", - "global_id": 15767, - "bbox": [ - 268.88, - 458.43, - 490.38, - 472.93 - ], - "text": "zn \r⇒H[z]zn\n(5.30)", - "type": "text" - }, - { - "block_id": "p546-b8", - "global_id": 15768, - "bbox": [ - 101.85, - 479.39, - 263.55, - 492.96 - ], - "text": "Setting z = ej in this relationship yields", - "type": "text" - }, - { - "block_id": "p546-b9", - "global_id": 15769, - "bbox": [ - 256.76, - 498.5, - 490.38, - 512.99 - ], - "text": "ejn \r⇒H[ej]ejn\n(5.31)", - "type": "text" - }, - { - "block_id": "p546-b10", - "global_id": 15770, - "bbox": [ - 101.85, - 521.42, - 363.88, - 533.03 - ], - "text": "Noting that cos n is the real part of ejn, use of Eq. (3.34) yields", - "type": "text" - }, - { - "block_id": "p546-b11", - "global_id": 15771, - "bbox": [ - 240.77, - 540.96, - 490.38, - 553.06 - ], - "text": "cos n \r⇒Re{H[ej]ejn}\n(5.32)", - "type": "text" - }, - { - "block_id": "p546-b12", - "global_id": 15772, - "bbox": [ - 101.85, - 561.49, - 244.91, - 573.09 - ], - "text": "Expressing H[ej] in the polar form", - "type": "text" - }, - { - "block_id": "p546-b13", - "global_id": 15773, - "bbox": [ - 244.58, - 581.15, - 347.16, - 594.88 - ], - "text": "H[ej] = |H[ej]|ej̸\nH[ej]", - "type": "text" - }, - { - "block_id": "p546-b14", - "global_id": 15774, - "bbox": [ - 101.84, - 605.05, - 222.59, - 615.02 - ], - "text": "Eq. (5.32) can be expressed as", - "type": "text" - }, - { - "block_id": "p546-b15", - "global_id": 15775, - "bbox": [ - 216.22, - 622.95, - 376.03, - 635.05 - ], - "text": "cos n \r⇒|H[ej]|cos(n +̸ H[ej])", - "type": "text" - } - ] - }, - { - "page_num": 547, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p547-b0", - "global_id": 15776, - "bbox": [ - 269.55, - 62.89, - 516.12, - 71.98 - ], - "text": "5.5\nFrequency Response of Discrete-Time Systems\n527", - "type": "text" - }, - { - "block_id": "p547-b1", - "global_id": 15777, - "bbox": [ - 127.59, - 85.4, - 448.36, - 95.78 - ], - "text": "In other words, the system response y[n] to a sinusoidal input cos n is given by", - "type": "text" - }, - { - "block_id": "p547-b2", - "global_id": 15778, - "bbox": [ - 252.42, - 105.28, - 391.31, - 117.38 - ], - "text": "y[n] = |H[ej]|cos(n +̸ H[ej])", - "type": "text" - }, - { - "block_id": "p547-b3", - "global_id": 15779, - "bbox": [ - 127.59, - 128.59, - 445.7, - 138.97 - ], - "text": "Following the same argument, the system response to a sinusoid cos(n + θ) is", - "type": "text" - }, - { - "block_id": "p547-b4", - "global_id": 15780, - "bbox": [ - 244.29, - 148.47, - 516.12, - 160.57 - ], - "text": "y[n] = |H[ej]|cos(n + θ +̸ H[ej])\n(5.33)", - "type": "text" - }, - { - "block_id": "p547-b5", - "global_id": 15781, - "bbox": [ - 127.59, - 172.21, - 516.15, - 253.9 - ], - "text": "This result is valid only for BIBO-stable or asymptotically stable systems. The frequency response\nis meaningless for BIBO-unstable systems (which include marginally stable and asymptotically\nunstable systems). This follows from the fact that the frequency response in Eq. (5.31) is obtained\nby setting z = ej in Eq. (5.30). But, as shown in Sec. 3.8-2 [Eqs. (3.38) and (3.39)], the relationship\nof Eq. (5.30) applies only for values of z for which H[z] exists. For BIBO-unstable systems, the\nROC for H[z] does not include the unit circle where z = ej. This means, for BIBO-unstable\nsystems, that H[z] is meaningless when z = ej.†", - "type": "text" - }, - { - "block_id": "p547-b6", - "global_id": 15782, - "bbox": [ - 127.59, - 255.89, - 516.14, - 361.49 - ], - "text": "This important result shows that the response of an asymptotically or BIBO-stable LTID\nsystem to a discrete-time sinusoidal input of frequency is also a discrete-time sinusoid of the\nsame frequency. The amplitude of the output sinusoid is |H[ej]| times the input amplitude, and\nthe phase of the output sinusoid is shifted by̸\nH[ej] with respect to the input phase. Clearly,\n|H[ej]| is the amplitude gain, and a plot of |H[ej]| versus is the amplitude response of the\ndiscrete-time system. Similarly,̸ H[ej] is the phase response of the system, and a plot of̸ H[ej]\nversus shows how the system modifies or shifts the phase of the input sinusoid. Note that\nH[ej] incorporates the information of both amplitude and phase responses and therefore is called\nthe frequency responses of the system.", - "type": "text" - }, - { - "block_id": "p547-b7", - "global_id": 15783, - "bbox": [ - 127.89, - 376.15, - 438.05, - 388.27 - ], - "text": "STEADY-STATE RESPONSE TO CAUSAL SINUSOIDAL INPUT", - "type": "text" - }, - { - "block_id": "p547-b8", - "global_id": 15784, - "bbox": [ - 127.59, - 392.3, - 516.14, - 462.03 - ], - "text": "As in the case of continuous-time systems, we can show that the response of an LTID system to a\ncausal sinusoidal input cos nu[n] is y[n] in Eq. (5.33), plus a natural component consisting of the\ncharacteristic modes (see Prob. 5.5-9). For a stable system, all the modes decay exponentially, and\nonly the sinusoidal component in Eq. (5.33) persists. For this reason, this component is called the\nsinusoidal steady-state response of the system. Thus, yss[n], the steady-state response of a system\nto a causal sinusoidal input cos nu[n], is", - "type": "text" - }, - { - "block_id": "p547-b9", - "global_id": 15785, - "bbox": [ - 241.16, - 471.53, - 402.57, - 484.34 - ], - "text": "yss[n] = |H[ej]|cos(n +̸ H[ej])u[n]", - "type": "text" - }, - { - "block_id": "p547-b10", - "global_id": 15786, - "bbox": [ - 127.89, - 498.28, - 471.21, - 510.41 - ], - "text": "SYSTEM RESPONSE TO SAMPLED CONTINUOUS-TIME SINUSOIDS", - "type": "text" - }, - { - "block_id": "p547-b11", - "global_id": 15787, - "bbox": [ - 127.59, - 514.03, - 516.13, - 548.31 - ], - "text": "So far we have considered the response of a discrete-time system to a discrete-time sinusoid cos n\n(or exponential ejn). In practice, the input may be a sampled continuous-time sinusoid cos ωt (or\nan exponential ejωt). When a sinusoid cos ωt is sampled with sampling interval T, the resulting", - "type": "text" - }, - { - "block_id": "p547-b12", - "global_id": 15788, - "bbox": [ - 127.59, - 566.4, - 516.13, - 633.41 - ], - "text": "† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains\nnondecaying natural mode terms of the form cos0n or γ n cos0n (γ > 1). Hence, the response of such\na system to a sinusoid cosn will contain not just the sinusoid of frequency but also nondecaying natural\nmodes, rendering the concept of frequency response meaningless. Alternately, we can argue that when z = ej,\na BIBO-unstable system violates the dominance condition |γi| < |ej| for all i, where γi represents ith\ncharacteristic root of the system.", - "type": "text" - } - ] - }, - { - "page_num": 548, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p548-b0", - "global_id": 15789, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "528\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p548-b1", - "global_id": 15790, - "bbox": [ - 101.84, - 85.4, - 490.38, - 107.74 - ], - "text": "signal is a discrete-time sinusoid cos ωnT, obtained by setting t = nT in cosωt. Therefore, all the\nresults developed in this section apply if we substitute ωT for :", - "type": "text" - }, - { - "block_id": "p548-b2", - "global_id": 15791, - "bbox": [ - 279.79, - 121.01, - 490.37, - 131.38 - ], - "text": "= ωT\n(5.34)", - "type": "text" - }, - { - "block_id": "p548-b3", - "global_id": 15792, - "bbox": [ - 76.77, - 166.57, - 475.59, - 178.52 - ], - "text": "EXAMPLE 5.10\nSinusoidal Response of a Difference Equation System", - "type": "text" - }, - { - "block_id": "p548-b4", - "global_id": 15793, - "bbox": [ - 103.16, - 195.19, - 256.09, - 205.15 - ], - "text": "For a system specified by the equation", - "type": "text" - }, - { - "block_id": "p548-b5", - "global_id": 15794, - "bbox": [ - 232.57, - 216.69, - 347.6, - 227.07 - ], - "text": "y[n + 1] −0.8y[n] = x[n + 1]", - "type": "text" - }, - { - "block_id": "p548-b6", - "global_id": 15795, - "bbox": [ - 103.17, - 239.02, - 252.95, - 248.98 - ], - "text": "find the system response to the inputs", - "type": "text" - }, - { - "block_id": "p548-b7", - "global_id": 15796, - "bbox": [ - 121.1, - 252.93, - 162.4, - 266.92 - ], - "text": "(a) 1n = 1", - "type": "text" - }, - { - "block_id": "p548-b8", - "global_id": 15797, - "bbox": [ - 121.09, - 274.05, - 151.53, - 284.09 - ], - "text": "(b) cos", - "type": "text" - }, - { - "block_id": "p548-b9", - "global_id": 15798, - "bbox": [ - 152.64, - 262.72, - 164.48, - 276.69 - ], - "text": "'π", - "type": "text" - }, - { - "block_id": "p548-b10", - "global_id": 15799, - "bbox": [ - 159.5, - 262.72, - 199.64, - 291.17 - ], - "text": "6 n −0.2\n(", - "type": "text" - }, - { - "block_id": "p548-b11", - "global_id": 15800, - "bbox": [ - 121.65, - 291.79, - 388.24, - 302.16 - ], - "text": "(c) a sampled sinusoid cos1500t with sampling interval T = 0.001", - "type": "text" - }, - { - "block_id": "p548-b12", - "global_id": 15801, - "bbox": [ - 103.16, - 331.06, - 266.25, - 341.02 - ], - "text": "The system equation can be expressed as", - "type": "text" - }, - { - "block_id": "p548-b13", - "global_id": 15802, - "bbox": [ - 246.53, - 352.57, - 333.65, - 362.94 - ], - "text": "(E −0.8)y[n] = Ex[n]", - "type": "text" - }, - { - "block_id": "p548-b14", - "global_id": 15803, - "bbox": [ - 103.17, - 374.89, - 292.97, - 384.86 - ], - "text": "Therefore, the transfer function of the system is", - "type": "text" - }, - { - "block_id": "p548-b15", - "global_id": 15804, - "bbox": [ - 232.35, - 394.69, - 346.11, - 418.81 - ], - "text": "H[z] =\nz\nz −0.8 =\n1\n1 −0.8z−1", - "type": "text" - }, - { - "block_id": "p548-b16", - "global_id": 15805, - "bbox": [ - 103.16, - 427.45, - 207.33, - 437.41 - ], - "text": "The frequency response is", - "type": "text" - }, - { - "block_id": "p548-b17", - "global_id": 15806, - "bbox": [ - 185.1, - 447.34, - 393.88, - 471.36 - ], - "text": "H[ej] =\n1\n1 −0.8e−j =\n1\n(1 −0.8cos ) + j0.8sin", - "type": "text" - }, - { - "block_id": "p548-b18", - "global_id": 15807, - "bbox": [ - 103.17, - 481.18, - 144.91, - 491.15 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p548-b19", - "global_id": 15808, - "bbox": [ - 158.05, - 500.36, - 270.75, - 517.21 - ], - "text": "|H[ej]| =\n1", - "type": "text" - }, - { - "block_id": "p548-b20", - "global_id": 15809, - "bbox": [ - 211.63, - 500.36, - 386.66, - 526.2 - ], - "text": "(1 −0.8cos )2 + (0.8sin )2 =\n1\n√", - "type": "text" - }, - { - "block_id": "p548-b21", - "global_id": 15810, - "bbox": [ - 355.83, - 507.35, - 477.01, - 525.37 - ], - "text": "1.64 −1.6cos \n(5.35)", - "type": "text" - }, - { - "block_id": "p548-b22", - "global_id": 15811, - "bbox": [ - 103.16, - 536.5, - 117.54, - 546.46 - ], - "text": "and̸", - "type": "text" - }, - { - "block_id": "p548-b23", - "global_id": 15812, - "bbox": [ - 224.94, - 549.91, - 293.35, - 562.0 - ], - "text": "H[ej] = −tan−1", - "type": "text" - }, - { - "block_id": "p548-b24", - "global_id": 15813, - "bbox": [ - 294.96, - 537.64, - 354.24, - 569.08 - ], - "text": "0.8sin \n1 −0.8cos", - "type": "text" - }, - { - "block_id": "p548-b25", - "global_id": 15814, - "bbox": [ - 355.43, - 537.64, - 360.86, - 547.6 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p548-b26", - "global_id": 15815, - "bbox": [ - 452.95, - 552.04, - 477.01, - 562.0 - ], - "text": "(5.36)", - "type": "text" - }, - { - "block_id": "p548-b27", - "global_id": 15816, - "bbox": [ - 103.17, - 573.81, - 477.0, - 599.76 - ], - "text": "The amplitude response |H[ej]| can also be obtained by observing that |H|2 = HH∗. Since\nour system is real, we therefore see that", - "type": "text" - }, - { - "block_id": "p548-b28", - "global_id": 15817, - "bbox": [ - 202.19, - 607.18, - 477.01, - 621.67 - ], - "text": "|H[ej]|2 = H[ej]H∗[ej] = H[ej]H[e−j]\n(5.37)", - "type": "text" - } - ] - }, - { - "page_num": 549, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p549-b0", - "global_id": 15818, - "bbox": [ - 269.55, - 62.89, - 516.12, - 71.98 - ], - "text": "5.5\nFrequency Response of Discrete-Time Systems\n529", - "type": "text" - }, - { - "block_id": "p549-b1", - "global_id": 15819, - "bbox": [ - 128.9, - 86.08, - 280.32, - 97.68 - ], - "text": "Substituting for H[ej], it follows that", - "type": "text" - }, - { - "block_id": "p549-b2", - "global_id": 15820, - "bbox": [ - 192.39, - 110.75, - 238.46, - 125.14 - ], - "text": "|H[ej]|2 =", - "type": "text" - }, - { - "block_id": "p549-b3", - "global_id": 15821, - "bbox": [ - 240.5, - 100.88, - 293.93, - 132.32 - ], - "text": "1\n1 −0.8e−j", - "type": "text" - }, - { - "block_id": "p549-b4", - "global_id": 15822, - "bbox": [ - 295.65, - 100.88, - 351.46, - 132.32 - ], - "text": "1\n1 −0.8ej", - "type": "text" - }, - { - "block_id": "p549-b6", - "global_id": 15823, - "bbox": [ - 361.96, - 108.3, - 438.08, - 132.32 - ], - "text": "=\n1\n1.64 −1.6cos", - "type": "text" - }, - { - "block_id": "p549-b7", - "global_id": 15824, - "bbox": [ - 128.9, - 142.49, - 283.29, - 152.45 - ], - "text": "which matches the result found earlier.", - "type": "text" - }, - { - "block_id": "p549-b8", - "global_id": 15825, - "bbox": [ - 128.9, - 154.03, - 502.75, - 176.36 - ], - "text": "Figure 5.14 shows plots of amplitude and phase response as functions of . We now\ncompute the amplitude and the phase response for the various inputs.", - "type": "text" - }, - { - "block_id": "p549-b9", - "global_id": 15826, - "bbox": [ - 165.49, - 300.87, - 485.72, - 311.48 - ], - "text": "p\n2p\n3p\np\n\t\n2p", - "type": "text" - }, - { - "block_id": "p549-b10", - "global_id": 15827, - "bbox": [ - 275.75, - 210.04, - 279.75, - 218.04 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p549-b11", - "global_id": 15828, - "bbox": [ - 178.32, - 222.68, - 205.34, - 232.37 - ], - "text": "H [ej\t]", - "type": "text" - }, - { - "block_id": "p549-b12", - "global_id": 15829, - "bbox": [ - 282.32, - 303.48, - 286.32, - 311.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p549-b14", - "global_id": 15830, - "bbox": [ - 253.13, - 423.26, - 280.45, - 431.56 - ], - "text": "53.13", - "type": "text" - }, - { - "block_id": "p549-b15", - "global_id": 15831, - "bbox": [ - 178.93, - 337.94, - 308.89, - 353.07 - ], - "text": "53.13\nH [e j\t]", - "type": "text" - }, - { - "block_id": "p549-b16", - "global_id": 15832, - "bbox": [ - 154.96, - 385.39, - 453.89, - 395.75 - ], - "text": "p\n2p\n3p\np\n2p\n0", - "type": "text" - }, - { - "block_id": "p549-b17", - "global_id": 15833, - "bbox": [ - 307.84, - 437.14, - 317.48, - 445.14 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p549-b18", - "global_id": 15834, - "bbox": [ - 308.0, - 317.04, - 316.88, - 325.04 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p549-b19", - "global_id": 15835, - "bbox": [ - 136.48, - 451.83, - 332.01, - 461.07 - ], - "text": "Figure 5.14 Frequency response of the LTID system.", - "type": "text" - }, - { - "block_id": "p549-b20", - "global_id": 15836, - "bbox": [ - 128.91, - 484.71, - 502.76, - 510.66 - ], - "text": "(a) Since 1n = (ej)n with = 0, the amplitude response is H[ej0]. From Eq. (5.35) we\nobtain", - "type": "text" - }, - { - "block_id": "p549-b21", - "global_id": 15837, - "bbox": [ - 215.63, - 508.58, - 360.09, - 533.17 - ], - "text": "H[ej0] =\n1\n√1.64 −1.6cos(0) =\n1\n√", - "type": "text" - }, - { - "block_id": "p549-b22", - "global_id": 15838, - "bbox": [ - 353.09, - 515.15, - 416.04, - 533.95 - ], - "text": "0.04\n= 5 = 5̸ 0", - "type": "text" - }, - { - "block_id": "p549-b23", - "global_id": 15839, - "bbox": [ - 128.91, - 539.86, - 170.65, - 549.83 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p549-b24", - "global_id": 15840, - "bbox": [ - 241.58, - 549.9, - 390.09, - 561.78 - ], - "text": "|H[ej0]| = 5\nand̸\nH[ej0] = 0", - "type": "text" - }, - { - "block_id": "p549-b25", - "global_id": 15841, - "bbox": [ - 128.91, - 570.74, - 502.78, - 592.67 - ], - "text": "These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding\nto = 0. Therefore, the system response to input 1 is", - "type": "text" - }, - { - "block_id": "p549-b26", - "global_id": 15842, - "bbox": [ - 257.25, - 602.7, - 374.41, - 614.58 - ], - "text": "y[n] = 5(1n) = 5\nfor all n", - "type": "text" - } - ] - }, - { - "page_num": 550, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p550-b0", - "global_id": 15843, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "530\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p550-b1", - "global_id": 15844, - "bbox": [ - 121.09, - 85.96, - 443.22, - 96.33 - ], - "text": "(b) For x[n] = cos[(π/6)n −0.2], = π/6. According to Eqs. (5.35) and (5.36),", - "type": "text" - }, - { - "block_id": "p550-b2", - "global_id": 15845, - "bbox": [ - 185.29, - 106.27, - 276.47, - 123.11 - ], - "text": "|H[ejπ/6]| =\n1", - "type": "text" - }, - { - "block_id": "p550-b3", - "global_id": 15846, - "bbox": [ - 244.74, - 119.99, - 309.29, - 144.43 - ], - "text": "1.64 −1.6cos π\n6", - "type": "text" - }, - { - "block_id": "p550-b4", - "global_id": 15847, - "bbox": [ - 314.72, - 112.84, - 346.94, - 123.21 - ], - "text": "= 1.983̸", - "type": "text" - }, - { - "block_id": "p550-b5", - "global_id": 15848, - "bbox": [ - 191.0, - 161.99, - 265.27, - 174.09 - ], - "text": "H[ejπ/6] = −tan−1", - "type": "text" - }, - { - "block_id": "p550-b6", - "global_id": 15849, - "bbox": [ - 266.87, - 140.77, - 274.14, - 150.73 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p550-b7", - "global_id": 15850, - "bbox": [ - 266.87, - 158.23, - 274.14, - 174.64 - ], - "text": "⎢⎣", - "type": "text" - }, - { - "block_id": "p550-b8", - "global_id": 15851, - "bbox": [ - 275.33, - 145.46, - 327.41, - 191.65 - ], - "text": "0.8sin π\n6\n1 −0.8cos π\n6", - "type": "text" - }, - { - "block_id": "p550-b9", - "global_id": 15852, - "bbox": [ - 330.8, - 140.77, - 338.06, - 150.73 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p550-b10", - "global_id": 15853, - "bbox": [ - 330.8, - 158.23, - 394.88, - 174.64 - ], - "text": "⎥⎦= −0.916 rad", - "type": "text" - }, - { - "block_id": "p550-b11", - "global_id": 15854, - "bbox": [ - 103.16, - 200.31, - 476.97, - 222.22 - ], - "text": "These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding\nto = π/6. Therefore,", - "type": "text" - }, - { - "block_id": "p550-b12", - "global_id": 15855, - "bbox": [ - 159.55, - 239.26, - 224.24, - 249.64 - ], - "text": "y[n] = 1.983cos", - "type": "text" - }, - { - "block_id": "p550-b13", - "global_id": 15856, - "bbox": [ - 225.37, - 225.28, - 239.26, - 242.24 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p550-b14", - "global_id": 15857, - "bbox": [ - 234.29, - 225.28, - 309.74, - 256.71 - ], - "text": "6 n −0.2 −0.916", - "type": "text" - }, - { - "block_id": "p550-b15", - "global_id": 15858, - "bbox": [ - 311.82, - 239.26, - 358.42, - 249.64 - ], - "text": "= 1.983cos", - "type": "text" - }, - { - "block_id": "p550-b16", - "global_id": 15859, - "bbox": [ - 359.55, - 225.28, - 373.45, - 242.24 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p550-b17", - "global_id": 15860, - "bbox": [ - 368.47, - 225.28, - 420.61, - 256.71 - ], - "text": "6 n −1.116", - "type": "text" - }, - { - "block_id": "p550-b18", - "global_id": 15861, - "bbox": [ - 103.17, - 266.47, - 393.46, - 276.84 - ], - "text": "Figure 5.15 shows the input x[n] and the corresponding system response.", - "type": "text" - }, - { - "block_id": "p550-b19", - "global_id": 15862, - "bbox": [ - 210.47, - 402.74, - 214.47, - 410.74 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p550-b20", - "global_id": 15863, - "bbox": [ - 257.47, - 389.24, - 261.47, - 397.24 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p550-b21", - "global_id": 15864, - "bbox": [ - 308.47, - 405.24, - 316.47, - 413.24 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p550-b22", - "global_id": 15865, - "bbox": [ - 361.97, - 390.74, - 369.97, - 398.74 - ], - "text": "18", - "type": "text" - }, - { - "block_id": "p550-b23", - "global_id": 15866, - "bbox": [ - 415.47, - 404.24, - 423.47, - 412.24 - ], - "text": "24", - "type": "text" - }, - { - "block_id": "p550-b24", - "global_id": 15867, - "bbox": [ - 147.47, - 390.97, - 158.13, - 399.27 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p550-b25", - "global_id": 15868, - "bbox": [ - 101.84, - 405.08, - 116.5, - 413.37 - ], - "text": "11", - "type": "text" - }, - { - "block_id": "p550-b26", - "global_id": 15869, - "bbox": [ - 197.17, - 310.89, - 201.17, - 318.89 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p550-b27", - "global_id": 15870, - "bbox": [ - 275.41, - 368.51, - 288.29, - 376.59 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p550-b28", - "global_id": 15871, - "bbox": [ - 357.22, - 330.95, - 370.1, - 339.03 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p550-b29", - "global_id": 15872, - "bbox": [ - 415.11, - 414.55, - 419.11, - 422.55 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p550-b30", - "global_id": 15873, - "bbox": [ - 101.77, - 499.32, - 391.58, - 508.56 - ], - "text": "Figure 5.15 Sinusoidal input and the corresponding output of the LTID system.", - "type": "text" - }, - { - "block_id": "p550-b31", - "global_id": 15874, - "bbox": [ - 103.16, - 535.82, - 477.03, - 558.15 - ], - "text": "(c) A sinusoid cos 1500t sampled every T seconds (t = nT) results in a discrete-time\nsinusoid", - "type": "text" - }, - { - "block_id": "p550-b32", - "global_id": 15875, - "bbox": [ - 252.77, - 559.72, - 326.62, - 570.1 - ], - "text": "x[n] = cos 1500nT", - "type": "text" - }, - { - "block_id": "p550-b33", - "global_id": 15876, - "bbox": [ - 103.16, - 578.66, - 209.2, - 589.04 - ], - "text": "For T = 0.001, the input is", - "type": "text" - }, - { - "block_id": "p550-b34", - "global_id": 15877, - "bbox": [ - 256.5, - 590.61, - 323.67, - 600.99 - ], - "text": "x[n] = cos(1.5n)", - "type": "text" - } - ] - }, - { - "page_num": 551, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p551-b0", - "global_id": 15878, - "bbox": [ - 269.55, - 62.89, - 516.12, - 71.98 - ], - "text": "5.5\nFrequency Response of Discrete-Time Systems\n531", - "type": "text" - }, - { - "block_id": "p551-b1", - "global_id": 15879, - "bbox": [ - 128.9, - 85.83, - 360.45, - 96.2 - ], - "text": "In this case, = 1.5. According to Eqs. (5.35) and (5.36),", - "type": "text" - }, - { - "block_id": "p551-b2", - "global_id": 15880, - "bbox": [ - 209.45, - 105.35, - 378.1, - 129.93 - ], - "text": "|H[ej1.5]| =\n1\n√1.64 −1.6cos(1.5) = 0.809̸", - "type": "text" - }, - { - "block_id": "p551-b3", - "global_id": 15881, - "bbox": [ - 215.15, - 138.17, - 286.85, - 150.27 - ], - "text": "H[ej1.5] = −tan−1", - "type": "text" - }, - { - "block_id": "p551-b4", - "global_id": 15882, - "bbox": [ - 288.46, - 125.9, - 358.76, - 157.34 - ], - "text": "0.8sin(1.5)\n1 −0.8cos(1.5)", - "type": "text" - }, - { - "block_id": "p551-b5", - "global_id": 15883, - "bbox": [ - 359.96, - 125.9, - 365.39, - 135.87 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p551-b6", - "global_id": 15884, - "bbox": [ - 367.44, - 139.89, - 422.21, - 150.26 - ], - "text": "= −0.702 rad", - "type": "text" - }, - { - "block_id": "p551-b7", - "global_id": 15885, - "bbox": [ - 128.91, - 166.29, - 502.75, - 188.63 - ], - "text": "These values also could be read directly from Fig. 5.14 corresponding to = 1.5.\nTherefore,", - "type": "text" - }, - { - "block_id": "p551-b8", - "global_id": 15886, - "bbox": [ - 253.85, - 190.21, - 377.79, - 200.58 - ], - "text": "y[n] = 0.809cos(1.5n −0.702)", - "type": "text" - }, - { - "block_id": "p551-b9", - "global_id": 15887, - "bbox": [ - 128.9, - 215.12, - 502.78, - 265.14 - ], - "text": "FREQUENCY RESPONSE PLOTS USING MATLAB\nMATLAB makes it easy to compute and plot magnitude and phase responses directly using\na system’s transfer function. As the following code demonstrates, there is no need to derive\nseparate expressions for the magnitude and phase responses.", - "type": "text" - }, - { - "block_id": "p551-b10", - "global_id": 15888, - "bbox": [ - 128.9, - 274.45, - 463.22, - 322.27 - ], - "text": ">>\nOmega = linspace(-pi,pi,400); H = @(z) z./(z-0.8);\n>>\nsubplot(1,2,1); plot(Omega,abs(H(exp(1j*Omega))),’k’); axis tight;\n>>\nxlabel(’\\Omega’); ylabel(’|H[e^{j \\Omega}]|’);\n>>\nsubplot(1,2,2); plot(Omega,angle(H(exp(1j*Omega))*180/pi),’k’); axis tight;\n>>\nxlabel(’\\Omega’); ylabel(’\\angle H[e^{j \\Omega}] [deg]’);", - "type": "text" - }, - { - "block_id": "p551-b11", - "global_id": 15889, - "bbox": [ - 128.9, - 332.09, - 441.68, - 342.06 - ], - "text": "The resulting plots, shown in Fig. 5.16, confirm the earlier results of Fig. 5.14.", - "type": "text" - }, - { - "block_id": "p551-b12", - "global_id": 15890, - "bbox": [ - 244.21, - 435.27, - 247.81, - 442.47 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p551-b13", - "global_id": 15891, - "bbox": [ - 244.21, - 413.69, - 247.81, - 420.89 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p551-b14", - "global_id": 15892, - "bbox": [ - 244.21, - 392.11, - 247.81, - 399.31 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p551-b15", - "global_id": 15893, - "bbox": [ - 244.21, - 370.53, - 247.81, - 377.73 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p551-b16", - "global_id": 15894, - "bbox": [ - 232.32, - 399.64, - 241.05, - 423.43 - ], - "text": "|H[ej Ω]|", - "type": "text" - }, - { - "block_id": "p551-b17", - "global_id": 15895, - "bbox": [ - 250.68, - 453.17, - 402.41, - 470.3 - ], - "text": "–3\n–2\n–1\n0\n1\n2\n3\nΩ", - "type": "text" - }, - { - "block_id": "p551-b18", - "global_id": 15896, - "bbox": [ - 251.31, - 584.73, - 403.56, - 601.86 - ], - "text": "Ω\n–3\n–2\n–1\n0\n1\n2\n3", - "type": "text" - }, - { - "block_id": "p551-b19", - "global_id": 15897, - "bbox": [ - 238.4, - 573.6, - 249.2, - 580.8 - ], - "text": "–50", - "type": "text" - }, - { - "block_id": "p551-b20", - "global_id": 15898, - "bbox": [ - 245.6, - 528.49, - 249.2, - 535.69 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p551-b21", - "global_id": 15899, - "bbox": [ - 241.54, - 483.4, - 248.74, - 490.6 - ], - "text": "50", - "type": "text" - }, - { - "block_id": "p551-b22", - "global_id": 15900, - "bbox": [ - 227.63, - 510.52, - 236.36, - 548.42 - ], - "text": "H[ej Ω] [deg]", - "type": "text" - }, - { - "block_id": "p551-b23", - "global_id": 15901, - "bbox": [ - 128.9, - 612.06, - 416.05, - 621.3 - ], - "text": "Figure 5.16 MATLAB-generated magnitude and phase responses for Ex. 5.10.", - "type": "text" - } - ] - }, - { - "page_num": 552, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p552-b0", - "global_id": 15902, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "532\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p552-b1", - "global_id": 15903, - "bbox": [ - 101.84, - 85.87, - 490.39, - 155.68 - ], - "text": "Comment: Figures 5.14 and 5.16 show amplitude and phase response plots as functions of\n. These plots as well as Eqs. (5.35) and (5.36) indicate that the frequency response of a\ndiscrete-time system is a continuous (rather than discrete) function of frequency . There is no\ncontradiction here. This behavior is merely an indication of the fact that the frequency variable\n is continuous (takes on all possible values) and therefore the system response exists at every\nvalue of .", - "type": "text" - }, - { - "block_id": "p552-b2", - "global_id": 15904, - "bbox": [ - 107.82, - 192.37, - 429.68, - 204.33 - ], - "text": "DRILL 5.18\nFrequency Response of Difference Equation", - "type": "text" - }, - { - "block_id": "p552-b3", - "global_id": 15905, - "bbox": [ - 107.82, - 213.45, - 260.75, - 223.41 - ], - "text": "For a system specified by the equation", - "type": "text" - }, - { - "block_id": "p552-b4", - "global_id": 15906, - "bbox": [ - 246.52, - 234.95, - 345.71, - 245.32 - ], - "text": "y[n + 1] −0.5y[n] = x[n]", - "type": "text" - }, - { - "block_id": "p552-b5", - "global_id": 15907, - "bbox": [ - 107.82, - 257.28, - 484.39, - 279.2 - ], - "text": "find the amplitude and the phase response. Find the system response to sinusoidal input\ncos[1000t −(π/3)] sampled every T = 0.5 ms.", - "type": "text" - }, - { - "block_id": "p552-b6", - "global_id": 15908, - "bbox": [ - 108.09, - 292.72, - 162.68, - 303.68 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p552-b7", - "global_id": 15909, - "bbox": [ - 158.93, - 310.89, - 236.73, - 327.73 - ], - "text": "|H[ej]| =\n1\n√", - "type": "text" - }, - { - "block_id": "p552-b8", - "global_id": 15910, - "bbox": [ - 212.68, - 325.51, - 264.23, - 335.89 - ], - "text": "1.25 −cos ̸", - "type": "text" - }, - { - "block_id": "p552-b9", - "global_id": 15911, - "bbox": [ - 164.63, - 343.71, - 233.03, - 355.81 - ], - "text": "H[ej] = −tan−1", - "type": "text" - }, - { - "block_id": "p552-b10", - "global_id": 15912, - "bbox": [ - 234.64, - 331.44, - 287.83, - 362.88 - ], - "text": "sin \ncos −0.5", - "type": "text" - }, - { - "block_id": "p552-b11", - "global_id": 15913, - "bbox": [ - 289.03, - 331.44, - 294.46, - 341.41 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p552-b12", - "global_id": 15914, - "bbox": [ - 175.15, - 373.33, - 239.84, - 383.71 - ], - "text": "y[n] = 1.639cos", - "type": "text" - }, - { - "block_id": "p552-b14", - "global_id": 15915, - "bbox": [ - 247.69, - 359.34, - 325.35, - 390.77 - ], - "text": "0.5n −π\n3 −0.904", - "type": "text" - }, - { - "block_id": "p552-b15", - "global_id": 15916, - "bbox": [ - 327.42, - 373.33, - 433.27, - 383.71 - ], - "text": "= 1.639cos(0.5n −1.951)", - "type": "text" - }, - { - "block_id": "p552-b16", - "global_id": 15917, - "bbox": [ - 107.82, - 449.55, - 441.2, - 461.5 - ], - "text": "DRILL 5.19\nFrequency Response of an Ideal Delay System", - "type": "text" - }, - { - "block_id": "p552-b17", - "global_id": 15918, - "bbox": [ - 107.82, - 468.98, - 484.42, - 528.4 - ], - "text": "Show that for an ideal delay (H[z] = 1/z), the amplitude response |H[ej]| = 1, and the\nphase response̸\nH[ej] = −. Thus, a pure time delay does not affect the amplitude gain\nof sinusoidal input, but it causes a phase shift (delay) of radians in a discrete sinusoid of\nfrequency . Thus, for an ideal delay, the phase shift of the output sinusoid is proportional to\nthe frequency of the input sinusoid (linear phase shift).", - "type": "text" - }, - { - "block_id": "p552-b18", - "global_id": 15919, - "bbox": [ - 101.84, - 582.83, - 372.33, - 594.78 - ], - "text": "5.5-1 The Periodic Nature of Frequency Response", - "type": "text" - }, - { - "block_id": "p552-b19", - "global_id": 15920, - "bbox": [ - 101.84, - 599.27, - 490.4, - 634.79 - ], - "text": "In Ex. 5.10 and Fig. 5.14, we saw that the frequency response H[ej] is a periodic function of\n. This is not a coincidence. Unlike continuous-time systems, all LTID systems have periodic\nfrequency response. This is seen clearly from the nature of the expression of the frequency", - "type": "text" - } - ] - }, - { - "page_num": 553, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p553-b0", - "global_id": 15921, - "bbox": [ - 269.55, - 62.89, - 516.12, - 71.98 - ], - "text": "5.5\nFrequency Response of Discrete-Time Systems\n533", - "type": "text" - }, - { - "block_id": "p553-b1", - "global_id": 15922, - "bbox": [ - 127.59, - 83.37, - 496.13, - 97.36 - ], - "text": "response of an LTID system. Because e±j2πm = 1 for all integer values of m [see Eq. (B.10)],", - "type": "text" - }, - { - "block_id": "p553-b2", - "global_id": 15923, - "bbox": [ - 248.18, - 107.85, - 293.79, - 119.85 - ], - "text": "H[ej] = H", - "type": "text" - }, - { - "block_id": "p553-b4", - "global_id": 15924, - "bbox": [ - 298.17, - 101.57, - 338.43, - 119.85 - ], - "text": "ej(+2πm)", - "type": "text" - }, - { - "block_id": "p553-b5", - "global_id": 15925, - "bbox": [ - 358.35, - 109.89, - 395.54, - 119.95 - ], - "text": "m integer", - "type": "text" - }, - { - "block_id": "p553-b6", - "global_id": 15926, - "bbox": [ - 127.59, - 130.95, - 516.13, - 166.46 - ], - "text": "Therefore, the frequency response H[ej] is a periodic function of with a period 2π. This is the\nmathematical explanation of the periodic behavior. The physical explanation that follows provides\na much better insight into the periodic behavior.", - "type": "text" - }, - { - "block_id": "p553-b7", - "global_id": 15927, - "bbox": [ - 127.59, - 181.53, - 516.15, - 231.55 - ], - "text": "NON-UNIQUENESS OF DISCRETE-TIME SINUSOID WAVEFORMS\nA continuous-time sinusoid cos ωt has a unique waveform for every real value of ω in the range 0\nto ∞. Increasing ω results in a sinusoid of ever-increasing frequency. Such is not the case for the\ndiscrete-time sinusoid cos n because", - "type": "text" - }, - { - "block_id": "p553-b8", - "global_id": 15928, - "bbox": [ - 237.64, - 243.77, - 406.06, - 254.14 - ], - "text": "cos[( ± 2πm)n] = cos n\nm integer", - "type": "text" - }, - { - "block_id": "p553-b9", - "global_id": 15929, - "bbox": [ - 127.59, - 266.78, - 141.97, - 276.74 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p553-b10", - "global_id": 15930, - "bbox": [ - 259.57, - 275.21, - 384.14, - 289.71 - ], - "text": "ej(±2πm)n = ejn\nm integer", - "type": "text" - }, - { - "block_id": "p553-b11", - "global_id": 15931, - "bbox": [ - 127.59, - 297.71, - 516.14, - 357.13 - ], - "text": "This shows that the discrete-time sinusoids cos n (and exponentials ejn) separated by values of\n in integral multiples of 2π are identical. The reason for the periodic nature of the frequency\nresponse of an LTID system is now clear. Since the sinusoids (or exponentials) with frequencies\nseparated by interval 2π are identical, the system response to such sinusoids is also identical and,\nhence, is periodic with period 2π.", - "type": "text" - }, - { - "block_id": "p553-b12", - "global_id": 15932, - "bbox": [ - 127.59, - 358.71, - 516.13, - 406.04 - ], - "text": "This discussion shows that the discrete-time sinusoid cos n has a unique waveform only\nfor the values of in the range −π to π. This band is called the fundamental band. Every\nfrequency , no matter how large, is identical to some frequency, a, in the fundamental band\n(−π ≤a < π), where", - "type": "text" - }, - { - "block_id": "p553-b13", - "global_id": 15933, - "bbox": [ - 216.55, - 417.17, - 516.12, - 428.63 - ], - "text": "a = −2πm\n−π ≤a < π\nand\nm integer\n(5.38)", - "type": "text" - }, - { - "block_id": "p553-b14", - "global_id": 15934, - "bbox": [ - 127.59, - 440.07, - 516.13, - 486.0 - ], - "text": "The integer m can be positive or negative. We use Eq. (5.38) to plot the fundamental band\nfrequency a versus the frequency of a sinusoid (Fig. 5.17a). The frequency a is modulo\n2π value of .\nAll these conclusions are also valid for exponential ejn.", - "type": "text" - }, - { - "block_id": "p553-b15", - "global_id": 15935, - "bbox": [ - 127.59, - 501.07, - 516.12, - 575.01 - ], - "text": "ALL DISCRETE-TIME SIGNALS ARE INHERENTLY BANDLIMITED\nThis discussion leads to the surprising conclusion that all discrete-time signals are inherently\nbandlimited, with frequencies lying in the range −π to π radians per sample. In terms of frequency\nF = /2π, where F is in cycles per sample, all frequencies F separated by an integer number\nare identical. For instance, all discrete-time sinusoids of frequencies 0.3, 1.3, 2.3, . . . cycles per\nsample are identical. The fundamental range of frequencies is −0.5 to 0.5 cycles per sample.", - "type": "text" - }, - { - "block_id": "p553-b16", - "global_id": 15936, - "bbox": [ - 127.59, - 577.0, - 516.17, - 634.79 - ], - "text": "Any discrete-time sinusoid of frequency beyond the fundamental band, when plotted, appears\nand behaves, in every way, like a sinusoid having its frequency in the fundamental band. It is\nimpossible to distinguish between the two signals. Thus, in a basic sense, discrete-time frequencies\nbeyond || = π or |F| = 1/2 do not exist. Yet, in a “mathematical” sense, we must admit the\nexistence of sinusoids of frequencies beyond = π. What does this mean?", - "type": "text" - } - ] - }, - { - "page_num": 554, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p554-b0", - "global_id": 15937, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "534\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p554-b1", - "global_id": 15938, - "bbox": [ - 304.43, - 247.42, - 353.78, - 255.52 - ], - "text": "2.4p\n1.6p", - "type": "text" - }, - { - "block_id": "p554-b2", - "global_id": 15939, - "bbox": [ - 282.18, - 172.53, - 291.07, - 180.53 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p554-b3", - "global_id": 15940, - "bbox": [ - 281.8, - 269.74, - 291.45, - 277.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p554-b4", - "global_id": 15941, - "bbox": [ - 408.69, - 132.7, - 418.03, - 140.8 - ], - "text": "4p", - "type": "text" - }, - { - "block_id": "p554-b5", - "global_id": 15942, - "bbox": [ - 127.81, - 99.05, - 137.04, - 108.81 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p554-b7", - "global_id": 15943, - "bbox": [ - 125.76, - 206.48, - 139.09, - 216.25 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p554-b8", - "global_id": 15944, - "bbox": [ - 388.04, - 247.42, - 403.37, - 255.52 - ], - "text": "3.6p", - "type": "text" - }, - { - "block_id": "p554-b9", - "global_id": 15945, - "bbox": [ - 155.05, - 132.51, - 374.73, - 140.8 - ], - "text": "3p\n2p\n2p\n0\np\np", - "type": "text" - }, - { - "block_id": "p554-b10", - "global_id": 15946, - "bbox": [ - 231.03, - 154.04, - 243.03, - 162.24 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p554-b11", - "global_id": 15947, - "bbox": [ - 237.69, - 96.64, - 243.03, - 104.64 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p554-b12", - "global_id": 15948, - "bbox": [ - 155.05, - 241.14, - 434.16, - 249.43 - ], - "text": "4p\n\t\n3p\n2p\n2p", - "type": "text" - }, - { - "block_id": "p554-b13", - "global_id": 15949, - "bbox": [ - 142.18, - 223.63, - 157.51, - 231.73 - ], - "text": "0.4p", - "type": "text" - }, - { - "block_id": "p554-b14", - "global_id": 15950, - "bbox": [ - 244.47, - 241.33, - 289.49, - 249.43 - ], - "text": "0\np", - "type": "text" - }, - { - "block_id": "p554-b15", - "global_id": 15951, - "bbox": [ - 152.18, - 205.39, - 157.51, - 213.39 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p554-b16", - "global_id": 15952, - "bbox": [ - 197.99, - 241.14, - 209.99, - 249.33 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p554-b17", - "global_id": 15953, - "bbox": [ - 125.76, - 284.36, - 361.31, - 293.67 - ], - "text": "Figure 5.17 (a) Actual frequency versus (b) apparent frequency.", - "type": "text" - }, - { - "block_id": "p554-b18", - "global_id": 15954, - "bbox": [ - 102.14, - 328.21, - 234.16, - 340.33 - ], - "text": "A MAN NAMED ROBERT", - "type": "text" - }, - { - "block_id": "p554-b19", - "global_id": 15955, - "bbox": [ - 101.84, - 344.37, - 490.41, - 426.06 - ], - "text": "To give an analogy, consider a fictitious person Mr. Robert Thompson. His mother calls him\nRobby; his acquaintances call him Bob, his close friends call him by his nickname, Shorty. Yet,\nRobert, Robby, Bob, and Shorty are one and the same person. However, we cannot say that only\nMr. Robert Thompson exists, or only Robby exists, or only Shorty exists, or only Bob exists. All\nthese four persons exist, although they are one and the same individual. In a same way, we cannot\nsay that the frequency π/2 exists and frequency 5π/2 does not exist; they are both the same entity,\ncalled by different names.", - "type": "text" - }, - { - "block_id": "p554-b20", - "global_id": 15956, - "bbox": [ - 101.84, - 428.05, - 490.39, - 545.62 - ], - "text": "It is in this sense that we have to admit the existence of frequencies beyond the fundamental\nband. Indeed, mathematical expressions in the frequency domain automatically cater to this\nneed by their built-in periodicity. As seen earlier, the very structure of the frequency response\nis 2π-periodic. We shall also see later, in Ch. 9, that discrete-time signal spectra are also\n2π-periodic.\nAdmitting the existence of frequencies beyond π also serves mathematical and computational\nconvenience in digital signal-processing applications. Values of frequencies beyond π may also\noriginate naturally in the process of sampling continuous-time sinusoids. Because there is no upper\nlimit on the value of ω, there is no upper limit on the value of the resulting discrete-time frequency\n = ωT either.†", - "type": "text" - }, - { - "block_id": "p554-b21", - "global_id": 15957, - "bbox": [ - 101.84, - 547.19, - 490.39, - 581.48 - ], - "text": "The highest possible frequency is π and the lowest frequency is 0 (dc or constant). Clearly,\nthe high frequencies are those in the vicinity of = (2m + 1)π and the low frequencies are those\nin the vicinity of = 2πm for all positive or negative integer values of m.", - "type": "text" - }, - { - "block_id": "p554-b22", - "global_id": 15958, - "bbox": [ - 101.84, - 621.19, - 448.59, - 634.38 - ], - "text": "† However, if goes beyond π, the resulting aliasing reduces the apparent frequency to a < π.", - "type": "text" - } - ] - }, - { - "page_num": 555, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p555-b0", - "global_id": 15959, - "bbox": [ - 269.55, - 62.89, - 516.12, - 71.98 - ], - "text": "5.5\nFrequency Response of Discrete-Time Systems\n535", - "type": "text" - }, - { - "block_id": "p555-b1", - "global_id": 15960, - "bbox": [ - 127.89, - 86.19, - 397.28, - 98.32 - ], - "text": "FURTHER REDUCTION IN THE FREQUENCY RANGE", - "type": "text" - }, - { - "block_id": "p555-b2", - "global_id": 15961, - "bbox": [ - 127.59, - 101.93, - 516.14, - 160.13 - ], - "text": "Because cos(−n + θ) = cos(n −θ), a frequency in the range −π to 0 is identical to the\nfrequency (of the same magnitude) in the range 0 to π (but with a change in phase sign).\nConsequently the apparent frequency for a discrete-time sinusoid of any frequency is equal to some\nvalue in the range 0 to π. Thus, cos(8.7πn + θ) = cos(0.7πn + θ), and the apparent frequency is\n0.7π. Similarly,", - "type": "text" - }, - { - "block_id": "p555-b3", - "global_id": 15962, - "bbox": [ - 212.39, - 165.42, - 431.32, - 175.8 - ], - "text": "cos(9.6πn + θ) = cos(−0.4πn + θ) = cos(0.4πn −θ)", - "type": "text" - }, - { - "block_id": "p555-b4", - "global_id": 15963, - "bbox": [ - 127.59, - 186.82, - 516.16, - 292.84 - ], - "text": "Hence, the frequency 9.6π is identical (in every respect) to frequency −0.4π, which, in turn, is\nequal (within the sign of its phase) to frequency 0.4π. In this case, the apparent frequency reduces\nto |a| = 0.4π. We can generalize the result to say that the apparent frequency of a discrete-time\nsinusoid is |a|, as found from Eq. (5.38), and if a <0, there is a phase reversal. Figure 5.17b\nplots versus the apparent frequency |a|. The shaded bands represent the ranges of for\nwhich there is a phase reversal, when represented in terms of |a|. For example, the apparent\nfrequency for both the sinusoids cos(2.4π + θ) and cos(3.6π + θ) is |a| = 0.4π, as seen from\nFig. 5.17b. But 2.4π is in a clear band and 3.6π is in a shaded band. Hence, these sinusoids appear\nas cos(0.4π + θ) and cos(0.4π −θ), respectively.", - "type": "text" - }, - { - "block_id": "p555-b5", - "global_id": 15964, - "bbox": [ - 127.59, - 294.83, - 516.16, - 364.57 - ], - "text": "Although every discrete-time sinusoid can be expressed as having frequency in the range from\n0 to π, we generally use the frequency range from −π to π instead of 0 to π for two reasons. First,\nexponential representation of sinusoids with frequencies in the range 0 to π requires a frequency\nrange −π to π. Second, even when we are using a trigonometric representation, we generally need\nthe frequency range −π to π to have exact identity (without phase reversal) of a higher-frequency\nsinusoid.", - "type": "text" - }, - { - "block_id": "p555-b6", - "global_id": 15965, - "bbox": [ - 127.59, - 366.16, - 516.14, - 412.4 - ], - "text": "For certain practical advantages, in place of the range −π to π, we often use other contiguous\nranges of width 2π. The range 0 to 2π, for instance, is used in many applications. It is left as an\nexercise for the reader to show that the frequencies in the range from π to 2π are identical to those\nin the range from −π to 0.", - "type": "text" - }, - { - "block_id": "p555-b7", - "global_id": 15966, - "bbox": [ - 102.51, - 442.36, - 318.61, - 454.32 - ], - "text": "EXAMPLE 5.11\nApparent Frequency", - "type": "text" - }, - { - "block_id": "p555-b8", - "global_id": 15967, - "bbox": [ - 128.9, - 470.56, - 502.76, - 492.89 - ], - "text": "Express the following signals in terms of their apparent frequencies: (a) cos(0.5πn + θ), (b)\ncos(1.6πn + θ), (c) sin(1.6πn + θ), (d) cos(2.3πn + θ), and (e) cos(34.699n + θ).", - "type": "text" - }, - { - "block_id": "p555-b9", - "global_id": 15968, - "bbox": [ - 128.9, - 515.39, - 502.74, - 538.81 - ], - "text": "(a) = 0.5π is in the reduced range already. This is also apparent from Fig. 5.17a or 5.17b.\nBecause a = 0.5π, there is no phase reversal, and the apparent sinusoid is cos(0.5πn + θ).", - "type": "text" - }, - { - "block_id": "p555-b10", - "global_id": 15969, - "bbox": [ - 128.91, - 539.3, - 502.76, - 573.59 - ], - "text": "(b) We express 1.6π = −0.4π + 2π so that a = −0.4π and |a| = 0.4. Also, a is\nnegative, implying sign change for the phase. Hence, the apparent sinusoid is cos(0.4πn−θ).\nThis fact is also apparent from Fig. 5.17b.", - "type": "text" - }, - { - "block_id": "p555-b11", - "global_id": 15970, - "bbox": [ - 128.9, - 575.17, - 502.76, - 609.46 - ], - "text": "(c) We first convert the sine form to cosine form as sin(1.6πn+θ) = cos(1.6πn −\n(π/2) + θ). In part (b), we found a = −0.4π. Hence, the apparent sinusoid is cos(0.4πn +\n(π/2)−θ) = −sin(0.4πn −θ). In this case, both the phase and the amplitude change signs.", - "type": "text" - }, - { - "block_id": "p555-b12", - "global_id": 15971, - "bbox": [ - 146.84, - 611.03, - 502.75, - 622.49 - ], - "text": "(d) 2.3π = 0.3π +2π so that a = 0.3π. Hence, the apparent sinusoid is cos(0.3πn+θ).", - "type": "text" - } - ] - }, - { - "page_num": 556, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p556-b0", - "global_id": 15972, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "536\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p556-b1", - "global_id": 15973, - "bbox": [ - 103.16, - 85.96, - 477.02, - 120.25 - ], - "text": "(e) We have 34.699 = −3+6(2π). Hence, a = −3, and the apparent frequency |a| = 3\nrad/sample. Because a is negative, there is a sign change of the phase. Hence, the apparent\nsinusoid is cos(3n −θ).", - "type": "text" - }, - { - "block_id": "p556-b2", - "global_id": 15974, - "bbox": [ - 107.82, - 171.37, - 300.75, - 183.32 - ], - "text": "DRILL 5.20\nApparent Frequency", - "type": "text" - }, - { - "block_id": "p556-b3", - "global_id": 15975, - "bbox": [ - 107.82, - 192.03, - 484.42, - 226.32 - ], - "text": "Show that the sinusoids having frequencies of (a) 2π, (b) 3π, (c) 5π, (d) 3.2π, (e) 22.1327,\nand (f) π + 2 can be expressed, respectively, as sinusoids of frequencies (a) 0, (b) π, (c) π,\n(d) 0.8π, (e) 3, and (f) π −2. Show that in cases (d), (e), and (f), phase changes sign.", - "type": "text" - }, - { - "block_id": "p556-b4", - "global_id": 15976, - "bbox": [ - 101.84, - 259.67, - 285.94, - 271.63 - ], - "text": "5.5-2 Aliasing and Sampling Rate", - "type": "text" - }, - { - "block_id": "p556-b5", - "global_id": 15977, - "bbox": [ - 101.84, - 277.76, - 490.4, - 383.37 - ], - "text": "The non-uniqueness of discrete-time sinusoids and the periodic repetition of the same waveforms\nat intervals of 2π may seem innocuous, but in reality it leads to a serious problem for processing\ncontinuous-time signals by digital filters. A continuous-time sinusoid cosωt sampled every T\nseconds (t = nT) results in a discrete-time sinusoid cosωnT, which is cosn with = ωT.\nThe discrete-time sinusoids cosn have unique waveforms only for the values of frequencies\nin the range < π or ωT < π. Therefore, samples of continuous-time sinusoids of two (or\nmore) different frequencies can generate the same discrete-time signal, as shown in Fig. 5.18.\nThis phenomenon is known as aliasing because through sampling, two entirely different analog\nsinusoids take on the same “discrete-time” identity.†", - "type": "text" - }, - { - "block_id": "p556-b6", - "global_id": 15978, - "bbox": [ - 101.84, - 385.35, - 490.4, - 407.28 - ], - "text": "Aliasing causes ambiguity in digital signal processing, which makes it impossible to\ndetermine the true frequency of the sampled signal. Consider, for instance, digitally processing", - "type": "text" - }, - { - "block_id": "p556-b7", - "global_id": 15979, - "bbox": [ - 106.7, - 474.4, - 467.39, - 493.03 - ], - "text": "0\n0.2\n0.4\n0.6\n0.8\n1", - "type": "text" - }, - { - "block_id": "p556-b8", - "global_id": 15980, - "bbox": [ - 469.35, - 487.13, - 471.58, - 495.13 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p556-b9", - "global_id": 15981, - "bbox": [ - 212.81, - 422.31, - 241.19, - 439.41 - ], - "text": "cos 2pt\nf 1 Hz", - "type": "text" - }, - { - "block_id": "p556-b10", - "global_id": 15982, - "bbox": [ - 303.74, - 422.31, - 332.6, - 439.41 - ], - "text": "cos 12pt\nf 6 Hz", - "type": "text" - }, - { - "block_id": "p556-b11", - "global_id": 15983, - "bbox": [ - 228.55, - 527.14, - 259.74, - 536.9 - ], - "text": "fs 5 Hz", - "type": "text" - }, - { - "block_id": "p556-b12", - "global_id": 15984, - "bbox": [ - 101.84, - 544.13, - 282.91, - 553.37 - ], - "text": "Figure 5.18 Demonstration of the aliasing effect.", - "type": "text" - }, - { - "block_id": "p556-b13", - "global_id": 15985, - "bbox": [ - 101.84, - 577.36, - 490.39, - 633.41 - ], - "text": "† Figure 5.18 shows samples of two sinusoids cos12πt and cos2πt taken every 0.2 second. The corresponding\ndiscrete-time frequencies ( = ωT = 0.2ω) are cos2.4π and cos0.4π. The apparent frequency of 2.4π\nis 0.4π, identical to the discrete-time frequency corresponding to the lower sinusoid. This shows that the\nsamples of both these continuous-time sinusoids at 0.2-second intervals are identical, as verified from\nFig. 5.18.", - "type": "text" - } - ] - }, - { - "page_num": 557, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p557-b0", - "global_id": 15986, - "bbox": [ - 269.55, - 62.89, - 516.12, - 71.98 - ], - "text": "5.5\nFrequency Response of Discrete-Time Systems\n537", - "type": "text" - }, - { - "block_id": "p557-b1", - "global_id": 15987, - "bbox": [ - 127.59, - 85.4, - 516.13, - 131.64 - ], - "text": "a continuous-time signal that contains two distinct components of frequencies ω1 and ω2. The\nsamples of these components appear as discrete-time sinusoids of frequencies 1 = ω1T and\n2 = ω2T. If 1 and 2 happen to differ by an integer multiple of 2π (if ω2 −ω1 = 2kπ/T),\nthe two frequencies will be read as the same (lower of the two) frequency by the digital processor.‡", - "type": "text" - }, - { - "block_id": "p557-b2", - "global_id": 15988, - "bbox": [ - 127.59, - 133.23, - 516.15, - 228.78 - ], - "text": "As a result, the higher-frequency component ω2 not only is lost for good (by losing its identity to\nω1), but also it reincarnates as a component of frequency ω1, thus distorting the true amplitude of\nthe original component of frequency ω1. Hence, the resulting processed signal will be distorted.\nClearly, aliasing is highly undesirable and should be avoided. To avoid aliasing, the frequencies of\nthe continuous-time sinusoids to be processed should be kept within the fundamental band ωT ≤π\nor ω ≤π/T. Under this condition the question of ambiguity or aliasing does not arise because any\ncontinuous-time sinusoid of frequency in this range has a unique waveform when it is sampled.\nTherefore, if ωh is the highest frequency to be processed, then, to avoid aliasing,", - "type": "text" - }, - { - "block_id": "p557-b3", - "global_id": 15989, - "bbox": [ - 305.99, - 234.76, - 335.54, - 253.2 - ], - "text": "ωh < π", - "type": "text" - }, - { - "block_id": "p557-b4", - "global_id": 15990, - "bbox": [ - 329.89, - 249.13, - 335.43, - 259.09 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p557-b5", - "global_id": 15991, - "bbox": [ - 127.59, - 266.9, - 423.88, - 278.78 - ], - "text": "If fh is the highest frequency in hertz, fh = ωh/2π, and we avoid aliasing if", - "type": "text" - }, - { - "block_id": "p557-b6", - "global_id": 15992, - "bbox": [ - 265.48, - 287.56, - 293.44, - 305.59 - ], - "text": "fh < 1", - "type": "text" - }, - { - "block_id": "p557-b7", - "global_id": 15993, - "bbox": [ - 285.3, - 287.56, - 373.67, - 311.58 - ], - "text": "2T\nor\nT < 1", - "type": "text" - }, - { - "block_id": "p557-b8", - "global_id": 15994, - "bbox": [ - 365.3, - 294.54, - 516.13, - 312.28 - ], - "text": "2fh\n(5.39)", - "type": "text" - }, - { - "block_id": "p557-b9", - "global_id": 15995, - "bbox": [ - 127.59, - 321.65, - 516.13, - 369.07 - ], - "text": "This shows that discrete-time signal processing places the limit on the highest frequency fh that\ncan be processed for a given value of the sampling interval T. Fortunately, we can process a signal\nof any frequency (without aliasing) by choosing a suitably small value of T. Since the sampling\nfrequency fs is the reciprocal of the sampling interval T, we can also express Eq. (5.39) as", - "type": "text" - }, - { - "block_id": "p557-b10", - "global_id": 15996, - "bbox": [ - 259.22, - 377.86, - 283.9, - 395.88 - ], - "text": "fs = 1", - "type": "text" - }, - { - "block_id": "p557-b11", - "global_id": 15997, - "bbox": [ - 278.26, - 377.76, - 382.81, - 401.78 - ], - "text": "T > 2fh\nor\nfh < fs", - "type": "text" - }, - { - "block_id": "p557-b12", - "global_id": 15998, - "bbox": [ - 377.83, - 384.84, - 516.12, - 401.88 - ], - "text": "2\n(5.40)", - "type": "text" - }, - { - "block_id": "p557-b13", - "global_id": 15999, - "bbox": [ - 127.59, - 409.9, - 516.14, - 467.79 - ], - "text": "This result is a special case of the well-known sampling theorem (to be proved in Ch. 8). It states\nthat for a discrete-time system to process a continuous-time sinusoid, the sampling rate must be\ngreater than twice the frequency (in hertz) of the sinusoid. In short, a sampled sinusoid must have\na minimum of two samples per cycle.† For sampling rates below this minimum value, the output\nsignal will be aliased, which means it will be mistaken for a sinusoid of lower frequency.", - "type": "text" - }, - { - "block_id": "p557-b14", - "global_id": 16000, - "bbox": [ - 127.59, - 482.7, - 516.13, - 580.55 - ], - "text": "ANTI-ALIASING FILTER\nIf the sampling rate fails to satisfy Eq. (5.40), aliasing occurs, causing the frequencies beyond fs/2\nHz to masquerade as lower frequencies to corrupt the spectrum at frequencies below fs/2. To avoid\nsuch a corruption, a signal to be sampled is passed through an anti-aliasing filter of bandwidth fs/2\nprior to sampling. This operation ensures the condition of Eq. (5.40). The drawback of such a filter\nis that we lose the spectral components of the signal beyond frequency fs/2, which is preferable\nto the aliasing corruption of the signal at frequencies below fs/2. Chapter 8 presents a detailed\nanalysis of the aliasing problem.", - "type": "text" - }, - { - "block_id": "p557-b15", - "global_id": 16001, - "bbox": [ - 127.59, - 598.98, - 516.13, - 633.41 - ], - "text": "‡ In the case shown in Fig. 5.18, ω1 = 12π, ω2 = 2π, and T = 0.2. Hence, ω2 −ω1 = 10πT = 2π, and the\ntwo frequencies are read as the same frequency = 0.4π by the digital processor.\n† Strictly speaking, we must have more than two samples per cycle.", - "type": "text" - } - ] - }, - { - "page_num": 558, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p558-b0", - "global_id": 16002, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "538\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p558-b1", - "global_id": 16003, - "bbox": [ - 76.77, - 93.92, - 339.68, - 105.87 - ], - "text": "EXAMPLE 5.12\nMaximum Sampling Interval", - "type": "text" - }, - { - "block_id": "p558-b2", - "global_id": 16004, - "bbox": [ - 103.16, - 122.43, - 477.01, - 144.45 - ], - "text": "Determine the maximum sampling interval T that can be used in a discrete-time oscillator that\ngenerates a sinusoid of 50 kHz.", - "type": "text" - }, - { - "block_id": "p558-b3", - "global_id": 16005, - "bbox": [ - 103.16, - 166.96, - 416.34, - 178.41 - ], - "text": "Here the highest significant frequency fh = 50 kHz. Therefore from Eq. (5.39),", - "type": "text" - }, - { - "block_id": "p558-b4", - "global_id": 16006, - "bbox": [ - 257.14, - 187.27, - 284.86, - 204.21 - ], - "text": "T < 1", - "type": "text" - }, - { - "block_id": "p558-b5", - "global_id": 16007, - "bbox": [ - 276.5, - 193.84, - 323.03, - 211.99 - ], - "text": "2fh\n= 10µs", - "type": "text" - }, - { - "block_id": "p558-b6", - "global_id": 16008, - "bbox": [ - 103.17, - 220.69, - 477.0, - 232.15 - ], - "text": "The sampling interval must be less than 10 µs. The sampling frequency is fs = 1/T > 100 kHz.", - "type": "text" - }, - { - "block_id": "p558-b7", - "global_id": 16009, - "bbox": [ - 76.77, - 293.26, - 394.66, - 305.22 - ], - "text": "EXAMPLE 5.13\nMaximum Frequency Without Aliasing", - "type": "text" - }, - { - "block_id": "p558-b8", - "global_id": 16010, - "bbox": [ - 103.16, - 321.46, - 477.02, - 343.8 - ], - "text": "A discrete-time amplifier uses a sampling interval T = 25µs. What is the highest frequency of\na signal that can be processed with this amplifier without aliasing?", - "type": "text" - }, - { - "block_id": "p558-b9", - "global_id": 16011, - "bbox": [ - 103.16, - 366.72, - 167.35, - 376.68 - ], - "text": "From Eq. (5.39)", - "type": "text" - }, - { - "block_id": "p558-b10", - "global_id": 16012, - "bbox": [ - 253.71, - 376.65, - 281.66, - 394.68 - ], - "text": "fh < 1", - "type": "text" - }, - { - "block_id": "p558-b11", - "global_id": 16013, - "bbox": [ - 273.52, - 383.22, - 326.47, - 400.67 - ], - "text": "2T = 20 kHz", - "type": "text" - }, - { - "block_id": "p558-b12", - "global_id": 16014, - "bbox": [ - 102.2, - 451.65, - 477.42, - 465.6 - ], - "text": "5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS", - "type": "text" - }, - { - "block_id": "p558-b13", - "global_id": 16015, - "bbox": [ - 101.84, - 471.58, - 490.43, - 529.36 - ], - "text": "The frequency responses (amplitude and phase responses) of a system are determined by pole-zero\nlocations of the transfer function H[z]. Just as in continuous-time systems, it is possible to\ndetermine quickly the amplitude and the phase response and to obtain physical insight into the filter\ncharacteristics of a discrete-time system by using a graphical technique. The general Nth-order\ntransfer function H[z] in Eq. (5.26) can be expressed in factored form as", - "type": "text" - }, - { - "block_id": "p558-b14", - "global_id": 16016, - "bbox": [ - 220.25, - 546.83, - 258.74, - 557.98 - ], - "text": "H[z] = b0", - "type": "text" - }, - { - "block_id": "p558-b15", - "global_id": 16017, - "bbox": [ - 260.43, - 539.85, - 370.79, - 565.05 - ], - "text": "(z −z1)(z −z2)· · ·(z −zN)\n(z −γ1)(z −γ2)· · ·(z −γN)", - "type": "text" - }, - { - "block_id": "p558-b16", - "global_id": 16018, - "bbox": [ - 101.84, - 574.6, - 490.39, - 645.52 - ], - "text": "We can compute H[z] graphically by using the concepts discussed in Sec. 4.10. The directed line\nsegment from zi to z in the complex plane (Fig. 5.19a) represents the complex number z −zi. The\nlength of this segment is |z −zi| and its angle with the horizontal axis is̸\n(z −zi).\nTo compute the frequency response H[ej] we evaluate H[z] at z = ej. But for z = ej, |z| = 1\nand̸ z = so that z = ej represents a point on the unit circle at an angle with the horizontal. We\nnow connect all zeros (z1, z2,. . .,zN) and all poles (γ1, γ2, . . . , γN) to the point ej, as indicated in", - "type": "text" - } - ] - }, - { - "page_num": 559, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p559-b0", - "global_id": 16019, - "bbox": [ - 268.69, - 62.89, - 516.14, - 71.98 - ], - "text": "5.6\nFrequency Response from Pole-Zero Locations\n539", - "type": "text" - }, - { - "block_id": "p559-b1", - "global_id": 16020, - "bbox": [ - 259.28, - 113.14, - 281.72, - 121.22 - ], - "text": "z plane", - "type": "text" - }, - { - "block_id": "p559-b2", - "global_id": 16021, - "bbox": [ - 252.75, - 194.31, - 255.86, - 202.31 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p559-b3", - "global_id": 16022, - "bbox": [ - 191.02, - 136.14, - 195.8, - 145.68 - ], - "text": "zi", - "type": "text" - }, - { - "block_id": "p559-b4", - "global_id": 16023, - "bbox": [ - 188.15, - 307.74, - 389.03, - 315.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p559-b5", - "global_id": 16024, - "bbox": [ - 135.15, - 147.32, - 144.04, - 155.32 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p559-b6", - "global_id": 16025, - "bbox": [ - 372.08, - 100.69, - 380.97, - 108.69 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p559-b7", - "global_id": 16026, - "bbox": [ - 232.96, - 242.95, - 241.85, - 250.95 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p559-b8", - "global_id": 16027, - "bbox": [ - 476.75, - 193.39, - 485.64, - 201.39 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p559-b9", - "global_id": 16028, - "bbox": [ - 162.13, - 178.05, - 173.58, - 187.81 - ], - "text": "zi", - "type": "text" - }, - { - "block_id": "p559-b10", - "global_id": 16029, - "bbox": [ - 228.13, - 164.58, - 246.69, - 174.34 - ], - "text": "z zi", - "type": "text" - }, - { - "block_id": "p559-b11", - "global_id": 16030, - "bbox": [ - 205.18, - 217.74, - 208.29, - 225.74 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p559-b12", - "global_id": 16031, - "bbox": [ - 280.71, - 202.78, - 481.21, - 212.39 - ], - "text": "1\n1\nz2", - "type": "text" - }, - { - "block_id": "p559-b13", - "global_id": 16032, - "bbox": [ - 367.08, - 185.9, - 374.08, - 195.51 - ], - "text": "d1", - "type": "text" - }, - { - "block_id": "p559-b14", - "global_id": 16033, - "bbox": [ - 399.71, - 136.78, - 406.71, - 146.39 - ], - "text": "d2", - "type": "text" - }, - { - "block_id": "p559-b15", - "global_id": 16034, - "bbox": [ - 393.83, - 161.15, - 399.95, - 170.76 - ], - "text": "r2", - "type": "text" - }, - { - "block_id": "p559-b16", - "global_id": 16035, - "bbox": [ - 446.56, - 203.28, - 452.67, - 212.88 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p559-b17", - "global_id": 16036, - "bbox": [ - 446.47, - 123.06, - 457.31, - 132.67 - ], - "text": "e j", - "type": "text" - }, - { - "block_id": "p559-b18", - "global_id": 16037, - "bbox": [ - 448.41, - 171.15, - 454.52, - 180.76 - ], - "text": "r1", - "type": "text" - }, - { - "block_id": "p559-b19", - "global_id": 16038, - "bbox": [ - 332.8, - 191.2, - 403.99, - 200.82 - ], - "text": "f2", - "type": "text" - }, - { - "block_id": "p559-b20", - "global_id": 16039, - "bbox": [ - 358.49, - 165.51, - 365.49, - 175.14 - ], - "text": "u1", - "type": "text" - }, - { - "block_id": "p559-b21", - "global_id": 16040, - "bbox": [ - 455.36, - 189.7, - 463.7, - 199.33 - ], - "text": "f1", - "type": "text" - }, - { - "block_id": "p559-b22", - "global_id": 16041, - "bbox": [ - 328.78, - 218.23, - 360.1, - 235.32 - ], - "text": "u2\ng2", - "type": "text" - }, - { - "block_id": "p559-b23", - "global_id": 16042, - "bbox": [ - 329.16, - 170.57, - 336.61, - 180.2 - ], - "text": "g1", - "type": "text" - }, - { - "block_id": "p559-b24", - "global_id": 16043, - "bbox": [ - 142.82, - 242.38, - 146.82, - 250.38 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p559-b25", - "global_id": 16044, - "bbox": [ - 127.59, - 322.06, - 431.56, - 331.67 - ], - "text": "Figure 5.19 Vector representations of (a) complex numbers and (b) factors of H[z].", - "type": "text" - }, - { - "block_id": "p559-b26", - "global_id": 16045, - "bbox": [ - 127.59, - 352.85, - 516.13, - 388.63 - ], - "text": "Fig. 5.19b. Let r1, r2, . . . , rN be the lengths and φ1, φ2, . . . , φN be the angles, respectively, of the\nstraight lines connecting z1, z2, . . . , zN to the point ej. Similarly, let d1, d2, . . . , dN be the lengths\nand θ1, θ2, . . . , θN be the angles, respectively, of the lines connecting γ1, γ2, . . . , γN to ej. Then", - "type": "text" - }, - { - "block_id": "p559-b27", - "global_id": 16046, - "bbox": [ - 188.73, - 404.48, - 288.24, - 418.13 - ], - "text": "H[ej] = H[z]|z=ej = b0", - "type": "text" - }, - { - "block_id": "p559-b28", - "global_id": 16047, - "bbox": [ - 289.93, - 398.0, - 397.6, - 424.42 - ], - "text": "(r1ejφ1)(r2ejφ2)· · ·(rNejφN)\n(d1ejθ1)(d2ejθ2)· · ·(dNejθN)", - "type": "text" - }, - { - "block_id": "p559-b29", - "global_id": 16048, - "bbox": [ - 269.96, - 430.37, - 288.24, - 441.52 - ], - "text": "= b0", - "type": "text" - }, - { - "block_id": "p559-b30", - "global_id": 16049, - "bbox": [ - 289.93, - 423.39, - 332.78, - 448.9 - ], - "text": "r1r2 · · ·rN\nd1d2 · · ·dN", - "type": "text" - }, - { - "block_id": "p559-b31", - "global_id": 16050, - "bbox": [ - 334.9, - 428.65, - 454.5, - 440.65 - ], - "text": "ej[(φ1+φ2+···+φN)−(θ1+θ2+···+θN)]", - "type": "text" - }, - { - "block_id": "p559-b32", - "global_id": 16051, - "bbox": [ - 127.59, - 457.85, - 244.42, - 469.31 - ], - "text": "Therefore (assuming b0 > 0),", - "type": "text" - }, - { - "block_id": "p559-b33", - "global_id": 16052, - "bbox": [ - 183.51, - 485.57, - 236.1, - 498.44 - ], - "text": "|H[ej]| = b0", - "type": "text" - }, - { - "block_id": "p559-b34", - "global_id": 16053, - "bbox": [ - 237.8, - 480.3, - 280.65, - 505.82 - ], - "text": "r1r2 · · ·rN\nd1d2 · · ·dN", - "type": "text" - }, - { - "block_id": "p559-b35", - "global_id": 16054, - "bbox": [ - 284.8, - 487.29, - 303.09, - 498.44 - ], - "text": "= b0", - "type": "text" - }, - { - "block_id": "p559-b36", - "global_id": 16055, - "bbox": [ - 304.79, - 479.09, - 458.5, - 490.68 - ], - "text": "product of the distances of zeros to ej", - "type": "text" - }, - { - "block_id": "p559-b37", - "global_id": 16056, - "bbox": [ - 312.12, - 487.7, - 516.13, - 505.04 - ], - "text": "product of distances of poles to ej\n(5.41)", - "type": "text" - }, - { - "block_id": "p559-b38", - "global_id": 16057, - "bbox": [ - 127.59, - 516.06, - 141.97, - 526.02 - ], - "text": "and̸", - "type": "text" - }, - { - "block_id": "p559-b39", - "global_id": 16058, - "bbox": [ - 200.02, - 536.92, - 413.02, - 550.1 - ], - "text": "H[ej] = (φ1 + φ2 + · · · + φN) −(θ1 + θ2 + · · · + θN)", - "type": "text" - }, - { - "block_id": "p559-b40", - "global_id": 16059, - "bbox": [ - 228.64, - 549.97, - 448.82, - 563.97 - ], - "text": "= sum of zero angles to ej −sum of pole angles to ej", - "type": "text" - }, - { - "block_id": "p559-b41", - "global_id": 16060, - "bbox": [ - 127.59, - 575.36, - 516.14, - 634.79 - ], - "text": "In this manner, we can compute the frequency response H[ej] for any value of by selecting the\npoint on the unit circle at an angle . This point is ej. To compute the frequency response H[ej],\nwe connect all poles and zeros to this point and use the foregoing equations to determine |H[ej]|\nand̸\nH[ej]. We repeat this procedure for all values of from 0 to π to obtain the frequency\nresponse.", - "type": "text" - } - ] - }, - { - "page_num": 560, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p560-b0", - "global_id": 16061, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "540\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p560-b1", - "global_id": 16062, - "bbox": [ - 102.14, - 86.19, - 422.18, - 98.32 - ], - "text": "CONTROLLING GAIN BY PLACEMENT OF POLES AND ZEROS", - "type": "text" - }, - { - "block_id": "p560-b2", - "global_id": 16063, - "bbox": [ - 101.84, - 102.35, - 490.41, - 231.86 - ], - "text": "The nature of the influence of pole and zero locations on the frequency response is similar to that\nobserved in continuous-time systems, with minor differences. In place of the imaginary axis of the\ncontinuous-time systems, we have the unit circle in the discrete-time case. The nearer the pole (or\nzero) is to a point ej (on the unit circle) representing some frequency , the more influence that\npole (or zero) wields on the amplitude response at that frequency because the length of the vector\njoining that pole (or zero) to the point ej is small. The proximity of a pole (or a zero) has a similar\neffect on the phase response. From Eq. (5.41), it is clear that to enhance the amplitude response at\na frequency , we should place a pole as close as possible to the point ej (which is on the unit\ncircle).† Similarly, to suppress the amplitude response at a frequency , we should place a zero\nas close as possible to the point ej on the unit circle. Placing repeated poles or zeros will further\nenhance their influence.", - "type": "text" - }, - { - "block_id": "p560-b3", - "global_id": 16064, - "bbox": [ - 101.84, - 233.86, - 490.38, - 267.72 - ], - "text": "Total suppression of signal transmission at any frequency can be achieved by placing a zero\non the unit circle at a point corresponding to that frequency. This observation is used in the notch\n(bandstop) filter design.", - "type": "text" - }, - { - "block_id": "p560-b4", - "global_id": 16065, - "bbox": [ - 101.84, - 269.72, - 490.39, - 339.45 - ], - "text": "Placing a pole or a zero at the origin does not influence the amplitude response because the\nlength of the vector connecting the origin to any point on the unit circle is unity. However, a pole\n(or a zero) at the origin adds angle − (or ) to̸\nH[ej]. Hence, the phase spectrum − (or )\nis a linear function of frequency and therefore represents a pure time delay (or time advance) of\nT seconds (see Drill 5.19). Therefore, a pole (a zero) at the origin causes a time delay (or a time\nadvance) of T seconds in the response. There is no change in the amplitude response.", - "type": "text" - }, - { - "block_id": "p560-b5", - "global_id": 16066, - "bbox": [ - 101.85, - 341.45, - 490.42, - 399.23 - ], - "text": "For a stable system, all the poles must be located inside the unit circle. The zeros may lie\nanywhere. Also, for a physically realizable system, H[z] must be a proper fraction, that is, N ≥M.\nIf, to achieve a certain amplitude response, we require M > N, we can still make the system\nrealizable by placing a sufficient number of poles at the origin to make N = M. This will not\nchange the amplitude response, but it will increase the time delay of the response.", - "type": "text" - }, - { - "block_id": "p560-b6", - "global_id": 16067, - "bbox": [ - 101.85, - 401.23, - 490.38, - 423.15 - ], - "text": "In general, a pole at a point has the opposite effect of a zero at that point. Placing a zero closer\nto a pole tends to cancel the effect of that pole on the frequency response.", - "type": "text" - }, - { - "block_id": "p560-b7", - "global_id": 16068, - "bbox": [ - 101.84, - 437.38, - 490.4, - 559.13 - ], - "text": "LOWPASS FILTERS\nA lowpass filter generally has a maximum gain at or near = 0, which corresponds to point\nej0 = 1 on the unit circle. Clearly, placing a pole inside the unit circle near the point z = 1\n(Fig. 5.20a) would result in a lowpass response.‡ The corresponding amplitude and phase response\nappear in Fig. 5.20a. For smaller values of , the point ej (a point on the unit circle at an angle\n) is closer to the pole, and consequently the gain is higher. As increases, the distance of\nthe point ej from the pole increases. Consequently the gain decreases, resulting in a lowpass\ncharacteristic. Placing a zero at the origin does not change the amplitude response but it does\nmodify the phase response, as illustrated in Fig. 5.20b. Placing a zero at z = −1, however, changes\nboth the amplitude and the phase response (Fig. 5.20c). The point z = −1 corresponds to frequency", - "type": "text" - }, - { - "block_id": "p560-b8", - "global_id": 16069, - "bbox": [ - 101.84, - 577.22, - 490.4, - 633.58 - ], - "text": "† The closest we can place a pole is on the unit circle at the point representing . This choice would lead to\ninfinite gain, but should be avoided because it will render the system marginally stable (BIBO-unstable). The\ncloser the point to the unit circle, the more sensitive the system gain to parameter variations.\n‡ Placing the pole at z = 1 results in maximum (infinite) gain but renders the system BIBO-unstable, hence\nshould be avoided.", - "type": "text" - } - ] - }, - { - "page_num": 561, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p561-b0", - "global_id": 16070, - "bbox": [ - 268.69, - 62.89, - 516.14, - 71.98 - ], - "text": "5.6\nFrequency Response from Pole-Zero Locations\n541", - "type": "text" - }, - { - "block_id": "p561-b1", - "global_id": 16071, - "bbox": [ - 301.5, - 163.74, - 310.38, - 171.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p561-b2", - "global_id": 16072, - "bbox": [ - 301.5, - 256.74, - 311.15, - 264.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p561-b3", - "global_id": 16073, - "bbox": [ - 301.5, - 374.74, - 310.38, - 382.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p561-b4", - "global_id": 16074, - "bbox": [ - 301.5, - 472.74, - 310.83, - 480.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p561-b5", - "global_id": 16075, - "bbox": [ - 301.5, - 587.74, - 310.38, - 595.74 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p561-b6", - "global_id": 16076, - "bbox": [ - 424.51, - 501.74, - 440.5, - 509.74 - ], - "text": "Ideal", - "type": "text" - }, - { - "block_id": "p561-b7", - "global_id": 16077, - "bbox": [ - 422.14, - 408.29, - 438.14, - 416.29 - ], - "text": "Ideal", - "type": "text" - }, - { - "block_id": "p561-b8", - "global_id": 16078, - "bbox": [ - 380.97, - 89.45, - 391.26, - 97.66 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b9", - "global_id": 16079, - "bbox": [ - 434.27, - 163.59, - 447.19, - 171.81 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b10", - "global_id": 16080, - "bbox": [ - 355.22, - 138.56, - 474.55, - 158.61 - ], - "text": "(vT)\n0\np", - "type": "text" - }, - { - "block_id": "p561-b11", - "global_id": 16081, - "bbox": [ - 347.22, - 155.72, - 359.22, - 163.91 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p561-b12", - "global_id": 16082, - "bbox": [ - 337.23, - 355.95, - 356.09, - 364.25 - ], - "text": "p/2", - "type": "text" - }, - { - "block_id": "p561-b13", - "global_id": 16083, - "bbox": [ - 380.97, - 187.45, - 391.26, - 195.66 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b14", - "global_id": 16084, - "bbox": [ - 413.34, - 253.79, - 426.26, - 262.01 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b15", - "global_id": 16085, - "bbox": [ - 354.63, - 236.56, - 474.39, - 256.61 - ], - "text": "(vT)\n0\np", - "type": "text" - }, - { - "block_id": "p561-b16", - "global_id": 16086, - "bbox": [ - 380.97, - 295.45, - 391.26, - 303.66 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b17", - "global_id": 16087, - "bbox": [ - 383.6, - 391.29, - 393.89, - 399.51 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b18", - "global_id": 16088, - "bbox": [ - 379.29, - 509.17, - 389.58, - 517.38 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b19", - "global_id": 16089, - "bbox": [ - 434.3, - 361.79, - 447.22, - 370.01 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b20", - "global_id": 16090, - "bbox": [ - 354.63, - 344.56, - 474.39, - 364.61 - ], - "text": "(vT)\n0\np", - "type": "text" - }, - { - "block_id": "p561-b21", - "global_id": 16091, - "bbox": [ - 333.22, - 456.96, - 447.22, - 471.01 - ], - "text": "3p/2\nH", - "type": "text" - }, - { - "block_id": "p561-b22", - "global_id": 16092, - "bbox": [ - 354.63, - 445.56, - 474.39, - 465.61 - ], - "text": "(vT)\n0\np", - "type": "text" - }, - { - "block_id": "p561-b23", - "global_id": 16093, - "bbox": [ - 340.89, - 528.91, - 357.09, - 537.01 - ], - "text": "3p/2", - "type": "text" - }, - { - "block_id": "p561-b24", - "global_id": 16094, - "bbox": [ - 420.55, - 577.79, - 433.47, - 586.01 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p561-b25", - "global_id": 16095, - "bbox": [ - 354.63, - 560.56, - 474.39, - 580.61 - ], - "text": "(vT)\n0\np", - "type": "text" - }, - { - "block_id": "p561-b26", - "global_id": 16096, - "bbox": [ - 151.5, - 602.44, - 478.12, - 611.68 - ], - "text": "Figure 5.20 Various pole-zero configurations and the corresponding frequency responses.", - "type": "text" - } - ] - }, - { - "page_num": 562, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p562-b0", - "global_id": 16097, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "542\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p562-b1", - "global_id": 16098, - "bbox": [ - 101.84, - 83.27, - 490.39, - 168.99 - ], - "text": "= π (z = ej = ejπ = −1). Consequently, the amplitude response now becomes more attenuated\nat higher frequencies, with a zero gain at = π. We can approach ideal lowpass characteristics\nby using more poles staggered near z = 1 (but within the unit circle). Figure 5.20d shows a\nthird-order lowpass filter with three poles near z = 1 and a third-order zero at z = −1, with\ncorresponding amplitude and phase response. For an ideal lowpass filter, we need an enhanced\ngain at every frequency in the band (0, c). This can be achieved by placing a continuous wall of\npoles (requiring an infinite number of poles) opposite this band.", - "type": "text" - }, - { - "block_id": "p562-b2", - "global_id": 16099, - "bbox": [ - 102.14, - 183.87, - 202.38, - 195.99 - ], - "text": "HIGHPASS FILTERS", - "type": "text" - }, - { - "block_id": "p562-b3", - "global_id": 16100, - "bbox": [ - 101.84, - 200.03, - 490.4, - 257.81 - ], - "text": "A highpass filter has a small gain at lower frequencies and a high gain at higher frequencies. Such\na characteristic can be realized by placing a pole or poles near z = −1 because we want the gain\nat = π to be the highest. Placing a zero at z = 1 further enhances suppression of gain at lower\nfrequencies. Figure 5.20e shows a possible pole-zero configuration of the third-order highpass\nfilter with corresponding amplitude and phase responses.", - "type": "text" - }, - { - "block_id": "p562-b4", - "global_id": 16101, - "bbox": [ - 101.85, - 259.8, - 490.38, - 305.63 - ], - "text": "In the following two examples, we shall realize analog filters by using digital processors and\nsuitable interface devices (C/D and D/C), as shown in Fig. 3.2. At this point, we shall examine the\ndesign of a digital processor with the transfer function H[z] for the purpose of realizing bandpass\nand bandstop filters in the following examples.", - "type": "text" - }, - { - "block_id": "p562-b5", - "global_id": 16102, - "bbox": [ - 101.85, - 307.21, - 490.39, - 353.45 - ], - "text": "As Fig. 3.2 shows, the C/D device samples the continuous-time input x(t) to yield a\ndiscrete-time signal x[n], which serves as the input to H[z]. The output y[n] of H[z] is converted\nto a continuous-time signal y(t) by a D/C device. We also saw in Eq. (5.34) that a continuous-time\nsinusoid of frequency ω, when sampled, results in a discrete-time sinusoid = ωT.", - "type": "text" - }, - { - "block_id": "p562-b6", - "global_id": 16103, - "bbox": [ - 76.77, - 376.61, - 398.1, - 388.57 - ], - "text": "EXAMPLE 5.14\nBandpass Filter by Pole-Zero Placement", - "type": "text" - }, - { - "block_id": "p562-b7", - "global_id": 16104, - "bbox": [ - 103.16, - 405.24, - 477.02, - 428.23 - ], - "text": "By trial and error, design a tuned (bandpass) analog filter with zero transmission at 0 Hz and\nalso at the highest frequency fh = 500 Hz. The resonant frequency is required to be 125 Hz.", - "type": "text" - }, - { - "block_id": "p562-b8", - "global_id": 16105, - "bbox": [ - 103.16, - 446.45, - 477.02, - 567.62 - ], - "text": "Because fh = 500, we require T < 1/1000 [see Eq. (5.39)]. Let us select T = 10−3.† Recall that\nthe analog frequencies ω correspond to digital frequencies = ωT. Hence, analog frequencies\nω = 0 and 1000π correspond to = 0 and π, respectively. The gain is required to be zero at\nthese frequencies. Hence, we need to place zeros at ej corresponding to = 0 and = π. For\n = 0, z = ej = 1; for = π, ej = −1. Hence, there must be zeros at z = ±1. Moreover, we\nneed enhanced response at the resonant frequency ω = 250π, which corresponds to = π/4,\nwhich, in turn, corresponds to z = ej = ejπ/4. Therefore, to enhance the frequency response\nat ω = 250π, we place a pole in the vicinity of ejπ/4. Because this is a complex pole, we also\nneed its conjugate near e−jπ/4, as indicated in Fig. 5.21a. Let us choose these poles γ1 and γ2\nas", - "type": "text" - }, - { - "block_id": "p562-b9", - "global_id": 16106, - "bbox": [ - 208.95, - 567.48, - 370.73, - 580.65 - ], - "text": "γ1 = |γ |ejπ/4\nand\nγ2 = |γ |e−jπ/4", - "type": "text" - }, - { - "block_id": "p562-b10", - "global_id": 16107, - "bbox": [ - 101.84, - 611.23, - 490.4, - 645.37 - ], - "text": "† Strictly speaking, we need T < 0.001. However, we shall show in Ch. 8 that if the input does not contain\na finite amplitude component of 500 Hz, T = 0.001 is adequate. Generally, practical signals satisfy this\ncondition.", - "type": "text" - } - ] - }, - { - "page_num": 563, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p563-b0", - "global_id": 16108, - "bbox": [ - 268.69, - 62.89, - 516.14, - 71.98 - ], - "text": "5.6\nFrequency Response from Pole-Zero Locations\n543", - "type": "text" - }, - { - "block_id": "p563-b1", - "global_id": 16109, - "bbox": [ - 128.9, - 85.83, - 502.76, - 108.15 - ], - "text": "where |γ | < 1 for stability. The closer γ is to the unit circle, the more sharply peaked is the\nresponse around ω = 250π. We also have zeros at ±1. Hence,", - "type": "text" - }, - { - "block_id": "p563-b2", - "global_id": 16110, - "bbox": [ - 192.23, - 115.64, - 412.36, - 143.59 - ], - "text": "H[z] = K\n(z −1)(z + 1)\n(z −|γ |ejπ/4)(z −|γ |e−jπ/4) = K\nz2 −1", - "type": "text" - }, - { - "block_id": "p563-b3", - "global_id": 16111, - "bbox": [ - 362.79, - 131.84, - 379.96, - 145.0 - ], - "text": "z2 −", - "type": "text" - }, - { - "block_id": "p563-b4", - "global_id": 16112, - "bbox": [ - 381.51, - 126.3, - 389.94, - 136.26 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p563-b5", - "global_id": 16113, - "bbox": [ - 389.94, - 134.52, - 437.76, - 145.1 - ], - "text": "2|γ |z + |γ |2", - "type": "text" - }, - { - "block_id": "p563-b6", - "global_id": 16114, - "bbox": [ - 128.9, - 154.51, - 434.91, - 164.89 - ], - "text": "For convenience, we shall choose K = 1. The amplitude response is given by", - "type": "text" - }, - { - "block_id": "p563-b7", - "global_id": 16115, - "bbox": [ - 230.55, - 172.38, - 399.93, - 200.33 - ], - "text": "|H[ej]| =\n|ej2 −1|\n|ej −|γ |ejπ/4||ej −|γ |e−jπ/4|", - "type": "text" - }, - { - "block_id": "p563-b8", - "global_id": 16116, - "bbox": [ - 128.91, - 210.27, - 272.99, - 220.23 - ], - "text": "Now, by using Eq. (5.37), we obtain", - "type": "text" - }, - { - "block_id": "p563-b9", - "global_id": 16117, - "bbox": [ - 161.96, - 229.81, - 368.13, - 247.07 - ], - "text": "|H[ej]|2 =\n2(1 −cos 2)", - "type": "text" - }, - { - "block_id": "p563-b10", - "global_id": 16118, - "bbox": [ - 216.7, - 236.37, - 299.2, - 260.73 - ], - "text": "1 + |γ |2 −2|γ |cos", - "type": "text" - }, - { - "block_id": "p563-b11", - "global_id": 16119, - "bbox": [ - 299.19, - 243.37, - 324.99, - 260.32 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p563-b12", - "global_id": 16120, - "bbox": [ - 320.01, - 257.84, - 324.99, - 267.81 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p563-b13", - "global_id": 16121, - "bbox": [ - 327.18, - 236.37, - 345.87, - 246.33 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p563-b14", - "global_id": 16122, - "bbox": [ - 345.88, - 236.37, - 428.37, - 260.73 - ], - "text": "1 + |γ |2 −2|γ |cos", - "type": "text" - }, - { - "block_id": "p563-b15", - "global_id": 16123, - "bbox": [ - 428.37, - 243.37, - 454.17, - 260.32 - ], - "text": "+ π", - "type": "text" - }, - { - "block_id": "p563-b16", - "global_id": 16124, - "bbox": [ - 449.19, - 257.84, - 454.17, - 267.81 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p563-b17", - "global_id": 16125, - "bbox": [ - 456.35, - 236.37, - 468.5, - 246.33 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p563-b18", - "global_id": 16126, - "bbox": [ - 212.91, - 475.26, - 356.92, - 483.77 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p563-b19", - "global_id": 16127, - "bbox": [ - 174.09, - 458.2, - 182.97, - 466.2 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p563-b20", - "global_id": 16128, - "bbox": [ - 183.03, - 346.91, - 191.51, - 355.1 - ], - "text": "g", - "type": "text" - }, - { - "block_id": "p563-b21", - "global_id": 16129, - "bbox": [ - 408.25, - 372.88, - 441.71, - 381.18 - ], - "text": "g 0.83", - "type": "text" - }, - { - "block_id": "p563-b22", - "global_id": 16130, - "bbox": [ - 404.34, - 333.67, - 437.81, - 341.97 - ], - "text": "g 0.96", - "type": "text" - }, - { - "block_id": "p563-b23", - "global_id": 16131, - "bbox": [ - 404.34, - 296.12, - 427.81, - 304.42 - ], - "text": "g 1", - "type": "text" - }, - { - "block_id": "p563-b24", - "global_id": 16132, - "bbox": [ - 188.07, - 362.86, - 200.45, - 371.16 - ], - "text": "p4", - "type": "text" - }, - { - "block_id": "p563-b25", - "global_id": 16133, - "bbox": [ - 383.98, - 450.4, - 396.36, - 458.69 - ], - "text": "p4", - "type": "text" - }, - { - "block_id": "p563-b26", - "global_id": 16134, - "bbox": [ - 232.03, - 375.91, - 257.36, - 393.2 - ], - "text": "z 1\n(v 0)", - "type": "text" - }, - { - "block_id": "p563-b27", - "global_id": 16135, - "bbox": [ - 219.78, - 323.06, - 258.45, - 341.75 - ], - "text": "z e jp4\n(v 250p)", - "type": "text" - }, - { - "block_id": "p563-b28", - "global_id": 16136, - "bbox": [ - 409.03, - 458.2, - 418.67, - 466.2 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p563-b29", - "global_id": 16137, - "bbox": [ - 301.14, - 289.51, - 328.48, - 299.2 - ], - "text": "H [ej\t]", - "type": "text" - }, - { - "block_id": "p563-b30", - "global_id": 16138, - "bbox": [ - 298.03, - 433.38, - 461.29, - 443.2 - ], - "text": "v\n0\n250p", - "type": "text" - }, - { - "block_id": "p563-b31", - "global_id": 16139, - "bbox": [ - 281.89, - 586.87, - 290.77, - 594.87 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p563-b33", - "global_id": 16140, - "bbox": [ - 278.03, - 337.67, - 292.03, - 345.67 - ], - "text": "25.5", - "type": "text" - }, - { - "block_id": "p563-b34", - "global_id": 16141, - "bbox": [ - 278.03, - 406.85, - 292.03, - 414.85 - ], - "text": "6.41", - "type": "text" - }, - { - "block_id": "p563-b35", - "global_id": 16142, - "bbox": [ - 300.69, - 570.76, - 311.35, - 579.06 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p563-b37", - "global_id": 16143, - "bbox": [ - 285.09, - 502.37, - 289.09, - 517.41 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p563-b39", - "global_id": 16144, - "bbox": [ - 285.09, - 541.62, - 289.09, - 556.66 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p563-b40", - "global_id": 16145, - "bbox": [ - 260.05, - 517.55, - 274.21, - 525.84 - ], - "text": "2 g", - "type": "text" - }, - { - "block_id": "p563-b41", - "global_id": 16146, - "bbox": [ - 251.04, - 570.25, - 270.86, - 579.72 - ], - "text": "g2", - "type": "text" - }, - { - "block_id": "p563-b42", - "global_id": 16147, - "bbox": [ - 128.9, - 604.25, - 278.41, - 613.49 - ], - "text": "Figure 5.21 Designing a bandpass filter.", - "type": "text" - } - ] - }, - { - "page_num": 564, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p564-b0", - "global_id": 16148, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "544\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p564-b1", - "global_id": 16149, - "bbox": [ - 103.16, - 86.18, - 477.01, - 145.6 - ], - "text": "Figure 5.21b shows the amplitude response as a function of ω, as well as = ωT = 10−3ω\nfor values of |γ | = 0.83, 0.96, and 1. As expected, the gain is zero at ω = 0 and at 500 Hz\n(ω = 1000π). The gain peaks at about 125 Hz (ω = 250π). The resonance (peaking) becomes\npronounced as |γ | approaches 1. Figure 5.21c shows a canonical realization of this filter, which\nfollows from the transfer function H[z].", - "type": "text" - }, - { - "block_id": "p564-b2", - "global_id": 16150, - "bbox": [ - 103.46, - 160.34, - 431.17, - 172.46 - ], - "text": "MULTIPLE MAGNITUDE RESPONSE CURVES USING MATLAB", - "type": "text" - }, - { - "block_id": "p564-b3", - "global_id": 16151, - "bbox": [ - 103.16, - 176.08, - 477.03, - 210.36 - ], - "text": "By defining an anonymous function of the two variables z and γ in MATLAB, it is\nstraightforward to duplicate the three magnitude response curves in Fig. 5.21b, corresponding\nto the cases γ = 0.83, 0.96, and 1.", - "type": "text" - }, - { - "block_id": "p564-b4", - "global_id": 16152, - "bbox": [ - 103.16, - 230.57, - 464.06, - 348.13 - ], - "text": ">>\nOmega = linspace(0,pi,400);\n>>\nH = @(z,gamma_m) (z.^2-1)./(z.^2-sqrt(2)*gamma_m*z+gamma_m^2);\n>>\nplot(Omega,abs(H(exp(1j*Omega),0.83)),...\n>>\nOmega,abs(H(exp(1j*Omega),0.96)),...\n>>\nOmega,abs(H(exp(1j*Omega),0.99)));\n>>\ntext(.27*pi,35,’|\\gamma|=1’);\n>>\ntext(.28*pi,25.5,’|\\gamma|=0.96’);\n>>\ntext(.35*pi,6.41,’|\\gamma|=0.83’);\n>>\nset(gca,’xtick’,0:pi/4:pi,’ytick’,[0 6.41,25.5]);\n>>\naxis([0 pi 0 40]); xlabel(’\\Omega’); ylabel(’|H[e^{j \\Omega}]|’);", - "type": "text" - }, - { - "block_id": "p564-b5", - "global_id": 16153, - "bbox": [ - 103.16, - 367.77, - 477.03, - 389.69 - ], - "text": "The result, shown in Fig. 5.22, confirms the earlier result of Fig. 5.21b. Phase response curves\ncan be generated with minor modification to the MATLAB code.", - "type": "text" - }, - { - "block_id": "p564-b6", - "global_id": 16154, - "bbox": [ - 138.86, - 542.06, - 317.99, - 561.55 - ], - "text": "0\n0.7854\n1.5708\nΩ", - "type": "text" - }, - { - "block_id": "p564-b7", - "global_id": 16155, - "bbox": [ - 132.64, - 532.84, - 485.24, - 550.06 - ], - "text": "2.3562\n3.1416\n0", - "type": "text" - }, - { - "block_id": "p564-b8", - "global_id": 16156, - "bbox": [ - 121.39, - 515.77, - 135.39, - 523.77 - ], - "text": "6.41", - "type": "text" - }, - { - "block_id": "p564-b9", - "global_id": 16157, - "bbox": [ - 121.39, - 464.93, - 135.39, - 472.93 - ], - "text": "25.5", - "type": "text" - }, - { - "block_id": "p564-b10", - "global_id": 16158, - "bbox": [ - 108.19, - 469.53, - 117.88, - 495.8 - ], - "text": "|H[ej Ω]|", - "type": "text" - }, - { - "block_id": "p564-b11", - "global_id": 16159, - "bbox": [ - 231.49, - 439.9, - 247.13, - 448.2 - ], - "text": "|γ|=1", - "type": "text" - }, - { - "block_id": "p564-b12", - "global_id": 16160, - "bbox": [ - 235.24, - 464.66, - 260.88, - 472.96 - ], - "text": "|γ|=0.96", - "type": "text" - }, - { - "block_id": "p564-b13", - "global_id": 16161, - "bbox": [ - 258.49, - 515.67, - 284.13, - 523.96 - ], - "text": "|γ|=0.83", - "type": "text" - }, - { - "block_id": "p564-b14", - "global_id": 16162, - "bbox": [ - 103.16, - 571.83, - 374.47, - 581.07 - ], - "text": "Figure 5.22 MATLAB-generated magnitude response curves for Ex. 5.14.", - "type": "text" - } - ] - }, - { - "page_num": 565, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p565-b0", - "global_id": 16163, - "bbox": [ - 268.69, - 62.89, - 516.14, - 71.98 - ], - "text": "5.6\nFrequency Response from Pole-Zero Locations\n545", - "type": "text" - }, - { - "block_id": "p565-b1", - "global_id": 16164, - "bbox": [ - 102.51, - 93.91, - 423.19, - 105.87 - ], - "text": "EXAMPLE 5.15\nBandstop Filter by Pole-Zero Placement", - "type": "text" - }, - { - "block_id": "p565-b2", - "global_id": 16165, - "bbox": [ - 128.9, - 122.54, - 502.76, - 157.49 - ], - "text": "Design a second-order notch filter to have zero transmission at 250 Hz and a sharp recovery\nof gain to unity on both sides of 250 Hz. The highest significant frequency to be processed is\nfh = 400 Hz.", - "type": "text" - }, - { - "block_id": "p565-b3", - "global_id": 16166, - "bbox": [ - 128.9, - 177.69, - 502.78, - 284.92 - ], - "text": "In this case, T < 1/2fh = 1.25 × 10−3. Let us choose T = 10−3. For the frequency 250 Hz,\n = 2π(250)T = π/2. Thus, the frequency 250 Hz is represented by a point ej = ejπ/2 = j on\nthe unit circle, as depicted in Fig. 5.23a. Since we need zero transmission at this frequency, we\nmust place a zero at z = ejπ/2 = j and its conjugate at z = e−jπ/2 = −j. We also require a sharp\nrecovery of gain on both sides of frequency 250 Hz. To accomplish this goal, we place two\npoles close to the two zeros, to cancel out the effect of the two zeros as we move away from\nthe point j (corresponding to frequency 250 Hz). For this reason, let us use poles at ±ja with\na < 1 for stability. The closer the poles are to zeros (the closer the a to 1), the faster is the gain\nrecovery on either side of 250 Hz. The resulting transfer function is", - "type": "text" - }, - { - "block_id": "p565-b4", - "global_id": 16167, - "bbox": [ - 241.53, - 304.0, - 334.81, - 321.26 - ], - "text": "H[z] = K (z −j)(z + j)", - "type": "text" - }, - { - "block_id": "p565-b5", - "global_id": 16168, - "bbox": [ - 279.93, - 300.39, - 386.95, - 328.34 - ], - "text": "(z −ja)(z + ja) = K z2 + 1", - "type": "text" - }, - { - "block_id": "p565-b6", - "global_id": 16169, - "bbox": [ - 361.25, - 315.18, - 388.43, - 328.34 - ], - "text": "z2 + a2", - "type": "text" - }, - { - "block_id": "p565-b7", - "global_id": 16170, - "bbox": [ - 128.9, - 345.81, - 336.35, - 356.18 - ], - "text": "The dc gain (gain at = 0, or z = 1 ) of this filter is", - "type": "text" - }, - { - "block_id": "p565-b8", - "global_id": 16171, - "bbox": [ - 283.07, - 374.1, - 346.89, - 398.12 - ], - "text": "H[1] = K\n2\n1 + a2", - "type": "text" - }, - { - "block_id": "p565-b9", - "global_id": 16172, - "bbox": [ - 128.9, - 414.93, - 502.76, - 438.19 - ], - "text": "Because we require a dc gain of unity, we must select K = (1 + a2)/2. The transfer function is\ntherefore", - "type": "text" - }, - { - "block_id": "p565-b10", - "global_id": 16173, - "bbox": [ - 267.93, - 451.59, - 362.53, - 472.46 - ], - "text": "H[z] = (1 + a2)(z2 + 1)", - "type": "text" - }, - { - "block_id": "p565-b11", - "global_id": 16174, - "bbox": [ - 310.79, - 466.39, - 350.9, - 479.64 - ], - "text": "2(z2 + a2)", - "type": "text" - }, - { - "block_id": "p565-b12", - "global_id": 16175, - "bbox": [ - 128.9, - 497.13, - 240.41, - 507.09 - ], - "text": "and according to Eq. (5.37),", - "type": "text" - }, - { - "block_id": "p565-b13", - "global_id": 16176, - "bbox": [ - 225.51, - 525.23, - 310.56, - 543.42 - ], - "text": "|H[ej]|2 = (1 + a2)2", - "type": "text" - }, - { - "block_id": "p565-b14", - "global_id": 16177, - "bbox": [ - 290.45, - 522.55, - 404.96, - 550.59 - ], - "text": "4\n(ej2 + 1)(e−j2 + 1)\n(ej2 + a2)(e−j2 + a2)", - "type": "text" - }, - { - "block_id": "p565-b15", - "global_id": 16178, - "bbox": [ - 263.81, - 553.37, - 364.37, - 571.25 - ], - "text": "= (1 + a2)2(1 + cos 2)", - "type": "text" - }, - { - "block_id": "p565-b16", - "global_id": 16179, - "bbox": [ - 274.81, - 565.48, - 366.19, - 578.74 - ], - "text": "2(1 + a4 + 2a2 cos 2)", - "type": "text" - }, - { - "block_id": "p565-b17", - "global_id": 16180, - "bbox": [ - 128.9, - 596.07, - 502.76, - 619.62 - ], - "text": "Figure 5.23b shows |H[ej]| for values of a = 0.3, 0.6, and 0.95. Figure 5.23c shows a\nrealization of this filter.", - "type": "text" - } - ] - }, - { - "page_num": 566, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p566-b0", - "global_id": 16181, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "546\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p566-b1", - "global_id": 16182, - "bbox": [ - 235.2, - 99.98, - 274.34, - 117.28 - ], - "text": "p2\n(v 500p)", - "type": "text" - }, - { - "block_id": "p566-b2", - "global_id": 16183, - "bbox": [ - 271.87, - 150.48, - 292.76, - 158.78 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p566-b3", - "global_id": 16184, - "bbox": [ - 270.08, - 162.49, - 295.41, - 170.79 - ], - "text": "(v 0)", - "type": "text" - }, - { - "block_id": "p566-b4", - "global_id": 16185, - "bbox": [ - 216.22, - 290.64, - 244.89, - 298.93 - ], - "text": "a 0.95", - "type": "text" - }, - { - "block_id": "p566-b5", - "global_id": 16186, - "bbox": [ - 168.08, - 150.48, - 190.3, - 158.78 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p566-b6", - "global_id": 16187, - "bbox": [ - 147.14, - 162.49, - 191.81, - 170.79 - ], - "text": "(v 1000 p)", - "type": "text" - }, - { - "block_id": "p566-b7", - "global_id": 16188, - "bbox": [ - 233.2, - 181.48, - 246.09, - 189.7 - ], - "text": "ja", - "type": "text" - }, - { - "block_id": "p566-b8", - "global_id": 16189, - "bbox": [ - 233.2, - 198.08, - 242.09, - 206.29 - ], - "text": "j", - "type": "text" - }, - { - "block_id": "p566-b9", - "global_id": 16190, - "bbox": [ - 233.2, - 130.7, - 239.42, - 138.7 - ], - "text": "ja", - "type": "text" - }, - { - "block_id": "p566-b10", - "global_id": 16191, - "bbox": [ - 221.99, - 227.98, - 230.87, - 235.98 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p566-b12", - "global_id": 16192, - "bbox": [ - 221.6, - 444.99, - 231.25, - 452.99 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p566-b13", - "global_id": 16193, - "bbox": [ - 103.2, - 404.76, - 292.87, - 414.27 - ], - "text": "0\np", - "type": "text" - }, - { - "block_id": "p566-b14", - "global_id": 16194, - "bbox": [ - 95.25, - 296.71, - 99.25, - 304.71 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p566-b15", - "global_id": 16195, - "bbox": [ - 105.35, - 259.21, - 135.36, - 268.76 - ], - "text": "H [ejvT]", - "type": "text" - }, - { - "block_id": "p566-b16", - "global_id": 16196, - "bbox": [ - 103.2, - 417.65, - 370.56, - 439.39 - ], - "text": "v\n1000p\n500p\n0\n(vT)", - "type": "text" - }, - { - "block_id": "p566-b17", - "global_id": 16197, - "bbox": [ - 221.99, - 592.13, - 230.87, - 600.13 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p566-b18", - "global_id": 16198, - "bbox": [ - 144.92, - 474.76, - 166.59, - 484.24 - ], - "text": "1 a2", - "type": "text" - }, - { - "block_id": "p566-b19", - "global_id": 16199, - "bbox": [ - 191.4, - 577.03, - 205.07, - 586.43 - ], - "text": "a2", - "type": "text" - }, - { - "block_id": "p566-b20", - "global_id": 16200, - "bbox": [ - 153.75, - 484.94, - 157.75, - 492.94 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p566-b21", - "global_id": 16201, - "bbox": [ - 134.78, - 498.67, - 147.66, - 506.75 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p566-b22", - "global_id": 16202, - "bbox": [ - 174.77, - 481.94, - 292.08, - 498.78 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p566-b23", - "global_id": 16203, - "bbox": [ - 220.72, - 548.19, - 224.72, - 563.23 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p566-b24", - "global_id": 16204, - "bbox": [ - 220.72, - 508.94, - 224.72, - 523.98 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p566-b25", - "global_id": 16205, - "bbox": [ - 216.9, - 345.85, - 241.57, - 354.14 - ], - "text": "a 0.3", - "type": "text" - }, - { - "block_id": "p566-b26", - "global_id": 16206, - "bbox": [ - 209.45, - 320.78, - 234.11, - 329.08 - ], - "text": "a 0.6", - "type": "text" - }, - { - "block_id": "p566-b27", - "global_id": 16207, - "bbox": [ - 193.53, - 406.27, - 198.87, - 414.27 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p566-b28", - "global_id": 16208, - "bbox": [ - 194.2, - 414.91, - 198.2, - 422.91 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p566-b29", - "global_id": 16209, - "bbox": [ - 94.2, - 606.84, - 271.34, - 616.07 - ], - "text": "Figure 5.23 Designing a notch (bandstop) filter.", - "type": "text" - } - ] - }, - { - "page_num": 567, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p567-b0", - "global_id": 16210, - "bbox": [ - 309.15, - 62.89, - 516.14, - 71.98 - ], - "text": "5.7\nDigital Processing of Analog Signals\n547", - "type": "text" - }, - { - "block_id": "p567-b1", - "global_id": 16211, - "bbox": [ - 133.57, - 97.81, - 431.04, - 109.77 - ], - "text": "DRILL 5.21\nHighpass Filter by Pole-Zero Placement", - "type": "text" - }, - { - "block_id": "p567-b2", - "global_id": 16212, - "bbox": [ - 133.57, - 118.89, - 411.57, - 128.85 - ], - "text": "Use the graphical argument to show that a filter with transfer function", - "type": "text" - }, - { - "block_id": "p567-b3", - "global_id": 16213, - "bbox": [ - 292.06, - 150.16, - 350.46, - 167.42 - ], - "text": "H[z] = z −0.9", - "type": "text" - }, - { - "block_id": "p567-b4", - "global_id": 16214, - "bbox": [ - 334.93, - 164.53, - 338.8, - 174.49 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p567-b5", - "global_id": 16215, - "bbox": [ - 133.57, - 192.19, - 427.48, - 202.15 - ], - "text": "acts like a highpass filter. Make a rough sketch of the amplitude response.", - "type": "text" - }, - { - "block_id": "p567-b6", - "global_id": 16216, - "bbox": [ - 127.94, - 251.67, - 440.68, - 265.62 - ], - "text": "5.7 DIGITAL PROCESSING OF ANALOG SIGNALS", - "type": "text" - }, - { - "block_id": "p567-b7", - "global_id": 16217, - "bbox": [ - 127.59, - 271.6, - 516.16, - 329.38 - ], - "text": "An analog (meaning continuous-time) signal can be processed digitally by sampling the analog\nsignal and processing the samples by a digital (meaning discrete-time) processor. The output of\nthe processor is then converted back to analog signal, as shown in Fig. 5.24a. We saw some simple\ncases of such processing in Exs. 3.8, 3.9, 5.14, and 5.15. In this section, we shall derive a criterion\nfor designing such a digital processor for a general LTIC system.", - "type": "text" - }, - { - "block_id": "p567-b8", - "global_id": 16218, - "bbox": [ - 127.59, - 331.37, - 516.16, - 377.2 - ], - "text": "Suppose that we wish to realize an equivalent of an analog system with transfer function\nHa(s), shown in Fig. 5.24b. Let the digital processor transfer function in Fig. 5.24a that realizes\nthis desired Ha(s) be H[z]. In other words, we wish to make the two systems in Fig. 5.24 equivalent\n(at least approximately).", - "type": "text" - }, - { - "block_id": "p567-b9", - "global_id": 16219, - "bbox": [ - 127.59, - 378.78, - 516.13, - 413.07 - ], - "text": "By “equivalence” we mean that for a given input x(t), the systems in Fig. 5.24 yield the same\noutput y(t). Therefore, y(nT), the samples of the output in Fig. 5.24b, are identical to y[n], the\noutput of H[z] in Fig. 5.24a.", - "type": "text" - }, - { - "block_id": "p567-b10", - "global_id": 16220, - "bbox": [ - 151.71, - 468.02, - 438.83, - 476.1 - ], - "text": "x[n]\nx(t)\ny[n]\ny(t)", - "type": "text" - }, - { - "block_id": "p567-b11", - "global_id": 16221, - "bbox": [ - 151.71, - 566.55, - 438.83, - 574.63 - ], - "text": "x(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p567-b12", - "global_id": 16222, - "bbox": [ - 187.47, - 469.86, - 232.59, - 477.86 - ], - "text": "Continuous to", - "type": "text" - }, - { - "block_id": "p567-b13", - "global_id": 16223, - "bbox": [ - 196.59, - 478.86, - 223.47, - 486.86 - ], - "text": "discrete,", - "type": "text" - }, - { - "block_id": "p567-b14", - "global_id": 16224, - "bbox": [ - 202.96, - 487.86, - 217.1, - 495.86 - ], - "text": "C/D", - "type": "text" - }, - { - "block_id": "p567-b15", - "global_id": 16225, - "bbox": [ - 272.79, - 469.86, - 316.33, - 477.86 - ], - "text": "Discrete-time", - "type": "text" - }, - { - "block_id": "p567-b16", - "global_id": 16226, - "bbox": [ - 283.45, - 478.86, - 305.67, - 486.86 - ], - "text": "system", - "type": "text" - }, - { - "block_id": "p567-b17", - "global_id": 16227, - "bbox": [ - 287.45, - 487.77, - 301.67, - 495.86 - ], - "text": "H[z]", - "type": "text" - }, - { - "block_id": "p567-b18", - "global_id": 16228, - "bbox": [ - 362.67, - 469.86, - 399.79, - 486.86 - ], - "text": "Discrete to\ncontinuous,", - "type": "text" - }, - { - "block_id": "p567-b19", - "global_id": 16229, - "bbox": [ - 374.56, - 487.86, - 387.9, - 495.86 - ], - "text": "D/C", - "type": "text" - }, - { - "block_id": "p567-b20", - "global_id": 16230, - "bbox": [ - 296.0, - 538.85, - 304.88, - 546.85 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p567-b21", - "global_id": 16231, - "bbox": [ - 248.87, - 486.02, - 339.01, - 495.09 - ], - "text": "a\nb", - "type": "text" - }, - { - "block_id": "p567-b22", - "global_id": 16232, - "bbox": [ - 286.49, - 523.37, - 303.7, - 532.92 - ], - "text": "Ha(s)", - "type": "text" - }, - { - "block_id": "p567-b23", - "global_id": 16233, - "bbox": [ - 295.62, - 605.26, - 305.27, - 613.26 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p567-b24", - "global_id": 16234, - "bbox": [ - 291.84, - 574.7, - 309.05, - 584.24 - ], - "text": "Ha(s)", - "type": "text" - }, - { - "block_id": "p567-b25", - "global_id": 16235, - "bbox": [ - 151.5, - 619.96, - 358.35, - 629.19 - ], - "text": "Figure 5.24 Analog filter realization with a digital filter.", - "type": "text" - } - ] - }, - { - "page_num": 568, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p568-b0", - "global_id": 16236, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "548\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p568-b1", - "global_id": 16237, - "bbox": [ - 101.84, - 85.82, - 490.38, - 107.74 - ], - "text": "For the sake of generality, we are assuming a noncausal system. The argument and the results\nare also valid for causal systems. The output y(t) of the system in Fig. 5.24b is", - "type": "text" - }, - { - "block_id": "p568-b2", - "global_id": 16238, - "bbox": [ - 143.06, - 127.95, - 167.68, - 138.22 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p568-b3", - "global_id": 16239, - "bbox": [ - 169.72, - 114.39, - 186.67, - 126.45 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p568-b4", - "global_id": 16240, - "bbox": [ - 174.99, - 139.26, - 187.54, - 146.24 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p568-b5", - "global_id": 16241, - "bbox": [ - 189.15, - 127.94, - 322.12, - 139.02 - ], - "text": "x(τ)ha(t −τ)dτ\n= lim", - "type": "text" - }, - { - "block_id": "p568-b6", - "global_id": 16242, - "bbox": [ - 305.16, - 136.86, - 325.79, - 144.12 - ], - "text": "τ→0", - "type": "text" - }, - { - "block_id": "p568-b7", - "global_id": 16243, - "bbox": [ - 332.46, - 117.77, - 346.56, - 128.45 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p568-b8", - "global_id": 16244, - "bbox": [ - 328.0, - 141.79, - 351.02, - 148.99 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p568-b9", - "global_id": 16245, - "bbox": [ - 352.13, - 127.94, - 447.98, - 139.02 - ], - "text": "x(mτ)ha(t −mτ)τ", - "type": "text" - }, - { - "block_id": "p568-b10", - "global_id": 16246, - "bbox": [ - 101.85, - 159.37, - 490.39, - 181.71 - ], - "text": "For our purpose, it is convenient to use the notation T for τ. Assuming T (the sampling interval)\nto be small enough, such a change of notation yields", - "type": "text" - }, - { - "block_id": "p568-b11", - "global_id": 16247, - "bbox": [ - 232.45, - 201.92, - 264.67, - 212.19 - ], - "text": "y(t) = T", - "type": "text" - }, - { - "block_id": "p568-b12", - "global_id": 16248, - "bbox": [ - 271.01, - 191.74, - 285.11, - 202.41 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p568-b13", - "global_id": 16249, - "bbox": [ - 266.54, - 215.76, - 289.56, - 222.96 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p568-b14", - "global_id": 16250, - "bbox": [ - 290.68, - 201.92, - 359.77, - 212.99 - ], - "text": "x(mT)ha(t −mT)", - "type": "text" - }, - { - "block_id": "p568-b15", - "global_id": 16251, - "bbox": [ - 101.84, - 233.41, - 474.88, - 243.78 - ], - "text": "The response at the nth sampling instant is y(nT) obtained by setting t = nT in the equation is", - "type": "text" - }, - { - "block_id": "p568-b16", - "global_id": 16252, - "bbox": [ - 223.95, - 263.99, - 264.5, - 274.26 - ], - "text": "y(nT) = T", - "type": "text" - }, - { - "block_id": "p568-b17", - "global_id": 16253, - "bbox": [ - 270.85, - 253.82, - 284.94, - 264.49 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p568-b18", - "global_id": 16254, - "bbox": [ - 266.37, - 277.84, - 289.4, - 285.03 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p568-b19", - "global_id": 16255, - "bbox": [ - 290.51, - 263.99, - 490.38, - 275.07 - ], - "text": "x(mT)ha[(n −m)T]\n(5.42)", - "type": "text" - }, - { - "block_id": "p568-b20", - "global_id": 16256, - "bbox": [ - 101.85, - 295.48, - 490.39, - 317.81 - ], - "text": "In Fig. 5.24a, the input to H[z] is x(nT) = x[n]. If h[n] is the unit impulse response of H[z], then\ny[n], the output of H[z], is given by", - "type": "text" - }, - { - "block_id": "p568-b21", - "global_id": 16257, - "bbox": [ - 243.63, - 338.02, - 269.49, - 348.3 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p568-b22", - "global_id": 16258, - "bbox": [ - 275.99, - 327.85, - 290.09, - 338.52 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p568-b23", - "global_id": 16259, - "bbox": [ - 271.53, - 351.87, - 294.55, - 359.06 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p568-b24", - "global_id": 16260, - "bbox": [ - 295.67, - 338.02, - 490.38, - 348.4 - ], - "text": "x[m]h[n −m]\n(5.43)", - "type": "text" - }, - { - "block_id": "p568-b25", - "global_id": 16261, - "bbox": [ - 101.85, - 369.51, - 490.39, - 391.84 - ], - "text": "If the two systems are to be equivalent, y(nT) in Eq. (5.42) must be equal to y[n] in Eq. (5.43).\nTherefore,", - "type": "text" - }, - { - "block_id": "p568-b26", - "global_id": 16262, - "bbox": [ - 265.28, - 394.24, - 490.38, - 405.32 - ], - "text": "h[n] = Tha(nT)\n(5.44)", - "type": "text" - }, - { - "block_id": "p568-b27", - "global_id": 16263, - "bbox": [ - 101.84, - 410.5, - 490.39, - 459.95 - ], - "text": "This is the time-domain criterion for equivalence of the two systems.† According to this criterion,\nh[n], the unit impulse response of H[z] in Fig. 5.24a, should be T times the samples of ha(t),\nthe unit impulse response of the system in Fig. 5.24b. This is known as the impulse invariance\ncriterion of filter design.", - "type": "text" - }, - { - "block_id": "p568-b28", - "global_id": 16264, - "bbox": [ - 101.84, - 461.94, - 490.4, - 520.43 - ], - "text": "Strictly speaking, this realization guarantees the output equivalence only at the sampling\ninstants, that is, y(nT) = y[n], and that also requires the assumption that T →0. Clearly, this\ncriterion leads to an approximate realization of Ha(s). However, it can be shown that when the\nfrequency response of |Ha(jω)| is bandlimited, the realization is exact [2], provided the sampling\nrate is high enough to avoid any aliasing (T < 1/2fh).", - "type": "text" - }, - { - "block_id": "p568-b29", - "global_id": 16265, - "bbox": [ - 102.14, - 533.92, - 279.35, - 546.85 - ], - "text": "REALIZATION OF RATIONAL H(s)", - "type": "text" - }, - { - "block_id": "p568-b30", - "global_id": 16266, - "bbox": [ - 101.84, - 550.88, - 334.62, - 560.84 - ], - "text": "If we wish to realize an analog filter with transfer function", - "type": "text" - }, - { - "block_id": "p568-b31", - "global_id": 16267, - "bbox": [ - 267.65, - 568.81, - 323.38, - 592.83 - ], - "text": "Ha(s) =\nc\ns −λ", - "type": "text" - }, - { - "block_id": "p568-b32", - "global_id": 16268, - "bbox": [ - 101.84, - 610.24, - 490.4, - 633.41 - ], - "text": "† Because T is a constant, some authors ignore the factor T, which yields a simplified criterion h[n] = ha(nT).\nIgnoring T merely scales the amplitude response of the resulting filter.", - "type": "text" - } - ] - }, - { - "page_num": 569, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p569-b0", - "global_id": 16269, - "bbox": [ - 309.15, - 62.89, - 516.14, - 71.98 - ], - "text": "5.7\nDigital Processing of Analog Signals\n549", - "type": "text" - }, - { - "block_id": "p569-b1", - "global_id": 16270, - "bbox": [ - 127.59, - 85.46, - 439.52, - 96.54 - ], - "text": "The impulse response h(t), given by the inverse Laplace transform of Ha(s), is", - "type": "text" - }, - { - "block_id": "p569-b2", - "global_id": 16271, - "bbox": [ - 290.96, - 106.48, - 352.76, - 119.28 - ], - "text": "ha(t) = ceλtu(t)", - "type": "text" - }, - { - "block_id": "p569-b3", - "global_id": 16272, - "bbox": [ - 127.59, - 130.94, - 432.6, - 141.32 - ], - "text": "The corresponding digital filter unit impulse response h[n], per Eq. (5.44), is", - "type": "text" - }, - { - "block_id": "p569-b4", - "global_id": 16273, - "bbox": [ - 271.78, - 151.95, - 370.89, - 164.75 - ], - "text": "h[n] = Tha(nT) = TcenλT", - "type": "text" - }, - { - "block_id": "p569-b5", - "global_id": 16274, - "bbox": [ - 127.59, - 176.42, - 516.13, - 198.75 - ], - "text": "Figure 5.25 shows ha(t) and h[n]. The corresponding H[z], the z-transform of h[n], as found from\nTable 5.1, is", - "type": "text" - }, - { - "block_id": "p569-b6", - "global_id": 16275, - "bbox": [ - 291.71, - 198.99, - 516.13, - 223.01 - ], - "text": "H[z] =\nTcz\nz −eλT\n(5.45)", - "type": "text" - }, - { - "block_id": "p569-b7", - "global_id": 16276, - "bbox": [ - 301.56, - 311.79, - 490.59, - 319.79 - ], - "text": "n\nt", - "type": "text" - }, - { - "block_id": "p569-b8", - "global_id": 16277, - "bbox": [ - 181.95, - 240.68, - 376.52, - 251.98 - ], - "text": "h[n]\nha(t)", - "type": "text" - }, - { - "block_id": "p569-b9", - "global_id": 16278, - "bbox": [ - 231.71, - 327.46, - 428.02, - 335.46 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p569-b10", - "global_id": 16279, - "bbox": [ - 158.52, - 241.43, - 351.32, - 250.27 - ], - "text": "c\ncT", - "type": "text" - }, - { - "block_id": "p569-b11", - "global_id": 16280, - "bbox": [ - 151.5, - 341.91, - 516.12, - 363.83 - ], - "text": "Figure 5.25 Impulse response for analog and digital systems in the impulse invariance\nmethod of filter design.", - "type": "text" - }, - { - "block_id": "p569-b12", - "global_id": 16281, - "bbox": [ - 127.59, - 385.3, - 294.13, - 394.54 - ], - "text": "TABLE 5.3\nSelect Impulse-Invariance Pairs", - "type": "text" - }, - { - "block_id": "p569-b13", - "global_id": 16282, - "bbox": [ - 127.59, - 405.25, - 420.85, - 415.06 - ], - "text": "No.\nHa(s)\nha(t)\nh[n]\nH[z]", - "type": "text" - }, - { - "block_id": "p569-b14", - "global_id": 16283, - "bbox": [ - 127.59, - 423.58, - 414.91, - 432.92 - ], - "text": "1\nK\nKδ(t)\nTKδ[n]\nTK", - "type": "text" - }, - { - "block_id": "p569-b15", - "global_id": 16284, - "bbox": [ - 127.59, - 442.22, - 422.9, - 463.93 - ], - "text": "2\n1\ns\nu(t)\nTu[n]\nTz\nz −1", - "type": "text" - }, - { - "block_id": "p569-b16", - "global_id": 16285, - "bbox": [ - 127.59, - 467.15, - 432.83, - 490.13 - ], - "text": "3\n1\ns2\nt\nnT2\nT2z\n(z −1)2", - "type": "text" - }, - { - "block_id": "p569-b17", - "global_id": 16286, - "bbox": [ - 127.59, - 494.72, - 170.0, - 516.25 - ], - "text": "4\n1\ns3", - "type": "text" - }, - { - "block_id": "p569-b18", - "global_id": 16287, - "bbox": [ - 226.18, - 493.35, - 232.07, - 503.6 - ], - "text": "t2", - "type": "text" - }, - { - "block_id": "p569-b19", - "global_id": 16288, - "bbox": [ - 227.13, - 493.35, - 325.99, - 516.34 - ], - "text": "2\nk2T3", - "type": "text" - }, - { - "block_id": "p569-b20", - "global_id": 16289, - "bbox": [ - 315.6, - 493.35, - 442.5, - 516.34 - ], - "text": "2\nT3z(z + 1)", - "type": "text" - }, - { - "block_id": "p569-b21", - "global_id": 16290, - "bbox": [ - 407.48, - 506.67, - 439.66, - 516.34 - ], - "text": "2(z −1)3", - "type": "text" - }, - { - "block_id": "p569-b22", - "global_id": 16291, - "bbox": [ - 127.59, - 519.27, - 429.54, - 540.89 - ], - "text": "5\n1\ns −λ\neλt\nTeλnT\nTz\nz −eλT", - "type": "text" - }, - { - "block_id": "p569-b23", - "global_id": 16292, - "bbox": [ - 127.59, - 543.93, - 434.57, - 567.1 - ], - "text": "6\n1\n(s −λ)2\nteλt\nnT2eλnT\nT2zeλT", - "type": "text" - }, - { - "block_id": "p569-b24", - "global_id": 16293, - "bbox": [ - 405.14, - 557.25, - 440.47, - 567.1 - ], - "text": "(z −eλT)2", - "type": "text" - }, - { - "block_id": "p569-b25", - "global_id": 16294, - "bbox": [ - 127.59, - 568.42, - 514.94, - 593.3 - ], - "text": "7\nAs+B\ns2+2as+c\nTre−at cos(bt+θ)\nTre−anT cos(bnT+θ)\nTrz[zcos θ −e−aT cos(bT−θ)]", - "type": "text" - }, - { - "block_id": "p569-b26", - "global_id": 16295, - "bbox": [ - 410.27, - 581.74, - 508.82, - 593.3 - ], - "text": "z2−(2e−aT cos bT)z+e−2aT", - "type": "text" - }, - { - "block_id": "p569-b27", - "global_id": 16296, - "bbox": [ - 200.1, - 607.59, - 212.63, - 616.84 - ], - "text": "r =", - "type": "text" - }, - { - "block_id": "p569-b29", - "global_id": 16297, - "bbox": [ - 223.1, - 598.05, - 285.0, - 610.64 - ], - "text": "A2c + B2 −2ABa", - "type": "text" - }, - { - "block_id": "p569-b30", - "global_id": 16298, - "bbox": [ - 243.05, - 607.59, - 312.96, - 623.21 - ], - "text": "c −a2\n,\nb =", - "type": "text" - }, - { - "block_id": "p569-b31", - "global_id": 16299, - "bbox": [ - 314.8, - 599.93, - 322.39, - 608.89 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p569-b32", - "global_id": 16300, - "bbox": [ - 322.39, - 606.33, - 392.59, - 616.93 - ], - "text": "c −a2,\nθ = tan−1", - "type": "text" - }, - { - "block_id": "p569-b33", - "global_id": 16301, - "bbox": [ - 394.08, - 595.0, - 431.47, - 610.55 - ], - "text": "Aa −B", - "type": "text" - }, - { - "block_id": "p569-b34", - "global_id": 16302, - "bbox": [ - 401.33, - 615.57, - 406.81, - 624.54 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p569-b35", - "global_id": 16303, - "bbox": [ - 406.81, - 607.63, - 414.39, - 616.6 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p569-b36", - "global_id": 16304, - "bbox": [ - 414.4, - 614.96, - 435.88, - 624.54 - ], - "text": "c −a2", - "type": "text" - } - ] - }, - { - "page_num": 570, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p570-b0", - "global_id": 16305, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "550\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p570-b1", - "global_id": 16306, - "bbox": [ - 101.84, - 85.4, - 490.38, - 108.44 - ], - "text": "The procedure of finding H[z] can be systematized for any Nth-order system. First we express\nan Nth-order analog transfer function Ha(s) as a sum of partial fractions as‡", - "type": "text" - }, - { - "block_id": "p570-b2", - "global_id": 16307, - "bbox": [ - 258.83, - 137.93, - 291.14, - 149.01 - ], - "text": "Ha(s) =", - "type": "text" - }, - { - "block_id": "p570-b3", - "global_id": 16308, - "bbox": [ - 293.18, - 127.98, - 307.28, - 138.43 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p570-b4", - "global_id": 16309, - "bbox": [ - 294.8, - 152.32, - 305.65, - 159.59 - ], - "text": "i=1", - "type": "text" - }, - { - "block_id": "p570-b5", - "global_id": 16310, - "bbox": [ - 309.58, - 131.26, - 331.72, - 156.09 - ], - "text": "ci\ns −λi", - "type": "text" - }, - { - "block_id": "p570-b6", - "global_id": 16311, - "bbox": [ - 101.84, - 179.59, - 262.93, - 189.96 - ], - "text": "Then the corresponding H[z] is given by", - "type": "text" - }, - { - "block_id": "p570-b7", - "global_id": 16312, - "bbox": [ - 253.58, - 220.16, - 289.13, - 230.43 - ], - "text": "H[z] = T", - "type": "text" - }, - { - "block_id": "p570-b8", - "global_id": 16313, - "bbox": [ - 291.01, - 210.21, - 305.1, - 220.66 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p570-b9", - "global_id": 16314, - "bbox": [ - 292.63, - 234.56, - 303.48, - 241.82 - ], - "text": "i=1", - "type": "text" - }, - { - "block_id": "p570-b10", - "global_id": 16315, - "bbox": [ - 307.4, - 213.5, - 336.43, - 237.51 - ], - "text": "ciz\nz −eλiT", - "type": "text" - }, - { - "block_id": "p570-b11", - "global_id": 16316, - "bbox": [ - 101.84, - 262.23, - 490.41, - 308.06 - ], - "text": "This transfer function can be readily realized, as explained in Sec. 5.4. Table 5.3 lists several pairs\nof Ha(s) and their corresponding H[z]. For instance, to realize a digital integrator, we examine its\nHa(s) = 1/s. From Table 5.3, corresponding to Ha(s) = 1/s (pair 2), we find H[z] = Tz/(z −1).\nThis is exactly the result we obtained in Ex. 3.9 using another approach.", - "type": "text" - }, - { - "block_id": "p570-b12", - "global_id": 16317, - "bbox": [ - 101.84, - 309.65, - 490.38, - 331.97 - ], - "text": "Note that the frequency response Ha(jω) of a practical analog filter cannot be bandlimited.\nConsequently, all these realizations are approximate.", - "type": "text" - }, - { - "block_id": "p570-b13", - "global_id": 16318, - "bbox": [ - 101.84, - 351.65, - 490.42, - 582.91 - ], - "text": "CHOOSING THE SAMPLING INTERVAL T\nThe impulse-invariance criterion (5.44) was derived under the assumption that T →0. Such an\nassumption is neither practical nor necessary for satisfactory design. Avoiding of aliasing is the\nmost important consideration for the choice of T. In Eq. (5.39), we showed that for a sampling\ninterval T seconds, the highest frequency that can be sampled without aliasing is 1/2T Hz or\nπ/T radians per second. This implies that Ha(jω), the frequency response of the analog filter in\nFig. 5.24b should not have spectral components beyond frequency π/T radians per second. In\nother words, to avoid aliasing, the frequency response of the system Ha(s) must be bandlimited to\nπ/T radians per second. We shall see later in Ch. 7 that frequency response of a realizable LTIC\nsystem cannot be bandlimited; that is, the response generally exists for all frequencies up to ∞.\nTherefore, it is impossible to digitally realize an LTIC system exactly without aliasing. The saving\ngrace is that the frequency response of every realizable LTIC system decays with frequency. This\nallows for a compromise in digitally realizing an LTIC system with an acceptable level of aliasing.\nThe smaller the value of T, the smaller the aliasing, and the better the approximation. Since it is\nimpossible to make |Ha(jω)| zero, we are satisfied with making it negligible beyond the frequency\nπ/T. As a rule of thumb [3], we choose T such that |Ha(jω)| at the frequency ω = π/T is less than\na certain fraction (often taken as 1%) of the peak value of |Ha(jω)|. This ensures that aliasing is\nnegligible. The peak |Ha(jω)| usually occurs at ω = 0 for lowpass filters and at the band center\nfrequency ωc for bandpass filters.", - "type": "text" - }, - { - "block_id": "p570-b14", - "global_id": 16319, - "bbox": [ - 101.84, - 610.24, - 490.37, - 633.41 - ], - "text": "‡ Assuming Ha(s) has simple poles. For repeated poles, the form changes accordingly. Entry 6 in Table 5.3 is\nsuitable for repeated poles.", - "type": "text" - } - ] - }, - { - "page_num": 571, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p571-b0", - "global_id": 16320, - "bbox": [ - 309.15, - 62.89, - 516.14, - 71.98 - ], - "text": "5.7\nDigital Processing of Analog Signals\n551", - "type": "text" - }, - { - "block_id": "p571-b1", - "global_id": 16321, - "bbox": [ - 102.51, - 93.92, - 494.56, - 119.82 - ], - "text": "EXAMPLE 5.16\nButterworth Filter Design by the Impulse-Invariance\nMethod", - "type": "text" - }, - { - "block_id": "p571-b2", - "global_id": 16322, - "bbox": [ - 128.9, - 133.56, - 502.77, - 155.47 - ], - "text": "Design a digital filter to realize a first-order lowpass Butterworth filter with the transfer\nfunction", - "type": "text" - }, - { - "block_id": "p571-b3", - "global_id": 16323, - "bbox": [ - 257.1, - 150.07, - 317.01, - 175.21 - ], - "text": "Ha(s) =\nωc\ns + ωc", - "type": "text" - }, - { - "block_id": "p571-b4", - "global_id": 16324, - "bbox": [ - 338.63, - 155.63, - 502.75, - 168.51 - ], - "text": "ωc = 105\n(5.46)", - "type": "text" - }, - { - "block_id": "p571-b5", - "global_id": 16325, - "bbox": [ - 128.9, - 196.91, - 502.75, - 219.24 - ], - "text": "For this filter, we find the corresponding H[z] according to Eq. (5.45) (or pair 5 in Table 5.3)\nas", - "type": "text" - }, - { - "block_id": "p571-b6", - "global_id": 16326, - "bbox": [ - 281.0, - 216.6, - 502.75, - 240.93 - ], - "text": "H[z] =\nωcTz\nz −e−ωcT\n(5.47)", - "type": "text" - }, - { - "block_id": "p571-b7", - "global_id": 16327, - "bbox": [ - 128.9, - 246.26, - 502.74, - 304.45 - ], - "text": "Next, we select the value of T by means of the criterion according to which the gain at ω = π/T\ndrops to 1% of the maximum filter gain. However, this choice results in such a good design that\naliasing is imperceptible. The resulting amplitude response is so close to the desired response\nthat we can hardly notice the aliasing effect in our plot. For the sake of demonstrating the\naliasing effect, we shall deliberately select a 10% criterion (instead of 1%). We have", - "type": "text" - }, - { - "block_id": "p571-b8", - "global_id": 16328, - "bbox": [ - 268.29, - 324.63, - 311.91, - 335.71 - ], - "text": "|Ha(jω)| =", - "type": "text" - }, - { - "block_id": "p571-b10", - "global_id": 16329, - "bbox": [ - 318.39, - 317.64, - 343.23, - 334.5 - ], - "text": "ωc", - "type": "text" - }, - { - "block_id": "p571-b11", - "global_id": 16330, - "bbox": [ - 326.65, - 330.13, - 358.44, - 345.27 - ], - "text": "ω2 + ω2c", - "type": "text" - }, - { - "block_id": "p571-b13", - "global_id": 16331, - "bbox": [ - 128.91, - 354.89, - 502.76, - 389.87 - ], - "text": "In this case |Ha(jω)|max = 1, which occurs at ω = 0. Use of 10% criterion leads to |Ha(π/T)| =\n0.1. Observe that\n|Ha(jω)| ≈ωc", - "type": "text" - }, - { - "block_id": "p571-b14", - "global_id": 16332, - "bbox": [ - 309.98, - 378.79, - 369.76, - 395.83 - ], - "text": "ω\nω ≫ωc", - "type": "text" - }, - { - "block_id": "p571-b15", - "global_id": 16333, - "bbox": [ - 146.84, - 400.86, - 174.76, - 410.82 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p571-b16", - "global_id": 16334, - "bbox": [ - 209.05, - 407.27, - 277.18, - 425.32 - ], - "text": "|Ha(π/T)| ≈ωc", - "type": "text" - }, - { - "block_id": "p571-b17", - "global_id": 16335, - "bbox": [ - 263.89, - 412.81, - 422.11, - 431.59 - ], - "text": "π/T = 0.1\n\r⇒\nπ/T = 10ωc = 106", - "type": "text" - }, - { - "block_id": "p571-b18", - "global_id": 16336, - "bbox": [ - 128.9, - 438.44, - 502.76, - 462.0 - ], - "text": "Thus, the 10% criterion yields T = 10−6π. The 1% criterion would have given T = 10−7π.\nSubstitution of T = 10−6π in Eq. (5.47) yields", - "type": "text" - }, - { - "block_id": "p571-b19", - "global_id": 16337, - "bbox": [ - 278.56, - 471.84, - 502.75, - 495.96 - ], - "text": "H[z] =\n0.3142z\nz −0.7304\n(5.48)", - "type": "text" - }, - { - "block_id": "p571-b20", - "global_id": 16338, - "bbox": [ - 128.9, - 504.58, - 363.36, - 514.55 - ], - "text": "A canonical realization of this filter is shown in Fig. 5.26a.", - "type": "text" - }, - { - "block_id": "p571-b21", - "global_id": 16339, - "bbox": [ - 146.84, - 516.12, - 425.01, - 526.5 - ], - "text": "To find the frequency response of this digital filter, we rewrite H[z] as", - "type": "text" - }, - { - "block_id": "p571-b22", - "global_id": 16340, - "bbox": [ - 271.36, - 536.44, - 358.59, - 560.46 - ], - "text": "H[z] =\n0.3142\n1 −0.7304z−1", - "type": "text" - }, - { - "block_id": "p571-b23", - "global_id": 16341, - "bbox": [ - 128.9, - 569.09, - 170.65, - 579.05 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p571-b24", - "global_id": 16342, - "bbox": [ - 180.0, - 588.28, - 449.7, - 612.3 - ], - "text": "H[ejωT] =\n0.3142\n1 −0.7304e−jωT =\n0.3142\n(1 −0.7304cosωT) + j0.7304 sinωT", - "type": "text" - } - ] - }, - { - "page_num": 572, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p572-b0", - "global_id": 16343, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "552\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p572-b1", - "global_id": 16344, - "bbox": [ - 138.15, - 192.36, - 148.43, - 200.57 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p572-b2", - "global_id": 16345, - "bbox": [ - 315.01, - 302.87, - 320.35, - 310.87 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p572-b3", - "global_id": 16346, - "bbox": [ - 281.81, - 342.95, - 287.15, - 350.95 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p572-b4", - "global_id": 16347, - "bbox": [ - 117.75, - 190.73, - 121.75, - 198.73 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p572-b5", - "global_id": 16348, - "bbox": [ - 108.31, - 338.04, - 121.23, - 346.26 - ], - "text": "H", - "type": "text" - }, - { - "block_id": "p572-b6", - "global_id": 16349, - "bbox": [ - 205.08, - 317.69, - 214.73, - 325.69 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p572-b7", - "global_id": 16350, - "bbox": [ - 205.46, - 473.11, - 214.34, - 481.11 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p572-b8", - "global_id": 16351, - "bbox": [ - 143.25, - 340.94, - 224.93, - 350.41 - ], - "text": "105\n5 105", - "type": "text" - }, - { - "block_id": "p572-b9", - "global_id": 16352, - "bbox": [ - 101.77, - 458.0, - 121.77, - 466.3 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p572-b10", - "global_id": 16353, - "bbox": [ - 143.25, - 303.14, - 299.75, - 312.61 - ], - "text": "105\n5 105\n106", - "type": "text" - }, - { - "block_id": "p572-b11", - "global_id": 16354, - "bbox": [ - 288.75, - 358.51, - 299.75, - 367.98 - ], - "text": "106", - "type": "text" - }, - { - "block_id": "p572-b12", - "global_id": 16355, - "bbox": [ - 254.2, - 264.67, - 283.02, - 274.22 - ], - "text": "H[e jvT]", - "type": "text" - }, - { - "block_id": "p572-b13", - "global_id": 16356, - "bbox": [ - 196.12, - 379.48, - 227.72, - 389.03 - ], - "text": "H[e jvT]", - "type": "text" - }, - { - "block_id": "p572-b14", - "global_id": 16357, - "bbox": [ - 196.12, - 441.46, - 226.2, - 451.22 - ], - "text": "Ha( jv)", - "type": "text" - }, - { - "block_id": "p572-b15", - "global_id": 16358, - "bbox": [ - 165.65, - 287.45, - 192.47, - 297.21 - ], - "text": "Ha( jv)", - "type": "text" - }, - { - "block_id": "p572-b16", - "global_id": 16359, - "bbox": [ - 205.46, - 175.48, - 214.34, - 183.48 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p572-b17", - "global_id": 16360, - "bbox": [ - 157.13, - 101.97, - 170.01, - 110.06 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p572-b18", - "global_id": 16361, - "bbox": [ - 252.58, - 115.77, - 265.46, - 123.85 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p572-b19", - "global_id": 16362, - "bbox": [ - 193.65, - 161.97, - 215.65, - 169.97 - ], - "text": "0.7304", - "type": "text" - }, - { - "block_id": "p572-b20", - "global_id": 16363, - "bbox": [ - 181.82, - 102.42, - 251.3, - 118.36 - ], - "text": "0.3142", - "type": "text" - }, - { - "block_id": "p572-b21", - "global_id": 16364, - "bbox": [ - 227.21, - 128.03, - 231.21, - 143.07 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p572-b22", - "global_id": 16365, - "bbox": [ - 101.77, - 487.6, - 475.63, - 509.61 - ], - "text": "Figure 5.26 An example of filter design by the impulse-invariance method: (a) filter\nrealization, (b) amplitude response, and (c) phase response.", - "type": "text" - }, - { - "block_id": "p572-b23", - "global_id": 16366, - "bbox": [ - 103.16, - 537.17, - 269.18, - 547.13 - ], - "text": "The corresponding magnitude response is", - "type": "text" - }, - { - "block_id": "p572-b24", - "global_id": 16367, - "bbox": [ - 180.43, - 557.07, - 327.68, - 573.91 - ], - "text": "|H[ejωT]| =\n0.3142", - "type": "text" - }, - { - "block_id": "p572-b25", - "global_id": 16368, - "bbox": [ - 237.71, - 569.65, - 398.05, - 582.9 - ], - "text": "(1 −0.7304 cosωT)2 + (0.7304sinωT)2", - "type": "text" - }, - { - "block_id": "p572-b26", - "global_id": 16369, - "bbox": [ - 218.43, - 586.23, - 291.94, - 602.81 - ], - "text": "=\n0.3142\n√", - "type": "text" - }, - { - "block_id": "p572-b27", - "global_id": 16370, - "bbox": [ - 237.88, - 593.21, - 477.01, - 611.24 - ], - "text": "1.533 −1.4608cosωT\n(5.49)", - "type": "text" - } - ] - }, - { - "page_num": 573, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p573-b0", - "global_id": 16371, - "bbox": [ - 309.15, - 62.89, - 516.14, - 71.98 - ], - "text": "5.7\nDigital Processing of Analog Signals\n553", - "type": "text" - }, - { - "block_id": "p573-b1", - "global_id": 16372, - "bbox": [ - 128.9, - 86.24, - 229.62, - 96.2 - ], - "text": "and the phase response is̸", - "type": "text" - }, - { - "block_id": "p573-b2", - "global_id": 16373, - "bbox": [ - 237.44, - 114.65, - 309.54, - 126.75 - ], - "text": "H[ejωT] = −tan−1", - "type": "text" - }, - { - "block_id": "p573-b3", - "global_id": 16374, - "bbox": [ - 311.15, - 102.39, - 391.18, - 133.82 - ], - "text": "0.7304 sinωT\n1 −0.7304 cosωT", - "type": "text" - }, - { - "block_id": "p573-b5", - "global_id": 16375, - "bbox": [ - 478.69, - 116.79, - 502.75, - 126.75 - ], - "text": "(5.50)", - "type": "text" - }, - { - "block_id": "p573-b6", - "global_id": 16376, - "bbox": [ - 128.9, - 146.63, - 502.76, - 264.6 - ], - "text": "This frequency response differs from the desired response Ha(jω) because aliasing causes\nfrequencies above π/T to appear as frequencies below π/T. This generally results in increased\ngain for frequencies below π/T. For instance, the realized filter gain at ω = 0 is H[ej0] = H[1].\nThis value, as obtained from Eq. (5.48), is 1.1654 instead of the desired value 1. We can\npartly compensate for this distortion by multiplying H[z] or H[ejωT] by a normalizing constant\nK = Ha(0)/H[1] = 1/1.1654 = 0.858. This forces the resulting gain of H[ejωT] to be equal to\n1 at ω = 0. The normalized Hn[z] = 0.858H[z] = 0.858(0.1πz/(z −0.7304)). The amplitude\nresponse in Eq. (5.49) is multiplied by K = 0.858 and plotted in Fig. 5.26b over the frequency\nrange 0 ≤ω ≤π/T = 106. The multiplying constant K has no effect on the phase response in\nEq. (5.50), which is shown in Fig. 5.26c.", - "type": "text" - }, - { - "block_id": "p573-b7", - "global_id": 16377, - "bbox": [ - 146.84, - 265.25, - 460.97, - 277.64 - ], - "text": "Also, the desired frequency response, according to Eq. (5.46) with ωc = 105, is", - "type": "text" - }, - { - "block_id": "p573-b8", - "global_id": 16378, - "bbox": [ - 255.13, - 290.66, - 326.27, - 315.79 - ], - "text": "Ha(jω) =\nωc\njω + ωc", - "type": "text" - }, - { - "block_id": "p573-b9", - "global_id": 16379, - "bbox": [ - 330.02, - 289.72, - 364.65, - 307.6 - ], - "text": "=\n105", - "type": "text" - }, - { - "block_id": "p573-b10", - "global_id": 16380, - "bbox": [ - 341.03, - 304.52, - 374.83, - 315.09 - ], - "text": "jω + 105", - "type": "text" - }, - { - "block_id": "p573-b11", - "global_id": 16381, - "bbox": [ - 128.9, - 327.9, - 170.65, - 337.86 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p573-b12", - "global_id": 16382, - "bbox": [ - 194.2, - 350.3, - 271.25, - 369.3 - ], - "text": "|Ha(jω)| =\n105\n√", - "type": "text" - }, - { - "block_id": "p573-b13", - "global_id": 16383, - "bbox": [ - 249.5, - 351.24, - 432.46, - 377.07 - ], - "text": "ω2 + 1010\nand̸\nHa(jω) = −tan−1 ω", - "type": "text" - }, - { - "block_id": "p573-b14", - "global_id": 16384, - "bbox": [ - 422.32, - 365.09, - 435.76, - 375.67 - ], - "text": "105", - "type": "text" - }, - { - "block_id": "p573-b15", - "global_id": 16385, - "bbox": [ - 128.9, - 388.91, - 502.8, - 458.65 - ], - "text": "This desired amplitude and phase response are plotted (dotted) in Figs. 5.26b and 5.26c for\ncomparison with realized digital filter response. Observe that the amplitude response behavior\nof the analog and the digital filter is very close over the range ω ≤ωc = 105. However, for\nhigher frequencies, there is considerable aliasing, especially in the phase spectrum. Had we\nused the 1% rule, the realized frequency response would have been closer over another decade\nof the frequency range.", - "type": "text" - }, - { - "block_id": "p573-b16", - "global_id": 16386, - "bbox": [ - 128.9, - 473.1, - 502.74, - 511.17 - ], - "text": "IMPULSE INVARIANCE BY MATLAB\nWe can readily use the MATLAB impinvar command to confirm our digital filter designed\nby the impulse-invariance method.", - "type": "text" - }, - { - "block_id": "p573-b17", - "global_id": 16387, - "bbox": [ - 128.9, - 520.87, - 463.64, - 566.7 - ], - "text": ">>\nomegac = 10^5; Ba = [omegac]; Aa = [1 omegac]; Fs = 10^6/pi;\n>>\n[B,A] = impinvar(Ba,Aa,Fs)\nB = 0.3142\nA = 1.0000\n-0.7304", - "type": "text" - }, - { - "block_id": "p573-b18", - "global_id": 16388, - "bbox": [ - 128.9, - 575.81, - 465.54, - 585.78 - ], - "text": "This confirms our earlier result of Eq. (5.48) that the digital filter transfer function is", - "type": "text" - }, - { - "block_id": "p573-b19", - "global_id": 16389, - "bbox": [ - 278.56, - 598.6, - 351.89, - 622.72 - ], - "text": "H[z] =\n0.3142z\nz −0.7304", - "type": "text" - } - ] - }, - { - "page_num": 574, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p574-b0", - "global_id": 16390, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "554\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p574-b1", - "global_id": 16391, - "bbox": [ - 107.82, - 97.81, - 453.44, - 109.76 - ], - "text": "DRILL 5.22\nFilter Design by the Impulse-Invariance Method", - "type": "text" - }, - { - "block_id": "p574-b2", - "global_id": 16392, - "bbox": [ - 107.82, - 118.88, - 341.42, - 128.84 - ], - "text": "Design a digital filter to realize an analog transfer function", - "type": "text" - }, - { - "block_id": "p574-b3", - "global_id": 16393, - "bbox": [ - 265.39, - 138.78, - 325.63, - 162.8 - ], - "text": "Ha(s) =\n20\ns + 20", - "type": "text" - }, - { - "block_id": "p574-b4", - "global_id": 16394, - "bbox": [ - 108.09, - 183.96, - 162.68, - 194.92 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p574-b5", - "global_id": 16395, - "bbox": [ - 107.82, - 197.85, - 238.7, - 222.28 - ], - "text": "H[z] =\n20Tz\nz −e−20T with T =\nπ\n2000", - "type": "text" - }, - { - "block_id": "p574-b6", - "global_id": 16396, - "bbox": [ - 102.2, - 260.25, - 327.05, - 275.2 - ], - "text": "5.8 THE BILATERAL z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p574-b7", - "global_id": 16397, - "bbox": [ - 101.84, - 281.09, - 490.41, - 315.05 - ], - "text": "Situations involving noncausal signals or systems cannot be handled by the (unilateral) z-transform\ndiscussed so far. Such cases can be analyzed by the bilateral (or two-sided) z-transform defined in\nEq. (5.1) as", - "type": "text" - }, - { - "block_id": "p574-b8", - "global_id": 16398, - "bbox": [ - 255.68, - 324.77, - 282.54, - 335.04 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p574-b9", - "global_id": 16399, - "bbox": [ - 288.28, - 314.59, - 302.38, - 325.26 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p574-b10", - "global_id": 16400, - "bbox": [ - 284.59, - 338.61, - 306.06, - 345.81 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p574-b11", - "global_id": 16401, - "bbox": [ - 307.18, - 323.03, - 336.05, - 335.04 - ], - "text": "x[n]z−n", - "type": "text" - }, - { - "block_id": "p574-b12", - "global_id": 16402, - "bbox": [ - 101.84, - 352.41, - 302.89, - 362.48 - ], - "text": "As in Eq. (5.2), the inverse z-transform is given by", - "type": "text" - }, - { - "block_id": "p574-b13", - "global_id": 16403, - "bbox": [ - 245.68, - 371.3, - 289.54, - 395.32 - ], - "text": "x[n] =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p574-b14", - "global_id": 16404, - "bbox": [ - 291.83, - 364.3, - 346.55, - 388.14 - ], - "text": "5\nX[z]zn−1 dz", - "type": "text" - }, - { - "block_id": "p574-b15", - "global_id": 16405, - "bbox": [ - 101.84, - 403.91, - 390.11, - 413.97 - ], - "text": "These equations define the bilateral z-transform. Earlier, we showed that", - "type": "text" - }, - { - "block_id": "p574-b16", - "global_id": 16406, - "bbox": [ - 232.23, - 420.28, - 490.38, - 444.3 - ], - "text": "γ nu[n] ⇐⇒\nz\nz −γ\n|z| > |γ |\n(5.51)", - "type": "text" - }, - { - "block_id": "p574-b17", - "global_id": 16407, - "bbox": [ - 101.85, - 451.76, - 439.32, - 463.14 - ], - "text": "In contrast, the z-transform of the signal −γ nu[−(n + 1)], illustrated in Fig. 5.27a, is", - "type": "text" - }, - { - "block_id": "p574-b18", - "global_id": 16408, - "bbox": [ - 173.14, - 481.68, - 264.35, - 493.56 - ], - "text": "Z{−γ nu[−(n + 1)]} =", - "type": "text" - }, - { - "block_id": "p574-b19", - "global_id": 16409, - "bbox": [ - 266.4, - 473.01, - 280.5, - 483.68 - ], - "text": "−1\n\"", - "type": "text" - }, - { - "block_id": "p574-b20", - "global_id": 16410, - "bbox": [ - 267.17, - 497.03, - 279.72, - 504.01 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p574-b21", - "global_id": 16411, - "bbox": [ - 281.6, - 479.07, - 323.08, - 493.46 - ], - "text": "−γ nz−n =", - "type": "text" - }, - { - "block_id": "p574-b22", - "global_id": 16412, - "bbox": [ - 325.12, - 473.01, - 339.21, - 483.68 - ], - "text": "−1\n\"", - "type": "text" - }, - { - "block_id": "p574-b23", - "global_id": 16413, - "bbox": [ - 325.89, - 497.03, - 338.45, - 504.01 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p574-b24", - "global_id": 16414, - "bbox": [ - 340.33, - 483.18, - 348.1, - 493.15 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p574-b25", - "global_id": 16415, - "bbox": [ - 348.09, - 469.2, - 361.01, - 486.16 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p574-b26", - "global_id": 16416, - "bbox": [ - 357.37, - 490.57, - 361.25, - 500.53 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p574-b27", - "global_id": 16417, - "bbox": [ - 363.81, - 469.2, - 374.02, - 481.01 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p574-b28", - "global_id": 16418, - "bbox": [ - 256.58, - 517.74, - 274.17, - 527.71 - ], - "text": "= −", - "type": "text" - }, - { - "block_id": "p574-b30", - "global_id": 16419, - "bbox": [ - 282.66, - 511.07, - 299.77, - 534.78 - ], - "text": "z\nγ +", - "type": "text" - }, - { - "block_id": "p574-b31", - "global_id": 16420, - "bbox": [ - 301.31, - 503.76, - 314.47, - 521.04 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p574-b32", - "global_id": 16421, - "bbox": [ - 309.24, - 524.82, - 314.24, - 534.78 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p574-b33", - "global_id": 16422, - "bbox": [ - 317.03, - 503.76, - 327.24, - 515.64 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p574-b34", - "global_id": 16423, - "bbox": [ - 329.29, - 517.74, - 337.06, - 527.71 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p574-b35", - "global_id": 16424, - "bbox": [ - 338.61, - 503.76, - 351.77, - 521.04 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p574-b36", - "global_id": 16425, - "bbox": [ - 346.53, - 524.82, - 351.53, - 534.78 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p574-b37", - "global_id": 16426, - "bbox": [ - 354.32, - 503.76, - 364.53, - 515.64 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p574-b38", - "global_id": 16427, - "bbox": [ - 366.58, - 517.74, - 388.97, - 527.71 - ], - "text": "+ · · ·", - "type": "text" - }, - { - "block_id": "p574-b40", - "global_id": 16428, - "bbox": [ - 256.58, - 551.62, - 280.7, - 562.0 - ], - "text": "= 1 −", - "type": "text" - }, - { - "block_id": "p574-b42", - "global_id": 16429, - "bbox": [ - 288.44, - 544.95, - 322.59, - 568.66 - ], - "text": "1 + z\nγ +", - "type": "text" - }, - { - "block_id": "p574-b43", - "global_id": 16430, - "bbox": [ - 324.14, - 537.62, - 337.3, - 554.91 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p574-b44", - "global_id": 16431, - "bbox": [ - 332.06, - 558.69, - 337.07, - 568.66 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p574-b45", - "global_id": 16432, - "bbox": [ - 339.86, - 537.62, - 350.07, - 549.52 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p574-b46", - "global_id": 16433, - "bbox": [ - 352.12, - 551.62, - 359.89, - 561.58 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p574-b47", - "global_id": 16434, - "bbox": [ - 361.43, - 537.63, - 374.59, - 554.91 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p574-b48", - "global_id": 16435, - "bbox": [ - 369.36, - 558.69, - 374.36, - 568.66 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p574-b49", - "global_id": 16436, - "bbox": [ - 377.14, - 537.62, - 387.35, - 549.52 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p574-b50", - "global_id": 16437, - "bbox": [ - 389.4, - 551.62, - 411.79, - 561.58 - ], - "text": "+ · · ·", - "type": "text" - }, - { - "block_id": "p574-b52", - "global_id": 16438, - "bbox": [ - 256.58, - 575.93, - 298.36, - 592.88 - ], - "text": "= 1 −\n1", - "type": "text" - }, - { - "block_id": "p574-b53", - "global_id": 16439, - "bbox": [ - 283.45, - 586.3, - 305.73, - 610.01 - ], - "text": "1 −z\nγ", - "type": "text" - }, - { - "block_id": "p574-b55", - "global_id": 16440, - "bbox": [ - 334.94, - 575.83, - 340.18, - 599.54 - ], - "text": "z\nγ", - "type": "text" - }, - { - "block_id": "p574-b56", - "global_id": 16441, - "bbox": [ - 342.74, - 568.05, - 362.82, - 595.95 - ], - "text": "< 1", - "type": "text" - }, - { - "block_id": "p574-b57", - "global_id": 16442, - "bbox": [ - 256.58, - 610.76, - 343.8, - 634.79 - ], - "text": "=\nz\nz −γ\n|z| < |γ |", - "type": "text" - } - ] - }, - { - "page_num": 575, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p575-b0", - "global_id": 16443, - "bbox": [ - 353.04, - 62.57, - 516.13, - 71.98 - ], - "text": "5.8\nThe Bilateral z-Transform\n555", - "type": "text" - }, - { - "block_id": "p575-b1", - "global_id": 16444, - "bbox": [ - 284.7, - 103.33, - 339.59, - 112.87 - ], - "text": "gnu[(n 1)]", - "type": "text" - }, - { - "block_id": "p575-b2", - "global_id": 16445, - "bbox": [ - 387.88, - 169.24, - 462.08, - 177.24 - ], - "text": "Region of convergence", - "type": "text" - }, - { - "block_id": "p575-b3", - "global_id": 16446, - "bbox": [ - 180.73, - 126.94, - 289.52, - 146.51 - ], - "text": "n\n1\n3\n5\n7", - "type": "text" - }, - { - "block_id": "p575-b4", - "global_id": 16447, - "bbox": [ - 436.33, - 133.46, - 444.49, - 141.65 - ], - "text": "g", - "type": "text" - }, - { - "block_id": "p575-b5", - "global_id": 16448, - "bbox": [ - 452.23, - 92.41, - 474.67, - 100.49 - ], - "text": "z plane", - "type": "text" - }, - { - "block_id": "p575-b6", - "global_id": 16449, - "bbox": [ - 151.5, - 222.01, - 489.56, - 232.67 - ], - "text": "Figure 5.27 (a) −γ nu[−(n + 1)] and (b) the region of convergence (ROC) of its z-transform.", - "type": "text" - }, - { - "block_id": "p575-b7", - "global_id": 16450, - "bbox": [ - 127.59, - 256.1, - 169.34, - 266.07 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p575-b8", - "global_id": 16451, - "bbox": [ - 236.54, - 267.21, - 516.13, - 291.23 - ], - "text": "Z{−γ nu[−(n + 1)]} =\nz\nz −γ\n|z| < |γ |\n(5.52)", - "type": "text" - }, - { - "block_id": "p575-b9", - "global_id": 16452, - "bbox": [ - 127.59, - 306.69, - 516.14, - 389.8 - ], - "text": "A comparison of Eqs. (5.51) and (5.52) shows that the z-transform of γ nu[n] is identical to\nthat of −γ nu[−(n + 1)]. The regions of convergence, however, are different. In the former case,\nX[z] converges for |z| > |γ |; in the latter, X[z] converges for |z| < |γ | (see Fig. 5.27b). Clearly,\nthe inverse transform of X[z] is not unique unless the region of convergence is specified. If we\nadd the restriction that all our signals be causal, however, this ambiguity does not arise. The\ninverse transform of z/(z −γ ) is γ nu[n] even without specifying the ROC. Thus, in the unilateral\ntransform, we can ignore the ROC in determining the inverse z-transform of X[z].", - "type": "text" - }, - { - "block_id": "p575-b10", - "global_id": 16453, - "bbox": [ - 145.52, - 383.91, - 421.25, - 401.75 - ], - "text": "As in the case of the bilateral Laplace transform, if x[n] = %k", - "type": "text" - }, - { - "block_id": "p575-b11", - "global_id": 16454, - "bbox": [ - 127.59, - 391.37, - 516.13, - 426.44 - ], - "text": "i=1 xi[n], then the ROC\nfor X[z] is the intersection of the ROCs (region common to all ROCs) for the transforms\nX1[z],X2[z],. . .,Xk[z].", - "type": "text" - }, - { - "block_id": "p575-b12", - "global_id": 16455, - "bbox": [ - 127.59, - 427.65, - 516.15, - 461.53 - ], - "text": "The preceding results lead to the conclusion (similar to that for the Laplace transform) that\nif z = β is the largest magnitude pole for a causal sequence, its ROC is |z| > |β|. If z = α is the\nsmallest magnitude nonzero pole for an anticausal sequence, its ROC is |z| < |α|.", - "type": "text" - }, - { - "block_id": "p575-b13", - "global_id": 16456, - "bbox": [ - 127.59, - 477.72, - 516.13, - 531.23 - ], - "text": "REGION OF CONVERGENCE FOR LEFT-SIDED\nAND RIGHT-SIDED SEQUENCES\nLet us first consider a finite duration sequence xf [n], defined as a sequence that is nonzero for\nN1 ≤n ≤N2, where both N1 and N2 are finite numbers and N2 > N1. Also,", - "type": "text" - }, - { - "block_id": "p575-b14", - "global_id": 16457, - "bbox": [ - 280.41, - 553.63, - 310.29, - 565.08 - ], - "text": "Xf [z] =", - "type": "text" - }, - { - "block_id": "p575-b15", - "global_id": 16458, - "bbox": [ - 313.82, - 543.11, - 327.92, - 554.12 - ], - "text": "N2\n\"", - "type": "text" - }, - { - "block_id": "p575-b16", - "global_id": 16459, - "bbox": [ - 312.34, - 567.92, - 328.9, - 576.44 - ], - "text": "n=N1", - "type": "text" - }, - { - "block_id": "p575-b17", - "global_id": 16460, - "bbox": [ - 330.51, - 551.9, - 362.81, - 565.08 - ], - "text": "xf [n]z−n", - "type": "text" - }, - { - "block_id": "p575-b18", - "global_id": 16461, - "bbox": [ - 127.59, - 588.81, - 295.16, - 600.26 - ], - "text": "For example, if N1 = −2 and N2 = 1, then", - "type": "text" - }, - { - "block_id": "p575-b19", - "global_id": 16462, - "bbox": [ - 235.77, - 611.63, - 406.75, - 630.1 - ], - "text": "Xf [z] = xf [−2]z2 + xf [−1]z + xf [0] + xf [1]", - "type": "text" - }, - { - "block_id": "p575-b20", - "global_id": 16463, - "bbox": [ - 395.07, - 626.03, - 398.95, - 635.99 - ], - "text": "z", - "type": "text" - } - ] - }, - { - "page_num": 576, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p576-b0", - "global_id": 16464, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "556\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p576-b1", - "global_id": 16465, - "bbox": [ - 101.84, - 85.4, - 490.39, - 131.64 - ], - "text": "Assuming all the elements in xf [n] are finite, we observe that Xf [z] has two poles at z = ∞because\nof terms xf [−2]z2 + xf [−1]z and one pole at z = 0 because of term xf [1]/z. Thus, a finite-duration\nsequence could have poles at z = 0 and z = ∞. Observe that Xf [z] converges for all values of z\nexcept possibly z = 0 and z = ∞.", - "type": "text" - }, - { - "block_id": "p576-b2", - "global_id": 16466, - "bbox": [ - 101.85, - 133.23, - 490.39, - 155.56 - ], - "text": "This means that the ROC of a general signal x[n] + xf [n] is the same as the ROC of x[n] with\nthe possible exception of z = 0 and z = ∞.", - "type": "text" - }, - { - "block_id": "p576-b3", - "global_id": 16467, - "bbox": [ - 101.85, - 157.13, - 490.39, - 203.37 - ], - "text": "A right-sided sequence is zero for n < N2 < ∞and a left-sided sequence is zero for n > N1 >\n−∞. A causal sequence is always a right-sided sequence, but the converse is not necessarily true.\nAn anticausal sequence is always a left-sided sequence, but the converse is not necessarily true. A\ntwo-sided sequence is of infinite duration and is neither right-sided nor left-sided.", - "type": "text" - }, - { - "block_id": "p576-b4", - "global_id": 16468, - "bbox": [ - 101.84, - 204.96, - 490.39, - 287.06 - ], - "text": "A right-sided sequence xr[n] can be expressed as xr[n] = xc[n]+xf [n], where xc[n] is a causal\nsignal and xf [n] is a finite-duration signal. Therefore, the ROC for xr[n] is the same as the ROC for\nxc[n] except possibly z = ∞. If z = β is the largest magnitude pole for a right-sided sequence xr[n],\nits ROC is |β| < |z| ≤∞. Similarly, a left-sided sequence can be expressed as xl[n] = xa[n]+xf [n],\nwhere xa[n] is an anticausal sequence and xf [n] is a finite-duration signal. Therefore, the ROC for\nxl[n] is the same as the ROC for xa[n] except possibly z = 0. Thus, if z = α is the smallest magnitude\nnonzero pole for a left-sided sequence, its ROC is 0 ≤|z| < |α|.", - "type": "text" - }, - { - "block_id": "p576-b5", - "global_id": 16469, - "bbox": [ - 76.77, - 345.79, - 296.85, - 357.75 - ], - "text": "EXAMPLE 5.17\nBilateral z-Transform", - "type": "text" - }, - { - "block_id": "p576-b6", - "global_id": 16470, - "bbox": [ - 103.16, - 371.4, - 254.22, - 381.46 - ], - "text": "Determine the bilateral z-transform of", - "type": "text" - }, - { - "block_id": "p576-b7", - "global_id": 16471, - "bbox": [ - 215.13, - 391.5, - 283.53, - 419.06 - ], - "text": "x[n] = (0.9)nu[n]\n\n\n\nx1[n]", - "type": "text" - }, - { - "block_id": "p576-b8", - "global_id": 16472, - "bbox": [ - 284.65, - 391.5, - 361.73, - 419.06 - ], - "text": "+(1.2)nu[−(n + 1)\n\n\n\nx2[n]", - "type": "text" - }, - { - "block_id": "p576-b9", - "global_id": 16473, - "bbox": [ - 361.73, - 393.0, - 365.04, - 402.96 - ], - "text": "]", - "type": "text" - }, - { - "block_id": "p576-b10", - "global_id": 16474, - "bbox": [ - 103.16, - 443.55, - 305.62, - 453.51 - ], - "text": "From the results in Eqs. (5.51) and (5.52), we have", - "type": "text" - }, - { - "block_id": "p576-b11", - "global_id": 16475, - "bbox": [ - 232.16, - 460.95, - 348.01, - 485.06 - ], - "text": "X1[z] =\nz\nz −0.9\n|z| > 0.9", - "type": "text" - }, - { - "block_id": "p576-b12", - "global_id": 16476, - "bbox": [ - 232.16, - 485.32, - 348.01, - 509.75 - ], - "text": "X2[z] =\n−z\nz −1.2\n|z| < 1.2", - "type": "text" - }, - { - "block_id": "p576-b13", - "global_id": 16477, - "bbox": [ - 103.16, - 517.96, - 477.0, - 529.11 - ], - "text": "The common region where both X1[z] and X2[z] converge is 0.9 < |z| < 1.2 (Fig. 5.28b). Hence,", - "type": "text" - }, - { - "block_id": "p576-b14", - "global_id": 16478, - "bbox": [ - 200.74, - 539.88, - 281.69, - 551.03 - ], - "text": "X[z] = X1[z] + X2[z]", - "type": "text" - }, - { - "block_id": "p576-b15", - "global_id": 16479, - "bbox": [ - 219.82, - 552.14, - 298.5, - 576.27 - ], - "text": "=\nz\nz −0.9 −\nz\nz −1.2", - "type": "text" - }, - { - "block_id": "p576-b16", - "global_id": 16480, - "bbox": [ - 219.82, - 577.5, - 379.43, - 601.93 - ], - "text": "=\n−0.3z\n(z −0.9)(z −1.2)\n0.9 < |z| < 1.2", - "type": "text" - }, - { - "block_id": "p576-b17", - "global_id": 16481, - "bbox": [ - 103.16, - 611.03, - 362.47, - 621.41 - ], - "text": "The sequence x[n] and the ROC of X[z] are depicted in Fig. 5.28.", - "type": "text" - } - ] - }, - { - "page_num": 577, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p577-b0", - "global_id": 16482, - "bbox": [ - 353.04, - 62.57, - 516.13, - 71.98 - ], - "text": "5.8\nThe Bilateral z-Transform\n557", - "type": "text" - }, - { - "block_id": "p577-b1", - "global_id": 16483, - "bbox": [ - 381.91, - 102.43, - 464.85, - 112.93 - ], - "text": "Im\nz plane", - "type": "text" - }, - { - "block_id": "p577-b2", - "global_id": 16484, - "bbox": [ - 413.91, - 165.93, - 479.3, - 177.43 - ], - "text": "Re\n0.9", - "type": "text" - }, - { - "block_id": "p577-b3", - "global_id": 16485, - "bbox": [ - 412.91, - 184.93, - 422.91, - 192.93 - ], - "text": "1.2", - "type": "text" - }, - { - "block_id": "p577-b4", - "global_id": 16486, - "bbox": [ - 393.69, - 262.11, - 403.34, - 270.11 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p577-b5", - "global_id": 16487, - "bbox": [ - 130.25, - 227.09, - 279.84, - 235.39 - ], - "text": "0\n15\n25\n15", - "type": "text" - }, - { - "block_id": "p577-b6", - "global_id": 16488, - "bbox": [ - 181.27, - 124.51, - 208.34, - 141.16 - ], - "text": "1\nx[n]", - "type": "text" - }, - { - "block_id": "p577-b7", - "global_id": 16489, - "bbox": [ - 276.21, - 235.75, - 280.21, - 243.75 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p577-b8", - "global_id": 16490, - "bbox": [ - 190.37, - 262.11, - 199.25, - 270.11 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p577-b9", - "global_id": 16491, - "bbox": [ - 412.91, - 184.93, - 422.91, - 192.93 - ], - "text": "1.2", - "type": "text" - }, - { - "block_id": "p577-b10", - "global_id": 16492, - "bbox": [ - 413.91, - 165.93, - 423.91, - 173.93 - ], - "text": "0.9", - "type": "text" - }, - { - "block_id": "p577-b11", - "global_id": 16493, - "bbox": [ - 105.99, - 276.43, - 286.62, - 286.04 - ], - "text": "Figure 5.28 (a) Signal x[n] and (b) ROC of X[z].", - "type": "text" - }, - { - "block_id": "p577-b12", - "global_id": 16494, - "bbox": [ - 102.51, - 358.06, - 366.09, - 370.01 - ], - "text": "EXAMPLE 5.18\nInverse Bilateral z-Transform", - "type": "text" - }, - { - "block_id": "p577-b13", - "global_id": 16495, - "bbox": [ - 128.9, - 383.66, - 286.9, - 393.73 - ], - "text": "Find the inverse bilateral z-transform of", - "type": "text" - }, - { - "block_id": "p577-b14", - "global_id": 16496, - "bbox": [ - 269.3, - 407.98, - 361.17, - 432.41 - ], - "text": "X[z] =\n−z(z + 0.4)\n(z −0.8)(z −2)", - "type": "text" - }, - { - "block_id": "p577-b15", - "global_id": 16497, - "bbox": [ - 128.9, - 447.62, - 366.57, - 458.0 - ], - "text": "if the ROC is (a) |z| > 2, (b) |z| < 0.8, and (c) 0.8 < |z| < 2.", - "type": "text" - }, - { - "block_id": "p577-b16", - "global_id": 16498, - "bbox": [ - 146.84, - 480.83, - 158.46, - 490.79 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p577-b17", - "global_id": 16499, - "bbox": [ - 232.07, - 490.07, - 249.11, - 500.35 - ], - "text": "X[z]", - "type": "text" - }, - { - "block_id": "p577-b18", - "global_id": 16500, - "bbox": [ - 238.65, - 490.07, - 399.59, - 514.5 - ], - "text": "z\n=\n−(z + 0.4)\n(z −0.8)(z −2) =\n1\nz −0.8 −\n2\nz −2", - "type": "text" - }, - { - "block_id": "p577-b19", - "global_id": 16501, - "bbox": [ - 128.9, - 521.03, - 143.28, - 531.0 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p577-b20", - "global_id": 16502, - "bbox": [ - 267.6, - 525.9, - 362.85, - 550.02 - ], - "text": "X[z] =\nz\nz −0.8 −2\nz\nz −2", - "type": "text" - }, - { - "block_id": "p577-b21", - "global_id": 16503, - "bbox": [ - 128.9, - 555.24, - 417.8, - 565.62 - ], - "text": "Since the ROC is |z| > 2, both terms correspond to causal sequences and", - "type": "text" - }, - { - "block_id": "p577-b22", - "global_id": 16504, - "bbox": [ - 262.19, - 579.03, - 369.48, - 593.52 - ], - "text": "x[n] = [(0.8)n −2(2)n]u[n]", - "type": "text" - }, - { - "block_id": "p577-b23", - "global_id": 16505, - "bbox": [ - 128.91, - 611.45, - 273.9, - 621.41 - ], - "text": "This sequence appears in Fig. 5.29a.", - "type": "text" - } - ] - }, - { - "page_num": 578, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p578-b0", - "global_id": 16506, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "558\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p578-b1", - "global_id": 16507, - "bbox": [ - 103.17, - 85.95, - 477.03, - 108.29 - ], - "text": "(b) In this case, |z| < 0.8, which is less than the magnitudes of both poles. Hence, both\nterms correspond to anticausal sequences, and", - "type": "text" - }, - { - "block_id": "p578-b2", - "global_id": 16508, - "bbox": [ - 217.03, - 115.72, - 363.14, - 130.2 - ], - "text": "x[n] = [−(0.8)n + 2(2)n]u[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p578-b3", - "global_id": 16509, - "bbox": [ - 103.17, - 142.16, - 248.31, - 152.12 - ], - "text": "This sequence appears in Fig. 5.29b.", - "type": "text" - }, - { - "block_id": "p578-b4", - "global_id": 16510, - "bbox": [ - 103.17, - 153.7, - 477.02, - 176.03 - ], - "text": "(c) In this case, 0.8 < |z| < 2; the part of X[z] corresponding to the pole at 0.8 is a causal\nsequence, and the part corresponding to the pole at 2 is an anticausal sequence:", - "type": "text" - }, - { - "block_id": "p578-b5", - "global_id": 16511, - "bbox": [ - 215.94, - 186.07, - 364.23, - 197.95 - ], - "text": "x[n] = (0.8)nu[n] + 2(2)nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p578-b6", - "global_id": 16512, - "bbox": [ - 103.17, - 209.91, - 248.15, - 219.87 - ], - "text": "This sequence appears in Fig. 5.29c.", - "type": "text" - }, - { - "block_id": "p578-b7", - "global_id": 16513, - "bbox": [ - 156.25, - 275.17, - 226.57, - 283.17 - ], - "text": "5\n10\n15", - "type": "text" - }, - { - "block_id": "p578-b8", - "global_id": 16514, - "bbox": [ - 87.22, - 318.3, - 119.55, - 327.77 - ], - "text": "5 105", - "type": "text" - }, - { - "block_id": "p578-b9", - "global_id": 16515, - "bbox": [ - 101.89, - 353.35, - 119.55, - 362.82 - ], - "text": "106", - "type": "text" - }, - { - "block_id": "p578-b10", - "global_id": 16516, - "bbox": [ - 240.09, - 268.37, - 244.09, - 276.37 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p578-b11", - "global_id": 16517, - "bbox": [ - 119.9, - 275.17, - 123.9, - 283.17 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p578-b12", - "global_id": 16518, - "bbox": [ - 185.27, - 438.96, - 194.15, - 446.96 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p578-b13", - "global_id": 16519, - "bbox": [ - 349.58, - 274.88, - 426.26, - 283.17 - ], - "text": "15\n10\n5", - "type": "text" - }, - { - "block_id": "p578-b14", - "global_id": 16520, - "bbox": [ - 450.01, - 306.18, - 464.67, - 314.47 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p578-b15", - "global_id": 16521, - "bbox": [ - 450.01, - 336.99, - 464.67, - 345.28 - ], - "text": "40", - "type": "text" - }, - { - "block_id": "p578-b16", - "global_id": 16522, - "bbox": [ - 450.01, - 367.8, - 464.67, - 376.09 - ], - "text": "60", - "type": "text" - }, - { - "block_id": "p578-b17", - "global_id": 16523, - "bbox": [ - 432.1, - 268.37, - 436.1, - 276.37 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p578-b18", - "global_id": 16524, - "bbox": [ - 450.91, - 275.17, - 454.91, - 283.17 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p578-b19", - "global_id": 16525, - "bbox": [ - 105.4, - 289.2, - 474.46, - 297.28 - ], - "text": "x[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p578-b20", - "global_id": 16526, - "bbox": [ - 295.76, - 476.16, - 308.64, - 484.24 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p578-b21", - "global_id": 16527, - "bbox": [ - 393.08, - 569.51, - 397.08, - 577.51 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p578-b22", - "global_id": 16528, - "bbox": [ - 376.42, - 438.96, - 386.07, - 446.96 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p578-b23", - "global_id": 16529, - "bbox": [ - 202.71, - 569.3, - 345.14, - 577.6 - ], - "text": "10\n5\n5\n0\n10", - "type": "text" - }, - { - "block_id": "p578-b24", - "global_id": 16530, - "bbox": [ - 279.33, - 460.32, - 283.33, - 468.32 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p578-b25", - "global_id": 16531, - "bbox": [ - 274.33, - 586.59, - 283.21, - 594.59 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p578-b26", - "global_id": 16532, - "bbox": [ - 87.23, - 600.92, - 287.6, - 610.53 - ], - "text": "Figure 5.29 Three possible inverse transforms of X[z].", - "type": "text" - } - ] - }, - { - "page_num": 579, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p579-b0", - "global_id": 16533, - "bbox": [ - 353.04, - 62.57, - 516.13, - 71.98 - ], - "text": "5.8\nThe Bilateral z-Transform\n559", - "type": "text" - }, - { - "block_id": "p579-b1", - "global_id": 16534, - "bbox": [ - 133.57, - 97.81, - 373.96, - 109.77 - ], - "text": "DRILL 5.23\nInverse Bilateral z-Transform", - "type": "text" - }, - { - "block_id": "p579-b2", - "global_id": 16535, - "bbox": [ - 133.57, - 118.78, - 291.56, - 128.85 - ], - "text": "Find the inverse bilateral z-transform of", - "type": "text" - }, - { - "block_id": "p579-b3", - "global_id": 16536, - "bbox": [ - 251.1, - 134.97, - 305.76, - 159.8 - ], - "text": "X[z] =\nz\nz2 + 5", - "type": "text" - }, - { - "block_id": "p579-b4", - "global_id": 16537, - "bbox": [ - 301.12, - 148.18, - 325.24, - 162.71 - ], - "text": "6z + 1\n6", - "type": "text" - }, - { - "block_id": "p579-b5", - "global_id": 16538, - "bbox": [ - 348.75, - 140.29, - 391.42, - 155.14 - ], - "text": "1\n2 > |z| > 1\n3", - "type": "text" - }, - { - "block_id": "p579-b6", - "global_id": 16539, - "bbox": [ - 133.57, - 183.83, - 188.43, - 203.73 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p579-b7", - "global_id": 16540, - "bbox": [ - 139.13, - 200.42, - 153.13, - 211.74 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p579-b8", - "global_id": 16541, - "bbox": [ - 149.64, - 193.77, - 194.77, - 214.97 - ], - "text": "3\nnu[n] + 6", - "type": "text" - }, - { - "block_id": "p579-b10", - "global_id": 16542, - "bbox": [ - 200.34, - 200.42, - 214.35, - 211.74 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p579-b11", - "global_id": 16543, - "bbox": [ - 210.86, - 193.77, - 271.18, - 214.97 - ], - "text": "2\nnu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p579-b12", - "global_id": 16544, - "bbox": [ - 127.89, - 244.89, - 488.86, - 257.81 - ], - "text": "INVERSE TRANSFORM BY EXPANSION OF X[z] IN POWER SERIES OF z", - "type": "text" - }, - { - "block_id": "p579-b13", - "global_id": 16545, - "bbox": [ - 127.59, - 261.85, - 161.55, - 271.81 - ], - "text": "We have", - "type": "text" - }, - { - "block_id": "p579-b14", - "global_id": 16546, - "bbox": [ - 285.12, - 277.16, - 311.98, - 287.44 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p579-b15", - "global_id": 16547, - "bbox": [ - 314.03, - 267.7, - 328.13, - 277.66 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p579-b16", - "global_id": 16548, - "bbox": [ - 319.33, - 291.23, - 322.82, - 298.21 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p579-b17", - "global_id": 16549, - "bbox": [ - 329.23, - 275.44, - 358.1, - 287.44 - ], - "text": "x[n]z−n", - "type": "text" - }, - { - "block_id": "p579-b18", - "global_id": 16550, - "bbox": [ - 127.59, - 307.09, - 448.53, - 317.47 - ], - "text": "For an anticausal sequence, which exists only for n ≤−1, this equation becomes", - "type": "text" - }, - { - "block_id": "p579-b19", - "global_id": 16551, - "bbox": [ - 239.01, - 327.41, - 404.71, - 341.9 - ], - "text": "X[z] = x[−1]z + x[−2]z2 + x[−3]z3 + · · ·", - "type": "text" - }, - { - "block_id": "p579-b20", - "global_id": 16552, - "bbox": [ - 127.59, - 355.96, - 516.13, - 414.15 - ], - "text": "We can find the inverse z-transform of X[z] by dividing the numerator polynomial by the\ndenominator polynomial, both in ascending powers of z, to obtain a polynomial in ascending\npowers of z. Thus, to find the inverse transform of z/(z −0.5) (when the ROC is |z| < 0.5), we\ndivide z by −0.5+z to obtain −2z−4z2−8z3−· · ·. Hence, x[−1] = −2, x[−2] = −4, x[−3] = −8,\nand so on.", - "type": "text" - }, - { - "block_id": "p579-b21", - "global_id": 16553, - "bbox": [ - 127.59, - 440.57, - 365.86, - 453.38 - ], - "text": "5.8-1 Properties of the Bilateral z-Transform", - "type": "text" - }, - { - "block_id": "p579-b22", - "global_id": 16554, - "bbox": [ - 127.59, - 459.41, - 516.15, - 482.13 - ], - "text": "Properties of the bilateral z-transform are similar to those of the unilateral transform. We shall\nmerely state the properties here, without proofs, for xi[n] ⇐⇒Xi[z].", - "type": "text" - }, - { - "block_id": "p579-b23", - "global_id": 16555, - "bbox": [ - 127.89, - 497.42, - 186.25, - 509.54 - ], - "text": "LINEARITY", - "type": "text" - }, - { - "block_id": "p579-b24", - "global_id": 16556, - "bbox": [ - 240.84, - 510.96, - 402.89, - 522.11 - ], - "text": "a1x1[n] + a2x2[n] ⇐⇒a1X1[z] + a2X2[z]", - "type": "text" - }, - { - "block_id": "p579-b25", - "global_id": 16557, - "bbox": [ - 127.59, - 532.41, - 516.13, - 555.51 - ], - "text": "The ROC for a1X1[z] + a2X2[z] is the region common to (intersection of) the ROCs for X1[z] and\nX2[z].", - "type": "text" - }, - { - "block_id": "p579-b26", - "global_id": 16558, - "bbox": [ - 127.89, - 570.73, - 158.9, - 582.86 - ], - "text": "SHIFT", - "type": "text" - }, - { - "block_id": "p579-b27", - "global_id": 16559, - "bbox": [ - 205.82, - 580.69, - 271.56, - 597.54 - ], - "text": "x[n −m] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p579-b28", - "global_id": 16560, - "bbox": [ - 264.37, - 587.26, - 437.88, - 604.61 - ], - "text": "zm X[z]\nm is positive or negative integer", - "type": "text" - }, - { - "block_id": "p579-b29", - "global_id": 16561, - "bbox": [ - 127.59, - 609.25, - 516.14, - 634.79 - ], - "text": "The ROC for X[z]/zm is the ROC for X[z] except for the addition or deletion of z = 0 or z = ∞\ncaused by the factor 1/zm.", - "type": "text" - } - ] - }, - { - "page_num": 580, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p580-b0", - "global_id": 16562, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "560\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p580-b1", - "global_id": 16563, - "bbox": [ - 102.14, - 86.19, - 184.26, - 98.32 - ], - "text": "CONVOLUTION", - "type": "text" - }, - { - "block_id": "p580-b2", - "global_id": 16564, - "bbox": [ - 239.76, - 108.29, - 352.48, - 119.44 - ], - "text": "x1[n] ∗x2[n] ⇐⇒X1[z]X2[z]", - "type": "text" - }, - { - "block_id": "p580-b3", - "global_id": 16565, - "bbox": [ - 101.85, - 135.45, - 484.03, - 146.6 - ], - "text": "The ROC for X1[z]X2[z] is the region common to (intersection of) the ROCs for X1[z] and X2[z].", - "type": "text" - }, - { - "block_id": "p580-b4", - "global_id": 16566, - "bbox": [ - 102.14, - 162.2, - 227.8, - 176.8 - ], - "text": "MULTIPLICATION BY γ n", - "type": "text" - }, - { - "block_id": "p580-b5", - "global_id": 16567, - "bbox": [ - 257.45, - 191.22, - 313.4, - 202.99 - ], - "text": "γ nx[n] ⇐⇒X", - "type": "text" - }, - { - "block_id": "p580-b6", - "global_id": 16568, - "bbox": [ - 314.95, - 178.73, - 326.8, - 196.01 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p580-b7", - "global_id": 16569, - "bbox": [ - 321.57, - 199.79, - 326.57, - 209.75 - ], - "text": "γ", - "type": "text" - }, - { - "block_id": "p580-b8", - "global_id": 16570, - "bbox": [ - 329.36, - 178.73, - 334.79, - 188.69 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p580-b9", - "global_id": 16571, - "bbox": [ - 101.84, - 225.16, - 490.38, - 247.49 - ], - "text": "If the ROC for X[z] is |γ1| < |z| < |γ2|, then the ROC for X[z/γ ] is |γ γ1| < |z| < |γ γ2|, indicating\nthat the ROC is scaled by the factor |γ |.", - "type": "text" - }, - { - "block_id": "p580-b10", - "global_id": 16572, - "bbox": [ - 102.14, - 265.91, - 221.37, - 278.46 - ], - "text": "MULTIPLICATION BY n", - "type": "text" - }, - { - "block_id": "p580-b11", - "global_id": 16573, - "bbox": [ - 245.72, - 284.61, - 326.21, - 301.55 - ], - "text": "nx[n]u[n] ⇐⇒−z d", - "type": "text" - }, - { - "block_id": "p580-b12", - "global_id": 16574, - "bbox": [ - 319.43, - 291.28, - 346.52, - 308.63 - ], - "text": "dzX[z]", - "type": "text" - }, - { - "block_id": "p580-b13", - "global_id": 16575, - "bbox": [ - 101.85, - 322.24, - 331.57, - 332.61 - ], - "text": "The ROC for −z(dX/dz) is the same as the ROC for X[z].", - "type": "text" - }, - { - "block_id": "p580-b14", - "global_id": 16576, - "bbox": [ - 102.14, - 351.46, - 188.36, - 363.58 - ], - "text": "TIME REVERSAL", - "type": "text" - }, - { - "block_id": "p580-b15", - "global_id": 16577, - "bbox": [ - 259.24, - 373.55, - 333.0, - 383.93 - ], - "text": "x[−n] ⇐⇒X[1/z]", - "type": "text" - }, - { - "block_id": "p580-b16", - "global_id": 16578, - "bbox": [ - 101.85, - 400.71, - 452.45, - 411.86 - ], - "text": "If the ROC for X[z] is |γ1| < |z| < |γ2|, then the ROC for X[1/z] is 1/|γ1| > |z| > |1/γ2|.", - "type": "text" - }, - { - "block_id": "p580-b17", - "global_id": 16579, - "bbox": [ - 102.14, - 429.94, - 237.24, - 442.06 - ], - "text": "COMPLEX CONJUGATION", - "type": "text" - }, - { - "block_id": "p580-b18", - "global_id": 16580, - "bbox": [ - 261.76, - 450.3, - 330.48, - 462.31 - ], - "text": "x∗[n] ⇐⇒X∗[z∗]", - "type": "text" - }, - { - "block_id": "p580-b19", - "global_id": 16581, - "bbox": [ - 101.84, - 477.96, - 312.74, - 489.56 - ], - "text": "The ROC for X∗[z∗] is the same as the ROC for X[z].", - "type": "text" - }, - { - "block_id": "p580-b20", - "global_id": 16582, - "bbox": [ - 101.84, - 521.68, - 466.73, - 534.5 - ], - "text": "5.8-2 Using the Bilateral z-Transform for Analysis of LTID Systems", - "type": "text" - }, - { - "block_id": "p580-b21", - "global_id": 16583, - "bbox": [ - 101.84, - 540.53, - 490.4, - 562.55 - ], - "text": "Because the bilateral z-transform can handle noncausal signals, we can use this transform to\nanalyze noncausal linear systems. The zero-state response y[n] is given by", - "type": "text" - }, - { - "block_id": "p580-b22", - "global_id": 16584, - "bbox": [ - 252.16, - 580.59, - 340.09, - 592.59 - ], - "text": "y[n] = Z−1{X[z]H[z]}", - "type": "text" - }, - { - "block_id": "p580-b23", - "global_id": 16585, - "bbox": [ - 101.85, - 612.45, - 490.39, - 634.79 - ], - "text": "provided X[z]H[z] exists. The ROC of X[z]H[z] is the region in which both X[z] and H[z] exist,\nwhich means that the region is the common part of the ROC of both X[z] and H[z].", - "type": "text" - } - ] - }, - { - "page_num": 581, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p581-b0", - "global_id": 16586, - "bbox": [ - 353.04, - 62.57, - 516.13, - 71.98 - ], - "text": "5.8\nThe Bilateral z-Transform\n561", - "type": "text" - }, - { - "block_id": "p581-b1", - "global_id": 16587, - "bbox": [ - 102.51, - 93.92, - 454.41, - 105.87 - ], - "text": "EXAMPLE 5.19\nZero-State Response by Bilateral z-Transform", - "type": "text" - }, - { - "block_id": "p581-b2", - "global_id": 16588, - "bbox": [ - 128.9, - 122.54, - 341.02, - 132.5 - ], - "text": "For a causal system specified by the transfer function", - "type": "text" - }, - { - "block_id": "p581-b3", - "global_id": 16589, - "bbox": [ - 286.03, - 144.9, - 344.44, - 169.02 - ], - "text": "H[z] =\nz\nz −0.5", - "type": "text" - }, - { - "block_id": "p581-b4", - "global_id": 16590, - "bbox": [ - 128.9, - 182.63, - 271.18, - 192.59 - ], - "text": "find the zero-state response to input", - "type": "text" - }, - { - "block_id": "p581-b5", - "global_id": 16591, - "bbox": [ - 233.03, - 207.62, - 381.33, - 219.5 - ], - "text": "x[n] = (0.8)nu[n] + 2(2)nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p581-b6", - "global_id": 16592, - "bbox": [ - 232.07, - 223.54, - 398.39, - 247.97 - ], - "text": "X[z] =\nz\nz −0.8 −\n2z\nz −2 =\n−z(z + 0.4)\n(z −0.8)(z −2)", - "type": "text" - }, - { - "block_id": "p581-b7", - "global_id": 16593, - "bbox": [ - 128.9, - 278.34, - 502.75, - 300.68 - ], - "text": "The ROC corresponding to the causal term is |z| > 0.8, and that corresponding to the anticausal\nterm is |z| < 2. Hence, the ROC for X[z] is the common region, given by 0.8 < |z| < 2. Hence,", - "type": "text" - }, - { - "block_id": "p581-b8", - "global_id": 16594, - "bbox": [ - 233.96, - 315.24, - 397.71, - 339.67 - ], - "text": "X[z] =\n−z(z + 0.4)\n(z −0.8)(z −2)\n0.8 < |z| < 2", - "type": "text" - }, - { - "block_id": "p581-b9", - "global_id": 16595, - "bbox": [ - 128.9, - 354.16, - 170.65, - 364.13 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p581-b10", - "global_id": 16596, - "bbox": [ - 228.56, - 363.62, - 401.89, - 388.98 - ], - "text": "Y[z] = X[z]H[z] =\n−z2(z + 0.4)\n(z −0.5)(z −0.8)(z −2)", - "type": "text" - }, - { - "block_id": "p581-b11", - "global_id": 16597, - "bbox": [ - 128.9, - 395.1, - 502.76, - 417.43 - ], - "text": "Since the system is causal, the ROC of H[z] is |z| > 0.5. The ROC of X[z] is 0.8 < |z| < 2. The\ncommon region of convergence for X[z] and H[z] is 0.8 < |z| < 2. Therefore,", - "type": "text" - }, - { - "block_id": "p581-b12", - "global_id": 16598, - "bbox": [ - 216.76, - 432.58, - 414.91, - 457.95 - ], - "text": "Y[z] =\n−z2(z + 0.4)\n(z −0.5)(z −0.8)(z −2)\n0.8 < |z| < 2", - "type": "text" - }, - { - "block_id": "p581-b13", - "global_id": 16599, - "bbox": [ - 128.91, - 472.04, - 339.23, - 482.41 - ], - "text": "Expanding Y[z] into modified partial fractions yields", - "type": "text" - }, - { - "block_id": "p581-b14", - "global_id": 16600, - "bbox": [ - 188.27, - 497.9, - 271.36, - 522.02 - ], - "text": "Y[z] = −\nz\nz −0.5 + 8", - "type": "text" - }, - { - "block_id": "p581-b15", - "global_id": 16601, - "bbox": [ - 266.38, - 512.06, - 271.36, - 522.02 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p581-b16", - "global_id": 16602, - "bbox": [ - 273.67, - 490.59, - 308.78, - 522.02 - ], - "text": "z\nz −0.8", - "type": "text" - }, - { - "block_id": "p581-b18", - "global_id": 16603, - "bbox": [ - 318.25, - 498.0, - 333.75, - 514.53 - ], - "text": "−8", - "type": "text" - }, - { - "block_id": "p581-b19", - "global_id": 16604, - "bbox": [ - 328.77, - 512.06, - 333.75, - 522.02 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p581-b20", - "global_id": 16605, - "bbox": [ - 336.06, - 490.59, - 363.7, - 522.02 - ], - "text": "z\nz −2", - "type": "text" - }, - { - "block_id": "p581-b22", - "global_id": 16606, - "bbox": [ - 392.66, - 504.57, - 443.4, - 514.95 - ], - "text": "0.8 < |z| < 2", - "type": "text" - }, - { - "block_id": "p581-b23", - "global_id": 16607, - "bbox": [ - 128.91, - 537.18, - 502.76, - 571.05 - ], - "text": "Since the ROC extends outward from the pole at 0.8, both poles at 0.5 and 0.8 correspond to\ncausal sequence. The ROC extends inward from the pole at 2. Hence, the pole at 2 corresponds\nto anticausal sequence. Therefore,", - "type": "text" - }, - { - "block_id": "p581-b24", - "global_id": 16608, - "bbox": [ - 212.58, - 587.58, - 238.43, - 597.85 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p581-b26", - "global_id": 16609, - "bbox": [ - 244.4, - 583.46, - 291.6, - 597.96 - ], - "text": "−(0.5)n + 8", - "type": "text" - }, - { - "block_id": "p581-b27", - "global_id": 16610, - "bbox": [ - 288.11, - 579.57, - 353.85, - 600.77 - ], - "text": "3(0.8)n\nu[n] + 8", - "type": "text" - }, - { - "block_id": "p581-b28", - "global_id": 16611, - "bbox": [ - 350.36, - 586.57, - 419.08, - 600.77 - ], - "text": "3(2)nu[−(n + 1)]", - "type": "text" - } - ] - }, - { - "page_num": 582, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p582-b0", - "global_id": 16612, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "562\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p582-b1", - "global_id": 16613, - "bbox": [ - 76.77, - 93.92, - 478.8, - 105.87 - ], - "text": "EXAMPLE 5.20\nZero-State Response for an Input with No z-Transform", - "type": "text" - }, - { - "block_id": "p582-b2", - "global_id": 16614, - "bbox": [ - 103.16, - 122.54, - 355.15, - 132.5 - ], - "text": "For the system in Ex. 5.19, find the zero-state response to input", - "type": "text" - }, - { - "block_id": "p582-b3", - "global_id": 16615, - "bbox": [ - 215.13, - 142.54, - 283.53, - 170.1 - ], - "text": "x[n] = (0.8)nu[n]\n\n\n\nx1[n]", - "type": "text" - }, - { - "block_id": "p582-b4", - "global_id": 16616, - "bbox": [ - 284.65, - 142.54, - 365.06, - 170.1 - ], - "text": "+(0.6)nu[−(n + 1)]\n\n\n\nx2[n]", - "type": "text" - }, - { - "block_id": "p582-b5", - "global_id": 16617, - "bbox": [ - 103.16, - 194.18, - 463.61, - 205.33 - ], - "text": "The z-transforms of the causal and anticausal components x1[n] and x2[n] of the output are", - "type": "text" - }, - { - "block_id": "p582-b6", - "global_id": 16618, - "bbox": [ - 232.16, - 211.98, - 348.01, - 236.1 - ], - "text": "X1[z] =\nz\nz −0.8\n|z| > 0.8", - "type": "text" - }, - { - "block_id": "p582-b7", - "global_id": 16619, - "bbox": [ - 232.16, - 236.35, - 348.01, - 260.79 - ], - "text": "X2[z] =\n−z\nz −0.6\n|z| < 0.6", - "type": "text" - }, - { - "block_id": "p582-b8", - "global_id": 16620, - "bbox": [ - 103.16, - 285.53, - 477.01, - 332.54 - ], - "text": "Observe that a common ROC for X1[z] and X2[z] does not exist. Therefore, X[z] does not exist.\nIn such a case we take advantage of the superposition principle and find y1[n] and y2[n], the\nsystem responses to x1[n] and x2[n], separately. The desired response y[n] is the sum of y1[n]\nand y2[n]. Now", - "type": "text" - }, - { - "block_id": "p582-b9", - "global_id": 16621, - "bbox": [ - 175.82, - 339.08, - 289.25, - 363.2 - ], - "text": "H[z] =\nz\nz −0.5\n|z| > 0.5", - "type": "text" - }, - { - "block_id": "p582-b10", - "global_id": 16622, - "bbox": [ - 173.94, - 365.09, - 295.71, - 384.16 - ], - "text": "Y1[z] = X1[z]H[z] =\nz2", - "type": "text" - }, - { - "block_id": "p582-b11", - "global_id": 16623, - "bbox": [ - 257.64, - 373.01, - 381.92, - 390.45 - ], - "text": "(z −0.5)(z −0.8)\n|z| > 0.8", - "type": "text" - }, - { - "block_id": "p582-b12", - "global_id": 16624, - "bbox": [ - 173.94, - 393.23, - 299.59, - 412.29 - ], - "text": "Y2[z] = X2[z]H[z] =\n−z2", - "type": "text" - }, - { - "block_id": "p582-b13", - "global_id": 16625, - "bbox": [ - 257.64, - 401.14, - 406.23, - 418.59 - ], - "text": "(z −0.5)(z −0.6)\n0.5 < |z| < 0.6", - "type": "text" - }, - { - "block_id": "p582-b14", - "global_id": 16626, - "bbox": [ - 103.17, - 427.69, - 356.1, - 438.84 - ], - "text": "Expanding Y1[z] and Y2[z] into modified partial fractions yields", - "type": "text" - }, - { - "block_id": "p582-b15", - "global_id": 16627, - "bbox": [ - 179.95, - 448.69, - 225.8, - 466.41 - ], - "text": "Y1[z] = −5", - "type": "text" - }, - { - "block_id": "p582-b16", - "global_id": 16628, - "bbox": [ - 220.82, - 462.75, - 225.8, - 472.71 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p582-b17", - "global_id": 16629, - "bbox": [ - 228.1, - 441.27, - 263.22, - 472.71 - ], - "text": "z\nz −0.5", - "type": "text" - }, - { - "block_id": "p582-b19", - "global_id": 16630, - "bbox": [ - 272.7, - 448.69, - 288.19, - 465.64 - ], - "text": "+ 8", - "type": "text" - }, - { - "block_id": "p582-b20", - "global_id": 16631, - "bbox": [ - 283.21, - 462.75, - 288.19, - 472.71 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p582-b21", - "global_id": 16632, - "bbox": [ - 290.49, - 441.27, - 325.6, - 472.71 - ], - "text": "z\nz −0.8", - "type": "text" - }, - { - "block_id": "p582-b23", - "global_id": 16633, - "bbox": [ - 354.57, - 455.26, - 388.46, - 465.64 - ], - "text": "|z| > 0.8", - "type": "text" - }, - { - "block_id": "p582-b24", - "global_id": 16634, - "bbox": [ - 179.95, - 483.15, - 216.84, - 494.3 - ], - "text": "Y2[z] = 5", - "type": "text" - }, - { - "block_id": "p582-b25", - "global_id": 16635, - "bbox": [ - 217.94, - 469.17, - 253.05, - 500.6 - ], - "text": "z\nz −0.5", - "type": "text" - }, - { - "block_id": "p582-b27", - "global_id": 16636, - "bbox": [ - 262.54, - 483.15, - 276.83, - 493.53 - ], - "text": "−6", - "type": "text" - }, - { - "block_id": "p582-b28", - "global_id": 16637, - "bbox": [ - 277.93, - 469.17, - 313.06, - 500.6 - ], - "text": "z\nz −0.6", - "type": "text" - }, - { - "block_id": "p582-b30", - "global_id": 16638, - "bbox": [ - 342.01, - 483.15, - 400.22, - 493.53 - ], - "text": "0.5 < |z| < 0.6", - "type": "text" - }, - { - "block_id": "p582-b31", - "global_id": 16639, - "bbox": [ - 103.17, - 510.77, - 144.91, - 520.73 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p582-b32", - "global_id": 16640, - "bbox": [ - 207.74, - 532.28, - 237.57, - 543.43 - ], - "text": "y1[n] =", - "type": "text" - }, - { - "block_id": "p582-b34", - "global_id": 16641, - "bbox": [ - 243.55, - 530.93, - 256.01, - 542.25 - ], - "text": "−5", - "type": "text" - }, - { - "block_id": "p582-b35", - "global_id": 16642, - "bbox": [ - 252.52, - 524.27, - 342.2, - 545.47 - ], - "text": "3(0.5)n + 8\n3(0.8)n\nu[n]", - "type": "text" - }, - { - "block_id": "p582-b36", - "global_id": 16643, - "bbox": [ - 207.74, - 546.74, - 372.43, - 559.39 - ], - "text": "y2[n] = 5(0.5)nu[n] + 6(0.6)nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p582-b37", - "global_id": 16644, - "bbox": [ - 103.17, - 570.57, - 117.54, - 580.53 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p582-b38", - "global_id": 16645, - "bbox": [ - 151.91, - 582.12, - 240.55, - 593.26 - ], - "text": "y[n] = y1[n] + y2[n] =", - "type": "text" - }, - { - "block_id": "p582-b39", - "global_id": 16646, - "bbox": [ - 242.6, - 574.11, - 254.69, - 587.74 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p582-b40", - "global_id": 16647, - "bbox": [ - 249.46, - 574.11, - 428.26, - 595.62 - ], - "text": "3 (0.5)n + 8\n3(0.8)n\nu[n] + 6(0.6)nu[−(n + 1)]", - "type": "text" - } - ] - }, - { - "page_num": 583, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p583-b0", - "global_id": 16648, - "bbox": [ - 288.77, - 62.57, - 516.13, - 71.98 - ], - "text": "5.9\nConnecting the Laplace and z-Transforms\n563", - "type": "text" - }, - { - "block_id": "p583-b1", - "global_id": 16649, - "bbox": [ - 133.57, - 97.81, - 462.29, - 109.76 - ], - "text": "DRILL 5.24\nZero-State Response by Bilateral z-Transform", - "type": "text" - }, - { - "block_id": "p583-b2", - "global_id": 16650, - "bbox": [ - 133.57, - 118.88, - 412.94, - 128.85 - ], - "text": "For the causal system in Ex. 5.19, find the zero-state response to input", - "type": "text" - }, - { - "block_id": "p583-b3", - "global_id": 16651, - "bbox": [ - 250.14, - 140.38, - 276.03, - 150.66 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p583-b4", - "global_id": 16652, - "bbox": [ - 278.07, - 132.37, - 286.77, - 146.02 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p583-b5", - "global_id": 16653, - "bbox": [ - 283.29, - 132.37, - 393.56, - 153.58 - ], - "text": "4\nn u[n] + 5(3)nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p583-b6", - "global_id": 16654, - "bbox": [ - 133.57, - 162.3, - 188.43, - 182.19 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p583-b7", - "global_id": 16655, - "bbox": [ - 139.04, - 180.24, - 146.81, - 190.2 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p583-b8", - "global_id": 16656, - "bbox": [ - 148.36, - 172.23, - 157.06, - 185.87 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p583-b9", - "global_id": 16657, - "bbox": [ - 153.57, - 172.23, - 182.1, - 193.43 - ], - "text": "4\nn + 3", - "type": "text" - }, - { - "block_id": "p583-b10", - "global_id": 16658, - "bbox": [ - 182.1, - 172.23, - 190.81, - 185.87 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p583-b11", - "global_id": 16659, - "bbox": [ - 187.32, - 172.23, - 203.92, - 193.43 - ], - "text": "2\nn", - "type": "text" - }, - { - "block_id": "p583-b12", - "global_id": 16660, - "bbox": [ - 203.92, - 179.23, - 300.41, - 190.61 - ], - "text": "u[n] + 6(3)nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p583-b13", - "global_id": 16661, - "bbox": [ - 127.94, - 232.19, - 471.94, - 247.14 - ], - "text": "5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS", - "type": "text" - }, - { - "block_id": "p583-b14", - "global_id": 16662, - "bbox": [ - 127.59, - 253.13, - 516.14, - 286.99 - ], - "text": "We now show that discrete-time systems also can be analyzed by means of the Laplace transform.\nIn fact, we shall see that the z-transform is the Laplace transform in disguise and that discrete-time\nsystems can be analyzed as if they were continuous-time systems.", - "type": "text" - }, - { - "block_id": "p583-b15", - "global_id": 16663, - "bbox": [ - 127.59, - 288.99, - 516.16, - 370.68 - ], - "text": "So far we have considered the discrete-time signal as a sequence of numbers and not as\nan electrical signal (voltage or current). Similarly, we considered a discrete-time system as a\nmechanism that processes a sequence of numbers (input) to yield another sequence of numbers\n(output). The system was built by using delays (along with adders and multipliers) that delay\nsequences of numbers. A digital computer is a perfect example: every signal is a sequence of\nnumbers, and the processing involves delaying sequences of numbers (along with addition and\nmultiplication).", - "type": "text" - }, - { - "block_id": "p583-b16", - "global_id": 16664, - "bbox": [ - 127.6, - 372.26, - 516.14, - 394.6 - ], - "text": "Now suppose we have a discrete-time system with transfer function H[z] and input x[n].\nConsider a continuous-time signal x(t) such that its nth sample value is x[n], as shown in Fig. 5.30.†", - "type": "text" - }, - { - "block_id": "p583-b17", - "global_id": 16665, - "bbox": [ - 127.59, - 396.17, - 516.14, - 418.5 - ], - "text": "Let the sampled signal be x(t), consisting of impulses spaced T seconds apart with the nth impulse\nof strength x[n]. Thus,", - "type": "text" - }, - { - "block_id": "p583-b18", - "global_id": 16666, - "bbox": [ - 274.19, - 428.21, - 298.85, - 438.49 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p583-b19", - "global_id": 16667, - "bbox": [ - 300.9, - 418.04, - 314.99, - 428.71 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p583-b20", - "global_id": 16668, - "bbox": [ - 301.73, - 442.6, - 314.14, - 449.86 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p583-b21", - "global_id": 16669, - "bbox": [ - 316.1, - 428.21, - 369.51, - 438.49 - ], - "text": "x[n]δ(t −nT)", - "type": "text" - }, - { - "block_id": "p583-b22", - "global_id": 16670, - "bbox": [ - 127.59, - 461.22, - 516.14, - 603.1 - ], - "text": "Figure 5.30 shows x[n] and the corresponding x(t). The signal x[n] is applied to the input of\na discrete-time system with transfer function H[z], which is generally made up of delays, adders,\nand scalar multipliers. Hence, processing x[n] through H[z] amounts to operating on the sequence\nx[n] by means of delays, adders, and scalar multipliers. Suppose for x(t) samples, we perform\noperations identical to those performed on the samples of x[n] by H[z]. For this purpose, we need a\ncontinuous-time system with transfer function H(s) that is identical in structure to the discrete-time\nsystem H[z] except that the delays in H[z] are replaced by elements that delay continuous-time\nsignals (such as voltages or currents). There is no other difference between realizations of H[z]\nand H(s). If a continuous-time impulse δ(t) is applied to such a delay of T seconds, the output will\nbe δ(t −T). The continuous-time transfer function of such a delay is e−sT [see Eq. (4.30)]. Hence,\nthe delay elements with transfer function 1/z in the realization of H[z] will be replaced by the delay\nelements with transfer function e−sT in the realization of the corresponding H(s). This is the same", - "type": "text" - }, - { - "block_id": "p583-b23", - "global_id": 16671, - "bbox": [ - 127.59, - 621.19, - 423.25, - 633.41 - ], - "text": "† We can construct such x(t) from the sample values, as will be explained in Ch. 8.", - "type": "text" - } - ] - }, - { - "page_num": 584, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p584-b0", - "global_id": 16672, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "564\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p584-b1", - "global_id": 16673, - "bbox": [ - 112.96, - 125.38, - 125.84, - 133.46 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p584-b2", - "global_id": 16674, - "bbox": [ - 152.11, - 236.69, - 261.2, - 254.21 - ], - "text": "H[z]\nx[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p584-b3", - "global_id": 16675, - "bbox": [ - 384.94, - 246.17, - 407.75, - 255.71 - ], - "text": "H [esT]", - "type": "text" - }, - { - "block_id": "p584-b4", - "global_id": 16676, - "bbox": [ - 113.15, - 298.41, - 126.03, - 306.49 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p584-b5", - "global_id": 16677, - "bbox": [ - 185.9, - 397.33, - 194.78, - 405.33 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p584-b6", - "global_id": 16678, - "bbox": [ - 226.08, - 379.58, - 445.74, - 387.58 - ], - "text": "t\nn", - "type": "text" - }, - { - "block_id": "p584-b7", - "global_id": 16679, - "bbox": [ - 226.08, - 187.48, - 445.74, - 195.48 - ], - "text": "n\nt", - "type": "text" - }, - { - "block_id": "p584-b8", - "global_id": 16680, - "bbox": [ - 401.02, - 397.33, - 410.67, - 405.33 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p584-b9", - "global_id": 16681, - "bbox": [ - 381.49, - 104.65, - 392.59, - 112.73 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p584-b10", - "global_id": 16682, - "bbox": [ - 329.41, - 298.41, - 340.51, - 306.49 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p584-b11", - "global_id": 16683, - "bbox": [ - 340.79, - 236.69, - 451.12, - 244.86 - ], - "text": "y(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p584-b12", - "global_id": 16684, - "bbox": [ - 332.63, - 125.38, - 343.73, - 133.46 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p584-b13", - "global_id": 16685, - "bbox": [ - 101.84, - 418.35, - 380.51, - 427.68 - ], - "text": "Figure 5.30 Connection between the Laplace transform and the z-transform.", - "type": "text" - }, - { - "block_id": "p584-b14", - "global_id": 16686, - "bbox": [ - 101.84, - 449.8, - 490.4, - 556.83 - ], - "text": "as z being replaced by esT. Therefore, H(s) = H[esT]. Let us now apply x[n] to the input of H[z] and\napply x(t) at the input of H[esT]. Whatever operations are performed by the discrete-time system\nH[z] on x[n] (Fig. 5.30a) are also performed by the corresponding continuous-time system H[esT]\non the impulse sequence x(t) (Fig. 5.30b). The delaying of a sequence in H[z] would amount to\ndelaying of an impulse train in H[esT]. Adding and multiplying operations are the same in both\ncases. In other words, one-to-one correspondence of the two systems is preserved in every aspect.\nTherefore if y[n] is the output of the discrete-time system in Fig. 5.30a, then y(t), the output of the\ncontinuous-time system in Fig. 5.30b, would be a sequence of impulse whose nth impulse strength\nis y[n]. Thus,", - "type": "text" - }, - { - "block_id": "p584-b15", - "global_id": 16687, - "bbox": [ - 248.48, - 568.76, - 273.1, - 579.04 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p584-b16", - "global_id": 16688, - "bbox": [ - 275.15, - 558.59, - 289.25, - 569.26 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p584-b17", - "global_id": 16689, - "bbox": [ - 275.99, - 583.15, - 288.4, - 590.42 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p584-b18", - "global_id": 16690, - "bbox": [ - 290.36, - 568.76, - 343.74, - 579.04 - ], - "text": "y[n]δ(t −nT)", - "type": "text" - }, - { - "block_id": "p584-b19", - "global_id": 16691, - "bbox": [ - 101.84, - 598.68, - 490.38, - 620.6 - ], - "text": "The system in Fig. 5.30b, being a continuous-time system, can be analyzed via the Laplace\ntransform. If", - "type": "text" - }, - { - "block_id": "p584-b20", - "global_id": 16692, - "bbox": [ - 213.24, - 624.53, - 379.0, - 634.91 - ], - "text": "x(t) ⇐⇒X(s)\nand\ny(t) ⇐⇒Y(s)", - "type": "text" - } - ] - }, - { - "page_num": 585, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p585-b0", - "global_id": 16693, - "bbox": [ - 309.61, - 62.89, - 516.13, - 71.98 - ], - "text": "5.10\nMATLAB: Discrete-Time IIR Filters\n565", - "type": "text" - }, - { - "block_id": "p585-b1", - "global_id": 16694, - "bbox": [ - 127.59, - 85.82, - 144.74, - 95.78 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p585-b2", - "global_id": 16695, - "bbox": [ - 285.04, - 95.86, - 516.13, - 107.74 - ], - "text": "Y(s) = H[esT]X(s)\n(5.53)", - "type": "text" - }, - { - "block_id": "p585-b3", - "global_id": 16696, - "bbox": [ - 127.59, - 115.8, - 148.9, - 125.77 - ], - "text": "Also,", - "type": "text" - }, - { - "block_id": "p585-b4", - "global_id": 16697, - "bbox": [ - 262.52, - 135.26, - 299.09, - 145.54 - ], - "text": "X(s) = L", - "type": "text" - }, - { - "block_id": "p585-b5", - "global_id": 16698, - "bbox": [ - 300.2, - 118.28, - 320.48, - 135.76 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b6", - "global_id": 16699, - "bbox": [ - 307.22, - 149.65, - 319.63, - 156.91 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b7", - "global_id": 16700, - "bbox": [ - 321.59, - 135.26, - 375.0, - 145.54 - ], - "text": "x[n]δ(t −nT)", - "type": "text" - }, - { - "block_id": "p585-b9", - "global_id": 16701, - "bbox": [ - 127.59, - 162.56, - 356.38, - 174.16 - ], - "text": "Now because the Laplace transform of δ(t −nT) is e−snT,", - "type": "text" - }, - { - "block_id": "p585-b10", - "global_id": 16702, - "bbox": [ - 212.92, - 192.02, - 240.58, - 202.3 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p585-b11", - "global_id": 16703, - "bbox": [ - 242.62, - 181.85, - 256.72, - 192.52 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b12", - "global_id": 16704, - "bbox": [ - 243.46, - 206.41, - 255.87, - 213.68 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b13", - "global_id": 16705, - "bbox": [ - 257.82, - 190.29, - 376.54, - 202.4 - ], - "text": "x[n]e−snT\nand\nY(s) =", - "type": "text" - }, - { - "block_id": "p585-b14", - "global_id": 16706, - "bbox": [ - 378.58, - 181.85, - 392.68, - 192.52 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b15", - "global_id": 16707, - "bbox": [ - 379.42, - 206.41, - 391.83, - 213.68 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b16", - "global_id": 16708, - "bbox": [ - 393.79, - 190.29, - 429.75, - 202.3 - ], - "text": "y[n]e−snT", - "type": "text" - }, - { - "block_id": "p585-b17", - "global_id": 16709, - "bbox": [ - 127.59, - 221.56, - 345.24, - 231.53 - ], - "text": "Substitution of these expressions into Eq. (5.53) yields", - "type": "text" - }, - { - "block_id": "p585-b18", - "global_id": 16710, - "bbox": [ - 243.79, - 239.73, - 257.89, - 250.4 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b19", - "global_id": 16711, - "bbox": [ - 244.63, - 264.29, - 257.04, - 271.56 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b20", - "global_id": 16712, - "bbox": [ - 258.99, - 245.78, - 334.2, - 260.18 - ], - "text": "y[n]e−snT = H[esT]", - "type": "text" - }, - { - "block_id": "p585-b21", - "global_id": 16713, - "bbox": [ - 335.31, - 232.92, - 355.59, - 250.4 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b22", - "global_id": 16714, - "bbox": [ - 342.33, - 264.29, - 354.74, - 271.56 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b23", - "global_id": 16715, - "bbox": [ - 356.7, - 248.17, - 392.69, - 260.18 - ], - "text": "x[n]e−snT", - "type": "text" - }, - { - "block_id": "p585-b25", - "global_id": 16716, - "bbox": [ - 127.59, - 279.51, - 414.82, - 290.89 - ], - "text": "By introducing a new variable z = esT, this equation can be expressed as", - "type": "text" - }, - { - "block_id": "p585-b26", - "global_id": 16717, - "bbox": [ - 261.74, - 298.58, - 275.84, - 309.25 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b27", - "global_id": 16718, - "bbox": [ - 262.58, - 323.14, - 274.99, - 330.41 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b28", - "global_id": 16719, - "bbox": [ - 276.94, - 304.63, - 336.31, - 319.03 - ], - "text": "y[n]z−n = H[z]", - "type": "text" - }, - { - "block_id": "p585-b29", - "global_id": 16720, - "bbox": [ - 337.42, - 298.58, - 351.52, - 309.25 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b30", - "global_id": 16721, - "bbox": [ - 338.25, - 323.14, - 350.65, - 330.41 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b31", - "global_id": 16722, - "bbox": [ - 352.61, - 307.02, - 381.48, - 319.03 - ], - "text": "x[n]z−n", - "type": "text" - }, - { - "block_id": "p585-b32", - "global_id": 16723, - "bbox": [ - 127.59, - 336.08, - 135.89, - 346.04 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p585-b33", - "global_id": 16724, - "bbox": [ - 289.93, - 347.62, - 353.82, - 357.89 - ], - "text": "Y[z] = H[z]X[z]", - "type": "text" - }, - { - "block_id": "p585-b34", - "global_id": 16725, - "bbox": [ - 127.6, - 366.06, - 151.93, - 376.02 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p585-b35", - "global_id": 16726, - "bbox": [ - 221.39, - 383.67, - 248.25, - 393.95 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p585-b36", - "global_id": 16727, - "bbox": [ - 250.29, - 373.5, - 264.39, - 384.17 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b37", - "global_id": 16728, - "bbox": [ - 251.14, - 398.06, - 263.54, - 405.32 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b38", - "global_id": 16729, - "bbox": [ - 265.5, - 381.95, - 375.74, - 394.05 - ], - "text": "x[n]z−n\nand\nY[z] =", - "type": "text" - }, - { - "block_id": "p585-b39", - "global_id": 16730, - "bbox": [ - 377.78, - 373.5, - 391.88, - 384.17 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p585-b40", - "global_id": 16731, - "bbox": [ - 378.63, - 398.06, - 391.04, - 405.32 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p585-b41", - "global_id": 16732, - "bbox": [ - 392.99, - 381.95, - 421.83, - 393.95 - ], - "text": "y[n]z−n", - "type": "text" - }, - { - "block_id": "p585-b42", - "global_id": 16733, - "bbox": [ - 127.59, - 411.03, - 516.14, - 468.91 - ], - "text": "It is clear from this discussion that the z-transform can be considered to be the Laplace transform\nwith a change of variable z = esT or s = (1/T)lnz. Note that the transformation z = esT transforms\nthe imaginary axis in the s plane (s = jω) into a unit circle in the z plane (z = esT = ejωT, or |z| = 1).\nThe LHP and RHP in the s-plane map into the inside and the outside, respectively, of the unit circle\nin the z plane.", - "type": "text" - }, - { - "block_id": "p585-b43", - "global_id": 16734, - "bbox": [ - 127.94, - 497.29, - 420.57, - 511.24 - ], - "text": "5.10 MATLAB: DISCRETE-TIME IIR FILTERS", - "type": "text" - }, - { - "block_id": "p585-b44", - "global_id": 16735, - "bbox": [ - 127.59, - 517.22, - 516.13, - 575.01 - ], - "text": "Recent technological advancements have dramatically increased the popularity of discrete-time\nfilters. Unlike their continuous-time counterparts, the performance of discrete-time filters is not\naffected by component variations, temperature, humidity, or age. Furthermore, digital hardware is\neasily reprogrammed, which allows convenient change of device function. For example, certain\ndigital hearing aids are individually programmed to match the required response of a user.", - "type": "text" - }, - { - "block_id": "p585-b45", - "global_id": 16736, - "bbox": [ - 127.59, - 577.0, - 516.15, - 634.79 - ], - "text": "Typically, discrete-time filters are categorized as infinite-impulse response (IIR) or\nfinite-impulse response (FIR). A popular method to obtain a discrete-time IIR filter is by\ntransformation of a corresponding continuous-time filter design. MATLAB greatly assists this\nprocess. Although discrete-time IIR filter design is the emphasis of this section, methods for\ndiscrete-time FIR filter design are considered in Sec. 9.7.", - "type": "text" - } - ] - }, - { - "page_num": 586, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p586-b0", - "global_id": 16737, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "566\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p586-b1", - "global_id": 16738, - "bbox": [ - 101.84, - 86.52, - 360.71, - 98.48 - ], - "text": "5.10-1 Frequency Response and Pole-Zero Plots", - "type": "text" - }, - { - "block_id": "p586-b2", - "global_id": 16739, - "bbox": [ - 101.84, - 104.61, - 490.39, - 138.48 - ], - "text": "Frequency\nresponse\nand\npole-zero\nplots\nhelp\ncharacterize\nfilter\nbehavior.\nSimilar\nto\ncontinuous-time systems, rational transfer functions for realizable LTID systems are represented\nin the z-domain as", - "type": "text" - }, - { - "block_id": "p586-b3", - "global_id": 16740, - "bbox": [ - 193.29, - 162.07, - 241.44, - 179.33 - ], - "text": "H[z] = Y[z]", - "type": "text" - }, - { - "block_id": "p586-b4", - "global_id": 16741, - "bbox": [ - 224.51, - 162.07, - 272.4, - 186.41 - ], - "text": "X[z] = B[z]", - "type": "text" - }, - { - "block_id": "p586-b5", - "global_id": 16742, - "bbox": [ - 255.8, - 169.06, - 283.41, - 186.41 - ], - "text": "A[z] =", - "type": "text" - }, - { - "block_id": "p586-b6", - "global_id": 16743, - "bbox": [ - 286.66, - 154.57, - 301.72, - 166.84 - ], - "text": "%N", - "type": "text" - }, - { - "block_id": "p586-b7", - "global_id": 16744, - "bbox": [ - 286.66, - 160.82, - 331.92, - 182.4 - ], - "text": "k=0 bkz−k\n%N", - "type": "text" - }, - { - "block_id": "p586-b8", - "global_id": 16745, - "bbox": [ - 297.07, - 169.06, - 343.55, - 190.72 - ], - "text": "k=0 akz−k =", - "type": "text" - }, - { - "block_id": "p586-b9", - "global_id": 16746, - "bbox": [ - 346.8, - 154.57, - 361.85, - 166.84 - ], - "text": "%N", - "type": "text" - }, - { - "block_id": "p586-b10", - "global_id": 16747, - "bbox": [ - 346.8, - 160.82, - 397.13, - 182.4 - ], - "text": "k=0 bkzN−k\n%N", - "type": "text" - }, - { - "block_id": "p586-b11", - "global_id": 16748, - "bbox": [ - 357.2, - 169.47, - 490.38, - 190.72 - ], - "text": "k=0 akzN−k\n(5.54)", - "type": "text" - }, - { - "block_id": "p586-b12", - "global_id": 16749, - "bbox": [ - 101.85, - 211.03, - 490.38, - 233.36 - ], - "text": "When only the first (N1 + 1) numerator coefficients are nonzero and only the first (N2 + 1)\ndenominator coefficients are nonzero, Eq. (5.54) simplifies to", - "type": "text" - }, - { - "block_id": "p586-b13", - "global_id": 16750, - "bbox": [ - 178.71, - 258.73, - 226.86, - 275.99 - ], - "text": "H[z] = Y[z]", - "type": "text" - }, - { - "block_id": "p586-b14", - "global_id": 16751, - "bbox": [ - 209.93, - 258.73, - 257.82, - 283.07 - ], - "text": "X[z] = B[z]", - "type": "text" - }, - { - "block_id": "p586-b15", - "global_id": 16752, - "bbox": [ - 241.22, - 265.72, - 268.83, - 283.07 - ], - "text": "A[z] =", - "type": "text" - }, - { - "block_id": "p586-b16", - "global_id": 16753, - "bbox": [ - 272.08, - 251.15, - 290.13, - 264.37 - ], - "text": "%N1", - "type": "text" - }, - { - "block_id": "p586-b17", - "global_id": 16754, - "bbox": [ - 272.08, - 257.4, - 317.34, - 280.4 - ], - "text": "k=0 bkz−k\n%N2", - "type": "text" - }, - { - "block_id": "p586-b18", - "global_id": 16755, - "bbox": [ - 282.49, - 265.72, - 328.97, - 287.84 - ], - "text": "k=0 akz−k =", - "type": "text" - }, - { - "block_id": "p586-b19", - "global_id": 16756, - "bbox": [ - 332.22, - 251.15, - 350.27, - 264.37 - ], - "text": "%N1", - "type": "text" - }, - { - "block_id": "p586-b20", - "global_id": 16757, - "bbox": [ - 332.22, - 257.4, - 385.61, - 280.4 - ], - "text": "k=0 bkzN1−k\n%N2", - "type": "text" - }, - { - "block_id": "p586-b21", - "global_id": 16758, - "bbox": [ - 342.63, - 263.99, - 490.38, - 287.84 - ], - "text": "k=0 akzN2−k zN2−N1\n(5.55)", - "type": "text" - }, - { - "block_id": "p586-b22", - "global_id": 16759, - "bbox": [ - 101.84, - 308.51, - 490.41, - 342.38 - ], - "text": "The form of Eq. (5.55) has many advantages. It can be more efficient than Eq. (5.54); it still\nworks when N1 = N2 = N; and it more closely conforms to the notation of built-in MATLAB\ndiscrete-time signal-processing functions.", - "type": "text" - }, - { - "block_id": "p586-b23", - "global_id": 16760, - "bbox": [ - 101.84, - 344.38, - 490.39, - 414.11 - ], - "text": "The right-hand side of Eq. (5.55) is a form that is convenient for MATLAB computations.\nThe frequency response H[ej] is obtained by letting z = ej, where has units of radians.\nOften, = ωT, where ω is the continuous-time frequency in radians per second and T is the\nsampling period in seconds. Defining length-(N2 + 1) coefficient vector A = [a0,a1,...,aN2] and\nlength-(N1+1) coefficient vector B = [b0,b1,...,bN1], program CH5MP1 computes H[ej] by using\nEq. (5.55) for each frequency in the input vector .", - "type": "text" - }, - { - "block_id": "p586-b24", - "global_id": 16761, - "bbox": [ - 101.85, - 437.07, - 483.65, - 518.76 - ], - "text": "function [H] = CH5MP1(B,A,Omega);\n% CH5MP1.m : Chapter 5, MATLAB Program 1\n% Function M-file computes frequency response for LTID systems\n% INPUTS:\nB = vector of feedforward coefficients\n%\nA = vector of feedback coefficients\n%\nOmega = vector of frequencies [rad], typically -pi<=Omega<=pi\n% OUTPUTS:\nH =\nfrequency response", - "type": "text" - }, - { - "block_id": "p586-b25", - "global_id": 16762, - "bbox": [ - 101.85, - 532.71, - 410.44, - 554.63 - ], - "text": "N_1 = length(B)-1; N_2 = length(A)-1;\nH = polyval(B,exp(1j*Omega))./polyval(A,exp(1j*Omega)).*...", - "type": "text" - }, - { - "block_id": "p586-b26", - "global_id": 16763, - "bbox": [ - 122.76, - 556.62, - 248.29, - 566.58 - ], - "text": "exp(1j*Omega*(N_2-N_1));", - "type": "text" - }, - { - "block_id": "p586-b27", - "global_id": 16764, - "bbox": [ - 101.84, - 588.86, - 490.4, - 635.08 - ], - "text": "Note that owing to MATLAB’s indexing scheme, A(k) corresponds to coefficient ak−1 and B(k)\ncorresponds to coefficient bk−1. It is also possible to use the signal-processing toolbox function\nfreqz to evaluate the frequency response of a system described by Eq. (5.55). Under special\ncircumstances, the control system toolbox function bode can also be used.", - "type": "text" - } - ] - }, - { - "page_num": 587, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p587-b0", - "global_id": 16765, - "bbox": [ - 309.61, - 62.89, - 516.13, - 71.98 - ], - "text": "5.10\nMATLAB: Discrete-Time IIR Filters\n567", - "type": "text" - }, - { - "block_id": "p587-b1", - "global_id": 16766, - "bbox": [ - 127.59, - 85.82, - 516.13, - 107.74 - ], - "text": "Program CH5MP2 computes and plots the poles and zeros of an LTID system described by\nEq. (5.55), again using vectors B and A.", - "type": "text" - }, - { - "block_id": "p587-b2", - "global_id": 16767, - "bbox": [ - 127.59, - 118.87, - 419.59, - 166.69 - ], - "text": "function [p,z] = CH5MP2(B,A);\n% CH5MP2.m : Chapter 5, MATLAB Program 2\n% Function M-file computes and plots poles and zeros for LTID systems\n% INPUTS:\nB = vector of feedforward coefficients\n%\nA = vector of feedback coefficients", - "type": "text" - }, - { - "block_id": "p587-b3", - "global_id": 16768, - "bbox": [ - 127.59, - 178.65, - 453.22, - 246.4 - ], - "text": "N_1 = length(B)-1; N_2 = length(A)-1;\np = roots([A,zeros(1,N_1-N_2)]); z = roots([B,zeros(1,N_2-N_1)]);\nucirc = exp(1j*linspace(0,2*pi,200)); % Compute unit circle for plot\nplot(real(p),imag(p),’xk’,real(z),imag(z),’ok’,real(ucirc),imag(ucirc),’k:’);\nxlabel(’Real’); ylabel(’Imag’);\nax = axis; dx = 0.05*(ax(2)-ax(1)); dy = 0.05*(ax(4)-ax(3));\naxis(ax+[-dx,dx,-dy,dy]); axis equal;", - "type": "text" - }, - { - "block_id": "p587-b4", - "global_id": 16769, - "bbox": [ - 127.59, - 257.63, - 516.15, - 363.65 - ], - "text": "The right-hand side of Eq. (5.55) helps explain how the roots are computed. When N1̸ = N2, the\nterm zN2−N1 implies additional roots at the origin. If N1 > N2, the roots are poles, which are added\nby concatenating A with zeros(N_1-N_2,1); since N2 −N1 ≤0, zeros(N_2-N_1,1) produces\nthe empty set and B is unchanged. If N2 > N1, the roots are zeros, which are added by concatenating\nB with zeros(N_2-N_1,1); since N1 −N2 ≤0, zeros(N_1-N_2,1) produces the empty set and\nA is unchanged. Poles and zeros are indicated with black x’s and o’s, respectively. For visual\nreference, the unit circle is also plotted. The last two lines in CH5MP2 expand the plot axis box so\nthat root locations are not obscured and also ensure that the real and imaginary axes are drawn to\nthe same scale.", - "type": "text" - }, - { - "block_id": "p587-b5", - "global_id": 16770, - "bbox": [ - 127.59, - 389.83, - 285.14, - 401.78 - ], - "text": "5.10-2 Transformation Basics", - "type": "text" - }, - { - "block_id": "p587-b6", - "global_id": 16771, - "bbox": [ - 127.59, - 407.92, - 516.14, - 429.83 - ], - "text": "Transformation of a continuous-time filter to a discrete-time filter begins with the desired\ncontinuous-time transfer function", - "type": "text" - }, - { - "block_id": "p587-b7", - "global_id": 16772, - "bbox": [ - 236.38, - 441.41, - 286.12, - 458.67 - ], - "text": "H(s) = Y(s)", - "type": "text" - }, - { - "block_id": "p587-b8", - "global_id": 16773, - "bbox": [ - 268.4, - 441.41, - 317.88, - 465.74 - ], - "text": "X(s) = B(s)", - "type": "text" - }, - { - "block_id": "p587-b9", - "global_id": 16774, - "bbox": [ - 300.48, - 448.39, - 328.89, - 465.74 - ], - "text": "A(s) =", - "type": "text" - }, - { - "block_id": "p587-b10", - "global_id": 16775, - "bbox": [ - 332.14, - 433.91, - 348.36, - 446.18 - ], - "text": "%M", - "type": "text" - }, - { - "block_id": "p587-b11", - "global_id": 16776, - "bbox": [ - 342.55, - 440.15, - 405.52, - 454.49 - ], - "text": "k=0 bk+N−MsM−k", - "type": "text" - }, - { - "block_id": "p587-b12", - "global_id": 16777, - "bbox": [ - 343.67, - 449.46, - 358.73, - 461.74 - ], - "text": "%N", - "type": "text" - }, - { - "block_id": "p587-b13", - "global_id": 16778, - "bbox": [ - 354.08, - 456.45, - 393.99, - 470.05 - ], - "text": "k=0 aksN−k", - "type": "text" - }, - { - "block_id": "p587-b14", - "global_id": 16779, - "bbox": [ - 127.59, - 479.09, - 395.43, - 489.47 - ], - "text": "As a matter of convenience, H(s) is represented in factored form as", - "type": "text" - }, - { - "block_id": "p587-b15", - "global_id": 16780, - "bbox": [ - 266.59, - 503.41, - 319.88, - 520.36 - ], - "text": "H(s) = bN−M", - "type": "text" - }, - { - "block_id": "p587-b16", - "global_id": 16781, - "bbox": [ - 305.15, - 517.47, - 313.62, - 528.31 - ], - "text": "a0", - "type": "text" - }, - { - "block_id": "p587-b17", - "global_id": 16782, - "bbox": [ - 323.06, - 495.59, - 375.94, - 531.12 - ], - "text": "7M\nk=1(s −zk)\n7N\nk=1(s −pk)", - "type": "text" - }, - { - "block_id": "p587-b18", - "global_id": 16783, - "bbox": [ - 492.07, - 510.5, - 516.12, - 520.46 - ], - "text": "(5.56)", - "type": "text" - }, - { - "block_id": "p587-b19", - "global_id": 16784, - "bbox": [ - 127.59, - 541.03, - 368.0, - 552.59 - ], - "text": "where zk and pk are the system poles and zeros, respectively.", - "type": "text" - }, - { - "block_id": "p587-b20", - "global_id": 16785, - "bbox": [ - 127.59, - 552.68, - 516.14, - 634.79 - ], - "text": "A mapping rule converts the rational function H(s) to a rational function H[z]. Requiring\nthat the result be rational ensures that the system realization can proceed with only delay,\nsum, and multiplier blocks. There are many possible mapping rules. For obvious reasons, good\ntransformations tend to map the ω axis to the unit circle, ω = 0 to z = 1, ω = ∞to z = −1, and the\nleft half-plane to the interior of the unit circle. Put another way, sinusoids map to sinusoids, zero\nfrequency maps to zero frequency, high frequency maps to high frequency, and stable systems map\nto stable systems.", - "type": "text" - } - ] - }, - { - "page_num": 588, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p588-b0", - "global_id": 16786, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "568\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p588-b1", - "global_id": 16787, - "bbox": [ - 101.84, - 85.72, - 490.4, - 167.51 - ], - "text": "Section 5.9 suggests that the z-transform can be considered to be a Laplace transform with\na change of variable z = esT or s = (1/T)lnz, where T is the sampling interval. It is tempting,\ntherefore, to convert a continuous-time filter to a discrete-time filter by substituting s = (1/T)lnz\ninto H(s), or H[z] = H(s)|s=(1/T)lnz. This approach is impractical, however, since the resulting\nH[z] is not rational and therefore cannot be implemented by using standard blocks. Although\nnot considered here, the so-called matched-z transformation relies on the relationship z = esT to\ntransform system poles and zeros, so the connection is not completely without merit.", - "type": "text" - }, - { - "block_id": "p588-b2", - "global_id": 16788, - "bbox": [ - 101.84, - 192.24, - 422.27, - 204.2 - ], - "text": "5.10-3 Transformation by First-Order Backward Difference", - "type": "text" - }, - { - "block_id": "p588-b3", - "global_id": 16789, - "bbox": [ - 101.84, - 209.92, - 490.4, - 232.25 - ], - "text": "Consider the transfer function H(s) = Y(s)/X(s) = s, which corresponds to the first-order\ncontinuous-time differentiator", - "type": "text" - }, - { - "block_id": "p588-b4", - "global_id": 16790, - "bbox": [ - 270.21, - 229.91, - 304.4, - 246.86 - ], - "text": "y(t) = d", - "type": "text" - }, - { - "block_id": "p588-b5", - "global_id": 16791, - "bbox": [ - 298.07, - 236.58, - 322.03, - 253.93 - ], - "text": "dtx(t)", - "type": "text" - }, - { - "block_id": "p588-b6", - "global_id": 16792, - "bbox": [ - 101.84, - 258.92, - 490.36, - 280.85 - ], - "text": "An approximation that resembles the fundamental theorem of calculus is the first-order backward\ndifference", - "type": "text" - }, - { - "block_id": "p588-b7", - "global_id": 16793, - "bbox": [ - 252.73, - 278.41, - 338.3, - 295.67 - ], - "text": "y(t) = x(t) −x(t −T)", - "type": "text" - }, - { - "block_id": "p588-b8", - "global_id": 16794, - "bbox": [ - 101.84, - 292.78, - 442.4, - 317.56 - ], - "text": "T\nFor sampling interval T and t = nT, the corresponding discrete-time approximation is", - "type": "text" - }, - { - "block_id": "p588-b9", - "global_id": 16795, - "bbox": [ - 251.54, - 327.04, - 339.47, - 344.3 - ], - "text": "y[n] = x[n] −x[n −1]", - "type": "text" - }, - { - "block_id": "p588-b10", - "global_id": 16796, - "bbox": [ - 306.92, - 341.41, - 312.46, - 351.38 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p588-b11", - "global_id": 16797, - "bbox": [ - 101.84, - 359.21, - 210.56, - 369.17 - ], - "text": "which has transfer function", - "type": "text" - }, - { - "block_id": "p588-b12", - "global_id": 16798, - "bbox": [ - 240.14, - 367.03, - 350.39, - 385.61 - ], - "text": "H[z] = Y[z]/X[z] = 1 −z−1", - "type": "text" - }, - { - "block_id": "p588-b13", - "global_id": 16799, - "bbox": [ - 333.17, - 382.63, - 338.71, - 392.59 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p588-b14", - "global_id": 16800, - "bbox": [ - 101.84, - 397.4, - 490.41, - 444.87 - ], - "text": "This implies a transformation rule that uses the change of variable s = (1 −z−1)/T or z =\n1/(1 −sT). This transformation rule is appealing since the resulting H[z] is rational and has the\nsame number of poles and zeros as H(s). Section 3.4 discusses this transformation strategy in a\ndifferent way in describing the kinship of difference equations to differential equations.", - "type": "text" - }, - { - "block_id": "p588-b15", - "global_id": 16801, - "bbox": [ - 119.79, - 445.22, - 400.31, - 456.82 - ], - "text": "After some algebra, substituting s = (1 −z−1)/T into Eq. (5.56) yields", - "type": "text" - }, - { - "block_id": "p588-b16", - "global_id": 16802, - "bbox": [ - 185.2, - 481.54, - 213.18, - 491.81 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p588-b17", - "global_id": 16803, - "bbox": [ - 215.23, - 464.56, - 244.62, - 485.6 - ], - "text": "8\nbN−M", - "type": "text" - }, - { - "block_id": "p588-b18", - "global_id": 16804, - "bbox": [ - 246.51, - 467.05, - 310.37, - 487.02 - ], - "text": "7M\nk=1(1/T −zk)", - "type": "text" - }, - { - "block_id": "p588-b19", - "global_id": 16805, - "bbox": [ - 229.33, - 490.39, - 237.79, - 501.23 - ], - "text": "a0", - "type": "text" - }, - { - "block_id": "p588-b20", - "global_id": 16806, - "bbox": [ - 239.4, - 482.61, - 304.37, - 502.58 - ], - "text": "7N\nk=1(1/T −pk)", - "type": "text" - }, - { - "block_id": "p588-b21", - "global_id": 16807, - "bbox": [ - 311.57, - 463.49, - 342.66, - 483.46 - ], - "text": "9 7M\nk=1", - "type": "text" - }, - { - "block_id": "p588-b23", - "global_id": 16808, - "bbox": [ - 349.7, - 459.97, - 405.44, - 486.21 - ], - "text": "1 −\n1\n1−Tzk z−1", - "type": "text" - }, - { - "block_id": "p588-b24", - "global_id": 16809, - "bbox": [ - 320.78, - 484.63, - 342.27, - 504.6 - ], - "text": "7N\nk=1", - "type": "text" - }, - { - "block_id": "p588-b26", - "global_id": 16810, - "bbox": [ - 349.32, - 481.11, - 490.38, - 507.35 - ], - "text": "1 −\n1\n1−Tpk z−1\n\n(5.57)", - "type": "text" - }, - { - "block_id": "p588-b27", - "global_id": 16811, - "bbox": [ - 101.84, - 516.34, - 490.39, - 550.62 - ], - "text": "The discrete-time system has M zeros at 1/(1 −Tzk) and N poles at 1/(1 −Tpk). This\ntransformation rule preserves system stability but does not map the ω axis to the unit circle (see\nProb. 5.7-10).", - "type": "text" - }, - { - "block_id": "p588-b28", - "global_id": 16812, - "bbox": [ - 101.84, - 552.62, - 490.38, - 598.45 - ], - "text": "MATLAB program CH5MP3 uses the first-order backward difference method of Eq. (5.57)\nto convert a continuous-time filter described by coefficient vectors A = [a0,a1,...,aN] and\nB = [bN−M,bN−M+1,...,bN] into a discrete-time filter. The form of the discrete-time filter follows\nEq. (5.55).", - "type": "text" - }, - { - "block_id": "p588-b29", - "global_id": 16813, - "bbox": [ - 101.84, - 608.13, - 372.66, - 636.03 - ], - "text": "function [Bd,Ad] = CH5MP3(B,A,T);\n% CH5MP3.m : Chapter 5, MATLAB Program 3\n% Function M-file first-order backward difference transformation", - "type": "text" - } - ] - }, - { - "page_num": 589, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p589-b0", - "global_id": 16814, - "bbox": [ - 309.61, - 62.89, - 516.13, - 71.98 - ], - "text": "5.10\nMATLAB: Discrete-Time IIR Filters\n569", - "type": "text" - }, - { - "block_id": "p589-b1", - "global_id": 16815, - "bbox": [ - 127.59, - 86.3, - 461.91, - 144.08 - ], - "text": "% of a continuous-time filter described by B and A into a discrete-time filter.\n% INPUTS:\nB = vector of continuous-time filter feedforward coefficients\n%\nA = vector of continuous-time filter feedback coefficients\n%\nT = sampling interval\n% OUTPUTS:\nBd = vector of discrete-time filter feedforward coefficients\n%\nAd = vector of discrete-time filter feedback coefficients", - "type": "text" - }, - { - "block_id": "p589-b2", - "global_id": 16816, - "bbox": [ - 127.59, - 156.04, - 339.19, - 193.9 - ], - "text": "z = roots(B); p = roots(A); % s-domain roots\ngain = B(1)/A(1)*prod(1/T-z)/prod(1/T-p);\nzd = 1./(1-T*z); pd = 1./(1-T*p); % z-domain roots\nBd = gain*poly(zd); Ad = poly(pd);", - "type": "text" - }, - { - "block_id": "p589-b3", - "global_id": 16817, - "bbox": [ - 127.59, - 218.46, - 295.14, - 230.42 - ], - "text": "5.10-4 Bilinear Transformation", - "type": "text" - }, - { - "block_id": "p589-b4", - "global_id": 16818, - "bbox": [ - 127.59, - 236.55, - 516.13, - 258.46 - ], - "text": "The bilinear transformation is based on a better approximation than first-order backward\ndifferences. Again, consider the continuous-time integrator", - "type": "text" - }, - { - "block_id": "p589-b5", - "global_id": 16819, - "bbox": [ - 295.95, - 267.25, - 330.13, - 284.2 - ], - "text": "y(t) = d", - "type": "text" - }, - { - "block_id": "p589-b6", - "global_id": 16820, - "bbox": [ - 323.81, - 273.92, - 347.77, - 291.27 - ], - "text": "dtx(t)", - "type": "text" - }, - { - "block_id": "p589-b7", - "global_id": 16821, - "bbox": [ - 127.59, - 298.03, - 221.83, - 308.4 - ], - "text": "Represent signal x(t) as", - "type": "text" - }, - { - "block_id": "p589-b8", - "global_id": 16822, - "bbox": [ - 256.47, - 315.77, - 281.13, - 326.04 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p589-b9", - "global_id": 16823, - "bbox": [ - 283.18, - 302.21, - 294.93, - 314.49 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p589-b10", - "global_id": 16824, - "bbox": [ - 288.43, - 327.08, - 299.81, - 334.28 - ], - "text": "t−T", - "type": "text" - }, - { - "block_id": "p589-b11", - "global_id": 16825, - "bbox": [ - 303.16, - 309.1, - 387.24, - 333.12 - ], - "text": "d\ndτ x(τ)dτ + x(t −T)", - "type": "text" - }, - { - "block_id": "p589-b12", - "global_id": 16826, - "bbox": [ - 127.6, - 339.58, - 445.72, - 349.96 - ], - "text": "Letting t = nT and replacing the integral with a trapezoidal approximation yield", - "type": "text" - }, - { - "block_id": "p589-b13", - "global_id": 16827, - "bbox": [ - 220.88, - 359.57, - 262.65, - 376.51 - ], - "text": "x(nT) = T", - "type": "text" - }, - { - "block_id": "p589-b14", - "global_id": 16828, - "bbox": [ - 257.78, - 373.73, - 262.76, - 383.69 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p589-b15", - "global_id": 16829, - "bbox": [ - 265.73, - 352.25, - 278.67, - 369.53 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p589-b16", - "global_id": 16830, - "bbox": [ - 272.35, - 359.57, - 323.03, - 383.59 - ], - "text": "dtx(nT) + d", - "type": "text" - }, - { - "block_id": "p589-b17", - "global_id": 16831, - "bbox": [ - 316.72, - 366.24, - 366.18, - 383.59 - ], - "text": "dtx(nT −T)", - "type": "text" - }, - { - "block_id": "p589-b18", - "global_id": 16832, - "bbox": [ - 366.18, - 352.25, - 371.61, - 362.21 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p589-b19", - "global_id": 16833, - "bbox": [ - 373.17, - 366.24, - 422.83, - 376.51 - ], - "text": "+ x(nT −T)", - "type": "text" - }, - { - "block_id": "p589-b20", - "global_id": 16834, - "bbox": [ - 127.59, - 392.61, - 404.25, - 402.98 - ], - "text": "Substituting y(t) for (d/dt)x(t), the equivalent discrete-time system is", - "type": "text" - }, - { - "block_id": "p589-b21", - "global_id": 16835, - "bbox": [ - 249.03, - 411.78, - 283.7, - 428.72 - ], - "text": "x[n] = T", - "type": "text" - }, - { - "block_id": "p589-b22", - "global_id": 16836, - "bbox": [ - 278.83, - 418.45, - 394.65, - 435.9 - ], - "text": "2 (y[n] + y[n −1]) + x[n −1]", - "type": "text" - }, - { - "block_id": "p589-b23", - "global_id": 16837, - "bbox": [ - 127.6, - 442.67, - 296.39, - 452.74 - ], - "text": "From z-transforms, the transfer function is", - "type": "text" - }, - { - "block_id": "p589-b24", - "global_id": 16838, - "bbox": [ - 268.56, - 462.22, - 316.71, - 479.48 - ], - "text": "H[z] = Y[z]", - "type": "text" - }, - { - "block_id": "p589-b25", - "global_id": 16839, - "bbox": [ - 299.78, - 461.0, - 373.3, - 486.56 - ], - "text": "X[z] = 2(1 −z−1)", - "type": "text" - }, - { - "block_id": "p589-b26", - "global_id": 16840, - "bbox": [ - 331.07, - 475.79, - 373.96, - 486.66 - ], - "text": "T(1 + z−1)", - "type": "text" - }, - { - "block_id": "p589-b27", - "global_id": 16841, - "bbox": [ - 127.59, - 495.21, - 516.14, - 530.73 - ], - "text": "The implied change of variable s = 2(1 −z−1)/T(1 + z−1) or z = (1 + sT/2)/(1 −sT/2) is called\nthe bilinear transformation. Not only does the bilinear transformation result in a rational function\nH[z], the ω axis is correctly mapped to the unit circle (see Prob. 5.6-18a).", - "type": "text" - }, - { - "block_id": "p589-b28", - "global_id": 16842, - "bbox": [ - 145.53, - 531.08, - 467.62, - 542.68 - ], - "text": "After some algebra, substituting s = 2(1 −z−1)/T(1 + z−1) into Eq. (5.56) yields", - "type": "text" - }, - { - "block_id": "p589-b29", - "global_id": 16843, - "bbox": [ - 180.46, - 566.55, - 208.43, - 576.83 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p589-b30", - "global_id": 16844, - "bbox": [ - 210.48, - 549.58, - 239.86, - 570.61 - ], - "text": "8\nbN−M", - "type": "text" - }, - { - "block_id": "p589-b31", - "global_id": 16845, - "bbox": [ - 241.76, - 552.06, - 305.63, - 572.03 - ], - "text": "7M\nk=1(2/T −zk)", - "type": "text" - }, - { - "block_id": "p589-b32", - "global_id": 16846, - "bbox": [ - 224.58, - 575.41, - 233.04, - 586.24 - ], - "text": "a0", - "type": "text" - }, - { - "block_id": "p589-b33", - "global_id": 16847, - "bbox": [ - 234.65, - 567.62, - 299.63, - 587.58 - ], - "text": "7N\nk=1(2/T −pk)", - "type": "text" - }, - { - "block_id": "p589-b34", - "global_id": 16848, - "bbox": [ - 306.82, - 548.51, - 337.92, - 568.47 - ], - "text": "9 7M\nk=1", - "type": "text" - }, - { - "block_id": "p589-b36", - "global_id": 16849, - "bbox": [ - 344.95, - 544.98, - 408.0, - 570.29 - ], - "text": "1 −1+zkT/2\n1−zkT/2z−1", - "type": "text" - }, - { - "block_id": "p589-b37", - "global_id": 16850, - "bbox": [ - 316.03, - 569.65, - 337.53, - 589.62 - ], - "text": "7N\nk=1", - "type": "text" - }, - { - "block_id": "p589-b39", - "global_id": 16851, - "bbox": [ - 344.56, - 564.83, - 516.13, - 591.43 - ], - "text": "1 −1+pkT/2\n1−pkT/2z−1\n(1 + z−1)N−M\n(5.58)", - "type": "text" - }, - { - "block_id": "p589-b40", - "global_id": 16852, - "bbox": [ - 127.59, - 600.5, - 516.13, - 634.79 - ], - "text": "In addition to the M zeros at (1+zkT/2)/(1−zkT/2) and N poles at (1 + pkT/2)/(1 −pkT/2),\nthere are N−M zeros at minus 1. Since practical continuous-time filters require M ≤N for stability,\nthe number of added zeros is thankfully always nonnegative.", - "type": "text" - } - ] - }, - { - "page_num": 590, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p590-b0", - "global_id": 16853, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "570\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p590-b1", - "global_id": 16854, - "bbox": [ - 101.84, - 85.82, - 490.39, - 143.6 - ], - "text": "MATLAB program CH5MP4 converts a continuous-time filter described by coefficient vectors\nA = [a0,a1,...,aN] and B = [bN−M,bN−M+1,...,bN] into a discrete-time filter by using the bilinear\ntransformation of Eq. (5.58). The form of the discrete-time filter follows Eq. (5.55). If available,\nit is also possible to use the signal-processing toolbox function bilinear to perform the bilinear\ntransformation.", - "type": "text" - }, - { - "block_id": "p590-b2", - "global_id": 16855, - "bbox": [ - 101.84, - 159.18, - 483.66, - 276.73 - ], - "text": "function [Bd,Ad] = CH5MP4(B,A,T);\n% CH5MP4.m : Chapter 5, MATLAB Program 4\n% Function M-file bilinear transformation of a continuous-time filter\n% described by vectors B and A into a discrete-time filter.\n% Length of B must not exceed A.\n% INPUTS:\nB = vector of continuous-time filter feedforward coefficients\n%\nA = vector of continuous-time filter feedback coefficients\n%\nT = sampling interval\n% OUTPUTS:\nBd = vector of discrete-time filter feedforward coefficients\n%\nAd = vector of discrete-time filter feedback coefficients", - "type": "text" - }, - { - "block_id": "p590-b3", - "global_id": 16856, - "bbox": [ - 101.84, - 290.69, - 232.6, - 300.65 - ], - "text": "if (length(B)>length(A)),", - "type": "text" - }, - { - "block_id": "p590-b4", - "global_id": 16857, - "bbox": [ - 101.84, - 302.64, - 467.97, - 384.33 - ], - "text": "disp(’Numerator order must not exceed denominator order.’);\nreturn\nend\nz = roots(B); p = roots(A); % s-domain roots\ngain = real(B(1)/A(1)*prod(2/T-z)/prod(2/T-p));\nzd = (1+z*T/2)./(1-z*T/2); pd = (1+p*T/2)./(1-p*T/2); % z-domain roots\nBd = gain*poly([zd;-ones(length(A)-length(B),1)]); Ad = poly(pd);", - "type": "text" - }, - { - "block_id": "p590-b5", - "global_id": 16858, - "bbox": [ - 101.84, - 399.33, - 407.23, - 409.3 - ], - "text": "As with most high-level languages, MATLAB supports general if-structures:", - "type": "text" - }, - { - "block_id": "p590-b6", - "global_id": 16859, - "bbox": [ - 119.78, - 410.87, - 193.0, - 421.54 - ], - "text": "if expression,", - "type": "text" - }, - { - "block_id": "p590-b7", - "global_id": 16860, - "bbox": [ - 119.78, - 422.82, - 213.92, - 445.45 - ], - "text": "statements;\nelseif expression,", - "type": "text" - }, - { - "block_id": "p590-b8", - "global_id": 16861, - "bbox": [ - 119.78, - 446.74, - 210.73, - 469.35 - ], - "text": "statements;\nelse,", - "type": "text" - }, - { - "block_id": "p590-b9", - "global_id": 16862, - "bbox": [ - 101.85, - 470.65, - 490.37, - 517.18 - ], - "text": "statements;\nend\nIn the program CH5MP4, the if statement tests M > N. When true, an error message is displayed\nand the return command terminates program execution to prevent errors.", - "type": "text" - }, - { - "block_id": "p590-b10", - "global_id": 16863, - "bbox": [ - 101.84, - 546.96, - 364.35, - 558.92 - ], - "text": "5.10-5 Bilinear Transformation with Prewarping", - "type": "text" - }, - { - "block_id": "p590-b11", - "global_id": 16864, - "bbox": [ - 101.84, - 564.63, - 490.39, - 610.88 - ], - "text": "The bilinear transformation maps the entire infinite-length ω axis onto the finite-length unit circle\n(z = ej) according to ω = (2/T)tan(/2) (see Prob. 5.6-18b). Equivalently, = 2arctan(ωT/2).\nThe nonlinearity of the tangent function causes a frequency compression, commonly called\nfrequency warping, that distorts the transformation.", - "type": "text" - }, - { - "block_id": "p590-b12", - "global_id": 16865, - "bbox": [ - 101.85, - 612.87, - 490.38, - 635.87 - ], - "text": "To illustrate the warping effect, consider the bilinear transformation of a continuous-time\nlowpass filter with cutoff frequency ωc = 2π3000 rad/s. If the target digital system uses a sampling", - "type": "text" - } - ] - }, - { - "page_num": 591, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p591-b0", - "global_id": 16866, - "bbox": [ - 309.61, - 62.89, - 516.13, - 71.98 - ], - "text": "5.10\nMATLAB: Discrete-Time IIR Filters\n571", - "type": "text" - }, - { - "block_id": "p591-b1", - "global_id": 16867, - "bbox": [ - 127.59, - 85.46, - 516.13, - 108.87 - ], - "text": "rate of 10 kHz, then T = 1/(10,000) and ωc maps to c = 2arctan(ωcT/2) = 1.5116. Thus, the\ntransformed cutoff frequency is short of the desired c = ωcT = 0.6π = 1.8850.", - "type": "text" - }, - { - "block_id": "p591-b2", - "global_id": 16868, - "bbox": [ - 127.59, - 109.79, - 516.14, - 168.65 - ], - "text": "Cutoff frequencies are important and need to be as accurate as possible. By adjusting the\nparameter T used in the bilinear transform, one continuous-time frequency can be exactly mapped\nto one discrete-time frequency; the process is called prewarping. Continuing the last example,\nadjusting T = (2/ωc)tan(c/2) ≈1/6848 achieves the appropriate prewarping to ensure ωc =\n2π3000 maps to c = 0.6π.", - "type": "text" - }, - { - "block_id": "p591-b3", - "global_id": 16869, - "bbox": [ - 127.59, - 201.43, - 403.05, - 213.39 - ], - "text": "5.10-6 Example: Butterworth Filter Transformation", - "type": "text" - }, - { - "block_id": "p591-b4", - "global_id": 16870, - "bbox": [ - 127.59, - 219.51, - 516.12, - 253.39 - ], - "text": "To illustrate the transformation techniques, consider a continuous-time 10th-order Butterworth\nlowpass filter with cutoff frequency ωc = 2π3000, as designed in Sec. 4.12. First, we determine\ncontinuous-time coefficient vectors A and B.", - "type": "text" - }, - { - "block_id": "p591-b5", - "global_id": 16871, - "bbox": [ - 127.59, - 272.19, - 372.92, - 310.05 - ], - "text": ">>\nomega_c = 2*pi*3000; N=10;\n>>\npoles = roots([(1j*omega_c)^(-2*N),zeros(1,2*N-1),1]);\n>>\npoles = poles(find(poles<0));\n>>\nB = 1; A = poly(poles); A = A/A(end);", - "type": "text" - }, - { - "block_id": "p591-b6", - "global_id": 16872, - "bbox": [ - 127.59, - 329.38, - 516.11, - 351.3 - ], - "text": "Programs CH5MP3 and CH5MP4 are used to perform first-order forward difference and bilinear\ntransformations, respectively.", - "type": "text" - }, - { - "block_id": "p591-b7", - "global_id": 16873, - "bbox": [ - 127.59, - 370.11, - 457.67, - 407.96 - ], - "text": ">>\nOmega = linspace(0,pi,200); T = 1/10000; Omega_c = omega_c*T;\n>>\n[B1,A1] = CH5MP3(B,A,T); % First-order backward difference transformation\n>>\n[B2,A2] = CH5MP4(B,A,T); % Bilinear transformation\n>>\n[B3,A3] = CH5MP4(B,A,2/omega_c*tan(Omega_c/2)); % Bilinear with prewarping", - "type": "text" - }, - { - "block_id": "p591-b8", - "global_id": 16874, - "bbox": [ - 145.52, - 427.28, - 413.47, - 437.54 - ], - "text": "Magnitude responses are computed using CH5MP1 and then plotted.", - "type": "text" - }, - { - "block_id": "p591-b9", - "global_id": 16875, - "bbox": [ - 127.59, - 456.06, - 402.61, - 533.76 - ], - "text": ">>\nH1mag = abs(CH5MP1(B1,A1,Omega));\n>>\nH2mag = abs(CH5MP1(B2,A2,Omega));\n>>\nH3mag = abs(CH5MP1(B3,A3,Omega));\n>>\nplot(Omega,(Omega<=Omega_c),’k’,Omega,H1mag,’k-.’,...\n>>\nOmega,H2mag,’k--’,Omega,H3mag,’k:’);\n>>\naxis([0 pi -.05 1.5]);\n>>\nxlabel(’\\Omega [rad]’); ylabel(’Magnitude Response’);\n>>\nlegend(’Ideal’,’FOBD’,’BLT’,’Prewarp BLT’,’location’,’best’);", - "type": "text" - }, - { - "block_id": "p591-b10", - "global_id": 16876, - "bbox": [ - 127.59, - 553.1, - 516.11, - 575.01 - ], - "text": "The result of each transformation method is shown in Fig. 5.31, where FOBD and BLT stand for\nfirst-order backward difference and bilinear transformation, respectively.", - "type": "text" - }, - { - "block_id": "p591-b11", - "global_id": 16877, - "bbox": [ - 127.59, - 577.0, - 516.15, - 634.79 - ], - "text": "Although the first-order backward difference results in a lowpass filter, the method causes\nsignificant distortion that makes the resulting filter unacceptable with regard to cutoff frequency.\nThe bilinear transformation is better, but, as predicted, the cutoff frequency falls short of the\ndesired value. Bilinear transformation with prewarping properly locates the cutoff frequency and\nproduces a very acceptable filter response.", - "type": "text" - } - ] - }, - { - "page_num": 592, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p592-b0", - "global_id": 16878, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "572\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p592-b1", - "global_id": 16879, - "bbox": [ - 151.47, - 199.98, - 475.46, - 207.98 - ], - "text": "0\n0.5\n1\n1.5\n2\n2.5\n3", - "type": "text" - }, - { - "block_id": "p592-b2", - "global_id": 16880, - "bbox": [ - 309.5, - 211.84, - 333.19, - 220.14 - ], - "text": "Ω [rad]", - "type": "text" - }, - { - "block_id": "p592-b3", - "global_id": 16881, - "bbox": [ - 145.24, - 185.92, - 149.24, - 193.92 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p592-b4", - "global_id": 16882, - "bbox": [ - 138.49, - 166.55, - 149.17, - 174.55 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p592-b5", - "global_id": 16883, - "bbox": [ - 138.49, - 147.19, - 149.17, - 155.19 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p592-b6", - "global_id": 16884, - "bbox": [ - 138.49, - 127.83, - 149.17, - 135.83 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p592-b7", - "global_id": 16885, - "bbox": [ - 138.49, - 108.47, - 149.17, - 116.47 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p592-b8", - "global_id": 16886, - "bbox": [ - 145.24, - 89.11, - 149.24, - 97.11 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p592-b9", - "global_id": 16887, - "bbox": [ - 126.52, - 108.93, - 135.32, - 182.99 - ], - "text": "Magnitude Response", - "type": "text" - }, - { - "block_id": "p592-b10", - "global_id": 16888, - "bbox": [ - 423.35, - 94.95, - 462.75, - 134.5 - ], - "text": "Ideal\nFOBD\nBLT\nPrewarp BLT", - "type": "text" - }, - { - "block_id": "p592-b11", - "global_id": 16889, - "bbox": [ - 125.76, - 227.44, - 355.65, - 236.68 - ], - "text": "Figure 5.31 Comparison of various transformation techniques.", - "type": "text" - }, - { - "block_id": "p592-b12", - "global_id": 16890, - "bbox": [ - 101.84, - 262.16, - 336.48, - 274.11 - ], - "text": "5.10-7 Problems Finding Polynomial Roots", - "type": "text" - }, - { - "block_id": "p592-b13", - "global_id": 16891, - "bbox": [ - 101.84, - 280.24, - 490.39, - 314.41 - ], - "text": "Numerically, it is difficult to accurately determine the roots of a polynomial. Consider, for\nexample, a simple polynomial that has a root at minus 1 repeated four times, (s + 1)4 = s4 +\n4s3 + 6s2 + 4s + 1. The MATLAB roots command returns a surprising result:", - "type": "text" - }, - { - "block_id": "p592-b14", - "global_id": 16892, - "bbox": [ - 101.85, - 328.65, - 415.66, - 350.56 - ], - "text": ">>\nroots([1 4 6 4 1])’\nans = -1.0002\n-1.0000-0.0002i\n-1.0000+0.0002i\n-0.9998", - "type": "text" - }, - { - "block_id": "p592-b15", - "global_id": 16893, - "bbox": [ - 101.85, - 364.51, - 413.73, - 374.47 - ], - "text": "Even for this low-degree polynomial, MATLAB does not return the true roots.", - "type": "text" - }, - { - "block_id": "p592-b16", - "global_id": 16894, - "bbox": [ - 101.85, - 376.46, - 490.4, - 410.63 - ], - "text": "The problem worsens as polynomial degree increases. The bilinear transformation of the\n10th-order Butterworth filter, for example, should have 10 zeros at minus 1. Figure 5.32 shows\nthat the zeros, computed by CH5MP2 with the roots command, are not correctly located.", - "type": "text" - }, - { - "block_id": "p592-b17", - "global_id": 16895, - "bbox": [ - 101.85, - 412.33, - 490.39, - 458.16 - ], - "text": "When possible, programs should avoid root computations that may limit accuracy. For\nexample, results from the transformation programs CH5MP3 and CH5MP4 are more accurate if the\ntrue transfer function poles and zeros are passed directly as inputs rather than the polynomial\ncoefficient vectors. When roots must be computed, result accuracy should always be verified.", - "type": "text" - }, - { - "block_id": "p592-b18", - "global_id": 16896, - "bbox": [ - 101.84, - 487.18, - 458.99, - 499.14 - ], - "text": "5.10-8 Using Cascaded Second-Order Sections to Improve Design", - "type": "text" - }, - { - "block_id": "p592-b19", - "global_id": 16897, - "bbox": [ - 101.84, - 505.27, - 490.4, - 539.15 - ], - "text": "The dynamic range of high-degree polynomial coefficients is often large. Adding the difficulties\nassociated with factoring a high-degree polynomial, it is little surprise that high-order designs are\ndifficult.", - "type": "text" - }, - { - "block_id": "p592-b20", - "global_id": 16898, - "bbox": [ - 101.84, - 541.14, - 490.4, - 586.96 - ], - "text": "As with continuous-time filters, performance is improved by using a cascade of second-order\nsections to design and realize a discrete-time filter. Cascades of second-order sections are also\nmore robust to the coefficient quantization that occurs when discrete-time filters are implemented\non fixed-point digital hardware.", - "type": "text" - }, - { - "block_id": "p592-b21", - "global_id": 16899, - "bbox": [ - 101.84, - 588.96, - 490.39, - 634.79 - ], - "text": "To illustrate the performance possible with a cascade of second-order sections, consider\na 180th-order transformed Butterworth discrete-time filter with cutoff frequency c = 0.6π ≈\n1.8850. Program CH5MP5 completes this design, taking care to initially locate poles and zeros\nwithout root computations.", - "type": "text" - } - ] - }, - { - "page_num": 593, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p593-b0", - "global_id": 16900, - "bbox": [ - 309.61, - 62.89, - 516.13, - 71.98 - ], - "text": "5.10\nMATLAB: Discrete-Time IIR Filters\n573", - "type": "text" - }, - { - "block_id": "p593-b1", - "global_id": 16901, - "bbox": [ - 208.94, - 265.57, - 372.47, - 286.76 - ], - "text": "–1\n–0.5\n0\n0.5\n1\nReal", - "type": "text" - }, - { - "block_id": "p593-b2", - "global_id": 16902, - "bbox": [ - 169.23, - 248.55, - 177.23, - 256.55 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p593-b3", - "global_id": 16903, - "bbox": [ - 163.23, - 209.51, - 177.23, - 217.51 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p593-b4", - "global_id": 16904, - "bbox": [ - 173.23, - 170.48, - 177.23, - 178.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p593-b5", - "global_id": 16905, - "bbox": [ - 166.48, - 131.45, - 176.48, - 139.45 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p593-b6", - "global_id": 16906, - "bbox": [ - 173.23, - 92.4, - 177.23, - 100.4 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p593-b7", - "global_id": 16907, - "bbox": [ - 152.26, - 165.23, - 161.06, - 183.32 - ], - "text": "Imag", - "type": "text" - }, - { - "block_id": "p593-b8", - "global_id": 16908, - "bbox": [ - 151.5, - 293.44, - 350.62, - 302.67 - ], - "text": "Figure 5.32 Pole-zero plot computed by using roots.", - "type": "text" - }, - { - "block_id": "p593-b9", - "global_id": 16909, - "bbox": [ - 127.59, - 325.67, - 457.68, - 353.57 - ], - "text": "% CH5MP5.m : Chapter 5, MATLAB Program 5\n% Script M-file designs a 180th-order Butterworth lowpass discrete-time filter\n% with cutoff Omega_c = 0.6*pi using 90 cascaded second-order filter sections.", - "type": "text" - }, - { - "block_id": "p593-b10", - "global_id": 16910, - "bbox": [ - 127.59, - 365.52, - 444.96, - 423.31 - ], - "text": "omega_0 = 1; % Use normalized cutoff frequency for analog prototype\npsi = [0.5:1:90]*pi/180; % Butterworth pole angles\nOmega_c = 0.6*pi; % Discrete-time cutoff frequency\nOmega = linspace(0,pi,1000); % Frequency range for magnitude response\nHmag = zeros(90,1000); p = zeros(1,180); z = zeros(1,180); % Pre-allocation\nfor stage = 1:90,", - "type": "text" - }, - { - "block_id": "p593-b11", - "global_id": 16911, - "bbox": [ - 127.59, - 425.3, - 453.46, - 552.81 - ], - "text": "Q = 1/(2*cos(psi(stage))); % Compute Q for stage\nB = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute stage coefficients\n[B1,A1] = CH5MP4(B,A,2/omega_0*tan(0.6*pi/2)); % Transform stage to DT\np(stage*2-1:stage*2) = roots(A1); % Compute z-domain poles for stage\nz(stage*2-1:stage*2) = roots(B1); % Compute z-domain zeros for stage\nHmag(stage,:) = abs(CH5MP1(B1,A1,Omega)); % Compute stage mag response\nend\nucirc = exp(j*linspace(0,2*pi,200)); % Compute unit circle for pole-zero plot\nfigure;\nplot(real(p),imag(p),’kx’,real(z),imag(z),’ok’,real(ucirc),imag(ucirc),’k:’);\naxis equal; xlabel(’Real’); ylabel(’Imag’);\nfigure; plot(Omega,prod(Hmag),’k’); axis([0 pi -0.05 1.05]);\nxlabel(’\\Omega [rad]’); ylabel(’Magnitude Response’);", - "type": "text" - }, - { - "block_id": "p593-b12", - "global_id": 16912, - "bbox": [ - 127.59, - 565.05, - 503.88, - 575.3 - ], - "text": "The figure command preceding each plot command opens a separate window for each plot.", - "type": "text" - }, - { - "block_id": "p593-b13", - "global_id": 16913, - "bbox": [ - 127.59, - 577.0, - 516.15, - 634.79 - ], - "text": "The filter’s pole-zero plot is shown in Fig. 5.33, along with the unit circle, for reference. All\n180 zeros of the cascaded design are properly located at minus 1. The wall of poles provides an\namazing approximation to the desired brick-wall response, as shown by the magnitude response\nin Fig. 5.34. It is virtually impossible to realize such high-order designs with continuous-time\nfilters, which adds another reason for the popularity of discrete-time filters. Still, the design is not", - "type": "text" - } - ] - }, - { - "page_num": 594, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p594-b0", - "global_id": 16914, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "574\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p594-b1", - "global_id": 16915, - "bbox": [ - 173.61, - 265.57, - 352.47, - 286.76 - ], - "text": "–1\n–0.5\n0\n0.5\n1\nReal", - "type": "text" - }, - { - "block_id": "p594-b2", - "global_id": 16916, - "bbox": [ - 137.49, - 239.18, - 151.49, - 247.18 - ], - "text": "–0.8", - "type": "text" - }, - { - "block_id": "p594-b3", - "global_id": 16917, - "bbox": [ - 137.49, - 222.0, - 151.49, - 230.0 - ], - "text": "–0.6", - "type": "text" - }, - { - "block_id": "p594-b4", - "global_id": 16918, - "bbox": [ - 137.49, - 204.82, - 151.49, - 212.82 - ], - "text": "–0.4", - "type": "text" - }, - { - "block_id": "p594-b5", - "global_id": 16919, - "bbox": [ - 137.49, - 187.66, - 151.49, - 195.66 - ], - "text": "–0.2", - "type": "text" - }, - { - "block_id": "p594-b6", - "global_id": 16920, - "bbox": [ - 147.49, - 170.48, - 151.49, - 178.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p594-b7", - "global_id": 16921, - "bbox": [ - 140.74, - 153.3, - 150.74, - 161.3 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p594-b8", - "global_id": 16922, - "bbox": [ - 140.74, - 136.12, - 150.74, - 144.12 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p594-b9", - "global_id": 16923, - "bbox": [ - 140.74, - 118.95, - 150.74, - 126.95 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p594-b10", - "global_id": 16924, - "bbox": [ - 140.74, - 101.77, - 150.74, - 109.78 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p594-b11", - "global_id": 16925, - "bbox": [ - 126.52, - 165.23, - 135.32, - 183.32 - ], - "text": "Imag", - "type": "text" - }, - { - "block_id": "p594-b12", - "global_id": 16926, - "bbox": [ - 125.76, - 293.44, - 399.44, - 302.67 - ], - "text": "Figure 5.33 Pole-zero plot for 180th-order discrete-time Butterworth filter.", - "type": "text" - }, - { - "block_id": "p594-b13", - "global_id": 16927, - "bbox": [ - 151.47, - 437.47, - 474.89, - 445.47 - ], - "text": "0\n0.5\n1\n1.5\n2\n2.5\n3", - "type": "text" - }, - { - "block_id": "p594-b14", - "global_id": 16928, - "bbox": [ - 314.6, - 449.34, - 338.28, - 457.63 - ], - "text": "Ω [rad]", - "type": "text" - }, - { - "block_id": "p594-b15", - "global_id": 16929, - "bbox": [ - 145.24, - 423.41, - 149.24, - 431.41 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p594-b16", - "global_id": 16930, - "bbox": [ - 138.49, - 404.04, - 149.17, - 412.04 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p594-b17", - "global_id": 16931, - "bbox": [ - 138.49, - 384.69, - 149.17, - 392.69 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p594-b18", - "global_id": 16932, - "bbox": [ - 138.49, - 365.32, - 149.17, - 373.32 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p594-b19", - "global_id": 16933, - "bbox": [ - 138.49, - 345.96, - 149.17, - 353.96 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p594-b20", - "global_id": 16934, - "bbox": [ - 145.24, - 326.6, - 149.24, - 334.6 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p594-b21", - "global_id": 16935, - "bbox": [ - 126.52, - 346.42, - 135.32, - 420.48 - ], - "text": "Magnitude Response", - "type": "text" - }, - { - "block_id": "p594-b22", - "global_id": 16936, - "bbox": [ - 125.76, - 464.92, - 427.57, - 474.16 - ], - "text": "Figure 5.34 Magnitude response for a 180th-order discrete-time Butterworth filter.", - "type": "text" - }, - { - "block_id": "p594-b23", - "global_id": 16937, - "bbox": [ - 101.84, - 494.78, - 490.4, - 516.71 - ], - "text": "trivial; even functions from the MATLAB signal-processing toolbox fail to properly design such a\nhigh-order discrete-time Butterworth filter.", - "type": "text" - }, - { - "block_id": "p594-b24", - "global_id": 16938, - "bbox": [ - 102.2, - 546.1, - 202.15, - 560.05 - ], - "text": "5.11 SUMMARY", - "type": "text" - }, - { - "block_id": "p594-b25", - "global_id": 16939, - "bbox": [ - 101.84, - 566.05, - 490.4, - 611.87 - ], - "text": "In this chapter we discussed the analysis of linear, time-invariant, discrete-time (LTID) systems by\nmeans of the z-transform. The z-transform changes the difference equations of LTID systems into\nalgebraic equations. Therefore, solving these difference equations reduces to solving algebraic\nequations.", - "type": "text" - }, - { - "block_id": "p594-b26", - "global_id": 16940, - "bbox": [ - 101.84, - 613.45, - 490.39, - 647.74 - ], - "text": "The transfer function H[z] of an LTID system is equal to the ratio of the z-transform of the\noutput to the z-transform of the input when all initial conditions are zero. Therefore, if X[z] is\nthe z-transform of the input x[n] and Y[z] is the z-transform of the corresponding output y[n]", - "type": "text" - } - ] - }, - { - "page_num": 595, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p595-b0", - "global_id": 16941, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n575", - "type": "text" - }, - { - "block_id": "p595-b1", - "global_id": 16942, - "bbox": [ - 127.59, - 85.4, - 516.14, - 131.64 - ], - "text": "(when all initial conditions are zero), then Y[z] = H[z]X[z]. For an LTID system specified by\nthe difference equation Q[E]y[n] = P[E]x[n], the transfer function H[z] = P[z]/Q[z]. Moreover,\nH[z] is the z-transform of the system impulse response h[n]. We showed in Ch. 3 that the system\nresponse to an everlasting exponential zn is H[z]zn.", - "type": "text" - }, - { - "block_id": "p595-b2", - "global_id": 16943, - "bbox": [ - 127.59, - 133.23, - 516.14, - 179.47 - ], - "text": "We may also view the z-transform as a tool that expresses a signal x[n] as a sum of\nexponentials of the form zn over a continuum of the values of z. Using the fact that an LTID\nsystem response to zn is H[z]zn, we find the system response to x[n] as a sum of the system’s\nresponses to all the components of the form zn over the continuum of values of z.", - "type": "text" - }, - { - "block_id": "p595-b3", - "global_id": 16944, - "bbox": [ - 127.59, - 181.46, - 516.13, - 227.29 - ], - "text": "LTID systems can be realized by scalar multipliers, adders, and time delays. A given transfer\nfunction can be synthesized in many different ways. We discussed canonical, transposed canonical,\ncascade, and parallel forms of realization. The realization procedure is identical to that for\ncontinuous-time systems with 1/s (integrator) replaced by 1/z (unit delay).", - "type": "text" - }, - { - "block_id": "p595-b4", - "global_id": 16945, - "bbox": [ - 127.6, - 229.29, - 516.16, - 322.93 - ], - "text": "The majority of the input signals and practical systems are causal. Consequently, we are\nrequired to deal with causal signals most of the time. Restricting all signals to the causal type\ngreatly simplifies z-transform analysis; the ROC of a signal becomes irrelevant to the analysis\nprocess. This special case of z-transform (which is restricted to causal signals) is called the\nunilateral z-transform. Much of the chapter deals with this transform. Section 5.8 discusses the\ngeneral variety of the z-transform (bilateral z-transform), which can handle causal and noncausal\nsignals and systems. In the bilateral transform, the inverse transform of X[z] is not unique, but\ndepends on the ROC of X[z]. Thus, the ROC plays a crucial role in the bilateral z-transform.", - "type": "text" - }, - { - "block_id": "p595-b5", - "global_id": 16946, - "bbox": [ - 127.6, - 324.92, - 516.13, - 358.8 - ], - "text": "In Sec. 5.9, we showed that discrete-time systems can be analyzed by the Laplace transform\nas if they were continuous-time systems. In fact, we showed that the z-transform is the Laplace\ntransform with a change in variable.", - "type": "text" - }, - { - "block_id": "p595-b6", - "global_id": 16947, - "bbox": [ - 127.86, - 380.07, - 215.04, - 391.03 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p595-b7", - "global_id": 16948, - "bbox": [ - 127.59, - 398.1, - 481.58, - 407.15 - ], - "text": "1.\nLyons, R. G. Understanding Digital Signal Processing. Addison-Wesley, Reading, MA, 1997.", - "type": "text" - }, - { - "block_id": "p595-b8", - "global_id": 16949, - "bbox": [ - 127.59, - 412.05, - 516.12, - 432.07 - ], - "text": "2.\nOppenheim, A. V., and R. W. Schafer. Discrete-Time Signal Processing, 2nd ed. Prentice-Hall, Upper\nSaddle River, NJ, 1999.", - "type": "text" - }, - { - "block_id": "p595-b9", - "global_id": 16950, - "bbox": [ - 127.59, - 436.95, - 428.68, - 446.01 - ], - "text": "3.\nMitra, S. K. Digital Signal Processing, 2nd ed. McGraw-Hill, New York, 2001.", - "type": "text" - }, - { - "block_id": "p595-b10", - "global_id": 16951, - "bbox": [ - 106.67, - 469.63, - 217.26, - 486.57 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p595-b11", - "global_id": 16952, - "bbox": [ - 87.82, - 497.08, - 288.98, - 539.01 - ], - "text": "5.1-1\nUsing the definition, compute the z-transform of\nx[n] = (−1)n(u[n] −u[n −8]). Sketch the poles\nand zeros of X[z] in the z plane. No calculator is\nneeded to do this problem!", - "type": "text" - }, - { - "block_id": "p595-b12", - "global_id": 16953, - "bbox": [ - 87.82, - 543.86, - 288.99, - 586.08 - ], - "text": "5.1-2\nDetermine the unilateral z-transform X[z] of the\nsignal x[n] shown in Fig. P5.1-2. As the picture\nsuggests, x[n] = −3 for all n ≥9 and x[n] = 0\nfor all n < 3.", - "type": "text" - }, - { - "block_id": "p595-b13", - "global_id": 16954, - "bbox": [ - 87.82, - 591.19, - 288.98, - 600.25 - ], - "text": "5.1-3\n(a) A causal signal has z-transform given by", - "type": "text" - }, - { - "block_id": "p595-b14", - "global_id": 16955, - "bbox": [ - 131.9, - 600.6, - 172.16, - 612.81 - ], - "text": "X[z] =\nz2", - "type": "text" - }, - { - "block_id": "p595-b15", - "global_id": 16956, - "bbox": [ - 131.89, - 603.93, - 288.98, - 645.77 - ], - "text": "z3−1. Determine the time-domain\nsignal x[n] and sketch x[n] over −4 ≤n ≤\n11. [Hint: No complex arithmetic is needed\nto solve this problem!]", - "type": "text" - }, - { - "block_id": "p595-b16", - "global_id": 16957, - "bbox": [ - 386.13, - 500.44, - 398.96, - 508.66 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p595-b17", - "global_id": 16958, - "bbox": [ - 506.55, - 566.55, - 510.53, - 574.52 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p595-b18", - "global_id": 16959, - "bbox": [ - 374.22, - 517.53, - 378.21, - 525.5 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p595-b19", - "global_id": 16960, - "bbox": [ - 372.92, - 546.93, - 376.9, - 554.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p595-b20", - "global_id": 16961, - "bbox": [ - 366.36, - 575.71, - 376.57, - 584.02 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p595-b21", - "global_id": 16962, - "bbox": [ - 367.67, - 605.23, - 377.87, - 613.53 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p595-b22", - "global_id": 16963, - "bbox": [ - 452.92, - 553.41, - 497.39, - 561.39 - ], - "text": "15\n10", - "type": "text" - }, - { - "block_id": "p595-b23", - "global_id": 16964, - "bbox": [ - 342.22, - 569.87, - 422.11, - 578.17 - ], - "text": "5\n−5", - "type": "text" - }, - { - "block_id": "p595-b24", - "global_id": 16965, - "bbox": [ - 497.6, - 582.44, - 508.05, - 590.41 - ], - "text": "· · ·", - "type": "text" - }, - { - "block_id": "p595-b25", - "global_id": 16966, - "bbox": [ - 339.12, - 624.82, - 390.77, - 633.79 - ], - "text": "Figure P5.1-2", - "type": "text" - } - ] - }, - { - "page_num": 596, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p596-b0", - "global_id": 16967, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "576\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p596-b1", - "global_id": 16968, - "bbox": [ - 90.72, - 85.53, - 263.24, - 94.86 - ], - "text": "(b) Consider the causal semiperiodic signal y[n]", - "type": "text" - }, - { - "block_id": "p596-b2", - "global_id": 16969, - "bbox": [ - 106.16, - 96.49, - 263.25, - 149.68 - ], - "text": "shown in Fig. P5.1-3. Notice, y[n] continu-\nally repeats the sequence [1,2,3] for n ≥0.\nDetermine the unilateral z-transform Y[z] of\nthis signal. If possible, express your result\nas a rational function in standard form.", - "type": "text" - }, - { - "block_id": "p596-b3", - "global_id": 16970, - "bbox": [ - 62.09, - 155.36, - 263.24, - 186.34 - ], - "text": "5.1-4\nUsing the definition of the z-transform, find\nthe z-transform and the ROC for each of the\nfollowing signals.", - "type": "text" - }, - { - "block_id": "p596-b4", - "global_id": 16971, - "bbox": [ - 90.72, - 187.97, - 154.98, - 208.27 - ], - "text": "(a) u[n −m]\n(b) γ n sinπnu[n]", - "type": "text" - }, - { - "block_id": "p596-b5", - "global_id": 16972, - "bbox": [ - 91.23, - 206.97, - 156.47, - 219.19 - ], - "text": "(c) γ n cosπnu[n]", - "type": "text" - }, - { - "block_id": "p596-b6", - "global_id": 16973, - "bbox": [ - 133.98, - 228.17, - 146.16, - 235.74 - ], - "text": "y[ ]", - "type": "text" - }, - { - "block_id": "p596-b7", - "global_id": 16974, - "bbox": [ - 123.47, - 262.59, - 127.26, - 270.16 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p596-b8", - "global_id": 16975, - "bbox": [ - 123.47, - 278.4, - 127.26, - 285.97 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p596-b9", - "global_id": 16976, - "bbox": [ - 100.7, - 302.13, - 104.96, - 309.7 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p596-b10", - "global_id": 16977, - "bbox": [ - 237.22, - 269.59, - 246.5, - 277.16 - ], - "text": "· · ·", - "type": "text" - }, - { - "block_id": "p596-b11", - "global_id": 16978, - "bbox": [ - 245.26, - 299.41, - 249.24, - 307.38 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p596-b12", - "global_id": 16979, - "bbox": [ - 139.46, - 228.26, - 143.45, - 236.23 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p596-b13", - "global_id": 16980, - "bbox": [ - 104.77, - 302.45, - 161.0, - 310.42 - ], - "text": "3\n3", - "type": "text" - }, - { - "block_id": "p596-b14", - "global_id": 16981, - "bbox": [ - 123.27, - 247.15, - 127.26, - 255.12 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p596-b15", - "global_id": 16982, - "bbox": [ - 184.46, - 302.39, - 215.75, - 310.44 - ], - "text": "6\n9", - "type": "text" - }, - { - "block_id": "p596-b16", - "global_id": 16983, - "bbox": [ - 86.23, - 318.38, - 137.86, - 327.35 - ], - "text": "Figure P5.1-3", - "type": "text" - }, - { - "block_id": "p596-b17", - "global_id": 16984, - "bbox": [ - 90.72, - 334.74, - 140.24, - 350.38 - ], - "text": "(d) γ n sin πn", - "type": "text" - }, - { - "block_id": "p596-b18", - "global_id": 16985, - "bbox": [ - 132.62, - 341.03, - 157.36, - 356.74 - ], - "text": "2 u[n]", - "type": "text" - }, - { - "block_id": "p596-b19", - "global_id": 16986, - "bbox": [ - 91.23, - 357.32, - 141.74, - 372.94 - ], - "text": "(e) γ n cos πn", - "type": "text" - }, - { - "block_id": "p596-b20", - "global_id": 16987, - "bbox": [ - 134.1, - 363.6, - 158.86, - 379.31 - ], - "text": "2 u[n]", - "type": "text" - }, - { - "block_id": "p596-b21", - "global_id": 16988, - "bbox": [ - 92.21, - 386.67, - 101.17, - 395.64 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p596-b22", - "global_id": 16989, - "bbox": [ - 106.15, - 376.98, - 118.84, - 386.75 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p596-b23", - "global_id": 16990, - "bbox": [ - 106.87, - 399.22, - 118.13, - 405.97 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p596-b24", - "global_id": 16991, - "bbox": [ - 119.84, - 382.6, - 165.1, - 395.63 - ], - "text": "22k δ[n −2k]", - "type": "text" - }, - { - "block_id": "p596-b25", - "global_id": 16992, - "bbox": [ - 90.71, - 405.7, - 153.31, - 416.31 - ], - "text": "(g) γ n−1u[n −1]", - "type": "text" - }, - { - "block_id": "p596-b26", - "global_id": 16993, - "bbox": [ - 90.72, - 418.94, - 137.24, - 431.24 - ], - "text": "(h) nγ n u[n]", - "type": "text" - }, - { - "block_id": "p596-b27", - "global_id": 16994, - "bbox": [ - 92.71, - 432.87, - 126.58, - 442.21 - ], - "text": "(i) nu[n]", - "type": "text" - }, - { - "block_id": "p596-b28", - "global_id": 16995, - "bbox": [ - 92.71, - 444.5, - 116.52, - 461.19 - ], - "text": "(j) γ n", - "type": "text" - }, - { - "block_id": "p596-b29", - "global_id": 16996, - "bbox": [ - 108.7, - 451.84, - 133.15, - 467.45 - ], - "text": "n! u[n]", - "type": "text" - }, - { - "block_id": "p596-b30", - "global_id": 16997, - "bbox": [ - 90.72, - 467.71, - 183.54, - 480.3 - ], - "text": "(k) [2n−1 −(−2)n−1]u[n]", - "type": "text" - }, - { - "block_id": "p596-b31", - "global_id": 16998, - "bbox": [ - 92.7, - 482.65, - 130.73, - 499.33 - ], - "text": "(l) (lnα)n", - "type": "text" - }, - { - "block_id": "p596-b32", - "global_id": 16999, - "bbox": [ - 115.8, - 489.99, - 147.36, - 505.6 - ], - "text": "n!\nu[n]", - "type": "text" - }, - { - "block_id": "p596-b33", - "global_id": 17000, - "bbox": [ - 62.09, - 501.76, - 206.29, - 517.84 - ], - "text": "5.1-5\nShowing all work, evaluate $∞", - "type": "text" - }, - { - "block_id": "p596-b34", - "global_id": 17001, - "bbox": [ - 199.68, - 507.23, - 255.08, - 520.31 - ], - "text": "n=0 n(−3/2)−n.", - "type": "text" - }, - { - "block_id": "p596-b35", - "global_id": 17002, - "bbox": [ - 62.09, - 523.52, - 263.22, - 554.5 - ], - "text": "5.1-6\nUsing only the z-transforms of Table 5.1, deter-\nmine the z-transform of each of the following\nsignals.", - "type": "text" - }, - { - "block_id": "p596-b36", - "global_id": 17003, - "bbox": [ - 90.71, - 556.12, - 160.06, - 576.42 - ], - "text": "(a) u[n] −u[n −2]\n(b) γ n−2u[n −2]", - "type": "text" - }, - { - "block_id": "p596-b37", - "global_id": 17004, - "bbox": [ - 90.72, - 576.77, - 192.59, - 600.28 - ], - "text": "(c) 2n+1u[n −1] + en−1u[n]\n(d)", - "type": "text" - }, - { - "block_id": "p596-b39", - "global_id": 17005, - "bbox": [ - 110.36, - 581.05, - 149.04, - 600.28 - ], - "text": "2−n cos\nπ", - "type": "text" - }, - { - "block_id": "p596-b40", - "global_id": 17006, - "bbox": [ - 144.56, - 581.04, - 164.71, - 606.65 - ], - "text": "3 n", - "type": "text" - }, - { - "block_id": "p596-b41", - "global_id": 17007, - "bbox": [ - 164.71, - 590.94, - 193.91, - 600.28 - ], - "text": "u[n −1]", - "type": "text" - }, - { - "block_id": "p596-b42", - "global_id": 17008, - "bbox": [ - 91.23, - 603.44, - 149.52, - 613.84 - ], - "text": "(e) nγ nu[n −1]", - "type": "text" - }, - { - "block_id": "p596-b43", - "global_id": 17009, - "bbox": [ - 92.21, - 614.2, - 206.82, - 624.8 - ], - "text": "(f) n(n−1)(n−2)2n−3u[n−m]", - "type": "text" - }, - { - "block_id": "p596-b44", - "global_id": 17010, - "bbox": [ - 90.72, - 626.42, - 163.24, - 646.72 - ], - "text": "for m = 0,1,2,3\n(g) (−1)nnu[n]", - "type": "text" - }, - { - "block_id": "p596-b45", - "global_id": 17011, - "bbox": [ - 317.86, - 95.12, - 328.32, - 104.09 - ], - "text": "(h)", - "type": "text" - }, - { - "block_id": "p596-b46", - "global_id": 17012, - "bbox": [ - 333.31, - 85.43, - 345.99, - 95.19 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p596-b47", - "global_id": 17013, - "bbox": [ - 334.02, - 107.68, - 345.28, - 114.42 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p596-b48", - "global_id": 17014, - "bbox": [ - 346.99, - 94.75, - 399.3, - 104.09 - ], - "text": "kδ(n −2k + 1)", - "type": "text" - }, - { - "block_id": "p596-b49", - "global_id": 17015, - "bbox": [ - 289.23, - 117.24, - 490.39, - 137.26 - ], - "text": "5.1-7\nFind the inverse unilateral z-transform of each of\nthe following:", - "type": "text" - }, - { - "block_id": "p596-b50", - "global_id": 17016, - "bbox": [ - 318.37, - 137.02, - 373.74, - 159.01 - ], - "text": "(a)\nz(z −4)\nz2 −5z + 6", - "type": "text" - }, - { - "block_id": "p596-b51", - "global_id": 17017, - "bbox": [ - 317.86, - 157.53, - 373.74, - 179.53 - ], - "text": "(b)\nz −4\nz2 −5z + 6", - "type": "text" - }, - { - "block_id": "p596-b52", - "global_id": 17018, - "bbox": [ - 318.36, - 180.34, - 391.68, - 205.58 - ], - "text": "(c)\n(e−2 −2)z\n(z −e−2)(z −2)", - "type": "text" - }, - { - "block_id": "p596-b53", - "global_id": 17019, - "bbox": [ - 317.87, - 209.46, - 362.19, - 226.08 - ], - "text": "(d) (z −1)2", - "type": "text" - }, - { - "block_id": "p596-b54", - "global_id": 17020, - "bbox": [ - 344.98, - 222.78, - 351.71, - 232.36 - ], - "text": "z3", - "type": "text" - }, - { - "block_id": "p596-b55", - "global_id": 17021, - "bbox": [ - 318.37, - 235.0, - 404.87, - 257.0 - ], - "text": "(e)\nz(2z + 3)\n(z −1)(z2 −5z + 6)", - "type": "text" - }, - { - "block_id": "p596-b56", - "global_id": 17022, - "bbox": [ - 319.36, - 260.36, - 386.65, - 282.35 - ], - "text": "(f)\nz(−5z + 22)\n(z + 1)(z −2)2", - "type": "text" - }, - { - "block_id": "p596-b57", - "global_id": 17023, - "bbox": [ - 317.86, - 285.71, - 400.09, - 307.7 - ], - "text": "(g)\nz(1.4z + 0.08)\n(z −0.2)(z −0.8)2", - "type": "text" - }, - { - "block_id": "p596-b58", - "global_id": 17024, - "bbox": [ - 317.86, - 311.06, - 369.26, - 333.06 - ], - "text": "(h)\nz(z −2)\nz2 −z + 1", - "type": "text" - }, - { - "block_id": "p596-b59", - "global_id": 17025, - "bbox": [ - 319.86, - 333.87, - 396.15, - 352.75 - ], - "text": "(i) 2z2 −0.3z + 0.25", - "type": "text" - }, - { - "block_id": "p596-b60", - "global_id": 17026, - "bbox": [ - 336.73, - 347.19, - 393.91, - 359.11 - ], - "text": "z2 + 0.6z + 0.25", - "type": "text" - }, - { - "block_id": "p596-b61", - "global_id": 17027, - "bbox": [ - 319.86, - 361.68, - 409.36, - 383.67 - ], - "text": "(j)\n2z(3z −23)\n(z −1)(z2 −6z + 25)", - "type": "text" - }, - { - "block_id": "p596-b62", - "global_id": 17028, - "bbox": [ - 317.85, - 387.03, - 409.36, - 409.03 - ], - "text": "(k)\nz(3.83z + 11.34)\n(z −2)(z2 −5z + 25)", - "type": "text" - }, - { - "block_id": "p596-b63", - "global_id": 17029, - "bbox": [ - 319.85, - 410.64, - 399.12, - 429.52 - ], - "text": "(l) z2(−2z2 + 8z −7)", - "type": "text" - }, - { - "block_id": "p596-b64", - "global_id": 17030, - "bbox": [ - 340.51, - 426.21, - 392.65, - 435.88 - ], - "text": "(z −1)(z −2)3", - "type": "text" - }, - { - "block_id": "p596-b65", - "global_id": 17031, - "bbox": [ - 289.22, - 438.49, - 490.39, - 449.09 - ], - "text": "5.1-8\n(a) Expanding X[z] as a power series in z−1,", - "type": "text" - }, - { - "block_id": "p596-b66", - "global_id": 17032, - "bbox": [ - 333.31, - 450.71, - 453.75, - 460.05 - ], - "text": "find the first three terms of x[n] if", - "type": "text" - }, - { - "block_id": "p596-b67", - "global_id": 17033, - "bbox": [ - 317.86, - 461.84, - 490.38, - 499.27 - ], - "text": "X[z] =\n2z3 + 13z2 + z\nz3 + 7z2 + 2z + 1\n(b) Extend the procedure used in part (a) to find", - "type": "text" - }, - { - "block_id": "p596-b68", - "global_id": 17034, - "bbox": [ - 333.3, - 500.9, - 434.24, - 510.23 - ], - "text": "the first four terms of x[n] if", - "type": "text" - }, - { - "block_id": "p596-b69", - "global_id": 17035, - "bbox": [ - 356.93, - 512.02, - 465.55, - 530.9 - ], - "text": "X[z] = 2z4 + 16z3 + 17z2 + 3z", - "type": "text" - }, - { - "block_id": "p596-b70", - "global_id": 17036, - "bbox": [ - 289.22, - 525.33, - 490.38, - 562.2 - ], - "text": "z3 + 7z2 + 2z + 1\n5.1-9\nA right-sided signal x[n] has z-transform given\nby X[z] = z6+2z5+3z4+4z3", - "type": "text" - }, - { - "block_id": "p596-b71", - "global_id": 17037, - "bbox": [ - 317.85, - 553.24, - 490.38, - 584.13 - ], - "text": "z4−1\n. Using a power series\nexpansion of X[z], determine x[n] over −5 ≤\nn ≤5.", - "type": "text" - }, - { - "block_id": "p596-b72", - "global_id": 17038, - "bbox": [ - 284.73, - 588.74, - 401.67, - 598.08 - ], - "text": "5.1-10\nFind x[n] by expanding", - "type": "text" - }, - { - "block_id": "p596-b73", - "global_id": 17039, - "bbox": [ - 375.11, - 604.92, - 431.46, - 626.82 - ], - "text": "X[z] =\nγ z\n(z −γ )2", - "type": "text" - }, - { - "block_id": "p596-b74", - "global_id": 17040, - "bbox": [ - 317.86, - 636.11, - 404.61, - 646.72 - ], - "text": "as a power series in z−1.", - "type": "text" - } - ] - }, - { - "page_num": 597, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p597-b0", - "global_id": 17041, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n577", - "type": "text" - }, - { - "block_id": "p597-b1", - "global_id": 17042, - "bbox": [ - 83.34, - 85.94, - 288.98, - 94.98 - ], - "text": "5.1-11\n(a) In Table 5.1, if the numerator and the", - "type": "text" - }, - { - "block_id": "p597-b2", - "global_id": 17043, - "bbox": [ - 116.46, - 96.6, - 288.99, - 149.77 - ], - "text": "denominator powers of X[z] are M and\nN, respectively, explain why in some cases\nN −M = 0, while in others N −M = 1 or\nN −M = m (m any positive integer).\n(b) Without actually finding the z-transform,", - "type": "text" - }, - { - "block_id": "p597-b3", - "global_id": 17044, - "bbox": [ - 131.89, - 151.4, - 288.98, - 171.69 - ], - "text": "state what is N −M for X[z] corresponding\nto x[n] = γ nu[n −4].", - "type": "text" - }, - { - "block_id": "p597-b4", - "global_id": 17045, - "bbox": [ - 87.82, - 176.99, - 288.97, - 197.0 - ], - "text": "5.2-1\nFor a discrete-time signal shown in Fig. P5.2-1,\nshow that", - "type": "text" - }, - { - "block_id": "p597-b5", - "global_id": 17046, - "bbox": [ - 174.53, - 200.46, - 229.22, - 217.34 - ], - "text": "X[z] = 1 −z−m", - "type": "text" - }, - { - "block_id": "p597-b6", - "global_id": 17047, - "bbox": [ - 202.46, - 213.78, - 228.5, - 223.72 - ], - "text": "1 −z−1", - "type": "text" - }, - { - "block_id": "p597-b7", - "global_id": 17048, - "bbox": [ - 116.46, - 233.56, - 288.98, - 264.44 - ], - "text": "Find your answer by using the definition\nin Eq. (5.1) and by using Table 5.1 and an\nappropriate property of the z-transform.", - "type": "text" - }, - { - "block_id": "p597-b8", - "global_id": 17049, - "bbox": [ - 150.32, - 347.99, - 238.49, - 358.29 - ], - "text": "0\nm 1", - "type": "text" - }, - { - "block_id": "p597-b9", - "global_id": 17050, - "bbox": [ - 158.82, - 281.99, - 171.7, - 290.07 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p597-b10", - "global_id": 17051, - "bbox": [ - 277.52, - 348.15, - 281.52, - 356.15 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p597-b11", - "global_id": 17052, - "bbox": [ - 142.62, - 300.22, - 146.62, - 308.22 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p597-b12", - "global_id": 17053, - "bbox": [ - 116.46, - 366.21, - 168.11, - 375.17 - ], - "text": "Figure P5.2-1", - "type": "text" - }, - { - "block_id": "p597-b13", - "global_id": 17054, - "bbox": [ - 87.82, - 393.07, - 288.97, - 413.09 - ], - "text": "5.2-2\nDetermine the unilateral z-transform of signal\nx[n] = (1 −n)cos", - "type": "text" - }, - { - "block_id": "p597-b14", - "global_id": 17055, - "bbox": [ - 180.99, - 396.55, - 189.68, - 408.6 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p597-b15", - "global_id": 17056, - "bbox": [ - 186.44, - 396.55, - 220.58, - 415.9 - ], - "text": "2 (n −1)", - "type": "text" - }, - { - "block_id": "p597-b16", - "global_id": 17057, - "bbox": [ - 221.57, - 403.75, - 253.02, - 413.09 - ], - "text": "u[n −1].", - "type": "text" - }, - { - "block_id": "p597-b17", - "global_id": 17058, - "bbox": [ - 87.82, - 418.1, - 287.78, - 427.44 - ], - "text": "5.2-3\nSuppose a DT signal x[n] = 2(u[n −10] −u[n−", - "type": "text" - }, - { - "block_id": "p597-b18", - "global_id": 17059, - "bbox": [ - 116.46, - 429.05, - 288.98, - 460.32 - ], - "text": "6]) has a transform X(z). Define Y(z) =\n1\n2z−3\nd\ndzX(2z). Using graphic plot or vector nota-\ntion, determine the corresponding signal y[n].", - "type": "text" - }, - { - "block_id": "p597-b19", - "global_id": 17060, - "bbox": [ - 87.82, - 465.32, - 288.99, - 499.09 - ], - "text": "5.2-4\nSuppose a DT signal x[n] = 3(u[n] −u[n −5])\nhas\na\ntransform\nX(z).\nDefine\nY(z)\n=\n2z−4 d\ndzX", - "type": "text" - }, - { - "block_id": "p597-b20", - "global_id": 17061, - "bbox": [ - 147.24, - 480.03, - 154.93, - 492.28 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p597-b21", - "global_id": 17062, - "bbox": [ - 152.05, - 480.03, - 160.1, - 499.15 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p597-b22", - "global_id": 17063, - "bbox": [ - 116.46, - 487.61, - 288.98, - 507.53 - ], - "text": ". Using graphic plot or vector nota-\ntion, determine the corresponding signal y[n].", - "type": "text" - }, - { - "block_id": "p597-b23", - "global_id": 17064, - "bbox": [ - 87.82, - 512.82, - 288.98, - 554.76 - ], - "text": "5.2-5\nFind the z-transform of the signal illustrated in\nFig. P5.2-5. Solve this problem in two ways, as\nin Exs. 5.2d and 5.4. Verify that the two answers\nare equivalent.", - "type": "text" - }, - { - "block_id": "p597-b24", - "global_id": 17065, - "bbox": [ - 87.82, - 560.04, - 288.98, - 591.02 - ], - "text": "5.2-6\nUsing z-transform techniques and properties (no\ntime-domain convolution sum!), determine the\nconvolution y[n] = ( 1", - "type": "text" - }, - { - "block_id": "p597-b25", - "global_id": 17066, - "bbox": [ - 116.46, - 580.32, - 288.98, - 613.68 - ], - "text": "2)nu[n−3]∗( 1\n3)n−6u[n−4].\nExpress your answer in the form y[n] =\nc1γ n−N1", - "type": "text" - }, - { - "block_id": "p597-b26", - "global_id": 17067, - "bbox": [ - 128.68, - 600.94, - 220.95, - 615.09 - ], - "text": "1\nu[n −N1] + c2γ n−N2", - "type": "text" - }, - { - "block_id": "p597-b27", - "global_id": 17068, - "bbox": [ - 116.46, - 603.6, - 288.98, - 635.59 - ], - "text": "2\nu[n −N2], making\nsure to clearly identify the constants c1, c2, γ1,\nγ2, N1, and N2.", - "type": "text" - }, - { - "block_id": "p597-b28", - "global_id": 17069, - "bbox": [ - 348.37, - 155.74, - 352.37, - 163.74 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p597-b29", - "global_id": 17070, - "bbox": [ - 354.37, - 98.57, - 367.25, - 106.65 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p597-b30", - "global_id": 17071, - "bbox": [ - 488.02, - 155.6, - 492.02, - 163.6 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p597-b31", - "global_id": 17072, - "bbox": [ - 343.37, - 105.98, - 347.37, - 113.98 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p597-b32", - "global_id": 17073, - "bbox": [ - 409.15, - 155.74, - 473.83, - 163.74 - ], - "text": "8\n4", - "type": "text" - }, - { - "block_id": "p597-b33", - "global_id": 17074, - "bbox": [ - 323.68, - 171.66, - 375.32, - 180.62 - ], - "text": "Figure P5.2-5", - "type": "text" - }, - { - "block_id": "p597-b34", - "global_id": 17075, - "bbox": [ - 314.97, - 197.52, - 516.13, - 217.81 - ], - "text": "5.2-7\nDetermine the inverse unilateral z-transform x[n]\nof the signal", - "type": "text" - }, - { - "block_id": "p597-b35", - "global_id": 17076, - "bbox": [ - 377.54, - 233.04, - 414.33, - 249.56 - ], - "text": "X[z] = d7", - "type": "text" - }, - { - "block_id": "p597-b36", - "global_id": 17077, - "bbox": [ - 404.75, - 246.35, - 415.96, - 255.93 - ], - "text": "dz7", - "type": "text" - }, - { - "block_id": "p597-b38", - "global_id": 17078, - "bbox": [ - 444.29, - 232.77, - 456.06, - 243.27 - ], - "text": "z−4", - "type": "text" - }, - { - "block_id": "p597-b39", - "global_id": 17079, - "bbox": [ - 425.41, - 246.19, - 446.46, - 256.79 - ], - "text": "(z −1", - "type": "text" - }, - { - "block_id": "p597-b40", - "global_id": 17080, - "bbox": [ - 443.22, - 247.54, - 475.45, - 259.47 - ], - "text": "2)(z + 3)", - "type": "text" - }, - { - "block_id": "p597-b42", - "global_id": 17081, - "bbox": [ - 314.97, - 270.89, - 516.13, - 303.2 - ], - "text": "5.2-8\nUsing only the fact that γ nu[n]⇐⇒z/(z −γ )\nand properties of the z-transform, find the\nz-transform of each of the following:", - "type": "text" - }, - { - "block_id": "p597-b43", - "global_id": 17082, - "bbox": [ - 343.61, - 303.83, - 391.88, - 325.12 - ], - "text": "(a) n2u[n]\n(b) n2γ nu[n]", - "type": "text" - }, - { - "block_id": "p597-b44", - "global_id": 17083, - "bbox": [ - 343.61, - 325.75, - 429.15, - 347.03 - ], - "text": "(c) n3u[n]\n(d) an[u[n] −u[n −m]]", - "type": "text" - }, - { - "block_id": "p597-b45", - "global_id": 17084, - "bbox": [ - 344.12, - 347.39, - 410.72, - 358.0 - ], - "text": "(e) ne−2nu[n −m]", - "type": "text" - }, - { - "block_id": "p597-b46", - "global_id": 17085, - "bbox": [ - 345.11, - 356.64, - 444.61, - 368.96 - ], - "text": "(f) (n −2)(0.5)n−3 u[n −4]", - "type": "text" - }, - { - "block_id": "p597-b47", - "global_id": 17086, - "bbox": [ - 314.97, - 374.18, - 516.14, - 427.06 - ], - "text": "5.2-9\nUsing only pair 1 in Table 5.1 and appropriate\nproperties of the z-transform, derive iteratively\npairs 2 through 9. In other words, first derive\npair 2. Then use pair 2 (and pair 1, if needed) to\nderive pair 3, and so on.", - "type": "text" - }, - { - "block_id": "p597-b48", - "global_id": 17087, - "bbox": [ - 310.48, - 432.0, - 516.13, - 463.25 - ], - "text": "5.2-10\nFind the z-transform of cos(πn/4)u[n] using\nonly pairs 1 and 11b in Table 5.1 and a suitable\nproperty of the z-transform.", - "type": "text" - }, - { - "block_id": "p597-b49", - "global_id": 17088, - "bbox": [ - 310.48, - 468.48, - 516.13, - 499.44 - ], - "text": "5.2-11\nApply the time-reversal property to pair 6 of\nTable 5.1 to show that γ nu[−(n + 1)] ⇐⇒\n−z/(z −γ ) and the ROC is given by |z| < |γ |.", - "type": "text" - }, - { - "block_id": "p597-b50", - "global_id": 17089, - "bbox": [ - 310.48, - 504.37, - 516.13, - 535.63 - ], - "text": "5.2-12\n(a) If\nx[n] ⇐⇒X[z],\nthen\nshow\nthat\n(−1)nx[n] ⇐⇒X[−z].\n(b) Use this result to show that (−γ )nu[n] ⇐⇒", - "type": "text" - }, - { - "block_id": "p597-b51", - "global_id": 17090, - "bbox": [ - 344.11, - 537.25, - 516.13, - 557.55 - ], - "text": "z/(z + γ ).\n(c) Use these results to find the z-transforms of", - "type": "text" - }, - { - "block_id": "p597-b52", - "global_id": 17091, - "bbox": [ - 359.04, - 555.92, - 516.13, - 579.47 - ], - "text": "xi[n] = [2n−1 −(−2)n−1]u[n] and x−ii[n] =\nγ n cosπnu[n]", - "type": "text" - }, - { - "block_id": "p597-b53", - "global_id": 17092, - "bbox": [ - 310.48, - 584.4, - 474.5, - 593.74 - ], - "text": "5.2-13\n(a) If x[n] ⇐⇒X[z], then show that", - "type": "text" - }, - { - "block_id": "p597-b54", - "global_id": 17093, - "bbox": [ - 402.63, - 606.25, - 415.32, - 615.8 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p597-b55", - "global_id": 17094, - "bbox": [ - 403.34, - 628.28, - 414.61, - 635.03 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p597-b56", - "global_id": 17095, - "bbox": [ - 416.32, - 609.07, - 471.36, - 624.6 - ], - "text": "x[k] ⇐⇒zX[z]", - "type": "text" - }, - { - "block_id": "p597-b57", - "global_id": 17096, - "bbox": [ - 453.06, - 621.72, - 470.82, - 631.06 - ], - "text": "z −1", - "type": "text" - } - ] - }, - { - "page_num": 598, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p598-b0", - "global_id": 17097, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "578\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p598-b1", - "global_id": 17098, - "bbox": [ - 90.72, - 85.9, - 263.24, - 94.86 - ], - "text": "(b) Use this result to derive pair 2 from pair 1", - "type": "text" - }, - { - "block_id": "p598-b2", - "global_id": 17099, - "bbox": [ - 106.16, - 96.86, - 150.76, - 105.82 - ], - "text": "in Table 5.1.", - "type": "text" - }, - { - "block_id": "p598-b3", - "global_id": 17100, - "bbox": [ - 57.59, - 126.69, - 263.24, - 190.53 - ], - "text": "5.2-14\nA number of causal time-domain functions are\nshown in Fig. P5.2-14. List the function of\ntime that corresponds to each of the following\nfunctions of z. Few or no calculations are\nnecessary! Be careful, the graphs may be scaled\ndifferently.", - "type": "text" - }, - { - "block_id": "p598-b4", - "global_id": 17101, - "bbox": [ - 91.22, - 190.8, - 130.16, - 207.42 - ], - "text": "(a)\nz2", - "type": "text" - }, - { - "block_id": "p598-b5", - "global_id": 17102, - "bbox": [ - 107.35, - 204.12, - 146.25, - 213.78 - ], - "text": "(z −0.75)2", - "type": "text" - }, - { - "block_id": "p598-b6", - "global_id": 17103, - "bbox": [ - 90.72, - 216.16, - 153.89, - 235.04 - ], - "text": "(b)\nz2 −0.9z/", - "type": "text" - }, - { - "block_id": "p598-b7", - "global_id": 17104, - "bbox": [ - 153.89, - 211.83, - 161.48, - 220.8 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p598-b8", - "global_id": 17105, - "bbox": [ - 161.48, - 219.79, - 165.96, - 228.76 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p598-b9", - "global_id": 17106, - "bbox": [ - 107.35, - 230.74, - 135.56, - 242.66 - ], - "text": "z2 −0.9", - "type": "text" - }, - { - "block_id": "p598-b10", - "global_id": 17107, - "bbox": [ - 135.56, - 225.74, - 143.14, - 234.71 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p598-b11", - "global_id": 17108, - "bbox": [ - 143.15, - 233.32, - 176.6, - 242.66 - ], - "text": "2z + 0.81", - "type": "text" - }, - { - "block_id": "p598-b12", - "global_id": 17109, - "bbox": [ - 91.22, - 256.36, - 101.18, - 265.32 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p598-b13", - "global_id": 17110, - "bbox": [ - 106.15, - 246.94, - 118.84, - 256.43 - ], - "text": "4\n\"", - "type": "text" - }, - { - "block_id": "p598-b14", - "global_id": 17111, - "bbox": [ - 106.86, - 268.91, - 118.12, - 275.66 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p598-b15", - "global_id": 17112, - "bbox": [ - 119.84, - 254.27, - 134.49, - 265.23 - ], - "text": "z−2k", - "type": "text" - }, - { - "block_id": "p598-b16", - "global_id": 17113, - "bbox": [ - 90.72, - 279.38, - 126.26, - 296.26 - ], - "text": "(d)\nz−5", - "type": "text" - }, - { - "block_id": "p598-b17", - "global_id": 17114, - "bbox": [ - 107.35, - 292.69, - 133.4, - 302.63 - ], - "text": "1 −z−1", - "type": "text" - }, - { - "block_id": "p598-b18", - "global_id": 17115, - "bbox": [ - 91.22, - 305.7, - 121.21, - 322.32 - ], - "text": "(e)\nz2", - "type": "text" - }, - { - "block_id": "p598-b19", - "global_id": 17116, - "bbox": [ - 107.35, - 316.76, - 128.84, - 328.69 - ], - "text": "z4 −1", - "type": "text" - }, - { - "block_id": "p598-b20", - "global_id": 17117, - "bbox": [ - 92.21, - 331.48, - 146.25, - 353.19 - ], - "text": "(f)\n0.75z\n(z −0.75)2", - "type": "text" - }, - { - "block_id": "p598-b21", - "global_id": 17118, - "bbox": [ - 90.72, - 355.56, - 137.09, - 374.44 - ], - "text": "(g)\nz2 −z/", - "type": "text" - }, - { - "block_id": "p598-b22", - "global_id": 17119, - "bbox": [ - 137.09, - 351.23, - 144.67, - 360.2 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p598-b23", - "global_id": 17120, - "bbox": [ - 144.67, - 359.19, - 149.15, - 368.16 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p598-b24", - "global_id": 17121, - "bbox": [ - 107.35, - 370.14, - 122.96, - 381.98 - ], - "text": "z2 −", - "type": "text" - }, - { - "block_id": "p598-b25", - "global_id": 17122, - "bbox": [ - 124.35, - 365.14, - 131.94, - 374.11 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p598-b26", - "global_id": 17123, - "bbox": [ - 131.94, - 372.73, - 154.18, - 382.07 - ], - "text": "2z + 1", - "type": "text" - }, - { - "block_id": "p598-b27", - "global_id": 17124, - "bbox": [ - 90.72, - 383.08, - 172.21, - 401.96 - ], - "text": "(h) z−1 −5z−5 + 4z−6", - "type": "text" - }, - { - "block_id": "p598-b28", - "global_id": 17125, - "bbox": [ - 119.3, - 398.39, - 160.26, - 408.33 - ], - "text": "5(1 −z−1)2", - "type": "text" - }, - { - "block_id": "p598-b29", - "global_id": 17126, - "bbox": [ - 92.71, - 409.75, - 131.83, - 431.46 - ], - "text": "(i)\nz\nz −1.1", - "type": "text" - }, - { - "block_id": "p598-b30", - "global_id": 17127, - "bbox": [ - 319.86, - 85.82, - 389.06, - 102.71 - ], - "text": "(j)\n0.25z−1", - "type": "text" - }, - { - "block_id": "p598-b31", - "global_id": 17128, - "bbox": [ - 284.74, - 99.14, - 490.4, - 153.79 - ], - "text": "(1 −z−1)(1 −0.75z−1)\n5.2-15\nSuppose we upsample a causal signal x[n] by\nfactor N to produce signal y[n]. Express Y(z)\nin terms of X(z), taking care to mathematically\njustify your result.", - "type": "text" - }, - { - "block_id": "p598-b32", - "global_id": 17129, - "bbox": [ - 289.23, - 158.74, - 490.39, - 189.62 - ], - "text": "5.3-1\nUsing z-transform techniques, find the output\ny[n] of an LTID system specified by the equation\ny[n]−1", - "type": "text" - }, - { - "block_id": "p598-b33", - "global_id": 17130, - "bbox": [ - 317.86, - 180.37, - 490.38, - 200.68 - ], - "text": "3y[n−1] = x[n−1] if the initial condition\nis y[−1] = 2 and the input is x[n] = −u[n].", - "type": "text" - }, - { - "block_id": "p598-b34", - "global_id": 17131, - "bbox": [ - 289.22, - 205.35, - 490.39, - 225.65 - ], - "text": "5.3-2\nConsider an LTID system y[n] −y[n −2] = x[n]\nwith y[−1] = 0, y[−2] = 1, and x[n] = u[n].", - "type": "text" - }, - { - "block_id": "p598-b35", - "global_id": 17132, - "bbox": [ - 318.36, - 227.27, - 490.38, - 236.6 - ], - "text": "(a) Determine Y[z], expressed as a rational", - "type": "text" - }, - { - "block_id": "p598-b36", - "global_id": 17133, - "bbox": [ - 317.86, - 238.6, - 490.38, - 258.53 - ], - "text": "function in standard factored form.\n(b) Use Y[z] from part (a) to solve for the", - "type": "text" - }, - { - "block_id": "p598-b37", - "global_id": 17134, - "bbox": [ - 333.3, - 260.14, - 402.29, - 269.48 - ], - "text": "system output y[n].", - "type": "text" - }, - { - "block_id": "p598-b38", - "global_id": 17135, - "bbox": [ - 289.22, - 274.45, - 483.97, - 283.5 - ], - "text": "5.3-3\nSolve Prob. 3.8-23 by the z-transform method.", - "type": "text" - }, - { - "block_id": "p598-b39", - "global_id": 17136, - "bbox": [ - 289.22, - 288.48, - 490.38, - 331.13 - ], - "text": "5.3-4\nConsider a DT system with transfer function\nH[z] =\n2z−2\nz−0.5. Assuming the system is both\ncontrollable and observable, determine the ZIR\nyzir[n] given y[−1] = 1.", - "type": "text" - }, - { - "block_id": "p598-b40", - "global_id": 17137, - "bbox": [ - 289.22, - 335.36, - 353.59, - 344.41 - ], - "text": "5.3-5\n(a) Solve", - "type": "text" - }, - { - "block_id": "p598-b41", - "global_id": 17138, - "bbox": [ - 363.44, - 353.0, - 460.25, - 362.34 - ], - "text": "y[n + 1] + 2y[n] = x[n + 1]", - "type": "text" - }, - { - "block_id": "p598-b42", - "global_id": 17139, - "bbox": [ - 317.86, - 366.35, - 490.38, - 387.92 - ], - "text": "when y[0] = 1 and x[n] = e−(n−1)u[n]\n(b) Find the zero-input and the zero-state", - "type": "text" - }, - { - "block_id": "p598-b43", - "global_id": 17140, - "bbox": [ - 333.31, - 389.91, - 435.4, - 398.87 - ], - "text": "components of the response.", - "type": "text" - }, - { - "block_id": "p598-b44", - "global_id": 17141, - "bbox": [ - 289.22, - 403.85, - 490.39, - 423.85 - ], - "text": "5.3-6\nConsider a LTID system that is described by\nthe difference equation y[n] −1", - "type": "text" - }, - { - "block_id": "p598-b45", - "global_id": 17142, - "bbox": [ - 317.86, - 414.51, - 490.39, - 434.81 - ], - "text": "4y[n −2] =\nx[n −1].", - "type": "text" - }, - { - "block_id": "p598-b46", - "global_id": 17143, - "bbox": [ - 111.06, - 501.33, - 115.06, - 509.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b47", - "global_id": 17144, - "bbox": [ - 104.83, - 488.22, - 108.83, - 496.22 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b48", - "global_id": 17145, - "bbox": [ - 152.4, - 454.54, - 156.4, - 462.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p598-b49", - "global_id": 17146, - "bbox": [ - 168.81, - 501.33, - 172.81, - 509.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b50", - "global_id": 17147, - "bbox": [ - 162.58, - 488.22, - 166.58, - 496.22 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b51", - "global_id": 17148, - "bbox": [ - 210.15, - 454.54, - 214.15, - 462.54 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p598-b52", - "global_id": 17149, - "bbox": [ - 227.32, - 501.33, - 231.32, - 509.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b53", - "global_id": 17150, - "bbox": [ - 221.08, - 470.72, - 225.08, - 478.72 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b54", - "global_id": 17151, - "bbox": [ - 268.65, - 453.62, - 272.65, - 461.62 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p598-b55", - "global_id": 17152, - "bbox": [ - 285.06, - 501.33, - 289.06, - 509.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b56", - "global_id": 17153, - "bbox": [ - 278.83, - 470.72, - 282.83, - 478.72 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b57", - "global_id": 17154, - "bbox": [ - 326.39, - 453.62, - 330.4, - 461.62 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p598-b58", - "global_id": 17155, - "bbox": [ - 343.56, - 501.33, - 347.56, - 509.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b59", - "global_id": 17156, - "bbox": [ - 337.34, - 488.22, - 341.34, - 496.22 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b60", - "global_id": 17157, - "bbox": [ - 384.9, - 454.54, - 388.9, - 462.54 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p598-b61", - "global_id": 17158, - "bbox": [ - 401.31, - 501.33, - 405.31, - 509.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b62", - "global_id": 17159, - "bbox": [ - 395.08, - 488.22, - 399.08, - 496.22 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b63", - "global_id": 17160, - "bbox": [ - 442.65, - 454.54, - 446.65, - 462.54 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p598-b64", - "global_id": 17161, - "bbox": [ - 111.06, - 566.58, - 115.06, - 574.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b65", - "global_id": 17162, - "bbox": [ - 104.83, - 553.4, - 108.83, - 561.4 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b66", - "global_id": 17163, - "bbox": [ - 152.4, - 519.12, - 156.4, - 527.12 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p598-b67", - "global_id": 17164, - "bbox": [ - 168.81, - 566.58, - 172.81, - 574.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b68", - "global_id": 17165, - "bbox": [ - 162.58, - 540.18, - 166.58, - 548.18 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b69", - "global_id": 17166, - "bbox": [ - 210.15, - 518.43, - 214.15, - 526.43 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p598-b70", - "global_id": 17167, - "bbox": [ - 227.32, - 566.58, - 231.32, - 574.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b71", - "global_id": 17168, - "bbox": [ - 221.08, - 540.18, - 225.08, - 548.18 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b72", - "global_id": 17169, - "bbox": [ - 268.65, - 518.43, - 272.65, - 526.43 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p598-b73", - "global_id": 17170, - "bbox": [ - 285.06, - 566.58, - 289.06, - 574.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b74", - "global_id": 17171, - "bbox": [ - 278.83, - 553.4, - 282.83, - 561.4 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b75", - "global_id": 17172, - "bbox": [ - 320.87, - 519.12, - 328.87, - 527.12 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p598-b76", - "global_id": 17173, - "bbox": [ - 343.56, - 566.58, - 347.56, - 574.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b77", - "global_id": 17174, - "bbox": [ - 337.34, - 553.4, - 341.34, - 561.4 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b78", - "global_id": 17175, - "bbox": [ - 379.37, - 518.6, - 387.37, - 526.6 - ], - "text": "11", - "type": "text" - }, - { - "block_id": "p598-b79", - "global_id": 17176, - "bbox": [ - 401.31, - 566.58, - 405.31, - 574.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b80", - "global_id": 17177, - "bbox": [ - 395.08, - 553.4, - 399.08, - 561.4 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b81", - "global_id": 17178, - "bbox": [ - 437.12, - 519.12, - 445.12, - 527.12 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p598-b82", - "global_id": 17179, - "bbox": [ - 111.06, - 631.08, - 115.06, - 639.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b83", - "global_id": 17180, - "bbox": [ - 104.83, - 617.9, - 108.83, - 625.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b84", - "global_id": 17181, - "bbox": [ - 146.87, - 583.1, - 154.87, - 591.1 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p598-b85", - "global_id": 17182, - "bbox": [ - 168.81, - 631.08, - 172.81, - 639.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b86", - "global_id": 17183, - "bbox": [ - 162.58, - 617.9, - 166.58, - 625.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b87", - "global_id": 17184, - "bbox": [ - 204.62, - 583.62, - 212.62, - 591.62 - ], - "text": "14", - "type": "text" - }, - { - "block_id": "p598-b88", - "global_id": 17185, - "bbox": [ - 227.32, - 631.08, - 231.32, - 639.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b89", - "global_id": 17186, - "bbox": [ - 221.08, - 617.9, - 225.08, - 625.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b90", - "global_id": 17187, - "bbox": [ - 263.12, - 583.62, - 271.12, - 591.62 - ], - "text": "15", - "type": "text" - }, - { - "block_id": "p598-b91", - "global_id": 17188, - "bbox": [ - 285.06, - 631.08, - 289.06, - 639.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b92", - "global_id": 17189, - "bbox": [ - 278.83, - 617.9, - 282.83, - 625.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b93", - "global_id": 17190, - "bbox": [ - 290.45, - 582.71, - 298.45, - 590.71 - ], - "text": "16", - "type": "text" - }, - { - "block_id": "p598-b94", - "global_id": 17191, - "bbox": [ - 343.56, - 631.08, - 347.56, - 639.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b95", - "global_id": 17192, - "bbox": [ - 337.34, - 617.9, - 341.34, - 625.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b96", - "global_id": 17193, - "bbox": [ - 348.95, - 582.71, - 356.95, - 590.71 - ], - "text": "17", - "type": "text" - }, - { - "block_id": "p598-b97", - "global_id": 17194, - "bbox": [ - 401.31, - 631.08, - 405.31, - 639.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b98", - "global_id": 17195, - "bbox": [ - 395.08, - 617.9, - 399.08, - 625.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p598-b99", - "global_id": 17196, - "bbox": [ - 437.12, - 583.62, - 445.12, - 591.62 - ], - "text": "18", - "type": "text" - }, - { - "block_id": "p598-b100", - "global_id": 17197, - "bbox": [ - 104.83, - 645.24, - 160.95, - 654.21 - ], - "text": "Figure P5.2-14", - "type": "text" - } - ] - }, - { - "page_num": 599, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p599-b0", - "global_id": 17198, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n579", - "type": "text" - }, - { - "block_id": "p599-b1", - "global_id": 17199, - "bbox": [ - 116.46, - 85.9, - 288.99, - 150.39 - ], - "text": "(a) Use\ntransform-domain\ntechniques\nto\ndetermine the zero-state response yzsr[n]\nto input x[n] = 3u[n −5].\n(b) Use\ntransform-domain\ntechniques\nto\ndetermine the zero-input response yzir[n]\ngiven yzir[−2] = yzir[−1] = 1.", - "type": "text" - }, - { - "block_id": "p599-b2", - "global_id": 17200, - "bbox": [ - 87.82, - 154.27, - 288.99, - 163.6 - ], - "text": "5.3-7\n(a) Find the output y[n] of an LTID system", - "type": "text" - }, - { - "block_id": "p599-b3", - "global_id": 17201, - "bbox": [ - 131.89, - 165.6, - 221.61, - 174.57 - ], - "text": "specified by the equation", - "type": "text" - }, - { - "block_id": "p599-b4", - "global_id": 17202, - "bbox": [ - 158.03, - 185.59, - 258.39, - 194.93 - ], - "text": "2y[n + 2] −3y[n + 1] + y[n]", - "type": "text" - }, - { - "block_id": "p599-b5", - "global_id": 17203, - "bbox": [ - 177.8, - 199.53, - 262.84, - 208.87 - ], - "text": "= 4x[n + 2] −3x[n + 1]", - "type": "text" - }, - { - "block_id": "p599-b6", - "global_id": 17204, - "bbox": [ - 116.46, - 218.63, - 288.98, - 251.15 - ], - "text": "for input x[n] = (4)−nu[n] and initial\nconditions y[−1] = 0 and y[−2] = 1.\n(b) Find the zero-input and the zero-state", - "type": "text" - }, - { - "block_id": "p599-b7", - "global_id": 17205, - "bbox": [ - 116.96, - 253.14, - 288.97, - 273.06 - ], - "text": "components of the response.\n(c) Find the transient and the steady-state", - "type": "text" - }, - { - "block_id": "p599-b8", - "global_id": 17206, - "bbox": [ - 131.89, - 275.06, - 234.0, - 284.03 - ], - "text": "components of the response.", - "type": "text" - }, - { - "block_id": "p599-b9", - "global_id": 17207, - "bbox": [ - 87.82, - 288.64, - 288.98, - 319.89 - ], - "text": "5.3-8\nSolve Prob. 5.3-7 if initial conditions y[−1]\nand y[−2] are instead replaced with auxiliary\nconditions y[0] = 3/2 and y[1] = 35/4.", - "type": "text" - }, - { - "block_id": "p599-b10", - "global_id": 17208, - "bbox": [ - 87.82, - 324.8, - 152.18, - 333.84 - ], - "text": "5.3-9\n(a) Solve", - "type": "text" - }, - { - "block_id": "p599-b11", - "global_id": 17209, - "bbox": [ - 140.55, - 344.86, - 280.33, - 354.2 - ], - "text": "4y[n + 2] + 4y[n + 1] + y[n] = x[n + 1]", - "type": "text" - }, - { - "block_id": "p599-b12", - "global_id": 17210, - "bbox": [ - 116.46, - 365.22, - 288.98, - 385.51 - ], - "text": "with y[−1] = 0, y[−2] = 1, and x[n] = u[n].\n(b) Find the zero-input and the zero-state", - "type": "text" - }, - { - "block_id": "p599-b13", - "global_id": 17211, - "bbox": [ - 116.96, - 387.51, - 288.97, - 407.44 - ], - "text": "components of the response.\n(c) Find the transient and the steady-state", - "type": "text" - }, - { - "block_id": "p599-b14", - "global_id": 17212, - "bbox": [ - 131.89, - 409.42, - 234.0, - 418.39 - ], - "text": "components of the response.", - "type": "text" - }, - { - "block_id": "p599-b15", - "global_id": 17213, - "bbox": [ - 83.34, - 423.3, - 136.75, - 432.34 - ], - "text": "5.3-10\nSolve", - "type": "text" - }, - { - "block_id": "p599-b16", - "global_id": 17214, - "bbox": [ - 132.83, - 443.35, - 272.61, - 452.69 - ], - "text": "y[n + 2] −3y[n + 1] + 2y[n] = x[n + 1]", - "type": "text" - }, - { - "block_id": "p599-b17", - "global_id": 17215, - "bbox": [ - 116.46, - 462.66, - 278.76, - 473.05 - ], - "text": "if y[−1] = 2, y[−2] = 3, and x[n] = (3)nu[n].", - "type": "text" - }, - { - "block_id": "p599-b18", - "global_id": 17216, - "bbox": [ - 83.34, - 477.96, - 136.75, - 487.0 - ], - "text": "5.3-11\nSolve", - "type": "text" - }, - { - "block_id": "p599-b19", - "global_id": 17217, - "bbox": [ - 139.97, - 498.02, - 265.48, - 507.36 - ], - "text": "y[n + 2] −2y[n + 1] + 2y[n] = x[n]", - "type": "text" - }, - { - "block_id": "p599-b20", - "global_id": 17218, - "bbox": [ - 116.46, - 518.38, - 274.31, - 527.72 - ], - "text": "with y[−1] = 1, y[−2] = 0, and x[n] = u[n].", - "type": "text" - }, - { - "block_id": "p599-b21", - "global_id": 17219, - "bbox": [ - 83.34, - 532.63, - 288.96, - 561.29 - ], - "text": "5.3-12\nConsider a causal LTID system described as\nH(z) =\n21(z2+1)\n16(z2+ 1\n4 z−3\n8 ).", - "type": "text" - }, - { - "block_id": "p599-b22", - "global_id": 17220, - "bbox": [ - 116.96, - 560.15, - 288.98, - 569.12 - ], - "text": "(a) Determine the standard delay-form differ-", - "type": "text" - }, - { - "block_id": "p599-b23", - "global_id": 17221, - "bbox": [ - 116.46, - 571.1, - 288.99, - 591.03 - ], - "text": "ence equation description of this system.\n(b) Using transform-domain techniques, deter-", - "type": "text" - }, - { - "block_id": "p599-b24", - "global_id": 17222, - "bbox": [ - 116.96, - 592.66, - 288.98, - 634.87 - ], - "text": "mine the system impulse response h[n].\n(c) Using\ntransform-domain\ntechniques,\ndetermine yzir[n] given y[−1] = 16 and\ny[−2] = 8.", - "type": "text" - }, - { - "block_id": "p599-b25", - "global_id": 17223, - "bbox": [ - 310.48, - 85.94, - 516.11, - 105.85 - ], - "text": "5.3-13\nConsider a causal LTID system described as\ny[n] −5", - "type": "text" - }, - { - "block_id": "p599-b26", - "global_id": 17224, - "bbox": [ - 343.61, - 95.24, - 516.14, - 183.97 - ], - "text": "6y[n −1] + 1\n6y[n −2] = 3\n2x[n −1] +\n3\n2x[n −2].\n(a) Determine\nthe\n(standard-form)\nsystem\ntransfer function H(z) and sketch the system\npole-zero plot.\n(b) Using\ntransform-domain\ntechniques,\ndetermine yzir[n] given y[−1] = 2 and\ny[−2] = −2.", - "type": "text" - }, - { - "block_id": "p599-b27", - "global_id": 17225, - "bbox": [ - 310.48, - 188.87, - 363.9, - 197.91 - ], - "text": "5.3-14\nSolve", - "type": "text" - }, - { - "block_id": "p599-b28", - "global_id": 17226, - "bbox": [ - 381.27, - 209.24, - 478.46, - 218.58 - ], - "text": "y[n]+2y[n−1]+2y[n−2]", - "type": "text" - }, - { - "block_id": "p599-b29", - "global_id": 17227, - "bbox": [ - 392.09, - 223.19, - 470.26, - 232.53 - ], - "text": "= x[n−1]+2x[n−2]", - "type": "text" - }, - { - "block_id": "p599-b30", - "global_id": 17228, - "bbox": [ - 343.61, - 242.79, - 495.19, - 253.19 - ], - "text": "with y[0] = 0, y[1] = 1, and x[n] = enu[n].", - "type": "text" - }, - { - "block_id": "p599-b31", - "global_id": 17229, - "bbox": [ - 310.48, - 257.8, - 516.13, - 300.02 - ], - "text": "5.3-15\nA\nsystem\nwith\nimpulse\nresponse\nh[n] =\n2(1/3)nu[n −1] produces an output y[n] =\n(−2)nu[n −1]. Determine the corresponding\ninput x[n].", - "type": "text" - }, - { - "block_id": "p599-b32", - "global_id": 17230, - "bbox": [ - 310.48, - 304.92, - 516.13, - 412.59 - ], - "text": "5.3-16\nA professor recently received an unexpected\n$10 (a futile bribe attached to a test). Being\nthe savvy investor that she is, the professor\ndecides to invest the $10 into a savings account\nthat earns 0.5% interest compounded monthly\n(6.17% APY). Furthermore, she decides to\nsupplement this initial investment with an\nadditional\n$5\ndeposit\nmade\nevery\nmonth,\nbeginning the month immediately following her\ninitial investment.", - "type": "text" - }, - { - "block_id": "p599-b33", - "global_id": 17231, - "bbox": [ - 344.12, - 414.59, - 516.12, - 423.56 - ], - "text": "(a) Model the professor’s savings account as", - "type": "text" - }, - { - "block_id": "p599-b34", - "global_id": 17232, - "bbox": [ - 343.61, - 425.55, - 516.14, - 489.31 - ], - "text": "a\nconstant\ncoefficient\nlinear\ndifference\nequation. Designate y[n] as the account\nbalance\nat\nmonth\nn,\nwhere\nn = 0\ncorresponds to the first month that interest\nis awarded (and that her $5 deposits begin).\n(b) Determine a closed-form solution for y[n].", - "type": "text" - }, - { - "block_id": "p599-b35", - "global_id": 17233, - "bbox": [ - 344.12, - 490.93, - 516.13, - 522.19 - ], - "text": "That is, you should express y[n] as a\nfunction only of n.\n(c) If we consider the professor’s bank account", - "type": "text" - }, - { - "block_id": "p599-b36", - "global_id": 17234, - "bbox": [ - 343.61, - 524.19, - 516.13, - 566.03 - ], - "text": "as a system, what is the system impulse\nresponse h[n]? What is the system transfer\nfunction H[z]?\n(d) Explain this fact: if the input to the", - "type": "text" - }, - { - "block_id": "p599-b37", - "global_id": 17235, - "bbox": [ - 359.05, - 568.02, - 516.13, - 598.9 - ], - "text": "professor’s bank account is the everlasting\nexponential x[n] = 1n = 1, then the output\nis not y[n] = 1nH[1] = H[1].", - "type": "text" - }, - { - "block_id": "p599-b38", - "global_id": 17236, - "bbox": [ - 310.48, - 603.8, - 516.13, - 634.76 - ], - "text": "5.3-17\nSally deposits $100 into her savings account\non the first day of every month except for each\nDecember, when she uses her money to buy", - "type": "text" - } - ] - }, - { - "page_num": 600, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p600-b0", - "global_id": 17237, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "580\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p600-b1", - "global_id": 17238, - "bbox": [ - 90.72, - 85.52, - 263.24, - 204.45 - ], - "text": "holiday gifts. Define b[m] as the balance in\nSally’s account on the first day of month m.\nAssume Sally opens her account in January\n(m = 0), continues making monthly payments\nforever (except each December!), and that her\nmonthly interest rate is 1%. Sally’s account\nbalance satisfies a simple difference equation\nb[m] = (1.01)b[m −1] + p[m], where p[m]\ndesignates Sally’s monthly deposits. Determine\na closed-form expression for b[m] that is only a\nfunction of the month m.", - "type": "text" - }, - { - "block_id": "p600-b2", - "global_id": 17239, - "bbox": [ - 57.59, - 210.22, - 263.24, - 252.14 - ], - "text": "5.3-18\nFor each impulse response, determine the\nnumber of system poles, whether the poles\nare real or complex, and whether the system\nis BIBO-stable.", - "type": "text" - }, - { - "block_id": "p600-b3", - "global_id": 17240, - "bbox": [ - 90.72, - 252.7, - 213.53, - 274.79 - ], - "text": "(a) h1[n] = (−1 + (0.5)n)u[n]\n(b) h2[n] = (j)n(u[n] −u[n −10])", - "type": "text" - }, - { - "block_id": "p600-b4", - "global_id": 17241, - "bbox": [ - 57.59, - 279.83, - 180.42, - 288.87 - ], - "text": "5.3-19\nFind the following sums:", - "type": "text" - }, - { - "block_id": "p600-b5", - "global_id": 17242, - "bbox": [ - 91.22, - 297.98, - 101.18, - 306.95 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p600-b6", - "global_id": 17243, - "bbox": [ - 106.15, - 288.5, - 118.84, - 298.06 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p600-b7", - "global_id": 17244, - "bbox": [ - 106.86, - 310.54, - 118.12, - 317.28 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p600-b8", - "global_id": 17245, - "bbox": [ - 119.84, - 297.89, - 123.82, - 306.86 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p600-b9", - "global_id": 17246, - "bbox": [ - 90.72, - 326.13, - 101.18, - 335.1 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p600-b10", - "global_id": 17247, - "bbox": [ - 106.15, - 316.65, - 118.84, - 326.21 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p600-b11", - "global_id": 17248, - "bbox": [ - 106.86, - 338.69, - 118.12, - 345.43 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p600-b12", - "global_id": 17249, - "bbox": [ - 119.84, - 324.32, - 127.2, - 335.01 - ], - "text": "k2", - "type": "text" - }, - { - "block_id": "p600-b13", - "global_id": 17250, - "bbox": [ - 90.72, - 344.88, - 263.25, - 387.09 - ], - "text": "[Hint: Consider a system whose output y[n]\nis the desired sum. Examine the relationship\nbetween y[n] and y[n −1]. Note also that\ny[0] = 0.]", - "type": "text" - }, - { - "block_id": "p600-b14", - "global_id": 17251, - "bbox": [ - 57.59, - 392.86, - 176.93, - 401.9 - ], - "text": "5.3-20\nFind the following sum:", - "type": "text" - }, - { - "block_id": "p600-b15", - "global_id": 17252, - "bbox": [ - 166.2, - 420.11, - 178.89, - 429.67 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p600-b16", - "global_id": 17253, - "bbox": [ - 166.9, - 442.15, - 178.17, - 448.9 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p600-b17", - "global_id": 17254, - "bbox": [ - 179.88, - 427.78, - 187.25, - 438.47 - ], - "text": "k3", - "type": "text" - }, - { - "block_id": "p600-b18", - "global_id": 17255, - "bbox": [ - 90.72, - 467.2, - 221.59, - 476.26 - ], - "text": "[Hint: See the hint for Prob. 5.3-19.]", - "type": "text" - }, - { - "block_id": "p600-b19", - "global_id": 17256, - "bbox": [ - 57.59, - 482.02, - 176.93, - 491.06 - ], - "text": "5.3-21\nFind the following sum:", - "type": "text" - }, - { - "block_id": "p600-b20", - "global_id": 17257, - "bbox": [ - 144.12, - 509.28, - 156.8, - 518.84 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p600-b21", - "global_id": 17258, - "bbox": [ - 144.83, - 531.32, - 156.09, - 538.06 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p600-b22", - "global_id": 17259, - "bbox": [ - 157.81, - 516.89, - 207.59, - 527.73 - ], - "text": "kak\na̸ = 1", - "type": "text" - }, - { - "block_id": "p600-b23", - "global_id": 17260, - "bbox": [ - 90.72, - 556.36, - 221.59, - 565.42 - ], - "text": "[Hint: See the hint for Prob. 5.3-19.]", - "type": "text" - }, - { - "block_id": "p600-b24", - "global_id": 17261, - "bbox": [ - 57.59, - 571.19, - 263.24, - 591.19 - ], - "text": "5.3-22\nRedo\nProb.\n5.3-19\nusing\nthe\nresult\nin\nProb. 5.2-13a.", - "type": "text" - }, - { - "block_id": "p600-b25", - "global_id": 17262, - "bbox": [ - 57.59, - 596.96, - 263.24, - 616.95 - ], - "text": "5.3-23\nRedo\nProb.\n5.3-20\nusing\nthe\nresult\nin\nProb. 5.2-13a.", - "type": "text" - }, - { - "block_id": "p600-b26", - "global_id": 17263, - "bbox": [ - 57.59, - 622.73, - 263.24, - 642.72 - ], - "text": "5.3-24\nRedo\nProb.\n5.3-21\nusing\nthe\nresult\nin\nProb. 5.2-13a.", - "type": "text" - }, - { - "block_id": "p600-b27", - "global_id": 17264, - "bbox": [ - 284.74, - 85.94, - 490.39, - 94.98 - ], - "text": "5.3-25\n(a) Find the zero-state response of an LTID", - "type": "text" - }, - { - "block_id": "p600-b28", - "global_id": 17265, - "bbox": [ - 333.31, - 96.98, - 438.14, - 105.94 - ], - "text": "system with transfer function", - "type": "text" - }, - { - "block_id": "p600-b29", - "global_id": 17266, - "bbox": [ - 365.98, - 113.19, - 456.53, - 134.9 - ], - "text": "H[z] =\nz\n(z + 0.2)(z −0.8)", - "type": "text" - }, - { - "block_id": "p600-b30", - "global_id": 17267, - "bbox": [ - 317.86, - 144.0, - 490.38, - 165.57 - ], - "text": "and the input x[n] = e(n+1)u[n].\n(b) Write the difference equation relating the", - "type": "text" - }, - { - "block_id": "p600-b31", - "global_id": 17268, - "bbox": [ - 333.31, - 167.18, - 421.74, - 176.52 - ], - "text": "output y[n] to input x[n].", - "type": "text" - }, - { - "block_id": "p600-b32", - "global_id": 17269, - "bbox": [ - 284.74, - 181.13, - 459.43, - 190.47 - ], - "text": "5.3-26\nRepeat Prob. 5.3-25 for x[n] = u[n] and", - "type": "text" - }, - { - "block_id": "p600-b33", - "global_id": 17270, - "bbox": [ - 364.98, - 199.62, - 442.09, - 221.61 - ], - "text": "H[z] =\n2z + 3\n(z −2)(z −3)", - "type": "text" - }, - { - "block_id": "p600-b34", - "global_id": 17271, - "bbox": [ - 284.74, - 230.79, - 401.92, - 239.83 - ], - "text": "5.3-27\nRepeat Prob. 5.3-25 for", - "type": "text" - }, - { - "block_id": "p600-b35", - "global_id": 17272, - "bbox": [ - 367.56, - 249.03, - 439.5, - 271.01 - ], - "text": "H[z] =\n6(5z −1)\n6z2 −5z + 1", - "type": "text" - }, - { - "block_id": "p600-b36", - "global_id": 17273, - "bbox": [ - 317.86, - 278.98, - 389.61, - 288.32 - ], - "text": "and the input x[n] is", - "type": "text" - }, - { - "block_id": "p600-b37", - "global_id": 17274, - "bbox": [ - 317.86, - 288.68, - 395.58, - 310.24 - ], - "text": "(a) (4)−nu[n]\n(b) (4)−(n−2)u[n −2]", - "type": "text" - }, - { - "block_id": "p600-b38", - "global_id": 17275, - "bbox": [ - 317.86, - 310.6, - 382.46, - 332.16 - ], - "text": "(c) (4)−(n−2)u[n]\n(d) (4)−nu[n −2]", - "type": "text" - }, - { - "block_id": "p600-b39", - "global_id": 17276, - "bbox": [ - 284.74, - 336.77, - 459.43, - 346.11 - ], - "text": "5.3-28\nRepeat Prob. 5.3-25 for x[n] = u[n] and", - "type": "text" - }, - { - "block_id": "p600-b40", - "global_id": 17277, - "bbox": [ - 363.08, - 355.24, - 443.98, - 377.24 - ], - "text": "H[z] =\n2z −1\nz2 −1.6z + 0.8", - "type": "text" - }, - { - "block_id": "p600-b41", - "global_id": 17278, - "bbox": [ - 284.74, - 385.62, - 490.4, - 427.54 - ], - "text": "5.3-29\nFind the transfer functions corresponding to\neach of the systems specified by difference\nequations in Probs. 5.3-5, 5.3-7, 5.3-9, and\n5.3-14.", - "type": "text" - }, - { - "block_id": "p600-b42", - "global_id": 17279, - "bbox": [ - 284.74, - 432.15, - 490.41, - 452.44 - ], - "text": "5.3-30\nFind h[n], the unit impulse response of the sys-\ntems described by the following equations:", - "type": "text" - }, - { - "block_id": "p600-b43", - "global_id": 17280, - "bbox": [ - 318.37, - 454.07, - 490.38, - 463.41 - ], - "text": "(a) y[n]+3y[n−1]+2y[n−2] = x[n]+3x[n−", - "type": "text" - }, - { - "block_id": "p600-b44", - "global_id": 17281, - "bbox": [ - 317.87, - 465.03, - 490.39, - 485.32 - ], - "text": "1] + 3x[n −2]\n(b) y[n + 2] + 2y[n + 1] + y[n] = 2x[n + 2] −", - "type": "text" - }, - { - "block_id": "p600-b45", - "global_id": 17282, - "bbox": [ - 318.37, - 486.94, - 490.39, - 507.25 - ], - "text": "x[n + 1]\n(c) y[n]−y[n−1]+0.5y[n−2] = x[n]+2x[n−", - "type": "text" - }, - { - "block_id": "p600-b46", - "global_id": 17283, - "bbox": [ - 333.31, - 508.86, - 340.78, - 518.2 - ], - "text": "1]", - "type": "text" - }, - { - "block_id": "p600-b47", - "global_id": 17284, - "bbox": [ - 284.74, - 522.81, - 490.39, - 543.11 - ], - "text": "5.3-31\nFind h[n], the unit impulse response of the\nsystems in Probs. 5.3-25, 5.3-26, and 5.3-28.", - "type": "text" - }, - { - "block_id": "p600-b48", - "global_id": 17285, - "bbox": [ - 284.74, - 547.71, - 490.39, - 578.97 - ], - "text": "5.3-32\nA system has impulse response h[n] = u[n −\n3].\n(a) Determine the impulse response of the", - "type": "text" - }, - { - "block_id": "p600-b49", - "global_id": 17286, - "bbox": [ - 317.86, - 579.33, - 485.94, - 600.89 - ], - "text": "inverse system h−1[n].\n(b) Is the inverse stable? Is the inverse causal?", - "type": "text" - }, - { - "block_id": "p600-b50", - "global_id": 17287, - "bbox": [ - 318.37, - 601.25, - 490.39, - 611.85 - ], - "text": "(c) Your boss asks you to implement h−1[n]", - "type": "text" - }, - { - "block_id": "p600-b51", - "global_id": 17288, - "bbox": [ - 333.31, - 613.84, - 490.38, - 644.73 - ], - "text": "to the best of your ability. Describe your\nrealizable design, taking care to identify\nany deficiencies.", - "type": "text" - } - ] - }, - { - "page_num": 601, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p601-b0", - "global_id": 17289, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n581", - "type": "text" - }, - { - "block_id": "p601-b1", - "global_id": 17290, - "bbox": [ - 87.82, - 85.94, - 261.54, - 94.98 - ], - "text": "5.4-1\nA system has impulse response given by", - "type": "text" - }, - { - "block_id": "p601-b2", - "global_id": 17291, - "bbox": [ - 137.18, - 109.95, - 160.95, - 119.2 - ], - "text": "h[n] =", - "type": "text" - }, - { - "block_id": "p601-b3", - "global_id": 17292, - "bbox": [ - 162.79, - 97.36, - 191.69, - 118.96 - ], - "text": "1 + j\n√", - "type": "text" - }, - { - "block_id": "p601-b4", - "global_id": 17293, - "bbox": [ - 184.86, - 117.9, - 189.34, - 126.87 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p601-b5", - "global_id": 17294, - "bbox": [ - 192.89, - 97.36, - 202.17, - 108.12 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p601-b6", - "global_id": 17295, - "bbox": [ - 204.07, - 109.95, - 211.06, - 118.92 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p601-b7", - "global_id": 17296, - "bbox": [ - 212.45, - 97.36, - 236.47, - 118.96 - ], - "text": "1 −j\n√", - "type": "text" - }, - { - "block_id": "p601-b8", - "global_id": 17297, - "bbox": [ - 229.63, - 117.9, - 234.12, - 126.87 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p601-b9", - "global_id": 17298, - "bbox": [ - 237.65, - 97.36, - 252.33, - 108.12 - ], - "text": "n!", - "type": "text" - }, - { - "block_id": "p601-b10", - "global_id": 17299, - "bbox": [ - 253.33, - 109.95, - 268.27, - 119.2 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p601-b11", - "global_id": 17300, - "bbox": [ - 116.46, - 134.04, - 288.98, - 153.96 - ], - "text": "This system can be implemented according to\nFig. P5.4-1.", - "type": "text" - }, - { - "block_id": "p601-b12", - "global_id": 17301, - "bbox": [ - 119.71, - 169.92, - 303.44, - 178.58 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p601-b13", - "global_id": 17302, - "bbox": [ - 217.05, - 189.5, - 226.16, - 198.9 - ], - "text": "z–1", - "type": "text" - }, - { - "block_id": "p601-b14", - "global_id": 17303, - "bbox": [ - 217.05, - 250.7, - 226.16, - 260.1 - ], - "text": "z–1", - "type": "text" - }, - { - "block_id": "p601-b17", - "global_id": 17304, - "bbox": [ - 192.07, - 201.1, - 203.96, - 210.7 - ], - "text": "–A1", - "type": "text" - }, - { - "block_id": "p601-b18", - "global_id": 17305, - "bbox": [ - 191.81, - 262.3, - 203.7, - 271.9 - ], - "text": "–A2", - "type": "text" - }, - { - "block_id": "p601-b19", - "global_id": 17306, - "bbox": [ - 116.46, - 283.79, - 168.11, - 292.75 - ], - "text": "Figure P5.4-1", - "type": "text" - }, - { - "block_id": "p601-b20", - "global_id": 17307, - "bbox": [ - 116.96, - 305.71, - 288.98, - 316.11 - ], - "text": "(a) Determine the coefficients A1 and A2 to", - "type": "text" - }, - { - "block_id": "p601-b21", - "global_id": 17308, - "bbox": [ - 116.46, - 316.39, - 288.98, - 348.39 - ], - "text": "implement h[n] using the structure shown in\nFig. P5.4-1.\n(b) What is the zero-state response y0[n] of", - "type": "text" - }, - { - "block_id": "p601-b22", - "global_id": 17309, - "bbox": [ - 131.89, - 349.64, - 288.97, - 369.57 - ], - "text": "this system, given a shifted unit step input\nx[n] = u[n + 3]?", - "type": "text" - }, - { - "block_id": "p601-b23", - "global_id": 17310, - "bbox": [ - 87.82, - 374.47, - 288.98, - 383.51 - ], - "text": "5.4-2\n(a) Show the canonic direct form, a cascade,", - "type": "text" - }, - { - "block_id": "p601-b24", - "global_id": 17311, - "bbox": [ - 131.9, - 385.51, - 229.99, - 394.47 - ], - "text": "and a parallel realization of", - "type": "text" - }, - { - "block_id": "p601-b25", - "global_id": 17312, - "bbox": [ - 172.76, - 402.25, - 243.53, - 417.79 - ], - "text": "H[z] = z(3z −1.8)", - "type": "text" - }, - { - "block_id": "p601-b26", - "global_id": 17313, - "bbox": [ - 200.96, - 412.31, - 246.93, - 424.24 - ], - "text": "z2 −z + 0.16", - "type": "text" - }, - { - "block_id": "p601-b27", - "global_id": 17314, - "bbox": [ - 116.46, - 431.17, - 288.96, - 451.09 - ], - "text": "(b) Find\nthe\ntranspose\nof\nthe\nrealizations\nobtained in part (a).", - "type": "text" - }, - { - "block_id": "p601-b28", - "global_id": 17315, - "bbox": [ - 87.82, - 456.0, - 196.03, - 465.04 - ], - "text": "5.4-3\nRepeat Prob. 5.4-2 for", - "type": "text" - }, - { - "block_id": "p601-b29", - "global_id": 17316, - "bbox": [ - 165.03, - 472.77, - 239.21, - 494.76 - ], - "text": "H[z] =\n5z + 2.2\nz2 + z + 0.16", - "type": "text" - }, - { - "block_id": "p601-b30", - "global_id": 17317, - "bbox": [ - 87.82, - 501.72, - 196.03, - 510.76 - ], - "text": "5.4-4\nRepeat Prob. 5.4-2 for", - "type": "text" - }, - { - "block_id": "p601-b31", - "global_id": 17318, - "bbox": [ - 140.5, - 518.5, - 263.74, - 540.49 - ], - "text": "H[z] =\n3.8z −1.1\n(z −0.2)(z2 −0.6z + 0.25)", - "type": "text" - }, - { - "block_id": "p601-b32", - "global_id": 17319, - "bbox": [ - 87.82, - 548.25, - 196.03, - 557.29 - ], - "text": "5.4-5\nRepeat Prob. 5.4-2 for", - "type": "text" - }, - { - "block_id": "p601-b33", - "global_id": 17320, - "bbox": [ - 148.35, - 565.07, - 255.9, - 587.06 - ], - "text": "H[z] =\nz(1.6z −1.8)\n(z −0.2)(z2 + z + 0.5)", - "type": "text" - }, - { - "block_id": "p601-b34", - "global_id": 17321, - "bbox": [ - 87.82, - 594.83, - 196.03, - 603.87 - ], - "text": "5.4-6\nRepeat Prob. 5.4-2 for", - "type": "text" - }, - { - "block_id": "p601-b35", - "global_id": 17322, - "bbox": [ - 152.1, - 609.9, - 252.15, - 628.69 - ], - "text": "H[z] = z(2z2 + 1.3z + 0.96)", - "type": "text" - }, - { - "block_id": "p601-b36", - "global_id": 17323, - "bbox": [ - 183.2, - 625.47, - 248.78, - 635.14 - ], - "text": "(z + 0.5)(z −0.4)2", - "type": "text" - }, - { - "block_id": "p601-b37", - "global_id": 17324, - "bbox": [ - 314.97, - 85.94, - 516.13, - 94.98 - ], - "text": "5.4-7\nConsider the LTID system shown in Fig. P5.4-7.", - "type": "text" - }, - { - "block_id": "p601-b38", - "global_id": 17325, - "bbox": [ - 344.88, - 121.73, - 507.94, - 138.25 - ], - "text": "3\nz−1\n\nz−1\n2\nx[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p601-b39", - "global_id": 17326, - "bbox": [ - 343.61, - 150.33, - 395.25, - 159.29 - ], - "text": "Figure P5.4-7", - "type": "text" - }, - { - "block_id": "p601-b40", - "global_id": 17327, - "bbox": [ - 344.11, - 177.83, - 516.12, - 186.8 - ], - "text": "(a) Determine the standard delay-form differ-", - "type": "text" - }, - { - "block_id": "p601-b41", - "global_id": 17328, - "bbox": [ - 343.61, - 188.78, - 516.13, - 208.71 - ], - "text": "ence equation description of this system.\n(b) Determine the impulse response h[n] of this", - "type": "text" - }, - { - "block_id": "p601-b42", - "global_id": 17329, - "bbox": [ - 343.61, - 210.71, - 493.77, - 241.59 - ], - "text": "system.\n(c) Is this realization canonical? Explain.\n(d) Is this system stable? Explain.", - "type": "text" - }, - { - "block_id": "p601-b43", - "global_id": 17330, - "bbox": [ - 344.12, - 243.59, - 468.87, - 252.55 - ], - "text": "(e) Is this system causal? Explain.", - "type": "text" - }, - { - "block_id": "p601-b44", - "global_id": 17331, - "bbox": [ - 314.97, - 257.45, - 499.01, - 266.49 - ], - "text": "5.4-8\nRealize a system whose transfer function is", - "type": "text" - }, - { - "block_id": "p601-b45", - "global_id": 17332, - "bbox": [ - 370.69, - 273.53, - 487.86, - 292.41 - ], - "text": "H[z] = 2z4 + z3 + 0.8z2 + 2z + 8", - "type": "text" - }, - { - "block_id": "p601-b46", - "global_id": 17333, - "bbox": [ - 439.77, - 289.11, - 446.5, - 298.69 - ], - "text": "z4", - "type": "text" - }, - { - "block_id": "p601-b47", - "global_id": 17334, - "bbox": [ - 314.97, - 306.75, - 516.13, - 326.75 - ], - "text": "5.4-9\nRealize a system whose transfer function is\ngiven by", - "type": "text" - }, - { - "block_id": "p601-b48", - "global_id": 17335, - "bbox": [ - 401.14, - 337.06, - 426.31, - 346.3 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p601-b49", - "global_id": 17336, - "bbox": [ - 428.16, - 328.01, - 440.84, - 337.51 - ], - "text": "6\n\"", - "type": "text" - }, - { - "block_id": "p601-b50", - "global_id": 17337, - "bbox": [ - 428.74, - 349.98, - 440.27, - 356.73 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p601-b51", - "global_id": 17338, - "bbox": [ - 441.85, - 335.35, - 458.11, - 346.3 - ], - "text": "nz−n", - "type": "text" - }, - { - "block_id": "p601-b52", - "global_id": 17339, - "bbox": [ - 310.48, - 363.09, - 516.13, - 394.05 - ], - "text": "5.4-10\nConsider\nthe\nLTID\nsystem\nshown\nin\nFig. P5.4-10, where parameter c is an arbitrary,\nreal constant.", - "type": "text" - }, - { - "block_id": "p601-b53", - "global_id": 17340, - "bbox": [ - 314.5, - 411.71, - 416.29, - 433.95 - ], - "text": "x[n] \nz−1\nc", - "type": "text" - }, - { - "block_id": "p601-b54", - "global_id": 17341, - "bbox": [ - 453.9, - 418.8, - 514.54, - 433.95 - ], - "text": "z−1\n y[n]", - "type": "text" - }, - { - "block_id": "p601-b55", - "global_id": 17342, - "bbox": [ - 369.55, - 440.51, - 459.49, - 448.48 - ], - "text": "c\nc", - "type": "text" - }, - { - "block_id": "p601-b56", - "global_id": 17343, - "bbox": [ - 313.72, - 463.8, - 369.84, - 472.76 - ], - "text": "Figure P5.4-10", - "type": "text" - }, - { - "block_id": "p601-b57", - "global_id": 17344, - "bbox": [ - 344.12, - 491.3, - 516.11, - 500.27 - ], - "text": "(a) Determine the system transfer function", - "type": "text" - }, - { - "block_id": "p601-b58", - "global_id": 17345, - "bbox": [ - 343.61, - 501.88, - 516.12, - 522.18 - ], - "text": "H[z], expressed in standard rational form.\n(b) Determine all system poles and all system", - "type": "text" - }, - { - "block_id": "p601-b59", - "global_id": 17346, - "bbox": [ - 344.11, - 524.18, - 516.11, - 544.1 - ], - "text": "zeros.\n(c) Is the system of Fig. P5.4-10 canonical?", - "type": "text" - }, - { - "block_id": "p601-b60", - "global_id": 17347, - "bbox": [ - 343.61, - 546.09, - 516.11, - 566.02 - ], - "text": "Explain.\n(d) What constraints, if any, exist on parameter", - "type": "text" - }, - { - "block_id": "p601-b61", - "global_id": 17348, - "bbox": [ - 359.05, - 567.92, - 489.28, - 576.98 - ], - "text": "c to ensure that the system is stable?", - "type": "text" - }, - { - "block_id": "p601-b62", - "global_id": 17349, - "bbox": [ - 310.48, - 581.89, - 516.13, - 634.77 - ], - "text": "5.4-11\nThis problem demonstrates the enormous num-\nber of ways of implementing even a relatively\nlow-order transfer function. A second-order\ntransfer function has two real zeros and two\nreal poles. Discuss various ways of realizing", - "type": "text" - } - ] - }, - { - "page_num": 602, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p602-b0", - "global_id": 17350, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "582\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p602-b1", - "global_id": 17351, - "bbox": [ - 90.72, - 85.9, - 263.25, - 127.74 - ], - "text": "such a transfer function. Consider canonic\ndirect, cascade, parallel, and the corresponding\ntransposed forms. Note also that interchange of\ncascaded sections yields a different realization.", - "type": "text" - }, - { - "block_id": "p602-b2", - "global_id": 17352, - "bbox": [ - 57.59, - 132.83, - 263.24, - 229.54 - ], - "text": "5.4-12\nConsider a digital audio system: an input\nanalog-to-digital converter (ADC) is used to\ncollect input samples at a CD-quality rate\nFs = 44 kHz. Input samples are processed with\na digital filter to generate output samples, which\nare sent to a digital-to-analog converter (DAC)\nat the same rate Fs. Every sample interval\nT = 1/Fs, the digital processor executes the\nfollowing MATLAB-compatible code:", - "type": "text" - }, - { - "block_id": "p602-b3", - "global_id": 17353, - "bbox": [ - 90.72, - 241.59, - 250.76, - 360.14 - ], - "text": "% Read input sample from the ADC\nx = read_ADC;\n% Process input and...\nmem(1) = x - mem(3)*9/16;\n% ...compute output sample\ny = mem(1)*7/16 - mem(3)*7/16;\n% Send output sample to the DAC\nwrite_DAC = y;\n% Update memory for next iteration\nmem(3) = mem(2);\nmem(2) = mem(1);", - "type": "text" - }, - { - "block_id": "p602-b4", - "global_id": 17354, - "bbox": [ - 91.22, - 371.68, - 263.23, - 380.64 - ], - "text": "(a) Does the code implement DFI, DFII, TDFI,", - "type": "text" - }, - { - "block_id": "p602-b5", - "global_id": 17355, - "bbox": [ - 90.72, - 382.64, - 263.24, - 424.48 - ], - "text": "or TDFII? Support your answer by drawing\nthe appropriate block diagram labeled in a\nmanner that is consistent with the code.\n(b) Determine the transfer function H[z] of this", - "type": "text" - }, - { - "block_id": "p602-b6", - "global_id": 17356, - "bbox": [ - 91.22, - 426.48, - 263.22, - 446.4 - ], - "text": "system.\n(c) What is the basic filtering function of this", - "type": "text" - }, - { - "block_id": "p602-b7", - "global_id": 17357, - "bbox": [ - 90.72, - 448.4, - 263.24, - 479.28 - ], - "text": "system: LP, HP, BP, or BS? Justify your\nanswer.\n(d) Determine the transfer function H−1[z] of", - "type": "text" - }, - { - "block_id": "p602-b8", - "global_id": 17358, - "bbox": [ - 106.15, - 480.89, - 263.24, - 512.15 - ], - "text": "the inverse system to H[z] and draw its DFI\nblock implementation. How well will the\ninverse system operate?", - "type": "text" - }, - { - "block_id": "p602-b9", - "global_id": 17359, - "bbox": [ - 57.59, - 517.24, - 252.28, - 526.28 - ], - "text": "5.4-13\nRepeat Prob. 5.4-12 but instead use the code:", - "type": "text" - }, - { - "block_id": "p602-b10", - "global_id": 17360, - "bbox": [ - 90.72, - 538.33, - 250.76, - 634.96 - ], - "text": "% Read input sample from the ADC\nx = read_ADC;\n% Compute output sample\ny = x*7/32+mem(1);\n% Send output sample to the DAC\nwrite_DAC = y;\n% Update memory for next iteration\nmem(1) = mem(2);\nmem(2) = x*7/32 + y*9/16;", - "type": "text" - }, - { - "block_id": "p602-b11", - "global_id": 17361, - "bbox": [ - 289.22, - 85.64, - 490.39, - 127.86 - ], - "text": "5.5-1\nA CT sinusoid x(t) = cos(ωt) is sampled at\na greater-than-Nyquist rate Fs = 1000 Hz\nto produce a DT sinusoid x(t) = cos(n).\nDetermine the analog frequency ω if", - "type": "text" - }, - { - "block_id": "p602-b12", - "global_id": 17362, - "bbox": [ - 318.37, - 127.85, - 356.04, - 138.82 - ], - "text": "(a) = π", - "type": "text" - }, - { - "block_id": "p602-b13", - "global_id": 17363, - "bbox": [ - 352.8, - 134.92, - 356.04, - 141.39 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p602-b14", - "global_id": 17364, - "bbox": [ - 317.86, - 145.04, - 359.28, - 156.01 - ], - "text": "(b) = 2π", - "type": "text" - }, - { - "block_id": "p602-b15", - "global_id": 17365, - "bbox": [ - 354.42, - 152.12, - 357.66, - 158.59 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p602-b16", - "global_id": 17366, - "bbox": [ - 318.37, - 162.39, - 355.39, - 173.08 - ], - "text": "(c) = 7", - "type": "text" - }, - { - "block_id": "p602-b17", - "global_id": 17367, - "bbox": [ - 289.22, - 169.19, - 490.39, - 197.99 - ], - "text": "8\n5.5-2\nFind the amplitude and phase response of the\ndigital filters depicted in Fig. P5.5-2.", - "type": "text" - }, - { - "block_id": "p602-b18", - "global_id": 17368, - "bbox": [ - 289.22, - 200.4, - 490.39, - 218.38 - ], - "text": "5.5-3\nA causal LTID system H(z) =\n21(z−j)(z+j)\n16(z−1\n2 )(z+ 3\n4 ) has", - "type": "text" - }, - { - "block_id": "p602-b19", - "global_id": 17369, - "bbox": [ - 317.86, - 216.99, - 490.38, - 226.33 - ], - "text": "a periodic input x[n] that toggles between the", - "type": "text" - }, - { - "block_id": "p602-b20", - "global_id": 17370, - "bbox": [ - 317.86, - 232.81, - 483.91, - 242.15 - ], - "text": "values 1 and 2. That is, x[n] = [. . ., 1, 2, 1,", - "type": "text" - }, - { - "block_id": "p602-b21", - "global_id": 17371, - "bbox": [ - 317.86, - 227.23, - 490.38, - 253.11 - ], - "text": "↓\n2\n, 1, 2, 1, . . .], where x[0] = 2.", - "type": "text" - }, - { - "block_id": "p602-b22", - "global_id": 17372, - "bbox": [ - 318.37, - 253.47, - 490.38, - 264.06 - ], - "text": "(a) Plot the magnitude response |H(ej)| over", - "type": "text" - }, - { - "block_id": "p602-b23", - "global_id": 17373, - "bbox": [ - 317.86, - 265.69, - 490.39, - 318.86 - ], - "text": "−2π ≤ ≤2π.\n(b) Plot the phase response̸\nH(ej) over\n−2π ≤ ≤2π.\n(c) Determine\nthe\nsystem\noutput\ny[n]\nin\nresponse to the periodic input x[n].", - "type": "text" - }, - { - "block_id": "p602-b24", - "global_id": 17374, - "bbox": [ - 289.22, - 321.47, - 490.39, - 358.16 - ], - "text": "5.5-4\nA causal LTID system H(z) =\n−7(z+1)\n32(z−j 3\n4 )(z+j 3\n4 )\nhas a periodic input x[n] that cycles through\nthe 4 values 3, 2, 1, and 2. That is, x[n] =", - "type": "text" - }, - { - "block_id": "p602-b25", - "global_id": 17375, - "bbox": [ - 317.86, - 364.15, - 370.84, - 373.49 - ], - "text": "[. . ., 3, 2, 1, 2,", - "type": "text" - }, - { - "block_id": "p602-b26", - "global_id": 17376, - "bbox": [ - 317.86, - 358.52, - 490.39, - 395.4 - ], - "text": "↓\n3, 2, 1, 2, . . .],\nwhere\nx[0] =\n3.\n(a) Plot the magnitude response |H(ej)| over", - "type": "text" - }, - { - "block_id": "p602-b27", - "global_id": 17377, - "bbox": [ - 317.86, - 397.02, - 490.39, - 450.19 - ], - "text": "−2π ≤ ≤2π.\n(b) Plot the phase response̸\nH(ej) over\n−2π ≤ ≤2π.\n(c) Determine\nthe\nsystem\noutput\ny[n]\nin\nresponse to the periodic input x[n].", - "type": "text" - }, - { - "block_id": "p602-b28", - "global_id": 17378, - "bbox": [ - 289.22, - 455.1, - 490.4, - 486.73 - ], - "text": "5.5-5\nFind the amplitude and the phase response of\nthe filters shown in Fig. P5.5-5. [Hint: Express\nH[ej] as e−j2.5Ha[ej].]", - "type": "text" - }, - { - "block_id": "p602-b29", - "global_id": 17379, - "bbox": [ - 289.22, - 490.97, - 490.39, - 521.92 - ], - "text": "5.5-6\nFind\nthe\nfrequency\nresponse\nfor\nthe\nmoving-average system in Prob. 3.4-3. The\ninput–output equation of this system is given by", - "type": "text" - }, - { - "block_id": "p602-b30", - "global_id": 17380, - "bbox": [ - 367.23, - 540.46, - 396.77, - 551.06 - ], - "text": "y[n] = 1", - "type": "text" - }, - { - "block_id": "p602-b31", - "global_id": 17381, - "bbox": [ - 393.53, - 547.26, - 396.77, - 553.74 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p602-b32", - "global_id": 17382, - "bbox": [ - 398.97, - 532.77, - 411.66, - 542.27 - ], - "text": "4\n\"", - "type": "text" - }, - { - "block_id": "p602-b33", - "global_id": 17383, - "bbox": [ - 399.67, - 554.74, - 410.94, - 561.49 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p602-b34", - "global_id": 17384, - "bbox": [ - 412.65, - 541.81, - 441.02, - 551.06 - ], - "text": "x[n −k]", - "type": "text" - }, - { - "block_id": "p602-b35", - "global_id": 17385, - "bbox": [ - 289.22, - 570.92, - 490.37, - 579.96 - ], - "text": "5.5-7\n(a) Input–output relationships of two filters are", - "type": "text" - }, - { - "block_id": "p602-b36", - "global_id": 17386, - "bbox": [ - 333.3, - 581.96, - 490.4, - 634.77 - ], - "text": "described by\n(i) y[n] = −0.9y[n −1] + x[n]\n(ii) y[n] = 0.9y[n −1] + x[n]\nFor each case, find the transfer function, the\namplitude response, and the phase response.", - "type": "text" - } - ] - }, - { - "page_num": 603, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p603-b0", - "global_id": 17387, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n583", - "type": "text" - }, - { - "block_id": "p603-b1", - "global_id": 17388, - "bbox": [ - 116.46, - 85.9, - 288.98, - 116.78 - ], - "text": "Sketch the amplitude response, and state the\ntype (highpass, lowpass, etc.) of each filter.\n(b) Find the response of each of these filters to", - "type": "text" - }, - { - "block_id": "p603-b2", - "global_id": 17389, - "bbox": [ - 131.89, - 118.4, - 288.99, - 172.32 - ], - "text": "a sinusoid x[n] = cosn for = 0.01π\nand 0.99π. In general, show that the\ngain (amplitude response) of filter (i) at\nfrequency 0 is the same as the gain of\nfilter (ii) at frequency π −0.", - "type": "text" - }, - { - "block_id": "p603-b3", - "global_id": 17390, - "bbox": [ - 87.82, - 176.48, - 280.42, - 185.52 - ], - "text": "5.5-8\nFor an LTID system specified by the equation", - "type": "text" - }, - { - "block_id": "p603-b4", - "global_id": 17391, - "bbox": [ - 133.23, - 195.62, - 272.21, - 204.96 - ], - "text": "y[n + 1] −0.5y[n] = x[n + 1] + 0.8x[n]", - "type": "text" - }, - { - "block_id": "p603-b5", - "global_id": 17392, - "bbox": [ - 116.96, - 215.43, - 286.54, - 224.4 - ], - "text": "(a) Find the amplitude and the phase response.", - "type": "text" - }, - { - "block_id": "p603-b6", - "global_id": 17393, - "bbox": [ - 343.61, - 85.52, - 516.13, - 94.86 - ], - "text": "(b) Find the system response y[n] for the input", - "type": "text" - }, - { - "block_id": "p603-b7", - "global_id": 17394, - "bbox": [ - 359.05, - 96.49, - 452.41, - 105.82 - ], - "text": "x[n] = cos(0.5k −(π/3)).", - "type": "text" - }, - { - "block_id": "p603-b8", - "global_id": 17395, - "bbox": [ - 314.97, - 112.4, - 516.13, - 165.28 - ], - "text": "5.5-9\nFor an asymptotically stable LTID system, show\nthat the steady-state response to input ejnu[n]\nis H[ej]ejnu[n]. The steady-state response is\nthat part of the response which does not decay\nwith time and persists forever.", - "type": "text" - }, - { - "block_id": "p603-b9", - "global_id": 17396, - "bbox": [ - 310.48, - 171.86, - 516.13, - 191.87 - ], - "text": "5.5-10\nExpress the following signals in terms of\napparent frequencies:", - "type": "text" - }, - { - "block_id": "p603-b10", - "global_id": 17397, - "bbox": [ - 343.61, - 193.48, - 415.28, - 213.78 - ], - "text": "(a) cos(0.8πn + θ)\n(b) sin(1.2πn + θ)", - "type": "text" - }, - { - "block_id": "p603-b11", - "global_id": 17398, - "bbox": [ - 344.12, - 215.4, - 409.0, - 224.73 - ], - "text": "(c) cos(6.9n + θ)", - "type": "text" - }, - { - "block_id": "p603-b12", - "global_id": 17399, - "bbox": [ - 255.09, - 287.25, - 267.97, - 295.33 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p603-b13", - "global_id": 17400, - "bbox": [ - 174.64, - 244.43, - 187.52, - 252.51 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p603-b14", - "global_id": 17401, - "bbox": [ - 219.41, - 308.91, - 228.29, - 316.91 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p603-b15", - "global_id": 17402, - "bbox": [ - 184.37, - 275.81, - 194.37, - 283.81 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p603-b17", - "global_id": 17403, - "bbox": [ - 241.36, - 270.26, - 245.36, - 285.3 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b18", - "global_id": 17404, - "bbox": [ - 327.19, - 244.43, - 419.0, - 252.51 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p603-b19", - "global_id": 17405, - "bbox": [ - 369.23, - 308.91, - 378.88, - 316.91 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p603-b20", - "global_id": 17406, - "bbox": [ - 339.23, - 275.48, - 349.23, - 283.48 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p603-b22", - "global_id": 17407, - "bbox": [ - 396.43, - 270.26, - 400.43, - 285.3 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b23", - "global_id": 17408, - "bbox": [ - 224.28, - 329.95, - 364.39, - 338.03 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p603-b24", - "global_id": 17409, - "bbox": [ - 259.12, - 433.85, - 279.78, - 442.14 - ], - "text": "0.16", - "type": "text" - }, - { - "block_id": "p603-b25", - "global_id": 17410, - "bbox": [ - 312.59, - 388.82, - 322.59, - 396.82 - ], - "text": "1.8", - "type": "text" - }, - { - "block_id": "p603-b26", - "global_id": 17411, - "bbox": [ - 292.3, - 447.56, - 301.18, - 455.56 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p603-b27", - "global_id": 17412, - "bbox": [ - 246.19, - 331.07, - 314.54, - 346.48 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p603-b30", - "global_id": 17413, - "bbox": [ - 291.59, - 356.0, - 295.59, - 371.04 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b31", - "global_id": 17414, - "bbox": [ - 291.59, - 399.75, - 295.59, - 414.79 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b32", - "global_id": 17415, - "bbox": [ - 436.37, - 447.53, - 488.01, - 456.5 - ], - "text": "Figure P5.5-2", - "type": "text" - }, - { - "block_id": "p603-b33", - "global_id": 17416, - "bbox": [ - 271.08, - 532.22, - 279.96, - 540.22 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p603-b34", - "global_id": 17417, - "bbox": [ - 270.7, - 614.19, - 280.35, - 622.19 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p603-b35", - "global_id": 17418, - "bbox": [ - 133.46, - 496.75, - 418.59, - 510.9 - ], - "text": "x[n]\ny[n]\n2\n0.5\n2\n0.5", - "type": "text" - }, - { - "block_id": "p603-b38", - "global_id": 17419, - "bbox": [ - 176.19, - 508.02, - 180.19, - 523.06 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b39", - "global_id": 17420, - "bbox": [ - 220.62, - 508.02, - 224.62, - 523.06 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b40", - "global_id": 17421, - "bbox": [ - 265.93, - 508.02, - 269.93, - 523.06 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b41", - "global_id": 17422, - "bbox": [ - 309.93, - 508.02, - 313.93, - 523.06 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b42", - "global_id": 17423, - "bbox": [ - 355.93, - 508.02, - 359.93, - 523.06 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b43", - "global_id": 17424, - "bbox": [ - 133.46, - 578.25, - 418.59, - 592.7 - ], - "text": "x[n]\ny[n]\n2\n0.5\n1\n2\n0.5", - "type": "text" - }, - { - "block_id": "p603-b46", - "global_id": 17425, - "bbox": [ - 176.19, - 589.82, - 180.19, - 604.85 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b47", - "global_id": 17426, - "bbox": [ - 220.62, - 589.82, - 224.62, - 604.85 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b48", - "global_id": 17427, - "bbox": [ - 265.93, - 589.82, - 269.93, - 604.85 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b49", - "global_id": 17428, - "bbox": [ - 309.93, - 589.82, - 313.93, - 604.85 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b50", - "global_id": 17429, - "bbox": [ - 355.93, - 589.82, - 359.93, - 604.85 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p603-b51", - "global_id": 17430, - "bbox": [ - 130.58, - 628.36, - 182.22, - 637.33 - ], - "text": "Figure P5.5-5", - "type": "text" - } - ] - }, - { - "page_num": 604, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p604-b0", - "global_id": 17431, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "584\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p604-b1", - "global_id": 17432, - "bbox": [ - 199.74, - 191.73, - 346.01, - 199.73 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p604-b2", - "global_id": 17433, - "bbox": [ - 189.47, - 144.95, - 357.13, - 152.95 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p604-b3", - "global_id": 17434, - "bbox": [ - 413.63, - 191.71, - 465.27, - 200.67 - ], - "text": "Figure P5.6-1", - "type": "text" - }, - { - "block_id": "p604-b4", - "global_id": 17435, - "bbox": [ - 90.72, - 208.53, - 232.39, - 217.87 - ], - "text": "(d) cos(2.8πn + θ) + 2sin(3.7πn + θ)", - "type": "text" - }, - { - "block_id": "p604-b5", - "global_id": 17436, - "bbox": [ - 91.22, - 219.49, - 147.74, - 228.83 - ], - "text": "(e) sinc(πn/2)", - "type": "text" - }, - { - "block_id": "p604-b6", - "global_id": 17437, - "bbox": [ - 90.72, - 230.44, - 152.22, - 250.74 - ], - "text": "(f) sinc(3πn/2)\n(g) sinc(2πn)", - "type": "text" - }, - { - "block_id": "p604-b7", - "global_id": 17438, - "bbox": [ - 57.59, - 256.87, - 263.24, - 269.25 - ], - "text": "5.5-11\nShow\nthat\ncos(0.6πn\n+\n(π/6))\n+\n√", - "type": "text" - }, - { - "block_id": "p604-b8", - "global_id": 17439, - "bbox": [ - 90.72, - 267.83, - 263.24, - 288.12 - ], - "text": "3 cos(1.4πn + (π/3))\n=\n2 cos(0.6πn −\n(π/6)).", - "type": "text" - }, - { - "block_id": "p604-b9", - "global_id": 17440, - "bbox": [ - 57.59, - 294.54, - 263.23, - 303.58 - ], - "text": "5.5-12\n(a) A digital filter has the sampling interval", - "type": "text" - }, - { - "block_id": "p604-b10", - "global_id": 17441, - "bbox": [ - 90.72, - 305.21, - 263.25, - 347.42 - ], - "text": "T = 50µs. Determine the highest frequency\nthat can be processed by this filter without\naliasing.\n(b) If the highest frequency to be processed", - "type": "text" - }, - { - "block_id": "p604-b11", - "global_id": 17442, - "bbox": [ - 106.15, - 349.42, - 263.23, - 391.26 - ], - "text": "is 50 kHz, determine the minimum value\nof the sampling frequency Fs and the\nmaximum value of the sampling interval\nT that can be used.", - "type": "text" - }, - { - "block_id": "p604-b12", - "global_id": 17443, - "bbox": [ - 57.59, - 397.68, - 263.21, - 417.68 - ], - "text": "5.5-13\nConsider the discrete-time system represented\nby", - "type": "text" - }, - { - "block_id": "p604-b13", - "global_id": 17444, - "bbox": [ - 132.7, - 449.33, - 155.97, - 458.58 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p604-b14", - "global_id": 17445, - "bbox": [ - 157.82, - 440.01, - 170.5, - 449.77 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p604-b15", - "global_id": 17446, - "bbox": [ - 158.53, - 462.26, - 169.79, - 469.0 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p604-b16", - "global_id": 17447, - "bbox": [ - 171.49, - 447.82, - 221.25, - 458.67 - ], - "text": "(0.5)kx[n −k]", - "type": "text" - }, - { - "block_id": "p604-b17", - "global_id": 17448, - "bbox": [ - 91.22, - 495.78, - 263.24, - 504.75 - ], - "text": "(a) Determine and plot the magnitude response", - "type": "text" - }, - { - "block_id": "p604-b18", - "global_id": 17449, - "bbox": [ - 90.72, - 505.11, - 263.25, - 526.67 - ], - "text": "|H[ej]| of the system.\n(b) Determine and plot the phase response̸", - "type": "text" - }, - { - "block_id": "p604-b19", - "global_id": 17450, - "bbox": [ - 91.22, - 527.02, - 263.24, - 548.59 - ], - "text": "H[ej] of the system.\n(c) Find an efficient block representation that", - "type": "text" - }, - { - "block_id": "p604-b20", - "global_id": 17451, - "bbox": [ - 106.15, - 550.58, - 192.58, - 559.54 - ], - "text": "implements this system.", - "type": "text" - }, - { - "block_id": "p604-b21", - "global_id": 17452, - "bbox": [ - 62.08, - 565.97, - 263.23, - 596.93 - ], - "text": "5.6-1\nPole-zero configurations of certain filters are\nshown in Fig. P5.6-1. Sketch roughly the\namplitude response of these filters.", - "type": "text" - }, - { - "block_id": "p604-b22", - "global_id": 17453, - "bbox": [ - 62.08, - 603.35, - 263.26, - 634.31 - ], - "text": "5.6-2\nFigure P5.6-2 displays the pole-zero plot of a\nsecond-order real, causal LTID system that has\nH[−1] = −1.", - "type": "text" - }, - { - "block_id": "p604-b23", - "global_id": 17454, - "bbox": [ - 373.61, - 235.84, - 376.6, - 247.88 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p604-b24", - "global_id": 17455, - "bbox": [ - 366.52, - 274.16, - 376.93, - 283.5 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p604-b25", - "global_id": 17456, - "bbox": [ - 373.94, - 280.21, - 376.93, - 286.19 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p604-b26", - "global_id": 17457, - "bbox": [ - 423.91, - 264.88, - 427.9, - 272.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p604-b27", - "global_id": 17458, - "bbox": [ - 441.58, - 296.65, - 493.21, - 305.61 - ], - "text": "Figure P5.6-2", - "type": "text" - }, - { - "block_id": "p604-b28", - "global_id": 17459, - "bbox": [ - 318.37, - 315.08, - 490.39, - 324.88 - ], - "text": "(a) Determine the five constants b0, b1, b2, a1,", - "type": "text" - }, - { - "block_id": "p604-b29", - "global_id": 17460, - "bbox": [ - 333.31, - 326.04, - 490.37, - 348.29 - ], - "text": "and a2 that specify the transfer function\nH[z] = b0z2+b1z+b2", - "type": "text" - }, - { - "block_id": "p604-b30", - "global_id": 17461, - "bbox": [ - 317.86, - 339.42, - 490.38, - 360.81 - ], - "text": "z2+a1z+a2 .\n(b) Using the techniques of Sec. 5.6, accurately", - "type": "text" - }, - { - "block_id": "p604-b31", - "global_id": 17462, - "bbox": [ - 318.37, - 362.8, - 490.41, - 415.61 - ], - "text": "hand-sketch the system magnitude response\n|H[ej]| over the range (−2π ≤ ≤0).\n(c) Determine\nthe\noutput\ny[n]\nof\nthis\nsystem\nif\nthe\ninput\nis\nx[n]\n=\nsin", - "type": "text" - }, - { - "block_id": "p604-b32", - "global_id": 17463, - "bbox": [ - 344.77, - 399.06, - 357.34, - 411.32 - ], - "text": "πn", - "type": "text" - }, - { - "block_id": "p604-b33", - "global_id": 17464, - "bbox": [ - 351.84, - 399.06, - 362.15, - 418.19 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p604-b34", - "global_id": 17465, - "bbox": [ - 362.15, - 406.64, - 364.39, - 415.6 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p604-b35", - "global_id": 17466, - "bbox": [ - 289.22, - 431.39, - 490.38, - 462.65 - ], - "text": "5.6-3\nRepeat Prob. 5.6-2 if the zero at z = 1 is moved\nto z = −1 and H[1] = −1 is specified rather than\nH[−1] = −1.", - "type": "text" - }, - { - "block_id": "p604-b36", - "global_id": 17467, - "bbox": [ - 289.22, - 467.78, - 490.38, - 498.74 - ], - "text": "5.6-4\nFigure P5.6-4 displays the pole-zero plot of a\nsecond-order real, causal LTID system that has\nH[−1] = 1.", - "type": "text" - }, - { - "block_id": "p604-b37", - "global_id": 17468, - "bbox": [ - 318.37, - 500.64, - 490.39, - 510.44 - ], - "text": "(a) Determine the five constants k, b1, b2, a1,", - "type": "text" - }, - { - "block_id": "p604-b38", - "global_id": 17469, - "bbox": [ - 333.31, - 511.6, - 490.37, - 533.79 - ], - "text": "and a2 that specify the transfer function\nH[z] = k z2+b1z+b2", - "type": "text" - }, - { - "block_id": "p604-b39", - "global_id": 17470, - "bbox": [ - 365.64, - 524.92, - 400.41, - 539.21 - ], - "text": "z2+a1z+a2 .", - "type": "text" - }, - { - "block_id": "p604-b40", - "global_id": 17471, - "bbox": [ - 404.85, - 572.53, - 407.84, - 584.56 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p604-b41", - "global_id": 17472, - "bbox": [ - 403.96, - 610.85, - 414.37, - 620.2 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p604-b42", - "global_id": 17473, - "bbox": [ - 411.37, - 616.9, - 414.36, - 622.88 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p604-b43", - "global_id": 17474, - "bbox": [ - 439.35, - 601.57, - 443.33, - 609.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p604-b44", - "global_id": 17475, - "bbox": [ - 457.01, - 633.34, - 508.66, - 642.3 - ], - "text": "Figure P5.6-4", - "type": "text" - } - ] - }, - { - "page_num": 605, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p605-b0", - "global_id": 17476, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n585", - "type": "text" - }, - { - "block_id": "p605-b1", - "global_id": 17477, - "bbox": [ - 116.46, - 85.9, - 288.98, - 94.86 - ], - "text": "(b) Using the techniques of Sec. 5.6, accurately", - "type": "text" - }, - { - "block_id": "p605-b2", - "global_id": 17478, - "bbox": [ - 116.96, - 96.86, - 288.99, - 127.74 - ], - "text": "hand-sketch the system magnitude response\n|H[ej]| over the range (−π ≤ ≤π).\n(c) A signal x(t) = cos(100πt) + cos(500πt) is", - "type": "text" - }, - { - "block_id": "p605-b3", - "global_id": 17479, - "bbox": [ - 116.46, - 129.65, - 288.99, - 193.49 - ], - "text": "sampled at a greater than Nyquist rate Fs Hz\nand then input into the above LTID system\nto produce DT output y[n] = β cos(0n +\nθ). Determine Fs and 0. You do not need\nto find constants β and θ.\n(d) Is the inpulse response h[n] of this system", - "type": "text" - }, - { - "block_id": "p605-b4", - "global_id": 17480, - "bbox": [ - 131.9, - 195.48, - 286.03, - 204.45 - ], - "text": "absolutely summable? Justify your answer.", - "type": "text" - }, - { - "block_id": "p605-b5", - "global_id": 17481, - "bbox": [ - 87.82, - 211.44, - 289.0, - 242.4 - ], - "text": "5.6-5\nFigure P5.6-5 displays the pole-zero plot of a\nsecond-order real, causal LTID system that has\nH[1] = −1.", - "type": "text" - }, - { - "block_id": "p605-b6", - "global_id": 17482, - "bbox": [ - 172.21, - 301.67, - 175.19, - 313.69 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p605-b7", - "global_id": 17483, - "bbox": [ - 165.12, - 339.99, - 175.52, - 349.34 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p605-b8", - "global_id": 17484, - "bbox": [ - 172.54, - 346.04, - 175.52, - 352.02 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p605-b9", - "global_id": 17485, - "bbox": [ - 222.5, - 330.7, - 226.49, - 338.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p605-b10", - "global_id": 17486, - "bbox": [ - 240.17, - 362.47, - 291.81, - 371.44 - ], - "text": "Figure P5.6-5", - "type": "text" - }, - { - "block_id": "p605-b11", - "global_id": 17487, - "bbox": [ - 116.96, - 405.46, - 288.98, - 415.26 - ], - "text": "(a) Determine the five constants k, b1, b2, a1,", - "type": "text" - }, - { - "block_id": "p605-b12", - "global_id": 17488, - "bbox": [ - 131.89, - 416.42, - 288.97, - 438.61 - ], - "text": "and a2 that specify the transfer function\nH[z] = k z2+b1z+b2", - "type": "text" - }, - { - "block_id": "p605-b13", - "global_id": 17489, - "bbox": [ - 116.46, - 429.73, - 288.98, - 451.14 - ], - "text": "z2+a1z+a2 .\n(b) Using the techniques of Sec. 5.6, accurately", - "type": "text" - }, - { - "block_id": "p605-b14", - "global_id": 17490, - "bbox": [ - 116.96, - 453.12, - 288.99, - 484.01 - ], - "text": "hand-sketch the system magnitude response\n|H[ej]| over the range (−π ≤ ≤π).\n(c) A signal x(t) = cos(2πft) is sampled at a", - "type": "text" - }, - { - "block_id": "p605-b15", - "global_id": 17491, - "bbox": [ - 131.89, - 485.63, - 288.99, - 538.81 - ], - "text": "rate Fs = 1 kHz and then input into the\nabove LTID system to produce DT output\ny[n]. Determine, if possible, the frequency\nor frequencies f that will produce zero\noutput, y[n] = 0.", - "type": "text" - }, - { - "block_id": "p605-b16", - "global_id": 17492, - "bbox": [ - 87.82, - 545.5, - 288.98, - 587.71 - ], - "text": "5.6-6\nThe system y[n] −y[n −1] = x[n] −x[n −1] is\nan all-pass system that has zero phase response.\nIs there any difference between this system and\nthe system y[n] = x[n]? Justify your answer.", - "type": "text" - }, - { - "block_id": "p605-b17", - "global_id": 17493, - "bbox": [ - 87.82, - 594.69, - 289.0, - 636.61 - ], - "text": "5.6-7\nFigure P5.6-7 displays the pole-zero plot of a\nsecond-order real, causal LTID system that has\na repeated zero and H[1] = 4. The solid circle is\nthe unit circle.", - "type": "text" - }, - { - "block_id": "p605-b18", - "global_id": 17494, - "bbox": [ - 454.92, - 131.29, - 458.91, - 139.26 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p605-b19", - "global_id": 17495, - "bbox": [ - 343.61, - 205.66, - 395.25, - 214.62 - ], - "text": "Figure P5.6-7", - "type": "text" - }, - { - "block_id": "p605-b20", - "global_id": 17496, - "bbox": [ - 344.11, - 231.53, - 516.13, - 241.33 - ], - "text": "(a) Determine the five constants k, b1, b2, a1,", - "type": "text" - }, - { - "block_id": "p605-b21", - "global_id": 17497, - "bbox": [ - 359.05, - 242.5, - 516.11, - 264.69 - ], - "text": "and a2 that specify the transfer function\nH[z] = k z2+b1z+b2", - "type": "text" - }, - { - "block_id": "p605-b22", - "global_id": 17498, - "bbox": [ - 343.61, - 255.82, - 516.12, - 277.21 - ], - "text": "z2+a1z+a2 .\n(b) Using the techniques of Sec. 5.6, accurately", - "type": "text" - }, - { - "block_id": "p605-b23", - "global_id": 17499, - "bbox": [ - 344.12, - 279.21, - 516.15, - 310.83 - ], - "text": "hand-sketch the system magnitude response\n|H[ej]| over the range (−π ≤ ≤π).\n(c) Determine the steady-state output yss[n]", - "type": "text" - }, - { - "block_id": "p605-b24", - "global_id": 17500, - "bbox": [ - 359.05, - 311.71, - 516.13, - 332.0 - ], - "text": "of this system if the input is x[n] =\ncos( 3πn", - "type": "text" - }, - { - "block_id": "p605-b25", - "global_id": 17501, - "bbox": [ - 343.61, - 322.66, - 516.13, - 342.97 - ], - "text": "4 )u[n].\n(d) State whether this system is LP, HP, BP,", - "type": "text" - }, - { - "block_id": "p605-b26", - "global_id": 17502, - "bbox": [ - 359.05, - 344.95, - 516.13, - 386.8 - ], - "text": "BS, or other. If the digital system operates\nat Fs = 8 kHz, what is the approximate\nhertzian cutoff frequency (or frequencies)\nof this system?", - "type": "text" - }, - { - "block_id": "p605-b27", - "global_id": 17503, - "bbox": [ - 314.97, - 392.0, - 516.13, - 412.01 - ], - "text": "5.6-8\nThe magnitude and phase responses of a real,\nstable, LTI system are shown in Fig. P5.6-8.", - "type": "text" - }, - { - "block_id": "p605-b28", - "global_id": 17504, - "bbox": [ - 344.11, - 414.0, - 516.13, - 422.97 - ], - "text": "(a) What type of system is this: lowpass,", - "type": "text" - }, - { - "block_id": "p605-b29", - "global_id": 17505, - "bbox": [ - 343.61, - 424.95, - 516.14, - 444.88 - ], - "text": "highpass, bandpass, or bandstop?\n(b) What is the output of this system in response", - "type": "text" - }, - { - "block_id": "p605-b30", - "global_id": 17506, - "bbox": [ - 359.05, - 446.88, - 366.02, - 455.85 - ], - "text": "to", - "type": "text" - }, - { - "block_id": "p605-b31", - "global_id": 17507, - "bbox": [ - 392.84, - 465.05, - 437.64, - 475.13 - ], - "text": "x1[n] = 2sin", - "type": "text" - }, - { - "block_id": "p605-b32", - "global_id": 17508, - "bbox": [ - 438.63, - 452.46, - 451.26, - 467.74 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p605-b33", - "global_id": 17509, - "bbox": [ - 446.78, - 458.77, - 474.19, - 480.76 - ], - "text": "2 n + π\n4", - "type": "text" - }, - { - "block_id": "p605-b35", - "global_id": 17510, - "bbox": [ - 344.11, - 491.0, - 516.14, - 499.97 - ], - "text": "(c) What is the output of this system in response", - "type": "text" - }, - { - "block_id": "p605-b36", - "global_id": 17511, - "bbox": [ - 359.04, - 501.96, - 366.02, - 510.92 - ], - "text": "to", - "type": "text" - }, - { - "block_id": "p605-b37", - "global_id": 17512, - "bbox": [ - 401.82, - 520.14, - 442.61, - 530.22 - ], - "text": "x2[n] = cos", - "type": "text" - }, - { - "block_id": "p605-b38", - "global_id": 17513, - "bbox": [ - 443.62, - 507.55, - 460.73, - 523.2 - ], - "text": "7π", - "type": "text" - }, - { - "block_id": "p605-b39", - "global_id": 17514, - "bbox": [ - 454.01, - 507.55, - 473.36, - 535.84 - ], - "text": "4 n", - "type": "text" - }, - { - "block_id": "p605-b40", - "global_id": 17515, - "bbox": [ - 314.97, - 546.13, - 516.13, - 581.55 - ], - "text": "5.6-9\nConsider an LTID system with system function\nH[z] = b0\nz2+1\nz2−9/16.\n(a) Determine the constant b0 so that the system", - "type": "text" - }, - { - "block_id": "p605-b41", - "global_id": 17516, - "bbox": [ - 343.61, - 581.82, - 516.12, - 602.12 - ], - "text": "frequency response at = −π is −1.\n(b) Accurately sketch the system poles and", - "type": "text" - }, - { - "block_id": "p605-b42", - "global_id": 17517, - "bbox": [ - 344.11, - 604.12, - 516.12, - 624.04 - ], - "text": "zeros.\n(c) Using the locations of the system poles and", - "type": "text" - }, - { - "block_id": "p605-b43", - "global_id": 17518, - "bbox": [ - 359.05, - 624.39, - 502.31, - 635.0 - ], - "text": "zeros, sketch |H[ej]| over 0 ≤ ≤2π.", - "type": "text" - } - ] - }, - { - "page_num": 606, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p606-b0", - "global_id": 17519, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "586\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p606-b1", - "global_id": 17520, - "bbox": [ - 133.01, - 200.99, - 212.01, - 208.99 - ], - "text": "0\n0.25\n0.5", - "type": "text" - }, - { - "block_id": "p606-b2", - "global_id": 17521, - "bbox": [ - 200.39, - 212.84, - 403.77, - 221.14 - ], - "text": "Ω/π\nΩ/π", - "type": "text" - }, - { - "block_id": "p606-b3", - "global_id": 17522, - "bbox": [ - 120.03, - 90.33, - 281.76, - 208.99 - ], - "text": "0.75\n1\n0\n0.1\n0.2\n0.3\n0.4\n0.5\n0.6\n0.7\n0.8\n0.9\n1", - "type": "text" - }, - { - "block_id": "p606-b4", - "global_id": 17523, - "bbox": [ - 106.83, - 129.54, - 116.53, - 154.75 - ], - "text": "|H(ejΩ)|", - "type": "text" - }, - { - "block_id": "p606-b5", - "global_id": 17524, - "bbox": [ - 323.51, - 200.99, - 471.51, - 208.99 - ], - "text": "0\n0.25\n0.5\n0.75\n1", - "type": "text" - }, - { - "block_id": "p606-b6", - "global_id": 17525, - "bbox": [ - 305.28, - 188.96, - 321.28, - 196.96 - ], - "text": "–180", - "type": "text" - }, - { - "block_id": "p606-b7", - "global_id": 17526, - "bbox": [ - 305.28, - 176.35, - 321.28, - 184.35 - ], - "text": "–135", - "type": "text" - }, - { - "block_id": "p606-b8", - "global_id": 17527, - "bbox": [ - 309.28, - 163.73, - 321.28, - 171.73 - ], - "text": "–90", - "type": "text" - }, - { - "block_id": "p606-b9", - "global_id": 17528, - "bbox": [ - 309.28, - 151.12, - 321.28, - 159.12 - ], - "text": "–45", - "type": "text" - }, - { - "block_id": "p606-b10", - "global_id": 17529, - "bbox": [ - 317.28, - 138.5, - 321.28, - 146.5 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p606-b11", - "global_id": 17530, - "bbox": [ - 312.79, - 125.9, - 320.79, - 133.9 - ], - "text": "45", - "type": "text" - }, - { - "block_id": "p606-b12", - "global_id": 17531, - "bbox": [ - 312.79, - 113.3, - 320.79, - 121.3 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p606-b13", - "global_id": 17532, - "bbox": [ - 308.28, - 100.67, - 320.28, - 108.67 - ], - "text": "135", - "type": "text" - }, - { - "block_id": "p606-b14", - "global_id": 17533, - "bbox": [ - 308.28, - 88.06, - 320.28, - 96.06 - ], - "text": "180", - "type": "text" - }, - { - "block_id": "p606-b15", - "global_id": 17534, - "bbox": [ - 292.08, - 117.85, - 301.78, - 160.74 - ], - "text": "H(ejΩ) [deg]", - "type": "text" - }, - { - "block_id": "p606-b16", - "global_id": 17535, - "bbox": [ - 104.83, - 227.91, - 156.48, - 236.88 - ], - "text": "Figure P5.6-8", - "type": "text" - }, - { - "block_id": "p606-b17", - "global_id": 17536, - "bbox": [ - 90.72, - 247.96, - 263.24, - 257.3 - ], - "text": "(d) Determine the response y[n] to the input", - "type": "text" - }, - { - "block_id": "p606-b18", - "global_id": 17537, - "bbox": [ - 91.22, - 255.67, - 263.22, - 290.18 - ], - "text": "x[n] = (−1 + j) + jn + (1 −j)sin(πn + 1).\n(e) Draw\nan\nappropriate\nblock\ndiagram\nrepresentation of this system.", - "type": "text" - }, - { - "block_id": "p606-b19", - "global_id": 17538, - "bbox": [ - 57.59, - 295.29, - 263.24, - 326.25 - ], - "text": "5.6-10\nDo Prob. 5.10-3 by graphical procedure. Do\nthe\nsketches\napproximately,\nwithout\nusing\nMATLAB.", - "type": "text" - }, - { - "block_id": "p606-b20", - "global_id": 17539, - "bbox": [ - 57.59, - 331.37, - 263.24, - 362.33 - ], - "text": "5.6-11\nDo Prob. 5.10-8 by graphical procedure. Do\nthe\nsketches\napproximately,\nwithout\nusing\nMATLAB.", - "type": "text" - }, - { - "block_id": "p606-b21", - "global_id": 17540, - "bbox": [ - 57.59, - 367.44, - 263.23, - 376.48 - ], - "text": "5.6-12\n(a) Realize a digital filter whose transfer", - "type": "text" - }, - { - "block_id": "p606-b22", - "global_id": 17541, - "bbox": [ - 106.15, - 378.47, - 177.27, - 387.44 - ], - "text": "function is given by", - "type": "text" - }, - { - "block_id": "p606-b23", - "global_id": 17542, - "bbox": [ - 157.38, - 399.19, - 210.82, - 414.72 - ], - "text": "H[z] = K z + 1", - "type": "text" - }, - { - "block_id": "p606-b24", - "global_id": 17543, - "bbox": [ - 193.06, - 411.83, - 210.82, - 421.08 - ], - "text": "z −a", - "type": "text" - }, - { - "block_id": "p606-b25", - "global_id": 17544, - "bbox": [ - 90.72, - 432.1, - 263.23, - 441.07 - ], - "text": "(b) Sketch the amplitude response of this filter,", - "type": "text" - }, - { - "block_id": "p606-b26", - "global_id": 17545, - "bbox": [ - 91.22, - 442.69, - 263.24, - 462.99 - ], - "text": "assuming |a| < 1.\n(c) The amplitude response of this lowpass", - "type": "text" - }, - { - "block_id": "p606-b27", - "global_id": 17546, - "bbox": [ - 106.15, - 464.61, - 263.23, - 495.87 - ], - "text": "filter is maximum at = 0. The 3 dB band-\nwidth is the frequency at which the ampli-\ntude response drops to 0.707 (or 1/", - "type": "text" - }, - { - "block_id": "p606-b28", - "global_id": 17547, - "bbox": [ - 248.18, - 478.94, - 255.77, - 487.91 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p606-b29", - "global_id": 17548, - "bbox": [ - 106.15, - 486.9, - 263.24, - 517.78 - ], - "text": "2)\ntimes its maximum value. Determine the 3\ndB bandwidth of this filter when a = 0.2.", - "type": "text" - }, - { - "block_id": "p606-b30", - "global_id": 17549, - "bbox": [ - 57.59, - 522.9, - 263.25, - 630.57 - ], - "text": "5.6-13\nDesign a digital notch filter to reject frequency\n5000 Hz completely and to have a sharp\nrecovery on either side of 5000 Hz to a gain of\nunity. The highest frequency to be processed is\n20 kHz (Fh = 20,000). [Hint: See Ex. 5.15. The\nzeros should be at e±jωT for ω corresponding\nto 5000 Hz, and the poles are at ae±jωT with\na < 1. Leave your answer in terms of a. Realize\nthis filter using the canonical form. Find the\namplitude response of the filter.]", - "type": "text" - }, - { - "block_id": "p606-b31", - "global_id": 17550, - "bbox": [ - 284.74, - 248.38, - 490.36, - 268.37 - ], - "text": "5.6-14\nConsider the desired DT system magnitude\nresponse |H[ej]| in Fig. P5.6-14.", - "type": "text" - }, - { - "block_id": "p606-b32", - "global_id": 17551, - "bbox": [ - 291.99, - 280.96, - 321.3, - 291.48 - ], - "text": "|H[ej]|", - "type": "text" - }, - { - "block_id": "p606-b33", - "global_id": 17552, - "bbox": [ - 302.54, - 345.13, - 483.7, - 362.04 - ], - "text": "14π\n4\n15π\n4\n17π\n4\n19π\n4\n21π\n4\n22π\n4", - "type": "text" - }, - { - "block_id": "p606-b34", - "global_id": 17553, - "bbox": [ - 302.26, - 303.04, - 306.74, - 312.0 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p606-b35", - "global_id": 17554, - "bbox": [ - 287.98, - 365.33, - 344.1, - 374.29 - ], - "text": "Figure P5.6-14", - "type": "text" - }, - { - "block_id": "p606-b36", - "global_id": 17555, - "bbox": [ - 318.37, - 388.47, - 490.38, - 397.43 - ], - "text": "(a) Is the filter LP, HP, BP, BS, or other?", - "type": "text" - }, - { - "block_id": "p606-b37", - "global_id": 17556, - "bbox": [ - 317.86, - 399.42, - 490.37, - 419.35 - ], - "text": "Explain.\n(b) Sketch the pole-zero plot of a 2nd-order", - "type": "text" - }, - { - "block_id": "p606-b38", - "global_id": 17557, - "bbox": [ - 333.31, - 421.35, - 490.39, - 452.97 - ], - "text": "system\nthat\nbehaves\nas\na\nreasonable\napproximation of Fig. P5.6-14. What is the\ncoefficient b0?", - "type": "text" - }, - { - "block_id": "p606-b39", - "global_id": 17558, - "bbox": [ - 284.74, - 457.14, - 490.39, - 510.02 - ], - "text": "5.6-15\nShow that a first-order LTID system with a pole\nat z = r and a zero at z = 1/r (r ≤1) is an allpass\nfilter. In other words, show that the amplitude\nresponse |H[ej]| of a system with the transfer\nfunction", - "type": "text" - }, - { - "block_id": "p606-b40", - "global_id": 17559, - "bbox": [ - 361.07, - 534.33, - 386.24, - 543.58 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p606-b41", - "global_id": 17560, - "bbox": [ - 389.29, - 518.36, - 408.23, - 533.52 - ], - "text": "z −1", - "type": "text" - }, - { - "block_id": "p606-b42", - "global_id": 17561, - "bbox": [ - 390.87, - 530.93, - 447.18, - 549.94 - ], - "text": "r\nz −r\nr ≤1", - "type": "text" - }, - { - "block_id": "p606-b43", - "global_id": 17562, - "bbox": [ - 317.86, - 559.04, - 490.4, - 633.77 - ], - "text": "is constant with frequency. This is a first-order\nallpass filter. [Hint: Show that the ratio of the\ndistances of any point on the unit circle from\nthe zero (at z = 1/r) and the pole (at z = r) is a\nconstant 1/r.]\nGeneralize this result to show that an LTID\nsystem with two poles at z = re±jθ and two zeros", - "type": "text" - } - ] - }, - { - "page_num": 607, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p607-b0", - "global_id": 17563, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n587", - "type": "text" - }, - { - "block_id": "p607-b1", - "global_id": 17564, - "bbox": [ - 116.46, - 83.73, - 288.99, - 118.25 - ], - "text": "at z = (1/r)e±jθ (r ≤1) is an allpass filter. In\nother words, show that the amplitude response\nof a system with the transfer function", - "type": "text" - }, - { - "block_id": "p607-b2", - "global_id": 17565, - "bbox": [ - 130.81, - 147.21, - 155.98, - 156.46 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p607-b4", - "global_id": 17566, - "bbox": [ - 165.07, - 129.09, - 184.02, - 144.26 - ], - "text": "z −1", - "type": "text" - }, - { - "block_id": "p607-b5", - "global_id": 17567, - "bbox": [ - 179.94, - 133.29, - 193.85, - 150.62 - ], - "text": "r ejθ", - "type": "text" - }, - { - "block_id": "p607-b7", - "global_id": 17568, - "bbox": [ - 207.1, - 129.09, - 226.04, - 144.25 - ], - "text": "z −1", - "type": "text" - }, - { - "block_id": "p607-b8", - "global_id": 17569, - "bbox": [ - 221.96, - 133.29, - 240.92, - 150.62 - ], - "text": "r e−jθ", - "type": "text" - }, - { - "block_id": "p607-b10", - "global_id": 17570, - "bbox": [ - 168.16, - 152.97, - 238.99, - 162.82 - ], - "text": "(z −rejθ)(z −re−jθ)", - "type": "text" - }, - { - "block_id": "p607-b11", - "global_id": 17571, - "bbox": [ - 148.99, - 184.33, - 155.98, - 193.29 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p607-b12", - "global_id": 17572, - "bbox": [ - 159.02, - 168.86, - 174.63, - 181.37 - ], - "text": "z2 −", - "type": "text" - }, - { - "block_id": "p607-b13", - "global_id": 17573, - "bbox": [ - 176.02, - 159.53, - 187.76, - 175.17 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p607-b14", - "global_id": 17574, - "bbox": [ - 183.67, - 172.12, - 206.84, - 187.82 - ], - "text": "r cosθ", - "type": "text" - }, - { - "block_id": "p607-b16", - "global_id": 17575, - "bbox": [ - 213.79, - 166.2, - 234.22, - 181.46 - ], - "text": "z + 1", - "type": "text" - }, - { - "block_id": "p607-b17", - "global_id": 17576, - "bbox": [ - 228.26, - 178.15, - 235.19, - 187.73 - ], - "text": "r2", - "type": "text" - }, - { - "block_id": "p607-b18", - "global_id": 17577, - "bbox": [ - 162.27, - 184.33, - 274.64, - 200.03 - ], - "text": "z2 −(2rcosθ)z + r2\nr ≤1", - "type": "text" - }, - { - "block_id": "p607-b19", - "global_id": 17578, - "bbox": [ - 116.46, - 209.88, - 212.35, - 218.85 - ], - "text": "is constant with frequency.", - "type": "text" - }, - { - "block_id": "p607-b20", - "global_id": 17579, - "bbox": [ - 83.34, - 223.48, - 288.98, - 233.55 - ], - "text": "5.6-16\n(a) If h1[n] and h2[n], the impulse responses of", - "type": "text" - }, - { - "block_id": "p607-b21", - "global_id": 17580, - "bbox": [ - 131.89, - 234.44, - 288.98, - 255.47 - ], - "text": "two LTID systems are related by h2[n] =\n(−1)nh1[n], then show that", - "type": "text" - }, - { - "block_id": "p607-b22", - "global_id": 17581, - "bbox": [ - 170.4, - 264.8, - 250.49, - 276.59 - ], - "text": "H2[ej] = H1[ej(±π)]", - "type": "text" - }, - { - "block_id": "p607-b23", - "global_id": 17582, - "bbox": [ - 116.46, - 288.0, - 288.99, - 319.62 - ], - "text": "How is the frequency response spectrum\nH2[ej] related to the H1[ej]?\n(b) If H1[z] represents an ideal lowpass filter", - "type": "text" - }, - { - "block_id": "p607-b24", - "global_id": 17583, - "bbox": [ - 131.89, - 319.24, - 288.98, - 341.55 - ], - "text": "with cutoff frequency c, sketch H2[ej].\nWhat type of filter is H2[ej]?", - "type": "text" - }, - { - "block_id": "p607-b25", - "global_id": 17584, - "bbox": [ - 83.34, - 345.73, - 288.99, - 420.53 - ], - "text": "5.6-17\nMappings such as the bilinear transformation\nare useful in the conversion of continuous-time\nfilters to discrete-time filters. Another useful\ntype of transformation is one that converts\na discrete-time filter into a different type of\ndiscrete-time filter. Consider a transformation\nthat replaces z with −z.", - "type": "text" - }, - { - "block_id": "p607-b26", - "global_id": 17585, - "bbox": [ - 116.96, - 422.52, - 288.98, - 431.48 - ], - "text": "(a) Show that this transformation converts", - "type": "text" - }, - { - "block_id": "p607-b27", - "global_id": 17586, - "bbox": [ - 116.46, - 433.48, - 288.99, - 464.36 - ], - "text": "lowpass filters into highpass filters and\nhighpass filters into lowpass filters.\n(b) If the original filter is an FIR filter with", - "type": "text" - }, - { - "block_id": "p607-b28", - "global_id": 17587, - "bbox": [ - 131.89, - 465.98, - 288.98, - 486.29 - ], - "text": "impulse response h[n], what is the impulse\nresponse of the transformed filter?", - "type": "text" - }, - { - "block_id": "p607-b29", - "global_id": 17588, - "bbox": [ - 83.34, - 491.2, - 288.97, - 511.21 - ], - "text": "5.6-18\nThe bilinear transformation is defined by the\nrule s = 2(1 −z−1)/T(1 + z−1).", - "type": "text" - }, - { - "block_id": "p607-b30", - "global_id": 17589, - "bbox": [ - 116.96, - 512.83, - 288.8, - 522.16 - ], - "text": "(a) Show that this transformation maps the ω", - "type": "text" - }, - { - "block_id": "p607-b31", - "global_id": 17590, - "bbox": [ - 131.89, - 522.52, - 288.48, - 533.12 - ], - "text": "axis in the s plane to the unit circle z = ej", - "type": "text" - }, - { - "block_id": "p607-b32", - "global_id": 17591, - "bbox": [ - 116.46, - 535.02, - 288.98, - 555.04 - ], - "text": "in the z plane.\n(b) Show that this transformation maps to", - "type": "text" - }, - { - "block_id": "p607-b33", - "global_id": 17592, - "bbox": [ - 131.89, - 556.66, - 188.64, - 566.0 - ], - "text": "2arctan(ωT/2).", - "type": "text" - }, - { - "block_id": "p607-b34", - "global_id": 17593, - "bbox": [ - 87.82, - 570.92, - 288.99, - 634.77 - ], - "text": "5.7-1\nIn Ch. 3, we used another approximation to find\na digital system to realize an analog system.\nWe showed that an analog system specified\nby Eq. (3.12) can be realized by using the\ndigital system specified by Eq. (3.13). Compare\nthat solution with the one resulting from the", - "type": "text" - }, - { - "block_id": "p607-b35", - "global_id": 17594, - "bbox": [ - 343.61, - 85.9, - 516.13, - 116.78 - ], - "text": "impulse-invariance method. Show that one result\nis a close approximation of the other and that the\napproximation improves as T →0.", - "type": "text" - }, - { - "block_id": "p607-b36", - "global_id": 17595, - "bbox": [ - 314.97, - 121.43, - 516.14, - 163.65 - ], - "text": "5.7-2\nA CT system has impulse response hct(t) =\ne−tu(t).\nDraw\nthe\nDFI\nrealization\nof\nthe\ncorresponding DT system designed by the\nimpulse-invariance method with T = 0.1.", - "type": "text" - }, - { - "block_id": "p607-b37", - "global_id": 17596, - "bbox": [ - 314.97, - 168.6, - 516.14, - 199.55 - ], - "text": "5.7-3\n(a) Using\nthe\nimpulse-invariance\ncriterion,\ndesign a digital filter to realize an analog\nfilter with transfer function", - "type": "text" - }, - { - "block_id": "p607-b38", - "global_id": 17597, - "bbox": [ - 393.41, - 207.79, - 480.57, - 229.78 - ], - "text": "Ha(s) =\n7s + 20\n2(s2 + 7s + 10)", - "type": "text" - }, - { - "block_id": "p607-b39", - "global_id": 17598, - "bbox": [ - 343.61, - 239.86, - 516.11, - 248.83 - ], - "text": "(b) Show a canonical and a parallel realization", - "type": "text" - }, - { - "block_id": "p607-b40", - "global_id": 17599, - "bbox": [ - 359.05, - 250.81, - 516.14, - 270.74 - ], - "text": "of the filter. Use a 1% criterion for the\nchoice of T.", - "type": "text" - }, - { - "block_id": "p607-b41", - "global_id": 17600, - "bbox": [ - 314.97, - 275.69, - 516.14, - 306.65 - ], - "text": "5.7-4\nUse the impulse-invariance criterion to design a\ndigital filter to realize the second-order analog\nButterworth filter with transfer function", - "type": "text" - }, - { - "block_id": "p607-b42", - "global_id": 17601, - "bbox": [ - 389.73, - 315.25, - 447.64, - 331.17 - ], - "text": "Ha(s) =\n1", - "type": "text" - }, - { - "block_id": "p607-b43", - "global_id": 17602, - "bbox": [ - 421.99, - 326.2, - 437.6, - 338.04 - ], - "text": "s2 +", - "type": "text" - }, - { - "block_id": "p607-b44", - "global_id": 17603, - "bbox": [ - 438.99, - 321.21, - 446.58, - 330.17 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p607-b45", - "global_id": 17604, - "bbox": [ - 446.58, - 328.79, - 468.82, - 338.13 - ], - "text": "2s + 1", - "type": "text" - }, - { - "block_id": "p607-b46", - "global_id": 17605, - "bbox": [ - 343.61, - 347.52, - 481.99, - 356.57 - ], - "text": "Use a 1% criterion for the choice of T.", - "type": "text" - }, - { - "block_id": "p607-b47", - "global_id": 17606, - "bbox": [ - 314.97, - 361.53, - 516.13, - 436.32 - ], - "text": "5.7-5\nDesign\na\ndigital\nintegrator\nusing\nthe\nimpulse-invariance method. Find and give a\nrough sketch of the amplitude response, and\ncompare it with that of the ideal integrator. If\nthis integrator is used primarily for integrating\naudio signals (whose bandwidth is 20 kHz),\ndetermine a suitable value for T.", - "type": "text" - }, - { - "block_id": "p607-b48", - "global_id": 17607, - "bbox": [ - 314.97, - 441.27, - 516.14, - 537.98 - ], - "text": "5.7-6\nAn oscillator by definition is a source (no input)\nthat generates a sinusoid of a certain frequency\nω0. Therefore, an oscillator is a system whose\nzero-input response is a sinusoid of the desired\nfrequency. Find the transfer function of a digital\noscillator to oscillate at 10 kHz by the methods\ndescribed in parts (a) and (b). In both methods,\nselect T so that there are 10 samples in each\ncycle of the sinusoid.", - "type": "text" - }, - { - "block_id": "p607-b49", - "global_id": 17608, - "bbox": [ - 344.11, - 539.6, - 516.11, - 548.94 - ], - "text": "(a) Choose H[z] directly so that its zero-input", - "type": "text" - }, - { - "block_id": "p607-b50", - "global_id": 17609, - "bbox": [ - 343.61, - 550.94, - 516.14, - 582.49 - ], - "text": "response is a discrete-time sinusoid of\nfrequency = ωT corresponding to 10 kHz.\n(b) Choose Ha(s) whose zero-input response", - "type": "text" - }, - { - "block_id": "p607-b51", - "global_id": 17610, - "bbox": [ - 344.11, - 583.81, - 516.13, - 636.61 - ], - "text": "is an analog sinusoid of 10 kHz. Now use\nthe impulse invariance method to determine\nH[z].\n(c) Show\na\ncanonical\nrealization\nof\nthe\noscillator.", - "type": "text" - } - ] - }, - { - "page_num": 608, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p608-b0", - "global_id": 17611, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "588\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p608-b1", - "global_id": 17612, - "bbox": [ - 62.08, - 85.94, - 263.24, - 149.77 - ], - "text": "5.7-7\nA variant of the impulse invariance method\nis the step-invariance method of digital filter\nsynthesis. In this method, for a given Ha(s), we\ndesign H[z] in Fig. 5.24a such that y(nT) in\nFig. 5.24b is identical to y[n] in Fig. 5.24a when\nx(t) = u(t).", - "type": "text" - }, - { - "block_id": "p608-b2", - "global_id": 17613, - "bbox": [ - 91.22, - 151.77, - 184.37, - 160.74 - ], - "text": "(a) Show that, in general,", - "type": "text" - }, - { - "block_id": "p608-b3", - "global_id": 17614, - "bbox": [ - 120.32, - 171.21, - 166.28, - 186.74 - ], - "text": "H[z] = z −1", - "type": "text" - }, - { - "block_id": "p608-b4", - "global_id": 17615, - "bbox": [ - 155.66, - 177.49, - 174.13, - 193.12 - ], - "text": "z\nZ", - "type": "text" - }, - { - "block_id": "p608-b6", - "global_id": 17616, - "bbox": [ - 185.79, - 171.21, - 222.51, - 186.46 - ], - "text": "L−1 Ha(s)", - "type": "text" - }, - { - "block_id": "p608-b7", - "global_id": 17617, - "bbox": [ - 210.56, - 184.15, - 214.05, - 193.12 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p608-b9", - "global_id": 17618, - "bbox": [ - 229.75, - 188.33, - 243.19, - 195.01 - ], - "text": "t=kT", - "type": "text" - }, - { - "block_id": "p608-b10", - "global_id": 17619, - "bbox": [ - 244.19, - 164.91, - 249.07, - 173.88 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p608-b11", - "global_id": 17620, - "bbox": [ - 90.72, - 203.69, - 230.58, - 213.02 - ], - "text": "(b) Use this method to design H[z] for", - "type": "text" - }, - { - "block_id": "p608-b12", - "global_id": 17621, - "bbox": [ - 156.71, - 220.73, - 211.0, - 243.4 - ], - "text": "Ha(s) =\nωc\ns + ωc", - "type": "text" - }, - { - "block_id": "p608-b13", - "global_id": 17622, - "bbox": [ - 91.22, - 252.45, - 263.24, - 294.29 - ], - "text": "(c) Use\nthe\nstep-invariance\nmethod\nto\nsynthesize a discrete-time integrator and\ncompare its amplitude response with that of\nthe ideal integrator.", - "type": "text" - }, - { - "block_id": "p608-b14", - "global_id": 17623, - "bbox": [ - 62.08, - 299.21, - 263.25, - 352.1 - ], - "text": "5.7-8\nUse the ramp-invariance method to synthesize\na discrete-time differentiator and integrator. In\nthis method, for a given Ha(s), we design H[z]\nsuch that y(nT) in Fig. 5.24b is identical to y[n]\nin Fig. 5.24a when x(t) = tu(t).", - "type": "text" - }, - { - "block_id": "p608-b15", - "global_id": 17624, - "bbox": [ - 62.08, - 357.03, - 263.24, - 398.94 - ], - "text": "5.7-9\nIn an impulse-invariance design, show that if\nHa(s) is a transfer function of a stable system,\nthe corresponding H[z] is also a transfer function\nof a stable system.", - "type": "text" - }, - { - "block_id": "p608-b16", - "global_id": 17625, - "bbox": [ - 57.59, - 403.86, - 263.22, - 423.87 - ], - "text": "5.7-10\nFirst-order backward differences provide the\ntransformation rule s = (1 −z−1)/T.", - "type": "text" - }, - { - "block_id": "p608-b17", - "global_id": 17626, - "bbox": [ - 91.22, - 425.86, - 263.23, - 434.83 - ], - "text": "(a) Show that this transformation maps the", - "type": "text" - }, - { - "block_id": "p608-b18", - "global_id": 17627, - "bbox": [ - 90.72, - 436.44, - 263.23, - 467.71 - ], - "text": "ω axis in the s plane to a circle of radius\n1/2 centered at (1/2,0) in the z plane.\n(b) Show that this transformation maps the", - "type": "text" - }, - { - "block_id": "p608-b19", - "global_id": 17628, - "bbox": [ - 106.15, - 469.6, - 263.23, - 500.58 - ], - "text": "left-half s plane to the interior of the unit\ncircle in the z plane, which ensures that\nstability is preserved.", - "type": "text" - }, - { - "block_id": "p608-b20", - "global_id": 17629, - "bbox": [ - 62.08, - 505.5, - 263.24, - 536.47 - ], - "text": "5.8-1\nFind the z-transform (if it exists) and the\ncorresponding ROC for each of the following\nsignals:", - "type": "text" - }, - { - "block_id": "p608-b21", - "global_id": 17630, - "bbox": [ - 90.72, - 537.03, - 203.62, - 558.38 - ], - "text": "(a) (0.8)nu[n] + 2nu[−(n + 1)]\n(b) 2nu[n] −3nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p608-b22", - "global_id": 17631, - "bbox": [ - 91.22, - 553.28, - 184.04, - 569.34 - ], - "text": "(c) (−2)n+3u[−n] + %∞", - "type": "text" - }, - { - "block_id": "p608-b23", - "global_id": 17632, - "bbox": [ - 90.72, - 558.74, - 252.78, - 580.3 - ], - "text": "k=0(0.5)k−1δ(n −2k)\n(d) (0.8)nu[n] + (0.9)nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p608-b24", - "global_id": 17633, - "bbox": [ - 91.22, - 578.67, - 212.55, - 591.26 - ], - "text": "(e) [(0.8)n + 3(0.4)n]u[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p608-b25", - "global_id": 17634, - "bbox": [ - 90.72, - 589.62, - 221.51, - 624.14 - ], - "text": "(f) [(0.8)n + 3(0.4)n]u[n]\n(g) (0.8)nu[n] + 3(0.4)nu[−(n + 1)]\n(h) (0.5)|n|", - "type": "text" - }, - { - "block_id": "p608-b26", - "global_id": 17635, - "bbox": [ - 92.71, - 625.76, - 154.53, - 635.1 - ], - "text": "(i) nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p608-b27", - "global_id": 17636, - "bbox": [ - 289.22, - 85.94, - 490.38, - 105.94 - ], - "text": "5.8-2\nUsing the definition, compute the bilateral\nz-transform X(z) of", - "type": "text" - }, - { - "block_id": "p608-b28", - "global_id": 17637, - "bbox": [ - 318.37, - 106.3, - 382.01, - 116.89 - ], - "text": "(a) x[n] = 3nu[−n]", - "type": "text" - }, - { - "block_id": "p608-b29", - "global_id": 17638, - "bbox": [ - 317.86, - 117.15, - 366.23, - 127.86 - ], - "text": "(b) x[n] = ( 1", - "type": "text" - }, - { - "block_id": "p608-b30", - "global_id": 17639, - "bbox": [ - 362.99, - 117.25, - 384.8, - 130.44 - ], - "text": "3)nu[n]", - "type": "text" - }, - { - "block_id": "p608-b31", - "global_id": 17640, - "bbox": [ - 317.86, - 129.85, - 489.43, - 138.82 - ], - "text": "Express your answers in standard rational form.", - "type": "text" - }, - { - "block_id": "p608-b32", - "global_id": 17641, - "bbox": [ - 289.22, - 143.42, - 490.39, - 152.76 - ], - "text": "5.8-3\nDetermine the inverse z-transform x[n] of", - "type": "text" - }, - { - "block_id": "p608-b33", - "global_id": 17642, - "bbox": [ - 317.86, - 157.64, - 342.04, - 166.89 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p608-b34", - "global_id": 17643, - "bbox": [ - 351.49, - 152.57, - 366.22, - 160.84 - ], - "text": "z2−1", - "type": "text" - }, - { - "block_id": "p608-b35", - "global_id": 17644, - "bbox": [ - 345.07, - 154.37, - 459.47, - 170.17 - ], - "text": "3 z\n(z−1)(z+2) with ROC 1 < |z| < 2.", - "type": "text" - }, - { - "block_id": "p608-b36", - "global_id": 17645, - "bbox": [ - 289.22, - 171.87, - 428.44, - 180.93 - ], - "text": "5.8-4\nFind the inverse z-transform of", - "type": "text" - }, - { - "block_id": "p608-b37", - "global_id": 17646, - "bbox": [ - 361.34, - 187.65, - 445.73, - 212.89 - ], - "text": "X[z] =\n(e−2 −2)z\n(z −e−2)(z −2)", - "type": "text" - }, - { - "block_id": "p608-b38", - "global_id": 17647, - "bbox": [ - 317.86, - 222.47, - 381.27, - 231.44 - ], - "text": "when the ROC is,", - "type": "text" - }, - { - "block_id": "p608-b39", - "global_id": 17648, - "bbox": [ - 317.86, - 233.06, - 380.53, - 253.36 - ], - "text": "(a) |z| > 2\n(b) e−2 < |z| < 2", - "type": "text" - }, - { - "block_id": "p608-b40", - "global_id": 17649, - "bbox": [ - 318.37, - 253.72, - 364.87, - 264.32 - ], - "text": "(c) |z| < e−2", - "type": "text" - }, - { - "block_id": "p608-b41", - "global_id": 17650, - "bbox": [ - 289.22, - 269.21, - 490.4, - 300.18 - ], - "text": "5.8-5\nUse partial fraction expansions, z-transform\ntables, and a region of convergence (|z| < 1/2)\nto determine the inverse z-transform of", - "type": "text" - }, - { - "block_id": "p608-b42", - "global_id": 17651, - "bbox": [ - 349.82, - 308.97, - 419.73, - 324.13 - ], - "text": "X(z) =\n1", - "type": "text" - }, - { - "block_id": "p608-b43", - "global_id": 17652, - "bbox": [ - 377.74, - 322.12, - 431.12, - 331.45 - ], - "text": "(2z + 1)(z + 1)", - "type": "text" - }, - { - "block_id": "p608-b45", - "global_id": 17653, - "bbox": [ - 434.73, - 320.76, - 452.43, - 331.36 - ], - "text": "z + 1", - "type": "text" - }, - { - "block_id": "p608-b46", - "global_id": 17654, - "bbox": [ - 449.19, - 314.91, - 457.24, - 334.04 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p608-b47", - "global_id": 17655, - "bbox": [ - 289.22, - 342.69, - 490.38, - 373.67 - ], - "text": "5.8-6\nUsing z-transform techniques and properties (no\ntime-domain convolution sum!), determine the\nconvolution y[n] = ( 1", - "type": "text" - }, - { - "block_id": "p608-b48", - "global_id": 17656, - "bbox": [ - 317.86, - 363.06, - 490.39, - 396.32 - ], - "text": "3)n−3u[n −2] ∗(2)nu[−n].\nExpress your answer in the form y[n] =\nc1γ n", - "type": "text" - }, - { - "block_id": "p608-b49", - "global_id": 17657, - "bbox": [ - 317.86, - 385.06, - 490.39, - 418.23 - ], - "text": "1 u[n + N1] + c2γ n\n2 u[−n + N2], making sure\nto clearly identify the constants c1, c2, γ1, γ2,\nN1, and N2.", - "type": "text" - }, - { - "block_id": "p608-b50", - "global_id": 17658, - "bbox": [ - 289.22, - 422.39, - 490.4, - 453.37 - ], - "text": "5.8-7\nUsing partial fraction expansions, z-transform\ntables, and the fact that h[n] is stable, determine\nthe inverse z-transform of", - "type": "text" - }, - { - "block_id": "p608-b51", - "global_id": 17659, - "bbox": [ - 363.28, - 460.09, - 428.14, - 478.88 - ], - "text": "H[z] =\nz4 + z3", - "type": "text" - }, - { - "block_id": "p608-b52", - "global_id": 17660, - "bbox": [ - 391.5, - 475.5, - 436.99, - 486.19 - ], - "text": "(z −2)(z + 1", - "type": "text" - }, - { - "block_id": "p608-b53", - "global_id": 17661, - "bbox": [ - 433.76, - 470.0, - 444.97, - 488.78 - ], - "text": "2)\n.", - "type": "text" - }, - { - "block_id": "p608-b54", - "global_id": 17662, - "bbox": [ - 289.22, - 497.45, - 390.59, - 506.49 - ], - "text": "5.8-8\nConsider the system", - "type": "text" - }, - { - "block_id": "p608-b55", - "global_id": 17663, - "bbox": [ - 372.87, - 518.28, - 406.32, - 534.13 - ], - "text": "H[z] = z", - "type": "text" - }, - { - "block_id": "p608-b57", - "global_id": 17664, - "bbox": [ - 409.93, - 516.64, - 427.63, - 527.25 - ], - "text": "z −1", - "type": "text" - }, - { - "block_id": "p608-b58", - "global_id": 17665, - "bbox": [ - 424.39, - 510.79, - 432.45, - 529.92 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p608-b60", - "global_id": 17666, - "bbox": [ - 404.69, - 529.51, - 429.38, - 541.35 - ], - "text": "z3 −27", - "type": "text" - }, - { - "block_id": "p608-b61", - "global_id": 17667, - "bbox": [ - 424.52, - 524.89, - 434.18, - 544.03 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p608-b62", - "global_id": 17668, - "bbox": [ - 318.37, - 556.68, - 490.38, - 566.02 - ], - "text": "(a) Draw the pole-zero diagram for H[z] and", - "type": "text" - }, - { - "block_id": "p608-b63", - "global_id": 17669, - "bbox": [ - 317.86, - 568.01, - 490.38, - 587.94 - ], - "text": "identify all possible regions of convergence.\n(b) Draw the pole-zero diagram for H−1[z] and", - "type": "text" - }, - { - "block_id": "p608-b64", - "global_id": 17670, - "bbox": [ - 333.3, - 589.93, - 490.36, - 598.9 - ], - "text": "identify all possible regions of convergence.", - "type": "text" - }, - { - "block_id": "p608-b65", - "global_id": 17671, - "bbox": [ - 289.22, - 603.5, - 490.38, - 635.5 - ], - "text": "5.8-9\nA discrete-time signal x[n] has a rational\nz-transform that contains a pole at z = 0.5. Given\nx1[n] = (1/3)nx[n] is absolutely summable and", - "type": "text" - } - ] - }, - { - "page_num": 609, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p609-b0", - "global_id": 17672, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n589", - "type": "text" - }, - { - "block_id": "p609-b1", - "global_id": 17673, - "bbox": [ - 116.46, - 84.52, - 288.98, - 116.83 - ], - "text": "x2[n] = (1/4)nx[n] is not absolutely summable,\ndetermine whether x[n] is left-sided, right-sided,\nor two-sided. Justify your answer!", - "type": "text" - }, - { - "block_id": "p609-b2", - "global_id": 17674, - "bbox": [ - 83.34, - 121.56, - 288.98, - 174.74 - ], - "text": "5.8-10\nLet x[n] be an absolutely summable signal with\nrational z-transform X[z]. X[z] is known to have\na pole at z = (0.75+0.75j), and other poles may\nbe present. Recall that an absolutely summable\nsignal satisfies %∞", - "type": "text" - }, - { - "block_id": "p609-b3", - "global_id": 17675, - "bbox": [ - 116.46, - 165.4, - 251.21, - 196.65 - ], - "text": "−∞|x[n]| < ∞.\n(a) Can x[n] be left-sided? Explain.\n(b) Can x[n] be right-sided? Explain.", - "type": "text" - }, - { - "block_id": "p609-b4", - "global_id": 17676, - "bbox": [ - 116.46, - 198.27, - 272.93, - 218.57 - ], - "text": "(c) Can x[n] be two-sided? Explain.\n(d) Can x[n] be of finite duration? Explain.", - "type": "text" - }, - { - "block_id": "p609-b5", - "global_id": 17677, - "bbox": [ - 83.34, - 223.6, - 288.97, - 243.59 - ], - "text": "5.8-11\nConsider a causal system that has transfer\nfunction", - "type": "text" - }, - { - "block_id": "p609-b6", - "global_id": 17678, - "bbox": [ - 175.78, - 243.19, - 228.48, - 258.73 - ], - "text": "H[z] = z −0.5", - "type": "text" - }, - { - "block_id": "p609-b7", - "global_id": 17679, - "bbox": [ - 203.99, - 255.84, - 228.47, - 265.18 - ], - "text": "z + 0.5", - "type": "text" - }, - { - "block_id": "p609-b8", - "global_id": 17680, - "bbox": [ - 116.46, - 272.21, - 288.97, - 292.13 - ], - "text": "When appropriate, assume initial conditions of\nzero.", - "type": "text" - }, - { - "block_id": "p609-b9", - "global_id": 17681, - "bbox": [ - 116.96, - 293.76, - 288.98, - 303.84 - ], - "text": "(a) Determine the output y1[n] of this system in", - "type": "text" - }, - { - "block_id": "p609-b10", - "global_id": 17682, - "bbox": [ - 116.46, - 303.66, - 288.98, - 325.75 - ], - "text": "response to x1[n] = (3/4)nu[n].\n(b) Determine the output y2[n] of this system in", - "type": "text" - }, - { - "block_id": "p609-b11", - "global_id": 17683, - "bbox": [ - 116.96, - 325.57, - 288.98, - 347.67 - ], - "text": "response to x2[n] = (3/4)n.\n(c) Determine the output y3[n] of this system in", - "type": "text" - }, - { - "block_id": "p609-b12", - "global_id": 17684, - "bbox": [ - 131.89, - 347.5, - 265.63, - 358.63 - ], - "text": "response to x3[n] = (3/4)nu[−n −1].", - "type": "text" - }, - { - "block_id": "p609-b13", - "global_id": 17685, - "bbox": [ - 83.34, - 361.55, - 288.98, - 404.83 - ], - "text": "5.8-12\nLet x[n] = (−1)nu[n−n0]+αnu[−n]. Determine\nthe constraints on the complex number α and the\ninteger n0 so that the z-transform X[z] exists with\nregion of convergence 1 < |z| < 2.", - "type": "text" - }, - { - "block_id": "p609-b14", - "global_id": 17686, - "bbox": [ - 83.34, - 409.86, - 288.98, - 451.78 - ], - "text": "5.8-13\nUsing the definition, compute the bilateral\nz-transform, including the region of convergence\n(ROC),\nof\nthe\nfollowing\ncomplex-valued\nfunctions:", - "type": "text" - }, - { - "block_id": "p609-b15", - "global_id": 17687, - "bbox": [ - 116.46, - 452.13, - 239.09, - 474.43 - ], - "text": "(a) x1[n] = (−j)−nu[−n] + δ[−n]\n(b) x2[n] = (j)n cos(n + 1)u[n]", - "type": "text" - }, - { - "block_id": "p609-b16", - "global_id": 17688, - "bbox": [ - 116.46, - 475.31, - 227.55, - 496.35 - ], - "text": "(c) x3[n] = jsinh(n)u[−n + 1]\n(d) x4[n] = %0", - "type": "text" - }, - { - "block_id": "p609-b17", - "global_id": 17689, - "bbox": [ - 170.12, - 485.21, - 240.78, - 497.5 - ], - "text": "k=−∞(2j)nδ[n −2k]", - "type": "text" - }, - { - "block_id": "p609-b18", - "global_id": 17690, - "bbox": [ - 83.34, - 500.62, - 289.0, - 531.59 - ], - "text": "5.8-14\nUse partial fraction expansions, z-transform\ntables, and a region of convergence (0.5 < |z| <\n2) to determine the inverse z-transform of", - "type": "text" - }, - { - "block_id": "p609-b19", - "global_id": 17691, - "bbox": [ - 116.97, - 531.29, - 210.08, - 547.28 - ], - "text": "(a) X1[z] =\n1", - "type": "text" - }, - { - "block_id": "p609-b20", - "global_id": 17692, - "bbox": [ - 162.45, - 541.84, - 252.74, - 556.87 - ], - "text": "1 + 13\n6 z−1 + 1\n6z−2 −1\n3z−3", - "type": "text" - }, - { - "block_id": "p609-b21", - "global_id": 17693, - "bbox": [ - 83.34, - 555.16, - 289.0, - 610.57 - ], - "text": "(b) X2[z] =\n1\nz−3(2 −z−1)(1 + 2z−1)\n5.8-15\nUse partial fraction expansions, z-transform\ntables, and the fact that the systems are stable\nto determine the inverse z-transform of", - "type": "text" - }, - { - "block_id": "p609-b22", - "global_id": 17694, - "bbox": [ - 116.97, - 609.4, - 201.84, - 627.03 - ], - "text": "(a) H1[z] =\nz−1", - "type": "text" - }, - { - "block_id": "p609-b23", - "global_id": 17695, - "bbox": [ - 167.06, - 622.82, - 184.76, - 633.43 - ], - "text": "z −1", - "type": "text" - }, - { - "block_id": "p609-b24", - "global_id": 17696, - "bbox": [ - 181.52, - 616.97, - 193.18, - 636.1 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p609-b25", - "global_id": 17697, - "bbox": [ - 193.18, - 616.97, - 228.96, - 636.1 - ], - "text": "1 + 1\n2z−1", - "type": "text" - }, - { - "block_id": "p609-b26", - "global_id": 17698, - "bbox": [ - 343.61, - 85.52, - 428.36, - 101.88 - ], - "text": "(b) H2[z] =\nz + 1", - "type": "text" - }, - { - "block_id": "p609-b27", - "global_id": 17699, - "bbox": [ - 390.59, - 98.71, - 422.27, - 108.38 - ], - "text": "z3(z −2)", - "type": "text" - }, - { - "block_id": "p609-b29", - "global_id": 17700, - "bbox": [ - 425.87, - 97.69, - 443.57, - 108.29 - ], - "text": "z + 1", - "type": "text" - }, - { - "block_id": "p609-b30", - "global_id": 17701, - "bbox": [ - 440.33, - 91.83, - 448.38, - 110.97 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p609-b31", - "global_id": 17702, - "bbox": [ - 310.48, - 112.51, - 516.13, - 132.81 - ], - "text": "5.8-16\nBy inserting N −1 zeros between every sample\nof a unit step, we obtain a signal", - "type": "text" - }, - { - "block_id": "p609-b32", - "global_id": 17703, - "bbox": [ - 392.94, - 152.46, - 416.71, - 161.71 - ], - "text": "h[n] =", - "type": "text" - }, - { - "block_id": "p609-b33", - "global_id": 17704, - "bbox": [ - 418.55, - 143.15, - 431.24, - 152.91 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p609-b34", - "global_id": 17705, - "bbox": [ - 419.27, - 165.38, - 430.53, - 172.13 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p609-b35", - "global_id": 17706, - "bbox": [ - 432.23, - 152.46, - 466.8, - 161.71 - ], - "text": "δ[n −Nk]", - "type": "text" - }, - { - "block_id": "p609-b36", - "global_id": 17707, - "bbox": [ - 343.61, - 182.3, - 516.13, - 213.55 - ], - "text": "Determine H[z], the bilateral z-transform of\nh[n]. Identify the number and location(s) of the\npoles of H[z].", - "type": "text" - }, - { - "block_id": "p609-b37", - "global_id": 17708, - "bbox": [ - 310.48, - 218.54, - 516.14, - 249.41 - ], - "text": "5.8-17\nUsing transform-domain techniques, determine\nthe zero-state response yzsr[n] of LTID system\ny[n] −1", - "type": "text" - }, - { - "block_id": "p609-b38", - "global_id": 17709, - "bbox": [ - 343.61, - 240.16, - 516.13, - 260.45 - ], - "text": "4y[n −2] = x[n] to the noncausal input\nx[n] = 2nu[2 −n].", - "type": "text" - }, - { - "block_id": "p609-b39", - "global_id": 17710, - "bbox": [ - 310.48, - 265.43, - 516.13, - 285.44 - ], - "text": "5.8-18\nDetermine the zero-state response of a system\nhaving a transfer function", - "type": "text" - }, - { - "block_id": "p609-b40", - "global_id": 17711, - "bbox": [ - 359.78, - 293.98, - 499.96, - 315.69 - ], - "text": "H[z] =\nz\n(z + 0.2)(z −0.8)\n|z| > 0.8", - "type": "text" - }, - { - "block_id": "p609-b41", - "global_id": 17712, - "bbox": [ - 343.61, - 325.74, - 437.65, - 335.08 - ], - "text": "and an input x[n] given by", - "type": "text" - }, - { - "block_id": "p609-b42", - "global_id": 17713, - "bbox": [ - 343.61, - 335.64, - 435.3, - 357.0 - ], - "text": "(a) x[n] = enu[n]\n(b) x[n] = 2nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p609-b43", - "global_id": 17714, - "bbox": [ - 343.61, - 357.56, - 467.74, - 378.91 - ], - "text": "(c) x[n] = enu[n] + 2nu[−(n + 1)]\n(d) x[n] = 2nu[n] + u[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p609-b44", - "global_id": 17715, - "bbox": [ - 344.12, - 379.27, - 443.09, - 389.87 - ], - "text": "(e) x[n] = e−2nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p609-b45", - "global_id": 17716, - "bbox": [ - 310.48, - 394.85, - 516.11, - 414.86 - ], - "text": "5.8-19\nThe discrete cross-correlation between real\nsignal x[n] and real signal y[n] is", - "type": "text" - }, - { - "block_id": "p609-b46", - "global_id": 17717, - "bbox": [ - 382.61, - 434.51, - 412.13, - 444.52 - ], - "text": "cxy[n] =", - "type": "text" - }, - { - "block_id": "p609-b47", - "global_id": 17718, - "bbox": [ - 417.47, - 425.2, - 430.16, - 434.96 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p609-b48", - "global_id": 17719, - "bbox": [ - 413.97, - 447.35, - 433.66, - 454.02 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p609-b49", - "global_id": 17720, - "bbox": [ - 434.66, - 434.51, - 477.13, - 443.76 - ], - "text": "x[k]y[k −n]", - "type": "text" - }, - { - "block_id": "p609-b50", - "global_id": 17721, - "bbox": [ - 343.61, - 464.71, - 516.14, - 517.89 - ], - "text": "Let signal x[n] have z-transform X[z] with\nROC Rx, and let signal y[n] have z-transform\nY[z] with ROC Ry. Determine CXY[z] (the\nbilateral z-transform of cxy[n]) in terms of the\nz-transforms of x[n] and y[n].", - "type": "text" - }, - { - "block_id": "p609-b51", - "global_id": 17722, - "bbox": [ - 310.48, - 522.87, - 516.14, - 542.87 - ], - "text": "5.8-20\nTransform properties can be very useful. The\naccumulation property states", - "type": "text" - }, - { - "block_id": "p609-b52", - "global_id": 17723, - "bbox": [ - 387.78, - 553.23, - 400.47, - 562.79 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p609-b53", - "global_id": 17724, - "bbox": [ - 384.28, - 575.17, - 403.97, - 581.85 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p609-b54", - "global_id": 17725, - "bbox": [ - 404.97, - 556.34, - 475.47, - 578.05 - ], - "text": "x[k] ⇐⇒\nz\nz −1X[z]", - "type": "text" - }, - { - "block_id": "p609-b55", - "global_id": 17726, - "bbox": [ - 343.61, - 592.92, - 516.13, - 634.77 - ], - "text": "This property is the DT version of the Laplace\ntransform “integration in time” property. Given\nx[n] ⇐⇒X[z], prove the accumulation property.\n[Hint: Polynomial long division can be helpful.]", - "type": "text" - } - ] - }, - { - "page_num": 610, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p610-b0", - "global_id": 17727, - "bbox": [ - 60.0, - 59.76, - 433.19, - 69.45 - ], - "text": "590\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p610-b1", - "global_id": 17728, - "bbox": [ - 57.59, - 85.94, - 263.23, - 116.9 - ], - "text": "5.10-1\nUse MATLAB to generate pole-zero plots for\nthe causal systems with the following transfer\nfunctions:", - "type": "text" - }, - { - "block_id": "p610-b2", - "global_id": 17729, - "bbox": [ - 91.22, - 115.56, - 148.14, - 128.6 - ], - "text": "(a) Ha[z] = z4−", - "type": "text" - }, - { - "block_id": "p610-b3", - "global_id": 17730, - "bbox": [ - 148.14, - 111.41, - 153.62, - 117.89 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p610-b4", - "global_id": 17731, - "bbox": [ - 139.82, - 115.56, - 170.91, - 131.23 - ], - "text": "2z2+1\nz4+0.4096", - "type": "text" - }, - { - "block_id": "p610-b5", - "global_id": 17732, - "bbox": [ - 90.72, - 140.79, - 134.66, - 150.87 - ], - "text": "(b) Hb[z] =", - "type": "text" - }, - { - "block_id": "p610-b6", - "global_id": 17733, - "bbox": [ - 139.32, - 135.77, - 166.62, - 144.11 - ], - "text": "−3z−1+ 3", - "type": "text" - }, - { - "block_id": "p610-b7", - "global_id": 17734, - "bbox": [ - 163.88, - 135.82, - 177.36, - 146.6 - ], - "text": "4 z−3", - "type": "text" - }, - { - "block_id": "p610-b8", - "global_id": 17735, - "bbox": [ - 90.72, - 145.92, - 263.24, - 175.48 - ], - "text": "3+ 3\n2 z−2+2z−4\nIn each case, determine whether the system is\nstable.", - "type": "text" - }, - { - "block_id": "p610-b9", - "global_id": 17736, - "bbox": [ - 57.59, - 188.5, - 263.25, - 219.46 - ], - "text": "5.10-2\nFor each of the following stable LTID systems,\nuse MATLAB to generate magnitude and phase\nresponse plots over −π ≤ < π.", - "type": "text" - }, - { - "block_id": "p610-b10", - "global_id": 17737, - "bbox": [ - 91.22, - 219.09, - 153.32, - 231.16 - ], - "text": "(a) Ha[z] = cos(z)", - "type": "text" - }, - { - "block_id": "p610-b11", - "global_id": 17738, - "bbox": [ - 137.49, - 226.25, - 153.16, - 233.0 - ], - "text": "z−0.5", - "type": "text" - }, - { - "block_id": "p610-b12", - "global_id": 17739, - "bbox": [ - 90.72, - 235.58, - 174.15, - 248.53 - ], - "text": "(b) Hb[z] = z3 sin(z−1)", - "type": "text" - }, - { - "block_id": "p610-b13", - "global_id": 17740, - "bbox": [ - 57.59, - 260.82, - 263.25, - 335.61 - ], - "text": "5.10-3\nConsider an LTID system described by the\ndifference equation 4y[n + 2] −y[n] = x[n +\n2] + x[n].\n(a) Plot the pole-zero diagram for this system.\n(b) Plot\nthe\nsystem’s\nmagnitude\nresponse\n|H[ej]| over −π ≤ ≤π.\n(c) What type of system is this: lowpass,", - "type": "text" - }, - { - "block_id": "p610-b14", - "global_id": 17741, - "bbox": [ - 90.72, - 337.6, - 256.31, - 357.53 - ], - "text": "highpass, bandpass, or bandstop?\n(d) Is this system stable? Justify your answer.", - "type": "text" - }, - { - "block_id": "p610-b15", - "global_id": 17742, - "bbox": [ - 91.22, - 359.52, - 248.83, - 368.49 - ], - "text": "(e) Is this system real? Justify your answer.", - "type": "text" - }, - { - "block_id": "p610-b16", - "global_id": 17743, - "bbox": [ - 92.21, - 370.11, - 263.24, - 379.44 - ], - "text": "(f) If the system input is of the form x[n] =", - "type": "text" - }, - { - "block_id": "p610-b17", - "global_id": 17744, - "bbox": [ - 90.72, - 381.07, - 263.24, - 423.28 - ], - "text": "cos(n), what is the greatest possible\namplitude of the output? Justify your\nanswer.\n(g) Draw an efficient, causal implementation of", - "type": "text" - }, - { - "block_id": "p610-b18", - "global_id": 17745, - "bbox": [ - 106.16, - 425.28, - 263.22, - 445.2 - ], - "text": "this system using only add, scale, and delay\nblocks.", - "type": "text" - }, - { - "block_id": "p610-b19", - "global_id": 17746, - "bbox": [ - 57.59, - 458.23, - 263.23, - 480.64 - ], - "text": "5.10-4\nConsider an LTID system with system function\nH(z) = b0 z2+1", - "type": "text" - }, - { - "block_id": "p610-b20", - "global_id": 17747, - "bbox": [ - 91.22, - 470.93, - 263.24, - 493.63 - ], - "text": "z2−4/9.\n(a) Determine the constant b0 so that the system", - "type": "text" - }, - { - "block_id": "p610-b21", - "global_id": 17748, - "bbox": [ - 90.72, - 493.92, - 258.83, - 514.21 - ], - "text": "frequency response at = −π is −1.\n(b) Plot the pole-zero diagram for this system.", - "type": "text" - }, - { - "block_id": "p610-b22", - "global_id": 17749, - "bbox": [ - 90.72, - 516.21, - 263.25, - 569.01 - ], - "text": "(c) Plot\nthe\nsystem’s\nmagnitude\nresponse\n|H[ej]| over −π ≤ ≤π.\n(d) Plot the system’s phase response̸\nH[ej]\nover −π ≤ ≤π.\n(e) What type of system is this: lowpass,", - "type": "text" - }, - { - "block_id": "p610-b23", - "global_id": 17750, - "bbox": [ - 92.21, - 571.0, - 263.24, - 590.93 - ], - "text": "highpass, bandpass, or bandstop?\n(f) Determine the response y[n] to the input", - "type": "text" - }, - { - "block_id": "p610-b24", - "global_id": 17751, - "bbox": [ - 106.16, - 589.29, - 263.24, - 601.89 - ], - "text": "x[n] = (−1 −j) + (−j)n + (1 −j)cos(πn +", - "type": "text" - }, - { - "block_id": "p610-b25", - "global_id": 17752, - "bbox": [ - 90.72, - 602.14, - 263.22, - 623.8 - ], - "text": "1\n3).\n(g) Draw a TDFII block diagram representation", - "type": "text" - }, - { - "block_id": "p610-b26", - "global_id": 17753, - "bbox": [ - 106.15, - 625.8, - 158.21, - 634.76 - ], - "text": "of this system.", - "type": "text" - }, - { - "block_id": "p610-b27", - "global_id": 17754, - "bbox": [ - 284.74, - 85.94, - 490.39, - 116.9 - ], - "text": "5.10-5\nConsider\nthe\nLTID\nsystem\nshown\nin\nFig. P5.10-5, where parameters c1 and c2 are\narbitrary constants.", - "type": "text" - }, - { - "block_id": "p610-b28", - "global_id": 17755, - "bbox": [ - 299.32, - 138.96, - 482.81, - 149.58 - ], - "text": "x[n]\nz−1\n\nz−1\ny[n]", - "type": "text" - }, - { - "block_id": "p610-b29", - "global_id": 17756, - "bbox": [ - 350.89, - 158.99, - 430.11, - 168.79 - ], - "text": "c1\nc2", - "type": "text" - }, - { - "block_id": "p610-b30", - "global_id": 17757, - "bbox": [ - 297.94, - 184.53, - 354.06, - 193.49 - ], - "text": "Figure P5.10-5", - "type": "text" - }, - { - "block_id": "p610-b31", - "global_id": 17758, - "bbox": [ - 317.86, - 212.57, - 490.39, - 243.83 - ], - "text": "(a) Determine\nthe\nsystem\nfunction\nH[z],\nexpressed in standard rational form.\n(b) What is the order N of this system?", - "type": "text" - }, - { - "block_id": "p610-b32", - "global_id": 17759, - "bbox": [ - 318.37, - 245.73, - 490.38, - 254.78 - ], - "text": "(c) Determine the N poles and N zeros of", - "type": "text" - }, - { - "block_id": "p610-b33", - "global_id": 17760, - "bbox": [ - 317.86, - 256.78, - 490.39, - 287.66 - ], - "text": "this system. Use MATLAB to create the\ncorresponding pole-zero plot.\n(d) What constraints, if any, exist on parameters", - "type": "text" - }, - { - "block_id": "p610-b34", - "global_id": 17761, - "bbox": [ - 318.37, - 289.57, - 490.39, - 364.38 - ], - "text": "c1 and c2 to ensure that the system is stable?\n(e) Determine\nc1\nand\nc2\nso\nthat\nthis\nsystem functions as an LPF with narrow\npassband. Use MATLAB to generate the\ncorresponding magnitude response |H[ej]|\nover −π ≤ ≤π.\n(f) Ms. Zeroine, the heroine of DT systems,", - "type": "text" - }, - { - "block_id": "p610-b35", - "global_id": 17762, - "bbox": [ - 317.86, - 365.99, - 490.39, - 409.18 - ], - "text": "believes that if x[n] = 0, then y[n] = 0 also.\nIs Ms. Zeroine correct? Fully justify your\nanswer.\n(g) Dr. Strange suggests that by setting c1 = −1", - "type": "text" - }, - { - "block_id": "p610-b36", - "global_id": 17763, - "bbox": [ - 333.31, - 409.83, - 490.4, - 441.08 - ], - "text": "and c2 = −2, the system will act as a\nhighpass filter. Is Dr. Strange right? Fully\njustify your answer.", - "type": "text" - }, - { - "block_id": "p610-b37", - "global_id": 17764, - "bbox": [ - 284.74, - 446.53, - 490.39, - 554.21 - ], - "text": "5.10-6\nOne\ninteresting\nand\nuseful\napplication\nof\ndiscrete systems is the implementation of\ncomplex (rather than real) systems. A complex\nsystem is one in which a real-valued input can\nproduce a complex-valued output. Complex\nsystems\nthat\nare\ndescribed\nby\nconstant\ncoefficient difference equations require at least\none complex-valued coefficient, and they are\ncapable of operating on complex-valued inputs.\nConsider the complex discrete-time system", - "type": "text" - }, - { - "block_id": "p610-b38", - "global_id": 17765, - "bbox": [ - 367.18, - 567.61, - 439.38, - 592.86 - ], - "text": "H[z] =\nz2 −j\nz −0.9ej3π/4", - "type": "text" - }, - { - "block_id": "p610-b39", - "global_id": 17766, - "bbox": [ - 318.37, - 614.84, - 490.38, - 623.8 - ], - "text": "(a) Determine and plot the system zeros and", - "type": "text" - }, - { - "block_id": "p610-b40", - "global_id": 17767, - "bbox": [ - 333.31, - 625.8, - 354.48, - 634.76 - ], - "text": "poles.", - "type": "text" - } - ] - }, - { - "page_num": 611, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p611-b0", - "global_id": 17768, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n591", - "type": "text" - }, - { - "block_id": "p611-b1", - "global_id": 17769, - "bbox": [ - 171.64, - 104.26, - 414.52, - 113.74 - ], - "text": "x[n]\ny[n]\nz–1\nz–1\nz–1", - "type": "text" - }, - { - "block_id": "p611-b2", - "global_id": 17770, - "bbox": [ - 326.64, - 132.71, - 485.49, - 141.68 - ], - "text": "1/4\nFigure P5.10-9", - "type": "text" - }, - { - "block_id": "p611-b3", - "global_id": 17771, - "bbox": [ - 116.46, - 155.66, - 288.98, - 166.25 - ], - "text": "(b) Sketch the magnitude response |H[ejω]| of", - "type": "text" - }, - { - "block_id": "p611-b4", - "global_id": 17772, - "bbox": [ - 131.89, - 167.87, - 288.98, - 188.18 - ], - "text": "this system over −2π ≤ω ≤2π. Comment\non the system’s behavior.", - "type": "text" - }, - { - "block_id": "p611-b5", - "global_id": 17773, - "bbox": [ - 83.34, - 193.08, - 222.17, - 202.13 - ], - "text": "5.10-7\nConsider the complex system", - "type": "text" - }, - { - "block_id": "p611-b6", - "global_id": 17774, - "bbox": [ - 164.83, - 203.17, - 239.4, - 228.41 - ], - "text": "H[z] =\nz4 −1\n2(z2 + 0.81j)", - "type": "text" - }, - { - "block_id": "p611-b7", - "global_id": 17775, - "bbox": [ - 116.46, - 235.01, - 288.97, - 254.94 - ], - "text": "Refer to Prob. 5.10-6 for an introduction to\ncomplex systems.", - "type": "text" - }, - { - "block_id": "p611-b8", - "global_id": 17776, - "bbox": [ - 116.46, - 256.56, - 288.98, - 298.78 - ], - "text": "(a) Plot the pole-zero diagram for H[z].\n(b) Plot\nthe\nsystem’s\nmagnitude\nresponse\n|H[ej]| over −π ≤ ≤π.\n(c) Explain why H[z] is a noncausal system. Do", - "type": "text" - }, - { - "block_id": "p611-b9", - "global_id": 17777, - "bbox": [ - 116.46, - 300.76, - 288.98, - 342.61 - ], - "text": "not give a general definition of causality;\nspecifically identify what makes this system\nnoncausal.\n(d) One way to make this system causal is to", - "type": "text" - }, - { - "block_id": "p611-b10", - "global_id": 17778, - "bbox": [ - 131.89, - 344.23, - 240.8, - 353.57 - ], - "text": "add two poles to H[z]. That is,", - "type": "text" - }, - { - "block_id": "p611-b11", - "global_id": 17779, - "bbox": [ - 154.99, - 356.68, - 264.72, - 378.21 - ], - "text": "Hcausal[z] = H[z]\n1\n(z −a)(z −b)", - "type": "text" - }, - { - "block_id": "p611-b12", - "global_id": 17780, - "bbox": [ - 116.96, - 384.45, - 288.98, - 416.97 - ], - "text": "Find poles a and b such that |Hcausal[ej]| =\n|H[ej]|.\n(e) Draw an efficient block implementation of", - "type": "text" - }, - { - "block_id": "p611-b13", - "global_id": 17781, - "bbox": [ - 131.89, - 418.6, - 166.76, - 428.67 - ], - "text": "Hcausal[z].", - "type": "text" - }, - { - "block_id": "p611-b14", - "global_id": 17782, - "bbox": [ - 83.34, - 432.84, - 288.97, - 452.84 - ], - "text": "5.10-8\nA\ndiscrete-time\nLTI\nsystem\nis\nshown\nin\nFig. P5.10-8.", - "type": "text" - }, - { - "block_id": "p611-b15", - "global_id": 17783, - "bbox": [ - 116.96, - 454.83, - 288.98, - 463.8 - ], - "text": "(a) Determine the difference equation that", - "type": "text" - }, - { - "block_id": "p611-b16", - "global_id": 17784, - "bbox": [ - 116.46, - 465.79, - 288.99, - 485.72 - ], - "text": "describes this system.\n(b) Determine the magnitude response |H[ej]|", - "type": "text" - }, - { - "block_id": "p611-b17", - "global_id": 17785, - "bbox": [ - 116.96, - 487.71, - 288.99, - 551.47 - ], - "text": "for this system and simplify your answer.\nPlot the magnitude response over −π ≤ ≤\nπ. What type of standard filter (lowpass,\nhighpass,\nbandpass,\nor\nbandstop)\nbest\ndescribes this system?\n(c) Determine the impulse response h[n] of this", - "type": "text" - }, - { - "block_id": "p611-b18", - "global_id": 17786, - "bbox": [ - 131.89, - 553.47, - 159.05, - 562.43 - ], - "text": "system.", - "type": "text" - }, - { - "block_id": "p611-b19", - "global_id": 17787, - "bbox": [ - 89.77, - 580.1, - 286.65, - 588.18 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p611-b20", - "global_id": 17788, - "bbox": [ - 156.77, - 599.64, - 224.21, - 616.95 - ], - "text": "z–1\nz–1\n1/2", - "type": "text" - }, - { - "block_id": "p611-b22", - "global_id": 17789, - "bbox": [ - 86.57, - 624.88, - 142.69, - 633.84 - ], - "text": "Figure P5.10-8", - "type": "text" - }, - { - "block_id": "p611-b23", - "global_id": 17790, - "bbox": [ - 310.48, - 155.57, - 516.13, - 186.82 - ], - "text": "5.10-9\nDetermine the impulse response h[n] for the\nsystem shown in Fig. P5.10-9. Is the system\nstable? Is the system causal?", - "type": "text" - }, - { - "block_id": "p611-b24", - "global_id": 17791, - "bbox": [ - 306.0, - 193.61, - 516.14, - 257.45 - ], - "text": "5.10-10\nAn LTID filter has an impulse response function\ngiven by h[n] = δ[n −1] + δ[n + 1]. Determine\nand carefully sketch the magnitude response\n|H[ej]| over the range −π ≤ ≤π. For\nthis range of frequencies, is this filter lowpass,\nhighpass, bandpass, or bandstop?", - "type": "text" - }, - { - "block_id": "p611-b25", - "global_id": 17792, - "bbox": [ - 306.0, - 264.24, - 516.11, - 284.24 - ], - "text": "5.10-11\nA causal, stable discrete system has the rather\nstrange transfer function H[z] = cos(z−1).", - "type": "text" - }, - { - "block_id": "p611-b26", - "global_id": 17793, - "bbox": [ - 344.11, - 286.24, - 516.13, - 295.2 - ], - "text": "(a) Write MATLAB code that will compute", - "type": "text" - }, - { - "block_id": "p611-b27", - "global_id": 17794, - "bbox": [ - 343.61, - 297.19, - 516.13, - 339.03 - ], - "text": "and plot the magnitude response of this\nsystem over an appropriate range of digital\nfrequencies . Comment on the system.\n(b) Determine the impulse response h[n]. Plot", - "type": "text" - }, - { - "block_id": "p611-b28", - "global_id": 17795, - "bbox": [ - 344.12, - 340.66, - 516.12, - 360.96 - ], - "text": "h[n] over (0 ≤n ≤10).\n(c) Determine a difference equation description", - "type": "text" - }, - { - "block_id": "p611-b29", - "global_id": 17796, - "bbox": [ - 359.05, - 362.95, - 516.15, - 415.75 - ], - "text": "for an FIR filter that closely approximates\nthe system H[z] = cos(z−1). To verify\nproper\nbehavior,\nplot\nthe\nFIR\nfilter’s\nmagnitude response and compare it with the\nmagnitude response computed in part (a).", - "type": "text" - }, - { - "block_id": "p611-b30", - "global_id": 17797, - "bbox": [ - 306.0, - 422.54, - 516.15, - 499.23 - ], - "text": "5.10-12\nThe MATLAB signal-processing toolbox func-\ntion butter helps design digital Butterworth\nfilters. Use MATLAB help to learn how butter\nworks. For each of the following cases, design\nthe filter, plot the filter’s poles and zeros in the\ncomplex z plane, and plot the decibel magnitude\nresponse 20log10 |H[ej]|.", - "type": "text" - }, - { - "block_id": "p611-b31", - "global_id": 17798, - "bbox": [ - 344.12, - 499.33, - 516.14, - 508.3 - ], - "text": "(a) Design an eighth-order digital lowpass filter", - "type": "text" - }, - { - "block_id": "p611-b32", - "global_id": 17799, - "bbox": [ - 343.61, - 509.91, - 516.14, - 530.21 - ], - "text": "with c = π/3.\n(b) Design an eighth-order digital highpass", - "type": "text" - }, - { - "block_id": "p611-b33", - "global_id": 17800, - "bbox": [ - 344.12, - 531.83, - 516.14, - 552.13 - ], - "text": "filter with c = π/3.\n(c) Design an eighth-order digital bandpass", - "type": "text" - }, - { - "block_id": "p611-b34", - "global_id": 17801, - "bbox": [ - 343.61, - 553.75, - 516.13, - 585.0 - ], - "text": "filter with passband between 5π/24 and\n11π/24.\n(d) Design an eighth-order digital bandstop", - "type": "text" - }, - { - "block_id": "p611-b35", - "global_id": 17802, - "bbox": [ - 359.05, - 586.63, - 516.12, - 606.93 - ], - "text": "filter with stopband between 5π/24 and\n11π/24.", - "type": "text" - }, - { - "block_id": "p611-b36", - "global_id": 17803, - "bbox": [ - 306.0, - 613.72, - 516.13, - 633.97 - ], - "text": "5.10-13\nThe\nMATLAB\nsignal-processing\ntoolbox\nfunction cheby1 helps design digital Chebyshev", - "type": "text" - } - ] - }, - { - "page_num": 612, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p612-b0", - "global_id": 17804, - "bbox": [ - 60.0, - 62.29, - 433.19, - 71.98 - ], - "text": "592\nCHAPTER 5\nDISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p612-b1", - "global_id": 17805, - "bbox": [ - 90.72, - 85.9, - 263.24, - 160.62 - ], - "text": "type I filters. A Chebyshev type I filter has\npassband ripple and smooth stopband. Setting\nthe passband ripple to Rp = 3 dB, repeat\nProb. 5.10-12 using the cheby1 command. With\nall other parameters held constant, what is the\ngeneral effect of reducing Rp, the allowable\npassband ripple?", - "type": "text" - }, - { - "block_id": "p612-b2", - "global_id": 17806, - "bbox": [ - 53.11, - 165.53, - 263.24, - 219.37 - ], - "text": "5.10-14\nThe\nMATLAB\nsignal-processing\ntoolbox\nfunction cheby2 helps design digital Chebyshev\ntype II filters. A Chebyshev type II filter has\nsmooth passband and ripple in the stopband.\nSetting\nthe\nstopband\nripple\nRs = 20 dB", - "type": "text" - }, - { - "block_id": "p612-b3", - "global_id": 17807, - "bbox": [ - 317.86, - 85.9, - 490.4, - 128.41 - ], - "text": "down, repeat Prob. 5.10-12 using the cheby2\ncommand. With all other parameters held\nconstant, what is the general effect of increasing\nRs, the minimum stopband attenuation?", - "type": "text" - }, - { - "block_id": "p612-b4", - "global_id": 17808, - "bbox": [ - 280.26, - 132.65, - 490.39, - 207.7 - ], - "text": "5.10-15\nThe\nMATLAB\nsignal-processing\ntoolbox\nfunction ellip helps design digital elliptic\nfilters. An elliptic filter has ripple in both the\npassband and the stopband. Setting the passband\nripple to Rp = 3 dB and the stopband ripple\nRs = 20 dB down, repeat Prob. 5.10-12 using\nthe ellip command.", - "type": "text" - } - ] - }, - { - "page_num": 613, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p613-b0", - "global_id": 17809, - "bbox": [ - 90.21, - 67.04, - 159.28, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p613-b1", - "global_id": 17810, - "bbox": [ - 172.76, - 125.73, - 506.35, - 173.26 - ], - "text": "CONTINUOUS-TIME SIGNAL\nANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p613-b2", - "global_id": 17811, - "bbox": [ - 140.04, - 79.92, - 159.47, - 118.77 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p613-b3", - "global_id": 17812, - "bbox": [ - 127.59, - 266.14, - 516.16, - 431.52 - ], - "text": "Electrical engineers instinctively think of signals in terms of their frequency spectra and think\nof systems in terms of their frequency response. Most teenagers know about the audible\nportion of audio signals having a bandwidth of about 20 kHz and the need for good-quality\nspeakers to respond up to 20 kHz. This is basically thinking in the frequency domain. In\nChs. 4 and 5 we discussed extensively the frequency-domain representation of systems and their\nspectral response (system response to signals of various frequencies). In Chs. 6 through 9, we\ndiscuss spectral representation of signals, where signals are expressed as a sum of sinusoids\nor exponentials. Actually, we touched on this topic in Chs. 4 and 5. Recall that the Laplace\ntransform of a continuous-time signal is its spectral representation in terms of exponentials (or\nsinusoids) of complex frequencies. Similarly the z-transform of a discrete-time signal is its spectral\nrepresentation in terms of discrete-time exponentials. However, in the earlier chapters we were\nconcerned mainly with system representation; the spectral representation of signals was incidental\nto the system analysis. Spectral analysis of signals is an important topic in its own right, and now\nwe turn to this subject.", - "type": "text" - }, - { - "block_id": "p613-b4", - "global_id": 17813, - "bbox": [ - 127.59, - 433.51, - 516.15, - 491.3 - ], - "text": "In this chapter we show that a periodic signal can be represented as a sum of sinusoids (or\nexponentials) of various frequencies. These results are extended to aperiodic signals in Ch. 7 and\nto discrete-time signals in Ch. 9. The fascinating subject of sampling of continuous-time signals\nis discussed in Ch. 8, leading to A/D (analog-to-digital) and D/A conversion. Chapter 8 forms the\nbridge between the continuous-time and the discrete-time worlds.", - "type": "text" - }, - { - "block_id": "p613-b5", - "global_id": 17814, - "bbox": [ - 127.94, - 520.79, - 390.27, - 550.67 - ], - "text": "6.1 PERIODIC SIGNAL REPRESENTATION\nBY TRIGONOMETRIC FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p613-b6", - "global_id": 17815, - "bbox": [ - 127.59, - 556.25, - 510.82, - 568.12 - ], - "text": "As seen in Sec. 1.3-3 [Eq. (1.7)], a periodic signal x(t) with period T0 (Fig. 6.1) has the property", - "type": "text" - }, - { - "block_id": "p613-b7", - "global_id": 17816, - "bbox": [ - 266.18, - 578.38, - 377.37, - 589.53 - ], - "text": "x(t) = x(t + T0)\nfor all t", - "type": "text" - }, - { - "block_id": "p613-b8", - "global_id": 17817, - "bbox": [ - 127.59, - 600.5, - 516.13, - 636.28 - ], - "text": "The smallest value of T0 that satisfies this periodicity condition is the fundamental period of x(t).\nAs argued in Sec. 1.3-3, this equation implies that x(t) starts at −∞and continues to ∞. Moreover,\nthe area under a periodic signal x(t) over any interval of duration T0 is the same; that is, for any", - "type": "text" - }, - { - "block_id": "p613-b9", - "global_id": 17818, - "bbox": [ - 501.19, - 656.12, - 516.13, - 666.22 - ], - "text": "593", - "type": "text" - } - ] - }, - { - "page_num": 614, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p614-b0", - "global_id": 17819, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "594\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p614-b1", - "global_id": 17820, - "bbox": [ - 471.53, - 132.09, - 473.76, - 140.09 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p614-b2", - "global_id": 17821, - "bbox": [ - 309.53, - 164.68, - 316.98, - 174.29 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p614-b3", - "global_id": 17822, - "bbox": [ - 298.76, - 87.86, - 310.18, - 95.94 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p614-b4", - "global_id": 17823, - "bbox": [ - 125.76, - 184.35, - 281.17, - 194.14 - ], - "text": "Figure 6.1 A periodic signal of period T0.", - "type": "text" - }, - { - "block_id": "p614-b5", - "global_id": 17824, - "bbox": [ - 101.84, - 220.22, - 185.4, - 230.28 - ], - "text": "real numbers a and b", - "type": "text" - }, - { - "block_id": "p614-b6", - "global_id": 17825, - "bbox": [ - 238.59, - 232.19, - 264.2, - 245.8 - ], - "text": "# a+T0", - "type": "text" - }, - { - "block_id": "p614-b7", - "global_id": 17826, - "bbox": [ - 243.86, - 257.28, - 247.34, - 264.26 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p614-b8", - "global_id": 17827, - "bbox": [ - 266.31, - 245.75, - 300.0, - 256.02 - ], - "text": "x(t)dt =", - "type": "text" - }, - { - "block_id": "p614-b9", - "global_id": 17828, - "bbox": [ - 302.05, - 232.19, - 327.66, - 245.8 - ], - "text": "# b+T0", - "type": "text" - }, - { - "block_id": "p614-b10", - "global_id": 17829, - "bbox": [ - 307.31, - 257.28, - 310.8, - 264.26 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p614-b11", - "global_id": 17830, - "bbox": [ - 329.77, - 245.75, - 353.46, - 256.02 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p614-b12", - "global_id": 17831, - "bbox": [ - 101.84, - 275.97, - 490.39, - 311.44 - ], - "text": "This result follows from the fact that a periodic signal takes the same values at intervals of T0.\nHence, the values over any segment of duration T0 are repeated in any other interval of the same\nduration. For convenience, the area under x(t) over any interval of duration T0 will be denoted by", - "type": "text" - }, - { - "block_id": "p614-b13", - "global_id": 17832, - "bbox": [ - 277.07, - 318.82, - 282.33, - 328.78 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p614-b14", - "global_id": 17833, - "bbox": [ - 282.33, - 343.92, - 289.2, - 352.22 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p614-b15", - "global_id": 17834, - "bbox": [ - 291.29, - 332.38, - 314.98, - 342.66 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p614-b16", - "global_id": 17835, - "bbox": [ - 101.84, - 366.75, - 490.38, - 413.76 - ], - "text": "The frequency of a sinusoid cos2πf0t or sin2πf0t is f0, and the period is T0 = 1/f0. These\nsinusoids can also be expressed as cosω0t or sinω0t, where ω0 = 2πf0 is the radian frequency,\nalthough for brevity, it is often referred to as frequency (see Sec. B.2). A sinusoid of frequency nf0\nis said to be the nth harmonic of the sinusoid of frequency f0.", - "type": "text" - }, - { - "block_id": "p614-b17", - "global_id": 17836, - "bbox": [ - 101.85, - 414.58, - 490.4, - 436.91 - ], - "text": "Let us consider a signal x(t) made up of a sines and cosines of frequency ω0 and all of its\nharmonics (including the zeroth harmonic; i.e., dc) with arbitrary amplitudes†:", - "type": "text" - }, - { - "block_id": "p614-b18", - "global_id": 17837, - "bbox": [ - 217.74, - 462.71, - 262.72, - 474.17 - ], - "text": "x(t) = a0 +", - "type": "text" - }, - { - "block_id": "p614-b19", - "global_id": 17838, - "bbox": [ - 264.28, - 452.54, - 278.37, - 463.21 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p614-b20", - "global_id": 17839, - "bbox": [ - 265.11, - 477.11, - 277.52, - 484.37 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p614-b21", - "global_id": 17840, - "bbox": [ - 279.48, - 462.71, - 490.39, - 474.58 - ], - "text": "an cosnω0t + bn sinnω0t\n(6.1)", - "type": "text" - }, - { - "block_id": "p614-b22", - "global_id": 17841, - "bbox": [ - 101.84, - 499.79, - 318.22, - 511.66 - ], - "text": "The frequency ω0 is called the fundamental frequency.", - "type": "text" - }, - { - "block_id": "p614-b23", - "global_id": 17842, - "bbox": [ - 101.85, - 511.74, - 490.39, - 547.52 - ], - "text": "We now prove an extremely important property: x(t) in Eq. (6.1) is a periodic signal with the\nsame period as that of the fundamental, regardless of the values of the amplitudes an and bn. Note\nthat the period T0 of the fundamental satisfies", - "type": "text" - }, - { - "block_id": "p614-b24", - "global_id": 17843, - "bbox": [ - 218.7, - 562.1, - 247.15, - 580.13 - ], - "text": "T0 = 1", - "type": "text" - }, - { - "block_id": "p614-b25", - "global_id": 17844, - "bbox": [ - 241.28, - 576.06, - 247.54, - 586.9 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p614-b26", - "global_id": 17845, - "bbox": [ - 251.28, - 561.69, - 273.25, - 579.05 - ], - "text": "= 2π", - "type": "text" - }, - { - "block_id": "p614-b27", - "global_id": 17846, - "bbox": [ - 263.02, - 575.75, - 273.03, - 586.9 - ], - "text": "ω0", - "type": "text" - }, - { - "block_id": "p614-b28", - "global_id": 17847, - "bbox": [ - 295.37, - 568.67, - 490.39, - 580.13 - ], - "text": "and\nω0T0 = 2π\n(6.2)", - "type": "text" - }, - { - "block_id": "p614-b29", - "global_id": 17848, - "bbox": [ - 101.84, - 610.24, - 490.38, - 634.15 - ], - "text": "† In Eq. (6.1), the constant term a0 corresponds to the cosine term for n = 0 because cos(0 × ω0)t = 1.\nHowever, sin(0 × ω0)t = 0. Hence, the sine term for n = 0 is nonexistent.", - "type": "text" - } - ] - }, - { - "page_num": 615, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p615-b0", - "global_id": 17849, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n595", - "type": "text" - }, - { - "block_id": "p615-b1", - "global_id": 17850, - "bbox": [ - 127.59, - 85.46, - 492.73, - 96.61 - ], - "text": "To prove the periodicity of x(t), all we need is to show that x(t) = x(t + T0). From Eq. (6.1),", - "type": "text" - }, - { - "block_id": "p615-b2", - "global_id": 17851, - "bbox": [ - 190.52, - 113.52, - 255.9, - 124.97 - ], - "text": "x(t + T0) = a0 +", - "type": "text" - }, - { - "block_id": "p615-b3", - "global_id": 17852, - "bbox": [ - 257.45, - 103.34, - 271.55, - 114.01 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p615-b4", - "global_id": 17853, - "bbox": [ - 258.29, - 127.91, - 270.7, - 135.17 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p615-b5", - "global_id": 17854, - "bbox": [ - 272.65, - 113.52, - 423.32, - 125.39 - ], - "text": "an cosnω0(t + T0) + bn sinnω0(t + T0)", - "type": "text" - }, - { - "block_id": "p615-b6", - "global_id": 17855, - "bbox": [ - 227.8, - 147.62, - 255.9, - 159.08 - ], - "text": "= a0 +", - "type": "text" - }, - { - "block_id": "p615-b7", - "global_id": 17856, - "bbox": [ - 257.45, - 137.45, - 271.55, - 148.11 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p615-b8", - "global_id": 17857, - "bbox": [ - 258.29, - 162.01, - 270.7, - 169.27 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p615-b9", - "global_id": 17858, - "bbox": [ - 272.65, - 147.62, - 453.2, - 159.49 - ], - "text": "an cos(nω0t + nω0T0) + bn sin(nω0t + nω0T0)", - "type": "text" - }, - { - "block_id": "p615-b10", - "global_id": 17859, - "bbox": [ - 127.59, - 176.56, - 300.03, - 188.02 - ], - "text": "From Eq. (6.2), we have nω0T0 = 2πn, and", - "type": "text" - }, - { - "block_id": "p615-b11", - "global_id": 17860, - "bbox": [ - 198.05, - 204.61, - 263.42, - 216.06 - ], - "text": "x(t + T0) = a0 +", - "type": "text" - }, - { - "block_id": "p615-b12", - "global_id": 17861, - "bbox": [ - 264.98, - 194.43, - 279.08, - 205.11 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p615-b13", - "global_id": 17862, - "bbox": [ - 265.82, - 219.0, - 278.22, - 226.27 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p615-b14", - "global_id": 17863, - "bbox": [ - 280.18, - 204.61, - 445.66, - 216.48 - ], - "text": "an cos(nω0t + 2πn) + bn sin(nω0t + 2πn)", - "type": "text" - }, - { - "block_id": "p615-b15", - "global_id": 17864, - "bbox": [ - 235.32, - 238.72, - 263.42, - 250.17 - ], - "text": "= a0 +", - "type": "text" - }, - { - "block_id": "p615-b16", - "global_id": 17865, - "bbox": [ - 264.98, - 228.54, - 279.08, - 239.21 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p615-b17", - "global_id": 17866, - "bbox": [ - 265.82, - 253.1, - 278.22, - 260.37 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p615-b18", - "global_id": 17867, - "bbox": [ - 280.18, - 238.72, - 401.89, - 250.59 - ], - "text": "an cosnω0t + bn sinnω0t = x(t)", - "type": "text" - }, - { - "block_id": "p615-b19", - "global_id": 17868, - "bbox": [ - 127.59, - 267.97, - 516.15, - 326.63 - ], - "text": "We could also infer this result intuitively. In one fundamental period T0, the nth harmonic executes\nn complete cycles. Hence, every sinusoid on the right-hand side of Eq. (6.1) executes a complete\nnumber of cycles in one fundamental period T0. Therefore, at t = T0, every sinusoid starts as if\nit were the origin and repeats the same drama over the next T0 seconds, and so on, ad infinitum.\nHence, the sum of such harmonics results in a periodic signal of period T0.", - "type": "text" - }, - { - "block_id": "p615-b20", - "global_id": 17869, - "bbox": [ - 127.59, - 327.43, - 516.14, - 375.17 - ], - "text": "This result shows that any combination of sinusoids of frequencies 0, f0, 2f0, . . ., kf0 is a\nperiodic signal of period T0 = 1/f0 regardless of the values of amplitudes ak and bk of these\nsinusoids. By changing the values of ak and bk in Eq. (6.1), we can construct a variety of periodic\nsignals, all of the same period T0 (T0 = 1/f0 = 2π/ω0).", - "type": "text" - }, - { - "block_id": "p615-b21", - "global_id": 17870, - "bbox": [ - 127.59, - 375.26, - 516.14, - 421.5 - ], - "text": "The converse of this result is also true. We shall show in Sec. 6.5-4 that a periodic signal x(t)\nwith a period T0 can be expressed as a sum of a sinusoid of frequency f0 (f0 = 1/T0) and all its\nharmonics, as shown in Eq. (6.1).† The infinite series on the right-hand side of Eq. (6.1) is known\nas the trigonometric Fourier series of a periodic signal x(t).", - "type": "text" - }, - { - "block_id": "p615-b22", - "global_id": 17871, - "bbox": [ - 127.89, - 435.73, - 416.64, - 447.85 - ], - "text": "COMPUTING THE COEFFICIENTS OF A FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p615-b23", - "global_id": 17872, - "bbox": [ - 127.59, - 451.78, - 452.27, - 461.84 - ], - "text": "To determine the coefficients of a Fourier series, consider an integral I defined by", - "type": "text" - }, - { - "block_id": "p615-b24", - "global_id": 17873, - "bbox": [ - 267.89, - 476.15, - 281.54, - 486.42 - ], - "text": "I =", - "type": "text" - }, - { - "block_id": "p615-b25", - "global_id": 17874, - "bbox": [ - 283.59, - 462.59, - 288.85, - 472.55 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p615-b26", - "global_id": 17875, - "bbox": [ - 288.85, - 487.68, - 295.71, - 495.99 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p615-b27", - "global_id": 17876, - "bbox": [ - 297.82, - 476.15, - 375.65, - 487.29 - ], - "text": "cosnω0tcosmω0tdt", - "type": "text" - }, - { - "block_id": "p615-b28", - "global_id": 17877, - "bbox": [ - 127.59, - 504.05, - 151.92, - 514.01 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p615-b29", - "global_id": 17878, - "bbox": [ - 157.44, - 495.6, - 162.0, - 505.57 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p615-b30", - "global_id": 17879, - "bbox": [ - 127.59, - 503.95, - 516.14, - 525.96 - ], - "text": "T0 stands for integration over any contiguous interval of T0 seconds. By using a\ntrigonometric identity (see Sec. B.8-6), this integral can be expressed as", - "type": "text" - }, - { - "block_id": "p615-b31", - "global_id": 17880, - "bbox": [ - 218.62, - 539.81, - 239.0, - 551.43 - ], - "text": "I = 1", - "type": "text" - }, - { - "block_id": "p615-b32", - "global_id": 17881, - "bbox": [ - 235.51, - 547.37, - 239.0, - 554.34 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p615-b33", - "global_id": 17882, - "bbox": [ - 240.19, - 527.17, - 250.88, - 537.56 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p615-b34", - "global_id": 17883, - "bbox": [ - 250.88, - 552.69, - 257.75, - 561.0 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p615-b35", - "global_id": 17884, - "bbox": [ - 259.85, - 541.15, - 336.53, - 552.3 - ], - "text": "cos(n + m)ω0tdt +", - "type": "text" - }, - { - "block_id": "p615-b36", - "global_id": 17885, - "bbox": [ - 338.08, - 527.6, - 343.34, - 537.56 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p615-b37", - "global_id": 17886, - "bbox": [ - 343.34, - 552.69, - 350.21, - 561.0 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p615-b38", - "global_id": 17887, - "bbox": [ - 352.31, - 541.15, - 419.49, - 552.3 - ], - "text": "cos(n −m)ω0tdt", - "type": "text" - }, - { - "block_id": "p615-b39", - "global_id": 17888, - "bbox": [ - 419.67, - 527.17, - 425.1, - 537.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p615-b40", - "global_id": 17889, - "bbox": [ - 497.04, - 541.57, - 516.13, - 551.53 - ], - "text": "(6.3)", - "type": "text" - }, - { - "block_id": "p615-b41", - "global_id": 17890, - "bbox": [ - 127.59, - 577.95, - 516.12, - 634.01 - ], - "text": "† Strictly speaking, this statement applies only if a periodic signal x(t) is a continuous function of t. However,\nSec. 6.5-4 shows that it can be applied even for discontinuous signals, if we interpret the equality in Eq. (6.1)\nin the mean-square sense instead of in the ordinary sense. This means that the power of the difference between\nthe periodic signal x(t) and its Fourier series on the right-hand side of Eq. (6.1) approaches zero as the number\nof terms in the series approaches infinity.", - "type": "text" - } - ] - }, - { - "page_num": 616, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p616-b0", - "global_id": 17891, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "596\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p616-b1", - "global_id": 17892, - "bbox": [ - 101.84, - 85.46, - 490.39, - 155.62 - ], - "text": "Because cos ω0t executes one complete cycle during any interval of duration T0, cos(n + m)ω0t\nexecutes (n + m) complete cycles during any interval of duration T0. Therefore, the first integral\nin Eq. (6.3), which represents the area under n+m complete cycles of a sinusoid, equals zero. The\nsame argument shows that the second integral in Eq. (6.3) is also zero, except when n = m. Hence,\nI in Eq. (6.3) is zero for all n̸ = m. When n = m, the first integral in Eq. (6.3) is still zero, but the\nsecond integral yields", - "type": "text" - }, - { - "block_id": "p616-b2", - "global_id": 17893, - "bbox": [ - 261.25, - 161.32, - 281.63, - 172.95 - ], - "text": "I = 1", - "type": "text" - }, - { - "block_id": "p616-b3", - "global_id": 17894, - "bbox": [ - 278.14, - 168.88, - 281.63, - 175.86 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p616-b4", - "global_id": 17895, - "bbox": [ - 283.93, - 149.11, - 289.19, - 159.08 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b5", - "global_id": 17896, - "bbox": [ - 289.19, - 174.21, - 296.06, - 182.51 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b6", - "global_id": 17897, - "bbox": [ - 299.27, - 156.01, - 329.28, - 172.95 - ], - "text": "dt = T0", - "type": "text" - }, - { - "block_id": "p616-b7", - "global_id": 17898, - "bbox": [ - 322.53, - 170.15, - 327.51, - 180.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p616-b8", - "global_id": 17899, - "bbox": [ - 101.84, - 188.77, - 124.25, - 198.73 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p616-b9", - "global_id": 17900, - "bbox": [ - 203.52, - 196.7, - 208.78, - 206.66 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b10", - "global_id": 17901, - "bbox": [ - 208.77, - 221.8, - 215.64, - 230.1 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b11", - "global_id": 17902, - "bbox": [ - 217.75, - 210.26, - 305.57, - 221.41 - ], - "text": "cosnω0tcosmω0tdt =", - "type": "text" - }, - { - "block_id": "p616-b12", - "global_id": 17903, - "bbox": [ - 307.62, - 193.28, - 370.68, - 221.18 - ], - "text": ")0\nn̸ = m\nT0", - "type": "text" - }, - { - "block_id": "p616-b13", - "global_id": 17904, - "bbox": [ - 317.96, - 210.67, - 490.39, - 234.47 - ], - "text": "2\nm = n̸ = 0\n(6.4)", - "type": "text" - }, - { - "block_id": "p616-b14", - "global_id": 17905, - "bbox": [ - 101.84, - 239.7, - 273.25, - 249.66 - ], - "text": "Using similar arguments, we can show that", - "type": "text" - }, - { - "block_id": "p616-b15", - "global_id": 17906, - "bbox": [ - 205.16, - 258.09, - 210.42, - 268.06 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b16", - "global_id": 17907, - "bbox": [ - 210.43, - 283.2, - 217.3, - 291.5 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b17", - "global_id": 17908, - "bbox": [ - 219.4, - 271.66, - 303.93, - 282.81 - ], - "text": "sinnω0tsinmω0tdt =", - "type": "text" - }, - { - "block_id": "p616-b18", - "global_id": 17909, - "bbox": [ - 305.96, - 254.67, - 369.02, - 282.58 - ], - "text": ")0\nn̸ = m\nT0", - "type": "text" - }, - { - "block_id": "p616-b19", - "global_id": 17910, - "bbox": [ - 316.31, - 272.07, - 490.39, - 295.87 - ], - "text": "2\nn = m̸ = 0\n(6.5)", - "type": "text" - }, - { - "block_id": "p616-b20", - "global_id": 17911, - "bbox": [ - 101.84, - 304.08, - 208.64, - 315.38 - ], - "text": "and\n#", - "type": "text" - }, - { - "block_id": "p616-b21", - "global_id": 17912, - "bbox": [ - 208.65, - 330.52, - 215.51, - 338.82 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b22", - "global_id": 17913, - "bbox": [ - 217.61, - 318.98, - 490.39, - 330.13 - ], - "text": "sinnω0tcosmω0tdt = 0\nfor all n and m\n(6.6)", - "type": "text" - }, - { - "block_id": "p616-b23", - "global_id": 17914, - "bbox": [ - 119.78, - 344.97, - 485.93, - 356.53 - ], - "text": "To determine a0 in Eq. (6.1), we integrate both sides of Eq. (6.1) over one period T0 to yield", - "type": "text" - }, - { - "block_id": "p616-b24", - "global_id": 17915, - "bbox": [ - 165.22, - 361.58, - 170.48, - 371.54 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b25", - "global_id": 17916, - "bbox": [ - 170.48, - 386.68, - 177.35, - 394.98 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b26", - "global_id": 17917, - "bbox": [ - 179.45, - 375.14, - 223.65, - 386.29 - ], - "text": "x(t)dt = a0", - "type": "text" - }, - { - "block_id": "p616-b27", - "global_id": 17918, - "bbox": [ - 225.26, - 361.58, - 230.52, - 371.54 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b28", - "global_id": 17919, - "bbox": [ - 230.52, - 386.68, - 237.38, - 394.98 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b29", - "global_id": 17920, - "bbox": [ - 240.6, - 375.14, - 257.85, - 385.41 - ], - "text": "dt +", - "type": "text" - }, - { - "block_id": "p616-b30", - "global_id": 17921, - "bbox": [ - 259.4, - 364.97, - 273.5, - 375.63 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p616-b31", - "global_id": 17922, - "bbox": [ - 260.24, - 389.53, - 272.65, - 396.79 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p616-b33", - "global_id": 17923, - "bbox": [ - 280.03, - 375.45, - 288.5, - 386.22 - ], - "text": "an", - "type": "text" - }, - { - "block_id": "p616-b34", - "global_id": 17924, - "bbox": [ - 290.11, - 361.58, - 295.37, - 371.54 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b35", - "global_id": 17925, - "bbox": [ - 295.36, - 386.68, - 302.23, - 394.98 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b36", - "global_id": 17926, - "bbox": [ - 304.33, - 375.14, - 365.53, - 386.29 - ], - "text": "cosnω0tdt + bn", - "type": "text" - }, - { - "block_id": "p616-b37", - "global_id": 17927, - "bbox": [ - 367.15, - 361.58, - 372.41, - 371.54 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b38", - "global_id": 17928, - "bbox": [ - 372.4, - 386.68, - 379.27, - 394.98 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b39", - "global_id": 17929, - "bbox": [ - 381.38, - 375.14, - 421.4, - 386.29 - ], - "text": "sinnω0tdt", - "type": "text" - }, - { - "block_id": "p616-b40", - "global_id": 17930, - "bbox": [ - 421.59, - 361.15, - 427.02, - 371.12 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p616-b41", - "global_id": 17931, - "bbox": [ - 101.84, - 406.51, - 490.39, - 452.75 - ], - "text": "Recall that T0 is the period of a sinusoid of frequency ω0. Therefore, functions cos nω0t and\nsin nω0t execute n complete cycles over any interval of T0 seconds so that the area under these\nfunctions over an interval T0 is zero, and the last two integrals on the right-hand side of the\nforegoing equation are zero. This yields", - "type": "text" - }, - { - "block_id": "p616-b42", - "global_id": 17932, - "bbox": [ - 176.18, - 456.14, - 181.44, - 466.1 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b43", - "global_id": 17933, - "bbox": [ - 181.45, - 481.23, - 188.32, - 489.54 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b44", - "global_id": 17934, - "bbox": [ - 190.42, - 469.69, - 234.62, - 480.84 - ], - "text": "x(t)dt = a0", - "type": "text" - }, - { - "block_id": "p616-b45", - "global_id": 17935, - "bbox": [ - 236.23, - 456.14, - 241.49, - 466.1 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b46", - "global_id": 17936, - "bbox": [ - 241.49, - 481.23, - 248.36, - 489.54 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b47", - "global_id": 17937, - "bbox": [ - 251.57, - 463.12, - 373.37, - 481.15 - ], - "text": "dt = a0T0\nand\na0 = 1", - "type": "text" - }, - { - "block_id": "p616-b48", - "global_id": 17938, - "bbox": [ - 366.12, - 477.08, - 375.14, - 487.92 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b49", - "global_id": 17939, - "bbox": [ - 377.95, - 456.14, - 383.21, - 466.1 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b50", - "global_id": 17940, - "bbox": [ - 383.2, - 481.23, - 390.07, - 489.54 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b51", - "global_id": 17941, - "bbox": [ - 392.17, - 469.69, - 415.86, - 479.97 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p616-b52", - "global_id": 17942, - "bbox": [ - 101.84, - 498.37, - 490.4, - 521.47 - ], - "text": "Next we multiply both sides of Eq. (6.1) by cosmω0t and integrate the resulting equation over an\ninterval T0:", - "type": "text" - }, - { - "block_id": "p616-b53", - "global_id": 17943, - "bbox": [ - 148.6, - 526.64, - 153.86, - 536.61 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b54", - "global_id": 17944, - "bbox": [ - 153.85, - 551.75, - 160.72, - 560.05 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b55", - "global_id": 17945, - "bbox": [ - 162.83, - 540.21, - 244.28, - 551.36 - ], - "text": "x(t)cos mω0tdt = a0", - "type": "text" - }, - { - "block_id": "p616-b56", - "global_id": 17946, - "bbox": [ - 245.89, - 526.64, - 251.15, - 536.61 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b57", - "global_id": 17947, - "bbox": [ - 251.15, - 551.75, - 258.02, - 560.05 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b58", - "global_id": 17948, - "bbox": [ - 260.12, - 540.21, - 314.63, - 551.36 - ], - "text": "cos mω0tdt +", - "type": "text" - }, - { - "block_id": "p616-b59", - "global_id": 17949, - "bbox": [ - 316.17, - 530.03, - 330.27, - 540.7 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p616-b60", - "global_id": 17950, - "bbox": [ - 317.02, - 554.6, - 329.42, - 561.86 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p616-b62", - "global_id": 17951, - "bbox": [ - 336.81, - 540.52, - 345.27, - 551.29 - ], - "text": "an", - "type": "text" - }, - { - "block_id": "p616-b63", - "global_id": 17952, - "bbox": [ - 346.88, - 526.64, - 352.14, - 536.61 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b64", - "global_id": 17953, - "bbox": [ - 352.14, - 551.75, - 359.01, - 560.05 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b65", - "global_id": 17954, - "bbox": [ - 361.11, - 540.21, - 442.27, - 551.36 - ], - "text": "cos nω0t cos mω0tdt", - "type": "text" - }, - { - "block_id": "p616-b66", - "global_id": 17955, - "bbox": [ - 306.86, - 571.83, - 325.83, - 582.9 - ], - "text": "+ bn", - "type": "text" - }, - { - "block_id": "p616-b67", - "global_id": 17956, - "bbox": [ - 327.45, - 558.27, - 332.71, - 568.23 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p616-b68", - "global_id": 17957, - "bbox": [ - 332.7, - 583.36, - 339.57, - 591.67 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p616-b69", - "global_id": 17958, - "bbox": [ - 341.68, - 571.82, - 421.17, - 582.97 - ], - "text": "sin nω0t cos mω0tdt", - "type": "text" - }, - { - "block_id": "p616-b70", - "global_id": 17959, - "bbox": [ - 421.35, - 557.84, - 426.78, - 567.8 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p616-b71", - "global_id": 17960, - "bbox": [ - 101.84, - 600.81, - 490.41, - 634.79 - ], - "text": "The first integral on the right-hand side is zero because it is an area under m integral number of\ncycles of a sinusoid. Also, the last integral on the right-hand side vanishes because of Eq. (6.6).\nThis leaves only the middle integral, which is also zero for all n̸ = m because of Eq. (6.4). But", - "type": "text" - } - ] - }, - { - "page_num": 617, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p617-b0", - "global_id": 17961, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n597", - "type": "text" - }, - { - "block_id": "p617-b1", - "global_id": 17962, - "bbox": [ - 127.59, - 85.4, - 516.15, - 120.46 - ], - "text": "n takes on all values from 1 to ∞, including m. When n = m, this integral is T0/2, according\nto Eq. (6.4). Therefore, from the infinite number of terms on the right-hand side, only one term\nsurvives to yield anT0/2 = amT0/2 (recall that n = m). Therefore,", - "type": "text" - }, - { - "block_id": "p617-b2", - "global_id": 17963, - "bbox": [ - 184.54, - 124.31, - 189.8, - 134.27 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p617-b3", - "global_id": 17964, - "bbox": [ - 189.8, - 149.4, - 196.67, - 157.71 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b4", - "global_id": 17965, - "bbox": [ - 198.76, - 131.19, - 292.49, - 149.01 - ], - "text": "x(t)cos mω0tdt = amT0", - "type": "text" - }, - { - "block_id": "p617-b5", - "global_id": 17966, - "bbox": [ - 280.48, - 131.29, - 379.26, - 155.31 - ], - "text": "2\nand\nam = 2", - "type": "text" - }, - { - "block_id": "p617-b6", - "global_id": 17967, - "bbox": [ - 372.01, - 145.25, - 381.03, - 156.09 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b7", - "global_id": 17968, - "bbox": [ - 383.83, - 124.31, - 389.09, - 134.27 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p617-b8", - "global_id": 17969, - "bbox": [ - 389.09, - 149.4, - 395.96, - 157.71 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b9", - "global_id": 17970, - "bbox": [ - 398.06, - 137.86, - 459.01, - 149.01 - ], - "text": "x(t)cos mω0tdt", - "type": "text" - }, - { - "block_id": "p617-b10", - "global_id": 17971, - "bbox": [ - 127.59, - 167.77, - 516.13, - 190.87 - ], - "text": "Similarly, by multiplying both sides of Eq. (6.1) by sin nω0t and then integrating over an interval\nT0, we obtain", - "type": "text" - }, - { - "block_id": "p617-b11", - "global_id": 17972, - "bbox": [ - 267.31, - 191.98, - 298.14, - 210.01 - ], - "text": "bm = 2", - "type": "text" - }, - { - "block_id": "p617-b12", - "global_id": 17973, - "bbox": [ - 290.89, - 205.94, - 299.91, - 216.77 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b13", - "global_id": 17974, - "bbox": [ - 302.71, - 185.0, - 307.98, - 194.96 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p617-b14", - "global_id": 17975, - "bbox": [ - 307.97, - 210.09, - 314.84, - 218.39 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b15", - "global_id": 17976, - "bbox": [ - 316.94, - 198.55, - 376.23, - 209.7 - ], - "text": "x(t)sin mω0tdt", - "type": "text" - }, - { - "block_id": "p617-b16", - "global_id": 17977, - "bbox": [ - 127.59, - 225.52, - 516.12, - 249.35 - ], - "text": "To sum up our discussion, which applies to real or complex x(t), we have shown that a periodic\nsignal x(t) with period T0 can be expressed as a sum of a sinusoid of period T0 and its harmonics:", - "type": "text" - }, - { - "block_id": "p617-b17", - "global_id": 17978, - "bbox": [ - 242.37, - 269.19, - 287.36, - 280.65 - ], - "text": "x(t) = a0 +", - "type": "text" - }, - { - "block_id": "p617-b18", - "global_id": 17979, - "bbox": [ - 288.91, - 259.02, - 303.01, - 269.69 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p617-b19", - "global_id": 17980, - "bbox": [ - 289.76, - 283.58, - 302.16, - 290.84 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p617-b20", - "global_id": 17981, - "bbox": [ - 304.12, - 269.19, - 516.13, - 281.06 - ], - "text": "an cos nω0t + bn sin nω0t\n(6.7)", - "type": "text" - }, - { - "block_id": "p617-b21", - "global_id": 17982, - "bbox": [ - 127.59, - 302.15, - 216.23, - 315.23 - ], - "text": "where ω0 = 2πf0 = 2π", - "type": "text" - }, - { - "block_id": "p617-b22", - "global_id": 17983, - "bbox": [ - 209.07, - 304.19, - 235.0, - 319.19 - ], - "text": "T0 and", - "type": "text" - }, - { - "block_id": "p617-b23", - "global_id": 17984, - "bbox": [ - 144.43, - 328.71, - 173.71, - 346.73 - ], - "text": "a0 = 1", - "type": "text" - }, - { - "block_id": "p617-b24", - "global_id": 17985, - "bbox": [ - 166.46, - 342.67, - 175.49, - 353.5 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b25", - "global_id": 17986, - "bbox": [ - 178.29, - 321.71, - 183.55, - 331.68 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p617-b26", - "global_id": 17987, - "bbox": [ - 183.55, - 346.82, - 190.41, - 355.12 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b27", - "global_id": 17988, - "bbox": [ - 192.52, - 328.71, - 259.23, - 346.73 - ], - "text": "x(t)dt,\nan = 2", - "type": "text" - }, - { - "block_id": "p617-b28", - "global_id": 17989, - "bbox": [ - 251.97, - 342.67, - 260.99, - 353.5 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b29", - "global_id": 17990, - "bbox": [ - 263.81, - 321.71, - 269.07, - 331.68 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p617-b30", - "global_id": 17991, - "bbox": [ - 269.06, - 346.82, - 275.93, - 355.12 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b31", - "global_id": 17992, - "bbox": [ - 278.04, - 328.71, - 404.14, - 346.73 - ], - "text": "x(t)cos nω0tdt,\nand\nbn = 2", - "type": "text" - }, - { - "block_id": "p617-b32", - "global_id": 17993, - "bbox": [ - 396.88, - 342.67, - 405.9, - 353.5 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b33", - "global_id": 17994, - "bbox": [ - 408.71, - 321.71, - 413.97, - 331.68 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p617-b34", - "global_id": 17995, - "bbox": [ - 413.97, - 346.82, - 420.84, - 355.12 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p617-b35", - "global_id": 17996, - "bbox": [ - 422.94, - 335.28, - 516.13, - 346.43 - ], - "text": "x(t)sin nω0tdt\n(6.8)", - "type": "text" - }, - { - "block_id": "p617-b36", - "global_id": 17997, - "bbox": [ - 127.59, - 367.13, - 516.14, - 417.15 - ], - "text": "COMPACT FORM OF FOURIER SERIES\nThe results derived so far are general and apply whether x(t) is a real or a complex function of t.\nHowever, when x(t) is real, coefficients an and bn are real for all n, and the trigonometric Fourier\nseries can be expressed in a compact form, using the results in Eq. (B.16):", - "type": "text" - }, - { - "block_id": "p617-b37", - "global_id": 17998, - "bbox": [ - 254.54, - 438.49, - 301.19, - 449.95 - ], - "text": "x(t) = C0 +", - "type": "text" - }, - { - "block_id": "p617-b38", - "global_id": 17999, - "bbox": [ - 302.74, - 428.32, - 316.84, - 438.98 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p617-b39", - "global_id": 18000, - "bbox": [ - 303.58, - 452.88, - 315.99, - 460.14 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p617-b40", - "global_id": 18001, - "bbox": [ - 317.94, - 438.49, - 516.13, - 450.36 - ], - "text": "Cn cos(nω0t + θn)\n(6.9)", - "type": "text" - }, - { - "block_id": "p617-b41", - "global_id": 18002, - "bbox": [ - 127.59, - 471.09, - 366.47, - 482.96 - ], - "text": "where Cn and θn are related to an and bn, as [see Eq. (B.17)]", - "type": "text" - }, - { - "block_id": "p617-b42", - "global_id": 18003, - "bbox": [ - 208.87, - 500.32, - 274.34, - 511.78 - ], - "text": "C0 = a0,\nCn =", - "type": "text" - }, - { - "block_id": "p617-b44", - "global_id": 18004, - "bbox": [ - 284.65, - 497.44, - 400.67, - 511.78 - ], - "text": "an2 + bn2,\nand\nθn = tan−1", - "type": "text" - }, - { - "block_id": "p617-b45", - "global_id": 18005, - "bbox": [ - 402.27, - 486.32, - 426.42, - 504.41 - ], - "text": "−bn", - "type": "text" - }, - { - "block_id": "p617-b46", - "global_id": 18006, - "bbox": [ - 414.08, - 507.7, - 422.54, - 518.46 - ], - "text": "an", - "type": "text" - }, - { - "block_id": "p617-b48", - "global_id": 18007, - "bbox": [ - 492.07, - 500.73, - 516.13, - 510.7 - ], - "text": "(6.10)", - "type": "text" - }, - { - "block_id": "p617-b49", - "global_id": 18008, - "bbox": [ - 127.59, - 529.6, - 297.78, - 539.57 - ], - "text": "These results are summarized in Table 6.1.", - "type": "text" - }, - { - "block_id": "p617-b50", - "global_id": 18009, - "bbox": [ - 127.59, - 541.56, - 516.14, - 587.39 - ], - "text": "The compact form in Eq. (6.9) uses the cosine form. We could just as well have used the sine\nform, with terms sin(nω0t + θn) instead of cos(nω0t + θn). The literature overwhelmingly favors\nthe cosine form, for no apparent reason except possibly that the cosine phasor is represented by\nthe horizontal axis, which happens to be the reference axis in phasor representation.", - "type": "text" - }, - { - "block_id": "p617-b51", - "global_id": 18010, - "bbox": [ - 127.59, - 588.97, - 516.13, - 611.3 - ], - "text": "Equation (6.8) shows that a0 (or C0) is the average value of x(t) (averaged over one period).\nThis value can often be determined by inspection of x(t).", - "type": "text" - }, - { - "block_id": "p617-b52", - "global_id": 18011, - "bbox": [ - 127.59, - 612.87, - 516.11, - 635.21 - ], - "text": "Because an and bn are real, Cn and θn are also real. In the following discussion of trigonometric\nFourier series, we shall assume real x(t), unless mentioned otherwise.", - "type": "text" - } - ] - }, - { - "page_num": 618, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p618-b0", - "global_id": 18012, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "598\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p618-b1", - "global_id": 18013, - "bbox": [ - 101.84, - 85.69, - 432.82, - 96.37 - ], - "text": "TABLE 6.1\nFourier Series Representation of a Periodic Signal of Period T0 (ω0 = 2π/T0)", - "type": "text" - }, - { - "block_id": "p618-b2", - "global_id": 18014, - "bbox": [ - 101.84, - 106.31, - 477.69, - 115.28 - ], - "text": "Series Form\nCoefficient Computation\nConversion Formulas", - "type": "text" - }, - { - "block_id": "p618-b3", - "global_id": 18015, - "bbox": [ - 101.84, - 123.16, - 296.13, - 139.37 - ], - "text": "Trigonometric\na0 = 1", - "type": "text" - }, - { - "block_id": "p618-b4", - "global_id": 18016, - "bbox": [ - 289.53, - 135.72, - 297.75, - 145.52 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b5", - "global_id": 18017, - "bbox": [ - 300.45, - 116.86, - 305.18, - 125.83 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p618-b6", - "global_id": 18018, - "bbox": [ - 305.18, - 139.31, - 311.52, - 146.96 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b7", - "global_id": 18019, - "bbox": [ - 313.51, - 129.07, - 444.44, - 139.37 - ], - "text": "f(t)dt\na0 = C0 = D0", - "type": "text" - }, - { - "block_id": "p618-b8", - "global_id": 18020, - "bbox": [ - 101.84, - 153.84, - 141.28, - 163.92 - ], - "text": "f(t) = a0+", - "type": "text" - }, - { - "block_id": "p618-b9", - "global_id": 18021, - "bbox": [ - 141.68, - 144.53, - 154.37, - 154.3 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p618-b10", - "global_id": 18022, - "bbox": [ - 142.26, - 166.77, - 153.79, - 173.52 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p618-b11", - "global_id": 18023, - "bbox": [ - 155.37, - 147.93, - 296.13, - 164.52 - ], - "text": "an cosnω0t+bn sinnω0t\nan = 2", - "type": "text" - }, - { - "block_id": "p618-b12", - "global_id": 18024, - "bbox": [ - 289.53, - 160.49, - 297.75, - 170.28 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b13", - "global_id": 18025, - "bbox": [ - 300.45, - 141.64, - 305.18, - 150.61 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p618-b14", - "global_id": 18026, - "bbox": [ - 305.18, - 164.09, - 311.52, - 171.74 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b15", - "global_id": 18027, - "bbox": [ - 313.51, - 151.56, - 479.79, - 164.15 - ], - "text": "f(t)cosnω0tdt\nan−jbn = Cnejθn = 2Dn", - "type": "text" - }, - { - "block_id": "p618-b16", - "global_id": 18028, - "bbox": [ - 269.43, - 172.33, - 296.13, - 188.55 - ], - "text": "bn = 2", - "type": "text" - }, - { - "block_id": "p618-b17", - "global_id": 18029, - "bbox": [ - 289.53, - 184.9, - 297.75, - 194.69 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b18", - "global_id": 18030, - "bbox": [ - 300.45, - 166.04, - 305.18, - 175.0 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p618-b19", - "global_id": 18031, - "bbox": [ - 305.18, - 188.48, - 311.52, - 196.13 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b20", - "global_id": 18032, - "bbox": [ - 313.51, - 175.96, - 489.89, - 188.55 - ], - "text": "f(t)sinnω0tdt\nan+jbn = Cne−jθn = 2D−n", - "type": "text" - }, - { - "block_id": "p618-b21", - "global_id": 18033, - "bbox": [ - 101.84, - 195.36, - 425.54, - 205.67 - ], - "text": "Compact trigonometric\nC0 = a0\nC0 = D0", - "type": "text" - }, - { - "block_id": "p618-b22", - "global_id": 18034, - "bbox": [ - 101.84, - 213.98, - 143.78, - 224.29 - ], - "text": "f(t) = C0 +", - "type": "text" - }, - { - "block_id": "p618-b23", - "global_id": 18035, - "bbox": [ - 145.17, - 204.66, - 157.86, - 214.42 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p618-b24", - "global_id": 18036, - "bbox": [ - 145.75, - 226.91, - 157.27, - 233.65 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p618-b25", - "global_id": 18037, - "bbox": [ - 158.85, - 213.98, - 287.99, - 224.66 - ], - "text": "Cn cos(nω0t + θn)\nCn =", - "type": "text" - }, - { - "block_id": "p618-b27", - "global_id": 18038, - "bbox": [ - 297.26, - 211.39, - 473.0, - 224.29 - ], - "text": "an2 + bn2\nCn = 2|Dn|\nn ≥1", - "type": "text" - }, - { - "block_id": "p618-b28", - "global_id": 18039, - "bbox": [ - 269.43, - 237.73, - 307.04, - 249.3 - ], - "text": "θn = tan−1", - "type": "text" - }, - { - "block_id": "p618-b29", - "global_id": 18040, - "bbox": [ - 308.54, - 226.4, - 330.5, - 242.71 - ], - "text": "−bn", - "type": "text" - }, - { - "block_id": "p618-b30", - "global_id": 18041, - "bbox": [ - 319.28, - 245.64, - 327.0, - 255.37 - ], - "text": "an", - "type": "text" - }, - { - "block_id": "p618-b32", - "global_id": 18042, - "bbox": [ - 395.43, - 238.99, - 428.73, - 249.3 - ], - "text": "θn ≠ Dn", - "type": "text" - }, - { - "block_id": "p618-b33", - "global_id": 18043, - "bbox": [ - 101.84, - 255.48, - 148.18, - 264.45 - ], - "text": "Exponential", - "type": "text" - }, - { - "block_id": "p618-b34", - "global_id": 18044, - "bbox": [ - 101.84, - 273.8, - 123.82, - 283.04 - ], - "text": "f(t) =", - "type": "text" - }, - { - "block_id": "p618-b35", - "global_id": 18045, - "bbox": [ - 129.3, - 264.48, - 141.99, - 274.25 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p618-b36", - "global_id": 18046, - "bbox": [ - 125.67, - 286.1, - 145.62, - 292.78 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p618-b37", - "global_id": 18047, - "bbox": [ - 146.62, - 267.89, - 298.13, - 284.1 - ], - "text": "Dnejnω0t\nDn = 1", - "type": "text" - }, - { - "block_id": "p618-b38", - "global_id": 18048, - "bbox": [ - 291.52, - 280.44, - 299.74, - 290.24 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b39", - "global_id": 18049, - "bbox": [ - 302.44, - 261.59, - 307.17, - 270.56 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p618-b40", - "global_id": 18050, - "bbox": [ - 307.17, - 284.04, - 313.51, - 291.69 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p618-b41", - "global_id": 18051, - "bbox": [ - 315.5, - 270.38, - 360.58, - 283.04 - ], - "text": "f(t)e−jnω0t dt", - "type": "text" - }, - { - "block_id": "p618-b42", - "global_id": 18052, - "bbox": [ - 101.84, - 318.92, - 251.8, - 330.87 - ], - "text": "6.1-1 The Fourier Spectrum", - "type": "text" - }, - { - "block_id": "p618-b43", - "global_id": 18053, - "bbox": [ - 101.84, - 336.59, - 490.39, - 384.33 - ], - "text": "The compact trigonometric Fourier series in Eq. (6.9) indicates that a periodic signal x(t) can be\nexpressed as a sum of sinusoids of frequencies 0 (dc), ω0, 2ω0, . . ., nω0, . . ., whose amplitudes\nare C0, C1, C2, . . ., Cn, . . ., and whose phases are 0, θ1, θ2, . . ., θn, . . ., respectively. We can\nreadily plot amplitude Cn versus n (the amplitude spectrum) and θn versus n (the phase spectrum).†", - "type": "text" - }, - { - "block_id": "p618-b44", - "global_id": 18054, - "bbox": [ - 101.84, - 384.42, - 490.39, - 478.48 - ], - "text": "Because n is proportional to the frequency nω0, these plots are scaled plots of Cn versus ω and\nθn versus ω. The two plots together are the frequency spectra of x(t). These spectra show at\na glance the frequency contents of the signal x(t) with their amplitudes and phases. Knowing\nthese spectra, we can reconstruct or synthesize the signal x(t) according to Eq. (6.9). Therefore,\nfrequency spectra, which are an alternative way of describing a periodic signal x(t), are in every\nway equivalent to the plot of x(t) as a function of t. The frequency spectra of a signal constitute\nthe frequency-domain description of x(t), in contrast to the time-domain description, where x(t) is\nspecified as a function of time.", - "type": "text" - }, - { - "block_id": "p618-b45", - "global_id": 18055, - "bbox": [ - 101.85, - 480.05, - 490.39, - 514.34 - ], - "text": "In computing θn, the phase of the nth harmonic from Eq. (6.10), the quadrant in which θn lies\nshould be determined from the signs of an and bn. For example, if an = −1 and bn = 1, θn lies in\nthe third quadrant, and", - "type": "text" - }, - { - "block_id": "p618-b46", - "global_id": 18056, - "bbox": [ - 246.24, - 507.95, - 302.21, - 527.42 - ], - "text": "θn = tan−1 −1", - "type": "text" - }, - { - "block_id": "p618-b47", - "global_id": 18057, - "bbox": [ - 293.29, - 521.89, - 302.21, - 529.15 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p618-b49", - "global_id": 18058, - "bbox": [ - 309.47, - 514.73, - 345.48, - 526.34 - ], - "text": "= −135◦", - "type": "text" - }, - { - "block_id": "p618-b50", - "global_id": 18059, - "bbox": [ - 101.84, - 535.33, - 152.33, - 545.29 - ], - "text": "Observe that", - "type": "text" - }, - { - "block_id": "p618-b51", - "global_id": 18060, - "bbox": [ - 239.65, - 538.89, - 275.38, - 557.28 - ], - "text": "tan−1 −1", - "type": "text" - }, - { - "block_id": "p618-b52", - "global_id": 18061, - "bbox": [ - 266.45, - 552.83, - 275.38, - 560.1 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p618-b53", - "global_id": 18062, - "bbox": [ - 276.58, - 538.89, - 280.59, - 548.86 - ], - "text": "̸", - "type": "text" - }, - { - "block_id": "p618-b54", - "global_id": 18063, - "bbox": [ - 282.94, - 545.69, - 352.08, - 557.28 - ], - "text": "= tan−1(1) = 45◦", - "type": "text" - }, - { - "block_id": "p618-b55", - "global_id": 18064, - "bbox": [ - 101.84, - 577.36, - 490.39, - 633.41 - ], - "text": "† The amplitude Cn, by definition here, is nonnegative. Some authors define amplitude An that can take\npositive or negative values and magnitude Cn = |An| that can only be nonnegative. Thus, what we call\namplitude spectrum becomes magnitude spectrum. The distinction between amplitude and magnitude,\nalthough useful, is avoided in this book in the interest of keeping definitions of essentially similar entities\nto a minimum.", - "type": "text" - } - ] - }, - { - "page_num": 619, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p619-b0", - "global_id": 18065, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n599", - "type": "text" - }, - { - "block_id": "p619-b1", - "global_id": 18066, - "bbox": [ - 127.59, - 85.72, - 516.15, - 119.69 - ], - "text": "Although Cn, the amplitude of the nth harmonic as defined in Eq. (6.10), is positive, we shall\nfind it convenient to allow Cn to take on negative values when bn = 0. This will become clear in\nlater examples.", - "type": "text" - }, - { - "block_id": "p619-b2", - "global_id": 18067, - "bbox": [ - 102.51, - 173.44, - 470.67, - 199.35 - ], - "text": "EXAMPLE 6.1\nCompact Trigonometric Fourier Series of Periodic\nExponential Wave", - "type": "text" - }, - { - "block_id": "p619-b3", - "global_id": 18068, - "bbox": [ - 128.9, - 215.59, - 502.75, - 237.93 - ], - "text": "Find the compact trigonometric Fourier series for the periodic signal x(t) shown in Fig. 6.2a.\nSketch the amplitude and phase spectra for x(t).", - "type": "text" - }, - { - "block_id": "p619-b4", - "global_id": 18069, - "bbox": [ - 314.77, - 288.01, - 318.77, - 296.01 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p619-b5", - "global_id": 18070, - "bbox": [ - 306.46, - 329.95, - 472.98, - 341.48 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p619-b6", - "global_id": 18071, - "bbox": [ - 334.51, - 296.0, - 350.49, - 305.61 - ], - "text": "et2", - "type": "text" - }, - { - "block_id": "p619-b7", - "global_id": 18072, - "bbox": [ - 180.04, - 333.19, - 433.02, - 341.48 - ], - "text": "2p\np\np\n2p", - "type": "text" - }, - { - "block_id": "p619-b8", - "global_id": 18073, - "bbox": [ - 290.85, - 294.32, - 301.95, - 302.4 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p619-b9", - "global_id": 18074, - "bbox": [ - 304.03, - 346.94, - 312.91, - 354.94 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p619-b10", - "global_id": 18075, - "bbox": [ - 214.4, - 421.97, - 218.4, - 429.97 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p619-b11", - "global_id": 18076, - "bbox": [ - 198.59, - 474.84, - 205.59, - 484.41 - ], - "text": "un", - "type": "text" - }, - { - "block_id": "p619-b12", - "global_id": 18077, - "bbox": [ - 193.04, - 506.78, - 212.4, - 515.08 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p619-b13", - "global_id": 18078, - "bbox": [ - 252.4, - 421.97, - 406.39, - 429.97 - ], - "text": "2\n4\n6\n8\n10", - "type": "text" - }, - { - "block_id": "p619-b14", - "global_id": 18079, - "bbox": [ - 214.4, - 456.36, - 406.39, - 464.36 - ], - "text": "0\n2\n4\n6\n8\n10", - "type": "text" - }, - { - "block_id": "p619-b15", - "global_id": 18080, - "bbox": [ - 222.59, - 368.27, - 240.59, - 376.27 - ], - "text": "0.504", - "type": "text" - }, - { - "block_id": "p619-b16", - "global_id": 18081, - "bbox": [ - 303.64, - 435.49, - 313.29, - 443.49 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p619-b17", - "global_id": 18082, - "bbox": [ - 304.03, - 526.48, - 312.91, - 534.48 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p619-b18", - "global_id": 18083, - "bbox": [ - 199.59, - 378.64, - 207.92, - 388.19 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p619-b19", - "global_id": 18084, - "bbox": [ - 411.07, - 418.95, - 416.4, - 426.95 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p619-b20", - "global_id": 18085, - "bbox": [ - 411.07, - 458.09, - 416.4, - 466.09 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p619-b21", - "global_id": 18086, - "bbox": [ - 258.63, - 384.78, - 313.01, - 400.9 - ], - "text": "0.244\n0.125", - "type": "text" - }, - { - "block_id": "p619-b22", - "global_id": 18087, - "bbox": [ - 127.51, - 541.1, - 353.5, - 550.41 - ], - "text": "Figure 6.2 (a) A periodic signal and (b, c) its Fourier spectra.", - "type": "text" - }, - { - "block_id": "p619-b23", - "global_id": 18088, - "bbox": [ - 146.84, - 577.67, - 493.41, - 589.13 - ], - "text": "In this case the period T0 = π and the fundamental frequency f0 = 1/T0 = 1/π Hz, and", - "type": "text" - }, - { - "block_id": "p619-b24", - "global_id": 18089, - "bbox": [ - 278.81, - 597.57, - 313.33, - 616.01 - ], - "text": "ω0 = 2π", - "type": "text" - }, - { - "block_id": "p619-b25", - "global_id": 18090, - "bbox": [ - 303.59, - 611.94, - 312.62, - 622.77 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p619-b26", - "global_id": 18091, - "bbox": [ - 317.58, - 604.55, - 352.85, - 614.93 - ], - "text": "= 2rad/s", - "type": "text" - } - ] - }, - { - "page_num": 620, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p620-b0", - "global_id": 18092, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "600\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p620-b1", - "global_id": 18093, - "bbox": [ - 103.16, - 86.24, - 144.91, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p620-b2", - "global_id": 18094, - "bbox": [ - 216.13, - 105.19, - 261.12, - 116.64 - ], - "text": "x(t) = a0 +", - "type": "text" - }, - { - "block_id": "p620-b3", - "global_id": 18095, - "bbox": [ - 262.67, - 95.01, - 276.77, - 105.69 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p620-b4", - "global_id": 18096, - "bbox": [ - 263.51, - 119.58, - 275.91, - 126.85 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p620-b5", - "global_id": 18097, - "bbox": [ - 277.87, - 105.19, - 363.86, - 117.06 - ], - "text": "an cos 2nt + bn sin 2nt", - "type": "text" - }, - { - "block_id": "p620-b6", - "global_id": 18098, - "bbox": [ - 103.17, - 132.8, - 127.5, - 142.76 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p620-b7", - "global_id": 18099, - "bbox": [ - 255.39, - 140.67, - 283.39, - 158.7 - ], - "text": "a0 = 1", - "type": "text" - }, - { - "block_id": "p620-b8", - "global_id": 18100, - "bbox": [ - 277.41, - 154.32, - 283.39, - 164.28 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p620-b9", - "global_id": 18101, - "bbox": [ - 286.69, - 133.69, - 291.95, - 143.65 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p620-b10", - "global_id": 18102, - "bbox": [ - 291.95, - 158.78, - 298.82, - 167.09 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p620-b11", - "global_id": 18103, - "bbox": [ - 300.92, - 147.24, - 324.61, - 157.52 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p620-b12", - "global_id": 18104, - "bbox": [ - 103.16, - 171.74, - 450.31, - 182.12 - ], - "text": "In this example the obvious choice for the interval of integration is from 0 to π. Hence,", - "type": "text" - }, - { - "block_id": "p620-b13", - "global_id": 18105, - "bbox": [ - 194.96, - 190.72, - 222.97, - 208.74 - ], - "text": "a0 = 1", - "type": "text" - }, - { - "block_id": "p620-b14", - "global_id": 18106, - "bbox": [ - 216.99, - 204.36, - 222.96, - 214.32 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p620-b15", - "global_id": 18107, - "bbox": [ - 226.27, - 183.73, - 240.27, - 195.78 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p620-b16", - "global_id": 18108, - "bbox": [ - 231.52, - 193.48, - 305.07, - 215.86 - ], - "text": "0\ne−t/2 dt = 0.504", - "type": "text" - }, - { - "block_id": "p620-b17", - "global_id": 18109, - "bbox": [ - 194.97, - 218.14, - 222.97, - 236.17 - ], - "text": "an = 2", - "type": "text" - }, - { - "block_id": "p620-b18", - "global_id": 18110, - "bbox": [ - 216.99, - 231.79, - 222.96, - 241.75 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p620-b19", - "global_id": 18111, - "bbox": [ - 226.27, - 211.15, - 240.27, - 223.21 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p620-b20", - "global_id": 18112, - "bbox": [ - 231.52, - 221.02, - 333.48, - 243.3 - ], - "text": "0\ne−t/2 cos2ntdt = 0.504", - "type": "text" - }, - { - "block_id": "p620-b21", - "global_id": 18113, - "bbox": [ - 334.58, - 210.73, - 376.78, - 242.16 - ], - "text": "2\n1 + 16n2", - "type": "text" - }, - { - "block_id": "p620-b23", - "global_id": 18114, - "bbox": [ - 103.16, - 250.83, - 117.54, - 260.79 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p620-b24", - "global_id": 18115, - "bbox": [ - 195.79, - 259.39, - 223.8, - 277.42 - ], - "text": "bn = 2", - "type": "text" - }, - { - "block_id": "p620-b25", - "global_id": 18116, - "bbox": [ - 217.82, - 273.04, - 223.79, - 283.0 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p620-b26", - "global_id": 18117, - "bbox": [ - 227.09, - 252.4, - 241.1, - 264.45 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p620-b27", - "global_id": 18118, - "bbox": [ - 232.35, - 262.26, - 332.65, - 284.54 - ], - "text": "0\ne−t/2 sin2ntdt = 0.504", - "type": "text" - }, - { - "block_id": "p620-b28", - "global_id": 18119, - "bbox": [ - 333.75, - 251.98, - 375.95, - 283.41 - ], - "text": "8n\n1 + 16n2", - "type": "text" - }, - { - "block_id": "p620-b30", - "global_id": 18120, - "bbox": [ - 103.17, - 289.84, - 144.91, - 299.8 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p620-b31", - "global_id": 18121, - "bbox": [ - 183.52, - 309.3, - 232.62, - 319.67 - ], - "text": "x(t) = 0.504", - "type": "text" - }, - { - "block_id": "p620-b33", - "global_id": 18122, - "bbox": [ - 239.93, - 309.3, - 254.23, - 319.67 - ], - "text": "1 +", - "type": "text" - }, - { - "block_id": "p620-b34", - "global_id": 18123, - "bbox": [ - 255.79, - 299.12, - 269.88, - 309.79 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p620-b35", - "global_id": 18124, - "bbox": [ - 256.62, - 323.68, - 269.03, - 330.95 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p620-b36", - "global_id": 18125, - "bbox": [ - 272.18, - 302.73, - 390.47, - 326.75 - ], - "text": "2\n1 + 16n2 (cos2nt + 4nsin2nt)", - "type": "text" - }, - { - "block_id": "p620-b38", - "global_id": 18126, - "bbox": [ - 103.17, - 336.9, - 188.95, - 346.86 - ], - "text": "Also from Eq. (6.10),", - "type": "text" - }, - { - "block_id": "p620-b39", - "global_id": 18127, - "bbox": [ - 137.8, - 356.91, - 203.53, - 368.36 - ], - "text": "C0 = a0 = 0.504", - "type": "text" - }, - { - "block_id": "p620-b40", - "global_id": 18128, - "bbox": [ - 137.8, - 384.18, - 158.25, - 395.63 - ], - "text": "Cn =", - "type": "text" - }, - { - "block_id": "p620-b42", - "global_id": 18129, - "bbox": [ - 168.55, - 383.98, - 231.62, - 396.82 - ], - "text": "a2n + b2n = 0.504", - "type": "text" - }, - { - "block_id": "p620-b44", - "global_id": 18130, - "bbox": [ - 241.1, - 376.25, - 332.61, - 401.63 - ], - "text": "4\n(1 + 16n2)2 +\n64n2", - "type": "text" - }, - { - "block_id": "p620-b45", - "global_id": 18131, - "bbox": [ - 300.55, - 384.18, - 382.21, - 401.63 - ], - "text": "(1 + 16n2)2 = 0.504", - "type": "text" - }, - { - "block_id": "p620-b46", - "global_id": 18132, - "bbox": [ - 383.32, - 370.19, - 415.35, - 394.18 - ], - "text": "2\n√", - "type": "text" - }, - { - "block_id": "p620-b47", - "global_id": 18133, - "bbox": [ - 399.68, - 392.46, - 433.94, - 403.03 - ], - "text": "1 + 16n2", - "type": "text" - }, - { - "block_id": "p620-b49", - "global_id": 18134, - "bbox": [ - 140.06, - 411.52, - 181.39, - 424.69 - ], - "text": "θn = tan−1", - "type": "text" - }, - { - "block_id": "p620-b50", - "global_id": 18135, - "bbox": [ - 183.0, - 399.25, - 207.15, - 417.34 - ], - "text": "−bn", - "type": "text" - }, - { - "block_id": "p620-b51", - "global_id": 18136, - "bbox": [ - 194.8, - 420.62, - 203.27, - 431.39 - ], - "text": "an", - "type": "text" - }, - { - "block_id": "p620-b53", - "global_id": 18137, - "bbox": [ - 217.63, - 409.54, - 327.62, - 423.61 - ], - "text": "= tan−1(−4n) = −tan−1 4n", - "type": "text" - }, - { - "block_id": "p620-b54", - "global_id": 18138, - "bbox": [ - 103.17, - 439.35, - 476.98, - 461.28 - ], - "text": "Amplitude and phases of the dc and the first seven harmonics are computed from the above\nequations as", - "type": "text" - }, - { - "block_id": "p620-b55", - "global_id": 18139, - "bbox": [ - 116.74, - 477.05, - 451.67, - 487.12 - ], - "text": "n\n0\n1\n2\n3\n4\n5\n6\n7", - "type": "text" - }, - { - "block_id": "p620-b56", - "global_id": 18140, - "bbox": [ - 113.92, - 497.77, - 460.37, - 508.54 - ], - "text": "Cn\n0.504\n0.244\n0.125\n0.084\n0.063\n0.0504\n0.042\n0.036", - "type": "text" - }, - { - "block_id": "p620-b57", - "global_id": 18141, - "bbox": [ - 115.04, - 516.96, - 465.75, - 529.26 - ], - "text": "θn\n0◦\n−75.96◦\n−82.87◦\n−85.24◦\n−86.42◦\n−87.14◦\n−87.61◦\n−87.95◦", - "type": "text" - }, - { - "block_id": "p620-b58", - "global_id": 18142, - "bbox": [ - 103.16, - 540.58, - 311.85, - 550.95 - ], - "text": "We can use these numerical values to express x(t) as", - "type": "text" - }, - { - "block_id": "p620-b59", - "global_id": 18143, - "bbox": [ - 164.12, - 569.12, - 246.48, - 579.49 - ], - "text": "x(t) = 0.504 + 0.504", - "type": "text" - }, - { - "block_id": "p620-b60", - "global_id": 18144, - "bbox": [ - 247.59, - 558.94, - 261.68, - 569.61 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p620-b61", - "global_id": 18145, - "bbox": [ - 248.47, - 583.51, - 260.88, - 590.77 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p620-b62", - "global_id": 18146, - "bbox": [ - 264.03, - 562.55, - 288.12, - 579.13 - ], - "text": "2\n√", - "type": "text" - }, - { - "block_id": "p620-b63", - "global_id": 18147, - "bbox": [ - 272.45, - 565.42, - 387.8, - 587.98 - ], - "text": "1 + 16n2 cos(2nt −tan−1 4n)", - "type": "text" - }, - { - "block_id": "p620-b64", - "global_id": 18148, - "bbox": [ - 181.0, - 592.94, - 416.07, - 605.04 - ], - "text": "= 0.504 + 0.244cos(2t −75.96◦) + 0.125cos(4t −82.87◦)", - "type": "text" - }, - { - "block_id": "p620-b65", - "global_id": 18149, - "bbox": [ - 190.47, - 607.89, - 477.01, - 619.98 - ], - "text": "+ 0.084cos(6t −85.24◦) + 0.063cos(8t −86.42◦) + · · ·\n(6.11)", - "type": "text" - } - ] - }, - { - "page_num": 621, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p621-b0", - "global_id": 18150, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n601", - "type": "text" - }, - { - "block_id": "p621-b1", - "global_id": 18151, - "bbox": [ - 128.9, - 86.62, - 502.75, - 138.13 - ], - "text": "PLOTTING FOURIER SERIES SPECTRA USING MATLAB\nMATLAB is well suited to compute and plot Fourier series spectra. The results in Fig. 6.3,\nwhich plot Cn and θn as functions of n, match Figs. 6.2b and 6.2c, which plot Cn and θn as\nfunctions of ω = nω0 = 2n. Plots of an and bn are similarly simple to generate.", - "type": "text" - }, - { - "block_id": "p621-b2", - "global_id": 18152, - "bbox": [ - 128.91, - 146.89, - 474.11, - 216.63 - ], - "text": ">>\nn = 0:10; theta_n = atan(-4*n);\n>>\nC_n(n==0) = 0.504; C_n(n~=0) = 0.504*2./sqrt(1+16*n(n~=0).^2);\n>>\nsubplot(1,2,1); stem(n,C_n,’.k’);\n>>\naxis([-.5 10.5 0 .6]); xlabel(’n’); ylabel(’C_n’);\n>>\nsubplot(1,2,2); stem(n,theta_n,’.k’);\n>>\naxis([-.5 10.5 -1.6 0]); xlabel(’n’); ylabel(’\\theta_n’);", - "type": "text" - }, - { - "block_id": "p621-b3", - "global_id": 18153, - "bbox": [ - 220.49, - 322.81, - 292.04, - 344.0 - ], - "text": "10\nn", - "type": "text" - }, - { - "block_id": "p621-b4", - "global_id": 18154, - "bbox": [ - 141.9, - 313.58, - 145.9, - 321.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p621-b5", - "global_id": 18155, - "bbox": [ - 135.14, - 289.58, - 145.81, - 297.58 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p621-b6", - "global_id": 18156, - "bbox": [ - 135.14, - 265.58, - 145.81, - 273.59 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p621-b7", - "global_id": 18157, - "bbox": [ - 135.14, - 241.58, - 145.81, - 249.58 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p621-b8", - "global_id": 18158, - "bbox": [ - 120.64, - 277.23, - 130.16, - 285.56 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p621-b9", - "global_id": 18159, - "bbox": [ - 154.69, - 322.81, - 481.82, - 344.0 - ], - "text": "0\n2\n4\n6\n8\n0\n2\n4\n6\n8\n10\nn", - "type": "text" - }, - { - "block_id": "p621-b10", - "global_id": 18160, - "bbox": [ - 323.39, - 309.08, - 337.39, - 317.08 - ], - "text": "–1.5", - "type": "text" - }, - { - "block_id": "p621-b11", - "global_id": 18161, - "bbox": [ - 329.39, - 286.58, - 337.39, - 294.58 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p621-b12", - "global_id": 18162, - "bbox": [ - 323.39, - 264.08, - 337.39, - 272.08 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p621-b13", - "global_id": 18163, - "bbox": [ - 332.4, - 241.58, - 336.4, - 249.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p621-b14", - "global_id": 18164, - "bbox": [ - 312.33, - 276.73, - 322.15, - 283.9 - ], - "text": "θn", - "type": "text" - }, - { - "block_id": "p621-b15", - "global_id": 18165, - "bbox": [ - 119.94, - 350.69, - 347.2, - 359.92 - ], - "text": "Figure 6.3 Fourier Series spectra for Ex. 6.1 using MATLAB.", - "type": "text" - }, - { - "block_id": "p621-b16", - "global_id": 18166, - "bbox": [ - 127.59, - 402.26, - 516.14, - 496.23 - ], - "text": "The amplitude and phase spectra for x(t), in Figs. 6.2b and 6.2c, tell us at a glance the\nfrequency composition of x(t), that is, the amplitudes and phases of various sinusoidal components\nof x(t). Knowing the frequency spectra, we can reconstruct x(t), as shown on the right-hand\nside of Eq. (6.11). Therefore the frequency spectra (Figs. 6.2b, 6.2c) provide an alternative\ndescription—the frequency-domain description of x(t). The time-domain description of x(t) is\nshown in Fig. 6.2a. A signal, therefore, has a dual identity: the time-domain identity x(t) and\nthe frequency-domain identity (Fourier spectra). The two identities complement each other; taken\ntogether, they provide a better understanding of a signal.", - "type": "text" - }, - { - "block_id": "p621-b17", - "global_id": 18167, - "bbox": [ - 127.59, - 497.91, - 516.13, - 568.06 - ], - "text": "An interesting aspect of Fourier series is that whenever there is a jump discontinuity in x(t),\nthe series at the point of discontinuity converges to an average of the left-hand and right-hand limits\nof x(t) at the instant of discontinuity.† In the present example, for instance, x(t) is discontinuous\nat t = 0 with x(0+) = 1 and x(0−) = x(π) = e−π/2 = 0.208. The corresponding Fourier series\nconverges to a value (1 + 0.208)/2 = 0.604 at t = 0. This is easily verified from Eq. (6.11) by\nsetting t = 0.", - "type": "text" - }, - { - "block_id": "p621-b18", - "global_id": 18168, - "bbox": [ - 127.59, - 610.24, - 516.13, - 633.41 - ], - "text": "† This behavior of the Fourier series is dictated by its convergence in the mean, discussed later in Secs. 6.2\nand 6.5.", - "type": "text" - } - ] - }, - { - "page_num": 622, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p622-b0", - "global_id": 18169, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "602\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p622-b1", - "global_id": 18170, - "bbox": [ - 76.77, - 93.92, - 453.93, - 119.82 - ], - "text": "EXAMPLE 6.2\nCompact Trigonometric Fourier Series of a Periodic\nTriangle Wave", - "type": "text" - }, - { - "block_id": "p622-b2", - "global_id": 18171, - "bbox": [ - 103.16, - 136.07, - 477.02, - 158.4 - ], - "text": "Find the compact trigonometric Fourier series for the triangular periodic signal x(t) shown in\nFig. 6.4a, and sketch the amplitude and phase spectra for x(t).", - "type": "text" - }, - { - "block_id": "p622-b3", - "global_id": 18172, - "bbox": [ - 432.55, - 223.34, - 434.77, - 231.34 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p622-b4", - "global_id": 18173, - "bbox": [ - 285.1, - 243.79, - 296.66, - 252.01 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p622-b5", - "global_id": 18174, - "bbox": [ - 273.38, - 216.01, - 278.27, - 224.01 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p622-b6", - "global_id": 18175, - "bbox": [ - 233.44, - 239.13, - 324.77, - 247.42 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p622-b7", - "global_id": 18176, - "bbox": [ - 277.33, - 262.01, - 286.21, - 270.01 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p622-b8", - "global_id": 18177, - "bbox": [ - 284.77, - 207.68, - 296.35, - 215.76 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p622-b9", - "global_id": 18178, - "bbox": [ - 192.44, - 239.13, - 366.77, - 247.42 - ], - "text": "2\n2", - "type": "text" - }, - { - "block_id": "p622-b10", - "global_id": 18179, - "bbox": [ - 381.73, - 423.27, - 387.07, - 431.27 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p622-b11", - "global_id": 18180, - "bbox": [ - 381.73, - 336.31, - 387.07, - 344.31 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p622-b12", - "global_id": 18181, - "bbox": [ - 169.27, - 397.61, - 176.27, - 407.18 - ], - "text": "un", - "type": "text" - }, - { - "block_id": "p622-b13", - "global_id": 18182, - "bbox": [ - 280.33, - 464.96, - 289.21, - 472.96 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p622-b14", - "global_id": 18183, - "bbox": [ - 175.21, - 424.27, - 372.44, - 443.81 - ], - "text": "0\np\n5p\n3p\n7p\n9p", - "type": "text" - }, - { - "block_id": "p622-b15", - "global_id": 18184, - "bbox": [ - 279.87, - 363.99, - 289.67, - 371.99 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p622-b16", - "global_id": 18185, - "bbox": [ - 171.6, - 296.75, - 179.94, - 306.29 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p622-b17", - "global_id": 18186, - "bbox": [ - 176.73, - 349.04, - 372.44, - 357.41 - ], - "text": "0\np\n5p\n3p\n7p\n9p", - "type": "text" - }, - { - "block_id": "p622-b18", - "global_id": 18187, - "bbox": [ - 252.63, - 219.8, - 349.27, - 228.09 - ], - "text": "0.5\n1.5", - "type": "text" - }, - { - "block_id": "p622-b19", - "global_id": 18188, - "bbox": [ - 277.6, - 322.97, - 293.94, - 340.41 - ], - "text": "8A\n25p2", - "type": "text" - }, - { - "block_id": "p622-b20", - "global_id": 18189, - "bbox": [ - 209.73, - 281.9, - 219.1, - 299.24 - ], - "text": "8A\np2", - "type": "text" - }, - { - "block_id": "p622-b21", - "global_id": 18190, - "bbox": [ - 240.2, - 317.85, - 249.56, - 335.3 - ], - "text": "8A\n9p", - "type": "text" - }, - { - "block_id": "p622-b22", - "global_id": 18191, - "bbox": [ - 171.78, - 407.58, - 177.12, - 415.58 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p622-b23", - "global_id": 18192, - "bbox": [ - 172.45, - 416.33, - 176.45, - 424.33 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p622-b24", - "global_id": 18193, - "bbox": [ - 172.02, - 442.58, - 177.35, - 450.58 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p622-b25", - "global_id": 18194, - "bbox": [ - 164.38, - 446.65, - 176.69, - 459.33 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p622-b26", - "global_id": 18195, - "bbox": [ - 101.77, - 479.58, - 364.85, - 488.89 - ], - "text": "Figure 6.4 (a) A triangular periodic signal and (b, c) its Fourier spectra.", - "type": "text" - }, - { - "block_id": "p622-b27", - "global_id": 18196, - "bbox": [ - 121.09, - 516.15, - 270.29, - 527.6 - ], - "text": "In this case the period T0 = 2. Hence,", - "type": "text" - }, - { - "block_id": "p622-b28", - "global_id": 18197, - "bbox": [ - 262.31, - 536.05, - 296.84, - 554.49 - ], - "text": "ω0 = 2π", - "type": "text" - }, - { - "block_id": "p622-b29", - "global_id": 18198, - "bbox": [ - 289.37, - 543.03, - 316.88, - 560.48 - ], - "text": "2 = π", - "type": "text" - }, - { - "block_id": "p622-b30", - "global_id": 18199, - "bbox": [ - 103.17, - 568.26, - 117.54, - 578.22 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p622-b31", - "global_id": 18200, - "bbox": [ - 215.25, - 585.87, - 260.24, - 597.33 - ], - "text": "x(t) = a0 +", - "type": "text" - }, - { - "block_id": "p622-b32", - "global_id": 18201, - "bbox": [ - 261.78, - 575.7, - 275.88, - 586.36 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p622-b33", - "global_id": 18202, - "bbox": [ - 262.63, - 600.26, - 275.03, - 607.52 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p622-b34", - "global_id": 18203, - "bbox": [ - 276.99, - 585.87, - 364.75, - 597.74 - ], - "text": "an cosnπt + bn sinnπt", - "type": "text" - } - ] - }, - { - "page_num": 623, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p623-b0", - "global_id": 18204, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n603", - "type": "text" - }, - { - "block_id": "p623-b1", - "global_id": 18205, - "bbox": [ - 128.9, - 86.24, - 153.23, - 96.21 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p623-b2", - "global_id": 18206, - "bbox": [ - 250.97, - 101.78, - 275.63, - 112.05 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p623-b3", - "global_id": 18207, - "bbox": [ - 277.67, - 87.78, - 366.26, - 106.07 - ], - "text": "2At\n|t| < 1", - "type": "text" - }, - { - "block_id": "p623-b4", - "global_id": 18208, - "bbox": [ - 128.9, - 101.91, - 502.76, - 161.76 - ], - "text": "2\n2A(1 −t)\n1\n2 < t < 3\n2\nHere it will be advantageous to choose the interval of integration from −1/2 to 3/2 rather than\n0 to 2.\nA glance at Fig. 6.4a shows that the average value (dc) of x(t) is zero so that a0 = 0. Also,", - "type": "text" - }, - { - "block_id": "p623-b5", - "global_id": 18209, - "bbox": [ - 211.66, - 178.45, - 238.67, - 196.47 - ], - "text": "an = 2", - "type": "text" - }, - { - "block_id": "p623-b6", - "global_id": 18210, - "bbox": [ - 233.69, - 192.51, - 238.67, - 202.47 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p623-b7", - "global_id": 18211, - "bbox": [ - 240.97, - 171.46, - 261.05, - 183.8 - ], - "text": "# 3/2", - "type": "text" - }, - { - "block_id": "p623-b8", - "global_id": 18212, - "bbox": [ - 246.24, - 196.34, - 261.92, - 203.6 - ], - "text": "−1/2", - "type": "text" - }, - { - "block_id": "p623-b9", - "global_id": 18213, - "bbox": [ - 263.54, - 185.02, - 318.72, - 195.39 - ], - "text": "x(t)cos nπtdt", - "type": "text" - }, - { - "block_id": "p623-b10", - "global_id": 18214, - "bbox": [ - 222.68, - 214.95, - 230.45, - 224.91 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p623-b11", - "global_id": 18215, - "bbox": [ - 232.49, - 201.4, - 252.57, - 213.74 - ], - "text": "# 1/2", - "type": "text" - }, - { - "block_id": "p623-b12", - "global_id": 18216, - "bbox": [ - 237.76, - 226.27, - 253.44, - 233.53 - ], - "text": "−1/2", - "type": "text" - }, - { - "block_id": "p623-b13", - "global_id": 18217, - "bbox": [ - 255.05, - 201.4, - 340.55, - 225.33 - ], - "text": "2Atcos nπtdt +\n# 3/2", - "type": "text" - }, - { - "block_id": "p623-b14", - "global_id": 18218, - "bbox": [ - 325.74, - 214.95, - 419.82, - 233.53 - ], - "text": "1/2\n2A(1 −t)cos nπtdt", - "type": "text" - }, - { - "block_id": "p623-b15", - "global_id": 18219, - "bbox": [ - 128.91, - 249.01, - 502.75, - 271.35 - ], - "text": "Detailed evaluation of these integrals shows that both have a value of zero. Therefore an = 0.\nNext,", - "type": "text" - }, - { - "block_id": "p623-b16", - "global_id": 18220, - "bbox": [ - 213.32, - 279.01, - 232.1, - 290.47 - ], - "text": "bn =", - "type": "text" - }, - { - "block_id": "p623-b17", - "global_id": 18221, - "bbox": [ - 234.15, - 265.46, - 254.22, - 277.81 - ], - "text": "# 1/2", - "type": "text" - }, - { - "block_id": "p623-b18", - "global_id": 18222, - "bbox": [ - 239.41, - 290.33, - 255.09, - 297.6 - ], - "text": "−1/2", - "type": "text" - }, - { - "block_id": "p623-b19", - "global_id": 18223, - "bbox": [ - 256.7, - 265.46, - 340.55, - 289.39 - ], - "text": "2Atsin nπtdt +\n# 3/2", - "type": "text" - }, - { - "block_id": "p623-b20", - "global_id": 18224, - "bbox": [ - 325.74, - 279.01, - 418.17, - 297.6 - ], - "text": "1/2\n2A(1 −t)sin nπtdt", - "type": "text" - }, - { - "block_id": "p623-b21", - "global_id": 18225, - "bbox": [ - 128.91, - 304.52, - 336.61, - 314.48 - ], - "text": "Detailed evaluation of these integrals yields, in turn,", - "type": "text" - }, - { - "block_id": "p623-b22", - "global_id": 18226, - "bbox": [ - 200.24, - 345.99, - 237.76, - 364.12 - ], - "text": "bn = 8A", - "type": "text" - }, - { - "block_id": "p623-b23", - "global_id": 18227, - "bbox": [ - 222.27, - 353.08, - 256.13, - 370.01 - ], - "text": "n2π2 sin", - "type": "text" - }, - { - "block_id": "p623-b24", - "global_id": 18228, - "bbox": [ - 258.34, - 338.68, - 277.22, - 355.95 - ], - "text": "nπ", - "type": "text" - }, - { - "block_id": "p623-b25", - "global_id": 18229, - "bbox": [ - 269.75, - 360.15, - 274.73, - 370.11 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p623-b27", - "global_id": 18230, - "bbox": [ - 288.19, - 352.66, - 295.96, - 362.62 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p623-b28", - "global_id": 18231, - "bbox": [ - 298.0, - 323.27, - 305.89, - 351.17 - ], - "text": "⎧\n⎪⎪⎪⎨", - "type": "text" - }, - { - "block_id": "p623-b29", - "global_id": 18232, - "bbox": [ - 298.0, - 359.13, - 305.89, - 378.07 - ], - "text": "⎪⎪⎪⎩", - "type": "text" - }, - { - "block_id": "p623-b30", - "global_id": 18233, - "bbox": [ - 305.89, - 331.89, - 425.25, - 366.28 - ], - "text": "0\nn even\n8A\nn2π2\nn = 1,5,9,13,. . .", - "type": "text" - }, - { - "block_id": "p623-b31", - "global_id": 18234, - "bbox": [ - 305.89, - 363.24, - 330.35, - 379.88 - ], - "text": "−8A", - "type": "text" - }, - { - "block_id": "p623-b32", - "global_id": 18235, - "bbox": [ - 314.85, - 369.91, - 430.23, - 387.26 - ], - "text": "n2π2\nn = 3,7,11,15,. . .", - "type": "text" - }, - { - "block_id": "p623-b33", - "global_id": 18236, - "bbox": [ - 128.91, - 401.27, - 170.65, - 411.23 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p623-b34", - "global_id": 18237, - "bbox": [ - 191.01, - 411.07, - 229.97, - 428.12 - ], - "text": "x(t) = 8A", - "type": "text" - }, - { - "block_id": "p623-b35", - "global_id": 18238, - "bbox": [ - 218.96, - 424.61, - 229.42, - 434.78 - ], - "text": "π2", - "type": "text" - }, - { - "block_id": "p623-b37", - "global_id": 18239, - "bbox": [ - 236.6, - 411.17, - 276.3, - 428.12 - ], - "text": "sinπt −1", - "type": "text" - }, - { - "block_id": "p623-b38", - "global_id": 18240, - "bbox": [ - 271.32, - 403.75, - 440.66, - 435.19 - ], - "text": "9 sin3πt + 1\n25 sin5πt −1\n49 sin7πt + · · ·\n!", - "type": "text" - }, - { - "block_id": "p623-b39", - "global_id": 18241, - "bbox": [ - 478.69, - 418.15, - 502.75, - 428.12 - ], - "text": "(6.12)", - "type": "text" - }, - { - "block_id": "p623-b40", - "global_id": 18242, - "bbox": [ - 128.91, - 442.37, - 502.78, - 476.25 - ], - "text": "To plot Fourier spectra, the series must be converted into compact trigonometric form as in\nEq. (6.9). In this case this is readily done by converting sine terms into cosine terms with a\nsuitable phase shift. For example,", - "type": "text" - }, - { - "block_id": "p623-b41", - "global_id": 18243, - "bbox": [ - 197.3, - 492.04, - 434.37, - 504.14 - ], - "text": "sinkt = cos(kt −90◦)\nand\n−sinkt = cos(kt + 90◦)", - "type": "text" - }, - { - "block_id": "p623-b42", - "global_id": 18244, - "bbox": [ - 128.91, - 522.08, - 352.31, - 532.04 - ], - "text": "By using these identities, Eq. (6.12) can be expressed as", - "type": "text" - }, - { - "block_id": "p623-b43", - "global_id": 18245, - "bbox": [ - 128.91, - 548.53, - 166.91, - 565.57 - ], - "text": "x(t) = 8A", - "type": "text" - }, - { - "block_id": "p623-b44", - "global_id": 18246, - "bbox": [ - 155.9, - 562.07, - 166.36, - 572.23 - ], - "text": "π2", - "type": "text" - }, - { - "block_id": "p623-b46", - "global_id": 18247, - "bbox": [ - 174.64, - 548.63, - 245.36, - 565.57 - ], - "text": "cos(πt−90◦)+ 1", - "type": "text" - }, - { - "block_id": "p623-b47", - "global_id": 18248, - "bbox": [ - 240.38, - 541.21, - 502.76, - 572.65 - ], - "text": "9 cos(3πt+90◦)+ 1\n25 cos(5πt−90◦)+ 1\n49 cos(7πt+90◦)+· · ·\n!", - "type": "text" - }, - { - "block_id": "p623-b48", - "global_id": 18249, - "bbox": [ - 128.9, - 588.8, - 502.72, - 610.72 - ], - "text": "In this series all the even harmonics are missing. The phases of the odd harmonics alternate\nfrom −90◦to 90◦. Figure 6.4 shows amplitude and phase spectra for x(t).", - "type": "text" - } - ] - }, - { - "page_num": 624, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p624-b0", - "global_id": 18250, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "604\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p624-b1", - "global_id": 18251, - "bbox": [ - 76.77, - 93.64, - 416.74, - 119.55 - ], - "text": "EXAMPLE 6.3\nConverting a Trigonometric FS to a Compact\nTrigonometric FS", - "type": "text" - }, - { - "block_id": "p624-b2", - "global_id": 18252, - "bbox": [ - 103.16, - 135.03, - 379.6, - 145.41 - ], - "text": "A periodic signal x(t) is represented by a trigonometric Fourier series", - "type": "text" - }, - { - "block_id": "p624-b3", - "global_id": 18253, - "bbox": [ - 165.65, - 153.69, - 414.53, - 165.79 - ], - "text": "x(t) = 2 + 3cos2t + 4sin2t + 2sin(3t + 30◦) −cos(7t + 150◦)", - "type": "text" - }, - { - "block_id": "p624-b4", - "global_id": 18254, - "bbox": [ - 103.17, - 176.21, - 477.04, - 198.13 - ], - "text": "Express this series as a compact trigonometric Fourier series, and sketch amplitude and phase\nspectra for x(t).", - "type": "text" - }, - { - "block_id": "p624-b5", - "global_id": 18255, - "bbox": [ - 103.16, - 221.04, - 477.03, - 254.92 - ], - "text": "In compact trigonometric Fourier series, the sine and cosine terms of the same frequency\nare combined into a single term and all terms are expressed as cosine terms with positive\namplitudes. Using Eqs. (6.9) and (6.10), we have", - "type": "text" - }, - { - "block_id": "p624-b6", - "global_id": 18256, - "bbox": [ - 214.58, - 263.2, - 365.59, - 275.3 - ], - "text": "3cos2t + 4sin2t = 5cos(2t −53.13◦)", - "type": "text" - }, - { - "block_id": "p624-b7", - "global_id": 18257, - "bbox": [ - 103.17, - 285.72, - 124.48, - 295.69 - ], - "text": "Also,", - "type": "text" - }, - { - "block_id": "p624-b8", - "global_id": 18258, - "bbox": [ - 184.8, - 295.54, - 395.39, - 307.64 - ], - "text": "sin(3t + 30◦) = cos(3t + 30◦−90◦) = cos(3t −60◦)", - "type": "text" - }, - { - "block_id": "p624-b9", - "global_id": 18259, - "bbox": [ - 103.17, - 315.84, - 117.54, - 325.8 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p624-b10", - "global_id": 18260, - "bbox": [ - 172.05, - 325.66, - 408.13, - 337.75 - ], - "text": "−cos(7t + 150◦) = cos(7t + 150◦−180◦) = cos(7t −30◦)", - "type": "text" - }, - { - "block_id": "p624-b11", - "global_id": 18261, - "bbox": [ - 103.17, - 345.95, - 144.91, - 355.92 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p624-b12", - "global_id": 18262, - "bbox": [ - 163.78, - 364.2, - 416.4, - 376.3 - ], - "text": "x(t) = 2 + 5cos(2t −53.13◦) + 2cos(3t −60◦) + cos(7t −30◦)", - "type": "text" - }, - { - "block_id": "p624-b13", - "global_id": 18263, - "bbox": [ - 214.47, - 490.11, - 223.35, - 498.11 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p624-b14", - "global_id": 18264, - "bbox": [ - 123.72, - 446.07, - 127.72, - 454.07 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p624-b15", - "global_id": 18265, - "bbox": [ - 156.38, - 519.01, - 208.19, - 527.01 - ], - "text": "3\n2\n1", - "type": "text" - }, - { - "block_id": "p624-b16", - "global_id": 18266, - "bbox": [ - 214.01, - 591.89, - 223.82, - 599.89 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p624-b17", - "global_id": 18267, - "bbox": [ - 230.38, - 519.01, - 307.02, - 527.01 - ], - "text": "7\n6\n5\n4", - "type": "text" - }, - { - "block_id": "p624-b18", - "global_id": 18268, - "bbox": [ - 111.77, - 552.14, - 129.1, - 560.44 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p624-b19", - "global_id": 18269, - "bbox": [ - 101.77, - 568.24, - 129.1, - 585.28 - ], - "text": "60\n53.13", - "type": "text" - }, - { - "block_id": "p624-b20", - "global_id": 18270, - "bbox": [ - 156.38, - 474.99, - 307.69, - 482.99 - ], - "text": "3\n2\n1\n7\n6\n5\n4", - "type": "text" - }, - { - "block_id": "p624-b21", - "global_id": 18271, - "bbox": [ - 124.02, - 412.22, - 128.02, - 420.22 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p624-b22", - "global_id": 18272, - "bbox": [ - 124.97, - 521.0, - 128.97, - 529.0 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p624-b23", - "global_id": 18273, - "bbox": [ - 123.72, - 457.07, - 127.72, - 465.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p624-b24", - "global_id": 18274, - "bbox": [ - 109.35, - 422.86, - 117.69, - 432.41 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p624-b25", - "global_id": 18275, - "bbox": [ - 105.14, - 524.95, - 326.38, - 539.79 - ], - "text": "un\nv", - "type": "text" - }, - { - "block_id": "p624-b26", - "global_id": 18276, - "bbox": [ - 321.05, - 471.81, - 326.38, - 479.81 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p624-b27", - "global_id": 18277, - "bbox": [ - 101.77, - 606.58, - 250.06, - 615.82 - ], - "text": "Figure 6.5 Fourier spectra of the signal.", - "type": "text" - } - ] - }, - { - "page_num": 625, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p625-b0", - "global_id": 18278, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n605", - "type": "text" - }, - { - "block_id": "p625-b1", - "global_id": 18279, - "bbox": [ - 128.9, - 86.24, - 502.76, - 132.07 - ], - "text": "In this case only four components (including dc) are present. The amplitude of dc is 2. The\nremaining three components are of frequencies ω = 2, 3, and 7 with amplitudes 5, 2, and 1\nand phases −53.13◦, −60◦, and −30◦, respectively. The amplitude and phase spectra for this\nsignal are shown in Figs. 6.5a and 6.5b, respectively.", - "type": "text" - }, - { - "block_id": "p625-b2", - "global_id": 18280, - "bbox": [ - 102.51, - 224.85, - 479.67, - 250.75 - ], - "text": "EXAMPLE 6.4\nCompact Trigonometric Fourier Series of a Periodic\nSquare Wave", - "type": "text" - }, - { - "block_id": "p625-b3", - "global_id": 18281, - "bbox": [ - 128.9, - 267.42, - 502.79, - 289.33 - ], - "text": "Find the compact trigonometric Fourier series for the square-pulse periodic signal shown in\nFig. 6.6a and sketch its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p625-b4", - "global_id": 18282, - "bbox": [ - 234.75, - 554.79, - 442.94, - 562.79 - ], - "text": "10\n2\n4\n6\n8", - "type": "text" - }, - { - "block_id": "p625-b5", - "global_id": 18283, - "bbox": [ - 179.36, - 432.17, - 183.36, - 440.17 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p625-b6", - "global_id": 18284, - "bbox": [ - 172.61, - 495.17, - 182.61, - 503.17 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p625-b7", - "global_id": 18285, - "bbox": [ - 157.77, - 466.7, - 166.11, - 476.25 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p625-b8", - "global_id": 18286, - "bbox": [ - 425.88, - 562.99, - 431.21, - 570.99 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p625-b9", - "global_id": 18287, - "bbox": [ - 304.32, - 337.58, - 315.9, - 345.66 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p625-b10", - "global_id": 18288, - "bbox": [ - 164.61, - 378.95, - 467.76, - 388.8 - ], - "text": "p\n2p\n3p\n2p\np\n3p\nt", - "type": "text" - }, - { - "block_id": "p625-b11", - "global_id": 18289, - "bbox": [ - 295.04, - 591.89, - 304.84, - 599.89 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p625-b12", - "global_id": 18290, - "bbox": [ - 295.5, - 401.04, - 304.38, - 409.04 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p625-b13", - "global_id": 18291, - "bbox": [ - 179.61, - 554.79, - 183.61, - 562.79 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p625-b14", - "global_id": 18292, - "bbox": [ - 292.48, - 344.8, - 296.48, - 352.8 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p625-b15", - "global_id": 18293, - "bbox": [ - 276.27, - 378.81, - 281.61, - 386.81 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p625-b16", - "global_id": 18294, - "bbox": [ - 276.94, - 387.56, - 280.94, - 395.56 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p625-b17", - "global_id": 18295, - "bbox": [ - 216.66, - 468.05, - 222.0, - 482.86 - ], - "text": "2\np", - "type": "text" - }, - { - "block_id": "p625-b18", - "global_id": 18296, - "bbox": [ - 295.24, - 520.85, - 304.57, - 537.39 - ], - "text": "2\n5p", - "type": "text" - }, - { - "block_id": "p625-b19", - "global_id": 18297, - "bbox": [ - 239.14, - 569.7, - 248.47, - 586.27 - ], - "text": "2\n3p", - "type": "text" - }, - { - "block_id": "p625-b20", - "global_id": 18298, - "bbox": [ - 317.64, - 378.81, - 322.97, - 386.81 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p625-b21", - "global_id": 18299, - "bbox": [ - 268.6, - 382.95, - 322.3, - 395.56 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p625-b23", - "global_id": 18300, - "bbox": [ - 119.94, - 606.51, - 391.75, - 615.82 - ], - "text": "Figure 6.6 (a) A square pulse periodic signal and (b) its Fourier spectrum.", - "type": "text" - } - ] - }, - { - "page_num": 626, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p626-b0", - "global_id": 18301, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "606\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p626-b1", - "global_id": 18302, - "bbox": [ - 121.09, - 85.82, - 359.02, - 97.28 - ], - "text": "Here the period is T0 = 2π and ω0 = 2π/T0 = 1. Therefore,", - "type": "text" - }, - { - "block_id": "p626-b2", - "global_id": 18303, - "bbox": [ - 221.11, - 110.89, - 266.1, - 122.34 - ], - "text": "x(t) = a0 +", - "type": "text" - }, - { - "block_id": "p626-b3", - "global_id": 18304, - "bbox": [ - 267.65, - 100.72, - 281.75, - 111.38 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p626-b4", - "global_id": 18305, - "bbox": [ - 268.49, - 125.28, - 280.9, - 132.54 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p626-b5", - "global_id": 18306, - "bbox": [ - 282.86, - 110.89, - 358.88, - 122.76 - ], - "text": "an cos nt + bn sin nt", - "type": "text" - }, - { - "block_id": "p626-b6", - "global_id": 18307, - "bbox": [ - 103.17, - 137.26, - 127.5, - 147.22 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p626-b7", - "global_id": 18308, - "bbox": [ - 254.11, - 145.13, - 283.39, - 163.16 - ], - "text": "a0 = 1", - "type": "text" - }, - { - "block_id": "p626-b8", - "global_id": 18309, - "bbox": [ - 276.14, - 159.09, - 285.17, - 169.93 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p626-b9", - "global_id": 18310, - "bbox": [ - 287.97, - 138.15, - 293.23, - 148.11 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p626-b10", - "global_id": 18311, - "bbox": [ - 293.23, - 163.24, - 300.09, - 171.55 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p626-b11", - "global_id": 18312, - "bbox": [ - 302.2, - 151.7, - 325.89, - 161.98 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p626-b12", - "global_id": 18313, - "bbox": [ - 103.16, - 175.98, - 477.02, - 198.32 - ], - "text": "From Fig. 6.6a, it is clear that a proper choice of region of integration is from −π to π. But\nsince x(t) = 1 only over (−π/2, π/2), and x(t) = 0 over the remaining segment,", - "type": "text" - }, - { - "block_id": "p626-b13", - "global_id": 18314, - "bbox": [ - 246.01, - 205.13, - 276.51, - 223.15 - ], - "text": "a0 = 1", - "type": "text" - }, - { - "block_id": "p626-b14", - "global_id": 18315, - "bbox": [ - 268.04, - 218.77, - 278.99, - 229.15 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p626-b15", - "global_id": 18316, - "bbox": [ - 282.29, - 198.13, - 303.42, - 210.48 - ], - "text": "# π/2", - "type": "text" - }, - { - "block_id": "p626-b16", - "global_id": 18317, - "bbox": [ - 287.56, - 223.01, - 304.29, - 230.27 - ], - "text": "−π/2", - "type": "text" - }, - { - "block_id": "p626-b17", - "global_id": 18318, - "bbox": [ - 307.01, - 205.13, - 332.97, - 222.07 - ], - "text": "dt = 1", - "type": "text" - }, - { - "block_id": "p626-b18", - "global_id": 18319, - "bbox": [ - 327.99, - 219.18, - 332.97, - 229.15 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p626-b19", - "global_id": 18320, - "bbox": [ - 103.17, - 234.85, - 477.02, - 257.18 - ], - "text": "We could have found a0, the average value of x(t), to be 1/2 merely by inspection of x(t) in\nFig. 6.6a. Also,", - "type": "text" - }, - { - "block_id": "p626-b20", - "global_id": 18321, - "bbox": [ - 213.04, - 263.99, - 241.04, - 282.01 - ], - "text": "an = 1", - "type": "text" - }, - { - "block_id": "p626-b21", - "global_id": 18322, - "bbox": [ - 235.06, - 277.63, - 241.04, - 287.59 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p626-b22", - "global_id": 18323, - "bbox": [ - 244.34, - 256.99, - 265.47, - 269.35 - ], - "text": "# π/2", - "type": "text" - }, - { - "block_id": "p626-b23", - "global_id": 18324, - "bbox": [ - 249.6, - 281.88, - 266.34, - 289.14 - ], - "text": "−π/2", - "type": "text" - }, - { - "block_id": "p626-b24", - "global_id": 18325, - "bbox": [ - 267.95, - 263.99, - 320.82, - 280.93 - ], - "text": "cosntdt = 2", - "type": "text" - }, - { - "block_id": "p626-b25", - "global_id": 18326, - "bbox": [ - 312.36, - 270.97, - 338.25, - 288.01 - ], - "text": "nπ sin", - "type": "text" - }, - { - "block_id": "p626-b26", - "global_id": 18327, - "bbox": [ - 339.35, - 256.57, - 358.23, - 273.85 - ], - "text": "nπ", - "type": "text" - }, - { - "block_id": "p626-b27", - "global_id": 18328, - "bbox": [ - 350.75, - 278.04, - 355.74, - 288.01 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p626-b29", - "global_id": 18329, - "bbox": [ - 224.05, - 314.33, - 231.82, - 324.3 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p626-b30", - "global_id": 18330, - "bbox": [ - 233.87, - 284.95, - 241.76, - 312.84 - ], - "text": "⎧\n⎪⎪⎪⎨", - "type": "text" - }, - { - "block_id": "p626-b31", - "global_id": 18331, - "bbox": [ - 233.87, - 320.81, - 241.76, - 339.74 - ], - "text": "⎪⎪⎪⎩", - "type": "text" - }, - { - "block_id": "p626-b32", - "global_id": 18332, - "bbox": [ - 241.76, - 293.56, - 353.14, - 327.95 - ], - "text": "0\nn even\n2\nπn\nn = 1,5,9,13,. . .", - "type": "text" - }, - { - "block_id": "p626-b33", - "global_id": 18333, - "bbox": [ - 241.75, - 325.02, - 259.19, - 341.56 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p626-b34", - "global_id": 18334, - "bbox": [ - 250.72, - 331.59, - 358.13, - 348.94 - ], - "text": "πn\nn = 3,7,11,15,. . .", - "type": "text" - }, - { - "block_id": "p626-b35", - "global_id": 18335, - "bbox": [ - 213.04, - 351.84, - 241.04, - 369.87 - ], - "text": "bn = 1", - "type": "text" - }, - { - "block_id": "p626-b36", - "global_id": 18336, - "bbox": [ - 235.06, - 365.49, - 241.04, - 375.45 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p626-b37", - "global_id": 18337, - "bbox": [ - 244.34, - 344.86, - 265.47, - 357.21 - ], - "text": "# π/2", - "type": "text" - }, - { - "block_id": "p626-b38", - "global_id": 18338, - "bbox": [ - 249.6, - 369.73, - 266.34, - 377.0 - ], - "text": "−π/2", - "type": "text" - }, - { - "block_id": "p626-b39", - "global_id": 18339, - "bbox": [ - 267.95, - 358.42, - 315.6, - 368.79 - ], - "text": "sin ntdt = 0", - "type": "text" - }, - { - "block_id": "p626-b40", - "global_id": 18340, - "bbox": [ - 103.17, - 381.93, - 142.43, - 391.89 - ], - "text": "Therefore", - "type": "text" - }, - { - "block_id": "p626-b41", - "global_id": 18341, - "bbox": [ - 171.41, - 390.48, - 204.29, - 407.43 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p626-b42", - "global_id": 18342, - "bbox": [ - 199.31, - 390.48, - 223.53, - 414.5 - ], - "text": "2 + 2\nπ", - "type": "text" - }, - { - "block_id": "p626-b44", - "global_id": 18343, - "bbox": [ - 233.56, - 390.48, - 269.04, - 407.43 - ], - "text": "cos t −1", - "type": "text" - }, - { - "block_id": "p626-b45", - "global_id": 18344, - "bbox": [ - 264.06, - 383.07, - 408.77, - 414.5 - ], - "text": "3 cos3t + 1\n5 cos5t −1\n7 cos7t + · · ·", - "type": "text" - }, - { - "block_id": "p626-b46", - "global_id": 18345, - "bbox": [ - 452.95, - 397.47, - 477.01, - 407.43 - ], - "text": "(6.13)", - "type": "text" - }, - { - "block_id": "p626-b47", - "global_id": 18346, - "bbox": [ - 103.16, - 420.31, - 477.03, - 478.51 - ], - "text": "Observe that bn = 0 and all the sine terms are zero. Only the cosine terms appear in the\ntrigonometric series. The series is therefore already in the compact form except that the\namplitudes of alternating harmonics are negative. Now by definition, amplitudes Cn are\npositive [see Eq. (6.10)]. The negative sign can be accommodated by associating a proper\nphase, as seen from the trigonometric identity†", - "type": "text" - }, - { - "block_id": "p626-b48", - "global_id": 18347, - "bbox": [ - 248.24, - 485.07, - 331.93, - 495.45 - ], - "text": "−cosx = cos(x −π)", - "type": "text" - }, - { - "block_id": "p626-b49", - "global_id": 18348, - "bbox": [ - 103.16, - 502.42, - 329.78, - 512.39 - ], - "text": "Using this fact, we can express the series in Eq. (6.13) as", - "type": "text" - }, - { - "block_id": "p626-b50", - "global_id": 18349, - "bbox": [ - 179.22, - 518.02, - 212.1, - 534.96 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p626-b51", - "global_id": 18350, - "bbox": [ - 207.12, - 518.02, - 231.34, - 542.03 - ], - "text": "2 + 2\nπ", - "type": "text" - }, - { - "block_id": "p626-b53", - "global_id": 18351, - "bbox": [ - 240.07, - 518.02, - 286.07, - 535.73 - ], - "text": "cos ω0t + 1", - "type": "text" - }, - { - "block_id": "p626-b54", - "global_id": 18352, - "bbox": [ - 281.09, - 518.02, - 399.58, - 542.03 - ], - "text": "3 cos(3ω0t −π) + 1\n5 cos 5ω0t", - "type": "text" - }, - { - "block_id": "p626-b55", - "global_id": 18353, - "bbox": [ - 214.85, - 545.91, - 231.54, - 562.86 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p626-b56", - "global_id": 18354, - "bbox": [ - 226.55, - 538.5, - 374.6, - 569.93 - ], - "text": "7 cos(7ω0t −π) + 1\n9 cos9ω0t + · · ·\n!", - "type": "text" - }, - { - "block_id": "p626-b57", - "global_id": 18355, - "bbox": [ - 103.16, - 586.89, - 476.99, - 621.39 - ], - "text": "† Because cos(x±π) = −cos x, we could have chosen the phase π or −π. In fact, cos(x±Nπ) = −cos x\nfor any odd integral value of N. Therefore the phase can be chosen as ±Nπ, where N is any convenient\nodd integer.", - "type": "text" - } - ] - }, - { - "page_num": 627, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p627-b0", - "global_id": 18356, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n607", - "type": "text" - }, - { - "block_id": "p627-b1", - "global_id": 18357, - "bbox": [ - 128.9, - 86.24, - 480.17, - 96.21 - ], - "text": "This is the desired form of the compact trigonometric Fourier series. The amplitudes are", - "type": "text" - }, - { - "block_id": "p627-b2", - "global_id": 18358, - "bbox": [ - 228.42, - 111.34, - 257.1, - 129.36 - ], - "text": "C0 = 1", - "type": "text" - }, - { - "block_id": "p627-b3", - "global_id": 18359, - "bbox": [ - 252.11, - 117.91, - 332.97, - 135.36 - ], - "text": "2\nand\nCn =", - "type": "text" - }, - { - "block_id": "p627-b4", - "global_id": 18360, - "bbox": [ - 335.02, - 100.93, - 402.03, - 142.01 - ], - "text": ")0\nn even\n2\nπn\nn odd", - "type": "text" - }, - { - "block_id": "p627-b5", - "global_id": 18361, - "bbox": [ - 128.91, - 150.04, - 188.11, - 160.01 - ], - "text": "The phases are", - "type": "text" - }, - { - "block_id": "p627-b6", - "global_id": 18362, - "bbox": [ - 234.12, - 167.23, - 252.3, - 178.69 - ], - "text": "θn =", - "type": "text" - }, - { - "block_id": "p627-b7", - "global_id": 18363, - "bbox": [ - 254.35, - 153.24, - 396.36, - 183.49 - ], - "text": "0\nfor all n̸ = 3,7,11,15,. . .\n−π\nn = 3,7,11,15,. . .", - "type": "text" - }, - { - "block_id": "p627-b8", - "global_id": 18364, - "bbox": [ - 128.9, - 191.87, - 502.78, - 297.46 - ], - "text": "We might use these values to plot amplitude and phase spectra. However, we can simplify our\ntask in this special case if we allow amplitude Cn to take on negative values. If this is allowed,\nwe do not need a phase of −π to account for the sign as seen from Eq. (6.13). This means that\nphases of all components are zero, and we can discard the phase spectrum and manage with\nonly the amplitude spectrum, as shown in Fig. 6.6b. Observe that there is no loss of information\nin doing so and that the amplitude spectrum in Fig. 6.6b has the complete information about the\nFourier series in Eq. (6.13). Therefore, whenever all sine terms vanish (bn = 0), it is convenient\nto allow Cn to take on negative values. This permits the spectral information to be conveyed\nby a single spectrum.†", - "type": "text" - }, - { - "block_id": "p627-b9", - "global_id": 18365, - "bbox": [ - 128.9, - 299.46, - 502.77, - 345.29 - ], - "text": "Let us investigate the behavior of the series at the points of discontinuities. For the\ndiscontinuity at t = π/2, the values of x(t) on either sides of the discontinuity are x((π/2)−) = 1\nand x((π/2)+) = 0. We can verify by setting t = π/2 in Eq. (6.13) that x(π/2) = 0.5, which is\na value midway between the values of x(t) on either side of the discontinuity at t = π/2.", - "type": "text" - }, - { - "block_id": "p627-b10", - "global_id": 18366, - "bbox": [ - 127.59, - 391.01, - 287.62, - 402.96 - ], - "text": "6.1-2 The Effect of Symmetry", - "type": "text" - }, - { - "block_id": "p627-b11", - "global_id": 18367, - "bbox": [ - 127.59, - 408.68, - 516.14, - 526.65 - ], - "text": "The Fourier series for the signal x(t) in Fig. 6.2a (Ex. 6.1) consists of sine and cosine terms, but\nthe series for the signal x(t) in Fig. 6.4a (Ex. 6.2) consists of sine terms only, and the series for\nthe signal x(t) in Fig. 6.6a (Ex. 6.4) consists of cosine terms only. This is no accident. We can\nshow that the Fourier series of any even periodic function x(t) consists of cosine terms only and\nthe series for any odd periodic function x(t) consists of sine terms only. Moreover, because of\nsymmetry (even or odd), the information of one period of x(t) is implicit in only half the period, as\nseen in Figs. 6.4a and 6.6a. In these cases, knowing the signal over a half-period and knowing the\nkind of symmetry (even or odd), we can determine the signal waveform over a complete period.\nFor this reason, the Fourier coefficients in these cases can be computed by integrating over only\nhalf the period rather than a complete period. To prove this result, recall that", - "type": "text" - }, - { - "block_id": "p627-b12", - "global_id": 18368, - "bbox": [ - 136.77, - 536.9, - 166.05, - 554.93 - ], - "text": "a0 = 1", - "type": "text" - }, - { - "block_id": "p627-b13", - "global_id": 18369, - "bbox": [ - 158.8, - 550.86, - 167.83, - 561.7 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p627-b14", - "global_id": 18370, - "bbox": [ - 170.63, - 529.92, - 194.58, - 543.53 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p627-b15", - "global_id": 18371, - "bbox": [ - 175.9, - 554.79, - 195.46, - 563.32 - ], - "text": "−T0/2", - "type": "text" - }, - { - "block_id": "p627-b16", - "global_id": 18372, - "bbox": [ - 197.06, - 536.9, - 263.78, - 554.93 - ], - "text": "x(t)dt,\nan = 2", - "type": "text" - }, - { - "block_id": "p627-b17", - "global_id": 18373, - "bbox": [ - 256.53, - 550.86, - 265.55, - 561.7 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p627-b18", - "global_id": 18374, - "bbox": [ - 268.35, - 529.92, - 292.31, - 543.53 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p627-b19", - "global_id": 18375, - "bbox": [ - 273.61, - 554.79, - 293.18, - 563.32 - ], - "text": "−T0/2", - "type": "text" - }, - { - "block_id": "p627-b20", - "global_id": 18376, - "bbox": [ - 294.79, - 536.9, - 419.78, - 554.93 - ], - "text": "x(t)cosnω0tdt,\nand\nbn = 2", - "type": "text" - }, - { - "block_id": "p627-b21", - "global_id": 18377, - "bbox": [ - 412.53, - 550.86, - 421.55, - 561.7 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p627-b22", - "global_id": 18378, - "bbox": [ - 424.36, - 529.92, - 448.31, - 543.53 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p627-b23", - "global_id": 18379, - "bbox": [ - 429.62, - 554.79, - 449.19, - 563.32 - ], - "text": "−T0/2", - "type": "text" - }, - { - "block_id": "p627-b24", - "global_id": 18380, - "bbox": [ - 450.79, - 543.47, - 506.76, - 554.62 - ], - "text": "x(t)sinnω0tdt", - "type": "text" - }, - { - "block_id": "p627-b25", - "global_id": 18381, - "bbox": [ - 127.59, - 570.24, - 516.11, - 593.34 - ], - "text": "Recall also that cosnω0t is an even function and sinnω0t is an odd function of t. If x(t) is an even\nfunction of t, then x(t)cosnω0t is also an even function and x(t)sin nω0t is an odd function of t", - "type": "text" - }, - { - "block_id": "p627-b26", - "global_id": 18382, - "bbox": [ - 127.59, - 610.24, - 516.14, - 633.41 - ], - "text": "† Here, the distinction between amplitude An and magnitude Cn = |An| would have been useful. But, for the\nreasons mentioned in the footnote on page 598, we refrain from this distinction formally.", - "type": "text" - } - ] - }, - { - "page_num": 628, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p628-b0", - "global_id": 18383, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "608\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p628-b1", - "global_id": 18384, - "bbox": [ - 101.84, - 85.82, - 318.5, - 95.78 - ], - "text": "(see Sec. 1.5-1). Therefore, following from Eq. (1.16),", - "type": "text" - }, - { - "block_id": "p628-b2", - "global_id": 18385, - "bbox": [ - 156.67, - 109.28, - 185.95, - 127.31 - ], - "text": "a0 = 2", - "type": "text" - }, - { - "block_id": "p628-b3", - "global_id": 18386, - "bbox": [ - 178.7, - 123.24, - 187.72, - 134.07 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p628-b4", - "global_id": 18387, - "bbox": [ - 190.53, - 102.29, - 214.47, - 115.9 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p628-b5", - "global_id": 18388, - "bbox": [ - 195.79, - 109.28, - 282.79, - 134.43 - ], - "text": "0\nx(t)dt,\nan = 4", - "type": "text" - }, - { - "block_id": "p628-b6", - "global_id": 18389, - "bbox": [ - 275.54, - 123.24, - 284.57, - 134.07 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p628-b7", - "global_id": 18390, - "bbox": [ - 287.37, - 102.29, - 311.32, - 115.9 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p628-b8", - "global_id": 18391, - "bbox": [ - 292.63, - 115.85, - 490.38, - 134.43 - ], - "text": "0\nx(t)cos nω0tdt,\nand\nbn = 0\n(6.14)", - "type": "text" - }, - { - "block_id": "p628-b9", - "global_id": 18392, - "bbox": [ - 101.85, - 144.36, - 490.21, - 166.7 - ], - "text": "Similarly, if x(t) is an odd function of t, then x(t)cos nω0t is an odd function of t and x(t)sin nω0t\nis an even function of t. Therefore,", - "type": "text" - }, - { - "block_id": "p628-b10", - "global_id": 18393, - "bbox": [ - 207.72, - 179.48, - 297.12, - 197.51 - ], - "text": "an = 0\nand\nbn = 4", - "type": "text" - }, - { - "block_id": "p628-b11", - "global_id": 18394, - "bbox": [ - 289.87, - 193.44, - 298.89, - 204.27 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p628-b12", - "global_id": 18395, - "bbox": [ - 301.7, - 172.49, - 325.65, - 186.09 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p628-b13", - "global_id": 18396, - "bbox": [ - 306.96, - 186.05, - 490.38, - 204.63 - ], - "text": "0\nx(t)sin nω0tdt\n(6.15)", - "type": "text" - }, - { - "block_id": "p628-b14", - "global_id": 18397, - "bbox": [ - 101.85, - 214.91, - 490.4, - 236.84 - ], - "text": "Observe that because of symmetry, the integration required to compute the coefficients need be\nperformed over only half the period.", - "type": "text" - }, - { - "block_id": "p628-b15", - "global_id": 18398, - "bbox": [ - 101.85, - 238.41, - 490.39, - 260.74 - ], - "text": "If a periodic signal x(t) shifted by half the period remains unchanged except for a sign—that\nis, if", - "type": "text" - }, - { - "block_id": "p628-b16", - "global_id": 18399, - "bbox": [ - 256.51, - 270.13, - 260.93, - 280.1 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p628-b18", - "global_id": 18400, - "bbox": [ - 268.79, - 263.15, - 292.83, - 280.1 - ], - "text": "t −T0", - "type": "text" - }, - { - "block_id": "p628-b19", - "global_id": 18401, - "bbox": [ - 286.08, - 277.3, - 291.06, - 287.26 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p628-b21", - "global_id": 18402, - "bbox": [ - 303.3, - 269.82, - 335.72, - 280.1 - ], - "text": "= −x(t)", - "type": "text" - }, - { - "block_id": "p628-b22", - "global_id": 18403, - "bbox": [ - 101.85, - 296.06, - 490.43, - 365.9 - ], - "text": "then the signal is said to have a half-wave symmetry. It can be shown that for a signal with a\nhalf-wave symmetry, all the even-numbered harmonics vanish (see Prob. 6.1-6). The signal in\nFig. 6.4a is an example of such a symmetry. The signal in Fig. 6.6a also has this symmetry,\nalthough it is not obvious owing to a dc component. If we subtract the dc component of 0.5 from\nthis signal, the remaining signal has half-wave symmetry. For this reason, this signal has only odd\nharmonics and a dc component of 0.5.", - "type": "text" - }, - { - "block_id": "p628-b23", - "global_id": 18404, - "bbox": [ - 107.82, - 399.64, - 390.4, - 411.6 - ], - "text": "DRILL 6.1\nCompact Trigonometric Fourier Series", - "type": "text" - }, - { - "block_id": "p628-b24", - "global_id": 18405, - "bbox": [ - 107.82, - 420.72, - 484.42, - 466.55 - ], - "text": "Find the compact trigonometric Fourier series for periodic signals shown in Fig. 6.7. Sketch\ntheir amplitude and phase spectra. Allow Cn to take on negative values if bn = 0 so that the\nphase spectrum can be eliminated. [Hint: Use Eqs. (6.14) and (6.15) for appropriate symmetry\nconditions.]", - "type": "text" - }, - { - "block_id": "p628-b25", - "global_id": 18406, - "bbox": [ - 108.09, - 480.07, - 170.31, - 491.03 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p628-b26", - "global_id": 18407, - "bbox": [ - 125.76, - 495.05, - 175.23, - 512.0 - ], - "text": "(a) x(t) = 1", - "type": "text" - }, - { - "block_id": "p628-b27", - "global_id": 18408, - "bbox": [ - 170.25, - 495.05, - 198.95, - 519.07 - ], - "text": "3 −4\nπ2", - "type": "text" - }, - { - "block_id": "p628-b29", - "global_id": 18409, - "bbox": [ - 208.48, - 495.05, - 250.94, - 512.0 - ], - "text": "cos πt −1", - "type": "text" - }, - { - "block_id": "p628-b30", - "global_id": 18410, - "bbox": [ - 245.96, - 487.64, - 419.89, - 519.07 - ], - "text": "4 cos 2πt + 1\n9 cos 3πt −1\n16 cos 4πt + · · ·", - "type": "text" - }, - { - "block_id": "p628-b31", - "global_id": 18411, - "bbox": [ - 159.24, - 522.44, - 175.23, - 539.39 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p628-b32", - "global_id": 18412, - "bbox": [ - 170.25, - 522.44, - 198.95, - 546.46 - ], - "text": "3 + 4\nπ2", - "type": "text" - }, - { - "block_id": "p628-b33", - "global_id": 18413, - "bbox": [ - 201.75, - 518.84, - 215.85, - 529.51 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p628-b34", - "global_id": 18414, - "bbox": [ - 202.59, - 543.41, - 215.0, - 550.67 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p628-b35", - "global_id": 18415, - "bbox": [ - 218.15, - 521.03, - 241.82, - 532.4 - ], - "text": "(−1)n", - "type": "text" - }, - { - "block_id": "p628-b36", - "global_id": 18416, - "bbox": [ - 225.76, - 529.01, - 274.84, - 546.36 - ], - "text": "n2\ncos nπt", - "type": "text" - }, - { - "block_id": "p628-b37", - "global_id": 18417, - "bbox": [ - 125.76, - 553.29, - 181.87, - 570.34 - ], - "text": "(b) x(t) = 2A", - "type": "text" - }, - { - "block_id": "p628-b38", - "global_id": 18418, - "bbox": [ - 172.86, - 567.02, - 178.83, - 576.99 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p628-b39", - "global_id": 18419, - "bbox": [ - 183.07, - 548.95, - 187.74, - 558.91 - ], - "text": "'", - "type": "text" - }, - { - "block_id": "p628-b40", - "global_id": 18420, - "bbox": [ - 188.85, - 553.39, - 229.66, - 570.33 - ], - "text": "sin πt −1", - "type": "text" - }, - { - "block_id": "p628-b41", - "global_id": 18421, - "bbox": [ - 224.68, - 548.95, - 386.62, - 577.4 - ], - "text": "2 sin 2πt + 1\n3 sin 3πt −1\n4 sin 4πt + · · ·\n(", - "type": "text" - }, - { - "block_id": "p628-b42", - "global_id": 18422, - "bbox": [ - 159.24, - 578.26, - 181.31, - 595.31 - ], - "text": "= 2A", - "type": "text" - }, - { - "block_id": "p628-b43", - "global_id": 18423, - "bbox": [ - 172.3, - 592.01, - 178.28, - 601.97 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p628-b44", - "global_id": 18424, - "bbox": [ - 182.51, - 573.94, - 187.18, - 583.9 - ], - "text": "'", - "type": "text" - }, - { - "block_id": "p628-b45", - "global_id": 18425, - "bbox": [ - 188.29, - 578.36, - 261.9, - 595.31 - ], - "text": "cos(πt −90◦) + 1", - "type": "text" - }, - { - "block_id": "p628-b46", - "global_id": 18426, - "bbox": [ - 256.92, - 578.36, - 406.64, - 602.38 - ], - "text": "2 cos(2πt + 90◦) + 1\n3 cos(3πt −90◦)", - "type": "text" - }, - { - "block_id": "p628-b47", - "global_id": 18427, - "bbox": [ - 182.21, - 611.77, - 194.66, - 623.08 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p628-b48", - "global_id": 18428, - "bbox": [ - 191.17, - 602.12, - 288.22, - 627.03 - ], - "text": "4 cos(4πt + 90◦) + · · ·\n(", - "type": "text" - } - ] - }, - { - "page_num": 629, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p629-b0", - "global_id": 18429, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n609", - "type": "text" - }, - { - "block_id": "p629-b1", - "global_id": 18430, - "bbox": [ - 299.84, - 93.28, - 311.42, - 101.36 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p629-b2", - "global_id": 18431, - "bbox": [ - 176.32, - 123.69, - 445.55, - 131.98 - ], - "text": "1\n2\n3\n4\n5\n1\n2\n3\n4\n5\n0", - "type": "text" - }, - { - "block_id": "p629-b3", - "global_id": 18432, - "bbox": [ - 309.83, - 137.38, - 318.71, - 145.38 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p629-b4", - "global_id": 18433, - "bbox": [ - 324.55, - 104.89, - 330.09, - 113.28 - ], - "text": "t2", - "type": "text" - }, - { - "block_id": "p629-b5", - "global_id": 18434, - "bbox": [ - 272.9, - 184.43, - 283.57, - 192.73 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p629-b6", - "global_id": 18435, - "bbox": [ - 317.92, - 158.14, - 329.51, - 166.22 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p629-b7", - "global_id": 18436, - "bbox": [ - 481.18, - 186.73, - 483.4, - 194.73 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p629-b8", - "global_id": 18437, - "bbox": [ - 481.18, - 123.94, - 483.4, - 131.94 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p629-b9", - "global_id": 18438, - "bbox": [ - 196.22, - 184.43, - 431.55, - 197.23 - ], - "text": "2\n4\n4\n1\n2\nA", - "type": "text" - }, - { - "block_id": "p629-b10", - "global_id": 18439, - "bbox": [ - 306.05, - 165.73, - 310.94, - 173.73 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p629-b11", - "global_id": 18440, - "bbox": [ - 309.44, - 202.72, - 319.09, - 210.72 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p629-b12", - "global_id": 18441, - "bbox": [ - 133.57, - 220.1, - 315.86, - 229.34 - ], - "text": "Figure 6.7 Periodic signals for Drills 6.1 and 6.6.", - "type": "text" - }, - { - "block_id": "p629-b13", - "global_id": 18442, - "bbox": [ - 127.59, - 269.45, - 446.56, - 281.41 - ], - "text": "6.1-3 Determining the Fundamental Frequency and Period", - "type": "text" - }, - { - "block_id": "p629-b14", - "global_id": 18443, - "bbox": [ - 127.59, - 287.54, - 516.11, - 333.37 - ], - "text": "We have seen that every periodic signal can be expressed as a sum of sinusoids of a fundamental\nfrequency ω0 and its harmonics. One may ask whether a sum of sinusoids of any frequencies\nrepresents a periodic signal. If so, how does one determine the period? Consider the following\nthree functions:", - "type": "text" - }, - { - "block_id": "p629-b15", - "global_id": 18444, - "bbox": [ - 202.88, - 347.0, - 268.73, - 358.14 - ], - "text": "x1(t) = 2 + 7cos", - "type": "text" - }, - { - "block_id": "p629-b16", - "global_id": 18445, - "bbox": [ - 269.87, - 338.99, - 278.56, - 352.62 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p629-b17", - "global_id": 18446, - "bbox": [ - 275.08, - 338.99, - 305.96, - 360.19 - ], - "text": "2t + θ1", - "type": "text" - }, - { - "block_id": "p629-b18", - "global_id": 18447, - "bbox": [ - 307.51, - 347.0, - 336.18, - 357.37 - ], - "text": "+ 3cos", - "type": "text" - }, - { - "block_id": "p629-b19", - "global_id": 18448, - "bbox": [ - 337.3, - 338.99, - 346.0, - 352.62 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p629-b20", - "global_id": 18449, - "bbox": [ - 342.51, - 338.99, - 373.4, - 360.19 - ], - "text": "3t + θ2", - "type": "text" - }, - { - "block_id": "p629-b21", - "global_id": 18450, - "bbox": [ - 374.95, - 347.0, - 403.62, - 357.37 - ], - "text": "+ 5cos", - "type": "text" - }, - { - "block_id": "p629-b22", - "global_id": 18451, - "bbox": [ - 404.75, - 338.99, - 413.45, - 352.62 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p629-b23", - "global_id": 18452, - "bbox": [ - 409.96, - 338.99, - 440.84, - 360.19 - ], - "text": "6t + θ3", - "type": "text" - }, - { - "block_id": "p629-b24", - "global_id": 18453, - "bbox": [ - 202.88, - 362.95, - 354.9, - 374.1 - ], - "text": "x2(t) = 2cos(2t + θ1) + 5sin(πt + θ2)", - "type": "text" - }, - { - "block_id": "p629-b25", - "global_id": 18454, - "bbox": [ - 202.88, - 380.4, - 251.25, - 391.55 - ], - "text": "x3(t) = 3sin", - "type": "text" - }, - { - "block_id": "p629-b27", - "global_id": 18455, - "bbox": [ - 256.37, - 371.46, - 269.78, - 390.78 - ], - "text": "3\n√", - "type": "text" - }, - { - "block_id": "p629-b28", - "global_id": 18456, - "bbox": [ - 269.78, - 372.39, - 297.97, - 390.78 - ], - "text": "2t + θ", - "type": "text" - }, - { - "block_id": "p629-b29", - "global_id": 18457, - "bbox": [ - 299.53, - 380.4, - 328.2, - 390.78 - ], - "text": "+ 7cos", - "type": "text" - }, - { - "block_id": "p629-b31", - "global_id": 18458, - "bbox": [ - 333.34, - 371.46, - 346.75, - 390.78 - ], - "text": "6\n√", - "type": "text" - }, - { - "block_id": "p629-b32", - "global_id": 18459, - "bbox": [ - 346.75, - 372.39, - 375.8, - 390.78 - ], - "text": "2t + φ", - "type": "text" - }, - { - "block_id": "p629-b33", - "global_id": 18460, - "bbox": [ - 127.59, - 404.82, - 516.13, - 450.65 - ], - "text": "Recall that every frequency in a periodic signal is an integer multiple of the fundamental\nfrequency ω0. Therefore the ratio of any two frequencies is of the form m/n, where m and n are\nintegers. This means that the ratio of any two frequencies is a rational number. When the ratio of\ntwo frequencies is a rational number, the frequencies are said to be harmonically related.", - "type": "text" - }, - { - "block_id": "p629-b34", - "global_id": 18461, - "bbox": [ - 127.59, - 452.64, - 516.15, - 570.2 - ], - "text": "The largest number of which all the frequencies are integer multiples is the fundamental\nfrequency. In other words, the fundamental frequency is the greatest common factor (GCF) of\nall the frequencies in the series. The frequencies in the spectrum of x1(t) are 1/2, 2/3, and 7/6 (we\ndo not consider dc). The ratios of the successive frequencies are 3:4 and 4:7, respectively. Because\nboth these numbers are rational, all the three frequencies in the spectrum are harmonically related,\nand the signal x1(t) is periodic. The GCF, that is, the greatest number of which 1/2, 2/3, and\n7/6 are integer multiples, is 1/6.† Moreover, 3(1/6) = 1/2, 4(1/6) = 2/3, and 7(1/6) = 7/6.\nTherefore the fundamental frequency is 1/6, and the three frequencies in the spectrum are the\nthird, fourth, and seventh harmonics. Observe that the fundamental frequency component is absent\nin this Fourier series.", - "type": "text" - }, - { - "block_id": "p629-b35", - "global_id": 18462, - "bbox": [ - 127.59, - 588.31, - 516.14, - 633.41 - ], - "text": "† The greatest common factor of a1/b1, a2/b2, . . ., am/bm is the ratio of the GCF of the numerators set\n(a1,a2,. . .,am) to the LCM (least common multiple) of the denominator set (b1,b2,. . .,bm). For instance, for\nthe set (2/3,6/7,2), the GCF of the numerator set (2,6,2) is 2; the LCM of the denominator set (3,7,1) is 21.\nTherefore, 2/21 is the largest number of which 2/3,6/7, and 2 are integer multiples.", - "type": "text" - } - ] - }, - { - "page_num": 630, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p630-b0", - "global_id": 18463, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "610\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p630-b1", - "global_id": 18464, - "bbox": [ - 101.84, - 85.46, - 490.38, - 119.75 - ], - "text": "The signal x2(t) is not periodic because the ratio of two frequencies in the spectrum is 2/π,\nwhich is not a rational number. The signal x3(t) is periodic because the ratio of frequencies\n3\n√", - "type": "text" - }, - { - "block_id": "p630-b2", - "global_id": 18465, - "bbox": [ - 115.25, - 100.94, - 154.49, - 119.75 - ], - "text": "2 and 6\n√", - "type": "text" - }, - { - "block_id": "p630-b3", - "global_id": 18466, - "bbox": [ - 154.49, - 100.94, - 412.19, - 119.75 - ], - "text": "2 is 1/2, a rational number. The greatest common factor of 3\n√", - "type": "text" - }, - { - "block_id": "p630-b4", - "global_id": 18467, - "bbox": [ - 412.19, - 100.94, - 451.42, - 119.75 - ], - "text": "2 and 6\n√", - "type": "text" - }, - { - "block_id": "p630-b5", - "global_id": 18468, - "bbox": [ - 451.42, - 100.94, - 482.91, - 119.75 - ], - "text": "2 is 3\n√", - "type": "text" - }, - { - "block_id": "p630-b6", - "global_id": 18469, - "bbox": [ - 101.85, - 109.79, - 490.39, - 132.79 - ], - "text": "2.\nTherefore, the fundamental frequency ω0 = 3", - "type": "text" - }, - { - "block_id": "p630-b7", - "global_id": 18470, - "bbox": [ - 282.58, - 112.91, - 291.01, - 122.87 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p630-b8", - "global_id": 18471, - "bbox": [ - 291.02, - 121.74, - 357.97, - 131.71 - ], - "text": "2, and the period", - "type": "text" - }, - { - "block_id": "p630-b9", - "global_id": 18472, - "bbox": [ - 254.0, - 149.97, - 294.47, - 168.41 - ], - "text": "T0 =\n2π", - "type": "text" - }, - { - "block_id": "p630-b10", - "global_id": 18473, - "bbox": [ - 276.58, - 165.42, - 285.28, - 175.8 - ], - "text": "(3", - "type": "text" - }, - { - "block_id": "p630-b11", - "global_id": 18474, - "bbox": [ - 285.27, - 157.0, - 293.7, - 166.96 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p630-b12", - "global_id": 18475, - "bbox": [ - 293.71, - 156.95, - 313.42, - 175.8 - ], - "text": "2)\n=", - "type": "text" - }, - { - "block_id": "p630-b13", - "global_id": 18476, - "bbox": [ - 316.66, - 141.53, - 325.09, - 151.5 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p630-b14", - "global_id": 18477, - "bbox": [ - 320.88, - 150.38, - 337.24, - 174.4 - ], - "text": "2\n3 π", - "type": "text" - }, - { - "block_id": "p630-b15", - "global_id": 18478, - "bbox": [ - 107.82, - 227.46, - 478.79, - 253.37 - ], - "text": "DRILL 6.2\nDetermining Periodicity, Fundamental Frequency, and\nHarmonic Content", - "type": "text" - }, - { - "block_id": "p630-b16", - "global_id": 18479, - "bbox": [ - 107.82, - 262.49, - 225.4, - 272.45 - ], - "text": "Determine whether the signal", - "type": "text" - }, - { - "block_id": "p630-b17", - "global_id": 18480, - "bbox": [ - 222.09, - 283.99, - 262.07, - 294.37 - ], - "text": "x(t) = cos", - "type": "text" - }, - { - "block_id": "p630-b18", - "global_id": 18481, - "bbox": [ - 263.19, - 275.98, - 271.88, - 289.62 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p630-b19", - "global_id": 18482, - "bbox": [ - 268.39, - 275.98, - 327.35, - 297.18 - ], - "text": "3t + 30◦\n+ sin", - "type": "text" - }, - { - "block_id": "p630-b20", - "global_id": 18483, - "bbox": [ - 328.47, - 275.98, - 337.16, - 289.62 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p630-b21", - "global_id": 18484, - "bbox": [ - 333.68, - 275.98, - 370.14, - 297.18 - ], - "text": "5t + 45◦", - "type": "text" - }, - { - "block_id": "p630-b22", - "global_id": 18485, - "bbox": [ - 107.82, - 306.32, - 484.41, - 328.24 - ], - "text": "is periodic. If it is periodic, find the fundamental frequency and the period. What harmonics are\npresent in x(t)?", - "type": "text" - }, - { - "block_id": "p630-b23", - "global_id": 18486, - "bbox": [ - 108.09, - 341.76, - 170.31, - 352.72 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p630-b24", - "global_id": 18487, - "bbox": [ - 107.82, - 359.71, - 484.4, - 371.16 - ], - "text": "Periodic with ω0 = 2/15 and period T0 = 15π. Signal x(t) contains the fifth and sixth harmonics.", - "type": "text" - }, - { - "block_id": "p630-b25", - "global_id": 18488, - "bbox": [ - 101.84, - 404.0, - 490.42, - 503.84 - ], - "text": "A HISTORICAL NOTE: BARON JEAN-BAPTISTE-JOSEPH FOURIER\n(1768–1830)\nThe Fourier series and integral comprise a most beautiful and fruitful development, which serves\nas an indispensable instrument in the treatment of many problems in mathematics, science, and\nengineering. Maxwell was so taken by the beauty of the Fourier series that he called it a great\nmathematical poem. In electrical engineering, it is central to the areas of communication, signal\nprocessing, and several other fields, including antennas, but its initial reception by the scientific\nworld was not enthusiastic. In fact, Fourier could not get his results published as a paper.", - "type": "text" - }, - { - "block_id": "p630-b26", - "global_id": 18489, - "bbox": [ - 101.84, - 505.83, - 490.43, - 635.35 - ], - "text": "Fourier, a tailor’s son, was orphaned at age 8 and educated at a local military college (run\nby Benedictine monks), where he excelled in mathematics. The Benedictines prevailed upon the\nyoung genius to choose the priesthood as his vocation, but revolution broke out before he could\ntake his vows. Fourier joined the people’s party. But in its early days, the French Revolution,\nlike most such upheavals, liquidated a large segment of the intelligentsia, including prominent\nscientists such as Lavoisier. Observing this trend, many intellectuals decided to leave France to\nsave themselves from a rapidly rising tide of barbarism. Fourier, despite his early enthusiasm for\nthe Revolution, narrowly escaped the guillotine twice. It was to the everlasting credit of Napoleon\nthat he stopped the persecution of the intelligentsia and founded new schools to replenish their\nranks. The 26-year-old Fourier was appointed chair of mathematics at the newly created École\nNormale in 1794 [1].", - "type": "text" - } - ] - }, - { - "page_num": 631, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p631-b0", - "global_id": 18490, - "bbox": [ - 203.95, - 62.89, - 516.11, - 71.98 - ], - "text": "6.1\nPeriodic Signal Representation by Trigonometric Fourier Series\n611", - "type": "text" - }, - { - "block_id": "p631-b1", - "global_id": 18491, - "bbox": [ - 234.92, - 318.99, - 408.77, - 328.95 - ], - "text": "Jean-Baptiste-Joseph Fourier and Napoleon", - "type": "text" - }, - { - "block_id": "p631-b2", - "global_id": 18492, - "bbox": [ - 127.59, - 349.86, - 516.15, - 515.23 - ], - "text": "Napoleon was the first modern ruler with a scientific education, and he was one of the rare\npersons who are equally comfortable with soldiers and scientists. The age of Napoleon was\none of the most fruitful in the history of science. Napoleon liked to sign himself as “member\nof Institut de France” (a fraternity of scientists), and he once expressed to Laplace his regret\nthat “force of circumstances has led me so far from the career of a scientist” [2]. Many great\nfigures in science and mathematics, including Fourier and Laplace, were honored and promoted\nby Napoleon. In 1798 he took a group of scientists, artists, and scholars—Fourier among them—on\nhis Egyptian expedition, with the promise of an exciting and historic union of adventure and\nresearch. Fourier proved to be a capable administrator of the newly formed Institut d’Égypte,\nwhich, incidentally, was responsible for the discovery of the Rosetta Stone. The inscription on\nthis stone in two languages and three scripts (hieroglyphic, demotic, and Greek) enabled Thomas\nYoung and Jean-François Champollion, a protégé of Fourier, to invent a method of translating\nhieroglyphic writings of ancient Egypt—the only significant result of Napoleon’s Egyptian\nexpedition.", - "type": "text" - }, - { - "block_id": "p631-b3", - "global_id": 18493, - "bbox": [ - 127.59, - 517.22, - 516.17, - 634.79 - ], - "text": "Back in France in 1801, Fourier briefly served in his former position as professor of\nmathematics at the École Polytechnique in Paris. In 1802 Napoleon appointed him the prefect\nof Isère (with its headquarters in Grenoble), a position in which Fourier served with distinction.\nFourier was named Baron of the Empire by Napoleon in 1809. Later, when Napoleon was exiled to\nElba, his route was to take him through Grenoble. Fourier had the route changed to avoid meeting\nNapoleon, which would have displeased Fourier’s new master, King Louis XVIII. Within a year,\nNapoleon escaped from Elba and returned to France. At Grenoble, Fourier was brought before him\nin chains. Napoleon scolded Fourier for his ungrateful behavior but reappointed him the prefect\nof Rhône at Lyons. Within four months Napoleon was defeated at Waterloo and was exiled to\nSt. Helena, where he died in 1821. Fourier once again was in disgrace as a Bonapartist and had", - "type": "text" - } - ] - }, - { - "page_num": 632, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p632-b0", - "global_id": 18494, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "612\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p632-b1", - "global_id": 18495, - "bbox": [ - 101.84, - 85.82, - 490.4, - 143.6 - ], - "text": "to pawn his possessions to keep himself alive. But through the intercession of a former student,\nwho was now a prefect of Paris, he was appointed director of the statistical bureau of the Seine, a\nposition that allowed him ample time for scholarly pursuits. Later, in 1827, he was elected to the\npowerful position of perpetual secretary of the Paris Academy of Science, a section of the Institut\nde France [3].", - "type": "text" - }, - { - "block_id": "p632-b2", - "global_id": 18496, - "bbox": [ - 101.84, - 145.59, - 490.42, - 382.71 - ], - "text": "While serving as the prefect of Grenoble, Fourier carried on his elaborate investigation of the\npropagation of heat in solid bodies, which led him to the Fourier series and the Fourier integral.\nOn December 21, 1807, he announced these results in a prize paper on the theory of heat. Fourier\nclaimed that an arbitrary function (continuous or with discontinuities) defined in a finite interval\nby an arbitrarily capricious graph can always be expressed as a sum of sinusoids (Fourier series).\nThe judges, who included the great French mathematicians Laplace, Lagrange, Legendre, Monge,\nand LaCroix, admitted the novelty and importance of Fourier’s work but criticized it for lack of\nmathematical rigor and generality. Lagrange thought it incredible that a sum of sines and cosines\ncould add up to anything but an infinitely differentiable function. Moreover, one of the properties\nof an infinitely differentiable function is that if we know its behavior over an arbitrarily small\ninterval, we can determine its behavior over the entire range (the Taylor–Maclaurin series). Such a\nfunction is far from an arbitrary or a capriciously drawn graph [4]. Laplace had additional reason\nto criticize Fourier’s work. Laplace and his students had already approached the problem of heat\nconduction from a different angle, and Laplace was reluctant to accept the superiority of Fourier’s\nmethod [5]. Fourier thought the criticism unjustified but was unable to prove his claim because the\ntools required for operations with infinite series were not available at the time. However, posterity\nhas proved Fourier to be closer to the truth than his critics. This is the classic conflict between\npure mathematicians and physicists or engineers, as we saw earlier (Ch. 4) in the life of Oliver\nHeaviside. In 1829 Dirichlet proved Fourier’s claim concerning capriciously drawn functions with\na few restrictions (Dirichlet conditions).", - "type": "text" - }, - { - "block_id": "p632-b3", - "global_id": 18497, - "bbox": [ - 101.84, - 384.7, - 490.38, - 430.53 - ], - "text": "Although three of the four judges were in favor of publication, Fourier’s paper was rejected\nbecause of vehement opposition by Lagrange. Fifteen years later, after several attempts and\ndisappointments, Fourier published the results in expanded form as a text, Théorie analytique\nde la chaleur, which is now a classic.", - "type": "text" - }, - { - "block_id": "p632-b4", - "global_id": 18498, - "bbox": [ - 102.2, - 474.47, - 355.06, - 504.36 - ], - "text": "6.2 EXISTENCE AND CONVERGENCE OF\nTHE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p632-b5", - "global_id": 18499, - "bbox": [ - 101.84, - 510.25, - 490.4, - 544.21 - ], - "text": "For the existence of the Fourier series, coefficients a0,an, and bn in Eq. (6.8) must be finite. It\nfollows from Eq. (6.8) that the existence of these coefficients is guaranteed if x(t) is absolutely\nintegrable over one period; that is,", - "type": "text" - }, - { - "block_id": "p632-b6", - "global_id": 18500, - "bbox": [ - 262.09, - 558.99, - 267.35, - 568.95 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p632-b7", - "global_id": 18501, - "bbox": [ - 267.35, - 584.08, - 274.22, - 592.39 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p632-b8", - "global_id": 18502, - "bbox": [ - 276.32, - 572.54, - 490.38, - 582.92 - ], - "text": "|x(t)|dt < ∞\n(6.16)", - "type": "text" - }, - { - "block_id": "p632-b9", - "global_id": 18503, - "bbox": [ - 101.85, - 612.86, - 490.39, - 634.79 - ], - "text": "However, existence, by itself, does not inform us about the nature and the manner in which the\nseries converges. We shall first discuss the notion of convergence.", - "type": "text" - } - ] - }, - { - "page_num": 633, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p633-b0", - "global_id": 18504, - "bbox": [ - 262.43, - 62.89, - 516.13, - 71.98 - ], - "text": "6.2\nExistence and Convergence of the Fourier Series\n613", - "type": "text" - }, - { - "block_id": "p633-b1", - "global_id": 18505, - "bbox": [ - 127.59, - 86.52, - 288.49, - 98.48 - ], - "text": "6.2-1 Convergence of a Series", - "type": "text" - }, - { - "block_id": "p633-b2", - "global_id": 18506, - "bbox": [ - 127.59, - 104.61, - 516.11, - 138.48 - ], - "text": "The key to many puzzles lies in the nature of the convergence of the Fourier series. Convergence\nof infinite series is a complex problem. It took mathematicians several decades to understand the\nconvergence aspect of the Fourier series. We shall barely scratch the surface here.", - "type": "text" - }, - { - "block_id": "p633-b3", - "global_id": 18507, - "bbox": [ - 127.59, - 140.47, - 516.15, - 329.76 - ], - "text": "Nothing annoys a student more than the discussion of convergence. “Have we not proved,”\nthey ask, “that a periodic signal x(t) can be expressed as a Fourier series”? Then why spoil the fun\nby this annoying discussion? All we have shown so far is that a signal represented by a Fourier\nseries in Eq. (6.1) is periodic. We have not proved the converse, that every periodic signal can\nbe expressed as a Fourier series. This issue will be tackled later, in Sec. 6.5-4, where it will be\nshown that a periodic signal can be represented by a Fourier series, as in Eq. (6.1), where the\nequality of the two sides of the equation is not in the ordinary sense, but in the mean-square sense\n(explained later in this discussion). But the astute reader should have been skeptical of the claims\nof the Fourier series to represent discontinuous functions in Figs. 6.2a and 6.6a. If x(t) has a jump\ndiscontinuity, say, at t = 0, then x(0+),x(0), and x(0−) are generally different. How could a series\nconsisting of the sum of continuous functions of the smoothest type (sinusoids) add to one value\nat t = 0−and a different value at t = 0 and yet another value at t = 0+? The demand is impossible\nto satisfy unless the math involved executes some spectacular acrobatics. How does a Fourier\nseries act under such conditions? Precisely for this reason, the great mathematicians Lagrange and\nLaplace, two of the judges examining Fourier’s paper, were skeptical of Fourier’s claims and voted\nagainst publication of the paper that later became a classic.", - "type": "text" - }, - { - "block_id": "p633-b4", - "global_id": 18508, - "bbox": [ - 127.59, - 331.76, - 516.16, - 401.49 - ], - "text": "There are also other issues. In any practical application, we can use only a finite number of\nterms in a series. If, with a fixed number of terms, the series guarantees convergence within an\narbitrarily small error at every value of t, such a series is highly desirable and is called a uniformly\nconvergent series. If a series converges at every value of t, but to guarantee convergence within a\ngiven error requires a different number of terms at different t, then the series is still convergent,\nbut less desirable. It goes under the name pointwise convergent series.", - "type": "text" - }, - { - "block_id": "p633-b5", - "global_id": 18509, - "bbox": [ - 127.59, - 403.39, - 516.13, - 461.27 - ], - "text": "Finally, we have the case of a series that refuses to converge at some t, no matter how many\nterms are added. But the series may converge in the mean; that is, the energy of the difference\nbetween x(t) and the corresponding finite term series approaches zero as the number of terms\napproaches infinity.† To explain this concept, let us consider representation of a function x(t) by\nan infinite series", - "type": "text" - }, - { - "block_id": "p633-b6", - "global_id": 18510, - "bbox": [ - 291.79, - 477.45, - 316.44, - 487.73 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p633-b7", - "global_id": 18511, - "bbox": [ - 318.49, - 467.28, - 332.59, - 477.95 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p633-b8", - "global_id": 18512, - "bbox": [ - 319.33, - 491.84, - 331.74, - 499.1 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p633-b9", - "global_id": 18513, - "bbox": [ - 333.69, - 477.45, - 351.93, - 488.53 - ], - "text": "zn(t)", - "type": "text" - }, - { - "block_id": "p633-b10", - "global_id": 18514, - "bbox": [ - 127.59, - 511.15, - 516.14, - 533.48 - ], - "text": "Let the partial sum of the first N terms of the series on the right-hand side be denoted by xN(t),\nthat is,", - "type": "text" - }, - { - "block_id": "p633-b11", - "global_id": 18515, - "bbox": [ - 289.02, - 552.38, - 319.21, - 563.46 - ], - "text": "xN(t) =", - "type": "text" - }, - { - "block_id": "p633-b12", - "global_id": 18516, - "bbox": [ - 321.26, - 542.43, - 335.35, - 552.88 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p633-b13", - "global_id": 18517, - "bbox": [ - 322.1, - 566.77, - 334.51, - 574.04 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p633-b14", - "global_id": 18518, - "bbox": [ - 336.46, - 552.38, - 354.71, - 563.46 - ], - "text": "zn(t)", - "type": "text" - }, - { - "block_id": "p633-b15", - "global_id": 18519, - "bbox": [ - 127.59, - 610.24, - 516.14, - 633.41 - ], - "text": "† The behavior is called “convergence in the mean” because minimizing the error energy over a certain\ninterval is equivalent to minimizing the mean-square value of the error over the same interval.", - "type": "text" - } - ] - }, - { - "page_num": 634, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p634-b0", - "global_id": 18520, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "614\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p634-b1", - "global_id": 18521, - "bbox": [ - 101.84, - 85.46, - 490.39, - 120.53 - ], - "text": "If we approximate x(t) by xN(t) (the partial sum of the first N terms of the series), the error in\nthe approximation is the difference x(t) −xN(t). The series converges in the mean to x(t) in the\ninterval (0, T0) if", - "type": "text" - }, - { - "block_id": "p634-b2", - "global_id": 18522, - "bbox": [ - 208.99, - 131.64, - 225.68, - 145.25 - ], - "text": "# T0", - "type": "text" - }, - { - "block_id": "p634-b3", - "global_id": 18523, - "bbox": [ - 214.24, - 141.4, - 383.25, - 163.78 - ], - "text": "0\n|x(t) −xN(t)|2 dt →0\nas\nN →∞", - "type": "text" - }, - { - "block_id": "p634-b4", - "global_id": 18524, - "bbox": [ - 101.84, - 180.7, - 490.41, - 298.67 - ], - "text": "Hence, the energy of the error x(t)−xN(t) approaches zero as N →∞. This form of convergence\ndoes not require the series to be equal to x(t) for all t. It just requires the energy of the difference\n(area under |x(t)−xN(t)|2) to vanish as N →∞. Superficially it may appear that if the energy of a\nsignal over an interval is zero, the signal (the error) must be zero everywhere. This is not true. The\nsignal energy can be zero even if there are nonzero values at a finite number of isolated points.\nThis is because although the signal is nonzero at a point (and zero everywhere else), the area under\nits square is still zero. Thus, a series that converges in the mean to x(t) need not converge to x(t) at\na finite number of points. This is precisely what happens to the Fourier series when x(t) has jump\ndiscontinuities. This is also what makes Fourier series convergence compatible with the Gibbs\nphenomenon, to be discussed later in this section.", - "type": "text" - }, - { - "block_id": "p634-b5", - "global_id": 18525, - "bbox": [ - 101.84, - 300.25, - 490.4, - 334.54 - ], - "text": "There is a simple criterion for ensuring that a periodic signal x(t) has a Fourier series that\nconverges in the mean. The Fourier series for x(t) converges to x(t) in the mean if x(t) has a finite\nenergy over one period, that is,", - "type": "text" - }, - { - "block_id": "p634-b6", - "global_id": 18526, - "bbox": [ - 261.21, - 344.85, - 266.47, - 354.81 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p634-b7", - "global_id": 18527, - "bbox": [ - 266.46, - 369.95, - 273.33, - 378.25 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p634-b8", - "global_id": 18528, - "bbox": [ - 275.44, - 354.61, - 490.38, - 368.78 - ], - "text": "|x(t)|2 dt < ∞\n(6.17)", - "type": "text" - }, - { - "block_id": "p634-b9", - "global_id": 18529, - "bbox": [ - 101.85, - 395.4, - 490.41, - 441.64 - ], - "text": "Thus, the periodic signal x(t), having a finite energy over one period, guarantees the convergence in\nthe mean of its Fourier series. In all the examples discussed so far, Eq. (6.17) is satisfied; hence the\ncorresponding Fourier series converges in the mean. Equation (6.17), like Eq. (6.16), guarantees\nthat the Fourier coefficients are finite.", - "type": "text" - }, - { - "block_id": "p634-b10", - "global_id": 18530, - "bbox": [ - 101.85, - 443.64, - 490.4, - 465.55 - ], - "text": "We shall now discuss an alternate set of criteria, due to Dirichlet, for convergence of the\nFourier series.", - "type": "text" - }, - { - "block_id": "p634-b11", - "global_id": 18531, - "bbox": [ - 101.84, - 486.13, - 490.39, - 548.11 - ], - "text": "DIRICHLET CONDITIONS\nDirichlet showed that if x(t) satisfies certain conditions (Dirichlet conditions), its Fourier series is\nguaranteed to converge pointwise at all points where x(t) is continuous. Moreover, at the points of\ndiscontinuities, x(t) converges to the value midway between the two values of x(t) on either side\nof the discontinuity. These conditions are:", - "type": "text" - }, - { - "block_id": "p634-b12", - "global_id": 18532, - "bbox": [ - 118.78, - 555.67, - 490.4, - 601.91 - ], - "text": "1. The function x(t) must be absolutely integrable; that is, it must satisfy Eq. (6.16).\n2. The function x(t) must have only a finite number of finite discontinuities in one\nperiod.\n3. The function x(t) must contain only a finite number of maxima and minima in one period.", - "type": "text" - }, - { - "block_id": "p634-b13", - "global_id": 18533, - "bbox": [ - 101.85, - 609.88, - 462.41, - 619.84 - ], - "text": "All practical signals, including those in Exs. 6.1, 6.2, 6.3, and 6.4, satisfy these conditions.", - "type": "text" - } - ] - }, - { - "page_num": 635, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p635-b0", - "global_id": 18534, - "bbox": [ - 262.43, - 62.89, - 516.13, - 71.98 - ], - "text": "6.2\nExistence and Convergence of the Fourier Series\n615", - "type": "text" - }, - { - "block_id": "p635-b1", - "global_id": 18535, - "bbox": [ - 127.59, - 86.52, - 477.22, - 98.48 - ], - "text": "6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping", - "type": "text" - }, - { - "block_id": "p635-b2", - "global_id": 18536, - "bbox": [ - 127.59, - 104.19, - 516.15, - 162.39 - ], - "text": "The trigonometric Fourier series of a signal x(t) shows explicitly the sinusoidal components of\nx(t). We can synthesize x(t) by adding the sinusoids in the spectrum of x(t). Let us synthesize\nthe square-pulse periodic signal x(t) of Fig. 6.6a by adding successive harmonics in its spectrum\nstep by step and observing the similarity of the resulting signal to x(t). The Fourier series for this\nfunction as found in Eq. (6.13) is", - "type": "text" - }, - { - "block_id": "p635-b3", - "global_id": 18537, - "bbox": [ - 201.52, - 174.28, - 234.4, - 191.22 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p635-b4", - "global_id": 18538, - "bbox": [ - 229.42, - 174.28, - 253.64, - 198.3 - ], - "text": "2 + 2\nπ", - "type": "text" - }, - { - "block_id": "p635-b6", - "global_id": 18539, - "bbox": [ - 263.66, - 174.28, - 299.15, - 191.22 - ], - "text": "cos t −1", - "type": "text" - }, - { - "block_id": "p635-b7", - "global_id": 18540, - "bbox": [ - 294.17, - 166.86, - 442.2, - 198.3 - ], - "text": "3 cos 3t + 1\n5 cos 5t −1\n7 cos 7t + · · ·", - "type": "text" - }, - { - "block_id": "p635-b8", - "global_id": 18541, - "bbox": [ - 127.59, - 209.32, - 516.16, - 327.3 - ], - "text": "We start the synthesis with only the first term in the series (n = 0), a constant 1/2 (dc); this is\na gross approximation of the square wave, as shown in Fig. 6.8a. In the next step we add the dc\n(n = 0) and the first harmonic (fundamental), which results in a signal shown in Fig. 6.8b. Observe\nthat the synthesized signal somewhat resembles x(t). It is a smoothed-out version of x(t). The\nsharp corners in x(t) are not reproduced in this signal because sharp corners mean rapid changes,\nand their reproduction requires rapidly varying (i.e., higher-frequency) components, which are\nexcluded. Figure 6.8c shows the sum of dc, first, and third harmonics (even harmonics are absent).\nAs we increase the number of harmonics progressively, as shown in Figs. 6.8d (sum up to the fifth\nharmonic) and 6.8e (sum up to the nineteenth harmonic), the edges of the pulses become sharper\nand the signal resembles x(t) more closely.", - "type": "text" - }, - { - "block_id": "p635-b9", - "global_id": 18542, - "bbox": [ - 127.89, - 342.67, - 416.41, - 354.79 - ], - "text": "ASYMPTOTIC RATE OF AMPLITUDE SPECTRUM DECAY", - "type": "text" - }, - { - "block_id": "p635-b10", - "global_id": 18543, - "bbox": [ - 127.59, - 358.82, - 516.15, - 416.6 - ], - "text": "Figure 6.8 brings out one interesting aspect of the Fourier series. Lower frequencies in the Fourier\nseries affect the large-scale behavior of x(t), whereas the higher frequencies determine the fine\nstructure such as rapid wiggling. Hence, sharp changes in x(t), being a part of fine structure,\nnecessitate higher frequencies in the Fourier series. The sharper the change [the higher the time\nderivative ˙x(t)], the higher are the frequencies needed in the series.", - "type": "text" - }, - { - "block_id": "p635-b11", - "global_id": 18544, - "bbox": [ - 127.59, - 418.6, - 516.17, - 572.02 - ], - "text": "The amplitude spectrum indicates the amounts (amplitudes) of various frequency components\nof x(t). If x(t) is a smooth function, its variations are less rapid. Synthesis of such a function\nrequires predominantly lower-frequency sinusoids and relatively small amounts of rapidly varying\n(higher-frequency) sinusoids. The amplitude spectrum of such a function would decay swiftly\nwith frequency. To synthesize such a function, we require fewer terms in the Fourier series\nfor a good approximation. On the other hand, a signal with sharp changes, such as jump\ndiscontinuities, contains rapid variations, and its synthesis requires a relatively large amount\nof high-frequency components. The amplitude spectrum of such a signal would decay slowly\nwith frequency, and to synthesize such a function, we require many terms in its Fourier\nseries for a good approximation. The square wave x(t) is a discontinuous function with jump\ndiscontinuities, and therefore its amplitude spectrum decays rather slowly, as 1/n [see Eq. (6.13)].\nOn the other hand, the triangular-pulse periodic signal in Fig. 6.4a is smoother because it is a\ncontinuous function (no jump discontinuities). Its spectrum decays rapidly with frequency as 1/n2", - "type": "text" - }, - { - "block_id": "p635-b12", - "global_id": 18545, - "bbox": [ - 127.59, - 574.01, - 192.05, - 583.97 - ], - "text": "[see Eq. (6.12)].", - "type": "text" - }, - { - "block_id": "p635-b13", - "global_id": 18546, - "bbox": [ - 127.59, - 585.55, - 516.14, - 619.84 - ], - "text": "We can show that if the first k −1 derivatives of a periodic signal x(t) are continuous and the\nkth derivative is discontinuous, then its amplitude spectrum Cn decays with frequency at least as\nrapidly as 1/nk+1 [6]. This result provides a simple and useful means for predicting the asymptotic", - "type": "text" - } - ] - }, - { - "page_num": 636, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p636-b0", - "global_id": 18547, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "616\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p636-b1", - "global_id": 18548, - "bbox": [ - 398.18, - 410.47, - 400.41, - 418.47 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p636-b2", - "global_id": 18549, - "bbox": [ - 398.18, - 322.67, - 400.41, - 330.67 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p636-b3", - "global_id": 18550, - "bbox": [ - 398.18, - 234.17, - 400.41, - 242.17 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p636-b4", - "global_id": 18551, - "bbox": [ - 398.18, - 143.92, - 400.41, - 151.92 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p636-b5", - "global_id": 18552, - "bbox": [ - 267.88, - 159.06, - 276.76, - 167.06 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p636-b6", - "global_id": 18553, - "bbox": [ - 267.42, - 247.69, - 277.23, - 255.69 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p636-b7", - "global_id": 18554, - "bbox": [ - 267.88, - 336.15, - 276.76, - 344.15 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p636-b8", - "global_id": 18555, - "bbox": [ - 267.66, - 424.05, - 276.99, - 432.05 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p636-b9", - "global_id": 18556, - "bbox": [ - 267.88, - 513.74, - 276.76, - 521.74 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p636-b10", - "global_id": 18557, - "bbox": [ - 275.02, - 88.18, - 279.02, - 96.18 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p636-b11", - "global_id": 18558, - "bbox": [ - 198.04, - 143.21, - 345.51, - 151.5 - ], - "text": "p\np2\np2\np", - "type": "text" - }, - { - "block_id": "p636-b12", - "global_id": 18559, - "bbox": [ - 280.48, - 111.65, - 283.48, - 125.1 - ], - "text": "2\n1", - "type": "text" - }, - { - "block_id": "p636-b13", - "global_id": 18560, - "bbox": [ - 125.76, - 528.44, - 469.34, - 537.67 - ], - "text": "Figure 6.8 Synthesis of a square-pulse periodic signal by successive addition of its harmonics.", - "type": "text" - }, - { - "block_id": "p636-b14", - "global_id": 18561, - "bbox": [ - 101.84, - 564.08, - 490.41, - 609.91 - ], - "text": "rate of convergence of the Fourier series. In the case of the square-wave signal (Fig. 6.6a), the\nzeroth derivative of the signal (the signal itself) is discontinuous so that k = 0. For the triangular\nperiodic signal in Fig. 6.4a, the first derivative is discontinuous; that is, k = 1. For this reason, the\nspectra of these signals decay as 1/n and 1/n2, respectively.", - "type": "text" - } - ] - }, - { - "page_num": 637, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p637-b0", - "global_id": 18562, - "bbox": [ - 262.43, - 62.89, - 516.13, - 71.98 - ], - "text": "6.2\nExistence and Convergence of the Fourier Series\n617", - "type": "text" - }, - { - "block_id": "p637-b1", - "global_id": 18563, - "bbox": [ - 102.51, - 93.91, - 482.37, - 119.82 - ], - "text": "EXAMPLE 6.5\nSquare-Wave Synthesis by Truncated Fourier Series\nUsing MATLAB", - "type": "text" - }, - { - "block_id": "p637-b2", - "global_id": 18564, - "bbox": [ - 128.9, - 136.49, - 502.73, - 158.4 - ], - "text": "Use MATLAB to synthesize and plot the square wave of Fig. 6.8a using a Fourier series that\nis truncated to the 19th harmonic. The result should match Fig. 6.8e.", - "type": "text" - }, - { - "block_id": "p637-b3", - "global_id": 18565, - "bbox": [ - 128.9, - 181.32, - 400.04, - 191.28 - ], - "text": "To synthesize the waveform, we use the Fourier series of Eq. (6.13).", - "type": "text" - }, - { - "block_id": "p637-b4", - "global_id": 18566, - "bbox": [ - 128.9, - 201.53, - 442.73, - 271.27 - ], - "text": ">>\nx = @(t) 1.0*(mod(t+pi/2,2*pi)<=pi);\n>>\nt = linspace(-2*pi,2*pi,10001);\n>>\nx19 = 0.5*ones(size(t));\n>>\nfor n=1:19, x19 = x19+2/(pi*n)*sin(pi*n/2)*cos(n*t); end\n>>\nplot(t,x19,’k-’); axis([-2*pi 2*pi -0.2 1.2]);\n>>\nxlabel(’t’); ylabel(’x_{19}(t)’);", - "type": "text" - }, - { - "block_id": "p637-b5", - "global_id": 18567, - "bbox": [ - 128.9, - 280.94, - 338.78, - 290.9 - ], - "text": "As expected, the result of Fig. 6.9 matches Fig. 6.8e.", - "type": "text" - }, - { - "block_id": "p637-b6", - "global_id": 18568, - "bbox": [ - 159.83, - 436.61, - 487.57, - 457.7 - ], - "text": "–6\n–4\n–2\n0\n2\n4\n6\nt", - "type": "text" - }, - { - "block_id": "p637-b7", - "global_id": 18569, - "bbox": [ - 149.46, - 412.27, - 153.46, - 420.27 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p637-b8", - "global_id": 18570, - "bbox": [ - 142.71, - 374.51, - 153.39, - 382.51 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p637-b9", - "global_id": 18571, - "bbox": [ - 149.46, - 336.74, - 153.46, - 344.74 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p637-b10", - "global_id": 18572, - "bbox": [ - 128.2, - 369.05, - 137.73, - 386.61 - ], - "text": "x19(t)", - "type": "text" - }, - { - "block_id": "p637-b11", - "global_id": 18573, - "bbox": [ - 127.51, - 464.38, - 437.25, - 473.62 - ], - "text": "Figure 6.9 Using MATLAB to synthesize a square wave via truncated Fourier series.", - "type": "text" - }, - { - "block_id": "p637-b12", - "global_id": 18574, - "bbox": [ - 127.59, - 516.7, - 516.16, - 602.59 - ], - "text": "PHASE SPECTRUM: THE WOMAN BEHIND A SUCCESSFUL MAN\nThe role of the amplitude spectrum in shaping the waveform x(t) is quite clear. However, the role\nof the phase spectrum in shaping this waveform is less obvious. Yet, the phase spectrum, like\nthe woman behind a successful man,† plays an equally important role in waveshaping. We can\nexplain this role by considering a signal x(t) that has rapid changes, such as jump discontinuities.\nTo synthesize an instantaneous change at a jump discontinuity, the phases of the various sinusoidal\ncomponents in its spectrum must be such that all (or most) of the harmonic components will have", - "type": "text" - }, - { - "block_id": "p637-b13", - "global_id": 18575, - "bbox": [ - 127.59, - 621.19, - 372.83, - 633.41 - ], - "text": "† Or, to keep up with the times, the man behind a successful woman.", - "type": "text" - } - ] - }, - { - "page_num": 638, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p638-b0", - "global_id": 18576, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "618\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p638-b1", - "global_id": 18577, - "bbox": [ - 101.84, - 85.82, - 490.39, - 131.64 - ], - "text": "one sign before the discontinuity and the opposite sign after the discontinuity. This will result in\na sharp change in x(t) at the point of discontinuity. We can verify this fact in any waveform with\njump discontinuity. Consider, for example, the sawtooth waveform in Fig. 6.7b. This waveform\nhas a discontinuity at t = 1. The Fourier series for this waveform, as given in Drill 6.1b, is", - "type": "text" - }, - { - "block_id": "p638-b2", - "global_id": 18578, - "bbox": [ - 107.92, - 160.34, - 146.87, - 177.38 - ], - "text": "x(t) = 2A", - "type": "text" - }, - { - "block_id": "p638-b3", - "global_id": 18579, - "bbox": [ - 137.86, - 174.08, - 143.84, - 184.04 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p638-b5", - "global_id": 18580, - "bbox": [ - 154.61, - 160.43, - 228.22, - 177.38 - ], - "text": "cos(πt −90◦) + 1", - "type": "text" - }, - { - "block_id": "p638-b6", - "global_id": 18581, - "bbox": [ - 223.24, - 153.02, - 484.32, - 184.46 - ], - "text": "2 cos(2πt + 90◦) + 1\n3 cos(3πt −90◦) + 1\n4 cos(4πt + 90◦) + · · ·\n!", - "type": "text" - }, - { - "block_id": "p638-b7", - "global_id": 18582, - "bbox": [ - 101.84, - 212.81, - 490.42, - 330.26 - ], - "text": "Figure 6.10 shows the first three components of this series. The phases of all the (infinite)\ncomponents are such that all the components are positive just before t = 1 and turn negative\njust after t = 1, the point of discontinuity. The same behavior is also observed at t = −1, where\na similar discontinuity occurs. This sign change in all the harmonics adds up to produce very\nnearly a jump discontinuity. The role of the phase spectrum is crucial in achieving a sharp change\nin the waveform. If we ignore the phase spectrum when trying to reconstruct this signal, the\nresult will be a smeared and spread-out waveform. In general, the phase spectrum is just as\ncrucial as the amplitude spectrum in determining the waveform. The synthesis of any signal x(t)\nis achieved by using a proper combination of amplitudes and phases of various sinusoids. This\nunique combination is the Fourier spectrum of x(t).", - "type": "text" - }, - { - "block_id": "p638-b8", - "global_id": 18583, - "bbox": [ - 290.93, - 396.47, - 294.93, - 404.47 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p638-b9", - "global_id": 18584, - "bbox": [ - 291.24, - 364.79, - 302.82, - 372.87 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p638-b10", - "global_id": 18585, - "bbox": [ - 280.88, - 547.38, - 284.88, - 555.38 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p638-b11", - "global_id": 18586, - "bbox": [ - 290.9, - 473.58, - 294.9, - 481.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p638-b12", - "global_id": 18587, - "bbox": [ - 290.86, - 592.76, - 294.86, - 600.76 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p638-b13", - "global_id": 18588, - "bbox": [ - 220.36, - 397.13, - 353.57, - 405.52 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p638-b14", - "global_id": 18589, - "bbox": [ - 449.48, - 596.77, - 451.7, - 604.77 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p638-b15", - "global_id": 18590, - "bbox": [ - 449.48, - 473.41, - 451.7, - 481.41 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p638-b16", - "global_id": 18591, - "bbox": [ - 449.48, - 396.05, - 451.7, - 404.05 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p638-b17", - "global_id": 18592, - "bbox": [ - 449.48, - 548.32, - 451.7, - 556.32 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p638-b18", - "global_id": 18593, - "bbox": [ - 427.1, - 530.05, - 482.86, - 538.05 - ], - "text": "Second harmonic", - "type": "text" - }, - { - "block_id": "p638-b19", - "global_id": 18594, - "bbox": [ - 421.47, - 457.85, - 463.25, - 465.85 - ], - "text": "Fundamental", - "type": "text" - }, - { - "block_id": "p638-b20", - "global_id": 18595, - "bbox": [ - 427.1, - 571.85, - 477.09, - 579.85 - ], - "text": "Third harmonic", - "type": "text" - }, - { - "block_id": "p638-b21", - "global_id": 18596, - "bbox": [ - 125.76, - 619.96, - 376.56, - 629.19 - ], - "text": "Figure 6.10 Role of the phase spectrum in shaping a periodic signal.", - "type": "text" - } - ] - }, - { - "page_num": 639, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p639-b0", - "global_id": 18597, - "bbox": [ - 262.43, - 62.89, - 516.13, - 71.98 - ], - "text": "6.2\nExistence and Convergence of the Fourier Series\n619", - "type": "text" - }, - { - "block_id": "p639-b1", - "global_id": 18598, - "bbox": [ - 127.89, - 86.19, - 418.92, - 112.27 - ], - "text": "FOURIER SYNTHESIS OF DISCONTINUOUS FUNCTIONS:\nTHE GIBBS PHENOMENON", - "type": "text" - }, - { - "block_id": "p639-b2", - "global_id": 18599, - "bbox": [ - 127.59, - 115.88, - 516.17, - 233.86 - ], - "text": "Figure 6.8 showed the square function x(t) and its approximation by a truncated trigonometric\nFourier series that includes only the first N harmonics for N = 1, 3, 5, and 19. The plot of the\ntruncated series approximates closely the function x(t) as N increases, and we expect that the series\nwill converge exactly to x(t) as N →∞. Yet the curious fact, as seen from Fig. 6.8, is that even for\nlarge N, the truncated series exhibits an oscillatory behavior and an overshoot approaching a value\nof about 9% in the vicinity of the discontinuity at the nearest peak of oscillation.† Regardless of the\nvalue of N, the overshoot remains at about 9%. Such strange behavior certainly would undermine\nanyone’s faith in the Fourier series. In fact, this behavior puzzled many scholars at the turn of the\ncentury. Josiah Willard Gibbs, an eminent mathematical physicist who was the inventor of vector\nanalysis, gave a mathematical explanation of this behavior (now called the Gibbs phenomenon).", - "type": "text" - }, - { - "block_id": "p639-b3", - "global_id": 18600, - "bbox": [ - 127.59, - 235.85, - 516.15, - 377.32 - ], - "text": "We can reconcile the apparent aberration in the behavior of the Fourier series by observing\nfrom Fig. 6.8 that the frequency of oscillation of the synthesized signal is Nf0, so the width of the\nspike with 9% overshoot is approximately 1/2Nf0. As we increase N, the frequency of oscillation\nincreases and the spike width 1/2Nf0 diminishes. As N →∞, the error power →0 because the\nerror consists mostly of the spikes, whose widths →0. Therefore, as N →∞, the corresponding\nFourier series differs from x(t) by about 9% at the immediate left and right of the points of\ndiscontinuity, and yet the error power →0. The reason for all this confusion is that in this case, the\nFourier series converges in the mean. When this happens, all we promise is that the error energy\n(over one period) →0 as N →∞. Thus, the series may differ from x(t) at some points and yet\nhave the error signal power zero, as verified earlier. Note that the series, in this case, also converges\npointwise at all points except the points of discontinuity. It is precisely at the discontinuities that\nthe series differs from x(t) by 9%.‡", - "type": "text" - }, - { - "block_id": "p639-b4", - "global_id": 18601, - "bbox": [ - 127.59, - 379.21, - 516.16, - 437.09 - ], - "text": "When we use only the first N terms in the Fourier series to synthesize a signal, we are\nabruptly terminating the series, giving a unit weight to the first N harmonics and zero weight\nto all the remaining harmonics beyond N. This abrupt termination of the series causes the Gibbs\nphenomenon in synthesis of discontinuous functions. Section 7.8 offers more discussion on the\nGibbs phenomenon, its ramifications, and cure.", - "type": "text" - }, - { - "block_id": "p639-b5", - "global_id": 18602, - "bbox": [ - 127.59, - 438.67, - 516.15, - 496.86 - ], - "text": "The Gibbs phenomenon is present only when there is a jump discontinuity in x(t). When\na continuous function x(t) is synthesized by using the first N terms of the Fourier series, the\nsynthesized function approaches x(t) for all t as N →∞. No Gibbs phenomenon appears. This\ncan be seen in Fig. 6.11, which shows one cycle of a continuous periodic signal being synthesized\nfrom the first 19 harmonics. Compare the similar situation for a discontinuous signal in Fig. 6.8.", - "type": "text" - }, - { - "block_id": "p639-b6", - "global_id": 18603, - "bbox": [ - 133.57, - 530.14, - 331.77, - 542.1 - ], - "text": "DRILL 6.3\nRate of Spectral Decay", - "type": "text" - }, - { - "block_id": "p639-b7", - "global_id": 18604, - "bbox": [ - 133.57, - 551.22, - 510.14, - 573.13 - ], - "text": "By inspection of signals in Figs. 6.2a, 6.7a, and 6.7b, determine the asymptotic rate of decay of\ntheir amplitude spectra.", - "type": "text" - }, - { - "block_id": "p639-b8", - "global_id": 18605, - "bbox": [ - 127.59, - 598.98, - 516.13, - 633.41 - ], - "text": "† There is also an undershoot of 9% at the other side [at t = (π/2)+] of the discontinuity.\n‡ Actually, at discontinuities, the series converges to a value midway between the values on either side of the\ndiscontinuity. The 9% overshoot occurs at t = (π/2)−and 9% undershoot occurs at t = (π/2)+.", - "type": "text" - } - ] - }, - { - "page_num": 640, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p640-b0", - "global_id": 18606, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "620\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p640-b1", - "global_id": 18607, - "bbox": [ - 308.54, - 195.52, - 472.68, - 203.55 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p640-b2", - "global_id": 18608, - "bbox": [ - 299.3, - 89.68, - 303.3, - 97.68 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p640-b3", - "global_id": 18609, - "bbox": [ - 125.76, - 212.44, - 409.63, - 221.67 - ], - "text": "Figure 6.11 Fourier synthesis of a continuous signal using first 19 harmonics.", - "type": "text" - }, - { - "block_id": "p640-b4", - "global_id": 18610, - "bbox": [ - 108.09, - 251.18, - 170.31, - 262.13 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p640-b5", - "global_id": 18611, - "bbox": [ - 107.82, - 268.19, - 235.74, - 279.5 - ], - "text": "1/n,1/n2, and 1/n, respectively.", - "type": "text" - }, - { - "block_id": "p640-b6", - "global_id": 18612, - "bbox": [ - 102.14, - 311.54, - 378.72, - 323.67 - ], - "text": "A HISTORICAL NOTE ON THE GIBBS PHENOMENON", - "type": "text" - }, - { - "block_id": "p640-b7", - "global_id": 18613, - "bbox": [ - 101.84, - 327.7, - 490.41, - 385.48 - ], - "text": "Normally speaking, troublesome functions with strange behavior are invented by mathematicians;\nwe rarely see such oddities in practice. In the case of the Gibbs phenomenon, however, the tables\nwere turned. A rather puzzling behavior was observed in a mundane object, a mechanical wave\nsynthesizer, and then well-known mathematicians of the day were dispatched on the scent of it to\ndiscover its hideout.", - "type": "text" - }, - { - "block_id": "p640-b8", - "global_id": 18614, - "bbox": [ - 101.84, - 387.48, - 490.42, - 600.68 - ], - "text": "Albert Michelson (of Michelson–Morley fame) was an intense, practical man who developed\ningenious physical instruments of extraordinary precision, mostly in the field of optics. His\nharmonic analyzer, developed in 1898, could compute the first 80 coefficients of the Fourier\nseries of a signal x(t) specified by any graphical description. The instrument could also be used\nas a harmonic synthesizer, which could plot a function x(t) generated by summing the first 80\nharmonics (Fourier components) of arbitrary amplitudes and phases. This analyzer, therefore, had\nthe ability of self-checking its operation by analyzing a signal x(t) and then adding the resulting\n80 components to see whether the sum yielded a close approximation of x(t).\nMichelson found that the instrument checked very well with most of signals analyzed.\nHowever, when he tried a discontinuous function, such as a square wave,† a curious behavior was\nobserved. The sum of 80 components showed oscillatory behavior (ringing), with an overshoot\nof 9% in the vicinity of the points of discontinuity. Moreover, this behavior was a constant\nfeature regardless of the number of terms added. A larger number of terms made the oscillations\nproportionately faster, but regardless of the number of terms added, the overshoot remained 9%.\nThis puzzling behavior caused Michelson to suspect some mechanical defect in his synthesizer. He\nwrote about his observation in a letter to Nature (December 1898). Josiah Willard Gibbs, who was\na professor at Yale, investigated and clarified this behavior for a sawtooth periodic signal in a letter\nto Nature [7]. Later, in 1906, Bôcher generalized the result for any function with discontinuity [8].", - "type": "text" - }, - { - "block_id": "p640-b9", - "global_id": 18615, - "bbox": [ - 101.84, - 621.19, - 261.42, - 633.41 - ], - "text": "† Actually, it was a periodic sawtooth signal.", - "type": "text" - } - ] - }, - { - "page_num": 641, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p641-b0", - "global_id": 18616, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n621", - "type": "text" - }, - { - "block_id": "p641-b1", - "global_id": 18617, - "bbox": [ - 235.31, - 306.95, - 408.39, - 316.91 - ], - "text": "Albert Michelson and Josiah Willard Gibbs", - "type": "text" - }, - { - "block_id": "p641-b2", - "global_id": 18618, - "bbox": [ - 127.59, - 333.53, - 516.12, - 367.51 - ], - "text": "It was Bôcher who gave the name Gibbs phenomenon to this behavior. Gibbs showed that the\npeculiar behavior in the synthesis of a square wave was inherent in the behavior of the Fourier\nseries because of nonuniform convergence at the points of discontinuity.", - "type": "text" - }, - { - "block_id": "p641-b3", - "global_id": 18619, - "bbox": [ - 127.59, - 369.5, - 516.14, - 439.24 - ], - "text": "This, however, is not the end of the story. Both Bôcher and Gibbs were under the impression\nthat this property had remained undiscovered until Gibbs’s work published in 1899. It is now\nknown that what is called the Gibbs phenomenon had been observed in 1848 by Wilbraham\nof Trinity College, Cambridge, who clearly saw the behavior of the sum of the Fourier series\ncomponents in the periodic sawtooth signal later investigated by Gibbs [9]. Apparently, this work\nwas not known to most people, including Gibbs and Bôcher.", - "type": "text" - }, - { - "block_id": "p641-b4", - "global_id": 18620, - "bbox": [ - 127.94, - 472.38, - 354.97, - 486.33 - ], - "text": "6.3 EXPONENTIAL FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p641-b5", - "global_id": 18621, - "bbox": [ - 127.59, - 490.68, - 515.51, - 503.05 - ], - "text": "By using Euler’s equality, we can express cos nω0t and sin nω0t in terms of exponentials ejnω0t", - "type": "text" - }, - { - "block_id": "p641-b6", - "global_id": 18622, - "bbox": [ - 127.59, - 502.63, - 516.15, - 574.01 - ], - "text": "and e−jnω0t. Clearly, we should be able to express the trigonometric Fourier series in Eq. (6.7)\nin terms of exponentials of the form ejnω0t with the index n taking on all integer values from\n−∞to ∞, including zero. Derivation of the exponential Fourier series from the results already\nderived for the trigonometric Fourier series is straightforward, involving conversion of sinusoids\nto exponentials. We shall, however, derive them here independently, without using the prior results\nof the trigonometric series.", - "type": "text" - }, - { - "block_id": "p641-b7", - "global_id": 18623, - "bbox": [ - 127.59, - 575.59, - 516.13, - 597.92 - ], - "text": "This discussion shows that the exponential Fourier series for a periodic signal x(t) can be\nexpressed as", - "type": "text" - }, - { - "block_id": "p641-b8", - "global_id": 18624, - "bbox": [ - 281.39, - 613.42, - 306.04, - 623.7 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p641-b9", - "global_id": 18625, - "bbox": [ - 311.79, - 603.25, - 325.88, - 613.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p641-b10", - "global_id": 18626, - "bbox": [ - 308.09, - 627.27, - 329.56, - 634.46 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p641-b11", - "global_id": 18627, - "bbox": [ - 330.69, - 611.7, - 361.7, - 624.5 - ], - "text": "Dnejnω0t", - "type": "text" - } - ] - }, - { - "page_num": 642, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p642-b0", - "global_id": 18628, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "622\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p642-b1", - "global_id": 18629, - "bbox": [ - 101.84, - 83.68, - 490.38, - 109.22 - ], - "text": "To derive the coefficients Dn, we multiply both sides of this equation by e−jmω0t (m integer) and\nintegrate over one period. This yields", - "type": "text" - }, - { - "block_id": "p642-b2", - "global_id": 18630, - "bbox": [ - 209.15, - 116.34, - 214.41, - 126.31 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b3", - "global_id": 18631, - "bbox": [ - 214.42, - 141.44, - 221.28, - 149.74 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b4", - "global_id": 18632, - "bbox": [ - 223.39, - 126.1, - 285.64, - 140.17 - ], - "text": "x(t)e−jmω0t dt =", - "type": "text" - }, - { - "block_id": "p642-b5", - "global_id": 18633, - "bbox": [ - 291.38, - 119.72, - 305.47, - 130.4 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p642-b6", - "global_id": 18634, - "bbox": [ - 287.68, - 143.75, - 309.16, - 150.94 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p642-b7", - "global_id": 18635, - "bbox": [ - 310.27, - 130.21, - 320.95, - 140.98 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p642-b8", - "global_id": 18636, - "bbox": [ - 322.56, - 116.34, - 327.82, - 126.31 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b9", - "global_id": 18637, - "bbox": [ - 327.83, - 141.44, - 334.69, - 149.74 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b10", - "global_id": 18638, - "bbox": [ - 337.9, - 126.1, - 382.89, - 140.17 - ], - "text": "ej(n−m)ω0t dt", - "type": "text" - }, - { - "block_id": "p642-b11", - "global_id": 18639, - "bbox": [ - 101.85, - 162.46, - 487.45, - 173.02 - ], - "text": "To simplify this expression, we use the orthogonality property of exponentials, which states that†", - "type": "text" - }, - { - "block_id": "p642-b12", - "global_id": 18640, - "bbox": [ - 223.47, - 177.66, - 228.73, - 187.62 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b13", - "global_id": 18641, - "bbox": [ - 228.73, - 202.76, - 235.59, - 211.07 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b14", - "global_id": 18642, - "bbox": [ - 237.7, - 187.42, - 305.58, - 201.5 - ], - "text": "ejnω0t e−jmω0t dt =", - "type": "text" - }, - { - "block_id": "p642-b15", - "global_id": 18643, - "bbox": [ - 307.63, - 177.24, - 490.38, - 208.25 - ], - "text": "0\nm̸ = n\nT0\nm = n\n(6.18)", - "type": "text" - }, - { - "block_id": "p642-b16", - "global_id": 18644, - "bbox": [ - 101.85, - 220.89, - 250.44, - 234.4 - ], - "text": "Thus,\n#", - "type": "text" - }, - { - "block_id": "p642-b17", - "global_id": 18645, - "bbox": [ - 250.43, - 249.54, - 257.3, - 257.84 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b18", - "global_id": 18646, - "bbox": [ - 260.51, - 234.2, - 346.56, - 249.15 - ], - "text": "x(t)e−jmω0t dt = DmT0", - "type": "text" - }, - { - "block_id": "p642-b19", - "global_id": 18647, - "bbox": [ - 101.84, - 264.67, - 189.55, - 274.64 - ], - "text": "from which we obtain", - "type": "text" - }, - { - "block_id": "p642-b20", - "global_id": 18648, - "bbox": [ - 242.73, - 274.08, - 275.77, - 292.1 - ], - "text": "Dm = 1", - "type": "text" - }, - { - "block_id": "p642-b21", - "global_id": 18649, - "bbox": [ - 268.52, - 288.04, - 277.55, - 298.87 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b22", - "global_id": 18650, - "bbox": [ - 280.35, - 267.09, - 285.61, - 277.06 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b23", - "global_id": 18651, - "bbox": [ - 285.61, - 292.19, - 292.47, - 300.49 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b24", - "global_id": 18652, - "bbox": [ - 294.58, - 276.85, - 349.5, - 291.02 - ], - "text": "x(t)e−jmω0t dt.", - "type": "text" - }, - { - "block_id": "p642-b25", - "global_id": 18653, - "bbox": [ - 119.78, - 307.32, - 379.7, - 317.29 - ], - "text": "To summarize, the exponential Fourier series can be expressed as", - "type": "text" - }, - { - "block_id": "p642-b26", - "global_id": 18654, - "bbox": [ - 172.96, - 337.97, - 197.62, - 348.24 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p642-b27", - "global_id": 18655, - "bbox": [ - 203.35, - 327.8, - 217.45, - 338.47 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p642-b28", - "global_id": 18656, - "bbox": [ - 199.66, - 351.82, - 221.14, - 359.01 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p642-b29", - "global_id": 18657, - "bbox": [ - 222.26, - 331.41, - 349.58, - 349.43 - ], - "text": "Dnejnω0t\nwhere\nDn = 1", - "type": "text" - }, - { - "block_id": "p642-b30", - "global_id": 18658, - "bbox": [ - 342.33, - 345.36, - 351.35, - 356.19 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b31", - "global_id": 18659, - "bbox": [ - 354.16, - 324.41, - 359.42, - 334.38 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b32", - "global_id": 18660, - "bbox": [ - 359.42, - 349.51, - 366.28, - 357.81 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b33", - "global_id": 18661, - "bbox": [ - 368.39, - 334.17, - 490.38, - 348.35 - ], - "text": "x(t)e−jnω0t dt\n(6.19)", - "type": "text" - }, - { - "block_id": "p642-b34", - "global_id": 18662, - "bbox": [ - 101.84, - 370.29, - 490.4, - 440.03 - ], - "text": "Observe the compactness of Eq. (6.19) and compare it with the trigonometric Fourier series\nexpression. Such a comparison demonstrates very clearly the principal virtue of the exponential\nFourier series. First, the form of the series is most compact. Second, the mathematical expression\nfor deriving the coefficients of the series is also compact. It is much more convenient to handle\nthe exponential series than the trigonometric one. For these reasons we shall use the exponential\n(rather than trigonometric) representation of signals in the rest of the book.", - "type": "text" - }, - { - "block_id": "p642-b35", - "global_id": 18663, - "bbox": [ - 101.84, - 441.6, - 490.38, - 463.94 - ], - "text": "We can now relate Dn to trigonometric series coefficients an and bn. Setting n = 0 in Eq. (6.19),\nwe obtain", - "type": "text" - }, - { - "block_id": "p642-b36", - "global_id": 18664, - "bbox": [ - 280.12, - 467.05, - 311.62, - 478.5 - ], - "text": "D0 = a0", - "type": "text" - }, - { - "block_id": "p642-b37", - "global_id": 18665, - "bbox": [ - 101.84, - 486.99, - 184.14, - 497.37 - ], - "text": "Moreover, for n̸ = 0,", - "type": "text" - }, - { - "block_id": "p642-b38", - "global_id": 18666, - "bbox": [ - 165.25, - 508.19, - 196.74, - 526.21 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p642-b39", - "global_id": 18667, - "bbox": [ - 189.49, - 522.14, - 198.52, - 532.98 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b40", - "global_id": 18668, - "bbox": [ - 201.32, - 501.2, - 206.58, - 511.16 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b41", - "global_id": 18669, - "bbox": [ - 206.58, - 526.29, - 213.44, - 534.6 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b42", - "global_id": 18670, - "bbox": [ - 215.55, - 508.09, - 291.57, - 525.9 - ], - "text": "x(t)cosnω0tdt −j", - "type": "text" - }, - { - "block_id": "p642-b43", - "global_id": 18671, - "bbox": [ - 285.43, - 522.14, - 294.45, - 532.98 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b44", - "global_id": 18672, - "bbox": [ - 297.25, - 501.2, - 302.51, - 511.16 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b45", - "global_id": 18673, - "bbox": [ - 302.51, - 526.29, - 309.38, - 534.6 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b46", - "global_id": 18674, - "bbox": [ - 311.48, - 508.19, - 385.67, - 525.9 - ], - "text": "x(t)sinnω0tdt = 1", - "type": "text" - }, - { - "block_id": "p642-b47", - "global_id": 18675, - "bbox": [ - 380.69, - 514.76, - 490.38, - 532.21 - ], - "text": "2 (an −jbn)\n(6.20)", - "type": "text" - }, - { - "block_id": "p642-b48", - "global_id": 18676, - "bbox": [ - 101.84, - 552.57, - 490.37, - 576.5 - ], - "text": "† We can readily prove this property as follows. For the case of m = n, the integrand in Eq. (6.18) is unity and\nthe integral is T0. When m̸ = n, the integral on the left-hand side of Eq. (6.18) can be expressed as", - "type": "text" - }, - { - "block_id": "p642-b49", - "global_id": 18677, - "bbox": [ - 183.71, - 576.86, - 188.44, - 585.83 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b50", - "global_id": 18678, - "bbox": [ - 188.44, - 599.3, - 194.78, - 606.95 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b51", - "global_id": 18679, - "bbox": [ - 196.78, - 585.64, - 247.21, - 598.31 - ], - "text": "ej(n−m)ω0t dt =", - "type": "text" - }, - { - "block_id": "p642-b52", - "global_id": 18680, - "bbox": [ - 249.06, - 576.86, - 253.79, - 585.83 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b53", - "global_id": 18681, - "bbox": [ - 253.79, - 599.3, - 260.13, - 606.95 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b54", - "global_id": 18682, - "bbox": [ - 262.12, - 589.06, - 335.18, - 599.14 - ], - "text": "cos(n −m)ω0tdt + j", - "type": "text" - }, - { - "block_id": "p642-b55", - "global_id": 18683, - "bbox": [ - 336.18, - 576.86, - 340.91, - 585.83 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p642-b56", - "global_id": 18684, - "bbox": [ - 340.91, - 599.3, - 347.25, - 606.95 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p642-b57", - "global_id": 18685, - "bbox": [ - 349.24, - 589.06, - 408.37, - 599.14 - ], - "text": "sin(n −m)ω0tdt", - "type": "text" - }, - { - "block_id": "p642-b58", - "global_id": 18686, - "bbox": [ - 101.84, - 613.11, - 490.38, - 633.41 - ], - "text": "Both the integrals on the right-hand side represent area under n −m number of cycles. Because n −m is an\ninteger, both the areas are zero. Hence, Eq. (6.18) follows.", - "type": "text" - } - ] - }, - { - "page_num": 643, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p643-b0", - "global_id": 18687, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n623", - "type": "text" - }, - { - "block_id": "p643-b1", - "global_id": 18688, - "bbox": [ - 127.59, - 85.82, - 141.97, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p643-b2", - "global_id": 18689, - "bbox": [ - 188.28, - 95.16, - 225.21, - 113.18 - ], - "text": "D−n = 1", - "type": "text" - }, - { - "block_id": "p643-b3", - "global_id": 18690, - "bbox": [ - 217.95, - 109.12, - 226.97, - 119.95 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p643-b4", - "global_id": 18691, - "bbox": [ - 229.78, - 88.17, - 235.04, - 98.14 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p643-b5", - "global_id": 18692, - "bbox": [ - 235.04, - 113.27, - 241.91, - 121.57 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p643-b6", - "global_id": 18693, - "bbox": [ - 244.02, - 95.06, - 320.03, - 112.88 - ], - "text": "x(t)cosnω0tdt + j", - "type": "text" - }, - { - "block_id": "p643-b7", - "global_id": 18694, - "bbox": [ - 313.88, - 109.12, - 322.91, - 119.95 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p643-b8", - "global_id": 18695, - "bbox": [ - 325.71, - 88.17, - 330.97, - 98.14 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p643-b9", - "global_id": 18696, - "bbox": [ - 330.98, - 113.27, - 337.84, - 121.57 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p643-b10", - "global_id": 18697, - "bbox": [ - 339.94, - 95.16, - 414.14, - 112.88 - ], - "text": "x(t)sinnω0tdt = 1", - "type": "text" - }, - { - "block_id": "p643-b11", - "global_id": 18698, - "bbox": [ - 409.16, - 101.73, - 516.13, - 119.18 - ], - "text": "2 (an + jbn)\n(6.21)", - "type": "text" - }, - { - "block_id": "p643-b12", - "global_id": 18699, - "bbox": [ - 127.59, - 128.01, - 516.13, - 151.83 - ], - "text": "These results are valid for general x(t), real or complex. When x(t) is real, an and bn are real, and\nEqs. (6.20) and (6.21) show that Dn and D−n are conjugates.", - "type": "text" - }, - { - "block_id": "p643-b13", - "global_id": 18700, - "bbox": [ - 301.96, - 161.12, - 341.27, - 174.31 - ], - "text": "D−n = D∗", - "type": "text" - }, - { - "block_id": "p643-b14", - "global_id": 18701, - "bbox": [ - 337.64, - 168.14, - 341.12, - 175.11 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p643-b15", - "global_id": 18702, - "bbox": [ - 127.59, - 186.16, - 300.48, - 196.12 - ], - "text": "Moreover, from Eq. (6.10), we observe that", - "type": "text" - }, - { - "block_id": "p643-b16", - "global_id": 18703, - "bbox": [ - 239.83, - 213.53, - 281.22, - 224.99 - ], - "text": "an −jbn =", - "type": "text" - }, - { - "block_id": "p643-b18", - "global_id": 18704, - "bbox": [ - 291.53, - 213.34, - 326.97, - 226.48 - ], - "text": "a2n + b2n e", - "type": "text" - }, - { - "block_id": "p643-b19", - "global_id": 18705, - "bbox": [ - 326.96, - 202.54, - 361.99, - 220.58 - ], - "text": "jtan−1 −bn\nan", - "type": "text" - }, - { - "block_id": "p643-b21", - "global_id": 18706, - "bbox": [ - 370.03, - 211.8, - 402.89, - 224.61 - ], - "text": "= Cnejθn", - "type": "text" - }, - { - "block_id": "p643-b22", - "global_id": 18707, - "bbox": [ - 127.59, - 237.89, - 155.51, - 247.85 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p643-b23", - "global_id": 18708, - "bbox": [ - 294.61, - 250.88, - 348.61, - 262.34 - ], - "text": "D0 = a0 = C0", - "type": "text" - }, - { - "block_id": "p643-b24", - "global_id": 18709, - "bbox": [ - 127.59, - 271.2, - 141.97, - 281.16 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p643-b25", - "global_id": 18710, - "bbox": [ - 253.49, - 282.86, - 281.22, - 295.66 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p643-b26", - "global_id": 18711, - "bbox": [ - 277.73, - 282.86, - 359.55, - 297.39 - ], - "text": "2Cnejθn\nD−n = 1", - "type": "text" - }, - { - "block_id": "p643-b27", - "global_id": 18712, - "bbox": [ - 356.07, - 282.98, - 389.22, - 297.39 - ], - "text": "2Cne−jθn", - "type": "text" - }, - { - "block_id": "p643-b28", - "global_id": 18713, - "bbox": [ - 127.59, - 304.1, - 210.56, - 314.48 - ], - "text": "Therefore, for n̸ = 0,", - "type": "text" - }, - { - "block_id": "p643-b29", - "global_id": 18714, - "bbox": [ - 217.74, - 325.66, - 285.35, - 338.08 - ], - "text": "|Dn| = |D−n| = 1", - "type": "text" - }, - { - "block_id": "p643-b30", - "global_id": 18715, - "bbox": [ - 281.86, - 327.0, - 516.13, - 340.19 - ], - "text": "2Cn,̸\nDn = θn,\nand̸ D−n = −θn\n(6.22)", - "type": "text" - }, - { - "block_id": "p643-b31", - "global_id": 18716, - "bbox": [ - 127.59, - 349.88, - 516.13, - 397.62 - ], - "text": "Note that |Dn| are the amplitudes and̸\nDn are the angles of various exponential components.\nFrom Eq. (6.22) it follows that when x(t) is real, the amplitude spectrum (|Dn| versus ω) is an even\nfunction of ω and the angle spectrum (̸ Dn versus ω) is an odd function of ω. For complex x(t),\nDn and D−n are generally not conjugates.", - "type": "text" - }, - { - "block_id": "p643-b32", - "global_id": 18717, - "bbox": [ - 102.51, - 425.61, - 509.28, - 437.56 - ], - "text": "EXAMPLE 6.6\nExponential Fourier Series of Periodic Exponential Wave", - "type": "text" - }, - { - "block_id": "p643-b33", - "global_id": 18718, - "bbox": [ - 128.9, - 454.23, - 426.34, - 464.19 - ], - "text": "Find the exponential Fourier series for the signal of Fig. 6.2a from Ex. 6.1.", - "type": "text" - }, - { - "block_id": "p643-b34", - "global_id": 18719, - "bbox": [ - 128.9, - 486.7, - 293.82, - 498.15 - ], - "text": "In this case T0 = π, ω0 = 2π/T0 = 2, and", - "type": "text" - }, - { - "block_id": "p643-b35", - "global_id": 18720, - "bbox": [ - 277.65, - 516.59, - 302.3, - 526.87 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p643-b36", - "global_id": 18721, - "bbox": [ - 308.04, - 506.42, - 322.14, - 517.09 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p643-b37", - "global_id": 18722, - "bbox": [ - 304.35, - 530.44, - 325.83, - 537.63 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p643-b38", - "global_id": 18723, - "bbox": [ - 326.94, - 515.08, - 353.38, - 527.67 - ], - "text": "Dnej2nt", - "type": "text" - }, - { - "block_id": "p643-b39", - "global_id": 18724, - "bbox": [ - 128.9, - 547.89, - 153.23, - 557.85 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p643-b40", - "global_id": 18725, - "bbox": [ - 178.46, - 565.9, - 209.95, - 583.93 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p643-b41", - "global_id": 18726, - "bbox": [ - 202.7, - 579.86, - 211.73, - 590.7 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p643-b42", - "global_id": 18727, - "bbox": [ - 214.53, - 558.91, - 219.79, - 568.87 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p643-b43", - "global_id": 18728, - "bbox": [ - 219.79, - 584.0, - 226.65, - 592.31 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p643-b44", - "global_id": 18729, - "bbox": [ - 228.76, - 565.9, - 293.01, - 582.85 - ], - "text": "x(t)e−j2nt dt = 1", - "type": "text" - }, - { - "block_id": "p643-b45", - "global_id": 18730, - "bbox": [ - 287.03, - 579.55, - 293.01, - 589.51 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p643-b46", - "global_id": 18731, - "bbox": [ - 296.3, - 558.91, - 310.32, - 570.97 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p643-b47", - "global_id": 18732, - "bbox": [ - 301.57, - 565.9, - 382.33, - 591.05 - ], - "text": "0\ne−t/2 e−j2nt dt = 1", - "type": "text" - }, - { - "block_id": "p643-b48", - "global_id": 18733, - "bbox": [ - 376.35, - 579.55, - 382.33, - 589.51 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p643-b49", - "global_id": 18734, - "bbox": [ - 385.63, - 558.91, - 399.63, - 570.97 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p643-b50", - "global_id": 18735, - "bbox": [ - 390.89, - 568.67, - 453.02, - 591.05 - ], - "text": "0\ne−(1/2+j2n)t dt", - "type": "text" - }, - { - "block_id": "p643-b51", - "global_id": 18736, - "bbox": [ - 191.69, - 594.79, - 231.87, - 619.62 - ], - "text": "=\n−1\nπ", - "type": "text" - }, - { - "block_id": "p643-b52", - "global_id": 18737, - "bbox": [ - 210.79, - 601.64, - 219.48, - 615.28 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p643-b53", - "global_id": 18738, - "bbox": [ - 216.0, - 600.05, - 291.06, - 623.15 - ], - "text": "2 + j2n\ne−(1/2+j2n)t", - "type": "text" - }, - { - "block_id": "p643-b55", - "global_id": 18739, - "bbox": [ - 294.95, - 592.41, - 299.14, - 599.38 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p643-b56", - "global_id": 18740, - "bbox": [ - 294.95, - 595.2, - 338.89, - 621.23 - ], - "text": "0\n= 0.504", - "type": "text" - }, - { - "block_id": "p643-b57", - "global_id": 18741, - "bbox": [ - 313.4, - 608.84, - 341.97, - 619.22 - ], - "text": "1 + j4n", - "type": "text" - } - ] - }, - { - "page_num": 644, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p644-b0", - "global_id": 18742, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "624\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p644-b1", - "global_id": 18743, - "bbox": [ - 103.16, - 86.24, - 117.54, - 96.21 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p644-b2", - "global_id": 18744, - "bbox": [ - 163.35, - 113.81, - 212.46, - 124.19 - ], - "text": "x(t) = 0.504", - "type": "text" - }, - { - "block_id": "p644-b3", - "global_id": 18745, - "bbox": [ - 217.27, - 103.64, - 231.36, - 114.31 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p644-b4", - "global_id": 18746, - "bbox": [ - 213.56, - 127.66, - 235.03, - 134.85 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p644-b5", - "global_id": 18747, - "bbox": [ - 237.35, - 107.24, - 282.39, - 131.26 - ], - "text": "1\n1 + j4nej2nt", - "type": "text" - }, - { - "block_id": "p644-b6", - "global_id": 18748, - "bbox": [ - 180.23, - 145.38, - 212.45, - 155.76 - ], - "text": "= 0.504", - "type": "text" - }, - { - "block_id": "p644-b8", - "global_id": 18749, - "bbox": [ - 219.0, - 138.81, - 400.73, - 162.83 - ], - "text": "1 +\n1\n1 + j4ej2t +\n1\n1 + j8ej4t +\n1\n1 + j12ej6t + · · ·", - "type": "text" - }, - { - "block_id": "p644-b9", - "global_id": 18750, - "bbox": [ - 217.59, - 159.3, - 416.83, - 190.73 - ], - "text": "+\n1\n1 −j4e−j2t +\n1\n1 −j8e−j4t +\n1\n1 −j12e−j6t + · · ·\n!", - "type": "text" - }, - { - "block_id": "p644-b10", - "global_id": 18751, - "bbox": [ - 103.17, - 200.8, - 477.01, - 222.82 - ], - "text": "Observe that the coefficients Dn are complex. Moreover, Dn and D−n are conjugates, as\nexpected.", - "type": "text" - }, - { - "block_id": "p644-b11", - "global_id": 18752, - "bbox": [ - 101.84, - 264.46, - 284.3, - 276.41 - ], - "text": "6.3-1 Exponential Fourier Spectra", - "type": "text" - }, - { - "block_id": "p644-b12", - "global_id": 18753, - "bbox": [ - 101.84, - 282.12, - 490.41, - 379.04 - ], - "text": "In exponential spectra, we plot coefficients Dn as a function of ω. But since Dn is complex in\ngeneral, we need both parts of one of two sets of plots: the real and the imaginary parts of Dn,\nor the magnitude and the angle of Dn. We prefer the latter because of its close connection to\nthe amplitudes and phases of corresponding components of the trigonometric Fourier series. We\ntherefore plot |Dn| versus ω and̸\nDn versus ω. This requires that the coefficients Dn be expressed\nin polar form as |Dn|ej̸\nDn, where |Dn| are the amplitudes and̸\nDn are the angles of various\nexponential components. Equation (6.22) shows that for real x(t), the amplitude spectrum (|Dn|\nversus ω) is an even function of ω and the angle spectrum (̸ Dn versus ω) is an odd function of ω.", - "type": "text" - }, - { - "block_id": "p644-b13", - "global_id": 18754, - "bbox": [ - 119.78, - 379.54, - 268.74, - 389.5 - ], - "text": "For the series in Ex. 6.6, for instance,", - "type": "text" - }, - { - "block_id": "p644-b14", - "global_id": 18755, - "bbox": [ - 163.31, - 399.41, - 208.77, - 410.87 - ], - "text": "D0 = 0.504", - "type": "text" - }, - { - "block_id": "p644-b15", - "global_id": 18756, - "bbox": [ - 163.31, - 414.18, - 210.55, - 432.21 - ], - "text": "D1 = 0.504", - "type": "text" - }, - { - "block_id": "p644-b16", - "global_id": 18757, - "bbox": [ - 187.55, - 417.29, - 433.84, - 438.2 - ], - "text": "1 + j4 = 0.122e−j75.96◦\n\r⇒\n|D1| = 0.122,̸\nD1 = −75.96◦", - "type": "text" - }, - { - "block_id": "p644-b17", - "global_id": 18758, - "bbox": [ - 157.88, - 441.05, - 210.55, - 459.07 - ], - "text": "D−1 = 0.504", - "type": "text" - }, - { - "block_id": "p644-b18", - "global_id": 18759, - "bbox": [ - 187.55, - 444.15, - 431.52, - 465.07 - ], - "text": "1 −j4 = 0.122ej75.96◦\n\r⇒\n|D−1| = 0.122,̸\nD−1 = 75.96◦", - "type": "text" - }, - { - "block_id": "p644-b19", - "global_id": 18760, - "bbox": [ - 101.84, - 473.25, - 116.22, - 483.21 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p644-b20", - "global_id": 18761, - "bbox": [ - 158.34, - 489.46, - 205.57, - 507.48 - ], - "text": "D2 = 0.504", - "type": "text" - }, - { - "block_id": "p644-b21", - "global_id": 18762, - "bbox": [ - 182.57, - 492.56, - 438.83, - 513.47 - ], - "text": "1 + j8 = 0.0625e−j82.87◦\n\r⇒\n|D2| = 0.0625,̸\nD2 = −82.87◦", - "type": "text" - }, - { - "block_id": "p644-b22", - "global_id": 18763, - "bbox": [ - 152.89, - 516.31, - 205.57, - 534.34 - ], - "text": "D−2 = 0.504", - "type": "text" - }, - { - "block_id": "p644-b23", - "global_id": 18764, - "bbox": [ - 182.57, - 519.43, - 436.49, - 540.33 - ], - "text": "1 −j8 = 0.0625ej82.87◦\n\r⇒\n|D−2| = 0.0625,̸\nD−2 = 82.87◦", - "type": "text" - }, - { - "block_id": "p644-b24", - "global_id": 18765, - "bbox": [ - 101.84, - 564.95, - 409.17, - 576.5 - ], - "text": "and so on. Note that Dn and D−n are conjugates, as expected [see Eq. (6.22)].", - "type": "text" - }, - { - "block_id": "p644-b25", - "global_id": 18766, - "bbox": [ - 101.84, - 577.0, - 490.37, - 598.92 - ], - "text": "Figure 6.12 shows the frequency spectra (amplitude and angle) of the exponential Fourier\nseries for the periodic signal x(t) in Fig. 6.2a.", - "type": "text" - }, - { - "block_id": "p644-b26", - "global_id": 18767, - "bbox": [ - 101.84, - 600.91, - 490.39, - 634.79 - ], - "text": "We notice some interesting features of these spectra. First, the spectra exist for positive as\nwell as negative values of ω (the frequency). Second, the amplitude spectrum is an even function\nof ω and the angle spectrum is an odd function of ω.", - "type": "text" - } - ] - }, - { - "page_num": 645, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p645-b0", - "global_id": 18768, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n625", - "type": "text" - }, - { - "block_id": "p645-b1", - "global_id": 18769, - "bbox": [ - 268.02, - 97.57, - 286.02, - 105.57 - ], - "text": "0.504", - "type": "text" - }, - { - "block_id": "p645-b2", - "global_id": 18770, - "bbox": [ - 163.4, - 164.84, - 416.14, - 173.14 - ], - "text": "2\n4\n6\n8\n10\n2\n4\n6\n8\n10", - "type": "text" - }, - { - "block_id": "p645-b3", - "global_id": 18771, - "bbox": [ - 285.52, - 178.75, - 294.4, - 186.75 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p645-b4", - "global_id": 18772, - "bbox": [ - 295.71, - 87.33, - 308.88, - 97.09 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p645-b5", - "global_id": 18773, - "bbox": [ - 307.07, - 135.29, - 351.09, - 149.79 - ], - "text": "0.122\n0.0625", - "type": "text" - }, - { - "block_id": "p645-b6", - "global_id": 18774, - "bbox": [ - 428.19, - 262.1, - 433.53, - 270.1 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p645-b7", - "global_id": 18775, - "bbox": [ - 428.19, - 165.04, - 433.53, - 173.04 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p645-b8", - "global_id": 18776, - "bbox": [ - 314.9, - 262.02, - 416.99, - 270.02 - ], - "text": "2\n4\n6\n8\n10", - "type": "text" - }, - { - "block_id": "p645-b9", - "global_id": 18777, - "bbox": [ - 284.64, - 306.74, - 294.44, - 314.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p645-b10", - "global_id": 18778, - "bbox": [ - 272.63, - 199.82, - 288.55, - 209.58 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p645-b11", - "global_id": 18779, - "bbox": [ - 163.51, - 260.75, - 272.26, - 269.05 - ], - "text": "2\n4\n6\n8\n10", - "type": "text" - }, - { - "block_id": "p645-b12", - "global_id": 18780, - "bbox": [ - 296.71, - 216.17, - 302.04, - 224.17 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p645-b13", - "global_id": 18781, - "bbox": [ - 297.37, - 224.92, - 301.37, - 232.92 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p645-b14", - "global_id": 18782, - "bbox": [ - 281.5, - 283.76, - 286.83, - 291.76 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p645-b15", - "global_id": 18783, - "bbox": [ - 273.86, - 287.83, - 286.17, - 300.51 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p645-b16", - "global_id": 18784, - "bbox": [ - 151.5, - 321.44, - 395.92, - 330.67 - ], - "text": "Figure 6.12 Exponential Fourier spectra for the signal in Fig. 6.2a.", - "type": "text" - }, - { - "block_id": "p645-b17", - "global_id": 18785, - "bbox": [ - 127.59, - 368.28, - 516.14, - 414.11 - ], - "text": "At times it may appear that the phase spectrum of a real periodic signal fails to satisfy the\nodd symmetry: for example, when Dk = D−k = −10. In this case, Dk = 10ejπ, and therefore,\nD−k = 10e−jπ. Recall that e±jπ = −1. Here, although Dk = D−k, their phases should be taken as π\nand −π.", - "type": "text" - }, - { - "block_id": "p645-b18", - "global_id": 18786, - "bbox": [ - 102.51, - 450.33, - 450.96, - 462.28 - ], - "text": "EXAMPLE 6.7\nPlotting Fourier Series Spectra with MATLAB", - "type": "text" - }, - { - "block_id": "p645-b19", - "global_id": 18787, - "bbox": [ - 128.9, - 478.95, - 502.76, - 500.86 - ], - "text": "Using MATLAB and the results of Ex. 6.6, compute and plot the exponential Fourier spectra\nfor the periodic signal x(t) shown in Fig. 6.2a. The result should match Fig. 6.12.", - "type": "text" - }, - { - "block_id": "p645-b20", - "global_id": 18788, - "bbox": [ - 128.9, - 523.68, - 301.96, - 535.23 - ], - "text": "The expression for Dn is derived in Ex. 6.6.", - "type": "text" - }, - { - "block_id": "p645-b21", - "global_id": 18789, - "bbox": [ - 128.9, - 543.99, - 359.05, - 601.77 - ], - "text": ">>\nclf; n = (-5:5); D_n = 0.504./(1+4j*n);\n>>\nsubplot(1,2,1); stem(n,abs(D_n),’.k’);\n>>\nxlabel(’n’); ylabel(’|D_n|’);\n>>\nsubplot(1,2,2); stem(n,angle(D_n),’.k’);\n>>\nxlabel(’n’); ylabel(’\\angle D_n [rad]’);", - "type": "text" - }, - { - "block_id": "p645-b22", - "global_id": 18790, - "bbox": [ - 128.9, - 611.03, - 502.75, - 621.41 - ], - "text": "Except that it is plotted as a function of n rather than ω, the result in Fig. 6.13 matches Fig. 6.12.", - "type": "text" - } - ] - }, - { - "page_num": 646, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p646-b0", - "global_id": 18791, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "626\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p646-b1", - "global_id": 18792, - "bbox": [ - 137.78, - 182.93, - 145.78, - 190.93 - ], - "text": "–5", - "type": "text" - }, - { - "block_id": "p646-b2", - "global_id": 18793, - "bbox": [ - 211.28, - 195.32, - 215.68, - 204.12 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p646-b3", - "global_id": 18794, - "bbox": [ - 132.69, - 173.71, - 136.69, - 181.71 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p646-b4", - "global_id": 18795, - "bbox": [ - 125.93, - 149.7, - 136.6, - 157.7 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p646-b5", - "global_id": 18796, - "bbox": [ - 125.93, - 125.71, - 136.6, - 133.71 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p646-b6", - "global_id": 18797, - "bbox": [ - 125.93, - 101.7, - 136.6, - 109.7 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p646-b7", - "global_id": 18798, - "bbox": [ - 111.43, - 135.96, - 120.95, - 147.94 - ], - "text": "|Dn|", - "type": "text" - }, - { - "block_id": "p646-b8", - "global_id": 18799, - "bbox": [ - 211.28, - 182.93, - 476.96, - 204.12 - ], - "text": "–5\n0\n5\n0\n5\nn", - "type": "text" - }, - { - "block_id": "p646-b9", - "global_id": 18800, - "bbox": [ - 319.18, - 173.71, - 327.18, - 181.71 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p646-b10", - "global_id": 18801, - "bbox": [ - 323.18, - 137.71, - 327.18, - 145.71 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p646-b11", - "global_id": 18802, - "bbox": [ - 323.18, - 101.7, - 327.18, - 109.7 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p646-b12", - "global_id": 18803, - "bbox": [ - 306.43, - 124.64, - 315.95, - 153.19 - ], - "text": "Dn [rad]", - "type": "text" - }, - { - "block_id": "p646-b13", - "global_id": 18804, - "bbox": [ - 110.73, - 210.81, - 325.76, - 220.05 - ], - "text": "Figure 6.13 Exponential Fourier series spectra for Ex. 6.7.", - "type": "text" - }, - { - "block_id": "p646-b14", - "global_id": 18805, - "bbox": [ - 101.84, - 268.67, - 490.39, - 332.14 - ], - "text": "WHAT IS A NEGATIVE FREQUENCY?\nThe existence of the spectrum at negative frequencies is somewhat disturbing because, by\ndefinition, the frequency (number of repetitions per second) is a positive quantity. How do we\ninterpret a negative frequency? We can use a trigonometric identity to express a sinusoid of a\nnegative frequency −ω0 as", - "type": "text" - }, - { - "block_id": "p646-b15", - "global_id": 18806, - "bbox": [ - 234.77, - 349.61, - 357.46, - 360.75 - ], - "text": "cos(−ω0t + θ) = cos(ω0t −θ)", - "type": "text" - }, - { - "block_id": "p646-b16", - "global_id": 18807, - "bbox": [ - 101.84, - 380.12, - 490.39, - 402.45 - ], - "text": "This equation clearly shows that the frequency of a sinusoid cos(ω0t + θ) is |ω0|, which is a\npositive quantity. The same conclusion is reached by observing that", - "type": "text" - }, - { - "block_id": "p646-b17", - "global_id": 18808, - "bbox": [ - 243.48, - 421.47, - 348.58, - 436.73 - ], - "text": "e±jω0t = cos ω0t ± jsin ω0t", - "type": "text" - }, - { - "block_id": "p646-b18", - "global_id": 18809, - "bbox": [ - 101.84, - 455.89, - 490.39, - 529.25 - ], - "text": "Thus, the frequency of exponentials e±jω0t is indeed |ω0|. How do we then interpret the spectral\nplots for negative values of ω? A more satisfying way of looking at the situation is to say that\nexponential spectra are a graphical representation of coefficients Dn as a function of ω. Existence\nof the spectrum at ω = −nω0 is merely an indication that an exponential component e−jnω0t exists\nin the series. We know that a sinusoid of frequency nω0 can be expressed in terms of a pair of\nexponentials ejnω0t and e−jnω0t.", - "type": "text" - }, - { - "block_id": "p646-b19", - "global_id": 18810, - "bbox": [ - 101.84, - 531.23, - 490.41, - 636.84 - ], - "text": "We see a close connection between the exponential spectra in Fig. 6.12 and the spectra of\nthe corresponding trigonometric Fourier series for x(t) (Figs. 6.2b, 6.2c). Equation (6.22) explains\nthe reason for the close connection, for real x(t), between the trigonometric spectra (Cn and θn)\nwith exponential spectra (|Dn| and̸\nDn). The dc components D0 and C0 are identical in both\nspectra. Moreover, the exponential amplitude spectrum |Dn| is half the trigonometric amplitude\nspectrum Cn for n ≥1. The exponential angle spectrum̸\nDn is identical to the trigonometric\nphase spectrum θn for n ≥0. We can therefore produce the exponential spectra merely by\ninspection of trigonometric spectra, and vice versa. The following example demonstrates this\nfeature.", - "type": "text" - } - ] - }, - { - "page_num": 647, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p647-b0", - "global_id": 18811, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n627", - "type": "text" - }, - { - "block_id": "p647-b1", - "global_id": 18812, - "bbox": [ - 102.51, - 93.92, - 488.59, - 119.82 - ], - "text": "EXAMPLE 6.8\nRelating Exponential to Trigonometric Fourier Series\nSpectra", - "type": "text" - }, - { - "block_id": "p647-b2", - "global_id": 18813, - "bbox": [ - 128.9, - 136.07, - 502.8, - 170.36 - ], - "text": "The trigonometric Fourier spectra of a certain periodic signal x(t) are shown in Fig. 6.14a.\nAfter inspecting these spectra, sketch the corresponding exponential Fourier spectra and verify\nyour results analytically.", - "type": "text" - }, - { - "block_id": "p647-b3", - "global_id": 18814, - "bbox": [ - 127.51, - 394.49, - 372.26, - 402.82 - ], - "text": "12\n12 9 6 3\n6\n9\n3\n0\n12\n3\n6\n9", - "type": "text" - }, - { - "block_id": "p647-b4", - "global_id": 18815, - "bbox": [ - 204.94, - 338.05, - 212.94, - 346.05 - ], - "text": "16", - "type": "text" - }, - { - "block_id": "p647-b5", - "global_id": 18816, - "bbox": [ - 205.53, - 361.78, - 209.53, - 369.78 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p647-b6", - "global_id": 18817, - "bbox": [ - 293.44, - 420.91, - 303.25, - 428.91 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p647-b7", - "global_id": 18818, - "bbox": [ - 385.51, - 381.37, - 455.51, - 389.37 - ], - "text": "3\n6\n9\n12\n0", - "type": "text" - }, - { - "block_id": "p647-b8", - "global_id": 18819, - "bbox": [ - 293.9, - 316.97, - 302.78, - 324.97 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p647-b9", - "global_id": 18820, - "bbox": [ - 385.84, - 276.46, - 455.08, - 284.46 - ], - "text": "3\n6\n9\n12\n0", - "type": "text" - }, - { - "block_id": "p647-b10", - "global_id": 18821, - "bbox": [ - 208.84, - 232.33, - 216.84, - 240.33 - ], - "text": "16", - "type": "text" - }, - { - "block_id": "p647-b11", - "global_id": 18822, - "bbox": [ - 217.84, - 289.56, - 272.84, - 297.56 - ], - "text": "3\n6\n9\n12", - "type": "text" - }, - { - "block_id": "p647-b12", - "global_id": 18823, - "bbox": [ - 188.67, - 230.66, - 197.01, - 240.21 - ], - "text": "Cn", - "type": "text" - }, - { - "block_id": "p647-b13", - "global_id": 18824, - "bbox": [ - 182.4, - 346.6, - 195.29, - 356.36 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p647-b14", - "global_id": 18825, - "bbox": [ - 365.79, - 357.52, - 381.23, - 367.28 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p647-b15", - "global_id": 18826, - "bbox": [ - 370.34, - 243.74, - 377.34, - 253.31 - ], - "text": "un", - "type": "text" - }, - { - "block_id": "p647-b16", - "global_id": 18827, - "bbox": [ - 453.86, - 290.51, - 459.2, - 298.51 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p647-b17", - "global_id": 18828, - "bbox": [ - 453.86, - 396.51, - 459.2, - 404.51 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p647-b18", - "global_id": 18829, - "bbox": [ - 264.97, - 297.51, - 270.31, - 305.51 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p647-b19", - "global_id": 18830, - "bbox": [ - 264.97, - 403.51, - 270.31, - 411.51 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p647-b20", - "global_id": 18831, - "bbox": [ - 388.57, - 360.45, - 393.9, - 368.45 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p647-b21", - "global_id": 18832, - "bbox": [ - 389.24, - 369.46, - 393.24, - 377.46 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p647-b22", - "global_id": 18833, - "bbox": [ - 372.86, - 302.54, - 378.19, - 310.54 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p647-b23", - "global_id": 18834, - "bbox": [ - 364.29, - 306.64, - 377.52, - 319.29 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p647-b24", - "global_id": 18835, - "bbox": [ - 372.86, - 406.54, - 378.19, - 414.54 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p647-b25", - "global_id": 18836, - "bbox": [ - 364.29, - 410.64, - 377.52, - 423.29 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p647-b26", - "global_id": 18837, - "bbox": [ - 127.51, - 435.6, - 296.96, - 444.84 - ], - "text": "Figure 6.14 Fourier series spectra for Ex. 6.8.", - "type": "text" - }, - { - "block_id": "p647-b27", - "global_id": 18838, - "bbox": [ - 128.9, - 472.51, - 502.76, - 542.25 - ], - "text": "The trigonometric spectral components exist at frequencies 0,3,6, and 9. The exponential\nspectral components exist at 0,3,6,9, and −3, −6, −9. Consider first the amplitude spectrum.\nThe dc component remains unchanged: that is, D0 = C0 = 16. Now |Dn| is an even function\nof ω and |Dn| = |D−n| = Cn/2. Thus, all the remaining spectrum |Dn| for positive n is half\nthe trigonometric amplitude spectrum Cn, and the spectrum |Dn| for negative n is a reflection\nabout the vertical axis of the spectrum for positive n, as shown in Fig. 6.14b.", - "type": "text" - }, - { - "block_id": "p647-b28", - "global_id": 18839, - "bbox": [ - 128.91, - 543.83, - 502.76, - 566.16 - ], - "text": "The angle spectrum is̸\nDn = θn for positive n and is −θn for negative n, as depicted in\nFig. 6.14b. We shall now verify that both sets of spectra represent the same signal.", - "type": "text" - }, - { - "block_id": "p647-b29", - "global_id": 18840, - "bbox": [ - 128.91, - 567.74, - 502.78, - 613.99 - ], - "text": "Signal x(t), whose trigonometric spectra are shown in Fig. 6.14a, has four spectral\ncomponents of frequencies 0, 3, 6, and 9. The dc component is 16. The amplitude and the phase\nof the component of frequency 3 are 12 and −π/4, respectively. Therefore, this component can\nbe expressed as 12cos(3t −π/4). Proceeding in this manner, we can write the Fourier series", - "type": "text" - } - ] - }, - { - "page_num": 648, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p648-b0", - "global_id": 18841, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "628\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p648-b1", - "global_id": 18842, - "bbox": [ - 103.16, - 85.88, - 142.89, - 96.26 - ], - "text": "for x(t) as", - "type": "text" - }, - { - "block_id": "p648-b2", - "global_id": 18843, - "bbox": [ - 159.83, - 113.03, - 231.68, - 123.4 - ], - "text": "x(t) = 16 + 12cos", - "type": "text" - }, - { - "block_id": "p648-b4", - "global_id": 18844, - "bbox": [ - 239.54, - 106.04, - 265.52, - 130.48 - ], - "text": "3t −π\n4", - "type": "text" - }, - { - "block_id": "p648-b6", - "global_id": 18845, - "bbox": [ - 275.98, - 113.03, - 304.65, - 123.4 - ], - "text": "+ 8cos", - "type": "text" - }, - { - "block_id": "p648-b8", - "global_id": 18846, - "bbox": [ - 312.5, - 106.04, - 338.48, - 130.48 - ], - "text": "6t −π\n2", - "type": "text" - }, - { - "block_id": "p648-b10", - "global_id": 18847, - "bbox": [ - 348.95, - 113.03, - 377.62, - 123.4 - ], - "text": "+ 4cos", - "type": "text" - }, - { - "block_id": "p648-b12", - "global_id": 18848, - "bbox": [ - 385.47, - 106.04, - 411.44, - 130.48 - ], - "text": "9t −π\n4", - "type": "text" - }, - { - "block_id": "p648-b14", - "global_id": 18849, - "bbox": [ - 103.17, - 140.65, - 477.04, - 163.64 - ], - "text": "Consider now the exponential spectra in Fig. 6.14b. They contain components of\nfrequencies 0 (dc), ±3, ±6, and ±9. The dc component is D0 = 16. The component ej3t", - "type": "text" - }, - { - "block_id": "p648-b15", - "global_id": 18850, - "bbox": [ - 103.16, - 164.14, - 477.03, - 210.38 - ], - "text": "(frequency 3) has magnitude 6 and angle −π/4. Therefore, this component strength is\n6e−jπ/4, and it can be expressed as (6e−jπ/4)ej3t. Similarly, the component of frequency −3\nis (6ejπ/4)e−j3t. Proceeding in this manner, ˆx(t), the signal corresponding to the spectra in\nFig. 6.14b, is", - "type": "text" - }, - { - "block_id": "p648-b16", - "global_id": 18851, - "bbox": [ - 159.83, - 221.93, - 205.81, - 232.3 - ], - "text": "ˆx(t) = 16 +", - "type": "text" - }, - { - "block_id": "p648-b18", - "global_id": 18852, - "bbox": [ - 211.29, - 213.92, - 317.39, - 232.3 - ], - "text": "6e−jπ/4ej3t + 6ejπ/4e−j3t\n+", - "type": "text" - }, - { - "block_id": "p648-b20", - "global_id": 18853, - "bbox": [ - 322.86, - 213.92, - 419.63, - 232.3 - ], - "text": "4e−jπ/2ej6t + 4ejπ/2e−j6t", - "type": "text" - }, - { - "block_id": "p648-b21", - "global_id": 18854, - "bbox": [ - 176.21, - 238.36, - 183.98, - 248.32 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p648-b23", - "global_id": 18855, - "bbox": [ - 190.56, - 230.35, - 287.34, - 248.74 - ], - "text": "2e−jπ/4ej9t + 2ejπ/4e−j9t", - "type": "text" - }, - { - "block_id": "p648-b24", - "global_id": 18856, - "bbox": [ - 176.71, - 254.84, - 212.33, - 265.22 - ], - "text": "= 16 + 6", - "type": "text" - }, - { - "block_id": "p648-b26", - "global_id": 18857, - "bbox": [ - 216.27, - 246.83, - 305.21, - 265.12 - ], - "text": "ej(3t−π/4) + e−j(3t−π/4)", - "type": "text" - }, - { - "block_id": "p648-b27", - "global_id": 18858, - "bbox": [ - 306.75, - 254.84, - 321.05, - 265.22 - ], - "text": "+ 4", - "type": "text" - }, - { - "block_id": "p648-b29", - "global_id": 18859, - "bbox": [ - 324.98, - 246.83, - 413.92, - 265.12 - ], - "text": "ej(6t−π/2) + e−j(6t−π/2)", - "type": "text" - }, - { - "block_id": "p648-b30", - "global_id": 18860, - "bbox": [ - 176.21, - 271.32, - 191.62, - 281.7 - ], - "text": "+ 2", - "type": "text" - }, - { - "block_id": "p648-b32", - "global_id": 18861, - "bbox": [ - 195.55, - 263.31, - 284.48, - 281.6 - ], - "text": "ej(9t−π/4) + e−j(9t−π/4)", - "type": "text" - }, - { - "block_id": "p648-b33", - "global_id": 18862, - "bbox": [ - 176.71, - 293.33, - 231.68, - 303.71 - ], - "text": "= 16 + 12cos", - "type": "text" - }, - { - "block_id": "p648-b35", - "global_id": 18863, - "bbox": [ - 239.54, - 286.35, - 265.52, - 310.78 - ], - "text": "3t −π\n4", - "type": "text" - }, - { - "block_id": "p648-b37", - "global_id": 18864, - "bbox": [ - 275.98, - 293.33, - 304.65, - 303.71 - ], - "text": "+ 8cos", - "type": "text" - }, - { - "block_id": "p648-b39", - "global_id": 18865, - "bbox": [ - 312.5, - 286.35, - 338.48, - 310.78 - ], - "text": "6t −π\n2", - "type": "text" - }, - { - "block_id": "p648-b41", - "global_id": 18866, - "bbox": [ - 348.95, - 293.33, - 377.62, - 303.71 - ], - "text": "+ 4cos", - "type": "text" - }, - { - "block_id": "p648-b43", - "global_id": 18867, - "bbox": [ - 385.47, - 286.35, - 411.44, - 310.78 - ], - "text": "9t −π\n4", - "type": "text" - }, - { - "block_id": "p648-b45", - "global_id": 18868, - "bbox": [ - 103.17, - 320.95, - 352.19, - 330.92 - ], - "text": "Clearly both sets of spectra represent the same periodic signal.", - "type": "text" - }, - { - "block_id": "p648-b46", - "global_id": 18869, - "bbox": [ - 102.14, - 368.67, - 243.6, - 380.79 - ], - "text": "BANDWIDTH OF A SIGNAL", - "type": "text" - }, - { - "block_id": "p648-b47", - "global_id": 18870, - "bbox": [ - 101.84, - 384.82, - 490.41, - 466.52 - ], - "text": "The difference between the highest and the lowest frequencies of the spectral components of a\nsignal is the bandwidth of the signal. The bandwidth of the signal whose exponential spectra are\nshown in Fig. 6.14b is 9 (in radians). The highest and lowest frequencies are 9 and 0, respectively.\nNote that the component of frequency 12 has zero amplitude and is nonexistent. Moreover, the\nlowest frequency is 0, not −9. Recall that the frequencies (in the conventional sense) of the spectral\ncomponents at ω = −3, −6, and −9 in reality are 3, 6, and 9.† The bandwidth can be more readily\nseen from the trigonometric spectra in Fig. 6.14a.", - "type": "text" - }, - { - "block_id": "p648-b48", - "global_id": 18871, - "bbox": [ - 76.77, - 494.53, - 404.76, - 506.49 - ], - "text": "EXAMPLE 6.9\nFourier Series Spectra of an Impulse Train", - "type": "text" - }, - { - "block_id": "p648-b49", - "global_id": 18872, - "bbox": [ - 103.16, - 523.14, - 477.05, - 559.06 - ], - "text": "Find the exponential Fourier series and sketch the corresponding spectra for the impulse train\nδT0(t) depicted in Fig. 6.15a. From this result, sketch the trigonometric spectrum and write the\ntrigonometric Fourier series for δT0(t).", - "type": "text" - }, - { - "block_id": "p648-b50", - "global_id": 18873, - "bbox": [ - 101.84, - 601.26, - 490.39, - 646.37 - ], - "text": "† Some authors do define bandwidth as the difference between the highest and the lowest (negative) frequency\nin the exponential spectrum. The bandwidth according to this definition is twice that defined here. In reality,\nthis phrasing defines not the signal bandwidth but the spectral width (width of the exponential spectrum of\nthe signal).", - "type": "text" - } - ] - }, - { - "page_num": 649, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p649-b0", - "global_id": 18874, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n629", - "type": "text" - }, - { - "block_id": "p649-b1", - "global_id": 18875, - "bbox": [ - 119.94, - 368.77, - 336.21, - 378.08 - ], - "text": "Figure 6.15 (a) Impulse train and (b, c) its Fourier spectra.", - "type": "text" - }, - { - "block_id": "p649-b2", - "global_id": 18876, - "bbox": [ - 146.84, - 395.79, - 396.85, - 405.75 - ], - "text": "The unit impulse train shown in Fig. 6.15a can be expressed as", - "type": "text" - }, - { - "block_id": "p649-b3", - "global_id": 18877, - "bbox": [ - 287.94, - 415.24, - 302.04, - 425.91 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p649-b4", - "global_id": 18878, - "bbox": [ - 284.24, - 439.27, - 305.72, - 446.46 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p649-b5", - "global_id": 18879, - "bbox": [ - 306.84, - 425.42, - 347.42, - 436.57 - ], - "text": "δ(t −nT0)", - "type": "text" - }, - { - "block_id": "p649-b6", - "global_id": 18880, - "bbox": [ - 128.9, - 456.36, - 495.78, - 468.77 - ], - "text": "Following Papoulis, we shall denote this function as δT0(t) for the sake of notational brevity.", - "type": "text" - }, - { - "block_id": "p649-b7", - "global_id": 18881, - "bbox": [ - 146.84, - 468.73, - 313.83, - 478.69 - ], - "text": "The exponential Fourier series is given by", - "type": "text" - }, - { - "block_id": "p649-b8", - "global_id": 18882, - "bbox": [ - 243.2, - 498.36, - 275.54, - 510.77 - ], - "text": "δT0(t) =", - "type": "text" - }, - { - "block_id": "p649-b9", - "global_id": 18883, - "bbox": [ - 281.28, - 488.19, - 295.38, - 498.86 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p649-b10", - "global_id": 18884, - "bbox": [ - 277.58, - 512.21, - 299.06, - 519.4 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p649-b11", - "global_id": 18885, - "bbox": [ - 300.18, - 491.38, - 386.28, - 509.82 - ], - "text": "Dnejnω0t\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p649-b12", - "global_id": 18886, - "bbox": [ - 376.54, - 505.75, - 385.56, - 516.58 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p649-b13", - "global_id": 18887, - "bbox": [ - 478.69, - 498.78, - 502.75, - 508.74 - ], - "text": "(6.23)", - "type": "text" - }, - { - "block_id": "p649-b14", - "global_id": 18888, - "bbox": [ - 128.91, - 529.66, - 153.24, - 539.62 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p649-b15", - "global_id": 18889, - "bbox": [ - 261.95, - 537.54, - 293.45, - 555.57 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p649-b16", - "global_id": 18890, - "bbox": [ - 286.19, - 551.5, - 295.21, - 562.33 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p649-b17", - "global_id": 18891, - "bbox": [ - 298.01, - 530.55, - 303.27, - 540.51 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p649-b18", - "global_id": 18892, - "bbox": [ - 303.28, - 555.65, - 310.15, - 563.95 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p649-b19", - "global_id": 18893, - "bbox": [ - 312.24, - 540.31, - 369.54, - 556.52 - ], - "text": "δT0(t)e−jnω0t dt", - "type": "text" - }, - { - "block_id": "p649-b20", - "global_id": 18894, - "bbox": [ - 128.91, - 569.35, - 502.74, - 593.72 - ], - "text": "Choosing the interval of integration (−T0/2,T0/2) and recognizing that over this interval\nδT0(t) = δ(t), we get", - "type": "text" - }, - { - "block_id": "p649-b21", - "global_id": 18895, - "bbox": [ - 259.51, - 594.41, - 291.01, - 612.44 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p649-b22", - "global_id": 18896, - "bbox": [ - 283.75, - 608.37, - 292.77, - 619.2 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p649-b23", - "global_id": 18897, - "bbox": [ - 295.58, - 587.43, - 319.53, - 601.03 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p649-b24", - "global_id": 18898, - "bbox": [ - 300.84, - 612.3, - 320.4, - 620.83 - ], - "text": "−T0/2", - "type": "text" - }, - { - "block_id": "p649-b25", - "global_id": 18899, - "bbox": [ - 322.01, - 597.18, - 371.97, - 611.26 - ], - "text": "δ(t)e−jnω0t dt", - "type": "text" - } - ] - }, - { - "page_num": 650, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p650-b0", - "global_id": 18900, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "630\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p650-b1", - "global_id": 18901, - "bbox": [ - 103.16, - 85.83, - 477.01, - 120.11 - ], - "text": "In this integral, the impulse is located at t = 0. From the sampling property of Eq. (1.11), the\nintegral on the right-hand side is the value of e−jnω0t at t = 0 (where the impulse is located).\nTherefore,", - "type": "text" - }, - { - "block_id": "p650-b2", - "global_id": 18902, - "bbox": [ - 272.61, - 119.37, - 304.1, - 137.4 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p650-b3", - "global_id": 18903, - "bbox": [ - 296.85, - 133.33, - 305.88, - 144.16 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p650-b4", - "global_id": 18904, - "bbox": [ - 452.95, - 126.36, - 477.01, - 136.32 - ], - "text": "(6.24)", - "type": "text" - }, - { - "block_id": "p650-b5", - "global_id": 18905, - "bbox": [ - 103.17, - 149.75, - 477.04, - 171.66 - ], - "text": "From this result, we see that the exponential spectrum is constant for all frequencies, as shown\nin Fig. 6.15b. The spectrum, being real, requires only the amplitude plot. All phases are zero.", - "type": "text" - }, - { - "block_id": "p650-b6", - "global_id": 18906, - "bbox": [ - 121.09, - 171.89, - 201.42, - 184.7 - ], - "text": "Substituting Dn = 1", - "type": "text" - }, - { - "block_id": "p650-b7", - "global_id": 18907, - "bbox": [ - 196.0, - 173.66, - 443.04, - 188.66 - ], - "text": "T0 into Eq. (6.23) yields the desired exponential Fourier series", - "type": "text" - }, - { - "block_id": "p650-b8", - "global_id": 18908, - "bbox": [ - 216.53, - 199.66, - 259.37, - 218.64 - ], - "text": "δT0(t) = 1", - "type": "text" - }, - { - "block_id": "p650-b9", - "global_id": 18909, - "bbox": [ - 252.11, - 213.62, - 261.14, - 224.45 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p650-b10", - "global_id": 18910, - "bbox": [ - 267.64, - 196.06, - 281.74, - 206.73 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p650-b11", - "global_id": 18911, - "bbox": [ - 263.94, - 220.08, - 285.42, - 227.27 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p650-b12", - "global_id": 18912, - "bbox": [ - 286.54, - 199.25, - 361.46, - 217.69 - ], - "text": "ejnω0t\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p650-b13", - "global_id": 18913, - "bbox": [ - 351.72, - 213.62, - 360.74, - 224.45 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p650-b14", - "global_id": 18914, - "bbox": [ - 121.09, - 237.53, - 382.24, - 247.49 - ], - "text": "To sketch the trigonometric spectrum, we use Eq. (6.22) to obtain", - "type": "text" - }, - { - "block_id": "p650-b15", - "global_id": 18915, - "bbox": [ - 218.17, - 257.43, - 272.15, - 275.46 - ], - "text": "C0 = D0 = 1", - "type": "text" - }, - { - "block_id": "p650-b16", - "global_id": 18916, - "bbox": [ - 264.9, - 271.39, - 273.93, - 282.22 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p650-b17", - "global_id": 18917, - "bbox": [ - 218.17, - 283.82, - 282.84, - 301.85 - ], - "text": "Cn = 2|Dn| = 2", - "type": "text" - }, - { - "block_id": "p650-b18", - "global_id": 18918, - "bbox": [ - 275.59, - 297.78, - 284.61, - 308.61 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p650-b19", - "global_id": 18919, - "bbox": [ - 306.23, - 290.39, - 362.02, - 300.77 - ], - "text": "n = 1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p650-b20", - "global_id": 18920, - "bbox": [ - 220.43, - 311.39, - 245.65, - 322.84 - ], - "text": "θn = 0", - "type": "text" - }, - { - "block_id": "p650-b21", - "global_id": 18921, - "bbox": [ - 103.16, - 333.71, - 477.02, - 357.66 - ], - "text": "Figure 6.15c shows the trigonometric Fourier spectrum. From this spectrum we can express\nδT0(t) as", - "type": "text" - }, - { - "block_id": "p650-b22", - "global_id": 18922, - "bbox": [ - 136.21, - 366.48, - 179.06, - 385.46 - ], - "text": "δT0(t) = 1", - "type": "text" - }, - { - "block_id": "p650-b23", - "global_id": 18923, - "bbox": [ - 171.81, - 380.42, - 180.83, - 391.26 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p650-b24", - "global_id": 18924, - "bbox": [ - 183.63, - 366.06, - 417.7, - 384.5 - ], - "text": "[1 + 2(cos ω0t + cos 2ω0t + cos 3ω0t + · · ·)]\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p650-b25", - "global_id": 18925, - "bbox": [ - 407.96, - 380.42, - 416.98, - 391.26 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p650-b26", - "global_id": 18926, - "bbox": [ - 452.95, - 373.46, - 477.01, - 383.42 - ], - "text": "(6.25)", - "type": "text" - }, - { - "block_id": "p650-b27", - "global_id": 18927, - "bbox": [ - 102.14, - 419.66, - 406.16, - 431.78 - ], - "text": "EFFECT OF SYMMETRY IN EXPONENTIAL FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p650-b28", - "global_id": 18928, - "bbox": [ - 101.84, - 435.4, - 490.42, - 530.96 - ], - "text": "When x(t) has an even symmetry, bn = 0, and from Eq. (6.20), Dn = an/2, which is real (positive\nor negative). Hence,̸\nDn can only be 0 or ±π. Moreover, we may compute Dn = an/2 by using\nEq. (6.14), which requires integration over a half-period only. Similarly, when x(t) has an odd\nsymmetry, an = 0, and Dn = −jbn/2 is imaginary (positive or negative). Hence,̸\nDn can only be 0\nor ±π/2. Moreover, we may compute Dn = −jbn/2 by using Eq. (6.15), which requires integration\nover a half-period only. Note, however, that in the exponential case, we are using the symmetry\nproperty indirectly by finding the trigonometric coefficients. We cannot apply it directly in finding\nDn from Eq. (6.19) since the function ejnω0t is neither even nor odd.", - "type": "text" - }, - { - "block_id": "p650-b29", - "global_id": 18929, - "bbox": [ - 107.82, - 550.4, - 470.73, - 576.3 - ], - "text": "DRILL 6.4\nRelating Trigonometric to Exponential Fourier Series\nSpectra", - "type": "text" - }, - { - "block_id": "p650-b30", - "global_id": 18930, - "bbox": [ - 107.82, - 585.0, - 484.4, - 619.29 - ], - "text": "The exponential Fourier spectra of a certain periodic signal x(t) are shown in Fig. 6.16.\nDetermine and sketch the trigonometric Fourier spectra of x(t) by inspection of Fig. 6.16. Now\nwrite the (compact) trigonometric Fourier series for x(t).", - "type": "text" - } - ] - }, - { - "page_num": 651, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p651-b0", - "global_id": 18931, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n631", - "type": "text" - }, - { - "block_id": "p651-b1", - "global_id": 18932, - "bbox": [ - 133.84, - 92.39, - 188.43, - 103.34 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p651-b2", - "global_id": 18933, - "bbox": [ - 133.57, - 113.93, - 195.46, - 124.31 - ], - "text": "x(t) = 4 + 6cos", - "type": "text" - }, - { - "block_id": "p651-b4", - "global_id": 18934, - "bbox": [ - 203.32, - 112.3, - 227.5, - 127.12 - ], - "text": "3t −π\n6", - "type": "text" - }, - { - "block_id": "p651-b6", - "global_id": 18935, - "bbox": [ - 237.66, - 113.93, - 266.34, - 124.31 - ], - "text": "+ 2cos", - "type": "text" - }, - { - "block_id": "p651-b8", - "global_id": 18936, - "bbox": [ - 274.19, - 112.3, - 298.37, - 127.12 - ], - "text": "6t −π\n4", - "type": "text" - }, - { - "block_id": "p651-b10", - "global_id": 18937, - "bbox": [ - 308.54, - 113.93, - 337.21, - 124.31 - ], - "text": "+ 4cos", - "type": "text" - }, - { - "block_id": "p651-b12", - "global_id": 18938, - "bbox": [ - 345.05, - 112.3, - 369.23, - 127.12 - ], - "text": "9t −π\n2", - "type": "text" - }, - { - "block_id": "p651-b14", - "global_id": 18939, - "bbox": [ - 227.67, - 209.83, - 462.77, - 218.41 - ], - "text": "3\n6\n9\n0", - "type": "text" - }, - { - "block_id": "p651-b15", - "global_id": 18940, - "bbox": [ - 218.29, - 187.47, - 222.29, - 195.47 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p651-b16", - "global_id": 18941, - "bbox": [ - 218.29, - 170.49, - 222.29, - 178.49 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p651-b17", - "global_id": 18942, - "bbox": [ - 243.37, - 210.41, - 286.37, - 218.41 - ], - "text": "3\n6\n9", - "type": "text" - }, - { - "block_id": "p651-b18", - "global_id": 18943, - "bbox": [ - 218.59, - 178.96, - 222.59, - 186.96 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p651-b19", - "global_id": 18944, - "bbox": [ - 218.94, - 195.97, - 222.94, - 203.97 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p651-b20", - "global_id": 18945, - "bbox": [ - 162.04, - 210.11, - 368.61, - 218.41 - ], - "text": "3\n6\n6\n9\n9", - "type": "text" - }, - { - "block_id": "p651-b21", - "global_id": 18946, - "bbox": [ - 238.88, - 167.29, - 398.45, - 178.36 - ], - "text": "Dn\nDn", - "type": "text" - }, - { - "block_id": "p651-b22", - "global_id": 18947, - "bbox": [ - 379.22, - 214.03, - 399.22, - 231.98 - ], - "text": "p6\np4", - "type": "text" - }, - { - "block_id": "p651-b23", - "global_id": 18948, - "bbox": [ - 379.22, - 238.93, - 399.22, - 247.23 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p651-b24", - "global_id": 18949, - "bbox": [ - 291.58, - 210.3, - 296.91, - 218.3 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p651-b25", - "global_id": 18950, - "bbox": [ - 466.18, - 198.33, - 471.51, - 206.33 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p651-b26", - "global_id": 18951, - "bbox": [ - 157.47, - 256.63, - 377.23, - 265.87 - ], - "text": "Figure 6.16 Exponential Fourier series spectra for Drill 6.4.", - "type": "text" - }, - { - "block_id": "p651-b27", - "global_id": 18952, - "bbox": [ - 133.57, - 318.29, - 500.88, - 344.19 - ], - "text": "DRILL 6.5\nFourier Series Spectrum of a Full-Wave Rectified Sine\nWave", - "type": "text" - }, - { - "block_id": "p651-b28", - "global_id": 18953, - "bbox": [ - 133.57, - 352.89, - 509.95, - 375.23 - ], - "text": "Find the exponential Fourier series and sketch the corresponding Fourier spectrum Dn versus ω\nfor the full-wave rectified sine wave depicted in Fig. 6.17.", - "type": "text" - }, - { - "block_id": "p651-b29", - "global_id": 18954, - "bbox": [ - 133.84, - 388.75, - 188.43, - 399.71 - ], - "text": "ANSWER", - "type": "text" - }, - { - "block_id": "p651-b30", - "global_id": 18955, - "bbox": [ - 133.57, - 405.35, - 165.65, - 416.97 - ], - "text": "x(t) = 2", - "type": "text" - }, - { - "block_id": "p651-b31", - "global_id": 18956, - "bbox": [ - 161.46, - 412.62, - 165.65, - 419.6 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p651-b32", - "global_id": 18957, - "bbox": [ - 168.64, - 399.22, - 186.18, - 411.28 - ], - "text": "%∞", - "type": "text" - }, - { - "block_id": "p651-b33", - "global_id": 18958, - "bbox": [ - 179.06, - 411.92, - 200.53, - 419.11 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p651-b34", - "global_id": 18959, - "bbox": [ - 203.35, - 405.35, - 239.19, - 420.63 - ], - "text": "1\n1−4n2 ej2nt", - "type": "text" - }, - { - "block_id": "p651-b35", - "global_id": 18960, - "bbox": [ - 294.24, - 463.76, - 298.24, - 471.76 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p651-b36", - "global_id": 18961, - "bbox": [ - 190.75, - 488.39, - 433.4, - 496.69 - ], - "text": "0\np\np\n2p\n2p\nt", - "type": "text" - }, - { - "block_id": "p651-b37", - "global_id": 18962, - "bbox": [ - 305.9, - 449.39, - 362.85, - 458.81 - ], - "text": "sin t\nx(t)", - "type": "text" - }, - { - "block_id": "p651-b38", - "global_id": 18963, - "bbox": [ - 157.47, - 503.39, - 360.43, - 512.63 - ], - "text": "Figure 6.17 Full-wave rectified sine wave for Drill 6.5.", - "type": "text" - }, - { - "block_id": "p651-b39", - "global_id": 18964, - "bbox": [ - 133.57, - 565.04, - 420.44, - 577.0 - ], - "text": "DRILL 6.6\nExponential Fourier Series and Spectra", - "type": "text" - }, - { - "block_id": "p651-b40", - "global_id": 18965, - "bbox": [ - 133.57, - 586.12, - 510.13, - 608.04 - ], - "text": "Find the exponential Fourier series and sketch the corresponding Fourier spectra for the periodic\nsignals shown in Fig. 6.7.", - "type": "text" - } - ] - }, - { - "page_num": 652, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p652-b0", - "global_id": 18966, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "632\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p652-b1", - "global_id": 18967, - "bbox": [ - 108.09, - 92.38, - 170.31, - 103.34 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p652-b2", - "global_id": 18968, - "bbox": [ - 125.76, - 109.84, - 175.23, - 126.79 - ], - "text": "(a) x(t) = 1", - "type": "text" - }, - { - "block_id": "p652-b3", - "global_id": 18969, - "bbox": [ - 170.25, - 109.84, - 198.95, - 133.86 - ], - "text": "3 + 2\nπ2", - "type": "text" - }, - { - "block_id": "p652-b4", - "global_id": 18970, - "bbox": [ - 214.75, - 106.24, - 228.85, - 116.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p652-b5", - "global_id": 18971, - "bbox": [ - 201.75, - 130.85, - 241.82, - 138.11 - ], - "text": "n=−∞(n̸=0)", - "type": "text" - }, - { - "block_id": "p652-b6", - "global_id": 18972, - "bbox": [ - 244.14, - 108.42, - 267.81, - 119.81 - ], - "text": "(−1)n", - "type": "text" - }, - { - "block_id": "p652-b7", - "global_id": 18973, - "bbox": [ - 251.74, - 114.69, - 286.16, - 133.76 - ], - "text": "n2\nejnπt", - "type": "text" - }, - { - "block_id": "p652-b8", - "global_id": 18974, - "bbox": [ - 125.76, - 148.21, - 179.66, - 165.17 - ], - "text": "(b) x(t) = jA", - "type": "text" - }, - { - "block_id": "p652-b9", - "global_id": 18975, - "bbox": [ - 171.75, - 161.95, - 177.73, - 171.92 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p652-b10", - "global_id": 18976, - "bbox": [ - 194.96, - 144.71, - 209.06, - 155.38 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p652-b11", - "global_id": 18977, - "bbox": [ - 181.96, - 169.31, - 222.02, - 176.58 - ], - "text": "n=−∞(n̸=0)", - "type": "text" - }, - { - "block_id": "p652-b12", - "global_id": 18978, - "bbox": [ - 224.35, - 146.89, - 248.02, - 158.27 - ], - "text": "(−1)n", - "type": "text" - }, - { - "block_id": "p652-b13", - "global_id": 18979, - "bbox": [ - 233.95, - 153.16, - 266.38, - 172.23 - ], - "text": "n\nejnπt", - "type": "text" - }, - { - "block_id": "p652-b14", - "global_id": 18980, - "bbox": [ - 101.84, - 227.49, - 238.18, - 239.44 - ], - "text": "6.3-2 Parseval’s Theorem", - "type": "text" - }, - { - "block_id": "p652-b15", - "global_id": 18981, - "bbox": [ - 101.84, - 245.16, - 373.18, - 255.54 - ], - "text": "The trigonometric Fourier series of a periodic signal x(t) is given by", - "type": "text" - }, - { - "block_id": "p652-b16", - "global_id": 18982, - "bbox": [ - 228.8, - 283.83, - 275.45, - 295.29 - ], - "text": "x(t) = C0 +", - "type": "text" - }, - { - "block_id": "p652-b17", - "global_id": 18983, - "bbox": [ - 277.0, - 273.66, - 291.1, - 284.33 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p652-b18", - "global_id": 18984, - "bbox": [ - 277.84, - 298.22, - 290.25, - 305.49 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p652-b19", - "global_id": 18985, - "bbox": [ - 292.2, - 283.83, - 363.44, - 295.7 - ], - "text": "Cn cos(nω0t + θn)", - "type": "text" - }, - { - "block_id": "p652-b20", - "global_id": 18986, - "bbox": [ - 101.85, - 323.81, - 490.4, - 357.68 - ], - "text": "Every term on the right-hand side of this equation is a power signal. As shown in Ex. 1.2, Eq. (1.3),\nthe power of x(t) is equal to the sum of the powers of all the sinusoidal components on the\nright-hand side.", - "type": "text" - }, - { - "block_id": "p652-b21", - "global_id": 18987, - "bbox": [ - 255.32, - 380.33, - 287.02, - 391.79 - ], - "text": "Px = C0", - "type": "text" - }, - { - "block_id": "p652-b22", - "global_id": 18988, - "bbox": [ - 287.52, - 373.76, - 308.55, - 397.78 - ], - "text": "2 + 1\n2", - "type": "text" - }, - { - "block_id": "p652-b23", - "global_id": 18989, - "bbox": [ - 310.86, - 370.16, - 324.96, - 380.83 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p652-b24", - "global_id": 18990, - "bbox": [ - 311.7, - 394.72, - 324.1, - 401.98 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p652-b25", - "global_id": 18991, - "bbox": [ - 326.06, - 378.9, - 336.41, - 390.61 - ], - "text": "C2", - "type": "text" - }, - { - "block_id": "p652-b26", - "global_id": 18992, - "bbox": [ - 332.7, - 380.75, - 490.38, - 392.6 - ], - "text": "n\n(6.26)", - "type": "text" - }, - { - "block_id": "p652-b27", - "global_id": 18993, - "bbox": [ - 101.85, - 417.22, - 490.38, - 439.24 - ], - "text": "This result is one form of Parseval’s theorem, as applied to power signals. It states that the power\nof a periodic signal is equal to the sum of the powers of its Fourier components.", - "type": "text" - }, - { - "block_id": "p652-b28", - "global_id": 18994, - "bbox": [ - 101.85, - 441.23, - 490.38, - 499.01 - ], - "text": "We can apply the same argument to the exponential Fourier series (see Prob. 1.1-11). The\npower of a periodic signal x(t) can be expressed as a sum of the powers of its exponential\ncomponents. In Eq. (1.4), we showed that the power of an exponential Dejω0t is |D2|. We can\nuse this result to express the power of a periodic signal x(t) in terms of its exponential Fourier\nseries coefficients as", - "type": "text" - }, - { - "block_id": "p652-b29", - "global_id": 18995, - "bbox": [ - 263.6, - 519.61, - 283.13, - 531.06 - ], - "text": "Px =", - "type": "text" - }, - { - "block_id": "p652-b30", - "global_id": 18996, - "bbox": [ - 288.87, - 509.43, - 302.97, - 520.1 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p652-b31", - "global_id": 18997, - "bbox": [ - 285.17, - 533.45, - 306.65, - 540.65 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p652-b32", - "global_id": 18998, - "bbox": [ - 307.76, - 518.17, - 490.38, - 530.69 - ], - "text": "|Dn|2\n(6.27)", - "type": "text" - }, - { - "block_id": "p652-b33", - "global_id": 18999, - "bbox": [ - 101.85, - 556.19, - 259.97, - 567.27 - ], - "text": "For a real x(t), |D−n| = |Dn|. Therefore,", - "type": "text" - }, - { - "block_id": "p652-b34", - "global_id": 19000, - "bbox": [ - 251.23, - 594.78, - 283.49, - 606.24 - ], - "text": "Px = D0", - "type": "text" - }, - { - "block_id": "p652-b35", - "global_id": 19001, - "bbox": [ - 283.99, - 590.67, - 303.82, - 605.16 - ], - "text": "2 + 2", - "type": "text" - }, - { - "block_id": "p652-b36", - "global_id": 19002, - "bbox": [ - 304.92, - 584.61, - 319.02, - 595.28 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p652-b37", - "global_id": 19003, - "bbox": [ - 305.77, - 609.17, - 318.18, - 616.44 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p652-b38", - "global_id": 19004, - "bbox": [ - 320.13, - 593.35, - 490.38, - 605.86 - ], - "text": "|Dn|2\n(6.28)", - "type": "text" - } - ] - }, - { - "page_num": 653, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p653-b0", - "global_id": 19005, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n633", - "type": "text" - }, - { - "block_id": "p653-b1", - "global_id": 19006, - "bbox": [ - 102.51, - 93.91, - 434.44, - 105.87 - ], - "text": "EXAMPLE 6.10\nHarmonic Distortion of Clipped Sinusoid", - "type": "text" - }, - { - "block_id": "p653-b2", - "global_id": 19007, - "bbox": [ - 128.9, - 121.13, - 502.77, - 167.37 - ], - "text": "The input signal to an audio amplifier of gain 100 is given by x(t) = 0.1cosω0t. Hence, the\noutput is a sinusoid 10 cos ω0t. However, the amplifier, being nonlinear at higher amplitude\nlevels, clips all amplitudes beyond ±8 volts, as shown in Fig. 6.18a. We shall determine the\nharmonic distortion incurred in this operation.", - "type": "text" - }, - { - "block_id": "p653-b3", - "global_id": 19008, - "bbox": [ - 243.72, - 204.48, - 251.72, - 212.48 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p653-b4", - "global_id": 19009, - "bbox": [ - 249.33, - 221.0, - 253.33, - 229.0 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p653-b5", - "global_id": 19010, - "bbox": [ - 291.24, - 199.95, - 302.35, - 208.03 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p653-b6", - "global_id": 19011, - "bbox": [ - 339.05, - 342.97, - 387.3, - 352.8 - ], - "text": "10 cos v0t 8", - "type": "text" - }, - { - "block_id": "p653-b7", - "global_id": 19012, - "bbox": [ - 191.64, - 316.47, - 239.89, - 326.3 - ], - "text": "10 cos v0t 8", - "type": "text" - }, - { - "block_id": "p653-b8", - "global_id": 19013, - "bbox": [ - 131.8, - 342.21, - 180.05, - 352.03 - ], - "text": "10 cos v0t 8", - "type": "text" - }, - { - "block_id": "p653-b9", - "global_id": 19014, - "bbox": [ - 254.27, - 359.0, - 264.07, - 367.0 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p653-b10", - "global_id": 19015, - "bbox": [ - 374.65, - 251.25, - 382.09, - 260.86 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p653-b11", - "global_id": 19016, - "bbox": [ - 310.64, - 236.71, - 328.09, - 246.32 - ], - "text": "0.5T0", - "type": "text" - }, - { - "block_id": "p653-b12", - "global_id": 19017, - "bbox": [ - 291.85, - 322.62, - 321.3, - 332.23 - ], - "text": "0.3976T0", - "type": "text" - }, - { - "block_id": "p653-b13", - "global_id": 19018, - "bbox": [ - 187.51, - 236.5, - 211.63, - 246.32 - ], - "text": "0.5T0", - "type": "text" - }, - { - "block_id": "p653-b14", - "global_id": 19019, - "bbox": [ - 261.14, - 251.37, - 290.59, - 260.98 - ], - "text": "0.1024T0", - "type": "text" - }, - { - "block_id": "p653-b15", - "global_id": 19020, - "bbox": [ - 261.14, - 338.51, - 407.84, - 349.87 - ], - "text": "0.1024T0\nt", - "type": "text" - }, - { - "block_id": "p653-b16", - "global_id": 19021, - "bbox": [ - 405.61, - 250.85, - 407.84, - 258.85 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p653-b17", - "global_id": 19022, - "bbox": [ - 269.6, - 317.74, - 283.99, - 327.29 - ], - "text": "yd(t)", - "type": "text" - }, - { - "block_id": "p653-b18", - "global_id": 19023, - "bbox": [ - 254.73, - 292.99, - 263.61, - 300.99 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p653-b19", - "global_id": 19024, - "bbox": [ - 119.94, - 373.33, - 478.44, - 383.41 - ], - "text": "Figure 6.18 (a) A clipped sinusoid cos ω0t. (b) The distortion component xd(t) of the signal in (a).", - "type": "text" - }, - { - "block_id": "p653-b20", - "global_id": 19025, - "bbox": [ - 128.91, - 396.25, - 502.76, - 442.5 - ], - "text": "The output y(t) is the clipped signal in Fig. 6.18a. The distortion signal yd(t), shown in\nFig. 6.18b, is the difference between the undistorted sinusoid 10cosω0t and the output signal\ny(t). The signal yd(t), whose period is T0 [the same as that of y(t)], can be described over the\nfirst cycle as", - "type": "text" - }, - { - "block_id": "p653-b21", - "global_id": 19026, - "bbox": [ - 179.29, - 468.06, - 208.08, - 479.13 - ], - "text": "yd(t) =", - "type": "text" - }, - { - "block_id": "p653-b22", - "global_id": 19027, - "bbox": [ - 210.13, - 444.64, - 218.02, - 466.56 - ], - "text": "⎧\n⎪⎨", - "type": "text" - }, - { - "block_id": "p653-b23", - "global_id": 19028, - "bbox": [ - 210.13, - 474.53, - 218.02, - 487.48 - ], - "text": "⎪⎩", - "type": "text" - }, - { - "block_id": "p653-b24", - "global_id": 19029, - "bbox": [ - 218.02, - 451.63, - 351.59, - 462.78 - ], - "text": "10 cos ω0t −8\n|t| ≤0.1024T0", - "type": "text" - }, - { - "block_id": "p653-b25", - "global_id": 19030, - "bbox": [ - 218.02, - 462.17, - 305.15, - 479.99 - ], - "text": "10 cos ω0t + 8\nT0", - "type": "text" - }, - { - "block_id": "p653-b26", - "global_id": 19031, - "bbox": [ - 218.02, - 462.17, - 445.68, - 494.66 - ], - "text": "2 −0.1024T0 ≤|t| ≤T0\n2 + 0.1024T0\n0\neverywhere else", - "type": "text" - }, - { - "block_id": "p653-b27", - "global_id": 19032, - "bbox": [ - 128.9, - 503.7, - 502.75, - 527.52 - ], - "text": "Observe that yd(t) is an even function of t and its mean value is zero. Hence, a0 = C0 = 0, and\nbn = 0. Thus, Cn = an and the Fourier series for yd(t) can be expressed as", - "type": "text" - }, - { - "block_id": "p653-b28", - "global_id": 19033, - "bbox": [ - 269.97, - 543.72, - 298.77, - 554.8 - ], - "text": "yd(t) =", - "type": "text" - }, - { - "block_id": "p653-b29", - "global_id": 19034, - "bbox": [ - 300.81, - 533.55, - 314.91, - 544.21 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p653-b30", - "global_id": 19035, - "bbox": [ - 301.66, - 558.11, - 314.07, - 565.37 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p653-b31", - "global_id": 19036, - "bbox": [ - 316.02, - 543.72, - 361.52, - 555.59 - ], - "text": "Cn cos nω0t", - "type": "text" - }, - { - "block_id": "p653-b32", - "global_id": 19037, - "bbox": [ - 128.9, - 572.99, - 502.76, - 608.46 - ], - "text": "As usual, we can compute the coefficients Cn (which is equal to an) by integrating\nyd(t)cos nω0t over one cycle (and then dividing by 2/T0). Because yd(t) has even symmetry,\nwe can find an by integrating the expression over a half-cycle only using Eq. (6.14). The", - "type": "text" - } - ] - }, - { - "page_num": 654, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p654-b0", - "global_id": 19038, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "634\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p654-b1", - "global_id": 19039, - "bbox": [ - 103.16, - 86.48, - 343.35, - 97.04 - ], - "text": "straightforward evaluation of the appropriate integral yields†", - "type": "text" - }, - { - "block_id": "p654-b2", - "global_id": 19040, - "bbox": [ - 111.53, - 120.21, - 131.99, - 131.66 - ], - "text": "Cn =", - "type": "text" - }, - { - "block_id": "p654-b3", - "global_id": 19041, - "bbox": [ - 134.03, - 99.78, - 141.92, - 118.71 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p654-b4", - "global_id": 19042, - "bbox": [ - 134.03, - 126.68, - 141.92, - 136.64 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p654-b5", - "global_id": 19043, - "bbox": [ - 148.1, - 107.65, - 158.06, - 131.26 - ], - "text": "20\nπ", - "type": "text" - }, - { - "block_id": "p654-b6", - "global_id": 19044, - "bbox": [ - 160.37, - 100.24, - 242.0, - 117.61 - ], - "text": "sin[0.6435(n + 1)]", - "type": "text" - }, - { - "block_id": "p654-b7", - "global_id": 19045, - "bbox": [ - 194.09, - 107.24, - 330.29, - 131.67 - ], - "text": "n + 1\n+ sin[0.6435(n −1)]", - "type": "text" - }, - { - "block_id": "p654-b8", - "global_id": 19046, - "bbox": [ - 282.38, - 121.29, - 303.2, - 131.67 - ], - "text": "n −1", - "type": "text" - }, - { - "block_id": "p654-b9", - "global_id": 19047, - "bbox": [ - 331.51, - 100.24, - 336.94, - 110.2 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p654-b10", - "global_id": 19048, - "bbox": [ - 338.49, - 107.65, - 358.97, - 124.18 - ], - "text": "−32", - "type": "text" - }, - { - "block_id": "p654-b11", - "global_id": 19049, - "bbox": [ - 350.5, - 121.29, - 356.48, - 131.26 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p654-b12", - "global_id": 19050, - "bbox": [ - 361.27, - 100.24, - 420.41, - 117.61 - ], - "text": "sin(0.6435n)", - "type": "text" - }, - { - "block_id": "p654-b13", - "global_id": 19051, - "bbox": [ - 391.68, - 121.61, - 396.66, - 131.57 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p654-b14", - "global_id": 19052, - "bbox": [ - 421.64, - 100.24, - 427.07, - 110.2 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p654-b15", - "global_id": 19053, - "bbox": [ - 438.76, - 114.53, - 460.72, - 124.6 - ], - "text": "n odd", - "type": "text" - }, - { - "block_id": "p654-b16", - "global_id": 19054, - "bbox": [ - 146.9, - 132.37, - 462.45, - 142.44 - ], - "text": "0\nn even", - "type": "text" - }, - { - "block_id": "p654-b17", - "global_id": 19055, - "bbox": [ - 103.16, - 153.39, - 413.29, - 164.54 - ], - "text": "Computing the coefficients C1, C2, C3, . . . from this expression, we can write", - "type": "text" - }, - { - "block_id": "p654-b18", - "global_id": 19056, - "bbox": [ - 170.62, - 175.3, - 409.56, - 186.45 - ], - "text": "yd(t) = 1.04cos ω0t + 0.733cos 3ω0t + 0.311cos 5ω0t + · · ·", - "type": "text" - }, - { - "block_id": "p654-b19", - "global_id": 19057, - "bbox": [ - 103.46, - 200.41, - 307.0, - 212.54 - ], - "text": "COMPUTING HARMONIC DISTORTION", - "type": "text" - }, - { - "block_id": "p654-b20", - "global_id": 19058, - "bbox": [ - 103.16, - 216.57, - 477.02, - 262.4 - ], - "text": "We can compute the amount of harmonic distortion in the output signal by computing the\npower of the distortion component yd(t). Because yd(t) is an even function of t and because the\nenergy in the first half-cycle is identical to the energy in the second half-cycle, we can compute\nthe power by averaging the energy over a quarter-cycle. Thus,", - "type": "text" - }, - { - "block_id": "p654-b21", - "global_id": 19059, - "bbox": [ - 195.71, - 274.18, - 229.36, - 293.63 - ], - "text": "Pyd = 1", - "type": "text" - }, - { - "block_id": "p654-b22", - "global_id": 19060, - "bbox": [ - 222.1, - 288.14, - 231.12, - 298.97 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p654-b23", - "global_id": 19061, - "bbox": [ - 233.93, - 267.19, - 257.88, - 280.8 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p654-b24", - "global_id": 19062, - "bbox": [ - 239.19, - 292.07, - 258.75, - 300.59 - ], - "text": "−T0/2", - "type": "text" - }, - { - "block_id": "p654-b25", - "global_id": 19063, - "bbox": [ - 260.36, - 281.06, - 268.26, - 291.83 - ], - "text": "yd", - "type": "text" - }, - { - "block_id": "p654-b26", - "global_id": 19064, - "bbox": [ - 268.96, - 274.18, - 324.61, - 298.97 - ], - "text": "2(t)dt =\n1\nT0/4", - "type": "text" - }, - { - "block_id": "p654-b27", - "global_id": 19065, - "bbox": [ - 326.91, - 267.19, - 350.87, - 280.8 - ], - "text": "# T0/4", - "type": "text" - }, - { - "block_id": "p654-b28", - "global_id": 19066, - "bbox": [ - 332.18, - 281.06, - 360.38, - 299.33 - ], - "text": "0\nyd", - "type": "text" - }, - { - "block_id": "p654-b29", - "global_id": 19067, - "bbox": [ - 361.07, - 279.32, - 384.3, - 291.02 - ], - "text": "2(t)dt", - "type": "text" - }, - { - "block_id": "p654-b30", - "global_id": 19068, - "bbox": [ - 211.08, - 304.27, - 229.36, - 321.22 - ], - "text": "= 4", - "type": "text" - }, - { - "block_id": "p654-b31", - "global_id": 19069, - "bbox": [ - 222.1, - 318.23, - 231.12, - 329.07 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p654-b32", - "global_id": 19070, - "bbox": [ - 233.93, - 297.28, - 269.78, - 310.89 - ], - "text": "# 0.1024T0", - "type": "text" - }, - { - "block_id": "p654-b33", - "global_id": 19071, - "bbox": [ - 239.19, - 307.04, - 382.5, - 329.43 - ], - "text": "0\n(10 cos ω0t −8)2 dt = 0.865", - "type": "text" - }, - { - "block_id": "p654-b34", - "global_id": 19072, - "bbox": [ - 103.17, - 338.23, - 477.01, - 361.5 - ], - "text": "The power of the desired signal 10cos ω0t is (10)2/2 = 50. Hence, the total harmonic distortion\nis‡", - "type": "text" - }, - { - "block_id": "p654-b35", - "global_id": 19073, - "bbox": [ - 232.49, - 359.41, - 283.02, - 377.43 - ], - "text": "Dtot = 0.865", - "type": "text" - }, - { - "block_id": "p654-b36", - "global_id": 19074, - "bbox": [ - 266.85, - 365.98, - 347.67, - 383.43 - ], - "text": "50\n× 100 = 1.73%", - "type": "text" - }, - { - "block_id": "p654-b37", - "global_id": 19075, - "bbox": [ - 103.16, - 388.56, - 477.02, - 411.82 - ], - "text": "The power of the third harmonic components of yd(t) is (0.733)2/2 = 0.2686. The third\nharmonic distortion is", - "type": "text" - }, - { - "block_id": "p654-b38", - "global_id": 19076, - "bbox": [ - 226.97, - 409.74, - 278.59, - 427.76 - ], - "text": "D3 = 0.2686", - "type": "text" - }, - { - "block_id": "p654-b39", - "global_id": 19077, - "bbox": [ - 259.93, - 416.31, - 353.2, - 433.76 - ], - "text": "50\n× 100 = 0.5372%", - "type": "text" - }, - { - "block_id": "p654-b40", - "global_id": 19078, - "bbox": [ - 101.84, - 470.02, - 490.42, - 538.37 - ], - "text": "† In addition, yd(t) exhibits half-wave symmetry (see Prob. 6.1-6), where the second half-cycle is the negative\nof the first. Because of this property, all the even harmonics vanish, and the odd harmonics can be computed\nby integrating the appropriate expressions over the first half-cycle only (from −T0/4 to T0/4) and doubling\nthe resulting values. Moreover, because of even symmetry, we can integrate the appropriate expressions over 0\nto T0/4 (instead of from −T0/4 to T0/4) and double the resulting values. In essence, this allows us to compute\nCn by integrating the expression over the quarter-cycle only and then quadrupling the resulting values. Thus,", - "type": "text" - }, - { - "block_id": "p654-b41", - "global_id": 19079, - "bbox": [ - 204.31, - 544.35, - 251.41, - 560.56 - ], - "text": "Cn = an = 8", - "type": "text" - }, - { - "block_id": "p654-b42", - "global_id": 19080, - "bbox": [ - 244.8, - 556.9, - 253.02, - 566.7 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p654-b43", - "global_id": 19081, - "bbox": [ - 255.72, - 538.05, - 288.71, - 550.4 - ], - "text": "# 0.1024T0", - "type": "text" - }, - { - "block_id": "p654-b44", - "global_id": 19082, - "bbox": [ - 260.45, - 550.26, - 387.76, - 567.04 - ], - "text": "0\n[10 cos ω0t −8] cos nω0tdt", - "type": "text" - }, - { - "block_id": "p654-b45", - "global_id": 19083, - "bbox": [ - 101.84, - 577.36, - 490.41, - 633.41 - ], - "text": "‡ In the literature, the harmonic distortion often refers to the rms distortion rather than the power distortion.\nThe rms values are the square-root values of the corresponding powers. Thus, the third harmonic distortion\nin this sense is √(0.2686/50) × 100 = 7.33%. Alternately, we may also compute this value directly from\nthe amplitudes of the third harmonic 0.733 and that of the fundamental as 10. The ratio of the rms values is\n(0.733/", - "type": "text" - }, - { - "block_id": "p654-b46", - "global_id": 19084, - "bbox": [ - 129.57, - 616.49, - 137.16, - 625.45 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p654-b47", - "global_id": 19085, - "bbox": [ - 137.16, - 616.49, - 175.28, - 633.41 - ], - "text": "2) : (10/\n√", - "type": "text" - }, - { - "block_id": "p654-b48", - "global_id": 19086, - "bbox": [ - 175.28, - 624.07, - 360.38, - 633.41 - ], - "text": "2) = 0.0733 and the percentage distortion is 7.33%.", - "type": "text" - } - ] - }, - { - "page_num": 655, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p655-b0", - "global_id": 19087, - "bbox": [ - 348.36, - 62.89, - 516.12, - 71.98 - ], - "text": "6.3\nExponential Fourier Series\n635", - "type": "text" - }, - { - "block_id": "p655-b1", - "global_id": 19088, - "bbox": [ - 127.59, - 86.52, - 327.32, - 98.48 - ], - "text": "6.3-3 Properties of the Fourier Series", - "type": "text" - }, - { - "block_id": "p655-b2", - "global_id": 19089, - "bbox": [ - 127.59, - 104.51, - 516.14, - 174.35 - ], - "text": "As with the Laplace and z-transforms, the Fourier series has a variety of properties that can\nsimplify work and help provide a more intuitive understanding of signals. Table 6.2 provides\nthe most important properties of the Fourier series for a periodic signal x(t) and its spectrum\nDn. Properties that involve two signals require that the two signals have a common fundamental\nfrequency ω0. While not given here, the proofs of these properties are straightforward and parallel\nthe proofs of the Fourier transform properties given in Ch. 7.", - "type": "text" - }, - { - "block_id": "p655-b3", - "global_id": 19090, - "bbox": [ - 127.59, - 176.34, - 516.14, - 210.21 - ], - "text": "To demonstrate the utility of Fourier series properties, let us consider an example where\nwe use a selection of properties to simplify the work of finding a piecewise polynomial signal’s\nspectrum.", - "type": "text" - }, - { - "block_id": "p655-b4", - "global_id": 19091, - "bbox": [ - 127.59, - 227.82, - 302.45, - 237.06 - ], - "text": "TABLE 6.2\nSelected Fourier Series Properties", - "type": "text" - }, - { - "block_id": "p655-b5", - "global_id": 19092, - "bbox": [ - 127.59, - 247.78, - 477.32, - 257.59 - ], - "text": "Operation\nx(t)\nDn", - "type": "text" - }, - { - "block_id": "p655-b6", - "global_id": 19093, - "bbox": [ - 127.59, - 266.1, - 506.68, - 308.32 - ], - "text": "Scalar multiplication\nkx(t)\nkDn\nAddition\nx1(t) + x2(t)\nD1,n + D2,n\nx1(t), x2(t) require same ω0\nConjugation\nx∗(t)\nD∗", - "type": "text" - }, - { - "block_id": "p655-b7", - "global_id": 19094, - "bbox": [ - 127.59, - 303.38, - 504.36, - 363.86 - ], - "text": "−n\nReversal\nx(−t)\nD−n\nTime shifting\nx(t −t0)\nDne−jnω0t0\nFrequency shifting\nx(t)ejn0ω0t\nDn−n0\nFrequency convolution\nx1(t)x2(t)\nD1,n ∗D2,n\nx1(t), x2(t) require same ω0", - "type": "text" - }, - { - "block_id": "p655-b8", - "global_id": 19095, - "bbox": [ - 127.59, - 363.57, - 339.36, - 380.25 - ], - "text": "Time differentiation\ndkx(t)", - "type": "text" - }, - { - "block_id": "p655-b9", - "global_id": 19096, - "bbox": [ - 323.26, - 369.86, - 503.72, - 386.53 - ], - "text": "dtk\n(jnω0)kDn", - "type": "text" - }, - { - "block_id": "p655-b10", - "global_id": 19097, - "bbox": [ - 102.51, - 442.36, - 377.04, - 454.32 - ], - "text": "EXAMPLE 6.11\nUsing Fourier Series Properties", - "type": "text" - }, - { - "block_id": "p655-b11", - "global_id": 19098, - "bbox": [ - 128.9, - 470.98, - 502.76, - 504.85 - ], - "text": "Use properties rather than integration to compute the exponential Fourier series coefficients\nDn of the triangular signal x(t) shown in Fig. 6.4. Verify the correctness of Dn for A = 1 by\nsynthesizing x(t) with a suitable truncation of Eq. (6.19).", - "type": "text" - }, - { - "block_id": "p655-b12", - "global_id": 19099, - "bbox": [ - 128.9, - 527.35, - 502.76, - 586.62 - ], - "text": "From Fig. 6.4, we see that x(t) is a piecewise linear function that is T0 = 2 periodic. To\ncompute Dn directly using Eq. (6.19) would therefore require tedious integration by parts.\nFortunately, we can compute Dn without integration by instead using Fourier series properties.\nFirst, however, we must compute the dc component D0 separately from other Dn. By simple\ninspection of Fig. 6.4, we see that x(t) has no dc component, so D0 = 0.", - "type": "text" - }, - { - "block_id": "p655-b13", - "global_id": 19100, - "bbox": [ - 128.9, - 587.13, - 502.76, - 621.41 - ], - "text": "To determine the remaining Dn, we begin by noting that x(t) has a constant slope of\neither 2A or −2A. Thus, differentiating x(t) once yields a square wave with amplitudes ±2A.\nHere, differentiation reduces x(t) from a piecewise linear to a piecewise constant function,", - "type": "text" - } - ] - }, - { - "page_num": 656, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p656-b0", - "global_id": 19101, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "636\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p656-b1", - "global_id": 19102, - "bbox": [ - 103.16, - 83.46, - 477.02, - 144.86 - ], - "text": "thereby eliminating the need for integration by parts.† Differentiating x(t) twice yields a pair\nof shifted impulse trains, weighted by ±4A. This second differentiation eliminates the need for\nany integration whatsoever, since the Fourier series coefficients of an impulse train are known\nto be\n1\nT0 (see Ex. 6.9).\nStated mathematically, we see that", - "type": "text" - }, - { - "block_id": "p656-b2", - "global_id": 19103, - "bbox": [ - 220.22, - 152.04, - 228.95, - 163.25 - ], - "text": "d2", - "type": "text" - }, - { - "block_id": "p656-b3", - "global_id": 19104, - "bbox": [ - 218.88, - 158.62, - 300.25, - 177.31 - ], - "text": "dt2 x(t) = 4Aδ2(t + 1", - "type": "text" - }, - { - "block_id": "p656-b4", - "global_id": 19105, - "bbox": [ - 296.76, - 158.62, - 362.5, - 173.15 - ], - "text": "2) −4Aδ2(t −1\n2)", - "type": "text" - }, - { - "block_id": "p656-b5", - "global_id": 19106, - "bbox": [ - 103.16, - 182.32, - 477.01, - 204.24 - ], - "text": "Transforming this expression and using the Fourier series properties of scalar multiplication,\naddition, frequency shifting, and time differentiation yield (for n̸ = 0)", - "type": "text" - }, - { - "block_id": "p656-b6", - "global_id": 19107, - "bbox": [ - 218.72, - 211.13, - 304.83, - 230.8 - ], - "text": "(jnπ)2Dn = 4Aejnω0/2", - "type": "text" - }, - { - "block_id": "p656-b7", - "global_id": 19108, - "bbox": [ - 290.26, - 211.13, - 359.77, - 236.79 - ], - "text": "2\n−4Ae−jnω0/2", - "type": "text" - }, - { - "block_id": "p656-b8", - "global_id": 19109, - "bbox": [ - 103.17, - 226.83, - 347.46, - 253.04 - ], - "text": "2\nSubstituting ω0 = 2π/T0 = π and solving for Dn yield", - "type": "text" - }, - { - "block_id": "p656-b9", - "global_id": 19110, - "bbox": [ - 242.03, - 256.04, - 336.43, - 278.1 - ], - "text": "Dn = 4Aejnπ/2 −e−jnπ/2", - "type": "text" - }, - { - "block_id": "p656-b10", - "global_id": 19111, - "bbox": [ - 290.81, - 273.52, - 322.98, - 284.1 - ], - "text": "−2n2π2", - "type": "text" - }, - { - "block_id": "p656-b11", - "global_id": 19112, - "bbox": [ - 103.16, - 289.71, - 427.17, - 299.67 - ], - "text": "Combining with the dc component and simplifying expressions, the final result is", - "type": "text" - }, - { - "block_id": "p656-b12", - "global_id": 19113, - "bbox": [ - 230.66, - 314.79, - 251.65, - 326.24 - ], - "text": "Dn =", - "type": "text" - }, - { - "block_id": "p656-b13", - "global_id": 19114, - "bbox": [ - 253.7, - 300.8, - 343.2, - 325.83 - ], - "text": "0\nn = 0\n−A4jsin(nπ/2)", - "type": "text" - }, - { - "block_id": "p656-b14", - "global_id": 19115, - "bbox": [ - 280.37, - 320.78, - 343.35, - 334.64 - ], - "text": "n2π2\nn̸ = 0", - "type": "text" - }, - { - "block_id": "p656-b15", - "global_id": 19116, - "bbox": [ - 103.16, - 340.33, - 477.04, - 410.84 - ], - "text": "Overall, this is a neat way to determine the signal’s spectrum. Through the selective use\nof properties, we have determined Dn without any integration. With a little care, this basic\napproach can yield the spectrum of any piecewise polynomial periodic function. Since all\nperiodic functions of practical interest to engineers can, to an arbitrary level of accuracy,\nbe represented as piecewise polynomial functions, we can always find their spectra without\nintegration except for the dc term D0.", - "type": "text" - }, - { - "block_id": "p656-b16", - "global_id": 19117, - "bbox": [ - 103.16, - 424.8, - 477.01, - 489.85 - ], - "text": "USING A TRUNCATED FOURIER SERIES TO VERIFY\nSPECTRUM CORRECTNESS\nWhile a signal’s spectrum Dn provides useful insight into signal character, it can be difficult\nto look at Dn and know that it is correct for a particular signal x(t). For example, is it at all\nobvious that Dn = 4Ajsin(nπ/2)", - "type": "text" - }, - { - "block_id": "p656-b17", - "global_id": 19118, - "bbox": [ - 103.16, - 478.81, - 477.03, - 584.42 - ], - "text": "−n2π2\nis the spectrum for the triangle wave of Fig. 6.4? Probably not.\nThere is an easy way, however, to verify a signal’s spectrum: synthesize x(t) using a\nsuitable truncation of Eq. (6.19). If the synthesized signal closely matches the original, we\ncan be relatively certain that the spectrum Dn is correct. What is a suitable truncation? Well,\nit depends. All significant Dn terms need to be included, but not so many as to make the\nreconstruction impractical to compute. Since in the present case Dn decays at a rate 1/n2,\na good approximation is possible with a relatively few number of terms; a 10-harmonic\ntruncation should be just fine. Let us use MATLAB to synthesize x(t) using a 10-harmonic\ntruncated Fourier series.", - "type": "text" - }, - { - "block_id": "p656-b18", - "global_id": 19119, - "bbox": [ - 103.16, - 603.6, - 477.0, - 627.13 - ], - "text": "† Differentiation also destroys the dc component of the signal, providing further justification as to why\nthe dc component needs to be separately computed.", - "type": "text" - } - ] - }, - { - "page_num": 657, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p657-b0", - "global_id": 19120, - "bbox": [ - 289.5, - 62.89, - 516.12, - 71.98 - ], - "text": "6.4\nLTIC System Response to Periodic Inputs\n637", - "type": "text" - }, - { - "block_id": "p657-b1", - "global_id": 19121, - "bbox": [ - 128.9, - 85.83, - 502.76, - 108.15 - ], - "text": "To begin, we define A, Dn, T0, ω0, and a time vector that spans two periods of the\nwaveform.", - "type": "text" - }, - { - "block_id": "p657-b2", - "global_id": 19122, - "bbox": [ - 128.9, - 118.41, - 395.65, - 140.33 - ], - "text": ">>\nA = 1; D = @(n) -A*4j*sin(n*pi/2)./(n.^2*pi^2);\n>>\nT0 = 2; omega0 = 2*pi/T0; t = (-T0:.001:T0);", - "type": "text" - }, - { - "block_id": "p657-b3", - "global_id": 19123, - "bbox": [ - 128.9, - 150.0, - 290.33, - 159.96 - ], - "text": "Next, we set the dc portion of the signal.", - "type": "text" - }, - { - "block_id": "p657-b4", - "global_id": 19124, - "bbox": [ - 128.9, - 170.21, - 311.97, - 180.17 - ], - "text": ">>\nD0 = 0; x10 = D0*ones(size(t));", - "type": "text" - }, - { - "block_id": "p657-b5", - "global_id": 19125, - "bbox": [ - 128.9, - 189.44, - 502.77, - 224.01 - ], - "text": "To add the desired 10 harmonics, we enter a loop for 1 ≤n ≤10 and add in the Dn and\nD−n terms. Although x(t) should be real, small round-off errors cause the reconstruction to be\ncomplex. These small imaginary parts are removed using the real command.", - "type": "text" - }, - { - "block_id": "p657-b6", - "global_id": 19126, - "bbox": [ - 128.9, - 233.98, - 515.94, - 267.85 - ], - "text": ">>\nfor n = 1:10,\n>>\nx10 = x10+real(D(n)*exp(1j*omega0*n*t)+D(-n)*exp(-1j*omega0*n*t));\n>>\nend", - "type": "text" - }, - { - "block_id": "p657-b7", - "global_id": 19127, - "bbox": [ - 128.9, - 277.11, - 405.82, - 287.49 - ], - "text": "Lastly, we plot the resulting truncated Fourier series synthesis of x(t).", - "type": "text" - }, - { - "block_id": "p657-b8", - "global_id": 19128, - "bbox": [ - 128.91, - 297.73, - 411.35, - 307.7 - ], - "text": ">>\nplot(t,x10,’k’); xlabel(’t’); ylabel(’x_{10}(t)’);", - "type": "text" - }, - { - "block_id": "p657-b9", - "global_id": 19129, - "bbox": [ - 128.91, - 317.37, - 502.77, - 340.79 - ], - "text": "Since the synthesized waveform shown in Fig. 6.19 closely matches the original waveform in\nFig. 6.4, we have high confidence that the computed Dn are correct.", - "type": "text" - }, - { - "block_id": "p657-b10", - "global_id": 19130, - "bbox": [ - 140.5, - 439.89, - 482.12, - 460.97 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2\nt", - "type": "text" - }, - { - "block_id": "p657-b11", - "global_id": 19131, - "bbox": [ - 135.14, - 430.38, - 143.14, - 438.38 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p657-b12", - "global_id": 19132, - "bbox": [ - 137.39, - 394.67, - 141.39, - 402.67 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p657-b13", - "global_id": 19133, - "bbox": [ - 137.39, - 358.66, - 141.39, - 366.66 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p657-b14", - "global_id": 19134, - "bbox": [ - 120.64, - 388.34, - 131.91, - 407.15 - ], - "text": "x10(t)", - "type": "text" - }, - { - "block_id": "p657-b15", - "global_id": 19135, - "bbox": [ - 119.94, - 467.28, - 338.85, - 476.89 - ], - "text": "Figure 6.19 A 10-harmonic truncated Fourier series of x(t).", - "type": "text" - }, - { - "block_id": "p657-b16", - "global_id": 19136, - "bbox": [ - 127.94, - 536.68, - 456.38, - 550.63 - ], - "text": "6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS", - "type": "text" - }, - { - "block_id": "p657-b17", - "global_id": 19137, - "bbox": [ - 127.59, - 556.61, - 516.15, - 603.94 - ], - "text": "A periodic signal can be expressed as a sum of everlasting exponentials (or sinusoids). We also\nknow how to find the response of an LTIC system to an everlasting exponential. From this\ninformation, we can readily determine the response of an LTIC system to periodic inputs. A\nperiodic signal x(t) with period T0 can be expressed as an exponential Fourier series", - "type": "text" - }, - { - "block_id": "p657-b18", - "global_id": 19138, - "bbox": [ - 253.07, - 626.37, - 277.72, - 636.65 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p657-b19", - "global_id": 19139, - "bbox": [ - 283.46, - 616.2, - 297.56, - 626.87 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p657-b20", - "global_id": 19140, - "bbox": [ - 279.77, - 640.22, - 301.24, - 647.42 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p657-b21", - "global_id": 19141, - "bbox": [ - 302.36, - 619.39, - 388.46, - 637.83 - ], - "text": "Dnejnω0t\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p657-b22", - "global_id": 19142, - "bbox": [ - 378.72, - 633.76, - 387.74, - 644.6 - ], - "text": "T0", - "type": "text" - } - ] - }, - { - "page_num": 658, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p658-b0", - "global_id": 19143, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "638\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p658-b1", - "global_id": 19144, - "bbox": [ - 101.84, - 85.46, - 490.39, - 119.75 - ], - "text": "In Sec. 4.8, we showed that the response of an LTIC system with transfer function H(s) to an\neverlasting exponential input ejωt is an everlasting exponential H(jω)ejωt. This input–output pair\ncan be displayed as†", - "type": "text" - }, - { - "block_id": "p658-b2", - "global_id": 19145, - "bbox": [ - 257.75, - 126.99, - 275.68, - 142.8 - ], - "text": "ejωt", - "type": "text" - }, - { - "block_id": "p658-b3", - "global_id": 19146, - "bbox": [ - 260.58, - 145.24, - 272.87, - 151.22 - ], - "text": "input", - "type": "text" - }, - { - "block_id": "p658-b4", - "global_id": 19147, - "bbox": [ - 277.73, - 126.99, - 333.85, - 138.99 - ], - "text": "⇒H(jω)ejωt", - "type": "text" - }, - { - "block_id": "p658-b5", - "global_id": 19148, - "bbox": [ - 296.28, - 134.76, - 334.48, - 153.14 - ], - "text": "output", - "type": "text" - }, - { - "block_id": "p658-b6", - "global_id": 19149, - "bbox": [ - 101.84, - 166.21, - 253.9, - 176.18 - ], - "text": "Therefore, from the linearity property,", - "type": "text" - }, - { - "block_id": "p658-b7", - "global_id": 19150, - "bbox": [ - 217.58, - 190.59, - 231.68, - 201.27 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p658-b8", - "global_id": 19151, - "bbox": [ - 213.88, - 214.61, - 235.36, - 221.81 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p658-b9", - "global_id": 19152, - "bbox": [ - 236.48, - 199.04, - 267.5, - 211.84 - ], - "text": "Dnejnω0t", - "type": "text" - }, - { - "block_id": "p658-b11", - "global_id": 19153, - "bbox": [ - 228.43, - 229.86, - 253.58, - 237.05 - ], - "text": "input x(t)", - "type": "text" - }, - { - "block_id": "p658-b12", - "global_id": 19154, - "bbox": [ - 270.17, - 200.76, - 286.67, - 210.73 - ], - "text": "⇒", - "type": "text" - }, - { - "block_id": "p658-b13", - "global_id": 19155, - "bbox": [ - 294.47, - 190.59, - 308.57, - 201.27 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p658-b14", - "global_id": 19156, - "bbox": [ - 290.77, - 214.61, - 312.25, - 221.81 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p658-b15", - "global_id": 19157, - "bbox": [ - 313.36, - 199.04, - 377.72, - 211.91 - ], - "text": "DnH(jnω0)ejnω0t", - "type": "text" - }, - { - "block_id": "p658-b17", - "global_id": 19158, - "bbox": [ - 317.67, - 229.86, - 351.44, - 237.05 - ], - "text": "response y(t)", - "type": "text" - }, - { - "block_id": "p658-b18", - "global_id": 19159, - "bbox": [ - 466.32, - 201.18, - 490.38, - 211.14 - ], - "text": "(6.29)", - "type": "text" - }, - { - "block_id": "p658-b19", - "global_id": 19160, - "bbox": [ - 101.85, - 252.72, - 490.4, - 275.06 - ], - "text": "The response y(t) is obtained in the form of an exponential Fourier series and is therefore a periodic\nsignal of the same period as that of the input.", - "type": "text" - }, - { - "block_id": "p658-b20", - "global_id": 19161, - "bbox": [ - 119.78, - 277.05, - 415.2, - 287.01 - ], - "text": "We shall demonstrate the utility of these results by the following example.", - "type": "text" - }, - { - "block_id": "p658-b21", - "global_id": 19162, - "bbox": [ - 76.77, - 318.46, - 285.29, - 330.42 - ], - "text": "EXAMPLE 6.12\nFull-Wave Rectifier", - "type": "text" - }, - { - "block_id": "p658-b22", - "global_id": 19163, - "bbox": [ - 103.16, - 344.06, - 477.02, - 389.99 - ], - "text": "A full-wave rectifier (Fig. 6.20a) is used to obtain a dc signal from a sinusoid sin t. The rectified\nsignal x(t), depicted in Fig. 6.17, is applied to the input of a lowpass RC filter, which suppresses\nthe time-varying component and yields a dc component with some residual ripple. Find the\nfilter output y(t). Find also the dc output and the rms value of the ripple voltage.", - "type": "text" - }, - { - "block_id": "p658-b23", - "global_id": 19164, - "bbox": [ - 103.16, - 412.49, - 477.02, - 435.9 - ], - "text": "First, we shall find the Fourier series for the rectified signal x(t), whose period is T0 = π.\nConsequently, ω0 = 2, and", - "type": "text" - }, - { - "block_id": "p658-b24", - "global_id": 19165, - "bbox": [ - 251.91, - 444.53, - 276.56, - 454.8 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p658-b25", - "global_id": 19166, - "bbox": [ - 282.3, - 434.35, - 296.4, - 445.02 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p658-b26", - "global_id": 19167, - "bbox": [ - 278.6, - 458.38, - 300.08, - 465.57 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p658-b27", - "global_id": 19168, - "bbox": [ - 301.2, - 443.02, - 327.64, - 455.61 - ], - "text": "Dnej2nt", - "type": "text" - }, - { - "block_id": "p658-b28", - "global_id": 19169, - "bbox": [ - 103.16, - 472.84, - 127.49, - 482.8 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p658-b29", - "global_id": 19170, - "bbox": [ - 212.5, - 480.89, - 242.72, - 498.91 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p658-b30", - "global_id": 19171, - "bbox": [ - 236.74, - 494.53, - 242.72, - 504.49 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p658-b31", - "global_id": 19172, - "bbox": [ - 246.02, - 473.89, - 260.02, - 485.95 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p658-b32", - "global_id": 19173, - "bbox": [ - 251.28, - 480.89, - 477.01, - 506.04 - ], - "text": "0\nsinte−j2nt dt =\n2\nπ(1 −4n2)\n(6.30)", - "type": "text" - }, - { - "block_id": "p658-b33", - "global_id": 19174, - "bbox": [ - 103.16, - 511.62, - 144.91, - 521.58 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p658-b34", - "global_id": 19175, - "bbox": [ - 234.34, - 530.57, - 258.99, - 540.84 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p658-b35", - "global_id": 19176, - "bbox": [ - 264.73, - 520.39, - 278.83, - 531.07 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p658-b36", - "global_id": 19177, - "bbox": [ - 261.03, - 544.42, - 282.51, - 551.61 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p658-b37", - "global_id": 19178, - "bbox": [ - 284.83, - 524.0, - 345.21, - 548.02 - ], - "text": "2\nπ(1 −4n2)ej2nt", - "type": "text" - }, - { - "block_id": "p658-b38", - "global_id": 19179, - "bbox": [ - 101.84, - 588.31, - 490.39, - 633.41 - ], - "text": "† This result applies only to asymptotically stable systems. This is because when s = jω, the integral on\nthe right-hand side of Eq. (2.39) does not converge for unstable systems. Moreover, for marginally stable\nsystems also, that integral does not converge in the ordinary sense, and H(jω) cannot be obtained from H(s)\nby replacing s with jω.", - "type": "text" - } - ] - }, - { - "page_num": 659, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p659-b0", - "global_id": 19180, - "bbox": [ - 289.5, - 62.89, - 516.12, - 71.98 - ], - "text": "6.4\nLTIC System Response to Periodic Inputs\n639", - "type": "text" - }, - { - "block_id": "p659-b1", - "global_id": 19181, - "bbox": [ - 294.25, - 187.38, - 303.13, - 195.38 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p659-b2", - "global_id": 19182, - "bbox": [ - 392.63, - 290.59, - 394.86, - 298.59 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p659-b3", - "global_id": 19183, - "bbox": [ - 273.47, - 211.95, - 284.89, - 220.04 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p659-b4", - "global_id": 19184, - "bbox": [ - 293.78, - 305.14, - 303.59, - 313.14 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p659-b5", - "global_id": 19185, - "bbox": [ - 133.55, - 290.64, - 460.49, - 299.67 - ], - "text": "3p\n3p\n0", - "type": "text" - }, - { - "block_id": "p659-b6", - "global_id": 19186, - "bbox": [ - 303.97, - 208.52, - 307.97, - 216.52 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p659-b7", - "global_id": 19187, - "bbox": [ - 360.05, - 94.06, - 376.28, - 102.35 - ], - "text": "15", - "type": "text" - }, - { - "block_id": "p659-b8", - "global_id": 19188, - "bbox": [ - 322.81, - 137.57, - 448.09, - 145.9 - ], - "text": "y(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p659-b13", - "global_id": 19189, - "bbox": [ - 146.56, - 131.96, - 277.45, - 145.67 - ], - "text": "sin t\nFull-wave", - "type": "text" - }, - { - "block_id": "p659-b14", - "global_id": 19190, - "bbox": [ - 248.57, - 140.96, - 273.89, - 148.96 - ], - "text": "rectifier", - "type": "text" - }, - { - "block_id": "p659-b15", - "global_id": 19191, - "bbox": [ - 394.25, - 132.87, - 404.32, - 146.51 - ], - "text": "1\n5 F", - "type": "text" - }, - { - "block_id": "p659-b16", - "global_id": 19192, - "bbox": [ - 119.94, - 319.77, - 390.41, - 329.08 - ], - "text": "Figure 6.20 (a) Full-wave rectifier with a lowpass filter and (b) its output.", - "type": "text" - }, - { - "block_id": "p659-b17", - "global_id": 19193, - "bbox": [ - 128.9, - 346.69, - 502.75, - 380.65 - ], - "text": "Next, we find the transfer function of the RC filter in Fig. 6.20a. This filter is identical\nto the RC circuit in Ex. 1.17 (Fig. 1.35) for which the differential equation relating the output\n(capacitor voltage) to the input x(t) was found to be [Eq. (1.31)]:", - "type": "text" - }, - { - "block_id": "p659-b18", - "global_id": 19194, - "bbox": [ - 277.35, - 392.2, - 354.31, - 402.58 - ], - "text": "(3D + 1)y(t) = x(t)", - "type": "text" - }, - { - "block_id": "p659-b19", - "global_id": 19195, - "bbox": [ - 128.91, - 414.12, - 405.73, - 424.49 - ], - "text": "The transfer function H(s) for this system is found from Eq. (2.41) as", - "type": "text" - }, - { - "block_id": "p659-b20", - "global_id": 19196, - "bbox": [ - 286.87, - 434.43, - 343.59, - 458.45 - ], - "text": "H(s) =\n1\n3s + 1", - "type": "text" - }, - { - "block_id": "p659-b21", - "global_id": 19197, - "bbox": [ - 128.9, - 467.06, - 143.28, - 477.03 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p659-b22", - "global_id": 19198, - "bbox": [ - 281.26, - 474.94, - 502.75, - 498.96 - ], - "text": "H(jω) =\n1\n3jω + 1\n(6.31)", - "type": "text" - }, - { - "block_id": "p659-b23", - "global_id": 19199, - "bbox": [ - 128.9, - 505.43, - 414.48, - 516.89 - ], - "text": "From Eq. (6.29), the filter output y(t) can be expressed as (with ω0 = 2)", - "type": "text" - }, - { - "block_id": "p659-b24", - "global_id": 19200, - "bbox": [ - 214.04, - 535.48, - 238.66, - 545.76 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p659-b25", - "global_id": 19201, - "bbox": [ - 244.41, - 525.31, - 258.51, - 535.98 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p659-b26", - "global_id": 19202, - "bbox": [ - 240.71, - 549.33, - 262.19, - 556.52 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p659-b27", - "global_id": 19203, - "bbox": [ - 263.3, - 531.36, - 338.1, - 546.63 - ], - "text": "DnH(jnω0)ejnω0t =", - "type": "text" - }, - { - "block_id": "p659-b28", - "global_id": 19204, - "bbox": [ - 343.84, - 525.31, - 357.94, - 535.98 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p659-b29", - "global_id": 19205, - "bbox": [ - 340.15, - 549.33, - 361.63, - 556.52 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p659-b30", - "global_id": 19206, - "bbox": [ - 362.74, - 533.97, - 416.98, - 546.56 - ], - "text": "DnH(j2n)ej2nt", - "type": "text" - }, - { - "block_id": "p659-b31", - "global_id": 19207, - "bbox": [ - 128.9, - 566.42, - 496.46, - 578.29 - ], - "text": "Substituting Dn and H(j2n) from Eqs. (6.30) and (6.31) in the foregoing equation, we obtain", - "type": "text" - }, - { - "block_id": "p659-b32", - "global_id": 19208, - "bbox": [ - 242.09, - 596.47, - 266.71, - 606.74 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p659-b33", - "global_id": 19209, - "bbox": [ - 272.45, - 586.29, - 286.55, - 596.97 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p659-b34", - "global_id": 19210, - "bbox": [ - 268.76, - 610.31, - 290.23, - 617.51 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p659-b35", - "global_id": 19211, - "bbox": [ - 292.54, - 589.89, - 388.95, - 613.92 - ], - "text": "2\nπ(1 −4n2)(j6n + 1)ej2nt", - "type": "text" - } - ] - }, - { - "page_num": 660, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p660-b0", - "global_id": 19212, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "640\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p660-b1", - "global_id": 19213, - "bbox": [ - 103.16, - 85.88, - 477.0, - 108.22 - ], - "text": "Note that the output y(t) is also a periodic signal given by the exponential Fourier series on the\nright-hand side. The output is shown in Fig. 6.20b.", - "type": "text" - }, - { - "block_id": "p660-b2", - "global_id": 19214, - "bbox": [ - 103.16, - 109.8, - 477.02, - 179.95 - ], - "text": "The output Fourier series coefficient corresponding to n = 0 is the dc component of\nthe output, given by 2/π. The remaining terms in the Fourier series constitute the unwanted\ncomponent called the ripple. We can determine the rms value of the ripple voltage by using\nEq. (6.27) to find the power of the ripple component. The power of the ripple is the power of\nall the components except the dc (n = 0). Note that ˆDn, the exponential Fourier coefficient for\nthe output y(t), is", - "type": "text" - }, - { - "block_id": "p660-b3", - "global_id": 19215, - "bbox": [ - 237.4, - 189.89, - 341.57, - 213.91 - ], - "text": "ˆDn =\n2\nπ(1 −4n2)(j6n + 1)", - "type": "text" - }, - { - "block_id": "p660-b4", - "global_id": 19216, - "bbox": [ - 103.17, - 223.73, - 246.93, - 233.69 - ], - "text": "Therefore, from Eq. (6.28), we have", - "type": "text" - }, - { - "block_id": "p660-b5", - "global_id": 19217, - "bbox": [ - 125.0, - 253.36, - 164.7, - 264.81 - ], - "text": "Pripple = 2", - "type": "text" - }, - { - "block_id": "p660-b6", - "global_id": 19218, - "bbox": [ - 165.8, - 243.18, - 179.9, - 253.85 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p660-b7", - "global_id": 19219, - "bbox": [ - 166.65, - 267.74, - 179.06, - 275.01 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p660-b8", - "global_id": 19220, - "bbox": [ - 181.01, - 249.24, - 218.73, - 264.43 - ], - "text": "|Dn|2 = 2", - "type": "text" - }, - { - "block_id": "p660-b9", - "global_id": 19221, - "bbox": [ - 219.83, - 243.18, - 233.93, - 253.85 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p660-b10", - "global_id": 19222, - "bbox": [ - 220.68, - 267.74, - 233.08, - 275.01 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p660-b12", - "global_id": 19223, - "bbox": [ - 239.47, - 246.78, - 319.4, - 270.81 - ], - "text": "2\nπ(1 −4n2)(j6n + 1)", - "type": "text" - }, - { - "block_id": "p660-b14", - "global_id": 19224, - "bbox": [ - 323.84, - 244.28, - 348.86, - 263.73 - ], - "text": "2\n= 8", - "type": "text" - }, - { - "block_id": "p660-b15", - "global_id": 19225, - "bbox": [ - 340.89, - 260.23, - 351.35, - 270.39 - ], - "text": "π2", - "type": "text" - }, - { - "block_id": "p660-b16", - "global_id": 19226, - "bbox": [ - 354.16, - 243.18, - 368.25, - 253.85 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p660-b17", - "global_id": 19227, - "bbox": [ - 354.99, - 267.74, - 367.4, - 275.01 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p660-b18", - "global_id": 19228, - "bbox": [ - 370.55, - 246.79, - 453.97, - 270.81 - ], - "text": "1\n(1 −4n2)2(36n2 + 1)", - "type": "text" - }, - { - "block_id": "p660-b19", - "global_id": 19229, - "bbox": [ - 103.16, - 284.29, - 477.03, - 318.58 - ], - "text": "Numerical computation of the right-hand side yields Pripple = 0.0025, and the ripple rms\nvalue = Pripple = 0.05. This shows that the rms ripple voltage is 5% of the amplitude of\nthe input sinusoid.", - "type": "text" - }, - { - "block_id": "p660-b20", - "global_id": 19230, - "bbox": [ - 102.14, - 369.57, - 246.68, - 381.69 - ], - "text": "WHY USE EXPONENTIALS?", - "type": "text" - }, - { - "block_id": "p660-b21", - "global_id": 19231, - "bbox": [ - 101.84, - 385.72, - 490.41, - 503.28 - ], - "text": "The exponential Fourier series is just another way of representing trigonometric Fourier series\n(or vice versa). The two forms carry identical information—no more, no less. The reasons for\npreferring the exponential form have already been mentioned: this form is more compact, and\nthe expression for deriving the exponential coefficients is also more compact than those in the\ntrigonometric series. Furthermore, the LTIC system response to exponential signals is also simpler\n(more compact) than the system response to sinusoids. In addition, the exponential form proves to\nbe much easier than the trigonometric form to manipulate mathematically and otherwise handle\nin the area of signals as well as systems. Moreover, exponential representation proves much more\nconvenient for analysis of complex x(t). For these reasons, in our future discussion we shall use\nthe exponential form exclusively.", - "type": "text" - }, - { - "block_id": "p660-b22", - "global_id": 19232, - "bbox": [ - 101.84, - 505.27, - 490.4, - 634.79 - ], - "text": "A minor disadvantage of the exponential form is that it cannot be visualized as easily\nas sinusoids. For intuitive and qualitative understanding, the sinusoids have the edge over\nexponentials. Fortunately, this difficulty can be overcome readily because of the close connection\nbetween exponential and Fourier spectra. For the purpose of mathematical analysis, we shall\ncontinue to use exponential signals and spectra; but to understand the physical situation intuitively\nor qualitatively, we shall speak in terms of sinusoids and trigonometric spectra. Thus, although\nall mathematical manipulation will be in terms of exponential spectra, we shall now speak of\nexponential and sinusoids interchangeably when we discuss intuitive and qualitative insights\nin attempting to arrive at an understanding of physical situations. This is an important point;\nreaders should make an extra effort to familiarize themselves with the two forms of spectra, their\nrelationships, and their convertibility.", - "type": "text" - } - ] - }, - { - "page_num": 661, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p661-b0", - "global_id": 19233, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n641", - "type": "text" - }, - { - "block_id": "p661-b1", - "global_id": 19234, - "bbox": [ - 127.89, - 86.19, - 310.43, - 98.32 - ], - "text": "DUAL PERSONALITY OF A SIGNAL", - "type": "text" - }, - { - "block_id": "p661-b2", - "global_id": 19235, - "bbox": [ - 127.59, - 102.35, - 516.15, - 184.04 - ], - "text": "The discussion so far shows that a periodic signal has a dual personality—the time domain and\nthe frequency domain. It can be described by its waveform or by its Fourier spectra. The time-\nand frequency-domain descriptions provide complementary insights into a signal. For in-depth\nperspective, we need to understand both these identities. It is important to learn to think of a signal\nfrom both perspectives. In the next chapter, we shall see that aperiodic signals also have this dual\npersonality. Moreover, we shall show that even LTI systems have this dual personality, which offers\ncomplementary insights into the system behavior.", - "type": "text" - }, - { - "block_id": "p661-b3", - "global_id": 19236, - "bbox": [ - 127.59, - 202.61, - 516.15, - 264.58 - ], - "text": "LIMITATIONS OF THE FOURIER SERIES METHOD OF ANALYSIS\nWe have developed here a method of representing a periodic signal as a weighted sum of\neverlasting exponentials whose frequencies lie along the ω axis in the s plane. This representation\n(Fourier series) is valuable in many applications. However, as a tool for analyzing linear systems,\nit has serious limitations and consequently has limited utility for the following reasons:", - "type": "text" - }, - { - "block_id": "p661-b4", - "global_id": 19237, - "bbox": [ - 144.52, - 272.56, - 516.16, - 318.38 - ], - "text": "1. The Fourier series can be used only for periodic inputs. All practical inputs are aperiodic\n(remember that a periodic signal starts at t = −∞).\n2. The Fourier methods can be applied readily to BIBO-stable (or asymptotically stable)\nsystems. It cannot handle unstable or even marginally stable systems.", - "type": "text" - }, - { - "block_id": "p661-b5", - "global_id": 19238, - "bbox": [ - 127.59, - 326.35, - 516.17, - 408.05 - ], - "text": "The first limitation can be overcome by representing aperiodic signals in terms of everlasting\nexponentials. This representation can be achieved through the Fourier integral, which may be\nconsidered to be an extension of the Fourier series. We shall therefore use the Fourier series as a\nstepping-stone to the Fourier integral developed in the next chapter. The second limitation can be\novercome by using exponentials est, where s is not restricted to the imaginary axis but is free to\ntake on complex values. This generalization leads to the Laplace integral, discussed in Ch. 4 (the\nLaplace transform).", - "type": "text" - }, - { - "block_id": "p661-b6", - "global_id": 19239, - "bbox": [ - 127.94, - 444.99, - 360.37, - 474.88 - ], - "text": "6.5 GENERALIZED FOURIER SERIES:\nSIGNALS AS VECTORS", - "type": "text" - }, - { - "block_id": "p661-b7", - "global_id": 19240, - "bbox": [ - 127.59, - 480.87, - 516.15, - 562.56 - ], - "text": "We now consider a very general approach to signal representation with far-reaching\nconsequences.† There is a perfect analogy between signals and vectors; the analogy is so strong\nthat the term analogy understates the reality. Signals are not just like vectors. Signals are vectors!\nA vector can be represented as a sum of its components in a variety of ways, depending on the\nchoice of coordinate system. A signal can also be represented as a sum of its components in a\nvariety of ways. Let us begin with some basic vector concepts and then apply these concepts to\nsignals.", - "type": "text" - }, - { - "block_id": "p661-b8", - "global_id": 19241, - "bbox": [ - 127.59, - 588.31, - 516.11, - 633.41 - ], - "text": "† This section closely follows the material from the author’s earlier book [10]. Omission of this section will\nnot cause any discontinuity in understanding the rest of the book. Derivation of Fourier series through the\nsignal-vector analogy provides an interesting insight into signal representation and other topics such as signal\ncorrelation, data truncation, and signal detection.", - "type": "text" - } - ] - }, - { - "page_num": 662, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p662-b0", - "global_id": 19242, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "642\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p662-b1", - "global_id": 19243, - "bbox": [ - 101.84, - 86.52, - 257.44, - 98.48 - ], - "text": "6.5-1 Component of a Vector", - "type": "text" - }, - { - "block_id": "p662-b2", - "global_id": 19244, - "bbox": [ - 101.84, - 104.61, - 490.38, - 138.48 - ], - "text": "A vector is specified by its magnitude and its direction. We shall denote all vectors by boldface.\nFor example, x is a certain vector with magnitude or length |x|. For the two vectors x and y shown\nin Fig. 6.21, we define their dot (inner or scalar) product as", - "type": "text" - }, - { - "block_id": "p662-b3", - "global_id": 19245, - "bbox": [ - 260.6, - 151.24, - 330.64, - 161.61 - ], - "text": "x · y = |x||y|cos θ", - "type": "text" - }, - { - "block_id": "p662-b4", - "global_id": 19246, - "bbox": [ - 101.85, - 174.38, - 490.4, - 196.7 - ], - "text": "where θ is the angle between these vectors. Using this definition, we can express |x|, the length of\na vector x, as", - "type": "text" - }, - { - "block_id": "p662-b5", - "global_id": 19247, - "bbox": [ - 274.93, - 195.99, - 317.3, - 210.41 - ], - "text": "|x|2 = x · x", - "type": "text" - }, - { - "block_id": "p662-b6", - "global_id": 19248, - "bbox": [ - 101.84, - 220.57, - 490.39, - 278.46 - ], - "text": "Let the component of x along y be cy as depicted in Fig. 6.21. Geometrically, the component\nof x along y is the projection of x on y and is obtained by drawing a perpendicular from the tip of\nx on the vector y, as illustrated in Fig. 6.21. What is the mathematical significance of a component\nof a vector along another vector? As seen from Fig. 6.21, the vector x can be expressed in terms\nof vector y as", - "type": "text" - }, - { - "block_id": "p662-b7", - "global_id": 19249, - "bbox": [ - 275.35, - 281.86, - 316.89, - 292.16 - ], - "text": "x = cy + e", - "type": "text" - }, - { - "block_id": "p662-b8", - "global_id": 19250, - "bbox": [ - 101.85, - 302.34, - 490.39, - 324.34 - ], - "text": "However, this is not the only way to express x in terms of y. From Fig. 6.22, which shows two of\nthe infinite other possibilities, we have", - "type": "text" - }, - { - "block_id": "p662-b9", - "global_id": 19251, - "bbox": [ - 249.09, - 337.1, - 342.64, - 348.55 - ], - "text": "x = c1y + e1 = c2y + e2", - "type": "text" - }, - { - "block_id": "p662-b10", - "global_id": 19252, - "bbox": [ - 101.84, - 360.57, - 490.38, - 382.56 - ], - "text": "In each of these three representations, x is represented in terms of y plus another vector called the\nerror vector. If we approximate x by cy,", - "type": "text" - }, - { - "block_id": "p662-b11", - "global_id": 19253, - "bbox": [ - 282.99, - 385.98, - 309.24, - 396.27 - ], - "text": "x ≃cy", - "type": "text" - }, - { - "block_id": "p662-b12", - "global_id": 19254, - "bbox": [ - 101.84, - 406.12, - 490.4, - 476.27 - ], - "text": "the error in the approximation is the vector e = x −cy. Similarly, the errors in approximations in\nthese drawings are e1 (Fig. 6.22a) and e2 (Fig. 6.22b). What is unique about the approximation in\nFig. 6.21 is that the error vector is the smallest. We can now define mathematically the component\nof a vector x along vector y to be cy where c is chosen to minimize the length of the error vector\ne = x −cy. Now, the length of the component of x along y is |x|cosθ. But it is also c|y|, as seen\nfrom Fig. 6.21. Therefore,", - "type": "text" - }, - { - "block_id": "p662-b13", - "global_id": 19255, - "bbox": [ - 266.86, - 479.67, - 324.38, - 490.05 - ], - "text": "c|y| = |x|cosθ", - "type": "text" - }, - { - "block_id": "p662-b14", - "global_id": 19256, - "bbox": [ - 101.85, - 499.83, - 243.99, - 510.2 - ], - "text": "Multiplying both sides by |y| yields", - "type": "text" - }, - { - "block_id": "p662-b15", - "global_id": 19257, - "bbox": [ - 245.13, - 518.84, - 347.12, - 533.33 - ], - "text": "c|y|2 = |x||y|cos θ = x · y", - "type": "text" - }, - { - "block_id": "p662-b16", - "global_id": 19258, - "bbox": [ - 139.04, - 571.36, - 143.04, - 579.36 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p662-b17", - "global_id": 19259, - "bbox": [ - 170.06, - 617.13, - 213.17, - 625.13 - ], - "text": "cy\ny", - "type": "text" - }, - { - "block_id": "p662-b18", - "global_id": 19260, - "bbox": [ - 180.04, - 580.95, - 183.59, - 588.95 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p662-b19", - "global_id": 19261, - "bbox": [ - 128.4, - 603.41, - 132.4, - 611.41 - ], - "text": "u", - "type": "text" - }, - { - "block_id": "p662-b20", - "global_id": 19262, - "bbox": [ - 223.8, - 617.27, - 478.29, - 626.5 - ], - "text": "Figure 6.21 Component (projection) of a vector along another vector.", - "type": "text" - } - ] - }, - { - "page_num": 663, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p663-b0", - "global_id": 19263, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n643", - "type": "text" - }, - { - "block_id": "p663-b1", - "global_id": 19264, - "bbox": [ - 205.34, - 166.74, - 348.92, - 174.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p663-b2", - "global_id": 19265, - "bbox": [ - 221.96, - 114.01, - 228.51, - 123.62 - ], - "text": "e1", - "type": "text" - }, - { - "block_id": "p663-b3", - "global_id": 19266, - "bbox": [ - 204.82, - 146.02, - 266.73, - 160.76 - ], - "text": "y\nc1y", - "type": "text" - }, - { - "block_id": "p663-b4", - "global_id": 19267, - "bbox": [ - 183.38, - 108.61, - 321.55, - 116.61 - ], - "text": "x\nx", - "type": "text" - }, - { - "block_id": "p663-b5", - "global_id": 19268, - "bbox": [ - 371.55, - 146.02, - 401.47, - 160.76 - ], - "text": "y\nc2y", - "type": "text" - }, - { - "block_id": "p663-b6", - "global_id": 19269, - "bbox": [ - 374.24, - 114.12, - 380.79, - 123.73 - ], - "text": "e2", - "type": "text" - }, - { - "block_id": "p663-b7", - "global_id": 19270, - "bbox": [ - 151.5, - 181.44, - 393.28, - 190.68 - ], - "text": "Figure 6.22 Approximation of a vector in terms of another vector.", - "type": "text" - }, - { - "block_id": "p663-b8", - "global_id": 19271, - "bbox": [ - 127.59, - 210.85, - 169.34, - 220.82 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p663-b9", - "global_id": 19272, - "bbox": [ - 281.67, - 219.66, - 314.99, - 236.93 - ], - "text": "c = x · y", - "type": "text" - }, - { - "block_id": "p663-b10", - "global_id": 19273, - "bbox": [ - 299.16, - 220.07, - 516.13, - 244.01 - ], - "text": "y · y =\n1\n|y|2 x · y\n(6.32)", - "type": "text" - }, - { - "block_id": "p663-b11", - "global_id": 19274, - "bbox": [ - 127.59, - 257.52, - 516.14, - 291.47 - ], - "text": "From Fig. 6.21, it is apparent that when x and y are perpendicular, or orthogonal, then x has a\nzero component along y; consequently, c = 0. Keeping an eye on Eq. (6.32), we therefore define x\nand y to be orthogonal if the inner (scalar or dot) product of the two vectors is zero, that is, if", - "type": "text" - }, - { - "block_id": "p663-b12", - "global_id": 19275, - "bbox": [ - 305.52, - 302.35, - 338.2, - 312.72 - ], - "text": "x · y = 0", - "type": "text" - }, - { - "block_id": "p663-b13", - "global_id": 19276, - "bbox": [ - 127.59, - 337.14, - 415.67, - 349.1 - ], - "text": "6.5-2 Signal Comparison and Component of a Signal", - "type": "text" - }, - { - "block_id": "p663-b14", - "global_id": 19277, - "bbox": [ - 127.59, - 355.23, - 516.15, - 389.88 - ], - "text": "The concept of a vector component and orthogonality can be extended to signals. Consider the\nproblem of approximating a real signal x(t) in terms of another real signal y(t) over an interval\n(t1, t2):", - "type": "text" - }, - { - "block_id": "p663-b15", - "global_id": 19278, - "bbox": [ - 268.84, - 390.68, - 374.39, - 402.14 - ], - "text": "x(t) ≃cy(t)\nt1 < t < t2", - "type": "text" - }, - { - "block_id": "p663-b16", - "global_id": 19279, - "bbox": [ - 127.59, - 409.28, - 279.1, - 419.65 - ], - "text": "The error e(t) in this approximation is", - "type": "text" - }, - { - "block_id": "p663-b17", - "global_id": 19280, - "bbox": [ - 249.83, - 436.17, - 274.46, - 446.44 - ], - "text": "e(t) =", - "type": "text" - }, - { - "block_id": "p663-b18", - "global_id": 19281, - "bbox": [ - 276.51, - 422.18, - 387.21, - 452.42 - ], - "text": "x(t) −cy(t)\nt1 < t < t2\n0\notherwise", - "type": "text" - }, - { - "block_id": "p663-b19", - "global_id": 19282, - "bbox": [ - 127.59, - 463.12, - 516.14, - 498.49 - ], - "text": "We now select a criterion for the “best approximation.” We know that the signal energy is one\npossible measure of a signal size. For best approximation, we shall use the criterion that minimizes\nthe size or energy of the error signal e(t) over the interval (t1,t2). This energy Ee is given by", - "type": "text" - }, - { - "block_id": "p663-b20", - "global_id": 19283, - "bbox": [ - 242.65, - 513.66, - 262.15, - 525.11 - ], - "text": "Ee =", - "type": "text" - }, - { - "block_id": "p663-b21", - "global_id": 19284, - "bbox": [ - 264.2, - 500.1, - 278.94, - 513.71 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p663-b22", - "global_id": 19285, - "bbox": [ - 269.45, - 525.2, - 274.37, - 533.5 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p663-b23", - "global_id": 19286, - "bbox": [ - 281.04, - 512.23, - 318.69, - 523.93 - ], - "text": "e2(t)dt =", - "type": "text" - }, - { - "block_id": "p663-b24", - "global_id": 19287, - "bbox": [ - 320.74, - 500.1, - 335.48, - 513.71 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p663-b25", - "global_id": 19288, - "bbox": [ - 326.0, - 525.2, - 330.92, - 533.5 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p663-b26", - "global_id": 19289, - "bbox": [ - 336.48, - 509.86, - 400.89, - 523.93 - ], - "text": "[x(t) −cy(t)]2 dt", - "type": "text" - }, - { - "block_id": "p663-b27", - "global_id": 19290, - "bbox": [ - 127.59, - 541.47, - 516.14, - 575.45 - ], - "text": "Note that the right-hand side is a definite integral with t as the dummy variable. Hence, Ee is a\nfunction of the parameter c (not t) and Ee is minimum for some choice of c. To minimize Ee, a\nnecessary condition is", - "type": "text" - }, - { - "block_id": "p663-b28", - "global_id": 19291, - "bbox": [ - 306.11, - 575.18, - 320.27, - 585.95 - ], - "text": "dEe", - "type": "text" - }, - { - "block_id": "p663-b29", - "global_id": 19292, - "bbox": [ - 308.74, - 581.85, - 338.81, - 599.2 - ], - "text": "dc = 0", - "type": "text" - }, - { - "block_id": "p663-b30", - "global_id": 19293, - "bbox": [ - 127.59, - 600.86, - 135.89, - 610.82 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p663-b31", - "global_id": 19294, - "bbox": [ - 262.58, - 609.46, - 271.98, - 633.48 - ], - "text": "d\ndc", - "type": "text" - }, - { - "block_id": "p663-b32", - "global_id": 19295, - "bbox": [ - 274.29, - 602.15, - 294.46, - 616.19 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p663-b33", - "global_id": 19296, - "bbox": [ - 284.98, - 627.67, - 289.9, - 635.98 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p663-b34", - "global_id": 19297, - "bbox": [ - 295.47, - 612.33, - 359.87, - 626.41 - ], - "text": "[x(t) −cy(t)]2 dt", - "type": "text" - }, - { - "block_id": "p663-b35", - "global_id": 19298, - "bbox": [ - 360.05, - 602.15, - 365.48, - 612.11 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p663-b36", - "global_id": 19299, - "bbox": [ - 367.53, - 616.13, - 382.33, - 626.51 - ], - "text": "= 0", - "type": "text" - } - ] - }, - { - "page_num": 664, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p664-b0", - "global_id": 19300, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "644\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p664-b1", - "global_id": 19301, - "bbox": [ - 101.84, - 85.82, - 331.65, - 95.78 - ], - "text": "Expanding the squared term inside the integral, we obtain", - "type": "text" - }, - { - "block_id": "p664-b2", - "global_id": 19302, - "bbox": [ - 159.9, - 106.71, - 169.3, - 130.73 - ], - "text": "d\ndc", - "type": "text" - }, - { - "block_id": "p664-b3", - "global_id": 19303, - "bbox": [ - 171.61, - 99.4, - 191.78, - 113.43 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p664-b4", - "global_id": 19304, - "bbox": [ - 182.29, - 124.92, - 187.22, - 133.23 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b5", - "global_id": 19305, - "bbox": [ - 193.9, - 111.95, - 221.57, - 123.66 - ], - "text": "x2(t)dt", - "type": "text" - }, - { - "block_id": "p664-b6", - "global_id": 19306, - "bbox": [ - 221.75, - 99.4, - 227.18, - 109.36 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p664-b7", - "global_id": 19307, - "bbox": [ - 228.73, - 106.71, - 246.3, - 123.35 - ], - "text": "−d", - "type": "text" - }, - { - "block_id": "p664-b8", - "global_id": 19308, - "bbox": [ - 239.25, - 120.77, - 248.65, - 130.73 - ], - "text": "dc", - "type": "text" - }, - { - "block_id": "p664-b10", - "global_id": 19309, - "bbox": [ - 256.38, - 99.82, - 281.64, - 123.76 - ], - "text": "2c\n# t2", - "type": "text" - }, - { - "block_id": "p664-b11", - "global_id": 19310, - "bbox": [ - 272.15, - 124.92, - 277.07, - 133.23 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b12", - "global_id": 19311, - "bbox": [ - 283.74, - 113.38, - 322.23, - 123.66 - ], - "text": "x(t)y(t)dt", - "type": "text" - }, - { - "block_id": "p664-b13", - "global_id": 19312, - "bbox": [ - 322.42, - 99.4, - 327.85, - 109.36 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p664-b14", - "global_id": 19313, - "bbox": [ - 329.4, - 106.71, - 346.97, - 123.66 - ], - "text": "+ d", - "type": "text" - }, - { - "block_id": "p664-b15", - "global_id": 19314, - "bbox": [ - 339.92, - 120.77, - 349.32, - 130.73 - ], - "text": "dc", - "type": "text" - }, - { - "block_id": "p664-b17", - "global_id": 19315, - "bbox": [ - 357.06, - 111.95, - 364.96, - 123.66 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p664-b18", - "global_id": 19316, - "bbox": [ - 366.57, - 99.82, - 381.31, - 113.43 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p664-b19", - "global_id": 19317, - "bbox": [ - 371.83, - 124.92, - 376.75, - 133.23 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b20", - "global_id": 19318, - "bbox": [ - 383.43, - 111.95, - 411.07, - 123.66 - ], - "text": "y2(t)dt", - "type": "text" - }, - { - "block_id": "p664-b21", - "global_id": 19319, - "bbox": [ - 411.26, - 99.4, - 416.68, - 109.36 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p664-b22", - "global_id": 19320, - "bbox": [ - 418.73, - 113.38, - 433.53, - 123.76 - ], - "text": "= 0", - "type": "text" - }, - { - "block_id": "p664-b23", - "global_id": 19321, - "bbox": [ - 101.85, - 142.24, - 176.81, - 152.2 - ], - "text": "from which we get", - "type": "text" - }, - { - "block_id": "p664-b24", - "global_id": 19322, - "bbox": [ - 219.97, - 159.98, - 232.72, - 170.35 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p664-b25", - "global_id": 19323, - "bbox": [ - 233.82, - 146.41, - 248.57, - 160.03 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p664-b26", - "global_id": 19324, - "bbox": [ - 239.08, - 171.51, - 244.01, - 179.82 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b27", - "global_id": 19325, - "bbox": [ - 250.68, - 159.98, - 309.62, - 170.35 - ], - "text": "x(t)y(t)dt + 2c", - "type": "text" - }, - { - "block_id": "p664-b28", - "global_id": 19326, - "bbox": [ - 310.73, - 146.41, - 325.49, - 160.03 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p664-b29", - "global_id": 19327, - "bbox": [ - 316.0, - 171.51, - 320.92, - 179.82 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b30", - "global_id": 19328, - "bbox": [ - 327.59, - 158.54, - 372.26, - 170.35 - ], - "text": "y2(t)dt = 0", - "type": "text" - }, - { - "block_id": "p664-b31", - "global_id": 19329, - "bbox": [ - 218.17, - 208.17, - 232.41, - 218.45 - ], - "text": "c =", - "type": "text" - }, - { - "block_id": "p664-b32", - "global_id": 19330, - "bbox": [ - 235.66, - 180.08, - 250.4, - 193.69 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p664-b33", - "global_id": 19331, - "bbox": [ - 240.91, - 205.18, - 245.84, - 213.49 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b34", - "global_id": 19332, - "bbox": [ - 241.08, - 193.64, - 291.0, - 221.92 - ], - "text": "x(t)y(t)dt\n# t2", - "type": "text" - }, - { - "block_id": "p664-b35", - "global_id": 19333, - "bbox": [ - 246.34, - 233.41, - 251.26, - 241.71 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b36", - "global_id": 19334, - "bbox": [ - 257.93, - 220.44, - 285.59, - 232.14 - ], - "text": "y2(t)dt", - "type": "text" - }, - { - "block_id": "p664-b37", - "global_id": 19335, - "bbox": [ - 294.43, - 201.6, - 312.77, - 218.55 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p664-b38", - "global_id": 19336, - "bbox": [ - 305.44, - 215.56, - 314.62, - 226.32 - ], - "text": "Ey", - "type": "text" - }, - { - "block_id": "p664-b39", - "global_id": 19337, - "bbox": [ - 318.54, - 194.61, - 333.28, - 208.22 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p664-b40", - "global_id": 19338, - "bbox": [ - 323.79, - 219.71, - 328.72, - 228.02 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b41", - "global_id": 19339, - "bbox": [ - 335.39, - 208.17, - 490.38, - 218.55 - ], - "text": "x(t)y(t)dt\n(6.33)", - "type": "text" - }, - { - "block_id": "p664-b42", - "global_id": 19340, - "bbox": [ - 101.85, - 247.74, - 490.39, - 317.48 - ], - "text": "We observe a remarkable similarity between the behavior of vectors and signals, as indicated by\nEqs. (6.32) and (6.33). It is evident from these two parallel expressions that the area under the\nproduct of two signals corresponds to the inner (scalar or dot) product of two vectors. In fact, the\narea under the product of x(t) and y(t) is called the inner product of x(t) and y(t), and is denoted\nby (x, y). The energy of a signal is the inner product of a signal with itself, and corresponds to the\nvector length square (which is the inner product of the vector with itself).", - "type": "text" - }, - { - "block_id": "p664-b43", - "global_id": 19341, - "bbox": [ - 119.78, - 319.06, - 463.24, - 329.44 - ], - "text": "To summarize our discussion, if a signal x(t) is approximated by another signal y(t) as", - "type": "text" - }, - { - "block_id": "p664-b44", - "global_id": 19342, - "bbox": [ - 273.15, - 341.25, - 319.08, - 351.52 - ], - "text": "x(t) ≃cy(t)", - "type": "text" - }, - { - "block_id": "p664-b45", - "global_id": 19343, - "bbox": [ - 101.85, - 363.75, - 490.37, - 385.77 - ], - "text": "then the optimum value of c that minimizes the energy of the error signal in this approximation is\ngiven by Eq. (6.33).", - "type": "text" - }, - { - "block_id": "p664-b46", - "global_id": 19344, - "bbox": [ - 101.85, - 387.34, - 490.4, - 446.32 - ], - "text": "Taking our clue from vectors, we say that a signal x(t) contains a component cy(t), where\nc is given by Eq. (6.33). Note that in vector terminology, cy(t) is the projection of x(t) on y(t).\nContinuing with the analogy, we say that if the component of a signal x(t) of the form y(t) is zero\n(i.e., c = 0), the signals x(t) and y(t) are orthogonal over the interval (t1, t2). Therefore, we define\nthe real signals x(t) and y(t) to be orthogonal over the interval (t1, t2) if†", - "type": "text" - }, - { - "block_id": "p664-b47", - "global_id": 19345, - "bbox": [ - 259.93, - 449.59, - 274.67, - 463.19 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p664-b48", - "global_id": 19346, - "bbox": [ - 265.18, - 474.68, - 270.11, - 482.99 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p664-b49", - "global_id": 19347, - "bbox": [ - 276.79, - 463.14, - 490.38, - 473.52 - ], - "text": "x(t)y(t)dt = 0\n(6.34)", - "type": "text" - }, - { - "block_id": "p664-b50", - "global_id": 19348, - "bbox": [ - 76.77, - 501.23, - 423.27, - 513.18 - ], - "text": "EXAMPLE 6.13\nSine-Wave Approximation of a Square Wave", - "type": "text" - }, - { - "block_id": "p664-b51", - "global_id": 19349, - "bbox": [ - 103.16, - 529.44, - 477.02, - 551.76 - ], - "text": "For the square signal x(t) shown in Fig. 6.23, find the component in x(t) of the form sint. In\nother words, approximate x(t) in terms of sin t", - "type": "text" - }, - { - "block_id": "p664-b52", - "global_id": 19350, - "bbox": [ - 234.36, - 563.31, - 344.82, - 573.69 - ], - "text": "x(t) ≃csint\n0 < t < 2π", - "type": "text" - }, - { - "block_id": "p664-b53", - "global_id": 19351, - "bbox": [ - 103.17, - 585.64, - 299.98, - 595.6 - ], - "text": "so that the energy of the error signal is minimum.", - "type": "text" - }, - { - "block_id": "p664-b54", - "global_id": 19352, - "bbox": [ - 101.84, - 633.15, - 384.13, - 645.37 - ], - "text": "† For complex signals, the definition is modified as in Eq. (6.37), in Sec. 6.5-3.", - "type": "text" - } - ] - }, - { - "page_num": 665, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p665-b0", - "global_id": 19353, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n645", - "type": "text" - }, - { - "block_id": "p665-b1", - "global_id": 19354, - "bbox": [ - 119.94, - 171.41, - 130.6, - 179.71 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p665-b2", - "global_id": 19355, - "bbox": [ - 126.6, - 94.08, - 156.35, - 109.21 - ], - "text": "1\nx(t)", - "type": "text" - }, - { - "block_id": "p665-b3", - "global_id": 19356, - "bbox": [ - 243.92, - 141.41, - 382.15, - 150.56 - ], - "text": "p\n2p", - "type": "text" - }, - { - "block_id": "p665-b4", - "global_id": 19357, - "bbox": [ - 376.03, - 128.99, - 378.25, - 136.99 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p665-b5", - "global_id": 19358, - "bbox": [ - 126.6, - 142.43, - 130.6, - 150.43 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p665-b6", - "global_id": 19359, - "bbox": [ - 119.94, - 193.84, - 392.51, - 203.08 - ], - "text": "Figure 6.23 Approximation of a square wave in terms of a single sinusoid.", - "type": "text" - }, - { - "block_id": "p665-b7", - "global_id": 19360, - "bbox": [ - 146.84, - 230.75, - 194.15, - 240.71 - ], - "text": "In this case,", - "type": "text" - }, - { - "block_id": "p665-b8", - "global_id": 19361, - "bbox": [ - 219.93, - 248.39, - 336.03, - 259.84 - ], - "text": "y(t) = sint\nand\nEy =", - "type": "text" - }, - { - "block_id": "p665-b9", - "global_id": 19362, - "bbox": [ - 338.07, - 234.82, - 355.56, - 247.17 - ], - "text": "# 2π", - "type": "text" - }, - { - "block_id": "p665-b10", - "global_id": 19363, - "bbox": [ - 343.33, - 246.95, - 410.74, - 266.97 - ], - "text": "0\nsin2(t)dt = π", - "type": "text" - }, - { - "block_id": "p665-b11", - "global_id": 19364, - "bbox": [ - 128.9, - 272.55, - 228.05, - 282.51 - ], - "text": "From Eq. (6.33), we find", - "type": "text" - }, - { - "block_id": "p665-b12", - "global_id": 19365, - "bbox": [ - 192.36, - 294.3, - 215.82, - 311.24 - ], - "text": "c = 1", - "type": "text" - }, - { - "block_id": "p665-b13", - "global_id": 19366, - "bbox": [ - 209.85, - 307.94, - 215.83, - 317.9 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p665-b14", - "global_id": 19367, - "bbox": [ - 219.12, - 287.31, - 236.62, - 299.66 - ], - "text": "# 2π", - "type": "text" - }, - { - "block_id": "p665-b15", - "global_id": 19368, - "bbox": [ - 224.39, - 294.3, - 298.62, - 319.45 - ], - "text": "0\nx(t)sintdt = 1", - "type": "text" - }, - { - "block_id": "p665-b16", - "global_id": 19369, - "bbox": [ - 292.64, - 307.94, - 298.62, - 317.9 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p665-b17", - "global_id": 19370, - "bbox": [ - 301.92, - 286.88, - 321.36, - 299.37 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p665-b18", - "global_id": 19371, - "bbox": [ - 312.61, - 300.87, - 357.7, - 319.45 - ], - "text": "0\nsintdt +", - "type": "text" - }, - { - "block_id": "p665-b19", - "global_id": 19372, - "bbox": [ - 359.24, - 287.31, - 376.74, - 299.66 - ], - "text": "# 2π", - "type": "text" - }, - { - "block_id": "p665-b20", - "global_id": 19373, - "bbox": [ - 364.51, - 312.19, - 368.69, - 319.16 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p665-b21", - "global_id": 19374, - "bbox": [ - 379.05, - 300.87, - 412.46, - 311.24 - ], - "text": "−sintdt", - "type": "text" - }, - { - "block_id": "p665-b22", - "global_id": 19375, - "bbox": [ - 412.65, - 286.88, - 418.08, - 296.84 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p665-b23", - "global_id": 19376, - "bbox": [ - 420.12, - 294.3, - 437.11, - 311.24 - ], - "text": "= 4", - "type": "text" - }, - { - "block_id": "p665-b24", - "global_id": 19377, - "bbox": [ - 431.13, - 307.94, - 437.11, - 317.9 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p665-b25", - "global_id": 19378, - "bbox": [ - 128.91, - 328.49, - 151.31, - 338.45 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p665-b26", - "global_id": 19379, - "bbox": [ - 289.41, - 337.71, - 323.28, - 354.56 - ], - "text": "x(t) ≃4", - "type": "text" - }, - { - "block_id": "p665-b27", - "global_id": 19380, - "bbox": [ - 317.3, - 344.6, - 342.09, - 361.73 - ], - "text": "π sint", - "type": "text" - }, - { - "block_id": "p665-b28", - "global_id": 19381, - "bbox": [ - 128.91, - 366.32, - 502.78, - 412.56 - ], - "text": "represents the best approximation of x(t) by the function sint, which will minimize the error\nenergy. This sinusoidal component of x(t) is shaded in Fig. 6.23. By analogy with vectors, we\nsay that the square function x(t) depicted in Fig. 6.23 has a component of signal sint and that\nthe magnitude of this component is 4/π.", - "type": "text" - }, - { - "block_id": "p665-b29", - "global_id": 19382, - "bbox": [ - 133.57, - 483.58, - 462.76, - 495.54 - ], - "text": "DRILL 6.7\nSine Wave Approximation of a Ramp Function", - "type": "text" - }, - { - "block_id": "p665-b30", - "global_id": 19383, - "bbox": [ - 133.57, - 504.25, - 510.16, - 550.49 - ], - "text": "Show that over an interval (−π < t < π), the “best” approximation of the signal x(t) = t in\nterms of the function sint is 2sint. Verify that the error signal e(t) = t −2sint is orthogonal to\nthe signal sint over the interval −π < t < π. Sketch the signals t and 2sint over the interval\n−π < t < π.", - "type": "text" - }, - { - "block_id": "p665-b31", - "global_id": 19384, - "bbox": [ - 127.59, - 594.78, - 321.02, - 606.73 - ], - "text": "6.5-3 Extension to Complex Signals", - "type": "text" - }, - { - "block_id": "p665-b32", - "global_id": 19385, - "bbox": [ - 127.59, - 612.76, - 516.13, - 634.79 - ], - "text": "So far we have restricted ourselves to real functions of t. To generalize the results to complex\nfunctions of t, consider again the problem of approximating a signal x(t) by a signal y(t) over an", - "type": "text" - } - ] - }, - { - "page_num": 666, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p666-b0", - "global_id": 19386, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "646\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p666-b1", - "global_id": 19387, - "bbox": [ - 101.84, - 85.46, - 184.92, - 96.92 - ], - "text": "interval (t1 < t < t2):", - "type": "text" - }, - { - "block_id": "p666-b2", - "global_id": 19388, - "bbox": [ - 273.15, - 97.78, - 319.08, - 108.05 - ], - "text": "x(t) ≃cy(t)", - "type": "text" - }, - { - "block_id": "p666-b3", - "global_id": 19389, - "bbox": [ - 101.84, - 116.95, - 490.37, - 140.06 - ], - "text": "where x(t) and y(t) now can be complex functions of t. Recall that the energy Ey of the complex\nsignal y(t) over an interval (t1, t2) is", - "type": "text" - }, - { - "block_id": "p666-b4", - "global_id": 19390, - "bbox": [ - 260.15, - 156.85, - 279.65, - 168.31 - ], - "text": "Ey =", - "type": "text" - }, - { - "block_id": "p666-b5", - "global_id": 19391, - "bbox": [ - 281.7, - 143.29, - 296.45, - 156.9 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b6", - "global_id": 19392, - "bbox": [ - 286.96, - 168.39, - 291.89, - 176.7 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p666-b7", - "global_id": 19393, - "bbox": [ - 298.56, - 153.05, - 331.9, - 167.13 - ], - "text": "|y(t)|2 dt", - "type": "text" - }, - { - "block_id": "p666-b8", - "global_id": 19394, - "bbox": [ - 101.84, - 185.58, - 288.96, - 195.64 - ], - "text": "In this case, both the coefficient c and the error", - "type": "text" - }, - { - "block_id": "p666-b9", - "global_id": 19395, - "bbox": [ - 260.32, - 207.42, - 331.9, - 217.7 - ], - "text": "e(t) = x(t) −cy(t)", - "type": "text" - }, - { - "block_id": "p666-b10", - "global_id": 19396, - "bbox": [ - 101.84, - 229.9, - 490.39, - 251.91 - ], - "text": "are complex (in general). For the “best” approximation, we choose c to minimize the energy Ee of\nthe error signal e(t). Now,", - "type": "text" - }, - { - "block_id": "p666-b11", - "global_id": 19397, - "bbox": [ - 245.09, - 259.64, - 264.59, - 271.1 - ], - "text": "Ee =", - "type": "text" - }, - { - "block_id": "p666-b12", - "global_id": 19398, - "bbox": [ - 266.63, - 246.08, - 281.39, - 259.7 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b13", - "global_id": 19399, - "bbox": [ - 271.9, - 271.18, - 276.82, - 279.49 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p666-b14", - "global_id": 19400, - "bbox": [ - 283.49, - 255.84, - 490.38, - 270.02 - ], - "text": "|x(t) −cy(t)|2 dt\n(6.35)", - "type": "text" - }, - { - "block_id": "p666-b15", - "global_id": 19401, - "bbox": [ - 101.85, - 285.48, - 163.27, - 295.44 - ], - "text": "Recall also that", - "type": "text" - }, - { - "block_id": "p666-b16", - "global_id": 19402, - "bbox": [ - 193.2, - 293.27, - 490.38, - 307.76 - ], - "text": "|u + v|2 = (u + v)(u∗+ v∗) = |u|2 + |v|2 + u∗v + uv∗\n(6.36)", - "type": "text" - }, - { - "block_id": "p666-b17", - "global_id": 19403, - "bbox": [ - 101.85, - 316.96, - 395.4, - 326.93 - ], - "text": "After some manipulation, we can use this result to rearrange Eq. (6.35) as", - "type": "text" - }, - { - "block_id": "p666-b18", - "global_id": 19404, - "bbox": [ - 141.09, - 349.4, - 160.59, - 360.86 - ], - "text": "Ee =", - "type": "text" - }, - { - "block_id": "p666-b19", - "global_id": 19405, - "bbox": [ - 162.64, - 335.84, - 177.38, - 349.45 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b20", - "global_id": 19406, - "bbox": [ - 167.9, - 360.94, - 172.82, - 369.24 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p666-b21", - "global_id": 19407, - "bbox": [ - 179.5, - 345.6, - 222.37, - 359.67 - ], - "text": "|x(t)|2 dt −", - "type": "text" - }, - { - "block_id": "p666-b23", - "global_id": 19408, - "bbox": [ - 228.35, - 342.83, - 245.79, - 368.28 - ], - "text": "1\nEy", - "type": "text" - }, - { - "block_id": "p666-b24", - "global_id": 19409, - "bbox": [ - 248.6, - 335.84, - 263.34, - 349.45 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b25", - "global_id": 19410, - "bbox": [ - 253.85, - 360.94, - 257.34, - 367.91 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p666-b26", - "global_id": 19411, - "bbox": [ - 265.45, - 347.68, - 308.08, - 359.67 - ], - "text": "x(t)y∗(t)dt", - "type": "text" - }, - { - "block_id": "p666-b28", - "global_id": 19412, - "bbox": [ - 311.49, - 337.34, - 324.8, - 359.36 - ], - "text": "2\n+", - "type": "text" - }, - { - "block_id": "p666-b29", - "global_id": 19413, - "bbox": [ - 326.34, - 331.97, - 334.01, - 365.84 - ], - "text": "c", - "type": "text" - }, - { - "block_id": "p666-b31", - "global_id": 19414, - "bbox": [ - 342.27, - 342.83, - 381.46, - 368.28 - ], - "text": "Ey −\n1\nEy", - "type": "text" - }, - { - "block_id": "p666-b32", - "global_id": 19415, - "bbox": [ - 384.26, - 335.84, - 399.01, - 349.45 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b33", - "global_id": 19416, - "bbox": [ - 389.52, - 360.94, - 394.44, - 369.24 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p666-b34", - "global_id": 19417, - "bbox": [ - 401.11, - 347.68, - 443.74, - 359.67 - ], - "text": "x(t)y∗(t)dt", - "type": "text" - }, - { - "block_id": "p666-b36", - "global_id": 19418, - "bbox": [ - 447.16, - 337.34, - 450.64, - 344.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p666-b37", - "global_id": 19419, - "bbox": [ - 101.84, - 380.15, - 490.38, - 402.17 - ], - "text": "Since the first two terms on the right-hand side are independent of c, it is clear that Ee is minimized\nby choosing c so that the third term on the right-hand side is zero. This yields", - "type": "text" - }, - { - "block_id": "p666-b38", - "global_id": 19420, - "bbox": [ - 251.55, - 413.17, - 276.37, - 430.12 - ], - "text": "c = 1", - "type": "text" - }, - { - "block_id": "p666-b39", - "global_id": 19421, - "bbox": [ - 269.04, - 427.13, - 278.22, - 437.9 - ], - "text": "Ey", - "type": "text" - }, - { - "block_id": "p666-b40", - "global_id": 19422, - "bbox": [ - 281.02, - 406.18, - 295.77, - 419.79 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b41", - "global_id": 19423, - "bbox": [ - 286.28, - 431.28, - 291.2, - 439.59 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p666-b42", - "global_id": 19424, - "bbox": [ - 297.87, - 418.02, - 340.5, - 430.02 - ], - "text": "x(t)y∗(t)dt", - "type": "text" - }, - { - "block_id": "p666-b43", - "global_id": 19425, - "bbox": [ - 101.84, - 448.57, - 490.4, - 471.56 - ], - "text": "In light of this result, we need to redefine orthogonality for the complex case as follows: two\ncomplex functions x1(t) and x2(t) are orthogonal over an interval (t1 < t < t2) if", - "type": "text" - }, - { - "block_id": "p666-b44", - "global_id": 19426, - "bbox": [ - 191.0, - 474.5, - 205.74, - 488.11 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b45", - "global_id": 19427, - "bbox": [ - 196.25, - 499.59, - 201.18, - 507.9 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p666-b46", - "global_id": 19428, - "bbox": [ - 207.85, - 486.33, - 234.73, - 499.21 - ], - "text": "x1(t)x∗", - "type": "text" - }, - { - "block_id": "p666-b47", - "global_id": 19429, - "bbox": [ - 231.07, - 488.06, - 299.71, - 500.39 - ], - "text": "2(t)dt = 0\nor", - "type": "text" - }, - { - "block_id": "p666-b48", - "global_id": 19430, - "bbox": [ - 320.74, - 474.5, - 335.49, - 488.11 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p666-b49", - "global_id": 19431, - "bbox": [ - 326.01, - 499.59, - 330.93, - 507.9 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p666-b50", - "global_id": 19432, - "bbox": [ - 337.6, - 486.33, - 345.68, - 498.34 - ], - "text": "x∗", - "type": "text" - }, - { - "block_id": "p666-b51", - "global_id": 19433, - "bbox": [ - 342.02, - 488.06, - 490.38, - 500.39 - ], - "text": "1(t)x2(t)dt = 0\n(6.37)", - "type": "text" - }, - { - "block_id": "p666-b52", - "global_id": 19434, - "bbox": [ - 101.84, - 516.89, - 490.36, - 538.8 - ], - "text": "Either equality suffices. This is a general definition of orthogonality, which reduces to Eq. (6.34)\nwhen the functions are real.", - "type": "text" - }, - { - "block_id": "p666-b53", - "global_id": 19435, - "bbox": [ - 107.82, - 569.75, - 484.42, - 581.71 - ], - "text": "DRILL 6.8\nComplex Exponential Approximation of a Square Wave", - "type": "text" - }, - { - "block_id": "p666-b54", - "global_id": 19436, - "bbox": [ - 107.82, - 590.41, - 484.41, - 624.7 - ], - "text": "Show that over an interval (0 < t < 2π), the “best” approximation of the square signal x(t)\nin Fig. 6.23 in terms of the signal ejt is given by (2/jπ)ejt. Verify that the error signal\ne(t) = x(t) −(2/jπ)ejt is orthogonal to the signal ejt.", - "type": "text" - } - ] - }, - { - "page_num": 667, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p667-b0", - "global_id": 19437, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n647", - "type": "text" - }, - { - "block_id": "p667-b1", - "global_id": 19438, - "bbox": [ - 127.89, - 86.19, - 389.87, - 98.32 - ], - "text": "ENERGY OF THE SUM OF ORTHOGONAL SIGNALS", - "type": "text" - }, - { - "block_id": "p667-b2", - "global_id": 19439, - "bbox": [ - 127.59, - 102.35, - 516.15, - 136.22 - ], - "text": "We know that the square of the length of a sum of two orthogonal vectors is equal to the sum of the\nsquares of the lengths of the two vectors. Thus, if vectors x and y are orthogonal, and if z = x+y,\nthen", - "type": "text" - }, - { - "block_id": "p667-b3", - "global_id": 19440, - "bbox": [ - 288.77, - 135.33, - 354.45, - 149.74 - ], - "text": "|z|2 = |x|2 + |y|2", - "type": "text" - }, - { - "block_id": "p667-b4", - "global_id": 19441, - "bbox": [ - 127.59, - 159.89, - 516.13, - 194.53 - ], - "text": "We have a similar result for signals. The energy of the sum of two orthogonal signals is equal to\nthe sum of the energies of the two signals. Thus, if signals x(t) and y(t) are orthogonal over an\ninterval (t1, t2), and if z(t) = x(t) + y(t), then", - "type": "text" - }, - { - "block_id": "p667-b5", - "global_id": 19442, - "bbox": [ - 296.15, - 206.4, - 347.07, - 217.85 - ], - "text": "Ez = Ex + Ey", - "type": "text" - }, - { - "block_id": "p667-b6", - "global_id": 19443, - "bbox": [ - 127.59, - 229.83, - 516.14, - 251.75 - ], - "text": "We now prove this result for complex signals, of which real signals are a special case. From\nEq. (6.36), it follows that", - "type": "text" - }, - { - "block_id": "p667-b7", - "global_id": 19444, - "bbox": [ - 151.52, - 256.62, - 166.28, - 270.23 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p667-b8", - "global_id": 19445, - "bbox": [ - 156.79, - 281.72, - 161.71, - 290.03 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p667-b9", - "global_id": 19446, - "bbox": [ - 168.38, - 266.38, - 237.42, - 280.46 - ], - "text": "|x(t) + y(t)|2 dt =", - "type": "text" - }, - { - "block_id": "p667-b10", - "global_id": 19447, - "bbox": [ - 239.47, - 256.62, - 254.22, - 270.23 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p667-b11", - "global_id": 19448, - "bbox": [ - 244.74, - 281.72, - 249.66, - 290.03 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p667-b12", - "global_id": 19449, - "bbox": [ - 256.33, - 266.38, - 299.21, - 280.46 - ], - "text": "|x(t)|2 dt +", - "type": "text" - }, - { - "block_id": "p667-b13", - "global_id": 19450, - "bbox": [ - 300.75, - 256.62, - 315.5, - 270.23 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p667-b14", - "global_id": 19451, - "bbox": [ - 306.02, - 281.72, - 310.94, - 290.03 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p667-b15", - "global_id": 19452, - "bbox": [ - 317.61, - 266.38, - 360.45, - 280.46 - ], - "text": "|y(t)|2 dt +", - "type": "text" - }, - { - "block_id": "p667-b16", - "global_id": 19453, - "bbox": [ - 362.0, - 256.62, - 376.75, - 270.23 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p667-b17", - "global_id": 19454, - "bbox": [ - 367.26, - 281.72, - 372.19, - 290.03 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p667-b18", - "global_id": 19455, - "bbox": [ - 378.86, - 268.46, - 430.99, - 280.46 - ], - "text": "x(t)y∗(t)dt +", - "type": "text" - }, - { - "block_id": "p667-b19", - "global_id": 19456, - "bbox": [ - 432.53, - 256.62, - 447.28, - 270.23 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p667-b20", - "global_id": 19457, - "bbox": [ - 437.8, - 281.72, - 442.72, - 290.03 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p667-b21", - "global_id": 19458, - "bbox": [ - 449.39, - 268.46, - 492.02, - 280.46 - ], - "text": "x∗(t)y(t)dt", - "type": "text" - }, - { - "block_id": "p667-b22", - "global_id": 19459, - "bbox": [ - 229.65, - 299.19, - 237.42, - 309.15 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p667-b23", - "global_id": 19460, - "bbox": [ - 239.47, - 285.62, - 254.22, - 299.23 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p667-b24", - "global_id": 19461, - "bbox": [ - 244.74, - 310.73, - 249.66, - 319.03 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p667-b25", - "global_id": 19462, - "bbox": [ - 256.33, - 295.39, - 299.21, - 309.46 - ], - "text": "|x(t)|2 dt +", - "type": "text" - }, - { - "block_id": "p667-b26", - "global_id": 19463, - "bbox": [ - 300.75, - 285.62, - 315.5, - 299.23 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p667-b27", - "global_id": 19464, - "bbox": [ - 306.02, - 310.73, - 310.94, - 319.03 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p667-b28", - "global_id": 19465, - "bbox": [ - 317.61, - 295.39, - 350.95, - 309.46 - ], - "text": "|y(t)|2 dt", - "type": "text" - }, - { - "block_id": "p667-b29", - "global_id": 19466, - "bbox": [ - 127.59, - 328.88, - 516.15, - 362.75 - ], - "text": "The last result follows from the fact that because of orthogonality, the two integrals of the products\nx(t)y∗(t) and x∗(t)y(t) are zero [see Eq. (6.37)]. This result can be extended to the sum of any\nnumber of mutually orthogonal signals.", - "type": "text" - }, - { - "block_id": "p667-b30", - "global_id": 19467, - "bbox": [ - 127.59, - 388.6, - 436.89, - 400.55 - ], - "text": "6.5-4 Signal Representation by an Orthogonal Signal Set", - "type": "text" - }, - { - "block_id": "p667-b31", - "global_id": 19468, - "bbox": [ - 127.59, - 406.68, - 516.17, - 464.47 - ], - "text": "In this section we show a way of representing a signal as a sum of orthogonal signals. Here again\nwe can benefit from the insight gained from a similar problem in vectors. We know that a vector\ncan be represented as a sum of orthogonal vectors, which form the coordinate system of a vector\nspace. The problem in signals is analogous, and the results for signals are parallel to those for\nvectors. So, let us review the case of vector representation.", - "type": "text" - }, - { - "block_id": "p667-b32", - "global_id": 19469, - "bbox": [ - 127.59, - 479.75, - 516.15, - 531.27 - ], - "text": "ORTHOGONAL VECTOR SPACE\nLet us investigate a three-dimensional Cartesian vector space described by three mutually\northogonal vectors x1, x2, and x3, as illustrated in Fig. 6.24. First, we shall seek to approximate a\nthree-dimensional vector x in terms of two mutually orthogonal vectors x1 and x2:", - "type": "text" - }, - { - "block_id": "p667-b33", - "global_id": 19470, - "bbox": [ - 290.63, - 542.41, - 352.6, - 553.87 - ], - "text": "x ≃c1x1 + c2x2", - "type": "text" - }, - { - "block_id": "p667-b34", - "global_id": 19471, - "bbox": [ - 127.59, - 565.77, - 268.71, - 575.81 - ], - "text": "The error e in this approximation is", - "type": "text" - }, - { - "block_id": "p667-b35", - "global_id": 19472, - "bbox": [ - 279.26, - 588.45, - 364.45, - 599.9 - ], - "text": "e = x −(c1x1 + c2x2)", - "type": "text" - }, - { - "block_id": "p667-b36", - "global_id": 19473, - "bbox": [ - 127.59, - 611.88, - 135.89, - 621.85 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p667-b37", - "global_id": 19474, - "bbox": [ - 282.98, - 625.07, - 360.73, - 636.53 - ], - "text": "x = c1x1 + c2x2 + e", - "type": "text" - } - ] - }, - { - "page_num": 668, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p668-b0", - "global_id": 19475, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "648\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p668-b1", - "global_id": 19476, - "bbox": [ - 214.37, - 122.71, - 218.37, - 130.71 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p668-b2", - "global_id": 19477, - "bbox": [ - 135.68, - 198.35, - 149.23, - 207.96 - ], - "text": "c1x1", - "type": "text" - }, - { - "block_id": "p668-b3", - "global_id": 19478, - "bbox": [ - 178.35, - 218.37, - 216.12, - 228.19 - ], - "text": "c1x1 c2x2", - "type": "text" - }, - { - "block_id": "p668-b4", - "global_id": 19479, - "bbox": [ - 251.77, - 168.6, - 265.32, - 178.2 - ], - "text": "c2x2", - "type": "text" - }, - { - "block_id": "p668-b5", - "global_id": 19480, - "bbox": [ - 161.54, - 97.04, - 175.09, - 106.65 - ], - "text": "c3x3", - "type": "text" - }, - { - "block_id": "p668-b6", - "global_id": 19481, - "bbox": [ - 229.6, - 156.11, - 233.15, - 164.11 - ], - "text": "e", - "type": "text" - }, - { - "block_id": "p668-b7", - "global_id": 19482, - "bbox": [ - 322.8, - 213.08, - 486.42, - 234.99 - ], - "text": "Figure 6.24 Representation of a vector in\nthree-dimensional space.", - "type": "text" - }, - { - "block_id": "p668-b8", - "global_id": 19483, - "bbox": [ - 101.84, - 262.43, - 490.39, - 309.83 - ], - "text": "As in the earlier geometrical argument, we see from Fig. 6.24 that the length of e is minimum\nwhen e is perpendicular to the x1–x2 plane, and c1x1 and c2x2 are the projections (components) of\nx on x1 and x2, respectively. Therefore, the constants c1 and c2 are given by Eq. (6.32). Observe\nthat the error vector is orthogonal to both the vectors x1 and x2.", - "type": "text" - }, - { - "block_id": "p668-b9", - "global_id": 19484, - "bbox": [ - 101.85, - 310.25, - 490.37, - 333.03 - ], - "text": "Now, let us determine the “best” approximation to x in terms of all three mutually orthogonal\nvectors x1, x2, and x3:", - "type": "text" - }, - { - "block_id": "p668-b10", - "global_id": 19485, - "bbox": [ - 250.76, - 340.47, - 490.38, - 351.93 - ], - "text": "x ≃c1x1 + c2x2 + c3x3\n(6.38)", - "type": "text" - }, - { - "block_id": "p668-b11", - "global_id": 19486, - "bbox": [ - 101.85, - 364.15, - 490.35, - 386.16 - ], - "text": "Figure 6.24 shows that a unique choice of c1, c2, and c3 exists, for which Eq. (6.38) is no longer\nan approximation but an equality", - "type": "text" - }, - { - "block_id": "p668-b12", - "global_id": 19487, - "bbox": [ - 250.76, - 402.14, - 340.98, - 413.6 - ], - "text": "x = c1x1 + c2x2 + c3x3", - "type": "text" - }, - { - "block_id": "p668-b13", - "global_id": 19488, - "bbox": [ - 101.84, - 428.8, - 490.39, - 450.82 - ], - "text": "In this case, c1x1,c2x2, and c3x3 are the projections (components) of x on x1,x2, and x3,\nrespectively; that is,", - "type": "text" - }, - { - "block_id": "p668-b14", - "global_id": 19489, - "bbox": [ - 221.49, - 457.02, - 260.4, - 475.46 - ], - "text": "ci = x · xi", - "type": "text" - }, - { - "block_id": "p668-b15", - "global_id": 19490, - "bbox": [ - 241.41, - 471.08, - 261.62, - 482.54 - ], - "text": "xi · xi", - "type": "text" - }, - { - "block_id": "p668-b16", - "global_id": 19491, - "bbox": [ - 265.36, - 457.43, - 490.38, - 482.9 - ], - "text": "=\n1\n|xi|2 x · xi\ni = 1,2,3\n(6.39)", - "type": "text" - }, - { - "block_id": "p668-b17", - "global_id": 19492, - "bbox": [ - 101.84, - 493.23, - 490.4, - 598.92 - ], - "text": "Note that the error in the approximation is zero when x is approximated in terms of three mutually\northogonal vectors: x1, x2, and x3. The reason is that x is a three-dimensional vector, and the\nvectors x1, x2, and x3 represent a complete set of orthogonal vectors in three-dimensional space.\nCompleteness here means that it is impossible to find another vector x4 in this space, which is\northogonal to all three vectors, x1,x2, and x3. Any vector in this space can then be represented\n(with zero error) in terms of these three vectors. Such vectors are known as basis vectors. If a set\nof vectors {xi} is not complete, the error in the approximation will generally not be zero. Thus, in\nthe three-dimensional case discussed earlier, it is generally not possible to represent a vector x in\nterms of only two basis vectors without an error.", - "type": "text" - }, - { - "block_id": "p668-b18", - "global_id": 19493, - "bbox": [ - 101.85, - 600.91, - 490.4, - 634.79 - ], - "text": "The choice of basis vectors is not unique. In fact, a set of basis vectors corresponds to a\nparticular choice of coordinate system. Thus, a three-dimensional vector x may be represented in\nmany different ways, depending on the coordinate system used.", - "type": "text" - } - ] - }, - { - "page_num": 669, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p669-b0", - "global_id": 19494, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n649", - "type": "text" - }, - { - "block_id": "p669-b1", - "global_id": 19495, - "bbox": [ - 127.89, - 86.19, - 287.79, - 98.32 - ], - "text": "ORTHOGONAL SIGNAL SPACE", - "type": "text" - }, - { - "block_id": "p669-b2", - "global_id": 19496, - "bbox": [ - 127.59, - 102.35, - 516.12, - 136.99 - ], - "text": "We start with real signals and then extend the discussion to complex signals. We proceed with our\nsignal approximation problem, using clues and insights developed for vector approximation. As\nbefore, we define orthogonality of a real signal set x1(t), x2(t), . . ., xN(t) over interval (t1,t2) as", - "type": "text" - }, - { - "block_id": "p669-b3", - "global_id": 19497, - "bbox": [ - 253.17, - 147.0, - 267.92, - 160.61 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p669-b4", - "global_id": 19498, - "bbox": [ - 258.43, - 172.09, - 263.36, - 180.4 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p669-b5", - "global_id": 19499, - "bbox": [ - 270.03, - 160.55, - 328.01, - 171.63 - ], - "text": "xm(t)xn(t)dt =", - "type": "text" - }, - { - "block_id": "p669-b6", - "global_id": 19500, - "bbox": [ - 330.06, - 146.57, - 516.13, - 177.51 - ], - "text": "0\nm̸ = n\nEn\nm = n\n(6.40)", - "type": "text" - }, - { - "block_id": "p669-b7", - "global_id": 19501, - "bbox": [ - 127.59, - 194.97, - 516.13, - 218.79 - ], - "text": "If the energies En = 1 for all n, then the set is normalized and is called an orthonormal set. An\northogonal set can always be normalized by dividing xn(t) by √En for all n.", - "type": "text" - }, - { - "block_id": "p669-b8", - "global_id": 19502, - "bbox": [ - 127.59, - 218.87, - 516.13, - 241.98 - ], - "text": "Now, consider approximating a signal x(t) over the interval (t1, t2) by a set of N real, mutually\northogonal signals x1(t), x2(t),. . .,xN(t) as", - "type": "text" - }, - { - "block_id": "p669-b9", - "global_id": 19503, - "bbox": [ - 216.15, - 268.5, - 383.12, - 279.65 - ], - "text": "x(t) ≃c1x1(t) + c2x2(t) + · · · + cNxN(t) ≃", - "type": "text" - }, - { - "block_id": "p669-b10", - "global_id": 19504, - "bbox": [ - 385.16, - 258.54, - 399.26, - 269.0 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p669-b11", - "global_id": 19505, - "bbox": [ - 386.0, - 282.89, - 398.41, - 290.16 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p669-b12", - "global_id": 19506, - "bbox": [ - 400.36, - 268.5, - 516.13, - 279.58 - ], - "text": "cnxn(t)\n(6.41)", - "type": "text" - }, - { - "block_id": "p669-b13", - "global_id": 19507, - "bbox": [ - 127.59, - 305.73, - 329.99, - 316.11 - ], - "text": "In the approximation of Eq. (6.41), the error e(t) is", - "type": "text" - }, - { - "block_id": "p669-b14", - "global_id": 19508, - "bbox": [ - 274.47, - 343.4, - 325.29, - 353.68 - ], - "text": "e(t) = x(t) −", - "type": "text" - }, - { - "block_id": "p669-b15", - "global_id": 19509, - "bbox": [ - 326.84, - 333.45, - 340.94, - 343.9 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p669-b16", - "global_id": 19510, - "bbox": [ - 327.69, - 357.79, - 340.1, - 365.06 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p669-b17", - "global_id": 19511, - "bbox": [ - 342.05, - 343.4, - 369.25, - 354.48 - ], - "text": "cnxn(t)", - "type": "text" - }, - { - "block_id": "p669-b18", - "global_id": 19512, - "bbox": [ - 127.59, - 380.89, - 259.84, - 391.65 - ], - "text": "and Ee, the error signal energy, is", - "type": "text" - }, - { - "block_id": "p669-b19", - "global_id": 19513, - "bbox": [ - 227.08, - 420.3, - 246.59, - 431.76 - ], - "text": "Ee =", - "type": "text" - }, - { - "block_id": "p669-b20", - "global_id": 19514, - "bbox": [ - 248.63, - 406.74, - 263.38, - 420.35 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p669-b21", - "global_id": 19515, - "bbox": [ - 253.89, - 431.84, - 258.81, - 440.15 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p669-b22", - "global_id": 19516, - "bbox": [ - 265.48, - 418.87, - 303.13, - 430.58 - ], - "text": "e2(t)dt =", - "type": "text" - }, - { - "block_id": "p669-b23", - "global_id": 19517, - "bbox": [ - 305.18, - 406.74, - 319.92, - 420.35 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p669-b24", - "global_id": 19518, - "bbox": [ - 310.44, - 431.84, - 315.36, - 440.15 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p669-b26", - "global_id": 19519, - "bbox": [ - 328.22, - 420.3, - 352.37, - 430.58 - ], - "text": "x(t) −", - "type": "text" - }, - { - "block_id": "p669-b27", - "global_id": 19520, - "bbox": [ - 353.92, - 410.35, - 368.02, - 420.8 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p669-b28", - "global_id": 19521, - "bbox": [ - 354.76, - 434.69, - 367.17, - 441.95 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p669-b29", - "global_id": 19522, - "bbox": [ - 369.13, - 420.3, - 396.33, - 431.38 - ], - "text": "cnxn(t)", - "type": "text" - }, - { - "block_id": "p669-b30", - "global_id": 19523, - "bbox": [ - 396.33, - 403.33, - 405.99, - 414.33 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p669-b31", - "global_id": 19524, - "bbox": [ - 408.71, - 420.61, - 516.13, - 430.68 - ], - "text": "dt\n(6.42)", - "type": "text" - }, - { - "block_id": "p669-b32", - "global_id": 19525, - "bbox": [ - 127.59, - 457.78, - 516.13, - 480.88 - ], - "text": "According to our criterion for best approximation, we select the values of ci that minimize Ee.\nHence, the necessary condition is ∂Ee/dci = 0 for i = 1,2,. . .,N, that is,", - "type": "text" - }, - { - "block_id": "p669-b33", - "global_id": 19526, - "bbox": [ - 251.04, - 502.17, - 262.76, - 527.31 - ], - "text": "∂\n∂ci", - "type": "text" - }, - { - "block_id": "p669-b34", - "global_id": 19527, - "bbox": [ - 265.56, - 495.59, - 280.31, - 509.2 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p669-b35", - "global_id": 19528, - "bbox": [ - 270.83, - 520.7, - 275.75, - 529.0 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p669-b37", - "global_id": 19529, - "bbox": [ - 288.6, - 509.16, - 312.75, - 519.43 - ], - "text": "x(t) −", - "type": "text" - }, - { - "block_id": "p669-b38", - "global_id": 19530, - "bbox": [ - 314.31, - 499.2, - 328.4, - 509.65 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p669-b39", - "global_id": 19531, - "bbox": [ - 315.15, - 523.55, - 327.56, - 530.81 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p669-b40", - "global_id": 19532, - "bbox": [ - 329.52, - 509.16, - 356.71, - 520.24 - ], - "text": "cnxn(t)", - "type": "text" - }, - { - "block_id": "p669-b41", - "global_id": 19533, - "bbox": [ - 356.71, - 492.17, - 366.38, - 503.18 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p669-b42", - "global_id": 19534, - "bbox": [ - 369.1, - 509.16, - 393.87, - 519.53 - ], - "text": "dt = 0", - "type": "text" - }, - { - "block_id": "p669-b43", - "global_id": 19535, - "bbox": [ - 127.59, - 546.74, - 516.13, - 568.66 - ], - "text": "When we expand the integrand, we find that all the cross-multiplication terms arising from the\northogonal signals are zero by virtue of orthogonality: that is, all terms of the form", - "type": "text" - }, - { - "block_id": "p669-b44", - "global_id": 19536, - "bbox": [ - 460.37, - 550.25, - 464.93, - 560.21 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p669-b45", - "global_id": 19537, - "bbox": [ - 127.59, - 558.28, - 516.14, - 592.57 - ], - "text": "xm(t)xn(t)dt\nwith m̸ = n vanish. Similarly, the derivative with respect to ci of all terms that do not contain ci is\nzero. For each i, this leaves only two nonzero terms:", - "type": "text" - }, - { - "block_id": "p669-b46", - "global_id": 19538, - "bbox": [ - 242.59, - 609.15, - 254.31, - 634.28 - ], - "text": "∂\n∂ci", - "type": "text" - }, - { - "block_id": "p669-b47", - "global_id": 19539, - "bbox": [ - 257.11, - 602.58, - 271.85, - 616.19 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p669-b48", - "global_id": 19540, - "bbox": [ - 262.37, - 627.67, - 267.29, - 635.98 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p669-b50", - "global_id": 19541, - "bbox": [ - 277.89, - 616.13, - 346.8, - 627.21 - ], - "text": "−2cix(t)xi(t) + ci", - "type": "text" - }, - { - "block_id": "p669-b51", - "global_id": 19542, - "bbox": [ - 347.31, - 608.12, - 376.45, - 627.21 - ], - "text": "2xi\n2(t)", - "type": "text" - }, - { - "block_id": "p669-b52", - "global_id": 19543, - "bbox": [ - 377.55, - 616.13, - 402.33, - 626.51 - ], - "text": "dt = 0", - "type": "text" - } - ] - }, - { - "page_num": 670, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p670-b0", - "global_id": 19544, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "650\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p670-b1", - "global_id": 19545, - "bbox": [ - 101.84, - 83.6, - 110.14, - 93.56 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p670-b2", - "global_id": 19546, - "bbox": [ - 177.88, - 101.7, - 190.62, - 112.08 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p670-b3", - "global_id": 19547, - "bbox": [ - 191.73, - 88.14, - 206.49, - 101.75 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b4", - "global_id": 19548, - "bbox": [ - 197.0, - 113.24, - 201.92, - 121.55 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b5", - "global_id": 19549, - "bbox": [ - 208.59, - 101.7, - 271.9, - 112.78 - ], - "text": "x(t)xi(t)dt + 2ci", - "type": "text" - }, - { - "block_id": "p670-b6", - "global_id": 19550, - "bbox": [ - 273.52, - 88.14, - 288.27, - 101.75 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b7", - "global_id": 19551, - "bbox": [ - 278.78, - 113.24, - 283.71, - 121.55 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b8", - "global_id": 19552, - "bbox": [ - 290.38, - 102.01, - 296.73, - 112.78 - ], - "text": "xi", - "type": "text" - }, - { - "block_id": "p670-b9", - "global_id": 19553, - "bbox": [ - 297.23, - 100.27, - 413.76, - 112.08 - ], - "text": "2(t)dt = 0\ni = 1,2,. . .,N", - "type": "text" - }, - { - "block_id": "p670-b10", - "global_id": 19554, - "bbox": [ - 101.84, - 129.18, - 143.59, - 139.14 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p670-b11", - "global_id": 19555, - "bbox": [ - 177.21, - 163.14, - 193.89, - 174.6 - ], - "text": "ci =", - "type": "text" - }, - { - "block_id": "p670-b12", - "global_id": 19556, - "bbox": [ - 197.14, - 135.06, - 211.88, - 148.66 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b13", - "global_id": 19557, - "bbox": [ - 202.39, - 160.16, - 207.32, - 168.46 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b14", - "global_id": 19558, - "bbox": [ - 202.56, - 148.62, - 254.92, - 176.9 - ], - "text": "x(t)xi(t)dt\n# t2", - "type": "text" - }, - { - "block_id": "p670-b15", - "global_id": 19559, - "bbox": [ - 207.82, - 188.38, - 212.74, - 196.69 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b16", - "global_id": 19560, - "bbox": [ - 219.41, - 177.16, - 225.77, - 187.93 - ], - "text": "xi", - "type": "text" - }, - { - "block_id": "p670-b17", - "global_id": 19561, - "bbox": [ - 226.28, - 156.57, - 276.11, - 187.13 - ], - "text": "2(t)dt\n= 1", - "type": "text" - }, - { - "block_id": "p670-b18", - "global_id": 19562, - "bbox": [ - 269.36, - 170.53, - 277.38, - 181.3 - ], - "text": "Ei", - "type": "text" - }, - { - "block_id": "p670-b19", - "global_id": 19563, - "bbox": [ - 280.19, - 149.58, - 294.93, - 163.19 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b20", - "global_id": 19564, - "bbox": [ - 285.44, - 174.68, - 290.37, - 182.99 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b21", - "global_id": 19565, - "bbox": [ - 297.05, - 163.14, - 490.38, - 174.22 - ], - "text": "x(t)xi(t)dt\ni = 1,2,. . .,N\n(6.43)", - "type": "text" - }, - { - "block_id": "p670-b22", - "global_id": 19566, - "bbox": [ - 101.85, - 204.33, - 490.37, - 226.25 - ], - "text": "A comparison of Eq. (6.43) with Eq. (6.39) forcefully brings out the analogy of signals with\nvectors.", - "type": "text" - }, - { - "block_id": "p670-b23", - "global_id": 19567, - "bbox": [ - 101.84, - 238.0, - 490.39, - 331.74 - ], - "text": "Finality Property. Equation (6.43) shows one interesting property of the coefficients of c1, c2,\n. . ., cN: the optimum value of any coefficient in Eq. (6.41) is independent of the number of terms\nused in the approximation. For example, if we used only one term (N = 1) or two terms (N = 2)\nor any number of terms, the optimum value of the coefficient c1 would be the same [as given by\nEq. (6.43)]. The advantage of this approximation of a signal x(t) by a set of mutually orthogonal\nsignals is that we can continue to add terms to the approximation without disturbing the previous\nterms. This property of finality of the values of the coefficients is very important from a practical\npoint of view.†", - "type": "text" - }, - { - "block_id": "p670-b24", - "global_id": 19568, - "bbox": [ - 101.84, - 347.42, - 490.39, - 386.98 - ], - "text": "ENERGY OF THE ERROR SIGNAL\nWhen the coefficients ci in Eq. (6.41) are chosen according to Eq. (6.43), the error signal energy\nis minimized. This minimum value of Ee is given by Eq. (6.42):", - "type": "text" - }, - { - "block_id": "p670-b25", - "global_id": 19569, - "bbox": [ - 172.24, - 410.48, - 191.74, - 421.94 - ], - "text": "Ee =", - "type": "text" - }, - { - "block_id": "p670-b26", - "global_id": 19570, - "bbox": [ - 193.79, - 396.92, - 208.54, - 410.54 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b27", - "global_id": 19571, - "bbox": [ - 199.05, - 422.02, - 203.98, - 430.33 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b29", - "global_id": 19572, - "bbox": [ - 216.83, - 410.48, - 240.98, - 420.76 - ], - "text": "x(t) −", - "type": "text" - }, - { - "block_id": "p670-b30", - "global_id": 19573, - "bbox": [ - 242.53, - 400.53, - 256.63, - 410.98 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p670-b31", - "global_id": 19574, - "bbox": [ - 243.38, - 424.87, - 255.78, - 432.13 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p670-b32", - "global_id": 19575, - "bbox": [ - 257.74, - 410.48, - 284.94, - 421.56 - ], - "text": "cnxn(t)", - "type": "text" - }, - { - "block_id": "p670-b33", - "global_id": 19576, - "bbox": [ - 284.94, - 393.51, - 294.61, - 404.51 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p670-b34", - "global_id": 19577, - "bbox": [ - 297.32, - 410.8, - 305.07, - 420.76 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p670-b35", - "global_id": 19578, - "bbox": [ - 183.97, - 445.98, - 191.74, - 455.94 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p670-b36", - "global_id": 19579, - "bbox": [ - 193.79, - 432.41, - 208.54, - 446.02 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b37", - "global_id": 19580, - "bbox": [ - 199.05, - 457.52, - 203.98, - 465.82 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b38", - "global_id": 19581, - "bbox": [ - 210.65, - 444.55, - 247.82, - 456.25 - ], - "text": "x2(t)dt +", - "type": "text" - }, - { - "block_id": "p670-b39", - "global_id": 19582, - "bbox": [ - 249.37, - 436.02, - 263.47, - 446.47 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p670-b40", - "global_id": 19583, - "bbox": [ - 250.21, - 460.37, - 262.62, - 467.63 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p670-b41", - "global_id": 19584, - "bbox": [ - 264.57, - 444.55, - 272.48, - 456.25 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p670-b42", - "global_id": 19585, - "bbox": [ - 268.99, - 451.27, - 272.48, - 458.24 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p670-b43", - "global_id": 19586, - "bbox": [ - 274.09, - 432.41, - 288.84, - 446.02 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b44", - "global_id": 19587, - "bbox": [ - 279.35, - 457.52, - 284.27, - 465.82 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b45", - "global_id": 19588, - "bbox": [ - 290.94, - 444.55, - 298.88, - 456.25 - ], - "text": "x2", - "type": "text" - }, - { - "block_id": "p670-b46", - "global_id": 19589, - "bbox": [ - 295.36, - 445.98, - 334.64, - 458.24 - ], - "text": "n(t)dt −2", - "type": "text" - }, - { - "block_id": "p670-b47", - "global_id": 19590, - "bbox": [ - 335.74, - 436.02, - 349.84, - 446.47 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p670-b48", - "global_id": 19591, - "bbox": [ - 336.6, - 460.37, - 349.01, - 467.63 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p670-b49", - "global_id": 19592, - "bbox": [ - 350.96, - 446.29, - 358.87, - 457.06 - ], - "text": "cn", - "type": "text" - }, - { - "block_id": "p670-b50", - "global_id": 19593, - "bbox": [ - 360.48, - 432.41, - 375.23, - 446.02 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p670-b51", - "global_id": 19594, - "bbox": [ - 365.74, - 457.52, - 370.67, - 465.82 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p670-b52", - "global_id": 19595, - "bbox": [ - 377.33, - 445.98, - 419.81, - 457.06 - ], - "text": "x(t)xn(t)dt", - "type": "text" - }, - { - "block_id": "p670-b53", - "global_id": 19596, - "bbox": [ - 101.84, - 487.36, - 490.38, - 522.47 - ], - "text": "† Contrast this situation with a polynomial approximation of x(t). Suppose we wish to find a two-point\napproximation of x(t) by a polynomial in t; that is, the polynomial is to be equal to x(t) at two points t1\nand t2. This can be done by choosing a first-order polynomial a0 + a1t with", - "type": "text" - }, - { - "block_id": "p670-b54", - "global_id": 19597, - "bbox": [ - 211.65, - 531.91, - 380.09, - 542.22 - ], - "text": "x(t1) = a0 + a1t1\nand\nx(t2) = a0 + a1t2", - "type": "text" - }, - { - "block_id": "p670-b55", - "global_id": 19598, - "bbox": [ - 101.84, - 551.96, - 490.39, - 572.95 - ], - "text": "Solution of these equations yields the desired values of a0 and a1. For a three-point approximation, we must\nchoose the polynomial a0 + a1t + a2t2 with", - "type": "text" - }, - { - "block_id": "p670-b56", - "global_id": 19599, - "bbox": [ - 220.53, - 580.96, - 371.7, - 592.7 - ], - "text": "x(ti) = a0 + a1ti + a2ti2\ni = 1,2, and 3", - "type": "text" - }, - { - "block_id": "p670-b57", - "global_id": 19600, - "bbox": [ - 101.84, - 602.53, - 490.38, - 633.41 - ], - "text": "The approximation improves with a larger number of points (higher-order polynomial), but the coefficients\na0, a1, a2, . . . do not have the finality property. Every time we increase the number of terms in the polynomial,\nwe need to recalculate the coefficients.", - "type": "text" - } - ] - }, - { - "page_num": 671, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p671-b0", - "global_id": 19601, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n651", - "type": "text" - }, - { - "block_id": "p671-b1", - "global_id": 19602, - "bbox": [ - 127.59, - 85.82, - 366.7, - 95.78 - ], - "text": "Substitution of Eqs. (6.40) and (6.43) in this equation yields", - "type": "text" - }, - { - "block_id": "p671-b2", - "global_id": 19603, - "bbox": [ - 190.57, - 121.76, - 210.06, - 133.22 - ], - "text": "Ee =", - "type": "text" - }, - { - "block_id": "p671-b3", - "global_id": 19604, - "bbox": [ - 212.11, - 108.2, - 226.86, - 121.81 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p671-b4", - "global_id": 19605, - "bbox": [ - 217.38, - 133.3, - 222.3, - 141.61 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p671-b5", - "global_id": 19606, - "bbox": [ - 228.97, - 120.33, - 266.14, - 132.04 - ], - "text": "x2(t)dt +", - "type": "text" - }, - { - "block_id": "p671-b6", - "global_id": 19607, - "bbox": [ - 267.7, - 111.81, - 281.79, - 122.26 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p671-b7", - "global_id": 19608, - "bbox": [ - 268.53, - 136.15, - 280.94, - 143.41 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p671-b8", - "global_id": 19609, - "bbox": [ - 282.9, - 120.33, - 290.8, - 132.04 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p671-b9", - "global_id": 19610, - "bbox": [ - 287.32, - 121.76, - 317.23, - 134.03 - ], - "text": "nEn −2", - "type": "text" - }, - { - "block_id": "p671-b10", - "global_id": 19611, - "bbox": [ - 318.33, - 111.81, - 332.43, - 122.26 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p671-b11", - "global_id": 19612, - "bbox": [ - 319.18, - 136.15, - 331.58, - 143.41 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p671-b12", - "global_id": 19613, - "bbox": [ - 333.54, - 120.33, - 341.45, - 132.04 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p671-b13", - "global_id": 19614, - "bbox": [ - 337.96, - 121.76, - 361.84, - 134.03 - ], - "text": "nEn =", - "type": "text" - }, - { - "block_id": "p671-b14", - "global_id": 19615, - "bbox": [ - 363.89, - 108.2, - 378.64, - 121.81 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p671-b15", - "global_id": 19616, - "bbox": [ - 369.15, - 133.3, - 374.08, - 141.61 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p671-b16", - "global_id": 19617, - "bbox": [ - 380.74, - 120.33, - 417.92, - 132.04 - ], - "text": "x2(t)dt −", - "type": "text" - }, - { - "block_id": "p671-b17", - "global_id": 19618, - "bbox": [ - 419.47, - 111.81, - 433.57, - 122.26 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p671-b18", - "global_id": 19619, - "bbox": [ - 420.31, - 136.15, - 432.72, - 143.41 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p671-b19", - "global_id": 19620, - "bbox": [ - 434.67, - 120.33, - 442.58, - 132.04 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p671-b20", - "global_id": 19621, - "bbox": [ - 439.09, - 122.08, - 516.13, - 134.03 - ], - "text": "nEn\n(6.44)", - "type": "text" - }, - { - "block_id": "p671-b21", - "global_id": 19622, - "bbox": [ - 127.59, - 158.26, - 259.37, - 169.62 - ], - "text": "Observe that because the term c2", - "type": "text" - }, - { - "block_id": "p671-b22", - "global_id": 19623, - "bbox": [ - 127.59, - 159.56, - 516.13, - 205.49 - ], - "text": "kEk is nonnegative, the error energy Ee generally decreases as N,\nthe number of terms, is increased. Hence, it is possible that the error energy →0 as N →∞. When\nthis happens, the orthogonal signal set is said to be complete. In this case, Eq. (6.41) is no more an\napproximation but an equality", - "type": "text" - }, - { - "block_id": "p671-b23", - "global_id": 19624, - "bbox": [ - 163.68, - 230.08, - 351.41, - 241.23 - ], - "text": "x(t) = c1x1(t) + c2x2(t) + · · · + cnxn(t) + · · · =", - "type": "text" - }, - { - "block_id": "p671-b24", - "global_id": 19625, - "bbox": [ - 353.46, - 219.9, - 367.55, - 230.57 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p671-b25", - "global_id": 19626, - "bbox": [ - 354.29, - 244.46, - 366.7, - 251.73 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p671-b26", - "global_id": 19627, - "bbox": [ - 368.66, - 230.08, - 516.13, - 241.53 - ], - "text": "cnxn(t)\nt1 < t < t2\n(6.45)", - "type": "text" - }, - { - "block_id": "p671-b27", - "global_id": 19628, - "bbox": [ - 127.59, - 266.25, - 516.13, - 300.99 - ], - "text": "where the coefficients cn are given by Eq. (6.43). Because the error signal energy approaches\nzero, it follows that the energy of x(t) is now equal to the sum of the energies of its orthogonal\ncomponents c1x1(t), c2x2(t), c3x3(t), . . ..", - "type": "text" - }, - { - "block_id": "p671-b28", - "global_id": 19629, - "bbox": [ - 127.59, - 301.8, - 516.14, - 359.99 - ], - "text": "The series on the right-hand side of Eq. (6.45) is called the generalized Fourier series of x(t)\nwith respect to the set {xn(t)}. When the set {xn(t)} is such that the error energy Ee →0 as N →∞\nfor every member of some particular class, we say that the set {xn(t)} is complete on (t1, t2) for that\nclass of x(t), and the set {xn(t)} is called a set of basis functions or basis signals. Unless otherwise\nmentioned, in the future we shall consider only the class of energy signals.", - "type": "text" - }, - { - "block_id": "p671-b29", - "global_id": 19630, - "bbox": [ - 127.59, - 361.57, - 516.14, - 467.59 - ], - "text": "Thus, when the set {xn(t)} is complete, we have the equality of Eq. (6.45). One subtle point\nthat must be understood clearly is the meaning of equality in Eq. (6.45). The equality here is not\nan equality in the ordinary sense, but in the sense that the error energy, that is, the energy of\nthe difference between the two sides of Eq. (6.45), approaches zero. If the equality exists in the\nordinary sense, the error energy is always zero, but the converse is not necessarily true. The error\nenergy can approach zero even though e(t), the difference between the two sides, is nonzero at\nsome isolated instants. The reason is that even if e(t) is nonzero at such instants, the area under\ne2(t) is still zero; thus the Fourier series on the right-hand side of Eq. (6.45) may differ from x(t)\nat a finite number of points.", - "type": "text" - }, - { - "block_id": "p671-b30", - "global_id": 19631, - "bbox": [ - 127.59, - 469.48, - 516.14, - 491.5 - ], - "text": "In Eq. (6.45), the energy of the left-hand side is Ex, and the energy of the right-hand side is\nthe sum of the energies of all the orthogonal components.† Thus,", - "type": "text" - }, - { - "block_id": "p671-b31", - "global_id": 19632, - "bbox": [ - 234.91, - 502.52, - 249.66, - 516.13 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p671-b32", - "global_id": 19633, - "bbox": [ - 240.17, - 527.63, - 245.1, - 535.93 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p671-b33", - "global_id": 19634, - "bbox": [ - 251.77, - 514.66, - 299.4, - 526.36 - ], - "text": "x2(t)dt = c2", - "type": "text" - }, - { - "block_id": "p671-b34", - "global_id": 19635, - "bbox": [ - 295.91, - 514.66, - 373.08, - 528.42 - ], - "text": "1E1 + c2\n2E2 + · · · =", - "type": "text" - }, - { - "block_id": "p671-b35", - "global_id": 19636, - "bbox": [ - 375.13, - 505.91, - 389.23, - 516.58 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p671-b36", - "global_id": 19637, - "bbox": [ - 375.97, - 530.48, - 388.37, - 537.74 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p671-b37", - "global_id": 19638, - "bbox": [ - 390.33, - 514.66, - 398.24, - 526.36 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p671-b38", - "global_id": 19639, - "bbox": [ - 394.75, - 516.4, - 516.13, - 528.35 - ], - "text": "nEn\n(6.46)", - "type": "text" - }, - { - "block_id": "p671-b39", - "global_id": 19640, - "bbox": [ - 127.59, - 552.26, - 516.14, - 598.18 - ], - "text": "This is Parseval’s theorem expressed for energy signals. In Eqs. (6.26) and (6.27), we have already\nencountered Parseval’s theorem for power signals. Recall that the signal energy (area under the\nsquared value of a signal) is analogous to the square of the length of a vector in the vector-signal\nanalogy. In vector space, we know that the square of the length of a vector is equal to the sum of", - "type": "text" - }, - { - "block_id": "p671-b40", - "global_id": 19641, - "bbox": [ - 127.59, - 621.19, - 294.33, - 634.09 - ], - "text": "† Note that the energy of a signal cx(t) is c2Ex.", - "type": "text" - } - ] - }, - { - "page_num": 672, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p672-b0", - "global_id": 19642, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "652\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p672-b1", - "global_id": 19643, - "bbox": [ - 101.84, - 85.82, - 490.4, - 107.74 - ], - "text": "the squares of the lengths of its orthogonal components. Parseval’s theorem of Eq. (6.46) is the\nstatement of this fact as it applies to signals.", - "type": "text" - }, - { - "block_id": "p672-b2", - "global_id": 19644, - "bbox": [ - 102.14, - 124.88, - 322.72, - 137.01 - ], - "text": "GENERALIZATION TO COMPLEX SIGNALS", - "type": "text" - }, - { - "block_id": "p672-b3", - "global_id": 19645, - "bbox": [ - 101.84, - 140.63, - 490.39, - 163.73 - ], - "text": "The foregoing results can be generalized to complex signals as follows: a set of functions x1(t),\nx2(t), . . ., xN(t) is mutually orthogonal over the interval (t1, t2) if", - "type": "text" - }, - { - "block_id": "p672-b4", - "global_id": 19646, - "bbox": [ - 226.75, - 171.56, - 241.49, - 185.17 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p672-b5", - "global_id": 19647, - "bbox": [ - 232.0, - 196.65, - 236.93, - 204.96 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p672-b6", - "global_id": 19648, - "bbox": [ - 243.59, - 183.39, - 272.01, - 196.19 - ], - "text": "xm(t)x∗", - "type": "text" - }, - { - "block_id": "p672-b7", - "global_id": 19649, - "bbox": [ - 268.35, - 185.11, - 301.76, - 197.38 - ], - "text": "n(t)dt =", - "type": "text" - }, - { - "block_id": "p672-b8", - "global_id": 19650, - "bbox": [ - 303.8, - 171.13, - 364.3, - 202.07 - ], - "text": "0\nm̸ = n\nEn\nm = n", - "type": "text" - }, - { - "block_id": "p672-b9", - "global_id": 19651, - "bbox": [ - 101.84, - 218.18, - 490.39, - 240.51 - ], - "text": "If this set is complete for a certain class of functions, then a function x(t) in this class can be\nexpressed as", - "type": "text" - }, - { - "block_id": "p672-b10", - "global_id": 19652, - "bbox": [ - 208.71, - 249.34, - 383.52, - 260.49 - ], - "text": "x(t) = c1x1(t) + c2x2(t) + · · · + cixi(t) + · · ·", - "type": "text" - }, - { - "block_id": "p672-b11", - "global_id": 19653, - "bbox": [ - 101.84, - 273.51, - 126.18, - 283.47 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p672-b12", - "global_id": 19654, - "bbox": [ - 249.34, - 289.46, - 278.35, - 307.49 - ], - "text": "cn = 1", - "type": "text" - }, - { - "block_id": "p672-b13", - "global_id": 19655, - "bbox": [ - 270.82, - 303.42, - 280.39, - 314.18 - ], - "text": "En", - "type": "text" - }, - { - "block_id": "p672-b14", - "global_id": 19656, - "bbox": [ - 283.19, - 282.47, - 297.94, - 296.08 - ], - "text": "# t2", - "type": "text" - }, - { - "block_id": "p672-b15", - "global_id": 19657, - "bbox": [ - 288.45, - 307.57, - 293.37, - 315.87 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p672-b16", - "global_id": 19658, - "bbox": [ - 300.05, - 294.31, - 322.96, - 306.3 - ], - "text": "x(t)x∗", - "type": "text" - }, - { - "block_id": "p672-b17", - "global_id": 19659, - "bbox": [ - 319.3, - 296.03, - 490.38, - 308.3 - ], - "text": "n(t)dt\n(6.47)", - "type": "text" - }, - { - "block_id": "p672-b18", - "global_id": 19660, - "bbox": [ - 76.77, - 357.89, - 476.94, - 383.8 - ], - "text": "EXAMPLE 6.14\nApproximating a Square Wave with a Set of Harmonic\nSine Waves", - "type": "text" - }, - { - "block_id": "p672-b19", - "global_id": 19661, - "bbox": [ - 103.16, - 397.12, - 477.02, - 431.41 - ], - "text": "In Ex. 6.13, the square signal x(t) in Fig. 6.23 is approximated by a single sinusoid sin t. In\nthis example, we approximate x(t) using the set of harmonic sine waves sin t, sin 2t, . . ., sin nt,\n. . ., and see how the approximation improves with the number of terms.", - "type": "text" - }, - { - "block_id": "p672-b20", - "global_id": 19662, - "bbox": [ - 103.16, - 453.91, - 477.01, - 494.66 - ], - "text": "To begin, we note that the set of harmonic sine waves sin t, sin 2t,. . .,sin nt,. . . is orthogonal\nover any interval of duration 2π.† The reader can verify this fact by showing that for any real\nnumber a,\n# a+2π", - "type": "text" - }, - { - "block_id": "p672-b21", - "global_id": 19663, - "bbox": [ - 217.26, - 507.41, - 220.74, - 514.38 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p672-b22", - "global_id": 19664, - "bbox": [ - 240.72, - 495.87, - 307.54, - 506.25 - ], - "text": "sin mt sin ntdt =", - "type": "text" - }, - { - "block_id": "p672-b23", - "global_id": 19665, - "bbox": [ - 309.59, - 481.88, - 477.01, - 512.03 - ], - "text": "0\nm̸ = n\nπ\nm = n\n(6.48)", - "type": "text" - }, - { - "block_id": "p672-b24", - "global_id": 19666, - "bbox": [ - 103.16, - 520.15, - 254.42, - 530.53 - ], - "text": "Using this set, we approximate x(t) as", - "type": "text" - }, - { - "block_id": "p672-b25", - "global_id": 19667, - "bbox": [ - 208.63, - 542.07, - 371.37, - 553.94 - ], - "text": "x(t) ≃c1 sin t + c2 sin 2t + · · · + cn sin Nt", - "type": "text" - }, - { - "block_id": "p672-b26", - "global_id": 19668, - "bbox": [ - 101.84, - 588.31, - 490.38, - 633.41 - ], - "text": "† This sine set, along with the cosine set cos0t, cos t, cos 2t,. . .,cosnt,. . ., forms a complete set. In this\ncase, however, the coefficients ci corresponding to the cosine terms are zero. For this reason, we have omitted\ncosine terms in this example. This composite sine and cosine set is the basis set for the trigonometric Fourier\nseries.", - "type": "text" - } - ] - }, - { - "page_num": 673, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p673-b0", - "global_id": 19669, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n653", - "type": "text" - }, - { - "block_id": "p673-b1", - "global_id": 19670, - "bbox": [ - 128.9, - 86.24, - 153.23, - 96.2 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p673-b2", - "global_id": 19671, - "bbox": [ - 236.29, - 117.68, - 254.52, - 129.13 - ], - "text": "cn =", - "type": "text" - }, - { - "block_id": "p673-b3", - "global_id": 19672, - "bbox": [ - 257.77, - 91.02, - 275.25, - 103.37 - ], - "text": "# 2π", - "type": "text" - }, - { - "block_id": "p673-b4", - "global_id": 19673, - "bbox": [ - 263.02, - 104.58, - 324.13, - 131.19 - ], - "text": "0\nx(t)sin ntdt\n# 2π", - "type": "text" - }, - { - "block_id": "p673-b5", - "global_id": 19674, - "bbox": [ - 269.56, - 128.6, - 317.6, - 150.98 - ], - "text": "0\nsin2 ntdt", - "type": "text" - }, - { - "block_id": "p673-b6", - "global_id": 19675, - "bbox": [ - 246.75, - 154.43, - 263.75, - 171.38 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p673-b7", - "global_id": 19676, - "bbox": [ - 257.77, - 168.08, - 263.74, - 178.04 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p673-b8", - "global_id": 19677, - "bbox": [ - 267.05, - 147.02, - 286.48, - 159.5 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p673-b9", - "global_id": 19678, - "bbox": [ - 277.73, - 161.0, - 328.91, - 179.59 - ], - "text": "0\nsin ntdt +", - "type": "text" - }, - { - "block_id": "p673-b10", - "global_id": 19679, - "bbox": [ - 330.45, - 147.44, - 347.95, - 159.79 - ], - "text": "# 2π", - "type": "text" - }, - { - "block_id": "p673-b11", - "global_id": 19680, - "bbox": [ - 335.72, - 172.32, - 339.91, - 179.3 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p673-b12", - "global_id": 19681, - "bbox": [ - 350.26, - 161.0, - 389.76, - 171.38 - ], - "text": "−sin ntdt", - "type": "text" - }, - { - "block_id": "p673-b13", - "global_id": 19682, - "bbox": [ - 389.94, - 147.02, - 395.37, - 156.98 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p673-b14", - "global_id": 19683, - "bbox": [ - 246.75, - 193.41, - 254.52, - 203.37 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p673-b15", - "global_id": 19684, - "bbox": [ - 256.57, - 176.44, - 273.11, - 191.61 - ], - "text": ") 4", - "type": "text" - }, - { - "block_id": "p673-b16", - "global_id": 19685, - "bbox": [ - 264.64, - 188.53, - 320.12, - 205.57 - ], - "text": "πn\nn odd", - "type": "text" - }, - { - "block_id": "p673-b17", - "global_id": 19686, - "bbox": [ - 263.44, - 204.12, - 323.58, - 214.18 - ], - "text": "0\nn even", - "type": "text" - }, - { - "block_id": "p673-b18", - "global_id": 19687, - "bbox": [ - 144.87, - 226.92, - 155.97, - 235.0 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p673-b19", - "global_id": 19688, - "bbox": [ - 134.17, - 427.52, - 138.17, - 435.52 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b20", - "global_id": 19689, - "bbox": [ - 127.51, - 591.72, - 138.17, - 600.02 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b21", - "global_id": 19690, - "bbox": [ - 127.51, - 494.12, - 138.17, - 502.42 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b22", - "global_id": 19691, - "bbox": [ - 127.51, - 396.32, - 138.17, - 404.62 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b23", - "global_id": 19692, - "bbox": [ - 127.51, - 298.53, - 138.17, - 306.82 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b24", - "global_id": 19693, - "bbox": [ - 134.17, - 524.62, - 138.17, - 532.62 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b25", - "global_id": 19694, - "bbox": [ - 349.16, - 551.26, - 351.38, - 559.26 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p673-b26", - "global_id": 19695, - "bbox": [ - 349.16, - 454.12, - 351.38, - 462.12 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p673-b27", - "global_id": 19696, - "bbox": [ - 349.16, - 356.42, - 351.38, - 364.42 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p673-b28", - "global_id": 19697, - "bbox": [ - 134.17, - 330.02, - 138.17, - 338.02 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b29", - "global_id": 19698, - "bbox": [ - 134.17, - 236.52, - 138.17, - 244.52 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p673-b30", - "global_id": 19699, - "bbox": [ - 235.45, - 564.09, - 359.63, - 572.19 - ], - "text": "p\n2p", - "type": "text" - }, - { - "block_id": "p673-b31", - "global_id": 19700, - "bbox": [ - 235.45, - 466.75, - 359.63, - 474.85 - ], - "text": "p\n2p", - "type": "text" - }, - { - "block_id": "p673-b32", - "global_id": 19701, - "bbox": [ - 235.45, - 369.43, - 359.63, - 377.54 - ], - "text": "p\n2p", - "type": "text" - }, - { - "block_id": "p673-b33", - "global_id": 19702, - "bbox": [ - 235.45, - 273.65, - 359.63, - 281.75 - ], - "text": "p\n2p", - "type": "text" - }, - { - "block_id": "p673-b34", - "global_id": 19703, - "bbox": [ - 286.08, - 246.71, - 306.08, - 255.0 - ], - "text": "N 1", - "type": "text" - }, - { - "block_id": "p673-b35", - "global_id": 19704, - "bbox": [ - 286.08, - 342.6, - 306.08, - 350.9 - ], - "text": "N 3", - "type": "text" - }, - { - "block_id": "p673-b36", - "global_id": 19705, - "bbox": [ - 286.08, - 439.9, - 306.08, - 448.2 - ], - "text": "N 5", - "type": "text" - }, - { - "block_id": "p673-b37", - "global_id": 19706, - "bbox": [ - 284.08, - 537.3, - 308.08, - 545.6 - ], - "text": "N 19", - "type": "text" - }, - { - "block_id": "p673-b38", - "global_id": 19707, - "bbox": [ - 127.51, - 610.26, - 413.03, - 619.49 - ], - "text": "Figure 6.25 Approximation of a square wave by a sum of harmonic sinusoids.", - "type": "text" - } - ] - }, - { - "page_num": 674, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p674-b0", - "global_id": 19708, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "654\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p674-b1", - "global_id": 19709, - "bbox": [ - 103.16, - 86.24, - 144.91, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p674-b2", - "global_id": 19710, - "bbox": [ - 180.77, - 96.14, - 214.64, - 112.98 - ], - "text": "x(t) ≃4", - "type": "text" - }, - { - "block_id": "p674-b3", - "global_id": 19711, - "bbox": [ - 208.67, - 109.78, - 214.65, - 119.74 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p674-b5", - "global_id": 19712, - "bbox": [ - 224.67, - 96.14, - 258.51, - 113.08 - ], - "text": "sin t + 1", - "type": "text" - }, - { - "block_id": "p674-b6", - "global_id": 19713, - "bbox": [ - 253.53, - 96.14, - 392.5, - 120.16 - ], - "text": "3 sin 3t + 1\n5 sin 5t + · · · + 1\nN sin Nt", - "type": "text" - }, - { - "block_id": "p674-b8", - "global_id": 19714, - "bbox": [ - 452.95, - 103.12, - 477.01, - 113.08 - ], - "text": "(6.49)", - "type": "text" - }, - { - "block_id": "p674-b9", - "global_id": 19715, - "bbox": [ - 103.16, - 127.24, - 477.01, - 149.26 - ], - "text": "Note that coefficients of terms sinkt are zero for even values of k. Figure 6.25 shows how the\napproximation improves as we increase the number of terms in the series.", - "type": "text" - }, - { - "block_id": "p674-b10", - "global_id": 19716, - "bbox": [ - 121.09, - 150.84, - 400.42, - 161.22 - ], - "text": "Let us investigate the error signal energy as N →∞. From Eq. (6.44),", - "type": "text" - }, - { - "block_id": "p674-b11", - "global_id": 19717, - "bbox": [ - 233.21, - 182.87, - 252.71, - 194.33 - ], - "text": "Ee =", - "type": "text" - }, - { - "block_id": "p674-b12", - "global_id": 19718, - "bbox": [ - 254.76, - 169.32, - 272.25, - 181.66 - ], - "text": "# 2π", - "type": "text" - }, - { - "block_id": "p674-b13", - "global_id": 19719, - "bbox": [ - 260.02, - 181.44, - 311.73, - 201.45 - ], - "text": "0\nx2(t)dt −", - "type": "text" - }, - { - "block_id": "p674-b14", - "global_id": 19720, - "bbox": [ - 313.29, - 172.7, - 327.38, - 183.37 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p674-b15", - "global_id": 19721, - "bbox": [ - 314.12, - 197.27, - 326.53, - 204.53 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p674-b16", - "global_id": 19722, - "bbox": [ - 328.49, - 181.44, - 336.39, - 193.15 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p674-b17", - "global_id": 19723, - "bbox": [ - 332.91, - 183.18, - 346.47, - 195.14 - ], - "text": "nEn", - "type": "text" - }, - { - "block_id": "p674-b18", - "global_id": 19724, - "bbox": [ - 103.16, - 216.22, - 139.96, - 226.18 - ], - "text": "Note that", - "type": "text" - }, - { - "block_id": "p674-b19", - "global_id": 19725, - "bbox": [ - 203.05, - 230.9, - 220.53, - 243.26 - ], - "text": "# 2π", - "type": "text" - }, - { - "block_id": "p674-b20", - "global_id": 19726, - "bbox": [ - 208.3, - 243.03, - 260.51, - 263.05 - ], - "text": "0\nx2(t)dt =", - "type": "text" - }, - { - "block_id": "p674-b21", - "global_id": 19727, - "bbox": [ - 262.56, - 230.9, - 276.57, - 242.97 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p674-b22", - "global_id": 19728, - "bbox": [ - 267.82, - 230.9, - 325.24, - 263.05 - ], - "text": "0\n12 dt +\n# 2π", - "type": "text" - }, - { - "block_id": "p674-b23", - "global_id": 19729, - "bbox": [ - 313.01, - 255.79, - 317.19, - 262.76 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p674-b24", - "global_id": 19730, - "bbox": [ - 327.54, - 240.67, - 376.14, - 254.84 - ], - "text": "−12 dt = 2π", - "type": "text" - }, - { - "block_id": "p674-b25", - "global_id": 19731, - "bbox": [ - 242.29, - 275.01, - 250.19, - 286.72 - ], - "text": "c2", - "type": "text" - }, - { - "block_id": "p674-b26", - "global_id": 19732, - "bbox": [ - 246.71, - 276.44, - 260.52, - 289.09 - ], - "text": "n =", - "type": "text" - }, - { - "block_id": "p674-b27", - "global_id": 19733, - "bbox": [ - 262.56, - 259.47, - 285.57, - 274.64 - ], - "text": ") 16", - "type": "text" - }, - { - "block_id": "p674-b28", - "global_id": 19734, - "bbox": [ - 270.63, - 271.56, - 334.08, - 288.6 - ], - "text": "n2π2\nn odd", - "type": "text" - }, - { - "block_id": "p674-b29", - "global_id": 19735, - "bbox": [ - 269.44, - 287.15, - 337.54, - 297.21 - ], - "text": "0\nn even", - "type": "text" - }, - { - "block_id": "p674-b30", - "global_id": 19736, - "bbox": [ - 103.17, - 310.58, - 184.5, - 320.54 - ], - "text": "and from Eq. (6.48),", - "type": "text" - }, - { - "block_id": "p674-b31", - "global_id": 19737, - "bbox": [ - 275.63, - 322.12, - 303.55, - 333.58 - ], - "text": "En = π", - "type": "text" - }, - { - "block_id": "p674-b32", - "global_id": 19738, - "bbox": [ - 103.17, - 341.46, - 144.91, - 351.42 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p674-b33", - "global_id": 19739, - "bbox": [ - 187.99, - 361.81, - 230.81, - 373.27 - ], - "text": "Ee = 2π −", - "type": "text" - }, - { - "block_id": "p674-b34", - "global_id": 19740, - "bbox": [ - 242.24, - 351.85, - 256.34, - 362.3 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p674-b35", - "global_id": 19741, - "bbox": [ - 232.35, - 376.2, - 266.22, - 383.46 - ], - "text": "n=1,3,5,. . .", - "type": "text" - }, - { - "block_id": "p674-b36", - "global_id": 19742, - "bbox": [ - 268.52, - 355.24, - 342.46, - 379.15 - ], - "text": "16\nn2π2 π = 2π −16", - "type": "text" - }, - { - "block_id": "p674-b37", - "global_id": 19743, - "bbox": [ - 334.0, - 368.87, - 339.97, - 378.84 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p674-b38", - "global_id": 19744, - "bbox": [ - 355.75, - 351.85, - 369.85, - 362.3 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p674-b39", - "global_id": 19745, - "bbox": [ - 345.88, - 376.2, - 379.74, - 383.46 - ], - "text": "n=1,3,5,. . .", - "type": "text" - }, - { - "block_id": "p674-b40", - "global_id": 19746, - "bbox": [ - 382.03, - 355.24, - 390.49, - 379.15 - ], - "text": "1\nn2", - "type": "text" - }, - { - "block_id": "p674-b41", - "global_id": 19747, - "bbox": [ - 121.09, - 390.67, - 285.88, - 401.05 - ], - "text": "For a single-term approximation (N = 1),", - "type": "text" - }, - { - "block_id": "p674-b42", - "global_id": 19748, - "bbox": [ - 242.1, - 412.98, - 297.62, - 431.01 - ], - "text": "Ee = 2π −16", - "type": "text" - }, - { - "block_id": "p674-b43", - "global_id": 19749, - "bbox": [ - 289.16, - 419.55, - 338.07, - 436.59 - ], - "text": "π = 1.1938", - "type": "text" - }, - { - "block_id": "p674-b44", - "global_id": 19750, - "bbox": [ - 103.17, - 446.51, - 258.99, - 456.88 - ], - "text": "For a two-term approximation (N = 3),", - "type": "text" - }, - { - "block_id": "p674-b45", - "global_id": 19751, - "bbox": [ - 223.21, - 469.49, - 278.73, - 487.52 - ], - "text": "Ee = 2π −16", - "type": "text" - }, - { - "block_id": "p674-b46", - "global_id": 19752, - "bbox": [ - 270.27, - 483.14, - 276.24, - 493.1 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p674-b48", - "global_id": 19753, - "bbox": [ - 287.76, - 469.49, - 309.79, - 493.51 - ], - "text": "1 + 1\n9", - "type": "text" - }, - { - "block_id": "p674-b50", - "global_id": 19754, - "bbox": [ - 319.76, - 476.06, - 356.96, - 486.44 - ], - "text": "= 0.6243", - "type": "text" - }, - { - "block_id": "p674-b51", - "global_id": 19755, - "bbox": [ - 103.17, - 505.57, - 433.45, - 517.13 - ], - "text": "Continuing this process, we compute the error energy Ee for various values of N as", - "type": "text" - }, - { - "block_id": "p674-b52", - "global_id": 19756, - "bbox": [ - 179.09, - 528.74, - 401.08, - 552.18 - ], - "text": "N\n1\n3\n5\n7\n99\n∞\nEe\n1.1938\n0.6243\n0.4206\n0.3166\n0.02545\n0", - "type": "text" - }, - { - "block_id": "p674-b53", - "global_id": 19757, - "bbox": [ - 121.09, - 564.48, - 329.56, - 574.86 - ], - "text": "Clearly, x(t) can be represented by the infinite series", - "type": "text" - }, - { - "block_id": "p674-b54", - "global_id": 19758, - "bbox": [ - 160.07, - 589.94, - 193.94, - 606.89 - ], - "text": "x(t) = 4", - "type": "text" - }, - { - "block_id": "p674-b55", - "global_id": 19759, - "bbox": [ - 187.97, - 603.58, - 193.95, - 613.55 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p674-b57", - "global_id": 19760, - "bbox": [ - 203.97, - 589.94, - 237.81, - 606.89 - ], - "text": "sin t + 1", - "type": "text" - }, - { - "block_id": "p674-b58", - "global_id": 19761, - "bbox": [ - 232.83, - 582.52, - 332.56, - 613.96 - ], - "text": "3 sin3t + 1\n5 sin5t + · · ·", - "type": "text" - }, - { - "block_id": "p674-b59", - "global_id": 19762, - "bbox": [ - 334.6, - 589.94, - 351.6, - 606.89 - ], - "text": "= 4", - "type": "text" - }, - { - "block_id": "p674-b60", - "global_id": 19763, - "bbox": [ - 345.62, - 603.58, - 351.59, - 613.55 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p674-b61", - "global_id": 19764, - "bbox": [ - 365.88, - 586.34, - 379.98, - 597.01 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p674-b62", - "global_id": 19765, - "bbox": [ - 356.0, - 610.91, - 389.86, - 618.17 - ], - "text": "n=1,3,5,. . .", - "type": "text" - }, - { - "block_id": "p674-b63", - "global_id": 19766, - "bbox": [ - 392.16, - 589.94, - 419.93, - 613.96 - ], - "text": "1\nn sinnt", - "type": "text" - } - ] - }, - { - "page_num": 675, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p675-b0", - "global_id": 19767, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n655", - "type": "text" - }, - { - "block_id": "p675-b1", - "global_id": 19768, - "bbox": [ - 128.9, - 85.83, - 502.76, - 132.07 - ], - "text": "The equality exists in the sense that the error signal energy →0 as N →∞. In this case, the\nerror energy decreases rather slowly with N, indicating that the series converges slowly. This is\nto be expected because x(t) has jump discontinuities and consequently, according to discussion\nin Sec. 6.2-2, the series converges asymptotically as 1/n.", - "type": "text" - }, - { - "block_id": "p675-b2", - "global_id": 19769, - "bbox": [ - 133.57, - 209.32, - 504.11, - 235.21 - ], - "text": "DRILL 6.9\nApproximating a Ramp Signal with a Set of Harmonic\nSine Waves", - "type": "text" - }, - { - "block_id": "p675-b3", - "global_id": 19770, - "bbox": [ - 133.57, - 243.92, - 510.16, - 278.21 - ], - "text": "Approximate the signal x(t) = t −π (Fig. 6.26) over the interval (0,2π) in terms of the set of\nsinusoids {sin nt}, n = 0,1,2,. . ., used in Ex. 6.14. Find Ee, the error energy. Show that Ee →0\nas N →∞.", - "type": "text" - }, - { - "block_id": "p675-b4", - "global_id": 19771, - "bbox": [ - 133.84, - 291.74, - 196.06, - 302.69 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p675-b5", - "global_id": 19772, - "bbox": [ - 133.57, - 302.2, - 189.19, - 320.05 - ], - "text": "x(t) ≃−2%N", - "type": "text" - }, - { - "block_id": "p675-b6", - "global_id": 19773, - "bbox": [ - 184.54, - 314.91, - 196.94, - 322.17 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p675-b7", - "global_id": 19774, - "bbox": [ - 199.75, - 308.33, - 294.41, - 323.59 - ], - "text": "1\nN sin nt\nand\nEe = 2", - "type": "text" - }, - { - "block_id": "p675-b8", - "global_id": 19775, - "bbox": [ - 290.93, - 302.2, - 340.26, - 322.87 - ], - "text": "3π3 −%N\nn=1", - "type": "text" - }, - { - "block_id": "p675-b9", - "global_id": 19776, - "bbox": [ - 343.06, - 308.04, - 350.73, - 323.54 - ], - "text": "4π\nn2", - "type": "text" - }, - { - "block_id": "p675-b10", - "global_id": 19777, - "bbox": [ - 157.69, - 365.17, - 179.19, - 373.72 - ], - "text": "x(t) p", - "type": "text" - }, - { - "block_id": "p675-b11", - "global_id": 19778, - "bbox": [ - 167.19, - 435.48, - 179.19, - 443.67 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p675-b12", - "global_id": 19779, - "bbox": [ - 262.47, - 405.66, - 394.08, - 415.3 - ], - "text": "p\nt\n2p", - "type": "text" - }, - { - "block_id": "p675-b13", - "global_id": 19780, - "bbox": [ - 157.47, - 458.35, - 300.19, - 467.59 - ], - "text": "Figure 6.26 Ramp signal for Drill 6.9.", - "type": "text" - }, - { - "block_id": "p675-b14", - "global_id": 19781, - "bbox": [ - 127.89, - 513.03, - 406.21, - 525.15 - ], - "text": "SOME EXAMPLES OF GENERALIZED FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p675-b15", - "global_id": 19782, - "bbox": [ - 127.59, - 529.18, - 516.16, - 634.79 - ], - "text": "Signals are vectors in every sense. Like a vector, a signal can be represented as a sum of its\ncomponents in a variety of ways. Just as vector coordinate systems are formed by mutually\northogonal vectors (rectangular, cylindrical, spherical), we also have signal coordinate systems\n(basis signals) formed by a variety of sets of mutually orthogonal signals. There exist a large\nnumber of orthogonal signal sets that can be used as basis signals for generalized Fourier series.\nSome well-known signal sets are trigonometric (sinusoid) functions, exponential functions, Walsh\nfunctions, Bessel functions, Legendre polynomials, Laguerre functions, Jacobi polynomials,\nHermite polynomials, and Chebyshev polynomials. The functions that concern us most in this\nbook are the trigonometric and the exponential sets discussed earlier in this chapter.", - "type": "text" - } - ] - }, - { - "page_num": 676, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p676-b0", - "global_id": 19783, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "656\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p676-b1", - "global_id": 19784, - "bbox": [ - 102.14, - 86.19, - 249.88, - 98.32 - ], - "text": "LEGENDRE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p676-b2", - "global_id": 19785, - "bbox": [ - 101.84, - 101.93, - 490.4, - 136.22 - ], - "text": "A set of Legendre polynomials Pn(t) (n = 0,1,2,3,. . .) forms a complete set of mutually\northogonal functions over an interval (−1 < t < 1). These polynomials can be defined by the\nRodrigues formula:", - "type": "text" - }, - { - "block_id": "p676-b3", - "global_id": 19786, - "bbox": [ - 208.3, - 136.32, - 270.98, - 161.77 - ], - "text": "Pn(t) =\n1\n2nn!\ndn", - "type": "text" - }, - { - "block_id": "p676-b4", - "global_id": 19787, - "bbox": [ - 260.91, - 140.2, - 383.94, - 161.67 - ], - "text": "dtn (t2 −1)n\nn = 0,1,2,. . .", - "type": "text" - }, - { - "block_id": "p676-b5", - "global_id": 19788, - "bbox": [ - 101.84, - 167.72, - 233.03, - 177.68 - ], - "text": "It follows from this equation that", - "type": "text" - }, - { - "block_id": "p676-b6", - "global_id": 19789, - "bbox": [ - 157.27, - 190.25, - 267.3, - 201.4 - ], - "text": "P0(t) = 1,P1(t) = t,P2(t) =", - "type": "text" - }, - { - "block_id": "p676-b7", - "global_id": 19790, - "bbox": [ - 269.34, - 182.24, - 278.04, - 195.88 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p676-b8", - "global_id": 19791, - "bbox": [ - 274.55, - 182.24, - 306.94, - 203.44 - ], - "text": "2t2 −1\n2", - "type": "text" - }, - { - "block_id": "p676-b9", - "global_id": 19792, - "bbox": [ - 308.04, - 190.25, - 341.91, - 201.4 - ], - "text": ",P3(t) =", - "type": "text" - }, - { - "block_id": "p676-b10", - "global_id": 19793, - "bbox": [ - 343.96, - 182.24, - 352.66, - 195.88 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p676-b11", - "global_id": 19794, - "bbox": [ - 349.17, - 182.24, - 384.5, - 203.44 - ], - "text": "2t3 −3\n2t", - "type": "text" - }, - { - "block_id": "p676-b12", - "global_id": 19795, - "bbox": [ - 385.6, - 190.67, - 434.97, - 200.63 - ], - "text": ", and so on", - "type": "text" - }, - { - "block_id": "p676-b13", - "global_id": 19796, - "bbox": [ - 119.78, - 213.62, - 400.81, - 223.59 - ], - "text": "We may verify the orthogonality of these polynomials by showing that", - "type": "text" - }, - { - "block_id": "p676-b14", - "global_id": 19797, - "bbox": [ - 214.66, - 234.23, - 227.96, - 246.57 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p676-b15", - "global_id": 19798, - "bbox": [ - 219.92, - 259.1, - 228.84, - 266.36 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p676-b16", - "global_id": 19799, - "bbox": [ - 230.45, - 247.79, - 291.76, - 258.87 - ], - "text": "Pm(t)Pn(t)dt =", - "type": "text" - }, - { - "block_id": "p676-b17", - "global_id": 19800, - "bbox": [ - 293.81, - 227.36, - 301.7, - 246.29 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p676-b18", - "global_id": 19801, - "bbox": [ - 293.81, - 254.26, - 301.7, - 264.22 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p676-b19", - "global_id": 19802, - "bbox": [ - 301.7, - 236.85, - 376.39, - 271.66 - ], - "text": "0\nm̸ = n\n2\n2m + 1\nm = n", - "type": "text" - }, - { - "block_id": "p676-b20", - "global_id": 19803, - "bbox": [ - 101.85, - 282.07, - 489.52, - 292.45 - ], - "text": "We can express a function x(t) in terms of Legendre polynomials over an interval (−1 < t < 1) as", - "type": "text" - }, - { - "block_id": "p676-b21", - "global_id": 19804, - "bbox": [ - 205.28, - 305.02, - 490.38, - 316.17 - ], - "text": "x(t) = c0P0(t) + c1P1(t) + · · · + crPr(t) + · · ·\n(6.50)", - "type": "text" - }, - { - "block_id": "p676-b22", - "global_id": 19805, - "bbox": [ - 101.85, - 328.38, - 126.18, - 338.35 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p676-b23", - "global_id": 19806, - "bbox": [ - 204.89, - 359.84, - 222.49, - 371.3 - ], - "text": "cr =", - "type": "text" - }, - { - "block_id": "p676-b24", - "global_id": 19807, - "bbox": [ - 225.74, - 332.68, - 239.05, - 345.03 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p676-b25", - "global_id": 19808, - "bbox": [ - 231.0, - 357.56, - 239.93, - 364.82 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p676-b26", - "global_id": 19809, - "bbox": [ - 231.17, - 346.24, - 285.06, - 373.35 - ], - "text": "x(t)Pr(t)dt\n# 1", - "type": "text" - }, - { - "block_id": "p676-b27", - "global_id": 19810, - "bbox": [ - 236.42, - 385.89, - 245.34, - 393.15 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p676-b28", - "global_id": 19811, - "bbox": [ - 246.96, - 374.89, - 255.76, - 385.65 - ], - "text": "Pr", - "type": "text" - }, - { - "block_id": "p676-b29", - "global_id": 19812, - "bbox": [ - 256.42, - 352.86, - 324.44, - 384.85 - ], - "text": "2(t)dt\n= 2r + 1", - "type": "text" - }, - { - "block_id": "p676-b30", - "global_id": 19813, - "bbox": [ - 309.48, - 367.33, - 314.46, - 377.29 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p676-b31", - "global_id": 19814, - "bbox": [ - 327.85, - 346.29, - 341.16, - 358.63 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p676-b32", - "global_id": 19815, - "bbox": [ - 333.11, - 371.16, - 342.03, - 378.42 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p676-b33", - "global_id": 19816, - "bbox": [ - 343.64, - 359.84, - 490.38, - 370.92 - ], - "text": "x(t)Pr(t)dt\n(6.51)", - "type": "text" - }, - { - "block_id": "p676-b34", - "global_id": 19817, - "bbox": [ - 101.84, - 399.92, - 490.39, - 422.25 - ], - "text": "Note that although the series representation is valid over the interval (−1, 1), it can be extended to\nany interval by the appropriate time scaling (see Prob. 6.5-8).", - "type": "text" - }, - { - "block_id": "p676-b35", - "global_id": 19818, - "bbox": [ - 76.77, - 463.72, - 311.12, - 475.68 - ], - "text": "EXAMPLE 6.15\nLegendre Fourier Series", - "type": "text" - }, - { - "block_id": "p676-b36", - "global_id": 19819, - "bbox": [ - 103.16, - 492.35, - 416.09, - 502.31 - ], - "text": "Determine the Legendre Fourier series of the square signal shown in Fig. 6.27.", - "type": "text" - }, - { - "block_id": "p676-b37", - "global_id": 19820, - "bbox": [ - 116.32, - 583.78, - 253.28, - 592.93 - ], - "text": "t\n1\n0", - "type": "text" - }, - { - "block_id": "p676-b38", - "global_id": 19821, - "bbox": [ - 187.45, - 560.09, - 191.45, - 568.09 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p676-b39", - "global_id": 19822, - "bbox": [ - 168.79, - 597.01, - 179.45, - 605.31 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p676-b40", - "global_id": 19823, - "bbox": [ - 211.01, - 565.24, - 222.43, - 573.33 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p676-b41", - "global_id": 19824, - "bbox": [ - 241.9, - 573.37, - 245.9, - 581.37 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p676-b42", - "global_id": 19825, - "bbox": [ - 278.12, - 603.89, - 424.07, - 613.13 - ], - "text": "Figure 6.27 Square signal for Ex. 6.15.", - "type": "text" - } - ] - }, - { - "page_num": 677, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p677-b0", - "global_id": 19826, - "bbox": [ - 273.38, - 62.89, - 516.13, - 71.98 - ], - "text": "6.5\nGeneralized Fourier Series: Signals as Vectors\n657", - "type": "text" - }, - { - "block_id": "p677-b1", - "global_id": 19827, - "bbox": [ - 146.84, - 86.24, - 440.32, - 96.21 - ], - "text": "From Eq. (6.50), we know that the Legendre Fourier series takes the form", - "type": "text" - }, - { - "block_id": "p677-b2", - "global_id": 19828, - "bbox": [ - 225.0, - 107.74, - 406.67, - 118.89 - ], - "text": "x(t) = c0P0(t) + c1P1(t) + · · · + crPr(t) + · · ·", - "type": "text" - }, - { - "block_id": "p677-b3", - "global_id": 19829, - "bbox": [ - 128.91, - 129.66, - 417.82, - 141.53 - ], - "text": "The coefficients c0,c1,c2,. . .,cr may be found from Eq. (6.51). We have", - "type": "text" - }, - { - "block_id": "p677-b4", - "global_id": 19830, - "bbox": [ - 252.18, - 157.23, - 276.83, - 167.5 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p677-b5", - "global_id": 19831, - "bbox": [ - 278.88, - 143.24, - 378.3, - 173.48 - ], - "text": "1\n· · · −1 < t < 0\n−1\n· · ·0 < t < 1", - "type": "text" - }, - { - "block_id": "p677-b6", - "global_id": 19832, - "bbox": [ - 128.9, - 184.85, - 143.28, - 194.81 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p677-b7", - "global_id": 19833, - "bbox": [ - 217.06, - 204.54, - 243.51, - 222.57 - ], - "text": "c0 = 1", - "type": "text" - }, - { - "block_id": "p677-b8", - "global_id": 19834, - "bbox": [ - 238.53, - 218.6, - 243.51, - 228.56 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p677-b9", - "global_id": 19835, - "bbox": [ - 245.81, - 197.55, - 259.13, - 209.89 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p677-b10", - "global_id": 19836, - "bbox": [ - 251.08, - 222.43, - 260.0, - 229.69 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p677-b11", - "global_id": 19837, - "bbox": [ - 261.61, - 211.11, - 302.33, - 221.49 - ], - "text": "x(t)dt = 0", - "type": "text" - }, - { - "block_id": "p677-b12", - "global_id": 19838, - "bbox": [ - 217.06, - 233.64, - 243.51, - 251.67 - ], - "text": "c1 = 3", - "type": "text" - }, - { - "block_id": "p677-b13", - "global_id": 19839, - "bbox": [ - 238.53, - 247.7, - 243.51, - 257.67 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p677-b14", - "global_id": 19840, - "bbox": [ - 245.81, - 226.65, - 259.13, - 239.0 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p677-b15", - "global_id": 19841, - "bbox": [ - 251.08, - 251.53, - 260.0, - 258.8 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p677-b16", - "global_id": 19842, - "bbox": [ - 261.61, - 233.64, - 306.29, - 250.59 - ], - "text": "tx(t)dt = 3", - "type": "text" - }, - { - "block_id": "p677-b17", - "global_id": 19843, - "bbox": [ - 301.31, - 247.7, - 306.29, - 257.67 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p677-b18", - "global_id": 19844, - "bbox": [ - 308.6, - 226.23, - 328.63, - 239.0 - ], - "text": "# 0", - "type": "text" - }, - { - "block_id": "p677-b19", - "global_id": 19845, - "bbox": [ - 320.58, - 251.53, - 329.5, - 258.8 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p677-b20", - "global_id": 19846, - "bbox": [ - 331.12, - 240.21, - 352.42, - 250.49 - ], - "text": "tdt −", - "type": "text" - }, - { - "block_id": "p677-b21", - "global_id": 19847, - "bbox": [ - 353.97, - 226.65, - 367.28, - 239.0 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p677-b22", - "global_id": 19848, - "bbox": [ - 359.23, - 240.53, - 380.69, - 258.8 - ], - "text": "0\ntdt", - "type": "text" - }, - { - "block_id": "p677-b24", - "global_id": 19849, - "bbox": [ - 389.64, - 233.64, - 413.41, - 250.18 - ], - "text": "= −3", - "type": "text" - }, - { - "block_id": "p677-b25", - "global_id": 19850, - "bbox": [ - 408.43, - 247.7, - 413.41, - 257.67 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p677-b26", - "global_id": 19851, - "bbox": [ - 217.06, - 262.75, - 243.51, - 280.78 - ], - "text": "c2 = 5", - "type": "text" - }, - { - "block_id": "p677-b27", - "global_id": 19852, - "bbox": [ - 238.53, - 276.81, - 243.51, - 286.77 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p677-b28", - "global_id": 19853, - "bbox": [ - 245.81, - 255.76, - 259.13, - 268.1 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p677-b29", - "global_id": 19854, - "bbox": [ - 251.08, - 280.64, - 260.0, - 287.9 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p677-b30", - "global_id": 19855, - "bbox": [ - 261.61, - 269.32, - 276.44, - 279.6 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p677-b31", - "global_id": 19856, - "bbox": [ - 277.55, - 255.34, - 290.45, - 272.71 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p677-b32", - "global_id": 19857, - "bbox": [ - 285.47, - 262.75, - 315.63, - 286.77 - ], - "text": "2t2 −1\n2", - "type": "text" - }, - { - "block_id": "p677-b34", - "global_id": 19858, - "bbox": [ - 324.65, - 269.32, - 349.44, - 279.7 - ], - "text": "dt = 0", - "type": "text" - }, - { - "block_id": "p677-b35", - "global_id": 19859, - "bbox": [ - 128.9, - 296.89, - 502.77, - 320.4 - ], - "text": "This result follows immediately from the fact that the integrand is an odd function of t. In fact,\nthis is true of all cr for even values of r, that is,", - "type": "text" - }, - { - "block_id": "p677-b36", - "global_id": 19860, - "bbox": [ - 260.32, - 330.44, - 371.34, - 341.9 - ], - "text": "c0 = c2 = c4 = c6 = · · · = 0", - "type": "text" - }, - { - "block_id": "p677-b37", - "global_id": 19861, - "bbox": [ - 128.91, - 352.77, - 150.22, - 362.73 - ], - "text": "Also,", - "type": "text" - }, - { - "block_id": "p677-b38", - "global_id": 19862, - "bbox": [ - 152.6, - 373.8, - 179.06, - 391.83 - ], - "text": "c3 = 7", - "type": "text" - }, - { - "block_id": "p677-b39", - "global_id": 19863, - "bbox": [ - 174.08, - 387.86, - 179.06, - 397.82 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p677-b40", - "global_id": 19864, - "bbox": [ - 181.36, - 366.82, - 194.67, - 379.16 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p677-b41", - "global_id": 19865, - "bbox": [ - 186.62, - 391.69, - 195.54, - 398.95 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p677-b42", - "global_id": 19866, - "bbox": [ - 197.15, - 380.37, - 211.98, - 390.65 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p677-b43", - "global_id": 19867, - "bbox": [ - 213.09, - 366.39, - 225.99, - 383.76 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p677-b44", - "global_id": 19868, - "bbox": [ - 221.01, - 366.39, - 262.04, - 397.82 - ], - "text": "2t3 −3\n2t", - "type": "text" - }, - { - "block_id": "p677-b45", - "global_id": 19869, - "bbox": [ - 263.15, - 373.8, - 289.12, - 390.75 - ], - "text": "dt = 7", - "type": "text" - }, - { - "block_id": "p677-b46", - "global_id": 19870, - "bbox": [ - 284.14, - 387.86, - 289.12, - 397.82 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p677-b47", - "global_id": 19871, - "bbox": [ - 291.42, - 366.39, - 310.16, - 379.16 - ], - "text": "# 0", - "type": "text" - }, - { - "block_id": "p677-b48", - "global_id": 19872, - "bbox": [ - 302.11, - 391.69, - 311.03, - 398.95 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p677-b49", - "global_id": 19873, - "bbox": [ - 312.64, - 366.39, - 325.54, - 383.76 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p677-b50", - "global_id": 19874, - "bbox": [ - 320.56, - 366.39, - 361.59, - 397.82 - ], - "text": "2t3 −3\n2t", - "type": "text" - }, - { - "block_id": "p677-b51", - "global_id": 19875, - "bbox": [ - 362.7, - 380.37, - 379.94, - 390.65 - ], - "text": "dt −", - "type": "text" - }, - { - "block_id": "p677-b52", - "global_id": 19876, - "bbox": [ - 381.49, - 366.82, - 394.8, - 379.16 - ], - "text": "# 1", - "type": "text" - }, - { - "block_id": "p677-b53", - "global_id": 19877, - "bbox": [ - 386.75, - 391.98, - 390.24, - 398.95 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p677-b54", - "global_id": 19878, - "bbox": [ - 396.41, - 366.39, - 409.31, - 383.76 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p677-b55", - "global_id": 19879, - "bbox": [ - 404.33, - 366.39, - 445.36, - 397.82 - ], - "text": "2t3 −3\n2t", - "type": "text" - }, - { - "block_id": "p677-b56", - "global_id": 19880, - "bbox": [ - 446.47, - 380.68, - 454.22, - 390.65 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p677-b57", - "global_id": 19881, - "bbox": [ - 454.4, - 366.39, - 459.83, - 376.35 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p677-b58", - "global_id": 19882, - "bbox": [ - 461.88, - 373.8, - 477.86, - 390.75 - ], - "text": "= 7", - "type": "text" - }, - { - "block_id": "p677-b59", - "global_id": 19883, - "bbox": [ - 472.88, - 387.86, - 477.87, - 397.82 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p677-b60", - "global_id": 19884, - "bbox": [ - 128.91, - 407.62, - 410.07, - 418.77 - ], - "text": "In a similar way, coefficients c5,c7,. . . can be evaluated. We now have", - "type": "text" - }, - { - "block_id": "p677-b61", - "global_id": 19885, - "bbox": [ - 253.57, - 428.2, - 292.73, - 439.81 - ], - "text": "x(t) = −3", - "type": "text" - }, - { - "block_id": "p677-b62", - "global_id": 19886, - "bbox": [ - 289.24, - 421.53, - 322.32, - 442.73 - ], - "text": "2t + 7\n8\n 5", - "type": "text" - }, - { - "block_id": "p677-b63", - "global_id": 19887, - "bbox": [ - 318.83, - 421.53, - 354.16, - 442.73 - ], - "text": "2t3 −3\n2t", - "type": "text" - }, - { - "block_id": "p677-b64", - "global_id": 19888, - "bbox": [ - 355.7, - 429.54, - 378.09, - 439.5 - ], - "text": "+ · · ·", - "type": "text" - }, - { - "block_id": "p677-b65", - "global_id": 19889, - "bbox": [ - 127.89, - 478.76, - 309.07, - 490.89 - ], - "text": "TRIGONOMETRIC FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p677-b66", - "global_id": 19890, - "bbox": [ - 127.59, - 494.92, - 480.3, - 504.88 - ], - "text": "We have already proved [see Eqs. (6.4), (6.5), and (6.6)] that the trigonometric signal set", - "type": "text" - }, - { - "block_id": "p677-b67", - "global_id": 19891, - "bbox": [ - 243.32, - 516.31, - 400.4, - 527.46 - ], - "text": "{1, cosω0t, cos2ω0t, . . ., cosnω0t, . . .;", - "type": "text" - }, - { - "block_id": "p677-b68", - "global_id": 19892, - "bbox": [ - 257.56, - 528.26, - 396.12, - 539.41 - ], - "text": "sinω0t, sin2ω0t, . . ., sinnω0t, . . .}", - "type": "text" - }, - { - "block_id": "p677-b69", - "global_id": 19893, - "bbox": [ - 127.59, - 549.93, - 516.15, - 596.17 - ], - "text": "is orthogonal over any interval of duration T0, where T0 = 1/f0 is the period of the sinusoid of\nfrequency f0. This is a complete set for a class of signals with finite energies [11, 12]. Therefore,\nwe can express a signal x(t) by a trigonometric Fourier series over any interval of duration T0\nseconds as", - "type": "text" - }, - { - "block_id": "p677-b70", - "global_id": 19894, - "bbox": [ - 240.78, - 608.03, - 402.95, - 619.9 - ], - "text": "x(t) = a0 + a1 cosω0t + a2 cos2ω0t + · · ·", - "type": "text" - }, - { - "block_id": "p677-b71", - "global_id": 19895, - "bbox": [ - 269.03, - 622.98, - 391.78, - 634.85 - ], - "text": "+ b1 sinω0t + b2 sin2ω0t + · · ·", - "type": "text" - } - ] - }, - { - "page_num": 678, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p678-b0", - "global_id": 19896, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "658\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p678-b1", - "global_id": 19897, - "bbox": [ - 101.84, - 83.6, - 110.14, - 93.56 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p678-b2", - "global_id": 19898, - "bbox": [ - 177.48, - 101.85, - 222.47, - 113.31 - ], - "text": "x(t) = a0 +", - "type": "text" - }, - { - "block_id": "p678-b3", - "global_id": 19899, - "bbox": [ - 224.02, - 91.68, - 238.12, - 102.35 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p678-b4", - "global_id": 19900, - "bbox": [ - 224.86, - 116.24, - 237.27, - 123.51 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p678-b5", - "global_id": 19901, - "bbox": [ - 239.22, - 101.85, - 414.25, - 113.72 - ], - "text": "an cosnω0t + bn sinnω0t\nt1 < t < t1 + T0", - "type": "text" - }, - { - "block_id": "p678-b6", - "global_id": 19902, - "bbox": [ - 101.84, - 130.64, - 126.17, - 140.6 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p678-b7", - "global_id": 19903, - "bbox": [ - 262.47, - 138.74, - 327.57, - 157.18 - ], - "text": "ω0 = 2πf0 = 2π", - "type": "text" - }, - { - "block_id": "p678-b8", - "global_id": 19904, - "bbox": [ - 317.83, - 153.11, - 326.86, - 163.94 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p678-b9", - "global_id": 19905, - "bbox": [ - 101.84, - 169.86, - 420.62, - 180.69 - ], - "text": "We can use Eq. (6.43) to determine the Fourier coefficients a0, an, and bn. Thus,", - "type": "text" - }, - { - "block_id": "p678-b10", - "global_id": 19906, - "bbox": [ - 240.78, - 213.12, - 259.55, - 224.58 - ], - "text": "an =", - "type": "text" - }, - { - "block_id": "p678-b11", - "global_id": 19907, - "bbox": [ - 262.8, - 185.04, - 290.35, - 198.65 - ], - "text": "# t1+T0", - "type": "text" - }, - { - "block_id": "p678-b12", - "global_id": 19908, - "bbox": [ - 268.07, - 210.13, - 272.99, - 218.44 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p678-b13", - "global_id": 19909, - "bbox": [ - 292.45, - 198.59, - 350.09, - 209.74 - ], - "text": "x(t)cosnω0tdt", - "type": "text" - }, - { - "block_id": "p678-b14", - "global_id": 19910, - "bbox": [ - 268.78, - 214.19, - 296.33, - 227.8 - ], - "text": "# t1+T0", - "type": "text" - }, - { - "block_id": "p678-b15", - "global_id": 19911, - "bbox": [ - 274.04, - 239.29, - 278.96, - 247.6 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p678-b16", - "global_id": 19912, - "bbox": [ - 298.44, - 223.96, - 344.1, - 238.9 - ], - "text": "cos2 nω0tdt", - "type": "text" - }, - { - "block_id": "p678-b17", - "global_id": 19913, - "bbox": [ - 466.32, - 213.53, - 490.38, - 223.5 - ], - "text": "(6.52)", - "type": "text" - }, - { - "block_id": "p678-b18", - "global_id": 19914, - "bbox": [ - 101.85, - 256.36, - 490.39, - 279.46 - ], - "text": "The integral in the denominator of Eq. (6.52) has already been found to be T0/2 when n̸ = 0\n[Eq. (6.4) with m = n]. For n = 0, the denominator is T0. Hence,", - "type": "text" - }, - { - "block_id": "p678-b19", - "global_id": 19915, - "bbox": [ - 124.72, - 290.79, - 154.0, - 308.82 - ], - "text": "a0 = 1", - "type": "text" - }, - { - "block_id": "p678-b20", - "global_id": 19916, - "bbox": [ - 146.74, - 304.75, - 155.77, - 315.58 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p678-b21", - "global_id": 19917, - "bbox": [ - 158.57, - 283.81, - 186.13, - 297.41 - ], - "text": "# t1+T0", - "type": "text" - }, - { - "block_id": "p678-b22", - "global_id": 19918, - "bbox": [ - 163.84, - 308.9, - 168.76, - 317.21 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p678-b23", - "global_id": 19919, - "bbox": [ - 188.24, - 290.79, - 275.69, - 308.82 - ], - "text": "x(t)dt\nand\nan = 2", - "type": "text" - }, - { - "block_id": "p678-b24", - "global_id": 19920, - "bbox": [ - 268.44, - 304.75, - 277.46, - 315.58 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p678-b25", - "global_id": 19921, - "bbox": [ - 280.27, - 283.81, - 307.82, - 297.41 - ], - "text": "# t1+T0", - "type": "text" - }, - { - "block_id": "p678-b26", - "global_id": 19922, - "bbox": [ - 285.53, - 308.9, - 290.46, - 317.21 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p678-b27", - "global_id": 19923, - "bbox": [ - 309.92, - 297.36, - 490.38, - 308.51 - ], - "text": "x(t)cosnω0tdt\nn = 1,2,3,. . .\n(6.53)", - "type": "text" - }, - { - "block_id": "p678-b28", - "global_id": 19924, - "bbox": [ - 101.84, - 326.38, - 190.65, - 336.34 - ], - "text": "Similarly, we find that", - "type": "text" - }, - { - "block_id": "p678-b29", - "global_id": 19925, - "bbox": [ - 198.43, - 348.45, - 227.71, - 366.48 - ], - "text": "bn = 2", - "type": "text" - }, - { - "block_id": "p678-b30", - "global_id": 19926, - "bbox": [ - 220.45, - 362.41, - 229.48, - 373.25 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p678-b31", - "global_id": 19927, - "bbox": [ - 232.28, - 341.46, - 259.84, - 355.07 - ], - "text": "# t1+T0", - "type": "text" - }, - { - "block_id": "p678-b32", - "global_id": 19928, - "bbox": [ - 237.55, - 366.56, - 242.47, - 374.87 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p678-b33", - "global_id": 19929, - "bbox": [ - 261.95, - 355.02, - 490.38, - 366.17 - ], - "text": "x(t)sinnω0tdt\nn = 1,2,3,. . .\n(6.54)", - "type": "text" - }, - { - "block_id": "p678-b34", - "global_id": 19930, - "bbox": [ - 101.84, - 384.04, - 490.39, - 429.87 - ], - "text": "Note that the Fourier series in Eq. (6.49) of Ex. 6.14 is indeed the trigonometric Fourier series\nwith T0 = 2π and ω0 = 2π/T0. In this particular example, it is easy to verify from Eq. (6.53) that\nan = 0 for all n, including n = 0. Hence, the Fourier series in that example consisted only of sine\nterms.", - "type": "text" - }, - { - "block_id": "p678-b35", - "global_id": 19931, - "bbox": [ - 102.14, - 444.81, - 268.86, - 456.93 - ], - "text": "EXPONENTIAL FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p678-b36", - "global_id": 19932, - "bbox": [ - 101.84, - 457.34, - 490.39, - 495.91 - ], - "text": "As shown in the footnote on page 622, the set of exponentials ejnω0t (n = 0,±1,±2,. . .) is a set of\nfunctions orthogonal over any interval of duration T0 = 2π/ω0. An arbitrary signal x(t) can now\nbe expressed over an interval (t1,t1 + T0) as", - "type": "text" - }, - { - "block_id": "p678-b37", - "global_id": 19933, - "bbox": [ - 215.39, - 514.93, - 240.04, - 525.2 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p678-b38", - "global_id": 19934, - "bbox": [ - 245.79, - 504.75, - 259.89, - 515.42 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p678-b39", - "global_id": 19935, - "bbox": [ - 242.09, - 528.77, - 263.57, - 535.97 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p678-b40", - "global_id": 19936, - "bbox": [ - 264.69, - 513.2, - 376.33, - 526.38 - ], - "text": "Dnejnω0t\nt1 < t < t1 + T0", - "type": "text" - }, - { - "block_id": "p678-b41", - "global_id": 19937, - "bbox": [ - 101.84, - 546.65, - 190.64, - 556.61 - ], - "text": "where [see Eq. (6.47)]", - "type": "text" - }, - { - "block_id": "p678-b42", - "global_id": 19938, - "bbox": [ - 238.36, - 558.97, - 269.86, - 577.0 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p678-b43", - "global_id": 19939, - "bbox": [ - 262.6, - 572.93, - 271.62, - 583.77 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p678-b44", - "global_id": 19940, - "bbox": [ - 274.44, - 551.99, - 301.99, - 565.6 - ], - "text": "# t1+T0", - "type": "text" - }, - { - "block_id": "p678-b45", - "global_id": 19941, - "bbox": [ - 279.69, - 577.08, - 284.62, - 585.39 - ], - "text": "t1", - "type": "text" - }, - { - "block_id": "p678-b46", - "global_id": 19942, - "bbox": [ - 304.09, - 561.74, - 353.69, - 575.82 - ], - "text": "x(t)e−jnω0t dt", - "type": "text" - }, - { - "block_id": "p678-b47", - "global_id": 19943, - "bbox": [ - 101.84, - 596.72, - 490.41, - 634.79 - ], - "text": "WHY USE THE EXPONENTIAL SET?\nIf x(t) can be represented in terms of hundreds of different orthogonal sets, why do we exclusively\nuse the exponential (or trigonometric) set for the representation of signals or LTI systems? It so", - "type": "text" - } - ] - }, - { - "page_num": 679, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p679-b0", - "global_id": 19944, - "bbox": [ - 332.99, - 62.57, - 516.13, - 72.3 - ], - "text": "6.6\nNumerical Computation of Dn\n659", - "type": "text" - }, - { - "block_id": "p679-b1", - "global_id": 19945, - "bbox": [ - 127.59, - 85.82, - 516.14, - 143.6 - ], - "text": "happens that the exponential signal is an eigenfunction of LTI systems. In other words, for an\nLTI system, only an exponential input est yields the response that is also an exponential of the\nsame form, given by H(s)est. The same is true of the trigonometric set. This fact makes the use of\nexponential signals natural for LTI systems in the sense that the system analysis using exponentials\nas the basis signals is greatly simplified.", - "type": "text" - }, - { - "block_id": "p679-b2", - "global_id": 19946, - "bbox": [ - 127.94, - 171.83, - 374.44, - 187.2 - ], - "text": "6.6 NUMERICAL COMPUTATION OF Dn", - "type": "text" - }, - { - "block_id": "p679-b3", - "global_id": 19947, - "bbox": [ - 127.59, - 192.66, - 516.14, - 240.08 - ], - "text": "We can compute Dn numerically by using the DFT (the discrete Fourier transform discussed in\nSec. 8.5), which uses the samples of a periodic signal x(t) over one period. The sampling interval is\nT seconds. Hence, there are N0 = T0/T number of samples in one period T0. To find the relationship\nbetween Dn and the samples of x(t), consider Eq. (6.19) and write", - "type": "text" - }, - { - "block_id": "p679-b4", - "global_id": 19948, - "bbox": [ - 250.98, - 248.43, - 282.48, - 266.45 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p679-b5", - "global_id": 19949, - "bbox": [ - 275.23, - 262.39, - 284.25, - 273.22 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p679-b6", - "global_id": 19950, - "bbox": [ - 287.05, - 241.43, - 292.32, - 251.4 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p679-b7", - "global_id": 19951, - "bbox": [ - 292.31, - 266.54, - 299.18, - 274.84 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p679-b8", - "global_id": 19952, - "bbox": [ - 301.28, - 251.2, - 350.87, - 265.27 - ], - "text": "x(t)e−jnω0t dt", - "type": "text" - }, - { - "block_id": "p679-b9", - "global_id": 19953, - "bbox": [ - 264.21, - 288.51, - 288.46, - 298.89 - ], - "text": "= lim", - "type": "text" - }, - { - "block_id": "p679-b10", - "global_id": 19954, - "bbox": [ - 274.03, - 297.43, - 289.61, - 304.69 - ], - "text": "T→0", - "type": "text" - }, - { - "block_id": "p679-b11", - "global_id": 19955, - "bbox": [ - 293.95, - 281.94, - 310.12, - 306.74 - ], - "text": "1\nN0T", - "type": "text" - }, - { - "block_id": "p679-b12", - "global_id": 19956, - "bbox": [ - 313.18, - 277.72, - 330.24, - 289.02 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p679-b13", - "global_id": 19957, - "bbox": [ - 315.65, - 302.9, - 327.78, - 310.17 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p679-b14", - "global_id": 19958, - "bbox": [ - 331.35, - 284.71, - 391.97, - 298.79 - ], - "text": "x(kT)e−jnω0kT T", - "type": "text" - }, - { - "block_id": "p679-b15", - "global_id": 19959, - "bbox": [ - 264.21, - 324.73, - 288.46, - 335.11 - ], - "text": "= lim", - "type": "text" - }, - { - "block_id": "p679-b16", - "global_id": 19960, - "bbox": [ - 274.03, - 333.65, - 289.61, - 340.91 - ], - "text": "T→0", - "type": "text" - }, - { - "block_id": "p679-b17", - "global_id": 19961, - "bbox": [ - 293.95, - 318.16, - 304.08, - 342.95 - ], - "text": "1\nN0", - "type": "text" - }, - { - "block_id": "p679-b18", - "global_id": 19962, - "bbox": [ - 306.87, - 313.94, - 323.94, - 325.23 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p679-b19", - "global_id": 19963, - "bbox": [ - 309.34, - 339.13, - 321.47, - 346.39 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p679-b20", - "global_id": 19964, - "bbox": [ - 325.05, - 323.01, - 516.13, - 335.11 - ], - "text": "x(kT)e−jn0k\n(6.55)", - "type": "text" - }, - { - "block_id": "p679-b21", - "global_id": 19965, - "bbox": [ - 127.59, - 355.63, - 291.37, - 366.0 - ], - "text": "where x(kT) is the kth sample of x(t) and", - "type": "text" - }, - { - "block_id": "p679-b22", - "global_id": 19966, - "bbox": [ - 244.22, - 375.73, - 276.93, - 393.86 - ], - "text": "N0 = T0", - "type": "text" - }, - { - "block_id": "p679-b23", - "global_id": 19967, - "bbox": [ - 269.52, - 375.42, - 397.31, - 399.75 - ], - "text": "T\nand\n0 = ω0T = 2π", - "type": "text" - }, - { - "block_id": "p679-b24", - "global_id": 19968, - "bbox": [ - 387.02, - 389.79, - 397.15, - 400.62 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p679-b25", - "global_id": 19969, - "bbox": [ - 127.59, - 408.68, - 516.14, - 478.83 - ], - "text": "In practice, it is impossible to make T →0 in computing the right-hand side of Eq. (6.55). We\ncan make T small, but not zero, which will cause the data to increase without limit. Thus, we\nshall ignore the limit on T in Eq. (6.55) with the implicit understanding that T is reasonably small.\nNonzero T will result in some computational error, which is inevitable in any numerical evaluation\nof an integral. The error resulting from nonzero T is called the aliasing error, which is discussed\nin more detail in Ch. 8. Thus, we can express Eq. (6.55) as", - "type": "text" - }, - { - "block_id": "p679-b26", - "global_id": 19970, - "bbox": [ - 268.92, - 493.94, - 300.97, - 511.97 - ], - "text": "Dn ≈1", - "type": "text" - }, - { - "block_id": "p679-b27", - "global_id": 19971, - "bbox": [ - 293.17, - 507.9, - 303.29, - 518.73 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p679-b28", - "global_id": 19972, - "bbox": [ - 306.1, - 489.71, - 323.16, - 501.0 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p679-b29", - "global_id": 19973, - "bbox": [ - 308.56, - 514.9, - 320.69, - 522.16 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p679-b30", - "global_id": 19974, - "bbox": [ - 324.27, - 498.79, - 516.13, - 510.89 - ], - "text": "x(kT)e−jn0k\n(6.56)", - "type": "text" - }, - { - "block_id": "p679-b31", - "global_id": 19975, - "bbox": [ - 127.59, - 529.34, - 411.86, - 544.42 - ], - "text": "Since 0N0 = 2π, we know that ejn0(k+N0) = ejn0k, and it follows that", - "type": "text" - }, - { - "block_id": "p679-b32", - "global_id": 19976, - "bbox": [ - 297.96, - 554.78, - 345.26, - 567.74 - ], - "text": "Dn+N0 = Dn", - "type": "text" - }, - { - "block_id": "p679-b33", - "global_id": 19977, - "bbox": [ - 127.59, - 576.59, - 516.13, - 610.88 - ], - "text": "The periodicity property Dn+N0 = Dn means that beyond n = N0/2, the coefficients represent the\nvalues for negative n. For instance, when N0 = 32, D17 = D−15, D18 = D−14,. . .,D31 = D−1. The\ncycle repeats again from n = 32 on.", - "type": "text" - }, - { - "block_id": "p679-b34", - "global_id": 19978, - "bbox": [ - 127.59, - 612.76, - 516.12, - 634.79 - ], - "text": "We can use the efficient FFT (the fast Fourier transform discussed in Sec. 8.6) to compute\nthe right-hand side of Eq. (6.56). We shall use MATLAB to implement the FFT algorithm. For", - "type": "text" - } - ] - }, - { - "page_num": 680, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p680-b0", - "global_id": 19979, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "660\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p680-b1", - "global_id": 19980, - "bbox": [ - 101.84, - 85.46, - 490.39, - 109.29 - ], - "text": "this purpose, we need samples of x(t) over one period starting at t = 0. In this algorithm, it is also\npreferable (although not necessary) that N0 be a power of 2, (i.e., N0 = 2m where m is an integer).", - "type": "text" - }, - { - "block_id": "p680-b2", - "global_id": 19981, - "bbox": [ - 76.77, - 124.33, - 414.73, - 136.29 - ], - "text": "EXAMPLE 6.16\nNumerical Computation of Fourier Spectra", - "type": "text" - }, - { - "block_id": "p680-b3", - "global_id": 19982, - "bbox": [ - 103.16, - 152.95, - 477.02, - 174.87 - ], - "text": "Numerically compute and then plot the exponential Fourier spectra for the periodic signal in\nFig. 6.2a (Ex. 6.1).", - "type": "text" - }, - { - "block_id": "p680-b4", - "global_id": 19983, - "bbox": [ - 103.16, - 197.37, - 477.01, - 280.56 - ], - "text": "The samples of x(t) start at t = 0 and the last (N0th) sample is at t = T0 −T. At the points of\ndiscontinuity, the sample value is taken as the average of the values of the function on two sides\nof the discontinuity. Thus, the sample at t = 0 is not 1 but (e−π/2 +1)/2 = 0.604. To determine\nN0, we require that Dn for n ≥N0/2 be negligible. Because x(t) has a jump discontinuity, Dn\ndecays rather slowly as 1/n. Hence, a choice of N0 = 200 is acceptable because the (N0/2)nd\n(100th) harmonic is about 1% of the fundamental. However, we also require N0 to be a power\nof 2. Hence, we shall take N0 = 256 = 28.", - "type": "text" - }, - { - "block_id": "p680-b5", - "global_id": 19984, - "bbox": [ - 121.09, - 281.47, - 289.57, - 291.43 - ], - "text": "First, the basic parameters are established.", - "type": "text" - }, - { - "block_id": "p680-b6", - "global_id": 19985, - "bbox": [ - 103.17, - 301.68, - 411.71, - 323.6 - ], - "text": ">>\nT_0 = pi; N_0 = 256; T = T_0/N_0; t = (0:T:T*(N_0-1))’;\n>>\nx = exp(-t/2); x(1) = (exp(-pi/2)+1)/2;", - "type": "text" - }, - { - "block_id": "p680-b7", - "global_id": 19986, - "bbox": [ - 103.17, - 333.27, - 477.01, - 367.15 - ], - "text": "Next, the DFT, computed by means of the fft function, is used to approximate the exponential\nFourier spectra up to n = N0/2. To facilitate comparison with previous plots of Dn, we only\nplot the results over −5 ≤n ≤5.", - "type": "text" - }, - { - "block_id": "p680-b8", - "global_id": 19987, - "bbox": [ - 103.16, - 377.4, - 432.68, - 435.18 - ], - "text": ">>\nD_n = fft(x)/N_0; n = [-N_0/2:N_0/2-1]’;\n>>\nclf; subplot(1,2,1); stem(n,abs(fftshift(D_n)),’.k’);\n>>\naxis([-5 5 0 .6]); xlabel(’n’); ylabel(’|D_n|’);\n>>\nsubplot(1,2,2); stem(n,angle(fftshift(D_n)),’.k’);\n>>\naxis([-5 5 -2 2]); xlabel(’n’); ylabel(’\\angle D_n [rad]’);", - "type": "text" - }, - { - "block_id": "p680-b9", - "global_id": 19988, - "bbox": [ - 103.16, - 444.85, - 477.01, - 466.78 - ], - "text": "As shown in Fig. 6.28, the resulting approximation is visually indistinguishable from the true\nFourier series spectra shown in Fig. 6.12 or Fig. 6.13.", - "type": "text" - }, - { - "block_id": "p680-b10", - "global_id": 19989, - "bbox": [ - 203.71, - 588.62, - 208.11, - 597.42 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p680-b11", - "global_id": 19990, - "bbox": [ - 125.12, - 567.01, - 129.12, - 575.01 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p680-b12", - "global_id": 19991, - "bbox": [ - 118.36, - 543.0, - 129.04, - 551.0 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p680-b13", - "global_id": 19992, - "bbox": [ - 118.36, - 519.01, - 129.04, - 527.01 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p680-b14", - "global_id": 19993, - "bbox": [ - 118.36, - 495.0, - 129.04, - 503.0 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p680-b15", - "global_id": 19994, - "bbox": [ - 103.86, - 529.26, - 113.39, - 541.24 - ], - "text": "|Dn|", - "type": "text" - }, - { - "block_id": "p680-b16", - "global_id": 19995, - "bbox": [ - 129.36, - 576.23, - 469.39, - 597.42 - ], - "text": "–5\n–5\n0\n5\n0\n5\nn", - "type": "text" - }, - { - "block_id": "p680-b17", - "global_id": 19996, - "bbox": [ - 311.61, - 567.01, - 319.61, - 575.01 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p680-b18", - "global_id": 19997, - "bbox": [ - 315.61, - 531.01, - 319.61, - 539.01 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p680-b19", - "global_id": 19998, - "bbox": [ - 315.61, - 495.0, - 319.61, - 503.0 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p680-b20", - "global_id": 19999, - "bbox": [ - 298.86, - 519.07, - 308.39, - 546.49 - ], - "text": "Dn [rad]", - "type": "text" - }, - { - "block_id": "p680-b21", - "global_id": 20000, - "bbox": [ - 103.16, - 604.1, - 433.7, - 613.34 - ], - "text": "Figure 6.28 Numerical approximation of exponential Fourier series spectra using the DFT.", - "type": "text" - } - ] - }, - { - "page_num": 681, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p681-b0", - "global_id": 20001, - "bbox": [ - 302.52, - 62.89, - 516.12, - 71.98 - ], - "text": "6.7\nMATLAB: Fourier Series Applications\n661", - "type": "text" - }, - { - "block_id": "p681-b1", - "global_id": 20002, - "bbox": [ - 127.94, - 94.37, - 432.83, - 108.31 - ], - "text": "6.7 MATLAB: FOURIER SERIES APPLICATIONS", - "type": "text" - }, - { - "block_id": "p681-b2", - "global_id": 20003, - "bbox": [ - 127.59, - 114.3, - 516.14, - 148.18 - ], - "text": "Computational packages such as MATLAB simplify the Fourier-based analysis, design, and\nsynthesis of periodic signals. MATLAB permits rapid and sophisticated calculations, which\npromote practical application and intuitive understanding of the Fourier series.", - "type": "text" - }, - { - "block_id": "p681-b3", - "global_id": 20004, - "bbox": [ - 127.59, - 173.27, - 415.98, - 185.23 - ], - "text": "6.7-1 Periodic Functions and the Gibbs Phenomenon", - "type": "text" - }, - { - "block_id": "p681-b4", - "global_id": 20005, - "bbox": [ - 127.59, - 190.94, - 516.11, - 213.27 - ], - "text": "It is sufficient to define any T0-periodic function over the interval (0 ≤t < T0). For example,\nconsider the 2π-periodic function given by", - "type": "text" - }, - { - "block_id": "p681-b5", - "global_id": 20006, - "bbox": [ - 250.01, - 242.75, - 274.66, - 253.02 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p681-b6", - "global_id": 20007, - "bbox": [ - 276.71, - 216.35, - 284.6, - 241.25 - ], - "text": "⎧\n⎪⎪⎨", - "type": "text" - }, - { - "block_id": "p681-b7", - "global_id": 20008, - "bbox": [ - 276.71, - 249.23, - 284.6, - 265.17 - ], - "text": "⎪⎪⎩", - "type": "text" - }, - { - "block_id": "p681-b8", - "global_id": 20009, - "bbox": [ - 284.6, - 224.72, - 386.54, - 270.96 - ], - "text": "t/A\n0 ≤t < A\n1\nA ≤t < π\n0\nπ ≤t < 2π\nx(t + 2π)\notherwise", - "type": "text" - }, - { - "block_id": "p681-b9", - "global_id": 20010, - "bbox": [ - 127.6, - 282.31, - 516.15, - 304.64 - ], - "text": "Although similar to a square wave, x(t) has a linearly rising edge of width A, where (0 < A < π).\nAs A →0, x(t) approaches a square wave; as A →π, x(t) approaches a type of sawtooth wave.", - "type": "text" - }, - { - "block_id": "p681-b10", - "global_id": 20011, - "bbox": [ - 145.53, - 306.23, - 462.98, - 316.89 - ], - "text": "In MATLAB, the mod command helps represent periodic functions such as x(t).", - "type": "text" - }, - { - "block_id": "p681-b11", - "global_id": 20012, - "bbox": [ - 127.59, - 326.64, - 474.42, - 334.62 - ], - "text": ">>\nx = @(t,A) mod(t,2*pi)/A.*(mod(t,2*pi)=A)&(mod(t,2*pi)>\nA = pi/2; [x_20,t] = CH6MP1(A,20);\n>>\nplot(t,x_20,’k’,t,x(t,A),’k:’); axis([-pi/4,2*pi+pi/4,-0.1,1.1]);\n>>\nxlabel(’t’); ylabel(’x_{20}(t)’);", - "type": "text" - }, - { - "block_id": "p682-b6", - "global_id": 20039, - "bbox": [ - 101.85, - 303.58, - 490.37, - 325.49 - ], - "text": "As expected, the falling edge is accompanied by the overshoot that is characteristic of the Gibbs\nphenomenon.", - "type": "text" - }, - { - "block_id": "p682-b7", - "global_id": 20040, - "bbox": [ - 101.85, - 327.38, - 490.39, - 349.41 - ], - "text": "Increasing N to 100, as shown in Fig. 6.30, improves the approximation but does not reduce\nthe overshoot.", - "type": "text" - }, - { - "block_id": "p682-b8", - "global_id": 20041, - "bbox": [ - 101.85, - 360.47, - 467.98, - 394.35 - ], - "text": ">>\n[x_100,t] = CH6MP1(A,100);\n>>\nplot(t,x_100,’k’,t,x(t,A),’k:’); axis([-pi/4,2*pi+pi/4,-0.1,1.1]);\n>>\nxlabel(’t’); ylabel(’x_{100}(t)’);", - "type": "text" - }, - { - "block_id": "p682-b9", - "global_id": 20042, - "bbox": [ - 101.84, - 416.37, - 490.4, - 462.61 - ], - "text": "Reducing A to π/64 produces a curious result. For N = 20, both the rising and falling edges\nare accompanied by roughly 9% of overshoot, as shown in Fig. 6.31. As the number of terms is\nincreased, overshoot persists only in the vicinity of jump discontinuities. For xN(t), increasing N\ndecreases the overshoot near the rising edge but not near the falling edge. Remember that it is a", - "type": "text" - }, - { - "block_id": "p682-b10", - "global_id": 20043, - "bbox": [ - 187.95, - 592.19, - 490.05, - 613.28 - ], - "text": "0\n1\n2\n3\n4\n5\n6\n7\nt", - "type": "text" - }, - { - "block_id": "p682-b11", - "global_id": 20044, - "bbox": [ - 147.71, - 574.15, - 151.71, - 582.15 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p682-b12", - "global_id": 20045, - "bbox": [ - 140.96, - 556.53, - 151.64, - 564.53 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p682-b13", - "global_id": 20046, - "bbox": [ - 140.96, - 538.9, - 151.64, - 546.9 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p682-b14", - "global_id": 20047, - "bbox": [ - 140.96, - 521.28, - 151.64, - 529.28 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p682-b15", - "global_id": 20048, - "bbox": [ - 140.96, - 503.65, - 151.64, - 511.65 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p682-b16", - "global_id": 20049, - "bbox": [ - 147.71, - 486.03, - 151.71, - 494.03 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p682-b17", - "global_id": 20050, - "bbox": [ - 126.45, - 523.39, - 137.73, - 542.19 - ], - "text": "x20(t)", - "type": "text" - }, - { - "block_id": "p682-b18", - "global_id": 20051, - "bbox": [ - 125.76, - 619.58, - 339.53, - 629.66 - ], - "text": "Figure 6.29 Comparison of x20(t) and x(t) when A = π/2.", - "type": "text" - } - ] - }, - { - "page_num": 683, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p683-b0", - "global_id": 20052, - "bbox": [ - 302.52, - 62.89, - 516.12, - 71.98 - ], - "text": "6.7\nMATLAB: Fourier Series Applications\n663", - "type": "text" - }, - { - "block_id": "p683-b1", - "global_id": 20053, - "bbox": [ - 213.13, - 199.67, - 515.52, - 220.75 - ], - "text": "0\n1\n2\n3\n4\n5\n6\n7\nt", - "type": "text" - }, - { - "block_id": "p683-b2", - "global_id": 20054, - "bbox": [ - 173.45, - 181.62, - 177.45, - 189.62 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p683-b3", - "global_id": 20055, - "bbox": [ - 166.7, - 164.0, - 177.38, - 172.0 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p683-b4", - "global_id": 20056, - "bbox": [ - 166.7, - 146.37, - 177.38, - 154.37 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p683-b5", - "global_id": 20057, - "bbox": [ - 166.7, - 128.76, - 177.38, - 136.76 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p683-b6", - "global_id": 20058, - "bbox": [ - 166.7, - 111.13, - 177.38, - 119.13 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p683-b7", - "global_id": 20059, - "bbox": [ - 173.45, - 93.5, - 177.45, - 101.5 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p683-b8", - "global_id": 20060, - "bbox": [ - 152.19, - 129.36, - 163.94, - 151.92 - ], - "text": "x100 (t)", - "type": "text" - }, - { - "block_id": "p683-b9", - "global_id": 20061, - "bbox": [ - 151.5, - 227.07, - 368.51, - 237.15 - ], - "text": "Figure 6.30 Comparison of x100(t) and x(t) when A = π/2.", - "type": "text" - }, - { - "block_id": "p683-b10", - "global_id": 20062, - "bbox": [ - 213.13, - 367.38, - 514.67, - 388.47 - ], - "text": "0\n1\n2\n3\n4\n5\n6\n7\nt", - "type": "text" - }, - { - "block_id": "p683-b11", - "global_id": 20063, - "bbox": [ - 173.45, - 349.33, - 177.45, - 357.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p683-b12", - "global_id": 20064, - "bbox": [ - 166.7, - 331.72, - 177.38, - 339.72 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p683-b13", - "global_id": 20065, - "bbox": [ - 166.7, - 314.08, - 177.38, - 322.08 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p683-b14", - "global_id": 20066, - "bbox": [ - 166.7, - 296.47, - 177.38, - 304.47 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p683-b15", - "global_id": 20067, - "bbox": [ - 166.7, - 278.84, - 177.38, - 286.84 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p683-b16", - "global_id": 20068, - "bbox": [ - 173.45, - 261.22, - 177.45, - 269.22 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p683-b17", - "global_id": 20069, - "bbox": [ - 152.19, - 298.58, - 163.47, - 317.38 - ], - "text": "x20(t)", - "type": "text" - }, - { - "block_id": "p683-b18", - "global_id": 20070, - "bbox": [ - 151.5, - 394.78, - 369.75, - 404.86 - ], - "text": "Figure 6.31 Comparison of x20(t) and x(t) when A = π/64.", - "type": "text" - }, - { - "block_id": "p683-b19", - "global_id": 20071, - "bbox": [ - 213.13, - 535.1, - 515.52, - 556.18 - ], - "text": "0\n1\n2\n3\n4\n5\n6\n7\nt", - "type": "text" - }, - { - "block_id": "p683-b20", - "global_id": 20072, - "bbox": [ - 173.45, - 517.05, - 177.45, - 525.05 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p683-b21", - "global_id": 20073, - "bbox": [ - 166.7, - 499.43, - 177.38, - 507.43 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p683-b22", - "global_id": 20074, - "bbox": [ - 166.7, - 481.8, - 177.38, - 489.8 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p683-b23", - "global_id": 20075, - "bbox": [ - 166.7, - 464.19, - 177.38, - 472.19 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p683-b24", - "global_id": 20076, - "bbox": [ - 166.7, - 446.56, - 177.38, - 454.56 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p683-b25", - "global_id": 20077, - "bbox": [ - 173.45, - 428.93, - 177.45, - 436.93 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p683-b26", - "global_id": 20078, - "bbox": [ - 152.19, - 464.79, - 163.94, - 487.35 - ], - "text": "x100 (t)", - "type": "text" - }, - { - "block_id": "p683-b27", - "global_id": 20079, - "bbox": [ - 151.5, - 562.49, - 372.99, - 572.57 - ], - "text": "Figure 6.32 Comparison of x100(t) and x(t) when A = π/64.", - "type": "text" - }, - { - "block_id": "p683-b28", - "global_id": 20080, - "bbox": [ - 127.59, - 588.96, - 516.15, - 634.79 - ], - "text": "true jump discontinuity that causes the Gibbs phenomenon. A continuous signal, no matter how\nsharply it rises, can always be represented by a Fourier series at every point within any small\nerror by increasing N. This is not the case when a true jump discontinuity is present. Figure 6.32\nillustrates this behavior using N = 100.", - "type": "text" - } - ] - }, - { - "page_num": 684, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p684-b0", - "global_id": 20081, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "664\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p684-b1", - "global_id": 20082, - "bbox": [ - 101.84, - 86.52, - 307.86, - 98.48 - ], - "text": "6.7-2 Optimization and Phase Spectra", - "type": "text" - }, - { - "block_id": "p684-b2", - "global_id": 20083, - "bbox": [ - 101.84, - 104.61, - 490.4, - 186.3 - ], - "text": "Although magnitude spectra typically receive the most attention, phase spectra are critically\nimportant in some applications. Consider the problem of characterizing the frequency response\nof an unknown system. By applying sinusoids one at a time, the frequency response is empirically\nmeasured one point at a time. This process is tedious at best. Applying a superposition of many\nsinusoids, however, allows simultaneous measurement of many points of the frequency response.\nSuch measurements can be taken by a spectrum analyzer equipped with a transfer function mode\nor by applying Fourier analysis techniques, which are discussed in later chapters.", - "type": "text" - }, - { - "block_id": "p684-b3", - "global_id": 20084, - "bbox": [ - 119.78, - 187.89, - 439.31, - 198.26 - ], - "text": "A multitone test signal m(t) is constructed as a superposition of N real sinusoids", - "type": "text" - }, - { - "block_id": "p684-b4", - "global_id": 20085, - "bbox": [ - 239.84, - 218.74, - 267.23, - 229.01 - ], - "text": "m(t) =", - "type": "text" - }, - { - "block_id": "p684-b5", - "global_id": 20086, - "bbox": [ - 269.28, - 208.78, - 283.38, - 219.23 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p684-b6", - "global_id": 20087, - "bbox": [ - 270.13, - 233.12, - 282.53, - 240.39 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p684-b7", - "global_id": 20088, - "bbox": [ - 284.48, - 218.74, - 352.39, - 230.61 - ], - "text": "Mn cos(ωnt + θn)", - "type": "text" - }, - { - "block_id": "p684-b8", - "global_id": 20089, - "bbox": [ - 101.84, - 249.08, - 490.4, - 295.33 - ], - "text": "where Mn and θn establish the relative magnitude and phase of each sinusoidal component. It is\nsensible to constrain all gains to be equal, Mn = M for all n. This ensures equal treatment at each\npoint of the measured frequency response. Although the value M is normally chosen to set the\ndesired signal power, we set M = 1 for convenience.", - "type": "text" - }, - { - "block_id": "p684-b9", - "global_id": 20090, - "bbox": [ - 101.85, - 297.32, - 490.4, - 319.24 - ], - "text": "While not required, it is also sensible to space the sinusoidal components uniformly in\nfrequency.", - "type": "text" - }, - { - "block_id": "p684-b10", - "global_id": 20091, - "bbox": [ - 244.04, - 330.34, - 271.44, - 340.61 - ], - "text": "m(t) =", - "type": "text" - }, - { - "block_id": "p684-b11", - "global_id": 20092, - "bbox": [ - 273.48, - 320.38, - 287.58, - 330.83 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p684-b12", - "global_id": 20093, - "bbox": [ - 274.33, - 344.72, - 286.74, - 351.99 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p684-b13", - "global_id": 20094, - "bbox": [ - 288.69, - 330.34, - 490.38, - 341.49 - ], - "text": "cos(nω0t + θn)\n(6.58)", - "type": "text" - }, - { - "block_id": "p684-b14", - "global_id": 20095, - "bbox": [ - 101.84, - 358.41, - 490.43, - 380.32 - ], - "text": "Another sensible alternative, which spaces components logarithmically in frequency, is treated in\nProb. 6.7-4.", - "type": "text" - }, - { - "block_id": "p684-b15", - "global_id": 20096, - "bbox": [ - 101.84, - 382.31, - 490.4, - 428.84 - ], - "text": "Equation (6.58) is now a truncated compact-form Fourier series with a flat magnitude\nspectrum. Frequency resolution and range are set by ω0 and N, respectively. For example, a 2 kHz\nrange with a resolution of 100 Hz requires ω0 = 2π100 and N = 20. The only remaining unknowns\nare the θn.", - "type": "text" - }, - { - "block_id": "p684-b16", - "global_id": 20097, - "bbox": [ - 101.84, - 429.72, - 490.38, - 464.0 - ], - "text": "While it is tempting to set θn = 0 for all n, the results are quite unsatisfactory. MATLAB helps\ndemonstrate the problem by using ω0 = 2π100 and N = 20 sinusoids, each with a peak-to-peak\nvoltage of 1 volt.", - "type": "text" - }, - { - "block_id": "p684-b17", - "global_id": 20098, - "bbox": [ - 101.84, - 473.97, - 499.35, - 519.8 - ], - "text": ">>\nm = @(theta,t,omega) sum(cos(omega*t+theta*ones(size(t))));\n>>\nN = 20; omega = 2*pi*100*[1:N]’; theta = zeros(size(omega));\n>>\nt = linspace(-0.01,0.01,10000);\n>>\nplot(t,m(theta,t,omega),’k’); xlabel(’t [sec]’); ylabel(’m(t) [volts]’);", - "type": "text" - }, - { - "block_id": "p684-b18", - "global_id": 20099, - "bbox": [ - 101.84, - 528.77, - 490.4, - 563.06 - ], - "text": "As shown in Fig. 6.33, θn = 0 causes each sinusoid to constructively add. The resulting 20 volt\npeak can saturate system components, such as operational amplifiers operating with ±12 volt rails.\nTo improve signal performance, the maximum amplitude of m(t) over t needs to be reduced.", - "type": "text" - }, - { - "block_id": "p684-b19", - "global_id": 20100, - "bbox": [ - 101.84, - 564.63, - 490.38, - 634.79 - ], - "text": "One way to reduce maxt (|m(t)|) is to reduce M, the strength of each component.\nUnfortunately, this approach reduces the system’s signal-to-noise ratio and ultimately degrades\nmeasurement quality. Therefore, reducing M is not a smart decision. The phases θn, however, can\nbe adjusted to reduce maxt (|m(t)|) while preserving signal power. In fact, since θn = 0 maximizes\nmaxt (|m(t)|), just about any other choice of θn will improve the situation. Even a random choice\nshould improve performance.", - "type": "text" - } - ] - }, - { - "page_num": 685, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p685-b0", - "global_id": 20101, - "bbox": [ - 302.52, - 62.89, - 516.12, - 71.98 - ], - "text": "6.7\nMATLAB: Fourier Series Applications\n665", - "type": "text" - }, - { - "block_id": "p685-b1", - "global_id": 20102, - "bbox": [ - 143.08, - 202.67, - 493.94, - 223.76 - ], - "text": "–0.01\n–0.008\n–0.006\n–0.004\n–0.002\n0\n0.002\n0.004\n0.006\n0.008\n0.01\nt [sec]", - "type": "text" - }, - { - "block_id": "p685-b2", - "global_id": 20103, - "bbox": [ - 140.83, - 193.44, - 148.83, - 201.44 - ], - "text": "–5", - "type": "text" - }, - { - "block_id": "p685-b3", - "global_id": 20104, - "bbox": [ - 144.83, - 172.3, - 148.83, - 180.3 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p685-b4", - "global_id": 20105, - "bbox": [ - 144.83, - 151.15, - 148.83, - 159.15 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p685-b5", - "global_id": 20106, - "bbox": [ - 140.34, - 130.0, - 148.34, - 138.0 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p685-b6", - "global_id": 20107, - "bbox": [ - 140.34, - 108.84, - 148.34, - 116.84 - ], - "text": "15", - "type": "text" - }, - { - "block_id": "p685-b7", - "global_id": 20108, - "bbox": [ - 140.34, - 87.7, - 148.34, - 95.7 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p685-b8", - "global_id": 20109, - "bbox": [ - 128.36, - 123.97, - 137.16, - 164.3 - ], - "text": "m(t) [volts]", - "type": "text" - }, - { - "block_id": "p685-b9", - "global_id": 20110, - "bbox": [ - 127.59, - 230.06, - 279.7, - 240.37 - ], - "text": "Figure 6.33 Test signal m(t) with θn = 0.", - "type": "text" - }, - { - "block_id": "p685-b10", - "global_id": 20111, - "bbox": [ - 127.59, - 260.91, - 516.15, - 366.52 - ], - "text": "As with any computer, MATLAB cannot generate truly random numbers. Rather, it generates\npseudo-random numbers. Pseudo-random numbers are deterministic sequences that appear to be\nrandom. The particular sequence of numbers that is realized depends entirely on the initial state of\nthe pseudo-random number generator. Setting the generator’s initial state to a known value allows\na “random” experiment with reproducible results. The command rng(0) initializes the state of\nthe pseudo-random number generator to a known condition of zero, and the MATLAB command\nrand(a,b) generates an a-by-b matrix of pseudo-random numbers that are uniformly distributed\nover the interval (0,1). Radian phases occupy the wider interval (0,2π), so the results from rand\nneed to be appropriately scaled.", - "type": "text" - }, - { - "block_id": "p685-b11", - "global_id": 20112, - "bbox": [ - 127.59, - 377.5, - 342.03, - 387.46 - ], - "text": ">>\nrng(0); theta_rand0 = 2*pi*rand(N,1);", - "type": "text" - }, - { - "block_id": "p685-b12", - "global_id": 20113, - "bbox": [ - 145.52, - 397.45, - 403.75, - 408.53 - ], - "text": "Next, we recompute and plot m(t) using the randomly chosen θn.", - "type": "text" - }, - { - "block_id": "p685-b13", - "global_id": 20114, - "bbox": [ - 127.59, - 418.81, - 430.95, - 464.63 - ], - "text": ">>\nm_rand0 = m(theta_rand0,t,omega);\n>>\nplot(t,m_rand0,’k’); axis([-0.01,0.01,-10,10]);\n>>\nxlabel(’t [sec]’); ylabel(’m(t) [volts]’);\n>>\nset(gca,’ytick’,[min(m_rand0),max(m_rand0)]); grid on;", - "type": "text" - }, - { - "block_id": "p685-b14", - "global_id": 20115, - "bbox": [ - 127.59, - 475.04, - 516.14, - 521.95 - ], - "text": "For a vector input, the min and max commands return the minimum and maximum values of the\nvector. Using these values to set y axis tick marks makes it easy to identify the extreme values of\nthe m(t). As seen from Fig. 6.34, the maximum amplitude is now 7.6307, which is significantly\nsmaller than the maximum of 20 when θn = 0.", - "type": "text" - }, - { - "block_id": "p685-b15", - "global_id": 20116, - "bbox": [ - 127.59, - 522.86, - 516.13, - 616.51 - ], - "text": "Randomly chosen phases suffer a fatal fault: there is little guarantee of optimal performance.\nFor example, repeating the experiment with rng(5) produces a maximum magnitude of 8.2399\nvolts, as shown in Fig. 6.35. This value is significantly higher than the previous maximum of\n7.6307 volts. Clearly, it is better to replace a random solution with an optimal solution.\nWhat constitutes “optimal”? Many choices exist, but desired signal criteria naturally suggest\nthat optimal phases minimize the maximum magnitude of m(t) over all t. To find these optimal\nphases, MATLAB’s fminsearch command is useful. First, the function to be minimized, called\nthe objective function, is defined.", - "type": "text" - }, - { - "block_id": "p685-b16", - "global_id": 20117, - "bbox": [ - 127.59, - 626.95, - 461.78, - 634.92 - ], - "text": ">>\nmaxmagm = @(theta,t,omega) max(abs(sum(cos(omega*t+theta*ones(size(t))))));", - "type": "text" - } - ] - }, - { - "page_num": 686, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p686-b0", - "global_id": 20118, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "666\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p686-b1", - "global_id": 20119, - "bbox": [ - 136.93, - 199.67, - 486.93, - 220.75 - ], - "text": "–0.01\n–0.008\n–0.006\n–0.004\n–0.002\n0\n0.002\n0.004\n0.006\n0.008\n0.01\nt [sec]", - "type": "text" - }, - { - "block_id": "p686-b2", - "global_id": 20120, - "bbox": [ - 114.58, - 176.94, - 140.58, - 184.94 - ], - "text": "–7.4460", - "type": "text" - }, - { - "block_id": "p686-b3", - "global_id": 20121, - "bbox": [ - 117.59, - 97.22, - 139.59, - 105.22 - ], - "text": "7.6307", - "type": "text" - }, - { - "block_id": "p686-b4", - "global_id": 20122, - "bbox": [ - 102.61, - 120.96, - 111.41, - 161.29 - ], - "text": "m(t) [volts]", - "type": "text" - }, - { - "block_id": "p686-b5", - "global_id": 20123, - "bbox": [ - 101.84, - 227.07, - 354.95, - 237.75 - ], - "text": "Figure 6.34 Test signal m(t) with random θn found by using rng(0).", - "type": "text" - }, - { - "block_id": "p686-b6", - "global_id": 20124, - "bbox": [ - 136.07, - 370.19, - 486.93, - 391.27 - ], - "text": "–0.01\n–0.008\n–0.006\n–0.004\n–0.002\n0\n0.002\n0.004\n0.006\n0.008\n0.01\nt [sec]", - "type": "text" - }, - { - "block_id": "p686-b7", - "global_id": 20125, - "bbox": [ - 113.59, - 351.66, - 139.59, - 359.66 - ], - "text": "–8.2399", - "type": "text" - }, - { - "block_id": "p686-b8", - "global_id": 20126, - "bbox": [ - 117.59, - 266.7, - 139.59, - 274.7 - ], - "text": "7.8268", - "type": "text" - }, - { - "block_id": "p686-b9", - "global_id": 20127, - "bbox": [ - 102.61, - 291.47, - 111.41, - 331.8 - ], - "text": "m(t) [volts]", - "type": "text" - }, - { - "block_id": "p686-b10", - "global_id": 20128, - "bbox": [ - 101.84, - 397.58, - 397.31, - 408.26 - ], - "text": "Figure 6.35 Test signal m(t) with random θn found by using rand(’state’,1).", - "type": "text" - }, - { - "block_id": "p686-b11", - "global_id": 20129, - "bbox": [ - 101.84, - 427.27, - 490.39, - 461.43 - ], - "text": "The anonymous function argument order is important; fminsearch uses the first input argument\nas the variable of minimization. To minimize over θ, as desired, θ must be the first argument of\nthe objective function maxmagm.", - "type": "text" - }, - { - "block_id": "p686-b12", - "global_id": 20130, - "bbox": [ - 119.78, - 462.72, - 394.4, - 473.09 - ], - "text": "Next, the time vector is shortened to include only one period of m(t).", - "type": "text" - }, - { - "block_id": "p686-b13", - "global_id": 20131, - "bbox": [ - 101.84, - 482.92, - 253.52, - 492.89 - ], - "text": ">>\nt = linspace(0,0.01,401);", - "type": "text" - }, - { - "block_id": "p686-b14", - "global_id": 20132, - "bbox": [ - 101.84, - 501.72, - 490.38, - 524.05 - ], - "text": "A full period ensures that all values of m(t) are considered; the short length of t helps ensure that\nfunctions execute quickly. An initial value of θ is randomly chosen to begin the search.", - "type": "text" - }, - { - "block_id": "p686-b15", - "global_id": 20133, - "bbox": [ - 101.84, - 533.88, - 405.2, - 555.8 - ], - "text": ">>\nrng(0); theta_init = 2*pi*rand(N,1);\n>>\ntheta_opt = fminsearch(maxmagm,theta_init,[],t,omega);", - "type": "text" - }, - { - "block_id": "p686-b16", - "global_id": 20134, - "bbox": [ - 101.84, - 564.63, - 490.39, - 635.08 - ], - "text": "Notice that fminsearch finds the minimizer to maxmagm over θ by using an initial value\ntheta_init. Most numerical minimization techniques are capable of finding only local minima,\nand fminsearch is no exception. As a result, fminsearch does not always produce a unique\nsolution. The empty square brackets indicate no special options are requested, and the remaining\nordered arguments are secondary inputs for the objective function. Full format details for\nfminsearch are available from MATLAB’s help facilities.", - "type": "text" - } - ] - }, - { - "page_num": 687, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p687-b0", - "global_id": 20135, - "bbox": [ - 414.16, - 62.89, - 516.14, - 71.98 - ], - "text": "6.8\nSummary\n667", - "type": "text" - }, - { - "block_id": "p687-b1", - "global_id": 20136, - "bbox": [ - 161.26, - 199.67, - 512.68, - 220.75 - ], - "text": "–0.01\n–0.008\n–0.006\n–0.004\n–0.002\n0\n0.002\n0.004\n0.006\n0.008\n0.01\nt [sec]", - "type": "text" - }, - { - "block_id": "p687-b2", - "global_id": 20137, - "bbox": [ - 140.33, - 165.92, - 166.33, - 173.92 - ], - "text": "–5.3632", - "type": "text" - }, - { - "block_id": "p687-b3", - "global_id": 20138, - "bbox": [ - 143.34, - 109.32, - 165.34, - 117.32 - ], - "text": "5.3414", - "type": "text" - }, - { - "block_id": "p687-b4", - "global_id": 20139, - "bbox": [ - 128.36, - 120.96, - 137.16, - 161.29 - ], - "text": "m(t) [volts]", - "type": "text" - }, - { - "block_id": "p687-b5", - "global_id": 20140, - "bbox": [ - 127.59, - 227.07, - 318.87, - 236.68 - ], - "text": "Figure 6.36 Test signal m(t) with optimized phases.", - "type": "text" - }, - { - "block_id": "p687-b6", - "global_id": 20141, - "bbox": [ - 127.59, - 256.59, - 516.12, - 278.51 - ], - "text": "Figure 6.36 shows the phase-optimized test signal. The maximum magnitude is reduced to a\nvalue of 5.3632 volts, which is a significant improvement over the original peak of 20 volts.", - "type": "text" - }, - { - "block_id": "p687-b7", - "global_id": 20142, - "bbox": [ - 127.59, - 280.5, - 516.15, - 338.57 - ], - "text": "Although the signals shown in Figs. 6.33 through 6.36 look different, they all possess the\nsame magnitude spectra. The signals differ only in phase spectra. It is interesting to investigate\nthe similarities and differences of these signals in ways other than graphs and mathematics. For\nexample, is there an audible difference between the signals? For computers equipped with sound\ncapability, the MATLAB sound command can be used to find out.", - "type": "text" - }, - { - "block_id": "p687-b8", - "global_id": 20143, - "bbox": [ - 127.59, - 347.4, - 491.6, - 365.33 - ], - "text": ">>\nFs = 8000; t = [0:1/Fs:2];\n% Two second records at a sampling rate of 8kHz\n>>\nsound(m(theta,t,omega)/20,Fs);\n% Play (scaled) m(t) constructed using zero phases", - "type": "text" - }, - { - "block_id": "p687-b9", - "global_id": 20144, - "bbox": [ - 127.59, - 374.54, - 516.14, - 432.74 - ], - "text": "Since the sound command clips magnitudes that exceed 1, the input vector is scaled by 1/20 to\navoid clipping and the resulting sound distortion. The signals using other phase assignments are\ncreated and played in a similar fashion. How well does the human ear discern the differences in\nphase spectra? If you are like most people, you will not be able to discern any differences in how\nthese waveforms sound.", - "type": "text" - }, - { - "block_id": "p687-b10", - "global_id": 20145, - "bbox": [ - 127.94, - 461.42, - 220.22, - 475.37 - ], - "text": "6.8 SUMMARY", - "type": "text" - }, - { - "block_id": "p687-b11", - "global_id": 20146, - "bbox": [ - 127.59, - 481.36, - 516.13, - 539.15 - ], - "text": "In this chapter we showed how a periodic signal can be represented as a sum of sinusoids or\nexponentials. If the frequency of a periodic signal is f0, then it can be expressed as a weighted\nsum of a sinusoid of frequency f0 and its harmonics (the trigonometric Fourier series). We can\nreconstruct the periodic signal from a knowledge of the amplitudes and phases of these sinusoidal\ncomponents (amplitude and phase spectra).", - "type": "text" - }, - { - "block_id": "p687-b12", - "global_id": 20147, - "bbox": [ - 127.59, - 540.72, - 516.16, - 575.01 - ], - "text": "If a periodic signal x(t) has an even symmetry, its Fourier series contains only cosine terms\n(including dc). In contrast, if x(t) has an odd symmetry, its Fourier series contains only sine terms.\nIf x(t) has neither type of symmetry, its Fourier series contains both sine and cosine terms.", - "type": "text" - }, - { - "block_id": "p687-b13", - "global_id": 20148, - "bbox": [ - 127.59, - 576.59, - 516.14, - 634.79 - ], - "text": "At points of discontinuity, the Fourier series for x(t) converges to the mean of the values of x(t)\non either side of the discontinuity. For signals with discontinuities, the Fourier series converges in\nthe mean and exhibits Gibbs phenomenon at the points of discontinuity. The amplitude spectrum\nof the Fourier series for a periodic signal x(t) with jump discontinuities decays slowly (as 1/n)\nwith frequency. We need a large number of terms in the Fourier series to approximate x(t) within", - "type": "text" - } - ] - }, - { - "page_num": 688, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p688-b0", - "global_id": 20149, - "bbox": [ - 60.0, - 62.89, - 428.39, - 71.98 - ], - "text": "668\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p688-b1", - "global_id": 20150, - "bbox": [ - 101.84, - 85.82, - 490.41, - 119.69 - ], - "text": "a given error. In contrast, the amplitude spectrum of a smoother periodic signal decays faster with\nfrequency and we require a smaller number of terms in the series to approximate x(t) within a\ngiven error.", - "type": "text" - }, - { - "block_id": "p688-b2", - "global_id": 20151, - "bbox": [ - 101.84, - 121.68, - 490.43, - 227.29 - ], - "text": "A sinusoid can be expressed in terms of exponentials. Therefore, the Fourier series of a\nperiodic signal can also be expressed as a sum of exponentials (the exponential Fourier series).\nThe exponential form of the Fourier series and the expressions for the series coefficients are more\ncompact than those of the trigonometric Fourier series. Also, the response of LTIC systems to an\nexponential input is much simpler than that for a sinusoidal input. Moreover, the exponential form\nof representation lends itself better to mathematical manipulations than does the trigonometric\nform. This includes the establishment of useful Fourier series properties that simplify work and\nhelp provide a more intuitive understanding of signals. For these reasons, the exponential form of\nthe series is preferred in modern practice in the areas of signals and systems.", - "type": "text" - }, - { - "block_id": "p688-b3", - "global_id": 20152, - "bbox": [ - 101.84, - 229.29, - 490.41, - 322.93 - ], - "text": "The plots of amplitudes and angles of various exponential components of the Fourier series as\nfunctions of the frequency are the exponential Fourier spectra (amplitude and angle spectra) of the\nsignal. Because a sinusoid cosω0t can be represented as a sum of two exponentials, ejω0t and e−jω0t,\nthe frequencies in the exponential spectra range from ω = −∞to ∞. By definition, frequency of\na signal is always a positive quantity. Presence of a spectral component of a negative frequency\n−nω0 merely indicates that the Fourier series contains terms of the form e−jnω0t. The spectra of\nthe trigonometric and exponential Fourier series are closely related, and one can be found by the\ninspection of the other.", - "type": "text" - }, - { - "block_id": "p688-b4", - "global_id": 20153, - "bbox": [ - 101.85, - 324.92, - 490.42, - 466.39 - ], - "text": "In Sec. 6.5 we discuss a method of representing signals by the generalized Fourier series, of\nwhich the trigonometric and exponential Fourier series are special cases. Signals are vectors in\nevery sense. Just as a vector can be represented as a sum of its components in a variety of ways,\ndepending on the choice of the coordinate system, a signal can be represented as a sum of its\ncomponents in a variety of ways, of which the trigonometric and exponential Fourier series are\nonly two examples. Just as we have vector coordinate systems formed by mutually orthogonal\nvectors, we also have signal coordinate systems (basis signals) formed by mutually orthogonal\nsignals. Any signal in this signal space can be represented as a sum of the basis signals. Each set\nof basis signals yields a particular Fourier series representation of the signal. The signal is equal to\nits Fourier series, not in the ordinary sense, but in the special sense that the energy of the difference\nbetween the signal and its Fourier series approaches zero. This allows for the signal to differ from\nits Fourier series at some isolated points.", - "type": "text" - }, - { - "block_id": "p688-b5", - "global_id": 20154, - "bbox": [ - 102.11, - 496.21, - 189.29, - 507.16 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p688-b6", - "global_id": 20155, - "bbox": [ - 101.84, - 514.23, - 366.33, - 523.29 - ], - "text": "1.\nBell, E. T. Men of Mathematics. Simon & Schuster, New York, 1937.", - "type": "text" - }, - { - "block_id": "p688-b7", - "global_id": 20156, - "bbox": [ - 101.84, - 528.17, - 490.39, - 548.19 - ], - "text": "2.\nDurant, W., and Durant, A. The Age of Napoleon, Part XI in The Story of Civilization Series. Simon &\nSchuster, New York, 1975.", - "type": "text" - }, - { - "block_id": "p688-b8", - "global_id": 20157, - "bbox": [ - 101.84, - 553.09, - 422.89, - 562.14 - ], - "text": "3.\nCalinger, R. Classics of Mathematics, 4th ed. Moore Publishing, Oak Park, IL, 1982.", - "type": "text" - }, - { - "block_id": "p688-b9", - "global_id": 20158, - "bbox": [ - 101.84, - 567.03, - 369.82, - 576.08 - ], - "text": "4.\nLanczos, C. Discourse on Fourier Series. Oliver Boyd, London, 1966.", - "type": "text" - }, - { - "block_id": "p688-b10", - "global_id": 20159, - "bbox": [ - 101.84, - 580.98, - 420.4, - 590.03 - ], - "text": "5.\nKörner, T. W. Fourier Analysis. Cambridge University Press, Cambridge, UK, 1989.", - "type": "text" - }, - { - "block_id": "p688-b11", - "global_id": 20160, - "bbox": [ - 101.84, - 594.93, - 395.66, - 603.98 - ], - "text": "6.\nGuillemin, E. A. Theory of Linear Physical Systems. Wiley, New York, 1963.", - "type": "text" - }, - { - "block_id": "p688-b12", - "global_id": 20161, - "bbox": [ - 101.84, - 608.88, - 290.95, - 617.93 - ], - "text": "7.\nGibbs, W. J. Nature, vol. 59, p. 606, April 1899.", - "type": "text" - }, - { - "block_id": "p688-b13", - "global_id": 20162, - "bbox": [ - 101.84, - 622.82, - 315.5, - 631.87 - ], - "text": "8.\nBôcher, M. Annals of Mathematics, vol. 7, no. 2, 1906.", - "type": "text" - } - ] - }, - { - "page_num": 689, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p689-b0", - "global_id": 20163, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n669", - "type": "text" - }, - { - "block_id": "p689-b1", - "global_id": 20164, - "bbox": [ - 127.59, - 85.81, - 499.77, - 94.86 - ], - "text": "9.\nCarslaw, H. S. Bulletin of the American Mathematical Society, vol. 31, pp. 420–424, October 1925.", - "type": "text" - }, - { - "block_id": "p689-b2", - "global_id": 20165, - "bbox": [ - 123.11, - 99.76, - 414.2, - 108.81 - ], - "text": "10.\nLathi, B. P. Signals, Systems, and Communication. Wiley, New York, 1965.", - "type": "text" - }, - { - "block_id": "p689-b3", - "global_id": 20166, - "bbox": [ - 123.11, - 113.7, - 434.85, - 122.75 - ], - "text": "11.\nWalker P. L. The Theory of Fourier Series and Integrals. Wiley, New York, 1986.", - "type": "text" - }, - { - "block_id": "p689-b4", - "global_id": 20167, - "bbox": [ - 123.11, - 127.65, - 516.12, - 147.67 - ], - "text": "12.\nChurchill, R. V., and Brown, J. W. Fourier Series and Boundary Value Problems, 3rd ed. McGraw-Hill,\nNew York, 1978.", - "type": "text" - }, - { - "block_id": "p689-b5", - "global_id": 20168, - "bbox": [ - 106.67, - 167.7, - 217.26, - 184.64 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p689-b6", - "global_id": 20169, - "bbox": [ - 86.99, - 195.01, - 288.98, - 247.89 - ], - "text": "6.1-1\nFor each of the periodic signals shown in\nFig. P6.1-1, find the compact trigonometric\nFourier series and sketch the amplitude and\nphase spectra. If either the sine or cosine terms\nare absent in the Fourier series, explain why.", - "type": "text" - }, - { - "block_id": "p689-b7", - "global_id": 20170, - "bbox": [ - 86.99, - 257.27, - 288.98, - 266.61 - ], - "text": "6.1-2\n(a) Find the trigonometric Fourier series for y(t)", - "type": "text" - }, - { - "block_id": "p689-b8", - "global_id": 20171, - "bbox": [ - 115.63, - 268.6, - 288.98, - 288.53 - ], - "text": "shown in Fig. P6.1-2.\n(b) The signal y(t) can be obtained by time", - "type": "text" - }, - { - "block_id": "p689-b9", - "global_id": 20172, - "bbox": [ - 116.13, - 290.15, - 288.99, - 354.28 - ], - "text": "reversal of x(t) shown in Fig. 6.2a. Use this\nfact to obtain the Fourier series for y(t) from\nthe results in Ex. 6.1. Verify that the Fourier\nseries thus obtained is identical to that found\nin part (a).\n(c) Show that, in general, time reversal of a", - "type": "text" - }, - { - "block_id": "p689-b10", - "global_id": 20173, - "bbox": [ - 131.07, - 356.28, - 288.99, - 387.16 - ], - "text": "periodic signal does not affect the amplitude\nspectrum, and the phase spectrum is also\nunchanged except for the change of sign.", - "type": "text" - }, - { - "block_id": "p689-b11", - "global_id": 20174, - "bbox": [ - 86.99, - 396.84, - 288.97, - 405.88 - ], - "text": "6.1-3\n(a) Find the trigonometric Fourier series for the", - "type": "text" - }, - { - "block_id": "p689-b12", - "global_id": 20175, - "bbox": [ - 115.63, - 407.5, - 288.97, - 427.81 - ], - "text": "periodic signal y(t) depicted in Fig. P6.1-3.\n(b) The signal y(t) can be obtained by time", - "type": "text" - }, - { - "block_id": "p689-b13", - "global_id": 20176, - "bbox": [ - 116.13, - 429.42, - 288.98, - 493.55 - ], - "text": "compression of x(t) shown in Fig. 6.2a by a\nfactor 2. Use this fact to obtain the Fourier\nseries for y(t) from the results in Ex. 6.1.\nVerify that the Fourier series thus obtained\nis identical to that found in part (a).\n(c) Show that, in general, time compression of", - "type": "text" - }, - { - "block_id": "p689-b14", - "global_id": 20177, - "bbox": [ - 131.06, - 495.46, - 288.98, - 592.18 - ], - "text": "a periodic signal by a factor a expands the\nFourier spectra along the ω axis by the\nsame factor a. In other words C0,Cn, and\nθn remain unchanged, but the fundamental\nfrequency is increased by the factor a,\nthus expanding the spectrum. Similarly, time\nexpansion of a periodic signal by a factor a\ncompresses its Fourier spectra along the ω\naxis by the factor a.", - "type": "text" - }, - { - "block_id": "p689-b15", - "global_id": 20178, - "bbox": [ - 87.0, - 601.87, - 288.97, - 610.91 - ], - "text": "6.1-4\n(a) Find the trigonometric Fourier series for the", - "type": "text" - }, - { - "block_id": "p689-b16", - "global_id": 20179, - "bbox": [ - 131.07, - 612.52, - 288.98, - 632.82 - ], - "text": "periodic signal g(t) in Fig. P6.1-4. Take\nadvantage of the symmetry.", - "type": "text" - }, - { - "block_id": "p689-b17", - "global_id": 20180, - "bbox": [ - 342.79, - 194.65, - 516.13, - 203.99 - ], - "text": "(b) Observe that g(t) is identical to x(t) in", - "type": "text" - }, - { - "block_id": "p689-b18", - "global_id": 20181, - "bbox": [ - 343.29, - 205.98, - 516.14, - 269.74 - ], - "text": "Fig. 6.4a left-shifted by 0.5 second. Use this\nfact to obtain the Fourier series for g(t) from\nthe results in Ex. 6.2. Verify that the Fourier\nseries thus obtained is identical to that found\nin part (a).\n(c) Show that, in general, a time shift of T sec-", - "type": "text" - }, - { - "block_id": "p689-b19", - "global_id": 20182, - "bbox": [ - 358.22, - 271.74, - 516.15, - 324.53 - ], - "text": "onds of a periodic signal does not affect the\namplitude spectrum. However, the phase of\nthe nth harmonic is increased or decreased\nnω0T depending on whether the signal is\nadvanced or delayed by T seconds.", - "type": "text" - }, - { - "block_id": "p689-b20", - "global_id": 20183, - "bbox": [ - 314.15, - 329.74, - 516.13, - 382.61 - ], - "text": "6.1-5\nDetermine the trigonometric Fourier series coef-\nficients an and bn for the following signals. In\neach case, also determine the signals’ funda-\nmental radian frequency ω0. No integration is\nrequired to solve this problem.", - "type": "text" - }, - { - "block_id": "p689-b21", - "global_id": 20184, - "bbox": [ - 342.78, - 384.23, - 417.66, - 405.27 - ], - "text": "(a) xa(t) = cos(3πt)\n(b) xb(t) = sin(7πt)", - "type": "text" - }, - { - "block_id": "p689-b22", - "global_id": 20185, - "bbox": [ - 342.78, - 406.15, - 503.28, - 427.19 - ], - "text": "(c) xc(t) = 2 + 4cos(3πt) −2jsin(7πt)\n(d) xd(t) = (1+j)sin(3πt) + (2−j)cos(7πt)", - "type": "text" - }, - { - "block_id": "p689-b23", - "global_id": 20186, - "bbox": [ - 343.29, - 428.07, - 492.01, - 438.15 - ], - "text": "(e) xe(t) = sin(3πt + 1) + 2cos(7πt −2)", - "type": "text" - }, - { - "block_id": "p689-b24", - "global_id": 20187, - "bbox": [ - 344.28, - 439.03, - 467.25, - 449.1 - ], - "text": "(f) xf(t) = sin(6πt) + 2cos(14πt)", - "type": "text" - }, - { - "block_id": "p689-b25", - "global_id": 20188, - "bbox": [ - 314.15, - 453.57, - 516.14, - 517.4 - ], - "text": "6.1-6\nIf the two halves of one period of a periodic\nsignal are identical in shape except that one is the\nnegative of the other, the periodic signal is said to\nhave a half-wave symmetry. If a periodic signal\nx(t) with a period T0 satisfies the half-wave\nsymmetry condition, then", - "type": "text" - }, - { - "block_id": "p689-b26", - "global_id": 20189, - "bbox": [ - 393.61, - 537.19, - 397.6, - 546.16 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p689-b28", - "global_id": 20190, - "bbox": [ - 404.68, - 530.91, - 426.53, - 546.16 - ], - "text": "t −T0", - "type": "text" - }, - { - "block_id": "p689-b29", - "global_id": 20191, - "bbox": [ - 420.43, - 543.65, - 424.91, - 552.61 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p689-b31", - "global_id": 20192, - "bbox": [ - 436.11, - 536.91, - 465.3, - 546.16 - ], - "text": "= −x(t)", - "type": "text" - }, - { - "block_id": "p689-b32", - "global_id": 20193, - "bbox": [ - 342.78, - 565.81, - 516.12, - 596.69 - ], - "text": "In this case, show that all the even-numbered\nharmonics vanish and that the odd-numbered\nharmonic coefficients are given by", - "type": "text" - }, - { - "block_id": "p689-b33", - "global_id": 20194, - "bbox": [ - 375.64, - 611.4, - 402.34, - 627.62 - ], - "text": "an = 4", - "type": "text" - }, - { - "block_id": "p689-b34", - "global_id": 20195, - "bbox": [ - 395.74, - 623.96, - 403.96, - 633.76 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p689-b35", - "global_id": 20196, - "bbox": [ - 406.65, - 605.11, - 428.6, - 617.45 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p689-b36", - "global_id": 20197, - "bbox": [ - 411.38, - 617.32, - 483.12, - 634.09 - ], - "text": "0\nx(t)cos nω0tdt", - "type": "text" - } - ] - }, - { - "page_num": 690, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p690-b0", - "global_id": 20198, - "bbox": [ - 60.0, - 60.36, - 428.39, - 69.45 - ], - "text": "670\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p690-b1", - "global_id": 20199, - "bbox": [ - 143.55, - 125.66, - 258.97, - 133.96 - ], - "text": "7\n5\n3\n1", - "type": "text" - }, - { - "block_id": "p690-b2", - "global_id": 20200, - "bbox": [ - 264.81, - 143.59, - 275.47, - 151.89 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p690-b3", - "global_id": 20201, - "bbox": [ - 282.43, - 91.45, - 286.43, - 99.45 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p690-b4", - "global_id": 20202, - "bbox": [ - 272.43, - 125.96, - 398.65, - 133.96 - ], - "text": "0\n1\n3\n5\n7", - "type": "text" - }, - { - "block_id": "p690-b5", - "global_id": 20203, - "bbox": [ - 274.99, - 160.53, - 283.87, - 168.53 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p690-b6", - "global_id": 20204, - "bbox": [ - 129.41, - 217.27, - 425.78, - 225.57 - ], - "text": "20p\n10p\n10p\n20p", - "type": "text" - }, - { - "block_id": "p690-b7", - "global_id": 20205, - "bbox": [ - 129.81, - 288.77, - 180.73, - 297.06 - ], - "text": "8p\n6p", - "type": "text" - }, - { - "block_id": "p690-b8", - "global_id": 20206, - "bbox": [ - 186.8, - 433.62, - 197.46, - 441.91 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p690-b9", - "global_id": 20207, - "bbox": [ - 199.65, - 288.77, - 250.57, - 297.06 - ], - "text": "4p\n2p", - "type": "text" - }, - { - "block_id": "p690-b10", - "global_id": 20208, - "bbox": [ - 249.09, - 346.1, - 268.45, - 354.4 - ], - "text": "p4", - "type": "text" - }, - { - "block_id": "p690-b11", - "global_id": 20209, - "bbox": [ - 291.51, - 358.03, - 304.21, - 366.32 - ], - "text": "p4", - "type": "text" - }, - { - "block_id": "p690-b12", - "global_id": 20210, - "bbox": [ - 309.69, - 288.96, - 319.02, - 297.06 - ], - "text": "2p", - "type": "text" - }, - { - "block_id": "p690-b13", - "global_id": 20211, - "bbox": [ - 198.74, - 360.31, - 431.72, - 368.61 - ], - "text": "p\np\n2p", - "type": "text" - }, - { - "block_id": "p690-b14", - "global_id": 20212, - "bbox": [ - 344.61, - 288.96, - 423.78, - 297.06 - ], - "text": "4p\n6p\n8p", - "type": "text" - }, - { - "block_id": "p690-b15", - "global_id": 20213, - "bbox": [ - 263.47, - 217.27, - 288.59, - 225.47 - ], - "text": "p\np", - "type": "text" - }, - { - "block_id": "p690-b16", - "global_id": 20214, - "bbox": [ - 282.43, - 182.19, - 286.43, - 190.19 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p690-b17", - "global_id": 20215, - "bbox": [ - 274.53, - 233.09, - 284.34, - 241.09 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p690-b18", - "global_id": 20216, - "bbox": [ - 281.94, - 256.89, - 285.94, - 264.89 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p690-b19", - "global_id": 20217, - "bbox": [ - 272.43, - 289.06, - 276.43, - 297.06 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p690-b20", - "global_id": 20218, - "bbox": [ - 277.53, - 303.4, - 286.41, - 311.4 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p690-b21", - "global_id": 20219, - "bbox": [ - 270.85, - 339.0, - 274.85, - 347.0 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p690-b22", - "global_id": 20220, - "bbox": [ - 274.77, - 380.8, - 284.1, - 388.8 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p690-b23", - "global_id": 20221, - "bbox": [ - 270.84, - 467.72, - 274.84, - 475.72 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p690-b24", - "global_id": 20222, - "bbox": [ - 294.89, - 506.14, - 421.32, - 514.14 - ], - "text": "1\n2\n4\n6\n8", - "type": "text" - }, - { - "block_id": "p690-b25", - "global_id": 20223, - "bbox": [ - 274.99, - 447.03, - 283.87, - 455.03 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p690-b26", - "global_id": 20224, - "bbox": [ - 274.86, - 519.74, - 283.65, - 527.74 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p690-b27", - "global_id": 20225, - "bbox": [ - 293.92, - 433.91, - 388.13, - 441.91 - ], - "text": "4\n6\n1\n3", - "type": "text" - }, - { - "block_id": "p690-b28", - "global_id": 20226, - "bbox": [ - 270.84, - 403.81, - 274.84, - 411.81 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p690-b29", - "global_id": 20227, - "bbox": [ - 440.27, - 506.14, - 442.5, - 514.14 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p690-b30", - "global_id": 20228, - "bbox": [ - 440.27, - 433.64, - 442.5, - 441.64 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p690-b31", - "global_id": 20229, - "bbox": [ - 440.27, - 359.54, - 442.5, - 367.54 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p690-b32", - "global_id": 20230, - "bbox": [ - 440.27, - 288.84, - 442.5, - 296.84 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p690-b33", - "global_id": 20231, - "bbox": [ - 440.27, - 216.12, - 442.5, - 224.12 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p690-b34", - "global_id": 20232, - "bbox": [ - 440.27, - 126.21, - 442.5, - 134.21 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p690-b35", - "global_id": 20233, - "bbox": [ - 221.72, - 433.62, - 248.87, - 441.91 - ], - "text": "3 2", - "type": "text" - }, - { - "block_id": "p690-b36", - "global_id": 20234, - "bbox": [ - 169.34, - 505.85, - 249.84, - 514.14 - ], - "text": "2\n6", - "type": "text" - }, - { - "block_id": "p690-b37", - "global_id": 20235, - "bbox": [ - 104.83, - 533.91, - 156.48, - 542.88 - ], - "text": "Figure P6.1-1", - "type": "text" - }, - { - "block_id": "p690-b38", - "global_id": 20236, - "bbox": [ - 236.6, - 569.0, - 250.22, - 578.62 - ], - "text": "et2", - "type": "text" - }, - { - "block_id": "p690-b39", - "global_id": 20237, - "bbox": [ - 163.78, - 606.89, - 390.04, - 615.19 - ], - "text": "2p\n2p\np\np", - "type": "text" - }, - { - "block_id": "p690-b40", - "global_id": 20238, - "bbox": [ - 281.88, - 567.67, - 309.18, - 575.76 - ], - "text": "y(t)\n1", - "type": "text" - }, - { - "block_id": "p690-b41", - "global_id": 20239, - "bbox": [ - 271.22, - 606.6, - 445.02, - 615.13 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p690-b42", - "global_id": 20240, - "bbox": [ - 104.83, - 621.36, - 156.48, - 630.32 - ], - "text": "Figure P6.1-2", - "type": "text" - } - ] - }, - { - "page_num": 691, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p691-b0", - "global_id": 20241, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n671", - "type": "text" - }, - { - "block_id": "p691-b1", - "global_id": 20242, - "bbox": [ - 305.75, - 87.5, - 309.75, - 95.5 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p691-b2", - "global_id": 20243, - "bbox": [ - 136.0, - 129.85, - 486.84, - 138.43 - ], - "text": "t\n0\np2\np\np2\np\n2p\n2p", - "type": "text" - }, - { - "block_id": "p691-b3", - "global_id": 20244, - "bbox": [ - 325.66, - 87.87, - 337.09, - 95.95 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p691-b4", - "global_id": 20245, - "bbox": [ - 333.76, - 100.93, - 343.98, - 110.55 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p691-b5", - "global_id": 20246, - "bbox": [ - 130.58, - 144.92, - 182.22, - 153.88 - ], - "text": "Figure P6.1-3", - "type": "text" - }, - { - "block_id": "p691-b6", - "global_id": 20247, - "bbox": [ - 248.08, - 168.65, - 259.95, - 176.73 - ], - "text": "g(t)", - "type": "text" - }, - { - "block_id": "p691-b7", - "global_id": 20248, - "bbox": [ - 320.63, - 189.35, - 489.35, - 198.72 - ], - "text": "1\n2\n3\n4\nt", - "type": "text" - }, - { - "block_id": "p691-b8", - "global_id": 20249, - "bbox": [ - 287.94, - 167.27, - 292.83, - 175.27 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p691-b9", - "global_id": 20250, - "bbox": [ - 265.07, - 202.11, - 276.62, - 210.32 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p691-b10", - "global_id": 20251, - "bbox": [ - 154.59, - 189.05, - 245.98, - 197.35 - ], - "text": "1\n2\n3", - "type": "text" - }, - { - "block_id": "p691-b11", - "global_id": 20252, - "bbox": [ - 130.58, - 216.57, - 182.22, - 225.54 - ], - "text": "Figure P6.1-4", - "type": "text" - }, - { - "block_id": "p691-b12", - "global_id": 20253, - "bbox": [ - 165.15, - 268.78, - 393.03, - 277.08 - ], - "text": "2\n4\n6\n8\n10\n2\n4\n6", - "type": "text" - }, - { - "block_id": "p691-b13", - "global_id": 20254, - "bbox": [ - 288.8, - 301.09, - 297.68, - 309.09 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p691-b14", - "global_id": 20255, - "bbox": [ - 245.32, - 325.71, - 249.32, - 333.71 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p691-b15", - "global_id": 20256, - "bbox": [ - 238.65, - 366.44, - 249.32, - 374.74 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p691-b16", - "global_id": 20257, - "bbox": [ - 238.65, - 284.48, - 249.32, - 292.78 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p691-b17", - "global_id": 20258, - "bbox": [ - 220.28, - 349.88, - 281.96, - 358.08 - ], - "text": "p\np", - "type": "text" - }, - { - "block_id": "p691-b18", - "global_id": 20259, - "bbox": [ - 288.34, - 385.06, - 298.14, - 393.06 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p691-b19", - "global_id": 20260, - "bbox": [ - 263.94, - 325.87, - 282.8, - 335.48 - ], - "text": "et10", - "type": "text" - }, - { - "block_id": "p691-b20", - "global_id": 20261, - "bbox": [ - 245.32, - 241.89, - 249.32, - 249.89 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p691-b21", - "global_id": 20262, - "bbox": [ - 428.23, - 352.14, - 430.46, - 360.14 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p691-b22", - "global_id": 20263, - "bbox": [ - 428.23, - 256.23, - 430.46, - 264.23 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p691-b23", - "global_id": 20264, - "bbox": [ - 130.58, - 399.23, - 182.22, - 408.2 - ], - "text": "Figure P6.1-6", - "type": "text" - }, - { - "block_id": "p691-b24", - "global_id": 20265, - "bbox": [ - 115.63, - 427.23, - 128.57, - 436.2 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p691-b25", - "global_id": 20266, - "bbox": [ - 149.24, - 437.78, - 175.93, - 454.0 - ], - "text": "bn = 4", - "type": "text" - }, - { - "block_id": "p691-b26", - "global_id": 20267, - "bbox": [ - 169.34, - 450.33, - 177.56, - 460.13 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p691-b27", - "global_id": 20268, - "bbox": [ - 180.24, - 431.48, - 202.2, - 443.83 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p691-b28", - "global_id": 20269, - "bbox": [ - 184.98, - 443.69, - 255.22, - 460.47 - ], - "text": "0\nx(t)sin nω0tdt", - "type": "text" - }, - { - "block_id": "p691-b29", - "global_id": 20270, - "bbox": [ - 115.63, - 467.33, - 288.99, - 487.25 - ], - "text": "Using these results, find the Fourier series for the\nperiodic signals in Fig. P6.1-6.", - "type": "text" - }, - { - "block_id": "p691-b30", - "global_id": 20271, - "bbox": [ - 86.99, - 492.27, - 288.99, - 632.82 - ], - "text": "6.1-7\nOver a finite interval, a signal can be represented\nby more than one trigonometric (or exponential)\nFourier series. For instance, if we wish to\nrepresent x(t) = t over an interval 0 < t < 1 by a\nFourier series with fundamental frequency ω0 =\n2, we can draw a pulse x(t) = t over the interval\n0 < t < 1 and repeat the pulse every π seconds so\nthat T0 = π and ω0 = 2 (Fig. P6.1-7a). If we want\nthe fundamental frequency ω0 to be 4, we repeat\nthe pulse every π/2 seconds. If we want the\nseries to contain only cosine terms with ω0 = 2,\nwe construct a pulse x(t) = |t| over −1 < t < 1,\nand repeat it every π seconds (Fig. P6.1-7b). The", - "type": "text" - }, - { - "block_id": "p691-b31", - "global_id": 20272, - "bbox": [ - 342.78, - 427.23, - 516.14, - 523.88 - ], - "text": "resulting signal is an even function with period\nπ. Hence, its Fourier series will have only cosine\nterms with ω0 = 2. The resulting Fourier series\nrepresents x(t) = t over 0 < t < 1, as desired.\nWe do not care what it represents outside this\ninterval.\nSketch the periodic signal x(t) such that x(t) = t\nfor 0 < t < 1 and the Fourier series for x(t)\nsatisfies the following conditions.", - "type": "text" - }, - { - "block_id": "p691-b32", - "global_id": 20273, - "bbox": [ - 343.29, - 525.49, - 516.13, - 535.8 - ], - "text": "(a) ω0 = π/2 and contains all harmonics, but", - "type": "text" - }, - { - "block_id": "p691-b33", - "global_id": 20274, - "bbox": [ - 342.78, - 536.82, - 516.13, - 557.72 - ], - "text": "cosine terms only\n(b) ω0 = 2 and contains all harmonics, but sine", - "type": "text" - }, - { - "block_id": "p691-b34", - "global_id": 20275, - "bbox": [ - 343.29, - 558.74, - 516.13, - 579.64 - ], - "text": "terms only\n(c) ω0 = π/2 and contains all harmonics, which", - "type": "text" - }, - { - "block_id": "p691-b35", - "global_id": 20276, - "bbox": [ - 342.78, - 580.65, - 516.13, - 601.55 - ], - "text": "are exclusively neither sine nor cosine\n(d) ω0 = 1 and contains only odd harmonics and", - "type": "text" - }, - { - "block_id": "p691-b36", - "global_id": 20277, - "bbox": [ - 343.29, - 602.58, - 516.13, - 623.47 - ], - "text": "cosine terms\n(e) ω0 = π/2 and contains only odd harmonics", - "type": "text" - }, - { - "block_id": "p691-b37", - "global_id": 20278, - "bbox": [ - 358.22, - 624.49, - 410.02, - 633.46 - ], - "text": "and sine terms", - "type": "text" - } - ] - }, - { - "page_num": 692, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p692-b0", - "global_id": 20279, - "bbox": [ - 60.0, - 60.36, - 428.39, - 69.45 - ], - "text": "672\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p692-b1", - "global_id": 20280, - "bbox": [ - 365.16, - 89.04, - 369.16, - 97.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p692-b2", - "global_id": 20281, - "bbox": [ - 354.58, - 127.24, - 358.58, - 135.24 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p692-b3", - "global_id": 20282, - "bbox": [ - 175.06, - 140.74, - 366.58, - 148.74 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p692-b4", - "global_id": 20283, - "bbox": [ - 173.17, - 89.04, - 177.17, - 97.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p692-b5", - "global_id": 20284, - "bbox": [ - 104.83, - 126.94, - 436.14, - 135.24 - ], - "text": "t\nt\n1\n1\n1\n0\np\np\np\np", - "type": "text" - }, - { - "block_id": "p692-b6", - "global_id": 20285, - "bbox": [ - 104.83, - 154.92, - 156.48, - 163.88 - ], - "text": "Figure P6.1-7", - "type": "text" - }, - { - "block_id": "p692-b7", - "global_id": 20286, - "bbox": [ - 91.39, - 186.92, - 263.24, - 197.23 - ], - "text": "(f) ω0 = 1 and contains only odd harmon-", - "type": "text" - }, - { - "block_id": "p692-b8", - "global_id": 20287, - "bbox": [ - 89.89, - 198.25, - 263.25, - 283.93 - ], - "text": "ics, which are exclusively neither sine nor\ncosine.\n[Hint: For parts (d), (e), and (f), you need to use\nhalf-wave symmetry discussed in Prob. 6.1-6.\nCosine terms imply a possible dc component.]\nYou are asked only to sketch the periodic signal\nx(t) satisfying the given conditions. Do not find\nthe values of the Fourier coefficients.", - "type": "text" - }, - { - "block_id": "p692-b9", - "global_id": 20288, - "bbox": [ - 61.25, - 290.2, - 263.23, - 332.12 - ], - "text": "6.1-8\nState with reasons whether the following signals\nare periodic or aperiodic. For periodic signals,\nfind the period and state which harmonics are\npresent in the series.", - "type": "text" - }, - { - "block_id": "p692-b10", - "global_id": 20289, - "bbox": [ - 89.89, - 333.74, - 180.85, - 354.04 - ], - "text": "(a) 3sin t + 2sin 3t\n(b) 2 + 5sin 4t + 4cos 7t", - "type": "text" - }, - { - "block_id": "p692-b11", - "global_id": 20290, - "bbox": [ - 89.89, - 355.65, - 174.66, - 375.96 - ], - "text": "(c) 2sin 3t + 7cos πt\n(d) 7cos πt + 5sin 2πt", - "type": "text" - }, - { - "block_id": "p692-b12", - "global_id": 20291, - "bbox": [ - 90.4, - 377.95, - 122.76, - 386.92 - ], - "text": "(e) 3cos", - "type": "text" - }, - { - "block_id": "p692-b13", - "global_id": 20292, - "bbox": [ - 124.75, - 369.99, - 132.34, - 378.96 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p692-b14", - "global_id": 20293, - "bbox": [ - 132.34, - 377.57, - 175.65, - 386.91 - ], - "text": "2t + 5cos 2t", - "type": "text" - }, - { - "block_id": "p692-b15", - "global_id": 20294, - "bbox": [ - 91.39, - 387.17, - 124.96, - 402.51 - ], - "text": "(f) sin 5t", - "type": "text" - }, - { - "block_id": "p692-b16", - "global_id": 20295, - "bbox": [ - 119.3, - 380.58, - 201.43, - 408.87 - ], - "text": "2 + 3cos 6t\n5 + 3sin\n t", - "type": "text" - }, - { - "block_id": "p692-b17", - "global_id": 20296, - "bbox": [ - 198.03, - 380.58, - 232.24, - 408.87 - ], - "text": "7 + 30◦", - "type": "text" - }, - { - "block_id": "p692-b18", - "global_id": 20297, - "bbox": [ - 89.89, - 409.17, - 156.82, - 424.42 - ], - "text": "(g) sin3t + cos 15", - "type": "text" - }, - { - "block_id": "p692-b19", - "global_id": 20298, - "bbox": [ - 150.09, - 415.36, - 160.51, - 430.78 - ], - "text": "4 t", - "type": "text" - }, - { - "block_id": "p692-b20", - "global_id": 20299, - "bbox": [ - 89.89, - 429.11, - 169.7, - 439.45 - ], - "text": "(h) (3sin 2t + sin 5t)2", - "type": "text" - }, - { - "block_id": "p692-b21", - "global_id": 20300, - "bbox": [ - 91.88, - 440.07, - 140.33, - 450.4 - ], - "text": "(i) (5sin 2t)3", - "type": "text" - }, - { - "block_id": "p692-b22", - "global_id": 20301, - "bbox": [ - 61.25, - 456.68, - 263.23, - 487.63 - ], - "text": "6.3-1\nFor each of the periodic signals in Fig. P6.1-1,\nfind exponential Fourier series and sketch the\ncorresponding spectra.", - "type": "text" - }, - { - "block_id": "p692-b23", - "global_id": 20302, - "bbox": [ - 61.25, - 493.6, - 263.23, - 513.9 - ], - "text": "6.3-2\nA 2π-periodic signal x(t) is specified over one\nperiod as", - "type": "text" - }, - { - "block_id": "p692-b24", - "global_id": 20303, - "bbox": [ - 125.77, - 553.67, - 147.95, - 562.92 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p692-b25", - "global_id": 20304, - "bbox": [ - 149.8, - 532.6, - 156.9, - 552.32 - ], - "text": "⎧\n⎪⎨", - "type": "text" - }, - { - "block_id": "p692-b26", - "global_id": 20305, - "bbox": [ - 149.8, - 559.49, - 156.9, - 571.15 - ], - "text": "⎪⎩", - "type": "text" - }, - { - "block_id": "p692-b27", - "global_id": 20306, - "bbox": [ - 158.09, - 537.56, - 219.09, - 559.09 - ], - "text": "1\nAt\n0 ≤t < A", - "type": "text" - }, - { - "block_id": "p692-b28", - "global_id": 20307, - "bbox": [ - 156.9, - 557.52, - 225.28, - 577.81 - ], - "text": "1\nA ≤t < π\n0\nπ ≤t < 2π", - "type": "text" - }, - { - "block_id": "p692-b29", - "global_id": 20308, - "bbox": [ - 89.9, - 602.51, - 263.24, - 635.11 - ], - "text": "Sketch x(t) over two periods from t = 0 to\n4π. Show that the exponential Fourier series\ncoefficients Dn for this series are given by", - "type": "text" - }, - { - "block_id": "p692-b30", - "global_id": 20309, - "bbox": [ - 321.55, - 211.52, - 340.6, - 221.83 - ], - "text": "Dn =", - "type": "text" - }, - { - "block_id": "p692-b31", - "global_id": 20310, - "bbox": [ - 342.44, - 190.45, - 349.54, - 210.19 - ], - "text": "⎧\n⎪⎨", - "type": "text" - }, - { - "block_id": "p692-b32", - "global_id": 20311, - "bbox": [ - 342.44, - 217.36, - 349.54, - 229.01 - ], - "text": "⎪⎩", - "type": "text" - }, - { - "block_id": "p692-b33", - "global_id": 20312, - "bbox": [ - 350.74, - 194.76, - 484.42, - 216.75 - ], - "text": "2π −A\n4π\nn = 0", - "type": "text" - }, - { - "block_id": "p692-b34", - "global_id": 20313, - "bbox": [ - 350.74, - 215.49, - 365.99, - 237.11 - ], - "text": "1\n2πn", - "type": "text" - }, - { - "block_id": "p692-b35", - "global_id": 20314, - "bbox": [ - 368.17, - 208.81, - 408.21, - 224.45 - ], - "text": "e−jAn −1", - "type": "text" - }, - { - "block_id": "p692-b36", - "global_id": 20315, - "bbox": [ - 386.83, - 219.69, - 439.63, - 237.02 - ], - "text": "An\n+ je−jnπ", - "type": "text" - }, - { - "block_id": "p692-b38", - "global_id": 20316, - "bbox": [ - 464.77, - 221.41, - 484.68, - 230.74 - ], - "text": "n̸ = 0", - "type": "text" - }, - { - "block_id": "p692-b39", - "global_id": 20317, - "bbox": [ - 288.4, - 246.92, - 490.38, - 267.22 - ], - "text": "6.3-3\nA periodic signal x(t) is expressed by the\nfollowing Fourier series:", - "type": "text" - }, - { - "block_id": "p692-b40", - "global_id": 20318, - "bbox": [ - 323.12, - 283.6, - 388.68, - 292.94 - ], - "text": "x(t) = 3cost+ sin", - "type": "text" - }, - { - "block_id": "p692-b42", - "global_id": 20319, - "bbox": [ - 395.73, - 277.31, - 414.75, - 292.85 - ], - "text": "t −π", - "type": "text" - }, - { - "block_id": "p692-b43", - "global_id": 20320, - "bbox": [ - 410.27, - 290.34, - 414.75, - 299.31 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p692-b45", - "global_id": 20321, - "bbox": [ - 424.28, - 283.6, - 450.09, - 292.94 - ], - "text": "−2cos", - "type": "text" - }, - { - "block_id": "p692-b47", - "global_id": 20322, - "bbox": [ - 457.15, - 277.31, - 476.16, - 292.85 - ], - "text": "t −π", - "type": "text" - }, - { - "block_id": "p692-b48", - "global_id": 20323, - "bbox": [ - 471.68, - 290.34, - 476.16, - 299.31 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p692-b50", - "global_id": 20324, - "bbox": [ - 317.55, - 309.38, - 490.38, - 318.34 - ], - "text": "(a) Sketch the amplitude and phase spectra for", - "type": "text" - }, - { - "block_id": "p692-b51", - "global_id": 20325, - "bbox": [ - 317.04, - 320.34, - 490.39, - 340.27 - ], - "text": "the trigonometric series.\n(b) By inspection of spectra in part (a), sketch", - "type": "text" - }, - { - "block_id": "p692-b52", - "global_id": 20326, - "bbox": [ - 317.55, - 342.25, - 490.39, - 362.18 - ], - "text": "the exponential Fourier series spectra.\n(c) By inspection of spectra in part (b), write the", - "type": "text" - }, - { - "block_id": "p692-b53", - "global_id": 20327, - "bbox": [ - 317.04, - 363.81, - 490.38, - 384.1 - ], - "text": "exponential Fourier series for x(t).\n(d) Show that the series found in part (c) is", - "type": "text" - }, - { - "block_id": "p692-b54", - "global_id": 20328, - "bbox": [ - 332.48, - 386.09, - 490.38, - 406.02 - ], - "text": "equivalent to the trigonometric series for\nx(t).", - "type": "text" - }, - { - "block_id": "p692-b55", - "global_id": 20329, - "bbox": [ - 288.4, - 410.92, - 490.37, - 430.93 - ], - "text": "6.3-4\nThe trigonometric Fourier series of a certain\nperiodic signal is given by", - "type": "text" - }, - { - "block_id": "p692-b56", - "global_id": 20330, - "bbox": [ - 343.53, - 442.33, - 380.43, - 451.67 - ], - "text": "x(t) = 3 +", - "type": "text" - }, - { - "block_id": "p692-b57", - "global_id": 20331, - "bbox": [ - 381.82, - 434.33, - 389.41, - 443.3 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p692-b58", - "global_id": 20332, - "bbox": [ - 389.41, - 442.33, - 445.18, - 451.67 - ], - "text": "3cos 2t + sin 2t", - "type": "text" - }, - { - "block_id": "p692-b59", - "global_id": 20333, - "bbox": [ - 368.95, - 456.08, - 412.4, - 471.33 - ], - "text": "+ sin 3t −1", - "type": "text" - }, - { - "block_id": "p692-b60", - "global_id": 20334, - "bbox": [ - 407.92, - 452.09, - 433.42, - 477.7 - ], - "text": "2 cos", - "type": "text" - }, - { - "block_id": "p692-b61", - "global_id": 20335, - "bbox": [ - 433.42, - 455.7, - 456.92, - 477.7 - ], - "text": "5t + π\n3", - "type": "text" - }, - { - "block_id": "p692-b63", - "global_id": 20336, - "bbox": [ - 317.04, - 485.7, - 490.39, - 505.63 - ], - "text": "(a) Sketch the trigonometric Fourier spectra.\n(b) By inspection of spectra in part (a), sketch", - "type": "text" - }, - { - "block_id": "p692-b64", - "global_id": 20337, - "bbox": [ - 317.55, - 507.62, - 490.39, - 527.55 - ], - "text": "the exponential Fourier series spectra.\n(c) By inspection of spectra in part (b), write the", - "type": "text" - }, - { - "block_id": "p692-b65", - "global_id": 20338, - "bbox": [ - 317.04, - 529.17, - 490.38, - 549.46 - ], - "text": "exponential Fourier series for x(t).\n(d) Show that the series found in part (c) is", - "type": "text" - }, - { - "block_id": "p692-b66", - "global_id": 20339, - "bbox": [ - 332.48, - 551.46, - 490.38, - 571.39 - ], - "text": "equivalent to the trigonometric series for\nx(t).", - "type": "text" - }, - { - "block_id": "p692-b67", - "global_id": 20340, - "bbox": [ - 288.4, - 576.29, - 490.38, - 596.29 - ], - "text": "6.3-5\nThe exponential Fourier series of a certain\nfunction is given as", - "type": "text" - }, - { - "block_id": "p692-b68", - "global_id": 20341, - "bbox": [ - 347.01, - 603.99, - 444.84, - 617.04 - ], - "text": "x(t) = (2 + j2)e−j3t + j2e−jt", - "type": "text" - }, - { - "block_id": "p692-b69", - "global_id": 20342, - "bbox": [ - 372.44, - 619.38, - 459.8, - 632.41 - ], - "text": "+ 3 −j2ejt + (2 −j2)ej3t", - "type": "text" - } - ] - }, - { - "page_num": 693, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p693-b0", - "global_id": 20343, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n673", - "type": "text" - }, - { - "block_id": "p693-b1", - "global_id": 20344, - "bbox": [ - 115.63, - 90.75, - 288.97, - 110.68 - ], - "text": "(a) Sketch the exponential Fourier spectra.\n(b) By inspection of the spectra in part (a),", - "type": "text" - }, - { - "block_id": "p693-b2", - "global_id": 20345, - "bbox": [ - 116.14, - 112.66, - 288.97, - 154.51 - ], - "text": "sketch the trigonometric Fourier spectra for\nx(t). Find the compact trigonometric Fourier\nseries from these spectra.\n(c) Show that the trigonometric series found", - "type": "text" - }, - { - "block_id": "p693-b3", - "global_id": 20346, - "bbox": [ - 115.63, - 156.5, - 288.97, - 187.39 - ], - "text": "in part (b) is equivalent to the exponential\nseries for x(t).\n(d) Find the signal bandwidth.", - "type": "text" - }, - { - "block_id": "p693-b4", - "global_id": 20347, - "bbox": [ - 87.0, - 199.6, - 288.97, - 219.61 - ], - "text": "6.3-6\nFigure P6.3-6 shows the trigonometric Fourier\nspectra of a periodic signal x(t).", - "type": "text" - }, - { - "block_id": "p693-b5", - "global_id": 20348, - "bbox": [ - 116.14, - 221.59, - 288.98, - 230.56 - ], - "text": "(a) By inspection of Fig. P6.3-6, find the", - "type": "text" - }, - { - "block_id": "p693-b6", - "global_id": 20349, - "bbox": [ - 115.63, - 232.56, - 289.0, - 263.44 - ], - "text": "trigonometric Fourier series representing\nx(t).\n(b) By inspection of Fig. P6.3-6, sketch the", - "type": "text" - }, - { - "block_id": "p693-b7", - "global_id": 20350, - "bbox": [ - 116.14, - 265.06, - 288.98, - 285.36 - ], - "text": "exponential Fourier spectra of x(t).\n(c) By inspection of the exponential Fourier", - "type": "text" - }, - { - "block_id": "p693-b8", - "global_id": 20351, - "bbox": [ - 115.64, - 287.35, - 288.98, - 318.23 - ], - "text": "spectra obtained in part (b), find the expo-\nnential Fourier series for x(t).\n(d) Show that the series found in parts (a) and", - "type": "text" - }, - { - "block_id": "p693-b9", - "global_id": 20352, - "bbox": [ - 131.07, - 320.23, - 195.59, - 329.19 - ], - "text": "(c) are equivalent.", - "type": "text" - }, - { - "block_id": "p693-b10", - "global_id": 20353, - "bbox": [ - 87.0, - 341.41, - 288.98, - 361.41 - ], - "text": "6.3-7\nFigure P6.3-7 shows the exponential Fourier\nspectra of a periodic signal x(t).", - "type": "text" - }, - { - "block_id": "p693-b11", - "global_id": 20354, - "bbox": [ - 343.29, - 85.9, - 516.12, - 94.87 - ], - "text": "(a) By inspection of Fig. P6.3-7, find the expo-", - "type": "text" - }, - { - "block_id": "p693-b12", - "global_id": 20355, - "bbox": [ - 342.79, - 96.49, - 516.15, - 116.79 - ], - "text": "nential Fourier series representing x(t).\n(b) By inspection of Fig. P6.3-7, sketch the", - "type": "text" - }, - { - "block_id": "p693-b13", - "global_id": 20356, - "bbox": [ - 343.29, - 118.41, - 516.12, - 138.71 - ], - "text": "trigonometric Fourier spectra for x(t).\n(c) By inspection of the trigonometric Fourier", - "type": "text" - }, - { - "block_id": "p693-b14", - "global_id": 20357, - "bbox": [ - 342.79, - 140.7, - 516.14, - 171.59 - ], - "text": "spectra found in part (b), find the trigono-\nmetric Fourier series for x(t).\n(d) Show that the series found in parts (a) and", - "type": "text" - }, - { - "block_id": "p693-b15", - "global_id": 20358, - "bbox": [ - 358.22, - 173.58, - 422.73, - 182.54 - ], - "text": "(c) are equivalent.", - "type": "text" - }, - { - "block_id": "p693-b16", - "global_id": 20359, - "bbox": [ - 314.15, - 191.4, - 516.13, - 222.65 - ], - "text": "6.3-8\nLet periodic signal x(t) have exponential Fourier\nseries spectrum Dn. Prove the following proper-\nties.", - "type": "text" - }, - { - "block_id": "p693-b17", - "global_id": 20360, - "bbox": [ - 343.29, - 224.27, - 516.13, - 234.95 - ], - "text": "(a) If x(t) has even symmetry, then Dn also has", - "type": "text" - }, - { - "block_id": "p693-b18", - "global_id": 20361, - "bbox": [ - 342.78, - 235.6, - 516.13, - 256.87 - ], - "text": "even symmetry.\n(b) If x(t) has odd symmetry, then Dn also has", - "type": "text" - }, - { - "block_id": "p693-b19", - "global_id": 20362, - "bbox": [ - 343.29, - 257.52, - 516.13, - 278.79 - ], - "text": "odd symmetry.\n(c) If x(t) is real, then Dn is conjugate symmet-", - "type": "text" - }, - { - "block_id": "p693-b20", - "global_id": 20363, - "bbox": [ - 358.22, - 277.8, - 403.64, - 289.38 - ], - "text": "ric (Dn = D∗", - "type": "text" - }, - { - "block_id": "p693-b21", - "global_id": 20364, - "bbox": [ - 342.78, - 279.44, - 516.13, - 300.7 - ], - "text": "−n).\n(d) If x(t) is imaginary, then Dn is conjugate", - "type": "text" - }, - { - "block_id": "p693-b22", - "global_id": 20365, - "bbox": [ - 358.22, - 299.72, - 452.48, - 311.29 - ], - "text": "antisymmetric (Dn = −D∗", - "type": "text" - }, - { - "block_id": "p693-b23", - "global_id": 20366, - "bbox": [ - 449.11, - 301.36, - 463.12, - 312.07 - ], - "text": "−n).", - "type": "text" - }, - { - "block_id": "p693-b24", - "global_id": 20367, - "bbox": [ - 314.14, - 319.48, - 516.13, - 328.52 - ], - "text": "6.3-9\n(a) Find the exponential Fourier series for the", - "type": "text" - }, - { - "block_id": "p693-b25", - "global_id": 20368, - "bbox": [ - 342.78, - 330.52, - 516.13, - 350.45 - ], - "text": "signal in Fig. P6.3-9a.\n(b) Using the results in part (a), find the Fourier", - "type": "text" - }, - { - "block_id": "p693-b26", - "global_id": 20369, - "bbox": [ - 358.22, - 352.06, - 516.13, - 361.4 - ], - "text": "series for the signal ˆx(t) in Fig. P6.3-9b,", - "type": "text" - }, - { - "block_id": "p693-b27", - "global_id": 20370, - "bbox": [ - 228.01, - 469.55, - 236.89, - 477.55 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p693-b28", - "global_id": 20371, - "bbox": [ - 182.28, - 398.17, - 186.28, - 406.17 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p693-b29", - "global_id": 20372, - "bbox": [ - 182.28, - 419.34, - 186.28, - 427.34 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p693-b30", - "global_id": 20373, - "bbox": [ - 239.88, - 446.61, - 243.88, - 454.61 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p693-b31", - "global_id": 20374, - "bbox": [ - 371.92, - 396.42, - 375.92, - 404.42 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p693-b32", - "global_id": 20375, - "bbox": [ - 307.6, - 441.52, - 319.6, - 449.72 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p693-b33", - "global_id": 20376, - "bbox": [ - 333.72, - 396.42, - 363.18, - 404.42 - ], - "text": "3\n2\n1", - "type": "text" - }, - { - "block_id": "p693-b34", - "global_id": 20377, - "bbox": [ - 356.08, - 469.55, - 365.89, - 477.55 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p693-b35", - "global_id": 20378, - "bbox": [ - 266.08, - 444.11, - 271.41, - 452.11 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p693-b36", - "global_id": 20379, - "bbox": [ - 171.42, - 401.07, - 396.09, - 416.62 - ], - "text": "v\nCn", - "type": "text" - }, - { - "block_id": "p693-b37", - "global_id": 20380, - "bbox": [ - 308.87, - 394.74, - 315.39, - 404.31 - ], - "text": "un", - "type": "text" - }, - { - "block_id": "p693-b38", - "global_id": 20381, - "bbox": [ - 313.47, - 417.24, - 318.81, - 425.24 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p693-b39", - "global_id": 20382, - "bbox": [ - 305.83, - 421.31, - 318.14, - 433.99 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p693-b40", - "global_id": 20383, - "bbox": [ - 427.37, - 469.53, - 479.01, - 478.49 - ], - "text": "Figure P6.3-6", - "type": "text" - }, - { - "block_id": "p693-b41", - "global_id": 20384, - "bbox": [ - 204.55, - 566.66, - 213.43, - 574.66 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p693-b42", - "global_id": 20385, - "bbox": [ - 154.1, - 548.91, - 342.99, - 557.21 - ], - "text": "3\n3\n1\n3", - "type": "text" - }, - { - "block_id": "p693-b43", - "global_id": 20386, - "bbox": [ - 200.85, - 503.12, - 204.85, - 511.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p693-b44", - "global_id": 20387, - "bbox": [ - 224.01, - 549.21, - 228.01, - 557.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p693-b45", - "global_id": 20388, - "bbox": [ - 385.86, - 614.19, - 395.67, - 622.19 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p693-b46", - "global_id": 20389, - "bbox": [ - 405.79, - 549.21, - 442.98, - 557.21 - ], - "text": "3\n1", - "type": "text" - }, - { - "block_id": "p693-b47", - "global_id": 20390, - "bbox": [ - 234.35, - 494.19, - 431.58, - 503.95 - ], - "text": "Dn\nDn", - "type": "text" - }, - { - "block_id": "p693-b48", - "global_id": 20391, - "bbox": [ - 394.97, - 582.39, - 406.97, - 590.58 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p693-b49", - "global_id": 20392, - "bbox": [ - 380.18, - 502.13, - 385.51, - 510.13 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p693-b50", - "global_id": 20393, - "bbox": [ - 270.04, - 549.1, - 457.15, - 557.1 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p693-b51", - "global_id": 20394, - "bbox": [ - 130.58, - 628.36, - 182.22, - 637.33 - ], - "text": "Figure P6.3-7", - "type": "text" - } - ] - }, - { - "page_num": 694, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p694-b0", - "global_id": 20395, - "bbox": [ - 60.0, - 60.36, - 428.39, - 69.45 - ], - "text": "674\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p694-b1", - "global_id": 20396, - "bbox": [ - 280.39, - 287.74, - 289.27, - 295.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p694-b2", - "global_id": 20397, - "bbox": [ - 276.25, - 237.02, - 293.64, - 249.12 - ], - "text": "1\n˜", - "type": "text" - }, - { - "block_id": "p694-b3", - "global_id": 20398, - "bbox": [ - 280.01, - 215.88, - 289.66, - 223.88 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p694-b4", - "global_id": 20399, - "bbox": [ - 277.13, - 162.59, - 293.6, - 178.16 - ], - "text": "1\nˆ", - "type": "text" - }, - { - "block_id": "p694-b5", - "global_id": 20400, - "bbox": [ - 449.56, - 116.62, - 451.78, - 124.62 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p694-b6", - "global_id": 20401, - "bbox": [ - 449.56, - 261.9, - 451.78, - 269.9 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p694-b7", - "global_id": 20402, - "bbox": [ - 449.56, - 190.03, - 451.78, - 198.03 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p694-b8", - "global_id": 20403, - "bbox": [ - 288.46, - 124.28, - 299.12, - 132.57 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p694-b9", - "global_id": 20404, - "bbox": [ - 288.46, - 197.87, - 299.12, - 206.16 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p694-b10", - "global_id": 20405, - "bbox": [ - 288.46, - 269.74, - 299.12, - 278.03 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p694-b11", - "global_id": 20406, - "bbox": [ - 276.25, - 96.57, - 280.25, - 104.57 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p694-b12", - "global_id": 20407, - "bbox": [ - 237.5, - 117.56, - 327.83, - 126.85 - ], - "text": "4\n4", - "type": "text" - }, - { - "block_id": "p694-b13", - "global_id": 20408, - "bbox": [ - 280.39, - 142.28, - 289.27, - 150.28 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p694-b14", - "global_id": 20409, - "bbox": [ - 195.5, - 117.85, - 413.5, - 126.85 - ], - "text": "8\n8\n12", - "type": "text" - }, - { - "block_id": "p694-b15", - "global_id": 20410, - "bbox": [ - 150.9, - 193.1, - 413.5, - 201.4 - ], - "text": "4\n4\n8\n8\n12\n12", - "type": "text" - }, - { - "block_id": "p694-b16", - "global_id": 20411, - "bbox": [ - 150.9, - 264.93, - 413.5, - 273.23 - ], - "text": "4\n4\n8\n8\n12\n12", - "type": "text" - }, - { - "block_id": "p694-b17", - "global_id": 20412, - "bbox": [ - 290.19, - 88.05, - 301.29, - 96.13 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p694-b18", - "global_id": 20413, - "bbox": [ - 290.19, - 163.01, - 301.29, - 171.09 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p694-b19", - "global_id": 20414, - "bbox": [ - 290.19, - 237.44, - 301.29, - 245.52 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p694-b20", - "global_id": 20415, - "bbox": [ - 150.9, - 117.56, - 165.16, - 125.85 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p694-b21", - "global_id": 20416, - "bbox": [ - 104.83, - 301.91, - 156.48, - 310.88 - ], - "text": "Figure P6.3-9", - "type": "text" - }, - { - "block_id": "p694-b22", - "global_id": 20417, - "bbox": [ - 90.4, - 322.34, - 263.24, - 353.22 - ], - "text": "which is a time-shifted version of the signal\nx(t).\n(c) Using the results in part (a), find the Fourier", - "type": "text" - }, - { - "block_id": "p694-b23", - "global_id": 20418, - "bbox": [ - 105.33, - 354.84, - 263.24, - 386.1 - ], - "text": "series for the signal ˜x(t) in Fig. P6.3-9c,\nwhich is a time-scaled version of the signal\nx(t).", - "type": "text" - }, - { - "block_id": "p694-b24", - "global_id": 20419, - "bbox": [ - 56.77, - 391.03, - 263.23, - 411.33 - ], - "text": "6.3-10\nA periodic signal x(t) is expressed as an expo-\nnential Fourier series", - "type": "text" - }, - { - "block_id": "p694-b25", - "global_id": 20420, - "bbox": [ - 139.52, - 431.61, - 161.71, - 440.86 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p694-b26", - "global_id": 20421, - "bbox": [ - 167.18, - 422.3, - 179.86, - 432.06 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p694-b27", - "global_id": 20422, - "bbox": [ - 163.55, - 443.92, - 183.5, - 450.6 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p694-b28", - "global_id": 20423, - "bbox": [ - 184.49, - 429.9, - 213.0, - 441.62 - ], - "text": "Dnejnω0t", - "type": "text" - }, - { - "block_id": "p694-b29", - "global_id": 20424, - "bbox": [ - 90.4, - 468.99, - 263.24, - 477.96 - ], - "text": "(a) Show that the exponential Fourier series for", - "type": "text" - }, - { - "block_id": "p694-b30", - "global_id": 20425, - "bbox": [ - 105.33, - 479.58, - 199.41, - 488.92 - ], - "text": "ˆx(t) = x(t −T) is given by", - "type": "text" - }, - { - "block_id": "p694-b31", - "global_id": 20426, - "bbox": [ - 147.23, - 511.05, - 169.42, - 520.3 - ], - "text": "ˆx(t) =", - "type": "text" - }, - { - "block_id": "p694-b32", - "global_id": 20427, - "bbox": [ - 174.9, - 501.74, - 187.58, - 511.51 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p694-b33", - "global_id": 20428, - "bbox": [ - 171.26, - 523.36, - 191.21, - 530.04 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p694-b34", - "global_id": 20429, - "bbox": [ - 192.21, - 509.2, - 220.72, - 521.07 - ], - "text": "ˆDnejnω0t", - "type": "text" - }, - { - "block_id": "p694-b35", - "global_id": 20430, - "bbox": [ - 105.33, - 543.58, - 136.45, - 552.55 - ], - "text": "in which", - "type": "text" - }, - { - "block_id": "p694-b36", - "global_id": 20431, - "bbox": [ - 112.61, - 565.53, - 255.27, - 577.69 - ], - "text": "| ˆDn| = |Dn|\nand̸\nˆDn ≠ Dn −nω0T", - "type": "text" - }, - { - "block_id": "p694-b37", - "global_id": 20432, - "bbox": [ - 105.33, - 591.92, - 263.24, - 633.76 - ], - "text": "This result shows that time shifting of a\nperiodic signal by T seconds merely changes\nthe phase spectrum by nω0T. The amplitude\nspectrum is unchanged.", - "type": "text" - }, - { - "block_id": "p694-b38", - "global_id": 20433, - "bbox": [ - 317.04, - 322.33, - 490.39, - 331.3 - ], - "text": "(b) Show that the exponential Fourier series for", - "type": "text" - }, - { - "block_id": "p694-b39", - "global_id": 20434, - "bbox": [ - 332.48, - 332.92, - 415.58, - 342.26 - ], - "text": "˜x(t) = x(at) is given by", - "type": "text" - }, - { - "block_id": "p694-b40", - "global_id": 20435, - "bbox": [ - 370.35, - 367.95, - 392.54, - 377.2 - ], - "text": "˜x(t) =", - "type": "text" - }, - { - "block_id": "p694-b41", - "global_id": 20436, - "bbox": [ - 398.01, - 358.63, - 410.69, - 368.4 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p694-b42", - "global_id": 20437, - "bbox": [ - 394.38, - 380.25, - 414.33, - 386.93 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p694-b43", - "global_id": 20438, - "bbox": [ - 415.32, - 366.24, - 451.9, - 377.97 - ], - "text": "Dnejn(aω0)t", - "type": "text" - }, - { - "block_id": "p694-b44", - "global_id": 20439, - "bbox": [ - 332.48, - 404.04, - 490.39, - 478.76 - ], - "text": "This result shows that time compression of\na periodic signal by a factor a expands its\nFourier spectra along the ω axis by the\nsame factor a. Similarly, time expansion of a\nperiodic signal by a factor a compresses its\nFourier spectra along the ω axis by the factor\na. Intuitively explain this result.", - "type": "text" - }, - { - "block_id": "p694-b45", - "global_id": 20440, - "bbox": [ - 283.92, - 484.34, - 490.39, - 493.38 - ], - "text": "6.3-11\n(a) The Fourier series for the periodic signal", - "type": "text" - }, - { - "block_id": "p694-b46", - "global_id": 20441, - "bbox": [ - 332.48, - 495.38, - 490.38, - 515.3 - ], - "text": "in Fig. 6.7a is given in Drill 6.1. Verify\nParseval’s theorem for this series, given that", - "type": "text" - }, - { - "block_id": "p694-b47", - "global_id": 20442, - "bbox": [ - 387.75, - 531.68, - 400.43, - 541.44 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p694-b48", - "global_id": 20443, - "bbox": [ - 388.32, - 553.91, - 399.85, - 560.66 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p694-b49", - "global_id": 20444, - "bbox": [ - 402.62, - 533.71, - 433.43, - 556.61 - ], - "text": "1\nn4 = π4", - "type": "text" - }, - { - "block_id": "p694-b50", - "global_id": 20445, - "bbox": [ - 424.43, - 547.73, - 433.4, - 556.7 - ], - "text": "90", - "type": "text" - }, - { - "block_id": "p694-b51", - "global_id": 20446, - "bbox": [ - 317.04, - 576.92, - 490.39, - 586.26 - ], - "text": "(b) If x(t) is approximated by the first N terms", - "type": "text" - }, - { - "block_id": "p694-b52", - "global_id": 20447, - "bbox": [ - 332.48, - 588.16, - 490.39, - 608.85 - ], - "text": "in this series, find N so that the power of the\nerror signal is less than 1% of Px.", - "type": "text" - }, - { - "block_id": "p694-b53", - "global_id": 20448, - "bbox": [ - 283.91, - 613.76, - 490.39, - 622.8 - ], - "text": "6.3-12\n(a) The Fourier series for the periodic signal", - "type": "text" - }, - { - "block_id": "p694-b54", - "global_id": 20449, - "bbox": [ - 332.48, - 624.8, - 490.38, - 633.76 - ], - "text": "in Fig. 6.7b is given in Drill 6.1. Verify", - "type": "text" - } - ] - }, - { - "page_num": 695, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p695-b0", - "global_id": 20450, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n675", - "type": "text" - }, - { - "block_id": "p695-b1", - "global_id": 20451, - "bbox": [ - 131.07, - 85.9, - 288.74, - 94.86 - ], - "text": "Parseval’s theorem for this series, given that", - "type": "text" - }, - { - "block_id": "p695-b2", - "global_id": 20452, - "bbox": [ - 186.34, - 103.73, - 199.02, - 113.5 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p695-b3", - "global_id": 20453, - "bbox": [ - 186.92, - 125.98, - 198.45, - 132.73 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p695-b4", - "global_id": 20454, - "bbox": [ - 201.21, - 105.78, - 232.03, - 128.67 - ], - "text": "1\nn2 = π2", - "type": "text" - }, - { - "block_id": "p695-b5", - "global_id": 20455, - "bbox": [ - 225.27, - 119.79, - 229.75, - 128.76 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p695-b6", - "global_id": 20456, - "bbox": [ - 115.63, - 141.49, - 288.98, - 150.82 - ], - "text": "(b) If x(t) is approximated by the first N terms", - "type": "text" - }, - { - "block_id": "p695-b7", - "global_id": 20457, - "bbox": [ - 131.07, - 152.72, - 288.98, - 173.41 - ], - "text": "in this series, find N so that the power of the\nerror signal is less than 10% of Px.", - "type": "text" - }, - { - "block_id": "p695-b8", - "global_id": 20458, - "bbox": [ - 82.51, - 177.35, - 288.99, - 241.48 - ], - "text": "6.3-13\nThe signal x(t) in Fig. 6.17 is approximated by\nthe first 2N + 1 terms (from n = −N to N) in\nits exponential Fourier series given in Drill 6.5.\nDetermine the value of N if this (2N + 1)-term\nFourier series power is to be no less than 99.75%\nof the power of x(t).", - "type": "text" - }, - { - "block_id": "p695-b9", - "global_id": 20459, - "bbox": [ - 82.51, - 246.09, - 288.98, - 256.17 - ], - "text": "6.3-14\n(a) A 2 rad/s periodic signal x1(t) has Fourier", - "type": "text" - }, - { - "block_id": "p695-b10", - "global_id": 20460, - "bbox": [ - 131.07, - 257.05, - 288.98, - 278.09 - ], - "text": "series\nspectrum\nD1[n].\nDetermine\nthe\nFourier series spectrum X2[n] of x2(t) =", - "type": "text" - }, - { - "block_id": "p695-b11", - "global_id": 20461, - "bbox": [ - 115.63, - 277.6, - 288.98, - 300.0 - ], - "text": "1\n3x1(−t −5) in terms of X1[n].\n(b) A 2 rad/s periodic signal x1(t) has Fourier", - "type": "text" - }, - { - "block_id": "p695-b12", - "global_id": 20462, - "bbox": [ - 116.14, - 300.88, - 288.98, - 343.84 - ], - "text": "series\nspectrum\nD1[n].\nDetermine\nthe\nFourier series spectrum X2[n] of x2(t) =\ncos(10t)x1(t) in terms of X1[n].\n(c) A 3 rad/s periodic signal x1(t) has Fourier", - "type": "text" - }, - { - "block_id": "p695-b13", - "global_id": 20463, - "bbox": [ - 131.07, - 344.72, - 288.98, - 376.72 - ], - "text": "series\nspectrum\nD1[n].\nDetermine\nthe\nFourier series spectrum X2[n] of x2(t) =\nx1(−t) −3x1(t + 2) in terms of X1[n].", - "type": "text" - }, - { - "block_id": "p695-b14", - "global_id": 20464, - "bbox": [ - 82.51, - 380.59, - 247.33, - 389.93 - ], - "text": "6.3-15\nA 2-periodic signal x(t) is defined as", - "type": "text" - }, - { - "block_id": "p695-b15", - "global_id": 20465, - "bbox": [ - 130.66, - 411.47, - 152.85, - 420.72 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p695-b16", - "global_id": 20466, - "bbox": [ - 154.69, - 393.09, - 161.79, - 410.12 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p695-b17", - "global_id": 20467, - "bbox": [ - 154.69, - 417.29, - 161.79, - 426.26 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p695-b18", - "global_id": 20468, - "bbox": [ - 166.78, - 397.2, - 267.78, - 431.71 - ], - "text": "−t2 −t + 0.25\n−1 ≤t < 0\nt2 −t + 0.25\n0 ≤t < 1\nx(t + 2)\n∀t", - "type": "text" - }, - { - "block_id": "p695-b19", - "global_id": 20469, - "bbox": [ - 115.64, - 446.88, - 263.7, - 467.92 - ], - "text": "(a) Plot x(t) over −2 ≤t ≤2.\n(b) Determine D0, the dc content of x(t).", - "type": "text" - }, - { - "block_id": "p695-b20", - "global_id": 20470, - "bbox": [ - 116.14, - 469.17, - 288.97, - 478.14 - ], - "text": "(c) Similar to Ex. 6.11, use properties and not", - "type": "text" - }, - { - "block_id": "p695-b21", - "global_id": 20471, - "bbox": [ - 115.63, - 479.76, - 288.98, - 500.73 - ], - "text": "integration to determine Dn for n̸ = 0.\n(d) Plot the magnitude spectrum |Dn| over a", - "type": "text" - }, - { - "block_id": "p695-b22", - "global_id": 20472, - "bbox": [ - 116.14, - 501.96, - 288.98, - 522.65 - ], - "text": "suitable range of n.\n(e) How does the magnitude spectrum |Dn|", - "type": "text" - }, - { - "block_id": "p695-b23", - "global_id": 20473, - "bbox": [ - 131.07, - 523.6, - 288.99, - 554.85 - ], - "text": "compare to that of signal y(t) = cos(πt)?\nNote similarities as well as major differ-\nences.", - "type": "text" - }, - { - "block_id": "p695-b24", - "global_id": 20474, - "bbox": [ - 82.51, - 559.46, - 247.33, - 568.8 - ], - "text": "6.3-16\nA 3-periodic signal x(t) is defined as", - "type": "text" - }, - { - "block_id": "p695-b25", - "global_id": 20475, - "bbox": [ - 140.08, - 590.34, - 162.26, - 599.59 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p695-b26", - "global_id": 20476, - "bbox": [ - 164.11, - 571.97, - 171.21, - 589.0 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p695-b27", - "global_id": 20477, - "bbox": [ - 164.11, - 596.17, - 171.21, - 605.13 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p695-b28", - "global_id": 20478, - "bbox": [ - 176.19, - 579.34, - 258.37, - 610.59 - ], - "text": "|t|\n−1 ≤t ≤1\n0\n1 < |t| ≤1.5\nx(t + 3)\n∀t", - "type": "text" - }, - { - "block_id": "p695-b29", - "global_id": 20479, - "bbox": [ - 116.14, - 625.76, - 223.02, - 635.1 - ], - "text": "(a) Plot x(t) over −3 ≤t ≤3.", - "type": "text" - }, - { - "block_id": "p695-b30", - "global_id": 20480, - "bbox": [ - 342.79, - 85.58, - 516.12, - 105.88 - ], - "text": "(b) Determine D0, the dc content of x(t).\n(c) Similar to Ex. 6.11, use properties and not", - "type": "text" - }, - { - "block_id": "p695-b31", - "global_id": 20481, - "bbox": [ - 342.78, - 107.49, - 516.14, - 128.47 - ], - "text": "integration to determine Dn for n̸ = 0.\n(d) Plot the magnitude spectrum |Dn| over a", - "type": "text" - }, - { - "block_id": "p695-b32", - "global_id": 20482, - "bbox": [ - 358.22, - 129.7, - 516.13, - 149.71 - ], - "text": "suitable range of n. What is the most domi-\nnant frequency component of this signal?", - "type": "text" - }, - { - "block_id": "p695-b33", - "global_id": 20483, - "bbox": [ - 314.15, - 154.62, - 516.13, - 174.62 - ], - "text": "6.4-1\nFind the response of an LTIC system with\ntransfer function", - "type": "text" - }, - { - "block_id": "p695-b34", - "global_id": 20484, - "bbox": [ - 394.78, - 179.55, - 462.95, - 201.26 - ], - "text": "H(s) =\ns\ns2 + 2s + 3", - "type": "text" - }, - { - "block_id": "p695-b35", - "global_id": 20485, - "bbox": [ - 342.78, - 209.46, - 486.0, - 218.42 - ], - "text": "to the periodic input shown in Fig. 6.2a.", - "type": "text" - }, - { - "block_id": "p695-b36", - "global_id": 20486, - "bbox": [ - 314.15, - 223.03, - 516.14, - 254.29 - ], - "text": "6.4-2\nA periodic signal x(t) = 1 + 2cos(5πt) +\n3sin(14πt) is applied to an LTIC system to\nproduce output y(t).", - "type": "text" - }, - { - "block_id": "p695-b37", - "global_id": 20487, - "bbox": [ - 343.29, - 255.9, - 516.13, - 265.98 - ], - "text": "(a) Determine ω0, the fundamental radian fre-", - "type": "text" - }, - { - "block_id": "p695-b38", - "global_id": 20488, - "bbox": [ - 342.78, - 266.87, - 516.13, - 309.08 - ], - "text": "quency of x(t).\n(b) Determine\nDn, the exponential\nFourier\nseries spectrum of x(t).\n(c) If the system is an ideal lowpass filter with", - "type": "text" - }, - { - "block_id": "p695-b39", - "global_id": 20489, - "bbox": [ - 342.78, - 310.7, - 516.14, - 341.96 - ], - "text": "cutoff frequency fc = 2 Hz, what is the\noutput y(t)?\n(d) If the system is an ideal highpass filter with", - "type": "text" - }, - { - "block_id": "p695-b40", - "global_id": 20490, - "bbox": [ - 343.29, - 343.58, - 516.14, - 374.84 - ], - "text": "cutoff frequency fc = 2 Hz, what is the\noutput y(t)?\n(e) If the system is an ideal bandpass filter with", - "type": "text" - }, - { - "block_id": "p695-b41", - "global_id": 20491, - "bbox": [ - 344.28, - 376.83, - 516.14, - 407.71 - ], - "text": "a 4 Hz passband centered at 4 Hz, what is\nthe output y(t)?\n(f) If the system is an ideal bandstop filter with", - "type": "text" - }, - { - "block_id": "p695-b42", - "global_id": 20492, - "bbox": [ - 342.78, - 409.71, - 516.13, - 440.59 - ], - "text": "a 5 Hz stopband centered at 10 Hz, what is\nthe output y(t)?\n(g) Describe the frequency response of a filter", - "type": "text" - }, - { - "block_id": "p695-b43", - "global_id": 20493, - "bbox": [ - 358.22, - 442.21, - 516.13, - 462.51 - ], - "text": "that, in response to x(t), would produce the\noutput y(t) = 4cos(5πt) −9sin(14πt).", - "type": "text" - }, - { - "block_id": "p695-b44", - "global_id": 20494, - "bbox": [ - 314.14, - 467.11, - 516.13, - 477.42 - ], - "text": "6.4-3\nConsider a T0 = 1 periodic signal x(t) defined as", - "type": "text" - }, - { - "block_id": "p695-b45", - "global_id": 20495, - "bbox": [ - 373.93, - 492.57, - 396.12, - 501.82 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p695-b46", - "global_id": 20496, - "bbox": [ - 397.96, - 479.98, - 478.81, - 507.33 - ], - "text": "1 −t2\n0 < t ≤1\nx(t + 1)\n∀t", - "type": "text" - }, - { - "block_id": "p695-b47", - "global_id": 20497, - "bbox": [ - 342.79, - 523.81, - 516.14, - 566.02 - ], - "text": "(a) Sketch x(t) for −2 ≤t ≤2.\n(b) Determine\nDn, the exponential\nFourier\nseries spectrum of x(t).\n(c) If x(t) is applied to an ideal bandpass filter", - "type": "text" - }, - { - "block_id": "p695-b48", - "global_id": 20498, - "bbox": [ - 358.22, - 568.01, - 516.11, - 587.94 - ], - "text": "with a 1 Hz passband centered at 3 Hz,\ndetermine the output y(t).", - "type": "text" - }, - { - "block_id": "p695-b49", - "global_id": 20499, - "bbox": [ - 314.15, - 592.85, - 516.14, - 601.89 - ], - "text": "6.4-4\n(a) Find the exponential Fourier series for a", - "type": "text" - }, - { - "block_id": "p695-b50", - "global_id": 20500, - "bbox": [ - 342.79, - 603.5, - 516.13, - 634.77 - ], - "text": "signal x(t) = cos 5t sin3t. You can do this\nwithout evaluating any integrals.\n(b) Sketch the Fourier spectra.", - "type": "text" - } - ] - }, - { - "page_num": 696, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p696-b0", - "global_id": 20501, - "bbox": [ - 60.0, - 60.36, - 428.39, - 69.45 - ], - "text": "676\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p696-b1", - "global_id": 20502, - "bbox": [ - 90.4, - 85.58, - 263.24, - 94.92 - ], - "text": "(c) The signal x(t) is applied at the input of", - "type": "text" - }, - { - "block_id": "p696-b2", - "global_id": 20503, - "bbox": [ - 105.33, - 96.91, - 263.22, - 116.83 - ], - "text": "an LTIC system with frequency response, as\nshown in Fig. P6.4-4. Find the output y(t).", - "type": "text" - }, - { - "block_id": "p696-b3", - "global_id": 20504, - "bbox": [ - 137.82, - 141.89, - 154.26, - 149.99 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p696-b4", - "global_id": 20505, - "bbox": [ - 163.18, - 151.98, - 167.18, - 159.98 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p696-b5", - "global_id": 20506, - "bbox": [ - 93.15, - 176.8, - 251.99, - 186.11 - ], - "text": "6\n2.5\n2.5\n6\nv", - "type": "text" - }, - { - "block_id": "p696-b6", - "global_id": 20507, - "bbox": [ - 75.78, - 198.81, - 127.43, - 207.77 - ], - "text": "Figure P6.4-4", - "type": "text" - }, - { - "block_id": "p696-b7", - "global_id": 20508, - "bbox": [ - 61.25, - 222.96, - 263.24, - 232.01 - ], - "text": "6.4-5\n(a) Find the exponential Fourier series for a", - "type": "text" - }, - { - "block_id": "p696-b8", - "global_id": 20509, - "bbox": [ - 89.89, - 233.62, - 263.24, - 253.92 - ], - "text": "periodic signal x(t) shown in Fig. P6.4-5a.\n(b) The signal x(t) is applied at the input of an", - "type": "text" - }, - { - "block_id": "p696-b9", - "global_id": 20510, - "bbox": [ - 105.33, - 255.91, - 263.23, - 275.84 - ], - "text": "LTIC system shown in Fig. P6.4-5b. Find the\nexpression for the output y(t).", - "type": "text" - }, - { - "block_id": "p696-b10", - "global_id": 20511, - "bbox": [ - 61.25, - 280.64, - 263.24, - 300.94 - ], - "text": "6.4-6\nA T-periodic τ/T duty-cycle square wave p(t) is\ndefined as", - "type": "text" - }, - { - "block_id": "p696-b11", - "global_id": 20512, - "bbox": [ - 114.76, - 323.41, - 137.42, - 332.65 - ], - "text": "p(t) =", - "type": "text" - }, - { - "block_id": "p696-b12", - "global_id": 20513, - "bbox": [ - 139.27, - 305.03, - 146.37, - 322.06 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p696-b13", - "global_id": 20514, - "bbox": [ - 139.27, - 329.24, - 146.37, - 338.2 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p696-b14", - "global_id": 20515, - "bbox": [ - 163.75, - 310.64, - 221.64, - 321.6 - ], - "text": "1\n|t| < τ", - "type": "text" - }, - { - "block_id": "p696-b15", - "global_id": 20516, - "bbox": [ - 151.34, - 317.71, - 230.57, - 343.69 - ], - "text": "2\n0\nτ\n2 < |t| < T\n2\np(t + T)\n∀t", - "type": "text" - }, - { - "block_id": "p696-b16", - "global_id": 20517, - "bbox": [ - 89.89, - 356.74, - 263.24, - 411.26 - ], - "text": "where 0 < τ < T. Also consider the frequency\nresponse H(ω) of a lowpass communications\nchannel with 10 rad/s bandwidth (e.g., |H(ω)| ≈\n0 for ω > 10). If T and τ are properly chosen,\nwe can estimate H(ω) at points ω = nω0 as", - "type": "text" - }, - { - "block_id": "p696-b17", - "global_id": 20518, - "bbox": [ - 89.89, - 409.69, - 263.24, - 431.83 - ], - "text": "ˆH(nω0) =\n1\nP0 Yn, where Yn is the exponential FS\nspectrum of the channel output y(t) in response", - "type": "text" - }, - { - "block_id": "p696-b18", - "global_id": 20519, - "bbox": [ - 317.04, - 85.58, - 490.39, - 105.88 - ], - "text": "to input p(t) and P0 is the dc component of\np(t).", - "type": "text" - }, - { - "block_id": "p696-b19", - "global_id": 20520, - "bbox": [ - 317.55, - 107.87, - 490.37, - 116.83 - ], - "text": "(a) Using direct integration, determine the", - "type": "text" - }, - { - "block_id": "p696-b20", - "global_id": 20521, - "bbox": [ - 317.04, - 118.74, - 490.39, - 149.71 - ], - "text": "exponential Fourier series coefficients Pn of\nsignal p(t).\n(b) Determine a suitable value T so that p(t)", - "type": "text" - }, - { - "block_id": "p696-b21", - "global_id": 20522, - "bbox": [ - 317.55, - 151.33, - 490.38, - 193.55 - ], - "text": "applied to system H(ω) has 21 component\nfrequencies over the system bandwidth 0 ≤\nω ≤10.\n(c) Assuming T is properly chosen, determine a", - "type": "text" - }, - { - "block_id": "p696-b22", - "global_id": 20523, - "bbox": [ - 317.04, - 195.16, - 490.39, - 226.43 - ], - "text": "suitable duty cycle τ/T so that H(nω0) ≈Yn.\nCarefully justify your result.\n(d) From the perspective of using p(t) to help", - "type": "text" - }, - { - "block_id": "p696-b23", - "global_id": 20524, - "bbox": [ - 317.55, - 228.41, - 490.4, - 270.26 - ], - "text": "measure the system frequency response\nH(ω), what happens if T is properly chosen\nbut τ/T is chosen too small?\n(e) From the perspective of using p(t) to help", - "type": "text" - }, - { - "block_id": "p696-b24", - "global_id": 20525, - "bbox": [ - 332.48, - 272.25, - 490.4, - 303.13 - ], - "text": "measure the system frequency response\nH(ω), what happens if T is properly chosen\nbut τ/T is chosen too large?", - "type": "text" - }, - { - "block_id": "p696-b25", - "global_id": 20526, - "bbox": [ - 288.4, - 308.74, - 490.39, - 339.69 - ], - "text": "6.5-1\nDerive Eq. (6.32) in an alternate way by observ-\ning that e = (x−cy) and |e|2 = (x−cy)·(x −\ncy) = |x|2 + c2|y|2 −2cx · y.", - "type": "text" - }, - { - "block_id": "p696-b26", - "global_id": 20527, - "bbox": [ - 288.4, - 344.98, - 490.39, - 366.03 - ], - "text": "6.5-2\nA signal x(t) is approximated in terms of a signal\ny(t) over an interval (t1, t2):", - "type": "text" - }, - { - "block_id": "p696-b27", - "global_id": 20528, - "bbox": [ - 355.84, - 383.74, - 451.09, - 394.05 - ], - "text": "x(t) ≃cy(t)\nt1 < t < t2", - "type": "text" - }, - { - "block_id": "p696-b28", - "global_id": 20529, - "bbox": [ - 317.04, - 411.82, - 490.38, - 431.83 - ], - "text": "where c is chosen to minimize the error\nenergy.", - "type": "text" - }, - { - "block_id": "p696-b29", - "global_id": 20530, - "bbox": [ - 259.62, - 626.64, - 269.43, - 634.64 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p696-b30", - "global_id": 20531, - "bbox": [ - 238.1, - 533.41, - 264.55, - 541.7 - ], - "text": "C 1 F", - "type": "text" - }, - { - "block_id": "p696-b31", - "global_id": 20532, - "bbox": [ - 187.24, - 577.06, - 341.81, - 598.77 - ], - "text": "R 1 \t\nx(t)\ny(t)", - "type": "text" - }, - { - "block_id": "p696-b34", - "global_id": 20533, - "bbox": [ - 260.08, - 512.83, - 268.96, - 520.83 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p696-b35", - "global_id": 20534, - "bbox": [ - 258.62, - 455.41, - 262.62, - 463.41 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p696-b36", - "global_id": 20535, - "bbox": [ - 257.52, - 498.14, - 376.58, - 507.33 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p696-b37", - "global_id": 20536, - "bbox": [ - 269.18, - 451.31, - 280.29, - 459.39 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p696-b38", - "global_id": 20537, - "bbox": [ - 279.86, - 462.79, - 290.08, - 472.41 - ], - "text": "et", - "type": "text" - }, - { - "block_id": "p696-b39", - "global_id": 20538, - "bbox": [ - 159.88, - 499.04, - 365.72, - 507.33 - ], - "text": "1\n2\n3\n3\n2\n1", - "type": "text" - }, - { - "block_id": "p696-b40", - "global_id": 20539, - "bbox": [ - 405.63, - 626.61, - 457.27, - 635.58 - ], - "text": "Figure P6.4-5", - "type": "text" - } - ] - }, - { - "page_num": 697, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p697-b0", - "global_id": 20540, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n677", - "type": "text" - }, - { - "block_id": "p697-b1", - "global_id": 20541, - "bbox": [ - 116.14, - 85.58, - 288.98, - 94.92 - ], - "text": "(a) Show that y(t) and the error e(t) = x(t) −", - "type": "text" - }, - { - "block_id": "p697-b2", - "global_id": 20542, - "bbox": [ - 115.63, - 96.54, - 288.99, - 116.83 - ], - "text": "cy(t) are orthogonal over the interval (t1, t2).\n(b) If possible, explain the result in terms of a", - "type": "text" - }, - { - "block_id": "p697-b3", - "global_id": 20543, - "bbox": [ - 116.14, - 118.83, - 288.98, - 138.76 - ], - "text": "signal-vector analogy.\n(c) Verify this result for the square signal x(t) in", - "type": "text" - }, - { - "block_id": "p697-b4", - "global_id": 20544, - "bbox": [ - 131.07, - 140.74, - 288.97, - 160.67 - ], - "text": "Fig. 6.23 and its approximation in terms of\nsignal sin t.", - "type": "text" - }, - { - "block_id": "p697-b5", - "global_id": 20545, - "bbox": [ - 86.99, - 165.81, - 288.99, - 262.81 - ], - "text": "6.5-3\nIf x(t) and y(t) are orthogonal, then show that\nthe energy of the signal x(t) + y(t) is identical\nto the energy of the signal x(t) −y(t) and is\ngiven by Ex + Ey. Explain this result by using\nthe vector analogy. In general, show that for\northogonal signals x(t) and y(t) and for any pair\nof arbitrary real constants c1 and c2, the energies\nof c1x(t) + c2y(t) and c1x(t) −c2y(t) are both\ngiven by c2", - "type": "text" - }, - { - "block_id": "p697-b6", - "global_id": 20546, - "bbox": [ - 152.62, - 252.44, - 193.83, - 264.97 - ], - "text": "1Ex + c2\n2Ey.", - "type": "text" - }, - { - "block_id": "p697-b7", - "global_id": 20547, - "bbox": [ - 86.99, - 267.95, - 288.98, - 277.29 - ], - "text": "6.5-4\n(a) For the signals x(t) and y(t) depicted in", - "type": "text" - }, - { - "block_id": "p697-b8", - "global_id": 20548, - "bbox": [ - 115.63, - 279.28, - 288.98, - 343.71 - ], - "text": "Fig. P6.5-4, find the component of the form\ny(t) contained in x(t). In other words, find\nthe optimum value of c in the approximation\nx(t) ≈cy(t) so that the error signal energy is\nminimum.\n(b) Find the error signal e(t) and its energy Ee.", - "type": "text" - }, - { - "block_id": "p697-b9", - "global_id": 20549, - "bbox": [ - 131.07, - 345.03, - 288.98, - 375.92 - ], - "text": "Show that the error signal is orthogonal to\ny(t), and that Ex = c2Ey + Ee. Explain this\nresult in terms of vectors.", - "type": "text" - }, - { - "block_id": "p697-b10", - "global_id": 20550, - "bbox": [ - 89.81, - 439.0, - 163.62, - 447.6 - ], - "text": "1\nt\n0", - "type": "text" - }, - { - "block_id": "p697-b11", - "global_id": 20551, - "bbox": [ - 90.45, - 393.64, - 124.13, - 411.98 - ], - "text": "1\nx(t)", - "type": "text" - }, - { - "block_id": "p697-b12", - "global_id": 20552, - "bbox": [ - 128.54, - 458.2, - 137.42, - 466.2 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p697-b13", - "global_id": 20553, - "bbox": [ - 200.66, - 439.0, - 274.47, - 447.6 - ], - "text": "1\nt\n0", - "type": "text" - }, - { - "block_id": "p697-b14", - "global_id": 20554, - "bbox": [ - 201.3, - 393.64, - 234.97, - 411.98 - ], - "text": "1\ny(t)", - "type": "text" - }, - { - "block_id": "p697-b15", - "global_id": 20555, - "bbox": [ - 238.92, - 458.2, - 248.73, - 466.2 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p697-b16", - "global_id": 20556, - "bbox": [ - 85.74, - 474.13, - 137.39, - 483.09 - ], - "text": "Figure P6.5-4", - "type": "text" - }, - { - "block_id": "p697-b17", - "global_id": 20557, - "bbox": [ - 86.99, - 502.01, - 288.99, - 566.14 - ], - "text": "6.5-5\nFor\nthe\nsignals\nx(t)\nand\ny(t)\nshown\nin\nFig. P6.5-4, find the component of the form\nx(t) contained in y(t). In other words, find the\noptimum value of c in the approximation y(t) ≈\ncx(t) so that the error signal energy is minimum.\nWhat is the error signal energy?", - "type": "text" - }, - { - "block_id": "p697-b18", - "global_id": 20558, - "bbox": [ - 87.0, - 571.27, - 288.99, - 635.4 - ], - "text": "6.5-6\nRepresent the signal x(t) shown in Fig. P6.5-4a\nover the interval from 0 to 1 by a trigonometric\nFourier series of fundamental frequency ω0 =\n2π. Compute the error energy in the representa-\ntion of x(t) by only the first N terms of this series\nfor N = 1, 2, 3, and 4.", - "type": "text" - }, - { - "block_id": "p697-b19", - "global_id": 20559, - "bbox": [ - 314.15, - 85.64, - 516.14, - 105.94 - ], - "text": "6.5-7\nRepresent x(t) = t over the interval (0,1) by a\ntrigonometric Fourier series that has", - "type": "text" - }, - { - "block_id": "p697-b20", - "global_id": 20560, - "bbox": [ - 342.78, - 107.56, - 461.49, - 128.83 - ], - "text": "(a) ω0 = 2π and only sine terms\n(b) ω0 = π and only sine terms", - "type": "text" - }, - { - "block_id": "p697-b21", - "global_id": 20561, - "bbox": [ - 342.78, - 129.48, - 516.12, - 160.73 - ], - "text": "(c) ω0 = π and only cosine terms\nYou may use a dc term in these series if\nnecessary.", - "type": "text" - }, - { - "block_id": "p697-b22", - "global_id": 20562, - "bbox": [ - 314.15, - 166.07, - 516.12, - 186.07 - ], - "text": "6.5-8\nIn Ex. 6.15, we represented the function in\nFig. 6.27 by Legendre polynomials.", - "type": "text" - }, - { - "block_id": "p697-b23", - "global_id": 20563, - "bbox": [ - 343.29, - 188.07, - 516.13, - 197.03 - ], - "text": "(a) Use the results in Ex. 6.15 to represent", - "type": "text" - }, - { - "block_id": "p697-b24", - "global_id": 20564, - "bbox": [ - 342.78, - 198.65, - 516.13, - 229.91 - ], - "text": "the signal g(t) in Fig. P6.5-8 by Legendre\npolynomials.\n(b) Compute the error energy for the approx-", - "type": "text" - }, - { - "block_id": "p697-b25", - "global_id": 20565, - "bbox": [ - 358.22, - 231.9, - 516.13, - 251.83 - ], - "text": "imations having one and two (nonzero)\nterms.", - "type": "text" - }, - { - "block_id": "p697-b26", - "global_id": 20566, - "bbox": [ - 357.27, - 295.15, - 494.92, - 305.26 - ], - "text": "t\np", - "type": "text" - }, - { - "block_id": "p697-b27", - "global_id": 20567, - "bbox": [ - 416.94, - 285.25, - 488.19, - 305.33 - ], - "text": "p\n0", - "type": "text" - }, - { - "block_id": "p697-b28", - "global_id": 20568, - "bbox": [ - 428.21, - 272.4, - 432.21, - 280.4 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p697-b29", - "global_id": 20569, - "bbox": [ - 410.51, - 309.32, - 421.18, - 317.61 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p697-b30", - "global_id": 20570, - "bbox": [ - 452.16, - 271.87, - 463.27, - 279.95 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p697-b31", - "global_id": 20571, - "bbox": [ - 342.78, - 332.07, - 394.42, - 341.04 - ], - "text": "Figure P6.5-8", - "type": "text" - }, - { - "block_id": "p697-b32", - "global_id": 20572, - "bbox": [ - 314.15, - 359.29, - 516.14, - 565.6 - ], - "text": "6.5-9\nWalsh functions, which can take on only two\namplitude values, form a complete set of\northonormal functions and are of great practi-\ncal importance in digital applications because\nthey can be easily generated by logic circuitry\nand because multiplication with these func-\ntions can be implemented by simply using a\npolarity-reversing switch. Figure P6.5-9 shows\nthe first eight functions in this set. Represent\nx(t) in Fig. P6.5-4a over the interval (0, 1) by\nusing a Walsh Fourier series with these eight\nbasis functions. Compute the energy of e(t), the\nerror in the approximation, using the first N\nnonzero terms in the series for N = 1, 2, 3, and\n4. In Prob. 6.5-6 we found the trigonometric\nFourier series for x(t). How does the Walsh\nseries compare with the trigonometric series in\nProb. 6.5-6 from the viewpoint of the error\nenergy for a given N?", - "type": "text" - }, - { - "block_id": "p697-b33", - "global_id": 20573, - "bbox": [ - 309.66, - 569.64, - 516.13, - 634.76 - ], - "text": "6.5-10\nFor\nthe\nfour-dimensional\nreal\nspace\nR4,\nthe\nso-called\nWalsh\nbasis\nis\ngiven\nby:\nφ1 = [1,1,1,1], φ2 = [1,1,−1,−1], φ3 =\n[1,−1,−1,1], and φ4 = [1,−1,1,−1]. Denoting\nelements x = [x1,x2,x3,x4] and y = [y1,y2,y3,y4]\n(x,y ∈R4), we can define orthogonality as", - "type": "text" - } - ] - }, - { - "page_num": 698, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p698-b0", - "global_id": 20574, - "bbox": [ - 60.0, - 60.36, - 428.39, - 69.45 - ], - "text": "678\nCHAPTER 6\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES", - "type": "text" - }, - { - "block_id": "p698-b1", - "global_id": 20575, - "bbox": [ - 264.72, - 127.84, - 459.23, - 135.84 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p698-b2", - "global_id": 20576, - "bbox": [ - 264.72, - 203.82, - 459.23, - 211.82 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p698-b3", - "global_id": 20577, - "bbox": [ - 264.72, - 283.5, - 459.23, - 291.5 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p698-b4", - "global_id": 20578, - "bbox": [ - 264.72, - 359.9, - 459.23, - 367.9 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p698-b5", - "global_id": 20579, - "bbox": [ - 252.62, - 123.92, - 453.61, - 131.92 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p698-b6", - "global_id": 20580, - "bbox": [ - 252.62, - 187.9, - 256.62, - 195.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p698-b7", - "global_id": 20581, - "bbox": [ - 252.55, - 281.18, - 256.55, - 289.18 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p698-b8", - "global_id": 20582, - "bbox": [ - 252.62, - 344.07, - 256.62, - 352.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p698-b9", - "global_id": 20583, - "bbox": [ - 449.61, - 187.91, - 453.61, - 195.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p698-b10", - "global_id": 20584, - "bbox": [ - 449.61, - 279.58, - 453.61, - 287.58 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p698-b11", - "global_id": 20585, - "bbox": [ - 449.61, - 343.69, - 453.61, - 351.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p698-b12", - "global_id": 20586, - "bbox": [ - 176.58, - 201.5, - 186.58, - 209.5 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p698-b13", - "global_id": 20587, - "bbox": [ - 138.6, - 281.18, - 224.48, - 289.18 - ], - "text": "0.25\n0.75", - "type": "text" - }, - { - "block_id": "p698-b14", - "global_id": 20588, - "bbox": [ - 138.58, - 357.59, - 224.58, - 365.59 - ], - "text": "0.5\n0.75\n0.25", - "type": "text" - }, - { - "block_id": "p698-b15", - "global_id": 20589, - "bbox": [ - 336.57, - 126.4, - 422.65, - 134.4 - ], - "text": "0.5\n0.75\n0.25", - "type": "text" - }, - { - "block_id": "p698-b16", - "global_id": 20590, - "bbox": [ - 114.11, - 87.89, - 128.21, - 97.5 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p698-b17", - "global_id": 20591, - "bbox": [ - 114.11, - 163.87, - 128.21, - 173.48 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p698-b18", - "global_id": 20592, - "bbox": [ - 114.11, - 238.74, - 128.21, - 248.35 - ], - "text": "x3(t)", - "type": "text" - }, - { - "block_id": "p698-b19", - "global_id": 20593, - "bbox": [ - 114.11, - 319.46, - 128.21, - 329.07 - ], - "text": "x4(t)", - "type": "text" - }, - { - "block_id": "p698-b20", - "global_id": 20594, - "bbox": [ - 311.95, - 87.89, - 326.06, - 97.5 - ], - "text": "x5(t)", - "type": "text" - }, - { - "block_id": "p698-b21", - "global_id": 20595, - "bbox": [ - 336.57, - 202.38, - 422.65, - 210.38 - ], - "text": "0.5\n0.75\n0.25", - "type": "text" - }, - { - "block_id": "p698-b22", - "global_id": 20596, - "bbox": [ - 311.95, - 163.87, - 326.06, - 173.48 - ], - "text": "x6(t)", - "type": "text" - }, - { - "block_id": "p698-b23", - "global_id": 20597, - "bbox": [ - 336.57, - 282.06, - 422.65, - 290.06 - ], - "text": "0.5\n0.75\n0.25", - "type": "text" - }, - { - "block_id": "p698-b24", - "global_id": 20598, - "bbox": [ - 311.95, - 238.74, - 326.06, - 248.35 - ], - "text": "x7(t)", - "type": "text" - }, - { - "block_id": "p698-b25", - "global_id": 20599, - "bbox": [ - 336.57, - 358.46, - 422.65, - 366.46 - ], - "text": "0.5\n0.75\n0.25", - "type": "text" - }, - { - "block_id": "p698-b26", - "global_id": 20600, - "bbox": [ - 311.95, - 319.46, - 326.06, - 329.07 - ], - "text": "x8(t)", - "type": "text" - }, - { - "block_id": "p698-b27", - "global_id": 20601, - "bbox": [ - 104.83, - 388.92, - 156.48, - 397.88 - ], - "text": "Figure P6.5-9", - "type": "text" - }, - { - "block_id": "p698-b28", - "global_id": 20602, - "bbox": [ - 89.89, - 404.76, - 102.5, - 416.0 - ], - "text": "%4", - "type": "text" - }, - { - "block_id": "p698-b29", - "global_id": 20603, - "bbox": [ - 99.26, - 410.18, - 126.84, - 423.31 - ], - "text": "k=1 xky∗", - "type": "text" - }, - { - "block_id": "p698-b30", - "global_id": 20604, - "bbox": [ - 89.89, - 411.49, - 263.24, - 453.71 - ], - "text": "k = 0. In linear algebra terminology,\northogonality here means that the inner product\nof vectors x and y is zero. Lastly, define a vector\nz = [−4,0,1,−7].", - "type": "text" - }, - { - "block_id": "p698-b31", - "global_id": 20605, - "bbox": [ - 90.4, - 455.7, - 263.22, - 464.67 - ], - "text": "(a) Show that the Walsh basis functions are", - "type": "text" - }, - { - "block_id": "p698-b32", - "global_id": 20606, - "bbox": [ - 89.89, - 466.66, - 263.25, - 497.54 - ], - "text": "mutually orthogonal. This requires a total of\nsix calculations.\n(b) Are the Walsh basis functions normal? That", - "type": "text" - }, - { - "block_id": "p698-b33", - "global_id": 20607, - "bbox": [ - 90.4, - 499.54, - 263.24, - 531.16 - ], - "text": "is, does the inner product of each Walsh\nbasis function with itself evaluate to 1?\n(c) Determine the coefficients [c1,c2,c3,c4] to", - "type": "text" - }, - { - "block_id": "p698-b34", - "global_id": 20608, - "bbox": [ - 105.33, - 532.32, - 263.22, - 552.24 - ], - "text": "represent z using Walsh basis functions as\nˆz = %4", - "type": "text" - }, - { - "block_id": "p698-b35", - "global_id": 20609, - "bbox": [ - 89.89, - 542.99, - 263.23, - 629.79 - ], - "text": "k=1 ckφk.\n(d) Determine\nthe\nbest\nthree-dimensional\napproximation ˆz3D to z in terms of the\nWalsh basis functions. That is, your estimate\ncan only be a linear combination of three\nfunctions\nfrom\n[φ1,φ2,φ3,φ4].\nEvaluate\nthe three-term sum to determine the four\nelements of vector ˆz3D.", - "type": "text" - }, - { - "block_id": "p698-b36", - "global_id": 20610, - "bbox": [ - 283.91, - 409.38, - 490.41, - 517.05 - ], - "text": "6.5-11\nA function can be expanded in terms of many\ndifferent types of basis functions, not just\nthe complex exponentials of Fourier analysis.\nFor example, Walsh functions are explored in\nProb. 6.5-9. Laguerre polynomials Lk(t), which\nhave support on the interval [0,∞), are another\npossible set of basis functions. The Laguerre\nexpansion using any number of terms we choose.\nFor the Laguerre expansion, we define orthogo-\nnality a little differently, as", - "type": "text" - }, - { - "block_id": "p698-b37", - "global_id": 20611, - "bbox": [ - 363.91, - 521.36, - 379.36, - 532.32 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p698-b38", - "global_id": 20612, - "bbox": [ - 368.64, - 531.85, - 443.52, - 550.34 - ], - "text": "0\ne−tx(t)y(t)dt = 0.", - "type": "text" - }, - { - "block_id": "p698-b39", - "global_id": 20613, - "bbox": [ - 317.04, - 556.98, - 490.39, - 591.12 - ], - "text": "Notice the presence of the e−t term in the inte-\ngral. Using this definition, Laguerre polynomials\nare orthonormal.", - "type": "text" - }, - { - "block_id": "p698-b40", - "global_id": 20614, - "bbox": [ - 317.55, - 592.75, - 490.38, - 602.82 - ], - "text": "(a) Show that L0(t) is normal. That is, show that", - "type": "text" - }, - { - "block_id": "p698-b41", - "global_id": 20615, - "bbox": [ - 367.94, - 606.06, - 383.39, - 617.02 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p698-b42", - "global_id": 20616, - "bbox": [ - 372.67, - 616.55, - 422.79, - 635.04 - ], - "text": "0\ne−tL0(t)L∗", - "type": "text" - }, - { - "block_id": "p698-b43", - "global_id": 20617, - "bbox": [ - 419.39, - 618.26, - 454.92, - 629.41 - ], - "text": "0(t)dt = 1", - "type": "text" - } - ] - }, - { - "page_num": 699, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p699-b0", - "global_id": 20618, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n679", - "type": "text" - }, - { - "block_id": "p699-b1", - "global_id": 20619, - "bbox": [ - 115.63, - 85.58, - 288.97, - 95.66 - ], - "text": "(b) Show that L1(t) is normal. That is, show that", - "type": "text" - }, - { - "block_id": "p699-b2", - "global_id": 20620, - "bbox": [ - 166.54, - 96.68, - 181.99, - 107.65 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p699-b3", - "global_id": 20621, - "bbox": [ - 171.27, - 107.17, - 221.39, - 125.67 - ], - "text": "0\ne−tL1(t)L∗", - "type": "text" - }, - { - "block_id": "p699-b4", - "global_id": 20622, - "bbox": [ - 217.99, - 108.89, - 253.52, - 120.04 - ], - "text": "1(t)dt = 1", - "type": "text" - }, - { - "block_id": "p699-b5", - "global_id": 20623, - "bbox": [ - 115.63, - 132.13, - 288.98, - 153.78 - ], - "text": "(c) Show that L0(t) is orthogonal to L1(t).\n(d) Compute the coefficient c0 that produces the", - "type": "text" - }, - { - "block_id": "p699-b6", - "global_id": 20624, - "bbox": [ - 116.14, - 154.05, - 288.98, - 186.04 - ], - "text": "best approximation ˆx0(t) = c0L0(t) of the\nfunction x(t) = e−tu(t).\n(e) For the best estimate ˆx1(t) = c0L0(t) +", - "type": "text" - }, - { - "block_id": "p699-b7", - "global_id": 20625, - "bbox": [ - 131.07, - 185.66, - 288.98, - 240.1 - ], - "text": "c1L1(t) of the function x(t) = e−tu(t), the\ncoefficient c0 remains unchanged from part\n()(d) and c1 =\n1\n4. Confirm that c1 =\n1\n4\nand explain why the coefficient c0 does not\nchange.", - "type": "text" - }, - { - "block_id": "p699-b8", - "global_id": 20626, - "bbox": [ - 86.99, - 243.35, - 288.98, - 286.92 - ], - "text": "6.7-1\nA periodic signal has ω0 =\n2\n3π\nand ex-\nponential Fourier series spectrum Dn = jcos\n(πn/10)(u[n + 10] −u[n −11]). [Hint: Refer to\nProb. 6.3-8 for some useful properties.]", - "type": "text" - }, - { - "block_id": "p699-b9", - "global_id": 20627, - "bbox": [ - 116.14, - 288.83, - 288.98, - 299.23 - ], - "text": "(a) Determine the period T0 of the correspond-", - "type": "text" - }, - { - "block_id": "p699-b10", - "global_id": 20628, - "bbox": [ - 115.63, - 299.51, - 288.97, - 319.8 - ], - "text": "ing signal x(t).\n(b) Is the time-domain signal real, imaginary, or", - "type": "text" - }, - { - "block_id": "p699-b11", - "global_id": 20629, - "bbox": [ - 116.14, - 321.8, - 288.97, - 341.72 - ], - "text": "neither? Justify your answer.\n(c) Is the time-domain signal even, odd, or", - "type": "text" - }, - { - "block_id": "p699-b12", - "global_id": 20630, - "bbox": [ - 115.63, - 343.71, - 288.98, - 363.64 - ], - "text": "neither? Justify your answer.\n(d) Use MATLAB to synthesize the time-", - "type": "text" - }, - { - "block_id": "p699-b13", - "global_id": 20631, - "bbox": [ - 131.07, - 365.26, - 288.98, - 386.29 - ], - "text": "domain signal x(t) and plot it over the\ninterval [−T0,T0].", - "type": "text" - }, - { - "block_id": "p699-b14", - "global_id": 20632, - "bbox": [ - 86.99, - 390.46, - 288.97, - 411.43 - ], - "text": "6.7-2\nRepeat Prob. 6.7-1 for a periodic signal with\nω0 = 3", - "type": "text" - }, - { - "block_id": "p699-b15", - "global_id": 20633, - "bbox": [ - 115.63, - 401.13, - 288.98, - 421.42 - ], - "text": "2π and Dn = 2sin(πn/10)(u[n + 10] −\nu[n −11]).", - "type": "text" - }, - { - "block_id": "p699-b16", - "global_id": 20634, - "bbox": [ - 86.99, - 426.03, - 268.15, - 436.34 - ], - "text": "6.7-3\nConsider the (T0 = 1)-periodic signal x(t):", - "type": "text" - }, - { - "block_id": "p699-b17", - "global_id": 20635, - "bbox": [ - 145.16, - 450.15, - 167.35, - 459.4 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p699-b18", - "global_id": 20636, - "bbox": [ - 169.19, - 437.56, - 259.46, - 464.92 - ], - "text": "2t −t2\n0 < t ≤1\nx(t + 1)\n∀t\n.", - "type": "text" - }, - { - "block_id": "p699-b19", - "global_id": 20637, - "bbox": [ - 115.63, - 480.05, - 288.96, - 500.35 - ], - "text": "(a) Sketch x(t) for −2 ≤t ≤2.\n(b) Using properties and minimal integration,", - "type": "text" - }, - { - "block_id": "p699-b20", - "global_id": 20638, - "bbox": [ - 131.07, - 502.26, - 288.97, - 522.26 - ], - "text": "determine Dn, the exponential Fourier spec-\ntrum of x(t).", - "type": "text" - }, - { - "block_id": "p699-b21", - "global_id": 20639, - "bbox": [ - 343.29, - 85.8, - 516.13, - 96.2 - ], - "text": "(c) Verify the correctness of Dn by using MAT-", - "type": "text" - }, - { - "block_id": "p699-b22", - "global_id": 20640, - "bbox": [ - 342.78, - 96.49, - 516.13, - 127.74 - ], - "text": "LAB to synthesize x(t) with a suitable\ntruncation of Eq. (6.19).\n(d) Suppose x(t) is applied to an ideal bandpass", - "type": "text" - }, - { - "block_id": "p699-b23", - "global_id": 20641, - "bbox": [ - 342.79, - 129.74, - 516.13, - 171.58 - ], - "text": "filter with passband between 2.5 and 3.5 Hz.\nDetermine the filter output y(t). Simplify\nyour answer.\n[Hint: Refer to Ex. 6.11.]", - "type": "text" - }, - { - "block_id": "p699-b24", - "global_id": 20642, - "bbox": [ - 314.15, - 176.66, - 516.14, - 251.46 - ], - "text": "6.7-4\nSection 6.7 discusses the construction of a\nphase-optimized multitone test signal with lin-\nearly spaced frequency components. This prob-\nlem investigates a similar signal with logarithmi-\ncally spaced frequency components.\nA multitone test signal m(t) is constructed by\nusing a superposition of N real sinusoids", - "type": "text" - }, - { - "block_id": "p699-b25", - "global_id": 20643, - "bbox": [ - 385.18, - 273.45, - 409.84, - 282.7 - ], - "text": "m(t) =", - "type": "text" - }, - { - "block_id": "p699-b26", - "global_id": 20644, - "bbox": [ - 411.68, - 264.34, - 424.37, - 273.9 - ], - "text": "N\n\"", - "type": "text" - }, - { - "block_id": "p699-b27", - "global_id": 20645, - "bbox": [ - 412.26, - 286.37, - 423.79, - 293.12 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p699-b28", - "global_id": 20646, - "bbox": [ - 425.37, - 273.44, - 473.73, - 283.46 - ], - "text": "cos(ωnt + θn)", - "type": "text" - }, - { - "block_id": "p699-b29", - "global_id": 20647, - "bbox": [ - 342.78, - 304.32, - 516.12, - 324.62 - ], - "text": "where θn establishes the relative phase of each\nsinusoidal component.", - "type": "text" - }, - { - "block_id": "p699-b30", - "global_id": 20648, - "bbox": [ - 343.29, - 326.25, - 516.13, - 335.59 - ], - "text": "(a) Determine a suitable set of N = 10 frequen-", - "type": "text" - }, - { - "block_id": "p699-b31", - "global_id": 20649, - "bbox": [ - 342.78, - 337.2, - 516.14, - 402.68 - ], - "text": "cies ωn that logarithmically spans [(2π) ≤\nω ≤100(2π)] yet still results in a periodic\ntest signal m(t). Determine the period T0 of\nyour signal. Using θn = 0, plot the resulting\n(T0)-periodic signal over −T0/2 ≤t ≤T0/2.\n(b) Determine a suitable set of phases θn that", - "type": "text" - }, - { - "block_id": "p699-b32", - "global_id": 20650, - "bbox": [ - 343.29, - 402.95, - 516.14, - 521.89 - ], - "text": "minimize the maximum magnitude of m(t).\nPlot the resulting signal and identify the\nmaximum magnitude that results.\n(c) Many\nsystems\nsuffer\nfrom\nwhat\nis\ncalled one-over-f\nnoise. The power of\nthis\nundesirable\nnoise\nis\nproportional\nto\n1/f.\nThus,\nlow-frequency\nnoise\nis\nstronger than high-frequency noise. What\nmodifications to m(t) are appropriate for\nuse in environments with 1/f noise? Justify\nyour answer.", - "type": "text" - } - ] - }, - { - "page_num": 700, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p700-b0", - "global_id": 20651, - "bbox": [ - 64.47, - 67.04, - 133.54, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p700-b1", - "global_id": 20652, - "bbox": [ - 147.02, - 100.33, - 438.42, - 173.26 - ], - "text": "CONTINUOUS-TIME SIGNAL\nANALYSIS: THE FOURIER\nTRANSFORM", - "type": "text" - }, - { - "block_id": "p700-b2", - "global_id": 20653, - "bbox": [ - 114.3, - 79.92, - 133.73, - 118.77 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p700-b3", - "global_id": 20654, - "bbox": [ - 101.84, - 266.14, - 490.42, - 395.66 - ], - "text": "We can analyze linear systems in many different ways by taking advantage of the property of\nlinearity, whereby the input is expressed as a sum of simpler components. The system response to\nany complex input can be found by summing the system’s response to these simpler components\nof the input. In time-domain analysis, we separated the input into impulse components. In the\nfrequency-domain analysis in Ch. 4, we separated the input into exponentials of the form est (the\nLaplace transform), where the complex frequency s = σ + jω. The Laplace transform, although\nvery valuable for system analysis, proves somewhat awkward for signal analysis, where we prefer\nto represent signals in terms of exponentials ejωt instead of est. This is accomplished by the Fourier\ntransform. In a sense, the Fourier transform may be considered to be a special case of the Laplace\ntransform with s = jω. Although this view is true most of the time, it does not always hold because\nof the nature of convergence of the Laplace and Fourier integrals.", - "type": "text" - }, - { - "block_id": "p700-b4", - "global_id": 20655, - "bbox": [ - 101.84, - 397.65, - 490.41, - 431.52 - ], - "text": "In Ch. 6, we succeeded in representing periodic signals as a sum of (everlasting) sinusoids or\nexponentials of the form ejωt. The Fourier integral developed in this chapter extends this spectral\nrepresentation to aperiodic signals.", - "type": "text" - }, - { - "block_id": "p700-b5", - "global_id": 20656, - "bbox": [ - 102.2, - 474.58, - 374.36, - 504.47 - ], - "text": "7.1 APERIODIC SIGNAL REPRESENTATION\nBY THE FOURIER INTEGRAL", - "type": "text" - }, - { - "block_id": "p700-b6", - "global_id": 20657, - "bbox": [ - 101.84, - 510.46, - 490.39, - 592.16 - ], - "text": "Applying a limiting process, we now show that an aperiodic signal can be expressed as a\ncontinuous sum (integral) of everlasting exponentials. To represent an aperiodic signal x(t) such\nas the one depicted in Fig. 7.1a by everlasting exponentials, let us construct a new periodic signal\nxT0(t) formed by repeating the signal x(t) at intervals of T0 seconds, as illustrated in Fig. 7.1b. The\nperiod T0 is made long enough to avoid overlap between the repeating pulses. The periodic signal\nxT0(t) can be represented by an exponential Fourier series. If we let T0 →∞, the pulses in the\nperiodic signal repeat after an infinite interval and, therefore,", - "type": "text" - }, - { - "block_id": "p700-b7", - "global_id": 20658, - "bbox": [ - 259.8, - 617.47, - 332.42, - 634.81 - ], - "text": "lim\nT0→∞xT0(t) = x(t)", - "type": "text" - }, - { - "block_id": "p700-b8", - "global_id": 20659, - "bbox": [ - 60.0, - 656.12, - 74.94, - 666.22 - ], - "text": "680", - "type": "text" - } - ] - }, - { - "page_num": 701, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p701-b0", - "global_id": 20660, - "bbox": [ - 232.84, - 62.89, - 516.12, - 71.98 - ], - "text": "7.1\nAperiodic Signal Representation by the Fourier Integral\n681", - "type": "text" - }, - { - "block_id": "p701-b1", - "global_id": 20661, - "bbox": [ - 151.5, - 273.07, - 492.11, - 282.68 - ], - "text": "Figure 7.1 Construction of a periodic signal: (a) signal x(t) and (b) periodic extension of x(t).", - "type": "text" - }, - { - "block_id": "p701-b2", - "global_id": 20662, - "bbox": [ - 127.59, - 305.93, - 516.13, - 330.29 - ], - "text": "Thus, the Fourier series representing xT0(t) will also represent x(t) in the limit T0 →∞. The\nexponential Fourier series for xT0(t) is given by", - "type": "text" - }, - { - "block_id": "p701-b3", - "global_id": 20663, - "bbox": [ - 277.48, - 351.93, - 309.96, - 364.34 - ], - "text": "xT0(t) =", - "type": "text" - }, - { - "block_id": "p701-b4", - "global_id": 20664, - "bbox": [ - 315.7, - 341.75, - 329.8, - 352.42 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p701-b5", - "global_id": 20665, - "bbox": [ - 312.01, - 365.78, - 333.49, - 372.97 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p701-b6", - "global_id": 20666, - "bbox": [ - 334.6, - 350.21, - 516.13, - 363.01 - ], - "text": "Dnejnω0t\n(7.1)", - "type": "text" - }, - { - "block_id": "p701-b7", - "global_id": 20667, - "bbox": [ - 127.59, - 386.27, - 185.66, - 399.36 - ], - "text": "where ω0 = 2π", - "type": "text" - }, - { - "block_id": "p701-b8", - "global_id": 20668, - "bbox": [ - 178.5, - 388.31, - 204.42, - 403.32 - ], - "text": "T0 and", - "type": "text" - }, - { - "block_id": "p701-b9", - "global_id": 20669, - "bbox": [ - 262.36, - 407.69, - 293.85, - 425.72 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p701-b10", - "global_id": 20670, - "bbox": [ - 286.6, - 421.65, - 295.62, - 432.48 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p701-b11", - "global_id": 20671, - "bbox": [ - 298.42, - 400.7, - 322.38, - 414.31 - ], - "text": "# T0/2", - "type": "text" - }, - { - "block_id": "p701-b12", - "global_id": 20672, - "bbox": [ - 303.69, - 425.58, - 323.26, - 434.11 - ], - "text": "−T0/2", - "type": "text" - }, - { - "block_id": "p701-b13", - "global_id": 20673, - "bbox": [ - 324.86, - 412.54, - 516.13, - 426.67 - ], - "text": "xT0(t)e−jnω0tdt\n(7.2)", - "type": "text" - }, - { - "block_id": "p701-b14", - "global_id": 20674, - "bbox": [ - 127.59, - 442.67, - 516.13, - 464.99 - ], - "text": "Observe that integrating xT0(t) over (−T0/2,T0/2) is the same as integrating x(t) over (−∞,∞).\nTherefore, Eq. (7.2) can be expressed as", - "type": "text" - }, - { - "block_id": "p701-b15", - "global_id": 20675, - "bbox": [ - 269.77, - 478.39, - 301.27, - 496.41 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p701-b16", - "global_id": 20676, - "bbox": [ - 294.02, - 492.34, - 303.04, - 503.18 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p701-b17", - "global_id": 20677, - "bbox": [ - 305.85, - 471.4, - 322.79, - 483.46 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p701-b18", - "global_id": 20678, - "bbox": [ - 311.1, - 496.28, - 323.66, - 503.25 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p701-b19", - "global_id": 20679, - "bbox": [ - 325.27, - 483.23, - 516.13, - 495.33 - ], - "text": "x(t)e−jnω0tdt\n(7.3)", - "type": "text" - }, - { - "block_id": "p701-b20", - "global_id": 20680, - "bbox": [ - 127.59, - 515.62, - 516.13, - 537.63 - ], - "text": "It is interesting to see how the nature of the spectrum changes as T0 increases. To understand this\nfascinating behavior, let us define X(ω), a continuous function of ω, as", - "type": "text" - }, - { - "block_id": "p701-b21", - "global_id": 20681, - "bbox": [ - 274.96, - 557.6, - 305.45, - 567.87 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p701-b22", - "global_id": 20682, - "bbox": [ - 307.5, - 544.04, - 324.45, - 556.1 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p701-b23", - "global_id": 20683, - "bbox": [ - 312.77, - 568.91, - 325.32, - 575.89 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p701-b24", - "global_id": 20684, - "bbox": [ - 326.93, - 555.87, - 516.13, - 567.97 - ], - "text": "x(t)e−jωtdt\n(7.4)", - "type": "text" - }, - { - "block_id": "p701-b25", - "global_id": 20685, - "bbox": [ - 127.59, - 588.36, - 297.49, - 598.32 - ], - "text": "A glance at Eqs. (7.3) and (7.4) shows that", - "type": "text" - }, - { - "block_id": "p701-b26", - "global_id": 20686, - "bbox": [ - 289.66, - 611.36, - 321.15, - 629.39 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p701-b27", - "global_id": 20687, - "bbox": [ - 313.89, - 625.31, - 322.91, - 636.15 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p701-b28", - "global_id": 20688, - "bbox": [ - 324.61, - 617.93, - 516.13, - 629.08 - ], - "text": "X(nω0)\n(7.5)", - "type": "text" - } - ] - }, - { - "page_num": 702, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p702-b0", - "global_id": 20689, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "682\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p702-b1", - "global_id": 20690, - "bbox": [ - 214.97, - 169.67, - 218.97, - 177.67 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p702-b2", - "global_id": 20691, - "bbox": [ - 207.4, - 183.22, - 216.28, - 191.22 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p702-b3", - "global_id": 20692, - "bbox": [ - 302.8, - 169.7, - 308.14, - 177.7 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p702-b4", - "global_id": 20693, - "bbox": [ - 302.8, - 247.09, - 308.14, - 255.09 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p702-b5", - "global_id": 20694, - "bbox": [ - 195.84, - 205.22, - 204.62, - 214.77 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p702-b6", - "global_id": 20695, - "bbox": [ - 214.84, - 246.97, - 218.84, - 254.97 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p702-b7", - "global_id": 20696, - "bbox": [ - 207.02, - 260.73, - 216.67, - 268.73 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p702-b8", - "global_id": 20697, - "bbox": [ - 195.84, - 89.97, - 280.27, - 99.52 - ], - "text": "Dn\nEnvelope", - "type": "text" - }, - { - "block_id": "p702-b9", - "global_id": 20698, - "bbox": [ - 248.0, - 99.73, - 274.02, - 117.84 - ], - "text": "X(v)\n1\nT0", - "type": "text" - }, - { - "block_id": "p702-b10", - "global_id": 20699, - "bbox": [ - 244.51, - 205.71, - 274.72, - 213.71 - ], - "text": "Envelope", - "type": "text" - }, - { - "block_id": "p702-b11", - "global_id": 20700, - "bbox": [ - 245.28, - 214.96, - 271.31, - 233.07 - ], - "text": "X(v)\n1\nT0", - "type": "text" - }, - { - "block_id": "p702-b12", - "global_id": 20701, - "bbox": [ - 329.8, - 236.12, - 486.4, - 269.99 - ], - "text": "Figure 7.2 Change in the Fourier spec-\ntrum when the period T0 in Fig. 7.1 is\ndoubled.", - "type": "text" - }, - { - "block_id": "p702-b13", - "global_id": 20702, - "bbox": [ - 101.84, - 292.24, - 490.42, - 446.08 - ], - "text": "This means that the Fourier coefficients Dn are 1/T0 times the samples of X(ω) uniformly\nspaced at intervals of ω0, as depicted in Fig. 7.2a.† Therefore, (1/T0)X(ω) is the envelope for\nthe coefficients Dn. We now let T0 →∞by doubling T0 repeatedly. Doubling T0 halves the\nfundamental frequency ω0 so that there are now twice as many components (samples) in the\nspectrum. However, by doubling T0, the envelope (1/T0)X(ω) is halved, as shown in Fig. 7.2b. If\nwe continue this process of doubling T0 repeatedly, the spectrum progressively becomes denser\nwhile its magnitude becomes smaller. Note, however, that the relative shape of the envelope\nremains the same [proportional to X(ω) in Eq. (7.4)]. In the limit as T0 →∞, ω0 →0 and\nDn →0. This result makes for a spectrum so dense that the spectral components are spaced\nat zero (infinitesimal) intervals. At the same time, the amplitude of each component is zero\n(infinitesimal). We have nothing of everything, yet we have something! This paradox sounds like\nAlice in Wonderland, but as we shall see, these are the classic characteristics of a very familiar\nphenomenon.‡", - "type": "text" - }, - { - "block_id": "p702-b14", - "global_id": 20703, - "bbox": [ - 119.78, - 448.07, - 290.52, - 458.04 - ], - "text": "Substitution of Eq. (7.5) in Eq. (7.1) yields", - "type": "text" - }, - { - "block_id": "p702-b15", - "global_id": 20704, - "bbox": [ - 241.4, - 478.38, - 273.88, - 490.79 - ], - "text": "xT0(t) =", - "type": "text" - }, - { - "block_id": "p702-b16", - "global_id": 20705, - "bbox": [ - 279.63, - 468.2, - 293.73, - 478.87 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p702-b17", - "global_id": 20706, - "bbox": [ - 275.93, - 492.23, - 297.41, - 499.42 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p702-b18", - "global_id": 20707, - "bbox": [ - 299.73, - 471.39, - 329.17, - 482.54 - ], - "text": "X(nω0)", - "type": "text" - }, - { - "block_id": "p702-b19", - "global_id": 20708, - "bbox": [ - 309.68, - 485.77, - 318.7, - 496.6 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p702-b20", - "global_id": 20709, - "bbox": [ - 330.37, - 476.66, - 490.39, - 488.75 - ], - "text": "ejnω0t\n(7.6)", - "type": "text" - }, - { - "block_id": "p702-b21", - "global_id": 20710, - "bbox": [ - 101.84, - 509.94, - 490.38, - 533.35 - ], - "text": "As T0 →∞, ω0 becomes infinitesimal (ω0 →0). Hence, we shall replace ω0 by a more appropriate\nnotation, ω. In terms of this new notation, ω0 = 2π", - "type": "text" - }, - { - "block_id": "p702-b22", - "global_id": 20711, - "bbox": [ - 302.85, - 522.31, - 349.23, - 537.31 - ], - "text": "T0 becomes", - "type": "text" - }, - { - "block_id": "p702-b23", - "global_id": 20712, - "bbox": [ - 275.64, - 545.41, - 314.4, - 562.77 - ], - "text": "ω = 2π", - "type": "text" - }, - { - "block_id": "p702-b24", - "global_id": 20713, - "bbox": [ - 304.66, - 559.78, - 313.68, - 570.62 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p702-b25", - "global_id": 20714, - "bbox": [ - 101.84, - 588.02, - 490.38, - 633.41 - ], - "text": "† For the sake of simplicity, we assume Dn, and therefore X(ω), in Fig. 7.2, to be real. The argument, however,\nis also valid for complex Dn [or X(ω)].\n‡ If nothing else, the reader now has irrefutable proof of the proposition that 0% ownership of everything is\nbetter than 100% ownership of nothing.", - "type": "text" - } - ] - }, - { - "page_num": 703, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p703-b0", - "global_id": 20715, - "bbox": [ - 232.84, - 62.89, - 516.12, - 71.98 - ], - "text": "7.1\nAperiodic Signal Representation by the Fourier Integral\n683", - "type": "text" - }, - { - "block_id": "p703-b1", - "global_id": 20716, - "bbox": [ - 335.93, - 153.35, - 346.21, - 161.55 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p703-b2", - "global_id": 20717, - "bbox": [ - 281.49, - 176.18, - 395.36, - 187.67 - ], - "text": "0\nv\nnv", - "type": "text" - }, - { - "block_id": "p703-b3", - "global_id": 20718, - "bbox": [ - 247.08, - 90.2, - 274.16, - 99.75 - ], - "text": "X(v)ejvt", - "type": "text" - }, - { - "block_id": "p703-b4", - "global_id": 20719, - "bbox": [ - 338.96, - 105.04, - 419.26, - 114.74 - ], - "text": "Area X(nv)ejnvtv", - "type": "text" - }, - { - "block_id": "p703-b5", - "global_id": 20720, - "bbox": [ - 151.5, - 194.06, - 454.21, - 204.37 - ], - "text": "Figure 7.3 The Fourier series becomes the Fourier integral in the limit as T0 →∞.", - "type": "text" - }, - { - "block_id": "p703-b6", - "global_id": 20721, - "bbox": [ - 127.59, - 229.55, - 216.94, - 239.52 - ], - "text": "and Eq. (7.6) becomes", - "type": "text" - }, - { - "block_id": "p703-b7", - "global_id": 20722, - "bbox": [ - 247.56, - 257.28, - 280.05, - 269.69 - ], - "text": "xT0(t) =", - "type": "text" - }, - { - "block_id": "p703-b8", - "global_id": 20723, - "bbox": [ - 285.8, - 247.11, - 299.89, - 257.78 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p703-b9", - "global_id": 20724, - "bbox": [ - 282.1, - 271.13, - 303.57, - 278.33 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p703-b10", - "global_id": 20725, - "bbox": [ - 304.69, - 243.3, - 359.52, - 260.58 - ], - "text": "X(nω)ω", - "type": "text" - }, - { - "block_id": "p703-b11", - "global_id": 20726, - "bbox": [ - 329.54, - 264.36, - 340.5, - 274.73 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p703-b12", - "global_id": 20727, - "bbox": [ - 360.92, - 243.3, - 366.35, - 253.26 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p703-b13", - "global_id": 20728, - "bbox": [ - 367.45, - 255.56, - 395.52, - 267.56 - ], - "text": "e(jnω)t", - "type": "text" - }, - { - "block_id": "p703-b14", - "global_id": 20729, - "bbox": [ - 127.59, - 290.61, - 516.14, - 336.85 - ], - "text": "This equation shows that xT0(t) can be expressed as a sum of everlasting exponentials of\nfrequencies 0,±ω,±2ω,±3ω,. . . (the Fourier series). The amount of the component of\nfrequency nω is [X(nω)ω]/2π. In the limit as T0 →∞, ω →0 and xT0(t) →x(t).\nTherefore,", - "type": "text" - }, - { - "block_id": "p703-b15", - "global_id": 20730, - "bbox": [ - 210.75, - 353.9, - 255.18, - 364.28 - ], - "text": "x(t) = lim", - "type": "text" - }, - { - "block_id": "p703-b16", - "global_id": 20731, - "bbox": [ - 237.45, - 353.9, - 312.63, - 371.23 - ], - "text": "T0→∞xT0(t) = lim", - "type": "text" - }, - { - "block_id": "p703-b17", - "global_id": 20732, - "bbox": [ - 295.24, - 362.81, - 316.71, - 370.08 - ], - "text": "ω→0", - "type": "text" - }, - { - "block_id": "p703-b18", - "global_id": 20733, - "bbox": [ - 319.01, - 347.33, - 329.96, - 371.35 - ], - "text": "1\n2π", - "type": "text" - }, - { - "block_id": "p703-b19", - "global_id": 20734, - "bbox": [ - 336.96, - 343.73, - 351.06, - 354.4 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p703-b20", - "global_id": 20735, - "bbox": [ - 333.27, - 367.75, - 354.75, - 374.94 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p703-b21", - "global_id": 20736, - "bbox": [ - 355.86, - 352.18, - 516.13, - 364.28 - ], - "text": "X(nω)e(jnω)tω\n(7.7)", - "type": "text" - }, - { - "block_id": "p703-b22", - "global_id": 20737, - "bbox": [ - 127.59, - 387.43, - 516.13, - 410.99 - ], - "text": "The sum on the right-hand side of Eq. (7.7) can be viewed as the area under the function X(ω)ejωt,\nas illustrated in Fig. 7.3. Therefore,", - "type": "text" - }, - { - "block_id": "p703-b23", - "global_id": 20738, - "bbox": [ - 267.93, - 426.65, - 304.29, - 443.59 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p703-b24", - "global_id": 20739, - "bbox": [ - 295.82, - 440.29, - 306.78, - 450.67 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p703-b25", - "global_id": 20740, - "bbox": [ - 310.08, - 419.66, - 327.03, - 431.72 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p703-b26", - "global_id": 20741, - "bbox": [ - 315.34, - 444.53, - 327.9, - 451.51 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p703-b27", - "global_id": 20742, - "bbox": [ - 329.51, - 431.49, - 516.13, - 443.59 - ], - "text": "X(ω)ejωtdω\n(7.8)", - "type": "text" - }, - { - "block_id": "p703-b28", - "global_id": 20743, - "bbox": [ - 127.59, - 466.15, - 516.15, - 524.03 - ], - "text": "The integral on the right-hand side is called the Fourier integral. We have now succeeded in\nrepresenting an aperiodic signal x(t) by a Fourier integral (rather than a Fourier series).† This\nintegral is basically a Fourier series (in the limit) with fundamental frequency ω →0, as seen\nfrom Eq. (7.7). The amount of the exponential ejnωt is X(nω)ω/2π. Thus, the function X(ω)\ngiven by Eq. (7.4) acts as a spectral function.", - "type": "text" - }, - { - "block_id": "p703-b29", - "global_id": 20744, - "bbox": [ - 127.59, - 525.62, - 516.14, - 559.91 - ], - "text": "We call X(ω) the direct Fourier transform of x(t), and x(t) the inverse Fourier transform\nof X(ω). The same information is conveyed by the statement that x(t) and X(ω) are a Fourier\ntransform pair. Symbolically, this statement is expressed as", - "type": "text" - }, - { - "block_id": "p703-b30", - "global_id": 20745, - "bbox": [ - 227.31, - 573.12, - 416.41, - 587.2 - ], - "text": "X(ω) = F[x(t)]\nand\nx(t) = F−1 [X(ω)]", - "type": "text" - }, - { - "block_id": "p703-b31", - "global_id": 20746, - "bbox": [ - 127.59, - 610.24, - 516.11, - 633.41 - ], - "text": "† This derivation should not be considered to be a rigorous proof of Eq. (7.8). The situation is not as simple\nas we have made it appear [1].", - "type": "text" - } - ] - }, - { - "page_num": 704, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p704-b0", - "global_id": 20747, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "684\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p704-b1", - "global_id": 20748, - "bbox": [ - 101.84, - 83.6, - 110.14, - 93.56 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p704-b2", - "global_id": 20749, - "bbox": [ - 266.75, - 103.73, - 325.48, - 114.0 - ], - "text": "x(t) ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p704-b3", - "global_id": 20750, - "bbox": [ - 101.84, - 128.79, - 163.56, - 138.75 - ], - "text": "To recapitulate,", - "type": "text" - }, - { - "block_id": "p704-b4", - "global_id": 20751, - "bbox": [ - 249.21, - 154.23, - 279.72, - 164.51 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p704-b5", - "global_id": 20752, - "bbox": [ - 281.76, - 140.68, - 298.7, - 152.74 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p704-b6", - "global_id": 20753, - "bbox": [ - 287.02, - 165.55, - 299.58, - 172.53 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p704-b7", - "global_id": 20754, - "bbox": [ - 301.18, - 152.51, - 490.39, - 164.61 - ], - "text": "x(t)e−jωtdt\n(7.9)", - "type": "text" - }, - { - "block_id": "p704-b8", - "global_id": 20755, - "bbox": [ - 101.84, - 184.63, - 116.22, - 194.59 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p704-b9", - "global_id": 20756, - "bbox": [ - 242.18, - 201.45, - 278.55, - 218.39 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p704-b10", - "global_id": 20757, - "bbox": [ - 270.08, - 215.09, - 281.04, - 225.47 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p704-b11", - "global_id": 20758, - "bbox": [ - 284.34, - 194.46, - 301.28, - 206.52 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p704-b12", - "global_id": 20759, - "bbox": [ - 289.59, - 219.34, - 302.15, - 226.31 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p704-b13", - "global_id": 20760, - "bbox": [ - 303.76, - 206.3, - 490.38, - 218.39 - ], - "text": "X(ω)ejωtdω\n(7.10)", - "type": "text" - }, - { - "block_id": "p704-b14", - "global_id": 20761, - "bbox": [ - 101.84, - 238.41, - 490.4, - 284.14 - ], - "text": "It is helpful to keep in mind that the Fourier integral in Eq. (7.10) is of the nature of a\nFourier series with fundamental frequency ω approaching zero [Eq. (7.7)]. Therefore, most of the\ndiscussion and properties of Fourier series apply to the Fourier transform as well. The transform\nX(ω) is the frequency-domain specification of x(t).", - "type": "text" - }, - { - "block_id": "p704-b15", - "global_id": 20762, - "bbox": [ - 101.85, - 285.82, - 490.39, - 308.15 - ], - "text": "We can plot the spectrum X(ω) as a function of ω. Since X(ω) is complex, we have both\namplitude and angle (or phase) spectra", - "type": "text" - }, - { - "block_id": "p704-b16", - "global_id": 20763, - "bbox": [ - 253.17, - 325.55, - 338.56, - 337.55 - ], - "text": "X(ω) = |X(ω)|ej̸\nX(ω)", - "type": "text" - }, - { - "block_id": "p704-b17", - "global_id": 20764, - "bbox": [ - 101.84, - 354.92, - 490.4, - 365.29 - ], - "text": "in which |X(ω)| is the amplitude and̸ X(ω) is the angle (or phase) of X(ω). According to Eq. (7.9),", - "type": "text" - }, - { - "block_id": "p704-b18", - "global_id": 20765, - "bbox": [ - 248.05, - 387.87, - 286.32, - 398.15 - ], - "text": "X(−ω) =", - "type": "text" - }, - { - "block_id": "p704-b19", - "global_id": 20766, - "bbox": [ - 288.37, - 374.31, - 305.31, - 386.37 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p704-b20", - "global_id": 20767, - "bbox": [ - 293.63, - 399.19, - 306.18, - 406.17 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p704-b21", - "global_id": 20768, - "bbox": [ - 307.79, - 386.15, - 344.0, - 398.15 - ], - "text": "x(t)ejωtdt", - "type": "text" - }, - { - "block_id": "p704-b22", - "global_id": 20769, - "bbox": [ - 101.84, - 421.25, - 268.63, - 431.22 - ], - "text": "Taking the conjugates of both sides yields", - "type": "text" - }, - { - "block_id": "p704-b23", - "global_id": 20770, - "bbox": [ - 258.73, - 446.76, - 490.38, - 458.86 - ], - "text": "x∗(t) ⇐⇒X∗(−ω)\n(7.11)", - "type": "text" - }, - { - "block_id": "p704-b24", - "global_id": 20771, - "bbox": [ - 101.85, - 476.12, - 490.38, - 498.46 - ], - "text": "This property is known as the conjugation property. Now, if x(t) is a real function of t, then\nx(t) = x∗(t), and from the conjugation property, we find that", - "type": "text" - }, - { - "block_id": "p704-b25", - "global_id": 20772, - "bbox": [ - 263.55, - 514.0, - 328.68, - 526.0 - ], - "text": "X(−ω) = X∗(ω)", - "type": "text" - }, - { - "block_id": "p704-b26", - "global_id": 20773, - "bbox": [ - 101.84, - 543.36, - 490.38, - 565.69 - ], - "text": "This is the conjugate symmetry property of the Fourier transform, applicable to real x(t). Therefore,\nfor real x(t),", - "type": "text" - }, - { - "block_id": "p704-b27", - "global_id": 20774, - "bbox": [ - 192.78, - 575.85, - 490.38, - 586.22 - ], - "text": "|X(−ω)| = |X(ω)|\nand̸\nX(−ω) = −̸ X(ω)\n(7.12)", - "type": "text" - }, - { - "block_id": "p704-b28", - "global_id": 20775, - "bbox": [ - 101.85, - 600.5, - 490.38, - 610.87 - ], - "text": "Thus, for real x(t), the amplitude spectrum |X(ω)| is an even function, and the phase spectrum̸", - "type": "text" - }, - { - "block_id": "p704-b29", - "global_id": 20776, - "bbox": [ - 101.84, - 612.45, - 490.4, - 634.79 - ], - "text": "X(ω) is an odd function of ω. These results were derived earlier for the Fourier spectrum of a\nperiodic signal [Eq. (6.22)] and should come as no surprise.", - "type": "text" - } - ] - }, - { - "page_num": 705, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p705-b0", - "global_id": 20777, - "bbox": [ - 232.84, - 62.89, - 516.12, - 71.98 - ], - "text": "7.1\nAperiodic Signal Representation by the Fourier Integral\n685", - "type": "text" - }, - { - "block_id": "p705-b1", - "global_id": 20778, - "bbox": [ - 102.51, - 93.91, - 432.18, - 105.87 - ], - "text": "EXAMPLE 7.1\nFourier Transform of a Causal Exponential", - "type": "text" - }, - { - "block_id": "p705-b2", - "global_id": 20779, - "bbox": [ - 128.9, - 120.9, - 281.79, - 132.5 - ], - "text": "Find the Fourier transform of e−atu(t).", - "type": "text" - }, - { - "block_id": "p705-b3", - "global_id": 20780, - "bbox": [ - 128.9, - 148.44, - 228.33, - 158.4 - ], - "text": "By definition [Eq. (7.9)],", - "type": "text" - }, - { - "block_id": "p705-b4", - "global_id": 20781, - "bbox": [ - 183.1, - 176.15, - 213.61, - 186.43 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p705-b5", - "global_id": 20782, - "bbox": [ - 215.66, - 162.59, - 232.59, - 174.65 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p705-b6", - "global_id": 20783, - "bbox": [ - 220.91, - 187.47, - 233.47, - 194.44 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p705-b7", - "global_id": 20784, - "bbox": [ - 235.08, - 174.43, - 303.18, - 186.43 - ], - "text": "e−atu(t)e−jωtdt =", - "type": "text" - }, - { - "block_id": "p705-b8", - "global_id": 20785, - "bbox": [ - 305.22, - 162.59, - 322.16, - 174.65 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p705-b9", - "global_id": 20786, - "bbox": [ - 310.48, - 169.17, - 437.06, - 194.73 - ], - "text": "0\ne−(a+jω)tdt =\n−1\na + jωe−(a+jω)t", - "type": "text" - }, - { - "block_id": "p705-b11", - "global_id": 20787, - "bbox": [ - 440.94, - 166.79, - 448.06, - 173.76 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p705-b12", - "global_id": 20788, - "bbox": [ - 440.94, - 188.64, - 444.43, - 195.61 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p705-b13", - "global_id": 20789, - "bbox": [ - 128.9, - 201.78, - 502.76, - 227.73 - ], - "text": "But |e−jωt| = 1. Therefore, as t →∞, e−(a+jω)t = e−ate−jωt = ∞if a < 0, but it is equal to 0 if\na > 0. Therefore,", - "type": "text" - }, - { - "block_id": "p705-b14", - "global_id": 20790, - "bbox": [ - 264.81, - 226.99, - 366.85, - 250.91 - ], - "text": "X(ω) =\n1\na + jω\na > 0", - "type": "text" - }, - { - "block_id": "p705-b15", - "global_id": 20791, - "bbox": [ - 128.9, - 260.86, - 281.53, - 271.23 - ], - "text": "Expressing a + jω in the polar form as", - "type": "text" - }, - { - "block_id": "p705-b16", - "global_id": 20792, - "bbox": [ - 284.03, - 252.42, - 292.46, - 262.39 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p705-b17", - "global_id": 20793, - "bbox": [ - 292.45, - 257.9, - 409.17, - 271.23 - ], - "text": "a2 + ω2 ejtan−1(ω/a), we obtain", - "type": "text" - }, - { - "block_id": "p705-b18", - "global_id": 20794, - "bbox": [ - 255.24, - 281.17, - 310.95, - 298.02 - ], - "text": "X(ω) =\n1\n√", - "type": "text" - }, - { - "block_id": "p705-b19", - "global_id": 20795, - "bbox": [ - 297.4, - 283.49, - 375.94, - 306.5 - ], - "text": "a2 + ω2 e−jtan−1 (ω/a)", - "type": "text" - }, - { - "block_id": "p705-b20", - "global_id": 20796, - "bbox": [ - 128.9, - 315.44, - 170.65, - 325.41 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p705-b21", - "global_id": 20797, - "bbox": [ - 204.04, - 324.67, - 265.46, - 341.51 - ], - "text": "|X(ω)| =\n1\n√", - "type": "text" - }, - { - "block_id": "p705-b22", - "global_id": 20798, - "bbox": [ - 251.91, - 320.24, - 420.8, - 349.99 - ], - "text": "a2 + ω2\nand̸\nX(ω) = −tan−1 ω", - "type": "text" - }, - { - "block_id": "p705-b23", - "global_id": 20799, - "bbox": [ - 415.15, - 338.63, - 420.13, - 348.59 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p705-b25", - "global_id": 20800, - "bbox": [ - 128.9, - 355.59, - 502.75, - 377.93 - ], - "text": "The amplitude spectrum |X(ω)| and the phase spectrum̸\nX(ω) are depicted in Fig. 7.4b.\nObserve that |X(ω)| is an even function of ω, and̸\nX(ω) is an odd function of ω, as expected.", - "type": "text" - }, - { - "block_id": "p705-b26", - "global_id": 20801, - "bbox": [ - 151.72, - 472.2, - 236.75, - 481.17 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p705-b27", - "global_id": 20802, - "bbox": [ - 143.1, - 416.89, - 147.1, - 424.89 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p705-b28", - "global_id": 20803, - "bbox": [ - 131.94, - 404.42, - 143.36, - 412.5 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p705-b29", - "global_id": 20804, - "bbox": [ - 184.95, - 440.11, - 210.24, - 449.8 - ], - "text": "eatu(t)", - "type": "text" - }, - { - "block_id": "p705-b30", - "global_id": 20805, - "bbox": [ - 344.31, - 402.34, - 363.73, - 410.64 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p705-b31", - "global_id": 20806, - "bbox": [ - 329.25, - 518.58, - 338.9, - 526.58 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p705-b32", - "global_id": 20807, - "bbox": [ - 322.73, - 491.22, - 401.77, - 504.24 - ], - "text": "X(v)\np", - "type": "text" - }, - { - "block_id": "p705-b33", - "global_id": 20808, - "bbox": [ - 314.25, - 500.34, - 327.4, - 513.09 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p705-b34", - "global_id": 20809, - "bbox": [ - 340.9, - 429.35, - 346.24, - 437.35 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p705-b35", - "global_id": 20810, - "bbox": [ - 341.57, - 438.2, - 345.57, - 446.2 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p705-b36", - "global_id": 20811, - "bbox": [ - 327.1, - 471.51, - 389.03, - 481.27 - ], - "text": "v\n0", - "type": "text" - }, - { - "block_id": "p705-b37", - "global_id": 20812, - "bbox": [ - 180.41, - 518.58, - 189.29, - 526.58 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p705-b38", - "global_id": 20813, - "bbox": [ - 300.64, - 402.05, - 312.0, - 410.35 - ], - "text": "1a", - "type": "text" - }, - { - "block_id": "p705-b39", - "global_id": 20814, - "bbox": [ - 119.94, - 532.39, - 304.21, - 543.25 - ], - "text": "Figure 7.4 (a) e−atu(t) and (b) its Fourier spectra.", - "type": "text" - }, - { - "block_id": "p705-b40", - "global_id": 20815, - "bbox": [ - 127.59, - 597.72, - 516.14, - 647.74 - ], - "text": "EXISTENCE OF THE FOURIER TRANSFORM\nIn Ex. 7.1 we observed that when a < 0, the Fourier integral for e−atu(t) does not converge. Hence,\nthe Fourier transform for e−atu(t) does not exist if a < 0 (growing exponential). Clearly, not all\nsignals are Fourier transformable.", - "type": "text" - } - ] - }, - { - "page_num": 706, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p706-b0", - "global_id": 20816, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "686\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p706-b1", - "global_id": 20817, - "bbox": [ - 101.84, - 85.82, - 490.4, - 131.64 - ], - "text": "Because the Fourier transform is derived here as a limiting case of the Fourier series, it follows\nthat the basic qualifications of the Fourier series, such as equality in the mean and convergence\nconditions in suitably modified form, apply to the Fourier transform as well. It can be shown that\nif x(t) has a finite energy, that is, if", - "type": "text" - }, - { - "block_id": "p706-b2", - "global_id": 20818, - "bbox": [ - 258.6, - 133.37, - 275.55, - 145.43 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p706-b3", - "global_id": 20819, - "bbox": [ - 263.87, - 158.24, - 276.43, - 165.22 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p706-b4", - "global_id": 20820, - "bbox": [ - 278.04, - 143.13, - 333.62, - 157.2 - ], - "text": "|x(t)|2 dt < ∞", - "type": "text" - }, - { - "block_id": "p706-b5", - "global_id": 20821, - "bbox": [ - 101.84, - 172.66, - 486.04, - 183.03 - ], - "text": "then the Fourier transform X(ω) is finite and converges to x(t) in the mean. This means, if we let", - "type": "text" - }, - { - "block_id": "p706-b6", - "global_id": 20822, - "bbox": [ - 231.48, - 199.7, - 275.39, - 210.08 - ], - "text": "ˆx(t) = lim", - "type": "text" - }, - { - "block_id": "p706-b7", - "global_id": 20823, - "bbox": [ - 258.18, - 208.52, - 279.29, - 215.71 - ], - "text": "W→∞", - "type": "text" - }, - { - "block_id": "p706-b8", - "global_id": 20824, - "bbox": [ - 281.58, - 193.13, - 292.54, - 217.15 - ], - "text": "1\n2π", - "type": "text" - }, - { - "block_id": "p706-b9", - "global_id": 20825, - "bbox": [ - 295.84, - 186.15, - 311.48, - 198.42 - ], - "text": "# W", - "type": "text" - }, - { - "block_id": "p706-b10", - "global_id": 20826, - "bbox": [ - 301.1, - 211.02, - 312.35, - 218.21 - ], - "text": "−W", - "type": "text" - }, - { - "block_id": "p706-b11", - "global_id": 20827, - "bbox": [ - 314.47, - 197.98, - 360.55, - 209.98 - ], - "text": "X(ω)ejωtdω", - "type": "text" - }, - { - "block_id": "p706-b12", - "global_id": 20828, - "bbox": [ - 101.85, - 225.78, - 264.91, - 241.15 - ], - "text": "then Eq. (7.10) implies\n# ∞", - "type": "text" - }, - { - "block_id": "p706-b13", - "global_id": 20829, - "bbox": [ - 253.22, - 253.97, - 265.78, - 260.94 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p706-b14", - "global_id": 20830, - "bbox": [ - 267.39, - 234.18, - 311.17, - 252.93 - ], - "text": "x(t) −ˆx(t)", - "type": "text" - }, - { - "block_id": "p706-b15", - "global_id": 20831, - "bbox": [ - 311.17, - 234.18, - 490.38, - 253.03 - ], - "text": "2 dt = 0\n(7.13)", - "type": "text" - }, - { - "block_id": "p706-b16", - "global_id": 20832, - "bbox": [ - 101.85, - 266.18, - 490.39, - 300.47 - ], - "text": "In other words, x(t) and its Fourier integral [the right-hand side of Eq. (7.10)] can differ at some\nvalues of t without contradicting Eq. (7.13). We shall now discuss an alternate set of criteria due\nto Dirichlet for convergence of the Fourier transform.", - "type": "text" - }, - { - "block_id": "p706-b17", - "global_id": 20833, - "bbox": [ - 101.84, - 302.04, - 490.39, - 348.29 - ], - "text": "As with the Fourier series, if x(t) satisfies certain conditions (Dirichlet conditions), its Fourier\ntransform is guaranteed to converge pointwise at all points where x(t) is continuous. Moreover, at\nthe points of discontinuity, x(t) converges to the value midway between the two values of x(t) on\neither side of the discontinuity. The Dirichlet conditions are as follows:", - "type": "text" - }, - { - "block_id": "p706-b18", - "global_id": 20834, - "bbox": [ - 118.78, - 355.84, - 305.61, - 380.0 - ], - "text": "1. x(t) should be absolutely integrable, that is,\n# ∞", - "type": "text" - }, - { - "block_id": "p706-b19", - "global_id": 20835, - "bbox": [ - 281.25, - 392.82, - 293.81, - 399.79 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p706-b20", - "global_id": 20836, - "bbox": [ - 295.42, - 381.5, - 490.38, - 391.88 - ], - "text": "|x(t)|dt < ∞\n(7.14)", - "type": "text" - }, - { - "block_id": "p706-b21", - "global_id": 20837, - "bbox": [ - 118.78, - 407.58, - 490.4, - 453.41 - ], - "text": "If this condition is satisfied, we see that the integral on the right-hand side of Eq. (7.9) is\nguaranteed to have a finite value.\n2. x(t) must have only a finite number of finite discontinuities within any finite interval.\n3. x(t) must contain only a finite number of maxima and minima within any finite interval.", - "type": "text" - }, - { - "block_id": "p706-b22", - "global_id": 20838, - "bbox": [ - 101.85, - 461.39, - 490.42, - 519.17 - ], - "text": "We stress here that although the Dirichlet conditions are sufficient for the existence and\npointwise convergence of the Fourier transform, they are not necessary. For example, we saw in\nEx. 7.1 that a growing exponential, which violates Dirichlet’s first condition in Eq. (7.14), does not\nhave a Fourier transform. But the signal of the form (sinat)/t, which does violate this condition,\ndoes have a Fourier transform.", - "type": "text" - }, - { - "block_id": "p706-b23", - "global_id": 20839, - "bbox": [ - 101.85, - 521.15, - 490.37, - 555.03 - ], - "text": "Any signal that can be generated in practice satisfies the Dirichlet conditions and therefore\nhas a Fourier transform. Thus, the physical existence of a signal is a sufficient condition for the\nexistence of its transform.", - "type": "text" - }, - { - "block_id": "p706-b24", - "global_id": 20840, - "bbox": [ - 102.14, - 569.37, - 323.23, - 581.49 - ], - "text": "LINEARITY OF THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p706-b25", - "global_id": 20841, - "bbox": [ - 101.84, - 585.52, - 264.95, - 595.49 - ], - "text": "The Fourier transform is linear; that is, if", - "type": "text" - }, - { - "block_id": "p706-b26", - "global_id": 20842, - "bbox": [ - 202.76, - 605.44, - 389.47, - 616.59 - ], - "text": "x1(t) ⇐⇒X1(ω)\nand\nx2(t) ⇐⇒X2(ω)", - "type": "text" - }, - { - "block_id": "p706-b27", - "global_id": 20843, - "bbox": [ - 101.84, - 626.19, - 118.99, - 636.16 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p706-b28", - "global_id": 20844, - "bbox": [ - 212.69, - 637.73, - 490.38, - 648.88 - ], - "text": "a1x1(t) + a2x2(t) ⇐⇒a1X1(ω) + a2X2(ω)\n(7.15)", - "type": "text" - } - ] - }, - { - "page_num": 707, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p707-b0", - "global_id": 20845, - "bbox": [ - 232.84, - 62.89, - 516.12, - 71.98 - ], - "text": "7.1\nAperiodic Signal Representation by the Fourier Integral\n687", - "type": "text" - }, - { - "block_id": "p707-b1", - "global_id": 20846, - "bbox": [ - 127.59, - 85.82, - 516.14, - 119.69 - ], - "text": "The proof is trivial and follows directly from Eq. (7.9). This result can be extended to any finite\nnumber of terms. It can be extended to an infinite number of terms only if the conditions required\nfor interchangeability of the operations of summation and integration are satisfied.", - "type": "text" - }, - { - "block_id": "p707-b2", - "global_id": 20847, - "bbox": [ - 127.59, - 145.03, - 415.36, - 156.98 - ], - "text": "7.1-1 Physical Appreciation of the Fourier Transform", - "type": "text" - }, - { - "block_id": "p707-b3", - "global_id": 20848, - "bbox": [ - 127.59, - 163.12, - 516.15, - 280.67 - ], - "text": "In understanding any aspect of the Fourier transform, we should remember that Fourier\nrepresentation is a way of expressing a signal in terms of everlasting sinusoids (or exponentials).\nThe Fourier spectrum of a signal indicates the relative amplitudes and phases of sinusoids that\nare required to synthesize that signal. A periodic signal Fourier spectrum has finite amplitudes\nand exists at discrete frequencies (ω0 and its multiples). Such a spectrum is easy to visualize, but\nthe spectrum of an aperiodic signal is not easy to visualize because it has a continuous spectrum.\nThe continuous spectrum concept can be appreciated by considering an analogous, more tangible\nphenomenon. One familiar example of a continuous distribution is the loading of a beam. Consider\na beam loaded with weights D1,D2,D3,. . .,Dn units at the uniformly spaced points y1,y2,. . .,yn,\nas shown in Fig. 7.5a.", - "type": "text" - }, - { - "block_id": "p707-b4", - "global_id": 20849, - "bbox": [ - 145.52, - 282.56, - 497.59, - 294.12 - ], - "text": "The total load WT on the beam is given by the sum of these loads at each of the n points:", - "type": "text" - }, - { - "block_id": "p707-b5", - "global_id": 20850, - "bbox": [ - 296.9, - 312.69, - 319.94, - 324.14 - ], - "text": "WT =", - "type": "text" - }, - { - "block_id": "p707-b6", - "global_id": 20851, - "bbox": [ - 321.98, - 302.73, - 336.08, - 313.18 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p707-b7", - "global_id": 20852, - "bbox": [ - 323.6, - 327.08, - 334.45, - 334.34 - ], - "text": "i=1", - "type": "text" - }, - { - "block_id": "p707-b8", - "global_id": 20853, - "bbox": [ - 337.18, - 313.0, - 346.31, - 323.77 - ], - "text": "Di", - "type": "text" - }, - { - "block_id": "p707-b9", - "global_id": 20854, - "bbox": [ - 127.59, - 344.62, - 516.14, - 439.77 - ], - "text": "Consider now the case of a continuously loaded beam, as depicted in Fig. 7.5b. In this case,\nalthough there appears to be a load at every point, the load at any one point is zero. This does not\nmean that there is no load on the beam. A meaningful measure of load in this situation is not the\nload at a point, but rather the loading density per unit length at that point. Let X(y) be the loading\ndensity per unit length of beam. It then follows that the load over a beam length y(y →0), at\nsome point y, is X(y)y. To find the total load on the beam, we divide the beam into segments of\ninterval y(y →0). The load over the nth such segment of length y is X(ny)y. The total\nload WT is given by", - "type": "text" - }, - { - "block_id": "p707-b10", - "global_id": 20855, - "bbox": [ - 239.4, - 449.14, - 281.05, - 460.59 - ], - "text": "WT = lim", - "type": "text" - }, - { - "block_id": "p707-b11", - "global_id": 20856, - "bbox": [ - 264.47, - 458.05, - 284.33, - 465.32 - ], - "text": "y→0", - "type": "text" - }, - { - "block_id": "p707-b12", - "global_id": 20857, - "bbox": [ - 285.43, - 438.71, - 299.53, - 449.64 - ], - "text": "yn\n\"", - "type": "text" - }, - { - "block_id": "p707-b13", - "global_id": 20858, - "bbox": [ - 289.2, - 463.2, - 295.28, - 471.51 - ], - "text": "y1", - "type": "text" - }, - { - "block_id": "p707-b14", - "global_id": 20859, - "bbox": [ - 300.64, - 449.14, - 355.37, - 459.41 - ], - "text": "X(ny)y =", - "type": "text" - }, - { - "block_id": "p707-b15", - "global_id": 20860, - "bbox": [ - 357.42, - 435.58, - 373.33, - 448.67 - ], - "text": "# yn", - "type": "text" - }, - { - "block_id": "p707-b16", - "global_id": 20861, - "bbox": [ - 362.68, - 460.68, - 368.77, - 468.98 - ], - "text": "y1", - "type": "text" - }, - { - "block_id": "p707-b17", - "global_id": 20862, - "bbox": [ - 375.43, - 449.14, - 404.32, - 459.41 - ], - "text": "X(y)dy", - "type": "text" - }, - { - "block_id": "p707-b18", - "global_id": 20863, - "bbox": [ - 127.59, - 478.86, - 516.14, - 512.84 - ], - "text": "The load now exists at every point, and y is now a continuous variable. In the case of discrete\nloading (Fig. 7.5a), the load exists only at n discrete points. At other points, there is no load. On\nthe other hand, in the continuously loaded case, the load exists at every point, but at any specific", - "type": "text" - }, - { - "block_id": "p707-b19", - "global_id": 20864, - "bbox": [ - 151.56, - 540.06, - 303.96, - 549.67 - ], - "text": "D1\nD2\nD3\nD4\nDn", - "type": "text" - }, - { - "block_id": "p707-b20", - "global_id": 20865, - "bbox": [ - 152.57, - 585.81, - 302.85, - 596.29 - ], - "text": "y1\ny2\ny3\ny4\nyn", - "type": "text" - }, - { - "block_id": "p707-b21", - "global_id": 20866, - "bbox": [ - 223.48, - 605.26, - 232.36, - 613.26 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p707-b22", - "global_id": 20867, - "bbox": [ - 434.03, - 546.73, - 442.92, - 554.94 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p707-b23", - "global_id": 20868, - "bbox": [ - 321.82, - 584.68, - 468.38, - 598.06 - ], - "text": "ny\ny1\nyn", - "type": "text" - }, - { - "block_id": "p707-b24", - "global_id": 20869, - "bbox": [ - 389.0, - 605.26, - 398.65, - 613.26 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p707-b25", - "global_id": 20870, - "bbox": [ - 368.82, - 533.68, - 385.07, - 541.76 - ], - "text": "X(y)", - "type": "text" - }, - { - "block_id": "p707-b26", - "global_id": 20871, - "bbox": [ - 151.5, - 619.96, - 377.04, - 629.19 - ], - "text": "Figure 7.5 Weight-loading analogy for the Fourier transform.", - "type": "text" - } - ] - }, - { - "page_num": 708, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p708-b0", - "global_id": 20872, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "688\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p708-b1", - "global_id": 20873, - "bbox": [ - 101.84, - 85.46, - 490.38, - 107.79 - ], - "text": "point y, the load is zero. The load over a small interval y, however, is [X(ny)]y (Fig. 7.5b).\nThus, even though the load at a point y is zero, the relative load at that point is X(y).", - "type": "text" - }, - { - "block_id": "p708-b2", - "global_id": 20874, - "bbox": [ - 101.85, - 109.38, - 490.41, - 143.66 - ], - "text": "An exactly analogous situation exists in the case of a signal spectrum. When x(t) is periodic,\nthe spectrum is discrete, and x(t) can be expressed as a sum of discrete exponentials with finite\namplitudes:", - "type": "text" - }, - { - "block_id": "p708-b3", - "global_id": 20875, - "bbox": [ - 259.35, - 146.9, - 284.0, - 157.18 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p708-b4", - "global_id": 20876, - "bbox": [ - 286.05, - 137.43, - 300.15, - 147.39 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p708-b5", - "global_id": 20877, - "bbox": [ - 291.35, - 160.97, - 294.83, - 167.94 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p708-b6", - "global_id": 20878, - "bbox": [ - 301.25, - 145.18, - 332.27, - 157.98 - ], - "text": "Dnejnω0t", - "type": "text" - }, - { - "block_id": "p708-b7", - "global_id": 20879, - "bbox": [ - 101.84, - 174.17, - 490.4, - 255.86 - ], - "text": "For an aperiodic signal, the spectrum becomes continuous; that is, the spectrum exists for every\nvalue of ω, but the amplitude of each component in the spectrum is zero. The meaningful measure\nhere is not the amplitude of a component of some frequency but the spectral density per unit\nbandwidth. From Eq. (7.7), it is clear that x(t) is synthesized by adding exponentials of the form\nejnωt, in which the contribution by any one exponential component is zero. But the contribution\nby exponentials in an infinitesimal band ω located at ω = nω is (1/2π)X(nω)ω, and the\naddition of all these components yields x(t) in the integral form:", - "type": "text" - }, - { - "block_id": "p708-b8", - "global_id": 20880, - "bbox": [ - 166.83, - 274.43, - 210.92, - 284.81 - ], - "text": "x(t) = lim", - "type": "text" - }, - { - "block_id": "p708-b9", - "global_id": 20881, - "bbox": [ - 193.54, - 283.34, - 215.01, - 290.61 - ], - "text": "ω→0", - "type": "text" - }, - { - "block_id": "p708-b10", - "global_id": 20882, - "bbox": [ - 217.3, - 267.86, - 228.25, - 291.88 - ], - "text": "1\n2π", - "type": "text" - }, - { - "block_id": "p708-b11", - "global_id": 20883, - "bbox": [ - 235.25, - 264.26, - 249.35, - 274.93 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p708-b12", - "global_id": 20884, - "bbox": [ - 231.56, - 288.28, - 253.04, - 295.47 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p708-b13", - "global_id": 20885, - "bbox": [ - 254.15, - 267.86, - 352.79, - 284.81 - ], - "text": "X(nω)e(jnω)tω = 1", - "type": "text" - }, - { - "block_id": "p708-b14", - "global_id": 20886, - "bbox": [ - 344.32, - 281.5, - 355.28, - 291.88 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p708-b15", - "global_id": 20887, - "bbox": [ - 358.58, - 260.88, - 375.53, - 272.93 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p708-b16", - "global_id": 20888, - "bbox": [ - 363.84, - 285.75, - 376.4, - 292.72 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p708-b17", - "global_id": 20889, - "bbox": [ - 378.01, - 270.63, - 425.2, - 284.7 - ], - "text": "X(ω)ejωt dω", - "type": "text" - }, - { - "block_id": "p708-b18", - "global_id": 20890, - "bbox": [ - 101.84, - 304.23, - 490.41, - 386.33 - ], - "text": "Thus, nω approaches a continuous variable ω. The spectrum now exists at every ω. The\ncontribution by components within a band dω is (1/2π)X(ω)dω = X(ω)df, where df is the\nbandwidth in hertz. Clearly, X(ω) is the spectral density per unit bandwidth (in hertz).† It also\nfollows that even if the amplitude of any one component is infinitesimal, the relative amount\nof a component of frequency ω is X(ω). Although X(ω) is a spectral density, in practice, it is\ncustomarily called the spectrum of x(t) rather than the spectral density of x(t). Deferring to this\nconvention, we shall call X(ω) the Fourier spectrum (or Fourier transform) of x(t).", - "type": "text" - }, - { - "block_id": "p708-b19", - "global_id": 20891, - "bbox": [ - 102.14, - 400.79, - 279.2, - 412.91 - ], - "text": "A MARVELOUS BALANCING ACT", - "type": "text" - }, - { - "block_id": "p708-b20", - "global_id": 20892, - "bbox": [ - 101.84, - 416.53, - 490.4, - 510.59 - ], - "text": "An important point to remember here is that x(t) is represented (or synthesized) by exponentials\nor sinusoids that are everlasting (not causal). Such conceptualization leads to a rather fascinating\npicture when we try to visualize the synthesis of a timelimited pulse signal x(t) [Fig. 7.6] by the\nsinusoidal components in its Fourier spectrum. The signal x(t) exists only over an interval (a,b)\nand is zero outside this interval. The spectrum of x(t) contains an infinite number of exponentials\n(or sinusoids), which start at t = −∞and continue forever. The amplitudes and phases of these\ncomponents add up exactly to x(t) over the finite interval (a,b) and to zero everywhere outside\nthis interval. Juggling the amplitudes and phases of an infinite number of components to achieve", - "type": "text" - }, - { - "block_id": "p708-b21", - "global_id": 20893, - "bbox": [ - 166.08, - 575.81, - 286.21, - 583.81 - ], - "text": "a\nb", - "type": "text" - }, - { - "block_id": "p708-b22", - "global_id": 20894, - "bbox": [ - 204.07, - 529.46, - 215.37, - 537.54 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p708-b23", - "global_id": 20895, - "bbox": [ - 234.51, - 574.16, - 236.73, - 582.16 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p708-b24", - "global_id": 20896, - "bbox": [ - 330.8, - 563.22, - 486.39, - 585.14 - ], - "text": "Figure 7.6 The marvel of the Fourier\ntransform.", - "type": "text" - }, - { - "block_id": "p708-b25", - "global_id": 20897, - "bbox": [ - 101.84, - 610.24, - 490.37, - 633.41 - ], - "text": "† To stress that the signal spectrum is a density function, we shall shade the plot of |X(ω)| (as in Fig. 7.4b).\nThe representation of̸\nX(ω), however, will be a line plot, primarily to avoid visual confusion.", - "type": "text" - } - ] - }, - { - "page_num": 709, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p709-b0", - "global_id": 20898, - "bbox": [ - 304.47, - 62.89, - 516.12, - 71.98 - ], - "text": "7.2\nTransforms of Some Useful Functions\n689", - "type": "text" - }, - { - "block_id": "p709-b1", - "global_id": 20899, - "bbox": [ - 127.59, - 85.82, - 516.14, - 119.69 - ], - "text": "such a perfect and delicate balance boggles the human imagination. Yet the Fourier transform\naccomplishes it routinely, without much thinking on our part. Indeed, we become so involved in\nmathematical manipulations that we fail to notice this marvel.", - "type": "text" - }, - { - "block_id": "p709-b2", - "global_id": 20900, - "bbox": [ - 127.94, - 152.18, - 439.82, - 166.13 - ], - "text": "7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS", - "type": "text" - }, - { - "block_id": "p709-b3", - "global_id": 20901, - "bbox": [ - 127.59, - 172.12, - 516.14, - 194.03 - ], - "text": "For convenience, we now introduce a compact notation for the useful gate, triangle, and\ninterpolation functions.", - "type": "text" - }, - { - "block_id": "p709-b4", - "global_id": 20902, - "bbox": [ - 127.89, - 210.37, - 250.02, - 222.49 - ], - "text": "UNIT GATE FUNCTION", - "type": "text" - }, - { - "block_id": "p709-b5", - "global_id": 20903, - "bbox": [ - 127.59, - 226.11, - 516.14, - 248.44 - ], - "text": "We define a unit gate function rect (x) as a gate pulse of unit height and unit width, centered at the\norigin, as illustrated in Fig. 7.7a†:", - "type": "text" - }, - { - "block_id": "p709-b6", - "global_id": 20904, - "bbox": [ - 265.31, - 279.41, - 303.61, - 289.78 - ], - "text": "rect (x) =", - "type": "text" - }, - { - "block_id": "p709-b7", - "global_id": 20905, - "bbox": [ - 305.66, - 255.99, - 313.55, - 277.91 - ], - "text": "⎧\n⎪⎨", - "type": "text" - }, - { - "block_id": "p709-b8", - "global_id": 20906, - "bbox": [ - 305.66, - 285.88, - 313.55, - 298.83 - ], - "text": "⎪⎩", - "type": "text" - }, - { - "block_id": "p709-b9", - "global_id": 20907, - "bbox": [ - 318.53, - 261.82, - 371.04, - 273.54 - ], - "text": "0\n|x| > 1", - "type": "text" - }, - { - "block_id": "p709-b10", - "global_id": 20908, - "bbox": [ - 319.73, - 269.38, - 371.04, - 292.7 - ], - "text": "2\n1\n2\n|x| = 1", - "type": "text" - }, - { - "block_id": "p709-b11", - "global_id": 20909, - "bbox": [ - 318.53, - 285.72, - 371.04, - 306.23 - ], - "text": "2\n1\n|x| < 1", - "type": "text" - }, - { - "block_id": "p709-b12", - "global_id": 20910, - "bbox": [ - 367.55, - 302.07, - 371.04, - 309.04 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p709-b13", - "global_id": 20911, - "bbox": [ - 492.07, - 279.82, - 516.13, - 289.78 - ], - "text": "(7.16)", - "type": "text" - }, - { - "block_id": "p709-b14", - "global_id": 20912, - "bbox": [ - 127.59, - 320.45, - 516.13, - 354.74 - ], - "text": "The gate pulse in Fig. 7.7b is the unit gate pulse rect (x) expanded by a factor τ along the\nhorizontal axis and therefore can be expressed as rect (x/τ) (see Sec. 1.2-2). Observe that τ, the\ndenominator of the argument of rect (x/τ), indicates the width of the pulse.", - "type": "text" - }, - { - "block_id": "p709-b15", - "global_id": 20913, - "bbox": [ - 127.89, - 371.07, - 276.13, - 383.19 - ], - "text": "UNIT TRIANGLE FUNCTION", - "type": "text" - }, - { - "block_id": "p709-b16", - "global_id": 20914, - "bbox": [ - 127.59, - 386.81, - 516.13, - 409.14 - ], - "text": "We define a unit triangle function (x) as a triangular pulse of unit height and unit width, centered\nat the origin, as shown in Fig. 7.8a", - "type": "text" - }, - { - "block_id": "p709-b17", - "global_id": 20915, - "bbox": [ - 257.56, - 432.52, - 287.27, - 442.8 - ], - "text": "(x) =", - "type": "text" - }, - { - "block_id": "p709-b18", - "global_id": 20916, - "bbox": [ - 289.32, - 415.55, - 296.2, - 425.51 - ], - "text": ")", - "type": "text" - }, - { - "block_id": "p709-b19", - "global_id": 20917, - "bbox": [ - 301.18, - 423.11, - 378.53, - 434.83 - ], - "text": "0\n|x| ≥1", - "type": "text" - }, - { - "block_id": "p709-b20", - "global_id": 20918, - "bbox": [ - 301.18, - 430.67, - 378.79, - 451.18 - ], - "text": "2\n1 −2|x|\n|x| < 1", - "type": "text" - }, - { - "block_id": "p709-b21", - "global_id": 20919, - "bbox": [ - 375.3, - 432.94, - 516.13, - 453.99 - ], - "text": "2\n(7.17)", - "type": "text" - }, - { - "block_id": "p709-b22", - "global_id": 20920, - "bbox": [ - 198.41, - 559.42, - 383.08, - 567.42 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p709-b23", - "global_id": 20921, - "bbox": [ - 239.77, - 524.42, - 243.33, - 532.42 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p709-b24", - "global_id": 20922, - "bbox": [ - 195.87, - 537.65, - 199.87, - 545.65 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p709-b25", - "global_id": 20923, - "bbox": [ - 196.51, - 490.11, - 231.41, - 501.89 - ], - "text": "1\nrect (x)", - "type": "text" - }, - { - "block_id": "p709-b26", - "global_id": 20924, - "bbox": [ - 171.52, - 535.84, - 467.13, - 553.97 - ], - "text": "1\n2\n1\n2\nx\nt\n2\nt\n2\n0", - "type": "text" - }, - { - "block_id": "p709-b27", - "global_id": 20925, - "bbox": [ - 371.88, - 486.35, - 410.81, - 501.92 - ], - "text": "1\nrect ( )\nx\nt", - "type": "text" - }, - { - "block_id": "p709-b29", - "global_id": 20926, - "bbox": [ - 151.5, - 574.11, - 244.09, - 583.35 - ], - "text": "Figure 7.7 A gate pulse.", - "type": "text" - }, - { - "block_id": "p709-b30", - "global_id": 20927, - "bbox": [ - 127.59, - 610.24, - 516.14, - 633.41 - ], - "text": "† At |x| = 0.5, we require rect(x) = 0.5 because the inverse Fourier transform of a discontinuous signal\nconverges to the mean of its two values at the discontinuity.", - "type": "text" - } - ] - }, - { - "page_num": 710, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p710-b0", - "global_id": 20928, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "690\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p710-b1", - "global_id": 20929, - "bbox": [ - 170.08, - 159.74, - 351.92, - 167.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p710-b2", - "global_id": 20930, - "bbox": [ - 177.48, - 137.59, - 181.48, - 145.59 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p710-b3", - "global_id": 20931, - "bbox": [ - 167.12, - 98.09, - 195.98, - 106.38 - ], - "text": "(x)\n1", - "type": "text" - }, - { - "block_id": "p710-b4", - "global_id": 20932, - "bbox": [ - 215.14, - 136.63, - 353.89, - 145.59 - ], - "text": "x\n0", - "type": "text" - }, - { - "block_id": "p710-b5", - "global_id": 20933, - "bbox": [ - 339.61, - 97.11, - 343.61, - 105.11 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p710-b6", - "global_id": 20934, - "bbox": [ - 144.82, - 136.63, - 431.97, - 154.33 - ], - "text": "x\n1\n2\n1\n2\nt\n2\nt\n2", - "type": "text" - }, - { - "block_id": "p710-b7", - "global_id": 20935, - "bbox": [ - 361.01, - 93.86, - 378.46, - 108.81 - ], - "text": "( )\nx\nt", - "type": "text" - }, - { - "block_id": "p710-b9", - "global_id": 20936, - "bbox": [ - 125.76, - 174.44, - 230.86, - 183.68 - ], - "text": "Figure 7.8 A triangle pulse.", - "type": "text" - }, - { - "block_id": "p710-b10", - "global_id": 20937, - "bbox": [ - 101.84, - 204.47, - 490.39, - 226.8 - ], - "text": "The pulse in Fig. 7.8b is (x/τ). Observe that here, as for the gate pulse, the denominator τ of the\nargument of (x/τ) indicates the pulse width.", - "type": "text" - }, - { - "block_id": "p710-b11", - "global_id": 20938, - "bbox": [ - 102.14, - 241.06, - 295.03, - 253.98 - ], - "text": "INTERPOLATION FUNCTION SINC(x)", - "type": "text" - }, - { - "block_id": "p710-b12", - "global_id": 20939, - "bbox": [ - 101.84, - 254.39, - 490.39, - 291.89 - ], - "text": "The function sinx/x is the “sine over argument” function denoted by sinc(x).† This function plays\nan important role in signal processing. It is also known as the filtering or interpolating function.\nWe define", - "type": "text" - }, - { - "block_id": "p710-b13", - "global_id": 20940, - "bbox": [ - 265.88, - 290.67, - 325.14, - 307.72 - ], - "text": "sinc(x) = sinx", - "type": "text" - }, - { - "block_id": "p710-b14", - "global_id": 20941, - "bbox": [ - 314.34, - 297.76, - 490.38, - 314.68 - ], - "text": "x\n(7.18)", - "type": "text" - }, - { - "block_id": "p710-b15", - "global_id": 20942, - "bbox": [ - 101.84, - 320.35, - 282.32, - 330.31 - ], - "text": "Inspection of Eq. (7.18) shows the following:", - "type": "text" - }, - { - "block_id": "p710-b16", - "global_id": 20943, - "bbox": [ - 118.78, - 337.86, - 490.4, - 419.97 - ], - "text": "1. sinc(x) is an even function of x.\n2. sinc(x) = 0 when sin x = 0 except at x = 0, where it appears to be indeterminate. This\nmeans that sincx = 0 for x = ±π,±2π,±3π,. . ..\n3. Using L’Hôpital’s rule, we find sinc(0) = 1.\n4. sinc(x) is the product of an oscillating signal sinx (of period 2π) and a monotonically\ndecreasing function 1/x. Therefore, sinc(x) exhibits damped oscillations of period 2π,\nwith amplitude decreasing continuously as 1/x.", - "type": "text" - }, - { - "block_id": "p710-b17", - "global_id": 20944, - "bbox": [ - 101.85, - 427.53, - 490.39, - 461.82 - ], - "text": "Figure 7.9a shows sinc(x). Observe that sinc(x) = 0 for values of x that are positive and\nnegative integer multiples of π. Figure 7.9b shows sinc(3ω/7). The argument 3ω/7 = π when\nω = 7π/3. Therefore, the first zero of this function occurs at ω = 7π/3.", - "type": "text" - }, - { - "block_id": "p710-b18", - "global_id": 20945, - "bbox": [ - 107.82, - 494.88, - 326.28, - 506.84 - ], - "text": "DRILL 7.1\nSketching Basic Functions", - "type": "text" - }, - { - "block_id": "p710-b19", - "global_id": 20946, - "bbox": [ - 107.82, - 515.54, - 430.57, - 525.92 - ], - "text": "Sketch: (a) rect(x/8), (b) (ω/10), (c) sinc(3πω/2), and (d) sinc(t)rect(t/4π).", - "type": "text" - }, - { - "block_id": "p710-b20", - "global_id": 20947, - "bbox": [ - 101.84, - 592.97, - 395.74, - 605.19 - ], - "text": "† sinc(x) is also denoted by Sa (x) in the literature. Some authors define sinc(x) as", - "type": "text" - }, - { - "block_id": "p710-b21", - "global_id": 20948, - "bbox": [ - 265.64, - 611.84, - 325.38, - 627.46 - ], - "text": "sinc(x) = sinπx", - "type": "text" - }, - { - "block_id": "p710-b22", - "global_id": 20949, - "bbox": [ - 309.38, - 624.49, - 319.65, - 633.74 - ], - "text": "πx", - "type": "text" - } - ] - }, - { - "page_num": 711, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p711-b0", - "global_id": 20950, - "bbox": [ - 304.47, - 62.89, - 516.12, - 71.98 - ], - "text": "7.2\nTransforms of Some Useful Functions\n691", - "type": "text" - }, - { - "block_id": "p711-b1", - "global_id": 20951, - "bbox": [ - 281.75, - 244.52, - 290.63, - 252.52 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p711-b2", - "global_id": 20952, - "bbox": [ - 280.59, - 349.75, - 290.24, - 357.75 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p711-b3", - "global_id": 20953, - "bbox": [ - 288.52, - 325.76, - 292.52, - 333.76 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p711-b4", - "global_id": 20954, - "bbox": [ - 276.87, - 267.82, - 280.87, - 275.82 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p711-b5", - "global_id": 20955, - "bbox": [ - 306.04, - 326.75, - 396.09, - 343.57 - ], - "text": "v\n14p\n3\n7p\n3", - "type": "text" - }, - { - "block_id": "p711-b6", - "global_id": 20956, - "bbox": [ - 190.07, - 179.1, - 419.61, - 189.31 - ], - "text": "x\n0\n2p\np\np\n2p\n3p", - "type": "text" - }, - { - "block_id": "p711-b7", - "global_id": 20957, - "bbox": [ - 291.26, - 106.94, - 295.26, - 114.94 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p711-b8", - "global_id": 20958, - "bbox": [ - 318.35, - 134.03, - 342.11, - 142.11 - ], - "text": "sinc (x)", - "type": "text" - }, - { - "block_id": "p711-b9", - "global_id": 20959, - "bbox": [ - 368.08, - 179.22, - 377.41, - 187.33 - ], - "text": "3p", - "type": "text" - }, - { - "block_id": "p711-b10", - "global_id": 20960, - "bbox": [ - 302.77, - 94.98, - 306.77, - 109.71 - ], - "text": "1\nx", - "type": "text" - }, - { - "block_id": "p711-b11", - "global_id": 20961, - "bbox": [ - 212.85, - 326.62, - 234.06, - 343.57 - ], - "text": "14p\n 3", - "type": "text" - }, - { - "block_id": "p711-b12", - "global_id": 20962, - "bbox": [ - 248.37, - 326.62, - 266.3, - 343.57 - ], - "text": "7p\n 3", - "type": "text" - }, - { - "block_id": "p711-b13", - "global_id": 20963, - "bbox": [ - 313.31, - 270.52, - 345.16, - 287.47 - ], - "text": "sinc \n)\n3v\n7(", - "type": "text" - }, - { - "block_id": "p711-b14", - "global_id": 20964, - "bbox": [ - 302.77, - 216.53, - 314.73, - 231.35 - ], - "text": "1\nx", - "type": "text" - }, - { - "block_id": "p711-b15", - "global_id": 20965, - "bbox": [ - 151.5, - 364.44, - 243.66, - 373.68 - ], - "text": "Figure 7.9 A sinc pulse.", - "type": "text" - }, - { - "block_id": "p711-b16", - "global_id": 20966, - "bbox": [ - 102.51, - 397.29, - 424.87, - 409.25 - ], - "text": "EXAMPLE 7.2\nFourier Transform of a Rectangular Pulse", - "type": "text" - }, - { - "block_id": "p711-b17", - "global_id": 20967, - "bbox": [ - 128.9, - 425.01, - 363.08, - 435.39 - ], - "text": "Find the Fourier transform of x(t) = rect(t/τ) (Fig. 7.10a).", - "type": "text" - }, - { - "block_id": "p711-b18", - "global_id": 20968, - "bbox": [ - 258.4, - 451.25, - 288.89, - 461.53 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p711-b19", - "global_id": 20969, - "bbox": [ - 290.94, - 437.69, - 307.89, - 449.75 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p711-b20", - "global_id": 20970, - "bbox": [ - 296.21, - 462.57, - 308.76, - 469.55 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p711-b21", - "global_id": 20971, - "bbox": [ - 310.36, - 451.67, - 325.29, - 461.63 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p711-b22", - "global_id": 20972, - "bbox": [ - 326.41, - 440.26, - 337.08, - 454.55 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p711-b23", - "global_id": 20973, - "bbox": [ - 333.03, - 458.33, - 337.35, - 468.29 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p711-b25", - "global_id": 20974, - "bbox": [ - 346.27, - 449.53, - 373.09, - 461.53 - ], - "text": "e−jωtdt", - "type": "text" - }, - { - "block_id": "p711-b26", - "global_id": 20975, - "bbox": [ - 128.9, - 483.13, - 393.07, - 493.51 - ], - "text": "Since rect (t/τ) = 1 for |t| < τ/2, and since it is zero for |t| > τ/2,", - "type": "text" - }, - { - "block_id": "p711-b27", - "global_id": 20976, - "bbox": [ - 225.32, - 510.73, - 255.83, - 521.0 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p711-b28", - "global_id": 20977, - "bbox": [ - 257.87, - 497.17, - 278.08, - 509.52 - ], - "text": "# τ/2", - "type": "text" - }, - { - "block_id": "p711-b29", - "global_id": 20978, - "bbox": [ - 263.13, - 522.04, - 278.96, - 529.31 - ], - "text": "−τ/2", - "type": "text" - }, - { - "block_id": "p711-b30", - "global_id": 20979, - "bbox": [ - 280.57, - 509.0, - 307.39, - 521.0 - ], - "text": "e−jωtdt", - "type": "text" - }, - { - "block_id": "p711-b31", - "global_id": 20980, - "bbox": [ - 248.06, - 542.5, - 274.08, - 559.03 - ], - "text": "= −1", - "type": "text" - }, - { - "block_id": "p711-b32", - "global_id": 20981, - "bbox": [ - 266.84, - 544.95, - 355.01, - 566.42 - ], - "text": "jω(e−jωτ/2 −ejωτ/2) =", - "type": "text" - }, - { - "block_id": "p711-b33", - "global_id": 20982, - "bbox": [ - 358.25, - 521.53, - 396.03, - 545.89 - ], - "text": "2sin\nωτ", - "type": "text" - }, - { - "block_id": "p711-b34", - "global_id": 20983, - "bbox": [ - 388.62, - 543.0, - 393.6, - 552.96 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p711-b36", - "global_id": 20984, - "bbox": [ - 378.34, - 556.14, - 384.86, - 566.1 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p711-b37", - "global_id": 20985, - "bbox": [ - 248.06, - 590.17, - 262.19, - 600.13 - ], - "text": "= τ", - "type": "text" - }, - { - "block_id": "p711-b38", - "global_id": 20986, - "bbox": [ - 264.58, - 577.03, - 276.21, - 586.99 - ], - "text": "sin", - "type": "text" - }, - { - "block_id": "p711-b39", - "global_id": 20987, - "bbox": [ - 277.31, - 562.62, - 296.27, - 579.59 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p711-b40", - "global_id": 20988, - "bbox": [ - 288.86, - 584.09, - 293.84, - 594.06 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p711-b42", - "global_id": 20989, - "bbox": [ - 270.94, - 589.74, - 289.9, - 606.71 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p711-b43", - "global_id": 20990, - "bbox": [ - 282.49, - 611.22, - 287.47, - 621.18 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p711-b44", - "global_id": 20991, - "bbox": [ - 292.3, - 589.74, - 342.05, - 600.55 - ], - "text": "= τ sinc", - "type": "text" - }, - { - "block_id": "p711-b45", - "global_id": 20992, - "bbox": [ - 342.05, - 576.18, - 361.01, - 593.15 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p711-b46", - "global_id": 20993, - "bbox": [ - 353.6, - 597.66, - 358.58, - 607.62 - ], - "text": "2", - "type": "text" - } - ] - }, - { - "page_num": 712, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p712-b0", - "global_id": 20994, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "692\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p712-b1", - "global_id": 20995, - "bbox": [ - 193.56, - 220.09, - 197.56, - 228.09 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p712-b2", - "global_id": 20996, - "bbox": [ - 157.42, - 182.57, - 201.59, - 194.06 - ], - "text": "X(v)\nt", - "type": "text" - }, - { - "block_id": "p712-b3", - "global_id": 20997, - "bbox": [ - 261.86, - 220.65, - 391.52, - 231.17 - ], - "text": "v\n0", - "type": "text" - }, - { - "block_id": "p712-b4", - "global_id": 20998, - "bbox": [ - 457.16, - 207.5, - 462.49, - 215.5 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p712-b5", - "global_id": 20999, - "bbox": [ - 389.44, - 195.14, - 394.77, - 203.14 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p712-b6", - "global_id": 21000, - "bbox": [ - 368.71, - 233.38, - 380.71, - 241.58 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p712-b7", - "global_id": 21001, - "bbox": [ - 388.58, - 136.13, - 392.58, - 144.13 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p712-b8", - "global_id": 21002, - "bbox": [ - 357.76, - 97.6, - 396.61, - 108.29 - ], - "text": "X(v)\nt", - "type": "text" - }, - { - "block_id": "p712-b9", - "global_id": 21003, - "bbox": [ - 187.98, - 157.03, - 196.86, - 165.03 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p712-b10", - "global_id": 21004, - "bbox": [ - 195.47, - 136.22, - 199.47, - 144.22 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p712-b11", - "global_id": 21005, - "bbox": [ - 186.27, - 94.03, - 230.15, - 105.76 - ], - "text": "1\nx(t)", - "type": "text" - }, - { - "block_id": "p712-b12", - "global_id": 21006, - "bbox": [ - 223.31, - 134.4, - 225.54, - 142.4 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p712-b13", - "global_id": 21007, - "bbox": [ - 380.72, - 157.03, - 390.36, - 165.03 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p712-b14", - "global_id": 21008, - "bbox": [ - 456.4, - 141.88, - 461.74, - 149.88 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p712-b15", - "global_id": 21009, - "bbox": [ - 187.08, - 248.14, - 390.41, - 256.14 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p712-b16", - "global_id": 21010, - "bbox": [ - 407.85, - 191.45, - 430.07, - 199.74 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p712-b17", - "global_id": 21011, - "bbox": [ - 172.5, - 134.68, - 406.03, - 151.6 - ], - "text": "t\n2\nt\n2\n2p\nt", - "type": "text" - }, - { - "block_id": "p712-b18", - "global_id": 21012, - "bbox": [ - 400.84, - 203.56, - 410.17, - 218.46 - ], - "text": "2p\nt", - "type": "text" - }, - { - "block_id": "p712-b19", - "global_id": 21013, - "bbox": [ - 425.19, - 135.84, - 434.52, - 150.74 - ], - "text": "4p\nt", - "type": "text" - }, - { - "block_id": "p712-b20", - "global_id": 21014, - "bbox": [ - 206.69, - 204.06, - 430.49, - 235.62 - ], - "text": "4p\nt\n2p\nt", - "type": "text" - }, - { - "block_id": "p712-b21", - "global_id": 21015, - "bbox": [ - 226.86, - 220.72, - 236.19, - 235.62 - ], - "text": "4p\nt", - "type": "text" - }, - { - "block_id": "p712-b22", - "global_id": 21016, - "bbox": [ - 164.25, - 135.88, - 347.76, - 150.84 - ], - "text": "4p\n t", - "type": "text" - }, - { - "block_id": "p712-b23", - "global_id": 21017, - "bbox": [ - 358.25, - 135.88, - 375.19, - 150.84 - ], - "text": "2p\n t", - "type": "text" - }, - { - "block_id": "p712-b24", - "global_id": 21018, - "bbox": [ - 142.33, - 220.7, - 159.26, - 235.67 - ], - "text": "4p\n t", - "type": "text" - }, - { - "block_id": "p712-b25", - "global_id": 21019, - "bbox": [ - 162.22, - 220.7, - 352.83, - 235.67 - ], - "text": "4p\n t\n2p\n t", - "type": "text" - }, - { - "block_id": "p712-b26", - "global_id": 21020, - "bbox": [ - 356.21, - 220.7, - 373.15, - 235.67 - ], - "text": "2p\n t", - "type": "text" - }, - { - "block_id": "p712-b27", - "global_id": 21021, - "bbox": [ - 103.16, - 262.3, - 477.01, - 284.64 - ], - "text": "Figure 7.10 (a) A gate pulse x(t), (b) its Fourier spectrum X(ω), (c) its amplitude spectrum\n|X(ω)|, and (d) its phase spectrum̸\nX(ω).", - "type": "text" - }, - { - "block_id": "p712-b28", - "global_id": 21022, - "bbox": [ - 103.17, - 312.2, - 144.91, - 322.16 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p712-b29", - "global_id": 21023, - "bbox": [ - 234.28, - 329.08, - 249.21, - 339.04 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p712-b30", - "global_id": 21024, - "bbox": [ - 249.21, - 314.68, - 261.19, - 331.96 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p712-b31", - "global_id": 21025, - "bbox": [ - 257.13, - 335.74, - 261.44, - 345.7 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p712-b33", - "global_id": 21026, - "bbox": [ - 272.61, - 328.67, - 317.82, - 339.04 - ], - "text": "⇐⇒τ sinc", - "type": "text" - }, - { - "block_id": "p712-b34", - "global_id": 21027, - "bbox": [ - 317.83, - 314.68, - 336.79, - 331.65 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p712-b35", - "global_id": 21028, - "bbox": [ - 329.38, - 336.16, - 334.36, - 346.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p712-b37", - "global_id": 21029, - "bbox": [ - 452.95, - 329.08, - 477.01, - 339.04 - ], - "text": "(7.19)", - "type": "text" - }, - { - "block_id": "p712-b38", - "global_id": 21030, - "bbox": [ - 103.17, - 352.95, - 477.04, - 447.01 - ], - "text": "Recall that sinc(x) = 0 when x = ±nπ. Hence, sinc(ωτ/2) = 0 when ωτ/2 = ±nπ;\nthat is, when ω = ±2nπ/τ,(n = 1,2,3,. . .), as depicted in Fig. 7.10b. The Fourier transform\nX(ω) shown in Fig. 7.10b exhibits positive and negative values. A negative amplitude can be\nconsidered to be a positive amplitude with a phase of −π or π. We use this observation to plot\nthe amplitude spectrum |X(ω)| = |sinc(ωτ/2)| (Fig. 7.10c) and the phase spectrum̸\nX(ω)\n(Fig. 7.10d). The phase spectrum, which is required to be an odd function of ω, may be drawn\nin several other ways because a negative sign can be accounted for by a phase of ±nπ, where\nn is any odd integer. All such representations are equivalent.", - "type": "text" - }, - { - "block_id": "p712-b39", - "global_id": 21031, - "bbox": [ - 102.14, - 477.36, - 230.37, - 499.72 - ], - "text": "BANDWIDTH OF RECT t", - "type": "text" - }, - { - "block_id": "p712-b40", - "global_id": 21032, - "bbox": [ - 226.47, - 493.43, - 230.74, - 502.4 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p712-b42", - "global_id": 21033, - "bbox": [ - 101.84, - 503.33, - 490.41, - 585.44 - ], - "text": "The spectrum X(ω) in Fig. 7.10 peaks at ω = 0 and decays at higher frequencies. Therefore,\nrect(t/τ) is a lowpass signal with most of the signal energy in lower-frequency components.\nStrictly speaking, because the spectrum extends from 0 to ∞, the bandwidth is ∞. However,\nmuch of the spectrum is concentrated within the first lobe (from ω = 0 to ω = 2π/τ). Therefore,\na rough estimate of the bandwidth of a rectangular pulse of width τ seconds is 2π/τ rad/s, or 1/τ\nHz.† Note the reciprocal relationship of the pulse width with its bandwidth. We shall observe later\nthat this result is true, in general.", - "type": "text" - }, - { - "block_id": "p712-b43", - "global_id": 21034, - "bbox": [ - 101.84, - 610.24, - 490.39, - 633.41 - ], - "text": "† To compute bandwidth, we must consider the spectrum for positive values of ω only. See the discussion in\nSec. 6.3.", - "type": "text" - } - ] - }, - { - "page_num": 713, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p713-b0", - "global_id": 21035, - "bbox": [ - 304.47, - 62.89, - 516.12, - 71.98 - ], - "text": "7.2\nTransforms of Some Useful Functions\n693", - "type": "text" - }, - { - "block_id": "p713-b1", - "global_id": 21036, - "bbox": [ - 102.51, - 93.91, - 451.77, - 105.86 - ], - "text": "EXAMPLE 7.3\nFourier Transform of the Dirac Delta Function", - "type": "text" - }, - { - "block_id": "p713-b2", - "global_id": 21037, - "bbox": [ - 128.9, - 119.2, - 332.42, - 129.57 - ], - "text": "Find the Fourier transform of the unit impulse δ(t).", - "type": "text" - }, - { - "block_id": "p713-b3", - "global_id": 21038, - "bbox": [ - 128.9, - 152.49, - 395.37, - 162.45 - ], - "text": "Using the sampling property of the impulse [Eq. (1.11)], we obtain", - "type": "text" - }, - { - "block_id": "p713-b4", - "global_id": 21039, - "bbox": [ - 206.84, - 179.32, - 246.65, - 189.59 - ], - "text": "F[δ(t)] =", - "type": "text" - }, - { - "block_id": "p713-b5", - "global_id": 21040, - "bbox": [ - 248.7, - 165.75, - 265.64, - 177.81 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p713-b6", - "global_id": 21041, - "bbox": [ - 253.96, - 190.63, - 266.52, - 197.6 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p713-b7", - "global_id": 21042, - "bbox": [ - 268.13, - 177.59, - 424.82, - 189.69 - ], - "text": "δ(t)e−jωtdt = 1\nand\nδ(t) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p713-b8", - "global_id": 21043, - "bbox": [ - 128.9, - 206.62, - 289.97, - 217.0 - ], - "text": "Figure 7.11 shows δ(t) and its spectrum.", - "type": "text" - }, - { - "block_id": "p713-b9", - "global_id": 21044, - "bbox": [ - 177.94, - 302.66, - 213.77, - 311.09 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p713-b10", - "global_id": 21045, - "bbox": [ - 324.24, - 263.56, - 328.24, - 271.56 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p713-b11", - "global_id": 21046, - "bbox": [ - 370.71, - 302.48, - 376.05, - 310.48 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p713-b12", - "global_id": 21047, - "bbox": [ - 336.24, - 242.13, - 366.46, - 250.43 - ], - "text": "X(v) 1", - "type": "text" - }, - { - "block_id": "p713-b13", - "global_id": 21048, - "bbox": [ - 169.5, - 316.64, - 335.57, - 324.64 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p713-b14", - "global_id": 21049, - "bbox": [ - 333.24, - 303.09, - 337.24, - 311.09 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p713-b15", - "global_id": 21050, - "bbox": [ - 180.94, - 242.13, - 214.23, - 250.43 - ], - "text": "x(t) d(t)", - "type": "text" - }, - { - "block_id": "p713-b16", - "global_id": 21051, - "bbox": [ - 119.94, - 331.27, - 334.24, - 340.58 - ], - "text": "Figure 7.11 (a) Unit impulse and (b) its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p713-b17", - "global_id": 21052, - "bbox": [ - 102.51, - 402.53, - 495.26, - 414.48 - ], - "text": "EXAMPLE 7.4\nInverse Fourier Transform of the Dirac Delta Function", - "type": "text" - }, - { - "block_id": "p713-b18", - "global_id": 21053, - "bbox": [ - 128.9, - 427.81, - 300.21, - 438.18 - ], - "text": "Find the inverse Fourier transform of δ(ω).", - "type": "text" - }, - { - "block_id": "p713-b19", - "global_id": 21054, - "bbox": [ - 128.9, - 461.1, - 436.03, - 471.06 - ], - "text": "On the basis of Eq. (7.10) and the sampling property of the impulse function,", - "type": "text" - }, - { - "block_id": "p713-b20", - "global_id": 21055, - "bbox": [ - 235.46, - 481.36, - 300.19, - 498.3 - ], - "text": "F−1[δ(ω)] = 1", - "type": "text" - }, - { - "block_id": "p713-b21", - "global_id": 21056, - "bbox": [ - 291.71, - 495.0, - 302.67, - 505.38 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p713-b22", - "global_id": 21057, - "bbox": [ - 305.98, - 474.37, - 322.92, - 486.42 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p713-b23", - "global_id": 21058, - "bbox": [ - 311.24, - 499.25, - 323.79, - 506.22 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p713-b24", - "global_id": 21059, - "bbox": [ - 325.4, - 481.36, - 391.52, - 498.3 - ], - "text": "δ(ω)ejωtdω = 1", - "type": "text" - }, - { - "block_id": "p713-b25", - "global_id": 21060, - "bbox": [ - 383.05, - 495.0, - 394.0, - 505.38 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p713-b26", - "global_id": 21061, - "bbox": [ - 128.91, - 515.59, - 170.65, - 525.55 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p713-b27", - "global_id": 21062, - "bbox": [ - 232.07, - 524.81, - 502.75, - 548.83 - ], - "text": "1\n2π ⇐⇒δ(ω)\nand\n1 ⇐⇒2πδ(ω)\n(7.20)", - "type": "text" - }, - { - "block_id": "p713-b28", - "global_id": 21063, - "bbox": [ - 128.91, - 553.42, - 502.76, - 575.75 - ], - "text": "This result shows that the spectrum of a constant signal x(t) = 1 is an impulse 2πδ(ω), as\nillustrated in Fig. 7.12.", - "type": "text" - }, - { - "block_id": "p713-b29", - "global_id": 21064, - "bbox": [ - 128.91, - 577.74, - 502.77, - 611.61 - ], - "text": "The result [Eq. (7.20)] could have been anticipated on qualitative grounds. Recall that the\nFourier transform of x(t) is a spectral representation of x(t) in terms of everlasting exponential\ncomponents of the form ejωt. Now, to represent a constant signal x(t) = 1, we need a single", - "type": "text" - } - ] - }, - { - "page_num": 714, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p714-b0", - "global_id": 21065, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "694\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p714-b1", - "global_id": 21066, - "bbox": [ - 146.2, - 114.19, - 150.2, - 122.19 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p714-b2", - "global_id": 21067, - "bbox": [ - 146.2, - 152.88, - 354.31, - 161.36 - ], - "text": "0\n0\nt\nv", - "type": "text" - }, - { - "block_id": "p714-b3", - "global_id": 21068, - "bbox": [ - 148.26, - 172.14, - 318.49, - 180.14 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p714-b4", - "global_id": 21069, - "bbox": [ - 159.23, - 93.41, - 369.71, - 103.22 - ], - "text": "X(v) 2pd(v)\nx(t) 1", - "type": "text" - }, - { - "block_id": "p714-b5", - "global_id": 21070, - "bbox": [ - 94.2, - 186.77, - 340.85, - 196.08 - ], - "text": "Figure 7.12 (a) A constant (dc) signal and (b) its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p714-b6", - "global_id": 21071, - "bbox": [ - 103.16, - 220.13, - 477.02, - 257.62 - ], - "text": "everlasting exponential ejωt with ω = 0.† This results in a spectrum at a single frequency ω = 0.\nAnother way of looking at the situation is that x(t) = 1 is a dc signal that has a single frequency\nω = 0 (dc).", - "type": "text" - }, - { - "block_id": "p714-b7", - "global_id": 21072, - "bbox": [ - 101.84, - 296.46, - 490.38, - 318.79 - ], - "text": "If an impulse at ω = 0 is a spectrum of a dc signal, what does an impulse at ω = ω0 represent?\nWe shall answer this question in the next example.", - "type": "text" - }, - { - "block_id": "p714-b8", - "global_id": 21073, - "bbox": [ - 76.77, - 350.96, - 450.25, - 376.87 - ], - "text": "EXAMPLE 7.5\nInverse Fourier Transform of a Shifted Dirac Delta\nFunction", - "type": "text" - }, - { - "block_id": "p714-b9", - "global_id": 21074, - "bbox": [ - 103.16, - 390.19, - 295.84, - 401.34 - ], - "text": "Find the inverse Fourier transform of δ(ω −ω0).", - "type": "text" - }, - { - "block_id": "p714-b10", - "global_id": 21075, - "bbox": [ - 103.16, - 423.48, - 356.09, - 433.44 - ], - "text": "Using the sampling property of the impulse function, we obtain", - "type": "text" - }, - { - "block_id": "p714-b11", - "global_id": 21076, - "bbox": [ - 179.3, - 443.73, - 265.4, - 461.45 - ], - "text": "F−1[δ(ω −ω0)] = 1", - "type": "text" - }, - { - "block_id": "p714-b12", - "global_id": 21077, - "bbox": [ - 256.93, - 457.38, - 267.88, - 467.75 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p714-b13", - "global_id": 21078, - "bbox": [ - 271.19, - 436.75, - 288.13, - 448.8 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p714-b14", - "global_id": 21079, - "bbox": [ - 276.45, - 461.62, - 289.01, - 468.6 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p714-b15", - "global_id": 21080, - "bbox": [ - 290.62, - 443.73, - 379.22, - 461.45 - ], - "text": "δ(ω −ω0)ejωt dω = 1", - "type": "text" - }, - { - "block_id": "p714-b16", - "global_id": 21081, - "bbox": [ - 370.75, - 448.58, - 400.26, - 467.75 - ], - "text": "2π ejω0t", - "type": "text" - }, - { - "block_id": "p714-b17", - "global_id": 21082, - "bbox": [ - 103.16, - 477.97, - 144.91, - 487.93 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p714-b18", - "global_id": 21083, - "bbox": [ - 170.46, - 487.18, - 477.01, - 511.2 - ], - "text": "1\n2π ejω0t ⇐⇒δ(ω −ω0)\nand\nejω0t ⇐⇒2πδ(ω −ω0)\n(7.21)", - "type": "text" - }, - { - "block_id": "p714-b19", - "global_id": 21084, - "bbox": [ - 103.17, - 514.01, - 477.03, - 540.32 - ], - "text": "This result shows that the spectrum of an everlasting exponential ejω0t is a single impulse at\nω = ω0. We reach the same conclusion by qualitative reasoning. To represent the everlasting", - "type": "text" - }, - { - "block_id": "p714-b20", - "global_id": 21085, - "bbox": [ - 101.84, - 577.36, - 490.38, - 633.41 - ], - "text": "† The constant multiplier 2π in the spectrum [X(ω) = 2πδ(ω)] may be a bit puzzling. Since 1 = ejωt with\nω = 0, it appears that the Fourier transform of x(t) = 1 should be an impulse of strength unity rather than\n2π. Recall, however, that in the Fourier transform x(t) is synthesized by exponentials not of amplitude\nX(nω)ω but of amplitude 1/2π times X(nω)ω, as seen from Eq. (7.7). Had we used variable f (hertz)\ninstead of ω, the spectrum would have been the unit impulse.", - "type": "text" - } - ] - }, - { - "page_num": 715, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p715-b0", - "global_id": 21086, - "bbox": [ - 304.47, - 62.89, - 516.12, - 71.98 - ], - "text": "7.2\nTransforms of Some Useful Functions\n695", - "type": "text" - }, - { - "block_id": "p715-b1", - "global_id": 21087, - "bbox": [ - 128.9, - 84.1, - 502.76, - 110.41 - ], - "text": "exponential ejω0t, we need a single everlasting exponential ejωt with ω = ω0. Therefore, the\nspectrum consists of a single component at frequency ω = ω0.", - "type": "text" - }, - { - "block_id": "p715-b2", - "global_id": 21088, - "bbox": [ - 146.84, - 111.63, - 268.64, - 121.6 - ], - "text": "From Eq. (7.21) it follows that", - "type": "text" - }, - { - "block_id": "p715-b3", - "global_id": 21089, - "bbox": [ - 266.86, - 129.02, - 364.81, - 144.28 - ], - "text": "e−jω0t ⇐⇒2πδ(ω + ω0)", - "type": "text" - }, - { - "block_id": "p715-b4", - "global_id": 21090, - "bbox": [ - 102.51, - 198.3, - 375.39, - 210.26 - ], - "text": "EXAMPLE 7.6\nFourier Transform of a Sinusoid", - "type": "text" - }, - { - "block_id": "p715-b5", - "global_id": 21091, - "bbox": [ - 128.9, - 223.58, - 424.89, - 234.73 - ], - "text": "Find the Fourier transform of the everlasting sinusoid cos ω0t (Fig. 7.13a).", - "type": "text" - }, - { - "block_id": "p715-b6", - "global_id": 21092, - "bbox": [ - 307.38, - 302.93, - 411.63, - 312.75 - ], - "text": "0\nv0\nv0", - "type": "text" - }, - { - "block_id": "p715-b7", - "global_id": 21093, - "bbox": [ - 171.27, - 270.87, - 417.16, - 280.5 - ], - "text": "x(t)\nX(v)\np\ncos v0t", - "type": "text" - }, - { - "block_id": "p715-b8", - "global_id": 21094, - "bbox": [ - 183.31, - 301.92, - 422.85, - 317.46 - ], - "text": "t\nv\n0", - "type": "text" - }, - { - "block_id": "p715-b9", - "global_id": 21095, - "bbox": [ - 180.95, - 328.56, - 365.96, - 336.56 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p715-b10", - "global_id": 21096, - "bbox": [ - 119.94, - 343.19, - 342.95, - 352.5 - ], - "text": "Figure 7.13 (a) A cosine signal and (b) its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p715-b11", - "global_id": 21097, - "bbox": [ - 146.84, - 368.21, - 237.05, - 378.18 - ], - "text": "Recall Euler’s formula", - "type": "text" - }, - { - "block_id": "p715-b12", - "global_id": 21098, - "bbox": [ - 263.63, - 378.41, - 309.13, - 390.9 - ], - "text": "cos ω0t = 1", - "type": "text" - }, - { - "block_id": "p715-b13", - "global_id": 21099, - "bbox": [ - 305.65, - 376.13, - 368.03, - 392.94 - ], - "text": "2(ejω0t + e−jω0t)", - "type": "text" - }, - { - "block_id": "p715-b14", - "global_id": 21100, - "bbox": [ - 128.91, - 399.1, - 253.15, - 409.06 - ], - "text": "Applying Eq. (7.21), we obtain", - "type": "text" - }, - { - "block_id": "p715-b15", - "global_id": 21101, - "bbox": [ - 237.15, - 420.6, - 394.52, - 431.75 - ], - "text": "cos ω0t ⇐⇒π[δ(ω + ω0) + δ(ω −ω0)]", - "type": "text" - }, - { - "block_id": "p715-b16", - "global_id": 21102, - "bbox": [ - 128.91, - 442.52, - 502.76, - 490.25 - ], - "text": "The spectrum of cos ω0t consists of two impulses at ω0 and −ω0, as shown in Fig. 7.13b.\nThe result also follows from qualitative reasoning. An everlasting sinusoid cos ω0t can be\nsynthesized by two everlasting exponentials, ejω0t and e−jω0t. Therefore, the Fourier spectrum\nconsists of only two components of frequencies ω0 and −ω0.", - "type": "text" - }, - { - "block_id": "p715-b17", - "global_id": 21103, - "bbox": [ - 102.51, - 543.98, - 409.6, - 555.93 - ], - "text": "EXAMPLE 7.7\nFourier Transform of a Periodic Signal", - "type": "text" - }, - { - "block_id": "p715-b18", - "global_id": 21104, - "bbox": [ - 128.9, - 572.18, - 502.75, - 582.56 - ], - "text": "Determine the Fourier transform of a periodic signal x(t) using its Fourier series representation.", - "type": "text" - }, - { - "block_id": "p715-b19", - "global_id": 21105, - "bbox": [ - 128.9, - 599.5, - 502.77, - 633.37 - ], - "text": "We can use a Fourier series to express a periodic signal as a sum of exponentials of the form\nejnω0t, whose Fourier transform is found in Eq. (7.21). Hence, we can readily find the Fourier\ntransform of a periodic signal by using the linearity property in Eq. (7.15).", - "type": "text" - } - ] - }, - { - "page_num": 716, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p716-b0", - "global_id": 21106, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "696\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p716-b1", - "global_id": 21107, - "bbox": [ - 121.09, - 85.89, - 395.88, - 97.76 - ], - "text": "The Fourier series of a periodic signal x(t) with period T0 is given by", - "type": "text" - }, - { - "block_id": "p716-b2", - "global_id": 21108, - "bbox": [ - 221.3, - 115.93, - 245.95, - 126.2 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p716-b3", - "global_id": 21109, - "bbox": [ - 251.69, - 105.76, - 265.79, - 116.42 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p716-b4", - "global_id": 21110, - "bbox": [ - 248.0, - 129.78, - 269.48, - 136.97 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p716-b5", - "global_id": 21111, - "bbox": [ - 270.59, - 108.94, - 356.69, - 127.38 - ], - "text": "Dnejnω0t\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p716-b6", - "global_id": 21112, - "bbox": [ - 346.95, - 123.32, - 355.97, - 134.15 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p716-b7", - "global_id": 21113, - "bbox": [ - 103.16, - 147.47, - 318.54, - 158.03 - ], - "text": "Taking the Fourier transform of both sides, we obtain†", - "type": "text" - }, - { - "block_id": "p716-b8", - "global_id": 21114, - "bbox": [ - 227.72, - 177.7, - 271.23, - 188.07 - ], - "text": "X(ω) = 2π", - "type": "text" - }, - { - "block_id": "p716-b9", - "global_id": 21115, - "bbox": [ - 277.04, - 167.52, - 291.13, - 178.19 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p716-b10", - "global_id": 21116, - "bbox": [ - 273.34, - 191.54, - 294.81, - 198.74 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p716-b11", - "global_id": 21117, - "bbox": [ - 295.94, - 177.7, - 477.01, - 188.84 - ], - "text": "Dnδ(ω −nω0)\n(7.22)", - "type": "text" - }, - { - "block_id": "p716-b12", - "global_id": 21118, - "bbox": [ - 76.77, - 271.78, - 394.79, - 283.73 - ], - "text": "EXAMPLE 7.8\nFourier Transform of a Dirac Delta Train", - "type": "text" - }, - { - "block_id": "p716-b13", - "global_id": 21119, - "bbox": [ - 103.16, - 297.07, - 477.0, - 319.4 - ], - "text": "Determine the Fourier transform of a Dirac delta train δT0(t) (Fig. 7.14a) using its Fourier\nseries representation.", - "type": "text" - }, - { - "block_id": "p716-b14", - "global_id": 21120, - "bbox": [ - 103.16, - 456.57, - 367.27, - 465.88 - ], - "text": "Figure 7.14 (a) The uniform impulse train and (b) its Fourier transform.", - "type": "text" - }, - { - "block_id": "p716-b15", - "global_id": 21121, - "bbox": [ - 103.17, - 489.16, - 477.0, - 513.53 - ], - "text": "As shown in Eq. (6.24) from Ex. 6.9, the Fourier coefficients Dn for δT0(t) are constant\nDn = 1/T0. From Eq. (7.22), the Fourier transform of δT0(t) is therefore", - "type": "text" - }, - { - "block_id": "p716-b16", - "global_id": 21122, - "bbox": [ - 164.12, - 525.07, - 208.82, - 542.43 - ], - "text": "X(ω) = 2π", - "type": "text" - }, - { - "block_id": "p716-b17", - "global_id": 21123, - "bbox": [ - 199.08, - 539.45, - 208.1, - 550.28 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p716-b18", - "global_id": 21124, - "bbox": [ - 215.82, - 521.89, - 229.91, - 532.55 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p716-b19", - "global_id": 21125, - "bbox": [ - 212.12, - 545.91, - 233.59, - 553.1 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p716-b20", - "global_id": 21126, - "bbox": [ - 234.72, - 525.07, - 413.86, - 544.47 - ], - "text": "δ(ω −nω0) = ω0δω0(ω),\nwhere ω0 = 2π", - "type": "text" - }, - { - "block_id": "p716-b21", - "global_id": 21127, - "bbox": [ - 404.12, - 539.45, - 413.15, - 550.28 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p716-b22", - "global_id": 21128, - "bbox": [ - 103.16, - 563.36, - 310.04, - 573.32 - ], - "text": "The corresponding spectrum is shown in Fig. 7.14b.", - "type": "text" - }, - { - "block_id": "p716-b23", - "global_id": 21129, - "bbox": [ - 101.84, - 621.19, - 386.17, - 633.41 - ], - "text": "† We assume here that the linearity property can be extended to an infinite sum.", - "type": "text" - } - ] - }, - { - "page_num": 717, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p717-b0", - "global_id": 21130, - "bbox": [ - 304.47, - 62.89, - 516.12, - 71.98 - ], - "text": "7.2\nTransforms of Some Useful Functions\n697", - "type": "text" - }, - { - "block_id": "p717-b1", - "global_id": 21131, - "bbox": [ - 102.51, - 93.92, - 441.14, - 105.87 - ], - "text": "EXAMPLE 7.9\nFourier Transform of the Unit Step Function", - "type": "text" - }, - { - "block_id": "p717-b2", - "global_id": 21132, - "bbox": [ - 128.9, - 122.12, - 352.77, - 132.49 - ], - "text": "Find the Fourier transform of the unit step function u(t).", - "type": "text" - }, - { - "block_id": "p717-b3", - "global_id": 21133, - "bbox": [ - 128.9, - 154.99, - 502.76, - 177.33 - ], - "text": "Trying to find the Fourier transform of u(t) by direct integration leads to an indeterminate\nresult because", - "type": "text" - }, - { - "block_id": "p717-b4", - "global_id": 21134, - "bbox": [ - 210.34, - 183.06, - 241.94, - 193.33 - ], - "text": "U(ω) =", - "type": "text" - }, - { - "block_id": "p717-b5", - "global_id": 21135, - "bbox": [ - 243.99, - 169.5, - 260.93, - 181.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p717-b6", - "global_id": 21136, - "bbox": [ - 249.24, - 194.38, - 261.8, - 201.35 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p717-b7", - "global_id": 21137, - "bbox": [ - 263.41, - 181.33, - 315.6, - 193.33 - ], - "text": "u(t)e−jωtdt =", - "type": "text" - }, - { - "block_id": "p717-b8", - "global_id": 21138, - "bbox": [ - 317.64, - 169.5, - 334.58, - 181.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p717-b9", - "global_id": 21139, - "bbox": [ - 322.9, - 176.07, - 390.2, - 201.64 - ], - "text": "0\ne−jωtdt = −1", - "type": "text" - }, - { - "block_id": "p717-b10", - "global_id": 21140, - "bbox": [ - 379.08, - 181.33, - 409.83, - 200.41 - ], - "text": "jω e−jωt", - "type": "text" - }, - { - "block_id": "p717-b12", - "global_id": 21141, - "bbox": [ - 413.71, - 173.69, - 420.83, - 180.67 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p717-b13", - "global_id": 21142, - "bbox": [ - 413.71, - 195.56, - 417.2, - 202.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p717-b14", - "global_id": 21143, - "bbox": [ - 128.9, - 205.98, - 502.75, - 243.46 - ], - "text": "The upper limit of e−jωt as t →∞yields an indeterminate answer. So we approach this problem\nby considering u(t) to be a decaying exponential e−atu(t) in the limit as a →0 (Fig. 7.15a).\nThus,", - "type": "text" - }, - { - "block_id": "p717-b15", - "global_id": 21144, - "bbox": [ - 278.71, - 245.04, - 319.9, - 255.42 - ], - "text": "u(t) = lim", - "type": "text" - }, - { - "block_id": "p717-b16", - "global_id": 21145, - "bbox": [ - 305.93, - 243.32, - 352.96, - 261.22 - ], - "text": "a→0e−atu(t)", - "type": "text" - }, - { - "block_id": "p717-b17", - "global_id": 21146, - "bbox": [ - 128.91, - 267.93, - 143.28, - 277.89 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p717-b18", - "global_id": 21147, - "bbox": [ - 239.05, - 282.37, - 286.66, - 292.75 - ], - "text": "U(ω) = lim", - "type": "text" - }, - { - "block_id": "p717-b19", - "global_id": 21148, - "bbox": [ - 272.69, - 280.65, - 361.06, - 298.55 - ], - "text": "a→0F{e−atu(t)} = lim", - "type": "text" - }, - { - "block_id": "p717-b20", - "global_id": 21149, - "bbox": [ - 347.09, - 291.29, - 361.73, - 298.55 - ], - "text": "a→0", - "type": "text" - }, - { - "block_id": "p717-b21", - "global_id": 21150, - "bbox": [ - 366.08, - 275.8, - 391.22, - 299.72 - ], - "text": "1\na + jω", - "type": "text" - }, - { - "block_id": "p717-b22", - "global_id": 21151, - "bbox": [ - 128.91, - 306.65, - 433.28, - 316.62 - ], - "text": "Expressing the right-hand side in terms of its real and imaginary parts yields", - "type": "text" - }, - { - "block_id": "p717-b23", - "global_id": 21152, - "bbox": [ - 198.87, - 333.8, - 246.48, - 344.17 - ], - "text": "U(ω) = lim", - "type": "text" - }, - { - "block_id": "p717-b24", - "global_id": 21153, - "bbox": [ - 232.51, - 342.71, - 247.15, - 349.98 - ], - "text": "a→0", - "type": "text" - }, - { - "block_id": "p717-b25", - "global_id": 21154, - "bbox": [ - 248.26, - 319.81, - 331.51, - 351.15 - ], - "text": "a\na2 + ω2 −j\nω\na2 + ω2", - "type": "text" - }, - { - "block_id": "p717-b26", - "global_id": 21155, - "bbox": [ - 333.2, - 319.81, - 338.63, - 329.77 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p717-b27", - "global_id": 21156, - "bbox": [ - 340.68, - 333.8, - 364.47, - 344.17 - ], - "text": "= lim", - "type": "text" - }, - { - "block_id": "p717-b28", - "global_id": 21157, - "bbox": [ - 350.5, - 342.71, - 365.14, - 349.98 - ], - "text": "a→0", - "type": "text" - }, - { - "block_id": "p717-b29", - "global_id": 21158, - "bbox": [ - 366.24, - 319.81, - 402.91, - 351.15 - ], - "text": "a\na2 + ω2", - "type": "text" - }, - { - "block_id": "p717-b30", - "global_id": 21159, - "bbox": [ - 404.61, - 319.81, - 410.04, - 329.77 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p717-b31", - "global_id": 21160, - "bbox": [ - 411.59, - 327.23, - 429.35, - 344.17 - ], - "text": "+ 1", - "type": "text" - }, - { - "block_id": "p717-b32", - "global_id": 21161, - "bbox": [ - 422.11, - 340.87, - 431.4, - 351.15 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p717-b33", - "global_id": 21162, - "bbox": [ - 409.41, - 399.91, - 413.41, - 414.79 - ], - "text": "1\na", - "type": "text" - }, - { - "block_id": "p717-b34", - "global_id": 21163, - "bbox": [ - 456.09, - 477.64, - 461.43, - 485.64 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p717-b35", - "global_id": 21164, - "bbox": [ - 415.33, - 434.3, - 441.33, - 453.02 - ], - "text": "a\na2 v2", - "type": "text" - }, - { - "block_id": "p717-b36", - "global_id": 21165, - "bbox": [ - 127.89, - 404.93, - 131.89, - 412.93 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p717-b37", - "global_id": 21166, - "bbox": [ - 128.25, - 477.66, - 301.36, - 486.77 - ], - "text": "t\n0", - "type": "text" - }, - { - "block_id": "p717-b38", - "global_id": 21167, - "bbox": [ - 217.82, - 499.27, - 411.02, - 507.27 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p717-b39", - "global_id": 21168, - "bbox": [ - 404.34, - 478.77, - 408.34, - 486.77 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p717-b40", - "global_id": 21169, - "bbox": [ - 188.67, - 438.93, - 212.96, - 448.62 - ], - "text": "eatu(t)", - "type": "text" - }, - { - "block_id": "p717-b41", - "global_id": 21170, - "bbox": [ - 138.3, - 389.82, - 149.85, - 397.9 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p717-b42", - "global_id": 21171, - "bbox": [ - 127.52, - 513.96, - 377.94, - 523.2 - ], - "text": "Figure 7.15 Derivation of the Fourier transform of the step function.", - "type": "text" - }, - { - "block_id": "p717-b43", - "global_id": 21172, - "bbox": [ - 128.91, - 546.84, - 502.75, - 572.78 - ], - "text": "The function a/(a2 + ω2) has interesting properties. First, the area under this function\n(Fig. 7.15b) is π regardless of the value of a:", - "type": "text" - }, - { - "block_id": "p717-b44", - "global_id": 21173, - "bbox": [ - 243.7, - 576.98, - 260.64, - 589.03 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p717-b45", - "global_id": 21174, - "bbox": [ - 248.96, - 601.85, - 261.51, - 608.82 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p717-b46", - "global_id": 21175, - "bbox": [ - 264.32, - 583.55, - 351.44, - 607.88 - ], - "text": "a\na2 + ω2 dω = tan−1 ω", - "type": "text" - }, - { - "block_id": "p717-b47", - "global_id": 21176, - "bbox": [ - 345.79, - 597.92, - 350.77, - 607.88 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p717-b49", - "global_id": 21177, - "bbox": [ - 356.07, - 581.17, - 363.19, - 588.14 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p717-b50", - "global_id": 21178, - "bbox": [ - 356.07, - 602.74, - 368.63, - 609.71 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p717-b51", - "global_id": 21179, - "bbox": [ - 371.18, - 590.53, - 386.97, - 600.49 - ], - "text": "= π", - "type": "text" - } - ] - }, - { - "page_num": 718, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p718-b0", - "global_id": 21180, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "698\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p718-b1", - "global_id": 21181, - "bbox": [ - 103.16, - 85.83, - 477.02, - 120.11 - ], - "text": "Second, when a →0, this function approaches zero for all ω̸ = 0, and all its area (π) is\nconcentrated at a single point ω = 0. Clearly, as a →0, this function approaches an impulse of\nstrength π. Thus,", - "type": "text" - }, - { - "block_id": "p718-b2", - "global_id": 21182, - "bbox": [ - 248.91, - 120.08, - 327.81, - 137.03 - ], - "text": "U(ω) = πδ(ω) + 1", - "type": "text" - }, - { - "block_id": "p718-b3", - "global_id": 21183, - "bbox": [ - 320.58, - 127.07, - 477.01, - 144.0 - ], - "text": "jω\n(7.23)", - "type": "text" - }, - { - "block_id": "p718-b4", - "global_id": 21184, - "bbox": [ - 103.16, - 150.59, - 477.03, - 220.74 - ], - "text": "Note that u(t) is not a “true” dc signal because it is not constant over the interval −∞to ∞.\nTo synthesize “true” dc, we require only one everlasting exponential with ω = 0 (impulse at\nω = 0). The signal u(t) has a jump discontinuity at t = 0. It is impossible to synthesize such\na signal with a single everlasting exponential ejωt. To synthesize this signal from everlasting\nexponentials, we need, in addition to an impulse at ω = 0, all the frequency components, as\nindicated by the term 1/jω in Eq. (7.23).", - "type": "text" - }, - { - "block_id": "p718-b5", - "global_id": 21185, - "bbox": [ - 76.77, - 298.63, - 394.49, - 310.59 - ], - "text": "EXAMPLE 7.10\nFourier Transform of the Sign Function", - "type": "text" - }, - { - "block_id": "p718-b6", - "global_id": 21186, - "bbox": [ - 103.16, - 326.83, - 477.02, - 349.17 - ], - "text": "Find the Fourier transform of the sign function sgn(t) [pronounced signum (t)], depicted in\nFig. 7.16.", - "type": "text" - }, - { - "block_id": "p718-b7", - "global_id": 21187, - "bbox": [ - 256.38, - 448.99, - 258.6, - 456.99 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p718-b8", - "global_id": 21188, - "bbox": [ - 211.3, - 403.37, - 231.96, - 411.46 - ], - "text": "sgn (t)", - "type": "text" - }, - { - "block_id": "p718-b9", - "global_id": 21189, - "bbox": [ - 186.37, - 419.89, - 190.37, - 427.89 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p718-b10", - "global_id": 21190, - "bbox": [ - 186.37, - 449.07, - 190.37, - 457.07 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p718-b11", - "global_id": 21191, - "bbox": [ - 196.29, - 467.6, - 433.76, - 481.0 - ], - "text": "1\nFigure 7.16 The signum function sgn(t).", - "type": "text" - }, - { - "block_id": "p718-b12", - "global_id": 21192, - "bbox": [ - 121.09, - 511.36, - 171.58, - 521.32 - ], - "text": "Observe that", - "type": "text" - }, - { - "block_id": "p718-b13", - "global_id": 21193, - "bbox": [ - 198.49, - 522.91, - 381.68, - 533.28 - ], - "text": "sgn(t) + 1 = 2u(t)\n\r⇒\nsgn(t) = 2u(t) −1", - "type": "text" - }, - { - "block_id": "p718-b14", - "global_id": 21194, - "bbox": [ - 103.16, - 542.25, - 363.72, - 552.21 - ], - "text": "Using Eqs. (7.20) and (7.23) and the linearity property, we obtain", - "type": "text" - }, - { - "block_id": "p718-b15", - "global_id": 21195, - "bbox": [ - 259.87, - 562.15, - 316.85, - 579.09 - ], - "text": "sgn(t) ⇐⇒2", - "type": "text" - }, - { - "block_id": "p718-b16", - "global_id": 21196, - "bbox": [ - 309.62, - 575.79, - 318.91, - 586.07 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p718-b17", - "global_id": 21197, - "bbox": [ - 119.78, - 624.83, - 352.36, - 634.79 - ], - "text": "Table 7.1 provides many common Fourier transform pairs.", - "type": "text" - } - ] - }, - { - "page_num": 719, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p719-b0", - "global_id": 21198, - "bbox": [ - 304.47, - 62.89, - 516.12, - 71.98 - ], - "text": "7.2\nTransforms of Some Useful Functions\n699", - "type": "text" - }, - { - "block_id": "p719-b1", - "global_id": 21199, - "bbox": [ - 127.59, - 86.07, - 291.05, - 95.3 - ], - "text": "TABLE 7.1\nSelect Fourier Transform Pairs", - "type": "text" - }, - { - "block_id": "p719-b2", - "global_id": 21200, - "bbox": [ - 127.59, - 106.02, - 314.0, - 115.28 - ], - "text": "No.\nx(t)\nX(ω)", - "type": "text" - }, - { - "block_id": "p719-b3", - "global_id": 21201, - "bbox": [ - 136.3, - 123.16, - 502.34, - 144.69 - ], - "text": "1\ne−atu(t)\n1\na + jω\na > 0", - "type": "text" - }, - { - "block_id": "p719-b4", - "global_id": 21202, - "bbox": [ - 136.3, - 148.3, - 502.34, - 169.82 - ], - "text": "2\neatu(−t)\n1\na −jω\na > 0", - "type": "text" - }, - { - "block_id": "p719-b5", - "global_id": 21203, - "bbox": [ - 136.3, - 173.35, - 502.34, - 194.97 - ], - "text": "3\ne−a|t|\n2a\na2 + ω2\na > 0", - "type": "text" - }, - { - "block_id": "p719-b6", - "global_id": 21204, - "bbox": [ - 136.3, - 198.59, - 502.34, - 220.12 - ], - "text": "4\nte−atu(t)\n1\n(a + jω)2\na > 0", - "type": "text" - }, - { - "block_id": "p719-b7", - "global_id": 21205, - "bbox": [ - 136.3, - 223.46, - 502.34, - 245.36 - ], - "text": "5\ntne−atu(t)\nn!\n(a + jω)n+1\na > 0", - "type": "text" - }, - { - "block_id": "p719-b8", - "global_id": 21206, - "bbox": [ - 136.3, - 250.17, - 299.45, - 259.51 - ], - "text": "6\nδ(t)\n1", - "type": "text" - }, - { - "block_id": "p719-b9", - "global_id": 21207, - "bbox": [ - 136.3, - 270.59, - 322.82, - 279.93 - ], - "text": "7\n1\n2πδ(ω)", - "type": "text" - }, - { - "block_id": "p719-b10", - "global_id": 21208, - "bbox": [ - 136.3, - 289.76, - 342.2, - 301.1 - ], - "text": "8\nejω0t\n2πδ(ω −ω0)", - "type": "text" - }, - { - "block_id": "p719-b11", - "global_id": 21209, - "bbox": [ - 136.3, - 311.44, - 389.95, - 321.52 - ], - "text": "9\ncos ω0t\nπ[δ(ω −ω0) + δ(ω + ω0)]", - "type": "text" - }, - { - "block_id": "p719-b12", - "global_id": 21210, - "bbox": [ - 131.82, - 331.86, - 392.44, - 341.94 - ], - "text": "10\nsin ω0t\njπ[δ(ω + ω0) −δ(ω −ω0)]", - "type": "text" - }, - { - "block_id": "p719-b13", - "global_id": 21211, - "bbox": [ - 131.82, - 351.1, - 335.82, - 366.35 - ], - "text": "11\nu(t)\nπδ(ω) + 1", - "type": "text" - }, - { - "block_id": "p719-b14", - "global_id": 21212, - "bbox": [ - 329.31, - 363.37, - 337.68, - 372.62 - ], - "text": "jω", - "type": "text" - }, - { - "block_id": "p719-b15", - "global_id": 21213, - "bbox": [ - 131.82, - 376.24, - 304.53, - 397.77 - ], - "text": "12\nsgnt\n2\njω", - "type": "text" - }, - { - "block_id": "p719-b16", - "global_id": 21214, - "bbox": [ - 131.82, - 400.88, - 301.54, - 417.24 - ], - "text": "13\ncos ω0tu(t)\nπ", - "type": "text" - }, - { - "block_id": "p719-b17", - "global_id": 21215, - "bbox": [ - 297.06, - 400.88, - 422.09, - 423.04 - ], - "text": "2 [δ(ω −ω0) + δ(ω + ω0)] +\njω\nω2", - "type": "text" - }, - { - "block_id": "p719-b18", - "global_id": 21216, - "bbox": [ - 409.19, - 413.75, - 432.19, - 425.81 - ], - "text": "0 −ω2", - "type": "text" - }, - { - "block_id": "p719-b19", - "global_id": 21217, - "bbox": [ - 131.82, - 423.86, - 423.01, - 446.03 - ], - "text": "14\nsin ω0tu(t)\nπ\n2j[δ(ω −ω0) −δ(ω + ω0)] +\nω0\nω2", - "type": "text" - }, - { - "block_id": "p719-b20", - "global_id": 21218, - "bbox": [ - 409.9, - 436.73, - 432.88, - 448.78 - ], - "text": "0 −ω2", - "type": "text" - }, - { - "block_id": "p719-b21", - "global_id": 21219, - "bbox": [ - 131.82, - 446.85, - 348.48, - 469.29 - ], - "text": "15\ne−at sin ω0tu(t)\nω0\n(a + jω)2 + ω2", - "type": "text" - }, - { - "block_id": "p719-b22", - "global_id": 21220, - "bbox": [ - 345.06, - 453.13, - 502.34, - 471.54 - ], - "text": "0\na > 0", - "type": "text" - }, - { - "block_id": "p719-b23", - "global_id": 21221, - "bbox": [ - 131.82, - 471.86, - 348.48, - 494.3 - ], - "text": "16\ne−at cos ω0tu(t)\na + jω\n(a + jω)2 + ω2", - "type": "text" - }, - { - "block_id": "p719-b24", - "global_id": 21222, - "bbox": [ - 345.06, - 478.15, - 502.34, - 496.55 - ], - "text": "0\na > 0", - "type": "text" - }, - { - "block_id": "p719-b25", - "global_id": 21223, - "bbox": [ - 131.82, - 502.34, - 202.34, - 511.31 - ], - "text": "17\nrect", - "type": "text" - }, - { - "block_id": "p719-b26", - "global_id": 21224, - "bbox": [ - 203.35, - 492.07, - 213.07, - 504.93 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p719-b27", - "global_id": 21225, - "bbox": [ - 209.43, - 508.33, - 213.31, - 517.3 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p719-b29", - "global_id": 21226, - "bbox": [ - 294.97, - 501.97, - 315.37, - 511.31 - ], - "text": "τ sinc", - "type": "text" - }, - { - "block_id": "p719-b30", - "global_id": 21227, - "bbox": [ - 316.37, - 492.07, - 332.39, - 504.65 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p719-b31", - "global_id": 21228, - "bbox": [ - 325.71, - 508.71, - 330.19, - 517.67 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p719-b33", - "global_id": 21229, - "bbox": [ - 131.82, - 520.98, - 197.58, - 536.32 - ], - "text": "18\nW", - "type": "text" - }, - { - "block_id": "p719-b34", - "global_id": 21230, - "bbox": [ - 191.03, - 526.98, - 308.4, - 542.68 - ], - "text": "π sinc(Wt)\nrect", - "type": "text" - }, - { - "block_id": "p719-b35", - "global_id": 21231, - "bbox": [ - 309.41, - 517.08, - 324.64, - 529.67 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p719-b36", - "global_id": 21232, - "bbox": [ - 315.49, - 533.63, - 327.44, - 542.68 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p719-b38", - "global_id": 21233, - "bbox": [ - 131.82, - 550.81, - 196.12, - 560.14 - ], - "text": "19", - "type": "text" - }, - { - "block_id": "p719-b39", - "global_id": 21234, - "bbox": [ - 197.12, - 540.92, - 206.84, - 553.77 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p719-b40", - "global_id": 21235, - "bbox": [ - 203.19, - 557.18, - 207.08, - 566.14 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p719-b41", - "global_id": 21236, - "bbox": [ - 209.35, - 540.91, - 338.51, - 566.52 - ], - "text": "τ\n2 sinc2 ωτ\n4", - "type": "text" - }, - { - "block_id": "p719-b43", - "global_id": 21237, - "bbox": [ - 131.82, - 563.97, - 239.44, - 592.27 - ], - "text": "20\nW\n2π sinc2\nWt", - "type": "text" - }, - { - "block_id": "p719-b44", - "global_id": 21238, - "bbox": [ - 232.3, - 583.3, - 236.78, - 592.27 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p719-b47", - "global_id": 21239, - "bbox": [ - 303.17, - 566.66, - 318.41, - 579.25 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p719-b48", - "global_id": 21240, - "bbox": [ - 309.26, - 583.21, - 321.21, - 592.27 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p719-b50", - "global_id": 21241, - "bbox": [ - 131.82, - 605.02, - 140.79, - 613.98 - ], - "text": "21", - "type": "text" - }, - { - "block_id": "p719-b51", - "global_id": 21242, - "bbox": [ - 192.54, - 595.33, - 205.22, - 605.09 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p719-b52", - "global_id": 21243, - "bbox": [ - 188.91, - 616.95, - 208.86, - 623.63 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p719-b53", - "global_id": 21244, - "bbox": [ - 209.85, - 604.64, - 304.08, - 614.72 - ], - "text": "δ(t −nT)\nω0", - "type": "text" - }, - { - "block_id": "p719-b54", - "global_id": 21245, - "bbox": [ - 309.2, - 595.33, - 321.89, - 605.09 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p719-b55", - "global_id": 21246, - "bbox": [ - 305.58, - 616.95, - 325.53, - 623.63 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p719-b56", - "global_id": 21247, - "bbox": [ - 326.52, - 598.36, - 514.04, - 614.95 - ], - "text": "δ(ω −nω0)\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p719-b57", - "global_id": 21248, - "bbox": [ - 506.72, - 611.29, - 511.7, - 620.26 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p719-b58", - "global_id": 21249, - "bbox": [ - 131.82, - 624.14, - 299.75, - 636.07 - ], - "text": "22\ne−t2/2σ 2\nσ", - "type": "text" - }, - { - "block_id": "p719-b59", - "global_id": 21250, - "bbox": [ - 299.93, - 619.22, - 307.51, - 628.18 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p719-b60", - "global_id": 21251, - "bbox": [ - 307.51, - 624.14, - 348.73, - 636.07 - ], - "text": "2πe−σ 2ω2/2", - "type": "text" - } - ] - }, - { - "page_num": 720, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p720-b0", - "global_id": 21252, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "700\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p720-b1", - "global_id": 21253, - "bbox": [ - 107.82, - 97.81, - 450.49, - 109.76 - ], - "text": "DRILL 7.2\nInverse Fourier Transform of a Rectangular Pulse", - "type": "text" - }, - { - "block_id": "p720-b2", - "global_id": 21254, - "bbox": [ - 107.82, - 118.47, - 484.41, - 141.58 - ], - "text": "Show that the inverse Fourier transform of X(ω) illustrated in Fig. 7.17 is x(t) =\n(ω0/π)sinc(ω0t). Sketch x(t).", - "type": "text" - }, - { - "block_id": "p720-b3", - "global_id": 21255, - "bbox": [ - 201.01, - 176.37, - 227.94, - 191.93 - ], - "text": "X(v)\n1", - "type": "text" - }, - { - "block_id": "p720-b4", - "global_id": 21256, - "bbox": [ - 127.89, - 232.2, - 284.56, - 242.02 - ], - "text": "v\nv0\nv0", - "type": "text" - }, - { - "block_id": "p720-b5", - "global_id": 21257, - "bbox": [ - 308.35, - 218.98, - 487.97, - 243.68 - ], - "text": "Figure 7.17 A frequency-domain rectangular\npulse X(ω) = rect", - "type": "text" - }, - { - "block_id": "p720-b7", - "global_id": 21258, - "bbox": [ - 387.08, - 231.68, - 398.13, - 247.76 - ], - "text": "ω\n2ω0", - "type": "text" - }, - { - "block_id": "p720-b9", - "global_id": 21259, - "bbox": [ - 405.25, - 233.72, - 407.74, - 243.69 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p720-b10", - "global_id": 21260, - "bbox": [ - 107.82, - 319.16, - 404.35, - 331.11 - ], - "text": "DRILL 7.3\nFourier Transform of a General Sinusoid", - "type": "text" - }, - { - "block_id": "p720-b11", - "global_id": 21261, - "bbox": [ - 107.82, - 336.2, - 359.32, - 350.97 - ], - "text": "Show that cos(ω0t + θ) ⇐⇒π[δ(ω + ω0)e−jθ + δ(ω −ω0)ejθ].", - "type": "text" - }, - { - "block_id": "p720-b12", - "global_id": 21262, - "bbox": [ - 101.84, - 392.93, - 441.74, - 404.88 - ], - "text": "7.2-1 Connection Between the Fourier and Laplace Transforms", - "type": "text" - }, - { - "block_id": "p720-b13", - "global_id": 21263, - "bbox": [ - 101.84, - 410.6, - 431.25, - 420.98 - ], - "text": "The general (bilateral) Laplace transform of a signal x(t), according to Eq. (4.1), is", - "type": "text" - }, - { - "block_id": "p720-b14", - "global_id": 21264, - "bbox": [ - 252.05, - 442.03, - 279.71, - 452.3 - ], - "text": "X(s) =", - "type": "text" - }, - { - "block_id": "p720-b15", - "global_id": 21265, - "bbox": [ - 281.75, - 428.46, - 298.69, - 440.53 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p720-b16", - "global_id": 21266, - "bbox": [ - 287.01, - 453.35, - 299.57, - 460.32 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p720-b17", - "global_id": 21267, - "bbox": [ - 301.18, - 438.23, - 490.38, - 452.4 - ], - "text": "x(t)e−st dt\n(7.24)", - "type": "text" - }, - { - "block_id": "p720-b18", - "global_id": 21268, - "bbox": [ - 101.84, - 473.46, - 248.02, - 483.84 - ], - "text": "Setting s = jω in this equation yields", - "type": "text" - }, - { - "block_id": "p720-b19", - "global_id": 21269, - "bbox": [ - 246.72, - 504.89, - 281.1, - 515.17 - ], - "text": "X(jω) =", - "type": "text" - }, - { - "block_id": "p720-b20", - "global_id": 21270, - "bbox": [ - 283.15, - 491.34, - 300.09, - 503.39 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p720-b21", - "global_id": 21271, - "bbox": [ - 288.41, - 516.21, - 300.96, - 523.18 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p720-b22", - "global_id": 21272, - "bbox": [ - 302.57, - 501.09, - 345.33, - 515.17 - ], - "text": "x(t)e−jωt dt", - "type": "text" - }, - { - "block_id": "p720-b23", - "global_id": 21273, - "bbox": [ - 101.84, - 536.28, - 490.39, - 606.55 - ], - "text": "where X(jω) = X(s)|s=jω. But, the right-hand-side integral defines X(ω), the Fourier transform of\nx(t). Does this mean that the Fourier transform can be obtained from the corresponding Laplace\ntransform by setting s = jω? In other words, is it true that X(jω) = X(ω)? Yes and no. Yes, it\nis true in most cases. For example, when x(t) = e−atu(t), its Laplace transform is 1/(s + a), and\nX(jω) = 1/(jω +a), which is equal to X(ω) (assuming a < 0). However, for the unit step function\nu(t), the Laplace transform is", - "type": "text" - }, - { - "block_id": "p720-b24", - "global_id": 21274, - "bbox": [ - 246.73, - 612.8, - 291.49, - 629.65 - ], - "text": "u(t) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p720-b25", - "global_id": 21275, - "bbox": [ - 287.05, - 619.37, - 345.51, - 636.72 - ], - "text": "s\nRes > 0", - "type": "text" - } - ] - }, - { - "page_num": 721, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p721-b0", - "global_id": 21276, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n701", - "type": "text" - }, - { - "block_id": "p721-b1", - "global_id": 21277, - "bbox": [ - 127.59, - 85.82, - 261.79, - 95.78 - ], - "text": "The Fourier transform is given by", - "type": "text" - }, - { - "block_id": "p721-b2", - "global_id": 21278, - "bbox": [ - 278.22, - 109.91, - 325.23, - 126.75 - ], - "text": "u(t) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p721-b3", - "global_id": 21279, - "bbox": [ - 317.99, - 116.48, - 365.49, - 133.83 - ], - "text": "jω + πδ(ω)", - "type": "text" - }, - { - "block_id": "p721-b4", - "global_id": 21280, - "bbox": [ - 127.59, - 147.58, - 267.91, - 157.96 - ], - "text": "Clearly, X(jω)̸ = X(ω) in this case.", - "type": "text" - }, - { - "block_id": "p721-b5", - "global_id": 21281, - "bbox": [ - 127.54, - 159.54, - 516.15, - 241.65 - ], - "text": "To understand this puzzle, consider the fact that we obtain X(jω) by setting s = jω in\nEq. (7.24). This implies that the integral on the right-hand side of Eq. (7.24) converges for s = jω,\nmeaning that s = jω (the imaginary axis) lies in the ROC for X(s). The general rule is that only\nwhen the ROC for X(s) includes the ω axis, does setting s = jω in X(s) yield the Fourier transform\nX(ω), that is, X(jω) = X(ω). This is the case of absolutely integrable x(t). If the ROC of X(s)\nexcludes the ω axis, X(jω)̸ = X(ω). This is the case for exponentially growing x(t) and also x(t)\nthat is constant or is oscillating with constant amplitude.", - "type": "text" - }, - { - "block_id": "p721-b6", - "global_id": 21282, - "bbox": [ - 127.59, - 243.64, - 516.14, - 265.56 - ], - "text": "The reason for this peculiar behavior has something to do with the nature of convergence of\nthe Laplace and the Fourier integrals when x(t) is not absolutely integrable.†", - "type": "text" - }, - { - "block_id": "p721-b7", - "global_id": 21283, - "bbox": [ - 127.59, - 267.55, - 516.15, - 301.43 - ], - "text": "This discussion shows that although the Fourier transform may be considered as a special case\nof the Laplace transform, we need to circumscribe such a view. This fact can also be confirmed by\nnoting that a periodic signal has the Fourier transform, but the Laplace transform does not exist.", - "type": "text" - }, - { - "block_id": "p721-b8", - "global_id": 21284, - "bbox": [ - 127.94, - 334.89, - 471.01, - 348.84 - ], - "text": "7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p721-b9", - "global_id": 21285, - "bbox": [ - 127.59, - 354.83, - 516.16, - 388.71 - ], - "text": "We now study some of the important properties of the Fourier transform and their implications as\nwell as applications. We have already encountered two important properties, linearity [Eq. (7.15)]\nand the conjugation property [Eq. (7.11)].", - "type": "text" - }, - { - "block_id": "p721-b10", - "global_id": 21286, - "bbox": [ - 127.59, - 390.7, - 516.13, - 412.63 - ], - "text": "Before embarking on this study, we shall explain an important and pervasive aspect of the\nFourier transform: the time-frequency duality.", - "type": "text" - }, - { - "block_id": "p721-b11", - "global_id": 21287, - "bbox": [ - 127.56, - 434.89, - 516.13, - 545.78 - ], - "text": "† To explain this point, consider the unit step function and its transforms. Both the Laplace and the Fourier\ntransform synthesize x(t), using everlasting exponentials of the form est. The frequency s can be anywhere\nin the complex plane for the Laplace transform, but it must be restricted to the ω axis in the case of the\nFourier transform. The unit step function is readily synthesized in the Laplace transform by a relatively simple\nspectrum X(s) = 1/s, in which the frequencies s are chosen in the RHP [the region of convergence for u(t) is\nRe s > 0]. In the Fourier transform, however, we are restricted to values of s on the ω axis only. The function\nu(t) can still be synthesized by frequencies along the ω axis, but the spectrum is more complicated than it is\nwhen we are free to choose the frequencies in the RHP. In contrast, when x(t) is absolutely integrable, the\nregion of convergence for the Laplace transform includes the ω axis, and we can synthesize x(t) by using\nfrequencies along the ω axis in both transforms. This leads to X(jω) = X(ω).", - "type": "text" - }, - { - "block_id": "p721-b12", - "global_id": 21288, - "bbox": [ - 127.54, - 547.77, - 516.1, - 633.48 - ], - "text": "We may explain this concept by an example of two countries, X and Y. Suppose these countries want\nto construct similar dams in their respective territories. Country X has financial resources but not much\nmanpower. In contrast, Y has considerable manpower but few financial resources. The dams will still be\nconstructed in both countries, although the methods used will be different. Country X will use expensive\nbut efficient equipment to compensate for its lack of manpower, whereas Y will use the cheapest possible\nequipment in a labor-intensive approach to the project. Similarly, both Fourier and Laplace integrals converge\nfor u(t), but the makeup of the components used to synthesize u(t) will be very different for two cases because\nof the constraints of the Fourier transform, which are not present for the Laplace transform.", - "type": "text" - } - ] - }, - { - "page_num": 722, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p722-b0", - "global_id": 21289, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "702\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p722-b1", - "global_id": 21290, - "bbox": [ - 102.14, - 86.19, - 435.74, - 98.32 - ], - "text": "TIME-FREQUENCY DUALITY IN THE TRANSFORM OPERATIONS", - "type": "text" - }, - { - "block_id": "p722-b2", - "global_id": 21291, - "bbox": [ - 101.84, - 102.35, - 490.42, - 184.04 - ], - "text": "Equations (7.9) and (7.10) show an interesting fact: the direct and the inverse transform operations\nare remarkably similar. These operations, required to go from x(t) to X(ω) and then from X(ω) to\nx(t), are depicted graphically in Fig. 7.18. The inverse transform equation can be obtained from the\ndirect transform equation by replacing x(t) with X(ω), t with ω, and ω with t. In a similar way, we\ncan obtain the direct from the inverse. There are only two minor differences in these operations:\nthe factor 2π appears only in the inverse operator, and the exponential indices in the two operations\nhave opposite signs. Otherwise the two equations are duals of each other.†", - "type": "text" - }, - { - "block_id": "p722-b3", - "global_id": 21292, - "bbox": [ - 101.84, - 186.03, - 490.4, - 279.69 - ], - "text": "This observation has far-reaching consequences in the study of the Fourier transform. It is the\nbasis of the so-called duality of time and frequency. The duality principle may be compared with\na photograph and its negative. A photograph can be obtained from its negative, and by using an\nidentical procedure, a negative can be obtained from the photograph. For any result or relationship\nbetween x(t) and X(ω), there exists a dual result or relationship, obtained by interchanging the\nroles of x(t) and X(ω) in the original result (along with some minor modifications arising because\nof the factor 2π and a sign change). For example, the time-shifting property, to be proved later,\nstates that if x(t) ⇐⇒X(ω), then", - "type": "text" - }, - { - "block_id": "p722-b4", - "global_id": 21293, - "bbox": [ - 246.72, - 293.06, - 344.5, - 305.93 - ], - "text": "x(t −t0) ⇐⇒X(ω)e−jωt0", - "type": "text" - }, - { - "block_id": "p722-b5", - "global_id": 21294, - "bbox": [ - 125.76, - 508.83, - 433.41, - 518.07 - ], - "text": "Figure 7.18 A near symmetry between the direct and the inverse Fourier transforms.", - "type": "text" - }, - { - "block_id": "p722-b6", - "global_id": 21295, - "bbox": [ - 101.84, - 545.3, - 490.38, - 568.48 - ], - "text": "† Of the two differences, the former can be eliminated by change of variable from ω to f (in hertz). In this\ncase ω = 2πf and dω = 2π df.", - "type": "text" - }, - { - "block_id": "p722-b7", - "global_id": 21296, - "bbox": [ - 110.81, - 570.47, - 327.06, - 579.44 - ], - "text": "Therefore, the direct and the inverse transforms are given by", - "type": "text" - }, - { - "block_id": "p722-b8", - "global_id": 21297, - "bbox": [ - 173.66, - 597.08, - 209.63, - 606.41 - ], - "text": "X(2πf) =", - "type": "text" - }, - { - "block_id": "p722-b9", - "global_id": 21298, - "bbox": [ - 211.47, - 584.87, - 226.92, - 595.84 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p722-b10", - "global_id": 21299, - "bbox": [ - 216.2, - 607.11, - 227.86, - 613.59 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p722-b11", - "global_id": 21300, - "bbox": [ - 229.36, - 593.66, - 344.66, - 606.41 - ], - "text": "x(t)e−j2πft dt\nand\nx(t) =", - "type": "text" - }, - { - "block_id": "p722-b12", - "global_id": 21301, - "bbox": [ - 346.51, - 584.87, - 361.96, - 595.84 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p722-b13", - "global_id": 21302, - "bbox": [ - 351.24, - 607.11, - 362.9, - 613.59 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p722-b14", - "global_id": 21303, - "bbox": [ - 364.4, - 593.66, - 417.26, - 606.41 - ], - "text": "X(2πf)ej2πft df", - "type": "text" - }, - { - "block_id": "p722-b15", - "global_id": 21304, - "bbox": [ - 101.84, - 624.45, - 416.87, - 633.41 - ], - "text": "This leaves only one significant difference, that of sign change in the exponential index.", - "type": "text" - } - ] - }, - { - "page_num": 723, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p723-b0", - "global_id": 21305, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n703", - "type": "text" - }, - { - "block_id": "p723-b1", - "global_id": 21306, - "bbox": [ - 127.59, - 85.82, - 402.2, - 95.78 - ], - "text": "The dual of this property (the frequency-shifting property) states that", - "type": "text" - }, - { - "block_id": "p723-b2", - "global_id": 21307, - "bbox": [ - 273.31, - 107.05, - 370.41, - 122.32 - ], - "text": "x(t)ejω0t ⇐⇒X(ω −ω0)", - "type": "text" - }, - { - "block_id": "p723-b3", - "global_id": 21308, - "bbox": [ - 127.59, - 137.34, - 516.13, - 183.17 - ], - "text": "Observe the role reversal of time and frequency in these two equations (with the minor difference\nof the sign change in the exponential index). The value of this principle lies in the fact that\nwhenever we derive any result, we can be sure that it has a dual. This possibility can give valuable\ninsights about many unsuspected properties or results in signal processing.", - "type": "text" - }, - { - "block_id": "p723-b4", - "global_id": 21309, - "bbox": [ - 127.59, - 185.16, - 516.14, - 219.03 - ], - "text": "The properties of the Fourier transform are useful not only in deriving the direct and inverse\ntransforms of many functions, but also in obtaining several valuable results in signal processing.\nThe reader should not fail to observe the ever-present duality in this discussion.", - "type": "text" - }, - { - "block_id": "p723-b5", - "global_id": 21310, - "bbox": [ - 127.59, - 235.69, - 516.13, - 274.53 - ], - "text": "LINEARITY\nThe linearity property, already introduced as Eq. (7.15), states that if x1(t) ⇐⇒X1(ω) and\nx2(t) ⇐⇒X2(ω), then a1x1(t) + a2x2(t) ⇐⇒a1X1(ω) + a2X2(ω).", - "type": "text" - }, - { - "block_id": "p723-b6", - "global_id": 21311, - "bbox": [ - 127.59, - 290.41, - 507.97, - 316.52 - ], - "text": "CONJUGATION AND CONJUGATE SYMMETRY\nThe conjugation property, which has already been introduced, states that if x(t) ⇐⇒X(ω), then", - "type": "text" - }, - { - "block_id": "p723-b7", - "global_id": 21312, - "bbox": [ - 284.48, - 330.18, - 359.24, - 342.19 - ], - "text": "x∗(t) ⇐⇒X∗(−ω)", - "type": "text" - }, - { - "block_id": "p723-b8", - "global_id": 21313, - "bbox": [ - 127.59, - 358.08, - 516.14, - 380.0 - ], - "text": "From this property follows the conjugate symmetry property, also introduced earlier, which states\nthat if x(t) is real, then", - "type": "text" - }, - { - "block_id": "p723-b9", - "global_id": 21314, - "bbox": [ - 289.3, - 385.62, - 354.43, - 397.62 - ], - "text": "X(−ω) = X∗(ω)", - "type": "text" - }, - { - "block_id": "p723-b10", - "global_id": 21315, - "bbox": [ - 127.59, - 414.38, - 260.13, - 440.49 - ], - "text": "DUALITY\nThe duality property states that if", - "type": "text" - }, - { - "block_id": "p723-b11", - "global_id": 21316, - "bbox": [ - 292.49, - 447.83, - 351.22, - 458.11 - ], - "text": "x(t) ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p723-b12", - "global_id": 21317, - "bbox": [ - 127.59, - 471.01, - 144.74, - 480.98 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p723-b13", - "global_id": 21318, - "bbox": [ - 282.63, - 488.32, - 516.13, - 498.7 - ], - "text": "X(t) ⇐⇒2πx(−ω)\n(7.25)", - "type": "text" - }, - { - "block_id": "p723-b14", - "global_id": 21319, - "bbox": [ - 127.59, - 512.42, - 276.11, - 522.47 - ], - "text": "Proof. From Eq. (7.10) we can write", - "type": "text" - }, - { - "block_id": "p723-b15", - "global_id": 21320, - "bbox": [ - 269.87, - 536.6, - 306.23, - 553.54 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p723-b16", - "global_id": 21321, - "bbox": [ - 297.76, - 550.24, - 308.71, - 560.62 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p723-b17", - "global_id": 21322, - "bbox": [ - 312.01, - 529.62, - 328.96, - 541.67 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p723-b18", - "global_id": 21323, - "bbox": [ - 317.28, - 554.49, - 329.83, - 561.46 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p723-b19", - "global_id": 21324, - "bbox": [ - 331.44, - 539.37, - 373.86, - 553.44 - ], - "text": "X(u)ejut du", - "type": "text" - }, - { - "block_id": "p723-b20", - "global_id": 21325, - "bbox": [ - 127.59, - 574.4, - 155.51, - 584.36 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p723-b21", - "global_id": 21326, - "bbox": [ - 265.01, - 582.75, - 328.38, - 606.68 - ], - "text": "2πx(−t) =\n# ∞", - "type": "text" - }, - { - "block_id": "p723-b22", - "global_id": 21327, - "bbox": [ - 316.69, - 607.62, - 329.25, - 614.6 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p723-b23", - "global_id": 21328, - "bbox": [ - 330.86, - 592.51, - 378.72, - 606.58 - ], - "text": "X(u)e−jut du", - "type": "text" - }, - { - "block_id": "p723-b24", - "global_id": 21329, - "bbox": [ - 127.59, - 624.41, - 262.62, - 634.79 - ], - "text": "Changing t to ω yields Eq. (7.25).", - "type": "text" - } - ] - }, - { - "page_num": 724, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p724-b0", - "global_id": 21330, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "704\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p724-b1", - "global_id": 21331, - "bbox": [ - 78.18, - 95.83, - 486.33, - 107.78 - ], - "text": "EXAMPLE 7.11\nApplying the Duality Property of the Fourier Transform", - "type": "text" - }, - { - "block_id": "p724-b2", - "global_id": 21332, - "bbox": [ - 103.16, - 125.53, - 454.98, - 135.49 - ], - "text": "Apply the duality property [Eq. (7.25)] of the Fourier transform to the pair in Fig. 7.19a.", - "type": "text" - }, - { - "block_id": "p724-b3", - "global_id": 21333, - "bbox": [ - 177.61, - 227.58, - 181.61, - 235.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p724-b4", - "global_id": 21334, - "bbox": [ - 168.84, - 178.42, - 172.84, - 186.42 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p724-b5", - "global_id": 21335, - "bbox": [ - 210.59, - 226.79, - 212.81, - 234.79 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p724-b6", - "global_id": 21336, - "bbox": [ - 251.17, - 251.22, - 260.05, - 259.22 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p724-b7", - "global_id": 21337, - "bbox": [ - 178.83, - 170.31, - 189.93, - 178.4 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p724-b8", - "global_id": 21338, - "bbox": [ - 149.45, - 283.85, - 160.55, - 291.93 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p724-b9", - "global_id": 21339, - "bbox": [ - 353.31, - 173.19, - 368.86, - 181.29 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p724-b10", - "global_id": 21340, - "bbox": [ - 381.23, - 272.5, - 396.78, - 280.61 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p724-b11", - "global_id": 21341, - "bbox": [ - 380.11, - 339.16, - 384.11, - 347.16 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p724-b12", - "global_id": 21342, - "bbox": [ - 383.7, - 180.49, - 387.25, - 188.49 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p724-b13", - "global_id": 21343, - "bbox": [ - 456.26, - 228.08, - 461.6, - 236.08 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p724-b14", - "global_id": 21344, - "bbox": [ - 417.9, - 339.01, - 423.23, - 347.01 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p724-b15", - "global_id": 21345, - "bbox": [ - 250.62, - 367.77, - 260.59, - 375.77 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p724-b16", - "global_id": 21346, - "bbox": [ - 181.36, - 291.53, - 184.91, - 299.53 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p724-b17", - "global_id": 21347, - "bbox": [ - 187.62, - 338.72, - 266.3, - 353.8 - ], - "text": "t\nt\n2p", - "type": "text" - }, - { - "block_id": "p724-b18", - "global_id": 21348, - "bbox": [ - 364.57, - 282.74, - 373.91, - 290.85 - ], - "text": "2p", - "type": "text" - }, - { - "block_id": "p724-b19", - "global_id": 21349, - "bbox": [ - 219.67, - 338.81, - 229.01, - 353.8 - ], - "text": "t\n4p", - "type": "text" - }, - { - "block_id": "p724-b20", - "global_id": 21350, - "bbox": [ - 144.44, - 225.91, - 202.51, - 242.83 - ], - "text": "t\n2\nt\n2", - "type": "text" - }, - { - "block_id": "p724-b21", - "global_id": 21351, - "bbox": [ - 345.84, - 337.39, - 404.71, - 354.31 - ], - "text": "t\n2\nt\n2", - "type": "text" - }, - { - "block_id": "p724-b22", - "global_id": 21352, - "bbox": [ - 352.31, - 225.89, - 369.25, - 240.86 - ], - "text": "2p\n t", - "type": "text" - }, - { - "block_id": "p724-b23", - "global_id": 21353, - "bbox": [ - 115.51, - 338.83, - 132.45, - 353.8 - ], - "text": "4p\n t", - "type": "text" - }, - { - "block_id": "p724-b24", - "global_id": 21354, - "bbox": [ - 146.09, - 338.83, - 163.03, - 353.8 - ], - "text": "2p\n t", - "type": "text" - }, - { - "block_id": "p724-b25", - "global_id": 21355, - "bbox": [ - 390.95, - 225.89, - 400.29, - 240.86 - ], - "text": "2p\nt", - "type": "text" - }, - { - "block_id": "p724-b26", - "global_id": 21356, - "bbox": [ - 103.16, - 397.1, - 317.71, - 406.34 - ], - "text": "Figure 7.19 The duality property of the Fourier transform.", - "type": "text" - }, - { - "block_id": "p724-b27", - "global_id": 21357, - "bbox": [ - 121.09, - 425.9, - 220.34, - 435.87 - ], - "text": "From Eq. (7.19) we have", - "type": "text" - }, - { - "block_id": "p724-b28", - "global_id": 21358, - "bbox": [ - 235.16, - 452.8, - 250.09, - 462.76 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p724-b29", - "global_id": 21359, - "bbox": [ - 251.2, - 441.38, - 261.88, - 455.67 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p724-b30", - "global_id": 21360, - "bbox": [ - 257.82, - 459.46, - 262.13, - 469.42 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p724-b32", - "global_id": 21361, - "bbox": [ - 235.16, - 463.56, - 269.97, - 482.69 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p724-b33", - "global_id": 21362, - "bbox": [ - 272.0, - 452.38, - 315.83, - 462.76 - ], - "text": "⇐⇒τ sinc", - "type": "text" - }, - { - "block_id": "p724-b34", - "global_id": 21363, - "bbox": [ - 316.95, - 438.4, - 335.91, - 455.36 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p724-b35", - "global_id": 21364, - "bbox": [ - 328.5, - 459.87, - 333.48, - 469.83 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p724-b37", - "global_id": 21365, - "bbox": [ - 293.17, - 465.85, - 345.03, - 484.99 - ], - "text": "X(ω)", - "type": "text" - }, - { - "block_id": "p724-b38", - "global_id": 21366, - "bbox": [ - 103.16, - 494.85, - 476.84, - 517.18 - ], - "text": "Also, X(t) is the same as X(ω) with ω replaced by t, and x(−ω) is the same as x(t) with t\nreplaced by −ω. Therefore, the duality property of Eq. (7.25) yields", - "type": "text" - }, - { - "block_id": "p724-b39", - "global_id": 21367, - "bbox": [ - 194.25, - 533.7, - 216.92, - 544.07 - ], - "text": "τ sinc", - "type": "text" - }, - { - "block_id": "p724-b40", - "global_id": 21368, - "bbox": [ - 219.13, - 522.7, - 234.04, - 536.99 - ], - "text": "τt", - "type": "text" - }, - { - "block_id": "p724-b41", - "global_id": 21369, - "bbox": [ - 227.5, - 541.18, - 232.48, - 551.15 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p724-b43", - "global_id": 21370, - "bbox": [ - 194.25, - 544.78, - 240.84, - 563.91 - ], - "text": "X(t)", - "type": "text" - }, - { - "block_id": "p724-b44", - "global_id": 21371, - "bbox": [ - 242.89, - 533.7, - 292.05, - 544.07 - ], - "text": "⇐⇒2π rect", - "type": "text" - }, - { - "block_id": "p724-b45", - "global_id": 21372, - "bbox": [ - 293.16, - 519.71, - 315.38, - 536.67 - ], - "text": "−ω", - "type": "text" - }, - { - "block_id": "p724-b46", - "global_id": 21373, - "bbox": [ - 305.57, - 540.77, - 309.89, - 550.73 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p724-b48", - "global_id": 21374, - "bbox": [ - 264.06, - 547.16, - 323.49, - 566.37 - ], - "text": "2πx(−ω)", - "type": "text" - }, - { - "block_id": "p724-b49", - "global_id": 21375, - "bbox": [ - 325.55, - 533.7, - 363.35, - 544.07 - ], - "text": "= 2π rect", - "type": "text" - }, - { - "block_id": "p724-b50", - "global_id": 21376, - "bbox": [ - 363.36, - 519.71, - 377.81, - 536.67 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p724-b51", - "global_id": 21377, - "bbox": [ - 371.9, - 540.77, - 376.21, - 550.73 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p724-b53", - "global_id": 21378, - "bbox": [ - 103.16, - 576.17, - 477.04, - 622.41 - ], - "text": "In this result, we used the fact that rect(−x) = rect(x) because rect is an even function.\nFigure 7.19b shows this pair graphically. Observe the interchange of the roles of t and ω\n(with the minor adjustment of the factor 2π). This result appears as pair 18 in Table 7.1 (with\nτ/2 = W).", - "type": "text" - }, - { - "block_id": "p724-b54", - "global_id": 21379, - "bbox": [ - 103.16, - 624.4, - 477.0, - 646.32 - ], - "text": "As an interesting exercise, the reader should generate the dual of every pair in Table 7.1\nby applying the duality property.", - "type": "text" - } - ] - }, - { - "page_num": 725, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p725-b0", - "global_id": 21380, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n705", - "type": "text" - }, - { - "block_id": "p725-b1", - "global_id": 21381, - "bbox": [ - 133.57, - 97.81, - 453.64, - 123.72 - ], - "text": "DRILL 7.4\nApplying the Duality Property of the Fourier\nTransform", - "type": "text" - }, - { - "block_id": "p725-b2", - "global_id": 21382, - "bbox": [ - 133.57, - 132.84, - 412.82, - 142.8 - ], - "text": "Apply the duality property to pairs 1, 3, and 9 (Table 7.1) to show that", - "type": "text" - }, - { - "block_id": "p725-b3", - "global_id": 21383, - "bbox": [ - 151.5, - 149.12, - 281.95, - 160.73 - ], - "text": "(a) 1/(jt + a) ⇐⇒2πeaωu(−ω)", - "type": "text" - }, - { - "block_id": "p725-b4", - "global_id": 21384, - "bbox": [ - 151.5, - 161.68, - 274.71, - 175.68 - ], - "text": "(b) 2a/(t2 + a2) ⇐⇒2πe−a|ω|", - "type": "text" - }, - { - "block_id": "p725-b5", - "global_id": 21385, - "bbox": [ - 152.06, - 180.24, - 302.7, - 191.39 - ], - "text": "(c) δ(t + t0) + δ(t −t0) ⇐⇒2cos t0ω", - "type": "text" - }, - { - "block_id": "p725-b6", - "global_id": 21386, - "bbox": [ - 127.59, - 220.48, - 260.9, - 246.59 - ], - "text": "THE SCALING PROPERTY\nIf", - "type": "text" - }, - { - "block_id": "p725-b7", - "global_id": 21387, - "bbox": [ - 292.49, - 248.18, - 351.22, - 258.45 - ], - "text": "x(t) ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p725-b8", - "global_id": 21388, - "bbox": [ - 127.59, - 266.87, - 241.15, - 276.93 - ], - "text": "then, for any real constant a,", - "type": "text" - }, - { - "block_id": "p725-b9", - "global_id": 21389, - "bbox": [ - 280.01, - 276.9, - 332.07, - 293.75 - ], - "text": "x(at) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p725-b10", - "global_id": 21390, - "bbox": [ - 324.23, - 283.79, - 342.19, - 300.82 - ], - "text": "|a|X", - "type": "text" - }, - { - "block_id": "p725-b11", - "global_id": 21391, - "bbox": [ - 343.74, - 272.48, - 356.9, - 286.45 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p725-b12", - "global_id": 21392, - "bbox": [ - 351.24, - 290.86, - 356.22, - 300.82 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p725-b14", - "global_id": 21393, - "bbox": [ - 492.07, - 283.89, - 516.13, - 293.85 - ], - "text": "(7.26)", - "type": "text" - }, - { - "block_id": "p725-b15", - "global_id": 21394, - "bbox": [ - 127.59, - 308.09, - 275.85, - 318.15 - ], - "text": "Proof. For a positive real constant a,", - "type": "text" - }, - { - "block_id": "p725-b16", - "global_id": 21395, - "bbox": [ - 191.95, - 333.91, - 236.37, - 344.19 - ], - "text": "F[x(at)] =", - "type": "text" - }, - { - "block_id": "p725-b17", - "global_id": 21396, - "bbox": [ - 238.42, - 320.36, - 255.37, - 332.41 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p725-b18", - "global_id": 21397, - "bbox": [ - 243.68, - 345.23, - 256.24, - 352.2 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p725-b19", - "global_id": 21398, - "bbox": [ - 257.84, - 327.35, - 322.71, - 344.29 - ], - "text": "x(at)e−jωtdt = 1", - "type": "text" - }, - { - "block_id": "p725-b20", - "global_id": 21399, - "bbox": [ - 317.73, - 341.3, - 322.71, - 351.26 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p725-b21", - "global_id": 21400, - "bbox": [ - 325.01, - 320.36, - 341.96, - 332.41 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p725-b22", - "global_id": 21401, - "bbox": [ - 330.27, - 345.23, - 342.83, - 352.2 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p725-b23", - "global_id": 21402, - "bbox": [ - 344.43, - 327.35, - 422.97, - 344.29 - ], - "text": "x(u)e(−jω/a)u du = 1", - "type": "text" - }, - { - "block_id": "p725-b24", - "global_id": 21403, - "bbox": [ - 417.99, - 334.22, - 430.25, - 351.26 - ], - "text": "aX", - "type": "text" - }, - { - "block_id": "p725-b25", - "global_id": 21404, - "bbox": [ - 431.79, - 322.91, - 444.94, - 336.9 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p725-b26", - "global_id": 21405, - "bbox": [ - 439.29, - 341.3, - 444.27, - 351.26 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p725-b28", - "global_id": 21406, - "bbox": [ - 127.59, - 360.06, - 300.96, - 370.44 - ], - "text": "Similarly, we can demonstrate that if a < 0,", - "type": "text" - }, - { - "block_id": "p725-b29", - "global_id": 21407, - "bbox": [ - 278.97, - 378.86, - 335.95, - 396.12 - ], - "text": "x(at) ⇐⇒−1", - "type": "text" - }, - { - "block_id": "p725-b30", - "global_id": 21408, - "bbox": [ - 327.09, - 386.16, - 343.24, - 403.2 - ], - "text": "a X", - "type": "text" - }, - { - "block_id": "p725-b31", - "global_id": 21409, - "bbox": [ - 344.78, - 374.85, - 357.93, - 388.83 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p725-b32", - "global_id": 21410, - "bbox": [ - 352.28, - 393.24, - 357.26, - 403.2 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p725-b34", - "global_id": 21411, - "bbox": [ - 127.59, - 410.1, - 230.26, - 420.07 - ], - "text": "Hence follows Eq. (7.26).", - "type": "text" - }, - { - "block_id": "p725-b35", - "global_id": 21412, - "bbox": [ - 127.59, - 434.53, - 516.14, - 592.15 - ], - "text": "SIGNIFICANCE OF THE SCALING PROPERTY\nThe function x(at) represents the function x(t) compressed in time by a factor a (see Sec. 1.2-2).\nSimilarly, a function X(ω/a) represents the function X(ω) expanded in frequency by the same\nfactor a. The scaling property states that time compression of a signal results in its spectral\nexpansion, and time expansion of the signal results in its spectral compression. Intuitively,\ncompression in time by factor a means that the signal is varying faster by factor a.† To synthesize\nsuch a signal, the frequencies of its sinusoidal components must be increased by the factor a,\nimplying that its frequency spectrum is expanded by the factor a. Similarly, a signal expanded\nin time varies more slowly; hence the frequencies of its components are lowered, implying that\nits frequency spectrum is compressed. For instance, the signal cos2ω0t is the same as the signal\ncosω0t time-compressed by a factor of 2. Clearly, the spectrum of the former (impulse at ±2ω0)\nis an expanded version of the spectrum of the latter (impulse at ±ω0). The effect of this scaling is\ndemonstrated in Fig. 7.20.", - "type": "text" - }, - { - "block_id": "p725-b36", - "global_id": 21413, - "bbox": [ - 127.59, - 610.24, - 516.11, - 633.41 - ], - "text": "† We are assuming a > 1, although the argument still holds if a < 1. In the latter case, compression becomes\nexpansion by factor 1/a, and vice versa.", - "type": "text" - } - ] - }, - { - "page_num": 726, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p726-b0", - "global_id": 21414, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "706\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p726-b1", - "global_id": 21415, - "bbox": [ - 186.46, - 229.72, - 190.46, - 237.72 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p726-b2", - "global_id": 21416, - "bbox": [ - 177.73, - 180.53, - 181.73, - 188.53 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p726-b3", - "global_id": 21417, - "bbox": [ - 204.39, - 234.11, - 206.61, - 242.11 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p726-b4", - "global_id": 21418, - "bbox": [ - 186.41, - 144.56, - 190.41, - 152.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p726-b5", - "global_id": 21419, - "bbox": [ - 177.64, - 95.4, - 181.64, - 103.4 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p726-b6", - "global_id": 21420, - "bbox": [ - 219.38, - 143.77, - 221.61, - 151.77 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p726-b7", - "global_id": 21421, - "bbox": [ - 187.63, - 87.29, - 198.73, - 95.37 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p726-b8", - "global_id": 21422, - "bbox": [ - 187.63, - 170.65, - 198.73, - 178.73 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p726-b9", - "global_id": 21423, - "bbox": [ - 356.34, - 96.54, - 371.9, - 104.64 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p726-b10", - "global_id": 21424, - "bbox": [ - 356.34, - 172.43, - 371.9, - 180.53 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p726-b11", - "global_id": 21425, - "bbox": [ - 387.45, - 114.02, - 391.0, - 122.02 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p726-b12", - "global_id": 21426, - "bbox": [ - 469.8, - 143.36, - 475.13, - 151.36 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p726-b13", - "global_id": 21427, - "bbox": [ - 460.29, - 230.64, - 465.63, - 238.64 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p726-b14", - "global_id": 21428, - "bbox": [ - 207.11, - 142.86, - 211.11, - 159.87 - ], - "text": "2\nt", - "type": "text" - }, - { - "block_id": "p726-b15", - "global_id": 21429, - "bbox": [ - 128.85, - 227.52, - 236.93, - 235.71 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p726-b16", - "global_id": 21430, - "bbox": [ - 412.62, - 144.49, - 460.73, - 159.47 - ], - "text": "t\n2p\nt\n4p", - "type": "text" - }, - { - "block_id": "p726-b17", - "global_id": 21431, - "bbox": [ - 384.15, - 183.5, - 391.7, - 191.61 - ], - "text": "2t", - "type": "text" - }, - { - "block_id": "p726-b18", - "global_id": 21432, - "bbox": [ - 394.11, - 228.03, - 399.45, - 243.02 - ], - "text": "t\np", - "type": "text" - }, - { - "block_id": "p726-b19", - "global_id": 21433, - "bbox": [ - 153.23, - 142.89, - 165.48, - 159.81 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p726-b20", - "global_id": 21434, - "bbox": [ - 351.47, - 228.05, - 364.21, - 243.01 - ], - "text": "p\n t", - "type": "text" - }, - { - "block_id": "p726-b21", - "global_id": 21435, - "bbox": [ - 287.13, - 144.51, - 304.07, - 159.47 - ], - "text": "4p\n t", - "type": "text" - }, - { - "block_id": "p726-b22", - "global_id": 21436, - "bbox": [ - 328.46, - 144.51, - 345.4, - 159.47 - ], - "text": "2p\n t", - "type": "text" - }, - { - "block_id": "p726-b23", - "global_id": 21437, - "bbox": [ - 125.76, - 258.44, - 340.8, - 267.67 - ], - "text": "Figure 7.20 The scaling property of the Fourier transform.", - "type": "text" - }, - { - "block_id": "p726-b24", - "global_id": 21438, - "bbox": [ - 102.14, - 287.56, - 417.18, - 299.68 - ], - "text": "RECIPROCITY OF SIGNAL DURATION AND ITS BANDWIDTH", - "type": "text" - }, - { - "block_id": "p726-b25", - "global_id": 21439, - "bbox": [ - 101.84, - 303.3, - 490.39, - 361.49 - ], - "text": "The scaling property implies that if x(t) is wider, its spectrum is narrower, and vice versa. Doubling\nthe signal duration halves its bandwidth, and vice versa. This suggests that the bandwidth of a\nsignal is inversely proportional to the signal duration or width (in seconds).† We have already\nverified this fact for the gate pulse, where we found that the bandwidth of a gate pulse of width τ\nseconds is 1/τ Hz. More discussion of this interesting topic can be found in the literature [2].", - "type": "text" - }, - { - "block_id": "p726-b26", - "global_id": 21440, - "bbox": [ - 101.84, - 363.08, - 490.36, - 385.31 - ], - "text": "By letting a = −1 in Eq. (7.26), we obtain the inversion (or reflection) property of time and\nfrequency:", - "type": "text" - }, - { - "block_id": "p726-b27", - "global_id": 21441, - "bbox": [ - 258.97, - 386.98, - 490.38, - 397.36 - ], - "text": "x(−t) ⇐⇒X(−ω)\n(7.27)", - "type": "text" - }, - { - "block_id": "p726-b28", - "global_id": 21442, - "bbox": [ - 76.77, - 430.97, - 388.52, - 442.92 - ], - "text": "EXAMPLE 7.12\nFourier Transform Reflection Property", - "type": "text" - }, - { - "block_id": "p726-b29", - "global_id": 21443, - "bbox": [ - 103.16, - 459.51, - 477.02, - 481.42 - ], - "text": "Using the reflection property of the Fourier transform and Table 7.1, find the Fourier transforms\nof eatu(−t) and e−a|t|.", - "type": "text" - }, - { - "block_id": "p726-b30", - "global_id": 21444, - "bbox": [ - 103.16, - 504.34, - 315.68, - 514.3 - ], - "text": "Application of Eq. (7.27) to pair 1 of Table 7.1 yields", - "type": "text" - }, - { - "block_id": "p726-b31", - "global_id": 21445, - "bbox": [ - 226.93, - 524.09, - 353.25, - 548.0 - ], - "text": "eatu(−t) ⇐⇒\n1\na −jω\na > 0", - "type": "text" - }, - { - "block_id": "p726-b32", - "global_id": 21446, - "bbox": [ - 103.16, - 557.77, - 124.47, - 567.73 - ], - "text": "Also,", - "type": "text" - }, - { - "block_id": "p726-b33", - "global_id": 21447, - "bbox": [ - 236.33, - 565.21, - 343.84, - 579.6 - ], - "text": "e−a|t| = e−atu(t) + eatu(−t)", - "type": "text" - }, - { - "block_id": "p726-b34", - "global_id": 21448, - "bbox": [ - 101.84, - 610.73, - 490.37, - 633.91 - ], - "text": "† When a signal has infinite duration, we must consider its effective or equivalent duration. There is no unique\ndefinition of effective signal duration. One possible definition is given in Eq. (2.47).", - "type": "text" - } - ] - }, - { - "page_num": 727, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p727-b0", - "global_id": 21449, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n707", - "type": "text" - }, - { - "block_id": "p727-b1", - "global_id": 21450, - "bbox": [ - 128.9, - 86.24, - 170.65, - 96.2 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p727-b2", - "global_id": 21451, - "bbox": [ - 217.82, - 95.36, - 502.75, - 119.38 - ], - "text": "e−a|t| ⇐⇒\n1\na + jω +\n1\na −jω =\n2a\na2 + ω2\na > 0\n(7.28)", - "type": "text" - }, - { - "block_id": "p727-b3", - "global_id": 21452, - "bbox": [ - 128.91, - 124.17, - 368.81, - 137.76 - ], - "text": "The signal e−a|t| and its spectrum are illustrated in Fig. 7.21.", - "type": "text" - }, - { - "block_id": "p727-b4", - "global_id": 21453, - "bbox": [ - 212.84, - 177.16, - 216.84, - 185.16 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p727-b5", - "global_id": 21454, - "bbox": [ - 231.76, - 188.54, - 270.73, - 198.23 - ], - "text": "x(t) eat", - "type": "text" - }, - { - "block_id": "p727-b6", - "global_id": 21455, - "bbox": [ - 214.15, - 236.83, - 495.15, - 245.38 - ], - "text": "0\nv\nt\n0", - "type": "text" - }, - { - "block_id": "p727-b7", - "global_id": 21456, - "bbox": [ - 449.59, - 176.38, - 502.65, - 194.7 - ], - "text": "X(v) \n2a\na2 v2", - "type": "text" - }, - { - "block_id": "p727-b8", - "global_id": 21457, - "bbox": [ - 221.12, - 251.53, - 434.55, - 259.53 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p727-b9", - "global_id": 21458, - "bbox": [ - 136.48, - 264.28, - 322.58, - 276.76 - ], - "text": "Figure 7.21 (a) e−a|t| and (b) its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p727-b10", - "global_id": 21459, - "bbox": [ - 127.59, - 320.44, - 296.17, - 346.55 - ], - "text": "THE TIME-SHIFTING PROPERTY\nIf", - "type": "text" - }, - { - "block_id": "p727-b11", - "global_id": 21460, - "bbox": [ - 292.49, - 350.66, - 351.22, - 360.93 - ], - "text": "x(t) ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p727-b12", - "global_id": 21461, - "bbox": [ - 127.59, - 371.68, - 144.74, - 381.64 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p727-b13", - "global_id": 21462, - "bbox": [ - 272.47, - 384.04, - 516.13, - 396.91 - ], - "text": "x(t −t0) ⇐⇒X(ω)e−jωt0\n(7.29)", - "type": "text" - }, - { - "block_id": "p727-b14", - "global_id": 21463, - "bbox": [ - 127.59, - 407.7, - 213.66, - 417.74 - ], - "text": "Proof. By definition,", - "type": "text" - }, - { - "block_id": "p727-b15", - "global_id": 21464, - "bbox": [ - 252.31, - 427.18, - 309.37, - 438.33 - ], - "text": "F[x(t −t0)] =", - "type": "text" - }, - { - "block_id": "p727-b16", - "global_id": 21465, - "bbox": [ - 311.42, - 413.62, - 328.37, - 425.68 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p727-b17", - "global_id": 21466, - "bbox": [ - 316.68, - 438.5, - 329.24, - 445.47 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p727-b18", - "global_id": 21467, - "bbox": [ - 330.85, - 423.38, - 391.23, - 438.33 - ], - "text": "x(t −t0)e−jωt dt", - "type": "text" - }, - { - "block_id": "p727-b19", - "global_id": 21468, - "bbox": [ - 127.59, - 453.12, - 233.82, - 464.58 - ], - "text": "Letting t −t0 = u, we have", - "type": "text" - }, - { - "block_id": "p727-b20", - "global_id": 21469, - "bbox": [ - 171.03, - 482.05, - 228.09, - 493.2 - ], - "text": "F[x(t −t0)] =", - "type": "text" - }, - { - "block_id": "p727-b21", - "global_id": 21470, - "bbox": [ - 230.14, - 468.5, - 247.08, - 480.55 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p727-b22", - "global_id": 21471, - "bbox": [ - 235.4, - 493.37, - 247.95, - 500.34 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p727-b23", - "global_id": 21472, - "bbox": [ - 249.56, - 478.25, - 347.74, - 492.33 - ], - "text": "x(u)e−jω(u+t0) du = e−jωt0", - "type": "text" - }, - { - "block_id": "p727-b24", - "global_id": 21473, - "bbox": [ - 349.86, - 468.5, - 366.81, - 480.55 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p727-b25", - "global_id": 21474, - "bbox": [ - 355.12, - 493.37, - 367.68, - 500.34 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p727-b26", - "global_id": 21475, - "bbox": [ - 369.28, - 478.25, - 471.68, - 492.33 - ], - "text": "x(u)e−jωu du = X(ω)e−jωt0", - "type": "text" - }, - { - "block_id": "p727-b27", - "global_id": 21476, - "bbox": [ - 127.59, - 511.3, - 516.14, - 534.09 - ], - "text": "This result shows that delaying a signal by t0 seconds does not change its amplitude spectrum. The\nphase spectrum, however, is changed by −ωt0.", - "type": "text" - }, - { - "block_id": "p727-b28", - "global_id": 21477, - "bbox": [ - 127.89, - 548.89, - 388.93, - 561.01 - ], - "text": "PHYSICAL EXPLANATION OF THE LINEAR PHASE", - "type": "text" - }, - { - "block_id": "p727-b29", - "global_id": 21478, - "bbox": [ - 127.59, - 565.05, - 516.16, - 636.28 - ], - "text": "Time delay in a signal causes a linear phase shift in its spectrum. This result can also be derived by\nheuristic reasoning. Imagine x(t) being synthesized by its Fourier components, which are sinusoids\nof certain amplitudes and phases. The delayed signal x(t −t0) can be synthesized by the same\nsinusoidal components, each delayed by t0 seconds. The amplitudes of the components remain\nunchanged. Therefore, the amplitude spectrum of x(t −t0) is identical to that of x(t). The time\ndelay of t0 in each sinusoid, however, does change the phase of each component. Now, a sinusoid", - "type": "text" - } - ] - }, - { - "page_num": 728, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p728-b0", - "global_id": 21479, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "708\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p728-b1", - "global_id": 21480, - "bbox": [ - 402.33, - 224.46, - 404.55, - 232.46 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p728-b2", - "global_id": 21481, - "bbox": [ - 147.65, - 168.37, - 152.87, - 177.97 - ], - "text": "t0", - "type": "text" - }, - { - "block_id": "p728-b3", - "global_id": 21482, - "bbox": [ - 402.33, - 128.18, - 404.55, - 136.18 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p728-b4", - "global_id": 21483, - "bbox": [ - 125.76, - 268.43, - 355.87, - 277.67 - ], - "text": "Figure 7.22 Physical explanation of the time-shifting property.", - "type": "text" - }, - { - "block_id": "p728-b5", - "global_id": 21484, - "bbox": [ - 101.84, - 299.87, - 226.89, - 311.74 - ], - "text": "cosωt delayed by t0 is given by", - "type": "text" - }, - { - "block_id": "p728-b6", - "global_id": 21485, - "bbox": [ - 236.6, - 323.9, - 355.63, - 335.05 - ], - "text": "cos ω(t −t0) = cos(ωt −ωt0)", - "type": "text" - }, - { - "block_id": "p728-b7", - "global_id": 21486, - "bbox": [ - 101.84, - 347.92, - 490.39, - 441.98 - ], - "text": "Therefore a time delay t0 in a sinusoid of frequency ω manifests as a phase delay of ωt0. This is\na linear function of ω, meaning that higher-frequency components must undergo proportionately\nhigher phase shifts to achieve the same time delay. This effect is depicted in Fig. 7.22 with two\nsinusoids, the frequency of the lower sinusoid being twice that of the upper. The same time delay t0\namounts to a phase shift of π/2 in the upper sinusoid and a phase shift of π in the lower sinusoid.\nThis verifies the fact that to achieve the same time delay, higher-frequency sinusoids must undergo\nproportionately higher phase shifts. The principle of linear phase shift is very important, and we\nshall encounter it again in distortionless signal transmission and filtering applications.", - "type": "text" - }, - { - "block_id": "p728-b8", - "global_id": 21487, - "bbox": [ - 76.77, - 472.03, - 410.44, - 483.99 - ], - "text": "EXAMPLE 7.13\nFourier Transform Time-Shifting Property", - "type": "text" - }, - { - "block_id": "p728-b9", - "global_id": 21488, - "bbox": [ - 103.16, - 499.02, - 386.21, - 510.61 - ], - "text": "Use the time-shifting property to find the Fourier transform of e−a|t−t0|.", - "type": "text" - }, - { - "block_id": "p728-b10", - "global_id": 21489, - "bbox": [ - 103.16, - 529.91, - 477.0, - 555.45 - ], - "text": "This function, shown in Fig. 7.23a, is a time-shifted version of e−a|t| (depicted in Fig. 7.21a).\nFrom Eqs. (7.28) and (7.29), we have", - "type": "text" - }, - { - "block_id": "p728-b11", - "global_id": 21490, - "bbox": [ - 235.41, - 565.28, - 343.75, - 589.31 - ], - "text": "e−a|t−t0| ⇐⇒\n2a\na2 + ω2 e−jωt0", - "type": "text" - }, - { - "block_id": "p728-b12", - "global_id": 21491, - "bbox": [ - 103.17, - 595.88, - 477.01, - 622.18 - ], - "text": "The spectrum of e−a|t−t0| (Fig. 7.23b) is the same as that of e−a|t| (Fig. 7.21b), except for\nan added phase shift of −ωt0.", - "type": "text" - } - ] - }, - { - "page_num": 729, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p729-b0", - "global_id": 21492, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n709", - "type": "text" - }, - { - "block_id": "p729-b1", - "global_id": 21493, - "bbox": [ - 128.91, - 85.83, - 502.75, - 108.15 - ], - "text": "Observe that the time delay t0 causes a linear phase spectrum −ωt0. This example clearly\ndemonstrates the effect of time shift.", - "type": "text" - }, - { - "block_id": "p729-b2", - "global_id": 21494, - "bbox": [ - 120.73, - 267.57, - 380.63, - 276.81 - ], - "text": "Figure 7.23 Effect of time-shifting on the Fourier spectrum of a signal.", - "type": "text" - }, - { - "block_id": "p729-b3", - "global_id": 21495, - "bbox": [ - 102.51, - 347.68, - 506.19, - 359.64 - ], - "text": "EXAMPLE 7.14\nFourier Transform of a Time-Shifted Rectangular Pulse", - "type": "text" - }, - { - "block_id": "p729-b4", - "global_id": 21496, - "bbox": [ - 128.9, - 375.88, - 496.65, - 386.26 - ], - "text": "Find the Fourier transform of the time-shifted rectangular pulse x(t) illustrated in Fig. 7.24a.", - "type": "text" - }, - { - "block_id": "p729-b5", - "global_id": 21497, - "bbox": [ - 128.9, - 408.76, - 502.76, - 443.05 - ], - "text": "The pulse x(t) is the gate pulse rect(t/τ) in Fig. 7.10a delayed by 3τ/4 seconds. Hence,\naccording to Eq. (7.29), its Fourier transform is the Fourier transform of rect(t/τ) multiplied\nby e−jω(3τ/4). Therefore,", - "type": "text" - }, - { - "block_id": "p729-b6", - "global_id": 21498, - "bbox": [ - 255.28, - 450.27, - 310.5, - 460.64 - ], - "text": "X(ω) = τ sinc", - "type": "text" - }, - { - "block_id": "p729-b7", - "global_id": 21499, - "bbox": [ - 311.61, - 436.28, - 330.56, - 453.25 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p729-b8", - "global_id": 21500, - "bbox": [ - 323.15, - 457.75, - 328.13, - 467.72 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p729-b10", - "global_id": 21501, - "bbox": [ - 339.68, - 448.54, - 375.88, - 460.54 - ], - "text": "e−jω(3τ/4)", - "type": "text" - }, - { - "block_id": "p729-b11", - "global_id": 21502, - "bbox": [ - 128.9, - 474.55, - 502.78, - 520.79 - ], - "text": "The amplitude spectrum |X(ω)| (depicted in Fig. 7.24b) of this pulse is the same as that\nindicated in Fig. 7.10c. But the phase spectrum has an added linear term −3ωτ/4. Hence,\nthe phase spectrum of x(t) (Fig. 7.24a) is identical to that in Fig. 7.10d plus a linear term\n−3ωτ/4, as shown in Fig. 7.24c.", - "type": "text" - }, - { - "block_id": "p729-b12", - "global_id": 21503, - "bbox": [ - 129.2, - 535.52, - 371.83, - 547.64 - ], - "text": "PHASE SPECTRUM USING PRINCIPAL VALUES", - "type": "text" - }, - { - "block_id": "p729-b13", - "global_id": 21504, - "bbox": [ - 128.9, - 551.26, - 502.78, - 621.41 - ], - "text": "There is an alternate way of spectral representation of̸\nX(ω). The phase angle computed on a\ncalculator or by using a computer subroutine is generally the principal value (modulo 2π value)\nof the phase angle, which always lies in the range −π to π. For instance, the principal value\nof angle 3π/2 is −π/2, and so on. The principal value differs from the actual value by ±2π\nradians (and its integer multiples) in a way that ensures that the principal value remains within\n−π to π. Thus, the principal value will show jump discontinuities of ±2π whenever the actual", - "type": "text" - } - ] - }, - { - "page_num": 730, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p730-b0", - "global_id": 21505, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "710\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p730-b1", - "global_id": 21506, - "bbox": [ - 103.16, - 85.83, - 476.99, - 120.12 - ], - "text": "phase crosses ±π. The phase plot in Fig. 7.24c is redrawn in Fig. 7.24d using the principal\nvalue for the phase. This phase pattern, which contains phase discontinuities of magnitudes 2π\nand π, becomes repetitive at intervals of ω = 8π/τ.", - "type": "text" - }, - { - "block_id": "p730-b2", - "global_id": 21507, - "bbox": [ - 336.41, - 187.39, - 338.64, - 195.39 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p730-b3", - "global_id": 21508, - "bbox": [ - 350.56, - 259.8, - 433.75, - 274.55 - ], - "text": "8p\nt\n6p\nt\n4p\nt", - "type": "text" - }, - { - "block_id": "p730-b4", - "global_id": 21509, - "bbox": [ - 244.29, - 186.29, - 321.92, - 203.07 - ], - "text": "t\n4\n3t\n4\n5t\n4", - "type": "text" - }, - { - "block_id": "p730-b5", - "global_id": 21510, - "bbox": [ - 300.5, - 436.84, - 334.49, - 453.52 - ], - "text": "4p\n3t\n8p\n3t", - "type": "text" - }, - { - "block_id": "p730-b6", - "global_id": 21511, - "bbox": [ - 372.47, - 426.43, - 385.8, - 443.11 - ], - "text": "16p\n3t", - "type": "text" - }, - { - "block_id": "p730-b7", - "global_id": 21512, - "bbox": [ - 282.37, - 257.14, - 286.37, - 265.14 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p730-b8", - "global_id": 21513, - "bbox": [ - 238.38, - 219.06, - 257.97, - 227.36 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p730-b9", - "global_id": 21514, - "bbox": [ - 288.1, - 211.25, - 291.65, - 219.25 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p730-b10", - "global_id": 21515, - "bbox": [ - 272.36, - 332.36, - 276.36, - 340.36 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p730-b11", - "global_id": 21516, - "bbox": [ - 349.97, - 318.67, - 355.3, - 326.67 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p730-b12", - "global_id": 21517, - "bbox": [ - 429.85, - 237.79, - 435.18, - 245.79 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p730-b13", - "global_id": 21518, - "bbox": [ - 328.03, - 334.56, - 351.55, - 342.86 - ], - "text": "3p2", - "type": "text" - }, - { - "block_id": "p730-b14", - "global_id": 21519, - "bbox": [ - 273.46, - 435.62, - 277.46, - 443.62 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p730-b15", - "global_id": 21520, - "bbox": [ - 289.74, - 476.66, - 305.29, - 484.85 - ], - "text": "vt", - "type": "text" - }, - { - "block_id": "p730-b16", - "global_id": 21521, - "bbox": [ - 213.49, - 157.0, - 224.59, - 165.08 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p730-b17", - "global_id": 21522, - "bbox": [ - 102.09, - 181.32, - 225.36, - 194.71 - ], - "text": "(a)\n0", - "type": "text" - }, - { - "block_id": "p730-b18", - "global_id": 21523, - "bbox": [ - 228.88, - 154.04, - 232.88, - 162.04 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p730-b19", - "global_id": 21524, - "bbox": [ - 101.77, - 252.95, - 111.42, - 260.95 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p730-b20", - "global_id": 21525, - "bbox": [ - 102.09, - 327.78, - 110.97, - 335.78 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p730-b21", - "global_id": 21526, - "bbox": [ - 102.09, - 430.87, - 111.42, - 438.87 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p730-b22", - "global_id": 21527, - "bbox": [ - 313.26, - 259.8, - 322.59, - 267.91 - ], - "text": "2p", - "type": "text" - }, - { - "block_id": "p730-b23", - "global_id": 21528, - "bbox": [ - 268.12, - 303.01, - 277.46, - 311.11 - ], - "text": "4p", - "type": "text" - }, - { - "block_id": "p730-b24", - "global_id": 21529, - "bbox": [ - 260.6, - 312.33, - 301.59, - 326.14 - ], - "text": "3p2\n5p2", - "type": "text" - }, - { - "block_id": "p730-b25", - "global_id": 21530, - "bbox": [ - 253.94, - 342.68, - 277.46, - 350.98 - ], - "text": "5p2", - "type": "text" - }, - { - "block_id": "p730-b26", - "global_id": 21531, - "bbox": [ - 257.94, - 444.41, - 277.46, - 452.71 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p730-b27", - "global_id": 21532, - "bbox": [ - 265.46, - 457.65, - 277.46, - 465.85 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p730-b28", - "global_id": 21533, - "bbox": [ - 284.28, - 415.47, - 297.14, - 423.77 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p730-b29", - "global_id": 21534, - "bbox": [ - 255.46, - 350.67, - 277.46, - 367.21 - ], - "text": "4p\n5.5p", - "type": "text" - }, - { - "block_id": "p730-b30", - "global_id": 21535, - "bbox": [ - 261.46, - 367.67, - 277.46, - 375.96 - ], - "text": "7p", - "type": "text" - }, - { - "block_id": "p730-b31", - "global_id": 21536, - "bbox": [ - 272.12, - 401.23, - 277.46, - 409.23 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p730-b32", - "global_id": 21537, - "bbox": [ - 316.15, - 266.55, - 319.7, - 274.55 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p730-b33", - "global_id": 21538, - "bbox": [ - 379.34, - 406.61, - 457.13, - 414.91 - ], - "text": "X(v) (principal value)", - "type": "text" - }, - { - "block_id": "p730-b34", - "global_id": 21539, - "bbox": [ - 309.8, - 301.47, - 332.5, - 309.77 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p730-b35", - "global_id": 21540, - "bbox": [ - 120.19, - 259.59, - 137.12, - 274.55 - ], - "text": "8p\n t", - "type": "text" - }, - { - "block_id": "p730-b36", - "global_id": 21541, - "bbox": [ - 157.33, - 259.58, - 174.26, - 274.55 - ], - "text": "6p\n t", - "type": "text" - }, - { - "block_id": "p730-b37", - "global_id": 21542, - "bbox": [ - 193.65, - 259.58, - 210.59, - 274.55 - ], - "text": "4p\n t", - "type": "text" - }, - { - "block_id": "p730-b38", - "global_id": 21543, - "bbox": [ - 231.7, - 259.58, - 248.63, - 274.55 - ], - "text": "2p\n t", - "type": "text" - }, - { - "block_id": "p730-b39", - "global_id": 21544, - "bbox": [ - 101.77, - 491.59, - 359.37, - 500.83 - ], - "text": "Figure 7.24 A time-shifted rectangular pulse and its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p730-b40", - "global_id": 21545, - "bbox": [ - 107.82, - 571.82, - 412.34, - 583.77 - ], - "text": "DRILL 7.5\nFourier Transform Time-Shifting Property", - "type": "text" - }, - { - "block_id": "p730-b41", - "global_id": 21546, - "bbox": [ - 107.82, - 592.89, - 484.41, - 626.77 - ], - "text": "Use pair 18 of Table 7.1 and the time-shifting property to show that the Fourier transform of\nsinc[ω0(t −T)] is (π/ω0)rect (ω/2ω0)e−jωT. Sketch the amplitude and phase spectra of the\nFourier transform.", - "type": "text" - } - ] - }, - { - "page_num": 731, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p731-b0", - "global_id": 21547, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n711", - "type": "text" - }, - { - "block_id": "p731-b1", - "global_id": 21548, - "bbox": [ - 127.89, - 86.19, - 333.49, - 98.32 - ], - "text": "THE FREQUENCY-SHIFTING PROPERTY", - "type": "text" - }, - { - "block_id": "p731-b2", - "global_id": 21549, - "bbox": [ - 127.59, - 102.35, - 134.23, - 112.31 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p731-b3", - "global_id": 21550, - "bbox": [ - 292.49, - 120.09, - 351.22, - 130.36 - ], - "text": "x(t) ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p731-b4", - "global_id": 21551, - "bbox": [ - 127.59, - 143.56, - 144.74, - 153.52 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p731-b5", - "global_id": 21552, - "bbox": [ - 273.31, - 157.19, - 516.13, - 172.45 - ], - "text": "x(t)ejω0t ⇐⇒X(ω −ω0)\n(7.30)", - "type": "text" - }, - { - "block_id": "p731-b6", - "global_id": 21553, - "bbox": [ - 127.59, - 185.7, - 213.66, - 195.74 - ], - "text": "Proof. By definition,", - "type": "text" - }, - { - "block_id": "p731-b7", - "global_id": 21554, - "bbox": [ - 180.08, - 215.01, - 236.5, - 227.01 - ], - "text": "F[x(t)ejω0t] =", - "type": "text" - }, - { - "block_id": "p731-b8", - "global_id": 21555, - "bbox": [ - 238.55, - 203.18, - 255.48, - 215.23 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p731-b9", - "global_id": 21556, - "bbox": [ - 243.8, - 228.05, - 256.36, - 235.02 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p731-b10", - "global_id": 21557, - "bbox": [ - 257.97, - 212.93, - 327.71, - 227.01 - ], - "text": "x(t)ejω0te−jωt dt =", - "type": "text" - }, - { - "block_id": "p731-b11", - "global_id": 21558, - "bbox": [ - 329.76, - 203.18, - 346.7, - 215.23 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p731-b12", - "global_id": 21559, - "bbox": [ - 335.02, - 228.05, - 347.57, - 235.02 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p731-b13", - "global_id": 21560, - "bbox": [ - 349.18, - 215.01, - 463.64, - 227.88 - ], - "text": "x(t)e−j(ω−ω0)tdt = X(ω −ω0)", - "type": "text" - }, - { - "block_id": "p731-b14", - "global_id": 21561, - "bbox": [ - 127.59, - 246.39, - 516.13, - 283.88 - ], - "text": "According to this property, the multiplication of a signal by a factor ejω0t shifts the spectrum\nof that signal by ω = ω0. Note the duality between the time-shifting and the frequency-shifting\nproperties.", - "type": "text" - }, - { - "block_id": "p731-b15", - "global_id": 21562, - "bbox": [ - 145.52, - 285.46, - 307.42, - 297.33 - ], - "text": "Changing ω0 to −ω0 in Eq. (7.30) yields", - "type": "text" - }, - { - "block_id": "p731-b16", - "global_id": 21563, - "bbox": [ - 270.59, - 307.4, - 516.13, - 322.66 - ], - "text": "x(t)e−jω0t ⇐⇒X(ω + ω0)\n(7.31)", - "type": "text" - }, - { - "block_id": "p731-b17", - "global_id": 21564, - "bbox": [ - 127.59, - 334.36, - 516.14, - 359.89 - ], - "text": "Because ejω0t is not a real function that can be generated, frequency shifting in practice is\nachieved by multiplying x(t) by a sinusoid. Observe that", - "type": "text" - }, - { - "block_id": "p731-b18", - "global_id": 21565, - "bbox": [ - 247.26, - 374.22, - 308.69, - 386.71 - ], - "text": "x(t)cos ω0t = 1", - "type": "text" - }, - { - "block_id": "p731-b19", - "global_id": 21566, - "bbox": [ - 305.21, - 371.95, - 396.46, - 388.75 - ], - "text": "2[x(t)ejω0t + x(t)e−jω0t]", - "type": "text" - }, - { - "block_id": "p731-b20", - "global_id": 21567, - "bbox": [ - 127.59, - 402.03, - 299.19, - 411.99 - ], - "text": "From Eqs. (7.30) and (7.31), it follows that", - "type": "text" - }, - { - "block_id": "p731-b21", - "global_id": 21568, - "bbox": [ - 234.05, - 426.32, - 306.84, - 438.81 - ], - "text": "x(t)cos ω0t ⇐⇒1", - "type": "text" - }, - { - "block_id": "p731-b22", - "global_id": 21569, - "bbox": [ - 303.35, - 427.66, - 516.12, - 440.85 - ], - "text": "2[X(ω −ω0) + X(ω + ω0)]\n(7.32)", - "type": "text" - }, - { - "block_id": "p731-b23", - "global_id": 21570, - "bbox": [ - 127.59, - 453.72, - 516.13, - 476.82 - ], - "text": "This result shows that the multiplication of a signal x(t) by a sinusoid of frequency ω0 shifts the\nspectrum X(ω) by ±ω0, as depicted in Fig. 7.25.", - "type": "text" - }, - { - "block_id": "p731-b24", - "global_id": 21571, - "bbox": [ - 127.59, - 477.63, - 516.13, - 523.87 - ], - "text": "Multiplication of a sinusoid cosω0t by x(t) amounts to modulating the sinusoid amplitude.\nThis type of modulation is known as amplitude modulation. The sinusoid cosω0t is called the\ncarrier, the signal x(t) is the modulating signal, and the signal x(t)cosω0t is the modulated signal.\nFurther discussion of modulation and demodulation appears in Sec. 7.7.", - "type": "text" - }, - { - "block_id": "p731-b25", - "global_id": 21572, - "bbox": [ - 145.52, - 525.45, - 331.58, - 536.6 - ], - "text": "To sketch a signal x(t)cos ω0t, we observe that", - "type": "text" - }, - { - "block_id": "p731-b26", - "global_id": 21573, - "bbox": [ - 224.87, - 557.15, - 279.59, - 568.29 - ], - "text": "x(t)cos ω0t =", - "type": "text" - }, - { - "block_id": "p731-b27", - "global_id": 21574, - "bbox": [ - 281.64, - 543.15, - 412.66, - 574.17 - ], - "text": "x(t)\nwhen cos ω0t = 1\n−x(t)\nwhen cos ω0t = −1", - "type": "text" - }, - { - "block_id": "p731-b28", - "global_id": 21575, - "bbox": [ - 127.59, - 588.55, - 516.13, - 635.56 - ], - "text": "Therefore, x(t)cos ω0t touches x(t) when the sinusoid cos ω0t is at its positive peaks and touches\n−x(t) when cos ω0t is at its negative peaks. This means that x(t) and −x(t) act as envelopes for the\nsignal x(t)cos ω0t (see Fig. 7.25). The signal −x(t) is a mirror image of x(t) about the horizontal\naxis. Figure 7.25 shows the signals x(t) and x(t)cos ω0t and their spectra.", - "type": "text" - } - ] - }, - { - "page_num": 732, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p732-b0", - "global_id": 21576, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "712\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p732-b1", - "global_id": 21577, - "bbox": [ - 107.22, - 128.1, - 118.33, - 136.18 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p732-b2", - "global_id": 21578, - "bbox": [ - 101.84, - 304.71, - 119.61, - 313.01 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p732-b3", - "global_id": 21579, - "bbox": [ - 106.35, - 220.74, - 117.46, - 228.83 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p732-b4", - "global_id": 21580, - "bbox": [ - 355.96, - 161.88, - 441.07, - 170.54 - ], - "text": "0\nW\nW\nv", - "type": "text" - }, - { - "block_id": "p732-b5", - "global_id": 21581, - "bbox": [ - 247.93, - 266.25, - 474.66, - 277.27 - ], - "text": "0\nv\nt", - "type": "text" - }, - { - "block_id": "p732-b6", - "global_id": 21582, - "bbox": [ - 247.93, - 165.09, - 250.15, - 173.09 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p732-b7", - "global_id": 21583, - "bbox": [ - 336.72, - 266.06, - 453.35, - 275.88 - ], - "text": "v0\nv0", - "type": "text" - }, - { - "block_id": "p732-b8", - "global_id": 21584, - "bbox": [ - 339.04, - 281.86, - 349.7, - 289.94 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p732-b9", - "global_id": 21585, - "bbox": [ - 373.33, - 97.52, - 388.88, - 105.62 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p732-b10", - "global_id": 21586, - "bbox": [ - 399.65, - 231.9, - 404.54, - 239.9 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p732-b11", - "global_id": 21587, - "bbox": [ - 402.67, - 97.54, - 411.56, - 105.62 - ], - "text": "2A", - "type": "text" - }, - { - "block_id": "p732-b12", - "global_id": 21588, - "bbox": [ - 198.17, - 224.69, - 235.34, - 234.32 - ], - "text": "x(t) cos v0t", - "type": "text" - }, - { - "block_id": "p732-b13", - "global_id": 21589, - "bbox": [ - 180.73, - 192.12, - 401.07, - 200.12 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p732-b14", - "global_id": 21590, - "bbox": [ - 180.73, - 322.74, - 400.91, - 330.74 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p732-b15", - "global_id": 21591, - "bbox": [ - 101.84, - 337.44, - 360.62, - 346.68 - ], - "text": "Figure 7.25 Amplitude modulation of a signal causes spectral shifting.", - "type": "text" - }, - { - "block_id": "p732-b16", - "global_id": 21592, - "bbox": [ - 76.77, - 426.41, - 421.33, - 438.37 - ], - "text": "EXAMPLE 7.15\nSpectral Shifting by Amplitude Modulation", - "type": "text" - }, - { - "block_id": "p732-b17", - "global_id": 21593, - "bbox": [ - 103.16, - 454.62, - 477.02, - 476.95 - ], - "text": "Find and sketch the Fourier transform of the modulated signal x(t)cos10t in which x(t) is a\ngate pulse rect(t/4), as illustrated in Fig. 7.26a.", - "type": "text" - }, - { - "block_id": "p732-b18", - "global_id": 21594, - "bbox": [ - 103.16, - 499.45, - 477.0, - 521.79 - ], - "text": "From pair 17 of Table 7.1, we find rect(t/4) ⇐⇒4 sinc(2ω), which is depicted in Fig. 7.26b.\nFrom Eq. (7.32) it follows that", - "type": "text" - }, - { - "block_id": "p732-b19", - "global_id": 21595, - "bbox": [ - 203.66, - 531.98, - 274.8, - 543.7 - ], - "text": "x(t)cos10t ⇐⇒1", - "type": "text" - }, - { - "block_id": "p732-b20", - "global_id": 21596, - "bbox": [ - 271.31, - 533.33, - 376.51, - 546.52 - ], - "text": "2[X(ω + 10) + X(ω −10)]", - "type": "text" - }, - { - "block_id": "p732-b21", - "global_id": 21597, - "bbox": [ - 103.16, - 555.24, - 276.01, - 565.62 - ], - "text": "In this case, X(ω) = 4 sinc(2ω). Therefore,", - "type": "text" - }, - { - "block_id": "p732-b22", - "global_id": 21598, - "bbox": [ - 179.64, - 577.17, - 400.53, - 587.54 - ], - "text": "x(t)cos 10t ⇐⇒2 sinc[2(ω + 10)] + 2 sinc[2(ω −10)]", - "type": "text" - }, - { - "block_id": "p732-b23", - "global_id": 21599, - "bbox": [ - 103.17, - 599.08, - 477.02, - 621.41 - ], - "text": "The spectrum (Fig. 7.26c) of x(t)cos 10t is obtained by shifting X(ω) in Fig. 7.26b to the\nleft by 10 and also to the right by 10, and then multiplying it by 0.5, as depicted in Fig. 7.26d.", - "type": "text" - } - ] - }, - { - "page_num": 733, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p733-b0", - "global_id": 21600, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n713", - "type": "text" - }, - { - "block_id": "p733-b1", - "global_id": 21601, - "bbox": [ - 417.6, - 157.15, - 422.93, - 165.15 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p733-b2", - "global_id": 21602, - "bbox": [ - 125.58, - 253.12, - 483.98, - 264.2 - ], - "text": "v\n10\n0\n10\n2\n2", - "type": "text" - }, - { - "block_id": "p733-b3", - "global_id": 21603, - "bbox": [ - 125.09, - 158.83, - 216.92, - 167.13 - ], - "text": "2\n2", - "type": "text" - }, - { - "block_id": "p733-b4", - "global_id": 21604, - "bbox": [ - 176.62, - 110.9, - 180.62, - 118.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p733-b5", - "global_id": 21605, - "bbox": [ - 169.18, - 185.37, - 178.06, - 193.37 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p733-b6", - "global_id": 21606, - "bbox": [ - 169.08, - 289.11, - 177.96, - 297.11 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p733-b7", - "global_id": 21607, - "bbox": [ - 385.33, - 226.25, - 389.33, - 234.25 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p733-b8", - "global_id": 21608, - "bbox": [ - 152.82, - 102.39, - 163.92, - 110.48 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p733-b9", - "global_id": 21609, - "bbox": [ - 191.6, - 206.15, - 228.07, - 214.23 - ], - "text": "x(t) cos 10t", - "type": "text" - }, - { - "block_id": "p733-b10", - "global_id": 21610, - "bbox": [ - 377.45, - 185.37, - 387.09, - 193.37 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p733-b11", - "global_id": 21611, - "bbox": [ - 377.61, - 289.11, - 386.93, - 297.11 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p733-b12", - "global_id": 21612, - "bbox": [ - 386.44, - 105.92, - 390.44, - 113.92 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p733-b13", - "global_id": 21613, - "bbox": [ - 471.64, - 268.83, - 476.97, - 276.83 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p733-b14", - "global_id": 21614, - "bbox": [ - 230.93, - 159.05, - 233.15, - 167.05 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p733-b15", - "global_id": 21615, - "bbox": [ - 233.05, - 242.97, - 235.27, - 250.97 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p733-b16", - "global_id": 21616, - "bbox": [ - 361.91, - 106.99, - 377.46, - 115.09 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p733-b17", - "global_id": 21617, - "bbox": [ - 399.93, - 163.27, - 405.27, - 171.27 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p733-b18", - "global_id": 21618, - "bbox": [ - 358.15, - 163.02, - 404.6, - 179.92 - ], - "text": "2\np", - "type": "text" - }, - { - "block_id": "p733-b19", - "global_id": 21619, - "bbox": [ - 350.57, - 167.15, - 362.82, - 179.94 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p733-b20", - "global_id": 21620, - "bbox": [ - 111.22, - 303.81, - 368.85, - 313.04 - ], - "text": "Figure 7.26 An example of spectral shifting by amplitude modulation.", - "type": "text" - }, - { - "block_id": "p733-b21", - "global_id": 21621, - "bbox": [ - 133.57, - 372.94, - 503.77, - 384.9 - ], - "text": "DRILL 7.6\nFourier Transform of an Amplitude-Modulated Signal", - "type": "text" - }, - { - "block_id": "p733-b22", - "global_id": 21622, - "bbox": [ - 133.57, - 390.4, - 497.09, - 419.8 - ], - "text": "Sketch signal e−|t| cos10t. Find the Fourier transform of this signal and sketch its spectrum.\nAnswer: X(ω) =\n1\n(ω−10)2+1 +\n1\n(ω+10)2+1. See Fig. 7.21b for the spectrum of e−a|t|.", - "type": "text" - }, - { - "block_id": "p733-b23", - "global_id": 21623, - "bbox": [ - 133.57, - 467.18, - 501.76, - 479.13 - ], - "text": "DRILL 7.7\nAmplitude Modulation Using a Phase-Shifted Carrier", - "type": "text" - }, - { - "block_id": "p733-b24", - "global_id": 21624, - "bbox": [ - 133.57, - 488.25, - 173.44, - 498.21 - ], - "text": "Show that", - "type": "text" - }, - { - "block_id": "p733-b25", - "global_id": 21625, - "bbox": [ - 208.81, - 498.45, - 304.18, - 510.94 - ], - "text": "x(t)cos(ω0t + θ) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p733-b26", - "global_id": 21626, - "bbox": [ - 300.69, - 491.78, - 309.3, - 512.98 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p733-b27", - "global_id": 21627, - "bbox": [ - 309.3, - 491.78, - 434.91, - 510.94 - ], - "text": "X(ω −ω0)ejθ + X(ω + ω0)e−jθ", - "type": "text" - }, - { - "block_id": "p733-b28", - "global_id": 21628, - "bbox": [ - 127.89, - 542.92, - 304.15, - 555.04 - ], - "text": "APPLICATIONS OF MODULATION", - "type": "text" - }, - { - "block_id": "p733-b29", - "global_id": 21629, - "bbox": [ - 127.59, - 559.07, - 516.14, - 580.99 - ], - "text": "Modulation is used to shift signal spectra. Some of the situations that call for spectrum shifting are\npresented next.", - "type": "text" - }, - { - "block_id": "p733-b30", - "global_id": 21630, - "bbox": [ - 144.52, - 588.96, - 516.15, - 634.79 - ], - "text": "1. If several signals, all occupying the same frequency band, are transmitted simultaneously\nover the same transmission medium, they will all interfere; it will be impossible to separate\nor retrieve them at a receiver. For example, if all radio stations decide to broadcast audio\nsignals simultaneously, a receiver will not be able to separate them. This problem is solved", - "type": "text" - } - ] - }, - { - "page_num": 734, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p734-b0", - "global_id": 21631, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "714\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p734-b1", - "global_id": 21632, - "bbox": [ - 210.14, - 308.99, - 382.09, - 318.95 - ], - "text": "Old is gold, but sometimes it is fool’s gold.", - "type": "text" - }, - { - "block_id": "p734-b2", - "global_id": 21633, - "bbox": [ - 132.63, - 338.14, - 490.41, - 431.78 - ], - "text": "by using modulation, whereby each radio station is assigned a distinct carrier frequency.\nEach station transmits a modulated signal. This procedure shifts the signal spectrum to its\nallocated band, which is not occupied by any other station. A radio receiver can pick up\nany station by tuning to the band of the desired station. The receiver must now demodulate\nthe received signal (undo the effect of modulation). Demodulation therefore consists of\nanother spectral shift required to restore the signal to its original band. Note that both\nmodulation and demodulation implement spectral shifting; consequently, demodulation\noperation is similar to modulation (see Sec. 7.7).", - "type": "text" - }, - { - "block_id": "p734-b3", - "global_id": 21634, - "bbox": [ - 118.78, - 433.78, - 490.41, - 515.48 - ], - "text": "This method of transmitting several signals simultaneously over a channel by sharing\nits frequency band is known as frequency-division multiplexing (FDM).\n2. For effective radiation of power over a radio link, the antenna size must be of the\norder of the wavelength of the signal to be radiated. Audio signal frequencies are so\nlow (wavelengths are so large) that impracticably large antennas would be required for\nradiation. Here, shifting the spectrum to a higher frequency (a smaller wavelength) by\nmodulation solves the problem.", - "type": "text" - }, - { - "block_id": "p734-b4", - "global_id": 21635, - "bbox": [ - 102.14, - 538.32, - 184.26, - 550.44 - ], - "text": "CONVOLUTION", - "type": "text" - }, - { - "block_id": "p734-b5", - "global_id": 21636, - "bbox": [ - 101.84, - 554.47, - 467.06, - 564.44 - ], - "text": "The time-convolution property and its dual, the frequency-convolution property, state that if", - "type": "text" - }, - { - "block_id": "p734-b6", - "global_id": 21637, - "bbox": [ - 202.76, - 580.24, - 389.47, - 591.39 - ], - "text": "x1(t) ⇐⇒X1(ω)\nand\nx2(t) ⇐⇒X2(ω)", - "type": "text" - }, - { - "block_id": "p734-b7", - "global_id": 21638, - "bbox": [ - 101.84, - 606.84, - 118.99, - 616.81 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p734-b8", - "global_id": 21639, - "bbox": [ - 195.89, - 624.78, - 490.38, - 635.93 - ], - "text": "x1(t)∗x2(t) ⇐⇒X1(ω)X2(ω)\n(time convolution)\n(7.33)", - "type": "text" - } - ] - }, - { - "page_num": 735, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p735-b0", - "global_id": 21640, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n715", - "type": "text" - }, - { - "block_id": "p735-b1", - "global_id": 21641, - "bbox": [ - 127.59, - 85.82, - 141.97, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p735-b2", - "global_id": 21642, - "bbox": [ - 203.47, - 93.7, - 273.93, - 111.42 - ], - "text": "x1(t)x2(t) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p735-b3", - "global_id": 21643, - "bbox": [ - 265.46, - 100.27, - 516.12, - 117.72 - ], - "text": "2π X1(ω)∗X2(ω)\n(frequency convolution)\n(7.34)", - "type": "text" - }, - { - "block_id": "p735-b4", - "global_id": 21644, - "bbox": [ - 127.59, - 123.04, - 213.66, - 133.08 - ], - "text": "Proof. By definition,", - "type": "text" - }, - { - "block_id": "p735-b5", - "global_id": 21645, - "bbox": [ - 212.86, - 148.94, - 281.51, - 160.09 - ], - "text": "F|x1(t)∗x2(t)| =", - "type": "text" - }, - { - "block_id": "p735-b6", - "global_id": 21646, - "bbox": [ - 283.55, - 135.38, - 300.5, - 147.43 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p735-b7", - "global_id": 21647, - "bbox": [ - 288.82, - 160.26, - 301.38, - 167.23 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b8", - "global_id": 21648, - "bbox": [ - 302.98, - 147.22, - 321.41, - 159.22 - ], - "text": "e−jωt", - "type": "text" - }, - { - "block_id": "p735-b9", - "global_id": 21649, - "bbox": [ - 323.15, - 134.96, - 345.53, - 147.43 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p735-b10", - "global_id": 21650, - "bbox": [ - 333.85, - 160.26, - 346.4, - 167.23 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b11", - "global_id": 21651, - "bbox": [ - 348.01, - 148.94, - 415.19, - 160.09 - ], - "text": "x1(τ)x2(t −τ)dτ", - "type": "text" - }, - { - "block_id": "p735-b12", - "global_id": 21652, - "bbox": [ - 416.39, - 134.96, - 421.81, - 144.92 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p735-b13", - "global_id": 21653, - "bbox": [ - 422.94, - 149.26, - 430.68, - 159.22 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p735-b14", - "global_id": 21654, - "bbox": [ - 273.74, - 176.88, - 281.51, - 186.84 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p735-b15", - "global_id": 21655, - "bbox": [ - 283.55, - 163.31, - 300.5, - 175.38 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p735-b16", - "global_id": 21656, - "bbox": [ - 288.82, - 188.2, - 301.38, - 195.17 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b17", - "global_id": 21657, - "bbox": [ - 302.98, - 176.88, - 324.33, - 188.03 - ], - "text": "x1(τ)", - "type": "text" - }, - { - "block_id": "p735-b18", - "global_id": 21658, - "bbox": [ - 325.44, - 162.89, - 347.81, - 175.38 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p735-b19", - "global_id": 21659, - "bbox": [ - 336.12, - 188.2, - 348.68, - 195.17 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b20", - "global_id": 21660, - "bbox": [ - 350.29, - 175.16, - 413.38, - 188.03 - ], - "text": "e−jωtx2(t −τ)dt", - "type": "text" - }, - { - "block_id": "p735-b21", - "global_id": 21661, - "bbox": [ - 413.56, - 162.89, - 418.99, - 172.85 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p735-b22", - "global_id": 21662, - "bbox": [ - 420.1, - 176.88, - 429.66, - 187.15 - ], - "text": "dτ", - "type": "text" - }, - { - "block_id": "p735-b23", - "global_id": 21663, - "bbox": [ - 127.59, - 202.86, - 516.14, - 225.96 - ], - "text": "The inner integral is the Fourier transform of x2(t −τ), given by [time-shifting property in\nEq. (7.29)] X2(ω)e−jωτ. Hence,", - "type": "text" - }, - { - "block_id": "p735-b24", - "global_id": 21664, - "bbox": [ - 152.22, - 240.72, - 221.8, - 251.87 - ], - "text": "F[x1(t)∗x2(t)] =", - "type": "text" - }, - { - "block_id": "p735-b25", - "global_id": 21665, - "bbox": [ - 223.85, - 227.17, - 240.79, - 239.23 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p735-b26", - "global_id": 21666, - "bbox": [ - 229.1, - 252.04, - 241.66, - 259.02 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b27", - "global_id": 21667, - "bbox": [ - 243.27, - 239.0, - 357.68, - 251.87 - ], - "text": "x1(τ)e−jωτX2(ω)dτ = X2(ω)", - "type": "text" - }, - { - "block_id": "p735-b28", - "global_id": 21668, - "bbox": [ - 358.78, - 227.17, - 375.72, - 239.23 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p735-b29", - "global_id": 21669, - "bbox": [ - 364.04, - 252.04, - 376.6, - 259.02 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b30", - "global_id": 21670, - "bbox": [ - 378.21, - 239.0, - 491.51, - 251.87 - ], - "text": "x1(τ)e−jωτdτ = X1(ω)X2(ω)", - "type": "text" - }, - { - "block_id": "p735-b31", - "global_id": 21671, - "bbox": [ - 145.52, - 266.71, - 446.8, - 277.08 - ], - "text": "Let H(ω) be the Fourier transform of the unit impulse response h(t), that is,", - "type": "text" - }, - { - "block_id": "p735-b32", - "global_id": 21672, - "bbox": [ - 291.67, - 287.29, - 352.05, - 297.57 - ], - "text": "h(t) ⇐⇒H(ω)", - "type": "text" - }, - { - "block_id": "p735-b33", - "global_id": 21673, - "bbox": [ - 127.6, - 307.88, - 516.14, - 330.22 - ], - "text": "Application of the time-convolution property to y(t) = x(t) ∗h(t) yields [assuming that both x(t)\nand h(t) are Fourier transformable]", - "type": "text" - }, - { - "block_id": "p735-b34", - "global_id": 21674, - "bbox": [ - 284.46, - 340.42, - 516.13, - 350.8 - ], - "text": "Y(ω) = X(ω)H(ω)\n(7.35)", - "type": "text" - }, - { - "block_id": "p735-b35", - "global_id": 21675, - "bbox": [ - 127.6, - 361.43, - 516.13, - 383.34 - ], - "text": "The frequency-convolution property of Eq. (7.34) can be proved in exactly the same way by\nreversing the roles of x(t) and X(ω).", - "type": "text" - }, - { - "block_id": "p735-b36", - "global_id": 21676, - "bbox": [ - 102.51, - 411.68, - 425.87, - 437.58 - ], - "text": "EXAMPLE 7.16\nTime-Convolution Property to Show the\nTime-Integration Property", - "type": "text" - }, - { - "block_id": "p735-b37", - "global_id": 21677, - "bbox": [ - 128.9, - 453.54, - 325.35, - 463.5 - ], - "text": "Use the time-convolution property to show that if", - "type": "text" - }, - { - "block_id": "p735-b38", - "global_id": 21678, - "bbox": [ - 286.47, - 473.63, - 345.2, - 483.9 - ], - "text": "x(t) ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p735-b39", - "global_id": 21679, - "bbox": [ - 128.91, - 494.53, - 252.23, - 508.53 - ], - "text": "then\n# t", - "type": "text" - }, - { - "block_id": "p735-b40", - "global_id": 21680, - "bbox": [ - 245.72, - 521.12, - 258.28, - 528.1 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b41", - "global_id": 21681, - "bbox": [ - 259.89, - 502.82, - 334.25, - 520.08 - ], - "text": "x(τ)dτ ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p735-b42", - "global_id": 21682, - "bbox": [ - 319.16, - 509.81, - 391.2, - 527.16 - ], - "text": "jω\n+ πX(0)δ(ω)", - "type": "text" - }, - { - "block_id": "p735-b43", - "global_id": 21683, - "bbox": [ - 128.9, - 549.36, - 162.08, - 559.32 - ], - "text": "Because", - "type": "text" - }, - { - "block_id": "p735-b44", - "global_id": 21684, - "bbox": [ - 262.75, - 564.49, - 304.32, - 574.77 - ], - "text": "u(t −τ) =", - "type": "text" - }, - { - "block_id": "p735-b45", - "global_id": 21685, - "bbox": [ - 306.37, - 550.5, - 362.55, - 580.74 - ], - "text": "1\nτ ≤t\n0\nτ > t", - "type": "text" - }, - { - "block_id": "p735-b46", - "global_id": 21686, - "bbox": [ - 128.9, - 588.41, - 184.0, - 598.37 - ], - "text": "it follows that", - "type": "text" - }, - { - "block_id": "p735-b47", - "global_id": 21687, - "bbox": [ - 220.6, - 603.69, - 268.03, - 613.96 - ], - "text": "x(t)∗u(t) =", - "type": "text" - }, - { - "block_id": "p735-b48", - "global_id": 21688, - "bbox": [ - 270.08, - 590.13, - 287.02, - 602.19 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p735-b49", - "global_id": 21689, - "bbox": [ - 275.33, - 615.0, - 287.89, - 621.98 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b50", - "global_id": 21690, - "bbox": [ - 289.5, - 603.68, - 360.33, - 613.96 - ], - "text": "x(τ)u(t −τ)dτ =", - "type": "text" - }, - { - "block_id": "p735-b51", - "global_id": 21691, - "bbox": [ - 362.37, - 590.13, - 374.13, - 602.41 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p735-b52", - "global_id": 21692, - "bbox": [ - 367.63, - 615.0, - 380.19, - 621.98 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p735-b53", - "global_id": 21693, - "bbox": [ - 381.8, - 603.68, - 409.85, - 613.96 - ], - "text": "x(τ)dτ", - "type": "text" - } - ] - }, - { - "page_num": 736, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p736-b0", - "global_id": 21694, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "716\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p736-b1", - "global_id": 21695, - "bbox": [ - 103.16, - 86.24, - 376.72, - 96.21 - ], - "text": "Now, from the time-convolution property [Eq. (7.33)], it follows that", - "type": "text" - }, - { - "block_id": "p736-b2", - "global_id": 21696, - "bbox": [ - 188.73, - 113.53, - 236.14, - 123.81 - ], - "text": "x(t)∗u(t) =", - "type": "text" - }, - { - "block_id": "p736-b3", - "global_id": 21697, - "bbox": [ - 238.19, - 99.98, - 249.96, - 112.25 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p736-b4", - "global_id": 21698, - "bbox": [ - 243.46, - 124.85, - 256.01, - 131.82 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p736-b5", - "global_id": 21699, - "bbox": [ - 257.62, - 113.53, - 330.78, - 123.81 - ], - "text": "x(τ)dτ ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p736-b6", - "global_id": 21700, - "bbox": [ - 331.89, - 99.54, - 345.75, - 116.92 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p736-b7", - "global_id": 21701, - "bbox": [ - 338.51, - 113.53, - 386.02, - 130.88 - ], - "text": "jω + πδ(ω)", - "type": "text" - }, - { - "block_id": "p736-b8", - "global_id": 21702, - "bbox": [ - 386.02, - 99.54, - 391.45, - 109.51 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p736-b9", - "global_id": 21703, - "bbox": [ - 228.37, - 133.87, - 260.07, - 151.13 - ], - "text": "= X(ω)", - "type": "text" - }, - { - "block_id": "p736-b10", - "global_id": 21704, - "bbox": [ - 244.99, - 140.86, - 317.02, - 158.2 - ], - "text": "jω\n+ πX(0)δ(ω)", - "type": "text" - }, - { - "block_id": "p736-b11", - "global_id": 21705, - "bbox": [ - 103.16, - 184.65, - 284.98, - 194.61 - ], - "text": "In deriving the last result, we used Eq. (1.10).", - "type": "text" - }, - { - "block_id": "p736-b12", - "global_id": 21706, - "bbox": [ - 107.82, - 261.3, - 436.25, - 273.25 - ], - "text": "DRILL 7.8\nFourier Transform Time-Convolution Property", - "type": "text" - }, - { - "block_id": "p736-b13", - "global_id": 21707, - "bbox": [ - 107.82, - 282.37, - 298.46, - 292.34 - ], - "text": "Use the time-convolution property to show that:", - "type": "text" - }, - { - "block_id": "p736-b14", - "global_id": 21708, - "bbox": [ - 125.76, - 299.89, - 206.5, - 310.18 - ], - "text": "(a) x(t)∗δ(t) = x(t)", - "type": "text" - }, - { - "block_id": "p736-b15", - "global_id": 21709, - "bbox": [ - 125.76, - 311.22, - 304.23, - 327.96 - ], - "text": "(b) e−atu(t)∗e−btu(t) =\n1\nb−a[e−at −e−bt]u(t)", - "type": "text" - }, - { - "block_id": "p736-b16", - "global_id": 21710, - "bbox": [ - 102.14, - 358.76, - 366.64, - 370.88 - ], - "text": "TIME DIFFERENTIATION AND TIME INTEGRATION", - "type": "text" - }, - { - "block_id": "p736-b17", - "global_id": 21711, - "bbox": [ - 101.84, - 374.91, - 108.48, - 384.88 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p736-b18", - "global_id": 21712, - "bbox": [ - 266.75, - 388.7, - 325.48, - 398.98 - ], - "text": "x(t) ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p736-b19", - "global_id": 21713, - "bbox": [ - 101.84, - 408.95, - 121.98, - 419.5 - ], - "text": "then†", - "type": "text" - }, - { - "block_id": "p736-b20", - "global_id": 21714, - "bbox": [ - 212.74, - 419.31, - 232.55, - 429.59 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p736-b21", - "global_id": 21715, - "bbox": [ - 218.68, - 426.29, - 490.37, - 443.64 - ], - "text": "dt\n⇐⇒jωX(ω)\n(time differentiation)\n(7.36)", - "type": "text" - }, - { - "block_id": "p736-b22", - "global_id": 21716, - "bbox": [ - 101.84, - 450.16, - 192.59, - 466.4 - ], - "text": "and\n# t", - "type": "text" - }, - { - "block_id": "p736-b23", - "global_id": 21717, - "bbox": [ - 186.09, - 479.0, - 198.65, - 485.97 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p736-b24", - "global_id": 21718, - "bbox": [ - 200.26, - 460.69, - 274.62, - 477.95 - ], - "text": "x(τ)dτ ⇐⇒X(ω)", - "type": "text" - }, - { - "block_id": "p736-b25", - "global_id": 21719, - "bbox": [ - 259.53, - 467.68, - 490.37, - 485.03 - ], - "text": "jω\n+ πX(0)δ(ω)\n(time integration)\n(7.37)", - "type": "text" - }, - { - "block_id": "p736-b26", - "global_id": 21720, - "bbox": [ - 101.84, - 494.9, - 324.83, - 504.94 - ], - "text": "Proof. Differentiation of both sides of Eq. (7.10) yields", - "type": "text" - }, - { - "block_id": "p736-b27", - "global_id": 21721, - "bbox": [ - 234.39, - 516.32, - 254.19, - 526.6 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p736-b28", - "global_id": 21722, - "bbox": [ - 240.34, - 516.74, - 276.93, - 540.66 - ], - "text": "dt\n= 1", - "type": "text" - }, - { - "block_id": "p736-b29", - "global_id": 21723, - "bbox": [ - 268.46, - 530.38, - 279.42, - 540.76 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p736-b30", - "global_id": 21724, - "bbox": [ - 282.73, - 509.74, - 299.67, - 521.8 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p736-b31", - "global_id": 21725, - "bbox": [ - 287.98, - 534.62, - 300.54, - 541.6 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p736-b32", - "global_id": 21726, - "bbox": [ - 302.15, - 519.51, - 358.84, - 533.58 - ], - "text": "jωX(ω)ejωt dω", - "type": "text" - }, - { - "block_id": "p736-b33", - "global_id": 21727, - "bbox": [ - 101.84, - 560.62, - 490.38, - 589.91 - ], - "text": "† Valid only if the transform of dx/dt exists. In other words, dx/dt must satisfy the Dirichlet conditions. The\nfirst Dirichlet condition implies\n# ∞", - "type": "text" - }, - { - "block_id": "p736-b34", - "global_id": 21728, - "bbox": [ - 264.39, - 601.18, - 276.05, - 607.66 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p736-b36", - "global_id": 21729, - "bbox": [ - 281.66, - 584.86, - 299.5, - 594.11 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p736-b37", - "global_id": 21730, - "bbox": [ - 287.01, - 597.79, - 293.98, - 606.76 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p736-b38", - "global_id": 21731, - "bbox": [ - 300.69, - 578.15, - 332.57, - 603.53 - ], - "text": "dt < ∞", - "type": "text" - }, - { - "block_id": "p736-b39", - "global_id": 21732, - "bbox": [ - 101.84, - 613.11, - 490.39, - 633.41 - ], - "text": "We also require that x(t) →0 as t →±∞. Otherwise, x(t) has a dc component, which gets lost in\ndifferentiation, and there is no one-to-one relationship between x(t) and dx/dt.", - "type": "text" - } - ] - }, - { - "page_num": 737, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p737-b0", - "global_id": 21733, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n717", - "type": "text" - }, - { - "block_id": "p737-b1", - "global_id": 21734, - "bbox": [ - 184.37, - 86.07, - 342.24, - 95.3 - ], - "text": "TABLE 7.2\nFourier Transform Properties", - "type": "text" - }, - { - "block_id": "p737-b2", - "global_id": 21735, - "bbox": [ - 184.37, - 106.02, - 407.18, - 115.28 - ], - "text": "Operation\nx(t)\nX(ω)", - "type": "text" - }, - { - "block_id": "p737-b3", - "global_id": 21736, - "bbox": [ - 184.37, - 124.35, - 410.76, - 133.69 - ], - "text": "Scalar multiplication\nkx(t)\nkX(ω)", - "type": "text" - }, - { - "block_id": "p737-b4", - "global_id": 21737, - "bbox": [ - 184.37, - 143.28, - 441.86, - 153.36 - ], - "text": "Addition\nx1(t) + x2(t)\nX1(ω) + X2(ω)", - "type": "text" - }, - { - "block_id": "p737-b5", - "global_id": 21738, - "bbox": [ - 184.37, - 160.95, - 417.64, - 171.54 - ], - "text": "Conjugation\nx∗(t)\nX∗(−ω)", - "type": "text" - }, - { - "block_id": "p737-b6", - "global_id": 21739, - "bbox": [ - 184.37, - 181.14, - 422.66, - 190.48 - ], - "text": "Duality\nX(t)\n2πx(−ω)", - "type": "text" - }, - { - "block_id": "p737-b7", - "global_id": 21740, - "bbox": [ - 184.37, - 198.87, - 405.64, - 220.4 - ], - "text": "Scaling (a real)\nx(at)\n1\n|a|X", - "type": "text" - }, - { - "block_id": "p737-b8", - "global_id": 21741, - "bbox": [ - 407.04, - 194.88, - 418.98, - 207.47 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p737-b9", - "global_id": 21742, - "bbox": [ - 413.89, - 211.44, - 418.38, - 220.4 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p737-b11", - "global_id": 21743, - "bbox": [ - 184.37, - 222.45, - 426.51, - 233.79 - ], - "text": "Time shifting\nx(t −t0)\nX(ω)e−jωt0", - "type": "text" - }, - { - "block_id": "p737-b12", - "global_id": 21744, - "bbox": [ - 184.37, - 241.38, - 426.16, - 253.33 - ], - "text": "Frequency shifting (ω0 real)\nx(t)ejω0t\nX(ω −ω0)", - "type": "text" - }, - { - "block_id": "p737-b13", - "global_id": 21745, - "bbox": [ - 184.37, - 261.57, - 432.08, - 271.65 - ], - "text": "Time convolution\nx1(t)∗x2(t)\nX1(ω)X2(ω)", - "type": "text" - }, - { - "block_id": "p737-b14", - "global_id": 21746, - "bbox": [ - 184.37, - 279.31, - 451.89, - 300.93 - ], - "text": "Frequency convolution\nx1(t)x2(t)\n1\n2π X1(ω)∗X2(ω)", - "type": "text" - }, - { - "block_id": "p737-b15", - "global_id": 21747, - "bbox": [ - 184.37, - 301.58, - 337.75, - 318.26 - ], - "text": "Time differentiation\ndnx(t)", - "type": "text" - }, - { - "block_id": "p737-b16", - "global_id": 21748, - "bbox": [ - 321.41, - 307.87, - 426.75, - 324.54 - ], - "text": "dtn\n(jω)nX(ω)", - "type": "text" - }, - { - "block_id": "p737-b17", - "global_id": 21749, - "bbox": [ - 184.37, - 333.72, - 243.94, - 342.69 - ], - "text": "Time integration", - "type": "text" - }, - { - "block_id": "p737-b18", - "global_id": 21750, - "bbox": [ - 314.74, - 321.14, - 325.39, - 332.31 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p737-b19", - "global_id": 21751, - "bbox": [ - 319.48, - 343.38, - 331.14, - 349.85 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p737-b20", - "global_id": 21752, - "bbox": [ - 332.63, - 327.06, - 407.98, - 342.6 - ], - "text": "x(u)du\nX(ω)", - "type": "text" - }, - { - "block_id": "p737-b21", - "global_id": 21753, - "bbox": [ - 394.38, - 333.35, - 459.36, - 348.96 - ], - "text": "jω\n+ πX(0)δ(ω)", - "type": "text" - }, - { - "block_id": "p737-b22", - "global_id": 21754, - "bbox": [ - 127.59, - 401.69, - 214.51, - 411.65 - ], - "text": "This result shows that", - "type": "text" - }, - { - "block_id": "p737-b23", - "global_id": 21755, - "bbox": [ - 285.25, - 450.02, - 305.06, - 460.29 - ], - "text": "dx(t)", - "type": "text" - }, - { - "block_id": "p737-b24", - "global_id": 21756, - "bbox": [ - 291.2, - 457.0, - 359.66, - 474.35 - ], - "text": "dt\n⇐⇒jωX(ω)", - "type": "text" - }, - { - "block_id": "p737-b25", - "global_id": 21757, - "bbox": [ - 127.59, - 506.57, - 301.62, - 516.54 - ], - "text": "Repeated application of this property yields", - "type": "text" - }, - { - "block_id": "p737-b26", - "global_id": 21758, - "bbox": [ - 276.87, - 552.32, - 300.93, - 563.61 - ], - "text": "dnx(t)", - "type": "text" - }, - { - "block_id": "p737-b27", - "global_id": 21759, - "bbox": [ - 282.94, - 558.8, - 368.05, - 577.66 - ], - "text": "dtn\n⇐⇒(jω)nX(ω)", - "type": "text" - }, - { - "block_id": "p737-b28", - "global_id": 21760, - "bbox": [ - 127.59, - 612.86, - 442.57, - 622.83 - ], - "text": "The time-integration property [Eq. (7.37)] has already been proved in Ex. 7.16.", - "type": "text" - }, - { - "block_id": "p737-b29", - "global_id": 21761, - "bbox": [ - 145.52, - 624.83, - 453.39, - 634.79 - ], - "text": "Table 7.2 summarizes the most important properties of the Fourier transform.", - "type": "text" - } - ] - }, - { - "page_num": 738, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p738-b0", - "global_id": 21762, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "718\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p738-b1", - "global_id": 21763, - "bbox": [ - 76.77, - 93.92, - 448.05, - 105.87 - ], - "text": "EXAMPLE 7.17\nFourier Transform Time-Differentiation Property", - "type": "text" - }, - { - "block_id": "p738-b2", - "global_id": 21764, - "bbox": [ - 103.16, - 122.12, - 477.05, - 156.41 - ], - "text": "Use the time-differentiation property to find the Fourier transform of the triangle pulse (t/τ)\nillustrated in Fig. 7.27a. Verify the correctness of the spectrum by using it to synthesize a\nperiodic replication of the original time-domain signal with τ = 1.", - "type": "text" - }, - { - "block_id": "p738-b3", - "global_id": 21765, - "bbox": [ - 236.62, - 243.43, - 238.84, - 251.43 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p738-b4", - "global_id": 21766, - "bbox": [ - 239.2, - 317.82, - 241.42, - 325.82 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p738-b5", - "global_id": 21767, - "bbox": [ - 236.62, - 405.56, - 238.84, - 413.56 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p738-b6", - "global_id": 21768, - "bbox": [ - 175.94, - 243.12, - 179.94, - 251.12 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p738-b7", - "global_id": 21769, - "bbox": [ - 160.29, - 202.75, - 227.73, - 212.54 - ], - "text": "1\nx(t) ( )", - "type": "text" - }, - { - "block_id": "p738-b8", - "global_id": 21770, - "bbox": [ - 169.62, - 254.89, - 408.6, - 271.56 - ], - "text": "sinc2( )\n(a)", - "type": "text" - }, - { - "block_id": "p738-b9", - "global_id": 21771, - "bbox": [ - 176.88, - 316.43, - 180.88, - 324.43 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p738-b10", - "global_id": 21772, - "bbox": [ - 169.24, - 351.52, - 178.89, - 359.52 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p738-b11", - "global_id": 21773, - "bbox": [ - 176.88, - 404.27, - 180.88, - 412.27 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p738-b12", - "global_id": 21774, - "bbox": [ - 169.62, - 455.99, - 178.5, - 463.99 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p738-b13", - "global_id": 21775, - "bbox": [ - 397.62, - 329.41, - 402.96, - 337.41 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p738-b14", - "global_id": 21776, - "bbox": [ - 344.3, - 314.08, - 348.3, - 322.08 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p738-b15", - "global_id": 21777, - "bbox": [ - 333.62, - 224.73, - 378.15, - 243.7 - ], - "text": "t\n2\nX(v)", - "type": "text" - }, - { - "block_id": "p738-b16", - "global_id": 21778, - "bbox": [ - 396.34, - 251.96, - 405.22, - 259.96 - ], - "text": "vt", - "type": "text" - }, - { - "block_id": "p738-b17", - "global_id": 21779, - "bbox": [ - 398.78, - 260.6, - 402.78, - 268.6 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p738-b18", - "global_id": 21780, - "bbox": [ - 341.64, - 351.52, - 350.97, - 359.52 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p738-b19", - "global_id": 21781, - "bbox": [ - 116.17, - 240.66, - 127.73, - 257.78 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p738-b20", - "global_id": 21782, - "bbox": [ - 116.17, - 315.08, - 127.73, - 332.21 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p738-b21", - "global_id": 21783, - "bbox": [ - 116.17, - 402.79, - 127.73, - 419.91 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p738-b22", - "global_id": 21784, - "bbox": [ - 177.81, - 284.56, - 181.81, - 299.2 - ], - "text": "2\nt", - "type": "text" - }, - { - "block_id": "p738-b23", - "global_id": 21785, - "bbox": [ - 230.92, - 372.47, - 234.92, - 387.11 - ], - "text": "2\nt", - "type": "text" - }, - { - "block_id": "p738-b24", - "global_id": 21786, - "bbox": [ - 223.76, - 403.07, - 227.76, - 419.78 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p738-b25", - "global_id": 21787, - "bbox": [ - 157.36, - 329.58, - 169.01, - 344.22 - ], - "text": "2\nt", - "type": "text" - }, - { - "block_id": "p738-b26", - "global_id": 21788, - "bbox": [ - 157.36, - 431.45, - 169.01, - 446.09 - ], - "text": "4\nt", - "type": "text" - }, - { - "block_id": "p738-b27", - "global_id": 21789, - "bbox": [ - 224.7, - 297.7, - 228.7, - 314.41 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p738-b28", - "global_id": 21790, - "bbox": [ - 370.04, - 251.81, - 374.04, - 268.51 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p738-b29", - "global_id": 21791, - "bbox": [ - 127.34, - 372.47, - 131.34, - 387.11 - ], - "text": "2\nt", - "type": "text" - }, - { - "block_id": "p738-b30", - "global_id": 21792, - "bbox": [ - 203.79, - 281.96, - 211.34, - 289.96 - ], - "text": "dx", - "type": "text" - }, - { - "block_id": "p738-b31", - "global_id": 21793, - "bbox": [ - 204.45, - 290.67, - 210.68, - 298.67 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p738-b32", - "global_id": 21794, - "bbox": [ - 186.12, - 373.43, - 197.15, - 382.83 - ], - "text": "d2x", - "type": "text" - }, - { - "block_id": "p738-b33", - "global_id": 21795, - "bbox": [ - 186.78, - 383.75, - 196.48, - 393.15 - ], - "text": "dt2", - "type": "text" - }, - { - "block_id": "p738-b34", - "global_id": 21796, - "bbox": [ - 368.71, - 313.84, - 378.05, - 328.59 - ], - "text": "4p\nt", - "type": "text" - }, - { - "block_id": "p738-b35", - "global_id": 21797, - "bbox": [ - 223.59, - 241.01, - 227.59, - 258.2 - ], - "text": "t\n2", - "type": "text" - }, - { - "block_id": "p738-b36", - "global_id": 21798, - "bbox": [ - 220.36, - 206.6, - 223.91, - 214.6 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p738-b37", - "global_id": 21799, - "bbox": [ - 221.03, - 199.95, - 223.25, - 207.95 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p738-b38", - "global_id": 21800, - "bbox": [ - 280.62, - 313.39, - 405.79, - 328.59 - ], - "text": "8p\nt\n4p\n t\n8p\n t", - "type": "text" - }, - { - "block_id": "p738-b39", - "global_id": 21801, - "bbox": [ - 94.2, - 470.44, - 468.05, - 492.35 - ], - "text": "Figure 7.27 Finding\nthe\nFourier\ntransform\nof\na\npiecewise-linear\nsignal\nusing\nthe\ntime-differentiation property.", - "type": "text" - }, - { - "block_id": "p738-b40", - "global_id": 21802, - "bbox": [ - 103.16, - 509.95, - 477.04, - 591.65 - ], - "text": "To find the Fourier transform of this pulse, we differentiate the pulse successively, as\nillustrated in Fig. 7.27b and 7.27c. Because dx/dt is constant everywhere, its derivative,\nd2x/dt2, is zero everywhere. But dx/dt has jump discontinuities with a positive jump of 2/τ\nat t = ±τ/2, and a negative jump of 4/τ at t = 0. Recall that the derivative of a signal at a\njump discontinuity is an impulse at that point of strength equal to the amount of jump. Hence,\nd2x/dt2, the derivative of dx/dt, consists of a sequence of impulses, as depicted in Fig. 7.27c;\nthat is,", - "type": "text" - }, - { - "block_id": "p738-b41", - "global_id": 21803, - "bbox": [ - 201.93, - 591.13, - 226.0, - 602.35 - ], - "text": "d2x(t)", - "type": "text" - }, - { - "block_id": "p738-b42", - "global_id": 21804, - "bbox": [ - 208.01, - 592.48, - 245.5, - 616.4 - ], - "text": "dt2\n= 2", - "type": "text" - }, - { - "block_id": "p738-b43", - "global_id": 21805, - "bbox": [ - 240.25, - 606.13, - 244.57, - 616.09 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p738-b45", - "global_id": 21806, - "bbox": [ - 252.39, - 599.05, - 256.68, - 609.02 - ], - "text": "δ", - "type": "text" - }, - { - "block_id": "p738-b47", - "global_id": 21807, - "bbox": [ - 263.8, - 592.07, - 283.12, - 609.33 - ], - "text": "t + τ", - "type": "text" - }, - { - "block_id": "p738-b48", - "global_id": 21808, - "bbox": [ - 279.07, - 606.54, - 284.05, - 616.5 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p738-b50", - "global_id": 21809, - "bbox": [ - 293.79, - 599.05, - 338.44, - 609.43 - ], - "text": "−2δ(t) + δ", - "type": "text" - }, - { - "block_id": "p738-b52", - "global_id": 21810, - "bbox": [ - 345.57, - 592.07, - 364.9, - 609.33 - ], - "text": "t −τ", - "type": "text" - }, - { - "block_id": "p738-b53", - "global_id": 21811, - "bbox": [ - 360.85, - 606.54, - 365.83, - 616.5 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p738-b54", - "global_id": 21812, - "bbox": [ - 367.3, - 585.07, - 379.45, - 595.03 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 739, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p739-b0", - "global_id": 21813, - "bbox": [ - 288.88, - 62.89, - 516.13, - 71.98 - ], - "text": "7.3\nSome Properties of the Fourier Transform\n719", - "type": "text" - }, - { - "block_id": "p739-b1", - "global_id": 21814, - "bbox": [ - 128.9, - 86.24, - 333.4, - 96.21 - ], - "text": "From the time-differentiation property [Eq. (7.36)],", - "type": "text" - }, - { - "block_id": "p739-b2", - "global_id": 21815, - "bbox": [ - 245.33, - 106.37, - 269.39, - 117.58 - ], - "text": "d2x(t)", - "type": "text" - }, - { - "block_id": "p739-b3", - "global_id": 21816, - "bbox": [ - 251.4, - 112.85, - 387.54, - 131.64 - ], - "text": "dt2\n⇐⇒(jω)2X(ω) = −ω2X(ω)", - "type": "text" - }, - { - "block_id": "p739-b4", - "global_id": 21817, - "bbox": [ - 128.91, - 139.63, - 328.69, - 149.59 - ], - "text": "Also, from the time-shifting property [Eq. (7.29)],", - "type": "text" - }, - { - "block_id": "p739-b5", - "global_id": 21818, - "bbox": [ - 276.6, - 159.4, - 354.06, - 172.27 - ], - "text": "δ(t −t0) ⇐⇒e−jωt0", - "type": "text" - }, - { - "block_id": "p739-b6", - "global_id": 21819, - "bbox": [ - 128.9, - 183.45, - 269.21, - 193.41 - ], - "text": "Combining these results, we obtain", - "type": "text" - }, - { - "block_id": "p739-b7", - "global_id": 21820, - "bbox": [ - 162.66, - 204.03, - 220.13, - 220.97 - ], - "text": "−ω2X(ω) = 2", - "type": "text" - }, - { - "block_id": "p739-b8", - "global_id": 21821, - "bbox": [ - 214.88, - 217.67, - 219.2, - 227.63 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p739-b10", - "global_id": 21822, - "bbox": [ - 225.51, - 202.59, - 316.38, - 220.97 - ], - "text": "ej(ωτ/2) −2 + e−j(ωτ/2)", - "type": "text" - }, - { - "block_id": "p739-b11", - "global_id": 21823, - "bbox": [ - 318.43, - 204.03, - 334.69, - 220.97 - ], - "text": "= 4", - "type": "text" - }, - { - "block_id": "p739-b12", - "global_id": 21824, - "bbox": [ - 329.44, - 217.67, - 333.76, - 227.63 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p739-b14", - "global_id": 21825, - "bbox": [ - 343.99, - 203.61, - 370.6, - 220.97 - ], - "text": "cos ωτ", - "type": "text" - }, - { - "block_id": "p739-b15", - "global_id": 21826, - "bbox": [ - 363.19, - 196.61, - 395.56, - 228.05 - ], - "text": "2 −1", - "type": "text" - }, - { - "block_id": "p739-b16", - "global_id": 21827, - "bbox": [ - 397.62, - 204.03, - 421.64, - 220.56 - ], - "text": "= −8", - "type": "text" - }, - { - "block_id": "p739-b17", - "global_id": 21828, - "bbox": [ - 416.4, - 209.16, - 439.33, - 228.05 - ], - "text": "τ sin2", - "type": "text" - }, - { - "block_id": "p739-b18", - "global_id": 21829, - "bbox": [ - 440.93, - 196.61, - 459.89, - 213.57 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p739-b19", - "global_id": 21830, - "bbox": [ - 452.48, - 218.08, - 457.46, - 228.05 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p739-b21", - "global_id": 21831, - "bbox": [ - 128.91, - 238.21, - 143.28, - 248.17 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p739-b22", - "global_id": 21832, - "bbox": [ - 192.27, - 262.38, - 259.64, - 286.4 - ], - "text": "X(ω) =\n8\nω2τ sin2", - "type": "text" - }, - { - "block_id": "p739-b23", - "global_id": 21833, - "bbox": [ - 261.25, - 254.97, - 280.21, - 271.93 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p739-b24", - "global_id": 21834, - "bbox": [ - 272.8, - 276.44, - 277.78, - 286.4 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p739-b26", - "global_id": 21835, - "bbox": [ - 291.37, - 261.97, - 306.69, - 278.91 - ], - "text": "= τ", - "type": "text" - }, - { - "block_id": "p739-b27", - "global_id": 21836, - "bbox": [ - 302.64, - 276.44, - 307.62, - 286.4 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p739-b28", - "global_id": 21837, - "bbox": [ - 310.19, - 243.01, - 317.45, - 252.98 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p739-b29", - "global_id": 21838, - "bbox": [ - 310.19, - 260.48, - 317.45, - 282.86 - ], - "text": "⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p739-b30", - "global_id": 21839, - "bbox": [ - 318.65, - 255.81, - 330.28, - 265.77 - ], - "text": "sin", - "type": "text" - }, - { - "block_id": "p739-b31", - "global_id": 21840, - "bbox": [ - 332.49, - 241.4, - 351.45, - 258.38 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p739-b32", - "global_id": 21841, - "bbox": [ - 344.04, - 262.88, - 349.02, - 272.84 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p739-b34", - "global_id": 21842, - "bbox": [ - 333.49, - 272.45, - 344.53, - 282.42 - ], - "text": "ωτ", - "type": "text" - }, - { - "block_id": "p739-b35", - "global_id": 21843, - "bbox": [ - 337.12, - 286.93, - 342.1, - 296.89 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p739-b36", - "global_id": 21844, - "bbox": [ - 361.76, - 243.01, - 369.02, - 252.98 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p739-b37", - "global_id": 21845, - "bbox": [ - 361.76, - 260.48, - 369.02, - 282.86 - ], - "text": "⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p739-b38", - "global_id": 21846, - "bbox": [ - 369.02, - 246.32, - 372.51, - 253.3 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p739-b39", - "global_id": 21847, - "bbox": [ - 375.06, - 261.97, - 390.38, - 278.91 - ], - "text": "= τ", - "type": "text" - }, - { - "block_id": "p739-b40", - "global_id": 21848, - "bbox": [ - 386.33, - 257.95, - 431.59, - 286.4 - ], - "text": "2sinc2 ωτ\n4", - "type": "text" - }, - { - "block_id": "p739-b42", - "global_id": 21849, - "bbox": [ - 128.91, - 305.19, - 502.8, - 375.34 - ], - "text": "The spectrum X(ω) is depicted in Fig. 7.27d. This procedure of finding the Fourier transform\ncan be applied to any function x(t) made up of straight-line segments with x(t) →0 as\n|t| →∞. The second derivative of such a signal yields a sequence of impulses whose Fourier\ntransform can be found by inspection. This example suggests a numerical method of finding\nthe Fourier transform of an arbitrary signal x(t) by approximating the signal by straight-line\nsegments.", - "type": "text" - }, - { - "block_id": "p739-b43", - "global_id": 21850, - "bbox": [ - 129.2, - 390.06, - 411.38, - 416.14 - ], - "text": "SYNTHESIZING A PERIODIC REPLICATION TO VERIFY\nSPECTRUM CORRECTNESS", - "type": "text" - }, - { - "block_id": "p739-b44", - "global_id": 21851, - "bbox": [ - 128.9, - 419.76, - 502.75, - 454.04 - ], - "text": "While a signal’s spectrum X(ω) provides useful insight into signal character, it can be difficult\nto look at X(ω) and know that it is correct for a particular signal x(t). Is it obvious, for example,\nthat X(ω) = τ", - "type": "text" - }, - { - "block_id": "p739-b45", - "global_id": 21852, - "bbox": [ - 128.91, - 442.25, - 502.78, - 477.95 - ], - "text": "2sinc2(ωτ/4) is really the spectrum of a τ-duration rectangle function? Or is it\npossible that a mathematical error was made in the determination of X(ω)? It is difficult to be\ncertain by simple inspection of the spectrum.", - "type": "text" - }, - { - "block_id": "p739-b46", - "global_id": 21853, - "bbox": [ - 128.91, - 479.94, - 502.77, - 561.64 - ], - "text": "The same uncertainties exist when we are looking at a periodic signal’s Fourier series\nspectrum. In the Fourier series case, we can verify the correctness of a signal’s spectrum\nby synthesizing x(t) with a truncated Fourier series; the synthesized signal will match the\noriginal only if the computed spectrum is correct. This is exactly the approach that was\ntaken in Ex. 6.11. And since a truncated Fourier series involves a simple sum, tools like\nMATLAB make waveform synthesis relatively simple, at least in the case of the Fourier\nseries.", - "type": "text" - }, - { - "block_id": "p739-b47", - "global_id": 21854, - "bbox": [ - 128.91, - 563.22, - 502.77, - 621.41 - ], - "text": "In the case of the Fourier transform, however, synthesis of x(t) using Eq. (7.10) requires\nintegration, a task not well suited to numerical packages such as MATLAB. All is not lost,\nhowever. Consider Eq. (7.5). By scaling and sampling the spectrum X(ω) of an aperiodic signal\nx(t), we obtain the Fourier series coefficient of a signal that is the periodic replication of x(t).\nSimilar to Ex. 6.11, we can then synthesize a periodic replication of x(t) with a truncated", - "type": "text" - } - ] - }, - { - "page_num": 740, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p740-b0", - "global_id": 21855, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "720\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p740-b1", - "global_id": 21856, - "bbox": [ - 103.16, - 86.24, - 477.02, - 108.15 - ], - "text": "Fourier series to verify spectrum correctness. Let us demonstrate the idea for the current\nexample with τ = 1.", - "type": "text" - }, - { - "block_id": "p740-b2", - "global_id": 21857, - "bbox": [ - 121.09, - 108.1, - 250.26, - 120.12 - ], - "text": "To begin, we represent X(ω) = τ", - "type": "text" - }, - { - "block_id": "p740-b3", - "global_id": 21858, - "bbox": [ - 103.16, - 108.32, - 477.01, - 144.02 - ], - "text": "2sinc2(ωτ/4) using an anonymous function in MATLAB.\nSince MATLAB computes sinc(x) as (sin(πx))/πx, we must scale the input by 1/π to match\nthe notation of sinc in this book.", - "type": "text" - }, - { - "block_id": "p740-b4", - "global_id": 21859, - "bbox": [ - 103.16, - 154.27, - 416.98, - 164.23 - ], - "text": ">>\ntau = 1; X = @(omega) tau/2*(sinc(omega*tau/(4*pi))).^2;", - "type": "text" - }, - { - "block_id": "p740-b5", - "global_id": 21860, - "bbox": [ - 103.16, - 173.5, - 477.02, - 208.49 - ], - "text": "For our periodic replication, let us pick T0 = 2, which is comfortably wide enough to\naccommodate our (τ = 1)-width function without overlap. We use Eq. (7.5) to define the\nneeded Fourier series coefficients Dn.", - "type": "text" - }, - { - "block_id": "p740-b6", - "global_id": 21861, - "bbox": [ - 103.17, - 218.04, - 385.56, - 228.0 - ], - "text": ">>\nT0 = 2; omega0 = 2*pi/T0; D = @(n) X(n*omega0)/T0;", - "type": "text" - }, - { - "block_id": "p740-b7", - "global_id": 21862, - "bbox": [ - 103.17, - 237.26, - 477.01, - 259.59 - ], - "text": "Let us use 25 harmonics to synthesize the periodic replication x25(t) of our triangular signal\nx(t). To begin waveform synthesis, we set the dc portion of the signal.", - "type": "text" - }, - { - "block_id": "p740-b8", - "global_id": 21863, - "bbox": [ - 103.17, - 269.84, - 354.22, - 279.8 - ], - "text": ">>\nt = (-T0:.001:T0); x25 = D(0)*ones(size(t));", - "type": "text" - }, - { - "block_id": "p740-b9", - "global_id": 21864, - "bbox": [ - 103.16, - 289.06, - 477.03, - 323.64 - ], - "text": "To add the desired 25 harmonics, we enter a loop for 1 ≤n ≤25 and add in the Dn and D−n\nterms. Although the result should be real, small round-off errors cause the reconstruction to be\ncomplex. These small imaginary parts are removed by using the real command.", - "type": "text" - }, - { - "block_id": "p740-b10", - "global_id": 21865, - "bbox": [ - 103.16, - 333.61, - 490.2, - 367.47 - ], - "text": ">>\nfor n = 1:25,\n>>\nx25 = x25+real(D(n)*exp(1j*omega0*n*t)+D(-n)*exp(-1j*omega0*n*t));\n>>\nend", - "type": "text" - }, - { - "block_id": "p740-b11", - "global_id": 21866, - "bbox": [ - 103.16, - 376.73, - 380.08, - 387.11 - ], - "text": "Lastly, we plot the resulting truncated Fourier series synthesis of x(t).", - "type": "text" - }, - { - "block_id": "p740-b12", - "global_id": 21867, - "bbox": [ - 103.17, - 397.36, - 385.6, - 407.33 - ], - "text": ">>\nplot(t,x25,’k’); xlabel(’t’); ylabel(’x_{25}(t)’);", - "type": "text" - }, - { - "block_id": "p740-b13", - "global_id": 21868, - "bbox": [ - 103.17, - 417.0, - 477.03, - 450.87 - ], - "text": "Since the synthesized waveform shown in Fig. 7.28 closely matches a 2-periodic replication\nof the triangle wave in Fig. 7.27a, we have high confidence that both the computed Dn and, by\nextension, the Fourier spectrum X(ω) are correct.", - "type": "text" - }, - { - "block_id": "p740-b14", - "global_id": 21869, - "bbox": [ - 135.8, - 562.57, - 477.41, - 583.66 - ], - "text": "–2\n–1.5\n–1\n–0.5\n0\n0.5\n1\n1.5\n2\nt", - "type": "text" - }, - { - "block_id": "p740-b15", - "global_id": 21870, - "bbox": [ - 132.69, - 547.35, - 136.69, - 555.35 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p740-b16", - "global_id": 21871, - "bbox": [ - 125.93, - 517.35, - 136.6, - 525.35 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p740-b17", - "global_id": 21872, - "bbox": [ - 132.69, - 487.35, - 136.69, - 495.35 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p740-b18", - "global_id": 21873, - "bbox": [ - 111.43, - 511.02, - 122.7, - 529.84 - ], - "text": "x25(t)", - "type": "text" - }, - { - "block_id": "p740-b19", - "global_id": 21874, - "bbox": [ - 110.73, - 589.98, - 435.23, - 599.59 - ], - "text": "Figure 7.28 Synthesizing a 2-periodic replication of x(t) using a truncated Fourier series.", - "type": "text" - } - ] - }, - { - "page_num": 741, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p741-b0", - "global_id": 21875, - "bbox": [ - 280.23, - 62.89, - 516.13, - 71.98 - ], - "text": "7.4\nSignal Transmission Through LTIC Systems\n721", - "type": "text" - }, - { - "block_id": "p741-b1", - "global_id": 21876, - "bbox": [ - 133.57, - 97.81, - 475.68, - 109.77 - ], - "text": "DRILL 7.9\nFourier Transform Time-Differentiation Property", - "type": "text" - }, - { - "block_id": "p741-b2", - "global_id": 21877, - "bbox": [ - 133.57, - 118.47, - 449.33, - 128.85 - ], - "text": "Use the time-differentiation property to find the Fourier transform of rect (t/τ).", - "type": "text" - }, - { - "block_id": "p741-b3", - "global_id": 21878, - "bbox": [ - 127.94, - 174.69, - 482.83, - 188.64 - ], - "text": "7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS", - "type": "text" - }, - { - "block_id": "p741-b4", - "global_id": 21879, - "bbox": [ - 127.59, - 194.21, - 516.13, - 216.54 - ], - "text": "If x(t) and y(t) are the input and output of an LTIC system with impulse response h(t), then, as\ndemonstrated in Eq. (7.35),", - "type": "text" - }, - { - "block_id": "p741-b5", - "global_id": 21880, - "bbox": [ - 284.46, - 222.56, - 359.26, - 232.84 - ], - "text": "Y(ω) = H(ω)X(ω)", - "type": "text" - }, - { - "block_id": "p741-b6", - "global_id": 21881, - "bbox": [ - 127.6, - 244.46, - 516.13, - 278.75 - ], - "text": "This equation does not apply to (asymptotically) unstable systems because h(t) for such systems\nis not Fourier transformable. It applies to BIBO-stable as well as most of the marginally stable\nsystems.† Similarly, this equation does not apply if x(t) is not Fourier transformable.", - "type": "text" - }, - { - "block_id": "p741-b7", - "global_id": 21882, - "bbox": [ - 127.59, - 280.74, - 516.15, - 350.48 - ], - "text": "In Ch. 4, we saw that the Laplace transform is more versatile and capable of analyzing all kinds\nof LTIC systems whether stable, unstable, or marginally stable. Laplace transform can also handle\nexponentially growing inputs. In comparison to the Laplace transform, the Fourier transform in\nsystem analysis is not just clumsier, but also very restrictive. Hence, the Laplace transform is\npreferable to the Fourier transform in LTIC system analysis. We shall not belabor the application\nof the Fourier transform to LTIC system analysis. We consider just one example here.", - "type": "text" - }, - { - "block_id": "p741-b8", - "global_id": 21883, - "bbox": [ - 102.51, - 380.95, - 460.7, - 406.85 - ], - "text": "EXAMPLE 7.18\nFourier Transform to Determine the Zero-State\nResponse", - "type": "text" - }, - { - "block_id": "p741-b9", - "global_id": 21884, - "bbox": [ - 128.9, - 423.52, - 502.78, - 445.44 - ], - "text": "Use the Fourier transform to find the zero-state response of a stable LTIC system with\nfrequency response", - "type": "text" - }, - { - "block_id": "p741-b10", - "global_id": 21885, - "bbox": [ - 289.36, - 445.41, - 341.1, - 469.43 - ], - "text": "H(s) =\n1\ns + 2", - "type": "text" - }, - { - "block_id": "p741-b11", - "global_id": 21886, - "bbox": [ - 128.9, - 474.3, - 502.74, - 497.86 - ], - "text": "and the input is x(t) = e−tu(t). Stability implies that the region of convergence of H(s) includes\nthe ω axis.", - "type": "text" - }, - { - "block_id": "p741-b12", - "global_id": 21887, - "bbox": [ - 128.9, - 520.78, - 176.21, - 530.74 - ], - "text": "In this case,", - "type": "text" - }, - { - "block_id": "p741-b13", - "global_id": 21888, - "bbox": [ - 285.69, - 529.99, - 344.77, - 554.01 - ], - "text": "X(ω) =\n1\njω + 1", - "type": "text" - }, - { - "block_id": "p741-b14", - "global_id": 21889, - "bbox": [ - 127.59, - 588.31, - 516.13, - 633.41 - ], - "text": "† For marginally stable systems, if the input x(t) contains a finite-amplitude sinusoid of the system’s natural\nfrequency, which leads to resonance, the output is not Fourier transformable. It does, however, apply to\nmarginally stable systems if the input does not contain a finite-amplitude sinusoid of the system’s natural\nfrequency.", - "type": "text" - } - ] - }, - { - "page_num": 742, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p742-b0", - "global_id": 21890, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "722\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p742-b1", - "global_id": 21891, - "bbox": [ - 103.16, - 85.89, - 449.37, - 96.26 - ], - "text": "Moreover, because the system is stable, the frequency response H(jω) = H(ω). Hence,", - "type": "text" - }, - { - "block_id": "p742-b2", - "global_id": 21892, - "bbox": [ - 234.9, - 106.2, - 344.07, - 130.22 - ], - "text": "H(ω) = H(s)|s=jω =\n1\njω + 2", - "type": "text" - }, - { - "block_id": "p742-b3", - "global_id": 21893, - "bbox": [ - 103.16, - 140.04, - 144.91, - 150.0 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p742-b4", - "global_id": 21894, - "bbox": [ - 211.67, - 149.25, - 367.3, - 173.28 - ], - "text": "Y(ω) = H(ω)X(ω) =\n1\n(jω + 2)(jω + 1)", - "type": "text" - }, - { - "block_id": "p742-b5", - "global_id": 21895, - "bbox": [ - 103.16, - 180.11, - 325.33, - 190.07 - ], - "text": "Expanding the right-hand side in partial fractions yields", - "type": "text" - }, - { - "block_id": "p742-b6", - "global_id": 21896, - "bbox": [ - 240.76, - 200.01, - 338.22, - 224.03 - ], - "text": "Y(ω) =\n1\njω + 1 −\n1\njω + 2", - "type": "text" - }, - { - "block_id": "p742-b7", - "global_id": 21897, - "bbox": [ - 103.16, - 233.85, - 117.54, - 243.81 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p742-b8", - "global_id": 21898, - "bbox": [ - 245.75, - 241.27, - 334.43, - 255.66 - ], - "text": "y(t) = (e−t −e−2t)u(t)", - "type": "text" - }, - { - "block_id": "p742-b9", - "global_id": 21899, - "bbox": [ - 107.82, - 309.17, - 442.83, - 335.07 - ], - "text": "DRILL 7.10\nFourier Transform to Determine the Zero-State\nResponse", - "type": "text" - }, - { - "block_id": "p742-b10", - "global_id": 21900, - "bbox": [ - 107.82, - 342.77, - 484.41, - 354.15 - ], - "text": "For the system in Ex. 7.18, show that the zero-input response to the input etu(−t) is y(t) =", - "type": "text" - }, - { - "block_id": "p742-b11", - "global_id": 21901, - "bbox": [ - 109.02, - 354.39, - 476.67, - 368.92 - ], - "text": "1\n3[etu(−t) + e−2tu(t)]. [Hint: Use pair 2 (Table 7.1) to find the Fourier transform of etu(−t).]", - "type": "text" - }, - { - "block_id": "p742-b12", - "global_id": 21902, - "bbox": [ - 101.84, - 404.67, - 490.42, - 526.42 - ], - "text": "HEURISTIC UNDERSTANDING OF LINEAR SYSTEM RESPONSE\nIn finding the linear system response to arbitrary input, the time-domain method uses convolution\nintegral and the frequency-domain method uses the Fourier integral. Despite the apparent\ndissimilarities of the two methods, their philosophies are amazingly similar. In the time-domain\ncase, we express the input x(t) as a sum of its impulse components; in the frequency-domain case,\nthe input is expressed as a sum of everlasting exponentials (or sinusoids). In the former case, the\nresponse y(t) obtained by summing the system’s responses to impulse components results in the\nconvolution integral; in the latter case, the response obtained by summing the system’s response to\neverlasting exponential components results in the Fourier integral. These ideas can be expressed\nmathematically as follows:", - "type": "text" - }, - { - "block_id": "p742-b13", - "global_id": 21903, - "bbox": [ - 118.78, - 534.39, - 236.51, - 544.36 - ], - "text": "1. For the time-domain case,", - "type": "text" - }, - { - "block_id": "p742-b14", - "global_id": 21904, - "bbox": [ - 226.4, - 555.45, - 430.99, - 577.37 - ], - "text": "δ(t) \r⇒h(t)\nshows the system response\nto δ(t) is the impulse response h(t)", - "type": "text" - }, - { - "block_id": "p742-b15", - "global_id": 21905, - "bbox": [ - 171.29, - 589.11, - 195.93, - 599.38 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p742-b16", - "global_id": 21906, - "bbox": [ - 197.98, - 581.08, - 211.6, - 593.14 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p742-b17", - "global_id": 21907, - "bbox": [ - 202.54, - 583.03, - 385.45, - 605.36 - ], - "text": "−∞x(τ)δ(t −τ)dτ\nexpresses x(t) as a sum\nof impulse components", - "type": "text" - }, - { - "block_id": "p742-b18", - "global_id": 21908, - "bbox": [ - 171.15, - 617.1, - 195.78, - 627.38 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p742-b19", - "global_id": 21909, - "bbox": [ - 197.83, - 609.08, - 211.44, - 621.13 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p742-b20", - "global_id": 21910, - "bbox": [ - 202.39, - 611.02, - 446.88, - 633.36 - ], - "text": "−∞x(τ)h(t −τ)dτ\nexpresses y(t) as a sum of responses to\nthe impulse components of input x(t)", - "type": "text" - } - ] - }, - { - "page_num": 743, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p743-b0", - "global_id": 21911, - "bbox": [ - 280.23, - 62.89, - 516.13, - 71.98 - ], - "text": "7.4\nSignal Transmission Through LTIC Systems\n723", - "type": "text" - }, - { - "block_id": "p743-b1", - "global_id": 21912, - "bbox": [ - 144.52, - 85.82, - 284.22, - 95.78 - ], - "text": "2. For the frequency-domain case,", - "type": "text" - }, - { - "block_id": "p743-b2", - "global_id": 21913, - "bbox": [ - 238.94, - 107.72, - 430.37, - 129.64 - ], - "text": "ejωt \r⇒H(ω)ejωt\nshows the system response\nto ejωt is H(ω)ejωt", - "type": "text" - }, - { - "block_id": "p743-b3", - "global_id": 21914, - "bbox": [ - 205.02, - 133.35, - 257.2, - 154.57 - ], - "text": "x(t) =\n1\n2π\n$ ∞", - "type": "text" - }, - { - "block_id": "p743-b4", - "global_id": 21915, - "bbox": [ - 248.15, - 135.3, - 477.09, - 157.64 - ], - "text": "−∞X(ω)ejωtdω\nexpresses x(t) as a sum\nof everlasting exponential components", - "type": "text" - }, - { - "block_id": "p743-b5", - "global_id": 21916, - "bbox": [ - 183.25, - 161.35, - 235.41, - 182.56 - ], - "text": "y(t) =\n1\n2π\n$ ∞", - "type": "text" - }, - { - "block_id": "p743-b6", - "global_id": 21917, - "bbox": [ - 226.35, - 163.3, - 486.27, - 185.63 - ], - "text": "−∞X(ω)H(ω)ejωtdω\nexpresses y(t) as a sum of responses to\nthe exponential components of input x(t)", - "type": "text" - }, - { - "block_id": "p743-b7", - "global_id": 21918, - "bbox": [ - 127.59, - 197.42, - 516.15, - 243.24 - ], - "text": "The frequency-domain view sees a system in terms of its frequency response (system response\nto various sinusoidal components). It views a signal as a sum of various sinusoidal components.\nTransmission of an input signal through a (linear) system is viewed as transmission of various\nsinusoidal components of the input through the system.", - "type": "text" - }, - { - "block_id": "p743-b8", - "global_id": 21919, - "bbox": [ - 127.59, - 245.23, - 516.14, - 303.02 - ], - "text": "It was not by coincidence that we used the impulse function in time-domain analysis and the\nexponential ejωt in studying the frequency domain. The two functions happen to be duals of each\nother. Thus, the Fourier transform of an impulse δ(t −τ) is e−jωτ, and the Fourier transform of\nejω0t is an impulse 2πδ(ω −ω0). This time-frequency duality is a constant theme in the Fourier\ntransform and linear systems.", - "type": "text" - }, - { - "block_id": "p743-b9", - "global_id": 21920, - "bbox": [ - 127.59, - 328.19, - 371.46, - 340.14 - ], - "text": "7.4-1 Signal Distortion During Transmission", - "type": "text" - }, - { - "block_id": "p743-b10", - "global_id": 21921, - "bbox": [ - 127.59, - 345.86, - 516.13, - 368.19 - ], - "text": "For a system with frequency response H(ω), if X(ω) and Y(ω) are the spectra of the input and the\noutput signals, respectively, then", - "type": "text" - }, - { - "block_id": "p743-b11", - "global_id": 21922, - "bbox": [ - 284.45, - 370.4, - 516.13, - 380.77 - ], - "text": "Y(ω) = X(ω)H(ω)\n(7.38)", - "type": "text" - }, - { - "block_id": "p743-b12", - "global_id": 21923, - "bbox": [ - 127.59, - 389.75, - 516.14, - 459.9 - ], - "text": "The transmission of the input signal x(t) through the system changes it into the output signal y(t).\nEquation (7.38) shows the nature of this change or modification. Here, X(ω) and Y(ω) are the\nspectra of the input and the output, respectively. Therefore, H(ω) is the spectral response of the\nsystem. The output spectrum is obtained by the input spectrum multiplied by the spectral response\nof the system. Equation (7.38), which clearly brings out the spectral shaping (or modification) of\nthe signal by the system, can be expressed in polar form as", - "type": "text" - }, - { - "block_id": "p743-b13", - "global_id": 21924, - "bbox": [ - 232.16, - 469.6, - 411.05, - 483.99 - ], - "text": "|Y(ω)|ej̸\nY(ω) = |X(ω)||H(ω)|ej[̸\nX(ω)+̸\nH(jω)]", - "type": "text" - }, - { - "block_id": "p743-b14", - "global_id": 21925, - "bbox": [ - 145.52, - 496.47, - 187.27, - 506.43 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p743-b15", - "global_id": 21926, - "bbox": [ - 196.95, - 518.38, - 446.78, - 528.76 - ], - "text": "|Y(ω)| = |X(ω)||H(ω)|\nand̸\nY(ω) ≠ X(ω) +̸ H(ω)", - "type": "text" - }, - { - "block_id": "p743-b16", - "global_id": 21927, - "bbox": [ - 127.59, - 540.72, - 516.15, - 634.79 - ], - "text": "During transmission, the input signal amplitude spectrum |X(ω)| is changed to |X(ω)||H(ω)|.\nSimilarly, the input signal phase spectrum̸\nX(ω) is changed to̸\nX(ω) +̸ H(ω). An input signal\nspectral component of frequency ω is modified in amplitude by a factor |H(ω)| and is shifted\nin phase by an angle̸\nH(ω). Clearly, |H(ω)| is the amplitude response, and̸\nH(ω) is the phase\nresponse of the system. The plots of |H(ω)| and̸\nH(ω) as functions of ω show at a glance how\nthe system modifies the amplitudes and phases of various sinusoidal inputs. This is the reason why\nH(ω) is also called the frequency response of the system. During transmission through the system,\nsome frequency components may be boosted in amplitude, while others may be attenuated. The", - "type": "text" - } - ] - }, - { - "page_num": 744, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p744-b0", - "global_id": 21928, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "724\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p744-b1", - "global_id": 21929, - "bbox": [ - 101.84, - 85.82, - 490.39, - 107.74 - ], - "text": "relative phases of the various components also change. In general, the output waveform will be\ndifferent from the input waveform.", - "type": "text" - }, - { - "block_id": "p744-b2", - "global_id": 21930, - "bbox": [ - 102.14, - 122.25, - 278.11, - 134.37 - ], - "text": "DISTORTIONLESS TRANSMISSION", - "type": "text" - }, - { - "block_id": "p744-b3", - "global_id": 21931, - "bbox": [ - 101.84, - 138.4, - 490.39, - 196.18 - ], - "text": "In several applications, such as signal amplification or message signal transmission over a\ncommunication channel, we require that the output waveform be a replica of the input waveform.\nIn such cases we need to minimize the distortion caused by the amplifier or the communication\nchannel. It is, therefore, of practical interest to determine the characteristics of a system that allows\na signal to pass without distortion (distortionless transmission).", - "type": "text" - }, - { - "block_id": "p744-b4", - "global_id": 21932, - "bbox": [ - 101.84, - 198.18, - 490.4, - 244.01 - ], - "text": "Transmission is said to be distortionless if the input and the output have identical waveshapes\nwithin a multiplicative constant. A delayed output that retains the input waveform is also\nconsidered to be distortionless. Thus, in distortionless transmission, the input x(t) and the output\ny(t) satisfy the condition", - "type": "text" - }, - { - "block_id": "p744-b5", - "global_id": 21933, - "bbox": [ - 260.86, - 245.58, - 331.36, - 256.73 - ], - "text": "y(t) = G0x(t −td)", - "type": "text" - }, - { - "block_id": "p744-b6", - "global_id": 21934, - "bbox": [ - 101.84, - 264.49, - 281.01, - 274.45 - ], - "text": "The Fourier transform of this equation yields", - "type": "text" - }, - { - "block_id": "p744-b7", - "global_id": 21935, - "bbox": [ - 252.73, - 283.39, - 338.34, - 296.27 - ], - "text": "Y(ω) = G0X(ω)e−jωtd", - "type": "text" - }, - { - "block_id": "p744-b8", - "global_id": 21936, - "bbox": [ - 101.84, - 306.58, - 116.23, - 316.54 - ], - "text": "But", - "type": "text" - }, - { - "block_id": "p744-b9", - "global_id": 21937, - "bbox": [ - 258.71, - 318.12, - 333.52, - 328.39 - ], - "text": "Y(ω) = X(ω)H(ω)", - "type": "text" - }, - { - "block_id": "p744-b10", - "global_id": 21938, - "bbox": [ - 101.85, - 337.01, - 143.59, - 346.98 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p744-b11", - "global_id": 21939, - "bbox": [ - 261.85, - 346.84, - 329.21, - 360.33 - ], - "text": "H(ω) = G0 e−jωtd", - "type": "text" - }, - { - "block_id": "p744-b12", - "global_id": 21940, - "bbox": [ - 101.84, - 367.46, - 490.37, - 389.39 - ], - "text": "This is the frequency response required of a system for distortionless transmission. From this\nequation, it follows that", - "type": "text" - }, - { - "block_id": "p744-b13", - "global_id": 21941, - "bbox": [ - 213.36, - 400.05, - 490.38, - 411.2 - ], - "text": "|H(ω)| = G0\nand̸\nH(ω) = −ωtd\n(7.39)", - "type": "text" - }, - { - "block_id": "p744-b14", - "global_id": 21942, - "bbox": [ - 101.85, - 421.09, - 490.39, - 455.37 - ], - "text": "This result shows that for distortionless transmission, the amplitude response |H(ω)| must be a\nconstant, and the phase response̸\nH(ω) must be a linear function of ω with slope −td, where td is\nthe delay of the output with respect to input (Fig. 7.29).", - "type": "text" - }, - { - "block_id": "p744-b15", - "global_id": 21943, - "bbox": [ - 102.14, - 469.89, - 409.2, - 482.01 - ], - "text": "MEASURE OF TIME-DELAY VARIATION WITH FREQUENCY", - "type": "text" - }, - { - "block_id": "p744-b16", - "global_id": 21944, - "bbox": [ - 101.84, - 485.63, - 490.39, - 509.45 - ], - "text": "The gain |H(ω)| = G0 means that every spectral component is multiplied by a constant G0. Also,\nas seen in connection with Fig. 7.22, a linear phase̸\nH(ω) = −ωtd means that every spectral", - "type": "text" - }, - { - "block_id": "p744-b17", - "global_id": 21945, - "bbox": [ - 268.61, - 583.63, - 273.94, - 591.63 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p744-b18", - "global_id": 21946, - "bbox": [ - 276.85, - 598.65, - 300.27, - 606.94 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p744-b19", - "global_id": 21947, - "bbox": [ - 184.2, - 548.84, - 225.5, - 558.47 - ], - "text": "H(v)\nG0", - "type": "text" - }, - { - "block_id": "p744-b20", - "global_id": 21948, - "bbox": [ - 188.62, - 585.93, - 192.62, - 593.93 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p744-b21", - "global_id": 21949, - "bbox": [ - 310.8, - 604.7, - 486.4, - 626.62 - ], - "text": "Figure 7.29 LTIC system frequency res-\nponse for distortionless transmission.", - "type": "text" - } - ] - }, - { - "page_num": 745, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p745-b0", - "global_id": 21950, - "bbox": [ - 280.23, - 62.89, - 516.13, - 71.98 - ], - "text": "7.4\nSignal Transmission Through LTIC Systems\n725", - "type": "text" - }, - { - "block_id": "p745-b1", - "global_id": 21951, - "bbox": [ - 127.59, - 85.72, - 516.13, - 133.14 - ], - "text": "component is delayed by td seconds. This results in the output equal to G0 times the input delayed\nby td seconds. Because each spectral component is attenuated by the same factor (G0) and delayed\nby exactly the same amount (td), the output signal is an exact replica of the input (except for\nattenuating factor G0 and delay td).", - "type": "text" - }, - { - "block_id": "p745-b2", - "global_id": 21952, - "bbox": [ - 127.59, - 133.54, - 516.14, - 191.42 - ], - "text": "For distortionless transmission, we require a linear phase characteristic. The phase is not only\na linear function of ω, it should also pass through the origin ω = 0. In practice, many systems have\na phase characteristic that may be only approximately linear. A convenient way of judging phase\nlinearity is to plot the slope of̸\nH(ω) as a function of frequency. This slope, which is constant for\nan ideal linear phase (ILP) system, is a function of ω in the general case and can be expressed as", - "type": "text" - }, - { - "block_id": "p745-b3", - "global_id": 21953, - "bbox": [ - 280.68, - 202.75, - 330.77, - 220.5 - ], - "text": "tg(ω) = −d", - "type": "text" - }, - { - "block_id": "p745-b4", - "global_id": 21954, - "bbox": [ - 322.42, - 209.42, - 516.13, - 226.76 - ], - "text": "dω̸ H(ω)\n(7.40)", - "type": "text" - }, - { - "block_id": "p745-b5", - "global_id": 21955, - "bbox": [ - 127.59, - 236.04, - 516.16, - 318.15 - ], - "text": "If tg(ω) is constant, all the components are delayed by the same time interval tg. But if the slope is\nnot constant, the time delay tg varies with frequency. This variation means that different frequency\ncomponents undergo different amounts of time delay, and consequently, the output waveform will\nnot be a replica of the input waveform. As we shall see, tg(ω) plays an important role in bandpass\nsystems and is called the group delay or envelope delay. Observe that constant td [Eq. (7.39)]\nimplies constant tg. Note that̸\nH(ω) = φ0 −ωtg also has a constant tg. Thus, constant group delay\nis a more relaxed condition.", - "type": "text" - }, - { - "block_id": "p745-b6", - "global_id": 21956, - "bbox": [ - 127.59, - 319.74, - 516.13, - 355.51 - ], - "text": "It is often thought (erroneously) that flatness of amplitude response |H(ω)| alone can\nguarantee signal quality. However, a system that has a flat amplitude response may yet distort\na signal beyond recognition if the phase response is not linear (td not constant).", - "type": "text" - }, - { - "block_id": "p745-b7", - "global_id": 21957, - "bbox": [ - 127.89, - 369.57, - 458.65, - 381.69 - ], - "text": "THE NATURE OF DISTORTION IN AUDIO AND VIDEO SIGNALS", - "type": "text" - }, - { - "block_id": "p745-b8", - "global_id": 21958, - "bbox": [ - 127.59, - 385.72, - 516.15, - 515.23 - ], - "text": "Generally speaking, the human ear can readily perceive amplitude distortion but is relatively\ninsensitive to phase distortion. For the phase distortion to become noticeable, the variation in\ndelay [variation in the slope of̸\nH(ω)] should be comparable to the signal duration (or the\nphysically perceptible duration, in case the signal itself is long). In the case of audio signals, each\nspoken syllable can be considered to be an individual signal. The average duration of a spoken\nsyllable is of a magnitude of the order of 0.01 to 0.1 second. Audio systems may have nonlinear\nphases, yet no noticeable signal distortion results because in practical audio systems, maximum\nvariation in the slope of̸\nH(ω) is only a small fraction of a millisecond. This is the real truth\nunderlying the statement that “the human ear is relatively insensitive to phase distortion” [3]. As a\nresult, the manufacturers of audio equipment make available only |H(ω)|, the amplitude response\ncharacteristic of their systems.", - "type": "text" - }, - { - "block_id": "p745-b9", - "global_id": 21959, - "bbox": [ - 127.59, - 517.22, - 516.16, - 634.79 - ], - "text": "For video signals, in contrast, the situation is exactly the opposite. The human eye is sensitive\nto phase distortion but is relatively insensitive to amplitude distortion. Amplitude distortion\nin television signals manifests itself as a partial destruction of the relative half-tone values of\nthe resulting picture, but this effect is not readily apparent to the human eye. Phase distortion\n(nonlinear phase), on the other hand, causes different time delays in different picture elements. The\nresult is a smeared picture, and this effect is readily perceived by the human eye. Phase distortion\nis also very important in digital communication systems because the nonlinear phase characteristic\nof a channel causes pulse dispersion (spreading out), which in turn causes pulses to interfere with\nneighboring pulses. Such interference between pulses can cause an error in the pulse amplitude at\nthe receiver: a binary 1 may read as 0, and vice versa.", - "type": "text" - } - ] - }, - { - "page_num": 746, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p746-b0", - "global_id": 21960, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "726\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p746-b1", - "global_id": 21961, - "bbox": [ - 101.84, - 86.52, - 330.8, - 98.48 - ], - "text": "7.4-2 Bandpass Systems and Group Delay", - "type": "text" - }, - { - "block_id": "p746-b2", - "global_id": 21962, - "bbox": [ - 101.84, - 104.61, - 490.4, - 150.44 - ], - "text": "The distortionless transmission conditions [Eq. (7.39)] can be relaxed slightly for bandpass\nsystems. For lowpass systems, the phase characteristics not only should be linear over the band of\ninterest but also should pass through the origin. For bandpass systems, the phase characteristics\nmust be linear over the band of interest but need not pass through the origin.", - "type": "text" - }, - { - "block_id": "p746-b3", - "global_id": 21963, - "bbox": [ - 101.84, - 152.43, - 490.38, - 187.0 - ], - "text": "Consider an LTI system with amplitude and phase characteristics as shown in Fig. 7.30, where\nthe amplitude spectrum is a constant G0 and the phase is φ0 −ωtg over a band 2W centered at\nfrequency ωc. Over this band, we can describe H(ω) as†", - "type": "text" - }, - { - "block_id": "p746-b4", - "global_id": 21964, - "bbox": [ - 234.5, - 198.63, - 490.38, - 211.5 - ], - "text": "H(ω) = G0ej(φ0−ωtg)\nω ≥0\n(7.41)", - "type": "text" - }, - { - "block_id": "p746-b5", - "global_id": 21965, - "bbox": [ - 101.85, - 224.79, - 490.39, - 247.11 - ], - "text": "The phase of H(ω) in Eq. (7.41), shown dotted in Fig. 7.30b, is linear but does not pass through\nthe origin.", - "type": "text" - }, - { - "block_id": "p746-b6", - "global_id": 21966, - "bbox": [ - 101.84, - 248.69, - 490.39, - 319.55 - ], - "text": "Consider a modulated input signal z(t) = x(t) cosωct. This is a bandpass signal, whose\nspectrum is centered at ω = ωc. The signal cosωct is the carrier, and the signal x(t), which is a\nlowpass signal of bandwidth W (see Fig. 7.25), is the envelope of z(t).‡ We shall now show that\nthe transmission of z(t) through H(ω) results in distortionless transmission of the envelope x(t).\nHowever, the carrier phase changes by φ0. To show this, consider an input ˆz(t) = x(t)ejωct and\nthe corresponding output ˆy(t). From Eq. (7.30), ˆZ(ω) = X(ω −ωc), and the corresponding output", - "type": "text" - }, - { - "block_id": "p746-b7", - "global_id": 21967, - "bbox": [ - 210.1, - 430.49, - 218.98, - 438.49 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p746-b8", - "global_id": 21968, - "bbox": [ - 190.61, - 498.57, - 198.95, - 508.2 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p746-b9", - "global_id": 21969, - "bbox": [ - 190.61, - 388.9, - 198.95, - 398.53 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p746-b10", - "global_id": 21970, - "bbox": [ - 246.64, - 455.01, - 257.31, - 463.09 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p746-b11", - "global_id": 21971, - "bbox": [ - 258.83, - 350.06, - 269.5, - 358.14 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p746-b12", - "global_id": 21972, - "bbox": [ - 209.64, - 540.19, - 219.44, - 548.19 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p746-b13", - "global_id": 21973, - "bbox": [ - 155.63, - 387.37, - 304.57, - 397.13 - ], - "text": "v\nvc\nvc", - "type": "text" - }, - { - "block_id": "p746-b14", - "global_id": 21974, - "bbox": [ - 178.73, - 339.8, - 198.72, - 348.09 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p746-b15", - "global_id": 21975, - "bbox": [ - 177.26, - 451.07, - 200.37, - 459.37 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p746-b16", - "global_id": 21976, - "bbox": [ - 291.73, - 495.2, - 297.07, - 503.2 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p746-b17", - "global_id": 21977, - "bbox": [ - 205.73, - 354.43, - 214.51, - 364.03 - ], - "text": "G0", - "type": "text" - }, - { - "block_id": "p746-b18", - "global_id": 21978, - "bbox": [ - 247.07, - 408.99, - 270.49, - 417.28 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p746-b19", - "global_id": 21979, - "bbox": [ - 190.93, - 518.56, - 220.49, - 528.39 - ], - "text": "f0 vtg", - "type": "text" - }, - { - "block_id": "p746-b20", - "global_id": 21980, - "bbox": [ - 208.04, - 478.93, - 223.04, - 488.75 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p746-b21", - "global_id": 21981, - "bbox": [ - 208.04, - 367.47, - 223.04, - 377.29 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p746-b22", - "global_id": 21982, - "bbox": [ - 327.8, - 527.54, - 486.39, - 549.45 - ], - "text": "Figure 7.30 Generalized linear phase\ncharacteristics.", - "type": "text" - }, - { - "block_id": "p746-b23", - "global_id": 21983, - "bbox": [ - 101.84, - 577.06, - 490.39, - 634.75 - ], - "text": "† Because the phase function is an odd function of ω, if̸\nH(ω) = φ0 −ωtg for ω ≥0, over the band 2W\n(centered at ωc), then̸\nH(ω) = −φ0 −ωtg for ω < 0 over the band 2W (centered at −ωc), as shown in\nFig. 7.30a.\n‡ The envelope of a bandpass signal is well defined only when the bandwidth of the envelope is well below\nthe carrier ωc (W ≪ωc).", - "type": "text" - } - ] - }, - { - "page_num": 747, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p747-b0", - "global_id": 21984, - "bbox": [ - 280.23, - 62.89, - 516.13, - 71.98 - ], - "text": "7.4\nSignal Transmission Through LTIC Systems\n727", - "type": "text" - }, - { - "block_id": "p747-b1", - "global_id": 21985, - "bbox": [ - 127.59, - 86.09, - 232.88, - 98.51 - ], - "text": "spectrum ˆY(ω) is given by", - "type": "text" - }, - { - "block_id": "p747-b2", - "global_id": 21986, - "bbox": [ - 247.03, - 108.59, - 396.69, - 121.73 - ], - "text": "ˆY(ω) = H(ω)ˆZ(ω) = H(ω)X(ω −ωc)", - "type": "text" - }, - { - "block_id": "p747-b3", - "global_id": 21987, - "bbox": [ - 127.59, - 133.16, - 516.13, - 155.48 - ], - "text": "Recall that the bandwidth of X(ω) is W so that the bandwidth of X(ω −ωc) is 2W, centered at ωc.\nOver this range, H(ω) is given by Eq. (7.41). Hence,", - "type": "text" - }, - { - "block_id": "p747-b4", - "global_id": 21988, - "bbox": [ - 211.91, - 163.5, - 430.8, - 178.77 - ], - "text": "ˆY(ω) = G0X(ω −ωc)ej(φ0−ωtg) = G0ejφ0X(ω −ωc)e−ωtg", - "type": "text" - }, - { - "block_id": "p747-b5", - "global_id": 21989, - "bbox": [ - 127.59, - 190.13, - 298.17, - 200.5 - ], - "text": "Use of Eqs. (7.29) and (7.30) yields ˆy(t) as", - "type": "text" - }, - { - "block_id": "p747-b6", - "global_id": 21990, - "bbox": [ - 210.26, - 208.51, - 432.96, - 223.77 - ], - "text": "ˆy(t) = G0ejφ0x(t −tg)ejωc(t−tg) = G0x(t −tg)ej[ωc(t−tg)+φ0]", - "type": "text" - }, - { - "block_id": "p747-b7", - "global_id": 21991, - "bbox": [ - 127.59, - 233.91, - 516.15, - 282.08 - ], - "text": "This is the system response to input ˆz(t) = x(t)ejωct, which is a complex signal. We are really\ninterested in finding the response to the input z(t) = x(t)cosωct, which is the real part of\nˆz(t) = x(t)ejωct. Hence, we use Eq. (2.31) to obtain y(t), the system response to the input\nz(t) = x(t)cosωct, as", - "type": "text" - }, - { - "block_id": "p747-b8", - "global_id": 21992, - "bbox": [ - 244.42, - 283.84, - 516.13, - 294.99 - ], - "text": "y(t) = G0x(t −tg)cos[ωc(t −tg) + φ0)]\n(7.42)", - "type": "text" - }, - { - "block_id": "p747-b9", - "global_id": 21993, - "bbox": [ - 127.59, - 300.16, - 516.13, - 373.51 - ], - "text": "where tg, the group (or envelope) delay, is the negative slope of̸\nH(ω) at ωc.† The output y(t)\nis basically the delayed input z(t −tg), except that the output carrier acquires an extra phase\nφ0. The output envelope x(t −tg) is the delayed version of the input envelope x(t) and is not\naffected by extra phase φ0 of the carrier. In a modulated signal, such as x(t)cosωct, the information\ngenerally resides in the envelope x(t). Hence, the transmission is considered to be distortionless if\nthe envelope x(t) remains undistorted.", - "type": "text" - }, - { - "block_id": "p747-b10", - "global_id": 21994, - "bbox": [ - 127.59, - 375.51, - 516.13, - 398.12 - ], - "text": "Most practical systems satisfy Eq. (7.41), at least over a very small band. Figure 7.30b shows\na typical case in which this condition is satisfied for a small band W centered at frequency ωc.", - "type": "text" - }, - { - "block_id": "p747-b11", - "global_id": 21995, - "bbox": [ - 127.59, - 399.31, - 516.15, - 445.24 - ], - "text": "A system in Eq. (7.41) is said to have a generalized linear phase (GLP), as illustrated in\nFig. 7.30. The ideal linear phase (ILP) characteristics is shown in Fig. 7.29. For distortionless\ntransmission of bandpass signals, the system need satisfy Eq. (7.41) only over the bandwidth of\nthe bandpass signal.", - "type": "text" - }, - { - "block_id": "p747-b12", - "global_id": 21996, - "bbox": [ - 127.59, - 455.71, - 516.12, - 501.62 - ], - "text": "Caution. Recall that the phase response associated with the amplitude response may have jump\ndiscontinuities when the amplitude response goes negative. Jump discontinuities also arise because\nof the use of the principal value for phase. Under such conditions, to compute the group delay\n[Eq. (7.40)], we should ignore the jump discontinuities.", - "type": "text" - }, - { - "block_id": "p747-b13", - "global_id": 21997, - "bbox": [ - 127.59, - 542.51, - 278.59, - 554.74 - ], - "text": "† Equation (7.42) can also be expressed as", - "type": "text" - }, - { - "block_id": "p747-b14", - "global_id": 21998, - "bbox": [ - 264.18, - 563.22, - 379.54, - 573.3 - ], - "text": "y(t) = Gox(t −tg)cosωc(t −tph)", - "type": "text" - }, - { - "block_id": "p747-b15", - "global_id": 21999, - "bbox": [ - 127.59, - 581.04, - 516.13, - 601.34 - ], - "text": "where tph, called the phase delay at ωc, is given by tph(ωc) = (ωctg −φ0)/ωc. Generally, tph varies with ω,\nand we can write", - "type": "text" - }, - { - "block_id": "p747-b16", - "global_id": 22000, - "bbox": [ - 288.74, - 599.29, - 353.29, - 615.73 - ], - "text": "tph(ω) = ωtg −φ0", - "type": "text" - }, - { - "block_id": "p747-b17", - "global_id": 22001, - "bbox": [ - 127.59, - 612.01, - 341.15, - 634.75 - ], - "text": "ω\nRecall also that tg itself may vary with ω.", - "type": "text" - } - ] - }, - { - "page_num": 748, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p748-b0", - "global_id": 22002, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "728\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p748-b1", - "global_id": 22003, - "bbox": [ - 76.77, - 93.91, - 389.47, - 105.87 - ], - "text": "EXAMPLE 7.19\nDistortionless Bandpass Transmission", - "type": "text" - }, - { - "block_id": "p748-b2", - "global_id": 22004, - "bbox": [ - 121.09, - 128.1, - 320.4, - 138.47 - ], - "text": "(a) A signal z(t), shown in Fig. 7.31b, is given by", - "type": "text" - }, - { - "block_id": "p748-b3", - "global_id": 22005, - "bbox": [ - 272.6, - 157.99, - 341.93, - 169.07 - ], - "text": "z(t) = x(t)cosωct", - "type": "text" - }, - { - "block_id": "p748-b4", - "global_id": 22006, - "bbox": [ - 121.09, - 187.88, - 477.04, - 250.14 - ], - "text": "where ωc = 2000π. The pulse x(t) (Fig. 7.31a) is a lowpass pulse of duration 0.1\nsecond and has a bandwidth of about 10 Hz. This signal is passed through a filter\nwhose frequency response is shown in Fig. 7.31c (shown only for positive ω). Find\nand sketch the filter output y(t).\n(b) Find the filter response if ωc = 4000π.", - "type": "text" - }, - { - "block_id": "p748-b5", - "global_id": 22007, - "bbox": [ - 103.17, - 277.54, - 477.02, - 347.69 - ], - "text": "(a) The spectrum Z(ω) is a narrow band of width 20 Hz, centered at frequency f0 = 1 kHz.\nThe gain at the center frequency (1 kHz) is 2. The group delay, which is the negative of the\nslope of the phase plot, can be found by drawing tangents at ωc, as shown in Fig. 7.31c. The\nnegative of the slope of the tangent represents tg, and the intercept along the vertical axis by\nthe tangent represents φ0 at that frequency. From the tangents at ωc, we find tg, the group\ndelay, as", - "type": "text" - }, - { - "block_id": "p748-b6", - "global_id": 22008, - "bbox": [ - 239.09, - 365.19, - 307.63, - 383.63 - ], - "text": "tg = 2.4π −0.4π", - "type": "text" - }, - { - "block_id": "p748-b7", - "global_id": 22009, - "bbox": [ - 270.32, - 370.45, - 340.58, - 389.62 - ], - "text": "2000π\n= 10−3", - "type": "text" - }, - { - "block_id": "p748-b8", - "global_id": 22010, - "bbox": [ - 103.16, - 405.11, - 477.01, - 427.44 - ], - "text": "The vertical axis intercept is φ0 = −0.4π. Hence, by using Eq. (7.42) with gain G0 = 2, we\nobtain", - "type": "text" - }, - { - "block_id": "p748-b9", - "global_id": 22011, - "bbox": [ - 154.37, - 445.23, - 425.31, - 458.41 - ], - "text": "y(t) = 2x(t −tg)cos[ωc(t −tg) −0.4π]\nωc = 2000π\ntg = 10−3", - "type": "text" - }, - { - "block_id": "p748-b10", - "global_id": 22012, - "bbox": [ - 103.16, - 476.84, - 477.02, - 523.08 - ], - "text": "Figure 7.31d shows the output y(t), which consists of the modulated pulse envelope x(t)\ndelayed by 1 ms and the phase of the carrier changed by −0.4π. The output shows no distortion\nof the envelope x(t), only the delay. The carrier phase change does not affect the shape of\nenvelope. Hence, the transmission is considered distortionless.", - "type": "text" - }, - { - "block_id": "p748-b11", - "global_id": 22013, - "bbox": [ - 103.17, - 524.66, - 477.01, - 558.95 - ], - "text": "(b) Figure 7.31c shows that when ωc = 4000π, the slope of̸\nH(ω) is zero so that tg = 0.\nAlso, the gain G0 = 1.5, and the intercept of the tangent with the vertical axis is φ0 = −3.1π.\nHence,", - "type": "text" - }, - { - "block_id": "p748-b12", - "global_id": 22014, - "bbox": [ - 229.97, - 578.46, - 350.2, - 589.54 - ], - "text": "y(t) = 1.5x(t)cos(ωct −3.1π)", - "type": "text" - }, - { - "block_id": "p748-b13", - "global_id": 22015, - "bbox": [ - 103.16, - 608.76, - 410.29, - 618.72 - ], - "text": "This, too, is a distortionless transmission for the same reasons as for case (a).", - "type": "text" - } - ] - }, - { - "page_num": 749, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p749-b0", - "global_id": 22016, - "bbox": [ - 280.23, - 62.89, - 516.13, - 71.98 - ], - "text": "7.4\nSignal Transmission Through LTIC Systems\n729", - "type": "text" - }, - { - "block_id": "p749-b1", - "global_id": 22017, - "bbox": [ - 149.95, - 354.22, - 153.95, - 362.22 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p749-b2", - "global_id": 22018, - "bbox": [ - 148.64, - 265.93, - 152.64, - 273.93 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p749-b3", - "global_id": 22019, - "bbox": [ - 133.6, - 229.19, - 144.27, - 237.27 - ], - "text": "z(t)", - "type": "text" - }, - { - "block_id": "p749-b4", - "global_id": 22020, - "bbox": [ - 249.59, - 323.14, - 259.4, - 331.14 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p749-b5", - "global_id": 22021, - "bbox": [ - 149.78, - 147.0, - 153.78, - 155.0 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p749-b6", - "global_id": 22022, - "bbox": [ - 250.06, - 191.47, - 258.94, - 199.47 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p749-b7", - "global_id": 22023, - "bbox": [ - 133.14, - 109.72, - 144.73, - 117.8 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p749-b8", - "global_id": 22024, - "bbox": [ - 287.86, - 146.91, - 290.08, - 154.91 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p749-b9", - "global_id": 22025, - "bbox": [ - 287.86, - 268.15, - 290.08, - 276.15 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p749-b10", - "global_id": 22026, - "bbox": [ - 289.44, - 546.67, - 291.67, - 554.67 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p749-b11", - "global_id": 22027, - "bbox": [ - 342.41, - 398.99, - 347.75, - 406.99 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p749-b12", - "global_id": 22028, - "bbox": [ - 263.64, - 147.46, - 273.64, - 155.46 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p749-b13", - "global_id": 22029, - "bbox": [ - 263.64, - 252.18, - 273.64, - 260.18 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p749-b14", - "global_id": 22030, - "bbox": [ - 149.73, - 100.96, - 153.73, - 108.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p749-b15", - "global_id": 22031, - "bbox": [ - 149.73, - 221.18, - 153.73, - 229.18 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p749-b16", - "global_id": 22032, - "bbox": [ - 148.95, - 546.35, - 152.95, - 554.35 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p749-b17", - "global_id": 22033, - "bbox": [ - 133.22, - 506.35, - 144.65, - 514.43 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p749-b18", - "global_id": 22034, - "bbox": [ - 249.83, - 602.14, - 259.16, - 610.14 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p749-b19", - "global_id": 22035, - "bbox": [ - 250.06, - 470.31, - 258.94, - 478.31 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p749-b20", - "global_id": 22036, - "bbox": [ - 270.98, - 531.18, - 288.98, - 539.18 - ], - "text": "0.101", - "type": "text" - }, - { - "block_id": "p749-b21", - "global_id": 22037, - "bbox": [ - 148.95, - 500.2, - 152.95, - 508.2 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p749-b22", - "global_id": 22038, - "bbox": [ - 156.4, - 529.32, - 174.4, - 537.32 - ], - "text": "0.001", - "type": "text" - }, - { - "block_id": "p749-b23", - "global_id": 22039, - "bbox": [ - 119.94, - 360.76, - 139.93, - 369.06 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p749-b24", - "global_id": 22040, - "bbox": [ - 131.95, - 388.2, - 290.72, - 409.97 - ], - "text": "0\n0.4p\n2000p\n4000p", - "type": "text" - }, - { - "block_id": "p749-b25", - "global_id": 22041, - "bbox": [ - 130.95, - 426.95, - 153.95, - 445.58 - ], - "text": "2.4p\n3.1p", - "type": "text" - }, - { - "block_id": "p749-b26", - "global_id": 22042, - "bbox": [ - 308.27, - 422.74, - 331.69, - 431.04 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p749-b27", - "global_id": 22043, - "bbox": [ - 143.95, - 364.47, - 153.95, - 372.47 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p749-b28", - "global_id": 22044, - "bbox": [ - 119.94, - 616.84, - 235.26, - 626.08 - ], - "text": "Figure 7.31 Plots for Ex. 7.19.", - "type": "text" - } - ] - }, - { - "page_num": 750, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p750-b0", - "global_id": 22045, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "730\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p750-b1", - "global_id": 22046, - "bbox": [ - 102.2, - 94.37, - 333.77, - 108.31 - ], - "text": "7.5 IDEAL AND PRACTICAL FILTERS", - "type": "text" - }, - { - "block_id": "p750-b2", - "global_id": 22047, - "bbox": [ - 101.84, - 114.3, - 490.38, - 160.13 - ], - "text": "Ideal filters allow distortionless transmission of a certain band of frequencies and completely\nsuppress the remaining frequencies. The ideal lowpass filter (Fig. 7.32), for example, allows all\ncomponents below ω = W rad/s to pass without distortion and suppresses all components above\nω = W. Figure 7.33 illustrates ideal highpass and bandpass filter characteristics.", - "type": "text" - }, - { - "block_id": "p750-b3", - "global_id": 22048, - "bbox": [ - 101.84, - 161.71, - 490.4, - 196.69 - ], - "text": "The ideal lowpass filter in Fig. 7.32a has a linear phase of slope −td, which results in a time\ndelay of td seconds for all its input components of frequencies below W rad/s. Therefore, if the\ninput is a signal x(t) bandlimited to W rad/s, the output y(t) is x(t) delayed by td: that is,", - "type": "text" - }, - { - "block_id": "p750-b4", - "global_id": 22049, - "bbox": [ - 266.45, - 211.1, - 325.77, - 222.17 - ], - "text": "y(t) = x(t −td)", - "type": "text" - }, - { - "block_id": "p750-b5", - "global_id": 22050, - "bbox": [ - 101.85, - 236.58, - 490.38, - 258.91 - ], - "text": "The signal x(t) is transmitted by this system without distortion, but with time delay td. For\nthis filter, |H(ω)| = rect (ω/2W) and̸\nH(ω) = e−jωtd so that", - "type": "text" - }, - { - "block_id": "p750-b6", - "global_id": 22051, - "bbox": [ - 245.78, - 276.53, - 294.38, - 286.9 - ], - "text": "H(ω) = rect", - "type": "text" - }, - { - "block_id": "p750-b7", - "global_id": 22052, - "bbox": [ - 295.49, - 265.53, - 312.28, - 279.5 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p750-b8", - "global_id": 22053, - "bbox": [ - 302.11, - 283.91, - 315.39, - 293.98 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p750-b10", - "global_id": 22054, - "bbox": [ - 323.86, - 274.79, - 345.28, - 286.8 - ], - "text": "e−jωtd", - "type": "text" - }, - { - "block_id": "p750-b11", - "global_id": 22055, - "bbox": [ - 323.3, - 370.2, - 327.3, - 378.2 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p750-b12", - "global_id": 22056, - "bbox": [ - 302.9, - 312.62, - 314.94, - 320.7 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p750-b13", - "global_id": 22057, - "bbox": [ - 381.15, - 412.34, - 390.96, - 420.34 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p750-b14", - "global_id": 22058, - "bbox": [ - 399.99, - 370.82, - 405.21, - 380.37 - ], - "text": "td", - "type": "text" - }, - { - "block_id": "p750-b15", - "global_id": 22059, - "bbox": [ - 197.64, - 412.34, - 206.52, - 420.34 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p750-b16", - "global_id": 22060, - "bbox": [ - 471.08, - 368.01, - 473.3, - 376.01 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p750-b17", - "global_id": 22061, - "bbox": [ - 199.76, - 314.51, - 219.75, - 322.8 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p750-b18", - "global_id": 22062, - "bbox": [ - 195.51, - 396.54, - 239.84, - 406.3 - ], - "text": "H(v) vtd", - "type": "text" - }, - { - "block_id": "p750-b19", - "global_id": 22063, - "bbox": [ - 132.6, - 369.81, - 256.59, - 378.02 - ], - "text": "W\nW", - "type": "text" - }, - { - "block_id": "p750-b20", - "global_id": 22064, - "bbox": [ - 257.23, - 357.6, - 262.56, - 365.6 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p750-b21", - "global_id": 22065, - "bbox": [ - 190.26, - 333.93, - 194.26, - 341.93 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p750-b22", - "global_id": 22066, - "bbox": [ - 456.47, - 377.4, - 463.13, - 394.12 - ], - "text": "p\nW", - "type": "text" - }, - { - "block_id": "p750-b23", - "global_id": 22067, - "bbox": [ - 125.76, - 426.96, - 425.79, - 436.27 - ], - "text": "Figure 7.32 Ideal lowpass filter: (a) frequency response and (b) impulse response.", - "type": "text" - }, - { - "block_id": "p750-b24", - "global_id": 22068, - "bbox": [ - 176.59, - 461.87, - 199.69, - 470.17 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p750-b25", - "global_id": 22069, - "bbox": [ - 178.03, - 539.85, - 201.13, - 548.15 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p750-b26", - "global_id": 22070, - "bbox": [ - 262.29, - 474.6, - 282.76, - 482.89 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p750-b27", - "global_id": 22071, - "bbox": [ - 263.48, - 553.61, - 283.95, - 561.9 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p750-b28", - "global_id": 22072, - "bbox": [ - 250.3, - 506.37, - 376.77, - 515.95 - ], - "text": "v\n0", - "type": "text" - }, - { - "block_id": "p750-b29", - "global_id": 22073, - "bbox": [ - 252.92, - 526.94, - 261.8, - 534.94 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p750-b30", - "global_id": 22074, - "bbox": [ - 252.53, - 605.26, - 262.18, - 613.26 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p750-b31", - "global_id": 22075, - "bbox": [ - 203.8, - 582.12, - 365.85, - 591.94 - ], - "text": "0\nv0\nv0\nv", - "type": "text" - }, - { - "block_id": "p750-b32", - "global_id": 22076, - "bbox": [ - 125.76, - 619.88, - 400.38, - 629.19 - ], - "text": "Figure 7.33 Ideal (a) highpass and (b) bandpass filter frequency responses.", - "type": "text" - } - ] - }, - { - "page_num": 751, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p751-b0", - "global_id": 22077, - "bbox": [ - 352.54, - 62.89, - 516.13, - 71.98 - ], - "text": "7.5\nIdeal and Practical Filters\n731", - "type": "text" - }, - { - "block_id": "p751-b1", - "global_id": 22078, - "bbox": [ - 127.59, - 85.46, - 516.11, - 107.79 - ], - "text": "The unit impulse response h(t) of this filter is obtained from pair 18 (Table 7.1) and the\ntime-shifting property", - "type": "text" - }, - { - "block_id": "p751-b2", - "global_id": 22079, - "bbox": [ - 218.03, - 112.87, - 268.62, - 134.14 - ], - "text": "h(t) = F−1 '", - "type": "text" - }, - { - "block_id": "p751-b3", - "global_id": 22080, - "bbox": [ - 268.61, - 124.28, - 283.54, - 134.24 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p751-b4", - "global_id": 22081, - "bbox": [ - 284.65, - 112.87, - 301.45, - 126.85 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p751-b5", - "global_id": 22082, - "bbox": [ - 291.28, - 131.26, - 304.55, - 141.32 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p751-b7", - "global_id": 22083, - "bbox": [ - 313.02, - 122.15, - 334.44, - 134.14 - ], - "text": "e−jωtd", - "type": "text" - }, - { - "block_id": "p751-b8", - "global_id": 22084, - "bbox": [ - 335.61, - 112.87, - 340.28, - 122.83 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p751-b9", - "global_id": 22085, - "bbox": [ - 342.33, - 117.2, - 361.65, - 134.14 - ], - "text": "= W", - "type": "text" - }, - { - "block_id": "p751-b10", - "global_id": 22086, - "bbox": [ - 354.38, - 123.87, - 425.69, - 141.32 - ], - "text": "π sinc[W(t −td)]", - "type": "text" - }, - { - "block_id": "p751-b11", - "global_id": 22087, - "bbox": [ - 127.59, - 148.61, - 516.13, - 206.81 - ], - "text": "Recall that h(t) is the system response to impulse input δ(t), which is applied at t = 0. Figure 7.32b\nshows a curious fact: the response h(t) begins even before the input is applied (at t = 0). Clearly,\nthe filter is noncausal and therefore physically unrealizable. Similarly, one can show that other\nideal filters (such as the ideal highpass or ideal bandpass filters depicted in Fig. 7.33) are also\nphysically unrealizable.", - "type": "text" - }, - { - "block_id": "p751-b12", - "global_id": 22088, - "bbox": [ - 145.52, - 208.39, - 394.24, - 218.76 - ], - "text": "For a physically realizable system, h(t) must be causal; that is,", - "type": "text" - }, - { - "block_id": "p751-b13", - "global_id": 22089, - "bbox": [ - 277.6, - 230.01, - 366.13, - 240.39 - ], - "text": "h(t) = 0\nfor t < 0", - "type": "text" - }, - { - "block_id": "p751-b14", - "global_id": 22090, - "bbox": [ - 127.6, - 251.95, - 516.14, - 289.27 - ], - "text": "In the frequency domain, this condition is equivalent to the well-known Paley–Wiener criterion,\nwhich states that the necessary and sufficient condition for the amplitude response |H(ω)| to be\nrealizable is†\n# ∞", - "type": "text" - }, - { - "block_id": "p751-b15", - "global_id": 22091, - "bbox": [ - 278.18, - 302.09, - 290.74, - 309.07 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p751-b16", - "global_id": 22092, - "bbox": [ - 293.54, - 283.79, - 334.49, - 294.17 - ], - "text": "|ln|H(ω)||", - "type": "text" - }, - { - "block_id": "p751-b17", - "global_id": 22093, - "bbox": [ - 300.73, - 290.77, - 516.13, - 308.22 - ], - "text": "1 + ω2\ndω < ∞\n(7.43)", - "type": "text" - }, - { - "block_id": "p751-b18", - "global_id": 22094, - "bbox": [ - 127.59, - 314.96, - 516.14, - 409.02 - ], - "text": "If H(ω) does not satisfy this condition, it is unrealizable. Note that if |H(ω)| = 0 over any finite\nband, |ln|H(ω)|| = ∞over that band, and Eq. (7.43) is violated. If, however, H(ω) = 0 at a single\nfrequency (or a set of discrete frequencies), the integral in Eq. (7.43) may still be finite even\nthough the integrand is infinite at those discrete frequencies. Therefore, for a physically realizable\nsystem, H(ω) may be zero at some discrete frequencies, but it cannot be zero over any finite band.\nIn addition, if |H(ω)| decays exponentially (or at a higher rate) with ω, the integral in Eq. (7.43)\ngoes to infinity, and |H(ω)| cannot be realized. Clearly, |H(ω)| cannot decay too fast with ω.\nAccording to this criterion, ideal filter characteristics (Figs. 7.32 and 7.33) are unrealizable.", - "type": "text" - }, - { - "block_id": "p751-b19", - "global_id": 22095, - "bbox": [ - 127.6, - 410.59, - 516.11, - 432.93 - ], - "text": "The impulse response h(t) in Fig. 7.32 is not realizable. One practical approach to filter design\nis to cut off the tail of h(t) for t < 0. The resulting causal impulse response:h(t), given by", - "type": "text" - }, - { - "block_id": "p751-b20", - "global_id": 22096, - "bbox": [ - 291.68, - 442.14, - 350.84, - 454.45 - ], - "text": ":h(t) = h(t)u(t)", - "type": "text" - }, - { - "block_id": "p751-b21", - "global_id": 22097, - "bbox": [ - 127.59, - 463.77, - 516.13, - 547.91 - ], - "text": "is physically realizable because it is causal (Fig. 7.34). If td is sufficiently large, :h(t) will be a\nclose approximation of h(t), and the resulting filter :H(ω) will be a good approximation of an ideal\nfilter. This close realization of the ideal filter is achieved because of the increased value of time\ndelay td. This observation means that the price of close realization is higher delay in the output;\nthis situation is common in noncausal systems. Of course, theoretically, a delay td = ∞is needed\nto realize the ideal characteristics. But a glance at Fig. 7.32b shows that a delay td of three or\nfour times π", - "type": "text" - }, - { - "block_id": "p751-b22", - "global_id": 22098, - "bbox": [ - 127.59, - 535.5, - 516.13, - 562.39 - ], - "text": "W will make :h(t) a reasonably close version of h(t −td). For instance, an audio filter\nis required to handle frequencies of up to 20 kHz (W = 40,000π). In this case, a td of about 10−4", - "type": "text" - }, - { - "block_id": "p751-b23", - "global_id": 22099, - "bbox": [ - 127.59, - 578.99, - 337.02, - 591.2 - ], - "text": "† We are assuming that |H(ω)| is square integrable, that is,", - "type": "text" - }, - { - "block_id": "p751-b24", - "global_id": 22100, - "bbox": [ - 282.87, - 590.73, - 298.32, - 601.7 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p751-b25", - "global_id": 22101, - "bbox": [ - 287.6, - 612.97, - 299.26, - 619.45 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p751-b26", - "global_id": 22102, - "bbox": [ - 300.76, - 599.52, - 360.85, - 612.18 - ], - "text": "|H(ω)|2 dω < ∞", - "type": "text" - }, - { - "block_id": "p751-b27", - "global_id": 22103, - "bbox": [ - 127.59, - 624.07, - 500.84, - 633.41 - ], - "text": "Note that the Paley–Wiener criterion is a criterion for the realizability of the amplitude response |H(ω)|.", - "type": "text" - } - ] - }, - { - "page_num": 752, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p752-b0", - "global_id": 22104, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "732\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p752-b1", - "global_id": 22105, - "bbox": [ - 133.17, - 143.06, - 257.4, - 153.0 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p752-b2", - "global_id": 22106, - "bbox": [ - 125.3, - 90.33, - 137.11, - 100.83 - ], - "text": "hˆ(t)", - "type": "text" - }, - { - "block_id": "p752-b3", - "global_id": 22107, - "bbox": [ - 198.2, - 142.98, - 203.43, - 152.53 - ], - "text": "td", - "type": "text" - }, - { - "block_id": "p752-b4", - "global_id": 22108, - "bbox": [ - 281.8, - 134.07, - 486.42, - 156.0 - ], - "text": "Figure 7.34 Approximate realization of an ideal\nlowpass filter by truncation of its impulse response.", - "type": "text" - }, - { - "block_id": "p752-b5", - "global_id": 22109, - "bbox": [ - 101.84, - 184.72, - 490.42, - 219.01 - ], - "text": "(0.1 ms) would be a reasonable choice. The truncation operation [cutting the tail of h(t) to make\nit causal], however, creates some unsuspected problems. We discuss these problems and their cure\nin Sec. 7.8.", - "type": "text" - }, - { - "block_id": "p752-b6", - "global_id": 22110, - "bbox": [ - 101.84, - 221.0, - 490.39, - 242.91 - ], - "text": "In practice, we can realize a variety of filter characteristics that approach the ideal. Practical\n(realizable) filter characteristics are gradual, without jump discontinuities in amplitude response.", - "type": "text" - }, - { - "block_id": "p752-b7", - "global_id": 22111, - "bbox": [ - 107.82, - 278.79, - 393.0, - 290.75 - ], - "text": "DRILL 7.11\nThe Unrealizable Gaussian Response", - "type": "text" - }, - { - "block_id": "p752-b8", - "global_id": 22112, - "bbox": [ - 107.82, - 293.72, - 484.42, - 333.74 - ], - "text": "Show that a filter with Gaussian frequency response H(ω) = e−αω2 is unrealizable. Demonstrate\nthis fact in two ways: first by showing that its impulse response is noncausal, and then by\nshowing that |H(ω)| violates the Paley–Wiener criterion. [Hint: Use pair 22 in Table 7.1.]", - "type": "text" - }, - { - "block_id": "p752-b9", - "global_id": 22113, - "bbox": [ - 102.14, - 367.58, - 401.97, - 393.64 - ], - "text": "THINKING IN THE TIME AND FREQUENCY DOMAINS:\nA TWO-DIMENSIONAL VIEW OF SIGNALS AND SYSTEMS", - "type": "text" - }, - { - "block_id": "p752-b10", - "global_id": 22114, - "bbox": [ - 101.84, - 397.67, - 490.4, - 515.23 - ], - "text": "Both signals and systems have dual personalities, the time domain and the frequency domain. For\na deeper perspective, we should examine and understand both these identities because they offer\ncomplementary insights. An exponential signal, for instance, can be specified by its time-domain\ndescription such as e−2tu(t) or by its Fourier transform (its frequency-domain description)\n1/(jω + 2). The time-domain description depicts the waveform of a signal. The frequency-domain\ndescription portrays its spectral composition [relative amplitudes of its sinusoidal (or exponential)\ncomponents and their phases]. For the signal e−2t, for instance, the time-domain description\nportrays the exponentially decaying signal with a time constant 0.5. The frequency-domain\ndescription characterizes it as a lowpass signal, which can be synthesized by sinusoids with\namplitudes decaying with frequency roughly as 1/ω.", - "type": "text" - }, - { - "block_id": "p752-b11", - "global_id": 22115, - "bbox": [ - 101.85, - 517.22, - 490.43, - 610.88 - ], - "text": "An LTIC system can also be described or specified in the time domain by its impulse response\nh(t) or in the frequency domain by its frequency response H(ω). In Sec. 2.6, we studied intuitive\ninsights in the system behavior offered by the impulse response, which consists of characteristic\nmodes of the system. By purely qualitative reasoning, we saw that the system responds well to\nsignals that are similar to the characteristic modes and responds poorly to signals that are very\ndifferent from those modes. We also saw that the shape of the impulse response h(t) determines\nthe system time constant (speed of response), and pulse dispersion (spreading), which, in turn,\ndetermines the rate of pulse transmission.", - "type": "text" - }, - { - "block_id": "p752-b12", - "global_id": 22116, - "bbox": [ - 101.84, - 612.45, - 490.38, - 634.79 - ], - "text": "The frequency response H(ω) specifies the system response to exponential or sinusoidal input\nof various frequencies. This is precisely the filtering characteristic of the system.", - "type": "text" - } - ] - }, - { - "page_num": 753, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p753-b0", - "global_id": 22117, - "bbox": [ - 397.93, - 62.89, - 516.13, - 71.98 - ], - "text": "7.6\nSignal Energy\n733", - "type": "text" - }, - { - "block_id": "p753-b1", - "global_id": 22118, - "bbox": [ - 127.59, - 85.82, - 516.15, - 239.25 - ], - "text": "Experienced electrical engineers instinctively think in both domains (time and frequency)\nwhenever possible. When they look at a signal, they consider its waveform, the signal width\n(duration), and the rate at which the waveform decays. This is basically a time-domain perspective.\nThey also think of the signal in terms of its frequency spectrum, that is, in terms of its sinusoidal\ncomponents and their relative amplitudes and phases, whether the spectrum is lowpass, bandpass,\nhighpass, and so on. This is a frequency-domain perspective. Experienced electrical engineers\nthink of a system in terms of its impulse response h(t). The width of h(t) indicates the time\nconstant (response time): that is, how quickly the system is capable of responding to an input,\nand how much dispersion (spreading) it will cause. This is a time-domain perspective. From\nthe frequency-domain perspective, these engineers view a system as a filter, which selectively\ntransmits certain frequency components and suppresses the others [frequency response H(ω)].\nKnowing the input signal spectrum and the frequency response of the system, they create a mental\nimage of the output signal spectrum. This concept is precisely expressed by Y(ω) = X(ω)H(ω).", - "type": "text" - }, - { - "block_id": "p753-b2", - "global_id": 22119, - "bbox": [ - 127.59, - 241.24, - 516.16, - 322.93 - ], - "text": "We can analyze LTI systems by time-domain techniques or by frequency-domain techniques.\nThen why learn both? The reason is that the two domains offer complementary insights into system\nbehavior. Some aspects are easily grasped in one domain; other aspects may be easier to see in the\nother domain. Both time-domain and frequency-domain methods are as essential for the study of\nsignals and systems as two eyes are essential to a human being for correct visual perception of\nreality. A person can see with either eye, but for proper perception of three-dimensional reality,\nboth eyes are essential.", - "type": "text" - }, - { - "block_id": "p753-b3", - "global_id": 22120, - "bbox": [ - 127.59, - 324.92, - 516.15, - 418.58 - ], - "text": "It is important to keep the two domains separate, and not to mix the entities in the two domains.\nIf we are using the frequency domain to determine the system response, we must deal with all\nsignals in terms of their spectra (Fourier transforms) and all systems in terms of their frequency\nresponses. For example, to determine the system response y(t) to an input x(t), we must first\nconvert the input signal into its frequency-domain description X(ω). The system description also\nmust be in the frequency domain, that is, the frequency response H(ω). The output signal spectrum\nY(ω) = X(ω)H(ω). Thus, the result (output) is also in the frequency domain. To determine the final\nanswer y(t), we must take the inverse transform of Y(ω).", - "type": "text" - }, - { - "block_id": "p753-b4", - "global_id": 22121, - "bbox": [ - 127.94, - 464.74, - 260.47, - 478.69 - ], - "text": "7.6 SIGNAL ENERGY", - "type": "text" - }, - { - "block_id": "p753-b5", - "global_id": 22122, - "bbox": [ - 127.59, - 484.26, - 367.46, - 496.13 - ], - "text": "The signal energy Ex of a signal x(t) was defined in Ch. 1 as", - "type": "text" - }, - { - "block_id": "p753-b6", - "global_id": 22123, - "bbox": [ - 284.59, - 528.4, - 304.1, - 539.86 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p753-b7", - "global_id": 22124, - "bbox": [ - 306.15, - 514.85, - 323.1, - 526.9 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p753-b8", - "global_id": 22125, - "bbox": [ - 311.42, - 539.72, - 323.97, - 546.69 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p753-b9", - "global_id": 22126, - "bbox": [ - 325.58, - 524.6, - 516.12, - 538.78 - ], - "text": "|x(t)|2 dt\n(7.44)", - "type": "text" - }, - { - "block_id": "p753-b10", - "global_id": 22127, - "bbox": [ - 127.59, - 572.61, - 513.35, - 582.98 - ], - "text": "Signal energy can be related to the signal spectrum X(ω) by substituting Eq. (7.10) in Eq. (7.44):", - "type": "text" - }, - { - "block_id": "p753-b11", - "global_id": 22128, - "bbox": [ - 199.91, - 617.06, - 219.42, - 628.52 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p753-b12", - "global_id": 22129, - "bbox": [ - 221.47, - 603.5, - 238.42, - 615.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p753-b13", - "global_id": 22130, - "bbox": [ - 226.74, - 628.38, - 239.29, - 635.35 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p753-b14", - "global_id": 22131, - "bbox": [ - 240.89, - 615.34, - 293.56, - 627.33 - ], - "text": "x(t)x∗(t)dt =", - "type": "text" - }, - { - "block_id": "p753-b15", - "global_id": 22132, - "bbox": [ - 295.6, - 603.5, - 312.54, - 615.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p753-b16", - "global_id": 22133, - "bbox": [ - 300.86, - 628.38, - 313.42, - 635.35 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p753-b17", - "global_id": 22134, - "bbox": [ - 315.03, - 617.06, - 329.86, - 627.33 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p753-b18", - "global_id": 22135, - "bbox": [ - 330.97, - 603.07, - 346.06, - 620.45 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p753-b19", - "global_id": 22136, - "bbox": [ - 337.59, - 624.13, - 348.54, - 634.51 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p753-b20", - "global_id": 22137, - "bbox": [ - 351.86, - 603.51, - 368.79, - 615.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p753-b21", - "global_id": 22138, - "bbox": [ - 357.11, - 628.38, - 369.67, - 635.35 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p753-b22", - "global_id": 22139, - "bbox": [ - 371.28, - 613.26, - 428.04, - 627.33 - ], - "text": "X∗(ω)e−jωt dω", - "type": "text" - }, - { - "block_id": "p753-b23", - "global_id": 22140, - "bbox": [ - 428.24, - 603.07, - 433.67, - 613.04 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p753-b24", - "global_id": 22141, - "bbox": [ - 435.89, - 617.37, - 443.64, - 627.33 - ], - "text": "dt", - "type": "text" - } - ] - }, - { - "page_num": 754, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p754-b0", - "global_id": 22142, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "734\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p754-b1", - "global_id": 22143, - "bbox": [ - 101.84, - 84.26, - 490.4, - 107.83 - ], - "text": "Here, we used the fact that x∗(t), being the conjugate of x(t), can be expressed as the conjugate of\nthe right-hand side of Eq. (7.10). Now, interchanging the order of integration yields", - "type": "text" - }, - { - "block_id": "p754-b2", - "global_id": 22144, - "bbox": [ - 193.51, - 123.98, - 224.74, - 142.0 - ], - "text": "Ex = 1", - "type": "text" - }, - { - "block_id": "p754-b3", - "global_id": 22145, - "bbox": [ - 216.27, - 137.62, - 227.22, - 148.0 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p754-b4", - "global_id": 22146, - "bbox": [ - 230.53, - 116.99, - 247.47, - 129.05 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p754-b5", - "global_id": 22147, - "bbox": [ - 235.79, - 141.86, - 248.35, - 148.84 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p754-b6", - "global_id": 22148, - "bbox": [ - 249.96, - 128.82, - 274.77, - 140.82 - ], - "text": "X∗(ω)", - "type": "text" - }, - { - "block_id": "p754-b7", - "global_id": 22149, - "bbox": [ - 275.88, - 116.56, - 298.25, - 129.05 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p754-b8", - "global_id": 22150, - "bbox": [ - 286.57, - 141.86, - 299.13, - 148.84 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p754-b9", - "global_id": 22151, - "bbox": [ - 300.73, - 126.75, - 343.49, - 140.82 - ], - "text": "x(t)e−jωt dt", - "type": "text" - }, - { - "block_id": "p754-b10", - "global_id": 22152, - "bbox": [ - 343.67, - 116.56, - 349.1, - 126.52 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p754-b11", - "global_id": 22153, - "bbox": [ - 351.32, - 130.55, - 363.09, - 140.82 - ], - "text": "dω", - "type": "text" - }, - { - "block_id": "p754-b12", - "global_id": 22154, - "bbox": [ - 205.26, - 151.6, - 224.74, - 168.54 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p754-b13", - "global_id": 22155, - "bbox": [ - 216.27, - 165.24, - 227.22, - 175.62 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p754-b14", - "global_id": 22156, - "bbox": [ - 230.53, - 144.6, - 247.47, - 156.66 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p754-b15", - "global_id": 22157, - "bbox": [ - 235.79, - 169.49, - 248.35, - 176.46 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p754-b16", - "global_id": 22158, - "bbox": [ - 249.96, - 151.6, - 330.06, - 168.54 - ], - "text": "X(ω)X∗(ω)dω = 1", - "type": "text" - }, - { - "block_id": "p754-b17", - "global_id": 22159, - "bbox": [ - 321.6, - 165.24, - 332.55, - 175.62 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p754-b18", - "global_id": 22160, - "bbox": [ - 335.85, - 144.6, - 352.79, - 156.66 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p754-b19", - "global_id": 22161, - "bbox": [ - 341.11, - 169.49, - 353.67, - 176.46 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p754-b20", - "global_id": 22162, - "bbox": [ - 355.27, - 154.37, - 398.52, - 168.44 - ], - "text": "|X(ω)|2 dω", - "type": "text" - }, - { - "block_id": "p754-b21", - "global_id": 22163, - "bbox": [ - 101.85, - 191.37, - 158.46, - 201.33 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p754-b22", - "global_id": 22164, - "bbox": [ - 213.75, - 216.54, - 233.27, - 228.0 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p754-b23", - "global_id": 22165, - "bbox": [ - 235.31, - 202.98, - 252.26, - 215.03 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p754-b24", - "global_id": 22166, - "bbox": [ - 240.58, - 227.86, - 253.14, - 234.83 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p754-b25", - "global_id": 22167, - "bbox": [ - 254.74, - 209.97, - 309.83, - 226.92 - ], - "text": "|x(t)|2 dt = 1", - "type": "text" - }, - { - "block_id": "p754-b26", - "global_id": 22168, - "bbox": [ - 301.36, - 223.61, - 312.31, - 233.99 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p754-b27", - "global_id": 22169, - "bbox": [ - 315.61, - 202.98, - 332.56, - 215.03 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p754-b28", - "global_id": 22170, - "bbox": [ - 320.88, - 227.86, - 333.43, - 234.83 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p754-b29", - "global_id": 22171, - "bbox": [ - 335.04, - 212.74, - 490.38, - 226.92 - ], - "text": "|X(ω)|2 dω\n(7.45)", - "type": "text" - }, - { - "block_id": "p754-b30", - "global_id": 22172, - "bbox": [ - 101.85, - 246.65, - 490.41, - 292.58 - ], - "text": "This is Parseval’s theorem (for the Fourier transform). A similar result was obtained in Eqs. (6.26)\nand (6.27) for a periodic signal and its Fourier series. This result allows us to determine the signal\nenergy from either the time-domain specification x(t) or the corresponding frequency-domain\nspecification X(ω).", - "type": "text" - }, - { - "block_id": "p754-b31", - "global_id": 22173, - "bbox": [ - 101.84, - 294.16, - 490.39, - 352.36 - ], - "text": "The right-hand side of Eq. (7.45) can be interpreted to mean that the energy of a signal x(t)\nresults from energies contributed by all the spectral components of the signal x(t). The total signal\nenergy is the area under |X(ω)2| (divided by 2π). If we consider a small band ω (ω →0),\nas illustrated in Fig. 7.35, the energy Ex of the spectral components in this band is the area of\n|X(ω)|2 under this band (divided by 2π):", - "type": "text" - }, - { - "block_id": "p754-b32", - "global_id": 22174, - "bbox": [ - 185.77, - 367.83, - 225.01, - 385.86 - ], - "text": "Ex = 1", - "type": "text" - }, - { - "block_id": "p754-b33", - "global_id": 22175, - "bbox": [ - 216.55, - 367.42, - 367.14, - 391.85 - ], - "text": "2π |X(ω)|2 ω = |X(ω)|2 f\nω", - "type": "text" - }, - { - "block_id": "p754-b34", - "global_id": 22176, - "bbox": [ - 353.99, - 374.4, - 406.46, - 391.85 - ], - "text": "2π = f Hz", - "type": "text" - }, - { - "block_id": "p754-b35", - "global_id": 22177, - "bbox": [ - 101.84, - 405.56, - 490.4, - 452.74 - ], - "text": "Therefore, the energy contributed by the components in this band of f (in hertz) is |X(ω)|2f.\nThe total signal energy is the sum of energies of all such bands and is indicated by the area under\n|X(ω)|2 as in Eq. (7.45). Therefore, |X(ω)|2 is the energy spectral density (per unit bandwidth in\nhertz).", - "type": "text" - }, - { - "block_id": "p754-b36", - "global_id": 22178, - "bbox": [ - 119.78, - 451.11, - 490.37, - 464.69 - ], - "text": "For real signals, X(ω) and X(−ω) are conjugates, and |X(ω)|2 is an even function of ω because", - "type": "text" - }, - { - "block_id": "p754-b37", - "global_id": 22179, - "bbox": [ - 221.75, - 477.66, - 370.48, - 492.05 - ], - "text": "|X(ω)|2 = X(ω)X∗(ω) = X(ω)X(−ω)", - "type": "text" - }, - { - "block_id": "p754-b38", - "global_id": 22180, - "bbox": [ - 264.97, - 516.59, - 287.87, - 526.06 - ], - "text": "X(v)2", - "type": "text" - }, - { - "block_id": "p754-b39", - "global_id": 22181, - "bbox": [ - 252.16, - 598.93, - 307.48, - 608.56 - ], - "text": "0\nv0", - "type": "text" - }, - { - "block_id": "p754-b40", - "global_id": 22182, - "bbox": [ - 318.71, - 581.24, - 329.38, - 589.44 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p754-b41", - "global_id": 22183, - "bbox": [ - 375.05, - 598.8, - 380.39, - 606.8 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p754-b42", - "global_id": 22184, - "bbox": [ - 125.76, - 619.96, - 360.92, - 629.19 - ], - "text": "Figure 7.35 Interpretation of energy spectral density of a signal.", - "type": "text" - } - ] - }, - { - "page_num": 755, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p755-b0", - "global_id": 22185, - "bbox": [ - 397.93, - 62.89, - 516.13, - 71.98 - ], - "text": "7.6\nSignal Energy\n735", - "type": "text" - }, - { - "block_id": "p755-b1", - "global_id": 22186, - "bbox": [ - 127.59, - 86.06, - 368.88, - 96.62 - ], - "text": "Therefore, the energy of real signal x(t) can be expressed as†", - "type": "text" - }, - { - "block_id": "p755-b2", - "global_id": 22187, - "bbox": [ - 274.84, - 104.92, - 303.58, - 122.95 - ], - "text": "Ex = 1", - "type": "text" - }, - { - "block_id": "p755-b3", - "global_id": 22188, - "bbox": [ - 297.6, - 118.57, - 303.58, - 128.53 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p755-b4", - "global_id": 22189, - "bbox": [ - 306.88, - 97.93, - 323.83, - 109.99 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p755-b5", - "global_id": 22190, - "bbox": [ - 312.15, - 107.68, - 516.13, - 130.07 - ], - "text": "0\n|X(ω)|2 dω\n(7.46)", - "type": "text" - }, - { - "block_id": "p755-b6", - "global_id": 22191, - "bbox": [ - 127.59, - 136.55, - 516.15, - 172.02 - ], - "text": "The signal energy Ex, which results from contributions from all the frequency components from\nω = 0 to ∞, is given by (1/π times) the area under |X(ω)|2 from ω = 0 to ∞. It follows that the\nenergy contributed by spectral components of frequencies between ω1 and ω2 is", - "type": "text" - }, - { - "block_id": "p755-b7", - "global_id": 22192, - "bbox": [ - 270.37, - 178.63, - 307.12, - 196.66 - ], - "text": "Ex = 1", - "type": "text" - }, - { - "block_id": "p755-b8", - "global_id": 22193, - "bbox": [ - 301.14, - 192.28, - 307.12, - 202.24 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p755-b9", - "global_id": 22194, - "bbox": [ - 310.42, - 171.65, - 327.81, - 185.26 - ], - "text": "# ω2", - "type": "text" - }, - { - "block_id": "p755-b10", - "global_id": 22195, - "bbox": [ - 315.68, - 196.52, - 323.24, - 205.05 - ], - "text": "ω1", - "type": "text" - }, - { - "block_id": "p755-b11", - "global_id": 22196, - "bbox": [ - 329.9, - 181.4, - 516.13, - 195.58 - ], - "text": "|X(ω)|2 dω\n(7.47)", - "type": "text" - }, - { - "block_id": "p755-b12", - "global_id": 22197, - "bbox": [ - 102.51, - 221.89, - 415.92, - 233.85 - ], - "text": "EXAMPLE 7.20\nSignal Energy and Parseval’s Theorem", - "type": "text" - }, - { - "block_id": "p755-b13", - "global_id": 22198, - "bbox": [ - 128.9, - 247.89, - 502.76, - 284.1 - ], - "text": "Find the energy of signal x(t) = e−atu(t). Determine the frequency W (rad/s) so that the energy\ncontributed by the spectral components of all the frequencies below W is 95% of the signal\nenergy Ex.", - "type": "text" - }, - { - "block_id": "p755-b14", - "global_id": 22199, - "bbox": [ - 128.9, - 306.31, - 162.87, - 316.27 - ], - "text": "We have", - "type": "text" - }, - { - "block_id": "p755-b15", - "global_id": 22200, - "bbox": [ - 239.87, - 330.58, - 259.39, - 342.04 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p755-b16", - "global_id": 22201, - "bbox": [ - 261.44, - 317.03, - 278.39, - 329.08 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p755-b17", - "global_id": 22202, - "bbox": [ - 266.7, - 341.9, - 279.26, - 348.87 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p755-b18", - "global_id": 22203, - "bbox": [ - 280.87, - 329.15, - 318.54, - 340.86 - ], - "text": "x2(t)dt =", - "type": "text" - }, - { - "block_id": "p755-b19", - "global_id": 22204, - "bbox": [ - 320.59, - 317.03, - 337.53, - 329.08 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p755-b20", - "global_id": 22205, - "bbox": [ - 325.85, - 324.01, - 388.11, - 349.16 - ], - "text": "0\ne−2at dt = 1", - "type": "text" - }, - { - "block_id": "p755-b21", - "global_id": 22206, - "bbox": [ - 380.63, - 337.97, - 390.59, - 348.03 - ], - "text": "2a", - "type": "text" - }, - { - "block_id": "p755-b22", - "global_id": 22207, - "bbox": [ - 128.91, - 357.75, - 380.09, - 367.71 - ], - "text": "We can verify this result by Parseval’s theorem. For this signal,", - "type": "text" - }, - { - "block_id": "p755-b23", - "global_id": 22208, - "bbox": [ - 285.69, - 377.16, - 344.77, - 401.08 - ], - "text": "X(ω) =\n1\njω + a", - "type": "text" - }, - { - "block_id": "p755-b24", - "global_id": 22209, - "bbox": [ - 128.9, - 410.5, - 143.28, - 420.46 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p755-b25", - "global_id": 22210, - "bbox": [ - 177.66, - 429.1, - 206.4, - 447.12 - ], - "text": "Ex = 1", - "type": "text" - }, - { - "block_id": "p755-b26", - "global_id": 22211, - "bbox": [ - 200.42, - 442.74, - 206.4, - 452.7 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p755-b27", - "global_id": 22212, - "bbox": [ - 209.7, - 422.1, - 226.64, - 434.16 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p755-b28", - "global_id": 22213, - "bbox": [ - 214.95, - 429.1, - 290.73, - 454.24 - ], - "text": "0\n|X(ω)|2 dω = 1", - "type": "text" - }, - { - "block_id": "p755-b29", - "global_id": 22214, - "bbox": [ - 284.76, - 442.74, - 290.73, - 452.7 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p755-b30", - "global_id": 22215, - "bbox": [ - 294.03, - 422.1, - 310.98, - 434.16 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p755-b31", - "global_id": 22216, - "bbox": [ - 299.29, - 447.26, - 302.78, - 454.24 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p755-b32", - "global_id": 22217, - "bbox": [ - 313.77, - 429.09, - 381.32, - 453.02 - ], - "text": "1\nω2 + a2 dω = 1", - "type": "text" - }, - { - "block_id": "p755-b33", - "global_id": 22218, - "bbox": [ - 372.86, - 428.68, - 417.54, - 453.12 - ], - "text": "πa tan−1 ω", - "type": "text" - }, - { - "block_id": "p755-b34", - "global_id": 22219, - "bbox": [ - 411.89, - 443.05, - 416.87, - 453.02 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p755-b36", - "global_id": 22220, - "bbox": [ - 422.17, - 426.3, - 429.29, - 433.28 - ], - "text": "∞", - "type": "text" - }, - { - "block_id": "p755-b37", - "global_id": 22221, - "bbox": [ - 422.17, - 429.1, - 450.33, - 455.13 - ], - "text": "0\n= 1", - "type": "text" - }, - { - "block_id": "p755-b38", - "global_id": 22222, - "bbox": [ - 442.85, - 443.05, - 452.81, - 453.12 - ], - "text": "2a", - "type": "text" - }, - { - "block_id": "p755-b39", - "global_id": 22223, - "bbox": [ - 128.91, - 462.79, - 502.74, - 486.21 - ], - "text": "The band ω = 0 to ω = W contains 95% of the signal energy, that is, 0.95/2a. Therefore,\nfrom Eq. (7.47) with ω1 = 0 and ω2 = W, we obtain", - "type": "text" - }, - { - "block_id": "p755-b40", - "global_id": 22224, - "bbox": [ - 204.68, - 496.32, - 242.36, - 520.34 - ], - "text": "0.95\n2a = 1\nπ", - "type": "text" - }, - { - "block_id": "p755-b41", - "global_id": 22225, - "bbox": [ - 245.65, - 489.32, - 261.29, - 501.6 - ], - "text": "# W", - "type": "text" - }, - { - "block_id": "p755-b42", - "global_id": 22226, - "bbox": [ - 250.92, - 514.49, - 254.4, - 521.47 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p755-b43", - "global_id": 22227, - "bbox": [ - 264.6, - 495.9, - 317.87, - 520.24 - ], - "text": "dω\nω2 + a2 = 1", - "type": "text" - }, - { - "block_id": "p755-b44", - "global_id": 22228, - "bbox": [ - 309.4, - 495.9, - 355.28, - 520.34 - ], - "text": "πa tan−1 ω", - "type": "text" - }, - { - "block_id": "p755-b45", - "global_id": 22229, - "bbox": [ - 349.63, - 510.27, - 354.61, - 520.24 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p755-b47", - "global_id": 22230, - "bbox": [ - 359.92, - 496.73, - 365.73, - 503.7 - ], - "text": "W", - "type": "text" - }, - { - "block_id": "p755-b48", - "global_id": 22231, - "bbox": [ - 359.92, - 496.22, - 426.26, - 520.39 - ], - "text": "0 = 1\nπa tan−1 W", - "type": "text" - }, - { - "block_id": "p755-b49", - "global_id": 22232, - "bbox": [ - 419.97, - 510.27, - 424.95, - 520.24 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p755-b50", - "global_id": 22233, - "bbox": [ - 128.9, - 527.33, - 137.2, - 537.29 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p755-b51", - "global_id": 22234, - "bbox": [ - 225.08, - 544.26, - 294.75, - 568.69 - ], - "text": "0.95π\n2\n= tan−1 W", - "type": "text" - }, - { - "block_id": "p755-b52", - "global_id": 22235, - "bbox": [ - 288.47, - 551.24, - 407.78, - 568.59 - ], - "text": "a\n\r⇒\nW = 12.706a rad/s", - "type": "text" - }, - { - "block_id": "p755-b53", - "global_id": 22236, - "bbox": [ - 127.59, - 611.73, - 516.12, - 634.91 - ], - "text": "† In Eq. (7.46), it is assumed that X(ω) does not contain an impulse at ω = 0. If such an impulse exists, it\nshould be integrated separately with a multiplying factor of 1/2π rather than 1/π.", - "type": "text" - } - ] - }, - { - "page_num": 756, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p756-b0", - "global_id": 22237, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "736\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p756-b1", - "global_id": 22238, - "bbox": [ - 103.16, - 85.88, - 477.02, - 132.13 - ], - "text": "This result indicates that the spectral components of x(t) in the band from 0 (dc) to\n12.706a rad/s (2.02a Hz) contribute 95% of the total signal energy; all the remaining\nspectral components (in the band from 12.706a rad/s to ∞) contribute only 5% of the signal\nenergy.", - "type": "text" - }, - { - "block_id": "p756-b2", - "global_id": 22239, - "bbox": [ - 107.82, - 197.2, - 398.05, - 209.15 - ], - "text": "DRILL 7.12\nSignal Energy and Parseval’s Theorem", - "type": "text" - }, - { - "block_id": "p756-b3", - "global_id": 22240, - "bbox": [ - 107.82, - 214.24, - 484.41, - 240.2 - ], - "text": "Use Parseval’s theorem to show that the energy of the signal x(t) = 2a/(t2 + a2) is 2π/a. [Hint:\nFind X(ω) using pair 3 of Table 7.1 and the duality property.]", - "type": "text" - }, - { - "block_id": "p756-b4", - "global_id": 22241, - "bbox": [ - 102.14, - 276.95, - 331.33, - 289.07 - ], - "text": "THE ESSENTIAL BANDWIDTH OF A SIGNAL", - "type": "text" - }, - { - "block_id": "p756-b5", - "global_id": 22242, - "bbox": [ - 101.84, - 293.1, - 490.39, - 410.66 - ], - "text": "The spectra of all practical signals extend to infinity. However, because the energy of any practical\nsignal is finite, the signal spectrum must approach 0 as ω →∞. Most of the signal energy is\ncontained within a certain band of B Hz, and the energy contributed by the components beyond\nB Hz is negligible. We can therefore suppress the signal spectrum beyond B Hz with little effect\non the signal shape and energy. The bandwidth B is called the essential bandwidth of the signal.\nThe criterion for selecting B depends on the error tolerance in a particular application. We may,\nfor example, select B to be that band which contains 95% of the signal energy.† This figure may\nbe higher or lower than 95%, depending on the precision needed. Using such a criterion, we can\ndetermine the essential bandwidth of a signal. The essential bandwidth B for the signal e−atu(t),\nusing 95% energy criterion, was determined in Ex. 7.20 to be 2.02a Hz.", - "type": "text" - }, - { - "block_id": "p756-b6", - "global_id": 22243, - "bbox": [ - 101.85, - 412.24, - 490.4, - 447.23 - ], - "text": "Suppression of all the spectral components of x(t) beyond the essential bandwidth results in\na signal ˆx(t), which is a close approximation of x(t). If we use the 95% criterion for the essential\nbandwidth, the energy of the error (the difference) x(t) −ˆx(t) is 5% of Ex.", - "type": "text" - }, - { - "block_id": "p756-b7", - "global_id": 22244, - "bbox": [ - 102.19, - 487.65, - 371.8, - 517.54 - ], - "text": "7.7 APPLICATION TO COMMUNICATIONS:\nAMPLITUDE MODULATION", - "type": "text" - }, - { - "block_id": "p756-b8", - "global_id": 22245, - "bbox": [ - 101.84, - 523.42, - 490.38, - 569.35 - ], - "text": "Modulation causes a spectral shift in a signal and is used to gain certain advantages mentioned\nin our discussion of the frequency-shifting property. Broadly speaking, there are two classes of\nmodulation: amplitude (linear) modulation and angle (nonlinear) modulation. In this section, we\nshall discuss some practical forms of amplitude modulation.", - "type": "text" - }, - { - "block_id": "p756-b9", - "global_id": 22246, - "bbox": [ - 101.84, - 599.27, - 490.37, - 633.41 - ], - "text": "† For lowpass signals, the essential bandwidth may also be defined as a frequency at which the value of the\namplitude spectrum is a small fraction (about 1%) of its peak value. In Ex. 7.20, for instance, the peak value,\nwhich occurs at ω = 0, is 1/a.", - "type": "text" - } - ] - }, - { - "page_num": 757, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p757-b0", - "global_id": 22247, - "bbox": [ - 228.18, - 62.89, - 516.11, - 71.98 - ], - "text": "7.7\nApplication to Communications: Amplitude Modulation\n737", - "type": "text" - }, - { - "block_id": "p757-b1", - "global_id": 22248, - "bbox": [ - 127.59, - 86.52, - 491.95, - 98.48 - ], - "text": "7.7-1 Double-Sideband, Suppressed-Carrier (DSB-SC) Modulation", - "type": "text" - }, - { - "block_id": "p757-b2", - "global_id": 22249, - "bbox": [ - 127.59, - 104.19, - 516.15, - 174.35 - ], - "text": "In amplitude modulation, the amplitude A of the carrier Acos(ωct + θc) is varied in some manner\nwith the baseband (message)† signal m(t) (known as the modulating signal). The frequency ωc\nand the phase θc are constant. We can assume θc = 0 without loss of generality. If the carrier\namplitude A is made directly proportional to the modulating signal m(t), the modulated signal is\nm(t)cos ωct (Fig. 7.36). As was indicated earlier [Eq. (7.32)], this type of modulation simply shifts\nthe spectrum of m(t) to the carrier frequency (Fig. 7.36c). Thus, if", - "type": "text" - }, - { - "block_id": "p757-b3", - "global_id": 22250, - "bbox": [ - 290.03, - 187.2, - 353.68, - 197.48 - ], - "text": "m(t) ⇐⇒M(ω)", - "type": "text" - }, - { - "block_id": "p757-b4", - "global_id": 22251, - "bbox": [ - 268.12, - 398.69, - 270.35, - 406.69 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p757-b5", - "global_id": 22252, - "bbox": [ - 274.93, - 509.23, - 277.15, - 517.23 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p757-b6", - "global_id": 22253, - "bbox": [ - 430.19, - 395.05, - 434.19, - 403.05 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p757-b7", - "global_id": 22254, - "bbox": [ - 430.68, - 505.29, - 434.68, - 513.29 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p757-b8", - "global_id": 22255, - "bbox": [ - 430.29, - 468.93, - 435.18, - 476.93 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p757-b9", - "global_id": 22256, - "bbox": [ - 433.46, - 328.42, - 442.35, - 336.5 - ], - "text": "2A", - "type": "text" - }, - { - "block_id": "p757-b10", - "global_id": 22257, - "bbox": [ - 383.26, - 394.76, - 404.15, - 403.05 - ], - "text": "2pB", - "type": "text" - }, - { - "block_id": "p757-b11", - "global_id": 22258, - "bbox": [ - 366.77, - 503.18, - 485.86, - 512.95 - ], - "text": "vc\nvc", - "type": "text" - }, - { - "block_id": "p757-b12", - "global_id": 22259, - "bbox": [ - 453.19, - 394.95, - 467.41, - 403.05 - ], - "text": "2pB", - "type": "text" - }, - { - "block_id": "p757-b13", - "global_id": 22260, - "bbox": [ - 367.14, - 520.64, - 381.36, - 528.74 - ], - "text": "4pB", - "type": "text" - }, - { - "block_id": "p757-b14", - "global_id": 22261, - "bbox": [ - 471.58, - 394.95, - 476.92, - 402.95 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p757-b15", - "global_id": 22262, - "bbox": [ - 503.88, - 506.42, - 509.22, - 514.42 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p757-b16", - "global_id": 22263, - "bbox": [ - 226.77, - 232.6, - 410.85, - 240.6 - ], - "text": "(modulating signal)\n(modulated signal)", - "type": "text" - }, - { - "block_id": "p757-b17", - "global_id": 22264, - "bbox": [ - 251.55, - 216.59, - 264.87, - 224.68 - ], - "text": "m(t)", - "type": "text" - }, - { - "block_id": "p757-b18", - "global_id": 22265, - "bbox": [ - 405.91, - 319.9, - 423.24, - 328.0 - ], - "text": "M(v)", - "type": "text" - }, - { - "block_id": "p757-b19", - "global_id": 22266, - "bbox": [ - 131.83, - 455.66, - 145.16, - 463.74 - ], - "text": "m(t)", - "type": "text" - }, - { - "block_id": "p757-b20", - "global_id": 22267, - "bbox": [ - 127.59, - 541.29, - 147.59, - 549.59 - ], - "text": "m(t)", - "type": "text" - }, - { - "block_id": "p757-b21", - "global_id": 22268, - "bbox": [ - 135.78, - 358.17, - 149.11, - 366.26 - ], - "text": "m(t)", - "type": "text" - }, - { - "block_id": "p757-b22", - "global_id": 22269, - "bbox": [ - 361.77, - 216.57, - 399.71, - 226.14 - ], - "text": "m(t) cos vct", - "type": "text" - }, - { - "block_id": "p757-b23", - "global_id": 22270, - "bbox": [ - 226.01, - 459.42, - 263.95, - 468.98 - ], - "text": "m(t) cos vct", - "type": "text" - }, - { - "block_id": "p757-b24", - "global_id": 22271, - "bbox": [ - 301.7, - 275.93, - 327.9, - 293.03 - ], - "text": "cos vct\n(carrier)", - "type": "text" - }, - { - "block_id": "p757-b25", - "global_id": 22272, - "bbox": [ - 328.92, - 467.43, - 515.36, - 485.54 - ], - "text": "USB\nLSB\nLSB\nUSB", - "type": "text" - }, - { - "block_id": "p757-b26", - "global_id": 22273, - "bbox": [ - 310.36, - 299.91, - 319.24, - 307.91 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p757-b27", - "global_id": 22274, - "bbox": [ - 309.98, - 427.42, - 319.63, - 435.42 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p757-b28", - "global_id": 22275, - "bbox": [ - 310.36, - 561.31, - 319.24, - 569.31 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p757-b29", - "global_id": 22276, - "bbox": [ - 127.59, - 576.01, - 254.38, - 585.25 - ], - "text": "Figure 7.36 DSB-SC modulation.", - "type": "text" - }, - { - "block_id": "p757-b30", - "global_id": 22277, - "bbox": [ - 127.59, - 610.24, - 516.13, - 633.41 - ], - "text": "† The term baseband is used to designate the band of frequencies of the signal delivered by the source or the\ninput transducer.", - "type": "text" - } - ] - }, - { - "page_num": 758, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p758-b0", - "global_id": 22278, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "738\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p758-b1", - "global_id": 22279, - "bbox": [ - 101.84, - 85.82, - 118.99, - 95.78 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p758-b2", - "global_id": 22280, - "bbox": [ - 204.79, - 109.46, - 279.94, - 121.88 - ], - "text": "m(t)cos ωct ⇐⇒1", - "type": "text" - }, - { - "block_id": "p758-b3", - "global_id": 22281, - "bbox": [ - 276.45, - 110.8, - 490.38, - 124.71 - ], - "text": "2 [M(ω + ωc) + M(ω −ωc)]\n(7.48)", - "type": "text" - }, - { - "block_id": "p758-b4", - "global_id": 22282, - "bbox": [ - 101.84, - 138.69, - 490.39, - 232.75 - ], - "text": "Recall that M(ω −ωc) is M(ω)-shifted to the right by ωc and M(ω + ωc) is M(ω)-shifted to\nthe left by ωc. Thus, the process of modulation shifts the spectrum of the modulating signal to\nthe left and the right by ωc. Note also that if the bandwidth of m(t) is B Hz, then, as indicated in\nFig. 7.36c, the bandwidth of the modulated signal is 2B Hz. We also observe that the modulated\nsignal spectrum centered at ωc is composed of two parts: a portion that lies above ωc, known as\nthe upper sideband (USB), and a portion that lies below ωc, known as the lower sideband (LSB).\nSimilarly, the spectrum centered at −ωc has upper and lower sidebands. This form of modulation\nis called double sideband (DSB) modulation for the obvious reason.", - "type": "text" - }, - { - "block_id": "p758-b5", - "global_id": 22283, - "bbox": [ - 101.84, - 234.32, - 490.4, - 281.28 - ], - "text": "The relationship of B to ωc is of interest. Figure 7.36c shows that ωc ≥2πB to avoid the\noverlap of the spectra centered at ±ωc. If ωc < 2πB, the spectra overlap and the information of\nm(t) are lost in the process of modulation, a loss that makes it impossible to get back m(t) from\nthe modulated signal m(t)cos ωct.†", - "type": "text" - }, - { - "block_id": "p758-b6", - "global_id": 22284, - "bbox": [ - 76.77, - 314.05, - 455.88, - 326.0 - ], - "text": "EXAMPLE 7.21\nDouble-Sideband Suppressed-Carrier Modulation", - "type": "text" - }, - { - "block_id": "p758-b7", - "global_id": 22285, - "bbox": [ - 103.16, - 342.25, - 477.02, - 364.59 - ], - "text": "For a baseband signal m(t) = cos ωmt, find the DSB-SC signal and sketch its spectrum. Identify\nthe upper and lower sidebands.", - "type": "text" - }, - { - "block_id": "p758-b8", - "global_id": 22286, - "bbox": [ - 103.16, - 387.5, - 477.02, - 422.07 - ], - "text": "We shall work this problem in the frequency domain as well as the time domain to clarify the\nbasic concepts of DSB-SC modulation. In the frequency-domain approach, we work with the\nsignal spectra. The spectrum of the baseband signal m(t) = cos ωmt is given by", - "type": "text" - }, - { - "block_id": "p758-b9", - "global_id": 22287, - "bbox": [ - 218.58, - 432.91, - 361.6, - 443.99 - ], - "text": "M(ω) = π[δ(ω −ωm) + δ(ω + ωm)]", - "type": "text" - }, - { - "block_id": "p758-b10", - "global_id": 22288, - "bbox": [ - 103.17, - 454.83, - 425.89, - 465.91 - ], - "text": "The spectrum consists of two impulses located at ±ωm, as depicted in Fig. 7.37a.", - "type": "text" - }, - { - "block_id": "p758-b11", - "global_id": 22289, - "bbox": [ - 103.17, - 467.2, - 477.03, - 536.94 - ], - "text": "The DSB-SC (modulated) spectrum, as indicated by Eq. (7.48), is the baseband spectrum\nin Fig. 7.37a shifted to the right and the left by ωc (times 0.5), as depicted in Fig. 7.37b. This\nspectrum consists of impulses at ±(ωc −ωm) and ±(ωc + ωm). The spectrum beyond ωc is\nthe upper sideband (USB), and the one below ωc is the lower sideband (LSB). Observe that\nthe DSB-SC spectrum does not have as a component the carrier frequency ωc. This is why the\nterm double-sideband, suppressed carrier (DSB-SC) is used for this type of modulation.", - "type": "text" - }, - { - "block_id": "p758-b12", - "global_id": 22290, - "bbox": [ - 101.84, - 577.36, - 490.38, - 633.41 - ], - "text": "† Practical factors may impose additional restrictions on ωc. For instance, in broadcast applications, a\nradiating antenna can radiate only a narrow band without distortion. This restriction implies that avoiding\ndistortion caused by the radiating antenna calls for ωc/2πB ≫1. The broadcast band AM radio, for instance,\nwith B = 5 kHz and the band of 550–1600 kHz for carrier frequency gives a ratio of ωc/2πB roughly in the\nrange of 100–300.", - "type": "text" - } - ] - }, - { - "page_num": 759, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p759-b0", - "global_id": 22291, - "bbox": [ - 228.18, - 62.89, - 516.11, - 71.98 - ], - "text": "7.7\nApplication to Communications: Amplitude Modulation\n739", - "type": "text" - }, - { - "block_id": "p759-b1", - "global_id": 22292, - "bbox": [ - 271.05, - 140.99, - 275.05, - 148.99 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p759-b2", - "global_id": 22293, - "bbox": [ - 271.05, - 226.32, - 275.05, - 234.32 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p759-b3", - "global_id": 22294, - "bbox": [ - 351.71, - 196.95, - 365.04, - 205.24 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p759-b4", - "global_id": 22295, - "bbox": [ - 245.8, - 177.38, - 292.69, - 185.38 - ], - "text": "DSB spectrum", - "type": "text" - }, - { - "block_id": "p759-b5", - "global_id": 22296, - "bbox": [ - 264.9, - 156.49, - 273.78, - 164.49 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p759-b6", - "global_id": 22297, - "bbox": [ - 264.51, - 242.14, - 274.16, - 250.14 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p759-b7", - "global_id": 22298, - "bbox": [ - 160.5, - 214.67, - 377.39, - 222.67 - ], - "text": "USB\nLSB\nLSB\nUSB", - "type": "text" - }, - { - "block_id": "p759-b8", - "global_id": 22299, - "bbox": [ - 243.38, - 140.7, - 259.71, - 150.46 - ], - "text": "vm", - "type": "text" - }, - { - "block_id": "p759-b9", - "global_id": 22300, - "bbox": [ - 143.79, - 226.56, - 377.14, - 236.32 - ], - "text": "vc\nvc\n(vc vm)\nvc vm\n(vc vm)\nvc vm", - "type": "text" - }, - { - "block_id": "p759-b10", - "global_id": 22301, - "bbox": [ - 281.81, - 140.89, - 326.54, - 150.46 - ], - "text": "vm\nv", - "type": "text" - }, - { - "block_id": "p759-b11", - "global_id": 22302, - "bbox": [ - 427.13, - 227.0, - 432.47, - 235.0 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p759-b12", - "global_id": 22303, - "bbox": [ - 243.05, - 94.56, - 260.38, - 102.66 - ], - "text": "M(v)", - "type": "text" - }, - { - "block_id": "p759-b13", - "global_id": 22304, - "bbox": [ - 295.12, - 111.05, - 300.46, - 119.05 - ], - "text": "p", - "type": "text" - }, - { - "block_id": "p759-b14", - "global_id": 22305, - "bbox": [ - 119.94, - 256.84, - 302.11, - 266.08 - ], - "text": "Figure 7.37 An example of DSB-SC modulation.", - "type": "text" - }, - { - "block_id": "p759-b15", - "global_id": 22306, - "bbox": [ - 128.9, - 293.75, - 502.77, - 319.8 - ], - "text": "In the time-domain approach, we work directly with signals in the time domain. For the\nbaseband signal m(t) = cos ωmt, the DSB-SC signal ϕDSB-SC(t) is", - "type": "text" - }, - { - "block_id": "p759-b16", - "global_id": 22307, - "bbox": [ - 216.37, - 328.85, - 323.35, - 343.37 - ], - "text": "ϕDSB-SC(t) = m(t)cos ωct", - "type": "text" - }, - { - "block_id": "p759-b17", - "global_id": 22308, - "bbox": [ - 266.46, - 345.44, - 337.38, - 356.52 - ], - "text": "= cos ωmt cos ωct", - "type": "text" - }, - { - "block_id": "p759-b18", - "global_id": 22309, - "bbox": [ - 266.46, - 360.44, - 280.96, - 371.75 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p759-b19", - "global_id": 22310, - "bbox": [ - 277.48, - 361.79, - 502.75, - 374.98 - ], - "text": "2[cos(ωc + ωm)t + cos(ωc −ωm)t]\n(7.49)", - "type": "text" - }, - { - "block_id": "p759-b20", - "global_id": 22311, - "bbox": [ - 128.9, - 384.12, - 502.78, - 467.31 - ], - "text": "This result shows that when the baseband (message) signal is a single sinusoid of frequency\nωm, the modulated signal consists of two sinusoids: the component of frequency ωc + ωm (the\nupper sideband), and the component of frequency ωc −ωm (the lower sideband). Figure 7.37b\nillustrates precisely the spectrum of ϕDSB-SC(t). Thus, each component of frequency ωm\nin the modulating signal results in two components of frequencies ωc + ωm and ωc −ωm\nin the modulated signal. This being a DSB-SC (suppressed-carrier) modulation, there is no\ncomponent of the carrier frequency ωc on the right-hand side of Eq. (7.49).†", - "type": "text" - }, - { - "block_id": "p759-b21", - "global_id": 22312, - "bbox": [ - 127.59, - 518.1, - 516.14, - 557.67 - ], - "text": "DEMODULATION OF DSB-SC SIGNALS\nThe DSB-SC modulation translates or shifts the frequency spectrum to the left and the right by\nωc (i.e., at +ωc and −ωc), as seen from Eq. (7.48). To recover the original signal m(t) from", - "type": "text" - }, - { - "block_id": "p759-b22", - "global_id": 22313, - "bbox": [ - 127.59, - 577.36, - 516.13, - 633.41 - ], - "text": "† The term suppressed carrier does not necessarily mean absence of the spectrum at the carrier frequency.\n“Suppressed carrier” merely implies that there is no discrete component of the carrier frequency. Since no\ndiscrete component exists, the DSB-SC spectrum does not have impulses at ±ωc, which further implies that\nthe modulated signal m(t)cos ωct does not contain a term of the form kcos ωct [assuming that m(t) has a zero\nmean value].", - "type": "text" - } - ] - }, - { - "page_num": 760, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p760-b0", - "global_id": 22314, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "740\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p760-b1", - "global_id": 22315, - "bbox": [ - 329.76, - 100.51, - 375.75, - 108.51 - ], - "text": "Lowpass filter", - "type": "text" - }, - { - "block_id": "p760-b2", - "global_id": 22316, - "bbox": [ - 281.76, - 224.21, - 285.76, - 232.21 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p760-b3", - "global_id": 22317, - "bbox": [ - 289.76, - 188.47, - 294.64, - 196.47 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p760-b4", - "global_id": 22318, - "bbox": [ - 158.84, - 90.98, - 300.99, - 101.09 - ], - "text": "e(t)\nm(t) cos vct", - "type": "text" - }, - { - "block_id": "p760-b5", - "global_id": 22319, - "bbox": [ - 236.16, - 136.43, - 262.36, - 153.53 - ], - "text": "cos vct\n(carrier)", - "type": "text" - }, - { - "block_id": "p760-b6", - "global_id": 22320, - "bbox": [ - 273.23, - 160.4, - 282.11, - 168.4 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p760-b7", - "global_id": 22321, - "bbox": [ - 273.23, - 239.74, - 282.88, - 247.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p760-b8", - "global_id": 22322, - "bbox": [ - 159.72, - 223.92, - 424.24, - 233.68 - ], - "text": "2vc\n2vc\nv", - "type": "text" - }, - { - "block_id": "p760-b9", - "global_id": 22323, - "bbox": [ - 400.42, - 87.5, - 420.08, - 100.99 - ], - "text": "m(t)\n1\n2", - "type": "text" - }, - { - "block_id": "p760-b10", - "global_id": 22324, - "bbox": [ - 125.76, - 254.07, - 424.25, - 263.68 - ], - "text": "Figure 7.38 Demodulation of DSB-SC: (a) demodulator and (b) spectrum of e(t).", - "type": "text" - }, - { - "block_id": "p760-b11", - "global_id": 22325, - "bbox": [ - 101.84, - 299.39, - 490.41, - 416.94 - ], - "text": "the modulated signal, we must retranslate the spectrum to its original position. The process\nof recovering the signal from the modulated signal (retranslating the spectrum to its original\nposition) is referred to as demodulation, or detection. Observe that if the modulated signal\nspectrum in Fig. 7.36c is shifted to the left and to the right by ωc (and halved), we obtain the\nspectrum illustrated in Fig. 7.38b, which contains the desired baseband spectrum in addition to an\nunwanted spectrum at ±2ωc. The latter can be suppressed by a lowpass filter. Thus, demodulation,\nwhich is almost identical to modulation, consists of multiplication of the incoming modulated\nsignal m(t)cos ωct by a carrier cos ωct followed by a lowpass filter, as depicted in Fig. 7.38a.\nWe can verify this conclusion directly in the time domain by observing that the signal e(t) in\nFig. 7.38a is", - "type": "text" - }, - { - "block_id": "p760-b12", - "global_id": 22326, - "bbox": [ - 205.2, - 422.26, - 298.53, - 437.45 - ], - "text": "e(t) = m(t)cos2 ωct = 1", - "type": "text" - }, - { - "block_id": "p760-b13", - "global_id": 22327, - "bbox": [ - 295.05, - 426.38, - 387.04, - 439.57 - ], - "text": "2[m(t) + m(t)cos 2ωct]", - "type": "text" - }, - { - "block_id": "p760-b14", - "global_id": 22328, - "bbox": [ - 101.85, - 450.54, - 309.36, - 460.92 - ], - "text": "Therefore, the Fourier transform of the signal e(t) is", - "type": "text" - }, - { - "block_id": "p760-b15", - "global_id": 22329, - "bbox": [ - 199.63, - 476.36, - 236.65, - 487.98 - ], - "text": "E(ω) = 1", - "type": "text" - }, - { - "block_id": "p760-b16", - "global_id": 22330, - "bbox": [ - 233.16, - 476.36, - 392.6, - 490.89 - ], - "text": "2M(ω) + 1\n4[M(ω + 2ωc) + M(ω −2ωc)]", - "type": "text" - }, - { - "block_id": "p760-b17", - "global_id": 22331, - "bbox": [ - 101.84, - 504.86, - 490.39, - 563.06 - ], - "text": "Hence, e(t) consists of two components (1/2)m(t) and (1/2)m(t)cos 2ωct, with their spectra, as\nillustrated in Fig. 7.38b. The spectrum of the second component, being a modulated signal with\ncarrier frequency 2ωc, is centered at ±2ωc. Hence, this component is suppressed by the lowpass\nfilter in Fig. 7.38a. The desired component (1/2)M(ω), being a lowpass spectrum (centered at\nω = 0), passes through the filter unharmed, resulting in the output (1/2)m(t).", - "type": "text" - }, - { - "block_id": "p760-b18", - "global_id": 22332, - "bbox": [ - 101.85, - 565.05, - 490.41, - 634.79 - ], - "text": "A possible form of lowpass filter characteristics is depicted (dotted) in Fig. 7.38b. In this\nmethod of recovering the baseband signal, called synchronous detection, or coherent detection,\nwe use a carrier of exactly the same frequency (and phase) as the carrier used for modulation.\nThus, for demodulation, we need to generate a local carrier at the receiver in frequency and phase\ncoherence (synchronism) with the carrier used at the modulator. We shall demonstrate in Ex. 7.22\nthat both phase and frequency synchronism are extremely critical.", - "type": "text" - } - ] - }, - { - "page_num": 761, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p761-b0", - "global_id": 22333, - "bbox": [ - 228.18, - 62.89, - 516.11, - 71.98 - ], - "text": "7.7\nApplication to Communications: Amplitude Modulation\n741", - "type": "text" - }, - { - "block_id": "p761-b1", - "global_id": 22334, - "bbox": [ - 103.92, - 95.83, - 455.47, - 107.78 - ], - "text": "EXAMPLE 7.22\nFrequency and Phase Incoherence in DSB-SC", - "type": "text" - }, - { - "block_id": "p761-b2", - "global_id": 22335, - "bbox": [ - 128.9, - 125.71, - 502.76, - 147.63 - ], - "text": "Discuss the effect of lack of frequency and phase coherence (synchronism) between the carriers\nat the modulator (transmitter) and the demodulator (receiver) in DSB-SC.", - "type": "text" - }, - { - "block_id": "p761-b3", - "global_id": 22336, - "bbox": [ - 128.9, - 170.13, - 502.75, - 204.41 - ], - "text": "Let the modulator carrier be cos ωct (Fig. 7.36a). For the demodulator in Fig. 7.38a, we shall\nconsider two cases: with carrier cos(ωct+θ) (phase error of θ) and with carrier cos(ωc +ω)t\n(frequency error ω).", - "type": "text" - }, - { - "block_id": "p761-b4", - "global_id": 22337, - "bbox": [ - 128.91, - 206.0, - 502.76, - 240.28 - ], - "text": "(a) With the demodulator carrier cos(ωct + θ) (instead of cos ωct) in Fig. 7.38a,\nthe multiplier output is e(t) = m(t)cos ωctcos(ωct + θ) instead of m(t)cos2 ωct. From the\ntrigonometric identity, we obtain", - "type": "text" - }, - { - "block_id": "p761-b5", - "global_id": 22338, - "bbox": [ - 243.52, - 251.51, - 370.79, - 262.59 - ], - "text": "e(t) = m(t)cos ωct cos(ωct + θ)", - "type": "text" - }, - { - "block_id": "p761-b6", - "global_id": 22339, - "bbox": [ - 260.38, - 266.51, - 274.88, - 277.82 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p761-b7", - "global_id": 22340, - "bbox": [ - 271.39, - 267.85, - 388.15, - 281.05 - ], - "text": "2m(t)[cos θ + cos(2ωct + θ)]", - "type": "text" - }, - { - "block_id": "p761-b8", - "global_id": 22341, - "bbox": [ - 128.9, - 289.46, - 502.77, - 347.66 - ], - "text": "The spectrum of the component (1/2)m(t)cos(2ωct + θ) is centered at ±2ωc. Consequently,\nit will be filtered out by the lowpass filter at the output. The component (1/2)m(t)cos θ is the\nsignal m(t) multiplied by a constant (1/2)cos θ. The spectrum of this component is centered\nat ω = 0 (lowpass spectrum) and will pass through the lowpass filter at the output, yielding the\noutput (1/2)m(t)cos θ.", - "type": "text" - }, - { - "block_id": "p761-b9", - "global_id": 22342, - "bbox": [ - 128.9, - 349.24, - 502.78, - 395.48 - ], - "text": "If θ is constant, the phase asynchronism merely yields an output that is attenuated (by a\nfactor cos θ). Unfortunately, in practice, θ is often the phase difference between the carriers\ngenerated by two distant generators and varies randomly with time. This variation would result\nin an output whose gain varies randomly with time.", - "type": "text" - }, - { - "block_id": "p761-b10", - "global_id": 22343, - "bbox": [ - 128.9, - 397.06, - 502.76, - 431.35 - ], - "text": "(b) In the case of frequency error, the demodulator carrier is cos(ωc +ω)t. This situation\nis very similar to the phase error case in part (a) with θ replaced by (ω)t. Following the\nanalysis in part (a), we can express the demodulator product e(t) as", - "type": "text" - }, - { - "block_id": "p761-b11", - "global_id": 22344, - "bbox": [ - 229.53, - 442.57, - 365.97, - 454.03 - ], - "text": "e(t) = m(t)cos ωct cos(ωc + ω)t", - "type": "text" - }, - { - "block_id": "p761-b12", - "global_id": 22345, - "bbox": [ - 246.38, - 457.57, - 260.88, - 468.89 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p761-b13", - "global_id": 22346, - "bbox": [ - 257.4, - 458.92, - 402.13, - 472.11 - ], - "text": "2m(t)[cos(ω)t + cos(2ωc + ω)t]", - "type": "text" - }, - { - "block_id": "p761-b14", - "global_id": 22347, - "bbox": [ - 128.9, - 480.52, - 502.78, - 646.32 - ], - "text": "The spectrum of the component (1/2)m(t)cos(2ωc + ω)t is centered at ±(2ωc + ω).\nConsequently, this component will be filtered out by the lowpass filter at the output. The\ncomponent (1/2)m(t)cos(ω)t is the signal m(t) multiplied by a low-frequency carrier of\nfrequency ω. The spectrum of this component is centered at ±ω. In practice, the frequency\nerror (ω) is usually very small. Hence, the signal (1/2)m(t)cos(ω)t (whose spectrum is\ncentered at ±ω) is a lowpass signal and passes through the lowpass filter at the output,\nresulting in the output (1/2)m(t)cos(ω)t. The output is the desired signal m(t) multiplied by\na very-low-frequency sinusoid cos(ω)t. The output in this case is not merely an attenuated\nreplica of the desired signal m(t), but represents m(t) multiplied by a time-varying gain\ncos(ω)t. If, for instance, the transmitter and the receiver carrier frequencies differ just by\n1 Hz, the output will be the desired signal m(t) multiplied by a time-varying signal whose\ngain goes from the maximum to 0 every half-second. This is like a restless child fiddling with\nthe volume control knob of a receiver, going from maximum volume to zero volume every\nhalf-second. This kind of distortion (called the beat effect) is beyond repair.", - "type": "text" - } - ] - }, - { - "page_num": 762, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p762-b0", - "global_id": 22348, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "742\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p762-b1", - "global_id": 22349, - "bbox": [ - 101.84, - 86.52, - 292.27, - 98.48 - ], - "text": "7.7-2 Amplitude Modulation (AM)", - "type": "text" - }, - { - "block_id": "p762-b2", - "global_id": 22350, - "bbox": [ - 101.84, - 104.61, - 490.42, - 259.33 - ], - "text": "For the suppressed-carrier scheme just discussed, a receiver must generate a carrier in frequency\nand phase synchronism with the carrier at a transmitter that may be located hundreds or thousands\nof miles away. This situation calls for a sophisticated receiver, which could be quite costly. The\nother alternative is for the transmitter to transmit a carrier A cosωct [along with the modulated\nsignal m(t) cosωct] so that there is no need to generate a carrier at the receiver. In this case, the\ntransmitter needs to transmit much larger power, a rather expensive procedure. In point-to-point\ncommunications, where there is one transmitter for each receiver, substantial complexity in the\nreceiver system can be justified, provided there is a large enough saving in expensive high-power\ntransmitting equipment. On the other hand, for a broadcast system with a multitude of receivers for\neach transmitter, it is more economical to have one expensive high-power transmitter and simpler,\nless expensive receivers. The second option (transmitting a carrier along with the modulated\nsignal) is the obvious choice in this case. This is amplitude modulation (AM), in which the\ntransmitted signal ϕAM(t) is given by", - "type": "text" - }, - { - "block_id": "p762-b3", - "global_id": 22351, - "bbox": [ - 190.37, - 269.04, - 490.38, - 280.7 - ], - "text": "ϕAM(t) = Acos ωct + m(t)cosωct = [A + m(t)]cosωct\n(7.50)", - "type": "text" - }, - { - "block_id": "p762-b4", - "global_id": 22352, - "bbox": [ - 101.84, - 290.42, - 490.41, - 444.26 - ], - "text": "Recall that the DSB-SC signal is m(t) cos ωct. From Eq. (7.50) it follows that the AM signal is\nidentical to the DSB-SC signal with A+m(t) as the modulating signal [instead of m(t)]. Therefore,\nto sketch ϕAM(t), we sketch A + m(t) and −[A + m(t)] as the envelopes and fill in between with\nthe sinusoid of the carrier frequency. Two cases are considered in Fig. 7.39. In the first case, A\nis large enough so that A + m(t) ≥0 (is nonnegative) for all values of t. In the second case, A is\nnot large enough to satisfy this condition. In the first case, the envelope (Fig. 7.39d) has the same\nshape as m(t) (although riding on a dc of magnitude A). In the second case, the envelope shape\nis not m(t), for some parts get rectified (Fig. 7.39e). Thus, we can detect the desired signal m(t)\nby detecting the envelope in the first case. In the second case, such a detection is not possible. We\nshall see that envelope detection is an extremely simple and inexpensive operation, which does\nnot require generation of a local carrier for the demodulation. But as just noted, the envelope of\nAM has the information about m(t) only if the AM signal [A+m(t)]cos ωct satisfies the condition\nA + m(t) > 0 for all t. Thus, the condition for envelope detection of an AM signal is", - "type": "text" - }, - { - "block_id": "p762-b5", - "global_id": 22353, - "bbox": [ - 246.53, - 455.26, - 490.38, - 465.63 - ], - "text": "A + m(t) ≥0\nfor allt\n(7.51)", - "type": "text" - }, - { - "block_id": "p762-b6", - "global_id": 22354, - "bbox": [ - 101.85, - 476.64, - 454.8, - 488.51 - ], - "text": "If mp is the peak amplitude (positive or negative) of m(t), then Eq. (7.51) is equivalent to", - "type": "text" - }, - { - "block_id": "p762-b7", - "global_id": 22355, - "bbox": [ - 281.68, - 498.02, - 310.05, - 509.1 - ], - "text": "A ≥mp", - "type": "text" - }, - { - "block_id": "p762-b8", - "global_id": 22356, - "bbox": [ - 101.84, - 519.71, - 490.38, - 541.74 - ], - "text": "Thus, the minimum carrier amplitude required for the viability of envelope detection is mp. This\npoint is clearly illustrated in Fig. 7.39.", - "type": "text" - }, - { - "block_id": "p762-b9", - "global_id": 22357, - "bbox": [ - 119.78, - 543.31, - 266.53, - 553.69 - ], - "text": "We define the modulation index μ as", - "type": "text" - }, - { - "block_id": "p762-b10", - "global_id": 22358, - "bbox": [ - 279.98, - 560.3, - 310.56, - 577.29 - ], - "text": "μ = mp", - "type": "text" - }, - { - "block_id": "p762-b11", - "global_id": 22359, - "bbox": [ - 302.43, - 567.42, - 490.38, - 584.36 - ], - "text": "A\n(7.52)", - "type": "text" - }, - { - "block_id": "p762-b12", - "global_id": 22360, - "bbox": [ - 101.85, - 591.35, - 490.38, - 613.67 - ], - "text": "where A is the carrier amplitude. Note that mp is a constant of the signal m(t). Because A ≥mp and\nbecause there is no upper bound on A, it follows that", - "type": "text" - }, - { - "block_id": "p762-b13", - "global_id": 22361, - "bbox": [ - 276.11, - 624.68, - 316.12, - 635.06 - ], - "text": "0 ≤μ ≤1", - "type": "text" - } - ] - }, - { - "page_num": 763, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p763-b0", - "global_id": 22362, - "bbox": [ - 228.18, - 62.89, - 516.11, - 71.98 - ], - "text": "7.7\nApplication to Communications: Amplitude Modulation\n743", - "type": "text" - }, - { - "block_id": "p763-b1", - "global_id": 22363, - "bbox": [ - 151.5, - 398.37, - 487.0, - 407.68 - ], - "text": "Figure 7.39 An AM signal (a) for two values of A (b, c) and the respective envelopes (d, e).", - "type": "text" - }, - { - "block_id": "p763-b2", - "global_id": 22364, - "bbox": [ - 127.59, - 439.0, - 487.01, - 448.97 - ], - "text": "as the required condition for the viability of demodulation of AM by an envelope detector.", - "type": "text" - }, - { - "block_id": "p763-b3", - "global_id": 22365, - "bbox": [ - 127.59, - 450.54, - 516.14, - 508.74 - ], - "text": "When A < mp, Eq. (7.52) shows that μ > 1 (overmodulation, shown in Fig. 7.39e). In this\ncase, the option of envelope detection is no longer viable. We then need to use synchronous\ndemodulation. Note that synchronous demodulation can be used for any value of μ (see\nProb. 7.7-7). The envelope detector, which is considerably simpler and less expensive than the\nsynchronous detector, can be used only when μ ≤1.", - "type": "text" - }, - { - "block_id": "p763-b4", - "global_id": 22366, - "bbox": [ - 102.51, - 543.15, - 334.52, - 555.1 - ], - "text": "EXAMPLE 7.23\nAmplitude Modulation", - "type": "text" - }, - { - "block_id": "p763-b5", - "global_id": 22367, - "bbox": [ - 128.9, - 571.35, - 502.76, - 605.64 - ], - "text": "Sketch ϕAM(t) for modulation indices of μ = 0.5 (50% modulation) and μ = 1 (100%\nmodulation), when m(t) = Bcos ωmt. This case is referred to as tone modulation because the\nmodulating signal is a pure sinusoid (or tone).", - "type": "text" - } - ] - }, - { - "page_num": 764, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p764-b0", - "global_id": 22368, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "744\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p764-b1", - "global_id": 22369, - "bbox": [ - 103.16, - 85.83, - 389.77, - 97.28 - ], - "text": "In this case, mp = B and the modulation index according to Eq. (7.52) is", - "type": "text" - }, - { - "block_id": "p764-b2", - "global_id": 22370, - "bbox": [ - 276.5, - 106.66, - 302.48, - 123.61 - ], - "text": "μ = B", - "type": "text" - }, - { - "block_id": "p764-b3", - "global_id": 22371, - "bbox": [ - 296.39, - 120.72, - 302.48, - 130.68 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p764-b4", - "global_id": 22372, - "bbox": [ - 103.16, - 138.14, - 181.34, - 148.52 - ], - "text": "Hence, B = μA and", - "type": "text" - }, - { - "block_id": "p764-b5", - "global_id": 22373, - "bbox": [ - 229.43, - 150.1, - 350.57, - 161.18 - ], - "text": "m(t) = Bcosωmt = μAcosωmt", - "type": "text" - }, - { - "block_id": "p764-b6", - "global_id": 22374, - "bbox": [ - 103.17, - 169.44, - 144.91, - 179.41 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p764-b7", - "global_id": 22375, - "bbox": [ - 182.82, - 180.98, - 397.18, - 193.35 - ], - "text": "ϕAM(t) = [A + m(t)]cosωct = A[1 + μcosωmt]cosωct", - "type": "text" - }, - { - "block_id": "p764-b8", - "global_id": 22376, - "bbox": [ - 103.17, - 199.91, - 477.02, - 222.25 - ], - "text": "The modulated signals corresponding to μ = 0.5 and μ = 1 appear in Figs. 7.40a and\n7.40b, respectively.", - "type": "text" - }, - { - "block_id": "p764-b9", - "global_id": 22377, - "bbox": [ - 231.5, - 318.6, - 435.21, - 327.6 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p764-b10", - "global_id": 22378, - "bbox": [ - 150.77, - 260.77, - 398.21, - 278.61 - ], - "text": "1 0.5 cos vmt\n1 cos vmt\n3A2", - "type": "text" - }, - { - "block_id": "p764-b11", - "global_id": 22379, - "bbox": [ - 284.49, - 284.59, - 289.37, - 292.59 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p764-b12", - "global_id": 22380, - "bbox": [ - 285.37, - 312.8, - 289.37, - 320.8 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p764-b13", - "global_id": 22381, - "bbox": [ - 276.49, - 298.44, - 289.37, - 306.74 - ], - "text": "A2", - "type": "text" - }, - { - "block_id": "p764-b14", - "global_id": 22382, - "bbox": [ - 280.49, - 256.46, - 289.37, - 264.54 - ], - "text": "2A", - "type": "text" - }, - { - "block_id": "p764-b15", - "global_id": 22383, - "bbox": [ - 159.71, - 364.39, - 383.32, - 372.69 - ], - "text": "m 0.5\nm 1", - "type": "text" - }, - { - "block_id": "p764-b16", - "global_id": 22384, - "bbox": [ - 101.77, - 379.35, - 326.72, - 388.96 - ], - "text": "Figure 7.40 Tone-modulated AM: (a) μ = 0.5 and (b) μ = 1.", - "type": "text" - }, - { - "block_id": "p764-b17", - "global_id": 22385, - "bbox": [ - 102.14, - 433.03, - 385.14, - 445.15 - ], - "text": "DEMODULATION OF AM: THE ENVELOPE DETECTOR", - "type": "text" - }, - { - "block_id": "p764-b18", - "global_id": 22386, - "bbox": [ - 101.84, - 449.18, - 490.39, - 495.01 - ], - "text": "The AM signal can be demodulated coherently by a locally generated carrier (see Prob. 7.7-7).\nSince, however, coherent, or synchronous, demodulation of AM (with μ ≤1) will defeat the very\npurpose of AM, it is rarely used in practice. We shall consider here one of the noncoherent methods\nof AM demodulation, envelope detection.†", - "type": "text" - }, - { - "block_id": "p764-b19", - "global_id": 22387, - "bbox": [ - 101.84, - 497.0, - 490.39, - 578.69 - ], - "text": "In an envelope detector, the output of the detector follows the envelope of the (modulated)\ninput signal. The circuit illustrated in Fig. 7.41a functions as an envelope detector. During the\npositive cycle of the input signal, the diode conducts and the capacitor C charges up to the peak\nvoltage of the input signal (Fig. 7.41b). As the input signal falls below this peak value, the diode\nis cut off, because the capacitor voltage (which is very nearly the peak voltage) is greater than\nthe input signal voltage, a circumstance causing the diode to open. The capacitor now discharges\nthrough the resistor R at a slow rate (with a time constant RC). During the next positive cycle,", - "type": "text" - }, - { - "block_id": "p764-b20", - "global_id": 22388, - "bbox": [ - 101.84, - 599.27, - 490.41, - 633.41 - ], - "text": "† There are also other methods of noncoherent detection. The rectifier detector consists of a rectifier followed\nby a lowpass filter. This method is also simple and almost as inexpensive as the envelope detector [4]. The\nnonlinear detector, although simple and inexpensive, results in a distorted output.", - "type": "text" - } - ] - }, - { - "page_num": 765, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p765-b0", - "global_id": 22389, - "bbox": [ - 228.18, - 62.89, - 516.11, - 71.98 - ], - "text": "7.7\nApplication to Communications: Amplitude Modulation\n745", - "type": "text" - }, - { - "block_id": "p765-b1", - "global_id": 22390, - "bbox": [ - 222.22, - 210.86, - 392.27, - 224.85 - ], - "text": "RC too large\nEnvelope", - "type": "text" - }, - { - "block_id": "p765-b2", - "global_id": 22391, - "bbox": [ - 399.79, - 194.03, - 480.22, - 202.03 - ], - "text": "Envelope detector output", - "type": "text" - }, - { - "block_id": "p765-b3", - "global_id": 22392, - "bbox": [ - 314.69, - 398.74, - 324.5, - 406.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p765-b4", - "global_id": 22393, - "bbox": [ - 341.15, - 124.79, - 447.59, - 134.33 - ], - "text": "C\nvC(t)", - "type": "text" - }, - { - "block_id": "p765-b7", - "global_id": 22394, - "bbox": [ - 365.43, - 96.73, - 396.54, - 113.73 - ], - "text": "Capacitor\ndischarge", - "type": "text" - }, - { - "block_id": "p765-b8", - "global_id": 22395, - "bbox": [ - 396.31, - 141.94, - 401.2, - 149.94 - ], - "text": "R", - "type": "text" - }, - { - "block_id": "p765-b9", - "global_id": 22396, - "bbox": [ - 191.6, - 126.65, - 225.6, - 134.65 - ], - "text": "AM signal", - "type": "text" - }, - { - "block_id": "p765-b10", - "global_id": 22397, - "bbox": [ - 499.32, - 296.25, - 501.55, - 304.25 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p765-b11", - "global_id": 22398, - "bbox": [ - 315.15, - 173.03, - 324.03, - 181.03 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p765-b12", - "global_id": 22399, - "bbox": [ - 127.59, - 413.44, - 342.0, - 422.68 - ], - "text": "Figure 7.41 Demodulation by means of envelope detector.", - "type": "text" - }, - { - "block_id": "p765-b13", - "global_id": 22400, - "bbox": [ - 127.59, - 457.45, - 516.16, - 503.28 - ], - "text": "the same drama repeats. When the input signal becomes greater than the capacitor voltage, the\ndiode conducts again. The capacitor again charges to the peak value of this (new) cycle. As the\ninput voltage falls below the new peak value, the diode cuts off again and the capacitor discharges\nslowly during the cutoff period, a process that changes the capacitor voltage very slightly.", - "type": "text" - }, - { - "block_id": "p765-b14", - "global_id": 22401, - "bbox": [ - 127.59, - 505.27, - 516.14, - 611.96 - ], - "text": "In this manner, during each positive cycle, the capacitor charges up to the peak voltage of\nthe input signal and then decays slowly until the next positive cycle. Thus, the output voltage\nvC(t) follows closely the envelope of the input. The capacitor discharge between positive peaks,\nhowever, causes a ripple signal of frequency ωc in the output. This ripple can be reduced by\nincreasing the time constant RC so that the capacitor discharges very little between the positive\npeaks (RC ≫1/ωc). Making RC too large, however, would make it impossible for the capacitor\nvoltage to follow the envelope (see Fig. 7.41b). Thus, RC should be large in comparison to 1/ωc\nbut small in comparison to 1/2πB, where B is the highest frequency in m(t). Incidentally, these\ntwo conditions also require that ωc ≫2πB, a condition necessary for a well-defined envelope.", - "type": "text" - }, - { - "block_id": "p765-b15", - "global_id": 22402, - "bbox": [ - 127.59, - 612.45, - 516.14, - 634.79 - ], - "text": "The envelope-detector output vC(t) is A + m(t) plus a ripple of frequency ωc. The dc term A\ncan be blocked out by a capacitor or a simple RC highpass filter. The ripple is reduced further by", - "type": "text" - } - ] - }, - { - "page_num": 766, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p766-b0", - "global_id": 22403, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "746\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p766-b1", - "global_id": 22404, - "bbox": [ - 101.84, - 85.72, - 490.39, - 107.74 - ], - "text": "another (lowpass) RC filter. In the case of audio signals, the speakers also act as lowpass filters,\nwhich further enhances suppression of the high-frequency ripple.", - "type": "text" - }, - { - "block_id": "p766-b2", - "global_id": 22405, - "bbox": [ - 101.84, - 139.3, - 325.47, - 151.25 - ], - "text": "7.7-3 Single-Sideband Modulation (SSB)", - "type": "text" - }, - { - "block_id": "p766-b3", - "global_id": 22406, - "bbox": [ - 101.84, - 156.97, - 490.43, - 215.16 - ], - "text": "Now consider the baseband spectrum M(ω) (Fig. 7.42a) and the spectrum of the DSB-SC\nmodulated signal m(t)cos ωct (Fig. 7.42b). The DSB spectrum in Fig. 7.42b has two sidebands:\nthe upper and the lower (USB and LSB), both containing complete information on M(ω) [see\nEq. (7.12)]. Clearly, it is redundant to transmit both sidebands, a process that requires twice the\nbandwidth of the baseband signal. A scheme in which only one sideband is transmitted is known", - "type": "text" - }, - { - "block_id": "p766-b4", - "global_id": 22407, - "bbox": [ - 304.81, - 576.53, - 453.62, - 585.72 - ], - "text": "0\nv", - "type": "text" - }, - { - "block_id": "p766-b5", - "global_id": 22408, - "bbox": [ - 389.4, - 506.37, - 394.74, - 514.37 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p766-b6", - "global_id": 22409, - "bbox": [ - 390.93, - 432.41, - 396.26, - 440.41 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p766-b7", - "global_id": 22410, - "bbox": [ - 389.4, - 357.98, - 394.74, - 365.98 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p766-b8", - "global_id": 22411, - "bbox": [ - 352.25, - 275.47, - 357.59, - 283.47 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p766-b9", - "global_id": 22412, - "bbox": [ - 304.42, - 506.93, - 308.42, - 514.93 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p766-b10", - "global_id": 22413, - "bbox": [ - 304.42, - 431.36, - 308.42, - 439.36 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p766-b11", - "global_id": 22414, - "bbox": [ - 304.42, - 356.69, - 308.42, - 364.69 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p766-b12", - "global_id": 22415, - "bbox": [ - 304.52, - 321.49, - 317.09, - 329.79 - ], - "text": "A2", - "type": "text" - }, - { - "block_id": "p766-b13", - "global_id": 22416, - "bbox": [ - 304.52, - 395.95, - 317.09, - 404.25 - ], - "text": "A2", - "type": "text" - }, - { - "block_id": "p766-b14", - "global_id": 22417, - "bbox": [ - 304.52, - 471.52, - 317.09, - 479.81 - ], - "text": "A2", - "type": "text" - }, - { - "block_id": "p766-b15", - "global_id": 22418, - "bbox": [ - 313.47, - 541.41, - 326.04, - 549.71 - ], - "text": "A2", - "type": "text" - }, - { - "block_id": "p766-b16", - "global_id": 22419, - "bbox": [ - 241.69, - 357.78, - 360.12, - 367.54 - ], - "text": "vc\nvc", - "type": "text" - }, - { - "block_id": "p766-b17", - "global_id": 22420, - "bbox": [ - 241.69, - 432.22, - 360.12, - 441.98 - ], - "text": "vc\nvc", - "type": "text" - }, - { - "block_id": "p766-b18", - "global_id": 22421, - "bbox": [ - 241.69, - 506.18, - 256.36, - 515.94 - ], - "text": "vc", - "type": "text" - }, - { - "block_id": "p766-b19", - "global_id": 22422, - "bbox": [ - 186.18, - 577.53, - 415.51, - 587.29 - ], - "text": "2vc\n2vc", - "type": "text" - }, - { - "block_id": "p766-b20", - "global_id": 22423, - "bbox": [ - 352.12, - 506.37, - 360.12, - 515.94 - ], - "text": "vc", - "type": "text" - }, - { - "block_id": "p766-b21", - "global_id": 22424, - "bbox": [ - 204.93, - 319.51, - 390.28, - 335.86 - ], - "text": "USB\nLSB\nUSB\nLSB", - "type": "text" - }, - { - "block_id": "p766-b22", - "global_id": 22425, - "bbox": [ - 306.52, - 275.96, - 310.52, - 283.96 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p766-b23", - "global_id": 22426, - "bbox": [ - 274.72, - 238.91, - 316.91, - 248.42 - ], - "text": "M(v)\nA", - "type": "text" - }, - { - "block_id": "p766-b24", - "global_id": 22427, - "bbox": [ - 258.61, - 276.07, - 342.76, - 284.36 - ], - "text": "2pB\n2pB", - "type": "text" - }, - { - "block_id": "p766-b25", - "global_id": 22428, - "bbox": [ - 298.47, - 290.43, - 307.35, - 298.43 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p766-b26", - "global_id": 22429, - "bbox": [ - 298.08, - 373.01, - 307.73, - 381.01 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p766-b27", - "global_id": 22430, - "bbox": [ - 298.47, - 448.79, - 307.35, - 456.79 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p766-b28", - "global_id": 22431, - "bbox": [ - 298.24, - 520.45, - 307.57, - 528.45 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p766-b29", - "global_id": 22432, - "bbox": [ - 298.47, - 592.81, - 307.35, - 600.81 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p766-b30", - "global_id": 22433, - "bbox": [ - 125.76, - 607.3, - 490.39, - 629.31 - ], - "text": "Figure 7.42 Spectra for single-sideband transmission: (a) baseband, (b) DSB, (c) USB, (d)\nLSB, and (e) synchronously demodulated signal.", - "type": "text" - } - ] - }, - { - "page_num": 767, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p767-b0", - "global_id": 22434, - "bbox": [ - 228.18, - 62.89, - 516.11, - 71.98 - ], - "text": "7.7\nApplication to Communications: Amplitude Modulation\n747", - "type": "text" - }, - { - "block_id": "p767-b1", - "global_id": 22435, - "bbox": [ - 127.59, - 85.72, - 516.13, - 107.74 - ], - "text": "as single-sideband (SSB) transmission, which requires only half the bandwidth of the DSB signal.\nThus, we transmit only the upper sidebands (Fig. 7.42c) or only the lower sidebands (Fig. 7.42d).", - "type": "text" - }, - { - "block_id": "p767-b2", - "global_id": 22436, - "bbox": [ - 127.59, - 109.73, - 516.15, - 191.42 - ], - "text": "An SSB signal can be coherently (synchronously) demodulated. For example, multiplication\nof a USB signal (Fig. 7.42c) by 2cos ωct shifts its spectrum to the left and to the right by ωc,\nyielding the spectrum in Fig. 7.42e. Lowpass filtering of this signal yields the desired baseband\nsignal. The case is similar with an LSB signal. Hence, demodulation of SSB signals is identical\nto that of DSB-SC signals, and the synchronous demodulator in Fig. 7.38a can demodulate SSB\nsignals. Note that we are talking of SSB signals without an additional carrier. Hence, they are\nsuppressed-carrier signals (SSB-SC).", - "type": "text" - }, - { - "block_id": "p767-b3", - "global_id": 22437, - "bbox": [ - 102.51, - 247.09, - 366.38, - 259.04 - ], - "text": "EXAMPLE 7.24\nSingle-Sideband Modulation", - "type": "text" - }, - { - "block_id": "p767-b4", - "global_id": 22438, - "bbox": [ - 128.9, - 275.3, - 502.74, - 309.58 - ], - "text": "Find the USB (upper sideband) and LSB (lower sideband) signals when m(t) = cos ωmt. Sketch\ntheir spectra, and show that these SSB signals can be demodulated using the synchronous\ndemodulator in Fig. 7.38a.", - "type": "text" - }, - { - "block_id": "p767-b5", - "global_id": 22439, - "bbox": [ - 128.9, - 332.49, - 267.82, - 342.45 - ], - "text": "The DSB-SC signal for this case is", - "type": "text" - }, - { - "block_id": "p767-b6", - "global_id": 22440, - "bbox": [ - 148.33, - 352.66, - 349.01, - 365.95 - ], - "text": "ϕDSB-SC(t) = m(t)cos ωct = cos ωmtcos ωct\n= 1", - "type": "text" - }, - { - "block_id": "p767-b7", - "global_id": 22441, - "bbox": [ - 345.53, - 354.0, - 483.35, - 367.19 - ], - "text": "2[cos(ωc −ωm)t + cos(ωc + ωm)t]", - "type": "text" - }, - { - "block_id": "p767-b8", - "global_id": 22442, - "bbox": [ - 128.91, - 375.92, - 502.77, - 470.68 - ], - "text": "As pointed out in Ex. 7.21, the terms (1/2)cos(ωc + ωm)t and (1/2)cos(ωc −ωm)t\nrepresent the upper and lower sidebands, respectively. The spectra of the upper and lower\nsidebands are given in Figs. 7.43a and 7.43b. Observe that these spectra can be obtained\nfrom the DSB-SC spectrum in Fig. 7.37b by using a proper filter to suppress the undesired\nsidebands. For instance, the USB signal in Fig. 7.43a can be obtained by passing the DSB-SC\nsignal (Fig. 7.37b) through a highpass filter of cutoff frequency ωc. Similarly, the LSB signal\nin Fig. 7.43b can be obtained by passing the DSB-SC signal through a lowpass filter of cutoff\nfrequency ωc.", - "type": "text" - }, - { - "block_id": "p767-b9", - "global_id": 22443, - "bbox": [ - 128.91, - 471.56, - 502.77, - 493.89 - ], - "text": "If we apply the LSB signal (1/2)cos(ωc −ωm)t to the synchronous demodulator in\nFig. 7.38a, the multiplier output is", - "type": "text" - }, - { - "block_id": "p767-b10", - "global_id": 22444, - "bbox": [ - 190.99, - 504.08, - 222.34, - 515.71 - ], - "text": "e(t) = 1", - "type": "text" - }, - { - "block_id": "p767-b11", - "global_id": 22445, - "bbox": [ - 218.85, - 504.08, - 440.69, - 519.34 - ], - "text": "2 cos(ωc −ωm)tcos ωct = 1\n4[cos ωmt + cos(2ωc −ωm)t]", - "type": "text" - }, - { - "block_id": "p767-b12", - "global_id": 22446, - "bbox": [ - 128.9, - 527.35, - 502.76, - 573.59 - ], - "text": "The term (1/4)cos(2ωc−ωm)t is suppressed by the lowpass filter, producing the desired output\n(1/4)cos ωmt [which is m(t)/4]. The spectrum of this term is π[δ(ω +ωm)+δ(ω −ωm)]/4, as\ndepicted in Fig. 7.43c. In the same way, we can show that the USB signal can be demodulated\nby the synchronous demodulator.", - "type": "text" - }, - { - "block_id": "p767-b13", - "global_id": 22447, - "bbox": [ - 128.91, - 575.17, - 502.78, - 621.41 - ], - "text": "In the frequency domain, demodulation (multiplication by cos ωct) amounts to shifting\nthe LSB spectrum (Fig. 7.43b) to the left and the right by ωc (times 0.5) and then suppressing\nthe high frequency, as illustrated in Fig. 7.43c. The resulting spectrum represents the desired\nsignal (1/4)m(t).", - "type": "text" - } - ] - }, - { - "page_num": 768, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p768-b0", - "global_id": 22448, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "748\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p768-b1", - "global_id": 22449, - "bbox": [ - 262.1, - 209.5, - 266.1, - 217.5 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p768-b2", - "global_id": 22450, - "bbox": [ - 255.2, - 225.67, - 265.0, - 233.67 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p768-b3", - "global_id": 22451, - "bbox": [ - 262.1, - 132.95, - 447.05, - 141.24 - ], - "text": "0\nv", - "type": "text" - }, - { - "block_id": "p768-b4", - "global_id": 22452, - "bbox": [ - 441.71, - 209.79, - 447.05, - 217.79 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p768-b5", - "global_id": 22453, - "bbox": [ - 441.71, - 284.95, - 447.05, - 292.95 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p768-b6", - "global_id": 22454, - "bbox": [ - 125.37, - 133.13, - 344.25, - 143.01 - ], - "text": "vc\nvc\n(vc vm)", - "type": "text" - }, - { - "block_id": "p768-b7", - "global_id": 22455, - "bbox": [ - 191.6, - 210.56, - 232.04, - 220.32 - ], - "text": "(vc vm)", - "type": "text" - }, - { - "block_id": "p768-b8", - "global_id": 22456, - "bbox": [ - 354.96, - 133.25, - 383.29, - 143.01 - ], - "text": "vc vm", - "type": "text" - }, - { - "block_id": "p768-b9", - "global_id": 22457, - "bbox": [ - 172.25, - 209.61, - 186.92, - 219.37 - ], - "text": "vc", - "type": "text" - }, - { - "block_id": "p768-b10", - "global_id": 22458, - "bbox": [ - 172.55, - 285.53, - 343.88, - 295.29 - ], - "text": "vc\nvm\nvm\nvc", - "type": "text" - }, - { - "block_id": "p768-b11", - "global_id": 22459, - "bbox": [ - 335.88, - 209.8, - 343.88, - 219.37 - ], - "text": "vc", - "type": "text" - }, - { - "block_id": "p768-b12", - "global_id": 22460, - "bbox": [ - 98.17, - 285.49, - 417.0, - 296.24 - ], - "text": "(2vc vm)\n2vc vm", - "type": "text" - }, - { - "block_id": "p768-b13", - "global_id": 22461, - "bbox": [ - 300.25, - 209.73, - 328.58, - 219.49 - ], - "text": "vc vm", - "type": "text" - }, - { - "block_id": "p768-b14", - "global_id": 22462, - "bbox": [ - 255.66, - 149.15, - 264.54, - 157.15 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p768-b15", - "global_id": 22463, - "bbox": [ - 262.1, - 284.72, - 266.1, - 292.72 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p768-b16", - "global_id": 22464, - "bbox": [ - 255.66, - 301.14, - 264.54, - 309.14 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p768-b17", - "global_id": 22465, - "bbox": [ - 113.88, - 254.49, - 407.5, - 262.79 - ], - "text": "p4\np4\np4", - "type": "text" - }, - { - "block_id": "p768-b18", - "global_id": 22466, - "bbox": [ - 324.38, - 170.63, - 337.72, - 178.92 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p768-b19", - "global_id": 22467, - "bbox": [ - 364.02, - 93.94, - 377.36, - 102.24 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p768-b20", - "global_id": 22468, - "bbox": [ - 94.2, - 315.31, - 468.05, - 337.64 - ], - "text": "Figure 7.43 Single-sideband spectra for m(t) = cos ωmt: (a) USB, (b) LSB, and (c)\nsynchronously demodulated LSB signal.", - "type": "text" - }, - { - "block_id": "p768-b21", - "global_id": 22469, - "bbox": [ - 101.84, - 382.71, - 490.42, - 432.73 - ], - "text": "GENERATION OF SSB SIGNALS\nTwo methods are commonly used to generate SSB signals. The selective-filtering method uses\nsharp cutoff filters to eliminate the undesired sideband, and the second method uses phase-shifting\nnetworks to achieve the same goal [4].† We shall consider here only the first method.", - "type": "text" - }, - { - "block_id": "p768-b22", - "global_id": 22470, - "bbox": [ - 101.85, - 434.72, - 490.4, - 468.59 - ], - "text": "Selective filtering is the most commonly used method of generating SSB signals. In this\nmethod, a DSB-SC signal is passed through a sharp cutoff filter to eliminate the undesired\nsideband.", - "type": "text" - }, - { - "block_id": "p768-b23", - "global_id": 22471, - "bbox": [ - 101.84, - 470.17, - 490.41, - 576.89 - ], - "text": "To obtain the USB, the filter should pass all components above ωc unattenuated and\ncompletely suppress all components below ωc. Such an operation requires an ideal filter, which is\nunrealizable. It can, however, be realized closely if there is some separation between the passband\nand the stopband. Fortunately, the voice signal provides this condition, because its spectrum shows\nlittle power content at the origin (Fig. 7.44). Moreover, articulation tests show that for speech\nsignals, frequency components below 300 Hz are not important. In other words, we may suppress\nall speech components below 300 Hz without appreciably affecting intelligibility.‡ Thus, filtering\nof the unwanted sideband becomes relatively easy for speech signals because we have a 600 Hz\ntransition region around the cutoff frequency ωc. For some signals, which have considerable power", - "type": "text" - }, - { - "block_id": "p768-b24", - "global_id": 22472, - "bbox": [ - 101.84, - 598.98, - 490.38, - 633.41 - ], - "text": "† Yet another method, known as Weaver’s method, is also used to generate SSB signals.\n‡ Similarly, suppression of components of a speech signal above 3500 Hz causes no appreciable change in\nintelligibility.", - "type": "text" - } - ] - }, - { - "page_num": 769, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p769-b0", - "global_id": 22473, - "bbox": [ - 309.24, - 62.89, - 516.12, - 71.98 - ], - "text": "7.8\nData Truncation: Window Functions\n749", - "type": "text" - }, - { - "block_id": "p769-b1", - "global_id": 22474, - "bbox": [ - 136.49, - 163.74, - 306.68, - 175.74 - ], - "text": "200\n3200\n0\nf(Hz)", - "type": "text" - }, - { - "block_id": "p769-b2", - "global_id": 22475, - "bbox": [ - 130.28, - 101.21, - 138.28, - 149.86 - ], - "text": "Relative power", - "type": "text" - }, - { - "block_id": "p769-b3", - "global_id": 22476, - "bbox": [ - 326.55, - 167.79, - 434.22, - 177.03 - ], - "text": "Figure 7.44 Voice spectrum.", - "type": "text" - }, - { - "block_id": "p769-b4", - "global_id": 22477, - "bbox": [ - 127.59, - 200.4, - 516.16, - 282.51 - ], - "text": "at low frequencies (around ω = 0), SSB techniques cause considerable distortion. Such is the case\nwith video signals. Consequently, for video signals, instead of SSB, we use another technique, the\nvestigial sideband (VSB), which is a compromise between SSB and DSB. It inherits the advantages\nof SSB and DSB but avoids their disadvantages at a cost of slightly increased bandwidth. VSB\nsignals are relatively easy to generate, and their bandwidth is only slightly (typically 25%) greater\nthan that of SSB signals. In VSB signals, instead of rejecting one sideband completely (as in SSB),\nwe accept a gradual cutoff from one sideband [4].", - "type": "text" - }, - { - "block_id": "p769-b5", - "global_id": 22478, - "bbox": [ - 127.59, - 307.86, - 342.24, - 319.81 - ], - "text": "7.7-4 Frequency-Division Multiplexing", - "type": "text" - }, - { - "block_id": "p769-b6", - "global_id": 22479, - "bbox": [ - 127.59, - 325.94, - 516.15, - 455.46 - ], - "text": "Signal multiplexing allows transmission of several signals on the same channel. Later, in Ch. 8\n(Sec. 8.2-2), we shall discuss time-division multiplexing (TDM), where several signals time-share\nthe same channel, such as a cable or an optical fiber. In frequency-division multiplexing (FDM),\nthe use of modulation, as illustrated in Fig. 7.45, makes several signals share the band of the\nsame channel. Each signal is modulated by a different carrier frequency. The various carriers\nare adequately separated to avoid overlap (or interference) between the spectra of various\nmodulated signals. These carriers are referred to as subcarriers. Each signal may use a different\nkind of modulation, for example, DSB-SC, AM, SSB-SC, VSB-SC, or even other forms of\nmodulation, not discussed here [such as FM (frequency modulation) or PM (phase modulation)].\nThe modulated-signal spectra may be separated by a small guard band to avoid interference and to\nfacilitate signal separation at the receiver.", - "type": "text" - }, - { - "block_id": "p769-b7", - "global_id": 22480, - "bbox": [ - 127.59, - 457.45, - 516.12, - 491.33 - ], - "text": "When all the modulated spectra are added, we have a composite signal that may be considered\nto be a new baseband signal. Sometimes, this composite baseband signal may be used to further\nmodulate a high-frequency (radio frequency, or RF) carrier for the purpose of transmission.", - "type": "text" - }, - { - "block_id": "p769-b8", - "global_id": 22481, - "bbox": [ - 127.59, - 493.32, - 516.16, - 539.15 - ], - "text": "At the receiver, the incoming signal is first demodulated by the RF carrier to retrieve the\ncomposite baseband, which is then bandpass-filtered to separate the modulated signals. Then each\nmodulated signal is individually demodulated by an appropriate subcarrier to obtain all the basic\nbaseband signals.", - "type": "text" - }, - { - "block_id": "p769-b9", - "global_id": 22482, - "bbox": [ - 127.94, - 569.02, - 431.16, - 582.97 - ], - "text": "7.8 DATA TRUNCATION: WINDOW FUNCTIONS", - "type": "text" - }, - { - "block_id": "p769-b10", - "global_id": 22483, - "bbox": [ - 127.59, - 588.96, - 516.14, - 634.79 - ], - "text": "We often need to truncate data in diverse situations from numerical computations to filter design.\nFor example, if we need to compute numerically the Fourier transform of some signal, say, e−tu(t),\nwe will have to truncate the signal e−tu(t) beyond a sufficiently large value of t (typically five\ntime constants and above). The reason is that in numerical computations, we have to deal with", - "type": "text" - } - ] - }, - { - "page_num": 770, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p770-b0", - "global_id": 22484, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "750\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p770-b1", - "global_id": 22485, - "bbox": [ - 129.7, - 345.82, - 146.02, - 355.37 - ], - "text": "mn(t)", - "type": "text" - }, - { - "block_id": "p770-b2", - "global_id": 22486, - "bbox": [ - 129.7, - 176.17, - 146.02, - 185.78 - ], - "text": "m1(t)", - "type": "text" - }, - { - "block_id": "p770-b3", - "global_id": 22487, - "bbox": [ - 129.7, - 257.23, - 146.02, - 266.84 - ], - "text": "m2(t)", - "type": "text" - }, - { - "block_id": "p770-b4", - "global_id": 22488, - "bbox": [ - 470.04, - 399.55, - 486.37, - 409.16 - ], - "text": "m2(t)", - "type": "text" - }, - { - "block_id": "p770-b5", - "global_id": 22489, - "bbox": [ - 470.04, - 328.56, - 486.37, - 338.16 - ], - "text": "m1(t)", - "type": "text" - }, - { - "block_id": "p770-b6", - "global_id": 22490, - "bbox": [ - 470.04, - 467.85, - 486.37, - 477.39 - ], - "text": "mn(t)", - "type": "text" - }, - { - "block_id": "p770-b7", - "global_id": 22491, - "bbox": [ - 158.26, - 181.18, - 192.03, - 189.18 - ], - "text": "Modulator", - "type": "text" - }, - { - "block_id": "p770-b8", - "global_id": 22492, - "bbox": [ - 173.15, - 190.18, - 177.15, - 198.18 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p770-b9", - "global_id": 22493, - "bbox": [ - 158.26, - 262.5, - 192.03, - 270.5 - ], - "text": "Modulator", - "type": "text" - }, - { - "block_id": "p770-b10", - "global_id": 22494, - "bbox": [ - 173.15, - 271.5, - 177.15, - 279.5 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p770-b11", - "global_id": 22495, - "bbox": [ - 158.26, - 350.78, - 192.03, - 358.78 - ], - "text": "Modulator", - "type": "text" - }, - { - "block_id": "p770-b12", - "global_id": 22496, - "bbox": [ - 173.15, - 359.7, - 177.15, - 367.7 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p770-b13", - "global_id": 22497, - "bbox": [ - 415.42, - 333.01, - 457.64, - 341.01 - ], - "text": "Demodulator", - "type": "text" - }, - { - "block_id": "p770-b14", - "global_id": 22498, - "bbox": [ - 434.53, - 342.01, - 438.53, - 350.01 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p770-b15", - "global_id": 22499, - "bbox": [ - 415.42, - 403.8, - 457.64, - 411.8 - ], - "text": "Demodulator", - "type": "text" - }, - { - "block_id": "p770-b16", - "global_id": 22500, - "bbox": [ - 434.53, - 412.8, - 438.53, - 420.8 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p770-b17", - "global_id": 22501, - "bbox": [ - 415.42, - 473.23, - 457.64, - 481.23 - ], - "text": "Demodulator", - "type": "text" - }, - { - "block_id": "p770-b18", - "global_id": 22502, - "bbox": [ - 434.53, - 482.15, - 438.53, - 490.15 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p770-b19", - "global_id": 22503, - "bbox": [ - 170.24, - 404.83, - 180.05, - 412.83 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p770-b20", - "global_id": 22504, - "bbox": [ - 361.18, - 308.81, - 391.85, - 316.81 - ], - "text": "Bandpass", - "type": "text" - }, - { - "block_id": "p770-b21", - "global_id": 22505, - "bbox": [ - 368.74, - 317.81, - 384.29, - 325.81 - ], - "text": "filter", - "type": "text" - }, - { - "block_id": "p770-b22", - "global_id": 22506, - "bbox": [ - 261.01, - 262.05, - 293.9, - 279.05 - ], - "text": "RF\nmodulator", - "type": "text" - }, - { - "block_id": "p770-b23", - "global_id": 22507, - "bbox": [ - 279.19, - 404.06, - 319.63, - 421.06 - ], - "text": "RF\ndemodulator", - "type": "text" - }, - { - "block_id": "p770-b25", - "global_id": 22508, - "bbox": [ - 233.25, - 121.26, - 241.58, - 130.89 - ], - "text": "v1", - "type": "text" - }, - { - "block_id": "p770-b26", - "global_id": 22509, - "bbox": [ - 185.68, - 216.84, - 194.02, - 226.47 - ], - "text": "v1", - "type": "text" - }, - { - "block_id": "p770-b27", - "global_id": 22510, - "bbox": [ - 185.68, - 297.78, - 194.02, - 307.41 - ], - "text": "v2", - "type": "text" - }, - { - "block_id": "p770-b28", - "global_id": 22511, - "bbox": [ - 372.35, - 335.96, - 380.68, - 345.59 - ], - "text": "v1", - "type": "text" - }, - { - "block_id": "p770-b29", - "global_id": 22512, - "bbox": [ - 372.35, - 406.76, - 380.68, - 416.39 - ], - "text": "v2", - "type": "text" - }, - { - "block_id": "p770-b30", - "global_id": 22513, - "bbox": [ - 185.68, - 384.61, - 194.02, - 394.17 - ], - "text": "vn", - "type": "text" - }, - { - "block_id": "p770-b31", - "global_id": 22514, - "bbox": [ - 446.76, - 368.91, - 455.1, - 378.54 - ], - "text": "v1", - "type": "text" - }, - { - "block_id": "p770-b32", - "global_id": 22515, - "bbox": [ - 446.76, - 437.62, - 455.1, - 447.25 - ], - "text": "v2", - "type": "text" - }, - { - "block_id": "p770-b33", - "global_id": 22516, - "bbox": [ - 446.76, - 506.78, - 455.1, - 516.35 - ], - "text": "vn", - "type": "text" - }, - { - "block_id": "p770-b34", - "global_id": 22517, - "bbox": [ - 372.35, - 476.18, - 380.68, - 485.75 - ], - "text": "vn", - "type": "text" - }, - { - "block_id": "p770-b35", - "global_id": 22518, - "bbox": [ - 279.91, - 121.26, - 399.36, - 130.89 - ], - "text": "v2\nv3\nvn", - "type": "text" - }, - { - "block_id": "p770-b36", - "global_id": 22519, - "bbox": [ - 290.95, - 153.72, - 299.83, - 161.72 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p770-b37", - "global_id": 22520, - "bbox": [ - 152.59, - 122.32, - 430.39, - 130.62 - ], - "text": "v\n0", - "type": "text" - }, - { - "block_id": "p770-b38", - "global_id": 22521, - "bbox": [ - 382.95, - 526.74, - 391.83, - 534.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p770-b39", - "global_id": 22522, - "bbox": [ - 373.98, - 442.1, - 379.98, - 451.4 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p770-b40", - "global_id": 22523, - "bbox": [ - 172.61, - 322.95, - 178.61, - 332.25 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p770-b41", - "global_id": 22524, - "bbox": [ - 433.99, - 454.27, - 439.99, - 463.57 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p770-b42", - "global_id": 22525, - "bbox": [ - 373.98, - 371.82, - 379.98, - 381.12 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p770-b43", - "global_id": 22526, - "bbox": [ - 433.99, - 385.02, - 439.99, - 394.32 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p770-b44", - "global_id": 22527, - "bbox": [ - 125.76, - 541.37, - 481.15, - 550.68 - ], - "text": "Figure 7.45 Frequency-division multiplexing: (a) FDM spectrum (b) transmitter, and (c) receiver.", - "type": "text" - }, - { - "block_id": "p770-b45", - "global_id": 22528, - "bbox": [ - 101.84, - 576.59, - 490.38, - 635.49 - ], - "text": "data of finite duration. Similarly, the impulse response h(t) of an ideal lowpass filter is noncausal\nand approaches zero asymptotically as |t| →∞. For a practical design, we may want to truncate\nh(t) beyond a sufficiently large value of |t| to make h(t) causal and of finite duration. In signal\nsampling, to eliminate aliasing, we must use an antialiasing filter to truncate the signal spectrum\nbeyond the half-sampling frequency ωs/2. Again, we may want to synthesize a periodic signal", - "type": "text" - } - ] - }, - { - "page_num": 771, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p771-b0", - "global_id": 22529, - "bbox": [ - 309.24, - 62.89, - 516.12, - 71.98 - ], - "text": "7.8\nData Truncation: Window Functions\n751", - "type": "text" - }, - { - "block_id": "p771-b1", - "global_id": 22530, - "bbox": [ - 127.59, - 85.72, - 516.14, - 143.6 - ], - "text": "by adding the first n harmonics and truncating all the higher harmonics. These examples show\nthat data truncation can occur in both time and frequency domains. On the surface, truncation\nappears to be a simple problem of cutting off the data at a point at which values are deemed\nto be sufficiently small. Unfortunately, this is not the case. Simple truncation can cause some\nunsuspected problems.", - "type": "text" - }, - { - "block_id": "p771-b2", - "global_id": 22531, - "bbox": [ - 127.89, - 158.15, - 245.17, - 170.27 - ], - "text": "WINDOW FUNCTIONS", - "type": "text" - }, - { - "block_id": "p771-b3", - "global_id": 22532, - "bbox": [ - 127.59, - 174.3, - 516.14, - 256.0 - ], - "text": "Truncation operation may be regarded as multiplying a signal of a large width by a window\nfunction of a smaller (finite) width. Simple truncation amounts to using a rectangular window\nwR(t) (shown later in Fig. 7.48a) in which we assign unit weight to all the data within the window\nwidth (|t| < T/2), and assign zero weight to all the data lying outside the window (|t| > T/2). It\nis also possible to use a window in which the weight assigned to the data within the window may\nnot be constant. In a triangular window wT(t), for example, the weight assigned to data decreases\nlinearly over the window width (shown later in Fig. 7.48b).", - "type": "text" - }, - { - "block_id": "p771-b4", - "global_id": 22533, - "bbox": [ - 127.6, - 257.57, - 516.13, - 280.61 - ], - "text": "Consider a signal x(t) and a window function w(t). If x(t) ⇐⇒X(ω) and w(t) ⇐⇒W(ω), and\nif the windowed function xw(t) ⇐⇒Xw(ω), then", - "type": "text" - }, - { - "block_id": "p771-b5", - "global_id": 22534, - "bbox": [ - 211.46, - 288.68, - 376.3, - 306.33 - ], - "text": "xw(t) = x(t)w(t)\nand\nXw(ω) = 1", - "type": "text" - }, - { - "block_id": "p771-b6", - "global_id": 22535, - "bbox": [ - 367.84, - 295.25, - 432.26, - 312.71 - ], - "text": "2π X(ω)∗W(ω)", - "type": "text" - }, - { - "block_id": "p771-b7", - "global_id": 22536, - "bbox": [ - 127.59, - 319.55, - 516.14, - 473.39 - ], - "text": "According to the width property of convolution, it follows that the width of Xw(ω) equals the sum\nof the widths of X(ω) and W(ω). Thus, truncation of a signal increases its bandwidth by the amount\nof bandwidth of w(t). Clearly, the truncation of a signal causes its spectrum to spread (or smear)\nby the amount of the bandwidth of w(t). Recall that the signal bandwidth is inversely proportional\nto the signal duration (width). Hence, the wider the window, the smaller its bandwidth, and the\nsmaller the spectral spreading. This result is predictable because a wider window means that we\nare accepting more data (closer approximation), which should cause smaller distortion (smaller\nspectral spreading). Smaller window width (poorer approximation) causes more spectral spreading\n(more distortion). In addition, since W(ω) is really not strictly bandlimited and its spectrum →0\nonly asymptotically, the spectrum of Xw(ω) →0 asymptotically also at the same rate as that of\nW(ω), even if X(ω) is, in fact, strictly bandlimited. Thus, windowing causes the spectrum of X(ω)\nto spread into the band where it is supposed to be zero. This effect is called leakage. The following\nexample clarifies these twin effects of spectral spreading and leakage.", - "type": "text" - }, - { - "block_id": "p771-b8", - "global_id": 22537, - "bbox": [ - 127.59, - 474.97, - 516.14, - 580.99 - ], - "text": "Let us consider x(t) = cos ω0t and a rectangular window wR(t) = rect(t/T), illustrated in\nFig. 7.46b. The reason for selecting a sinusoid for x(t) is that its spectrum consists of spectral\nlines of zero width (Fig. 7.46a). Hence, this choice will make the effect of spectral spreading and\nleakage easily discernible. The spectrum of the truncated signal xw(t) is the convolution of the\ntwo impulses of X(ω) with the sinc spectrum of the window function. Because the convolution\nof any function with an impulse is the function itself (shifted at the location of the impulse), the\nresulting spectrum of the truncated signal is 1/2π times the two sinc pulses at ±ω0, as depicted\nin Fig. 7.46c (also see Fig. 7.26). Comparison of spectra X(ω) and Xw(ω) reveals the effects of\ntruncation. These are:", - "type": "text" - }, - { - "block_id": "p771-b9", - "global_id": 22538, - "bbox": [ - 144.52, - 588.55, - 516.14, - 634.79 - ], - "text": "1. The spectral lines of X(ω) have zero width. But the truncated signal is spread out by 2π/T\nabout each spectral line. The amount of spread is equal to the width of the mainlobe of\nthe window spectrum. One effect of this spectral spreading (or smearing) is that if x(t) has\ntwo spectral components of frequencies differing by less than 4π/T rad/s (2/T Hz), they", - "type": "text" - } - ] - }, - { - "page_num": 772, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p772-b0", - "global_id": 22539, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "752\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p772-b1", - "global_id": 22540, - "bbox": [ - 277.47, - 352.67, - 286.35, - 360.67 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p772-b2", - "global_id": 22541, - "bbox": [ - 277.47, - 155.37, - 286.35, - 163.37 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p772-b3", - "global_id": 22542, - "bbox": [ - 166.99, - 87.85, - 178.42, - 95.93 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p772-b4", - "global_id": 22543, - "bbox": [ - 252.07, - 123.64, - 254.3, - 131.64 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p772-b5", - "global_id": 22544, - "bbox": [ - 205.82, - 88.41, - 229.4, - 98.04 - ], - "text": "cos v0t", - "type": "text" - }, - { - "block_id": "p772-b6", - "global_id": 22545, - "bbox": [ - 178.93, - 133.5, - 182.93, - 141.5 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p772-b7", - "global_id": 22546, - "bbox": [ - 358.15, - 90.46, - 441.12, - 102.17 - ], - "text": "X(v)\np", - "type": "text" - }, - { - "block_id": "p772-b8", - "global_id": 22547, - "bbox": [ - 313.21, - 124.31, - 450.12, - 134.31 - ], - "text": "v0\nv0\n0\nv", - "type": "text" - }, - { - "block_id": "p772-b9", - "global_id": 22548, - "bbox": [ - 162.84, - 285.15, - 178.18, - 294.7 - ], - "text": "xw(t)", - "type": "text" - }, - { - "block_id": "p772-b10", - "global_id": 22549, - "bbox": [ - 118.78, - 322.0, - 182.68, - 338.81 - ], - "text": "0\n T", - "type": "text" - }, - { - "block_id": "p772-b11", - "global_id": 22550, - "bbox": [ - 127.72, - 321.07, - 248.23, - 338.85 - ], - "text": "2\nT\n2\nt", - "type": "text" - }, - { - "block_id": "p772-b12", - "global_id": 22551, - "bbox": [ - 189.79, - 506.79, - 204.45, - 515.09 - ], - "text": "40", - "type": "text" - }, - { - "block_id": "p772-b13", - "global_id": 22552, - "bbox": [ - 183.79, - 414.95, - 204.45, - 432.51 - ], - "text": "10\n13.3", - "type": "text" - }, - { - "block_id": "p772-b14", - "global_id": 22553, - "bbox": [ - 207.21, - 514.65, - 211.21, - 522.65 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p772-b15", - "global_id": 22554, - "bbox": [ - 220.88, - 393.03, - 251.54, - 401.03 - ], - "text": "Mainlobe", - "type": "text" - }, - { - "block_id": "p772-b16", - "global_id": 22555, - "bbox": [ - 313.31, - 459.42, - 350.41, - 467.42 - ], - "text": "Rolloff rate", - "type": "text" - }, - { - "block_id": "p772-b17", - "global_id": 22556, - "bbox": [ - 149.34, - 387.5, - 174.12, - 397.26 - ], - "text": "WR(v)", - "type": "text" - }, - { - "block_id": "p772-b18", - "global_id": 22557, - "bbox": [ - 154.32, - 396.8, - 169.14, - 404.8 - ], - "text": "(dB)", - "type": "text" - }, - { - "block_id": "p772-b19", - "global_id": 22558, - "bbox": [ - 416.59, - 512.61, - 421.93, - 520.61 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p772-b20", - "global_id": 22559, - "bbox": [ - 189.79, - 476.72, - 204.45, - 485.01 - ], - "text": "30", - "type": "text" - }, - { - "block_id": "p772-b21", - "global_id": 22560, - "bbox": [ - 189.79, - 446.02, - 204.45, - 454.32 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p772-b22", - "global_id": 22561, - "bbox": [ - 254.34, - 413.19, - 285.45, - 421.19 - ], - "text": "Sidelobes", - "type": "text" - }, - { - "block_id": "p772-b23", - "global_id": 22562, - "bbox": [ - 200.45, - 383.49, - 204.45, - 391.49 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p772-b24", - "global_id": 22563, - "bbox": [ - 372.0, - 469.39, - 422.43, - 477.68 - ], - "text": "20 dB/decade", - "type": "text" - }, - { - "block_id": "p772-b25", - "global_id": 22564, - "bbox": [ - 219.62, - 514.56, - 228.96, - 531.21 - ], - "text": "2p\nT", - "type": "text" - }, - { - "block_id": "p772-b26", - "global_id": 22565, - "bbox": [ - 277.95, - 514.56, - 291.28, - 531.21 - ], - "text": "10p\nT", - "type": "text" - }, - { - "block_id": "p772-b27", - "global_id": 22566, - "bbox": [ - 353.01, - 514.56, - 366.34, - 531.21 - ], - "text": "20p\nT", - "type": "text" - }, - { - "block_id": "p772-b28", - "global_id": 22567, - "bbox": [ - 277.25, - 536.74, - 286.58, - 544.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p772-b29", - "global_id": 22568, - "bbox": [ - 277.01, - 264.26, - 286.82, - 272.26 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p772-b30", - "global_id": 22569, - "bbox": [ - 183.94, - 232.97, - 187.94, - 240.97 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p772-b31", - "global_id": 22570, - "bbox": [ - 174.5, - 189.17, - 252.77, - 201.32 - ], - "text": "wR(t)\n1", - "type": "text" - }, - { - "block_id": "p772-b32", - "global_id": 22571, - "bbox": [ - 122.89, - 232.91, - 136.06, - 245.21 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p772-b33", - "global_id": 22572, - "bbox": [ - 131.83, - 232.05, - 250.65, - 249.77 - ], - "text": "2\nT\n2\nt", - "type": "text" - }, - { - "block_id": "p772-b34", - "global_id": 22573, - "bbox": [ - 363.69, - 176.23, - 404.97, - 186.01 - ], - "text": "WR(v)\nT", - "type": "text" - }, - { - "block_id": "p772-b35", - "global_id": 22574, - "bbox": [ - 411.61, - 214.78, - 448.72, - 222.78 - ], - "text": "Rolloff rate", - "type": "text" - }, - { - "block_id": "p772-b36", - "global_id": 22575, - "bbox": [ - 430.65, - 233.79, - 435.99, - 241.79 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p772-b37", - "global_id": 22576, - "bbox": [ - 371.83, - 237.25, - 425.47, - 258.77 - ], - "text": "T\n4p\n0.217T", - "type": "text" - }, - { - "block_id": "p772-b38", - "global_id": 22577, - "bbox": [ - 369.48, - 322.07, - 373.48, - 330.07 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p772-b39", - "global_id": 22578, - "bbox": [ - 454.32, - 334.31, - 485.44, - 342.31 - ], - "text": "Sidelobes", - "type": "text" - }, - { - "block_id": "p772-b40", - "global_id": 22579, - "bbox": [ - 436.42, - 285.29, - 467.08, - 293.29 - ], - "text": "Mainlobe", - "type": "text" - }, - { - "block_id": "p772-b41", - "global_id": 22580, - "bbox": [ - 424.78, - 321.78, - 433.11, - 331.41 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p772-b42", - "global_id": 22581, - "bbox": [ - 379.75, - 289.67, - 399.31, - 299.24 - ], - "text": "Xw(v)", - "type": "text" - }, - { - "block_id": "p772-b43", - "global_id": 22582, - "bbox": [ - 464.32, - 320.1, - 469.66, - 328.1 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p772-b44", - "global_id": 22583, - "bbox": [ - 316.82, - 321.59, - 411.42, - 345.77 - ], - "text": "T\n4p\nv0", - "type": "text" - }, - { - "block_id": "p772-b45", - "global_id": 22584, - "bbox": [ - 368.43, - 287.11, - 372.88, - 303.78 - ], - "text": "2\nT", - "type": "text" - }, - { - "block_id": "p772-b46", - "global_id": 22585, - "bbox": [ - 101.84, - 551.44, - 247.72, - 560.68 - ], - "text": "Figure 7.46 Windowing and its effects.", - "type": "text" - }, - { - "block_id": "p772-b47", - "global_id": 22586, - "bbox": [ - 118.78, - 588.96, - 490.39, - 635.56 - ], - "text": "will be indistinguishable in the truncated signal. The result is loss of spectral resolution.\nWe would like the spectral spreading [mainlobe width of W(ω)] to be as small as possible.\n2. In addition to the mainlobe spreading, the truncated signal has sidelobes, which decay\nslowly with frequency. The spectrum of x(t) is zero everywhere except at ±ω0. On the", - "type": "text" - } - ] - }, - { - "page_num": 773, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p773-b0", - "global_id": 22587, - "bbox": [ - 309.24, - 62.89, - 516.12, - 71.98 - ], - "text": "7.8\nData Truncation: Window Functions\n753", - "type": "text" - }, - { - "block_id": "p773-b1", - "global_id": 22588, - "bbox": [ - 158.37, - 85.46, - 516.14, - 191.48 - ], - "text": "other hand, the truncated signal spectrum Xw(ω) is zero nowhere because of the sidelobes.\nThese sidelobes decay asymptotically as 1/ω. Thus, the truncation causes spectral leakage\nin the band where the spectrum of the signal x(t) is zero. The peak sidelobe magnitude is\n0.217 times the mainlobe magnitude (13.3 dB below the peak mainlobe magnitude). Also,\nthe sidelobes decay at a rate 1/ω, which is −6 dB/octave (or −20 dB/decade). This is the\nsidelobe’s rolloff rate. We want smaller sidelobes with a faster rate of decay (high rolloff\nrate). Figure 7.46d, which plots |WR(ω)| as a function of ω, clearly shows the mainlobe\nand sidelobe features, with the first sidelobe amplitude −13.3 dB below the mainlobe\namplitude and the sidelobes decaying at a rate of −6 dB/octave (or −20 dB/decade).", - "type": "text" - }, - { - "block_id": "p773-b2", - "global_id": 22589, - "bbox": [ - 127.59, - 199.45, - 516.11, - 233.33 - ], - "text": "So far, we have discussed the effect on the signal spectrum of signal truncation (truncation\nin the time domain). Because of the time-frequency duality, the effect of spectral truncation\n(truncation in frequency domain) on the signal shape is similar.", - "type": "text" - }, - { - "block_id": "p773-b3", - "global_id": 22590, - "bbox": [ - 127.59, - 248.35, - 516.12, - 286.42 - ], - "text": "REMEDIES FOR SIDE EFFECTS OF TRUNCATION\nFor better results, we must try to minimize the twin side effects of truncations: spectral spreading\n(mainlobe width) and leakage (sidelobe). Let us consider each of these ills.", - "type": "text" - }, - { - "block_id": "p773-b4", - "global_id": 22591, - "bbox": [ - 144.52, - 294.38, - 516.16, - 495.63 - ], - "text": "1. The spectral spread (mainlobe width) of the truncated signal is equal to the bandwidth of\nthe window function w(t). We know that the signal bandwidth is inversely proportional\nto the signal width (duration). Hence, to reduce the spectral spread (mainlobe width), we\nneed to increase the window width.\n2. To improve the leakage behavior, we must search for the cause of the slow decay of\nsidelobes. In Ch. 6, we saw that the Fourier spectrum decays as 1/ω for a signal with\njump discontinuity, decays as 1/ω2 for a continuous signal whose first derivative is\ndiscontinuous, and so on.† Smoothness of a signal is measured by the number of continuous\nderivatives it possesses. The smoother the signal, the faster the decay of its spectrum. Thus,\nwe can achieve a given leakage behavior by selecting a suitably smooth (tapered) window.\n3. For a given window width, the remedies for the two effects are incompatible. If we try to\nimprove one, the other deteriorates. For instance, among all the windows of a given width,\nthe rectangular window has the smallest spectral spread (mainlobe width), but its sidelobes\nhave high level and they decay slowly. A tapered (smooth) window of the same width has\nsmaller and faster decaying sidelobes, but it has a wider mainlobe.‡ But we can compensate\nfor the increased mainlobe width by widening the window. Thus, we can remedy both the\nside effects of truncation by selecting a suitably smooth window of sufficient width.", - "type": "text" - }, - { - "block_id": "p773-b5", - "global_id": 22592, - "bbox": [ - 127.59, - 503.6, - 516.14, - 525.52 - ], - "text": "There are several well-known tapered-window functions, such as Bartlett (triangular),\nHanning (von Hann), Hamming, Blackman, and Kaiser, which truncate the data gradually. These", - "type": "text" - }, - { - "block_id": "p773-b6", - "global_id": 22593, - "bbox": [ - 127.59, - 544.19, - 516.14, - 633.41 - ], - "text": "† This result was demonstrated for periodic signals. However, it applies to aperiodic signals also. This is\nbecause we showed in the beginning of this chapter that if xT0(t) is a periodic signal formed by periodic\nextension of an aperiodic signal x(t), then the spectrum of xT0(t) is (1/T0 times) the samples of X(ω). Thus,\nwhat is true of the decay rate of the spectrum of xT0(t) is also true of the rate of decay of X(ω).\n‡ A tapered window yields a higher mainlobe width because the effective width of a tapered window is\nsmaller than that of the rectangular window; see Sec. 2.6-2 [Eq. (2.47)] for the definition of effective width.\nTherefore, from the reciprocity of the signal width and its bandwidth, it follows that the rectangular window\nmainlobe is narrower than a tapered window.", - "type": "text" - } - ] - }, - { - "page_num": 774, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p774-b0", - "global_id": 22594, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "754\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p774-b1", - "global_id": 22595, - "bbox": [ - 101.84, - 86.07, - 338.02, - 95.3 - ], - "text": "TABLE 7.3\nSome Window Functions and Their Characteristics", - "type": "text" - }, - { - "block_id": "p774-b2", - "global_id": 22596, - "bbox": [ - 101.84, - 106.31, - 490.39, - 137.2 - ], - "text": "Rolloff\nPeak\nMainlobe\nRate\nSidelobe\nNo.\nWindow w(t)\nWidth\nLevel (dB)", - "type": "text" - }, - { - "block_id": "p774-b3", - "global_id": 22597, - "bbox": [ - 101.84, - 151.97, - 196.48, - 160.93 - ], - "text": "1\nRectangular: rect", - "type": "text" - }, - { - "block_id": "p774-b4", - "global_id": 22598, - "bbox": [ - 197.98, - 139.01, - 209.23, - 154.56 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p774-b5", - "global_id": 22599, - "bbox": [ - 205.23, - 158.25, - 210.21, - 167.21 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p774-b6", - "global_id": 22600, - "bbox": [ - 212.09, - 139.01, - 480.68, - 167.21 - ], - "text": "4π\nT\n−6\n−13.3", - "type": "text" - }, - { - "block_id": "p774-b7", - "global_id": 22601, - "bbox": [ - 101.84, - 175.86, - 173.82, - 185.2 - ], - "text": "2\nBartlett:", - "type": "text" - }, - { - "block_id": "p774-b8", - "global_id": 22602, - "bbox": [ - 173.83, - 163.27, - 187.32, - 178.81 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p774-b9", - "global_id": 22603, - "bbox": [ - 181.07, - 182.5, - 190.54, - 191.56 - ], - "text": "2T", - "type": "text" - }, - { - "block_id": "p774-b10", - "global_id": 22604, - "bbox": [ - 192.43, - 163.27, - 480.68, - 191.47 - ], - "text": "8π\nT\n−12\n−26.5", - "type": "text" - }, - { - "block_id": "p774-b11", - "global_id": 22605, - "bbox": [ - 101.84, - 200.48, - 181.8, - 209.45 - ], - "text": "3\nHanning: 0.5", - "type": "text" - }, - { - "block_id": "p774-b13", - "global_id": 22606, - "bbox": [ - 187.68, - 187.53, - 235.41, - 209.45 - ], - "text": "1 + cos\n2πt", - "type": "text" - }, - { - "block_id": "p774-b14", - "global_id": 22607, - "bbox": [ - 226.02, - 206.77, - 231.0, - 215.73 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p774-b15", - "global_id": 22608, - "bbox": [ - 236.76, - 187.53, - 480.68, - 215.73 - ], - "text": "!\n8π\nT\n−18\n−31.5", - "type": "text" - }, - { - "block_id": "p774-b16", - "global_id": 22609, - "bbox": [ - 101.84, - 224.38, - 229.69, - 233.71 - ], - "text": "4\nHamming: 0.54 + 0.46cos", - "type": "text" - }, - { - "block_id": "p774-b17", - "global_id": 22610, - "bbox": [ - 230.7, - 211.78, - 251.2, - 227.42 - ], - "text": "2πt", - "type": "text" - }, - { - "block_id": "p774-b18", - "global_id": 22611, - "bbox": [ - 241.81, - 231.02, - 246.8, - 239.99 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p774-b19", - "global_id": 22612, - "bbox": [ - 252.55, - 211.78, - 480.68, - 239.99 - ], - "text": "8π\nT\n−6\n−42.7", - "type": "text" - }, - { - "block_id": "p774-b20", - "global_id": 22613, - "bbox": [ - 101.84, - 248.63, - 225.7, - 257.97 - ], - "text": "5\nBlackman: 0.42 + 0.5cos", - "type": "text" - }, - { - "block_id": "p774-b21", - "global_id": 22614, - "bbox": [ - 226.7, - 236.04, - 247.21, - 251.69 - ], - "text": "2πt", - "type": "text" - }, - { - "block_id": "p774-b22", - "global_id": 22615, - "bbox": [ - 237.83, - 255.28, - 242.81, - 264.25 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p774-b24", - "global_id": 22616, - "bbox": [ - 256.01, - 248.63, - 293.03, - 257.97 - ], - "text": "+ 0.08cos", - "type": "text" - }, - { - "block_id": "p774-b25", - "global_id": 22617, - "bbox": [ - 294.03, - 236.04, - 314.54, - 251.69 - ], - "text": "4πt", - "type": "text" - }, - { - "block_id": "p774-b26", - "global_id": 22618, - "bbox": [ - 305.14, - 255.28, - 310.13, - 264.25 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p774-b27", - "global_id": 22619, - "bbox": [ - 315.89, - 236.04, - 480.68, - 264.25 - ], - "text": "12π\nT\n−18\n−58.1", - "type": "text" - }, - { - "block_id": "p774-b28", - "global_id": 22620, - "bbox": [ - 101.84, - 290.84, - 160.87, - 299.81 - ], - "text": "6\nKaiser:", - "type": "text" - }, - { - "block_id": "p774-b29", - "global_id": 22621, - "bbox": [ - 164.31, - 275.87, - 170.54, - 285.66 - ], - "text": "I0", - "type": "text" - }, - { - "block_id": "p774-b31", - "global_id": 22622, - "bbox": [ - 177.61, - 275.59, - 182.73, - 284.55 - ], - "text": "α", - "type": "text" - }, - { - "block_id": "p774-b33", - "global_id": 22623, - "bbox": [ - 190.51, - 265.69, - 220.35, - 284.92 - ], - "text": "1 −4\n t", - "type": "text" - }, - { - "block_id": "p774-b34", - "global_id": 22624, - "bbox": [ - 216.34, - 282.23, - 221.33, - 291.2 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p774-b35", - "global_id": 22625, - "bbox": [ - 223.21, - 260.3, - 237.41, - 275.75 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p774-b36", - "global_id": 22626, - "bbox": [ - 191.41, - 284.19, - 480.68, - 306.92 - ], - "text": "I0(α)\n0 ≤α ≤10\n11.2π\nT\n−6\n−59.9", - "type": "text" - }, - { - "block_id": "p774-b37", - "global_id": 22627, - "bbox": [ - 391.64, - 309.41, - 434.65, - 318.75 - ], - "text": "(α = 8.168)", - "type": "text" - }, - { - "block_id": "p774-b38", - "global_id": 22628, - "bbox": [ - 199.24, - 350.9, - 203.24, - 358.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p774-b39", - "global_id": 22629, - "bbox": [ - 199.89, - 429.09, - 203.89, - 437.09 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p774-b40", - "global_id": 22630, - "bbox": [ - 125.76, - 427.59, - 387.52, - 456.03 - ], - "text": "(a)\n(b)\n2\nxo\n\n2\nxo\n\n2\nxo", - "type": "text" - }, - { - "block_id": "p774-b41", - "global_id": 22631, - "bbox": [ - 226.84, - 354.98, - 251.05, - 364.58 - ], - "text": "wHan(x)", - "type": "text" - }, - { - "block_id": "p774-b42", - "global_id": 22632, - "bbox": [ - 236.51, - 433.59, - 240.06, - 441.59 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p774-b43", - "global_id": 22633, - "bbox": [ - 376.21, - 350.9, - 380.21, - 358.9 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p774-b44", - "global_id": 22634, - "bbox": [ - 376.86, - 427.59, - 457.59, - 446.56 - ], - "text": "0\n2\nxo", - "type": "text" - }, - { - "block_id": "p774-b45", - "global_id": 22635, - "bbox": [ - 403.81, - 354.98, - 429.69, - 364.58 - ], - "text": "wHam(x)", - "type": "text" - }, - { - "block_id": "p774-b46", - "global_id": 22636, - "bbox": [ - 413.48, - 433.59, - 417.03, - 441.59 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p774-b47", - "global_id": 22637, - "bbox": [ - 125.76, - 462.64, - 322.52, - 471.96 - ], - "text": "Figure 7.47 (a) Hanning and (b) Hamming windows.", - "type": "text" - }, - { - "block_id": "p774-b48", - "global_id": 22638, - "bbox": [ - 101.84, - 505.27, - 490.4, - 563.06 - ], - "text": "windows offer different trade-offs with respect to spectral spread (mainlobe width), the peak\nsidelobe magnitude, and the leakage rolloff rate, as indicated in Table 7.3 [5, 6]. Observe that all\nwindows are symmetrical about the origin (i.e., are even functions of t). Because of this feature,\nW(ω) is a real function of ω; that is,̸\nW(ω) is either 0 or π. Hence, the phase function of the\ntruncated signal has a minimal amount of distortion.", - "type": "text" - }, - { - "block_id": "p774-b49", - "global_id": 22639, - "bbox": [ - 101.84, - 565.05, - 490.37, - 610.88 - ], - "text": "Figure 7.47 shows two well-known tapered-window functions, the von Hann (or Hanning)\nwindow wHan(x) and the Hamming window wHam(x). We have intentionally used the independent\nvariable x because windowing can be performed in the time domain as well as in the frequency\ndomain, so x could be t or ω, depending on the application.", - "type": "text" - }, - { - "block_id": "p774-b50", - "global_id": 22640, - "bbox": [ - 101.85, - 612.87, - 490.38, - 634.79 - ], - "text": "There are hundreds of windows, all with different characteristics. But the choice depends\non a particular application. The rectangular window has the narrowest mainlobe. The Bartlett", - "type": "text" - } - ] - }, - { - "page_num": 775, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p775-b0", - "global_id": 22641, - "bbox": [ - 310.23, - 62.89, - 516.12, - 71.98 - ], - "text": "7.9\nMATLAB: Fourier Transform Topics\n755", - "type": "text" - }, - { - "block_id": "p775-b1", - "global_id": 22642, - "bbox": [ - 127.59, - 85.82, - 516.14, - 227.29 - ], - "text": "(triangle) window (also called the Fejer or Cesaro) is inferior in all respects to the\nHanning window. For this reason, it is rarely used in practice. Hanning is preferred over\nHamming in spectral analysis because it has faster sidelobe decay. For filtering applications,\non the other hand, the Hamming window is chosen because it has the smallest sidelobe\nmagnitude for a given mainlobe width. The Hamming window is the most widely used\ngeneral-purpose window. The Kaiser window, which uses I0(α), the modified zero-order Bessel\nfunction, is more versatile and adjustable. Selecting a proper value of α (0 ≤α ≤10)\nallows the designer to tailor the window to suit a particular application. The parameter α\ncontrols the mainlobe-sidelobe trade-off. When α = 0, the Kaiser window is the rectangular\nwindow. For α = 5.4414, it is the Hamming window, and when α = 8.885, it is the\nBlackman window. As α increases, the mainlobe width increases and the sidelobe level\ndecreases.", - "type": "text" - }, - { - "block_id": "p775-b2", - "global_id": 22643, - "bbox": [ - 127.59, - 254.97, - 334.16, - 266.93 - ], - "text": "7.8-1 Using Windows in Filter Design", - "type": "text" - }, - { - "block_id": "p775-b3", - "global_id": 22644, - "bbox": [ - 127.59, - 272.65, - 516.14, - 378.67 - ], - "text": "We shall design an ideal lowpass filter of bandwidth W rad/s, with frequency response H(ω), as\nshown in Fig. 7.48e or Fig. 7.48f. For this filter, the impulse response h(t) = (W/π)sinc(Wt)\n(Fig. 7.48c) is noncausal and, therefore, unrealizable. Truncation of h(t) by a suitable window\n(Fig. 7.48a) makes it realizable, although the resulting filter is now an approximation to the\ndesired ideal filter.† We shall use a rectangular window wR(t) and a triangular (Bartlett) window\nwT(t) to truncate h(t), and then examine the resulting filters. The truncated impulse responses\nhR(t) = h(t)wR(t) and hT(t) = h(t)wT(t) are depicted in Fig. 7.48d. Hence, the windowed filter\nfrequency response is the convolution of H(ω) with the Fourier transform of the window, as\nillustrated in Figs. 7.48e and 7.48f. We make the following observations.", - "type": "text" - }, - { - "block_id": "p775-b4", - "global_id": 22645, - "bbox": [ - 144.52, - 386.53, - 516.14, - 492.24 - ], - "text": "1. The windowed filter spectra show spectral spreading at the edges, and instead of a sudden\nswitch there is a gradual transition from the passband to the stopband of the filter. The\ntransition band is smaller (2π/T rad/s) for the rectangular case than for the triangular case\n(4π/T rad/s).\n2. Although H(ω) is bandlimited, the windowed filters are not. But the stopband behavior\nof the triangular case is superior to that of the rectangular case. For the rectangular\nwindow, the leakage in the stopband decreases slowly (as 1/ω) in comparison to that of\nthe triangular window (as 1/ω2). Moreover, the rectangular case has a higher peak sidelobe\namplitude than that of the triangular window.", - "type": "text" - }, - { - "block_id": "p775-b5", - "global_id": 22646, - "bbox": [ - 127.94, - 524.45, - 422.32, - 538.4 - ], - "text": "7.9 MATLAB: FOURIER TRANSFORM TOPICS", - "type": "text" - }, - { - "block_id": "p775-b6", - "global_id": 22647, - "bbox": [ - 127.59, - 544.38, - 516.15, - 578.26 - ], - "text": "MATLAB is useful for investigating a variety of Fourier transform topics. In this section,\na rectangular pulse is used to investigate the scaling property, Parseval’s theorem, essential\nbandwidth, and spectral sampling. Kaiser window functions are also investigated.", - "type": "text" - }, - { - "block_id": "p775-b7", - "global_id": 22648, - "bbox": [ - 127.59, - 599.27, - 516.12, - 633.41 - ], - "text": "† In addition to truncation, we need to delay the truncated function by T/2 to render it causal. However,\nthe time delay only adds a linear phase to the spectrum without changing the amplitude spectrum. Thus, to\nsimplify our discussion, we shall ignore the delay.", - "type": "text" - } - ] - }, - { - "page_num": 776, - "width": 720.0, - "height": 576.0, - "blocks": [ - { - "block_id": "p776-b0", - "global_id": 22649, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "756\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p776-b1", - "global_id": 22650, - "bbox": [ - 289.77, - 291.29, - 299.14, - 311.86 - ], - "text": "WR(v)", - "type": "text" - }, - { - "block_id": "p776-b2", - "global_id": 22651, - "bbox": [ - 189.5, - 291.65, - 198.88, - 312.61 - ], - "text": "WT(v)", - "type": "text" - }, - { - "block_id": "p776-b3", - "global_id": 22652, - "bbox": [ - 299.52, - 141.71, - 307.46, - 157.97 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p776-b4", - "global_id": 22653, - "bbox": [ - 199.77, - 141.27, - 207.71, - 159.34 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p776-b5", - "global_id": 22654, - "bbox": [ - 168.72, - 208.52, - 277.86, - 218.32 - ], - "text": "*\n*", - "type": "text" - }, - { - "block_id": "p776-b8", - "global_id": 22655, - "bbox": [ - 241.94, - 166.47, - 249.78, - 173.0 - ], - "text": "W", - "type": "text" - }, - { - "block_id": "p776-b9", - "global_id": 22656, - "bbox": [ - 141.25, - 102.97, - 150.15, - 499.03 - ], - "text": "W\nW\nW\nW", - "type": "text" - }, - { - "block_id": "p776-b10", - "global_id": 22657, - "bbox": [ - 241.87, - 141.4, - 250.07, - 500.67 - ], - "text": "W\nW\n0", - "type": "text" - }, - { - "block_id": "p776-b11", - "global_id": 22658, - "bbox": [ - 141.9, - 142.21, - 149.99, - 480.73 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p776-b12", - "global_id": 22659, - "bbox": [ - 241.95, - 476.89, - 249.79, - 480.81 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p776-b13", - "global_id": 22660, - "bbox": [ - 216.5, - 326.49, - 224.33, - 335.19 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p776-b14", - "global_id": 22661, - "bbox": [ - 116.02, - 326.69, - 123.86, - 334.99 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p776-b15", - "global_id": 22662, - "bbox": [ - 303.71, - 477.57, - 313.08, - 497.27 - ], - "text": "HR(v)", - "type": "text" - }, - { - "block_id": "p776-b16", - "global_id": 22663, - "bbox": [ - 197.99, - 476.98, - 207.37, - 496.94 - ], - "text": "HT(v)", - "type": "text" - }, - { - "block_id": "p776-b17", - "global_id": 22664, - "bbox": [ - 142.99, - 279.07, - 150.83, - 282.99 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p776-b18", - "global_id": 22665, - "bbox": [ - 481.76, - 448.16, - 491.11, - 464.65 - ], - "text": "wT(t)", - "type": "text" - }, - { - "block_id": "p776-b19", - "global_id": 22666, - "bbox": [ - 444.14, - 198.33, - 452.12, - 450.41 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p776-b20", - "global_id": 22667, - "bbox": [ - 425.4, - 191.09, - 433.24, - 453.15 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p776-b21", - "global_id": 22668, - "bbox": [ - 481.76, - 206.66, - 491.11, - 222.88 - ], - "text": "wR(t)", - "type": "text" - }, - { - "block_id": "p776-b22", - "global_id": 22669, - "bbox": [ - 445.17, - 277.96, - 453.01, - 533.52 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p776-b23", - "global_id": 22670, - "bbox": [ - 343.25, - 293.38, - 351.74, - 541.9 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p776-b24", - "global_id": 22671, - "bbox": [ - 481.72, - 189.07, - 490.37, - 439.81 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p776-b25", - "global_id": 22672, - "bbox": [ - 389.79, - 467.79, - 399.15, - 483.08 - ], - "text": "hT(t)", - "type": "text" - }, - { - "block_id": "p776-b26", - "global_id": 22673, - "bbox": [ - 403.24, - 465.03, - 412.59, - 479.94 - ], - "text": "hR(t)", - "type": "text" - }, - { - "block_id": "p776-b27", - "global_id": 22674, - "bbox": [ - 342.09, - 199.41, - 350.77, - 449.7 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p776-b28", - "global_id": 22675, - "bbox": [ - 321.55, - 192.09, - 329.39, - 451.93 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p776-b29", - "global_id": 22676, - "bbox": [ - 408.73, - 199.72, - 416.65, - 211.04 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p776-b30", - "global_id": 22677, - "bbox": [ - 243.65, - 549.51, - 251.48, - 554.74 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p776-b31", - "global_id": 22678, - "bbox": [ - 142.99, - 177.69, - 152.6, - 329.4 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p776-b32", - "global_id": 22679, - "bbox": [ - 242.89, - 177.69, - 252.13, - 330.77 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p776-b33", - "global_id": 22680, - "bbox": [ - 144.23, - 548.68, - 152.07, - 553.9 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p776-b34", - "global_id": 22681, - "bbox": [ - 241.94, - 103.3, - 249.99, - 116.36 - ], - "text": "W", - "type": "text" - }, - { - "block_id": "p776-b35", - "global_id": 22682, - "bbox": [ - 127.58, - 294.04, - 135.42, - 298.4 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p776-b36", - "global_id": 22683, - "bbox": [ - 135.92, - 291.65, - 143.86, - 300.79 - ], - "text": "4p", - "type": "text" - }, - { - "block_id": "p776-b37", - "global_id": 22684, - "bbox": [ - 226.78, - 284.53, - 234.62, - 288.89 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p776-b38", - "global_id": 22685, - "bbox": [ - 235.12, - 282.14, - 243.06, - 291.29 - ], - "text": "4p", - "type": "text" - }, - { - "block_id": "p776-b39", - "global_id": 22686, - "bbox": [ - 126.32, - 525.18, - 134.16, - 529.54 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p776-b40", - "global_id": 22687, - "bbox": [ - 134.66, - 522.78, - 142.6, - 531.93 - ], - "text": "4p", - "type": "text" - }, - { - "block_id": "p776-b41", - "global_id": 22688, - "bbox": [ - 435.58, - 114.68, - 452.23, - 517.83 - ], - "text": "T\n2\nT\n2", - "type": "text" - }, - { - "block_id": "p776-b42", - "global_id": 22689, - "bbox": [ - 331.74, - 118.91, - 348.58, - 514.49 - ], - "text": "T\n2\n T\n2\nT\n2", - "type": "text" - }, - { - "block_id": "p776-b43", - "global_id": 22690, - "bbox": [ - 226.93, - 528.9, - 234.76, - 533.26 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p776-b44", - "global_id": 22691, - "bbox": [ - 235.26, - 526.51, - 243.2, - 535.66 - ], - "text": "2p", - "type": "text" - }, - { - "block_id": "p776-b45", - "global_id": 22692, - "bbox": [ - 434.72, - 364.27, - 451.23, - 377.18 - ], - "text": "T\n2", - "type": "text" - }, - { - "block_id": "p776-b46", - "global_id": 22693, - "bbox": [ - 331.69, - 364.27, - 348.21, - 377.18 - ], - "text": "T\n2", - "type": "text" - }, - { - "block_id": "p776-b47", - "global_id": 22694, - "bbox": [ - 434.9, - 264.92, - 451.42, - 269.28 - ], - "text": "T\n2", - "type": "text" - }, - { - "block_id": "p776-b48", - "global_id": 22695, - "bbox": [ - 100.09, - 86.59, - 109.33, - 237.73 - ], - "text": "Figure 7.48 Window-based filter design.", - "type": "text" - } - ] - }, - { - "page_num": 777, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p777-b0", - "global_id": 22696, - "bbox": [ - 310.23, - 62.89, - 516.12, - 71.98 - ], - "text": "7.9\nMATLAB: Fourier Transform Topics\n757", - "type": "text" - }, - { - "block_id": "p777-b1", - "global_id": 22697, - "bbox": [ - 127.59, - 86.52, - 395.07, - 98.48 - ], - "text": "7.9-1 The Sinc Function and the Scaling Property", - "type": "text" - }, - { - "block_id": "p777-b2", - "global_id": 22698, - "bbox": [ - 127.59, - 104.19, - 516.13, - 150.44 - ], - "text": "As shown in Ex. 7.2, the Fourier transform of x(t) = rect(t/τ) is X(ω) = τ sinc(ωτ/2). To\nrepresent X(ω) in MATLAB, a sinc function is first required. As an alternative to the signal\nprocessing toolbox function sinc, which computes sinc(x) as sin(πx)/πx, we create our own\nfunction that follows the conventions of this book and defines sinc(x) = sin(x)/x.", - "type": "text" - }, - { - "block_id": "p777-b3", - "global_id": 22699, - "bbox": [ - 127.59, - 161.87, - 436.12, - 195.74 - ], - "text": "function [y] = CH7MP1(x)\n% CH7MP1.m : Chapter 7, MATLAB Program 1\n% Function M-file computes the sinc function, y = sin(x)/x.", - "type": "text" - }, - { - "block_id": "p777-b4", - "global_id": 22700, - "bbox": [ - 127.59, - 209.68, - 294.96, - 231.61 - ], - "text": "y(x==0) = 1;\ny(x~=0) = sin(x(x~=0))./x(x~=0);", - "type": "text" - }, - { - "block_id": "p777-b5", - "global_id": 22701, - "bbox": [ - 127.59, - 242.05, - 516.15, - 312.48 - ], - "text": "The computational simplicity of sinc (x) = sin(x)/x is somewhat deceptive: sin(0)/0 results\nin a divide-by-zero error. Thus, program CH7MP1 assigns sinc (0) = 1 and computes the\nremaining values according to the definition. Notice that CH7MP1 cannot be directly replaced\nby an anonymous function. Anonymous functions cannot have multiple lines or contain certain\ncommands such as =, if, or for. M-files, however, can be used to define an anonymous function.\nFor example, we can represent X(ω) as an anonymous function that is defined in terms of CH7MP1.", - "type": "text" - }, - { - "block_id": "p777-b6", - "global_id": 22702, - "bbox": [ - 127.6, - 323.62, - 362.96, - 333.59 - ], - "text": ">>\nX = @(omega,tau) tau*CH7MP1(omega*tau/2);", - "type": "text" - }, - { - "block_id": "p777-b7", - "global_id": 22703, - "bbox": [ - 127.59, - 344.03, - 516.13, - 366.35 - ], - "text": "Once we have defined X(ω), it is simple to investigate the effects of scaling the pulse width τ.\nConsider the three cases τ = 1.0, τ = 0.5, and τ = 2.0.", - "type": "text" - }, - { - "block_id": "p777-b8", - "global_id": 22704, - "bbox": [ - 127.59, - 377.24, - 465.98, - 425.06 - ], - "text": ">>\nomega = linspace(-4*pi,4*pi,200);\n>>\nplot(omega,X(omega,1),’k-’,omega,X(omega,0.5),’k-.’,omega,X(omega,2),’k--’);\n>>\ngrid; axis tight; xlabel(’\\omega’); ylabel(’X(\\omega)’);\n>>\nlegend(’Baseline (\\tau = 1)’,’Compressed (\\tau = 0.5)’,...\n>>\n’Expanded (\\tau = 2.0)’);", - "type": "text" - }, - { - "block_id": "p777-b9", - "global_id": 22705, - "bbox": [ - 127.59, - 436.46, - 516.16, - 494.24 - ], - "text": "Figure 7.49 confirms the reciprocal relationship between signal duration and spectral bandwidth:\ntime compression causes spectral expansion, and time expansion causes spectral compression.\nAdditionally, spectral amplitudes are directly related to signal energy. As a signal is compressed,\nsignal energy and thus spectral magnitude decrease. The opposite effect occurs when the signal is\nexpanded.", - "type": "text" - }, - { - "block_id": "p777-b10", - "global_id": 22706, - "bbox": [ - 198.97, - 594.33, - 275.76, - 602.33 - ], - "text": "–10\n–5", - "type": "text" - }, - { - "block_id": "p777-b11", - "global_id": 22707, - "bbox": [ - 336.65, - 604.37, - 342.14, - 612.37 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p777-b12", - "global_id": 22708, - "bbox": [ - 337.41, - 594.33, - 476.25, - 602.33 - ], - "text": "0\n5\n10", - "type": "text" - }, - { - "block_id": "p777-b13", - "global_id": 22709, - "bbox": [ - 163.94, - 572.12, - 167.94, - 580.12 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p777-b14", - "global_id": 22710, - "bbox": [ - 163.94, - 542.24, - 167.94, - 550.24 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p777-b15", - "global_id": 22711, - "bbox": [ - 163.94, - 512.36, - 167.94, - 520.36 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p777-b16", - "global_id": 22712, - "bbox": [ - 152.14, - 542.75, - 160.43, - 559.34 - ], - "text": "X(ω)", - "type": "text" - }, - { - "block_id": "p777-b17", - "global_id": 22713, - "bbox": [ - 478.04, - 520.68, - 500.09, - 547.49 - ], - "text": "τ = 1 \nτ = 0.5\nτ = 2.0)", - "type": "text" - }, - { - "block_id": "p777-b18", - "global_id": 22714, - "bbox": [ - 151.5, - 619.58, - 430.15, - 629.19 - ], - "text": "Figure 7.49 Spectra X(ω) = τ sinc(ωτ/2) for τ = 1.0, τ = 0.5, and τ = 2.0.", - "type": "text" - } - ] - }, - { - "page_num": 778, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p778-b0", - "global_id": 22715, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "758\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p778-b1", - "global_id": 22716, - "bbox": [ - 101.84, - 86.52, - 377.94, - 98.48 - ], - "text": "7.9-2 Parseval’s Theorem and Essential Bandwidth", - "type": "text" - }, - { - "block_id": "p778-b2", - "global_id": 22717, - "bbox": [ - 101.84, - 104.61, - 486.55, - 114.57 - ], - "text": "Parseval’s theorem concisely relates energy between the time domain and the frequency domain:", - "type": "text" - }, - { - "block_id": "p778-b3", - "global_id": 22718, - "bbox": [ - 224.53, - 117.46, - 241.47, - 129.52 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p778-b4", - "global_id": 22719, - "bbox": [ - 229.79, - 142.34, - 242.34, - 149.32 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p778-b5", - "global_id": 22720, - "bbox": [ - 243.95, - 124.45, - 299.04, - 141.4 - ], - "text": "|x(t)|2 dt = 1", - "type": "text" - }, - { - "block_id": "p778-b6", - "global_id": 22721, - "bbox": [ - 290.57, - 138.1, - 301.53, - 148.47 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p778-b7", - "global_id": 22722, - "bbox": [ - 304.83, - 117.46, - 321.78, - 129.52 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p778-b8", - "global_id": 22723, - "bbox": [ - 310.1, - 142.34, - 322.65, - 149.32 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p778-b9", - "global_id": 22724, - "bbox": [ - 324.25, - 127.22, - 367.5, - 141.3 - ], - "text": "|X(ω)|2 dω", - "type": "text" - }, - { - "block_id": "p778-b10", - "global_id": 22725, - "bbox": [ - 101.85, - 157.91, - 490.39, - 185.65 - ], - "text": "This too is easily verified with MATLAB. For example, a unit amplitude pulse x(t) with duration\nτ has energy Ex = τ. Thus,\n# ∞", - "type": "text" - }, - { - "block_id": "p778-b11", - "global_id": 22726, - "bbox": [ - 255.28, - 198.47, - 267.83, - 205.44 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p778-b12", - "global_id": 22727, - "bbox": [ - 269.44, - 183.35, - 341.02, - 197.52 - ], - "text": "|X(ω)|2 dω = 2πτ", - "type": "text" - }, - { - "block_id": "p778-b13", - "global_id": 22728, - "bbox": [ - 101.85, - 211.27, - 399.72, - 221.93 - ], - "text": "Letting τ = 1, the energy of X(ω) is computed by using the quad function.", - "type": "text" - }, - { - "block_id": "p778-b14", - "global_id": 22729, - "bbox": [ - 101.85, - 231.68, - 410.44, - 265.55 - ], - "text": ">>\nX_squared = @(omega, tau) (tau*CH7MP1(omega*tau/2)).^2;\n>>\nquad(X_squared,-1e6,1e6,[],[],1)\nans = 6.2817", - "type": "text" - }, - { - "block_id": "p778-b15", - "global_id": 22730, - "bbox": [ - 101.85, - 275.02, - 490.42, - 333.1 - ], - "text": "Although not perfect, the result of the numerical integration is consistent with the expected value of\n2π ≈6.2832. For quad, the first argument is the function to be integrated, the next two arguments\nare the limits of integration, the empty square brackets indicate default values for special options,\nand the last argument is the secondary input τ for the anonymous function X_squared. Full format\ndetails for quad are available from MATLAB’s help facilities.", - "type": "text" - }, - { - "block_id": "p778-b16", - "global_id": 22731, - "bbox": [ - 101.85, - 334.8, - 490.41, - 368.68 - ], - "text": "A more interesting problem involves computing a signal’s essential bandwidth. Consider, for\nexample, finding the essential bandwidth W, in radians per second, that contains fraction β of the\nenergy of the square pulse x(t). That is, we want to find W such that", - "type": "text" - }, - { - "block_id": "p778-b17", - "global_id": 22732, - "bbox": [ - 246.71, - 379.95, - 257.66, - 403.97 - ], - "text": "1\n2π", - "type": "text" - }, - { - "block_id": "p778-b18", - "global_id": 22733, - "bbox": [ - 260.96, - 372.95, - 276.59, - 385.24 - ], - "text": "# W", - "type": "text" - }, - { - "block_id": "p778-b19", - "global_id": 22734, - "bbox": [ - 266.22, - 397.83, - 277.47, - 405.03 - ], - "text": "−W", - "type": "text" - }, - { - "block_id": "p778-b20", - "global_id": 22735, - "bbox": [ - 279.58, - 382.72, - 345.52, - 396.79 - ], - "text": "|X(ω)|2 dω = βτ", - "type": "text" - }, - { - "block_id": "p778-b21", - "global_id": 22736, - "bbox": [ - 101.85, - 413.67, - 340.43, - 424.02 - ], - "text": "Program CH7MP2 uses a guess-and-check method to find W.", - "type": "text" - }, - { - "block_id": "p778-b22", - "global_id": 22737, - "bbox": [ - 101.84, - 433.77, - 452.27, - 587.2 - ], - "text": "function [W,E_W] = CH7MP2(tau,beta,tol)\n% CH7MP2.m : Chapter 7, MATLAB Program 2\n% Function M-file computes essential bandwidth W for square pulse.\n% INPUTS:\ntau = pulse width\n%\nbeta = fraction of signal energy desired in W\n%\ntol = tolerance of relative energy error\n% OUTPUTS:\nW = essential bandwidth [rad/s]\n%\nE_W = Energy contained in bandwidth W\nW = 0; step = 2*pi/tau;\n% Initial guess and step values\nX_squared = @(omega,tau) (tau*CH7MP1(omega*tau/2)).^2;\nE = beta*tau;\n% Desired energy in W\nrelerr = (E-0)/E;\n% Initial relative error is 100 percent\nwhile(abs(relerr) > tol),", - "type": "text" - }, - { - "block_id": "p778-b23", - "global_id": 22738, - "bbox": [ - 122.76, - 589.19, - 405.19, - 635.02 - ], - "text": "if (relerr>0),\n% W too small, so...\nW=W+step;\n% ... increase W by step\nelseif (relerr<0),\n% W too large, so...\nstep = step/2;\n% ... decrease step and then W", - "type": "text" - } - ] - }, - { - "page_num": 779, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p779-b0", - "global_id": 22739, - "bbox": [ - 310.23, - 62.89, - 516.12, - 71.98 - ], - "text": "7.9\nMATLAB: Fourier Transform Topics\n759", - "type": "text" - }, - { - "block_id": "p779-b1", - "global_id": 22740, - "bbox": [ - 127.59, - 85.41, - 389.1, - 143.19 - ], - "text": "W = W-step;\nend\nE_W = 1/(2*pi)*quad(X_squared,-W,W,[],[],tau);\nrelerr = (E - E_W)/E;\nend", - "type": "text" - }, - { - "block_id": "p779-b2", - "global_id": 22741, - "bbox": [ - 127.59, - 152.05, - 516.17, - 198.17 - ], - "text": "Although this guess-and-check method is not the most efficient, it is relatively simple to\nunderstand: CH7MP2 sensibly adjusts W until the relative error is within tolerance. The number\nof iterations needed to converge to a solution depends on a variety of factors and is not known\nbeforehand. The while command is ideal for such situations:", - "type": "text" - }, - { - "block_id": "p779-b3", - "global_id": 22742, - "bbox": [ - 145.52, - 199.46, - 234.43, - 210.13 - ], - "text": "while expression,", - "type": "text" - }, - { - "block_id": "p779-b4", - "global_id": 22743, - "bbox": [ - 127.6, - 211.41, - 421.13, - 245.7 - ], - "text": "statements;\nend\nWhile the expression is true, the statements are continually repeated.", - "type": "text" - }, - { - "block_id": "p779-b5", - "global_id": 22744, - "bbox": [ - 127.59, - 247.59, - 516.15, - 317.43 - ], - "text": "To demonstrate CH7MP2, consider the 90% essential bandwidth W for a pulse of 1 second\nduration. Typing [W,E_W]=CH7MP2(1,0.9,0.001) returns an essential bandwidth W = 5.3014\nthat contains 89.97% of the energy. Reducing the error tolerance improves the estimate.\nCH7MP2(1,0.9,0.00005) returns an essential bandwidth W = 5.3321 that contains 90.00% of\nthe energy. These essential bandwidth calculations are consistent with estimates presented after\nEx. 7.2.", - "type": "text" - }, - { - "block_id": "p779-b6", - "global_id": 22745, - "bbox": [ - 127.59, - 341.37, - 258.58, - 353.33 - ], - "text": "7.9-3 Spectral Sampling", - "type": "text" - }, - { - "block_id": "p779-b7", - "global_id": 22746, - "bbox": [ - 127.59, - 359.04, - 516.14, - 405.28 - ], - "text": "Consider a signal with finite duration τ. A periodic signal xT0(t) is constructed by repeating x(t)\nevery T0 seconds, where T0 ≥τ. From Eq. (7.5), we can write the Fourier series coefficients of\nxT0(t) as Dn = (1/T0)X(n2π/T0). Put another way, the Fourier series coefficients are obtained by\nsampling the spectrum X(ω).", - "type": "text" - }, - { - "block_id": "p779-b8", - "global_id": 22747, - "bbox": [ - 127.59, - 407.28, - 516.15, - 465.06 - ], - "text": "By using spectral sampling, it is simple to determine the Fourier series coefficients for an\narbitrary duty-cycle, square-pulse periodic signal. The square pulse x(t) = rect(t/τ) has spectrum\nX(ω) = τ sinc(ωτ/2). Thus, the nth Fourier coefficient of the periodic extension xT0(t) is Dn =\n(τ/T0)sinc(nπτ/T0). As in Ex. 6.4, τ = π and T0 = 2π provide a square-pulse periodic signal.\nThe Fourier coefficients are determined by", - "type": "text" - }, - { - "block_id": "p779-b9", - "global_id": 22748, - "bbox": [ - 127.59, - 474.49, - 357.73, - 520.33 - ], - "text": ">>\ntau = pi; T_0 = 2*pi; n = [0:10];\n>>\nD_n = tau/T_0*MS7P1(n*pi*tau/T_0);\n>>\nstem(n,D_n); xlabel(’n’); ylabel(’D_n’);\n>>\naxis([-0.5 10.5 -0.2 0.55]);", - "type": "text" - }, - { - "block_id": "p779-b10", - "global_id": 22749, - "bbox": [ - 127.59, - 528.77, - 516.12, - 551.1 - ], - "text": "The results, shown in Fig. 7.50, agree with Fig. 6.6b. Doubling the period to T0 = 4π effectively\ndoubles the density of spectral samples and halves the spectral amplitude, as shown in Fig. 7.51.", - "type": "text" - }, - { - "block_id": "p779-b11", - "global_id": 22750, - "bbox": [ - 127.59, - 553.0, - 516.14, - 588.46 - ], - "text": "As T0 increases, the spectral sampling becomes progressively finer while the amplitude\nbecomes infinitesimal. An evolution of the Fourier series toward the Fourier integral is seen by\nallowing the period T0 to become large. Figure 7.52 shows the result for T0 = 40π.", - "type": "text" - }, - { - "block_id": "p779-b12", - "global_id": 22751, - "bbox": [ - 127.59, - 588.55, - 516.14, - 634.79 - ], - "text": "If T0 = τ, the signal xT0 is a constant and the spectrum should concentrate energy at dc. In this\ncase, the sinc function is sampled at the zero crossings and Dn = 0 for all n not equal to 0. Only\nthe sample corresponding to n = 0 is nonzero, indicating a dc signal, as expected. It is a simple\nmatter to modify the previous code to verify this case.", - "type": "text" - } - ] - }, - { - "page_num": 780, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p780-b0", - "global_id": 22752, - "bbox": [ - 60.0, - 62.89, - 456.14, - 71.98 - ], - "text": "760\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p780-b1", - "global_id": 22753, - "bbox": [ - 130.59, - 167.57, - 471.69, - 188.76 - ], - "text": "0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nn", - "type": "text" - }, - { - "block_id": "p780-b2", - "global_id": 22754, - "bbox": [ - 123.79, - 145.74, - 127.79, - 153.74 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p780-b3", - "global_id": 22755, - "bbox": [ - 117.05, - 121.98, - 127.72, - 129.98 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p780-b4", - "global_id": 22756, - "bbox": [ - 117.05, - 98.23, - 127.72, - 106.23 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p780-b5", - "global_id": 22757, - "bbox": [ - 102.54, - 121.34, - 113.82, - 130.33 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p780-b6", - "global_id": 22758, - "bbox": [ - 101.84, - 195.06, - 292.5, - 205.37 - ], - "text": "Figure 7.50 Fourier spectra for τ = π and T0 = 2π.", - "type": "text" - }, - { - "block_id": "p780-b7", - "global_id": 22759, - "bbox": [ - 130.02, - 315.74, - 470.27, - 336.93 - ], - "text": "0\n2\n4\n6\n8\n10\n12\n14\n16\n18\n20\nn", - "type": "text" - }, - { - "block_id": "p780-b8", - "global_id": 22760, - "bbox": [ - 123.79, - 293.91, - 127.79, - 301.91 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p780-b9", - "global_id": 22761, - "bbox": [ - 117.05, - 270.15, - 127.72, - 278.15 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p780-b10", - "global_id": 22762, - "bbox": [ - 117.05, - 246.4, - 127.72, - 254.4 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p780-b11", - "global_id": 22763, - "bbox": [ - 102.54, - 269.51, - 113.82, - 278.5 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p780-b12", - "global_id": 22764, - "bbox": [ - 101.84, - 343.24, - 292.5, - 353.55 - ], - "text": "Figure 7.51 Fourier spectra for τ = π and T0 = 4π.", - "type": "text" - }, - { - "block_id": "p780-b13", - "global_id": 22765, - "bbox": [ - 127.77, - 469.9, - 469.77, - 491.1 - ], - "text": "0\n20\n40\n60\n80\n100\n120\n140\n160\n180\n200\nn", - "type": "text" - }, - { - "block_id": "p780-b14", - "global_id": 22766, - "bbox": [ - 121.54, - 448.64, - 125.54, - 456.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p780-b15", - "global_id": 22767, - "bbox": [ - 117.54, - 426.46, - 125.54, - 434.46 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p780-b16", - "global_id": 22768, - "bbox": [ - 117.54, - 404.27, - 125.54, - 412.27 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p780-b17", - "global_id": 22769, - "bbox": [ - 102.54, - 425.91, - 113.82, - 434.92 - ], - "text": "Dn", - "type": "text" - }, - { - "block_id": "p780-b18", - "global_id": 22770, - "bbox": [ - 130.39, - 386.09, - 151.15, - 395.56 - ], - "text": "× 10–3", - "type": "text" - }, - { - "block_id": "p780-b19", - "global_id": 22771, - "bbox": [ - 101.84, - 497.41, - 296.98, - 507.72 - ], - "text": "Figure 7.52 Fourier spectra for τ = π and T0 = 40π.", - "type": "text" - }, - { - "block_id": "p780-b20", - "global_id": 22772, - "bbox": [ - 101.84, - 537.03, - 273.92, - 548.99 - ], - "text": "7.9-4 Kaiser Window Functions", - "type": "text" - }, - { - "block_id": "p780-b21", - "global_id": 22773, - "bbox": [ - 101.84, - 555.12, - 490.37, - 577.04 - ], - "text": "A window function is useful only if it can be easily computed and applied to a signal. The Kaiser\nwindow, for example, is flexible but appears rather intimidating:", - "type": "text" - }, - { - "block_id": "p780-b22", - "global_id": 22774, - "bbox": [ - 200.93, - 610.07, - 233.3, - 621.14 - ], - "text": "wK(t) =", - "type": "text" - }, - { - "block_id": "p780-b23", - "global_id": 22775, - "bbox": [ - 235.35, - 589.64, - 243.24, - 608.57 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p780-b24", - "global_id": 22776, - "bbox": [ - 235.35, - 616.54, - 243.24, - 626.5 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p780-b25", - "global_id": 22777, - "bbox": [ - 249.42, - 598.53, - 256.22, - 609.37 - ], - "text": "I0", - "type": "text" - }, - { - "block_id": "p780-b27", - "global_id": 22778, - "bbox": [ - 260.73, - 598.22, - 266.42, - 608.18 - ], - "text": "α", - "type": "text" - }, - { - "block_id": "p780-b29", - "global_id": 22779, - "bbox": [ - 275.08, - 590.21, - 325.28, - 608.6 - ], - "text": "1 −4(t/T)2", - "type": "text" - }, - { - "block_id": "p780-b30", - "global_id": 22780, - "bbox": [ - 276.94, - 605.8, - 382.87, - 624.02 - ], - "text": "I0(α)\n|t| < T/2", - "type": "text" - }, - { - "block_id": "p780-b31", - "global_id": 22781, - "bbox": [ - 284.86, - 623.39, - 385.14, - 633.36 - ], - "text": "0\notherwise", - "type": "text" - } - ] - }, - { - "page_num": 781, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p781-b0", - "global_id": 22782, - "bbox": [ - 310.23, - 62.89, - 516.12, - 71.98 - ], - "text": "7.9\nMATLAB: Fourier Transform Topics\n761", - "type": "text" - }, - { - "block_id": "p781-b1", - "global_id": 22783, - "bbox": [ - 127.59, - 85.46, - 516.12, - 107.79 - ], - "text": "Fortunately, the bark of a Kaiser window is worse than its bite! The function I0(x), a zero-order\nmodified Bessel function of the first kind, can be computed according to", - "type": "text" - }, - { - "block_id": "p781-b2", - "global_id": 22784, - "bbox": [ - 280.79, - 129.63, - 309.8, - 140.78 - ], - "text": "I0(x) =", - "type": "text" - }, - { - "block_id": "p781-b3", - "global_id": 22785, - "bbox": [ - 311.85, - 119.45, - 325.95, - 130.13 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p781-b4", - "global_id": 22786, - "bbox": [ - 312.83, - 144.02, - 324.97, - 151.29 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p781-b5", - "global_id": 22787, - "bbox": [ - 327.05, - 115.64, - 346.46, - 132.92 - ], - "text": "xk", - "type": "text" - }, - { - "block_id": "p781-b6", - "global_id": 22788, - "bbox": [ - 334.97, - 136.43, - 351.01, - 147.08 - ], - "text": "2kk!", - "type": "text" - }, - { - "block_id": "p781-b7", - "global_id": 22789, - "bbox": [ - 352.21, - 115.64, - 362.42, - 126.73 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p781-b8", - "global_id": 22790, - "bbox": [ - 127.59, - 162.9, - 516.14, - 209.02 - ], - "text": "or, more simply, by using the MATLAB function besseli(0,x). In fact, MATLAB supports a\nwide range of Bessel functions, including Bessel functions of the first and second kinds (besselj\nand bessely), modified Bessel functions of the first and second kinds (besseli and besselk),\nHankel functions (besselh), and Airy functions (airy).", - "type": "text" - }, - { - "block_id": "p781-b9", - "global_id": 22791, - "bbox": [ - 145.52, - 210.31, - 479.76, - 220.98 - ], - "text": "Program CH7MP3 computes Kaiser windows at times t by using parameters T and α.", - "type": "text" - }, - { - "block_id": "p781-b10", - "global_id": 22792, - "bbox": [ - 127.59, - 232.73, - 509.4, - 350.29 - ], - "text": "function [w_K] = CH7MP3(t,T,alpha)\n% CH7MP3.m : Chapter 7, MATLAB Program 3\n% Function M-file computes a width-T Kaiser window using parameter alpha.\n% Alpha can also be a string identifier: ’rectangular’, ’Hamming’, or\n% ’Blackman’.\n% INPUTS:\nt = independent variable of the window function\n%\nT = window width\n%\nalpha = Kaiser parameter or string identifier\n% OUTPUTS:\nw_K = Kaiser window function\nif strncmpi(alpha,’rectangular’,1),", - "type": "text" - }, - { - "block_id": "p781-b11", - "global_id": 22793, - "bbox": [ - 127.59, - 352.28, - 310.66, - 374.2 - ], - "text": "alpha = 0;\nelseif strncmpi(alpha,’Hamming’,3),", - "type": "text" - }, - { - "block_id": "p781-b12", - "global_id": 22794, - "bbox": [ - 127.59, - 376.19, - 315.89, - 398.11 - ], - "text": "alpha = 5.4414;\nelseif strncmpi(alpha,’Blackman’,1),", - "type": "text" - }, - { - "block_id": "p781-b13", - "global_id": 22795, - "bbox": [ - 127.59, - 400.1, - 253.12, - 422.02 - ], - "text": "alpha = 8.885;\nelseif isa(alpha,’char’)", - "type": "text" - }, - { - "block_id": "p781-b14", - "global_id": 22796, - "bbox": [ - 127.59, - 424.01, - 478.03, - 469.84 - ], - "text": "disp(’Unrecognized string identifier.’); return\nend\nw_K = zeros(size(t)); i = find(abs(t)>\nt = [-0.6:.001:0.6]; T = 1;\n>>\nplot(t,CH7MP3(t,T,’r’),’k-’,t,CH7MP3(t,T,’ham’),’k-.’,t,CH7MP3(t,T,’b’),’k--’);\n>>\naxis([-0.6 0.6 -.1 1.1]); xlabel(’t’); ylabel(’w_K(t)’);\n>>\nlegend(’Rectangular’,’Hamming’,’Blackman’,’Location’,’EastOutside’);", - "type": "text" - }, - { - "block_id": "p782-b12", - "global_id": 22811, - "bbox": [ - 102.2, - 329.91, - 202.15, - 343.86 - ], - "text": "7.10 SUMMARY", - "type": "text" - }, - { - "block_id": "p782-b13", - "global_id": 22812, - "bbox": [ - 101.84, - 349.86, - 490.39, - 443.5 - ], - "text": "In Ch. 6, we represented periodic signals as a sum of (everlasting) sinusoids or exponentials\n(Fourier series). In this chapter we extended this result to aperiodic signals, which are represented\nby the Fourier integral (instead of the Fourier series). An aperiodic signal x(t) may be regarded as\na periodic signal with period T0 →∞so that the Fourier integral is basically a Fourier series\nwith a fundamental frequency approaching zero. Therefore, for aperiodic signals, the Fourier\nspectra are continuous. This continuity means that a signal is represented as a sum of sinusoids (or\nexponentials) of all frequencies over a continuous frequency interval. The Fourier transform X(ω),\ntherefore, is the spectral density (per unit bandwidth in hertz).", - "type": "text" - }, - { - "block_id": "p782-b14", - "global_id": 22813, - "bbox": [ - 101.85, - 445.49, - 490.39, - 503.28 - ], - "text": "An ever-present aspect of the Fourier transform is the duality between time and frequency,\nwhich also implies duality between the signal x(t) and its transform X(ω). This duality\narises because of near-symmetrical equations for direct and inverse Fourier transforms. The\nduality principle has far-reaching consequences and yields many valuable insights into signal\nanalysis.", - "type": "text" - }, - { - "block_id": "p782-b15", - "global_id": 22814, - "bbox": [ - 101.84, - 505.27, - 490.42, - 598.92 - ], - "text": "The scaling property of the Fourier transform leads to the conclusion that the signal bandwidth\nis inversely proportional to signal duration (signal width). Time shifting of a signal does not change\nits amplitude spectrum, but it does add a linear phase component to its spectrum. Multiplication of\na signal by an exponential ejω0t shifts the spectrum to the right by ω0. In practice, spectral shifting\nis achieved by multiplying a signal by a sinusoid such as cosω0t (rather than the exponential\nejω0t). This process is known as amplitude modulation. Multiplication of two signals results in\nconvolution of their spectra, whereas convolution of two signals results in multiplication of their\nspectra.", - "type": "text" - }, - { - "block_id": "p782-b16", - "global_id": 22815, - "bbox": [ - 101.84, - 600.5, - 490.4, - 634.79 - ], - "text": "For an LTIC system with the frequency response H(ω), the input and output spectra X(ω) and\nY(ω) are related by the equation Y(ω) = X(ω)H(ω). This is valid only for asymptotically stable\nsystems. It also applies to marginally stable systems if the input does not contain a finite-amplitude", - "type": "text" - } - ] - }, - { - "page_num": 783, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p783-b0", - "global_id": 22816, - "bbox": [ - 409.68, - 62.89, - 516.14, - 71.98 - ], - "text": "7.10\nSummary\n763", - "type": "text" - }, - { - "block_id": "p783-b1", - "global_id": 22817, - "bbox": [ - 127.59, - 85.82, - 516.15, - 179.47 - ], - "text": "sinusoid of the natural frequency of the system. For asymptotically unstable systems, the frequency\nresponse H(ω) does not exist. For distortionless transmission of a signal through an LTIC system,\nthe amplitude response |H(ω)| of the system must be constant, and the phase response̸\nH(ω)\nshould be a linear function of ω over a band of interest. Ideal filters, which allow distortionless\ntransmission of a certain band of frequencies and suppress all the remaining frequencies, are\nphysically unrealizable (noncausal). In fact, it is impossible to build a physical system with zero\ngain [H(ω) = 0] over a finite band of frequencies. Such systems (which include ideal filters) can\nbe realized only with infinite time delay in the response.", - "type": "text" - }, - { - "block_id": "p783-b2", - "global_id": 22818, - "bbox": [ - 127.59, - 180.11, - 516.14, - 215.33 - ], - "text": "The energy of a signal x(t) is equal to 1/2π times the area under |X(ω)2| (Parseval’s theorem).\nThe energy contributed by spectral components within a band f (in hertz) is given by |X(ω)|2f.\nTherefore, |X(ω)|2 is the energy spectral density per unit bandwidth (in hertz).", - "type": "text" - }, - { - "block_id": "p783-b3", - "global_id": 22819, - "bbox": [ - 127.59, - 217.32, - 516.16, - 299.02 - ], - "text": "The process of modulation shifts the signal spectrum to different frequencies. Modulation is\nused for many reasons: to transmit several messages simultaneously over the same channel for the\nsake of utilizing channel’s high bandwidth, to effectively radiate power over a radio link, to shift a\nsignal spectrum at higher frequencies to overcome the difficulties associated with signal processing\nat lower frequencies, and to effect the exchange of transmission bandwidth and transmission power\nrequired to transmit data at a certain rate. Broadly speaking, there are two types of modulation,\namplitude and angle modulation. Each class has several subclasses.", - "type": "text" - }, - { - "block_id": "p783-b4", - "global_id": 22820, - "bbox": [ - 127.59, - 301.02, - 516.15, - 478.35 - ], - "text": "In practice, we often need to truncate data. Truncating is like viewing data through a window,\nwhich permits only certain portions of the data to be seen and hides (suppresses) the remainder.\nAbrupt truncation of data amounts to a rectangular window, which assigns a unit weight to data\nseen from the window and zero weight to the remaining data. Tapered windows, on the other hand,\nreduce the weight gradually from 1 to 0. Data truncation can cause some unsuspected problems.\nFor example, in computation of the Fourier transform, windowing (data truncation) causes spectral\nspreading (spectral smearing) that is characteristic of the window function used. A rectangular\nwindow results in the least spreading, but it does so at the cost of a high and oscillatory spectral\nleakage outside the signal band, which decays slowly as 1/ω. In comparison to a rectangular\nwindow, tapered windows, in general, have larger spectral spreading (smearing), but the spectral\nleakage is smaller and decays faster with frequency. If we try to reduce spectral leakage by using a\nsmoother window, the spectral spreading increases. Fortunately, spectral spreading can be reduced\nby increasing the window width. Therefore, we can achieve a given combination of spectral spread\n(transition bandwidth) and leakage characteristics by choosing a suitable tapered window function\nof a sufficiently long width T.", - "type": "text" - }, - { - "block_id": "p783-b5", - "global_id": 22821, - "bbox": [ - 127.86, - 505.06, - 215.04, - 516.02 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p783-b6", - "global_id": 22822, - "bbox": [ - 127.59, - 523.09, - 516.12, - 543.11 - ], - "text": "1.\nChurchill, R. V., and Brown, J. W. Fourier Series and Boundary Value Problems, 3rd ed. McGraw-Hill,\nNew York, 1978.", - "type": "text" - }, - { - "block_id": "p783-b7", - "global_id": 22823, - "bbox": [ - 127.59, - 547.99, - 509.65, - 557.05 - ], - "text": "2.\nBracewell, R. N. Fourier Transform and Its Applications, rev. 2nd ed. McGraw-Hill, New York, 1986.", - "type": "text" - }, - { - "block_id": "p783-b8", - "global_id": 22824, - "bbox": [ - 127.59, - 561.94, - 421.41, - 571.0 - ], - "text": "3.\nGuillemin, E. A. Theory of Linear Physical Systems. Wiley, New York, 1963.", - "type": "text" - }, - { - "block_id": "p783-b9", - "global_id": 22825, - "bbox": [ - 127.59, - 575.89, - 516.12, - 595.9 - ], - "text": "4.\nLathi, B. P. Modern Digital and Analog Communication Systems, 3rd ed. Oxford University Press, New\nYork, 1998.", - "type": "text" - }, - { - "block_id": "p783-b10", - "global_id": 22826, - "bbox": [ - 127.59, - 600.8, - 445.77, - 609.85 - ], - "text": "5.\nHamming, R. W. Digital Filters, 2nd ed. Prentice-Hall, Englewood Cliffs, NJ, 1983.", - "type": "text" - }, - { - "block_id": "p783-b11", - "global_id": 22827, - "bbox": [ - 127.59, - 614.84, - 516.14, - 634.76 - ], - "text": "6.\nHarris, F. J. On the use of windows for harmonic analysis with the discrete Fourier transform.\nProceedings of the IEEE, vol. 66, no. 1, pp. 51–83, January 1978.", - "type": "text" - } - ] - }, - { - "page_num": 784, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p784-b0", - "global_id": 22828, - "bbox": [ - 60.0, - 60.36, - 456.14, - 69.45 - ], - "text": "764\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p784-b1", - "global_id": 22829, - "bbox": [ - 80.93, - 91.05, - 191.52, - 107.98 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p784-b2", - "global_id": 22830, - "bbox": [ - 62.08, - 117.1, - 263.24, - 151.24 - ], - "text": "7.1-1\nSuppose signal x(t) = t2 [u(t) −u(t −2)] has\nFourier transform X(ω). Define a 3-periodic\nreplication of x(t) as y(t) = %∞", - "type": "text" - }, - { - "block_id": "p784-b3", - "global_id": 22831, - "bbox": [ - 90.72, - 141.9, - 263.24, - 173.16 - ], - "text": "n=−∞2x(t −1 −\n3n). Determine Yk, the Fourier series of y(t), in\nterms of the Fourier transform X(·).", - "type": "text" - }, - { - "block_id": "p784-b4", - "global_id": 22832, - "bbox": [ - 62.08, - 177.77, - 263.24, - 198.07 - ], - "text": "7.1-2\nShow that for a real x(t), Eq. (7.10) can be\nexpressed as", - "type": "text" - }, - { - "block_id": "p784-b5", - "global_id": 22833, - "bbox": [ - 102.28, - 207.7, - 132.89, - 222.95 - ], - "text": "x(t) = 1", - "type": "text" - }, - { - "block_id": "p784-b6", - "global_id": 22834, - "bbox": [ - 127.5, - 219.97, - 132.88, - 228.94 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p784-b7", - "global_id": 22835, - "bbox": [ - 135.97, - 201.41, - 151.42, - 212.38 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p784-b8", - "global_id": 22836, - "bbox": [ - 140.71, - 213.61, - 251.52, - 230.39 - ], - "text": "0\n|X(ω)| cos[ωt +̸ X(ω)]dω", - "type": "text" - }, - { - "block_id": "p784-b9", - "global_id": 22837, - "bbox": [ - 90.72, - 238.42, - 263.24, - 269.3 - ], - "text": "This is the trigonometric form of the Fourier\nintegral. Compare this with the compact trigono-\nmetric Fourier series.", - "type": "text" - }, - { - "block_id": "p784-b10", - "global_id": 22838, - "bbox": [ - 62.08, - 273.91, - 254.31, - 283.25 - ], - "text": "7.1-3\nShow that if x(t) is an even function of t, then", - "type": "text" - }, - { - "block_id": "p784-b11", - "global_id": 22839, - "bbox": [ - 128.56, - 298.41, - 162.33, - 307.75 - ], - "text": "X(ω) = 2", - "type": "text" - }, - { - "block_id": "p784-b12", - "global_id": 22840, - "bbox": [ - 163.33, - 286.21, - 178.78, - 297.18 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p784-b13", - "global_id": 22841, - "bbox": [ - 168.06, - 298.41, - 225.25, - 315.19 - ], - "text": "0\nx(t)cos ωtdt", - "type": "text" - }, - { - "block_id": "p784-b14", - "global_id": 22842, - "bbox": [ - 90.72, - 322.9, - 228.25, - 332.24 - ], - "text": "and if x(t) is an odd function of t, then", - "type": "text" - }, - { - "block_id": "p784-b15", - "global_id": 22843, - "bbox": [ - 125.05, - 347.41, - 168.32, - 356.74 - ], - "text": "X(ω) = −2j", - "type": "text" - }, - { - "block_id": "p784-b16", - "global_id": 22844, - "bbox": [ - 169.32, - 335.2, - 184.77, - 346.17 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p784-b17", - "global_id": 22845, - "bbox": [ - 174.04, - 347.41, - 228.75, - 364.18 - ], - "text": "0\nx(t)sinωtdt", - "type": "text" - }, - { - "block_id": "p784-b18", - "global_id": 22846, - "bbox": [ - 90.72, - 371.89, - 263.23, - 392.19 - ], - "text": "Hence, prove that if x(t) is a real and even\nfunction of t, then X(ω) is a real and even", - "type": "text" - }, - { - "block_id": "p784-b19", - "global_id": 22847, - "bbox": [ - 317.86, - 118.48, - 490.38, - 149.74 - ], - "text": "function of ω. In addition, if x(t) is a real and\nodd function of t, then X(ω) is an imaginary and\nodd function of ω.", - "type": "text" - }, - { - "block_id": "p784-b20", - "global_id": 22848, - "bbox": [ - 289.22, - 154.35, - 490.38, - 174.64 - ], - "text": "7.1-4\nA signal x(t) can be expressed as the sum of even\nand odd components (see Sec. 1.5-2):", - "type": "text" - }, - { - "block_id": "p784-b21", - "global_id": 22849, - "bbox": [ - 370.34, - 185.56, - 437.92, - 195.57 - ], - "text": "x(t) = xe(t) + xo(t)", - "type": "text" - }, - { - "block_id": "p784-b22", - "global_id": 22850, - "bbox": [ - 318.37, - 205.81, - 479.47, - 215.15 - ], - "text": "(a) If x(t) ⇐⇒X(ω), show that for real x(t),", - "type": "text" - }, - { - "block_id": "p784-b23", - "global_id": 22851, - "bbox": [ - 374.65, - 226.06, - 449.05, - 236.07 - ], - "text": "xe(t) ⇐⇒Re[X(ω)]", - "type": "text" - }, - { - "block_id": "p784-b24", - "global_id": 22852, - "bbox": [ - 333.3, - 246.68, - 346.25, - 255.65 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p784-b25", - "global_id": 22853, - "bbox": [ - 373.23, - 257.27, - 450.48, - 267.28 - ], - "text": "xo(t) ⇐⇒j Im[X(ω)]", - "type": "text" - }, - { - "block_id": "p784-b26", - "global_id": 22854, - "bbox": [ - 317.86, - 275.24, - 490.39, - 284.2 - ], - "text": "(b) Verify these results by finding the Fourier", - "type": "text" - }, - { - "block_id": "p784-b27", - "global_id": 22855, - "bbox": [ - 333.31, - 286.2, - 490.39, - 317.08 - ], - "text": "transforms of the even and odd components\nof the following signals: (i) u(t) and (ii)\ne−atu(t).", - "type": "text" - }, - { - "block_id": "p784-b28", - "global_id": 22856, - "bbox": [ - 289.22, - 321.99, - 490.39, - 341.99 - ], - "text": "7.1-5\nUsing Eq. (7.9), find the Fourier transforms of\nthe signals x(t) in Fig. P7.1-5.", - "type": "text" - }, - { - "block_id": "p784-b29", - "global_id": 22857, - "bbox": [ - 289.22, - 346.9, - 490.39, - 366.9 - ], - "text": "7.1-6\nUsing Eq. (7.9), find the Fourier transforms of\nthe signals depicted in Fig. P7.1-6.", - "type": "text" - }, - { - "block_id": "p784-b30", - "global_id": 22858, - "bbox": [ - 289.22, - 371.81, - 490.39, - 391.81 - ], - "text": "7.1-7\nUse Eq. (7.10) to find the inverse Fourier\ntransforms of the spectra in Fig. P7.1-7.", - "type": "text" - }, - { - "block_id": "p784-b31", - "global_id": 22859, - "bbox": [ - 111.9, - 485.64, - 168.84, - 493.72 - ], - "text": "0\nT", - "type": "text" - }, - { - "block_id": "p784-b32", - "global_id": 22860, - "bbox": [ - 101.84, - 434.69, - 105.84, - 442.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p784-b33", - "global_id": 22861, - "bbox": [ - 135.44, - 454.96, - 148.65, - 464.58 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p784-b34", - "global_id": 22862, - "bbox": [ - 121.26, - 441.46, - 132.37, - 449.54 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p784-b35", - "global_id": 22863, - "bbox": [ - 193.65, - 485.64, - 320.03, - 493.72 - ], - "text": "0\nt\nt\nT", - "type": "text" - }, - { - "block_id": "p784-b36", - "global_id": 22864, - "bbox": [ - 224.62, - 434.69, - 228.62, - 442.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p784-b37", - "global_id": 22865, - "bbox": [ - 258.88, - 455.13, - 267.1, - 464.58 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p784-b38", - "global_id": 22866, - "bbox": [ - 246.59, - 440.51, - 257.7, - 448.59 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p784-b39", - "global_id": 22867, - "bbox": [ - 150.37, - 499.2, - 283.63, - 507.2 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p784-b40", - "global_id": 22868, - "bbox": [ - 101.84, - 516.11, - 153.49, - 525.08 - ], - "text": "Figure P7.1-5", - "type": "text" - }, - { - "block_id": "p784-b41", - "global_id": 22869, - "bbox": [ - 113.08, - 549.94, - 117.08, - 557.94 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p784-b42", - "global_id": 22870, - "bbox": [ - 113.08, - 567.32, - 117.08, - 575.32 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p784-b43", - "global_id": 22871, - "bbox": [ - 169.36, - 590.77, - 223.26, - 598.77 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p784-b44", - "global_id": 22872, - "bbox": [ - 166.92, - 606.22, - 175.8, - 614.22 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p784-b45", - "global_id": 22873, - "bbox": [ - 226.14, - 577.19, - 393.88, - 585.19 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p784-b46", - "global_id": 22874, - "bbox": [ - 125.78, - 541.5, - 316.77, - 549.58 - ], - "text": "x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p784-b47", - "global_id": 22875, - "bbox": [ - 326.73, - 552.39, - 330.73, - 560.39 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p784-b48", - "global_id": 22876, - "bbox": [ - 268.03, - 589.45, - 374.7, - 598.27 - ], - "text": "t\nt\n0", - "type": "text" - }, - { - "block_id": "p784-b49", - "global_id": 22877, - "bbox": [ - 317.96, - 606.22, - 328.09, - 614.22 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p784-b50", - "global_id": 22878, - "bbox": [ - 104.83, - 620.4, - 156.48, - 629.36 - ], - "text": "Figure P7.1-6", - "type": "text" - } - ] - }, - { - "page_num": 785, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p785-b0", - "global_id": 22879, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n765", - "type": "text" - }, - { - "block_id": "p785-b1", - "global_id": 22880, - "bbox": [ - 346.56, - 108.66, - 350.56, - 116.66 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p785-b2", - "global_id": 22881, - "bbox": [ - 313.72, - 126.91, - 370.14, - 135.21 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p785-b3", - "global_id": 22882, - "bbox": [ - 345.01, - 92.39, - 349.01, - 100.39 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p785-b4", - "global_id": 22883, - "bbox": [ - 346.8, - 140.74, - 356.45, - 148.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p785-b5", - "global_id": 22884, - "bbox": [ - 251.82, - 113.75, - 404.78, - 133.35 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p785-b6", - "global_id": 22885, - "bbox": [ - 185.94, - 140.74, - 194.82, - 148.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p785-b7", - "global_id": 22886, - "bbox": [ - 197.79, - 127.26, - 201.79, - 135.26 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p785-b8", - "global_id": 22887, - "bbox": [ - 175.44, - 87.37, - 340.18, - 106.04 - ], - "text": "v2\nX(v)\nX(v)", - "type": "text" - }, - { - "block_id": "p785-b9", - "global_id": 22888, - "bbox": [ - 140.04, - 125.16, - 394.23, - 135.21 - ], - "text": "v0\nv0\n2\n2", - "type": "text" - }, - { - "block_id": "p785-b10", - "global_id": 22889, - "bbox": [ - 130.58, - 154.92, - 182.22, - 163.88 - ], - "text": "Figure P7.1-7", - "type": "text" - }, - { - "block_id": "p785-b11", - "global_id": 22890, - "bbox": [ - 200.51, - 181.15, - 204.51, - 189.15 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p785-b12", - "global_id": 22891, - "bbox": [ - 201.52, - 222.89, - 205.52, - 230.89 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p785-b13", - "global_id": 22892, - "bbox": [ - 228.86, - 186.74, - 246.86, - 194.84 - ], - "text": "cos v", - "type": "text" - }, - { - "block_id": "p785-b14", - "global_id": 22893, - "bbox": [ - 178.15, - 177.6, - 352.27, - 185.7 - ], - "text": "X(v)\nX(v)", - "type": "text" - }, - { - "block_id": "p785-b15", - "global_id": 22894, - "bbox": [ - 360.86, - 184.18, - 364.86, - 192.18 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p785-b16", - "global_id": 22895, - "bbox": [ - 360.08, - 222.9, - 364.08, - 230.9 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p785-b17", - "global_id": 22896, - "bbox": [ - 194.18, - 243.2, - 361.98, - 251.2 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p785-b18", - "global_id": 22897, - "bbox": [ - 253.57, - 209.91, - 433.12, - 217.91 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p785-b19", - "global_id": 22898, - "bbox": [ - 148.08, - 220.68, - 416.44, - 230.5 - ], - "text": "v0\nv0\np", - "type": "text" - }, - { - "block_id": "p785-b20", - "global_id": 22899, - "bbox": [ - 140.86, - 220.9, - 252.1, - 237.62 - ], - "text": "2\n\np", - "type": "text" - }, - { - "block_id": "p785-b21", - "global_id": 22900, - "bbox": [ - 247.43, - 229.62, - 251.43, - 237.62 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p785-b22", - "global_id": 22901, - "bbox": [ - 130.58, - 257.38, - 182.22, - 266.35 - ], - "text": "Figure P7.1-8", - "type": "text" - }, - { - "block_id": "p785-b23", - "global_id": 22902, - "bbox": [ - 87.82, - 278.35, - 288.98, - 298.35 - ], - "text": "7.1-8\nUse Eq. (7.10) to find the inverse Fourier\ntransforms of the spectra in Fig. P7.1-8.", - "type": "text" - }, - { - "block_id": "p785-b24", - "global_id": 22903, - "bbox": [ - 87.82, - 303.28, - 234.09, - 312.62 - ], - "text": "7.1-9\nIf x(t) ⇐⇒X(ω), then show that", - "type": "text" - }, - { - "block_id": "p785-b25", - "global_id": 22904, - "bbox": [ - 169.17, - 331.7, - 195.06, - 341.04 - ], - "text": "X(0) =", - "type": "text" - }, - { - "block_id": "p785-b26", - "global_id": 22905, - "bbox": [ - 196.89, - 319.5, - 212.35, - 330.47 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p785-b27", - "global_id": 22906, - "bbox": [ - 201.63, - 341.74, - 213.29, - 348.22 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p785-b28", - "global_id": 22907, - "bbox": [ - 214.79, - 331.7, - 236.12, - 340.95 - ], - "text": "x(t)dt", - "type": "text" - }, - { - "block_id": "p785-b29", - "global_id": 22908, - "bbox": [ - 116.46, - 360.9, - 129.4, - 369.87 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p785-b30", - "global_id": 22909, - "bbox": [ - 158.58, - 373.19, - 193.26, - 388.44 - ], - "text": "x(0) = 1", - "type": "text" - }, - { - "block_id": "p785-b31", - "global_id": 22910, - "bbox": [ - 185.63, - 385.47, - 195.5, - 394.81 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p785-b32", - "global_id": 22911, - "bbox": [ - 198.59, - 366.9, - 214.03, - 377.87 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p785-b33", - "global_id": 22912, - "bbox": [ - 203.32, - 389.14, - 214.98, - 395.62 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p785-b34", - "global_id": 22913, - "bbox": [ - 216.48, - 379.1, - 246.7, - 388.35 - ], - "text": "X(ω)dω", - "type": "text" - }, - { - "block_id": "p785-b35", - "global_id": 22914, - "bbox": [ - 116.46, - 405.31, - 170.03, - 414.27 - ], - "text": "Also show that", - "type": "text" - }, - { - "block_id": "p785-b36", - "global_id": 22915, - "bbox": [ - 133.51, - 419.68, - 148.97, - 430.66 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p785-b37", - "global_id": 22916, - "bbox": [ - 138.26, - 441.93, - 149.92, - 448.4 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p785-b38", - "global_id": 22917, - "bbox": [ - 151.41, - 431.89, - 195.88, - 441.23 - ], - "text": "sinc(x)dx =", - "type": "text" - }, - { - "block_id": "p785-b39", - "global_id": 22918, - "bbox": [ - 197.71, - 419.68, - 213.17, - 430.66 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p785-b40", - "global_id": 22919, - "bbox": [ - 202.45, - 441.93, - 214.11, - 448.4 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p785-b41", - "global_id": 22920, - "bbox": [ - 215.6, - 428.19, - 271.03, - 441.23 - ], - "text": "sinc2 (x)dx = π", - "type": "text" - }, - { - "block_id": "p785-b42", - "global_id": 22921, - "bbox": [ - 314.97, - 278.35, - 456.11, - 287.39 - ], - "text": "7.2-1\nSketch the following functions:", - "type": "text" - }, - { - "block_id": "p785-b43", - "global_id": 22922, - "bbox": [ - 343.61, - 289.01, - 401.15, - 309.31 - ], - "text": "(a) rect (t/2)\n(b) (3ω/100)", - "type": "text" - }, - { - "block_id": "p785-b44", - "global_id": 22923, - "bbox": [ - 343.61, - 310.93, - 417.46, - 331.23 - ], - "text": "(c) rect ((t −10)/8)\n(d) sinc(πω/5)", - "type": "text" - }, - { - "block_id": "p785-b45", - "global_id": 22924, - "bbox": [ - 344.12, - 332.85, - 423.29, - 342.18 - ], - "text": "(e) sinc((ω/5) −2π)", - "type": "text" - }, - { - "block_id": "p785-b46", - "global_id": 22925, - "bbox": [ - 345.11, - 343.81, - 437.26, - 353.15 - ], - "text": "(f) sinc(t/5)rect (t/10π)", - "type": "text" - }, - { - "block_id": "p785-b47", - "global_id": 22926, - "bbox": [ - 314.98, - 361.07, - 516.15, - 392.03 - ], - "text": "7.2-2\nUsing Eq. (7.9), show that the Fourier transform\nof rect (t −5) is sinc(ω/2)e−j5ω. Sketch the\nresulting amplitude and phase spectra.", - "type": "text" - }, - { - "block_id": "p785-b48", - "global_id": 22927, - "bbox": [ - 314.97, - 399.95, - 516.13, - 419.95 - ], - "text": "7.2-3\nUsing Eq. (7.10), show that the inverse Fourier\ntransform of rect ((ω −10)/2π) is sinc(πt)ej10t.", - "type": "text" - }, - { - "block_id": "p785-b49", - "global_id": 22928, - "bbox": [ - 314.97, - 427.57, - 516.13, - 447.87 - ], - "text": "7.2-4\nFind the inverse Fourier transform of X(ω) for\nthe spectra illustrated in Fig. P7.2-4. [Hint:", - "type": "text" - }, - { - "block_id": "p785-b50", - "global_id": 22929, - "bbox": [ - 204.83, - 479.41, - 208.83, - 487.41 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p785-b51", - "global_id": 22930, - "bbox": [ - 204.91, - 513.56, - 208.91, - 521.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p785-b52", - "global_id": 22931, - "bbox": [ - 260.51, - 501.25, - 265.84, - 509.25 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p785-b53", - "global_id": 22932, - "bbox": [ - 260.51, - 579.16, - 265.85, - 587.16 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p785-b54", - "global_id": 22933, - "bbox": [ - 432.34, - 501.25, - 437.67, - 509.25 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p785-b55", - "global_id": 22934, - "bbox": [ - 432.34, - 579.16, - 437.67, - 587.16 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p785-b56", - "global_id": 22935, - "bbox": [ - 197.43, - 614.19, - 206.31, - 622.19 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p785-b57", - "global_id": 22936, - "bbox": [ - 176.12, - 465.69, - 368.62, - 473.99 - ], - "text": "X(v)\nX(v)", - "type": "text" - }, - { - "block_id": "p785-b58", - "global_id": 22937, - "bbox": [ - 208.99, - 547.23, - 231.53, - 555.53 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p785-b59", - "global_id": 22938, - "bbox": [ - 391.72, - 558.66, - 414.26, - 566.96 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p785-b60", - "global_id": 22939, - "bbox": [ - 141.07, - 513.46, - 259.9, - 523.28 - ], - "text": "v0\nv0", - "type": "text" - }, - { - "block_id": "p785-b61", - "global_id": 22940, - "bbox": [ - 251.56, - 564.32, - 259.9, - 573.95 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p785-b62", - "global_id": 22941, - "bbox": [ - 424.41, - 513.65, - 432.74, - 523.28 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p785-b63", - "global_id": 22942, - "bbox": [ - 424.41, - 564.32, - 432.74, - 573.95 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p785-b64", - "global_id": 22943, - "bbox": [ - 313.82, - 513.46, - 328.82, - 523.28 - ], - "text": "v0", - "type": "text" - }, - { - "block_id": "p785-b65", - "global_id": 22944, - "bbox": [ - 377.58, - 477.14, - 381.58, - 485.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p785-b66", - "global_id": 22945, - "bbox": [ - 377.58, - 513.56, - 381.58, - 521.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p785-b67", - "global_id": 22946, - "bbox": [ - 369.72, - 614.19, - 379.53, - 622.19 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p785-b68", - "global_id": 22947, - "bbox": [ - 377.58, - 549.94, - 390.92, - 558.24 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p785-b69", - "global_id": 22948, - "bbox": [ - 351.84, - 597.66, - 371.84, - 605.96 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p785-b70", - "global_id": 22949, - "bbox": [ - 141.07, - 579.42, - 328.82, - 589.25 - ], - "text": "v0\nv0", - "type": "text" - }, - { - "block_id": "p785-b71", - "global_id": 22950, - "bbox": [ - 211.88, - 595.91, - 229.1, - 605.74 - ], - "text": "vt0", - "type": "text" - }, - { - "block_id": "p785-b72", - "global_id": 22951, - "bbox": [ - 130.58, - 628.36, - 182.22, - 637.33 - ], - "text": "Figure P7.2-4", - "type": "text" - } - ] - }, - { - "page_num": 786, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p786-b0", - "global_id": 22952, - "bbox": [ - 60.0, - 60.36, - 456.14, - 69.45 - ], - "text": "766\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p786-b1", - "global_id": 22953, - "bbox": [ - 90.72, - 87.53, - 263.25, - 131.0 - ], - "text": "X(ω) = |X(ω)|ej̸\nX(ω). This problem illustrates\nhow different phase spectra (both with the same\namplitude spectrum) represent entirely different\nsignals.]", - "type": "text" - }, - { - "block_id": "p786-b2", - "global_id": 22954, - "bbox": [ - 62.08, - 136.07, - 263.24, - 145.11 - ], - "text": "7.2-5\n(a) Can you find the Fourier transform of", - "type": "text" - }, - { - "block_id": "p786-b3", - "global_id": 22955, - "bbox": [ - 90.72, - 145.67, - 263.24, - 177.99 - ], - "text": "eatu(t) when a>1 by setting s = jω in the\nLaplace transform of eatu(t)? Explain.\n(b) Find the Laplace transform of x(t) shown", - "type": "text" - }, - { - "block_id": "p786-b4", - "global_id": 22956, - "bbox": [ - 106.16, - 179.97, - 263.25, - 232.78 - ], - "text": "in Fig. P7.2-5. Can you find the Fourier\ntransform of x(t) by setting s = jω in its\nLaplace transform? Explain. Verify your\nanswer by finding the Fourier and the\nLaplace transforms of x(t).", - "type": "text" - }, - { - "block_id": "p786-b5", - "global_id": 22957, - "bbox": [ - 135.22, - 275.1, - 143.43, - 284.55 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p786-b6", - "global_id": 22958, - "bbox": [ - 168.47, - 305.09, - 197.16, - 313.16 - ], - "text": "T\nt", - "type": "text" - }, - { - "block_id": "p786-b7", - "global_id": 22959, - "bbox": [ - 94.91, - 253.91, - 106.34, - 261.99 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p786-b8", - "global_id": 22960, - "bbox": [ - 106.15, - 305.17, - 110.15, - 313.17 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p786-b9", - "global_id": 22961, - "bbox": [ - 90.72, - 321.15, - 142.36, - 330.11 - ], - "text": "Figure P7.2-5", - "type": "text" - }, - { - "block_id": "p786-b10", - "global_id": 22962, - "bbox": [ - 62.08, - 351.63, - 263.23, - 371.63 - ], - "text": "7.3-1\nApply the duality property to the appropriate\npair in Table 7.1 to show that", - "type": "text" - }, - { - "block_id": "p786-b11", - "global_id": 22963, - "bbox": [ - 90.72, - 371.89, - 226.1, - 393.55 - ], - "text": "(a)\n1\n2[δ(t) + j/πt] ⇐⇒u(ω)\n(b) δ(t + T) + δ(t −T) ⇐⇒2cos Tω", - "type": "text" - }, - { - "block_id": "p786-b12", - "global_id": 22964, - "bbox": [ - 91.23, - 395.17, - 227.1, - 404.51 - ], - "text": "(c) δ(t + T) −δ(t −T) ⇐⇒2jsin Tω", - "type": "text" - }, - { - "block_id": "p786-b13", - "global_id": 22965, - "bbox": [ - 289.23, - 85.64, - 490.39, - 116.89 - ], - "text": "7.3-2\nA signal x(t) has Fourier transform X(ω). Deter-\nmine the Fourier transform Y(ω) in terms of\nX(ω) for each of the following signals y(t):", - "type": "text" - }, - { - "block_id": "p786-b14", - "global_id": 22966, - "bbox": [ - 318.37, - 117.15, - 361.73, - 127.86 - ], - "text": "(a) y(t) = 1", - "type": "text" - }, - { - "block_id": "p786-b15", - "global_id": 22967, - "bbox": [ - 317.86, - 118.52, - 411.7, - 138.82 - ], - "text": "5x(−2t + 3)\n(b) y(t) = ej2tx∗(−3t −6)", - "type": "text" - }, - { - "block_id": "p786-b16", - "global_id": 22968, - "bbox": [ - 289.22, - 144.55, - 490.4, - 186.76 - ], - "text": "7.3-3\nA signal x(t) has Fourier transform X(ω).\nDetermine the inverse Fourier transform y(t) in\nterms of x(t) for each of the following spectra\nY(ω),", - "type": "text" - }, - { - "block_id": "p786-b17", - "global_id": 22969, - "bbox": [ - 318.37, - 187.02, - 366.83, - 197.72 - ], - "text": "(a) Y(ω) = 4", - "type": "text" - }, - { - "block_id": "p786-b18", - "global_id": 22970, - "bbox": [ - 317.86, - 187.12, - 427.77, - 209.99 - ], - "text": "3e−j2ω/3X(−ω/3)\n(b) Y(ω) = 1", - "type": "text" - }, - { - "block_id": "p786-b19", - "global_id": 22971, - "bbox": [ - 363.59, - 193.44, - 428.05, - 212.58 - ], - "text": "3ej2(ω−2)X∗ ω−2\n3", - "type": "text" - }, - { - "block_id": "p786-b20", - "global_id": 22972, - "bbox": [ - 289.22, - 216.02, - 490.4, - 236.02 - ], - "text": "7.3-4\nThe Fourier transform of the triangular pulse\nx(t) in Fig. P7.3-4 is expressed as", - "type": "text" - }, - { - "block_id": "p786-b21", - "global_id": 22973, - "bbox": [ - 353.1, - 257.25, - 390.73, - 272.5 - ], - "text": "X(ω) = 1", - "type": "text" - }, - { - "block_id": "p786-b22", - "global_id": 22974, - "bbox": [ - 383.59, - 259.46, - 455.15, - 278.5 - ], - "text": "ω2 (ejω −jωejω −1)", - "type": "text" - }, - { - "block_id": "p786-b23", - "global_id": 22975, - "bbox": [ - 317.86, - 298.24, - 490.39, - 340.09 - ], - "text": "Use this information, and the time-shifting and\ntime-scaling properties, to find the Fourier trans-\nforms of the signals xi(t)(i = 1,2,3,4,5) shown\nin Fig. P7.3-4.", - "type": "text" - }, - { - "block_id": "p786-b24", - "global_id": 22976, - "bbox": [ - 289.22, - 346.11, - 490.39, - 377.07 - ], - "text": "7.3-5\nUsing only the time-shifting property and\nTable 7.1, find the Fourier transforms of the\nsignals depicted in Fig. P7.3-5.", - "type": "text" - }, - { - "block_id": "p786-b25", - "global_id": 22977, - "bbox": [ - 289.22, - 382.8, - 490.39, - 403.1 - ], - "text": "7.3-6\nConsider the fact that the τ-duration triangle\nfunction", - "type": "text" - }, - { - "block_id": "p786-b26", - "global_id": 22978, - "bbox": [ - 360.3, - 386.55, - 369.34, - 398.6 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p786-b27", - "global_id": 22979, - "bbox": [ - 365.5, - 398.93, - 368.3, - 405.41 - ], - "text": "τ", - "type": "text" - }, - { - "block_id": "p786-b29", - "global_id": 22980, - "bbox": [ - 378.68, - 394.13, - 490.39, - 403.09 - ], - "text": "has inverse Fourier transform", - "type": "text" - }, - { - "block_id": "p786-b30", - "global_id": 22981, - "bbox": [ - 125.41, - 477.42, - 155.92, - 485.42 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p786-b31", - "global_id": 22982, - "bbox": [ - 115.45, - 451.29, - 119.45, - 459.29 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p786-b32", - "global_id": 22983, - "bbox": [ - 164.72, - 477.35, - 447.31, - 485.35 - ], - "text": "t\nt\nt", - "type": "text" - }, - { - "block_id": "p786-b33", - "global_id": 22984, - "bbox": [ - 327.19, - 612.28, - 329.42, - 620.28 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p786-b34", - "global_id": 22985, - "bbox": [ - 220.3, - 547.77, - 222.52, - 555.77 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p786-b35", - "global_id": 22986, - "bbox": [ - 421.14, - 530.45, - 423.36, - 538.45 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p786-b36", - "global_id": 22987, - "bbox": [ - 193.83, - 547.85, - 197.83, - 555.85 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p786-b37", - "global_id": 22988, - "bbox": [ - 183.83, - 522.42, - 187.83, - 530.42 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p786-b38", - "global_id": 22989, - "bbox": [ - 444.45, - 536.02, - 448.45, - 544.02 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p786-b39", - "global_id": 22990, - "bbox": [ - 372.09, - 522.41, - 376.09, - 530.41 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p786-b40", - "global_id": 22991, - "bbox": [ - 382.07, - 542.07, - 386.07, - 550.07 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p786-b41", - "global_id": 22992, - "bbox": [ - 183.83, - 500.31, - 187.83, - 508.31 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p786-b42", - "global_id": 22993, - "bbox": [ - 202.2, - 612.07, - 218.86, - 620.36 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p786-b43", - "global_id": 22994, - "bbox": [ - 118.04, - 535.73, - 316.94, - 555.85 - ], - "text": "2\n2\n2", - "type": "text" - }, - { - "block_id": "p786-b44", - "global_id": 22995, - "bbox": [ - 232.07, - 612.07, - 276.73, - 620.36 - ], - "text": "0.5\n0.5", - "type": "text" - }, - { - "block_id": "p786-b45", - "global_id": 22996, - "bbox": [ - 365.06, - 477.13, - 434.45, - 485.42 - ], - "text": "1\n0\n1", - "type": "text" - }, - { - "block_id": "p786-b46", - "global_id": 22997, - "bbox": [ - 394.2, - 451.29, - 398.2, - 459.29 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p786-b47", - "global_id": 22998, - "bbox": [ - 228.79, - 477.13, - 274.53, - 485.42 - ], - "text": "1\n0", - "type": "text" - }, - { - "block_id": "p786-b48", - "global_id": 22999, - "bbox": [ - 253.95, - 451.29, - 257.95, - 459.29 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p786-b49", - "global_id": 23000, - "bbox": [ - 297.04, - 477.42, - 301.04, - 485.42 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p786-b50", - "global_id": 23001, - "bbox": [ - 143.26, - 443.84, - 154.69, - 451.92 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p786-b51", - "global_id": 23002, - "bbox": [ - 202.28, - 506.9, - 216.39, - 516.51 - ], - "text": "x3(t)", - "type": "text" - }, - { - "block_id": "p786-b52", - "global_id": 23003, - "bbox": [ - 285.26, - 455.28, - 299.37, - 464.89 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p786-b53", - "global_id": 23004, - "bbox": [ - 418.46, - 442.53, - 432.56, - 452.14 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p786-b54", - "global_id": 23005, - "bbox": [ - 394.32, - 520.15, - 408.43, - 529.75 - ], - "text": "x4(t)", - "type": "text" - }, - { - "block_id": "p786-b55", - "global_id": 23006, - "bbox": [ - 273.22, - 578.17, - 287.33, - 587.78 - ], - "text": "x5(t)", - "type": "text" - }, - { - "block_id": "p786-b56", - "global_id": 23007, - "bbox": [ - 297.44, - 612.36, - 307.44, - 620.36 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p786-b57", - "global_id": 23008, - "bbox": [ - 248.74, - 587.37, - 252.74, - 595.37 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p786-b58", - "global_id": 23009, - "bbox": [ - 357.75, - 548.05, - 376.09, - 556.35 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p786-b59", - "global_id": 23010, - "bbox": [ - 104.83, - 629.31, - 156.48, - 638.27 - ], - "text": "Figure P7.3-4", - "type": "text" - } - ] - }, - { - "page_num": 787, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p787-b0", - "global_id": 23011, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n767", - "type": "text" - }, - { - "block_id": "p787-b1", - "global_id": 23012, - "bbox": [ - 200.78, - 88.14, - 204.78, - 96.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p787-b2", - "global_id": 23013, - "bbox": [ - 181.79, - 130.35, - 192.46, - 138.64 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p787-b3", - "global_id": 23014, - "bbox": [ - 142.03, - 114.67, - 202.58, - 123.59 - ], - "text": "0\nT", - "type": "text" - }, - { - "block_id": "p787-b4", - "global_id": 23015, - "bbox": [ - 244.36, - 103.48, - 248.81, - 111.48 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p787-b5", - "global_id": 23016, - "bbox": [ - 192.14, - 147.28, - 201.02, - 155.28 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p787-b6", - "global_id": 23017, - "bbox": [ - 302.2, - 133.67, - 409.87, - 141.77 - ], - "text": "0\np", - "type": "text" - }, - { - "block_id": "p787-b7", - "global_id": 23018, - "bbox": [ - 188.76, - 178.12, - 192.76, - 186.12 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p787-b8", - "global_id": 23019, - "bbox": [ - 199.58, - 227.53, - 203.58, - 235.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p787-b9", - "global_id": 23020, - "bbox": [ - 236.58, - 190.54, - 251.47, - 198.62 - ], - "text": "cos t", - "type": "text" - }, - { - "block_id": "p787-b10", - "global_id": 23021, - "bbox": [ - 253.97, - 213.86, - 256.2, - 221.86 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p787-b11", - "global_id": 23022, - "bbox": [ - 253.97, - 115.36, - 256.2, - 123.36 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p787-b12", - "global_id": 23023, - "bbox": [ - 411.06, - 227.45, - 413.29, - 235.45 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p787-b13", - "global_id": 23024, - "bbox": [ - 411.06, - 121.12, - 413.29, - 129.12 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p787-b14", - "global_id": 23025, - "bbox": [ - 302.13, - 227.45, - 375.42, - 235.53 - ], - "text": "0\nT", - "type": "text" - }, - { - "block_id": "p787-b15", - "global_id": 23026, - "bbox": [ - 292.2, - 167.62, - 296.2, - 175.62 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p787-b16", - "global_id": 23027, - "bbox": [ - 332.2, - 190.0, - 345.42, - 199.62 - ], - "text": "eat", - "type": "text" - }, - { - "block_id": "p787-b17", - "global_id": 23028, - "bbox": [ - 352.38, - 147.28, - 362.19, - 155.28 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p787-b18", - "global_id": 23029, - "bbox": [ - 192.14, - 242.74, - 361.95, - 250.74 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p787-b19", - "global_id": 23030, - "bbox": [ - 244.15, - 227.23, - 257.01, - 235.53 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p787-b20", - "global_id": 23031, - "bbox": [ - 386.76, - 92.47, - 400.32, - 100.55 - ], - "text": "sin t", - "type": "text" - }, - { - "block_id": "p787-b21", - "global_id": 23032, - "bbox": [ - 130.58, - 256.91, - 182.22, - 265.88 - ], - "text": "Figure P7.3-5", - "type": "text" - }, - { - "block_id": "p787-b22", - "global_id": 23033, - "bbox": [ - 276.42, - 302.85, - 466.94, - 321.06 - ], - "text": "1\nt\nt", - "type": "text" - }, - { - "block_id": "p787-b23", - "global_id": 23034, - "bbox": [ - 200.8, - 339.81, - 391.6, - 347.81 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p787-b24", - "global_id": 23035, - "bbox": [ - 209.23, - 302.85, - 213.23, - 310.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p787-b25", - "global_id": 23036, - "bbox": [ - 131.86, - 326.12, - 456.74, - 334.52 - ], - "text": "0\n4 3 2\n2\n3\n4\n0\n4 3 2\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p787-b26", - "global_id": 23037, - "bbox": [ - 130.58, - 353.98, - 182.22, - 362.94 - ], - "text": "Figure P7.3-7", - "type": "text" - }, - { - "block_id": "p787-b27", - "global_id": 23038, - "bbox": [ - 117.66, - 389.0, - 160.92, - 408.74 - ], - "text": "τ\n4π sinc2 tτ\n4", - "type": "text" - }, - { - "block_id": "p787-b28", - "global_id": 23039, - "bbox": [ - 116.46, - 396.59, - 288.98, - 427.47 - ], - "text": ". Use the duality property to deter-\nmine the Fourier transform Y(ω) of signal y(t) =\n(t).", - "type": "text" - }, - { - "block_id": "p787-b29", - "global_id": 23040, - "bbox": [ - 87.82, - 432.93, - 288.99, - 452.92 - ], - "text": "7.3-7\nUse the time-shifting property to show that if\nx(t) ⇐⇒X(ω), then", - "type": "text" - }, - { - "block_id": "p787-b30", - "global_id": 23041, - "bbox": [ - 134.19, - 469.95, - 271.09, - 479.29 - ], - "text": "x(t + T) + x(t −T) ⇐⇒2X(ω)cosTω", - "type": "text" - }, - { - "block_id": "p787-b31", - "global_id": 23042, - "bbox": [ - 116.46, - 496.68, - 288.99, - 527.58 - ], - "text": "This is the dual of Eq. (7.32). Use this result and\nTable 7.1 to find the Fourier transforms of the\nsignals shown in Fig. P7.3-7.", - "type": "text" - }, - { - "block_id": "p787-b32", - "global_id": 23043, - "bbox": [ - 87.83, - 533.02, - 288.96, - 553.02 - ], - "text": "7.3-8\nProve the following results, which are duals of\neach other:", - "type": "text" - }, - { - "block_id": "p787-b33", - "global_id": 23044, - "bbox": [ - 126.14, - 566.85, - 191.43, - 578.28 - ], - "text": "x(t)sin ω0t ⇐⇒1", - "type": "text" - }, - { - "block_id": "p787-b34", - "global_id": 23045, - "bbox": [ - 187.29, - 568.2, - 279.31, - 580.13 - ], - "text": "2j[X(ω−ω0)−X(ω+ω0)]", - "type": "text" - }, - { - "block_id": "p787-b35", - "global_id": 23046, - "bbox": [ - 134.16, - 584.14, - 272.31, - 597.42 - ], - "text": "1\n2j[x(t+T)−x(t−T)] ⇐⇒X(ω)sin Tω", - "type": "text" - }, - { - "block_id": "p787-b36", - "global_id": 23047, - "bbox": [ - 116.46, - 612.89, - 288.98, - 632.82 - ], - "text": "Use the latter result and Table 7.1 to find the\nFourier transform of the signal in Fig. P7.3-8.", - "type": "text" - }, - { - "block_id": "p787-b37", - "global_id": 23048, - "bbox": [ - 434.45, - 401.21, - 438.45, - 409.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p787-b38", - "global_id": 23049, - "bbox": [ - 358.9, - 424.26, - 502.23, - 432.55 - ], - "text": "0\nt\n4 3 2", - "type": "text" - }, - { - "block_id": "p787-b39", - "global_id": 23050, - "bbox": [ - 417.01, - 434.33, - 427.68, - 442.63 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p787-b40", - "global_id": 23051, - "bbox": [ - 462.02, - 411.33, - 499.36, - 419.33 - ], - "text": "2\n3\n4", - "type": "text" - }, - { - "block_id": "p787-b41", - "global_id": 23052, - "bbox": [ - 343.61, - 454.44, - 395.25, - 463.4 - ], - "text": "Figure P7.3-8", - "type": "text" - }, - { - "block_id": "p787-b42", - "global_id": 23053, - "bbox": [ - 314.97, - 485.89, - 516.14, - 549.72 - ], - "text": "7.3-9\nThe signals in Fig. P7.3-9 are modulated signals\nwith carrier cos 10t. Find the Fourier transforms\nof these signals by using the appropriate prop-\nerties of the Fourier transform and Table 7.1.\nSketch the amplitude and phase spectra for\nFigs. P7.3-9a and P7.3-9b.", - "type": "text" - }, - { - "block_id": "p787-b43", - "global_id": 23054, - "bbox": [ - 310.48, - 554.89, - 516.14, - 585.85 - ], - "text": "7.3-10\nUse\nthe\nfrequency-shifting\nproperty\nand\nTable 7.1 to find the inverse Fourier transform\nof the spectra depicted in Fig. P7.3-10.", - "type": "text" - }, - { - "block_id": "p787-b44", - "global_id": 23055, - "bbox": [ - 310.48, - 590.71, - 516.13, - 611.02 - ], - "text": "7.3-11\nLet X(ω) = rect(ω) be the Fourier transform of\na signal x(t).", - "type": "text" - }, - { - "block_id": "p787-b45", - "global_id": 23056, - "bbox": [ - 343.61, - 612.63, - 497.47, - 633.67 - ], - "text": "(a) For ya(t) = x(t) ∗x(t), sketch Ya(ω).\n(b) For yb(t) = x(t) ∗x(t/2), sketch Yb(ω).", - "type": "text" - } - ] - }, - { - "page_num": 788, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p788-b0", - "global_id": 23057, - "bbox": [ - 60.0, - 60.36, - 456.14, - 69.45 - ], - "text": "768\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p788-b1", - "global_id": 23058, - "bbox": [ - 217.42, - 204.55, - 356.7, - 213.21 - ], - "text": "p\n3p", - "type": "text" - }, - { - "block_id": "p788-b2", - "global_id": 23059, - "bbox": [ - 276.82, - 232.75, - 285.7, - 240.75 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p788-b3", - "global_id": 23060, - "bbox": [ - 186.06, - 179.08, - 190.06, - 187.08 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p788-b4", - "global_id": 23061, - "bbox": [ - 376.26, - 193.04, - 378.48, - 201.04 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p788-b5", - "global_id": 23062, - "bbox": [ - 190.31, - 93.96, - 194.31, - 101.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p788-b6", - "global_id": 23063, - "bbox": [ - 115.16, - 119.75, - 453.2, - 129.93 - ], - "text": "p\np\np\n3p", - "type": "text" - }, - { - "block_id": "p788-b7", - "global_id": 23064, - "bbox": [ - 181.96, - 158.21, - 380.38, - 166.21 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p788-b8", - "global_id": 23065, - "bbox": [ - 184.17, - 120.47, - 188.17, - 128.47 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p788-b9", - "global_id": 23066, - "bbox": [ - 256.85, - 96.19, - 455.93, - 116.05 - ], - "text": "1\nt\nt", - "type": "text" - }, - { - "block_id": "p788-b10", - "global_id": 23067, - "bbox": [ - 104.83, - 246.91, - 156.48, - 255.88 - ], - "text": "Figure P7.3-9", - "type": "text" - }, - { - "block_id": "p788-b11", - "global_id": 23068, - "bbox": [ - 373.83, - 284.21, - 377.83, - 292.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p788-b12", - "global_id": 23069, - "bbox": [ - 295.78, - 308.98, - 442.79, - 317.27 - ], - "text": "0\n2\n4\n6\n4 2\n6", - "type": "text" - }, - { - "block_id": "p788-b13", - "global_id": 23070, - "bbox": [ - 177.02, - 322.84, - 375.86, - 330.84 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p788-b14", - "global_id": 23071, - "bbox": [ - 185.34, - 284.21, - 189.34, - 292.21 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p788-b15", - "global_id": 23072, - "bbox": [ - 105.49, - 308.98, - 253.3, - 317.27 - ], - "text": "0\n3\n5\n5\n3", - "type": "text" - }, - { - "block_id": "p788-b16", - "global_id": 23073, - "bbox": [ - 162.91, - 278.66, - 367.44, - 286.91 - ], - "text": "X(v)\nX(v)", - "type": "text" - }, - { - "block_id": "p788-b17", - "global_id": 23074, - "bbox": [ - 255.7, - 296.0, - 449.15, - 304.0 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p788-b18", - "global_id": 23075, - "bbox": [ - 104.83, - 337.01, - 160.95, - 345.98 - ], - "text": "Figure P7.3-10", - "type": "text" - }, - { - "block_id": "p788-b19", - "global_id": 23076, - "bbox": [ - 90.72, - 367.43, - 218.83, - 388.47 - ], - "text": "(c) For yc(t) = 2x(t), sketch Yc(ω).\n(d) For yd(t) = x2(t), sketch Yd(ω).", - "type": "text" - }, - { - "block_id": "p788-b20", - "global_id": 23077, - "bbox": [ - 91.22, - 388.36, - 232.34, - 399.42 - ], - "text": "(e) For ye(t) = 1 −x2(t), sketch Ye(ω).", - "type": "text" - }, - { - "block_id": "p788-b21", - "global_id": 23078, - "bbox": [ - 57.59, - 409.92, - 263.25, - 495.67 - ], - "text": "7.3-12\nUse the time-convolution property to prove pairs\n2, 4, 13, and 14 in Table 2.1 (assume λ < 0\nin pair 2, λ1 and λ2 < 0 in pair 4, λ1 < 0 and\nλ2 > 0 in pair 13, and λ1 and λ2 > 0 in pair\n14). These restrictions are placed because of\nthe Fourier transformability issue for the signals\nconcerned. For pair 2, you need to apply the\nresult in Eq. (1.10).", - "type": "text" - }, - { - "block_id": "p788-b22", - "global_id": 23079, - "bbox": [ - 57.59, - 506.61, - 263.24, - 526.9 - ], - "text": "7.3-13\nA signal x(t) is bandlimited to B Hz. Show that\nthe signal xn(t) is bandlimited to nB Hz.", - "type": "text" - }, - { - "block_id": "p788-b23", - "global_id": 23080, - "bbox": [ - 57.59, - 538.13, - 263.25, - 558.14 - ], - "text": "7.3-14\nFind the Fourier transform of the signal in\nFig. P7.3-5a by three different methods:", - "type": "text" - }, - { - "block_id": "p788-b24", - "global_id": 23081, - "bbox": [ - 90.72, - 560.12, - 263.23, - 580.05 - ], - "text": "(a) By direct integration using Eq. (7.9).\n(b) Using only pair 17 (Table 7.1) and the", - "type": "text" - }, - { - "block_id": "p788-b25", - "global_id": 23082, - "bbox": [ - 91.22, - 582.04, - 263.22, - 623.88 - ], - "text": "time-shifting property.\n(c) Using\nthe\ntime-differentiation\nand\ntime-shifting properties, along with the fact\nthat δ(t) ⇐⇒1.", - "type": "text" - }, - { - "block_id": "p788-b26", - "global_id": 23083, - "bbox": [ - 284.74, - 367.79, - 490.37, - 376.83 - ], - "text": "7.3-15\n(a) Prove the frequency-differentiation property", - "type": "text" - }, - { - "block_id": "p788-b27", - "global_id": 23084, - "bbox": [ - 333.31, - 378.83, - 484.96, - 387.8 - ], - "text": "(dual of the time-differentiation property):", - "type": "text" - }, - { - "block_id": "p788-b28", - "global_id": 23085, - "bbox": [ - 372.84, - 401.2, - 427.77, - 416.45 - ], - "text": "−jtx(t) ⇐⇒d", - "type": "text" - }, - { - "block_id": "p788-b29", - "global_id": 23086, - "bbox": [ - 420.26, - 407.2, - 450.86, - 422.81 - ], - "text": "dω X(ω)", - "type": "text" - }, - { - "block_id": "p788-b30", - "global_id": 23087, - "bbox": [ - 317.86, - 434.7, - 490.39, - 443.67 - ], - "text": "(b) Use this property and pair 1 (Table 7.1)", - "type": "text" - }, - { - "block_id": "p788-b31", - "global_id": 23088, - "bbox": [ - 333.31, - 445.66, - 490.39, - 465.59 - ], - "text": "to determine the Fourier transform of\nte−atu(t).", - "type": "text" - }, - { - "block_id": "p788-b32", - "global_id": 23089, - "bbox": [ - 284.74, - 470.85, - 490.4, - 523.73 - ], - "text": "7.3-16\nAdapt the method of Ex. 7.17 and use the\nfrequency-differentiation (see Prob. 7.3-15) and\nother properties to find the inverse Fourier\ntransform x(t) of the triangular spectrum X(ω) =\n(ω/2).", - "type": "text" - }, - { - "block_id": "p788-b33", - "global_id": 23090, - "bbox": [ - 284.74, - 528.99, - 490.4, - 570.91 - ], - "text": "7.3-17\nAdapt the method of Ex. 7.17 and use the\nfrequency-differentiation (see Prob. 7.3-15) and\nother properties to find the inverse Fourier trans-\nform x(t) of the spectrum X(ω) = π", - "type": "text" - }, - { - "block_id": "p788-b34", - "global_id": 23091, - "bbox": [ - 446.14, - 553.96, - 454.48, - 568.3 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p788-b35", - "global_id": 23092, - "bbox": [ - 450.81, - 553.96, - 473.15, - 573.49 - ], - "text": "2\nrect", - "type": "text" - }, - { - "block_id": "p788-b36", - "global_id": 23093, - "bbox": [ - 474.16, - 554.36, - 483.21, - 566.42 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p788-b37", - "global_id": 23094, - "bbox": [ - 479.54, - 554.36, - 488.15, - 573.49 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p788-b38", - "global_id": 23095, - "bbox": [ - 488.15, - 561.94, - 490.39, - 570.91 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p788-b39", - "global_id": 23096, - "bbox": [ - 289.22, - 576.18, - 487.99, - 585.22 - ], - "text": "7.4-1\nFor a stable LTIC system with transfer function", - "type": "text" - }, - { - "block_id": "p788-b40", - "global_id": 23097, - "bbox": [ - 380.19, - 598.84, - 426.87, - 620.46 - ], - "text": "H(s) =\n1\ns + 1", - "type": "text" - } - ] - }, - { - "page_num": 789, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p789-b0", - "global_id": 23098, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n769", - "type": "text" - }, - { - "block_id": "p789-b1", - "global_id": 23099, - "bbox": [ - 116.46, - 85.9, - 288.96, - 105.83 - ], - "text": "find the (zero-state) response if the input\nx(t) is", - "type": "text" - }, - { - "block_id": "p789-b2", - "global_id": 23100, - "bbox": [ - 116.46, - 106.18, - 160.41, - 127.74 - ], - "text": "(a) e−2tu(t)\n(b) e−tu(t)", - "type": "text" - }, - { - "block_id": "p789-b3", - "global_id": 23101, - "bbox": [ - 116.46, - 128.3, - 159.12, - 149.65 - ], - "text": "(c) etu(−t)\n(d) u(t)", - "type": "text" - }, - { - "block_id": "p789-b4", - "global_id": 23102, - "bbox": [ - 87.82, - 154.83, - 288.98, - 174.84 - ], - "text": "7.4-2\nA stable LTIC system is specified by the fre-\nquency response", - "type": "text" - }, - { - "block_id": "p789-b5", - "global_id": 23103, - "bbox": [ - 174.97, - 187.21, - 229.28, - 209.2 - ], - "text": "H(ω) =\n−1\njω −2", - "type": "text" - }, - { - "block_id": "p789-b6", - "global_id": 23104, - "bbox": [ - 116.46, - 221.83, - 288.98, - 263.68 - ], - "text": "Find the impulse response of this system and\nshow that this is a noncausal system. Find the\n(zero-state) response of this system if the input\nx(t) is", - "type": "text" - }, - { - "block_id": "p789-b7", - "global_id": 23105, - "bbox": [ - 116.46, - 264.03, - 159.12, - 285.59 - ], - "text": "(a) e−tu(t)\n(b) etu(−t)", - "type": "text" - }, - { - "block_id": "p789-b8", - "global_id": 23106, - "bbox": [ - 87.82, - 290.47, - 288.99, - 332.69 - ], - "text": "7.4-3\nA periodic signal x(t) = 1 + 2cos(5πt) +\n3sin(8πt) is applied to an LTIC system with\nimpulse response h(t) = 8sinc(4t)cos(2πt) to\nproduce output y(t) = x(t) ∗h(t).", - "type": "text" - }, - { - "block_id": "p789-b9", - "global_id": 23107, - "bbox": [ - 116.97, - 334.31, - 288.99, - 344.39 - ], - "text": "(a) Determine ω0, the fundamental radian fre-", - "type": "text" - }, - { - "block_id": "p789-b10", - "global_id": 23108, - "bbox": [ - 116.46, - 345.26, - 288.98, - 365.57 - ], - "text": "quency of x(t).\n(b) Determine X(ω), the Fourier transform of", - "type": "text" - }, - { - "block_id": "p789-b11", - "global_id": 23109, - "bbox": [ - 116.96, - 367.19, - 288.99, - 387.48 - ], - "text": "x(t).\n(c) Sketch the system’s magnitude response", - "type": "text" - }, - { - "block_id": "p789-b12", - "global_id": 23110, - "bbox": [ - 116.46, - 389.1, - 281.21, - 409.41 - ], - "text": "|H(ω)| over −10π ≤ω ≤10π.\n(d) Is the system h(t) distortionless? Explain.", - "type": "text" - }, - { - "block_id": "p789-b13", - "global_id": 23111, - "bbox": [ - 116.97, - 411.02, - 187.56, - 420.36 - ], - "text": "(e) Determine y(t).", - "type": "text" - }, - { - "block_id": "p789-b14", - "global_id": 23112, - "bbox": [ - 87.82, - 418.52, - 239.12, - 434.58 - ], - "text": "7.4-4\nA periodic delta train x(t) = %∞", - "type": "text" - }, - { - "block_id": "p789-b15", - "global_id": 23113, - "bbox": [ - 116.46, - 425.24, - 288.99, - 456.49 - ], - "text": "n=−∞δ(t −πn)\nis applied to an LTIC system with impulse\nresponse h(t) = sin(3t)sinc2 t", - "type": "text" - }, - { - "block_id": "p789-b16", - "global_id": 23114, - "bbox": [ - 231.97, - 452.33, - 235.85, - 458.81 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p789-b18", - "global_id": 23115, - "bbox": [ - 116.46, - 447.53, - 288.97, - 467.45 - ], - "text": "to produce\nzero-state output y(t) = x(t) ∗h(t).", - "type": "text" - }, - { - "block_id": "p789-b19", - "global_id": 23116, - "bbox": [ - 116.96, - 469.08, - 288.99, - 479.16 - ], - "text": "(a) Determine ω0, the fundamental radian fre-", - "type": "text" - }, - { - "block_id": "p789-b20", - "global_id": 23117, - "bbox": [ - 116.46, - 480.03, - 288.98, - 500.33 - ], - "text": "quency of x(t).\n(b) Determine X(ω), the Fourier transform of", - "type": "text" - }, - { - "block_id": "p789-b21", - "global_id": 23118, - "bbox": [ - 116.96, - 501.95, - 288.99, - 522.25 - ], - "text": "x(t).\n(c) Sketch the system’s magnitude response", - "type": "text" - }, - { - "block_id": "p789-b22", - "global_id": 23119, - "bbox": [ - 116.46, - 523.87, - 281.21, - 544.17 - ], - "text": "|H(ω)| over −10π ≤ω ≤10π.\n(d) Is the system h(t) distortionless? Explain.", - "type": "text" - }, - { - "block_id": "p789-b23", - "global_id": 23120, - "bbox": [ - 116.97, - 545.79, - 187.56, - 555.12 - ], - "text": "(e) Determine y(t).", - "type": "text" - }, - { - "block_id": "p789-b24", - "global_id": 23121, - "bbox": [ - 87.82, - 559.01, - 288.99, - 624.87 - ], - "text": "7.4-5\nSignals x1(t)=104rect (104t) and x2(t) = δ(t)\nare applied at the inputs of the ideal lowpass\nfilters H1(ω) = rect (ω/40,000π) and H2(ω) =\nrect (ω/20,000π) (Fig. P7.4-5). The outputs\ny1(t) and y2(t) of these filters are multiplied to\nobtain the signal y(t) = y1(t)y2(t).", - "type": "text" - }, - { - "block_id": "p789-b25", - "global_id": 23122, - "bbox": [ - 116.96, - 625.76, - 222.05, - 635.84 - ], - "text": "(a) Sketch X1(ω) and X2(ω).", - "type": "text" - }, - { - "block_id": "p789-b26", - "global_id": 23123, - "bbox": [ - 343.61, - 85.58, - 451.19, - 95.66 - ], - "text": "(b) Sketch H1(ω) and H2(ω).", - "type": "text" - }, - { - "block_id": "p789-b27", - "global_id": 23124, - "bbox": [ - 343.61, - 96.54, - 516.14, - 127.79 - ], - "text": "(c) Sketch Y1(ω) and Y2(ω).\n(d) Find\nthe\nbandwidths\nof\ny1(t),\ny2(t),\nand y(t).", - "type": "text" - }, - { - "block_id": "p789-b28", - "global_id": 23125, - "bbox": [ - 475.98, - 169.25, - 526.59, - 179.07 - ], - "text": "y(t) y1(t)y2(t)", - "type": "text" - }, - { - "block_id": "p789-b29", - "global_id": 23126, - "bbox": [ - 389.7, - 152.64, - 446.47, - 169.49 - ], - "text": "y1(t)\nH1(v)", - "type": "text" - }, - { - "block_id": "p789-b30", - "global_id": 23127, - "bbox": [ - 389.7, - 196.99, - 409.14, - 206.62 - ], - "text": "H2(v)", - "type": "text" - }, - { - "block_id": "p789-b31", - "global_id": 23128, - "bbox": [ - 432.37, - 189.87, - 446.47, - 199.48 - ], - "text": "y2(t)", - "type": "text" - }, - { - "block_id": "p789-b32", - "global_id": 23129, - "bbox": [ - 349.14, - 152.64, - 363.24, - 162.24 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p789-b33", - "global_id": 23130, - "bbox": [ - 349.14, - 189.87, - 363.24, - 199.48 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p789-b34", - "global_id": 23131, - "bbox": [ - 333.65, - 217.92, - 385.29, - 226.88 - ], - "text": "Figure P7.4-5", - "type": "text" - }, - { - "block_id": "p789-b35", - "global_id": 23132, - "bbox": [ - 314.97, - 252.27, - 516.15, - 403.77 - ], - "text": "7.4-6\nA lowpass system time constant is often defined\nas the width of its unit impulse response h(t) (see\nSec. 2.6-2). An input pulse p(t) to this system\nacts like an impulse of strength equal to the\narea of p(t) if the width of p(t) is much smaller\nthan the system time constant, and provided p(t)\nis a lowpass pulse, implying that its spectrum\nis concentrated at low frequencies. Verify this\nbehavior by considering a system whose unit\nimpulse response is h(t) = rect(t/10−3). The\ninput pulse is a triangle pulse p(t) = (t/10−6).\nShow that the system response to this pulse is\nvery nearly the system response to the input\nAδ(t), where A is the area under the pulse p(t).", - "type": "text" - }, - { - "block_id": "p789-b36", - "global_id": 23133, - "bbox": [ - 314.97, - 409.23, - 516.14, - 549.77 - ], - "text": "7.4-7\nA lowpass system time constant is often defined\nas the width of its unit impulse response h(t) (see\nSec. 2.6-2). An input pulse p(t) to this system\npasses practically without distortion if the width\nof p(t) is much greater than the system time\nconstant, and provided p(t) is a lowpass pulse,\nimplying that its spectrum is concentrated at low\nfrequencies. Verify this behavior by considering\na system whose unit impulse response is h(t) =\nrect(t/10−3). The input pulse is a triangle pulse\np(t)=(t). Show that the system output to this\npulse is very nearly kp(t), where k is the system\ngain to a dc signal, that is, k = H(0).", - "type": "text" - }, - { - "block_id": "p789-b37", - "global_id": 23134, - "bbox": [ - 314.98, - 554.93, - 516.15, - 597.15 - ], - "text": "7.4-8\nA causal signal h(t) has a Fourier transform\nH(ω). If R(ω) and X(ω) are the real and the\nimaginary parts of H(ω), that is, H(ω) = R(ω)+\njX(ω), then show that", - "type": "text" - }, - { - "block_id": "p789-b38", - "global_id": 23135, - "bbox": [ - 384.66, - 612.87, - 420.14, - 628.12 - ], - "text": "R(ω) = 1", - "type": "text" - }, - { - "block_id": "p789-b39", - "global_id": 23136, - "bbox": [ - 414.75, - 625.15, - 420.13, - 634.11 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p789-b40", - "global_id": 23137, - "bbox": [ - 423.22, - 606.58, - 438.67, - 617.55 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p789-b41", - "global_id": 23138, - "bbox": [ - 427.95, - 628.82, - 439.61, - 635.3 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p789-b42", - "global_id": 23139, - "bbox": [ - 442.3, - 612.49, - 474.91, - 634.39 - ], - "text": "X(ω)\nω −y dω", - "type": "text" - } - ] - }, - { - "page_num": 790, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p790-b0", - "global_id": 23140, - "bbox": [ - 60.0, - 60.36, - 456.14, - 69.45 - ], - "text": "770\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p790-b1", - "global_id": 23141, - "bbox": [ - 90.72, - 85.9, - 103.66, - 94.86 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p790-b2", - "global_id": 23142, - "bbox": [ - 128.07, - 94.65, - 170.93, - 109.82 - ], - "text": "X(ω) = −1", - "type": "text" - }, - { - "block_id": "p790-b3", - "global_id": 23143, - "bbox": [ - 165.55, - 106.93, - 170.93, - 115.9 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p790-b4", - "global_id": 23144, - "bbox": [ - 174.01, - 88.37, - 189.47, - 99.34 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p790-b5", - "global_id": 23145, - "bbox": [ - 178.75, - 110.61, - 190.41, - 117.09 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p790-b6", - "global_id": 23146, - "bbox": [ - 193.1, - 94.28, - 225.71, - 116.18 - ], - "text": "R(ω)\nω −y dω", - "type": "text" - }, - { - "block_id": "p790-b7", - "global_id": 23147, - "bbox": [ - 90.72, - 124.11, - 263.25, - 188.91 - ], - "text": "assuming that h(t) has no impulse at the ori-\ngin. This pair of integrals defines the Hilbert\ntransform. [Hint: Let he(t) and ho(t) be the even\nand odd components of h(t). Use the results in\nProb. 7.1-4. See Fig. 1.24 for the relationship\nbetween he(t) and ho(t).]", - "type": "text" - }, - { - "block_id": "p790-b8", - "global_id": 23148, - "bbox": [ - 90.72, - 190.23, - 263.25, - 308.79 - ], - "text": "This problem states one of the important\nproperties of causal systems: that the real and\nimaginary parts of the frequency response of\na causal system are related. If one specifies\nthe real part, the imaginary part cannot be\nspecified independently. The imaginary part is\npredetermined by the real part, and vice versa.\nThis result also leads to the conclusion that\nthe magnitude and angle of H(ω) are related,\nprovided all the poles and zeros of H(ω) lie in\nthe LHP.", - "type": "text" - }, - { - "block_id": "p790-b9", - "global_id": 23149, - "bbox": [ - 62.08, - 313.78, - 251.77, - 322.82 - ], - "text": "7.5-1\nConsider a filter with the frequency response", - "type": "text" - }, - { - "block_id": "p790-b10", - "global_id": 23150, - "bbox": [ - 141.22, - 333.94, - 212.24, - 346.24 - ], - "text": "H(ω) = e−(kω2+jωt0)", - "type": "text" - }, - { - "block_id": "p790-b11", - "global_id": 23151, - "bbox": [ - 90.72, - 359.1, - 263.24, - 444.77 - ], - "text": "Show that this filter is physically unrealizable\nby using the time-domain criterion [noncausal\nh(t)] and the frequency-domain (Paley–Wiener)\ncriterion. Can this filter be made approximately\nrealizable by choosing t0 sufficiently large? Use\nyour own (reasonable) criterion of approximate\nrealizability to determine t0. [Hint: Use pair 22\nin Table 7.1.]", - "type": "text" - }, - { - "block_id": "p790-b12", - "global_id": 23152, - "bbox": [ - 62.08, - 449.76, - 242.07, - 458.8 - ], - "text": "7.5-2\nShow that a filter with frequency response", - "type": "text" - }, - { - "block_id": "p790-b13", - "global_id": 23153, - "bbox": [ - 132.51, - 469.86, - 220.44, - 492.84 - ], - "text": "H(ω) =\n2(105)\nω2 + 1010 e−jωt0", - "type": "text" - }, - { - "block_id": "p790-b14", - "global_id": 23154, - "bbox": [ - 90.72, - 502.52, - 263.24, - 545.09 - ], - "text": "is unrealizable. Can this filter be made approx-\nimately realizable by choosing a sufficiently\nlarge t0? Use your own (reasonable) criterion of\napproximate realizability to determine t0.", - "type": "text" - }, - { - "block_id": "p790-b15", - "global_id": 23155, - "bbox": [ - 62.08, - 549.34, - 263.24, - 602.22 - ], - "text": "7.5-3\nDetermine whether the filters with the following\nfrequency response H(ω) are physically real-\nizable. If they are not realizable, can they be\nrealized approximately by allowing a finite time\ndelay in the response?", - "type": "text" - }, - { - "block_id": "p790-b16", - "global_id": 23156, - "bbox": [ - 90.72, - 600.96, - 181.11, - 624.14 - ], - "text": "(a) 10−6 sinc(10−6ω)\n(b) 10−4 (ω/40,000π)", - "type": "text" - }, - { - "block_id": "p790-b17", - "global_id": 23157, - "bbox": [ - 91.22, - 625.76, - 135.0, - 635.1 - ], - "text": "(c) 2π δ(ω)", - "type": "text" - }, - { - "block_id": "p790-b18", - "global_id": 23158, - "bbox": [ - 289.22, - 85.64, - 490.4, - 138.82 - ], - "text": "7.5-4\nConsider signal x1(t), its Fourier transform\nX1(f), and several other signals, as shown in\nFig. P7.5-4. Notice, spectra are drawn as a\nfunction of hertzian frequency f rather than\nradian frequency ω.", - "type": "text" - }, - { - "block_id": "p790-b19", - "global_id": 23159, - "bbox": [ - 318.37, - 140.43, - 490.39, - 150.51 - ], - "text": "(a) Accurately sketch X2(f), the Fourier trans-", - "type": "text" - }, - { - "block_id": "p790-b20", - "global_id": 23160, - "bbox": [ - 317.86, - 151.39, - 490.39, - 172.43 - ], - "text": "form of x2(t).\n(b) Accurately sketch x3(t), the inverse Fourier", - "type": "text" - }, - { - "block_id": "p790-b21", - "global_id": 23161, - "bbox": [ - 318.37, - 173.31, - 490.39, - 194.35 - ], - "text": "transform of X3(f).\n(c) The signal x4(t) = x1(t) + x2(t) is passed", - "type": "text" - }, - { - "block_id": "p790-b22", - "global_id": 23162, - "bbox": [ - 333.31, - 195.6, - 490.38, - 227.23 - ], - "text": "through an ideal lowpass filter with 3 Hz\ncutoff to produce output y4(t). Accurately\nsketch y4(t).", - "type": "text" - }, - { - "block_id": "p790-b34", - "global_id": 23163, - "bbox": [ - 292.55, - 264.22, - 305.12, - 578.0 - ], - "text": "x1(t)\nX1(f)\nx2(t)\nX3(f)\nx4(t)", - "type": "text" - }, - { - "block_id": "p790-b35", - "global_id": 23164, - "bbox": [ - 404.02, - 607.57, - 406.24, - 615.54 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p790-b36", - "global_id": 23165, - "bbox": [ - 404.02, - 458.32, - 406.24, - 466.29 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p790-b37", - "global_id": 23166, - "bbox": [ - 404.02, - 309.08, - 406.24, - 317.05 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p790-b38", - "global_id": 23167, - "bbox": [ - 395.24, - 532.98, - 415.21, - 541.03 - ], - "text": "f [Hz]", - "type": "text" - }, - { - "block_id": "p790-b39", - "global_id": 23168, - "bbox": [ - 395.24, - 384.48, - 415.21, - 392.53 - ], - "text": "f [Hz]", - "type": "text" - }, - { - "block_id": "p790-b40", - "global_id": 23169, - "bbox": [ - 337.18, - 518.57, - 347.39, - 526.87 - ], - "text": "−4", - "type": "text" - }, - { - "block_id": "p790-b41", - "global_id": 23170, - "bbox": [ - 337.18, - 370.07, - 347.39, - 378.37 - ], - "text": "−4", - "type": "text" - }, - { - "block_id": "p790-b42", - "global_id": 23171, - "bbox": [ - 347.65, - 593.57, - 357.86, - 601.87 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b43", - "global_id": 23172, - "bbox": [ - 368.75, - 518.57, - 378.96, - 526.87 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b44", - "global_id": 23173, - "bbox": [ - 347.65, - 444.32, - 357.86, - 452.62 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b45", - "global_id": 23174, - "bbox": [ - 368.75, - 370.07, - 378.96, - 378.37 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b46", - "global_id": 23175, - "bbox": [ - 347.65, - 295.07, - 357.86, - 303.37 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b47", - "global_id": 23176, - "bbox": [ - 403.38, - 593.9, - 407.36, - 601.87 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b48", - "global_id": 23177, - "bbox": [ - 403.38, - 518.9, - 407.36, - 526.87 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b49", - "global_id": 23178, - "bbox": [ - 403.38, - 444.65, - 407.36, - 452.62 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b50", - "global_id": 23179, - "bbox": [ - 403.38, - 370.4, - 407.36, - 378.37 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b51", - "global_id": 23180, - "bbox": [ - 403.38, - 295.4, - 407.36, - 303.37 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b52", - "global_id": 23181, - "bbox": [ - 456.01, - 593.9, - 459.99, - 601.87 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b53", - "global_id": 23182, - "bbox": [ - 434.91, - 518.9, - 438.89, - 526.87 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b54", - "global_id": 23183, - "bbox": [ - 456.01, - 444.65, - 459.99, - 452.62 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b55", - "global_id": 23184, - "bbox": [ - 434.91, - 370.4, - 438.89, - 378.37 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b56", - "global_id": 23185, - "bbox": [ - 456.01, - 295.4, - 459.99, - 303.37 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b57", - "global_id": 23186, - "bbox": [ - 466.63, - 518.9, - 470.62, - 526.87 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p790-b58", - "global_id": 23187, - "bbox": [ - 466.63, - 370.4, - 470.62, - 378.37 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p790-b59", - "global_id": 23188, - "bbox": [ - 311.58, - 582.77, - 321.79, - 591.07 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b60", - "global_id": 23189, - "bbox": [ - 311.58, - 433.52, - 321.79, - 441.82 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b61", - "global_id": 23190, - "bbox": [ - 311.58, - 284.27, - 321.79, - 292.57 - ], - "text": "−2", - "type": "text" - }, - { - "block_id": "p790-b62", - "global_id": 23191, - "bbox": [ - 318.33, - 567.06, - 322.32, - 575.03 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b63", - "global_id": 23192, - "bbox": [ - 318.33, - 507.7, - 322.32, - 515.67 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b64", - "global_id": 23193, - "bbox": [ - 318.33, - 417.81, - 322.32, - 425.78 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b65", - "global_id": 23194, - "bbox": [ - 318.33, - 359.17, - 322.32, - 367.14 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b66", - "global_id": 23195, - "bbox": [ - 318.33, - 268.56, - 322.32, - 276.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p790-b67", - "global_id": 23196, - "bbox": [ - 316.73, - 491.57, - 320.72, - 499.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p790-b68", - "global_id": 23197, - "bbox": [ - 316.73, - 342.68, - 320.72, - 350.65 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p790-b69", - "global_id": 23198, - "bbox": [ - 318.21, - 550.57, - 322.2, - 558.54 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b70", - "global_id": 23199, - "bbox": [ - 318.21, - 475.5, - 322.2, - 483.47 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b71", - "global_id": 23200, - "bbox": [ - 318.21, - 401.32, - 322.2, - 409.29 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b72", - "global_id": 23201, - "bbox": [ - 318.21, - 326.25, - 322.2, - 334.22 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b73", - "global_id": 23202, - "bbox": [ - 318.21, - 252.07, - 322.2, - 260.04 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p790-b74", - "global_id": 23203, - "bbox": [ - 287.98, - 625.88, - 339.62, - 634.84 - ], - "text": "Figure P7.5-4", - "type": "text" - } - ] - }, - { - "page_num": 791, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p791-b0", - "global_id": 23204, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n771", - "type": "text" - }, - { - "block_id": "p791-b1", - "global_id": 23205, - "bbox": [ - 87.82, - 86.08, - 288.99, - 118.7 - ], - "text": "7.6-1\nDefine x(t) =\n1\n2π sinc(t/2) with Fourier trans-\nform X(ω) = rect(ω). Use Parseval’s theorem to\ndetermine", - "type": "text" - }, - { - "block_id": "p791-b2", - "global_id": 23206, - "bbox": [ - 154.56, - 102.15, - 167.02, - 113.12 - ], - "text": "$ ∞", - "type": "text" - }, - { - "block_id": "p791-b3", - "global_id": 23207, - "bbox": [ - 158.66, - 108.03, - 223.99, - 120.9 - ], - "text": "−∞sinc2(t −2)dt.", - "type": "text" - }, - { - "block_id": "p791-b4", - "global_id": 23208, - "bbox": [ - 87.82, - 123.87, - 264.24, - 132.91 - ], - "text": "7.6-2\nShow that the energy of a Gaussian pulse", - "type": "text" - }, - { - "block_id": "p791-b5", - "global_id": 23209, - "bbox": [ - 162.61, - 145.46, - 202.23, - 160.62 - ], - "text": "x(t) =\n1", - "type": "text" - }, - { - "block_id": "p791-b6", - "global_id": 23210, - "bbox": [ - 187.84, - 158.94, - 192.63, - 167.9 - ], - "text": "σ", - "type": "text" - }, - { - "block_id": "p791-b7", - "global_id": 23211, - "bbox": [ - 193.8, - 151.41, - 201.38, - 160.38 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p791-b8", - "global_id": 23212, - "bbox": [ - 201.38, - 148.34, - 241.84, - 168.28 - ], - "text": "2π\ne−t2/2σ 2", - "type": "text" - }, - { - "block_id": "p791-b9", - "global_id": 23213, - "bbox": [ - 116.46, - 174.27, - 289.0, - 218.65 - ], - "text": "is 1/(2σ√π). Verify this result by using Parse-\nval’s theorem to derive the energy Ex from X(ω).\n[Hint: See pair 22 in Table 7.1. Use the fact that\n$ ∞", - "type": "text" - }, - { - "block_id": "p791-b10", - "global_id": 23214, - "bbox": [ - 120.56, - 211.93, - 173.99, - 226.44 - ], - "text": "−∞e−x2/2 dx =", - "type": "text" - }, - { - "block_id": "p791-b11", - "global_id": 23215, - "bbox": [ - 175.83, - 207.39, - 183.42, - 216.36 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p791-b12", - "global_id": 23216, - "bbox": [ - 183.42, - 214.91, - 199.41, - 224.25 - ], - "text": "2π.]", - "type": "text" - }, - { - "block_id": "p791-b13", - "global_id": 23217, - "bbox": [ - 87.82, - 229.4, - 288.98, - 256.42 - ], - "text": "7.6-3\nUse Parseval’s theorem of Eq. (7.45) to show\nthat\n# ∞", - "type": "text" - }, - { - "block_id": "p791-b14", - "global_id": 23218, - "bbox": [ - 167.17, - 267.69, - 178.83, - 274.16 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p791-b15", - "global_id": 23219, - "bbox": [ - 180.32, - 251.36, - 240.92, - 266.99 - ], - "text": "sinc2 (kx)dx = π", - "type": "text" - }, - { - "block_id": "p791-b16", - "global_id": 23220, - "bbox": [ - 236.61, - 264.29, - 240.59, - 273.26 - ], - "text": "k", - "type": "text" - }, - { - "block_id": "p791-b17", - "global_id": 23221, - "bbox": [ - 87.82, - 282.94, - 288.98, - 369.66 - ], - "text": "7.6-4\nA lowpass signal x(t) is applied to a squaring\ndevice. The squarer output x2(t) is applied to\na lowpass filter of bandwidth f (in hertz)\n(Fig. P7.6-4). Show that if f is very small\n(f →0), then the filter output is a dc signal\ny(t) ≈2Exf. [Hint: If x2(t) ⇐⇒A(ω), then\nshow that Y(ω) ≈[4πA(0)f]δ(ω) if f →0.\nNow, show that A(0) = Ex.]", - "type": "text" - }, - { - "block_id": "p791-b18", - "global_id": 23222, - "bbox": [ - 87.82, - 374.15, - 288.98, - 405.84 - ], - "text": "7.6-5\nGeneralize Parseval’s theorem to show that for\nreal, Fourier-transformable signals x1(t) and\nx2(t)", - "type": "text" - }, - { - "block_id": "p791-b19", - "global_id": 23223, - "bbox": [ - 146.54, - 411.32, - 161.99, - 422.29 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p791-b20", - "global_id": 23224, - "bbox": [ - 151.27, - 433.56, - 162.93, - 440.03 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p791-b21", - "global_id": 23225, - "bbox": [ - 164.43, - 423.53, - 206.53, - 433.61 - ], - "text": "x1(t)x2(t)dt", - "type": "text" - }, - { - "block_id": "p791-b22", - "global_id": 23226, - "bbox": [ - 156.35, - 442.91, - 174.01, - 458.16 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p791-b23", - "global_id": 23227, - "bbox": [ - 166.38, - 455.18, - 176.25, - 464.52 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p791-b24", - "global_id": 23228, - "bbox": [ - 179.33, - 436.61, - 194.78, - 447.58 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p791-b25", - "global_id": 23229, - "bbox": [ - 184.07, - 458.86, - 195.73, - 465.33 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p791-b26", - "global_id": 23230, - "bbox": [ - 197.23, - 448.82, - 259.74, - 458.9 - ], - "text": "X1(−ω)X2(ω)dω", - "type": "text" - }, - { - "block_id": "p791-b27", - "global_id": 23231, - "bbox": [ - 156.35, - 468.2, - 174.01, - 483.46 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p791-b28", - "global_id": 23232, - "bbox": [ - 166.38, - 480.48, - 176.25, - 489.82 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p791-b29", - "global_id": 23233, - "bbox": [ - 179.33, - 461.91, - 194.78, - 472.87 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p791-b30", - "global_id": 23234, - "bbox": [ - 184.07, - 484.15, - 195.73, - 490.62 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p791-b31", - "global_id": 23235, - "bbox": [ - 197.23, - 474.12, - 259.74, - 484.2 - ], - "text": "X1(ω)X2(−ω)dω", - "type": "text" - }, - { - "block_id": "p791-b32", - "global_id": 23236, - "bbox": [ - 87.82, - 502.68, - 152.35, - 511.72 - ], - "text": "7.6-6\nShow that", - "type": "text" - }, - { - "block_id": "p791-b33", - "global_id": 23237, - "bbox": [ - 135.92, - 516.5, - 151.37, - 527.46 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p791-b34", - "global_id": 23238, - "bbox": [ - 140.65, - 538.73, - 152.31, - 545.21 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p791-b35", - "global_id": 23239, - "bbox": [ - 153.81, - 528.7, - 270.37, - 538.04 - ], - "text": "sinc(Wt −mπ)sinc(Wt −nπ)dt", - "type": "text" - }, - { - "block_id": "p791-b36", - "global_id": 23240, - "bbox": [ - 145.73, - 557.34, - 152.72, - 566.31 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p791-b37", - "global_id": 23241, - "bbox": [ - 154.57, - 542.06, - 216.09, - 578.07 - ], - "text": ") 0\nm̸ = n\nπ\nW\nm = n", - "type": "text" - }, - { - "block_id": "p791-b38", - "global_id": 23242, - "bbox": [ - 343.61, - 85.8, - 421.29, - 94.86 - ], - "text": "[Hint: Recognize that", - "type": "text" - }, - { - "block_id": "p791-b39", - "global_id": 23243, - "bbox": [ - 367.21, - 109.6, - 444.78, - 118.94 - ], - "text": "sinc(Wt −kπ) = sinc", - "type": "text" - }, - { - "block_id": "p791-b41", - "global_id": 23244, - "bbox": [ - 450.31, - 109.88, - 457.78, - 118.85 - ], - "text": "W", - "type": "text" - }, - { - "block_id": "p791-b43", - "global_id": 23245, - "bbox": [ - 463.04, - 107.96, - 483.54, - 118.85 - ], - "text": "t −kπ", - "type": "text" - }, - { - "block_id": "p791-b44", - "global_id": 23246, - "bbox": [ - 477.5, - 114.97, - 482.9, - 121.45 - ], - "text": "W", - "type": "text" - }, - { - "block_id": "p791-b46", - "global_id": 23247, - "bbox": [ - 382.03, - 118.93, - 406.83, - 129.52 - ], - "text": "⇐⇒π", - "type": "text" - }, - { - "block_id": "p791-b47", - "global_id": 23248, - "bbox": [ - 402.28, - 120.92, - 423.77, - 133.08 - ], - "text": "W rect", - "type": "text" - }, - { - "block_id": "p791-b48", - "global_id": 23249, - "bbox": [ - 425.77, - 113.34, - 437.19, - 125.4 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p791-b49", - "global_id": 23250, - "bbox": [ - 430.59, - 113.34, - 444.5, - 132.47 - ], - "text": "2W", - "type": "text" - }, - { - "block_id": "p791-b50", - "global_id": 23251, - "bbox": [ - 445.49, - 119.3, - 476.72, - 129.8 - ], - "text": "e−jkπω/W", - "type": "text" - }, - { - "block_id": "p791-b51", - "global_id": 23252, - "bbox": [ - 343.61, - 144.51, - 495.31, - 153.48 - ], - "text": "Use this fact and the result in Prob. 7.6-5.]", - "type": "text" - }, - { - "block_id": "p791-b52", - "global_id": 23253, - "bbox": [ - 314.97, - 158.71, - 516.12, - 167.75 - ], - "text": "7.6-7\n(a) What does it mean to compute the 95%", - "type": "text" - }, - { - "block_id": "p791-b53", - "global_id": 23254, - "bbox": [ - 343.61, - 169.37, - 516.13, - 200.63 - ], - "text": "essential bandwidth B of a signal x(t) with\nFourier transform X(ω)?\n(b) Determine the 95% essential bandwidth B", - "type": "text" - }, - { - "block_id": "p791-b54", - "global_id": 23255, - "bbox": [ - 344.12, - 202.25, - 516.13, - 222.55 - ], - "text": "of a signal with spectrum X(ω) = rect(ω).\n(c) Determine the 95% essential bandwidth B", - "type": "text" - }, - { - "block_id": "p791-b55", - "global_id": 23256, - "bbox": [ - 359.05, - 224.17, - 503.43, - 233.51 - ], - "text": "of a signal with spectrum X(ω) = (ω).", - "type": "text" - }, - { - "block_id": "p791-b56", - "global_id": 23257, - "bbox": [ - 314.97, - 238.73, - 516.13, - 269.7 - ], - "text": "7.6-8\nUsing a 95% energy criterion, determine the\nessential bandwidth B of a signal that has a\nFourier transform given by X(ω) = e−|ω|.", - "type": "text" - }, - { - "block_id": "p791-b57", - "global_id": 23258, - "bbox": [ - 314.97, - 274.92, - 392.79, - 283.96 - ], - "text": "7.6-9\nFor the signal", - "type": "text" - }, - { - "block_id": "p791-b58", - "global_id": 23259, - "bbox": [ - 404.46, - 297.15, - 453.58, - 318.76 - ], - "text": "x(t) =\n2a\nt2 + a2", - "type": "text" - }, - { - "block_id": "p791-b59", - "global_id": 23260, - "bbox": [ - 343.61, - 330.84, - 516.13, - 373.45 - ], - "text": "determine the essential bandwidth B (in hertz) of\nx(t) such that the energy contained in the spec-\ntral components of x(t) of frequencies below B\nHz is 99% of the signal energy Ex.", - "type": "text" - }, - { - "block_id": "p791-b60", - "global_id": 23261, - "bbox": [ - 314.97, - 378.01, - 516.14, - 419.93 - ], - "text": "7.7-1\nFor\neach\nof\nthe\nfollowing\nbaseband\nsig-\nnals\n(i)\nm(t) = cos 1000t,\n(ii)\nm(t) =\n2cos 1000t + cos 2000t,\nand\n(iii)\nm(t) =\ncos 1000tcos 3000t:", - "type": "text" - }, - { - "block_id": "p791-b61", - "global_id": 23262, - "bbox": [ - 343.61, - 421.55, - 516.13, - 441.84 - ], - "text": "(a) Sketch the spectrum of m(t).\n(b) Sketch the spectrum of the DSB-SC signal", - "type": "text" - }, - { - "block_id": "p791-b62", - "global_id": 23263, - "bbox": [ - 344.11, - 443.46, - 516.13, - 463.77 - ], - "text": "m(t)cos 10,000t.\n(c) Identify the upper sideband (USB) and the", - "type": "text" - }, - { - "block_id": "p791-b63", - "global_id": 23264, - "bbox": [ - 343.61, - 465.75, - 516.12, - 485.68 - ], - "text": "lower sideband (LSB) spectra.\n(d) Identify the frequencies in the baseband,", - "type": "text" - }, - { - "block_id": "p791-b64", - "global_id": 23265, - "bbox": [ - 359.04, - 487.68, - 516.13, - 518.56 - ], - "text": "and the corresponding frequencies in the\nDSB-SC, USB, and LSB spectra. Explain\nthe nature of frequency shifting in each case.", - "type": "text" - }, - { - "block_id": "p791-b65", - "global_id": 23266, - "bbox": [ - 314.97, - 523.49, - 516.13, - 532.83 - ], - "text": "7.7-2\nA message signal m(t) with spectrum M(ω) =", - "type": "text" - }, - { - "block_id": "p791-b66", - "global_id": 23267, - "bbox": [ - 344.81, - 527.23, - 397.5, - 546.37 - ], - "text": "1\n1000\n\nω\n2π6000", - "type": "text" - }, - { - "block_id": "p791-b67", - "global_id": 23268, - "bbox": [ - 343.61, - 534.82, - 516.13, - 565.71 - ], - "text": "is to be transmitted using a com-\nmunication system. Assume all single-sideband\nsystems have suppressed carriers.", - "type": "text" - }, - { - "block_id": "p791-b68", - "global_id": 23269, - "bbox": [ - 344.12, - 567.32, - 502.86, - 576.66 - ], - "text": "(a) What is the hertzian bandwidth of m(t)?", - "type": "text" - }, - { - "block_id": "p791-b69", - "global_id": 23270, - "bbox": [ - 193.86, - 600.7, - 343.17, - 614.7 - ], - "text": "( )2\nLowpass", - "type": "text" - }, - { - "block_id": "p791-b70", - "global_id": 23271, - "bbox": [ - 321.1, - 609.7, - 336.81, - 617.7 - ], - "text": "filter", - "type": "text" - }, - { - "block_id": "p791-b71", - "global_id": 23272, - "bbox": [ - 144.09, - 595.76, - 413.06, - 605.57 - ], - "text": "y(t) \t 2Exf\nx(t)\nx2(t)", - "type": "text" - }, - { - "block_id": "p791-b72", - "global_id": 23273, - "bbox": [ - 130.58, - 629.31, - 182.22, - 638.27 - ], - "text": "Figure P7.6-4", - "type": "text" - } - ] - }, - { - "page_num": 792, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p792-b0", - "global_id": 23274, - "bbox": [ - 60.0, - 60.36, - 456.14, - 69.45 - ], - "text": "772\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p792-b1", - "global_id": 23275, - "bbox": [ - 90.72, - 85.9, - 263.23, - 94.86 - ], - "text": "(b) Sketch the spectrum of the transmitted sig-", - "type": "text" - }, - { - "block_id": "p792-b2", - "global_id": 23276, - "bbox": [ - 91.22, - 96.86, - 263.24, - 127.74 - ], - "text": "nal if the communication system is DSB-SC\nwith ωc = 2π 100,000.\n(c) Sketch the spectrum of the transmitted sig-", - "type": "text" - }, - { - "block_id": "p792-b3", - "global_id": 23277, - "bbox": [ - 90.72, - 129.74, - 263.24, - 182.53 - ], - "text": "nal if the communication system is AM with\nωc = 2π 100,000 and a modulation index\nof μ = 1. What is the corresponding carrier\namplitude A?\n(d) Sketch the spectrum of the transmitted sig-", - "type": "text" - }, - { - "block_id": "p792-b4", - "global_id": 23278, - "bbox": [ - 91.22, - 184.53, - 263.24, - 215.41 - ], - "text": "nal if the communication system is USB\nwith ωc = 2π 100,000.\n(e) Sketch the spectrum of the transmitted sig-", - "type": "text" - }, - { - "block_id": "p792-b5", - "global_id": 23279, - "bbox": [ - 92.21, - 217.4, - 263.25, - 248.29 - ], - "text": "nal if the communication system is LSB\nwith ωc = 2π 100,000.\n(f) Suppose we want to transmit m(t) on each", - "type": "text" - }, - { - "block_id": "p792-b6", - "global_id": 23280, - "bbox": [ - 106.15, - 250.28, - 263.24, - 346.92 - ], - "text": "of an FDM system’s four channels: DSB-SC\nat carrier ω1, AM (μ = 1) at carrier ω2,\nUSB at carrier ω3, and LSB at carrier\nω4. Determine carrier frequencies ω1 < ω2\n< ω3 < ω4 so that the FDM spectrum begins\nat a frequency of 100,000 Hz with 5,000 Hz\ndeadbands separating adjacent messages.\nWhat is the end hertzian frequency of the\nFDM signal?", - "type": "text" - }, - { - "block_id": "p792-b7", - "global_id": 23281, - "bbox": [ - 62.08, - 382.7, - 263.24, - 469.13 - ], - "text": "7.7-3\nYou are asked to design a DSB-SC modulator\nto generate a modulated signal km(t)cosωct,\nwhere m(t) is a signal bandlimited to B Hz\n(Fig. P7.7-3a). Figure P7.7-3b shows a DSB-SC\nmodulator available in the stockroom. The band-\npass filter is tuned to ωc and has a bandwidth of\n2B Hz. The carrier generator available generates\nnot cosωct, but cos3 ωct.", - "type": "text" - }, - { - "block_id": "p792-b8", - "global_id": 23282, - "bbox": [ - 91.22, - 470.44, - 263.24, - 479.41 - ], - "text": "(a) Explain whether you would be able to", - "type": "text" - }, - { - "block_id": "p792-b9", - "global_id": 23283, - "bbox": [ - 106.15, - 481.4, - 263.22, - 501.32 - ], - "text": "generate the desired signal using only this\nequipment. If so, what is the value of k?", - "type": "text" - }, - { - "block_id": "p792-b10", - "global_id": 23284, - "bbox": [ - 317.86, - 85.81, - 490.38, - 94.86 - ], - "text": "(b) Determine the signal spectra at points b", - "type": "text" - }, - { - "block_id": "p792-b11", - "global_id": 23285, - "bbox": [ - 317.86, - 96.77, - 490.38, - 138.7 - ], - "text": "and c, and indicate the frequency bands\noccupied by these spectra.\n(c) What is the minimum usable value of ωc?\n(d) Would this scheme work if the carrier", - "type": "text" - }, - { - "block_id": "p792-b12", - "global_id": 23286, - "bbox": [ - 318.37, - 137.07, - 490.38, - 160.62 - ], - "text": "generator output were cos2 ωct? Explain.\n(e) Would this scheme work if the carrier gen-", - "type": "text" - }, - { - "block_id": "p792-b13", - "global_id": 23287, - "bbox": [ - 333.31, - 158.98, - 490.38, - 182.53 - ], - "text": "erator output were cosn ωct for any integer\nn ≥2?", - "type": "text" - }, - { - "block_id": "p792-b14", - "global_id": 23288, - "bbox": [ - 289.22, - 188.85, - 490.4, - 285.56 - ], - "text": "7.7-4\nIn practice, the analog multiplication operation\nis difficult and expensive. For this reason, in\namplitude modulators, it is necessary to find\nsome alternative to multiplication of m(t) with\ncosωct. Fortunately, for this purpose, we can\nreplace multiplication with a switching opera-\ntion. A similar observation applies to demodu-\nlators. In the scheme depicted in Fig. P7.7-4a,\nthe period of the rectangular periodic pulse x(t)", - "type": "text" - }, - { - "block_id": "p792-b15", - "global_id": 23289, - "bbox": [ - 439.31, - 452.37, - 441.53, - 460.37 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p792-b16", - "global_id": 23290, - "bbox": [ - 362.12, - 424.32, - 366.12, - 432.32 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p792-b17", - "global_id": 23291, - "bbox": [ - 295.32, - 315.63, - 428.28, - 324.09 - ], - "text": "m(t)\nBandpass", - "type": "text" - }, - { - "block_id": "p792-b18", - "global_id": 23292, - "bbox": [ - 405.17, - 324.63, - 420.72, - 343.09 - ], - "text": "filter\nvc", - "type": "text" - }, - { - "block_id": "p792-b19", - "global_id": 23293, - "bbox": [ - 447.82, - 314.99, - 489.75, - 324.55 - ], - "text": "km(t) cos vct", - "type": "text" - }, - { - "block_id": "p792-b20", - "global_id": 23294, - "bbox": [ - 353.01, - 375.31, - 364.43, - 383.39 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p792-b21", - "global_id": 23295, - "bbox": [ - 390.17, - 412.46, - 401.27, - 420.54 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p792-b22", - "global_id": 23296, - "bbox": [ - 341.4, - 451.8, - 418.98, - 470.97 - ], - "text": "4\nT0\nT0\nT0\n 4", - "type": "text" - }, - { - "block_id": "p792-b23", - "global_id": 23297, - "bbox": [ - 371.43, - 452.01, - 378.87, - 461.62 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p792-b24", - "global_id": 23298, - "bbox": [ - 377.21, - 390.22, - 386.09, - 398.22 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p792-b25", - "global_id": 23299, - "bbox": [ - 376.75, - 476.52, - 386.56, - 484.52 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p792-b26", - "global_id": 23300, - "bbox": [ - 291.96, - 492.43, - 343.61, - 501.4 - ], - "text": "Figure P7.7-4", - "type": "text" - }, - { - "block_id": "p792-b27", - "global_id": 23301, - "bbox": [ - 173.33, - 587.87, - 216.89, - 596.89 - ], - "text": "0\nv", - "type": "text" - }, - { - "block_id": "p792-b28", - "global_id": 23302, - "bbox": [ - 140.5, - 524.0, - 181.13, - 534.52 - ], - "text": "M(v)\nA", - "type": "text" - }, - { - "block_id": "p792-b29", - "global_id": 23303, - "bbox": [ - 126.86, - 589.0, - 327.77, - 608.27 - ], - "text": "2pB\n2pB\ncos3 vct\n(Carrier)", - "type": "text" - }, - { - "block_id": "p792-b30", - "global_id": 23304, - "bbox": [ - 283.18, - 554.51, - 287.18, - 562.51 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p792-b31", - "global_id": 23305, - "bbox": [ - 257.45, - 537.18, - 445.73, - 546.75 - ], - "text": "m(t)\nkm(t) cos vct", - "type": "text" - }, - { - "block_id": "p792-b32", - "global_id": 23306, - "bbox": [ - 341.13, - 555.51, - 345.13, - 563.51 - ], - "text": "b", - "type": "text" - }, - { - "block_id": "p792-b33", - "global_id": 23307, - "bbox": [ - 363.17, - 542.3, - 393.84, - 550.3 - ], - "text": "Bandpass", - "type": "text" - }, - { - "block_id": "p792-b34", - "global_id": 23308, - "bbox": [ - 370.73, - 551.3, - 422.91, - 562.51 - ], - "text": "filter\nc", - "type": "text" - }, - { - "block_id": "p792-b35", - "global_id": 23309, - "bbox": [ - 164.39, - 615.14, - 352.96, - 623.14 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p792-b36", - "global_id": 23310, - "bbox": [ - 104.83, - 629.31, - 156.48, - 638.27 - ], - "text": "Figure P7.7-3", - "type": "text" - } - ] - }, - { - "page_num": 793, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p793-b0", - "global_id": 23311, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n773", - "type": "text" - }, - { - "block_id": "p793-b1", - "global_id": 23312, - "bbox": [ - 116.46, - 85.52, - 288.99, - 237.33 - ], - "text": "shown in Fig. P7.7-4b is T0 = 2π/ωc. The\nbandpass filter is centered at ±ωc and has a\nbandwidth of 2B Hz. Note that multiplication\nby a square periodic pulse x(t) in Fig. P7.7-4b\namounts to periodic on-off switching of m(t),\nwhich is bandlimited to B Hz. Such a switching\noperation is relatively simple and inexpensive.\nShow\nthat\nthis\nscheme\ncan\ngenerate\nan\namplitude-modulated signal kcos ωct. Deter-\nmine the value of k. Show that the same\nscheme can also be used for demodulation,\nprovided the bandpass filter in Fig. P7.7-4a\nis\nreplaced\nby\na\nlowpass\n(or\nbaseband)\nfilter.", - "type": "text" - }, - { - "block_id": "p793-b2", - "global_id": 23313, - "bbox": [ - 87.82, - 249.36, - 288.99, - 291.28 - ], - "text": "7.7-5\nFigure P7.7-5a shows a scheme to transmit\ntwo signals m1(t) and m2(t) simultaneously\non the same channel (without causing spectral\ninterference). Such a scheme, which transmits", - "type": "text" - }, - { - "block_id": "p793-b3", - "global_id": 23314, - "bbox": [ - 343.61, - 85.81, - 516.13, - 182.53 - ], - "text": "more than one signal, is known as signal mul-\ntiplexing. In this case, we transmit multiple\nsignals by sharing an available spectral band\non the channel; hence, this is an example of\nthe frequency-division multiplexing. The signal\nat point b is the multiplexed signal, which\nnow modulates a carrier of frequency 20,000\nrad/s. The modulated signal at point c is now\ntransmitted over the channel.", - "type": "text" - }, - { - "block_id": "p793-b4", - "global_id": 23315, - "bbox": [ - 343.61, - 184.44, - 516.13, - 204.45 - ], - "text": "(a) Sketch the spectra at points a, b, and c.\n(b) What must be the minimum bandwidth of", - "type": "text" - }, - { - "block_id": "p793-b5", - "global_id": 23316, - "bbox": [ - 344.12, - 206.45, - 516.14, - 227.11 - ], - "text": "the channel?\n(c) Design a receiver to recover signals m1(t)", - "type": "text" - }, - { - "block_id": "p793-b6", - "global_id": 23317, - "bbox": [ - 359.05, - 227.98, - 516.13, - 248.29 - ], - "text": "and m2(t) from the modulated signal at\npoint c.", - "type": "text" - }, - { - "block_id": "p793-b7", - "global_id": 23318, - "bbox": [ - 314.97, - 259.98, - 516.13, - 290.95 - ], - "text": "7.7-6\nThe system shown in Fig. P7.7-6 is used for\nscrambling audio signals. The output y(t) is the\nscrambled version of the input m(t).", - "type": "text" - }, - { - "block_id": "p793-b8", - "global_id": 23319, - "bbox": [ - 154.81, - 385.94, - 175.14, - 395.57 - ], - "text": "M2(v)", - "type": "text" - }, - { - "block_id": "p793-b9", - "global_id": 23320, - "bbox": [ - 154.81, - 325.59, - 175.14, - 335.22 - ], - "text": "M1(v)", - "type": "text" - }, - { - "block_id": "p793-b10", - "global_id": 23321, - "bbox": [ - 359.83, - 439.06, - 369.64, - 447.06 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p793-b11", - "global_id": 23322, - "bbox": [ - 345.71, - 393.69, - 349.71, - 401.69 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p793-b12", - "global_id": 23323, - "bbox": [ - 308.37, - 335.92, - 324.7, - 345.52 - ], - "text": "m1(t)", - "type": "text" - }, - { - "block_id": "p793-b13", - "global_id": 23324, - "bbox": [ - 268.83, - 378.14, - 285.16, - 387.74 - ], - "text": "m2(t)", - "type": "text" - }, - { - "block_id": "p793-b14", - "global_id": 23325, - "bbox": [ - 394.92, - 360.12, - 455.12, - 368.12 - ], - "text": "b\nc", - "type": "text" - }, - { - "block_id": "p793-b15", - "global_id": 23326, - "bbox": [ - 402.64, - 400.26, - 445.53, - 408.34 - ], - "text": "2 cos 20,000t", - "type": "text" - }, - { - "block_id": "p793-b16", - "global_id": 23327, - "bbox": [ - 289.32, - 424.43, - 332.21, - 432.51 - ], - "text": "2 cos 10,000t", - "type": "text" - }, - { - "block_id": "p793-b18", - "global_id": 23328, - "bbox": [ - 218.48, - 351.01, - 223.82, - 359.01 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p793-b19", - "global_id": 23329, - "bbox": [ - 180.6, - 419.58, - 184.6, - 427.58 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p793-b20", - "global_id": 23330, - "bbox": [ - 180.6, - 362.43, - 184.6, - 370.43 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p793-b21", - "global_id": 23331, - "bbox": [ - 130.58, - 419.28, - 222.68, - 427.58 - ], - "text": "5000\n5000", - "type": "text" - }, - { - "block_id": "p793-b22", - "global_id": 23332, - "bbox": [ - 130.58, - 362.13, - 222.68, - 370.43 - ], - "text": "5000\n5000", - "type": "text" - }, - { - "block_id": "p793-b23", - "global_id": 23333, - "bbox": [ - 217.23, - 408.26, - 222.57, - 416.26 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p793-b24", - "global_id": 23334, - "bbox": [ - 173.79, - 439.0, - 182.67, - 447.0 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p793-b25", - "global_id": 23335, - "bbox": [ - 130.58, - 453.23, - 182.22, - 462.2 - ], - "text": "Figure P7.7-5", - "type": "text" - }, - { - "block_id": "p793-b26", - "global_id": 23336, - "bbox": [ - 253.11, - 488.59, - 257.11, - 496.59 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p793-b27", - "global_id": 23337, - "bbox": [ - 224.28, - 526.43, - 271.61, - 534.72 - ], - "text": "0\n15\n15", - "type": "text" - }, - { - "block_id": "p793-b28", - "global_id": 23338, - "bbox": [ - 245.16, - 540.31, - 254.04, - 548.31 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p793-b29", - "global_id": 23339, - "bbox": [ - 273.56, - 487.88, - 290.89, - 495.98 - ], - "text": "M(v)", - "type": "text" - }, - { - "block_id": "p793-b30", - "global_id": 23340, - "bbox": [ - 181.08, - 609.5, - 229.3, - 617.6 - ], - "text": "2 cos 30,000pt", - "type": "text" - }, - { - "block_id": "p793-b31", - "global_id": 23341, - "bbox": [ - 244.69, - 624.15, - 254.5, - 632.15 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p793-b32", - "global_id": 23342, - "bbox": [ - 378.71, - 561.91, - 435.37, - 569.91 - ], - "text": "Scrambled output", - "type": "text" - }, - { - "block_id": "p793-b33", - "global_id": 23343, - "bbox": [ - 301.57, - 526.73, - 323.25, - 534.81 - ], - "text": "f(kHz)", - "type": "text" - }, - { - "block_id": "p793-b34", - "global_id": 23344, - "bbox": [ - 154.26, - 563.03, - 361.54, - 575.76 - ], - "text": "m(t)\ny(t)\nLowpass filter", - "type": "text" - }, - { - "block_id": "p793-b35", - "global_id": 23345, - "bbox": [ - 282.3, - 577.56, - 318.08, - 586.56 - ], - "text": "0–15 kHz", - "type": "text" - }, - { - "block_id": "p793-b36", - "global_id": 23346, - "bbox": [ - 130.58, - 638.32, - 182.22, - 647.29 - ], - "text": "Figure P7.7-6", - "type": "text" - } - ] - }, - { - "page_num": 794, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p794-b0", - "global_id": 23347, - "bbox": [ - 60.0, - 60.36, - 456.14, - 69.45 - ], - "text": "774\nCHAPTER 7\nCONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p794-b1", - "global_id": 23348, - "bbox": [ - 145.68, - 89.83, - 204.78, - 99.59 - ], - "text": "[A m(t)] cos vct", - "type": "text" - }, - { - "block_id": "p794-b2", - "global_id": 23349, - "bbox": [ - 208.56, - 144.63, - 230.89, - 154.2 - ], - "text": "cos vct", - "type": "text" - }, - { - "block_id": "p794-b3", - "global_id": 23350, - "bbox": [ - 254.47, - 94.97, - 282.91, - 102.97 - ], - "text": "Lowpass", - "type": "text" - }, - { - "block_id": "p794-b4", - "global_id": 23351, - "bbox": [ - 260.92, - 103.97, - 276.47, - 111.97 - ], - "text": "filter", - "type": "text" - }, - { - "block_id": "p794-b5", - "global_id": 23352, - "bbox": [ - 317.14, - 94.97, - 334.92, - 111.97 - ], - "text": "dc\nblock", - "type": "text" - }, - { - "block_id": "p794-b6", - "global_id": 23353, - "bbox": [ - 349.91, - 90.43, - 372.13, - 98.43 - ], - "text": "Output", - "type": "text" - }, - { - "block_id": "p794-b7", - "global_id": 23354, - "bbox": [ - 387.63, - 146.71, - 439.27, - 155.67 - ], - "text": "Figure P7.7-7", - "type": "text" - }, - { - "block_id": "p794-b8", - "global_id": 23355, - "bbox": [ - 251.23, - 175.49, - 311.26, - 185.19 - ], - "text": "10\n103", - "type": "text" - }, - { - "block_id": "p794-b9", - "global_id": 23356, - "bbox": [ - 244.56, - 224.72, - 259.23, - 233.01 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p794-b10", - "global_id": 23357, - "bbox": [ - 374.85, - 206.68, - 377.08, - 214.68 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p794-b11", - "global_id": 23358, - "bbox": [ - 220.71, - 173.21, - 234.03, - 181.29 - ], - "text": "m(t)", - "type": "text" - }, - { - "block_id": "p794-b12", - "global_id": 23359, - "bbox": [ - 405.63, - 228.34, - 457.27, - 237.3 - ], - "text": "Figure P7.7-8", - "type": "text" - }, - { - "block_id": "p794-b13", - "global_id": 23360, - "bbox": [ - 91.22, - 251.8, - 263.23, - 260.77 - ], - "text": "(a) Find the spectrum of the scrambled signal", - "type": "text" - }, - { - "block_id": "p794-b14", - "global_id": 23361, - "bbox": [ - 90.72, - 262.39, - 263.24, - 282.68 - ], - "text": "y(t).\n(b) Suggest a method of descrambling y(t) to", - "type": "text" - }, - { - "block_id": "p794-b15", - "global_id": 23362, - "bbox": [ - 90.72, - 284.31, - 263.25, - 337.49 - ], - "text": "obtain m(t).\nA slightly modified version of this scrambler\nwas first used commercially on the 25-mile\nradio-telephone circuit connecting Los Angeles\nand Santa Catalina Island.", - "type": "text" - }, - { - "block_id": "p794-b16", - "global_id": 23363, - "bbox": [ - 62.08, - 343.99, - 263.24, - 386.58 - ], - "text": "7.7-7\nFigure P7.7-7 presents a scheme for coherent\n(synchronous) demodulation. Show that this\nscheme can demodulate the AM signal [A +\nm(t)]cos ωct regardless of the value of A.", - "type": "text" - }, - { - "block_id": "p794-b17", - "global_id": 23364, - "bbox": [ - 62.08, - 392.12, - 263.25, - 434.33 - ], - "text": "7.7-8\nSketch the AM signal [A + m(t)]cos ωct for\nthe periodic triangle signal m(t) illustrated in\nFig. P7.7-8 corresponding to the following mod-\nulation indices:", - "type": "text" - }, - { - "block_id": "p794-b18", - "global_id": 23365, - "bbox": [ - 90.72, - 435.95, - 134.19, - 456.25 - ], - "text": "(a) μ = 0.5\n(b) μ = 1", - "type": "text" - }, - { - "block_id": "p794-b19", - "global_id": 23366, - "bbox": [ - 90.72, - 457.87, - 230.52, - 489.13 - ], - "text": "(c) μ = 2\n(d) μ = ∞\nHow do you interpret the case μ = ∞?", - "type": "text" - }, - { - "block_id": "p794-b20", - "global_id": 23367, - "bbox": [ - 62.08, - 495.34, - 214.2, - 504.68 - ], - "text": "7.9-1\nConsider the signal x(t) defined as", - "type": "text" - }, - { - "block_id": "p794-b21", - "global_id": 23368, - "bbox": [ - 118.09, - 537.62, - 140.28, - 546.87 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p794-b23", - "global_id": 23369, - "bbox": [ - 152.64, - 530.91, - 196.08, - 541.62 - ], - "text": "1 −|t|\n−1", - "type": "text" - }, - { - "block_id": "p794-b24", - "global_id": 23370, - "bbox": [ - 161.42, - 530.91, - 235.87, - 552.58 - ], - "text": "2 ≤t ≤1\n2\n0\notherwise\n.", - "type": "text" - }, - { - "block_id": "p794-b25", - "global_id": 23371, - "bbox": [ - 90.72, - 579.64, - 263.23, - 599.94 - ], - "text": "(a) Sketch the signal x(t) over −2 ≤t ≤2.\n(b) Use time-differentiation and other Fourier", - "type": "text" - }, - { - "block_id": "p794-b26", - "global_id": 23372, - "bbox": [ - 106.15, - 601.57, - 263.26, - 632.82 - ], - "text": "transform properties to determine X(ω).\nThe only integration you should use is to\ndetermine the dc component X(0).", - "type": "text" - }, - { - "block_id": "p794-b27", - "global_id": 23373, - "bbox": [ - 318.37, - 251.8, - 490.4, - 260.77 - ], - "text": "(c) Using MATLAB, verify the correctness of", - "type": "text" - }, - { - "block_id": "p794-b28", - "global_id": 23374, - "bbox": [ - 317.86, - 262.39, - 490.4, - 304.61 - ], - "text": "X(ω) by synthesizing a 3-periodic repli-\ncation of the original time-domain signal\nx(t).\n[Hint: Follow the approach taken in Ex. 7.17.]", - "type": "text" - }, - { - "block_id": "p794-b29", - "global_id": 23375, - "bbox": [ - 289.22, - 316.91, - 434.6, - 326.25 - ], - "text": "7.9-2\nConsider the signal x(t) = |t|rect", - "type": "text" - }, - { - "block_id": "p794-b30", - "global_id": 23376, - "bbox": [ - 435.6, - 309.7, - 450.61, - 322.02 - ], - "text": "t−1", - "type": "text" - }, - { - "block_id": "p794-b31", - "global_id": 23377, - "bbox": [ - 443.89, - 309.7, - 455.42, - 328.83 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p794-b32", - "global_id": 23378, - "bbox": [ - 317.86, - 317.28, - 490.38, - 348.17 - ], - "text": ".\n(a) Sketch the signal x(t) over −5 ≤t ≤5.\n(b) Use time-differentiation and other Fourier", - "type": "text" - }, - { - "block_id": "p794-b33", - "global_id": 23379, - "bbox": [ - 318.36, - 349.78, - 490.4, - 392.0 - ], - "text": "transform properties to determine X(ω).\nThe only integration you should use is to\ndetermine the dc component X(0).\n(c) Use MATLAB to plot the magnitude spec-", - "type": "text" - }, - { - "block_id": "p794-b34", - "global_id": 23380, - "bbox": [ - 317.86, - 393.62, - 490.4, - 424.88 - ], - "text": "trum |X(ω)| and the phase spectrum̸\nX(ω)\nover suitable ranges of ω.\n(d) Using MATLAB, verify the correctness of", - "type": "text" - }, - { - "block_id": "p794-b35", - "global_id": 23381, - "bbox": [ - 317.86, - 426.5, - 490.39, - 468.72 - ], - "text": "X(ω) by synthesizing a 10-periodic repli-\ncation of the original time-domain signal\nx(t).\n[Hint: Follow the approach taken in Ex. 7.17.]", - "type": "text" - }, - { - "block_id": "p794-b36", - "global_id": 23382, - "bbox": [ - 289.22, - 481.31, - 490.4, - 534.19 - ], - "text": "7.9-3\nConsider the continuous-time aperiodic signal\nx(t) = rect(t) with Fourier transform X(ω) =\nsinc(ω/2). Furthermore, let y(t) = (1 −|t −\n1|)(u(t) −u(t −2)) with Fourier transform\nY(ω).", - "type": "text" - }, - { - "block_id": "p794-b37", - "global_id": 23383, - "bbox": [ - 318.37, - 535.82, - 490.39, - 545.15 - ], - "text": "(a) Express the Fourier transform Y(ω) in terms", - "type": "text" - }, - { - "block_id": "p794-b38", - "global_id": 23384, - "bbox": [ - 317.87, - 546.77, - 490.38, - 567.07 - ], - "text": "of X(ω).\n(b) Suppose we create Fourier series coeffi-", - "type": "text" - }, - { - "block_id": "p794-b39", - "global_id": 23385, - "bbox": [ - 318.37, - 568.69, - 490.39, - 621.86 - ], - "text": "cients Vk by sampling Y(ω) according to\nVk = Y(2πk/3). Sketch the corresponding\ntime-domain signal v(t) over a suitable\nrange of time t.\n(c) Use MATLAB to synthesize and plot v(t)", - "type": "text" - }, - { - "block_id": "p794-b40", - "global_id": 23386, - "bbox": [ - 333.31, - 623.48, - 490.39, - 633.79 - ], - "text": "using the Fourier series coefficients Vk =", - "type": "text" - } - ] - }, - { - "page_num": 795, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p795-b0", - "global_id": 23387, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n775", - "type": "text" - }, - { - "block_id": "p795-b1", - "global_id": 23388, - "bbox": [ - 116.46, - 85.58, - 288.98, - 116.83 - ], - "text": "Y(2πk/3). Verify that the synthesized wave-\nform matches the result of part (b).\n(d) Suppose we again create Fourier series", - "type": "text" - }, - { - "block_id": "p795-b2", - "global_id": 23389, - "bbox": [ - 116.96, - 118.45, - 288.99, - 171.63 - ], - "text": "coefficients Vk according to Vk = Y(2πk/3).\nNext, we upsample Vk by factor 2 to create\nWk. Sketch the time domain signal p(t) that\nhas Fourier series coefficients Pk = Vk +Wk.\n(e) Use MATLAB to synthesize and plot p(t)", - "type": "text" - }, - { - "block_id": "p795-b3", - "global_id": 23390, - "bbox": [ - 131.89, - 173.25, - 288.98, - 215.46 - ], - "text": "using the Fourier series coefficients Pk =\nVk + Wk defined in part (d). Verify that the\nsynthesized waveform matches the result of\npart (d).", - "type": "text" - }, - { - "block_id": "p795-b4", - "global_id": 23391, - "bbox": [ - 87.82, - 219.59, - 288.99, - 252.11 - ], - "text": "7.9-4\nConsider the signal x(t) = e−atu(t). Modify\nCH7MP2 to compute the following essential\nbandwidths.", - "type": "text" - }, - { - "block_id": "p795-b5", - "global_id": 23392, - "bbox": [ - 116.96, - 253.73, - 288.98, - 263.07 - ], - "text": "(a) Setting a=1, determine the essential band-", - "type": "text" - }, - { - "block_id": "p795-b6", - "global_id": 23393, - "bbox": [ - 116.46, - 264.97, - 288.99, - 306.9 - ], - "text": "width W1 that contains 95% of the signal\nenergy. Compare this value with the theo-\nretical value presented in Ex. 7.20.\n(b) Setting a=2, determine the essential band-", - "type": "text" - }, - { - "block_id": "p795-b7", - "global_id": 23394, - "bbox": [ - 116.96, - 308.81, - 288.99, - 339.78 - ], - "text": "width W2 that contains 90% of the signal\nenergy.\n(c) Setting a=3, determine the essential band-", - "type": "text" - }, - { - "block_id": "p795-b8", - "global_id": 23395, - "bbox": [ - 131.89, - 341.69, - 288.99, - 361.7 - ], - "text": "width W3 that contains 75% of the signal\nenergy.", - "type": "text" - }, - { - "block_id": "p795-b9", - "global_id": 23396, - "bbox": [ - 87.82, - 367.09, - 288.97, - 387.39 - ], - "text": "7.9-5\nA unit amplitude pulse with duration τ is defined\nas", - "type": "text" - }, - { - "block_id": "p795-b10", - "global_id": 23397, - "bbox": [ - 153.72, - 404.08, - 175.91, - 413.33 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p795-b11", - "global_id": 23398, - "bbox": [ - 177.75, - 391.49, - 245.56, - 418.85 - ], - "text": "1\n|t| ≤τ/2\n0\notherwise", - "type": "text" - }, - { - "block_id": "p795-b12", - "global_id": 23399, - "bbox": [ - 116.96, - 434.7, - 288.98, - 445.38 - ], - "text": "(a) Determine the duration τ1 that results in a", - "type": "text" - }, - { - "block_id": "p795-b13", - "global_id": 23400, - "bbox": [ - 116.46, - 446.03, - 288.98, - 467.3 - ], - "text": "95% essential bandwidth of 5 Hz.\n(b) Determine the duration τ2 that results in a", - "type": "text" - }, - { - "block_id": "p795-b14", - "global_id": 23401, - "bbox": [ - 116.96, - 467.95, - 288.98, - 489.22 - ], - "text": "90% essential bandwidth of 10 Hz.\n(c) Determine the duration τ3 that results in a", - "type": "text" - }, - { - "block_id": "p795-b15", - "global_id": 23402, - "bbox": [ - 131.89, - 489.87, - 247.69, - 498.84 - ], - "text": "75% essential bandwidth 20 Hz.", - "type": "text" - }, - { - "block_id": "p795-b16", - "global_id": 23403, - "bbox": [ - 87.82, - 502.97, - 242.73, - 513.56 - ], - "text": "7.9-6\nConsider the signal x(t) = e−atu(t).", - "type": "text" - }, - { - "block_id": "p795-b17", - "global_id": 23404, - "bbox": [ - 344.12, - 85.81, - 516.13, - 96.2 - ], - "text": "(a) Determine the decay parameter a1 that", - "type": "text" - }, - { - "block_id": "p795-b18", - "global_id": 23405, - "bbox": [ - 343.61, - 96.86, - 516.13, - 129.08 - ], - "text": "results in a 95% essential bandwidth of\n5 Hz.\n(b) Determine the decay parameter a2 that", - "type": "text" - }, - { - "block_id": "p795-b19", - "global_id": 23406, - "bbox": [ - 344.12, - 129.74, - 516.13, - 161.96 - ], - "text": "results in a 90% essential bandwidth of\n10 Hz.\n(c) Determine the decay parameter a3 that", - "type": "text" - }, - { - "block_id": "p795-b20", - "global_id": 23407, - "bbox": [ - 359.05, - 162.61, - 515.93, - 171.58 - ], - "text": "results in a 75% essential bandwidth 20 Hz.", - "type": "text" - }, - { - "block_id": "p795-b21", - "global_id": 23408, - "bbox": [ - 314.97, - 176.49, - 516.13, - 229.36 - ], - "text": "7.9-7\nUse MATLAB to determine the 95, 90, and 75%\nessential bandwidths of a one-second triangle\nfunction with a peak amplitude of 1. Recall that\na triangle function can be constructed by the\nconvolution of two rectangular pulses.", - "type": "text" - }, - { - "block_id": "p795-b22", - "global_id": 23409, - "bbox": [ - 314.97, - 234.25, - 516.13, - 254.27 - ], - "text": "7.9-8\nA 1/3 duty-cycle square-pulse T0-periodic signal\nx(t) is described as", - "type": "text" - }, - { - "block_id": "p795-b23", - "global_id": 23410, - "bbox": [ - 351.14, - 276.15, - 373.33, - 285.4 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p795-b24", - "global_id": 23411, - "bbox": [ - 375.18, - 257.77, - 382.28, - 274.81 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p795-b25", - "global_id": 23412, - "bbox": [ - 375.18, - 281.98, - 382.28, - 290.95 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p795-b26", - "global_id": 23413, - "bbox": [ - 387.25, - 265.15, - 502.42, - 297.14 - ], - "text": "1\n−T0/6 ≤t ≤T0/6\n0\nT0/t ≤|t| ≤T0/2\nx(t + T0)\n∀t", - "type": "text" - }, - { - "block_id": "p795-b27", - "global_id": 23414, - "bbox": [ - 344.12, - 307.81, - 516.15, - 316.78 - ], - "text": "(a) Use spectral sampling to determine the", - "type": "text" - }, - { - "block_id": "p795-b28", - "global_id": 23415, - "bbox": [ - 343.61, - 318.4, - 516.14, - 360.62 - ], - "text": "Fourier series coefficients Dn of x(t) for\nT0 = 2π. Evaluate and plot Dn for (0 ≤n ≤\n10).\n(b) Use spectral sampling to determine the", - "type": "text" - }, - { - "block_id": "p795-b29", - "global_id": 23416, - "bbox": [ - 359.05, - 362.24, - 516.13, - 426.37 - ], - "text": "Fourier series coefficients Dn of x(t) for\nT0 = π. Evaluate and plot Dn for (0 ≤n ≤\n10). How does this result compare with your\nanswer to part (a)? What can be said about\nthe relation of T0 to Dn for signal x(t), which\nhas fixed duty cycle of 1/3?", - "type": "text" - }, - { - "block_id": "p795-b30", - "global_id": 23417, - "bbox": [ - 314.97, - 431.28, - 516.14, - 462.23 - ], - "text": "7.9-9\nDetermine the Fourier transform of a Gaussian\npulse defined as x(t) = e−t2. Plot both x(t) and\nX(ω). How do the two curves compare? [Hint:", - "type": "text" - }, - { - "block_id": "p795-b31", - "global_id": 23418, - "bbox": [ - 383.06, - 472.64, - 394.47, - 487.56 - ], - "text": "1\n√", - "type": "text" - }, - { - "block_id": "p795-b32", - "global_id": 23419, - "bbox": [ - 390.64, - 486.1, - 400.51, - 495.44 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p795-b33", - "global_id": 23420, - "bbox": [ - 403.6, - 466.34, - 419.05, - 477.31 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p795-b34", - "global_id": 23421, - "bbox": [ - 408.33, - 488.58, - 419.99, - 495.06 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p795-b35", - "global_id": 23422, - "bbox": [ - 421.49, - 475.51, - 477.87, - 487.88 - ], - "text": "e−(t−a)2/2dt = 1", - "type": "text" - }, - { - "block_id": "p795-b36", - "global_id": 23423, - "bbox": [ - 343.61, - 504.5, - 445.07, - 513.56 - ], - "text": "for any real or imaginary a.]", - "type": "text" - } - ] - }, - { - "page_num": 796, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p796-b0", - "global_id": 23424, - "bbox": [ - 64.47, - 67.04, - 133.54, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p796-b1", - "global_id": 23425, - "bbox": [ - 147.02, - 125.73, - 465.19, - 173.26 - ], - "text": "SAMPLING: THE BRIDGE FROM\nCONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p796-b2", - "global_id": 23426, - "bbox": [ - 114.3, - 79.92, - 133.73, - 118.77 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p796-b3", - "global_id": 23427, - "bbox": [ - 101.84, - 266.14, - 490.41, - 323.93 - ], - "text": "A continuous-time signal can be processed by applying its samples through a discrete-time system.\nFor this purpose, it is important to maintain the signal sampling rate high enough to permit the\nreconstruction of the original signal from these samples without error (or with an error within\na given tolerance). The necessary quantitative framework for this purpose is provided by the\nsampling theorem derived in Sec. 8.1.", - "type": "text" - }, - { - "block_id": "p796-b4", - "global_id": 23428, - "bbox": [ - 101.84, - 325.92, - 490.41, - 407.61 - ], - "text": "Sampling theory is the bridge between the continuous-time and discrete-time worlds. The\ninformation inherent in a sampled continuous-time signal is equivalent to that of a discrete-time\nsignal. A sampled continuous-time signal is a sequence of impulses, while a discrete-time signal\npresents the same information as a sequence of numbers. These are basically two different ways\nof presenting the same data. Clearly, all the concepts in the analysis of sampled signals apply to\ndiscrete-time signals. We should not be surprised to see that the Fourier spectra of the two kinds\nof signal are also the same (within a multiplicative constant).", - "type": "text" - }, - { - "block_id": "p796-b5", - "global_id": 23429, - "bbox": [ - 102.2, - 427.69, - 297.06, - 441.64 - ], - "text": "8.1 THE SAMPLING THEOREM", - "type": "text" - }, - { - "block_id": "p796-b6", - "global_id": 23430, - "bbox": [ - 101.84, - 447.21, - 490.39, - 482.58 - ], - "text": "We now show that a real signal whose spectrum is bandlimited to B Hz [X(ω) = 0 for |ω| > 2πB]\ncan be reconstructed exactly (without any error) from its samples taken uniformly at a rate fs > 2B\nsamples per second. In other words, the minimum sampling frequency is fs = 2B Hz.†", - "type": "text" - }, - { - "block_id": "p796-b7", - "global_id": 23431, - "bbox": [ - 101.84, - 483.08, - 490.39, - 553.23 - ], - "text": "To prove the sampling theorem, consider a signal x(t) (Fig. 8.1a) whose spectrum is\nbandlimited to B Hz (Fig. 8.1b).‡ For convenience, spectra are shown as functions of ω as well\nas of f (hertz). Sampling x(t) at a rate of fs Hz ( fs samples per second) can be accomplished\nby multiplying x(t) by an impulse train δT(t) (Fig. 8.1c), consisting of unit impulses repeating\nperiodically every T seconds, where T = 1/fs. The schematic of a sampler is shown in Fig. 8.1d.\nThe resulting sampled signal x(t) is shown in Fig. 8.1e. The sampled signal consists of impulses", - "type": "text" - }, - { - "block_id": "p796-b8", - "global_id": 23432, - "bbox": [ - 101.84, - 572.08, - 490.41, - 639.39 - ], - "text": "† The theorem stated here (and proved subsequently) applies to lowpass signals. A bandpass signal whose\nspectrum exists over a frequency band fc −(B/2) < |f| < fc + (B/2) has a bandwidth of B Hz. Such a signal\nis uniquely determined by 2B samples per second. In general, the sampling scheme is a bit more complex in\nthis case. It uses two interlaced sampling trains, each at a rate of B samples per second. See, for example, [1].\n‡ The spectrum X(ω) in Fig. 8.1b is shown as real, for convenience. However, our arguments are valid for\ncomplex X(ω) as well.", - "type": "text" - }, - { - "block_id": "p796-b9", - "global_id": 23433, - "bbox": [ - 60.0, - 656.12, - 74.94, - 666.22 - ], - "text": "776", - "type": "text" - } - ] - }, - { - "page_num": 797, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p797-b0", - "global_id": 23434, - "bbox": [ - 360.21, - 62.89, - 516.13, - 71.98 - ], - "text": "8.1\nThe Sampling Theorem\n777", - "type": "text" - }, - { - "block_id": "p797-b1", - "global_id": 23435, - "bbox": [ - 154.39, - 217.9, - 170.0, - 227.47 - ], - "text": "dT(t)", - "type": "text" - }, - { - "block_id": "p797-b2", - "global_id": 23436, - "bbox": [ - 368.76, - 267.88, - 384.73, - 277.45 - ], - "text": "dT(t)", - "type": "text" - }, - { - "block_id": "p797-b3", - "global_id": 23437, - "bbox": [ - 204.15, - 416.74, - 213.03, - 424.74 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p797-b4", - "global_id": 23438, - "bbox": [ - 427.87, - 196.98, - 437.52, - 204.98 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p797-b5", - "global_id": 23439, - "bbox": [ - 426.09, - 162.87, - 430.09, - 170.87 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p797-b6", - "global_id": 23440, - "bbox": [ - 437.09, - 91.86, - 452.65, - 99.96 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p797-b7", - "global_id": 23441, - "bbox": [ - 410.22, - 149.21, - 452.21, - 157.5 - ], - "text": "2pB\n2pB", - "type": "text" - }, - { - "block_id": "p797-b8", - "global_id": 23442, - "bbox": [ - 442.65, - 162.85, - 496.89, - 175.34 - ], - "text": "B\nf (Hz)", - "type": "text" - }, - { - "block_id": "p797-b9", - "global_id": 23443, - "bbox": [ - 477.31, - 388.1, - 497.39, - 396.18 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p797-b10", - "global_id": 23444, - "bbox": [ - 439.57, - 115.62, - 444.46, - 123.62 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p797-b11", - "global_id": 23445, - "bbox": [ - 491.49, - 149.4, - 496.83, - 157.4 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p797-b12", - "global_id": 23446, - "bbox": [ - 204.15, - 196.98, - 213.03, - 204.98 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p797-b13", - "global_id": 23447, - "bbox": [ - 196.69, - 94.39, - 207.8, - 102.47 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p797-b14", - "global_id": 23448, - "bbox": [ - 327.48, - 233.08, - 504.83, - 241.16 - ], - "text": "x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p797-b15", - "global_id": 23449, - "bbox": [ - 266.74, - 137.76, - 268.96, - 145.76 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p797-b16", - "global_id": 23450, - "bbox": [ - 266.74, - 257.56, - 268.96, - 265.56 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p797-b17", - "global_id": 23451, - "bbox": [ - 266.34, - 356.31, - 268.56, - 364.31 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p797-b18", - "global_id": 23452, - "bbox": [ - 204.15, - 283.94, - 213.03, - 291.94 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p797-b19", - "global_id": 23453, - "bbox": [ - 159.3, - 263.41, - 163.75, - 271.41 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p797-b20", - "global_id": 23454, - "bbox": [ - 436.73, - 371.65, - 450.96, - 379.75 - ], - "text": "2pB", - "type": "text" - }, - { - "block_id": "p797-b21", - "global_id": 23455, - "bbox": [ - 369.78, - 316.51, - 398.22, - 324.51 - ], - "text": "Lowpass", - "type": "text" - }, - { - "block_id": "p797-b22", - "global_id": 23456, - "bbox": [ - 376.22, - 325.51, - 391.78, - 333.51 - ], - "text": "filter", - "type": "text" - }, - { - "block_id": "p797-b23", - "global_id": 23457, - "bbox": [ - 425.89, - 388.09, - 446.29, - 400.06 - ], - "text": "B\n0", - "type": "text" - }, - { - "block_id": "p797-b24", - "global_id": 23458, - "bbox": [ - 428.22, - 416.74, - 437.17, - 424.74 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p797-b25", - "global_id": 23459, - "bbox": [ - 354.95, - 237.79, - 483.15, - 254.79 - ], - "text": "Ideal lowpass\nfilter, cutoff B Hz\nSampler", - "type": "text" - }, - { - "block_id": "p797-b26", - "global_id": 23460, - "bbox": [ - 428.03, - 283.94, - 437.36, - 291.94 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p797-b27", - "global_id": 23461, - "bbox": [ - 409.58, - 368.09, - 497.33, - 384.92 - ], - "text": "v\np\nT", - "type": "text" - }, - { - "block_id": "p797-b28", - "global_id": 23462, - "bbox": [ - 393.35, - 392.16, - 468.37, - 411.24 - ], - "text": "fs\nfs\nfs\n2", - "type": "text" - }, - { - "block_id": "p797-b29", - "global_id": 23463, - "bbox": [ - 388.93, - 368.07, - 469.93, - 384.92 - ], - "text": "2p\n T\nvs", - "type": "text" - }, - { - "block_id": "p797-b30", - "global_id": 23464, - "bbox": [ - 398.87, - 230.13, - 410.42, - 241.13 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p797-b31", - "global_id": 23465, - "bbox": [ - 195.03, - 303.35, - 451.65, - 320.05 - ], - "text": "x–(t)\nX(v)", - "type": "text" - }, - { - "block_id": "p797-b32", - "global_id": 23466, - "bbox": [ - 440.19, - 334.62, - 453.53, - 342.84 - ], - "text": "AT", - "type": "text" - }, - { - "block_id": "p797-b33", - "global_id": 23467, - "bbox": [ - 127.59, - 431.44, - 319.97, - 440.68 - ], - "text": "Figure 8.1 Sampled signal and its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p797-b34", - "global_id": 23468, - "bbox": [ - 127.59, - 451.97, - 516.13, - 474.3 - ], - "text": "spaced every T seconds (the sampling interval). The nth impulse, located at t = nT, has a strength\nx(nT), the value of x(t) at t = nT.", - "type": "text" - }, - { - "block_id": "p797-b35", - "global_id": 23469, - "bbox": [ - 247.5, - 485.42, - 318.43, - 496.5 - ], - "text": "x(t) = x(t)δT(t) =", - "type": "text" - }, - { - "block_id": "p797-b36", - "global_id": 23470, - "bbox": [ - 320.48, - 475.97, - 334.58, - 485.93 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p797-b37", - "global_id": 23471, - "bbox": [ - 325.78, - 499.49, - 329.27, - 506.47 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p797-b38", - "global_id": 23472, - "bbox": [ - 335.68, - 485.42, - 396.21, - 495.7 - ], - "text": "x(nT)δ(t −nT)", - "type": "text" - }, - { - "block_id": "p797-b39", - "global_id": 23473, - "bbox": [ - 127.59, - 514.22, - 516.14, - 536.56 - ], - "text": "Because the impulse train δT(t) is a periodic signal of period T, it can be expressed as a\ntrigonometric Fourier series like that already obtained in Ex. 6.9 [Eq. (6.25)],", - "type": "text" - }, - { - "block_id": "p797-b40", - "global_id": 23474, - "bbox": [ - 169.6, - 544.84, - 207.88, - 562.49 - ], - "text": "δT(t) = 1", - "type": "text" - }, - { - "block_id": "p797-b41", - "global_id": 23475, - "bbox": [ - 202.23, - 544.43, - 442.13, - 568.76 - ], - "text": "T [1 + 2cosωst + 2cos2ωst + 2cos3ωst + · · ·]\nωs = 2π", - "type": "text" - }, - { - "block_id": "p797-b42", - "global_id": 23476, - "bbox": [ - 434.0, - 551.41, - 473.63, - 568.76 - ], - "text": "T = 2πfs", - "type": "text" - }, - { - "block_id": "p797-b43", - "global_id": 23477, - "bbox": [ - 127.59, - 574.99, - 169.34, - 584.96 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p797-b44", - "global_id": 23478, - "bbox": [ - 154.94, - 592.52, - 234.76, - 610.17 - ], - "text": "x(t) = x(t)δT(t) = 1", - "type": "text" - }, - { - "block_id": "p797-b45", - "global_id": 23479, - "bbox": [ - 229.11, - 599.09, - 516.13, - 616.44 - ], - "text": "T [x(t) + 2x(t)cosωst + 2x(t)cos2ωst + 2x(t)cos3ωst + · · ·]\n(8.1)", - "type": "text" - }, - { - "block_id": "p797-b46", - "global_id": 23480, - "bbox": [ - 127.59, - 624.41, - 516.14, - 646.74 - ], - "text": "To find X(ω), the Fourier transform of x(t), we take the Fourier transform of the right-hand side of\nEq. (8.1), term by term. The transform of the first term in the brackets is X(ω). The transform", - "type": "text" - } - ] - }, - { - "page_num": 798, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p798-b0", - "global_id": 23481, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "778\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p798-b1", - "global_id": 23482, - "bbox": [ - 101.84, - 85.46, - 490.4, - 155.62 - ], - "text": "of the second term 2x(t)cos ωst is X(ω −ωs) + X(ω + ωs) [see Eq. (7.32)]. This represents\nspectrum X(ω) shifted to ωs and −ωs. Similarly, the transform of the third term 2x(t)cos2ωst\nis X(ω −2ωs) + X(ω + 2ωs), which represents the spectrum X(ω) shifted to 2ωs and −2ωs, and\nso on to infinity. This result means that the spectrum X(ω) consists of X(ω) repeating periodically\nwith period ωs = 2π/T rad/s, or fs = 1/T Hz, as depicted in Fig. 8.1f. There is also a constant\nmultiplier 1/T in Eq. (8.1). Therefore,", - "type": "text" - }, - { - "block_id": "p798-b2", - "global_id": 23483, - "bbox": [ - 240.51, - 168.84, - 279.9, - 185.79 - ], - "text": "X(ω) = 1", - "type": "text" - }, - { - "block_id": "p798-b3", - "global_id": 23484, - "bbox": [ - 274.26, - 182.8, - 279.8, - 192.76 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p798-b4", - "global_id": 23485, - "bbox": [ - 286.56, - 165.24, - 300.66, - 175.91 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p798-b5", - "global_id": 23486, - "bbox": [ - 282.86, - 189.26, - 304.34, - 196.45 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p798-b6", - "global_id": 23487, - "bbox": [ - 305.45, - 175.41, - 490.39, - 186.49 - ], - "text": "X(ω −nωs)\n(8.2)", - "type": "text" - }, - { - "block_id": "p798-b7", - "global_id": 23488, - "bbox": [ - 101.85, - 208.57, - 490.38, - 242.86 - ], - "text": "If we are to reconstruct x(t) from x(t), we should be able to recover X(ω) from X(ω). This\nrecovery is possible if there is no overlap between successive cycles of X(ω). Figure 8.1f indicates\nthat this requires", - "type": "text" - }, - { - "block_id": "p798-b8", - "global_id": 23489, - "bbox": [ - 281.66, - 244.63, - 490.39, - 256.09 - ], - "text": "fs > 2B\n(8.3)", - "type": "text" - }, - { - "block_id": "p798-b9", - "global_id": 23490, - "bbox": [ - 101.85, - 263.68, - 292.56, - 274.76 - ], - "text": "Also, the sampling interval T = 1/fs. Therefore,", - "type": "text" - }, - { - "block_id": "p798-b10", - "global_id": 23491, - "bbox": [ - 280.3, - 284.12, - 307.7, - 301.07 - ], - "text": "T < 1", - "type": "text" - }, - { - "block_id": "p798-b11", - "global_id": 23492, - "bbox": [ - 299.67, - 298.08, - 310.74, - 308.14 - ], - "text": "2B", - "type": "text" - }, - { - "block_id": "p798-b12", - "global_id": 23493, - "bbox": [ - 101.84, - 315.95, - 490.4, - 409.7 - ], - "text": "Thus, as long as the sampling frequency fs is greater than twice the signal bandwidth B (in hertz),\nX(ω) consists of nonoverlapping repetitions of X(ω). Figure 8.1f shows that the gap between the\ntwo adjacent spectral repetitions is fs −2B Hz, and x(t) can be recovered from its samples x(t) by\npassing the sampled signal x(t) through an ideal lowpass filter having a bandwidth of any value\nbetween B and fs −B Hz. The minimum sampling rate fs = 2B required to recover x(t) from its\nsamples x(t) is called the Nyquist rate for x(t), and the corresponding sampling interval T = 1/2B\nis called the Nyquist interval for x(t). Samples of a signal taken at its Nyquist rate are the Nyquist\nsamples of that signal.", - "type": "text" - }, - { - "block_id": "p798-b13", - "global_id": 23494, - "bbox": [ - 101.84, - 411.6, - 490.41, - 493.39 - ], - "text": "We are saying that the Nyquist rate 2B Hz is the minimum sampling rate required to\npreserve the information of x(t). This contradicts Eq. (8.3), where we showed that to preserve the\ninformation of x(t), the sampling rate fs needs to be greater than 2B Hz. Strictly speaking, Eq. (8.3)\nis the correct statement. However, if the spectrum X(ω) contains no impulse or its derivatives at\nthe highest frequency B Hz, then the minimum sampling rate 2B Hz is adequate. In practice, it is\nrare to observe X(ω) with an impulse or its derivatives at the highest frequency. If the contrary\nsituation were to occur, we should use Eq. (8.3).†", - "type": "text" - }, - { - "block_id": "p798-b14", - "global_id": 23495, - "bbox": [ - 101.84, - 511.61, - 490.4, - 633.41 - ], - "text": "† An interesting observation is that if the impulse is because of a cosine term, the sampling rate of 2B Hz is\nadequate. However, if the impulse is because of a sine term, then the rate must be greater than 2B Hz. This may\nbe seen from the fact that samples of sin2πBt using T = 1/2B are all zero because sin2πBnT = sinπn = 0.\nBut, samples of cos2πBt are cos2πBnT = cosπn = (−1)n. We can reconstruct cos2πBt from these samples.\nThis peculiar behavior occurs because in the sampled signal spectrum corresponding to the signal cos2πBt,\nthe impulses, which occur at frequencies (2n ± 1)B Hz (n = 0,±1,±2,. . .), interact constructively, whereas\nin the case of sin2πBt, the impulses, because of their opposite phases (e±jπ/2), interact destructively and\ncancel out in the sampled signal spectrum. Hence, sin2πBt cannot be reconstructed from its samples at a rate\n2B Hz. A similar situation exists for signal cos(2πBt+θ), which contains a component of the form sin2πBt.\nFor this reason, it is advisable to maintain sampling rate above 2B Hz if a finite-amplitude component of a\nsinusoid of frequency B Hz is present in the signal.", - "type": "text" - } - ] - }, - { - "page_num": 799, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p799-b0", - "global_id": 23496, - "bbox": [ - 360.21, - 62.89, - 516.13, - 71.98 - ], - "text": "8.1\nThe Sampling Theorem\n779", - "type": "text" - }, - { - "block_id": "p799-b1", - "global_id": 23497, - "bbox": [ - 127.59, - 85.82, - 516.15, - 167.51 - ], - "text": "The sampling theorem proved here uses samples taken at uniform intervals. This condition\nis not necessary. Samples can be taken arbitrarily at any instants as long as the sampling instants\nare recorded and there are, on average, 2B samples per second [2]. The essence of the sampling\ntheorem was known to mathematicians for a long time in the form of the interpolation formula\n[see later, Eq. (8.6)]. The origin of the sampling theorem was attributed by H. S. Black to Cauchy\nin 1841. The essential idea of the sampling theorem was rediscovered in the 1920s by Carson,\nNyquist, and Hartley.", - "type": "text" - }, - { - "block_id": "p799-b2", - "global_id": 23498, - "bbox": [ - 102.51, - 260.72, - 466.89, - 272.67 - ], - "text": "EXAMPLE 8.1\nSampling at, Below, and Above the Nyquist Rate", - "type": "text" - }, - { - "block_id": "p799-b3", - "global_id": 23499, - "bbox": [ - 128.9, - 289.33, - 502.77, - 359.07 - ], - "text": "In this example, we examine the effects of sampling a signal at the Nyquist rate, below\nthe Nyquist rate (undersampling), and above the Nyquist rate (oversampling). Consider a\nsignal x(t) = sinc2 (5πt) (Fig. 8.2a) whose spectrum is X(ω) = 0.2(ω/20π) (Fig. 8.2b).\nThe bandwidth of this signal is 5 Hz (10π rad/s). Consequently, the Nyquist rate is 10 Hz;\nthat is, we must sample the signal at a rate no less than 10 samples/s. The Nyquist interval is\nT = 1/2B = 0.1 second.", - "type": "text" - }, - { - "block_id": "p799-b4", - "global_id": 23500, - "bbox": [ - 128.9, - 381.58, - 502.76, - 416.94 - ], - "text": "Recall that the sampled signal spectrum consists of (1/T)X(ω) = (0.2/T)(ω/20π) repeating\nperiodically with a period equal to the sampling frequency fs Hz. For the three sampling rates\nfs = 5 Hz (undersampling), 10 Hz (Nyquist rate), and 20 Hz (oversampling), we see that", - "type": "text" - }, - { - "block_id": "p799-b5", - "global_id": 23501, - "bbox": [ - 218.63, - 443.93, - 280.28, - 457.12 - ], - "text": "fs (Hz)\nT = 1", - "type": "text" - }, - { - "block_id": "p799-b6", - "global_id": 23502, - "bbox": [ - 275.96, - 443.93, - 397.51, - 460.21 - ], - "text": "fs (s)\n1\nTX(ω)\nComments", - "type": "text" - }, - { - "block_id": "p799-b7", - "global_id": 23503, - "bbox": [ - 218.63, - 466.13, - 313.28, - 476.51 - ], - "text": "5\n0.2", - "type": "text" - }, - { - "block_id": "p799-b8", - "global_id": 23504, - "bbox": [ - 314.38, - 458.12, - 327.74, - 471.47 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p799-b9", - "global_id": 23505, - "bbox": [ - 319.6, - 458.12, - 336.66, - 479.32 - ], - "text": "20π", - "type": "text" - }, - { - "block_id": "p799-b10", - "global_id": 23506, - "bbox": [ - 351.6, - 466.54, - 413.02, - 476.51 - ], - "text": "Undersampling", - "type": "text" - }, - { - "block_id": "p799-b11", - "global_id": 23507, - "bbox": [ - 218.63, - 473.17, - 332.71, - 491.55 - ], - "text": "10\n0.1\n2\n ω", - "type": "text" - }, - { - "block_id": "p799-b12", - "global_id": 23508, - "bbox": [ - 324.57, - 473.17, - 341.64, - 494.37 - ], - "text": "20π", - "type": "text" - }, - { - "block_id": "p799-b13", - "global_id": 23509, - "bbox": [ - 351.6, - 481.59, - 400.59, - 491.55 - ], - "text": "Nyquist rate", - "type": "text" - }, - { - "block_id": "p799-b14", - "global_id": 23510, - "bbox": [ - 218.63, - 488.21, - 332.71, - 506.59 - ], - "text": "20\n0.05\n4\n ω", - "type": "text" - }, - { - "block_id": "p799-b15", - "global_id": 23511, - "bbox": [ - 324.57, - 488.21, - 341.64, - 509.41 - ], - "text": "20π", - "type": "text" - }, - { - "block_id": "p799-b16", - "global_id": 23512, - "bbox": [ - 351.6, - 496.63, - 407.88, - 506.59 - ], - "text": "Oversampling", - "type": "text" - }, - { - "block_id": "p799-b17", - "global_id": 23513, - "bbox": [ - 128.9, - 541.87, - 502.8, - 611.61 - ], - "text": "In the first case (undersampling), the sampling rate is 5 Hz (5 samples/s), and the\nspectrum (1/T)X(ω) repeats every 5 Hz (10π rad/s). The successive spectra overlap, as\ndepicted in Fig. 8.2d, and the spectrum X(ω) are not recoverable from X(ω); that is, x(t)\ncannot be reconstructed from its samples x(t) in Fig. 8.2c. In the second case, we use the\nNyquist sampling rate of 10 Hz (Fig. 8.2e). The spectrum X(ω) consists of back-to-back,\nnonoverlapping repetitions of (1/T)X(ω) repeating every 10 Hz. Hence, X(ω) can be recovered", - "type": "text" - } - ] - }, - { - "page_num": 800, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p800-b0", - "global_id": 23514, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "780\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p800-b1", - "global_id": 23515, - "bbox": [ - 139.74, - 174.43, - 211.36, - 183.17 - ], - "text": "0\n0.2\n0.2", - "type": "text" - }, - { - "block_id": "p800-b2", - "global_id": 23516, - "bbox": [ - 169.03, - 511.01, - 173.03, - 519.01 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p800-b3", - "global_id": 23517, - "bbox": [ - 125.78, - 598.58, - 226.42, - 606.88 - ], - "text": "0\n0.2\n0.3\n0.1\nt", - "type": "text" - }, - { - "block_id": "p800-b4", - "global_id": 23518, - "bbox": [ - 224.2, - 456.78, - 226.42, - 464.78 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p800-b5", - "global_id": 23519, - "bbox": [ - 224.2, - 316.02, - 226.42, - 324.02 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p800-b6", - "global_id": 23520, - "bbox": [ - 224.2, - 173.33, - 226.42, - 181.33 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p800-b7", - "global_id": 23521, - "bbox": [ - 169.03, - 368.44, - 173.03, - 376.44 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p800-b8", - "global_id": 23522, - "bbox": [ - 175.6, - 456.86, - 211.1, - 464.86 - ], - "text": "0\n0.2", - "type": "text" - }, - { - "block_id": "p800-b9", - "global_id": 23523, - "bbox": [ - 169.03, - 227.93, - 173.03, - 235.93 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p800-b10", - "global_id": 23524, - "bbox": [ - 112.98, - 316.05, - 210.94, - 324.35 - ], - "text": "0\n0.2\n0.4", - "type": "text" - }, - { - "block_id": "p800-b11", - "global_id": 23525, - "bbox": [ - 125.49, - 456.56, - 171.79, - 464.86 - ], - "text": "0.3\n0.1", - "type": "text" - }, - { - "block_id": "p800-b12", - "global_id": 23526, - "bbox": [ - 140.85, - 316.05, - 157.51, - 324.35 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p800-b13", - "global_id": 23527, - "bbox": [ - 352.76, - 121.62, - 362.76, - 129.62 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p800-b14", - "global_id": 23528, - "bbox": [ - 192.47, - 97.32, - 203.58, - 105.4 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p800-b15", - "global_id": 23529, - "bbox": [ - 376.06, - 252.31, - 391.62, - 260.41 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p800-b16", - "global_id": 23530, - "bbox": [ - 373.61, - 397.26, - 389.16, - 405.36 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p800-b17", - "global_id": 23531, - "bbox": [ - 373.61, - 540.07, - 389.16, - 548.18 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p800-b18", - "global_id": 23532, - "bbox": [ - 373.61, - 122.64, - 389.16, - 130.74 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p800-b19", - "global_id": 23533, - "bbox": [ - 456.47, - 172.21, - 461.8, - 180.21 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p800-b20", - "global_id": 23534, - "bbox": [ - 456.47, - 314.24, - 461.8, - 322.24 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p800-b21", - "global_id": 23535, - "bbox": [ - 456.47, - 454.87, - 461.8, - 462.87 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p800-b22", - "global_id": 23536, - "bbox": [ - 458.97, - 597.45, - 464.31, - 605.45 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p800-b23", - "global_id": 23537, - "bbox": [ - 358.49, - 403.44, - 362.49, - 411.44 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p800-b24", - "global_id": 23538, - "bbox": [ - 287.77, - 598.58, - 444.09, - 606.88 - ], - "text": "40p\n10p\n10p\n40p", - "type": "text" - }, - { - "block_id": "p800-b25", - "global_id": 23539, - "bbox": [ - 315.43, - 391.32, - 348.98, - 399.32 - ], - "text": "Ideal filter", - "type": "text" - }, - { - "block_id": "p800-b26", - "global_id": 23540, - "bbox": [ - 308.09, - 533.64, - 354.12, - 541.64 - ], - "text": "Practical filter", - "type": "text" - }, - { - "block_id": "p800-b27", - "global_id": 23541, - "bbox": [ - 290.44, - 481.03, - 441.12, - 489.45 - ], - "text": "5\n5\n20\n20", - "type": "text" - }, - { - "block_id": "p800-b28", - "global_id": 23542, - "bbox": [ - 173.16, - 211.94, - 372.15, - 221.44 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p800-b29", - "global_id": 23543, - "bbox": [ - 173.16, - 349.46, - 182.04, - 357.46 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p800-b30", - "global_id": 23544, - "bbox": [ - 173.16, - 495.18, - 371.72, - 503.18 - ], - "text": "(e)\n(f)", - "type": "text" - }, - { - "block_id": "p800-b31", - "global_id": 23545, - "bbox": [ - 344.63, - 337.96, - 441.19, - 348.16 - ], - "text": "5\n5\n20", - "type": "text" - }, - { - "block_id": "p800-b32", - "global_id": 23546, - "bbox": [ - 344.63, - 200.22, - 386.5, - 208.52 - ], - "text": "5\n5", - "type": "text" - }, - { - "block_id": "p800-b33", - "global_id": 23547, - "bbox": [ - 290.44, - 337.96, - 305.1, - 346.26 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p800-b34", - "global_id": 23548, - "bbox": [ - 362.58, - 351.78, - 371.91, - 359.78 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p800-b35", - "global_id": 23549, - "bbox": [ - 172.94, - 612.2, - 372.07, - 620.2 - ], - "text": "(g)\n(h)", - "type": "text" - }, - { - "block_id": "p800-b36", - "global_id": 23550, - "bbox": [ - 169.03, - 86.31, - 173.03, - 94.31 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p800-b37", - "global_id": 23551, - "bbox": [ - 287.77, - 456.48, - 444.09, - 464.77 - ], - "text": "40p\n10p\n10p\n40p", - "type": "text" - }, - { - "block_id": "p800-b38", - "global_id": 23552, - "bbox": [ - 322.47, - 315.81, - 444.09, - 324.1 - ], - "text": "40p\n20p\n20p", - "type": "text" - }, - { - "block_id": "p800-b39", - "global_id": 23553, - "bbox": [ - 339.96, - 173.54, - 391.66, - 181.84 - ], - "text": "10p\n10p", - "type": "text" - }, - { - "block_id": "p800-b40", - "global_id": 23554, - "bbox": [ - 287.04, - 315.81, - 307.04, - 324.1 - ], - "text": "40p", - "type": "text" - }, - { - "block_id": "p800-b41", - "global_id": 23555, - "bbox": [ - 359.48, - 546.41, - 363.48, - 554.41 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p800-b42", - "global_id": 23556, - "bbox": [ - 360.3, - 263.47, - 364.3, - 271.47 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p800-b43", - "global_id": 23557, - "bbox": [ - 441.72, - 200.85, - 461.8, - 208.93 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p800-b44", - "global_id": 23558, - "bbox": [ - 448.72, - 336.35, - 468.8, - 344.43 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p800-b45", - "global_id": 23559, - "bbox": [ - 448.72, - 481.65, - 468.8, - 489.73 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p800-b46", - "global_id": 23560, - "bbox": [ - 190.47, - 525.53, - 202.02, - 536.53 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p800-b47", - "global_id": 23561, - "bbox": [ - 189.47, - 378.08, - 201.02, - 389.08 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p800-b48", - "global_id": 23562, - "bbox": [ - 188.47, - 234.12, - 200.02, - 245.12 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p800-b49", - "global_id": 23563, - "bbox": [ - 103.16, - 626.89, - 307.38, - 636.13 - ], - "text": "Figure 8.2 Effects of undersampling and oversampling.", - "type": "text" - } - ] - }, - { - "page_num": 801, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p801-b0", - "global_id": 23564, - "bbox": [ - 360.21, - 62.89, - 516.13, - 71.98 - ], - "text": "8.1\nThe Sampling Theorem\n781", - "type": "text" - }, - { - "block_id": "p801-b1", - "global_id": 23565, - "bbox": [ - 128.9, - 87.97, - 502.77, - 146.17 - ], - "text": "from X(ω) using an ideal lowpass filter of bandwidth 5 Hz (Fig. 8.2f). Finally, in the last\ncase of oversampling (sampling rate 20 Hz), the spectrum X(ω) consists of nonoverlapping\nrepetitions of (1/T)X(ω) (repeating every 20 Hz) with empty bands between successive cycles\n(Fig. 8.2h). Hence, X(ω) can be recovered from X(ω) by using an ideal lowpass filter or even\na practical lowpass filter (shown dashed in Fig. 8.2h).†", - "type": "text" - }, - { - "block_id": "p801-b2", - "global_id": 23566, - "bbox": [ - 133.57, - 213.33, - 309.18, - 225.28 - ], - "text": "DRILL 8.1\nNyquist Sampling", - "type": "text" - }, - { - "block_id": "p801-b3", - "global_id": 23567, - "bbox": [ - 133.57, - 233.99, - 510.15, - 256.32 - ], - "text": "Find the Nyquist rate and the Nyquist sampling interval for the signals sinc(100πt) and\nsinc(100πt) + sinc(50πt).", - "type": "text" - }, - { - "block_id": "p801-b4", - "global_id": 23568, - "bbox": [ - 133.84, - 269.84, - 196.06, - 280.8 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p801-b5", - "global_id": 23569, - "bbox": [ - 133.57, - 288.19, - 510.16, - 310.11 - ], - "text": "The Nyquist sampling interval is 0.01 s and the Nyquist sampling rate is 100 Hz for both\nsignals.", - "type": "text" - }, - { - "block_id": "p801-b6", - "global_id": 23570, - "bbox": [ - 127.89, - 341.93, - 237.14, - 354.05 - ], - "text": "FOR SKEPTICS ONLY", - "type": "text" - }, - { - "block_id": "p801-b7", - "global_id": 23571, - "bbox": [ - 127.59, - 358.08, - 516.15, - 415.86 - ], - "text": "Rare is the reader who, at first encounter, is not skeptical of the sampling theorem. It seems\nimpossible that Nyquist samples can define the one and the only signal that passes through\nthose sample values. We can easily picture infinite number of signals passing through a given\nset of samples. However, among all these (infinite number of) signals, only one has the minimum\nbandwidth B ≤1/2T Hz, where T is the sampling interval. See Prob. 8.2-15.", - "type": "text" - }, - { - "block_id": "p801-b8", - "global_id": 23572, - "bbox": [ - 127.59, - 417.76, - 516.15, - 463.69 - ], - "text": "To summarize, for a given set of samples taken at a rate fs Hz, there is only one signal of\nbandwidth B ≤fs/2 that passes through those samples. All other signals that pass through those\nsamples have bandwidth higher than fs/2, and the samples are sub-Nyquist rate samples for those\nsignals.", - "type": "text" - }, - { - "block_id": "p801-b9", - "global_id": 23573, - "bbox": [ - 127.59, - 490.4, - 260.57, - 502.36 - ], - "text": "8.1-1 Practical Sampling", - "type": "text" - }, - { - "block_id": "p801-b10", - "global_id": 23574, - "bbox": [ - 127.59, - 508.08, - 516.14, - 590.18 - ], - "text": "In proving the sampling theorem, we assumed ideal samples obtained by multiplying a signal x(t)\nby an impulse train that is physically unrealizable. In practice, we multiply a signal x(t) by a train\nof pulses of finite width, depicted in Fig. 8.3c. The sampler is shown in Fig. 8.3d. The sampled\nsignal x(t) is illustrated in Fig. 8.3e. We wonder whether it is possible to recover or reconstruct\nx(t) from this x(t). Surprisingly, the answer is affirmative, provided the sampling rate is not below\nthe Nyquist rate. The signal x(t) can be recovered by lowpass filtering x(t) as if it were sampled\nby impulse train.", - "type": "text" - }, - { - "block_id": "p801-b11", - "global_id": 23575, - "bbox": [ - 127.59, - 610.24, - 516.13, - 633.41 - ], - "text": "† The filter should have a constant gain between 0 and 5 Hz and zero gain beyond 10 Hz. In practice, the gain\nbeyond 10 Hz can be made negligibly small, but not zero.", - "type": "text" - } - ] - }, - { - "page_num": 802, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p802-b0", - "global_id": 23576, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "782\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p802-b1", - "global_id": 23577, - "bbox": [ - 178.0, - 389.74, - 395.8, - 397.74 - ], - "text": "(e)\n(f)", - "type": "text" - }, - { - "block_id": "p802-b2", - "global_id": 23578, - "bbox": [ - 427.28, - 292.26, - 455.72, - 300.26 - ], - "text": "Lowpass", - "type": "text" - }, - { - "block_id": "p802-b3", - "global_id": 23579, - "bbox": [ - 433.73, - 301.86, - 449.28, - 309.86 - ], - "text": "filter", - "type": "text" - }, - { - "block_id": "p802-b4", - "global_id": 23580, - "bbox": [ - 401.66, - 364.62, - 406.55, - 372.62 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p802-b5", - "global_id": 23581, - "bbox": [ - 401.66, - 178.2, - 406.55, - 186.2 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p802-b6", - "global_id": 23582, - "bbox": [ - 420.97, - 347.18, - 428.64, - 356.75 - ], - "text": "vs", - "type": "text" - }, - { - "block_id": "p802-b7", - "global_id": 23583, - "bbox": [ - 384.57, - 361.69, - 427.09, - 374.16 - ], - "text": "fs\n0", - "type": "text" - }, - { - "block_id": "p802-b8", - "global_id": 23584, - "bbox": [ - 399.28, - 121.18, - 414.83, - 129.28 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p802-b9", - "global_id": 23585, - "bbox": [ - 368.57, - 160.98, - 460.24, - 169.42 - ], - "text": "2pB\n2pB\nv", - "type": "text" - }, - { - "block_id": "p802-b10", - "global_id": 23586, - "bbox": [ - 456.61, - 174.82, - 458.83, - 182.82 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p802-b11", - "global_id": 23587, - "bbox": [ - 450.19, - 345.02, - 455.53, - 353.02 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p802-b12", - "global_id": 23588, - "bbox": [ - 451.89, - 362.51, - 454.12, - 370.51 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p802-b13", - "global_id": 23589, - "bbox": [ - 133.12, - 248.59, - 137.57, - 256.59 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p802-b14", - "global_id": 23590, - "bbox": [ - 396.99, - 347.24, - 411.22, - 355.34 - ], - "text": "2pB", - "type": "text" - }, - { - "block_id": "p802-b15", - "global_id": 23591, - "bbox": [ - 241.0, - 248.14, - 243.2, - 256.04 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p802-b16", - "global_id": 23592, - "bbox": [ - 241.0, - 337.33, - 243.2, - 345.23 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p802-b17", - "global_id": 23593, - "bbox": [ - 402.55, - 223.55, - 459.42, - 241.15 - ], - "text": "Ideal lowpass\nfilter, cutoff B Hz", - "type": "text" - }, - { - "block_id": "p802-b18", - "global_id": 23594, - "bbox": [ - 306.75, - 220.49, - 317.85, - 228.58 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p802-b19", - "global_id": 23595, - "bbox": [ - 334.22, - 228.3, - 360.89, - 236.3 - ], - "text": "Sampler", - "type": "text" - }, - { - "block_id": "p802-b20", - "global_id": 23596, - "bbox": [ - 167.05, - 91.85, - 178.15, - 99.93 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p802-b21", - "global_id": 23597, - "bbox": [ - 178.4, - 190.18, - 396.23, - 198.18 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p802-b22", - "global_id": 23598, - "bbox": [ - 117.99, - 210.96, - 133.71, - 220.51 - ], - "text": "pT(t)", - "type": "text" - }, - { - "block_id": "p802-b23", - "global_id": 23599, - "bbox": [ - 348.94, - 253.81, - 364.67, - 263.35 - ], - "text": "pT(t)", - "type": "text" - }, - { - "block_id": "p802-b24", - "global_id": 23600, - "bbox": [ - 175.9, - 270.55, - 184.78, - 278.55 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p802-b25", - "global_id": 23601, - "bbox": [ - 468.79, - 220.5, - 489.15, - 230.1 - ], - "text": "C0x(t)", - "type": "text" - }, - { - "block_id": "p802-b26", - "global_id": 23602, - "bbox": [ - 386.74, - 270.55, - 396.07, - 278.55 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p802-b27", - "global_id": 23603, - "bbox": [ - 241.05, - 137.46, - 243.25, - 145.37 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p802-b28", - "global_id": 23604, - "bbox": [ - 167.27, - 285.32, - 178.82, - 296.32 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p802-b29", - "global_id": 23605, - "bbox": [ - 377.43, - 217.58, - 388.98, - 228.58 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p802-b30", - "global_id": 23606, - "bbox": [ - 395.42, - 289.21, - 410.98, - 297.32 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p802-b31", - "global_id": 23607, - "bbox": [ - 101.84, - 404.44, - 249.29, - 413.68 - ], - "text": "Figure 8.3 Effect of practical sampling.", - "type": "text" - }, - { - "block_id": "p802-b32", - "global_id": 23608, - "bbox": [ - 101.84, - 435.9, - 490.41, - 493.69 - ], - "text": "The plausibility of this result becomes apparent when we consider the fact that reconstruction\nof x(t) requires the knowledge of the Nyquist sample values. This information is available or built\ninto the sampled signal x(t) in Fig. 8.3e because the nth sampled pulse strength is x(nT). To prove\nthe result analytically, we observe that the sampling pulse train pT(t) depicted in Fig. 8.3c, being\na periodic signal, can be expressed as a trigonometric Fourier series", - "type": "text" - }, - { - "block_id": "p802-b33", - "global_id": 23609, - "bbox": [ - 198.52, - 515.07, - 250.62, - 526.53 - ], - "text": "pT(t) = C0 +", - "type": "text" - }, - { - "block_id": "p802-b34", - "global_id": 23610, - "bbox": [ - 252.17, - 504.9, - 266.27, - 515.57 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p802-b35", - "global_id": 23611, - "bbox": [ - 253.01, - 529.46, - 265.41, - 536.72 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p802-b36", - "global_id": 23612, - "bbox": [ - 267.37, - 508.09, - 391.51, - 526.94 - ], - "text": "Cn cos(nωst + θn)\nωs = 2π", - "type": "text" - }, - { - "block_id": "p802-b37", - "global_id": 23613, - "bbox": [ - 383.38, - 522.46, - 388.92, - 532.42 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p802-b38", - "global_id": 23614, - "bbox": [ - 101.84, - 548.14, - 124.25, - 558.1 - ], - "text": "Thus,", - "type": "text" - }, - { - "block_id": "p802-b39", - "global_id": 23615, - "bbox": [ - 191.54, - 579.27, - 280.04, - 590.35 - ], - "text": "x(t) = x(t)pT(t) = x(t)", - "type": "text" - }, - { - "block_id": "p802-b41", - "global_id": 23616, - "bbox": [ - 287.34, - 579.27, - 307.29, - 590.72 - ], - "text": "C0 +", - "type": "text" - }, - { - "block_id": "p802-b42", - "global_id": 23617, - "bbox": [ - 308.84, - 569.09, - 322.94, - 579.77 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p802-b43", - "global_id": 23618, - "bbox": [ - 309.68, - 593.66, - 322.09, - 600.93 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p802-b44", - "global_id": 23619, - "bbox": [ - 324.05, - 579.27, - 394.51, - 591.14 - ], - "text": "Cn cos(nωst + θn)", - "type": "text" - }, - { - "block_id": "p802-b46", - "global_id": 23620, - "bbox": [ - 208.42, - 613.38, - 253.02, - 624.53 - ], - "text": "= C0x(t) +", - "type": "text" - }, - { - "block_id": "p802-b47", - "global_id": 23621, - "bbox": [ - 254.57, - 603.2, - 268.67, - 613.87 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p802-b48", - "global_id": 23622, - "bbox": [ - 255.42, - 627.77, - 267.83, - 635.03 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p802-b49", - "global_id": 23623, - "bbox": [ - 269.77, - 613.38, - 355.08, - 624.46 - ], - "text": "Cnx(t)cos(nωst + θn)", - "type": "text" - } - ] - }, - { - "page_num": 803, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p803-b0", - "global_id": 23624, - "bbox": [ - 360.21, - 62.89, - 516.13, - 71.98 - ], - "text": "8.1\nThe Sampling Theorem\n783", - "type": "text" - }, - { - "block_id": "p803-b1", - "global_id": 23625, - "bbox": [ - 127.59, - 85.46, - 516.15, - 144.74 - ], - "text": "The sampled signal x(t) consists of C0x(t), C1x(t)cos(ωst + θ1), C2x(t)cos(2ωst + θ2), . . . . Note\nthat the first term C0x(t) is the desired signal and all the other terms are modulated signals with\nspectra centered at ±ωs,±2ωs,±3ωs,. . ., as illustrated in Fig. 8.3f. Clearly the signal x(t) can\nbe recovered by lowpass filtering of x(t), as shown in Fig. 8.3d. As before, it is necessary that\nωs > 4πB (or fs > 2B).", - "type": "text" - }, - { - "block_id": "p803-b2", - "global_id": 23626, - "bbox": [ - 102.51, - 177.69, - 303.29, - 189.65 - ], - "text": "EXAMPLE 8.2\nPractical Sampling", - "type": "text" - }, - { - "block_id": "p803-b3", - "global_id": 23627, - "bbox": [ - 128.9, - 204.49, - 502.73, - 240.18 - ], - "text": "Demonstrate practical sampling by sampling signal x(t) = sinc2(5πt) with the rectangular\npulse sequence pT(t) illustrated in Fig. 8.4c. Sketch the original and sampled signals and their\nspectra, and discuss recovery of x(t) from its samples.", - "type": "text" - }, - { - "block_id": "p803-b4", - "global_id": 23628, - "bbox": [ - 211.41, - 361.95, - 215.41, - 369.95 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p803-b5", - "global_id": 23629, - "bbox": [ - 204.44, - 272.07, - 208.44, - 280.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p803-b6", - "global_id": 23630, - "bbox": [ - 177.11, - 360.77, - 263.06, - 370.7 - ], - "text": "0.2\n–0.2\nt", - "type": "text" - }, - { - "block_id": "p803-b7", - "global_id": 23631, - "bbox": [ - 203.79, - 486.75, - 207.79, - 494.75 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p803-b8", - "global_id": 23632, - "bbox": [ - 386.38, - 308.06, - 396.38, - 316.06 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p803-b9", - "global_id": 23633, - "bbox": [ - 383.95, - 526.68, - 397.95, - 534.68 - ], - "text": "0.05", - "type": "text" - }, - { - "block_id": "p803-b10", - "global_id": 23634, - "bbox": [ - 322.58, - 600.13, - 476.41, - 608.42 - ], - "text": "5\n5\n20\n20", - "type": "text" - }, - { - "block_id": "p803-b11", - "global_id": 23635, - "bbox": [ - 208.97, - 391.66, - 405.7, - 399.66 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p803-b12", - "global_id": 23636, - "bbox": [ - 396.36, - 613.81, - 405.24, - 621.81 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p803-b13", - "global_id": 23637, - "bbox": [ - 204.14, - 419.82, - 208.14, - 427.82 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p803-b14", - "global_id": 23638, - "bbox": [ - 161.96, - 450.01, - 247.36, - 458.31 - ], - "text": "0\n0.2\n0.3\n0.1", - "type": "text" - }, - { - "block_id": "p803-b15", - "global_id": 23639, - "bbox": [ - 208.97, - 463.82, - 217.85, - 471.82 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p803-b16", - "global_id": 23640, - "bbox": [ - 141.16, - 410.24, - 233.21, - 421.59 - ], - "text": "pT(t)\n0.025", - "type": "text" - }, - { - "block_id": "p803-b17", - "global_id": 23641, - "bbox": [ - 493.74, - 575.78, - 499.07, - 583.78 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p803-b18", - "global_id": 23642, - "bbox": [ - 489.06, - 360.94, - 494.39, - 368.94 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p803-b19", - "global_id": 23643, - "bbox": [ - 377.83, - 376.24, - 420.41, - 384.54 - ], - "text": "5\n5", - "type": "text" - }, - { - "block_id": "p803-b20", - "global_id": 23644, - "bbox": [ - 160.33, - 575.64, - 263.28, - 583.93 - ], - "text": "0\n0.2\n0.3\n0.1\nt", - "type": "text" - }, - { - "block_id": "p803-b21", - "global_id": 23645, - "bbox": [ - 406.93, - 308.4, - 422.48, - 316.5 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p803-b22", - "global_id": 23646, - "bbox": [ - 225.15, - 283.9, - 236.25, - 291.98 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p803-b23", - "global_id": 23647, - "bbox": [ - 208.59, - 613.81, - 218.23, - 621.81 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p803-b24", - "global_id": 23648, - "bbox": [ - 319.91, - 575.74, - 479.08, - 584.04 - ], - "text": "40p\n40p\n10p\n10p", - "type": "text" - }, - { - "block_id": "p803-b25", - "global_id": 23649, - "bbox": [ - 373.16, - 360.67, - 425.08, - 368.97 - ], - "text": "10p\n10p", - "type": "text" - }, - { - "block_id": "p803-b26", - "global_id": 23650, - "bbox": [ - 285.34, - 450.23, - 287.56, - 458.23 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p803-b27", - "global_id": 23651, - "bbox": [ - 133.74, - 432.5, - 292.51, - 438.5 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p803-b28", - "global_id": 23652, - "bbox": [ - 482.99, - 600.34, - 503.07, - 608.42 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p803-b29", - "global_id": 23653, - "bbox": [ - 474.31, - 376.46, - 494.39, - 384.54 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p803-b30", - "global_id": 23654, - "bbox": [ - 228.07, - 507.48, - 422.48, - 523.9 - ], - "text": "x–(t)\nX(v)", - "type": "text" - }, - { - "block_id": "p803-b31", - "global_id": 23655, - "bbox": [ - 128.9, - 628.51, - 298.12, - 637.75 - ], - "text": "Figure 8.4 An example of practical sampling.", - "type": "text" - } - ] - }, - { - "page_num": 804, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p804-b0", - "global_id": 23656, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "784\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p804-b1", - "global_id": 23657, - "bbox": [ - 103.17, - 85.89, - 477.01, - 121.25 - ], - "text": "The signal x(t) and its spectrum are shown in Figs. 8.4a and 8.4b, respectively.\nThe period of pT(t) is 0.1 second so that the fundamental frequency (which is the sampling\nfrequency) is 10 Hz. Hence, ωs = 20π. The Fourier series for pT(t) can be expressed as", - "type": "text" - }, - { - "block_id": "p804-b2", - "global_id": 23658, - "bbox": [ - 233.77, - 139.84, - 285.87, - 151.29 - ], - "text": "pT(t) = C0 +", - "type": "text" - }, - { - "block_id": "p804-b3", - "global_id": 23659, - "bbox": [ - 287.41, - 129.66, - 301.51, - 140.34 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p804-b4", - "global_id": 23660, - "bbox": [ - 288.26, - 154.23, - 300.66, - 161.5 - ], - "text": "n=1", - "type": "text" - }, - { - "block_id": "p804-b5", - "global_id": 23661, - "bbox": [ - 302.61, - 139.84, - 346.22, - 151.71 - ], - "text": "Cn cosnωst", - "type": "text" - }, - { - "block_id": "p804-b6", - "global_id": 23662, - "bbox": [ - 103.17, - 170.91, - 131.08, - 180.87 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p804-b7", - "global_id": 23663, - "bbox": [ - 144.47, - 192.41, - 206.29, - 203.49 - ], - "text": "x(t) = x(t)pT(t)", - "type": "text" - }, - { - "block_id": "p804-b8", - "global_id": 23664, - "bbox": [ - 161.35, - 207.35, - 435.71, - 218.5 - ], - "text": "= C1x(t) + C1x(t)cos20πt + C2x(t)cos40πt + C3x(t)cos60πt + · · ·", - "type": "text" - }, - { - "block_id": "p804-b9", - "global_id": 23665, - "bbox": [ - 103.17, - 229.33, - 228.02, - 242.13 - ], - "text": "Use of Eq. (6.14) yields C0 = 1", - "type": "text" - }, - { - "block_id": "p804-b10", - "global_id": 23666, - "bbox": [ - 224.53, - 229.33, - 294.57, - 244.59 - ], - "text": "4 and Cn =\n2\nnπ sin", - "type": "text" - }, - { - "block_id": "p804-b11", - "global_id": 23667, - "bbox": [ - 295.67, - 222.66, - 308.55, - 236.23 - ], - "text": "nπ", - "type": "text" - }, - { - "block_id": "p804-b12", - "global_id": 23668, - "bbox": [ - 303.32, - 222.66, - 314.47, - 243.86 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p804-b13", - "global_id": 23669, - "bbox": [ - 314.47, - 231.09, - 411.13, - 241.05 - ], - "text": ". Consequently, we have", - "type": "text" - }, - { - "block_id": "p804-b14", - "global_id": 23670, - "bbox": [ - 146.84, - 253.64, - 208.66, - 264.72 - ], - "text": "x(t) = x(t)pT(t)", - "type": "text" - }, - { - "block_id": "p804-b15", - "global_id": 23671, - "bbox": [ - 163.73, - 268.64, - 178.23, - 279.96 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p804-b16", - "global_id": 23672, - "bbox": [ - 174.74, - 269.99, - 433.34, - 283.18 - ], - "text": "4x(t) + C1x(t)cos20πt + C2x(t)cos40πt + C3x(t)cos60πt + · · ·", - "type": "text" - }, - { - "block_id": "p804-b17", - "global_id": 23673, - "bbox": [ - 103.17, - 292.32, - 117.54, - 302.28 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p804-b18", - "global_id": 23674, - "bbox": [ - 152.75, - 310.16, - 191.48, - 327.11 - ], - "text": "X(ω) = 1", - "type": "text" - }, - { - "block_id": "p804-b19", - "global_id": 23675, - "bbox": [ - 186.5, - 310.06, - 427.42, - 334.18 - ], - "text": "4X(ω) + C1\n2 [X(ω −20π) + X(ω + 20π)] + C2\n2 [X(ω −40π)", - "type": "text" - }, - { - "block_id": "p804-b20", - "global_id": 23676, - "bbox": [ - 184.95, - 334.75, - 268.26, - 351.79 - ], - "text": "+ X(ω + 40π)] + C3", - "type": "text" - }, - { - "block_id": "p804-b21", - "global_id": 23677, - "bbox": [ - 260.96, - 341.42, - 408.39, - 358.87 - ], - "text": "2 [X(ω −60π) + X(ω + 60π)] + · · ·", - "type": "text" - }, - { - "block_id": "p804-b22", - "global_id": 23678, - "bbox": [ - 103.16, - 366.29, - 477.02, - 388.62 - ], - "text": "where Cn = (2/nπ)sin(nπ/4). The sampled signal and its spectrum are shown in Figs. 8.4d\nand 8.4e, respectively.", - "type": "text" - }, - { - "block_id": "p804-b23", - "global_id": 23679, - "bbox": [ - 103.17, - 390.21, - 477.02, - 448.4 - ], - "text": "The spectrum X(ω) consists of X(ω) repeating periodically at the interval of 20π rad/s\n(10 Hz). Hence, there is no overlap between cycles, and X(ω) can be recovered by using an\nideal lowpass filter of bandwidth 5 Hz. An ideal lowpass filter of unit gain (and bandwidth 5\nHz) will allow the first term on the right-hand side of the foregoing equation to pass fully and\nsuppress all the other terms. Hence, the output y(t) is", - "type": "text" - }, - { - "block_id": "p804-b24", - "global_id": 23680, - "bbox": [ - 266.4, - 458.6, - 297.75, - 470.21 - ], - "text": "y(t) = 1", - "type": "text" - }, - { - "block_id": "p804-b25", - "global_id": 23681, - "bbox": [ - 294.26, - 459.94, - 313.78, - 473.13 - ], - "text": "4x(t)", - "type": "text" - }, - { - "block_id": "p804-b26", - "global_id": 23682, - "bbox": [ - 107.82, - 556.23, - 364.47, - 568.18 - ], - "text": "DRILL 8.2\nThe Role of Sampling Pulse Area", - "type": "text" - }, - { - "block_id": "p804-b27", - "global_id": 23683, - "bbox": [ - 107.82, - 576.89, - 484.43, - 599.22 - ], - "text": "Show that the basic pulse p(t) used in the sampling pulse train in Fig. 8.4c cannot have zero\narea if we wish to reconstruct x(t) by lowpass-filtering the sampled signal.", - "type": "text" - } - ] - }, - { - "page_num": 805, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p805-b0", - "global_id": 23684, - "bbox": [ - 366.66, - 62.89, - 516.14, - 71.98 - ], - "text": "8.2\nSignal Reconstruction\n785", - "type": "text" - }, - { - "block_id": "p805-b1", - "global_id": 23685, - "bbox": [ - 127.94, - 94.37, - 329.56, - 108.31 - ], - "text": "8.2 SIGNAL RECONSTRUCTION", - "type": "text" - }, - { - "block_id": "p805-b2", - "global_id": 23686, - "bbox": [ - 127.59, - 113.89, - 516.13, - 219.91 - ], - "text": "The process of reconstructing a continuous-time signal x(t) from its samples is also known as\ninterpolation. In Sec. 8.1, we saw that a signal x(t) bandlimited to B Hz can be reconstructed\n(interpolated) exactly from its samples if the sampling frequency fs exceeds 2B Hz or the sampling\ninterval T is less than 1/2B. This reconstruction is accomplished by passing the sampled signal\nthrough an ideal lowpass filter of gain T and having a bandwidth of any value between B and fs −B\nHz. From a practical viewpoint, a good choice is the middle value fs/2 = 1/2T Hz or π/T rad/s.\nThis value allows for small deviations in the ideal filter characteristics on either side of the cutoff\nfrequency. With this choice of cutoff frequency and gain T, the ideal lowpass filter required for\nsignal reconstruction (or interpolation) is", - "type": "text" - }, - { - "block_id": "p805-b3", - "global_id": 23687, - "bbox": [ - 244.31, - 235.92, - 300.33, - 246.29 - ], - "text": "H(ω) = T rect", - "type": "text" - }, - { - "block_id": "p805-b4", - "global_id": 23688, - "bbox": [ - 301.44, - 221.93, - 321.49, - 238.9 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p805-b5", - "global_id": 23689, - "bbox": [ - 309.36, - 242.99, - 326.79, - 254.07 - ], - "text": "2πfs", - "type": "text" - }, - { - "block_id": "p805-b7", - "global_id": 23690, - "bbox": [ - 337.26, - 235.92, - 369.42, - 246.29 - ], - "text": "= T rect", - "type": "text" - }, - { - "block_id": "p805-b8", - "global_id": 23691, - "bbox": [ - 370.53, - 221.93, - 390.71, - 239.21 - ], - "text": "ωT", - "type": "text" - }, - { - "block_id": "p805-b9", - "global_id": 23692, - "bbox": [ - 378.99, - 242.99, - 389.94, - 253.37 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p805-b11", - "global_id": 23693, - "bbox": [ - 497.04, - 236.33, - 516.13, - 246.29 - ], - "text": "(8.4)", - "type": "text" - }, - { - "block_id": "p805-b12", - "global_id": 23694, - "bbox": [ - 127.59, - 262.37, - 516.09, - 284.29 - ], - "text": "The interpolation process here is expressed in the frequency domain as a filtering operation. Now\nwe shall examine this process from the time-domain viewpoint.", - "type": "text" - }, - { - "block_id": "p805-b13", - "global_id": 23695, - "bbox": [ - 127.89, - 298.73, - 390.31, - 310.85 - ], - "text": "TIME-DOMAIN VIEW: A SIMPLE INTERPOLATION", - "type": "text" - }, - { - "block_id": "p805-b14", - "global_id": 23696, - "bbox": [ - 127.59, - 314.88, - 516.14, - 444.39 - ], - "text": "Consider the interpolation system shown in Fig. 8.5a. We start with a very simple interpolating\nfilter, whose impulse response is rect(t/T), depicted in Fig. 8.5b. This is a gate pulse centered at\nthe origin, having unit height, and width T (the sampling interval). We shall find the output of this\nfilter when the input is the sampled signal x(t) consisting of an impulse train with the nth impulse\nat t = nT with strength x(nT). Each sample in x(t), being an impulse, produces at the output a gate\npulse of height equal to the strength of the sample. For instance, the nth sample is an impulse of\nstrength x(nT) located at t = nT and can be expressed as x(nT)δ(t−nT). When this impulse passes\nthrough the filter, it produces at the output a gate pulse of height x(nT), centered at t = nT (shaded\nin Fig. 8.5c). Each sample in x(t) will generate a corresponding gate pulse, resulting in the filter\noutput that is a staircase approximation of x(t), shown dotted in Fig. 8.5c. This filter thus gives a\ncrude form of interpolation.", - "type": "text" - }, - { - "block_id": "p805-b15", - "global_id": 23697, - "bbox": [ - 127.59, - 445.98, - 516.15, - 468.31 - ], - "text": "The frequency response of this filter H(ω) is the Fourier transform of the impulse response\nrect(t/T). Thus,", - "type": "text" - }, - { - "block_id": "p805-b16", - "global_id": 23698, - "bbox": [ - 220.14, - 475.38, - 262.3, - 485.76 - ], - "text": "h(t) = rect", - "type": "text" - }, - { - "block_id": "p805-b17", - "global_id": 23699, - "bbox": [ - 262.31, - 461.4, - 274.67, - 478.67 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p805-b18", - "global_id": 23700, - "bbox": [ - 270.23, - 482.77, - 275.76, - 492.73 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p805-b20", - "global_id": 23701, - "bbox": [ - 304.38, - 475.38, - 394.7, - 485.76 - ], - "text": "and\nH(ω) = Tsinc", - "type": "text" - }, - { - "block_id": "p805-b21", - "global_id": 23702, - "bbox": [ - 394.7, - 461.39, - 414.9, - 478.67 - ], - "text": "ωT", - "type": "text" - }, - { - "block_id": "p805-b22", - "global_id": 23703, - "bbox": [ - 406.66, - 482.87, - 411.64, - 492.83 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p805-b24", - "global_id": 23704, - "bbox": [ - 497.04, - 475.79, - 516.13, - 485.76 - ], - "text": "(8.5)", - "type": "text" - }, - { - "block_id": "p805-b25", - "global_id": 23705, - "bbox": [ - 127.59, - 499.08, - 516.14, - 533.37 - ], - "text": "The amplitude response |H(ω)| for this filter, illustrated in Fig. 8.5d, explains the reason for the\ncrudeness of this interpolation. This filter, also known as the zero-order hold (ZOH) filter, is a\npoor form of the ideal lowpass filter (shaded in Fig. 8.5d) required for exact interpolation.†", - "type": "text" - }, - { - "block_id": "p805-b26", - "global_id": 23706, - "bbox": [ - 127.59, - 535.26, - 516.15, - 581.18 - ], - "text": "We can improve on the ZOH filter by using a first-order hold filter, which results in a linear\ninterpolation instead of a staircase interpolation. A linear interpolator, whose impulse response is\na triangle pulse (t/2T), results in an interpolation in which successive sample tops are connected\nby straight-line segments (see Prob. 8.2-3).", - "type": "text" - }, - { - "block_id": "p805-b27", - "global_id": 23707, - "bbox": [ - 127.59, - 599.27, - 516.12, - 633.41 - ], - "text": "† Figure 8.5b shows that the impulse response of this filter is noncausal, and this filter is not realizable. In\npractice, we make it realizable by delaying the impulse response by T/2. This merely delays the output of\nthe filter by T/2.", - "type": "text" - } - ] - }, - { - "page_num": 806, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p806-b0", - "global_id": 23708, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "786\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p806-b1", - "global_id": 23709, - "bbox": [ - 376.45, - 158.19, - 443.33, - 166.19 - ], - "text": "Reconstructed signal", - "type": "text" - }, - { - "block_id": "p806-b2", - "global_id": 23710, - "bbox": [ - 404.34, - 167.11, - 415.44, - 175.19 - ], - "text": "y(t)", - "type": "text" - }, - { - "block_id": "p806-b3", - "global_id": 23711, - "bbox": [ - 375.17, - 139.57, - 386.6, - 147.65 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p806-b4", - "global_id": 23712, - "bbox": [ - 330.41, - 206.44, - 338.86, - 214.44 - ], - "text": "nT", - "type": "text" - }, - { - "block_id": "p806-b5", - "global_id": 23713, - "bbox": [ - 367.2, - 246.94, - 376.08, - 254.94 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p806-b6", - "global_id": 23714, - "bbox": [ - 429.29, - 194.6, - 431.51, - 202.6 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p806-b7", - "global_id": 23715, - "bbox": [ - 197.64, - 147.88, - 303.12, - 163.27 - ], - "text": "Sampled signal\nh(t)", - "type": "text" - }, - { - "block_id": "p806-b8", - "global_id": 23716, - "bbox": [ - 211.81, - 176.52, - 216.26, - 184.52 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p806-b9", - "global_id": 23717, - "bbox": [ - 183.06, - 192.29, - 220.72, - 203.5 - ], - "text": "0\nt", - "type": "text" - }, - { - "block_id": "p806-b10", - "global_id": 23718, - "bbox": [ - 180.16, - 209.16, - 189.97, - 217.16 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p806-b11", - "global_id": 23719, - "bbox": [ - 165.81, - 151.69, - 169.81, - 159.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p806-b12", - "global_id": 23720, - "bbox": [ - 299.77, - 268.48, - 319.76, - 276.78 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p806-b13", - "global_id": 23721, - "bbox": [ - 322.22, - 278.07, - 326.67, - 286.07 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p806-b14", - "global_id": 23722, - "bbox": [ - 268.75, - 347.93, - 340.74, - 366.78 - ], - "text": "0\n2pB\n2pB\n4pB\n( fs)", - "type": "text" - }, - { - "block_id": "p806-b15", - "global_id": 23723, - "bbox": [ - 447.8, - 344.59, - 453.13, - 352.59 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p806-b16", - "global_id": 23724, - "bbox": [ - 193.88, - 281.53, - 252.76, - 289.53 - ], - "text": "Ideal interpolation", - "type": "text" - }, - { - "block_id": "p806-b17", - "global_id": 23725, - "bbox": [ - 215.54, - 290.53, - 231.09, - 298.53 - ], - "text": "filter", - "type": "text" - }, - { - "block_id": "p806-b18", - "global_id": 23726, - "bbox": [ - 291.04, - 99.17, - 302.92, - 107.25 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p806-b19", - "global_id": 23727, - "bbox": [ - 339.05, - 90.33, - 350.48, - 98.41 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p806-b20", - "global_id": 23728, - "bbox": [ - 292.34, - 372.74, - 301.67, - 380.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p806-b21", - "global_id": 23729, - "bbox": [ - 292.91, - 118.83, - 301.79, - 126.83 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p806-b22", - "global_id": 23730, - "bbox": [ - 339.42, - 308.27, - 359.42, - 316.57 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p806-b23", - "global_id": 23731, - "bbox": [ - 249.1, - 87.36, - 260.65, - 98.36 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p806-b24", - "global_id": 23732, - "bbox": [ - 272.61, - 161.34, - 284.16, - 172.34 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p806-b25", - "global_id": 23733, - "bbox": [ - 125.76, - 387.24, - 490.39, - 421.19 - ], - "text": "Figure 8.5 Simple interpolation by means of a zero-order hold (ZOH) circuit. (a) ZOH\ninterpolator. (b) Impulse response of a ZOH circuit. (c) Signal reconstruction by ZOH, as\nviewed in the time domain. (d) Frequency response of a ZOH.", - "type": "text" - }, - { - "block_id": "p806-b26", - "global_id": 23734, - "bbox": [ - 101.84, - 441.57, - 490.41, - 479.64 - ], - "text": "TIME-DOMAIN VIEW: AN IDEAL INTERPOLATION\nThe ideal interpolation filter frequency response obtained in Eq. (8.4) is illustrated in Fig. 8.6a.\nThe impulse response of this filter, the inverse Fourier transform of H(ω) is", - "type": "text" - }, - { - "block_id": "p806-b27", - "global_id": 23735, - "bbox": [ - 261.59, - 496.01, - 304.88, - 506.38 - ], - "text": "h(t) = sinc", - "type": "text" - }, - { - "block_id": "p806-b28", - "global_id": 23736, - "bbox": [ - 304.88, - 482.02, - 322.54, - 499.3 - ], - "text": "πt", - "type": "text" - }, - { - "block_id": "p806-b29", - "global_id": 23737, - "bbox": [ - 314.6, - 503.4, - 320.14, - 513.36 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p806-b31", - "global_id": 23738, - "bbox": [ - 101.84, - 522.41, - 283.94, - 532.79 - ], - "text": "For the Nyquist sampling rate, T = 1/2B, and", - "type": "text" - }, - { - "block_id": "p806-b32", - "global_id": 23739, - "bbox": [ - 258.47, - 543.52, - 333.77, - 553.9 - ], - "text": "h(t) = sinc(2πBt)", - "type": "text" - }, - { - "block_id": "p806-b33", - "global_id": 23740, - "bbox": [ - 101.84, - 576.58, - 490.39, - 634.79 - ], - "text": "This h(t) is depicted in Fig. 8.6b. Observe the interesting fact that h(t) = 0 at all Nyquist\nsampling instants (t = ±n/2B) except at t = 0. When the sampled signal ¯x(t) is applied at the input\nof this filter, the output is x(t). Each sample in x(t), being an impulse, generates a sinc pulse of\nheight equal to the strength of the sample, as illustrated in Fig. 8.6c. The process is identical to that\ndepicted in Fig. 8.5c, except that h(t) is a sinc pulse instead of a gate pulse. Addition of the sinc", - "type": "text" - } - ] - }, - { - "page_num": 807, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p807-b0", - "global_id": 23741, - "bbox": [ - 366.66, - 62.89, - 516.14, - 71.98 - ], - "text": "8.2\nSignal Reconstruction\n787", - "type": "text" - }, - { - "block_id": "p807-b1", - "global_id": 23742, - "bbox": [ - 306.49, - 313.75, - 315.37, - 321.75 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p807-b2", - "global_id": 23743, - "bbox": [ - 321.49, - 215.77, - 388.37, - 223.77 - ], - "text": "Reconstructed signal", - "type": "text" - }, - { - "block_id": "p807-b3", - "global_id": 23744, - "bbox": [ - 349.38, - 224.68, - 360.48, - 232.77 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p807-b4", - "global_id": 23745, - "bbox": [ - 206.72, - 92.47, - 223.16, - 100.57 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p807-b5", - "global_id": 23746, - "bbox": [ - 206.72, - 110.46, - 211.17, - 118.46 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p807-b6", - "global_id": 23747, - "bbox": [ - 165.77, - 162.79, - 455.08, - 175.67 - ], - "text": "2pB\n2pB\nv\nt", - "type": "text" - }, - { - "block_id": "p807-b7", - "global_id": 23748, - "bbox": [ - 428.97, - 275.8, - 431.19, - 283.8 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p807-b8", - "global_id": 23749, - "bbox": [ - 213.28, - 184.41, - 222.16, - 192.41 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p807-b9", - "global_id": 23750, - "bbox": [ - 389.24, - 164.27, - 393.24, - 172.27 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p807-b10", - "global_id": 23751, - "bbox": [ - 389.28, - 87.43, - 401.15, - 95.51 - ], - "text": "h(t)", - "type": "text" - }, - { - "block_id": "p807-b11", - "global_id": 23752, - "bbox": [ - 334.54, - 134.03, - 427.37, - 151.82 - ], - "text": "1\n2B\n1\n 2B", - "type": "text" - }, - { - "block_id": "p807-b12", - "global_id": 23753, - "bbox": [ - 374.44, - 92.86, - 378.44, - 100.86 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p807-b13", - "global_id": 23754, - "bbox": [ - 381.52, - 184.41, - 390.85, - 192.41 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p807-b14", - "global_id": 23755, - "bbox": [ - 171.65, - 216.52, - 220.76, - 224.52 - ], - "text": "Sampled signal", - "type": "text" - }, - { - "block_id": "p807-b15", - "global_id": 23756, - "bbox": [ - 189.65, - 223.16, - 202.76, - 233.52 - ], - "text": "x(t )\n–", - "type": "text" - }, - { - "block_id": "p807-b16", - "global_id": 23757, - "bbox": [ - 151.5, - 328.44, - 359.44, - 337.68 - ], - "text": "Figure 8.6 Ideal interpolation for Nyquist sampling rate.", - "type": "text" - }, - { - "block_id": "p807-b17", - "global_id": 23758, - "bbox": [ - 127.59, - 358.79, - 516.14, - 393.08 - ], - "text": "pulses generated by all the samples results in x(t). The nth sample of the input x(t) is the impulse\nx(nT)δ(t −nT); the filter output of this impulse is x(nT)h(t −nT). Hence, the filter output to x(t),\nwhich is x(t), can now be expressed as a sum", - "type": "text" - }, - { - "block_id": "p807-b18", - "global_id": 23759, - "bbox": [ - 210.48, - 411.22, - 235.13, - 421.5 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p807-b19", - "global_id": 23760, - "bbox": [ - 237.18, - 401.75, - 251.28, - 411.72 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p807-b20", - "global_id": 23761, - "bbox": [ - 242.49, - 425.29, - 245.97, - 432.26 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p807-b21", - "global_id": 23762, - "bbox": [ - 252.39, - 411.22, - 322.9, - 421.5 - ], - "text": "x(nT)h(t −nT) =", - "type": "text" - }, - { - "block_id": "p807-b22", - "global_id": 23763, - "bbox": [ - 324.94, - 401.75, - 339.04, - 411.71 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p807-b23", - "global_id": 23764, - "bbox": [ - 330.25, - 425.29, - 333.73, - 432.26 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p807-b24", - "global_id": 23765, - "bbox": [ - 340.15, - 411.22, - 380.46, - 421.6 - ], - "text": "x(nT)sinc", - "type": "text" - }, - { - "block_id": "p807-b25", - "global_id": 23766, - "bbox": [ - 380.47, - 397.23, - 393.08, - 414.2 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p807-b26", - "global_id": 23767, - "bbox": [ - 387.43, - 411.22, - 427.8, - 428.56 - ], - "text": "T (t −nT)", - "type": "text" - }, - { - "block_id": "p807-b27", - "global_id": 23768, - "bbox": [ - 427.8, - 397.23, - 433.23, - 407.19 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p807-b28", - "global_id": 23769, - "bbox": [ - 127.59, - 442.57, - 436.92, - 452.94 - ], - "text": "For the case of Nyquist sampling rate, T = 1/2B, this expression simplifies to", - "type": "text" - }, - { - "block_id": "p807-b29", - "global_id": 23770, - "bbox": [ - 255.12, - 467.1, - 279.77, - 477.38 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p807-b30", - "global_id": 23771, - "bbox": [ - 281.82, - 457.63, - 295.92, - 467.6 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p807-b31", - "global_id": 23772, - "bbox": [ - 287.13, - 481.17, - 290.61, - 488.14 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p807-b32", - "global_id": 23773, - "bbox": [ - 297.03, - 467.1, - 516.13, - 477.48 - ], - "text": "x(nT)sinc(2πBt −nπ)\n(8.6)", - "type": "text" - }, - { - "block_id": "p807-b33", - "global_id": 23774, - "bbox": [ - 127.59, - 498.51, - 516.14, - 520.84 - ], - "text": "Equation (8.6) is the interpolation formula, which yields values of x(t) between samples as a\nweighted sum of all the sample values.", - "type": "text" - }, - { - "block_id": "p807-b34", - "global_id": 23775, - "bbox": [ - 102.51, - 550.31, - 471.98, - 576.22 - ], - "text": "EXAMPLE 8.3\nBandlimited Interpolation of the Kronecker Delta\nFunction", - "type": "text" - }, - { - "block_id": "p807-b35", - "global_id": 23776, - "bbox": [ - 128.9, - 589.54, - 403.87, - 599.92 - ], - "text": "Find a signal x(t) that is bandlimited to B Hz, and whose samples are", - "type": "text" - }, - { - "block_id": "p807-b36", - "global_id": 23777, - "bbox": [ - 195.17, - 611.46, - 436.5, - 621.83 - ], - "text": "x(0) = 1\nand\nx(±T) = x(±2T) = x(±3T) = · · · = 0", - "type": "text" - } - ] - }, - { - "page_num": 808, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p808-b0", - "global_id": 23778, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "788\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p808-b1", - "global_id": 23779, - "bbox": [ - 103.16, - 85.89, - 420.05, - 96.26 - ], - "text": "where the sampling interval T is the Nyquist interval for x(t), that is, T = 1/2B.", - "type": "text" - }, - { - "block_id": "p808-b2", - "global_id": 23780, - "bbox": [ - 103.16, - 119.18, - 477.04, - 165.01 - ], - "text": "Because we are given the Nyquist sample values, we use the interpolation formula of Eq. (8.6)\nto construct x(t) from its samples. Since all but one of the Nyquist samples are zero, only one\nterm (corresponding to n = 0) in the summation on the right-hand side of Eq. (8.6) survives.\nThus,", - "type": "text" - }, - { - "block_id": "p808-b3", - "global_id": 23781, - "bbox": [ - 253.95, - 166.58, - 326.23, - 176.96 - ], - "text": "x(t) = sinc(2πBt)", - "type": "text" - }, - { - "block_id": "p808-b4", - "global_id": 23782, - "bbox": [ - 103.17, - 185.92, - 477.02, - 219.8 - ], - "text": "This signal is illustrated in Fig. 8.6b. Observe that this is the only signal that has a bandwidth\nB Hz and the sample values x(0) = 1 and x(nT) = 0(n̸ = 0). No other signal satisfies these\nconditions.", - "type": "text" - }, - { - "block_id": "p808-b5", - "global_id": 23783, - "bbox": [ - 101.84, - 283.44, - 381.0, - 295.4 - ], - "text": "8.2-1 Practical Difficulties in Signal Reconstruction", - "type": "text" - }, - { - "block_id": "p808-b6", - "global_id": 23784, - "bbox": [ - 101.84, - 301.11, - 490.41, - 502.78 - ], - "text": "Consider the signal reconstruction procedure illustrated in Fig. 8.7a. If x(t) is sampled at the\nNyquist rate fs = 2B Hz, the spectrum X(ω) consists of repetitions of X(ω) without any gap\nbetween successive cycles, as depicted in Fig. 8.7b. To recover x(t) from x(t), we need to pass\nthe sampled signal x(t) through an ideal lowpass filter, shown dotted in Fig. 8.7b. As seen in\nSec. 7.5, such a filter is unrealizable; it can be closely approximated only with infinite time delay\nin the response. In other words, we can recover the signal x(t) from its samples with infinite\ntime delay. A practical solution to this problem is to sample the signal at a rate higher than the\nNyquist rate ( fs > 2B or ωs > 4πB). The result is X(ω), consisting of repetitions of X(ω) with a\nfinite bandgap between successive cycles, as illustrated in Fig. 8.7c. Now, we can recover X(ω)\nfrom X(ω) using a lowpass filter with a gradual cutoff characteristic, shown dotted in Fig. 8.7c.\nBut even in this case, if the unwanted spectrum is to be suppressed, the filter gain must be zero\nbeyond some frequency (see Fig. 8.7c). According to the Paley–Wiener criterion [Eq. (7.43)], it is\nimpossible to realize even this filter. The only advantage in this case is that the required filter can\nbe closely approximated with a smaller time delay. All this means that it is impossible in practice\nto recover a bandlimited signal x(t) exactly from its samples, even if the sampling rate is higher\nthan the Nyquist rate. However, as the sampling rate increases, the recovered signal approaches\nthe desired signal more closely.", - "type": "text" - }, - { - "block_id": "p808-b7", - "global_id": 23785, - "bbox": [ - 101.84, - 524.99, - 490.41, - 636.28 - ], - "text": "THE TREACHERY OF ALIASING\nThere is another fundamental practical difficulty in reconstructing a signal from its samples. The\nsampling theorem was proved on the assumption that the signal x(t) is bandlimited. All practical\nsignals are timelimited; that is, they are of finite duration or width. We can demonstrate (see\nProb. 8.2-20) that a signal cannot be timelimited and bandlimited simultaneously. If a signal is\ntimelimited, it cannot be bandlimited, and vice versa (but it can be simultaneously nontimelimited\nand nonbandlimited). Clearly, all practical signals, which are necessarily timelimited, are\nnonbandlimited, as shown in Fig. 8.8a; they have infinite bandwidth, and the spectrum X(ω)\nconsists of overlapping cycles of X(ω) repeating every fs Hz (the sampling frequency), as", - "type": "text" - } - ] - }, - { - "page_num": 809, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p809-b0", - "global_id": 23786, - "bbox": [ - 366.66, - 62.89, - 516.14, - 71.98 - ], - "text": "8.2\nSignal Reconstruction\n789", - "type": "text" - }, - { - "block_id": "p809-b1", - "global_id": 23787, - "bbox": [ - 267.92, - 353.74, - 276.8, - 361.74 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p809-b2", - "global_id": 23788, - "bbox": [ - 285.15, - 338.07, - 333.7, - 347.64 - ], - "text": "2pB\nvs", - "type": "text" - }, - { - "block_id": "p809-b3", - "global_id": 23789, - "bbox": [ - 306.83, - 231.13, - 314.5, - 240.69 - ], - "text": "vs", - "type": "text" - }, - { - "block_id": "p809-b4", - "global_id": 23790, - "bbox": [ - 267.46, - 246.78, - 277.27, - 254.78 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p809-b5", - "global_id": 23791, - "bbox": [ - 373.2, - 338.07, - 378.54, - 346.07 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p809-b6", - "global_id": 23792, - "bbox": [ - 373.2, - 231.13, - 378.54, - 239.13 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p809-b7", - "global_id": 23793, - "bbox": [ - 267.92, - 138.9, - 276.8, - 146.9 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p809-b8", - "global_id": 23794, - "bbox": [ - 216.21, - 94.76, - 343.41, - 111.76 - ], - "text": "Ideal lowpass\nfilter cutoff B Hz\nSampler", - "type": "text" - }, - { - "block_id": "p809-b9", - "global_id": 23795, - "bbox": [ - 233.75, - 126.03, - 249.71, - 135.6 - ], - "text": "dT(t)", - "type": "text" - }, - { - "block_id": "p809-b10", - "global_id": 23796, - "bbox": [ - 184.43, - 90.25, - 368.99, - 98.34 - ], - "text": "x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p809-b11", - "global_id": 23797, - "bbox": [ - 275.15, - 161.47, - 290.7, - 169.57 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p809-b12", - "global_id": 23798, - "bbox": [ - 247.79, - 269.65, - 263.34, - 277.75 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p809-b13", - "global_id": 23799, - "bbox": [ - 262.73, - 87.34, - 274.28, - 98.34 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p809-b14", - "global_id": 23800, - "bbox": [ - 151.5, - 368.23, - 516.13, - 390.24 - ], - "text": "Figure 8.7 (a) Signal reconstruction from its samples. (b) Spectrum of a signal sampled at\nthe Nyquist rate. (c) Spectrum of a signal sampled above the Nyquist rate.", - "type": "text" - }, - { - "block_id": "p809-b15", - "global_id": 23801, - "bbox": [ - 127.59, - 420.48, - 516.15, - 505.79 - ], - "text": "illustrated in Fig. 8.8b.† Because of infinite bandwidth in this case, the spectral overlap is\nunavoidable, regardless of the sampling rate. Sampling at a higher rate reduces but does not\neliminate overlapping between repeating spectral cycles. Because of the overlapping tails, X(ω)\nno longer has complete information about X(ω), and it is no longer possible, even theoretically, to\nrecover x(t) exactly from the sampled signal x(t). If the sampled signal is passed through an ideal\nlowpass filter of cutoff frequency fs/2 Hz, the output is not X(ω) but Xa(ω) (Fig. 8.8c), which is a\nversion of X(ω) distorted as a result of two separate causes:", - "type": "text" - }, - { - "block_id": "p809-b16", - "global_id": 23802, - "bbox": [ - 144.52, - 513.35, - 516.12, - 548.34 - ], - "text": "1. The loss of the tail of X(ω) beyond |f| > fs/2 Hz.\n2. The reappearance of this tail inverted or folded onto the spectrum. Note that the spectra\ncross at frequency fs/2 = 1/2T Hz. This frequency is called the folding frequency.", - "type": "text" - }, - { - "block_id": "p809-b17", - "global_id": 23803, - "bbox": [ - 127.59, - 566.4, - 516.12, - 633.42 - ], - "text": "† Figure 8.8b shows that from the infinite number of repeating cycles, only the neighboring spectral cycles\noverlap. This is a somewhat simplified picture. In reality, all the cycles overlap and interact with every other\ncycle because of the infinite width of all practical signal spectra. Fortunately, all practical spectra also must\ndecay at higher frequencies. This results in insignificant amount of interference from cycles other than the\nimmediate neighbors. When such an assumption is not justified, aliasing computations become little more\ninvolved.", - "type": "text" - } - ] - }, - { - "page_num": 810, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p810-b0", - "global_id": 23804, - "bbox": [ - 369.55, - 241.45, - 395.99, - 249.45 - ], - "text": "Lost tail", - "type": "text" - }, - { - "block_id": "p810-b1", - "global_id": 23805, - "bbox": [ - 291.52, - 355.62, - 295.52, - 363.62 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p810-b2", - "global_id": 23806, - "bbox": [ - 291.52, - 122.99, - 295.52, - 130.99 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p810-b3", - "global_id": 23807, - "bbox": [ - 172.76, - 315.54, - 249.86, - 323.54 - ], - "text": "Reconstructed spectrum", - "type": "text" - }, - { - "block_id": "p810-b4", - "global_id": 23808, - "bbox": [ - 202.03, - 324.44, - 220.59, - 334.01 - ], - "text": "Xa(v)", - "type": "text" - }, - { - "block_id": "p810-b5", - "global_id": 23809, - "bbox": [ - 346.27, - 303.72, - 406.28, - 320.72 - ], - "text": "Folded tail distorts\nlower frequencies", - "type": "text" - }, - { - "block_id": "p810-b6", - "global_id": 23810, - "bbox": [ - 369.34, - 325.8, - 441.35, - 342.8 - ], - "text": "Lost tail results in loss\nof higher frequencies", - "type": "text" - }, - { - "block_id": "p810-b7", - "global_id": 23811, - "bbox": [ - 203.11, - 408.11, - 352.45, - 425.91 - ], - "text": "Sampler\nx(t)", - "type": "text" - }, - { - "block_id": "p810-b8", - "global_id": 23812, - "bbox": [ - 293.41, - 456.48, - 302.74, - 464.48 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p810-b9", - "global_id": 23813, - "bbox": [ - 344.31, - 443.85, - 360.16, - 453.41 - ], - "text": "dT(t)", - "type": "text" - }, - { - "block_id": "p810-b10", - "global_id": 23814, - "bbox": [ - 291.52, - 549.63, - 295.52, - 557.63 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p810-b11", - "global_id": 23815, - "bbox": [ - 396.81, - 489.6, - 441.92, - 497.6 - ], - "text": "Sample signal", - "type": "text" - }, - { - "block_id": "p810-b12", - "global_id": 23816, - "bbox": [ - 404.7, - 498.6, - 434.03, - 506.6 - ], - "text": "spectrum", - "type": "text" - }, - { - "block_id": "p810-b13", - "global_id": 23817, - "bbox": [ - 247.59, - 582.4, - 382.05, - 592.17 - ], - "text": "fs\nfs2\nfs2", - "type": "text" - }, - { - "block_id": "p810-b14", - "global_id": 23818, - "bbox": [ - 333.43, - 372.83, - 382.05, - 382.6 - ], - "text": "fs\nfs2", - "type": "text" - }, - { - "block_id": "p810-b15", - "global_id": 23819, - "bbox": [ - 333.43, - 256.71, - 382.05, - 267.07 - ], - "text": "fs\nfs2", - "type": "text" - }, - { - "block_id": "p810-b16", - "global_id": 23820, - "bbox": [ - 247.59, - 372.83, - 266.49, - 382.6 - ], - "text": "fs2", - "type": "text" - }, - { - "block_id": "p810-b17", - "global_id": 23821, - "bbox": [ - 247.59, - 256.71, - 266.49, - 266.47 - ], - "text": "fs2", - "type": "text" - }, - { - "block_id": "p810-b18", - "global_id": 23822, - "bbox": [ - 293.64, - 594.83, - 302.52, - 602.83 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p810-b19", - "global_id": 23823, - "bbox": [ - 294.13, - 385.66, - 303.01, - 393.66 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p810-b20", - "global_id": 23824, - "bbox": [ - 293.25, - 269.83, - 302.9, - 277.83 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p810-b21", - "global_id": 23825, - "bbox": [ - 293.64, - 133.68, - 302.52, - 141.68 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p810-b22", - "global_id": 23826, - "bbox": [ - 450.47, - 121.8, - 455.8, - 129.8 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p810-b23", - "global_id": 23827, - "bbox": [ - 452.26, - 257.1, - 454.48, - 265.1 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p810-b24", - "global_id": 23828, - "bbox": [ - 452.26, - 370.05, - 454.48, - 378.05 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p810-b25", - "global_id": 23829, - "bbox": [ - 450.47, - 223.98, - 455.8, - 231.98 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p810-b26", - "global_id": 23830, - "bbox": [ - 450.47, - 355.61, - 455.8, - 363.61 - ], - "text": "v", - "type": "text" - }, - { - "block_id": "p810-b27", - "global_id": 23831, - "bbox": [ - 452.26, - 582.62, - 454.48, - 590.62 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p810-b28", - "global_id": 23832, - "bbox": [ - 396.81, - 163.3, - 441.92, - 171.3 - ], - "text": "Sample signal", - "type": "text" - }, - { - "block_id": "p810-b29", - "global_id": 23833, - "bbox": [ - 404.7, - 172.3, - 434.03, - 180.3 - ], - "text": "spectrum", - "type": "text" - }, - { - "block_id": "p810-b30", - "global_id": 23834, - "bbox": [ - 331.88, - 549.34, - 455.8, - 559.1 - ], - "text": "v\nvs\nvs2", - "type": "text" - }, - { - "block_id": "p810-b31", - "global_id": 23835, - "bbox": [ - 246.04, - 355.95, - 383.9, - 365.71 - ], - "text": "vs\nvs2\nvs2", - "type": "text" - }, - { - "block_id": "p810-b32", - "global_id": 23836, - "bbox": [ - 207.82, - 226.06, - 383.9, - 236.16 - ], - "text": "0\nvs\nvs2\nvs2\nvs", - "type": "text" - }, - { - "block_id": "p810-b33", - "global_id": 23837, - "bbox": [ - 207.97, - 549.34, - 268.05, - 559.1 - ], - "text": "vs2\nvs", - "type": "text" - }, - { - "block_id": "p810-b34", - "global_id": 23838, - "bbox": [ - 306.25, - 479.36, - 327.8, - 488.93 - ], - "text": "Xaa(v)", - "type": "text" - }, - { - "block_id": "p810-b35", - "global_id": 23839, - "bbox": [ - 313.71, - 64.18, - 329.26, - 72.28 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p810-b36", - "global_id": 23840, - "bbox": [ - 241.87, - 416.8, - 264.31, - 426.37 - ], - "text": "Haa(v)", - "type": "text" - }, - { - "block_id": "p810-b37", - "global_id": 23841, - "bbox": [ - 287.57, - 408.36, - 304.67, - 417.9 - ], - "text": "xaa(t)", - "type": "text" - }, - { - "block_id": "p810-b38", - "global_id": 23842, - "bbox": [ - 235.47, - 290.39, - 251.91, - 298.49 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p810-b39", - "global_id": 23843, - "bbox": [ - 235.47, - 161.73, - 251.91, - 169.83 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p810-b40", - "global_id": 23844, - "bbox": [ - 235.47, - 482.38, - 251.91, - 490.48 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p810-b41", - "global_id": 23845, - "bbox": [ - 377.1, - 405.19, - 381.1, - 416.11 - ], - "text": "x–", - "type": "text" - }, - { - "block_id": "p810-b42", - "global_id": 23846, - "bbox": [ - 380.46, - 408.11, - 394.01, - 417.66 - ], - "text": "aa(t)", - "type": "text" - }, - { - "block_id": "p810-b43", - "global_id": 23847, - "bbox": [ - 368.09, - 561.22, - 440.1, - 578.22 - ], - "text": "Lost tail results in loss\nof higher frequencies", - "type": "text" - }, - { - "block_id": "p810-b44", - "global_id": 23848, - "bbox": [ - 135.79, - 506.02, - 249.54, - 523.02 - ], - "text": "Reconstructed spectrum\n(no distortion of lower frequencies)", - "type": "text" - }, - { - "block_id": "p810-b45", - "global_id": 23849, - "bbox": [ - 289.4, - 236.03, - 326.94, - 253.03 - ], - "text": "Lost tail is\nfolded back", - "type": "text" - }, - { - "block_id": "p810-b46", - "global_id": 23850, - "bbox": [ - 306.25, - 156.76, - 321.8, - 164.86 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p810-b47", - "global_id": 23851, - "bbox": [ - 125.76, - 608.99, - 490.4, - 655.24 - ], - "text": "Figure 8.8 Aliasing effect. (a) Spectrum of a practical signal x(t). (b) Spectrum of sampled\nx(t). (c) Reconstructed signal spectrum. (d) Sampling scheme using anti-aliasing filter.\n(e) Sampled signal spectrum (dotted) and the reconstructed signal spectrum (solid) when\nanti-aliasing filter is used.", - "type": "text" - } - ] - }, - { - "page_num": 811, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p811-b0", - "global_id": 23852, - "bbox": [ - 366.66, - 62.89, - 516.14, - 71.98 - ], - "text": "8.2\nSignal Reconstruction\n791", - "type": "text" - }, - { - "block_id": "p811-b1", - "global_id": 23853, - "bbox": [ - 158.37, - 85.82, - 516.14, - 204.07 - ], - "text": "The spectrum may be viewed as if the lost tail is folding back onto itself at the\nfolding frequency. For instance, a component of frequency (fs/2) + fz shows up as or\n“impersonates” a component of lower frequency (fs/2) −fz in the reconstructed signal.\nThus, the components of frequencies above fs/2 reappear as components of frequencies\nbelow fs/2. This tail inversion, known as spectral folding or aliasing, is shown shaded\nin Fig. 8.8b and also in Fig. 8.8c. In the process of aliasing, not only are we losing all the\ncomponents of frequencies above the folding frequency fs/2 Hz, but these very components\nreappear (aliased) as lower-frequency components, as shown in Figs. 8.8b and 8.8c. Such\naliasing destroys the integrity of the frequency components below the folding frequency\nfs/2, as depicted in Fig. 8.8c.", - "type": "text" - }, - { - "block_id": "p811-b2", - "global_id": 23854, - "bbox": [ - 127.59, - 211.35, - 516.13, - 281.09 - ], - "text": "The aliasing problem is analogous to that of an army with a platoon that has secretly defected\nto the enemy side. The platoon is, however, ostensibly loyal to the army. The army is in double\njeopardy. First, the army has lost this platoon as a fighting force. In addition, during actual fighting,\nthe army will have to contend with sabotage by the defectors and will have to find another\nloyal platoon to neutralize the defectors. Thus, the army has lost two platoons in nonproductive\nactivity.", - "type": "text" - }, - { - "block_id": "p811-b3", - "global_id": 23855, - "bbox": [ - 127.59, - 321.75, - 516.12, - 395.69 - ], - "text": "DEFECTORS ELIMINATED: THE ANTI-ALIASING FILTER\nIf you were the commander of the betrayed army, the solution to the problem would be obvious. As\nsoon as the commander got wind of the defection, he would incapacitate, by whatever means, the\ndefecting platoon before the fighting begins. This way he loses only one (the defecting) platoon.\nThis is a partial solution to the double jeopardy of betrayal and sabotage, a solution that partly\nrectifies the problem and cuts the losses to half.", - "type": "text" - }, - { - "block_id": "p811-b4", - "global_id": 23856, - "bbox": [ - 127.59, - 397.68, - 516.16, - 563.06 - ], - "text": "We follow exactly the same procedure. The potential defectors are all the frequency\ncomponents beyond the folding frequency fs/2 = 1/2T Hz. We should eliminate (suppress)\nthese components from x(t) before sampling x(t). Such suppression of higher frequencies can be\naccomplished by an ideal lowpass filter of cutoff fs/2 Hz, as shown in Fig. 8.8d. This is called the\nanti-aliasing filter. Figure 8.8d also shows that anti-aliasing filtering is performed before sampling.\nFigure 8.8e shows the sampled signal spectrum (dotted) and the reconstructed signal Xaa(ω) when\nan anti-aliasing scheme is used. An anti-aliasing filter essentially bandlimits the signal x(t) to\nfs/2 Hz. This way, we lose only the components beyond the folding frequency fs/2 Hz. These\nsuppressed components now cannot reappear to corrupt the components of frequencies below the\nfolding frequency. Clearly, use of an anti-aliasing filter results in the reconstructed signal spectrum\nXaa(ω) = X(ω) for |f| < fs/2. Thus, although we lost the spectrum beyond fs/2 Hz, the spectrum\nfor all the frequencies below fs/2 remains intact. The effective aliasing distortion is cut in half\nowing to elimination of folding. We stress again that the anti-aliasing operation must be performed\nbefore the signal is sampled.", - "type": "text" - }, - { - "block_id": "p811-b5", - "global_id": 23857, - "bbox": [ - 127.59, - 565.05, - 516.15, - 611.58 - ], - "text": "An anti-aliasing filter also helps to reduce noise. Noise, generally, has a wideband spectrum,\nand without anti-aliasing, the aliasing phenomenon itself will cause the noise lying outside the\ndesired band to appear in the signal band. Anti-aliasing suppresses the entire noise spectrum\nbeyond frequency fs/2.", - "type": "text" - }, - { - "block_id": "p811-b6", - "global_id": 23858, - "bbox": [ - 127.59, - 612.86, - 516.14, - 635.49 - ], - "text": "The anti-aliasing filter, being an ideal filter, is unrealizable. In practice, we use a steep cutoff\nfilter, which leaves a sharply attenuated spectrum beyond the folding frequency fs/2.", - "type": "text" - } - ] - }, - { - "page_num": 812, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p812-b0", - "global_id": 23859, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "792\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p812-b1", - "global_id": 23860, - "bbox": [ - 102.14, - 86.19, - 359.81, - 112.27 - ], - "text": "SAMPLING FORCES NONBANDLIMITED SIGNALS\nTO APPEAR BANDLIMITED", - "type": "text" - }, - { - "block_id": "p812-b2", - "global_id": 23861, - "bbox": [ - 101.84, - 115.88, - 490.4, - 221.9 - ], - "text": "Figure 8.8b shows that the spectrum of a signal x(t) consists of overlapping cycles of X(ω). This\nmeans that x(t) are sub-Nyquist samples of x(t). However, we may also view the spectrum in\nFig. 8.8b as the spectrum Xa(ω) (Fig. 8.8c), repeating periodically every fs Hz without overlap.\nThe spectrum Xa(ω) is bandlimited to fs/2 Hz. Hence, these (sub-Nyquist) samples of x(t) are\nactually the Nyquist samples for signal xa(t). In conclusion, sampling a nonbandlimited signal\nx(t) at a rate fs Hz makes the samples appear to be the Nyquist samples of some signal xa(t),\nbandlimited to fs/2 Hz. In other words, sampling makes a nonbandlimited signal appear to be a\nbandlimited signal xa(t) with bandwidth fs/2 Hz. A similar conclusion applies if x(t) is bandlimited\nbut sampled at a sub-Nyquist rate.", - "type": "text" - }, - { - "block_id": "p812-b3", - "global_id": 23862, - "bbox": [ - 102.14, - 237.46, - 329.51, - 249.58 - ], - "text": "VERIFICATION OF ALIASING IN SINUSOIDS", - "type": "text" - }, - { - "block_id": "p812-b4", - "global_id": 23863, - "bbox": [ - 101.84, - 253.61, - 490.38, - 299.44 - ], - "text": "We showed in Fig. 8.8b how sampling a signal below the Nyquist rate causes aliasing, which makes\na signal of higher frequency (fs/2) + fz Hz masquerade as a signal of lower frequency (fs/2) −fz\nHz. Figure 8.8b demonstrates this result in the frequency domain. Let us now verify it in the time\ndomain to gain a deeper appreciation of aliasing.", - "type": "text" - }, - { - "block_id": "p812-b5", - "global_id": 23864, - "bbox": [ - 101.84, - 301.01, - 489.87, - 324.84 - ], - "text": "We can prove our proposition by showing that samples of sinusoids of frequencies (ωs/2)+ωz\nand (ωs/2) −ωz are identical when the sampling frequency is fs = ωs/2π Hz.", - "type": "text" - }, - { - "block_id": "p812-b6", - "global_id": 23865, - "bbox": [ - 101.84, - 324.93, - 490.4, - 347.26 - ], - "text": "For a sinusoid x(t) = cos ωt, sampled at intervals of T seconds, x(nT), its nth sample (at\nt = nT) is", - "type": "text" - }, - { - "block_id": "p812-b7", - "global_id": 23866, - "bbox": [ - 234.94, - 351.32, - 357.28, - 361.69 - ], - "text": "x(nT) = cosωnT\nn integer", - "type": "text" - }, - { - "block_id": "p812-b8", - "global_id": 23867, - "bbox": [ - 101.84, - 371.72, - 353.86, - 383.77 - ], - "text": "Hence, samples of sinusoids of frequency ω = (ωs/2) ± ωz are†", - "type": "text" - }, - { - "block_id": "p812-b9", - "global_id": 23868, - "bbox": [ - 145.18, - 401.1, - 193.5, - 411.48 - ], - "text": "x(nT) = cos", - "type": "text" - }, - { - "block_id": "p812-b10", - "global_id": 23869, - "bbox": [ - 194.61, - 387.12, - 211.76, - 405.21 - ], - "text": "ωs", - "type": "text" - }, - { - "block_id": "p812-b11", - "global_id": 23870, - "bbox": [ - 204.91, - 401.1, - 233.56, - 418.55 - ], - "text": "2 ± ωz", - "type": "text" - }, - { - "block_id": "p812-b13", - "global_id": 23871, - "bbox": [ - 240.79, - 401.1, - 277.22, - 411.48 - ], - "text": "nT = cos", - "type": "text" - }, - { - "block_id": "p812-b14", - "global_id": 23872, - "bbox": [ - 278.33, - 387.12, - 295.49, - 405.21 - ], - "text": "ωs", - "type": "text" - }, - { - "block_id": "p812-b15", - "global_id": 23873, - "bbox": [ - 288.63, - 408.59, - 293.61, - 418.55 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p812-b17", - "global_id": 23874, - "bbox": [ - 303.91, - 401.1, - 374.22, - 412.18 - ], - "text": "nT cosωznT ∓sin", - "type": "text" - }, - { - "block_id": "p812-b18", - "global_id": 23875, - "bbox": [ - 375.32, - 387.12, - 392.48, - 405.21 - ], - "text": "ωs", - "type": "text" - }, - { - "block_id": "p812-b19", - "global_id": 23876, - "bbox": [ - 385.62, - 408.59, - 390.6, - 418.55 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p812-b21", - "global_id": 23877, - "bbox": [ - 400.9, - 401.1, - 446.28, - 412.18 - ], - "text": "nT sinωznT", - "type": "text" - }, - { - "block_id": "p812-b22", - "global_id": 23878, - "bbox": [ - 101.85, - 430.03, - 482.24, - 441.11 - ], - "text": "Recognizing that ωsT = 2πfsT = 2π, and sin(ωs/2)nT = sinπn = 0 for all integer n, we obtain", - "type": "text" - }, - { - "block_id": "p812-b23", - "global_id": 23879, - "bbox": [ - 234.71, - 459.24, - 283.02, - 469.62 - ], - "text": "x(nT) = cos", - "type": "text" - }, - { - "block_id": "p812-b24", - "global_id": 23880, - "bbox": [ - 284.14, - 445.26, - 301.3, - 463.34 - ], - "text": "ωs", - "type": "text" - }, - { - "block_id": "p812-b25", - "global_id": 23881, - "bbox": [ - 294.43, - 466.73, - 299.42, - 476.69 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p812-b27", - "global_id": 23882, - "bbox": [ - 309.72, - 459.24, - 356.75, - 470.32 - ], - "text": "nT cosωznT", - "type": "text" - }, - { - "block_id": "p812-b28", - "global_id": 23883, - "bbox": [ - 101.84, - 488.16, - 490.41, - 546.35 - ], - "text": "Clearly, the samples of a sinusoid of frequency (fs/2)+fz are identical to the samples of a sinusoid\n(fs/2)−fz.‡ For instance, when a sinusoid of frequency 100 Hz is sampled at a rate of 120 Hz, the\napparent frequency of the sinusoid that results from reconstruction of the samples is 20 Hz. This\nfollows from the fact that here, 100 = (fs/2)+fz = 60+fz so that fz = 40. Hence, (fs/2)−fz = 20.\nSuch would precisely be the conclusion arrived at from Fig. 8.8b.", - "type": "text" - }, - { - "block_id": "p812-b29", - "global_id": 23884, - "bbox": [ - 101.84, - 566.1, - 490.38, - 634.38 - ], - "text": "† Here we have ignored the phase aspect of the sinusoid. Sampled versions of a sinusoid x(t) = cos(ωt + θ)\nwith two different frequencies (ωs/2) ± ωz have identical frequency, but the phase signs may be reversed\ndepending on the value of ωz.\n‡ The reader is encouraged to verify this result graphically by plotting the spectrum of a sinusoid of frequency\n(ωs/2) + ωz (impulses at ±[(ωs/2) + ωz]) and its periodic repetition at intervals ωs. Although the result is\nvalid for all values of ωz, consider the case of ωz < ωs/2 to simplify the graphics.", - "type": "text" - } - ] - }, - { - "page_num": 813, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p813-b0", - "global_id": 23885, - "bbox": [ - 366.66, - 62.89, - 516.14, - 71.98 - ], - "text": "8.2\nSignal Reconstruction\n793", - "type": "text" - }, - { - "block_id": "p813-b1", - "global_id": 23886, - "bbox": [ - 127.59, - 85.72, - 516.12, - 108.82 - ], - "text": "This discussion again shows that sampling a sinusoid of frequency f aliasing can be avoided\nif the sampling rate fs > 2f Hz.", - "type": "text" - }, - { - "block_id": "p813-b2", - "global_id": 23887, - "bbox": [ - 254.99, - 118.23, - 386.54, - 142.67 - ], - "text": "0 ≤f < fs\n2\nor\n0 ≤ω < π\nT", - "type": "text" - }, - { - "block_id": "p813-b3", - "global_id": 23888, - "bbox": [ - 127.59, - 151.56, - 516.15, - 185.44 - ], - "text": "Violating this condition leads to aliasing, implying that the samples appear to be those of a\nlower-frequency signal. Because of this loss of identity, it is impossible to reconstruct the signal\nfaithfully from its samples.", - "type": "text" - }, - { - "block_id": "p813-b4", - "global_id": 23889, - "bbox": [ - 127.59, - 200.73, - 516.13, - 251.83 - ], - "text": "GENERAL CONDITION FOR ALIASING IN SINUSOIDS\nWe can generalize the foregoing result by showing that samples of a sinusoid of frequency f0 are\nidentical to those of a sinusoid of frequency f0 + mfs Hz (integer m), where fs is the sampling\nfrequency. The samples of cos2π(f0 + mfs)t are", - "type": "text" - }, - { - "block_id": "p813-b5", - "global_id": 23890, - "bbox": [ - 204.94, - 263.4, - 438.01, - 274.86 - ], - "text": "cos 2π(f0 + mfs)nT = cos(2πf0nT + 2πmn) = cos 2πf0nT", - "type": "text" - }, - { - "block_id": "p813-b6", - "global_id": 23891, - "bbox": [ - 127.59, - 286.43, - 516.14, - 369.24 - ], - "text": "The result follows because mn is an integer and fsT = 1. This result shows that sinusoids of\nfrequencies that differ by an integer multiple of fs result in identical set of samples. In other words,\nsamples of sinusoids separated by frequency fs Hz are identical. This implies that samples of\nsinusoids in any frequency band of fs Hz are unique; that is, no two sinusoids in that band have\nthe same samples (when sampled at a rate fs Hz). For instance, frequencies in the band from −fs/2\nto fs/2 have unique samples (at the sampling rate fs). This band is called the fundamental band.\nRecall also that fs/2 is the folding frequency.", - "type": "text" - }, - { - "block_id": "p813-b7", - "global_id": 23892, - "bbox": [ - 127.59, - 370.44, - 516.15, - 405.9 - ], - "text": "From the discussion thus far, we conclude that if a continuous-time sinusoid of frequency f\nHz is sampled at a rate of fs Hz (samples/s), the resulting samples would appear as samples of a\ncontinuous-time sinusoid of frequency fa in the fundamental band, where", - "type": "text" - }, - { - "block_id": "p813-b8", - "global_id": 23893, - "bbox": [ - 225.02, - 415.21, - 309.39, - 433.34 - ], - "text": "fa = f −mfs\n−fs", - "type": "text" - }, - { - "block_id": "p813-b9", - "global_id": 23894, - "bbox": [ - 304.41, - 415.21, - 516.12, - 439.34 - ], - "text": "2 ≤fa < fs\n2\nm an integer\n(8.7)", - "type": "text" - }, - { - "block_id": "p813-b10", - "global_id": 23895, - "bbox": [ - 127.59, - 447.82, - 516.11, - 494.06 - ], - "text": "The frequency fa lies in the fundamental band from −fs/2 to fs/2. Figure 8.9a shows the plot of fa\nversus f, where f is the actual frequency and fa is the corresponding fundamental band frequency,\nwhose samples are identical to those of the sinusoid of frequency f, when the sampling rate is fs\nHz.", - "type": "text" - }, - { - "block_id": "p813-b11", - "global_id": 23896, - "bbox": [ - 127.59, - 496.05, - 516.11, - 517.97 - ], - "text": "Recall, however, that the sign change of a frequency does not alter the actual frequency of the\nwaveform. This is because", - "type": "text" - }, - { - "block_id": "p813-b12", - "global_id": 23897, - "bbox": [ - 261.06, - 521.23, - 382.65, - 532.31 - ], - "text": "cos(−ωat + θ) = cos(ωat −θ)", - "type": "text" - }, - { - "block_id": "p813-b13", - "global_id": 23898, - "bbox": [ - 127.59, - 541.27, - 516.17, - 611.42 - ], - "text": "Clearly the apparent frequency of a sinusoid of frequency −fa is also fa. However, its phase\nundergoes a sign change. This means the apparent frequency of any sampled sinusoid lies in\nthe range from 0 to fs/2 Hz. To summarize, if a continuous-time sinusoid of frequency f Hz is\nsampled at a rate of fs Hz (samples/second), the resulting samples would appear as samples of\na continuous-time sinusoid of frequency |fa| that lies in the band from 0 to fs/2. According to\nEq. (8.7),", - "type": "text" - }, - { - "block_id": "p813-b14", - "global_id": 23899, - "bbox": [ - 230.36, - 612.82, - 342.64, - 630.57 - ], - "text": "|fa| = |f −mfs|\n|fa| ≤fs", - "type": "text" - }, - { - "block_id": "p813-b15", - "global_id": 23900, - "bbox": [ - 337.66, - 619.8, - 413.34, - 636.94 - ], - "text": "2\nm an integer", - "type": "text" - } - ] - }, - { - "page_num": 814, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p814-b0", - "global_id": 23901, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "794\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p814-b1", - "global_id": 23902, - "bbox": [ - 152.84, - 193.94, - 157.4, - 212.72 - ], - "text": "2\nfs", - "type": "text" - }, - { - "block_id": "p814-b2", - "global_id": 23903, - "bbox": [ - 144.05, - 143.91, - 157.4, - 162.7 - ], - "text": "2\n fs", - "type": "text" - }, - { - "block_id": "p814-b3", - "global_id": 23904, - "bbox": [ - 430.1, - 235.86, - 432.33, - 243.86 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p814-b4", - "global_id": 23905, - "bbox": [ - 430.1, - 129.57, - 432.33, - 137.57 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p814-b5", - "global_id": 23906, - "bbox": [ - 152.84, - 87.39, - 157.4, - 106.18 - ], - "text": "2\nfs", - "type": "text" - }, - { - "block_id": "p814-b6", - "global_id": 23907, - "bbox": [ - 363.14, - 128.77, - 379.7, - 138.53 - ], - "text": "5fs2", - "type": "text" - }, - { - "block_id": "p814-b7", - "global_id": 23908, - "bbox": [ - 281.22, - 263.74, - 291.02, - 271.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p814-b8", - "global_id": 23909, - "bbox": [ - 131.05, - 209.47, - 141.26, - 219.23 - ], - "text": "fa", - "type": "text" - }, - { - "block_id": "p814-b9", - "global_id": 23910, - "bbox": [ - 281.68, - 168.52, - 290.56, - 176.52 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p814-b10", - "global_id": 23911, - "bbox": [ - 198.17, - 235.05, - 415.38, - 245.86 - ], - "text": "3fs\n5fs2\n3fs2\nfs2\nfs\n2fs", - "type": "text" - }, - { - "block_id": "p814-b11", - "global_id": 23912, - "bbox": [ - 198.17, - 128.77, - 415.38, - 139.58 - ], - "text": "3fs\n3fs2\nfs2\nfs\n2fs", - "type": "text" - }, - { - "block_id": "p814-b12", - "global_id": 23913, - "bbox": [ - 133.54, - 103.14, - 138.77, - 112.69 - ], - "text": "fa", - "type": "text" - }, - { - "block_id": "p814-b13", - "global_id": 23914, - "bbox": [ - 125.76, - 278.07, - 459.47, - 288.75 - ], - "text": "Figure 8.9 Apparent frequencies of a sampled sinusoid: (a) fa versus f and (b) |fa| versus f.", - "type": "text" - }, - { - "block_id": "p814-b14", - "global_id": 23915, - "bbox": [ - 101.84, - 307.94, - 490.39, - 357.39 - ], - "text": "The plot of the apparent frequency |fa| versus f is shown in Fig. 8.9b.† As expected, the apparent\nfrequency |fa| of any sampled sinusoid, regardless of its frequency, is always in the range of 0 to\nfs/2 Hz. However, when fa is negative, the phase of the apparent sinusoid undergoes a sign change.\nThe frequency belts in which such phase changes occur are shown shaded in Fig. 8.9b.", - "type": "text" - }, - { - "block_id": "p814-b15", - "global_id": 23916, - "bbox": [ - 101.85, - 358.97, - 490.39, - 406.7 - ], - "text": "Consider, for example, a sinusoid cos(2πft +θ) with f =8000 Hz sampled at a rate fs = 3000\nHz. Using Eq. (8.7), we obtain fa = 8000 −3 × 3000 = −1000. Hence, |fa| = 1000. The samples\nwould appear to have come from a sinusoid cos(2000πt −θ). Observe the sign change of the\nphase because fa is negative.‡", - "type": "text" - }, - { - "block_id": "p814-b16", - "global_id": 23917, - "bbox": [ - 101.84, - 406.78, - 490.39, - 430.61 - ], - "text": "In the light of the foregoing development, let us consider a sinusoid of frequency f =\n(fs/2) + fz, sampled at a rate of fs Hz. According to Eq. (8.7),", - "type": "text" - }, - { - "block_id": "p814-b17", - "global_id": 23918, - "bbox": [ - 231.68, - 441.35, - 256.98, - 459.48 - ], - "text": "fa = fs", - "type": "text" - }, - { - "block_id": "p814-b18", - "global_id": 23919, - "bbox": [ - 252.0, - 441.35, - 360.06, - 465.47 - ], - "text": "2 + fz −(1 × fs) = −fs\n2 + fz", - "type": "text" - }, - { - "block_id": "p814-b19", - "global_id": 23920, - "bbox": [ - 101.84, - 475.43, - 490.4, - 499.27 - ], - "text": "Hence, the apparent frequency is |fa| = (fs/2) −fz, confirming our earlier result. However, the\nphase of the sinusoid will suffer a sign change because fa is negative.", - "type": "text" - }, - { - "block_id": "p814-b20", - "global_id": 23921, - "bbox": [ - 101.85, - 499.76, - 490.4, - 546.67 - ], - "text": "Figure 8.10 shows how samples of sinusoids of two different frequencies (sampled at the same\nrate) generate identical sets of samples. Both the sinusoids are sampled at a rate fs = 5 Hz (T = 0.2\nsecond). The frequencies of the two sinusoids, 1 Hz (period 1) and 6 Hz (period 1/6), differ by\nfs = 5 Hz.", - "type": "text" - }, - { - "block_id": "p814-b21", - "global_id": 23922, - "bbox": [ - 101.84, - 566.1, - 490.38, - 633.41 - ], - "text": "† The plots in Figs. 8.9 and 5.17 are identical. This is because a sampled sinusoid is basically a discrete-time\nsinusoid.\n‡ For phase sign change, we are assuming that the signal has the form cos(2πft + θ). If the form is\nsin(2πft + θ), the rule changes slightly. It is left as an exercise for the reader to show that when fa < 0,\nthis sinusoid appears as −sin(2π|fa|t −θ). Thus, in addition to phase change, the amplitude also changes\nsign.", - "type": "text" - } - ] - }, - { - "page_num": 815, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p815-b0", - "global_id": 23923, - "bbox": [ - 366.66, - 62.89, - 516.14, - 71.98 - ], - "text": "8.2\nSignal Reconstruction\n795", - "type": "text" - }, - { - "block_id": "p815-b1", - "global_id": 23924, - "bbox": [ - 131.06, - 141.08, - 495.93, - 160.44 - ], - "text": "0\n0.2\n0.4\n0.6\n0.8\n1\nt", - "type": "text" - }, - { - "block_id": "p815-b2", - "global_id": 23925, - "bbox": [ - 256.42, - 188.45, - 286.96, - 198.21 - ], - "text": "fs 5 Hz", - "type": "text" - }, - { - "block_id": "p815-b3", - "global_id": 23926, - "bbox": [ - 229.83, - 87.81, - 257.72, - 104.91 - ], - "text": "cos 2pt\nf 1 Hz", - "type": "text" - }, - { - "block_id": "p815-b4", - "global_id": 23927, - "bbox": [ - 331.33, - 87.81, - 359.71, - 104.91 - ], - "text": "cos 12pt\nf 6 Hz", - "type": "text" - }, - { - "block_id": "p815-b5", - "global_id": 23928, - "bbox": [ - 127.59, - 210.44, - 273.02, - 219.68 - ], - "text": "Figure 8.10 Demonstration of aliasing.", - "type": "text" - }, - { - "block_id": "p815-b6", - "global_id": 23929, - "bbox": [ - 127.59, - 261.32, - 516.14, - 366.93 - ], - "text": "The reason for aliasing can be clearly seen in Fig. 8.10. The root of the problem is the\nsampling rate, which may be adequate for the lower-frequency sinusoid but is clearly inadequate\nfor the higher-frequency sinusoid. The figure clearly shows that between the successive samples\nof the higher-frequency sinusoid, there are wiggles, which are bypassed or ignored, and are\nunrepresented in the samples, indicating a sub-Nyquist rate of sampling. The frequency of the\napparent signal xa(t) is always the lowest possible frequency that lies within the band |f| ≤fs/2.\nThus, the apparent frequency of the samples in this example is 1 Hz. If these samples are chosen\nto reconstruct a signal using a lowpass filter of bandwidth fs/2, we shall obtain a sinusoid of\nfrequency 1 Hz.", - "type": "text" - }, - { - "block_id": "p815-b7", - "global_id": 23930, - "bbox": [ - 102.51, - 406.5, - 435.82, - 418.45 - ], - "text": "EXAMPLE 8.4\nApparent Frequency of Sampled Sinusoids", - "type": "text" - }, - { - "block_id": "p815-b8", - "global_id": 23931, - "bbox": [ - 128.9, - 434.7, - 502.76, - 468.99 - ], - "text": "A continuous-time sinusoid cos(2πft + θ) is sampled at a rate fs = 1000 Hz. Determine the\napparent (aliased) sinusoid of the resulting samples if the input signal frequency f is (a) 400\nHz, (b) 600 Hz, (c) 1000 Hz, and (d) 2400 Hz.", - "type": "text" - }, - { - "block_id": "p815-b9", - "global_id": 23932, - "bbox": [ - 128.91, - 470.56, - 502.77, - 504.85 - ], - "text": "The folding frequency is fs/2 = 500. Hence, sinusoids below 500 Hz (frequency within\nthe fundamental band) will not be aliased and sinusoids of frequency above 500 Hz will be\naliased.", - "type": "text" - }, - { - "block_id": "p815-b10", - "global_id": 23933, - "bbox": [ - 128.9, - 527.35, - 502.75, - 549.68 - ], - "text": "(a) Since f = 400 Hz is less than 500 Hz, there is no aliasing. The apparent sinusoid is\ncos(2πft + θ) with f = 400.", - "type": "text" - }, - { - "block_id": "p815-b11", - "global_id": 23934, - "bbox": [ - 128.91, - 551.26, - 502.78, - 585.54 - ], - "text": "(b) Since f = 600 Hz can be expressed as 600 = −400 + 1000, we see that fa = −400.\nHence, the aliased frequency is 400 Hz and the phase changes sign. The apparent (aliased)\nsinusoid is cos(2πft −θ) with f = 400.", - "type": "text" - }, - { - "block_id": "p815-b12", - "global_id": 23935, - "bbox": [ - 128.91, - 587.13, - 502.76, - 621.41 - ], - "text": "(c) Since f = 1000 Hz can be expressed as 1000 = 0 + 1000, we see that fa = 0. Hence,\nthe aliased frequency is 0 Hz (dc), and there is no phase sign change. The apparent sinusoid is\ny(t) = cos(0πt ± θ) = cos(θ). This is a dc signal with constant sample values for all n.", - "type": "text" - } - ] - }, - { - "page_num": 816, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p816-b0", - "global_id": 23936, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "796\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p816-b1", - "global_id": 23937, - "bbox": [ - 103.17, - 85.96, - 477.0, - 120.24 - ], - "text": "(d) Here, f = 2400 Hz can be expressed as 2400 = 400 + (2 × 1000) so that fa = 400.\nHence, the aliased frequency is 400 Hz and there is no sign change for the phase. The apparent\nsinusoid is cos(2πft + θ) with f = 400.", - "type": "text" - }, - { - "block_id": "p816-b2", - "global_id": 23938, - "bbox": [ - 103.16, - 122.23, - 477.03, - 156.11 - ], - "text": "We could have found these answers directly from Fig. 8.9b. For example, for case (b), we\nread |fa| = 400 corresponding to f = 600. Moreover, f = 600 lies in the shaded belt. Hence,\nthere is a phase sign change.", - "type": "text" - }, - { - "block_id": "p816-b3", - "global_id": 23939, - "bbox": [ - 107.82, - 223.75, - 395.67, - 235.71 - ], - "text": "DRILL 8.3\nA Case of Identical Sampled Sinusoids", - "type": "text" - }, - { - "block_id": "p816-b4", - "global_id": 23940, - "bbox": [ - 107.82, - 244.41, - 484.41, - 266.74 - ], - "text": "Show that samples of 90 Hz and 110 Hz sinusoids of the form cosωt are identical when sampled\nat a rate 200 Hz.", - "type": "text" - }, - { - "block_id": "p816-b5", - "global_id": 23941, - "bbox": [ - 107.82, - 325.93, - 417.95, - 337.89 - ], - "text": "DRILL 8.4\nApparent Frequency of Sampled Sinusoids", - "type": "text" - }, - { - "block_id": "p816-b6", - "global_id": 23942, - "bbox": [ - 107.82, - 346.91, - 484.38, - 370.42 - ], - "text": "A sinusoid of frequency f0 Hz is sampled at a rate of 100 Hz. Determine the apparent frequency\nof the samples if f0 is (a) 40 Hz, (b) 60 Hz, (c) 140 Hz, and (d) 160 Hz.", - "type": "text" - }, - { - "block_id": "p816-b7", - "global_id": 23943, - "bbox": [ - 108.09, - 382.45, - 170.31, - 393.41 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p816-b8", - "global_id": 23944, - "bbox": [ - 107.82, - 400.8, - 314.8, - 410.76 - ], - "text": "All four cases have an apparent frequency of 40 Hz.", - "type": "text" - }, - { - "block_id": "p816-b9", - "global_id": 23945, - "bbox": [ - 101.84, - 451.32, - 378.95, - 463.27 - ], - "text": "8.2-2 Some Applications of the Sampling Theorem", - "type": "text" - }, - { - "block_id": "p816-b10", - "global_id": 23946, - "bbox": [ - 101.84, - 469.4, - 490.42, - 634.79 - ], - "text": "The sampling theorem is very important in signal analysis, processing, and transmission because\nit allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing\na continuous-time signal is therefore equivalent to processing a discrete sequence of numbers.\nSuch processing leads us directly into the area of digital filtering. In the field of communication,\nthe transmission of a continuous-time message reduces to the transmission of a sequence of\nnumbers by means of pulse trains. The continuous-time signal x(t) is sampled, and sample values\nare used to modify certain parameters of a periodic pulse train. We may vary the amplitudes\n(Fig. 8.11b), widths (Fig. 8.11c), or positions (Fig. 8.11d) of the pulses in proportion to the\nsample values of the signal x(t). Accordingly, we may have pulse-amplitude modulation (PAM),\npulse-width modulation (PWM), or pulse-position modulation (PPM). The most important form\nof pulse modulation today is pulse-code modulation (PCM), discussed in Sec. 8.3 in connection\nwith Fig. 8.14b. In all these cases, instead of transmitting x(t), we transmit the corresponding\npulse-modulated signal. At the receiver, we read the information of the pulse-modulated signal\nand reconstruct the analog signal x(t).", - "type": "text" - } - ] - }, - { - "page_num": 817, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p817-b0", - "global_id": 23947, - "bbox": [ - 366.66, - 62.89, - 516.14, - 71.98 - ], - "text": "8.2\nSignal Reconstruction\n797", - "type": "text" - }, - { - "block_id": "p817-b1", - "global_id": 23948, - "bbox": [ - 351.81, - 124.11, - 354.04, - 132.11 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p817-b2", - "global_id": 23949, - "bbox": [ - 351.81, - 210.83, - 354.04, - 218.83 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p817-b3", - "global_id": 23950, - "bbox": [ - 351.81, - 300.47, - 354.04, - 308.47 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p817-b4", - "global_id": 23951, - "bbox": [ - 351.81, - 375.14, - 354.04, - 383.14 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p817-b5", - "global_id": 23952, - "bbox": [ - 250.61, - 153.91, - 259.49, - 161.91 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p817-b6", - "global_id": 23953, - "bbox": [ - 250.15, - 240.61, - 259.96, - 248.61 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p817-b7", - "global_id": 23954, - "bbox": [ - 250.61, - 313.96, - 259.49, - 321.96 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p817-b8", - "global_id": 23955, - "bbox": [ - 361.87, - 261.42, - 414.08, - 296.42 - ], - "text": "Pulse locations\nare the same but\ntheir widths\nchange.", - "type": "text" - }, - { - "block_id": "p817-b9", - "global_id": 23956, - "bbox": [ - 250.4, - 388.74, - 259.73, - 396.74 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p817-b10", - "global_id": 23957, - "bbox": [ - 361.87, - 334.76, - 414.31, - 369.76 - ], - "text": "Pulse widths are\nthe same but\ntheir locations\nchange.", - "type": "text" - }, - { - "block_id": "p817-b11", - "global_id": 23958, - "bbox": [ - 167.42, - 87.44, - 181.0, - 95.52 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p817-b12", - "global_id": 23959, - "bbox": [ - 151.5, - 403.24, - 516.13, - 425.24 - ], - "text": "Figure 8.11 Pulse-modulated signals. (a) The signal. (b) The PAM signal. (c) The PWM\n(PDM) signal. (d) The PAM signal.", - "type": "text" - }, - { - "block_id": "p817-b13", - "global_id": 23960, - "bbox": [ - 127.59, - 449.65, - 516.16, - 519.39 - ], - "text": "One advantage of using pulse modulation is that it permits the simultaneous transmission\nof several signals on a time-sharing basis—time-division multiplexing (TDM). Because a\npulse-modulated signal occupies only a part of the channel time, we can transmit several\npulse-modulated signals on the same channel by interweaving them. Figure 8.12 shows the TDM\nof two PAM signals. In this manner, we can multiplex several signals on the same channel by\nreducing pulse widths.†", - "type": "text" - }, - { - "block_id": "p817-b14", - "global_id": 23961, - "bbox": [ - 127.59, - 521.38, - 516.11, - 555.26 - ], - "text": "Digital signals also offer an advantage in the area of communications, where signals must\ntravel over distances. Transmission of digital signals is more rugged than that of analog signals\nbecause digital signals can withstand channel noise and distortion much better as long as the noise", - "type": "text" - }, - { - "block_id": "p817-b15", - "global_id": 23962, - "bbox": [ - 127.59, - 577.36, - 516.12, - 633.41 - ], - "text": "† Another method of transmitting several baseband signals simultaneously is frequency-division multiplexing\n(FDM) discussed in Sec. 7.7-4. In FDM, various signals are multiplexed by sharing the channel bandwidth.\nThe spectrum of each message is shifted to a specific band not occupied by any other signal. The information\nof various signals is located in nonoverlapping frequency bands of the channel (Fig. 7.45). In a way, TDM\nand FDM are duals of each other.", - "type": "text" - } - ] - }, - { - "page_num": 818, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p818-b0", - "global_id": 23963, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "798\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p818-b1", - "global_id": 23964, - "bbox": [ - 412.35, - 154.15, - 414.58, - 162.15 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p818-b2", - "global_id": 23965, - "bbox": [ - 137.14, - 87.47, - 151.25, - 97.08 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p818-b3", - "global_id": 23966, - "bbox": [ - 297.74, - 124.06, - 311.85, - 133.67 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p818-b4", - "global_id": 23967, - "bbox": [ - 125.76, - 208.44, - 328.75, - 217.68 - ], - "text": "Figure 8.12 Time-division multiplexing of two signals.", - "type": "text" - }, - { - "block_id": "p818-b5", - "global_id": 23968, - "bbox": [ - 126.53, - 254.15, - 135.41, - 262.15 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p818-b6", - "global_id": 23969, - "bbox": [ - 125.76, - 291.82, - 135.41, - 299.82 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p818-b7", - "global_id": 23970, - "bbox": [ - 126.53, - 329.48, - 135.41, - 337.48 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p818-b8", - "global_id": 23971, - "bbox": [ - 126.08, - 367.81, - 135.41, - 375.81 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p818-b9", - "global_id": 23972, - "bbox": [ - 144.53, - 265.2, - 164.09, - 273.49 - ], - "text": "A2", - "type": "text" - }, - { - "block_id": "p818-b10", - "global_id": 23973, - "bbox": [ - 151.2, - 242.07, - 375.84, - 255.35 - ], - "text": "A2\nt", - "type": "text" - }, - { - "block_id": "p818-b11", - "global_id": 23974, - "bbox": [ - 373.61, - 285.02, - 375.83, - 293.02 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p818-b12", - "global_id": 23975, - "bbox": [ - 373.61, - 322.68, - 375.83, - 330.68 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p818-b13", - "global_id": 23976, - "bbox": [ - 373.61, - 361.01, - 375.83, - 369.01 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p818-b14", - "global_id": 23977, - "bbox": [ - 125.76, - 390.9, - 490.39, - 424.85 - ], - "text": "Figure 8.13 Digital signal transmission: (a) at the transmitter, (b) received distorted signal\n(without noise), (c) received distorted signal (with noise), and (d) regenerated signal at the\nreceiver.", - "type": "text" - }, - { - "block_id": "p818-b15", - "global_id": 23978, - "bbox": [ - 101.84, - 445.49, - 490.4, - 539.15 - ], - "text": "and the distortion are within limits. An analog signal can be converted to digital binary form\nthrough sampling and quantization (rounding off), as explained in the next section. The digital\n(binary) message in Fig. 8.13a is distorted by the channel, as illustrated in Fig. 8.13b. Yet if the\ndistortion remains within a limit, we can recover the data without error because we need only\nmake a simple binary decision: Is the received pulse positive or negative? Figure 8.13c shows the\nsame data with channel distortion and noise. Here again, the data can be recovered correctly as\nlong as the distortion and the noise are within limits. Such is not the case with analog messages.\nAny distortion or noise, no matter how small, will distort the received signal.", - "type": "text" - }, - { - "block_id": "p818-b16", - "global_id": 23979, - "bbox": [ - 101.84, - 541.14, - 490.43, - 634.79 - ], - "text": "The greatest advantage of digital communication over the analog counterpart, however, is the\nviability of regenerative repeaters in the former. In an analog transmission system, a message signal\ngrows progressively weaker as it travels along the channel (transmission path), whereas the channel\nnoise and the signal distortion, being cumulative, become progressively stronger. Ultimately, the\nsignal, overwhelmed by noise and distortion, is mutilated. Amplification is of little help because\nit enhances the signal and the noise in the same proportion. Consequently, the distance over which\nan analog message can be transmitted is limited by the transmitted power. If a transmission path\nis long enough, the channel distortion and noise will accumulate sufficiently to overwhelm even a", - "type": "text" - } - ] - }, - { - "page_num": 819, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p819-b0", - "global_id": 23980, - "bbox": [ - 307.3, - 62.89, - 516.12, - 71.98 - ], - "text": "8.3\nAnalog-to-Digital (A/D) Conversion\n799", - "type": "text" - }, - { - "block_id": "p819-b1", - "global_id": 23981, - "bbox": [ - 127.59, - 85.82, - 516.15, - 143.6 - ], - "text": "digital signal. The trick is to set up repeaters along the transmission path at distances short enough\nto permit detection of signal pulses before the noise and distortion have a chance to accumulate\nsufficiently. At each repeater, the pulses are detected, and new, clean pulses are transmitted to the\nnext repeater, which, in turn, duplicates the same process. If the noise and distortion remain within\nlimits (which is possible because of the closely spaced repeaters), pulses can be detected correctly.†", - "type": "text" - }, - { - "block_id": "p819-b2", - "global_id": 23982, - "bbox": [ - 127.59, - 145.59, - 516.15, - 215.33 - ], - "text": "This way the digital messages can be transmitted over longer distances with greater reliability. In\ncontrast, analog messages cannot be cleaned up periodically, and their transmission is therefore\nless reliable. The most significant error in digitized signals comes from quantizing (rounding\noff). This error, discussed in Sec. 8.3, can be reduced as much as desired by increasing the\nnumber of quantization levels, at the cost of an increased bandwidth of the transmission medium\n(channel).", - "type": "text" - }, - { - "block_id": "p819-b3", - "global_id": 23983, - "bbox": [ - 127.94, - 247.51, - 431.09, - 261.46 - ], - "text": "8.3 ANALOG-TO-DIGITAL (A/D) CONVERSION", - "type": "text" - }, - { - "block_id": "p819-b4", - "global_id": 23984, - "bbox": [ - 127.59, - 267.35, - 516.15, - 420.88 - ], - "text": "The amplitude of an analog signal can take on any value over a continuous range. Hence, analog\nsignal amplitude can take on an infinite number of values. In contrast, a digital signal amplitude\ncan take on only a finite number of values. An analog signal can be converted into a digital signal\nby means of sampling and quantizing (rounding off). Sampling an analog signal alone will not\nyield a digital signal because a sample of analog signal can still take on any value in a continuous\nrange. It is digitized by rounding off its value to one of the closest permissible numbers (or\nquantized levels), as illustrated in Fig. 8.14a, which represents one possible quantizing scheme.\nThe amplitudes of the analog signal x(t) lie in the range (−V,V). This range is partitioned into\nL subintervals, each of magnitude = 2V/L. Next, each sample amplitude is approximated by\nthe midpoint value of the subinterval in which the sample falls (see Fig. 8.14a for L = 16). It is\nclear that each sample is approximated to one of the L numbers. Thus, the signal is digitized with\nquantized samples taking on any one of the L values. This is an L-ary digital signal (see Sec. 1.3-2).\nEach sample can now be represented by one of L distinct pulses.", - "type": "text" - }, - { - "block_id": "p819-b5", - "global_id": 23985, - "bbox": [ - 127.59, - 422.87, - 516.16, - 518.92 - ], - "text": "From a practical viewpoint, dealing with a large number of distinct pulses is difficult. We\nprefer to use the smallest possible number of distinct pulses, the very smallest number being 2.\nA digital signal using only two symbols or values is the binary signal. A binary digital signal (a\nsignal that can take on only two values) is very desirable because of its simplicity, economy,\nand ease of engineering. We can convert an L-ary signal into a binary signal by using pulse\ncoding. Figure 8.14b shows one such code for the case of L = 16. This code, formed by binary\nrepresentation of the 16 decimal digits from 0 to 15, is known as the natural binary code (NBC).\nFor L quantization levels, we need a minimum of b binary code digits, where 2b = L or b = log2 L.", - "type": "text" - }, - { - "block_id": "p819-b6", - "global_id": 23986, - "bbox": [ - 127.59, - 518.51, - 516.15, - 600.2 - ], - "text": "Each of the 16 levels is assigned one binary code word of four digits. Thus, each sample in\nthis example is encoded by four binary digits. To transmit or digitally process the binary data, we\nneed to assign a distinct electrical pulse to each of the two binary states. One possible way is to\nassign a negative pulse to a binary 0 and a positive pulse to a binary 1 so that each sample is now\nrepresented by a group of four binary pulses (pulse code), as depicted in Fig. 8.14b. The resulting\nbinary signal is a digital signal obtained from the analog signal x(t) through A/D conversion. In\ncommunications jargon, such a signal is known as a pulse-code-modulated (PCM) signal.", - "type": "text" - }, - { - "block_id": "p819-b7", - "global_id": 23987, - "bbox": [ - 127.59, - 621.19, - 319.43, - 633.41 - ], - "text": "† The error in pulse detection can be made negligible.", - "type": "text" - } - ] - }, - { - "page_num": 820, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p820-b0", - "global_id": 23988, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "800\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p820-b1", - "global_id": 23989, - "bbox": [ - 293.7, - 270.19, - 302.58, - 278.19 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p820-b2", - "global_id": 23990, - "bbox": [ - 293.23, - 607.74, - 303.04, - 615.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p820-b3", - "global_id": 23991, - "bbox": [ - 191.86, - 297.42, - 399.37, - 305.97 - ], - "text": "Digit\nBinary equivalent\nPulse code waveform", - "type": "text" - }, - { - "block_id": "p820-b4", - "global_id": 23992, - "bbox": [ - 258.89, - 315.49, - 274.89, - 323.49 - ], - "text": "0000", - "type": "text" - }, - { - "block_id": "p820-b5", - "global_id": 23993, - "bbox": [ - 258.89, - 333.31, - 274.89, - 341.31 - ], - "text": "0001", - "type": "text" - }, - { - "block_id": "p820-b6", - "global_id": 23994, - "bbox": [ - 258.89, - 351.12, - 274.89, - 359.12 - ], - "text": "0010", - "type": "text" - }, - { - "block_id": "p820-b7", - "global_id": 23995, - "bbox": [ - 258.89, - 368.95, - 274.89, - 376.95 - ], - "text": "0011", - "type": "text" - }, - { - "block_id": "p820-b8", - "global_id": 23996, - "bbox": [ - 258.89, - 386.77, - 274.89, - 394.77 - ], - "text": "0100", - "type": "text" - }, - { - "block_id": "p820-b9", - "global_id": 23997, - "bbox": [ - 258.89, - 404.58, - 274.89, - 412.58 - ], - "text": "0101", - "type": "text" - }, - { - "block_id": "p820-b10", - "global_id": 23998, - "bbox": [ - 258.89, - 422.4, - 274.89, - 430.4 - ], - "text": "0110", - "type": "text" - }, - { - "block_id": "p820-b11", - "global_id": 23999, - "bbox": [ - 258.89, - 440.22, - 274.89, - 448.22 - ], - "text": "0111", - "type": "text" - }, - { - "block_id": "p820-b12", - "global_id": 24000, - "bbox": [ - 258.89, - 458.04, - 274.89, - 466.04 - ], - "text": "1000", - "type": "text" - }, - { - "block_id": "p820-b13", - "global_id": 24001, - "bbox": [ - 258.89, - 475.86, - 274.89, - 483.86 - ], - "text": "1001", - "type": "text" - }, - { - "block_id": "p820-b14", - "global_id": 24002, - "bbox": [ - 258.89, - 493.68, - 274.89, - 501.68 - ], - "text": "1010", - "type": "text" - }, - { - "block_id": "p820-b15", - "global_id": 24003, - "bbox": [ - 258.89, - 511.5, - 274.89, - 519.5 - ], - "text": "1011", - "type": "text" - }, - { - "block_id": "p820-b16", - "global_id": 24004, - "bbox": [ - 258.89, - 529.32, - 274.89, - 537.32 - ], - "text": "1100", - "type": "text" - }, - { - "block_id": "p820-b17", - "global_id": 24005, - "bbox": [ - 258.89, - 547.14, - 274.89, - 555.14 - ], - "text": "1101", - "type": "text" - }, - { - "block_id": "p820-b18", - "global_id": 24006, - "bbox": [ - 258.89, - 564.96, - 274.89, - 572.96 - ], - "text": "1110", - "type": "text" - }, - { - "block_id": "p820-b19", - "global_id": 24007, - "bbox": [ - 258.89, - 582.78, - 274.89, - 590.78 - ], - "text": "1111", - "type": "text" - }, - { - "block_id": "p820-b20", - "global_id": 24008, - "bbox": [ - 196.09, - 315.57, - 200.09, - 323.57 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p820-b21", - "global_id": 24009, - "bbox": [ - 196.09, - 333.39, - 200.09, - 341.39 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p820-b22", - "global_id": 24010, - "bbox": [ - 196.09, - 351.2, - 200.09, - 359.2 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p820-b23", - "global_id": 24011, - "bbox": [ - 196.09, - 369.03, - 200.09, - 377.03 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p820-b24", - "global_id": 24012, - "bbox": [ - 196.09, - 386.85, - 200.09, - 394.85 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p820-b25", - "global_id": 24013, - "bbox": [ - 196.09, - 404.66, - 200.09, - 412.66 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p820-b26", - "global_id": 24014, - "bbox": [ - 196.09, - 422.48, - 200.09, - 430.48 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p820-b27", - "global_id": 24015, - "bbox": [ - 196.09, - 440.8, - 200.09, - 448.8 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p820-b28", - "global_id": 24016, - "bbox": [ - 196.09, - 458.12, - 200.09, - 466.12 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p820-b29", - "global_id": 24017, - "bbox": [ - 196.09, - 475.94, - 200.09, - 483.94 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p820-b30", - "global_id": 24018, - "bbox": [ - 196.09, - 493.76, - 204.09, - 501.76 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p820-b31", - "global_id": 24019, - "bbox": [ - 196.09, - 511.58, - 204.09, - 519.58 - ], - "text": "11", - "type": "text" - }, - { - "block_id": "p820-b32", - "global_id": 24020, - "bbox": [ - 196.09, - 529.4, - 204.09, - 537.4 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p820-b33", - "global_id": 24021, - "bbox": [ - 196.09, - 547.22, - 204.09, - 555.22 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p820-b34", - "global_id": 24022, - "bbox": [ - 196.09, - 565.04, - 204.09, - 573.04 - ], - "text": "14", - "type": "text" - }, - { - "block_id": "p820-b35", - "global_id": 24023, - "bbox": [ - 196.09, - 582.86, - 204.09, - 590.86 - ], - "text": "15", - "type": "text" - }, - { - "block_id": "p820-b36", - "global_id": 24024, - "bbox": [ - 144.13, - 101.83, - 149.02, - 109.83 - ], - "text": "V", - "type": "text" - }, - { - "block_id": "p820-b37", - "global_id": 24025, - "bbox": [ - 137.46, - 245.61, - 149.02, - 253.83 - ], - "text": "V", - "type": "text" - }, - { - "block_id": "p820-b38", - "global_id": 24026, - "bbox": [ - 406.88, - 161.73, - 409.1, - 169.73 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p820-b39", - "global_id": 24027, - "bbox": [ - 179.63, - 113.84, - 359.51, - 121.92 - ], - "text": "Quantized samples of x(t)\nx(t)", - "type": "text" - }, - { - "block_id": "p820-b40", - "global_id": 24028, - "bbox": [ - 126.45, - 134.99, - 134.45, - 224.76 - ], - "text": "Allowed quantization levels", - "type": "text" - }, - { - "block_id": "p820-b41", - "global_id": 24029, - "bbox": [ - 201.07, - 210.64, - 209.96, - 227.02 - ], - "text": "2V\nL", - "type": "text" - }, - { - "block_id": "p820-b42", - "global_id": 24030, - "bbox": [ - 125.76, - 622.37, - 476.16, - 631.68 - ], - "text": "Figure 8.14 Analog-to-digital (A/D) conversion of a signal: (a) quantizing and (b) pulse coding.", - "type": "text" - } - ] - }, - { - "page_num": 821, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p821-b0", - "global_id": 24031, - "bbox": [ - 307.3, - 62.89, - 516.12, - 71.98 - ], - "text": "8.3\nAnalog-to-Digital (A/D) Conversion\n801", - "type": "text" - }, - { - "block_id": "p821-b1", - "global_id": 24032, - "bbox": [ - 127.59, - 85.72, - 516.13, - 107.74 - ], - "text": "The convenient contraction of “binary digit” to bit has become an industry standard\nabbreviation.", - "type": "text" - }, - { - "block_id": "p821-b2", - "global_id": 24033, - "bbox": [ - 127.59, - 109.73, - 516.15, - 215.33 - ], - "text": "The audio signal bandwidth is about 15 kHz, but subjective tests show that signal articulation\n(intelligibility) is not affected if all the components above 3400 Hz are suppressed [3]. Since the\nobjective in telephone communication is intelligibility rather than high fidelity, the components\nabove 3400 Hz are eliminated by a lowpass filter.† The resulting signal is then sampled at a\nrate of 8000 samples/s (8 kHz). This rate is intentionally kept higher than the Nyquist sampling\nrate of 6.8 kHz to avoid unrealizable filters required for signal reconstruction. Each sample is\nfinally quantized into 256 levels (L = 256), which requires a group of eight binary pulses to\nencode each sample (28 = 256). Thus, a digitized telephone signal consists of data amounting\nto 8 × 8000 = 64,000 or 64 kbit/s, requiring 64,000 binary pulses per second for its transmission.", - "type": "text" - }, - { - "block_id": "p821-b3", - "global_id": 24034, - "bbox": [ - 127.59, - 217.32, - 516.15, - 275.11 - ], - "text": "The compact disc (CD), a high-fidelity application of A/D conversion, requires the audio\nsignal bandwidth of 20 kHz. Although the Nyquist sampling rate is only 40 kHz, an actual\nsampling rate of 44.1 kHz is used for the reason mentioned earlier. The signal is quantized into a\nrather large number of levels (L = 65,536) to reduce quantizing error. The binary-coded samples\nare now recorded on the CD.", - "type": "text" - }, - { - "block_id": "p821-b4", - "global_id": 24035, - "bbox": [ - 127.89, - 298.03, - 241.71, - 310.16 - ], - "text": "A HISTORICAL NOTE", - "type": "text" - }, - { - "block_id": "p821-b5", - "global_id": 24036, - "bbox": [ - 127.59, - 314.11, - 516.15, - 383.93 - ], - "text": "The binary system of representing any number by using 1s and 0s was invented in India by Pingala\n(ca. 200 BCE). It was again worked out independently in the West by Gottfried Wilhelm Leibniz\n(1646–1716). He felt a spiritual significance in this discovery, reasoning that 1 representing unity\nwas clearly a symbol for God, while 0 represented the nothingness. He reasoned that if all numbers\ncan be represented merely by the use of 1 and 0, this surely proves that God created the universe\nout of nothing!", - "type": "text" - }, - { - "block_id": "p821-b6", - "global_id": 24037, - "bbox": [ - 102.51, - 421.12, - 366.38, - 433.07 - ], - "text": "EXAMPLE 8.5\nADC Bit Number and Bit Rate", - "type": "text" - }, - { - "block_id": "p821-b7", - "global_id": 24038, - "bbox": [ - 128.9, - 445.05, - 359.82, - 456.77 - ], - "text": "A signal x(t) bandlimited to 3kHz is sampled at a rate 33 1", - "type": "text" - }, - { - "block_id": "p821-b8", - "global_id": 24039, - "bbox": [ - 128.91, - 446.81, - 502.76, - 504.6 - ], - "text": "3% higher than the Nyquist rate. The\nmaximum acceptable error in the sample amplitude (the maximum error due to quantization)\nis 0.5% of the peak amplitude V. The quantized samples are binary-coded. Find the required\nsampling rate, the number of bits required to encode each sample, and the bit rate of the\nresulting PCM signal.", - "type": "text" - }, - { - "block_id": "p821-b9", - "global_id": 24040, - "bbox": [ - 128.9, - 527.1, - 502.77, - 550.51 - ], - "text": "The Nyquist sampling rate is fNyq = 2×3000 = 6000 Hz (samples/s). The actual sampling rate\nis fA = 6000 × (1 1", - "type": "text" - }, - { - "block_id": "p821-b10", - "global_id": 24041, - "bbox": [ - 128.9, - 539.05, - 502.76, - 573.34 - ], - "text": "3) = 8000 Hz.\nThe quantization step is , and the maximum quantization error is ±/2, where =\n2V/L. The maximum error due to quantization, /2, should be no greater than 0.5% of the", - "type": "text" - }, - { - "block_id": "p821-b11", - "global_id": 24042, - "bbox": [ - 127.59, - 621.19, - 439.41, - 633.41 - ], - "text": "† Components below 300 Hz may also be suppressed without affecting the articulation.", - "type": "text" - } - ] - }, - { - "page_num": 822, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p822-b0", - "global_id": 24043, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "802\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p822-b1", - "global_id": 24044, - "bbox": [ - 103.16, - 86.14, - 246.68, - 96.2 - ], - "text": "signal peak amplitude V. Therefore,", - "type": "text" - }, - { - "block_id": "p822-b3", - "global_id": 24045, - "bbox": [ - 222.56, - 106.04, - 278.13, - 130.16 - ], - "text": "2 = V\nL = 0.5", - "type": "text" - }, - { - "block_id": "p822-b4", - "global_id": 24046, - "bbox": [ - 264.43, - 112.71, - 360.32, - 130.16 - ], - "text": "100V\n\r⇒\nL = 200", - "type": "text" - }, - { - "block_id": "p822-b5", - "global_id": 24047, - "bbox": [ - 103.16, - 137.94, - 477.02, - 171.91 - ], - "text": "For binary coding, L must be a power of 2. Hence, the next higher value of L that is a power\nof 2 is L = 256. Because log2 256 = 8, we need 8 bits to encode each sample. Therefore the bit\nrate of the PCM signal is", - "type": "text" - }, - { - "block_id": "p822-b6", - "global_id": 24048, - "bbox": [ - 240.78, - 173.5, - 339.4, - 183.87 - ], - "text": "8000 × 8 = 64,000 bits/s", - "type": "text" - }, - { - "block_id": "p822-b7", - "global_id": 24049, - "bbox": [ - 107.82, - 255.59, - 374.1, - 267.54 - ], - "text": "DRILL 8.5\nBit Number and Bit Rate for ASCII", - "type": "text" - }, - { - "block_id": "p822-b8", - "global_id": 24050, - "bbox": [ - 107.82, - 276.66, - 484.39, - 298.58 - ], - "text": "The American Standard Code for Information Interchange (ASCII) has 128 characters, which\nare binary-coded. A certain computer generates 100,000 characters per second. Show that", - "type": "text" - }, - { - "block_id": "p822-b9", - "global_id": 24051, - "bbox": [ - 125.76, - 306.46, - 374.46, - 316.51 - ], - "text": "(a) 7 bits (binary digits) are required to encode each character", - "type": "text" - }, - { - "block_id": "p822-b10", - "global_id": 24052, - "bbox": [ - 125.76, - 321.41, - 379.76, - 331.46 - ], - "text": "(b) 700,000 bits/s are required to transmit the computer output.", - "type": "text" - }, - { - "block_id": "p822-b11", - "global_id": 24053, - "bbox": [ - 102.2, - 383.73, - 445.65, - 397.67 - ], - "text": "8.4 DUAL OF TIME SAMPLING: SPECTRAL SAMPLING", - "type": "text" - }, - { - "block_id": "p822-b12", - "global_id": 24054, - "bbox": [ - 101.84, - 403.66, - 490.39, - 474.48 - ], - "text": "As in other cases, the sampling theorem has its dual. In Sec. 8.1, we discussed the time-sampling\ntheorem and showed that a signal bandlimited to B Hz can be reconstructed from the signal samples\ntaken at a rate of fs > 2B samples/s. Note that the signal spectrum exists over the frequency range\n(in hertz) of −B to B. Therefore, 2B is the spectral width (not the bandwidth, which is B) of the\nsignal. This fact means that a signal x(t) can be reconstructed from samples taken at a rate fs > the\nspectral width of X(ω) in hertz ( fs > 2B).", - "type": "text" - }, - { - "block_id": "p822-b13", - "global_id": 24055, - "bbox": [ - 101.84, - 475.29, - 490.4, - 534.67 - ], - "text": "We now prove the dual of the time-sampling theorem. This is the spectral sampling theorem,\nwhich applies to timelimited signals (the dual of bandlimited signals). A timelimited signal x(t)\nexists only over a finite interval of τ seconds, as shown in Fig. 8.15a. Generally, a timelimited\nsignal is characterized by x(t) = 0 for t < T1 and t > T2 (assuming T2 > T1). The signal width or\nduration is τ = T2 −T1 seconds.", - "type": "text" - }, - { - "block_id": "p822-b14", - "global_id": 24056, - "bbox": [ - 101.85, - 534.75, - 490.39, - 569.04 - ], - "text": "The spectral sampling theorem states that the spectrum X(ω) of a signal x(t) timelimited to a\nduration of τ seconds can be reconstructed from the samples of X(ω) taken at a rate R samples/Hz,\nwhere R > τ (the signal width or duration) in seconds.", - "type": "text" - }, - { - "block_id": "p822-b15", - "global_id": 24057, - "bbox": [ - 101.85, - 570.62, - 490.4, - 592.95 - ], - "text": "Figure 8.15a shows a timelimited signal x(t) and its Fourier transform X(ω). Although X(ω)\nis complex in general, it is adequate for our line of reasoning to show X(ω) as a real function.", - "type": "text" - }, - { - "block_id": "p822-b16", - "global_id": 24058, - "bbox": [ - 214.73, - 617.06, - 245.22, - 627.33 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p822-b17", - "global_id": 24059, - "bbox": [ - 247.27, - 603.5, - 264.21, - 615.56 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p822-b18", - "global_id": 24060, - "bbox": [ - 252.53, - 628.38, - 265.09, - 635.35 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p822-b19", - "global_id": 24061, - "bbox": [ - 266.69, - 615.34, - 318.34, - 627.33 - ], - "text": "x(t)e−jωtdt =", - "type": "text" - }, - { - "block_id": "p822-b20", - "global_id": 24062, - "bbox": [ - 320.39, - 603.5, - 333.24, - 615.56 - ], - "text": "# τ", - "type": "text" - }, - { - "block_id": "p822-b21", - "global_id": 24063, - "bbox": [ - 325.66, - 615.34, - 490.39, - 635.64 - ], - "text": "0\nx(t)e−jωtdt\n(8.8)", - "type": "text" - } - ] - }, - { - "page_num": 823, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p823-b0", - "global_id": 24064, - "bbox": [ - 284.76, - 62.89, - 516.15, - 71.98 - ], - "text": "8.4\nDual of Time Sampling: Spectral Sampling\n803", - "type": "text" - }, - { - "block_id": "p823-b1", - "global_id": 24065, - "bbox": [ - 127.59, - 264.44, - 408.77, - 273.68 - ], - "text": "Figure 8.15 Periodic repetition of a signal amounts to sampling its spectrum.", - "type": "text" - }, - { - "block_id": "p823-b2", - "global_id": 24066, - "bbox": [ - 127.59, - 293.36, - 516.13, - 315.69 - ], - "text": "We now construct xT0(t), a periodic signal formed by repeating x(t) every T0 seconds (T0 > τ), as\ndepicted in Fig. 8.15b. This periodic signal can be expressed by the exponential Fourier series", - "type": "text" - }, - { - "block_id": "p823-b3", - "global_id": 24067, - "bbox": [ - 249.15, - 334.43, - 281.63, - 346.84 - ], - "text": "xT0(t) =", - "type": "text" - }, - { - "block_id": "p823-b4", - "global_id": 24068, - "bbox": [ - 287.38, - 324.26, - 301.48, - 334.94 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p823-b5", - "global_id": 24069, - "bbox": [ - 283.68, - 348.28, - 305.16, - 355.47 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p823-b6", - "global_id": 24070, - "bbox": [ - 306.28, - 327.45, - 392.37, - 345.89 - ], - "text": "Dnejnω0t\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p823-b7", - "global_id": 24071, - "bbox": [ - 382.64, - 341.82, - 391.66, - 352.65 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p823-b8", - "global_id": 24072, - "bbox": [ - 127.59, - 364.39, - 228.08, - 375.85 - ], - "text": "where (assuming T0 > τ)", - "type": "text" - }, - { - "block_id": "p823-b9", - "global_id": 24073, - "bbox": [ - 225.68, - 385.52, - 257.17, - 403.55 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p823-b10", - "global_id": 24074, - "bbox": [ - 249.91, - 399.48, - 258.93, - 410.32 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p823-b11", - "global_id": 24075, - "bbox": [ - 261.74, - 378.53, - 278.43, - 392.14 - ], - "text": "# T0", - "type": "text" - }, - { - "block_id": "p823-b12", - "global_id": 24076, - "bbox": [ - 267.0, - 385.52, - 349.52, - 410.68 - ], - "text": "0\nx(t)e−jnω0tdt = 1", - "type": "text" - }, - { - "block_id": "p823-b13", - "global_id": 24077, - "bbox": [ - 342.26, - 399.48, - 351.28, - 410.32 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p823-b14", - "global_id": 24078, - "bbox": [ - 354.1, - 378.53, - 366.93, - 390.59 - ], - "text": "# τ", - "type": "text" - }, - { - "block_id": "p823-b15", - "global_id": 24079, - "bbox": [ - 359.35, - 390.37, - 417.87, - 410.68 - ], - "text": "0\nx(t)e−jnω0tdt", - "type": "text" - }, - { - "block_id": "p823-b16", - "global_id": 24080, - "bbox": [ - 127.59, - 418.32, - 246.89, - 428.28 - ], - "text": "From Eq. (8.8), it follows that", - "type": "text" - }, - { - "block_id": "p823-b17", - "global_id": 24081, - "bbox": [ - 289.66, - 428.26, - 321.15, - 446.28 - ], - "text": "Dn = 1", - "type": "text" - }, - { - "block_id": "p823-b18", - "global_id": 24082, - "bbox": [ - 313.89, - 442.21, - 322.91, - 453.05 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p823-b19", - "global_id": 24083, - "bbox": [ - 324.61, - 434.83, - 354.07, - 445.98 - ], - "text": "X(nω0)", - "type": "text" - }, - { - "block_id": "p823-b20", - "global_id": 24084, - "bbox": [ - 127.59, - 457.81, - 516.14, - 541.0 - ], - "text": "This result indicates that the coefficients of the Fourier series for xT0(t) are (1/T0) times the sample\nvalues of the spectrum X(ω) taken at intervals of ω0. This means that the spectrum of the periodic\nsignal xT0(t) is the sampled spectrum X(ω), as illustrated in Fig. 8.15b. Now as long as T0 > τ, the\nsuccessive cycles of x(t) appearing in xT0(t) do not overlap, and x(t) can be recovered from xT0(t).\nSuch recovery implies indirectly that X(ω) can be reconstructed from its samples. These samples\nare separated by the fundamental frequency f0 = 1/T0 Hz of the periodic signal xT0(t). Hence, the\ncondition for recovery is T0 > τ; that is,", - "type": "text" - }, - { - "block_id": "p823-b21", - "global_id": 24085, - "bbox": [ - 302.24, - 548.93, - 327.3, - 566.96 - ], - "text": "f0 < 1", - "type": "text" - }, - { - "block_id": "p823-b22", - "global_id": 24086, - "bbox": [ - 322.06, - 555.91, - 341.48, - 572.95 - ], - "text": "τ Hz", - "type": "text" - }, - { - "block_id": "p823-b23", - "global_id": 24087, - "bbox": [ - 127.59, - 579.55, - 516.12, - 602.96 - ], - "text": "Therefore, to be able to reconstruct the spectrum X(ω) from the samples of X(ω), the samples\nshould be taken at frequency intervals f0 < 1/τ Hz. If R is the sampling rate (samples/Hz), then", - "type": "text" - }, - { - "block_id": "p823-b24", - "global_id": 24088, - "bbox": [ - 274.58, - 610.88, - 300.7, - 627.83 - ], - "text": "R = 1", - "type": "text" - }, - { - "block_id": "p823-b25", - "global_id": 24089, - "bbox": [ - 294.84, - 624.84, - 301.1, - 635.68 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p823-b26", - "global_id": 24090, - "bbox": [ - 304.84, - 617.45, - 369.13, - 627.83 - ], - "text": "> τ samples/Hz", - "type": "text" - } - ] - }, - { - "page_num": 824, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p824-b0", - "global_id": 24091, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "804\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p824-b1", - "global_id": 24092, - "bbox": [ - 102.14, - 86.19, - 247.7, - 98.32 - ], - "text": "SPECTRAL INTERPOLATION", - "type": "text" - }, - { - "block_id": "p824-b2", - "global_id": 24093, - "bbox": [ - 101.84, - 101.93, - 490.39, - 148.18 - ], - "text": "Consider a signal timelimited to τ seconds and centered at Tc. We now show that the spectrum\nX(ω) of x(t) can be reconstructed from the samples of X(ω). For this case, using the dual of the\napproach employed to derive the signal interpolation formula in Eq. (8.6), we obtain the spectral\ninterpolation formula†", - "type": "text" - }, - { - "block_id": "p824-b3", - "global_id": 24094, - "bbox": [ - 142.96, - 181.58, - 173.47, - 191.86 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p824-b4", - "global_id": 24095, - "bbox": [ - 179.21, - 171.41, - 193.31, - 182.09 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p824-b5", - "global_id": 24096, - "bbox": [ - 175.52, - 195.43, - 196.99, - 202.63 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p824-b6", - "global_id": 24097, - "bbox": [ - 198.11, - 181.58, - 244.71, - 192.73 - ], - "text": "X(nω0)sinc", - "type": "text" - }, - { - "block_id": "p824-b7", - "global_id": 24098, - "bbox": [ - 244.71, - 167.6, - 268.38, - 185.75 - ], - "text": "ωT0", - "type": "text" - }, - { - "block_id": "p824-b8", - "global_id": 24099, - "bbox": [ - 258.27, - 181.58, - 291.9, - 199.04 - ], - "text": "2\n−nπ", - "type": "text" - }, - { - "block_id": "p824-b10", - "global_id": 24100, - "bbox": [ - 299.63, - 174.6, - 400.25, - 193.04 - ], - "text": "e−j(ω−nω0)Tc\nω0 = 2π", - "type": "text" - }, - { - "block_id": "p824-b11", - "global_id": 24101, - "bbox": [ - 390.51, - 188.97, - 399.53, - 199.81 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p824-b12", - "global_id": 24102, - "bbox": [ - 422.37, - 181.58, - 490.39, - 193.04 - ], - "text": "T0 > τ\n(8.9)", - "type": "text" - }, - { - "block_id": "p824-b13", - "global_id": 24103, - "bbox": [ - 101.84, - 226.28, - 490.4, - 260.57 - ], - "text": "For the case in Fig. 8.15, Tc = T0/2. If the pulse x(t) were to be centered at the origin, then Tc = 0,\nand the exponential term at the extreme right in Eq. (8.9) would vanish. In such a case, Eq. (8.9)\nwould be the exact dual of Eq. (8.6).", - "type": "text" - }, - { - "block_id": "p824-b14", - "global_id": 24104, - "bbox": [ - 76.77, - 302.41, - 374.51, - 314.37 - ], - "text": "EXAMPLE 8.6\nSpectral Sampling and Interpolation", - "type": "text" - }, - { - "block_id": "p824-b15", - "global_id": 24105, - "bbox": [ - 103.16, - 330.62, - 477.0, - 352.95 - ], - "text": "The spectrum X(ω) of a unit-duration signal x(t), centered at the origin, is sampled at the\nintervals of 1 Hz or 2π rad/s (the Nyquist rate). The samples are", - "type": "text" - }, - { - "block_id": "p824-b16", - "global_id": 24106, - "bbox": [ - 178.22, - 364.5, - 401.96, - 374.87 - ], - "text": "X(0) = 1\nand\nX(±2πn) = 0\nn = 1,2,3,. . .", - "type": "text" - }, - { - "block_id": "p824-b17", - "global_id": 24107, - "bbox": [ - 103.17, - 386.41, - 141.25, - 396.79 - ], - "text": "Find x(t).", - "type": "text" - }, - { - "block_id": "p824-b18", - "global_id": 24108, - "bbox": [ - 103.16, - 419.29, - 477.01, - 465.53 - ], - "text": "We use the interpolation formula Eq. (8.9) (with Tc = 0) to construct X(ω) from its samples.\nSince all but one of the Nyquist samples are zero, only one term (corresponding to n = 0) in the\nsummation on the right-hand side of Eq. (8.9) survives. Thus, with X(0) = 1 and τ = T0 = 1,\nwe obtain", - "type": "text" - }, - { - "block_id": "p824-b19", - "global_id": 24109, - "bbox": [ - 201.38, - 470.7, - 249.98, - 481.07 - ], - "text": "X(ω) = sinc", - "type": "text" - }, - { - "block_id": "p824-b20", - "global_id": 24110, - "bbox": [ - 249.98, - 456.71, - 264.43, - 473.67 - ], - "text": "ω", - "type": "text" - }, - { - "block_id": "p824-b21", - "global_id": 24111, - "bbox": [ - 258.77, - 478.18, - 263.76, - 488.15 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p824-b23", - "global_id": 24112, - "bbox": [ - 292.47, - 470.7, - 378.81, - 481.07 - ], - "text": "and\nx(t) = rect(t)", - "type": "text" - }, - { - "block_id": "p824-b24", - "global_id": 24113, - "bbox": [ - 103.16, - 494.97, - 477.02, - 517.31 - ], - "text": "For a signal of unit duration, this is the only spectrum with the sample values X(0) = 1 and\nX(2πn) = 0(n̸ = 0). No other spectrum satisfies these conditions.", - "type": "text" - }, - { - "block_id": "p824-b25", - "global_id": 24114, - "bbox": [ - 101.84, - 584.47, - 413.83, - 602.39 - ], - "text": "† This can be obtained by observing that the Fourier transform of xT0(t) is 2π %", - "type": "text" - }, - { - "block_id": "p824-b26", - "global_id": 24115, - "bbox": [ - 101.84, - 591.2, - 490.39, - 633.41 - ], - "text": "n Dnδ(ω −nω0) [see\nEq. (7.22)]. We can recover x(t) from xT0(t) by multiplying the latter with rect(t −Tc)/T0, whose Fourier\ntransform is T0 sinc(ωT0/2)e−jωTc. Hence, X(ω) is 1/2π times the convolution of these two Fourier\ntransforms, which yields Eq. (8.9).", - "type": "text" - } - ] - }, - { - "page_num": 825, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p825-b0", - "global_id": 24116, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n805", - "type": "text" - }, - { - "block_id": "p825-b1", - "global_id": 24117, - "bbox": [ - 127.94, - 94.37, - 383.77, - 140.19 - ], - "text": "8.5 NUMERICAL COMPUTATION OF THE\nFOURIER TRANSFORM: THE DISCRETE\nFOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p825-b2", - "global_id": 24118, - "bbox": [ - 127.59, - 145.76, - 516.15, - 203.97 - ], - "text": "Numerical computation of the Fourier transform of x(t) requires sample values of x(t) because a\ndigital computer can work only with discrete data (sequence of numbers). Moreover, a computer\ncan compute X(ω) only at some discrete values of ω [samples of X(ω)]. We therefore need to\nrelate the samples of X(ω) to samples of x(t). This task can be accomplished by using the results\nof the two sampling theorems developed in Secs. 8.1 and 8.4.", - "type": "text" - }, - { - "block_id": "p825-b3", - "global_id": 24119, - "bbox": [ - 127.59, - 205.54, - 516.13, - 301.1 - ], - "text": "We begin with a timelimited signal x(t) (Fig. 8.16a) and its spectrum X(ω) (Fig. 8.16b).\nSince x(t) is timelimited, X(ω) is nonbandlimited. For convenience, we shall show all spectra\nas functions of the frequency variable f (in hertz) rather than ω. According to the sampling\ntheorem, the spectrum X(ω) of the sampled signal x(t) consists of X(ω) repeating every fs Hz,\nwhere fs = 1/T, as depicted in Fig. 8.16d.† In the next step, the sampled signal in Fig. 8.16c\nis repeated periodically every T0 seconds, as illustrated in Fig. 8.16e. According to the spectral\nsampling theorem, such an operation results in sampling the spectrum at a rate of T0 samples/Hz.\nThis sampling rate means that the samples are spaced at f0 = 1/T0 Hz, as depicted in Fig. 8.16f.", - "type": "text" - }, - { - "block_id": "p825-b4", - "global_id": 24120, - "bbox": [ - 127.59, - 301.19, - 516.13, - 335.48 - ], - "text": "The foregoing discussion shows that when a signal x(t) is sampled and then periodically\nrepeated, the corresponding spectrum is also sampled and periodically repeated. Our goal is to\nrelate the samples of x(t) to the samples of X(ω).", - "type": "text" - }, - { - "block_id": "p825-b5", - "global_id": 24121, - "bbox": [ - 127.89, - 350.59, - 244.73, - 362.72 - ], - "text": "NUMBER OF SAMPLES", - "type": "text" - }, - { - "block_id": "p825-b6", - "global_id": 24122, - "bbox": [ - 127.59, - 366.65, - 516.13, - 389.44 - ], - "text": "One interesting observation from Figs. 8.16e and 8.16f is that N0, the number of samples of the\nsignal in Fig. 8.16e in one period T0, is identical to N′", - "type": "text" - }, - { - "block_id": "p825-b7", - "global_id": 24123, - "bbox": [ - 127.59, - 378.7, - 516.14, - 401.32 - ], - "text": "0, the number of samples of the spectrum in\nFig. 8.16f in one period fs. To see this, we notice that", - "type": "text" - }, - { - "block_id": "p825-b8", - "global_id": 24124, - "bbox": [ - 230.69, - 411.24, - 263.4, - 429.36 - ], - "text": "N0 = T0", - "type": "text" - }, - { - "block_id": "p825-b9", - "global_id": 24125, - "bbox": [ - 255.99, - 416.18, - 284.18, - 435.26 - ], - "text": "T\nN′", - "type": "text" - }, - { - "block_id": "p825-b10", - "global_id": 24126, - "bbox": [ - 281.7, - 411.24, - 305.01, - 436.13 - ], - "text": "0 = fs\nf0", - "type": "text" - }, - { - "block_id": "p825-b11", - "global_id": 24127, - "bbox": [ - 316.67, - 411.34, - 341.35, - 429.36 - ], - "text": "fs = 1", - "type": "text" - }, - { - "block_id": "p825-b12", - "global_id": 24128, - "bbox": [ - 335.71, - 411.34, - 409.57, - 435.26 - ], - "text": "T\nand\nf0 = 1", - "type": "text" - }, - { - "block_id": "p825-b13", - "global_id": 24129, - "bbox": [ - 402.32, - 425.29, - 411.34, - 436.13 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p825-b14", - "global_id": 24130, - "bbox": [ - 492.07, - 418.32, - 516.13, - 428.28 - ], - "text": "(8.10)", - "type": "text" - }, - { - "block_id": "p825-b15", - "global_id": 24131, - "bbox": [ - 127.59, - 445.96, - 260.4, - 455.92 - ], - "text": "Using these relations, we see that", - "type": "text" - }, - { - "block_id": "p825-b16", - "global_id": 24132, - "bbox": [ - 282.9, - 456.82, - 315.62, - 474.95 - ], - "text": "N0 = T0", - "type": "text" - }, - { - "block_id": "p825-b17", - "global_id": 24133, - "bbox": [ - 308.2, - 456.82, - 336.24, - 480.84 - ], - "text": "T = fs", - "type": "text" - }, - { - "block_id": "p825-b18", - "global_id": 24134, - "bbox": [ - 330.37, - 470.88, - 336.63, - 481.71 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p825-b19", - "global_id": 24135, - "bbox": [ - 340.37, - 461.77, - 359.31, - 473.77 - ], - "text": "= N′", - "type": "text" - }, - { - "block_id": "p825-b20", - "global_id": 24136, - "bbox": [ - 356.83, - 468.85, - 360.32, - 475.83 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p825-b21", - "global_id": 24137, - "bbox": [ - 127.59, - 493.52, - 516.13, - 591.36 - ], - "text": "ALIASING AND LEAKAGE IN NUMERICAL COMPUTATION\nFigure 8.16f shows the presence of aliasing in the samples of the spectrum X(ω). This aliasing\nerror can be reduced as much as desired by increasing the sampling frequency fs (decreasing the\nsampling interval T = 1/fs). The aliasing can never be eliminated for timelimited x(t), however,\nbecause its spectrum X(ω) is nonbandlimited. Had we started with a signal having a bandlimited\nspectrum X(ω), there would be no aliasing in the spectrum in Fig. 8.16f. Unfortunately, such\na signal is nontimelimited, and its repetition (in Fig. 8.16e) would result in signal overlapping\n(aliasing in the time domain). In this case, we shall have to contend with errors in signal", - "type": "text" - }, - { - "block_id": "p825-b22", - "global_id": 24138, - "bbox": [ - 127.59, - 610.24, - 516.13, - 633.41 - ], - "text": "† There is a multiplying constant 1/T for the spectrum in Fig. 8.16d [see Eq. (8.2)], but this is irrelevant to\nour discussion here.", - "type": "text" - } - ] - }, - { - "page_num": 826, - "width": 720.0, - "height": 576.0, - "blocks": [ - { - "block_id": "p826-b0", - "global_id": 24139, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "806\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p826-b1", - "global_id": 24140, - "bbox": [ - 435.48, - 180.73, - 443.56, - 192.16 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p826-b2", - "global_id": 24141, - "bbox": [ - 399.33, - 169.59, - 409.73, - 494.97 - ], - "text": "0\n0\nt", - "type": "text" - }, - { - "block_id": "p826-b3", - "global_id": 24142, - "bbox": [ - 383.93, - 200.89, - 391.93, - 502.17 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p826-b4", - "global_id": 24143, - "bbox": [ - 297.9, - 170.09, - 309.11, - 494.97 - ], - "text": "T\n0\n0", - "type": "text" - }, - { - "block_id": "p826-b5", - "global_id": 24144, - "bbox": [ - 275.64, - 197.45, - 283.64, - 502.01 - ], - "text": "(c)\n(d)", - "type": "text" - }, - { - "block_id": "p826-b6", - "global_id": 24145, - "bbox": [ - 194.54, - 170.09, - 203.64, - 494.77 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p826-b7", - "global_id": 24146, - "bbox": [ - 217.12, - 239.08, - 225.12, - 243.53 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p826-b8", - "global_id": 24147, - "bbox": [ - 191.72, - 210.49, - 201.33, - 217.94 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p826-b9", - "global_id": 24148, - "bbox": [ - 171.16, - 201.89, - 179.16, - 501.66 - ], - "text": "(e)\n(f)", - "type": "text" - }, - { - "block_id": "p826-b10", - "global_id": 24149, - "bbox": [ - 193.31, - 316.53, - 202.43, - 604.19 - ], - "text": "t\nf", - "type": "text" - }, - { - "block_id": "p826-b11", - "global_id": 24150, - "bbox": [ - 298.38, - 526.99, - 307.93, - 604.19 - ], - "text": "f\nfs\nfs", - "type": "text" - }, - { - "block_id": "p826-b12", - "global_id": 24151, - "bbox": [ - 399.63, - 601.97, - 407.63, - 604.19 - ], - "text": "f", - "type": "text" - }, - { - "block_id": "p826-b13", - "global_id": 24152, - "bbox": [ - 298.52, - 316.53, - 306.52, - 318.76 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p826-b14", - "global_id": 24153, - "bbox": [ - 398.18, - 316.53, - 406.18, - 318.76 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p826-b15", - "global_id": 24154, - "bbox": [ - 224.44, - 387.44, - 238.63, - 410.41 - ], - "text": "fo T0", - "type": "text" - }, - { - "block_id": "p826-b16", - "global_id": 24155, - "bbox": [ - 234.64, - 404.69, - 242.64, - 408.69 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p826-b17", - "global_id": 24156, - "bbox": [ - 184.75, - 518.78, - 197.33, - 538.8 - ], - "text": "fs T", - "type": "text" - }, - { - "block_id": "p826-b18", - "global_id": 24157, - "bbox": [ - 193.34, - 534.58, - 201.34, - 538.58 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p826-b19", - "global_id": 24158, - "bbox": [ - 289.09, - 526.71, - 297.09, - 530.71 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p826-b20", - "global_id": 24159, - "bbox": [ - 338.9, - 507.47, - 347.0, - 523.02 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p826-b21", - "global_id": 24160, - "bbox": [ - 437.85, - 507.47, - 445.95, - 523.02 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p826-b22", - "global_id": 24161, - "bbox": [ - 212.82, - 86.84, - 218.82, - 327.81 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p826-b23", - "global_id": 24162, - "bbox": [ - 335.53, - 180.73, - 346.53, - 192.28 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p826-b24", - "global_id": 24163, - "bbox": [ - 155.23, - 86.59, - 164.84, - 308.84 - ], - "text": "Figure 8.16 Relationship between samples of x(t) and X(ω).", - "type": "text" - } - ] - }, - { - "page_num": 827, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p827-b0", - "global_id": 24164, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n807", - "type": "text" - }, - { - "block_id": "p827-b1", - "global_id": 24165, - "bbox": [ - 127.59, - 85.82, - 516.16, - 251.2 - ], - "text": "samples. In other words, in computing the direct or inverse Fourier transform numerically, we\ncan reduce the error as much as we wish, but the error can never be eliminated. This is true\nof numerical computation of the direct and inverse Fourier transforms, regardless of the method\nused. For example, if we determine the Fourier transform by direct integration numerically, by\nusing Eq. (7.9), there will be an error because the interval of integration t can never be made\nzero. Similar remarks apply to numerical computation of the inverse transform. Therefore, we\nshould always keep in mind the nature of this error in our results. In our discussion (Fig. 8.16),\nwe assumed x(t) to be a timelimited signal. If x(t) were not timelimited, we would need to\ntimelimit it because numerical computations can work only with finite data. Furthermore, this\ndata truncation causes error because of spectral spreading (smearing) and leakage, as discussed in\nSec. 7.8. The leakage also causes aliasing. Leakage can be reduced by using a tapered window for\nsignal truncation. But this choice increases spectral spreading or smearing. Spectral spreading can\nbe reduced by increasing the window width (i.e., more data), which increases T0, and reduces f0\n(increases spectral or frequency resolution).", - "type": "text" - }, - { - "block_id": "p827-b2", - "global_id": 24166, - "bbox": [ - 127.59, - 269.04, - 516.15, - 378.84 - ], - "text": "PICKET FENCE EFFECT\nThe numerical computation method yields only the uniform sample values of X(ω). The major\npeaks or valleys of X(ω) can lie between two samples and may remain hidden, giving a false\npicture of reality. Viewing samples is like viewing the signal and its spectrum through a “picket\nfence” with upright posts that are very wide and placed close together. What is hidden behind the\npickets is much more than what we can see. Such misleading results can be avoided by using a\nsufficiently large N0, the number of samples, to increase resolution. We can also use zero padding\n(discussed later) or the spectral interpolation formula [Eq. (8.9)] to determine the values of X(ω)\nbetween samples.", - "type": "text" - }, - { - "block_id": "p827-b3", - "global_id": 24167, - "bbox": [ - 127.59, - 396.67, - 516.13, - 446.7 - ], - "text": "POINTS OF DISCONTINUITY\nIf x(t) or X(ω) has a jump discontinuity at a sampling point, the sample value should be taken as\nthe average of the values on the two sides of the discontinuity because the Fourier representation\nat a point of discontinuity converges to the average value.", - "type": "text" - }, - { - "block_id": "p827-b4", - "global_id": 24168, - "bbox": [ - 127.89, - 464.53, - 450.85, - 476.66 - ], - "text": "DERIVATION OF THE DISCRETE FOURIER TRANSFORM (DFT)", - "type": "text" - }, - { - "block_id": "p827-b5", - "global_id": 24169, - "bbox": [ - 127.59, - 480.28, - 516.12, - 504.1 - ], - "text": "If x(nT) and X(rω0) are the nth and rth samples of x(t) and X(ω), respectively, then we define new\nvariables xn and Xr as", - "type": "text" - }, - { - "block_id": "p827-b6", - "global_id": 24170, - "bbox": [ - 273.34, - 511.07, - 344.96, - 529.2 - ], - "text": "xn = Tx(nT) = T0", - "type": "text" - }, - { - "block_id": "p827-b7", - "global_id": 24171, - "bbox": [ - 335.39, - 525.13, - 345.51, - 535.97 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p827-b8", - "global_id": 24172, - "bbox": [ - 347.21, - 517.74, - 516.12, - 528.12 - ], - "text": "x(nT)\n(8.11)", - "type": "text" - }, - { - "block_id": "p827-b9", - "global_id": 24173, - "bbox": [ - 127.59, - 547.76, - 141.97, - 557.72 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p827-b10", - "global_id": 24174, - "bbox": [ - 296.91, - 568.61, - 346.81, - 580.07 - ], - "text": "Xr = X(rω0)", - "type": "text" - }, - { - "block_id": "p827-b11", - "global_id": 24175, - "bbox": [ - 127.59, - 594.16, - 151.92, - 604.12 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p827-b12", - "global_id": 24176, - "bbox": [ - 288.21, - 610.95, - 353.31, - 629.39 - ], - "text": "ω0 = 2πf0 = 2π", - "type": "text" - }, - { - "block_id": "p827-b13", - "global_id": 24177, - "bbox": [ - 343.58, - 625.31, - 352.6, - 636.15 - ], - "text": "T0", - "type": "text" - } - ] - }, - { - "page_num": 828, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p828-b0", - "global_id": 24178, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "808\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p828-b1", - "global_id": 24179, - "bbox": [ - 119.78, - 86.06, - 413.24, - 98.11 - ], - "text": "We shall now show that xn and Xr are related by the following equations†:", - "type": "text" - }, - { - "block_id": "p828-b2", - "global_id": 24180, - "bbox": [ - 258.38, - 119.86, - 277.65, - 131.32 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p828-b3", - "global_id": 24181, - "bbox": [ - 279.7, - 109.07, - 296.76, - 120.36 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p828-b4", - "global_id": 24182, - "bbox": [ - 282.03, - 134.25, - 294.44, - 141.51 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p828-b5", - "global_id": 24183, - "bbox": [ - 297.87, - 118.14, - 490.38, - 130.94 - ], - "text": "xne−jr0n\n(8.12)", - "type": "text" - }, - { - "block_id": "p828-b6", - "global_id": 24184, - "bbox": [ - 101.85, - 152.68, - 116.22, - 162.64 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p828-b7", - "global_id": 24185, - "bbox": [ - 254.03, - 168.02, - 283.31, - 186.05 - ], - "text": "xn = 1", - "type": "text" - }, - { - "block_id": "p828-b8", - "global_id": 24186, - "bbox": [ - 275.51, - 181.98, - 285.64, - 192.81 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p828-b9", - "global_id": 24187, - "bbox": [ - 288.44, - 163.8, - 305.5, - 175.09 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p828-b10", - "global_id": 24188, - "bbox": [ - 291.07, - 188.98, - 302.86, - 196.24 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p828-b11", - "global_id": 24189, - "bbox": [ - 306.61, - 172.87, - 490.38, - 185.67 - ], - "text": "Xrejr0n\n(8.13)", - "type": "text" - }, - { - "block_id": "p828-b12", - "global_id": 24190, - "bbox": [ - 101.85, - 204.4, - 126.18, - 214.36 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p828-b13", - "global_id": 24191, - "bbox": [ - 262.8, - 214.06, - 327.23, - 232.5 - ], - "text": "0 = ω0T = 2π", - "type": "text" - }, - { - "block_id": "p828-b14", - "global_id": 24192, - "bbox": [ - 316.94, - 228.43, - 327.07, - 239.26 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p828-b15", - "global_id": 24193, - "bbox": [ - 101.84, - 246.21, - 490.39, - 280.19 - ], - "text": "These equations define the direct and the inverse discrete Fourier transforms, with Xr the direct\ndiscrete Fourier transform (DFT) of xn, and xn the inverse discrete Fourier transform (IDFT) of Xr.\nThe notation", - "type": "text" - }, - { - "block_id": "p828-b16", - "global_id": 24194, - "bbox": [ - 275.57, - 283.96, - 315.99, - 295.42 - ], - "text": "xn ⇐⇒Xr", - "type": "text" - }, - { - "block_id": "p828-b17", - "global_id": 24195, - "bbox": [ - 101.84, - 304.34, - 490.39, - 398.41 - ], - "text": "is also used to indicate that xn and Xr are a DFT pair. Remember that xn is T0/N0 times the nth\nsample of x(t) and Xr is the rth sample of X(ω). Knowing the sample values of x(t), we can use the\nDFT to compute the sample values of X(ω)—and vice versa. Note, however, that xn is a function\nof n (n = 0,1,2,. . .,N0 −1) rather than of t and that Xr is a function of r (r = 0,1,2,. . .,N0 −1)\nrather than of ω. Moreover, both xn and Xr are periodic sequences of period N0 (Figs. 8.16e,\n8.16f). Such sequences are called N0-periodic sequences. The proof of the DFT relationships in\nEqs. (8.12) and (8.13) follows directly from the results of the sampling theorem. The sampled\nsignal x(t) (Fig. 8.16c) can be expressed as", - "type": "text" - }, - { - "block_id": "p828-b18", - "global_id": 24196, - "bbox": [ - 242.93, - 421.66, - 267.58, - 431.94 - ], - "text": "x(t) =", - "type": "text" - }, - { - "block_id": "p828-b19", - "global_id": 24197, - "bbox": [ - 269.63, - 410.86, - 286.69, - 422.15 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p828-b20", - "global_id": 24198, - "bbox": [ - 271.95, - 436.05, - 284.36, - 443.31 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p828-b21", - "global_id": 24199, - "bbox": [ - 287.8, - 421.66, - 349.29, - 431.94 - ], - "text": "x(nT)δ (t −nT)", - "type": "text" - }, - { - "block_id": "p828-b22", - "global_id": 24200, - "bbox": [ - 101.84, - 454.31, - 366.63, - 465.91 - ], - "text": "Since δ (t −nT) ⇐⇒e−jnωT, applying the Fourier transform yields", - "type": "text" - }, - { - "block_id": "p828-b23", - "global_id": 24201, - "bbox": [ - 246.72, - 489.15, - 277.21, - 499.43 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p828-b24", - "global_id": 24202, - "bbox": [ - 279.26, - 478.36, - 296.33, - 489.65 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p828-b25", - "global_id": 24203, - "bbox": [ - 281.59, - 503.55, - 294.0, - 510.81 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p828-b26", - "global_id": 24204, - "bbox": [ - 297.44, - 487.43, - 344.46, - 499.43 - ], - "text": "x(nT)e−jnωT", - "type": "text" - }, - { - "block_id": "p828-b27", - "global_id": 24205, - "bbox": [ - 101.84, - 523.69, - 490.37, - 546.03 - ], - "text": "But from Fig. 8.1f [or Eq. (8.2)], it is clear that over the interval |ω| ≤ωs/2, X(ω), the Fourier\ntransform of x(t) is X(ω)/T, assuming negligible aliasing. Hence,", - "type": "text" - }, - { - "block_id": "p828-b28", - "global_id": 24206, - "bbox": [ - 195.54, - 569.27, - 272.48, - 579.54 - ], - "text": "X(ω) = TX(ω) = T", - "type": "text" - }, - { - "block_id": "p828-b29", - "global_id": 24207, - "bbox": [ - 274.36, - 558.47, - 291.41, - 569.77 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p828-b30", - "global_id": 24208, - "bbox": [ - 276.69, - 583.67, - 289.09, - 590.93 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p828-b31", - "global_id": 24209, - "bbox": [ - 292.53, - 562.29, - 395.0, - 579.54 - ], - "text": "x(nT)e−jnωT\n|ω| ≤ωs", - "type": "text" - }, - { - "block_id": "p828-b32", - "global_id": 24210, - "bbox": [ - 388.14, - 576.76, - 393.12, - 586.72 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p828-b33", - "global_id": 24211, - "bbox": [ - 101.84, - 610.24, - 490.39, - 634.75 - ], - "text": "† In Eqs. (8.12) and (8.13), the summation is performed from 0 to N0 −1. It is shown in Sec. 9.1-2 [Eqs. (9.6)\nand (9.7)] that the summation may be performed over any successive N0 values of n or r.", - "type": "text" - } - ] - }, - { - "page_num": 829, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p829-b0", - "global_id": 24212, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n809", - "type": "text" - }, - { - "block_id": "p829-b1", - "global_id": 24213, - "bbox": [ - 127.59, - 85.82, - 141.97, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p829-b2", - "global_id": 24214, - "bbox": [ - 250.45, - 107.42, - 317.76, - 118.88 - ], - "text": "Xr = X(rω0) = T", - "type": "text" - }, - { - "block_id": "p829-b3", - "global_id": 24215, - "bbox": [ - 319.64, - 96.63, - 336.69, - 107.93 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b4", - "global_id": 24216, - "bbox": [ - 321.96, - 121.82, - 334.36, - 129.09 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p829-b5", - "global_id": 24217, - "bbox": [ - 337.81, - 105.7, - 516.13, - 117.8 - ], - "text": "x(nT)e−nkrω0T\n(8.14)", - "type": "text" - }, - { - "block_id": "p829-b6", - "global_id": 24218, - "bbox": [ - 127.59, - 136.63, - 292.78, - 147.78 - ], - "text": "If we let ω0T = 0, then from Eq. (8.10),", - "type": "text" - }, - { - "block_id": "p829-b7", - "global_id": 24219, - "bbox": [ - 270.1, - 157.79, - 371.42, - 176.23 - ], - "text": "0 = ω0T = 2πf0T = 2π", - "type": "text" - }, - { - "block_id": "p829-b8", - "global_id": 24220, - "bbox": [ - 361.13, - 172.16, - 371.26, - 183.0 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p829-b9", - "global_id": 24221, - "bbox": [ - 127.59, - 192.83, - 215.85, - 202.79 - ], - "text": "Also, from Eq. (8.11),", - "type": "text" - }, - { - "block_id": "p829-b10", - "global_id": 24222, - "bbox": [ - 297.37, - 206.25, - 345.86, - 217.32 - ], - "text": "Tx(nT) = xn", - "type": "text" - }, - { - "block_id": "p829-b11", - "global_id": 24223, - "bbox": [ - 127.59, - 226.85, - 249.28, - 236.81 - ], - "text": "Therefore, Eq. (8.14) becomes", - "type": "text" - }, - { - "block_id": "p829-b12", - "global_id": 24224, - "bbox": [ - 255.18, - 259.85, - 274.46, - 271.31 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p829-b13", - "global_id": 24225, - "bbox": [ - 276.51, - 249.05, - 293.57, - 260.34 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b14", - "global_id": 24226, - "bbox": [ - 278.84, - 274.24, - 291.24, - 281.5 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p829-b15", - "global_id": 24227, - "bbox": [ - 294.68, - 252.87, - 386.34, - 271.31 - ], - "text": "xne−jr0n\n0 = 2π", - "type": "text" - }, - { - "block_id": "p829-b16", - "global_id": 24228, - "bbox": [ - 376.06, - 267.24, - 386.18, - 278.07 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p829-b17", - "global_id": 24229, - "bbox": [ - 127.59, - 292.46, - 291.64, - 302.42 - ], - "text": "This is Eq. (8.12), which we set to prove.", - "type": "text" - }, - { - "block_id": "p829-b18", - "global_id": 24230, - "bbox": [ - 127.59, - 304.41, - 516.13, - 338.28 - ], - "text": "The inverse transform relationship of Eq. (8.13) can be derived by using a similar procedure\nwith the roles of t and ω reversed, but here we shall use a more direct proof. To prove Eq. (8.13),\nwe multiply both sides of Eq. (8.12) by ejm0r and sum over r as", - "type": "text" - }, - { - "block_id": "p829-b19", - "global_id": 24231, - "bbox": [ - 235.53, - 350.52, - 252.59, - 361.82 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b20", - "global_id": 24232, - "bbox": [ - 238.16, - 375.71, - 249.95, - 382.98 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p829-b21", - "global_id": 24233, - "bbox": [ - 253.7, - 357.21, - 296.66, - 372.4 - ], - "text": "Xrejm0r =", - "type": "text" - }, - { - "block_id": "p829-b22", - "global_id": 24234, - "bbox": [ - 298.71, - 350.52, - 315.77, - 361.82 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b23", - "global_id": 24235, - "bbox": [ - 301.34, - 375.71, - 313.13, - 382.98 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p829-b24", - "global_id": 24236, - "bbox": [ - 316.88, - 344.34, - 340.12, - 361.82 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b25", - "global_id": 24237, - "bbox": [ - 325.39, - 375.71, - 337.8, - 382.98 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p829-b26", - "global_id": 24238, - "bbox": [ - 341.23, - 359.6, - 376.72, - 372.4 - ], - "text": "xne−jr0n", - "type": "text" - }, - { - "block_id": "p829-b28", - "global_id": 24239, - "bbox": [ - 384.52, - 359.6, - 407.53, - 371.6 - ], - "text": "ejm0r", - "type": "text" - }, - { - "block_id": "p829-b29", - "global_id": 24240, - "bbox": [ - 127.59, - 393.92, - 419.41, - 403.88 - ], - "text": "By interchanging the order of summation on the right-hand side, we have", - "type": "text" - }, - { - "block_id": "p829-b30", - "global_id": 24241, - "bbox": [ - 242.25, - 416.13, - 259.31, - 427.43 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b31", - "global_id": 24242, - "bbox": [ - 244.88, - 441.32, - 256.68, - 448.59 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p829-b32", - "global_id": 24243, - "bbox": [ - 260.42, - 422.81, - 303.38, - 438.0 - ], - "text": "Xrejm0r =", - "type": "text" - }, - { - "block_id": "p829-b33", - "global_id": 24244, - "bbox": [ - 305.42, - 416.13, - 322.49, - 427.43 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b34", - "global_id": 24245, - "bbox": [ - 307.76, - 441.32, - 320.16, - 448.59 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p829-b35", - "global_id": 24246, - "bbox": [ - 323.6, - 427.24, - 331.51, - 438.0 - ], - "text": "xn", - "type": "text" - }, - { - "block_id": "p829-b36", - "global_id": 24247, - "bbox": [ - 333.12, - 409.95, - 356.37, - 427.43 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b37", - "global_id": 24248, - "bbox": [ - 341.94, - 441.32, - 353.73, - 448.59 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p829-b38", - "global_id": 24249, - "bbox": [ - 357.47, - 425.2, - 394.63, - 437.2 - ], - "text": "ej(m−n)0r", - "type": "text" - }, - { - "block_id": "p829-b40", - "global_id": 24250, - "bbox": [ - 127.59, - 459.12, - 516.14, - 482.94 - ], - "text": "As the footnote below readily shows, the inner sum on the right-hand side is zero for n̸ = m and\nis N0 when n = m.† Thus, the outer sum will have only one nonzero term when n = m, and it is", - "type": "text" - }, - { - "block_id": "p829-b41", - "global_id": 24251, - "bbox": [ - 127.59, - 500.8, - 181.43, - 513.01 - ], - "text": "† We show that", - "type": "text" - }, - { - "block_id": "p829-b42", - "global_id": 24252, - "bbox": [ - 239.62, - 512.79, - 255.47, - 523.11 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b43", - "global_id": 24253, - "bbox": [ - 241.79, - 535.59, - 253.31, - 542.34 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p829-b44", - "global_id": 24254, - "bbox": [ - 256.46, - 518.95, - 286.08, - 531.91 - ], - "text": "ejk0n =", - "type": "text" - }, - { - "block_id": "p829-b45", - "global_id": 24255, - "bbox": [ - 287.93, - 510.07, - 516.13, - 537.43 - ], - "text": "N0\nk = 0, ±N0, ±2N0, . . .\n0\notherwise\n(8.15)", - "type": "text" - }, - { - "block_id": "p829-b46", - "global_id": 24256, - "bbox": [ - 127.59, - 545.86, - 516.13, - 580.37 - ], - "text": "Recall that 0N0 = 2π. So ejk0n = 1 when k = 0,±N0,±2N0,. . .. Hence, the sum on the left-hand side of\nEq. (8.15) is N0. To compute the sum for other values of k, we note that the sum on the left-hand side of\nEq. (8.15) is a geometric progression with common ratio α = ejk0. Therefore, (see Sec. B.8-3)", - "type": "text" - }, - { - "block_id": "p829-b47", - "global_id": 24257, - "bbox": [ - 222.11, - 589.22, - 237.95, - 599.54 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p829-b48", - "global_id": 24258, - "bbox": [ - 224.26, - 612.03, - 235.79, - 618.77 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p829-b49", - "global_id": 24259, - "bbox": [ - 238.95, - 590.9, - 310.97, - 608.35 - ], - "text": "ejk0n = ejk0N0 −1", - "type": "text" - }, - { - "block_id": "p829-b50", - "global_id": 24260, - "bbox": [ - 275.38, - 595.39, - 421.61, - 614.8 - ], - "text": "ejk0 −1 = 0\n(ejk0N0 = ej2πm = 1)", - "type": "text" - } - ] - }, - { - "page_num": 830, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p830-b0", - "global_id": 24261, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "810\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p830-b1", - "global_id": 24262, - "bbox": [ - 101.84, - 85.4, - 200.06, - 96.86 - ], - "text": "N0xn = N0xm. Therefore,", - "type": "text" - }, - { - "block_id": "p830-b2", - "global_id": 24263, - "bbox": [ - 223.55, - 103.7, - 254.37, - 121.73 - ], - "text": "xm = 1", - "type": "text" - }, - { - "block_id": "p830-b3", - "global_id": 24264, - "bbox": [ - 246.56, - 117.66, - 256.69, - 128.49 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p830-b4", - "global_id": 24265, - "bbox": [ - 259.5, - 99.47, - 276.56, - 110.77 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p830-b5", - "global_id": 24266, - "bbox": [ - 262.13, - 124.67, - 273.93, - 131.93 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p830-b6", - "global_id": 24267, - "bbox": [ - 277.67, - 103.29, - 366.49, - 121.73 - ], - "text": "Xrejm0r\n0 = 2π", - "type": "text" - }, - { - "block_id": "p830-b7", - "global_id": 24268, - "bbox": [ - 356.2, - 117.66, - 366.33, - 128.49 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p830-b8", - "global_id": 24269, - "bbox": [ - 101.84, - 140.69, - 490.38, - 175.43 - ], - "text": "Because Xr is N0 periodic, we need to determine the values of Xr over any one period.\nIt is customary to determine Xr over the range (0, N0 −1), rather than over the range\n(−N0/2,(N0/2)−1).†", - "type": "text" - }, - { - "block_id": "p830-b9", - "global_id": 24270, - "bbox": [ - 101.84, - 190.05, - 490.38, - 241.99 - ], - "text": "CHOICE OF T AND T0\nIn DFT computation, we first need to select suitable values for N0 and T or T0. For this purpose, we\nbegin by deciding on B, the essential bandwidth (in hertz) of the signal. The sampling frequency\nfs must be at least 2B, that is,", - "type": "text" - }, - { - "block_id": "p830-b10", - "global_id": 24271, - "bbox": [ - 284.28, - 243.36, - 289.77, - 254.12 - ], - "text": "fs", - "type": "text" - }, - { - "block_id": "p830-b11", - "global_id": 24272, - "bbox": [ - 284.79, - 250.03, - 309.15, - 267.48 - ], - "text": "2 ≥B", - "type": "text" - }, - { - "block_id": "p830-b12", - "global_id": 24273, - "bbox": [ - 101.84, - 274.02, - 334.2, - 285.89 - ], - "text": "Moreover, the sampling interval T = 1/fs [Eq. (8.10)], and", - "type": "text" - }, - { - "block_id": "p830-b13", - "global_id": 24274, - "bbox": [ - 280.43, - 296.51, - 307.56, - 313.35 - ], - "text": "T ≤1", - "type": "text" - }, - { - "block_id": "p830-b14", - "global_id": 24275, - "bbox": [ - 299.54, - 303.49, - 490.38, - 320.53 - ], - "text": "2B\n(8.16)", - "type": "text" - }, - { - "block_id": "p830-b15", - "global_id": 24276, - "bbox": [ - 101.85, - 330.37, - 357.98, - 340.43 - ], - "text": "Once we pick B, we can choose T according to Eq. (8.16). Also,", - "type": "text" - }, - { - "block_id": "p830-b16", - "global_id": 24277, - "bbox": [ - 280.85, - 352.54, - 307.92, - 370.57 - ], - "text": "f0 = 1", - "type": "text" - }, - { - "block_id": "p830-b17", - "global_id": 24278, - "bbox": [ - 300.66, - 366.5, - 309.68, - 377.33 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p830-b18", - "global_id": 24279, - "bbox": [ - 466.32, - 359.52, - 490.38, - 369.49 - ], - "text": "(8.17)", - "type": "text" - }, - { - "block_id": "p830-b19", - "global_id": 24280, - "bbox": [ - 101.85, - 387.72, - 490.36, - 411.54 - ], - "text": "where f0 is the frequency resolution [separation between samples of X(ω)]. Hence, if f0 is given,\nwe can pick T0 according to Eq. (8.17). Knowing T0 and T, we determine N0 from", - "type": "text" - }, - { - "block_id": "p830-b20", - "global_id": 24281, - "bbox": [ - 278.91, - 421.91, - 311.63, - 440.04 - ], - "text": "N0 = T0", - "type": "text" - }, - { - "block_id": "p830-b21", - "global_id": 24282, - "bbox": [ - 304.21, - 435.97, - 309.75, - 445.93 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p830-b22", - "global_id": 24283, - "bbox": [ - 102.14, - 457.26, - 185.75, - 469.39 - ], - "text": "ZERO PADDING", - "type": "text" - }, - { - "block_id": "p830-b23", - "global_id": 24284, - "bbox": [ - 101.84, - 473.0, - 490.4, - 579.02 - ], - "text": "Recall that observing Xr is like observing the spectrum X(ω) through a picket fence. If the\nfrequency sampling interval f0 is not sufficiently small, we could miss out on some significant\ndetails and obtain a misleading picture. To obtain a higher number of samples, we need to reduce\nf0. Because f0 = 1/T0, a higher number of samples requires us to increase the value of T0, the period\nof repetition for x(t). This option increases N0, the number of samples of x(t), by adding dummy\nsamples of 0 value. This addition of dummy samples is known as zero padding. Thus, zero padding\nincreases the number of samples and may help in getting a better idea of the spectrum X(ω) from\nits samples Xr. To continue with our picket fence analogy, zero padding is like using more, and\nnarrower, pickets.", - "type": "text" - }, - { - "block_id": "p830-b24", - "global_id": 24285, - "bbox": [ - 101.84, - 599.27, - 490.4, - 633.41 - ], - "text": "† The DFT of Eq. (8.12) and the IDFT of Eq. (8.13) represent a transform in their own right, and they are\nexact. There is no approximation. However, xn and Xr, thus obtained, are only approximations to the actual\nsamples of a signal x(t) and of its Fourier transform X(ω).", - "type": "text" - } - ] - }, - { - "page_num": 831, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p831-b0", - "global_id": 24286, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n811", - "type": "text" - }, - { - "block_id": "p831-b1", - "global_id": 24287, - "bbox": [ - 127.89, - 86.19, - 481.43, - 98.32 - ], - "text": "ZERO PADDING DOES NOT IMPROVE ACCURACY OR RESOLUTION", - "type": "text" - }, - { - "block_id": "p831-b2", - "global_id": 24288, - "bbox": [ - 127.59, - 101.93, - 516.16, - 207.95 - ], - "text": "Actually, we are not observing X(ω) through a picket fence. We are observing a distorted version\nof X(ω) resulting from the truncation of x(t). Hence, we should keep in mind that even if the fence\nwere transparent, we would see a reality distorted by aliasing. Seeing through the picket fence\njust gives us an imperfect view of the imperfectly represented reality. Zero padding only allows us\nto look at more samples of that imperfect reality. It can never reduce the imperfection in what is\nbehind the fence. The imperfection, which is caused by aliasing, can be lessened only by reducing\nthe sampling interval T. Observe that reducing T also increases N0, the number of samples, and\nis like increasing the number of pickets while reducing their width. But in this case, the reality\nbehind the fence is also better dressed and we see more of it.", - "type": "text" - }, - { - "block_id": "p831-b3", - "global_id": 24289, - "bbox": [ - 102.51, - 233.89, - 456.73, - 245.85 - ], - "text": "EXAMPLE 8.7\nNumber of Samples and Frequency Resolution", - "type": "text" - }, - { - "block_id": "p831-b4", - "global_id": 24290, - "bbox": [ - 128.9, - 261.99, - 502.75, - 285.4 - ], - "text": "A signal x(t) has a duration of 2 ms and an essential bandwidth of 10 kHz. It is desirable to\nhave a frequency resolution of 100 Hz in the DFT (f0 = 100). Determine N0.", - "type": "text" - }, - { - "block_id": "p831-b5", - "global_id": 24291, - "bbox": [ - 128.9, - 306.82, - 371.9, - 318.69 - ], - "text": "To have f0 = 100 Hz, the effective signal duration T0 must be", - "type": "text" - }, - { - "block_id": "p831-b6", - "global_id": 24292, - "bbox": [ - 267.44, - 326.92, - 295.89, - 344.95 - ], - "text": "T0 = 1", - "type": "text" - }, - { - "block_id": "p831-b7", - "global_id": 24293, - "bbox": [ - 290.02, - 340.88, - 296.28, - 351.71 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p831-b8", - "global_id": 24294, - "bbox": [ - 300.02, - 326.92, - 364.22, - 350.94 - ], - "text": "=\n1\n100 = 10 ms", - "type": "text" - }, - { - "block_id": "p831-b9", - "global_id": 24295, - "bbox": [ - 128.91, - 360.13, - 502.75, - 383.55 - ], - "text": "Since the signal duration is only 2 ms, we need zero padding over 8 ms. Also, B = 10,000.\nHence, fs = 2B = 20,000 and T = 1/fs = 50 µs. Furthermore,", - "type": "text" - }, - { - "block_id": "p831-b10", - "global_id": 24296, - "bbox": [ - 265.78, - 391.97, - 295.35, - 410.1 - ], - "text": "N0 = fs", - "type": "text" - }, - { - "block_id": "p831-b11", - "global_id": 24297, - "bbox": [ - 289.47, - 406.03, - 295.73, - 416.86 - ], - "text": "f0", - "type": "text" - }, - { - "block_id": "p831-b12", - "global_id": 24298, - "bbox": [ - 299.47, - 392.07, - 337.87, - 409.02 - ], - "text": "= 20,000", - "type": "text" - }, - { - "block_id": "p831-b13", - "global_id": 24299, - "bbox": [ - 316.71, - 398.64, - 365.88, - 416.09 - ], - "text": "100\n= 200", - "type": "text" - }, - { - "block_id": "p831-b14", - "global_id": 24300, - "bbox": [ - 128.91, - 425.59, - 502.76, - 483.48 - ], - "text": "The fast Fourier transform (FFT) algorithm (discussed later; see Sec. 8.6) is used to compute\nDFT, where it proves convenient (although not necessary) to select N0 as a power of 2; that\nis, N0 = 2n (n, integer). Let us choose N0 = 256. Increasing N0 from 200 to 256 can be used\nto reduce aliasing error (by reducing T), to improve resolution (by increasing T0 using zero\npadding), or a combination of both.", - "type": "text" - }, - { - "block_id": "p831-b15", - "global_id": 24301, - "bbox": [ - 128.91, - 492.92, - 438.14, - 504.79 - ], - "text": "Reducing Aliasing Error. We maintain the same T0 so that f0 = 100. Hence,", - "type": "text" - }, - { - "block_id": "p831-b16", - "global_id": 24302, - "bbox": [ - 195.06, - 512.9, - 401.3, - 530.92 - ], - "text": "fs = N0f0 = 256 × 100 = 25,600\nand\nT = 1", - "type": "text" - }, - { - "block_id": "p831-b17", - "global_id": 24303, - "bbox": [ - 395.83, - 526.85, - 401.31, - 537.62 - ], - "text": "fs", - "type": "text" - }, - { - "block_id": "p831-b18", - "global_id": 24304, - "bbox": [ - 405.05, - 519.47, - 436.6, - 529.84 - ], - "text": "= 39µs", - "type": "text" - }, - { - "block_id": "p831-b19", - "global_id": 24305, - "bbox": [ - 128.91, - 546.11, - 502.76, - 569.52 - ], - "text": "Thus, increasing N0 from 200 to 256 permits us to reduce the sampling interval T from 50 µs\nto 39 µs while maintaining the same frequency resolution (f0 = 100).", - "type": "text" - }, - { - "block_id": "p831-b20", - "global_id": 24306, - "bbox": [ - 128.91, - 577.89, - 439.73, - 588.26 - ], - "text": "Improving Resolution. Here, we maintain the same T = 50 µs, which yields", - "type": "text" - }, - { - "block_id": "p831-b21", - "global_id": 24307, - "bbox": [ - 167.53, - 597.98, - 406.19, - 616.01 - ], - "text": "T0 = N0T = 256(50 × 10−6) = 12.8 ms\nand\nf0 = 1", - "type": "text" - }, - { - "block_id": "p831-b22", - "global_id": 24308, - "bbox": [ - 398.94, - 611.94, - 407.96, - 622.77 - ], - "text": "T0", - "type": "text" - }, - { - "block_id": "p831-b23", - "global_id": 24309, - "bbox": [ - 411.7, - 604.55, - 464.13, - 614.93 - ], - "text": "= 78.125 Hz", - "type": "text" - } - ] - }, - { - "page_num": 832, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p832-b0", - "global_id": 24310, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "812\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p832-b1", - "global_id": 24311, - "bbox": [ - 103.16, - 86.14, - 476.99, - 108.16 - ], - "text": "Thus, increasing N0 from 200 to 256 can improve the frequency resolution from 100 to 78.125\nHz while maintaining the same aliasing error (T = 50 µs).", - "type": "text" - }, - { - "block_id": "p832-b2", - "global_id": 24312, - "bbox": [ - 103.16, - 118.04, - 477.02, - 153.08 - ], - "text": "Combination of Reducing Aliasing Error and Improving Resolution. To simultaneously\nreduce alias error and improve resolution, we could choose T = 45 µs and T0 = 11.5 ms so\nthat f0 = 86.96 Hz. Many other combinations exist as well.", - "type": "text" - }, - { - "block_id": "p832-b3", - "global_id": 24313, - "bbox": [ - 76.77, - 215.78, - 487.71, - 227.73 - ], - "text": "EXAMPLE 8.8\nDFT to Compute the Fourier Transform of an Exponential", - "type": "text" - }, - { - "block_id": "p832-b4", - "global_id": 24314, - "bbox": [ - 103.16, - 242.76, - 477.01, - 266.31 - ], - "text": "Use the DFT to compute (samples of) the Fourier transform of e−2tu(t). Plot the resulting\nFourier spectra.", - "type": "text" - }, - { - "block_id": "p832-b5", - "global_id": 24315, - "bbox": [ - 103.16, - 287.59, - 477.03, - 358.97 - ], - "text": "We first determine T and T0. The Fourier transform of e−2tu(t) is 1/(jω + 2). This lowpass\nsignal is not bandlimited. In Sec. 7.6, we used the energy criterion to compute the essential\nbandwidth of a signal. Here, we shall present a simpler, but workable alternative to the energy\ncriterion. The essential bandwidth of a signal will be taken as the frequency at which |X(ω)|\ndrops to 1% of its peak value (see the footnote on page 736). In this case, the peak value occurs\nat ω = 0, where |X(0)| = 0.5. Observe that", - "type": "text" - }, - { - "block_id": "p832-b6", - "global_id": 24316, - "bbox": [ - 219.02, - 368.64, - 278.45, - 385.48 - ], - "text": "|X(ω)| =\n1\n√", - "type": "text" - }, - { - "block_id": "p832-b7", - "global_id": 24317, - "bbox": [ - 266.89, - 380.81, - 293.45, - 394.07 - ], - "text": "ω2 + 4", - "type": "text" - }, - { - "block_id": "p832-b8", - "global_id": 24318, - "bbox": [ - 296.7, - 368.64, - 313.56, - 385.17 - ], - "text": "≈1", - "type": "text" - }, - { - "block_id": "p832-b9", - "global_id": 24319, - "bbox": [ - 307.71, - 375.21, - 361.16, - 392.24 - ], - "text": "ω\nω ≫2", - "type": "text" - }, - { - "block_id": "p832-b10", - "global_id": 24320, - "bbox": [ - 103.17, - 402.49, - 477.02, - 424.83 - ], - "text": "Also, 1% of the peak value is 0.01 × 0.5 = 0.005. Hence, the essential bandwidth B is at\nω = 2πB, where", - "type": "text" - }, - { - "block_id": "p832-b11", - "global_id": 24321, - "bbox": [ - 201.93, - 424.08, - 363.39, - 448.11 - ], - "text": "|X(ω)| ≈\n1\n2πB = 0.005\n⇒\nB = 100", - "type": "text" - }, - { - "block_id": "p832-b12", - "global_id": 24322, - "bbox": [ - 352.44, - 431.07, - 378.25, - 447.69 - ], - "text": "π\nHz", - "type": "text" - }, - { - "block_id": "p832-b13", - "global_id": 24323, - "bbox": [ - 103.17, - 453.05, - 184.5, - 463.01 - ], - "text": "and from Eq. (8.16),", - "type": "text" - }, - { - "block_id": "p832-b14", - "global_id": 24324, - "bbox": [ - 235.19, - 462.98, - 262.32, - 479.83 - ], - "text": "T ≤1", - "type": "text" - }, - { - "block_id": "p832-b15", - "global_id": 24325, - "bbox": [ - 254.3, - 462.57, - 344.97, - 487.0 - ], - "text": "2B = π\n200 = 0.015708", - "type": "text" - }, - { - "block_id": "p832-b16", - "global_id": 24326, - "bbox": [ - 103.16, - 491.9, - 477.03, - 525.77 - ], - "text": "Had we used 1% energy criterion to determine the essential bandwidth, following the\nprocedure in Ex. 7.20, we would have obtained B = 20.26 Hz, which is somewhat smaller\nthan the value just obtained by using the 1% amplitude criterion.", - "type": "text" - }, - { - "block_id": "p832-b17", - "global_id": 24327, - "bbox": [ - 103.16, - 527.66, - 477.0, - 574.67 - ], - "text": "The second issue is to determine T0. Because the signal is not timelimited, we have\nto truncate it at T0 such that x(T0) ≪1. A reasonable choice would be T0 = 4 because\nx(4) = e−8 = 0.000335 ≪1. The result is N0 = T0/T = 254.6, which is not a power of 2. Hence,\nwe choose T0 = 4, and T = 0.015625 = 1/64, yielding N0 = 256, which is a power of 2.", - "type": "text" - }, - { - "block_id": "p832-b18", - "global_id": 24328, - "bbox": [ - 103.16, - 575.48, - 477.01, - 621.41 - ], - "text": "Note that there is a great deal of flexibility in determining T and T0, depending on the\naccuracy desired and the computational capacity available. We could just as well have chosen\nT = 0.03125, yielding N0 = 128, although this choice would have given a slightly higher\naliasing error.", - "type": "text" - } - ] - }, - { - "page_num": 833, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p833-b0", - "global_id": 24329, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n813", - "type": "text" - }, - { - "block_id": "p833-b1", - "global_id": 24330, - "bbox": [ - 128.9, - 85.83, - 502.76, - 133.15 - ], - "text": "Because the signal has a jump discontinuity at t = 0, the first sample (at t = 0) is 0.5, the\naverages of the values on the two sides of the discontinuity. We compute Xr (the DFT) from\nthe samples of e−2tu(t) according to Eq. (8.12). Note that Xr is the rth sample of X(ω), and\nthese samples are spaced at f0 = 1/T0 = 0.25 Hz (ω0 = π/2 rad/s).", - "type": "text" - }, - { - "block_id": "p833-b2", - "global_id": 24331, - "bbox": [ - 128.9, - 133.64, - 502.76, - 192.93 - ], - "text": "Because Xr is N0 periodic, Xr = X(r+256) so that X256 = X0. Hence, we need to plot Xr\nover the range r = 0 to 255 (not 256). Moreover, because of this periodicity, X−r = X(−r+256),\nand the values of Xr over the range r = −127 to −1 are identical to those over the range\nr = 129 to 255. Thus, X−127 = X129, X−126 = X130,. . .,X−1 = X255. In addition, because\nof the property of conjugate symmetry of the Fourier transform, X−r = X∗", - "type": "text" - }, - { - "block_id": "p833-b3", - "global_id": 24332, - "bbox": [ - 128.91, - 181.88, - 502.75, - 204.88 - ], - "text": "r , it follows that\nX−1 = X∗", - "type": "text" - }, - { - "block_id": "p833-b4", - "global_id": 24333, - "bbox": [ - 128.91, - 192.15, - 502.76, - 215.75 - ], - "text": "1, X−2 = X∗\n2,. . .,X−128 = X∗\n128. Thus, we need Xr only over the range r = 0 to N0/2\n(128 in this case).", - "type": "text" - }, - { - "block_id": "p833-b5", - "global_id": 24334, - "bbox": [ - 128.9, - 217.33, - 502.76, - 287.49 - ], - "text": "Figure 8.17 shows the computed plots of |Xr| and̸\nXr. The exact spectra are depicted by\ncontinuous curves for comparison. Note the nearly perfect agreement between the two sets of\nspectra. We have depicted the plot of only the first 28 points rather than all 128 points, which\nwould have made the figure very crowded, resulting in loss of clarity. The points are at the\nintervals of 1/T0 = 1/4 Hz or ω0 = 1.5708 rad/s. The 28 samples, therefore, exhibit the plots\nover the range ω = 0 to ω = 28(1.5708) ≈44 rad/s or 7 Hz.", - "type": "text" - }, - { - "block_id": "p833-b6", - "global_id": 24335, - "bbox": [ - 151.96, - 449.14, - 454.56, - 457.76 - ], - "text": "v\n10\n20\n30\n40\n0", - "type": "text" - }, - { - "block_id": "p833-b7", - "global_id": 24336, - "bbox": [ - 139.45, - 326.46, - 149.45, - 334.46 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p833-b8", - "global_id": 24337, - "bbox": [ - 139.45, - 346.3, - 149.45, - 354.3 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p833-b9", - "global_id": 24338, - "bbox": [ - 139.45, - 367.08, - 149.45, - 375.08 - ], - "text": "0.3", - "type": "text" - }, - { - "block_id": "p833-b10", - "global_id": 24339, - "bbox": [ - 139.45, - 387.85, - 149.45, - 395.85 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p833-b11", - "global_id": 24340, - "bbox": [ - 139.45, - 408.62, - 149.45, - 416.62 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p833-b12", - "global_id": 24341, - "bbox": [ - 175.87, - 334.67, - 195.78, - 342.97 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p833-b13", - "global_id": 24342, - "bbox": [ - 127.51, - 472.46, - 149.73, - 480.75 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p833-b14", - "global_id": 24343, - "bbox": [ - 133.3, - 493.0, - 149.97, - 501.29 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p833-b15", - "global_id": 24344, - "bbox": [ - 139.3, - 519.43, - 149.97, - 527.73 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p833-b16", - "global_id": 24345, - "bbox": [ - 197.56, - 360.9, - 215.78, - 368.9 - ], - "text": "Exact", - "type": "text" - }, - { - "block_id": "p833-b17", - "global_id": 24346, - "bbox": [ - 177.34, - 552.82, - 195.56, - 560.82 - ], - "text": "Exact", - "type": "text" - }, - { - "block_id": "p833-b18", - "global_id": 24347, - "bbox": [ - 197.56, - 376.07, - 234.91, - 384.07 - ], - "text": "FFT values", - "type": "text" - }, - { - "block_id": "p833-b19", - "global_id": 24348, - "bbox": [ - 129.64, - 550.09, - 149.64, - 558.39 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p833-b20", - "global_id": 24349, - "bbox": [ - 210.73, - 560.26, - 248.23, - 568.26 - ], - "text": "FFT values", - "type": "text" - }, - { - "block_id": "p833-b21", - "global_id": 24350, - "bbox": [ - 127.51, - 574.71, - 394.09, - 585.59 - ], - "text": "Figure 8.17 Discrete Fourier transform of an exponential signal e−2tu(t).", - "type": "text" - } - ] - }, - { - "page_num": 834, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p834-b0", - "global_id": 24351, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "814\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p834-b1", - "global_id": 24352, - "bbox": [ - 103.16, - 85.89, - 477.03, - 156.03 - ], - "text": "In this example, we knew X(ω) beforehand; hence we could make intelligent choices for B\n(or the sampling frequency fs). In practice, we generally do not know X(ω) beforehand. In fact,\nthat is the very thing we are trying to determine. In such a case, we must make an intelligent\nguess for B or fs from circumstantial evidence. We should then continue reducing the value\nof T and recomputing the transform until the result stabilizes within the desired number of\nsignificant digits.", - "type": "text" - }, - { - "block_id": "p834-b2", - "global_id": 24353, - "bbox": [ - 103.46, - 170.77, - 403.88, - 182.89 - ], - "text": "USING MATLAB TO COMPUTE AND PLOT THE RESULTS", - "type": "text" - }, - { - "block_id": "p834-b3", - "global_id": 24354, - "bbox": [ - 103.16, - 186.92, - 477.03, - 209.13 - ], - "text": "Let us now use MATLAB to confirm the results of this example. First, parameters are defined\nand MATLAB’s fft command is used to compute the DFT.", - "type": "text" - }, - { - "block_id": "p834-b4", - "global_id": 24355, - "bbox": [ - 103.16, - 218.55, - 365.52, - 246.44 - ], - "text": ">>\nT_0 = 4; N_0 = 256; T = T_0/N_0; t = (0:T:T*(N_0-1))’;\n>>\nx = T*exp(-2*t); x(1) = x(1)/2;\n>>\nX_r = fft(x); r = [-N_0/2:N_0/2-1]’; omega_r = r*2*pi/T_0;", - "type": "text" - }, - { - "block_id": "p834-b5", - "global_id": 24356, - "bbox": [ - 103.16, - 256.66, - 344.56, - 266.63 - ], - "text": "The true Fourier transform is also computed for comparison.", - "type": "text" - }, - { - "block_id": "p834-b6", - "global_id": 24357, - "bbox": [ - 103.16, - 276.33, - 348.59, - 284.3 - ], - "text": ">>\nomega = linspace(-pi/T,pi/T,5001); X = 1./(j*omega+2);", - "type": "text" - }, - { - "block_id": "p834-b7", - "global_id": 24358, - "bbox": [ - 103.16, - 294.52, - 366.67, - 304.48 - ], - "text": "For clarity, we display spectrum over a restricted frequency range.", - "type": "text" - }, - { - "block_id": "p834-b8", - "global_id": 24359, - "bbox": [ - 103.16, - 314.19, - 420.55, - 371.96 - ], - "text": ">>\nsubplot(1,2,1); stem(omega_r,fftshift(abs(X_r)),’k.’);\n>>\nline(omega,abs(X),’color’,[0 0 0]); axis([-0.01 44 -0.01 0.51]);\n>>\nxlabel(’\\omega’); ylabel(’|X(\\omega)|’);\n>>\nsubplot(1,2,2); stem(omega_r,fftshift(angle(X_r)),’k.’);\n>>\nline(omega,angle(X),’color’,[0 0 0]); axis([-0.01 44 -pi/2-0.01 0.01]);\n>>\nxlabel(’\\omega’); ylabel(’\\angle X(\\omega)’);", - "type": "text" - }, - { - "block_id": "p834-b9", - "global_id": 24360, - "bbox": [ - 103.16, - 382.19, - 405.96, - 392.15 - ], - "text": "The results, shown in Fig. 8.18, match the earlier results shown in Fig. 8.17.", - "type": "text" - }, - { - "block_id": "p834-b10", - "global_id": 24361, - "bbox": [ - 83.74, - 529.91, - 87.74, - 537.91 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p834-b11", - "global_id": 24362, - "bbox": [ - 76.99, - 509.43, - 87.67, - 517.43 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p834-b12", - "global_id": 24363, - "bbox": [ - 76.99, - 488.94, - 87.67, - 496.94 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p834-b13", - "global_id": 24364, - "bbox": [ - 76.99, - 468.47, - 87.67, - 476.47 - ], - "text": "0.3", - "type": "text" - }, - { - "block_id": "p834-b14", - "global_id": 24365, - "bbox": [ - 76.99, - 447.99, - 87.67, - 455.99 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p834-b15", - "global_id": 24366, - "bbox": [ - 76.99, - 427.5, - 87.67, - 435.5 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p834-b16", - "global_id": 24367, - "bbox": [ - 65.19, - 472.14, - 73.49, - 491.93 - ], - "text": "|X(ω)|", - "type": "text" - }, - { - "block_id": "p834-b17", - "global_id": 24368, - "bbox": [ - 90.01, - 541.18, - 471.58, - 560.04 - ], - "text": "0\n10\n20\n30\n40\nω\nω\n0\n10\n20\n30\n40", - "type": "text" - }, - { - "block_id": "p834-b18", - "global_id": 24369, - "bbox": [ - 295.74, - 526.55, - 309.74, - 534.55 - ], - "text": "–1.5", - "type": "text" - }, - { - "block_id": "p834-b19", - "global_id": 24370, - "bbox": [ - 301.74, - 493.07, - 309.74, - 501.07 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p834-b20", - "global_id": 24371, - "bbox": [ - 295.74, - 459.6, - 309.74, - 467.6 - ], - "text": "–0.5", - "type": "text" - }, - { - "block_id": "p834-b21", - "global_id": 24372, - "bbox": [ - 305.74, - 426.12, - 309.74, - 434.12 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p834-b22", - "global_id": 24373, - "bbox": [ - 284.94, - 469.59, - 293.24, - 488.18 - ], - "text": "X(ω)", - "type": "text" - }, - { - "block_id": "p834-b23", - "global_id": 24374, - "bbox": [ - 64.55, - 567.38, - 330.04, - 578.26 - ], - "text": "Figure 8.18 MATLAB-computed DFT of an exponential signal e−2tu(t).", - "type": "text" - } - ] - }, - { - "page_num": 835, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p835-b0", - "global_id": 24375, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n815", - "type": "text" - }, - { - "block_id": "p835-b1", - "global_id": 24376, - "bbox": [ - 102.51, - 93.91, - 438.18, - 119.82 - ], - "text": "EXAMPLE 8.9\nDFT to Compute the Fourier Transform of a\nRectangular Pulse", - "type": "text" - }, - { - "block_id": "p835-b2", - "global_id": 24377, - "bbox": [ - 128.9, - 136.07, - 362.16, - 146.45 - ], - "text": "Use the DFT to compute the Fourier transform of 8rect(t).", - "type": "text" - }, - { - "block_id": "p835-b3", - "global_id": 24378, - "bbox": [ - 128.9, - 169.36, - 502.79, - 286.92 - ], - "text": "This gate function and its Fourier transform are illustrated in Figs. 8.19a and 8.19b. To\ndetermine the value of the sampling interval T, we must first decide on the essential\nbandwidth B. In Fig. 8.19b, we see that X(ω) decays rather slowly with ω. Hence, the essential\nbandwidth B is rather large. For instance, at B = 15.5 Hz (97.39 rad/s), X(ω) = −0.1643, which\nis about 2% of the peak at X(0). Hence, the essential bandwidth is well above 16 Hz if we use\nthe 1% of the peak amplitude criterion for computing the essential bandwidth. However, we\nshall deliberately take B = 4 for two reasons: to show the effect of aliasing and because the\nuse of B > 4 would give an enormous number of samples, which could not be conveniently\ndisplayed on the page without losing sight of the essentials. Thus, we shall intentionally accept\napproximation to graphically clarify the concepts of the DFT.", - "type": "text" - }, - { - "block_id": "p835-b4", - "global_id": 24379, - "bbox": [ - 128.9, - 288.5, - 502.76, - 359.73 - ], - "text": "The choice of B = 4 results in the sampling interval T = 1/2B = 1/8. Looking again at\nthe spectrum in Fig. 8.19b, we see that the choice of the frequency resolution f0 = 1/4 Hz is\nreasonable. Such a choice gives us four samples in each lobe of X(ω). In this case T0 = 1/f0 = 4\nseconds and N0 = T0/T = 32. The duration of x(t) is only 1 second. We must repeat it every\n4 seconds (T0 = 4), as depicted in Fig. 8.19c, and take samples every 1/8 second. This choice\nyields 32 samples (N0 = 32). Also,", - "type": "text" - }, - { - "block_id": "p835-b5", - "global_id": 24380, - "bbox": [ - 270.88, - 368.85, - 336.42, - 381.65 - ], - "text": "xn = Tx(nT) = 1", - "type": "text" - }, - { - "block_id": "p835-b6", - "global_id": 24381, - "bbox": [ - 332.93, - 370.19, - 360.77, - 383.38 - ], - "text": "8x(nT)", - "type": "text" - }, - { - "block_id": "p835-b7", - "global_id": 24382, - "bbox": [ - 128.9, - 392.11, - 502.76, - 426.39 - ], - "text": "Since x(t) = 8rect (t), the values of xn are 1, 0, or 0.5 (at the points of discontinuity),\nas illustrated in Fig. 8.19c, where xn is depicted as a function of t as well as n, for\nconvenience.", - "type": "text" - }, - { - "block_id": "p835-b8", - "global_id": 24383, - "bbox": [ - 128.9, - 427.97, - 502.79, - 545.95 - ], - "text": "In the derivation of the DFT, we assumed that x(t) begins at t = 0 (Fig. 8.16a), and then\ntook N0 samples over the interval (0, T0). In the present case, however, x(t) begins at −1/2.\nThis difficulty is easily resolved when we realize that the DFT obtained by this procedure is\nactually the DFT of xn repeating periodically every T0 seconds. Figure 8.19c clearly indicates\nthat periodic repeating the segment of xn over the interval from −2 to 2 seconds yields the same\nsignal as the periodic repeating the segment of xn over the interval from 0 to 4 seconds. Hence,\nthe DFT of the samples taken from −2 to 2 seconds is the same as that of the samples taken\nfrom 0 to 4 seconds. Therefore, regardless of where x(t) starts, we can always take the samples\nof x(t) and its periodic extension over the interval from 0 to T0. In the present example, the 32\nsample values are", - "type": "text" - }, - { - "block_id": "p835-b9", - "global_id": 24384, - "bbox": [ - 224.66, - 575.08, - 242.89, - 586.54 - ], - "text": "xn =", - "type": "text" - }, - { - "block_id": "p835-b10", - "global_id": 24385, - "bbox": [ - 244.94, - 548.69, - 252.83, - 573.59 - ], - "text": "⎧\n⎪⎪⎨", - "type": "text" - }, - { - "block_id": "p835-b11", - "global_id": 24386, - "bbox": [ - 244.94, - 581.56, - 252.83, - 597.5 - ], - "text": "⎪⎪⎩", - "type": "text" - }, - { - "block_id": "p835-b12", - "global_id": 24387, - "bbox": [ - 252.83, - 557.06, - 405.8, - 567.43 - ], - "text": "1\n0 ≤n ≤3\nand\n29 ≤n ≤31", - "type": "text" - }, - { - "block_id": "p835-b13", - "global_id": 24388, - "bbox": [ - 252.83, - 574.98, - 328.35, - 585.36 - ], - "text": "0\n5 ≤n ≤27", - "type": "text" - }, - { - "block_id": "p835-b14", - "global_id": 24389, - "bbox": [ - 252.83, - 592.92, - 320.6, - 603.3 - ], - "text": "0.5\nn = 4,28", - "type": "text" - } - ] - }, - { - "page_num": 836, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p836-b0", - "global_id": 24390, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "816\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p836-b1", - "global_id": 24391, - "bbox": [ - 94.2, - 620.84, - 296.23, - 630.08 - ], - "text": "Figure 8.19 Discrete Fourier transform of a gate pulse.", - "type": "text" - } - ] - }, - { - "page_num": 837, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p837-b0", - "global_id": 24392, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n817", - "type": "text" - }, - { - "block_id": "p837-b1", - "global_id": 24393, - "bbox": [ - 128.9, - 85.83, - 502.76, - 121.2 - ], - "text": "Observe that the last sample is at t = 31/8, not at 4, because the signal repetition starts\nat t = 4, and the sample at t = 4 is the same as the sample at t = 0. Now, N0 = 32 and\n0 = 2π/32 = π/16. Therefore [see Eq. (8.12)],", - "type": "text" - }, - { - "block_id": "p837-b2", - "global_id": 24394, - "bbox": [ - 274.04, - 141.15, - 293.32, - 152.61 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p837-b3", - "global_id": 24395, - "bbox": [ - 295.37, - 131.27, - 309.47, - 141.64 - ], - "text": "31\n\"", - "type": "text" - }, - { - "block_id": "p837-b4", - "global_id": 24396, - "bbox": [ - 296.21, - 155.54, - 308.61, - 162.8 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p837-b5", - "global_id": 24397, - "bbox": [ - 310.57, - 139.43, - 357.1, - 152.23 - ], - "text": "xne−jr(π/16)n", - "type": "text" - }, - { - "block_id": "p837-b6", - "global_id": 24398, - "bbox": [ - 128.9, - 172.28, - 443.29, - 183.84 - ], - "text": "Values of Xr are computed according to this equation and plotted in Fig. 8.19d.", - "type": "text" - }, - { - "block_id": "p837-b7", - "global_id": 24399, - "bbox": [ - 128.91, - 183.92, - 502.77, - 242.11 - ], - "text": "The samples Xr are separated by f0 = 1/T0 Hz. In this case T0 = 4, so the frequency\nresolution f0 is 1/4 Hz, as desired. The folding frequency fs/2 = B = 4 Hz corresponds to\nr = N0/2 = 16. Because Xr is N0 periodic (N0 = 32), the values of Xr for r = −16 to n = −1\nare the same as those for r = 16 to n = 31. For instance, X17 = X−15, X18 = X−14, and so on.\nThe DFT gives us the samples of the spectrum X(ω).", - "type": "text" - }, - { - "block_id": "p837-b8", - "global_id": 24400, - "bbox": [ - 128.9, - 243.7, - 502.78, - 421.45 - ], - "text": "For the sake of comparison, Fig. 8.19d also shows the shaded curve 8sinc(ω/2), which is\nthe Fourier transform of 8rect(t). The values of Xr computed from the DFT equation show\naliasing error, which is clearly seen by comparing the two superimposed plots. The error\nin X2 is just about 1.3%. However, the aliasing error increases rapidly with r. For instance,\nthe error in X6 is about 12%, and the error in X10 is 33%. The error in X14 is a whopping\n72%. The percent error increases rapidly near the folding frequency (r = 16) because x(t)\nhas a jump discontinuity, which makes X(ω) decay slowly as 1/ω. Hence, near the folding\nfrequency, the inverted tail (due to aliasing) is very nearly equal to X(ω) itself. Moreover,\nthe final values are the difference between the exact and the folded values (which are very\nclose to the exact values). Hence, the percent error near the folding frequency (r = 16 in\nthis case) is very high, although the absolute error is very small. Clearly, for signals with\njump discontinuities, the aliasing error near the folding frequency will always be high (in\npercentage terms), regardless of the choice of N0. To ensure a negligible aliasing error at any\nvalue r, we must make sure that N0 ≫r. This observation is valid for all signals with jump\ndiscontinuities.", - "type": "text" - }, - { - "block_id": "p837-b9", - "global_id": 24401, - "bbox": [ - 128.9, - 448.1, - 502.77, - 486.45 - ], - "text": "USING MATLAB TO COMPUTE AND PLOT THE RESULTS\nOnce again, MATLAB lets us easily confirm the results of this example. First, parameters are\ndefined and MATLAB’s fft command is used to compute the DFT.", - "type": "text" - }, - { - "block_id": "p837-b10", - "global_id": 24402, - "bbox": [ - 128.9, - 496.36, - 463.65, - 530.23 - ], - "text": ">>\nT_0 = 4; N_0 = 32; T = T_0/N_0;\n>>\nx_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]’;\n>>\nX_r = fft(x_n); r = [-N_0/2:N_0/2-1]’; omega_r = r*2*pi/T_0;", - "type": "text" - }, - { - "block_id": "p837-b11", - "global_id": 24403, - "bbox": [ - 128.9, - 539.84, - 370.3, - 549.8 - ], - "text": "The true Fourier transform is also computed for comparison.", - "type": "text" - }, - { - "block_id": "p837-b12", - "global_id": 24404, - "bbox": [ - 128.9, - 560.0, - 463.65, - 569.96 - ], - "text": ">>\nomega = linspace(-pi/T,pi/T,5001); X = 8*sinc(omega/(2*pi));", - "type": "text" - }, - { - "block_id": "p837-b13", - "global_id": 24405, - "bbox": [ - 128.9, - 579.58, - 418.63, - 589.54 - ], - "text": "Since it is real, we can display the resulting spectrum using a single plot.", - "type": "text" - }, - { - "block_id": "p837-b14", - "global_id": 24406, - "bbox": [ - 128.9, - 599.73, - 379.96, - 621.64 - ], - "text": ">>\nclf; stem(omega_r,fftshift(real(X_r)),’k.’);\n>>\nline(omega,X,’color’,[0 0 0]);", - "type": "text" - } - ] - }, - { - "page_num": 838, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p838-b0", - "global_id": 24407, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "818\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p838-b1", - "global_id": 24408, - "bbox": [ - 103.16, - 86.66, - 380.38, - 96.63 - ], - "text": ">>\nxlabel(’\\omega’); ylabel(’X(\\omega)’); axis tight", - "type": "text" - }, - { - "block_id": "p838-b2", - "global_id": 24409, - "bbox": [ - 103.16, - 106.3, - 477.02, - 140.17 - ], - "text": "The result, shown in Fig. 8.20, matches the earlier result shown in Fig. 8.19d. The DFT\napproximation does not perfectly follow the true Fourier transform, especially at high\nfrequencies, because the parameter B is deliberately set too small.", - "type": "text" - }, - { - "block_id": "p838-b3", - "global_id": 24410, - "bbox": [ - 115.11, - 289.53, - 258.45, - 297.53 - ], - "text": "–25\n–20\n–15\n–10\n–5", - "type": "text" - }, - { - "block_id": "p838-b4", - "global_id": 24411, - "bbox": [ - 287.56, - 289.53, - 459.8, - 307.05 - ], - "text": "ω\n0\n5\n10\n15\n20\n25", - "type": "text" - }, - { - "block_id": "p838-b5", - "global_id": 24412, - "bbox": [ - 114.21, - 261.29, - 118.21, - 269.29 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p838-b6", - "global_id": 24413, - "bbox": [ - 114.21, - 239.43, - 118.21, - 247.43 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p838-b7", - "global_id": 24414, - "bbox": [ - 114.21, - 217.54, - 118.21, - 225.54 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p838-b8", - "global_id": 24415, - "bbox": [ - 114.21, - 195.67, - 118.21, - 203.67 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p838-b9", - "global_id": 24416, - "bbox": [ - 114.21, - 173.81, - 118.21, - 181.81 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p838-b10", - "global_id": 24417, - "bbox": [ - 102.41, - 221.45, - 110.7, - 238.04 - ], - "text": "X(ω)", - "type": "text" - }, - { - "block_id": "p838-b11", - "global_id": 24418, - "bbox": [ - 101.77, - 314.64, - 302.7, - 323.88 - ], - "text": "Figure 8.20 MATLAB-computed DFT of a gate pulse.", - "type": "text" - }, - { - "block_id": "p838-b12", - "global_id": 24419, - "bbox": [ - 101.84, - 376.28, - 283.67, - 388.24 - ], - "text": "8.5-1 Some Properties of the DFT", - "type": "text" - }, - { - "block_id": "p838-b13", - "global_id": 24420, - "bbox": [ - 101.84, - 394.37, - 490.42, - 428.24 - ], - "text": "The discrete Fourier transform is basically the Fourier transform of a sampled signal repeated\nperiodically. Hence, the properties derived earlier for the Fourier transform apply to the DFT as\nwell.", - "type": "text" - }, - { - "block_id": "p838-b14", - "global_id": 24421, - "bbox": [ - 102.14, - 444.15, - 160.5, - 456.27 - ], - "text": "LINEARITY", - "type": "text" - }, - { - "block_id": "p838-b15", - "global_id": 24422, - "bbox": [ - 101.84, - 459.89, - 236.3, - 471.76 - ], - "text": "If xn ⇐⇒Xr and gn ⇐⇒Gr, then", - "type": "text" - }, - { - "block_id": "p838-b16", - "global_id": 24423, - "bbox": [ - 237.01, - 484.16, - 354.57, - 495.62 - ], - "text": "a1xn + a2gn ⇐⇒a1Xr + a2Gr", - "type": "text" - }, - { - "block_id": "p838-b17", - "global_id": 24424, - "bbox": [ - 101.84, - 508.86, - 179.08, - 518.82 - ], - "text": "The proof is trivial.", - "type": "text" - }, - { - "block_id": "p838-b18", - "global_id": 24425, - "bbox": [ - 102.14, - 534.73, - 229.79, - 546.86 - ], - "text": "CONJUGATE SYMMETRY", - "type": "text" - }, - { - "block_id": "p838-b19", - "global_id": 24426, - "bbox": [ - 101.84, - 549.25, - 338.62, - 560.85 - ], - "text": "From the conjugation property x∗(t) ⇐⇒X∗(−ω), we have", - "type": "text" - }, - { - "block_id": "p838-b20", - "global_id": 24427, - "bbox": [ - 272.2, - 573.03, - 280.28, - 585.03 - ], - "text": "x∗", - "type": "text" - }, - { - "block_id": "p838-b21", - "global_id": 24428, - "bbox": [ - 276.62, - 573.03, - 315.29, - 587.39 - ], - "text": "n ←→X∗", - "type": "text" - }, - { - "block_id": "p838-b22", - "global_id": 24429, - "bbox": [ - 311.21, - 579.82, - 319.36, - 587.02 - ], - "text": "−r", - "type": "text" - }, - { - "block_id": "p838-b23", - "global_id": 24430, - "bbox": [ - 101.84, - 599.45, - 342.91, - 609.41 - ], - "text": "From this equation and the time-reversal property, we obtain", - "type": "text" - }, - { - "block_id": "p838-b24", - "global_id": 24431, - "bbox": [ - 271.7, - 621.58, - 279.78, - 633.58 - ], - "text": "x∗", - "type": "text" - }, - { - "block_id": "p838-b25", - "global_id": 24432, - "bbox": [ - 276.12, - 621.58, - 320.04, - 635.95 - ], - "text": "−n ←→X∗", - "type": "text" - }, - { - "block_id": "p838-b26", - "global_id": 24433, - "bbox": [ - 315.96, - 628.6, - 318.68, - 635.57 - ], - "text": "r", - "type": "text" - } - ] - }, - { - "page_num": 839, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p839-b0", - "global_id": 24434, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n819", - "type": "text" - }, - { - "block_id": "p839-b1", - "global_id": 24435, - "bbox": [ - 127.6, - 84.26, - 516.11, - 108.53 - ], - "text": "When x(t) is real, then the conjugate-symmetry property states that X∗(ω) = X(−ω). Hence,\nfor real xn,", - "type": "text" - }, - { - "block_id": "p839-b2", - "global_id": 24436, - "bbox": [ - 303.15, - 109.55, - 313.31, - 121.54 - ], - "text": "X∗", - "type": "text" - }, - { - "block_id": "p839-b3", - "global_id": 24437, - "bbox": [ - 309.23, - 111.27, - 339.91, - 123.91 - ], - "text": "r = X−r", - "type": "text" - }, - { - "block_id": "p839-b4", - "global_id": 24438, - "bbox": [ - 127.59, - 131.76, - 270.9, - 143.32 - ], - "text": "Moreover, Xr is N0 periodic. Hence,", - "type": "text" - }, - { - "block_id": "p839-b5", - "global_id": 24439, - "bbox": [ - 299.08, - 143.54, - 309.24, - 155.54 - ], - "text": "X∗", - "type": "text" - }, - { - "block_id": "p839-b6", - "global_id": 24440, - "bbox": [ - 305.16, - 145.26, - 343.97, - 157.91 - ], - "text": "r = XN0−r", - "type": "text" - }, - { - "block_id": "p839-b7", - "global_id": 24441, - "bbox": [ - 127.59, - 165.75, - 516.12, - 187.77 - ], - "text": "Because of this property, we need compute only half the DFTs for real xn. The other half are the\nconjugates.", - "type": "text" - }, - { - "block_id": "p839-b8", - "global_id": 24442, - "bbox": [ - 127.89, - 203.12, - 209.93, - 215.24 - ], - "text": "TIME SHIFTING", - "type": "text" - }, - { - "block_id": "p839-b9", - "global_id": 24443, - "bbox": [ - 127.59, - 218.68, - 333.4, - 229.24 - ], - "text": "The time-shifting (circular shifting) property states†", - "type": "text" - }, - { - "block_id": "p839-b10", - "global_id": 24444, - "bbox": [ - 283.34, - 240.3, - 359.77, - 253.48 - ], - "text": "xn−k ⇐⇒Xre−jr0k", - "type": "text" - }, - { - "block_id": "p839-b11", - "global_id": 24445, - "bbox": [ - 127.59, - 261.98, - 386.03, - 276.26 - ], - "text": "Proof. We use Eq. (8.13) to find the inverse DFT of Xre−jr0k as", - "type": "text" - }, - { - "block_id": "p839-b12", - "global_id": 24446, - "bbox": [ - 218.62, - 292.01, - 228.75, - 316.81 - ], - "text": "1\nN0", - "type": "text" - }, - { - "block_id": "p839-b13", - "global_id": 24447, - "bbox": [ - 231.55, - 287.79, - 248.61, - 299.09 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p839-b14", - "global_id": 24448, - "bbox": [ - 234.18, - 312.98, - 245.97, - 320.25 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p839-b15", - "global_id": 24449, - "bbox": [ - 249.72, - 292.01, - 328.62, - 309.66 - ], - "text": "Xre−jr0kejrω0n = 1", - "type": "text" - }, - { - "block_id": "p839-b16", - "global_id": 24450, - "bbox": [ - 320.81, - 305.97, - 330.94, - 316.81 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p839-b17", - "global_id": 24451, - "bbox": [ - 333.75, - 287.79, - 350.81, - 299.09 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p839-b18", - "global_id": 24452, - "bbox": [ - 336.38, - 312.98, - 348.18, - 320.25 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p839-b19", - "global_id": 24453, - "bbox": [ - 351.92, - 294.47, - 425.67, - 309.66 - ], - "text": "Xrejr0(n−k) = xn−k", - "type": "text" - }, - { - "block_id": "p839-b20", - "global_id": 24454, - "bbox": [ - 127.89, - 332.91, - 248.24, - 345.03 - ], - "text": "FREQUENCY SHIFTING", - "type": "text" - }, - { - "block_id": "p839-b21", - "global_id": 24455, - "bbox": [ - 127.59, - 349.07, - 423.39, - 359.03 - ], - "text": "A dual of the time-shifting property, the frequency-shifting property states", - "type": "text" - }, - { - "block_id": "p839-b22", - "global_id": 24456, - "bbox": [ - 283.93, - 367.7, - 359.28, - 382.89 - ], - "text": "xnejn0m ⇐⇒Xr−m", - "type": "text" - }, - { - "block_id": "p839-b23", - "global_id": 24457, - "bbox": [ - 127.59, - 395.31, - 516.13, - 417.3 - ], - "text": "Proof. This proof is identical to that of the time-shifting property except that we start with\nEq. (8.12).", - "type": "text" - }, - { - "block_id": "p839-b24", - "global_id": 24458, - "bbox": [ - 127.89, - 432.66, - 269.73, - 444.78 - ], - "text": "CIRCULAR CONVOLUTION", - "type": "text" - }, - { - "block_id": "p839-b25", - "global_id": 24459, - "bbox": [ - 127.59, - 448.81, - 338.93, - 458.78 - ], - "text": "The circular (or periodic) convolution property states", - "type": "text" - }, - { - "block_id": "p839-b26", - "global_id": 24460, - "bbox": [ - 286.57, - 471.56, - 516.13, - 483.01 - ], - "text": "xn ∗⃝gn ⇐⇒XrGr\n(8.18)", - "type": "text" - }, - { - "block_id": "p839-b27", - "global_id": 24461, - "bbox": [ - 127.59, - 495.14, - 141.97, - 505.1 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p839-b28", - "global_id": 24462, - "bbox": [ - 280.06, - 504.88, - 329.66, - 522.91 - ], - "text": "xngn ⇐⇒1", - "type": "text" - }, - { - "block_id": "p839-b29", - "global_id": 24463, - "bbox": [ - 321.85, - 518.84, - 331.98, - 529.68 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p839-b30", - "global_id": 24464, - "bbox": [ - 333.67, - 511.45, - 516.13, - 522.91 - ], - "text": "Xr ∗⃝Gr\n(8.19)", - "type": "text" - }, - { - "block_id": "p839-b31", - "global_id": 24465, - "bbox": [ - 127.59, - 536.4, - 481.02, - 547.96 - ], - "text": "For two N0-periodic sequences xn and gn, circular (or periodic) convolution is defined by", - "type": "text" - }, - { - "block_id": "p839-b32", - "global_id": 24466, - "bbox": [ - 252.4, - 569.49, - 289.55, - 580.95 - ], - "text": "xn ∗⃝gn =", - "type": "text" - }, - { - "block_id": "p839-b33", - "global_id": 24467, - "bbox": [ - 291.6, - 558.69, - 308.66, - 569.99 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p839-b34", - "global_id": 24468, - "bbox": [ - 294.07, - 583.88, - 306.2, - 591.15 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p839-b35", - "global_id": 24469, - "bbox": [ - 309.77, - 569.49, - 345.35, - 580.95 - ], - "text": "xkgn−k =", - "type": "text" - }, - { - "block_id": "p839-b36", - "global_id": 24470, - "bbox": [ - 347.4, - 558.69, - 364.46, - 569.99 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p839-b37", - "global_id": 24471, - "bbox": [ - 349.86, - 583.88, - 361.99, - 591.15 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p839-b38", - "global_id": 24472, - "bbox": [ - 365.57, - 569.81, - 516.13, - 580.57 - ], - "text": "gkxn−k\n(8.20)", - "type": "text" - }, - { - "block_id": "p839-b39", - "global_id": 24473, - "bbox": [ - 127.59, - 610.24, - 516.13, - 634.75 - ], - "text": "† Time shifting is also known as circular shifting because such a shift can be interpreted as a circular shift of\nthe N0 samples in the first cycle 0 ≤n ≤N0 −1.", - "type": "text" - } - ] - }, - { - "page_num": 840, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p840-b0", - "global_id": 24474, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "820\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p840-b1", - "global_id": 24475, - "bbox": [ - 173.59, - 180.73, - 234.71, - 190.48 - ], - "text": "n 0\ng3", - "type": "text" - }, - { - "block_id": "p840-b2", - "global_id": 24476, - "bbox": [ - 218.6, - 135.05, - 225.6, - 144.65 - ], - "text": "g0", - "type": "text" - }, - { - "block_id": "p840-b3", - "global_id": 24477, - "bbox": [ - 173.59, - 88.64, - 180.59, - 98.25 - ], - "text": "g1", - "type": "text" - }, - { - "block_id": "p840-b4", - "global_id": 24478, - "bbox": [ - 173.81, - 165.22, - 180.36, - 174.83 - ], - "text": "x1", - "type": "text" - }, - { - "block_id": "p840-b5", - "global_id": 24479, - "bbox": [ - 173.81, - 105.55, - 180.36, - 115.16 - ], - "text": "x3", - "type": "text" - }, - { - "block_id": "p840-b6", - "global_id": 24480, - "bbox": [ - 322.38, - 181.67, - 383.51, - 191.27 - ], - "text": "n 1\ng0", - "type": "text" - }, - { - "block_id": "p840-b7", - "global_id": 24481, - "bbox": [ - 127.91, - 135.05, - 374.3, - 144.65 - ], - "text": "g3\ng1\ng2", - "type": "text" - }, - { - "block_id": "p840-b8", - "global_id": 24482, - "bbox": [ - 322.38, - 88.64, - 329.38, - 98.25 - ], - "text": "g2", - "type": "text" - }, - { - "block_id": "p840-b9", - "global_id": 24483, - "bbox": [ - 322.6, - 165.69, - 329.16, - 175.3 - ], - "text": "x1", - "type": "text" - }, - { - "block_id": "p840-b10", - "global_id": 24484, - "bbox": [ - 143.9, - 135.05, - 299.61, - 144.65 - ], - "text": "x2\nx2", - "type": "text" - }, - { - "block_id": "p840-b11", - "global_id": 24485, - "bbox": [ - 322.6, - 105.08, - 329.16, - 114.69 - ], - "text": "x3", - "type": "text" - }, - { - "block_id": "p840-b12", - "global_id": 24486, - "bbox": [ - 204.06, - 135.05, - 358.85, - 144.65 - ], - "text": "x0\nx0", - "type": "text" - }, - { - "block_id": "p840-b13", - "global_id": 24487, - "bbox": [ - 125.76, - 198.44, - 336.4, - 207.68 - ], - "text": "Figure 8.21 Graphical depictions of circular convolution.", - "type": "text" - }, - { - "block_id": "p840-b14", - "global_id": 24488, - "bbox": [ - 101.84, - 232.53, - 398.99, - 244.4 - ], - "text": "To prove Eq. (8.18), we find the DFT of the circular convolution xn ∗⃝gn as", - "type": "text" - }, - { - "block_id": "p840-b15", - "global_id": 24489, - "bbox": [ - 186.45, - 258.54, - 203.51, - 269.84 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p840-b16", - "global_id": 24490, - "bbox": [ - 188.77, - 283.73, - 201.18, - 291.0 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p840-b17", - "global_id": 24491, - "bbox": [ - 204.62, - 252.36, - 228.59, - 269.84 - ], - "text": "8N0−1\n\"", - "type": "text" - }, - { - "block_id": "p840-b18", - "global_id": 24492, - "bbox": [ - 213.99, - 283.73, - 226.13, - 291.0 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p840-b19", - "global_id": 24493, - "bbox": [ - 229.7, - 269.66, - 254.83, - 280.42 - ], - "text": "xkgn−k", - "type": "text" - }, - { - "block_id": "p840-b20", - "global_id": 24494, - "bbox": [ - 255.46, - 252.36, - 262.36, - 262.32 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p840-b21", - "global_id": 24495, - "bbox": [ - 263.47, - 265.22, - 300.0, - 279.62 - ], - "text": "e−jrω0n =", - "type": "text" - }, - { - "block_id": "p840-b22", - "global_id": 24496, - "bbox": [ - 302.05, - 258.54, - 319.11, - 269.84 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p840-b23", - "global_id": 24497, - "bbox": [ - 304.52, - 283.73, - 316.65, - 291.0 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p840-b24", - "global_id": 24498, - "bbox": [ - 320.22, - 269.66, - 327.74, - 280.42 - ], - "text": "xk", - "type": "text" - }, - { - "block_id": "p840-b25", - "global_id": 24499, - "bbox": [ - 329.46, - 252.36, - 353.43, - 269.84 - ], - "text": "8N0−1\n\"", - "type": "text" - }, - { - "block_id": "p840-b26", - "global_id": 24500, - "bbox": [ - 338.7, - 283.73, - 351.1, - 291.0 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p840-b27", - "global_id": 24501, - "bbox": [ - 354.55, - 267.61, - 398.38, - 280.42 - ], - "text": "gn−ke−jrω0n", - "type": "text" - }, - { - "block_id": "p840-b28", - "global_id": 24502, - "bbox": [ - 398.88, - 252.36, - 405.78, - 262.32 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p840-b29", - "global_id": 24503, - "bbox": [ - 292.23, - 305.56, - 300.0, - 315.52 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p840-b30", - "global_id": 24504, - "bbox": [ - 302.05, - 294.77, - 319.11, - 306.05 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p840-b31", - "global_id": 24505, - "bbox": [ - 304.52, - 319.95, - 316.65, - 327.21 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p840-b32", - "global_id": 24506, - "bbox": [ - 320.22, - 303.84, - 404.89, - 316.64 - ], - "text": "xk(Gre−jr0k) = XrGr", - "type": "text" - }, - { - "block_id": "p840-b33", - "global_id": 24507, - "bbox": [ - 101.84, - 341.55, - 291.16, - 351.51 - ], - "text": "Equation (8.19) can be proved in the same way.", - "type": "text" - }, - { - "block_id": "p840-b34", - "global_id": 24508, - "bbox": [ - 101.84, - 353.51, - 490.4, - 448.65 - ], - "text": "For periodic sequences, the convolution can be visualized in terms of two sequences, with\none sequence fixed and the other inverted and moved past the fixed sequence, one digit at a\ntime. If the two sequences are N0 periodic, the same configuration will repeat after N0 shifts\nof the sequence. Clearly the convolution xn ∗⃝gn becomes N0 periodic. Such convolution can be\nconveniently visualized in terms of N0 sequences, as illustrated in Fig. 8.21, for the case of N0 = 4.\nThe inner N0-point sequence xn is clockwise and fixed. The outer N0-point sequence gn is inverted\nso that it becomes counterclockwise. This sequence is now rotated clockwise 1 unit at a time. We\nmultiply the overlapping numbers and add. For example, the value of xn ∗⃝gn at n = 0 (Fig. 8.21) is", - "type": "text" - }, - { - "block_id": "p840-b35", - "global_id": 24509, - "bbox": [ - 245.06, - 463.35, - 346.67, - 474.8 - ], - "text": "x0g0 + x1g3 + x2g2 + x3g1", - "type": "text" - }, - { - "block_id": "p840-b36", - "global_id": 24510, - "bbox": [ - 101.84, - 489.92, - 283.29, - 501.79 - ], - "text": "and the value of xn ∗⃝gn at n = 1 is (Fig. 8.21)", - "type": "text" - }, - { - "block_id": "p840-b37", - "global_id": 24511, - "bbox": [ - 245.06, - 516.48, - 346.67, - 527.94 - ], - "text": "x0g1 + x1g0 + x2g3 + x3g2", - "type": "text" - }, - { - "block_id": "p840-b38", - "global_id": 24512, - "bbox": [ - 101.84, - 543.47, - 142.52, - 553.43 - ], - "text": "and so on.", - "type": "text" - }, - { - "block_id": "p840-b39", - "global_id": 24513, - "bbox": [ - 101.84, - 582.83, - 298.28, - 594.78 - ], - "text": "8.5-2 Some Applications of the DFT", - "type": "text" - }, - { - "block_id": "p840-b40", - "global_id": 24514, - "bbox": [ - 101.84, - 600.91, - 490.39, - 634.79 - ], - "text": "The DFT is useful not only in the computation of direct and inverse Fourier transforms, but\nalso in other applications such as convolution, correlation, and filtering. Use of the efficient FFT\nalgorithm, discussed shortly (Sec. 8.6), makes it particularly appealing.", - "type": "text" - } - ] - }, - { - "page_num": 841, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p841-b0", - "global_id": 24515, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n821", - "type": "text" - }, - { - "block_id": "p841-b1", - "global_id": 24516, - "bbox": [ - 127.89, - 86.19, - 254.7, - 98.32 - ], - "text": "LINEAR CONVOLUTION", - "type": "text" - }, - { - "block_id": "p841-b2", - "global_id": 24517, - "bbox": [ - 127.59, - 101.93, - 516.13, - 148.18 - ], - "text": "Let x(t) and g(t) be the two signals to be convolved. In general, these signals may have different\ntime durations. To convolve them by using their samples, they must be sampled at the same rate\n(not below the Nyquist rate of either signal). Let xn (0 ≤n ≤N1 −1) and gn (0 ≤n ≤N2 −1) be\nthe corresponding discrete sequences representing these samples. Now,", - "type": "text" - }, - { - "block_id": "p841-b3", - "global_id": 24518, - "bbox": [ - 289.28, - 159.25, - 354.44, - 169.52 - ], - "text": "c(t) = x(t) ∗g(t)", - "type": "text" - }, - { - "block_id": "p841-b4", - "global_id": 24519, - "bbox": [ - 127.6, - 180.51, - 470.39, - 192.14 - ], - "text": "and if we define three sequences as xn = Tx(nT), gn = Tg(nT), and cn = Tc(nT), then†", - "type": "text" - }, - { - "block_id": "p841-b5", - "global_id": 24520, - "bbox": [ - 298.89, - 202.13, - 344.33, - 213.59 - ], - "text": "cn = xn ∗gn", - "type": "text" - }, - { - "block_id": "p841-b6", - "global_id": 24521, - "bbox": [ - 127.59, - 223.89, - 456.15, - 235.45 - ], - "text": "where we define the linear convolution sum of two discrete sequences xn and gn as", - "type": "text" - }, - { - "block_id": "p841-b7", - "global_id": 24522, - "bbox": [ - 268.92, - 253.15, - 324.68, - 264.61 - ], - "text": "cn = xn ∗gn =", - "type": "text" - }, - { - "block_id": "p841-b8", - "global_id": 24523, - "bbox": [ - 330.28, - 242.98, - 344.38, - 253.65 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p841-b9", - "global_id": 24524, - "bbox": [ - 326.73, - 267.44, - 347.93, - 274.64 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p841-b10", - "global_id": 24525, - "bbox": [ - 349.05, - 253.47, - 374.18, - 264.23 - ], - "text": "xkgn−k", - "type": "text" - }, - { - "block_id": "p841-b11", - "global_id": 24526, - "bbox": [ - 127.59, - 284.0, - 516.16, - 415.42 - ], - "text": "Because of the width property of the convolution, cn exists for 0 ≤n ≤N1+N2−1. To be able to use\nthe DFT circular convolution technique, we must make sure that the circular convolution will yield\nthe same result as does linear convolution. In other words, the signal resulting from the circular\nconvolution must have the same length (N1 + N2 −1) as that of the signal resulting from linear\nconvolution. This step can be accomplished by adding N2 −1 dummy samples of zero value to xn\nand N1 −1 dummy samples of zero value to gn (zero padding). This procedure changes the length\nof both xn and gn to N1 +N2 −1. The circular convolution now is identical to the linear convolution\nexcept that it repeats periodically with period N1 +N2 −1. A little reflection will show that in such\na case the circular convolution procedure in Fig. 8.21 over one cycle (0 ≤n ≤N1 + N2 −1) is\nidentical to the linear convolution of the two sequences xn and gn. We can use the DFT to find the\nconvolution xn ∗gn in three steps, as follows:", - "type": "text" - }, - { - "block_id": "p841-b12", - "global_id": 24527, - "bbox": [ - 144.52, - 421.8, - 516.14, - 467.73 - ], - "text": "1. Find the DFTs Xr and Gr corresponding to suitably padded xn and gn.\n2. Multiply Xr by Gr.\n3. Find the IDFT of XrGr. This procedure of convolution, when implemented by the fast\nFourier transform algorithm (discussed later), is known as fast convolution.", - "type": "text" - }, - { - "block_id": "p841-b13", - "global_id": 24528, - "bbox": [ - 127.59, - 482.35, - 516.14, - 592.15 - ], - "text": "FILTERING\nWe generally think of filtering in terms of a hardware-oriented solution (e.g., building a circuit\nwith RLC components and operational amplifiers). However, filtering also has a software-oriented\nsolution [a computer algorithm that yields the filtered output y(t) for a given input x(t)]. This goal\ncan be conveniently accomplished by using the DFT. If x(t) is the signal to be filtered, then Xr,\nthe DFT of xn, is found. The spectrum Xr is then shaped (filtered) as desired by multiplying Xr by\nHr, where Hr are the samples of H(ω) for the filter [Hr = H(rω0)]. Finally, we take the IDFT of\nXrHr to obtain the filtered output yn[yn =Ty(nT)]. This procedure is demonstrated in the following\nexample.", - "type": "text" - }, - { - "block_id": "p841-b14", - "global_id": 24529, - "bbox": [ - 127.59, - 610.24, - 516.12, - 633.41 - ], - "text": "† We can show that cn = limT→0 xn ∗gn; [see 4]. Error is inherent in any numerical method used to compute\nconvolution of continuous-time signals; since T̸ = 0 in practice, there will be some error in this equation.", - "type": "text" - } - ] - }, - { - "page_num": 842, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p842-b0", - "global_id": 24530, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "822\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p842-b1", - "global_id": 24531, - "bbox": [ - 76.77, - 93.92, - 355.6, - 105.87 - ], - "text": "EXAMPLE 8.10\nDFT to Determine Filter Output", - "type": "text" - }, - { - "block_id": "p842-b2", - "global_id": 24532, - "bbox": [ - 103.16, - 122.12, - 477.0, - 144.45 - ], - "text": "The signal x(t) in Fig. 8.22a is passed through an ideal lowpass filter of frequency response\nH(ω) depicted in Fig. 8.22b. Use the DFT to find the sampled version of the filter output.", - "type": "text" - }, - { - "block_id": "p842-b3", - "global_id": 24533, - "bbox": [ - 138.02, - 328.99, - 394.97, - 337.34 - ], - "text": "16\n32\n4\n16\n32", - "type": "text" - }, - { - "block_id": "p842-b4", - "global_id": 24534, - "bbox": [ - 266.97, - 353.35, - 392.16, - 361.35 - ], - "text": "0\n2\n4\n6\n8", - "type": "text" - }, - { - "block_id": "p842-b5", - "global_id": 24535, - "bbox": [ - 338.6, - 261.97, - 347.93, - 269.97 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p842-b6", - "global_id": 24536, - "bbox": [ - 262.29, - 283.49, - 266.29, - 291.49 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p842-b7", - "global_id": 24537, - "bbox": [ - 108.57, - 328.57, - 364.57, - 337.34 - ], - "text": "8\n24\n24\n40", - "type": "text" - }, - { - "block_id": "p842-b8", - "global_id": 24538, - "bbox": [ - 110.44, - 352.8, - 424.17, - 361.7 - ], - "text": "6\n8\n4\n2\n10\n10", - "type": "text" - }, - { - "block_id": "p842-b9", - "global_id": 24539, - "bbox": [ - 416.83, - 328.99, - 424.83, - 336.99 - ], - "text": "40", - "type": "text" - }, - { - "block_id": "p842-b10", - "global_id": 24540, - "bbox": [ - 272.79, - 282.66, - 280.9, - 292.2 - ], - "text": "Hr", - "type": "text" - }, - { - "block_id": "p842-b11", - "global_id": 24541, - "bbox": [ - 347.27, - 193.52, - 363.71, - 201.62 - ], - "text": "H(v)", - "type": "text" - }, - { - "block_id": "p842-b12", - "global_id": 24542, - "bbox": [ - 307.82, - 241.23, - 375.65, - 250.62 - ], - "text": "2\n2", - "type": "text" - }, - { - "block_id": "p842-b13", - "global_id": 24543, - "bbox": [ - 336.75, - 334.0, - 339.86, - 342.0 - ], - "text": "r", - "type": "text" - }, - { - "block_id": "p842-b14", - "global_id": 24544, - "bbox": [ - 264.53, - 376.91, - 273.41, - 384.91 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p842-b15", - "global_id": 24545, - "bbox": [ - 264.3, - 535.0, - 273.63, - 543.0 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p842-b16", - "global_id": 24546, - "bbox": [ - 337.85, - 199.67, - 341.85, - 207.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p842-b17", - "global_id": 24547, - "bbox": [ - 142.07, - 241.94, - 186.4, - 250.23 - ], - "text": "0.5\n0.5", - "type": "text" - }, - { - "block_id": "p842-b18", - "global_id": 24548, - "bbox": [ - 159.41, - 193.54, - 183.31, - 204.31 - ], - "text": "8\nx(t)", - "type": "text" - }, - { - "block_id": "p842-b19", - "global_id": 24549, - "bbox": [ - 160.83, - 262.39, - 169.71, - 270.39 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p842-b20", - "global_id": 24550, - "bbox": [ - 217.77, - 395.43, - 224.32, - 404.97 - ], - "text": "yn", - "type": "text" - }, - { - "block_id": "p842-b21", - "global_id": 24551, - "bbox": [ - 184.15, - 410.95, - 188.15, - 418.95 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p842-b22", - "global_id": 24552, - "bbox": [ - 105.94, - 515.59, - 415.95, - 524.99 - ], - "text": "8\n16\n24\n31\n16\n8", - "type": "text" - }, - { - "block_id": "p842-b23", - "global_id": 24553, - "bbox": [ - 232.57, - 328.27, - 258.49, - 336.57 - ], - "text": "8 4", - "type": "text" - }, - { - "block_id": "p842-b24", - "global_id": 24554, - "bbox": [ - 217.07, - 516.67, - 388.44, - 529.51 - ], - "text": "0\nn", - "type": "text" - }, - { - "block_id": "p842-b25", - "global_id": 24555, - "bbox": [ - 198.67, - 241.75, - 200.9, - 249.75 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p842-b26", - "global_id": 24556, - "bbox": [ - 332.43, - 361.24, - 352.51, - 369.32 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p842-b27", - "global_id": 24557, - "bbox": [ - 384.82, - 248.07, - 404.9, - 256.15 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p842-b28", - "global_id": 24558, - "bbox": [ - 101.77, - 549.33, - 313.4, - 558.94 - ], - "text": "Figure 8.22 DFT solution for filtering x(t) through H(ω).", - "type": "text" - }, - { - "block_id": "p842-b29", - "global_id": 24559, - "bbox": [ - 103.17, - 586.19, - 477.02, - 621.98 - ], - "text": "We have already found the 32-point DFT of x(t) (see Fig. 8.19d). Next we multiply Xr by\nHr. To find Hr, we recall using f0 = 1/4 in computing the 32-point DFT of x(t). Because Xr is\n32-periodic, Hr must also be 32-periodic with samples separated by 1/4 Hz. This fact means", - "type": "text" - } - ] - }, - { - "page_num": 843, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p843-b0", - "global_id": 24560, - "bbox": [ - 255.93, - 62.89, - 516.14, - 71.98 - ], - "text": "8.5\nNumerical Computation of the Fourier Transform\n823", - "type": "text" - }, - { - "block_id": "p843-b1", - "global_id": 24561, - "bbox": [ - 128.9, - 85.83, - 502.76, - 109.65 - ], - "text": "that Hr must be repeated every 8 Hz or 16π rad/s (see Fig. 8.22c). The resulting 32 samples of\nHr over (0 ≤ω ≤16π) are as follows:", - "type": "text" - }, - { - "block_id": "p843-b2", - "global_id": 24562, - "bbox": [ - 219.49, - 134.89, - 239.87, - 146.34 - ], - "text": "Hr =", - "type": "text" - }, - { - "block_id": "p843-b3", - "global_id": 24563, - "bbox": [ - 241.92, - 114.45, - 249.81, - 133.39 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p843-b4", - "global_id": 24564, - "bbox": [ - 241.92, - 141.36, - 249.81, - 151.32 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p843-b5", - "global_id": 24565, - "bbox": [ - 249.81, - 122.82, - 406.0, - 157.11 - ], - "text": "1\n0 ≤r ≤7\nand\n25 ≤r ≤31\n0\n9 ≤r ≤23\n0.5\nr = 8,24", - "type": "text" - }, - { - "block_id": "p843-b6", - "global_id": 24566, - "bbox": [ - 128.9, - 172.37, - 502.76, - 195.09 - ], - "text": "We multiply Xr with Hr. The desired output signal samples yn are found by taking the inverse\nDFT of XrHr. The resulting output signal is illustrated in Fig. 8.22d.", - "type": "text" - }, - { - "block_id": "p843-b7", - "global_id": 24567, - "bbox": [ - 128.9, - 196.37, - 502.78, - 219.0 - ], - "text": "It is quite simple to verify the results of this filtering example using MATLAB. First,\nparameters are defined, and MATLAB’s fft command is used to compute the DFT of xn.", - "type": "text" - }, - { - "block_id": "p843-b8", - "global_id": 24568, - "bbox": [ - 128.91, - 232.53, - 489.8, - 254.45 - ], - "text": ">>\nT_0 = 4; N_0 = 32; T = T_0/N_0; n = (0:N_0-1); r = n;\n>>\nx_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]’; X_r = fft(x_n);", - "type": "text" - }, - { - "block_id": "p843-b9", - "global_id": 24569, - "bbox": [ - 128.91, - 271.99, - 502.75, - 295.51 - ], - "text": "The DFT of the filter’s output is just the product of the filter response Hr and the input DFT\nXr. The output yn is obtained using the ifft command and then plotted.", - "type": "text" - }, - { - "block_id": "p843-b10", - "global_id": 24570, - "bbox": [ - 128.9, - 308.24, - 406.12, - 354.07 - ], - "text": ">>\nH_r = [ones(1,8) 0.5 zeros(1,15) 0.5 ones(1,7)]’;\n>>\nY_r = H_r.*X_r; y_n = ifft(Y_r);\n>>\nclf; stem(n,real(y_n),’k.’);\n>>\nxlabel(’n’); ylabel(’y_n’); axis([0 31 -.1 1.1]);", - "type": "text" - }, - { - "block_id": "p843-b11", - "global_id": 24571, - "bbox": [ - 128.9, - 371.72, - 502.79, - 405.6 - ], - "text": "The result, shown in Fig. 8.23, matches the earlier result shown in Fig. 8.22d. Recall, this\nDFT-based approach shows the samples yn of the filter output y(t) (sampled in this case at a\nrate T = 1", - "type": "text" - }, - { - "block_id": "p843-b12", - "global_id": 24572, - "bbox": [ - 128.91, - 395.22, - 502.76, - 419.04 - ], - "text": "8) over 0 ≤n ≤N0 −1 = 31 when the input pulse x(t) is periodically replicated to\nform samples xn (see Fig. 8.19c).", - "type": "text" - }, - { - "block_id": "p843-b13", - "global_id": 24573, - "bbox": [ - 164.66, - 564.16, - 494.11, - 585.34 - ], - "text": "0\n5\n10\n15\n20\n25\n30\nn", - "type": "text" - }, - { - "block_id": "p843-b14", - "global_id": 24574, - "bbox": [ - 158.43, - 546.12, - 162.43, - 554.12 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p843-b15", - "global_id": 24575, - "bbox": [ - 151.68, - 528.49, - 162.36, - 536.49 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p843-b16", - "global_id": 24576, - "bbox": [ - 151.68, - 510.87, - 162.36, - 518.87 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p843-b17", - "global_id": 24577, - "bbox": [ - 151.68, - 493.25, - 162.36, - 501.25 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p843-b18", - "global_id": 24578, - "bbox": [ - 151.68, - 475.62, - 162.36, - 483.62 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p843-b19", - "global_id": 24579, - "bbox": [ - 158.43, - 458.0, - 162.43, - 466.0 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p843-b20", - "global_id": 24580, - "bbox": [ - 137.18, - 501.43, - 148.45, - 508.92 - ], - "text": "yn", - "type": "text" - }, - { - "block_id": "p843-b21", - "global_id": 24581, - "bbox": [ - 136.48, - 592.02, - 388.85, - 601.26 - ], - "text": "Figure 8.23 Using MATLAB and the DFT to determine filter output.", - "type": "text" - } - ] - }, - { - "page_num": 844, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p844-b0", - "global_id": 24582, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "824\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p844-b1", - "global_id": 24583, - "bbox": [ - 102.2, - 94.37, - 379.26, - 108.31 - ], - "text": "8.6 THE FAST FOURIER TRANSFORM (FFT)", - "type": "text" - }, - { - "block_id": "p844-b2", - "global_id": 24584, - "bbox": [ - 101.84, - 114.3, - 490.4, - 148.18 - ], - "text": "The number of computations required in performing the DFT was dramatically reduced by an\nalgorithm developed by Cooley and Tukey in 1965 [5]. This algorithm, known as the fast Fourier\ntransform (FFT), reduces the number of computations from something on the order of N2", - "type": "text" - }, - { - "block_id": "p844-b3", - "global_id": 24585, - "bbox": [ - 101.84, - 138.21, - 490.39, - 184.04 - ], - "text": "0 to\nN0 logN0. To compute one sample Xr from Eq. (8.12), we require N0 complex multiplications\nand N0 −1 complex additions. To compute N0 such values (Xr for r = 0,1,. . .,N0 −1), we require\na total of N2", - "type": "text" - }, - { - "block_id": "p844-b4", - "global_id": 24586, - "bbox": [ - 101.85, - 173.66, - 490.39, - 207.95 - ], - "text": "0 complex multiplications and N0(N0 −1) complex additions. For a large N0, these\ncomputations can be prohibitively time-consuming, even for a high-speed computer. The FFT\nalgorithm is what made the use of Fourier transform accessible for digital signal processing.", - "type": "text" - }, - { - "block_id": "p844-b5", - "global_id": 24587, - "bbox": [ - 101.84, - 223.15, - 490.4, - 297.08 - ], - "text": "HOW DOES THE FFT REDUCE THE NUMBER OF COMPUTATIONS?\nIt is easy to understand the magic of the FFT. The secret is in the linearity of the Fourier transform\nand also of the DFT. Because of linearity, we can compute the Fourier transform of a signal x(t) as\na sum of the Fourier transforms of segments of x(t) of shorter duration. The same principle applies\nto the computation of the DFT. Consider a signal of length N0 = 16 samples. As seen earlier, DFT\ncomputation of this sequence requires N2", - "type": "text" - }, - { - "block_id": "p844-b6", - "global_id": 24588, - "bbox": [ - 101.84, - 286.7, - 490.4, - 453.27 - ], - "text": "0 = 256 multiplications and N0(N0 −1) = 240 additions.\nWe can split this sequence into two shorter sequences, each of length 8. To compute DFT of\neach of these segments, we need 64 multiplications and 56 additions. Thus, we need a total of\n128 multiplications and 112 additions. Suppose, we split the original sequence in four segments\nof length 4 each. To compute the DFT of each segment, we require 16 multiplications and 12\nadditions. Hence, we need a total of 64 multiplications and 48 additions. If we split the sequence\nin eight segments of length 2 each, we need 4 multiplications and 2 additions for each segment,\nresulting in a total of 32 multiplications and 8 additions. Thus, we have been able to reduce the\nnumber of multiplications from 256 to 32 and the number of additions from 240 to 8. Moreover,\nsome of these multiplications turn out to be multiplications by 1 or −1. All this fantastic economy\nin the number of computations is realized by the FFT without any approximation! The values\nobtained by the FFT are identical to those obtained by the DFT. In this example, we considered\na relatively small value of N0 = 16. The reduction in the number of computations is much more\ndramatic for higher values of N0.", - "type": "text" - }, - { - "block_id": "p844-b7", - "global_id": 24589, - "bbox": [ - 101.84, - 454.39, - 490.36, - 476.41 - ], - "text": "The FFT algorithm is simplified if we choose N0 to be a power of 2, although such a choice\nis not essential. For convenience, we define", - "type": "text" - }, - { - "block_id": "p844-b8", - "global_id": 24590, - "bbox": [ - 246.71, - 484.75, - 344.52, - 501.82 - ], - "text": "WN0 = e−(j2π/N0) = e−j0", - "type": "text" - }, - { - "block_id": "p844-b9", - "global_id": 24591, - "bbox": [ - 101.84, - 512.12, - 128.13, - 522.08 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p844-b10", - "global_id": 24592, - "bbox": [ - 224.34, - 533.22, - 243.62, - 544.68 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p844-b11", - "global_id": 24593, - "bbox": [ - 245.66, - 522.42, - 262.73, - 533.72 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p844-b12", - "global_id": 24594, - "bbox": [ - 247.99, - 547.61, - 260.4, - 554.88 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p844-b13", - "global_id": 24595, - "bbox": [ - 263.84, - 531.71, - 287.47, - 544.3 - ], - "text": "xnWnr", - "type": "text" - }, - { - "block_id": "p844-b14", - "global_id": 24596, - "bbox": [ - 280.55, - 533.22, - 490.38, - 546.81 - ], - "text": "N0\n0 ≤r ≤N0 −1\n(8.21)", - "type": "text" - }, - { - "block_id": "p844-b15", - "global_id": 24597, - "bbox": [ - 101.85, - 562.5, - 116.23, - 572.46 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p844-b16", - "global_id": 24598, - "bbox": [ - 214.65, - 577.04, - 243.92, - 595.06 - ], - "text": "xn = 1", - "type": "text" - }, - { - "block_id": "p844-b17", - "global_id": 24599, - "bbox": [ - 236.11, - 590.99, - 246.24, - 601.83 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p844-b18", - "global_id": 24600, - "bbox": [ - 249.05, - 572.8, - 266.11, - 584.1 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p844-b19", - "global_id": 24601, - "bbox": [ - 251.68, - 597.99, - 263.48, - 605.26 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p844-b20", - "global_id": 24602, - "bbox": [ - 267.22, - 581.75, - 297.34, - 594.68 - ], - "text": "XrW−nr", - "type": "text" - }, - { - "block_id": "p844-b21", - "global_id": 24603, - "bbox": [ - 284.98, - 583.61, - 490.38, - 597.33 - ], - "text": "N0\n0 ≤n ≤N0 −1\n(8.22)", - "type": "text" - }, - { - "block_id": "p844-b22", - "global_id": 24604, - "bbox": [ - 101.85, - 612.86, - 490.37, - 634.79 - ], - "text": "Although there are many variations of the Tukey–Cooley algorithm, these can be grouped into two\nbasic types: decimation in time and decimation in frequency.", - "type": "text" - } - ] - }, - { - "page_num": 845, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p845-b0", - "global_id": 24605, - "bbox": [ - 321.17, - 62.89, - 516.12, - 71.98 - ], - "text": "8.6\nThe Fast Fourier Transform (FFT)\n825", - "type": "text" - }, - { - "block_id": "p845-b1", - "global_id": 24606, - "bbox": [ - 127.89, - 86.19, - 344.61, - 98.32 - ], - "text": "THE DECIMATION-IN-TIME ALGORITHM", - "type": "text" - }, - { - "block_id": "p845-b2", - "global_id": 24607, - "bbox": [ - 127.59, - 101.93, - 516.11, - 124.26 - ], - "text": "Here we divide the N0-point data sequence xn into two (N0/2)-point sequences consisting of even-\nand odd-numbered samples, respectively, as follows:", - "type": "text" - }, - { - "block_id": "p845-b3", - "global_id": 24608, - "bbox": [ - 244.15, - 136.4, - 320.06, - 162.69 - ], - "text": "x0,x2,x4,. . .,xN0−2\n\n\n\nsequence gn", - "type": "text" - }, - { - "block_id": "p845-b4", - "global_id": 24609, - "bbox": [ - 320.06, - 136.4, - 399.57, - 162.69 - ], - "text": ",x1,x3,x5,. . .,xN0−1\n\n\n\nsequence hn", - "type": "text" - }, - { - "block_id": "p845-b5", - "global_id": 24610, - "bbox": [ - 127.59, - 173.58, - 217.51, - 183.55 - ], - "text": "Then, from Eq. (8.21),", - "type": "text" - }, - { - "block_id": "p845-b6", - "global_id": 24611, - "bbox": [ - 232.83, - 205.97, - 252.1, - 217.43 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p845-b7", - "global_id": 24612, - "bbox": [ - 254.15, - 195.17, - 283.18, - 206.47 - ], - "text": "(N0/2)−1\n\"", - "type": "text" - }, - { - "block_id": "p845-b8", - "global_id": 24613, - "bbox": [ - 262.47, - 220.36, - 274.87, - 227.63 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p845-b9", - "global_id": 24614, - "bbox": [ - 284.29, - 204.47, - 314.9, - 217.12 - ], - "text": "x2nW2nr", - "type": "text" - }, - { - "block_id": "p845-b10", - "global_id": 24615, - "bbox": [ - 304.49, - 205.97, - 324.88, - 220.11 - ], - "text": "N0 +", - "type": "text" - }, - { - "block_id": "p845-b11", - "global_id": 24616, - "bbox": [ - 326.42, - 195.17, - 355.45, - 206.47 - ], - "text": "(N0/2)−1\n\"", - "type": "text" - }, - { - "block_id": "p845-b12", - "global_id": 24617, - "bbox": [ - 334.74, - 220.36, - 347.14, - 227.63 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p845-b13", - "global_id": 24618, - "bbox": [ - 356.57, - 203.48, - 410.22, - 217.12 - ], - "text": "x2n+1W(2n+1)r", - "type": "text" - }, - { - "block_id": "p845-b14", - "global_id": 24619, - "bbox": [ - 385.69, - 211.39, - 393.33, - 219.7 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p845-b15", - "global_id": 24620, - "bbox": [ - 127.59, - 237.92, - 171.86, - 247.88 - ], - "text": "Also, since", - "type": "text" - }, - { - "block_id": "p845-b16", - "global_id": 24621, - "bbox": [ - 295.6, - 248.92, - 343.69, - 262.76 - ], - "text": "WN0/2 = W2", - "type": "text" - }, - { - "block_id": "p845-b17", - "global_id": 24622, - "bbox": [ - 339.47, - 255.65, - 347.11, - 263.95 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p845-b18", - "global_id": 24623, - "bbox": [ - 127.59, - 270.3, - 160.14, - 280.26 - ], - "text": "we have", - "type": "text" - }, - { - "block_id": "p845-b19", - "global_id": 24624, - "bbox": [ - 226.55, - 300.62, - 245.82, - 312.08 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p845-b20", - "global_id": 24625, - "bbox": [ - 247.87, - 289.82, - 276.89, - 301.12 - ], - "text": "(N0/2)−1\n\"", - "type": "text" - }, - { - "block_id": "p845-b21", - "global_id": 24626, - "bbox": [ - 256.18, - 315.01, - 268.58, - 322.27 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p845-b22", - "global_id": 24627, - "bbox": [ - 278.01, - 299.12, - 305.12, - 311.77 - ], - "text": "x2nWnr", - "type": "text" - }, - { - "block_id": "p845-b23", - "global_id": 24628, - "bbox": [ - 298.2, - 299.12, - 336.21, - 314.22 - ], - "text": "N0/2 + Wr", - "type": "text" - }, - { - "block_id": "p845-b24", - "global_id": 24629, - "bbox": [ - 332.77, - 305.91, - 340.41, - 314.22 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p845-b25", - "global_id": 24630, - "bbox": [ - 342.51, - 289.82, - 371.54, - 301.12 - ], - "text": "(N0/2)−1\n\"", - "type": "text" - }, - { - "block_id": "p845-b26", - "global_id": 24631, - "bbox": [ - 350.83, - 315.01, - 363.24, - 322.27 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p845-b27", - "global_id": 24632, - "bbox": [ - 372.65, - 299.12, - 408.7, - 311.77 - ], - "text": "x2n+1Wnr", - "type": "text" - }, - { - "block_id": "p845-b28", - "global_id": 24633, - "bbox": [ - 401.78, - 305.69, - 416.68, - 314.22 - ], - "text": "N0/2", - "type": "text" - }, - { - "block_id": "p845-b29", - "global_id": 24634, - "bbox": [ - 238.05, - 324.66, - 281.04, - 337.62 - ], - "text": "= Gr + Wr", - "type": "text" - }, - { - "block_id": "p845-b30", - "global_id": 24635, - "bbox": [ - 277.6, - 326.16, - 516.13, - 339.76 - ], - "text": "N0Hr\n0 ≤r ≤N0 −1\n(8.23)", - "type": "text" - }, - { - "block_id": "p845-b31", - "global_id": 24636, - "bbox": [ - 127.59, - 348.68, - 516.13, - 372.5 - ], - "text": "where Gr and Hr are the (N0/2)-point DFTs of the even- and odd-numbered sequences, gn and hn,\nrespectively. Also, Gr and Hr, being the (N0/2)-point DFTs, are (N0/2) periodic. Hence,", - "type": "text" - }, - { - "block_id": "p845-b32", - "global_id": 24637, - "bbox": [ - 236.21, - 383.15, - 516.13, - 395.56 - ], - "text": "Gr+(N0/2) = Gr\nand\nHr+(N0/2) = Hr\n(8.24)", - "type": "text" - }, - { - "block_id": "p845-b33", - "global_id": 24638, - "bbox": [ - 127.59, - 406.08, - 168.65, - 416.04 - ], - "text": "Moreover,", - "type": "text" - }, - { - "block_id": "p845-b34", - "global_id": 24639, - "bbox": [ - 235.0, - 415.67, - 272.44, - 428.79 - ], - "text": "Wr+(N0/2)", - "type": "text" - }, - { - "block_id": "p845-b35", - "global_id": 24640, - "bbox": [ - 243.3, - 415.67, - 308.73, - 432.24 - ], - "text": "N0\n= WN0/2", - "type": "text" - }, - { - "block_id": "p845-b36", - "global_id": 24641, - "bbox": [ - 293.1, - 417.01, - 320.97, - 432.24 - ], - "text": "N0\nWr", - "type": "text" - }, - { - "block_id": "p845-b37", - "global_id": 24642, - "bbox": [ - 317.53, - 416.79, - 366.95, - 432.66 - ], - "text": "N0 = e−jπWr", - "type": "text" - }, - { - "block_id": "p845-b38", - "global_id": 24643, - "bbox": [ - 363.51, - 417.01, - 403.53, - 432.66 - ], - "text": "N0 = −Wr", - "type": "text" - }, - { - "block_id": "p845-b39", - "global_id": 24644, - "bbox": [ - 400.09, - 418.92, - 516.13, - 432.11 - ], - "text": "N0\n(8.25)", - "type": "text" - }, - { - "block_id": "p845-b40", - "global_id": 24645, - "bbox": [ - 127.59, - 438.45, - 314.64, - 448.42 - ], - "text": "From Eqs. (8.23), (8.24), and (8.25), we obtain", - "type": "text" - }, - { - "block_id": "p845-b41", - "global_id": 24646, - "bbox": [ - 273.96, - 459.06, - 354.0, - 472.97 - ], - "text": "Xr+(N0/2) = Gr −Wr", - "type": "text" - }, - { - "block_id": "p845-b42", - "global_id": 24647, - "bbox": [ - 350.56, - 460.87, - 516.13, - 474.15 - ], - "text": "N0Hr\n(8.26)", - "type": "text" - }, - { - "block_id": "p845-b43", - "global_id": 24648, - "bbox": [ - 127.59, - 483.48, - 516.16, - 517.36 - ], - "text": "This property can be used to reduce the number of computations. We can compute the first\nN0/2 points (0 ≤n ≤(N0/2) −1) of Xr by using Eq. (8.23) and the last N0/2 points by using\nEq. (8.26) as", - "type": "text" - }, - { - "block_id": "p845-b44", - "global_id": 24649, - "bbox": [ - 260.2, - 519.36, - 314.7, - 531.87 - ], - "text": "Xr = Gr + Wr", - "type": "text" - }, - { - "block_id": "p845-b45", - "global_id": 24650, - "bbox": [ - 234.66, - 518.28, - 409.06, - 546.6 - ], - "text": "N0Hr\n0 ≤r ≤N0\n2 −1\nXr+(N0/2) = Gr −Wr", - "type": "text" - }, - { - "block_id": "p845-b46", - "global_id": 24651, - "bbox": [ - 311.26, - 527.71, - 516.13, - 547.92 - ], - "text": "N0Hr\n0 ≤r ≤N0\n2 −1\n(8.27)", - "type": "text" - }, - { - "block_id": "p845-b47", - "global_id": 24652, - "bbox": [ - 311.54, - 617.27, - 461.0, - 626.5 - ], - "text": "Figure 8.24 Butterfly signal flow graph.", - "type": "text" - } - ] - }, - { - "page_num": 846, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p846-b0", - "global_id": 24653, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "826\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - 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"block_id": "p846-b105", - "global_id": 24758, - "bbox": [ - 102.06, - 191.38, - 108.61, - 200.98 - ], - "text": "x6", - "type": "text" - }, - { - "block_id": "p846-b106", - "global_id": 24759, - "bbox": [ - 102.06, - 222.13, - 108.61, - 231.73 - ], - "text": "x1", - "type": "text" - }, - { - "block_id": "p846-b107", - "global_id": 24760, - "bbox": [ - 102.06, - 317.97, - 108.61, - 327.57 - ], - "text": "x7", - "type": "text" - }, - { - "block_id": "p846-b108", - "global_id": 24761, - "bbox": [ - 102.06, - 285.42, - 108.61, - 295.03 - ], - "text": "x5", - "type": "text" - }, - { - "block_id": "p846-b109", - "global_id": 24762, - "bbox": [ - 102.06, - 253.94, - 108.61, - 263.55 - ], - "text": "x3", - "type": "text" - }, - { - "block_id": "p846-b110", - "global_id": 24763, - "bbox": [ - 313.56, - 127.96, - 320.11, - 137.56 - ], - "text": "x4", - "type": "text" - }, - { - "block_id": "p846-b111", - "global_id": 24764, - "bbox": [ - 313.56, - 160.44, - 320.11, - 170.05 - ], - "text": "x2", - "type": "text" - }, - { - "block_id": "p846-b112", - "global_id": 24765, - "bbox": [ - 313.56, - 192.3, - 320.11, - 201.91 - ], - "text": "x6", - "type": "text" - }, - { - "block_id": "p846-b113", - "global_id": 24766, - "bbox": [ - 313.56, - 224.22, - 320.11, - 233.83 - ], - "text": "x1", - "type": "text" - }, - { - "block_id": "p846-b114", - "global_id": 24767, - "bbox": [ - 313.56, - 320.22, - 320.11, - 329.83 - ], - "text": "x7", - "type": "text" - }, - { - "block_id": "p846-b115", - "global_id": 24768, - "bbox": [ - 313.56, - 288.6, - 320.11, - 298.21 - ], - "text": "x3", - "type": "text" - }, - { - "block_id": "p846-b116", - "global_id": 24769, - "bbox": [ - 313.56, - 256.36, - 320.11, - 265.97 - ], - "text": "x5", - "type": "text" - }, - { - "block_id": "p846-b117", - "global_id": 24770, - "bbox": [ - 345.08, - 107.22, - 368.08, - 117.04 - ], - "text": "N0 2", - "type": "text" - }, - { - "block_id": "p846-b118", - "global_id": 24771, - "bbox": [ - 349.02, - 117.52, - 364.14, - 125.52 - ], - "text": "DFT", - "type": "text" - }, - { - "block_id": "p846-b119", - "global_id": 24772, - "bbox": [ - 345.08, - 171.72, - 368.08, - 181.54 - ], - "text": "N0 2", - "type": "text" - }, - { - "block_id": "p846-b120", - "global_id": 24773, - "bbox": [ - 349.02, - 182.02, - 364.14, - 190.02 - ], - "text": "DFT", - "type": "text" - }, - { - "block_id": "p846-b121", - "global_id": 24774, - "bbox": [ - 345.08, - 236.22, - 368.08, - 246.04 - ], - "text": "N0 2", - "type": "text" - }, - { - "block_id": "p846-b122", - "global_id": 24775, - "bbox": [ - 349.02, - 246.52, - 364.14, - 254.52 - ], - "text": "DFT", - "type": "text" - }, - { - "block_id": "p846-b123", - "global_id": 24776, - "bbox": [ - 345.08, - 300.72, - 368.08, - 310.54 - ], - "text": "N0 2", - "type": "text" - }, - { - "block_id": "p846-b124", - "global_id": 24777, - "bbox": [ - 349.02, - 311.02, - 364.14, - 319.02 - ], - "text": "DFT", - "type": "text" - }, - { - "block_id": "p846-b125", - "global_id": 24778, - "bbox": [ - 101.84, - 630.44, - 278.42, - 639.68 - ], - "text": "Figure 8.25 Successive steps in an 8-point FFT.", - "type": "text" - } - ] - }, - { - "page_num": 847, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p847-b0", - "global_id": 24779, - "bbox": [ - 286.1, - 62.89, - 516.13, - 71.98 - ], - "text": "8.7\nMATLAB: The Discrete Fourier Transform\n827", - "type": "text" - }, - { - "block_id": "p847-b1", - "global_id": 24780, - "bbox": [ - 127.59, - 85.4, - 516.14, - 132.72 - ], - "text": "Thus, an N0-point DFT can be computed by combining the two (N0/2)-point DFTs, as in\nEq. (8.27). These equations can be represented conveniently by the signal flow graph depicted\nin Fig. 8.24. This structure is known as a butterfly. Figure 8.25a shows the implementation of\nEq. (8.24) for the case of N0 = 8.", - "type": "text" - }, - { - "block_id": "p847-b2", - "global_id": 24781, - "bbox": [ - 127.59, - 133.23, - 516.14, - 191.42 - ], - "text": "The next step is to compute the (N0/2)-point DFTs Gr and Hr. We repeat the same\nprocedure by dividing gn and hn into two (N0/4)-point sequences corresponding to the even- and\nodd-numbered samples. Then we continue this process until we reach the one-point DFT. These\nsteps for the case of N0 = 8 are shown in Figs. 8.25a, 8.25b, and 8.25c. Figure 8.25c shows that\nthe two-point DFTs require no multiplication.", - "type": "text" - }, - { - "block_id": "p847-b3", - "global_id": 24782, - "bbox": [ - 127.59, - 193.31, - 516.13, - 228.06 - ], - "text": "To count the number of computations required in the first step, assume that Gr and Hr are\nknown. Equation (8.27) clearly shows that to compute all the N0 points of the Xr, we require N0\ncomplex additions and N0/2 complex multiplications† (corresponding to Wr", - "type": "text" - }, - { - "block_id": "p847-b4", - "global_id": 24783, - "bbox": [ - 127.59, - 217.22, - 516.16, - 322.93 - ], - "text": "N0Hr).\nIn the second step, to compute the (N0/2)-point DFT Gr from the (N0/4)-point DFT, we\nrequire N0/2 complex additions and N0/4 complex multiplications. We require an equal number\nof computations for Hr. Hence, in the second step, there are N0 complex additions and N0/2\ncomplex multiplications. The number of computations required remains the same in each step.\nSince a total of log2 N0 steps is needed to arrive at a one-point DFT, we require, conservatively, a\ntotal of N0 log2 N0 complex additions and (N0/2)log2 N0 complex multiplications, to compute the\nN0-point DFT. Actually, as Fig. 8.25c shows, many multiplications are multiplications by 1 or −1,\nwhich further reduces the number of computations.", - "type": "text" - }, - { - "block_id": "p847-b5", - "global_id": 24784, - "bbox": [ - 127.59, - 324.92, - 516.15, - 406.61 - ], - "text": "The procedure for obtaining IDFT is identical to that used to obtain the DFT except that\nWN0 = ej(2π/N0) instead of e−j(2π/N0) (in addition to the multiplier 1/N0). Another FFT algorithm,\nthe decimation-in-frequency algorithm, is similar to the decimation-in-time algorithm. The only\ndifference is that instead of dividing xn into two sequences of even- and odd-numbered samples, we\ndivide xn into two sequences formed by the first N0/2 and the last N0/2 samples, proceeding in the\nsame way until a single-point DFT is reached in log2 N0 steps. The total number of computations\nin this algorithm is the same as that in the decimation-in-time algorithm.", - "type": "text" - }, - { - "block_id": "p847-b6", - "global_id": 24785, - "bbox": [ - 127.94, - 436.5, - 470.03, - 450.45 - ], - "text": "8.7 MATLAB: THE DISCRETE FOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p847-b7", - "global_id": 24786, - "bbox": [ - 127.59, - 456.44, - 516.16, - 490.31 - ], - "text": "As an idea, the discrete Fourier transform (DFT) has been known for hundreds of years. Practical\ncomputing devices, however, are responsible for bringing the DFT into common use. MATLAB is\ncapable of DFT computations that would have been impractical just a few decades ago.", - "type": "text" - }, - { - "block_id": "p847-b8", - "global_id": 24787, - "bbox": [ - 127.59, - 515.67, - 391.11, - 527.62 - ], - "text": "8.7-1 Computing the Discrete Fourier Transform", - "type": "text" - }, - { - "block_id": "p847-b9", - "global_id": 24788, - "bbox": [ - 127.59, - 533.34, - 516.16, - 591.82 - ], - "text": "The MATLAB command fft(x) computes the DFT of a vector x that is defined over (0 ≤n ≤\nN0 −1) (Problem 8.7-1 considers how to scale the DFT to accommodate signals that do not begin\nat n = 0.) As its name suggests, the function fft uses the computationally more efficient fast\nFourier transform algorithm when it is appropriate to do so. The inverse DFT is easily computed\nby using the ifft function.", - "type": "text" - }, - { - "block_id": "p847-b10", - "global_id": 24789, - "bbox": [ - 127.59, - 610.24, - 502.24, - 623.19 - ], - "text": "† Actually, N0/2 is a conservative figure because some multiplications corresponding to the cases of Wr", - "type": "text" - }, - { - "block_id": "p847-b11", - "global_id": 24790, - "bbox": [ - 127.59, - 613.11, - 516.13, - 633.41 - ], - "text": "N0 =\n1,j, and so on, are eliminated.", - "type": "text" - } - ] - }, - { - "page_num": 848, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p848-b0", - "global_id": 24791, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "828\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p848-b1", - "global_id": 24792, - "bbox": [ - 101.84, - 85.82, - 490.39, - 108.82 - ], - "text": "To illustrate MATLAB’s DFT capabilities, consider 50 points of a 10 Hz sinusoid sampled at\nfs = 50 Hz and scaled by T = 1/fs.", - "type": "text" - }, - { - "block_id": "p848-b2", - "global_id": 24793, - "bbox": [ - 101.84, - 119.2, - 300.6, - 141.11 - ], - "text": ">>\nT = 1/50; N_0 = 50; n = (0:N_0-1);\n>>\nx = T*cos(2*pi*10*n*T);", - "type": "text" - }, - { - "block_id": "p848-b3", - "global_id": 24794, - "bbox": [ - 101.84, - 151.99, - 490.38, - 173.91 - ], - "text": "In this case, the vector x contains exactly 10 cycles of the sinusoid. The fft command computes\nthe DFT.", - "type": "text" - }, - { - "block_id": "p848-b4", - "global_id": 24795, - "bbox": [ - 101.84, - 185.36, - 180.3, - 195.33 - ], - "text": ">>\nX = fft(x);", - "type": "text" - }, - { - "block_id": "p848-b5", - "global_id": 24796, - "bbox": [ - 101.84, - 206.11, - 490.37, - 228.82 - ], - "text": "Since the DFT is both discrete and periodic, fft needs to return only the N0 discrete values\ncontained in the single period (0 ≤f < fs).", - "type": "text" - }, - { - "block_id": "p848-b6", - "global_id": 24797, - "bbox": [ - 101.84, - 230.02, - 490.39, - 253.53 - ], - "text": "While Xr can be plotted as a function of r, it is more convenient to plot the DFT as a function\nof frequency f. A frequency vector, in hertz, is created by using N0 and T.", - "type": "text" - }, - { - "block_id": "p848-b7", - "global_id": 24798, - "bbox": [ - 101.85, - 263.49, - 436.59, - 285.41 - ], - "text": ">>\nf = (0:N_0-1)/(T*N_0); stem(f,abs(X),’k.’);\n>>\naxis([0 50 -0.05 0.55]); xlabel(’f [Hz]’); ylabel(’|X(f)|’);", - "type": "text" - }, - { - "block_id": "p848-b8", - "global_id": 24799, - "bbox": [ - 101.85, - 296.29, - 490.4, - 330.16 - ], - "text": "As expected, Fig. 8.26 shows content at a frequency of 10 Hz. Since the time-domain signal is\nreal, X(f) is conjugate symmetric. Thus, content at 10 Hz implies equal content at −10 Hz. The\ncontent visible at 40 Hz is an alias of the −10 Hz content.", - "type": "text" - }, - { - "block_id": "p848-b9", - "global_id": 24800, - "bbox": [ - 101.85, - 331.74, - 490.38, - 354.36 - ], - "text": "Often, it is preferred to plot a DFT over the principal frequency range (−fs/2 ≤f < fs/2). The\nMATLAB function fftshift properly rearranges the output of fft to accomplish this task.", - "type": "text" - }, - { - "block_id": "p848-b10", - "global_id": 24801, - "bbox": [ - 101.85, - 365.53, - 447.05, - 387.45 - ], - "text": ">>\nstem(f-1/(T*2),fftshift(abs(X)),’k.’);\n>>\naxis([-25 25 -0.05 0.55]); xlabel(’f [Hz]’); ylabel(’|X(f)|’);", - "type": "text" - }, - { - "block_id": "p848-b11", - "global_id": 24802, - "bbox": [ - 101.85, - 398.33, - 490.37, - 420.25 - ], - "text": "When we use fftshift, the conjugate symmetry that accompanies the DFT of a real signal\nbecomes apparent, as shown in Fig. 8.27.", - "type": "text" - }, - { - "block_id": "p848-b12", - "global_id": 24803, - "bbox": [ - 101.85, - 422.24, - 490.37, - 444.16 - ], - "text": "Since DFTs are generally complex-valued, the magnitude plots of Figs. 8.26 and 8.27 offer\nonly half the picture; the signal’s phase spectrum, shown in Fig. 8.28, completes it.", - "type": "text" - }, - { - "block_id": "p848-b13", - "global_id": 24804, - "bbox": [ - 101.85, - 455.61, - 494.12, - 477.53 - ], - "text": ">>\nstem(f-1/(T*2),fftshift(angle(X)),’k.’);\n>>\naxis([-25 25 -1.1*pi 1.1*pi]); xlabel(’f [Hz]’); ylabel(’\\angle X(f)’);", - "type": "text" - }, - { - "block_id": "p848-b14", - "global_id": 24805, - "bbox": [ - 127.56, - 592.1, - 468.37, - 613.28 - ], - "text": "0\n5\n10\n15\n20\n25\n30\n35\n40\n45\n50\nf [Hz]", - "type": "text" - }, - { - "block_id": "p848-b15", - "global_id": 24806, - "bbox": [ - 121.33, - 576.86, - 125.33, - 584.86 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p848-b16", - "global_id": 24807, - "bbox": [ - 114.58, - 552.86, - 125.26, - 560.86 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p848-b17", - "global_id": 24808, - "bbox": [ - 114.58, - 528.87, - 125.26, - 536.87 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p848-b18", - "global_id": 24809, - "bbox": [ - 102.61, - 540.69, - 111.41, - 559.35 - ], - "text": "|X(f)|", - "type": "text" - }, - { - "block_id": "p848-b19", - "global_id": 24810, - "bbox": [ - 101.84, - 619.58, - 328.71, - 629.19 - ], - "text": "Figure 8.26 |X(f)| computed over (0 ≤f < 50) by using fft.", - "type": "text" - } - ] - }, - { - "page_num": 849, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p849-b0", - "global_id": 24811, - "bbox": [ - 286.1, - 62.89, - 516.13, - 71.98 - ], - "text": "8.7\nMATLAB: The Discrete Fourier Transform\n829", - "type": "text" - }, - { - "block_id": "p849-b1", - "global_id": 24812, - "bbox": [ - 149.08, - 164.57, - 493.84, - 185.75 - ], - "text": "–25\n–20\n–15\n–10\n–5\n0\n5\n10\n15\n20\n25\nf [Hz]", - "type": "text" - }, - { - "block_id": "p849-b2", - "global_id": 24813, - "bbox": [ - 147.08, - 150.47, - 151.08, - 158.47 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p849-b3", - "global_id": 24814, - "bbox": [ - 140.33, - 126.47, - 151.01, - 134.47 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p849-b4", - "global_id": 24815, - "bbox": [ - 140.33, - 102.48, - 151.01, - 110.48 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p849-b5", - "global_id": 24816, - "bbox": [ - 128.36, - 114.29, - 137.16, - 132.96 - ], - "text": "|X(f)|", - "type": "text" - }, - { - "block_id": "p849-b6", - "global_id": 24817, - "bbox": [ - 127.59, - 192.07, - 388.47, - 201.68 - ], - "text": "Figure 8.27 |X(f)| displayed over (−25 ≤f < 25) by using fftshift.", - "type": "text" - }, - { - "block_id": "p849-b7", - "global_id": 24818, - "bbox": [ - 167.1, - 303.98, - 513.55, - 325.16 - ], - "text": "–25\n–20\n–15\n–10\n–5\n0\n5\n10\n15\n20\n25\nf [Hz]", - "type": "text" - }, - { - "block_id": "p849-b8", - "global_id": 24819, - "bbox": [ - 163.7, - 279.58, - 171.7, - 287.58 - ], - "text": "–2", - "type": "text" - }, - { - "block_id": "p849-b9", - "global_id": 24820, - "bbox": [ - 165.95, - 258.75, - 169.95, - 266.75 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p849-b10", - "global_id": 24821, - "bbox": [ - 165.95, - 237.91, - 169.95, - 245.91 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p849-b11", - "global_id": 24822, - "bbox": [ - 152.19, - 250.87, - 160.19, - 265.98 - ], - "text": "X(f)", - "type": "text" - }, - { - "block_id": "p849-b12", - "global_id": 24823, - "bbox": [ - 151.5, - 331.48, - 335.59, - 341.09 - ], - "text": "Figure 8.28̸\nX(f) displayed over (−25≤f <25).", - "type": "text" - }, - { - "block_id": "p849-b13", - "global_id": 24824, - "bbox": [ - 127.59, - 363.63, - 516.14, - 445.32 - ], - "text": "Since the signal is real, the phase spectrum necessarily has odd symmetry. Additionally,\nthe phase at ±10 Hz is zero, as expected for a zero-phase cosine function. More interesting,\nhowever, are the phase values found at the remaining frequencies. Does a simple cosine really\nhave such complicated phase characteristics? The answer, of course, is no. The magnitude plot of\nFig. 8.27 helps identify the problem: there is zero content at frequencies other than ±10 Hz. Phase\ncomputations are not reliable at points where the magnitude response is zero. One way to remedy\nthis problem is to assign a phase of zero when the magnitude response is near or at zero.", - "type": "text" - }, - { - "block_id": "p849-b14", - "global_id": 24825, - "bbox": [ - 127.59, - 472.12, - 383.47, - 484.07 - ], - "text": "8.7-2 Improving the Picture with Zero Padding", - "type": "text" - }, - { - "block_id": "p849-b15", - "global_id": 24826, - "bbox": [ - 127.59, - 490.2, - 516.12, - 524.78 - ], - "text": "DFT magnitude and phase plots paint a picture of a signal’s spectrum. At times, however, the\npicture can be somewhat misleading. Given a sampling frequency fs = 50 Hz and a sampling\ninterval T = 1/fs, consider the signal", - "type": "text" - }, - { - "block_id": "p849-b16", - "global_id": 24827, - "bbox": [ - 284.05, - 542.56, - 321.0, - 552.83 - ], - "text": "y[n] = Te", - "type": "text" - }, - { - "block_id": "p849-b17", - "global_id": 24828, - "bbox": [ - 320.99, - 539.27, - 330.6, - 546.53 - ], - "text": "j2π", - "type": "text" - }, - { - "block_id": "p849-b19", - "global_id": 24829, - "bbox": [ - 335.12, - 537.5, - 346.28, - 549.1 - ], - "text": "10 1\n3", - "type": "text" - }, - { - "block_id": "p849-b21", - "global_id": 24830, - "bbox": [ - 351.27, - 539.49, - 358.63, - 546.46 - ], - "text": "nT", - "type": "text" - }, - { - "block_id": "p849-b22", - "global_id": 24831, - "bbox": [ - 127.59, - 565.17, - 465.64, - 576.89 - ], - "text": "This complex-valued, periodic signal contains a single positive frequency at 10 1", - "type": "text" - }, - { - "block_id": "p849-b23", - "global_id": 24832, - "bbox": [ - 127.59, - 566.93, - 516.13, - 588.85 - ], - "text": "3 Hz. Let us\ncompute the signal’s DFT using 50 samples.", - "type": "text" - }, - { - "block_id": "p849-b24", - "global_id": 24833, - "bbox": [ - 127.59, - 601.15, - 472.79, - 635.01 - ], - "text": ">>\ny = T*exp(j*2*pi*(10+1/3)*n*T); Y = fft(y);\n>>\nstem(f-25,fftshift(abs(Y)),’k.’);\n>>\naxis([-25 25 -0.05 1.05]); xlabel(’f [Hz]’); ylabel(’|Y(f)|’);", - "type": "text" - } - ] - }, - { - "page_num": 850, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p850-b0", - "global_id": 24834, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "830\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p850-b1", - "global_id": 24835, - "bbox": [ - 145.54, - 165.58, - 491.72, - 186.76 - ], - "text": "–25\n–20\n–15\n–10\n–5\n0\n5\n10\n15\n20\n25\nf [Hz]", - "type": "text" - }, - { - "block_id": "p850-b2", - "global_id": 24836, - "bbox": [ - 145.24, - 153.09, - 149.24, - 161.09 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p850-b3", - "global_id": 24837, - "bbox": [ - 138.49, - 120.35, - 149.17, - 128.35 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p850-b4", - "global_id": 24838, - "bbox": [ - 145.24, - 87.62, - 149.24, - 95.62 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p850-b5", - "global_id": 24839, - "bbox": [ - 126.52, - 114.17, - 135.32, - 132.83 - ], - "text": "|Y(f)|", - "type": "text" - }, - { - "block_id": "p850-b6", - "global_id": 24840, - "bbox": [ - 125.76, - 193.06, - 274.59, - 202.68 - ], - "text": "Figure 8.29 |Y(f)| using 50 data points.", - "type": "text" - }, - { - "block_id": "p850-b7", - "global_id": 24841, - "bbox": [ - 130.02, - 311.13, - 470.28, - 332.31 - ], - "text": "5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\nf [Hz]", - "type": "text" - }, - { - "block_id": "p850-b8", - "global_id": 24842, - "bbox": [ - 123.79, - 298.64, - 127.79, - 306.64 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p850-b9", - "global_id": 24843, - "bbox": [ - 117.05, - 265.9, - 127.72, - 273.9 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p850-b10", - "global_id": 24844, - "bbox": [ - 123.79, - 233.17, - 127.79, - 241.17 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p850-b11", - "global_id": 24845, - "bbox": [ - 102.54, - 256.79, - 113.82, - 281.38 - ], - "text": "|Yzp(f)|", - "type": "text" - }, - { - "block_id": "p850-b12", - "global_id": 24846, - "bbox": [ - 101.84, - 338.62, - 397.7, - 348.63 - ], - "text": "Figure 8.30 |Yzp(f)| over 5 ≤f ≤15 using 50 data points padded with 550 zeros.", - "type": "text" - }, - { - "block_id": "p850-b13", - "global_id": 24847, - "bbox": [ - 101.84, - 377.31, - 490.4, - 411.17 - ], - "text": "In this case, the vector y contains a noninteger number of cycles. Figure 8.29 shows the significant\nfrequency leakage that results. Also notice that since y[n] is not real, the DFT is not conjugate\nsymmetric.", - "type": "text" - }, - { - "block_id": "p850-b14", - "global_id": 24848, - "bbox": [ - 119.78, - 411.41, - 421.49, - 423.14 - ], - "text": "In this example, the discrete DFT frequencies do not include the actual 10 1", - "type": "text" - }, - { - "block_id": "p850-b15", - "global_id": 24849, - "bbox": [ - 101.85, - 413.17, - 490.41, - 447.04 - ], - "text": "3 Hz frequency of\nthe signal. Thus, it is difficult to determine the signal’s frequency from Fig. 8.29. To improve the\npicture, the signal is zero-padded to 12 times its original length.", - "type": "text" - }, - { - "block_id": "p850-b16", - "global_id": 24850, - "bbox": [ - 101.85, - 465.49, - 473.2, - 511.32 - ], - "text": ">>\ny_zp = [y,zeros(1,11*length(y))]; Y_zp = fft(y_zp);\n>>\nf_zp = (0:12*N_0-1)/(T*12*N_0);\n>>\nstem(f_zp-25,fftshift(abs(Y_zp)),’k.’);\n>>\naxis([-25 25 -0.05 1.05]); xlabel(’f [Hz]’); ylabel(’|Y_{zp}(f)|’);", - "type": "text" - }, - { - "block_id": "p850-b17", - "global_id": 24851, - "bbox": [ - 101.85, - 527.43, - 431.21, - 539.14 - ], - "text": "Figure 8.30, zoomed in to 5 ≤f ≤15, correctly shows the peak frequency at 10 1", - "type": "text" - }, - { - "block_id": "p850-b18", - "global_id": 24852, - "bbox": [ - 101.85, - 529.18, - 490.38, - 551.1 - ], - "text": "3 Hz and better\nrepresents the signal’s spectrum.", - "type": "text" - }, - { - "block_id": "p850-b19", - "global_id": 24853, - "bbox": [ - 101.85, - 553.1, - 490.41, - 634.79 - ], - "text": "It is important to keep in mind that zero padding does not increase the resolution or accuracy\nof the DFT. To return to the picket fence analogy, zero padding increases the number of pickets\nin our fence but cannot change what is behind the fence. More formally, the characteristics of\nthe sinc function, such as main beam width and sidelobe levels, depend on the fixed width of the\npulse, not on the number of zeros that follow. Adding zeros cannot change the characteristics of\nthe sinc function and thus cannot change the resolution or accuracy of the DFT. Adding zeros\nsimply allows the sinc function to be sampled more finely.", - "type": "text" - } - ] - }, - { - "page_num": 851, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p851-b0", - "global_id": 24854, - "bbox": [ - 286.1, - 62.89, - 516.12, - 71.98 - ], - "text": "8.7\nMATLAB: The Discrete Fourier Transform\n831", - "type": "text" - }, - { - "block_id": "p851-b1", - "global_id": 24855, - "bbox": [ - 127.59, - 86.25, - 230.33, - 98.21 - ], - "text": "8.7-3 Quantization", - "type": "text" - }, - { - "block_id": "p851-b2", - "global_id": 24856, - "bbox": [ - 127.59, - 104.24, - 516.12, - 150.17 - ], - "text": "A B-bit analog-to-digital converter (ADC) samples an analog signal and quantizes amplitudes by\nusing 2B discrete levels. This quantization results in signal distortion that is particularly noticeable\nfor small B. Typically, quantization is classified as symmetric or asymmetric and as either rounding\nor truncating. Let us investigate rounding-type quantizers.", - "type": "text" - }, - { - "block_id": "p851-b3", - "global_id": 24857, - "bbox": [ - 145.52, - 151.57, - 437.92, - 163.62 - ], - "text": "The quantized output xq of an asymmetric rounding converter is given as†", - "type": "text" - }, - { - "block_id": "p851-b4", - "global_id": 24858, - "bbox": [ - 268.66, - 173.04, - 307.43, - 191.17 - ], - "text": "xq = xmax", - "type": "text" - }, - { - "block_id": "p851-b5", - "global_id": 24859, - "bbox": [ - 290.12, - 173.04, - 332.03, - 197.94 - ], - "text": "2B−1 ⌊x\nxmax", - "type": "text" - }, - { - "block_id": "p851-b6", - "global_id": 24860, - "bbox": [ - 333.73, - 173.14, - 375.07, - 197.16 - ], - "text": "2B−1 + 1\n2⌋", - "type": "text" - }, - { - "block_id": "p851-b7", - "global_id": 24861, - "bbox": [ - 127.6, - 206.65, - 407.6, - 218.21 - ], - "text": "The quantized output xq of a symmetric rounding converter is given as", - "type": "text" - }, - { - "block_id": "p851-b8", - "global_id": 24862, - "bbox": [ - 261.37, - 228.31, - 300.15, - 246.44 - ], - "text": "xq = xmax", - "type": "text" - }, - { - "block_id": "p851-b9", - "global_id": 24863, - "bbox": [ - 282.84, - 241.57, - 300.99, - 252.43 - ], - "text": "2B−1", - "type": "text" - }, - { - "block_id": "p851-b11", - "global_id": 24864, - "bbox": [ - 310.55, - 228.31, - 326.83, - 244.95 - ], - "text": "⌊x", - "type": "text" - }, - { - "block_id": "p851-b12", - "global_id": 24865, - "bbox": [ - 316.16, - 242.37, - 332.59, - 253.2 - ], - "text": "xmax", - "type": "text" - }, - { - "block_id": "p851-b13", - "global_id": 24866, - "bbox": [ - 334.29, - 228.41, - 374.42, - 252.43 - ], - "text": "2B−1⌋+ 1\n2", - "type": "text" - }, - { - "block_id": "p851-b15", - "global_id": 24867, - "bbox": [ - 127.6, - 262.84, - 516.16, - 284.76 - ], - "text": "Program CH8MP1 quantizes a signal using one of these two rounding quantizer rules and also\nensures no more than 2B output levels.", - "type": "text" - }, - { - "block_id": "p851-b16", - "global_id": 24868, - "bbox": [ - 127.59, - 295.26, - 519.85, - 412.85 - ], - "text": "function [xq] = CH8MP1(x,xmax,B,method)\n% CH8MP1.m : Chapter 8, MATLAB Program 1\n% Function M-file quantizes x over (-xmax,xmax) using 2^b levels.\n% Uses rounding rule, supports symmetric and asymmetric quantization\n% INPUTS:\nx = input signal\n%\nxmax = maximum magnitude of signal to be quantized\n%\nB = number of quantization bits\n%\nmethod = default ’sym’ for symmetrical, ’asym’ for asymmetrical\n% OUTPUTS:\nxq = quantized signal\nif (nargin<3),", - "type": "text" - }, - { - "block_id": "p851-b17", - "global_id": 24869, - "bbox": [ - 127.59, - 414.84, - 389.1, - 436.76 - ], - "text": "disp(’Insufficient number of inputs.’); return\nelseif (nargin==3),", - "type": "text" - }, - { - "block_id": "p851-b18", - "global_id": 24870, - "bbox": [ - 127.59, - 438.76, - 226.96, - 460.68 - ], - "text": "method = ’sym’;\nelseif (nargin>4),", - "type": "text" - }, - { - "block_id": "p851-b19", - "global_id": 24871, - "bbox": [ - 127.59, - 462.68, - 315.88, - 484.6 - ], - "text": "disp(’Too many inputs.’); return\nend", - "type": "text" - }, - { - "block_id": "p851-b20", - "global_id": 24872, - "bbox": [ - 127.59, - 498.55, - 488.49, - 520.46 - ], - "text": "x(abs(x)>xmax)=xmax*sign(x(abs(x)>xmax));\n% Limit amplitude to xmax\nswitch lower(method)", - "type": "text" - }, - { - "block_id": "p851-b21", - "global_id": 24873, - "bbox": [ - 148.51, - 522.46, - 206.04, - 532.42 - ], - "text": "case ’asym’", - "type": "text" - }, - { - "block_id": "p851-b22", - "global_id": 24874, - "bbox": [ - 148.51, - 534.42, - 483.23, - 568.3 - ], - "text": "xq = xmax/(2^(B-1))*floor(x*2^(B-1)/xmax+1/2);\nxq(xq>=xmax)=xmax*(1-2^(1-B));\n% Ensure only 2^B levels\ncase ’sym’", - "type": "text" - }, - { - "block_id": "p851-b23", - "global_id": 24875, - "bbox": [ - 169.42, - 570.3, - 483.23, - 592.22 - ], - "text": "xq = xmax/(2^(B-1))*(floor(x*2^(B-1)/xmax)+1/2);\nxq(xq>=xmax)=xmax*(1-2^(1-B)/2);\n% Ensure only 2^B levels", - "type": "text" - }, - { - "block_id": "p851-b24", - "global_id": 24876, - "bbox": [ - 127.59, - 610.24, - 516.13, - 633.42 - ], - "text": "† Large values of x may return quantized values xq outside the 2B allowable levels. In such cases, xq should\nbe clamped to the nearest permitted level.", - "type": "text" - } - ] - }, - { - "page_num": 852, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p852-b0", - "global_id": 24877, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "832\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p852-b1", - "global_id": 24878, - "bbox": [ - 122.76, - 85.41, - 169.83, - 95.37 - ], - "text": "otherwise", - "type": "text" - }, - { - "block_id": "p852-b2", - "global_id": 24879, - "bbox": [ - 101.84, - 97.36, - 399.97, - 119.28 - ], - "text": "disp(’Unrecognized quantization method.’); return\nend", - "type": "text" - }, - { - "block_id": "p852-b3", - "global_id": 24880, - "bbox": [ - 101.84, - 130.43, - 490.38, - 188.22 - ], - "text": "Several MATLAB commands require discussion. First, the nargin function returns the number\nof input arguments. In this program, nargin is used to ensure that a correct number of inputs\nis supplied. If the number of inputs supplied is incorrect, an error message is displayed and\nthe function terminates. If only three input arguments are detected, the quantization type is not\nexplicitly specified and the program assigns the default symmetric method.", - "type": "text" - }, - { - "block_id": "p852-b4", - "global_id": 24881, - "bbox": [ - 101.84, - 190.21, - 490.38, - 212.13 - ], - "text": "As with many high-level languages such as C, MATLAB supports general switch/case\nstructures†:", - "type": "text" - }, - { - "block_id": "p852-b5", - "global_id": 24882, - "bbox": [ - 119.78, - 213.71, - 224.37, - 236.32 - ], - "text": "switch\nswitch_expr,\ncase\ncase_expr,", - "type": "text" - }, - { - "block_id": "p852-b6", - "global_id": 24883, - "bbox": [ - 119.78, - 237.62, - 210.73, - 272.19 - ], - "text": "statements;\n...\notherwise,", - "type": "text" - }, - { - "block_id": "p852-b7", - "global_id": 24884, - "bbox": [ - 101.84, - 273.48, - 490.4, - 343.92 - ], - "text": "statements;\nend\nCH8MP1 switches among cases of the string method. In this way, method-specific parameters are\neasily set. The command lower is used to convert a string to all lowercase characters. In this way,\nstrings such as SYM, Sym, and sym are all indistinguishable. Similar to lower, the MATLAB\ncommand upper converts a string to all uppercase.", - "type": "text" - }, - { - "block_id": "p852-b8", - "global_id": 24885, - "bbox": [ - 101.84, - 345.63, - 490.41, - 440.77 - ], - "text": "The floor command rounds input values to the nearest integer toward minus infinity.\nMathematically, it computes ⌊·⌋. To accommodate different types of rounding, MATLAB supplies\nthree other rounding commands: ceil, round, and fix. The ceil command rounds input values\nto the nearest integers toward infinity, (⌈·⌉); the round command rounds input values toward the\nnearest integer; the fix command rounds input values to the nearest integer toward zero. For\nexample, if x = [-0.5 0.5];, floor(x) yields [-1 0], ceil(x) yields [0 1], round(x)\nyields [-1 1], and fix(x) yields [0 0]. Finally, CH8MP1 checks and, if necessary, corrects large\nvalues of xq that may be outside the allowable 2B levels.", - "type": "text" - }, - { - "block_id": "p852-b9", - "global_id": 24886, - "bbox": [ - 101.85, - 441.27, - 490.38, - 463.19 - ], - "text": "To verify operation, CH8MP1 is used to determine the transfer characteristics of a symmetric\n3-bit quantizer operating over (−10,10).", - "type": "text" - }, - { - "block_id": "p852-b10", - "global_id": 24887, - "bbox": [ - 101.84, - 474.91, - 405.21, - 508.79 - ], - "text": ">>\nx = (-10:.0001:10); xsq = CH8MP1(x,10,3,’sym’);\n>>\nplot(x,xsq,’k’); axis([-10 10 -10.5 10.5]); grid on;\n>>\nxlabel(’Quantizer input’); ylabel(’Quantizer output’);", - "type": "text" - }, - { - "block_id": "p852-b11", - "global_id": 24888, - "bbox": [ - 101.84, - 515.91, - 490.43, - 565.76 - ], - "text": "Figure 8.31 shows the results. Clearly, the quantized output is limited to 2B = 8 levels. Zero is not\na quantization level for symmetric quantizers, so half of the levels occur above zero and half of\nthe levels occur below zero. In fact, symmetric quantizers get their name from the symmetry in\nquantization levels above and below zero.", - "type": "text" - }, - { - "block_id": "p852-b12", - "global_id": 24889, - "bbox": [ - 101.85, - 567.76, - 490.39, - 601.63 - ], - "text": "By changing the method in CH8MP1 from ’sym’ to ’asym’, we obtain the transfer\ncharacteristics of an asymmetric 3-bit quantizer, as shown in Fig. 8.32. Again, the quantized output\nis limited to 2B = 8 levels, and zero is now one of the included levels. With zero as a quantization", - "type": "text" - }, - { - "block_id": "p852-b13", - "global_id": 24890, - "bbox": [ - 101.84, - 621.19, - 443.25, - 633.67 - ], - "text": "† A functionally equivalent structure can be written by using if, elseif, and else statements.", - "type": "text" - } - ] - }, - { - "page_num": 853, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p853-b0", - "global_id": 24891, - "bbox": [ - 286.1, - 62.89, - 516.13, - 71.98 - ], - "text": "8.7\nMATLAB: The Discrete Fourier Transform\n833", - "type": "text" - }, - { - "block_id": "p853-b1", - "global_id": 24892, - "bbox": [ - 171.01, - 200.67, - 406.46, - 221.75 - ], - "text": "–10\n–5\n0\n5\n10\nQuantizer input", - "type": "text" - }, - { - "block_id": "p853-b2", - "global_id": 24893, - "bbox": [ - 164.74, - 188.92, - 176.74, - 196.92 - ], - "text": "–10", - "type": "text" - }, - { - "block_id": "p853-b3", - "global_id": 24894, - "bbox": [ - 168.74, - 163.74, - 176.74, - 171.74 - ], - "text": "–5", - "type": "text" - }, - { - "block_id": "p853-b4", - "global_id": 24895, - "bbox": [ - 172.74, - 138.56, - 176.74, - 146.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p853-b5", - "global_id": 24896, - "bbox": [ - 170.99, - 113.39, - 174.99, - 121.39 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p853-b6", - "global_id": 24897, - "bbox": [ - 166.49, - 88.2, - 174.49, - 96.2 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p853-b7", - "global_id": 24898, - "bbox": [ - 152.26, - 115.29, - 161.06, - 174.68 - ], - "text": "Quantizer output", - "type": "text" - }, - { - "block_id": "p853-b8", - "global_id": 24899, - "bbox": [ - 151.5, - 228.44, - 397.69, - 237.68 - ], - "text": "Figure 8.31 Transfer characteristics of a symmetric 3-bit quantizer.", - "type": "text" - }, - { - "block_id": "p853-b9", - "global_id": 24900, - "bbox": [ - 171.3, - 373.71, - 406.46, - 394.8 - ], - "text": "–10\n–5\n0\n5\n10\nQuantizer input", - "type": "text" - }, - { - "block_id": "p853-b10", - "global_id": 24901, - "bbox": [ - 164.74, - 361.96, - 176.74, - 369.97 - ], - "text": "–10", - "type": "text" - }, - { - "block_id": "p853-b11", - "global_id": 24902, - "bbox": [ - 168.74, - 336.79, - 176.74, - 344.79 - ], - "text": "–5", - "type": "text" - }, - { - "block_id": "p853-b12", - "global_id": 24903, - "bbox": [ - 172.74, - 311.61, - 176.74, - 319.61 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p853-b13", - "global_id": 24904, - "bbox": [ - 172.74, - 286.43, - 176.74, - 294.43 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p853-b14", - "global_id": 24905, - "bbox": [ - 168.74, - 261.25, - 176.74, - 269.25 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p853-b15", - "global_id": 24906, - "bbox": [ - 152.26, - 288.33, - 161.06, - 347.72 - ], - "text": "Quantizer output", - "type": "text" - }, - { - "block_id": "p853-b16", - "global_id": 24907, - "bbox": [ - 151.5, - 401.48, - 406.16, - 410.72 - ], - "text": "Figure 8.32 Transfer characteristics of an asymmetric 3-bit quantizer.", - "type": "text" - }, - { - "block_id": "p853-b17", - "global_id": 24908, - "bbox": [ - 127.59, - 431.9, - 516.15, - 465.77 - ], - "text": "level, we need one fewer quantization level above zero than there are levels below. Not surprisingly,\nasymmetric quantizers get their name from the asymmetry in quantization levels above and below\nzero.", - "type": "text" - }, - { - "block_id": "p853-b18", - "global_id": 24909, - "bbox": [ - 127.59, - 467.77, - 516.15, - 513.59 - ], - "text": "There is no doubt that quantization can change a signal. It follows that the spectrum of a\nquantized signal can also change. While these changes are difficult to characterize mathematically,\nthey are easy to investigate by using MATLAB. Consider a 1 Hz cosine sampled at fs = 50 Hz over\n1 second.", - "type": "text" - }, - { - "block_id": "p853-b19", - "global_id": 24910, - "bbox": [ - 127.59, - 524.52, - 488.42, - 534.48 - ], - "text": ">>\nx = cos(2*pi*n*T); X = fft(x); T = 1/50; N_0 = 50; n = (0:N_0-1);", - "type": "text" - }, - { - "block_id": "p853-b20", - "global_id": 24911, - "bbox": [ - 127.59, - 544.83, - 516.13, - 566.75 - ], - "text": "Upon quantizing by means of a 2-bit asymmetric rounding quantizer, both the signal and spectrum\nare substantially changed.", - "type": "text" - }, - { - "block_id": "p853-b21", - "global_id": 24912, - "bbox": [ - 127.59, - 577.13, - 436.52, - 634.92 - ], - "text": ">>\nxaq = CH8MP1(x,1,2,’asym’); Xaq = fft(xaq);\n>>\nsubplot(2,2,1); stem(n,x,’k’); axis([0 49 -1.1 1.1]);\n>>\nxlabel(’n’);ylabel(’x[n]’);\n>>\nsubplot(2,2,2); stem(f-25,fftshift(abs(X)),’k’); axis([-25,25 -1 26])\n>>\nxlabel(’f’);ylabel(’|X(f)|’);\n>>\nsubplot(2,2,3); stem(n,xaq,’k’);axis([0 49 -1.1 1.1]);", - "type": "text" - } - ] - }, - { - "page_num": 854, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p854-b0", - "global_id": 24913, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "834\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p854-b1", - "global_id": 24914, - "bbox": [ - 88.66, - 171.77, - 232.61, - 192.96 - ], - "text": "0\n10\n20\n30\n40\nn", - "type": "text" - }, - { - "block_id": "p854-b2", - "global_id": 24915, - "bbox": [ - 78.43, - 159.65, - 86.43, - 167.65 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p854-b3", - "global_id": 24916, - "bbox": [ - 82.43, - 130.66, - 86.43, - 138.66 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p854-b4", - "global_id": 24917, - "bbox": [ - 82.43, - 101.7, - 86.43, - 109.7 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p854-b5", - "global_id": 24918, - "bbox": [ - 68.21, - 126.23, - 77.01, - 140.89 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p854-b6", - "global_id": 24919, - "bbox": [ - 322.17, - 171.77, - 468.73, - 192.96 - ], - "text": "–20\n–10\n0\n10\n20\nf", - "type": "text" - }, - { - "block_id": "p854-b7", - "global_id": 24920, - "bbox": [ - 304.43, - 160.19, - 308.43, - 168.19 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p854-b8", - "global_id": 24921, - "bbox": [ - 299.93, - 136.58, - 307.93, - 144.58 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p854-b9", - "global_id": 24922, - "bbox": [ - 299.93, - 112.96, - 307.93, - 120.96 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p854-b10", - "global_id": 24923, - "bbox": [ - 287.95, - 124.47, - 296.75, - 143.14 - ], - "text": "|X(f)|", - "type": "text" - }, - { - "block_id": "p854-b11", - "global_id": 24924, - "bbox": [ - 88.66, - 274.52, - 232.61, - 295.71 - ], - "text": "0\n10\n20\n30\n40\nn", - "type": "text" - }, - { - "block_id": "p854-b12", - "global_id": 24925, - "bbox": [ - 78.43, - 262.36, - 86.43, - 270.36 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p854-b13", - "global_id": 24926, - "bbox": [ - 82.43, - 233.04, - 86.43, - 241.04 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p854-b14", - "global_id": 24927, - "bbox": [ - 82.43, - 203.73, - 86.43, - 211.73 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p854-b15", - "global_id": 24928, - "bbox": [ - 65.68, - 225.69, - 76.95, - 246.28 - ], - "text": "xaq[n]", - "type": "text" - }, - { - "block_id": "p854-b16", - "global_id": 24929, - "bbox": [ - 322.17, - 274.52, - 468.73, - 295.71 - ], - "text": "–20\n–10\n0\n10\n20\nf", - "type": "text" - }, - { - "block_id": "p854-b17", - "global_id": 24930, - "bbox": [ - 304.43, - 262.91, - 308.43, - 270.91 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p854-b18", - "global_id": 24931, - "bbox": [ - 299.93, - 239.02, - 307.93, - 247.02 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p854-b19", - "global_id": 24932, - "bbox": [ - 299.93, - 215.13, - 307.93, - 223.13 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p854-b20", - "global_id": 24933, - "bbox": [ - 285.43, - 223.93, - 296.7, - 248.53 - ], - "text": "|Xaq(f)|", - "type": "text" - }, - { - "block_id": "p854-b21", - "global_id": 24934, - "bbox": [ - 64.98, - 302.38, - 272.69, - 311.62 - ], - "text": "Figure 8.33 Signal and spectrum effects of quantization.", - "type": "text" - }, - { - "block_id": "p854-b22", - "global_id": 24935, - "bbox": [ - 101.84, - 333.83, - 444.63, - 361.72 - ], - "text": ">>\nxlabel(’n’);ylabel(’x_{aq}[n]’);\n>>\nsubplot(2,2,4); stem(f-25,fftshift(abs(fft(xaq))),’k’); axis([-25,25 -1 26]);\n>>\nxlabel(’f’);ylabel(’|X_{aq}(f)|’);", - "type": "text" - }, - { - "block_id": "p854-b23", - "global_id": 24936, - "bbox": [ - 101.84, - 372.74, - 490.4, - 407.73 - ], - "text": "The results are shown in Fig. 8.33. The original signal x[n] appears sinusoidal and has pure\nspectral content at ±1 Hz. The asymmetrically quantized signal xaq[n] is significantly distorted.\nThe corresponding magnitude spectrum |Xaq(f)| is spread over a broad range of frequencies.", - "type": "text" - }, - { - "block_id": "p854-b24", - "global_id": 24937, - "bbox": [ - 102.2, - 437.52, - 194.48, - 451.47 - ], - "text": "8.8 SUMMARY", - "type": "text" - }, - { - "block_id": "p854-b25", - "global_id": 24938, - "bbox": [ - 101.84, - 457.35, - 490.4, - 587.66 - ], - "text": "A signal bandlimited to B Hz can be reconstructed exactly from its samples if the sampling\nrate fs > 2B Hz (the sampling theorem). Such a reconstruction, although possible theoretically,\nposes practical problems such as the need for ideal filters, which are unrealizable or are realizable\nonly with infinite delay. Therefore, in practice, there is always an error in reconstructing a signal\nfrom its samples. Moreover, practical signals are not bandlimited, which causes an additional\nerror (aliasing error) in signal reconstruction from its samples. When a signal is sampled at\na frequency fs Hz, samples of a sinusoid of frequency (fs/2) + x Hz appear as samples of a\nlower frequency (fs/2) −x Hz. This phenomenon, in which higher frequencies appear as lower\nfrequencies, is known as aliasing. Aliasing error can be reduced by bandlimiting a signal to fs/2\nHz (half the sampling frequency). Such bandlimiting, done prior to sampling, is accomplished by\nan anti-aliasing filter that is an ideal lowpass filter of cutoff frequency fs/2 Hz.", - "type": "text" - }, - { - "block_id": "p854-b26", - "global_id": 24939, - "bbox": [ - 101.85, - 588.96, - 490.43, - 634.79 - ], - "text": "The sampling theorem is very important in signal analysis, processing, and transmission\nbecause it allows us to replace a continuous-time signal with a discrete sequence of numbers.\nProcessing a continuous-time signal is therefore equivalent to processing a discrete sequence of\nnumbers. This leads us directly into the area of digital filtering (discrete-time systems). In the field", - "type": "text" - } - ] - }, - { - "page_num": 855, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p855-b0", - "global_id": 24940, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n835", - "type": "text" - }, - { - "block_id": "p855-b1", - "global_id": 24941, - "bbox": [ - 127.59, - 85.82, - 516.15, - 119.69 - ], - "text": "of communication, the transmission of a continuous-time message reduces to the transmission\nof a sequence of numbers. This opens doors to many new techniques of communicating\ncontinuous-time signals by pulse trains.", - "type": "text" - }, - { - "block_id": "p855-b2", - "global_id": 24942, - "bbox": [ - 127.59, - 121.27, - 516.16, - 215.34 - ], - "text": "The dual of the sampling theorem states that for a signal timelimited to τ seconds, its spectrum\nX(ω) can be reconstructed from the samples of X(ω) taken at uniform intervals not greater than\n1/τ Hz. In other words, the spectrum should be sampled at a rate not less than τ samples/Hz.\nTo compute the direct or the inverse Fourier transform numerically, we need a relationship\nbetween the samples of x(t) and X(ω). The sampling theorem and its dual provide such a\nquantitative relationship in the form of a discrete Fourier transform (DFT). The DFT computations\nare greatly facilitated by a fast Fourier transform (FFT) algorithm, which reduces the number of\ncomputations from something on the order of N2", - "type": "text" - }, - { - "block_id": "p855-b3", - "global_id": 24943, - "bbox": [ - 317.69, - 205.27, - 373.7, - 218.25 - ], - "text": "0 to N0 logN0.", - "type": "text" - }, - { - "block_id": "p855-b4", - "global_id": 24944, - "bbox": [ - 127.86, - 239.64, - 215.04, - 250.6 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p855-b5", - "global_id": 24945, - "bbox": [ - 127.59, - 257.66, - 516.12, - 277.68 - ], - "text": "1.\nLinden, D. A. A discussion of sampling theorem. Proceedings of the IRE, vol. 47, pp. 1219–1226, July\n1959.", - "type": "text" - }, - { - "block_id": "p855-b6", - "global_id": 24946, - "bbox": [ - 127.59, - 282.57, - 444.36, - 291.63 - ], - "text": "2.\nSiebert, W. M. Circuits, Signals, and Systems. MIT/McGraw-Hill, New York, 1986.", - "type": "text" - }, - { - "block_id": "p855-b7", - "global_id": 24947, - "bbox": [ - 127.59, - 296.52, - 446.66, - 305.58 - ], - "text": "3.\nBennett, W. R. Introduction to Signal Transmission. McGraw-Hill, New York, 1970.", - "type": "text" - }, - { - "block_id": "p855-b8", - "global_id": 24948, - "bbox": [ - 127.59, - 310.46, - 473.55, - 319.52 - ], - "text": "4.\nLathi, B. P. Linear Systems and Signals. Berkeley-Cambridge Press, Carmichael, CA, 1992.", - "type": "text" - }, - { - "block_id": "p855-b9", - "global_id": 24949, - "bbox": [ - 127.59, - 324.5, - 516.12, - 344.43 - ], - "text": "5.\nCooley, J. W., and Tukey, J. W. An algorithm for the machine calculation of complex Fourier series.\nMathematics of Computation, vol. 19, pp. 297–301, April 1965.", - "type": "text" - }, - { - "block_id": "p855-b10", - "global_id": 24950, - "bbox": [ - 106.67, - 380.19, - 217.26, - 397.13 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p855-b11", - "global_id": 24951, - "bbox": [ - 87.57, - 407.76, - 289.83, - 449.7 - ], - "text": "[Note: In many problems, the plots of spectra are shown\nas functions of frequency f Hz for convenience, although\nwe have labeled them as functions of ω as X(ω), Y(ω),\netc.]", - "type": "text" - }, - { - "block_id": "p855-b12", - "global_id": 24952, - "bbox": [ - 87.82, - 461.97, - 288.99, - 493.22 - ], - "text": "8.1-1\nIf fs is the Nyquist rate for signal x(t), determine\nthe Nyquist rate for each of the following\nsignals:", - "type": "text" - }, - { - "block_id": "p855-b13", - "global_id": 24953, - "bbox": [ - 116.97, - 493.41, - 164.58, - 504.92 - ], - "text": "(a) ya(t) = d", - "type": "text" - }, - { - "block_id": "p855-b14", - "global_id": 24954, - "bbox": [ - 116.46, - 494.84, - 212.27, - 515.88 - ], - "text": "dtx(t)\n(b) yb(t) = x(t)cos(2πf0t)", - "type": "text" - }, - { - "block_id": "p855-b15", - "global_id": 24955, - "bbox": [ - 116.96, - 516.76, - 288.98, - 526.84 - ], - "text": "(c) yc(t) = x(t +a)+x(t −b), for real constants", - "type": "text" - }, - { - "block_id": "p855-b16", - "global_id": 24956, - "bbox": [ - 116.46, - 528.0, - 229.98, - 548.76 - ], - "text": "a and b\n(d) yd(t) = x(at), for real a > 0", - "type": "text" - }, - { - "block_id": "p855-b17", - "global_id": 24957, - "bbox": [ - 314.97, - 407.9, - 516.14, - 439.59 - ], - "text": "8.1-2\nFigure P8.1-2 shows Fourier spectra of signals\nx1(t) and x2(t). Determine the Nyquist sam-\npling rates for signals x1(t),x2(t),x2", - "type": "text" - }, - { - "block_id": "p855-b18", - "global_id": 24958, - "bbox": [ - 343.61, - 428.41, - 516.12, - 450.55 - ], - "text": "1(t),x3\n2(t), and\nx1(t)x2(t).", - "type": "text" - }, - { - "block_id": "p855-b19", - "global_id": 24959, - "bbox": [ - 314.97, - 461.69, - 516.14, - 492.95 - ], - "text": "8.1-3\nA signal x(t) has a bandwidth of B = 1000 Hz.\nFor a positive integer N, what is the Nyquist rate\nfor the signal y(t) = xN(t)?", - "type": "text" - }, - { - "block_id": "p855-b20", - "global_id": 24960, - "bbox": [ - 314.97, - 505.12, - 516.13, - 525.13 - ], - "text": "8.1-4\nDetermine the Nyquist sampling rate and the\nNyquist sampling interval for the signals:", - "type": "text" - }, - { - "block_id": "p855-b21", - "global_id": 24961, - "bbox": [ - 343.61, - 525.4, - 422.99, - 547.04 - ], - "text": "(a) sinc2(100πt)\n(b) 0.01sinc2(100πt)", - "type": "text" - }, - { - "block_id": "p855-b22", - "global_id": 24962, - "bbox": [ - 199.58, - 627.9, - 395.92, - 635.9 - ], - "text": "0\n0", - "type": "text" - }, - { - "block_id": "p855-b23", - "global_id": 24963, - "bbox": [ - 253.93, - 615.47, - 496.26, - 623.47 - ], - "text": "v\nv", - "type": "text" - }, - { - "block_id": "p855-b24", - "global_id": 24964, - "bbox": [ - 235.08, - 626.43, - 498.08, - 635.9 - ], - "text": "3p 105\n2p 105", - "type": "text" - }, - { - "block_id": "p855-b25", - "global_id": 24965, - "bbox": [ - 200.0, - 571.1, - 410.83, - 580.73 - ], - "text": "X1(v)\nX2(v)", - "type": "text" - }, - { - "block_id": "p855-b26", - "global_id": 24966, - "bbox": [ - 130.58, - 642.31, - 182.22, - 651.28 - ], - "text": "Figure P8.1-2", - "type": "text" - } - ] - }, - { - "page_num": 856, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p856-b0", - "global_id": 24967, - "bbox": [ - 60.0, - 60.36, - 421.85, - 69.45 - ], - "text": "836\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p856-b1", - "global_id": 24968, - "bbox": [ - 90.72, - 86.08, - 206.71, - 107.73 - ], - "text": "(c) sinc(100πt) + 3sinc2(60πt)\n(d) sinc(50πt)sinc(100πt)", - "type": "text" - }, - { - "block_id": "p856-b2", - "global_id": 24969, - "bbox": [ - 62.08, - 115.94, - 263.24, - 125.27 - ], - "text": "8.1-5\n(a) Sketch |X(ω)|, the amplitude spectrum", - "type": "text" - }, - { - "block_id": "p856-b3", - "global_id": 24970, - "bbox": [ - 90.72, - 126.89, - 263.24, - 191.03 - ], - "text": "of a signal x(t) = 3cos6πt + sin18πt +\n2cos(28 −ϵ)πt, where ϵ is a very small\nnumber →0. Determine the minimum sam-\npling rate required to be able to reconstruct\nx(t) from these samples.\n(b) Sketch the amplitude spectrum of the sam-", - "type": "text" - }, - { - "block_id": "p856-b4", - "global_id": 24971, - "bbox": [ - 106.16, - 193.02, - 263.25, - 245.82 - ], - "text": "pled signal when the sampling rate is 25%\nabove the Nyquist rate (show the spectrum\nover the frequency range ±50 Hz only).\nHow would you reconstruct x(t) from these\nsamples?", - "type": "text" - }, - { - "block_id": "p856-b5", - "global_id": 24972, - "bbox": [ - 62.08, - 254.32, - 263.23, - 263.36 - ], - "text": "8.1-6\n(a) Derive the sampling theorem by consid-", - "type": "text" - }, - { - "block_id": "p856-b6", - "global_id": 24973, - "bbox": [ - 90.72, - 265.36, - 263.25, - 307.2 - ], - "text": "ering the fact that the sampled signal\nx(t)\n=\nx(t)δT(t),\nand\nusing\nthe\nfrequency-convolutionpropertyinEq.(7.34).\n(b) For a sampling train consisting of shifted", - "type": "text" - }, - { - "block_id": "p856-b7", - "global_id": 24974, - "bbox": [ - 106.16, - 308.81, - 263.24, - 351.03 - ], - "text": "unit impulses at instants nT + τ instead\nof at nT (for all positive and negative\ninteger values of n), find the spectrum of the\nsampled signal.", - "type": "text" - }, - { - "block_id": "p856-b8", - "global_id": 24975, - "bbox": [ - 62.08, - 359.53, - 263.25, - 434.33 - ], - "text": "8.1-7\nA signal is bandlimited to 12 kHz. The band\nbetween 10 and 12 kHz has been so cor-\nrupted by excessive noise that the information\nin this band is nonrecoverable. Determine the\nminimum sampling rate for this signal so that\nthe uncorrupted portion of the band can be\nrecovered. If we were to filter out the corrupted", - "type": "text" - }, - { - "block_id": "p856-b9", - "global_id": 24976, - "bbox": [ - 317.86, - 85.9, - 490.38, - 105.83 - ], - "text": "spectrum prior to sampling, what would be the\nminimum sampling rate?", - "type": "text" - }, - { - "block_id": "p856-b10", - "global_id": 24977, - "bbox": [ - 289.22, - 111.01, - 490.4, - 218.98 - ], - "text": "8.1-8\nA continuous-time signal x(t) = ((t −1)/2)\nis sampled at three rates: 10, 2, and 1 Hz.\nSketch the resulting sampled signals. Because\nx(t) is timelimited, its bandwidth is infinite.\nHowever, most of its energy is concentrated in\na small band. Determine a reasonable minimum\nsampling rate that will allow reconstruction of\nthis signal with a small error. The answer is not\nunique. Make a reasonable assumption of what\nyou define as a “negligible” or “small” error.", - "type": "text" - }, - { - "block_id": "p856-b11", - "global_id": 24978, - "bbox": [ - 289.23, - 220.55, - 490.23, - 233.51 - ], - "text": "8.1-9\n(a) A signal x(t) = 5sinc2 (5πt) + cos20πt", - "type": "text" - }, - { - "block_id": "p856-b12", - "global_id": 24979, - "bbox": [ - 317.86, - 235.49, - 490.4, - 288.29 - ], - "text": "is sampled at a rate of 10 Hz. Find the\nspectrum of the sampled signal. Can x(t)\nbe reconstructed by lowpass filtering the\nsampled signal?\n(b) Repeat part (a) for a sampling frequency of", - "type": "text" - }, - { - "block_id": "p856-b13", - "global_id": 24980, - "bbox": [ - 318.37, - 290.29, - 490.39, - 321.17 - ], - "text": "20 Hz. Can you reconstruct the signal from\nthis sampled signal? Explain.\n(c) If x(t) = 5sinc2 (5πt) + sin20πt, can you", - "type": "text" - }, - { - "block_id": "p856-b14", - "global_id": 24981, - "bbox": [ - 317.86, - 322.79, - 490.39, - 365.01 - ], - "text": "reconstruct x(t) from the samples of x(t) at\na rate of 20 Hz? Explain your answer with\nspectral representation(s).\n(d) For x(t) = 5sinc2 (5πt) + sin20πt, can you", - "type": "text" - }, - { - "block_id": "p856-b15", - "global_id": 24982, - "bbox": [ - 333.31, - 366.63, - 490.39, - 408.85 - ], - "text": "reconstruct x(t) from the samples of x(t)\nat a rate of 21 Hz? Explain your answer\nwith spectral representation(s). Comment\non your results.", - "type": "text" - }, - { - "block_id": "p856-b16", - "global_id": 24983, - "bbox": [ - 284.74, - 414.02, - 490.4, - 423.36 - ], - "text": "8.1-10\n(a) The highest frequency in the spectrum X(ω)", - "type": "text" - }, - { - "block_id": "p856-b17", - "global_id": 24984, - "bbox": [ - 333.31, - 424.99, - 490.39, - 434.33 - ], - "text": "(Fig. P8.1-10a) of a bandpass signal x(t)", - "type": "text" - }, - { - "block_id": "p856-b18", - "global_id": 24985, - "bbox": [ - 113.5, - 505.38, - 290.94, - 513.68 - ], - "text": "0\n30", - "type": "text" - }, - { - "block_id": "p856-b19", - "global_id": 24986, - "bbox": [ - 290.44, - 451.83, - 305.99, - 459.93 - ], - "text": "X(v)", - "type": "text" - }, - { - "block_id": "p856-b20", - "global_id": 24987, - "bbox": [ - 168.1, - 505.38, - 449.83, - 513.68 - ], - "text": "20\n20\n30", - "type": "text" - }, - { - "block_id": "p856-b21", - "global_id": 24988, - "bbox": [ - 275.42, - 457.34, - 279.42, - 465.34 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p856-b22", - "global_id": 24989, - "bbox": [ - 279.0, - 523.43, - 287.88, - 531.43 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p856-b23", - "global_id": 24990, - "bbox": [ - 278.53, - 615.14, - 288.34, - 623.14 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p856-b24", - "global_id": 24991, - "bbox": [ - 140.5, - 599.35, - 290.94, - 608.53 - ], - "text": "0\n28", - "type": "text" - }, - { - "block_id": "p856-b25", - "global_id": 24992, - "bbox": [ - 290.44, - 544.78, - 305.55, - 552.88 - ], - "text": "Y(v)", - "type": "text" - }, - { - "block_id": "p856-b26", - "global_id": 24993, - "bbox": [ - 195.12, - 600.23, - 422.21, - 608.53 - ], - "text": "18\n18\n28", - "type": "text" - }, - { - "block_id": "p856-b27", - "global_id": 24994, - "bbox": [ - 275.42, - 550.28, - 279.42, - 558.28 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p856-b28", - "global_id": 24995, - "bbox": [ - 321.81, - 508.11, - 341.89, - 516.19 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p856-b29", - "global_id": 24996, - "bbox": [ - 429.13, - 600.03, - 449.21, - 608.11 - ], - "text": "f (Hz)", - "type": "text" - }, - { - "block_id": "p856-b30", - "global_id": 24997, - "bbox": [ - 104.83, - 629.31, - 160.95, - 638.27 - ], - "text": "Figure P8.1-10", - "type": "text" - } - ] - }, - { - "page_num": 857, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p857-b0", - "global_id": 24998, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n837", - "type": "text" - }, - { - "block_id": "p857-b1", - "global_id": 24999, - "bbox": [ - 116.46, - 85.9, - 288.99, - 149.66 - ], - "text": "is 30 Hz. Hence, the minimum sampling\nfrequency needed to sample x(t) is 60 Hz.\nShow the spectrum of the signal sampled at\na rate of 60 Hz. Can you reconstruct x(t)\nfrom these samples? How?\n(b) A certain busy student looks at X(ω), con-", - "type": "text" - }, - { - "block_id": "p857-b2", - "global_id": 25000, - "bbox": [ - 116.97, - 151.65, - 288.99, - 226.38 - ], - "text": "cludes that its bandwidth is really 10 Hz,\nand decides that the sampling rate 20 Hz\nis adequate for sampling x(t). Sketch the\nspectrum of the signal sampled at a rate of\n20 Hz. Can x(t) be reconstructed from these\nsamples?\n(c) The same student, using the same reasoning,", - "type": "text" - }, - { - "block_id": "p857-b3", - "global_id": 25001, - "bbox": [ - 131.9, - 227.99, - 288.99, - 303.08 - ], - "text": "looks at Y(ω) in Fig. P8.1-10b, the spectrum\nof another bandpass signal y(t), and con-\ncludes that the sampling rate of 20 Hz can\nbe used to sample y(t). Sketch the spectrum\nof the signal y(t) sampled at a rate of 20\nHz. Can y(t) be reconstructed from these\nsamples?", - "type": "text" - }, - { - "block_id": "p857-b4", - "global_id": 25002, - "bbox": [ - 83.34, - 307.7, - 288.99, - 349.9 - ], - "text": "8.1-11\nA signal x(t) whose spectrum X(ω), as shown\nin Fig. P8.1-11, is sampled at a frequency fs =\nf1 + f2 Hz. Find all the sample values of x(t)\nmerely by inspection of X(ω).", - "type": "text" - }, - { - "block_id": "p857-b5", - "global_id": 25003, - "bbox": [ - 83.34, - 354.81, - 288.98, - 407.69 - ], - "text": "8.1-12\nAs described in Sec. 8.1-1, practical sampling\ncan be achieved by multiplying a signal x(t) by\na periodic train of pulses pT(t). The pulse train\npT(t) can be created by the periodic replication\nof some pulse p(t) as", - "type": "text" - }, - { - "block_id": "p857-b6", - "global_id": 25004, - "bbox": [ - 161.19, - 425.38, - 188.45, - 435.39 - ], - "text": "pT(t) =", - "type": "text" - }, - { - "block_id": "p857-b7", - "global_id": 25005, - "bbox": [ - 193.8, - 416.06, - 206.48, - 425.83 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p857-b8", - "global_id": 25006, - "bbox": [ - 190.3, - 438.21, - 209.98, - 444.89 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p857-b9", - "global_id": 25007, - "bbox": [ - 210.99, - 425.38, - 244.26, - 434.63 - ], - "text": "p(t −kT)", - "type": "text" - }, - { - "block_id": "p857-b10", - "global_id": 25008, - "bbox": [ - 116.46, - 453.61, - 288.98, - 473.92 - ], - "text": "It is desired to sample a signal at a rate fs = 100\nHz, and two pulses are under consideration:", - "type": "text" - }, - { - "block_id": "p857-b11", - "global_id": 25009, - "bbox": [ - 140.96, - 482.94, - 181.2, - 494.37 - ], - "text": "pa(t) = −1", - "type": "text" - }, - { - "block_id": "p857-b12", - "global_id": 25010, - "bbox": [ - 177.97, - 482.87, - 251.97, - 496.45 - ], - "text": "4u(t) + 5\n4u(t −2T\n20 )+", - "type": "text" - }, - { - "block_id": "p857-b13", - "global_id": 25011, - "bbox": [ - 168.39, - 498.25, - 181.2, - 508.57 - ], - "text": "−5", - "type": "text" - }, - { - "block_id": "p857-b14", - "global_id": 25012, - "bbox": [ - 177.97, - 498.18, - 264.49, - 511.76 - ], - "text": "4u(t −3T\n20 ) + 1\n4u(t −5T\n20 )", - "type": "text" - }, - { - "block_id": "p857-b15", - "global_id": 25013, - "bbox": [ - 116.46, - 519.7, - 129.4, - 528.66 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p857-b16", - "global_id": 25014, - "bbox": [ - 143.77, - 535.72, - 261.67, - 549.13 - ], - "text": "pb(t) = e−t/T [u(t) −u(t −1.5T)]", - "type": "text" - }, - { - "block_id": "p857-b17", - "global_id": 25015, - "bbox": [ - 343.61, - 85.58, - 496.23, - 106.62 - ], - "text": "(a) Plot pT(t) using pa(t) over 0 ≤t ≤4T.\n(b) Plot pT(t) using pb(t) over 0 ≤t ≤4T.", - "type": "text" - }, - { - "block_id": "p857-b18", - "global_id": 25016, - "bbox": [ - 344.12, - 107.49, - 516.13, - 117.57 - ], - "text": "(c) Which pulse, pa(t) or pb(t), is more suitable", - "type": "text" - }, - { - "block_id": "p857-b19", - "global_id": 25017, - "bbox": [ - 359.05, - 118.83, - 516.13, - 138.76 - ], - "text": "as a sampling pulse? Carefully explain your\nanswer.", - "type": "text" - }, - { - "block_id": "p857-b20", - "global_id": 25018, - "bbox": [ - 310.48, - 153.66, - 516.14, - 436.67 - ], - "text": "8.1-13\nIn digital data transmission over a communica-\ntion channel, it is important to know the upper\ntheoretical limit on the rate of digital pulses that\ncan be transmitted over a channel of bandwidth\nB Hz. In digital transmission, the relative shape\nof the pulse is not important. We are interested\nin knowing only the amplitude represented by\nthe pulse. For instance, in binary communica-\ntion, we are interested in knowing whether the\nreceived pulse amplitude is 1 or −1 (positive\nor negative). Thus, each pulse represents one\npiece of information. Consider one independent\namplitude value (not necessarily binary) as one\npiece of information. Show that 2B indepen-\ndent pieces of information per second can be\ntransmitted correctly (assuming no noise) over\na channel of bandwidth B Hz. This important\nprinciple in communication theory states that\n1 Hz of bandwidth can transmit two independent\npieces of information per second. It represents\nthe upper rate of pulse transmission over a chan-\nnel without any error in reception in the absence\nof noise. [Hint: According to the interpolation\nformula [Eq. (8.6)], a continuous-time signal of\nbandwidth B Hz can be constructed from 2B\npieces of information/second.]", - "type": "text" - }, - { - "block_id": "p857-b21", - "global_id": 25019, - "bbox": [ - 310.48, - 451.57, - 516.14, - 548.29 - ], - "text": "8.1-14\nThis example is one of those interesting situa-\ntions leading to a curious result in the category\nof defying gravity. The sinc function can be\nrecovered from its samples taken at extremely\nlow frequencies in apparent defiance of the\nsampling theorem.\nConsider a sinc pulse x(t) = sinc(4πt) for which\nX(ω) = (1/4)rect(ω/8π). The bandwidth of x(t)\nis B = 2 Hz, and its Nyquist rate is 4 Hz.", - "type": "text" - }, - { - "block_id": "p857-b22", - "global_id": 25020, - "bbox": [ - 268.68, - 624.31, - 272.68, - 632.31 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p857-b23", - "global_id": 25021, - "bbox": [ - 257.58, - 566.46, - 285.64, - 575.0 - ], - "text": "X(v)\n1", - "type": "text" - }, - { - "block_id": "p857-b24", - "global_id": 25022, - "bbox": [ - 166.98, - 623.85, - 396.55, - 634.62 - ], - "text": "f2\nf1\nf1\nf2\nf (Hz)", - "type": "text" - }, - { - "block_id": "p857-b25", - "global_id": 25023, - "bbox": [ - 130.58, - 641.27, - 186.69, - 650.23 - ], - "text": "Figure P8.1-11", - "type": "text" - } - ] - }, - { - "page_num": 858, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p858-b0", - "global_id": 25024, - "bbox": [ - 60.0, - 60.36, - 421.85, - 69.45 - ], - "text": "838\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p858-b1", - "global_id": 25025, - "bbox": [ - 91.22, - 85.58, - 263.23, - 94.92 - ], - "text": "(a) Sample x(t) at a rate 4 Hz and sketch the", - "type": "text" - }, - { - "block_id": "p858-b2", - "global_id": 25026, - "bbox": [ - 90.72, - 96.91, - 263.23, - 116.83 - ], - "text": "spectrum of the sampled signal.\n(b) To recover x(t) from its samples, we pass the", - "type": "text" - }, - { - "block_id": "p858-b3", - "global_id": 25027, - "bbox": [ - 91.23, - 118.83, - 263.24, - 204.5 - ], - "text": "sampled signal through an ideal lowpass fil-\nter of bandwidth B = 2 Hz and gain G = T =\n1/4. Sketch this system and show that for\nthis system H(ω) = (1/4)rect(ω/8π). Show\nalso that when the input is the sampled x(t)\nat a rate 4 Hz, the output of this system is\nindeed x(t), as expected.\n(c) Now sample x(t) at half the Nyquist rate, at", - "type": "text" - }, - { - "block_id": "p858-b4", - "global_id": 25028, - "bbox": [ - 90.72, - 206.5, - 263.25, - 248.34 - ], - "text": "2 Hz. Apply this sampled signal at the input\nof the lowpass filter used in part (b). Find\nthe output.\n(d) Repeat part (c) for the sampling rate 1 Hz.", - "type": "text" - }, - { - "block_id": "p858-b5", - "global_id": 25029, - "bbox": [ - 91.23, - 250.34, - 263.24, - 259.31 - ], - "text": "(e) Show that the output of the lowpass filter", - "type": "text" - }, - { - "block_id": "p858-b6", - "global_id": 25030, - "bbox": [ - 92.22, - 260.92, - 263.25, - 325.06 - ], - "text": "in part (b) is x(t) to the sampled x(t) if\nthe sampling rate is 4/N, where N is any\npositive integer. This means that we can\nrecover x(t) from its samples taken at an\narbitrarily small rate by letting N →∞.\n(f) The mystery may be clarified a bit by", - "type": "text" - }, - { - "block_id": "p858-b7", - "global_id": 25031, - "bbox": [ - 106.16, - 327.05, - 263.24, - 357.94 - ], - "text": "examining the problem in the time domain.\nFind the samples of x(t) when the sampling\nrate is 2/N (N integer).", - "type": "text" - }, - { - "block_id": "p858-b8", - "global_id": 25032, - "bbox": [ - 62.08, - 372.5, - 263.24, - 480.47 - ], - "text": "8.2-1\nA signal x(t) = sinc(200πt) is sampled (multi-\nplied) by a periodic pulse train pT(t) represented\nin Fig. P8.2-1. Find and sketch the spectrum of\nthe sampled signal. Explain whether you will be\nable to reconstruct x(t) from these samples. Find\nthe filter output if the sampled signal is passed\nthrough an ideal lowpass filter of bandwidth 100\nHz and unit gain. What is the filter output if its\nbandwidth B Hz is between 100 and 150 Hz?\nWhat happens if the bandwidth exceeds 150 Hz?", - "type": "text" - }, - { - "block_id": "p858-b9", - "global_id": 25033, - "bbox": [ - 62.08, - 495.34, - 263.26, - 559.18 - ], - "text": "8.2-2\nShow that the circuit in Fig. P8.2-2 is a realiza-\ntion of the causal ZOH (zero-order hold) circuit.\nYou can do this by showing that the unit impulse\nresponse h(t) of this circuit is indeed equal to\nthat in Eq. (8.5) delayed by T/2 seconds to make\nit causal.", - "type": "text" - }, - { - "block_id": "p858-b10", - "global_id": 25034, - "bbox": [ - 336.74, - 94.48, - 400.39, - 107.11 - ], - "text": "Delay", - "type": "text" - }, - { - "block_id": "p858-b11", - "global_id": 25035, - "bbox": [ - 344.07, - 103.4, - 348.52, - 111.4 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p858-b14", - "global_id": 25036, - "bbox": [ - 285.91, - 89.27, - 488.21, - 107.81 - ], - "text": "Input\nOutput", - "type": "text" - }, - { - "block_id": "p858-b15", - "global_id": 25037, - "bbox": [ - 285.82, - 137.73, - 337.47, - 146.69 - ], - "text": "Figure P8.2-2", - "type": "text" - }, - { - "block_id": "p858-b16", - "global_id": 25038, - "bbox": [ - 289.22, - 160.94, - 490.4, - 169.98 - ], - "text": "8.2-3\n(a) A first-order hold circuit (FOH) can also", - "type": "text" - }, - { - "block_id": "p858-b17", - "global_id": 25039, - "bbox": [ - 317.86, - 171.6, - 490.39, - 290.52 - ], - "text": "be used to reconstruct a signal x(t) from\nits samples. The impulse response of this\ncircuit is h(t) = (t/2T), where T is the\nsampling interval. Consider a typical sam-\npled signal x(t) and show that this circuit\nperforms the linear interpolation. In other\nwords, the filter output consists of sample\ntops connected by straight-line segments.\nFollow the procedure discussed in Sec. 8.2\n(Fig. 8.5c).\n(b) Determine the frequency and magnitude", - "type": "text" - }, - { - "block_id": "p858-b18", - "global_id": 25040, - "bbox": [ - 318.37, - 292.52, - 490.39, - 334.36 - ], - "text": "responses of this filter, and compare it\nwith (i) the ideal filter required for signal\nreconstruction and (ii) a ZOH circuit.\n(c) This filter, being noncausal, is unrealiz-", - "type": "text" - }, - { - "block_id": "p858-b19", - "global_id": 25041, - "bbox": [ - 317.86, - 336.36, - 490.41, - 411.08 - ], - "text": "able. By delaying its impulse response,\nthe filter can be made realizable. What is\nthe minimum delay required to make it\nrealizable? How would this delay affect the\nreconstructed signal and the filter frequency\nresponse?\n(d) Show that the causal FOH circuit in part (c)", - "type": "text" - }, - { - "block_id": "p858-b20", - "global_id": 25042, - "bbox": [ - 333.31, - 413.07, - 490.39, - 443.96 - ], - "text": "can be realized by the ZOH circuit depicted\nin Fig. P8.2-2 followed by an identical filter\nin cascade.", - "type": "text" - }, - { - "block_id": "p858-b21", - "global_id": 25043, - "bbox": [ - 289.22, - 449.74, - 494.78, - 480.99 - ], - "text": "8.2-4\nSuppose signal x(t)=sin(2πt/8)(u(t) −u(t −8))\nis sampled at a rate fs = 1 Hz to generate signal\nx[n].", - "type": "text" - }, - { - "block_id": "p858-b22", - "global_id": 25044, - "bbox": [ - 317.86, - 482.61, - 490.39, - 502.91 - ], - "text": "(a) Sketch x(t) and x[n].\n(b) Has aliasing occurred in sampling x(t) to", - "type": "text" - }, - { - "block_id": "p858-b23", - "global_id": 25045, - "bbox": [ - 318.37, - 504.53, - 490.39, - 524.83 - ], - "text": "produce x[n]? Explain.\n(c) Sketch the output ˆx(t) produced when x[n]", - "type": "text" - }, - { - "block_id": "p858-b24", - "global_id": 25046, - "bbox": [ - 333.31, - 526.83, - 490.38, - 557.71 - ], - "text": "is applied to the causal ZOH reconstructor\nof Prob. 8.2-2. How does ˆx(t) compare with\nx(t)?", - "type": "text" - }, - { - "block_id": "p858-b25", - "global_id": 25047, - "bbox": [ - 409.37, - 575.47, - 430.7, - 583.47 - ], - "text": "0.8 ms", - "type": "text" - }, - { - "block_id": "p858-b26", - "global_id": 25048, - "bbox": [ - 284.51, - 612.21, - 393.37, - 623.14 - ], - "text": "4 ms\nt", - "type": "text" - }, - { - "block_id": "p858-b27", - "global_id": 25049, - "bbox": [ - 268.66, - 575.26, - 284.63, - 584.81 - ], - "text": "pT(t)", - "type": "text" - }, - { - "block_id": "p858-b28", - "global_id": 25050, - "bbox": [ - 314.18, - 615.14, - 329.51, - 623.14 - ], - "text": "8 ms", - "type": "text" - }, - { - "block_id": "p858-b29", - "global_id": 25051, - "bbox": [ - 104.83, - 595.71, - 419.68, - 601.71 - ], - "text": "• • •\n• • •", - "type": "text" - }, - { - "block_id": "p858-b30", - "global_id": 25052, - "bbox": [ - 104.83, - 629.31, - 156.48, - 638.27 - ], - "text": "Figure P8.2-1", - "type": "text" - } - ] - }, - { - "page_num": 859, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p859-b0", - "global_id": 25053, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n839", - "type": "text" - }, - { - "block_id": "p859-b1", - "global_id": 25054, - "bbox": [ - 116.46, - 86.14, - 288.98, - 95.47 - ], - "text": "(d) Sketch the output ˆx(t) produced when x[n]", - "type": "text" - }, - { - "block_id": "p859-b2", - "global_id": 25055, - "bbox": [ - 131.9, - 97.46, - 288.99, - 117.39 - ], - "text": "is applied to the FOH reconstructor of\nProb.8.2-3.Howdoes ˆx(t)comparewithx(t)?", - "type": "text" - }, - { - "block_id": "p859-b3", - "global_id": 25056, - "bbox": [ - 87.82, - 123.92, - 288.98, - 144.22 - ], - "text": "8.2-5\nRepeat Prob. 8.2-4 for the signal x(t) =\ncos(2πt/8)(u(t) −u(t −8)).", - "type": "text" - }, - { - "block_id": "p859-b4", - "global_id": 25057, - "bbox": [ - 87.83, - 151.05, - 288.99, - 203.93 - ], - "text": "8.2-6\nIs it possible to sample a physically realizable\n(nonzero) signal x(t) with a physically realizable\nsystem without aliasing? If possible, explain\nwhat conditions must be met. If not possible,\nexplain why not.", - "type": "text" - }, - { - "block_id": "p859-b5", - "global_id": 25058, - "bbox": [ - 87.83, - 210.75, - 289.0, - 307.47 - ], - "text": "8.2-7\nIn the text, for sampling purposes, we used\ntimelimited narrow pulses such as impulses\nor rectangular pulses of width less than the\nsampling interval T. Show that it is not necessary\nto restrict the sampling pulse width. We can use\nsampling pulses of arbitrarily large duration and\nstill be able to reconstruct the signal x(t) as long\nas the pulse rate is no less than the Nyquist rate\nfor x(t).", - "type": "text" - }, - { - "block_id": "p859-b6", - "global_id": 25059, - "bbox": [ - 116.46, - 309.09, - 288.99, - 417.06 - ], - "text": "Consider x(t) to be bandlimited to B Hz.\nThe sampling pulse to be used is an exponential\ne−atu(t). We multiply x(t) by a periodic train of\nexponential pulses of the form e−atu(t) spaced T\nseconds apart. Find the spectrum of the sampled\nsignal, and show that x(t) can be reconstructed\nfrom this sampled signal provided the sampling\nrate is no less than 2B Hz or T < 1/2B.\nExplain how you would reconstruct x(t) from the\nsampled signal.", - "type": "text" - }, - { - "block_id": "p859-b7", - "global_id": 25060, - "bbox": [ - 87.82, - 423.58, - 288.98, - 498.68 - ], - "text": "8.2-8\nIn Ex. 8.2, the sampling of a signal x(t) was\naccomplished by multiplying the signal by a\npulse train pT(t), resulting in the sampled signal\ndepicted in Fig. 8.4d. This procedure is known\nas the natural sampling. Figure P8.2-8 shows the\nso-called flat-top sampling of the same signal\nx(t) = sinc2 (5πt).", - "type": "text" - }, - { - "block_id": "p859-b8", - "global_id": 25061, - "bbox": [ - 116.96, - 500.3, - 288.97, - 509.64 - ], - "text": "(a) Show that the signal x(t) can be recovered", - "type": "text" - }, - { - "block_id": "p859-b9", - "global_id": 25062, - "bbox": [ - 116.46, - 511.63, - 288.99, - 542.52 - ], - "text": "from flat-top samples if the sampling rate is\nno less than the Nyquist rate.\n(b) Explain how you would recover x(t) from", - "type": "text" - }, - { - "block_id": "p859-b10", - "global_id": 25063, - "bbox": [ - 116.97, - 544.51, - 288.99, - 564.43 - ], - "text": "the flat-top samples.\n(c) Find the expression for the sampled signal", - "type": "text" - }, - { - "block_id": "p859-b11", - "global_id": 25064, - "bbox": [ - 131.9, - 566.05, - 264.89, - 575.39 - ], - "text": "spectrum X(ω) and sketch it roughly.", - "type": "text" - }, - { - "block_id": "p859-b12", - "global_id": 25065, - "bbox": [ - 87.82, - 582.2, - 288.99, - 614.52 - ], - "text": "8.2-9\nA sinusoid of frequency f0 Hz is sampled at a\nrate fs = 20 Hz. Find the apparent frequency of\nthe sampled signal if f0 is:", - "type": "text" - }, - { - "block_id": "p859-b13", - "global_id": 25066, - "bbox": [ - 116.46, - 615.17, - 153.56, - 635.1 - ], - "text": "(a) 8 Hz\n(b) 12 Hz", - "type": "text" - }, - { - "block_id": "p859-b14", - "global_id": 25067, - "bbox": [ - 412.82, - 88.06, - 416.82, - 96.06 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p859-b15", - "global_id": 25068, - "bbox": [ - 371.08, - 176.11, - 488.3, - 186.74 - ], - "text": "0\n0.2\n0.1\n0.3\nt", - "type": "text" - }, - { - "block_id": "p859-b16", - "global_id": 25069, - "bbox": [ - 438.57, - 106.62, - 450.12, - 117.62 - ], - "text": "x–(t)", - "type": "text" - }, - { - "block_id": "p859-b17", - "global_id": 25070, - "bbox": [ - 345.1, - 194.66, - 396.74, - 203.63 - ], - "text": "Figure P8.2-8", - "type": "text" - }, - { - "block_id": "p859-b18", - "global_id": 25071, - "bbox": [ - 343.61, - 222.75, - 380.7, - 242.67 - ], - "text": "(c) 20 Hz\n(d) 22 Hz", - "type": "text" - }, - { - "block_id": "p859-b19", - "global_id": 25072, - "bbox": [ - 344.12, - 244.67, - 380.7, - 253.63 - ], - "text": "(e) 32 Hz", - "type": "text" - }, - { - "block_id": "p859-b20", - "global_id": 25073, - "bbox": [ - 310.48, - 259.03, - 516.12, - 302.3 - ], - "text": "8.2-10\nA sinusoid of unknown frequency f0 is sampled\nat a rate 60 Hz. The apparent frequency of the\nsamples is 20 Hz. Determine f0 if it is known\nthat f0 lies in the range:", - "type": "text" - }, - { - "block_id": "p859-b21", - "global_id": 25074, - "bbox": [ - 343.61, - 302.96, - 394.16, - 322.89 - ], - "text": "(a) 0–30 Hz\n(b) 30–60 Hz", - "type": "text" - }, - { - "block_id": "p859-b22", - "global_id": 25075, - "bbox": [ - 343.61, - 324.87, - 398.64, - 344.8 - ], - "text": "(c) 60–90 Hz\n(d) 90–120 Hz", - "type": "text" - }, - { - "block_id": "p859-b23", - "global_id": 25076, - "bbox": [ - 310.48, - 349.92, - 516.14, - 468.85 - ], - "text": "8.2-11\nA signal x(t) = 3cos6πt+cos16πt+2cos20πt\nis sampled at a rate 25% above the Nyquist\nrate. Sketch the spectrum of the sampled signal.\nHow would you reconstruct x(t) from these\nsamples? If the sampling frequency is 25%\nbelow the Nyquist rate, what are the frequencies\nof the sinusoids present in the output of the\nfilter with cutoff frequency equal to the folding\nfrequency? Do not write the actual output; give\njust the frequencies of the sinusoids present in\nthe output.", - "type": "text" - }, - { - "block_id": "p859-b24", - "global_id": 25077, - "bbox": [ - 310.48, - 473.97, - 509.37, - 483.31 - ], - "text": "8.2-12\nA complex signal x(t) has a spectrum given as", - "type": "text" - }, - { - "block_id": "p859-b25", - "global_id": 25078, - "bbox": [ - 373.29, - 505.17, - 400.75, - 514.42 - ], - "text": "X(ω) =", - "type": "text" - }, - { - "block_id": "p859-b26", - "global_id": 25079, - "bbox": [ - 402.58, - 492.58, - 480.28, - 519.94 - ], - "text": "ω\n0 ≤ω ≤2π10\n0\notherwise", - "type": "text" - }, - { - "block_id": "p859-b27", - "global_id": 25080, - "bbox": [ - 343.61, - 536.1, - 516.13, - 556.4 - ], - "text": "Let x(t) be sampled at rate fs = 24 Hz to produce\nsignal x(t) with spectrum X(ω).", - "type": "text" - }, - { - "block_id": "p859-b28", - "global_id": 25081, - "bbox": [ - 343.61, - 558.02, - 516.13, - 578.32 - ], - "text": "(a) Sketch X(ω).\n(b) Has aliasing occurred in sampling x(t) to", - "type": "text" - }, - { - "block_id": "p859-b29", - "global_id": 25082, - "bbox": [ - 344.12, - 579.94, - 516.14, - 600.24 - ], - "text": "produce x(t)? Explain.\n(c) Can x(t) be exactly recovered from x(t)?", - "type": "text" - }, - { - "block_id": "p859-b30", - "global_id": 25083, - "bbox": [ - 359.05, - 602.23, - 389.18, - 611.19 - ], - "text": "Explain.", - "type": "text" - }, - { - "block_id": "p859-b31", - "global_id": 25084, - "bbox": [ - 310.48, - 616.32, - 516.13, - 636.61 - ], - "text": "8.2-13\nRepeat Prob. 8.2-12 for the sampling rate fs = 16\nHz.", - "type": "text" - } - ] - }, - { - "page_num": 860, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p860-b0", - "global_id": 25085, - "bbox": [ - 60.0, - 60.36, - 421.85, - 69.45 - ], - "text": "840\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p860-b1", - "global_id": 25086, - "bbox": [ - 57.59, - 85.64, - 263.24, - 105.94 - ], - "text": "8.2-14\nRepeat Prob. 8.2-12 for the sampling rate fs = 8\nHz.", - "type": "text" - }, - { - "block_id": "p860-b2", - "global_id": 25087, - "bbox": [ - 57.59, - 110.58, - 263.23, - 119.92 - ], - "text": "8.2-15\n(a) Show that the signal x(t), reconstructed", - "type": "text" - }, - { - "block_id": "p860-b3", - "global_id": 25088, - "bbox": [ - 90.72, - 121.54, - 263.24, - 152.8 - ], - "text": "from its samples x(nT), using Eq. (8.6) has\na bandwidth B ≤1/2T Hz.\n(b) Show that x(t) is the smallest bandwidth", - "type": "text" - }, - { - "block_id": "p860-b4", - "global_id": 25089, - "bbox": [ - 106.16, - 154.42, - 263.25, - 185.67 - ], - "text": "signal that passes through samples x(nT).\n[Hint:\nUse\nthe\nreductio\nad\nabsurdum\nmethod.]", - "type": "text" - }, - { - "block_id": "p860-b5", - "global_id": 25090, - "bbox": [ - 57.59, - 190.62, - 263.25, - 375.0 - ], - "text": "8.2-16\nIn digital communication systems, the effi-\ncient use of channel bandwidth is ensured by\ntransmitting digital data encoded by means of\nbandlimited pulses. Unfortunately, bandlimited\npulses are non-timelimited; that is, they have\ninfinite duration, which causes pulses repre-\nsenting successive digits to interfere and cause\nerrors in the reading of true pulse values. This\ndifficulty can be resolved by shaping a pulse\np(t) in such a way that it is bandlimited, yet\ncauses zero interference at the sampling instants.\nTo transmit R pulses per second, we require a\nminimum bandwidth R/2 Hz (see Prob. 8.1-13).\nThe bandwidth of p(t) should be R/2 Hz, and\nits samples, in order to cause no interference\nat all other sampling instants, must satisfy the\ncondition", - "type": "text" - }, - { - "block_id": "p860-b6", - "global_id": 25091, - "bbox": [ - 101.73, - 394.18, - 131.89, - 403.43 - ], - "text": "p(nT) =", - "type": "text" - }, - { - "block_id": "p860-b7", - "global_id": 25092, - "bbox": [ - 133.73, - 378.9, - 139.92, - 387.86 - ], - "text": ")", - "type": "text" - }, - { - "block_id": "p860-b8", - "global_id": 25093, - "bbox": [ - 139.92, - 383.55, - 231.41, - 398.81 - ], - "text": "1\nn = 0\nT = 1", - "type": "text" - }, - { - "block_id": "p860-b9", - "global_id": 25094, - "bbox": [ - 139.92, - 396.11, - 231.91, - 412.83 - ], - "text": "R\n0\nn̸ = 0", - "type": "text" - }, - { - "block_id": "p860-b10", - "global_id": 25095, - "bbox": [ - 90.72, - 424.49, - 263.24, - 499.31 - ], - "text": "Because the pulse rate is R pulses per second, the\nsampling instants are located at intervals of 1/R\nseconds. Hence, the foregoing condition ensures\nthat any given pulse will not interfere with the\namplitude of any other pulse at its center. Find\np(t). Is p(t) unique in the sense that no other\npulse satisfies the given requirements?", - "type": "text" - }, - { - "block_id": "p860-b11", - "global_id": 25096, - "bbox": [ - 284.74, - 85.94, - 490.41, - 226.49 - ], - "text": "8.2-17\nThe problem of pulse interference in digital data\ntransmission was outlined in Prob. 8.2-16, where\nwe found a pulse shape p(t) to eliminate the\ninterference. Unfortunately, the pulse found is\nnot only noncausal, (and unrealizable) but also\nhas a serious drawback: because of its slow\ndecay (as 1/t), it is prone to severe interference\ndue to small parameter deviation. To make the\npulse decay rapidly, Nyquist proposed relaxing\nthe bandwidth requirement from R/2 Hz to kR/2\nHz with 1 ≤k ≤2. The pulse must still have\na property of noninterference with other pulses,\nfor example,", - "type": "text" - }, - { - "block_id": "p860-b12", - "global_id": 25097, - "bbox": [ - 339.7, - 254.03, - 369.87, - 263.28 - ], - "text": "p(nT) =", - "type": "text" - }, - { - "block_id": "p860-b13", - "global_id": 25098, - "bbox": [ - 371.71, - 238.74, - 377.9, - 247.71 - ], - "text": ")", - "type": "text" - }, - { - "block_id": "p860-b14", - "global_id": 25099, - "bbox": [ - 377.89, - 243.4, - 460.69, - 258.65 - ], - "text": "1\nn = 0\nT = 1", - "type": "text" - }, - { - "block_id": "p860-b15", - "global_id": 25100, - "bbox": [ - 377.89, - 255.96, - 461.18, - 272.68 - ], - "text": "R\n0\nn̸ = 0", - "type": "text" - }, - { - "block_id": "p860-b16", - "global_id": 25101, - "bbox": [ - 317.86, - 290.94, - 490.38, - 332.79 - ], - "text": "Show that this condition is satisfied only if the\npulse spectrum P(ω) has an odd symmetry about\nthe set of dotted axes, as shown in Fig. P8.2-17.\nThe bandwidth of P(ω) is kR/2 Hz (1 ≤k ≤2).", - "type": "text" - }, - { - "block_id": "p860-b17", - "global_id": 25102, - "bbox": [ - 284.74, - 338.08, - 490.37, - 358.38 - ], - "text": "8.2-18\nThe Nyquist samples of a signal x(t) bandlimited\nto B Hz are", - "type": "text" - }, - { - "block_id": "p860-b18", - "global_id": 25103, - "bbox": [ - 321.75, - 384.06, - 351.44, - 393.31 - ], - "text": "x(nT) =", - "type": "text" - }, - { - "block_id": "p860-b19", - "global_id": 25104, - "bbox": [ - 353.29, - 368.78, - 409.26, - 383.97 - ], - "text": ")1\nn = 0,1", - "type": "text" - }, - { - "block_id": "p860-b20", - "global_id": 25105, - "bbox": [ - 359.47, - 384.4, - 463.44, - 399.65 - ], - "text": "0\nall n̸ = 0,1\nT = 1", - "type": "text" - }, - { - "block_id": "p860-b21", - "global_id": 25106, - "bbox": [ - 456.22, - 396.96, - 466.18, - 406.02 - ], - "text": "2B", - "type": "text" - }, - { - "block_id": "p860-b22", - "global_id": 25107, - "bbox": [ - 317.86, - 420.98, - 353.75, - 429.95 - ], - "text": "Show that", - "type": "text" - }, - { - "block_id": "p860-b23", - "global_id": 25108, - "bbox": [ - 370.9, - 444.7, - 436.16, - 460.32 - ], - "text": "x(t) = sinc(2πBt)", - "type": "text" - }, - { - "block_id": "p860-b24", - "global_id": 25109, - "bbox": [ - 402.7, - 457.35, - 429.43, - 466.69 - ], - "text": "1 −2Bt", - "type": "text" - }, - { - "block_id": "p860-b25", - "global_id": 25110, - "bbox": [ - 317.86, - 482.31, - 490.39, - 502.33 - ], - "text": "This pulse, known as the duobinary pulse, is\nused in digital transmission applications.", - "type": "text" - }, - { - "block_id": "p860-b26", - "global_id": 25111, - "bbox": [ - 292.49, - 604.09, - 297.38, - 620.98 - ], - "text": "R\n2", - "type": "text" - }, - { - "block_id": "p860-b27", - "global_id": 25112, - "bbox": [ - 122.26, - 523.89, - 127.15, - 540.45 - ], - "text": "R\n1", - "type": "text" - }, - { - "block_id": "p860-b28", - "global_id": 25113, - "bbox": [ - 120.26, - 558.75, - 129.15, - 575.39 - ], - "text": "2R\n1", - "type": "text" - }, - { - "block_id": "p860-b29", - "global_id": 25114, - "bbox": [ - 124.2, - 604.38, - 128.2, - 612.38 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p860-b30", - "global_id": 25115, - "bbox": [ - 269.27, - 527.68, - 284.82, - 535.78 - ], - "text": "P(v)", - "type": "text" - }, - { - "block_id": "p860-b31", - "global_id": 25116, - "bbox": [ - 371.6, - 604.37, - 387.15, - 612.45 - ], - "text": "f Hz", - "type": "text" - }, - { - "block_id": "p860-b32", - "global_id": 25117, - "bbox": [ - 104.83, - 629.31, - 160.95, - 638.27 - ], - "text": "Figure P8.2-17", - "type": "text" - } - ] - }, - { - "page_num": 861, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p861-b0", - "global_id": 25118, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n841", - "type": "text" - }, - { - "block_id": "p861-b1", - "global_id": 25119, - "bbox": [ - 83.34, - 85.92, - 288.97, - 106.91 - ], - "text": "8.2-19\nA signal bandlimited to B Hz is sampled at a rate\nfs = 2B Hz. Show that", - "type": "text" - }, - { - "block_id": "p861-b2", - "global_id": 25120, - "bbox": [ - 156.59, - 117.62, - 172.04, - 128.59 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p861-b3", - "global_id": 25121, - "bbox": [ - 161.32, - 139.85, - 172.99, - 146.33 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p861-b4", - "global_id": 25122, - "bbox": [ - 174.48, - 129.82, - 211.63, - 139.07 - ], - "text": "x(t)dt = T", - "type": "text" - }, - { - "block_id": "p861-b5", - "global_id": 25123, - "bbox": [ - 214.32, - 120.5, - 227.0, - 130.27 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p861-b6", - "global_id": 25124, - "bbox": [ - 214.83, - 142.12, - 226.49, - 148.6 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p861-b7", - "global_id": 25125, - "bbox": [ - 228.0, - 129.82, - 248.86, - 139.07 - ], - "text": "x(nT)", - "type": "text" - }, - { - "block_id": "p861-b8", - "global_id": 25126, - "bbox": [ - 147.73, - 148.78, - 163.18, - 159.75 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p861-b9", - "global_id": 25127, - "bbox": [ - 152.46, - 171.01, - 164.12, - 177.49 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p861-b10", - "global_id": 25128, - "bbox": [ - 165.62, - 157.55, - 211.63, - 170.23 - ], - "text": "|x(t)|2 dt = T", - "type": "text" - }, - { - "block_id": "p861-b11", - "global_id": 25129, - "bbox": [ - 214.32, - 151.66, - 227.0, - 161.42 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p861-b12", - "global_id": 25130, - "bbox": [ - 214.83, - 173.28, - 226.49, - 179.75 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p861-b13", - "global_id": 25131, - "bbox": [ - 228.0, - 159.53, - 257.23, - 170.23 - ], - "text": "|x(nT)|2", - "type": "text" - }, - { - "block_id": "p861-b14", - "global_id": 25132, - "bbox": [ - 116.46, - 195.27, - 288.99, - 215.28 - ], - "text": "[Hint: Use the orthogonality property of the sinc\nfunction in Prob. 7.6-6.]", - "type": "text" - }, - { - "block_id": "p861-b15", - "global_id": 25133, - "bbox": [ - 83.34, - 220.71, - 288.99, - 328.38 - ], - "text": "8.2-20\nProve that a signal cannot be simultaneously\ntimelimited and bandlimited. [Hint: Show that\na contrary assumption leads to contradiction.\nAssume a signal to be simultaneously timelim-\nited and bandlimited so that X(ω) = 0 for |ω| ≥\n2πB. In this case, X(ω) = X(ω)rect(ω/4πB′)\nfor B′ > B. This fact means that x(t) is equal\nto x(t) ∗2B′ sinc (2πB′t). The latter cannot\nbe timelimited because the sinc function tail\nextends to infinity.]", - "type": "text" - }, - { - "block_id": "p861-b16", - "global_id": 25134, - "bbox": [ - 87.82, - 333.79, - 288.98, - 375.71 - ], - "text": "8.3-1\nPhysically\nimplementable\ndigital\nsystems,\nsuch as smartphones and computers, require\nthat\nsignals\nbe\nboth\ntime-sampled\nand\namplitude-quantized.", - "type": "text" - }, - { - "block_id": "p861-b17", - "global_id": 25135, - "bbox": [ - 116.96, - 377.7, - 288.97, - 386.67 - ], - "text": "(a) Why is time sampling necessary? When", - "type": "text" - }, - { - "block_id": "p861-b18", - "global_id": 25136, - "bbox": [ - 116.46, - 388.66, - 288.97, - 419.55 - ], - "text": "does time sampling result in unrecoverable\nchanges to the signal?\n(b) Why is amplitude-quantization necessary?", - "type": "text" - }, - { - "block_id": "p861-b19", - "global_id": 25137, - "bbox": [ - 131.89, - 421.54, - 288.99, - 441.47 - ], - "text": "When does amplitude quantization result in\nunrecoverable changes to the signal?", - "type": "text" - }, - { - "block_id": "p861-b20", - "global_id": 25138, - "bbox": [ - 87.82, - 446.88, - 288.99, - 544.33 - ], - "text": "8.3-2\nTypical analog-to-digital converters (ADCs)\noperate over a range of input amplitudes\n[−Vref,Vref]. Why is it desirable to condition\nthe input x(t) to an ADC so that its maximum\nmagnitude is close to, but does not exceed, Vref?\nWhat happens if the maximum magnitude of\nx(t) is greater than Vref? What happens if the\nmaximum magnitude of x(t) is much smaller\nthan Vref?", - "type": "text" - }, - { - "block_id": "p861-b21", - "global_id": 25139, - "bbox": [ - 87.82, - 549.01, - 288.98, - 579.96 - ], - "text": "8.3-3\nA compact disc (CD) records audio signals\ndigitally by means of a binary code. Assume an\naudio signal bandwidth of 15 kHz.", - "type": "text" - }, - { - "block_id": "p861-b22", - "global_id": 25140, - "bbox": [ - 116.46, - 581.96, - 288.98, - 601.89 - ], - "text": "(a) What is the Nyquist rate?\n(b) If the Nyquist samples are quantized into", - "type": "text" - }, - { - "block_id": "p861-b23", - "global_id": 25141, - "bbox": [ - 131.89, - 603.5, - 288.99, - 634.77 - ], - "text": "65,536\nlevels\n(L = 65,536)\nand\nthen\nbinary-coded, what number of binary digits\nis required to encode a sample?", - "type": "text" - }, - { - "block_id": "p861-b24", - "global_id": 25142, - "bbox": [ - 344.11, - 85.9, - 516.13, - 94.86 - ], - "text": "(c) Determine the number of binary digits per", - "type": "text" - }, - { - "block_id": "p861-b25", - "global_id": 25143, - "bbox": [ - 343.61, - 96.86, - 516.11, - 127.74 - ], - "text": "second (bits/s) required to encode the audio\nsignal.\n(d) For practical reasons discussed in the text,", - "type": "text" - }, - { - "block_id": "p861-b26", - "global_id": 25144, - "bbox": [ - 359.04, - 129.74, - 516.14, - 182.53 - ], - "text": "signals are sampled at a rate well above\nthe Nyquist rate. Practical CDs use 44,100\nsamples/s. If L = 65,536, determine the\nnumber of pulses per second required to\nencode the signal.", - "type": "text" - }, - { - "block_id": "p861-b27", - "global_id": 25145, - "bbox": [ - 314.97, - 190.67, - 516.13, - 221.62 - ], - "text": "8.3-4\nA TV signal (video and audio) has a bandwidth\nof 4.5 MHz. This signal is sampled, quantized,\nand binary-coded.", - "type": "text" - }, - { - "block_id": "p861-b28", - "global_id": 25146, - "bbox": [ - 344.11, - 223.62, - 516.12, - 232.59 - ], - "text": "(a) Determine the sampling rate if the signal", - "type": "text" - }, - { - "block_id": "p861-b29", - "global_id": 25147, - "bbox": [ - 343.61, - 234.57, - 516.13, - 265.45 - ], - "text": "is to be sampled at a rate 20% above the\nNyquist rate.\n(b) If the samples are quantized into 1024", - "type": "text" - }, - { - "block_id": "p861-b30", - "global_id": 25148, - "bbox": [ - 344.11, - 267.45, - 516.13, - 298.33 - ], - "text": "levels, what number of binary pulses is\nrequired to encode each sample?\n(c) Determine the binary pulse rate (bits/s) of", - "type": "text" - }, - { - "block_id": "p861-b31", - "global_id": 25149, - "bbox": [ - 359.04, - 300.33, - 445.45, - 309.29 - ], - "text": "the binary-coded signal.", - "type": "text" - }, - { - "block_id": "p861-b32", - "global_id": 25150, - "bbox": [ - 314.97, - 317.42, - 516.13, - 326.46 - ], - "text": "8.3-5\n(a) In a certain A/D scheme, there are 16 quan-", - "type": "text" - }, - { - "block_id": "p861-b33", - "global_id": 25151, - "bbox": [ - 343.61, - 328.45, - 516.14, - 392.21 - ], - "text": "tization levels. Give one possible binary\ncode and one possible quaternary (4-ary)\ncode. For the quaternary code, use 0, 1, 2,\nand 3 as the four symbols. Use the minimum\nnumber of digits in your code.\n(b) To represent a given number of quantization", - "type": "text" - }, - { - "block_id": "p861-b34", - "global_id": 25152, - "bbox": [ - 359.04, - 394.12, - 516.13, - 447.97 - ], - "text": "levels L, we require a minimum of bM digits\nfor an M-ary code. Show that the ratio of\nthe number of digits in a binary code to\nthe number of digits in a quaternary (4-ary)\ncode is 2, that is, b2/b4 = 2.", - "type": "text" - }, - { - "block_id": "p861-b35", - "global_id": 25153, - "bbox": [ - 314.97, - 455.13, - 516.14, - 562.8 - ], - "text": "8.3-6\nFive telemetry signals, each of bandwidth 1 kHz,\nare quantized and binary-coded. These signals\nare time-division multiplexed (signal bits inter-\nleaved). Choose the number of quantization\nlevels so that the maximum error in sample\namplitudes is no greater than 0.2% of the peak\nsignal amplitude. The signals must be sampled\nat least 20% above the Nyquist rate. Determine\nthe data rate (bits per second) of the multiplexed\nsignal.", - "type": "text" - }, - { - "block_id": "p861-b36", - "global_id": 25154, - "bbox": [ - 314.97, - 570.63, - 516.13, - 612.84 - ], - "text": "8.4-1\nA triangle function x(t) = △(t/5) has spectrum\nX(ω). Sketch the corresponding time-domain\nsignal xT0(t) if X(ω) is sampled at the following\nrates:", - "type": "text" - }, - { - "block_id": "p861-b37", - "global_id": 25155, - "bbox": [ - 343.61, - 614.46, - 429.0, - 635.73 - ], - "text": "(a) f0 = 10 samples/Hz\n(b) f0 = 5 samples/Hz", - "type": "text" - } - ] - }, - { - "page_num": 862, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p862-b0", - "global_id": 25156, - "bbox": [ - 60.0, - 60.36, - 421.85, - 69.45 - ], - "text": "842\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p862-b1", - "global_id": 25157, - "bbox": [ - 90.72, - 85.52, - 178.34, - 106.79 - ], - "text": "(c) f0 = 4 samples/Hz\n(d) f0 = 2.5 samples/Hz", - "type": "text" - }, - { - "block_id": "p862-b2", - "global_id": 25158, - "bbox": [ - 62.08, - 110.43, - 263.23, - 152.64 - ], - "text": "8.4-2\nThe Fourier transform of a signal x(t), bandlim-\nited to B Hz, is X(ω). The signal x(t) is repeated\nperiodically at intervals T, where T = 1.25/B.\nThe resulting signal y(t) is", - "type": "text" - }, - { - "block_id": "p862-b3", - "global_id": 25159, - "bbox": [ - 141.49, - 170.74, - 163.65, - 179.99 - ], - "text": "y(t) =", - "type": "text" - }, - { - "block_id": "p862-b4", - "global_id": 25160, - "bbox": [ - 165.49, - 161.42, - 178.18, - 171.19 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p862-b5", - "global_id": 25161, - "bbox": [ - 166.01, - 183.04, - 177.67, - 189.52 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p862-b6", - "global_id": 25162, - "bbox": [ - 179.17, - 170.74, - 212.47, - 179.98 - ], - "text": "x(t −nT)", - "type": "text" - }, - { - "block_id": "p862-b7", - "global_id": 25163, - "bbox": [ - 90.72, - 198.91, - 214.73, - 208.24 - ], - "text": "Show that y(t) can be expressed as", - "type": "text" - }, - { - "block_id": "p862-b8", - "global_id": 25164, - "bbox": [ - 118.51, - 219.03, - 235.45, - 229.71 - ], - "text": "y(t) = C0 + C1 cos(1.6πBt + θ1)", - "type": "text" - }, - { - "block_id": "p862-b9", - "global_id": 25165, - "bbox": [ - 90.72, - 239.54, - 112.62, - 248.5 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p862-b10", - "global_id": 25166, - "bbox": [ - 143.27, - 255.89, - 169.95, - 272.11 - ], - "text": "C0 = 1", - "type": "text" - }, - { - "block_id": "p862-b11", - "global_id": 25167, - "bbox": [ - 164.86, - 261.81, - 188.79, - 277.42 - ], - "text": "T X(0)", - "type": "text" - }, - { - "block_id": "p862-b12", - "global_id": 25168, - "bbox": [ - 143.27, - 279.24, - 169.95, - 295.47 - ], - "text": "C1 = 2", - "type": "text" - }, - { - "block_id": "p862-b13", - "global_id": 25169, - "bbox": [ - 164.86, - 291.81, - 169.85, - 300.77 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p862-b14", - "global_id": 25170, - "bbox": [ - 172.73, - 272.16, - 181.12, - 297.26 - ], - "text": "X", - "type": "text" - }, - { - "block_id": "p862-b15", - "global_id": 25171, - "bbox": [ - 182.52, - 272.57, - 199.63, - 288.21 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p862-b16", - "global_id": 25172, - "bbox": [ - 192.3, - 291.81, - 197.29, - 300.77 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p862-b18", - "global_id": 25173, - "bbox": [ - 90.72, - 310.32, - 103.66, - 319.29 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p862-b19", - "global_id": 25174, - "bbox": [ - 149.12, - 324.03, - 178.19, - 334.34 - ], - "text": "θ1 ≠ X", - "type": "text" - }, - { - "block_id": "p862-b20", - "global_id": 25175, - "bbox": [ - 179.59, - 311.44, - 196.69, - 327.09 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p862-b21", - "global_id": 25176, - "bbox": [ - 189.37, - 330.68, - 194.35, - 339.64 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p862-b23", - "global_id": 25177, - "bbox": [ - 90.72, - 346.6, - 263.24, - 377.49 - ], - "text": "Recall that a bandlimited signal is not time-\nlimited, and hence has infinite duration. The\nperiodic repetitions are all overlapping.", - "type": "text" - }, - { - "block_id": "p862-b24", - "global_id": 25178, - "bbox": [ - 62.08, - 382.09, - 263.24, - 446.23 - ], - "text": "8.5-1\nFor a signal x(t) that is timelimited to 10 ms and\nhas an essential bandwidth of 10 kHz, determine\nN0, the number of signal samples necessary to\ncompute a power-of-2 FFT with a frequency\nresolution f0 of at least 50 Hz. Explain whether\nany zero padding is necessary.", - "type": "text" - }, - { - "block_id": "p862-b25", - "global_id": 25179, - "bbox": [ - 62.08, - 450.83, - 263.24, - 526.6 - ], - "text": "8.5-2\nTo\ncompute\nthe\nDFT\nof\nsignal\nx(t)\nin\nFig. P8.5-2, write the sequence xn (for n = 0 to\nN0 −1) if the frequency resolution f0 must be at\nleast 0.25 Hz. Assume the essential bandwidth\n(the folding frequency) of x(t) to be at least 3\nHz. Do not compute the DFT; just write the\nappropriate sequence xn.", - "type": "text" - }, - { - "block_id": "p862-b26", - "global_id": 25180, - "bbox": [ - 62.08, - 530.84, - 263.23, - 550.84 - ], - "text": "8.5-3\nSuppose we want to sample a finite-duration\nsignal x(t) that occupies 0 ≤t ≤T.", - "type": "text" - }, - { - "block_id": "p862-b27", - "global_id": 25181, - "bbox": [ - 318.37, - 85.9, - 490.39, - 94.86 - ], - "text": "(a) Devise a way to use the DFT to help select", - "type": "text" - }, - { - "block_id": "p862-b28", - "global_id": 25182, - "bbox": [ - 317.86, - 96.49, - 490.39, - 138.7 - ], - "text": "a suitable sampling rate fs for signal x(t).\n[Hint: Consider the characteristics of an\noversampled signal’s DFT spectrum.]\n(b) Test the method you devised in part (a)", - "type": "text" - }, - { - "block_id": "p862-b29", - "global_id": 25183, - "bbox": [ - 333.31, - 138.69, - 442.16, - 149.66 - ], - "text": "using the signal x(t) = ( t−1", - "type": "text" - }, - { - "block_id": "p862-b30", - "global_id": 25184, - "bbox": [ - 333.31, - 140.32, - 490.38, - 172.92 - ], - "text": "2 ). Use MAT-\nLAB to compute any needed DFTs. What\nvalue fs seems reasonable for this signal?", - "type": "text" - }, - { - "block_id": "p862-b31", - "global_id": 25185, - "bbox": [ - 289.22, - 176.46, - 490.4, - 262.24 - ], - "text": "8.5-4\nChoose appropriate values for N0 and T and\ncompute the DFT of the signal e−tu(t). Use two\ndifferent criteria for determining the effective\nbandwidth of e−tu(t). As the bandwidth, use the\nfrequency at which the amplitude response drops\nto 1% of its peak value (at ω = 0). Next, use\nthe 99% energy criterion for determining the\nbandwidth (see Ex. 7.20).", - "type": "text" - }, - { - "block_id": "p862-b32", - "global_id": 25186, - "bbox": [ - 289.22, - 267.15, - 434.3, - 276.19 - ], - "text": "8.5-5\nRepeat Prob. 8.5-4 for the signal", - "type": "text" - }, - { - "block_id": "p862-b33", - "global_id": 25187, - "bbox": [ - 380.59, - 286.18, - 426.47, - 307.79 - ], - "text": "x(t) =\n2\nt2 + 1", - "type": "text" - }, - { - "block_id": "p862-b34", - "global_id": 25188, - "bbox": [ - 289.22, - 316.34, - 490.39, - 369.52 - ], - "text": "8.5-6\nFor the signals x(t) and g(t) represented in\nFig. P8.5-6, write the appropriate sequences xn\nand gn necessary for the computation of the\nconvolution of x(t) and g(t) using DFT. Use\nT = 1/8.", - "type": "text" - }, - { - "block_id": "p862-b35", - "global_id": 25189, - "bbox": [ - 289.23, - 374.41, - 490.39, - 449.22 - ], - "text": "8.5-7\nFor this problem, interpret the N-point DFT as\nan N-periodic function of r. To stress this fact,\nwe shall change the notation Xr to X(r). Are\nthe following frequency-domain signals valid\nDFTs? Answer yes or no. For each valid DFT,\ndetermine the size N of the DFT and whether\nthe time-domain signal is real.", - "type": "text" - }, - { - "block_id": "p862-b36", - "global_id": 25190, - "bbox": [ - 317.86, - 450.84, - 395.26, - 471.14 - ], - "text": "(a) X(r) = j −π\n(b) X(r) = sin(r/10)", - "type": "text" - }, - { - "block_id": "p862-b37", - "global_id": 25191, - "bbox": [ - 317.86, - 472.76, - 401.55, - 493.05 - ], - "text": "(c) X(r) = sin(πr/10)\n(d) X(r) =", - "type": "text" - }, - { - "block_id": "p862-b39", - "global_id": 25192, - "bbox": [ - 363.77, - 483.71, - 391.44, - 493.05 - ], - "text": "(1 + j)/", - "type": "text" - }, - { - "block_id": "p862-b40", - "global_id": 25193, - "bbox": [ - 391.43, - 476.14, - 399.02, - 485.1 - ], - "text": "√", - "type": "text" - }, - { - "block_id": "p862-b41", - "global_id": 25194, - "bbox": [ - 399.02, - 476.5, - 410.15, - 493.05 - ], - "text": "2\n\nr", - "type": "text" - }, - { - "block_id": "p862-b42", - "global_id": 25195, - "bbox": [ - 318.37, - 494.67, - 490.38, - 505.35 - ], - "text": "(e) X(r) = ⟨r + π⟩10 where ⟨·⟩10 denotes the", - "type": "text" - }, - { - "block_id": "p862-b43", - "global_id": 25196, - "bbox": [ - 333.31, - 505.92, - 408.55, - 514.97 - ], - "text": "modulo-N operation.", - "type": "text" - }, - { - "block_id": "p862-b44", - "global_id": 25197, - "bbox": [ - 289.22, - 519.87, - 490.4, - 551.09 - ], - "text": "8.7-1\nMATLAB’s fft command computes the DFT of\na vector x assuming the first sample occurs at\ntime n = 0. Given that X = fft(x) has already", - "type": "text" - }, - { - "block_id": "p862-b45", - "global_id": 25198, - "bbox": [ - 257.26, - 575.96, - 261.26, - 583.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p862-b46", - "global_id": 25199, - "bbox": [ - 245.74, - 624.91, - 249.74, - 632.91 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p862-b47", - "global_id": 25200, - "bbox": [ - 238.82, - 570.59, - 249.92, - 578.67 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p862-b48", - "global_id": 25201, - "bbox": [ - 194.48, - 624.35, - 425.27, - 635.58 - ], - "text": "1\n1\nt\nFigure P8.5-2", - "type": "text" - } - ] - }, - { - "page_num": 863, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p863-b0", - "global_id": 25202, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n843", - "type": "text" - }, - { - "block_id": "p863-b1", - "global_id": 25203, - "bbox": [ - 171.86, - 96.57, - 300.44, - 105.51 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p863-b2", - "global_id": 25204, - "bbox": [ - 240.69, - 140.19, - 405.49, - 148.19 - ], - "text": "t\nt", - "type": "text" - }, - { - "block_id": "p863-b3", - "global_id": 25205, - "bbox": [ - 184.38, - 89.75, - 286.59, - 101.86 - ], - "text": "x(t)\ng(t)", - "type": "text" - }, - { - "block_id": "p863-b4", - "global_id": 25206, - "bbox": [ - 228.64, - 142.9, - 394.84, - 150.9 - ], - "text": "1\n1\n2", - "type": "text" - }, - { - "block_id": "p863-b5", - "global_id": 25207, - "bbox": [ - 208.81, - 156.36, - 347.34, - 167.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p863-b6", - "global_id": 25208, - "bbox": [ - 173.81, - 141.27, - 177.81, - 149.27 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p863-b7", - "global_id": 25209, - "bbox": [ - 432.37, - 159.71, - 484.02, - 168.68 - ], - "text": "Figure P8.5-6", - "type": "text" - }, - { - "block_id": "p863-b8", - "global_id": 25210, - "bbox": [ - 116.46, - 182.83, - 288.98, - 203.49 - ], - "text": "been computed, derive a method to correct X to\nreflect an arbitrary starting time n = n0.", - "type": "text" - }, - { - "block_id": "p863-b9", - "global_id": 25211, - "bbox": [ - 87.82, - 209.53, - 288.98, - 263.38 - ], - "text": "8.7-2\nConsider a complex signal composed of two\nclosely spaced complex exponentials: x1[n] =\nej2πn30/100 + ej2πn33/100. For each of the follow-\ning cases, plot the length-N DFT magnitude as a\nfunction of frequency fr, where fr = r/N.", - "type": "text" - }, - { - "block_id": "p863-b10", - "global_id": 25212, - "bbox": [ - 116.96, - 264.02, - 288.98, - 274.1 - ], - "text": "(a) Compute and plot the DFT of x1[n] using 10", - "type": "text" - }, - { - "block_id": "p863-b11", - "global_id": 25213, - "bbox": [ - 116.46, - 274.98, - 288.98, - 306.24 - ], - "text": "samples (0 ≤n ≤9). From the plot, can both\nexponentials be identified? Explain.\n(b) Zero-pad the signal from part (a) with", - "type": "text" - }, - { - "block_id": "p863-b12", - "global_id": 25214, - "bbox": [ - 116.96, - 308.23, - 288.99, - 350.82 - ], - "text": "490 zeros and then compute and plot the\n500-point DFT. Does this improve the pic-\nture of the DFT? Explain.\n(c) Compute and plot the DFT of x1[n] using", - "type": "text" - }, - { - "block_id": "p863-b13", - "global_id": 25215, - "bbox": [ - 116.46, - 351.69, - 288.97, - 382.95 - ], - "text": "100 samples (0 ≤n ≤99). From the plot, can\nboth exponentials be identified? Explain.\n(d) Zero-pad the signal from part (c) with", - "type": "text" - }, - { - "block_id": "p863-b14", - "global_id": 25216, - "bbox": [ - 131.9, - 384.94, - 288.99, - 415.83 - ], - "text": "400 zeros and then compute and plot the\n500-point DFT. Does this improve the pic-\nture of the DFT? Explain.", - "type": "text" - }, - { - "block_id": "p863-b15", - "global_id": 25217, - "bbox": [ - 87.82, - 422.6, - 288.97, - 443.34 - ], - "text": "8.7-3\nRepeat Prob. 8.7-2, using the complex signal\nx2[n] = ej2πn30/100 + ej2πn31.5/100.", - "type": "text" - }, - { - "block_id": "p863-b16", - "global_id": 25218, - "bbox": [ - 87.82, - 449.38, - 288.98, - 514.19 - ], - "text": "8.7-4\nConsider a complex signal composed of a dc\nterm and two complex exponentials: y1[n] =\n1 + ej2πn30/100 + 0.5 ∗ej2πn43/100. For each of\nthe following cases, plot the length-N DFT\nmagnitude as a function of frequency fr, where\nfr = r/N.", - "type": "text" - }, - { - "block_id": "p863-b17", - "global_id": 25219, - "bbox": [ - 116.96, - 515.21, - 288.98, - 524.17 - ], - "text": "(a) Use MATLAB to compute and plot the DFT", - "type": "text" - }, - { - "block_id": "p863-b18", - "global_id": 25220, - "bbox": [ - 116.46, - 525.79, - 288.98, - 611.85 - ], - "text": "of y1[n] with 20 samples (0 ≤n≤19). From\nthe plot, can the two non-dc exponentials\nbe identified? Given the amplitude rela-\ntion between the two, the lower-frequency\npeak should be twice as large as the\nhigher-frequency peak. Is this the case?\nExplain.\n(b) Zero-pad the signal from part (a) to a total", - "type": "text" - }, - { - "block_id": "p863-b19", - "global_id": 25221, - "bbox": [ - 131.9, - 613.84, - 289.0, - 633.77 - ], - "text": "length of 500. Does this improve locating\nthe two non-dc exponential components? Is", - "type": "text" - }, - { - "block_id": "p863-b20", - "global_id": 25222, - "bbox": [ - 344.12, - 182.83, - 516.13, - 213.71 - ], - "text": "the lower-frequency peak twice as large as\nthe higher-frequency peak? Explain.\n(c) MATLAB’s signal-processing toolbox func-", - "type": "text" - }, - { - "block_id": "p863-b21", - "global_id": 25223, - "bbox": [ - 359.05, - 215.71, - 516.14, - 279.46 - ], - "text": "tion window allows window functions to be\neasily generated. Generate a length-20 Han-\nning window and apply it to y1[n]. Using\nthis windowed function, repeat parts (a)\nand (b). Comment on whether the window\nfunction helps or hinders the analysis.", - "type": "text" - }, - { - "block_id": "p863-b22", - "global_id": 25224, - "bbox": [ - 314.97, - 288.67, - 516.12, - 309.41 - ], - "text": "8.7-5\nRepeat Prob. 8.7-4, using the complex signal\ny2[n] = 1 + ej2πn30/100 + 0.5ej2πn38/100.", - "type": "text" - }, - { - "block_id": "p863-b23", - "global_id": 25225, - "bbox": [ - 314.97, - 317.88, - 516.13, - 360.76 - ], - "text": "8.7-6\nThis problem investigates the idea of zero\npadding applied in the frequency domain. When\nasked, plot the length-N DFT magnitude as a\nfunction of frequency fr, where fr = r/N.", - "type": "text" - }, - { - "block_id": "p863-b24", - "global_id": 25226, - "bbox": [ - 344.11, - 361.78, - 516.13, - 371.0 - ], - "text": "(a) In MATLAB, create a vector x that con-", - "type": "text" - }, - { - "block_id": "p863-b25", - "global_id": 25227, - "bbox": [ - 343.61, - 372.37, - 516.14, - 414.84 - ], - "text": "tains one period of the sinusoid x[n] =\ncos((π/2)n). Plot the result. How “sinu-\nsoidal” does the signal appear to be?\n(b) Use the fft command to compute the DFT", - "type": "text" - }, - { - "block_id": "p863-b26", - "global_id": 25228, - "bbox": [ - 344.12, - 416.58, - 516.13, - 447.47 - ], - "text": "X of vector x. Plot the magnitude of the DFT\ncoefficients. Do they make sense?\n(c) Zero-pad the DFT vector to a total length", - "type": "text" - }, - { - "block_id": "p863-b27", - "global_id": 25229, - "bbox": [ - 343.61, - 449.45, - 516.14, - 589.93 - ], - "text": "of 100 by inserting the appropriate number\nof zeros in the middle of the vector X.\nCall this zero-padded DFT sequence Y.\nWhy are zeros inserted in the middle rather\nthan the end? Take the inverse DFT of Y\nand plot the result. What similarities exist\nbetween the new signal y and the original\nsignal x? What are the differences between\nx and y? What is the effect of zero padding\nin the frequency domain? How is this type\nof zero padding similar to zero padding in\nthe time domain?\n(d) Derive a general modification to the pro-", - "type": "text" - }, - { - "block_id": "p863-b28", - "global_id": 25230, - "bbox": [ - 359.05, - 591.93, - 516.13, - 633.77 - ], - "text": "cedure of zero padding in the frequency\ndomain to ensure that the amplitude of\nthe resulting time-domain signal is left\nunchanged.", - "type": "text" - } - ] - }, - { - "page_num": 864, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p864-b0", - "global_id": 25231, - "bbox": [ - 60.0, - 62.89, - 421.85, - 71.98 - ], - "text": "844\nCHAPTER 8\nSAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE", - "type": "text" - }, - { - "block_id": "p864-b1", - "global_id": 25232, - "bbox": [ - 91.22, - 85.9, - 263.23, - 94.86 - ], - "text": "(e) Consider one period of a square wave", - "type": "text" - }, - { - "block_id": "p864-b2", - "global_id": 25233, - "bbox": [ - 106.15, - 96.86, - 263.25, - 182.53 - ], - "text": "described\nby\nthe\nlength-8\nvector\n[1\n1\n1\n1\n−1\n−1\n−1\n−1].\nZero-pad the DFT of this vector to a\nlength of 100, and call the result S. Scale\nS according to part (d), take the inverse\nDFT, and plot the result. Does the new\ntime-domain signal s[n] look like a square\nwave? Explain.", - "type": "text" - }, - { - "block_id": "p864-b3", - "global_id": 25234, - "bbox": [ - 62.08, - 188.14, - 263.23, - 208.15 - ], - "text": "8.7-7\nThe quantized output xq of a truncating asym-\nmetric converter is given as", - "type": "text" - }, - { - "block_id": "p864-b4", - "global_id": 25235, - "bbox": [ - 136.53, - 225.16, - 169.07, - 237.16 - ], - "text": "xq = xmax", - "type": "text" - }, - { - "block_id": "p864-b5", - "global_id": 25236, - "bbox": [ - 156.12, - 225.42, - 212.25, - 240.35 - ], - "text": "2B−1 ⌊\nx\nxmax 2B−1 1", - "type": "text" - }, - { - "block_id": "p864-b6", - "global_id": 25237, - "bbox": [ - 209.02, - 226.85, - 217.43, - 238.77 - ], - "text": "2⌋", - "type": "text" - }, - { - "block_id": "p864-b7", - "global_id": 25238, - "bbox": [ - 90.72, - 253.83, - 263.23, - 277.0 - ], - "text": "Any values outside the 2B allowable levels\nshould be clamped to the nearest level.", - "type": "text" - }, - { - "block_id": "p864-b8", - "global_id": 25239, - "bbox": [ - 91.22, - 279.0, - 263.22, - 287.97 - ], - "text": "(a) Similar to Fig. 8.31, plot the transfer charac-", - "type": "text" - }, - { - "block_id": "p864-b9", - "global_id": 25240, - "bbox": [ - 90.72, - 289.95, - 263.25, - 309.88 - ], - "text": "teristics for a 3-bit version of this quantizer.\n(b) Apply 3-bit truncating asymmetric quanti-", - "type": "text" - }, - { - "block_id": "p864-b10", - "global_id": 25241, - "bbox": [ - 106.15, - 311.5, - 263.24, - 331.8 - ], - "text": "zation to a 1 Hz cosine sampled at fs = 50\nHz over 1 second. Plot the original signal", - "type": "text" - }, - { - "block_id": "p864-b11", - "global_id": 25242, - "bbox": [ - 333.31, - 85.58, - 490.38, - 138.76 - ], - "text": "x(t), the quantized signal xq(t), and the mag-\nnitude spectra of both. How does truncating\nasymmetric quantization compare to the\nresults of asymmetric rounding quantization\nshown in Fig. 8.33?", - "type": "text" - }, - { - "block_id": "p864-b12", - "global_id": 25243, - "bbox": [ - 289.22, - 143.64, - 490.38, - 163.66 - ], - "text": "8.7-8\nThe quantized output xq of a symmetric truncat-\ning converter is given as", - "type": "text" - }, - { - "block_id": "p864-b13", - "global_id": 25244, - "bbox": [ - 345.69, - 175.85, - 378.24, - 187.85 - ], - "text": "xq = xmax", - "type": "text" - }, - { - "block_id": "p864-b14", - "global_id": 25245, - "bbox": [ - 365.28, - 167.64, - 386.46, - 190.26 - ], - "text": "2B−1", - "type": "text" - }, - { - "block_id": "p864-b15", - "global_id": 25246, - "bbox": [ - 386.46, - 174.29, - 437.08, - 191.04 - ], - "text": "⌊\nx\nxmax 2B−1 −1", - "type": "text" - }, - { - "block_id": "p864-b16", - "global_id": 25247, - "bbox": [ - 433.85, - 167.64, - 462.56, - 189.46 - ], - "text": "2⌋+ 1\n2", - "type": "text" - }, - { - "block_id": "p864-b17", - "global_id": 25248, - "bbox": [ - 317.86, - 199.03, - 490.38, - 222.21 - ], - "text": "Any values outside the 2B allowable levels\nshould be clamped to the nearest level.", - "type": "text" - }, - { - "block_id": "p864-b18", - "global_id": 25249, - "bbox": [ - 318.37, - 224.2, - 490.37, - 233.17 - ], - "text": "(a) Similar to Fig. 8.31, plot the transfer charac-", - "type": "text" - }, - { - "block_id": "p864-b19", - "global_id": 25250, - "bbox": [ - 317.86, - 235.16, - 490.37, - 255.09 - ], - "text": "teristics for a 3-bit version of this quantizer.\n(b) Apply 3-bit truncating symmetric quanti-", - "type": "text" - }, - { - "block_id": "p864-b20", - "global_id": 25251, - "bbox": [ - 333.31, - 256.7, - 490.39, - 331.8 - ], - "text": "zation to a 1 Hz cosine sampled at fs =\n50 Hz over 1 second. Plot the original\nsignal x(t), the quantized signal xq(t), and\nthe magnitude spectra of both. How does\ntruncating symmetric quantization compare\nto the results of asymmetric rounding quan-\ntization shown in Fig. 8.33?", - "type": "text" - } - ] - }, - { - "page_num": 865, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p865-b0", - "global_id": 25252, - "bbox": [ - 90.21, - 67.04, - 159.28, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p865-b1", - "global_id": 25253, - "bbox": [ - 172.76, - 125.73, - 436.98, - 173.26 - ], - "text": "FOURIER ANALYSIS OF\nDISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p865-b2", - "global_id": 25254, - "bbox": [ - 140.04, - 79.92, - 159.47, - 118.77 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p865-b3", - "global_id": 25255, - "bbox": [ - 127.59, - 266.14, - 516.15, - 335.88 - ], - "text": "In Chs. 6 and 7, we studied the ways of representing a continuous-time signal as a sum of sinusoids\nor exponentials. In this chapter we shall discuss similar development for discrete-time signals. Our\napproach is parallel to that used for continuous-time signals. We first represent a periodic x[n] as\na Fourier series formed by a discrete-time exponential (or sinusoid) and its harmonics. Later we\nextend this representation to an aperiodic signal x[n] by considering x[n] as a limiting case of a\nperiodic signal with the period approaching infinity.", - "type": "text" - }, - { - "block_id": "p865-b4", - "global_id": 25256, - "bbox": [ - 127.94, - 366.24, - 421.98, - 380.19 - ], - "text": "9.1 DISCRETE-TIME FOURIER SERIES (DTFS)", - "type": "text" - }, - { - "block_id": "p865-b5", - "global_id": 25257, - "bbox": [ - 127.59, - 385.76, - 516.13, - 432.0 - ], - "text": "A continuous-time sinusoid cosωt is a periodic signal regardless of the value of ω. Such is not the\ncase for the discrete-time sinusoid cosn (or exponential ejn). A sinusoid cosn is periodic only\nif /2π is a rational number. This can be proved by observing that if this sinusoid is N0 periodic,\nthen", - "type": "text" - }, - { - "block_id": "p865-b6", - "global_id": 25258, - "bbox": [ - 274.0, - 435.2, - 369.72, - 446.34 - ], - "text": "cos(n + N0) = cosn", - "type": "text" - }, - { - "block_id": "p865-b7", - "global_id": 25259, - "bbox": [ - 127.59, - 455.61, - 218.38, - 465.57 - ], - "text": "This is possible only if", - "type": "text" - }, - { - "block_id": "p865-b8", - "global_id": 25260, - "bbox": [ - 268.6, - 468.77, - 375.11, - 480.22 - ], - "text": "N0 = 2πm\nm integer", - "type": "text" - }, - { - "block_id": "p865-b9", - "global_id": 25261, - "bbox": [ - 127.59, - 488.77, - 516.12, - 511.1 - ], - "text": "Here, both m and N0 are integers. Hence, /2π = m/N0 is a rational number. Thus, a sinusoid\ncosn (or exponential ejn) is periodic only if", - "type": "text" - }, - { - "block_id": "p865-b10", - "global_id": 25262, - "bbox": [ - 258.34, - 521.49, - 294.68, - 546.69 - ], - "text": "2π = m\nN0", - "type": "text" - }, - { - "block_id": "p865-b11", - "global_id": 25263, - "bbox": [ - 316.3, - 528.89, - 386.56, - 538.85 - ], - "text": "a rational number", - "type": "text" - }, - { - "block_id": "p865-b12", - "global_id": 25264, - "bbox": [ - 127.59, - 555.93, - 516.14, - 578.26 - ], - "text": "When this condition (/2π a rational number) is satisfied, the period N0 of the sinusoid cosn is\ngiven by", - "type": "text" - }, - { - "block_id": "p865-b13", - "global_id": 25265, - "bbox": [ - 292.56, - 587.1, - 322.26, - 598.55 - ], - "text": "N0 = m", - "type": "text" - }, - { - "block_id": "p865-b14", - "global_id": 25266, - "bbox": [ - 323.36, - 573.11, - 342.24, - 590.49 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p865-b17", - "global_id": 25267, - "bbox": [ - 497.04, - 587.51, - 516.13, - 597.47 - ], - "text": "(9.1)", - "type": "text" - }, - { - "block_id": "p865-b18", - "global_id": 25268, - "bbox": [ - 127.59, - 612.45, - 516.13, - 634.79 - ], - "text": "To compute N0, we must choose the smallest value of m that will make m(2π/) an integer. For\nexample, if = 4π/17, then the smallest value of m that will make m(2π/) = m(17/2) an", - "type": "text" - }, - { - "block_id": "p865-b19", - "global_id": 25269, - "bbox": [ - 501.19, - 656.12, - 516.13, - 666.22 - ], - "text": "845", - "type": "text" - } - ] - }, - { - "page_num": 866, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p866-b0", - "global_id": 25270, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "846\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p866-b1", - "global_id": 25271, - "bbox": [ - 101.84, - 85.82, - 192.71, - 95.78 - ], - "text": "integer is 2. Therefore,", - "type": "text" - }, - { - "block_id": "p866-b2", - "global_id": 25272, - "bbox": [ - 234.03, - 103.72, - 263.72, - 115.17 - ], - "text": "N0 = m", - "type": "text" - }, - { - "block_id": "p866-b3", - "global_id": 25273, - "bbox": [ - 264.83, - 89.73, - 283.7, - 107.11 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p866-b6", - "global_id": 25274, - "bbox": [ - 294.67, - 103.72, - 309.47, - 114.09 - ], - "text": "= 2", - "type": "text" - }, - { - "block_id": "p866-b7", - "global_id": 25275, - "bbox": [ - 310.57, - 89.73, - 328.45, - 107.11 - ], - "text": "17", - "type": "text" - }, - { - "block_id": "p866-b8", - "global_id": 25276, - "bbox": [ - 320.98, - 111.2, - 325.97, - 121.17 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p866-b10", - "global_id": 25277, - "bbox": [ - 338.43, - 103.72, - 358.2, - 114.09 - ], - "text": "= 17", - "type": "text" - }, - { - "block_id": "p866-b11", - "global_id": 25278, - "bbox": [ - 101.84, - 128.48, - 480.42, - 138.85 - ], - "text": "However, a sinusoid cos(0.8n) is not a periodic signal because 0.8/2π is not a rational number.", - "type": "text" - }, - { - "block_id": "p866-b12", - "global_id": 25279, - "bbox": [ - 101.84, - 164.08, - 318.51, - 189.99 - ], - "text": "9.1-1 Periodic Signal Representation by\nDiscrete-Time Fourier Series", - "type": "text" - }, - { - "block_id": "p866-b13", - "global_id": 25280, - "bbox": [ - 101.84, - 196.02, - 490.39, - 229.99 - ], - "text": "A continuous-time periodic signal of period T0 can be represented as a trigonometric Fourier\nseries consisting of a sinusoid of the fundamental frequency ω0 = 2π/T0, and all its harmonics.\nThe exponential form of the Fourier series consists of exponentials ej0t, e±jω0t, e±j2ω0t, e±j3ω0t,. . . .", - "type": "text" - }, - { - "block_id": "p866-b14", - "global_id": 25281, - "bbox": [ - 101.85, - 231.98, - 490.38, - 265.85 - ], - "text": "A discrete-time periodic signal can be represented by a discrete-time Fourier series using a\nparallel development. Recall that a periodic signal x[n] with period N0 is characterized by the fact\nthat", - "type": "text" - }, - { - "block_id": "p866-b15", - "global_id": 25282, - "bbox": [ - 263.37, - 268.15, - 328.87, - 279.3 - ], - "text": "x[n] = x[n + N0]", - "type": "text" - }, - { - "block_id": "p866-b16", - "global_id": 25283, - "bbox": [ - 101.84, - 287.87, - 490.41, - 369.67 - ], - "text": "The smallest value of N0 for which this equation holds is the fundamental period. The fundamental\nfrequency is 0 = 2π/N0 rad/sample. An N0-periodic signal x[n] can be represented by a\ndiscrete-time Fourier series made up of sinusoids of fundamental frequency 0 = 2π/N0 and\nits harmonics. As in the continuous-time case, we may use a trigonometric or an exponential\nform of the Fourier series. Because of its compactness and ease of mathematical manipulations,\nthe exponential form is preferable to the trigonometric. For this reason, we shall bypass the\ntrigonometric form and go directly to the exponential form of the discrete-time Fourier series.", - "type": "text" - }, - { - "block_id": "p866-b17", - "global_id": 25284, - "bbox": [ - 101.85, - 370.01, - 490.39, - 417.48 - ], - "text": "The exponential Fourier series consists of the exponentials ej0n, e±j0n, e±j20n, . . ., e±jn0n,\n. . ., and so on. There would be an infinite number of harmonics, except for the property proved\nin Sec. 5.5-1, that discrete-time exponentials whose frequencies are separated by 2π (or integer\nmultiples of 2π) are identical because", - "type": "text" - }, - { - "block_id": "p866-b18", - "global_id": 25285, - "bbox": [ - 206.38, - 425.39, - 385.84, - 439.88 - ], - "text": "ej(±2πm)n = ejne±2πmn = ejn\nm integer", - "type": "text" - }, - { - "block_id": "p866-b19", - "global_id": 25286, - "bbox": [ - 101.84, - 451.9, - 490.38, - 475.72 - ], - "text": "The consequence of this result is that the rth harmonic is identical to the (r + N0)th harmonic. To\ndemonstrate this, let gn denote the nth harmonic ejn0n. Then", - "type": "text" - }, - { - "block_id": "p866-b20", - "global_id": 25287, - "bbox": [ - 203.4, - 482.13, - 388.16, - 499.21 - ], - "text": "gr+N0 = ej(r+N0)0n = ej(r0n+2πn) = ejr0n = gr", - "type": "text" - }, - { - "block_id": "p866-b21", - "global_id": 25288, - "bbox": [ - 101.84, - 509.06, - 116.22, - 519.02 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p866-b22", - "global_id": 25289, - "bbox": [ - 195.96, - 521.32, - 396.27, - 534.28 - ], - "text": "gr = gr+N0 = gr+2N0 = · · · = gr+mN0\nm integer", - "type": "text" - }, - { - "block_id": "p866-b23", - "global_id": 25290, - "bbox": [ - 101.84, - 540.72, - 490.4, - 634.79 - ], - "text": "Thus, the first harmonic is identical to the (N0 +1)th harmonic, the second harmonic is identical to\nthe (N0 +2)th harmonic, and so on. In other words, there are only N0 independent harmonics, and\ntheir frequencies range over an interval 2π (because the harmonics are separated by 0 = 2π/N0).\nThis means that, unlike the continuous-time counterpart, the discrete-time Fourier series has only\na finite number (N0) of terms. This result is consistent with our observation in Sec. 5.5-1 that\nall discrete-time signals are bandlimited to a band from −π to π. Because the harmonics are\nseparated by 0 = 2π/N0, there can only be N0 harmonics in this band. We also saw that this\nband can be taken from 0 to 2π or any other contiguous band of width 2π. This means we may", - "type": "text" - } - ] - }, - { - "page_num": 867, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p867-b0", - "global_id": 25291, - "bbox": [ - 310.73, - 62.89, - 516.13, - 71.98 - ], - "text": "9.1\nDiscrete-Time Fourier Series (DTFS)\n847", - "type": "text" - }, - { - "block_id": "p867-b1", - "global_id": 25292, - "bbox": [ - 127.59, - 83.68, - 516.11, - 121.17 - ], - "text": "choose the N0 independent harmonics ejr0n over 0 ≤r ≤N0 −1, or over −1 ≤r ≤N0 −2, or over\n1 ≤r ≤N0, or over any other suitable choice for that matter. Every one of these sets will have the\nsame harmonics, although in different order.", - "type": "text" - }, - { - "block_id": "p867-b2", - "global_id": 25293, - "bbox": [ - 127.59, - 119.55, - 516.13, - 157.04 - ], - "text": "Let us consider the first choice, which corresponds to exponentials ejr0n for r = 0, 1, 2, . . . ,\nN0 −1. The Fourier series for an N0-periodic signal x[n] consists of only these N0 harmonics, and\ncan be expressed as", - "type": "text" - }, - { - "block_id": "p867-b3", - "global_id": 25294, - "bbox": [ - 253.13, - 168.99, - 279.02, - 179.27 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p867-b4", - "global_id": 25295, - "bbox": [ - 281.07, - 158.2, - 298.14, - 169.5 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b5", - "global_id": 25296, - "bbox": [ - 283.71, - 183.39, - 295.5, - 190.65 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p867-b6", - "global_id": 25297, - "bbox": [ - 299.24, - 162.01, - 388.39, - 180.45 - ], - "text": "Drejr0n\n0 = 2π", - "type": "text" - }, - { - "block_id": "p867-b7", - "global_id": 25298, - "bbox": [ - 378.1, - 176.38, - 388.23, - 187.22 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p867-b8", - "global_id": 25299, - "bbox": [ - 127.59, - 195.3, - 516.14, - 221.91 - ], - "text": "To compute coefficients Dr, we multiply both sides by e−jm0n and sum over n from n = 0 to\n(N0 −1).", - "type": "text" - }, - { - "block_id": "p867-b9", - "global_id": 25300, - "bbox": [ - 240.78, - 221.58, - 257.84, - 232.87 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b10", - "global_id": 25301, - "bbox": [ - 243.11, - 246.76, - 255.51, - 254.02 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p867-b11", - "global_id": 25302, - "bbox": [ - 258.95, - 228.26, - 314.57, - 242.65 - ], - "text": "x[n]e−jm0n =", - "type": "text" - }, - { - "block_id": "p867-b12", - "global_id": 25303, - "bbox": [ - 316.62, - 221.58, - 333.69, - 232.87 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b13", - "global_id": 25304, - "bbox": [ - 318.95, - 246.76, - 331.36, - 254.02 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p867-b14", - "global_id": 25305, - "bbox": [ - 335.9, - 221.58, - 352.97, - 232.87 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b15", - "global_id": 25306, - "bbox": [ - 338.54, - 246.76, - 350.33, - 254.02 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p867-b16", - "global_id": 25307, - "bbox": [ - 354.08, - 230.65, - 516.13, - 243.45 - ], - "text": "Drej(r−m)0n\n(9.2)", - "type": "text" - }, - { - "block_id": "p867-b17", - "global_id": 25308, - "bbox": [ - 127.59, - 260.82, - 423.69, - 270.78 - ], - "text": "The right-hand sum, after interchanging the order of summation, results in", - "type": "text" - }, - { - "block_id": "p867-b18", - "global_id": 25309, - "bbox": [ - 272.51, - 281.87, - 289.58, - 293.16 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b19", - "global_id": 25310, - "bbox": [ - 275.15, - 307.05, - 286.94, - 314.31 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p867-b20", - "global_id": 25311, - "bbox": [ - 290.69, - 292.66, - 301.09, - 303.74 - ], - "text": "Dr", - "type": "text" - }, - { - "block_id": "p867-b21", - "global_id": 25312, - "bbox": [ - 302.85, - 275.69, - 326.1, - 293.16 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b22", - "global_id": 25313, - "bbox": [ - 311.36, - 307.05, - 323.77, - 314.31 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p867-b23", - "global_id": 25314, - "bbox": [ - 327.21, - 290.94, - 364.52, - 302.94 - ], - "text": "ej(r−m)0n", - "type": "text" - }, - { - "block_id": "p867-b25", - "global_id": 25315, - "bbox": [ - 127.59, - 323.69, - 516.11, - 358.75 - ], - "text": "The inner sum, according to Eq. (8.15) in Sec. 8.5, is zero for all values of r̸ = m. It is nonzero\nwith a value N0 only when r = m. This fact means the outside sum has only one term DmN0\n(corresponding to r = m). Therefore, the right-hand side of Eq. (9.2) is equal to DmN0, and", - "type": "text" - }, - { - "block_id": "p867-b26", - "global_id": 25316, - "bbox": [ - 272.02, - 369.05, - 289.07, - 380.35 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b27", - "global_id": 25317, - "bbox": [ - 274.34, - 394.24, - 286.74, - 401.51 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p867-b28", - "global_id": 25318, - "bbox": [ - 290.19, - 375.73, - 371.21, - 391.0 - ], - "text": "x[n]e−jm0n = DmN0", - "type": "text" - }, - { - "block_id": "p867-b29", - "global_id": 25319, - "bbox": [ - 127.59, - 411.29, - 141.97, - 421.25 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p867-b30", - "global_id": 25320, - "bbox": [ - 270.27, - 424.58, - 304.35, - 442.61 - ], - "text": "Dm = 1", - "type": "text" - }, - { - "block_id": "p867-b31", - "global_id": 25321, - "bbox": [ - 296.55, - 438.54, - 306.68, - 449.38 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p867-b32", - "global_id": 25322, - "bbox": [ - 309.48, - 420.36, - 326.54, - 431.66 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b33", - "global_id": 25323, - "bbox": [ - 311.8, - 445.55, - 324.21, - 452.82 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p867-b34", - "global_id": 25324, - "bbox": [ - 327.65, - 429.43, - 372.95, - 441.43 - ], - "text": "x[n]e−jm0n", - "type": "text" - }, - { - "block_id": "p867-b35", - "global_id": 25325, - "bbox": [ - 127.59, - 459.2, - 516.13, - 470.35 - ], - "text": "We now have a discrete-time Fourier series (DTFS) representation of an N0-periodic signal x[n] as", - "type": "text" - }, - { - "block_id": "p867-b36", - "global_id": 25326, - "bbox": [ - 282.21, - 491.45, - 308.1, - 501.72 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p867-b37", - "global_id": 25327, - "bbox": [ - 310.15, - 480.65, - 327.21, - 491.95 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b38", - "global_id": 25328, - "bbox": [ - 312.78, - 505.84, - 324.57, - 513.11 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p867-b39", - "global_id": 25329, - "bbox": [ - 328.32, - 489.72, - 516.13, - 502.52 - ], - "text": "Drejr0n\n(9.3)", - "type": "text" - }, - { - "block_id": "p867-b40", - "global_id": 25330, - "bbox": [ - 127.59, - 522.87, - 151.93, - 532.84 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p867-b41", - "global_id": 25331, - "bbox": [ - 243.5, - 536.18, - 275.41, - 554.2 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p867-b42", - "global_id": 25332, - "bbox": [ - 267.61, - 550.13, - 277.73, - 560.97 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p867-b43", - "global_id": 25333, - "bbox": [ - 280.54, - 531.94, - 297.61, - 543.24 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p867-b44", - "global_id": 25334, - "bbox": [ - 282.87, - 557.13, - 295.28, - 564.4 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p867-b45", - "global_id": 25335, - "bbox": [ - 298.71, - 535.76, - 398.04, - 554.2 - ], - "text": "x[n]e−jr0n\n0 = 2π", - "type": "text" - }, - { - "block_id": "p867-b46", - "global_id": 25336, - "bbox": [ - 387.75, - 550.13, - 397.88, - 560.97 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p867-b47", - "global_id": 25337, - "bbox": [ - 497.04, - 543.16, - 516.13, - 553.12 - ], - "text": "(9.4)", - "type": "text" - }, - { - "block_id": "p867-b48", - "global_id": 25338, - "bbox": [ - 127.59, - 571.19, - 516.15, - 605.07 - ], - "text": "Observe that DTFS Eqs. (9.3) and (9.4) are identical (within a scaling constant) to the DFT\nEqs. (8.13) and (8.12).† Therefore, we can use the efficient FFT algorithm to compute the DTFS\ncoefficients.", - "type": "text" - }, - { - "block_id": "p867-b49", - "global_id": 25339, - "bbox": [ - 127.59, - 621.19, - 514.3, - 634.75 - ], - "text": "† If we let x[n] = N0xk and Dr = Xr, Eqs. (9.3) and (9.4) are identical to Eqs. (8.13) and (8.12), respectively.", - "type": "text" - } - ] - }, - { - "page_num": 868, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p868-b0", - "global_id": 25340, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "848\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p868-b1", - "global_id": 25341, - "bbox": [ - 101.84, - 85.66, - 349.09, - 98.48 - ], - "text": "9.1-2 Fourier Spectra of a Periodic Signal x[n]", - "type": "text" - }, - { - "block_id": "p868-b2", - "global_id": 25342, - "bbox": [ - 101.84, - 104.51, - 282.75, - 116.06 - ], - "text": "The Fourier series consists of N0 components", - "type": "text" - }, - { - "block_id": "p868-b3", - "global_id": 25343, - "bbox": [ - 209.87, - 126.26, - 381.86, - 140.39 - ], - "text": "D0,D1ej0n,D2ej20n,. . .,DN0−1ej(N0−1)0n", - "type": "text" - }, - { - "block_id": "p868-b4", - "global_id": 25344, - "bbox": [ - 101.84, - 151.77, - 490.38, - 198.01 - ], - "text": "The frequencies of these components are 0, 0, 20, . . ., (N0 −1)0, where 0 = 2π/N0. The\namount of the rth harmonic is Dr. We can plot this amount Dr (the Fourier coefficient) as a function\nof index r or frequency . Such a plot, called the Fourier spectrum of x[n], gives us, at a glance,\nthe graphical picture of the amounts of various harmonics of x[n].", - "type": "text" - }, - { - "block_id": "p868-b5", - "global_id": 25345, - "bbox": [ - 101.84, - 199.59, - 490.38, - 221.92 - ], - "text": "In general, the Fourier coefficients Dr are complex, and they can be represented in the polar\nform as", - "type": "text" - }, - { - "block_id": "p868-b6", - "global_id": 25346, - "bbox": [ - 265.85, - 224.57, - 325.25, - 237.75 - ], - "text": "Dr = |Dr|ej̸ Dr", - "type": "text" - }, - { - "block_id": "p868-b7", - "global_id": 25347, - "bbox": [ - 101.84, - 247.1, - 490.4, - 329.2 - ], - "text": "The plot of |Dr| versus is called the amplitude spectrum and that of̸\nDr versus is called the\nangle (or phase) spectrum. These two plots together are the frequency spectra of x[n]. Knowing\nthese spectra, we can reconstruct or synthesize x[n] according to Eq. (9.3). Therefore, the Fourier\n(or frequency) spectra, which are an alternative way of describing a periodic signal x[n], are in\nevery way equivalent (in terms of the information) to the plot of x[n] as a function of n. The\nFourier spectra of a signal constitute the frequency-domain description of x[n], in contrast to the\ntime-domain description, where x[n] is specified as a function of index n (representing time).", - "type": "text" - }, - { - "block_id": "p868-b8", - "global_id": 25348, - "bbox": [ - 101.85, - 331.2, - 490.41, - 378.52 - ], - "text": "The results are very similar to the representation of a continuous-time periodic signal by an\nexponential Fourier series except that, generally, the continuous-time signal spectrum bandwidth is\ninfinite and consists of an infinite number of exponential components (harmonics). The spectrum\nof the discrete-time periodic signal, in contrast, is bandlimited and has at most N0 components.", - "type": "text" - }, - { - "block_id": "p868-b9", - "global_id": 25349, - "bbox": [ - 101.84, - 392.7, - 348.37, - 419.58 - ], - "text": "PERIODIC EXTENSION OF FOURIER SPECTRUM\nWe now show that if φ[r] is an N0-periodic function of r, then", - "type": "text" - }, - { - "block_id": "p868-b10", - "global_id": 25350, - "bbox": [ - 252.77, - 431.66, - 269.83, - 442.95 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p868-b11", - "global_id": 25351, - "bbox": [ - 255.4, - 456.85, - 267.2, - 464.11 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p868-b12", - "global_id": 25352, - "bbox": [ - 270.95, - 442.46, - 297.75, - 452.73 - ], - "text": "φ[r] =", - "type": "text" - }, - { - "block_id": "p868-b13", - "global_id": 25353, - "bbox": [ - 303.53, - 432.99, - 317.63, - 442.95 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p868-b14", - "global_id": 25354, - "bbox": [ - 299.8, - 456.76, - 321.37, - 465.28 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p868-b15", - "global_id": 25355, - "bbox": [ - 322.48, - 442.46, - 490.39, - 452.84 - ], - "text": "φ[r]\n(9.5)", - "type": "text" - }, - { - "block_id": "p868-b16", - "global_id": 25356, - "bbox": [ - 101.84, - 476.64, - 490.4, - 522.89 - ], - "text": "where r = ⟨N0⟩indicates summation over any N0 consecutive values of r. Because φ[r] is N0\nperiodic, the same values repeat with period N0. Hence, the sum of any set of N0 consecutive\nvalues of φ[r] must be the same no matter the value of r at which we start summing. Basically, it\nrepresents the sum over one cycle.", - "type": "text" - }, - { - "block_id": "p868-b17", - "global_id": 25357, - "bbox": [ - 119.78, - 521.26, - 436.77, - 536.34 - ], - "text": "To apply this result to the DTFS, we observe that e−jr0n is N0 periodic because", - "type": "text" - }, - { - "block_id": "p868-b18", - "global_id": 25358, - "bbox": [ - 220.96, - 544.14, - 370.78, - 558.53 - ], - "text": "e−jr0(n+N0) = e−jr0ne−j2πr = e−jr0n", - "type": "text" - }, - { - "block_id": "p868-b19", - "global_id": 25359, - "bbox": [ - 101.84, - 568.84, - 490.38, - 606.32 - ], - "text": "Therefore, if x[n] is N0 periodic, x[n]e−jr0n is also N0 periodic. Hence, from Eq. (9.4), it follows\nthat Dr is also N0 periodic, as is Drejr0n. Now, because of Eq. (9.5), we can express Eqs. (9.3)\nand (9.4) as", - "type": "text" - }, - { - "block_id": "p868-b20", - "global_id": 25360, - "bbox": [ - 254.21, - 611.99, - 280.1, - 622.27 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p868-b21", - "global_id": 25361, - "bbox": [ - 285.89, - 602.52, - 299.98, - 612.49 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p868-b22", - "global_id": 25362, - "bbox": [ - 282.15, - 626.28, - 303.73, - 634.81 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p868-b23", - "global_id": 25363, - "bbox": [ - 304.83, - 610.27, - 490.39, - 623.07 - ], - "text": "Drejr0n\n(9.6)", - "type": "text" - } - ] - }, - { - "page_num": 869, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p869-b0", - "global_id": 25364, - "bbox": [ - 310.73, - 62.89, - 516.13, - 71.98 - ], - "text": "9.1\nDiscrete-Time Fourier Series (DTFS)\n849", - "type": "text" - }, - { - "block_id": "p869-b1", - "global_id": 25365, - "bbox": [ - 127.59, - 85.82, - 141.97, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p869-b2", - "global_id": 25366, - "bbox": [ - 269.86, - 95.21, - 301.79, - 113.24 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p869-b3", - "global_id": 25367, - "bbox": [ - 293.98, - 109.17, - 304.11, - 120.0 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p869-b4", - "global_id": 25368, - "bbox": [ - 310.96, - 92.31, - 325.06, - 102.28 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p869-b5", - "global_id": 25369, - "bbox": [ - 306.92, - 116.07, - 329.11, - 124.6 - ], - "text": "n=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p869-b6", - "global_id": 25370, - "bbox": [ - 330.21, - 100.05, - 516.13, - 112.16 - ], - "text": "x[n]e−jr0n\n(9.7)", - "type": "text" - }, - { - "block_id": "p869-b7", - "global_id": 25371, - "bbox": [ - 127.59, - 132.14, - 516.15, - 238.15 - ], - "text": "If we plot Dr for all values of r (rather than only 0 ≤r ≤N0 −1), then the spectrum Dr is N0\nperiodic. Moreover, Eq. (9.6) shows that x[n] can be synthesized not only by the N0 exponentials\ncorresponding to 0 ≤r ≤N0 −1, but also by any successive N0 exponentials in this spectrum,\nstarting at any value of r (positive or negative). For this reason, it is customary to show the\nspectrum Dr for all values of r (not just over the interval 0 ≤r ≤N0 −1). Yet we must remember\nthat to synthesize x[n] from this spectrum, we need to add only N0 consecutive components. All\nthese observations are consistent with our discussion in Ch. 5, where we showed that a sinusoid of\na given frequency is equivalent to multitudes of sinusoids, all separated by integer multiple of 2π\nin frequency.", - "type": "text" - }, - { - "block_id": "p869-b8", - "global_id": 25372, - "bbox": [ - 127.59, - 239.73, - 516.13, - 274.8 - ], - "text": "Along the scale, Dr repeats every 2π intervals, and along the r scale, Dr repeats at intervals\nof N0. Equations (9.6) and (9.7) show that both x[n] and its spectrum Dr are N0 periodic and both\nhave exactly the same number of components (N0) over one period.", - "type": "text" - }, - { - "block_id": "p869-b9", - "global_id": 25373, - "bbox": [ - 127.6, - 275.6, - 516.14, - 297.93 - ], - "text": "Equation (9.7) shows that Dr is complex in general, and D−r is the conjugate of Dr if x[n] is\nreal. Thus,", - "type": "text" - }, - { - "block_id": "p869-b10", - "global_id": 25374, - "bbox": [ - 240.1, - 301.01, - 402.95, - 312.47 - ], - "text": "|Dr| = |D−r|\nand̸\nDr = −̸ D−r", - "type": "text" - }, - { - "block_id": "p869-b11", - "global_id": 25375, - "bbox": [ - 127.59, - 320.95, - 516.14, - 355.24 - ], - "text": "so that the amplitude spectrum |Dr| is an even function , and̸\nDr is an odd function of r (or ).\nAll these concepts will be clarified by the examples to follow. The first example is rather trivial\nand serves mainly to familiarize the reader with the basic concepts of DTFS.", - "type": "text" - }, - { - "block_id": "p869-b12", - "global_id": 25376, - "bbox": [ - 102.51, - 384.73, - 432.14, - 396.69 - ], - "text": "EXAMPLE 9.1\nDiscrete-Time Fourier Series of a Sinusoid", - "type": "text" - }, - { - "block_id": "p869-b13", - "global_id": 25377, - "bbox": [ - 128.9, - 410.02, - 502.76, - 432.35 - ], - "text": "Find the discrete-time Fourier series (DTFS) for x[n] = sin 0.1πn (Fig. 9.1a). Sketch the\namplitude and phase spectra.", - "type": "text" - }, - { - "block_id": "p869-b14", - "global_id": 25378, - "bbox": [ - 128.9, - 454.85, - 502.75, - 478.68 - ], - "text": "In this case, the sinusoid sin 0.1πn is periodic because /2π = 1/20 is a rational number and\nthe period N0 is [see Eq. (9.1)]", - "type": "text" - }, - { - "block_id": "p869-b15", - "global_id": 25379, - "bbox": [ - 245.42, - 494.36, - 275.1, - 505.82 - ], - "text": "N0 = m", - "type": "text" - }, - { - "block_id": "p869-b16", - "global_id": 25380, - "bbox": [ - 275.1, - 480.38, - 293.98, - 497.76 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p869-b19", - "global_id": 25381, - "bbox": [ - 304.95, - 494.36, - 321.97, - 504.64 - ], - "text": "= m", - "type": "text" - }, - { - "block_id": "p869-b20", - "global_id": 25382, - "bbox": [ - 321.96, - 480.38, - 344.57, - 497.76 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p869-b21", - "global_id": 25383, - "bbox": [ - 329.88, - 501.44, - 348.31, - 511.81 - ], - "text": "0.1π", - "type": "text" - }, - { - "block_id": "p869-b23", - "global_id": 25384, - "bbox": [ - 359.28, - 494.36, - 386.24, - 504.74 - ], - "text": "= 20m", - "type": "text" - }, - { - "block_id": "p869-b24", - "global_id": 25385, - "bbox": [ - 128.9, - 521.58, - 502.76, - 544.99 - ], - "text": "The smallest value of m that makes 20m an integer is m = 1. Therefore, the period N0 = 20 so\nthat 0 = 2π/N0 = 0.1π, and from Eq. (9.6),", - "type": "text" - }, - { - "block_id": "p869-b25", - "global_id": 25386, - "bbox": [ - 272.25, - 557.11, - 298.14, - 567.39 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p869-b26", - "global_id": 25387, - "bbox": [ - 303.34, - 547.64, - 317.44, - 557.61 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p869-b27", - "global_id": 25388, - "bbox": [ - 300.19, - 571.5, - 320.59, - 578.76 - ], - "text": "r=⟨20⟩", - "type": "text" - }, - { - "block_id": "p869-b28", - "global_id": 25389, - "bbox": [ - 321.7, - 555.38, - 358.9, - 568.19 - ], - "text": "Drej0.1πrn", - "type": "text" - }, - { - "block_id": "p869-b29", - "global_id": 25390, - "bbox": [ - 128.9, - 589.6, - 502.76, - 611.61 - ], - "text": "where the sum is performed over any 20 consecutive values of r. We shall select the range\n−10 ≤r < 10 (values of r from −10 to 9). This choice corresponds to synthesizing x[n] using", - "type": "text" - } - ] - }, - { - "page_num": 870, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p870-b0", - "global_id": 25391, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "850\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p870-b1", - "global_id": 25392, - "bbox": [ - 94.2, - 426.51, - 341.66, - 436.12 - ], - "text": "Figure 9.1 Discrete-time sinusoid sin 0.1πn and its Fourier spectra.", - "type": "text" - }, - { - "block_id": "p870-b2", - "global_id": 25393, - "bbox": [ - 103.16, - 455.02, - 433.7, - 465.4 - ], - "text": "the spectral components in the fundamental frequency range (−π ≤ < π). Thus,", - "type": "text" - }, - { - "block_id": "p870-b3", - "global_id": 25394, - "bbox": [ - 246.34, - 479.82, - 272.23, - 490.1 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p870-b4", - "global_id": 25395, - "bbox": [ - 277.6, - 469.94, - 291.7, - 480.32 - ], - "text": "9\n\"", - "type": "text" - }, - { - "block_id": "p870-b5", - "global_id": 25396, - "bbox": [ - 274.28, - 494.22, - 294.99, - 501.48 - ], - "text": "r=−10", - "type": "text" - }, - { - "block_id": "p870-b6", - "global_id": 25397, - "bbox": [ - 296.12, - 478.1, - 333.32, - 490.9 - ], - "text": "Drej0.1πrn", - "type": "text" - }, - { - "block_id": "p870-b7", - "global_id": 25398, - "bbox": [ - 103.16, - 510.9, - 222.12, - 520.87 - ], - "text": "where, according to Eq. (9.7),", - "type": "text" - }, - { - "block_id": "p870-b8", - "global_id": 25399, - "bbox": [ - 192.78, - 528.73, - 224.36, - 546.75 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p870-b9", - "global_id": 25400, - "bbox": [ - 216.9, - 542.78, - 226.86, - 552.75 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p870-b10", - "global_id": 25401, - "bbox": [ - 232.78, - 525.41, - 246.88, - 535.79 - ], - "text": "9\n\"", - "type": "text" - }, - { - "block_id": "p870-b11", - "global_id": 25402, - "bbox": [ - 229.16, - 549.68, - 250.49, - 556.95 - ], - "text": "n=−10", - "type": "text" - }, - { - "block_id": "p870-b12", - "global_id": 25403, - "bbox": [ - 251.61, - 533.57, - 321.42, - 545.67 - ], - "text": "sin0.1πne−j0.1πrn", - "type": "text" - }, - { - "block_id": "p870-b13", - "global_id": 25404, - "bbox": [ - 205.88, - 564.82, - 224.36, - 581.76 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p870-b14", - "global_id": 25405, - "bbox": [ - 216.9, - 578.87, - 226.86, - 588.84 - ], - "text": "20", - "type": "text" - }, - { - "block_id": "p870-b15", - "global_id": 25406, - "bbox": [ - 232.78, - 561.5, - 246.88, - 571.89 - ], - "text": "9\n\"", - "type": "text" - }, - { - "block_id": "p870-b16", - "global_id": 25407, - "bbox": [ - 229.16, - 585.77, - 250.49, - 593.04 - ], - "text": "n=−10", - "type": "text" - }, - { - "block_id": "p870-b17", - "global_id": 25408, - "bbox": [ - 252.8, - 564.82, - 364.96, - 588.84 - ], - "text": "1\n2j(ej0.1πn −e−j0.1πn)e−j0.1πrn", - "type": "text" - }, - { - "block_id": "p870-b18", - "global_id": 25409, - "bbox": [ - 205.88, - 600.91, - 225.75, - 617.85 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p870-b19", - "global_id": 25410, - "bbox": [ - 216.9, - 614.86, - 229.62, - 624.93 - ], - "text": "40j", - "type": "text" - }, - { - "block_id": "p870-b20", - "global_id": 25411, - "bbox": [ - 231.93, - 590.5, - 255.83, - 607.98 - ], - "text": "9\n\"", - "type": "text" - }, - { - "block_id": "p870-b21", - "global_id": 25412, - "bbox": [ - 238.12, - 621.87, - 259.45, - 629.14 - ], - "text": "n=−10", - "type": "text" - }, - { - "block_id": "p870-b22", - "global_id": 25413, - "bbox": [ - 260.57, - 603.36, - 310.83, - 617.75 - ], - "text": "ej0.1πn(1−r) −", - "type": "text" - }, - { - "block_id": "p870-b23", - "global_id": 25414, - "bbox": [ - 316.01, - 597.59, - 330.1, - 607.98 - ], - "text": "9\n\"", - "type": "text" - }, - { - "block_id": "p870-b24", - "global_id": 25415, - "bbox": [ - 312.38, - 621.87, - 333.7, - 629.14 - ], - "text": "n=−10", - "type": "text" - }, - { - "block_id": "p870-b25", - "global_id": 25416, - "bbox": [ - 334.83, - 605.75, - 380.69, - 617.75 - ], - "text": "e−j0.1πn(1+r)", - "type": "text" - } - ] - }, - { - "page_num": 871, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p871-b0", - "global_id": 25417, - "bbox": [ - 310.73, - 62.89, - 516.13, - 71.98 - ], - "text": "9.1\nDiscrete-Time Fourier Series (DTFS)\n851", - "type": "text" - }, - { - "block_id": "p871-b1", - "global_id": 25418, - "bbox": [ - 128.9, - 85.83, - 502.77, - 133.15 - ], - "text": "In these sums, r takes on all values between −10 and 9. From Eq. (8.15), it follows that\nthe first sum on the right-hand side is zero for all values of r except r = 1, when the sum is\nequal to N0 = 20. Similarly, the second sum is zero for all values of r except r = −1, when it\nis equal to N0 = 20. Therefore,", - "type": "text" - }, - { - "block_id": "p871-b2", - "global_id": 25419, - "bbox": [ - 248.43, - 146.98, - 279.53, - 165.01 - ], - "text": "D1 = 1", - "type": "text" - }, - { - "block_id": "p871-b3", - "global_id": 25420, - "bbox": [ - 273.17, - 146.98, - 380.66, - 171.0 - ], - "text": "2j\nand\nD−1 = −1", - "type": "text" - }, - { - "block_id": "p871-b4", - "global_id": 25421, - "bbox": [ - 374.29, - 160.94, - 382.04, - 171.0 - ], - "text": "2j", - "type": "text" - }, - { - "block_id": "p871-b5", - "global_id": 25422, - "bbox": [ - 128.91, - 185.81, - 444.15, - 195.77 - ], - "text": "and all other coefficients are zero. The corresponding Fourier series is given by", - "type": "text" - }, - { - "block_id": "p871-b6", - "global_id": 25423, - "bbox": [ - 235.92, - 210.69, - 321.54, - 227.63 - ], - "text": "x[n] = sin 0.1πn = 1", - "type": "text" - }, - { - "block_id": "p871-b7", - "global_id": 25424, - "bbox": [ - 315.17, - 213.14, - 502.76, - 234.71 - ], - "text": "2j(ej0.1πn −e−j0.1πn)\n(9.8)", - "type": "text" - }, - { - "block_id": "p871-b8", - "global_id": 25425, - "bbox": [ - 128.91, - 249.09, - 484.85, - 260.55 - ], - "text": "Here the fundamental frequency 0 = 0.1π, and there are only two nonzero components:", - "type": "text" - }, - { - "block_id": "p871-b9", - "global_id": 25426, - "bbox": [ - 208.31, - 274.39, - 239.41, - 292.41 - ], - "text": "D1 = 1", - "type": "text" - }, - { - "block_id": "p871-b10", - "global_id": 25427, - "bbox": [ - 233.04, - 274.39, - 383.37, - 298.41 - ], - "text": "2j = 1\n2e−jπ/2\nand\nD−1 = −1", - "type": "text" - }, - { - "block_id": "p871-b11", - "global_id": 25428, - "bbox": [ - 377.01, - 274.39, - 422.85, - 298.41 - ], - "text": "2j = 1\n2ejπ/2", - "type": "text" - }, - { - "block_id": "p871-b12", - "global_id": 25429, - "bbox": [ - 128.9, - 313.21, - 170.65, - 323.17 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p871-b13", - "global_id": 25430, - "bbox": [ - 205.64, - 340.19, - 274.23, - 352.69 - ], - "text": "|D1| = |D−1| = 1", - "type": "text" - }, - { - "block_id": "p871-b14", - "global_id": 25431, - "bbox": [ - 270.74, - 334.55, - 373.77, - 354.73 - ], - "text": "2\nand̸\nD1 = −π", - "type": "text" - }, - { - "block_id": "p871-b15", - "global_id": 25432, - "bbox": [ - 368.79, - 334.55, - 423.84, - 358.99 - ], - "text": "2 ,̸\nD−1 = π", - "type": "text" - }, - { - "block_id": "p871-b16", - "global_id": 25433, - "bbox": [ - 418.86, - 349.03, - 423.84, - 358.99 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p871-b17", - "global_id": 25434, - "bbox": [ - 128.9, - 371.36, - 502.81, - 513.25 - ], - "text": "Sketches of Dr for the interval (−10 ≤r < 10) appear in Figs. 9.1b and 9.1c. According to\nEq. (9.8), there are only two components corresponding to r = 1 and −1. The remaining 18\ncoefficients are zero. The rth component Dr is the amplitude of the frequency r0 = 0.1rπ.\nTherefore, the frequency interval corresponding to −10 ≤r < 10 is −π ≤ < π, as depicted\nin Figs. 9.1b and 9.1c. This spectrum over the range −10 ≤r < 10 (or −π ≤ < π) is\nsufficient to specify the frequency-domain description (Fourier series), and we can synthesize\nx[n] by adding these spectral components. Because of the periodicity property discussed in\nthis section, the spectrum Dr is a periodic function of r with period N0 = 20. For this reason,\nwe repeat the spectrum with period N0 = 20 (or = 2π), as illustrated in Figs. 9.1b and 9.1c,\nwhich are periodic extensions of the spectrum in the range −10 ≤r < 10. Observe that the\namplitude spectrum is an even function and the angle or phase spectrum is an odd function of\nr (or ), as expected.", - "type": "text" - }, - { - "block_id": "p871-b18", - "global_id": 25435, - "bbox": [ - 128.9, - 515.24, - 502.78, - 596.93 - ], - "text": "The result [Eq. (9.8)] is a trigonometric identity and could have been obtained immediately\nwithout the formality of finding the Fourier coefficients. We have intentionally chosen this\ntrivial example to introduce the reader gently to the new concept of the discrete-time Fourier\nseries and its periodic nature. The Fourier series is a way of expressing a periodic signal x[n]\nin terms of exponentials of the form ejr0n and its harmonics. The result in Eq. (9.8) is merely\na statement of the (obvious) fact that sin 0.1πn can be expressed as a sum of two exponentials\nej0.1πn and e−j0.1πn.", - "type": "text" - }, - { - "block_id": "p871-b19", - "global_id": 25436, - "bbox": [ - 128.91, - 597.29, - 502.77, - 621.92 - ], - "text": "Because of the periodicity of the discrete-time exponentials ejr0n, the Fourier series\ncomponents can be selected in any range of length N0 = 20 (or = 2π). For example, if", - "type": "text" - } - ] - }, - { - "page_num": 872, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p872-b0", - "global_id": 25437, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "852\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p872-b1", - "global_id": 25438, - "bbox": [ - 103.16, - 85.83, - 460.79, - 96.21 - ], - "text": "we select the frequency range 0 ≤ < 2π (or 0 ≤r < 20), we obtain the Fourier series as", - "type": "text" - }, - { - "block_id": "p872-b2", - "global_id": 25439, - "bbox": [ - 212.89, - 111.12, - 298.51, - 128.07 - ], - "text": "x[n] = sin 0.1πn = 1", - "type": "text" - }, - { - "block_id": "p872-b3", - "global_id": 25440, - "bbox": [ - 292.15, - 113.58, - 367.29, - 135.14 - ], - "text": "2j(ej0.1πn −ej1.9πn)", - "type": "text" - }, - { - "block_id": "p872-b4", - "global_id": 25441, - "bbox": [ - 103.16, - 147.9, - 476.5, - 161.48 - ], - "text": "This series is equivalent to that in Eq. (9.8) because the two exponentials ej1.9πn and e−j0.1πn", - "type": "text" - }, - { - "block_id": "p872-b5", - "global_id": 25442, - "bbox": [ - 103.16, - 159.45, - 425.99, - 173.44 - ], - "text": "are equivalent. This follows from the fact that ej1.9πn = ej1.9πn × e−j2πn = e−j0.1πn.", - "type": "text" - }, - { - "block_id": "p872-b6", - "global_id": 25443, - "bbox": [ - 103.16, - 175.02, - 477.03, - 221.26 - ], - "text": "We could have selected the spectrum over any other range of width = 2π in Figs. 9.1b\nand 9.1c as a valid discrete-time Fourier series. The reader may verify this by proving that such\na spectrum starting anywhere (and of width = 2π) is equivalent to the same two components\non the right-hand side of Eq. (9.8).", - "type": "text" - }, - { - "block_id": "p872-b7", - "global_id": 25444, - "bbox": [ - 107.82, - 283.65, - 380.74, - 295.61 - ], - "text": "DRILL 9.1\nDTFS Spectra on Alternate Intervals", - "type": "text" - }, - { - "block_id": "p872-b8", - "global_id": 25445, - "bbox": [ - 107.82, - 304.31, - 484.41, - 326.65 - ], - "text": "From the spectra in Fig. 9.1, write the Fourier series corresponding to the interval −10 ≥r >\n−30 (or −π ≥ > −3π). Show that this Fourier is equivalent to that in Eq. (9.8).", - "type": "text" - }, - { - "block_id": "p872-b9", - "global_id": 25446, - "bbox": [ - 107.82, - 382.64, - 462.12, - 394.6 - ], - "text": "DRILL 9.2\nDiscrete-Time Fourier Series of a Sum of Sinusoids", - "type": "text" - }, - { - "block_id": "p872-b10", - "global_id": 25447, - "bbox": [ - 107.82, - 403.72, - 241.2, - 413.68 - ], - "text": "Find the period and the DTFS for", - "type": "text" - }, - { - "block_id": "p872-b11", - "global_id": 25448, - "bbox": [ - 232.66, - 425.22, - 359.58, - 435.6 - ], - "text": "x[n] = 4cos0.2πn + 6sin0.5πn", - "type": "text" - }, - { - "block_id": "p872-b12", - "global_id": 25449, - "bbox": [ - 107.82, - 447.13, - 338.51, - 458.21 - ], - "text": "over the interval 0 ≤r ≤19. Use Eq. (9.4) to compute Dr.", - "type": "text" - }, - { - "block_id": "p872-b13", - "global_id": 25450, - "bbox": [ - 108.09, - 471.04, - 170.31, - 482.0 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p872-b14", - "global_id": 25451, - "bbox": [ - 107.82, - 485.37, - 392.0, - 500.44 - ], - "text": "N0 = 20 and x[n] = 2ej0.2πn + (3e−jπ/2)ej0.5πn + (3ejπ/2)ej1.5πn + 2ej1.8πn", - "type": "text" - }, - { - "block_id": "p872-b15", - "global_id": 25452, - "bbox": [ - 107.82, - 554.93, - 445.84, - 566.88 - ], - "text": "DRILL 9.3\nFundamental Period of Discrete-Time Sinusoids", - "type": "text" - }, - { - "block_id": "p872-b16", - "global_id": 25453, - "bbox": [ - 107.82, - 575.59, - 426.17, - 586.74 - ], - "text": "Find the fundamental periods N0, if any, for: (a) sin(301πn/4) and (b) cos1.3n.", - "type": "text" - }, - { - "block_id": "p872-b17", - "global_id": 25454, - "bbox": [ - 108.09, - 599.49, - 170.31, - 610.45 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p872-b18", - "global_id": 25455, - "bbox": [ - 107.82, - 617.43, - 383.19, - 629.3 - ], - "text": "(a) N0 = 8, (b) N0 does not exist because the sinusoid is not periodic.", - "type": "text" - } - ] - }, - { - "page_num": 873, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p873-b0", - "global_id": 25456, - "bbox": [ - 310.73, - 62.89, - 516.13, - 71.98 - ], - "text": "9.1\nDiscrete-Time Fourier Series (DTFS)\n853", - "type": "text" - }, - { - "block_id": "p873-b1", - "global_id": 25457, - "bbox": [ - 103.92, - 95.83, - 509.93, - 107.78 - ], - "text": "EXAMPLE 9.2\nDiscrete-Time Fourier Series of a Periodic Gate Function", - "type": "text" - }, - { - "block_id": "p873-b2", - "global_id": 25458, - "bbox": [ - 128.9, - 122.94, - 502.77, - 144.87 - ], - "text": "Compute and plot the discrete-time Fourier series for the periodic sampled gate function shown\nin Fig. 9.2a.", - "type": "text" - }, - { - "block_id": "p873-b3", - "global_id": 25459, - "bbox": [ - 119.94, - 370.11, - 383.49, - 379.42 - ], - "text": "Figure 9.2 (a) Periodic sampled gate pulse and (b) its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p873-b4", - "global_id": 25460, - "bbox": [ - 146.84, - 396.71, - 377.9, - 408.17 - ], - "text": "In this case, N0 = 32 and 0 = 2π/32 = π/16. Therefore,", - "type": "text" - }, - { - "block_id": "p873-b5", - "global_id": 25461, - "bbox": [ - 268.98, - 420.14, - 294.87, - 430.42 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p873-b6", - "global_id": 25462, - "bbox": [ - 300.07, - 410.68, - 314.16, - 420.64 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p873-b7", - "global_id": 25463, - "bbox": [ - 296.92, - 434.54, - 317.31, - 441.8 - ], - "text": "r=⟨32⟩", - "type": "text" - }, - { - "block_id": "p873-b8", - "global_id": 25464, - "bbox": [ - 318.43, - 418.42, - 362.19, - 431.22 - ], - "text": "Drejr(π/16)n", - "type": "text" - }, - { - "block_id": "p873-b9", - "global_id": 25465, - "bbox": [ - 128.9, - 452.73, - 153.23, - 462.69 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p873-b10", - "global_id": 25466, - "bbox": [ - 259.22, - 460.61, - 290.81, - 478.64 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p873-b11", - "global_id": 25467, - "bbox": [ - 283.34, - 474.67, - 293.29, - 484.63 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p873-b12", - "global_id": 25468, - "bbox": [ - 299.07, - 457.72, - 313.16, - 467.68 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p873-b13", - "global_id": 25469, - "bbox": [ - 295.6, - 481.57, - 316.61, - 488.83 - ], - "text": "n=⟨32⟩", - "type": "text" - }, - { - "block_id": "p873-b14", - "global_id": 25470, - "bbox": [ - 317.73, - 465.46, - 371.95, - 477.46 - ], - "text": "x[n]e−jr(π/16)n", - "type": "text" - }, - { - "block_id": "p873-b15", - "global_id": 25471, - "bbox": [ - 128.9, - 496.37, - 502.74, - 518.7 - ], - "text": "For our convenience, we shall choose the interval −16 ≤n ≤15 for this summation, although\nany other interval of the same width (32 points) would give the same result.†", - "type": "text" - }, - { - "block_id": "p873-b16", - "global_id": 25472, - "bbox": [ - 259.06, - 533.29, - 290.65, - 551.31 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p873-b17", - "global_id": 25473, - "bbox": [ - 283.18, - 547.35, - 293.14, - 557.31 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p873-b18", - "global_id": 25474, - "bbox": [ - 299.07, - 529.97, - 313.16, - 540.36 - ], - "text": "15\n\"", - "type": "text" - }, - { - "block_id": "p873-b19", - "global_id": 25475, - "bbox": [ - 295.44, - 554.26, - 316.77, - 561.52 - ], - "text": "n=−16", - "type": "text" - }, - { - "block_id": "p873-b20", - "global_id": 25476, - "bbox": [ - 317.89, - 538.14, - 372.11, - 550.13 - ], - "text": "x[n]e−jr(π/16)n", - "type": "text" - }, - { - "block_id": "p873-b21", - "global_id": 25477, - "bbox": [ - 128.9, - 571.51, - 502.75, - 627.94 - ], - "text": "† In this example we have used the same equations as those for the DFT in Ex. 8.9, within a scaling\nconstant. In the present example, the values of x[n] at n = 4 and −4 are taken as 1 (full value), whereas\nin Ex. 8.9 these values are 0.5 (half the value). This is the reason for the slight difference in spectra\nin Figs. 9.2b and 8.19d. Unlike continuous-time signals, discontinuity is a meaningless concept in\ndiscrete-time signals.", - "type": "text" - } - ] - }, - { - "page_num": 874, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p874-b0", - "global_id": 25478, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "854\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p874-b1", - "global_id": 25479, - "bbox": [ - 103.16, - 85.83, - 409.92, - 96.2 - ], - "text": "Now, x[n] = 1 for −4 ≤n ≤4 and is zero for all other values of n. Therefore,", - "type": "text" - }, - { - "block_id": "p874-b2", - "global_id": 25480, - "bbox": [ - 243.1, - 108.68, - 274.68, - 126.7 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p874-b3", - "global_id": 25481, - "bbox": [ - 267.22, - 122.74, - 277.17, - 132.7 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p874-b4", - "global_id": 25482, - "bbox": [ - 281.36, - 105.36, - 295.45, - 115.74 - ], - "text": "4\n\"", - "type": "text" - }, - { - "block_id": "p874-b5", - "global_id": 25483, - "bbox": [ - 279.47, - 129.64, - 297.32, - 136.9 - ], - "text": "n=−4", - "type": "text" - }, - { - "block_id": "p874-b6", - "global_id": 25484, - "bbox": [ - 298.44, - 113.53, - 477.01, - 125.62 - ], - "text": "e−jr(π/16)n\n(9.9)", - "type": "text" - }, - { - "block_id": "p874-b7", - "global_id": 25485, - "bbox": [ - 103.17, - 145.56, - 466.97, - 157.16 - ], - "text": "This is a geometric progression with a common ratio e−j(π/16)r. Therefore (see Sec. B.8-3),†", - "type": "text" - }, - { - "block_id": "p874-b8", - "global_id": 25486, - "bbox": [ - 191.0, - 167.21, - 222.59, - 185.24 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p874-b9", - "global_id": 25487, - "bbox": [ - 215.12, - 181.26, - 225.07, - 191.23 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p874-b10", - "global_id": 25488, - "bbox": [ - 227.38, - 159.79, - 316.93, - 177.08 - ], - "text": "e−j(5πr/16) −ej(4πr/16)", - "type": "text" - }, - { - "block_id": "p874-b11", - "global_id": 25489, - "bbox": [ - 250.04, - 177.98, - 301.39, - 191.23 - ], - "text": "e−j(πr/16) −1", - "type": "text" - }, - { - "block_id": "p874-b12", - "global_id": 25490, - "bbox": [ - 318.62, - 159.79, - 324.05, - 169.75 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p874-b13", - "global_id": 25491, - "bbox": [ - 204.1, - 203.28, - 211.87, - 213.24 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p874-b14", - "global_id": 25492, - "bbox": [ - 213.92, - 189.29, - 229.31, - 206.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p874-b15", - "global_id": 25493, - "bbox": [ - 221.84, - 210.76, - 231.8, - 220.73 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p874-b16", - "global_id": 25494, - "bbox": [ - 233.0, - 187.69, - 290.18, - 205.98 - ], - "text": "e−j(0.5πr/16)", - "type": "text" - }, - { - "block_id": "p874-b17", - "global_id": 25495, - "bbox": [ - 290.18, - 187.69, - 387.99, - 205.98 - ], - "text": "e−j(4.5πr/16) −ej(4.5πr/16)", - "type": "text" - }, - { - "block_id": "p874-b18", - "global_id": 25496, - "bbox": [ - 240.92, - 202.85, - 290.18, - 221.14 - ], - "text": "e−j(0.5πr/16)", - "type": "text" - }, - { - "block_id": "p874-b19", - "global_id": 25497, - "bbox": [ - 290.18, - 202.85, - 387.99, - 221.14 - ], - "text": "e−j(0.5πr/16) −ej(0.5πr/16)", - "type": "text" - }, - { - "block_id": "p874-b20", - "global_id": 25498, - "bbox": [ - 204.1, - 246.34, - 211.87, - 256.3 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p874-b21", - "global_id": 25499, - "bbox": [ - 213.92, - 232.35, - 229.31, - 249.73 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p874-b22", - "global_id": 25500, - "bbox": [ - 221.84, - 253.82, - 231.8, - 263.78 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p874-b23", - "global_id": 25501, - "bbox": [ - 233.0, - 232.35, - 253.66, - 243.16 - ], - "text": "sin", - "type": "text" - }, - { - "block_id": "p874-b24", - "global_id": 25502, - "bbox": [ - 254.76, - 218.79, - 285.98, - 236.17 - ], - "text": "4.5πr", - "type": "text" - }, - { - "block_id": "p874-b25", - "global_id": 25503, - "bbox": [ - 269.47, - 240.27, - 279.42, - 250.23 - ], - "text": "16", - "type": "text" - }, - { - "block_id": "p874-b27", - "global_id": 25504, - "bbox": [ - 242.03, - 260.31, - 253.66, - 270.27 - ], - "text": "sin", - "type": "text" - }, - { - "block_id": "p874-b28", - "global_id": 25505, - "bbox": [ - 254.76, - 245.91, - 285.98, - 263.29 - ], - "text": "0.5πr", - "type": "text" - }, - { - "block_id": "p874-b29", - "global_id": 25506, - "bbox": [ - 269.47, - 267.38, - 279.42, - 277.35 - ], - "text": "16", - "type": "text" - }, - { - "block_id": "p874-b31", - "global_id": 25507, - "bbox": [ - 204.1, - 287.8, - 211.87, - 297.76 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p874-b32", - "global_id": 25508, - "bbox": [ - 213.92, - 273.8, - 229.31, - 291.19 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p874-b33", - "global_id": 25509, - "bbox": [ - 221.84, - 295.28, - 231.8, - 305.25 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p874-b34", - "global_id": 25510, - "bbox": [ - 233.0, - 273.8, - 290.49, - 291.96 - ], - "text": "sin(4.5r0)", - "type": "text" - }, - { - "block_id": "p874-b35", - "global_id": 25511, - "bbox": [ - 242.03, - 280.81, - 343.88, - 306.02 - ], - "text": "sin(0.5r0)\n0 = π", - "type": "text" - }, - { - "block_id": "p874-b36", - "global_id": 25512, - "bbox": [ - 336.41, - 288.21, - 477.01, - 305.25 - ], - "text": "16\n(9.10)", - "type": "text" - }, - { - "block_id": "p874-b37", - "global_id": 25513, - "bbox": [ - 103.16, - 313.9, - 370.21, - 323.87 - ], - "text": "This spectrum (with its periodic extension) is depicted in Fig. 9.2b.", - "type": "text" - }, - { - "block_id": "p874-b38", - "global_id": 25514, - "bbox": [ - 103.16, - 338.22, - 477.01, - 376.29 - ], - "text": "DISCRETE-TIME FOURIER SERIES USING MATLAB\nLet us confirm our results by using MATLAB to directly compute the DTFS according to\nEq. (9.4).", - "type": "text" - }, - { - "block_id": "p874-b39", - "global_id": 25515, - "bbox": [ - 103.16, - 385.78, - 380.37, - 467.48 - ], - "text": ">>\nN_0 = 32; n = (0:N_0-1); Omega_0 = 2*pi/N_0;\n>>\nx_n = [ones(1,5) zeros(1,23) ones(1,4)];\n>>\nfor r = 0:N_0-1,\n>>\nX_r(r+1) = sum(x_n.*exp(-j*r*Omega_0*n))/N_0;\n>>\nend\n>>\nr = n; stem(r,real(X_r),’k.’);\n>>\nxlabel(’r’); ylabel(’X_r’); axis([0 31 -.1 0.3]);", - "type": "text" - }, - { - "block_id": "p874-b40", - "global_id": 25516, - "bbox": [ - 103.16, - 476.39, - 338.84, - 486.35 - ], - "text": "The MATLAB result, shown in Fig. 9.3, matches Fig. 9.2b.", - "type": "text" - }, - { - "block_id": "p874-b41", - "global_id": 25517, - "bbox": [ - 121.09, - 488.25, - 432.15, - 499.81 - ], - "text": "Alternatively, scaling the FFT by N0 produces the exact same result (Fig. 9.3).", - "type": "text" - }, - { - "block_id": "p874-b42", - "global_id": 25518, - "bbox": [ - 103.16, - 507.81, - 380.38, - 529.73 - ], - "text": ">>\nX_r = fft(x_n)/N_0; stem(r,real(X_r),’k.’);\n>>\nxlabel(’r’); ylabel(’X_r’); axis([0 31 -.1 0.3]);", - "type": "text" - }, - { - "block_id": "p874-b43", - "global_id": 25519, - "bbox": [ - 103.16, - 548.44, - 477.04, - 582.94 - ], - "text": "† Strictly speaking, the geometric progression sum formula applies only if the common ratio\ne−j(π/16)r̸ = 1. When r = 0, this ratio is unity. Hence, Eq. (9.10) is valid for values of r̸ = 0. For the\ncase r = 0, the sum in Eq. (9.9) is given by", - "type": "text" - }, - { - "block_id": "p874-b44", - "global_id": 25520, - "bbox": [ - 260.57, - 594.94, - 267.04, - 608.22 - ], - "text": "1\n32", - "type": "text" - }, - { - "block_id": "p874-b45", - "global_id": 25521, - "bbox": [ - 271.18, - 587.25, - 283.86, - 596.75 - ], - "text": "4\n\"", - "type": "text" - }, - { - "block_id": "p874-b46", - "global_id": 25522, - "bbox": [ - 269.23, - 609.23, - 285.81, - 615.98 - ], - "text": "n=−4", - "type": "text" - }, - { - "block_id": "p874-b47", - "global_id": 25523, - "bbox": [ - 286.8, - 594.94, - 318.0, - 605.55 - ], - "text": "x[n] = 9", - "type": "text" - }, - { - "block_id": "p874-b48", - "global_id": 25524, - "bbox": [ - 313.14, - 601.74, - 319.62, - 608.22 - ], - "text": "32", - "type": "text" - }, - { - "block_id": "p874-b49", - "global_id": 25525, - "bbox": [ - 103.16, - 622.86, - 477.02, - 643.16 - ], - "text": "Fortunately, the value of D0, as computed from Eq. (9.10), also happens to be 9/32. Hence, Eq. (9.10) is\nvalid for all r.", - "type": "text" - } - ] - }, - { - "page_num": 875, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p875-b0", - "global_id": 25526, - "bbox": [ - 247.51, - 62.89, - 516.12, - 71.98 - ], - "text": "9.2\nAperiodic Signal Representation by Fourier Integral\n855", - "type": "text" - }, - { - "block_id": "p875-b1", - "global_id": 25527, - "bbox": [ - 159.32, - 168.97, - 488.79, - 190.16 - ], - "text": "0\n5\n10\n15\n20\n25\n30\nr", - "type": "text" - }, - { - "block_id": "p875-b2", - "global_id": 25528, - "bbox": [ - 143.25, - 159.76, - 157.25, - 167.76 - ], - "text": "–0.1", - "type": "text" - }, - { - "block_id": "p875-b3", - "global_id": 25529, - "bbox": [ - 153.25, - 141.75, - 157.25, - 149.75 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p875-b4", - "global_id": 25530, - "bbox": [ - 146.35, - 123.75, - 157.02, - 131.75 - ], - "text": "0.1", - "type": "text" - }, - { - "block_id": "p875-b5", - "global_id": 25531, - "bbox": [ - 146.35, - 105.76, - 157.02, - 113.76 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p875-b6", - "global_id": 25532, - "bbox": [ - 129.6, - 122.99, - 140.88, - 130.99 - ], - "text": "Xr", - "type": "text" - }, - { - "block_id": "p875-b7", - "global_id": 25533, - "bbox": [ - 128.9, - 196.85, - 456.1, - 206.09 - ], - "text": "Figure 9.3 MATLAB-computed DTFS spectra for periodic sampled gate pulse of Ex. 9.2.", - "type": "text" - }, - { - "block_id": "p875-b8", - "global_id": 25534, - "bbox": [ - 127.94, - 272.23, - 400.1, - 302.12 - ], - "text": "9.2 APERIODIC SIGNAL REPRESENTATION\nBY FOURIER INTEGRAL", - "type": "text" - }, - { - "block_id": "p875-b9", - "global_id": 25535, - "bbox": [ - 127.59, - 308.11, - 516.14, - 341.98 - ], - "text": "In Sec. 9.1 we succeeded in representing periodic signals as a sum of (everlasting) exponentials.\nIn this section we extend this representation to aperiodic signals. The procedure is identical\nconceptually to that used in Ch. 7 for continuous-time signals.", - "type": "text" - }, - { - "block_id": "p875-b10", - "global_id": 25536, - "bbox": [ - 127.59, - 343.56, - 516.13, - 415.75 - ], - "text": "Applying a limiting process, we now show that an aperiodic signal x[n] can be expressed as\na continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal x[n] such\nas the one illustrated in Fig. 9.4a by everlasting exponential signals, let us construct a new periodic\nsignal xN0[n] formed by repeating the signal x[n] every N0 units, as shown in Fig. 9.4b. The period\nN0 is made large enough to avoid overlap between the repeating cycles (N0 ≥2N +1). The periodic\nsignal xN0[n] can be represented by an exponential Fourier series. If we let N0 →∞, the signal", - "type": "text" - }, - { - "block_id": "p875-b11", - "global_id": 25537, - "bbox": [ - 151.5, - 619.58, - 445.98, - 629.19 - ], - "text": "Figure 9.4 Generation of a periodic signal by periodic extension of a signal x[n].", - "type": "text" - } - ] - }, - { - "page_num": 876, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p876-b0", - "global_id": 25538, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "856\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p876-b1", - "global_id": 25539, - "bbox": [ - 101.84, - 85.4, - 305.31, - 95.78 - ], - "text": "x[n] repeats after an infinite interval, and therefore,", - "type": "text" - }, - { - "block_id": "p876-b2", - "global_id": 25540, - "bbox": [ - 257.8, - 105.59, - 334.43, - 122.92 - ], - "text": "lim\nN0→∞xN0[n] = x[n]", - "type": "text" - }, - { - "block_id": "p876-b3", - "global_id": 25541, - "bbox": [ - 101.84, - 130.68, - 490.38, - 155.04 - ], - "text": "Thus, the Fourier series representing xN0[n] will also represent x[n] in the limit N0 →∞. The\nexponential Fourier series for xN0[n] is given by", - "type": "text" - }, - { - "block_id": "p876-b4", - "global_id": 25542, - "bbox": [ - 220.98, - 168.69, - 255.47, - 181.1 - ], - "text": "xN0[n] =", - "type": "text" - }, - { - "block_id": "p876-b5", - "global_id": 25543, - "bbox": [ - 261.25, - 159.22, - 275.35, - 169.18 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p876-b6", - "global_id": 25544, - "bbox": [ - 257.52, - 182.98, - 279.08, - 191.5 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p876-b7", - "global_id": 25545, - "bbox": [ - 280.2, - 161.7, - 369.06, - 180.14 - ], - "text": "Drejr0n\n0 = 2π", - "type": "text" - }, - { - "block_id": "p876-b8", - "global_id": 25546, - "bbox": [ - 358.78, - 176.07, - 368.9, - 186.9 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p876-b9", - "global_id": 25547, - "bbox": [ - 466.32, - 169.1, - 490.38, - 179.06 - ], - "text": "(9.11)", - "type": "text" - }, - { - "block_id": "p876-b10", - "global_id": 25548, - "bbox": [ - 101.85, - 199.67, - 126.18, - 209.64 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p876-b11", - "global_id": 25549, - "bbox": [ - 244.48, - 210.72, - 276.39, - 228.74 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p876-b12", - "global_id": 25550, - "bbox": [ - 268.6, - 224.68, - 278.72, - 235.51 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p876-b13", - "global_id": 25551, - "bbox": [ - 285.22, - 207.11, - 299.32, - 217.78 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p876-b14", - "global_id": 25552, - "bbox": [ - 281.53, - 231.14, - 303.0, - 238.33 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p876-b15", - "global_id": 25553, - "bbox": [ - 304.12, - 215.57, - 490.38, - 227.66 - ], - "text": "x[n]e−jr0n\n(9.12)", - "type": "text" - }, - { - "block_id": "p876-b16", - "global_id": 25554, - "bbox": [ - 101.85, - 244.31, - 490.37, - 266.64 - ], - "text": "The limits for the sum on the right-hand side of Eq. (9.12) should be from −N to N. But because\nx[n] = 0 for |n| > N, it does not matter if the limits are taken from −∞to ∞.", - "type": "text" - }, - { - "block_id": "p876-b17", - "global_id": 25555, - "bbox": [ - 101.85, - 268.54, - 490.38, - 290.56 - ], - "text": "It is interesting to see how the nature of the spectrum changes as N0 increases. To understand\nthis behavior, let us define X(), a continuous function of , as", - "type": "text" - }, - { - "block_id": "p876-b18", - "global_id": 25556, - "bbox": [ - 249.39, - 308.06, - 280.92, - 318.34 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p876-b19", - "global_id": 25557, - "bbox": [ - 286.66, - 297.89, - 300.76, - 308.56 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p876-b20", - "global_id": 25558, - "bbox": [ - 282.97, - 321.91, - 304.44, - 329.1 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p876-b21", - "global_id": 25559, - "bbox": [ - 305.56, - 306.34, - 490.38, - 318.44 - ], - "text": "x[n]e−jn\n(9.13)", - "type": "text" - }, - { - "block_id": "p876-b22", - "global_id": 25560, - "bbox": [ - 101.85, - 337.63, - 278.37, - 347.59 - ], - "text": "From this definition and Eq. (9.12), we have", - "type": "text" - }, - { - "block_id": "p876-b23", - "global_id": 25561, - "bbox": [ - 263.25, - 355.79, - 295.17, - 373.82 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p876-b24", - "global_id": 25562, - "bbox": [ - 287.36, - 369.75, - 297.49, - 380.58 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p876-b25", - "global_id": 25563, - "bbox": [ - 299.19, - 362.36, - 490.38, - 373.51 - ], - "text": "X(r0)\n(9.14)", - "type": "text" - }, - { - "block_id": "p876-b26", - "global_id": 25564, - "bbox": [ - 101.84, - 387.06, - 490.4, - 540.9 - ], - "text": "This result shows that the Fourier coefficients Dr are 1/N0 times the samples of X() taken every\n0 rad/s.† Therefore, (1/N0)X() is the envelope for the coefficients Dr. We now let N0 →∞by\ndoubling N0 repeatedly. Doubling N0 halves the fundamental frequency 0, with the result that the\nspacing between successive spectral components (harmonics) is halved, and there are now twice\nas many components (samples) in the spectrum. At the same time, by doubling N0, the envelope of\nthe coefficients Dr is halved, as seen from Eq. (9.14). If we continue this process of doubling N0\nrepeatedly, the number of components doubles in each step; the spectrum progressively becomes\ndenser, while its magnitude Dr becomes smaller. Note, however, that the relative shape of the\nenvelope remains the same [proportional to X() in Eq. (9.13)]. In the limit, as N0 →∞, the\nfundamental frequency 0 →0, and Dr →0. The separation between successive harmonics, which\nis 0, is approaching zero (infinitesimal), and the spectrum becomes so dense that it appears to be\ncontinuous. But as the number of harmonics increases indefinitely, the harmonic amplitudes Dr\nbecome vanishingly small (infinitesimal). We discussed an identical situation in Sec. 7.1.", - "type": "text" - }, - { - "block_id": "p876-b27", - "global_id": 25565, - "bbox": [ - 119.78, - 542.48, - 432.64, - 553.93 - ], - "text": "We follow the procedure in Sec. 7.1 and let N0 →∞. According to Eq. (9.13),", - "type": "text" - }, - { - "block_id": "p876-b28", - "global_id": 25566, - "bbox": [ - 242.16, - 570.78, - 281.79, - 581.93 - ], - "text": "X(r0) =", - "type": "text" - }, - { - "block_id": "p876-b29", - "global_id": 25567, - "bbox": [ - 287.52, - 560.61, - 301.62, - 571.28 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p876-b30", - "global_id": 25568, - "bbox": [ - 283.83, - 584.63, - 305.31, - 591.82 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p876-b31", - "global_id": 25569, - "bbox": [ - 306.42, - 569.06, - 349.57, - 581.06 - ], - "text": "x[n]e−jr0n", - "type": "text" - }, - { - "block_id": "p876-b32", - "global_id": 25570, - "bbox": [ - 101.84, - 610.24, - 490.38, - 634.75 - ], - "text": "† For the sake of simplicity we assume Dr and therefore X() to be real. The argument, however, is also valid\nfor complex Dr [or X()].", - "type": "text" - } - ] - }, - { - "page_num": 877, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p877-b0", - "global_id": 25571, - "bbox": [ - 247.51, - 62.89, - 516.12, - 71.98 - ], - "text": "9.2\nAperiodic Signal Representation by Fourier Integral\n857", - "type": "text" - }, - { - "block_id": "p877-b1", - "global_id": 25572, - "bbox": [ - 127.59, - 85.82, - 312.52, - 95.78 - ], - "text": "Using Eq. (9.14), we can express Eq. (9.11) as", - "type": "text" - }, - { - "block_id": "p877-b2", - "global_id": 25573, - "bbox": [ - 201.53, - 106.1, - 247.07, - 125.08 - ], - "text": "xN0[n] = 1", - "type": "text" - }, - { - "block_id": "p877-b3", - "global_id": 25574, - "bbox": [ - 239.27, - 120.06, - 249.4, - 130.89 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p877-b4", - "global_id": 25575, - "bbox": [ - 255.94, - 103.21, - 270.04, - 113.18 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p877-b5", - "global_id": 25576, - "bbox": [ - 252.21, - 126.97, - 273.77, - 135.5 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p877-b6", - "global_id": 25577, - "bbox": [ - 274.88, - 108.56, - 336.63, - 123.82 - ], - "text": "X(r0)ejr0n =", - "type": "text" - }, - { - "block_id": "p877-b7", - "global_id": 25578, - "bbox": [ - 342.41, - 103.21, - 356.51, - 113.18 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p877-b8", - "global_id": 25579, - "bbox": [ - 338.68, - 126.97, - 360.25, - 135.5 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p877-b9", - "global_id": 25580, - "bbox": [ - 361.36, - 110.95, - 412.79, - 123.82 - ], - "text": "X(r0)ejr0n", - "type": "text" - }, - { - "block_id": "p877-b10", - "global_id": 25581, - "bbox": [ - 414.4, - 98.69, - 433.66, - 116.84 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p877-b11", - "global_id": 25582, - "bbox": [ - 422.32, - 119.75, - 433.27, - 130.12 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p877-b13", - "global_id": 25583, - "bbox": [ - 127.59, - 144.7, - 379.62, - 157.11 - ], - "text": "In the limit as N0 →∞, 0 →0 and xN0[n] →x[n]. Therefore,", - "type": "text" - }, - { - "block_id": "p877-b14", - "global_id": 25584, - "bbox": [ - 246.96, - 172.87, - 291.58, - 183.25 - ], - "text": "x[n] = lim", - "type": "text" - }, - { - "block_id": "p877-b15", - "global_id": 25585, - "bbox": [ - 274.9, - 181.78, - 294.96, - 190.31 - ], - "text": "0→0", - "type": "text" - }, - { - "block_id": "p877-b16", - "global_id": 25586, - "bbox": [ - 299.8, - 163.4, - 313.9, - 173.37 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p877-b17", - "global_id": 25587, - "bbox": [ - 296.06, - 187.16, - 317.63, - 195.69 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p877-b18", - "global_id": 25588, - "bbox": [ - 318.74, - 158.89, - 366.4, - 177.04 - ], - "text": "X(r0)0", - "type": "text" - }, - { - "block_id": "p877-b19", - "global_id": 25589, - "bbox": [ - 340.15, - 179.95, - 351.11, - 190.32 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p877-b20", - "global_id": 25590, - "bbox": [ - 368.09, - 158.89, - 373.52, - 168.85 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p877-b21", - "global_id": 25591, - "bbox": [ - 374.63, - 171.15, - 516.13, - 183.25 - ], - "text": "ejr0n\n(9.15)", - "type": "text" - }, - { - "block_id": "p877-b22", - "global_id": 25592, - "bbox": [ - 127.59, - 204.89, - 516.13, - 216.77 - ], - "text": "Because 0 is infinitesimal, it will be appropriate to replace 0 with an infinitesimal notation :", - "type": "text" - }, - { - "block_id": "p877-b23", - "global_id": 25593, - "bbox": [ - 300.87, - 224.51, - 340.65, - 241.87 - ], - "text": "= 2π", - "type": "text" - }, - { - "block_id": "p877-b24", - "global_id": 25594, - "bbox": [ - 330.36, - 238.88, - 340.49, - 249.71 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p877-b25", - "global_id": 25595, - "bbox": [ - 492.07, - 231.9, - 516.13, - 241.87 - ], - "text": "(9.16)", - "type": "text" - }, - { - "block_id": "p877-b26", - "global_id": 25596, - "bbox": [ - 127.59, - 257.99, - 270.75, - 267.96 - ], - "text": "Equation (9.15) can be expressed as", - "type": "text" - }, - { - "block_id": "p877-b27", - "global_id": 25597, - "bbox": [ - 240.27, - 284.18, - 285.94, - 294.55 - ], - "text": "x[n] = lim", - "type": "text" - }, - { - "block_id": "p877-b28", - "global_id": 25598, - "bbox": [ - 268.2, - 293.08, - 290.39, - 300.35 - ], - "text": "→0", - "type": "text" - }, - { - "block_id": "p877-b29", - "global_id": 25599, - "bbox": [ - 292.69, - 277.61, - 303.64, - 301.63 - ], - "text": "1\n2π", - "type": "text" - }, - { - "block_id": "p877-b30", - "global_id": 25600, - "bbox": [ - 310.68, - 274.71, - 324.78, - 284.67 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p877-b31", - "global_id": 25601, - "bbox": [ - 306.95, - 298.46, - 328.51, - 306.99 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p877-b32", - "global_id": 25602, - "bbox": [ - 329.62, - 282.45, - 516.13, - 294.55 - ], - "text": "X(r)ejrn\n(9.17)", - "type": "text" - }, - { - "block_id": "p877-b33", - "global_id": 25603, - "bbox": [ - 127.59, - 316.2, - 515.13, - 338.54 - ], - "text": "The range r = ⟨N0⟩implies the interval of N0 number of harmonics, which is N0 = 2π\naccording to Eq. (9.16). In the limit, the right-hand side of Eq. (9.17) becomes the integral", - "type": "text" - }, - { - "block_id": "p877-b34", - "global_id": 25604, - "bbox": [ - 266.75, - 348.18, - 304.36, - 365.12 - ], - "text": "x[n] = 1", - "type": "text" - }, - { - "block_id": "p877-b35", - "global_id": 25605, - "bbox": [ - 295.88, - 361.82, - 306.84, - 372.2 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p877-b36", - "global_id": 25606, - "bbox": [ - 310.14, - 341.19, - 315.4, - 351.15 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p877-b37", - "global_id": 25607, - "bbox": [ - 315.4, - 350.95, - 516.13, - 373.33 - ], - "text": "2π\nX()ejn d\n(9.18)", - "type": "text" - }, - { - "block_id": "p877-b38", - "global_id": 25608, - "bbox": [ - 127.59, - 382.89, - 151.93, - 392.85 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p877-b39", - "global_id": 25609, - "bbox": [ - 154.93, - 374.45, - 159.49, - 384.41 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p877-b40", - "global_id": 25610, - "bbox": [ - 127.59, - 382.47, - 516.12, - 404.8 - ], - "text": "2π indicates integration over any continuous interval of 2π. The spectrum X() is given\nby [Eq. (9.13)]", - "type": "text" - }, - { - "block_id": "p877-b41", - "global_id": 25611, - "bbox": [ - 275.14, - 414.5, - 306.66, - 424.78 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p877-b42", - "global_id": 25612, - "bbox": [ - 312.41, - 404.33, - 326.51, - 415.01 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p877-b43", - "global_id": 25613, - "bbox": [ - 308.71, - 428.35, - 330.18, - 435.55 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p877-b44", - "global_id": 25614, - "bbox": [ - 331.31, - 412.78, - 516.13, - 424.88 - ], - "text": "x[n]e−jn\n(9.19)", - "type": "text" - }, - { - "block_id": "p877-b45", - "global_id": 25615, - "bbox": [ - 127.59, - 442.57, - 516.15, - 512.42 - ], - "text": "The integral on the right-hand side of Eq. (9.18) is called the Fourier integral. We have now\nsucceeded in representing an aperiodic signal x[n] by a Fourier integral (rather than a Fourier\nseries). This integral is basically a Fourier series (in the limit) with fundamental frequency\n →0, as seen in Eq. (9.17). The amount of the exponential ejrn is X(r)/2π. Thus, the\nfunction X() given by Eq. (9.19) acts as a spectral function, which indicates the relative amounts\nof various exponential components of x[n].", - "type": "text" - }, - { - "block_id": "p877-b46", - "global_id": 25616, - "bbox": [ - 127.6, - 513.99, - 516.14, - 536.32 - ], - "text": "We call X() the (direct) discrete-time Fourier transform (DTFT) of x[n], and x[n] the inverse\ndiscrete-time Fourier transform (IDTFT) of X(). This nomenclature can be represented as", - "type": "text" - }, - { - "block_id": "p877-b47", - "global_id": 25617, - "bbox": [ - 211.16, - 547.57, - 432.57, - 557.95 - ], - "text": "X() = DTFT{x[n]}\nand\nx[n] = IDTFT{X()}", - "type": "text" - }, - { - "block_id": "p877-b48", - "global_id": 25618, - "bbox": [ - 127.6, - 569.2, - 516.12, - 591.53 - ], - "text": "The same information is conveyed by the statement that x[n] and X() are a (discrete-time) Fourier\ntransform pair. Symbolically, this is expressed as", - "type": "text" - }, - { - "block_id": "p877-b49", - "global_id": 25619, - "bbox": [ - 291.37, - 602.78, - 352.37, - 613.06 - ], - "text": "x[n] ⇐⇒X()", - "type": "text" - }, - { - "block_id": "p877-b50", - "global_id": 25620, - "bbox": [ - 127.6, - 624.41, - 417.97, - 634.79 - ], - "text": "The Fourier transform X() is the frequency-domain description of x[n].", - "type": "text" - } - ] - }, - { - "page_num": 878, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p878-b0", - "global_id": 25621, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "858\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p878-b1", - "global_id": 25622, - "bbox": [ - 101.84, - 86.52, - 270.7, - 98.48 - ], - "text": "9.2-1 Nature of Fourier Spectra", - "type": "text" - }, - { - "block_id": "p878-b2", - "global_id": 25623, - "bbox": [ - 101.84, - 104.61, - 490.38, - 126.52 - ], - "text": "We now discuss several important features of the discrete-time Fourier transform and the spectra\nassociated with it.", - "type": "text" - }, - { - "block_id": "p878-b3", - "global_id": 25624, - "bbox": [ - 101.84, - 140.17, - 490.39, - 191.0 - ], - "text": "FOURIER SPECTRA ARE CONTINUOUS FUNCTIONS OF \nAlthough x[n] is a discrete-time signal, X(), its DTFT is a continuous function of for the\nsimple reason that is a continuous variable, which can take any value over a continuous interval\nfrom −∞to ∞.", - "type": "text" - }, - { - "block_id": "p878-b4", - "global_id": 25625, - "bbox": [ - 101.84, - 204.64, - 378.58, - 245.5 - ], - "text": "FOURIER SPECTRA ARE PERIODIC FUNCTIONS OF \nWITH PERIOD 2π\nFrom Eq. (9.19), it follows that", - "type": "text" - }, - { - "block_id": "p878-b5", - "global_id": 25626, - "bbox": [ - 163.74, - 264.02, - 218.09, - 274.4 - ], - "text": "X( + 2π) =", - "type": "text" - }, - { - "block_id": "p878-b6", - "global_id": 25627, - "bbox": [ - 223.83, - 253.85, - 237.93, - 264.52 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p878-b7", - "global_id": 25628, - "bbox": [ - 220.14, - 277.87, - 241.62, - 285.07 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p878-b8", - "global_id": 25629, - "bbox": [ - 242.73, - 259.91, - 308.84, - 274.3 - ], - "text": "x[n]e−j(+2π)n =", - "type": "text" - }, - { - "block_id": "p878-b9", - "global_id": 25630, - "bbox": [ - 314.59, - 253.85, - 328.68, - 264.52 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p878-b10", - "global_id": 25631, - "bbox": [ - 310.89, - 277.87, - 332.36, - 285.07 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p878-b11", - "global_id": 25632, - "bbox": [ - 333.49, - 259.91, - 428.5, - 274.3 - ], - "text": "x[n]e−jne−j2πn = X()", - "type": "text" - }, - { - "block_id": "p878-b12", - "global_id": 25633, - "bbox": [ - 101.85, - 293.82, - 490.4, - 352.02 - ], - "text": "Clearly, the spectrum X() is a continuous, periodic function of with period 2π. We must\nremember, however, that to synthesize x[n], we need to use the spectrum over a frequency interval\nof only 2π, starting at any value of [see Eq. (9.18)]. As a matter of convenience, we shall choose\nthis interval to be the fundamental frequency range (−π, π). It is, therefore, not necessary to show\ndiscrete-time-signal spectra beyond the fundamental range, although we often do so.", - "type": "text" - }, - { - "block_id": "p878-b13", - "global_id": 25634, - "bbox": [ - 101.85, - 353.59, - 490.39, - 399.84 - ], - "text": "The reason for the periodic behavior of X() was discussed in Ch. 5, where we showed that, in\na basic sense, the discrete-time frequency is bandlimited to || ≤π. However, all discrete-time\nsinusoids with frequencies separated by an integer multiple of 2π are identical. This is why the\nspectrum is 2π periodic.", - "type": "text" - }, - { - "block_id": "p878-b14", - "global_id": 25635, - "bbox": [ - 101.84, - 413.49, - 296.37, - 440.4 - ], - "text": "CONJUGATE SYMMETRY OF X()\nFrom Eq. (9.19), we obtain the DTFT of x∗[n] as", - "type": "text" - }, - { - "block_id": "p878-b15", - "global_id": 25636, - "bbox": [ - 209.21, - 457.2, - 271.5, - 469.3 - ], - "text": "DTFT{x∗[n]} =", - "type": "text" - }, - { - "block_id": "p878-b16", - "global_id": 25637, - "bbox": [ - 277.24, - 448.75, - 291.34, - 459.41 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p878-b17", - "global_id": 25638, - "bbox": [ - 273.55, - 472.77, - 295.02, - 479.96 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p878-b18", - "global_id": 25639, - "bbox": [ - 296.14, - 454.81, - 383.03, - 469.2 - ], - "text": "x∗[n]e−jn = X∗(−)", - "type": "text" - }, - { - "block_id": "p878-b19", - "global_id": 25640, - "bbox": [ - 101.84, - 489.07, - 162.35, - 499.03 - ], - "text": "In other words,", - "type": "text" - }, - { - "block_id": "p878-b20", - "global_id": 25641, - "bbox": [ - 257.6, - 498.89, - 490.38, - 510.98 - ], - "text": "x∗[n] ⇐⇒X∗(−)\n(9.20)", - "type": "text" - }, - { - "block_id": "p878-b21", - "global_id": 25642, - "bbox": [ - 101.84, - 517.74, - 443.85, - 529.34 - ], - "text": "For real x[n], Eq. (9.20) reduces to x[n] ⇐⇒X∗(−), which implies that for real x[n]", - "type": "text" - }, - { - "block_id": "p878-b22", - "global_id": 25643, - "bbox": [ - 262.52, - 538.01, - 329.7, - 550.02 - ], - "text": "X() = X∗(−)", - "type": "text" - }, - { - "block_id": "p878-b23", - "global_id": 25644, - "bbox": [ - 101.84, - 560.5, - 490.4, - 582.84 - ], - "text": "Therefore, for real x[n], X() and X(−) are conjugates. Since X() is generally complex, we\nhave both amplitude and angle (or phase) spectra", - "type": "text" - }, - { - "block_id": "p878-b24", - "global_id": 25645, - "bbox": [ - 251.79, - 593.37, - 339.95, - 605.37 - ], - "text": "X() = |X()|ej̸\nX()", - "type": "text" - }, - { - "block_id": "p878-b25", - "global_id": 25646, - "bbox": [ - 101.84, - 615.86, - 377.0, - 626.24 - ], - "text": "Because of conjugate symmetry of X(), it follows that for real x[n],", - "type": "text" - }, - { - "block_id": "p878-b26", - "global_id": 25647, - "bbox": [ - 190.73, - 636.62, - 401.5, - 647.0 - ], - "text": "|X()| = |X(−)|\nand̸\nX() = −̸ X(−)", - "type": "text" - } - ] - }, - { - "page_num": 879, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p879-b0", - "global_id": 25648, - "bbox": [ - 247.51, - 62.89, - 516.12, - 71.98 - ], - "text": "9.2\nAperiodic Signal Representation by Fourier Integral\n859", - "type": "text" - }, - { - "block_id": "p879-b1", - "global_id": 25649, - "bbox": [ - 127.59, - 85.46, - 516.13, - 107.79 - ], - "text": "Therefore, the amplitude spectrum |X()| is an even function of and the phase spectrum̸ X()\nis an odd function of for real x[n].", - "type": "text" - }, - { - "block_id": "p879-b2", - "global_id": 25650, - "bbox": [ - 127.89, - 122.34, - 396.13, - 148.4 - ], - "text": "PHYSICAL APPRECIATION OF THE DISCRETE-TIME\nFOURIER TRANSFORM", - "type": "text" - }, - { - "block_id": "p879-b3", - "global_id": 25651, - "bbox": [ - 127.59, - 152.43, - 516.12, - 198.26 - ], - "text": "In understanding any aspect of the Fourier transform, we should remember that Fourier\nrepresentation is a way of expressing a signal x[n] as a sum of everlasting exponentials (or\nsinusoids). The Fourier spectrum of a signal indicates the relative amplitudes and phases of the\nexponentials (or sinusoids) required to synthesize x[n].", - "type": "text" - }, - { - "block_id": "p879-b4", - "global_id": 25652, - "bbox": [ - 127.59, - 200.25, - 516.11, - 222.17 - ], - "text": "A detailed explanation of the nature of such sums over a continuum of frequencies is provided\nin Sec. 7.1-1.", - "type": "text" - }, - { - "block_id": "p879-b5", - "global_id": 25653, - "bbox": [ - 127.89, - 236.71, - 264.17, - 248.83 - ], - "text": "EXISTENCE OF THE DTFT", - "type": "text" - }, - { - "block_id": "p879-b6", - "global_id": 25654, - "bbox": [ - 127.59, - 251.22, - 516.14, - 274.78 - ], - "text": "Because |e−jn| = 1, from Eq. (9.19), it follows that the existence of X() is guaranteed if x[n] is\nabsolutely summable; that is,", - "type": "text" - }, - { - "block_id": "p879-b7", - "global_id": 25655, - "bbox": [ - 292.36, - 274.31, - 306.45, - 284.99 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p879-b8", - "global_id": 25656, - "bbox": [ - 288.66, - 298.33, - 310.13, - 305.52 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p879-b9", - "global_id": 25657, - "bbox": [ - 311.26, - 284.48, - 516.13, - 294.86 - ], - "text": "|x[n]| < ∞\n(9.21)", - "type": "text" - }, - { - "block_id": "p879-b10", - "global_id": 25658, - "bbox": [ - 127.59, - 312.41, - 516.11, - 334.32 - ], - "text": "This shows that the condition of absolute summability is a sufficient condition for the existence of\nthe DTFT representation. This condition also guarantees its uniform convergence. The inequality", - "type": "text" - }, - { - "block_id": "p879-b11", - "global_id": 25659, - "bbox": [ - 261.52, - 338.8, - 285.5, - 356.27 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p879-b12", - "global_id": 25660, - "bbox": [ - 267.71, - 369.62, - 289.19, - 376.81 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p879-b13", - "global_id": 25661, - "bbox": [ - 290.3, - 355.77, - 312.07, - 366.05 - ], - "text": "|x[n]|", - "type": "text" - }, - { - "block_id": "p879-b14", - "global_id": 25662, - "bbox": [ - 312.08, - 338.8, - 321.74, - 350.68 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p879-b15", - "global_id": 25663, - "bbox": [ - 324.29, - 355.77, - 331.8, - 365.73 - ], - "text": "≥", - "type": "text" - }, - { - "block_id": "p879-b16", - "global_id": 25664, - "bbox": [ - 337.55, - 345.6, - 351.64, - 356.27 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p879-b17", - "global_id": 25665, - "bbox": [ - 333.86, - 369.62, - 355.33, - 376.81 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p879-b18", - "global_id": 25666, - "bbox": [ - 356.45, - 354.34, - 381.7, - 366.05 - ], - "text": "|x[n]|2", - "type": "text" - }, - { - "block_id": "p879-b19", - "global_id": 25667, - "bbox": [ - 127.59, - 386.29, - 516.14, - 420.17 - ], - "text": "shows that the energy of an absolutely summable sequence is finite. However, not all finite-energy\nsignals are absolutely summable. Signal x[n] = sinc(n) is such an example. For such signals, the\nDTFT converges, not uniformly, but in the mean.†", - "type": "text" - }, - { - "block_id": "p879-b20", - "global_id": 25668, - "bbox": [ - 145.52, - 421.75, - 357.09, - 432.13 - ], - "text": "To summarize, X() exists under a weaker condition", - "type": "text" - }, - { - "block_id": "p879-b21", - "global_id": 25669, - "bbox": [ - 290.37, - 440.41, - 304.46, - 451.08 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p879-b22", - "global_id": 25670, - "bbox": [ - 286.67, - 464.43, - 308.14, - 471.63 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p879-b23", - "global_id": 25671, - "bbox": [ - 309.26, - 446.47, - 516.13, - 460.96 - ], - "text": "|x[n]|2 < ∞\n(9.22)", - "type": "text" - }, - { - "block_id": "p879-b24", - "global_id": 25672, - "bbox": [ - 127.59, - 481.1, - 516.15, - 574.76 - ], - "text": "The DTFT under this condition is guaranteed to converge in the mean. Thus, the DTFT of the\nexponentially growing signal γ nu[n] does not exist when |γ | > 1 because the signal violates\nEqs. (9.21) and (9.22). But the DTFT exists for the signal sinc(n), which violates Eq. (9.21) but\ndoes satisfy Eq. (9.22) (see later, Ex. 9.6). In addition, if the use of δ(), the continuous-time\nimpulse function, is permitted, we can even find the DTFT of some signals that violate both\nEq. (9.21) and Eq. (9.22). Such signals are not absolutely summable, nor do they have finite\nenergy. For example, as seen from pairs 11 and 12 of Table 9.1, the DTFT of x[n] = 1 for all\nn and x[n] = ej0n exist, although they violate Eqs. (9.21) and (9.22).", - "type": "text" - }, - { - "block_id": "p879-b25", - "global_id": 25673, - "bbox": [ - 127.59, - 592.84, - 174.16, - 605.06 - ], - "text": "† This means", - "type": "text" - }, - { - "block_id": "p879-b26", - "global_id": 25674, - "bbox": [ - 242.86, - 615.72, - 262.25, - 629.92 - ], - "text": "lim\nM→∞", - "type": "text" - }, - { - "block_id": "p879-b27", - "global_id": 25675, - "bbox": [ - 263.24, - 603.14, - 275.97, - 614.11 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p879-b28", - "global_id": 25676, - "bbox": [ - 267.98, - 625.37, - 276.91, - 631.85 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p879-b29", - "global_id": 25677, - "bbox": [ - 279.05, - 599.65, - 309.91, - 630.13 - ], - "text": "X() −", - "type": "text" - }, - { - "block_id": "p879-b30", - "global_id": 25678, - "bbox": [ - 314.44, - 606.23, - 327.13, - 615.79 - ], - "text": "M\n\"", - "type": "text" - }, - { - "block_id": "p879-b31", - "global_id": 25679, - "bbox": [ - 311.3, - 628.17, - 330.03, - 634.85 - ], - "text": "n=−M", - "type": "text" - }, - { - "block_id": "p879-b32", - "global_id": 25680, - "bbox": [ - 331.28, - 613.63, - 364.85, - 624.59 - ], - "text": "x[n]e−jn", - "type": "text" - }, - { - "block_id": "p879-b34", - "global_id": 25681, - "bbox": [ - 368.26, - 603.49, - 371.5, - 609.96 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p879-b35", - "global_id": 25682, - "bbox": [ - 373.99, - 615.34, - 400.85, - 624.68 - ], - "text": "d = 0", - "type": "text" - } - ] - }, - { - "page_num": 880, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p880-b0", - "global_id": 25683, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "860\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p880-b1", - "global_id": 25684, - "bbox": [ - 101.84, - 86.07, - 319.03, - 95.3 - ], - "text": "TABLE 9.1\nSelect Discrete-Time Fourier Transform Pairs", - "type": "text" - }, - { - "block_id": "p880-b2", - "global_id": 25685, - "bbox": [ - 101.84, - 106.02, - 247.72, - 115.28 - ], - "text": "No.\nx[n]\nX()", - "type": "text" - }, - { - "block_id": "p880-b3", - "global_id": 25686, - "bbox": [ - 106.33, - 123.08, - 490.23, - 133.69 - ], - "text": "1\nδ[n −k]\ne−jk\nIntegerk", - "type": "text" - }, - { - "block_id": "p880-b4", - "global_id": 25687, - "bbox": [ - 106.33, - 145.9, - 247.64, - 162.79 - ], - "text": "2\nγ nu[n]\nej", - "type": "text" - }, - { - "block_id": "p880-b5", - "global_id": 25688, - "bbox": [ - 228.97, - 153.45, - 490.39, - 169.06 - ], - "text": "ej −γ\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p880-b6", - "global_id": 25689, - "bbox": [ - 106.33, - 175.0, - 247.64, - 191.88 - ], - "text": "3\n−γ nu[−(n + 1)]\nej", - "type": "text" - }, - { - "block_id": "p880-b7", - "global_id": 25690, - "bbox": [ - 228.97, - 182.54, - 490.39, - 198.16 - ], - "text": "ej −γ\n|γ | > 1", - "type": "text" - }, - { - "block_id": "p880-b8", - "global_id": 25691, - "bbox": [ - 106.33, - 204.46, - 272.96, - 221.08 - ], - "text": "4\nγ |n|\n1 −γ 2", - "type": "text" - }, - { - "block_id": "p880-b9", - "global_id": 25692, - "bbox": [ - 228.97, - 211.74, - 490.38, - 227.44 - ], - "text": "1 −2γ cos + γ 2\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p880-b10", - "global_id": 25693, - "bbox": [ - 106.33, - 233.29, - 255.82, - 250.18 - ], - "text": "5\nnγ nu[n]\nγ ej", - "type": "text" - }, - { - "block_id": "p880-b11", - "global_id": 25694, - "bbox": [ - 228.97, - 240.84, - 490.38, - 256.46 - ], - "text": "(ej −γ )2\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p880-b12", - "global_id": 25695, - "bbox": [ - 106.33, - 263.76, - 339.05, - 283.01 - ], - "text": "6\nγ n cos(0n + θ)u[n]\nej[ej cos θ −γ cos(0 −θ)]", - "type": "text" - }, - { - "block_id": "p880-b13", - "global_id": 25696, - "bbox": [ - 235.57, - 272.93, - 490.38, - 289.37 - ], - "text": "ej2 −(2γ cos0)ej + γ 2\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p880-b14", - "global_id": 25697, - "bbox": [ - 106.33, - 297.32, - 269.87, - 312.94 - ], - "text": "7\nu[n] −u[n −M]\nsin(M/2)", - "type": "text" - }, - { - "block_id": "p880-b15", - "global_id": 25698, - "bbox": [ - 232.88, - 301.9, - 313.41, - 319.3 - ], - "text": "sin(/2) e−j(M−1)/2", - "type": "text" - }, - { - "block_id": "p880-b16", - "global_id": 25699, - "bbox": [ - 106.33, - 330.88, - 144.92, - 346.5 - ], - "text": "8\nc", - "type": "text" - }, - { - "block_id": "p880-b17", - "global_id": 25700, - "bbox": [ - 137.11, - 337.16, - 184.58, - 352.87 - ], - "text": "π sinc(cn)", - "type": "text" - }, - { - "block_id": "p880-b18", - "global_id": 25701, - "bbox": [ - 231.28, - 327.85, - 243.97, - 337.62 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b19", - "global_id": 25702, - "bbox": [ - 227.77, - 350.0, - 247.46, - 356.68 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b20", - "global_id": 25703, - "bbox": [ - 248.47, - 337.54, - 261.9, - 346.5 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p880-b21", - "global_id": 25704, - "bbox": [ - 262.91, - 324.57, - 301.66, - 340.22 - ], - "text": "−2πk", - "type": "text" - }, - { - "block_id": "p880-b22", - "global_id": 25705, - "bbox": [ - 278.56, - 343.53, - 292.89, - 353.54 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p880-b24", - "global_id": 25706, - "bbox": [ - 463.32, - 337.16, - 489.49, - 347.47 - ], - "text": "c ≤π", - "type": "text" - }, - { - "block_id": "p880-b25", - "global_id": 25707, - "bbox": [ - 106.33, - 358.13, - 189.28, - 386.43 - ], - "text": "9\nc\n2π sinc2\ncn", - "type": "text" - }, - { - "block_id": "p880-b26", - "global_id": 25708, - "bbox": [ - 179.62, - 377.46, - 184.11, - 386.43 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p880-b27", - "global_id": 25709, - "bbox": [ - 190.48, - 358.13, - 243.97, - 371.17 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b28", - "global_id": 25710, - "bbox": [ - 227.77, - 383.55, - 247.46, - 390.23 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b30", - "global_id": 25711, - "bbox": [ - 256.68, - 358.13, - 295.43, - 373.77 - ], - "text": "−2πk", - "type": "text" - }, - { - "block_id": "p880-b31", - "global_id": 25712, - "bbox": [ - 272.33, - 377.09, - 286.66, - 387.1 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p880-b33", - "global_id": 25713, - "bbox": [ - 463.32, - 370.72, - 489.49, - 381.03 - ], - "text": "c ≤π", - "type": "text" - }, - { - "block_id": "p880-b34", - "global_id": 25714, - "bbox": [ - 101.84, - 396.74, - 246.91, - 413.62 - ], - "text": "10\nu[n]\nej", - "type": "text" - }, - { - "block_id": "p880-b35", - "global_id": 25715, - "bbox": [ - 228.97, - 404.29, - 270.91, - 419.99 - ], - "text": "ej −1 + π", - "type": "text" - }, - { - "block_id": "p880-b36", - "global_id": 25716, - "bbox": [ - 276.31, - 394.97, - 289.0, - 404.73 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b37", - "global_id": 25717, - "bbox": [ - 272.81, - 417.11, - 292.5, - 423.79 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b38", - "global_id": 25718, - "bbox": [ - 293.5, - 404.29, - 336.18, - 413.62 - ], - "text": "δ( −2πk)", - "type": "text" - }, - { - "block_id": "p880-b39", - "global_id": 25719, - "bbox": [ - 101.84, - 434.85, - 237.63, - 444.19 - ], - "text": "11\n1\nfor all n\n2π", - "type": "text" - }, - { - "block_id": "p880-b40", - "global_id": 25720, - "bbox": [ - 243.04, - 425.54, - 255.72, - 435.3 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b41", - "global_id": 25721, - "bbox": [ - 239.53, - 447.69, - 259.21, - 454.37 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b42", - "global_id": 25722, - "bbox": [ - 260.23, - 434.85, - 302.91, - 444.19 - ], - "text": "δ( −2πk)", - "type": "text" - }, - { - "block_id": "p880-b43", - "global_id": 25723, - "bbox": [ - 101.84, - 464.17, - 237.63, - 474.76 - ], - "text": "12\nej0n\n2π", - "type": "text" - }, - { - "block_id": "p880-b44", - "global_id": 25724, - "bbox": [ - 243.04, - 456.11, - 255.72, - 465.88 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b45", - "global_id": 25725, - "bbox": [ - 239.53, - 478.26, - 259.21, - 484.94 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b46", - "global_id": 25726, - "bbox": [ - 260.23, - 465.42, - 323.4, - 475.73 - ], - "text": "δ( −0 −2πk)", - "type": "text" - }, - { - "block_id": "p880-b47", - "global_id": 25727, - "bbox": [ - 101.84, - 496.0, - 233.15, - 506.07 - ], - "text": "13\ncos0n\nπ", - "type": "text" - }, - { - "block_id": "p880-b48", - "global_id": 25728, - "bbox": [ - 238.55, - 486.68, - 251.24, - 496.45 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b49", - "global_id": 25729, - "bbox": [ - 235.04, - 508.83, - 254.73, - 515.51 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b50", - "global_id": 25730, - "bbox": [ - 255.74, - 496.0, - 391.87, - 506.3 - ], - "text": "δ( −0 −2πk) + δ( + 0 −2πk)", - "type": "text" - }, - { - "block_id": "p880-b51", - "global_id": 25731, - "bbox": [ - 101.84, - 526.57, - 235.64, - 536.65 - ], - "text": "14\nsin0n\njπ", - "type": "text" - }, - { - "block_id": "p880-b52", - "global_id": 25732, - "bbox": [ - 241.05, - 517.25, - 253.73, - 527.01 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b53", - "global_id": 25733, - "bbox": [ - 237.54, - 539.4, - 257.23, - 546.08 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b54", - "global_id": 25734, - "bbox": [ - 258.24, - 526.57, - 394.37, - 536.88 - ], - "text": "δ( + 0 −2πk) −δ( −0 −2πk)", - "type": "text" - }, - { - "block_id": "p880-b55", - "global_id": 25735, - "bbox": [ - 101.84, - 547.59, - 329.02, - 573.81 - ], - "text": "15\n(cos0n)u[n]\nej2 −ej cos0\nej2 −2ej cos0 + 1 + π", - "type": "text" - }, - { - "block_id": "p880-b56", - "global_id": 25736, - "bbox": [ - 324.54, - 563.88, - 329.02, - 572.84 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p880-b57", - "global_id": 25737, - "bbox": [ - 335.62, - 547.82, - 348.31, - 557.59 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b58", - "global_id": 25738, - "bbox": [ - 332.11, - 569.97, - 351.8, - 576.65 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b59", - "global_id": 25739, - "bbox": [ - 352.81, - 557.13, - 480.97, - 567.21 - ], - "text": "δ(−2πk−0) + δ(−2πk+0)", - "type": "text" - }, - { - "block_id": "p880-b60", - "global_id": 25740, - "bbox": [ - 101.84, - 578.54, - 329.52, - 604.39 - ], - "text": "16\n(sin0n)u[n]\nej sin0\nej2 −2ej cos0 + 1 + π", - "type": "text" - }, - { - "block_id": "p880-b61", - "global_id": 25741, - "bbox": [ - 323.79, - 594.36, - 330.77, - 603.42 - ], - "text": "2j", - "type": "text" - }, - { - "block_id": "p880-b62", - "global_id": 25742, - "bbox": [ - 335.47, - 578.39, - 348.15, - 588.16 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p880-b63", - "global_id": 25743, - "bbox": [ - 331.96, - 600.54, - 351.65, - 607.22 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p880-b64", - "global_id": 25744, - "bbox": [ - 352.66, - 587.71, - 480.82, - 597.78 - ], - "text": "δ(−2πk−0) −δ(−2πk+0)", - "type": "text" - } - ] - }, - { - "page_num": 881, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p881-b0", - "global_id": 25745, - "bbox": [ - 247.51, - 62.89, - 516.12, - 71.98 - ], - "text": "9.2\nAperiodic Signal Representation by Fourier Integral\n861", - "type": "text" - }, - { - "block_id": "p881-b1", - "global_id": 25746, - "bbox": [ - 102.51, - 93.91, - 365.42, - 105.87 - ], - "text": "EXAMPLE 9.3\nDTFT of a Causal Exponential", - "type": "text" - }, - { - "block_id": "p881-b2", - "global_id": 25747, - "bbox": [ - 128.9, - 121.11, - 260.12, - 132.5 - ], - "text": "Find the DTFT of x[n] = γ nu[n].", - "type": "text" - }, - { - "block_id": "p881-b3", - "global_id": 25748, - "bbox": [ - 128.9, - 155.41, - 262.09, - 165.38 - ], - "text": "Using the definition, the DTFT is", - "type": "text" - }, - { - "block_id": "p881-b4", - "global_id": 25749, - "bbox": [ - 244.7, - 190.02, - 276.22, - 200.29 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p881-b5", - "global_id": 25750, - "bbox": [ - 278.27, - 179.85, - 292.37, - 190.52 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p881-b6", - "global_id": 25751, - "bbox": [ - 279.11, - 204.42, - 291.52, - 211.68 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p881-b7", - "global_id": 25752, - "bbox": [ - 293.48, - 185.91, - 335.09, - 200.29 - ], - "text": "γ ne−jn =", - "type": "text" - }, - { - "block_id": "p881-b8", - "global_id": 25753, - "bbox": [ - 337.14, - 179.85, - 351.23, - 190.52 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p881-b9", - "global_id": 25754, - "bbox": [ - 337.97, - 204.42, - 350.38, - 211.68 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p881-b10", - "global_id": 25755, - "bbox": [ - 351.23, - 188.3, - 386.47, - 200.29 - ], - "text": "(γ e−j)n", - "type": "text" - }, - { - "block_id": "p881-b11", - "global_id": 25756, - "bbox": [ - 128.9, - 226.2, - 491.14, - 237.79 - ], - "text": "This is an infinite geometric series with a common ratio γ e−j. Therefore (see Sec. B.8-3),", - "type": "text" - }, - { - "block_id": "p881-b12", - "global_id": 25757, - "bbox": [ - 277.76, - 252.71, - 352.18, - 276.73 - ], - "text": "X() =\n1\n1 −γ e−j", - "type": "text" - }, - { - "block_id": "p881-b13", - "global_id": 25758, - "bbox": [ - 128.9, - 291.39, - 479.4, - 302.99 - ], - "text": "provided |γ e−j| < 1. But because |e−j| = 1, this condition implies |γ | < 1. Therefore,", - "type": "text" - }, - { - "block_id": "p881-b14", - "global_id": 25759, - "bbox": [ - 253.23, - 317.91, - 378.43, - 341.93 - ], - "text": "X() =\n1\n1 −γ e−j\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p881-b15", - "global_id": 25760, - "bbox": [ - 128.91, - 356.39, - 502.78, - 378.72 - ], - "text": "If |γ | > 1, X() does not converge. This result is in conformity with Eqs. (9.21) and (9.22).\nTo determine magnitude and phase responses, we note that", - "type": "text" - }, - { - "block_id": "p881-b16", - "global_id": 25761, - "bbox": [ - 254.1, - 393.64, - 502.75, - 417.66 - ], - "text": "X() =\n1\n1 −γ cos + jγ sin\n(9.23)", - "type": "text" - }, - { - "block_id": "p881-b17", - "global_id": 25762, - "bbox": [ - 128.9, - 430.25, - 137.77, - 440.22 - ], - "text": "so", - "type": "text" - }, - { - "block_id": "p881-b18", - "global_id": 25763, - "bbox": [ - 189.76, - 438.14, - 290.64, - 454.99 - ], - "text": "|X()| =\n1", - "type": "text" - }, - { - "block_id": "p881-b19", - "global_id": 25764, - "bbox": [ - 238.49, - 438.14, - 403.01, - 463.82 - ], - "text": "(1 −γ cos)2 + (γ sin)2 =\n1", - "type": "text" - }, - { - "block_id": "p881-b20", - "global_id": 25765, - "bbox": [ - 368.59, - 450.56, - 440.71, - 463.82 - ], - "text": "1 + γ 2 −2γ cos", - "type": "text" - }, - { - "block_id": "p881-b21", - "global_id": 25766, - "bbox": [ - 128.91, - 471.29, - 143.28, - 481.25 - ], - "text": "and̸", - "type": "text" - }, - { - "block_id": "p881-b22", - "global_id": 25767, - "bbox": [ - 256.6, - 484.68, - 320.15, - 496.79 - ], - "text": "X() = −tan−1", - "type": "text" - }, - { - "block_id": "p881-b23", - "global_id": 25768, - "bbox": [ - 321.76, - 472.42, - 374.08, - 503.87 - ], - "text": "γ sin\n1 −γ cos", - "type": "text" - }, - { - "block_id": "p881-b24", - "global_id": 25769, - "bbox": [ - 375.27, - 472.42, - 380.7, - 482.38 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p881-b25", - "global_id": 25770, - "bbox": [ - 128.91, - 509.63, - 502.76, - 556.88 - ], - "text": "Figure 9.5 shows x[n] = γ nu[n] and its spectra for γ = 0.8. Observe that the frequency\nspectra are continuous and periodic functions of with the period 2π. As explained earlier, we\nneed to use the spectrum only over the frequency interval of 2π. We often select this interval\nto be the fundamental frequency range (−π,π).", - "type": "text" - }, - { - "block_id": "p881-b26", - "global_id": 25771, - "bbox": [ - 128.91, - 558.45, - 502.76, - 580.78 - ], - "text": "The amplitude spectrum |X()| is an even function and the phase spectrum̸\nX() is an\nodd function of .", - "type": "text" - } - ] - }, - { - "page_num": 882, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p882-b0", - "global_id": 25772, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "862\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p882-b1", - "global_id": 25773, - "bbox": [ - 165.86, - 97.39, - 178.74, - 105.47 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p882-b2", - "global_id": 25774, - "bbox": [ - 346.46, - 171.83, - 350.46, - 179.83 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p882-b4", - "global_id": 25775, - "bbox": [ - 254.75, - 171.91, - 332.21, - 179.91 - ], - "text": "5\n10", - "type": "text" - }, - { - "block_id": "p882-b5", - "global_id": 25776, - "bbox": [ - 262.3, - 276.47, - 266.3, - 284.47 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p882-b6", - "global_id": 25777, - "bbox": [ - 255.38, - 292.9, - 265.02, - 300.9 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p882-b7", - "global_id": 25778, - "bbox": [ - 255.76, - 187.42, - 264.64, - 195.42 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p882-b8", - "global_id": 25779, - "bbox": [ - 255.76, - 393.14, - 264.64, - 401.14 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p882-b9", - "global_id": 25780, - "bbox": [ - 262.3, - 356.94, - 266.3, - 364.94 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p882-b10", - "global_id": 25781, - "bbox": [ - 251.7, - 215.88, - 255.7, - 223.88 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p882-b11", - "global_id": 25782, - "bbox": [ - 265.05, - 323.85, - 283.05, - 331.85 - ], - "text": "0.932", - "type": "text" - }, - { - "block_id": "p882-b12", - "global_id": 25783, - "bbox": [ - 267.65, - 209.39, - 288.12, - 217.68 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p882-b13", - "global_id": 25784, - "bbox": [ - 296.78, - 314.0, - 319.88, - 322.3 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p882-b14", - "global_id": 25785, - "bbox": [ - 211.77, - 93.69, - 243.79, - 103.22 - ], - "text": "(0.8)nu[n]", - "type": "text" - }, - { - "block_id": "p882-b15", - "global_id": 25786, - "bbox": [ - 111.4, - 279.16, - 405.97, - 287.45 - ], - "text": "2p\n3p\np\n2p\np\n3p", - "type": "text" - }, - { - "block_id": "p882-b16", - "global_id": 25787, - "bbox": [ - 111.4, - 359.58, - 421.52, - 368.21 - ], - "text": "2p\n3p\np\n2p\np\n3p", - "type": "text" - }, - { - "block_id": "p882-b17", - "global_id": 25788, - "bbox": [ - 94.2, - 406.41, - 303.7, - 417.08 - ], - "text": "Figure 9.5 Exponential γ nu[n] and its frequency spectra.", - "type": "text" - }, - { - "block_id": "p882-b18", - "global_id": 25789, - "bbox": [ - 76.77, - 462.05, - 368.21, - 474.0 - ], - "text": "EXAMPLE 9.4\nDTFT of an Anticausal Exponential", - "type": "text" - }, - { - "block_id": "p882-b19", - "global_id": 25790, - "bbox": [ - 103.16, - 488.26, - 317.2, - 499.65 - ], - "text": "Find the DTFT of γ nu[−(n + 1)] depicted in Fig. 9.6.", - "type": "text" - }, - { - "block_id": "p882-b20", - "global_id": 25791, - "bbox": [ - 196.6, - 534.01, - 300.69, - 543.54 - ], - "text": "x[n]\ng n u[(n 1)]", - "type": "text" - }, - { - "block_id": "p882-b21", - "global_id": 25792, - "bbox": [ - 127.13, - 597.24, - 364.57, - 605.53 - ], - "text": "0\n5\n10\nn", - "type": "text" - }, - { - "block_id": "p882-b22", - "global_id": 25793, - "bbox": [ - 94.2, - 614.55, - 240.16, - 625.22 - ], - "text": "Figure 9.6 Exponential γ nu[−(n + 1)].", - "type": "text" - } - ] - }, - { - "page_num": 883, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p883-b0", - "global_id": 25794, - "bbox": [ - 247.51, - 62.89, - 516.12, - 71.98 - ], - "text": "9.2\nAperiodic Signal Representation by Fourier Integral\n863", - "type": "text" - }, - { - "block_id": "p883-b1", - "global_id": 25795, - "bbox": [ - 146.84, - 86.24, - 280.02, - 96.2 - ], - "text": "Using the definition, the DTFT is", - "type": "text" - }, - { - "block_id": "p883-b2", - "global_id": 25796, - "bbox": [ - 177.81, - 116.67, - 209.34, - 126.95 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p883-b3", - "global_id": 25797, - "bbox": [ - 215.08, - 106.5, - 229.18, - 117.17 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p883-b4", - "global_id": 25798, - "bbox": [ - 211.39, - 130.52, - 232.87, - 137.71 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p883-b5", - "global_id": 25799, - "bbox": [ - 233.98, - 112.56, - 323.25, - 127.05 - ], - "text": "γ nu[−(n + 1)]e−jn =", - "type": "text" - }, - { - "block_id": "p883-b6", - "global_id": 25800, - "bbox": [ - 327.17, - 106.5, - 341.26, - 117.17 - ], - "text": "−∞\n\"", - "type": "text" - }, - { - "block_id": "p883-b7", - "global_id": 25801, - "bbox": [ - 325.29, - 131.07, - 343.14, - 138.33 - ], - "text": "n=−1", - "type": "text" - }, - { - "block_id": "p883-b8", - "global_id": 25802, - "bbox": [ - 343.14, - 112.56, - 388.7, - 126.94 - ], - "text": "(γ e−j)n =", - "type": "text" - }, - { - "block_id": "p883-b9", - "global_id": 25803, - "bbox": [ - 392.63, - 106.5, - 406.73, - 117.17 - ], - "text": "−∞\n\"", - "type": "text" - }, - { - "block_id": "p883-b10", - "global_id": 25804, - "bbox": [ - 390.75, - 131.07, - 408.59, - 138.33 - ], - "text": "n=−1", - "type": "text" - }, - { - "block_id": "p883-b11", - "global_id": 25805, - "bbox": [ - 409.71, - 102.68, - 423.42, - 120.06 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p883-b12", - "global_id": 25806, - "bbox": [ - 417.63, - 114.95, - 437.2, - 134.02 - ], - "text": "γ ej", - "type": "text" - }, - { - "block_id": "p883-b13", - "global_id": 25807, - "bbox": [ - 437.7, - 102.68, - 453.36, - 114.5 - ], - "text": "−n", - "type": "text" - }, - { - "block_id": "p883-b14", - "global_id": 25808, - "bbox": [ - 128.9, - 148.1, - 217.73, - 158.47 - ], - "text": "Setting n = −m yields", - "type": "text" - }, - { - "block_id": "p883-b15", - "global_id": 25809, - "bbox": [ - 196.02, - 178.15, - 221.9, - 188.42 - ], - "text": "x[n] =", - "type": "text" - }, - { - "block_id": "p883-b16", - "global_id": 25810, - "bbox": [ - 223.95, - 167.97, - 238.05, - 178.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p883-b17", - "global_id": 25811, - "bbox": [ - 224.03, - 192.54, - 237.98, - 199.8 - ], - "text": "m=1", - "type": "text" - }, - { - "block_id": "p883-b18", - "global_id": 25812, - "bbox": [ - 239.16, - 164.15, - 252.87, - 181.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p883-b19", - "global_id": 25813, - "bbox": [ - 247.08, - 176.42, - 266.65, - 195.49 - ], - "text": "γ ej", - "type": "text" - }, - { - "block_id": "p883-b20", - "global_id": 25814, - "bbox": [ - 267.15, - 164.15, - 278.92, - 175.98 - ], - "text": "m", - "type": "text" - }, - { - "block_id": "p883-b21", - "global_id": 25815, - "bbox": [ - 281.46, - 171.58, - 298.26, - 188.52 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p883-b22", - "global_id": 25816, - "bbox": [ - 292.48, - 174.03, - 321.88, - 195.49 - ], - "text": "γ ej +", - "type": "text" - }, - { - "block_id": "p883-b23", - "global_id": 25817, - "bbox": [ - 323.42, - 164.15, - 337.13, - 181.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p883-b24", - "global_id": 25818, - "bbox": [ - 331.34, - 176.42, - 350.92, - 195.49 - ], - "text": "γ ej", - "type": "text" - }, - { - "block_id": "p883-b25", - "global_id": 25819, - "bbox": [ - 351.42, - 164.15, - 361.63, - 176.05 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p883-b26", - "global_id": 25820, - "bbox": [ - 363.68, - 178.15, - 371.45, - 188.11 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p883-b27", - "global_id": 25821, - "bbox": [ - 372.99, - 164.15, - 386.71, - 181.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p883-b28", - "global_id": 25822, - "bbox": [ - 380.92, - 176.42, - 400.5, - 195.49 - ], - "text": "γ ej", - "type": "text" - }, - { - "block_id": "p883-b29", - "global_id": 25823, - "bbox": [ - 401.0, - 164.15, - 411.21, - 176.05 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p883-b30", - "global_id": 25824, - "bbox": [ - 413.26, - 178.15, - 435.65, - 188.11 - ], - "text": "+ · · ·", - "type": "text" - }, - { - "block_id": "p883-b31", - "global_id": 25825, - "bbox": [ - 128.91, - 209.33, - 456.82, - 220.93 - ], - "text": "This is a geometric series with a common ratio ej/γ . Therefore, from Sec. B.8-3,", - "type": "text" - }, - { - "block_id": "p883-b32", - "global_id": 25826, - "bbox": [ - 196.83, - 230.87, - 434.84, - 254.89 - ], - "text": "X() =\n1\nγ e−j −1 =\n1\n(γ cos −1) −jγ sin,\n|γ | > 1", - "type": "text" - }, - { - "block_id": "p883-b33", - "global_id": 25827, - "bbox": [ - 128.91, - 264.73, - 170.65, - 274.69 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p883-b34", - "global_id": 25828, - "bbox": [ - 167.26, - 284.58, - 250.42, - 301.43 - ], - "text": "|X()| =\n1", - "type": "text" - }, - { - "block_id": "p883-b35", - "global_id": 25829, - "bbox": [ - 215.99, - 289.43, - 403.85, - 310.27 - ], - "text": "1 + γ 2 −2γ cos\nand̸\nX() = tan−1", - "type": "text" - }, - { - "block_id": "p883-b36", - "global_id": 25830, - "bbox": [ - 405.47, - 277.17, - 457.77, - 308.6 - ], - "text": "γ sin\nγ cos −1", - "type": "text" - }, - { - "block_id": "p883-b37", - "global_id": 25831, - "bbox": [ - 458.97, - 277.17, - 464.4, - 287.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p883-b38", - "global_id": 25832, - "bbox": [ - 128.9, - 320.72, - 502.77, - 366.54 - ], - "text": "Except for the change of sign, this Fourier transform (and the corresponding frequency spectra)\nis identical to that of x[n] = γ nu[n]. Yet there is no ambiguity in determining the IDTFT of\nX() = 1/(γ e−j −1) because of the restrictions on the value of γ in each case. If |γ | < 1,\nthen the inverse transform is x[n] = −γ nu[n]. If |γ | > 1, it is x[n] = γ n[−(n + 1)].", - "type": "text" - }, - { - "block_id": "p883-b39", - "global_id": 25833, - "bbox": [ - 102.51, - 434.55, - 358.12, - 446.5 - ], - "text": "EXAMPLE 9.5\nDTFT of a Rectangular Pulse", - "type": "text" - }, - { - "block_id": "p883-b40", - "global_id": 25834, - "bbox": [ - 128.9, - 463.17, - 502.76, - 485.09 - ], - "text": "Find the DTFT of the discrete-time rectangular pulse illustrated in Fig. 9.7a. This pulse is also\nknown as the 9-point rectangular window function.", - "type": "text" - }, - { - "block_id": "p883-b41", - "global_id": 25835, - "bbox": [ - 205.2, - 514.17, - 236.72, - 524.44 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p883-b42", - "global_id": 25836, - "bbox": [ - 242.47, - 503.99, - 256.56, - 514.66 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p883-b43", - "global_id": 25837, - "bbox": [ - 238.77, - 528.02, - 260.24, - 535.21 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p883-b44", - "global_id": 25838, - "bbox": [ - 261.37, - 510.05, - 308.47, - 524.44 - ], - "text": "x[n]e−jn =", - "type": "text" - }, - { - "block_id": "p883-b45", - "global_id": 25839, - "bbox": [ - 317.7, - 503.53, - 344.67, - 514.66 - ], - "text": "(M−1)/2\n\"", - "type": "text" - }, - { - "block_id": "p883-b46", - "global_id": 25840, - "bbox": [ - 310.52, - 528.59, - 351.86, - 535.85 - ], - "text": "n=−(M−1)/2", - "type": "text" - }, - { - "block_id": "p883-b47", - "global_id": 25841, - "bbox": [ - 351.86, - 512.44, - 426.48, - 524.54 - ], - "text": "(e−j)n\nM = 9", - "type": "text" - }, - { - "block_id": "p883-b48", - "global_id": 25842, - "bbox": [ - 146.84, - 544.74, - 460.35, - 558.32 - ], - "text": "This is a geometric progression with a common ratio e−j and (see Sec. B.8-3)", - "type": "text" - }, - { - "block_id": "p883-b49", - "global_id": 25843, - "bbox": [ - 243.0, - 565.85, - 381.4, - 586.72 - ], - "text": "X() = e−j[(M+1)/2] −ej[(M−1)/2]", - "type": "text" - }, - { - "block_id": "p883-b50", - "global_id": 25844, - "bbox": [ - 313.05, - 580.64, - 346.62, - 593.9 - ], - "text": "e−j −1", - "type": "text" - }, - { - "block_id": "p883-b51", - "global_id": 25845, - "bbox": [ - 266.76, - 588.8, - 306.47, - 614.66 - ], - "text": "= e−j/2", - "type": "text" - }, - { - "block_id": "p883-b52", - "global_id": 25846, - "bbox": [ - 306.47, - 588.8, - 387.48, - 607.08 - ], - "text": "e−j(M/2) −ej(M/2)", - "type": "text" - }, - { - "block_id": "p883-b53", - "global_id": 25847, - "bbox": [ - 289.18, - 608.58, - 376.07, - 621.73 - ], - "text": "e−j/2(e−j/2 −ej/2)", - "type": "text" - } - ] - }, - { - "page_num": 884, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p884-b0", - "global_id": 25848, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "864\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p884-b1", - "global_id": 25849, - "bbox": [ - 241.02, - 107.05, - 248.79, - 117.01 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p884-b2", - "global_id": 25850, - "bbox": [ - 252.02, - 93.9, - 263.66, - 103.86 - ], - "text": "sin", - "type": "text" - }, - { - "block_id": "p884-b3", - "global_id": 25851, - "bbox": [ - 265.87, - 79.5, - 282.09, - 96.78 - ], - "text": "M", - "type": "text" - }, - { - "block_id": "p884-b4", - "global_id": 25852, - "bbox": [ - 275.65, - 79.5, - 298.15, - 110.94 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p884-b5", - "global_id": 25853, - "bbox": [ - 254.9, - 107.46, - 477.01, - 124.49 - ], - "text": "sin(0.5)\n(9.24)", - "type": "text" - }, - { - "block_id": "p884-b6", - "global_id": 25854, - "bbox": [ - 241.02, - 126.68, - 292.39, - 144.04 - ], - "text": "= sin(4.5)", - "type": "text" - }, - { - "block_id": "p884-b7", - "global_id": 25855, - "bbox": [ - 252.02, - 133.66, - 477.01, - 151.11 - ], - "text": "sin(0.5)\nfor M = 9\n(9.25)", - "type": "text" - }, - { - "block_id": "p884-b8", - "global_id": 25856, - "bbox": [ - 103.17, - 160.28, - 298.76, - 170.65 - ], - "text": "Figure 9.7b shows the spectrum X() for M = 9.", - "type": "text" - }, - { - "block_id": "p884-b9", - "global_id": 25857, - "bbox": [ - 267.33, - 204.23, - 280.2, - 212.31 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p884-b10", - "global_id": 25858, - "bbox": [ - 249.49, - 260.45, - 313.82, - 268.75 - ], - "text": "0\n4\n4", - "type": "text" - }, - { - "block_id": "p884-b11", - "global_id": 25859, - "bbox": [ - 278.62, - 278.69, - 287.5, - 286.69 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p884-b12", - "global_id": 25860, - "bbox": [ - 356.73, - 261.34, - 360.73, - 269.34 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p884-b14", - "global_id": 25861, - "bbox": [ - 285.32, - 213.33, - 289.32, - 221.33 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p884-b15", - "global_id": 25862, - "bbox": [ - 281.58, - 353.96, - 285.58, - 361.96 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p884-b16", - "global_id": 25863, - "bbox": [ - 278.02, - 367.43, - 287.66, - 375.43 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p884-b17", - "global_id": 25864, - "bbox": [ - 155.23, - 351.69, - 408.56, - 359.99 - ], - "text": "2p\n2p\np\np", - "type": "text" - }, - { - "block_id": "p884-b18", - "global_id": 25865, - "bbox": [ - 274.11, - 300.32, - 307.71, - 309.23 - ], - "text": "X(\t)\n9", - "type": "text" - }, - { - "block_id": "p884-b19", - "global_id": 25866, - "bbox": [ - 101.77, - 382.05, - 352.11, - 391.36 - ], - "text": "Figure 9.7 (a) Discrete-time gate pulse and (b) its Fourier spectrum.", - "type": "text" - }, - { - "block_id": "p884-b20", - "global_id": 25867, - "bbox": [ - 103.46, - 419.82, - 405.77, - 431.94 - ], - "text": "DISCRETE-TIME FOURIER TRANSFORM USING MATLAB", - "type": "text" - }, - { - "block_id": "p884-b21", - "global_id": 25868, - "bbox": [ - 103.16, - 435.97, - 477.02, - 495.25 - ], - "text": "Within a scale factor, the DTFS is identical to the DFT and, therefore, the FFT. That is, the\nDTFS is just the FFT scaled by\n1\nN0 . Combined with Eq. (9.14), we see that the DFT Xr of\nfinite-duration signal x[n] (repeated with period N0 large enough to avoid overlap) is just\nsamples of the DTFT X() taken at = r0. That is, the length-N0 DFT of signal x[n] yields\nN0 samples of its DTFT X() as", - "type": "text" - }, - { - "block_id": "p884-b22", - "global_id": 25869, - "bbox": [ - 222.17, - 503.66, - 356.11, - 516.76 - ], - "text": "Xr = X(r0),\nwhere 0 = 2π", - "type": "text" - }, - { - "block_id": "p884-b23", - "global_id": 25870, - "bbox": [ - 348.56, - 505.71, - 477.01, - 519.75 - ], - "text": "N0\n(9.26)", - "type": "text" - }, - { - "block_id": "p884-b24", - "global_id": 25871, - "bbox": [ - 103.17, - 527.63, - 477.02, - 574.95 - ], - "text": "This relationship provides a way to use MATLAB’s fft command to validate our DTFT\ncalculations. By appropriately zero-padding x[n], we can obtain as many samples of X() as\nare desired. Let us demonstrate the process for the current example using N0 = 64. Notice that\nin taking the DFT, we modulo-N0 shift our rectangular pulse signal to occupy 0 ≤n ≤N0 −1.", - "type": "text" - }, - { - "block_id": "p884-b25", - "global_id": 25872, - "bbox": [ - 103.16, - 583.16, - 416.3, - 621.02 - ], - "text": ">>\nOmega = linspace(0,2*pi,1000);\n>>\nX = sin(4.5*Omega)./sin(0.5*Omega); X(mod(Omega,2*pi)==0) = 4.5/0.5;\n>>\nN_0 = 64; M = 9; x = [ones(1,(M+1)/2) zeros(1,N_0-M) ones(1,(M-1)/2)];\n>>\nXr = fft(x); Omega_0 = 2*pi/N_0; r = 0:N_0-1;", - "type": "text" - } - ] - }, - { - "page_num": 885, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p885-b0", - "global_id": 25873, - "bbox": [ - 247.51, - 62.89, - 516.12, - 71.98 - ], - "text": "9.2\nAperiodic Signal Representation by Fourier Integral\n865", - "type": "text" - }, - { - "block_id": "p885-b1", - "global_id": 25874, - "bbox": [ - 128.9, - 86.72, - 437.83, - 104.66 - ], - "text": ">>\nplot(Omega,abs(X),’k-’,Omega_0*r,abs(Xr),’k.’); axis([0 2*pi 0 9.5]);\n>>\nxlabel(’\\Omega’); ylabel(’|X(\\Omega)|’);", - "type": "text" - }, - { - "block_id": "p885-b2", - "global_id": 25875, - "bbox": [ - 128.9, - 114.88, - 468.36, - 124.84 - ], - "text": "As shown in Fig. 9.8, the FFT samples align exactly with our analytical DTFT result.", - "type": "text" - }, - { - "block_id": "p885-b3", - "global_id": 25876, - "bbox": [ - 110.46, - 213.45, - 494.84, - 230.68 - ], - "text": "0\n1\n2\n3\n4\n5\n6\n0", - "type": "text" - }, - { - "block_id": "p885-b4", - "global_id": 25877, - "bbox": [ - 110.46, - 198.14, - 114.46, - 206.14 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p885-b5", - "global_id": 25878, - "bbox": [ - 110.46, - 182.82, - 114.46, - 190.82 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p885-b6", - "global_id": 25879, - "bbox": [ - 110.46, - 167.5, - 114.46, - 175.5 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p885-b7", - "global_id": 25880, - "bbox": [ - 110.46, - 152.19, - 114.46, - 160.19 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p885-b8", - "global_id": 25881, - "bbox": [ - 98.67, - 170.24, - 106.96, - 190.69 - ], - "text": "|X(Ω)|", - "type": "text" - }, - { - "block_id": "p885-b9", - "global_id": 25882, - "bbox": [ - 98.03, - 237.37, - 309.33, - 246.61 - ], - "text": "Figure 9.8 Using the FFT to verify the DTFT for Ex. 9.5.", - "type": "text" - }, - { - "block_id": "p885-b10", - "global_id": 25883, - "bbox": [ - 102.51, - 306.52, - 424.2, - 318.48 - ], - "text": "EXAMPLE 9.6\nInverse DTFT of a Rectangular Spectrum", - "type": "text" - }, - { - "block_id": "p885-b11", - "global_id": 25884, - "bbox": [ - 128.9, - 334.18, - 502.75, - 368.05 - ], - "text": "Find the inverse DTFT of the rectangular pulse spectrum described over the fundamental band\n(|| ≤π) by X() = rect(/2c) for c ≤π. Because of the periodicity property, X()\nrepeats at the intervals of 2π, as shown in Fig. 9.9a.", - "type": "text" - }, - { - "block_id": "p885-b12", - "global_id": 25885, - "bbox": [ - 295.04, - 578.66, - 304.84, - 586.66 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p885-b13", - "global_id": 25886, - "bbox": [ - 295.5, - 492.87, - 304.38, - 500.87 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p885-b14", - "global_id": 25887, - "bbox": [ - 302.04, - 562.13, - 306.04, - 570.13 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p885-b15", - "global_id": 25888, - "bbox": [ - 276.39, - 517.13, - 319.76, - 531.53 - ], - "text": "x[n]\n0.25", - "type": "text" - }, - { - "block_id": "p885-b16", - "global_id": 25889, - "bbox": [ - 319.94, - 563.15, - 400.3, - 572.49 - ], - "text": "4\n8\nn", - "type": "text" - }, - { - "block_id": "p885-b17", - "global_id": 25890, - "bbox": [ - 303.24, - 413.84, - 319.68, - 422.14 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p885-b18", - "global_id": 25891, - "bbox": [ - 301.74, - 470.89, - 319.34, - 487.33 - ], - "text": "0\n4", - "type": "text" - }, - { - "block_id": "p885-b19", - "global_id": 25892, - "bbox": [ - 294.34, - 421.96, - 298.34, - 429.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p885-b20", - "global_id": 25893, - "bbox": [ - 158.54, - 470.74, - 462.8, - 487.33 - ], - "text": "2p\n2p\np\np\np\n\t\n4\np", - "type": "text" - }, - { - "block_id": "p885-b21", - "global_id": 25894, - "bbox": [ - 119.94, - 593.35, - 418.11, - 602.59 - ], - "text": "Figure 9.9 Periodic gate spectrum and its inverse discrete-time Fourier transform.", - "type": "text" - } - ] - }, - { - "page_num": 886, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p886-b0", - "global_id": 25895, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "866\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p886-b1", - "global_id": 25896, - "bbox": [ - 121.09, - 86.24, - 218.49, - 96.21 - ], - "text": "According to Eq. (9.18),", - "type": "text" - }, - { - "block_id": "p886-b2", - "global_id": 25897, - "bbox": [ - 188.53, - 107.84, - 226.13, - 124.79 - ], - "text": "x[n] = 1", - "type": "text" - }, - { - "block_id": "p886-b3", - "global_id": 25898, - "bbox": [ - 217.65, - 121.49, - 228.61, - 131.86 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p886-b4", - "global_id": 25899, - "bbox": [ - 231.92, - 100.85, - 245.92, - 112.91 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p886-b5", - "global_id": 25900, - "bbox": [ - 237.17, - 125.73, - 246.79, - 132.71 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p886-b6", - "global_id": 25901, - "bbox": [ - 249.1, - 107.84, - 322.22, - 124.79 - ], - "text": "X()ejn d = 1", - "type": "text" - }, - { - "block_id": "p886-b7", - "global_id": 25902, - "bbox": [ - 313.76, - 121.49, - 324.71, - 131.86 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p886-b8", - "global_id": 25903, - "bbox": [ - 328.01, - 100.85, - 345.91, - 113.94 - ], - "text": "# c", - "type": "text" - }, - { - "block_id": "p886-b9", - "global_id": 25904, - "bbox": [ - 333.27, - 125.73, - 346.78, - 133.74 - ], - "text": "−c", - "type": "text" - }, - { - "block_id": "p886-b10", - "global_id": 25905, - "bbox": [ - 348.89, - 110.61, - 378.77, - 124.69 - ], - "text": "ejn d", - "type": "text" - }, - { - "block_id": "p886-b11", - "global_id": 25906, - "bbox": [ - 206.65, - 138.23, - 255.02, - 162.25 - ], - "text": "=\n1\nj2πnejn", - "type": "text" - }, - { - "block_id": "p886-b13", - "global_id": 25907, - "bbox": [ - 258.77, - 135.43, - 266.85, - 143.44 - ], - "text": "c", - "type": "text" - }, - { - "block_id": "p886-b14", - "global_id": 25908, - "bbox": [ - 258.77, - 157.0, - 272.28, - 165.0 - ], - "text": "−c", - "type": "text" - }, - { - "block_id": "p886-b15", - "global_id": 25909, - "bbox": [ - 275.33, - 137.81, - 322.83, - 155.17 - ], - "text": "= sin(cn)", - "type": "text" - }, - { - "block_id": "p886-b16", - "global_id": 25910, - "bbox": [ - 298.61, - 137.81, - 347.94, - 162.15 - ], - "text": "πn\n= c", - "type": "text" - }, - { - "block_id": "p886-b17", - "global_id": 25911, - "bbox": [ - 339.28, - 144.8, - 391.65, - 162.25 - ], - "text": "π sinc(cn)", - "type": "text" - }, - { - "block_id": "p886-b18", - "global_id": 25912, - "bbox": [ - 103.17, - 173.47, - 356.69, - 184.93 - ], - "text": "The signal x[n] is depicted in Fig. 9.9b (for the case c = π/4).", - "type": "text" - }, - { - "block_id": "p886-b19", - "global_id": 25913, - "bbox": [ - 107.82, - 247.27, - 281.81, - 259.23 - ], - "text": "DRILL 9.4\nFinding the DTFT", - "type": "text" - }, - { - "block_id": "p886-b20", - "global_id": 25914, - "bbox": [ - 107.82, - 268.35, - 416.23, - 278.31 - ], - "text": "Find the DTFT and sketch the corresponding amplitude and phase spectra for", - "type": "text" - }, - { - "block_id": "p886-b21", - "global_id": 25915, - "bbox": [ - 125.76, - 282.67, - 245.96, - 311.19 - ], - "text": "(a) x[n] = γ |k| with |γ | < 1\n(b) y[n] = δ[n + 1] −δ[n −1]", - "type": "text" - }, - { - "block_id": "p886-b22", - "global_id": 25916, - "bbox": [ - 108.09, - 330.69, - 170.31, - 341.65 - ], - "text": "ANSWERS", - "type": "text" - }, - { - "block_id": "p886-b23", - "global_id": 25917, - "bbox": [ - 125.76, - 345.23, - 226.45, - 363.44 - ], - "text": "(a) X() =\n1 −γ 2", - "type": "text" - }, - { - "block_id": "p886-b24", - "global_id": 25918, - "bbox": [ - 177.12, - 360.02, - 249.85, - 370.6 - ], - "text": "1 −2γ cos + γ 2", - "type": "text" - }, - { - "block_id": "p886-b25", - "global_id": 25919, - "bbox": [ - 125.76, - 373.17, - 360.49, - 383.55 - ], - "text": "(b) |Y()| = 2|sin| and̸\nY(ω) = (π/2)[1 −sgn(sin)]", - "type": "text" - }, - { - "block_id": "p886-b26", - "global_id": 25920, - "bbox": [ - 101.84, - 417.8, - 413.17, - 430.61 - ], - "text": "9.2-2 Connection Between the DTFT and the z-Transform", - "type": "text" - }, - { - "block_id": "p886-b27", - "global_id": 25921, - "bbox": [ - 101.84, - 436.64, - 490.4, - 458.66 - ], - "text": "The connection between the (bilateral) z-transform and the DTFT is similar to that between the\nLaplace transform and the Fourier transform. The z-transform of x[n], according to Eq. (5.1), is", - "type": "text" - }, - { - "block_id": "p886-b28", - "global_id": 25922, - "bbox": [ - 255.68, - 476.57, - 282.54, - 486.84 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p886-b29", - "global_id": 25923, - "bbox": [ - 288.28, - 466.4, - 302.38, - 477.06 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p886-b30", - "global_id": 25924, - "bbox": [ - 284.59, - 490.42, - 306.06, - 497.61 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p886-b31", - "global_id": 25925, - "bbox": [ - 307.18, - 474.84, - 490.38, - 486.95 - ], - "text": "x[n]z−n\n(9.27)", - "type": "text" - }, - { - "block_id": "p886-b32", - "global_id": 25926, - "bbox": [ - 101.85, - 503.96, - 250.81, - 517.55 - ], - "text": "Setting z = ej in this equation yields", - "type": "text" - }, - { - "block_id": "p886-b33", - "global_id": 25927, - "bbox": [ - 247.53, - 533.72, - 282.78, - 545.72 - ], - "text": "X[ej] =", - "type": "text" - }, - { - "block_id": "p886-b34", - "global_id": 25928, - "bbox": [ - 288.53, - 525.27, - 302.63, - 535.95 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p886-b35", - "global_id": 25929, - "bbox": [ - 284.83, - 549.29, - 306.31, - 556.49 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p886-b36", - "global_id": 25930, - "bbox": [ - 307.43, - 533.72, - 344.2, - 545.72 - ], - "text": "x[n]e−jn", - "type": "text" - }, - { - "block_id": "p886-b37", - "global_id": 25931, - "bbox": [ - 101.84, - 564.63, - 490.39, - 634.79 - ], - "text": "The right-hand side sum defines X(), the DTFT of x[n]. Does this mean that the DTFT can be\nobtained from the corresponding z-transform by setting z = ej? In other words, is it true that\nX[ej] = X()? Yes, it is true in most cases. For example, when x[n] = anu[n], its z-transform is\nz/(z −a), and X[ej] = ej/(ej −a), which is equal to X() (assuming |a| < 1). However, for\nthe unit step function u[n], the z-transform is z/(z −1), and X[ej] = ej/(ej −1). As seen from\nTable 9.1, pair 10, this is not equal to X() in this case.", - "type": "text" - } - ] - }, - { - "page_num": 887, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p887-b0", - "global_id": 25932, - "bbox": [ - 362.61, - 62.89, - 516.12, - 71.98 - ], - "text": "9.3\nProperties of the DTFT\n867", - "type": "text" - }, - { - "block_id": "p887-b1", - "global_id": 25933, - "bbox": [ - 127.59, - 83.68, - 516.14, - 168.99 - ], - "text": "We obtained X[ej] by setting z = ej in Eq. (9.27). This implies that the sum on the right-hand\nside of Eq. (9.27) converges for z = ej, which means the unit circle (characterized by z = ej) lies\nin the region of convergence for X[z]. Hence, the general rule is that setting z = ej in X[z] yields\nthe DTFT X() only when the ROC for X[z] includes the unit circle. This applies for all x[n] that\nare absolutely summable. If the ROC of X[z] excludes the unit circle, X[ej]̸ = X(). This applies\nto all exponentially growing x[n] and also x[n], which either is constant or oscillates with constant\namplitude.", - "type": "text" - }, - { - "block_id": "p887-b2", - "global_id": 25934, - "bbox": [ - 127.6, - 170.98, - 516.14, - 192.91 - ], - "text": "The reason for this peculiar behavior has something to do with the nature of convergence of\nthe z-transform and the DTFT.†", - "type": "text" - }, - { - "block_id": "p887-b3", - "global_id": 25935, - "bbox": [ - 127.59, - 194.9, - 516.15, - 228.77 - ], - "text": "This discussion shows that although the DTFT may be considered to be a special case of the\nz-transform, we need to circumscribe such a view. This cautionary note is supported by the fact\nthat a periodic signal has the DTFT, but its z-transform does not exist.", - "type": "text" - }, - { - "block_id": "p887-b4", - "global_id": 25936, - "bbox": [ - 127.94, - 261.12, - 325.26, - 275.07 - ], - "text": "9.3 PROPERTIES OF THE DTFT", - "type": "text" - }, - { - "block_id": "p887-b5", - "global_id": 25937, - "bbox": [ - 127.59, - 281.05, - 516.17, - 314.93 - ], - "text": "A close connection exists between the DTFT and the CTFT (continuous-time Fourier transform).\nFor this reason, which Sec. 9.4 discusses, the properties of the DTFT are very similar to those of\nthe CTFT, as the following discussion shows.", - "type": "text" - }, - { - "block_id": "p887-b6", - "global_id": 25938, - "bbox": [ - 127.89, - 331.19, - 263.58, - 343.31 - ], - "text": "LINEARITY OF THE DTFT", - "type": "text" - }, - { - "block_id": "p887-b7", - "global_id": 25939, - "bbox": [ - 127.59, - 347.35, - 134.23, - 357.31 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p887-b8", - "global_id": 25940, - "bbox": [ - 226.24, - 363.48, - 417.47, - 374.63 - ], - "text": "x1[n] ⇐⇒X1()\nand\nx2[n] ⇐⇒X2()", - "type": "text" - }, - { - "block_id": "p887-b9", - "global_id": 25941, - "bbox": [ - 127.59, - 385.89, - 144.74, - 395.86 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p887-b10", - "global_id": 25942, - "bbox": [ - 236.17, - 402.04, - 407.55, - 413.19 - ], - "text": "a1x1[n] + a2x2[n] ⇐⇒a1X1() + a2X2()", - "type": "text" - }, - { - "block_id": "p887-b11", - "global_id": 25943, - "bbox": [ - 127.59, - 424.45, - 389.71, - 434.41 - ], - "text": "The proof is trivial. The result can be extended to any finite sums.", - "type": "text" - }, - { - "block_id": "p887-b12", - "global_id": 25944, - "bbox": [ - 127.59, - 449.88, - 326.06, - 476.79 - ], - "text": "CONJUGATE SYMMETRY OF X()\nIn Eq. (9.20), we proved the conjugation property", - "type": "text" - }, - { - "block_id": "p887-b13", - "global_id": 25945, - "bbox": [ - 283.35, - 489.68, - 516.13, - 501.78 - ], - "text": "x∗[n] ⇐⇒X∗(−)\n(9.28)", - "type": "text" - }, - { - "block_id": "p887-b14", - "global_id": 25946, - "bbox": [ - 127.59, - 522.56, - 516.14, - 633.41 - ], - "text": "† To explain this point, consider the unit step function u[n] and its transforms. Both the z-transform and the\nDTFT synthesize x[n], using everlasting exponentials of the form zn. The value of z can be anywhere in\nthe complex z-plane for the z-transform, but it must be restricted to the unit circle (z = ej) in the case of\nthe DTFT. The unit step function is readily synthesized in the z-transform by a relatively simple spectrum\nX[z] = z/(z −1), by choosing z outside the unit circle (the ROC for u[n] is |z| > 1). In the DTFT, however,\nwe are restricted to values of z only on the unit circle (z = ej). The function u[n] can still be synthesized\nby values of z on the unit circle, but the spectrum is more complicated than when we are free to choose\nz anywhere, including the region outside the unit circle. In contrast, when x[n] is absolutely summable, the\nregion of convergence for the z-transform includes the unit circle, and we can synthesize x[n] by using z along\nthe unit circle in both the transforms. This leads to X[ej] = X().", - "type": "text" - } - ] - }, - { - "page_num": 888, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p888-b0", - "global_id": 25947, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "868\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p888-b1", - "global_id": 25948, - "bbox": [ - 101.84, - 85.46, - 490.36, - 107.79 - ], - "text": "We also showed that as a consequence of this, when x[n] is real, X() and X(−) are conjugates,\nthat is,", - "type": "text" - }, - { - "block_id": "p888-b2", - "global_id": 25949, - "bbox": [ - 262.52, - 109.93, - 329.7, - 121.93 - ], - "text": "X(−) = X∗()", - "type": "text" - }, - { - "block_id": "p888-b3", - "global_id": 25950, - "bbox": [ - 101.84, - 132.11, - 490.4, - 154.44 - ], - "text": "This is the conjugate symmetry property. Since X() is generally complex, we have both amplitude\nand angle (or phase) spectra", - "type": "text" - }, - { - "block_id": "p888-b4", - "global_id": 25951, - "bbox": [ - 252.18, - 156.58, - 339.55, - 168.58 - ], - "text": "X() = |X()|ej̸ X()", - "type": "text" - }, - { - "block_id": "p888-b5", - "global_id": 25952, - "bbox": [ - 101.84, - 178.75, - 239.96, - 189.13 - ], - "text": "Hence, for real x[n], it follows that", - "type": "text" - }, - { - "block_id": "p888-b6", - "global_id": 25953, - "bbox": [ - 190.73, - 202.19, - 401.5, - 212.57 - ], - "text": "|X()| = |X(−)|\nand̸\nX() = −̸ X(−)", - "type": "text" - }, - { - "block_id": "p888-b7", - "global_id": 25954, - "bbox": [ - 101.85, - 225.63, - 490.39, - 247.96 - ], - "text": "Therefore, for real x[n], the amplitude spectrum |X()| is an even function of and the phase\nspectrum̸\nX() is an odd function of .", - "type": "text" - }, - { - "block_id": "p888-b8", - "global_id": 25955, - "bbox": [ - 101.84, - 263.46, - 446.21, - 289.57 - ], - "text": "TIME AND FREQUENCY REVERSAL\nAlso called the reflection property, the time and frequency reversal property states that", - "type": "text" - }, - { - "block_id": "p888-b9", - "global_id": 25956, - "bbox": [ - 257.85, - 302.63, - 490.38, - 313.01 - ], - "text": "x[−n] ⇐⇒X(−)\n(9.29)", - "type": "text" - }, - { - "block_id": "p888-b10", - "global_id": 25957, - "bbox": [ - 101.85, - 326.07, - 457.93, - 336.45 - ], - "text": "Demonstration of this property is straightforward. From Eq. (9.19), the DTFT of x[−n] is", - "type": "text" - }, - { - "block_id": "p888-b11", - "global_id": 25958, - "bbox": [ - 171.82, - 357.64, - 237.76, - 368.02 - ], - "text": "DTFT{x[−n]} =", - "type": "text" - }, - { - "block_id": "p888-b12", - "global_id": 25959, - "bbox": [ - 243.5, - 347.47, - 257.6, - 358.14 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p888-b13", - "global_id": 25960, - "bbox": [ - 239.8, - 371.49, - 261.28, - 378.68 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p888-b14", - "global_id": 25961, - "bbox": [ - 262.4, - 353.52, - 317.27, - 367.92 - ], - "text": "x[−n]e−jn =", - "type": "text" - }, - { - "block_id": "p888-b15", - "global_id": 25962, - "bbox": [ - 323.79, - 347.47, - 337.89, - 358.14 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p888-b16", - "global_id": 25963, - "bbox": [ - 319.32, - 371.49, - 342.34, - 378.68 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p888-b17", - "global_id": 25964, - "bbox": [ - 343.47, - 353.52, - 420.41, - 367.92 - ], - "text": "x[m]ejm = X(−)", - "type": "text" - }, - { - "block_id": "p888-b18", - "global_id": 25965, - "bbox": [ - 76.77, - 420.24, - 336.02, - 432.2 - ], - "text": "EXAMPLE 9.7\nUsing the Reflection Property", - "type": "text" - }, - { - "block_id": "p888-b19", - "global_id": 25966, - "bbox": [ - 103.16, - 448.86, - 477.04, - 470.77 - ], - "text": "Use the time-frequency reversal property of Eq. (9.29) and pair 2 in Table 9.1 to derive pair 4\nin Table 9.1.", - "type": "text" - }, - { - "block_id": "p888-b20", - "global_id": 25967, - "bbox": [ - 103.16, - 493.69, - 168.6, - 503.65 - ], - "text": "Pair 2 states that", - "type": "text" - }, - { - "block_id": "p888-b21", - "global_id": 25968, - "bbox": [ - 229.97, - 501.42, - 290.71, - 519.89 - ], - "text": "γ nu[n] =\nej", - "type": "text" - }, - { - "block_id": "p888-b22", - "global_id": 25969, - "bbox": [ - 270.2, - 509.62, - 350.22, - 526.97 - ], - "text": "ej −γ\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p888-b23", - "global_id": 25970, - "bbox": [ - 103.17, - 533.92, - 198.06, - 543.88 - ], - "text": "Hence, from Eq. (9.29),", - "type": "text" - }, - { - "block_id": "p888-b24", - "global_id": 25971, - "bbox": [ - 220.64, - 543.7, - 300.04, - 562.18 - ], - "text": "γ −nu[−n] =\ne−j", - "type": "text" - }, - { - "block_id": "p888-b25", - "global_id": 25972, - "bbox": [ - 274.09, - 551.9, - 359.54, - 569.25 - ], - "text": "e−j −γ\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p888-b26", - "global_id": 25973, - "bbox": [ - 103.17, - 573.97, - 477.01, - 599.6 - ], - "text": "Moreover, γ |n| could be expressed as a sum of γ n u[n] and γ −n u[−n], except that the impulse\nat n = 0 is counted twice (once from each of the two exponentials). Hence,", - "type": "text" - }, - { - "block_id": "p888-b27", - "global_id": 25974, - "bbox": [ - 224.07, - 607.02, - 356.1, - 621.41 - ], - "text": "γ |n| = γ nu[n] + γ −nu[−n] −δ[n]", - "type": "text" - } - ] - }, - { - "page_num": 889, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p889-b0", - "global_id": 25975, - "bbox": [ - 362.61, - 62.89, - 516.12, - 71.98 - ], - "text": "9.3\nProperties of the DTFT\n869", - "type": "text" - }, - { - "block_id": "p889-b1", - "global_id": 25976, - "bbox": [ - 128.9, - 86.24, - 421.26, - 96.21 - ], - "text": "Combining these results and invoking the linearity property, we can write", - "type": "text" - }, - { - "block_id": "p889-b2", - "global_id": 25977, - "bbox": [ - 169.49, - 114.7, - 194.39, - 124.66 - ], - "text": "DTFT", - "type": "text" - }, - { - "block_id": "p889-b3", - "global_id": 25978, - "bbox": [ - 194.4, - 106.28, - 218.42, - 124.25 - ], - "text": "*\nγ |n|+", - "type": "text" - }, - { - "block_id": "p889-b4", - "global_id": 25979, - "bbox": [ - 220.46, - 106.09, - 251.99, - 124.25 - ], - "text": "=\nej", - "type": "text" - }, - { - "block_id": "p889-b5", - "global_id": 25980, - "bbox": [ - 231.48, - 106.09, - 300.43, - 131.64 - ], - "text": "ej −γ +\ne−j", - "type": "text" - }, - { - "block_id": "p889-b6", - "global_id": 25981, - "bbox": [ - 274.49, - 106.38, - 388.56, - 131.64 - ], - "text": "e−j −γ −1 =\n1 −γ 2", - "type": "text" - }, - { - "block_id": "p889-b7", - "global_id": 25982, - "bbox": [ - 339.79, - 114.29, - 462.18, - 131.74 - ], - "text": "1 −2γ cos + γ 2\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p889-b8", - "global_id": 25983, - "bbox": [ - 128.91, - 141.58, - 278.91, - 151.54 - ], - "text": "which agrees with pair 4 in Table 9.1.", - "type": "text" - }, - { - "block_id": "p889-b9", - "global_id": 25984, - "bbox": [ - 133.57, - 224.54, - 369.63, - 236.49 - ], - "text": "DRILL 9.5\nUsing the Reflection Property", - "type": "text" - }, - { - "block_id": "p889-b10", - "global_id": 25985, - "bbox": [ - 133.57, - 245.61, - 491.38, - 255.58 - ], - "text": "In Table 9.1, derive pair 13 from pair 15 by using the time-reversal property of Eq. (9.29).", - "type": "text" - }, - { - "block_id": "p889-b11", - "global_id": 25986, - "bbox": [ - 127.89, - 289.9, - 420.29, - 302.44 - ], - "text": "MULTIPLICATION BY n: FREQUENCY DIFFERENTIATION", - "type": "text" - }, - { - "block_id": "p889-b12", - "global_id": 25987, - "bbox": [ - 283.8, - 315.43, - 358.72, - 332.69 - ], - "text": "nx[n] ⇐⇒jdX()", - "type": "text" - }, - { - "block_id": "p889-b13", - "global_id": 25988, - "bbox": [ - 338.88, - 322.83, - 516.13, - 339.77 - ], - "text": "d\n(9.30)", - "type": "text" - }, - { - "block_id": "p889-b14", - "global_id": 25989, - "bbox": [ - 127.59, - 352.18, - 493.74, - 362.56 - ], - "text": "The result follows immediately by differentiating both sides of Eq. (9.19) with respect to .", - "type": "text" - }, - { - "block_id": "p889-b15", - "global_id": 25990, - "bbox": [ - 102.51, - 419.37, - 450.52, - 431.32 - ], - "text": "EXAMPLE 9.8\nUsing the Frequency-Differentiation Property", - "type": "text" - }, - { - "block_id": "p889-b16", - "global_id": 25991, - "bbox": [ - 128.9, - 447.99, - 502.78, - 469.9 - ], - "text": "Use the frequency-differentiation property of Eq. (9.30) and pair 2 in Table 9.1 to derive pair 5\nin Table 9.1.", - "type": "text" - }, - { - "block_id": "p889-b17", - "global_id": 25992, - "bbox": [ - 128.9, - 492.82, - 194.34, - 502.78 - ], - "text": "Pair 2 states that", - "type": "text" - }, - { - "block_id": "p889-b18", - "global_id": 25993, - "bbox": [ - 255.71, - 500.54, - 316.46, - 519.02 - ], - "text": "γ nu[n] =\nej", - "type": "text" - }, - { - "block_id": "p889-b19", - "global_id": 25994, - "bbox": [ - 295.94, - 508.75, - 375.96, - 526.1 - ], - "text": "ej −γ\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p889-b20", - "global_id": 25995, - "bbox": [ - 128.91, - 533.04, - 223.8, - 543.0 - ], - "text": "Hence, from Eq. (9.30),", - "type": "text" - }, - { - "block_id": "p889-b21", - "global_id": 25996, - "bbox": [ - 209.72, - 554.32, - 266.56, - 571.27 - ], - "text": "nγ nu[n] = j d", - "type": "text" - }, - { - "block_id": "p889-b22", - "global_id": 25997, - "bbox": [ - 257.7, - 568.07, - 270.7, - 578.34 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p889-b23", - "global_id": 25998, - "bbox": [ - 273.0, - 547.01, - 300.88, - 564.29 - ], - "text": "ej", - "type": "text" - }, - { - "block_id": "p889-b24", - "global_id": 25999, - "bbox": [ - 280.36, - 565.19, - 308.51, - 578.34 - ], - "text": "ej −γ", - "type": "text" - }, - { - "block_id": "p889-b26", - "global_id": 26000, - "bbox": [ - 319.51, - 552.78, - 360.04, - 570.96 - ], - "text": "=\nγ ej", - "type": "text" - }, - { - "block_id": "p889-b27", - "global_id": 26001, - "bbox": [ - 330.52, - 560.99, - 421.96, - 578.34 - ], - "text": "(ej −γ )2\n|γ | < 1", - "type": "text" - }, - { - "block_id": "p889-b28", - "global_id": 26002, - "bbox": [ - 128.91, - 588.61, - 278.91, - 598.58 - ], - "text": "which agrees with pair 5 in Table 9.1.", - "type": "text" - } - ] - }, - { - "page_num": 890, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p890-b0", - "global_id": 26003, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "870\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p890-b1", - "global_id": 26004, - "bbox": [ - 102.14, - 86.19, - 244.53, - 98.32 - ], - "text": "TIME-SHIFTING PROPERTY", - "type": "text" - }, - { - "block_id": "p890-b2", - "global_id": 26005, - "bbox": [ - 101.84, - 102.35, - 108.48, - 112.31 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p890-b3", - "global_id": 26006, - "bbox": [ - 265.62, - 115.8, - 326.61, - 126.08 - ], - "text": "x[n] ⇐⇒X()", - "type": "text" - }, - { - "block_id": "p890-b4", - "global_id": 26007, - "bbox": [ - 101.84, - 136.42, - 118.99, - 146.39 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p890-b5", - "global_id": 26008, - "bbox": [ - 213.1, - 148.16, - 490.38, - 160.25 - ], - "text": "x[n −k] ⇐⇒X()e−jk\nfor integer k\n(9.31)", - "type": "text" - }, - { - "block_id": "p890-b6", - "global_id": 26009, - "bbox": [ - 101.85, - 170.5, - 490.4, - 192.41 - ], - "text": "This property can be proved by direct substitution in the equation defining the direct transform.\nFrom Eq. (9.19), we obtain", - "type": "text" - }, - { - "block_id": "p890-b7", - "global_id": 26010, - "bbox": [ - 185.9, - 213.35, - 238.61, - 223.63 - ], - "text": "x[n −k] ⇐⇒", - "type": "text" - }, - { - "block_id": "p890-b8", - "global_id": 26011, - "bbox": [ - 244.34, - 203.18, - 258.44, - 213.85 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p890-b9", - "global_id": 26012, - "bbox": [ - 240.65, - 227.2, - 262.13, - 234.4 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p890-b10", - "global_id": 26013, - "bbox": [ - 263.24, - 209.24, - 325.8, - 223.63 - ], - "text": "x[n −k]e−jn =", - "type": "text" - }, - { - "block_id": "p890-b11", - "global_id": 26014, - "bbox": [ - 332.33, - 203.18, - 346.42, - 213.85 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p890-b12", - "global_id": 26015, - "bbox": [ - 327.85, - 227.2, - 350.88, - 234.4 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p890-b13", - "global_id": 26016, - "bbox": [ - 352.0, - 211.63, - 405.82, - 223.63 - ], - "text": "x[m]e−j[m+k]", - "type": "text" - }, - { - "block_id": "p890-b14", - "global_id": 26017, - "bbox": [ - 219.48, - 245.69, - 249.61, - 257.69 - ], - "text": "= e−jk", - "type": "text" - }, - { - "block_id": "p890-b15", - "global_id": 26018, - "bbox": [ - 255.04, - 237.24, - 269.14, - 247.91 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p890-b16", - "global_id": 26019, - "bbox": [ - 251.34, - 261.26, - 272.82, - 268.45 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p890-b17", - "global_id": 26020, - "bbox": [ - 273.94, - 243.3, - 369.49, - 257.69 - ], - "text": "x[m]e−jm = e−jkX()", - "type": "text" - }, - { - "block_id": "p890-b18", - "global_id": 26021, - "bbox": [ - 101.85, - 279.88, - 490.39, - 313.86 - ], - "text": "This result shows that delaying a signal by k samples does not change its amplitude spectrum. The\nphase spectrum, however, is changed by −k. This added phase is a linear function of with\nslope −k.", - "type": "text" - }, - { - "block_id": "p890-b19", - "global_id": 26022, - "bbox": [ - 102.14, - 329.23, - 338.76, - 341.35 - ], - "text": "PHYSICAL EXPLANATION OF LINEAR PHASE", - "type": "text" - }, - { - "block_id": "p890-b20", - "global_id": 26023, - "bbox": [ - 101.84, - 345.38, - 490.41, - 367.3 - ], - "text": "Time delay in a signal causes a linear phase shift in its spectrum. The heuristic explanation of this\nresult is exactly parallel to that for continuous-time signals given in Sec. 7.3 (see Fig. 7.22).", - "type": "text" - }, - { - "block_id": "p890-b21", - "global_id": 26024, - "bbox": [ - 76.77, - 416.85, - 330.35, - 428.81 - ], - "text": "EXAMPLE 9.9\nDemonstrating Linear Phase", - "type": "text" - }, - { - "block_id": "p890-b22", - "global_id": 26025, - "bbox": [ - 103.16, - 445.06, - 477.02, - 467.39 - ], - "text": "To demonstrate the linear phase associated with a time shift, find the DTFT of x[n] =\n(1/4)sinc(π(n −2)/4), shown in Fig. 9.10a.", - "type": "text" - }, - { - "block_id": "p890-b23", - "global_id": 26026, - "bbox": [ - 103.16, - 490.31, - 202.22, - 500.27 - ], - "text": "In Ex. 9.6, we found that", - "type": "text" - }, - { - "block_id": "p890-b24", - "global_id": 26027, - "bbox": [ - 206.62, - 505.23, - 249.83, - 536.67 - ], - "text": "1\n4 sinc\nπn", - "type": "text" - }, - { - "block_id": "p890-b25", - "global_id": 26028, - "bbox": [ - 241.36, - 526.7, - 246.34, - 536.67 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p890-b27", - "global_id": 26029, - "bbox": [ - 259.79, - 519.21, - 278.92, - 529.18 - ], - "text": "⇐⇒", - "type": "text" - }, - { - "block_id": "p890-b28", - "global_id": 26030, - "bbox": [ - 285.43, - 509.05, - 299.53, - 519.72 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p890-b29", - "global_id": 26031, - "bbox": [ - 280.97, - 533.07, - 303.99, - 540.27 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p890-b30", - "global_id": 26032, - "bbox": [ - 305.11, - 519.63, - 320.03, - 529.59 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p890-b31", - "global_id": 26033, - "bbox": [ - 321.14, - 505.23, - 366.82, - 522.62 - ], - "text": "−2πm", - "type": "text" - }, - { - "block_id": "p890-b32", - "global_id": 26034, - "bbox": [ - 339.88, - 526.29, - 356.01, - 536.67 - ], - "text": "π/2", - "type": "text" - }, - { - "block_id": "p890-b34", - "global_id": 26035, - "bbox": [ - 103.17, - 550.42, - 367.69, - 560.48 - ], - "text": "Use of the time-shifting property [Eq. (9.31)] yields (for integer k)", - "type": "text" - }, - { - "block_id": "p890-b35", - "global_id": 26036, - "bbox": [ - 182.85, - 566.16, - 250.16, - 597.6 - ], - "text": "1\n4 sinc\nπ(n −2)", - "type": "text" - }, - { - "block_id": "p890-b36", - "global_id": 26037, - "bbox": [ - 230.2, - 587.63, - 235.18, - 597.6 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p890-b38", - "global_id": 26038, - "bbox": [ - 260.14, - 580.14, - 279.27, - 590.11 - ], - "text": "⇐⇒", - "type": "text" - }, - { - "block_id": "p890-b39", - "global_id": 26039, - "bbox": [ - 285.77, - 569.97, - 299.87, - 580.65 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p890-b40", - "global_id": 26040, - "bbox": [ - 281.31, - 593.99, - 304.33, - 601.19 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p890-b41", - "global_id": 26041, - "bbox": [ - 305.45, - 580.56, - 320.38, - 590.52 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p890-b42", - "global_id": 26042, - "bbox": [ - 321.49, - 566.16, - 367.16, - 583.54 - ], - "text": "−2πm", - "type": "text" - }, - { - "block_id": "p890-b43", - "global_id": 26043, - "bbox": [ - 340.22, - 587.22, - 356.36, - 597.6 - ], - "text": "π/2", - "type": "text" - }, - { - "block_id": "p890-b45", - "global_id": 26044, - "bbox": [ - 377.31, - 578.42, - 398.01, - 590.42 - ], - "text": "e−j2", - "type": "text" - }, - { - "block_id": "p890-b46", - "global_id": 26045, - "bbox": [ - 103.16, - 611.45, - 331.93, - 621.41 - ], - "text": "The spectrum of the shifted signal is shown in Fig. 9.10b.", - "type": "text" - } - ] - }, - { - "page_num": 891, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p891-b0", - "global_id": 26046, - "bbox": [ - 362.61, - 62.89, - 516.12, - 71.98 - ], - "text": "9.3\nProperties of the DTFT\n871", - "type": "text" - }, - { - "block_id": "p891-b1", - "global_id": 26047, - "bbox": [ - 294.34, - 184.43, - 298.34, - 192.43 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p891-b2", - "global_id": 26048, - "bbox": [ - 295.04, - 265.15, - 304.84, - 273.15 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p891-b3", - "global_id": 26049, - "bbox": [ - 303.24, - 178.71, - 323.87, - 187.0 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p891-b4", - "global_id": 26050, - "bbox": [ - 300.74, - 235.76, - 319.34, - 249.2 - ], - "text": "0\n4", - "type": "text" - }, - { - "block_id": "p891-b5", - "global_id": 26051, - "bbox": [ - 295.5, - 154.29, - 304.38, - 162.29 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p891-b6", - "global_id": 26052, - "bbox": [ - 152.88, - 232.61, - 462.8, - 244.83 - ], - "text": "2p\n2p\np\np\np", - "type": "text" - }, - { - "block_id": "p891-b7", - "global_id": 26053, - "bbox": [ - 284.03, - 128.87, - 335.82, - 145.52 - ], - "text": "0\n2\n8", - "type": "text" - }, - { - "block_id": "p891-b8", - "global_id": 26054, - "bbox": [ - 270.53, - 92.76, - 306.27, - 109.34 - ], - "text": "x[n]\n0.25", - "type": "text" - }, - { - "block_id": "p891-b9", - "global_id": 26055, - "bbox": [ - 396.3, - 140.13, - 400.3, - 148.13 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p891-b10", - "global_id": 26056, - "bbox": [ - 277.12, - 232.8, - 290.06, - 249.2 - ], - "text": "4\np", - "type": "text" - }, - { - "block_id": "p891-b11", - "global_id": 26057, - "bbox": [ - 223.35, - 199.62, - 274.97, - 207.91 - ], - "text": "X(\t) 2", - "type": "text" - }, - { - "block_id": "p891-b12", - "global_id": 26058, - "bbox": [ - 119.94, - 279.47, - 458.3, - 289.08 - ], - "text": "Figure 9.10 Shifted signal x[n] possesses spectrum X() with linear phase component −2.", - "type": "text" - }, - { - "block_id": "p891-b13", - "global_id": 26059, - "bbox": [ - 133.57, - 362.06, - 391.55, - 374.01 - ], - "text": "DRILL 9.6\nUsing the Time-Shifting Property", - "type": "text" - }, - { - "block_id": "p891-b14", - "global_id": 26060, - "bbox": [ - 133.57, - 383.13, - 510.18, - 405.05 - ], - "text": "Verify the result in Eq. (9.24) from pair 7 in Table 9.1 and the time-shifting property of the\nDTFT.", - "type": "text" - }, - { - "block_id": "p891-b15", - "global_id": 26061, - "bbox": [ - 127.89, - 434.92, - 307.6, - 447.05 - ], - "text": "FREQUENCY-SHIFTING PROPERTY", - "type": "text" - }, - { - "block_id": "p891-b16", - "global_id": 26062, - "bbox": [ - 127.59, - 451.08, - 134.23, - 461.04 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p891-b17", - "global_id": 26063, - "bbox": [ - 291.37, - 465.92, - 352.36, - 476.2 - ], - "text": "x[n] ⇐⇒X()", - "type": "text" - }, - { - "block_id": "p891-b18", - "global_id": 26064, - "bbox": [ - 127.59, - 487.46, - 144.74, - 497.42 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p891-b19", - "global_id": 26065, - "bbox": [ - 270.79, - 498.19, - 516.13, - 513.38 - ], - "text": "x[n]ejcn ⇐⇒X( −c)\n(9.32)", - "type": "text" - }, - { - "block_id": "p891-b20", - "global_id": 26066, - "bbox": [ - 127.59, - 523.85, - 516.1, - 545.76 - ], - "text": "This property is the dual of the time-shifting property. To prove the frequency-shifting property,\nwe use Eq. (9.19) as", - "type": "text" - }, - { - "block_id": "p891-b21", - "global_id": 26067, - "bbox": [ - 179.98, - 563.52, - 236.15, - 577.91 - ], - "text": "x[n]ejcn ⇐⇒", - "type": "text" - }, - { - "block_id": "p891-b22", - "global_id": 26068, - "bbox": [ - 241.88, - 557.46, - 255.98, - 568.13 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p891-b23", - "global_id": 26069, - "bbox": [ - 238.19, - 581.48, - 259.67, - 588.67 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p891-b24", - "global_id": 26070, - "bbox": [ - 260.78, - 563.52, - 326.81, - 577.91 - ], - "text": "x[n]ejcne−jn =", - "type": "text" - }, - { - "block_id": "p891-b25", - "global_id": 26071, - "bbox": [ - 332.55, - 557.46, - 346.65, - 568.13 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p891-b26", - "global_id": 26072, - "bbox": [ - 328.86, - 581.48, - 350.34, - 588.67 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p891-b27", - "global_id": 26073, - "bbox": [ - 351.45, - 563.52, - 463.74, - 578.71 - ], - "text": "x[n]e−j(−c)n = X( −c)", - "type": "text" - }, - { - "block_id": "p891-b28", - "global_id": 26074, - "bbox": [ - 145.52, - 601.13, - 268.71, - 611.09 - ], - "text": "From this result, it follows that", - "type": "text" - }, - { - "block_id": "p891-b29", - "global_id": 26075, - "bbox": [ - 268.08, - 620.71, - 375.65, - 635.91 - ], - "text": "x[n]e−jcn ⇐⇒X( + c)", - "type": "text" - } - ] - }, - { - "page_num": 892, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p892-b0", - "global_id": 26076, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "872\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p892-b1", - "global_id": 26077, - "bbox": [ - 101.84, - 85.82, - 306.32, - 95.78 - ], - "text": "Adding this pair to the pair in Eq. (9.32), we obtain", - "type": "text" - }, - { - "block_id": "p892-b2", - "global_id": 26078, - "bbox": [ - 200.86, - 110.69, - 284.09, - 123.12 - ], - "text": "x[n]cos(cn) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p892-b3", - "global_id": 26079, - "bbox": [ - 280.6, - 112.04, - 490.38, - 125.23 - ], - "text": "2{X( −c) + X( + c)}\n(9.33)", - "type": "text" - }, - { - "block_id": "p892-b4", - "global_id": 26080, - "bbox": [ - 101.85, - 138.99, - 230.08, - 149.05 - ], - "text": "This is the modulation property.", - "type": "text" - }, - { - "block_id": "p892-b5", - "global_id": 26081, - "bbox": [ - 119.78, - 149.4, - 334.9, - 161.0 - ], - "text": "Multiplying both sides of pair (9.32) by ejθ, we obtain", - "type": "text" - }, - { - "block_id": "p892-b6", - "global_id": 26082, - "bbox": [ - 232.53, - 173.14, - 358.5, - 188.34 - ], - "text": "x[n]ej(cn+θ) ⇐⇒X( −c)ejθ", - "type": "text" - }, - { - "block_id": "p892-b7", - "global_id": 26083, - "bbox": [ - 101.84, - 204.31, - 346.03, - 214.27 - ], - "text": "Using this pair, we can generalize the modulation property as", - "type": "text" - }, - { - "block_id": "p892-b8", - "global_id": 26084, - "bbox": [ - 179.39, - 229.18, - 278.86, - 241.61 - ], - "text": "x[n]cos(cn + θ) ⇐⇒1", - "type": "text" - }, - { - "block_id": "p892-b9", - "global_id": 26085, - "bbox": [ - 275.37, - 226.91, - 412.84, - 243.72 - ], - "text": "2{X( −c)ejθ + X( + c)e−jθ}", - "type": "text" - }, - { - "block_id": "p892-b10", - "global_id": 26086, - "bbox": [ - 76.77, - 288.31, - 296.18, - 300.27 - ], - "text": "EXAMPLE 9.10\nModulation Property", - "type": "text" - }, - { - "block_id": "p892-b11", - "global_id": 26087, - "bbox": [ - 103.16, - 316.52, - 477.01, - 339.55 - ], - "text": "A signal x[n] = sinc(πn/4) modulates a carrier coscn. Find and sketch the spectrum of the\nmodulated signal x[n]coscn for", - "type": "text" - }, - { - "block_id": "p892-b12", - "global_id": 26088, - "bbox": [ - 121.09, - 346.41, - 222.83, - 372.81 - ], - "text": "(a) c = π/2\n(b) c = 7π/8 = 0.875π", - "type": "text" - }, - { - "block_id": "p892-b13", - "global_id": 26089, - "bbox": [ - 121.09, - 400.2, - 334.36, - 410.58 - ], - "text": "(a) For x[n] = sinc(πn/4), we find (Table 9.1, pair 8)", - "type": "text" - }, - { - "block_id": "p892-b14", - "global_id": 26090, - "bbox": [ - 223.36, - 430.25, - 261.92, - 440.63 - ], - "text": "X() = 4", - "type": "text" - }, - { - "block_id": "p892-b15", - "global_id": 26091, - "bbox": [ - 267.5, - 420.08, - 281.59, - 430.75 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p892-b16", - "global_id": 26092, - "bbox": [ - 263.02, - 444.1, - 286.05, - 451.29 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p892-b17", - "global_id": 26093, - "bbox": [ - 287.17, - 430.67, - 302.1, - 440.63 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p892-b18", - "global_id": 26094, - "bbox": [ - 303.21, - 416.26, - 348.88, - 433.65 - ], - "text": "−2πm", - "type": "text" - }, - { - "block_id": "p892-b19", - "global_id": 26095, - "bbox": [ - 321.95, - 437.33, - 338.08, - 447.7 - ], - "text": "π/2", - "type": "text" - }, - { - "block_id": "p892-b21", - "global_id": 26096, - "bbox": [ - 103.17, - 461.2, - 469.68, - 471.57 - ], - "text": "Figure 9.11a shows the DTFT X(). From the modulation property of Eq. (9.33), we obtain", - "type": "text" - }, - { - "block_id": "p892-b22", - "global_id": 26097, - "bbox": [ - 126.29, - 491.24, - 217.89, - 501.61 - ], - "text": "x[n]cos(0.5πn) ⇐⇒2", - "type": "text" - }, - { - "block_id": "p892-b23", - "global_id": 26098, - "bbox": [ - 223.46, - 481.06, - 237.56, - 491.74 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p892-b24", - "global_id": 26099, - "bbox": [ - 218.99, - 505.09, - 242.01, - 512.28 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p892-b25", - "global_id": 26100, - "bbox": [ - 243.13, - 491.65, - 258.06, - 501.61 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p892-b26", - "global_id": 26101, - "bbox": [ - 259.17, - 477.25, - 335.15, - 494.63 - ], - "text": "−0.5π −2πm", - "type": "text" - }, - { - "block_id": "p892-b27", - "global_id": 26102, - "bbox": [ - 291.42, - 498.31, - 309.84, - 508.69 - ], - "text": "0.5π", - "type": "text" - }, - { - "block_id": "p892-b29", - "global_id": 26103, - "bbox": [ - 344.63, - 491.24, - 368.88, - 501.61 - ], - "text": "+ rect", - "type": "text" - }, - { - "block_id": "p892-b30", - "global_id": 26104, - "bbox": [ - 370.0, - 477.25, - 445.97, - 494.63 - ], - "text": "+ 0.5π −2πm", - "type": "text" - }, - { - "block_id": "p892-b31", - "global_id": 26105, - "bbox": [ - 402.23, - 498.31, - 420.66, - 508.69 - ], - "text": "0.5π", - "type": "text" - }, - { - "block_id": "p892-b33", - "global_id": 26106, - "bbox": [ - 103.17, - 522.19, - 477.02, - 556.47 - ], - "text": "Figure 9.11b shows half the X() shifted by π/2 and Fig. 9.11c shows half the X() shifted by\n−π/2. The spectrum of the modulated signal is obtained by adding these two shifted spectra\nand multiplying by half, as shown in Fig. 9.11d.", - "type": "text" - }, - { - "block_id": "p892-b34", - "global_id": 26107, - "bbox": [ - 103.17, - 558.05, - 477.02, - 580.38 - ], - "text": "(b) Figure 9.12a shows X(), which is the same as that in part (a). For c = 7π/8 =\n0.875π, the modulation property of Eq. (9.33) yields", - "type": "text" - }, - { - "block_id": "p892-b35", - "global_id": 26108, - "bbox": [ - 110.79, - 600.05, - 212.35, - 610.42 - ], - "text": "x[n]cos(0.875πn) ⇐⇒2", - "type": "text" - }, - { - "block_id": "p892-b36", - "global_id": 26109, - "bbox": [ - 217.92, - 589.87, - 232.02, - 600.55 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p892-b37", - "global_id": 26110, - "bbox": [ - 213.45, - 613.9, - 236.47, - 621.09 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p892-b38", - "global_id": 26111, - "bbox": [ - 237.6, - 600.46, - 252.53, - 610.42 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p892-b39", - "global_id": 26112, - "bbox": [ - 253.64, - 586.06, - 339.58, - 603.44 - ], - "text": "−0.875π −2πm", - "type": "text" - }, - { - "block_id": "p892-b40", - "global_id": 26113, - "bbox": [ - 290.86, - 607.12, - 309.28, - 617.5 - ], - "text": "0.5π", - "type": "text" - }, - { - "block_id": "p892-b42", - "global_id": 26114, - "bbox": [ - 349.06, - 600.05, - 374.41, - 610.42 - ], - "text": "+ rect", - "type": "text" - }, - { - "block_id": "p892-b43", - "global_id": 26115, - "bbox": [ - 375.52, - 586.06, - 461.46, - 603.44 - ], - "text": "+ 0.875π −2πm", - "type": "text" - }, - { - "block_id": "p892-b44", - "global_id": 26116, - "bbox": [ - 412.75, - 607.12, - 431.17, - 617.5 - ], - "text": "0.5π", - "type": "text" - } - ] - }, - { - "page_num": 893, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p893-b0", - "global_id": 26117, - "bbox": [ - 362.61, - 62.89, - 516.12, - 71.98 - ], - "text": "9.3\nProperties of the DTFT\n873", - "type": "text" - }, - { - "block_id": "p893-b1", - "global_id": 26118, - "bbox": [ - 299.62, - 98.35, - 303.62, - 106.35 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p893-b2", - "global_id": 26119, - "bbox": [ - 301.18, - 169.94, - 310.06, - 177.94 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p893-b3", - "global_id": 26120, - "bbox": [ - 299.62, - 208.79, - 303.62, - 216.79 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p893-b4", - "global_id": 26121, - "bbox": [ - 300.03, - 273.12, - 309.84, - 281.12 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p893-b5", - "global_id": 26122, - "bbox": [ - 299.62, - 311.4, - 303.62, - 319.4 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p893-b6", - "global_id": 26123, - "bbox": [ - 301.18, - 375.64, - 310.06, - 383.64 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p893-b7", - "global_id": 26124, - "bbox": [ - 313.2, - 397.77, - 424.07, - 406.07 - ], - "text": "0.5[X(\t 0.5p) X(\t 0.5p)]", - "type": "text" - }, - { - "block_id": "p893-b8", - "global_id": 26125, - "bbox": [ - 313.2, - 295.45, - 355.64, - 303.74 - ], - "text": "X(\t 0.5p)", - "type": "text" - }, - { - "block_id": "p893-b9", - "global_id": 26126, - "bbox": [ - 313.2, - 198.46, - 355.64, - 206.75 - ], - "text": "X(\t 0.5p)", - "type": "text" - }, - { - "block_id": "p893-b10", - "global_id": 26127, - "bbox": [ - 313.2, - 90.2, - 329.64, - 98.49 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p893-b11", - "global_id": 26128, - "bbox": [ - 299.62, - 413.09, - 303.62, - 421.09 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p893-b12", - "global_id": 26129, - "bbox": [ - 307.22, - 455.13, - 311.22, - 463.13 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p893-b13", - "global_id": 26130, - "bbox": [ - 300.96, - 477.17, - 310.28, - 485.17 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p893-b14", - "global_id": 26131, - "bbox": [ - 299.62, - 430.44, - 303.62, - 438.44 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p893-b15", - "global_id": 26132, - "bbox": [ - 173.27, - 453.13, - 480.24, - 471.46 - ], - "text": "2p\n2p\np\n\t\n2\n5p", - "type": "text" - }, - { - "block_id": "p893-b16", - "global_id": 26133, - "bbox": [ - 456.39, - 250.42, - 465.72, - 266.81 - ], - "text": "2\n5p", - "type": "text" - }, - { - "block_id": "p893-b17", - "global_id": 26134, - "bbox": [ - 334.17, - 453.13, - 403.55, - 471.46 - ], - "text": "2\n3p\n2\np", - "type": "text" - }, - { - "block_id": "p893-b18", - "global_id": 26135, - "bbox": [ - 268.23, - 418.79, - 280.84, - 427.09 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p893-b19", - "global_id": 26136, - "bbox": [ - 141.24, - 452.94, - 323.64, - 471.46 - ], - "text": "4\np\n4\np\n2\n3p\n\n2\n5p\n\np", - "type": "text" - }, - { - "block_id": "p893-b20", - "global_id": 26137, - "bbox": [ - 203.41, - 248.37, - 249.45, - 266.84 - ], - "text": "2\n3p\n\np", - "type": "text" - }, - { - "block_id": "p893-b21", - "global_id": 26138, - "bbox": [ - 266.72, - 453.13, - 279.36, - 469.52 - ], - "text": "2\np", - "type": "text" - }, - { - "block_id": "p893-b22", - "global_id": 26139, - "bbox": [ - 173.27, - 145.17, - 480.24, - 161.74 - ], - "text": "0\n2p\n2p\np\n\t\n4\np\n4\np\np", - "type": "text" - }, - { - "block_id": "p893-b23", - "global_id": 26140, - "bbox": [ - 173.27, - 351.16, - 480.24, - 369.42 - ], - "text": "0\n2p\n2p\np\n\t\n2\n3p", - "type": "text" - }, - { - "block_id": "p893-b24", - "global_id": 26141, - "bbox": [ - 268.23, - 299.76, - 280.84, - 308.05 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p893-b25", - "global_id": 26142, - "bbox": [ - 141.24, - 350.91, - 323.64, - 369.42 - ], - "text": "4\np\n4\np\n2\n5p\n\np", - "type": "text" - }, - { - "block_id": "p893-b26", - "global_id": 26143, - "bbox": [ - 266.72, - 351.1, - 279.36, - 367.5 - ], - "text": "2\np", - "type": "text" - }, - { - "block_id": "p893-b27", - "global_id": 26144, - "bbox": [ - 173.27, - 248.54, - 480.24, - 259.37 - ], - "text": "0\n2p\n2p\np", - "type": "text" - }, - { - "block_id": "p893-b28", - "global_id": 26145, - "bbox": [ - 206.06, - 197.15, - 218.66, - 205.44 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p893-b29", - "global_id": 26146, - "bbox": [ - 174.97, - 96.05, - 187.58, - 104.35 - ], - "text": "p2", - "type": "text" - }, - { - "block_id": "p893-b30", - "global_id": 26147, - "bbox": [ - 283.27, - 248.54, - 323.64, - 264.93 - ], - "text": "4\np\n4\np", - "type": "text" - }, - { - "block_id": "p893-b31", - "global_id": 26148, - "bbox": [ - 111.17, - 491.87, - 295.2, - 501.11 - ], - "text": "Figure 9.11 Instance of modulation for Ex. 9.10a.", - "type": "text" - }, - { - "block_id": "p893-b32", - "global_id": 26149, - "bbox": [ - 128.91, - 542.32, - 502.76, - 613.56 - ], - "text": "Figure 9.12b shows X() shifted by 7π/8 and Fig. 9.12c shows X() shifted by −7π/8.\nThe spectrum of the modulated signal is obtained by adding these two shifted spectra and\nmultiplying by half, as shown in Fig. 9.12d. In this case, the two shifted spectra overlap.\nSince the operation of modulation thus causes aliasing, it does not achieve the desired effect\nof spectral shifting. In this example, to realize spectral shifting without aliasing requires\nc ≤3π/4.", - "type": "text" - } - ] - }, - { - "page_num": 894, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p894-b0", - "global_id": 26150, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "874\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p894-b1", - "global_id": 26151, - "bbox": [ - 266.34, - 111.51, - 270.34, - 119.51 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p894-b2", - "global_id": 26152, - "bbox": [ - 267.9, - 180.36, - 276.78, - 188.36 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p894-b3", - "global_id": 26153, - "bbox": [ - 266.34, - 216.78, - 270.34, - 224.78 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p894-b4", - "global_id": 26154, - "bbox": [ - 267.44, - 280.13, - 277.25, - 288.13 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p894-b5", - "global_id": 26155, - "bbox": [ - 266.34, - 316.73, - 270.34, - 324.73 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p894-b6", - "global_id": 26156, - "bbox": [ - 267.9, - 379.43, - 276.78, - 387.43 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p894-b7", - "global_id": 26157, - "bbox": [ - 266.34, - 409.69, - 270.34, - 417.69 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p894-b8", - "global_id": 26158, - "bbox": [ - 267.68, - 480.11, - 277.01, - 488.11 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p894-b9", - "global_id": 26159, - "bbox": [ - 266.34, - 433.91, - 270.34, - 441.91 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p894-b10", - "global_id": 26160, - "bbox": [ - 83.09, - 454.67, - 464.21, - 473.12 - ], - "text": "0\n2p\np\n3p\n3p\n2p\n\t\n4\np\n8\n5p\n\n8\n5p\n8\n11p", - "type": "text" - }, - { - "block_id": "p894-b11", - "global_id": 26161, - "bbox": [ - 174.74, - 257.98, - 194.83, - 274.38 - ], - "text": "8\n11p", - "type": "text" - }, - { - "block_id": "p894-b12", - "global_id": 26162, - "bbox": [ - 201.65, - 454.49, - 365.08, - 473.02 - ], - "text": "8\n11p\np", - "type": "text" - }, - { - "block_id": "p894-b13", - "global_id": 26163, - "bbox": [ - 79.29, - 356.05, - 464.21, - 372.44 - ], - "text": "0\n2p\n3p\n3p\n2p\n\t\n4\np", - "type": "text" - }, - { - "block_id": "p894-b14", - "global_id": 26164, - "bbox": [ - 188.57, - 355.98, - 365.08, - 374.31 - ], - "text": "4\np\n8\n5p\n\n8\n9p\n\n8\n7p\n8\n9p\n8\n11p", - "type": "text" - }, - { - "block_id": "p894-b15", - "global_id": 26165, - "bbox": [ - 79.29, - 256.02, - 464.21, - 272.42 - ], - "text": "0\n2p\n3p\n3p\n2p\n\t\n4\np", - "type": "text" - }, - { - "block_id": "p894-b16", - "global_id": 26166, - "bbox": [ - 207.75, - 256.07, - 347.03, - 274.38 - ], - "text": "4\np\n8\n7p\n\n8\n7p\n8\n5p\n8\n9p", - "type": "text" - }, - { - "block_id": "p894-b17", - "global_id": 26167, - "bbox": [ - 137.33, - 156.27, - 424.69, - 172.91 - ], - "text": "0\n2p\np\n2p\np\n\t\n4\np", - "type": "text" - }, - { - "block_id": "p894-b18", - "global_id": 26168, - "bbox": [ - 285.73, - 156.46, - 291.07, - 172.91 - ], - "text": "4\np", - "type": "text" - }, - { - "block_id": "p894-b19", - "global_id": 26169, - "bbox": [ - 279.65, - 201.03, - 314.33, - 217.42 - ], - "text": "8\n7p\nX \t \n(", - "type": "text" - }, - { - "block_id": "p894-b20", - "global_id": 26170, - "bbox": [ - 315.3, - 204.52, - 317.97, - 216.25 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p894-b21", - "global_id": 26171, - "bbox": [ - 279.65, - 301.37, - 314.33, - 317.76 - ], - "text": "8\n7p\nX \t \n(", - "type": "text" - }, - { - "block_id": "p894-b22", - "global_id": 26172, - "bbox": [ - 279.65, - 400.93, - 340.04, - 417.33 - ], - "text": "8\n7p\n0.5 X \t \n\n[", - "type": "text" - }, - { - "block_id": "p894-b23", - "global_id": 26173, - "bbox": [ - 380.97, - 403.85, - 383.63, - 415.58 - ], - "text": "[", - "type": "text" - }, - { - "block_id": "p894-b24", - "global_id": 26174, - "bbox": [ - 298.17, - 403.85, - 300.83, - 415.58 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p894-b25", - "global_id": 26175, - "bbox": [ - 328.99, - 403.85, - 331.66, - 415.58 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p894-b26", - "global_id": 26176, - "bbox": [ - 341.66, - 400.93, - 377.28, - 417.33 - ], - "text": "8\n7p\nX \t \n(", - "type": "text" - }, - { - "block_id": "p894-b27", - "global_id": 26177, - "bbox": [ - 378.35, - 403.85, - 381.01, - 415.58 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p894-b28", - "global_id": 26178, - "bbox": [ - 315.3, - 304.86, - 317.97, - 316.59 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p894-b29", - "global_id": 26179, - "bbox": [ - 279.65, - 102.12, - 296.09, - 110.42 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p894-b30", - "global_id": 26180, - "bbox": [ - 67.55, - 494.8, - 251.72, - 504.04 - ], - "text": "Figure 9.12 Instance of modulation for Ex. 9.10b.", - "type": "text" - }, - { - "block_id": "p894-b31", - "global_id": 26181, - "bbox": [ - 107.82, - 579.49, - 395.04, - 591.45 - ], - "text": "DRILL 9.7\nUsing the Frequency-Shifting Property", - "type": "text" - }, - { - "block_id": "p894-b32", - "global_id": 26182, - "bbox": [ - 107.82, - 600.57, - 484.4, - 622.49 - ], - "text": "In Table 9.1, derive pairs 12 and 13 from pair 11 and the frequency-shifting/modulation\nproperty.", - "type": "text" - } - ] - }, - { - "page_num": 895, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p895-b0", - "global_id": 26183, - "bbox": [ - 362.61, - 62.89, - 516.12, - 71.98 - ], - "text": "9.3\nProperties of the DTFT\n875", - "type": "text" - }, - { - "block_id": "p895-b1", - "global_id": 26184, - "bbox": [ - 127.89, - 86.19, - 401.39, - 98.32 - ], - "text": "TIME- AND FREQUENCY-CONVOLUTION PROPERTY", - "type": "text" - }, - { - "block_id": "p895-b2", - "global_id": 26185, - "bbox": [ - 127.59, - 102.35, - 134.23, - 112.31 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p895-b3", - "global_id": 26186, - "bbox": [ - 226.24, - 114.36, - 417.47, - 125.51 - ], - "text": "x1[n] ⇐⇒X1()\nand\nx2[n] ⇐⇒X2()", - "type": "text" - }, - { - "block_id": "p895-b4", - "global_id": 26187, - "bbox": [ - 127.59, - 134.02, - 144.74, - 143.98 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p895-b5", - "global_id": 26188, - "bbox": [ - 260.83, - 146.03, - 516.13, - 157.18 - ], - "text": "x1[n] ∗x2[n] ⇐⇒X1()X2()\n(9.34)", - "type": "text" - }, - { - "block_id": "p895-b6", - "global_id": 26189, - "bbox": [ - 127.59, - 165.69, - 141.97, - 175.65 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p895-b7", - "global_id": 26190, - "bbox": [ - 252.82, - 174.04, - 325.75, - 191.76 - ], - "text": "x1[n]x2[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p895-b8", - "global_id": 26191, - "bbox": [ - 317.28, - 180.61, - 516.12, - 198.06 - ], - "text": "2π X1() ∗⃝X2()\n(9.35)", - "type": "text" - }, - { - "block_id": "p895-b9", - "global_id": 26192, - "bbox": [ - 127.59, - 203.32, - 151.92, - 213.28 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p895-b10", - "global_id": 26193, - "bbox": [ - 249.52, - 221.41, - 307.69, - 232.56 - ], - "text": "x1[n] ∗x2[n] =", - "type": "text" - }, - { - "block_id": "p895-b11", - "global_id": 26194, - "bbox": [ - 314.2, - 211.24, - 328.3, - 221.9 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p895-b12", - "global_id": 26195, - "bbox": [ - 309.73, - 235.26, - 332.75, - 242.45 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p895-b13", - "global_id": 26196, - "bbox": [ - 333.87, - 221.41, - 394.19, - 232.56 - ], - "text": "x1[m]x2[n −m]", - "type": "text" - }, - { - "block_id": "p895-b14", - "global_id": 26197, - "bbox": [ - 127.59, - 250.03, - 516.13, - 271.95 - ], - "text": "For two continuous, periodic signals, we define the periodic convolution, denoted by\nsymbol ∗⃝as†", - "type": "text" - }, - { - "block_id": "p895-b15", - "global_id": 26198, - "bbox": [ - 233.02, - 272.4, - 315.02, - 290.12 - ], - "text": "X1() ∗⃝X2() = 1", - "type": "text" - }, - { - "block_id": "p895-b16", - "global_id": 26199, - "bbox": [ - 306.55, - 286.05, - 317.5, - 296.42 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p895-b17", - "global_id": 26200, - "bbox": [ - 320.81, - 265.41, - 326.07, - 275.37 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p895-b18", - "global_id": 26201, - "bbox": [ - 326.07, - 278.97, - 410.69, - 297.55 - ], - "text": "2π\nX1(u)X2( −u)du", - "type": "text" - }, - { - "block_id": "p895-b19", - "global_id": 26202, - "bbox": [ - 127.59, - 303.39, - 516.13, - 337.36 - ], - "text": "The convolution here is not the linear convolution used so far. This is a periodic (or circular)\nconvolution applicable to the convolution of two continuous, periodic functions with the same\nperiod. The limit of integration in the convolution extends only to one period.", - "type": "text" - }, - { - "block_id": "p895-b20", - "global_id": 26203, - "bbox": [ - 127.59, - 339.35, - 516.15, - 373.22 - ], - "text": "Proof of the time-convolution property is identical to that given in Sec. 5.2 [Eq. (5.19)]. All\nwe have to do is replace z with ej. To prove the frequency-convolution property of Eq. (9.35), we\nhave", - "type": "text" - }, - { - "block_id": "p895-b21", - "global_id": 26204, - "bbox": [ - 162.15, - 391.15, - 223.38, - 402.3 - ], - "text": "x1[n]x2[n] ⇐⇒", - "type": "text" - }, - { - "block_id": "p895-b22", - "global_id": 26205, - "bbox": [ - 229.12, - 380.98, - 243.22, - 391.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p895-b23", - "global_id": 26206, - "bbox": [ - 225.42, - 405.0, - 246.9, - 412.19 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p895-b24", - "global_id": 26207, - "bbox": [ - 248.01, - 387.04, - 319.1, - 402.3 - ], - "text": "x1[n]x2[n]e−jn =", - "type": "text" - }, - { - "block_id": "p895-b25", - "global_id": 26208, - "bbox": [ - 324.84, - 380.98, - 338.93, - 391.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p895-b26", - "global_id": 26209, - "bbox": [ - 321.15, - 405.0, - 342.62, - 412.19 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p895-b27", - "global_id": 26210, - "bbox": [ - 343.74, - 391.15, - 363.76, - 402.3 - ], - "text": "x2[n]", - "type": "text" - }, - { - "block_id": "p895-b28", - "global_id": 26211, - "bbox": [ - 364.87, - 377.16, - 379.97, - 394.54 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p895-b29", - "global_id": 26212, - "bbox": [ - 371.49, - 398.22, - 382.45, - 408.6 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p895-b30", - "global_id": 26213, - "bbox": [ - 385.75, - 377.59, - 391.01, - 387.55 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p895-b31", - "global_id": 26214, - "bbox": [ - 391.01, - 387.35, - 453.81, - 409.73 - ], - "text": "2π\nX1(u)e−jnu du", - "type": "text" - }, - { - "block_id": "p895-b32", - "global_id": 26215, - "bbox": [ - 453.81, - 377.16, - 459.24, - 387.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p895-b33", - "global_id": 26216, - "bbox": [ - 460.35, - 389.43, - 481.05, - 401.43 - ], - "text": "e−jn", - "type": "text" - }, - { - "block_id": "p895-b34", - "global_id": 26217, - "bbox": [ - 127.59, - 422.77, - 385.02, - 432.73 - ], - "text": "Interchanging the order of summation and integration, we obtain", - "type": "text" - }, - { - "block_id": "p895-b35", - "global_id": 26218, - "bbox": [ - 153.77, - 446.64, - 226.71, - 464.36 - ], - "text": "x1[n]x2[n] ⇐⇒1", - "type": "text" - }, - { - "block_id": "p895-b36", - "global_id": 26219, - "bbox": [ - 218.24, - 460.29, - 229.2, - 470.66 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p895-b37", - "global_id": 26220, - "bbox": [ - 232.5, - 439.66, - 237.76, - 449.62 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p895-b38", - "global_id": 26221, - "bbox": [ - 237.76, - 453.21, - 270.22, - 471.8 - ], - "text": "2π\nX1(u)", - "type": "text" - }, - { - "block_id": "p895-b39", - "global_id": 26222, - "bbox": [ - 271.32, - 436.24, - 295.3, - 453.72 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p895-b40", - "global_id": 26223, - "bbox": [ - 277.52, - 467.06, - 298.99, - 474.26 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p895-b41", - "global_id": 26224, - "bbox": [ - 300.11, - 451.49, - 354.96, - 464.36 - ], - "text": "x2[n]e−j(−u)n", - "type": "text" - }, - { - "block_id": "p895-b43", - "global_id": 26225, - "bbox": [ - 362.76, - 446.64, - 394.26, - 463.59 - ], - "text": "du = 1", - "type": "text" - }, - { - "block_id": "p895-b44", - "global_id": 26226, - "bbox": [ - 385.79, - 460.29, - 396.75, - 470.66 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p895-b45", - "global_id": 26227, - "bbox": [ - 400.05, - 439.66, - 405.31, - 449.62 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p895-b46", - "global_id": 26228, - "bbox": [ - 405.31, - 453.21, - 489.93, - 471.8 - ], - "text": "2π\nX1(u)X2( −u)du", - "type": "text" - }, - { - "block_id": "p895-b47", - "global_id": 26229, - "bbox": [ - 102.51, - 504.04, - 386.02, - 516.0 - ], - "text": "EXAMPLE 9.11\nDTFT of an Accumulator System", - "type": "text" - }, - { - "block_id": "p895-b48", - "global_id": 26230, - "bbox": [ - 128.9, - 538.22, - 261.87, - 548.6 - ], - "text": "If x[n] ⇐⇒X(), then show that", - "type": "text" - }, - { - "block_id": "p895-b49", - "global_id": 26231, - "bbox": [ - 267.92, - 528.27, - 282.02, - 538.72 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p895-b50", - "global_id": 26232, - "bbox": [ - 264.36, - 552.51, - 285.57, - 559.7 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p895-b51", - "global_id": 26233, - "bbox": [ - 286.68, - 538.22, - 351.49, - 548.6 - ], - "text": "x[k] ⇐⇒πX(0)", - "type": "text" - }, - { - "block_id": "p895-b52", - "global_id": 26234, - "bbox": [ - 356.16, - 528.05, - 370.26, - 538.72 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p895-b53", - "global_id": 26235, - "bbox": [ - 352.6, - 552.51, - 373.8, - 559.7 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p895-b54", - "global_id": 26236, - "bbox": [ - 374.93, - 530.01, - 454.12, - 548.6 - ], - "text": "δ( −2πk) +\nej", - "type": "text" - }, - { - "block_id": "p895-b55", - "global_id": 26237, - "bbox": [ - 434.41, - 538.22, - 487.95, - 555.67 - ], - "text": "ej −1X().", - "type": "text" - }, - { - "block_id": "p895-b56", - "global_id": 26238, - "bbox": [ - 127.59, - 599.27, - 516.11, - 633.41 - ], - "text": "† In Eq. (8.20), we defined periodic convolution for two discrete, periodic sequences in a different way.\nAlthough we are using the same symbol ∗⃝for both discrete and continuous cases, the meaning will be clear\nfrom the context.", - "type": "text" - } - ] - }, - { - "page_num": 896, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p896-b0", - "global_id": 26239, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "876\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p896-b1", - "global_id": 26240, - "bbox": [ - 103.16, - 86.24, - 198.77, - 96.2 - ], - "text": "To begin, we notice that", - "type": "text" - }, - { - "block_id": "p896-b2", - "global_id": 26241, - "bbox": [ - 203.71, - 115.86, - 254.49, - 126.14 - ], - "text": "x[n] ∗u[n] =", - "type": "text" - }, - { - "block_id": "p896-b3", - "global_id": 26242, - "bbox": [ - 260.1, - 105.69, - 274.19, - 116.37 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p896-b4", - "global_id": 26243, - "bbox": [ - 256.53, - 130.16, - 277.74, - 137.36 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p896-b5", - "global_id": 26244, - "bbox": [ - 278.85, - 115.86, - 336.42, - 126.14 - ], - "text": "x[k]u[n −k] =", - "type": "text" - }, - { - "block_id": "p896-b6", - "global_id": 26245, - "bbox": [ - 342.02, - 105.91, - 356.12, - 116.37 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p896-b7", - "global_id": 26246, - "bbox": [ - 338.47, - 130.16, - 359.67, - 137.36 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p896-b8", - "global_id": 26247, - "bbox": [ - 360.79, - 115.86, - 376.46, - 126.14 - ], - "text": "x[k]", - "type": "text" - }, - { - "block_id": "p896-b9", - "global_id": 26248, - "bbox": [ - 103.17, - 147.61, - 472.58, - 157.57 - ], - "text": "Applying the time-convolution property of Eq. (9.34) and pair 10 in Table 9.1, it follows that", - "type": "text" - }, - { - "block_id": "p896-b10", - "global_id": 26249, - "bbox": [ - 158.7, - 167.8, - 172.8, - 178.24 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p896-b11", - "global_id": 26250, - "bbox": [ - 155.15, - 192.04, - 176.35, - 199.23 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p896-b12", - "global_id": 26251, - "bbox": [ - 177.47, - 177.75, - 290.89, - 188.02 - ], - "text": "x[k] = x[n] ∗u[n] ⇐⇒X()", - "type": "text" - }, - { - "block_id": "p896-b13", - "global_id": 26252, - "bbox": [ - 292.0, - 160.77, - 298.91, - 170.74 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p896-b14", - "global_id": 26253, - "bbox": [ - 298.91, - 177.75, - 304.88, - 187.71 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p896-b15", - "global_id": 26254, - "bbox": [ - 310.55, - 167.58, - 324.65, - 178.24 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p896-b16", - "global_id": 26255, - "bbox": [ - 306.99, - 192.04, - 328.19, - 199.23 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p896-b17", - "global_id": 26256, - "bbox": [ - 329.31, - 169.54, - 408.51, - 188.12 - ], - "text": "δ( −2πk) +\nej", - "type": "text" - }, - { - "block_id": "p896-b18", - "global_id": 26257, - "bbox": [ - 388.8, - 181.94, - 416.93, - 195.2 - ], - "text": "ej −1", - "type": "text" - }, - { - "block_id": "p896-b19", - "global_id": 26258, - "bbox": [ - 418.13, - 160.77, - 425.03, - 170.74 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p896-b20", - "global_id": 26259, - "bbox": [ - 103.17, - 209.13, - 477.03, - 231.47 - ], - "text": "Because of 2π periodicity, X(0) = X(2πk). Moreover, X()δ( −2πk) = X(2πk)δ( −\n2πk) = X(0)δ( −2πk). Hence,", - "type": "text" - }, - { - "block_id": "p896-b21", - "global_id": 26260, - "bbox": [ - 183.1, - 240.76, - 197.2, - 251.21 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p896-b22", - "global_id": 26261, - "bbox": [ - 179.55, - 265.0, - 200.75, - 272.19 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p896-b23", - "global_id": 26262, - "bbox": [ - 201.87, - 250.71, - 266.68, - 261.09 - ], - "text": "x[k] ⇐⇒πX(0)", - "type": "text" - }, - { - "block_id": "p896-b24", - "global_id": 26263, - "bbox": [ - 271.34, - 240.54, - 285.44, - 251.21 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p896-b25", - "global_id": 26264, - "bbox": [ - 267.78, - 265.0, - 288.99, - 272.19 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p896-b26", - "global_id": 26265, - "bbox": [ - 290.1, - 242.5, - 369.3, - 261.09 - ], - "text": "δ( −2πk) +\nej", - "type": "text" - }, - { - "block_id": "p896-b27", - "global_id": 26266, - "bbox": [ - 349.59, - 250.71, - 400.63, - 268.16 - ], - "text": "ej −1X()", - "type": "text" - }, - { - "block_id": "p896-b28", - "global_id": 26267, - "bbox": [ - 107.82, - 346.89, - 418.96, - 358.84 - ], - "text": "DRILL 9.8\nUsing the Frequency-Convolution Property", - "type": "text" - }, - { - "block_id": "p896-b29", - "global_id": 26268, - "bbox": [ - 107.82, - 367.55, - 484.36, - 389.88 - ], - "text": "In Table 9.1, derive pair 9 from pair 8, assuming c ≤π/2. Use the frequency-convolution\nproperty.", - "type": "text" - }, - { - "block_id": "p896-b30", - "global_id": 26269, - "bbox": [ - 101.84, - 425.31, - 222.16, - 451.42 - ], - "text": "PARSEVAL’S THEOREM\nIf", - "type": "text" - }, - { - "block_id": "p896-b31", - "global_id": 26270, - "bbox": [ - 265.62, - 466.78, - 326.61, - 477.05 - ], - "text": "x[n] ⇐⇒X()", - "type": "text" - }, - { - "block_id": "p896-b32", - "global_id": 26271, - "bbox": [ - 101.84, - 494.89, - 255.43, - 505.97 - ], - "text": "then Ex, the energy of x[n], is given by", - "type": "text" - }, - { - "block_id": "p896-b33", - "global_id": 26272, - "bbox": [ - 217.13, - 534.12, - 236.65, - 545.58 - ], - "text": "Ex =", - "type": "text" - }, - { - "block_id": "p896-b34", - "global_id": 26273, - "bbox": [ - 242.4, - 523.95, - 256.49, - 534.62 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p896-b35", - "global_id": 26274, - "bbox": [ - 238.7, - 547.97, - 260.17, - 555.17 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p896-b36", - "global_id": 26275, - "bbox": [ - 261.3, - 527.55, - 308.58, - 544.5 - ], - "text": "|x[n]|2 = 1", - "type": "text" - }, - { - "block_id": "p896-b37", - "global_id": 26276, - "bbox": [ - 300.11, - 541.2, - 311.07, - 551.57 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p896-b38", - "global_id": 26277, - "bbox": [ - 314.37, - 520.56, - 319.63, - 530.52 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p896-b39", - "global_id": 26278, - "bbox": [ - 319.63, - 530.32, - 490.38, - 552.71 - ], - "text": "2π\n|X()|2 d\n(9.36)", - "type": "text" - }, - { - "block_id": "p896-b40", - "global_id": 26279, - "bbox": [ - 101.85, - 574.61, - 293.98, - 584.57 - ], - "text": "To prove this property, we have from Eq. (9.28),", - "type": "text" - }, - { - "block_id": "p896-b41", - "global_id": 26280, - "bbox": [ - 247.98, - 611.7, - 283.64, - 623.7 - ], - "text": "X∗() =", - "type": "text" - }, - { - "block_id": "p896-b42", - "global_id": 26281, - "bbox": [ - 289.38, - 603.25, - 303.47, - 613.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p896-b43", - "global_id": 26282, - "bbox": [ - 285.69, - 627.27, - 307.16, - 634.46 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p896-b44", - "global_id": 26283, - "bbox": [ - 308.28, - 611.7, - 343.75, - 623.7 - ], - "text": "x∗[n]ejn", - "type": "text" - } - ] - }, - { - "page_num": 897, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p897-b0", - "global_id": 26284, - "bbox": [ - 362.61, - 62.89, - 516.12, - 71.98 - ], - "text": "9.3\nProperties of the DTFT\n877", - "type": "text" - }, - { - "block_id": "p897-b1", - "global_id": 26285, - "bbox": [ - 127.59, - 85.54, - 148.54, - 95.5 - ], - "text": "Now,", - "type": "text" - }, - { - "block_id": "p897-b2", - "global_id": 26286, - "bbox": [ - 169.88, - 106.04, - 183.98, - 116.72 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p897-b3", - "global_id": 26287, - "bbox": [ - 166.19, - 130.07, - 187.67, - 137.26 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p897-b4", - "global_id": 26288, - "bbox": [ - 188.78, - 112.1, - 224.36, - 126.49 - ], - "text": "|x[n]|2 =", - "type": "text" - }, - { - "block_id": "p897-b5", - "global_id": 26289, - "bbox": [ - 230.09, - 106.04, - 244.19, - 116.72 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p897-b6", - "global_id": 26290, - "bbox": [ - 226.4, - 130.07, - 247.88, - 137.26 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p897-b7", - "global_id": 26291, - "bbox": [ - 248.99, - 114.5, - 295.09, - 126.49 - ], - "text": "x∗[n]x[n] =", - "type": "text" - }, - { - "block_id": "p897-b8", - "global_id": 26292, - "bbox": [ - 300.83, - 106.04, - 314.92, - 116.72 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p897-b9", - "global_id": 26293, - "bbox": [ - 297.14, - 130.07, - 318.61, - 137.26 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p897-b10", - "global_id": 26294, - "bbox": [ - 319.73, - 114.5, - 339.93, - 126.49 - ], - "text": "x∗[n]", - "type": "text" - }, - { - "block_id": "p897-b11", - "global_id": 26295, - "bbox": [ - 341.04, - 102.23, - 356.13, - 119.61 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p897-b12", - "global_id": 26296, - "bbox": [ - 347.66, - 123.29, - 358.62, - 133.67 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p897-b13", - "global_id": 26297, - "bbox": [ - 361.92, - 102.66, - 367.18, - 112.63 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p897-b14", - "global_id": 26298, - "bbox": [ - 367.18, - 112.42, - 428.74, - 134.8 - ], - "text": "2π\nX()ejn d", - "type": "text" - }, - { - "block_id": "p897-b15", - "global_id": 26299, - "bbox": [ - 428.73, - 102.23, - 434.16, - 112.2 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p897-b16", - "global_id": 26300, - "bbox": [ - 287.32, - 144.22, - 306.8, - 161.16 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p897-b17", - "global_id": 26301, - "bbox": [ - 298.33, - 157.86, - 309.29, - 168.24 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p897-b18", - "global_id": 26302, - "bbox": [ - 312.59, - 137.22, - 317.85, - 147.19 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p897-b19", - "global_id": 26303, - "bbox": [ - 317.84, - 150.79, - 349.53, - 169.37 - ], - "text": "2π\nX()", - "type": "text" - }, - { - "block_id": "p897-b20", - "global_id": 26304, - "bbox": [ - 350.64, - 133.8, - 374.62, - 151.28 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p897-b21", - "global_id": 26305, - "bbox": [ - 356.82, - 164.63, - 378.3, - 171.83 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p897-b22", - "global_id": 26306, - "bbox": [ - 379.41, - 149.06, - 414.89, - 161.06 - ], - "text": "x∗[n]ejn", - "type": "text" - }, - { - "block_id": "p897-b24", - "global_id": 26307, - "bbox": [ - 423.79, - 150.79, - 436.79, - 161.06 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p897-b25", - "global_id": 26308, - "bbox": [ - 287.32, - 175.11, - 306.8, - 192.06 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p897-b26", - "global_id": 26309, - "bbox": [ - 298.33, - 188.76, - 309.29, - 199.13 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p897-b27", - "global_id": 26310, - "bbox": [ - 312.59, - 168.12, - 317.85, - 178.08 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p897-b28", - "global_id": 26311, - "bbox": [ - 317.84, - 175.11, - 411.01, - 200.26 - ], - "text": "2π\nX()X∗()d = 1", - "type": "text" - }, - { - "block_id": "p897-b29", - "global_id": 26312, - "bbox": [ - 402.54, - 188.76, - 413.5, - 199.13 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p897-b30", - "global_id": 26313, - "bbox": [ - 416.8, - 168.12, - 422.06, - 178.08 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p897-b31", - "global_id": 26314, - "bbox": [ - 422.05, - 177.88, - 477.52, - 200.26 - ], - "text": "2π\n|X()|2 d", - "type": "text" - }, - { - "block_id": "p897-b32", - "global_id": 26315, - "bbox": [ - 145.52, - 227.17, - 502.45, - 237.13 - ], - "text": "Table 9.2 summarizes Parseval’s theorem and the other important properties of the DTFT.", - "type": "text" - }, - { - "block_id": "p897-b33", - "global_id": 26316, - "bbox": [ - 127.59, - 256.06, - 264.74, - 265.3 - ], - "text": "TABLE 9.2\nProperties of the DTFT", - "type": "text" - }, - { - "block_id": "p897-b34", - "global_id": 26317, - "bbox": [ - 127.59, - 276.01, - 441.88, - 285.27 - ], - "text": "Operation\nx[n]\nX()", - "type": "text" - }, - { - "block_id": "p897-b35", - "global_id": 26318, - "bbox": [ - 127.59, - 294.34, - 493.91, - 304.42 - ], - "text": "Linearity\na1x1[n] + a2x2[n]\na1X1() + a2X2()", - "type": "text" - }, - { - "block_id": "p897-b36", - "global_id": 26319, - "bbox": [ - 127.59, - 312.0, - 452.33, - 322.61 - ], - "text": "Conjugation\nx∗[n]\nX∗(−)", - "type": "text" - }, - { - "block_id": "p897-b37", - "global_id": 26320, - "bbox": [ - 127.59, - 332.2, - 445.95, - 341.53 - ], - "text": "Scalar multiplication\nax[n]\naX()", - "type": "text" - }, - { - "block_id": "p897-b38", - "global_id": 26321, - "bbox": [ - 127.59, - 349.62, - 449.65, - 365.24 - ], - "text": "Multiplication by n\nnx[n]\njdX()", - "type": "text" - }, - { - "block_id": "p897-b39", - "global_id": 26322, - "bbox": [ - 127.59, - 362.26, - 448.47, - 384.17 - ], - "text": "d\nTime reversal\nx[−n]\nX(−)", - "type": "text" - }, - { - "block_id": "p897-b40", - "global_id": 26323, - "bbox": [ - 127.59, - 392.5, - 509.88, - 403.09 - ], - "text": "Time shifting\nx[n −k]\nX()e−jk\nk integer", - "type": "text" - }, - { - "block_id": "p897-b41", - "global_id": 26324, - "bbox": [ - 127.59, - 411.43, - 461.6, - 422.7 - ], - "text": "Frequency shifting\nx[n]ejcn\nX( −c)", - "type": "text" - }, - { - "block_id": "p897-b42", - "global_id": 26325, - "bbox": [ - 127.59, - 431.62, - 467.69, - 441.7 - ], - "text": "Time convolution\nx1[n] ∗x2[n]\nX1()X2()", - "type": "text" - }, - { - "block_id": "p897-b43", - "global_id": 26326, - "bbox": [ - 127.59, - 449.35, - 432.98, - 470.97 - ], - "text": "Frequency convolution\nx1[n]x2[n]\n1\n2π", - "type": "text" - }, - { - "block_id": "p897-b44", - "global_id": 26327, - "bbox": [ - 436.07, - 443.07, - 440.8, - 452.03 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p897-b45", - "global_id": 26328, - "bbox": [ - 440.81, - 455.27, - 516.13, - 472.05 - ], - "text": "2π\nX1[u]X2[ −u]du", - "type": "text" - }, - { - "block_id": "p897-b46", - "global_id": 26329, - "bbox": [ - 127.59, - 481.85, - 301.68, - 492.16 - ], - "text": "Parseval’s theorem\nEx =", - "type": "text" - }, - { - "block_id": "p897-b47", - "global_id": 26330, - "bbox": [ - 307.15, - 472.54, - 319.84, - 482.3 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p897-b48", - "global_id": 26331, - "bbox": [ - 303.52, - 494.15, - 323.47, - 500.83 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p897-b49", - "global_id": 26332, - "bbox": [ - 324.47, - 475.94, - 450.29, - 492.16 - ], - "text": "|x[n]|2\nEx = 1", - "type": "text" - }, - { - "block_id": "p897-b50", - "global_id": 26333, - "bbox": [ - 442.67, - 488.22, - 452.54, - 497.56 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p897-b51", - "global_id": 26334, - "bbox": [ - 455.63, - 469.65, - 460.36, - 478.62 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p897-b52", - "global_id": 26335, - "bbox": [ - 460.36, - 478.43, - 510.72, - 498.63 - ], - "text": "2π\n|X()|2 d", - "type": "text" - }, - { - "block_id": "p897-b53", - "global_id": 26336, - "bbox": [ - 102.51, - 528.03, - 469.7, - 539.99 - ], - "text": "EXAMPLE 9.12\nUsing Parseval’s Theorem to Find Signal Energy", - "type": "text" - }, - { - "block_id": "p897-b54", - "global_id": 26337, - "bbox": [ - 128.9, - 556.24, - 351.18, - 567.7 - ], - "text": "Find the energy of x[n] = sinc(cn), assuming c < π.", - "type": "text" - }, - { - "block_id": "p897-b55", - "global_id": 26338, - "bbox": [ - 128.9, - 589.12, - 386.57, - 599.49 - ], - "text": "From pair 8, Table 9.1, the fundamental band spectrum of x[n] is", - "type": "text" - }, - { - "block_id": "p897-b56", - "global_id": 26339, - "bbox": [ - 226.27, - 609.69, - 298.65, - 627.75 - ], - "text": "sinc(cn) ⇐⇒π", - "type": "text" - }, - { - "block_id": "p897-b57", - "global_id": 26340, - "bbox": [ - 290.48, - 623.75, - 301.33, - 634.83 - ], - "text": "c", - "type": "text" - }, - { - "block_id": "p897-b58", - "global_id": 26341, - "bbox": [ - 304.13, - 617.09, - 319.06, - 627.05 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p897-b60", - "global_id": 26342, - "bbox": [ - 328.09, - 623.75, - 343.91, - 634.83 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p897-b62", - "global_id": 26343, - "bbox": [ - 373.37, - 616.68, - 404.4, - 626.64 - ], - "text": "|| ≤π", - "type": "text" - } - ] - }, - { - "page_num": 898, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p898-b0", - "global_id": 26344, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "878\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p898-b1", - "global_id": 26345, - "bbox": [ - 103.16, - 86.24, - 288.91, - 96.2 - ], - "text": "From Parseval’s theorem [Eq. (9.36)], we have", - "type": "text" - }, - { - "block_id": "p898-b2", - "global_id": 26346, - "bbox": [ - 216.23, - 108.88, - 247.47, - 126.9 - ], - "text": "Ex = 1", - "type": "text" - }, - { - "block_id": "p898-b3", - "global_id": 26347, - "bbox": [ - 239.0, - 122.52, - 249.96, - 132.9 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p898-b4", - "global_id": 26348, - "bbox": [ - 253.26, - 101.88, - 267.26, - 113.94 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p898-b5", - "global_id": 26349, - "bbox": [ - 258.51, - 126.76, - 268.13, - 133.74 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p898-b6", - "global_id": 26350, - "bbox": [ - 272.03, - 107.52, - 282.49, - 118.42 - ], - "text": "π2", - "type": "text" - }, - { - "block_id": "p898-b7", - "global_id": 26351, - "bbox": [ - 271.64, - 122.32, - 282.87, - 134.78 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p898-b9", - "global_id": 26352, - "bbox": [ - 291.11, - 115.86, - 306.04, - 125.82 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p898-b11", - "global_id": 26353, - "bbox": [ - 315.07, - 122.52, - 330.89, - 133.6 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p898-b12", - "global_id": 26354, - "bbox": [ - 332.59, - 101.46, - 348.23, - 113.35 - ], - "text": "!2", - "type": "text" - }, - { - "block_id": "p898-b13", - "global_id": 26355, - "bbox": [ - 350.94, - 115.45, - 363.94, - 125.72 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p898-b14", - "global_id": 26356, - "bbox": [ - 103.17, - 143.08, - 471.84, - 154.95 - ], - "text": "Because rect(/2c) = 1 over || ≤c and is zero otherwise, the preceding integral yields", - "type": "text" - }, - { - "block_id": "p898-b15", - "global_id": 26357, - "bbox": [ - 232.56, - 164.98, - 263.79, - 183.01 - ], - "text": "Ex = 1", - "type": "text" - }, - { - "block_id": "p898-b16", - "global_id": 26358, - "bbox": [ - 255.32, - 178.63, - 266.27, - 189.0 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p898-b17", - "global_id": 26359, - "bbox": [ - 269.57, - 157.57, - 288.35, - 174.53 - ], - "text": "π2", - "type": "text" - }, - { - "block_id": "p898-b18", - "global_id": 26360, - "bbox": [ - 277.5, - 178.43, - 288.74, - 190.89 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p898-b20", - "global_id": 26361, - "bbox": [ - 298.27, - 164.57, - 343.25, - 182.63 - ], - "text": "(2c) = π", - "type": "text" - }, - { - "block_id": "p898-b21", - "global_id": 26362, - "bbox": [ - 335.09, - 178.63, - 345.93, - 189.7 - ], - "text": "c", - "type": "text" - }, - { - "block_id": "p898-b22", - "global_id": 26363, - "bbox": [ - 102.2, - 255.91, - 448.43, - 269.86 - ], - "text": "9.4 LTI DISCRETE-TIME SYSTEM ANALYSIS BY DTFT", - "type": "text" - }, - { - "block_id": "p898-b23", - "global_id": 26364, - "bbox": [ - 101.84, - 275.43, - 490.39, - 297.77 - ], - "text": "Consider a linear, time-invariant, discrete-time system with the unit impulse response h[n]. We\nshall find the (zero-state) system response y[n] for the input x[n]. Let", - "type": "text" - }, - { - "block_id": "p898-b24", - "global_id": 26365, - "bbox": [ - 166.84, - 321.63, - 425.39, - 332.01 - ], - "text": "x[n] ⇐⇒X()\ny[n] ⇐⇒Y()\nand\nh[n] ⇐⇒H()", - "type": "text" - }, - { - "block_id": "p898-b25", - "global_id": 26366, - "bbox": [ - 101.85, - 355.89, - 330.96, - 366.26 - ], - "text": "Because y[n] = x[n] ∗h[n], it follows from Eq. (9.34) that", - "type": "text" - }, - { - "block_id": "p898-b26", - "global_id": 26367, - "bbox": [ - 257.18, - 390.14, - 490.38, - 400.52 - ], - "text": "Y() = X()H()\n(9.37)", - "type": "text" - }, - { - "block_id": "p898-b27", - "global_id": 26368, - "bbox": [ - 101.85, - 424.81, - 490.4, - 446.72 - ], - "text": "This result is similar to that obtained for continuous-time systems. Let us examine the role of\nH(), the DTFT of the unit impulse response h[n].", - "type": "text" - }, - { - "block_id": "p898-b28", - "global_id": 26369, - "bbox": [ - 101.85, - 448.72, - 490.4, - 494.55 - ], - "text": "Equation (9.37) holds for BIBO-stable systems and also for marginally stable systems if the\ninput does not contain the system’s natural mode(s). In other cases, the response grows with n and\nis not Fourier-transformable. Moreover, the input x[n] also has to be DTF-transformable. For cases\nwhere Eq. (9.37) does not apply, we use the z-transform for system analysis.", - "type": "text" - }, - { - "block_id": "p898-b29", - "global_id": 26370, - "bbox": [ - 101.85, - 496.54, - 490.41, - 518.45 - ], - "text": "Equation (9.37) shows that the output signal frequency spectrum is the product of the input\nsignal frequency spectrum and the frequency response of the system. From this equation, we obtain", - "type": "text" - }, - { - "block_id": "p898-b30", - "global_id": 26371, - "bbox": [ - 168.13, - 542.33, - 424.11, - 552.71 - ], - "text": "|Y()| = |X()||H()|\nand̸\nY() ≠ X() +̸ H()", - "type": "text" - }, - { - "block_id": "p898-b31", - "global_id": 26372, - "bbox": [ - 101.85, - 577.0, - 490.38, - 610.87 - ], - "text": "This result shows that the output amplitude spectrum is the product of the input amplitude spectrum\nand the amplitude response of the system. The output phase spectrum is the sum of the input phase\nspectrum and the phase response of the system.", - "type": "text" - }, - { - "block_id": "p898-b32", - "global_id": 26373, - "bbox": [ - 101.85, - 612.86, - 490.41, - 634.79 - ], - "text": "We can also interpret Eq. (9.37) in terms of the frequency-domain viewpoint, which sees a\nsystem in terms of its frequency response (system response to various exponential or sinusoidal", - "type": "text" - } - ] - }, - { - "page_num": 899, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p899-b0", - "global_id": 26374, - "bbox": [ - 277.57, - 62.89, - 516.14, - 71.98 - ], - "text": "9.4\nLTI Discrete-Time System Analysis by DTFT\n879", - "type": "text" - }, - { - "block_id": "p899-b1", - "global_id": 26375, - "bbox": [ - 127.59, - 85.82, - 516.14, - 131.64 - ], - "text": "components). The frequency domain views a signal as a sum of various exponential or sinusoidal\ncomponents. The transmission of a signal through a (linear) system is viewed as transmission of\nvarious exponential or sinusoidal components of the input signal through the system. This concept\ncan be understood by displaying the input–output relationships by a directed arrow as follows:", - "type": "text" - }, - { - "block_id": "p899-b2", - "global_id": 26376, - "bbox": [ - 282.33, - 143.06, - 360.89, - 157.46 - ], - "text": "ejn \r⇒H()ejn", - "type": "text" - }, - { - "block_id": "p899-b3", - "global_id": 26377, - "bbox": [ - 127.59, - 169.89, - 376.79, - 183.47 - ], - "text": "which shows that the system response to ejn is H()ejn, and", - "type": "text" - }, - { - "block_id": "p899-b4", - "global_id": 26378, - "bbox": [ - 266.75, - 197.41, - 304.36, - 214.35 - ], - "text": "x[n] = 1", - "type": "text" - }, - { - "block_id": "p899-b5", - "global_id": 26379, - "bbox": [ - 295.88, - 211.05, - 306.84, - 221.43 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p899-b6", - "global_id": 26380, - "bbox": [ - 310.14, - 190.41, - 315.4, - 200.38 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p899-b7", - "global_id": 26381, - "bbox": [ - 315.4, - 200.18, - 376.97, - 222.56 - ], - "text": "2π\nX()ejn d", - "type": "text" - }, - { - "block_id": "p899-b8", - "global_id": 26382, - "bbox": [ - 127.59, - 234.75, - 516.15, - 257.08 - ], - "text": "which shows x[n] as a sum of everlasting exponential components. Invoking the linearity property,\nwe obtain", - "type": "text" - }, - { - "block_id": "p899-b9", - "global_id": 26383, - "bbox": [ - 255.36, - 260.99, - 292.93, - 277.94 - ], - "text": "y[n] = 1", - "type": "text" - }, - { - "block_id": "p899-b10", - "global_id": 26384, - "bbox": [ - 284.46, - 274.63, - 295.42, - 285.0 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p899-b11", - "global_id": 26385, - "bbox": [ - 298.72, - 254.0, - 303.98, - 263.96 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p899-b12", - "global_id": 26386, - "bbox": [ - 303.98, - 263.75, - 388.36, - 286.14 - ], - "text": "2π\nX()H()ejn d", - "type": "text" - }, - { - "block_id": "p899-b13", - "global_id": 26387, - "bbox": [ - 127.59, - 295.34, - 516.16, - 317.67 - ], - "text": "which gives y[n] as a sum of responses to all input components and is equivalent to Eq. (9.37).\nThus, X() is the input spectrum and Y() is the output spectrum, given by X()H().", - "type": "text" - }, - { - "block_id": "p899-b14", - "global_id": 26388, - "bbox": [ - 102.51, - 364.6, - 402.06, - 376.56 - ], - "text": "EXAMPLE 9.13\nLTID System Analysis by the DTFT", - "type": "text" - }, - { - "block_id": "p899-b15", - "global_id": 26389, - "bbox": [ - 128.9, - 392.81, - 502.76, - 427.09 - ], - "text": "An LTID system is specified by the equation y[n] −0.5y[n −1] = x[n]. Find H(), the\nfrequency response of this system. Determine the (zero-state) response y[n] if the input\nx[n] = (0.8)nu[n].", - "type": "text" - }, - { - "block_id": "p899-b16", - "global_id": 26390, - "bbox": [ - 128.9, - 449.59, - 502.74, - 471.93 - ], - "text": "Let x[n] ⇐⇒X() and y[n] ⇐⇒Y(). Taking the DTFT of the system’s difference equation\nyields", - "type": "text" - }, - { - "block_id": "p899-b17", - "global_id": 26391, - "bbox": [ - 261.58, - 471.78, - 370.09, - 483.88 - ], - "text": "(1 −0.5e−j)Y() = X()", - "type": "text" - }, - { - "block_id": "p899-b18", - "global_id": 26392, - "bbox": [ - 128.91, - 492.85, - 226.3, - 502.81 - ], - "text": "According to Eq. (9.37),", - "type": "text" - }, - { - "block_id": "p899-b19", - "global_id": 26393, - "bbox": [ - 237.59, - 513.82, - 295.06, - 531.08 - ], - "text": "H() = Y()", - "type": "text" - }, - { - "block_id": "p899-b20", - "global_id": 26394, - "bbox": [ - 273.47, - 512.59, - 380.71, - 538.25 - ], - "text": "X() =\n1\n1 −e−j =\nej", - "type": "text" - }, - { - "block_id": "p899-b21", - "global_id": 26395, - "bbox": [ - 357.27, - 524.99, - 392.88, - 538.25 - ], - "text": "ej −0.5", - "type": "text" - }, - { - "block_id": "p899-b22", - "global_id": 26396, - "bbox": [ - 128.91, - 546.41, - 254.01, - 557.78 - ], - "text": "Also, x[n] = (0.8)nu[n]. Hence,", - "type": "text" - }, - { - "block_id": "p899-b23", - "global_id": 26397, - "bbox": [ - 280.05, - 557.18, - 338.26, - 575.66 - ], - "text": "X() =\nej", - "type": "text" - }, - { - "block_id": "p899-b24", - "global_id": 26398, - "bbox": [ - 314.82, - 569.58, - 350.42, - 582.84 - ], - "text": "ej −0.8", - "type": "text" - }, - { - "block_id": "p899-b25", - "global_id": 26399, - "bbox": [ - 128.91, - 588.45, - 143.28, - 598.41 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p899-b26", - "global_id": 26400, - "bbox": [ - 226.72, - 596.17, - 368.83, - 614.66 - ], - "text": "Y() = X()H() =\n2ej", - "type": "text" - }, - { - "block_id": "p899-b27", - "global_id": 26401, - "bbox": [ - 317.67, - 608.58, - 403.75, - 621.83 - ], - "text": "(ej −0.8)(ej −0.5)", - "type": "text" - } - ] - }, - { - "page_num": 900, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p900-b0", - "global_id": 26402, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "880\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p900-b1", - "global_id": 26403, - "bbox": [ - 103.16, - 86.24, - 477.01, - 108.15 - ], - "text": "We can express the right-hand side as a sum of two first-order terms (modified partial fraction\nexpansion as discussed in Sec. B.5-6) as follows†:", - "type": "text" - }, - { - "block_id": "p900-b2", - "global_id": 26404, - "bbox": [ - 179.81, - 120.39, - 201.29, - 130.66 - ], - "text": "Y()", - "type": "text" - }, - { - "block_id": "p900-b3", - "global_id": 26405, - "bbox": [ - 184.41, - 119.16, - 264.23, - 144.72 - ], - "text": "ej\n=\nej", - "type": "text" - }, - { - "block_id": "p900-b4", - "global_id": 26406, - "bbox": [ - 215.56, - 118.32, - 339.33, - 144.82 - ], - "text": "(ej −0.5)(ej −0.8) =\n−5", - "type": "text" - }, - { - "block_id": "p900-b5", - "global_id": 26407, - "bbox": [ - 315.9, - 125.89, - 362.02, - 144.82 - ], - "text": "3\nej −0.5 +", - "type": "text" - }, - { - "block_id": "p900-b6", - "global_id": 26408, - "bbox": [ - 364.77, - 118.32, - 400.37, - 144.82 - ], - "text": "8\n3\nej −0.8", - "type": "text" - }, - { - "block_id": "p900-b7", - "global_id": 26409, - "bbox": [ - 103.17, - 154.33, - 159.78, - 164.3 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p900-b8", - "global_id": 26410, - "bbox": [ - 193.19, - 182.29, - 234.32, - 192.56 - ], - "text": "Y() = −", - "type": "text" - }, - { - "block_id": "p900-b9", - "global_id": 26411, - "bbox": [ - 235.42, - 168.3, - 248.32, - 185.68 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p900-b10", - "global_id": 26412, - "bbox": [ - 243.34, - 189.78, - 248.32, - 199.74 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p900-b11", - "global_id": 26413, - "bbox": [ - 249.52, - 168.3, - 281.98, - 185.58 - ], - "text": "ej", - "type": "text" - }, - { - "block_id": "p900-b12", - "global_id": 26414, - "bbox": [ - 258.55, - 182.29, - 304.67, - 199.74 - ], - "text": "ej −0.5 +", - "type": "text" - }, - { - "block_id": "p900-b13", - "global_id": 26415, - "bbox": [ - 306.21, - 168.3, - 319.12, - 185.68 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p900-b14", - "global_id": 26416, - "bbox": [ - 314.14, - 189.78, - 319.12, - 199.74 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p900-b15", - "global_id": 26417, - "bbox": [ - 320.32, - 168.3, - 352.79, - 185.58 - ], - "text": "ej", - "type": "text" - }, - { - "block_id": "p900-b16", - "global_id": 26418, - "bbox": [ - 329.34, - 186.48, - 364.95, - 199.74 - ], - "text": "ej −0.8", - "type": "text" - }, - { - "block_id": "p900-b17", - "global_id": 26419, - "bbox": [ - 216.73, - 210.18, - 234.32, - 220.14 - ], - "text": "= −", - "type": "text" - }, - { - "block_id": "p900-b18", - "global_id": 26420, - "bbox": [ - 235.42, - 196.19, - 248.32, - 213.57 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p900-b19", - "global_id": 26421, - "bbox": [ - 243.34, - 217.67, - 248.32, - 227.63 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p900-b20", - "global_id": 26422, - "bbox": [ - 249.52, - 196.19, - 329.55, - 227.63 - ], - "text": "1\n1 −0.5e−j +\n8", - "type": "text" - }, - { - "block_id": "p900-b21", - "global_id": 26423, - "bbox": [ - 324.56, - 217.67, - 329.55, - 227.63 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p900-b22", - "global_id": 26424, - "bbox": [ - 330.74, - 196.19, - 385.27, - 227.63 - ], - "text": "1\n1 −0.8e−j", - "type": "text" - }, - { - "block_id": "p900-b23", - "global_id": 26425, - "bbox": [ - 103.16, - 237.8, - 354.2, - 247.76 - ], - "text": "From entry 2 of Table 9.1, the inverse DTFT of this equation is", - "type": "text" - }, - { - "block_id": "p900-b24", - "global_id": 26426, - "bbox": [ - 224.29, - 259.31, - 250.15, - 269.59 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p900-b26", - "global_id": 26427, - "bbox": [ - 256.12, - 257.97, - 268.57, - 269.27 - ], - "text": "−5", - "type": "text" - }, - { - "block_id": "p900-b27", - "global_id": 26428, - "bbox": [ - 265.09, - 251.3, - 355.88, - 272.5 - ], - "text": "3(0.5)n + 8\n3(0.8)n\nu[n]", - "type": "text" - }, - { - "block_id": "p900-b28", - "global_id": 26429, - "bbox": [ - 103.17, - 281.64, - 477.05, - 351.38 - ], - "text": "This example demonstrates the procedure for using the DTFT to determine an LTID\nsystem response. It is similar to the Fourier transform method in the analysis of LTIC systems.\nAs in the case of the Fourier transform, this method can be used only if the system is\nasymptotically or BIBO-stable and if the input signal is DTF-transformable.‡ We shall not\nbelabor this method further because it is clumsier and more restrictive than the z-transform\nmethod discussed in Ch. 5.", - "type": "text" - }, - { - "block_id": "p900-b29", - "global_id": 26430, - "bbox": [ - 101.84, - 410.45, - 285.28, - 422.4 - ], - "text": "9.4-1 Distortionless Transmission", - "type": "text" - }, - { - "block_id": "p900-b30", - "global_id": 26431, - "bbox": [ - 101.84, - 428.54, - 490.41, - 462.41 - ], - "text": "In several applications, digital signals are passed through LTI systems, and we require that the\noutput waveform be a replica of the input waveform. As in the continuous-time case, transmission\nis said to be distortionless if the input x[n] and the output y[n] satisfy the condition", - "type": "text" - }, - { - "block_id": "p900-b31", - "global_id": 26432, - "bbox": [ - 257.98, - 475.23, - 334.26, - 487.0 - ], - "text": "y[n] = G0 x[n −nd]", - "type": "text" - }, - { - "block_id": "p900-b32", - "global_id": 26433, - "bbox": [ - 101.85, - 498.74, - 474.32, - 509.51 - ], - "text": "Here, nd, the delay (in samples), is assumed to be integer. Taking the Fourier transform yields", - "type": "text" - }, - { - "block_id": "p900-b33", - "global_id": 26434, - "bbox": [ - 249.46, - 519.9, - 341.6, - 533.39 - ], - "text": "Y() = G0 X()e−jnd", - "type": "text" - }, - { - "block_id": "p900-b34", - "global_id": 26435, - "bbox": [ - 101.84, - 545.24, - 116.23, - 555.2 - ], - "text": "But", - "type": "text" - }, - { - "block_id": "p900-b35", - "global_id": 26436, - "bbox": [ - 256.62, - 558.69, - 335.61, - 568.97 - ], - "text": "Y() = X()H()", - "type": "text" - }, - { - "block_id": "p900-b36", - "global_id": 26437, - "bbox": [ - 101.84, - 588.02, - 490.39, - 633.41 - ], - "text": "† Here, Y() is a function of variable ej. Hence, x = ej for the purpose of comparison with the expression\nin Sec. B.5-6.\n‡ It can also be applied to marginally stable systems if the input does not contain natural mode(s) of the\nsystem.", - "type": "text" - } - ] - }, - { - "page_num": 901, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p901-b0", - "global_id": 26438, - "bbox": [ - 277.57, - 62.89, - 516.14, - 71.98 - ], - "text": "9.4\nLTI Discrete-Time System Analysis by DTFT\n881", - "type": "text" - }, - { - "block_id": "p901-b1", - "global_id": 26439, - "bbox": [ - 305.96, - 162.38, - 359.62, - 172.14 - ], - "text": "H(\t) vnd", - "type": "text" - }, - { - "block_id": "p901-b2", - "global_id": 26440, - "bbox": [ - 231.24, - 111.32, - 252.6, - 119.62 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p901-b3", - "global_id": 26441, - "bbox": [ - 216.11, - 146.69, - 219.79, - 154.04 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p901-b4", - "global_id": 26442, - "bbox": [ - 210.94, - 111.17, - 219.72, - 120.78 - ], - "text": "G0", - "type": "text" - }, - { - "block_id": "p901-b6", - "global_id": 26443, - "bbox": [ - 370.55, - 156.12, - 512.16, - 189.99 - ], - "text": "Figure 9.13 LTI system frequency\nresponse for distortionless trans-\nmission.", - "type": "text" - }, - { - "block_id": "p901-b7", - "global_id": 26444, - "bbox": [ - 127.59, - 213.47, - 169.34, - 223.43 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p901-b8", - "global_id": 26445, - "bbox": [ - 285.94, - 225.61, - 356.61, - 239.1 - ], - "text": "H() = G0 e−jnd", - "type": "text" - }, - { - "block_id": "p901-b9", - "global_id": 26446, - "bbox": [ - 127.59, - 248.22, - 516.15, - 270.14 - ], - "text": "This is the frequency response required for distortionless transmission. From this equation, it\nfollows that", - "type": "text" - }, - { - "block_id": "p901-b10", - "global_id": 26447, - "bbox": [ - 236.46, - 274.04, - 516.13, - 285.19 - ], - "text": "|H()| = G0\nand̸\nH() = −nd\n(9.38)", - "type": "text" - }, - { - "block_id": "p901-b11", - "global_id": 26448, - "bbox": [ - 127.59, - 294.52, - 516.14, - 342.25 - ], - "text": "Thus, for distortionless transmission, the amplitude response |H()| must be a constant, and the\nphase response̸\nH() must be a linear function of with slope −nd, where nd is the delay in the\nnumber of samples with respect to input (Fig. 9.13). These are precisely the characteristics of an\nideal delay of nd samples with a gain of G0 [see Eq. (9.31)].", - "type": "text" - }, - { - "block_id": "p901-b12", - "global_id": 26449, - "bbox": [ - 127.59, - 356.26, - 516.14, - 430.2 - ], - "text": "MEASURE OF DELAY VARIATION\nFor distortionless transmission, we require a linear phase characteristic. In practice, many systems\nhave a phase characteristic that may be only approximately linear. A convenient way of judging\nphase linearity is to plot the slope of̸\nH() as a function of frequency. This slope is constant for\nthe ideal linear phase (ILP) system, but it may vary with in the general case. The slope can be\nexpressed as", - "type": "text" - }, - { - "block_id": "p901-b13", - "global_id": 26450, - "bbox": [ - 278.05, - 432.25, - 331.88, - 450.0 - ], - "text": "ng() = −d", - "type": "text" - }, - { - "block_id": "p901-b14", - "global_id": 26451, - "bbox": [ - 323.02, - 438.92, - 516.13, - 456.27 - ], - "text": "d̸ H()\n(9.39)", - "type": "text" - }, - { - "block_id": "p901-b15", - "global_id": 26452, - "bbox": [ - 127.59, - 462.47, - 516.15, - 544.58 - ], - "text": "If ng() is constant, all the components are delayed by ng samples. But if the slope is not constant,\nthe delay ng varies with frequency. This variation means that different frequency components\nundergo different amounts of delay, and consequently, the output waveform will not be a replica\nof the input waveform. As in the case of LTIC systems, ng(), as defined in Eq. (9.39), plays\nan important role in bandpass systems and is called the group delay or envelope delay. Observe\nthat constant nd implies constant ng. Note that̸\nH() = φ0 −nd also has a constant ng. Thus,\nconstant group delay is a more relaxed condition.", - "type": "text" - }, - { - "block_id": "p901-b16", - "global_id": 26453, - "bbox": [ - 127.59, - 560.09, - 516.16, - 635.52 - ], - "text": "DISTORTIONLESS TRANSMISSION OVER BANDPASS SYSTEMS\nAs in the case of continuous-time systems, the distortionless transmission conditions can be\nrelaxed for discrete-time bandpass systems. For lowpass systems, the phase characteristic should\nnot only be linear over the band of interest, it should also pass through the origin [Eq. (9.38)].\nFor bandpass systems, the phase characteristic should be linear over the band of interest, but it\nneed not pass through the origin (ng should be constant). The amplitude response is required to", - "type": "text" - } - ] - }, - { - "page_num": 902, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p902-b0", - "global_id": 26454, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "882\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p902-b1", - "global_id": 26455, - "bbox": [ - 101.84, - 85.82, - 490.41, - 107.74 - ], - "text": "be constant over the passband. Thus, for distortionless transmission over a bandpass system, the\nfrequency response for positive range of is of the form†", - "type": "text" - }, - { - "block_id": "p902-b2", - "global_id": 26456, - "bbox": [ - 232.34, - 116.06, - 359.89, - 128.94 - ], - "text": "H() = G0ej(φ0−ng)\n ≥0", - "type": "text" - }, - { - "block_id": "p902-b3", - "global_id": 26457, - "bbox": [ - 101.85, - 138.64, - 490.36, - 161.25 - ], - "text": "The proof is identical to that for the continuous-time case in Sec. 7.4-2 and will not be repeated.\nIn using Eq. (9.39) to compute ng, we should ignore jump discontinuities in the phase function.", - "type": "text" - }, - { - "block_id": "p902-b4", - "global_id": 26458, - "bbox": [ - 101.84, - 184.56, - 270.68, - 196.51 - ], - "text": "9.4-2 Ideal and Practical Filters", - "type": "text" - }, - { - "block_id": "p902-b5", - "global_id": 26459, - "bbox": [ - 101.84, - 202.64, - 490.38, - 249.17 - ], - "text": "Ideal filters allow distortionless transmission of a certain band of frequencies and suppress all the\nremaining frequencies. The general ideal lowpass filter shown in Fig. 9.14 for || ≤π allows\nall components below the cutoff frequency = c to pass without distortion and suppresses all\ncomponents above c. Figure 9.15 illustrates ideal highpass and bandpass filter characteristics.", - "type": "text" - }, - { - "block_id": "p902-b6", - "global_id": 26460, - "bbox": [ - 101.85, - 250.05, - 490.39, - 285.04 - ], - "text": "The ideal lowpass filter in Fig. 9.14a has a linear phase of slope −nd, which results in a delay\nof nd samples for all its input components of frequencies below c rad/sample. Therefore, if the\ninput is a signal x[n] bandlimited to c, the output y[n] is x[n] delayed by nd; that is,", - "type": "text" - }, - { - "block_id": "p902-b7", - "global_id": 26461, - "bbox": [ - 264.11, - 294.39, - 328.12, - 305.47 - ], - "text": "y[n] = x[n −nd]", - "type": "text" - }, - { - "block_id": "p902-b8", - "global_id": 26462, - "bbox": [ - 101.85, - 314.82, - 490.39, - 337.15 - ], - "text": "The signal x[n] is transmitted by this system without distortion, but with delay of nd samples. For\nthis filter,", - "type": "text" - }, - { - "block_id": "p902-b9", - "global_id": 26463, - "bbox": [ - 218.89, - 346.14, - 251.54, - 356.42 - ], - "text": "H() =", - "type": "text" - }, - { - "block_id": "p902-b10", - "global_id": 26464, - "bbox": [ - 258.05, - 335.97, - 272.15, - 346.64 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p902-b11", - "global_id": 26465, - "bbox": [ - 253.58, - 359.99, - 276.61, - 367.18 - ], - "text": "m=−∞", - "type": "text" - }, - { - "block_id": "p902-b12", - "global_id": 26466, - "bbox": [ - 277.72, - 346.55, - 292.65, - 356.52 - ], - "text": "rect", - "type": "text" - }, - { - "block_id": "p902-b13", - "global_id": 26467, - "bbox": [ - 293.77, - 332.16, - 339.45, - 349.53 - ], - "text": "−2πm", - "type": "text" - }, - { - "block_id": "p902-b14", - "global_id": 26468, - "bbox": [ - 312.41, - 353.21, - 328.23, - 364.29 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p902-b16", - "global_id": 26469, - "bbox": [ - 348.48, - 344.42, - 372.17, - 356.42 - ], - "text": "e−jnd", - "type": "text" - }, - { - "block_id": "p902-b17", - "global_id": 26470, - "bbox": [ - 101.84, - 373.29, - 490.39, - 395.62 - ], - "text": "The unit impulse response h[n] of this filter is obtained from pair 8 (Table 9.1) and the time-shifting\nproperty", - "type": "text" - }, - { - "block_id": "p902-b18", - "global_id": 26471, - "bbox": [ - 240.68, - 394.98, - 281.18, - 412.24 - ], - "text": "h[n] = c", - "type": "text" - }, - { - "block_id": "p902-b19", - "global_id": 26472, - "bbox": [ - 272.52, - 401.96, - 351.56, - 419.41 - ], - "text": "π sinc[c(n −nd)]", - "type": "text" - }, - { - "block_id": "p902-b20", - "global_id": 26473, - "bbox": [ - 101.85, - 423.19, - 490.4, - 445.53 - ], - "text": "Because h[n] is the system response to impulse input δ[n], which is applied at n = 0, it must\nbe causal (i.e., it must not start before n = 0) for a realizable system. Figure 9.14b shows h[n] for", - "type": "text" - }, - { - "block_id": "p902-b21", - "global_id": 26474, - "bbox": [ - 167.8, - 562.62, - 378.67, - 570.62 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p902-b23", - "global_id": 26475, - "bbox": [ - 107.57, - 518.34, - 233.79, - 528.1 - ], - "text": "c\n\tc", - "type": "text" - }, - { - "block_id": "p902-b24", - "global_id": 26476, - "bbox": [ - 165.86, - 481.06, - 426.18, - 498.37 - ], - "text": "c \n1", - "type": "text" - }, - { - "block_id": "p902-b25", - "global_id": 26477, - "bbox": [ - 302.99, - 517.55, - 474.17, - 527.43 - ], - "text": "0\nn", - "type": "text" - }, - { - "block_id": "p902-b26", - "global_id": 26478, - "bbox": [ - 339.36, - 473.39, - 352.69, - 481.47 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p902-b27", - "global_id": 26479, - "bbox": [ - 298.24, - 476.62, - 434.59, - 501.33 - ], - "text": "4\np\n4\n1", - "type": "text" - }, - { - "block_id": "p902-b28", - "global_id": 26480, - "bbox": [ - 369.91, - 517.4, - 376.91, - 526.95 - ], - "text": "nd", - "type": "text" - }, - { - "block_id": "p902-b29", - "global_id": 26481, - "bbox": [ - 168.34, - 547.05, - 222.89, - 556.81 - ], - "text": "H(\t) \tnd", - "type": "text" - }, - { - "block_id": "p902-b30", - "global_id": 26482, - "bbox": [ - 445.77, - 488.61, - 470.48, - 498.37 - ], - "text": "nd 12", - "type": "text" - }, - { - "block_id": "p902-b31", - "global_id": 26483, - "bbox": [ - 175.09, - 461.14, - 196.45, - 469.44 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p902-b32", - "global_id": 26484, - "bbox": [ - 101.84, - 577.32, - 386.71, - 586.56 - ], - "text": "Figure 9.14 Ideal lowpass filter: its frequency response and impulse response.", - "type": "text" - }, - { - "block_id": "p902-b33", - "global_id": 26485, - "bbox": [ - 101.84, - 610.24, - 489.73, - 634.75 - ], - "text": "† Because the phase function is an odd function of , if̸\nH() = φ0 −ng for ≥0, over the band 2W\n(centered at c), then̸\nH() = −φ0 −ng for < 0 over the band 2W (centered at −c).", - "type": "text" - } - ] - }, - { - "page_num": 903, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p903-b0", - "global_id": 26486, - "bbox": [ - 322.62, - 62.89, - 516.13, - 71.98 - ], - "text": "9.5\nDTFT Connection with the CTFT\n883", - "type": "text" - }, - { - "block_id": "p903-b1", - "global_id": 26487, - "bbox": [ - 276.01, - 138.53, - 280.01, - 146.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p903-b2", - "global_id": 26488, - "bbox": [ - 276.01, - 244.46, - 280.01, - 252.46 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p903-b3", - "global_id": 26489, - "bbox": [ - 278.66, - 174.54, - 287.54, - 182.54 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p903-b4", - "global_id": 26490, - "bbox": [ - 278.27, - 271.74, - 287.92, - 279.74 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p903-b6", - "global_id": 26491, - "bbox": [ - 229.09, - 244.59, - 390.22, - 255.09 - ], - "text": "0\n\t0", - "type": "text" - }, - { - "block_id": "p903-b7", - "global_id": 26492, - "bbox": [ - 196.06, - 195.12, - 220.05, - 203.42 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p903-b8", - "global_id": 26493, - "bbox": [ - 286.87, - 104.31, - 308.23, - 112.6 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p903-b9", - "global_id": 26494, - "bbox": [ - 286.87, - 213.07, - 308.23, - 221.36 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p903-b10", - "global_id": 26495, - "bbox": [ - 196.06, - 88.22, - 220.05, - 96.51 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p903-b11", - "global_id": 26496, - "bbox": [ - 151.5, - 286.43, - 396.72, - 295.67 - ], - "text": "Figure 9.15 Ideal highpass and bandpass filter frequency response.", - "type": "text" - }, - { - "block_id": "p903-b12", - "global_id": 26497, - "bbox": [ - 373.33, - 371.51, - 377.33, - 379.51 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p903-b13", - "global_id": 26498, - "bbox": [ - 256.14, - 326.04, - 269.46, - 334.12 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p903-b14", - "global_id": 26499, - "bbox": [ - 205.84, - 372.13, - 209.84, - 380.13 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p903-b15", - "global_id": 26500, - "bbox": [ - 305.02, - 344.63, - 330.02, - 357.35 - ], - "text": "c 4", - "type": "text" - }, - { - "block_id": "p903-b16", - "global_id": 26501, - "bbox": [ - 200.23, - 330.33, - 367.47, - 354.39 - ], - "text": "p\nnd 12\n4\n1", - "type": "text" - }, - { - "block_id": "p903-b17", - "global_id": 26502, - "bbox": [ - 271.9, - 371.11, - 278.9, - 380.65 - ], - "text": "nd", - "type": "text" - }, - { - "block_id": "p903-b18", - "global_id": 26503, - "bbox": [ - 151.5, - 387.63, - 516.13, - 409.55 - ], - "text": "Figure 9.16 Approximate realization of an ideal lowpass filter by truncation of its impulse\nresponse.", - "type": "text" - }, - { - "block_id": "p903-b19", - "global_id": 26504, - "bbox": [ - 127.59, - 437.61, - 516.13, - 471.89 - ], - "text": "c = π/4 and nd = 12. This figure also shows that h[n] is noncausal, hence unrealizable. Similarly,\none can show that other ideal filters (such as the ideal highpass or and bandpass filters depicted in\nFig. 9.15) are also noncausal and therefore physically unrealizable.", - "type": "text" - }, - { - "block_id": "p903-b20", - "global_id": 26505, - "bbox": [ - 127.59, - 473.89, - 516.15, - 543.62 - ], - "text": "One practical approach to realize an ideal lowpass filter approximately is to truncate both\ntails (positive and negative) of h[n] so that it has a finite length and then delay sufficiently to\nmake it causal (Fig. 9.16). We now synthesize a system with this truncated (and delayed) impulse\nresponse. For closer approximation, the truncating window has to be correspondingly wider. The\ndelay required also increases correspondingly. Thus, the price of closer realization is higher delay\nin the output; this situation is common in noncausal systems.", - "type": "text" - }, - { - "block_id": "p903-b21", - "global_id": 26506, - "bbox": [ - 127.94, - 580.98, - 396.95, - 594.93 - ], - "text": "9.5 DTFT CONNECTION WITH THE CTFT", - "type": "text" - }, - { - "block_id": "p903-b22", - "global_id": 26507, - "bbox": [ - 127.59, - 600.5, - 516.14, - 635.49 - ], - "text": "Consider a continuous-time signal xc(t) (Fig. 9.17a) with the Fourier transform Xc(ω) bandlimited\nto B Hz (Fig. 9.17b). This signal is sampled with a sampling interval T. The sampling rate is at\nleast equal to the Nyquist rate; that is, T ≤1/2B. The sampled signal xc(t) (Fig. 9.17c) can be", - "type": "text" - } - ] - }, - { - "page_num": 904, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p904-b0", - "global_id": 26508, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "884\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p904-b1", - "global_id": 26509, - "bbox": [ - 159.22, - 329.82, - 163.22, - 337.82 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p904-b2", - "global_id": 26510, - "bbox": [ - 189.85, - 347.74, - 198.73, - 355.74 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p904-b3", - "global_id": 26511, - "bbox": [ - 128.58, - 329.52, - 204.69, - 337.82 - ], - "text": "4\n2", - "type": "text" - }, - { - "block_id": "p904-b4", - "global_id": 26512, - "bbox": [ - 189.85, - 157.75, - 198.73, - 165.75 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p904-b5", - "global_id": 26513, - "bbox": [ - 167.46, - 281.39, - 180.34, - 289.47 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p904-b6", - "global_id": 26514, - "bbox": [ - 189.85, - 251.81, - 198.73, - 259.81 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p904-b7", - "global_id": 26515, - "bbox": [ - 229.67, - 210.33, - 234.12, - 218.33 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p904-b8", - "global_id": 26516, - "bbox": [ - 159.92, - 136.97, - 163.92, - 144.97 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p904-b9", - "global_id": 26517, - "bbox": [ - 248.02, - 329.84, - 451.13, - 338.64 - ], - "text": "0\n2p\n2p\np\np\n\t\nn", - "type": "text" - }, - { - "block_id": "p904-b10", - "global_id": 26518, - "bbox": [ - 378.0, - 139.13, - 382.0, - 147.13 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p904-b11", - "global_id": 26519, - "bbox": [ - 371.0, - 157.75, - 380.8, - 165.75 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p904-b12", - "global_id": 26520, - "bbox": [ - 250.59, - 138.3, - 453.99, - 146.95 - ], - "text": "2pB\n2pB\nv\nt", - "type": "text" - }, - { - "block_id": "p904-b13", - "global_id": 26521, - "bbox": [ - 125.76, - 232.69, - 252.82, - 241.29 - ], - "text": "0\n2T\n4T\nt", - "type": "text" - }, - { - "block_id": "p904-b14", - "global_id": 26522, - "bbox": [ - 372.81, - 347.74, - 381.44, - 355.74 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p904-b15", - "global_id": 26523, - "bbox": [ - 362.41, - 277.29, - 367.29, - 293.73 - ], - "text": "A\nT", - "type": "text" - }, - { - "block_id": "p904-b16", - "global_id": 26524, - "bbox": [ - 362.4, - 179.5, - 367.29, - 195.94 - ], - "text": "A\nT", - "type": "text" - }, - { - "block_id": "p904-b17", - "global_id": 26525, - "bbox": [ - 205.6, - 294.69, - 219.37, - 304.23 - ], - "text": "xc(t)", - "type": "text" - }, - { - "block_id": "p904-b18", - "global_id": 26526, - "bbox": [ - 390.99, - 277.01, - 407.43, - 285.3 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p904-b19", - "global_id": 26527, - "bbox": [ - 388.08, - 182.71, - 406.29, - 192.28 - ], - "text": "Xc(v)", - "type": "text" - }, - { - "block_id": "p904-b20", - "global_id": 26528, - "bbox": [ - 168.81, - 88.14, - 402.31, - 98.76 - ], - "text": "A\nXc(v)\nxc(t)", - "type": "text" - }, - { - "block_id": "p904-b21", - "global_id": 26529, - "bbox": [ - 167.89, - 181.91, - 181.66, - 194.38 - ], - "text": "–xc(t)", - "type": "text" - }, - { - "block_id": "p904-b22", - "global_id": 26530, - "bbox": [ - 377.9, - 232.49, - 381.9, - 240.49 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p904-b23", - "global_id": 26531, - "bbox": [ - 371.24, - 252.34, - 380.56, - 260.34 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p904-b24", - "global_id": 26532, - "bbox": [ - 317.06, - 230.46, - 450.14, - 248.78 - ], - "text": "2pB\nv\n2p\nT\np\nT\n\n2p\nT", - "type": "text" - }, - { - "block_id": "p904-b25", - "global_id": 26533, - "bbox": [ - 125.76, - 362.44, - 383.37, - 371.68 - ], - "text": "Figure 9.17 Connection between the DTFT and the Fourier transform.", - "type": "text" - }, - { - "block_id": "p904-b26", - "global_id": 26534, - "bbox": [ - 101.84, - 393.34, - 151.76, - 403.3 - ], - "text": "expressed as", - "type": "text" - }, - { - "block_id": "p904-b27", - "global_id": 26535, - "bbox": [ - 237.07, - 414.73, - 265.32, - 425.81 - ], - "text": "xc(t) =", - "type": "text" - }, - { - "block_id": "p904-b28", - "global_id": 26536, - "bbox": [ - 271.06, - 404.56, - 285.16, - 415.22 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p904-b29", - "global_id": 26537, - "bbox": [ - 267.36, - 428.58, - 288.84, - 435.77 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p904-b30", - "global_id": 26538, - "bbox": [ - 289.96, - 414.73, - 355.15, - 425.81 - ], - "text": "xc(nT)δ(t −nT)", - "type": "text" - }, - { - "block_id": "p904-b31", - "global_id": 26539, - "bbox": [ - 101.84, - 444.19, - 387.11, - 454.15 - ], - "text": "The continuous-time Fourier transform of the foregoing equation yields", - "type": "text" - }, - { - "block_id": "p904-b32", - "global_id": 26540, - "bbox": [ - 240.37, - 474.97, - 274.47, - 486.05 - ], - "text": "Xc(ω) =", - "type": "text" - }, - { - "block_id": "p904-b33", - "global_id": 26541, - "bbox": [ - 280.21, - 464.79, - 294.3, - 475.47 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p904-b34", - "global_id": 26542, - "bbox": [ - 276.52, - 488.81, - 297.99, - 496.01 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p904-b35", - "global_id": 26543, - "bbox": [ - 299.11, - 473.24, - 490.38, - 486.05 - ], - "text": "xc(nT)e−jnTω\n(9.40)", - "type": "text" - }, - { - "block_id": "p904-b36", - "global_id": 26544, - "bbox": [ - 101.84, - 512.18, - 490.4, - 547.16 - ], - "text": "In Sec. 8.1 (Fig. 8.1f), we showed that Xc(ω) is Xc(ω)/T repeating periodically with a period\nωs = 2π/T, as illustrated in Fig. 9.17d. Let us construct a discrete-time signal x[n] such that its\nnth sample value is equal to the value of the nth sample of xc(t), as depicted in Fig. 9.17e, that is,", - "type": "text" - }, - { - "block_id": "p904-b37", - "global_id": 26545, - "bbox": [ - 268.78, - 559.16, - 490.38, - 570.24 - ], - "text": "x[n] = xc(nT)\n(9.41)", - "type": "text" - }, - { - "block_id": "p904-b38", - "global_id": 26546, - "bbox": [ - 101.84, - 582.23, - 269.21, - 592.61 - ], - "text": "Now, X(), the DTFT of x[n], is given by", - "type": "text" - }, - { - "block_id": "p904-b39", - "global_id": 26547, - "bbox": [ - 207.08, - 613.42, - 238.6, - 623.7 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p904-b40", - "global_id": 26548, - "bbox": [ - 244.35, - 603.25, - 258.45, - 613.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p904-b41", - "global_id": 26549, - "bbox": [ - 240.65, - 627.27, - 262.13, - 634.46 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p904-b42", - "global_id": 26550, - "bbox": [ - 263.25, - 609.31, - 311.46, - 623.7 - ], - "text": "x[n]e−jn =", - "type": "text" - }, - { - "block_id": "p904-b43", - "global_id": 26551, - "bbox": [ - 317.2, - 603.25, - 331.29, - 613.92 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p904-b44", - "global_id": 26552, - "bbox": [ - 313.51, - 627.27, - 334.98, - 634.46 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p904-b45", - "global_id": 26553, - "bbox": [ - 336.1, - 611.7, - 384.63, - 624.5 - ], - "text": "xc(nT)e−jn", - "type": "text" - } - ] - }, - { - "page_num": 905, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p905-b0", - "global_id": 26554, - "bbox": [ - 322.62, - 62.89, - 516.13, - 71.98 - ], - "text": "9.5\nDTFT Connection with the CTFT\n885", - "type": "text" - }, - { - "block_id": "p905-b1", - "global_id": 26555, - "bbox": [ - 127.59, - 87.55, - 516.13, - 109.88 - ], - "text": "Comparison of this equation with Eq. (9.40) shows that letting ωT = in Xc(ω) yields X(),\nthat is,", - "type": "text" - }, - { - "block_id": "p905-b2", - "global_id": 26556, - "bbox": [ - 281.27, - 114.14, - 361.96, - 125.22 - ], - "text": "X() = Xc(ω)|ωT=", - "type": "text" - }, - { - "block_id": "p905-b3", - "global_id": 26557, - "bbox": [ - 127.59, - 134.86, - 446.59, - 145.94 - ], - "text": "Alternately, X() can be obtained from Xc(ω) by replacing ω with /T, that is,", - "type": "text" - }, - { - "block_id": "p905-b4", - "global_id": 26558, - "bbox": [ - 287.67, - 164.2, - 330.86, - 175.27 - ], - "text": "X() = Xc", - "type": "text" - }, - { - "block_id": "p905-b6", - "global_id": 26559, - "bbox": [ - 341.11, - 171.58, - 346.65, - 181.55 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p905-b8", - "global_id": 26560, - "bbox": [ - 492.07, - 164.61, - 516.12, - 174.57 - ], - "text": "(9.42)", - "type": "text" - }, - { - "block_id": "p905-b9", - "global_id": 26561, - "bbox": [ - 127.59, - 195.34, - 516.12, - 217.67 - ], - "text": "Therefore, X() is identical to Xc(ω), frequency-scaled by factor T, as shown in Fig. 9.17f. Thus,\nω = 2π/T in Fig. 9.17d corresponds to = 2π in Fig. 9.17f.", - "type": "text" - }, - { - "block_id": "p905-b10", - "global_id": 26562, - "bbox": [ - 127.59, - 244.21, - 493.4, - 256.16 - ], - "text": "9.5-1 Use of DFT and FFT for Numerical Computation of the DTFT", - "type": "text" - }, - { - "block_id": "p905-b11", - "global_id": 26563, - "bbox": [ - 127.59, - 262.3, - 516.13, - 296.17 - ], - "text": "The discrete Fourier transform (DFT), as discussed in Ch. 8, is a tool for computing the samples\nof the continuous-time Fourier transform (CTFT). Because of the close connection between CTFT\nand DTFT, as seen in Eq. (9.42), we can also use this same DFT to compute DTFT samples.", - "type": "text" - }, - { - "block_id": "p905-b12", - "global_id": 26564, - "bbox": [ - 127.59, - 298.06, - 516.12, - 321.57 - ], - "text": "In Ch. 8, Eqs. (8.12) and (8.13) relate an N0-point sequence xn to another N0-point sequence\nXr. Changing the notation xn to x[n] in these equations, we obtain", - "type": "text" - }, - { - "block_id": "p905-b13", - "global_id": 26565, - "bbox": [ - 280.29, - 343.65, - 299.57, - 355.1 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p905-b14", - "global_id": 26566, - "bbox": [ - 301.62, - 332.85, - 318.68, - 344.14 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p905-b15", - "global_id": 26567, - "bbox": [ - 303.94, - 358.04, - 316.35, - 365.3 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p905-b16", - "global_id": 26568, - "bbox": [ - 319.79, - 341.93, - 516.13, - 354.02 - ], - "text": "x[n]e−jr0n\n(9.43)", - "type": "text" - }, - { - "block_id": "p905-b17", - "global_id": 26569, - "bbox": [ - 127.59, - 376.78, - 141.97, - 386.74 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p905-b18", - "global_id": 26570, - "bbox": [ - 275.95, - 392.62, - 312.88, - 409.56 - ], - "text": "x[n] = 1", - "type": "text" - }, - { - "block_id": "p905-b19", - "global_id": 26571, - "bbox": [ - 305.08, - 406.57, - 315.21, - 417.41 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p905-b20", - "global_id": 26572, - "bbox": [ - 318.02, - 388.38, - 335.07, - 399.68 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p905-b21", - "global_id": 26573, - "bbox": [ - 320.64, - 413.57, - 332.44, - 420.84 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p905-b22", - "global_id": 26574, - "bbox": [ - 336.19, - 397.46, - 516.13, - 410.27 - ], - "text": "Xrejr0n\n(9.44)", - "type": "text" - }, - { - "block_id": "p905-b23", - "global_id": 26575, - "bbox": [ - 127.59, - 429.25, - 186.16, - 442.34 - ], - "text": "where 0 = 2π", - "type": "text" - }, - { - "block_id": "p905-b24", - "global_id": 26576, - "bbox": [ - 127.59, - 430.88, - 516.13, - 453.99 - ], - "text": "N0 . Comparing Eq. (9.19) with Eq. (9.43), we recognize that Xr is the sample of X()\nat = r0, that is,", - "type": "text" - }, - { - "block_id": "p905-b25", - "global_id": 26577, - "bbox": [ - 267.36, - 454.87, - 374.16, - 473.31 - ], - "text": "Xr = X(r0)\n0 = 2π", - "type": "text" - }, - { - "block_id": "p905-b26", - "global_id": 26578, - "bbox": [ - 363.87, - 469.24, - 374.0, - 480.08 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p905-b27", - "global_id": 26579, - "bbox": [ - 127.59, - 487.03, - 516.14, - 522.81 - ], - "text": "Hence, DFT Eqs. (9.43) and (9.44) can be viewed to relate an N0-point sequence x[n] to the\nN0-point samples of corresponding X(). We can now use the efficient algorithm FFT (discussed\nin Ch. 8) to compute Xr from x[n], and vice versa.", - "type": "text" - }, - { - "block_id": "p905-b28", - "global_id": 26580, - "bbox": [ - 127.59, - 522.89, - 516.15, - 581.09 - ], - "text": "If x[n] is not timelimited, we can still find the approximate values of Xr by suitably windowing\nx[n]. To reduce the error, the window should be tapered and should have sufficient width to\nsatisfy error specifications. In practice, the numerical computation of signals, which are generally\nnon-timelimited, is performed in this manner because of the computational economy of the DFT,\nespecially for signals of long duration.", - "type": "text" - }, - { - "block_id": "p905-b29", - "global_id": 26581, - "bbox": [ - 127.59, - 596.72, - 516.15, - 636.28 - ], - "text": "COMPUTATION OF DISCRETE-TIME FOURIER SERIES (DTFS)\nThe discrete-time Fourier series (DTFS) equations [(9.3) and (9.4)] are identical to the DFT\nequations [(8.13) and (8.12)] within a scaling constant N0. If we let x[n] = N0xn and Dr = Xr", - "type": "text" - } - ] - }, - { - "page_num": 906, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p906-b0", - "global_id": 26582, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "886\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p906-b1", - "global_id": 26583, - "bbox": [ - 101.84, - 85.82, - 233.55, - 95.78 - ], - "text": "in Eqs. (9.4) and (9.3), we obtain", - "type": "text" - }, - { - "block_id": "p906-b2", - "global_id": 26584, - "bbox": [ - 189.18, - 124.52, - 208.45, - 135.97 - ], - "text": "Xr =", - "type": "text" - }, - { - "block_id": "p906-b3", - "global_id": 26585, - "bbox": [ - 210.5, - 113.72, - 227.56, - 125.01 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p906-b4", - "global_id": 26586, - "bbox": [ - 212.82, - 138.9, - 225.23, - 146.17 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p906-b5", - "global_id": 26587, - "bbox": [ - 228.67, - 117.95, - 348.17, - 135.97 - ], - "text": "xne−jr0n\nand\nxn = 1", - "type": "text" - }, - { - "block_id": "p906-b6", - "global_id": 26588, - "bbox": [ - 340.36, - 131.9, - 350.49, - 142.74 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p906-b7", - "global_id": 26589, - "bbox": [ - 353.3, - 113.72, - 370.35, - 125.01 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p906-b8", - "global_id": 26590, - "bbox": [ - 355.92, - 138.9, - 367.72, - 146.17 - ], - "text": "r=0", - "type": "text" - }, - { - "block_id": "p906-b9", - "global_id": 26591, - "bbox": [ - 371.47, - 122.79, - 402.56, - 135.59 - ], - "text": "Xrejr0n", - "type": "text" - }, - { - "block_id": "p906-b10", - "global_id": 26592, - "bbox": [ - 101.84, - 162.82, - 490.36, - 186.23 - ], - "text": "This is precisely the DFT and IDFT of Eqs. (8.12) and (8.13). For instance, to compute the DTFS\nfor the periodic signal in Fig. 9.2a, we use the values of xn = x[n]/N0 as", - "type": "text" - }, - { - "block_id": "p906-b11", - "global_id": 26593, - "bbox": [ - 204.88, - 209.09, - 223.1, - 220.55 - ], - "text": "xn =", - "type": "text" - }, - { - "block_id": "p906-b12", - "global_id": 26594, - "bbox": [ - 225.14, - 195.09, - 237.73, - 208.83 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p906-b13", - "global_id": 26595, - "bbox": [ - 231.3, - 203.2, - 381.18, - 225.55 - ], - "text": "32\n0 ≤n ≤4\nand\n28 ≤n ≤31\n0\n5 ≤n ≤27", - "type": "text" - }, - { - "block_id": "p906-b14", - "global_id": 26596, - "bbox": [ - 101.84, - 243.87, - 490.39, - 361.43 - ], - "text": "Numerical computations in modern digital signal processing are conveniently performed\nwith the discrete Fourier transform, introduced in Sec. 8.5. The DFT computations can be very\nefficiently executed by using the fast Fourier transform (FFT) algorithm discussed in Sec. 8.6.\nThe DFT is indeed the workhorse of modern digital signal processing. The discrete-time Fourier\ntransform (DTFT) and the inverse discrete-time Fourier transform (IDTFT) can be computed by\nusing the DFT. For an N0-point signal x[n], its DFT yields exactly N0 samples of X() at frequency\nintervals of 2π/N0. We can obtain a larger number of samples of X() by padding a sufficient\nnumber of zero-valued samples to x[n]. The N0-point DFT of x[n] gives exact values of the DTFT\nsamples if x[n] has a finite length N0. If the length of x[n] is infinite, we need to use the appropriate\nwindow function to truncate x[n].", - "type": "text" - }, - { - "block_id": "p906-b15", - "global_id": 26597, - "bbox": [ - 101.85, - 363.42, - 490.39, - 445.11 - ], - "text": "Because of the convolution property, we can use the DFT to compute the convolution of\ntwo signals x[n] and h[n], as discussed in Sec. 8.5. This procedure, known as fast convolution,\nrequires padding both signals by a suitable number of zeros, to make the linear convolution of\nthe two signals identical to the circular (or periodic) convolution of the padded signals. Large\nblocks of data may be processed by sectioning the data into smaller blocks and processing such\nsmaller blocks in sequence. Such a procedure requires smaller memory and reduces the processing\ntime [1].", - "type": "text" - }, - { - "block_id": "p906-b16", - "global_id": 26598, - "bbox": [ - 102.2, - 481.35, - 337.85, - 511.24 - ], - "text": "9.6 GENERALIZATION OF THE DTFT\nTO THE z-TRANSFORM", - "type": "text" - }, - { - "block_id": "p906-b17", - "global_id": 26599, - "bbox": [ - 101.84, - 517.22, - 490.38, - 539.15 - ], - "text": "LTID systems can be analyzed by using the DTFT. This method, however, has the following\nlimitations.", - "type": "text" - }, - { - "block_id": "p906-b18", - "global_id": 26600, - "bbox": [ - 118.78, - 547.11, - 490.4, - 604.9 - ], - "text": "1. Existence of the DTFT is guaranteed only for absolutely summable signals. The DTFT\ndoes not exist for exponentially or even linearly growing signals. This means that the DTFT\nmethod is applicable only for a limited class of inputs.\n2. Moreover, this method can be applied only to asymptotically or BIBO-stable systems; it\ncannot be used for unstable or even marginally stable systems.", - "type": "text" - }, - { - "block_id": "p906-b19", - "global_id": 26601, - "bbox": [ - 101.84, - 612.86, - 490.42, - 634.79 - ], - "text": "These are serious limitations in the study of LTID system analysis. Actually, it is the first\nlimitation that is also the cause of the second limitation. Because the DTFT is incapable of", - "type": "text" - } - ] - }, - { - "page_num": 907, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p907-b0", - "global_id": 26602, - "bbox": [ - 270.13, - 62.57, - 516.13, - 71.98 - ], - "text": "9.6\nGeneralization of the DTFT to the z-Transform\n887", - "type": "text" - }, - { - "block_id": "p907-b1", - "global_id": 26603, - "bbox": [ - 127.59, - 83.03, - 516.13, - 108.58 - ], - "text": "handling growing signals, it is incapable of handling unstable or marginally stable systems.† Our\ngoal is, therefore, to extend the DTFT concept so that it can handle exponentially growing signals.", - "type": "text" - }, - { - "block_id": "p907-b2", - "global_id": 26604, - "bbox": [ - 127.59, - 110.57, - 516.15, - 216.17 - ], - "text": "We may wonder what causes this limitation on DTFT so that it is incapable of handling\nexponentially growing signals. Recall that in the DTFT, we are using sinusoids or exponentials\nof the form ejn to synthesize an arbitrary signal x[n]. These signals are sinusoids with constant\namplitudes. They are incapable of synthesizing exponentially growing signals no matter how\nmany such components we add. Our hope, therefore, lies in trying to synthesize x[n] by using\nexponentially growing sinusoids or exponentials. This goal can be accomplished by generalizing\nthe frequency variable j to σ + j, that is, by using exponentials of the form e(σ+j)n instead of\nexponentials ejn. The procedure is almost identical to that used in extending the Fourier transform\nto the Laplace transform.", - "type": "text" - }, - { - "block_id": "p907-b3", - "global_id": 26605, - "bbox": [ - 145.52, - 215.69, - 350.87, - 228.12 - ], - "text": "Let us define a new variable ˆX(j) = X(). Hence,", - "type": "text" - }, - { - "block_id": "p907-b4", - "global_id": 26606, - "bbox": [ - 273.2, - 245.29, - 307.5, - 257.62 - ], - "text": "ˆX(j) =", - "type": "text" - }, - { - "block_id": "p907-b5", - "global_id": 26607, - "bbox": [ - 313.24, - 237.17, - 327.33, - 247.84 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p907-b6", - "global_id": 26608, - "bbox": [ - 309.55, - 261.2, - 331.02, - 268.39 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p907-b7", - "global_id": 26609, - "bbox": [ - 332.14, - 245.63, - 516.13, - 257.72 - ], - "text": "x[n]e−jn\n(9.45)", - "type": "text" - }, - { - "block_id": "p907-b8", - "global_id": 26610, - "bbox": [ - 127.59, - 278.32, - 141.97, - 288.28 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p907-b9", - "global_id": 26611, - "bbox": [ - 263.84, - 286.37, - 301.44, - 303.32 - ], - "text": "x[n] = 1", - "type": "text" - }, - { - "block_id": "p907-b10", - "global_id": 26612, - "bbox": [ - 292.97, - 300.02, - 303.92, - 310.39 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p907-b11", - "global_id": 26613, - "bbox": [ - 307.23, - 279.38, - 321.24, - 291.43 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p907-b12", - "global_id": 26614, - "bbox": [ - 312.49, - 304.25, - 322.11, - 311.23 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p907-b13", - "global_id": 26615, - "bbox": [ - 324.41, - 289.13, - 379.87, - 303.22 - ], - "text": "ˆX(j)ejn d", - "type": "text" - }, - { - "block_id": "p907-b14", - "global_id": 26616, - "bbox": [ - 127.59, - 314.72, - 309.9, - 328.3 - ], - "text": "Consider now the DTFT of x[n]e−σn (σ real):", - "type": "text" - }, - { - "block_id": "p907-b15", - "global_id": 26617, - "bbox": [ - 197.06, - 345.54, - 275.93, - 357.65 - ], - "text": "DTFT{x[n]e−σn} =", - "type": "text" - }, - { - "block_id": "p907-b16", - "global_id": 26618, - "bbox": [ - 281.66, - 337.1, - 295.76, - 347.77 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p907-b17", - "global_id": 26619, - "bbox": [ - 277.97, - 361.12, - 299.45, - 368.32 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p907-b18", - "global_id": 26620, - "bbox": [ - 300.56, - 343.15, - 368.36, - 357.55 - ], - "text": "x[n]e−σn e−jn =", - "type": "text" - }, - { - "block_id": "p907-b19", - "global_id": 26621, - "bbox": [ - 374.1, - 337.1, - 388.2, - 347.77 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p907-b20", - "global_id": 26622, - "bbox": [ - 370.4, - 361.12, - 391.88, - 368.32 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p907-b21", - "global_id": 26623, - "bbox": [ - 392.99, - 345.54, - 446.14, - 357.55 - ], - "text": "x[n]e−(σ+j)n", - "type": "text" - }, - { - "block_id": "p907-b22", - "global_id": 26624, - "bbox": [ - 127.59, - 378.5, - 364.67, - 390.94 - ], - "text": "It follows from Eq. (9.45) that this sum is ˆX(σ + j). Thus,", - "type": "text" - }, - { - "block_id": "p907-b23", - "global_id": 26625, - "bbox": [ - 216.36, - 408.56, - 295.22, - 420.66 - ], - "text": "DTFT{x[n]e−σn} =", - "type": "text" - }, - { - "block_id": "p907-b24", - "global_id": 26626, - "bbox": [ - 300.97, - 400.11, - 315.07, - 410.78 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p907-b25", - "global_id": 26627, - "bbox": [ - 297.27, - 424.13, - 318.75, - 431.33 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p907-b26", - "global_id": 26628, - "bbox": [ - 319.87, - 406.17, - 516.13, - 420.66 - ], - "text": "x[n]e−(σ+j)n = ˆX(σ + j)\n(9.46)", - "type": "text" - }, - { - "block_id": "p907-b27", - "global_id": 26629, - "bbox": [ - 127.59, - 441.52, - 377.56, - 453.95 - ], - "text": "Hence, the inverse DTFT of ˆX(σ + j) is x[n]e−σn. Therefore,", - "type": "text" - }, - { - "block_id": "p907-b28", - "global_id": 26630, - "bbox": [ - 245.3, - 463.62, - 302.48, - 480.56 - ], - "text": "x[n]e−σn = 1", - "type": "text" - }, - { - "block_id": "p907-b29", - "global_id": 26631, - "bbox": [ - 294.02, - 477.26, - 304.98, - 487.64 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p907-b30", - "global_id": 26632, - "bbox": [ - 308.28, - 456.63, - 322.29, - 468.68 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p907-b31", - "global_id": 26633, - "bbox": [ - 313.54, - 481.51, - 323.16, - 488.48 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p907-b32", - "global_id": 26634, - "bbox": [ - 325.47, - 466.39, - 398.41, - 480.46 - ], - "text": "ˆX(σ + j)ejn d", - "type": "text" - }, - { - "block_id": "p907-b33", - "global_id": 26635, - "bbox": [ - 127.59, - 493.9, - 272.08, - 507.48 - ], - "text": "Multiplying both sides by eσn yields", - "type": "text" - }, - { - "block_id": "p907-b34", - "global_id": 26636, - "bbox": [ - 247.46, - 517.27, - 285.06, - 534.21 - ], - "text": "x[n] = 1", - "type": "text" - }, - { - "block_id": "p907-b35", - "global_id": 26637, - "bbox": [ - 276.59, - 530.91, - 287.54, - 541.29 - ], - "text": "2π", - "type": "text" - }, - { - "block_id": "p907-b36", - "global_id": 26638, - "bbox": [ - 290.85, - 510.27, - 304.86, - 522.33 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p907-b37", - "global_id": 26639, - "bbox": [ - 296.11, - 535.15, - 305.73, - 542.12 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p907-b38", - "global_id": 26640, - "bbox": [ - 308.03, - 520.03, - 516.13, - 534.21 - ], - "text": "ˆX(σ + j)e(σ+j)n d\n(9.47)", - "type": "text" - }, - { - "block_id": "p907-b39", - "global_id": 26641, - "bbox": [ - 127.59, - 551.07, - 256.64, - 561.13 - ], - "text": "Let us define a new variable z as", - "type": "text" - }, - { - "block_id": "p907-b40", - "global_id": 26642, - "bbox": [ - 192.73, - 569.43, - 450.99, - 593.35 - ], - "text": "z = eσ+j\nso that\nlnz = σ + j\nand\n1\nz dz = jd", - "type": "text" - }, - { - "block_id": "p907-b41", - "global_id": 26643, - "bbox": [ - 127.59, - 610.24, - 516.11, - 633.41 - ], - "text": "† Recall that the output of an unstable system grows exponentially. Also, the output of a marginally stable\nsystem to characteristic mode input grows with time.", - "type": "text" - } - ] - }, - { - "page_num": 908, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p908-b0", - "global_id": 26644, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "888\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p908-b1", - "global_id": 26645, - "bbox": [ - 159.6, - 90.46, - 168.49, - 98.46 - ], - "text": "Im", - "type": "text" - }, - { - "block_id": "p908-b2", - "global_id": 26646, - "bbox": [ - 242.55, - 158.71, - 251.44, - 166.71 - ], - "text": "Re", - "type": "text" - }, - { - "block_id": "p908-b3", - "global_id": 26647, - "bbox": [ - 204.65, - 90.38, - 227.09, - 98.46 - ], - "text": "z plane", - "type": "text" - }, - { - "block_id": "p908-b4", - "global_id": 26648, - "bbox": [ - 185.18, - 136.0, - 275.74, - 147.89 - ], - "text": "Path of integration\nr", - "type": "text" - }, - { - "block_id": "p908-b5", - "global_id": 26649, - "bbox": [ - 285.8, - 189.07, - 453.59, - 211.0 - ], - "text": "Figure 9.18 Contour of integration for the\nz-transform.", - "type": "text" - }, - { - "block_id": "p908-b6", - "global_id": 26650, - "bbox": [ - 101.84, - 231.17, - 490.39, - 280.61 - ], - "text": "Because z = eσ+j is complex, we can express it as z = rej, where r = eσ. Thus, z lies on a circle\nof radius r, and as varies from −π to π, z circumambulates along this circle, completing exactly\none counterclockwise rotation, as illustrated in Fig. 9.18. Changing to variable z in Eq. (9.47)\nyields", - "type": "text" - }, - { - "block_id": "p908-b7", - "global_id": 26651, - "bbox": [ - 240.85, - 281.15, - 284.7, - 305.17 - ], - "text": "x[n] =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p908-b8", - "global_id": 26652, - "bbox": [ - 287.01, - 274.16, - 490.38, - 298.1 - ], - "text": "5\nˆX(lnz)zn−1dz\n(9.48)", - "type": "text" - }, - { - "block_id": "p908-b9", - "global_id": 26653, - "bbox": [ - 101.84, - 312.38, - 222.2, - 322.34 - ], - "text": "and from Eq. (9.46) we obtain", - "type": "text" - }, - { - "block_id": "p908-b10", - "global_id": 26654, - "bbox": [ - 250.3, - 330.56, - 286.82, - 342.99 - ], - "text": "ˆX(lnz) =", - "type": "text" - }, - { - "block_id": "p908-b11", - "global_id": 26655, - "bbox": [ - 292.55, - 322.44, - 306.65, - 333.12 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p908-b12", - "global_id": 26656, - "bbox": [ - 288.86, - 346.46, - 310.34, - 353.66 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p908-b13", - "global_id": 26657, - "bbox": [ - 311.45, - 330.89, - 490.38, - 342.99 - ], - "text": "x[n]z−n\n(9.49)", - "type": "text" - }, - { - "block_id": "p908-b14", - "global_id": 26658, - "bbox": [ - 101.85, - 362.55, - 172.88, - 372.51 - ], - "text": "where the integral", - "type": "text" - }, - { - "block_id": "p908-b15", - "global_id": 26659, - "bbox": [ - 101.84, - 354.11, - 490.37, - 384.46 - ], - "text": "6\nindicates a contour integral around a circle of radius r in the counterclockwise\ndirection.", - "type": "text" - }, - { - "block_id": "p908-b16", - "global_id": 26660, - "bbox": [ - 101.84, - 386.46, - 490.39, - 420.33 - ], - "text": "Equations (9.48) and (9.49) are the desired extensions. They are, however, in a clumsy form.\nFor the sake of convenience, we make another notational change by observing that ˆX(lnz) is a\nfunction of z. Let us denote it by a simpler notation X[z]. Thus, Eq. (9.48) becomes", - "type": "text" - }, - { - "block_id": "p908-b17", - "global_id": 26661, - "bbox": [ - 245.68, - 430.65, - 289.54, - 454.67 - ], - "text": "x[n] =\n1\n2πj", - "type": "text" - }, - { - "block_id": "p908-b18", - "global_id": 26662, - "bbox": [ - 291.83, - 423.66, - 490.38, - 447.6 - ], - "text": "5\nX[z]zn−1dz\n(9.50)", - "type": "text" - }, - { - "block_id": "p908-b19", - "global_id": 26663, - "bbox": [ - 101.84, - 464.87, - 196.18, - 474.83 - ], - "text": "and Eq. (9.49) becomes", - "type": "text" - }, - { - "block_id": "p908-b20", - "global_id": 26664, - "bbox": [ - 255.13, - 485.1, - 281.98, - 495.38 - ], - "text": "X[z] =", - "type": "text" - }, - { - "block_id": "p908-b21", - "global_id": 26665, - "bbox": [ - 287.73, - 474.93, - 301.83, - 485.6 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p908-b22", - "global_id": 26666, - "bbox": [ - 284.03, - 498.95, - 305.51, - 506.14 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p908-b23", - "global_id": 26667, - "bbox": [ - 306.63, - 483.38, - 490.38, - 495.48 - ], - "text": "x[n]z−n\n(9.51)", - "type": "text" - }, - { - "block_id": "p908-b24", - "global_id": 26668, - "bbox": [ - 101.85, - 513.38, - 490.39, - 547.67 - ], - "text": "This is the (bilateral) z-transform pair. Equation (9.50) expresses x[n] as a continuous sum of\nexponentials of the form zn = e(σ+j)n = rn ejn. Thus, by selecting a proper value for r (or σ), we\ncan make the exponential grow (or decay) at any exponential rate we desire.", - "type": "text" - }, - { - "block_id": "p908-b25", - "global_id": 26669, - "bbox": [ - 119.78, - 546.04, - 263.87, - 559.62 - ], - "text": "If we let σ = 0, we have z = ej and", - "type": "text" - }, - { - "block_id": "p908-b26", - "global_id": 26670, - "bbox": [ - 211.76, - 571.42, - 289.96, - 585.41 - ], - "text": "X[z]|z=ej = ˆX(lnz)", - "type": "text" - }, - { - "block_id": "p908-b28", - "global_id": 26671, - "bbox": [ - 293.19, - 571.42, - 381.67, - 589.89 - ], - "text": "z=ej = ˆX(j) = X()", - "type": "text" - }, - { - "block_id": "p908-b29", - "global_id": 26672, - "bbox": [ - 101.85, - 599.27, - 489.89, - 610.87 - ], - "text": "Thus, the familiar DTFT is just a special case of the z-transform X[z] obtained by letting z = ej", - "type": "text" - }, - { - "block_id": "p908-b30", - "global_id": 26673, - "bbox": [ - 101.84, - 611.22, - 490.39, - 634.79 - ], - "text": "and assuming that the sum on the right-hand side of Eq. (9.51) converges when z = ej. This also\nimplies that the ROC for X[z] includes the unit circle.", - "type": "text" - } - ] - }, - { - "page_num": 909, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p909-b0", - "global_id": 26674, - "bbox": [ - 259.25, - 62.89, - 516.14, - 71.98 - ], - "text": "9.7\nMATLAB: Working with the DTFS and the DTFT\n889", - "type": "text" - }, - { - "block_id": "p909-b1", - "global_id": 26675, - "bbox": [ - 127.94, - 92.21, - 404.46, - 122.09 - ], - "text": "9.7 MATLAB: WORKING WITH THE DTFS\nAND THE DTFT", - "type": "text" - }, - { - "block_id": "p909-b2", - "global_id": 26676, - "bbox": [ - 127.59, - 128.09, - 516.14, - 173.92 - ], - "text": "This section investigates various methods to compute the discrete-time Fourier series (DTFS).\nPerformance of these methods is assessed by using MATLAB’s stopwatch and profiling functions.\nAdditionally, the discrete-time Fourier transform (DTFT) is applied to the important topic of finite\nimpulse response (FIR) filter design.", - "type": "text" - }, - { - "block_id": "p909-b3", - "global_id": 26677, - "bbox": [ - 127.59, - 198.1, - 399.7, - 210.06 - ], - "text": "9.7-1 Computing the Discrete-Time Fourier Series", - "type": "text" - }, - { - "block_id": "p909-b4", - "global_id": 26678, - "bbox": [ - 127.59, - 216.19, - 516.13, - 250.06 - ], - "text": "Within a scale factor, the DTFS is identical to the DFT. Thus, methods to compute the DFT can\nbe readily used to compute the DTFS. Specifically, the DTFS is the DFT scaled by 1/N0. As an\nexample, consider a 50 Hz sinusoid sampled at 1000 Hz over one-tenth of a second.", - "type": "text" - }, - { - "block_id": "p909-b5", - "global_id": 26679, - "bbox": [ - 127.59, - 259.74, - 347.23, - 281.67 - ], - "text": ">>\nT = 1/1000; N_0 = 100; n = (0:N_0-1)’;\n>>\nx = cos(2*pi*50*n*T);", - "type": "text" - }, - { - "block_id": "p909-b6", - "global_id": 26680, - "bbox": [ - 127.59, - 290.77, - 296.74, - 300.74 - ], - "text": "The DTFS is obtained by scaling the DFT.", - "type": "text" - }, - { - "block_id": "p909-b7", - "global_id": 26681, - "bbox": [ - 127.59, - 310.42, - 483.25, - 344.3 - ], - "text": ">>\nX = fft(x)/N_0; f = (0:N_0-1)/(T*N_0);\n>>\nstem(f-1/(2*T),fftshift(abs(X)),’k.’);\n>>\naxis([-500 500 -0.05 0.55]); xlabel(’f [Hz]’); ylabel(’|X(f)|’);", - "type": "text" - }, - { - "block_id": "p909-b8", - "global_id": 26682, - "bbox": [ - 127.59, - 352.99, - 516.13, - 375.32 - ], - "text": "Figure 9.19 shows a peak magnitude of 0.5 at ±50 Hz. This result is consistent with Euler’s\nrepresentation", - "type": "text" - }, - { - "block_id": "p909-b9", - "global_id": 26683, - "bbox": [ - 242.78, - 375.29, - 315.85, - 392.24 - ], - "text": "cos(2π50nT) = 1", - "type": "text" - }, - { - "block_id": "p909-b10", - "global_id": 26684, - "bbox": [ - 310.87, - 375.29, - 399.87, - 399.31 - ], - "text": "2ej2π50nT + 1\n2e−j2π50nT", - "type": "text" - }, - { - "block_id": "p909-b11", - "global_id": 26685, - "bbox": [ - 127.59, - 403.13, - 471.59, - 415.0 - ], - "text": "Lacking the 1/N0 scale factor, the DFT would have a peak amplitude 100 times larger.", - "type": "text" - }, - { - "block_id": "p909-b12", - "global_id": 26686, - "bbox": [ - 145.52, - 415.39, - 402.44, - 426.23 - ], - "text": "The inverse DTFS is obtained by scaling the inverse DFT by N0.", - "type": "text" - }, - { - "block_id": "p909-b13", - "global_id": 26687, - "bbox": [ - 127.59, - 435.14, - 415.26, - 457.07 - ], - "text": ">>\nx = real(ifft(X)*N_0); stem(n,x,’k.’);\n>>\naxis([0 99 -1.1 1.1]); xlabel(’n’); ylabel(’x[n]’);", - "type": "text" - }, - { - "block_id": "p909-b14", - "global_id": 26688, - "bbox": [ - 127.59, - 465.76, - 516.15, - 500.33 - ], - "text": "Figure 9.20 confirms that the sinusoid x[n] is properly recovered. Although the result is\ntheoretically real, computer round-off errors produce a small imaginary component, which the\nreal command removes.", - "type": "text" - }, - { - "block_id": "p909-b15", - "global_id": 26689, - "bbox": [ - 144.75, - 592.1, - 495.3, - 613.28 - ], - "text": "–500\n–400\n–300\n–200\n–100\n0\n100\n200\n300\n400\n500\nf [Hz]", - "type": "text" - }, - { - "block_id": "p909-b16", - "global_id": 26690, - "bbox": [ - 147.07, - 576.86, - 151.07, - 584.86 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p909-b17", - "global_id": 26691, - "bbox": [ - 140.32, - 552.86, - 151.0, - 560.86 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p909-b18", - "global_id": 26692, - "bbox": [ - 140.32, - 528.87, - 151.0, - 536.87 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p909-b19", - "global_id": 26693, - "bbox": [ - 128.36, - 540.69, - 137.16, - 559.35 - ], - "text": "|X(f)|", - "type": "text" - }, - { - "block_id": "p909-b20", - "global_id": 26694, - "bbox": [ - 127.59, - 619.96, - 309.99, - 629.19 - ], - "text": "Figure 9.19 DTFS computed by scaling the DFT.", - "type": "text" - } - ] - }, - { - "page_num": 910, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p910-b0", - "global_id": 26695, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "890\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p910-b1", - "global_id": 26696, - "bbox": [ - 146.97, - 165.57, - 457.26, - 186.76 - ], - "text": "0\n10\n20\n30\n40\n50\n60\n70\n80\n90\nn", - "type": "text" - }, - { - "block_id": "p910-b2", - "global_id": 26697, - "bbox": [ - 136.74, - 153.08, - 144.74, - 161.08 - ], - "text": "–1", - "type": "text" - }, - { - "block_id": "p910-b3", - "global_id": 26698, - "bbox": [ - 140.74, - 120.34, - 144.74, - 128.34 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p910-b4", - "global_id": 26699, - "bbox": [ - 140.74, - 87.63, - 144.74, - 95.63 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p910-b5", - "global_id": 26700, - "bbox": [ - 126.52, - 115.91, - 135.32, - 130.57 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p910-b6", - "global_id": 26701, - "bbox": [ - 125.76, - 193.44, - 363.92, - 202.68 - ], - "text": "Figure 9.20 Inverse DTFS computed by scaling the inverse DFT.", - "type": "text" - }, - { - "block_id": "p910-b7", - "global_id": 26702, - "bbox": [ - 101.84, - 225.49, - 490.43, - 283.27 - ], - "text": "Although MATLAB’s fft command provides an efficient method to compute the DTFS,\nother important computational methods exist. A matrix-based approach is one popular way\nto implement Eq. (9.4). Although not as efficient as an FFT-based algorithm, matrix-based\napproaches provide insight into the DTFS and serve as an excellent model for solving similarly\nstructured problems.", - "type": "text" - }, - { - "block_id": "p910-b8", - "global_id": 26703, - "bbox": [ - 101.85, - 283.62, - 490.39, - 307.18 - ], - "text": "To begin, define WN0 = ej0, which is a constant for a given N0. Substituting WN0 into Eq. (9.4)\nyields", - "type": "text" - }, - { - "block_id": "p910-b9", - "global_id": 26704, - "bbox": [ - 249.81, - 315.88, - 281.73, - 333.91 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p910-b10", - "global_id": 26705, - "bbox": [ - 273.92, - 329.83, - 284.05, - 340.67 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p910-b11", - "global_id": 26706, - "bbox": [ - 286.86, - 311.65, - 303.92, - 322.95 - ], - "text": "N0−1\n\"", - "type": "text" - }, - { - "block_id": "p910-b12", - "global_id": 26707, - "bbox": [ - 289.19, - 336.84, - 301.59, - 344.1 - ], - "text": "n=0", - "type": "text" - }, - { - "block_id": "p910-b13", - "global_id": 26708, - "bbox": [ - 305.03, - 320.6, - 341.75, - 332.73 - ], - "text": "x[n]W−nr", - "type": "text" - }, - { - "block_id": "p910-b14", - "global_id": 26709, - "bbox": [ - 329.39, - 327.87, - 337.03, - 336.18 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p910-b15", - "global_id": 26710, - "bbox": [ - 101.84, - 352.72, - 285.86, - 363.8 - ], - "text": "An inner product of two vectors computes Dr.", - "type": "text" - }, - { - "block_id": "p910-b16", - "global_id": 26711, - "bbox": [ - 171.59, - 397.23, - 203.5, - 415.26 - ], - "text": "Dr = 1", - "type": "text" - }, - { - "block_id": "p910-b17", - "global_id": 26712, - "bbox": [ - 195.7, - 411.19, - 205.82, - 422.03 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p910-b18", - "global_id": 26713, - "bbox": [ - 208.63, - 392.81, - 213.3, - 402.77 - ], - "text": "'", - "type": "text" - }, - { - "block_id": "p910-b19", - "global_id": 26714, - "bbox": [ - 213.3, - 401.96, - 245.42, - 414.18 - ], - "text": "1\nW−r", - "type": "text" - }, - { - "block_id": "p910-b20", - "global_id": 26715, - "bbox": [ - 236.55, - 401.96, - 276.71, - 417.54 - ], - "text": "N0\nW−2r", - "type": "text" - }, - { - "block_id": "p910-b21", - "global_id": 26716, - "bbox": [ - 264.35, - 400.97, - 352.15, - 417.54 - ], - "text": "N0\n. . .\nW−(N0−1)r", - "type": "text" - }, - { - "block_id": "p910-b22", - "global_id": 26717, - "bbox": [ - 321.01, - 409.23, - 328.65, - 417.54 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p910-b23", - "global_id": 26718, - "bbox": [ - 352.82, - 392.81, - 357.49, - 402.77 - ], - "text": "(", - "type": "text" - }, - { - "block_id": "p910-b24", - "global_id": 26719, - "bbox": [ - 358.6, - 368.89, - 365.86, - 378.86 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p910-b25", - "global_id": 26720, - "bbox": [ - 358.6, - 386.37, - 365.86, - 426.68 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p910-b26", - "global_id": 26721, - "bbox": [ - 381.59, - 376.51, - 397.66, - 410.79 - ], - "text": "x[0]\nx[1]\nx[2]", - "type": "text" - }, - { - "block_id": "p910-b27", - "global_id": 26722, - "bbox": [ - 370.84, - 411.4, - 408.4, - 442.37 - ], - "text": "...\nx[N0 −1]", - "type": "text" - }, - { - "block_id": "p910-b28", - "global_id": 26723, - "bbox": [ - 413.39, - 368.89, - 420.65, - 378.86 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p910-b29", - "global_id": 26724, - "bbox": [ - 413.39, - 386.37, - 420.65, - 426.68 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p910-b30", - "global_id": 26725, - "bbox": [ - 101.85, - 454.85, - 239.33, - 464.91 - ], - "text": "Stacking the results for all r yields", - "type": "text" - }, - { - "block_id": "p910-b31", - "global_id": 26726, - "bbox": [ - 126.15, - 475.29, - 133.41, - 485.25 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p910-b32", - "global_id": 26727, - "bbox": [ - 126.15, - 492.77, - 133.41, - 533.08 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p910-b33", - "global_id": 26728, - "bbox": [ - 145.17, - 482.89, - 156.34, - 517.96 - ], - "text": "D0\nD1\nD2", - "type": "text" - }, - { - "block_id": "p910-b34", - "global_id": 26729, - "bbox": [ - 138.38, - 517.8, - 163.13, - 549.72 - ], - "text": "...\nDN0−1", - "type": "text" - }, - { - "block_id": "p910-b35", - "global_id": 26730, - "bbox": [ - 168.62, - 475.29, - 175.88, - 485.25 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p910-b36", - "global_id": 26731, - "bbox": [ - 168.62, - 492.77, - 175.88, - 533.08 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p910-b37", - "global_id": 26732, - "bbox": [ - 177.92, - 503.63, - 196.74, - 520.58 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p910-b38", - "global_id": 26733, - "bbox": [ - 188.94, - 517.59, - 199.07, - 528.42 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p910-b39", - "global_id": 26734, - "bbox": [ - 201.87, - 472.3, - 209.13, - 482.27 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p910-b40", - "global_id": 26735, - "bbox": [ - 201.87, - 489.78, - 209.13, - 536.07 - ], - "text": "⎢⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p910-b41", - "global_id": 26736, - "bbox": [ - 214.12, - 478.32, - 372.81, - 502.31 - ], - "text": "1\n1\n1\n· · ·\n1\n1\nW−1", - "type": "text" - }, - { - "block_id": "p910-b42", - "global_id": 26737, - "bbox": [ - 246.74, - 490.08, - 305.34, - 505.67 - ], - "text": "N0\nW−2", - "type": "text" - }, - { - "block_id": "p910-b43", - "global_id": 26738, - "bbox": [ - 295.69, - 489.09, - 388.43, - 505.67 - ], - "text": "N0\n· · ·\nW−(N0−1)", - "type": "text" - }, - { - "block_id": "p910-b44", - "global_id": 26739, - "bbox": [ - 214.12, - 497.36, - 367.65, - 516.71 - ], - "text": "N0\n1\nW−2", - "type": "text" - }, - { - "block_id": "p910-b45", - "global_id": 26740, - "bbox": [ - 246.74, - 504.48, - 305.34, - 520.07 - ], - "text": "N0\nW−4", - "type": "text" - }, - { - "block_id": "p910-b46", - "global_id": 26741, - "bbox": [ - 295.69, - 503.49, - 390.18, - 520.07 - ], - "text": "N0\n· · ·\nW−2(N0−1)", - "type": "text" - }, - { - "block_id": "p910-b47", - "global_id": 26742, - "bbox": [ - 215.36, - 511.76, - 371.57, - 536.04 - ], - "text": "N0\n...\n...\n...\n· · ·\n...", - "type": "text" - }, - { - "block_id": "p910-b48", - "global_id": 26743, - "bbox": [ - 214.12, - 538.24, - 265.79, - 551.47 - ], - "text": "1\nW−(N0−1)", - "type": "text" - }, - { - "block_id": "p910-b49", - "global_id": 26744, - "bbox": [ - 237.36, - 538.24, - 316.47, - 554.81 - ], - "text": "N0\nW−2(N0−1)", - "type": "text" - }, - { - "block_id": "p910-b50", - "global_id": 26745, - "bbox": [ - 284.55, - 536.76, - 389.68, - 554.81 - ], - "text": "N0\n· · ·\nW−(N0−1)2", - "type": "text" - }, - { - "block_id": "p910-b51", - "global_id": 26746, - "bbox": [ - 358.26, - 546.51, - 365.9, - 554.81 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p910-b52", - "global_id": 26747, - "bbox": [ - 395.67, - 472.3, - 402.93, - 482.27 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p910-b53", - "global_id": 26748, - "bbox": [ - 395.67, - 489.78, - 402.93, - 536.07 - ], - "text": "⎥⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p910-b54", - "global_id": 26749, - "bbox": [ - 404.04, - 475.29, - 411.3, - 485.25 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p910-b55", - "global_id": 26750, - "bbox": [ - 404.04, - 492.77, - 411.3, - 533.08 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p910-b56", - "global_id": 26751, - "bbox": [ - 427.03, - 482.89, - 443.1, - 517.18 - ], - "text": "x[0]\nx[1]\nx[2]", - "type": "text" - }, - { - "block_id": "p910-b57", - "global_id": 26752, - "bbox": [ - 416.28, - 517.79, - 453.84, - 548.77 - ], - "text": "...\nx[N0 −1]", - "type": "text" - }, - { - "block_id": "p910-b58", - "global_id": 26753, - "bbox": [ - 458.83, - 475.29, - 466.09, - 485.25 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p910-b59", - "global_id": 26754, - "bbox": [ - 458.83, - 492.77, - 466.09, - 533.08 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p910-b60", - "global_id": 26755, - "bbox": [ - 101.85, - 565.92, - 324.61, - 575.88 - ], - "text": "In matrix notation, this equation is compactly written as", - "type": "text" - }, - { - "block_id": "p910-b61", - "global_id": 26756, - "bbox": [ - 267.28, - 588.12, - 297.36, - 605.07 - ], - "text": "D = 1", - "type": "text" - }, - { - "block_id": "p910-b62", - "global_id": 26757, - "bbox": [ - 289.55, - 602.08, - 299.68, - 612.92 - ], - "text": "N0", - "type": "text" - }, - { - "block_id": "p910-b63", - "global_id": 26758, - "bbox": [ - 301.37, - 595.03, - 324.96, - 607.1 - ], - "text": "WN0x", - "type": "text" - }, - { - "block_id": "p910-b64", - "global_id": 26759, - "bbox": [ - 101.84, - 623.83, - 432.4, - 636.88 - ], - "text": "Since it is also used to compute the DFT, matrix WN0 is often called a DFT matrix.", - "type": "text" - } - ] - }, - { - "page_num": 911, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p911-b0", - "global_id": 26760, - "bbox": [ - 259.25, - 62.89, - 516.14, - 71.98 - ], - "text": "9.7\nMATLAB: Working with the DTFS and the DTFT\n891", - "type": "text" - }, - { - "block_id": "p911-b1", - "global_id": 26761, - "bbox": [ - 127.59, - 85.85, - 516.13, - 119.82 - ], - "text": "Let us create an anonymous function to compute the N0-by-N0 DFT matrix WN0. Although\nnot used here, the signal-processing toolbox function dftmtx computes the same DFT matrix,\nalthough in a less obvious but more efficient fashion.", - "type": "text" - }, - { - "block_id": "p911-b2", - "global_id": 26762, - "bbox": [ - 127.59, - 132.05, - 430.95, - 142.01 - ], - "text": ">>\nW = @(N_0) (exp(-j*2*pi/N_0)).^((0:N_0-1)’*(0:N_0-1));", - "type": "text" - }, - { - "block_id": "p911-b3", - "global_id": 26763, - "bbox": [ - 127.59, - 153.66, - 512.47, - 163.62 - ], - "text": "While less efficient than FFT-based methods, the matrix approach correctly computes the DTFS.", - "type": "text" - }, - { - "block_id": "p911-b4", - "global_id": 26764, - "bbox": [ - 127.59, - 175.85, - 483.25, - 197.77 - ], - "text": ">>\nX = W(N_0)*x/N_0; stem(f-1/(2*T),fftshift(abs(X)),’k.’);\n>>\naxis([-500 500 -0.05 0.55]); xlabel(’f [Hz]’); ylabel(’|X(f)|’);", - "type": "text" - }, - { - "block_id": "p911-b5", - "global_id": 26765, - "bbox": [ - 127.59, - 209.42, - 516.13, - 231.34 - ], - "text": "The resulting plot is indistinguishable from Fig. 9.19. Problem 9.7-1 investigates a matrix-based\napproach to compute Eq. (9.3), the inverse DTFS.", - "type": "text" - }, - { - "block_id": "p911-b6", - "global_id": 26766, - "bbox": [ - 127.59, - 258.07, - 320.71, - 270.02 - ], - "text": "9.7-2 Measuring Code Performance", - "type": "text" - }, - { - "block_id": "p911-b7", - "global_id": 26767, - "bbox": [ - 127.59, - 276.16, - 516.15, - 334.23 - ], - "text": "Writing efficient code is important, particularly if the code is frequently used, requires complicated\noperations, involves large data sets, or operates in real time. MATLAB provides several tools\nfor assessing code performance. When properly used, the profile function provides detailed\nstatistics that help assess code performance. MATLAB help thoroughly describes the use of the\nsophisticated profile command.", - "type": "text" - }, - { - "block_id": "p911-b8", - "global_id": 26768, - "bbox": [ - 127.59, - 335.93, - 516.13, - 381.76 - ], - "text": "A simpler method of assessing code efficiency is to measure execution time and compare it\nwith a reference. The MATLAB command tic starts a stopwatch timer. The toc command reads\nthe timer. Sandwiching instructions between tic and toc returns the elapsed time. For example,\nthe execution time of the 100-point matrix-based DTFS computation is", - "type": "text" - }, - { - "block_id": "p911-b9", - "global_id": 26769, - "bbox": [ - 127.59, - 393.99, - 321.1, - 415.9 - ], - "text": ">>\ntic; W(N_0)*x/N_0; toc\nElapsed time is 0.004417 seconds.", - "type": "text" - }, - { - "block_id": "p911-b10", - "global_id": 26770, - "bbox": [ - 127.59, - 427.56, - 516.18, - 485.34 - ], - "text": "Different machines operate at different speeds with different operating systems and with different\nbackground tasks. Therefore, elapsed-time measurements can vary considerably from machine to\nmachine and from execution to execution. For relatively simple and short events like the present\ncase, execution times can be so brief that MATLAB may report unreliable times or fail to register\nan elapsed time at all.", - "type": "text" - }, - { - "block_id": "p911-b11", - "global_id": 26771, - "bbox": [ - 127.59, - 487.33, - 516.12, - 509.25 - ], - "text": "To increase the elapsed time and therefore the accuracy of the time measurement, a loop is\nused to repeat the calculation.", - "type": "text" - }, - { - "block_id": "p911-b12", - "global_id": 26772, - "bbox": [ - 127.59, - 521.48, - 357.72, - 543.39 - ], - "text": ">>\ntic; for i=1:100, W(N_0)*x/N_0; end; toc\nElapsed time is 0.173388 seconds.", - "type": "text" - }, - { - "block_id": "p911-b13", - "global_id": 26773, - "bbox": [ - 127.59, - 555.04, - 516.16, - 600.87 - ], - "text": "This elapsed time suggests that each 100-point DTFS calculation takes a little under 2\nmilliseconds. What exactly does this mean, however? Elapsed time is only meaningful relative\nto some reference. Let us see what difference occurs by precomputing the DFT matrix, rather than\nrepeatedly using our anonymous function.", - "type": "text" - }, - { - "block_id": "p911-b14", - "global_id": 26774, - "bbox": [ - 127.59, - 613.1, - 425.71, - 635.02 - ], - "text": ">>\nW100 = W(100); tic; for i=1:100, W100*x/N_0; end; toc\nElapsed time is 0.001199 seconds.", - "type": "text" - } - ] - }, - { - "page_num": 912, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p912-b0", - "global_id": 26775, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "892\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p912-b1", - "global_id": 26776, - "bbox": [ - 101.84, - 85.82, - 490.4, - 107.74 - ], - "text": "Amazingly, this small change makes a hundredfold change in our computational efficiency!\nClearly, it is much better to precompute the DFT matrix.", - "type": "text" - }, - { - "block_id": "p912-b2", - "global_id": 26777, - "bbox": [ - 101.84, - 109.73, - 490.39, - 131.64 - ], - "text": "To provide another example, consider the time it takes to compute the same DTFS using the\nFFT-based approach.", - "type": "text" - }, - { - "block_id": "p912-b3", - "global_id": 26778, - "bbox": [ - 101.84, - 141.65, - 321.52, - 163.57 - ], - "text": ">>\ntic; for i=1:100, fft(x)/N_0; end; toc\nElapsed time is 0.000399 seconds.", - "type": "text" - }, - { - "block_id": "p912-b4", - "global_id": 26779, - "bbox": [ - 101.84, - 173.0, - 490.4, - 254.69 - ], - "text": "With this as a reference, our fastest matrix-based computations appear to be several times slower\nthan the FFT-based computations. This difference becomes more dramatic as N0 is increased. Since\nthe two methods provide identical results, there is little incentive to use the slower matrix-based\napproach, and the FFT-based algorithm is generally preferred. Even so, the FFT can exhibit curious\nbehavior: adding a few data points, even the artificial samples introduced by zero padding, can\ndramatically increase or decrease execution times. The tic and toc commands illustrate this\nstrange result. Consider computing the DTFS of 1015 random data points 100 times.", - "type": "text" - }, - { - "block_id": "p912-b5", - "global_id": 26780, - "bbox": [ - 101.85, - 264.7, - 457.51, - 286.62 - ], - "text": ">>\nx1 = rand(1015,1); tic; for i=1:100; fft(x1)/1015; end; T1 = toc\nT1 = 0.0067", - "type": "text" - }, - { - "block_id": "p912-b6", - "global_id": 26781, - "bbox": [ - 101.85, - 296.05, - 259.39, - 306.01 - ], - "text": "Next, pad the sequence with four zeros.", - "type": "text" - }, - { - "block_id": "p912-b7", - "global_id": 26782, - "bbox": [ - 101.85, - 316.02, - 473.2, - 337.93 - ], - "text": ">>\nx2 = [x1;zeros(4,1)]; tic; for i=1:100; fft(x2)/1019; end; T2 = toc\nT2 = 0.0134", - "type": "text" - }, - { - "block_id": "p912-b8", - "global_id": 26783, - "bbox": [ - 101.85, - 347.37, - 490.39, - 382.32 - ], - "text": "The ratio of the two elapsed times indicates that adding four points to an already long sequence\nincreases the computation time by a factor of 2. Next, the sequence is zero-padded to a length of\nN0 = 1024.", - "type": "text" - }, - { - "block_id": "p912-b9", - "global_id": 26784, - "bbox": [ - 101.84, - 391.24, - 473.2, - 413.16 - ], - "text": ">>\nx3 = [x2;zeros(5,1)]; tic; for i=1:100; fft(x3)/1024; end; T3 = toc\nT3 = 0.0017", - "type": "text" - }, - { - "block_id": "p912-b10", - "global_id": 26785, - "bbox": [ - 101.84, - 422.59, - 490.39, - 456.75 - ], - "text": "In this case, the added data decrease the original execution time by a factor of 4 and the second\nexecution time by a factor of 8! These results are particularly surprising when it is realized that the\nlengths of y1, y2, and y3 differ by less than 1%.", - "type": "text" - }, - { - "block_id": "p912-b11", - "global_id": 26786, - "bbox": [ - 101.85, - 458.35, - 490.42, - 516.24 - ], - "text": "As it turns out, the efficiency of the fft command depends on the factorability of N0. With\nthe factor command, 1015 = (5)(7)(29), 1019 is prime, and 1024 = (2)10. The most factorable\nlength, 1024, results in the fastest execution, while the least factorable length, 1019, results in\nthe slowest execution. To ensure the greatest factorability and fastest operation, vector lengths are\nideally a power of 2.", - "type": "text" - }, - { - "block_id": "p912-b12", - "global_id": 26787, - "bbox": [ - 101.84, - 540.75, - 359.7, - 552.71 - ], - "text": "9.7-3 FIR Filter Design by Frequency Sampling", - "type": "text" - }, - { - "block_id": "p912-b13", - "global_id": 26788, - "bbox": [ - 101.84, - 558.83, - 490.41, - 592.7 - ], - "text": "Finite impulse response (FIR) digital filters are flexible, always stable, and relatively easy to\nimplement. These qualities make FIR filters a popular choice among digital filter designers. The\ndifference equation of a length-N causal FIR filter is conveniently expressed as", - "type": "text" - }, - { - "block_id": "p912-b14", - "global_id": 26789, - "bbox": [ - 154.24, - 613.38, - 380.52, - 624.53 - ], - "text": "y[n] = h0x[n] + h1x[n −1] + · · · + hN−1x[n −(N −1)] =", - "type": "text" - }, - { - "block_id": "p912-b15", - "global_id": 26790, - "bbox": [ - 382.56, - 603.2, - 396.66, - 613.87 - ], - "text": "N−1\n\"", - "type": "text" - }, - { - "block_id": "p912-b16", - "global_id": 26791, - "bbox": [ - 383.54, - 627.77, - 395.67, - 635.03 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p912-b17", - "global_id": 26792, - "bbox": [ - 397.76, - 613.38, - 437.97, - 624.46 - ], - "text": "hkx[n −k]", - "type": "text" - } - ] - }, - { - "page_num": 913, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p913-b0", - "global_id": 26793, - "bbox": [ - 259.25, - 62.89, - 516.14, - 71.98 - ], - "text": "9.7\nMATLAB: Working with the DTFS and the DTFT\n893", - "type": "text" - }, - { - "block_id": "p913-b1", - "global_id": 26794, - "bbox": [ - 127.59, - 85.82, - 516.15, - 119.69 - ], - "text": "The filter coefficients, or tap weights as they are sometimes called, are expressed by using the\nvariable h to emphasize that the coefficients themselves represent the impulse response of the\nfilter.", - "type": "text" - }, - { - "block_id": "p913-b2", - "global_id": 26795, - "bbox": [ - 145.52, - 121.68, - 277.99, - 131.64 - ], - "text": "The filter’s frequency response is", - "type": "text" - }, - { - "block_id": "p913-b3", - "global_id": 26796, - "bbox": [ - 264.11, - 145.48, - 321.59, - 162.74 - ], - "text": "H() = Y()", - "type": "text" - }, - { - "block_id": "p913-b4", - "global_id": 26797, - "bbox": [ - 300.0, - 152.46, - 332.73, - 169.81 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p913-b5", - "global_id": 26798, - "bbox": [ - 334.77, - 142.29, - 348.87, - 152.96 - ], - "text": "N−1\n\"", - "type": "text" - }, - { - "block_id": "p913-b6", - "global_id": 26799, - "bbox": [ - 335.74, - 166.85, - 347.88, - 174.11 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p913-b7", - "global_id": 26800, - "bbox": [ - 349.97, - 150.74, - 378.98, - 163.54 - ], - "text": "hke−jk", - "type": "text" - }, - { - "block_id": "p913-b8", - "global_id": 26801, - "bbox": [ - 127.59, - 183.11, - 516.13, - 205.43 - ], - "text": "Since H() is a 2π-periodic function of the continuous variable , it is sufficient to specify H()\nover a single period (0 ≤ < 2π).", - "type": "text" - }, - { - "block_id": "p913-b9", - "global_id": 26802, - "bbox": [ - 127.59, - 207.02, - 516.13, - 241.31 - ], - "text": "In many filtering applications, the desired magnitude response |Hd()| is known but not the\nfilter coefficients h[n]. The question, then, is one of determining the filter coefficients from the\ndesired magnitude response.", - "type": "text" - }, - { - "block_id": "p913-b10", - "global_id": 26803, - "bbox": [ - 127.59, - 242.88, - 516.09, - 265.21 - ], - "text": "Consider the design of a lowpass filter with cutoff frequency c = π/4. An anonymous\nfunction represents the desired ideal frequency response.", - "type": "text" - }, - { - "block_id": "p913-b11", - "global_id": 26804, - "bbox": [ - 127.59, - 275.29, - 493.71, - 285.26 - ], - "text": ">>\nH_d = @(Omega) (mod(Omega,2*pi)2*pi-pi/4);", - "type": "text" - }, - { - "block_id": "p913-b12", - "global_id": 26805, - "bbox": [ - 127.59, - 294.35, - 516.16, - 364.49 - ], - "text": "Since the inverse DTFT of Hd() is a sampled sinc function, it is impossible to perfectly achieve\nthe desired response with a causal, finite-length FIR filter. A realizable FIR filter is necessarily\nan approximation, and an infinite number of possible solutions exist. Thought of another way,\nHd() specifies an infinite number of points, but the FIR filter only has N unknown tap weights.\nIn general, we expect a length-N filter to match only N points of the desired response over\n(0 ≤ < 2π). Which frequencies should be chosen?", - "type": "text" - }, - { - "block_id": "p913-b13", - "global_id": 26806, - "bbox": [ - 127.59, - 366.08, - 516.14, - 412.32 - ], - "text": "A simple and sensible method is to select N frequencies uniformly spaced on the interval (0 ≤\n < 2π), (0,2π/N,4π/N,6π/N,...,(N −1)2π/N). By choosing uniformly spaced frequency\nsamples, the N-point inverse DFT can be used to determine the tap weights h[n]. Program CH9MP1\nillustrates this procedure.", - "type": "text" - }, - { - "block_id": "p913-b14", - "global_id": 26807, - "bbox": [ - 127.59, - 421.85, - 466.15, - 489.59 - ], - "text": "function [h] = CH9MP1(N,H_d);\n% CH9MP1.m : Chapter 9, MATLAB Program 1\n% Function M-file designs a length-N FIR filter by sampling the desired\n% magnitude response H_d.\nPhase response is left as zero.\n% INPUTS:\nN = desired FIR filter length\n%\nH_d = anonymous function that defines the desired magnitude response\n% OUTPUTS:\nh = impulse response (FIR filter coefficients)", - "type": "text" - }, - { - "block_id": "p913-b15", - "global_id": 26808, - "bbox": [ - 127.59, - 501.55, - 364.58, - 539.41 - ], - "text": "% Create N equally spaced frequency samples:\nOmega = linspace(0,2*pi*(1-1/N),N)’;\n% Sample the desired magnitude response and create h[n]:\nH = 1.0*H_d(Omega); h = real(ifft(H));", - "type": "text" - }, - { - "block_id": "p913-b16", - "global_id": 26809, - "bbox": [ - 127.59, - 549.35, - 516.13, - 595.57 - ], - "text": "To complete the design, the filter length must be specified. Small values of N reduce the\nfilter’s complexity but also reduce the quality of the filter’s response. Large values of N improve\nthe approximation of Hd() but also increase complexity. A balance is needed. We choose an\nintermediate value of N = 21 and use CH9MP1 to design the filter.", - "type": "text" - }, - { - "block_id": "p913-b17", - "global_id": 26810, - "bbox": [ - 127.59, - 605.36, - 284.5, - 615.32 - ], - "text": ">>\nN = 21; h = CH9MP1(N,H_d);", - "type": "text" - }, - { - "block_id": "p913-b18", - "global_id": 26811, - "bbox": [ - 127.59, - 624.83, - 501.93, - 635.08 - ], - "text": "To assess the filter quality, the frequency response is computed by means of program CH5MP1.", - "type": "text" - } - ] - }, - { - "page_num": 914, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p914-b0", - "global_id": 26812, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "894\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p914-b1", - "global_id": 26813, - "bbox": [ - 101.84, - 86.24, - 494.12, - 167.93 - ], - "text": ">>\nOmega = linspace(0,2*pi,1000); samples = linspace(0,2*pi*(1-1/N),N)’;\n>>\nH = CH5MP1(h,1,Omega);\n>>\nsubplot(2,1,1); stem([0:N-1],h,’k.’); xlabel(’n’); ylabel(’h[n]’);\n>>\nsubplot(2,1,2);\n>>\nplot(samples,H_d(samples),’k.’,Omega,H_d(Omega),’k:’,Omega,abs(H),’k’);\n>>\naxis([0 2*pi -0.1 1.6]); xlabel(’\\Omega’); ylabel(’|H(\\Omega)|’);\n>>\nlegend(’Samples’,’Desired’,’Actual’,’Location’,’North’);", - "type": "text" - }, - { - "block_id": "p914-b2", - "global_id": 26814, - "bbox": [ - 101.84, - 189.33, - 490.4, - 235.15 - ], - "text": "As shown in Fig. 9.21, the filter’s frequency response intersects the desired response at the sampled\nvalues of Hd(). The overall response, however, has significant ripple between sample points\nthat renders the filter practically useless. Increasing the filter length does not alleviate the ripple\nproblems. Figure 9.22 shows the case N = 41.", - "type": "text" - }, - { - "block_id": "p914-b3", - "global_id": 26815, - "bbox": [ - 101.84, - 237.14, - 490.39, - 318.83 - ], - "text": "To understand the poor behavior of filters designed with CH9MP1, remember that the impulse\nresponse of an ideal lowpass filter is a sinc function with the peak centered at zero. Thought\nof another way, the peak of the sinc is centered at n = 0 because the phase of Hd() is zero.\nConstrained to be causal, the impulse response of the designed filter still has a peak at n = 0 but\ncannot include values for negative n. As a result, the sinc function is split in an unnatural way\nwith sharp discontinuities on both ends of h[n]. Sharp discontinuities in the time domain appear as\nhigh-frequency oscillations in the frequency domain, which is why H() has significant ripple.", - "type": "text" - }, - { - "block_id": "p914-b4", - "global_id": 26816, - "bbox": [ - 101.85, - 320.42, - 490.39, - 366.66 - ], - "text": "To improve the filter behavior, the peak of the sinc is moved to n = (N −1)/2, the center\nof the length-N filter response. In this way, the peak is not split, no large discontinuities are\npresent, and frequency response ripple is consequently reduced. From DFT properties, a cyclic\nshift of (N −1)/2 in the time domain requires a scale factor of e−j(N−1)/2 in the frequency", - "type": "text" - }, - { - "block_id": "p914-b5", - "global_id": 26817, - "bbox": [ - 129.81, - 484.17, - 470.06, - 505.36 - ], - "text": "0\n2\n4\n6\n8\n10\n12\n14\n16\n18\n20\nn", - "type": "text" - }, - { - "block_id": "p914-b6", - "global_id": 26818, - "bbox": [ - 297.06, - 615.45, - 303.82, - 624.25 - ], - "text": "Ω", - "type": "text" - }, - { - "block_id": "p914-b7", - "global_id": 26819, - "bbox": [ - 113.45, - 474.95, - 127.92, - 482.95 - ], - "text": "–0.2", - "type": "text" - }, - { - "block_id": "p914-b8", - "global_id": 26820, - "bbox": [ - 123.58, - 448.45, - 127.58, - 456.45 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p914-b9", - "global_id": 26821, - "bbox": [ - 116.84, - 421.95, - 127.5, - 429.95 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p914-b10", - "global_id": 26822, - "bbox": [ - 116.84, - 395.45, - 127.5, - 403.45 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p914-b11", - "global_id": 26823, - "bbox": [ - 102.61, - 431.14, - 111.41, - 445.8 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p914-b12", - "global_id": 26824, - "bbox": [ - 129.81, - 604.17, - 453.74, - 612.17 - ], - "text": "0\n1\n2\n3\n4\n5\n6", - "type": "text" - }, - { - "block_id": "p914-b13", - "global_id": 26825, - "bbox": [ - 123.58, - 590.24, - 127.58, - 598.24 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p914-b14", - "global_id": 26826, - "bbox": [ - 116.84, - 566.63, - 127.5, - 574.63 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p914-b15", - "global_id": 26827, - "bbox": [ - 123.58, - 543.03, - 127.58, - 551.03 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p914-b16", - "global_id": 26828, - "bbox": [ - 116.84, - 519.42, - 127.5, - 527.42 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p914-b17", - "global_id": 26829, - "bbox": [ - 104.39, - 547.37, - 113.33, - 568.43 - ], - "text": "|H(Ω)|", - "type": "text" - }, - { - "block_id": "p914-b18", - "global_id": 26830, - "bbox": [ - 297.43, - 531.43, - 321.83, - 538.63 - ], - "text": "Samples", - "type": "text" - }, - { - "block_id": "p914-b19", - "global_id": 26831, - "bbox": [ - 297.43, - 542.31, - 319.83, - 549.51 - ], - "text": "Desired", - "type": "text" - }, - { - "block_id": "p914-b20", - "global_id": 26832, - "bbox": [ - 297.43, - 553.18, - 316.62, - 560.38 - ], - "text": "Actual", - "type": "text" - }, - { - "block_id": "p914-b21", - "global_id": 26833, - "bbox": [ - 101.84, - 631.92, - 320.11, - 641.16 - ], - "text": "Figure 9.21 Length-21 FIR lowpass filter using zero phase.", - "type": "text" - } - ] - }, - { - "page_num": 915, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p915-b0", - "global_id": 26834, - "bbox": [ - 259.25, - 62.89, - 516.14, - 71.98 - ], - "text": "9.7\nMATLAB: Working with the DTFS and the DTFT\n895", - "type": "text" - }, - { - "block_id": "p915-b1", - "global_id": 26835, - "bbox": [ - 155.56, - 176.45, - 495.82, - 197.64 - ], - "text": "0\n5\n10\n15\n20\n25\n30\n35\n40\nn", - "type": "text" - }, - { - "block_id": "p915-b2", - "global_id": 26836, - "bbox": [ - 319.43, - 306.97, - 326.19, - 315.77 - ], - "text": "Ω", - "type": "text" - }, - { - "block_id": "p915-b3", - "global_id": 26837, - "bbox": [ - 139.2, - 167.24, - 153.67, - 175.24 - ], - "text": "–0.2", - "type": "text" - }, - { - "block_id": "p915-b4", - "global_id": 26838, - "bbox": [ - 149.33, - 140.74, - 153.33, - 148.74 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p915-b5", - "global_id": 26839, - "bbox": [ - 142.59, - 114.23, - 153.25, - 122.23 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p915-b6", - "global_id": 26840, - "bbox": [ - 142.59, - 87.74, - 153.25, - 95.74 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p915-b7", - "global_id": 26841, - "bbox": [ - 128.36, - 123.42, - 137.16, - 138.08 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p915-b8", - "global_id": 26842, - "bbox": [ - 155.56, - 296.45, - 476.28, - 304.45 - ], - "text": "0\n1\n2\n3\n4\n5\n6", - "type": "text" - }, - { - "block_id": "p915-b9", - "global_id": 26843, - "bbox": [ - 149.33, - 282.52, - 153.33, - 290.52 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p915-b10", - "global_id": 26844, - "bbox": [ - 142.59, - 258.91, - 153.25, - 266.91 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p915-b11", - "global_id": 26845, - "bbox": [ - 149.33, - 235.31, - 153.33, - 243.31 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p915-b12", - "global_id": 26846, - "bbox": [ - 142.59, - 211.7, - 153.25, - 219.7 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p915-b13", - "global_id": 26847, - "bbox": [ - 130.14, - 239.31, - 139.08, - 260.37 - ], - "text": "|H(Ω)|", - "type": "text" - }, - { - "block_id": "p915-b14", - "global_id": 26848, - "bbox": [ - 323.18, - 223.71, - 347.58, - 230.91 - ], - "text": "Samples", - "type": "text" - }, - { - "block_id": "p915-b15", - "global_id": 26849, - "bbox": [ - 323.18, - 234.59, - 345.57, - 241.79 - ], - "text": "Desired", - "type": "text" - }, - { - "block_id": "p915-b16", - "global_id": 26850, - "bbox": [ - 323.18, - 245.46, - 342.37, - 252.66 - ], - "text": "Actual", - "type": "text" - }, - { - "block_id": "p915-b17", - "global_id": 26851, - "bbox": [ - 127.59, - 323.44, - 345.85, - 332.68 - ], - "text": "Figure 9.22 Length-41 FIR lowpass filter using zero phase.", - "type": "text" - }, - { - "block_id": "p915-b18", - "global_id": 26852, - "bbox": [ - 127.59, - 352.28, - 516.14, - 378.1 - ], - "text": "domain.† Notice that the scale factor e−j(N−1)/2 affects only phase, not magnitude, and results\nin a linear phase filter. Program CH9MP2 implements the procedure.", - "type": "text" - }, - { - "block_id": "p915-b19", - "global_id": 26853, - "bbox": [ - 127.59, - 388.61, - 466.15, - 456.36 - ], - "text": "function [h] = CH9MP2(N,H_d);\n% CH9MP2.m : Chapter 9, MATLAB Program 2\n% Function M-file designs a length-N FIR filter by sampling the desired\n% magnitude response H_d.\nPhase is defined to shift h[n] by (N-1)/2.\n% INPUTS:\nN = desired FIR filter length\n%\nH_d = anonymous function that defines the desired magnitude response\n% OUTPUTS:\nh = impulse response (FIR filter coefficients)", - "type": "text" - }, - { - "block_id": "p915-b20", - "global_id": 26854, - "bbox": [ - 127.59, - 468.31, - 406.91, - 526.09 - ], - "text": "% Create N equally spaced frequency samples and use to sample H_d:\nOmega = linspace(0,2*pi*(1-1/N),N)’; H = H_d(Omega);\n% Define phase to shift h[n] by (N-1)/2:\nH = H.*exp(-j*Omega*((N-1)/2));\nH(fix(N/2)+2:N,1) = H(fix(N/2)+2:N,1)*((-1)^(N-1));\nh = real(ifft(H));", - "type": "text" - }, - { - "block_id": "p915-b21", - "global_id": 26855, - "bbox": [ - 127.59, - 544.48, - 516.13, - 633.42 - ], - "text": "† Technically, the shift property requires (N −1)/2 to be an integer, which occurs only for odd-length filters.\nThe next-to-last line of program CH9MP2 implements a correction factor, of sorts, required to accommodate\nthe fractional shifts desired for even-length filters. The mathematical derivation of this correction is nontrivial\nand is not included here. Those hesitant to use this correction factor have an alternative: simply round\n(N −1)/2 to the nearest integer. Although the rounded shift is slightly off-center for even-length filters,\nthere is usually little or no appreciable difference in the characteristics of the filter. Even so, true centering\nis desirable because the resulting impulse response is symmetric, which can reduce by half the number of\nmultiplies required to implement the filter.", - "type": "text" - } - ] - }, - { - "page_num": 916, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p916-b0", - "global_id": 26856, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "896\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p916-b1", - "global_id": 26857, - "bbox": [ - 129.81, - 176.49, - 470.06, - 197.68 - ], - "text": "0\n2\n4\n6\n8\n10\n12\n14\n16\n18\n20\nn", - "type": "text" - }, - { - "block_id": "p916-b2", - "global_id": 26858, - "bbox": [ - 113.73, - 167.27, - 127.92, - 175.27 - ], - "text": "–0.2", - "type": "text" - }, - { - "block_id": "p916-b3", - "global_id": 26859, - "bbox": [ - 123.58, - 140.77, - 127.58, - 148.77 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p916-b4", - "global_id": 26860, - "bbox": [ - 116.84, - 114.26, - 127.5, - 122.26 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p916-b5", - "global_id": 26861, - "bbox": [ - 116.84, - 87.77, - 127.5, - 95.77 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p916-b6", - "global_id": 26862, - "bbox": [ - 102.61, - 123.46, - 111.41, - 138.12 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p916-b7", - "global_id": 26863, - "bbox": [ - 130.09, - 296.49, - 347.22, - 304.49 - ], - "text": "0\n1\n2\n3\n4", - "type": "text" - }, - { - "block_id": "p916-b8", - "global_id": 26864, - "bbox": [ - 296.3, - 309.97, - 303.06, - 318.77 - ], - "text": "Ω", - "type": "text" - }, - { - "block_id": "p916-b9", - "global_id": 26865, - "bbox": [ - 397.15, - 296.49, - 453.94, - 304.49 - ], - "text": "5\n6", - "type": "text" - }, - { - "block_id": "p916-b10", - "global_id": 26866, - "bbox": [ - 123.58, - 282.56, - 127.58, - 290.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p916-b11", - "global_id": 26867, - "bbox": [ - 116.84, - 258.94, - 127.5, - 266.94 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p916-b12", - "global_id": 26868, - "bbox": [ - 123.58, - 235.34, - 127.58, - 243.34 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p916-b13", - "global_id": 26869, - "bbox": [ - 116.84, - 211.73, - 127.5, - 219.73 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p916-b14", - "global_id": 26870, - "bbox": [ - 104.39, - 239.69, - 113.33, - 260.75 - ], - "text": "|H(Ω)|", - "type": "text" - }, - { - "block_id": "p916-b15", - "global_id": 26871, - "bbox": [ - 297.43, - 223.75, - 321.83, - 230.95 - ], - "text": "Samples", - "type": "text" - }, - { - "block_id": "p916-b16", - "global_id": 26872, - "bbox": [ - 297.43, - 234.63, - 319.83, - 241.83 - ], - "text": "Desired", - "type": "text" - }, - { - "block_id": "p916-b17", - "global_id": 26873, - "bbox": [ - 297.43, - 245.5, - 316.62, - 252.7 - ], - "text": "Actual", - "type": "text" - }, - { - "block_id": "p916-b18", - "global_id": 26874, - "bbox": [ - 101.84, - 326.44, - 325.09, - 335.68 - ], - "text": "Figure 9.23 Length-21 FIR lowpass filter using linear phase.", - "type": "text" - }, - { - "block_id": "p916-b19", - "global_id": 26875, - "bbox": [ - 101.84, - 378.03, - 490.41, - 448.18 - ], - "text": "Figure 9.23 shows the results for the N = 21 case using CH9MP2 to compute h[n]. As hoped, the\nimpulse response looks like a sinc function with the peak centered at n = 10. Additionally, the\nfrequency response ripple is greatly reduced. With CH9MP2, increasing N improves the quality of\nthe filter, as shown in Fig. 9.24 for the case N = 41. While the magnitude response is needed to\nestablish the general shape of the filter response, it is the proper selection of phase that ensures the\nacceptability of the filter’s behavior.", - "type": "text" - }, - { - "block_id": "p916-b20", - "global_id": 26876, - "bbox": [ - 101.84, - 450.17, - 490.37, - 472.1 - ], - "text": "To illustrate the flexibility of the design method, consider a bandpass filter with passband\n(π/4 < || < π/2).", - "type": "text" - }, - { - "block_id": "p916-b21", - "global_id": 26877, - "bbox": [ - 101.84, - 489.32, - 457.5, - 511.24 - ], - "text": ">>\nH_d = @(Omega) (mod(Omega,2*pi)>pi/4)&(mod(Omega,2*pi)>\n(mod(Omega,2*pi)>3*pi/2)&(mod(Omega,2*pi)<7*pi/4);", - "type": "text" - }, - { - "block_id": "p916-b22", - "global_id": 26878, - "bbox": [ - 101.84, - 530.8, - 490.39, - 553.14 - ], - "text": "Figure 9.25 shows the results for N = 50. Notice that this even-length filter uses a fractional shift\nand is symmetric about n = 24.5.", - "type": "text" - }, - { - "block_id": "p916-b23", - "global_id": 26879, - "bbox": [ - 101.84, - 555.13, - 490.41, - 636.82 - ], - "text": "Although FIR filter design by means of frequency sampling is very flexible, it is not\nalways appropriate. Extreme care is needed for filters, such as digital differentiators and Hilbert\ntransformers, that require special phase characteristics for proper operation. Additionally, if\nfrequency samples occur near jump discontinuities of Hd(), rounding errors may, in rare cases,\ndisrupt the desired symmetry of the sampled magnitude response. Such cases are corrected\nby slightly adjusting the location of problematic jump discontinuities or by changing the\nvalue of N.", - "type": "text" - } - ] - }, - { - "page_num": 917, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p917-b0", - "global_id": 26880, - "bbox": [ - 259.25, - 62.89, - 516.14, - 71.98 - ], - "text": "9.7\nMATLAB: Working with the DTFS and the DTFT\n897", - "type": "text" - }, - { - "block_id": "p917-b1", - "global_id": 26881, - "bbox": [ - 155.56, - 176.11, - 328.56, - 184.11 - ], - "text": "0\n5\n10\n15\n20", - "type": "text" - }, - { - "block_id": "p917-b2", - "global_id": 26882, - "bbox": [ - 317.18, - 306.97, - 323.94, - 315.77 - ], - "text": "Ω", - "type": "text" - }, - { - "block_id": "p917-b3", - "global_id": 26883, - "bbox": [ - 322.81, - 176.11, - 495.82, - 197.3 - ], - "text": "25\n30\n35\n40\nn", - "type": "text" - }, - { - "block_id": "p917-b4", - "global_id": 26884, - "bbox": [ - 139.33, - 166.89, - 153.33, - 174.89 - ], - "text": "–0.2", - "type": "text" - }, - { - "block_id": "p917-b5", - "global_id": 26885, - "bbox": [ - 149.33, - 140.39, - 153.33, - 148.39 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p917-b6", - "global_id": 26886, - "bbox": [ - 142.59, - 113.88, - 153.25, - 121.88 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p917-b7", - "global_id": 26887, - "bbox": [ - 142.59, - 87.39, - 153.25, - 95.39 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p917-b8", - "global_id": 26888, - "bbox": [ - 128.36, - 123.08, - 137.16, - 137.74 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p917-b9", - "global_id": 26889, - "bbox": [ - 156.41, - 296.11, - 479.41, - 304.11 - ], - "text": "0\n1\n2\n3\n4\n5\n6", - "type": "text" - }, - { - "block_id": "p917-b10", - "global_id": 26890, - "bbox": [ - 149.33, - 282.18, - 153.33, - 290.18 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p917-b11", - "global_id": 26891, - "bbox": [ - 142.59, - 258.56, - 153.25, - 266.56 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p917-b12", - "global_id": 26892, - "bbox": [ - 149.33, - 234.96, - 153.33, - 242.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p917-b13", - "global_id": 26893, - "bbox": [ - 142.59, - 211.35, - 153.25, - 219.35 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p917-b14", - "global_id": 26894, - "bbox": [ - 130.14, - 239.31, - 139.08, - 260.37 - ], - "text": "|H(Ω)|", - "type": "text" - }, - { - "block_id": "p917-b15", - "global_id": 26895, - "bbox": [ - 323.18, - 223.37, - 347.58, - 230.57 - ], - "text": "Samples", - "type": "text" - }, - { - "block_id": "p917-b16", - "global_id": 26896, - "bbox": [ - 323.18, - 234.25, - 345.57, - 241.45 - ], - "text": "Desired", - "type": "text" - }, - { - "block_id": "p917-b17", - "global_id": 26897, - "bbox": [ - 323.18, - 245.12, - 342.37, - 252.32 - ], - "text": "Actual", - "type": "text" - }, - { - "block_id": "p917-b18", - "global_id": 26898, - "bbox": [ - 127.59, - 323.44, - 350.84, - 332.68 - ], - "text": "Figure 9.24 Length-41 FIR lowpass filter using linear phase.", - "type": "text" - }, - { - "block_id": "p917-b19", - "global_id": 26899, - "bbox": [ - 155.56, - 446.88, - 495.81, - 468.07 - ], - "text": "0\n5\n10\n15\n20\n25\n30\n35\n40\n45\n50\nn", - "type": "text" - }, - { - "block_id": "p917-b20", - "global_id": 26900, - "bbox": [ - 319.43, - 580.94, - 326.19, - 589.74 - ], - "text": "Ω", - "type": "text" - }, - { - "block_id": "p917-b21", - "global_id": 26901, - "bbox": [ - 139.33, - 429.7, - 153.33, - 437.7 - ], - "text": "–0.2", - "type": "text" - }, - { - "block_id": "p917-b22", - "global_id": 26902, - "bbox": [ - 149.33, - 397.91, - 153.33, - 405.91 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p917-b23", - "global_id": 26903, - "bbox": [ - 142.59, - 366.11, - 153.25, - 374.11 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p917-b24", - "global_id": 26904, - "bbox": [ - 128.36, - 393.84, - 137.16, - 408.5 - ], - "text": "h[n]", - "type": "text" - }, - { - "block_id": "p917-b25", - "global_id": 26905, - "bbox": [ - 156.41, - 566.87, - 479.12, - 574.87 - ], - "text": "0\n1\n2\n3\n4\n5\n6", - "type": "text" - }, - { - "block_id": "p917-b26", - "global_id": 26906, - "bbox": [ - 149.33, - 552.92, - 153.33, - 560.92 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p917-b27", - "global_id": 26907, - "bbox": [ - 142.59, - 529.33, - 153.25, - 537.33 - ], - "text": "0.5", - "type": "text" - }, - { - "block_id": "p917-b28", - "global_id": 26908, - "bbox": [ - 149.33, - 505.73, - 153.33, - 513.73 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p917-b29", - "global_id": 26909, - "bbox": [ - 142.59, - 482.13, - 153.25, - 490.13 - ], - "text": "1.5", - "type": "text" - }, - { - "block_id": "p917-b30", - "global_id": 26910, - "bbox": [ - 130.14, - 510.07, - 139.08, - 531.14 - ], - "text": "|H(Ω)|", - "type": "text" - }, - { - "block_id": "p917-b31", - "global_id": 26911, - "bbox": [ - 323.18, - 494.14, - 347.58, - 501.34 - ], - "text": "Samples", - "type": "text" - }, - { - "block_id": "p917-b32", - "global_id": 26912, - "bbox": [ - 323.18, - 505.01, - 345.57, - 512.21 - ], - "text": "Desired", - "type": "text" - }, - { - "block_id": "p917-b33", - "global_id": 26913, - "bbox": [ - 323.18, - 515.89, - 342.37, - 523.09 - ], - "text": "Actual", - "type": "text" - }, - { - "block_id": "p917-b34", - "global_id": 26914, - "bbox": [ - 127.59, - 597.41, - 355.02, - 606.65 - ], - "text": "Figure 9.25 Length-50 FIR bandpass filter using linear phase.", - "type": "text" - } - ] - }, - { - "page_num": 918, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p918-b0", - "global_id": 26915, - "bbox": [ - 60.0, - 62.89, - 373.81, - 71.98 - ], - "text": "898\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p918-b1", - "global_id": 26916, - "bbox": [ - 102.2, - 94.25, - 194.48, - 108.2 - ], - "text": "9.8 SUMMARY", - "type": "text" - }, - { - "block_id": "p918-b2", - "global_id": 26917, - "bbox": [ - 101.84, - 114.18, - 490.41, - 171.96 - ], - "text": "This chapter deals with the analysis and processing of discrete-time signals. For analysis, our\napproach is parallel to that used in continuous-time signals. We first represent a periodic x[n] as\na Fourier series formed by a discrete-time exponential and its harmonics. Later we extend this\nrepresentation to an aperiodic signal x[n] by considering x[n] to be a limiting case of a periodic\nsignal with period approaching infinity.", - "type": "text" - }, - { - "block_id": "p918-b3", - "global_id": 26918, - "bbox": [ - 101.84, - 173.96, - 490.42, - 269.1 - ], - "text": "Periodic signals are represented by discrete-time Fourier series (DTFS); aperiodic signals are\nrepresented by the discrete-time Fourier integral. The development, although similar to that of\ncontinuous-time signals, also reveals some significant differences. The basic difference in the two\ncases arises because a continuous-time exponential ejωt has a unique waveform for every value of\nω in the range −∞to ∞. In contrast, a discrete-time exponential ejn has a unique waveform only\nfor values of in a continuous interval of 2π. Therefore, if 0 is the fundamental frequency, then\nat most 2π/0 exponentials in the Fourier series are independent. Consequently, the discrete-time\nexponential Fourier series has only N0 = 2π/0 terms.", - "type": "text" - }, - { - "block_id": "p918-b4", - "global_id": 26919, - "bbox": [ - 101.85, - 269.6, - 490.4, - 315.42 - ], - "text": "The discrete-time Fourier transform (DTFT) of an aperiodic signal is a continuous function\nof and is periodic with period 2π. We can synthesize x[n] from spectral components of X() in\nany band of width 2π. In a basic sense, the DTFT has a finite spectral width of 2π, which makes\nit bandlimited to π radians.", - "type": "text" - }, - { - "block_id": "p918-b5", - "global_id": 26920, - "bbox": [ - 101.85, - 317.42, - 490.41, - 387.16 - ], - "text": "Linear, time-invariant, discrete-time (LTID) systems can be analyzed by means of the DTFT\nif the input signals are DTF-transformable and if the system is stable. Analysis of unstable (or\nmarginally stable) systems and/or exponentially growing inputs can be handled by the z-transform,\nwhich is a generalized DTFT. The relationship of the DTFT to the z-transform is similar to that of\nthe Fourier transform to the Laplace transform. Whereas the z-transform is superior to the DTFT\nfor analysis of LTID systems, the DTFT is preferable in signal analysis.", - "type": "text" - }, - { - "block_id": "p918-b6", - "global_id": 26921, - "bbox": [ - 101.84, - 388.74, - 490.39, - 446.93 - ], - "text": "If H() is the DTFT of the system’s impulse response h[n], then |H()| is the amplitude\nresponse, and̸\nH() is the phase response of the system. Moreover, if X() and Y() are\nthe DTFTs of the input x[n] and the corresponding output y[n], then Y() = H()X().\nTherefore, the output spectrum is the product of the input spectrum and the system’s frequency\nresponse.", - "type": "text" - }, - { - "block_id": "p918-b7", - "global_id": 26922, - "bbox": [ - 101.84, - 448.93, - 490.4, - 494.76 - ], - "text": "Because of the similarity between the DFT and DTFT relationships, numerical computations\nof the DTFT of finite-length signals can be handled by using the DFT and the FFT, introduced in\nSecs. 8.5 and 8.6. For signals of infinite length, we use a window of suitable length to truncate the\nsignal so that the final results are within a given error tolerance.", - "type": "text" - }, - { - "block_id": "p918-b8", - "global_id": 26923, - "bbox": [ - 102.11, - 531.19, - 180.95, - 542.15 - ], - "text": "REFERENCE", - "type": "text" - }, - { - "block_id": "p918-b9", - "global_id": 26924, - "bbox": [ - 101.84, - 549.22, - 490.38, - 569.23 - ], - "text": "1.\nMitra, S. K. Digital Signal Processing: A Computer-Based Approach, 2nd ed. McGraw-Hill, New York,\n2001.", - "type": "text" - } - ] - }, - { - "page_num": 919, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p919-b0", - "global_id": 26925, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n899", - "type": "text" - }, - { - "block_id": "p919-b1", - "global_id": 26926, - "bbox": [ - 106.67, - 91.05, - 217.26, - 107.98 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p919-b2", - "global_id": 26927, - "bbox": [ - 87.82, - 119.09, - 288.99, - 150.78 - ], - "text": "9.1-1\nFind the discrete-time Fourier series (DTFS)\nand sketch their spectra |Dr| and̸\nDr for\n0≤r≤N0−1 for the following periodic signal:", - "type": "text" - }, - { - "block_id": "p919-b3", - "global_id": 26928, - "bbox": [ - 145.61, - 164.06, - 259.84, - 173.4 - ], - "text": "x[n] = 4cos2.4πn + 2sin3.2πn", - "type": "text" - }, - { - "block_id": "p919-b4", - "global_id": 26929, - "bbox": [ - 87.82, - 187.41, - 293.35, - 196.75 - ], - "text": "9.1-2\nRepeat Prob. 9.1-1 for x[n]=cos2.2πncos3.3πn.", - "type": "text" - }, - { - "block_id": "p919-b5", - "global_id": 26930, - "bbox": [ - 87.82, - 201.61, - 282.29, - 210.94 - ], - "text": "9.1-3\nRepeat Prob. 9.1-1 for x[n]=2cos3.2π(n−3).", - "type": "text" - }, - { - "block_id": "p919-b6", - "global_id": 26931, - "bbox": [ - 87.82, - 215.8, - 288.98, - 247.05 - ], - "text": "9.1-4\nDetermine and sketch the DTFS spectrum Dr of\na 7-periodic signal x[n] that over 0 ≤n ≤6 is\ngiven by", - "type": "text" - }, - { - "block_id": "p919-b7", - "global_id": 26932, - "bbox": [ - 163.84, - 261.07, - 241.61, - 270.41 - ], - "text": "[0,1,−2,3,−4,5,−6]", - "type": "text" - }, - { - "block_id": "p919-b8", - "global_id": 26933, - "bbox": [ - 343.61, - 118.69, - 516.14, - 138.99 - ], - "text": "How does the spectrum Dr change if x[n] is time\nreversed?", - "type": "text" - }, - { - "block_id": "p919-b9", - "global_id": 26934, - "bbox": [ - 314.97, - 148.97, - 516.15, - 179.92 - ], - "text": "9.1-5\nFind the discrete-time Fourier series and the\ncorresponding amplitude and phase spectra for\nthe x[n] shown in Fig. P9.1-5.", - "type": "text" - }, - { - "block_id": "p919-b10", - "global_id": 26935, - "bbox": [ - 314.97, - 189.61, - 516.13, - 209.91 - ], - "text": "9.1-6\nRepeat Prob. 9.1-5 for the x[n] depicted in\nFig. P9.1-6.", - "type": "text" - }, - { - "block_id": "p919-b11", - "global_id": 26936, - "bbox": [ - 314.97, - 219.59, - 516.13, - 239.89 - ], - "text": "9.1-7\nRepeat Prob. 9.1-5 for the x[n] illustrated in\nFig. P9.1-7.", - "type": "text" - }, - { - "block_id": "p919-b12", - "global_id": 26937, - "bbox": [ - 314.98, - 249.57, - 516.13, - 269.87 - ], - "text": "9.1-8\nAn N0-periodic signal x[n] is represented by its\nDTFS, as in Eq. (9.3). Prove Parseval’s theorem", - "type": "text" - }, - { - "block_id": "p919-b13", - "global_id": 26938, - "bbox": [ - 250.53, - 301.11, - 263.73, - 309.19 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p919-b14", - "global_id": 26939, - "bbox": [ - 351.08, - 357.96, - 373.71, - 367.2 - ], - "text": "n\n12", - "type": "text" - }, - { - "block_id": "p919-b15", - "global_id": 26940, - "bbox": [ - 375.58, - 345.41, - 384.88, - 351.41 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p919-b16", - "global_id": 26941, - "bbox": [ - 160.85, - 358.9, - 330.08, - 367.21 - ], - "text": "9\n9\n6\n3\n3\n6\n0", - "type": "text" - }, - { - "block_id": "p919-b17", - "global_id": 26942, - "bbox": [ - 130.58, - 345.41, - 139.88, - 351.41 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p919-b18", - "global_id": 26943, - "bbox": [ - 239.58, - 315.21, - 243.58, - 323.21 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p919-b19", - "global_id": 26944, - "bbox": [ - 130.58, - 379.56, - 182.22, - 388.52 - ], - "text": "Figure P9.1-5", - "type": "text" - }, - { - "block_id": "p919-b20", - "global_id": 26945, - "bbox": [ - 284.03, - 410.02, - 297.23, - 418.11 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p919-b21", - "global_id": 26946, - "bbox": [ - 283.08, - 468.11, - 395.28, - 476.11 - ], - "text": "0\n12", - "type": "text" - }, - { - "block_id": "p919-b22", - "global_id": 26947, - "bbox": [ - 423.08, - 454.33, - 432.38, - 460.33 - ], - "text": "• • •", - "type": "text" - }, - { - "block_id": "p919-b23", - "global_id": 26948, - "bbox": [ - 273.58, - 425.77, - 277.58, - 433.77 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p919-b24", - "global_id": 26949, - "bbox": [ - 221.34, - 467.82, - 419.28, - 476.11 - ], - "text": "3\n6\n15\n6", - "type": "text" - }, - { - "block_id": "p919-b25", - "global_id": 26950, - "bbox": [ - 266.91, - 498.32, - 277.58, - 506.61 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p919-b26", - "global_id": 26951, - "bbox": [ - 193.34, - 467.82, - 204.01, - 476.11 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p919-b27", - "global_id": 26952, - "bbox": [ - 133.08, - 454.33, - 174.51, - 463.61 - ], - "text": "12\n• • •", - "type": "text" - }, - { - "block_id": "p919-b28", - "global_id": 26953, - "bbox": [ - 424.08, - 468.35, - 428.08, - 476.35 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p919-b29", - "global_id": 26954, - "bbox": [ - 130.58, - 517.47, - 182.22, - 526.44 - ], - "text": "Figure P9.1-6", - "type": "text" - }, - { - "block_id": "p919-b30", - "global_id": 26955, - "bbox": [ - 301.55, - 562.05, - 305.55, - 570.05 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p919-b31", - "global_id": 26956, - "bbox": [ - 306.67, - 602.8, - 310.67, - 610.8 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p919-b32", - "global_id": 26957, - "bbox": [ - 334.71, - 565.39, - 341.87, - 574.84 - ], - "text": "an", - "type": "text" - }, - { - "block_id": "p919-b33", - "global_id": 26958, - "bbox": [ - 180.41, - 603.92, - 434.08, - 613.74 - ], - "text": "2N0\n2N0\nN0\nN0", - "type": "text" - }, - { - "block_id": "p919-b34", - "global_id": 26959, - "bbox": [ - 318.67, - 548.12, - 331.87, - 556.2 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p919-b35", - "global_id": 26960, - "bbox": [ - 469.21, - 598.54, - 473.21, - 606.54 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p919-b36", - "global_id": 26961, - "bbox": [ - 130.58, - 620.4, - 182.22, - 629.36 - ], - "text": "Figure P9.1-7", - "type": "text" - } - ] - }, - { - "page_num": 920, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p920-b0", - "global_id": 26962, - "bbox": [ - 60.0, - 60.36, - 373.81, - 69.45 - ], - "text": "900\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p920-b1", - "global_id": 26963, - "bbox": [ - 90.72, - 85.9, - 208.74, - 94.86 - ], - "text": "(for the DTFS), which states that", - "type": "text" - }, - { - "block_id": "p920-b2", - "global_id": 26964, - "bbox": [ - 123.73, - 104.54, - 132.95, - 126.89 - ], - "text": "1\nN0", - "type": "text" - }, - { - "block_id": "p920-b3", - "global_id": 26965, - "bbox": [ - 139.6, - 101.92, - 152.29, - 110.89 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p920-b4", - "global_id": 26966, - "bbox": [ - 135.64, - 123.28, - 156.24, - 131.14 - ], - "text": "n=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p920-b5", - "global_id": 26967, - "bbox": [ - 157.24, - 106.74, - 189.41, - 119.69 - ], - "text": "|x[n]|2 =", - "type": "text" - }, - { - "block_id": "p920-b6", - "global_id": 26968, - "bbox": [ - 194.91, - 101.92, - 207.6, - 110.89 - ], - "text": "\"", - "type": "text" - }, - { - "block_id": "p920-b7", - "global_id": 26969, - "bbox": [ - 191.24, - 123.28, - 211.28, - 131.14 - ], - "text": "r=⟨N0⟩", - "type": "text" - }, - { - "block_id": "p920-b8", - "global_id": 26970, - "bbox": [ - 212.27, - 109.0, - 230.92, - 120.46 - ], - "text": "|Dr|2", - "type": "text" - }, - { - "block_id": "p920-b9", - "global_id": 26971, - "bbox": [ - 90.72, - 141.09, - 263.24, - 171.98 - ], - "text": "In the text [Eq. (9.36)], we obtain Parseval’s\ntheorem for the DTFT. [Hint: If w is complex,\nthen |w|2 = ww∗and use Eq. (8.15).]", - "type": "text" - }, - { - "block_id": "p920-b10", - "global_id": 26972, - "bbox": [ - 62.08, - 176.88, - 263.23, - 196.88 - ], - "text": "9.1-9\nAnswer yes or no, and justify your answers with\nan appropriate example or proof.", - "type": "text" - }, - { - "block_id": "p920-b11", - "global_id": 26973, - "bbox": [ - 90.72, - 198.87, - 263.24, - 229.76 - ], - "text": "(a) Is\na\nsum\nof\naperiodic\ndiscrete-time\nsequences ever periodic?\n(b) Is a sum of periodic discrete-time sequences", - "type": "text" - }, - { - "block_id": "p920-b12", - "global_id": 26974, - "bbox": [ - 106.15, - 231.75, - 160.8, - 240.72 - ], - "text": "ever aperiodic?", - "type": "text" - }, - { - "block_id": "p920-b13", - "global_id": 26975, - "bbox": [ - 62.08, - 245.33, - 263.24, - 265.62 - ], - "text": "9.2-1\nShow that for a real x[n], Eq. (9.18) can be\nexpressed as", - "type": "text" - }, - { - "block_id": "p920-b14", - "global_id": 26976, - "bbox": [ - 100.13, - 275.79, - 131.85, - 291.05 - ], - "text": "x[n] = 1", - "type": "text" - }, - { - "block_id": "p920-b15", - "global_id": 26977, - "bbox": [ - 126.48, - 288.07, - 131.86, - 297.04 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p920-b16", - "global_id": 26978, - "bbox": [ - 134.95, - 269.5, - 147.67, - 280.47 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p920-b17", - "global_id": 26979, - "bbox": [ - 139.68, - 281.71, - 253.82, - 298.48 - ], - "text": "0\n|X()|cos(n +̸ X())d", - "type": "text" - }, - { - "block_id": "p920-b18", - "global_id": 26980, - "bbox": [ - 90.72, - 307.22, - 247.95, - 316.18 - ], - "text": "This is the trigonometric form of the DTFT.", - "type": "text" - }, - { - "block_id": "p920-b19", - "global_id": 26981, - "bbox": [ - 62.08, - 320.79, - 263.24, - 341.09 - ], - "text": "9.2-2\nA signal x[n] can be expressed as the sum of\neven and odd components (Sec. 1.5-1):", - "type": "text" - }, - { - "block_id": "p920-b20", - "global_id": 26982, - "bbox": [ - 141.53, - 352.63, - 212.43, - 362.65 - ], - "text": "x[n] = xe[n] + xo[n]", - "type": "text" - }, - { - "block_id": "p920-b21", - "global_id": 26983, - "bbox": [ - 91.22, - 384.48, - 255.46, - 393.82 - ], - "text": "(a) If x[n] ⇐⇒X(), show that for real x[n],", - "type": "text" - }, - { - "block_id": "p920-b22", - "global_id": 26984, - "bbox": [ - 146.48, - 405.37, - 222.92, - 415.38 - ], - "text": "xe[n] ⇐⇒Re[X()]", - "type": "text" - }, - { - "block_id": "p920-b23", - "global_id": 26985, - "bbox": [ - 106.15, - 426.63, - 119.1, - 435.59 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p920-b24", - "global_id": 26986, - "bbox": [ - 145.06, - 437.22, - 224.35, - 447.23 - ], - "text": "xo[n] ⇐⇒j Im[X()]", - "type": "text" - }, - { - "block_id": "p920-b25", - "global_id": 26987, - "bbox": [ - 90.72, - 455.5, - 263.24, - 464.47 - ], - "text": "(b) Verify these results by finding the DTFT of", - "type": "text" - }, - { - "block_id": "p920-b26", - "global_id": 26988, - "bbox": [ - 106.15, - 466.46, - 263.24, - 486.38 - ], - "text": "the even and odd components of the signal\n(0.8)nu[n].", - "type": "text" - }, - { - "block_id": "p920-b27", - "global_id": 26989, - "bbox": [ - 62.08, - 491.29, - 263.25, - 522.26 - ], - "text": "9.2-3\nFor the following signals, find the DTFT\ndirectly, using the definition in Eq. (9.19).\nAssume |γ | < 1.", - "type": "text" - }, - { - "block_id": "p920-b28", - "global_id": 26990, - "bbox": [ - 91.22, - 523.87, - 120.82, - 533.21 - ], - "text": "(a) δ[n]", - "type": "text" - }, - { - "block_id": "p920-b29", - "global_id": 26991, - "bbox": [ - 317.86, - 85.52, - 361.89, - 94.86 - ], - "text": "(b) δ[n −k]", - "type": "text" - }, - { - "block_id": "p920-b30", - "global_id": 26992, - "bbox": [ - 317.86, - 95.42, - 372.17, - 116.78 - ], - "text": "(c) γ nu[n −1]\n(d) γ nu[n + 1]", - "type": "text" - }, - { - "block_id": "p920-b31", - "global_id": 26993, - "bbox": [ - 318.37, - 117.34, - 371.59, - 127.74 - ], - "text": "(e) (−γ )nu[n]", - "type": "text" - }, - { - "block_id": "p920-b32", - "global_id": 26994, - "bbox": [ - 319.37, - 128.1, - 346.18, - 138.7 - ], - "text": "(f) γ |n|", - "type": "text" - }, - { - "block_id": "p920-b33", - "global_id": 26995, - "bbox": [ - 289.22, - 143.86, - 490.38, - 176.17 - ], - "text": "9.2-4\nUse Eq. (9.18) to find the inverse DTFT for the\nfollowing spectra, given only over the interval\n|| ≤π. Assume c and 0 < π.", - "type": "text" - }, - { - "block_id": "p920-b34", - "global_id": 26996, - "bbox": [ - 317.86, - 175.19, - 396.07, - 207.71 - ], - "text": "(a) ejk\nintegerk\n(b) cosk\nintegerk\n(c) cos2(/2)", - "type": "text" - }, - { - "block_id": "p920-b35", - "global_id": 26997, - "bbox": [ - 317.86, - 214.17, - 340.51, - 223.51 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p920-b37", - "global_id": 26998, - "bbox": [ - 348.76, - 220.53, - 363.09, - 230.55 - ], - "text": "2c", - "type": "text" - }, - { - "block_id": "p920-b39", - "global_id": 26999, - "bbox": [ - 318.37, - 229.85, - 382.56, - 239.93 - ], - "text": "(e) 2πδ( −0)", - "type": "text" - }, - { - "block_id": "p920-b40", - "global_id": 27000, - "bbox": [ - 319.37, - 240.8, - 432.34, - 250.88 - ], - "text": "(f) π[δ( −0) + δ( + 0)]", - "type": "text" - }, - { - "block_id": "p920-b41", - "global_id": 27001, - "bbox": [ - 289.22, - 255.01, - 490.38, - 264.35 - ], - "text": "9.2-5\n(a) Determine and plot the DTFT X() of the", - "type": "text" - }, - { - "block_id": "p920-b42", - "global_id": 27002, - "bbox": [ - 317.86, - 265.97, - 490.38, - 286.27 - ], - "text": "triangular signal x[n] shown in Fig. P9.2-5.\n(b) Using Ex. 9.5 as a guide, use MATLAB and", - "type": "text" - }, - { - "block_id": "p920-b43", - "global_id": 27003, - "bbox": [ - 333.31, - 288.26, - 490.36, - 308.19 - ], - "text": "the FFT to validate the DTFT calculations\nand plot of part (a).", - "type": "text" - }, - { - "block_id": "p920-b44", - "global_id": 27004, - "bbox": [ - 407.79, - 326.46, - 420.65, - 334.68 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p920-b45", - "global_id": 27005, - "bbox": [ - 484.83, - 392.36, - 488.81, - 400.33 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p920-b46", - "global_id": 27006, - "bbox": [ - 394.53, - 337.31, - 398.52, - 345.28 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p920-b47", - "global_id": 27007, - "bbox": [ - 341.08, - 393.32, - 466.53, - 401.62 - ], - "text": "−6\n6", - "type": "text" - }, - { - "block_id": "p920-b48", - "global_id": 27008, - "bbox": [ - 378.31, - 408.6, - 429.95, - 417.56 - ], - "text": "Figure P9.2-5", - "type": "text" - }, - { - "block_id": "p920-b49", - "global_id": 27009, - "bbox": [ - 289.22, - 434.12, - 490.39, - 454.12 - ], - "text": "9.2-6\nUsing Eq. (9.18), show that the inverse DTFT of\nrect(( −π/4)/π) is 0.5sinc(πn/2)ejπn/4.", - "type": "text" - }, - { - "block_id": "p920-b50", - "global_id": 27010, - "bbox": [ - 289.22, - 459.29, - 490.4, - 479.29 - ], - "text": "9.2-7\nUsing Eq. (9.19), find the DTFT of the signals\nx[n] in Fig. P9.2-7.", - "type": "text" - }, - { - "block_id": "p920-b51", - "global_id": 27011, - "bbox": [ - 289.22, - 484.46, - 490.4, - 504.45 - ], - "text": "9.2-8\nUsing Eq. (9.19), find the DTFT of the signals\ndepicted in Fig. P9.2-8.", - "type": "text" - }, - { - "block_id": "p920-b52", - "global_id": 27012, - "bbox": [ - 289.22, - 509.62, - 490.39, - 529.63 - ], - "text": "9.2-9\nUse Eq. (9.18) to find the inverse DTFT of the\nspectra (shown only for || ≤π) in Fig. P9.2-9.", - "type": "text" - }, - { - "block_id": "p920-b53", - "global_id": 27013, - "bbox": [ - 165.05, - 599.9, - 350.89, - 609.26 - ], - "text": "0\nn", - "type": "text" - }, - { - "block_id": "p920-b54", - "global_id": 27014, - "bbox": [ - 185.1, - 558.7, - 323.99, - 568.23 - ], - "text": "an (a 1)\nan (a 1)", - "type": "text" - }, - { - "block_id": "p920-b55", - "global_id": 27015, - "bbox": [ - 230.44, - 599.9, - 285.39, - 609.26 - ], - "text": "0\nn", - "type": "text" - }, - { - "block_id": "p920-b56", - "global_id": 27016, - "bbox": [ - 194.63, - 616.68, - 322.38, - 624.68 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p920-b57", - "global_id": 27017, - "bbox": [ - 145.9, - 552.19, - 276.59, - 560.27 - ], - "text": "x[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p920-b58", - "global_id": 27018, - "bbox": [ - 214.1, - 600.72, - 338.67, - 610.33 - ], - "text": "N0\nN0", - "type": "text" - }, - { - "block_id": "p920-b59", - "global_id": 27019, - "bbox": [ - 375.63, - 616.65, - 427.27, - 625.62 - ], - "text": "Figure P9.2-7", - "type": "text" - } - ] - }, - { - "page_num": 921, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p921-b0", - "global_id": 27020, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n901", - "type": "text" - }, - { - "block_id": "p921-b1", - "global_id": 27021, - "bbox": [ - 198.26, - 93.44, - 202.26, - 101.44 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p921-b2", - "global_id": 27022, - "bbox": [ - 198.26, - 106.39, - 202.26, - 114.39 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p921-b3", - "global_id": 27023, - "bbox": [ - 227.86, - 130.75, - 264.14, - 138.75 - ], - "text": "6\n13", - "type": "text" - }, - { - "block_id": "p921-b4", - "global_id": 27024, - "bbox": [ - 265.83, - 113.95, - 269.83, - 121.95 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p921-b5", - "global_id": 27025, - "bbox": [ - 207.5, - 130.75, - 211.5, - 138.75 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p921-b6", - "global_id": 27026, - "bbox": [ - 220.65, - 146.73, - 353.87, - 154.73 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p921-b7", - "global_id": 27027, - "bbox": [ - 302.85, - 130.83, - 389.73, - 140.65 - ], - "text": "N0 1\n(N0 1)", - "type": "text" - }, - { - "block_id": "p921-b8", - "global_id": 27028, - "bbox": [ - 350.56, - 93.46, - 354.56, - 101.46 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p921-b9", - "global_id": 27029, - "bbox": [ - 384.9, - 113.95, - 388.9, - 121.95 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p921-b10", - "global_id": 27030, - "bbox": [ - 417.38, - 146.71, - 469.01, - 155.67 - ], - "text": "Figure P9.2-8", - "type": "text" - }, - { - "block_id": "p921-b11", - "global_id": 27031, - "bbox": [ - 338.9, - 193.06, - 342.9, - 201.06 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p921-b12", - "global_id": 27032, - "bbox": [ - 305.85, - 209.59, - 360.75, - 217.88 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p921-b13", - "global_id": 27033, - "bbox": [ - 338.05, - 177.81, - 342.05, - 185.81 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p921-b14", - "global_id": 27034, - "bbox": [ - 187.55, - 228.02, - 339.55, - 236.02 - ], - "text": "(b)\n(a)", - "type": "text" - }, - { - "block_id": "p921-b15", - "global_id": 27035, - "bbox": [ - 194.54, - 209.58, - 383.71, - 226.76 - ], - "text": "0\n2\n2\n3p\n4", - "type": "text" - }, - { - "block_id": "p921-b17", - "global_id": 27036, - "bbox": [ - 316.13, - 172.05, - 332.57, - 180.34 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p921-b18", - "global_id": 27037, - "bbox": [ - 216.19, - 186.95, - 225.41, - 196.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p921-b19", - "global_id": 27038, - "bbox": [ - 173.67, - 172.05, - 190.11, - 180.34 - ], - "text": "X(\t)", - "type": "text" - }, - { - "block_id": "p921-b20", - "global_id": 27039, - "bbox": [ - 138.03, - 209.89, - 154.76, - 226.76 - ], - "text": "3p\n4", - "type": "text" - }, - { - "block_id": "p921-b21", - "global_id": 27040, - "bbox": [ - 130.58, - 242.19, - 182.22, - 251.15 - ], - "text": "Figure P9.2-9", - "type": "text" - }, - { - "block_id": "p921-b22", - "global_id": 27041, - "bbox": [ - 83.34, - 263.33, - 288.99, - 294.29 - ], - "text": "9.2-10\nUse Eq. (9.18) to find the inverse DTFT\nof the spectra (shown only for || ≤π) in\nFig. P9.2-10.", - "type": "text" - }, - { - "block_id": "p921-b23", - "global_id": 27042, - "bbox": [ - 310.48, - 263.33, - 516.14, - 283.32 - ], - "text": "9.2-11\nFind the DTFT for the signals shown in\nFig. P9.2-11.", - "type": "text" - }, - { - "block_id": "p921-b24", - "global_id": 27043, - "bbox": [ - 353.36, - 367.72, - 357.36, - 375.72 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p921-b26", - "global_id": 27044, - "bbox": [ - 190.94, - 385.92, - 355.49, - 393.92 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p921-b27", - "global_id": 27045, - "bbox": [ - 174.94, - 320.27, - 371.47, - 339.62 - ], - "text": "cos \t \nX(\t)\nX(\t)", - "type": "text" - }, - { - "block_id": "p921-b28", - "global_id": 27046, - "bbox": [ - 241.05, - 363.56, - 246.38, - 380.46 - ], - "text": "2\np", - "type": "text" - }, - { - "block_id": "p921-b29", - "global_id": 27047, - "bbox": [ - 141.15, - 363.46, - 401.93, - 380.46 - ], - "text": "4\np\n2\np", - "type": "text" - }, - { - "block_id": "p921-b30", - "global_id": 27048, - "bbox": [ - 293.97, - 363.56, - 306.61, - 380.46 - ], - "text": "4\np", - "type": "text" - }, - { - "block_id": "p921-b31", - "global_id": 27049, - "bbox": [ - 130.58, - 400.08, - 186.69, - 409.05 - ], - "text": "Figure P9.2-10", - "type": "text" - }, - { - "block_id": "p921-b32", - "global_id": 27050, - "bbox": [ - 213.97, - 582.33, - 217.97, - 590.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p921-b33", - "global_id": 27051, - "bbox": [ - 203.51, - 539.98, - 207.51, - 547.98 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p921-b34", - "global_id": 27052, - "bbox": [ - 235.97, - 582.33, - 239.97, - 590.33 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p921-b35", - "global_id": 27053, - "bbox": [ - 179.14, - 564.48, - 207.51, - 577.67 - ], - "text": "3\n3", - "type": "text" - }, - { - "block_id": "p921-b36", - "global_id": 27054, - "bbox": [ - 214.92, - 424.34, - 227.8, - 432.42 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p921-b37", - "global_id": 27055, - "bbox": [ - 215.94, - 524.42, - 351.02, - 532.5 - ], - "text": "x[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p921-b38", - "global_id": 27056, - "bbox": [ - 323.44, - 424.34, - 336.32, - 432.42 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p921-b39", - "global_id": 27057, - "bbox": [ - 213.97, - 483.1, - 396.17, - 491.18 - ], - "text": "n\nn\n0\n3", - "type": "text" - }, - { - "block_id": "p921-b40", - "global_id": 27058, - "bbox": [ - 204.97, - 435.29, - 208.97, - 443.29 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p921-b41", - "global_id": 27059, - "bbox": [ - 321.47, - 483.18, - 348.17, - 491.18 - ], - "text": "0\n3", - "type": "text" - }, - { - "block_id": "p921-b42", - "global_id": 27060, - "bbox": [ - 312.19, - 439.93, - 316.19, - 447.93 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p921-b43", - "global_id": 27061, - "bbox": [ - 370.17, - 483.18, - 374.17, - 491.18 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p921-b44", - "global_id": 27062, - "bbox": [ - 336.17, - 582.33, - 340.17, - 590.33 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p921-b45", - "global_id": 27063, - "bbox": [ - 325.4, - 551.48, - 329.4, - 559.48 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p921-b46", - "global_id": 27064, - "bbox": [ - 310.84, - 582.04, - 353.17, - 590.33 - ], - "text": "2\n2", - "type": "text" - }, - { - "block_id": "p921-b47", - "global_id": 27065, - "bbox": [ - 221.73, - 503.64, - 230.61, - 511.64 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p921-b48", - "global_id": 27066, - "bbox": [ - 221.73, - 625.69, - 230.61, - 633.69 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p921-b49", - "global_id": 27067, - "bbox": [ - 355.35, - 503.64, - 364.99, - 511.64 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p921-b50", - "global_id": 27068, - "bbox": [ - 355.51, - 625.69, - 364.83, - 633.69 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p921-b51", - "global_id": 27069, - "bbox": [ - 251.97, - 582.25, - 378.17, - 590.25 - ], - "text": "n\nn", - "type": "text" - }, - { - "block_id": "p921-b52", - "global_id": 27070, - "bbox": [ - 427.37, - 625.67, - 483.49, - 634.64 - ], - "text": "Figure P9.2-11", - "type": "text" - } - ] - }, - { - "page_num": 922, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p922-b0", - "global_id": 27071, - "bbox": [ - 60.0, - 60.36, - 373.81, - 69.45 - ], - "text": "902\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p922-b1", - "global_id": 27072, - "bbox": [ - 57.59, - 85.64, - 263.25, - 151.12 - ], - "text": "9.2-12\nFind the inverse DTFT of X() (shown only\nfor ||≤π) for the spectra illustrated in\nFig. P9.2-12. [Hint: X() = |X()|ej̸\nX(). This\nproblem illustrates how different phase spec-\ntra (both with the same amplitude spectrum)\nrepresent entirely different signals.]", - "type": "text" - }, - { - "block_id": "p922-b2", - "global_id": 27073, - "bbox": [ - 57.59, - 155.93, - 263.24, - 165.94 - ], - "text": "9.2-13\n(a) Show that time-expanded signal xe[n] in", - "type": "text" - }, - { - "block_id": "p922-b3", - "global_id": 27074, - "bbox": [ - 106.15, - 167.27, - 227.03, - 176.23 - ], - "text": "Eq. (3.2) can also be expressed as", - "type": "text" - }, - { - "block_id": "p922-b4", - "global_id": 27075, - "bbox": [ - 136.26, - 197.17, - 162.91, - 207.18 - ], - "text": "xe[n] =", - "type": "text" - }, - { - "block_id": "p922-b5", - "global_id": 27076, - "bbox": [ - 168.26, - 187.86, - 180.94, - 197.62 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p922-b6", - "global_id": 27077, - "bbox": [ - 164.75, - 210.01, - 184.44, - 216.69 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p922-b7", - "global_id": 27078, - "bbox": [ - 185.45, - 197.17, - 233.12, - 206.42 - ], - "text": "x[k]δ[n −Lk]", - "type": "text" - }, - { - "block_id": "p922-b8", - "global_id": 27079, - "bbox": [ - 90.72, - 228.66, - 263.24, - 238.67 - ], - "text": "(b) Find the DTFT of xe[n] by finding the DTFT", - "type": "text" - }, - { - "block_id": "p922-b9", - "global_id": 27080, - "bbox": [ - 91.22, - 239.99, - 263.25, - 270.88 - ], - "text": "of the right-hand side of the equation in part\n(a).\n(c) Use the result in part (b) and Table 9.1", - "type": "text" - }, - { - "block_id": "p922-b10", - "global_id": 27081, - "bbox": [ - 106.16, - 272.5, - 263.24, - 292.79 - ], - "text": "to find the DTFT of z[n], shown in\nFig. P9.2-13.", - "type": "text" - }, - { - "block_id": "p922-b11", - "global_id": 27082, - "bbox": [ - 57.59, - 297.91, - 263.23, - 306.95 - ], - "text": "9.2-14\n(a) A glance at Eq. (9.18) shows that the", - "type": "text" - }, - { - "block_id": "p922-b12", - "global_id": 27083, - "bbox": [ - 106.16, - 308.95, - 263.24, - 361.74 - ], - "text": "inverse DTFT equation is identical to the\ninverse (continuous-time) Fourier transform\nEq. (7.10) for a signal x(t) bandlimited\nto π rad/s. Hence, we should be able to\nuse the continuous-time Fourier transform", - "type": "text" - }, - { - "block_id": "p922-b13", - "global_id": 27084, - "bbox": [ - 317.86, - 85.9, - 490.4, - 160.62 - ], - "text": "Table 7.1 to find DTFT pairs that corre-\nspond to continuous-time transform pairs\nfor bandlimited signals. Use this fact to\nderive DTFT pairs 8, 9, 11, 12, 13, and 14 in\nTable 9.1 by means of the appropriate pairs\nin Table 7.1.\n(b) Can this method be used to derive pairs 2, 3,", - "type": "text" - }, - { - "block_id": "p922-b14", - "global_id": 27085, - "bbox": [ - 333.31, - 162.62, - 490.39, - 182.54 - ], - "text": "4, 5, 6, 7, 10, 15, and 16 in Table 9.1? Justify\nyour answer with specific reason(s).", - "type": "text" - }, - { - "block_id": "p922-b15", - "global_id": 27086, - "bbox": [ - 284.74, - 191.11, - 490.4, - 222.07 - ], - "text": "9.2-15\nAre the following frequency-domain signals\nvalid DTFT’s? Answer yes or no, and justify\nyour answers.", - "type": "text" - }, - { - "block_id": "p922-b16", - "global_id": 27087, - "bbox": [ - 317.86, - 223.69, - 385.66, - 243.99 - ], - "text": "(a) X() = + π\n(b) X() = j + π", - "type": "text" - }, - { - "block_id": "p922-b17", - "global_id": 27088, - "bbox": [ - 317.86, - 245.61, - 402.06, - 265.91 - ], - "text": "(c) X() = sin(10)\n(d) X() = sin(/10)", - "type": "text" - }, - { - "block_id": "p922-b18", - "global_id": 27089, - "bbox": [ - 318.37, - 267.53, - 381.53, - 276.87 - ], - "text": "(e) X() = δ()", - "type": "text" - }, - { - "block_id": "p922-b19", - "global_id": 27090, - "bbox": [ - 289.23, - 285.44, - 490.39, - 327.36 - ], - "text": "9.3-1\nUsing only pairs 2 and 5 (Table 9.1) and\nthe time-shifting property of Eq. (9.31), find\nthe DTFT of the following signals, assuming\n|a| < 1.", - "type": "text" - }, - { - "block_id": "p922-b20", - "global_id": 27091, - "bbox": [ - 317.87, - 328.98, - 387.23, - 349.27 - ], - "text": "(a) u[n] −u[n −9]\n(b) an−mu[n −m]", - "type": "text" - }, - { - "block_id": "p922-b21", - "global_id": 27092, - "bbox": [ - 318.37, - 349.62, - 414.91, - 360.23 - ], - "text": "(c) an−3(u[n] −u[n −10])", - "type": "text" - }, - { - "block_id": "p922-b22", - "global_id": 27093, - "bbox": [ - 175.61, - 388.85, - 179.61, - 396.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p922-b23", - "global_id": 27094, - "bbox": [ - 175.61, - 423.67, - 179.61, - 431.67 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p922-b24", - "global_id": 27095, - "bbox": [ - 168.35, - 519.05, - 177.23, - 527.05 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p922-b25", - "global_id": 27096, - "bbox": [ - 359.85, - 388.85, - 363.85, - 396.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p922-b26", - "global_id": 27097, - "bbox": [ - 359.85, - 423.67, - 363.85, - 431.67 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p922-b27", - "global_id": 27098, - "bbox": [ - 352.12, - 519.05, - 361.93, - 527.05 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p922-b32", - "global_id": 27099, - "bbox": [ - 298.23, - 484.56, - 314.12, - 494.38 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p922-b33", - "global_id": 27100, - "bbox": [ - 298.23, - 423.75, - 412.4, - 433.57 - ], - "text": "0\n\t0", - "type": "text" - }, - { - "block_id": "p922-b34", - "global_id": 27101, - "bbox": [ - 115.55, - 470.77, - 412.4, - 494.38 - ], - "text": "0\n\t0", - "type": "text" - }, - { - "block_id": "p922-b35", - "global_id": 27102, - "bbox": [ - 115.55, - 423.75, - 229.72, - 433.57 - ], - "text": "0\n\t0", - "type": "text" - }, - { - "block_id": "p922-b36", - "global_id": 27103, - "bbox": [ - 220.5, - 470.77, - 229.72, - 480.59 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p922-b37", - "global_id": 27104, - "bbox": [ - 181.65, - 503.51, - 201.54, - 513.34 - ], - "text": "n0", - "type": "text" - }, - { - "block_id": "p922-b38", - "global_id": 27105, - "bbox": [ - 176.47, - 452.66, - 399.26, - 469.63 - ], - "text": "X(\t)\nX(\t)", - "type": "text" - }, - { - "block_id": "p922-b39", - "global_id": 27106, - "bbox": [ - 146.2, - 379.28, - 350.62, - 387.57 - ], - "text": "X(\t)\nX(\t)", - "type": "text" - }, - { - "block_id": "p922-b40", - "global_id": 27107, - "bbox": [ - 361.18, - 454.4, - 366.51, - 470.79 - ], - "text": "2\np", - "type": "text" - }, - { - "block_id": "p922-b41", - "global_id": 27108, - "bbox": [ - 339.87, - 496.72, - 352.51, - 513.12 - ], - "text": "2\np", - "type": "text" - }, - { - "block_id": "p922-b42", - "global_id": 27109, - "bbox": [ - 104.83, - 533.22, - 160.95, - 542.19 - ], - "text": "Figure P9.2-12", - "type": "text" - }, - { - "block_id": "p922-b43", - "global_id": 27110, - "bbox": [ - 240.11, - 564.36, - 244.11, - 572.36 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p922-b44", - "global_id": 27111, - "bbox": [ - 134.61, - 610.14, - 366.59, - 622.12 - ], - "text": "3\n6\n3\n6\n9\n9\n0\nn", - "type": "text" - }, - { - "block_id": "p922-b45", - "global_id": 27112, - "bbox": [ - 251.11, - 555.44, - 263.55, - 563.52 - ], - "text": "z[n]", - "type": "text" - }, - { - "block_id": "p922-b46", - "global_id": 27113, - "bbox": [ - 104.83, - 628.36, - 160.95, - 637.33 - ], - "text": "Figure P9.2-13", - "type": "text" - } - ] - }, - { - "page_num": 923, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p923-b0", - "global_id": 27114, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n903", - "type": "text" - }, - { - "block_id": "p923-b1", - "global_id": 27115, - "bbox": [ - 116.46, - 85.18, - 164.78, - 95.78 - ], - "text": "(d) an−mu[n]", - "type": "text" - }, - { - "block_id": "p923-b2", - "global_id": 27116, - "bbox": [ - 116.96, - 96.34, - 171.3, - 106.74 - ], - "text": "(e) anu[n −m]", - "type": "text" - }, - { - "block_id": "p923-b3", - "global_id": 27117, - "bbox": [ - 116.46, - 107.1, - 208.46, - 139.62 - ], - "text": "(f) (n −m)an−mu[n −m]\n(g) (n −m)anu[n]\n(h) nan−mu[n −m]", - "type": "text" - }, - { - "block_id": "p923-b4", - "global_id": 27118, - "bbox": [ - 87.82, - 144.65, - 288.98, - 164.96 - ], - "text": "9.3-2\nThe triangular pulse x[n] shown in Fig. P9.3-2a\nis given by", - "type": "text" - }, - { - "block_id": "p923-b5", - "global_id": 27119, - "bbox": [ - 151.94, - 177.24, - 251.81, - 196.12 - ], - "text": "X() = 4ej6 −5ej5 + ej", - "type": "text" - }, - { - "block_id": "p923-b6", - "global_id": 27120, - "bbox": [ - 199.83, - 190.55, - 235.34, - 202.48 - ], - "text": "(ej −1)2", - "type": "text" - }, - { - "block_id": "p923-b7", - "global_id": 27121, - "bbox": [ - 116.46, - 216.46, - 288.99, - 258.3 - ], - "text": "Use this information and the DTFT properties to\nfind the DTFT of the signals x1[n],x2[n],x3[n],\nand x4[n] shown in Figs. P9.3-2b, P9.3-2c,\nP9.3-2d, and P9.3-2e, respectively.", - "type": "text" - }, - { - "block_id": "p923-b8", - "global_id": 27122, - "bbox": [ - 87.82, - 262.0, - 288.98, - 295.34 - ], - "text": "9.3-3\nSuppose signal x[n] = sinc2(πn/2) modulates\na carrier cos(cn) to produce signal y[n] =\nx[n]cos(cn). Find and sketch the DTFT of:", - "type": "text" - }, - { - "block_id": "p923-b9", - "global_id": 27123, - "bbox": [ - 116.46, - 296.21, - 196.83, - 317.48 - ], - "text": "(a) x[n]\n(b) y[n] for c = π/2", - "type": "text" - }, - { - "block_id": "p923-b10", - "global_id": 27124, - "bbox": [ - 116.46, - 318.14, - 201.31, - 339.4 - ], - "text": "(c) y[n] for c = 3π/4\n(d) y[n] for c = π", - "type": "text" - }, - { - "block_id": "p923-b11", - "global_id": 27125, - "bbox": [ - 314.97, - 85.64, - 516.13, - 105.94 - ], - "text": "9.3-4\nShow that periodic convolution X() ∗⃝Y() =\n2πX() if", - "type": "text" - }, - { - "block_id": "p923-b12", - "global_id": 27126, - "bbox": [ - 393.76, - 125.6, - 422.14, - 134.85 - ], - "text": "X() =", - "type": "text" - }, - { - "block_id": "p923-b13", - "global_id": 27127, - "bbox": [ - 423.98, - 116.56, - 436.67, - 126.05 - ], - "text": "4\n\"", - "type": "text" - }, - { - "block_id": "p923-b14", - "global_id": 27128, - "bbox": [ - 424.68, - 138.53, - 435.95, - 145.27 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p923-b15", - "global_id": 27129, - "bbox": [ - 437.66, - 123.89, - 465.47, - 136.19 - ], - "text": "ak e−jk", - "type": "text" - }, - { - "block_id": "p923-b16", - "global_id": 27130, - "bbox": [ - 343.61, - 154.82, - 356.55, - 163.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p923-b17", - "global_id": 27131, - "bbox": [ - 384.59, - 161.69, - 453.35, - 177.32 - ], - "text": "Y() = sin(5/2)", - "type": "text" - }, - { - "block_id": "p923-b18", - "global_id": 27132, - "bbox": [ - 418.04, - 166.27, - 474.65, - 183.68 - ], - "text": "sin(/2) e−j2", - "type": "text" - }, - { - "block_id": "p923-b19", - "global_id": 27133, - "bbox": [ - 343.61, - 190.4, - 483.05, - 200.8 - ], - "text": "where ak is a set of arbitrary constants.", - "type": "text" - }, - { - "block_id": "p923-b20", - "global_id": 27134, - "bbox": [ - 314.97, - 204.36, - 516.14, - 236.29 - ], - "text": "9.3-5\nUsing only pair 2 (Table 9.1) and properties of\nDTFT, find the DTFT of the following signals,\nassuming |a| < 1 and 0 < π.", - "type": "text" - }, - { - "block_id": "p923-b21", - "global_id": 27135, - "bbox": [ - 343.61, - 234.06, - 411.34, - 257.24 - ], - "text": "(a) an cos0nu[n]\n(b) n2anu[n]", - "type": "text" - }, - { - "block_id": "p923-b22", - "global_id": 27136, - "bbox": [ - 344.12, - 257.8, - 426.79, - 268.2 - ], - "text": "(c) (n −k)a2nu[n −m]", - "type": "text" - }, - { - "block_id": "p923-b23", - "global_id": 27137, - "bbox": [ - 314.97, - 273.1, - 516.13, - 304.07 - ], - "text": "9.3-6\nUse pair 10 in Table 9.1, and suitable properties\nof the DTFT, to derive pairs 11, 12, 13, 14, 15,\nand 16.", - "type": "text" - }, - { - "block_id": "p923-b24", - "global_id": 27138, - "bbox": [ - 314.97, - 308.98, - 497.79, - 318.02 - ], - "text": "9.3-7\nUse the time-shifting property to show that", - "type": "text" - }, - { - "block_id": "p923-b25", - "global_id": 27139, - "bbox": [ - 361.12, - 329.37, - 498.62, - 338.71 - ], - "text": "x[n + k] + x[n −k] ⇐⇒2X() cosk", - "type": "text" - }, - { - "block_id": "p923-b26", - "global_id": 27140, - "bbox": [ - 223.55, - 428.48, - 232.43, - 436.48 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p923-b27", - "global_id": 27141, - "bbox": [ - 229.99, - 365.37, - 233.99, - 373.37 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p923-b28", - "global_id": 27142, - "bbox": [ - 172.28, - 410.67, - 182.95, - 418.96 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p923-b29", - "global_id": 27143, - "bbox": [ - 169.94, - 504.58, - 180.6, - 512.87 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p923-b30", - "global_id": 27144, - "bbox": [ - 231.04, - 410.59, - 285.39, - 421.7 - ], - "text": "0\nn", - "type": "text" - }, - { - "block_id": "p923-b31", - "global_id": 27145, - "bbox": [ - 293.8, - 506.62, - 501.14, - 514.62 - ], - "text": "n\nn", - "type": "text" - }, - { - "block_id": "p923-b32", - "global_id": 27146, - "bbox": [ - 429.72, - 600.26, - 433.72, - 608.26 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p923-b33", - "global_id": 27147, - "bbox": [ - 497.14, - 413.48, - 501.14, - 421.48 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p923-b34", - "global_id": 27148, - "bbox": [ - 204.59, - 356.44, - 217.47, - 364.52 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p923-b35", - "global_id": 27149, - "bbox": [ - 235.32, - 455.58, - 488.21, - 465.19 - ], - "text": "x2[n]\nx3[n]", - "type": "text" - }, - { - "block_id": "p923-b36", - "global_id": 27150, - "bbox": [ - 437.67, - 363.67, - 453.55, - 373.27 - ], - "text": "x1[n]", - "type": "text" - }, - { - "block_id": "p923-b37", - "global_id": 27151, - "bbox": [ - 421.66, - 428.48, - 431.31, - 436.48 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p923-b38", - "global_id": 27152, - "bbox": [ - 369.38, - 410.59, - 479.98, - 418.96 - ], - "text": "4\n4\n0", - "type": "text" - }, - { - "block_id": "p923-b39", - "global_id": 27153, - "bbox": [ - 417.44, - 365.12, - 421.44, - 373.12 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p923-b40", - "global_id": 27154, - "bbox": [ - 223.55, - 521.89, - 232.43, - 529.89 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p923-b41", - "global_id": 27155, - "bbox": [ - 231.04, - 504.11, - 280.6, - 512.47 - ], - "text": "4\n0", - "type": "text" - }, - { - "block_id": "p923-b42", - "global_id": 27156, - "bbox": [ - 421.83, - 522.01, - 431.15, - 530.01 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p923-b43", - "global_id": 27157, - "bbox": [ - 404.3, - 503.73, - 464.88, - 512.1 - ], - "text": "2\n2\n0", - "type": "text" - }, - { - "block_id": "p923-b44", - "global_id": 27158, - "bbox": [ - 428.03, - 457.33, - 432.03, - 465.33 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p923-b45", - "global_id": 27159, - "bbox": [ - 379.78, - 548.44, - 395.67, - 558.05 - ], - "text": "x4[n]", - "type": "text" - }, - { - "block_id": "p923-b46", - "global_id": 27160, - "bbox": [ - 323.79, - 615.14, - 332.67, - 623.14 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p923-b47", - "global_id": 27161, - "bbox": [ - 236.47, - 597.37, - 418.02, - 605.74 - ], - "text": "2\n7\n0\n7\n2", - "type": "text" - }, - { - "block_id": "p923-b48", - "global_id": 27162, - "bbox": [ - 221.19, - 457.83, - 225.19, - 465.83 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p923-b49", - "global_id": 27163, - "bbox": [ - 320.13, - 550.95, - 324.13, - 558.95 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p923-b50", - "global_id": 27164, - "bbox": [ - 130.58, - 629.31, - 182.22, - 638.27 - ], - "text": "Figure P9.3-2", - "type": "text" - } - ] - }, - { - "page_num": 924, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p924-b0", - "global_id": 27165, - "bbox": [ - 60.0, - 60.36, - 373.81, - 69.45 - ], - "text": "904\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p924-b1", - "global_id": 27166, - "bbox": [ - 173.86, - 144.74, - 182.74, - 152.74 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p924-b2", - "global_id": 27167, - "bbox": [ - 354.86, - 97.84, - 358.86, - 105.84 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p924-b3", - "global_id": 27168, - "bbox": [ - 383.83, - 123.22, - 413.54, - 131.22 - ], - "text": "4\n8", - "type": "text" - }, - { - "block_id": "p924-b4", - "global_id": 27169, - "bbox": [ - 358.13, - 141.19, - 367.78, - 149.19 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p924-b5", - "global_id": 27170, - "bbox": [ - 281.39, - 122.73, - 440.91, - 131.55 - ], - "text": "12\n12\n8\n4\n0", - "type": "text" - }, - { - "block_id": "p924-b6", - "global_id": 27171, - "bbox": [ - 173.19, - 96.91, - 177.19, - 104.91 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p924-b7", - "global_id": 27172, - "bbox": [ - 137.84, - 122.44, - 450.16, - 131.59 - ], - "text": "6\n2\n2\n6\nn\nn", - "type": "text" - }, - { - "block_id": "p924-b8", - "global_id": 27173, - "bbox": [ - 184.89, - 87.92, - 380.82, - 96.0 - ], - "text": "x[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p924-b9", - "global_id": 27174, - "bbox": [ - 104.83, - 158.91, - 156.48, - 167.88 - ], - "text": "Figure P9.3-7", - "type": "text" - }, - { - "block_id": "p924-b10", - "global_id": 27175, - "bbox": [ - 250.03, - 231.85, - 260.69, - 240.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p924-b11", - "global_id": 27176, - "bbox": [ - 256.69, - 189.19, - 260.69, - 197.19 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p924-b12", - "global_id": 27177, - "bbox": [ - 188.21, - 218.28, - 247.87, - 226.58 - ], - "text": "2\n4\n6", - "type": "text" - }, - { - "block_id": "p924-b13", - "global_id": 27178, - "bbox": [ - 285.04, - 206.34, - 338.04, - 214.34 - ], - "text": "2\n4\n6", - "type": "text" - }, - { - "block_id": "p924-b14", - "global_id": 27179, - "bbox": [ - 366.79, - 220.29, - 370.79, - 228.29 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p924-b15", - "global_id": 27180, - "bbox": [ - 269.28, - 182.42, - 282.16, - 190.5 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p924-b16", - "global_id": 27181, - "bbox": [ - 404.63, - 234.85, - 456.27, - 243.82 - ], - "text": "Figure P9.3-8", - "type": "text" - }, - { - "block_id": "p924-b17", - "global_id": 27182, - "bbox": [ - 90.72, - 258.32, - 263.24, - 278.24 - ], - "text": "Use this result to find the DTFT of the signals\nshown in Fig. P9.3-7.", - "type": "text" - }, - { - "block_id": "p924-b18", - "global_id": 27183, - "bbox": [ - 62.08, - 283.22, - 244.9, - 292.26 - ], - "text": "9.3-8\nUse the time-shifting property to show that", - "type": "text" - }, - { - "block_id": "p924-b19", - "global_id": 27184, - "bbox": [ - 107.72, - 304.53, - 246.23, - 313.87 - ], - "text": "x[n + k] −x[n −k] ⇐⇒2jX() sink", - "type": "text" - }, - { - "block_id": "p924-b20", - "global_id": 27185, - "bbox": [ - 90.72, - 326.52, - 263.25, - 346.44 - ], - "text": "Use this result to find the DTFT of the signal\nshown in Fig. P9.3-8.", - "type": "text" - }, - { - "block_id": "p924-b21", - "global_id": 27186, - "bbox": [ - 62.08, - 351.12, - 263.24, - 382.38 - ], - "text": "9.3-9\nSuppose signal x[n] has spectrum X() that\nis bandlimited to π/2 rad/sample. Next, define\nsignal y[n] as", - "type": "text" - }, - { - "block_id": "p924-b22", - "global_id": 27187, - "bbox": [ - 132.21, - 399.83, - 155.48, - 409.08 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p924-b23", - "global_id": 27188, - "bbox": [ - 157.33, - 387.24, - 215.56, - 414.61 - ], - "text": "x[n]\nn even\n0\nn odd", - "type": "text" - }, - { - "block_id": "p924-b24", - "global_id": 27189, - "bbox": [ - 90.72, - 426.36, - 263.23, - 446.67 - ], - "text": "Determine the spectrum of Y() in terms of\nX(). Sketch Y() if, over −π ≤ ≤π,", - "type": "text" - }, - { - "block_id": "p924-b25", - "global_id": 27190, - "bbox": [ - 103.02, - 463.73, - 131.19, - 472.98 - ], - "text": "Y() =", - "type": "text" - }, - { - "block_id": "p924-b26", - "global_id": 27191, - "bbox": [ - 133.04, - 451.14, - 244.77, - 478.51 - ], - "text": "|2/π|\n−π/2 ≤ ≤π/2\n0\notherwise", - "type": "text" - }, - { - "block_id": "p924-b27", - "global_id": 27192, - "bbox": [ - 57.59, - 490.33, - 256.52, - 499.67 - ], - "text": "9.3-10\nRepeat Prob. 9.3-9 if y[n] is instead defined as", - "type": "text" - }, - { - "block_id": "p924-b28", - "global_id": 27193, - "bbox": [ - 132.21, - 517.12, - 155.48, - 526.37 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p924-b29", - "global_id": 27194, - "bbox": [ - 157.33, - 504.53, - 215.56, - 531.89 - ], - "text": "x[n]\nn odd\n0\nn even", - "type": "text" - }, - { - "block_id": "p924-b30", - "global_id": 27195, - "bbox": [ - 57.59, - 544.01, - 263.24, - 574.97 - ], - "text": "9.3-11\nUsing only pair 2 in Table 9.1 and the convolu-\ntion property, find the inverse DTFT of X() =\ne2j/(ej −γ )2.", - "type": "text" - }, - { - "block_id": "p924-b31", - "global_id": 27196, - "bbox": [ - 57.59, - 579.94, - 263.24, - 632.82 - ], - "text": "9.3-12\nIn Table 9.1, you are given pair 1. From this\ninformation and using suitable properties of the\nDTFT, derive pairs 2, 3, 4, 5, 6, and 7 of\nTable 9.1. For example, starting with pair 1,\nderive pair 2. From pair 2, use suitable properties", - "type": "text" - }, - { - "block_id": "p924-b32", - "global_id": 27197, - "bbox": [ - 317.86, - 258.32, - 490.39, - 278.24 - ], - "text": "of the DTFT to derive pair 3. From pairs 2 and\n3, derive pair 4, and so on.", - "type": "text" - }, - { - "block_id": "p924-b33", - "global_id": 27198, - "bbox": [ - 284.74, - 281.58, - 490.39, - 326.03 - ], - "text": "9.3-13\nFrom\nthe\npair\nej(0/2)n\n⇐⇒\n2πδ( −\n(0/2)) over the fundamental band, and the\nfrequency-convolution property, find the DTFT\nof ej0n. Assume 0 <π/2.", - "type": "text" - }, - { - "block_id": "p924-b34", - "global_id": 27199, - "bbox": [ - 284.74, - 329.97, - 490.38, - 349.97 - ], - "text": "9.3-14\nFrom the definition and properties of the DTFT,\nshow that", - "type": "text" - }, - { - "block_id": "p924-b35", - "global_id": 27200, - "bbox": [ - 318.37, - 357.42, - 328.32, - 366.38 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p924-b36", - "global_id": 27201, - "bbox": [ - 336.93, - 347.73, - 349.62, - 357.5 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p924-b37", - "global_id": 27202, - "bbox": [ - 333.31, - 369.35, - 353.26, - 376.03 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p924-b38", - "global_id": 27203, - "bbox": [ - 354.25, - 350.76, - 410.51, - 367.06 - ], - "text": "sinc(cn) = π", - "type": "text" - }, - { - "block_id": "p924-b39", - "global_id": 27204, - "bbox": [ - 403.08, - 363.41, - 412.93, - 373.42 - ], - "text": "c", - "type": "text" - }, - { - "block_id": "p924-b40", - "global_id": 27205, - "bbox": [ - 423.59, - 357.04, - 450.01, - 367.35 - ], - "text": "c < π", - "type": "text" - }, - { - "block_id": "p924-b41", - "global_id": 27206, - "bbox": [ - 317.86, - 385.59, - 328.32, - 394.55 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p924-b42", - "global_id": 27207, - "bbox": [ - 336.93, - 375.9, - 349.62, - 385.67 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p924-b43", - "global_id": 27208, - "bbox": [ - 333.31, - 397.52, - 353.26, - 404.2 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p924-b44", - "global_id": 27209, - "bbox": [ - 353.25, - 383.71, - 462.65, - 395.52 - ], - "text": "(−1)nsinc(cn) = 0\nc < π", - "type": "text" - }, - { - "block_id": "p924-b45", - "global_id": 27210, - "bbox": [ - 318.37, - 413.76, - 328.32, - 422.72 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p924-b46", - "global_id": 27211, - "bbox": [ - 336.93, - 404.06, - 349.62, - 413.83 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p924-b47", - "global_id": 27212, - "bbox": [ - 333.31, - 425.69, - 353.26, - 432.37 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p924-b48", - "global_id": 27213, - "bbox": [ - 354.25, - 407.1, - 413.24, - 423.4 - ], - "text": "sinc2(cn) = π", - "type": "text" - }, - { - "block_id": "p924-b49", - "global_id": 27214, - "bbox": [ - 405.83, - 419.75, - 415.68, - 429.76 - ], - "text": "c", - "type": "text" - }, - { - "block_id": "p924-b50", - "global_id": 27215, - "bbox": [ - 426.34, - 413.38, - 461.89, - 423.69 - ], - "text": "c < π/2", - "type": "text" - }, - { - "block_id": "p924-b51", - "global_id": 27216, - "bbox": [ - 317.86, - 441.93, - 328.32, - 450.89 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p924-b52", - "global_id": 27217, - "bbox": [ - 336.93, - 432.23, - 349.62, - 442.0 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p924-b53", - "global_id": 27218, - "bbox": [ - 333.31, - 453.86, - 353.26, - 460.54 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p924-b54", - "global_id": 27219, - "bbox": [ - 353.25, - 440.04, - 472.85, - 451.86 - ], - "text": "(−1)nsinc2(cn) = 0\nc <π/2", - "type": "text" - }, - { - "block_id": "p924-b55", - "global_id": 27220, - "bbox": [ - 318.37, - 467.32, - 328.32, - 476.28 - ], - "text": "(e)", - "type": "text" - }, - { - "block_id": "p924-b56", - "global_id": 27221, - "bbox": [ - 333.31, - 454.74, - 346.03, - 465.71 - ], - "text": "# π", - "type": "text" - }, - { - "block_id": "p924-b57", - "global_id": 27222, - "bbox": [ - 338.04, - 476.98, - 346.97, - 483.46 - ], - "text": "−π", - "type": "text" - }, - { - "block_id": "p924-b58", - "global_id": 27223, - "bbox": [ - 350.31, - 460.66, - 391.21, - 470.0 - ], - "text": "sin(M/2)", - "type": "text" - }, - { - "block_id": "p924-b59", - "global_id": 27224, - "bbox": [ - 354.22, - 466.94, - 444.72, - 482.65 - ], - "text": "sin(/2) = 2π\noddM", - "type": "text" - }, - { - "block_id": "p924-b60", - "global_id": 27225, - "bbox": [ - 319.36, - 492.23, - 328.33, - 501.19 - ], - "text": "(f)", - "type": "text" - }, - { - "block_id": "p924-b61", - "global_id": 27226, - "bbox": [ - 336.93, - 482.54, - 349.62, - 492.3 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p924-b62", - "global_id": 27227, - "bbox": [ - 333.31, - 504.15, - 353.26, - 510.83 - ], - "text": "n=−∞", - "type": "text" - }, - { - "block_id": "p924-b63", - "global_id": 27228, - "bbox": [ - 354.25, - 488.15, - 482.94, - 502.16 - ], - "text": "|sinc(cn)|4 = 2π/3c\nc <π/2", - "type": "text" - }, - { - "block_id": "p924-b64", - "global_id": 27229, - "bbox": [ - 284.74, - 513.91, - 490.39, - 556.12 - ], - "text": "9.3-15\nShow that the energy of signal xc(t) specified in\nEq. (9.41) is identical to T times the energy of\nthe discrete-time signal x[n], assuming xc(t) is\nbandlimited to B ≤1/2T Hz. [Hint: Recall that", - "type": "text" - }, - { - "block_id": "p924-b65", - "global_id": 27230, - "bbox": [ - 338.08, - 559.91, - 353.53, - 570.87 - ], - "text": "# ∞", - "type": "text" - }, - { - "block_id": "p924-b66", - "global_id": 27231, - "bbox": [ - 342.81, - 582.14, - 354.47, - 588.61 - ], - "text": "−∞", - "type": "text" - }, - { - "block_id": "p924-b67", - "global_id": 27232, - "bbox": [ - 355.97, - 572.11, - 471.01, - 581.45 - ], - "text": "sinc[π(t −m)] sinc[π(t −n)] dt", - "type": "text" - }, - { - "block_id": "p924-b68", - "global_id": 27233, - "bbox": [ - 347.89, - 597.88, - 354.88, - 606.84 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p924-b69", - "global_id": 27234, - "bbox": [ - 356.72, - 585.29, - 406.59, - 612.64 - ], - "text": "0\nm̸ = n\n1\nm = n", - "type": "text" - }, - { - "block_id": "p924-b70", - "global_id": 27235, - "bbox": [ - 317.86, - 623.85, - 456.56, - 632.82 - ], - "text": "That is, sinc functions are orthogonal.]", - "type": "text" - } - ] - }, - { - "page_num": 925, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p925-b0", - "global_id": 27236, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n905", - "type": "text" - }, - { - "block_id": "p925-b1", - "global_id": 27237, - "bbox": [ - 87.82, - 85.94, - 288.99, - 116.9 - ], - "text": "9.4-1\nUse the DTFT method to find the zero-state\nresponse y[n] of a causal system with frequency\nresponse", - "type": "text" - }, - { - "block_id": "p925-b2", - "global_id": 27238, - "bbox": [ - 155.35, - 122.88, - 248.9, - 148.12 - ], - "text": "H() =\nej + 0.32\nej2 + ej + 0.16", - "type": "text" - }, - { - "block_id": "p925-b3", - "global_id": 27239, - "bbox": [ - 116.46, - 153.7, - 236.48, - 164.1 - ], - "text": "and the input x[n] = (−0.5)nu[n].", - "type": "text" - }, - { - "block_id": "p925-b4", - "global_id": 27240, - "bbox": [ - 87.82, - 169.01, - 196.03, - 178.05 - ], - "text": "9.4-2\nRepeat Prob. 9.4-1 for", - "type": "text" - }, - { - "block_id": "p925-b5", - "global_id": 27241, - "bbox": [ - 155.35, - 184.03, - 248.9, - 209.28 - ], - "text": "H() =\nej + 0.32\nej2 + ej + 0.16", - "type": "text" - }, - { - "block_id": "p925-b6", - "global_id": 27242, - "bbox": [ - 116.46, - 215.85, - 194.65, - 225.19 - ], - "text": "and input x[n] = u[n].", - "type": "text" - }, - { - "block_id": "p925-b7", - "global_id": 27243, - "bbox": [ - 87.82, - 230.1, - 196.03, - 239.14 - ], - "text": "9.4-3\nRepeat Prob. 9.4-1 for", - "type": "text" - }, - { - "block_id": "p925-b8", - "global_id": 27244, - "bbox": [ - 169.76, - 247.12, - 223.49, - 263.92 - ], - "text": "H() =\nej", - "type": "text" - }, - { - "block_id": "p925-b9", - "global_id": 27245, - "bbox": [ - 202.18, - 258.44, - 234.49, - 270.37 - ], - "text": "ej −0.5", - "type": "text" - }, - { - "block_id": "p925-b10", - "global_id": 27246, - "bbox": [ - 116.46, - 277.33, - 129.4, - 286.29 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p925-b11", - "global_id": 27247, - "bbox": [ - 139.18, - 294.48, - 266.27, - 305.32 - ], - "text": "x[n] = 0.8nu[n] + 2(2)nu[−(n + 1)]", - "type": "text" - }, - { - "block_id": "p925-b12", - "global_id": 27248, - "bbox": [ - 87.82, - 315.31, - 288.99, - 346.27 - ], - "text": "9.4-4\nDetermine and sketch the magnitude and phase\nresponse for an LTID system specified by the\nequation", - "type": "text" - }, - { - "block_id": "p925-b13", - "global_id": 27249, - "bbox": [ - 133.23, - 355.96, - 272.21, - 365.29 - ], - "text": "y[n] + 0.5y[n −1] = x[n] −0.9x[n −1]", - "type": "text" - }, - { - "block_id": "p925-b14", - "global_id": 27250, - "bbox": [ - 116.46, - 374.99, - 288.98, - 395.28 - ], - "text": "Determine the system output y[n] for the input\nx[n] = cos( πn", - "type": "text" - }, - { - "block_id": "p925-b15", - "global_id": 27251, - "bbox": [ - 160.36, - 385.94, - 193.64, - 398.1 - ], - "text": "3 + 0.5).", - "type": "text" - }, - { - "block_id": "p925-b16", - "global_id": 27252, - "bbox": [ - 87.82, - 400.19, - 288.98, - 420.19 - ], - "text": "9.4-5\nRepeat Prob. 9.4-4 if the LTID system is instead\nspecified by the equation", - "type": "text" - }, - { - "block_id": "p925-b17", - "global_id": 27253, - "bbox": [ - 133.23, - 429.88, - 272.21, - 439.22 - ], - "text": "y[n] −0.5y[n −1] = x[n] + 0.9x[n −1]", - "type": "text" - }, - { - "block_id": "p925-b18", - "global_id": 27254, - "bbox": [ - 87.82, - 449.21, - 288.98, - 469.21 - ], - "text": "9.4-6\nAn accumulator system has the property that an\ninput x[n] results in the output", - "type": "text" - }, - { - "block_id": "p925-b19", - "global_id": 27255, - "bbox": [ - 172.76, - 486.02, - 196.03, - 495.27 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p925-b20", - "global_id": 27256, - "bbox": [ - 201.38, - 476.92, - 214.07, - 486.48 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p925-b21", - "global_id": 27257, - "bbox": [ - 197.87, - 498.86, - 217.56, - 505.54 - ], - "text": "k=−∞", - "type": "text" - }, - { - "block_id": "p925-b22", - "global_id": 27258, - "bbox": [ - 218.57, - 486.02, - 232.68, - 495.27 - ], - "text": "x[k]", - "type": "text" - }, - { - "block_id": "p925-b23", - "global_id": 27259, - "bbox": [ - 116.96, - 518.43, - 288.98, - 527.77 - ], - "text": "(a) Find the unit impulse response h[n] and the", - "type": "text" - }, - { - "block_id": "p925-b24", - "global_id": 27260, - "bbox": [ - 116.46, - 529.39, - 288.99, - 560.65 - ], - "text": "frequency response H() for the accumula-\ntor.\n(b) Use the results of part (a) to find the DTFT", - "type": "text" - }, - { - "block_id": "p925-b25", - "global_id": 27261, - "bbox": [ - 131.9, - 562.26, - 158.79, - 571.6 - ], - "text": "of u[n].", - "type": "text" - }, - { - "block_id": "p925-b26", - "global_id": 27262, - "bbox": [ - 87.82, - 576.51, - 288.99, - 596.52 - ], - "text": "9.4-7\nA\nnoncausal\n7-point\nmoving\naverage\nis\ndescribed by the equation", - "type": "text" - }, - { - "block_id": "p925-b27", - "global_id": 27263, - "bbox": [ - 163.38, - 608.98, - 194.18, - 624.24 - ], - "text": "y[n] = 1", - "type": "text" - }, - { - "block_id": "p925-b28", - "global_id": 27264, - "bbox": [ - 189.69, - 621.63, - 194.18, - 630.6 - ], - "text": "7", - "type": "text" - }, - { - "block_id": "p925-b29", - "global_id": 27265, - "bbox": [ - 198.18, - 605.84, - 210.87, - 615.34 - ], - "text": "3\n\"", - "type": "text" - }, - { - "block_id": "p925-b30", - "global_id": 27266, - "bbox": [ - 196.36, - 627.82, - 212.68, - 634.57 - ], - "text": "k=−3", - "type": "text" - }, - { - "block_id": "p925-b31", - "global_id": 27267, - "bbox": [ - 213.69, - 614.89, - 242.06, - 624.14 - ], - "text": "x[n −k]", - "type": "text" - }, - { - "block_id": "p925-b32", - "global_id": 27268, - "bbox": [ - 344.11, - 90.75, - 516.13, - 99.71 - ], - "text": "(a) Find and sketch the magnitude and phase", - "type": "text" - }, - { - "block_id": "p925-b33", - "global_id": 27269, - "bbox": [ - 343.61, - 101.71, - 516.12, - 121.63 - ], - "text": "responses of the system.\n(b) How can this system be made causal?", - "type": "text" - }, - { - "block_id": "p925-b34", - "global_id": 27270, - "bbox": [ - 359.05, - 123.62, - 516.13, - 154.51 - ], - "text": "Plot the magnitude and phase responses of\nthe causal system, and comment on any\ndifferences from part (a).", - "type": "text" - }, - { - "block_id": "p925-b35", - "global_id": 27271, - "bbox": [ - 314.97, - 159.82, - 516.13, - 180.12 - ], - "text": "9.4-8\nAn LTID system frequency response over || ≤\nπ is", - "type": "text" - }, - { - "block_id": "p925-b36", - "global_id": 27272, - "bbox": [ - 387.0, - 195.44, - 431.66, - 204.78 - ], - "text": "H() = rect", - "type": "text" - }, - { - "block_id": "p925-b38", - "global_id": 27273, - "bbox": [ - 439.26, - 201.8, - 444.64, - 210.77 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p925-b40", - "global_id": 27274, - "bbox": [ - 453.13, - 193.73, - 472.24, - 204.69 - ], - "text": "e−j2", - "type": "text" - }, - { - "block_id": "p925-b41", - "global_id": 27275, - "bbox": [ - 343.61, - 225.05, - 516.13, - 245.35 - ], - "text": "Find the output y[n] of this system, if the input\nx[n] is given by", - "type": "text" - }, - { - "block_id": "p925-b42", - "global_id": 27276, - "bbox": [ - 343.61, - 246.98, - 400.64, - 267.27 - ], - "text": "(a) sinc(πn/2)\n(b) sinc(πn)", - "type": "text" - }, - { - "block_id": "p925-b43", - "global_id": 27277, - "bbox": [ - 344.12, - 265.28, - 404.38, - 278.23 - ], - "text": "(c) sinc2 (πn/4)", - "type": "text" - }, - { - "block_id": "p925-b44", - "global_id": 27278, - "bbox": [ - 314.97, - 283.54, - 516.13, - 314.8 - ], - "text": "9.4-9\n(a) If\nx[n] ⇐⇒X(),\nthen,\nshow\nthat\n(−1)nx[n] ⇐⇒X( −π).\n(b) Sketch γ nu[n] and (−γ )nu[n] for γ = 0.8;", - "type": "text" - }, - { - "block_id": "p925-b45", - "global_id": 27279, - "bbox": [ - 344.12, - 315.36, - 516.13, - 358.63 - ], - "text": "see the spectra for γ nu[n] in Figs. 9.5b and\n9.5c. From these spectra, sketch the spectra\nfor (−γ )nu[n].\n(c) An ideal lowpass filter of cutoff frequency", - "type": "text" - }, - { - "block_id": "p925-b46", - "global_id": 27280, - "bbox": [ - 359.05, - 360.25, - 516.15, - 424.39 - ], - "text": "c is specified by the frequency response\nH() = rect(/2c). Find its impulse\nresponse h[n]. Find the frequency response\nof a filter whose impulse response is\n(−1)nh[n]. Sketch the frequency response\nof this filter. What kind of filter is this?", - "type": "text" - }, - { - "block_id": "p925-b47", - "global_id": 27281, - "bbox": [ - 310.48, - 428.27, - 516.14, - 460.96 - ], - "text": "9.4-10\nAn analog differentiator y(t) =\nd\ndtx(t) can be\napproximated using a backward difference sys-\ntem described as", - "type": "text" - }, - { - "block_id": "p925-b48", - "global_id": 27282, - "bbox": [ - 389.63, - 475.79, - 468.92, - 491.32 - ], - "text": "y[n] = x[n] −x[n −1]", - "type": "text" - }, - { - "block_id": "p925-b49", - "global_id": 27283, - "bbox": [ - 439.59, - 488.72, - 444.57, - 497.68 - ], - "text": "T", - "type": "text" - }, - { - "block_id": "p925-b50", - "global_id": 27284, - "bbox": [ - 343.61, - 512.89, - 516.14, - 565.7 - ], - "text": "Find and sketch the magnitude and phase\nresponses of this DT system. For what fre-\nquencies does the system most behave as a\ndifferentiator? For what frequencies does the\nsystem least behave as a differentiator?", - "type": "text" - }, - { - "block_id": "p925-b51", - "global_id": 27285, - "bbox": [ - 310.48, - 571.01, - 516.13, - 635.88 - ], - "text": "9.4-11\nA filter with impulse response h[n] is modi-\nfied as shown in Fig. P9.4-11. Determine the\nresulting filter impulse response h1[n]. Find also\nthe resulting filter frequency response H1() in\nterms of the frequency response H(). How are\nH() and H1() related?", - "type": "text" - } - ] - }, - { - "page_num": 926, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p926-b0", - "global_id": 27286, - "bbox": [ - 60.0, - 60.36, - 373.81, - 69.45 - ], - "text": "906\nCHAPTER 9\nFOURIER ANALYSIS OF DISCRETE-TIME SIGNALS", - "type": "text" - }, - { - "block_id": "p926-b1", - "global_id": 27287, - "bbox": [ - 153.51, - 116.04, - 357.51, - 124.12 - ], - "text": "y[n]\nx[n]", - "type": "text" - }, - { - "block_id": "p926-b2", - "global_id": 27288, - "bbox": [ - 241.56, - 126.22, - 258.88, - 134.52 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p926-b3", - "global_id": 27289, - "bbox": [ - 200.7, - 89.22, - 363.02, - 105.38 - ], - "text": "H1(\t)\n(1)n\n(1)n", - "type": "text" - }, - { - "block_id": "p926-b4", - "global_id": 27290, - "bbox": [ - 379.63, - 144.71, - 435.75, - 153.68 - ], - "text": "Figure P9.4-11", - "type": "text" - }, - { - "block_id": "p926-b5", - "global_id": 27291, - "bbox": [ - 57.59, - 167.6, - 263.23, - 177.67 - ], - "text": "9.4-12\n(a) Consider an LTID system S1, specified", - "type": "text" - }, - { - "block_id": "p926-b6", - "global_id": 27292, - "bbox": [ - 90.72, - 178.93, - 263.24, - 276.91 - ], - "text": "by a difference equation of the form of\nEqs. (3.15) or (3.16) or (3.20) in Ch. 3. We\nconstruct another system S2 by replacing\ncoefficients ai (i=0,1,2,. . .,N) by coef-\nficients (−1)iai and replacing all coeffi-\ncients bi (i = 0,1,2,. . .,N) with coefficients\n(−1)ibi. How are the frequency responses of\nthe two systems related?\n(b) If S1 represents a lowpass filter, what kind", - "type": "text" - }, - { - "block_id": "p926-b7", - "global_id": 27293, - "bbox": [ - 91.22, - 277.18, - 263.24, - 297.48 - ], - "text": "of filter is specified by S2?\n(c) What type of filter (lowpass, highpass, etc.)", - "type": "text" - }, - { - "block_id": "p926-b8", - "global_id": 27294, - "bbox": [ - 106.15, - 299.48, - 242.45, - 308.44 - ], - "text": "is specified by the difference equation", - "type": "text" - }, - { - "block_id": "p926-b9", - "global_id": 27295, - "bbox": [ - 140.06, - 321.64, - 229.32, - 330.98 - ], - "text": "y[n] −0.8y[n −1] = x[n]", - "type": "text" - }, - { - "block_id": "p926-b10", - "global_id": 27296, - "bbox": [ - 106.15, - 344.56, - 263.23, - 364.48 - ], - "text": "What kind of filter is specified by the\nfollowing difference equation?", - "type": "text" - }, - { - "block_id": "p926-b11", - "global_id": 27297, - "bbox": [ - 140.06, - 377.68, - 229.32, - 387.02 - ], - "text": "y[n] + 0.8y[n −1] = x[n]", - "type": "text" - }, - { - "block_id": "p926-b12", - "global_id": 27298, - "bbox": [ - 57.59, - 400.52, - 263.21, - 409.56 - ], - "text": "9.4-13\n(a) The system shown in Fig. P9.4-13 contains", - "type": "text" - }, - { - "block_id": "p926-b13", - "global_id": 27299, - "bbox": [ - 106.15, - 411.55, - 263.25, - 464.36 - ], - "text": "two identical LTID filters with frequency\nresponse H0() and corresponding impulse\nresponse h0[n]. It is easy to see that the\nsystem is linear. Show that this system is\nalso time-invariant. Do this by finding the", - "type": "text" - }, - { - "block_id": "p926-b14", - "global_id": 27300, - "bbox": [ - 317.86, - 167.48, - 490.39, - 199.47 - ], - "text": "response of the system to input δ[n −k] in\nterms of h0[n].\n(b) If H0() = rect(/2W) over the funda-", - "type": "text" - }, - { - "block_id": "p926-b15", - "global_id": 27301, - "bbox": [ - 333.31, - 200.35, - 490.39, - 231.61 - ], - "text": "mental band, and c + W ≤π, find H(),\nthe frequency response of this system. What\nkind of filter is this?", - "type": "text" - }, - { - "block_id": "p926-b16", - "global_id": 27302, - "bbox": [ - 289.22, - 237.41, - 490.39, - 258.45 - ], - "text": "9.5-1\nDetermine the DTFT of x[n] = sin(0n) from\nthe CTFT of xc(t) = sin(ω0t).", - "type": "text" - }, - { - "block_id": "p926-b17", - "global_id": 27303, - "bbox": [ - 289.22, - 263.51, - 490.38, - 283.8 - ], - "text": "9.5-2\nA CT signal x(t), bandlimited to 25 kHz, is\nsampled at 50 kHz to produce", - "type": "text" - }, - { - "block_id": "p926-b18", - "global_id": 27304, - "bbox": [ - 324.65, - 307.28, - 483.58, - 316.62 - ], - "text": "x[n] = δ[n+4] −2δ[n+2] + δ[n+1] −3δ[n]", - "type": "text" - }, - { - "block_id": "p926-b19", - "global_id": 27305, - "bbox": [ - 351.19, - 321.23, - 464.43, - 330.57 - ], - "text": "−δ[n−1] −2δ[n−2] −δ[n−4]", - "type": "text" - }, - { - "block_id": "p926-b20", - "global_id": 27306, - "bbox": [ - 317.86, - 354.06, - 416.19, - 363.4 - ], - "text": "Determine the CTFT X(ω).", - "type": "text" - }, - { - "block_id": "p926-b21", - "global_id": 27307, - "bbox": [ - 289.23, - 369.49, - 490.39, - 400.45 - ], - "text": "9.7-1\nThis problem uses a matrix-based approach\nto investigate the computation of the inverse\nDTFS.", - "type": "text" - }, - { - "block_id": "p926-b22", - "global_id": 27308, - "bbox": [ - 318.37, - 402.45, - 490.39, - 411.41 - ], - "text": "(a) Implement Eq. (9.3), the inverse DTFS,", - "type": "text" - }, - { - "block_id": "p926-b23", - "global_id": 27309, - "bbox": [ - 317.86, - 413.41, - 490.39, - 466.21 - ], - "text": "using a matrix-based approach.\n(b) Compare\nthe\nexecution\nspeed\nof\nthe\nmatrix-based approach to the IFFT-based\napproach for input vectors of sizes 10, 100,\nand 1000.", - "type": "text" - }, - { - "block_id": "p926-b24", - "global_id": 27310, - "bbox": [ - 159.93, - 611.9, - 299.14, - 621.66 - ], - "text": "2 sin \tcn\nsin \tcn", - "type": "text" - }, - { - "block_id": "p926-b25", - "global_id": 27311, - "bbox": [ - 159.27, - 482.42, - 299.8, - 492.18 - ], - "text": "cos \tcn\n2 cos \tcn", - "type": "text" - }, - { - "block_id": "p926-b27", - "global_id": 27312, - "bbox": [ - 219.46, - 515.33, - 239.78, - 525.15 - ], - "text": "H0(\t)", - "type": "text" - }, - { - "block_id": "p926-b28", - "global_id": 27313, - "bbox": [ - 219.46, - 573.83, - 239.78, - 583.65 - ], - "text": "H0(\t)", - "type": "text" - }, - { - "block_id": "p926-b29", - "global_id": 27314, - "bbox": [ - 112.19, - 535.08, - 365.0, - 543.16 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p926-b30", - "global_id": 27315, - "bbox": [ - 104.83, - 628.36, - 160.95, - 637.33 - ], - "text": "Figure P9.4-13", - "type": "text" - } - ] - }, - { - "page_num": 927, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p927-b0", - "global_id": 27316, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n907", - "type": "text" - }, - { - "block_id": "p927-b1", - "global_id": 27317, - "bbox": [ - 116.96, - 85.9, - 288.97, - 94.86 - ], - "text": "(c) What is the result of multiplying the DFT", - "type": "text" - }, - { - "block_id": "p927-b2", - "global_id": 27318, - "bbox": [ - 131.89, - 96.78, - 288.98, - 116.78 - ], - "text": "matrix WN0 by the inverse DTFS matrix?\nDiscuss your result.", - "type": "text" - }, - { - "block_id": "p927-b3", - "global_id": 27319, - "bbox": [ - 87.82, - 122.01, - 288.99, - 142.0 - ], - "text": "9.7-2\nA stable, first-order highpass IIR digital filter\nhas transfer function", - "type": "text" - }, - { - "block_id": "p927-b4", - "global_id": 27320, - "bbox": [ - 148.34, - 160.89, - 173.51, - 170.14 - ], - "text": "H[z] =", - "type": "text" - }, - { - "block_id": "p927-b5", - "global_id": 27321, - "bbox": [ - 175.36, - 148.3, - 201.98, - 163.95 - ], - "text": "1 + α", - "type": "text" - }, - { - "block_id": "p927-b6", - "global_id": 27322, - "bbox": [ - 190.23, - 167.63, - 194.72, - 176.6 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p927-b7", - "global_id": 27323, - "bbox": [ - 203.54, - 148.3, - 246.62, - 163.95 - ], - "text": "1 −z−1", - "type": "text" - }, - { - "block_id": "p927-b8", - "global_id": 27324, - "bbox": [ - 217.84, - 166.66, - 249.36, - 176.6 - ], - "text": "1 −αz−1", - "type": "text" - }, - { - "block_id": "p927-b10", - "global_id": 27325, - "bbox": [ - 116.96, - 196.0, - 288.98, - 205.34 - ], - "text": "(a) Derive an expression relating α to the 3 dB", - "type": "text" - }, - { - "block_id": "p927-b11", - "global_id": 27326, - "bbox": [ - 116.46, - 206.95, - 288.99, - 227.25 - ], - "text": "cutoff frequency c.\n(b) Test your expression from part (a) in the fol-", - "type": "text" - }, - { - "block_id": "p927-b12", - "global_id": 27327, - "bbox": [ - 116.96, - 228.88, - 288.99, - 336.85 - ], - "text": "lowing manner. First, compute α to achieve\na 3 dB cutoff frequency of 1 kHz, assuming\na sampling rate of Fs = 5 kHz. Determine\na difference equation description of the\nsystem, and verify that the system is sta-\nble. Next, compute and plot the magnitude\nresponse of the resulting filter. Verify that\nthe filter is highpass and has the correct\ncutoff frequency.\n(c) Holding α constant, what happens to the", - "type": "text" - }, - { - "block_id": "p927-b13", - "global_id": 27328, - "bbox": [ - 116.46, - 338.46, - 288.99, - 380.68 - ], - "text": "cutoff frequency c as Fs is increased to 50\nkHz? What happens to the cutoff frequency\nfc as Fs is increased to 50 kHz?\n(d) Is there a well-behaved inverse filter to", - "type": "text" - }, - { - "block_id": "p927-b14", - "global_id": 27329, - "bbox": [ - 116.96, - 382.3, - 288.98, - 403.56 - ], - "text": "H[z]? Explain.\n(e) Determine α for c = π/2. Comment on the", - "type": "text" - }, - { - "block_id": "p927-b15", - "global_id": 27330, - "bbox": [ - 131.89, - 404.21, - 247.65, - 413.55 - ], - "text": "resulting filter, particularly h[n].", - "type": "text" - }, - { - "block_id": "p927-b16", - "global_id": 27331, - "bbox": [ - 87.82, - 418.79, - 289.0, - 471.66 - ], - "text": "9.7-3\nUsing the frequency sampling method, design\na length-35 linear phase FIR highstop filter\nthat has cutoff frequency c = 2π/3. Plot\nthe resulting filter’s impulse response h[n] and\nmagnitude response |H()|.", - "type": "text" - }, - { - "block_id": "p927-b17", - "global_id": 27332, - "bbox": [ - 87.82, - 476.89, - 288.98, - 529.76 - ], - "text": "9.7-4\nUsing the frequency-sampling method, design\na length-71 linear phase FIR bandstop filter\nthat has stopband (π/3 < || < π/2). Plot\nthe resulting filter’s impulse response h[n] and\nmagnitude response |H()|.", - "type": "text" - }, - { - "block_id": "p927-b18", - "global_id": 27333, - "bbox": [ - 87.83, - 535.0, - 288.99, - 555.0 - ], - "text": "9.7-5\nFigure P9.7-5 provides the desired magnitude\nresponse |H()| of a real filter. Mathematically,", - "type": "text" - }, - { - "block_id": "p927-b19", - "global_id": 27334, - "bbox": [ - 137.0, - 580.82, - 171.51, - 590.07 - ], - "text": "|H()| =", - "type": "text" - }, - { - "block_id": "p927-b20", - "global_id": 27335, - "bbox": [ - 173.35, - 562.44, - 180.46, - 579.47 - ], - "text": "⎧\n⎨", - "type": "text" - }, - { - "block_id": "p927-b21", - "global_id": 27336, - "bbox": [ - 173.35, - 586.64, - 180.46, - 595.61 - ], - "text": "⎩", - "type": "text" - }, - { - "block_id": "p927-b22", - "global_id": 27337, - "bbox": [ - 185.43, - 568.18, - 259.21, - 592.78 - ], - "text": "4\nπ\n0 ≤ < π\n4\n2 −4\nπ", - "type": "text" - }, - { - "block_id": "p927-b23", - "global_id": 27338, - "bbox": [ - 221.52, - 579.5, - 225.4, - 585.98 - ], - "text": "π", - "type": "text" - }, - { - "block_id": "p927-b24", - "global_id": 27339, - "bbox": [ - 195.65, - 579.5, - 260.43, - 601.43 - ], - "text": "4 ≤ < π\n2\n0\nπ", - "type": "text" - }, - { - "block_id": "p927-b25", - "global_id": 27340, - "bbox": [ - 222.61, - 592.09, - 260.94, - 604.24 - ], - "text": "2 ≤ ≤π", - "type": "text" - }, - { - "block_id": "p927-b26", - "global_id": 27341, - "bbox": [ - 343.61, - 85.58, - 516.14, - 105.88 - ], - "text": "Since the digital filter is real, |H()| = |H(−)|\nand |H()| = |H( + 2π)| for all .", - "type": "text" - }, - { - "block_id": "p927-b27", - "global_id": 27342, - "bbox": [ - 344.12, - 107.86, - 516.12, - 116.83 - ], - "text": "(a) Can a realizable filter have this exact mag-", - "type": "text" - }, - { - "block_id": "p927-b28", - "global_id": 27343, - "bbox": [ - 343.61, - 118.83, - 516.14, - 138.76 - ], - "text": "nitude response? Explain your answer.\n(b) Use the frequency-sampling method to", - "type": "text" - }, - { - "block_id": "p927-b29", - "global_id": 27344, - "bbox": [ - 359.05, - 140.74, - 516.13, - 182.59 - ], - "text": "design an FIR filter with this magni-\ntude response (or a reasonable approxima-\ntion). Use MATLAB to plot the magnitude\nresponse of your filter.", - "type": "text" - }, - { - "block_id": "p927-b30", - "global_id": 27345, - "bbox": [ - 352.98, - 320.23, - 512.03, - 328.52 - ], - "text": "p\np\np2\np2\n0", - "type": "text" - }, - { - "block_id": "p927-b31", - "global_id": 27346, - "bbox": [ - 356.36, - 305.22, - 360.36, - 313.22 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p927-b33", - "global_id": 27347, - "bbox": [ - 338.27, - 249.8, - 346.57, - 270.84 - ], - "text": "H(\t)", - "type": "text" - }, - { - "block_id": "p927-b34", - "global_id": 27348, - "bbox": [ - 350.36, - 285.9, - 360.36, - 293.9 - ], - "text": "0.2", - "type": "text" - }, - { - "block_id": "p927-b35", - "global_id": 27349, - "bbox": [ - 350.36, - 266.58, - 360.36, - 274.58 - ], - "text": "0.4", - "type": "text" - }, - { - "block_id": "p927-b36", - "global_id": 27350, - "bbox": [ - 350.36, - 247.26, - 360.36, - 255.26 - ], - "text": "0.6", - "type": "text" - }, - { - "block_id": "p927-b37", - "global_id": 27351, - "bbox": [ - 350.36, - 227.94, - 360.36, - 235.94 - ], - "text": "0.8", - "type": "text" - }, - { - "block_id": "p927-b38", - "global_id": 27352, - "bbox": [ - 356.36, - 208.63, - 360.36, - 216.63 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p927-b39", - "global_id": 27353, - "bbox": [ - 337.64, - 347.6, - 389.27, - 356.57 - ], - "text": "Figure P9.7-5", - "type": "text" - }, - { - "block_id": "p927-b40", - "global_id": 27354, - "bbox": [ - 314.97, - 376.74, - 516.15, - 429.62 - ], - "text": "9.7-6\nA real FIR comb filter is needed that has mag-\nnitude response |H()| = [0,3,0,3,0,3,0,3] for\n = [0,π/4,π/2,3π/4,π,5π/4,3π/2,7π/4],\nrespectively. Provide the impulse response h[n]\nof a filter that accomplishes these specifications.", - "type": "text" - }, - { - "block_id": "p927-b41", - "global_id": 27355, - "bbox": [ - 314.98, - 434.55, - 516.13, - 498.39 - ], - "text": "9.7-7\nA permutation matrix P has a single one in each\nrow and column with the remaining elements\nall zero. Permutation matrices are useful for\nreordering the elements of a vector; the opera-\ntion Px reorders the elements of a column vector\nx based on the form of P.", - "type": "text" - }, - { - "block_id": "p927-b42", - "global_id": 27356, - "bbox": [ - 344.12, - 500.0, - 516.12, - 510.68 - ], - "text": "(a) Fully describe an N0 × N0 permutation", - "type": "text" - }, - { - "block_id": "p927-b43", - "global_id": 27357, - "bbox": [ - 343.61, - 511.26, - 516.13, - 555.03 - ], - "text": "matrix named RN0 that reverses the order of\nthe elements of a column vector x.\n(b) Given\nDFT\nmatrix\nWN0,\nverify\nthat\n(WN0)(WN0) = W2", - "type": "text" - }, - { - "block_id": "p927-b44", - "global_id": 27358, - "bbox": [ - 359.05, - 544.21, - 516.12, - 565.95 - ], - "text": "N0 produces a scaled\npermutation\nmatrix.\nHow\ndoes\nW2", - "type": "text" - }, - { - "block_id": "p927-b45", - "global_id": 27359, - "bbox": [ - 344.11, - 556.91, - 516.13, - 587.87 - ], - "text": "N0x\nreorder the elements of x?\n(c) What is the result of (W2", - "type": "text" - }, - { - "block_id": "p927-b46", - "global_id": 27360, - "bbox": [ - 464.94, - 577.49, - 491.89, - 591.05 - ], - "text": "N0)(W2", - "type": "text" - }, - { - "block_id": "p927-b47", - "global_id": 27361, - "bbox": [ - 359.05, - 578.53, - 516.13, - 600.57 - ], - "text": "N0)x =\nW4", - "type": "text" - }, - { - "block_id": "p927-b48", - "global_id": 27362, - "bbox": [ - 368.01, - 591.6, - 384.53, - 603.82 - ], - "text": "N0x?", - "type": "text" - } - ] - }, - { - "page_num": 928, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p928-b0", - "global_id": 27363, - "bbox": [ - 64.47, - 67.04, - 133.54, - 79.04 - ], - "text": "C H A P T E R", - "type": "text" - }, - { - "block_id": "p928-b1", - "global_id": 27364, - "bbox": [ - 147.02, - 151.14, - 393.7, - 173.26 - ], - "text": "STATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p928-b2", - "global_id": 27365, - "bbox": [ - 94.87, - 79.92, - 133.73, - 118.77 - ], - "text": "10", - "type": "text" - }, - { - "block_id": "p928-b3", - "global_id": 27366, - "bbox": [ - 101.84, - 266.04, - 490.41, - 288.06 - ], - "text": "In Sec. 1.10, basic notions of state variables were introduced. In this chapter, we shall discuss\nstate variables in more depth.", - "type": "text" - }, - { - "block_id": "p928-b4", - "global_id": 27367, - "bbox": [ - 101.84, - 290.05, - 490.41, - 383.7 - ], - "text": "Most of this book deals with an external (input–output) description of systems. As noted in\nCh. 1, such a description may be inadequate in some cases, and we need a systematic way of\nfinding a system’s internal description. State-space analysis of systems meets this need. In this\nmethod, we first select a set of key variables, called the state variables, in the system. Every\npossible signal or variable in the system at any instant t can be expressed in terms of the state\nvariables and the input(s) at that instant t. If we know all the state variables as a function of t, we\ncan determine every possible signal or variable in the system at any instant with a relatively simple\nrelationship. The system description in this method consists of two parts:", - "type": "text" - }, - { - "block_id": "p928-b5", - "global_id": 27368, - "bbox": [ - 118.78, - 391.57, - 490.38, - 425.55 - ], - "text": "1. A set of equations relating the state variables to the inputs (the state equation).\n2. A set of equations relating outputs to the state variables and the inputs (the output\nequation).", - "type": "text" - }, - { - "block_id": "p928-b6", - "global_id": 27369, - "bbox": [ - 101.84, - 433.51, - 490.4, - 479.34 - ], - "text": "The analysis procedure, therefore, consists of solving the state equation first, and then solving\nthe output equation. The state-space description is capable of determining every possible system\nvariable (or output) from knowledge of the input and the initial state (conditions) of the system.\nFor this reason, it is an internal description of the system.", - "type": "text" - }, - { - "block_id": "p928-b7", - "global_id": 27370, - "bbox": [ - 101.84, - 481.34, - 490.38, - 527.17 - ], - "text": "By its nature, state variable analysis is eminently suited for multiple-input, multiple-output\n(MIMO) systems. A single-input, single output (SISO) system is a special case of MIMO systems.\nIn addition, the state-space techniques are useful for several other reasons, mentioned in Sec. 1.10,\nand repeated here.", - "type": "text" - }, - { - "block_id": "p928-b8", - "global_id": 27371, - "bbox": [ - 118.78, - 535.13, - 490.41, - 616.83 - ], - "text": "1. The state equations of a system provide a mathematical model of great generality that\ncan describe not just linear systems, but also nonlinear systems; not just time-invariant\nsystems, but also time-varying parameter systems; not just SISO systems, but also MIMO\nsystems. Indeed, state equations are ideally suited for analysis, synthesis, and optimization\nof MIMO systems.\n2. Compact matrix notation along with powerful techniques of linear algebra greatly\nfacilitates complex manipulations. Without such features, many important results of", - "type": "text" - }, - { - "block_id": "p928-b9", - "global_id": 27372, - "bbox": [ - 60.01, - 656.12, - 74.94, - 666.22 - ], - "text": "908", - "type": "text" - } - ] - }, - { - "page_num": 929, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p929-b0", - "global_id": 27373, - "bbox": [ - 339.2, - 62.89, - 516.13, - 71.98 - ], - "text": "10.1\nMathematical Preliminaries\n909", - "type": "text" - }, - { - "block_id": "p929-b1", - "global_id": 27374, - "bbox": [ - 144.52, - 85.82, - 516.16, - 155.56 - ], - "text": "modern system theory would have been difficult to obtain. State equations can yield a\ngreat deal of information about a system even when they are not solved explicitly.\n3. State equations lend themselves readily to digital computer simulation of complex systems\nof high order, with or without nonlinearities, and with multiple inputs and outputs.\n4. For second-order systems (N = 2), a graphical method called phase-plane analysis can be\nused on state equations, whether they are linear or nonlinear.", - "type": "text" - }, - { - "block_id": "p929-b2", - "global_id": 27375, - "bbox": [ - 127.94, - 186.5, - 374.15, - 200.45 - ], - "text": "10.1 MATHEMATICAL PRELIMINARIES", - "type": "text" - }, - { - "block_id": "p929-b3", - "global_id": 27376, - "bbox": [ - 127.59, - 206.43, - 516.1, - 228.36 - ], - "text": "This chapter requires some understanding of matrix algebra. Section B.6 introduces basic concepts\nof matrix algebra, but misses a few needed mathematical concepts, which we present next.", - "type": "text" - }, - { - "block_id": "p929-b4", - "global_id": 27377, - "bbox": [ - 127.59, - 254.77, - 364.84, - 266.72 - ], - "text": "10.1-1 Derivatives and Integrals of a Matrix", - "type": "text" - }, - { - "block_id": "p929-b5", - "global_id": 27378, - "bbox": [ - 127.59, - 272.85, - 510.11, - 282.82 - ], - "text": "Elements of a matrix need not be constants; they may be functions of a variable. For example, if", - "type": "text" - }, - { - "block_id": "p929-b6", - "global_id": 27379, - "bbox": [ - 269.38, - 301.66, - 286.39, - 311.96 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p929-b7", - "global_id": 27380, - "bbox": [ - 288.44, - 287.68, - 363.29, - 317.82 - ], - "text": "e−2t\nsin t\net\ne−t + e−2t", - "type": "text" - }, - { - "block_id": "p929-b8", - "global_id": 27381, - "bbox": [ - 368.91, - 287.68, - 374.34, - 297.64 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p929-b9", - "global_id": 27382, - "bbox": [ - 492.07, - 302.08, - 516.13, - 312.04 - ], - "text": "(10.1)", - "type": "text" - }, - { - "block_id": "p929-b10", - "global_id": 27383, - "bbox": [ - 127.59, - 330.67, - 516.13, - 353.0 - ], - "text": "then the matrix elements are functions of t. Here, it is helpful to denote A by A(t). Next, we define\nthe derivative and integral of A(t).", - "type": "text" - }, - { - "block_id": "p929-b11", - "global_id": 27384, - "bbox": [ - 127.59, - 354.57, - 516.15, - 376.91 - ], - "text": "The derivative of a matrix A(t) (with respect to t) is defined as a matrix whose ijth element is\nthe derivative (with respect to t) of the ijth element of the matrix A. Thus, if", - "type": "text" - }, - { - "block_id": "p929-b12", - "global_id": 27385, - "bbox": [ - 286.71, - 390.11, - 356.51, - 401.19 - ], - "text": "A(t) = [aij(t)]m×n", - "type": "text" - }, - { - "block_id": "p929-b13", - "global_id": 27386, - "bbox": [ - 127.59, - 414.11, - 144.74, - 424.07 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p929-b14", - "global_id": 27387, - "bbox": [ - 212.93, - 425.06, - 256.08, - 449.08 - ], - "text": "d\ndt[A(t)] =", - "type": "text" - }, - { - "block_id": "p929-b15", - "global_id": 27388, - "bbox": [ - 258.13, - 417.74, - 271.08, - 435.02 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p929-b16", - "global_id": 27389, - "bbox": [ - 264.75, - 431.73, - 293.62, - 449.08 - ], - "text": "dtaij(t)", - "type": "text" - }, - { - "block_id": "p929-b17", - "global_id": 27390, - "bbox": [ - 293.62, - 417.75, - 299.05, - 427.71 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p929-b18", - "global_id": 27391, - "bbox": [ - 299.05, - 443.94, - 313.03, - 451.13 - ], - "text": "m×n", - "type": "text" - }, - { - "block_id": "p929-b19", - "global_id": 27392, - "bbox": [ - 333.46, - 429.4, - 431.49, - 442.81 - ], - "text": "or\n˙A(t) = [˙aij(t)]m×n", - "type": "text" - }, - { - "block_id": "p929-b20", - "global_id": 27393, - "bbox": [ - 127.59, - 458.96, - 358.74, - 468.92 - ], - "text": "Thus, the derivative of the matrix in Eq. (10.1) is given by", - "type": "text" - }, - { - "block_id": "p929-b21", - "global_id": 27394, - "bbox": [ - 251.43, - 485.43, - 278.83, - 498.06 - ], - "text": "˙A(t) =", - "type": "text" - }, - { - "block_id": "p929-b22", - "global_id": 27395, - "bbox": [ - 280.87, - 473.78, - 381.23, - 504.02 - ], - "text": "−2e−2t\ncos t\net\n−e−t −2e−2t", - "type": "text" - }, - { - "block_id": "p929-b23", - "global_id": 27396, - "bbox": [ - 386.86, - 473.78, - 392.29, - 483.74 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p929-b24", - "global_id": 27397, - "bbox": [ - 127.59, - 516.77, - 516.14, - 539.1 - ], - "text": "Similarly, we define the integral of A(t) (with respect to t) as a matrix whose ijth element is\nthe integral (with respect to t) of the ijth element of the matrix A:", - "type": "text" - }, - { - "block_id": "p929-b25", - "global_id": 27398, - "bbox": [ - 263.33, - 544.39, - 268.59, - 554.36 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p929-b26", - "global_id": 27399, - "bbox": [ - 274.27, - 557.95, - 310.69, - 568.24 - ], - "text": "A(t)dt =", - "type": "text" - }, - { - "block_id": "p929-b27", - "global_id": 27400, - "bbox": [ - 312.74, - 543.96, - 324.72, - 554.36 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p929-b28", - "global_id": 27401, - "bbox": [ - 330.4, - 557.95, - 359.0, - 569.03 - ], - "text": "aij(t)dt", - "type": "text" - }, - { - "block_id": "p929-b30", - "global_id": 27402, - "bbox": [ - 365.89, - 570.15, - 379.88, - 577.34 - ], - "text": "m×n", - "type": "text" - }, - { - "block_id": "p929-b31", - "global_id": 27403, - "bbox": [ - 127.59, - 588.2, - 307.35, - 598.25 - ], - "text": "Thus, for the matrix A in Eq. (10.1), we have", - "type": "text" - }, - { - "block_id": "p929-b32", - "global_id": 27404, - "bbox": [ - 231.91, - 603.54, - 237.17, - 613.5 - ], - "text": "#", - "type": "text" - }, - { - "block_id": "p929-b33", - "global_id": 27405, - "bbox": [ - 242.84, - 617.1, - 279.27, - 627.4 - ], - "text": "A(t)dt =", - "type": "text" - }, - { - "block_id": "p929-b34", - "global_id": 27406, - "bbox": [ - 281.32, - 603.0, - 296.29, - 613.07 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p929-b35", - "global_id": 27407, - "bbox": [ - 299.33, - 607.72, - 324.1, - 621.3 - ], - "text": "e−2t dt", - "type": "text" - }, - { - "block_id": "p929-b36", - "global_id": 27408, - "bbox": [ - 353.13, - 603.0, - 357.7, - 612.96 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p929-b37", - "global_id": 27409, - "bbox": [ - 296.19, - 611.33, - 382.33, - 624.92 - ], - "text": "sin dt\n$", - "type": "text" - }, - { - "block_id": "p929-b38", - "global_id": 27410, - "bbox": [ - 303.8, - 619.68, - 319.63, - 633.26 - ], - "text": "et dt", - "type": "text" - }, - { - "block_id": "p929-b39", - "global_id": 27411, - "bbox": [ - 334.24, - 614.95, - 338.8, - 624.92 - ], - "text": "$", - "type": "text" - }, - { - "block_id": "p929-b40", - "global_id": 27412, - "bbox": [ - 340.74, - 619.36, - 401.22, - 633.36 - ], - "text": "(e−t + 2e−2t)dt", - "type": "text" - }, - { - "block_id": "p929-b41", - "global_id": 27413, - "bbox": [ - 406.38, - 603.11, - 411.81, - 613.07 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 930, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p930-b0", - "global_id": 27414, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "910\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p930-b1", - "global_id": 27415, - "bbox": [ - 119.78, - 85.82, - 373.28, - 95.78 - ], - "text": "Since differentiation is a linear operation, it is easy to show that", - "type": "text" - }, - { - "block_id": "p930-b2", - "global_id": 27416, - "bbox": [ - 192.79, - 105.96, - 259.55, - 129.98 - ], - "text": "d\ndt(A + B) = dA", - "type": "text" - }, - { - "block_id": "p930-b3", - "global_id": 27417, - "bbox": [ - 249.37, - 105.96, - 284.71, - 129.98 - ], - "text": "dt + dB", - "type": "text" - }, - { - "block_id": "p930-b4", - "global_id": 27418, - "bbox": [ - 274.81, - 105.96, - 399.44, - 129.98 - ], - "text": "dt\nand\nd\ndt(cA) = cdA", - "type": "text" - }, - { - "block_id": "p930-b5", - "global_id": 27419, - "bbox": [ - 389.26, - 120.01, - 397.01, - 129.98 - ], - "text": "dt", - "type": "text" - }, - { - "block_id": "p930-b6", - "global_id": 27420, - "bbox": [ - 101.84, - 138.19, - 282.05, - 148.15 - ], - "text": "The derivative of a matrix product is given as", - "type": "text" - }, - { - "block_id": "p930-b7", - "global_id": 27421, - "bbox": [ - 223.46, - 158.33, - 279.36, - 182.35 - ], - "text": "d\ndt(AB) = dA", - "type": "text" - }, - { - "block_id": "p930-b8", - "global_id": 27422, - "bbox": [ - 269.18, - 158.33, - 318.35, - 182.37 - ], - "text": "dt B + AdB", - "type": "text" - }, - { - "block_id": "p930-b9", - "global_id": 27423, - "bbox": [ - 308.45, - 162.67, - 490.38, - 182.35 - ], - "text": "dt = ˙AB + A ˙B\n(10.2)", - "type": "text" - }, - { - "block_id": "p930-b10", - "global_id": 27424, - "bbox": [ - 101.84, - 190.27, - 473.72, - 200.65 - ], - "text": "We can prove Eq. (10.2) as follows. Let A be an m × n matrix and B an n × p matrix. Then, if", - "type": "text" - }, - { - "block_id": "p930-b11", - "global_id": 27425, - "bbox": [ - 279.67, - 212.4, - 312.56, - 222.69 - ], - "text": "C = AB", - "type": "text" - }, - { - "block_id": "p930-b12", - "global_id": 27426, - "bbox": [ - 101.84, - 234.95, - 203.01, - 244.91 - ], - "text": "from Eq. (B.33), we have", - "type": "text" - }, - { - "block_id": "p930-b13", - "global_id": 27427, - "bbox": [ - 267.55, - 254.73, - 287.45, - 266.18 - ], - "text": "cik =", - "type": "text" - }, - { - "block_id": "p930-b14", - "global_id": 27428, - "bbox": [ - 289.49, - 244.77, - 303.59, - 255.22 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p930-b15", - "global_id": 27429, - "bbox": [ - 291.11, - 269.12, - 301.96, - 276.38 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p930-b16", - "global_id": 27430, - "bbox": [ - 304.7, - 255.04, - 324.06, - 265.81 - ], - "text": "aijbjk", - "type": "text" - }, - { - "block_id": "p930-b17", - "global_id": 27431, - "bbox": [ - 101.84, - 284.64, - 116.22, - 294.61 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p930-b18", - "global_id": 27432, - "bbox": [ - 193.57, - 302.36, - 213.47, - 313.82 - ], - "text": "˙cik =", - "type": "text" - }, - { - "block_id": "p930-b19", - "global_id": 27433, - "bbox": [ - 215.51, - 292.41, - 229.61, - 302.87 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p930-b20", - "global_id": 27434, - "bbox": [ - 217.12, - 316.76, - 227.98, - 324.03 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p930-b21", - "global_id": 27435, - "bbox": [ - 230.72, - 302.36, - 250.08, - 313.44 - ], - "text": "˙aijbjk", - "type": "text" - }, - { - "block_id": "p930-b22", - "global_id": 27436, - "bbox": [ - 215.51, - 320.9, - 250.71, - 341.09 - ], - "text": "dik", - "type": "text" - }, - { - "block_id": "p930-b23", - "global_id": 27437, - "bbox": [ - 251.81, - 302.36, - 259.58, - 312.33 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p930-b24", - "global_id": 27438, - "bbox": [ - 260.68, - 292.41, - 274.78, - 302.87 - ], - "text": "n\n\"", - "type": "text" - }, - { - "block_id": "p930-b25", - "global_id": 27439, - "bbox": [ - 262.3, - 316.76, - 273.16, - 324.03 - ], - "text": "j=1", - "type": "text" - }, - { - "block_id": "p930-b26", - "global_id": 27440, - "bbox": [ - 275.89, - 300.31, - 295.26, - 313.44 - ], - "text": "aij˙bjk", - "type": "text" - }, - { - "block_id": "p930-b27", - "global_id": 27441, - "bbox": [ - 260.68, - 320.9, - 295.88, - 340.64 - ], - "text": "eik", - "type": "text" - }, - { - "block_id": "p930-b28", - "global_id": 27442, - "bbox": [ - 316.92, - 302.36, - 490.38, - 313.82 - ], - "text": "or\n˙cik = dik + eik\n(10.3)", - "type": "text" - }, - { - "block_id": "p930-b29", - "global_id": 27443, - "bbox": [ - 101.85, - 348.19, - 490.39, - 371.7 - ], - "text": "Equation (10.3) along with the multiplication rule clearly indicates that dik is the ikth element of\nmatrix ˙AB and eik is the ikth element of matrix A ˙B. Equation (10.2) then follows.", - "type": "text" - }, - { - "block_id": "p930-b30", - "global_id": 27444, - "bbox": [ - 119.78, - 367.97, - 287.44, - 382.16 - ], - "text": "If we let B = A−1 in Eq. (10.2), we obtain", - "type": "text" - }, - { - "block_id": "p930-b31", - "global_id": 27445, - "bbox": [ - 232.37, - 392.34, - 298.24, - 416.36 - ], - "text": "d\ndt(AA−1) = dA", - "type": "text" - }, - { - "block_id": "p930-b32", - "global_id": 27446, - "bbox": [ - 288.05, - 392.34, - 341.63, - 416.37 - ], - "text": "dt A−1 + A d", - "type": "text" - }, - { - "block_id": "p930-b33", - "global_id": 27447, - "bbox": [ - 335.31, - 397.18, - 360.56, - 416.36 - ], - "text": "dtA−1", - "type": "text" - }, - { - "block_id": "p930-b34", - "global_id": 27448, - "bbox": [ - 101.84, - 424.57, - 139.2, - 434.53 - ], - "text": "But since", - "type": "text" - }, - { - "block_id": "p930-b35", - "global_id": 27449, - "bbox": [ - 255.08, - 432.52, - 314.82, - 456.54 - ], - "text": "d\ndt(AA−1) = d", - "type": "text" - }, - { - "block_id": "p930-b36", - "global_id": 27450, - "bbox": [ - 308.5, - 439.19, - 338.35, - 456.54 - ], - "text": "dtI = 0", - "type": "text" - }, - { - "block_id": "p930-b37", - "global_id": 27451, - "bbox": [ - 101.84, - 461.76, - 134.4, - 471.73 - ], - "text": "we have", - "type": "text" - }, - { - "block_id": "p930-b38", - "global_id": 27452, - "bbox": [ - 246.28, - 469.99, - 329.34, - 494.01 - ], - "text": "d\ndt(A−1) = −A−1 dA", - "type": "text" - }, - { - "block_id": "p930-b39", - "global_id": 27453, - "bbox": [ - 319.16, - 474.83, - 346.66, - 494.03 - ], - "text": "dt A−1", - "type": "text" - }, - { - "block_id": "p930-b40", - "global_id": 27454, - "bbox": [ - 101.84, - 514.88, - 356.02, - 540.78 - ], - "text": "10.1-2 The Characteristic Equation of a Matrix:\nThe Cayley–Hamilton Theorem", - "type": "text" - }, - { - "block_id": "p930-b41", - "global_id": 27455, - "bbox": [ - 101.84, - 546.49, - 413.35, - 556.87 - ], - "text": "For an (n × n) square matrix A, any vector x (x̸ = 0) that satisfies the equation", - "type": "text" - }, - { - "block_id": "p930-b42", - "global_id": 27456, - "bbox": [ - 278.88, - 568.62, - 490.37, - 579.0 - ], - "text": "Ax = λx\n(10.4)", - "type": "text" - }, - { - "block_id": "p930-b43", - "global_id": 27457, - "bbox": [ - 101.84, - 590.75, - 490.39, - 613.08 - ], - "text": "is an eigenvector (or characteristic vector), and λ is the corresponding eigenvalue (or\ncharacteristic value) of A. Equation (10.4) can be expressed as", - "type": "text" - }, - { - "block_id": "p930-b44", - "global_id": 27458, - "bbox": [ - 215.4, - 624.83, - 376.84, - 635.21 - ], - "text": "(A −λI)x = 0\nor\n(λI −A)x = 0", - "type": "text" - } - ] - }, - { - "page_num": 931, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p931-b0", - "global_id": 27459, - "bbox": [ - 339.2, - 62.89, - 516.13, - 71.98 - ], - "text": "10.1\nMathematical Preliminaries\n911", - "type": "text" - }, - { - "block_id": "p931-b1", - "global_id": 27460, - "bbox": [ - 127.59, - 85.82, - 410.81, - 95.78 - ], - "text": "The solution for this set of homogeneous equations exists if and only if", - "type": "text" - }, - { - "block_id": "p931-b2", - "global_id": 27461, - "bbox": [ - 218.11, - 127.21, - 261.01, - 137.5 - ], - "text": "|λI −A| =", - "type": "text" - }, - { - "block_id": "p931-b4", - "global_id": 27462, - "bbox": [ - 271.28, - 105.87, - 395.77, - 146.75 - ], - "text": "λ −a11\n−a12\n· · ·\n−a1n\n−a21\nλ −a22\n· · ·\n−a2n\n...", - "type": "text" - }, - { - "block_id": "p931-b5", - "global_id": 27463, - "bbox": [ - 275.55, - 128.82, - 400.05, - 159.48 - ], - "text": "...\n· · ·\n...\n−an1\n−an2\n· · ·\nλ −ann", - "type": "text" - }, - { - "block_id": "p931-b7", - "global_id": 27464, - "bbox": [ - 410.81, - 127.2, - 516.13, - 137.58 - ], - "text": "= 0\n(10.5)", - "type": "text" - }, - { - "block_id": "p931-b8", - "global_id": 27465, - "bbox": [ - 127.59, - 169.09, - 495.45, - 179.15 - ], - "text": "Equation (10.5) is known as the characteristic equation of matrix A and can be expressed as", - "type": "text" - }, - { - "block_id": "p931-b9", - "global_id": 27466, - "bbox": [ - 207.46, - 185.57, - 516.13, - 200.83 - ], - "text": "Q(λ) = |λI −A| = λn + an−1λn−1 + · · · + a1λ + a0λ0 = 0\n(10.6)", - "type": "text" - }, - { - "block_id": "p931-b10", - "global_id": 27467, - "bbox": [ - 127.59, - 210.6, - 516.13, - 244.88 - ], - "text": "Q(λ) is called the characteristic polynomial of matrix A. The n zeros of the characteristic\npolynomial are the eigenvalues of A and, corresponding to each eigenvalue, there is an eigenvector\nthat satisfies Eq. (10.4).", - "type": "text" - }, - { - "block_id": "p931-b11", - "global_id": 27468, - "bbox": [ - 127.59, - 246.47, - 516.13, - 268.79 - ], - "text": "The Cayley–Hamilton theorem states that every n×n matrix A satisfies its own characteristic\nequation. In other words, Eq. (10.6) is valid if λ is replaced by A:", - "type": "text" - }, - { - "block_id": "p931-b12", - "global_id": 27469, - "bbox": [ - 225.3, - 275.1, - 516.13, - 290.47 - ], - "text": "Q(A) = An + an−1An−1 + · · · + a1A + a0A0 = 0\n(10.7)", - "type": "text" - }, - { - "block_id": "p931-b13", - "global_id": 27470, - "bbox": [ - 127.89, - 304.18, - 265.49, - 316.3 - ], - "text": "FUNCTIONS OF A MATRIX", - "type": "text" - }, - { - "block_id": "p931-b14", - "global_id": 27471, - "bbox": [ - 127.59, - 320.34, - 516.11, - 342.25 - ], - "text": "We now demonstrate the use of the Cayley–Hamilton theorem [Eq. (10.7)] to evaluate functions\nof an n × n square matrix A.", - "type": "text" - }, - { - "block_id": "p931-b15", - "global_id": 27472, - "bbox": [ - 145.52, - 343.83, - 399.33, - 354.2 - ], - "text": "Consider a function f(λ) in the form of an infinite power series:", - "type": "text" - }, - { - "block_id": "p931-b16", - "global_id": 27473, - "bbox": [ - 241.04, - 371.43, - 335.16, - 384.32 - ], - "text": "f(λ) = α0 + α1λ + α2λ2", - "type": "text" - }, - { - "block_id": "p931-b17", - "global_id": 27474, - "bbox": [ - 331.68, - 372.86, - 369.42, - 385.51 - ], - "text": "2 + · · · =", - "type": "text" - }, - { - "block_id": "p931-b18", - "global_id": 27475, - "bbox": [ - 371.47, - 362.69, - 385.56, - 373.37 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p931-b19", - "global_id": 27476, - "bbox": [ - 373.08, - 387.26, - 383.94, - 394.52 - ], - "text": "i=0", - "type": "text" - }, - { - "block_id": "p931-b20", - "global_id": 27477, - "bbox": [ - 386.67, - 371.36, - 516.13, - 383.94 - ], - "text": "αiλi\n(10.8)", - "type": "text" - }, - { - "block_id": "p931-b21", - "global_id": 27478, - "bbox": [ - 127.59, - 402.92, - 516.12, - 425.25 - ], - "text": "Since λ, being an eigenvalue (characteristic root) of A, satisfies the characteristic equation\n[Eq. (10.6)], we can write", - "type": "text" - }, - { - "block_id": "p931-b22", - "global_id": 27479, - "bbox": [ - 231.1, - 431.67, - 516.13, - 446.93 - ], - "text": "λn = −an−1λn−1 −an−2λn−2 −· · · −a1λ −a0\n(10.9)", - "type": "text" - }, - { - "block_id": "p931-b23", - "global_id": 27480, - "bbox": [ - 127.59, - 455.47, - 516.14, - 526.84 - ], - "text": "If we multiply both sides by λ, the left-hand side is λn+1, and the right-hand side contains the\nterms λn, λn−1, . . . , λ. Using Eq. (10.9), we substitute λn in terms of λn−1, λn−2,. . ., λ so that\nthe highest power on the right-hand side is reduced to n −1. Continuing in this way, we see\nthat λn+k can be expressed in terms of λn−1, λn−2,. . .,λ for any k. Hence, the infinite series on\nthe right-hand side of Eq. (10.8) can always be expressed in terms of λn−1, λn−2,. . .,λ and a\nconstant as", - "type": "text" - }, - { - "block_id": "p931-b24", - "global_id": 27481, - "bbox": [ - 238.81, - 524.3, - 516.12, - 539.87 - ], - "text": "f(λ) = β0 + β1λ + β2λ2 + · · · + βn−1λn−1\n(10.10)", - "type": "text" - }, - { - "block_id": "p931-b25", - "global_id": 27482, - "bbox": [ - 127.59, - 546.85, - 516.12, - 569.18 - ], - "text": "If we assume that there are n distinct eigenvalues λ1, λ2, . . . , λn, then Eq. (10.10) holds for these\nn values of λ. The substitution of these values in Eq. (10.10) yields n simultaneous equations", - "type": "text" - }, - { - "block_id": "p931-b26", - "global_id": 27483, - "bbox": [ - 211.27, - 572.29, - 218.53, - 582.26 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p931-b27", - "global_id": 27484, - "bbox": [ - 211.27, - 589.77, - 218.53, - 618.12 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p931-b28", - "global_id": 27485, - "bbox": [ - 223.51, - 579.9, - 244.6, - 603.01 - ], - "text": "f(λ1)\nf(λ2)", - "type": "text" - }, - { - "block_id": "p931-b29", - "global_id": 27486, - "bbox": [ - 223.51, - 602.84, - 244.6, - 633.43 - ], - "text": "...\nf(λn)", - "type": "text" - }, - { - "block_id": "p931-b30", - "global_id": 27487, - "bbox": [ - 249.58, - 572.29, - 256.84, - 582.26 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p931-b31", - "global_id": 27488, - "bbox": [ - 249.58, - 589.77, - 266.67, - 618.12 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p931-b32", - "global_id": 27489, - "bbox": [ - 268.71, - 572.29, - 275.98, - 582.25 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p931-b33", - "global_id": 27490, - "bbox": [ - 268.71, - 589.77, - 275.98, - 618.12 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p931-b34", - "global_id": 27491, - "bbox": [ - 280.95, - 578.92, - 324.24, - 591.05 - ], - "text": "1\nλ1\nλ2", - "type": "text" - }, - { - "block_id": "p931-b35", - "global_id": 27492, - "bbox": [ - 320.75, - 578.05, - 375.58, - 592.47 - ], - "text": "1\n· · ·\nλn−1", - "type": "text" - }, - { - "block_id": "p931-b36", - "global_id": 27493, - "bbox": [ - 280.95, - 585.49, - 366.66, - 603.63 - ], - "text": "1\n1\nλ2\nλ2", - "type": "text" - }, - { - "block_id": "p931-b37", - "global_id": 27494, - "bbox": [ - 320.75, - 590.63, - 375.58, - 605.05 - ], - "text": "2\n· · ·\nλn−1", - "type": "text" - }, - { - "block_id": "p931-b38", - "global_id": 27495, - "bbox": [ - 280.95, - 598.08, - 368.15, - 634.06 - ], - "text": "2\n...\n...\n...\n· · ·\n...\n1\nλn\nλ2", - "type": "text" - }, - { - "block_id": "p931-b39", - "global_id": 27496, - "bbox": [ - 320.75, - 621.75, - 375.58, - 635.25 - ], - "text": "n\n· · ·\nλn−1", - "type": "text" - }, - { - "block_id": "p931-b40", - "global_id": 27497, - "bbox": [ - 363.18, - 628.27, - 366.66, - 635.25 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p931-b41", - "global_id": 27498, - "bbox": [ - 381.07, - 572.29, - 388.33, - 582.26 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p931-b42", - "global_id": 27499, - "bbox": [ - 381.07, - 589.77, - 388.33, - 618.12 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p931-b43", - "global_id": 27500, - "bbox": [ - 389.44, - 572.29, - 396.7, - 582.25 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p931-b44", - "global_id": 27501, - "bbox": [ - 389.44, - 589.77, - 396.7, - 618.12 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p931-b45", - "global_id": 27502, - "bbox": [ - 406.14, - 579.9, - 415.25, - 603.01 - ], - "text": "β0\nβ1", - "type": "text" - }, - { - "block_id": "p931-b46", - "global_id": 27503, - "bbox": [ - 401.68, - 602.84, - 419.7, - 633.5 - ], - "text": "...\nβn−1", - "type": "text" - }, - { - "block_id": "p931-b47", - "global_id": 27504, - "bbox": [ - 425.2, - 572.29, - 432.46, - 582.26 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p931-b48", - "global_id": 27505, - "bbox": [ - 425.2, - 589.77, - 432.46, - 618.12 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - } - ] - }, - { - "page_num": 932, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p932-b0", - "global_id": 27506, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "912\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p932-b1", - "global_id": 27507, - "bbox": [ - 101.84, - 85.4, - 245.23, - 95.78 - ], - "text": "Solving for the β coefficients yields", - "type": "text" - }, - { - "block_id": "p932-b2", - "global_id": 27508, - "bbox": [ - 180.81, - 102.44, - 188.07, - 112.4 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p932-b3", - "global_id": 27509, - "bbox": [ - 180.81, - 119.92, - 188.07, - 148.27 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p932-b4", - "global_id": 27510, - "bbox": [ - 197.51, - 110.05, - 206.62, - 133.15 - ], - "text": "β0\nβ1", - "type": "text" - }, - { - "block_id": "p932-b5", - "global_id": 27511, - "bbox": [ - 193.05, - 132.98, - 211.07, - 163.65 - ], - "text": "...\nβn−1", - "type": "text" - }, - { - "block_id": "p932-b6", - "global_id": 27512, - "bbox": [ - 216.57, - 102.44, - 223.83, - 112.4 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p932-b7", - "global_id": 27513, - "bbox": [ - 216.57, - 119.92, - 233.64, - 148.27 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p932-b8", - "global_id": 27514, - "bbox": [ - 235.69, - 102.44, - 242.95, - 112.4 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p932-b9", - "global_id": 27515, - "bbox": [ - 235.69, - 119.92, - 242.95, - 148.26 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p932-b10", - "global_id": 27516, - "bbox": [ - 247.94, - 109.06, - 291.22, - 121.2 - ], - "text": "1\nλ1\nλ2", - "type": "text" - }, - { - "block_id": "p932-b11", - "global_id": 27517, - "bbox": [ - 287.73, - 108.2, - 342.57, - 122.62 - ], - "text": "1\n· · ·\nλn−1", - "type": "text" - }, - { - "block_id": "p932-b12", - "global_id": 27518, - "bbox": [ - 247.94, - 115.65, - 333.65, - 133.77 - ], - "text": "1\n1\nλ2\nλ2", - "type": "text" - }, - { - "block_id": "p932-b13", - "global_id": 27519, - "bbox": [ - 287.73, - 120.78, - 342.57, - 135.2 - ], - "text": "2\n· · ·\nλn−1", - "type": "text" - }, - { - "block_id": "p932-b14", - "global_id": 27520, - "bbox": [ - 247.94, - 128.22, - 335.13, - 164.2 - ], - "text": "2\n...\n...\n...\n· · ·\n...\n1\nλn\nλ2", - "type": "text" - }, - { - "block_id": "p932-b15", - "global_id": 27521, - "bbox": [ - 287.73, - 151.9, - 342.57, - 165.39 - ], - "text": "n\n· · ·\nλn−1", - "type": "text" - }, - { - "block_id": "p932-b16", - "global_id": 27522, - "bbox": [ - 330.16, - 158.42, - 333.65, - 165.39 - ], - "text": "n", - "type": "text" - }, - { - "block_id": "p932-b17", - "global_id": 27523, - "bbox": [ - 348.05, - 102.44, - 355.31, - 112.4 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p932-b18", - "global_id": 27524, - "bbox": [ - 348.05, - 119.92, - 355.31, - 148.27 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p932-b19", - "global_id": 27525, - "bbox": [ - 355.31, - 102.44, - 373.1, - 113.71 - ], - "text": "−1 ⎡", - "type": "text" - }, - { - "block_id": "p932-b20", - "global_id": 27526, - "bbox": [ - 365.84, - 119.92, - 373.1, - 148.27 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p932-b21", - "global_id": 27527, - "bbox": [ - 378.09, - 110.05, - 399.18, - 133.15 - ], - "text": "f(λ1)\nf(λ2)", - "type": "text" - }, - { - "block_id": "p932-b22", - "global_id": 27528, - "bbox": [ - 378.09, - 132.98, - 399.18, - 163.57 - ], - "text": "...\nf(λn)", - "type": "text" - }, - { - "block_id": "p932-b23", - "global_id": 27529, - "bbox": [ - 404.16, - 102.44, - 411.43, - 112.4 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p932-b24", - "global_id": 27530, - "bbox": [ - 404.16, - 119.92, - 490.38, - 148.27 - ], - "text": "⎥⎥⎥⎦\n(10.11)", - "type": "text" - }, - { - "block_id": "p932-b25", - "global_id": 27531, - "bbox": [ - 101.85, - 175.06, - 490.39, - 197.39 - ], - "text": "Since A also satisfies Eq. (10.9), we may advance a similar argument to show that if f(A) is a\nfunction of a square matrix A expressed as an infinite power series in A, then", - "type": "text" - }, - { - "block_id": "p932-b26", - "global_id": 27532, - "bbox": [ - 209.87, - 213.31, - 347.35, - 228.69 - ], - "text": "f(A) = α0I + α1A + α2A2 + · · · =", - "type": "text" - }, - { - "block_id": "p932-b27", - "global_id": 27533, - "bbox": [ - 349.4, - 207.37, - 363.5, - 218.03 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p932-b28", - "global_id": 27534, - "bbox": [ - 351.02, - 231.93, - 361.88, - 239.19 - ], - "text": "i=0", - "type": "text" - }, - { - "block_id": "p932-b29", - "global_id": 27535, - "bbox": [ - 364.61, - 215.92, - 381.86, - 228.62 - ], - "text": "αiAi", - "type": "text" - }, - { - "block_id": "p932-b30", - "global_id": 27536, - "bbox": [ - 101.84, - 249.36, - 490.38, - 271.28 - ], - "text": "and, as argued earlier, the right-hand side can be expressed by using terms of power less than or\nequal to n −1,", - "type": "text" - }, - { - "block_id": "p932-b31", - "global_id": 27537, - "bbox": [ - 185.26, - 278.96, - 372.04, - 294.33 - ], - "text": "f(A) = β0I + β1A + β2A2 + · · · + βn−1An−1 =", - "type": "text" - }, - { - "block_id": "p932-b32", - "global_id": 27538, - "bbox": [ - 374.08, - 273.01, - 388.18, - 283.69 - ], - "text": "n−1\n\"", - "type": "text" - }, - { - "block_id": "p932-b33", - "global_id": 27539, - "bbox": [ - 375.7, - 297.58, - 386.56, - 304.85 - ], - "text": "i=0", - "type": "text" - }, - { - "block_id": "p932-b34", - "global_id": 27540, - "bbox": [ - 389.29, - 281.58, - 490.38, - 294.26 - ], - "text": "βiAi\n(10.12)", - "type": "text" - }, - { - "block_id": "p932-b35", - "global_id": 27541, - "bbox": [ - 101.84, - 311.61, - 490.36, - 333.95 - ], - "text": "in which the coefficients βis are found from Eq. (10.11). If some of the eigenvalues are repeated\n(multiple roots), the results are somewhat modified.", - "type": "text" - }, - { - "block_id": "p932-b36", - "global_id": 27542, - "bbox": [ - 119.78, - 335.93, - 434.24, - 345.9 - ], - "text": "We shall demonstrate the utility of this result with the following two examples.", - "type": "text" - }, - { - "block_id": "p932-b37", - "global_id": 27543, - "bbox": [ - 101.84, - 371.13, - 442.69, - 383.08 - ], - "text": "10.1-3 Computation of an Exponential and a Power of a Matrix", - "type": "text" - }, - { - "block_id": "p932-b38", - "global_id": 27544, - "bbox": [ - 101.84, - 385.6, - 222.4, - 399.18 - ], - "text": "Let us compute eAt defined by", - "type": "text" - }, - { - "block_id": "p932-b39", - "global_id": 27545, - "bbox": [ - 192.91, - 410.8, - 271.36, - 429.61 - ], - "text": "eAt = I + At + A2t2", - "type": "text" - }, - { - "block_id": "p932-b40", - "global_id": 27546, - "bbox": [ - 258.93, - 410.73, - 326.68, - 436.77 - ], - "text": "2! + · · · + Antn", - "type": "text" - }, - { - "block_id": "p932-b41", - "global_id": 27547, - "bbox": [ - 314.24, - 419.32, - 362.12, - 436.67 - ], - "text": "n! + · · · =", - "type": "text" - }, - { - "block_id": "p932-b42", - "global_id": 27548, - "bbox": [ - 364.17, - 409.15, - 378.26, - 419.81 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p932-b43", - "global_id": 27549, - "bbox": [ - 365.15, - 433.71, - 377.28, - 440.97 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p932-b44", - "global_id": 27550, - "bbox": [ - 380.57, - 410.73, - 397.52, - 422.63 - ], - "text": "Aktk", - "type": "text" - }, - { - "block_id": "p932-b45", - "global_id": 27551, - "bbox": [ - 385.68, - 426.39, - 393.03, - 436.67 - ], - "text": "k!", - "type": "text" - }, - { - "block_id": "p932-b46", - "global_id": 27552, - "bbox": [ - 101.84, - 451.14, - 236.15, - 461.11 - ], - "text": "From Eq. (10.12), we can express", - "type": "text" - }, - { - "block_id": "p932-b47", - "global_id": 27553, - "bbox": [ - 264.01, - 468.9, - 285.85, - 483.29 - ], - "text": "eAt =", - "type": "text" - }, - { - "block_id": "p932-b48", - "global_id": 27554, - "bbox": [ - 287.9, - 462.84, - 302.0, - 473.52 - ], - "text": "n−1\n\"", - "type": "text" - }, - { - "block_id": "p932-b49", - "global_id": 27555, - "bbox": [ - 289.51, - 487.41, - 300.37, - 494.67 - ], - "text": "i=1", - "type": "text" - }, - { - "block_id": "p932-b50", - "global_id": 27556, - "bbox": [ - 303.1, - 471.51, - 327.72, - 484.09 - ], - "text": "βi(A)i", - "type": "text" - }, - { - "block_id": "p932-b51", - "global_id": 27557, - "bbox": [ - 101.84, - 501.65, - 336.5, - 513.95 - ], - "text": "in which the βis are given by Eq. (10.11), with f(λi) = eλit.", - "type": "text" - }, - { - "block_id": "p932-b52", - "global_id": 27558, - "bbox": [ - 76.77, - 542.48, - 395.1, - 554.44 - ], - "text": "EXAMPLE 10.1\nComputing the Exponential of a Matrix", - "type": "text" - }, - { - "block_id": "p932-b53", - "global_id": 27559, - "bbox": [ - 103.16, - 567.49, - 202.61, - 581.06 - ], - "text": "Compute eAt for the case", - "type": "text" - }, - { - "block_id": "p932-b54", - "global_id": 27560, - "bbox": [ - 252.42, - 588.29, - 269.43, - 598.58 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p932-b55", - "global_id": 27561, - "bbox": [ - 271.47, - 574.3, - 317.35, - 604.54 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p932-b56", - "global_id": 27562, - "bbox": [ - 322.33, - 574.3, - 327.76, - 584.27 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 933, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p933-b0", - "global_id": 27563, - "bbox": [ - 341.95, - 62.89, - 516.13, - 71.98 - ], - "text": "10.2\nIntroduction to State Space\n913", - "type": "text" - }, - { - "block_id": "p933-b1", - "global_id": 27564, - "bbox": [ - 128.9, - 86.24, - 245.94, - 96.2 - ], - "text": "The characteristic equation is", - "type": "text" - }, - { - "block_id": "p933-b2", - "global_id": 27565, - "bbox": [ - 194.46, - 113.38, - 237.37, - 123.68 - ], - "text": "|λI −A| =", - "type": "text" - }, - { - "block_id": "p933-b4", - "global_id": 27566, - "bbox": [ - 247.63, - 107.31, - 284.35, - 129.64 - ], - "text": "λ\n−1\n2\nλ + 3", - "type": "text" - }, - { - "block_id": "p933-b5", - "global_id": 27567, - "bbox": [ - 289.33, - 98.94, - 437.2, - 126.84 - ], - "text": "= λ2 + 3λ + 2 = (λ + 1)(λ + 2) = 0", - "type": "text" - }, - { - "block_id": "p933-b6", - "global_id": 27568, - "bbox": [ - 128.9, - 140.59, - 329.98, - 152.05 - ], - "text": "Hence, the eigenvalues are λ1 = −1, λ2 = −2, and", - "type": "text" - }, - { - "block_id": "p933-b7", - "global_id": 27569, - "bbox": [ - 283.32, - 158.4, - 348.35, - 173.66 - ], - "text": "eAt = β0I + β1A", - "type": "text" - }, - { - "block_id": "p933-b8", - "global_id": 27570, - "bbox": [ - 128.91, - 184.84, - 163.49, - 194.81 - ], - "text": "in which", - "type": "text" - }, - { - "block_id": "p933-b9", - "global_id": 27571, - "bbox": [ - 158.7, - 198.0, - 178.2, - 217.06 - ], - "text": "β0", - "type": "text" - }, - { - "block_id": "p933-b10", - "global_id": 27572, - "bbox": [ - 169.1, - 217.87, - 178.2, - 229.02 - ], - "text": "β1", - "type": "text" - }, - { - "block_id": "p933-b11", - "global_id": 27573, - "bbox": [ - 183.69, - 198.0, - 189.12, - 207.96 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p933-b12", - "global_id": 27574, - "bbox": [ - 191.17, - 211.99, - 198.94, - 221.95 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p933-b13", - "global_id": 27575, - "bbox": [ - 200.99, - 198.0, - 239.09, - 228.25 - ], - "text": "1\n−1\n1\n−2", - "type": "text" - }, - { - "block_id": "p933-b14", - "global_id": 27576, - "bbox": [ - 244.07, - 198.0, - 283.98, - 216.19 - ], - "text": "!−1 e−t", - "type": "text" - }, - { - "block_id": "p933-b15", - "global_id": 27577, - "bbox": [ - 270.45, - 216.65, - 285.72, - 228.15 - ], - "text": "e−2t", - "type": "text" - }, - { - "block_id": "p933-b16", - "global_id": 27578, - "bbox": [ - 291.34, - 198.0, - 296.77, - 207.96 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p933-b17", - "global_id": 27579, - "bbox": [ - 299.92, - 211.99, - 307.69, - 221.95 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p933-b18", - "global_id": 27580, - "bbox": [ - 309.74, - 198.0, - 347.84, - 228.25 - ], - "text": "2\n−1\n1\n−1", - "type": "text" - }, - { - "block_id": "p933-b19", - "global_id": 27581, - "bbox": [ - 352.82, - 198.0, - 383.31, - 216.19 - ], - "text": "! e−t", - "type": "text" - }, - { - "block_id": "p933-b20", - "global_id": 27582, - "bbox": [ - 369.78, - 216.65, - 385.06, - 228.15 - ], - "text": "e−2t", - "type": "text" - }, - { - "block_id": "p933-b21", - "global_id": 27583, - "bbox": [ - 390.67, - 198.0, - 396.1, - 207.96 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p933-b22", - "global_id": 27584, - "bbox": [ - 398.15, - 211.99, - 405.92, - 221.95 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p933-b23", - "global_id": 27585, - "bbox": [ - 407.97, - 198.0, - 461.93, - 216.29 - ], - "text": "2e−t −e−2t", - "type": "text" - }, - { - "block_id": "p933-b24", - "global_id": 27586, - "bbox": [ - 420.86, - 214.26, - 459.43, - 228.15 - ], - "text": "e−t −e−2t", - "type": "text" - }, - { - "block_id": "p933-b25", - "global_id": 27587, - "bbox": [ - 467.55, - 198.0, - 472.98, - 207.96 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p933-b26", - "global_id": 27588, - "bbox": [ - 128.91, - 239.62, - 143.28, - 249.58 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p933-b27", - "global_id": 27589, - "bbox": [ - 199.7, - 260.59, - 275.21, - 275.08 - ], - "text": "eAt = (2e−t −e−2t)", - "type": "text" - }, - { - "block_id": "p933-b28", - "global_id": 27590, - "bbox": [ - 276.32, - 250.72, - 306.65, - 280.96 - ], - "text": "1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p933-b29", - "global_id": 27591, - "bbox": [ - 311.63, - 250.72, - 317.06, - 260.68 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p933-b30", - "global_id": 27592, - "bbox": [ - 318.61, - 260.59, - 374.58, - 274.98 - ], - "text": "+ (e−t −e−2t)", - "type": "text" - }, - { - "block_id": "p933-b31", - "global_id": 27593, - "bbox": [ - 375.68, - 250.72, - 421.55, - 280.96 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p933-b32", - "global_id": 27594, - "bbox": [ - 426.53, - 250.72, - 431.96, - 260.68 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p933-b33", - "global_id": 27595, - "bbox": [ - 213.77, - 292.6, - 221.54, - 302.57 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p933-b34", - "global_id": 27596, - "bbox": [ - 223.58, - 278.62, - 345.84, - 296.9 - ], - "text": "2e−t −e−2t\ne−t −e−2t", - "type": "text" - }, - { - "block_id": "p933-b35", - "global_id": 27597, - "bbox": [ - 234.0, - 294.86, - 352.22, - 308.85 - ], - "text": "−2e−t + 2e−2t\n−e−t + 2e−2t", - "type": "text" - }, - { - "block_id": "p933-b36", - "global_id": 27598, - "bbox": [ - 357.85, - 278.62, - 363.28, - 288.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p933-b37", - "global_id": 27599, - "bbox": [ - 473.71, - 293.01, - 502.75, - 302.98 - ], - "text": "(10.13)", - "type": "text" - }, - { - "block_id": "p933-b38", - "global_id": 27600, - "bbox": [ - 127.89, - 345.44, - 240.08, - 360.04 - ], - "text": "COMPUTATION OF Ak", - "type": "text" - }, - { - "block_id": "p933-b39", - "global_id": 27601, - "bbox": [ - 127.59, - 359.84, - 313.48, - 374.02 - ], - "text": "As Eq. (10.12) indicates, we can express Ak as", - "type": "text" - }, - { - "block_id": "p933-b40", - "global_id": 27602, - "bbox": [ - 253.18, - 382.12, - 390.04, - 397.49 - ], - "text": "Ak = β0I + β1A + · · · + βn−1An−1", - "type": "text" - }, - { - "block_id": "p933-b41", - "global_id": 27603, - "bbox": [ - 127.59, - 407.9, - 373.56, - 420.11 - ], - "text": "in which the βis are given by Eq. (10.11) with f(λi) = λk", - "type": "text" - }, - { - "block_id": "p933-b42", - "global_id": 27604, - "bbox": [ - 127.59, - 409.45, - 516.12, - 431.37 - ], - "text": "i . For a completed example of the\ncomputation of Ak by this method, see Ex. 10.13.", - "type": "text" - }, - { - "block_id": "p933-b43", - "global_id": 27605, - "bbox": [ - 127.94, - 461.42, - 374.59, - 475.37 - ], - "text": "10.2 INTRODUCTION TO STATE SPACE", - "type": "text" - }, - { - "block_id": "p933-b44", - "global_id": 27606, - "bbox": [ - 127.59, - 481.26, - 516.14, - 539.92 - ], - "text": "From the discussion in Ch. 1, we know that to determine a system’s response(s) at any instant t,\nwe need to know the system’s inputs during its entire past, from −∞to t. If the inputs are known\nonly for t > t0, we can still determine the system output(s) for any t > t0, provided we know certain\ninitial conditions in the system at t = t0. These initial conditions collectively are called the initial\nstate of the system (at t = t0).", - "type": "text" - }, - { - "block_id": "p933-b45", - "global_id": 27607, - "bbox": [ - 127.59, - 540.72, - 516.13, - 598.92 - ], - "text": "The state variables q1(t),q2(t),. . .,qN(t) are the minimum number of system variables such\nthat their initial values at any instant t0 are sufficient to determine the behavior of the system for\nall time t ≥t0 when the input(s) to the system is known for t ≥t0. This statement implies that an\noutput of a system at any instant is determined completely from a knowledge of the values of the\nsystem state and the input at that instant.", - "type": "text" - }, - { - "block_id": "p933-b46", - "global_id": 27608, - "bbox": [ - 127.59, - 600.91, - 516.14, - 634.79 - ], - "text": "Initial conditions of a system can be specified in many different ways. Consequently, the\nsystem state can also be specified in many different ways. This means that state variables are not\nunique.", - "type": "text" - } - ] - }, - { - "page_num": 934, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p934-b0", - "global_id": 27609, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "914\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p934-b1", - "global_id": 27610, - "bbox": [ - 101.84, - 85.82, - 490.39, - 131.64 - ], - "text": "This discussion is also valid for multiple-input, multiple-output (MIMO) systems, where every\npossible system output at any instant t is determined completely from a knowledge of the system\nstate and the input(s) at the instant t. These ideas should become clear from the following example\nof an RLC circuit.", - "type": "text" - }, - { - "block_id": "p934-b2", - "global_id": 27611, - "bbox": [ - 76.77, - 166.56, - 466.48, - 192.47 - ], - "text": "EXAMPLE 10.2\nState-Space Description and Output Equations of an\nRLC Circuit", - "type": "text" - }, - { - "block_id": "p934-b3", - "global_id": 27612, - "bbox": [ - 103.16, - 206.11, - 477.02, - 240.08 - ], - "text": "Find a state-space description of the RLC circuit shown in Fig. 10.1. Verify that all possible\nsystem outputs at some instant t can be determined from knowledge of the system state and\nthe input at that instant t.", - "type": "text" - }, - { - "block_id": "p934-b4", - "global_id": 27613, - "bbox": [ - 140.03, - 299.71, - 145.1, - 309.32 - ], - "text": "i1", - "type": "text" - }, - { - "block_id": "p934-b5", - "global_id": 27614, - "bbox": [ - 198.95, - 320.8, - 204.18, - 330.41 - ], - "text": "i2", - "type": "text" - }, - { - "block_id": "p934-b6", - "global_id": 27615, - "bbox": [ - 305.29, - 299.75, - 312.29, - 309.35 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p934-b7", - "global_id": 27616, - "bbox": [ - 163.82, - 318.43, - 321.1, - 330.41 - ], - "text": "i3\ni4\nv1", - "type": "text" - }, - { - "block_id": "p934-b8", - "global_id": 27617, - "bbox": [ - 178.39, - 346.33, - 184.78, - 355.94 - ], - "text": "v2", - "type": "text" - }, - { - "block_id": "p934-b11", - "global_id": 27618, - "bbox": [ - 179.35, - 315.66, - 306.65, - 325.27 - ], - "text": "v3", - "type": "text" - }, - { - "block_id": "p934-b13", - "global_id": 27619, - "bbox": [ - 237.52, - 348.03, - 244.36, - 357.63 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p934-b16", - "global_id": 27620, - "bbox": [ - 296.77, - 346.33, - 303.17, - 355.94 - ], - "text": "v4", - "type": "text" - }, - { - "block_id": "p934-b19", - "global_id": 27621, - "bbox": [ - 112.95, - 340.87, - 135.08, - 356.37 - ], - "text": "+\n_\nx", - "type": "text" - }, - { - "block_id": "p934-b21", - "global_id": 27622, - "bbox": [ - 161.84, - 291.08, - 294.38, - 305.44 - ], - "text": "1 H\n1\n2", - "type": "text" - }, - { - "block_id": "p934-b22", - "global_id": 27623, - "bbox": [ - 205.75, - 345.59, - 335.16, - 361.19 - ], - "text": "1\n3 \t\n2 \t\n1\n5 F", - "type": "text" - }, - { - "block_id": "p934-b23", - "global_id": 27624, - "bbox": [ - 345.69, - 383.68, - 467.98, - 392.92 - ], - "text": "Figure 10.1 Circuit for Ex. 10.2.", - "type": "text" - }, - { - "block_id": "p934-b24", - "global_id": 27625, - "bbox": [ - 103.16, - 423.18, - 477.01, - 458.64 - ], - "text": "It is known that inductor currents and capacitor voltages in an RLC circuit can be used\nas one possible choice of state variables. For this reason, we shall choose q1 (the capacitor\nvoltage) and q2 (the inductor current) as our state variables.", - "type": "text" - }, - { - "block_id": "p934-b25", - "global_id": 27626, - "bbox": [ - 121.09, - 459.14, - 302.9, - 469.11 - ], - "text": "The node equation at the intermediate node is", - "type": "text" - }, - { - "block_id": "p934-b26", - "global_id": 27627, - "bbox": [ - 258.68, - 480.64, - 321.0, - 492.1 - ], - "text": "i3 = i1 −i2 −q2", - "type": "text" - }, - { - "block_id": "p934-b27", - "global_id": 27628, - "bbox": [ - 103.16, - 502.56, - 288.99, - 514.02 - ], - "text": "but i3 = 0.2˙q1, i1 = 2(x −q1), i2 = 3q1. Hence,", - "type": "text" - }, - { - "block_id": "p934-b28", - "global_id": 27629, - "bbox": [ - 232.76, - 524.48, - 346.91, - 535.94 - ], - "text": "0.2˙q1 = 2(x −q1) −3q1 −q2", - "type": "text" - }, - { - "block_id": "p934-b29", - "global_id": 27630, - "bbox": [ - 103.16, - 546.81, - 111.46, - 556.77 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p934-b30", - "global_id": 27631, - "bbox": [ - 241.27, - 558.35, - 338.86, - 569.81 - ], - "text": "˙q1 = −25q1 −5q2 + 10x", - "type": "text" - }, - { - "block_id": "p934-b31", - "global_id": 27632, - "bbox": [ - 103.16, - 577.7, - 477.02, - 599.61 - ], - "text": "This is the first state equation. To obtain the second state equation, we sum the voltages in the\nextreme right loop formed by C, L, and the 2 resistor so that they are equal to zero:", - "type": "text" - }, - { - "block_id": "p934-b32", - "global_id": 27633, - "bbox": [ - 250.97, - 611.15, - 329.21, - 622.61 - ], - "text": "−q1 + ˙q2 + 2q2 = 0", - "type": "text" - } - ] - }, - { - "page_num": 935, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p935-b0", - "global_id": 27634, - "bbox": [ - 341.95, - 62.89, - 516.13, - 71.98 - ], - "text": "10.2\nIntroduction to State Space\n915", - "type": "text" - }, - { - "block_id": "p935-b1", - "global_id": 27635, - "bbox": [ - 128.9, - 84.02, - 137.2, - 93.98 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p935-b2", - "global_id": 27636, - "bbox": [ - 288.53, - 95.57, - 342.63, - 107.02 - ], - "text": "˙q2 = q1 −2q2", - "type": "text" - }, - { - "block_id": "p935-b3", - "global_id": 27637, - "bbox": [ - 128.9, - 114.91, - 259.42, - 124.87 - ], - "text": "Thus, the two state equations are", - "type": "text" - }, - { - "block_id": "p935-b4", - "global_id": 27638, - "bbox": [ - 267.02, - 135.98, - 364.61, - 147.43 - ], - "text": "˙q1 = −25q1 −5q2 + 10x", - "type": "text" - }, - { - "block_id": "p935-b5", - "global_id": 27639, - "bbox": [ - 288.53, - 147.93, - 342.63, - 159.39 - ], - "text": "˙q2 = q1 −2q2", - "type": "text" - }, - { - "block_id": "p935-b6", - "global_id": 27640, - "bbox": [ - 128.9, - 169.57, - 502.76, - 191.6 - ], - "text": "Every possible output can now be expressed as a linear combination of q1, q2, and x. From\nFig. 10.1, we have", - "type": "text" - }, - { - "block_id": "p935-b7", - "global_id": 27641, - "bbox": [ - 200.11, - 203.14, - 244.17, - 214.59 - ], - "text": "v1 = x −q1", - "type": "text" - }, - { - "block_id": "p935-b8", - "global_id": 27642, - "bbox": [ - 201.77, - 218.08, - 257.09, - 229.53 - ], - "text": "i1 = 2(x −q1)", - "type": "text" - }, - { - "block_id": "p935-b9", - "global_id": 27643, - "bbox": [ - 200.11, - 233.02, - 228.85, - 244.48 - ], - "text": "v2 = q1", - "type": "text" - }, - { - "block_id": "p935-b10", - "global_id": 27644, - "bbox": [ - 200.11, - 247.96, - 431.52, - 319.2 - ], - "text": "i2 = 3q1\ni3 = i1 −i2 −q2 = 2(x −q1) −3q1 −q2 = −5q1 −q2 + 2x\ni4 = q2\nv4 = 2i4 = 2q2\nv3 = q1 −v4 = q1 −2q2", - "type": "text" - }, - { - "block_id": "p935-b11", - "global_id": 27645, - "bbox": [ - 128.9, - 329.97, - 502.76, - 376.67 - ], - "text": "This set of equations is known as the output equation of the system. It is clear from this set that\nevery possible output at some instant t can be determined from knowledge of q1(t), q2(t), and\nx(t), the system state, and the input at the instant t. Once we have solved the state equations to\nobtain q1(t) and q2(t), we can determine every possible output for any given input x(t).", - "type": "text" - }, - { - "block_id": "p935-b12", - "global_id": 27646, - "bbox": [ - 127.59, - 416.64, - 516.12, - 440.15 - ], - "text": "For continuous-time systems, the state equations are N simultaneous first-order differential\nequations in N state variables q1, q2, . . . , qN of the form", - "type": "text" - }, - { - "block_id": "p935-b13", - "global_id": 27647, - "bbox": [ - 213.38, - 458.98, - 429.75, - 470.44 - ], - "text": "˙qi = gi(q1,q2,. . .,qN,x1,x2,. . .,xj)\ni = 1,2,. . .,N", - "type": "text" - }, - { - "block_id": "p935-b14", - "global_id": 27648, - "bbox": [ - 127.59, - 489.67, - 516.14, - 512.0 - ], - "text": "where x1, x2, . . . , xj are the j system inputs. For a linear system, these equations reduce to a simpler\nlinear form", - "type": "text" - }, - { - "block_id": "p935-b15", - "global_id": 27649, - "bbox": [ - 149.33, - 532.32, - 516.12, - 543.78 - ], - "text": "˙qi = ai1q1 + ai2q2 + · · · + aiNqN + bi1x1 + bi2x2 + · · · + bijxj\ni = 1,2,. . .,N\n(10.14)", - "type": "text" - }, - { - "block_id": "p935-b16", - "global_id": 27650, - "bbox": [ - 127.6, - 563.02, - 417.59, - 574.17 - ], - "text": "If there are k outputs y1,y2,. . .,yk, the k output equations are of the form", - "type": "text" - }, - { - "block_id": "p935-b17", - "global_id": 27651, - "bbox": [ - 138.73, - 593.71, - 516.12, - 605.17 - ], - "text": "ym = cm1q1 + cm2q2 + · · · + cmNqN + dm1x1 + dm2x2 + · · · + dmjxj\nm = 1,2,. . .,k\n(10.15)", - "type": "text" - }, - { - "block_id": "p935-b18", - "global_id": 27652, - "bbox": [ - 127.6, - 624.73, - 496.57, - 634.79 - ], - "text": "The N simultaneous first-order state equations are also known as the normal-form equations.", - "type": "text" - } - ] - }, - { - "page_num": 936, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p936-b0", - "global_id": 27653, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "916\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p936-b1", - "global_id": 27654, - "bbox": [ - 119.78, - 85.82, - 382.93, - 95.78 - ], - "text": "These equations can be written more conveniently in matrix form:", - "type": "text" - }, - { - "block_id": "p936-b2", - "global_id": 27655, - "bbox": [ - 148.36, - 101.16, - 155.62, - 111.12 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b3", - "global_id": 27656, - "bbox": [ - 148.36, - 118.63, - 155.62, - 146.99 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b4", - "global_id": 27657, - "bbox": [ - 156.41, - 108.76, - 164.87, - 131.87 - ], - "text": "˙q1\n˙q2", - "type": "text" - }, - { - "block_id": "p936-b5", - "global_id": 27658, - "bbox": [ - 155.62, - 131.71, - 165.25, - 162.3 - ], - "text": "...\n˙qN", - "type": "text" - }, - { - "block_id": "p936-b6", - "global_id": 27659, - "bbox": [ - 166.17, - 101.16, - 173.43, - 111.12 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b7", - "global_id": 27660, - "bbox": [ - 166.17, - 118.63, - 173.43, - 146.99 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b9", - "global_id": 27661, - "bbox": [ - 159.23, - 170.61, - 162.55, - 177.58 - ], - "text": "˙q", - "type": "text" - }, - { - "block_id": "p936-b10", - "global_id": 27662, - "bbox": [ - 175.47, - 130.09, - 183.24, - 140.05 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p936-b11", - "global_id": 27663, - "bbox": [ - 185.29, - 101.16, - 192.55, - 111.12 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b12", - "global_id": 27664, - "bbox": [ - 185.29, - 118.63, - 192.55, - 146.99 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b13", - "global_id": 27665, - "bbox": [ - 192.55, - 108.77, - 277.86, - 162.37 - ], - "text": "a11\na12\n· · ·\na1N\na21\na22\n· · ·\na2N\n...\n...\n· · ·\n...\naN1\naN2\n· · ·\naNN", - "type": "text" - }, - { - "block_id": "p936-b14", - "global_id": 27666, - "bbox": [ - 278.78, - 101.16, - 286.05, - 111.12 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b15", - "global_id": 27667, - "bbox": [ - 278.78, - 118.63, - 286.05, - 146.99 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b16", - "global_id": 27668, - "bbox": [ - 185.29, - 158.8, - 286.06, - 177.13 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p936-b17", - "global_id": 27669, - "bbox": [ - 287.15, - 101.16, - 294.42, - 111.12 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b18", - "global_id": 27670, - "bbox": [ - 287.15, - 118.63, - 294.42, - 146.99 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b19", - "global_id": 27671, - "bbox": [ - 295.21, - 109.08, - 303.67, - 131.87 - ], - "text": "q1\nq2", - "type": "text" - }, - { - "block_id": "p936-b20", - "global_id": 27672, - "bbox": [ - 294.42, - 131.71, - 304.05, - 162.3 - ], - "text": "...\nqN", - "type": "text" - }, - { - "block_id": "p936-b21", - "global_id": 27673, - "bbox": [ - 304.96, - 101.16, - 312.23, - 111.12 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b22", - "global_id": 27674, - "bbox": [ - 304.96, - 118.63, - 312.23, - 146.99 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b24", - "global_id": 27675, - "bbox": [ - 298.04, - 171.16, - 301.36, - 177.13 - ], - "text": "q", - "type": "text" - }, - { - "block_id": "p936-b25", - "global_id": 27676, - "bbox": [ - 313.33, - 130.09, - 321.11, - 140.05 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p936-b26", - "global_id": 27677, - "bbox": [ - 322.21, - 101.16, - 329.47, - 111.12 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b27", - "global_id": 27678, - "bbox": [ - 322.21, - 118.63, - 329.47, - 146.99 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b28", - "global_id": 27679, - "bbox": [ - 329.48, - 108.77, - 412.08, - 162.37 - ], - "text": "b11\nb12\n· · ·\nb1j\nb21\nb22\n· · ·\nb2j\n...\n...\n· · ·\n...\nbN1\nbN2\n· · ·\nbNj", - "type": "text" - }, - { - "block_id": "p936-b29", - "global_id": 27680, - "bbox": [ - 412.58, - 101.16, - 419.84, - 111.12 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b30", - "global_id": 27681, - "bbox": [ - 412.58, - 118.63, - 419.84, - 146.99 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b31", - "global_id": 27682, - "bbox": [ - 322.21, - 158.8, - 419.84, - 177.13 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p936-b32", - "global_id": 27683, - "bbox": [ - 420.95, - 101.16, - 428.21, - 111.12 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b33", - "global_id": 27684, - "bbox": [ - 420.95, - 118.63, - 428.21, - 146.99 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b34", - "global_id": 27685, - "bbox": [ - 428.21, - 109.08, - 436.12, - 131.87 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p936-b35", - "global_id": 27686, - "bbox": [ - 428.99, - 131.71, - 435.34, - 162.3 - ], - "text": "...\nxj", - "type": "text" - }, - { - "block_id": "p936-b36", - "global_id": 27687, - "bbox": [ - 436.62, - 101.16, - 443.88, - 111.12 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b37", - "global_id": 27688, - "bbox": [ - 436.62, - 118.63, - 443.88, - 146.99 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b39", - "global_id": 27689, - "bbox": [ - 430.92, - 171.16, - 433.91, - 177.13 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p936-b40", - "global_id": 27690, - "bbox": [ - 101.84, - 190.55, - 161.86, - 204.77 - ], - "text": "and\n⎡", - "type": "text" - }, - { - "block_id": "p936-b41", - "global_id": 27691, - "bbox": [ - 154.59, - 212.28, - 161.86, - 240.63 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b42", - "global_id": 27692, - "bbox": [ - 161.86, - 202.72, - 169.76, - 225.52 - ], - "text": "y1\ny2", - "type": "text" - }, - { - "block_id": "p936-b43", - "global_id": 27693, - "bbox": [ - 161.99, - 225.35, - 169.51, - 255.94 - ], - "text": "...\nyk", - "type": "text" - }, - { - "block_id": "p936-b44", - "global_id": 27694, - "bbox": [ - 170.26, - 194.81, - 177.52, - 204.77 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b45", - "global_id": 27695, - "bbox": [ - 170.26, - 212.28, - 177.52, - 240.63 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b47", - "global_id": 27696, - "bbox": [ - 164.57, - 264.81, - 167.55, - 270.79 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p936-b48", - "global_id": 27697, - "bbox": [ - 179.57, - 223.74, - 187.34, - 233.7 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p936-b49", - "global_id": 27698, - "bbox": [ - 189.39, - 194.81, - 196.65, - 204.77 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b50", - "global_id": 27699, - "bbox": [ - 189.39, - 212.28, - 196.65, - 240.63 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b51", - "global_id": 27700, - "bbox": [ - 196.65, - 202.41, - 275.96, - 243.29 - ], - "text": "c11\nc12\n· · ·\nc1N\nc21\nc22\n· · ·\nc2N\n...", - "type": "text" - }, - { - "block_id": "p936-b52", - "global_id": 27701, - "bbox": [ - 196.79, - 225.35, - 275.76, - 256.01 - ], - "text": "...\n· · ·\n...\nck1\nck2\n· · ·\nckN", - "type": "text" - }, - { - "block_id": "p936-b53", - "global_id": 27702, - "bbox": [ - 276.88, - 194.81, - 284.14, - 204.77 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b54", - "global_id": 27703, - "bbox": [ - 276.88, - 212.28, - 284.14, - 240.63 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b56", - "global_id": 27704, - "bbox": [ - 234.61, - 264.81, - 238.93, - 270.79 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p936-b57", - "global_id": 27705, - "bbox": [ - 285.25, - 194.81, - 292.51, - 204.77 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b58", - "global_id": 27706, - "bbox": [ - 285.25, - 212.28, - 292.51, - 240.63 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b59", - "global_id": 27707, - "bbox": [ - 293.3, - 202.72, - 301.76, - 225.52 - ], - "text": "q1\nq2", - "type": "text" - }, - { - "block_id": "p936-b60", - "global_id": 27708, - "bbox": [ - 292.51, - 225.35, - 302.14, - 255.94 - ], - "text": "...\nqN", - "type": "text" - }, - { - "block_id": "p936-b61", - "global_id": 27709, - "bbox": [ - 303.06, - 194.81, - 310.32, - 204.77 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b62", - "global_id": 27710, - "bbox": [ - 303.06, - 212.28, - 310.32, - 240.63 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b64", - "global_id": 27711, - "bbox": [ - 296.13, - 264.81, - 299.45, - 270.79 - ], - "text": "q", - "type": "text" - }, - { - "block_id": "p936-b65", - "global_id": 27712, - "bbox": [ - 311.43, - 223.74, - 319.2, - 233.7 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p936-b66", - "global_id": 27713, - "bbox": [ - 320.3, - 194.81, - 327.56, - 204.77 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b67", - "global_id": 27714, - "bbox": [ - 320.3, - 212.28, - 327.56, - 240.63 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b68", - "global_id": 27715, - "bbox": [ - 327.57, - 202.41, - 405.83, - 243.29 - ], - "text": "d11\nd12\n· · ·\nd1j\nd21\nd22\n· · ·\nd2j\n...", - "type": "text" - }, - { - "block_id": "p936-b69", - "global_id": 27716, - "bbox": [ - 327.71, - 225.35, - 405.64, - 256.01 - ], - "text": "...\n· · ·\n...\ndk1\ndk2\n· · ·\ndkj", - "type": "text" - }, - { - "block_id": "p936-b70", - "global_id": 27717, - "bbox": [ - 406.34, - 194.81, - 413.6, - 204.77 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b71", - "global_id": 27718, - "bbox": [ - 406.34, - 212.28, - 413.6, - 240.63 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b73", - "global_id": 27719, - "bbox": [ - 364.8, - 264.81, - 369.11, - 270.79 - ], - "text": "D", - "type": "text" - }, - { - "block_id": "p936-b74", - "global_id": 27720, - "bbox": [ - 414.71, - 194.81, - 421.97, - 204.77 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p936-b75", - "global_id": 27721, - "bbox": [ - 414.71, - 212.28, - 421.97, - 240.63 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p936-b76", - "global_id": 27722, - "bbox": [ - 421.97, - 202.72, - 429.88, - 225.52 - ], - "text": "x1\nx2", - "type": "text" - }, - { - "block_id": "p936-b77", - "global_id": 27723, - "bbox": [ - 422.75, - 225.35, - 429.11, - 255.94 - ], - "text": "...\nxj", - "type": "text" - }, - { - "block_id": "p936-b78", - "global_id": 27724, - "bbox": [ - 430.38, - 194.81, - 437.64, - 204.77 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p936-b79", - "global_id": 27725, - "bbox": [ - 430.38, - 212.28, - 437.64, - 240.63 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p936-b81", - "global_id": 27726, - "bbox": [ - 424.68, - 264.81, - 427.67, - 270.79 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p936-b82", - "global_id": 27727, - "bbox": [ - 101.84, - 278.55, - 110.14, - 288.51 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p936-b83", - "global_id": 27728, - "bbox": [ - 269.8, - 292.88, - 490.38, - 303.27 - ], - "text": "˙q = Aq + Bx\n(10.16)", - "type": "text" - }, - { - "block_id": "p936-b84", - "global_id": 27729, - "bbox": [ - 101.85, - 314.12, - 116.22, - 324.08 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p936-b85", - "global_id": 27730, - "bbox": [ - 269.8, - 328.47, - 490.38, - 338.85 - ], - "text": "y = Cq + Dx\n(10.17)", - "type": "text" - }, - { - "block_id": "p936-b86", - "global_id": 27731, - "bbox": [ - 101.85, - 349.6, - 490.39, - 371.61 - ], - "text": "Equation (10.16) is the state equation and Eq. (10.17) is the output equation; q, y, and x are the\nstate vector, the output vector, and the input vector, respectively.", - "type": "text" - }, - { - "block_id": "p936-b87", - "global_id": 27732, - "bbox": [ - 101.84, - 373.5, - 490.38, - 395.51 - ], - "text": "For discrete-time systems, the state equations are N simultaneous first-order difference\nequations. Discrete-time systems are discussed in Sec. 10.7.", - "type": "text" - }, - { - "block_id": "p936-b88", - "global_id": 27733, - "bbox": [ - 102.2, - 426.67, - 418.85, - 456.56 - ], - "text": "10.3 A SYSTEMATIC PROCEDURE TO DETERMINE\nSTATE EQUATIONS", - "type": "text" - }, - { - "block_id": "p936-b89", - "global_id": 27734, - "bbox": [ - 101.84, - 462.55, - 490.39, - 496.42 - ], - "text": "We shall discuss here a systematic procedure to determine the state-space description of linear\ntime-invariant systems. In particular, we shall consider systems of two types: (1) RLC networks\nand (2) systems specified by block diagrams or Nth-order transfer functions.", - "type": "text" - }, - { - "block_id": "p936-b90", - "global_id": 27735, - "bbox": [ - 101.84, - 523.05, - 234.78, - 535.0 - ], - "text": "10.3-1 Electrical Circuits", - "type": "text" - }, - { - "block_id": "p936-b91", - "global_id": 27736, - "bbox": [ - 101.84, - 541.13, - 490.41, - 563.06 - ], - "text": "The method used in Ex. 10.2 proves effective in most of the simple cases. The steps are as\nfollows:", - "type": "text" - }, - { - "block_id": "p936-b92", - "global_id": 27737, - "bbox": [ - 118.78, - 571.02, - 490.4, - 628.8 - ], - "text": "1. Choose all independent capacitor voltages and inductor currents to be the state variables.\n2. Choose a set of loop currents; express the state variables and their first derivatives in terms\nof these loop currents.\n3. Write loop equations, and eliminate all variables other than state variables (and their first\nderivatives) from the equations derived in steps 2 and 3.", - "type": "text" - } - ] - }, - { - "page_num": 937, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p937-b0", - "global_id": 27738, - "bbox": [ - 236.03, - 62.89, - 516.15, - 71.98 - ], - "text": "10.3\nA Systematic Procedure to Determine State Equations\n917", - "type": "text" - }, - { - "block_id": "p937-b1", - "global_id": 27739, - "bbox": [ - 102.51, - 93.92, - 390.62, - 105.87 - ], - "text": "EXAMPLE 10.3\nState Equations of an RLC Circuit", - "type": "text" - }, - { - "block_id": "p937-b2", - "global_id": 27740, - "bbox": [ - 128.9, - 122.54, - 370.67, - 132.5 - ], - "text": "Write the state equations for the network shown in Fig. 10.2.", - "type": "text" - }, - { - "block_id": "p937-b3", - "global_id": 27741, - "bbox": [ - 146.62, - 222.45, - 151.13, - 230.45 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p937-b4", - "global_id": 27742, - "bbox": [ - 127.73, - 229.53, - 381.91, - 243.94 - ], - "text": "_\nx\ni1\ni2\ni3\nq2", - "type": "text" - }, - { - "block_id": "p937-b5", - "global_id": 27743, - "bbox": [ - 253.45, - 185.17, - 260.45, - 194.78 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p937-b8", - "global_id": 27744, - "bbox": [ - 177.32, - 181.51, - 384.29, - 191.68 - ], - "text": "1 H\n2 \t\n1", - "type": "text" - }, - { - "block_id": "p937-b9", - "global_id": 27745, - "bbox": [ - 254.02, - 231.08, - 446.54, - 245.0 - ], - "text": "2 \t\n2 \t\n1\n2 F", - "type": "text" - }, - { - "block_id": "p937-b10", - "global_id": 27746, - "bbox": [ - 127.51, - 275.75, - 249.8, - 284.98 - ], - "text": "Figure 10.2 Circuit for Ex. 10.3.", - "type": "text" - }, - { - "block_id": "p937-b11", - "global_id": 27747, - "bbox": [ - 128.9, - 312.57, - 502.77, - 336.06 - ], - "text": "Step 1. There is one inductor and one capacitor in the network. Therefore, we shall choose\nthe inductor current q1 and the capacitor voltage q2 as the state variables.", - "type": "text" - }, - { - "block_id": "p937-b12", - "global_id": 27748, - "bbox": [ - 128.91, - 344.45, - 502.73, - 366.46 - ], - "text": "Step 2. The relationship between the loop currents and the state variables can be written by\ninspection:", - "type": "text" - }, - { - "block_id": "p937-b13", - "global_id": 27749, - "bbox": [ - 296.17, - 378.0, - 502.75, - 389.45 - ], - "text": "q1 = i2\n(10.18)", - "type": "text" - }, - { - "block_id": "p937-b14", - "global_id": 27750, - "bbox": [ - 291.48, - 393.0, - 502.75, - 407.84 - ], - "text": "1\n2 ˙q2 = i2 −i3\n(10.19)", - "type": "text" - }, - { - "block_id": "p937-b15", - "global_id": 27751, - "bbox": [ - 128.91, - 416.59, - 253.7, - 426.64 - ], - "text": "Step 3. The loop equations are", - "type": "text" - }, - { - "block_id": "p937-b16", - "global_id": 27752, - "bbox": [ - 269.18, - 438.18, - 502.75, - 464.58 - ], - "text": "4i1 −2i2 = x\n(10.20)\n2(i2 −i1) + ˙q1 + q2 = 0\n(10.21)", - "type": "text" - }, - { - "block_id": "p937-b17", - "global_id": 27753, - "bbox": [ - 306.3, - 468.07, - 502.75, - 479.52 - ], - "text": "−q2 + 3i3 = 0\n(10.22)", - "type": "text" - }, - { - "block_id": "p937-b18", - "global_id": 27754, - "bbox": [ - 128.91, - 490.3, - 502.76, - 512.31 - ], - "text": "Now we eliminate i1, i2, and i3 from the state and loop equations as follows. From Eq. (10.21),\nwe have", - "type": "text" - }, - { - "block_id": "p937-b19", - "global_id": 27755, - "bbox": [ - 277.1, - 513.9, - 354.05, - 525.35 - ], - "text": "˙q1 = 2(i1 −i2) −q2", - "type": "text" - }, - { - "block_id": "p937-b20", - "global_id": 27756, - "bbox": [ - 128.9, - 533.14, - 482.8, - 544.69 - ], - "text": "We can eliminate i1 and i2 from this equation by using Eqs. (10.18) and (10.20) to obtain", - "type": "text" - }, - { - "block_id": "p937-b21", - "global_id": 27757, - "bbox": [ - 276.53, - 553.4, - 349.49, - 566.19 - ], - "text": "˙q1 = −q1 −q2 + 1", - "type": "text" - }, - { - "block_id": "p937-b22", - "global_id": 27758, - "bbox": [ - 346.0, - 555.05, - 355.11, - 567.93 - ], - "text": "2x", - "type": "text" - }, - { - "block_id": "p937-b23", - "global_id": 27759, - "bbox": [ - 128.9, - 577.08, - 388.19, - 587.04 - ], - "text": "The substitution of Eqs. (10.18) and (10.22) in Eq. (10.19) yields", - "type": "text" - }, - { - "block_id": "p937-b24", - "global_id": 27760, - "bbox": [ - 285.59, - 597.23, - 335.92, - 610.03 - ], - "text": "˙q2 = 2q1 −2", - "type": "text" - }, - { - "block_id": "p937-b25", - "global_id": 27761, - "bbox": [ - 332.43, - 598.89, - 345.58, - 611.77 - ], - "text": "3q2", - "type": "text" - } - ] - }, - { - "page_num": 938, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p938-b0", - "global_id": 27762, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "918\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p938-b1", - "global_id": 27763, - "bbox": [ - 103.16, - 86.24, - 409.32, - 96.21 - ], - "text": "These are the desired state equations. We can express them in matrix form as", - "type": "text" - }, - { - "block_id": "p938-b2", - "global_id": 27764, - "bbox": [ - 223.58, - 99.61, - 237.48, - 118.67 - ], - "text": "˙q1", - "type": "text" - }, - { - "block_id": "p938-b3", - "global_id": 27765, - "bbox": [ - 229.01, - 119.48, - 237.48, - 130.63 - ], - "text": "˙q2", - "type": "text" - }, - { - "block_id": "p938-b4", - "global_id": 27766, - "bbox": [ - 237.98, - 99.61, - 243.4, - 109.57 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p938-b5", - "global_id": 27767, - "bbox": [ - 245.45, - 113.6, - 253.23, - 123.57 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p938-b6", - "global_id": 27768, - "bbox": [ - 255.27, - 99.61, - 296.61, - 130.05 - ], - "text": "−1\n−1\n2\n−2", - "type": "text" - }, - { - "block_id": "p938-b7", - "global_id": 27769, - "bbox": [ - 292.39, - 125.89, - 295.88, - 132.86 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p938-b8", - "global_id": 27770, - "bbox": [ - 297.07, - 99.61, - 317.49, - 118.67 - ], - "text": "! q1", - "type": "text" - }, - { - "block_id": "p938-b9", - "global_id": 27771, - "bbox": [ - 309.03, - 119.79, - 317.49, - 130.63 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p938-b10", - "global_id": 27772, - "bbox": [ - 318.0, - 99.61, - 323.43, - 109.57 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p938-b11", - "global_id": 27773, - "bbox": [ - 324.98, - 113.6, - 332.75, - 123.57 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p938-b12", - "global_id": 27774, - "bbox": [ - 334.29, - 99.61, - 344.41, - 113.36 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p938-b13", - "global_id": 27775, - "bbox": [ - 340.18, - 113.94, - 345.16, - 130.05 - ], - "text": "2\n0", - "type": "text" - }, - { - "block_id": "p938-b14", - "global_id": 27776, - "bbox": [ - 345.61, - 99.61, - 351.04, - 109.57 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p938-b15", - "global_id": 27777, - "bbox": [ - 352.14, - 113.92, - 477.0, - 123.98 - ], - "text": "x\n(10.23)", - "type": "text" - }, - { - "block_id": "p938-b16", - "global_id": 27778, - "bbox": [ - 103.16, - 141.43, - 477.02, - 175.31 - ], - "text": "The derivation of state equations from loop equations is facilitated considerably by\nchoosing loops in such a way that only one loop current passes through each of the inductors\nor capacitors.", - "type": "text" - }, - { - "block_id": "p938-b17", - "global_id": 27779, - "bbox": [ - 101.84, - 217.21, - 381.04, - 243.32 - ], - "text": "AN ALTERNATIVE PROCEDURE\nWe can also determine the state equations by the following procedure.", - "type": "text" - }, - { - "block_id": "p938-b18", - "global_id": 27780, - "bbox": [ - 118.78, - 251.3, - 490.39, - 332.99 - ], - "text": "1. Choose all independent capacitor voltages and inductor currents to be the state variables.\n2. Replace each capacitor by a voltage source equal to the capacitor voltage, and replace each\ninductor by a current source equal to the inductor current. This step will transform the RLC\nnetwork into a network consisting only of resistors, current sources, and voltage sources.\n3. Find the current through each capacitor and equate it to C˙qi, where qi is the capacitor\nvoltage. Similarly, find the voltage across each inductor and equate it to L˙qj, where qj is\nthe inductor current.", - "type": "text" - }, - { - "block_id": "p938-b19", - "global_id": 27781, - "bbox": [ - 76.77, - 354.79, - 453.54, - 366.74 - ], - "text": "EXAMPLE 10.4\nAlternate Procedure to Determine State Equations", - "type": "text" - }, - { - "block_id": "p938-b20", - "global_id": 27782, - "bbox": [ - 103.16, - 383.41, - 477.01, - 405.33 - ], - "text": "Use the three-step alternative procedure just outlined to write the state equations for the\nnetwork in Fig. 10.2.", - "type": "text" - }, - { - "block_id": "p938-b21", - "global_id": 27783, - "bbox": [ - 103.16, - 428.14, - 477.02, - 462.12 - ], - "text": "In the network in Fig. 10.2, we replace the inductor by a current source of current q1 and\nthe capacitor by a voltage source of voltage q2, as shown in Fig. 10.3. The resulting network\nconsists of four resistors, two voltage sources, and one current source.", - "type": "text" - }, - { - "block_id": "p938-b22", - "global_id": 27784, - "bbox": [ - 117.77, - 542.44, - 123.41, - 552.44 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p938-b23", - "global_id": 27785, - "bbox": [ - 101.99, - 542.79, - 302.13, - 563.24 - ], - "text": "_\nx\n+\n_", - "type": "text" - }, - { - "block_id": "p938-b24", - "global_id": 27786, - "bbox": [ - 253.49, - 520.83, - 310.14, - 533.73 - ], - "text": "iC\nvL", - "type": "text" - }, - { - "block_id": "p938-b25", - "global_id": 27787, - "bbox": [ - 279.76, - 558.5, - 286.76, - 568.11 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p938-b26", - "global_id": 27788, - "bbox": [ - 236.53, - 496.79, - 243.53, - 506.39 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p938-b28", - "global_id": 27789, - "bbox": [ - 149.37, - 495.65, - 356.34, - 503.95 - ], - "text": "2 \t\n1", - "type": "text" - }, - { - "block_id": "p938-b29", - "global_id": 27790, - "bbox": [ - 194.4, - 554.77, - 418.38, - 563.07 - ], - "text": "2 \t\n2", - "type": "text" - }, - { - "block_id": "p938-b30", - "global_id": 27791, - "bbox": [ - 101.77, - 606.58, - 315.66, - 615.82 - ], - "text": "Figure 10.3 Equivalent circuit of the network in Fig. 10.2.", - "type": "text" - } - ] - }, - { - "page_num": 939, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p939-b0", - "global_id": 27792, - "bbox": [ - 236.03, - 62.89, - 516.15, - 71.98 - ], - "text": "10.3\nA Systematic Procedure to Determine State Equations\n919", - "type": "text" - }, - { - "block_id": "p939-b1", - "global_id": 27793, - "bbox": [ - 128.9, - 86.14, - 502.76, - 193.33 - ], - "text": "We can determine the voltage vL across the inductor and the current ic through the\ncapacitor by using the principle of superposition. This step can be accomplished by inspection.\nFor example, vL has three components arising from three sources. To compute the component\ndue to x, we assume that q1 = 0 (open circuit) and q2 = 0 (short circuit). Under these conditions,\nthe entire network to the right of the 2 resistor is opened, and the component of vL due to\nx is the voltage across the 2 resistor. This voltage is clearly (1/2)x. Similarly, to find the\ncomponent of vL due to q1, we short x and q2. The source q1 sees an equivalent resistor of 1 \nacross it, and hence vL = −q1. Continuing the process, we find that the component of vL due\nto q2 is −q2. Hence,", - "type": "text" - }, - { - "block_id": "p939-b2", - "global_id": 27794, - "bbox": [ - 270.07, - 192.08, - 316.27, - 204.88 - ], - "text": "vL = ˙q1 = 1", - "type": "text" - }, - { - "block_id": "p939-b3", - "global_id": 27795, - "bbox": [ - 312.78, - 193.42, - 361.09, - 206.62 - ], - "text": "2x −q1 −q2", - "type": "text" - }, - { - "block_id": "p939-b4", - "global_id": 27796, - "bbox": [ - 128.9, - 212.77, - 267.6, - 222.73 - ], - "text": "Using the same procedure, we find", - "type": "text" - }, - { - "block_id": "p939-b5", - "global_id": 27797, - "bbox": [ - 276.02, - 232.92, - 298.94, - 245.72 - ], - "text": "ic = 1", - "type": "text" - }, - { - "block_id": "p939-b6", - "global_id": 27798, - "bbox": [ - 295.45, - 232.92, - 355.14, - 247.77 - ], - "text": "2 ˙q2 = q1 −1\n3q2", - "type": "text" - }, - { - "block_id": "p939-b7", - "global_id": 27799, - "bbox": [ - 128.9, - 256.01, - 455.07, - 266.57 - ], - "text": "These equations are identical to the state equations [Eq. (10.23)] obtained earlier.†", - "type": "text" - }, - { - "block_id": "p939-b8", - "global_id": 27800, - "bbox": [ - 127.59, - 332.15, - 387.42, - 344.1 - ], - "text": "10.3-2 State Equations from a Transfer Function", - "type": "text" - }, - { - "block_id": "p939-b9", - "global_id": 27801, - "bbox": [ - 127.59, - 349.64, - 515.62, - 360.2 - ], - "text": "It is relatively easy to determine the state equations of a system specified by its transfer function.‡", - "type": "text" - }, - { - "block_id": "p939-b10", - "global_id": 27802, - "bbox": [ - 127.59, - 362.19, - 401.58, - 372.15 - ], - "text": "Consider, for example, a first-order system with the transfer function", - "type": "text" - }, - { - "block_id": "p939-b11", - "global_id": 27803, - "bbox": [ - 295.4, - 386.33, - 347.12, - 410.25 - ], - "text": "H(s) =\n1\ns + a", - "type": "text" - }, - { - "block_id": "p939-b12", - "global_id": 27804, - "bbox": [ - 127.59, - 423.12, - 516.14, - 445.13 - ], - "text": "The system realization appears in Fig. 10.4. The integrator output q serves as a natural state\nvariable since, in practical realization, initial conditions are placed on the integrator output. The", - "type": "text" - }, - { - "block_id": "p939-b13", - "global_id": 27805, - "bbox": [ - 127.59, - 467.47, - 516.14, - 633.41 - ], - "text": "† This procedure requires modification if the system contains all-capacitor and voltage-source tie sets or\nall-inductor and current-source cut sets. In the case of all-capacitor and voltage-source tie sets, all capacitor\nvoltages cannot be independent. One capacitor voltage can be expressed in terms of the remaining capacitor\nvoltages and the voltage source(s) in that tie set. Consequently, one of the capacitor voltages should not be\nused as a state variable, and that capacitor should not be replaced by a voltage source. Similarly, in all-inductor\nand current-source tie sets, one inductor should not be replaced by a current source. If there are all-capacitor\ntie sets or all-inductor cut sets only, no further complications occur. In all-capacitor voltage-source tie\nsets and/or all-inductor current-source cut sets, we have additional difficulties in that the terms involving\nderivatives of the input may occur. This problem can be solved by redefining the state variables. The final\nstate variables will not be capacitor voltages and inductor currents.\n‡ We implicitly assume that the system is controllable and observable. This implies that there are no\npole-zero cancellations in the transfer function. If such cancellations are present, the state variable description\nrepresents only the part of the system that is controllable and observable (the part of the system that is coupled\nto the input and the output). In other words, the internal description represented by the state equations is no\nbetter than the external description represented by the input–output equation.", - "type": "text" - } - ] - }, - { - "page_num": 940, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p940-b0", - "global_id": 27806, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "920\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p940-b2", - "global_id": 27807, - "bbox": [ - 178.84, - 134.7, - 182.84, - 149.61 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p940-b3", - "global_id": 27808, - "bbox": [ - 116.54, - 90.17, - 182.5, - 103.99 - ], - "text": ".q\nx", - "type": "text" - }, - { - "block_id": "p940-b4", - "global_id": 27809, - "bbox": [ - 123.26, - 140.55, - 133.92, - 148.76 - ], - "text": "a", - "type": "text" - }, - { - "block_id": "p940-b5", - "global_id": 27810, - "bbox": [ - 178.81, - 179.66, - 182.81, - 187.66 - ], - "text": "q", - "type": "text" - }, - { - "block_id": "p940-b6", - "global_id": 27811, - "bbox": [ - 196.32, - 166.26, - 199.87, - 174.26 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p940-b7", - "global_id": 27812, - "bbox": [ - 218.8, - 178.06, - 385.51, - 191.27 - ], - "text": "Figure 10.4 Block realization of H(s) =\n1\ns+a.", - "type": "text" - }, - { - "block_id": "p940-b8", - "global_id": 27813, - "bbox": [ - 101.84, - 219.27, - 320.23, - 229.64 - ], - "text": "integrator input is naturally ˙q. From Fig. 10.4, we have", - "type": "text" - }, - { - "block_id": "p940-b9", - "global_id": 27814, - "bbox": [ - 233.42, - 247.22, - 358.83, - 257.6 - ], - "text": "˙q = −aq + x\nand\ny = q", - "type": "text" - }, - { - "block_id": "p940-b10", - "global_id": 27815, - "bbox": [ - 101.84, - 287.55, - 490.4, - 321.43 - ], - "text": "In Sec. 4.6 we saw that a given transfer function can be realized in several ways. Consequently,\nwe should be able to obtain different state-space descriptions of the same system by using different\nrealizations. This assertion will be clarified by the following example.", - "type": "text" - }, - { - "block_id": "p940-b11", - "global_id": 27816, - "bbox": [ - 76.77, - 353.44, - 449.59, - 365.39 - ], - "text": "EXAMPLE 10.5\nState-Space Description from a Transfer Function", - "type": "text" - }, - { - "block_id": "p940-b12", - "global_id": 27817, - "bbox": [ - 103.16, - 382.06, - 310.19, - 392.02 - ], - "text": "Consider a system specified by the transfer function", - "type": "text" - }, - { - "block_id": "p940-b13", - "global_id": 27818, - "bbox": [ - 124.56, - 404.25, - 234.88, - 442.3 - ], - "text": "H(s) =\n2s + 10\ns3 + 8s2 + 19s + 12\n\n\n\ndirect form", - "type": "text" - }, - { - "block_id": "p940-b14", - "global_id": 27819, - "bbox": [ - 236.92, - 411.23, - 244.69, - 421.19 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p940-b15", - "global_id": 27820, - "bbox": [ - 246.74, - 397.24, - 274.38, - 428.68 - ], - "text": "2\ns + 1", - "type": "text" - }, - { - "block_id": "p940-b16", - "global_id": 27821, - "bbox": [ - 275.59, - 397.24, - 311.06, - 414.62 - ], - "text": "s + 5", - "type": "text" - }, - { - "block_id": "p940-b17", - "global_id": 27822, - "bbox": [ - 291.34, - 418.3, - 311.06, - 428.68 - ], - "text": "s + 3", - "type": "text" - }, - { - "block_id": "p940-b18", - "global_id": 27823, - "bbox": [ - 312.26, - 397.24, - 347.73, - 428.68 - ], - "text": "1\ns + 4", - "type": "text" - }, - { - "block_id": "p940-b20", - "global_id": 27824, - "bbox": [ - 246.74, - 424.7, - 355.66, - 443.87 - ], - "text": "cascade", - "type": "text" - }, - { - "block_id": "p940-b21", - "global_id": 27825, - "bbox": [ - 357.71, - 411.23, - 365.48, - 421.19 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p940-b22", - "global_id": 27826, - "bbox": [ - 368.73, - 402.19, - 431.96, - 428.68 - ], - "text": "4\n3\ns + 1 −\n2\ns + 3 +", - "type": "text" - }, - { - "block_id": "p940-b23", - "global_id": 27827, - "bbox": [ - 367.53, - 402.19, - 455.62, - 442.3 - ], - "text": "2\n3\ns + 4\n\n\n\nparallel", - "type": "text" - }, - { - "block_id": "p940-b24", - "global_id": 27828, - "bbox": [ - 103.16, - 453.2, - 477.03, - 499.44 - ], - "text": "The procedure developed in Sec. 4.6 allows us to realize H(s) as, among others, direct form\nII (DFII), transpose DFII (TDFII), cascade, and parallel. These realizations are depicted in\nFig. 10.5. Determine state-space descriptions for each of these realizations. As mentioned\nearlier, the output of each integrator serves as a natural state variable.", - "type": "text" - }, - { - "block_id": "p940-b25", - "global_id": 27829, - "bbox": [ - 103.16, - 522.27, - 477.02, - 568.95 - ], - "text": "Direct Form II and Its Transpose\nHere we shall realize the system using the canonical form (direct form II and its transpose)\ndiscussed in Sec. 4.6. If we choose the state variables to be the three integrator outputs q1, q2,\nand q3, then, according to Fig. 10.5a,", - "type": "text" - }, - { - "block_id": "p940-b26", - "global_id": 27830, - "bbox": [ - 231.35, - 579.72, - 348.79, - 621.06 - ], - "text": "˙q1 = q2\n˙q2 = q3\n˙q3 = −12q1 −19q2 −8q3 + x", - "type": "text" - } - ] - }, - { - "page_num": 941, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p941-b0", - "global_id": 27831, - "bbox": [ - 236.03, - 62.89, - 516.15, - 71.98 - ], - "text": "10.3\nA Systematic Procedure to Determine State Equations\n921", - "type": "text" - }, - { - "block_id": "p941-b1", - "global_id": 27832, - "bbox": [ - 131.77, - 98.59, - 135.33, - 106.59 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p941-b2", - "global_id": 27833, - "bbox": [ - 168.83, - 145.02, - 179.49, - 153.32 - ], - "text": "8", - "type": "text" - }, - { - "block_id": "p941-b3", - "global_id": 27834, - "bbox": [ - 414.4, - 166.31, - 425.07, - 174.61 - 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"bbox": [ - 314.17, - 412.09, - 320.28, - 421.7 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p941-b42", - "global_id": 27870, - "bbox": [ - 315.12, - 407.5, - 317.12, - 415.5 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p941-b43", - "global_id": 27871, - "bbox": [ - 314.17, - 539.32, - 320.28, - 548.92 - ], - "text": "z3", - "type": "text" - }, - { - "block_id": "p941-b44", - "global_id": 27872, - "bbox": [ - 315.12, - 534.73, - 317.12, - 542.73 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p941-b45", - "global_id": 27873, - "bbox": [ - 314.17, - 476.43, - 320.28, - 486.03 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p941-b46", - "global_id": 27874, - "bbox": [ - 315.12, - 471.84, - 317.12, - 479.84 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p941-b47", - "global_id": 27875, - "bbox": [ - 320.75, - 461.93, - 326.86, - 471.54 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p941-b48", - "global_id": 27876, - "bbox": [ - 320.75, - 526.23, - 326.86, - 535.84 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p941-b49", - "global_id": 27877, - "bbox": [ - 320.75, - 590.53, - 326.86, - 600.13 - ], - "text": "z3", - "type": "text" - }, - { - "block_id": "p941-b50", - "global_id": 27878, - "bbox": [ - 312.48, - 317.9, - 320.81, - 327.5 - ], - "text": "w2", - "type": "text" - }, - { - "block_id": "p941-b51", - "global_id": 27879, - "bbox": [ - 314.4, - 313.31, - 447.91, - 327.5 - ], - "text": ".\nw3", - "type": "text" - }, - { - "block_id": "p941-b52", - "global_id": 27880, - "bbox": [ - 441.5, - 313.31, - 443.5, - 321.31 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p941-b53", - "global_id": 27881, - "bbox": [ - 207.54, - 375.74, - 447.37, - 385.35 - ], - "text": "w1\nw2\nw3", - "type": "text" - }, - { - "block_id": "p941-b54", - "global_id": 27882, - "bbox": [ - 312.5, - 391.25, - 321.38, - 399.25 - ], - "text": "(c)", - "type": "text" - }, - { - "block_id": "p941-b55", - "global_id": 27883, - "bbox": [ - 190.52, - 258.75, - 199.4, - 266.75 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p941-b56", - "global_id": 27884, - "bbox": [ - 333.29, - 412.9, - 345.29, - 421.19 - ], - "text": "43", - "type": "text" - }, - { - "block_id": "p941-b57", - "global_id": 27885, - "bbox": [ - 333.64, - 540.25, - 345.64, - 548.54 - ], - "text": "23", - "type": "text" - }, - { - "block_id": "p941-b58", - "global_id": 27886, - "bbox": [ - 312.24, - 606.15, - 321.57, - 614.15 - ], - "text": "(d)", - "type": "text" - }, - { - "block_id": "p941-b59", - "global_id": 27887, - "bbox": [ - 195.81, - 123.95, - 399.23, - 144.01 - ], - "text": "1\ns\n1\ns", - "type": "text" - }, - { - "block_id": "p941-b60", - "global_id": 27888, - "bbox": [ - 395.23, - 183.03, - 399.23, - 197.94 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b61", - "global_id": 27889, - "bbox": [ - 395.23, - 243.84, - 399.23, - 258.75 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b62", - "global_id": 27890, - "bbox": [ - 209.85, - 344.62, - 213.85, - 359.53 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b63", - "global_id": 27891, - "bbox": [ - 314.94, - 438.89, - 318.94, - 453.8 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b64", - "global_id": 27892, - "bbox": [ - 314.94, - 502.84, - 318.94, - 517.75 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b65", - "global_id": 27893, - "bbox": [ - 314.94, - 565.78, - 318.94, - 580.7 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b66", - "global_id": 27894, - "bbox": [ - 314.94, - 344.62, - 318.94, - 359.53 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b67", - "global_id": 27895, - "bbox": [ - 441.35, - 344.62, - 445.35, - 359.53 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b68", - "global_id": 27896, - "bbox": [ - 195.81, - 171.2, - 199.81, - 186.11 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b69", - "global_id": 27897, - "bbox": [ - 195.81, - 219.45, - 199.81, - 234.36 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p941-b78", - "global_id": 27898, - "bbox": [ - 119.94, - 620.48, - 419.87, - 630.09 - ], - "text": "Figure 10.5 (a) DFII, (b) TDFII, (c) cascade, and (d) parallel realizations of H(s).", - "type": "text" - } - ] - }, - { - "page_num": 942, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p942-b0", - "global_id": 27899, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "922\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p942-b1", - "global_id": 27900, - "bbox": [ - 103.16, - 86.14, - 219.82, - 96.21 - ], - "text": "Also, the output y is given by", - "type": "text" - }, - { - "block_id": "p942-b2", - "global_id": 27901, - "bbox": [ - 260.07, - 97.78, - 319.6, - 109.24 - ], - "text": "y = 10q1 + 2q2", - "type": "text" - }, - { - "block_id": "p942-b3", - "global_id": 27902, - "bbox": [ - 103.16, - 117.12, - 327.03, - 127.08 - ], - "text": "In matrix form, these state and output equations become", - "type": "text" - }, - { - "block_id": "p942-b4", - "global_id": 27903, - "bbox": [ - 200.8, - 130.28, - 208.07, - 140.24 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p942-b5", - "global_id": 27904, - "bbox": [ - 200.8, - 148.22, - 208.07, - 158.18 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p942-b6", - "global_id": 27905, - "bbox": [ - 208.07, - 138.19, - 216.53, - 173.25 - ], - "text": "˙q1\n˙q2\n˙q3", - "type": "text" - }, - { - "block_id": "p942-b7", - "global_id": 27906, - "bbox": [ - 217.04, - 130.28, - 224.3, - 140.24 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p942-b8", - "global_id": 27907, - "bbox": [ - 217.04, - 148.22, - 234.12, - 160.21 - ], - "text": "⎦=", - "type": "text" - }, - { - "block_id": "p942-b9", - "global_id": 27908, - "bbox": [ - 236.17, - 130.28, - 243.43, - 140.24 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p942-b10", - "global_id": 27909, - "bbox": [ - 236.17, - 148.22, - 243.43, - 158.18 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p942-b11", - "global_id": 27910, - "bbox": [ - 243.43, - 138.6, - 311.57, - 172.48 - ], - "text": "0\n1\n0\n0\n0\n1\n−12\n−19\n−8", - "type": "text" - }, - { - "block_id": "p942-b12", - "global_id": 27911, - "bbox": [ - 311.58, - 130.28, - 318.84, - 140.24 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p942-b13", - "global_id": 27912, - "bbox": [ - 311.58, - 148.22, - 318.84, - 158.18 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p942-b14", - "global_id": 27913, - "bbox": [ - 236.17, - 169.7, - 318.84, - 188.03 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p942-b15", - "global_id": 27914, - "bbox": [ - 319.95, - 130.28, - 327.21, - 140.24 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p942-b16", - "global_id": 27915, - "bbox": [ - 319.95, - 148.22, - 327.21, - 158.18 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p942-b17", - "global_id": 27916, - "bbox": [ - 327.21, - 138.5, - 335.67, - 173.25 - ], - "text": "q1\nq2\nq3", - "type": "text" - }, - { - "block_id": "p942-b18", - "global_id": 27917, - "bbox": [ - 336.17, - 130.28, - 343.44, - 140.24 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p942-b19", - "global_id": 27918, - "bbox": [ - 336.17, - 148.22, - 352.76, - 160.21 - ], - "text": "⎦+", - "type": "text" - }, - { - "block_id": "p942-b20", - "global_id": 27919, - "bbox": [ - 354.3, - 130.28, - 361.56, - 140.24 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p942-b21", - "global_id": 27920, - "bbox": [ - 354.3, - 148.22, - 361.56, - 158.18 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p942-b22", - "global_id": 27921, - "bbox": [ - 361.57, - 138.6, - 366.55, - 172.48 - ], - "text": "0\n0\n1", - "type": "text" - }, - { - "block_id": "p942-b23", - "global_id": 27922, - "bbox": [ - 366.55, - 130.28, - 373.81, - 140.24 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p942-b24", - "global_id": 27923, - "bbox": [ - 366.55, - 148.22, - 373.81, - 158.18 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p942-b26", - "global_id": 27924, - "bbox": [ - 362.07, - 182.05, - 366.05, - 188.03 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p942-b27", - "global_id": 27925, - "bbox": [ - 374.92, - 150.56, - 379.34, - 160.53 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p942-b28", - "global_id": 27926, - "bbox": [ - 103.17, - 197.92, - 117.54, - 207.89 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p942-b29", - "global_id": 27927, - "bbox": [ - 246.12, - 219.03, - 309.46, - 244.85 - ], - "text": "y = [10\n2\n0]\n\n\n\nC", - "type": "text" - }, - { - "block_id": "p942-b30", - "global_id": 27928, - "bbox": [ - 310.56, - 199.07, - 317.82, - 209.03 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p942-b31", - "global_id": 27929, - "bbox": [ - 310.56, - 217.0, - 317.82, - 226.96 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p942-b32", - "global_id": 27930, - "bbox": [ - 317.82, - 207.29, - 326.29, - 242.04 - ], - "text": "q1\nq2\nq3", - "type": "text" - }, - { - "block_id": "p942-b33", - "global_id": 27931, - "bbox": [ - 326.8, - 199.07, - 334.06, - 209.03 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p942-b34", - "global_id": 27932, - "bbox": [ - 326.8, - 217.0, - 334.06, - 226.96 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p942-b35", - "global_id": 27933, - "bbox": [ - 103.17, - 251.83, - 477.03, - 273.76 - ], - "text": "We can readily verify the state equations of the DFII structure by using MATLAB’s tf2ss\ncommand:", - "type": "text" - }, - { - "block_id": "p942-b36", - "global_id": 27934, - "bbox": [ - 103.17, - 291.97, - 291.45, - 409.53 - ], - "text": ">>\nnum = [2 10]; den = [1 8 19 12];\n>>\n[A,B,C,D] = tf2ss(num,den)\nA = -8\n-19\n-12\n1\n0\n0\n0\n1\n0\nB =\n1\n0\n0\nC =\n0\n2\n10\nD =\n0", - "type": "text" - }, - { - "block_id": "p942-b37", - "global_id": 27935, - "bbox": [ - 103.17, - 426.77, - 477.01, - 461.06 - ], - "text": "MATLAB’s convention for labeling state variables q1,q2,. . .,qn in a block diagram, such as\nshown in Fig. 10.5a, is reversed. That is, MATLAB labels q1 as qn, q2 and qn−1, and so on.\nKeeping this in mind, we see that MATLAB indeed confirms our earlier results.", - "type": "text" - }, - { - "block_id": "p942-b38", - "global_id": 27936, - "bbox": [ - 103.17, - 463.05, - 476.99, - 485.25 - ], - "text": "It is also possible to determine the transfer function from the state-space representation\nusing the ss2tf and tf commands:", - "type": "text" - }, - { - "block_id": "p942-b39", - "global_id": 27937, - "bbox": [ - 103.17, - 503.19, - 348.95, - 525.1 - ], - "text": ">>\n[num,den] = ss2tf(A,B,C,D); H = tf(num,den)\nH =", - "type": "text" - }, - { - "block_id": "p942-b40", - "global_id": 27938, - "bbox": [ - 145.0, - 527.09, - 265.3, - 560.97 - ], - "text": "2 s + 10\n-----------------------\ns^3 + 8 s^2 + 19 s + 12", - "type": "text" - }, - { - "block_id": "p942-b41", - "global_id": 27939, - "bbox": [ - 103.17, - 578.53, - 477.04, - 613.25 - ], - "text": "Transpose Direct Form II\nWe can also realize H(s) by using the transpose of the DFII form, as shown in Fig. 10.5b. If\nwe label the output of the three integrators as the state variables v1, v2, and v3, then, according", - "type": "text" - } - ] - }, - { - "page_num": 943, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p943-b0", - "global_id": 27940, - "bbox": [ - 236.03, - 62.89, - 516.15, - 71.98 - ], - "text": "10.3\nA Systematic Procedure to Determine State Equations\n923", - "type": "text" - }, - { - "block_id": "p943-b1", - "global_id": 27941, - "bbox": [ - 128.9, - 86.24, - 182.31, - 96.2 - ], - "text": "to Fig. 10.5b,", - "type": "text" - }, - { - "block_id": "p943-b2", - "global_id": 27942, - "bbox": [ - 276.72, - 107.74, - 354.91, - 134.14 - ], - "text": "˙v1 = −12v3 + 10x\n˙v2 = v1 −19v3 + 2x", - "type": "text" - }, - { - "block_id": "p943-b3", - "global_id": 27943, - "bbox": [ - 276.72, - 137.63, - 329.15, - 149.09 - ], - "text": "˙v3 = v2 −8v3", - "type": "text" - }, - { - "block_id": "p943-b4", - "global_id": 27944, - "bbox": [ - 128.9, - 159.86, - 238.65, - 169.92 - ], - "text": "and the output y is given by", - "type": "text" - }, - { - "block_id": "p943-b5", - "global_id": 27945, - "bbox": [ - 303.49, - 171.51, - 327.68, - 182.66 - ], - "text": "y = v3", - "type": "text" - }, - { - "block_id": "p943-b6", - "global_id": 27946, - "bbox": [ - 128.9, - 190.85, - 368.23, - 200.81 - ], - "text": "The matrix form of these state and output equations become", - "type": "text" - }, - { - "block_id": "p943-b7", - "global_id": 27947, - "bbox": [ - 234.88, - 204.01, - 242.14, - 213.97 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b8", - "global_id": 27948, - "bbox": [ - 234.88, - 221.94, - 242.14, - 231.91 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b9", - "global_id": 27949, - "bbox": [ - 242.14, - 211.92, - 250.05, - 246.98 - ], - "text": "˙v1\n˙v2\n˙v3", - "type": "text" - }, - { - "block_id": "p943-b10", - "global_id": 27950, - "bbox": [ - 250.55, - 204.01, - 257.81, - 213.97 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b11", - "global_id": 27951, - "bbox": [ - 250.55, - 221.94, - 267.63, - 233.93 - ], - "text": "⎦=", - "type": "text" - }, - { - "block_id": "p943-b12", - "global_id": 27952, - "bbox": [ - 269.68, - 204.01, - 276.95, - 213.97 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b13", - "global_id": 27953, - "bbox": [ - 269.68, - 221.94, - 276.95, - 231.91 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b14", - "global_id": 27954, - "bbox": [ - 276.95, - 211.92, - 324.56, - 246.21 - ], - "text": "0\n0\n−12\n1\n0\n−19\n0\n1\n−8", - "type": "text" - }, - { - "block_id": "p943-b15", - "global_id": 27955, - "bbox": [ - 324.56, - 204.01, - 331.83, - 213.97 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b16", - "global_id": 27956, - "bbox": [ - 324.56, - 221.94, - 331.83, - 231.91 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p943-b17", - "global_id": 27957, - "bbox": [ - 269.68, - 243.42, - 331.84, - 263.71 - ], - "text": "ˆA", - "type": "text" - }, - { - "block_id": "p943-b18", - "global_id": 27958, - "bbox": [ - 332.93, - 204.01, - 340.2, - 213.97 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b19", - "global_id": 27959, - "bbox": [ - 332.93, - 221.94, - 340.2, - 231.91 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b20", - "global_id": 27960, - "bbox": [ - 340.2, - 212.23, - 348.1, - 246.98 - ], - "text": "v1\nv2\nv3", - "type": "text" - }, - { - "block_id": "p943-b21", - "global_id": 27961, - "bbox": [ - 348.6, - 204.01, - 355.87, - 213.97 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b22", - "global_id": 27962, - "bbox": [ - 348.6, - 221.94, - 365.19, - 233.93 - ], - "text": "⎦+", - "type": "text" - }, - { - "block_id": "p943-b23", - "global_id": 27963, - "bbox": [ - 366.73, - 204.01, - 373.99, - 213.97 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b24", - "global_id": 27964, - "bbox": [ - 366.73, - 221.94, - 373.99, - 231.91 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b25", - "global_id": 27965, - "bbox": [ - 374.0, - 212.33, - 383.96, - 246.21 - ], - "text": "10\n2\n0", - "type": "text" - }, - { - "block_id": "p943-b26", - "global_id": 27966, - "bbox": [ - 383.96, - 204.01, - 391.23, - 213.97 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b27", - "global_id": 27967, - "bbox": [ - 383.96, - 221.94, - 391.23, - 231.91 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p943-b29", - "global_id": 27968, - "bbox": [ - 376.99, - 255.8, - 380.98, - 263.71 - ], - "text": "ˆB", - "type": "text" - }, - { - "block_id": "p943-b30", - "global_id": 27969, - "bbox": [ - 392.33, - 224.28, - 396.76, - 234.24 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p943-b31", - "global_id": 27970, - "bbox": [ - 128.91, - 273.6, - 143.28, - 283.57 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p943-b32", - "global_id": 27971, - "bbox": [ - 274.64, - 294.71, - 333.0, - 322.5 - ], - "text": "y = [0\n0\n1]\n\n\n\nˆC", - "type": "text" - }, - { - "block_id": "p943-b33", - "global_id": 27972, - "bbox": [ - 334.1, - 274.75, - 341.36, - 284.71 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b34", - "global_id": 27973, - "bbox": [ - 334.1, - 292.68, - 341.36, - 302.65 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b35", - "global_id": 27974, - "bbox": [ - 341.36, - 282.97, - 349.26, - 317.72 - ], - "text": "v1\nv2\nv3", - "type": "text" - }, - { - "block_id": "p943-b36", - "global_id": 27975, - "bbox": [ - 349.76, - 274.75, - 357.03, - 284.71 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b37", - "global_id": 27976, - "bbox": [ - 349.76, - 292.68, - 357.03, - 302.65 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p943-b38", - "global_id": 27977, - "bbox": [ - 128.91, - 329.12, - 502.76, - 363.41 - ], - "text": "Observe closely the relationship between the state-space descriptions of H(s) by means of\nthe DFII and TDFII realizations. The A matrices in these two cases are the transpose of each\nother; also, the B of one is the transpose of C in the other, and vice versa. Hence,", - "type": "text" - }, - { - "block_id": "p943-b39", - "global_id": 27978, - "bbox": [ - 217.81, - 370.83, - 413.86, - 385.33 - ], - "text": "(A)T = ˆA,\n(B)T = ˆC,\nand\n(C)T = ˆB", - "type": "text" - }, - { - "block_id": "p943-b40", - "global_id": 27979, - "bbox": [ - 128.91, - 397.28, - 387.37, - 407.24 - ], - "text": "This is no coincidence. This duality relation is generally true [1].", - "type": "text" - }, - { - "block_id": "p943-b41", - "global_id": 27980, - "bbox": [ - 128.9, - 421.11, - 502.75, - 455.06 - ], - "text": "Cascade Realization\nThe three integrator outputs w1, w2, and w3 in Fig. 10.5c are the state variables. Writing\nequations for the summer outputs yields", - "type": "text" - }, - { - "block_id": "p943-b42", - "global_id": 27981, - "bbox": [ - 171.71, - 466.61, - 459.45, - 478.07 - ], - "text": "˙w1 = −w1 + x,\n˙w2 = 2w1 −3w2,\nand\n˙w3 = 5w2 + ˙w2 −4w3", - "type": "text" - }, - { - "block_id": "p943-b43", - "global_id": 27982, - "bbox": [ - 128.9, - 488.52, - 502.75, - 511.62 - ], - "text": "Since ˙w2 = 2w1 −3w2, we see that ˙w3 = 2w1 +2w2 −4w3. From Fig. 10.5c, we further see that\ny = w3. Put into matrix form, the state and output equations are therefore", - "type": "text" - }, - { - "block_id": "p943-b44", - "global_id": 27983, - "bbox": [ - 229.87, - 514.06, - 237.13, - 524.02 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b45", - "global_id": 27984, - "bbox": [ - 229.87, - 531.98, - 237.13, - 541.95 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b46", - "global_id": 27985, - "bbox": [ - 237.13, - 521.97, - 247.26, - 557.02 - ], - "text": "˙w1\n˙w2\n˙w3", - "type": "text" - }, - { - "block_id": "p943-b47", - "global_id": 27986, - "bbox": [ - 247.76, - 514.06, - 255.02, - 524.02 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b48", - "global_id": 27987, - "bbox": [ - 247.76, - 531.98, - 264.84, - 543.98 - ], - "text": "⎦=", - "type": "text" - }, - { - "block_id": "p943-b49", - "global_id": 27988, - "bbox": [ - 266.89, - 514.06, - 274.16, - 524.02 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b50", - "global_id": 27989, - "bbox": [ - 266.89, - 531.98, - 274.16, - 541.95 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b51", - "global_id": 27990, - "bbox": [ - 274.16, - 521.97, - 332.33, - 556.25 - ], - "text": "−1\n0\n0\n2\n−3\n0\n2\n2\n−4", - "type": "text" - }, - { - "block_id": "p943-b52", - "global_id": 27991, - "bbox": [ - 332.33, - 514.06, - 339.59, - 524.02 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b53", - "global_id": 27992, - "bbox": [ - 332.33, - 531.98, - 339.59, - 541.95 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p943-b54", - "global_id": 27993, - "bbox": [ - 340.7, - 514.06, - 347.96, - 524.02 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b55", - "global_id": 27994, - "bbox": [ - 340.7, - 531.98, - 347.96, - 541.95 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b56", - "global_id": 27995, - "bbox": [ - 347.96, - 522.28, - 358.09, - 557.02 - ], - "text": "w1\nw2\nw3", - "type": "text" - }, - { - "block_id": "p943-b57", - "global_id": 27996, - "bbox": [ - 358.6, - 514.06, - 365.87, - 524.02 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b58", - "global_id": 27997, - "bbox": [ - 358.6, - 531.98, - 375.18, - 543.98 - ], - "text": "⎦+", - "type": "text" - }, - { - "block_id": "p943-b59", - "global_id": 27998, - "bbox": [ - 376.73, - 514.06, - 383.99, - 524.02 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b60", - "global_id": 27999, - "bbox": [ - 376.73, - 531.98, - 383.99, - 541.95 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b61", - "global_id": 28000, - "bbox": [ - 383.99, - 522.38, - 388.97, - 556.25 - ], - "text": "1\n0\n0", - "type": "text" - }, - { - "block_id": "p943-b62", - "global_id": 28001, - "bbox": [ - 388.98, - 514.06, - 396.24, - 524.02 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b63", - "global_id": 28002, - "bbox": [ - 388.98, - 531.98, - 401.77, - 544.29 - ], - "text": "⎦x", - "type": "text" - }, - { - "block_id": "p943-b64", - "global_id": 28003, - "bbox": [ - 128.9, - 567.62, - 143.28, - 577.58 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p943-b65", - "global_id": 28004, - "bbox": [ - 273.8, - 588.72, - 331.6, - 599.0 - ], - "text": "y = [0\n0\n1]", - "type": "text" - }, - { - "block_id": "p943-b66", - "global_id": 28005, - "bbox": [ - 332.71, - 568.76, - 339.97, - 578.72 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p943-b67", - "global_id": 28006, - "bbox": [ - 332.71, - 586.69, - 339.97, - 596.65 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p943-b68", - "global_id": 28007, - "bbox": [ - 339.97, - 576.98, - 350.1, - 611.73 - ], - "text": "w1\nw2\nw3", - "type": "text" - }, - { - "block_id": "p943-b69", - "global_id": 28008, - "bbox": [ - 350.6, - 568.76, - 357.86, - 578.72 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p943-b70", - "global_id": 28009, - "bbox": [ - 350.6, - 586.69, - 357.86, - 596.65 - ], - "text": "⎦", - "type": "text" - } - ] - }, - { - "page_num": 944, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p944-b0", - "global_id": 28010, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "924\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p944-b1", - "global_id": 28011, - "bbox": [ - 103.16, - 86.28, - 477.01, - 120.24 - ], - "text": "Parallel Realization (Diagonal Representation)\nThe three integrator outputs z1, z2, and z3 in Fig. 10.5d are the state variables. The state\nequations are", - "type": "text" - }, - { - "block_id": "p944-b2", - "global_id": 28012, - "bbox": [ - 262.26, - 131.78, - 312.9, - 143.24 - ], - "text": "˙z1 = −z1 + x", - "type": "text" - }, - { - "block_id": "p944-b3", - "global_id": 28013, - "bbox": [ - 262.26, - 146.73, - 317.89, - 173.13 - ], - "text": "˙z2 = −3z2 + x\n˙z3 = −4z3 + x", - "type": "text" - }, - { - "block_id": "p944-b4", - "global_id": 28014, - "bbox": [ - 103.16, - 184.01, - 206.11, - 193.97 - ], - "text": "and the output equation is", - "type": "text" - }, - { - "block_id": "p944-b5", - "global_id": 28015, - "bbox": [ - 250.92, - 194.2, - 271.89, - 205.82 - ], - "text": "y = 4", - "type": "text" - }, - { - "block_id": "p944-b6", - "global_id": 28016, - "bbox": [ - 268.4, - 194.2, - 328.77, - 208.74 - ], - "text": "3z1 −2z2 + 2\n3z3", - "type": "text" - }, - { - "block_id": "p944-b7", - "global_id": 28017, - "bbox": [ - 103.16, - 214.89, - 242.6, - 224.85 - ], - "text": "In matrix form, these equations are", - "type": "text" - }, - { - "block_id": "p944-b8", - "global_id": 28018, - "bbox": [ - 206.9, - 228.04, - 214.16, - 238.01 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p944-b9", - "global_id": 28019, - "bbox": [ - 206.9, - 245.98, - 214.16, - 255.94 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p944-b10", - "global_id": 28020, - "bbox": [ - 214.16, - 235.95, - 221.53, - 271.02 - ], - "text": "˙z1\n˙z2\n˙z3", - "type": "text" - }, - { - "block_id": "p944-b11", - "global_id": 28021, - "bbox": [ - 222.02, - 228.04, - 229.28, - 238.01 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p944-b12", - "global_id": 28022, - "bbox": [ - 222.02, - 245.98, - 239.1, - 257.98 - ], - "text": "⎦=", - "type": "text" - }, - { - "block_id": "p944-b13", - "global_id": 28023, - "bbox": [ - 241.14, - 228.04, - 248.41, - 238.01 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p944-b14", - "global_id": 28024, - "bbox": [ - 241.14, - 245.98, - 248.41, - 255.94 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p944-b15", - "global_id": 28025, - "bbox": [ - 248.41, - 235.95, - 306.59, - 270.24 - ], - "text": "−1\n0\n0\n0\n−3\n0\n0\n0\n−4", - "type": "text" - }, - { - "block_id": "p944-b16", - "global_id": 28026, - "bbox": [ - 306.59, - 228.04, - 313.85, - 238.01 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p944-b17", - "global_id": 28027, - "bbox": [ - 306.59, - 245.98, - 313.85, - 255.94 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p944-b18", - "global_id": 28028, - "bbox": [ - 314.96, - 228.04, - 322.22, - 238.01 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p944-b19", - "global_id": 28029, - "bbox": [ - 314.96, - 245.98, - 322.22, - 255.94 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p944-b20", - "global_id": 28030, - "bbox": [ - 322.22, - 236.27, - 329.59, - 271.02 - ], - "text": "z1\nz2\nz3", - "type": "text" - }, - { - "block_id": "p944-b21", - "global_id": 28031, - "bbox": [ - 330.08, - 228.04, - 337.34, - 238.01 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p944-b22", - "global_id": 28032, - "bbox": [ - 330.08, - 245.98, - 346.67, - 257.98 - ], - "text": "⎦+", - "type": "text" - }, - { - "block_id": "p944-b23", - "global_id": 28033, - "bbox": [ - 348.22, - 228.04, - 355.48, - 238.01 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p944-b24", - "global_id": 28034, - "bbox": [ - 348.22, - 245.98, - 355.48, - 255.94 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p944-b25", - "global_id": 28035, - "bbox": [ - 355.48, - 236.37, - 360.46, - 270.24 - ], - "text": "1\n1\n1", - "type": "text" - }, - { - "block_id": "p944-b26", - "global_id": 28036, - "bbox": [ - 360.47, - 228.04, - 367.73, - 238.01 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p944-b27", - "global_id": 28037, - "bbox": [ - 360.47, - 245.98, - 373.26, - 258.29 - ], - "text": "⎦x", - "type": "text" - }, - { - "block_id": "p944-b28", - "global_id": 28038, - "bbox": [ - 224.85, - 287.87, - 239.1, - 298.14 - ], - "text": "y =", - "type": "text" - }, - { - "block_id": "p944-b29", - "global_id": 28039, - "bbox": [ - 241.14, - 279.86, - 249.75, - 293.59 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p944-b30", - "global_id": 28040, - "bbox": [ - 246.26, - 279.86, - 293.43, - 301.15 - ], - "text": "3\n−2\n2\n3", - "type": "text" - }, - { - "block_id": "p944-b31", - "global_id": 28041, - "bbox": [ - 294.54, - 267.9, - 301.8, - 277.86 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p944-b32", - "global_id": 28042, - "bbox": [ - 294.54, - 285.83, - 301.8, - 295.8 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p944-b33", - "global_id": 28043, - "bbox": [ - 301.8, - 276.12, - 309.17, - 310.87 - ], - "text": "z1\nz2\nz3", - "type": "text" - }, - { - "block_id": "p944-b34", - "global_id": 28044, - "bbox": [ - 309.66, - 267.9, - 316.92, - 277.86 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p944-b35", - "global_id": 28045, - "bbox": [ - 309.66, - 285.83, - 316.92, - 295.8 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p944-b36", - "global_id": 28046, - "bbox": [ - 101.84, - 348.91, - 490.4, - 410.89 - ], - "text": "A GENERAL CASE\nIt is clear that a system has several state-space descriptions. Notable among these are the variables\nobtained from the DFII, its transpose, and the diagonalized variables (in the parallel realization).\nState equations in these forms can be written immediately by inspection of the transfer function.\nConsider the general Nth-order transfer function", - "type": "text" - }, - { - "block_id": "p944-b37", - "global_id": 28047, - "bbox": [ - 207.31, - 418.67, - 374.59, - 439.55 - ], - "text": "H(s) = b0sN + b1sN−1 + · · · + bN−1s + bN", - "type": "text" - }, - { - "block_id": "p944-b38", - "global_id": 28048, - "bbox": [ - 243.8, - 433.47, - 370.11, - 447.5 - ], - "text": "sN + a1sN−1 + · · · + aN−1s + aN", - "type": "text" - }, - { - "block_id": "p944-b39", - "global_id": 28049, - "bbox": [ - 461.34, - 429.69, - 490.38, - 439.65 - ], - "text": "(10.24)", - "type": "text" - }, - { - "block_id": "p944-b40", - "global_id": 28050, - "bbox": [ - 228.3, - 447.16, - 374.59, - 468.03 - ], - "text": "= b0sN + b1sN−1 + · · · + bN−1s + bN", - "type": "text" - }, - { - "block_id": "p944-b41", - "global_id": 28051, - "bbox": [ - 251.56, - 464.83, - 363.27, - 475.98 - ], - "text": "(s −λ1)(s −λ2)· · ·(s −λN)", - "type": "text" - }, - { - "block_id": "p944-b42", - "global_id": 28052, - "bbox": [ - 228.3, - 477.5, - 282.83, - 502.4 - ], - "text": "= b0 +\nk1\ns −λ1", - "type": "text" - }, - { - "block_id": "p944-b43", - "global_id": 28053, - "bbox": [ - 286.07, - 477.5, - 320.28, - 502.4 - ], - "text": "+\nk2\ns −λ2", - "type": "text" - }, - { - "block_id": "p944-b44", - "global_id": 28054, - "bbox": [ - 323.51, - 477.5, - 382.82, - 502.33 - ], - "text": "+ · · · +\nkN\ns −λN", - "type": "text" - }, - { - "block_id": "p944-b45", - "global_id": 28055, - "bbox": [ - 461.34, - 484.59, - 490.38, - 494.55 - ], - "text": "(10.25)", - "type": "text" - }, - { - "block_id": "p944-b46", - "global_id": 28056, - "bbox": [ - 101.85, - 510.92, - 490.4, - 533.26 - ], - "text": "The realizations of H(s) found by using direct form II [Eq. (10.24)] and the parallel form\n[Eq. (10.25)] appear in Figs. 10.6a and 10.6b, respectively.", - "type": "text" - }, - { - "block_id": "p944-b47", - "global_id": 28057, - "bbox": [ - 101.84, - 534.83, - 490.39, - 557.16 - ], - "text": "The N integrator outputs q1, q2, . . . , qN in Fig. 10.6a are the state variables. By inspection of\nthis figure, we obtain", - "type": "text" - }, - { - "block_id": "p944-b48", - "global_id": 28058, - "bbox": [ - 202.73, - 568.58, - 232.91, - 591.98 - ], - "text": "˙q1 = q2\n˙q2 = q3", - "type": "text" - }, - { - "block_id": "p944-b49", - "global_id": 28059, - "bbox": [ - 192.22, - 591.52, - 234.08, - 622.48 - ], - "text": "...\n˙qN−1 = qN", - "type": "text" - }, - { - "block_id": "p944-b50", - "global_id": 28060, - "bbox": [ - 201.15, - 622.98, - 399.97, - 634.44 - ], - "text": "˙qN = −aNq1 −aN−1q2 −· · · −a2qN−1 −a1qN + x", - "type": "text" - } - ] - }, - { - "page_num": 945, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p945-b0", - "global_id": 28061, - "bbox": [ - 236.03, - 62.89, - 516.15, - 71.98 - ], - "text": "10.3\nA Systematic Procedure to Determine State Equations\n925", - "type": "text" - }, - { - "block_id": "p945-b1", - "global_id": 28062, - "bbox": [ - 409.13, - 358.75, - 418.94, - 366.75 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p945-b2", - "global_id": 28063, - "bbox": [ - 237.06, - 149.72, - 244.06, - 159.32 - ], - "text": "b1", - "type": "text" - }, - { - "block_id": "p945-b3", - "global_id": 28064, - "bbox": [ - 237.06, - 306.22, - 252.34, - 315.82 - ], - "text": "bN1", - "type": "text" - }, - { - "block_id": "p945-b4", - "global_id": 28065, - "bbox": [ - 239.39, - 343.51, - 247.4, - 353.05 - ], - "text": "bN", - "type": "text" - }, - { - "block_id": "p945-b5", - "global_id": 28066, - "bbox": [ - 195.13, - 149.39, - 208.8, - 159.21 - ], - "text": "a1", - "type": "text" - }, - { - "block_id": "p945-b6", - "global_id": 28067, - "bbox": [ - 192.13, - 305.92, - 214.08, - 315.74 - ], - "text": "aN1", - "type": "text" - }, - { - "block_id": "p945-b7", - "global_id": 28068, - "bbox": [ - 195.13, - 343.29, - 209.8, - 353.05 - ], - "text": "aN", - "type": "text" - }, - { - "block_id": "p945-b8", - "global_id": 28069, - "bbox": [ - 159.46, - 92.53, - 285.48, - 100.53 - ], - "text": "x\ny", - "type": "text" - }, - { - "block_id": "p945-b9", - "global_id": 28070, - "bbox": [ - 195.13, - 187.0, - 244.06, - 196.82 - ], - "text": "a2\nb2", - "type": "text" - }, - { - "block_id": "p945-b10", - "global_id": 28071, - "bbox": [ - 224.35, - 292.75, - 231.35, - 302.36 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p945-b11", - "global_id": 28072, - "bbox": [ - 218.86, - 343.51, - 225.86, - 353.11 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p945-b12", - "global_id": 28073, - "bbox": [ - 195.13, - 248.75, - 245.4, - 258.51 - ], - "text": "bm\nam", - "type": "text" - }, - { - "block_id": "p945-b13", - "global_id": 28074, - "bbox": [ - 203.01, - 234.68, - 219.34, - 244.29 - ], - "text": "qm1", - "type": "text" - }, - { - "block_id": "p945-b14", - "global_id": 28075, - "bbox": [ - 204.06, - 174.69, - 219.34, - 184.29 - ], - "text": "qN1", - "type": "text" - }, - { - "block_id": "p945-b15", - "global_id": 28076, - "bbox": [ - 211.33, - 135.6, - 219.34, - 145.15 - ], - "text": "qN", - "type": "text" - }, - { - "block_id": "p945-b16", - "global_id": 28077, - "bbox": [ - 219.03, - 91.84, - 227.03, - 101.39 - ], - "text": "qN", - "type": "text" - }, - { - "block_id": "p945-b17", - "global_id": 28078, - "bbox": [ - 220.4, - 87.43, - 222.4, - 95.43 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p945-b18", - "global_id": 28079, - "bbox": [ - 221.08, - 119.41, - 225.08, - 134.45 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b19", - "global_id": 28080, - "bbox": [ - 221.08, - 219.66, - 225.08, - 234.7 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b20", - "global_id": 28081, - "bbox": [ - 221.08, - 277.16, - 225.08, - 292.2 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b21", - "global_id": 28082, - "bbox": [ - 221.08, - 316.16, - 225.08, - 331.2 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b22", - "global_id": 28083, - "bbox": [ - 221.08, - 158.66, - 225.08, - 173.7 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b33", - "global_id": 28084, - "bbox": [ - 241.11, - 91.92, - 396.17, - 101.53 - ], - "text": "b0\nb0", - "type": "text" - }, - { - "block_id": "p945-b34", - "global_id": 28085, - "bbox": [ - 411.64, - 228.37, - 417.75, - 237.98 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p945-b35", - "global_id": 28086, - "bbox": [ - 428.92, - 175.56, - 435.03, - 185.17 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p945-b36", - "global_id": 28087, - "bbox": [ - 428.11, - 112.06, - 434.66, - 121.67 - ], - "text": "k1", - "type": "text" - }, - { - "block_id": "p945-b37", - "global_id": 28088, - "bbox": [ - 413.34, - 162.66, - 419.46, - 172.27 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p945-b38", - "global_id": 28089, - "bbox": [ - 412.33, - 306.26, - 418.44, - 315.8 - ], - "text": "zn", - "type": "text" - }, - { - "block_id": "p945-b39", - "global_id": 28090, - "bbox": [ - 428.92, - 257.41, - 435.47, - 266.96 - ], - "text": "kn", - "type": "text" - }, - { - "block_id": "p945-b40", - "global_id": 28091, - "bbox": [ - 322.77, - 177.36, - 505.07, - 185.36 - ], - "text": "x\ny", - "type": "text" - }, - { - "block_id": "p945-b41", - "global_id": 28092, - "bbox": [ - 382.8, - 219.4, - 390.25, - 229.03 - ], - "text": "l2", - "type": "text" - }, - { - "block_id": "p945-b42", - "global_id": 28093, - "bbox": [ - 385.17, - 299.73, - 392.62, - 309.3 - ], - "text": "ln", - "type": "text" - }, - { - "block_id": "p945-b43", - "global_id": 28094, - "bbox": [ - 406.64, - 175.56, - 412.75, - 185.17 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p945-b44", - "global_id": 28095, - "bbox": [ - 407.49, - 171.15, - 409.49, - 179.15 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p945-b45", - "global_id": 28096, - "bbox": [ - 406.64, - 112.06, - 412.75, - 121.67 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p945-b46", - "global_id": 28097, - "bbox": [ - 407.49, - 107.64, - 409.49, - 115.64 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p945-b47", - "global_id": 28098, - "bbox": [ - 406.64, - 257.5, - 412.75, - 267.04 - ], - "text": "zn", - "type": "text" - }, - { - "block_id": "p945-b48", - "global_id": 28099, - "bbox": [ - 407.49, - 253.08, - 409.49, - 261.08 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p945-b49", - "global_id": 28100, - "bbox": [ - 407.75, - 138.83, - 411.75, - 153.87 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b50", - "global_id": 28101, - "bbox": [ - 407.75, - 202.77, - 411.75, - 217.81 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b51", - "global_id": 28102, - "bbox": [ - 407.75, - 282.72, - 411.75, - 297.76 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p945-b56", - "global_id": 28103, - "bbox": [ - 382.8, - 156.06, - 390.25, - 165.68 - ], - "text": "l1", - "type": "text" - }, - { - "block_id": "p945-b57", - "global_id": 28104, - "bbox": [ - 218.43, - 358.75, - 227.31, - 366.75 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p945-b58", - "global_id": 28105, - "bbox": [ - 151.5, - 373.35, - 477.43, - 382.68 - ], - "text": "Figure 10.6 (a) Direct form II and (b) parallel realizations for an Nth-order LTIC system.", - "type": "text" - }, - { - "block_id": "p945-b59", - "global_id": 28106, - "bbox": [ - 127.59, - 410.92, - 185.98, - 420.98 - ], - "text": "and output y is", - "type": "text" - }, - { - "block_id": "p945-b60", - "global_id": 28107, - "bbox": [ - 241.95, - 434.31, - 400.86, - 445.77 - ], - "text": "y = bNq1 + bN−1q2 + · · · + b1qN + b0˙qN", - "type": "text" - }, - { - "block_id": "p945-b61", - "global_id": 28108, - "bbox": [ - 127.59, - 461.08, - 459.18, - 472.95 - ], - "text": "We can eliminate ˙qN in this output equation by using the last state equation to yield", - "type": "text" - }, - { - "block_id": "p945-b62", - "global_id": 28109, - "bbox": [ - 184.6, - 490.83, - 459.09, - 502.29 - ], - "text": "y = (bN −b0aN)q1 + (bN−1 −b0aN−1)q2 + · · · + (b1 −b0a1)qN + b0x", - "type": "text" - }, - { - "block_id": "p945-b63", - "global_id": 28110, - "bbox": [ - 191.07, - 505.87, - 338.3, - 519.38 - ], - "text": "= ˆbNq1 + ˆbN−1q2 + · · · + ˆb1qN + b0x", - "type": "text" - }, - { - "block_id": "p945-b64", - "global_id": 28111, - "bbox": [ - 127.59, - 537.77, - 316.0, - 551.29 - ], - "text": "where ˆbi = bi −b0ai. In matrix form, we obtain", - "type": "text" - }, - { - "block_id": "p945-b65", - "global_id": 28112, - "bbox": [ - 171.56, - 560.97, - 178.82, - 570.93 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p945-b66", - "global_id": 28113, - "bbox": [ - 171.56, - 578.44, - 178.82, - 618.75 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p945-b67", - "global_id": 28114, - "bbox": [ - 184.08, - 568.57, - 192.54, - 591.68 - ], - "text": "˙q1\n˙q2", - "type": "text" - }, - { - "block_id": "p945-b68", - "global_id": 28115, - "bbox": [ - 178.82, - 591.52, - 197.79, - 622.17 - ], - "text": "...\n˙qN−1", - "type": "text" - }, - { - "block_id": "p945-b69", - "global_id": 28116, - "bbox": [ - 183.29, - 622.98, - 192.91, - 634.06 - ], - "text": "˙qN", - "type": "text" - }, - { - "block_id": "p945-b70", - "global_id": 28117, - "bbox": [ - 198.3, - 560.97, - 205.56, - 570.93 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p945-b71", - "global_id": 28118, - "bbox": [ - 198.3, - 578.44, - 205.56, - 618.75 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p945-b72", - "global_id": 28119, - "bbox": [ - 207.6, - 595.87, - 215.37, - 605.83 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p945-b73", - "global_id": 28120, - "bbox": [ - 217.42, - 560.97, - 224.68, - 570.93 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p945-b74", - "global_id": 28121, - "bbox": [ - 217.42, - 578.44, - 224.68, - 618.75 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p945-b75", - "global_id": 28122, - "bbox": [ - 224.69, - 568.57, - 393.35, - 634.13 - ], - "text": "0\n1\n0\n· · ·\n0\n0\n0\n0\n1\n· · ·\n0\n0\n...\n...\n...\n· · ·\n...\n...\n0\n0\n0\n· · ·\n0\n1\n−aN\n−aN−1\n−aN−2\n· · ·\n−a2\n−a1", - "type": "text" - }, - { - "block_id": "p945-b76", - "global_id": 28123, - "bbox": [ - 393.86, - 560.97, - 401.12, - 570.93 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p945-b77", - "global_id": 28124, - "bbox": [ - 393.86, - 578.44, - 401.12, - 618.75 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p945-b78", - "global_id": 28125, - "bbox": [ - 402.23, - 560.97, - 409.49, - 570.93 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p945-b79", - "global_id": 28126, - "bbox": [ - 402.23, - 578.44, - 409.49, - 618.75 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p945-b80", - "global_id": 28127, - "bbox": [ - 414.75, - 568.88, - 423.21, - 591.68 - ], - "text": "q1\nq2", - "type": "text" - }, - { - "block_id": "p945-b81", - "global_id": 28128, - "bbox": [ - 409.49, - 591.52, - 428.46, - 622.17 - ], - "text": "...\nqN−1", - "type": "text" - }, - { - "block_id": "p945-b82", - "global_id": 28129, - "bbox": [ - 413.95, - 623.29, - 423.58, - 634.06 - ], - "text": "qN", - "type": "text" - }, - { - "block_id": "p945-b83", - "global_id": 28130, - "bbox": [ - 428.97, - 560.97, - 436.23, - 570.93 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p945-b84", - "global_id": 28131, - "bbox": [ - 428.97, - 578.44, - 436.23, - 618.75 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p945-b85", - "global_id": 28132, - "bbox": [ - 437.78, - 595.87, - 445.55, - 605.83 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p945-b86", - "global_id": 28133, - "bbox": [ - 447.09, - 560.97, - 454.36, - 570.93 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p945-b87", - "global_id": 28134, - "bbox": [ - 447.09, - 578.44, - 454.36, - 618.75 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p945-b88", - "global_id": 28135, - "bbox": [ - 454.36, - 568.99, - 459.35, - 633.35 - ], - "text": "0\n0\n...\n0\n1", - "type": "text" - }, - { - "block_id": "p945-b89", - "global_id": 28136, - "bbox": [ - 459.34, - 560.97, - 466.6, - 570.93 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p945-b90", - "global_id": 28137, - "bbox": [ - 459.34, - 578.44, - 466.6, - 618.74 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p945-b91", - "global_id": 28138, - "bbox": [ - 467.71, - 596.18, - 472.13, - 606.14 - ], - "text": "x", - "type": "text" - } - ] - }, - { - "page_num": 946, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p946-b0", - "global_id": 28139, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "926\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p946-b1", - "global_id": 28140, - "bbox": [ - 101.84, - 85.82, - 116.22, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p946-b2", - "global_id": 28141, - "bbox": [ - 218.45, - 115.78, - 323.32, - 128.98 - ], - "text": "y = [ˆbN\nˆbN−1\n· · ·\nˆb1]", - "type": "text" - }, - { - "block_id": "p946-b3", - "global_id": 28142, - "bbox": [ - 324.43, - 88.43, - 331.7, - 98.39 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p946-b4", - "global_id": 28143, - "bbox": [ - 324.43, - 105.89, - 331.7, - 134.25 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p946-b5", - "global_id": 28144, - "bbox": [ - 332.48, - 96.34, - 340.94, - 119.13 - ], - "text": "q1\nq2", - "type": "text" - }, - { - "block_id": "p946-b6", - "global_id": 28145, - "bbox": [ - 331.69, - 118.97, - 341.31, - 149.56 - ], - "text": "...\nqN", - "type": "text" - }, - { - "block_id": "p946-b7", - "global_id": 28146, - "bbox": [ - 342.23, - 88.43, - 349.5, - 98.39 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p946-b8", - "global_id": 28147, - "bbox": [ - 342.23, - 105.89, - 373.76, - 134.25 - ], - "text": "⎥⎥⎥⎦+ b0x", - "type": "text" - }, - { - "block_id": "p946-b9", - "global_id": 28148, - "bbox": [ - 101.84, - 157.59, - 490.39, - 179.93 - ], - "text": "In Fig. 10.6b, the N integrator outputs z1, z2, . . . , zN are the state variables. By inspection of\nthis figure, we obtain", - "type": "text" - }, - { - "block_id": "p946-b10", - "global_id": 28149, - "bbox": [ - 268.71, - 182.23, - 321.91, - 205.64 - ], - "text": "˙z1 = λ1z1 + x\n˙z2 = λ2z2 + x", - "type": "text" - }, - { - "block_id": "p946-b11", - "global_id": 28150, - "bbox": [ - 267.13, - 205.17, - 325.08, - 236.14 - ], - "text": "...\n˙zN = λNzN + x", - "type": "text" - }, - { - "block_id": "p946-b12", - "global_id": 28151, - "bbox": [ - 101.84, - 244.21, - 116.22, - 254.17 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p946-b13", - "global_id": 28152, - "bbox": [ - 227.0, - 256.91, - 365.21, - 268.37 - ], - "text": "y = k1z1 + k2z2 + · · · + kNzN + b0x", - "type": "text" - }, - { - "block_id": "p946-b14", - "global_id": 28153, - "bbox": [ - 101.85, - 277.03, - 187.39, - 289.59 - ], - "text": "or\n⎡", - "type": "text" - }, - { - "block_id": "p946-b15", - "global_id": 28154, - "bbox": [ - 180.13, - 297.11, - 187.39, - 337.42 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p946-b16", - "global_id": 28155, - "bbox": [ - 192.65, - 287.23, - 200.01, - 310.35 - ], - "text": "˙z1\n˙z2", - "type": "text" - }, - { - "block_id": "p946-b17", - "global_id": 28156, - "bbox": [ - 187.39, - 310.18, - 205.26, - 340.84 - ], - "text": "...\n˙zN−1", - "type": "text" - }, - { - "block_id": "p946-b18", - "global_id": 28157, - "bbox": [ - 191.85, - 341.65, - 200.38, - 352.73 - ], - "text": "˙zN", - "type": "text" - }, - { - "block_id": "p946-b19", - "global_id": 28158, - "bbox": [ - 205.76, - 279.63, - 213.02, - 289.59 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p946-b20", - "global_id": 28159, - "bbox": [ - 205.76, - 297.11, - 213.02, - 337.42 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p946-b21", - "global_id": 28160, - "bbox": [ - 215.07, - 314.54, - 222.84, - 324.5 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p946-b22", - "global_id": 28161, - "bbox": [ - 224.89, - 279.63, - 232.15, - 289.59 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p946-b23", - "global_id": 28162, - "bbox": [ - 224.89, - 297.11, - 232.15, - 337.42 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p946-b24", - "global_id": 28163, - "bbox": [ - 232.15, - 287.23, - 331.89, - 328.11 - ], - "text": "λ1\n0\n· · ·\n0\n0\n0\nλ2\n· · ·\n0\n0\n...", - "type": "text" - }, - { - "block_id": "p946-b25", - "global_id": 28164, - "bbox": [ - 255.02, - 310.18, - 305.2, - 328.11 - ], - "text": "...\n· · ·\n...", - "type": "text" - }, - { - "block_id": "p946-b26", - "global_id": 28165, - "bbox": [ - 234.38, - 310.18, - 333.99, - 352.73 - ], - "text": "...\n0\n0\n· · ·\nλN−1\n0\n0\n0\n· · ·\n0\nλN", - "type": "text" - }, - { - "block_id": "p946-b27", - "global_id": 28166, - "bbox": [ - 334.9, - 279.63, - 342.16, - 289.59 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p946-b28", - "global_id": 28167, - "bbox": [ - 334.9, - 297.11, - 342.16, - 337.42 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p946-b29", - "global_id": 28168, - "bbox": [ - 343.27, - 279.63, - 350.53, - 289.59 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p946-b30", - "global_id": 28169, - "bbox": [ - 343.27, - 297.11, - 350.53, - 337.42 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p946-b31", - "global_id": 28170, - "bbox": [ - 355.79, - 287.55, - 363.15, - 310.35 - ], - "text": "z1\nz2", - "type": "text" - }, - { - "block_id": "p946-b32", - "global_id": 28171, - "bbox": [ - 350.53, - 310.18, - 368.4, - 340.84 - ], - "text": "...\nzN−1", - "type": "text" - }, - { - "block_id": "p946-b33", - "global_id": 28172, - "bbox": [ - 355.0, - 341.96, - 363.53, - 352.73 - ], - "text": "zN", - "type": "text" - }, - { - "block_id": "p946-b34", - "global_id": 28173, - "bbox": [ - 368.91, - 279.63, - 376.17, - 289.59 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p946-b35", - "global_id": 28174, - "bbox": [ - 368.91, - 297.11, - 376.17, - 337.42 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p946-b36", - "global_id": 28175, - "bbox": [ - 377.72, - 314.54, - 385.49, - 324.5 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p946-b37", - "global_id": 28176, - "bbox": [ - 387.04, - 279.63, - 394.3, - 289.59 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p946-b38", - "global_id": 28177, - "bbox": [ - 387.04, - 297.11, - 394.3, - 337.42 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p946-b39", - "global_id": 28178, - "bbox": [ - 394.3, - 287.65, - 399.28, - 352.02 - ], - "text": "1\n1\n...\n1\n1", - "type": "text" - }, - { - "block_id": "p946-b40", - "global_id": 28179, - "bbox": [ - 399.28, - 279.63, - 406.55, - 289.59 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p946-b41", - "global_id": 28180, - "bbox": [ - 399.28, - 297.11, - 406.55, - 337.42 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p946-b42", - "global_id": 28181, - "bbox": [ - 407.65, - 314.85, - 490.38, - 324.92 - ], - "text": "x\n(10.26)", - "type": "text" - }, - { - "block_id": "p946-b43", - "global_id": 28182, - "bbox": [ - 101.84, - 361.17, - 116.22, - 371.13 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p946-b44", - "global_id": 28183, - "bbox": [ - 205.58, - 398.69, - 219.82, - 408.96 - ], - "text": "y =", - "type": "text" - }, - { - "block_id": "p946-b45", - "global_id": 28184, - "bbox": [ - 221.87, - 390.68, - 323.52, - 409.74 - ], - "text": "k1\nk2\n· · ·\nkN−1\nkN", - "type": "text" - }, - { - "block_id": "p946-b47", - "global_id": 28185, - "bbox": [ - 329.47, - 363.78, - 336.73, - 373.74 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p946-b48", - "global_id": 28186, - "bbox": [ - 329.47, - 381.25, - 336.73, - 421.56 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p946-b49", - "global_id": 28187, - "bbox": [ - 341.99, - 371.7, - 349.36, - 394.49 - ], - "text": "z1\nz2", - "type": "text" - }, - { - "block_id": "p946-b50", - "global_id": 28188, - "bbox": [ - 336.73, - 394.33, - 354.6, - 424.99 - ], - "text": "...\nzN−1", - "type": "text" - }, - { - "block_id": "p946-b51", - "global_id": 28189, - "bbox": [ - 341.2, - 426.1, - 349.73, - 436.87 - ], - "text": "zN", - "type": "text" - }, - { - "block_id": "p946-b52", - "global_id": 28190, - "bbox": [ - 355.1, - 363.78, - 362.37, - 373.74 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p946-b53", - "global_id": 28191, - "bbox": [ - 355.1, - 381.25, - 362.37, - 421.56 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p946-b54", - "global_id": 28192, - "bbox": [ - 363.91, - 398.69, - 386.63, - 409.84 - ], - "text": "+ b0x", - "type": "text" - }, - { - "block_id": "p946-b55", - "global_id": 28193, - "bbox": [ - 101.85, - 445.32, - 490.41, - 479.2 - ], - "text": "Observe that the diagonalized form of the state matrix [Eq. (10.26)] has the transfer function poles\nas its diagonal elements. The presence of repeated poles in H(s) will modify the procedure slightly.\nThe handling of these cases is discussed in Sec. 4.6.", - "type": "text" - }, - { - "block_id": "p946-b56", - "global_id": 28194, - "bbox": [ - 101.85, - 481.19, - 490.4, - 527.01 - ], - "text": "It is clear from the foregoing discussion that a state-space description is not unique. For\nany realization of H(s) obtained from integrators, scalar multipliers, and adders, a corresponding\nstate-space description exists. Since there are uncountable possible realizations of H(s), there are\nuncountable possible state-space descriptions.", - "type": "text" - }, - { - "block_id": "p946-b57", - "global_id": 28195, - "bbox": [ - 119.78, - 529.0, - 474.38, - 538.97 - ], - "text": "The advantages and drawbacks of various types of realization were discussed in Sec. 4.6.", - "type": "text" - }, - { - "block_id": "p946-b58", - "global_id": 28196, - "bbox": [ - 102.2, - 569.02, - 348.68, - 582.97 - ], - "text": "10.4 SOLUTION OF STATE EQUATIONS", - "type": "text" - }, - { - "block_id": "p946-b59", - "global_id": 28197, - "bbox": [ - 101.84, - 588.86, - 490.42, - 634.79 - ], - "text": "The state equations of a linear system are N simultaneous linear differential equations of the first\norder. We studied the techniques of solving linear differential equations in Chs. 2 and 4. The\nsame techniques can be applied to state equations without any modification. However, it is more\nconvenient to carry out the solution in the framework of matrix notation.", - "type": "text" - } - ] - }, - { - "page_num": 947, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p947-b0", - "global_id": 28198, - "bbox": [ - 341.6, - 62.89, - 516.12, - 71.98 - ], - "text": "10.4\nSolution of State Equations\n927", - "type": "text" - }, - { - "block_id": "p947-b1", - "global_id": 28199, - "bbox": [ - 127.59, - 85.82, - 516.12, - 119.69 - ], - "text": "These equations can be solved in both the time and frequency domains (Laplace transform).\nThe latter is relatively easier to deal with than the time-domain solution. For this reason, we shall\nfirst consider the Laplace transform solution.", - "type": "text" - }, - { - "block_id": "p947-b2", - "global_id": 28200, - "bbox": [ - 127.59, - 144.17, - 417.96, - 156.13 - ], - "text": "10.4-1 Laplace Transform Solution of State Equations", - "type": "text" - }, - { - "block_id": "p947-b3", - "global_id": 28201, - "bbox": [ - 127.59, - 162.16, - 324.33, - 172.22 - ], - "text": "The ith state equation [Eq. (10.14)] is of the form", - "type": "text" - }, - { - "block_id": "p947-b4", - "global_id": 28202, - "bbox": [ - 202.29, - 183.23, - 516.12, - 194.69 - ], - "text": "˙qi = ai1q1 + ai2q2 + · · · + aiNqN + bi1x1 + bi2x2 + · · · + bijxj\n(10.27)", - "type": "text" - }, - { - "block_id": "p947-b5", - "global_id": 28203, - "bbox": [ - 127.59, - 205.04, - 353.55, - 215.0 - ], - "text": "We shall take the Laplace transform of this equation. Let", - "type": "text" - }, - { - "block_id": "p947-b6", - "global_id": 28204, - "bbox": [ - 290.88, - 226.01, - 352.84, - 237.09 - ], - "text": "qi(t) ⇐⇒Qi(s)", - "type": "text" - }, - { - "block_id": "p947-b7", - "global_id": 28205, - "bbox": [ - 127.59, - 247.81, - 153.87, - 257.77 - ], - "text": "so that", - "type": "text" - }, - { - "block_id": "p947-b8", - "global_id": 28206, - "bbox": [ - 273.59, - 259.34, - 370.12, - 270.42 - ], - "text": "˙qi(t) ⇐⇒sQi(s) −qi(0)", - "type": "text" - }, - { - "block_id": "p947-b9", - "global_id": 28207, - "bbox": [ - 127.59, - 278.42, - 161.35, - 288.39 - ], - "text": "Also, let", - "type": "text" - }, - { - "block_id": "p947-b10", - "global_id": 28208, - "bbox": [ - 291.72, - 289.96, - 352.01, - 301.04 - ], - "text": "xi(t) ⇐⇒Xi(s)", - "type": "text" - }, - { - "block_id": "p947-b11", - "global_id": 28209, - "bbox": [ - 127.59, - 309.04, - 302.99, - 319.01 - ], - "text": "The Laplace transform of Eq. (10.27) yields", - "type": "text" - }, - { - "block_id": "p947-b12", - "global_id": 28210, - "bbox": [ - 192.09, - 330.01, - 451.65, - 341.16 - ], - "text": "sQi(s) −qi(0) = ai1Q1(s) + ai2Q2(s) + · · · + aiNQN(s) + bi1X1(s)", - "type": "text" - }, - { - "block_id": "p947-b13", - "global_id": 28211, - "bbox": [ - 261.02, - 344.95, - 367.73, - 356.1 - ], - "text": "+ bi2X2(s) + · · · + bijXj(s)", - "type": "text" - }, - { - "block_id": "p947-b14", - "global_id": 28212, - "bbox": [ - 127.59, - 366.65, - 387.2, - 376.71 - ], - "text": "Taking the Laplace transforms of all N state equations, we obtain", - "type": "text" - }, - { - "block_id": "p947-b15", - "global_id": 28213, - "bbox": [ - 199.71, - 408.93, - 203.59, - 418.89 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p947-b16", - "global_id": 28214, - "bbox": [ - 204.7, - 379.68, - 211.96, - 389.65 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p947-b17", - "global_id": 28215, - "bbox": [ - 204.7, - 397.16, - 211.96, - 425.51 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p947-b18", - "global_id": 28216, - "bbox": [ - 212.75, - 387.29, - 235.24, - 410.4 - ], - "text": "Q1(s)\nQ2(s)", - "type": "text" - }, - { - "block_id": "p947-b19", - "global_id": 28217, - "bbox": [ - 211.96, - 410.23, - 236.03, - 440.82 - ], - "text": "...\nQN(s)", - "type": "text" - }, - { - "block_id": "p947-b20", - "global_id": 28218, - "bbox": [ - 236.03, - 379.68, - 243.29, - 389.65 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p947-b21", - "global_id": 28219, - "bbox": [ - 236.03, - 397.16, - 243.29, - 425.51 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p947-b23", - "global_id": 28220, - "bbox": [ - 217.71, - 449.28, - 230.27, - 456.47 - ], - "text": "Q(s)", - "type": "text" - }, - { - "block_id": "p947-b24", - "global_id": 28221, - "bbox": [ - 244.39, - 408.62, - 252.17, - 418.58 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p947-b25", - "global_id": 28222, - "bbox": [ - 253.27, - 379.68, - 260.53, - 389.65 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p947-b26", - "global_id": 28223, - "bbox": [ - 253.27, - 397.16, - 260.53, - 425.51 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p947-b27", - "global_id": 28224, - "bbox": [ - 261.32, - 387.29, - 282.71, - 410.4 - ], - "text": "q1(0)\nq2(0)", - "type": "text" - }, - { - "block_id": "p947-b28", - "global_id": 28225, - "bbox": [ - 260.53, - 410.23, - 283.5, - 440.82 - ], - "text": "...\nqN(0)", - "type": "text" - }, - { - "block_id": "p947-b29", - "global_id": 28226, - "bbox": [ - 283.5, - 379.68, - 290.76, - 389.65 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p947-b30", - "global_id": 28227, - "bbox": [ - 283.5, - 397.16, - 290.76, - 425.51 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p947-b31", - "global_id": 28228, - "bbox": [ - 253.27, - 437.33, - 290.76, - 457.18 - ], - "text": "q(0)", - "type": "text" - }, - { - "block_id": "p947-b32", - "global_id": 28229, - "bbox": [ - 292.81, - 408.62, - 300.58, - 418.58 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p947-b33", - "global_id": 28230, - "bbox": [ - 303.29, - 379.68, - 310.55, - 389.65 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p947-b34", - "global_id": 28231, - "bbox": [ - 303.29, - 397.16, - 310.55, - 425.51 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p947-b35", - "global_id": 28232, - "bbox": [ - 311.35, - 387.29, - 395.28, - 428.17 - ], - "text": "a11\na12\n· · ·\na1N\na21\na22\n· · ·\na2N\n...", - "type": "text" - }, - { - "block_id": "p947-b36", - "global_id": 28233, - "bbox": [ - 310.55, - 410.23, - 395.86, - 440.89 - ], - "text": "...\n· · ·\n...\naN1\naN2\n· · ·\naNN", - "type": "text" - }, - { - "block_id": "p947-b37", - "global_id": 28234, - "bbox": [ - 396.78, - 379.68, - 404.05, - 389.65 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p947-b38", - "global_id": 28235, - "bbox": [ - 396.78, - 397.16, - 404.05, - 425.51 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p947-b39", - "global_id": 28236, - "bbox": [ - 303.29, - 437.33, - 404.06, - 454.9 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p947-b40", - "global_id": 28237, - "bbox": [ - 404.71, - 379.68, - 411.97, - 389.65 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p947-b41", - "global_id": 28238, - "bbox": [ - 404.71, - 397.16, - 411.97, - 425.51 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p947-b42", - "global_id": 28239, - "bbox": [ - 412.77, - 387.29, - 435.26, - 410.4 - ], - "text": "Q1(s)\nQ2(s)", - "type": "text" - }, - { - "block_id": "p947-b43", - "global_id": 28240, - "bbox": [ - 411.98, - 410.23, - 436.05, - 440.82 - ], - "text": "...\nQN(s)", - "type": "text" - }, - { - "block_id": "p947-b44", - "global_id": 28241, - "bbox": [ - 436.04, - 379.68, - 443.3, - 389.65 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p947-b45", - "global_id": 28242, - "bbox": [ - 436.04, - 397.16, - 443.3, - 425.51 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p947-b46", - "global_id": 28243, - "bbox": [ - 404.71, - 437.33, - 443.31, - 454.9 - ], - "text": "Q(s)", - "type": "text" - }, - { - "block_id": "p947-b47", - "global_id": 28244, - "bbox": [ - 303.56, - 486.85, - 308.22, - 492.83 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p947-b48", - "global_id": 28245, - "bbox": [ - 309.82, - 454.73, - 317.08, - 464.69 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p947-b49", - "global_id": 28246, - "bbox": [ - 309.82, - 472.2, - 317.08, - 500.55 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p947-b50", - "global_id": 28247, - "bbox": [ - 317.87, - 462.33, - 399.09, - 503.21 - ], - "text": "b11\nb12\n· · ·\nb1j\nb21\nb22\n· · ·\nb2j\n...", - "type": "text" - }, - { - "block_id": "p947-b51", - "global_id": 28248, - "bbox": [ - 317.08, - 485.27, - 399.68, - 515.93 - ], - "text": "...\n· · ·\n...\nbN1\nbN2\n· · ·\nbNj", - "type": "text" - }, - { - "block_id": "p947-b52", - "global_id": 28249, - "bbox": [ - 400.18, - 454.73, - 407.44, - 464.69 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p947-b53", - "global_id": 28250, - "bbox": [ - 400.18, - 472.2, - 407.44, - 500.55 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p947-b54", - "global_id": 28251, - "bbox": [ - 309.82, - 512.37, - 407.45, - 529.94 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p947-b55", - "global_id": 28252, - "bbox": [ - 408.11, - 454.73, - 415.37, - 464.69 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p947-b56", - "global_id": 28253, - "bbox": [ - 408.11, - 472.2, - 415.37, - 500.55 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p947-b57", - "global_id": 28254, - "bbox": [ - 415.37, - 462.33, - 436.75, - 485.44 - ], - "text": "X1(s)\nX2(s)", - "type": "text" - }, - { - "block_id": "p947-b58", - "global_id": 28255, - "bbox": [ - 416.14, - 485.27, - 435.98, - 515.86 - ], - "text": "...\nXj(s)", - "type": "text" - }, - { - "block_id": "p947-b59", - "global_id": 28256, - "bbox": [ - 436.74, - 454.73, - 444.01, - 464.69 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p947-b60", - "global_id": 28257, - "bbox": [ - 436.74, - 472.2, - 444.01, - 500.55 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p947-b61", - "global_id": 28258, - "bbox": [ - 408.11, - 512.37, - 444.02, - 529.94 - ], - "text": "X(s)", - "type": "text" - }, - { - "block_id": "p947-b62", - "global_id": 28259, - "bbox": [ - 127.59, - 539.9, - 298.42, - 549.86 - ], - "text": "Defining the vectors, as indicated, we have", - "type": "text" - }, - { - "block_id": "p947-b63", - "global_id": 28260, - "bbox": [ - 259.14, - 560.87, - 384.59, - 571.25 - ], - "text": "sQ(s) −q(0) = AQ(s) + BX(s)", - "type": "text" - }, - { - "block_id": "p947-b64", - "global_id": 28261, - "bbox": [ - 127.6, - 582.67, - 135.89, - 592.63 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p947-b65", - "global_id": 28262, - "bbox": [ - 259.14, - 594.22, - 384.59, - 604.59 - ], - "text": "sQ(s) −AQ(s) = q(0) + BX(s)", - "type": "text" - }, - { - "block_id": "p947-b66", - "global_id": 28263, - "bbox": [ - 127.6, - 613.29, - 141.98, - 623.25 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p947-b67", - "global_id": 28264, - "bbox": [ - 263.08, - 624.83, - 380.66, - 635.21 - ], - "text": "(sI −A)Q(s) = x(0) + BX(s)", - "type": "text" - } - ] - }, - { - "page_num": 948, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p948-b0", - "global_id": 28265, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "928\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p948-b1", - "global_id": 28266, - "bbox": [ - 101.84, - 85.54, - 354.86, - 95.91 - ], - "text": "where I is the N × N identity matrix. Solving for Q(s), we have", - "type": "text" - }, - { - "block_id": "p948-b2", - "global_id": 28267, - "bbox": [ - 229.01, - 109.31, - 363.22, - 121.42 - ], - "text": "Q(s) = (sI −A)−1[q(0) + BX(s)]", - "type": "text" - }, - { - "block_id": "p948-b3", - "global_id": 28268, - "bbox": [ - 250.13, - 125.98, - 490.38, - 136.36 - ], - "text": "= (s)[q(0) + BX(s)]\n(10.28)", - "type": "text" - }, - { - "block_id": "p948-b4", - "global_id": 28269, - "bbox": [ - 101.85, - 151.9, - 126.18, - 161.87 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p948-b5", - "global_id": 28270, - "bbox": [ - 259.25, - 167.1, - 332.49, - 179.12 - ], - "text": "(s) = (sI −A)−1", - "type": "text" - }, - { - "block_id": "p948-b6", - "global_id": 28271, - "bbox": [ - 101.84, - 191.75, - 196.19, - 201.72 - ], - "text": "Thus, from Eq. (10.28),", - "type": "text" - }, - { - "block_id": "p948-b7", - "global_id": 28272, - "bbox": [ - 234.46, - 208.67, - 357.78, - 219.05 - ], - "text": "Q(s) = (s)q(0) + (s)BX(s)", - "type": "text" - }, - { - "block_id": "p948-b8", - "global_id": 28273, - "bbox": [ - 101.85, - 231.61, - 116.22, - 241.57 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p948-b9", - "global_id": 28274, - "bbox": [ - 213.55, - 246.8, - 301.43, - 264.48 - ], - "text": "q(t) = L−1[(s)]q(0)", - "type": "text" - }, - { - "block_id": "p948-b10", - "global_id": 28275, - "bbox": [ - 247.89, - 266.93, - 294.86, - 272.91 - ], - "text": "zero-input response", - "type": "text" - }, - { - "block_id": "p948-b11", - "global_id": 28276, - "bbox": [ - 302.53, - 246.8, - 378.7, - 272.65 - ], - "text": "+L−1[(s)BX(s)]\n\n\n\nzero-state response", - "type": "text" - }, - { - "block_id": "p948-b12", - "global_id": 28277, - "bbox": [ - 461.34, - 248.94, - 490.38, - 258.9 - ], - "text": "(10.29)", - "type": "text" - }, - { - "block_id": "p948-b13", - "global_id": 28278, - "bbox": [ - 101.85, - 284.64, - 490.39, - 330.47 - ], - "text": "Equation (10.29) gives the desired solution. Observe the two components of the solution.\nThe first component yields q(t) when the input x(t) = 0. Hence, the first component is the\nzero-input response. In a similar manner, we see that the second component is the zero-state\nresponse.", - "type": "text" - }, - { - "block_id": "p948-b14", - "global_id": 28279, - "bbox": [ - 76.77, - 373.22, - 434.28, - 385.17 - ], - "text": "EXAMPLE 10.6\nLaplace Transform Solution to State Equations", - "type": "text" - }, - { - "block_id": "p948-b15", - "global_id": 28280, - "bbox": [ - 103.16, - 401.42, - 477.03, - 423.76 - ], - "text": "Using the Laplace transform, find the state vector q(t) for the system whose state equation is\ngiven by", - "type": "text" - }, - { - "block_id": "p948-b16", - "global_id": 28281, - "bbox": [ - 263.77, - 425.32, - 316.41, - 435.63 - ], - "text": "˙q = Aq + Bx", - "type": "text" - }, - { - "block_id": "p948-b17", - "global_id": 28282, - "bbox": [ - 103.16, - 444.67, - 127.49, - 454.64 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p948-b18", - "global_id": 28283, - "bbox": [ - 194.95, - 460.02, - 211.97, - 470.31 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p948-b20", - "global_id": 28284, - "bbox": [ - 219.44, - 452.79, - 259.89, - 476.48 - ], - "text": "−12\n2\n3\n−36\n−1", - "type": "text" - }, - { - "block_id": "p948-b21", - "global_id": 28285, - "bbox": [ - 259.89, - 446.03, - 265.32, - 456.0 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p948-b22", - "global_id": 28286, - "bbox": [ - 286.35, - 460.02, - 302.81, - 470.31 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p948-b23", - "global_id": 28287, - "bbox": [ - 304.86, - 446.03, - 314.97, - 459.76 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p948-b24", - "global_id": 28288, - "bbox": [ - 310.74, - 460.35, - 315.72, - 476.48 - ], - "text": "3\n1", - "type": "text" - }, - { - "block_id": "p948-b25", - "global_id": 28289, - "bbox": [ - 316.18, - 446.03, - 321.61, - 456.0 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p948-b26", - "global_id": 28290, - "bbox": [ - 342.64, - 460.02, - 385.22, - 470.31 - ], - "text": "x(t) = u(t)", - "type": "text" - }, - { - "block_id": "p948-b27", - "global_id": 28291, - "bbox": [ - 103.17, - 484.51, - 303.04, - 495.66 - ], - "text": "and the initial conditions are q1(0) = 2, q2(0) = 1.", - "type": "text" - }, - { - "block_id": "p948-b28", - "global_id": 28292, - "bbox": [ - 103.16, - 517.8, - 209.88, - 527.76 - ], - "text": "From Eq. (10.28), we have", - "type": "text" - }, - { - "block_id": "p948-b29", - "global_id": 28293, - "bbox": [ - 234.73, - 544.29, - 345.47, - 554.66 - ], - "text": "Q(s) = (s)[q(0) + BX(s)]", - "type": "text" - }, - { - "block_id": "p948-b30", - "global_id": 28294, - "bbox": [ - 103.17, - 571.19, - 225.12, - 581.56 - ], - "text": "Let us first find (s). We have", - "type": "text" - }, - { - "block_id": "p948-b31", - "global_id": 28295, - "bbox": [ - 180.0, - 603.51, - 228.99, - 613.81 - ], - "text": "(sI −A) = s", - "type": "text" - }, - { - "block_id": "p948-b32", - "global_id": 28296, - "bbox": [ - 230.1, - 589.53, - 255.45, - 619.77 - ], - "text": "1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p948-b33", - "global_id": 28297, - "bbox": [ - 255.45, - 589.53, - 260.88, - 599.49 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p948-b34", - "global_id": 28298, - "bbox": [ - 262.43, - 603.52, - 270.2, - 613.48 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p948-b36", - "global_id": 28299, - "bbox": [ - 277.18, - 596.29, - 317.62, - 619.98 - ], - "text": "−12\n2\n3\n−36\n−1", - "type": "text" - }, - { - "block_id": "p948-b37", - "global_id": 28300, - "bbox": [ - 317.62, - 589.53, - 323.05, - 599.49 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p948-b38", - "global_id": 28301, - "bbox": [ - 325.1, - 603.52, - 332.87, - 613.48 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p948-b40", - "global_id": 28302, - "bbox": [ - 340.35, - 596.29, - 390.51, - 608.01 - ], - "text": "s + 12\n−2", - "type": "text" - }, - { - "block_id": "p948-b41", - "global_id": 28303, - "bbox": [ - 347.73, - 603.85, - 394.75, - 619.98 - ], - "text": "3\n36\ns + 1", - "type": "text" - }, - { - "block_id": "p948-b42", - "global_id": 28304, - "bbox": [ - 394.75, - 589.53, - 400.18, - 599.49 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 949, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p949-b0", - "global_id": 28305, - "bbox": [ - 341.6, - 62.89, - 516.12, - 71.98 - ], - "text": "10.4\nSolution of State Equations\n929", - "type": "text" - }, - { - "block_id": "p949-b1", - "global_id": 28306, - "bbox": [ - 128.9, - 86.24, - 143.28, - 96.21 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p949-b2", - "global_id": 28307, - "bbox": [ - 225.79, - 115.95, - 309.35, - 130.37 - ], - "text": "(s) = (sI −A)−1 =", - "type": "text" - }, - { - "block_id": "p949-b3", - "global_id": 28308, - "bbox": [ - 311.4, - 103.09, - 352.44, - 124.39 - ], - "text": "s+1\n(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b4", - "global_id": 28309, - "bbox": [ - 318.78, - 109.0, - 398.47, - 141.73 - ], - "text": "2/3\n(s+4)(s+9)\n−36\n(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b5", - "global_id": 28310, - "bbox": [ - 364.81, - 126.91, - 398.47, - 141.73 - ], - "text": "s+12\n(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b7", - "global_id": 28311, - "bbox": [ - 128.9, - 156.07, - 214.46, - 166.45 - ], - "text": "Now, q(0) is given as", - "type": "text" - }, - { - "block_id": "p949-b8", - "global_id": 28312, - "bbox": [ - 293.0, - 173.68, - 320.78, - 184.05 - ], - "text": "q(0) =", - "type": "text" - }, - { - "block_id": "p949-b9", - "global_id": 28313, - "bbox": [ - 322.83, - 159.69, - 333.24, - 177.98 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p949-b10", - "global_id": 28314, - "bbox": [ - 328.25, - 179.97, - 333.24, - 189.93 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p949-b11", - "global_id": 28315, - "bbox": [ - 333.23, - 159.69, - 338.66, - 169.66 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p949-b12", - "global_id": 28316, - "bbox": [ - 128.91, - 197.96, - 215.31, - 208.34 - ], - "text": "Also, X(s) = 1/s, and", - "type": "text" - }, - { - "block_id": "p949-b13", - "global_id": 28317, - "bbox": [ - 269.61, - 217.82, - 304.58, - 228.12 - ], - "text": "BX(s) =", - "type": "text" - }, - { - "block_id": "p949-b14", - "global_id": 28318, - "bbox": [ - 306.63, - 203.84, - 316.74, - 215.59 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p949-b15", - "global_id": 28319, - "bbox": [ - 312.5, - 216.18, - 317.49, - 236.27 - ], - "text": "3\n1", - "type": "text" - }, - { - "block_id": "p949-b16", - "global_id": 28320, - "bbox": [ - 317.93, - 203.84, - 329.54, - 221.22 - ], - "text": "!1", - "type": "text" - }, - { - "block_id": "p949-b17", - "global_id": 28321, - "bbox": [ - 325.1, - 217.82, - 340.55, - 235.17 - ], - "text": "s =", - "type": "text" - }, - { - "block_id": "p949-b18", - "global_id": 28322, - "bbox": [ - 342.6, - 203.84, - 354.07, - 215.37 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p949-b19", - "global_id": 28323, - "bbox": [ - 349.22, - 215.89, - 355.42, - 239.21 - ], - "text": "3s\n1\ns", - "type": "text" - }, - { - "block_id": "p949-b20", - "global_id": 28324, - "bbox": [ - 356.62, - 203.84, - 362.05, - 213.8 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p949-b21", - "global_id": 28325, - "bbox": [ - 128.91, - 244.88, - 170.65, - 254.84 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p949-b22", - "global_id": 28326, - "bbox": [ - 243.63, - 264.33, - 307.42, - 274.7 - ], - "text": "q(0) + BX(s) =", - "type": "text" - }, - { - "block_id": "p949-b24", - "global_id": 28327, - "bbox": [ - 315.66, - 254.9, - 338.9, - 285.71 - ], - "text": "2 + 1\n3s\n1 + 1\ns", - "type": "text" - }, - { - "block_id": "p949-b26", - "global_id": 28328, - "bbox": [ - 348.32, - 264.33, - 356.1, - 274.29 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p949-b27", - "global_id": 28329, - "bbox": [ - 358.14, - 247.35, - 380.65, - 261.88 - ], - "text": "6s+1", - "type": "text" - }, - { - "block_id": "p949-b28", - "global_id": 28330, - "bbox": [ - 367.27, - 262.39, - 378.9, - 278.22 - ], - "text": "3s\ns+1", - "type": "text" - }, - { - "block_id": "p949-b29", - "global_id": 28331, - "bbox": [ - 371.73, - 278.74, - 374.45, - 285.71 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p949-b31", - "global_id": 28332, - "bbox": [ - 128.9, - 291.96, - 143.28, - 301.92 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p949-b32", - "global_id": 28333, - "bbox": [ - 237.63, - 313.46, - 348.38, - 323.83 - ], - "text": "Q(s) = (s)[q(0) + BX(s)]", - "type": "text" - }, - { - "block_id": "p949-b33", - "global_id": 28334, - "bbox": [ - 258.74, - 338.46, - 266.51, - 348.42 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p949-b34", - "global_id": 28335, - "bbox": [ - 268.56, - 321.48, - 309.6, - 343.78 - ], - "text": "s+1\n(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b35", - "global_id": 28336, - "bbox": [ - 275.94, - 328.39, - 355.63, - 359.13 - ], - "text": "2/3\n(s+4)(s+9)\n−36\n(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b36", - "global_id": 28337, - "bbox": [ - 321.97, - 344.31, - 355.63, - 359.13 - ], - "text": "s+12\n(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b37", - "global_id": 28338, - "bbox": [ - 356.86, - 321.49, - 386.64, - 336.01 - ], - "text": "6s+1", - "type": "text" - }, - { - "block_id": "p949-b38", - "global_id": 28339, - "bbox": [ - 373.27, - 336.52, - 384.9, - 352.35 - ], - "text": "3s\ns+1", - "type": "text" - }, - { - "block_id": "p949-b39", - "global_id": 28340, - "bbox": [ - 377.74, - 352.87, - 380.45, - 359.84 - ], - "text": "s", - "type": "text" - }, - { - "block_id": "p949-b41", - "global_id": 28341, - "bbox": [ - 258.74, - 375.75, - 266.51, - 385.71 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p949-b43", - "global_id": 28342, - "bbox": [ - 275.94, - 365.39, - 312.31, - 381.71 - ], - "text": "2s2+3s+1\ns(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b44", - "global_id": 28343, - "bbox": [ - 277.3, - 382.23, - 310.96, - 397.06 - ], - "text": "s−59\n(s+4)(s+9)", - "type": "text" - }, - { - "block_id": "p949-b46", - "global_id": 28344, - "bbox": [ - 258.74, - 413.02, - 266.51, - 422.99 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p949-b47", - "global_id": 28345, - "bbox": [ - 268.56, - 396.05, - 289.67, - 410.26 - ], - "text": "1/36", - "type": "text" - }, - { - "block_id": "p949-b48", - "global_id": 28346, - "bbox": [ - 281.45, - 403.0, - 320.15, - 418.31 - ], - "text": "s\n−21/20", - "type": "text" - }, - { - "block_id": "p949-b49", - "global_id": 28347, - "bbox": [ - 305.73, - 403.0, - 354.12, - 418.68 - ], - "text": "s+4 + 136/45", - "type": "text" - }, - { - "block_id": "p949-b50", - "global_id": 28348, - "bbox": [ - 291.95, - 411.11, - 349.59, - 426.17 - ], - "text": "s+9\n−63/5", - "type": "text" - }, - { - "block_id": "p949-b51", - "global_id": 28349, - "bbox": [ - 295.72, - 418.91, - 338.11, - 434.6 - ], - "text": "s+4 + 68/5", - "type": "text" - }, - { - "block_id": "p949-b52", - "global_id": 28350, - "bbox": [ - 325.44, - 427.03, - 337.07, - 434.29 - ], - "text": "s+9", - "type": "text" - }, - { - "block_id": "p949-b54", - "global_id": 28351, - "bbox": [ - 128.91, - 443.63, - 341.13, - 453.6 - ], - "text": "The inverse Laplace transform of this equation yields", - "type": "text" - }, - { - "block_id": "p949-b55", - "global_id": 28352, - "bbox": [ - 233.79, - 459.79, - 258.57, - 478.84 - ], - "text": "q1(t)", - "type": "text" - }, - { - "block_id": "p949-b56", - "global_id": 28353, - "bbox": [ - 239.22, - 479.65, - 258.57, - 490.8 - ], - "text": "q2(t)", - "type": "text" - }, - { - "block_id": "p949-b57", - "global_id": 28354, - "bbox": [ - 258.57, - 459.79, - 264.0, - 469.75 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p949-b58", - "global_id": 28355, - "bbox": [ - 266.05, - 473.77, - 273.82, - 483.74 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p949-b59", - "global_id": 28356, - "bbox": [ - 275.86, - 456.79, - 292.49, - 471.32 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p949-b60", - "global_id": 28357, - "bbox": [ - 287.26, - 457.68, - 391.69, - 483.99 - ], - "text": "36 −21\n20e−4t + 136\n45 e−9t\nu(t)", - "type": "text" - }, - { - "block_id": "p949-b61", - "global_id": 28358, - "bbox": [ - 294.04, - 480.69, - 309.97, - 492.0 - ], - "text": "−63", - "type": "text" - }, - { - "block_id": "p949-b62", - "global_id": 28359, - "bbox": [ - 304.75, - 474.03, - 383.72, - 495.85 - ], - "text": "5 e−4t + 68\n5 e−9t\nu(t)", - "type": "text" - }, - { - "block_id": "p949-b64", - "global_id": 28360, - "bbox": [ - 128.9, - 504.38, - 424.53, - 514.35 - ], - "text": "This result is readily confirmed using MATLAB and its symbolic toolbox.", - "type": "text" - }, - { - "block_id": "p949-b65", - "global_id": 28361, - "bbox": [ - 128.9, - 524.6, - 442.67, - 570.43 - ], - "text": ">>\nsyms s\n>>\nA = [-12 2/3;-36 -1]; B = [1/3; 1]; q0 = [2;1]; X = 1/s;\n>>\nq = ilaplace(inv(s*eye(2)-A)*(q0+B*X))\nq = (136*exp(-9*t))/45 - (21*exp(-4*t))/20 + 1/36", - "type": "text" - }, - { - "block_id": "p949-b66", - "global_id": 28362, - "bbox": [ - 202.11, - 572.41, - 385.18, - 582.38 - ], - "text": "(68*exp(-9*t))/5 - (63*exp(-4*t))/5", - "type": "text" - }, - { - "block_id": "p949-b67", - "global_id": 28363, - "bbox": [ - 128.9, - 592.05, - 502.76, - 613.98 - ], - "text": "To create a plot of the state vector, we use MATLAB’s subs command to substitute the\nsymbolic variable t with a vector of desired values.", - "type": "text" - } - ] - }, - { - "page_num": 950, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p950-b0", - "global_id": 28364, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "930\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p950-b1", - "global_id": 28365, - "bbox": [ - 103.16, - 86.66, - 437.9, - 120.53 - ], - "text": ">>\nt = (0:.01:2); q = subs(q); q1 = q(1,:); q2 = q(2,:);\n>>\nplot(t,q1,’k’,t,q2,’k--’); xlabel(’t’); ylabel(’Amplitude’);\n>>\nlegend(’q_1(t)’,’q_2(t)’,’Location’,’SE’);", - "type": "text" - }, - { - "block_id": "p950-b2", - "global_id": 28366, - "bbox": [ - 103.16, - 130.21, - 262.31, - 140.17 - ], - "text": "The resulting plot is shown in Fig. 10.7.", - "type": "text" - }, - { - "block_id": "p950-b3", - "global_id": 28367, - "bbox": [ - 101.77, - 279.65, - 257.78, - 288.89 - ], - "text": "Figure 10.7 State vector plot for Ex. 10.6.", - "type": "text" - }, - { - "block_id": "p950-b4", - "global_id": 28368, - "bbox": [ - 102.14, - 330.95, - 172.78, - 343.07 - ], - "text": "THE OUTPUT", - "type": "text" - }, - { - "block_id": "p950-b5", - "global_id": 28369, - "bbox": [ - 101.84, - 347.1, - 227.91, - 357.07 - ], - "text": "The output equation is given by", - "type": "text" - }, - { - "block_id": "p950-b6", - "global_id": 28370, - "bbox": [ - 269.8, - 358.64, - 322.42, - 368.94 - ], - "text": "y = Cq + Dx", - "type": "text" - }, - { - "block_id": "p950-b7", - "global_id": 28371, - "bbox": [ - 101.84, - 377.24, - 116.22, - 387.2 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p950-b8", - "global_id": 28372, - "bbox": [ - 249.52, - 388.77, - 342.71, - 399.07 - ], - "text": "Y(s) = CQ(s) + DX(s)", - "type": "text" - }, - { - "block_id": "p950-b9", - "global_id": 28373, - "bbox": [ - 101.85, - 407.36, - 329.8, - 417.32 - ], - "text": "Upon substituting Eq. (10.28) into this equation, we have", - "type": "text" - }, - { - "block_id": "p950-b10", - "global_id": 28374, - "bbox": [ - 211.73, - 427.36, - 380.52, - 437.73 - ], - "text": "Y(s) = C{(s)[q(0) + Bx X(s)]} + DX(s)", - "type": "text" - }, - { - "block_id": "p950-b11", - "global_id": 28375, - "bbox": [ - 232.28, - 442.3, - 289.06, - 466.67 - ], - "text": "= C(s)q(0)\n\n\n\nzero-input response", - "type": "text" - }, - { - "block_id": "p950-b12", - "global_id": 28376, - "bbox": [ - 290.18, - 442.3, - 375.31, - 466.42 - ], - "text": "+[C(s)B + D]X(s)\n\n\n\nzero-state response", - "type": "text" - }, - { - "block_id": "p950-b13", - "global_id": 28377, - "bbox": [ - 461.34, - 442.71, - 490.38, - 452.67 - ], - "text": "(10.30)", - "type": "text" - }, - { - "block_id": "p950-b14", - "global_id": 28378, - "bbox": [ - 101.85, - 476.02, - 399.88, - 486.4 - ], - "text": "The zero-state response [i.e., the response Y(s) when q(0) = 0] is given by", - "type": "text" - }, - { - "block_id": "p950-b15", - "global_id": 28379, - "bbox": [ - 242.81, - 496.43, - 349.43, - 506.73 - ], - "text": "Y(s) = [C(s)B + D]X(s)", - "type": "text" - }, - { - "block_id": "p950-b16", - "global_id": 28380, - "bbox": [ - 101.85, - 517.25, - 490.41, - 552.62 - ], - "text": "Note that the transfer function of a system is defined under the zero-state condition [see Eq. (4.19)].\nThe matrix C(s)B + D is the transfer function matrix H(s) of the system, which relates the\nresponses y1, y2, . . . , yk to the inputs x1, x2, . . . , xj:", - "type": "text" - }, - { - "block_id": "p950-b17", - "global_id": 28381, - "bbox": [ - 255.1, - 561.15, - 490.38, - 571.53 - ], - "text": "H(s) = C(s)B + D\n(10.31)", - "type": "text" - }, - { - "block_id": "p950-b18", - "global_id": 28382, - "bbox": [ - 101.85, - 581.98, - 218.6, - 591.94 - ], - "text": "and the zero-state response is", - "type": "text" - }, - { - "block_id": "p950-b19", - "global_id": 28383, - "bbox": [ - 262.15, - 593.52, - 330.09, - 603.81 - ], - "text": "Y(s) = H(s)X(s)", - "type": "text" - }, - { - "block_id": "p950-b20", - "global_id": 28384, - "bbox": [ - 101.85, - 611.69, - 490.4, - 634.73 - ], - "text": "The matrix H(s) is a k × j matrix (k is the number of outputs and j is the number of inputs). The\nijth element Hij(s) of H(s) is the transfer function that relates the output yi(t) to the input xj(t).", - "type": "text" - } - ] - }, - { - "page_num": 951, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p951-b0", - "global_id": 28385, - "bbox": [ - 341.6, - 62.89, - 516.12, - 71.98 - ], - "text": "10.4\nSolution of State Equations\n931", - "type": "text" - }, - { - "block_id": "p951-b1", - "global_id": 28386, - "bbox": [ - 102.51, - 93.92, - 505.83, - 105.87 - ], - "text": "EXAMPLE 10.7\nTransfer Function Matrix from State-Space Description", - "type": "text" - }, - { - "block_id": "p951-b2", - "global_id": 28387, - "bbox": [ - 128.9, - 122.42, - 311.53, - 132.39 - ], - "text": "Let us consider a system with a state equation", - "type": "text" - }, - { - "block_id": "p951-b3", - "global_id": 28388, - "bbox": [ - 236.45, - 135.36, - 250.35, - 154.42 - ], - "text": "˙q1", - "type": "text" - }, - { - "block_id": "p951-b4", - "global_id": 28389, - "bbox": [ - 241.88, - 155.22, - 250.35, - 166.37 - ], - "text": "˙q2", - "type": "text" - }, - { - "block_id": "p951-b5", - "global_id": 28390, - "bbox": [ - 250.85, - 135.36, - 256.27, - 145.32 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b6", - "global_id": 28391, - "bbox": [ - 258.32, - 149.34, - 266.1, - 159.31 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p951-b7", - "global_id": 28392, - "bbox": [ - 268.14, - 135.36, - 309.03, - 165.6 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p951-b8", - "global_id": 28393, - "bbox": [ - 309.04, - 135.36, - 328.37, - 154.42 - ], - "text": "! q1", - "type": "text" - }, - { - "block_id": "p951-b9", - "global_id": 28394, - "bbox": [ - 319.9, - 155.53, - 328.37, - 166.37 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p951-b10", - "global_id": 28395, - "bbox": [ - 328.87, - 135.36, - 334.3, - 145.32 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b11", - "global_id": 28396, - "bbox": [ - 335.84, - 149.34, - 343.61, - 159.31 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p951-b12", - "global_id": 28397, - "bbox": [ - 345.16, - 135.36, - 370.52, - 165.6 - ], - "text": "1\n0\n1\n1", - "type": "text" - }, - { - "block_id": "p951-b13", - "global_id": 28398, - "bbox": [ - 370.52, - 135.36, - 389.29, - 154.42 - ], - "text": "! x1", - "type": "text" - }, - { - "block_id": "p951-b14", - "global_id": 28399, - "bbox": [ - 381.38, - 155.53, - 389.29, - 166.37 - ], - "text": "x2", - "type": "text" - }, - { - "block_id": "p951-b15", - "global_id": 28400, - "bbox": [ - 389.79, - 135.36, - 395.22, - 145.32 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b16", - "global_id": 28401, - "bbox": [ - 128.91, - 176.74, - 219.93, - 186.7 - ], - "text": "and an output equation", - "type": "text" - }, - { - "block_id": "p951-b17", - "global_id": 28402, - "bbox": [ - 237.9, - 189.66, - 245.16, - 199.63 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b18", - "global_id": 28403, - "bbox": [ - 237.9, - 207.6, - 245.16, - 217.56 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p951-b19", - "global_id": 28404, - "bbox": [ - 245.16, - 197.89, - 253.06, - 232.64 - ], - "text": "y1\ny2\ny3", - "type": "text" - }, - { - "block_id": "p951-b20", - "global_id": 28405, - "bbox": [ - 253.56, - 189.66, - 260.83, - 199.63 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b21", - "global_id": 28406, - "bbox": [ - 253.56, - 207.6, - 270.65, - 219.6 - ], - "text": "⎦=", - "type": "text" - }, - { - "block_id": "p951-b22", - "global_id": 28407, - "bbox": [ - 272.7, - 189.66, - 279.96, - 199.63 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b23", - "global_id": 28408, - "bbox": [ - 272.7, - 207.6, - 279.96, - 217.56 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p951-b24", - "global_id": 28409, - "bbox": [ - 279.96, - 197.99, - 299.88, - 231.86 - ], - "text": "1\n0\n1\n1\n0\n2", - "type": "text" - }, - { - "block_id": "p951-b25", - "global_id": 28410, - "bbox": [ - 299.89, - 189.66, - 307.15, - 199.63 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b26", - "global_id": 28411, - "bbox": [ - 299.89, - 207.6, - 307.15, - 217.56 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p951-b27", - "global_id": 28412, - "bbox": [ - 308.26, - 195.65, - 322.15, - 214.71 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p951-b28", - "global_id": 28413, - "bbox": [ - 313.68, - 215.82, - 322.15, - 226.66 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p951-b29", - "global_id": 28414, - "bbox": [ - 322.65, - 195.65, - 328.08, - 205.61 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b30", - "global_id": 28415, - "bbox": [ - 329.63, - 209.63, - 337.4, - 219.6 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p951-b31", - "global_id": 28416, - "bbox": [ - 338.95, - 189.66, - 346.21, - 199.63 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b32", - "global_id": 28417, - "bbox": [ - 338.95, - 207.6, - 346.21, - 217.56 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p951-b33", - "global_id": 28418, - "bbox": [ - 346.21, - 197.99, - 366.14, - 231.86 - ], - "text": "0\n0\n1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p951-b34", - "global_id": 28419, - "bbox": [ - 366.14, - 189.66, - 373.4, - 199.63 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b35", - "global_id": 28420, - "bbox": [ - 366.14, - 207.6, - 373.4, - 217.56 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p951-b36", - "global_id": 28421, - "bbox": [ - 374.5, - 195.65, - 387.84, - 214.71 - ], - "text": "x1", - "type": "text" - }, - { - "block_id": "p951-b37", - "global_id": 28422, - "bbox": [ - 379.93, - 215.82, - 387.84, - 226.66 - ], - "text": "x2", - "type": "text" - }, - { - "block_id": "p951-b38", - "global_id": 28423, - "bbox": [ - 388.35, - 195.65, - 393.78, - 205.61 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b39", - "global_id": 28424, - "bbox": [ - 128.91, - 243.01, - 340.85, - 252.97 - ], - "text": "Determine the transfer function matrix of the system.", - "type": "text" - }, - { - "block_id": "p951-b40", - "global_id": 28425, - "bbox": [ - 128.9, - 275.88, - 176.21, - 285.85 - ], - "text": "In this case,", - "type": "text" - }, - { - "block_id": "p951-b41", - "global_id": 28426, - "bbox": [ - 160.02, - 308.06, - 177.03, - 318.35 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p951-b42", - "global_id": 28427, - "bbox": [ - 179.08, - 294.07, - 219.97, - 324.31 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p951-b43", - "global_id": 28428, - "bbox": [ - 219.97, - 294.07, - 225.4, - 304.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b44", - "global_id": 28429, - "bbox": [ - 245.33, - 308.06, - 261.79, - 318.35 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p951-b45", - "global_id": 28430, - "bbox": [ - 263.84, - 294.07, - 289.19, - 324.31 - ], - "text": "1\n0\n1\n1", - "type": "text" - }, - { - "block_id": "p951-b46", - "global_id": 28431, - "bbox": [ - 289.19, - 294.07, - 294.62, - 304.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b47", - "global_id": 28432, - "bbox": [ - 314.55, - 308.06, - 331.56, - 318.35 - ], - "text": "C =", - "type": "text" - }, - { - "block_id": "p951-b48", - "global_id": 28433, - "bbox": [ - 333.61, - 288.1, - 340.87, - 298.06 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b49", - "global_id": 28434, - "bbox": [ - 333.61, - 306.02, - 340.87, - 315.99 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p951-b50", - "global_id": 28435, - "bbox": [ - 340.87, - 296.42, - 360.79, - 330.3 - ], - "text": "1\n0\n1\n1\n0\n2", - "type": "text" - }, - { - "block_id": "p951-b51", - "global_id": 28436, - "bbox": [ - 360.8, - 288.1, - 368.06, - 298.06 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b52", - "global_id": 28437, - "bbox": [ - 360.8, - 306.02, - 406.1, - 318.35 - ], - "text": "⎦\nD =", - "type": "text" - }, - { - "block_id": "p951-b53", - "global_id": 28438, - "bbox": [ - 408.15, - 288.1, - 415.41, - 298.06 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b54", - "global_id": 28439, - "bbox": [ - 408.15, - 306.02, - 415.41, - 315.99 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p951-b55", - "global_id": 28440, - "bbox": [ - 415.41, - 296.42, - 435.34, - 330.3 - ], - "text": "0\n0\n1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p951-b56", - "global_id": 28441, - "bbox": [ - 435.34, - 288.1, - 442.6, - 298.06 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b57", - "global_id": 28442, - "bbox": [ - 435.34, - 306.02, - 502.75, - 318.43 - ], - "text": "⎦\n(10.32)", - "type": "text" - }, - { - "block_id": "p951-b58", - "global_id": 28443, - "bbox": [ - 128.91, - 341.43, - 143.28, - 351.39 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p951-b59", - "global_id": 28444, - "bbox": [ - 192.37, - 355.43, - 275.94, - 369.84 - ], - "text": "(s) = (sI −A)−1 =", - "type": "text" - }, - { - "block_id": "p951-b60", - "global_id": 28445, - "bbox": [ - 277.99, - 345.56, - 318.08, - 375.8 - ], - "text": "s\n−1\n2\ns + 3", - "type": "text" - }, - { - "block_id": "p951-b61", - "global_id": 28446, - "bbox": [ - 318.08, - 345.56, - 332.44, - 357.45 - ], - "text": "!−1", - "type": "text" - }, - { - "block_id": "p951-b62", - "global_id": 28447, - "bbox": [ - 334.99, - 359.55, - 342.76, - 369.51 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p951-b63", - "global_id": 28448, - "bbox": [ - 344.81, - 342.57, - 385.86, - 364.09 - ], - "text": "s+3\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b64", - "global_id": 28449, - "bbox": [ - 352.19, - 349.56, - 431.89, - 380.43 - ], - "text": "1\n(s+1)(s+2)\n−2\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b65", - "global_id": 28450, - "bbox": [ - 398.23, - 365.83, - 431.89, - 380.43 - ], - "text": "s\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b67", - "global_id": 28451, - "bbox": [ - 473.71, - 359.96, - 502.75, - 369.92 - ], - "text": "(10.33)", - "type": "text" - }, - { - "block_id": "p951-b68", - "global_id": 28452, - "bbox": [ - 128.91, - 386.77, - 335.99, - 397.15 - ], - "text": "Hence, the transfer function matrix H(s) is given by", - "type": "text" - }, - { - "block_id": "p951-b69", - "global_id": 28453, - "bbox": [ - 196.75, - 408.46, - 278.78, - 418.76 - ], - "text": "H(s) = C(s)B + D", - "type": "text" - }, - { - "block_id": "p951-b70", - "global_id": 28454, - "bbox": [ - 217.85, - 436.45, - 225.62, - 446.41 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p951-b71", - "global_id": 28455, - "bbox": [ - 227.67, - 416.49, - 234.94, - 426.45 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b72", - "global_id": 28456, - "bbox": [ - 227.67, - 434.43, - 234.94, - 444.39 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p951-b73", - "global_id": 28457, - "bbox": [ - 234.94, - 424.81, - 254.86, - 458.69 - ], - "text": "1\n0\n1\n1\n0\n2", - "type": "text" - }, - { - "block_id": "p951-b74", - "global_id": 28458, - "bbox": [ - 254.86, - 416.49, - 262.12, - 426.45 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b75", - "global_id": 28459, - "bbox": [ - 254.86, - 434.43, - 262.12, - 444.39 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p951-b76", - "global_id": 28460, - "bbox": [ - 263.23, - 419.48, - 304.27, - 440.99 - ], - "text": "s+3\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b77", - "global_id": 28461, - "bbox": [ - 270.61, - 426.46, - 350.31, - 457.34 - ], - "text": "1\n(s+1)(s+2)\n−2\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b78", - "global_id": 28462, - "bbox": [ - 316.64, - 442.73, - 350.31, - 457.34 - ], - "text": "s\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b79", - "global_id": 28463, - "bbox": [ - 351.53, - 419.48, - 384.17, - 452.71 - ], - "text": "1\n0\n1\n1", - "type": "text" - }, - { - "block_id": "p951-b80", - "global_id": 28464, - "bbox": [ - 384.17, - 422.47, - 389.6, - 432.43 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p951-b81", - "global_id": 28465, - "bbox": [ - 391.15, - 436.45, - 398.93, - 446.41 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p951-b82", - "global_id": 28466, - "bbox": [ - 400.47, - 416.49, - 407.73, - 426.45 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b83", - "global_id": 28467, - "bbox": [ - 400.47, - 434.43, - 407.73, - 444.39 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p951-b84", - "global_id": 28468, - "bbox": [ - 407.73, - 424.81, - 427.66, - 458.69 - ], - "text": "0\n0\n1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p951-b85", - "global_id": 28469, - "bbox": [ - 427.66, - 416.49, - 434.92, - 426.45 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b86", - "global_id": 28470, - "bbox": [ - 427.66, - 434.43, - 434.92, - 444.39 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p951-b87", - "global_id": 28471, - "bbox": [ - 217.85, - 482.4, - 225.62, - 492.37 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p951-b88", - "global_id": 28472, - "bbox": [ - 227.67, - 456.47, - 234.93, - 466.43 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p951-b89", - "global_id": 28473, - "bbox": [ - 227.67, - 473.94, - 234.93, - 496.32 - ], - "text": "⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p951-b90", - "global_id": 28474, - "bbox": [ - 236.13, - 463.03, - 269.79, - 477.85 - ], - "text": "s+4\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b91", - "global_id": 28475, - "bbox": [ - 247.15, - 463.32, - 315.83, - 494.19 - ], - "text": "1\n(s+1)(s+2)\ns+4\ns+2", - "type": "text" - }, - { - "block_id": "p951-b92", - "global_id": 28476, - "bbox": [ - 236.13, - 479.66, - 304.82, - 512.39 - ], - "text": "1\ns+2\n2(s−2)\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b93", - "global_id": 28477, - "bbox": [ - 282.17, - 496.08, - 315.83, - 512.39 - ], - "text": "s2+5s+2\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p951-b94", - "global_id": 28478, - "bbox": [ - 317.04, - 456.47, - 324.3, - 466.43 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p951-b95", - "global_id": 28479, - "bbox": [ - 317.04, - 473.94, - 502.75, - 496.32 - ], - "text": "⎥⎥⎦\n(10.34)", - "type": "text" - }, - { - "block_id": "p951-b96", - "global_id": 28480, - "bbox": [ - 128.91, - 521.88, - 245.66, - 531.84 - ], - "text": "and the zero-state response is", - "type": "text" - }, - { - "block_id": "p951-b97", - "global_id": 28481, - "bbox": [ - 281.87, - 533.43, - 349.8, - 543.72 - ], - "text": "Y(s) = H(s)X(s)", - "type": "text" - }, - { - "block_id": "p951-b98", - "global_id": 28482, - "bbox": [ - 128.91, - 552.55, - 502.74, - 588.02 - ], - "text": "Remember that the ijth element of the transfer function matrix in Eq. (10.34) represents the\ntransfer function that relates the output yi(t) to the input xj(t). For instance, the transfer function\nthat relates the output y3 to the input x2 is H32(s), where", - "type": "text" - }, - { - "block_id": "p951-b99", - "global_id": 28483, - "bbox": [ - 268.56, - 593.78, - 361.9, - 621.83 - ], - "text": "H32(s) =\ns2 + 5s + 2\n(s + 1)(s + 2)", - "type": "text" - } - ] - }, - { - "page_num": 952, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p952-b0", - "global_id": 28484, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "932\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p952-b1", - "global_id": 28485, - "bbox": [ - 103.16, - 86.24, - 477.0, - 108.15 - ], - "text": "We can readily verify the transfer function matrix using MATLAB and its symbolic\ntoolbox functions.", - "type": "text" - }, - { - "block_id": "p952-b2", - "global_id": 28486, - "bbox": [ - 103.16, - 118.41, - 396.05, - 164.23 - ], - "text": ">>\nA = [0 1;-2 -3]; B = [1 0;1 1];\n>>\nC = [1 0;1 1;0 2]; D = [0 0;1 0;0 1];\n>>\nsyms s; H = collect(simplify(C*inv(s*eye(2)-A)*B+D))\nH =", - "type": "text" - }, - { - "block_id": "p952-b3", - "global_id": 28487, - "bbox": [ - 145.0, - 166.23, - 464.04, - 200.1 - ], - "text": "[\n(s + 4)/(s^2 + 3*s + 2),\n1/(s^2 + 3*s + 2)]\n[\n(s + 4)/(s + 2),\n1/(s + 2)]\n[ (2*s - 4)/(s^2 + 3*s + 2), (s^2 + 5*s + 2)/(s^2 + 3*s + 2)]", - "type": "text" - }, - { - "block_id": "p952-b4", - "global_id": 28488, - "bbox": [ - 103.16, - 209.36, - 477.01, - 231.99 - ], - "text": "Transfer functions relating particular inputs to particular outputs, such as H32(s), can be\nobtained using the ss2tf and tf functions.", - "type": "text" - }, - { - "block_id": "p952-b5", - "global_id": 28489, - "bbox": [ - 103.17, - 241.95, - 401.29, - 263.86 - ], - "text": ">>\n[num,den] = ss2tf(A,B,C,D,2); H_32 = tf(num(3,:),den)\nH_32 =", - "type": "text" - }, - { - "block_id": "p952-b6", - "global_id": 28490, - "bbox": [ - 160.68, - 265.85, - 228.68, - 299.73 - ], - "text": "s^2 + 5 s + 2\n-------------\ns^2 + 3 s + 2", - "type": "text" - }, - { - "block_id": "p952-b7", - "global_id": 28491, - "bbox": [ - 101.84, - 343.19, - 490.41, - 476.89 - ], - "text": "CHARACTERISTIC ROOTS (EIGENVALUES) OF A MATRIX\nIt is interesting to observe that the denominator of every transfer function in Eq. (10.34) is\n(s + 1)(s + 2) except for H21(s) and H22(s), where the factor (s + 1) is canceled. This is no\ncoincidence. We see that the denominator of every element of (s) is |sI −A| because (s) =\n(sI −A)−1, and the inverse of a matrix has its determinant in the denominator. Since C, B, and D\nare matrices with constant elements, we see from Eq. (10.31) that the denominator of (s) will also\nbe the denominator of H(s). Hence, the denominator of every element of H(s) is |sI −A|, except\nfor the possible cancellation of the common factors mentioned earlier. In other words, the zeros\nof the polynomial |sI −A| are also the poles of all transfer functions of the system. Therefore, the\nzeros of the polynomial |sI−A| are the characteristic roots of the system. Hence, the characteristic\nroots of the system are the roots of the equation", - "type": "text" - }, - { - "block_id": "p952-b8", - "global_id": 28492, - "bbox": [ - 271.94, - 489.83, - 490.38, - 500.2 - ], - "text": "|sI −A| = 0\n(10.35)", - "type": "text" - }, - { - "block_id": "p952-b9", - "global_id": 28493, - "bbox": [ - 101.85, - 513.14, - 490.39, - 535.47 - ], - "text": "Since |sI −A| is an Nth-order polynomial in s with N zeros λ1, λ2, . . . , λN, we can write\nEq. (10.35) as", - "type": "text" - }, - { - "block_id": "p952-b10", - "global_id": 28494, - "bbox": [ - 140.61, - 544.28, - 451.63, - 559.86 - ], - "text": "|sI −A| = sN + a1sN−1 + · · · + aN−1s + aN = (s −λ1)(s −λ2)· · ·(s −λN) = 0", - "type": "text" - }, - { - "block_id": "p952-b11", - "global_id": 28495, - "bbox": [ - 101.85, - 572.12, - 209.05, - 582.08 - ], - "text": "For the system in Ex. 10.7,", - "type": "text" - }, - { - "block_id": "p952-b12", - "global_id": 28496, - "bbox": [ - 208.32, - 600.67, - 249.65, - 610.96 - ], - "text": "|sI −A| =", - "type": "text" - }, - { - "block_id": "p952-b14", - "global_id": 28497, - "bbox": [ - 254.94, - 594.9, - 274.86, - 616.92 - ], - "text": "s\n0\n0\ns", - "type": "text" - }, - { - "block_id": "p952-b15", - "global_id": 28498, - "bbox": [ - 274.86, - 586.21, - 287.42, - 614.11 - ], - "text": "−", - "type": "text" - }, - { - "block_id": "p952-b17", - "global_id": 28499, - "bbox": [ - 292.2, - 595.0, - 327.67, - 616.92 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p952-b18", - "global_id": 28500, - "bbox": [ - 327.67, - 586.21, - 340.73, - 614.11 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p952-b20", - "global_id": 28501, - "bbox": [ - 346.01, - 594.58, - 380.68, - 616.92 - ], - "text": "s\n−1\n2\ns + 3", - "type": "text" - }, - { - "block_id": "p952-b22", - "global_id": 28502, - "bbox": [ - 241.87, - 618.87, - 361.3, - 633.36 - ], - "text": "= s2 + 3s + 2 = (s + 1)(s + 2)", - "type": "text" - } - ] - }, - { - "page_num": 953, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p953-b0", - "global_id": 28503, - "bbox": [ - 341.6, - 62.89, - 516.12, - 71.98 - ], - "text": "10.4\nSolution of State Equations\n933", - "type": "text" - }, - { - "block_id": "p953-b1", - "global_id": 28504, - "bbox": [ - 127.59, - 85.54, - 155.51, - 95.5 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p953-b2", - "global_id": 28505, - "bbox": [ - 260.68, - 99.68, - 383.03, - 111.14 - ], - "text": "λ1 = −1\nand\nλ2 = −2", - "type": "text" - }, - { - "block_id": "p953-b3", - "global_id": 28506, - "bbox": [ - 127.59, - 120.34, - 516.15, - 166.58 - ], - "text": "Equation (10.35) is known as the characteristic equation of the matrix A, and λ1, λ2, . . . , λN are\nthe characteristic roots of A. The term eigenvalue, meaning “characteristic value” in German, is\nalso commonly used in the literature. Thus, we have shown that the characteristic roots of a system\nare the eigenvalues (characteristic values) of the matrix A.", - "type": "text" - }, - { - "block_id": "p953-b4", - "global_id": 28507, - "bbox": [ - 127.59, - 168.17, - 516.11, - 190.49 - ], - "text": "At this point, the reader will recall that if λ1, λ2, . . . , λN are the poles of the transfer function,\nthen the zero-input response is of the form", - "type": "text" - }, - { - "block_id": "p953-b5", - "global_id": 28508, - "bbox": [ - 248.18, - 199.66, - 516.12, - 214.92 - ], - "text": "y0(t) = c1eλ1t + c2eλ2t + · · · + cNeλNt\n(10.36)", - "type": "text" - }, - { - "block_id": "p953-b6", - "global_id": 28509, - "bbox": [ - 127.59, - 227.84, - 516.13, - 273.66 - ], - "text": "This fact is also obvious from Eq. (10.30). The denominator of every element of the zero-input\nresponse matrix C(s)q(0) is |sI−A| = (s−λ1)(s−λ2)· · ·(s−λN). Therefore, the partial fraction\nexpansion and the subsequent inverse Laplace transform will yield a zero-input component of the\nform in Eq. (10.36).", - "type": "text" - }, - { - "block_id": "p953-b7", - "global_id": 28510, - "bbox": [ - 127.59, - 300.15, - 393.08, - 312.11 - ], - "text": "10.4-2 Time-Domain Solution of State Equations", - "type": "text" - }, - { - "block_id": "p953-b8", - "global_id": 28511, - "bbox": [ - 127.59, - 318.24, - 209.78, - 328.2 - ], - "text": "The state equation is", - "type": "text" - }, - { - "block_id": "p953-b9", - "global_id": 28512, - "bbox": [ - 294.3, - 332.36, - 516.12, - 342.76 - ], - "text": "˙q = Aq + Bx\n(10.37)", - "type": "text" - }, - { - "block_id": "p953-b10", - "global_id": 28513, - "bbox": [ - 127.59, - 353.46, - 405.91, - 363.42 - ], - "text": "We now show that the solution of the vector differential Eq. (10.37) is", - "type": "text" - }, - { - "block_id": "p953-b11", - "global_id": 28514, - "bbox": [ - 250.04, - 380.98, - 316.4, - 392.86 - ], - "text": "q(t) = eAtq(0) +", - "type": "text" - }, - { - "block_id": "p953-b12", - "global_id": 28515, - "bbox": [ - 317.94, - 368.92, - 329.72, - 381.2 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p953-b13", - "global_id": 28516, - "bbox": [ - 323.22, - 380.76, - 392.48, - 401.07 - ], - "text": "0\neA(t−τ)Bx(τ)dτ", - "type": "text" - }, - { - "block_id": "p953-b14", - "global_id": 28517, - "bbox": [ - 127.59, - 410.93, - 516.15, - 446.22 - ], - "text": "Before proceeding further, we must define the matrix exponential eAt. An exponential of a matrix\nis defined by an infinite series identical to that used in defining an exponential of a scalar. We shall\ndefine", - "type": "text" - }, - { - "block_id": "p953-b15", - "global_id": 28518, - "bbox": [ - 203.32, - 447.95, - 281.06, - 466.77 - ], - "text": "eAt = I + At + A2t2", - "type": "text" - }, - { - "block_id": "p953-b16", - "global_id": 28519, - "bbox": [ - 268.62, - 447.95, - 312.43, - 473.92 - ], - "text": "2! + A3t3", - "type": "text" - }, - { - "block_id": "p953-b17", - "global_id": 28520, - "bbox": [ - 300.0, - 447.88, - 367.74, - 473.92 - ], - "text": "3! + · · · + Antn", - "type": "text" - }, - { - "block_id": "p953-b18", - "global_id": 28521, - "bbox": [ - 355.31, - 456.47, - 403.19, - 473.82 - ], - "text": "n! + · · · =", - "type": "text" - }, - { - "block_id": "p953-b19", - "global_id": 28522, - "bbox": [ - 405.24, - 446.3, - 419.34, - 456.97 - ], - "text": "∞\n\"", - "type": "text" - }, - { - "block_id": "p953-b20", - "global_id": 28523, - "bbox": [ - 406.21, - 470.86, - 418.35, - 478.12 - ], - "text": "k=0", - "type": "text" - }, - { - "block_id": "p953-b21", - "global_id": 28524, - "bbox": [ - 421.64, - 447.88, - 438.6, - 459.78 - ], - "text": "Aktk", - "type": "text" - }, - { - "block_id": "p953-b22", - "global_id": 28525, - "bbox": [ - 426.74, - 456.89, - 516.12, - 473.82 - ], - "text": "k!\n(10.38)", - "type": "text" - }, - { - "block_id": "p953-b23", - "global_id": 28526, - "bbox": [ - 127.59, - 486.56, - 188.45, - 496.52 - ], - "text": "For example, if", - "type": "text" - }, - { - "block_id": "p953-b24", - "global_id": 28527, - "bbox": [ - 296.94, - 506.35, - 313.95, - 516.64 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p953-b25", - "global_id": 28528, - "bbox": [ - 316.0, - 492.35, - 341.36, - 522.6 - ], - "text": "0\n1\n2\n1", - "type": "text" - }, - { - "block_id": "p953-b26", - "global_id": 28529, - "bbox": [ - 341.35, - 492.35, - 346.78, - 502.32 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p953-b27", - "global_id": 28530, - "bbox": [ - 127.59, - 532.71, - 144.74, - 542.67 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p953-b28", - "global_id": 28531, - "bbox": [ - 272.21, - 550.44, - 292.17, - 560.73 - ], - "text": "At =", - "type": "text" - }, - { - "block_id": "p953-b29", - "global_id": 28532, - "bbox": [ - 294.21, - 536.46, - 319.57, - 566.7 - ], - "text": "0\n1\n2\n1", - "type": "text" - }, - { - "block_id": "p953-b30", - "global_id": 28533, - "bbox": [ - 319.57, - 536.46, - 325.0, - 546.42 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p953-b31", - "global_id": 28534, - "bbox": [ - 325.0, - 550.44, - 337.77, - 560.72 - ], - "text": "t =", - "type": "text" - }, - { - "block_id": "p953-b32", - "global_id": 28535, - "bbox": [ - 339.81, - 536.46, - 365.91, - 566.7 - ], - "text": "0\nt\n2t\nt", - "type": "text" - }, - { - "block_id": "p953-b33", - "global_id": 28536, - "bbox": [ - 366.08, - 536.46, - 371.51, - 546.42 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p953-b34", - "global_id": 28537, - "bbox": [ - 127.59, - 576.81, - 141.97, - 586.77 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p953-b35", - "global_id": 28538, - "bbox": [ - 219.15, - 589.0, - 236.77, - 600.84 - ], - "text": "A2t2", - "type": "text" - }, - { - "block_id": "p953-b36", - "global_id": 28539, - "bbox": [ - 224.34, - 597.53, - 248.28, - 614.97 - ], - "text": "2!\n=", - "type": "text" - }, - { - "block_id": "p953-b37", - "global_id": 28540, - "bbox": [ - 250.33, - 583.53, - 275.68, - 613.78 - ], - "text": "0\n1\n2\n1", - "type": "text" - }, - { - "block_id": "p953-b38", - "global_id": 28541, - "bbox": [ - 275.68, - 583.53, - 306.46, - 613.78 - ], - "text": "! 0\n1\n2\n1", - "type": "text" - }, - { - "block_id": "p953-b39", - "global_id": 28542, - "bbox": [ - 306.47, - 583.53, - 319.53, - 600.82 - ], - "text": "!t2", - "type": "text" - }, - { - "block_id": "p953-b40", - "global_id": 28543, - "bbox": [ - 314.06, - 583.53, - 358.45, - 614.97 - ], - "text": "2 =\n 2\n1\n2\n3", - "type": "text" - }, - { - "block_id": "p953-b41", - "global_id": 28544, - "bbox": [ - 358.44, - 583.53, - 371.51, - 600.82 - ], - "text": "!t2", - "type": "text" - }, - { - "block_id": "p953-b42", - "global_id": 28545, - "bbox": [ - 366.04, - 597.53, - 383.02, - 614.97 - ], - "text": "2 =", - "type": "text" - }, - { - "block_id": "p953-b44", - "global_id": 28546, - "bbox": [ - 391.25, - 588.33, - 414.4, - 601.72 - ], - "text": "t2\nt2", - "type": "text" - }, - { - "block_id": "p953-b45", - "global_id": 28547, - "bbox": [ - 391.25, - 597.66, - 417.88, - 618.86 - ], - "text": "2\nt2\n3t2\n2", - "type": "text" - }, - { - "block_id": "p953-b47", - "global_id": 28548, - "bbox": [ - 127.59, - 626.88, - 168.27, - 636.84 - ], - "text": "and so on.", - "type": "text" - } - ] - }, - { - "page_num": 954, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p954-b0", - "global_id": 28549, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "934\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p954-b1", - "global_id": 28550, - "bbox": [ - 101.84, - 85.82, - 490.38, - 119.69 - ], - "text": "We can show that the infinite series in Eq. (10.38) is absolutely and uniformly convergent\nfor all values of t. Consequently, it can be differentiated or integrated term by term. Thus, to find\n(d/dt)eAt, we differentiate the series on the right-hand side of Eq. (10.38) term by term:", - "type": "text" - }, - { - "block_id": "p954-b2", - "global_id": 28551, - "bbox": [ - 187.34, - 129.36, - 282.22, - 155.23 - ], - "text": "d\ndteAt = A + A2t + A3t2", - "type": "text" - }, - { - "block_id": "p954-b3", - "global_id": 28552, - "bbox": [ - 269.78, - 129.36, - 313.59, - 155.33 - ], - "text": "2! + A4t3", - "type": "text" - }, - { - "block_id": "p954-b4", - "global_id": 28553, - "bbox": [ - 301.16, - 137.88, - 339.23, - 155.33 - ], - "text": "3! + · · ·", - "type": "text" - }, - { - "block_id": "p954-b5", - "global_id": 28554, - "bbox": [ - 210.53, - 165.02, - 227.54, - 175.31 - ], - "text": "= A", - "type": "text" - }, - { - "block_id": "p954-b7", - "global_id": 28555, - "bbox": [ - 232.97, - 156.49, - 287.54, - 175.31 - ], - "text": "I + At + A2t2", - "type": "text" - }, - { - "block_id": "p954-b8", - "global_id": 28556, - "bbox": [ - 275.1, - 156.49, - 318.91, - 182.47 - ], - "text": "2! + A3t3", - "type": "text" - }, - { - "block_id": "p954-b9", - "global_id": 28557, - "bbox": [ - 306.48, - 151.03, - 351.07, - 182.47 - ], - "text": "3! + · · ·\n!", - "type": "text" - }, - { - "block_id": "p954-b10", - "global_id": 28558, - "bbox": [ - 353.12, - 163.51, - 381.52, - 175.31 - ], - "text": "= AeAt", - "type": "text" - }, - { - "block_id": "p954-b11", - "global_id": 28559, - "bbox": [ - 210.53, - 194.42, - 218.3, - 204.38 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p954-b13", - "global_id": 28560, - "bbox": [ - 225.77, - 185.89, - 280.34, - 204.71 - ], - "text": "I + At + A2t2", - "type": "text" - }, - { - "block_id": "p954-b14", - "global_id": 28561, - "bbox": [ - 267.91, - 185.89, - 311.72, - 211.87 - ], - "text": "2! + A3t3", - "type": "text" - }, - { - "block_id": "p954-b15", - "global_id": 28562, - "bbox": [ - 299.29, - 180.43, - 367.82, - 211.87 - ], - "text": "3! + · · · + · · ·\n!", - "type": "text" - }, - { - "block_id": "p954-b16", - "global_id": 28563, - "bbox": [ - 367.82, - 192.92, - 406.1, - 204.71 - ], - "text": "A = eAtA", - "type": "text" - }, - { - "block_id": "p954-b17", - "global_id": 28564, - "bbox": [ - 101.85, - 221.27, - 129.76, - 231.23 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p954-b18", - "global_id": 28565, - "bbox": [ - 255.06, - 230.24, - 338.37, - 254.26 - ], - "text": "d\ndteAt = AeAt = eAtA", - "type": "text" - }, - { - "block_id": "p954-b19", - "global_id": 28566, - "bbox": [ - 101.84, - 259.02, - 287.28, - 268.98 - ], - "text": "Also note that from Eq. (10.38), it follows that", - "type": "text" - }, - { - "block_id": "p954-b20", - "global_id": 28567, - "bbox": [ - 284.04, - 275.9, - 308.19, - 290.32 - ], - "text": "e0 = I", - "type": "text" - }, - { - "block_id": "p954-b21", - "global_id": 28568, - "bbox": [ - 101.85, - 300.43, - 489.76, - 311.81 - ], - "text": "where I is just the identity matrix. If we premultiply or postmultiply the infinite series for eAt", - "type": "text" - }, - { - "block_id": "p954-b22", - "global_id": 28569, - "bbox": [ - 101.84, - 312.17, - 320.6, - 323.78 - ], - "text": "[Eq. (10.38)] by an infinite series for e−At, we find that", - "type": "text" - }, - { - "block_id": "p954-b23", - "global_id": 28570, - "bbox": [ - 237.97, - 333.09, - 490.38, - 345.19 - ], - "text": "(e−At)(eAt) = (eAt)(e−At) = I\n(10.39)", - "type": "text" - }, - { - "block_id": "p954-b24", - "global_id": 28571, - "bbox": [ - 101.85, - 356.64, - 224.44, - 366.61 - ], - "text": "In Sec. 10.1-1, we showed that", - "type": "text" - }, - { - "block_id": "p954-b25", - "global_id": 28572, - "bbox": [ - 247.85, - 365.89, - 304.29, - 389.91 - ], - "text": "d\ndt(UV) = dU", - "type": "text" - }, - { - "block_id": "p954-b26", - "global_id": 28573, - "bbox": [ - 294.11, - 365.89, - 344.39, - 389.93 - ], - "text": "dt V + UdV", - "type": "text" - }, - { - "block_id": "p954-b27", - "global_id": 28574, - "bbox": [ - 101.85, - 379.95, - 341.94, - 404.63 - ], - "text": "dt\nUsing this relationship, we observe that", - "type": "text" - }, - { - "block_id": "p954-b28", - "global_id": 28575, - "bbox": [ - 231.08, - 414.65, - 279.66, - 438.67 - ], - "text": "d\ndt[e−Atq] =", - "type": "text" - }, - { - "block_id": "p954-b29", - "global_id": 28576, - "bbox": [ - 281.71, - 407.33, - 295.95, - 424.61 - ], - "text": "d", - "type": "text" - }, - { - "block_id": "p954-b30", - "global_id": 28577, - "bbox": [ - 289.63, - 419.6, - 315.58, - 438.67 - ], - "text": "dte−At", - "type": "text" - }, - { - "block_id": "p954-b32", - "global_id": 28578, - "bbox": [ - 322.94, - 417.2, - 362.34, - 431.61 - ], - "text": "q + e−At ˙q", - "type": "text" - }, - { - "block_id": "p954-b33", - "global_id": 28579, - "bbox": [ - 271.89, - 439.61, - 490.38, - 454.1 - ], - "text": "= −e−AtAq + e−At ˙q\n(10.40)", - "type": "text" - }, - { - "block_id": "p954-b34", - "global_id": 28580, - "bbox": [ - 101.85, - 463.02, - 349.32, - 476.6 - ], - "text": "We now premultiply both sides of Eq. (10.37) by e−At to yield", - "type": "text" - }, - { - "block_id": "p954-b35", - "global_id": 28581, - "bbox": [ - 243.61, - 483.53, - 348.63, - 497.93 - ], - "text": "e−At ˙q = e−AtAq + e−AtBx", - "type": "text" - }, - { - "block_id": "p954-b36", - "global_id": 28582, - "bbox": [ - 101.84, - 509.47, - 110.14, - 519.43 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p954-b37", - "global_id": 28583, - "bbox": [ - 239.72, - 516.9, - 352.51, - 531.3 - ], - "text": "−e−AtAq + e−At ˙q = e−AtBx", - "type": "text" - }, - { - "block_id": "p954-b38", - "global_id": 28584, - "bbox": [ - 101.85, - 540.1, - 283.4, - 550.06 - ], - "text": "Substituting this result into Eq. (10.40) yields", - "type": "text" - }, - { - "block_id": "p954-b39", - "global_id": 28585, - "bbox": [ - 256.86, - 559.26, - 336.57, - 583.28 - ], - "text": "d\ndt[e−Atq] = e−AtBx", - "type": "text" - }, - { - "block_id": "p954-b40", - "global_id": 28586, - "bbox": [ - 101.84, - 590.68, - 352.03, - 600.74 - ], - "text": "The integration of both sides of this equation from 0 to t yields", - "type": "text" - }, - { - "block_id": "p954-b41", - "global_id": 28587, - "bbox": [ - 241.08, - 615.85, - 264.08, - 627.87 - ], - "text": "e−Atq", - "type": "text" - }, - { - "block_id": "p954-b42", - "global_id": 28588, - "bbox": [ - 264.08, - 609.1, - 269.26, - 625.04 - ], - "text": "t", - "type": "text" - }, - { - "block_id": "p954-b43", - "global_id": 28589, - "bbox": [ - 267.32, - 604.01, - 294.93, - 631.37 - ], - "text": "0 =\n# t", - "type": "text" - }, - { - "block_id": "p954-b44", - "global_id": 28590, - "bbox": [ - 288.43, - 615.85, - 351.15, - 636.15 - ], - "text": "0\ne−AτBx(τ)dτ", - "type": "text" - } - ] - }, - { - "page_num": 955, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p955-b0", - "global_id": 28591, - "bbox": [ - 341.6, - 62.89, - 516.12, - 71.98 - ], - "text": "10.4\nSolution of State Equations\n935", - "type": "text" - }, - { - "block_id": "p955-b1", - "global_id": 28592, - "bbox": [ - 127.59, - 83.6, - 135.89, - 93.56 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p955-b2", - "global_id": 28593, - "bbox": [ - 250.24, - 102.36, - 322.26, - 114.46 - ], - "text": "e−Atq(t) −q(0) =", - "type": "text" - }, - { - "block_id": "p955-b3", - "global_id": 28594, - "bbox": [ - 324.31, - 90.53, - 336.06, - 102.81 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p955-b4", - "global_id": 28595, - "bbox": [ - 329.56, - 102.36, - 392.29, - 122.67 - ], - "text": "0\ne−AτBx(τ)dτ", - "type": "text" - }, - { - "block_id": "p955-b5", - "global_id": 28596, - "bbox": [ - 127.59, - 131.45, - 155.51, - 141.42 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p955-b6", - "global_id": 28597, - "bbox": [ - 255.43, - 151.56, - 317.56, - 163.66 - ], - "text": "e−Atq = q(0) +", - "type": "text" - }, - { - "block_id": "p955-b7", - "global_id": 28598, - "bbox": [ - 319.11, - 139.72, - 330.88, - 151.99 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p955-b8", - "global_id": 28599, - "bbox": [ - 324.38, - 151.56, - 387.09, - 171.86 - ], - "text": "0\ne−AτBx(τ)dτ", - "type": "text" - }, - { - "block_id": "p955-b9", - "global_id": 28600, - "bbox": [ - 127.59, - 178.97, - 381.93, - 192.56 - ], - "text": "Premultiplying this result by eAt and using Eq. (10.39), we have", - "type": "text" - }, - { - "block_id": "p955-b10", - "global_id": 28601, - "bbox": [ - 249.77, - 211.85, - 307.52, - 229.31 - ], - "text": "q(t) = eAtq(0)", - "type": "text" - }, - { - "block_id": "p955-b11", - "global_id": 28602, - "bbox": [ - 287.72, - 231.76, - 297.35, - 237.74 - ], - "text": "ZIR", - "type": "text" - }, - { - "block_id": "p955-b12", - "global_id": 28603, - "bbox": [ - 308.63, - 213.36, - 316.4, - 223.33 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p955-b13", - "global_id": 28604, - "bbox": [ - 317.5, - 199.8, - 329.27, - 212.08 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p955-b14", - "global_id": 28605, - "bbox": [ - 317.5, - 211.63, - 393.95, - 244.74 - ], - "text": "0\neA(t−τ)Bx(τ)dτ\n\n\n\nZSR", - "type": "text" - }, - { - "block_id": "p955-b15", - "global_id": 28606, - "bbox": [ - 487.08, - 213.78, - 516.12, - 223.74 - ], - "text": "(10.41)", - "type": "text" - }, - { - "block_id": "p955-b16", - "global_id": 28607, - "bbox": [ - 127.59, - 257.77, - 516.14, - 292.06 - ], - "text": "This is the desired solution. The first term on the right-hand side represents q(t) when the input\nx(t) = 0. Hence, it is the zero-input component. The second term, by a similar argument, is seen\nto be the zero-state component.", - "type": "text" - }, - { - "block_id": "p955-b17", - "global_id": 28608, - "bbox": [ - 127.6, - 294.05, - 516.17, - 339.89 - ], - "text": "The results of Eq. (10.41) can be expressed more conveniently in terms of the matrix\nconvolution. We can define the convolution of two matrices in a manner similar to the\nmultiplication of two matrices, except that the multiplication of two elements is replaced by their\nconvolution. For example,", - "type": "text" - }, - { - "block_id": "p955-b18", - "global_id": 28609, - "bbox": [ - 193.55, - 346.55, - 225.25, - 377.57 - ], - "text": "x1\nx2\nx3\nx4", - "type": "text" - }, - { - "block_id": "p955-b19", - "global_id": 28610, - "bbox": [ - 225.76, - 346.55, - 231.19, - 356.52 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p955-b20", - "global_id": 28611, - "bbox": [ - 232.73, - 360.54, - 237.92, - 370.5 - ], - "text": "∗", - "type": "text" - }, - { - "block_id": "p955-b21", - "global_id": 28612, - "bbox": [ - 239.47, - 346.55, - 272.29, - 377.57 - ], - "text": "g1\ng2\ng3\ng4", - "type": "text" - }, - { - "block_id": "p955-b22", - "global_id": 28613, - "bbox": [ - 272.8, - 346.55, - 278.23, - 356.52 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p955-b23", - "global_id": 28614, - "bbox": [ - 280.28, - 360.54, - 288.05, - 370.5 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p955-b24", - "global_id": 28615, - "bbox": [ - 290.1, - 346.55, - 444.75, - 377.87 - ], - "text": "(x1 ∗g1 + x2 ∗g3)\n(x1 ∗g2 + x2 ∗g4)\n(x3 ∗g1 + x4 ∗g3)\n(x3 ∗g2 + x4 ∗g4)", - "type": "text" - }, - { - "block_id": "p955-b25", - "global_id": 28616, - "bbox": [ - 444.75, - 346.55, - 450.18, - 356.52 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p955-b26", - "global_id": 28617, - "bbox": [ - 127.59, - 391.64, - 433.5, - 401.6 - ], - "text": "By using this definition of matrix convolution, we can express Eq. (10.41) as", - "type": "text" - }, - { - "block_id": "p955-b27", - "global_id": 28618, - "bbox": [ - 266.38, - 412.51, - 516.12, - 427.0 - ], - "text": "q(t) = eAtq(0) + eAt ∗Bx(t)\n(10.42)", - "type": "text" - }, - { - "block_id": "p955-b28", - "global_id": 28619, - "bbox": [ - 127.59, - 442.33, - 516.12, - 464.35 - ], - "text": "Note that the limits of the convolution integral [Eq. (10.41)] are from 0 to t. Hence, all the elements\nof eAt in the convolution term of Eq. (10.42) are implicitly assumed to be multiplied by u(t).", - "type": "text" - }, - { - "block_id": "p955-b29", - "global_id": 28620, - "bbox": [ - 127.59, - 466.24, - 516.14, - 488.26 - ], - "text": "The result of Eqs. (10.41) and (10.42) can be easily generalized for any initial value of t. It is\nleft as an exercise for the reader to show that the solution of the state equation can be expressed as", - "type": "text" - }, - { - "block_id": "p955-b30", - "global_id": 28621, - "bbox": [ - 240.4, - 507.34, - 325.33, - 520.21 - ], - "text": "q(t) = eA(t−t0)q(t0) +", - "type": "text" - }, - { - "block_id": "p955-b31", - "global_id": 28622, - "bbox": [ - 326.87, - 495.5, - 338.64, - 507.78 - ], - "text": "# t", - "type": "text" - }, - { - "block_id": "p955-b32", - "global_id": 28623, - "bbox": [ - 332.14, - 520.6, - 337.06, - 528.9 - ], - "text": "t0", - "type": "text" - }, - { - "block_id": "p955-b33", - "global_id": 28624, - "bbox": [ - 340.36, - 507.34, - 402.12, - 519.35 - ], - "text": "eA(t−τ)Bx(τ)dτ", - "type": "text" - }, - { - "block_id": "p955-b34", - "global_id": 28625, - "bbox": [ - 127.89, - 541.3, - 222.24, - 555.88 - ], - "text": "DETERMINING eAt", - "type": "text" - }, - { - "block_id": "p955-b35", - "global_id": 28626, - "bbox": [ - 127.59, - 556.3, - 516.14, - 617.7 - ], - "text": "The exponential eAt required in Eqs. (10.41) and (10.42) can be computed from the definition in\nEq. (10.38). Unfortunately, this is an infinite series, and its computation can be quite laborious.\nMoreover, we may not be able to recognize the closed-form expression for the answer. There are\nseveral efficient methods of determining eAt in closed form. It was shown in Sec. 10.1-3 that for\nan N × N matrix A,", - "type": "text" - }, - { - "block_id": "p955-b36", - "global_id": 28627, - "bbox": [ - 235.22, - 620.27, - 516.12, - 635.64 - ], - "text": "eAt = β0I + β1A + β2A2 + · · · + βN−1AN−1\n(10.43)", - "type": "text" - } - ] - }, - { - "page_num": 956, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p956-b0", - "global_id": 28628, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "936\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p956-b1", - "global_id": 28629, - "bbox": [ - 101.84, - 85.82, - 202.57, - 102.21 - ], - "text": "where\n⎡", - "type": "text" - }, - { - "block_id": "p956-b2", - "global_id": 28630, - "bbox": [ - 195.31, - 109.73, - 202.57, - 138.08 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p956-b3", - "global_id": 28631, - "bbox": [ - 207.83, - 99.86, - 216.93, - 122.97 - ], - "text": "β0\nβ1", - "type": "text" - }, - { - "block_id": "p956-b4", - "global_id": 28632, - "bbox": [ - 202.57, - 122.8, - 222.18, - 153.46 - ], - "text": "...\nβN−1", - "type": "text" - }, - { - "block_id": "p956-b5", - "global_id": 28633, - "bbox": [ - 222.69, - 92.25, - 229.95, - 102.21 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p956-b6", - "global_id": 28634, - "bbox": [ - 222.69, - 109.73, - 239.77, - 138.08 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p956-b7", - "global_id": 28635, - "bbox": [ - 241.81, - 89.26, - 249.07, - 99.22 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p956-b8", - "global_id": 28636, - "bbox": [ - 241.81, - 106.74, - 249.07, - 141.06 - ], - "text": "⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p956-b9", - "global_id": 28637, - "bbox": [ - 249.07, - 96.57, - 294.74, - 108.7 - ], - "text": "1\nλ1\nλ2", - "type": "text" - }, - { - "block_id": "p956-b10", - "global_id": 28638, - "bbox": [ - 291.25, - 95.69, - 348.46, - 110.12 - ], - "text": "1\n· · ·\nλN−1", - "type": "text" - }, - { - "block_id": "p956-b11", - "global_id": 28639, - "bbox": [ - 249.08, - 103.15, - 337.95, - 125.27 - ], - "text": "1\n1\nλ2\nλ2", - "type": "text" - }, - { - "block_id": "p956-b12", - "global_id": 28640, - "bbox": [ - 291.25, - 112.26, - 348.46, - 126.69 - ], - "text": "2\n· · ·\nλN−1", - "type": "text" - }, - { - "block_id": "p956-b13", - "global_id": 28641, - "bbox": [ - 250.33, - 119.72, - 337.95, - 143.04 - ], - "text": "2\n...", - "type": "text" - }, - { - "block_id": "p956-b14", - "global_id": 28642, - "bbox": [ - 268.29, - 125.1, - 270.78, - 143.04 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p956-b15", - "global_id": 28643, - "bbox": [ - 249.07, - 125.1, - 340.23, - 156.32 - ], - "text": "...\n· · ·\n...\n1\nλN\nλ2", - "type": "text" - }, - { - "block_id": "p956-b16", - "global_id": 28644, - "bbox": [ - 290.46, - 143.39, - 348.46, - 157.64 - ], - "text": "N\n· · ·\nλN−1", - "type": "text" - }, - { - "block_id": "p956-b17", - "global_id": 28645, - "bbox": [ - 334.47, - 150.66, - 339.12, - 157.64 - ], - "text": "N", - "type": "text" - }, - { - "block_id": "p956-b18", - "global_id": 28646, - "bbox": [ - 348.96, - 89.26, - 356.22, - 99.23 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p956-b19", - "global_id": 28647, - "bbox": [ - 348.96, - 106.74, - 356.22, - 141.07 - ], - "text": "⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p956-b20", - "global_id": 28648, - "bbox": [ - 356.22, - 91.25, - 374.01, - 102.21 - ], - "text": "−1 ⎡", - "type": "text" - }, - { - "block_id": "p956-b21", - "global_id": 28649, - "bbox": [ - 366.75, - 109.73, - 374.01, - 138.08 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p956-b22", - "global_id": 28650, - "bbox": [ - 374.7, - 98.63, - 388.36, - 110.13 - ], - "text": "eλ1t", - "type": "text" - }, - { - "block_id": "p956-b23", - "global_id": 28651, - "bbox": [ - 374.7, - 110.59, - 388.36, - 122.09 - ], - "text": "eλ2t", - "type": "text" - }, - { - "block_id": "p956-b24", - "global_id": 28652, - "bbox": [ - 374.01, - 122.8, - 389.04, - 152.59 - ], - "text": "...\neλNt", - "type": "text" - }, - { - "block_id": "p956-b25", - "global_id": 28653, - "bbox": [ - 389.66, - 92.25, - 396.93, - 102.21 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p956-b26", - "global_id": 28654, - "bbox": [ - 389.66, - 109.73, - 396.93, - 138.08 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p956-b27", - "global_id": 28655, - "bbox": [ - 101.85, - 162.71, - 374.53, - 174.58 - ], - "text": "and λ1,λ2,. . .,λN are the N characteristic values (eigenvalues) of A.", - "type": "text" - }, - { - "block_id": "p956-b28", - "global_id": 28656, - "bbox": [ - 119.78, - 171.47, - 441.44, - 185.05 - ], - "text": "We can also determine eAt by comparing Eqs. (10.41) and (10.29). It is clear that", - "type": "text" - }, - { - "block_id": "p956-b29", - "global_id": 28657, - "bbox": [ - 224.37, - 190.48, - 490.38, - 204.98 - ], - "text": "eAt = L−1[(s)] = L−1[(sI −A)−1]\n(10.44)", - "type": "text" - }, - { - "block_id": "p956-b30", - "global_id": 28658, - "bbox": [ - 101.85, - 211.32, - 490.38, - 236.85 - ], - "text": "Thus, eAt and (s) are a Laplace transform pair. To be consistent with Laplace transform notation,\neAt is often denoted by φ(t), the state transition matrix (STM):", - "type": "text" - }, - { - "block_id": "p956-b31", - "global_id": 28659, - "bbox": [ - 275.79, - 242.29, - 316.46, - 256.68 - ], - "text": "eAt = φ(t)", - "type": "text" - }, - { - "block_id": "p956-b32", - "global_id": 28660, - "bbox": [ - 76.77, - 270.66, - 438.95, - 282.62 - ], - "text": "EXAMPLE 10.8\nTime-Domain Method to Solve State Equations", - "type": "text" - }, - { - "block_id": "p956-b33", - "global_id": 28661, - "bbox": [ - 103.16, - 299.27, - 291.19, - 309.23 - ], - "text": "Use the time-domain method to solve Ex. 10.6.", - "type": "text" - }, - { - "block_id": "p956-b34", - "global_id": 28662, - "bbox": [ - 103.16, - 332.15, - 299.88, - 342.11 - ], - "text": "For this case, the characteristic roots are given by", - "type": "text" - }, - { - "block_id": "p956-b35", - "global_id": 28663, - "bbox": [ - 163.81, - 359.47, - 205.15, - 369.77 - ], - "text": "|sI −A| =", - "type": "text" - }, - { - "block_id": "p956-b37", - "global_id": 28664, - "bbox": [ - 210.43, - 352.24, - 260.59, - 363.96 - ], - "text": "s + 12\n−2", - "type": "text" - }, - { - "block_id": "p956-b38", - "global_id": 28665, - "bbox": [ - 217.81, - 359.8, - 264.83, - 375.93 - ], - "text": "3\n36\ns + 1", - "type": "text" - }, - { - "block_id": "p956-b39", - "global_id": 28666, - "bbox": [ - 264.83, - 345.03, - 416.37, - 372.92 - ], - "text": "= s2 + 13s + 36 = (s + 4)(s + 9) = 0", - "type": "text" - }, - { - "block_id": "p956-b40", - "global_id": 28667, - "bbox": [ - 103.16, - 386.85, - 259.54, - 398.31 - ], - "text": "The roots are λ1 = −4 and λ2 = −9, so", - "type": "text" - }, - { - "block_id": "p956-b41", - "global_id": 28668, - "bbox": [ - 194.06, - 401.78, - 208.59, - 420.84 - ], - "text": "β0", - "type": "text" - }, - { - "block_id": "p956-b42", - "global_id": 28669, - "bbox": [ - 199.49, - 421.65, - 208.59, - 432.8 - ], - "text": "β1", - "type": "text" - }, - { - "block_id": "p956-b43", - "global_id": 28670, - "bbox": [ - 209.09, - 401.78, - 214.52, - 411.74 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p956-b44", - "global_id": 28671, - "bbox": [ - 216.57, - 415.77, - 224.34, - 425.73 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p956-b45", - "global_id": 28672, - "bbox": [ - 226.39, - 401.78, - 259.52, - 432.03 - ], - "text": "1\n−4\n1\n−9", - "type": "text" - }, - { - "block_id": "p956-b46", - "global_id": 28673, - "bbox": [ - 259.52, - 401.78, - 296.19, - 419.97 - ], - "text": "!−1 e−4t", - "type": "text" - }, - { - "block_id": "p956-b47", - "global_id": 28674, - "bbox": [ - 280.91, - 420.43, - 296.19, - 431.93 - ], - "text": "e−9t", - "type": "text" - }, - { - "block_id": "p956-b48", - "global_id": 28675, - "bbox": [ - 296.83, - 401.78, - 302.26, - 411.74 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p956-b49", - "global_id": 28676, - "bbox": [ - 304.31, - 409.2, - 320.29, - 426.14 - ], - "text": "= 1", - "type": "text" - }, - { - "block_id": "p956-b50", - "global_id": 28677, - "bbox": [ - 315.31, - 423.25, - 320.29, - 433.22 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p956-b51", - "global_id": 28678, - "bbox": [ - 322.6, - 401.78, - 380.04, - 420.07 - ], - "text": "9e−4t −4e−9t", - "type": "text" - }, - { - "block_id": "p956-b52", - "global_id": 28679, - "bbox": [ - 333.01, - 418.03, - 375.07, - 431.93 - ], - "text": "e−4t −e−9t", - "type": "text" - }, - { - "block_id": "p956-b53", - "global_id": 28680, - "bbox": [ - 380.69, - 401.78, - 386.12, - 411.74 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p956-b54", - "global_id": 28681, - "bbox": [ - 103.17, - 443.35, - 117.54, - 453.31 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p956-b55", - "global_id": 28682, - "bbox": [ - 168.63, - 460.7, - 233.67, - 475.97 - ], - "text": "eAt = β0I + β1A", - "type": "text" - }, - { - "block_id": "p956-b56", - "global_id": 28683, - "bbox": [ - 182.7, - 487.05, - 190.47, - 497.02 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p956-b57", - "global_id": 28684, - "bbox": [ - 192.52, - 479.04, - 201.21, - 492.68 - ], - "text": "9", - "type": "text" - }, - { - "block_id": "p956-b58", - "global_id": 28685, - "bbox": [ - 197.73, - 473.06, - 280.35, - 503.31 - ], - "text": "5e−4t −4\n5e−9t 1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p956-b59", - "global_id": 28686, - "bbox": [ - 280.36, - 473.06, - 285.79, - 483.02 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p956-b60", - "global_id": 28687, - "bbox": [ - 287.33, - 487.06, - 295.1, - 497.02 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p956-b61", - "global_id": 28688, - "bbox": [ - 296.65, - 479.05, - 305.35, - 492.68 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p956-b62", - "global_id": 28689, - "bbox": [ - 301.87, - 473.06, - 406.12, - 503.52 - ], - "text": "5e−4t −1\n5e−9t \n−12\n2\n3\n−36\n−1", - "type": "text" - }, - { - "block_id": "p956-b63", - "global_id": 28690, - "bbox": [ - 406.12, - 473.06, - 411.55, - 483.02 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p956-b64", - "global_id": 28691, - "bbox": [ - 182.7, - 518.15, - 190.47, - 528.11 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p956-b65", - "global_id": 28692, - "bbox": [ - 192.52, - 501.18, - 212.84, - 515.7 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p956-b66", - "global_id": 28693, - "bbox": [ - 206.64, - 502.06, - 337.58, - 523.88 - ], - "text": "5 e−4t + 8\n5e−9t\n2\n15(e−4t −e−9t)", - "type": "text" - }, - { - "block_id": "p956-b67", - "global_id": 28694, - "bbox": [ - 200.23, - 518.4, - 285.28, - 539.91 - ], - "text": "36\n5 (−e−4t + e−9t)\n 8", - "type": "text" - }, - { - "block_id": "p956-b68", - "global_id": 28695, - "bbox": [ - 281.8, - 518.4, - 339.07, - 539.6 - ], - "text": "5e−4t −3\n5e−9t", - "type": "text" - }, - { - "block_id": "p956-b70", - "global_id": 28696, - "bbox": [ - 103.16, - 548.72, - 314.69, - 558.69 - ], - "text": "The zero-input response is given by [see Eq. (10.41)]", - "type": "text" - }, - { - "block_id": "p956-b71", - "global_id": 28697, - "bbox": [ - 183.17, - 577.32, - 222.96, - 589.2 - ], - "text": "eAtq(0) =", - "type": "text" - }, - { - "block_id": "p956-b73", - "global_id": 28698, - "bbox": [ - 235.2, - 569.39, - 247.65, - 580.71 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p956-b74", - "global_id": 28699, - "bbox": [ - 244.17, - 562.74, - 372.39, - 583.93 - ], - "text": "5e−4t + 8\n5e−9t\n2\n15(e−4t −e−9t)", - "type": "text" - }, - { - "block_id": "p956-b75", - "global_id": 28700, - "bbox": [ - 233.88, - 579.08, - 320.1, - 600.59 - ], - "text": "36\n5 (−e−4t + e−9t)\n 8", - "type": "text" - }, - { - "block_id": "p956-b76", - "global_id": 28701, - "bbox": [ - 316.62, - 579.08, - 373.89, - 600.28 - ], - "text": "5e−4t −3\n5e−9t", - "type": "text" - }, - { - "block_id": "p956-b77", - "global_id": 28702, - "bbox": [ - 373.88, - 561.84, - 391.59, - 583.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p956-b78", - "global_id": 28703, - "bbox": [ - 386.61, - 585.11, - 391.59, - 595.07 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p956-b79", - "global_id": 28704, - "bbox": [ - 391.59, - 564.83, - 397.02, - 574.8 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p956-b80", - "global_id": 28705, - "bbox": [ - 215.19, - 612.7, - 222.96, - 622.66 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p956-b81", - "global_id": 28706, - "bbox": [ - 225.01, - 595.72, - 248.81, - 610.24 - ], - "text": "−16", - "type": "text" - }, - { - "block_id": "p956-b82", - "global_id": 28707, - "bbox": [ - 231.19, - 596.6, - 321.45, - 626.59 - ], - "text": "15 e−4t + 46\n15e−9t\nu(t)\n −64", - "type": "text" - }, - { - "block_id": "p956-b83", - "global_id": 28708, - "bbox": [ - 240.86, - 612.95, - 321.45, - 634.77 - ], - "text": "5 e−4t + 69\n5 e−9t\nu(t)", - "type": "text" - } - ] - }, - { - "page_num": 957, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p957-b0", - "global_id": 28709, - "bbox": [ - 341.6, - 62.89, - 516.12, - 71.98 - ], - "text": "10.4\nSolution of State Equations\n937", - "type": "text" - }, - { - "block_id": "p957-b1", - "global_id": 28710, - "bbox": [ - 128.9, - 85.88, - 425.65, - 96.26 - ], - "text": "Note here the presence of u(t), indicating that the response begins at t = 0.", - "type": "text" - }, - { - "block_id": "p957-b2", - "global_id": 28711, - "bbox": [ - 146.84, - 94.23, - 392.63, - 108.22 - ], - "text": "The zero-state component is eAt ∗Bx [see Eq. (10.42)], where", - "type": "text" - }, - { - "block_id": "p957-b3", - "global_id": 28712, - "bbox": [ - 264.74, - 128.39, - 286.19, - 138.68 - ], - "text": "Bx =", - "type": "text" - }, - { - "block_id": "p957-b4", - "global_id": 28713, - "bbox": [ - 288.24, - 111.42, - 299.11, - 126.16 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p957-b5", - "global_id": 28714, - "bbox": [ - 294.87, - 126.74, - 299.85, - 146.84 - ], - "text": "3\n1", - "type": "text" - }, - { - "block_id": "p957-b7", - "global_id": 28715, - "bbox": [ - 307.6, - 128.39, - 332.78, - 138.67 - ], - "text": "u(t) =", - "type": "text" - }, - { - "block_id": "p957-b8", - "global_id": 28716, - "bbox": [ - 334.82, - 114.41, - 344.94, - 128.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p957-b9", - "global_id": 28717, - "bbox": [ - 341.45, - 122.5, - 361.49, - 144.76 - ], - "text": "3u(t)\nu(t)", - "type": "text" - }, - { - "block_id": "p957-b10", - "global_id": 28718, - "bbox": [ - 361.49, - 114.4, - 366.92, - 124.37 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p957-b11", - "global_id": 28719, - "bbox": [ - 128.91, - 159.0, - 143.28, - 168.96 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p957-b12", - "global_id": 28720, - "bbox": [ - 175.32, - 175.61, - 225.47, - 187.41 - ], - "text": "eAt∗Bx(t) =", - "type": "text" - }, - { - "block_id": "p957-b13", - "global_id": 28721, - "bbox": [ - 227.52, - 160.14, - 247.83, - 174.67 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p957-b14", - "global_id": 28722, - "bbox": [ - 241.63, - 161.02, - 403.3, - 182.85 - ], - "text": "5 e−4t + 8\n5e−9t\nu(t)\n2\n15(e−4t −e−9t)u(t)", - "type": "text" - }, - { - "block_id": "p957-b15", - "global_id": 28723, - "bbox": [ - 235.22, - 177.38, - 335.65, - 198.89 - ], - "text": "36\n5 (−e−4t + e−9tu(t))\n 8", - "type": "text" - }, - { - "block_id": "p957-b16", - "global_id": 28724, - "bbox": [ - 332.16, - 177.38, - 404.79, - 198.58 - ], - "text": "5e−4t −3\n5e−9t\nu(t)", - "type": "text" - }, - { - "block_id": "p957-b18", - "global_id": 28725, - "bbox": [ - 412.52, - 177.12, - 417.71, - 187.08 - ], - "text": "∗", - "type": "text" - }, - { - "block_id": "p957-b19", - "global_id": 28726, - "bbox": [ - 419.26, - 163.13, - 429.37, - 176.86 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p957-b20", - "global_id": 28727, - "bbox": [ - 425.89, - 171.23, - 445.94, - 193.49 - ], - "text": "3u(t)\nu(t)", - "type": "text" - }, - { - "block_id": "p957-b21", - "global_id": 28728, - "bbox": [ - 450.91, - 163.13, - 456.34, - 173.09 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p957-b22", - "global_id": 28729, - "bbox": [ - 128.9, - 204.99, - 502.75, - 228.33 - ], - "text": "Note again the presence of the term u(t) in every element of eAt. This is the case because the\nlimits of the convolution integral run from 0 to t [Eq. (10.41)]. Thus,", - "type": "text" - }, - { - "block_id": "p957-b23", - "global_id": 28730, - "bbox": [ - 168.7, - 244.39, - 221.94, - 258.8 - ], - "text": "eAt ∗Bx(t) =", - "type": "text" - }, - { - "block_id": "p957-b25", - "global_id": 28731, - "bbox": [ - 234.19, - 239.08, - 246.65, - 250.38 - ], - "text": "−3", - "type": "text" - }, - { - "block_id": "p957-b26", - "global_id": 28732, - "bbox": [ - 243.16, - 232.41, - 328.76, - 253.61 - ], - "text": "5e−4t + 8\n5e−9t\nu(t) ∗1", - "type": "text" - }, - { - "block_id": "p957-b27", - "global_id": 28733, - "bbox": [ - 325.28, - 236.8, - 455.29, - 253.61 - ], - "text": "3u(t)\n2\n15(e−4t −e−9t)u(t) ∗u(t)", - "type": "text" - }, - { - "block_id": "p957-b28", - "global_id": 28734, - "bbox": [ - 232.87, - 248.76, - 363.98, - 270.27 - ], - "text": "36\n5 (−e−4t + e−9t)u(t) ∗1\n3u(t)\n 8", - "type": "text" - }, - { - "block_id": "p957-b29", - "global_id": 28735, - "bbox": [ - 360.49, - 248.76, - 456.78, - 269.96 - ], - "text": "5e−4t −3\n5e−9t\nu(t) ∗u(t)", - "type": "text" - }, - { - "block_id": "p957-b31", - "global_id": 28736, - "bbox": [ - 214.17, - 282.38, - 221.94, - 292.34 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p957-b33", - "global_id": 28737, - "bbox": [ - 230.18, - 272.96, - 244.38, - 284.26 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p957-b34", - "global_id": 28738, - "bbox": [ - 239.15, - 272.96, - 373.92, - 287.49 - ], - "text": "15e−4tu(t) ∗u(t) + 2\n5e−9tu(t) ∗u(t)", - "type": "text" - }, - { - "block_id": "p957-b35", - "global_id": 28739, - "bbox": [ - 231.92, - 289.3, - 244.37, - 300.6 - ], - "text": "−4", - "type": "text" - }, - { - "block_id": "p957-b36", - "global_id": 28740, - "bbox": [ - 240.89, - 289.3, - 372.17, - 303.83 - ], - "text": "5e−4tu(t) ∗u(t) + 9\n5e−9tu(t) ∗u(t)", - "type": "text" - }, - { - "block_id": "p957-b38", - "global_id": 28741, - "bbox": [ - 128.91, - 312.99, - 502.76, - 334.9 - ], - "text": "Substitution for the preceding convolution integrals from the convolution table (Table 2.1)\nyields", - "type": "text" - }, - { - "block_id": "p957-b39", - "global_id": 28742, - "bbox": [ - 206.27, - 350.96, - 259.51, - 365.38 - ], - "text": "eAt ∗Bx(t) =", - "type": "text" - }, - { - "block_id": "p957-b41", - "global_id": 28743, - "bbox": [ - 267.75, - 345.66, - 281.94, - 356.96 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p957-b42", - "global_id": 28744, - "bbox": [ - 276.71, - 345.66, - 414.23, - 360.19 - ], - "text": "60(1 −e−4t)u(t) + 2\n45(1 −e−9t)u(t)", - "type": "text" - }, - { - "block_id": "p957-b43", - "global_id": 28745, - "bbox": [ - 271.23, - 362.0, - 283.68, - 373.31 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p957-b44", - "global_id": 28746, - "bbox": [ - 280.19, - 362.0, - 410.75, - 376.54 - ], - "text": "5(1 −e−4t)u(t) + 1\n5(1 −e−9t)u(t)", - "type": "text" - }, - { - "block_id": "p957-b46", - "global_id": 28747, - "bbox": [ - 251.74, - 388.95, - 259.51, - 398.91 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p957-b47", - "global_id": 28748, - "bbox": [ - 261.56, - 371.97, - 278.18, - 386.5 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p957-b48", - "global_id": 28749, - "bbox": [ - 272.96, - 372.87, - 372.79, - 394.37 - ], - "text": "36 + 1\n60e−4t −2\n45e−9t\nu(t)", - "type": "text" - }, - { - "block_id": "p957-b49", - "global_id": 28750, - "bbox": [ - 285.78, - 393.6, - 355.96, - 410.41 - ], - "text": "1\n5(e−4t −e−9t)u(t)", - "type": "text" - }, - { - "block_id": "p957-b51", - "global_id": 28751, - "bbox": [ - 128.91, - 419.21, - 415.27, - 429.58 - ], - "text": "The sum of the two components now gives the desired solution for q(t):", - "type": "text" - }, - { - "block_id": "p957-b52", - "global_id": 28752, - "bbox": [ - 215.47, - 449.76, - 241.2, - 460.06 - ], - "text": "q(t) =", - "type": "text" - }, - { - "block_id": "p957-b53", - "global_id": 28753, - "bbox": [ - 243.25, - 435.77, - 268.03, - 454.83 - ], - "text": "q1(t)", - "type": "text" - }, - { - "block_id": "p957-b54", - "global_id": 28754, - "bbox": [ - 248.68, - 455.64, - 268.03, - 466.79 - ], - "text": "q2(t)", - "type": "text" - }, - { - "block_id": "p957-b55", - "global_id": 28755, - "bbox": [ - 273.01, - 435.77, - 278.44, - 445.73 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p957-b56", - "global_id": 28756, - "bbox": [ - 280.49, - 449.76, - 288.26, - 459.72 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p957-b57", - "global_id": 28757, - "bbox": [ - 290.31, - 432.78, - 306.94, - 447.31 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p957-b58", - "global_id": 28758, - "bbox": [ - 301.7, - 433.67, - 405.03, - 463.66 - ], - "text": "36 −21\n20e−4t + 136\n45 e−9t\nu(t)\n −63", - "type": "text" - }, - { - "block_id": "p957-b59", - "global_id": 28759, - "bbox": [ - 315.31, - 450.01, - 395.89, - 471.84 - ], - "text": "5 e−4t + 68\n5 e−9t\nu(t)", - "type": "text" - }, - { - "block_id": "p957-b61", - "global_id": 28760, - "bbox": [ - 128.9, - 480.37, - 502.75, - 514.24 - ], - "text": "This result confirms the solution obtained by using the frequency-domain method [see\nEx. 10.6]. Once the state variables q1 and q2 have been found for t≥0, all the remaining\nvariables can be determined from the output equation.", - "type": "text" - }, - { - "block_id": "p957-b62", - "global_id": 28761, - "bbox": [ - 127.59, - 537.09, - 253.65, - 563.2 - ], - "text": "THE OUTPUT\nThe output equation is given by", - "type": "text" - }, - { - "block_id": "p957-b63", - "global_id": 28762, - "bbox": [ - 279.98, - 564.79, - 363.75, - 575.08 - ], - "text": "y(t) = Cq(t) + Dx(t)", - "type": "text" - }, - { - "block_id": "p957-b64", - "global_id": 28763, - "bbox": [ - 127.59, - 583.09, - 418.7, - 593.14 - ], - "text": "The substitution of the solution for q [Eq. (10.42)] in this equation yields", - "type": "text" - }, - { - "block_id": "p957-b65", - "global_id": 28764, - "bbox": [ - 242.49, - 598.64, - 401.24, - 613.13 - ], - "text": "y(t) = C[eAtq(0) + eAt ∗Bx(t)] + Dx(t)", - "type": "text" - }, - { - "block_id": "p957-b66", - "global_id": 28765, - "bbox": [ - 127.59, - 623.1, - 278.95, - 633.14 - ], - "text": "Since the elements of B are constants,", - "type": "text" - }, - { - "block_id": "p957-b67", - "global_id": 28766, - "bbox": [ - 273.06, - 638.65, - 370.67, - 653.06 - ], - "text": "eAt ∗Bx(t) = eAtB ∗x(t)", - "type": "text" - } - ] - }, - { - "page_num": 958, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p958-b0", - "global_id": 28767, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "938\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p958-b1", - "global_id": 28768, - "bbox": [ - 101.84, - 85.82, - 282.15, - 95.78 - ], - "text": "With this result, the output equation becomes", - "type": "text" - }, - { - "block_id": "p958-b2", - "global_id": 28769, - "bbox": [ - 217.3, - 109.12, - 374.94, - 121.0 - ], - "text": "y(t) = C[eAtq(0) + eAtB ∗x(t)] + Dx(t)", - "type": "text" - }, - { - "block_id": "p958-b3", - "global_id": 28770, - "bbox": [ - 101.84, - 135.84, - 490.39, - 170.13 - ], - "text": "Now recall that the convolution of x(t) with the unit impulse δ(t) yields x(t). Let us define a j × j\ndiagonal matrix δ(t) such that all its diagonal terms are unit impulse functions. It is then obvious\nthat", - "type": "text" - }, - { - "block_id": "p958-b4", - "global_id": 28771, - "bbox": [ - 263.07, - 176.65, - 329.17, - 186.95 - ], - "text": "δ(t) ∗x(t) = x(t)", - "type": "text" - }, - { - "block_id": "p958-b5", - "global_id": 28772, - "bbox": [ - 101.85, - 199.29, - 276.29, - 209.25 - ], - "text": "and the output equation can be expressed as", - "type": "text" - }, - { - "block_id": "p958-b6", - "global_id": 28773, - "bbox": [ - 205.54, - 222.59, - 386.69, - 234.47 - ], - "text": "y(t) = C[eAtq(0) + eAtB ∗x(t)] + Dδ(t) ∗x(t)", - "type": "text" - }, - { - "block_id": "p958-b7", - "global_id": 28774, - "bbox": [ - 222.96, - 239.11, - 370.23, - 250.99 - ], - "text": "= CeAtq(0) + [CeAtB + Dδ(t)] ∗x(t)", - "type": "text" - }, - { - "block_id": "p958-b8", - "global_id": 28775, - "bbox": [ - 101.85, - 265.92, - 384.94, - 277.29 - ], - "text": "With the notation φ(t) for eAt, the output equation may be expressed as", - "type": "text" - }, - { - "block_id": "p958-b9", - "global_id": 28776, - "bbox": [ - 206.93, - 292.13, - 281.13, - 316.51 - ], - "text": "y(t) = Cφ(t)q(0)\n\n\n\nzero-input response", - "type": "text" - }, - { - "block_id": "p958-b10", - "global_id": 28777, - "bbox": [ - 282.25, - 292.13, - 385.31, - 316.51 - ], - "text": "+[Cφ(t)B + Dδ(t)] ∗x(t)\n\n\n\nzero-state response", - "type": "text" - }, - { - "block_id": "p958-b11", - "global_id": 28778, - "bbox": [ - 101.84, - 330.68, - 352.47, - 341.06 - ], - "text": "The zero-state response, that is, the response when q(0) = 0, is", - "type": "text" - }, - { - "block_id": "p958-b12", - "global_id": 28779, - "bbox": [ - 209.69, - 355.89, - 382.55, - 366.18 - ], - "text": "y(t) = [Cφ(t)B + Dδ(t)] ∗x(t) = h(t) ∗x(t)", - "type": "text" - }, - { - "block_id": "p958-b13", - "global_id": 28780, - "bbox": [ - 101.84, - 381.52, - 126.18, - 391.48 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p958-b14", - "global_id": 28781, - "bbox": [ - 250.28, - 398.01, - 490.38, - 408.38 - ], - "text": "h(t) = Cφ(t)B + Dδ(t)\n(10.45)", - "type": "text" - }, - { - "block_id": "p958-b15", - "global_id": 28782, - "bbox": [ - 101.84, - 420.24, - 490.39, - 466.48 - ], - "text": "The matrix h(t) is a k × j matrix known as the impulse response matrix. The reason for this\ndesignation is obvious. The ijth element of h(t) is hij(t), which represents the zero-state response\nyi when the input xj(t) = δ(t) and when all other inputs (and all the initial conditions) are zero. Not\nsurprisingly, vectors h(t) and H(s) form a Laplace transform pair,", - "type": "text" - }, - { - "block_id": "p958-b16", - "global_id": 28783, - "bbox": [ - 265.94, - 481.32, - 326.29, - 491.62 - ], - "text": "L[h(t)] = H(s)", - "type": "text" - }, - { - "block_id": "p958-b17", - "global_id": 28784, - "bbox": [ - 76.77, - 535.17, - 471.16, - 547.13 - ], - "text": "EXAMPLE 10.9\nState Transition Matrix by Inverse Laplace Transform", - "type": "text" - }, - { - "block_id": "p958-b18", - "global_id": 28785, - "bbox": [ - 103.16, - 562.38, - 384.28, - 573.75 - ], - "text": "For the system described in Ex. 10.7, use Eq. (10.44) to determine eAt:", - "type": "text" - }, - { - "block_id": "p958-b19", - "global_id": 28786, - "bbox": [ - 246.08, - 581.19, - 334.11, - 595.58 - ], - "text": "φ(t) = eAt = L−1(s)", - "type": "text" - } - ] - }, - { - "page_num": 959, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p959-b0", - "global_id": 28787, - "bbox": [ - 293.79, - 62.89, - 516.13, - 71.98 - ], - "text": "10.5\nLinear Transformation of a State Vector\n939", - "type": "text" - }, - { - "block_id": "p959-b1", - "global_id": 28788, - "bbox": [ - 128.9, - 86.24, - 502.72, - 96.21 - ], - "text": "This problem was solved earlier with frequency-domain techniques. From Eq. (10.33), we have", - "type": "text" - }, - { - "block_id": "p959-b2", - "global_id": 28789, - "bbox": [ - 236.65, - 114.65, - 281.08, - 126.65 - ], - "text": "φ(t) = L−1", - "type": "text" - }, - { - "block_id": "p959-b3", - "global_id": 28790, - "bbox": [ - 282.7, - 99.4, - 323.74, - 120.92 - ], - "text": "s+3\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p959-b4", - "global_id": 28791, - "bbox": [ - 290.08, - 106.38, - 369.78, - 137.26 - ], - "text": "1\n(s+1)(s+2)\n−2\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p959-b5", - "global_id": 28792, - "bbox": [ - 336.12, - 122.66, - 369.78, - 137.26 - ], - "text": "s\n(s+1)(s+2)", - "type": "text" - }, - { - "block_id": "p959-b7", - "global_id": 28793, - "bbox": [ - 255.47, - 148.53, - 281.08, - 160.21 - ], - "text": "= L−1", - "type": "text" - }, - { - "block_id": "p959-b8", - "global_id": 28794, - "bbox": [ - 282.7, - 133.27, - 297.64, - 147.55 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p959-b9", - "global_id": 28795, - "bbox": [ - 290.08, - 140.57, - 326.61, - 155.42 - ], - "text": "s+1 −\n1\ns+2", - "type": "text" - }, - { - "block_id": "p959-b10", - "global_id": 28796, - "bbox": [ - 290.08, - 140.57, - 375.51, - 171.76 - ], - "text": "1\ns+1 −\n1\ns+2\n−2\ns+1 +\n2\ns+2", - "type": "text" - }, - { - "block_id": "p959-b11", - "global_id": 28797, - "bbox": [ - 338.97, - 156.63, - 375.51, - 171.76 - ], - "text": "−1\ns+1 +\n2\ns+2", - "type": "text" - }, - { - "block_id": "p959-b13", - "global_id": 28798, - "bbox": [ - 255.47, - 182.05, - 263.24, - 192.01 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p959-b14", - "global_id": 28799, - "bbox": [ - 265.29, - 168.05, - 382.57, - 186.34 - ], - "text": "2e−t −e−2t\ne−t −e−2t", - "type": "text" - }, - { - "block_id": "p959-b15", - "global_id": 28800, - "bbox": [ - 270.73, - 184.31, - 388.95, - 198.3 - ], - "text": "−2e−t + 2e−2t\n−e−t + 2e−2t", - "type": "text" - }, - { - "block_id": "p959-b16", - "global_id": 28801, - "bbox": [ - 389.59, - 168.05, - 395.02, - 178.02 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p959-b17", - "global_id": 28802, - "bbox": [ - 128.9, - 210.58, - 494.41, - 220.54 - ], - "text": "The same result is obtained in Ex. 10.1 (Sec. 10.1-3) by using Eq. (10.43) [see Eq. (10.13)].", - "type": "text" - }, - { - "block_id": "p959-b18", - "global_id": 28803, - "bbox": [ - 146.84, - 222.12, - 323.06, - 232.49 - ], - "text": "Also, δ(t) is a diagonal j × j or 2 × 2 matrix:", - "type": "text" - }, - { - "block_id": "p959-b19", - "global_id": 28804, - "bbox": [ - 276.68, - 249.68, - 301.71, - 259.96 - ], - "text": "δ(t) =", - "type": "text" - }, - { - "block_id": "p959-b20", - "global_id": 28805, - "bbox": [ - 303.76, - 235.7, - 349.56, - 265.94 - ], - "text": "δ(t)\n0\n0\nδ(t)", - "type": "text" - }, - { - "block_id": "p959-b21", - "global_id": 28806, - "bbox": [ - 349.56, - 235.7, - 354.99, - 245.66 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p959-b22", - "global_id": 28807, - "bbox": [ - 128.9, - 277.04, - 477.65, - 287.41 - ], - "text": "Substituting the matrices φ(t), δ(t), C, D, and B [Eq. (10.32)] into Eq. (10.45), we have", - "type": "text" - }, - { - "block_id": "p959-b23", - "global_id": 28808, - "bbox": [ - 155.07, - 310.57, - 180.82, - 320.86 - ], - "text": "h(t) =", - "type": "text" - }, - { - "block_id": "p959-b24", - "global_id": 28809, - "bbox": [ - 182.87, - 290.61, - 190.13, - 300.57 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p959-b25", - "global_id": 28810, - "bbox": [ - 182.87, - 308.54, - 190.13, - 318.51 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p959-b26", - "global_id": 28811, - "bbox": [ - 190.13, - 298.93, - 210.05, - 332.81 - ], - "text": "1\n0\n1\n1\n0\n2", - "type": "text" - }, - { - "block_id": "p959-b27", - "global_id": 28812, - "bbox": [ - 210.06, - 290.61, - 217.32, - 300.57 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p959-b28", - "global_id": 28813, - "bbox": [ - 210.06, - 308.54, - 217.32, - 318.51 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p959-b29", - "global_id": 28814, - "bbox": [ - 218.43, - 296.58, - 335.7, - 314.87 - ], - "text": "2e−t −e−2t\ne−t −e−2t", - "type": "text" - }, - { - "block_id": "p959-b30", - "global_id": 28815, - "bbox": [ - 223.85, - 312.84, - 342.07, - 326.83 - ], - "text": "−2e−t + 2e−2t\n−e−t + 2e−2t", - "type": "text" - }, - { - "block_id": "p959-b31", - "global_id": 28816, - "bbox": [ - 342.72, - 296.58, - 373.5, - 326.83 - ], - "text": "! 1\n0\n1\n1", - "type": "text" - }, - { - "block_id": "p959-b32", - "global_id": 28817, - "bbox": [ - 373.51, - 296.58, - 378.94, - 306.54 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p959-b33", - "global_id": 28818, - "bbox": [ - 380.48, - 310.57, - 388.25, - 320.53 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p959-b34", - "global_id": 28819, - "bbox": [ - 389.8, - 290.61, - 397.06, - 300.57 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p959-b35", - "global_id": 28820, - "bbox": [ - 389.8, - 308.54, - 397.06, - 318.51 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p959-b36", - "global_id": 28821, - "bbox": [ - 397.07, - 298.93, - 416.99, - 332.81 - ], - "text": "0\n0\n1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p959-b37", - "global_id": 28822, - "bbox": [ - 416.99, - 290.61, - 424.26, - 300.57 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p959-b38", - "global_id": 28823, - "bbox": [ - 416.99, - 308.54, - 424.26, - 318.51 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p959-b39", - "global_id": 28824, - "bbox": [ - 425.36, - 296.58, - 471.16, - 326.83 - ], - "text": "δ(t)\n0\n0\nδ(t)", - "type": "text" - }, - { - "block_id": "p959-b40", - "global_id": 28825, - "bbox": [ - 471.15, - 296.58, - 476.58, - 306.54 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p959-b41", - "global_id": 28826, - "bbox": [ - 173.05, - 350.42, - 180.82, - 360.38 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p959-b42", - "global_id": 28827, - "bbox": [ - 182.87, - 330.46, - 190.13, - 340.42 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p959-b43", - "global_id": 28828, - "bbox": [ - 182.87, - 348.39, - 190.13, - 358.36 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p959-b44", - "global_id": 28829, - "bbox": [ - 194.01, - 334.75, - 315.36, - 348.75 - ], - "text": "3e−t −2e−2t\ne−t −e−2t", - "type": "text" - }, - { - "block_id": "p959-b45", - "global_id": 28830, - "bbox": [ - 195.12, - 349.09, - 303.72, - 360.7 - ], - "text": "δ(t) + 2e−2t\ne−2t", - "type": "text" - }, - { - "block_id": "p959-b46", - "global_id": 28831, - "bbox": [ - 190.13, - 358.66, - 335.12, - 372.66 - ], - "text": "−6e−t + 8e−2t\nδ(t) −2e−2t + 4e−2t", - "type": "text" - }, - { - "block_id": "p959-b47", - "global_id": 28832, - "bbox": [ - 335.76, - 330.46, - 343.02, - 340.42 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p959-b48", - "global_id": 28833, - "bbox": [ - 335.76, - 348.39, - 343.02, - 358.36 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p959-b49", - "global_id": 28834, - "bbox": [ - 128.9, - 384.03, - 502.77, - 405.94 - ], - "text": "As the reader can verify, the Laplace transform of this equation yields the transfer function\nmatrix H(s) in Eq. (10.34).", - "type": "text" - }, - { - "block_id": "p959-b50", - "global_id": 28835, - "bbox": [ - 127.94, - 470.28, - 473.06, - 484.23 - ], - "text": "10.5 LINEAR TRANSFORMATION OF A STATE VECTOR", - "type": "text" - }, - { - "block_id": "p959-b51", - "global_id": 28836, - "bbox": [ - 127.59, - 490.21, - 516.15, - 548.0 - ], - "text": "In Sec. 10.2 we saw that the state of a system can be specified in several ways. The sets of all\npossible state variables are related—in other words, if we are given one set of state variables,\nwe should be able to relate it to any other set. We are particularly interested in a linear type\nof relationship. Let q1,q2,. . .,qN and w1,w2,. . .,wN be two different sets of state variables\nspecifying the same system. Let these sets be related by linear equations as", - "type": "text" - }, - { - "block_id": "p959-b52", - "global_id": 28837, - "bbox": [ - 252.68, - 570.56, - 387.37, - 596.96 - ], - "text": "w1 = p11q1 + p12q2 + · · · + p1NqN\nw2 = p21q1 + p22q2 + · · · + p2NqN", - "type": "text" - }, - { - "block_id": "p959-b53", - "global_id": 28838, - "bbox": [ - 263.31, - 600.48, - 265.8, - 618.41 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p959-b54", - "global_id": 28839, - "bbox": [ - 251.1, - 622.98, - 391.7, - 634.44 - ], - "text": "wN = pN1q1 + pN2q2 + · · · + pNNqN", - "type": "text" - } - ] - }, - { - "page_num": 960, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p960-b0", - "global_id": 28840, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "940\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p960-b1", - "global_id": 28841, - "bbox": [ - 101.84, - 83.6, - 215.56, - 95.5 - ], - "text": "or\n⎡", - "type": "text" - }, - { - "block_id": "p960-b2", - "global_id": 28842, - "bbox": [ - 208.3, - 103.02, - 215.56, - 131.36 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p960-b3", - "global_id": 28843, - "bbox": [ - 216.35, - 93.45, - 226.48, - 116.25 - ], - "text": "w1\nw2", - "type": "text" - }, - { - "block_id": "p960-b4", - "global_id": 28844, - "bbox": [ - 215.56, - 116.08, - 226.85, - 146.67 - ], - "text": "...\nwN", - "type": "text" - }, - { - "block_id": "p960-b5", - "global_id": 28845, - "bbox": [ - 232.76, - 85.54, - 240.02, - 95.5 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p960-b6", - "global_id": 28846, - "bbox": [ - 232.76, - 103.02, - 240.02, - 131.36 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p960-b8", - "global_id": 28847, - "bbox": [ - 221.64, - 154.78, - 226.68, - 161.75 - ], - "text": "w", - "type": "text" - }, - { - "block_id": "p960-b9", - "global_id": 28848, - "bbox": [ - 242.06, - 114.47, - 249.83, - 124.43 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p960-b10", - "global_id": 28849, - "bbox": [ - 251.88, - 85.54, - 259.15, - 95.5 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p960-b11", - "global_id": 28850, - "bbox": [ - 251.88, - 103.01, - 259.15, - 131.36 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p960-b12", - "global_id": 28851, - "bbox": [ - 259.94, - 93.14, - 344.01, - 134.02 - ], - "text": "p11\np12\n· · ·\np1N\np21\np22\n· · ·\np2N\n...", - "type": "text" - }, - { - "block_id": "p960-b13", - "global_id": 28852, - "bbox": [ - 288.92, - 116.08, - 320.35, - 134.02 - ], - "text": "...\n. . .", - "type": "text" - }, - { - "block_id": "p960-b14", - "global_id": 28853, - "bbox": [ - 259.15, - 116.08, - 344.59, - 146.74 - ], - "text": "...\npN1\npN2\n· · ·\npNN", - "type": "text" - }, - { - "block_id": "p960-b15", - "global_id": 28854, - "bbox": [ - 345.51, - 85.54, - 352.77, - 95.5 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p960-b16", - "global_id": 28855, - "bbox": [ - 345.51, - 103.02, - 352.77, - 131.36 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p960-b17", - "global_id": 28856, - "bbox": [ - 251.88, - 143.18, - 352.77, - 162.38 - ], - "text": "P", - "type": "text" - }, - { - "block_id": "p960-b18", - "global_id": 28857, - "bbox": [ - 353.88, - 85.54, - 361.14, - 95.5 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p960-b19", - "global_id": 28858, - "bbox": [ - 353.88, - 103.02, - 361.14, - 131.36 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p960-b20", - "global_id": 28859, - "bbox": [ - 361.93, - 93.45, - 370.4, - 116.25 - ], - "text": "q1\nq2", - "type": "text" - }, - { - "block_id": "p960-b21", - "global_id": 28860, - "bbox": [ - 361.14, - 116.08, - 370.77, - 146.67 - ], - "text": "...\nqN", - "type": "text" - }, - { - "block_id": "p960-b22", - "global_id": 28861, - "bbox": [ - 376.68, - 85.54, - 383.94, - 95.5 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p960-b23", - "global_id": 28862, - "bbox": [ - 376.68, - 103.02, - 383.94, - 131.36 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p960-b25", - "global_id": 28863, - "bbox": [ - 366.97, - 154.78, - 370.84, - 161.75 - ], - "text": "q", - "type": "text" - }, - { - "block_id": "p960-b26", - "global_id": 28864, - "bbox": [ - 101.84, - 170.19, - 480.89, - 180.23 - ], - "text": "Defining the vector w and matrix P as just shown, we obtain the compact matrix representation", - "type": "text" - }, - { - "block_id": "p960-b27", - "global_id": 28865, - "bbox": [ - 280.77, - 192.11, - 490.38, - 202.48 - ], - "text": "w = Pq\n(10.46)", - "type": "text" - }, - { - "block_id": "p960-b28", - "global_id": 28866, - "bbox": [ - 101.85, - 214.77, - 116.22, - 224.73 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p960-b29", - "global_id": 28867, - "bbox": [ - 276.07, - 224.97, - 490.38, - 237.18 - ], - "text": "q = P−1w\n(10.47)", - "type": "text" - }, - { - "block_id": "p960-b30", - "global_id": 28868, - "bbox": [ - 101.84, - 246.38, - 490.39, - 268.39 - ], - "text": "Thus, the state vector q is transformed into another state vector w through the linear transformation\nin Eq. (10.46).", - "type": "text" - }, - { - "block_id": "p960-b31", - "global_id": 28869, - "bbox": [ - 101.84, - 266.16, - 490.39, - 304.25 - ], - "text": "If we know w, we can determine q from q = P−1w, provided P−1 exists. This is equivalent to\nsaying that P is a nonsingular matrix† (|P|̸ = 0). Thus, if P is a nonsingular matrix, the vector w\ndefined by Eq. (10.46) is also a state vector. Consider the state equation of a system", - "type": "text" - }, - { - "block_id": "p960-b32", - "global_id": 28870, - "bbox": [ - 269.8, - 316.1, - 322.44, - 326.41 - ], - "text": "˙q = Aq + Bx", - "type": "text" - }, - { - "block_id": "p960-b33", - "global_id": 28871, - "bbox": [ - 101.85, - 338.78, - 108.48, - 348.74 - ], - "text": "If", - "type": "text" - }, - { - "block_id": "p960-b34", - "global_id": 28872, - "bbox": [ - 280.77, - 350.81, - 311.46, - 361.11 - ], - "text": "w = Pq", - "type": "text" - }, - { - "block_id": "p960-b35", - "global_id": 28873, - "bbox": [ - 101.85, - 370.48, - 118.99, - 380.45 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p960-b36", - "global_id": 28874, - "bbox": [ - 276.07, - 380.68, - 316.17, - 392.81 - ], - "text": "q = P−1w", - "type": "text" - }, - { - "block_id": "p960-b37", - "global_id": 28875, - "bbox": [ - 101.84, - 402.18, - 116.22, - 412.15 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p960-b38", - "global_id": 28876, - "bbox": [ - 276.07, - 410.0, - 316.17, - 424.51 - ], - "text": "˙q = P−1 ˙w", - "type": "text" - }, - { - "block_id": "p960-b39", - "global_id": 28877, - "bbox": [ - 101.84, - 433.89, - 258.74, - 443.85 - ], - "text": "Hence, the state equation now becomes", - "type": "text" - }, - { - "block_id": "p960-b40", - "global_id": 28878, - "bbox": [ - 252.63, - 451.49, - 339.6, - 466.01 - ], - "text": "P−1 ˙w = AP−1w + Bx", - "type": "text" - }, - { - "block_id": "p960-b41", - "global_id": 28879, - "bbox": [ - 101.85, - 478.38, - 110.14, - 488.34 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p960-b42", - "global_id": 28880, - "bbox": [ - 254.3, - 498.38, - 337.93, - 510.51 - ], - "text": "˙w = PAP−1w + PBx", - "type": "text" - }, - { - "block_id": "p960-b43", - "global_id": 28881, - "bbox": [ - 263.55, - 515.24, - 490.38, - 527.95 - ], - "text": "= ˆAw + ˆBx\n(10.48)", - "type": "text" - }, - { - "block_id": "p960-b44", - "global_id": 28882, - "bbox": [ - 101.85, - 540.23, - 126.18, - 550.2 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p960-b45", - "global_id": 28883, - "bbox": [ - 229.46, - 549.94, - 490.38, - 562.64 - ], - "text": "ˆA = PAP−1\nand\nˆB = PB\n(10.49)", - "type": "text" - }, - { - "block_id": "p960-b46", - "global_id": 28884, - "bbox": [ - 101.85, - 571.93, - 490.42, - 593.86 - ], - "text": "Equation (10.48) is a state equation for the same system, but now it is expressed in terms of the\nstate vector w.", - "type": "text" - }, - { - "block_id": "p960-b47", - "global_id": 28885, - "bbox": [ - 101.84, - 610.24, - 490.39, - 633.41 - ], - "text": "† This condition is equivalent to saying that all N equations in Eq. (10.46) are linearly independent; that is,\nnone of the N equations can be expressed as a linear combination of the remaining equations.", - "type": "text" - } - ] - }, - { - "page_num": 961, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p961-b0", - "global_id": 28886, - "bbox": [ - 293.79, - 62.89, - 516.13, - 71.98 - ], - "text": "10.5\nLinear Transformation of a State Vector\n941", - "type": "text" - }, - { - "block_id": "p961-b1", - "global_id": 28887, - "bbox": [ - 145.52, - 85.82, - 434.79, - 95.78 - ], - "text": "The output equation is also modified. Let the original output equation be", - "type": "text" - }, - { - "block_id": "p961-b2", - "global_id": 28888, - "bbox": [ - 295.55, - 105.33, - 348.17, - 115.62 - ], - "text": "y = Cq + Dx", - "type": "text" - }, - { - "block_id": "p961-b3", - "global_id": 28889, - "bbox": [ - 127.59, - 125.59, - 363.08, - 135.63 - ], - "text": "In terms of the new state variable w, this equation becomes", - "type": "text" - }, - { - "block_id": "p961-b4", - "global_id": 28890, - "bbox": [ - 283.25, - 143.35, - 360.47, - 155.47 - ], - "text": "y = C(P−1w) + Dx", - "type": "text" - }, - { - "block_id": "p961-b5", - "global_id": 28891, - "bbox": [ - 290.28, - 160.21, - 337.53, - 172.84 - ], - "text": "= ˆCw + Dx", - "type": "text" - }, - { - "block_id": "p961-b6", - "global_id": 28892, - "bbox": [ - 127.59, - 182.88, - 151.92, - 192.84 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p961-b7", - "global_id": 28893, - "bbox": [ - 300.98, - 192.1, - 516.12, - 204.8 - ], - "text": "ˆC = CP−1\n(10.50)", - "type": "text" - }, - { - "block_id": "p961-b8", - "global_id": 28894, - "bbox": [ - 102.51, - 238.4, - 437.46, - 250.35 - ], - "text": "EXAMPLE 10.10\nLinear Transformation of the State Vector", - "type": "text" - }, - { - "block_id": "p961-b9", - "global_id": 28895, - "bbox": [ - 128.9, - 263.87, - 334.62, - 273.83 - ], - "text": "The state equations of a certain system are given by", - "type": "text" - }, - { - "block_id": "p961-b10", - "global_id": 28896, - "bbox": [ - 246.14, - 276.56, - 260.03, - 295.62 - ], - "text": "˙q1", - "type": "text" - }, - { - "block_id": "p961-b11", - "global_id": 28897, - "bbox": [ - 251.57, - 296.43, - 260.03, - 307.58 - ], - "text": "˙q2", - "type": "text" - }, - { - "block_id": "p961-b12", - "global_id": 28898, - "bbox": [ - 260.54, - 276.56, - 265.97, - 286.52 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b13", - "global_id": 28899, - "bbox": [ - 268.01, - 290.54, - 275.78, - 300.51 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p961-b14", - "global_id": 28900, - "bbox": [ - 277.83, - 276.56, - 318.73, - 306.81 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p961-b15", - "global_id": 28901, - "bbox": [ - 318.72, - 276.56, - 338.05, - 295.62 - ], - "text": "! q1", - "type": "text" - }, - { - "block_id": "p961-b16", - "global_id": 28902, - "bbox": [ - 329.59, - 296.74, - 338.05, - 307.58 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p961-b17", - "global_id": 28903, - "bbox": [ - 338.55, - 276.56, - 343.98, - 286.52 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b18", - "global_id": 28904, - "bbox": [ - 345.53, - 290.54, - 353.3, - 300.51 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p961-b19", - "global_id": 28905, - "bbox": [ - 354.85, - 276.56, - 365.27, - 294.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p961-b20", - "global_id": 28906, - "bbox": [ - 360.28, - 296.84, - 365.27, - 306.81 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p961-b21", - "global_id": 28907, - "bbox": [ - 365.26, - 276.56, - 370.69, - 286.52 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b22", - "global_id": 28908, - "bbox": [ - 370.7, - 290.54, - 385.53, - 300.82 - ], - "text": "x(t)", - "type": "text" - }, - { - "block_id": "p961-b23", - "global_id": 28909, - "bbox": [ - 128.9, - 317.6, - 493.75, - 329.16 - ], - "text": "Find the state equations for this system when the new state variables w1 and w2 are given as", - "type": "text" - }, - { - "block_id": "p961-b24", - "global_id": 28910, - "bbox": [ - 269.96, - 330.4, - 285.53, - 349.46 - ], - "text": "w1", - "type": "text" - }, - { - "block_id": "p961-b25", - "global_id": 28911, - "bbox": [ - 275.4, - 350.57, - 285.53, - 361.41 - ], - "text": "w2", - "type": "text" - }, - { - "block_id": "p961-b26", - "global_id": 28912, - "bbox": [ - 286.03, - 330.4, - 291.46, - 340.36 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b27", - "global_id": 28913, - "bbox": [ - 293.51, - 344.38, - 301.28, - 354.34 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p961-b28", - "global_id": 28914, - "bbox": [ - 303.32, - 330.4, - 336.44, - 360.64 - ], - "text": "1\n1\n1\n−1", - "type": "text" - }, - { - "block_id": "p961-b29", - "global_id": 28915, - "bbox": [ - 336.44, - 330.4, - 355.77, - 349.46 - ], - "text": "! q1", - "type": "text" - }, - { - "block_id": "p961-b30", - "global_id": 28916, - "bbox": [ - 347.31, - 350.57, - 355.77, - 361.41 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p961-b31", - "global_id": 28917, - "bbox": [ - 356.27, - 330.4, - 361.7, - 340.36 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b32", - "global_id": 28918, - "bbox": [ - 473.71, - 344.79, - 502.75, - 354.76 - ], - "text": "(10.51)", - "type": "text" - }, - { - "block_id": "p961-b33", - "global_id": 28919, - "bbox": [ - 128.9, - 387.08, - 442.86, - 397.13 - ], - "text": "According to Eq. (10.48), the state equation for the state variable w is given by", - "type": "text" - }, - { - "block_id": "p961-b34", - "global_id": 28920, - "bbox": [ - 287.86, - 405.88, - 343.8, - 418.5 - ], - "text": "˙w = ˆAw + ˆBx", - "type": "text" - }, - { - "block_id": "p961-b35", - "global_id": 28921, - "bbox": [ - 128.91, - 430.07, - 274.97, - 440.03 - ], - "text": "where [see Eqs. (10.49) and (10.50)]", - "type": "text" - }, - { - "block_id": "p961-b36", - "global_id": 28922, - "bbox": [ - 218.44, - 454.59, - 276.1, - 469.1 - ], - "text": "ˆA = PAP−1 =", - "type": "text" - }, - { - "block_id": "p961-b37", - "global_id": 28923, - "bbox": [ - 278.15, - 444.82, - 311.28, - 475.06 - ], - "text": "1\n1\n1\n−1", - "type": "text" - }, - { - "block_id": "p961-b38", - "global_id": 28924, - "bbox": [ - 311.28, - 444.82, - 358.7, - 475.06 - ], - "text": "! 0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p961-b39", - "global_id": 28925, - "bbox": [ - 358.71, - 444.82, - 398.37, - 475.06 - ], - "text": "! 1\n1\n1\n−1", - "type": "text" - }, - { - "block_id": "p961-b40", - "global_id": 28926, - "bbox": [ - 398.37, - 444.82, - 412.73, - 456.71 - ], - "text": "!−1", - "type": "text" - }, - { - "block_id": "p961-b41", - "global_id": 28927, - "bbox": [ - 268.33, - 489.7, - 276.1, - 499.66 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p961-b42", - "global_id": 28928, - "bbox": [ - 278.15, - 475.71, - 311.28, - 505.95 - ], - "text": "1\n1\n1\n−1", - "type": "text" - }, - { - "block_id": "p961-b43", - "global_id": 28929, - "bbox": [ - 311.28, - 475.71, - 358.7, - 505.95 - ], - "text": "! 0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p961-b44", - "global_id": 28930, - "bbox": [ - 358.71, - 472.72, - 376.12, - 487.25 - ], - "text": "! 1", - "type": "text" - }, - { - "block_id": "p961-b45", - "global_id": 28931, - "bbox": [ - 372.63, - 480.28, - 399.73, - 511.16 - ], - "text": "2\n1\n2\n1\n2\n−1", - "type": "text" - }, - { - "block_id": "p961-b46", - "global_id": 28932, - "bbox": [ - 396.24, - 504.18, - 399.73, - 511.16 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p961-b48", - "global_id": 28933, - "bbox": [ - 268.33, - 520.57, - 276.1, - 530.54 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p961-b49", - "global_id": 28934, - "bbox": [ - 278.15, - 506.59, - 319.05, - 536.83 - ], - "text": "−2\n0\n3\n−1", - "type": "text" - }, - { - "block_id": "p961-b50", - "global_id": 28935, - "bbox": [ - 319.05, - 506.59, - 324.48, - 516.55 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b51", - "global_id": 28936, - "bbox": [ - 128.91, - 547.74, - 143.28, - 557.7 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p961-b52", - "global_id": 28937, - "bbox": [ - 249.34, - 560.53, - 290.41, - 573.15 - ], - "text": "ˆB = PB =", - "type": "text" - }, - { - "block_id": "p961-b53", - "global_id": 28938, - "bbox": [ - 292.45, - 548.87, - 325.58, - 579.12 - ], - "text": "1\n1\n1\n−1", - "type": "text" - }, - { - "block_id": "p961-b54", - "global_id": 28939, - "bbox": [ - 325.58, - 548.87, - 341.42, - 567.16 - ], - "text": "! 1", - "type": "text" - }, - { - "block_id": "p961-b55", - "global_id": 28940, - "bbox": [ - 336.44, - 569.16, - 341.42, - 579.12 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p961-b56", - "global_id": 28941, - "bbox": [ - 341.42, - 548.87, - 346.85, - 558.84 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b57", - "global_id": 28942, - "bbox": [ - 348.89, - 562.86, - 356.66, - 572.82 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p961-b58", - "global_id": 28943, - "bbox": [ - 358.71, - 548.87, - 373.01, - 567.16 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p961-b59", - "global_id": 28944, - "bbox": [ - 364.15, - 568.75, - 376.9, - 579.12 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p961-b60", - "global_id": 28945, - "bbox": [ - 376.9, - 548.88, - 382.33, - 558.84 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b61", - "global_id": 28946, - "bbox": [ - 128.91, - 587.27, - 256.15, - 608.8 - ], - "text": "Therefore,\n ˙w1", - "type": "text" - }, - { - "block_id": "p961-b62", - "global_id": 28947, - "bbox": [ - 246.02, - 609.6, - 256.15, - 620.75 - ], - "text": "˙w2", - "type": "text" - }, - { - "block_id": "p961-b63", - "global_id": 28948, - "bbox": [ - 256.65, - 589.74, - 262.08, - 599.7 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b64", - "global_id": 28949, - "bbox": [ - 264.13, - 603.73, - 271.9, - 613.69 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p961-b65", - "global_id": 28950, - "bbox": [ - 273.95, - 589.74, - 314.84, - 619.98 - ], - "text": "−2\n0\n3\n−1", - "type": "text" - }, - { - "block_id": "p961-b66", - "global_id": 28951, - "bbox": [ - 314.84, - 589.74, - 335.83, - 608.8 - ], - "text": "! w1", - "type": "text" - }, - { - "block_id": "p961-b67", - "global_id": 28952, - "bbox": [ - 325.7, - 609.92, - 335.83, - 620.75 - ], - "text": "w2", - "type": "text" - }, - { - "block_id": "p961-b68", - "global_id": 28953, - "bbox": [ - 336.33, - 589.74, - 341.76, - 599.7 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b69", - "global_id": 28954, - "bbox": [ - 343.31, - 603.73, - 351.08, - 613.69 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p961-b70", - "global_id": 28955, - "bbox": [ - 352.63, - 589.74, - 366.93, - 608.03 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p961-b71", - "global_id": 28956, - "bbox": [ - 358.06, - 609.6, - 370.81, - 619.98 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p961-b72", - "global_id": 28957, - "bbox": [ - 370.81, - 589.74, - 376.24, - 599.7 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p961-b73", - "global_id": 28958, - "bbox": [ - 376.24, - 603.73, - 391.07, - 614.0 - ], - "text": "x(t)", - "type": "text" - } - ] - }, - { - "page_num": 962, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p962-b0", - "global_id": 28959, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "942\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p962-b1", - "global_id": 28960, - "bbox": [ - 103.16, - 86.16, - 477.01, - 120.12 - ], - "text": "This is the desired state equation for the state vector w. The solution of this equation requires\na knowledge of the initial state w(0). This can be obtained from the given initial state q(0) by\nusing Eq. (10.51).", - "type": "text" - }, - { - "block_id": "p962-b2", - "global_id": 28961, - "bbox": [ - 121.09, - 122.11, - 372.11, - 132.07 - ], - "text": "We can obtain the same result with less effort using MATLAB.", - "type": "text" - }, - { - "block_id": "p962-b3", - "global_id": 28962, - "bbox": [ - 103.17, - 142.32, - 275.76, - 224.01 - ], - "text": ">>\nA = [0 1;-2 -3]; B = [1; 2];\n>>\nP = [1 1;1 -1];\n>>\nAhat = P*A*inv(P), Bhat = P*B\nAhat =\n-2\n0\n3\n-1\nBhat =\n3\n-1", - "type": "text" - }, - { - "block_id": "p962-b4", - "global_id": 28963, - "bbox": [ - 101.84, - 266.21, - 490.4, - 387.96 - ], - "text": "INVARIANCE OF EIGENVALUES\nWe have seen that the poles of all possible transfer functions of a system are the eigenvalues of\nthe matrix A. If we transform a state vector from q to w, the variables w1, w2, . . . , wN are linear\ncombinations of q1, q2, . . . , qN and therefore may be considered to be outputs. Hence, the poles\nof the transfer functions relating w1, w2, . . . , wN to the various inputs must also be the eigenvalues\nof matrix A. On the other hand, the system is also specified by Eq. (10.48). This means that the\npoles of the transfer functions must be the eigenvalues of ˆA. Therefore, the eigenvalues of matrix\nA remain unchanged for the linear transformation of variables represented by Eq. (10.46), and the\neigenvalues of matrix A and matrix ˆA( ˆA = PAP−1) are identical, implying that the characteristic\nequations of A and ˆA are also identical. This result also can be proved alternately as follows.", - "type": "text" - }, - { - "block_id": "p962-b5", - "global_id": 28964, - "bbox": [ - 119.78, - 387.71, - 295.21, - 399.92 - ], - "text": "Consider the matrix P(sI −A)P−1. We have", - "type": "text" - }, - { - "block_id": "p962-b6", - "global_id": 28965, - "bbox": [ - 179.78, - 406.11, - 412.45, - 420.62 - ], - "text": "P(sI −A)P−1 = PsIP−1 −PAP−1 = sPIP−1 −ˆA = sI −ˆA", - "type": "text" - }, - { - "block_id": "p962-b7", - "global_id": 28966, - "bbox": [ - 101.85, - 431.52, - 295.29, - 441.49 - ], - "text": "Taking the determinants of both sides, we obtain", - "type": "text" - }, - { - "block_id": "p962-b8", - "global_id": 28967, - "bbox": [ - 242.17, - 449.57, - 350.06, - 462.19 - ], - "text": "|P||sI −A||P−1| = |sI −ˆA|", - "type": "text" - }, - { - "block_id": "p962-b9", - "global_id": 28968, - "bbox": [ - 101.85, - 470.85, - 373.84, - 483.06 - ], - "text": "The determinants |P| and |P−1| are reciprocals of each other. Hence,", - "type": "text" - }, - { - "block_id": "p962-b10", - "global_id": 28969, - "bbox": [ - 258.67, - 491.14, - 333.56, - 503.76 - ], - "text": "|sI −A| = |sI −ˆA|", - "type": "text" - }, - { - "block_id": "p962-b11", - "global_id": 28970, - "bbox": [ - 101.85, - 511.92, - 490.39, - 536.59 - ], - "text": "This is the desired result. We have shown that the characteristic equations of A and ˆA are identical.\nHence, the eigenvalues of A and ˆA are identical.", - "type": "text" - }, - { - "block_id": "p962-b12", - "global_id": 28971, - "bbox": [ - 119.78, - 538.5, - 253.86, - 548.54 - ], - "text": "In Ex. 10.10, matrix A is given as", - "type": "text" - }, - { - "block_id": "p962-b13", - "global_id": 28972, - "bbox": [ - 263.42, - 564.6, - 280.44, - 574.89 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p962-b14", - "global_id": 28973, - "bbox": [ - 282.48, - 550.61, - 323.37, - 580.85 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p962-b15", - "global_id": 28974, - "bbox": [ - 323.38, - 550.61, - 328.81, - 560.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p962-b16", - "global_id": 28975, - "bbox": [ - 101.84, - 591.09, - 218.88, - 601.05 - ], - "text": "The characteristic equation is", - "type": "text" - }, - { - "block_id": "p962-b17", - "global_id": 28976, - "bbox": [ - 217.77, - 617.1, - 259.1, - 627.4 - ], - "text": "|sI −A| =", - "type": "text" - }, - { - "block_id": "p962-b19", - "global_id": 28977, - "bbox": [ - 264.39, - 611.02, - 299.06, - 633.36 - ], - "text": "s\n−1\n2\ns + 3", - "type": "text" - }, - { - "block_id": "p962-b20", - "global_id": 28978, - "bbox": [ - 299.07, - 602.66, - 374.46, - 630.55 - ], - "text": "= s2 + 3s + 2 = 0", - "type": "text" - } - ] - }, - { - "page_num": 963, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p963-b0", - "global_id": 28979, - "bbox": [ - 293.79, - 62.89, - 516.13, - 71.98 - ], - "text": "10.5\nLinear Transformation of a State Vector\n943", - "type": "text" - }, - { - "block_id": "p963-b1", - "global_id": 28980, - "bbox": [ - 127.59, - 85.82, - 148.9, - 95.78 - ], - "text": "Also,", - "type": "text" - }, - { - "block_id": "p963-b2", - "global_id": 28981, - "bbox": [ - 289.17, - 101.16, - 306.17, - 113.79 - ], - "text": "ˆA =", - "type": "text" - }, - { - "block_id": "p963-b3", - "global_id": 28982, - "bbox": [ - 308.22, - 89.51, - 349.12, - 119.75 - ], - "text": "−2\n0\n3\n−1", - "type": "text" - }, - { - "block_id": "p963-b4", - "global_id": 28983, - "bbox": [ - 349.13, - 89.51, - 354.56, - 99.47 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p963-b5", - "global_id": 28984, - "bbox": [ - 127.59, - 128.93, - 141.97, - 138.89 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p963-b6", - "global_id": 28985, - "bbox": [ - 233.95, - 142.93, - 275.29, - 155.56 - ], - "text": "|sI −ˆA| =", - "type": "text" - }, - { - "block_id": "p963-b7", - "global_id": 28986, - "bbox": [ - 277.33, - 131.28, - 332.18, - 161.52 - ], - "text": "s + 2\n0\n−3\ns + 1", - "type": "text" - }, - { - "block_id": "p963-b8", - "global_id": 28987, - "bbox": [ - 332.18, - 131.28, - 337.61, - 141.24 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p963-b9", - "global_id": 28988, - "bbox": [ - 339.66, - 141.15, - 409.77, - 155.64 - ], - "text": "= s2 + 3s + 2 = 0", - "type": "text" - }, - { - "block_id": "p963-b10", - "global_id": 28989, - "bbox": [ - 127.59, - 170.96, - 431.29, - 183.66 - ], - "text": "This result verifies that the characteristic equations of A and ˆA are identical.", - "type": "text" - }, - { - "block_id": "p963-b11", - "global_id": 28990, - "bbox": [ - 127.59, - 208.77, - 317.69, - 221.18 - ], - "text": "10.5-1 Diagonalization of Matrix A", - "type": "text" - }, - { - "block_id": "p963-b12", - "global_id": 28991, - "bbox": [ - 127.59, - 227.22, - 516.13, - 274.63 - ], - "text": "For several reasons, it is desirable to make matrix A diagonal. If A is not diagonal, we can\ntransform the state variables such that the resulting matrix ˆA is diagonal.† One can show that\nfor any diagonal matrix A, the diagonal elements of this matrix must necessarily be λ1, λ2, . . . ,\nλN (the eigenvalues) of the matrix. Consider the diagonal matrix A:", - "type": "text" - }, - { - "block_id": "p963-b13", - "global_id": 28992, - "bbox": [ - 261.87, - 306.37, - 278.89, - 316.66 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p963-b14", - "global_id": 28993, - "bbox": [ - 280.94, - 277.44, - 288.2, - 287.41 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p963-b15", - "global_id": 28994, - "bbox": [ - 280.94, - 294.91, - 288.2, - 323.27 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p963-b16", - "global_id": 28995, - "bbox": [ - 288.2, - 285.05, - 373.66, - 338.58 - ], - "text": "a1\n0\n0\n· · ·\n0\n0\na2\n0\n· · ·\n0\n...\n...\n...\n· · ·\n...\n0\n0\n0\n· · ·\naN", - "type": "text" - }, - { - "block_id": "p963-b17", - "global_id": 28996, - "bbox": [ - 374.58, - 277.44, - 381.84, - 287.41 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p963-b18", - "global_id": 28997, - "bbox": [ - 374.58, - 294.91, - 381.84, - 323.27 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p963-b19", - "global_id": 28998, - "bbox": [ - 127.59, - 350.04, - 281.29, - 360.01 - ], - "text": "The characteristic equation is given by", - "type": "text" - }, - { - "block_id": "p963-b20", - "global_id": 28999, - "bbox": [ - 208.03, - 393.25, - 249.36, - 403.54 - ], - "text": "|sI −A| =", - "type": "text" - }, - { - "block_id": "p963-b21", - "global_id": 29000, - "bbox": [ - 251.41, - 364.31, - 258.67, - 374.27 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p963-b22", - "global_id": 29001, - "bbox": [ - 251.41, - 381.79, - 258.67, - 410.14 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p963-b23", - "global_id": 29002, - "bbox": [ - 258.67, - 371.91, - 411.59, - 425.45 - ], - "text": "(s −a1)\n0\n0\n· · ·\n0\n0\n(s −a2)\n0\n· · ·\n0\n...\n...\n...\n· · ·\n...\n0\n0\n0\n· · ·\n(s −aN)", - "type": "text" - }, - { - "block_id": "p963-b24", - "global_id": 29003, - "bbox": [ - 411.59, - 364.31, - 418.85, - 374.27 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p963-b25", - "global_id": 29004, - "bbox": [ - 411.59, - 381.79, - 435.69, - 410.14 - ], - "text": "⎥⎥⎥⎦= 0", - "type": "text" - }, - { - "block_id": "p963-b26", - "global_id": 29005, - "bbox": [ - 127.59, - 434.7, - 135.89, - 444.66 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p963-b27", - "global_id": 29006, - "bbox": [ - 258.29, - 447.44, - 385.43, - 458.59 - ], - "text": "(s −a1)(s −a2)· · ·(s −aN) = 0", - "type": "text" - }, - { - "block_id": "p963-b28", - "global_id": 29007, - "bbox": [ - 127.59, - 467.17, - 516.13, - 489.51 - ], - "text": "The nonzero (diagonal) elements of a diagonal matrix are therefore its eigenvalues λ1, λ2, . . . , λN.\nWe shall denote the diagonal matrix by the symbol, A:", - "type": "text" - }, - { - "block_id": "p963-b29", - "global_id": 29008, - "bbox": [ - 260.88, - 522.74, - 278.47, - 532.7 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p963-b30", - "global_id": 29009, - "bbox": [ - 280.52, - 493.81, - 287.78, - 503.78 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p963-b31", - "global_id": 29010, - "bbox": [ - 280.52, - 511.28, - 287.78, - 539.64 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p963-b32", - "global_id": 29011, - "bbox": [ - 287.78, - 501.42, - 374.66, - 554.95 - ], - "text": "λ1\n0\n0\n· · ·\n0\n0\nλ2\n0\n· · ·\n0\n...\n...\n...\n· · ·\n...\n0\n0\n0\n· · ·\nλN", - "type": "text" - }, - { - "block_id": "p963-b33", - "global_id": 29012, - "bbox": [ - 375.57, - 493.81, - 382.83, - 503.78 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p963-b34", - "global_id": 29013, - "bbox": [ - 375.57, - 511.28, - 516.12, - 539.64 - ], - "text": "⎥⎥⎥⎦\n(10.52)", - "type": "text" - }, - { - "block_id": "p963-b35", - "global_id": 29014, - "bbox": [ - 127.59, - 566.68, - 516.14, - 591.34 - ], - "text": "Let us now consider the transformation of the state vector A such that the resulting matrix ˆA is a\ndiagonal matrix \n.", - "type": "text" - }, - { - "block_id": "p963-b36", - "global_id": 29015, - "bbox": [ - 127.59, - 610.24, - 516.1, - 633.41 - ], - "text": "† In this discussion we assume distinct eigenvalues. If the eigenvalues are not distinct, we can reduce the\nmatrix to a modified diagonalized (Jordan) form.", - "type": "text" - } - ] - }, - { - "page_num": 964, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p964-b0", - "global_id": 29016, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "944\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p964-b1", - "global_id": 29017, - "bbox": [ - 119.78, - 85.82, - 200.58, - 95.78 - ], - "text": "Consider the system", - "type": "text" - }, - { - "block_id": "p964-b2", - "global_id": 29018, - "bbox": [ - 269.8, - 103.55, - 322.44, - 113.86 - ], - "text": "˙q = Aq + Bx", - "type": "text" - }, - { - "block_id": "p964-b3", - "global_id": 29019, - "bbox": [ - 101.84, - 126.63, - 490.39, - 148.96 - ], - "text": "We shall assume that λ1, λ2, . . . , λN, the eigenvalues of A, are distinct (no repeated roots). Let us\ntransform the state vector q into the new state vector z, using the transformation", - "type": "text" - }, - { - "block_id": "p964-b4", - "global_id": 29020, - "bbox": [ - 282.16, - 164.64, - 490.38, - 175.01 - ], - "text": "z = Pq\n(10.53)", - "type": "text" - }, - { - "block_id": "p964-b5", - "global_id": 29021, - "bbox": [ - 101.84, - 191.11, - 309.69, - 201.07 - ], - "text": "Then, after the development of Eq. (10.48), we have", - "type": "text" - }, - { - "block_id": "p964-b6", - "global_id": 29022, - "bbox": [ - 257.08, - 214.92, - 335.16, - 227.04 - ], - "text": "˙z = PAP−1z + PBx", - "type": "text" - }, - { - "block_id": "p964-b7", - "global_id": 29023, - "bbox": [ - 101.84, - 239.0, - 490.38, - 265.14 - ], - "text": "We desire the transformation to be such that PAP−1 is a diagonal matrix \n given by Eq. (10.52),\nor", - "type": "text" - }, - { - "block_id": "p964-b8", - "global_id": 29024, - "bbox": [ - 270.63, - 270.59, - 490.38, - 283.29 - ], - "text": "˙z = \nz + ˆBx\n(10.54)", - "type": "text" - }, - { - "block_id": "p964-b9", - "global_id": 29025, - "bbox": [ - 101.84, - 296.4, - 129.76, - 306.36 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p964-b10", - "global_id": 29026, - "bbox": [ - 271.9, - 312.32, - 319.83, - 324.44 - ], - "text": "= PAP−1", - "type": "text" - }, - { - "block_id": "p964-b11", - "global_id": 29027, - "bbox": [ - 101.84, - 337.63, - 110.14, - 347.59 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p964-b12", - "global_id": 29028, - "bbox": [ - 276.61, - 355.37, - 490.38, - 365.75 - ], - "text": "P = PA\n(10.55)", - "type": "text" - }, - { - "block_id": "p964-b13", - "global_id": 29029, - "bbox": [ - 101.84, - 378.44, - 407.23, - 388.82 - ], - "text": "We know \n and A. Equation (10.55) therefore can be solved to determine P.", - "type": "text" - }, - { - "block_id": "p964-b14", - "global_id": 29030, - "bbox": [ - 76.77, - 419.88, - 391.82, - 431.83 - ], - "text": "EXAMPLE 10.11\nDiagonal Form of the State Equations", - "type": "text" - }, - { - "block_id": "p964-b15", - "global_id": 29031, - "bbox": [ - 103.16, - 448.49, - 410.84, - 458.46 - ], - "text": "Find the diagonalized form of the state equations for the system in Ex. 10.10.", - "type": "text" - }, - { - "block_id": "p964-b16", - "global_id": 29032, - "bbox": [ - 103.16, - 481.37, - 150.47, - 491.33 - ], - "text": "In this case,", - "type": "text" - }, - { - "block_id": "p964-b17", - "global_id": 29033, - "bbox": [ - 257.4, - 497.84, - 274.41, - 508.13 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p964-b18", - "global_id": 29034, - "bbox": [ - 276.46, - 483.85, - 317.35, - 514.09 - ], - "text": "0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p964-b19", - "global_id": 29035, - "bbox": [ - 317.35, - 483.86, - 322.78, - 493.82 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p964-b20", - "global_id": 29036, - "bbox": [ - 103.16, - 522.06, - 264.79, - 533.52 - ], - "text": "We found λ1 = −1 and λ2 = −2. Hence,", - "type": "text" - }, - { - "block_id": "p964-b21", - "global_id": 29037, - "bbox": [ - 257.11, - 548.96, - 274.7, - 558.92 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p964-b22", - "global_id": 29038, - "bbox": [ - 276.75, - 534.98, - 317.64, - 565.21 - ], - "text": "−1\n0\n0\n−2", - "type": "text" - }, - { - "block_id": "p964-b23", - "global_id": 29039, - "bbox": [ - 317.64, - 534.98, - 323.07, - 544.94 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p964-b24", - "global_id": 29040, - "bbox": [ - 103.16, - 576.58, - 202.46, - 586.54 - ], - "text": "and Eq. (10.55) becomes", - "type": "text" - }, - { - "block_id": "p964-b25", - "global_id": 29041, - "bbox": [ - 191.0, - 589.74, - 231.89, - 619.98 - ], - "text": "−1\n0\n0\n−2", - "type": "text" - }, - { - "block_id": "p964-b26", - "global_id": 29042, - "bbox": [ - 231.89, - 589.74, - 278.22, - 620.75 - ], - "text": "! p11\np12\np21\np22", - "type": "text" - }, - { - "block_id": "p964-b27", - "global_id": 29043, - "bbox": [ - 278.73, - 589.74, - 284.16, - 599.7 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p964-b28", - "global_id": 29044, - "bbox": [ - 286.21, - 603.73, - 293.98, - 613.69 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p964-b29", - "global_id": 29045, - "bbox": [ - 296.03, - 589.74, - 335.82, - 620.75 - ], - "text": "p11\np12\np21\np22", - "type": "text" - }, - { - "block_id": "p964-b30", - "global_id": 29046, - "bbox": [ - 336.32, - 589.74, - 383.75, - 619.98 - ], - "text": "! 0\n1\n−2\n−3", - "type": "text" - }, - { - "block_id": "p964-b31", - "global_id": 29047, - "bbox": [ - 383.76, - 589.74, - 389.18, - 599.7 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 965, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p965-b0", - "global_id": 29048, - "bbox": [ - 293.79, - 62.89, - 516.13, - 71.98 - ], - "text": "10.5\nLinear Transformation of a State Vector\n945", - "type": "text" - }, - { - "block_id": "p965-b1", - "global_id": 29049, - "bbox": [ - 128.9, - 86.24, - 332.72, - 96.2 - ], - "text": "Equating the four elements on two sides, we obtain", - "type": "text" - }, - { - "block_id": "p965-b2", - "global_id": 29050, - "bbox": [ - 276.92, - 105.75, - 354.23, - 162.04 - ], - "text": "−p11 = −2p12\n−p12 = p11 −3p12\n−2p21 = −2p22\n−2p22 = p21 −3p22", - "type": "text" - }, - { - "block_id": "p965-b3", - "global_id": 29051, - "bbox": [ - 128.9, - 170.92, - 502.76, - 241.74 - ], - "text": "The reader will immediately recognize that the first two equations are identical and that the last\ntwo equations are identical. Hence, two equations may be discarded, leaving us with only two\nequations [p11 = 2p12 and p21 = p22] and four unknowns. This observation means that there is\nno unique solution. There is, in fact, an infinite number of solutions. We can assign any value\nto p11 and p21 to yield one possible solution.† If p11 = k1 and p21 = k2, then we have p12 = k1/2\nand p22 = k2:", - "type": "text" - }, - { - "block_id": "p965-b4", - "global_id": 29052, - "bbox": [ - 287.75, - 256.21, - 303.66, - 266.5 - ], - "text": "P =", - "type": "text" - }, - { - "block_id": "p965-b6", - "global_id": 29053, - "bbox": [ - 311.14, - 250.93, - 319.04, - 261.77 - ], - "text": "k1", - "type": "text" - }, - { - "block_id": "p965-b7", - "global_id": 29054, - "bbox": [ - 330.7, - 248.54, - 336.79, - 256.85 - ], - "text": "k1", - "type": "text" - }, - { - "block_id": "p965-b8", - "global_id": 29055, - "bbox": [ - 311.14, - 256.84, - 337.7, - 273.72 - ], - "text": "2\nk2\nk2", - "type": "text" - }, - { - "block_id": "p965-b9", - "global_id": 29056, - "bbox": [ - 338.49, - 242.22, - 343.92, - 252.19 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b10", - "global_id": 29057, - "bbox": [ - 128.91, - 281.91, - 502.75, - 304.24 - ], - "text": "We may assign any values to k1 and k2. For convenience, let k1 = 2 and k2 = 1. This substitution\nyields", - "type": "text" - }, - { - "block_id": "p965-b11", - "global_id": 29058, - "bbox": [ - 291.46, - 319.44, - 307.37, - 329.73 - ], - "text": "P =", - "type": "text" - }, - { - "block_id": "p965-b12", - "global_id": 29059, - "bbox": [ - 309.42, - 305.45, - 334.78, - 335.69 - ], - "text": "2\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p965-b13", - "global_id": 29060, - "bbox": [ - 334.77, - 305.45, - 340.2, - 315.41 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b14", - "global_id": 29061, - "bbox": [ - 128.91, - 345.07, - 302.11, - 355.03 - ], - "text": "The transformed variables [Eq. (10.53)] are", - "type": "text" - }, - { - "block_id": "p965-b15", - "global_id": 29062, - "bbox": [ - 246.43, - 356.23, - 259.22, - 375.29 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p965-b16", - "global_id": 29063, - "bbox": [ - 251.85, - 376.41, - 259.22, - 387.25 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p965-b17", - "global_id": 29064, - "bbox": [ - 259.72, - 356.23, - 265.15, - 366.19 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b18", - "global_id": 29065, - "bbox": [ - 267.2, - 370.22, - 274.97, - 380.19 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p965-b19", - "global_id": 29066, - "bbox": [ - 277.02, - 356.23, - 302.37, - 386.48 - ], - "text": "2\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p965-b20", - "global_id": 29067, - "bbox": [ - 302.37, - 356.23, - 322.8, - 375.29 - ], - "text": "! q1", - "type": "text" - }, - { - "block_id": "p965-b21", - "global_id": 29068, - "bbox": [ - 314.33, - 376.41, - 322.8, - 387.25 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p965-b22", - "global_id": 29069, - "bbox": [ - 323.3, - 356.23, - 328.73, - 366.19 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b23", - "global_id": 29070, - "bbox": [ - 330.77, - 370.22, - 338.55, - 380.19 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p965-b24", - "global_id": 29071, - "bbox": [ - 340.59, - 356.23, - 379.31, - 375.6 - ], - "text": "2q1 + q2", - "type": "text" - }, - { - "block_id": "p965-b25", - "global_id": 29072, - "bbox": [ - 348.51, - 376.1, - 376.81, - 387.56 - ], - "text": "q1 + q2", - "type": "text" - }, - { - "block_id": "p965-b26", - "global_id": 29073, - "bbox": [ - 379.81, - 356.23, - 385.24, - 366.19 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b27", - "global_id": 29074, - "bbox": [ - 128.91, - 395.74, - 502.75, - 418.54 - ], - "text": "This expression relates the new state variables z1 and z2 to the original state variables q1 and\nq2. The system equation with z as the state vector is given by [see Eq. (10.54)]", - "type": "text" - }, - { - "block_id": "p965-b28", - "global_id": 29075, - "bbox": [ - 290.35, - 424.99, - 341.33, - 437.61 - ], - "text": "˙z = \nz + ˆBx", - "type": "text" - }, - { - "block_id": "p965-b29", - "global_id": 29076, - "bbox": [ - 128.9, - 447.66, - 153.23, - 457.62 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p965-b30", - "global_id": 29077, - "bbox": [ - 257.11, - 460.45, - 298.17, - 473.07 - ], - "text": "ˆB = PB =", - "type": "text" - }, - { - "block_id": "p965-b31", - "global_id": 29078, - "bbox": [ - 300.22, - 448.79, - 325.58, - 479.04 - ], - "text": "2\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p965-b32", - "global_id": 29079, - "bbox": [ - 325.58, - 448.79, - 341.42, - 467.08 - ], - "text": "! 1", - "type": "text" - }, - { - "block_id": "p965-b33", - "global_id": 29080, - "bbox": [ - 336.43, - 469.08, - 341.42, - 479.04 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p965-b34", - "global_id": 29081, - "bbox": [ - 341.42, - 448.79, - 346.85, - 458.76 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b35", - "global_id": 29082, - "bbox": [ - 348.89, - 462.78, - 356.66, - 472.74 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p965-b36", - "global_id": 29083, - "bbox": [ - 358.71, - 448.79, - 369.13, - 467.08 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p965-b37", - "global_id": 29084, - "bbox": [ - 364.15, - 469.08, - 369.13, - 479.04 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p965-b38", - "global_id": 29085, - "bbox": [ - 369.12, - 448.79, - 374.55, - 458.76 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b39", - "global_id": 29086, - "bbox": [ - 128.91, - 487.13, - 265.23, - 508.68 - ], - "text": "Hence,\n ˙z1", - "type": "text" - }, - { - "block_id": "p965-b40", - "global_id": 29087, - "bbox": [ - 257.87, - 509.48, - 265.23, - 520.63 - ], - "text": "˙z2", - "type": "text" - }, - { - "block_id": "p965-b41", - "global_id": 29088, - "bbox": [ - 265.72, - 489.62, - 271.15, - 499.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b42", - "global_id": 29089, - "bbox": [ - 273.2, - 503.6, - 280.97, - 513.56 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p965-b43", - "global_id": 29090, - "bbox": [ - 283.02, - 489.62, - 323.91, - 519.86 - ], - "text": "−1\n0\n0\n−2", - "type": "text" - }, - { - "block_id": "p965-b44", - "global_id": 29091, - "bbox": [ - 323.92, - 489.62, - 342.15, - 508.68 - ], - "text": "! z1", - "type": "text" - }, - { - "block_id": "p965-b45", - "global_id": 29092, - "bbox": [ - 334.78, - 509.79, - 342.15, - 520.63 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p965-b46", - "global_id": 29093, - "bbox": [ - 342.64, - 489.62, - 348.07, - 499.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b47", - "global_id": 29094, - "bbox": [ - 349.62, - 503.6, - 357.39, - 513.56 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p965-b48", - "global_id": 29095, - "bbox": [ - 358.94, - 489.62, - 369.35, - 507.9 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p965-b49", - "global_id": 29096, - "bbox": [ - 364.37, - 509.89, - 369.35, - 519.86 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p965-b50", - "global_id": 29097, - "bbox": [ - 369.35, - 489.62, - 374.78, - 499.58 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p965-b51", - "global_id": 29098, - "bbox": [ - 374.77, - 503.92, - 502.75, - 513.98 - ], - "text": "x\n(10.56)", - "type": "text" - }, - { - "block_id": "p965-b52", - "global_id": 29099, - "bbox": [ - 128.9, - 547.97, - 502.76, - 594.4 - ], - "text": "† If, however, we want the state equations in diagonalized form, as in Eq. (10.26), where all the elements\nof ˆB matrix are unity, there is a unique solution. The reason is that the equation ˆB = PB, where all the\nelements of ˆB are unity, imposes additional constraints. In the present example, this condition will yield\np11 = 1/2, p12 = 1/4, p21 = 1/3, and p22 = 1/3. The relationship between z and q is then", - "type": "text" - }, - { - "block_id": "p965-b53", - "global_id": 29100, - "bbox": [ - 245.01, - 603.65, - 267.35, - 615.32 - ], - "text": "z1 = 1", - "type": "text" - }, - { - "block_id": "p965-b54", - "global_id": 29101, - "bbox": [ - 264.11, - 603.65, - 353.61, - 616.93 - ], - "text": "2q1 + 1\n4q2\nand\nz2 = 1", - "type": "text" - }, - { - "block_id": "p965-b55", - "global_id": 29102, - "bbox": [ - 350.38, - 603.65, - 386.16, - 616.93 - ], - "text": "3q1 + 1\n3q2", - "type": "text" - } - ] - }, - { - "page_num": 966, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p966-b0", - "global_id": 29103, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "946\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p966-b1", - "global_id": 29104, - "bbox": [ - 103.16, - 84.03, - 111.46, - 93.99 - ], - "text": "or", - "type": "text" - }, - { - "block_id": "p966-b2", - "global_id": 29105, - "bbox": [ - 259.76, - 103.55, - 315.39, - 115.0 - ], - "text": "˙z1 = −z1 + 4x", - "type": "text" - }, - { - "block_id": "p966-b3", - "global_id": 29106, - "bbox": [ - 259.76, - 118.5, - 320.38, - 129.95 - ], - "text": "˙z2 = −2z2 + 3x", - "type": "text" - }, - { - "block_id": "p966-b4", - "global_id": 29107, - "bbox": [ - 103.17, - 138.85, - 477.04, - 208.59 - ], - "text": "Note the distinctive nature of these state equations. Each state equation involves only one\nvariable and therefore can be solved by itself. A general state equation has the derivative of\none state variable equal to a linear combination of all state variables. Such is not the case with\nthe diagonalized matrix \n. Each state variable zi is chosen so that it is uncoupled from the rest\nof the variables; hence, a system with N eigenvalues is split into N decoupled systems, each\nwith an equation of the form", - "type": "text" - }, - { - "block_id": "p966-b5", - "global_id": 29108, - "bbox": [ - 242.59, - 210.16, - 337.6, - 221.62 - ], - "text": "˙zi = λizi + (input terms)", - "type": "text" - }, - { - "block_id": "p966-b6", - "global_id": 29109, - "bbox": [ - 103.17, - 228.52, - 477.04, - 250.44 - ], - "text": "This fact also can be readily seen from Fig. 10.8a, which is a realization of the system\nrepresented by Eq. (10.56). In contrast, consider the original state equations [see Ex. 10.10]", - "type": "text" - }, - { - "block_id": "p966-b7", - "global_id": 29110, - "bbox": [ - 241.06, - 259.99, - 296.56, - 271.45 - ], - "text": "˙q1 = q2 + x(t)", - "type": "text" - }, - { - "block_id": "p966-b8", - "global_id": 29111, - "bbox": [ - 241.06, - 274.94, - 339.11, - 286.4 - ], - "text": "˙q2 = −2q1 −3q2 + 2x(t)", - "type": "text" - }, - { - "block_id": "p966-b9", - "global_id": 29112, - "bbox": [ - 103.16, - 295.29, - 477.01, - 329.17 - ], - "text": "A realization for these equations is shown in Fig. 10.8b. It can be seen from Fig. 10.8a that the\nstates z1 and z2 are decoupled, whereas the states q1 and q2 (Fig. 10.8b) are coupled. It should\nbe remembered that Figs. 10.8a and 10.8b are realizations of the same system.†", - "type": "text" - }, - { - "block_id": "p966-b10", - "global_id": 29113, - "bbox": [ - 157.96, - 526.74, - 341.69, - 534.74 - ], - "text": "(a)\n(b)", - "type": "text" - }, - { - "block_id": "p966-b11", - "global_id": 29114, - "bbox": [ - 357.19, - 445.17, - 367.86, - 453.46 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p966-b12", - "global_id": 29115, - "bbox": [ - 366.5, - 499.27, - 377.16, - 507.57 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p966-b13", - "global_id": 29116, - "bbox": [ - 264.43, - 418.44, - 267.98, - 426.44 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p966-b14", - "global_id": 29117, - "bbox": [ - 303.48, - 454.54, - 307.48, - 462.54 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p966-b15", - "global_id": 29118, - "bbox": [ - 351.08, - 385.92, - 358.08, - 395.52 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p966-b16", - "global_id": 29119, - "bbox": [ - 357.06, - 428.66, - 364.06, - 438.27 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p966-b17", - "global_id": 29120, - "bbox": [ - 352.67, - 381.5, - 354.67, - 389.5 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p966-b18", - "global_id": 29121, - "bbox": [ - 394.58, - 455.1, - 401.58, - 464.71 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p966-b19", - "global_id": 29122, - "bbox": [ - 393.95, - 511.21, - 400.95, - 520.82 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p966-b20", - "global_id": 29123, - "bbox": [ - 395.94, - 450.68, - 397.94, - 458.68 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p966-b21", - "global_id": 29124, - "bbox": [ - 395.51, - 480.29, - 399.51, - 495.33 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p966-b22", - "global_id": 29125, - "bbox": [ - 352.39, - 412.64, - 356.39, - 427.68 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p966-b25", - "global_id": 29126, - "bbox": [ - 171.98, - 431.72, - 182.64, - 440.01 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p966-b26", - "global_id": 29127, - "bbox": [ - 171.98, - 504.72, - 182.64, - 513.01 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p966-b27", - "global_id": 29128, - "bbox": [ - 138.44, - 460.25, - 142.44, - 468.25 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p966-b28", - "global_id": 29129, - "bbox": [ - 110.02, - 418.44, - 113.57, - 426.44 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p966-b29", - "global_id": 29130, - "bbox": [ - 136.77, - 387.51, - 140.77, - 395.51 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p966-b30", - "global_id": 29131, - "bbox": [ - 200.99, - 412.64, - 204.99, - 427.68 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p966-b31", - "global_id": 29132, - "bbox": [ - 200.99, - 485.39, - 204.99, - 500.43 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p966-b34", - "global_id": 29133, - "bbox": [ - 217.16, - 429.42, - 223.27, - 439.02 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p966-b35", - "global_id": 29134, - "bbox": [ - 217.16, - 502.92, - 223.27, - 512.52 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p966-b36", - "global_id": 29135, - "bbox": [ - 199.87, - 385.92, - 205.98, - 395.52 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p966-b37", - "global_id": 29136, - "bbox": [ - 200.73, - 381.5, - 202.73, - 389.5 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p966-b38", - "global_id": 29137, - "bbox": [ - 200.85, - 458.92, - 206.97, - 468.52 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p966-b39", - "global_id": 29138, - "bbox": [ - 201.71, - 454.5, - 203.71, - 462.5 - ], - "text": ".", - "type": "text" - }, - { - "block_id": "p966-b40", - "global_id": 29139, - "bbox": [ - 94.2, - 541.43, - 304.82, - 550.67 - ], - "text": "Figure 10.8 Two realizations of the second-order system.", - "type": "text" - }, - { - "block_id": "p966-b41", - "global_id": 29140, - "bbox": [ - 103.16, - 582.95, - 477.01, - 617.45 - ], - "text": "† Here we have only a simulated state equation; the outputs are not shown. The outputs are linear\ncombinations of state variables (and inputs). Hence, the output equation can be easily incorporated into\nthese diagrams.", - "type": "text" - } - ] - }, - { - "page_num": 967, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p967-b0", - "global_id": 29141, - "bbox": [ - 317.02, - 62.89, - 516.13, - 71.98 - ], - "text": "10.6\nControllability and Observability\n947", - "type": "text" - }, - { - "block_id": "p967-b1", - "global_id": 29142, - "bbox": [ - 128.9, - 86.62, - 502.77, - 136.64 - ], - "text": "MATRIX DIAGONALIZATION VIA MATLAB\nThe key to diagonalizing matrix A is to determine a matrix P that satisfies \nP = PA\n[Eq. (10.55)], where \n is a diagonal matrix of the eigenvalues of A. This problem is directly\nrelated to the classic eigenvalue problem, stated as", - "type": "text" - }, - { - "block_id": "p967-b2", - "global_id": 29143, - "bbox": [ - 295.94, - 148.18, - 335.72, - 158.48 - ], - "text": "AV = V", - "type": "text" - }, - { - "block_id": "p967-b3", - "global_id": 29144, - "bbox": [ - 128.9, - 170.43, - 502.75, - 192.43 - ], - "text": "where V is a matrix of eigenvectors for A. If we can find V, we can take its inverse to determine\nP. That is, P = V−1. This relationship is more fully developed in Sec. 10.8.", - "type": "text" - }, - { - "block_id": "p967-b4", - "global_id": 29145, - "bbox": [ - 128.9, - 194.42, - 502.76, - 228.3 - ], - "text": "MATLAB’s built-in function eig can determine the eigenvectors of a matrix and,\ntherefore, can help us determine a suitable matrix P. Let us demonstrate this approach for\nthe current case.", - "type": "text" - }, - { - "block_id": "p967-b5", - "global_id": 29146, - "bbox": [ - 128.9, - 238.55, - 306.74, - 344.16 - ], - "text": ">>\nA = [0 1;-2 -3]; B = [1; 2];\n>>\n[V, Lambda] = eig(A);\n>>\nP = inv(V), Lambda, Bhat = P*B\nP =\n2.8284\n1.4142\n2.2361\n2.2361\nLambda =\n-1\n0\n0\n-2\nBhat =\n5.6569\n6.7082", - "type": "text" - }, - { - "block_id": "p967-b6", - "global_id": 29147, - "bbox": [ - 128.9, - 353.82, - 170.65, - 363.79 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p967-b7", - "global_id": 29148, - "bbox": [ - 232.92, - 370.29, - 247.17, - 380.59 - ], - "text": "z =", - "type": "text" - }, - { - "block_id": "p967-b8", - "global_id": 29149, - "bbox": [ - 249.21, - 356.31, - 262.01, - 375.37 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p967-b9", - "global_id": 29150, - "bbox": [ - 254.64, - 376.48, - 262.01, - 387.32 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p967-b10", - "global_id": 29151, - "bbox": [ - 262.51, - 356.31, - 267.94, - 366.27 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p967-b11", - "global_id": 29152, - "bbox": [ - 269.99, - 370.29, - 277.76, - 380.25 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p967-b12", - "global_id": 29153, - "bbox": [ - 279.81, - 356.31, - 349.97, - 386.55 - ], - "text": "2.8284\n1.4142\n2.2361\n2.2361", - "type": "text" - }, - { - "block_id": "p967-b13", - "global_id": 29154, - "bbox": [ - 349.99, - 356.31, - 369.32, - 375.37 - ], - "text": "! q1", - "type": "text" - }, - { - "block_id": "p967-b14", - "global_id": 29155, - "bbox": [ - 360.85, - 376.48, - 369.32, - 387.32 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p967-b15", - "global_id": 29156, - "bbox": [ - 369.82, - 356.31, - 375.25, - 366.27 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p967-b16", - "global_id": 29157, - "bbox": [ - 377.3, - 370.29, - 398.74, - 380.59 - ], - "text": "= Pq", - "type": "text" - }, - { - "block_id": "p967-b17", - "global_id": 29158, - "bbox": [ - 128.91, - 394.92, - 143.29, - 404.89 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p967-b18", - "global_id": 29159, - "bbox": [ - 206.16, - 410.04, - 220.4, - 420.35 - ], - "text": "˙z =", - "type": "text" - }, - { - "block_id": "p967-b19", - "global_id": 29160, - "bbox": [ - 222.45, - 396.07, - 235.25, - 415.13 - ], - "text": "˙z1", - "type": "text" - }, - { - "block_id": "p967-b20", - "global_id": 29161, - "bbox": [ - 227.89, - 415.93, - 235.25, - 427.08 - ], - "text": "˙z2", - "type": "text" - }, - { - "block_id": "p967-b21", - "global_id": 29162, - "bbox": [ - 235.74, - 396.07, - 241.17, - 406.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p967-b22", - "global_id": 29163, - "bbox": [ - 243.22, - 410.05, - 250.99, - 420.02 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p967-b23", - "global_id": 29164, - "bbox": [ - 253.04, - 396.07, - 293.93, - 426.31 - ], - "text": "−1\n0\n0\n−2", - "type": "text" - }, - { - "block_id": "p967-b24", - "global_id": 29165, - "bbox": [ - 293.94, - 396.07, - 312.16, - 415.13 - ], - "text": "! z1", - "type": "text" - }, - { - "block_id": "p967-b25", - "global_id": 29166, - "bbox": [ - 304.79, - 416.24, - 312.16, - 427.08 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p967-b26", - "global_id": 29167, - "bbox": [ - 312.66, - 396.07, - 318.09, - 406.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p967-b27", - "global_id": 29168, - "bbox": [ - 319.63, - 410.05, - 327.4, - 420.02 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p967-b28", - "global_id": 29169, - "bbox": [ - 328.95, - 396.07, - 361.76, - 414.36 - ], - "text": "5.6569", - "type": "text" - }, - { - "block_id": "p967-b29", - "global_id": 29170, - "bbox": [ - 334.38, - 416.35, - 361.76, - 426.31 - ], - "text": "6.7082", - "type": "text" - }, - { - "block_id": "p967-b30", - "global_id": 29171, - "bbox": [ - 361.78, - 396.07, - 367.21, - 406.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p967-b31", - "global_id": 29172, - "bbox": [ - 367.21, - 407.72, - 425.5, - 420.35 - ], - "text": "x(t) = \nz+ ˆBx", - "type": "text" - }, - { - "block_id": "p967-b32", - "global_id": 29173, - "bbox": [ - 128.91, - 434.95, - 502.78, - 471.57 - ], - "text": "Recall that neither P nor ˆB are unique, which explains why the MATLAB output does not\nneed to match our previous solution. Still, the MATLAB results do their job and successfully\ndiagonalize matrix A.", - "type": "text" - }, - { - "block_id": "p967-b33", - "global_id": 29174, - "bbox": [ - 127.94, - 521.71, - 422.84, - 535.66 - ], - "text": "10.6 CONTROLLABILITY AND OBSERVABILITY", - "type": "text" - }, - { - "block_id": "p967-b34", - "global_id": 29175, - "bbox": [ - 127.59, - 541.65, - 364.72, - 551.61 - ], - "text": "Consider a diagonalized state-space description of a system", - "type": "text" - }, - { - "block_id": "p967-b35", - "global_id": 29176, - "bbox": [ - 242.39, - 562.39, - 516.12, - 575.1 - ], - "text": "˙z = \nz + ˆBx\nand\nY = ˆCz + Dx\n(10.57)", - "type": "text" - }, - { - "block_id": "p967-b36", - "global_id": 29177, - "bbox": [ - 127.59, - 588.2, - 516.13, - 610.53 - ], - "text": "We shall assume that all N eigenvalues λ1, λ2, . . . , λN are distinct. The state equations in\nEq. (10.57) are of the form", - "type": "text" - }, - { - "block_id": "p967-b37", - "global_id": 29178, - "bbox": [ - 200.04, - 621.59, - 443.08, - 635.1 - ], - "text": "˙zm = λmzm + ˆbm1x1 + ˆbm2x2 + · · · + ˆbmjxj\nm = 1,2,. . .,N", - "type": "text" - } - ] - }, - { - "page_num": 968, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p968-b0", - "global_id": 29179, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "948\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p968-b1", - "global_id": 29180, - "bbox": [ - 101.84, - 86.09, - 355.29, - 100.29 - ], - "text": "If ˆbm1, ˆbm2, . . . , ˆbmj (the mth row in matrix ˆB) are all zero, then", - "type": "text" - }, - { - "block_id": "p968-b2", - "global_id": 29181, - "bbox": [ - 275.28, - 116.78, - 316.45, - 128.23 - ], - "text": "˙zm = λmzm", - "type": "text" - }, - { - "block_id": "p968-b3", - "global_id": 29182, - "bbox": [ - 101.84, - 145.45, - 490.39, - 215.29 - ], - "text": "and the variable zm is uncontrollable because zm is not coupled to any of the inputs. Moreover, zm\nis decoupled from all the remaining (N −1) state variables because of the diagonalized nature of\nthe variables. Hence, there is no direct or indirect coupling of zm with any of the inputs, and the\nsystem is uncontrollable. In contrast, if at least one element in the mth row of ˆB is nonzero, zm is\ncoupled to at least one input and is therefore controllable. Thus, a system with a diagonalized state\n[Eq. (10.57)] is completely controllable if and only if the matrix ˆB has no row of zero elements.", - "type": "text" - }, - { - "block_id": "p968-b4", - "global_id": 29183, - "bbox": [ - 119.78, - 217.28, - 298.53, - 227.24 - ], - "text": "The outputs [see Eq. (10.57)] are of the form", - "type": "text" - }, - { - "block_id": "p968-b5", - "global_id": 29184, - "bbox": [ - 172.65, - 255.23, - 304.16, - 266.69 - ], - "text": "yi = ˆci1z1 + ˆci2z2 + · · · + ˆciNzN +", - "type": "text" - }, - { - "block_id": "p968-b6", - "global_id": 29185, - "bbox": [ - 305.71, - 244.79, - 319.81, - 255.73 - ], - "text": "j\n\"", - "type": "text" - }, - { - "block_id": "p968-b7", - "global_id": 29186, - "bbox": [ - 305.77, - 269.62, - 319.73, - 276.88 - ], - "text": "m=1", - "type": "text" - }, - { - "block_id": "p968-b8", - "global_id": 29187, - "bbox": [ - 320.91, - 255.23, - 419.42, - 266.31 - ], - "text": "dimxm\ni = 1,2,. . ., k", - "type": "text" - }, - { - "block_id": "p968-b9", - "global_id": 29188, - "bbox": [ - 101.84, - 293.28, - 490.39, - 399.3 - ], - "text": "If ˆcim = 0, then the state zm will not appear in the expression for yi. Since all the states are decoupled\nbecause of the diagonalized nature of the equations, the state zm cannot be observed directly or\nindirectly (through other states) at the output yi. Hence, the mth mode eλmt will not be observed at\nthe output yi. If ˆc1m, ˆc2m, . . . , ˆckm (the mth column in matrix ˆC) are all zero, the state zm will not\nbe observable at any of the k outputs, and the state zm is unobservable. In contrast, if at least one\nelement in the mth column of ˆC is nonzero, zm is observable at least at one output. Thus, a system\nwith diagonalized equations of the form in Eq. (10.57) is completely observable if and only if the\nmatrix ˆC has no column of zero elements. In this discussion, we assumed distinct eigenvalues; for\nrepeated eigenvalues, the modified criteria can be found in the literature [1, 2].", - "type": "text" - }, - { - "block_id": "p968-b10", - "global_id": 29189, - "bbox": [ - 101.84, - 401.29, - 490.37, - 435.17 - ], - "text": "If the state-space description is not in diagonalized form, it may be converted into diagonalized\nform using the procedure in Ex. 10.11. It is also possible to test for controllability and observability\neven if the state-space description is in undiagonalized form [1, 2].", - "type": "text" - }, - { - "block_id": "p968-b11", - "global_id": 29190, - "bbox": [ - 76.77, - 467.38, - 370.87, - 479.34 - ], - "text": "EXAMPLE 10.12\nControllability and Observability", - "type": "text" - }, - { - "block_id": "p968-b12", - "global_id": 29191, - "bbox": [ - 103.16, - 496.01, - 401.45, - 505.97 - ], - "text": "Investigate the controllability and observability of the systems in Fig. 10.9.", - "type": "text" - }, - { - "block_id": "p968-b13", - "global_id": 29192, - "bbox": [ - 103.16, - 528.78, - 477.01, - 550.8 - ], - "text": "In both cases, the state variables are identified as the two integrator outputs, q1 and q2. The\nstate equations for the system in Fig. 10.9a are", - "type": "text" - }, - { - "block_id": "p968-b14", - "global_id": 29193, - "bbox": [ - 265.27, - 562.34, - 310.36, - 573.79 - ], - "text": "˙q1 = q1 + x", - "type": "text" - }, - { - "block_id": "p968-b15", - "global_id": 29194, - "bbox": [ - 265.27, - 577.29, - 477.0, - 588.74 - ], - "text": "˙q2 = q1 −q2\n(10.58)", - "type": "text" - }, - { - "block_id": "p968-b16", - "global_id": 29195, - "bbox": [ - 103.17, - 599.61, - 117.54, - 609.58 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p968-b17", - "global_id": 29196, - "bbox": [ - 244.72, - 611.15, - 334.95, - 622.61 - ], - "text": "y = ˙q2 −q2 = q1 −2q2", - "type": "text" - } - ] - }, - { - "page_num": 969, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p969-b0", - "global_id": 29197, - "bbox": [ - 317.02, - 62.89, - 516.13, - 71.98 - ], - "text": "10.6\nControllability and Observability\n949", - "type": "text" - }, - { - "block_id": "p969-b1", - "global_id": 29198, - "bbox": [ - 147.91, - 192.82, - 151.46, - 200.82 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p969-b2", - "global_id": 29199, - "bbox": [ - 351.96, - 235.07, - 355.51, - 243.07 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p969-b3", - "global_id": 29200, - "bbox": [ - 243.43, - 264.15, - 253.08, - 272.15 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p969-b4", - "global_id": 29201, - "bbox": [ - 130.13, - 96.26, - 364.68, - 105.87 - ], - "text": "x\ny", - "type": "text" - }, - { - "block_id": "p969-b5", - "global_id": 29202, - "bbox": [ - 243.82, - 170.05, - 252.7, - 178.05 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p969-b6", - "global_id": 29203, - "bbox": [ - 268.48, - 155.55, - 325.37, - 163.85 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p969-b7", - "global_id": 29204, - "bbox": [ - 181.53, - 250.14, - 238.43, - 258.43 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p969-b11", - "global_id": 29205, - "bbox": [ - 188.35, - 91.87, - 195.35, - 106.01 - ], - "text": ".q1", - "type": "text" - }, - { - "block_id": "p969-b12", - "global_id": 29206, - "bbox": [ - 206.53, - 186.28, - 213.53, - 200.43 - ], - "text": ".q1", - "type": "text" - }, - { - "block_id": "p969-b13", - "global_id": 29207, - "bbox": [ - 332.95, - 186.28, - 339.95, - 200.43 - ], - "text": ".q2", - "type": "text" - }, - { - "block_id": "p969-b14", - "global_id": 29208, - "bbox": [ - 293.48, - 91.87, - 300.48, - 106.01 - ], - "text": ".q2", - "type": "text" - }, - { - "block_id": "p969-b15", - "global_id": 29209, - "bbox": [ - 188.28, - 154.53, - 195.28, - 164.14 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p969-b16", - "global_id": 29210, - "bbox": [ - 206.57, - 248.62, - 339.51, - 258.22 - ], - "text": "q1\nq2", - "type": "text" - }, - { - "block_id": "p969-b17", - "global_id": 29211, - "bbox": [ - 293.52, - 154.53, - 300.52, - 164.14 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p969-b18", - "global_id": 29212, - "bbox": [ - 189.93, - 122.54, - 193.93, - 137.45 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p969-b19", - "global_id": 29213, - "bbox": [ - 208.07, - 217.42, - 212.07, - 232.33 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p969-b20", - "global_id": 29214, - "bbox": [ - 334.48, - 217.42, - 338.48, - 232.33 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p969-b21", - "global_id": 29215, - "bbox": [ - 295.01, - 122.54, - 299.01, - 137.45 - ], - "text": "1\ns", - "type": "text" - }, - { - "block_id": "p969-b22", - "global_id": 29216, - "bbox": [ - 119.94, - 278.84, - 251.7, - 288.08 - ], - "text": "Figure 10.9 Systems for Ex. 10.12.", - "type": "text" - }, - { - "block_id": "p969-b23", - "global_id": 29217, - "bbox": [ - 128.9, - 315.75, - 156.82, - 325.71 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p969-b24", - "global_id": 29218, - "bbox": [ - 201.25, - 332.22, - 218.26, - 342.52 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p969-b25", - "global_id": 29219, - "bbox": [ - 220.31, - 318.23, - 253.42, - 348.48 - ], - "text": "1\n0\n1\n−1", - "type": "text" - }, - { - "block_id": "p969-b26", - "global_id": 29220, - "bbox": [ - 253.43, - 318.23, - 258.86, - 328.19 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p969-b27", - "global_id": 29221, - "bbox": [ - 278.79, - 332.22, - 295.25, - 342.52 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p969-b28", - "global_id": 29222, - "bbox": [ - 297.29, - 318.23, - 307.71, - 336.52 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p969-b29", - "global_id": 29223, - "bbox": [ - 302.73, - 338.51, - 307.71, - 348.48 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p969-b30", - "global_id": 29224, - "bbox": [ - 307.71, - 318.23, - 313.14, - 328.19 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p969-b31", - "global_id": 29225, - "bbox": [ - 333.06, - 332.12, - 430.42, - 342.6 - ], - "text": "C = [1\n−2]\nD = 0", - "type": "text" - }, - { - "block_id": "p969-b32", - "global_id": 29226, - "bbox": [ - 233.11, - 364.1, - 274.44, - 374.39 - ], - "text": "|sI −A| =", - "type": "text" - }, - { - "block_id": "p969-b34", - "global_id": 29227, - "bbox": [ - 279.73, - 358.02, - 329.13, - 380.35 - ], - "text": "s −1\n0\n−1\ns + 1", - "type": "text" - }, - { - "block_id": "p969-b35", - "global_id": 29228, - "bbox": [ - 329.13, - 349.65, - 398.55, - 377.54 - ], - "text": "= (s −1)(s + 1)", - "type": "text" - }, - { - "block_id": "p969-b36", - "global_id": 29229, - "bbox": [ - 128.9, - 388.73, - 170.65, - 398.69 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p969-b37", - "global_id": 29230, - "bbox": [ - 257.3, - 400.27, - 374.37, - 411.73 - ], - "text": "λ1 = 1\nand\nλ2 = −1", - "type": "text" - }, - { - "block_id": "p969-b38", - "global_id": 29231, - "bbox": [ - 128.91, - 419.61, - 143.28, - 429.57 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p969-b39", - "global_id": 29232, - "bbox": [ - 286.74, - 434.74, - 304.33, - 444.7 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p969-b40", - "global_id": 29233, - "bbox": [ - 306.38, - 420.76, - 339.5, - 451.0 - ], - "text": "1\n0\n0\n−1", - "type": "text" - }, - { - "block_id": "p969-b41", - "global_id": 29234, - "bbox": [ - 339.5, - 420.76, - 344.93, - 430.72 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p969-b42", - "global_id": 29235, - "bbox": [ - 128.91, - 459.38, - 502.75, - 481.3 - ], - "text": "We shall now use the procedure in Sec. 10.5-1 to diagonalize this system. According to\nEq. (10.55), we have", - "type": "text" - }, - { - "block_id": "p969-b43", - "global_id": 29236, - "bbox": [ - 225.62, - 484.49, - 258.74, - 514.73 - ], - "text": "1\n0\n0\n−1", - "type": "text" - }, - { - "block_id": "p969-b44", - "global_id": 29237, - "bbox": [ - 258.74, - 484.5, - 303.97, - 515.51 - ], - "text": "! p11\np12\np21\np22", - "type": "text" - }, - { - "block_id": "p969-b45", - "global_id": 29238, - "bbox": [ - 304.47, - 484.49, - 309.9, - 494.46 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p969-b46", - "global_id": 29239, - "bbox": [ - 311.95, - 498.48, - 319.72, - 508.44 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p969-b47", - "global_id": 29240, - "bbox": [ - 321.77, - 484.49, - 361.56, - 515.51 - ], - "text": "p11\np12\np21\np22", - "type": "text" - }, - { - "block_id": "p969-b48", - "global_id": 29241, - "bbox": [ - 362.07, - 484.49, - 400.62, - 514.73 - ], - "text": "! 1\n0\n1\n−1", - "type": "text" - }, - { - "block_id": "p969-b49", - "global_id": 29242, - "bbox": [ - 400.62, - 484.49, - 406.05, - 494.46 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p969-b50", - "global_id": 29243, - "bbox": [ - 128.91, - 526.1, - 269.76, - 536.06 - ], - "text": "The solution of this equation yields", - "type": "text" - }, - { - "block_id": "p969-b51", - "global_id": 29244, - "bbox": [ - 249.31, - 547.61, - 381.86, - 559.07 - ], - "text": "p12 = 0\nand\n−2p21 = p22", - "type": "text" - }, - { - "block_id": "p969-b52", - "global_id": 29245, - "bbox": [ - 128.9, - 569.53, - 285.1, - 580.98 - ], - "text": "Choosing p11 = 1 and p21 = 1, we have", - "type": "text" - }, - { - "block_id": "p969-b53", - "global_id": 29246, - "bbox": [ - 287.58, - 597.08, - 303.48, - 607.38 - ], - "text": "P =", - "type": "text" - }, - { - "block_id": "p969-b54", - "global_id": 29247, - "bbox": [ - 305.53, - 583.1, - 338.65, - 613.34 - ], - "text": "1\n0\n1\n−2", - "type": "text" - }, - { - "block_id": "p969-b55", - "global_id": 29248, - "bbox": [ - 338.66, - 583.1, - 344.09, - 593.06 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 970, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p970-b0", - "global_id": 29249, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "950\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p970-b1", - "global_id": 29250, - "bbox": [ - 103.16, - 86.24, - 117.54, - 96.21 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p970-b2", - "global_id": 29251, - "bbox": [ - 227.48, - 99.03, - 268.54, - 111.66 - ], - "text": "ˆB = PB =", - "type": "text" - }, - { - "block_id": "p970-b3", - "global_id": 29252, - "bbox": [ - 270.59, - 87.38, - 303.71, - 117.63 - ], - "text": "1\n0\n1\n−2", - "type": "text" - }, - { - "block_id": "p970-b4", - "global_id": 29253, - "bbox": [ - 303.72, - 87.38, - 319.56, - 105.67 - ], - "text": "! 1", - "type": "text" - }, - { - "block_id": "p970-b5", - "global_id": 29254, - "bbox": [ - 314.58, - 107.66, - 319.56, - 117.63 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p970-b6", - "global_id": 29255, - "bbox": [ - 319.56, - 87.38, - 324.99, - 97.34 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p970-b7", - "global_id": 29256, - "bbox": [ - 327.04, - 101.36, - 334.81, - 111.33 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p970-b8", - "global_id": 29257, - "bbox": [ - 336.86, - 87.38, - 347.27, - 105.67 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p970-b9", - "global_id": 29258, - "bbox": [ - 342.28, - 107.66, - 347.27, - 117.63 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p970-b10", - "global_id": 29259, - "bbox": [ - 347.27, - 87.38, - 352.7, - 97.34 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p970-b11", - "global_id": 29260, - "bbox": [ - 103.17, - 126.27, - 384.59, - 138.97 - ], - "text": "All the rows of ˆB are nonzero. Hence, the system is controllable. Also,", - "type": "text" - }, - { - "block_id": "p970-b12", - "global_id": 29261, - "bbox": [ - 242.95, - 148.18, - 337.22, - 160.81 - ], - "text": "Y = Cq = CP−1z = ˆCz", - "type": "text" - }, - { - "block_id": "p970-b13", - "global_id": 29262, - "bbox": [ - 103.17, - 172.84, - 117.54, - 182.8 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p970-b14", - "global_id": 29263, - "bbox": [ - 159.63, - 185.81, - 247.59, - 200.33 - ], - "text": "ˆC = CP−1 = [1\n−2]", - "type": "text" - }, - { - "block_id": "p970-b15", - "global_id": 29264, - "bbox": [ - 247.6, - 176.05, - 280.72, - 206.29 - ], - "text": "1\n0\n1\n−2", - "type": "text" - }, - { - "block_id": "p970-b16", - "global_id": 29265, - "bbox": [ - 280.72, - 176.05, - 295.08, - 187.93 - ], - "text": "!−1", - "type": "text" - }, - { - "block_id": "p970-b17", - "global_id": 29266, - "bbox": [ - 297.62, - 189.93, - 341.77, - 200.31 - ], - "text": "= [1\n−2]", - "type": "text" - }, - { - "block_id": "p970-b18", - "global_id": 29267, - "bbox": [ - 341.77, - 176.05, - 375.49, - 209.3 - ], - "text": "1\n0\n1\n2\n−1", - "type": "text" - }, - { - "block_id": "p970-b19", - "global_id": 29268, - "bbox": [ - 372.01, - 202.32, - 375.5, - 209.3 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p970-b20", - "global_id": 29269, - "bbox": [ - 376.69, - 176.05, - 382.12, - 186.01 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p970-b21", - "global_id": 29270, - "bbox": [ - 384.17, - 190.03, - 420.55, - 200.31 - ], - "text": "= [0\n1]", - "type": "text" - }, - { - "block_id": "p970-b22", - "global_id": 29271, - "bbox": [ - 103.17, - 215.83, - 477.0, - 253.95 - ], - "text": "The first column of ˆC is zero. Hence, the mode z1 (corresponding to λ1 =1) is unobservable.\nThe system is therefore controllable but not observable. We come to the same conclusion by\nrealizing the system with the diagonalized state variables z1 and z2, whose state equations are", - "type": "text" - }, - { - "block_id": "p970-b23", - "global_id": 29272, - "bbox": [ - 264.87, - 261.66, - 315.29, - 274.29 - ], - "text": "˙z = \nz + ˆBx", - "type": "text" - }, - { - "block_id": "p970-b24", - "global_id": 29273, - "bbox": [ - 264.87, - 279.03, - 292.77, - 291.66 - ], - "text": "y = ˆCz", - "type": "text" - }, - { - "block_id": "p970-b25", - "global_id": 29274, - "bbox": [ - 103.16, - 303.69, - 266.89, - 313.65 - ], - "text": "Using our previous calculations, we have", - "type": "text" - }, - { - "block_id": "p970-b26", - "global_id": 29275, - "bbox": [ - 264.75, - 325.19, - 307.62, - 336.65 - ], - "text": "˙z1 = z1 + x", - "type": "text" - }, - { - "block_id": "p970-b27", - "global_id": 29276, - "bbox": [ - 264.75, - 340.14, - 315.4, - 351.6 - ], - "text": "˙z2 = −z2 + x", - "type": "text" - }, - { - "block_id": "p970-b28", - "global_id": 29277, - "bbox": [ - 103.17, - 362.47, - 117.54, - 372.43 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p970-b29", - "global_id": 29278, - "bbox": [ - 278.02, - 374.02, - 301.67, - 385.17 - ], - "text": "y = z2", - "type": "text" - }, - { - "block_id": "p970-b30", - "global_id": 29279, - "bbox": [ - 103.16, - 393.36, - 477.02, - 415.27 - ], - "text": "Figure 10.10a shows a realization of these equations. It is clear that each of the two modes is\ncontrollable, but the first mode (corresponding to λ = 1) is not observable at the output.", - "type": "text" - }, - { - "block_id": "p970-b31", - "global_id": 29280, - "bbox": [ - 121.09, - 417.26, - 325.01, - 427.23 - ], - "text": "The state equations for the system in Fig. 10.9b are", - "type": "text" - }, - { - "block_id": "p970-b32", - "global_id": 29281, - "bbox": [ - 220.98, - 438.77, - 477.0, - 465.17 - ], - "text": "˙q1 = −q1 + x\n˙q2 = ˙q1 −q1 + q2 = −2q1 + q2 + x\n(10.59)", - "type": "text" - }, - { - "block_id": "p970-b33", - "global_id": 29282, - "bbox": [ - 103.17, - 476.05, - 117.54, - 486.01 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p970-b34", - "global_id": 29283, - "bbox": [ - 277.46, - 487.59, - 302.21, - 498.74 - ], - "text": "y = q2", - "type": "text" - }, - { - "block_id": "p970-b35", - "global_id": 29284, - "bbox": [ - 103.16, - 506.93, - 131.08, - 516.89 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p970-b36", - "global_id": 29285, - "bbox": [ - 177.12, - 523.4, - 194.13, - 533.69 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p970-b37", - "global_id": 29286, - "bbox": [ - 196.18, - 509.41, - 229.3, - 539.65 - ], - "text": "−1\n0\n−2\n1", - "type": "text" - }, - { - "block_id": "p970-b38", - "global_id": 29287, - "bbox": [ - 229.31, - 509.42, - 234.74, - 519.38 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p970-b39", - "global_id": 29288, - "bbox": [ - 255.77, - 523.4, - 272.23, - 533.7 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p970-b40", - "global_id": 29289, - "bbox": [ - 274.28, - 509.42, - 284.69, - 527.7 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p970-b41", - "global_id": 29290, - "bbox": [ - 279.71, - 529.69, - 284.69, - 539.66 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p970-b42", - "global_id": 29291, - "bbox": [ - 284.69, - 509.42, - 290.12, - 519.38 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p970-b43", - "global_id": 29292, - "bbox": [ - 311.15, - 523.4, - 328.16, - 533.7 - ], - "text": "C =", - "type": "text" - }, - { - "block_id": "p970-b44", - "global_id": 29293, - "bbox": [ - 330.21, - 515.39, - 403.06, - 533.78 - ], - "text": "0\n1\nD = 0", - "type": "text" - }, - { - "block_id": "p970-b45", - "global_id": 29294, - "bbox": [ - 207.37, - 555.28, - 248.69, - 565.57 - ], - "text": "|sI −A| =", - "type": "text" - }, - { - "block_id": "p970-b47", - "global_id": 29295, - "bbox": [ - 253.98, - 549.2, - 303.39, - 571.53 - ], - "text": "s + 1\n0\n−1\ns −1", - "type": "text" - }, - { - "block_id": "p970-b48", - "global_id": 29296, - "bbox": [ - 303.4, - 540.83, - 372.8, - 568.73 - ], - "text": "= (s + 1)(s −1)", - "type": "text" - }, - { - "block_id": "p970-b49", - "global_id": 29297, - "bbox": [ - 103.17, - 579.5, - 216.62, - 590.95 - ], - "text": "so that λ1 = −1, λ2 = 1, and", - "type": "text" - }, - { - "block_id": "p970-b50", - "global_id": 29298, - "bbox": [ - 261.0, - 596.44, - 278.59, - 606.4 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p970-b51", - "global_id": 29299, - "bbox": [ - 280.64, - 582.45, - 313.76, - 612.69 - ], - "text": "−1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p970-b52", - "global_id": 29300, - "bbox": [ - 313.75, - 582.45, - 319.18, - 592.41 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 971, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p971-b0", - "global_id": 29301, - "bbox": [ - 317.02, - 62.89, - 516.13, - 71.98 - ], - "text": "10.6\nControllability and Observability\n951", - "type": "text" - }, - { - "block_id": "p971-b1", - "global_id": 29302, - "bbox": [ - 138.35, - 123.73, - 141.9, - 131.73 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p971-b2", - "global_id": 29303, - "bbox": [ - 311.86, - 140.97, - 315.41, - 148.97 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p971-b3", - "global_id": 29304, - "bbox": [ - 136.88, - 264.84, - 315.41, - 274.41 - ], - "text": "x\ny", - "type": "text" - }, - { - "block_id": "p971-b4", - "global_id": 29305, - "bbox": [ - 222.05, - 107.47, - 239.18, - 124.36 - ], - "text": "1\ns 1", - "type": "text" - }, - { - "block_id": "p971-b5", - "global_id": 29306, - "bbox": [ - 222.05, - 144.98, - 239.18, - 161.87 - ], - "text": "1\ns 1", - "type": "text" - }, - { - "block_id": "p971-b7", - "global_id": 29307, - "bbox": [ - 224.28, - 307.14, - 233.61, - 315.14 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p971-b8", - "global_id": 29308, - "bbox": [ - 206.94, - 225.64, - 224.07, - 242.52 - ], - "text": "1\ns 1", - "type": "text" - }, - { - "block_id": "p971-b9", - "global_id": 29309, - "bbox": [ - 206.94, - 268.84, - 224.07, - 285.72 - ], - "text": "1\ns 1", - "type": "text" - }, - { - "block_id": "p971-b10", - "global_id": 29310, - "bbox": [ - 224.5, - 185.89, - 233.38, - 193.89 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p971-b11", - "global_id": 29311, - "bbox": [ - 283.79, - 110.83, - 289.91, - 120.43 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p971-b12", - "global_id": 29312, - "bbox": [ - 275.93, - 154.83, - 282.04, - 164.44 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p971-b13", - "global_id": 29313, - "bbox": [ - 245.02, - 220.75, - 251.13, - 230.36 - ], - "text": "z1", - "type": "text" - }, - { - "block_id": "p971-b14", - "global_id": 29314, - "bbox": [ - 245.21, - 278.64, - 251.32, - 288.24 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p971-b15", - "global_id": 29315, - "bbox": [ - 332.86, - 294.48, - 498.77, - 316.4 - ], - "text": "Figure 10.10 Equivalents of the systems\nin Fig. 10.9.", - "type": "text" - }, - { - "block_id": "p971-b16", - "global_id": 29316, - "bbox": [ - 128.9, - 334.54, - 266.05, - 344.5 - ], - "text": "Diagonalizing the matrix, we have", - "type": "text" - }, - { - "block_id": "p971-b17", - "global_id": 29317, - "bbox": [ - 224.51, - 347.38, - 257.64, - 377.62 - ], - "text": "1\n0\n0\n−1", - "type": "text" - }, - { - "block_id": "p971-b18", - "global_id": 29318, - "bbox": [ - 257.63, - 347.38, - 303.97, - 378.4 - ], - "text": "! p11\np12\np21\np22", - "type": "text" - }, - { - "block_id": "p971-b19", - "global_id": 29319, - "bbox": [ - 304.47, - 347.38, - 309.9, - 357.35 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p971-b20", - "global_id": 29320, - "bbox": [ - 311.95, - 361.37, - 319.72, - 371.33 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p971-b21", - "global_id": 29321, - "bbox": [ - 321.77, - 347.38, - 361.56, - 378.4 - ], - "text": "p11\np12\np21\np22", - "type": "text" - }, - { - "block_id": "p971-b22", - "global_id": 29322, - "bbox": [ - 362.07, - 347.38, - 401.73, - 377.62 - ], - "text": "! −1\n0\n−2\n1", - "type": "text" - }, - { - "block_id": "p971-b23", - "global_id": 29323, - "bbox": [ - 401.73, - 347.38, - 407.16, - 357.35 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p971-b24", - "global_id": 29324, - "bbox": [ - 128.91, - 388.27, - 502.75, - 410.6 - ], - "text": "The solution of this equation yields p11 = −p12 and p22 = 0. Choosing p11 = −1 and p21 = 1,\nwe obtain", - "type": "text" - }, - { - "block_id": "p971-b25", - "global_id": 29325, - "bbox": [ - 287.58, - 415.77, - 303.48, - 426.06 - ], - "text": "P =", - "type": "text" - }, - { - "block_id": "p971-b26", - "global_id": 29326, - "bbox": [ - 305.53, - 401.78, - 338.66, - 432.02 - ], - "text": "−1\n1\n1\n0", - "type": "text" - }, - { - "block_id": "p971-b27", - "global_id": 29327, - "bbox": [ - 338.66, - 401.78, - 344.09, - 411.74 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p971-b28", - "global_id": 29328, - "bbox": [ - 128.9, - 440.24, - 143.28, - 450.2 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p971-b29", - "global_id": 29329, - "bbox": [ - 239.91, - 462.68, - 280.97, - 475.31 - ], - "text": "ˆB = PB =", - "type": "text" - }, - { - "block_id": "p971-b30", - "global_id": 29330, - "bbox": [ - 283.02, - 451.03, - 316.14, - 481.27 - ], - "text": "−1\n1\n1\n0", - "type": "text" - }, - { - "block_id": "p971-b31", - "global_id": 29331, - "bbox": [ - 316.15, - 451.03, - 333.09, - 469.32 - ], - "text": "! 1", - "type": "text" - }, - { - "block_id": "p971-b32", - "global_id": 29332, - "bbox": [ - 328.11, - 471.31, - 333.09, - 481.27 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p971-b33", - "global_id": 29333, - "bbox": [ - 333.1, - 451.03, - 338.53, - 461.0 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p971-b34", - "global_id": 29334, - "bbox": [ - 340.57, - 465.02, - 348.35, - 474.98 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p971-b35", - "global_id": 29335, - "bbox": [ - 350.39, - 451.03, - 360.8, - 469.32 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p971-b36", - "global_id": 29336, - "bbox": [ - 355.82, - 471.31, - 360.8, - 481.27 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p971-b37", - "global_id": 29337, - "bbox": [ - 360.8, - 451.03, - 366.23, - 461.0 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p971-b38", - "global_id": 29338, - "bbox": [ - 239.36, - 488.69, - 290.94, - 503.21 - ], - "text": "ˆC = CP−1 =", - "type": "text" - }, - { - "block_id": "p971-b39", - "global_id": 29339, - "bbox": [ - 292.99, - 478.93, - 347.23, - 509.17 - ], - "text": "0\n1\n 0\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p971-b40", - "global_id": 29340, - "bbox": [ - 347.24, - 478.93, - 352.67, - 488.89 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p971-b41", - "global_id": 29341, - "bbox": [ - 354.71, - 492.92, - 362.48, - 502.88 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p971-b42", - "global_id": 29342, - "bbox": [ - 364.52, - 484.91, - 392.3, - 503.19 - ], - "text": "1\n1", - "type": "text" - }, - { - "block_id": "p971-b43", - "global_id": 29343, - "bbox": [ - 128.91, - 520.48, - 502.75, - 557.1 - ], - "text": "The first row of ˆB is zero. Hence, the mode corresponding to λ1 = 1 is not controllable.\nHowever, since none of the columns of ˆC vanish, both modes are observable at the output.\nHence the system is observable but not controllable.", - "type": "text" - }, - { - "block_id": "p971-b44", - "global_id": 29344, - "bbox": [ - 128.91, - 559.1, - 502.76, - 582.51 - ], - "text": "We reach the same conclusion by realizing the system with the diagonalized state variables\nz1 and z2. The two state equations are", - "type": "text" - }, - { - "block_id": "p971-b45", - "global_id": 29345, - "bbox": [ - 290.61, - 589.91, - 341.03, - 602.53 - ], - "text": "˙z = \nz + ˆBx", - "type": "text" - }, - { - "block_id": "p971-b46", - "global_id": 29346, - "bbox": [ - 290.61, - 607.28, - 318.51, - 619.9 - ], - "text": "y = ˆCz", - "type": "text" - } - ] - }, - { - "page_num": 972, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p972-b0", - "global_id": 29347, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "952\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p972-b1", - "global_id": 29348, - "bbox": [ - 103.16, - 86.24, - 266.89, - 96.21 - ], - "text": "Using our previous calculations, we have", - "type": "text" - }, - { - "block_id": "p972-b2", - "global_id": 29349, - "bbox": [ - 264.75, - 107.74, - 315.4, - 134.14 - ], - "text": "˙z1 = z1\n˙z2 = −z2 + x", - "type": "text" - }, - { - "block_id": "p972-b3", - "global_id": 29350, - "bbox": [ - 103.17, - 145.02, - 139.14, - 154.98 - ], - "text": "and thus,", - "type": "text" - }, - { - "block_id": "p972-b4", - "global_id": 29351, - "bbox": [ - 268.65, - 166.52, - 311.03, - 177.98 - ], - "text": "y = z1 + z2", - "type": "text" - }, - { - "block_id": "p972-b5", - "global_id": 29352, - "bbox": [ - 103.16, - 188.85, - 477.01, - 211.86 - ], - "text": "Figure 10.10b shows a realization of these equations. Clearly, each of the two modes is\nobservable at the output, but the mode corresponding to λ1 = 1 is not controllable.", - "type": "text" - }, - { - "block_id": "p972-b6", - "global_id": 29353, - "bbox": [ - 103.46, - 235.47, - 386.19, - 261.53 - ], - "text": "USING MATLAB TO DETERMINE CONTROLLABILITY\nAND OBSERVABILITY", - "type": "text" - }, - { - "block_id": "p972-b7", - "global_id": 29354, - "bbox": [ - 103.16, - 265.49, - 477.04, - 312.44 - ], - "text": "As demonstrated in Ex. 10.11, we can use MATLAB’s eig function to determine the matrix P\nthat will diagonalize A. We can then use P to determine ˆB and ˆC, from which we can determine\nthe controllability and observability of a system. Let us demonstrate the process for the two\npresent systems.", - "type": "text" - }, - { - "block_id": "p972-b8", - "global_id": 29355, - "bbox": [ - 121.09, - 311.69, - 421.89, - 324.4 - ], - "text": "First, let us use MATLAB to compute ˆB and ˆC for the system in Fig. 10.9a.", - "type": "text" - }, - { - "block_id": "p972-b9", - "global_id": 29356, - "bbox": [ - 103.17, - 346.61, - 432.67, - 380.48 - ], - "text": ">>\nA = [1 0;1 -1]; B = [1; 0]; C = [1 -2];\n>>\n[V, Lambda] = eig(A); P=inv(V); Bhat = P*B, Chat = C*inv(P)\nBhat =\n-0.5000", - "type": "text" - }, - { - "block_id": "p972-b10", - "global_id": 29357, - "bbox": [ - 124.08, - 382.47, - 212.99, - 404.39 - ], - "text": "1.1180\nChat =\n-2\n0", - "type": "text" - }, - { - "block_id": "p972-b11", - "global_id": 29358, - "bbox": [ - 103.17, - 418.29, - 477.02, - 442.96 - ], - "text": "Since all the rows of ˆB are nonzero, the system is controllable. However, one column of ˆC is\nzero, so one mode is unobservable.", - "type": "text" - }, - { - "block_id": "p972-b12", - "global_id": 29359, - "bbox": [ - 121.1, - 442.2, - 423.01, - 454.91 - ], - "text": "Next, let us use MATLAB to compute ˆB and ˆC for the system in Fig. 10.9b.", - "type": "text" - }, - { - "block_id": "p972-b13", - "global_id": 29360, - "bbox": [ - 103.17, - 477.12, - 432.68, - 534.9 - ], - "text": ">>\nA = [-1 0;-2 1]; B = [1; 1]; C = [0 1];\n>>\n[V, Lambda] = eig(A); P=inv(V); Bhat = P*B, Chat = C*inv(P)\nBhat =\n0\n1.4142\nChat =\n1.0000\n0.7071", - "type": "text" - }, - { - "block_id": "p972-b14", - "global_id": 29361, - "bbox": [ - 103.17, - 548.8, - 477.02, - 573.47 - ], - "text": "One of the rows of ˆB is zero, so one mode is uncontrollable. Since all of the columns of ˆC are\nnonzero, the system is observable.", - "type": "text" - }, - { - "block_id": "p972-b15", - "global_id": 29362, - "bbox": [ - 103.17, - 575.46, - 477.0, - 597.37 - ], - "text": "As expected, the MATLAB results confirm our earlier conclusions regarding the\ncontrollability and observability of the systems of Fig. 10.9.", - "type": "text" - } - ] - }, - { - "page_num": 973, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p973-b0", - "global_id": 29363, - "bbox": [ - 264.66, - 62.89, - 516.14, - 71.98 - ], - "text": "10.7\nState-Space Analysis of Discrete-Time Systems\n953", - "type": "text" - }, - { - "block_id": "p973-b1", - "global_id": 29364, - "bbox": [ - 127.59, - 86.52, - 495.37, - 98.48 - ], - "text": "10.6-1 Inadequacy of the Transfer Function Description of a System", - "type": "text" - }, - { - "block_id": "p973-b2", - "global_id": 29365, - "bbox": [ - 127.59, - 104.61, - 516.12, - 126.52 - ], - "text": "Example 10.12 demonstrates the inadequacy of the transfer function to describe an LTI system in\ngeneral. The systems in Figs. 10.9a and 10.9b both have the same transfer function", - "type": "text" - }, - { - "block_id": "p973-b3", - "global_id": 29366, - "bbox": [ - 295.4, - 135.82, - 347.12, - 159.84 - ], - "text": "H(s) =\n1\ns + 1", - "type": "text" - }, - { - "block_id": "p973-b4", - "global_id": 29367, - "bbox": [ - 127.59, - 167.81, - 516.17, - 225.59 - ], - "text": "Yet the two systems are very different. Their true nature is revealed in Figs. 10.10a and 10.10b,\nrespectively. Both the systems are unstable, but their transfer function H(s) = 1/(s + 1) does not\ngive any hint of it. Moreover, the systems are very different from the viewpoint of controllability\nand observability. The system in Fig. 10.9a is controllable but not observable, whereas the system\nin Fig. 10.9b is observable but not controllable.", - "type": "text" - }, - { - "block_id": "p973-b5", - "global_id": 29368, - "bbox": [ - 127.6, - 227.58, - 516.16, - 297.32 - ], - "text": "The transfer function description of a system looks at a system only from the input and output\nterminals. Consequently, the transfer function description can specify only the part of the system\nthat is coupled to the input and the output terminals. From Figs. 10.10a and 10.10b, we see that\nin both cases only a part of the system that has a transfer function H(s) = 1/(s + 1) is coupled\nto the input and the output terminals. This is why both systems have the same transfer function\nH(s) = 1/(s + 1).", - "type": "text" - }, - { - "block_id": "p973-b6", - "global_id": 29369, - "bbox": [ - 127.6, - 299.31, - 516.17, - 345.13 - ], - "text": "The state variable description [Eqs. (10.58) and (10.59)], on the other hand, contains all\nthe information about these systems to describe them completely. The reason is that the state\nvariable description is an internal description, not the external description obtained from the\nsystem behavior at external terminals.", - "type": "text" - }, - { - "block_id": "p973-b7", - "global_id": 29370, - "bbox": [ - 127.6, - 347.13, - 516.17, - 428.83 - ], - "text": "Apparently, the transfer function fails to describe these systems completely because the\ntransfer functions of these systems have a common factor s−1 in the numerator and denominator;\nthis common factor is canceled out in the systems in Fig. 10.9, with a consequent loss of the\ninformation. Such a situation occurs when a system is uncontrollable and/or unobservable. If a\nsystem is both controllable and observable (which is the case with most of the practical systems)\nthe transfer function describes the system completely. In such a case, the internal and external\ndescriptions are equivalent.", - "type": "text" - }, - { - "block_id": "p973-b8", - "global_id": 29371, - "bbox": [ - 127.94, - 457.78, - 336.55, - 487.66 - ], - "text": "10.7 STATE-SPACE ANALYSIS OF\nDISCRETE-TIME SYSTEMS", - "type": "text" - }, - { - "block_id": "p973-b9", - "global_id": 29372, - "bbox": [ - 127.59, - 493.56, - 516.15, - 527.53 - ], - "text": "We have shown that an Nth-order differential equation can be expressed in terms of N first-order\ndifferential equations. In the following analogous procedure, we show that a general Nth-order\ndifference equation can be expressed in terms of N first-order difference equations.", - "type": "text" - }, - { - "block_id": "p973-b10", - "global_id": 29373, - "bbox": [ - 145.52, - 529.42, - 271.95, - 539.49 - ], - "text": "Consider the z-transfer function", - "type": "text" - }, - { - "block_id": "p973-b11", - "global_id": 29374, - "bbox": [ - 237.56, - 545.01, - 404.04, - 565.89 - ], - "text": "H[z] = b0zN + b1zN−1 + · · · + bN−1z + bN", - "type": "text" - }, - { - "block_id": "p973-b12", - "global_id": 29375, - "bbox": [ - 273.26, - 559.81, - 399.56, - 573.84 - ], - "text": "zN + a1zN−1 + · · · + aN−1z + aN", - "type": "text" - }, - { - "block_id": "p973-b13", - "global_id": 29376, - "bbox": [ - 127.59, - 581.86, - 471.28, - 592.24 - ], - "text": "The input x[n] and the output y[n] of this system are related by the difference equation", - "type": "text" - }, - { - "block_id": "p973-b14", - "global_id": 29377, - "bbox": [ - 153.41, - 599.02, - 490.31, - 614.28 - ], - "text": "(EN + a1EN−1 + · · · + aN−1E + aN)y[n] = (b0EN + b1EN−1 + · · · + bN−1E + bN)x[n]", - "type": "text" - }, - { - "block_id": "p973-b15", - "global_id": 29378, - "bbox": [ - 127.6, - 624.83, - 381.35, - 634.79 - ], - "text": "The DFII realization of this equation is illustrated in Fig. 10.11.", - "type": "text" - } - ] - }, - { - "page_num": 974, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p974-b0", - "global_id": 29379, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "954\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p974-b1", - "global_id": 29380, - "bbox": [ - 253.46, - 89.48, - 260.46, - 99.09 - ], - "text": "b0", - "type": "text" - }, - { - "block_id": "p974-b2", - "global_id": 29381, - "bbox": [ - 151.93, - 145.56, - 256.41, - 157.11 - ], - "text": "b1\na1", - "type": "text" - }, - { - "block_id": "p974-b3", - "global_id": 29382, - "bbox": [ - 151.93, - 183.64, - 256.41, - 195.19 - ], - "text": "b2\na2", - "type": "text" - }, - { - "block_id": "p974-b4", - "global_id": 29383, - "bbox": [ - 151.26, - 244.62, - 256.92, - 256.1 - ], - "text": "bm\nam", - "type": "text" - }, - { - "block_id": "p974-b5", - "global_id": 29384, - "bbox": [ - 147.79, - 302.64, - 260.55, - 314.19 - ], - "text": "bN1\naN1", - "type": "text" - }, - { - "block_id": "p974-b6", - "global_id": 29385, - "bbox": [ - 151.42, - 339.72, - 256.91, - 351.21 - ], - "text": "bN\naN", - "type": "text" - }, - { - "block_id": "p974-b7", - "global_id": 29386, - "bbox": [ - 106.31, - 89.56, - 304.48, - 99.32 - ], - "text": "x[n]\ny[n]\nqN[n 1]", - "type": "text" - }, - { - "block_id": "p974-b8", - "global_id": 29387, - "bbox": [ - 179.97, - 133.19, - 197.66, - 142.74 - ], - "text": "qN[n]", - "type": "text" - }, - { - "block_id": "p974-b9", - "global_id": 29388, - "bbox": [ - 182.36, - 171.85, - 197.64, - 181.46 - ], - "text": "qN1", - "type": "text" - }, - { - "block_id": "p974-b10", - "global_id": 29389, - "bbox": [ - 172.73, - 231.78, - 197.67, - 241.38 - ], - "text": "qm1[n]", - "type": "text" - }, - { - "block_id": "p974-b11", - "global_id": 29390, - "bbox": [ - 180.68, - 289.46, - 197.01, - 299.06 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p974-b12", - "global_id": 29391, - "bbox": [ - 199.21, - 339.93, - 215.54, - 349.54 - ], - "text": "q1[n]", - "type": "text" - }, - { - "block_id": "p974-b18", - "global_id": 29392, - "bbox": [ - 205.41, - 116.93, - 209.41, - 131.84 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p974-b19", - "global_id": 29393, - "bbox": [ - 205.41, - 156.18, - 209.41, - 171.09 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p974-b20", - "global_id": 29394, - "bbox": [ - 205.41, - 217.18, - 209.41, - 232.09 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p974-b21", - "global_id": 29395, - "bbox": [ - 205.41, - 274.68, - 209.41, - 289.59 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p974-b22", - "global_id": 29396, - "bbox": [ - 205.41, - 313.68, - 209.41, - 328.59 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p974-b23", - "global_id": 29397, - "bbox": [ - 316.8, - 331.07, - 486.41, - 352.99 - ], - "text": "Figure 10.11 Direct form II realization of\nan Nth-order, discrete-time system.", - "type": "text" - }, - { - "block_id": "p974-b24", - "global_id": 29398, - "bbox": [ - 101.84, - 377.01, - 490.39, - 411.3 - ], - "text": "Signals appearing at the outputs of N delay elements are denoted by q1[n], q2[n], . . . , qN[n].\nThe input of the first delay is qN[n + 1]. We can now write N equations, one at the input of each\ndelay:", - "type": "text" - }, - { - "block_id": "p974-b25", - "global_id": 29399, - "bbox": [ - 185.42, - 414.45, - 255.19, - 437.56 - ], - "text": "q1[n + 1] = q2[n]\nq2[n + 1] = q3[n]", - "type": "text" - }, - { - "block_id": "p974-b26", - "global_id": 29400, - "bbox": [ - 174.92, - 437.39, - 256.78, - 468.05 - ], - "text": "...\nqN−1[n + 1] = qN[n]", - "type": "text" - }, - { - "block_id": "p974-b27", - "global_id": 29401, - "bbox": [ - 183.84, - 468.86, - 417.31, - 480.01 - ], - "text": "qN[n + 1] = −aNq1[n] −aN−1q2[n] −· · · −a1qN[n] + x[n]", - "type": "text" - }, - { - "block_id": "p974-b28", - "global_id": 29402, - "bbox": [ - 461.34, - 442.17, - 490.38, - 452.13 - ], - "text": "(10.60)", - "type": "text" - }, - { - "block_id": "p974-b29", - "global_id": 29403, - "bbox": [ - 101.85, - 488.96, - 116.22, - 498.92 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p974-b30", - "global_id": 29404, - "bbox": [ - 182.7, - 502.51, - 409.53, - 513.66 - ], - "text": "y[n] = bNq1[n] + bN−1q2[n] + · · · + b1qN[n] + b0qN+1[n]", - "type": "text" - }, - { - "block_id": "p974-b31", - "global_id": 29405, - "bbox": [ - 101.84, - 522.78, - 474.72, - 533.93 - ], - "text": "We can eliminate qN+1[n] from this equation by using the last equation in Eq. (10.60) to yield", - "type": "text" - }, - { - "block_id": "p974-b32", - "global_id": 29406, - "bbox": [ - 129.81, - 546.05, - 462.42, - 557.51 - ], - "text": "y[n] = (bN −b0aN)q1[n] + (bN−1 −b0aN−1)q2[n] + · · · + (b1 −b0a1)qN[n] + b0x[n]", - "type": "text" - }, - { - "block_id": "p974-b33", - "global_id": 29407, - "bbox": [ - 147.9, - 561.08, - 490.38, - 574.29 - ], - "text": "= ˆbNq1[n] + ˆbN−1q2[n] + · · · + ˆb1qN[n] + b0x[n]\n(10.61)", - "type": "text" - }, - { - "block_id": "p974-b34", - "global_id": 29408, - "bbox": [ - 101.84, - 586.49, - 185.13, - 600.0 - ], - "text": "where ˆbi = bi −b0ai.", - "type": "text" - }, - { - "block_id": "p974-b35", - "global_id": 29409, - "bbox": [ - 101.85, - 600.5, - 490.39, - 634.79 - ], - "text": "Equation (10.60) shows N first-order difference equations in N variables q1[n], q2[n], . . . ,\nqN[n]. These variables should immediately be recognized as state variables, since the specification\nof the initial values of these variables in Fig. 10.11 will uniquely determine the response y[n] for a", - "type": "text" - } - ] - }, - { - "page_num": 975, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p975-b0", - "global_id": 29410, - "bbox": [ - 264.66, - 62.89, - 516.14, - 71.98 - ], - "text": "10.7\nState-Space Analysis of Discrete-Time Systems\n955", - "type": "text" - }, - { - "block_id": "p975-b1", - "global_id": 29411, - "bbox": [ - 127.59, - 85.4, - 516.11, - 107.74 - ], - "text": "given x[n]. Thus, Eq. (10.60) represents the state equations, and Eq. (10.61) is the output equation.\nIn matrix form, we can write these equations as", - "type": "text" - }, - { - "block_id": "p975-b2", - "global_id": 29412, - "bbox": [ - 132.57, - 112.92, - 139.83, - 122.89 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p975-b3", - "global_id": 29413, - "bbox": [ - 132.57, - 130.4, - 139.83, - 170.7 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p975-b4", - "global_id": 29414, - "bbox": [ - 145.09, - 120.53, - 181.51, - 143.64 - ], - "text": "q1[n + 1]\nq2[n + 1]", - "type": "text" - }, - { - "block_id": "p975-b5", - "global_id": 29415, - "bbox": [ - 139.83, - 143.47, - 186.77, - 174.13 - ], - "text": "...\nqN−1[n + 1]", - "type": "text" - }, - { - "block_id": "p975-b6", - "global_id": 29416, - "bbox": [ - 144.3, - 174.93, - 182.3, - 186.01 - ], - "text": "qN[n + 1]", - "type": "text" - }, - { - "block_id": "p975-b7", - "global_id": 29417, - "bbox": [ - 186.78, - 112.92, - 194.04, - 122.89 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p975-b8", - "global_id": 29418, - "bbox": [ - 186.78, - 130.4, - 194.04, - 170.7 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p975-b10", - "global_id": 29419, - "bbox": [ - 153.12, - 194.42, - 173.49, - 201.69 - ], - "text": "q[n+1]", - "type": "text" - }, - { - "block_id": "p975-b11", - "global_id": 29420, - "bbox": [ - 195.64, - 147.83, - 203.41, - 157.8 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p975-b12", - "global_id": 29421, - "bbox": [ - 205.01, - 112.92, - 212.27, - 122.89 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p975-b13", - "global_id": 29422, - "bbox": [ - 205.01, - 130.4, - 212.27, - 170.7 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p975-b14", - "global_id": 29423, - "bbox": [ - 218.94, - 120.53, - 375.56, - 161.41 - ], - "text": "0\n1\n0\n· · ·\n0\n0\n0\n0\n1\n· · ·\n0\n0\n...", - "type": "text" - }, - { - "block_id": "p975-b15", - "global_id": 29424, - "bbox": [ - 252.94, - 143.47, - 255.43, - 161.41 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p975-b16", - "global_id": 29425, - "bbox": [ - 290.14, - 143.47, - 347.62, - 161.41 - ], - "text": "...\n· · ·\n...", - "type": "text" - }, - { - "block_id": "p975-b17", - "global_id": 29426, - "bbox": [ - 212.27, - 143.47, - 380.94, - 186.08 - ], - "text": "...\n0\n0\n0\n· · ·\n0\n1\n−aN\n−aN−1\n−aN−2\n· · ·\n−a2\n−a1", - "type": "text" - }, - { - "block_id": "p975-b18", - "global_id": 29427, - "bbox": [ - 381.45, - 112.92, - 388.71, - 122.89 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p975-b19", - "global_id": 29428, - "bbox": [ - 381.45, - 130.4, - 388.71, - 170.7 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p975-b20", - "global_id": 29429, - "bbox": [ - 205.01, - 182.53, - 388.72, - 200.86 - ], - "text": "A", - "type": "text" - }, - { - "block_id": "p975-b21", - "global_id": 29430, - "bbox": [ - 389.82, - 112.92, - 397.08, - 122.89 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p975-b22", - "global_id": 29431, - "bbox": [ - 389.82, - 130.4, - 397.08, - 170.7 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p975-b23", - "global_id": 29432, - "bbox": [ - 402.33, - 120.53, - 422.92, - 143.64 - ], - "text": "q1[n]\nq2[n]", - "type": "text" - }, - { - "block_id": "p975-b24", - "global_id": 29433, - "bbox": [ - 397.08, - 143.47, - 428.17, - 174.13 - ], - "text": "...\nqN−1[n]", - "type": "text" - }, - { - "block_id": "p975-b25", - "global_id": 29434, - "bbox": [ - 401.54, - 174.93, - 423.71, - 186.01 - ], - "text": "qN[n]", - "type": "text" - }, - { - "block_id": "p975-b26", - "global_id": 29435, - "bbox": [ - 428.17, - 112.92, - 435.43, - 122.89 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p975-b27", - "global_id": 29436, - "bbox": [ - 428.17, - 130.4, - 435.43, - 170.7 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p975-b29", - "global_id": 29437, - "bbox": [ - 406.9, - 194.33, - 418.35, - 201.53 - ], - "text": "q[n]", - "type": "text" - }, - { - "block_id": "p975-b30", - "global_id": 29438, - "bbox": [ - 436.54, - 147.83, - 444.31, - 157.8 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p975-b31", - "global_id": 29439, - "bbox": [ - 445.41, - 112.92, - 452.67, - 122.89 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p975-b32", - "global_id": 29440, - "bbox": [ - 445.41, - 130.4, - 452.67, - 170.7 - ], - "text": "⎢⎢⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p975-b33", - "global_id": 29441, - "bbox": [ - 452.68, - 120.94, - 457.66, - 185.31 - ], - "text": "0\n0\n...\n0\n1", - "type": "text" - }, - { - "block_id": "p975-b34", - "global_id": 29442, - "bbox": [ - 457.66, - 112.92, - 464.92, - 122.89 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p975-b35", - "global_id": 29443, - "bbox": [ - 457.66, - 130.4, - 464.92, - 170.7 - ], - "text": "⎥⎥⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p975-b37", - "global_id": 29444, - "bbox": [ - 453.18, - 194.88, - 457.16, - 200.86 - ], - "text": "B", - "type": "text" - }, - { - "block_id": "p975-b38", - "global_id": 29445, - "bbox": [ - 466.03, - 147.83, - 516.12, - 158.21 - ], - "text": "x[n] (10.62)", - "type": "text" - }, - { - "block_id": "p975-b39", - "global_id": 29446, - "bbox": [ - 127.6, - 214.18, - 141.97, - 224.14 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p975-b40", - "global_id": 29447, - "bbox": [ - 220.84, - 247.09, - 247.26, - 257.38 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p975-b41", - "global_id": 29448, - "bbox": [ - 249.3, - 239.08, - 334.68, - 258.71 - ], - "text": "ˆbN\nˆbN−1\n· · ·\nˆb1", - "type": "text" - }, - { - "block_id": "p975-b43", - "global_id": 29449, - "bbox": [ - 249.3, - 255.15, - 339.1, - 273.48 - ], - "text": "C", - "type": "text" - }, - { - "block_id": "p975-b44", - "global_id": 29450, - "bbox": [ - 340.21, - 218.15, - 347.47, - 228.11 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p975-b45", - "global_id": 29451, - "bbox": [ - 340.21, - 235.63, - 347.47, - 263.98 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p975-b46", - "global_id": 29452, - "bbox": [ - 348.26, - 225.76, - 368.84, - 248.87 - ], - "text": "q1[n]\nq2[n]", - "type": "text" - }, - { - "block_id": "p975-b47", - "global_id": 29453, - "bbox": [ - 347.47, - 248.7, - 369.64, - 279.29 - ], - "text": "...\nqN[n]", - "type": "text" - }, - { - "block_id": "p975-b48", - "global_id": 29454, - "bbox": [ - 369.64, - 218.15, - 376.9, - 228.11 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p975-b49", - "global_id": 29455, - "bbox": [ - 369.64, - 235.63, - 405.7, - 263.98 - ], - "text": "⎥⎥⎥⎦+ b0", - "type": "text" - }, - { - "block_id": "p975-b50", - "global_id": 29456, - "bbox": [ - 394.58, - 265.01, - 398.89, - 270.99 - ], - "text": "D", - "type": "text" - }, - { - "block_id": "p975-b51", - "global_id": 29457, - "bbox": [ - 406.81, - 247.09, - 516.12, - 257.46 - ], - "text": "x[n]\n(10.63)", - "type": "text" - }, - { - "block_id": "p975-b52", - "global_id": 29458, - "bbox": [ - 127.6, - 288.65, - 170.18, - 298.61 - ], - "text": "In general,", - "type": "text" - }, - { - "block_id": "p975-b53", - "global_id": 29459, - "bbox": [ - 269.92, - 311.84, - 373.26, - 322.22 - ], - "text": "q[n + 1] = Aq[n] + Bx[n]", - "type": "text" - }, - { - "block_id": "p975-b54", - "global_id": 29460, - "bbox": [ - 286.33, - 326.78, - 373.81, - 337.08 - ], - "text": "y[n] = Cq[n] + Dx[n]", - "type": "text" - }, - { - "block_id": "p975-b55", - "global_id": 29461, - "bbox": [ - 127.6, - 350.8, - 516.15, - 420.54 - ], - "text": "Here we have represented a discrete-time system with state equations for DFII form. There are\nseveral other possible representations, as discussed in Sec. 10.3. We may, for example, use the\ncascade, parallel, or transpose of DFII forms to realize the system, or we may use some linear\ntransformation of the state vector to realize other forms. In all cases, the output of each delay\nelement qualifies as a state variable. We then write the equation at the input of each delay element.\nThe N equations thus obtained are the N state equations.", - "type": "text" - }, - { - "block_id": "p975-b56", - "global_id": 29462, - "bbox": [ - 127.59, - 446.98, - 290.45, - 458.93 - ], - "text": "10.7-1 Solution in State Space", - "type": "text" - }, - { - "block_id": "p975-b57", - "global_id": 29463, - "bbox": [ - 127.59, - 465.07, - 235.76, - 475.03 - ], - "text": "Consider the state equation", - "type": "text" - }, - { - "block_id": "p975-b58", - "global_id": 29464, - "bbox": [ - 270.19, - 479.13, - 373.53, - 489.51 - ], - "text": "q[n + 1] = Aq[n] + Bx[n]", - "type": "text" - }, - { - "block_id": "p975-b59", - "global_id": 29465, - "bbox": [ - 127.6, - 500.17, - 262.95, - 510.13 - ], - "text": "From this equation, it follows that", - "type": "text" - }, - { - "block_id": "p975-b60", - "global_id": 29466, - "bbox": [ - 262.27, - 523.36, - 381.45, - 533.74 - ], - "text": "q[n] = Aq[n −1] + Bx[n −1]", - "type": "text" - }, - { - "block_id": "p975-b61", - "global_id": 29467, - "bbox": [ - 127.6, - 547.37, - 141.97, - 557.33 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p975-b62", - "global_id": 29468, - "bbox": [ - 254.34, - 570.56, - 389.37, - 595.88 - ], - "text": "q[n −1] = Aq[n −2] + Bx[n −2]\nq[n −2] = Aq[n −3] + Bx[n −3]", - "type": "text" - }, - { - "block_id": "p975-b63", - "global_id": 29469, - "bbox": [ - 293.03, - 600.48, - 295.52, - 618.41 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p975-b64", - "global_id": 29470, - "bbox": [ - 270.19, - 622.98, - 357.68, - 633.36 - ], - "text": "q[1] = Aq[0] + Bx[0]", - "type": "text" - } - ] - }, - { - "page_num": 976, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p976-b0", - "global_id": 29471, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "956\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p976-b1", - "global_id": 29472, - "bbox": [ - 101.84, - 85.4, - 372.41, - 95.78 - ], - "text": "Substituting the expression for q[n −1] into that for q[n], we obtain", - "type": "text" - }, - { - "block_id": "p976-b2", - "global_id": 29473, - "bbox": [ - 205.96, - 106.5, - 386.27, - 118.42 - ], - "text": "q[n] = A2q[n −2] + ABx[n −2] + Bx[n −1]", - "type": "text" - }, - { - "block_id": "p976-b3", - "global_id": 29474, - "bbox": [ - 101.85, - 130.68, - 367.17, - 141.06 - ], - "text": "Substituting the expression for q[n −2] in this equation, we obtain", - "type": "text" - }, - { - "block_id": "p976-b4", - "global_id": 29475, - "bbox": [ - 175.39, - 151.78, - 416.85, - 163.69 - ], - "text": "q[n] = A3q[n −3] + A2Bx[n −3] + ABx[n −2] + Bx[n −1]", - "type": "text" - }, - { - "block_id": "p976-b5", - "global_id": 29476, - "bbox": [ - 101.85, - 176.36, - 236.16, - 186.33 - ], - "text": "Continuing in this way, we obtain", - "type": "text" - }, - { - "block_id": "p976-b6", - "global_id": 29477, - "bbox": [ - 176.28, - 196.76, - 415.95, - 208.96 - ], - "text": "q[n] = Anq[0] + An−1Bx[0] + An−2Bx[1] + · · · + Bx[n −1]", - "type": "text" - }, - { - "block_id": "p976-b7", - "global_id": 29478, - "bbox": [ - 195.47, - 222.97, - 242.95, - 234.95 - ], - "text": "= Anq[0] +", - "type": "text" - }, - { - "block_id": "p976-b8", - "global_id": 29479, - "bbox": [ - 244.49, - 214.4, - 258.59, - 225.08 - ], - "text": "n−1\n\"", - "type": "text" - }, - { - "block_id": "p976-b9", - "global_id": 29480, - "bbox": [ - 244.57, - 238.97, - 258.52, - 246.24 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p976-b10", - "global_id": 29481, - "bbox": [ - 259.7, - 222.75, - 315.74, - 234.87 - ], - "text": "An−1−mBx[m]", - "type": "text" - }, - { - "block_id": "p976-b11", - "global_id": 29482, - "bbox": [ - 101.85, - 256.23, - 490.36, - 278.57 - ], - "text": "The upper limit of this summation is nonnegative. Hence, n ≥1, and the summation is recognized\nas the convolution sum", - "type": "text" - }, - { - "block_id": "p976-b12", - "global_id": 29483, - "bbox": [ - 254.07, - 279.39, - 338.16, - 291.6 - ], - "text": "An−1u[n −1] ∗Bx[n]", - "type": "text" - }, - { - "block_id": "p976-b13", - "global_id": 29484, - "bbox": [ - 101.85, - 301.28, - 158.46, - 311.25 - ], - "text": "Consequently,", - "type": "text" - }, - { - "block_id": "p976-b14", - "global_id": 29485, - "bbox": [ - 220.4, - 312.29, - 277.76, - 329.86 - ], - "text": "q[n] = Anq[0]", - "type": "text" - }, - { - "block_id": "p976-b15", - "global_id": 29486, - "bbox": [ - 251.56, - 332.3, - 275.63, - 338.27 - ], - "text": "zero input", - "type": "text" - }, - { - "block_id": "p976-b16", - "global_id": 29487, - "bbox": [ - 278.86, - 312.07, - 371.84, - 337.84 - ], - "text": "+An−1u[n −1] ∗Bx[n]\n\n\n\nzero state", - "type": "text" - }, - { - "block_id": "p976-b17", - "global_id": 29488, - "bbox": [ - 461.34, - 314.32, - 490.38, - 324.28 - ], - "text": "(10.64)", - "type": "text" - }, - { - "block_id": "p976-b18", - "global_id": 29489, - "bbox": [ - 101.85, - 347.16, - 116.22, - 357.12 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p976-b19", - "global_id": 29490, - "bbox": [ - 201.53, - 369.38, - 265.77, - 379.67 - ], - "text": "y[n] = Cq + Dx", - "type": "text" - }, - { - "block_id": "p976-b20", - "global_id": 29491, - "bbox": [ - 220.17, - 393.76, - 274.84, - 405.75 - ], - "text": "= CAnq[0] +", - "type": "text" - }, - { - "block_id": "p976-b21", - "global_id": 29492, - "bbox": [ - 276.39, - 385.2, - 290.49, - 395.87 - ], - "text": "n−1\n\"", - "type": "text" - }, - { - "block_id": "p976-b22", - "global_id": 29493, - "bbox": [ - 276.45, - 409.76, - 290.41, - 417.02 - ], - "text": "m=0", - "type": "text" - }, - { - "block_id": "p976-b23", - "global_id": 29494, - "bbox": [ - 291.59, - 393.54, - 377.86, - 405.66 - ], - "text": "CAn−1−mBx[m] + Dx", - "type": "text" - }, - { - "block_id": "p976-b24", - "global_id": 29495, - "bbox": [ - 220.17, - 419.69, - 490.38, - 431.89 - ], - "text": "= CAnq[0] + CAn−1u[n −1] ∗Bx[n] + Dx\n(10.65)", - "type": "text" - }, - { - "block_id": "p976-b25", - "global_id": 29496, - "bbox": [ - 101.85, - 444.56, - 224.44, - 454.53 - ], - "text": "In Sec. 10.1-3, we showed that", - "type": "text" - }, - { - "block_id": "p976-b26", - "global_id": 29497, - "bbox": [ - 209.89, - 462.56, - 490.38, - 477.93 - ], - "text": "An = β0I + β1A + β2A2 + · · · + βN−1AN−1\n(10.66)", - "type": "text" - }, - { - "block_id": "p976-b27", - "global_id": 29498, - "bbox": [ - 101.85, - 489.74, - 283.87, - 499.81 - ], - "text": "where (assuming N distinct eigenvalues of A)", - "type": "text" - }, - { - "block_id": "p976-b28", - "global_id": 29499, - "bbox": [ - 197.62, - 507.02, - 204.89, - 516.98 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p976-b29", - "global_id": 29500, - "bbox": [ - 197.62, - 524.49, - 204.89, - 552.85 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p976-b30", - "global_id": 29501, - "bbox": [ - 210.14, - 514.63, - 219.24, - 537.73 - ], - "text": "β0\nβ1", - "type": "text" - }, - { - "block_id": "p976-b31", - "global_id": 29502, - "bbox": [ - 204.89, - 537.56, - 224.49, - 568.22 - ], - "text": "...\nβN−1", - "type": "text" - }, - { - "block_id": "p976-b32", - "global_id": 29503, - "bbox": [ - 225.0, - 507.02, - 232.26, - 516.98 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p976-b33", - "global_id": 29504, - "bbox": [ - 225.0, - 524.49, - 242.09, - 552.85 - ], - "text": "⎥⎥⎥⎦=", - "type": "text" - }, - { - "block_id": "p976-b34", - "global_id": 29505, - "bbox": [ - 244.13, - 507.01, - 251.4, - 516.98 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p976-b35", - "global_id": 29506, - "bbox": [ - 244.13, - 524.49, - 251.4, - 552.85 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p976-b36", - "global_id": 29507, - "bbox": [ - 251.39, - 513.32, - 297.05, - 525.46 - ], - "text": "1\nλ1\nλ2", - "type": "text" - }, - { - "block_id": "p976-b37", - "global_id": 29508, - "bbox": [ - 293.56, - 512.46, - 350.77, - 526.89 - ], - "text": "1\n· · ·\nλN−1", - "type": "text" - }, - { - "block_id": "p976-b38", - "global_id": 29509, - "bbox": [ - 251.39, - 519.91, - 340.27, - 538.05 - ], - "text": "1\n1\nλ2\nλ2", - "type": "text" - }, - { - "block_id": "p976-b39", - "global_id": 29510, - "bbox": [ - 293.56, - 525.04, - 350.77, - 539.47 - ], - "text": "2\n· · ·\nλN−1", - "type": "text" - }, - { - "block_id": "p976-b40", - "global_id": 29511, - "bbox": [ - 253.88, - 532.49, - 340.27, - 555.81 - ], - "text": "2\n...", - "type": "text" - }, - { - "block_id": "p976-b41", - "global_id": 29512, - "bbox": [ - 270.6, - 537.88, - 273.09, - 555.81 - ], - "text": "...", - "type": "text" - }, - { - "block_id": "p976-b42", - "global_id": 29513, - "bbox": [ - 251.39, - 537.88, - 342.55, - 569.1 - ], - "text": "...\n· · ·\n...\n1\nλN\nλ2", - "type": "text" - }, - { - "block_id": "p976-b43", - "global_id": 29514, - "bbox": [ - 292.77, - 556.16, - 350.77, - 570.41 - ], - "text": "N\n· · ·\nλN−1", - "type": "text" - }, - { - "block_id": "p976-b44", - "global_id": 29515, - "bbox": [ - 336.78, - 563.44, - 341.43, - 570.41 - ], - "text": "N", - "type": "text" - }, - { - "block_id": "p976-b45", - "global_id": 29516, - "bbox": [ - 351.27, - 507.02, - 358.53, - 516.98 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p976-b46", - "global_id": 29517, - "bbox": [ - 351.27, - 524.49, - 358.53, - 552.85 - ], - "text": "⎥⎥⎥⎦", - "type": "text" - }, - { - "block_id": "p976-b47", - "global_id": 29518, - "bbox": [ - 358.53, - 507.02, - 376.33, - 517.98 - ], - "text": "−1 ⎡", - "type": "text" - }, - { - "block_id": "p976-b48", - "global_id": 29519, - "bbox": [ - 369.06, - 524.49, - 376.33, - 552.85 - ], - "text": "⎢⎢⎢⎣", - "type": "text" - }, - { - "block_id": "p976-b49", - "global_id": 29520, - "bbox": [ - 377.12, - 513.49, - 386.06, - 524.59 - ], - "text": "λn", - "type": "text" - }, - { - "block_id": "p976-b50", - "global_id": 29521, - "bbox": [ - 377.12, - 520.23, - 386.06, - 536.54 - ], - "text": "1\nλn", - "type": "text" - }, - { - "block_id": "p976-b51", - "global_id": 29522, - "bbox": [ - 376.33, - 532.18, - 386.06, - 567.04 - ], - "text": "2...\nλn", - "type": "text" - }, - { - "block_id": "p976-b52", - "global_id": 29523, - "bbox": [ - 381.78, - 562.5, - 386.43, - 569.47 - ], - "text": "N", - "type": "text" - }, - { - "block_id": "p976-b53", - "global_id": 29524, - "bbox": [ - 387.35, - 507.02, - 394.61, - 516.98 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p976-b54", - "global_id": 29525, - "bbox": [ - 387.35, - 524.49, - 490.38, - 552.85 - ], - "text": "⎥⎥⎥⎦\n(10.67)", - "type": "text" - }, - { - "block_id": "p976-b55", - "global_id": 29526, - "bbox": [ - 101.84, - 580.19, - 290.14, - 592.07 - ], - "text": "and λ1, λ2, . . . , λN are the N eigenvalues of A.", - "type": "text" - }, - { - "block_id": "p976-b56", - "global_id": 29527, - "bbox": [ - 101.84, - 588.34, - 490.41, - 614.48 - ], - "text": "We can also determine An from the z-transform formula, which will be derived later, in\nEq. (10.71):", - "type": "text" - }, - { - "block_id": "p976-b57", - "global_id": 29528, - "bbox": [ - 246.51, - 612.92, - 345.72, - 627.43 - ], - "text": "An = Z−1[(I −z−1A)−1]", - "type": "text" - } - ] - }, - { - "page_num": 977, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p977-b0", - "global_id": 29529, - "bbox": [ - 264.66, - 62.89, - 516.14, - 71.98 - ], - "text": "10.7\nState-Space Analysis of Discrete-Time Systems\n957", - "type": "text" - }, - { - "block_id": "p977-b1", - "global_id": 29530, - "bbox": [ - 102.51, - 93.91, - 472.64, - 105.87 - ], - "text": "EXAMPLE 10.13\nState-Space Analysis of a Discrete-Time System", - "type": "text" - }, - { - "block_id": "p977-b2", - "global_id": 29531, - "bbox": [ - 128.9, - 122.12, - 502.76, - 145.22 - ], - "text": "Give a state-space description of the system in Fig. 10.12. Find the output y[n] if the input\nx[n] = u[n] and the initial conditions are q1[0] = 2 and q2[0] = 3.", - "type": "text" - }, - { - "block_id": "p977-b5", - "global_id": 29532, - "bbox": [ - 144.45, - 195.67, - 157.33, - 203.75 - ], - "text": "x[n]", - "type": "text" - }, - { - "block_id": "p977-b6", - "global_id": 29533, - "bbox": [ - 270.9, - 242.0, - 283.78, - 250.08 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p977-b7", - "global_id": 29534, - "bbox": [ - 219.31, - 201.03, - 250.3, - 210.85 - ], - "text": "q2[n 1]", - "type": "text" - }, - { - "block_id": "p977-b8", - "global_id": 29535, - "bbox": [ - 225.12, - 256.96, - 241.45, - 266.57 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p977-b9", - "global_id": 29536, - "bbox": [ - 207.59, - 297.8, - 223.92, - 307.4 - ], - "text": "q1[n]", - "type": "text" - }, - { - "block_id": "p977-b10", - "global_id": 29537, - "bbox": [ - 231.29, - 243.59, - 235.29, - 251.59 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p977-b11", - "global_id": 29538, - "bbox": [ - 233.43, - 297.66, - 243.14, - 305.95 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p977-b12", - "global_id": 29539, - "bbox": [ - 213.79, - 222.85, - 217.79, - 237.76 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p977-b13", - "global_id": 29540, - "bbox": [ - 213.79, - 271.1, - 217.79, - 286.01 - ], - "text": "1\nz", - "type": "text" - }, - { - "block_id": "p977-b14", - "global_id": 29541, - "bbox": [ - 190.17, - 256.9, - 194.17, - 273.92 - ], - "text": "5\n6", - "type": "text" - }, - { - "block_id": "p977-b15", - "global_id": 29542, - "bbox": [ - 185.98, - 297.97, - 197.53, - 310.06 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p977-b16", - "global_id": 29543, - "bbox": [ - 193.53, - 307.0, - 431.18, - 316.28 - ], - "text": "6\nFigure 10.12 System for Ex. 10.13.", - "type": "text" - }, - { - "block_id": "p977-b17", - "global_id": 29544, - "bbox": [ - 146.84, - 346.23, - 500.97, - 357.38 - ], - "text": "Recognizing that q2[n] = q1[n + 1], the state equations are [see Eqs. (10.62) and (10.63)]", - "type": "text" - }, - { - "block_id": "p977-b18", - "global_id": 29545, - "bbox": [ - 234.22, - 360.72, - 276.08, - 379.78 - ], - "text": "q1[n + 1]", - "type": "text" - }, - { - "block_id": "p977-b19", - "global_id": 29546, - "bbox": [ - 239.65, - 380.59, - 276.08, - 391.73 - ], - "text": "q2[n + 1]", - "type": "text" - }, - { - "block_id": "p977-b20", - "global_id": 29547, - "bbox": [ - 276.09, - 360.72, - 281.52, - 370.69 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p977-b21", - "global_id": 29548, - "bbox": [ - 283.56, - 374.71, - 291.33, - 384.67 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p977-b22", - "global_id": 29549, - "bbox": [ - 293.38, - 360.72, - 327.85, - 390.75 - ], - "text": "0\n1\n−1", - "type": "text" - }, - { - "block_id": "p977-b23", - "global_id": 29550, - "bbox": [ - 307.78, - 379.43, - 327.11, - 393.97 - ], - "text": "6\n5\n6", - "type": "text" - }, - { - "block_id": "p977-b24", - "global_id": 29551, - "bbox": [ - 328.3, - 360.72, - 360.85, - 379.78 - ], - "text": "! q1[n]", - "type": "text" - }, - { - "block_id": "p977-b25", - "global_id": 29552, - "bbox": [ - 340.27, - 380.59, - 360.85, - 391.73 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p977-b26", - "global_id": 29553, - "bbox": [ - 360.85, - 360.72, - 366.28, - 370.69 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p977-b27", - "global_id": 29554, - "bbox": [ - 367.83, - 374.71, - 375.6, - 384.67 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p977-b28", - "global_id": 29555, - "bbox": [ - 377.14, - 360.72, - 387.56, - 379.01 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p977-b29", - "global_id": 29556, - "bbox": [ - 382.58, - 381.0, - 387.56, - 390.96 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p977-b30", - "global_id": 29557, - "bbox": [ - 387.56, - 360.72, - 392.99, - 370.69 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p977-b31", - "global_id": 29558, - "bbox": [ - 392.99, - 375.02, - 397.41, - 384.99 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p977-b32", - "global_id": 29559, - "bbox": [ - 128.91, - 403.24, - 143.28, - 413.2 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p977-b33", - "global_id": 29560, - "bbox": [ - 267.83, - 428.33, - 293.69, - 438.61 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p977-b34", - "global_id": 29561, - "bbox": [ - 295.74, - 414.35, - 358.4, - 445.36 - ], - "text": "−1\n5\n q1[n]\nq2[n]", - "type": "text" - }, - { - "block_id": "p977-b35", - "global_id": 29562, - "bbox": [ - 358.4, - 414.34, - 363.83, - 424.31 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p977-b36", - "global_id": 29563, - "bbox": [ - 128.9, - 454.43, - 502.76, - 478.37 - ], - "text": "To find the solution [Eq. (10.65)], we must first determine An. The characteristic equation of\nA is", - "type": "text" - }, - { - "block_id": "p977-b37", - "global_id": 29564, - "bbox": [ - 187.43, - 493.71, - 230.34, - 504.01 - ], - "text": "|λI −A| =", - "type": "text" - }, - { - "block_id": "p977-b39", - "global_id": 29565, - "bbox": [ - 235.83, - 487.42, - 272.47, - 512.97 - ], - "text": "λ\n−1\n1\n6\nλ −5", - "type": "text" - }, - { - "block_id": "p977-b40", - "global_id": 29566, - "bbox": [ - 268.98, - 506.0, - 272.47, - 512.97 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p977-b41", - "global_id": 29567, - "bbox": [ - 273.66, - 479.26, - 315.24, - 507.16 - ], - "text": "= λ2 −5", - "type": "text" - }, - { - "block_id": "p977-b42", - "global_id": 29568, - "bbox": [ - 310.26, - 479.72, - 358.72, - 511.16 - ], - "text": "6λ + 1\n6 =", - "type": "text" - }, - { - "block_id": "p977-b43", - "global_id": 29569, - "bbox": [ - 358.73, - 487.14, - 381.22, - 503.68 - ], - "text": "λ −1", - "type": "text" - }, - { - "block_id": "p977-b44", - "global_id": 29570, - "bbox": [ - 376.24, - 501.19, - 381.22, - 511.16 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p977-b46", - "global_id": 29571, - "bbox": [ - 396.97, - 487.14, - 419.47, - 503.68 - ], - "text": "λ −1", - "type": "text" - }, - { - "block_id": "p977-b47", - "global_id": 29572, - "bbox": [ - 414.49, - 501.19, - 419.47, - 511.16 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p977-b49", - "global_id": 29573, - "bbox": [ - 429.44, - 493.71, - 444.24, - 504.09 - ], - "text": "= 0", - "type": "text" - }, - { - "block_id": "p977-b50", - "global_id": 29574, - "bbox": [ - 128.91, - 521.26, - 435.56, - 532.72 - ], - "text": "Hence, λ1 = 1/3 and λ2 = 1/2 are the eigenvalues of A and [see Eq. (10.66)]", - "type": "text" - }, - { - "block_id": "p977-b51", - "global_id": 29575, - "bbox": [ - 283.74, - 538.96, - 347.93, - 554.33 - ], - "text": "An = β0I + β1A", - "type": "text" - }, - { - "block_id": "p977-b52", - "global_id": 29576, - "bbox": [ - 128.91, - 565.51, - 222.69, - 575.47 - ], - "text": "where [see Eq. (10.67)]", - "type": "text" - }, - { - "block_id": "p977-b53", - "global_id": 29577, - "bbox": [ - 169.93, - 581.66, - 184.46, - 600.72 - ], - "text": "β0", - "type": "text" - }, - { - "block_id": "p977-b54", - "global_id": 29578, - "bbox": [ - 175.35, - 601.53, - 184.46, - 612.67 - ], - "text": "β1", - "type": "text" - }, - { - "block_id": "p977-b55", - "global_id": 29579, - "bbox": [ - 184.97, - 581.66, - 190.39, - 591.63 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p977-b56", - "global_id": 29580, - "bbox": [ - 192.44, - 595.65, - 200.21, - 605.61 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p977-b57", - "global_id": 29581, - "bbox": [ - 202.25, - 581.66, - 227.31, - 615.13 - ], - "text": "1\n1\n3\n1\n1\n2", - "type": "text" - }, - { - "block_id": "p977-b58", - "global_id": 29582, - "bbox": [ - 228.51, - 578.68, - 259.35, - 593.44 - ], - "text": "!−1 1", - "type": "text" - }, - { - "block_id": "p977-b59", - "global_id": 29583, - "bbox": [ - 250.66, - 579.57, - 268.05, - 609.61 - ], - "text": "3\nn\n 1", - "type": "text" - }, - { - "block_id": "p977-b60", - "global_id": 29584, - "bbox": [ - 255.87, - 595.97, - 268.05, - 617.17 - ], - "text": "2\nn", - "type": "text" - }, - { - "block_id": "p977-b62", - "global_id": 29585, - "bbox": [ - 276.78, - 595.65, - 284.55, - 605.61 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p977-b63", - "global_id": 29586, - "bbox": [ - 286.6, - 581.66, - 327.49, - 611.9 - ], - "text": "3\n−2\n−6\n6", - "type": "text" - }, - { - "block_id": "p977-b64", - "global_id": 29587, - "bbox": [ - 327.5, - 581.66, - 360.8, - 599.95 - ], - "text": "! (3)−n", - "type": "text" - }, - { - "block_id": "p977-b65", - "global_id": 29588, - "bbox": [ - 339.47, - 600.3, - 360.8, - 611.9 - ], - "text": "(2)−n", - "type": "text" - }, - { - "block_id": "p977-b66", - "global_id": 29589, - "bbox": [ - 361.3, - 581.66, - 366.73, - 591.63 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p977-b67", - "global_id": 29590, - "bbox": [ - 368.78, - 595.65, - 376.55, - 605.61 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p977-b68", - "global_id": 29591, - "bbox": [ - 378.6, - 581.66, - 451.91, - 599.95 - ], - "text": "3(3)−n −2(2)−n", - "type": "text" - }, - { - "block_id": "p977-b69", - "global_id": 29592, - "bbox": [ - 384.03, - 597.91, - 455.8, - 611.9 - ], - "text": "−6(3)−n + 6(2)−n", - "type": "text" - }, - { - "block_id": "p977-b70", - "global_id": 29593, - "bbox": [ - 456.31, - 581.66, - 461.74, - 591.63 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 978, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p978-b0", - "global_id": 29594, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "958\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p978-b1", - "global_id": 29595, - "bbox": [ - 103.16, - 86.24, - 117.54, - 96.2 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p978-b2", - "global_id": 29596, - "bbox": [ - 161.43, - 105.64, - 255.62, - 120.16 - ], - "text": "An = [3(3)−n −2(2)−n]", - "type": "text" - }, - { - "block_id": "p978-b3", - "global_id": 29597, - "bbox": [ - 256.73, - 95.87, - 282.08, - 126.12 - ], - "text": "1\n0\n0\n1", - "type": "text" - }, - { - "block_id": "p978-b4", - "global_id": 29598, - "bbox": [ - 282.08, - 95.87, - 287.51, - 105.83 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p978-b5", - "global_id": 29599, - "bbox": [ - 289.06, - 106.15, - 377.29, - 120.14 - ], - "text": "+ [−6(3)−n + 6(2)−n]", - "type": "text" - }, - { - "block_id": "p978-b6", - "global_id": 29600, - "bbox": [ - 378.4, - 95.87, - 412.88, - 125.89 - ], - "text": "0\n1\n−1", - "type": "text" - }, - { - "block_id": "p978-b7", - "global_id": 29601, - "bbox": [ - 392.8, - 114.59, - 412.13, - 129.12 - ], - "text": "6\n5\n6", - "type": "text" - }, - { - "block_id": "p978-b8", - "global_id": 29602, - "bbox": [ - 413.32, - 95.87, - 418.75, - 105.83 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p978-b9", - "global_id": 29603, - "bbox": [ - 174.65, - 137.97, - 182.42, - 147.93 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p978-b10", - "global_id": 29604, - "bbox": [ - 184.47, - 123.99, - 336.14, - 142.27 - ], - "text": "3(3)−n −2(2)−n\n−6(3)−n + 6(2)−n", - "type": "text" - }, - { - "block_id": "p978-b11", - "global_id": 29605, - "bbox": [ - 194.88, - 140.23, - 336.14, - 154.23 - ], - "text": "(3)−n −(2)−n\n−2(3)−n + 3(2)−n", - "type": "text" - }, - { - "block_id": "p978-b12", - "global_id": 29606, - "bbox": [ - 336.65, - 123.99, - 342.08, - 133.95 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p978-b13", - "global_id": 29607, - "bbox": [ - 447.96, - 138.39, - 477.0, - 148.35 - ], - "text": "(10.68)", - "type": "text" - }, - { - "block_id": "p978-b14", - "global_id": 29608, - "bbox": [ - 103.16, - 163.5, - 477.02, - 185.82 - ], - "text": "We can now determine the state vector q[n] from Eq. (10.64). Since we are interested in the\noutput y[n], we shall use Eq. (10.65) directly. Note that", - "type": "text" - }, - { - "block_id": "p978-b15", - "global_id": 29609, - "bbox": [ - 169.6, - 191.47, - 410.59, - 205.98 - ], - "text": "CAn = [−1\n5]An = [2(3)−n −3(2)−n\n−4(3)−n + 9(2)−n]", - "type": "text" - }, - { - "block_id": "p978-b16", - "global_id": 29610, - "bbox": [ - 103.17, - 214.31, - 282.85, - 226.31 - ], - "text": "and the zero-input response is CAnq[0], with", - "type": "text" - }, - { - "block_id": "p978-b17", - "global_id": 29611, - "bbox": [ - 267.66, - 241.81, - 294.63, - 252.18 - ], - "text": "q[0] =", - "type": "text" - }, - { - "block_id": "p978-b18", - "global_id": 29612, - "bbox": [ - 296.68, - 227.82, - 307.09, - 246.11 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p978-b19", - "global_id": 29613, - "bbox": [ - 302.11, - 248.1, - 307.09, - 258.06 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p978-b20", - "global_id": 29614, - "bbox": [ - 307.09, - 227.82, - 312.52, - 237.78 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p978-b21", - "global_id": 29615, - "bbox": [ - 103.16, - 267.75, - 235.68, - 277.72 - ], - "text": "Hence, the zero-input response is", - "type": "text" - }, - { - "block_id": "p978-b22", - "global_id": 29616, - "bbox": [ - 227.76, - 283.46, - 351.9, - 297.95 - ], - "text": "CAnq[0] = −8(3)−n + 21(2)−n", - "type": "text" - }, - { - "block_id": "p978-b23", - "global_id": 29617, - "bbox": [ - 103.16, - 305.98, - 477.02, - 366.0 - ], - "text": "The zero-state component is given by the convolution sum of CAn−1u[n−1] and Bx[n]. We can\nuse the shifting property of the convolution sum [Eq. (3.32)] to obtain the zero-state component\nby finding the convolution sum of CAnu[n] and Bx[n] and then replacing n with n −1 in the\nresult. We use this procedure because the convolution sums are listed in Table 3.1 for functions\nof the type x[n]u[n], rather than x[n]u[n −1].", - "type": "text" - }, - { - "block_id": "p978-b24", - "global_id": 29618, - "bbox": [ - 160.04, - 377.8, - 391.14, - 391.81 - ], - "text": "CAnu[n] ∗Bx[n] = [2(3)−n −3(2)−n −4(3)−n + 9(2)−n] ∗", - "type": "text" - }, - { - "block_id": "p978-b25", - "global_id": 29619, - "bbox": [ - 392.69, - 367.53, - 408.91, - 385.81 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p978-b26", - "global_id": 29620, - "bbox": [ - 398.12, - 387.39, - 414.71, - 397.67 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p978-b27", - "global_id": 29621, - "bbox": [ - 414.72, - 367.53, - 420.15, - 377.49 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p978-b28", - "global_id": 29622, - "bbox": [ - 228.06, - 399.22, - 359.92, - 413.71 - ], - "text": "= −4(3)−n ∗u[n] + 9(2)−n ∗u[n]", - "type": "text" - }, - { - "block_id": "p978-b29", - "global_id": 29623, - "bbox": [ - 103.17, - 423.98, - 242.38, - 433.95 - ], - "text": "Using Table 3.1 (pair 4), we obtain", - "type": "text" - }, - { - "block_id": "p978-b30", - "global_id": 29624, - "bbox": [ - 158.9, - 450.82, - 249.49, - 462.81 - ], - "text": "CAnu[n] ∗Bx[n] = −4", - "type": "text" - }, - { - "block_id": "p978-b32", - "global_id": 29625, - "bbox": [ - 257.99, - 444.22, - 301.85, - 455.82 - ], - "text": "1 −3−(n+1)", - "type": "text" - }, - { - "block_id": "p978-b33", - "global_id": 29626, - "bbox": [ - 269.31, - 458.97, - 289.85, - 473.52 - ], - "text": "1 −1\n3", - "type": "text" - }, - { - "block_id": "p978-b35", - "global_id": 29627, - "bbox": [ - 310.87, - 452.43, - 343.31, - 462.81 - ], - "text": "u[n] + 9", - "type": "text" - }, - { - "block_id": "p978-b37", - "global_id": 29628, - "bbox": [ - 351.8, - 444.22, - 395.66, - 455.82 - ], - "text": "1 −2−(n+1)", - "type": "text" - }, - { - "block_id": "p978-b38", - "global_id": 29629, - "bbox": [ - 363.13, - 458.97, - 383.66, - 473.52 - ], - "text": "1 −1\n2", - "type": "text" - }, - { - "block_id": "p978-b40", - "global_id": 29630, - "bbox": [ - 404.67, - 452.43, - 421.27, - 462.71 - ], - "text": "u[n]", - "type": "text" - }, - { - "block_id": "p978-b41", - "global_id": 29631, - "bbox": [ - 226.92, - 477.78, - 234.69, - 487.74 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p978-b43", - "global_id": 29632, - "bbox": [ - 240.68, - 469.77, - 270.48, - 488.15 - ], - "text": "12 + 6", - "type": "text" - }, - { - "block_id": "p978-b44", - "global_id": 29633, - "bbox": [ - 270.5, - 469.77, - 323.88, - 488.15 - ], - "text": "3−(n+1)\n−18", - "type": "text" - }, - { - "block_id": "p978-b46", - "global_id": 29634, - "bbox": [ - 327.89, - 469.77, - 380.96, - 488.15 - ], - "text": "2−(n+1)\nu[n]", - "type": "text" - }, - { - "block_id": "p978-b47", - "global_id": 29635, - "bbox": [ - 103.17, - 498.02, - 427.52, - 508.39 - ], - "text": "Now the desired (zero-state) response is obtained by replacing n by n −1. Hence,", - "type": "text" - }, - { - "block_id": "p978-b48", - "global_id": 29636, - "bbox": [ - 178.54, - 514.14, - 401.63, - 528.63 - ], - "text": "CAnu[n] ∗Bx[n −1] = [12 + 6(3)−n −18(2)−n]u[n −1]", - "type": "text" - }, - { - "block_id": "p978-b49", - "global_id": 29637, - "bbox": [ - 103.17, - 538.9, - 158.81, - 548.87 - ], - "text": "It follows that", - "type": "text" - }, - { - "block_id": "p978-b50", - "global_id": 29638, - "bbox": [ - 157.41, - 554.61, - 422.76, - 569.1 - ], - "text": "y[n] = [−8(3)−n + 21(2)−nu[n] + [12 + 6(3)−n −18(2)−n]u[n −1]", - "type": "text" - }, - { - "block_id": "p978-b51", - "global_id": 29639, - "bbox": [ - 103.17, - 575.35, - 477.02, - 601.29 - ], - "text": "This is the desired answer. We can simplify this answer by observing that 12 + 6(3)−n −\n18(2)−n = 0 for n = 0. Hence, u[n −1] may be replaced by u[n], and", - "type": "text" - }, - { - "block_id": "p978-b52", - "global_id": 29640, - "bbox": [ - 221.85, - 607.04, - 477.0, - 621.53 - ], - "text": "y[n] = [12 −2(3)−n + 3(2)−n]u[n]\n(10.69)", - "type": "text" - } - ] - }, - { - "page_num": 979, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p979-b0", - "global_id": 29641, - "bbox": [ - 264.66, - 62.89, - 516.14, - 71.98 - ], - "text": "10.7\nState-Space Analysis of Discrete-Time Systems\n959", - "type": "text" - }, - { - "block_id": "p979-b1", - "global_id": 29642, - "bbox": [ - 128.9, - 86.62, - 502.78, - 136.64 - ], - "text": "USING MATLAB TO OBTAIN A GRAPHICAL SOLUTION\nMATLAB is equipped with tools to simulate digital systems, which makes it easy to obtain a\ngraphical solution to the system. Let us use MATLAB simulation to determine the total system\noutput over 0 ≤n ≤25.", - "type": "text" - }, - { - "block_id": "p979-b2", - "global_id": 29643, - "bbox": [ - 128.9, - 146.35, - 471.7, - 194.16 - ], - "text": ">>\nA = [0 1;-1/6 5/6]; B = [0; 1]; C = [-1 5]; D = 0;\n>>\nN = 25; n = (0:N); x = ones(1,N+1); q0 = [2;3];\n>>\nsys = ss(A,B,C,D,-1); % Discrete-time state space model\n>>\n[y,q] = lsim(sys,x,n,q0); % Simulate output and state vector\n>>\nclf; stem(n,y,’k.’); xlabel(’n’); ylabel(’y[n]’); axis([-.5 25.5 11.5 13.5]);", - "type": "text" - }, - { - "block_id": "p979-b3", - "global_id": 29644, - "bbox": [ - 128.9, - 204.38, - 502.77, - 238.55 - ], - "text": "The MATLAB results, shown in Fig. 10.13, exactly align with the analytical solution derived\nearlier. Also notice that the zero-input and zero-state responses can be separately obtained\nusing the same code and respectively setting either x or q0 to zero.", - "type": "text" - }, - { - "block_id": "p979-b4", - "global_id": 29645, - "bbox": [ - 173.13, - 352.9, - 500.52, - 372.11 - ], - "text": "0\n5\n10\n15\n20\n25\nn", - "type": "text" - }, - { - "block_id": "p979-b5", - "global_id": 29646, - "bbox": [ - 149.22, - 343.68, - 163.22, - 351.68 - ], - "text": "11.5", - "type": "text" - }, - { - "block_id": "p979-b6", - "global_id": 29647, - "bbox": [ - 155.97, - 325.68, - 163.97, - 333.68 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p979-b7", - "global_id": 29648, - "bbox": [ - 149.22, - 307.68, - 163.22, - 315.68 - ], - "text": "12.5", - "type": "text" - }, - { - "block_id": "p979-b8", - "global_id": 29649, - "bbox": [ - 155.97, - 289.68, - 163.97, - 297.68 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p979-b9", - "global_id": 29650, - "bbox": [ - 149.22, - 271.67, - 163.22, - 279.67 - ], - "text": "13.5", - "type": "text" - }, - { - "block_id": "p979-b10", - "global_id": 29651, - "bbox": [ - 137.25, - 303.24, - 146.05, - 317.9 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p979-b11", - "global_id": 29652, - "bbox": [ - 136.48, - 378.79, - 396.25, - 388.03 - ], - "text": "Figure 10.13 Graphical solution to Ex. 10.13 by MATLAB simulation.", - "type": "text" - }, - { - "block_id": "p979-b12", - "global_id": 29653, - "bbox": [ - 127.59, - 437.25, - 303.43, - 450.06 - ], - "text": "10.7-2 The z-Transform Solution", - "type": "text" - }, - { - "block_id": "p979-b13", - "global_id": 29654, - "bbox": [ - 127.59, - 456.09, - 295.69, - 466.16 - ], - "text": "The z-transform of Eq. (10.62) is given by", - "type": "text" - }, - { - "block_id": "p979-b14", - "global_id": 29655, - "bbox": [ - 258.57, - 478.5, - 385.16, - 488.88 - ], - "text": "zQ[z] −zq[0] = AQ[z] + BX[z]", - "type": "text" - }, - { - "block_id": "p979-b15", - "global_id": 29656, - "bbox": [ - 127.59, - 501.64, - 169.34, - 511.6 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p979-b16", - "global_id": 29657, - "bbox": [ - 262.05, - 514.4, - 381.68, - 524.78 - ], - "text": "(zI −A)Q[z] = zq[0] + BX[z]", - "type": "text" - }, - { - "block_id": "p979-b17", - "global_id": 29658, - "bbox": [ - 127.59, - 534.55, - 141.97, - 544.52 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p979-b18", - "global_id": 29659, - "bbox": [ - 233.23, - 555.14, - 404.95, - 567.24 - ], - "text": "Q[z] = (zI −A)−1zq[0] + (zI −A)−1BX[z]", - "type": "text" - }, - { - "block_id": "p979-b19", - "global_id": 29660, - "bbox": [ - 253.54, - 571.58, - 410.51, - 583.69 - ], - "text": "= (I −z−1A)−1q[0] + (zI −A)−1BX[z]", - "type": "text" - }, - { - "block_id": "p979-b20", - "global_id": 29661, - "bbox": [ - 127.59, - 596.45, - 155.51, - 606.41 - ], - "text": "Hence,", - "type": "text" - }, - { - "block_id": "p979-b21", - "global_id": 29662, - "bbox": [ - 210.15, - 607.48, - 332.51, - 633.57 - ], - "text": "q[n] = Z−1[(I −z−1A)−1]q[0]\n\n\n\nzero-input response", - "type": "text" - }, - { - "block_id": "p979-b22", - "global_id": 29663, - "bbox": [ - 333.6, - 607.48, - 433.58, - 633.32 - ], - "text": "+Z−1[(zI −A)−1BX[z]]\n\n\n\nzero-state response", - "type": "text" - }, - { - "block_id": "p979-b23", - "global_id": 29664, - "bbox": [ - 487.08, - 609.62, - 516.12, - 619.58 - ], - "text": "(10.70)", - "type": "text" - } - ] - }, - { - "page_num": 980, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p980-b0", - "global_id": 29665, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "960\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p980-b1", - "global_id": 29666, - "bbox": [ - 101.84, - 85.82, - 328.78, - 95.78 - ], - "text": "A comparison of Eq. (10.70) with Eq. (10.64) shows that", - "type": "text" - }, - { - "block_id": "p980-b2", - "global_id": 29667, - "bbox": [ - 246.51, - 105.57, - 490.38, - 120.17 - ], - "text": "An = Z−1[(I −z−1A)−1]\n(10.71)", - "type": "text" - }, - { - "block_id": "p980-b3", - "global_id": 29668, - "bbox": [ - 101.85, - 134.6, - 227.91, - 144.56 - ], - "text": "The output equation is given by", - "type": "text" - }, - { - "block_id": "p980-b4", - "global_id": 29669, - "bbox": [ - 182.96, - 158.57, - 273.76, - 168.87 - ], - "text": "Y[z] = CQ[z] + DX[z]", - "type": "text" - }, - { - "block_id": "p980-b5", - "global_id": 29670, - "bbox": [ - 202.72, - 173.28, - 409.28, - 185.38 - ], - "text": "= C[(I −z−1A)−1q[0] + (zI −A)−1BX[z]] + DX[z]", - "type": "text" - }, - { - "block_id": "p980-b6", - "global_id": 29671, - "bbox": [ - 202.72, - 189.72, - 398.77, - 201.83 - ], - "text": "= C(I −z−1A)−1q[0] + [C(zI −A)−1B + D]X[z]", - "type": "text" - }, - { - "block_id": "p980-b7", - "global_id": 29672, - "bbox": [ - 202.72, - 206.15, - 288.98, - 232.25 - ], - "text": "= C(I −z−1A)−1q[0]\n\n\n\nzero-input response", - "type": "text" - }, - { - "block_id": "p980-b8", - "global_id": 29673, - "bbox": [ - 290.09, - 207.88, - 339.78, - 223.4 - ], - "text": "+ H[z]X[z]", - "type": "text" - }, - { - "block_id": "p980-b9", - "global_id": 29674, - "bbox": [ - 298.96, - 225.85, - 344.6, - 231.82 - ], - "text": "zero-state response", - "type": "text" - }, - { - "block_id": "p980-b10", - "global_id": 29675, - "bbox": [ - 461.34, - 208.3, - 490.38, - 218.26 - ], - "text": "(10.72)", - "type": "text" - }, - { - "block_id": "p980-b11", - "global_id": 29676, - "bbox": [ - 101.85, - 245.87, - 126.18, - 255.83 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p980-b12", - "global_id": 29677, - "bbox": [ - 243.5, - 259.4, - 490.38, - 271.49 - ], - "text": "H[z] = C(zI −A)−1B + D\n(10.73)", - "type": "text" - }, - { - "block_id": "p980-b13", - "global_id": 29678, - "bbox": [ - 101.85, - 282.52, - 490.39, - 305.55 - ], - "text": "Note that H[z] is the transfer function matrix of the system, and Hij[z], the ijth element of H[z], is\nthe transfer function relating the output yi[n] to the input xj[n]. If we define h[n] as", - "type": "text" - }, - { - "block_id": "p980-b14", - "global_id": 29679, - "bbox": [ - 260.44, - 317.14, - 331.81, - 329.16 - ], - "text": "h[n] = Z−1[H[z]]", - "type": "text" - }, - { - "block_id": "p980-b15", - "global_id": 29680, - "bbox": [ - 101.85, - 343.25, - 490.4, - 377.54 - ], - "text": "then h[n] represents the unit impulse function response matrix of the system. Thus, hij[n], the ijth\nelement of h[n], represents the zero-state response yi[n] when the input xj[n] = δ[n] and all other\ninputs are zero.", - "type": "text" - }, - { - "block_id": "p980-b16", - "global_id": 29681, - "bbox": [ - 76.77, - 407.77, - 405.39, - 419.73 - ], - "text": "EXAMPLE 10.14\nz-Transform Solution to State Equations", - "type": "text" - }, - { - "block_id": "p980-b17", - "global_id": 29682, - "bbox": [ - 103.16, - 435.97, - 393.48, - 446.35 - ], - "text": "Use the z-transform to find the response y[n] for the system in Ex. 10.13.", - "type": "text" - }, - { - "block_id": "p980-b18", - "global_id": 29683, - "bbox": [ - 103.16, - 469.26, - 205.53, - 479.22 - ], - "text": "According to Eq. (10.72),", - "type": "text" - }, - { - "block_id": "p980-b19", - "global_id": 29684, - "bbox": [ - 140.55, - 485.44, - 247.39, - 509.72 - ], - "text": "Y[z] = [−1\n5]\n 1\n−1", - "type": "text" - }, - { - "block_id": "p980-b20", - "global_id": 29685, - "bbox": [ - 212.19, - 498.94, - 252.78, - 518.89 - ], - "text": "z\n1\n6z\n1 −5\n6z", - "type": "text" - }, - { - "block_id": "p980-b21", - "global_id": 29686, - "bbox": [ - 253.99, - 485.44, - 280.36, - 503.72 - ], - "text": "!−1 2", - "type": "text" - }, - { - "block_id": "p980-b22", - "global_id": 29687, - "bbox": [ - 275.38, - 505.71, - 280.36, - 515.67 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p980-b23", - "global_id": 29688, - "bbox": [ - 280.36, - 485.44, - 285.79, - 495.4 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p980-b24", - "global_id": 29689, - "bbox": [ - 287.34, - 485.44, - 372.8, - 518.68 - ], - "text": "+ [−1\n5]\n z\n−1\n1\n6\nz −5", - "type": "text" - }, - { - "block_id": "p980-b25", - "global_id": 29690, - "bbox": [ - 369.31, - 511.7, - 372.8, - 518.68 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p980-b26", - "global_id": 29691, - "bbox": [ - 373.99, - 485.44, - 404.88, - 503.44 - ], - "text": "!−1 0", - "type": "text" - }, - { - "block_id": "p980-b27", - "global_id": 29692, - "bbox": [ - 396.57, - 503.62, - 408.21, - 518.22 - ], - "text": "z\nz−1", - "type": "text" - }, - { - "block_id": "p980-b28", - "global_id": 29693, - "bbox": [ - 409.41, - 485.44, - 414.84, - 495.4 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p980-b29", - "global_id": 29694, - "bbox": [ - 160.31, - 535.86, - 204.46, - 546.24 - ], - "text": "= [−1\n5]", - "type": "text" - }, - { - "block_id": "p980-b30", - "global_id": 29695, - "bbox": [ - 205.57, - 518.98, - 241.91, - 533.13 - ], - "text": "z(6z−5)", - "type": "text" - }, - { - "block_id": "p980-b31", - "global_id": 29696, - "bbox": [ - 212.95, - 526.46, - 292.0, - 557.31 - ], - "text": "6z2 −5z+1\n6z\n6z2 −5z+1\n−z\n6z2 −5z+1\n6z2", - "type": "text" - }, - { - "block_id": "p980-b32", - "global_id": 29697, - "bbox": [ - 258.65, - 548.03, - 292.0, - 557.31 - ], - "text": "6z2 −5z+1", - "type": "text" - }, - { - "block_id": "p980-b33", - "global_id": 29698, - "bbox": [ - 293.2, - 518.98, - 310.9, - 540.25 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p980-b34", - "global_id": 29699, - "bbox": [ - 305.92, - 542.25, - 310.9, - 552.21 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p980-b35", - "global_id": 29700, - "bbox": [ - 310.9, - 521.97, - 316.33, - 531.93 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p980-b36", - "global_id": 29701, - "bbox": [ - 317.88, - 535.86, - 361.53, - 546.24 - ], - "text": "+ [−1\n5]", - "type": "text" - }, - { - "block_id": "p980-b37", - "global_id": 29702, - "bbox": [ - 362.64, - 515.99, - 369.91, - 525.95 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p980-b38", - "global_id": 29703, - "bbox": [ - 362.64, - 533.93, - 369.91, - 543.89 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p980-b39", - "global_id": 29704, - "bbox": [ - 399.78, - 520.77, - 402.49, - 527.74 - ], - "text": "z", - "type": "text" - }, - { - "block_id": "p980-b40", - "global_id": 29705, - "bbox": [ - 371.09, - 531.24, - 389.48, - 538.51 - ], - "text": "(z−1)", - "type": "text" - }, - { - "block_id": "p980-b42", - "global_id": 29706, - "bbox": [ - 393.5, - 529.23, - 410.88, - 538.44 - ], - "text": "z2 −5", - "type": "text" - }, - { - "block_id": "p980-b43", - "global_id": 29707, - "bbox": [ - 407.89, - 520.84, - 431.17, - 541.24 - ], - "text": "6 z+ 1\n6", - "type": "text" - }, - { - "block_id": "p980-b44", - "global_id": 29708, - "bbox": [ - 398.03, - 539.85, - 403.73, - 548.52 - ], - "text": "z2", - "type": "text" - }, - { - "block_id": "p980-b45", - "global_id": 29709, - "bbox": [ - 371.09, - 552.02, - 389.48, - 559.29 - ], - "text": "(z−1)", - "type": "text" - }, - { - "block_id": "p980-b47", - "global_id": 29710, - "bbox": [ - 393.5, - 550.01, - 410.88, - 559.22 - ], - "text": "z2 −5", - "type": "text" - }, - { - "block_id": "p980-b48", - "global_id": 29711, - "bbox": [ - 407.89, - 541.62, - 431.17, - 562.02 - ], - "text": "6 z+ 1\n6", - "type": "text" - }, - { - "block_id": "p980-b49", - "global_id": 29712, - "bbox": [ - 432.37, - 515.99, - 439.63, - 525.95 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p980-b50", - "global_id": 29713, - "bbox": [ - 432.37, - 533.93, - 439.63, - 543.89 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p980-b51", - "global_id": 29714, - "bbox": [ - 160.31, - 561.85, - 211.38, - 582.82 - ], - "text": "= 13z2 −3z", - "type": "text" - }, - { - "block_id": "p980-b52", - "global_id": 29715, - "bbox": [ - 171.32, - 577.46, - 193.84, - 590.62 - ], - "text": "z2 −5", - "type": "text" - }, - { - "block_id": "p980-b53", - "global_id": 29716, - "bbox": [ - 190.36, - 565.47, - 283.6, - 593.53 - ], - "text": "6z+ 1\n6\n+\n(5z−1)z\n(z−1)", - "type": "text" - }, - { - "block_id": "p980-b55", - "global_id": 29717, - "bbox": [ - 257.44, - 577.46, - 279.97, - 590.62 - ], - "text": "z2 −5", - "type": "text" - }, - { - "block_id": "p980-b56", - "global_id": 29718, - "bbox": [ - 276.49, - 572.33, - 304.92, - 593.53 - ], - "text": "6z+ 1\n6", - "type": "text" - }, - { - "block_id": "p980-b57", - "global_id": 29719, - "bbox": [ - 160.31, - 594.71, - 190.61, - 611.66 - ], - "text": "= −8z", - "type": "text" - }, - { - "block_id": "p980-b58", - "global_id": 29720, - "bbox": [ - 171.32, - 608.24, - 189.87, - 619.87 - ], - "text": "z−1", - "type": "text" - }, - { - "block_id": "p980-b59", - "global_id": 29721, - "bbox": [ - 186.38, - 595.03, - 220.21, - 622.78 - ], - "text": "3\n+ 21z", - "type": "text" - }, - { - "block_id": "p980-b60", - "global_id": 29722, - "bbox": [ - 203.43, - 608.24, - 221.98, - 619.87 - ], - "text": "z−1", - "type": "text" - }, - { - "block_id": "p980-b61", - "global_id": 29723, - "bbox": [ - 218.49, - 595.03, - 251.88, - 622.78 - ], - "text": "2\n+ 12z", - "type": "text" - }, - { - "block_id": "p980-b62", - "global_id": 29724, - "bbox": [ - 235.55, - 595.03, - 283.1, - 619.15 - ], - "text": "z−1 + 12z", - "type": "text" - }, - { - "block_id": "p980-b63", - "global_id": 29725, - "bbox": [ - 266.77, - 595.03, - 312.28, - 619.15 - ], - "text": "z−1 + 6z", - "type": "text" - }, - { - "block_id": "p980-b64", - "global_id": 29726, - "bbox": [ - 297.98, - 608.24, - 316.52, - 619.87 - ], - "text": "z−1", - "type": "text" - }, - { - "block_id": "p980-b65", - "global_id": 29727, - "bbox": [ - 313.04, - 595.03, - 346.87, - 622.78 - ], - "text": "3\n−18z", - "type": "text" - }, - { - "block_id": "p980-b66", - "global_id": 29728, - "bbox": [ - 330.09, - 608.24, - 348.63, - 619.87 - ], - "text": "z−1", - "type": "text" - }, - { - "block_id": "p980-b67", - "global_id": 29729, - "bbox": [ - 345.15, - 615.81, - 348.63, - 622.78 - ], - "text": "2", - "type": "text" - } - ] - }, - { - "page_num": 981, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p981-b0", - "global_id": 29730, - "bbox": [ - 264.72, - 62.89, - 516.13, - 71.98 - ], - "text": "10.8\nMATLAB: Toolboxes and State-Space Analysis\n961", - "type": "text" - }, - { - "block_id": "p981-b1", - "global_id": 29731, - "bbox": [ - 128.9, - 86.24, - 170.65, - 96.21 - ], - "text": "Therefore,", - "type": "text" - }, - { - "block_id": "p981-b2", - "global_id": 29732, - "bbox": [ - 203.14, - 93.67, - 311.1, - 108.16 - ], - "text": "y[n] = [−8(3)−n + 21(2)−n", - "type": "text" - }, - { - "block_id": "p981-b3", - "global_id": 29733, - "bbox": [ - 234.37, - 103.52, - 311.64, - 121.9 - ], - "text": "zero-input response", - "type": "text" - }, - { - "block_id": "p981-b4", - "global_id": 29734, - "bbox": [ - 312.73, - 93.67, - 411.41, - 108.16 - ], - "text": "+12 + 6(3)−n −18(2)−n", - "type": "text" - }, - { - "block_id": "p981-b5", - "global_id": 29735, - "bbox": [ - 321.6, - 103.52, - 411.93, - 121.9 - ], - "text": "zero-state response", - "type": "text" - }, - { - "block_id": "p981-b6", - "global_id": 29736, - "bbox": [ - 411.93, - 97.78, - 428.53, - 108.06 - ], - "text": "]u[n", - "type": "text" - }, - { - "block_id": "p981-b7", - "global_id": 29737, - "bbox": [ - 127.89, - 160.18, - 385.02, - 186.25 - ], - "text": "LINEAR TRANSFORMATION, CONTROLLABILITY,\nAND OBSERVABILITY", - "type": "text" - }, - { - "block_id": "p981-b8", - "global_id": 29738, - "bbox": [ - 127.59, - 190.28, - 516.15, - 212.19 - ], - "text": "The procedure for linear transformation is parallel to that in the continuous-time case (Sec. 10.5).\nIf w is the transformed-state vector given by", - "type": "text" - }, - { - "block_id": "p981-b9", - "global_id": 29739, - "bbox": [ - 306.52, - 223.41, - 337.21, - 233.7 - ], - "text": "w = Pq", - "type": "text" - }, - { - "block_id": "p981-b10", - "global_id": 29740, - "bbox": [ - 127.59, - 245.41, - 144.74, - 255.38 - ], - "text": "then", - "type": "text" - }, - { - "block_id": "p981-b11", - "global_id": 29741, - "bbox": [ - 260.5, - 255.12, - 383.22, - 267.33 - ], - "text": "w[n + 1] = PAP−1w[n] + PBx", - "type": "text" - }, - { - "block_id": "p981-b12", - "global_id": 29742, - "bbox": [ - 127.59, - 276.13, - 141.97, - 286.09 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p981-b13", - "global_id": 29743, - "bbox": [ - 277.44, - 285.84, - 366.28, - 297.96 - ], - "text": "y[n] = (CP−1)w + Dx", - "type": "text" - }, - { - "block_id": "p981-b14", - "global_id": 29744, - "bbox": [ - 127.59, - 306.85, - 516.15, - 328.77 - ], - "text": "Controllability and observability may be investigated by diagonalizing the matrix, as explained in\nSec. 10.5-1.", - "type": "text" - }, - { - "block_id": "p981-b15", - "global_id": 29745, - "bbox": [ - 127.94, - 357.89, - 344.86, - 387.77 - ], - "text": "10.8 MATLAB: TOOLBOXES AND\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p981-b16", - "global_id": 29746, - "bbox": [ - 127.59, - 393.76, - 516.13, - 463.5 - ], - "text": "The preceding MATLAB sections provide a comprehensive introduction to the basic MATLAB\nenvironment. However, MATLAB also offers a wide range of toolboxes that perform specialized\ntasks. Once installed, toolbox functions operate no differently from ordinary MATLAB functions.\nAlthough toolboxes are purchased at extra cost, they save time and offer the convenience of\npredefined functions. It would take significant effort to duplicate a toolbox’s functionality by using\ncustom user-defined programs.", - "type": "text" - }, - { - "block_id": "p981-b17", - "global_id": 29747, - "bbox": [ - 127.59, - 465.49, - 516.16, - 523.27 - ], - "text": "Three toolboxes are particularly appropriate in the study of signals and systems: the control\nsystem toolbox, the signal-processing toolbox, and the symbolic math toolbox. Functions from\nthese toolboxes have been utilized throughout the text in the MATLAB examples as well as\ncertain end-of-chapter problems. This section provides a more formal introduction to a selection\nof functions, both standard and toolbox, that are appropriate for state-space problems.", - "type": "text" - }, - { - "block_id": "p981-b18", - "global_id": 29748, - "bbox": [ - 127.59, - 547.0, - 495.3, - 559.82 - ], - "text": "10.8-1 z-Transform Solutions to Discrete-Time, State-Space Systems", - "type": "text" - }, - { - "block_id": "p981-b19", - "global_id": 29749, - "bbox": [ - 127.59, - 565.95, - 516.11, - 599.82 - ], - "text": "As with continuous-time systems, it is often more convenient to solve discrete-time systems in the\ntransform domain rather than in the time domain. As given in Ex. 10.13, consider the state-space\ndescription of the system shown in Fig. 10.12.", - "type": "text" - }, - { - "block_id": "p981-b20", - "global_id": 29750, - "bbox": [ - 233.89, - 602.9, - 275.74, - 621.96 - ], - "text": "q1[n + 1]", - "type": "text" - }, - { - "block_id": "p981-b21", - "global_id": 29751, - "bbox": [ - 239.32, - 622.76, - 275.74, - 633.91 - ], - "text": "q2[n + 1]", - "type": "text" - }, - { - "block_id": "p981-b22", - "global_id": 29752, - "bbox": [ - 275.75, - 602.9, - 281.18, - 612.86 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p981-b23", - "global_id": 29753, - "bbox": [ - 283.23, - 616.89, - 291.0, - 626.85 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p981-b24", - "global_id": 29754, - "bbox": [ - 293.05, - 602.9, - 327.52, - 632.93 - ], - "text": "0\n1\n−1", - "type": "text" - }, - { - "block_id": "p981-b25", - "global_id": 29755, - "bbox": [ - 307.44, - 621.61, - 326.78, - 636.15 - ], - "text": "6\n5\n6", - "type": "text" - }, - { - "block_id": "p981-b26", - "global_id": 29756, - "bbox": [ - 327.97, - 602.9, - 360.52, - 621.96 - ], - "text": "! q1[n]", - "type": "text" - }, - { - "block_id": "p981-b27", - "global_id": 29757, - "bbox": [ - 339.93, - 622.76, - 360.52, - 633.91 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p981-b28", - "global_id": 29758, - "bbox": [ - 360.51, - 602.9, - 365.94, - 612.86 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p981-b29", - "global_id": 29759, - "bbox": [ - 367.5, - 616.89, - 375.27, - 626.85 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p981-b30", - "global_id": 29760, - "bbox": [ - 376.81, - 602.9, - 387.23, - 621.19 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p981-b31", - "global_id": 29761, - "bbox": [ - 382.25, - 623.18, - 387.23, - 633.14 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p981-b32", - "global_id": 29762, - "bbox": [ - 387.23, - 602.9, - 392.65, - 612.87 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p981-b33", - "global_id": 29763, - "bbox": [ - 393.76, - 616.89, - 409.82, - 627.16 - ], - "text": "x[n]", - "type": "text" - } - ] - }, - { - "page_num": 982, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p982-b0", - "global_id": 29764, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "962\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p982-b1", - "global_id": 29765, - "bbox": [ - 101.84, - 85.82, - 116.22, - 95.78 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p982-b2", - "global_id": 29766, - "bbox": [ - 248.72, - 86.96, - 338.08, - 111.22 - ], - "text": "y[n] = [−1\n5]\n q1[n]", - "type": "text" - }, - { - "block_id": "p982-b3", - "global_id": 29767, - "bbox": [ - 317.5, - 106.82, - 338.08, - 117.97 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p982-b4", - "global_id": 29768, - "bbox": [ - 338.08, - 86.96, - 343.51, - 96.92 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p982-b5", - "global_id": 29769, - "bbox": [ - 101.84, - 124.44, - 490.37, - 147.54 - ], - "text": "We are interested in the output y[n] in response to the input x[n] = u[n] with initial conditions\nq1[0] = 2 and q2[0] = 3.", - "type": "text" - }, - { - "block_id": "p982-b6", - "global_id": 29770, - "bbox": [ - 119.78, - 148.69, - 417.12, - 158.73 - ], - "text": "To describe this system, the state matrices A, B, C, and D are first defined.", - "type": "text" - }, - { - "block_id": "p982-b7", - "global_id": 29771, - "bbox": [ - 101.84, - 168.26, - 384.24, - 178.22 - ], - "text": ">>\nA = [0 1;-1/6 5/6]; B = [0; 1]; C = [-1 5]; D = 0;", - "type": "text" - }, - { - "block_id": "p982-b8", - "global_id": 29772, - "bbox": [ - 101.84, - 187.17, - 320.27, - 197.14 - ], - "text": "Additionally, the vector of initial conditions is defined.", - "type": "text" - }, - { - "block_id": "p982-b9", - "global_id": 29773, - "bbox": [ - 101.84, - 206.66, - 185.52, - 216.62 - ], - "text": ">>\nq_0 = [2;3];", - "type": "text" - }, - { - "block_id": "p982-b10", - "global_id": 29774, - "bbox": [ - 119.78, - 225.58, - 359.7, - 235.54 - ], - "text": "In the transform domain, the solution to the state equation is", - "type": "text" - }, - { - "block_id": "p982-b11", - "global_id": 29775, - "bbox": [ - 207.49, - 243.91, - 490.38, - 256.01 - ], - "text": "Q[z] = (I −z−1A)−1q[0] + (zI −A)−1BX[z]\n(10.74)", - "type": "text" - }, - { - "block_id": "p982-b12", - "global_id": 29776, - "bbox": [ - 101.85, - 266.52, - 469.2, - 276.48 - ], - "text": "The solution is separated into two parts: the zero-input response and the zero-state response.", - "type": "text" - }, - { - "block_id": "p982-b13", - "global_id": 29777, - "bbox": [ - 101.85, - 278.48, - 490.41, - 300.39 - ], - "text": "MATLAB’s symbolic toolbox makes possible a symbolic representation of Eq. (10.74). First,\na symbolic variable z needs to be defined.", - "type": "text" - }, - { - "block_id": "p982-b14", - "global_id": 29778, - "bbox": [ - 101.85, - 309.93, - 190.76, - 319.89 - ], - "text": ">>\nz = sym(’z’);", - "type": "text" - }, - { - "block_id": "p982-b15", - "global_id": 29779, - "bbox": [ - 101.85, - 328.83, - 490.41, - 374.95 - ], - "text": "The sym command is used to construct symbolic variables, objects, and numbers. Typing whos\nconfirms that z is indeed a symbolic object. The syms command is a shorthand command for\nconstructing symbolic objects. For example, syms z s is equivalent to the two instructions z =\nsym(’z’); and s = sym(’s’);.", - "type": "text" - }, - { - "block_id": "p982-b16", - "global_id": 29780, - "bbox": [ - 101.85, - 376.25, - 490.39, - 398.86 - ], - "text": "Next, a symbolic expression for X[z] needs to be constructed for the unit step input, x[n] =\nu[n]. The z-transform is computed by means of the ztrans command.", - "type": "text" - }, - { - "block_id": "p982-b17", - "global_id": 29781, - "bbox": [ - 101.85, - 408.1, - 227.37, - 430.02 - ], - "text": ">>\nX = ztrans(sym(’1’))\nX = z/(z-1)", - "type": "text" - }, - { - "block_id": "p982-b18", - "global_id": 29782, - "bbox": [ - 101.85, - 438.55, - 490.4, - 497.04 - ], - "text": "Several comments are in order. First, the ztrans command assumes a causal signal. For n ≥0,\nu[n] has a constant value of 1. Second, the argument of ztrans needs to be a symbolic expression,\neven if the expression is a constant. Thus, a symbolic one sym(’1’) is required. Also note\nthat continuous-time systems use Laplace transforms rather than z-transforms. In such cases, the\nlaplace command replaces the ztrans command.", - "type": "text" - }, - { - "block_id": "p982-b19", - "global_id": 29783, - "bbox": [ - 119.78, - 498.33, - 259.87, - 508.71 - ], - "text": "Construction of Q[z] is now trivial.", - "type": "text" - }, - { - "block_id": "p982-b20", - "global_id": 29784, - "bbox": [ - 101.84, - 517.69, - 330.32, - 535.62 - ], - "text": ">>\nQ = inv(eye(2)-z^(-1)*A)*q_0 + inv(z*eye(2)-A)*B*X\nQ =", - "type": "text" - }, - { - "block_id": "p982-b21", - "global_id": 29785, - "bbox": [ - 123.01, - 537.61, - 461.5, - 545.58 - ], - "text": "(18*z)/(6*z^2-5*z+1) + (2*z*(6*z-5))/(6*z^2-5*z+1) + (6*z)/((z-1)*(6*z^2-5*z+1))", - "type": "text" - }, - { - "block_id": "p982-b22", - "global_id": 29786, - "bbox": [ - 148.41, - 547.58, - 469.96, - 555.55 - ], - "text": "(18*z^2)/(6*z^2-5*z+1) - (2*z)/(6*z^2-5*z+1) + (6*z^2)/((z-1)*(6*z^2-5*z+1))", - "type": "text" - }, - { - "block_id": "p982-b23", - "global_id": 29787, - "bbox": [ - 101.84, - 565.05, - 490.4, - 610.88 - ], - "text": "Unfortunately, not all MATLAB functions work with symbolic objects. Still, the symbolic toolbox\noverloads many standard MATLAB functions, such as inv, to work with symbolic objects. Recall\nthat overloaded functions have identical names but different behavior; proper function selection is\ntypically determined by context.", - "type": "text" - }, - { - "block_id": "p982-b24", - "global_id": 29788, - "bbox": [ - 101.84, - 612.87, - 490.38, - 634.79 - ], - "text": "The expression Q is somewhat unwieldy. The simplify command uses various algebraic\ntechniques to simplify the result.", - "type": "text" - } - ] - }, - { - "page_num": 983, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p983-b0", - "global_id": 29789, - "bbox": [ - 264.72, - 62.89, - 516.13, - 71.98 - ], - "text": "10.8\nMATLAB: Toolboxes and State-Space Analysis\n963", - "type": "text" - }, - { - "block_id": "p983-b1", - "global_id": 29790, - "bbox": [ - 127.59, - 86.24, - 446.63, - 108.15 - ], - "text": ">>\nQ = simplify(Q)\nQ = -(2*z*(- 6*z^2 + 2*z + 1))/(6*z^3 - 11*z^2 + 6*z - 1)", - "type": "text" - }, - { - "block_id": "p983-b2", - "global_id": 29791, - "bbox": [ - 185.11, - 110.15, - 446.62, - 120.12 - ], - "text": "(2*z*(9*z^2 - 7*z + 1))/(6*z^3 - 11*z^2 + 6*z - 1)", - "type": "text" - }, - { - "block_id": "p983-b3", - "global_id": 29792, - "bbox": [ - 127.59, - 130.39, - 516.14, - 152.31 - ], - "text": "The resulting expression is mathematically equivalent to the original but notationally more\ncompact.", - "type": "text" - }, - { - "block_id": "p983-b4", - "global_id": 29793, - "bbox": [ - 145.52, - 153.89, - 363.56, - 164.27 - ], - "text": "Since D = 0, the output Y[z] is given by Y[z] = CQ[z].", - "type": "text" - }, - { - "block_id": "p983-b5", - "global_id": 29794, - "bbox": [ - 127.59, - 175.12, - 441.4, - 197.04 - ], - "text": ">>\nY = simplify(C*Q)\nY = (6*z*(13*z^2 - 11*z + 2))/(6*z^3 - 11*z^2 + 6*z - 1)", - "type": "text" - }, - { - "block_id": "p983-b6", - "global_id": 29795, - "bbox": [ - 127.59, - 207.22, - 516.11, - 229.52 - ], - "text": "The corresponding time-domain expression is obtained by using the inverse z-transform command\niztrans.", - "type": "text" - }, - { - "block_id": "p983-b7", - "global_id": 29796, - "bbox": [ - 127.6, - 240.09, - 305.42, - 262.01 - ], - "text": ">>\ny = iztrans(Y)\ny = 3*(1/2)^n - 2*(1/3)^n + 12", - "type": "text" - }, - { - "block_id": "p983-b8", - "global_id": 29797, - "bbox": [ - 127.59, - 272.29, - 516.16, - 330.07 - ], - "text": "Like ztrans, the iztrans command assumes a causal signal, so the result implies multiplication\nby a unit step. That is, the system output is y[n] = (3(1/2)n−2(1/3)n+12)u[n], which is equivalent\nto Eq. (10.69) derived in Ex. 10.13. Continuous-time systems use inverse Laplace transforms rather\nthan inverse z-transforms. In such cases, the ilaplace command therefore replaces the iztrans\ncommand.", - "type": "text" - }, - { - "block_id": "p983-b9", - "global_id": 29798, - "bbox": [ - 145.52, - 331.65, - 516.14, - 342.8 - ], - "text": "Following a similar procedure, it is a simple matter to compute the zero-input response yzir[n]:", - "type": "text" - }, - { - "block_id": "p983-b10", - "global_id": 29799, - "bbox": [ - 127.59, - 352.88, - 425.72, - 374.8 - ], - "text": ">>\ny_zir = iztrans(simplify(C*inv(eye(2)-z^(-1)*A)*q_0))\ny_zir = 21*(1/2)^n - 8*(1/3)^n", - "type": "text" - }, - { - "block_id": "p983-b11", - "global_id": 29800, - "bbox": [ - 127.59, - 385.07, - 267.45, - 395.04 - ], - "text": "The zero-state response is given by", - "type": "text" - }, - { - "block_id": "p983-b12", - "global_id": 29801, - "bbox": [ - 127.59, - 405.89, - 331.57, - 427.82 - ], - "text": ">>\ny_zsr = y - y_zir\ny_zsr = 6*(1/3)^n - 18*(1/2)^n + 12", - "type": "text" - }, - { - "block_id": "p983-b13", - "global_id": 29802, - "bbox": [ - 127.59, - 438.09, - 470.44, - 448.34 - ], - "text": "Typing iztrans(simplify(C*inv(z*eye(2)-A)*B*X)) produces the same result.", - "type": "text" - }, - { - "block_id": "p983-b14", - "global_id": 29803, - "bbox": [ - 127.59, - 450.05, - 516.16, - 483.92 - ], - "text": "MATLAB plotting functions, such as plot and stem, do not directly support symbolic\nexpressions. By using the subs command, however, it is easy to replace a symbolic variable with\na vector of desired values.", - "type": "text" - }, - { - "block_id": "p983-b15", - "global_id": 29804, - "bbox": [ - 164.24, - 592.09, - 491.63, - 613.28 - ], - "text": "0\n5\n10\n15\n20\n25\nn", - "type": "text" - }, - { - "block_id": "p983-b16", - "global_id": 29805, - "bbox": [ - 140.33, - 582.86, - 154.33, - 590.86 - ], - "text": "11.5", - "type": "text" - }, - { - "block_id": "p983-b17", - "global_id": 29806, - "bbox": [ - 147.08, - 564.87, - 155.08, - 572.87 - ], - "text": "12", - "type": "text" - }, - { - "block_id": "p983-b18", - "global_id": 29807, - "bbox": [ - 140.33, - 546.86, - 154.33, - 554.86 - ], - "text": "12.5", - "type": "text" - }, - { - "block_id": "p983-b19", - "global_id": 29808, - "bbox": [ - 147.08, - 528.86, - 155.08, - 536.86 - ], - "text": "13", - "type": "text" - }, - { - "block_id": "p983-b20", - "global_id": 29809, - "bbox": [ - 140.33, - 510.86, - 154.33, - 518.86 - ], - "text": "13.5", - "type": "text" - }, - { - "block_id": "p983-b21", - "global_id": 29810, - "bbox": [ - 128.36, - 542.43, - 137.16, - 557.09 - ], - "text": "y[n]", - "type": "text" - }, - { - "block_id": "p983-b22", - "global_id": 29811, - "bbox": [ - 127.59, - 619.58, - 394.58, - 629.19 - ], - "text": "Figure 10.14 Output y[n] computed by using the symbolic math toolbox.", - "type": "text" - } - ] - }, - { - "page_num": 984, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p984-b0", - "global_id": 29812, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "964\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p984-b1", - "global_id": 29813, - "bbox": [ - 101.84, - 86.24, - 415.67, - 108.15 - ], - "text": ">>\nn = [0:25]; stem(n,subs(y,n),’k.’);\n>>\nxlabel(’n’); ylabel(’y[n]’); axis([-.5 25.5 11.5 13.5]);", - "type": "text" - }, - { - "block_id": "p984-b2", - "global_id": 29814, - "bbox": [ - 101.84, - 117.58, - 490.37, - 163.4 - ], - "text": "Figure 10.14 shows the results, which are equivalent to the results obtained in Ex. 10.13. Although\nthere are plotting commands in the symbolic math toolbox such as ezplot that plot symbolic\nexpression, these plotting routines lack the flexibility needed to satisfactorily plot discrete-time\nfunctions.", - "type": "text" - }, - { - "block_id": "p984-b3", - "global_id": 29815, - "bbox": [ - 101.84, - 187.91, - 427.05, - 199.86 - ], - "text": "10.8-2 Transfer Functions from State-Space Representations", - "type": "text" - }, - { - "block_id": "p984-b4", - "global_id": 29816, - "bbox": [ - 101.84, - 205.98, - 490.41, - 227.91 - ], - "text": "A system’s transfer function provides a wealth of useful information. From Eq. (10.73), the transfer\nfunction for the system described in Ex. 10.13 is", - "type": "text" - }, - { - "block_id": "p984-b5", - "global_id": 29817, - "bbox": [ - 101.84, - 237.9, - 352.9, - 259.82 - ], - "text": ">>\nH = collect(simplify(C*inv(z*eye(2)-A)*B+D))\nH = (30*z - 6)/(6*z^2 - 5*z + 1)", - "type": "text" - }, - { - "block_id": "p984-b6", - "global_id": 29818, - "bbox": [ - 101.84, - 269.24, - 490.43, - 291.44 - ], - "text": "It is also possible to determine the numerator and denominator transfer function coefficients from\na state-space model by using the signal-processing toolbox function ss2tf.", - "type": "text" - }, - { - "block_id": "p984-b7", - "global_id": 29819, - "bbox": [ - 101.84, - 301.15, - 290.13, - 335.03 - ], - "text": ">>\n[num,den] = ss2tf(A,B,C,D)\nnum =\n0\n5.0000\n-1.0000\nden = 1.0000\n-0.8333\n0.1667", - "type": "text" - }, - { - "block_id": "p984-b8", - "global_id": 29820, - "bbox": [ - 119.78, - 344.03, - 374.34, - 354.4 - ], - "text": "The denominator of H[z] provides the characteristic polynomial", - "type": "text" - }, - { - "block_id": "p984-b9", - "global_id": 29821, - "bbox": [ - 270.78, - 361.31, - 296.91, - 375.4 - ], - "text": "γ 2 −5", - "type": "text" - }, - { - "block_id": "p984-b10", - "global_id": 29822, - "bbox": [ - 293.43, - 364.09, - 320.26, - 378.63 - ], - "text": "6γ + 1\n6", - "type": "text" - }, - { - "block_id": "p984-b11", - "global_id": 29823, - "bbox": [ - 101.84, - 386.84, - 395.16, - 397.22 - ], - "text": "Equivalently, the characteristic polynomial is the determinant of (zI −A).", - "type": "text" - }, - { - "block_id": "p984-b12", - "global_id": 29824, - "bbox": [ - 101.84, - 407.22, - 399.98, - 429.13 - ], - "text": ">>\nsyms gamma; char_poly = subs(det(z*eye(2)-A),z,gamma)\nchar_poly = gamma^2 - (5*gamma)/6 + 1/6", - "type": "text" - }, - { - "block_id": "p984-b13", - "global_id": 29825, - "bbox": [ - 101.84, - 438.54, - 490.39, - 460.76 - ], - "text": "Here, the subs command replaces the symbolic variable z with the desired symbolic variable\ngamma.", - "type": "text" - }, - { - "block_id": "p984-b14", - "global_id": 29826, - "bbox": [ - 101.84, - 462.46, - 490.41, - 496.33 - ], - "text": "The roots command does not accommodate symbolic expressions. Thus, the sym2poly\ncommand converts the symbolic expression into a polynomial coefficient vector suitable for the\nroots command.", - "type": "text" - }, - { - "block_id": "p984-b15", - "global_id": 29827, - "bbox": [ - 101.84, - 506.32, - 258.75, - 528.25 - ], - "text": ">>\nroots(sym2poly(char_poly))\nans = 0.5000", - "type": "text" - }, - { - "block_id": "p984-b16", - "global_id": 29828, - "bbox": [ - 154.13, - 530.24, - 185.52, - 540.2 - ], - "text": "0.3333", - "type": "text" - }, - { - "block_id": "p984-b17", - "global_id": 29829, - "bbox": [ - 119.78, - 549.21, - 406.37, - 559.59 - ], - "text": "Taking the inverse z-transform of H[z] yields the impulse response h[n].", - "type": "text" - }, - { - "block_id": "p984-b18", - "global_id": 29830, - "bbox": [ - 101.85, - 569.58, - 394.74, - 591.49 - ], - "text": ">>\nh = iztrans(H)\nh = 18*(1/2)^n - 12*(1/3)^n - 6*kroneckerDelta(n, 0)", - "type": "text" - }, - { - "block_id": "p984-b19", - "global_id": 29831, - "bbox": [ - 101.84, - 597.29, - 490.39, - 635.08 - ], - "text": "As suggested by the characteristic roots, the characteristic modes of the system are (1/2)n and\n(1/3)n. Notice that the symbolic math toolbox represents δ[n] as kroneckerDelta(n, 0).\nIn general, δ[n −a] is represented as kroneckerDelta(n-a, 0). This notation is frequently", - "type": "text" - } - ] - }, - { - "page_num": 985, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p985-b0", - "global_id": 29832, - "bbox": [ - 264.72, - 62.89, - 516.13, - 71.98 - ], - "text": "10.8\nMATLAB: Toolboxes and State-Space Analysis\n965", - "type": "text" - }, - { - "block_id": "p985-b1", - "global_id": 29833, - "bbox": [ - 127.59, - 85.4, - 516.14, - 119.69 - ], - "text": "encountered. Consider, for example, delaying the input x[n] = u[n] by 2, x[n −2] = u[n −2].\nIn the transform domain, this is equivalent to z−2X[z]. Taking the inverse z-transform of z−2X[z]\nyields", - "type": "text" - }, - { - "block_id": "p985-b2", - "global_id": 29834, - "bbox": [ - 127.6, - 130.07, - 446.62, - 151.98 - ], - "text": ">>\niztrans(z^(-2)*X)\nans = 1 - kroneckerDelta(n, 0) - kroneckerDelta(n - 1, 0)", - "type": "text" - }, - { - "block_id": "p985-b3", - "global_id": 29835, - "bbox": [ - 127.6, - 161.36, - 501.97, - 171.74 - ], - "text": "That is, MATLAB represents the delayed unit step u[n −2] as (−δ[n −1] −δ[n −0] + 1)u[n].", - "type": "text" - }, - { - "block_id": "p985-b4", - "global_id": 29836, - "bbox": [ - 145.53, - 173.74, - 478.95, - 183.7 - ], - "text": "The transfer function also permits convenient calculation of the zero-state response.", - "type": "text" - }, - { - "block_id": "p985-b5", - "global_id": 29837, - "bbox": [ - 127.6, - 194.07, - 331.57, - 215.99 - ], - "text": ">>\ny_zsr = iztrans(H*X)\ny_zsr = 6*(1/3)^n - 18*(1/2)^n + 12", - "type": "text" - }, - { - "block_id": "p985-b6", - "global_id": 29838, - "bbox": [ - 127.6, - 225.78, - 304.98, - 235.75 - ], - "text": "The result agrees with previous calculations.", - "type": "text" - }, - { - "block_id": "p985-b7", - "global_id": 29839, - "bbox": [ - 127.59, - 260.63, - 490.34, - 272.58 - ], - "text": "10.8-3 Controllability and Observability of Discrete-Time Systems", - "type": "text" - }, - { - "block_id": "p985-b8", - "global_id": 29840, - "bbox": [ - 127.59, - 278.7, - 516.11, - 312.58 - ], - "text": "In their controllability and observability, discrete-time systems are analogous to continuous-time\nsystems. For example, consider the LTID system described by the constant coefficient difference\nequation", - "type": "text" - }, - { - "block_id": "p985-b9", - "global_id": 29841, - "bbox": [ - 226.9, - 313.0, - 258.49, - 324.62 - ], - "text": "y[n] + 5", - "type": "text" - }, - { - "block_id": "p985-b10", - "global_id": 29842, - "bbox": [ - 255.01, - 313.0, - 416.81, - 327.54 - ], - "text": "6y[n −1] + 1\n6y[n −2] = x[n] + 1\n2x[n −1]", - "type": "text" - }, - { - "block_id": "p985-b11", - "global_id": 29843, - "bbox": [ - 127.6, - 333.81, - 516.13, - 368.45 - ], - "text": "Figure 10.15 illustrates the direct form II (DFII) realization of this system. The system input is\nx[n], the system output is y[n], and the outputs of the delay blocks are designated as state variables\nq1[n] and q2[n].", - "type": "text" - }, - { - "block_id": "p985-b12", - "global_id": 29844, - "bbox": [ - 145.52, - 369.68, - 414.88, - 379.64 - ], - "text": "The corresponding state and output equations (see Prob. 10.7-1) are", - "type": "text" - }, - { - "block_id": "p985-b13", - "global_id": 29845, - "bbox": [ - 171.23, - 397.15, - 216.27, - 407.53 - ], - "text": "Q[n + 1] =", - "type": "text" - }, - { - "block_id": "p985-b14", - "global_id": 29846, - "bbox": [ - 218.31, - 383.17, - 260.17, - 402.23 - ], - "text": "q1[n + 1]", - "type": "text" - }, - { - "block_id": "p985-b15", - "global_id": 29847, - "bbox": [ - 223.75, - 403.03, - 260.17, - 414.18 - ], - "text": "q2[n + 1]", - "type": "text" - }, - { - "block_id": "p985-b16", - "global_id": 29848, - "bbox": [ - 260.18, - 383.17, - 265.61, - 393.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p985-b17", - "global_id": 29849, - "bbox": [ - 267.66, - 397.15, - 275.43, - 407.11 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p985-b18", - "global_id": 29850, - "bbox": [ - 277.48, - 383.17, - 315.83, - 413.19 - ], - "text": "0\n1\n−1", - "type": "text" - }, - { - "block_id": "p985-b19", - "global_id": 29851, - "bbox": [ - 291.87, - 401.89, - 318.97, - 416.42 - ], - "text": "6\n−5", - "type": "text" - }, - { - "block_id": "p985-b20", - "global_id": 29852, - "bbox": [ - 315.49, - 409.45, - 318.97, - 416.42 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p985-b21", - "global_id": 29853, - "bbox": [ - 320.17, - 383.17, - 352.72, - 402.23 - ], - "text": "! q1[n]", - "type": "text" - }, - { - "block_id": "p985-b22", - "global_id": 29854, - "bbox": [ - 332.14, - 403.03, - 352.72, - 414.18 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p985-b23", - "global_id": 29855, - "bbox": [ - 352.72, - 383.17, - 358.15, - 393.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p985-b24", - "global_id": 29856, - "bbox": [ - 359.69, - 397.15, - 367.47, - 407.11 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p985-b25", - "global_id": 29857, - "bbox": [ - 369.01, - 383.17, - 379.43, - 401.45 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p985-b26", - "global_id": 29858, - "bbox": [ - 374.44, - 403.44, - 379.43, - 413.41 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p985-b27", - "global_id": 29859, - "bbox": [ - 379.43, - 383.17, - 384.86, - 393.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p985-b28", - "global_id": 29860, - "bbox": [ - 384.86, - 397.15, - 472.48, - 407.45 - ], - "text": "x[n] = AQ[n] + Bx[n]", - "type": "text" - }, - { - "block_id": "p985-b29", - "global_id": 29861, - "bbox": [ - 127.6, - 425.11, - 141.97, - 435.08 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p985-b30", - "global_id": 29862, - "bbox": [ - 216.84, - 440.42, - 242.7, - 450.7 - ], - "text": "y[n] =", - "type": "text" - }, - { - "block_id": "p985-b32", - "global_id": 29863, - "bbox": [ - 248.67, - 439.17, - 261.12, - 450.48 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p985-b33", - "global_id": 29864, - "bbox": [ - 257.63, - 439.17, - 284.73, - 453.71 - ], - "text": "6\n−1", - "type": "text" - }, - { - "block_id": "p985-b34", - "global_id": 29865, - "bbox": [ - 281.25, - 426.44, - 316.98, - 453.71 - ], - "text": "3\n\n q1[n]", - "type": "text" - }, - { - "block_id": "p985-b35", - "global_id": 29866, - "bbox": [ - 296.39, - 446.3, - 316.98, - 457.45 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p985-b36", - "global_id": 29867, - "bbox": [ - 316.97, - 426.44, - 322.4, - 436.4 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p985-b37", - "global_id": 29868, - "bbox": [ - 323.95, - 440.42, - 426.88, - 450.8 - ], - "text": "+ 1x[n] = CQ[n] + Dx[n]", - "type": "text" - }, - { - "block_id": "p985-b38", - "global_id": 29869, - "bbox": [ - 127.59, - 465.23, - 478.62, - 475.27 - ], - "text": "To describe this system in MATLAB, the state matrices A, B, C, and D are first defined.", - "type": "text" - }, - { - "block_id": "p985-b39", - "global_id": 29870, - "bbox": [ - 127.59, - 485.64, - 441.34, - 495.6 - ], - "text": ">>\nA = [0 1;-1/6 -5/6]; B = [0; 1]; C = [-1/6 -1/3]; D = 1;", - "type": "text" - }, - { - "block_id": "p985-b40", - "global_id": 29871, - "bbox": [ - 129.8, - 518.15, - 316.69, - 526.23 - ], - "text": "x[n]\ny[n]", - "type": "text" - }, - { - "block_id": "p985-b41", - "global_id": 29872, - "bbox": [ - 219.02, - 581.53, - 228.14, - 590.93 - ], - "text": "z–1", - "type": "text" - }, - { - "block_id": "p985-b42", - "global_id": 29873, - "bbox": [ - 219.02, - 538.33, - 228.14, - 547.73 - ], - "text": "z–1", - "type": "text" - }, - { - "block_id": "p985-b45", - "global_id": 29874, - "bbox": [ - 239.78, - 524.85, - 243.78, - 532.85 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p985-b46", - "global_id": 29875, - "bbox": [ - 226.28, - 594.97, - 242.61, - 604.57 - ], - "text": "q1[n]", - "type": "text" - }, - { - "block_id": "p985-b47", - "global_id": 29876, - "bbox": [ - 226.28, - 551.77, - 242.61, - 561.37 - ], - "text": "q2[n]", - "type": "text" - }, - { - "block_id": "p985-b48", - "global_id": 29877, - "bbox": [ - 197.44, - 568.7, - 245.44, - 582.3 - ], - "text": "5\n6\n–\n1\n2", - "type": "text" - }, - { - "block_id": "p985-b49", - "global_id": 29878, - "bbox": [ - 197.44, - 611.87, - 206.31, - 625.47 - ], - "text": "1\n6\n–", - "type": "text" - }, - { - "block_id": "p985-b50", - "global_id": 29879, - "bbox": [ - 327.55, - 592.88, - 512.15, - 626.76 - ], - "text": "Figure 10.15 Direct form II realization of\ny[n] + (5/6)y[n −1] + (1/6)y[n −2] = x[n] +\n(1/2)x[n −1].", - "type": "text" - } - ] - }, - { - "page_num": 986, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p986-b0", - "global_id": 29880, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "966\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p986-b1", - "global_id": 29881, - "bbox": [ - 101.84, - 85.87, - 490.39, - 107.87 - ], - "text": "To assess the controllability and observability of this system, the state matrix A needs to be\ndiagonalized.† As shown in Eq. (10.55), this requires a transformation matrix P such that", - "type": "text" - }, - { - "block_id": "p986-b2", - "global_id": 29882, - "bbox": [ - 276.98, - 119.32, - 490.38, - 129.7 - ], - "text": "PA = \nP\n(10.75)", - "type": "text" - }, - { - "block_id": "p986-b3", - "global_id": 29883, - "bbox": [ - 101.84, - 141.16, - 490.38, - 163.49 - ], - "text": "where \n is a diagonal matrix containing the unique eigenvalues of A. Recall, the transformation\nmatrix P is not unique.", - "type": "text" - }, - { - "block_id": "p986-b4", - "global_id": 29884, - "bbox": [ - 101.84, - 165.4, - 490.4, - 187.41 - ], - "text": "To determine a matrix P, it is helpful to review the eigenvalue problem. Mathematically, an\neigendecomposition of A is expressed as", - "type": "text" - }, - { - "block_id": "p986-b5", - "global_id": 29885, - "bbox": [ - 276.23, - 198.87, - 316.0, - 209.16 - ], - "text": "AV = V", - "type": "text" - }, - { - "block_id": "p986-b6", - "global_id": 29886, - "bbox": [ - 101.84, - 220.7, - 490.37, - 243.03 - ], - "text": "where V is a matrix of eigenvectors and \n is a diagonal matrix of eigenvalues. Pre- and\npost-multiplying both sides of this equation by V−1 yields", - "type": "text" - }, - { - "block_id": "p986-b7", - "global_id": 29887, - "bbox": [ - 243.0, - 250.26, - 348.74, - 264.79 - ], - "text": "V−1AVV−1 = V−1V\nV−1", - "type": "text" - }, - { - "block_id": "p986-b8", - "global_id": 29888, - "bbox": [ - 101.84, - 276.74, - 184.94, - 286.7 - ], - "text": "Simplification yields", - "type": "text" - }, - { - "block_id": "p986-b9", - "global_id": 29889, - "bbox": [ - 266.08, - 286.45, - 490.38, - 298.66 - ], - "text": "V−1A = \nV−1\n(10.76)", - "type": "text" - }, - { - "block_id": "p986-b10", - "global_id": 29890, - "bbox": [ - 101.85, - 307.5, - 490.39, - 329.5 - ], - "text": "Comparing Eqs. (10.75) and (10.76), we see that a suitable transformation matrix P is given by an\ninverse eigenvector matrix V−1.", - "type": "text" - }, - { - "block_id": "p986-b11", - "global_id": 29891, - "bbox": [ - 101.84, - 331.41, - 490.38, - 353.41 - ], - "text": "The eig command is used to verify that A has the required distinct eigenvalues as well as\ncompute the needed eigenvector matrix V.", - "type": "text" - }, - { - "block_id": "p986-b12", - "global_id": 29892, - "bbox": [ - 101.84, - 371.59, - 222.14, - 393.51 - ], - "text": ">>\n[V,Lambda] = eig(A)\nV =", - "type": "text" - }, - { - "block_id": "p986-b13", - "global_id": 29893, - "bbox": [ - 122.76, - 395.5, - 227.36, - 429.37 - ], - "text": "0.9487\n-0.8944\n-0.3162\n0.4472\nLambda =", - "type": "text" - }, - { - "block_id": "p986-b14", - "global_id": 29894, - "bbox": [ - 138.45, - 431.37, - 227.35, - 453.29 - ], - "text": "-0.3333\n0\n0\n-0.5000", - "type": "text" - }, - { - "block_id": "p986-b15", - "global_id": 29895, - "bbox": [ - 101.84, - 462.83, - 468.18, - 473.17 - ], - "text": "Since the diagonal elements of Lambda are all unique, a transformation matrix P is given by", - "type": "text" - }, - { - "block_id": "p986-b16", - "global_id": 29896, - "bbox": [ - 101.84, - 486.08, - 180.29, - 496.05 - ], - "text": ">>\nP = inv(V);", - "type": "text" - }, - { - "block_id": "p986-b17", - "global_id": 29897, - "bbox": [ - 101.85, - 501.45, - 490.38, - 527.59 - ], - "text": "The transformed state matrices ˆA = PAP−1, ˆB = PB, and ˆC = CP−1 are easily computed by\nusing transformation matrix P. Notice that matrix D is unaffected by state variable transformations.", - "type": "text" - }, - { - "block_id": "p986-b18", - "global_id": 29898, - "bbox": [ - 101.85, - 537.81, - 363.36, - 559.72 - ], - "text": ">>\nAhat = P*A*inv(P), Bhat = P*B, Chat = C*inv(P)\nAhat =", - "type": "text" - }, - { - "block_id": "p986-b19", - "global_id": 29899, - "bbox": [ - 138.45, - 561.71, - 227.36, - 583.63 - ], - "text": "-0.3333\n-0.0000\n0.0000\n-0.5000", - "type": "text" - }, - { - "block_id": "p986-b20", - "global_id": 29900, - "bbox": [ - 101.84, - 599.27, - 490.4, - 633.67 - ], - "text": "† This approach requires that the state matrix A have unique eigenvalues. Systems with repeated roots require\nthat state matrix A be transformed into a modified diagonal form, also called the Jordan form. The MATLAB\nfunction jordan is used in these cases.", - "type": "text" - } - ] - }, - { - "page_num": 987, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p987-b0", - "global_id": 29901, - "bbox": [ - 264.72, - 62.89, - 516.13, - 71.98 - ], - "text": "10.8\nMATLAB: Toolboxes and State-Space Analysis\n967", - "type": "text" - }, - { - "block_id": "p987-b1", - "global_id": 29902, - "bbox": [ - 148.51, - 85.41, - 179.89, - 95.37 - ], - "text": "Bhat =", - "type": "text" - }, - { - "block_id": "p987-b2", - "global_id": 29903, - "bbox": [ - 148.51, - 97.36, - 200.81, - 131.24 - ], - "text": "6.3246\n6.7082\nChat =", - "type": "text" - }, - { - "block_id": "p987-b3", - "global_id": 29904, - "bbox": [ - 164.2, - 133.23, - 253.11, - 143.19 - ], - "text": "-0.0527\n-0.0000", - "type": "text" - }, - { - "block_id": "p987-b4", - "global_id": 29905, - "bbox": [ - 127.59, - 149.93, - 516.13, - 210.47 - ], - "text": "The proper operation of P is verified by the correct diagonalization of A, ˆA = \n. Since no row\nof ˆB is zero, the system is controllable. Since, however, at least one column of ˆC is zero, the\nsystem is not observable. These characteristics are no coincidence. The DFII realization, which\nis more descriptively called the controller canonical form, is always controllable but not always\nobservable.", - "type": "text" - }, - { - "block_id": "p987-b5", - "global_id": 29906, - "bbox": [ - 127.6, - 212.46, - 516.13, - 247.1 - ], - "text": "As a second example, consider the same system realized using the transposed direct form II\nstructure (TDFII), as shown in Fig. 10.16. The system input is x[n], the system output is y[n], and\nthe outputs of the delay blocks are designated as state variables v1[n] and v2[n].", - "type": "text" - }, - { - "block_id": "p987-b6", - "global_id": 29907, - "bbox": [ - 145.52, - 248.32, - 414.88, - 258.28 - ], - "text": "The corresponding state and output equations (see Prob. 10.7-2) are", - "type": "text" - }, - { - "block_id": "p987-b7", - "global_id": 29908, - "bbox": [ - 170.78, - 278.08, - 215.26, - 288.46 - ], - "text": "V[n + 1] =", - "type": "text" - }, - { - "block_id": "p987-b8", - "global_id": 29909, - "bbox": [ - 217.31, - 264.1, - 258.6, - 283.16 - ], - "text": "v1[n + 1]", - "type": "text" - }, - { - "block_id": "p987-b9", - "global_id": 29910, - "bbox": [ - 222.73, - 283.96, - 258.6, - 295.11 - ], - "text": "v2[n + 1]", - "type": "text" - }, - { - "block_id": "p987-b10", - "global_id": 29911, - "bbox": [ - 258.61, - 264.1, - 264.04, - 274.06 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p987-b11", - "global_id": 29912, - "bbox": [ - 266.08, - 278.08, - 273.86, - 288.05 - ], - "text": "=", - "type": "text" - }, - { - "block_id": "p987-b13", - "global_id": 29913, - "bbox": [ - 282.1, - 268.66, - 309.49, - 280.39 - ], - "text": "0\n−1", - "type": "text" - }, - { - "block_id": "p987-b14", - "global_id": 29914, - "bbox": [ - 282.1, - 276.23, - 309.49, - 296.73 - ], - "text": "6\n1\n−5", - "type": "text" - }, - { - "block_id": "p987-b15", - "global_id": 29915, - "bbox": [ - 306.0, - 292.57, - 309.49, - 299.55 - ], - "text": "6", - "type": "text" - }, - { - "block_id": "p987-b16", - "global_id": 29916, - "bbox": [ - 310.69, - 261.11, - 343.44, - 283.16 - ], - "text": "v1[n]", - "type": "text" - }, - { - "block_id": "p987-b17", - "global_id": 29917, - "bbox": [ - 323.41, - 283.96, - 343.44, - 295.11 - ], - "text": "v2[n]", - "type": "text" - }, - { - "block_id": "p987-b18", - "global_id": 29918, - "bbox": [ - 343.43, - 264.1, - 348.86, - 274.06 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p987-b19", - "global_id": 29919, - "bbox": [ - 350.41, - 278.08, - 358.19, - 288.05 - ], - "text": "+", - "type": "text" - }, - { - "block_id": "p987-b21", - "global_id": 29920, - "bbox": [ - 365.92, - 268.66, - 378.37, - 279.97 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p987-b22", - "global_id": 29921, - "bbox": [ - 365.92, - 276.23, - 378.37, - 296.32 - ], - "text": "6\n−1", - "type": "text" - }, - { - "block_id": "p987-b23", - "global_id": 29922, - "bbox": [ - 374.89, - 292.57, - 378.37, - 299.55 - ], - "text": "3", - "type": "text" - }, - { - "block_id": "p987-b25", - "global_id": 29923, - "bbox": [ - 386.86, - 278.08, - 472.94, - 288.38 - ], - "text": "x[n] = AV[n] + Bx[n]", - "type": "text" - }, - { - "block_id": "p987-b26", - "global_id": 29924, - "bbox": [ - 127.6, - 308.32, - 141.97, - 318.29 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p987-b27", - "global_id": 29925, - "bbox": [ - 226.68, - 309.47, - 307.7, - 333.73 - ], - "text": "y[n] = [0\n1]\n v1[n]", - "type": "text" - }, - { - "block_id": "p987-b28", - "global_id": 29926, - "bbox": [ - 287.68, - 329.33, - 307.7, - 340.48 - ], - "text": "v2[n]", - "type": "text" - }, - { - "block_id": "p987-b29", - "global_id": 29927, - "bbox": [ - 307.7, - 309.47, - 313.13, - 319.43 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p987-b30", - "global_id": 29928, - "bbox": [ - 314.68, - 323.45, - 417.04, - 333.83 - ], - "text": "+ 1x[n] = CV[n] + Dx[n]", - "type": "text" - }, - { - "block_id": "p987-b31", - "global_id": 29929, - "bbox": [ - 127.6, - 347.95, - 460.28, - 357.99 - ], - "text": "To describe this system in MATLAB, the state matrices A, B, C, and D are defined.", - "type": "text" - }, - { - "block_id": "p987-b32", - "global_id": 29930, - "bbox": [ - 127.6, - 368.06, - 441.36, - 378.02 - ], - "text": ">>\nA = [0 -1/6;1 -5/6]; B = [-1/6; -1/3]; C = [0 1]; D = 1;", - "type": "text" - }, - { - "block_id": "p987-b33", - "global_id": 29931, - "bbox": [ - 145.52, - 387.42, - 364.4, - 397.47 - ], - "text": "To diagonalize A, a transformation matrix P is created.", - "type": "text" - }, - { - "block_id": "p987-b34", - "global_id": 29932, - "bbox": [ - 127.6, - 407.54, - 247.89, - 429.45 - ], - "text": ">>\n[V,Lambda] = eig(A)\nV =", - "type": "text" - }, - { - "block_id": "p987-b35", - "global_id": 29933, - "bbox": [ - 148.51, - 431.44, - 253.11, - 465.32 - ], - "text": "0.4472\n0.3162\n0.8944\n0.9487\nLambda =", - "type": "text" - }, - { - "block_id": "p987-b36", - "global_id": 29934, - "bbox": [ - 164.2, - 467.31, - 253.1, - 489.23 - ], - "text": "-0.3333\n0\n0\n-0.5000", - "type": "text" - }, - { - "block_id": "p987-b37", - "global_id": 29935, - "bbox": [ - 129.8, - 509.97, - 268.31, - 525.21 - ], - "text": "x[n]\ny[n]\n1", - "type": "text" - }, - { - "block_id": "p987-b38", - "global_id": 29936, - "bbox": [ - 191.75, - 541.03, - 200.86, - 550.43 - ], - "text": "z–1", - "type": "text" - }, - { - "block_id": "p987-b39", - "global_id": 29937, - "bbox": [ - 191.75, - 605.83, - 200.86, - 615.23 - ], - "text": "z–1", - "type": "text" - }, - { - "block_id": "p987-b42", - "global_id": 29938, - "bbox": [ - 199.7, - 592.92, - 215.58, - 602.53 - ], - "text": "v1[n]", - "type": "text" - }, - { - "block_id": "p987-b43", - "global_id": 29939, - "bbox": [ - 199.01, - 528.02, - 214.89, - 537.63 - ], - "text": "v2[n]", - "type": "text" - }, - { - "block_id": "p987-b44", - "global_id": 29940, - "bbox": [ - 166.26, - 561.79, - 227.75, - 575.48 - ], - "text": "5\n6\n–\n1\n2", - "type": "text" - }, - { - "block_id": "p987-b45", - "global_id": 29941, - "bbox": [ - 218.87, - 616.77, - 512.14, - 638.69 - ], - "text": "1\n6\n–\nFigure 10.16 Transposed direct form II realization of\ny[n]+(5/6)y[n−1]+(1/6)y[n−2] = x[n]+(1/2)x[n−1].", - "type": "text" - } - ] - }, - { - "page_num": 988, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p988-b0", - "global_id": 29942, - "bbox": [ - 60.0, - 62.89, - 269.83, - 71.98 - ], - "text": "968\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p988-b1", - "global_id": 29943, - "bbox": [ - 101.84, - 85.82, - 490.39, - 131.64 - ], - "text": "The characteristic modes of a system do not depend on implementation, so the eigenvalues of\nthe DFII and TDFII realizations are the same. However, the eigenvectors of the two realizations\nare quite different. Since the transformation matrix P depends on the eigenvectors, different\nrealizations can possess different observability and controllability characteristics.", - "type": "text" - }, - { - "block_id": "p988-b2", - "global_id": 29944, - "bbox": [ - 101.85, - 130.9, - 490.38, - 155.56 - ], - "text": "Using transformation matrix P, the transformed state matrices ˆA = PAP−1, ˆB = PB, and\nˆC = CP−1 are computed.", - "type": "text" - }, - { - "block_id": "p988-b3", - "global_id": 29945, - "bbox": [ - 101.84, - 165.24, - 363.36, - 199.12 - ], - "text": ">>\nP = inv(V);\n>>\nAhat = P*A*inv(P), Bhat = P*B, Chat = C*inv(P)\nAhat =", - "type": "text" - }, - { - "block_id": "p988-b4", - "global_id": 29946, - "bbox": [ - 122.76, - 201.11, - 227.36, - 234.98 - ], - "text": "-0.3333\n0\n0.0000\n-0.5000\nBhat =", - "type": "text" - }, - { - "block_id": "p988-b5", - "global_id": 29947, - "bbox": [ - 122.76, - 236.97, - 175.06, - 270.85 - ], - "text": "-0.3727\n-0.0000\nChat =", - "type": "text" - }, - { - "block_id": "p988-b6", - "global_id": 29948, - "bbox": [ - 143.68, - 272.84, - 227.36, - 282.8 - ], - "text": "0.8944\n0.9487", - "type": "text" - }, - { - "block_id": "p988-b7", - "global_id": 29949, - "bbox": [ - 101.84, - 289.16, - 490.39, - 361.65 - ], - "text": "Again, the proper operation of P is verified by the correct diagonalization of A, ˆA = \n. Since\nno column of ˆC is zero, the system is observable. However, at least one row of ˆB is zero, and\ntherefore the system is not controllable. The TDFII realization, which is more descriptively called\nthe observer canonical form, is always observable but not always controllable. It is interesting\nto note that the properties of controllability and observability are influenced by the particular\nrealization of a system.", - "type": "text" - }, - { - "block_id": "p988-b8", - "global_id": 29950, - "bbox": [ - 101.84, - 385.83, - 413.47, - 397.79 - ], - "text": "10.8-4 Matrix Exponentiation and the Matrix Exponential", - "type": "text" - }, - { - "block_id": "p988-b9", - "global_id": 29951, - "bbox": [ - 101.84, - 403.92, - 490.42, - 449.75 - ], - "text": "Matrix exponentiation is important to many problems, including the solution of discrete-time\nstate-space equations. Equation (10.64), for example, shows that the state response requires matrix\nexponentiation, An. For a square A and specific n, MATLAB happily returns An by using the ^\noperator. From the system in Ex. 10.13 and n = 3, we have", - "type": "text" - }, - { - "block_id": "p988-b10", - "global_id": 29952, - "bbox": [ - 101.85, - 459.43, - 279.68, - 481.36 - ], - "text": ">>\nA = [0 1;-1/6 5/6]; n = 3; A^n\nans =", - "type": "text" - }, - { - "block_id": "p988-b11", - "global_id": 29953, - "bbox": [ - 138.45, - 483.35, - 227.36, - 505.26 - ], - "text": "-0.1389\n0.5278\n-0.0880\n0.3009", - "type": "text" - }, - { - "block_id": "p988-b12", - "global_id": 29954, - "bbox": [ - 101.85, - 514.37, - 300.94, - 524.62 - ], - "text": "The same result is also obtained by typing A*A*A.", - "type": "text" - }, - { - "block_id": "p988-b13", - "global_id": 29955, - "bbox": [ - 101.85, - 521.69, - 490.38, - 548.24 - ], - "text": "Often, it is useful to solve An symbolically. Noting An = Z−1[(I −z−1A)−1], the symbolic\ntoolbox can produce a symbolic expression for An.", - "type": "text" - }, - { - "block_id": "p988-b14", - "global_id": 29956, - "bbox": [ - 101.84, - 557.93, - 405.2, - 603.76 - ], - "text": ">>\nsyms z n; An = simplify(iztrans(inv(eye(2)-z^(-1)*A)))\nAn =\n[ 3*(1/3)^n - 2*(1/2)^n, 6*(1/2)^n - 6*(1/3)^n]\n[\n1/3^n - 1/2^n,\n3/2^n - 2/3^n]", - "type": "text" - }, - { - "block_id": "p988-b15", - "global_id": 29957, - "bbox": [ - 101.84, - 612.45, - 490.39, - 635.08 - ], - "text": "Notice that this result is identical to Eq. (10.68), derived earlier. Substituting the case n = 3 into\nAn provides a result that is identical to the one elicited by the previous A^n command.", - "type": "text" - } - ] - }, - { - "page_num": 989, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p989-b0", - "global_id": 29958, - "bbox": [ - 409.68, - 62.89, - 516.14, - 71.98 - ], - "text": "10.9\nSummary\n969", - "type": "text" - }, - { - "block_id": "p989-b1", - "global_id": 29959, - "bbox": [ - 127.59, - 86.24, - 253.12, - 108.15 - ], - "text": ">>\ndouble(subs(An,n,3))\nans =", - "type": "text" - }, - { - "block_id": "p989-b2", - "global_id": 29960, - "bbox": [ - 164.2, - 110.15, - 253.11, - 132.07 - ], - "text": "-0.1389\n0.5278\n-0.0880\n0.3009", - "type": "text" - }, - { - "block_id": "p989-b3", - "global_id": 29961, - "bbox": [ - 127.59, - 143.69, - 516.14, - 169.22 - ], - "text": "For continuous-time systems, the matrix exponential eAt is commonly encountered. The expm\ncommand can compute the matrix exponential symbolically. Using the system from Ex. 10.8 yields", - "type": "text" - }, - { - "block_id": "p989-b4", - "global_id": 29962, - "bbox": [ - 127.59, - 185.03, - 504.17, - 230.86 - ], - "text": ">>\nsyms t; A = [-12 2/3;-36 -1]; eAt = simplify(expm(A*t))\neAt =\n[\n-(exp(-9*t)*(3*exp(5*t) - 8))/5, (2*exp(-9*t)*(exp(5*t) - 1))/15]\n[ -(36*exp(-9*t)*(exp(5*t) - 1))/5,\n(exp(-9*t)*(8*exp(5*t) - 3))/5]", - "type": "text" - }, - { - "block_id": "p989-b5", - "global_id": 29963, - "bbox": [ - 127.6, - 246.1, - 516.14, - 268.3 - ], - "text": "This result is identical to the result computed in Ex. 10.8. Similar to the discrete-time case, an\nidentical result is obtained by typing syms s; simplify(ilaplace(inv(s*eye(2)-A))).", - "type": "text" - }, - { - "block_id": "p989-b6", - "global_id": 29964, - "bbox": [ - 127.6, - 269.91, - 516.15, - 291.92 - ], - "text": "For a specific t, the matrix exponential is also easy to compute, either through substitution or\ndirect computation. Consider the case t = 3.", - "type": "text" - }, - { - "block_id": "p989-b7", - "global_id": 29965, - "bbox": [ - 127.6, - 307.74, - 258.35, - 329.65 - ], - "text": ">>\ndouble(subs(eAt,t,3))\nans = 1.0e-004 *", - "type": "text" - }, - { - "block_id": "p989-b8", - "global_id": 29966, - "bbox": [ - 164.2, - 331.65, - 253.11, - 353.57 - ], - "text": "-0.0369\n0.0082\n-0.4424\n0.0983", - "type": "text" - }, - { - "block_id": "p989-b9", - "global_id": 29967, - "bbox": [ - 127.6, - 368.8, - 337.63, - 379.06 - ], - "text": "The command expm(A*3) produces the same result.", - "type": "text" - }, - { - "block_id": "p989-b10", - "global_id": 29968, - "bbox": [ - 127.94, - 413.61, - 227.89, - 427.55 - ], - "text": "10.9 SUMMARY", - "type": "text" - }, - { - "block_id": "p989-b11", - "global_id": 29969, - "bbox": [ - 127.59, - 433.44, - 516.15, - 551.1 - ], - "text": "An Nth-order system can be described in terms of N key variables—the state variables of the\nsystem. The state variables are not unique; rather, they can be selected in a variety of ways.\nEvery possible system output can be expressed as a linear combination of the state variables and\nthe inputs. Therefore, the state variables describe the entire system, not merely the relationship\nbetween certain input(s) and output(s). For this reason, the state variable description is an internal\ndescription of the system. Such a description is therefore the most general system description,\nand it contains the information of the external descriptions, such as the impulse response and the\ntransfer function. The state variable description can also be extended to time-varying parameter\nsystems and nonlinear systems. An external description of a system may not characterize the\nsystem completely.", - "type": "text" - }, - { - "block_id": "p989-b12", - "global_id": 29970, - "bbox": [ - 127.59, - 553.1, - 516.16, - 634.79 - ], - "text": "The state equations of a system can be written directly from knowledge of the system\nstructure, from the system equations, or from the block diagram representation of the system. State\nequations consist of a set of N first-order differential equations and can be solved by time-domain\nor frequency-domain (transform) methods. Suitable procedures exist to transform one given set of\nstate variables into another. Because a set of state variables is not unique, we can have an infinite\nvariety of state-space descriptions of the same system. The use of an appropriate transformation\nallows us to see clearly which of the system states are controllable and which are observable.", - "type": "text" - } - ] - }, - { - "page_num": 990, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p990-b0", - "global_id": 29971, - "bbox": [ - 60.0, - 60.36, - 269.83, - 69.45 - ], - "text": "970\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p990-b1", - "global_id": 29972, - "bbox": [ - 102.11, - 86.41, - 189.29, - 97.37 - ], - "text": "REFERENCES", - "type": "text" - }, - { - "block_id": "p990-b2", - "global_id": 29973, - "bbox": [ - 101.84, - 104.44, - 396.17, - 113.49 - ], - "text": "1.\nKailath, Thomas. Linear Systems. Prentice-Hall, Englewood Cliffs, NJ, 1980.", - "type": "text" - }, - { - "block_id": "p990-b3", - "global_id": 29974, - "bbox": [ - 101.84, - 118.38, - 409.53, - 127.43 - ], - "text": "2.\nZadeh, L., and C. Desoer. Linear System Theory. McGraw-Hill, New York, 1963.", - "type": "text" - }, - { - "block_id": "p990-b4", - "global_id": 29975, - "bbox": [ - 80.93, - 160.07, - 191.52, - 177.0 - ], - "text": "PROBLEMS", - "type": "text" - }, - { - "block_id": "p990-b5", - "global_id": 29976, - "bbox": [ - 62.58, - 187.67, - 263.23, - 240.54 - ], - "text": "10.1-1\nConvert each of the following second-order\ndifferential\nequations\ninto\na\nset\nof\ntwo\nfirst-order differential equations (state equa-\ntions). State which of the sets represent non-\nlinear equations.", - "type": "text" - }, - { - "block_id": "p990-b6", - "global_id": 29977, - "bbox": [ - 95.7, - 242.16, - 181.6, - 262.46 - ], - "text": "(a) ¨y + 10˙y + 2y = x\n(b) ¨y + 2ey˙y + logy = x", - "type": "text" - }, - { - "block_id": "p990-b7", - "global_id": 29978, - "bbox": [ - 96.2, - 264.07, - 196.46, - 274.15 - ], - "text": "(c) ¨y + φ1(y)˙y + φ2(y)y = x", - "type": "text" - }, - { - "block_id": "p990-b8", - "global_id": 29979, - "bbox": [ - 62.58, - 278.83, - 263.23, - 298.84 - ], - "text": "10.2-1\nWrite the state equations for the RLC network\nin Fig. P10.2-1.", - "type": "text" - }, - { - "block_id": "p990-b9", - "global_id": 29980, - "bbox": [ - 64.07, - 385.64, - 266.45, - 393.94 - ], - "text": "x\n3", - "type": "text" - }, - { - "block_id": "p990-b10", - "global_id": 29981, - "bbox": [ - 182.55, - 335.54, - 194.77, - 343.84 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p990-b11", - "global_id": 29982, - "bbox": [ - 80.13, - 372.91, - 202.92, - 396.72 - ], - "text": "1 \t\n1 H\n+\n_", - "type": "text" - }, - { - "block_id": "p990-b12", - "global_id": 29983, - "bbox": [ - 130.26, - 317.56, - 140.76, - 331.16 - ], - "text": "1\n2 F", - "type": "text" - }, - { - "block_id": "p990-b13", - "global_id": 29984, - "bbox": [ - 62.82, - 432.83, - 118.94, - 441.79 - ], - "text": "Figure P10.2-1", - "type": "text" - }, - { - "block_id": "p990-b14", - "global_id": 29985, - "bbox": [ - 62.58, - 460.93, - 263.23, - 480.94 - ], - "text": "10.2-2\nWrite the state and output equations for the\nnetwork in Fig. P10.2-2.", - "type": "text" - }, - { - "block_id": "p990-b15", - "global_id": 29986, - "bbox": [ - 62.58, - 486.36, - 263.23, - 506.36 - ], - "text": "10.2-3\nWrite the state and output equations for the\nnetwork in Fig. P10.2-3.", - "type": "text" - }, - { - "block_id": "p990-b16", - "global_id": 29987, - "bbox": [ - 310.07, - 244.18, - 314.59, - 259.75 - ], - "text": "+\n_", - "type": "text" - }, - { - "block_id": "p990-b17", - "global_id": 29988, - "bbox": [ - 431.03, - 265.69, - 435.54, - 281.26 - ], - "text": "+\n_", - "type": "text" - }, - { - "block_id": "p990-b18", - "global_id": 29989, - "bbox": [ - 292.18, - 247.39, - 298.73, - 257.0 - ], - "text": "x1", - "type": "text" - }, - { - "block_id": "p990-b19", - "global_id": 29990, - "bbox": [ - 447.84, - 268.9, - 454.4, - 278.51 - ], - "text": "x2", - "type": "text" - }, - { - "block_id": "p990-b20", - "global_id": 29991, - "bbox": [ - 316.25, - 189.73, - 429.48, - 198.39 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p990-b21", - "global_id": 29992, - "bbox": [ - 383.98, - 247.78, - 396.43, - 261.6 - ], - "text": "1\n5", - "type": "text" - }, - { - "block_id": "p990-b22", - "global_id": 29993, - "bbox": [ - 331.26, - 199.9, - 410.04, - 208.46 - ], - "text": "1 F\n1", - "type": "text" - }, - { - "block_id": "p990-b23", - "global_id": 29994, - "bbox": [ - 442.96, - 230.58, - 455.11, - 244.5 - ], - "text": "1\n2 H", - "type": "text" - }, - { - "block_id": "p990-b24", - "global_id": 29995, - "bbox": [ - 289.97, - 306.38, - 346.08, - 315.35 - ], - "text": "Figure P10.2-3", - "type": "text" - }, - { - "block_id": "p990-b25", - "global_id": 29996, - "bbox": [ - 289.73, - 331.11, - 490.38, - 351.1 - ], - "text": "10.2-4\nWrite the state and output equations for the\nelectrical network in Fig. P10.2-4.", - "type": "text" - }, - { - "block_id": "p990-b26", - "global_id": 29997, - "bbox": [ - 342.69, - 420.23, - 347.2, - 435.81 - ], - "text": "+\n_", - "type": "text" - }, - { - "block_id": "p990-b27", - "global_id": 29998, - "bbox": [ - 402.22, - 398.37, - 408.77, - 407.98 - ], - "text": "y1", - "type": "text" - }, - { - "block_id": "p990-b28", - "global_id": 29999, - "bbox": [ - 327.05, - 423.2, - 330.6, - 431.2 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p990-b31", - "global_id": 30000, - "bbox": [ - 463.17, - 398.37, - 469.73, - 407.98 - ], - "text": "y2", - "type": "text" - }, - { - "block_id": "p990-b34", - "global_id": 30001, - "bbox": [ - 375.4, - 437.36, - 440.67, - 450.8 - ], - "text": "1 H\n1 F", - "type": "text" - }, - { - "block_id": "p990-b35", - "global_id": 30002, - "bbox": [ - 362.46, - 366.11, - 374.69, - 374.41 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p990-b36", - "global_id": 30003, - "bbox": [ - 367.59, - 399.44, - 440.09, - 408.13 - ], - "text": "1 \t\n1", - "type": "text" - }, - { - "block_id": "p990-b37", - "global_id": 30004, - "bbox": [ - 322.85, - 483.71, - 378.96, - 492.68 - ], - "text": "Figure P10.2-4", - "type": "text" - }, - { - "block_id": "p990-b38", - "global_id": 30005, - "bbox": [ - 385.75, - 588.22, - 389.3, - 596.22 - ], - "text": "y", - "type": "text" - }, - { - "block_id": "p990-b43", - "global_id": 30006, - "bbox": [ - 213.32, - 583.46, - 220.32, - 593.07 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p990-b44", - "global_id": 30007, - "bbox": [ - 325.34, - 568.52, - 332.34, - 578.12 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p990-b45", - "global_id": 30008, - "bbox": [ - 145.9, - 588.57, - 149.45, - 596.57 - ], - "text": "x", - "type": "text" - }, - { - "block_id": "p990-b46", - "global_id": 30009, - "bbox": [ - 161.61, - 582.13, - 166.12, - 600.32 - ], - "text": "+\n_", - "type": "text" - }, - { - "block_id": "p990-b47", - "global_id": 30010, - "bbox": [ - 297.26, - 543.55, - 309.48, - 551.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p990-b48", - "global_id": 30011, - "bbox": [ - 243.84, - 582.85, - 358.55, - 596.45 - ], - "text": "1 H\n1\n2 F", - "type": "text" - }, - { - "block_id": "p990-b49", - "global_id": 30012, - "bbox": [ - 193.0, - 538.24, - 205.35, - 551.55 - ], - "text": "1\n3", - "type": "text" - }, - { - "block_id": "p990-b50", - "global_id": 30013, - "bbox": [ - 288.13, - 582.74, - 300.55, - 596.04 - ], - "text": "1\n2", - "type": "text" - }, - { - "block_id": "p990-b51", - "global_id": 30014, - "bbox": [ - 405.63, - 617.7, - 461.74, - 626.67 - ], - "text": "Figure P10.2-2", - "type": "text" - } - ] - }, - { - "page_num": 991, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p991-b0", - "global_id": 30015, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n971", - "type": "text" - }, - { - "block_id": "p991-b1", - "global_id": 30016, - "bbox": [ - 88.32, - 85.94, - 288.98, - 105.94 - ], - "text": "10.2-5\nWrite the state and output equations for the\nnetwork in Fig. P10.2-5.", - "type": "text" - }, - { - "block_id": "p991-b2", - "global_id": 30017, - "bbox": [ - 125.64, - 185.81, - 285.91, - 202.18 - ], - "text": "x\n+\n_\ny", - "type": "text" - }, - { - "block_id": "p991-b5", - "global_id": 30018, - "bbox": [ - 186.08, - 126.77, - 196.53, - 134.77 - ], - "text": "1 F", - "type": "text" - }, - { - "block_id": "p991-b6", - "global_id": 30019, - "bbox": [ - 185.81, - 185.68, - 272.98, - 208.8 - ], - "text": "2 \t\n1 \t\n1\n2 F", - "type": "text" - }, - { - "block_id": "p991-b7", - "global_id": 30020, - "bbox": [ - 121.45, - 237.68, - 177.56, - 246.65 - ], - "text": "Figure P10.2-5", - "type": "text" - }, - { - "block_id": "p991-b8", - "global_id": 30021, - "bbox": [ - 88.32, - 267.53, - 288.99, - 287.53 - ], - "text": "10.2-6\nWrite the state and output equations of the\nsystem shown in Fig. P10.2-6.", - "type": "text" - }, - { - "block_id": "p991-b10", - "global_id": 30022, - "bbox": [ - 96.49, - 308.53, - 258.7, - 318.1 - ], - "text": "y\nx\nw\nz", - "type": "text" - }, - { - "block_id": "p991-b12", - "global_id": 30023, - "bbox": [ - 163.33, - 312.78, - 233.9, - 329.7 - ], - "text": "5\ns 10\n1\ns 2", - "type": "text" - }, - { - "block_id": "p991-b13", - "global_id": 30024, - "bbox": [ - 190.25, - 354.63, - 208.02, - 371.56 - ], - "text": "1\ns 1", - "type": "text" - }, - { - "block_id": "p991-b14", - "global_id": 30025, - "bbox": [ - 85.58, - 383.27, - 141.69, - 392.24 - ], - "text": "Figure P10.2-6", - "type": "text" - }, - { - "block_id": "p991-b15", - "global_id": 30026, - "bbox": [ - 88.32, - 413.13, - 288.99, - 433.13 - ], - "text": "10.2-7\nWrite the state and output equations of the\nsystem shown in Fig. P10.2-7.", - "type": "text" - }, - { - "block_id": "p991-b16", - "global_id": 30027, - "bbox": [ - 288.43, - 488.92, - 294.98, - 498.52 - ], - "text": "y1", - "type": "text" - }, - { - "block_id": "p991-b17", - "global_id": 30028, - "bbox": [ - 288.43, - 528.7, - 294.98, - 538.31 - ], - "text": "y2", - "type": "text" - }, - { - "block_id": "p991-b18", - "global_id": 30029, - "bbox": [ - 205.66, - 457.73, - 212.66, - 467.33 - ], - "text": "q1", - "type": "text" - }, - { - "block_id": "p991-b19", - "global_id": 30030, - "bbox": [ - 203.72, - 500.63, - 210.72, - 510.23 - ], - "text": "q2", - "type": "text" - }, - { - "block_id": "p991-b20", - "global_id": 30031, - "bbox": [ - 203.72, - 543.63, - 210.72, - 553.23 - ], - "text": "q3", - "type": "text" - }, - { - "block_id": "p991-b21", - "global_id": 30032, - "bbox": [ - 204.84, - 586.75, - 211.84, - 596.36 - ], - "text": "q4", - "type": "text" - }, - { - "block_id": "p991-b22", - "global_id": 30033, - "bbox": [ - 98.19, - 565.03, - 104.74, - 574.64 - ], - "text": "x2", - "type": "text" - }, - { - "block_id": "p991-b23", - "global_id": 30034, - "bbox": [ - 98.19, - 502.24, - 104.74, - 511.84 - ], - "text": "x1", - "type": "text" - }, - { - "block_id": "p991-b26", - "global_id": 30035, - "bbox": [ - 164.76, - 590.62, - 185.98, - 608.56 - ], - "text": "1\ns l4", - "type": "text" - }, - { - "block_id": "p991-b27", - "global_id": 30036, - "bbox": [ - 164.76, - 547.6, - 185.98, - 565.54 - ], - "text": "1\ns l3", - "type": "text" - }, - { - "block_id": "p991-b28", - "global_id": 30037, - "bbox": [ - 164.76, - 504.57, - 185.98, - 522.51 - ], - "text": "1\ns l2", - "type": "text" - }, - { - "block_id": "p991-b29", - "global_id": 30038, - "bbox": [ - 164.76, - 461.42, - 185.98, - 479.37 - ], - "text": "1\ns l1", - "type": "text" - }, - { - "block_id": "p991-b30", - "global_id": 30039, - "bbox": [ - 85.58, - 625.88, - 141.69, - 634.84 - ], - "text": "Figure P10.2-7", - "type": "text" - }, - { - "block_id": "p991-b31", - "global_id": 30040, - "bbox": [ - 315.47, - 85.94, - 514.85, - 94.98 - ], - "text": "10.2-8\nFor a system specified by the transfer function", - "type": "text" - }, - { - "block_id": "p991-b32", - "global_id": 30041, - "bbox": [ - 395.43, - 105.48, - 468.09, - 127.48 - ], - "text": "H(s) =\n3s + 10\ns2 + 7s + 12", - "type": "text" - }, - { - "block_id": "p991-b33", - "global_id": 30042, - "bbox": [ - 348.59, - 137.17, - 516.14, - 168.05 - ], - "text": "write sets of state equations for DFII and its\ntranspose, cascade, and parallel forms. Also\nwrite the corresponding output equations.", - "type": "text" - }, - { - "block_id": "p991-b34", - "global_id": 30043, - "bbox": [ - 315.47, - 173.04, - 432.65, - 182.08 - ], - "text": "10.2-9\nRepeat Prob. 10.2-8 for", - "type": "text" - }, - { - "block_id": "p991-b35", - "global_id": 30044, - "bbox": [ - 349.09, - 184.08, - 359.04, - 193.04 - ], - "text": "(a)", - "type": "text" - }, - { - "block_id": "p991-b36", - "global_id": 30045, - "bbox": [ - 398.71, - 193.96, - 479.78, - 215.67 - ], - "text": "H(s) =\n4s\n(s + 1)(s + 2)2", - "type": "text" - }, - { - "block_id": "p991-b37", - "global_id": 30046, - "bbox": [ - 348.59, - 223.18, - 359.04, - 232.15 - ], - "text": "(b)", - "type": "text" - }, - { - "block_id": "p991-b38", - "global_id": 30047, - "bbox": [ - 398.71, - 231.08, - 479.42, - 249.87 - ], - "text": "H(s) = s3 + 7s2 + 12s", - "type": "text" - }, - { - "block_id": "p991-b39", - "global_id": 30048, - "bbox": [ - 427.64, - 246.66, - 480.26, - 256.33 - ], - "text": "(s + 1)3(s + 2)", - "type": "text" - }, - { - "block_id": "p991-b40", - "global_id": 30049, - "bbox": [ - 315.47, - 263.59, - 516.12, - 283.88 - ], - "text": "10.3-1\nFind the state vector q(t) by using the Laplace\ntransform method if", - "type": "text" - }, - { - "block_id": "p991-b41", - "global_id": 30050, - "bbox": [ - 408.67, - 296.3, - 456.05, - 305.58 - ], - "text": "˙q = Aq + Bx", - "type": "text" - }, - { - "block_id": "p991-b42", - "global_id": 30051, - "bbox": [ - 348.59, - 318.45, - 370.49, - 327.41 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p991-b43", - "global_id": 30052, - "bbox": [ - 382.86, - 343.17, - 398.17, - 352.43 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p991-b44", - "global_id": 30053, - "bbox": [ - 400.02, - 330.58, - 436.82, - 357.93 - ], - "text": "0\n2\n−1\n−3", - "type": "text" - }, - { - "block_id": "p991-b45", - "global_id": 30054, - "bbox": [ - 436.82, - 330.58, - 441.7, - 339.54 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b46", - "global_id": 30055, - "bbox": [ - 460.63, - 343.17, - 475.45, - 352.43 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p991-b47", - "global_id": 30056, - "bbox": [ - 477.3, - 330.58, - 486.67, - 346.98 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p991-b48", - "global_id": 30057, - "bbox": [ - 482.19, - 348.97, - 486.67, - 357.93 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p991-b49", - "global_id": 30058, - "bbox": [ - 486.67, - 330.58, - 491.55, - 339.54 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b50", - "global_id": 30059, - "bbox": [ - 373.18, - 369.07, - 398.17, - 378.41 - ], - "text": "q(0) =", - "type": "text" - }, - { - "block_id": "p991-b51", - "global_id": 30060, - "bbox": [ - 400.02, - 356.48, - 409.39, - 372.88 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p991-b52", - "global_id": 30061, - "bbox": [ - 404.9, - 374.87, - 409.39, - 383.84 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p991-b53", - "global_id": 30062, - "bbox": [ - 409.39, - 356.48, - 414.27, - 365.45 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b54", - "global_id": 30063, - "bbox": [ - 433.2, - 369.07, - 461.71, - 378.41 - ], - "text": "x(t) = 0", - "type": "text" - }, - { - "block_id": "p991-b55", - "global_id": 30064, - "bbox": [ - 315.47, - 396.12, - 432.65, - 405.16 - ], - "text": "10.3-2\nRepeat Prob. 10.3-1 for", - "type": "text" - }, - { - "block_id": "p991-b56", - "global_id": 30065, - "bbox": [ - 381.26, - 422.76, - 396.57, - 432.02 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p991-b57", - "global_id": 30066, - "bbox": [ - 398.42, - 410.17, - 435.22, - 437.53 - ], - "text": "−5\n−6\n1\n0", - "type": "text" - }, - { - "block_id": "p991-b58", - "global_id": 30067, - "bbox": [ - 435.22, - 410.17, - 440.1, - 419.14 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b59", - "global_id": 30068, - "bbox": [ - 459.03, - 422.76, - 473.85, - 432.02 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p991-b60", - "global_id": 30069, - "bbox": [ - 475.7, - 410.17, - 485.07, - 426.57 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p991-b61", - "global_id": 30070, - "bbox": [ - 480.59, - 428.56, - 485.07, - 437.53 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p991-b62", - "global_id": 30071, - "bbox": [ - 485.07, - 410.17, - 489.95, - 419.14 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b63", - "global_id": 30072, - "bbox": [ - 371.58, - 448.66, - 396.57, - 458.0 - ], - "text": "q(0) =", - "type": "text" - }, - { - "block_id": "p991-b64", - "global_id": 30073, - "bbox": [ - 398.42, - 436.08, - 407.79, - 452.47 - ], - "text": "5", - "type": "text" - }, - { - "block_id": "p991-b65", - "global_id": 30074, - "bbox": [ - 403.3, - 454.47, - 407.79, - 463.43 - ], - "text": "4", - "type": "text" - }, - { - "block_id": "p991-b66", - "global_id": 30075, - "bbox": [ - 407.79, - 436.08, - 412.67, - 445.04 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b67", - "global_id": 30076, - "bbox": [ - 440.56, - 448.66, - 492.99, - 458.0 - ], - "text": "x(t) = sin 100t", - "type": "text" - }, - { - "block_id": "p991-b68", - "global_id": 30077, - "bbox": [ - 315.47, - 475.7, - 432.66, - 484.74 - ], - "text": "10.3-3\nRepeat Prob. 10.3-1 for", - "type": "text" - }, - { - "block_id": "p991-b69", - "global_id": 30078, - "bbox": [ - 382.87, - 502.35, - 398.18, - 511.62 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p991-b70", - "global_id": 30079, - "bbox": [ - 400.02, - 489.76, - 436.82, - 517.12 - ], - "text": "−2\n0\n1\n−1", - "type": "text" - }, - { - "block_id": "p991-b71", - "global_id": 30080, - "bbox": [ - 436.82, - 489.76, - 441.71, - 498.73 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b72", - "global_id": 30081, - "bbox": [ - 460.64, - 502.35, - 475.46, - 511.62 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p991-b73", - "global_id": 30082, - "bbox": [ - 477.3, - 489.76, - 486.67, - 506.15 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p991-b74", - "global_id": 30083, - "bbox": [ - 482.19, - 508.15, - 486.67, - 517.12 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p991-b75", - "global_id": 30084, - "bbox": [ - 486.67, - 489.76, - 491.56, - 498.73 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b76", - "global_id": 30085, - "bbox": [ - 373.18, - 528.25, - 398.17, - 537.59 - ], - "text": "q(0) =", - "type": "text" - }, - { - "block_id": "p991-b77", - "global_id": 30086, - "bbox": [ - 400.01, - 515.66, - 412.88, - 532.06 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p991-b78", - "global_id": 30087, - "bbox": [ - 404.9, - 533.68, - 416.38, - 543.02 - ], - "text": "−1", - "type": "text" - }, - { - "block_id": "p991-b79", - "global_id": 30088, - "bbox": [ - 416.38, - 515.66, - 421.26, - 524.63 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b80", - "global_id": 30089, - "bbox": [ - 440.19, - 528.25, - 478.05, - 537.5 - ], - "text": "x(t) = u(t)", - "type": "text" - }, - { - "block_id": "p991-b81", - "global_id": 30090, - "bbox": [ - 315.47, - 555.29, - 432.65, - 564.33 - ], - "text": "10.3-4\nRepeat Prob. 10.3-1 for", - "type": "text" - }, - { - "block_id": "p991-b82", - "global_id": 30091, - "bbox": [ - 376.13, - 581.94, - 391.44, - 591.2 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p991-b83", - "global_id": 30092, - "bbox": [ - 393.28, - 569.35, - 430.1, - 596.7 - ], - "text": "−1\n1\n0\n−2", - "type": "text" - }, - { - "block_id": "p991-b84", - "global_id": 30093, - "bbox": [ - 430.09, - 569.35, - 434.98, - 578.31 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b85", - "global_id": 30094, - "bbox": [ - 453.91, - 581.94, - 468.72, - 591.2 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p991-b86", - "global_id": 30095, - "bbox": [ - 470.57, - 569.35, - 493.39, - 596.7 - ], - "text": "1\n1\n0\n1", - "type": "text" - }, - { - "block_id": "p991-b87", - "global_id": 30096, - "bbox": [ - 493.39, - 569.35, - 498.28, - 578.31 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b88", - "global_id": 30097, - "bbox": [ - 366.45, - 607.84, - 391.44, - 617.18 - ], - "text": "q(0) =", - "type": "text" - }, - { - "block_id": "p991-b89", - "global_id": 30098, - "bbox": [ - 393.28, - 595.25, - 402.65, - 611.65 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p991-b90", - "global_id": 30099, - "bbox": [ - 398.17, - 613.64, - 402.65, - 622.6 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p991-b91", - "global_id": 30100, - "bbox": [ - 402.65, - 595.25, - 407.54, - 604.21 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p991-b92", - "global_id": 30101, - "bbox": [ - 426.48, - 607.84, - 439.8, - 617.1 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p991-b93", - "global_id": 30102, - "bbox": [ - 441.63, - 595.25, - 460.35, - 611.56 - ], - "text": "u(t)", - "type": "text" - }, - { - "block_id": "p991-b94", - "global_id": 30103, - "bbox": [ - 446.59, - 613.27, - 460.28, - 622.51 - ], - "text": "δ(t)", - "type": "text" - }, - { - "block_id": "p991-b95", - "global_id": 30104, - "bbox": [ - 460.35, - 595.25, - 465.24, - 604.21 - ], - "text": "!", - "type": "text" - } - ] - }, - { - "page_num": 992, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p992-b0", - "global_id": 30105, - "bbox": [ - 60.0, - 60.36, - 269.83, - 69.45 - ], - "text": "972\nCHAPTER 10\nSTATE-SPACE ANALYSIS", - "type": "text" - }, - { - "block_id": "p992-b1", - "global_id": 30106, - "bbox": [ - 62.58, - 85.94, - 263.22, - 105.94 - ], - "text": "10.3-5\nUse the Laplace transform method to find the\nresponse y for", - "type": "text" - }, - { - "block_id": "p992-b2", - "global_id": 30107, - "bbox": [ - 150.87, - 116.51, - 207.58, - 125.78 - ], - "text": "˙q = Aq + Bx(t)", - "type": "text" - }, - { - "block_id": "p992-b3", - "global_id": 30108, - "bbox": [ - 151.87, - 130.47, - 208.08, - 139.73 - ], - "text": "y = Cq + Dx(t)", - "type": "text" - }, - { - "block_id": "p992-b4", - "global_id": 30109, - "bbox": [ - 95.7, - 150.77, - 117.6, - 159.74 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p992-b5", - "global_id": 30110, - "bbox": [ - 128.62, - 173.65, - 143.93, - 182.92 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p992-b6", - "global_id": 30111, - "bbox": [ - 145.77, - 161.06, - 175.58, - 188.42 - ], - "text": "−3\n1\n−2\n0", - "type": "text" - }, - { - "block_id": "p992-b7", - "global_id": 30112, - "bbox": [ - 175.59, - 161.06, - 180.48, - 170.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b8", - "global_id": 30113, - "bbox": [ - 199.4, - 173.65, - 214.22, - 182.92 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p992-b9", - "global_id": 30114, - "bbox": [ - 216.06, - 161.06, - 225.43, - 177.46 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p992-b10", - "global_id": 30115, - "bbox": [ - 220.95, - 179.45, - 225.43, - 188.42 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p992-b11", - "global_id": 30116, - "bbox": [ - 225.43, - 161.06, - 230.32, - 170.03 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b12", - "global_id": 30117, - "bbox": [ - 128.62, - 194.02, - 209.25, - 203.36 - ], - "text": "C = [0\n1]\nD = 0", - "type": "text" - }, - { - "block_id": "p992-b13", - "global_id": 30118, - "bbox": [ - 95.7, - 214.32, - 108.64, - 223.28 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p992-b14", - "global_id": 30119, - "bbox": [ - 131.03, - 228.24, - 211.82, - 237.58 - ], - "text": "x(t) = u(t)\nq(0) =", - "type": "text" - }, - { - "block_id": "p992-b15", - "global_id": 30120, - "bbox": [ - 213.66, - 215.65, - 223.03, - 232.04 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p992-b16", - "global_id": 30121, - "bbox": [ - 218.54, - 234.04, - 223.03, - 243.0 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p992-b17", - "global_id": 30122, - "bbox": [ - 223.03, - 215.65, - 227.92, - 224.61 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b18", - "global_id": 30123, - "bbox": [ - 62.58, - 250.95, - 179.76, - 259.99 - ], - "text": "10.3-6\nRepeat Prob. 10.3-5 for", - "type": "text" - }, - { - "block_id": "p992-b19", - "global_id": 30124, - "bbox": [ - 133.05, - 275.75, - 148.36, - 285.01 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p992-b20", - "global_id": 30125, - "bbox": [ - 150.21, - 263.17, - 187.01, - 290.53 - ], - "text": "−1\n1\n−1\n−1", - "type": "text" - }, - { - "block_id": "p992-b21", - "global_id": 30126, - "bbox": [ - 187.01, - 263.17, - 191.89, - 272.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b22", - "global_id": 30127, - "bbox": [ - 201.86, - 275.75, - 216.68, - 285.01 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p992-b23", - "global_id": 30128, - "bbox": [ - 218.52, - 263.17, - 227.89, - 279.56 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p992-b24", - "global_id": 30129, - "bbox": [ - 223.4, - 281.56, - 227.89, - 290.53 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p992-b25", - "global_id": 30130, - "bbox": [ - 227.89, - 263.17, - 232.78, - 272.13 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b26", - "global_id": 30131, - "bbox": [ - 133.05, - 296.13, - 213.68, - 305.47 - ], - "text": "C = [1\n1]\nD = 1", - "type": "text" - }, - { - "block_id": "p992-b27", - "global_id": 30132, - "bbox": [ - 126.18, - 316.6, - 206.96, - 325.94 - ], - "text": "x(t) = u(t)\nq(0) =", - "type": "text" - }, - { - "block_id": "p992-b28", - "global_id": 30133, - "bbox": [ - 208.79, - 304.01, - 218.16, - 320.4 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p992-b29", - "global_id": 30134, - "bbox": [ - 213.68, - 322.4, - 218.16, - 331.37 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p992-b30", - "global_id": 30135, - "bbox": [ - 218.16, - 304.01, - 223.05, - 312.97 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b31", - "global_id": 30136, - "bbox": [ - 62.58, - 341.5, - 263.24, - 373.5 - ], - "text": "10.3-7\nThe transfer function H(s) in Prob. 10.2-8 is\nrealized as a cascade of H1(s) followed by\nH2(s), where", - "type": "text" - }, - { - "block_id": "p992-b32", - "global_id": 30137, - "bbox": [ - 149.38, - 381.42, - 199.4, - 403.04 - ], - "text": "H1(s) =\n1\ns + 3", - "type": "text" - }, - { - "block_id": "p992-b33", - "global_id": 30138, - "bbox": [ - 149.38, - 404.54, - 208.36, - 420.9 - ], - "text": "H2(s) = 3s + 10", - "type": "text" - }, - { - "block_id": "p992-b34", - "global_id": 30139, - "bbox": [ - 186.13, - 417.18, - 203.89, - 426.52 - ], - "text": "s + 4", - "type": "text" - }, - { - "block_id": "p992-b35", - "global_id": 30140, - "bbox": [ - 95.7, - 434.37, - 263.24, - 476.21 - ], - "text": "Let the outputs of these subsystems be state\nvariables q1 and q2, respectively. Write the\nstate equations and the output equation for this\nsystem and verify that H(s) = Cφ(s)B + D.", - "type": "text" - }, - { - "block_id": "p992-b36", - "global_id": 30141, - "bbox": [ - 62.58, - 480.82, - 263.23, - 501.12 - ], - "text": "10.3-8\nFind the transfer function matrix H(s) for the\nsystem in Prob. 10.3-5.", - "type": "text" - }, - { - "block_id": "p992-b37", - "global_id": 30142, - "bbox": [ - 62.58, - 505.73, - 263.23, - 526.03 - ], - "text": "10.3-9\nFind the transfer function matrix H(s) for the\nsystem in Prob. 10.3-6.", - "type": "text" - }, - { - "block_id": "p992-b38", - "global_id": 30143, - "bbox": [ - 58.1, - 530.64, - 263.23, - 550.94 - ], - "text": "10.3-10\nFind the transfer function matrix H(s) for the\nsystem", - "type": "text" - }, - { - "block_id": "p992-b39", - "global_id": 30144, - "bbox": [ - 155.54, - 561.51, - 202.91, - 570.78 - ], - "text": "˙q = Aq + Bx", - "type": "text" - }, - { - "block_id": "p992-b40", - "global_id": 30145, - "bbox": [ - 156.54, - 575.47, - 203.4, - 584.73 - ], - "text": "y = Cq + Dx", - "type": "text" - }, - { - "block_id": "p992-b41", - "global_id": 30146, - "bbox": [ - 95.7, - 595.77, - 117.59, - 604.73 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p992-b42", - "global_id": 30147, - "bbox": [ - 96.9, - 618.65, - 112.21, - 627.91 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p992-b43", - "global_id": 30148, - "bbox": [ - 114.06, - 606.06, - 150.86, - 633.41 - ], - "text": "0\n1\n−1\n−2", - "type": "text" - }, - { - "block_id": "p992-b44", - "global_id": 30149, - "bbox": [ - 150.86, - 606.06, - 155.74, - 615.02 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b45", - "global_id": 30150, - "bbox": [ - 165.71, - 618.65, - 180.53, - 627.91 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p992-b46", - "global_id": 30151, - "bbox": [ - 182.38, - 606.06, - 205.19, - 633.41 - ], - "text": "0\n1\n1\n0", - "type": "text" - }, - { - "block_id": "p992-b47", - "global_id": 30152, - "bbox": [ - 205.19, - 606.06, - 210.08, - 615.02 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b48", - "global_id": 30153, - "bbox": [ - 220.04, - 618.65, - 233.36, - 627.91 - ], - "text": "x =", - "type": "text" - }, - { - "block_id": "p992-b49", - "global_id": 30154, - "bbox": [ - 235.21, - 606.06, - 257.15, - 623.19 - ], - "text": "x1(t)", - "type": "text" - }, - { - "block_id": "p992-b50", - "global_id": 30155, - "bbox": [ - 240.09, - 624.07, - 257.15, - 634.15 - ], - "text": "x2(t)", - "type": "text" - }, - { - "block_id": "p992-b51", - "global_id": 30156, - "bbox": [ - 257.15, - 606.06, - 262.03, - 615.02 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p992-b52", - "global_id": 30157, - "bbox": [ - 348.99, - 98.1, - 364.3, - 107.36 - ], - "text": "C =", - "type": "text" - }, - { - "block_id": "p992-b53", - "global_id": 30158, - "bbox": [ - 366.15, - 80.13, - 372.68, - 89.1 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p992-b54", - "global_id": 30159, - "bbox": [ - 366.15, - 96.27, - 372.68, - 105.24 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p992-b55", - "global_id": 30160, - "bbox": [ - 372.68, - 87.46, - 390.62, - 118.34 - ], - "text": "1\n2\n4\n1\n1\n1", - "type": "text" - }, - { - "block_id": "p992-b56", - "global_id": 30161, - "bbox": [ - 390.62, - 80.13, - 397.16, - 89.1 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p992-b57", - "global_id": 30162, - "bbox": [ - 390.62, - 96.27, - 431.39, - 107.36 - ], - "text": "⎦\nD =", - "type": "text" - }, - { - "block_id": "p992-b58", - "global_id": 30163, - "bbox": [ - 433.23, - 80.13, - 439.77, - 89.1 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p992-b59", - "global_id": 30164, - "bbox": [ - 433.23, - 96.27, - 439.77, - 105.24 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p992-b60", - "global_id": 30165, - "bbox": [ - 439.78, - 87.46, - 457.71, - 118.34 - ], - "text": "0\n0\n0\n0\n1\n0", - "type": "text" - }, - { - "block_id": "p992-b61", - "global_id": 30166, - "bbox": [ - 457.7, - 80.13, - 464.24, - 89.1 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p992-b62", - "global_id": 30167, - "bbox": [ - 457.7, - 96.27, - 464.24, - 105.24 - ], - "text": "⎦", - "type": "text" - }, - { - "block_id": "p992-b63", - "global_id": 30168, - "bbox": [ - 285.24, - 130.25, - 490.37, - 150.25 - ], - "text": "10.3-11\nRepeat Prob. 10.3-1, using the time-domain\nmethod.", - "type": "text" - }, - { - "block_id": "p992-b64", - "global_id": 30169, - "bbox": [ - 285.24, - 155.5, - 490.37, - 175.5 - ], - "text": "10.3-12\nRepeat Prob. 10.3-2, using the time-domain\nmethod.", - "type": "text" - }, - { - "block_id": "p992-b65", - "global_id": 30170, - "bbox": [ - 285.24, - 180.76, - 490.37, - 200.76 - ], - "text": "10.3-13\nRepeat Prob. 10.3-3, using the time-domain\nmethod.", - "type": "text" - }, - { - "block_id": "p992-b66", - "global_id": 30171, - "bbox": [ - 285.23, - 206.01, - 490.37, - 226.0 - ], - "text": "10.3-14\nRepeat Prob. 10.3-4, using the time-domain\nmethod.", - "type": "text" - }, - { - "block_id": "p992-b67", - "global_id": 30172, - "bbox": [ - 285.23, - 231.27, - 490.37, - 251.26 - ], - "text": "10.3-15\nRepeat Prob. 10.3-5, using the time-domain\nmethod.", - "type": "text" - }, - { - "block_id": "p992-b68", - "global_id": 30173, - "bbox": [ - 285.23, - 256.52, - 490.37, - 276.51 - ], - "text": "10.3-16\nRepeat Prob. 10.3-6, using the time-domain\nmethod.", - "type": "text" - }, - { - "block_id": "p992-b69", - "global_id": 30174, - "bbox": [ - 285.23, - 281.47, - 490.38, - 301.77 - ], - "text": "10.3-17\nFind the unit impulse response matrix h(t) for\nthe system in Prob. 10.3-7, using Eq. (10.45).", - "type": "text" - }, - { - "block_id": "p992-b70", - "global_id": 30175, - "bbox": [ - 285.23, - 306.72, - 490.38, - 327.02 - ], - "text": "10.3-18\nFind the unit impulse response matrix h(t) for\nthe system in Prob. 10.3-6.", - "type": "text" - }, - { - "block_id": "p992-b71", - "global_id": 30176, - "bbox": [ - 285.23, - 331.98, - 490.39, - 352.28 - ], - "text": "10.3-19\nFind the unit impulse response matrix h(t) for\nthe system in Prob. 10.3-10.", - "type": "text" - }, - { - "block_id": "p992-b72", - "global_id": 30177, - "bbox": [ - 289.73, - 357.53, - 490.37, - 377.53 - ], - "text": "10.4-1\nThe state equations of a certain system are\ngiven as", - "type": "text" - }, - { - "block_id": "p992-b73", - "global_id": 30178, - "bbox": [ - 373.66, - 392.57, - 419.03, - 402.88 - ], - "text": "˙q1 = q2 + 2x", - "type": "text" - }, - { - "block_id": "p992-b74", - "global_id": 30179, - "bbox": [ - 373.66, - 406.52, - 439.55, - 416.83 - ], - "text": "˙q2 = −q1 −q2 + x", - "type": "text" - }, - { - "block_id": "p992-b75", - "global_id": 30180, - "bbox": [ - 322.85, - 431.2, - 456.51, - 440.24 - ], - "text": "Define a new state vector w such that", - "type": "text" - }, - { - "block_id": "p992-b76", - "global_id": 30181, - "bbox": [ - 383.31, - 455.29, - 429.42, - 479.54 - ], - "text": "w1 = q2\nw2 = q2 −q1", - "type": "text" - }, - { - "block_id": "p992-b77", - "global_id": 30182, - "bbox": [ - 322.85, - 493.92, - 490.39, - 535.84 - ], - "text": "Find the state equations of the system with w\nas the state vector. Determine the characteristic\nroots (eigenvalues) of the matrix A in the\noriginal and the transformed state equations.", - "type": "text" - }, - { - "block_id": "p992-b78", - "global_id": 30183, - "bbox": [ - 289.73, - 541.09, - 474.98, - 550.13 - ], - "text": "10.4-2\nThe state equations of a certain system are", - "type": "text" - }, - { - "block_id": "p992-b79", - "global_id": 30184, - "bbox": [ - 366.94, - 565.17, - 446.27, - 589.43 - ], - "text": "˙q1 = q2\n˙q2 = −2q1 −3q2 + 2x", - "type": "text" - }, - { - "block_id": "p992-b80", - "global_id": 30185, - "bbox": [ - 323.35, - 603.8, - 490.39, - 612.84 - ], - "text": "(a) Determine a new state vector w (in terms", - "type": "text" - }, - { - "block_id": "p992-b81", - "global_id": 30186, - "bbox": [ - 338.28, - 614.76, - 490.38, - 634.76 - ], - "text": "of vector q) such that the resulting state\nequations are in diagonalized form.", - "type": "text" - } - ] - }, - { - "page_num": 993, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p993-b0", - "global_id": 30187, - "bbox": [ - 436.02, - 62.89, - 516.14, - 71.98 - ], - "text": "Problems\n973", - "type": "text" - }, - { - "block_id": "p993-b1", - "global_id": 30188, - "bbox": [ - 121.44, - 85.82, - 214.09, - 94.86 - ], - "text": "(b) For output y given by", - "type": "text" - }, - { - "block_id": "p993-b2", - "global_id": 30189, - "bbox": [ - 189.25, - 111.26, - 236.61, - 120.53 - ], - "text": "y = Cq + Dx", - "type": "text" - }, - { - "block_id": "p993-b3", - "global_id": 30190, - "bbox": [ - 136.88, - 137.38, - 158.78, - 146.34 - ], - "text": "where", - "type": "text" - }, - { - "block_id": "p993-b4", - "global_id": 30191, - "bbox": [ - 166.72, - 166.07, - 182.03, - 175.34 - ], - "text": "C =", - "type": "text" - }, - { - "block_id": "p993-b5", - "global_id": 30192, - "bbox": [ - 183.88, - 153.48, - 213.69, - 180.84 - ], - "text": "1\n1\n−1\n2", - "type": "text" - }, - { - "block_id": "p993-b6", - "global_id": 30193, - "bbox": [ - 213.69, - 153.48, - 218.57, - 162.45 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p993-b7", - "global_id": 30194, - "bbox": [ - 237.5, - 166.07, - 259.14, - 175.41 - ], - "text": "D = 0", - "type": "text" - }, - { - "block_id": "p993-b8", - "global_id": 30195, - "bbox": [ - 136.88, - 196.97, - 288.97, - 216.97 - ], - "text": "determine the output y in terms of the new\nstate vector w.", - "type": "text" - }, - { - "block_id": "p993-b9", - "global_id": 30196, - "bbox": [ - 88.32, - 222.35, - 176.37, - 231.39 - ], - "text": "10.4-3\nGiven a system", - "type": "text" - }, - { - "block_id": "p993-b10", - "global_id": 30197, - "bbox": [ - 149.0, - 258.45, - 162.82, - 267.72 - ], - "text": "˙q =", - "type": "text" - }, - { - "block_id": "p993-b11", - "global_id": 30198, - "bbox": [ - 164.66, - 240.49, - 171.2, - 249.45 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p993-b12", - "global_id": 30199, - "bbox": [ - 164.66, - 256.62, - 171.2, - 265.59 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p993-b13", - "global_id": 30200, - "bbox": [ - 171.2, - 247.83, - 216.57, - 278.71 - ], - "text": "0\n1\n0\n0\n0\n1\n0\n−2\n−3", - "type": "text" - }, - { - "block_id": "p993-b14", - "global_id": 30201, - "bbox": [ - 216.57, - 240.49, - 223.1, - 249.45 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p993-b15", - "global_id": 30202, - "bbox": [ - 216.57, - 256.62, - 237.47, - 267.73 - ], - "text": "⎦q +", - "type": "text" - }, - { - "block_id": "p993-b16", - "global_id": 30203, - "bbox": [ - 238.87, - 240.49, - 245.41, - 249.45 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p993-b17", - "global_id": 30204, - "bbox": [ - 238.87, - 256.62, - 245.41, - 265.59 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p993-b18", - "global_id": 30205, - "bbox": [ - 245.4, - 247.83, - 249.89, - 278.71 - ], - "text": "0\n0\n1", - "type": "text" - }, - { - "block_id": "p993-b19", - "global_id": 30206, - "bbox": [ - 249.89, - 240.49, - 256.42, - 249.46 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p993-b20", - "global_id": 30207, - "bbox": [ - 249.89, - 256.63, - 261.4, - 267.71 - ], - "text": "⎦x", - "type": "text" - }, - { - "block_id": "p993-b21", - "global_id": 30208, - "bbox": [ - 121.44, - 294.84, - 288.98, - 314.83 - ], - "text": "determine a new state vector w such that the\nstate equations are diagonalized.", - "type": "text" - }, - { - "block_id": "p993-b22", - "global_id": 30209, - "bbox": [ - 88.32, - 320.23, - 288.97, - 340.22 - ], - "text": "10.4-4\nThe state equations of a certain system are\ngiven in diagonalized form as", - "type": "text" - }, - { - "block_id": "p993-b23", - "global_id": 30210, - "bbox": [ - 145.5, - 367.27, - 159.32, - 376.55 - ], - "text": "˙q =", - "type": "text" - }, - { - "block_id": "p993-b24", - "global_id": 30211, - "bbox": [ - 161.16, - 349.32, - 167.7, - 358.28 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p993-b25", - "global_id": 30212, - "bbox": [ - 161.16, - 365.45, - 167.7, - 374.42 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p993-b26", - "global_id": 30213, - "bbox": [ - 167.7, - 356.27, - 220.06, - 387.54 - ], - "text": "−1\n0\n0\n0\n−3\n0\n0\n0\n−2", - "type": "text" - }, - { - "block_id": "p993-b27", - "global_id": 30214, - "bbox": [ - 220.07, - 349.32, - 226.6, - 358.28 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p993-b28", - "global_id": 30215, - "bbox": [ - 220.07, - 365.46, - 240.97, - 376.55 - ], - "text": "⎦q +", - "type": "text" - }, - { - "block_id": "p993-b29", - "global_id": 30216, - "bbox": [ - 242.37, - 349.32, - 248.9, - 358.28 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p993-b30", - "global_id": 30217, - "bbox": [ - 242.37, - 365.46, - 248.9, - 374.42 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p993-b31", - "global_id": 30218, - "bbox": [ - 248.9, - 356.65, - 253.39, - 387.54 - ], - "text": "1\n1\n1", - "type": "text" - }, - { - "block_id": "p993-b32", - "global_id": 30219, - "bbox": [ - 253.38, - 349.32, - 259.92, - 358.29 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p993-b33", - "global_id": 30220, - "bbox": [ - 253.38, - 365.46, - 264.9, - 376.53 - ], - "text": "⎦x", - "type": "text" - }, - { - "block_id": "p993-b34", - "global_id": 30221, - "bbox": [ - 121.44, - 403.74, - 234.9, - 412.7 - ], - "text": "The output equation is given by", - "type": "text" - }, - { - "block_id": "p993-b35", - "global_id": 30222, - "bbox": [ - 176.71, - 429.1, - 233.71, - 438.44 - ], - "text": "y = [1\n3\n1]q", - "type": "text" - }, - { - "block_id": "p993-b36", - "global_id": 30223, - "bbox": [ - 121.44, - 455.13, - 216.56, - 464.18 - ], - "text": "Determine the output y for", - "type": "text" - }, - { - "block_id": "p993-b37", - "global_id": 30224, - "bbox": [ - 154.62, - 491.24, - 179.61, - 500.58 - ], - "text": "q(0) =", - "type": "text" - }, - { - "block_id": "p993-b38", - "global_id": 30225, - "bbox": [ - 181.46, - 473.28, - 187.99, - 482.24 - ], - "text": "⎡", - "type": "text" - }, - { - "block_id": "p993-b39", - "global_id": 30226, - "bbox": [ - 181.46, - 489.41, - 187.99, - 498.38 - ], - "text": "⎣", - "type": "text" - }, - { - "block_id": "p993-b40", - "global_id": 30227, - "bbox": [ - 188.0, - 480.61, - 192.48, - 511.48 - ], - "text": "1\n2\n1", - "type": "text" - }, - { - "block_id": "p993-b41", - "global_id": 30228, - "bbox": [ - 192.48, - 473.27, - 199.02, - 482.24 - ], - "text": "⎤", - "type": "text" - }, - { - "block_id": "p993-b42", - "global_id": 30229, - "bbox": [ - 192.48, - 489.41, - 255.8, - 500.48 - ], - "text": "⎦\nx(t) = u(t)", - "type": "text" - }, - { - "block_id": "p993-b43", - "global_id": 30230, - "bbox": [ - 88.32, - 527.73, - 288.99, - 591.57 - ], - "text": "10.5-1\nWrite the state equations for the systems\ndepicted in Fig. P10.5-1. Determine a new state\nvector w such that the resulting state equations\nare in diagonalized form. Write the output y in\nterms of w. Determine in each case whether the\nsystem is controllable and observable.", - "type": "text" - }, - { - "block_id": "p993-b44", - "global_id": 30231, - "bbox": [ - 334.1, - 89.5, - 495.2, - 97.5 - ], - "text": "x\ny", - "type": "text" - }, - { - "block_id": "p993-b45", - "global_id": 30232, - "bbox": [ - 334.1, - 140.05, - 495.2, - 161.09 - ], - "text": "x\ny\ns a\ns b", - "type": "text" - }, - { - "block_id": "p993-b46", - "global_id": 30233, - "bbox": [ - 445.02, - 93.92, - 462.8, - 110.7 - ], - "text": "1\ns b", - "type": "text" - }, - { - "block_id": "p993-b47", - "global_id": 30234, - "bbox": [ - 373.02, - 144.38, - 390.8, - 161.15 - ], - "text": "1\ns b", - "type": "text" - }, - { - "block_id": "p993-b48", - "global_id": 30235, - "bbox": [ - 373.02, - 93.62, - 390.8, - 110.7 - ], - "text": "s a\ns b", - "type": "text" - }, - { - "block_id": "p993-b49", - "global_id": 30236, - "bbox": [ - 312.72, - 174.66, - 368.85, - 183.63 - ], - "text": "Figure P10.5-1", - "type": "text" - }, - { - "block_id": "p993-b50", - "global_id": 30237, - "bbox": [ - 315.47, - 197.61, - 504.97, - 206.65 - ], - "text": "10.6-1\nAn LTI discrete-time system is specified by", - "type": "text" - }, - { - "block_id": "p993-b51", - "global_id": 30238, - "bbox": [ - 385.01, - 223.46, - 400.32, - 232.72 - ], - "text": "A =", - "type": "text" - }, - { - "block_id": "p993-b52", - "global_id": 30239, - "bbox": [ - 402.17, - 210.87, - 424.98, - 238.22 - ], - "text": "2\n0\n1\n1", - "type": "text" - }, - { - "block_id": "p993-b53", - "global_id": 30240, - "bbox": [ - 424.98, - 210.87, - 429.87, - 219.83 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p993-b54", - "global_id": 30241, - "bbox": [ - 448.79, - 223.46, - 463.62, - 232.72 - ], - "text": "B =", - "type": "text" - }, - { - "block_id": "p993-b55", - "global_id": 30242, - "bbox": [ - 465.45, - 210.87, - 474.82, - 227.27 - ], - "text": "0", - "type": "text" - }, - { - "block_id": "p993-b56", - "global_id": 30243, - "bbox": [ - 470.34, - 229.26, - 474.82, - 238.22 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p993-b57", - "global_id": 30244, - "bbox": [ - 474.83, - 210.87, - 479.72, - 219.83 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p993-b58", - "global_id": 30245, - "bbox": [ - 385.01, - 243.83, - 476.09, - 253.17 - ], - "text": "C = [0\n1]\nD = [1]", - "type": "text" - }, - { - "block_id": "p993-b59", - "global_id": 30246, - "bbox": [ - 348.59, - 265.17, - 361.53, - 274.13 - ], - "text": "and", - "type": "text" - }, - { - "block_id": "p993-b60", - "global_id": 30247, - "bbox": [ - 382.31, - 279.14, - 407.31, - 288.48 - ], - "text": "q(0) =", - "type": "text" - }, - { - "block_id": "p993-b61", - "global_id": 30248, - "bbox": [ - 409.15, - 266.56, - 418.52, - 282.95 - ], - "text": "2", - "type": "text" - }, - { - "block_id": "p993-b62", - "global_id": 30249, - "bbox": [ - 414.04, - 284.95, - 418.52, - 293.92 - ], - "text": "1", - "type": "text" - }, - { - "block_id": "p993-b63", - "global_id": 30250, - "bbox": [ - 418.52, - 266.56, - 423.41, - 275.53 - ], - "text": "!", - "type": "text" - }, - { - "block_id": "p993-b64", - "global_id": 30251, - "bbox": [ - 442.34, - 279.14, - 482.42, - 288.39 - ], - "text": "x[n] = u[n]", - "type": "text" - }, - { - "block_id": "p993-b65", - "global_id": 30252, - "bbox": [ - 349.1, - 308.08, - 516.13, - 317.42 - ], - "text": "(a) Find the output y[n], using the time-", - "type": "text" - }, - { - "block_id": "p993-b66", - "global_id": 30253, - "bbox": [ - 348.59, - 319.41, - 516.13, - 339.34 - ], - "text": "domain method.\n(b) Find the output y[n], using the frequency-", - "type": "text" - }, - { - "block_id": "p993-b67", - "global_id": 30254, - "bbox": [ - 364.03, - 341.33, - 422.3, - 350.29 - ], - "text": "domain method.", - "type": "text" - }, - { - "block_id": "p993-b68", - "global_id": 30255, - "bbox": [ - 315.47, - 355.2, - 516.13, - 375.21 - ], - "text": "10.6-2\nAn LTI discrete-time system is specified by the\ndifference equation", - "type": "text" - }, - { - "block_id": "p993-b69", - "global_id": 30256, - "bbox": [ - 378.81, - 386.83, - 485.9, - 396.17 - ], - "text": "y[n + 2] + y[n + 1] + 0.16y[n]", - "type": "text" - }, - { - "block_id": "p993-b70", - "global_id": 30257, - "bbox": [ - 389.62, - 400.78, - 467.12, - 410.12 - ], - "text": "= x[n + 1] + 0.32x[n]", - "type": "text" - }, - { - "block_id": "p993-b71", - "global_id": 30258, - "bbox": [ - 349.09, - 433.07, - 516.12, - 442.04 - ], - "text": "(a) Show the DFII, its transpose, cascade, and", - "type": "text" - }, - { - "block_id": "p993-b72", - "global_id": 30259, - "bbox": [ - 348.59, - 444.04, - 516.12, - 463.96 - ], - "text": "parallel realizations of this system.\n(b) Write the state and the output equations", - "type": "text" - }, - { - "block_id": "p993-b73", - "global_id": 30260, - "bbox": [ - 364.03, - 465.95, - 516.14, - 485.88 - ], - "text": "from these realizations, using the output of\neach delay element as a state variable.", - "type": "text" - }, - { - "block_id": "p993-b74", - "global_id": 30261, - "bbox": [ - 315.47, - 490.79, - 432.65, - 499.83 - ], - "text": "10.6-3\nRepeat Prob. 10.6-2 for", - "type": "text" - }, - { - "block_id": "p993-b75", - "global_id": 30262, - "bbox": [ - 384.42, - 511.45, - 480.3, - 520.79 - ], - "text": "y[n + 2] + y[n + 1] −6y[n]", - "type": "text" - }, - { - "block_id": "p993-b76", - "global_id": 30263, - "bbox": [ - 395.23, - 525.4, - 475.78, - 534.74 - ], - "text": "= 2x[n + 2] + x[n + 1]", - "type": "text" - }, - { - "block_id": "p993-b77", - "global_id": 30264, - "bbox": [ - 315.47, - 546.66, - 516.13, - 566.66 - ], - "text": "10.7-1\nVerify the state and output equations for the\nLTID system shown in Fig. 10.15.", - "type": "text" - }, - { - "block_id": "p993-b78", - "global_id": 30265, - "bbox": [ - 315.47, - 571.57, - 516.13, - 591.56 - ], - "text": "10.7-2\nVerify the state and output equations for the\nLTID system shown in Fig. 10.16.", - "type": "text" - } - ] - }, - { - "page_num": 994, - "width": 576.0, - "height": 720.0, - "blocks": [] - }, - { - "page_num": 995, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p995-b0", - "global_id": 30266, - "bbox": [ - 86.26, - 100.75, - 160.16, - 121.67 - ], - "text": "INDEX", - "type": "text" - }, - { - "block_id": "p995-b1", - "global_id": 30267, - "bbox": [ - 85.74, - 207.51, - 202.42, - 225.68 - ], - "text": "Abscissa of convergence, 336\nAccumulator systems, 259, 295, 519", - "type": "text" - }, - { - "block_id": "p995-b2", - "global_id": 30268, - "bbox": [ - 85.74, - 227.91, - 230.18, - 266.46 - ], - "text": "discrete-time Fourier transform of, 875–76\nActive circuits, 382–85\nAdders, 388, 396, 399, 403\nAddition", - "type": "text" - }, - { - "block_id": "p995-b3", - "global_id": 30269, - "bbox": [ - 85.74, - 268.69, - 183.04, - 317.44 - ], - "text": "of complex numbers, 11–12\nof matrices, 38\nof sinusoids, 18–20\nAdditivity, 97–98\nAlgebra", - "type": "text" - }, - { - "block_id": "p995-b4", - "global_id": 30270, - "bbox": [ - 85.74, - 319.66, - 244.25, - 348.02 - ], - "text": "of complex numbers, 5–15\nmatrix, 38–42\nAliasing, 536–38, 788–95, 805, 811–12, 817, 834", - "type": "text" - }, - { - "block_id": "p995-b5", - "global_id": 30271, - "bbox": [ - 85.74, - 350.24, - 271.27, - 429.58 - ], - "text": "defined, 536\ngeneral condition for in sinusoids, 793–96\ntreachery of, 788–91\nverification of in sinusoids, 792–93\nAliasing error, 659, 811\nAmplitude, 16\nAmplitude modulation, 711–13, 736–49, 762\nAmplitude response, 413, 416–17, 421, 424, 435, 437–38,", - "type": "text" - }, - { - "block_id": "p995-b6", - "global_id": 30272, - "bbox": [ - 85.74, - 431.57, - 292.52, - 449.73 - ], - "text": "440–42\nAmplitude spectrum, 598, 607, 615–25, 667, 668, 707, 848, 870,", - "type": "text" - }, - { - "block_id": "p995-b7", - "global_id": 30273, - "bbox": [ - 85.74, - 451.73, - 149.29, - 480.08 - ], - "text": "878\nAnalog filters, 261\nAnalog signals, 133", - "type": "text" - }, - { - "block_id": "p995-b8", - "global_id": 30274, - "bbox": [ - 85.74, - 482.31, - 246.45, - 541.26 - ], - "text": "defined, 78\ndigital processing of, 547–54\nproperties of, 78\nAnalog systems, 109, 135, 261\nAnalog-to-digital (A/D) conversion, 799–802, 831\nAnalog-to-digital converters (ADC)", - "type": "text" - }, - { - "block_id": "p995-b9", - "global_id": 30275, - "bbox": [ - 85.74, - 543.48, - 175.42, - 582.03 - ], - "text": "bit number, 801–2\nbit rate, 801–2\nAngle modulation, 736, 763\nAngles", - "type": "text" - }, - { - "block_id": "p995-b10", - "global_id": 30276, - "bbox": [ - 85.74, - 584.26, - 225.87, - 633.01 - ], - "text": "electronic calculators in computing, 8–11\nprincipal value of, 9\nAngular acceleration, 116\nAngular position, 116\nAngular velocity, 116", - "type": "text" - }, - { - "block_id": "p995-b11", - "global_id": 30277, - "bbox": [ - 306.92, - 207.51, - 416.8, - 245.71 - ], - "text": "Anti-aliasing filters, 537, 791, 834\nAnti-causal exponentials, 862–63\nAnti-causal signals, 81\nAperiodic signals, 133", - "type": "text" - }, - { - "block_id": "p995-b12", - "global_id": 30278, - "bbox": [ - 306.92, - 247.81, - 449.47, - 346.47 - ], - "text": "discrete-time Fourier integral and, 855–67\nFourier integral and, 680–89, 762\nproperties of, 78–82\nApparent frequency, 534–36, 792–96\nArs Magna (Cardano), 2–5\nAssociative property, 171, 283\nAsymptotic stability. See Internal stability.\nAudio signals, 713–14, 725, 746\nAutomatic position control system, 406–8\nAuxiliary conditions, 153", - "type": "text" - }, - { - "block_id": "p995-b13", - "global_id": 30279, - "bbox": [ - 314.88, - 348.57, - 436.67, - 356.54 - ], - "text": "differential equation solution and, 161", - "type": "text" - }, - { - "block_id": "p995-b14", - "global_id": 30280, - "bbox": [ - 306.92, - 373.72, - 473.45, - 401.84 - ], - "text": "Backward difference system, 258, 295, 519, 568–69\nBandlimited signals, 533, 788, 792, 802\nBandpass filters, 441–43, 542–44, 749, 882–83, 896", - "type": "text" - }, - { - "block_id": "p995-b15", - "global_id": 30281, - "bbox": [ - 306.92, - 403.95, - 429.2, - 452.23 - ], - "text": "group delay and, 726–27\nideal, 730–31, 882–83\npoles and zeros of H(s) and, 443\nBandstop filters, 441–42, 445, 545–46\nBandwidth, 628", - "type": "text" - }, - { - "block_id": "p995-b16", - "global_id": 30282, - "bbox": [ - 306.92, - 454.33, - 482.22, - 563.05 - ], - "text": "continuous-time systems and, 208–10\ndata truncation and, 751–53\nessential, 736, 758–59\nFourier transform and, 692, 706, 762–63\nBartlett window, 753–55\nBaseband signals, 737–40, 746–47, 749\nBasis signals, 651, 655, 668\nBasis vectors, 648\nBeat effect, 741\nBhaskar, 2\nBilateral Laplace transform, 330, 335–37, 445–55, 467", - "type": "text" - }, - { - "block_id": "p995-b17", - "global_id": 30283, - "bbox": [ - 306.92, - 565.16, - 432.88, - 583.2 - ], - "text": "properties of, 451–55\nBilateral z-transform, 431, 490, 554–63", - "type": "text" - }, - { - "block_id": "p995-b18", - "global_id": 30284, - "bbox": [ - 306.92, - 585.31, - 432.45, - 633.58 - ], - "text": "in discrete-time system analysis, 563\nproperties of, 559–60\nBilinear transformation, 569–70\nBinary digital signals, 799\nBlack box, 95, 119, 120", - "type": "text" - }, - { - "block_id": "p995-b19", - "global_id": 30285, - "bbox": [ - 501.19, - 656.12, - 516.13, - 666.22 - ], - "text": "975", - "type": "text" - } - ] - }, - { - "page_num": 996, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p996-b0", - "global_id": 30286, - "bbox": [ - 60.0, - 60.36, - 124.49, - 69.45 - ], - "text": "976\nIndex", - "type": "text" - }, - { - "block_id": "p996-b1", - "global_id": 30287, - "bbox": [ - 60.0, - 85.98, - 201.91, - 123.83 - ], - "text": "Blackman window, 755, 761\nBlock diagrams, 386–88, 405, 407, 408, 519\nBôcher, M., 620–21\nBode plots, 419–35", - "type": "text" - }, - { - "block_id": "p996-b2", - "global_id": 30288, - "bbox": [ - 60.0, - 125.82, - 261.84, - 193.56 - ], - "text": "constant of, 421\nfirst-order pole and, 424–27\npole at the origin and, 422–23\nsecond-order pole and, 426–35\nBombelli, Raphael, 3\nBonaparte, Napoleon, 347, 610–11\nBounded-input/bounded-output (BIBO) stability, 110, 135, 263", - "type": "text" - }, - { - "block_id": "p996-b3", - "global_id": 30289, - "bbox": [ - 60.0, - 195.56, - 245.06, - 293.19 - ], - "text": "of continuous-time systems, 196–97, 199–203, 222–23\nof discrete-time systems, 298–99, 301–4, 314, 526, 527\nfrequency response and, 412–13\ninternal stability relationship to, 199–203, 301–4\nof the Laplace transform, 371–73\nsignal transmission and, 721\nsteady-state response and, 418\nof the z-transform, 518\nButterfly signal flow graph, 825\nButterworth filters, 440, 551–54", - "type": "text" - }, - { - "block_id": "p996-b4", - "global_id": 30290, - "bbox": [ - 67.98, - 295.18, - 237.03, - 303.15 - ], - "text": "cascaded second-order sections for Butterworth filter", - "type": "text" - }, - { - "block_id": "p996-b5", - "global_id": 30291, - "bbox": [ - 67.98, - 305.15, - 151.43, - 333.04 - ], - "text": "realization, 461–63\nMATLAB on, 459–63\ntransformation of, 571–72", - "type": "text" - }, - { - "block_id": "p996-b6", - "global_id": 30292, - "bbox": [ - 60.0, - 348.29, - 239.71, - 436.04 - ], - "text": "Canonic direct realization. See Direct form II realization\nCardano, Gerolamo, 2–5\nCartesian form, 8–15\nCascade realization, 394, 526, 920, 923\nCascade systems, 190, 192, 372–73\nCascaded RC filters, 461–62\nCausal exponentials, 861–62\nCausal signals, 81, 83, 134\nCausal sinusoidal input", - "type": "text" - }, - { - "block_id": "p996-b7", - "global_id": 30293, - "bbox": [ - 60.0, - 438.04, - 183.11, - 465.93 - ], - "text": "in continuous-time systems, 418–19\nin discrete-time systems, 527\nCausal systems, 104–6, 135, 263", - "type": "text" - }, - { - "block_id": "p996-b8", - "global_id": 30294, - "bbox": [ - 60.0, - 467.92, - 174.22, - 505.78 - ], - "text": "properties of, 104–6\nzero-state response and, 172, 283\nCayley–Hamilton theorem, 910–12\nCharacteristic equations", - "type": "text" - }, - { - "block_id": "p996-b9", - "global_id": 30295, - "bbox": [ - 60.0, - 507.77, - 193.28, - 555.6 - ], - "text": "of continuous-time systems, 153–55\nof discrete-time systems, 271, 273, 309\nof a matrix, 910–12, 933\nCharacteristic functions, 192\nCharacteristic modes", - "type": "text" - }, - { - "block_id": "p996-b10", - "global_id": 30296, - "bbox": [ - 67.98, - 557.59, - 261.26, - 565.56 - ], - "text": "of continuous-time systems, 153–55, 162–65, 167, 170, 196,", - "type": "text" - }, - { - "block_id": "p996-b11", - "global_id": 30297, - "bbox": [ - 60.0, - 567.55, - 265.01, - 595.44 - ], - "text": "198–99, 203–6\nof discrete-time systems, 271–74, 278–79, 297–301, 305, 313\nCharacteristic polynomials", - "type": "text" - }, - { - "block_id": "p996-b12", - "global_id": 30298, - "bbox": [ - 67.98, - 597.43, - 255.28, - 635.29 - ], - "text": "of continuous-time systems, 153–56, 164, 166, 202–3, 220\nof discrete-time systems, 271, 274–75, 279, 303–4\nof the Laplace transform, 371–72\nof the z-transform, 518", - "type": "text" - }, - { - "block_id": "p996-b13", - "global_id": 30299, - "bbox": [ - 281.17, - 85.96, - 343.37, - 93.93 - ], - "text": "Characteristic roots", - "type": "text" - }, - { - "block_id": "p996-b14", - "global_id": 30300, - "bbox": [ - 289.14, - 95.98, - 486.41, - 103.95 - ], - "text": "of continuous-time systems, 153–56, 162, 166, 198–203, 206,", - "type": "text" - }, - { - "block_id": "p996-b15", - "global_id": 30301, - "bbox": [ - 289.14, - 105.94, - 484.19, - 123.93 - ], - "text": "209, 211–12, 214–17, 222–24\nof discrete-time systems, 271, 273–75, 297, 299–301, 303–5,", - "type": "text" - }, - { - "block_id": "p996-b16", - "global_id": 30302, - "bbox": [ - 281.17, - 125.92, - 426.83, - 163.94 - ], - "text": "309, 314\ninvariance of, 942–43\nof a matrix, 911, 932–33\nCharacteristic values. See Characteristic roots", - "type": "text" - }, - { - "block_id": "p996-b17", - "global_id": 30303, - "bbox": [ - 281.17, - 165.93, - 424.17, - 213.98 - ], - "text": "Characteristic vectors, 910\nChebyshev filters, 440, 463–66\nCircular convolution, 819–20, 821\nClearing fractions, 26–27, 32–33, 342–43\nClosed Loop systems. See Feedback systems", - "type": "text" - }, - { - "block_id": "p996-b18", - "global_id": 30304, - "bbox": [ - 281.17, - 215.89, - 476.65, - 254.0 - ], - "text": "Coherent demodulation. See Synchronous demodulation\nCoefficients of Fourier series, computation, 595–98\nColumn vectors, 36\nCommutative property", - "type": "text" - }, - { - "block_id": "p996-b19", - "global_id": 30305, - "bbox": [ - 281.17, - 256.05, - 465.24, - 334.14 - ], - "text": "of the convolution integral, 170, 173, 181, 191–92\nof the convolution sum, 283\nCompact disc (CD), 801\nCompact form of Fourier series, 597–98, 599, 600, 604–7\nComplex factors of Q(x), 29\nComplex frequency, 89–91\nComplex inputs, 177, 297\nComplex numbers, 1–15, 54", - "type": "text" - }, - { - "block_id": "p996-b20", - "global_id": 30306, - "bbox": [ - 281.17, - 336.2, - 397.29, - 464.39 - ], - "text": "algebra of, 5–15\narithmetical operations for, 12–15\nconjugates of, 6–7\nhistorical note, 1–5\nlogarithms of, 15\norigins of, 2–5\nstandard forms of, 14–15\nuseful identities, 7–8\nworking with, 13–14\nComplex poles, 395, 432, 497, 542\nComplex roots, 154–56, 274–76\nComplex signals, 94–95\nConjugate symmetry", - "type": "text" - }, - { - "block_id": "p996-b21", - "global_id": 30307, - "bbox": [ - 281.17, - 466.44, - 465.22, - 564.57 - ], - "text": "of the discrete Fourier transform, 818–19\nof the discrete-time Fourier transform, 858–59, 867–68\nof the Fourier transform, 684, 703\nConjugation, 684, 703\nConstants, 54, 98, 100, 103, 130, 422\nConstant-parameter systems. See Time-invariant systems\nContinuous functions, 858\nContinuous-time filters, 455–63\nContinuous-time Fourier transform (CTFT), 867, 884–85\nContinuous-time signals, 107–8, 135", - "type": "text" - }, - { - "block_id": "p996-b22", - "global_id": 30308, - "bbox": [ - 281.17, - 566.62, - 422.74, - 614.67 - ], - "text": "defined, 78\ndiscrete-time systems and, 238\nFourier series and, 593–679\nFourier transform and, 678–769, 680–775\nContinuous-time systems, 135, 150–236", - "type": "text" - }, - { - "block_id": "p996-b23", - "global_id": 30309, - "bbox": [ - 289.14, - 616.71, - 413.36, - 634.71 - ], - "text": "analog systems compared with, 261\ndifferential equations of, 161, 196, 213", - "type": "text" - } - ] - }, - { - "page_num": 997, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p997-b0", - "global_id": 30310, - "bbox": [ - 450.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Index\n977", - "type": "text" - }, - { - "block_id": "p997-b1", - "global_id": 30311, - "bbox": [ - 85.74, - 85.98, - 228.08, - 203.77 - ], - "text": "discrete-time systems compared with, 261\nexternal input, response to, 168–96\nfrequency response of, 412–18, 732–33\ninternal conditions, response to, 151–63\nintuitive insights into, 189–90, 203–5\nLaplace transform, 330–487\nPeriodic inputs and, 637–641\nproperties of, 107–8\nsignal transmission through, 721–29\nstability of, 196–203, 222–23\nstate equations for, 915–16\nControl systems, 404–12", - "type": "text" - }, - { - "block_id": "p997-b2", - "global_id": 30312, - "bbox": [ - 85.74, - 205.78, - 203.71, - 243.71 - ], - "text": "analysis of, 406–12\ndesign specifications, 411\nstep input and, 407–9\nControllability/observability, 123–24", - "type": "text" - }, - { - "block_id": "p997-b3", - "global_id": 30313, - "bbox": [ - 85.74, - 245.72, - 237.17, - 283.65 - ], - "text": "of continuous-time systems, 197, 200–2, 223\nof discrete-time systems, 303, 965–68\nin state-space analysis, 947–53, 961\nConvergence", - "type": "text" - }, - { - "block_id": "p997-b4", - "global_id": 30314, - "bbox": [ - 85.74, - 285.66, - 210.19, - 333.57 - ], - "text": "abscissa of, 336\nof Fourier series, 613–14\nto the mean, 613, 614\nregion of. See region of convergence\nConvolution, 507–9", - "type": "text" - }, - { - "block_id": "p997-b5", - "global_id": 30315, - "bbox": [ - 85.74, - 335.58, - 258.93, - 443.39 - ], - "text": "with an impulse, 283\nof the bilateral z-transform, 560\ncircular, 819–20, 821\ndiscrete-time, 311–12\nfast, 821, 886\nfrequency. See Frequency convolution\nof the Fourier transform, 714–16\nlinear, 821\nperiodic, 886\ntime. See Time convolution\nConvolution integral, 170–93, 222, 282, 288, 313, 722", - "type": "text" - }, - { - "block_id": "p997-b6", - "global_id": 30316, - "bbox": [ - 85.74, - 445.41, - 234.95, - 483.33 - ], - "text": "explanation for use, 189–90\ngraphical understanding of, 178–90, 217–20\nproperties of, 170–72\nConvolution sum, 282–86, 313", - "type": "text" - }, - { - "block_id": "p997-b7", - "global_id": 30317, - "bbox": [ - 85.74, - 485.35, - 209.05, - 593.15 - ], - "text": "graphical procedure for, 288–93\nproperties of, 282–83\nfrom a table, 285–86\nConvolution table, 175–76\nCooley, J. W., 824\nCorner frequency, 424\nCramer’s rule, 23–25, 40, 51, 379, 385\nCritically damped systems, 409, 410\nCubic equations, 2–3, 58\nCustom filter function, 310–11\nCutoff frequency, 208, 209", - "type": "text" - }, - { - "block_id": "p997-b8", - "global_id": 30318, - "bbox": [ - 85.74, - 609.34, - 179.03, - 627.29 - ], - "text": "Damping coefficient, 115–18\nDashpots", - "type": "text" - }, - { - "block_id": "p997-b9", - "global_id": 30319, - "bbox": [ - 93.72, - 629.3, - 137.55, - 647.26 - ], - "text": "linear, 115\ntorsional, 116", - "type": "text" - }, - { - "block_id": "p997-b10", - "global_id": 30320, - "bbox": [ - 306.92, - 85.97, - 452.22, - 166.94 - ], - "text": "Data truncations, 749–55, 763\nDecades, 422\nDecibels, 421\nDecimation-in-frequency algorithm, 824, 827\nDecimation-in-time algorithm, 825–27\nDecomposition, 99–100, 151\nDelayed impulse, 168\nDemodulation, 714", - "type": "text" - }, - { - "block_id": "p997-b11", - "global_id": 30321, - "bbox": [ - 306.92, - 169.4, - 429.13, - 271.23 - ], - "text": "of amplitude modulation, 744–46\nof DSB-SC signals, 739–41\nsynchronous, 743–44\nDepressed cubic equation, 58\nDerivative formulas, 56\nDescartes, René, 2\nDetection. See Demodulation\nDeterministic signals, 82, 134\nDiagonal matrices, 37\nDifference equations, 259–60, 265–70", - "type": "text" - }, - { - "block_id": "p997-b12", - "global_id": 30322, - "bbox": [ - 306.92, - 273.7, - 492.43, - 375.53 - ], - "text": "causality condition in, 265–66\nclassical solution of, 298\ndifferential equation kinship with, 260\nfrequency response, 532\norder of, 260\nrecursive and non-recursive forms of, 259\nrecursive solution of, 266–70\nsinusoidal response of difference equation systems, 528\nz-transform solution of, 488, 510–19, 574\nDifferential equations, 161", - "type": "text" - }, - { - "block_id": "p997-b13", - "global_id": 30323, - "bbox": [ - 306.92, - 377.99, - 465.19, - 417.24 - ], - "text": "classical solution of, 196\ndifference equation kinship with, 260\nLaplace transform solution of, 346–48, 360–73\nDifferentiators", - "type": "text" - }, - { - "block_id": "p997-b14", - "global_id": 30324, - "bbox": [ - 306.92, - 419.7, - 449.07, - 490.25 - ], - "text": "digital, 256–58\nideal, 369–71, 373, 416–17\nDigital differentiator example, 258–59\nDigital filters, 108, 238, 261–62\nDigital integrators, 258–59\nDigital processing of analog signals, 547–53\nDigital signals, 135, 797–99", - "type": "text" - }, - { - "block_id": "p997-b15", - "global_id": 30325, - "bbox": [ - 306.92, - 492.71, - 466.19, - 615.41 - ], - "text": "advantages of, 261–62\nbinary, 799–801\ndefined, 78\nL-ary, 799\nproperties of, 78\nSee also Analog-to-digital conversion\nDigital systems, 109, 135, 261\nDirac definition of an impulse, 88, 134\nDirac delta train, 696–97\nDirac, P.A.M., 86\nDirect discrete Fourier transform (DFT), 808, 857\nDirect form I (DFI) realization", - "type": "text" - }, - { - "block_id": "p997-b16", - "global_id": 30326, - "bbox": [ - 314.88, - 617.86, - 460.23, - 646.69 - ], - "text": "Laplace transform and, 390–91, 394\nz-transform and, 521\nSee also Transposed direct form II realization", - "type": "text" - } - ] - }, - { - "page_num": 998, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p998-b0", - "global_id": 30327, - "bbox": [ - 60.0, - 60.36, - 124.49, - 69.45 - ], - "text": "978\nIndex", - "type": "text" - }, - { - "block_id": "p998-b1", - "global_id": 30328, - "bbox": [ - 60.0, - 85.98, - 238.85, - 93.95 - ], - "text": "Direct form II (DFII) realization, 920–25, 954, 965, 967", - "type": "text" - }, - { - "block_id": "p998-b2", - "global_id": 30329, - "bbox": [ - 60.0, - 95.98, - 254.89, - 154.03 - ], - "text": "Laplace transform and, 391, 398\nz-transform and, 520–22, 525\nDirect Fourier transform, 683, 702–3, 762\nDirect z-transform, 488–592\nDirichlet conditions, 612, 614, 686\nDiscrete Fourier transform (DFT), 659, 805–23, 827–34, 835", - "type": "text" - }, - { - "block_id": "p998-b3", - "global_id": 30330, - "bbox": [ - 60.0, - 156.08, - 225.23, - 334.29 - ], - "text": "aliasing and leakage and, 805–6\napplications of, 820–23\ncomputing Fourier transform, 812–18\nderivation of, 807–10\ndetermining filter output, 822–23\ndirect, 808, 857\ndiscrete-time Fourier transform and, 885–86, 898\ninverse, 808, 835, 857\nMATLAB on, 827–34\npicket fence effect and, 807\npoints of discontinuity, 807\nproperties of, 818–20\nzero padding and, 810–11, 829–30\nDiscrete-time complex exponentials, 252\nDiscrete-time convolution, 311–12\nDiscrete-time exponentials, 247–49\nDiscrete-time Fourier integral, 855–67\nDiscrete-time Fourier series (DTFS), 845–55", - "type": "text" - }, - { - "block_id": "p998-b4", - "global_id": 30331, - "bbox": [ - 60.0, - 336.34, - 216.82, - 394.38 - ], - "text": "computation of, 885–86\nMATLAB on, 889–97\nof periodic gate function, 853–55\nperiodic signals and, 846–47, 898\nof sinusoids, 849–52\nDiscrete-time Fourier transform (DTFT), 857–88", - "type": "text" - }, - { - "block_id": "p998-b5", - "global_id": 30332, - "bbox": [ - 60.0, - 396.43, - 265.51, - 534.58 - ], - "text": "of accumulator systems, 875–76\nof anti-causal exponentials, 862–63\nof causal exponentials, 861–62\ncontinuous-time Fourier transform and, 883–86\nexistence of, 859, 886\ninverse, 886\nlinear time-invariant discrete-time system analysis by, 879–80\nMATLAB on, 889–97\nphysical appreciation of, 859\nproperties of, 867–78\nof rectangular pulses, 863–65\ntable of, 860\nz-transform connection with, 866–67, 886–88, 898\nDiscrete-time signals, 78, 79, 107–8, 133, 237–53", - "type": "text" - }, - { - "block_id": "p998-b6", - "global_id": 30333, - "bbox": [ - 60.0, - 536.63, - 178.44, - 604.68 - ], - "text": "defined, 78\nFourier analysis of, 845–907\ninherently bandlimited, 533\nsize of, 238–40\nuseful models, 245–53\nuseful operations, 240–45\nDiscrete-time systems, 135, 237–329", - "type": "text" - }, - { - "block_id": "p998-b7", - "global_id": 30334, - "bbox": [ - 67.98, - 606.73, - 212.98, - 634.73 - ], - "text": "classification of, 262–64\ncontrollability/observability of, 303, 965–68\ndifference equations of, 259–60, 265–70, 298", - "type": "text" - }, - { - "block_id": "p998-b8", - "global_id": 30335, - "bbox": [ - 281.17, - 85.97, - 455.42, - 205.11 - ], - "text": "discrete-time Fourier transform analysis of, 878–83\nexamples of, 253–65\nexternal input, response to, 280–98\nfrequency response of, 526–38\ninternal conditions, response to, 270–76\nintuitive insights into, 305–6\nproperties of, 107–8, 264–65\nstability of, 263, 298–305, 314\nstate-space analysis of, 953–64\nz-transform analysis of, 488–592\nDistinct factors of Q(x), 27\nDistortionless transmission, 724–28, 730, 763, 880–82", - "type": "text" - }, - { - "block_id": "p998-b9", - "global_id": 30336, - "bbox": [ - 281.17, - 207.25, - 473.55, - 255.64 - ], - "text": "bandpass systems and, 726–27, 881–82\nmeasure of delay variation, 881\nDistributive property, 171, 283\nDivision of complex numbers, 12–14\nDouble-sideband, suppressed-carrier (DSB-SC) modulation,", - "type": "text" - }, - { - "block_id": "p998-b10", - "global_id": 30337, - "bbox": [ - 281.17, - 257.64, - 405.37, - 295.93 - ], - "text": "737–41, 742, 746–49\nDownsampling, 243–44\nDuality, 703–4\nDynamic systems, 103–4, 134–35, 263", - "type": "text" - }, - { - "block_id": "p998-b11", - "global_id": 30338, - "bbox": [ - 281.17, - 313.45, - 398.85, - 361.84 - ], - "text": "Eigenfunctions, 193\nEigenvalues. See Characteristic roots\nEigenvectors, 910\nEinstein, Albert, 348\nElectrical systems, 95–96, 111–14", - "type": "text" - }, - { - "block_id": "p998-b12", - "global_id": 30339, - "bbox": [ - 281.17, - 363.98, - 427.49, - 533.66 - ], - "text": "Laplace transform analysis of, 373–85, 467\nstate equations for, 916–19\nElectromechanical systems, 118–19\nElectronic calculators, 8–11\nEnergy signals, 67, 82, 134, 239–40\nEnergy spectral density, 734, 763\nEnvelope delay. See Group delay\nEnvelope detector, 743–45\nEquilibrium states, 196, 198\nError signals, 650–51\nError vectors, 642\nEssential bandwidth, 736, 758–59\nEuler, Leonhard, 2, 3\nEuler’s formula, 5–6, 45, 252\nEven component of a signal, 93–95\nEven functions, 92–93, 134\nEverlasting exponentials", - "type": "text" - }, - { - "block_id": "p998-b13", - "global_id": 30340, - "bbox": [ - 281.17, - 535.8, - 441.47, - 604.41 - ], - "text": "continuous-time systems and, 189, 193–95, 222\ndiscrete-time systems and, 296–97, 313\nFourier series and, 637, 638, 641\nFourier transform and, 687\nLaplace transform and, 367–68, 412, 419\nEverlasting signals, 81, 134\nExponential Fourier series, 621–37, 661, 803", - "type": "text" - }, - { - "block_id": "p998-b14", - "global_id": 30341, - "bbox": [ - 289.14, - 606.54, - 378.82, - 634.73 - ], - "text": "periodic inputs and, 637–41\nreasons for using, 640\nsymmetry effect on, 630–32", - "type": "text" - } - ] - }, - { - "page_num": 999, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p999-b0", - "global_id": 30342, - "bbox": [ - 450.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Index\n979", - "type": "text" - }, - { - "block_id": "p999-b1", - "global_id": 30343, - "bbox": [ - 85.74, - 85.98, - 249.89, - 123.83 - ], - "text": "Exponential Fourier spectra, 624–32, 664, 667, 668\nExponential functions, 89–91, 134\nExponential input, 193, 296\nExponentials", - "type": "text" - }, - { - "block_id": "p999-b2", - "global_id": 30344, - "bbox": [ - 85.74, - 125.82, - 238.26, - 233.42 - ], - "text": "computation of matrix, 922–913\ndiscrete-time, 247–49\ndiscrete-time complex, 252\neverlasting. See Everlasting exponentials\nmatrix, 968–69\nmonotonic, 20–22, 90, 91, 134\nsinusoid varying, 22–23, 90, 134\nsinusoids expressed in, 20\nExposition du système du monde (Laplace), 346\nExternal description of a system, 119–20, 135\nExternal input", - "type": "text" - }, - { - "block_id": "p999-b3", - "global_id": 30345, - "bbox": [ - 85.74, - 235.41, - 285.11, - 263.3 - ], - "text": "continuous-time system response to, 168–96\ndiscrete-time system response to, 280–98\nExternal stability. See Bounded-input/bounded-output stability", - "type": "text" - }, - { - "block_id": "p999-b4", - "global_id": 30346, - "bbox": [ - 85.74, - 278.64, - 269.93, - 296.58 - ], - "text": "Fast convolution, 821, 886\nFast Fourier transform (FFT), 659, 811, 821, 824–27, 835", - "type": "text" - }, - { - "block_id": "p999-b5", - "global_id": 30347, - "bbox": [ - 85.74, - 298.57, - 250.98, - 336.43 - ], - "text": "computations reduced by, 824\ndiscrete-time Fourier series and, 847\ndiscrete-time Fourier transform and, 885–86, 898\nFeedback systems", - "type": "text" - }, - { - "block_id": "p999-b6", - "global_id": 30348, - "bbox": [ - 93.72, - 338.41, - 239.15, - 346.38 - ], - "text": "Laplace transform and, 386–88, 392–95, 399,", - "type": "text" - }, - { - "block_id": "p999-b7", - "global_id": 30349, - "bbox": [ - 85.74, - 348.37, - 210.3, - 386.23 - ], - "text": "404–12\nz-transform and, 521\nFeedforward connections, 392–94, 403\nFiltering", - "type": "text" - }, - { - "block_id": "p999-b8", - "global_id": 30350, - "bbox": [ - 85.74, - 388.23, - 218.23, - 436.05 - ], - "text": "discrete Fourier transform and, 821–23\nMATLAB on, 308–10\nselective, 748–49\ntime constant and, 207–8\nFilters", - "type": "text" - }, - { - "block_id": "p999-b9", - "global_id": 30351, - "bbox": [ - 93.72, - 438.04, - 225.22, - 635.29 - ], - "text": "analog, 261\nanti-aliasing, 537, 791, 834\nbandpass, 441–43\nbandstop, 441–42, 445, 545–46\nButterworth. See Butterworth Filters\ncascaded RC, 461–62\nChebyshev, 440, 463–66\ncontinuous-time, 455–63\ncustom function, 310–11\ndigital, 108, 238, 261–62\nfinite impulse response, 524, 892–97\nfirst-order hold, 785\nfrequency response of, 412–18\nhighpass, 443, 445, 542, 730–31, 882–83\nIdeal. See Ideal filters\nimpulse invariance criterion of, 548\ninfinite impulse response, 524, 565–74\nlowpass, 439–41\nlowpass. See Lowpass filters\nnotch, 441–43, 540, 545–46", - "type": "text" - }, - { - "block_id": "p999-b10", - "global_id": 30352, - "bbox": [ - 306.92, - 85.68, - 466.61, - 224.17 - ], - "text": "poles and zeros of H(s) and, 436–45\npractical, 444–45, 882–83\nsharp cutoff, 748\nwindows in design of, 755\nzero-order hold, 785\nFinal value theorem, 359–61, 508\nFinite impulse response (FIR) filters, 524, 892–97\nFinite-duration signals, 333\nFinite-memory systems, 104\nFirst-order factors, method of, 497\nFirst-order hold filters, 785\nFolding frequency, 789–91, 793, 795, 817\nFor-loops, 216–18\nForced response", - "type": "text" - }, - { - "block_id": "p999-b11", - "global_id": 30353, - "bbox": [ - 306.92, - 226.21, - 412.09, - 264.23 - ], - "text": "difference equations and, 298\ndifferential equations and, 198\nForward amplifiers, 405–6\nFourier integral, 722", - "type": "text" - }, - { - "block_id": "p999-b12", - "global_id": 30354, - "bbox": [ - 306.92, - 266.27, - 422.92, - 294.27 - ], - "text": "aperiodic signal and, 680–89, 762\ndiscrete-time, 855–67\nFourier series, 593–679", - "type": "text" - }, - { - "block_id": "p999-b13", - "global_id": 30355, - "bbox": [ - 306.92, - 296.32, - 466.54, - 404.43 - ], - "text": "compact form of, 597–98, 599, 600, 604–7\ncomputing the coefficients of, 595–98\ndiscrete time. See Discrete-time Fourier series\nexistence of, 612–13\nexponential. See Exponential Fourier series\ngeneralized, 641–59, 668\nLegendre, 656–57\nlimitations of analysis method, 641\ntrigonometric. See Trigonometric Fourier series\nwaveshaping in, 615–17\nFourier spectrum, 598–607, 777", - "type": "text" - }, - { - "block_id": "p999-b14", - "global_id": 30356, - "bbox": [ - 306.92, - 406.48, - 434.97, - 444.49 - ], - "text": "exponential, 624–32, 664, 667, 668\nnature of, 858–59\nof a periodic signal, 848–55\nFourier transform, 680–755, 778, 802–3", - "type": "text" - }, - { - "block_id": "p999-b15", - "global_id": 30357, - "bbox": [ - 314.89, - 446.54, - 441.58, - 464.52 - ], - "text": "continuous-time, 867, 883–86\ndiscrete. See Discrete Fourier transform", - "type": "text" - }, - { - "block_id": "p999-b16", - "global_id": 30358, - "bbox": [ - 306.92, - 466.43, - 483.21, - 584.64 - ], - "text": "discrete-time. See Discrete-time Fourier transform\ndirect, 683, 702–3, 762\nexistence of, 685–86\nfast. See fast Fourier transform\ninterpolation and, 785\ninverse, 683, 693–95, 699, 762, 786–87\nphysical appreciation of, 687–89\nproperties of, 701–21\nuseful functions of, 689–701\nFourier transform pairs, 683, 700\nFourier, Baron Jean-Baptiste-Joseph, 610–12\nFractions, 1–2", - "type": "text" - }, - { - "block_id": "p999-b17", - "global_id": 30359, - "bbox": [ - 306.92, - 586.69, - 416.28, - 614.69 - ], - "text": "clearing, 26–27, 32–34, 342–43\npartial. See Partial fractions\nFrequency", - "type": "text" - }, - { - "block_id": "p999-b18", - "global_id": 30360, - "bbox": [ - 314.89, - 616.73, - 398.13, - 634.72 - ], - "text": "apparent, 534–36, 793–94\ncomplex, 89–91", - "type": "text" - } - ] - }, - { - "page_num": 1000, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1000-b0", - "global_id": 30361, - "bbox": [ - 60.0, - 60.36, - 124.49, - 69.45 - ], - "text": "980\nIndex", - "type": "text" - }, - { - "block_id": "p1000-b1", - "global_id": 30362, - "bbox": [ - 60.0, - 85.9, - 132.26, - 93.95 - ], - "text": "Frequency (continued)", - "type": "text" - }, - { - "block_id": "p1000-b2", - "global_id": 30363, - "bbox": [ - 60.0, - 96.17, - 175.6, - 206.18 - ], - "text": "corner, 424\ncutoff, 208, 209\nfolding, 789–91, 793, 795, 817\nfundamental, 594, 609–10, 846\nnegative, 626–28\nneper, 91\nradian, 16, 91, 594\nreduction in range, 535\nof sinusoids, 16\ntime delay variation with, 724–25\nFrequency convolution", - "type": "text" - }, - { - "block_id": "p1000-b3", - "global_id": 30364, - "bbox": [ - 60.0, - 208.41, - 216.17, - 267.4 - ], - "text": "of the bilateral Laplace transform, 452\nof the discrete-time Fourier transform, 875–76\nof the Fourier transform, 714–16\nof the Laplace transform, 357\nFrequency differentiation, 869\nFrequency domain analysis, 368, 722–23, 848", - "type": "text" - }, - { - "block_id": "p1000-b4", - "global_id": 30365, - "bbox": [ - 60.0, - 269.64, - 203.12, - 338.82 - ], - "text": "of electrical networks, 374–78\nof the Fourier series, 598, 601\ntwo-dimensional view and, 732–33\nSee also Laplace transform\nFrequency inversion, 706\nFrequency resolution, 807, 810–12, 815, 817\nFrequency response, 724", - "type": "text" - }, - { - "block_id": "p1000-b5", - "global_id": 30366, - "bbox": [ - 60.0, - 341.06, - 211.43, - 451.06 - ], - "text": "Bode plots and, 419–22\nof continuous-time systems, 412–18, 732–33\nof discrete-time systems, 526–38\nMATLAB on, 456–57, 531–32\nperiodic nature of, 532–36\nfrom pole-zero location, 538–47\npole-zero plots and, 566–68\npoles and zeros of H(s) and, 436–39\ntransfer function from, 435\nFrequency reversal, 868–69\nFrequency shifting", - "type": "text" - }, - { - "block_id": "p1000-b6", - "global_id": 30367, - "bbox": [ - 60.0, - 453.29, - 232.69, - 542.89 - ], - "text": "of the bilateral Laplace transform, 451\nof the discrete Fourier transform, 819\nof the discrete-time Fourier transform, 871–74\nof the Fourier transform, 711–13\nof the Laplace transform, 353–54\nFrequency spectra, 598, 601\nFrequency-division multiplexing (FDM), 714, 749–50\nFunction M-files, 214–15\nFunctions", - "type": "text" - }, - { - "block_id": "p1000-b7", - "global_id": 30368, - "bbox": [ - 67.98, - 545.13, - 144.9, - 634.73 - ], - "text": "characteristic, 193\ncontinuous, 858\neven, 92–93, 134\nexponential, 89–91, 134\nimproper, 25–26, 34\ninterpolation, 690\nMATLAB on, 126–33\nodd, 92–95, 134\nproper, 25–27", - "type": "text" - }, - { - "block_id": "p1000-b8", - "global_id": 30369, - "bbox": [ - 281.17, - 85.97, - 448.75, - 133.85 - ], - "text": "rational, 25–29, 338\nsingularity, 89\nFundamental band, 533, 534, 537, 793\nFundamental frequency, 594, 609–10, 846\nFundamental period, 79, 133, 239–40, 593, 595, 846", - "type": "text" - }, - { - "block_id": "p1000-b9", - "global_id": 30370, - "bbox": [ - 281.17, - 149.98, - 410.21, - 217.84 - ], - "text": "Gain enhancement by poles, 437–38\nGauss, Karl Friedrich, 3–4\nGeneralized Fourier series, 641–59, 668\nGeneralized linear phase (GLP), 726–27\nGibbs phenomenon, 619–21, 661–63\nGibbs, Josiah Willard, 620–21\nGraphical interpretation", - "type": "text" - }, - { - "block_id": "p1000-b10", - "global_id": 30371, - "bbox": [ - 281.17, - 219.85, - 433.39, - 257.76 - ], - "text": "of convolution integral, 178–90, 217–20\nof convolution sum, 288–93\nGreatest common factor of frequencies, 609–10\nGroup delay, 725–28, 881", - "type": "text" - }, - { - "block_id": "p1000-b11", - "global_id": 30372, - "bbox": [ - 281.17, - 273.56, - 296.32, - 281.78 - ], - "text": "H(s)", - "type": "text" - }, - { - "block_id": "p1000-b12", - "global_id": 30373, - "bbox": [ - 281.17, - 283.88, - 441.55, - 381.67 - ], - "text": "filter design and, 436–45\nrealization of, 548–49\nSee also Transfer functions\nHalf-wave symmetry, 608\nHamming window, 754–55, 761\nHanning window, 754–55, 761\nHardware realization, 64, 95, 133\nHarmonic distortion, 634\nHarmonically related frequencies, 609\nHeaviside “cover-up” method, 27–30, 33–35, 341,", - "type": "text" - }, - { - "block_id": "p1000-b13", - "global_id": 30374, - "bbox": [ - 281.17, - 383.66, - 438.87, - 421.57 - ], - "text": "342–43, 497\nHeaviside, Oliver, 347–48, 612\nHighpass filters, 443, 445, 542, 745, 747, 882–83\nHomogeneity, 97–98", - "type": "text" - }, - { - "block_id": "p1000-b14", - "global_id": 30375, - "bbox": [ - 281.17, - 437.7, - 443.78, - 635.29 - ], - "text": "Ideal delay, 369, 416\nIdeal differentiators, 369–71, 373, 416–17\nIdeal filters, 730–33, 763, 785, 791, 834, 882–83\nIdeal integrators, 369, 370, 373, 400, 416–18\nIdeal interpolation, 786–87\nIdeal linear phase (ILP), 725, 727\nIdeal masses, 114\nIdentity matrices, 37\nIdentity systems, 109, 192, 263\nImaginary numbers, 1–5\nImpedance, 374–77, 379, 380, 382, 384, 387, 399\nImproper functions, 25–26, 34\nImpulse invariance criterion of filter design, 548\nImpulse matching, 164–66\nImpulse response matrix, 938\nIndefinite integrals, 57\nIndicator function. See Relational operators\nInertia, moment of, 116–18\nInfinite impulse response (IIR) filters, 524, 565–74\nInformation transmission rate, 209–10", - "type": "text" - } - ] - }, - { - "page_num": 1001, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1001-b0", - "global_id": 30376, - "bbox": [ - 450.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Index\n981", - "type": "text" - }, - { - "block_id": "p1001-b1", - "global_id": 30377, - "bbox": [ - 85.74, - 85.98, - 231.65, - 93.95 - ], - "text": "Initial conditions, 97–100, 102, 122, 134, 335", - "type": "text" - }, - { - "block_id": "p1001-b2", - "global_id": 30378, - "bbox": [ - 85.74, - 94.37, - 214.15, - 144.05 - ], - "text": "at 0−and 0+, 363–64\ncontinuous-time systems and, 158–61\ngenerators of, 376–83\nInitial value theorem, 359–61, 508\nInput, 64", - "type": "text" - }, - { - "block_id": "p1001-b3", - "global_id": 30379, - "bbox": [ - 85.74, - 146.1, - 212.6, - 254.28 - ], - "text": "complex, 177, 297\nexponential, 193, 296\nexternal. See External input\nin linear systems, 97\nmultiple, 178, 287–88\nramp, 410–11\nsinusoidal. See Sinusoidal input\nstep, 407–10\nInput–output description, 111–19\nInstantaneous systems, 103–4, 134, 263\nIntegrals", - "type": "text" - }, - { - "block_id": "p1001-b4", - "global_id": 30380, - "bbox": [ - 85.74, - 256.25, - 219.71, - 314.41 - ], - "text": "convolution. See Convolutional integral\ndiscrete-time Fourier, 855–67\nFourier. See Fourier integral\nindefinite, 57\nof matrices, 909–10\nIntegrators", - "type": "text" - }, - { - "block_id": "p1001-b5", - "global_id": 30381, - "bbox": [ - 85.74, - 316.46, - 238.81, - 364.51 - ], - "text": "digital, 258–59\nideal, 369, 370, 373, 400, 416–18\nsystem realization and, 400\nIntegro-differential equations, 360–73, 466, 488\nInterconnected systems", - "type": "text" - }, - { - "block_id": "p1001-b6", - "global_id": 30382, - "bbox": [ - 85.74, - 366.56, - 173.42, - 394.58 - ], - "text": "continuous-time, 190–93\ndiscrete-time, 294–97\nInternal conditions", - "type": "text" - }, - { - "block_id": "p1001-b7", - "global_id": 30383, - "bbox": [ - 85.74, - 396.62, - 246.02, - 424.64 - ], - "text": "continuous-time system response to, 151–63\ndiscrete-time system response to, 270–76\nInternal description of a system, 119–21, 135, 908", - "type": "text" - }, - { - "block_id": "p1001-b8", - "global_id": 30384, - "bbox": [ - 85.74, - 426.61, - 234.04, - 444.68 - ], - "text": "See also State-space description of a system\nInternal stability, 110, 135, 263", - "type": "text" - }, - { - "block_id": "p1001-b9", - "global_id": 30385, - "bbox": [ - 93.72, - 446.73, - 241.16, - 474.75 - ], - "text": "BIBO relationship to, 199–203, 301–4\nof continuous-time systems, 196–203, 222–23\nof discrete-time systems, 298–302, 305, 314,", - "type": "text" - }, - { - "block_id": "p1001-b10", - "global_id": 30386, - "bbox": [ - 85.74, - 476.74, - 188.23, - 514.77 - ], - "text": "526, 527\nof the Laplace transform, 372\nof the z-transform, 518\nInterpolation, 785–88", - "type": "text" - }, - { - "block_id": "p1001-b11", - "global_id": 30387, - "bbox": [ - 85.74, - 516.82, - 196.01, - 584.91 - ], - "text": "of discrete-time signals, 243–44\nideal, 786–87\nsimple, 785–86\nspectral, 804\nInterpolation formula, 779, 787\nInterpolation function, 690\nIntuitive insights", - "type": "text" - }, - { - "block_id": "p1001-b12", - "global_id": 30388, - "bbox": [ - 85.74, - 586.97, - 266.7, - 635.03 - ], - "text": "into continuous-time systems, 189–90, 203–12\ninto discrete-time systems, 305–6\ninto the Laplace transform, 367–68\nInverse continuous-time systems, 192–93\nInverse discrete Fourier transform (IDFT), 808, 827, 857", - "type": "text" - }, - { - "block_id": "p1001-b13", - "global_id": 30389, - "bbox": [ - 306.92, - 85.97, - 477.67, - 93.94 - ], - "text": "Inverse discrete-time Fourier transform (IDTFT), 886", - "type": "text" - }, - { - "block_id": "p1001-b14", - "global_id": 30390, - "bbox": [ - 306.92, - 95.95, - 491.87, - 133.86 - ], - "text": "of rectangular spectrum, 865–66\nInverse discrete-time systems, 294–95\nInverse Fourier transform, 683, 693–95, 699, 762, 786–87\nInverse Laplace transform, 333, 335, 445, 549", - "type": "text" - }, - { - "block_id": "p1001-b15", - "global_id": 30391, - "bbox": [ - 306.92, - 135.87, - 510.15, - 153.82 - ], - "text": "finding, 338–46\nInverse z-transform, 488–89, 491, 499, 500, 501, 510, 554, 555,", - "type": "text" - }, - { - "block_id": "p1001-b16", - "global_id": 30392, - "bbox": [ - 306.92, - 155.82, - 353.7, - 183.76 - ], - "text": "559\nfinding, 495\nInversion", - "type": "text" - }, - { - "block_id": "p1001-b17", - "global_id": 30393, - "bbox": [ - 306.92, - 185.77, - 424.48, - 223.68 - ], - "text": "frequency, 706\nmatrix, 40–42\nInvertible systems, 109–10, 135, 263\nIrrational numbers, 1–2", - "type": "text" - }, - { - "block_id": "p1001-b18", - "global_id": 30394, - "bbox": [ - 306.92, - 236.85, - 442.6, - 274.76 - ], - "text": "Kaiser window, 755, 760–62\nKelvin, Lord, 348\nKennelly-Heaviside atmosphere layer, 348\nKirchhoff’s laws, 95", - "type": "text" - }, - { - "block_id": "p1001-b19", - "global_id": 30395, - "bbox": [ - 306.92, - 276.77, - 408.52, - 304.71 - ], - "text": "current (KCL), 111, 213, 374\nvoltage (KVL), 111, 374\nKronecker delta functions, 245", - "type": "text" - }, - { - "block_id": "p1001-b20", - "global_id": 30396, - "bbox": [ - 314.88, - 306.73, - 432.23, - 314.7 - ], - "text": "bandlimited interpolation of, 787–88", - "type": "text" - }, - { - "block_id": "p1001-b21", - "global_id": 30397, - "bbox": [ - 306.92, - 327.78, - 428.89, - 365.78 - ], - "text": "L-ary digital signals, 799\nL’Hôpital’s rule, 58, 211, 690\nLagrange, Louis de, 347, 612, 613\nLaplace transform, 167, 330–487, 721", - "type": "text" - }, - { - "block_id": "p1001-b22", - "global_id": 30398, - "bbox": [ - 306.92, - 367.71, - 492.41, - 605.35 - ], - "text": "bilateral. See Bilateral Laplace transform\ndifferential equation solutions and, 346–48, 360–73\nelectrical network analysis and, 373–85, 467\nexistence of, 336–37\nFourier transform connection with, 699–701, 866\nintuitive interpretation of, 367–69\ninverse, 549, 938–39\nproperties of, 349–62\nstability of, 371–74\nstate equation solutions by, 927–33\nsystem realization and, 388–404\nunilateral, 333–36, 337, 338, 345, 360, 445, 467\nz-transform connection with, 488, 489, 491, 563–65\nLaplace transform pairs, 333\nLaplace, Marquis Pierre-Simon de, 346–47, 611, 612, 613\nLeakage, 751, 753–55, 763, 805–6\nLeft half plane (LHP), 91, 198–99, 202, 211, 223, 435\nLeft shift, 71, 73, 130, 134, 503, 509, 510, 512\nLeft-sided sequences, 555–56\nLegendre Fourier series, 656–57\nLeibniz, Gottfried Wilhelm, 801\nLinear convolution, 821\nLinear dashpots, 115\nLinear phase", - "type": "text" - }, - { - "block_id": "p1001-b23", - "global_id": 30399, - "bbox": [ - 314.88, - 607.36, - 445.09, - 635.29 - ], - "text": "distortionless transmission and, 725, 881\ngeneralized, 726–27\nideal, 725, 727", - "type": "text" - } - ] - }, - { - "page_num": 1002, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1002-b0", - "global_id": 30400, - "bbox": [ - 60.0, - 60.36, - 124.49, - 69.45 - ], - "text": "982\nIndex", - "type": "text" - }, - { - "block_id": "p1002-b1", - "global_id": 30401, - "bbox": [ - 60.0, - 85.9, - 139.68, - 93.95 - ], - "text": "Linear phase (continued)", - "type": "text" - }, - { - "block_id": "p1002-b2", - "global_id": 30402, - "bbox": [ - 60.0, - 96.08, - 170.09, - 134.39 - ], - "text": "physical description of, 707–9\nphysical explanation of, 870–71\nLinear springs, 114\nLinear systems, 97–101, 134", - "type": "text" - }, - { - "block_id": "p1002-b3", - "global_id": 30403, - "bbox": [ - 60.0, - 136.53, - 247.8, - 164.72 - ], - "text": "heuristic understanding of, 722–23\nresponse of, 98–100\nLinear time-invariant continuous-time (LTIC) systems. See", - "type": "text" - }, - { - "block_id": "p1002-b4", - "global_id": 30404, - "bbox": [ - 60.0, - 166.71, - 238.04, - 184.8 - ], - "text": "Continuous-time systems\nLinear time-invariant discrete-time (LTID) systems. See", - "type": "text" - }, - { - "block_id": "p1002-b5", - "global_id": 30405, - "bbox": [ - 60.0, - 186.79, - 250.89, - 245.31 - ], - "text": "Discrete-time systems\nLinear time-invariant (LTI) systems, 103, 194–95\nLinear time-invariant discrete-time (LTID) systems, 879–80\nLinear time-varying systems, 103\nLinear transformation of vectors, 36, 939–47, 961\nLinearity", - "type": "text" - }, - { - "block_id": "p1002-b6", - "global_id": 30406, - "bbox": [ - 60.0, - 247.46, - 204.22, - 356.54 - ], - "text": "of the bilateral Laplace transform, 451\nof the bilateral z-transform, 559\nconcept of, 97–98\nof the discrete Fourier transform, 818, 824\nof the discrete-time Fourier transform, 867\nof discrete-time systems, 262\nof the Fourier transform, 686–87, 824\nof the Laplace transform, 331–32\nof the z-transform, 489\nLog magnitude, 27, 422–24\nLoop currents", - "type": "text" - }, - { - "block_id": "p1002-b7", - "global_id": 30407, - "bbox": [ - 60.0, - 358.68, - 204.36, - 396.99 - ], - "text": "continuous-time systems and, 159–63, 175\nLaplace transform and, 375\nLower sideband (LSB), 738–39, 747\nLowpass filters, 439–41, 540–42", - "type": "text" - }, - { - "block_id": "p1002-b8", - "global_id": 30408, - "bbox": [ - 67.98, - 399.12, - 184.3, - 417.21 - ], - "text": "ideal, 730, 784–85, 788–89, 882–83\npoles and zeros of H(s) and, 436–45", - "type": "text" - }, - { - "block_id": "p1002-b9", - "global_id": 30409, - "bbox": [ - 60.0, - 434.78, - 111.21, - 442.75 - ], - "text": "M-files, 212–20", - "type": "text" - }, - { - "block_id": "p1002-b10", - "global_id": 30410, - "bbox": [ - 60.0, - 444.9, - 205.43, - 493.31 - ], - "text": "function, 214–15\nscript, 213–14, 218\nMaclaurin series, 6, 55\nMagnitude response. See Amplitude response\nMarginally stable systems", - "type": "text" - }, - { - "block_id": "p1002-b11", - "global_id": 30411, - "bbox": [ - 60.01, - 495.46, - 211.44, - 564.1 - ], - "text": "continuous-time, 198–200, 203, 211, 222–24\ndiscrete-time, 301–2, 304, 314\nLaplace transform, 373\nsignal transmission and, 721\nz-transform, 519\nMathematical models of systems, 95–96, 125\nMATLAB", - "type": "text" - }, - { - "block_id": "p1002-b12", - "global_id": 30412, - "bbox": [ - 67.98, - 566.23, - 213.49, - 614.65 - ], - "text": "on Butterworth filters, 459–63\ncalculator operations in, 43–45\non continuous-time filters, 455–63\non discrete Fourier transform, 827–34\non discrete-time Fourier series and transform,", - "type": "text" - }, - { - "block_id": "p1002-b13", - "global_id": 30413, - "bbox": [ - 67.98, - 616.64, - 199.04, - 634.73 - ], - "text": "889–97\non discrete-time systems/signals, 306–12", - "type": "text" - }, - { - "block_id": "p1002-b14", - "global_id": 30414, - "bbox": [ - 281.17, - 85.97, - 428.07, - 294.43 - ], - "text": "elementary operations in, 42–53\non filtering, 308–10\nFourier series applications in, 661–67\nFourier transform topics in, 755–62\nfrequency response plots, 531–32\non functions, 126–33\nimpulse invariance, 553\nimpulse response and, 167\non infinite-impulse response filters, 565–74\nM-files in, 212–20\nmatrix operations in, 49–53\nmultiple magnitude response curves, 544\npartial fraction expansion in, 53\nperiodic functions, 661–63\nphase spectrum, 664–67\npolynomial roots and, 157\nsimple plotting in, 46–48\nstate-space analysis in, 961–69\nvector operations in, 45–46\nzero-input response and, 157–58\nMatrices, 36–42", - "type": "text" - }, - { - "block_id": "p1002-b15", - "global_id": 30415, - "bbox": [ - 281.17, - 296.49, - 435.9, - 605.21 - ], - "text": "algebra of, 38–42\ncharacteristic equation of, 909–10, 933\ncharacteristic roots of, 932–33\ncomputing exponential of, 912–13\ndefinitions and properties of, 37–38\nderivatives of, 909–10\ndiagonal, 37\ndiagonalization of, 943–44\nequal, 37\nfunctions of, 911–12\nidentity, 37\nimpulse response, 938\nintegrals of, 909–10\ninversion of, 40–42\nMATLAB operations, 49–53\nnonsingular, 41\nsquare, 36, 37, 41\nstate transition, 936\nsymmetric, 37\ntranspose of, 37–38\nzero, 37\nMatrix exponentials, 968–69\nMatrix exponentiation, 968–69\nMechanical systems, 114–18\nMemory, systems and, 104, 263\nMemoryless systems. See Instantaneous systems\nMethod of residues, 27\nMichelson, Albert, 620–21\nMinimum phase systems, 435, 436\nModified partial fractions, 35, 496\nModulation, 713–14, 736–49", - "type": "text" - }, - { - "block_id": "p1002-b16", - "global_id": 30416, - "bbox": [ - 289.14, - 607.26, - 425.39, - 635.28 - ], - "text": "amplitude, 711–13, 736, 742–46, 762\nangle, 736, 763\nof the discrete-time Fourier transform, 872", - "type": "text" - } - ] - }, - { - "page_num": 1003, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1003-b0", - "global_id": 30417, - "bbox": [ - 450.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Index\n983", - "type": "text" - }, - { - "block_id": "p1003-b1", - "global_id": 30418, - "bbox": [ - 93.72, - 85.98, - 254.31, - 93.95 - ], - "text": "double-sideband, suppressed-carrier, 737–41, 742,", - "type": "text" - }, - { - "block_id": "p1003-b2", - "global_id": 30419, - "bbox": [ - 85.74, - 95.94, - 256.88, - 194.13 - ], - "text": "746–49\npulse-amplitude, 796\npulse-code, 796, 799\npulse-position, 796\npulse-width, 796\nsingle-sideband, 746–49\nMoment of inertia, 116–18\nMonotonic exponentials, 20–22, 90, 91, 134\nMultiple inputs, 178, 287–88\nMultiple-input, multiple-output (MIMO) systems, 98,", - "type": "text" - }, - { - "block_id": "p1003-b3", - "global_id": 30420, - "bbox": [ - 85.74, - 196.12, - 131.35, - 214.12 - ], - "text": "125, 908\nMultiplication", - "type": "text" - }, - { - "block_id": "p1003-b4", - "global_id": 30421, - "bbox": [ - 93.72, - 216.09, - 223.09, - 284.29 - ], - "text": "bilateral z-transform and, 560\nof complex numbers, 12–14\ndiscrete-time Fourier transform and, 869\nof a function by an impulse, 87\nmatrix, 38–40\nscalar, 38, 400–1, 505\nz-transform and, 506–7", - "type": "text" - }, - { - "block_id": "p1003-b5", - "global_id": 30422, - "bbox": [ - 85.74, - 300.92, - 216.56, - 338.96 - ], - "text": "Natural binary code (NBC), 799\nNatural modes. See Characteristic modes\nNatural numbers, 1\nNatural response", - "type": "text" - }, - { - "block_id": "p1003-b6", - "global_id": 30423, - "bbox": [ - 85.74, - 341.02, - 229.54, - 469.29 - ], - "text": "difference equations and, 298\ndifferential equations and, 196\nNegative feedback, 406\nNegative frequency, 626–28\nNegative numbers, 1–3, 45\nNeper frequency, 91\nNeutral equilibrium, 197, 198\nNewton, Sir Isaac, 2, 346–47\nNoise, 66, 151, 371, 417, 791, 797–99\nNonanticipative systems. See Causal systems\nNon-bandlimited signals, 792\nNoncausal signals, 81\nNoncausal systems, 104–7, 135, 263", - "type": "text" - }, - { - "block_id": "p1003-b7", - "global_id": 30424, - "bbox": [ - 85.74, - 471.34, - 280.23, - 609.64 - ], - "text": "properties of, 104–6\nreasons for studying, 106–7\nNon-invertible systems, 109–10, 135, 263\nNon-inverting amplifiers, 382\nNonlinear systems, 97–101, 134\nNonsingular matrices, 41\nNon-uniqueness, 533\nNormal-form equations, 915\nNorton theorem, 375\nNotch filters, 441–43, 540, 545–46. See also Bandstop filters\nNumerical integration, 131–33\nNyquist interval, 778, 779\nNyquist rate, 778–81, 788–89, 792, 795, 821\nNyquist samples, 778, 781, 782, 788, 792", - "type": "text" - }, - { - "block_id": "p1003-b8", - "global_id": 30425, - "bbox": [ - 85.74, - 626.19, - 234.24, - 644.27 - ], - "text": "Observability. See controllability/observability\nOctave, 422", - "type": "text" - }, - { - "block_id": "p1003-b9", - "global_id": 30426, - "bbox": [ - 306.92, - 85.97, - 438.47, - 145.25 - ], - "text": "Odd component of a signals, 93–95\nOdd functions, 92–95, 134\nOperational amplifiers, 382–83, 399, 467\nOrdinary numbers, 1–5\nOrthogonal signal space, 649–50\nOrthogonal signals, 668", - "type": "text" - }, - { - "block_id": "p1003-b10", - "global_id": 30427, - "bbox": [ - 306.92, - 147.54, - 441.3, - 237.62 - ], - "text": "energy of the sum of, 647\nsignal representation by set, 647–59\nOrthogonal vector space, 647–48\nOrthogonality, 622\nOrthonormal sets, 649\nOscillators, 203\nOutput, 64, 97\nOutput equations, 122, 124, 908, 930, 941\nOverdamped systems, 409–10", - "type": "text" - }, - { - "block_id": "p1003-b11", - "global_id": 30428, - "bbox": [ - 306.92, - 256.87, - 484.39, - 295.63 - ], - "text": "Paley–Wiener criterion, 444, 731–32, 788\nParallel realization, 393–94, 525–26, 921, 924–25\nParallel systems, 190, 387\nParseval’s theorem, 632, 651–52, 734–35, 755, 758–59,", - "type": "text" - }, - { - "block_id": "p1003-b12", - "global_id": 30429, - "bbox": [ - 306.92, - 297.62, - 357.48, - 315.85 - ], - "text": "876–78\nPartial fractions", - "type": "text" - }, - { - "block_id": "p1003-b13", - "global_id": 30430, - "bbox": [ - 314.88, - 318.14, - 499.8, - 336.38 - ], - "text": "expansion of, 25–35, 53\ninverse transform by partial fraction expansion and tables,", - "type": "text" - }, - { - "block_id": "p1003-b14", - "global_id": 30431, - "bbox": [ - 314.88, - 338.38, - 496.18, - 356.61 - ], - "text": "495–98\nLaplace transform and, 338–39, 341, 344, 362, 394, 395,", - "type": "text" - }, - { - "block_id": "p1003-b15", - "global_id": 30432, - "bbox": [ - 306.92, - 358.6, - 432.62, - 428.15 - ], - "text": "419, 454\nmodified, 35\nz-transform, 499\nPassbands, 441, 444–45, 748, 755\nPeak time, 409–10\nPercent overshoot (PO), 409–10\nPeriodic (circular) convolution, 819–20", - "type": "text" - }, - { - "block_id": "p1003-b16", - "global_id": 30433, - "bbox": [ - 306.92, - 430.44, - 451.13, - 448.67 - ], - "text": "of the discrete-time Fourier transform, 875\nPeriodic extension", - "type": "text" - }, - { - "block_id": "p1003-b17", - "global_id": 30434, - "bbox": [ - 306.92, - 450.97, - 417.71, - 479.47 - ], - "text": "of the Fourier spectrum, 848–55\nproperties of, 80–81\nPeriodic functions", - "type": "text" - }, - { - "block_id": "p1003-b18", - "global_id": 30435, - "bbox": [ - 306.92, - 481.76, - 405.17, - 520.52 - ], - "text": "Fourier spectra as, 858\nMATLAB on, 661–63\nPeriodic gate function, 853–55\nPeriodic signals, 133, 637–40", - "type": "text" - }, - { - "block_id": "p1003-b19", - "global_id": 30436, - "bbox": [ - 306.92, - 522.81, - 465.08, - 582.1 - ], - "text": "discrete-time Fourier series and, 846–47\nFourier spectra of, 848–55\nFourier transform of, 695–96\nproperties of, 78–82\nand trigonometric Fourier series, 593–612, 661\nPeriods", - "type": "text" - }, - { - "block_id": "p1003-b20", - "global_id": 30437, - "bbox": [ - 306.92, - 584.4, - 458.33, - 623.15 - ], - "text": "fundamental, 79, 133, 239–40, 593, 595, 846\nsinusoid, 16\nPhase response, 413–25, 427–35, 439, 467\nPhase spectrum, 598, 607, 617–18, 707, 848", - "type": "text" - }, - { - "block_id": "p1003-b21", - "global_id": 30438, - "bbox": [ - 314.89, - 625.44, - 412.55, - 643.68 - ], - "text": "MATLAB on, 664–67\nusing principal values, 709–10", - "type": "text" - } - ] - }, - { - "page_num": 1004, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1004-b0", - "global_id": 30439, - "bbox": [ - 60.0, - 60.36, - 124.49, - 69.45 - ], - "text": "984\nIndex", - "type": "text" - }, - { - "block_id": "p1004-b1", - "global_id": 30440, - "bbox": [ - 60.0, - 85.98, - 180.17, - 176.14 - ], - "text": "Phase-plane analysis, 125, 909\nPhasors, 18–20\nPhysical systems. See Causal systems\nPicket fence effect, 807\nPickoff nodes, 190, 254–55, 396\nPingala, 801\nPointwise convergent series, 613\nPolar coordinates, 5–6\nPolar form, 8–15", - "type": "text" - }, - { - "block_id": "p1004-b2", - "global_id": 30441, - "bbox": [ - 60.0, - 178.44, - 173.36, - 227.52 - ], - "text": "arithmetical operations in, 12–15\nsinusoids and, 18\nPole-zero location, 538–47\nPole-zero plots, 566–68\nPoles", - "type": "text" - }, - { - "block_id": "p1004-b3", - "global_id": 30442, - "bbox": [ - 60.0, - 229.82, - 246.45, - 391.92 - ], - "text": "complex, 395, 432, 497, 542\ncontrolling gain by, 540\nfirst-order, 424–27\ngain enhancement by, 437–38\nH(s), filter design and, 436–45\nat the origin, 422–23\nrepeated, 395, 525, 926\nin the right half plane, 371, 435–36\nsecond-order, 426–35\nwall of, 439–41, 542\nPolynomial expansion, 458–59\nPolynomial roots, 157, 572\nPositive feedback, 406\nPower series, 55\nPower signals, 67, 82, 134, 239–40. See also Signal power\nPower, determining, 68–69", - "type": "text" - }, - { - "block_id": "p1004-b4", - "global_id": 30443, - "bbox": [ - 60.0, - 394.22, - 176.57, - 453.57 - ], - "text": "matrix, 912–13\nPowers, of complex numbers, 13–16\nPractical filters, 730–33, 882–83\nPreece, Sir William, 349\nPrewarping, 570–71\nPrincipal values", - "type": "text" - }, - { - "block_id": "p1004-b5", - "global_id": 30444, - "bbox": [ - 60.0, - 455.87, - 189.9, - 556.31 - ], - "text": "of the angle, 9\nphase spectrum using, 709–10\nProper functions, 25–27\nPulse-amplitude modulation (PAM), 796\nPulse-code modulation (PCM), 796, 799\nPulse dispersion, 209\nPulse-position modulation (PPM), 796\nPulse-width modulation (PWM), 796\nPupin, M., 348\nPythagoras, 2", - "type": "text" - }, - { - "block_id": "p1004-b6", - "global_id": 30445, - "bbox": [ - 60.0, - 575.69, - 138.95, - 593.94 - ], - "text": "Quadratic equations, 58\nQuadratic factors, 29–30", - "type": "text" - }, - { - "block_id": "p1004-b7", - "global_id": 30446, - "bbox": [ - 60.0, - 596.24, - 177.09, - 635.04 - ], - "text": "for the Laplace transform, 341–42\nfor the z-transform, 497\nQuantization, 799, 831–34\nQuantized levels, 799", - "type": "text" - }, - { - "block_id": "p1004-b8", - "global_id": 30447, - "bbox": [ - 281.17, - 85.98, - 410.03, - 184.9 - ], - "text": "Radian frequency, 16, 91, 594\nRandom signals, 82, 134\nRational functions, 25–29, 338\nReal numbers, 2–7, 43\nReal time, 105–6\nRectangular pulses, 863–65\nRectangular spectrum, 865–66\nRectangular windows, 751, 753–55, 763\nReflection property, 868–69\nRegion of convergence (ROC)", - "type": "text" - }, - { - "block_id": "p1004-b9", - "global_id": 30448, - "bbox": [ - 289.14, - 187.04, - 464.02, - 215.22 - ], - "text": "for continuous-time systems, 193\nfor finite-duration signals, 333\nfor the Laplace transform, 331–33, 337, 347, 448, 449,", - "type": "text" - }, - { - "block_id": "p1004-b10", - "global_id": 30449, - "bbox": [ - 281.17, - 217.22, - 420.63, - 275.72 - ], - "text": "454–55, 467\nfor the z-transform, 489–91, 555–58, 561\nRelational operators, 128–29\nRepeated factors of Q(x), 31–32\nRepeated poles, 395, 525, 926\nRepeated roots", - "type": "text" - }, - { - "block_id": "p1004-b11", - "global_id": 30450, - "bbox": [ - 281.17, - 277.86, - 470.22, - 386.9 - ], - "text": "of continuous-time systems, 154–56, 195, 198, 202, 223\nof discrete-time systems, 270, 273–74, 297, 301, 313–14\nResonance phenomenon, 163, 204, 205, 210–12, 305\nRight half plane (RHP), 91, 198, 200–3, 223, 371, 435–36\nRight shift, 71–72, 131, 134, 501–4, 509, 510\nRight-sided sequences, 555–56\nRise time, 206–7, 405, 409–10, 411\nRLC networks, 914, 916–18\nRMS value, 68–69, 70\nRolloff rate, 753, 754\nRoots", - "type": "text" - }, - { - "block_id": "p1004-b12", - "global_id": 30451, - "bbox": [ - 281.17, - 389.03, - 407.35, - 467.74 - ], - "text": "complex, 154–56, 274–76\nof complex numbers, 11–15\npolynomial, 157, 572\nrepeated. See Repeated roots\nunrepeated, 198, 202, 223, 301, 314\nRotational systems, 116–19\nRotational mass. See Moment of inertia\nRow vectors, 36, 45, 48–50", - "type": "text" - }, - { - "block_id": "p1004-b13", - "global_id": 30452, - "bbox": [ - 281.17, - 485.27, - 432.06, - 523.56 - ], - "text": "Sales estimate example, 255–56\nSallen–Key circuit, 383, 384, 461–62, 463, 466\nSampled continuous-time sinusoids, 527–31\nSampling, 776–844", - "type": "text" - }, - { - "block_id": "p1004-b14", - "global_id": 30453, - "bbox": [ - 281.17, - 525.7, - 478.69, - 604.42 - ], - "text": "practical, 781–84\nproperties of, 87–88, 134\nsignal reconstruction and, 785–99\nspectral, 759–60, 802–4\nSee also Discrete Fourier transform; Fast-Fourier transform\nSampling interval, 550–54\nSampling rate, 243–44, 536–37\nSampling theorem, 537, 776–84, 834–35", - "type": "text" - }, - { - "block_id": "p1004-b15", - "global_id": 30454, - "bbox": [ - 281.17, - 606.55, - 389.8, - 634.74 - ], - "text": "applications of, 796–99\nspectral, 802\nSavings account example, 253–55", - "type": "text" - } - ] - }, - { - "page_num": 1005, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1005-b0", - "global_id": 30455, - "bbox": [ - 450.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Index\n985", - "type": "text" - }, - { - "block_id": "p1005-b1", - "global_id": 30456, - "bbox": [ - 85.74, - 85.98, - 236.06, - 103.97 - ], - "text": "Scalar multiplication, 38, 400–1, 505, 509, 520\nScaling, 97–98, 130", - "type": "text" - }, - { - "block_id": "p1005-b2", - "global_id": 30457, - "bbox": [ - 85.74, - 106.03, - 242.14, - 174.16 - ], - "text": "of the Fourier transform, 705–6, 755, 757, 762\nof the Laplace transform, 357\nSee also Time scaling\nScript M-files, 213–14, 216, 218\nSelective-filtering method, 748–49\nSharp cutoff filters, 748\nShifting", - "type": "text" - }, - { - "block_id": "p1005-b3", - "global_id": 30458, - "bbox": [ - 85.74, - 176.14, - 266.63, - 264.41 - ], - "text": "of the bilateral z-transform, 559\nof the convolution integral, 171–72\nof the convolution sum, 283\nof discrete-time signals, 240\nSee also Frequency shifting; Time shifting\nSideband, 746–49\nsifting. See Sampling\nSignal distortion, 723–25\nSignal energy, 65–66, 70, 131–33, 733–36, 757, 877–78.", - "type": "text" - }, - { - "block_id": "p1005-b4", - "global_id": 30459, - "bbox": [ - 85.74, - 266.32, - 255.09, - 324.51 - ], - "text": "See also Energy signals\nSignal power, 65–67, 133. See also Power signals\nSignal reconstruction, 785–99. See also Interpolation\nSignal-to-noise power ratio, 66\nSignal transmission, 721–29\nSignals, 64–91, 133–34", - "type": "text" - }, - { - "block_id": "p1005-b5", - "global_id": 30460, - "bbox": [ - 93.72, - 326.49, - 237.84, - 585.23 - ], - "text": "analog. See Analog signals\nanti-causal, 81\naperiodic. See Aperiodic signals\naudio, 713–14, 725, 746\nbandlimited, 533, 788, 792, 802\nbaseband, 737–40, 746–47, 749\nbasis, 651, 655, 668\ncausal, 81, 83, 134\nclassification of, 78–82, 133–34\ncomparison and components of, 643–45\ncomplex, 94–95\ncontinuous time. See continuous-time signals\ndefined, 65\ndeterministic, 83, 134\ndigital. See Digital signals\ndiscrete time. See Discrete-time signals\nenergy, 82, 134, 239–40\nerror, 650–51\neven components of, 93–95\neverlasting, 81, 134\nfinite-duration, 333\nmodulating, 711, 737–39\nnon-bandlimited, 792\nnoncausal, 81\nodd components of, 93–95\northogonal. See Orthogonal signals periodic.", - "type": "text" - }, - { - "block_id": "p1005-b6", - "global_id": 30461, - "bbox": [ - 93.72, - 587.14, - 168.91, - 635.3 - ], - "text": "See Periodic signals\nphantoms of, 189\npower, 82, 134, 239–40\nrandom, 82, 134\nsize of, 64–70, 133", - "type": "text" - }, - { - "block_id": "p1005-b7", - "global_id": 30462, - "bbox": [ - 306.92, - 85.97, - 488.03, - 214.12 - ], - "text": "sketching, 20–23\ntime reversal of, 77\ntime limited, 802, 805, 807\ntwo-dimensional view of, 732–33\nuseful models, 82–91\nuseful operations, 71–78\nas vectors, 641–59\nvideo, 725, 749\nSinc function, 757\nSingle-input, single-output (SISO) systems, 98, 125, 908\nSingle-sideband (SSB) modulation, 746–49\nSingularity functions, 89\nSinusoidal input", - "type": "text" - }, - { - "block_id": "p1005-b8", - "global_id": 30463, - "bbox": [ - 314.88, - 216.09, - 446.81, - 264.2 - ], - "text": "causal. See Causal sinusoidal input\ncontinuous-time systems and, 208\ndiscrete-time systems and, 309\nfrequency response and, 413–17\nsteady-state response to causal sinusoidal", - "type": "text" - }, - { - "block_id": "p1005-b9", - "global_id": 30464, - "bbox": [ - 306.92, - 266.19, - 401.69, - 284.18 - ], - "text": "input, 418–19\nSinusoids, 16–20, 89–91, 134", - "type": "text" - }, - { - "block_id": "p1005-b10", - "global_id": 30465, - "bbox": [ - 306.92, - 286.23, - 451.63, - 514.53 - ], - "text": "addition of, 18–20\napparent frequency of sampled, 795–96\ncompression and expansion, 76\ncontinuous-time, 251–52, 533–37\ndiscrete-time, 251, 527, 528, 533–37\ndiscrete-time Fourier series of, 849–52\nin exponential terms, 20\nexponentially varying, 22–23, 80, 134\ngeneral condition for aliasing in, 793–96\npower of a sum of two equal-frequency, 70\nsampled continuous-time, 527–31\nverification of aliasing in, 792–93\nSketching signals, 20–23\nSliding-tape method, 290–93\nSoftware realization, 64, 95, 133\nSpectral density, 688\nSpectral folding. See Aliasing\nSpectral interpolation, 804\nSpectral resolution, 807\nSpectral sampling, 759–60, 802\nSpectral sampling theorem, 802\nSpectral spreading, 751–53, 755, 763, 807\nSprings", - "type": "text" - }, - { - "block_id": "p1005-b11", - "global_id": 30466, - "bbox": [ - 306.92, - 516.58, - 429.51, - 564.61 - ], - "text": "linear, 114\ntorsional, 116–17\nSquare matrices, 36, 37, 41\nSquare roots of negative numbers, 2–4\nStability", - "type": "text" - }, - { - "block_id": "p1005-b12", - "global_id": 30467, - "bbox": [ - 314.88, - 566.57, - 479.13, - 634.72 - ], - "text": "BIBO. See Bounded-input/bounded-output stability\nof continuous-time systems, 196–203, 222–23\nof discrete-time systems, 263, 298–305, 314\nof the Laplace transform, 371–74\nInternal. See Internal stability\nof the z-transform, 518–19\nmarginal. See marginally stable systems", - "type": "text" - } - ] - }, - { - "page_num": 1006, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1006-b0", - "global_id": 30468, - "bbox": [ - 60.0, - 60.36, - 124.49, - 69.45 - ], - "text": "986\nIndex", - "type": "text" - }, - { - "block_id": "p1006-b1", - "global_id": 30469, - "bbox": [ - 60.0, - 85.98, - 192.18, - 113.99 - ], - "text": "Stable equilibrium, 196–97\nStable systems, 110, 263\nState equations, 122–25, 135, 908–9, 969", - "type": "text" - }, - { - "block_id": "p1006-b2", - "global_id": 30470, - "bbox": [ - 60.0, - 116.04, - 215.18, - 204.22 - ], - "text": "alternative procedure to determine, 918–19\ndiagonal form of, 944–47\nsolution of, 926–39\nfor the state vector, 941–42\nsystematic procedure for determining, 913–26\ntime-domain method to solve, 936–37\nState transition matrix (STM), 936\nState variables, 121–25, 135, 908, 969\nState vectors, 927–30, 961", - "type": "text" - }, - { - "block_id": "p1006-b3", - "global_id": 30471, - "bbox": [ - 60.0, - 206.28, - 171.57, - 224.27 - ], - "text": "linear transformation of, 941–42\nState-space analysis, 908–73", - "type": "text" - }, - { - "block_id": "p1006-b4", - "global_id": 30472, - "bbox": [ - 60.0, - 226.32, - 208.8, - 304.47 - ], - "text": "controllability/observability in, 947–53, 961\nof discrete-time systems, 953–64\nin MATLAB, 961–69\ntransfer function and, 920–24\ntransfer function matrix, 931–32\nState-space description of a system, 121–25\nSteady-state error, 409–11\nSteady-state response", - "type": "text" - }, - { - "block_id": "p1006-b5", - "global_id": 30473, - "bbox": [ - 60.0, - 306.52, - 183.11, - 354.6 - ], - "text": "in continuous-time systems, 418–19\nin discrete-time systems, 527\nStem plots, 306–8\nStep input, 407–10\nStiffness", - "type": "text" - }, - { - "block_id": "p1006-b6", - "global_id": 30474, - "bbox": [ - 60.0, - 356.64, - 236.69, - 414.74 - ], - "text": "of linear springs, 114\nof torsional springs, 116–17\nStopbands, 441, 444, 445, 456, 457, 459, 460, 463, 755\nSubcarriers, 749\nSubtraction of complex numbers, 11–12\nSuperposition, 98, 99, 100, 123, 134", - "type": "text" - }, - { - "block_id": "p1006-b7", - "global_id": 30475, - "bbox": [ - 60.0, - 416.8, - 208.35, - 454.84 - ], - "text": "continuous-time systems and, 168, 170, 178\ndiscrete-time systems and, 287\nSymmetric matrices, 37\nSymmetry", - "type": "text" - }, - { - "block_id": "p1006-b8", - "global_id": 30476, - "bbox": [ - 60.0, - 456.82, - 194.82, - 504.96 - ], - "text": "conjugate. See Conjugate symmetry\nexponential Fourier series and, 630–32\ntrigonometric Fourier series and, 607–8\nSynchronous demodulation, 743–44, 747\nSystem realization, 388–404, 519–25, 567", - "type": "text" - }, - { - "block_id": "p1006-b9", - "global_id": 30477, - "bbox": [ - 60.0, - 507.02, - 265.38, - 585.17 - ], - "text": "cascade, 394, 525–26, 919–20, 923\nof complex conjugate poles, 395\ndirect. See Direct form I realization; Direct form II realization\ndifferences in performance, 525–26\nhardware, 64, 95, 133\nparallel. See Parallel realization\nsoftware, 64, 95, 129\nSystems, 95–133, 134–35", - "type": "text" - }, - { - "block_id": "p1006-b10", - "global_id": 30478, - "bbox": [ - 67.98, - 587.23, - 208.7, - 635.29 - ], - "text": "accumulator, 259, 295, 519\nanalog, 109, 135, 261\nbackward difference, 258, 295, 519, 568–69\nBIBO stability, assessing, 110\ncascade, 190, 192, 372, 373", - "type": "text" - }, - { - "block_id": "p1006-b11", - "global_id": 30479, - "bbox": [ - 289.14, - 85.88, - 464.71, - 124.28 - ], - "text": "causal. See causal systems\ncausality, assessing, 105\nclassification of, 97–110, 134–35\ncontinuous time. See Continuous-time systems control.", - "type": "text" - }, - { - "block_id": "p1006-b12", - "global_id": 30480, - "bbox": [ - 289.14, - 126.19, - 468.71, - 478.27 - ], - "text": "See control systems\ncritically damped, 409, 410\ndata for computing response, 96–97\ndefined, 64\ndigital, 78, 135, 261\ndiscrete time. See discrete time systems\ndynamic, 103–4, 134–35, 263\nelectrical, 95–96, 111–14\nelectrical. See Electrical systems\nelectromechanical, 118–19\nfeedback. See feedback systems\nfinite-memory, 104\nidentity, 109, 192, 263\ninput–output description, 111–19\ninstantaneous, 103–4, 263\ninterconnected. See interconnected systems\ninvertible, 109–10, 135, 263\nlinear. See Linear systems\nmathematical models of, 95–96, 125\nmechanical, 114–18\nmemory and, 104, 263\nminimum phase, 435, 436\nmultiple-input, multiple-output, 98, 125, 908\nnoncausal, 104–7, 263\nnon-invertible, 109–10, 135\nnonlinear, 97–101, 134\noverdamped, 409–10\nparallel, 190, 387\nphantoms of, 189\nproperties of, 264–65\nrotational, 116–19\nsingle-input, single-output, 98, 125, 908\nstable, 110, 263\ntranslational, 114–16\ntime invariant. See Time-invariant systems time varying.", - "type": "text" - }, - { - "block_id": "p1006-b13", - "global_id": 30481, - "bbox": [ - 289.14, - 480.18, - 396.46, - 518.59 - ], - "text": "See Time-varying systems\ntwo-dimensional view of, 732–33\nunderdamped, 409\nunstable, 110, 263", - "type": "text" - }, - { - "block_id": "p1006-b14", - "global_id": 30482, - "bbox": [ - 281.17, - 536.25, - 432.13, - 594.81 - ], - "text": "Tacoma Narrows Bridge failure, 212\nTapered windows, 753–54, 763, 807\nTaylor series, 55\nThéorie analytique de la chaleur (Fourier), 612\nThévenin’s theorem, 375, 378, 379\nTime constant", - "type": "text" - }, - { - "block_id": "p1006-b15", - "global_id": 30483, - "bbox": [ - 289.14, - 596.96, - 424.42, - 635.28 - ], - "text": "of continuous-time systems, 205–10, 223\nof the exponential, 21–22\nfiltering and, 207–9\ninformation transmission rate and, 209–10", - "type": "text" - } - ] - }, - { - "page_num": 1007, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1007-b0", - "global_id": 30484, - "bbox": [ - 450.66, - 62.89, - 516.13, - 71.98 - ], - "text": "Index\n987", - "type": "text" - }, - { - "block_id": "p1007-b1", - "global_id": 30485, - "bbox": [ - 85.74, - 85.98, - 174.74, - 113.97 - ], - "text": "pulse dispersion and, 209\nrise time and, 206–7\nTime convolution", - "type": "text" - }, - { - "block_id": "p1007-b2", - "global_id": 30486, - "bbox": [ - 85.74, - 116.02, - 241.92, - 184.07 - ], - "text": "of the bilateral Laplace transform, 452\nof the discrete-time Fourier transform, 875–76\nof the Fourier transform, 714–16\nof the Laplace transform, 357\nof the z-transform, 507–8\nTime delay, variation with frequency, 724–25\nTime differentiation", - "type": "text" - }, - { - "block_id": "p1007-b3", - "global_id": 30487, - "bbox": [ - 85.74, - 186.12, - 216.35, - 224.13 - ], - "text": "of the bilateral Laplace transform, 451\nof the Fourier transform, 716–18\nof the Laplace transform, 354–56\nTime integration", - "type": "text" - }, - { - "block_id": "p1007-b4", - "global_id": 30488, - "bbox": [ - 85.74, - 226.18, - 216.35, - 274.21 - ], - "text": "of the bilateral Laplace transform, 451\nof the Fourier transform, 716–18\nof the Laplace transform, 356–57\nTime inversion, 706\nTime reversal, 134", - "type": "text" - }, - { - "block_id": "p1007-b5", - "global_id": 30489, - "bbox": [ - 85.74, - 276.25, - 241.92, - 354.32 - ], - "text": "of the bilateral Laplace transform, 452\nof the bilateral z-transform, 560\nof the convolution integral, 178, 181\ndescribed, 76–77\nof the discrete-time Fourier transform, 868–69\nof discrete-time signals, 242\nof the z-transform, 506–7\nTime scaling, 77", - "type": "text" - }, - { - "block_id": "p1007-b6", - "global_id": 30490, - "bbox": [ - 85.74, - 356.37, - 216.35, - 384.36 - ], - "text": "of the bilateral Laplace transform, 452\ndescribed, 73–74\nTime shifting, 77, 79", - "type": "text" - }, - { - "block_id": "p1007-b7", - "global_id": 30491, - "bbox": [ - 85.74, - 386.41, - 229.96, - 484.51 - ], - "text": "of the bilateral Laplace transform, 451\nof the convolution integral, 178\ndescribed, 71–73\nof the discrete Fourier transform, 819\nof the discrete-time Fourier transform, 870\nof the Fourier transform, 707\nof the Laplace transform, 349–51\nof the z-transform, 501–5, 510\nTime-division multiplexing (TDM), 749, 797\nTime-domain analysis, 723", - "type": "text" - }, - { - "block_id": "p1007-b8", - "global_id": 30492, - "bbox": [ - 85.74, - 486.55, - 216.1, - 564.62 - ], - "text": "of continuous-time systems, 150–236\nof discrete-time systems, 237–329\nof the Fourier series, 598, 601\nof interpolation, 785–88\nstate equation solution in, 933–39\ntwo-dimensional view and, 732–33\nTime-frequency duality, 702–3, 723, 753\nTime invariant systems, 134", - "type": "text" - }, - { - "block_id": "p1007-b9", - "global_id": 30493, - "bbox": [ - 85.74, - 566.67, - 223.12, - 604.68 - ], - "text": "discrete-time, 262\nlinear. See Linear time-invariant systems\nproperties of, 102–3\nTime-varying systems, 134", - "type": "text" - }, - { - "block_id": "p1007-b10", - "global_id": 30494, - "bbox": [ - 93.72, - 606.73, - 158.14, - 634.73 - ], - "text": "discrete-time, 262\nlinear, 103\nproperties of, 102–3", - "type": "text" - }, - { - "block_id": "p1007-b11", - "global_id": 30495, - "bbox": [ - 306.92, - 85.97, - 420.66, - 134.37 - ], - "text": "Time-limited signals, 802, 805, 807\nTorque, 116–18\nTorsional dashpots, 116\nTorsional springs, 116, 117\nTotal response", - "type": "text" - }, - { - "block_id": "p1007-b12", - "global_id": 30496, - "bbox": [ - 306.92, - 136.51, - 444.03, - 174.81 - ], - "text": "of continuous-time systems, 195–96\nof discrete-time systems, 297–98\nTraité de mécanique céleste (Laplace), 346\nTransfer functions, 522", - "type": "text" - }, - { - "block_id": "p1007-b13", - "global_id": 30497, - "bbox": [ - 306.92, - 176.95, - 495.45, - 296.11 - ], - "text": "analog filter realization with, 548–49\nblock diagrams and, 386–88\nof continuous-time systems, 193–94, 222\nof discrete-time systems, 296–97, 314, 514–15, 567–68\nfrom the frequency response, 435\ninadequacy for system description, 953\nrealization of, 389–99, 401, 524–25\nstate equations from, 916, 919–26\nfrom state-space representations, 964–65\nTranslational systems, 114–16\nTranspose of a matrix, 37–38\nTransposed direct form II (TDFII) realization, 398, 967–69", - "type": "text" - }, - { - "block_id": "p1007-b14", - "global_id": 30498, - "bbox": [ - 306.92, - 298.25, - 489.83, - 336.55 - ], - "text": "state equations and, 920–24\nz-transform and, 520–22, 525–26\nTriangular windows, 751\nTrigonometric Fourier series, 640, 652, 657–58, 667, 668", - "type": "text" - }, - { - "block_id": "p1007-b15", - "global_id": 30499, - "bbox": [ - 306.92, - 338.69, - 426.46, - 397.21 - ], - "text": "exponential, 621–37, 661\nperiodic signals and, 593–612, 661\nsampling and, 777, 782\nsymmetry effect on, 607–8\nTrigonometric identities, 55–56\nTukey, J. W., 824", - "type": "text" - }, - { - "block_id": "p1007-b16", - "global_id": 30500, - "bbox": [ - 306.92, - 414.75, - 508.6, - 442.95 - ], - "text": "Underdamped systems, 409\nUniformly convergent series, 613\nUnilateral Laplace transform, 333–36, 337, 338, 345, 360, 445,", - "type": "text" - }, - { - "block_id": "p1007-b17", - "global_id": 30501, - "bbox": [ - 306.92, - 444.94, - 485.11, - 503.45 - ], - "text": "467\nUnilateral z-transform, 489, 491, 492, 495, 554–55, 559\nUniqueness, 335\nUnit delay, 517, 520, 521\nUnit-gate function, 689\nUnit-impulse function, 133", - "type": "text" - }, - { - "block_id": "p1007-b18", - "global_id": 30502, - "bbox": [ - 306.92, - 505.59, - 452.14, - 543.89 - ], - "text": "of discrete-time systems, 246–47, 280, 313\nas a generalized function, 88–89\nproperties of, 86–89\nUnit-impulse response", - "type": "text" - }, - { - "block_id": "p1007-b19", - "global_id": 30503, - "bbox": [ - 314.89, - 546.03, - 504.2, - 554.0 - ], - "text": "of continuous-time systems, 163–68, 170, 189–93, 220–21,", - "type": "text" - }, - { - "block_id": "p1007-b20", - "global_id": 30504, - "bbox": [ - 306.92, - 555.99, - 468.08, - 614.5 - ], - "text": "222, 731\nconvolution with, 171\ndetermining, 221\nof discrete-time systems, 277–80, 286, 295, 313\nUnit matrices, 37\nUnit-step function, 84–86, 88–89", - "type": "text" - }, - { - "block_id": "p1007-b21", - "global_id": 30505, - "bbox": [ - 314.89, - 616.64, - 420.27, - 634.72 - ], - "text": "of discrete-time systems, 246–47\nrelational operators and, 128–30", - "type": "text" - } - ] - }, - { - "page_num": 1008, - "width": 576.0, - "height": 720.0, - "blocks": [ - { - "block_id": "p1008-b0", - "global_id": 30506, - "bbox": [ - 60.0, - 62.89, - 124.49, - 71.98 - ], - "text": "988\nIndex", - "type": "text" - }, - { - "block_id": "p1008-b1", - "global_id": 30507, - "bbox": [ - 60.0, - 85.98, - 194.37, - 143.75 - ], - "text": "Unit-triangle function, 689–90\nUnrepeated roots, 198, 202, 223, 301, 314\nUnstable equilibrium, 196–97\nUnstable systems, 110, 263\nUpper sideband (USB), 737–39, 746–48\nUpsampling, 243–44", - "type": "text" - }, - { - "block_id": "p1008-b2", - "global_id": 30508, - "bbox": [ - 60.0, - 159.19, - 135.72, - 167.16 - ], - "text": "Vectors, 36–37, 641–59", - "type": "text" - }, - { - "block_id": "p1008-b3", - "global_id": 30509, - "bbox": [ - 60.0, - 169.15, - 159.86, - 296.67 - ], - "text": "basis, 648\ncharacteristic, 910\ncolumn, 36\ncomponents of, 642–43\nerror, 642\nMATLAB operations, 45–46\nmatrix multiplication by, 40\northogonal space, 647–48\nrow, 36, 45, 48–50\nsignals as, 641–59\nstate, 927–30, 961\nVestigial sideband (VSB), 749\nVideo signals, 725, 749", - "type": "text" - }, - { - "block_id": "p1008-b4", - "global_id": 30510, - "bbox": [ - 60.0, - 312.1, - 138.91, - 340.0 - ], - "text": "Waveshaping, 615–17\nWeber–Fechner law, 421\nWidth", - "type": "text" - }, - { - "block_id": "p1008-b5", - "global_id": 30511, - "bbox": [ - 60.0, - 341.99, - 184.05, - 369.89 - ], - "text": "of the convolution integral, 172, 187\nof the convolution sum, 283\nWindow functions, 749–55, 760–62", - "type": "text" - }, - { - "block_id": "p1008-b6", - "global_id": 30512, - "bbox": [ - 60.0, - 385.24, - 128.64, - 393.29 - ], - "text": "z-transform, 488–592", - "type": "text" - }, - { - "block_id": "p1008-b7", - "global_id": 30513, - "bbox": [ - 67.98, - 395.2, - 253.12, - 453.07 - ], - "text": "bilateral. See Bilateral z-transform\ndifference equation solutions of, 488, 510–19, 574\ndirect, 488–592\ndiscrete-time Fourier transform and, 866–67, 886–88, 898\nexistence of, 491–95\ninverse. See inverse z-transform", - "type": "text" - }, - { - "block_id": "p1008-b8", - "global_id": 30514, - "bbox": [ - 281.17, - 85.97, - 426.84, - 203.52 - ], - "text": "properties of, 501–9\nstability of, 518–19\nstate-space analysis and, 956, 959–65\nsystem realization and, 519–25, 567\ntime-reversal property, 506–7\ntime-shifting properties, 501–5\nunilateral, 489, 491, 492, 495, 554–55, 559\nz-domain differentiation property, 506\nz-domain scaling property, 505\nZero matrices, 37\nZero padding, 810–11, 829–30\nZero-input response, 119, 123", - "type": "text" - }, - { - "block_id": "p1008-b9", - "global_id": 30515, - "bbox": [ - 281.17, - 205.51, - 476.45, - 303.15 - ], - "text": "of continuous-time systems, 151–63, 195–96, 203, 220–22\ndescribed, 98–100\nof discrete-time systems, 270–76, 297–301, 309–11\ninsights into behavior of, 161–63\nof the Laplace transform, 363, 368\nin oscillators, 203\nof the z-transform, 512–13\nzero-state response independence from, 161\nZero-order hold (ZOH) filters, 785\nZero-state response, 119, 123", - "type": "text" - }, - { - "block_id": "p1008-b10", - "global_id": 30516, - "bbox": [ - 289.14, - 305.14, - 466.5, - 333.03 - ], - "text": "alternate interpretation, 515–18\ncausality and, 172–73\nof continuous-time systems, 151, 161, 168–96, 221–22,", - "type": "text" - }, - { - "block_id": "p1008-b11", - "global_id": 30517, - "bbox": [ - 281.17, - 335.03, - 466.24, - 392.8 - ], - "text": "512–16\ndescribed, 98–101\nof discrete-time systems, 280–98, 308–9, 311, 312, 313\nof the Laplace transform, 358, 363, 366–67, 369, 370\nzero-input response independence from, 161\nZeros", - "type": "text" - }, - { - "block_id": "p1008-b12", - "global_id": 30518, - "bbox": [ - 289.14, - 394.79, - 379.81, - 452.58 - ], - "text": "controlling gain by, 540\nfilter design, 436–45\nfirst-order, 424–27\ngain suppression by, 439–40\nat the origin, 422–23\nsecond-order, 426–35", - "type": "text" - } - ] - }, - { - "page_num": 1009, - "width": 576.0, - "height": 720.0, - "blocks": [] - }, - { - "page_num": 1010, - "width": 576.0, - "height": 720.0, - "blocks": [] - } - ] -} \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/llm.txt b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/llm.txt deleted file mode 100644 index 1dc928ce531481cf8865203908e424ce6e9bd8c4..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/llm.txt +++ /dev/null @@ -1,299 +0,0 @@ -# Book Title: LINEAR SYSTEMS AND SIGNALS -> Author: B. P. Lathi and R. A. Green | Category: engineering | Pages: 1010 | Year: 2017 - -## Overview & Metadata -- Document ID: linear-systems-and-signals-3rd-edition-bp-lathi -- Category: engineering -- Tags: 3rd, and, lathi, linear, signals, systems -- Primary Markdown HF Raw URL: https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.md -- Original PDF HF Download: https://huggingface.co/datasets/learner20011/CloverTexts-Data/resolve/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.pdf -- Layout Coordinates JSON: https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.layout.json -- Book LLM Text URL: https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/llm.txt -- Online Web Reader: /book?id=linear-systems-and-signals-3rd-edition-bp-lathi - -## Table of Contents -- Cover - Page 1 -- Half title - Page 3 -- Series page - Page 4 -- Title page - Page 5 -- Copyright page - Page 6 -- CONTENTS - Page 7 -- PREFACE - Page 17 -- B BACKGROUND - Page 21 - - B.1 COMPLEX NUMBERS - Page 21 - - B.1-1 A Historical Note - Page 21 - - B.1-2 Algebra of Complex Numbers - Page 25 - - B.2 SINUSOIDS - Page 36 - - B.2-1 Addition of Sinusoids - Page 38 - - B.2-2 Sinusoids in Terms of Exponentials - Page 40 - - B.3 SKETCHING SIGNALS - Page 40 - - B.3-1 Monotonic Exponentials - Page 40 - - B.3-2 The Exponentially Varying Sinusoid - Page 42 - - B.4 CRAMER’S RULE - Page 43 - - B.5 PARTIAL FRACTION EXPANSION - Page 45 - - B.5-1 Method of Clearing Fractions - Page 46 - - B.5-2 The Heaviside “Cover-Up” Method - Page 47 - - B.5-3 Repeated Factors of Q(x) - Page 51 - - B.5-4 A Combination of Heaviside “Cover-Up” and Clearing Fractions - Page 52 - - B.5-5 Improper F(x) with m = n - Page 54 - - B.5-6 Modified Partial Fractions - Page 55 - - B.6 VECTORS AND MATRICES - Page 56 - - B.6-1 Some Definitions and Properties - Page 57 - - B.6-2 Matrix Algebra - Page 58 - - B.7 MATLAB: ELEMENTARY OPERATIONS - Page 62 - - B.7-1 MATLAB Overview - Page 62 - - B.7-2 Calculator Operations - Page 63 - - B.7-3 Vector Operations - Page 65 - - B.7-4 Simple Plotting - Page 66 - - B.7-5 Element-by-Element Operations - Page 68 - - B.7-6 Matrix Operations - Page 69 - - B.7-7 Partial Fraction Expansions - Page 73 - - B.8 APPENDIX: USEFUL MATHEMATICAL FORMULAS - Page 74 - - B.8-1 Some Useful Constants - Page 74 - - B.8-2 Complex Numbers - Page 74 - - B.8-3 Sums - Page 74 - - B.8-4 Taylor and Maclaurin Series - Page 75 - - B.8-5 Power Series - Page 75 - - B.8-6 Trigonometric Identities - Page 75 - - B.8-7 Common Derivative Formulas - Page 76 - - B.8-8 Indefinite Integrals - Page 77 - - B.8-9 L’Hôpital’s Rule - Page 78 - - B.8-10 Solution of Quadratic and Cubic Equations - Page 78 - - REFERENCES - Page 78 - - PROBLEMS - Page 79 -- 1 SIGNALS AND SYSTEMS - Page 84 - - 1.1 SIZE OF A SIGNAL - Page 84 - - 1.1-1 Signal Energy - Page 85 - - 1.1-2 Signal Power - Page 85 - - 1.2 SOME USEFUL SIGNAL OPERATIONS - Page 91 - - 1.2-1 Time Shifting - Page 91 - - 1.2-2 Time Scaling - Page 93 - - 1.2-3 Time Reversal - Page 96 - - 1.2-4 Combined Operations - Page 97 - - 1.3 CLASSIFICATION OF SIGNALS - Page 98 - - 1.3-1 Continuous-Time and Discrete-Time Signals - Page 98 - - 1.3-2 Analog and Digital Signals - Page 98 - - 1.3-3 Periodic and Aperiodic Signals - Page 99 - - 1.3-4 Energy and Power Signals - Page 102 - - 1.3-5 Deterministic and Random Signals - Page 102 - - 1.4 SOME USEFUL SIGNAL MODELS - Page 102 - - 1.4-1 The Unit Step Function u(t) - Page 103 - - 1.4-2 The Unit Impulse Function δ(t) - Page 106 - - 1.4-3 The Exponential Function e^{st} - Page 109 - - 1.5 EVEN AND ODD FUNCTIONS - Page 112 - - 1.5-1 Some Properties of Even and Odd Functions - Page 112 - - 1.5-2 Even and Odd Components of a Signal - Page 113 - - 1.6 SYSTEMS - Page 115 - - 1.7 CLASSIFICATION OF SYSTEMS - Page 117 - - 1.7-1 Linear and Nonlinear Systems - Page 117 - - 1.7-2 Time-Invariant and Time-Varying Systems - Page 122 - - 1.7-3 Instantaneous and Dynamic Systems - Page 123 - - 1.7-4 Causal and Noncausal Systems - Page 124 - - 1.7-5 Continuous-Time and Discrete-Time Systems - Page 127 - - 1.7-6 Analog and Digital Systems - Page 129 - - 1.7-7 Invertible and Noninvertible Systems - Page 129 - - 1.7-8 Stable and Unstable Systems - Page 130 - - 1.8 SYSTEM MODEL: INPUT–OUTPUT DESCRIPTION - Page 131 - - 1.8-1 Electrical Systems - Page 131 - - 1.8-2 Mechanical Systems - Page 134 - - 1.8-3 Electromechanical Systems - Page 138 - - 1.9 INTERNAL AND EXTERNAL DESCRIPTIONS OF A SYSTEM - Page 139 - - 1.10 INTERNAL DESCRIPTION: THE STATE-SPACE DESCRIPTION - Page 141 - - 1.11 MATLAB: WORKING WITH FUNCTIONS - Page 146 - - 1.11-1 Anonymous Functions - Page 146 - - 1.11-2 Relational Operators and the Unit Step Function - Page 148 - - 1.11-3 Visualizing Operations on the Independent Variable - Page 150 - - 1.11-4 Numerical Integration and Estimating Signal Energy - Page 151 - - 1.12 SUMMARY - Page 153 - - REFERENCES - Page 155 - - PROBLEMS - Page 156 -- 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - Page 170 - - 2.1 INTRODUCTION - Page 170 - - 2.2 SYSTEM RESPONSE TO INTERNAL CONDITIONS: THE ZERO-INPUT RESPONSE - Page 171 - - 2.2-1 Some Insights into the Zero-Input Behavior of a System - Page 181 - - 2.3 THE UNIT IMPULSE RESPONSE h(t) - Page 183 - - 2.4 SYSTEM RESPONSE TO EXTERNAL INPUT: THE ZERO-STATE RESPONSE - Page 188 - - 2.4-1 The Convolution Integral - Page 190 - - 2.4-2 Graphical Understanding of Convolution Operation - Page 198 - - 2.4-3 Interconnected Systems - Page 210 - - 2.4-4 A Very Special Function for LTIC Systems: The Everlasting Exponential e^{st} - Page 213 - - 2.4-5 Total Response - Page 215 - - 2.5 SYSTEM STABILITY - Page 216 - - 2.5-1 External (BIBO) Stability - Page 216 - - 2.5-2 Internal (Asymptotic) Stability - Page 218 - - 2.5-3 Relationship Between BIBO and Asymptotic Stability - Page 219 - - 2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR - Page 223 - - 2.6-1 Dependence of System Behavior on Characteristic Modes - Page 223 - - 2.6-2 Response Time of a System: The System Time Constant - Page 225 - - 2.6-3 Time Constant and Rise Time of a System - Page 226 - - 2.6-4 Time Constant and Filtering - Page 227 - - 2.6-5 Time Constant and Pulse Dispersion (Spreading) - Page 229 - - 2.6-6 Time Constant and Rate of Information Transmission - Page 229 - - 2.6-7 The Resonance Phenomenon - Page 230 - - 2.7 MATLAB: M-FILES - Page 232 - - 2.7-1 Script M-Files - Page 233 - - 2.7-2 Function M-Files - Page 234 - - 2.7-3 For-Loops - Page 235 - - 2.7-4 Graphical Understanding of Convolution - Page 237 - - 2.8 APPENDIX: DETERMINING THE IMPULSE RESPONSE - Page 240 - - 2.9 SUMMARY - Page 241 - - REFERENCES - Page 243 - - PROBLEMS - Page 243 -- 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - Page 257 - - 3.1 INTRODUCTION - Page 257 - - 3.1-1 Size of a Discrete-Time Signal - Page 258 - - 3.2 USEFUL SIGNAL OPERATIONS - Page 260 - - 3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS - Page 265 - - 3.3-1 Discrete-Time Impulse Function δ[n] - Page 265 - - 3.3-2 Discrete-Time Unit Step Function u[n] - Page 266 - - 3.3-3 Discrete-Time Exponential γ^n - Page 267 - - 3.3-4 Discrete-Time Sinusoid cos(Omega n+θ) - Page 271 - - 3.3-5 Discrete-Time Complex Exponential e^{jOmega n} - Page 272 - - 3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS - Page 273 - - 3.4-1 Classification of Discrete-Time Systems - Page 282 - - 3.5 DISCRETE-TIME SYSTEM EQUATIONS - Page 285 - - 3.5-1 Recursive (Iterative) Solution of Difference Equation - Page 286 - - 3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS: THE ZERO-INPUT RESPONSE - Page 290 - - 3.7 THE UNIT IMPULSE RESPONSE h[n] - Page 297 - - 3.7-1 The Closed-Form Solution of h[n] - Page 298 - - 3.8 SYSTEM RESPONSE TO EXTERNAL INPUT: THE ZERO-STATE RESPONSE - Page 300 - - 3.8-1 Graphical Procedure for the Convolution Sum - Page 308 - - 3.8-2 Interconnected Systems - Page 314 - - 3.8-3 Total Response - Page 317 - - 3.9 SYSTEM STABILITY - Page 318 - - 3.9-1 External (BIBO) Stability - Page 318 - -## Available Raw Markdown Chunks (Direct Hugging Face Raw URLs - 0 Bot Protection) -- [001_Cover.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/001_Cover.md) — LINEAR SYSTEMS AND SIGNALS -- [002_Half title.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/002_Half%20title.md) — (c) -- [003_B.2 SINUSOIDS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/003_B.2%20SINUSOIDS.md) — [B.2 SINUSOIDS](#page-6-0) -- [004_B.3 SKETCHING SIGNALS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/004_B.3%20SKETCHING%20SIGNALS.md) — [B.3 SKETCHING](#page-6-0) SIGNALS -- [005_B.4 CRAMER’S RULE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/005_B.4%20CRAMER%E2%80%99S%20RULE.md) — EXAMPLE B.7 Using Cramer's Rule to Solve a System of Equations -- [006_B.5 PARTIAL FRACTION EXPANSION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/006_B.5%20PARTIAL%20FRACTION%20EXPANSION.md) — [B.5 PARTIAL](#page-6-0) FRACTION EXPANSION -- [007_B.6 VECTORS AND MATRICES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/007_B.6%20VECTORS%20AND%20MATRICES.md) — [B.6-1 Some Definitions and Properties](#page-6-0) -- [008_B.7 MATLAB - ELEMENTARY OPERATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/008_B.7%20MATLAB%20-%20ELEMENTARY%20OPERATIONS.md) — [B.7 MATLAB: ELEMENTARY](#page-6-0) OPERATIONS -- [009_B.8 APPENDIX - USEFUL MATHEMATICAL FORMULAS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/009_B.8%20APPENDIX%20-%20USEFUL%20MATHEMATICAL%20FORMULAS.md) — [B.8-4 Taylor and Maclaurin Series](#page-7-0) -- [010_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/010_REFERENCES.md) — [REFERENCES](#page-7-0) -- [011_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/011_PROBLEMS.md) — - (d) Express w1 + w2 in standard rectangular form. -- [012_1 SIGNALS AND SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/012_1%20SIGNALS%20AND%20SYSTEMS.md) — [SIGNALS AND](#page-7-0) SYSTEMS -- [013_1.1 SIZE OF A SIGNAL.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/013_1.1%20SIZE%20OF%20A%20SIGNAL.md) — [1.1 SIZE OF A](#page-7-0) SIGNAL -- [014_1.2 SOME USEFUL SIGNAL OPERATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/014_1.2%20SOME%20USEFUL%20SIGNAL%20OPERATIONS.md) — 1.2 SOME USEFUL SIGNAL [OPERATIONS](#page-7-0) -- [015_1.3 CLASSIFICATION OF SIGNALS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/015_1.3%20CLASSIFICATION%20OF%20SIGNALS.md) — [1.3 CLASSIFICATION OF](#page-7-0) SIGNALS -- [016_1.4 SOME USEFUL SIGNAL MODELS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/016_1.4%20SOME%20USEFUL%20SIGNAL%20MODELS.md) — [1.4 SOME](#page-7-0) USEFUL SIGNAL MODELS -- [017_1.5 EVEN AND ODD FUNCTIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/017_1.5%20EVEN%20AND%20ODD%20FUNCTIONS.md) — [1.5 EVEN AND](#page-7-0) ODD FUNCTIONS -- [018_1.6 SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/018_1.6%20SYSTEMS.md) — [1.6 SYSTEMS](#page-7-0) -- [019_1.7 CLASSIFICATION OF SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/019_1.7%20CLASSIFICATION%20OF%20SYSTEMS.md) — [1.7 CLASSIFICATION OF](#page-7-0) SYSTEMS -- [020_1.8 SYSTEM MODEL - INPUT–OUTPUT DESCRIPTION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/020_1.8%20SYSTEM%20MODEL%20-%20INPUT%E2%80%93OUTPUT%20DESCRIPTION.md) — 1.8 SYSTEM [MODEL: INPUT–OUTPUT](#page-8-0) DESCRIPTION -- [021_1.9 INTERNAL AND EXTERNAL DESCRIPTIONS OF A SYSTEM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/021_1.9%20INTERNAL%20AND%20EXTERNAL%20DESCRIPTIONS%20OF%20A%20SYSTEM.md) — 1.9 INTERNAL AND EXTERNAL [DESCRIPTIONS OF A](#page-8-0) SYSTEM -- [022_1.10 INTERNAL DESCRIPTION - THE STATE-SPACE DESCRIPTION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/022_1.10%20INTERNAL%20DESCRIPTION%20-%20THE%20STATE-SPACE%20DESCRIPTION.md) — 1.10 INTERNAL [DESCRIPTION: THE](#page-8-0) STATE-SPACE DESCRIPTION -- [023_1.11 MATLAB - WORKING WITH FUNCTIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/023_1.11%20MATLAB%20-%20WORKING%20WITH%20FUNCTIONS.md) — [1.11 MATLAB: WORKING WITH](#page-8-0) FUNCTIONS -- [024_1.12 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/024_1.12%20SUMMARY.md) — [1.12 SUMMARY](#page-8-0) -- [025_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/025_REFERENCES.md) — [REFERENCES](#page-8-0) -- [026_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/026_PROBLEMS.md) — TIME-DOMAIN ANALYSIS OF [CONTINUOUS-TIME](#page-8-0) SYSTEMS -- [027_2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/027_2%20TIME-DOMAIN%20ANALYSIS%20OF%20CONTINUOUS-TIME%20SYSTEMS.md) — 154 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS -- [028_2.2 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/028_2.2%20SYSTEM%20RESPONSE%20TO%20INTERNAL%20CONDITIONS%20-%20THE%20ZERO-INPUT%20RESPONSE.md) — (a) -- [029_2.3 THE UNIT IMPULSE RESPONSE h(t).md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/029_2.3%20THE%20UNIT%20IMPULSE%20RESPONSE%20h(t).md) — EXAMPLE 2.5 Impulse Response via Impulse Matching -- [030_2.4 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/030_2.4%20SYSTEM%20RESPONSE%20TO%20EXTERNAL%20INPUT%20-%20THE%20ZERO-STATE%20RESPONSE.md) — [2.4-1 The Convolution Integral](#page-8-0) -- [031_2.5 SYSTEM STABILITY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/031_2.5%20SYSTEM%20STABILITY.md) — [2.5 SYSTEM](#page-8-0) STABILITY -- [032_2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/032_2.6%20INTUITIVE%20INSIGHTS%20INTO%20SYSTEM%20BEHAVIOR.md) — 2.6 INTUITIVE [INSIGHTS INTO](#page-8-0) SYSTEM BEHAVIOR -- [033_2.7 MATLAB - M-FILES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/033_2.7%20MATLAB%20-%20M-FILES.md) — [2.7 MATLAB: M-FILES](#page-8-0) -- [034_2.8 APPENDIX - DETERMINING THE IMPULSE RESPONSE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/034_2.8%20APPENDIX%20-%20DETERMINING%20THE%20IMPULSE%20RESPONSE.md) — [2.8 APPENDIX: DETERMINING THE](#page-9-0) IMPULSE RESPONSE -- [035_2.9 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/035_2.9%20SUMMARY.md) — [2.9 SUMMARY](#page-9-0) -- [036_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/036_REFERENCES.md) — [REFERENCES](#page-9-0) -- [037_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/037_PROBLEMS.md) — [PROBLEMS](#page-9-0) -- [038_3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/038_3%20TIME-DOMAIN%20ANALYSIS%20OF%20DISCRETE-TIME%20SYSTEMS.md) — TIME-DOMAIN ANALYSIS OF [DISCRETE-TIME](#page-9-0) SYSTEMS -- [039_3.1 INTRODUCTION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/039_3.1%20INTRODUCTION.md) — [3.1 INTRODUCTION](#page-9-0) -- [040_3.2 USEFUL SIGNAL OPERATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/040_3.2%20USEFUL%20SIGNAL%20OPERATIONS.md) — 3.2 USEFUL SIGNAL [OPERATIONS](#page-9-0) -- [041_3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/041_3.3%20SOME%20USEFUL%20DISCRETE-TIME%20SIGNAL%20MODELS.md) — 3.3 SOME USEFUL [DISCRETE-TIME](#page-9-0) SIGNAL MODELS -- [042_3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/042_3.4%20EXAMPLES%20OF%20DISCRETE-TIME%20SYSTEMS.md) — [3.4 EXAMPLES OF](#page-9-0) DISCRETE-TIME SYSTEMS -- [043_3.5 DISCRETE-TIME SYSTEM EQUATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/043_3.5%20DISCRETE-TIME%20SYSTEM%20EQUATIONS.md) — [3.5 DISCRETE-TIME](#page-9-0) SYSTEM EQUATIONS -- [044_3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/044_3.6%20SYSTEM%20RESPONSE%20TO%20INTERNAL%20CONDITIONS%20-%20THE%20ZERO-INPUT%20RESPONSE.md) — 3.7 THE UNIT IMPULSE [RESPONSE](#page-9-0) h[n] -- [045_3.7 THE UNIT IMPULSE RESPONSE h[n].md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/045_3.7%20THE%20UNIT%20IMPULSE%20RESPONSE%20h%5Bn%5D.md) — EXAMPLE 3.17 Iterative Determination of the Impulse Response -- [046_3.8 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/046_3.8%20SYSTEM%20RESPONSE%20TO%20EXTERNAL%20INPUT%20-%20THE%20ZERO-STATE%20RESPONSE.md) — 3.8 SYSTEM [RESPONSE TO](#page-9-0) EXTERNAL INPUT: THE ZERO-STATE RESPONSE -- [047_3.9 SYSTEM STABILITY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/047_3.9%20SYSTEM%20STABILITY.md) — [3.9 SYSTEM](#page-9-0) STABILITY -- [048_3.10 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/048_3.10%20INTUITIVE%20INSIGHTS%20INTO%20SYSTEM%20BEHAVIOR.md) — [3.10 INTUITIVE](#page-9-0) INSIGHTS INTO SYSTEM BEHAVIOR -- [049_3.11 MATLAB - DISCRETE-TIME SIGNALS AND SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/049_3.11%20MATLAB%20-%20DISCRETE-TIME%20SIGNALS%20AND%20SYSTEMS.md) — [3.11 MATLAB: DISCRETE-TIME](#page-9-0) SIGNALS AND SYSTEMS -- [050_3.12 APPENDIX - IMPULSE RESPONSE FOR A SPECIAL CASE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/050_3.12%20APPENDIX%20-%20IMPULSE%20RESPONSE%20FOR%20A%20SPECIAL%20CASE.md) — [3.12 APPENDIX: IMPULSE](#page-9-0) RESPONSE FOR A SPECIAL CASE -- [051_3.13 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/051_3.13%20SUMMARY.md) — [3.13 SUMMARY](#page-9-0) -- [052_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/052_PROBLEMS.md) — [CONTINUOUS-TIME](#page-10-0) SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM -- [053_4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/053_4%20CONTINUOUS-TIME%20SYSTEM%20ANALYSIS%20USING%20THE%20LAPLACE%20TRANSFORM.md) — (a) -- [054_4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/054_4.2%20SOME%20PROPERTIES%20OF%20THE%20LAPLACE%20TRANSFORM.md) — 4.2 SOME [PROPERTIES OF THE](#page-10-0) LAPLACE TRANSFORM -- [055_4.3 SOLUTION OF DIFFERENTIAL AND INTEGRO-DIFFERENTIAL EQUATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/055_4.3%20SOLUTION%20OF%20DIFFERENTIAL%20AND%20INTEGRO-DIFFERENTIAL%20EQUATIONS.md) — 4.3 SOLUTION OF DIFFERENTIAL AND [INTEGRO-DIFFERENTIAL](#page-10-0) EQUATIONS -- [056_4.4 ANALYSIS OF ELECTRICAL NETWORKS - THE TRANSFORMED NETWORK.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/056_4.4%20ANALYSIS%20OF%20ELECTRICAL%20NETWORKS%20-%20THE%20TRANSFORMED%20NETWORK.md) — 4.4 ANALYSIS OF ELECTRICAL NETWORKS: THE TRANSFORMED NETWORK -- [057_4.5 BLOCK DIAGRAMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/057_4.5%20BLOCK%20DIAGRAMS.md) — 4.5 BLOCK [DIAGRAMS](#page-10-0) -- [058_4.6 SYSTEM REALIZATION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/058_4.6%20SYSTEM%20REALIZATION.md) — 4.6 SYSTEM [REALIZATION](#page-10-0) -- [059_4.7 APPLICATION TO FEEDBACK AND CONTROLS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/059_4.7%20APPLICATION%20TO%20FEEDBACK%20AND%20CONTROLS.md) — [4.7 APPLICATION TO](#page-10-0) FEEDBACK AND CONTROLS -- [060_4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/060_4.8%20FREQUENCY%20RESPONSE%20OF%20AN%20LTIC%20SYSTEM.md) — 4.8 FREQUENCY [RESPONSE OF AN](#page-10-0) LTIC SYSTEM -- [061_4.9 BODE PLOTS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/061_4.9%20BODE%20PLOTS.md) — [4.9 BODE](#page-10-0) PLOTS -- [062_4.10 FILTER DESIGN BY PLACEMENT OF POLES AND ZEROS OF H(s).md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/062_4.10%20FILTER%20DESIGN%20BY%20PLACEMENT%20OF%20POLES%20AND%20ZEROS%20OF%20H(s).md) — 4.10 FILTER DESIGN BY [PLACEMENT OF](#page-10-0) POLES AND ZEROS OF H(s) -- [063_4.11 THE BILATERAL LAPLACE TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/063_4.11%20THE%20BILATERAL%20LAPLACE%20TRANSFORM.md) — 4.11 THE BILATERAL LAPLACE [TRANSFORM](#page-10-0) -- [064_4.12 MATLAB - CONTINUOUS-TIME FILTERS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/064_4.12%20MATLAB%20-%20CONTINUOUS-TIME%20FILTERS.md) — [4.12 MATLAB: CONTINUOUS-TIME](#page-11-0) FILTERS -- [065_4.13 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/065_4.13%20SUMMARY.md) — [4.13 SUMMARY](#page-11-0) -- [066_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/066_REFERENCES.md) — [DISCRETE-TIME](#page-11-0) SYSTEM ANALYSIS USING THE z-TRANSFORM -- [067_5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/067_5%20DISCRETE-TIME%20SYSTEM%20ANALYSIS%20USING%20THE%20z-TRANSFORM.md) — 496 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM -- [068_5.2 SOME PROPERTIES OF THE z-TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/068_5.2%20SOME%20PROPERTIES%20OF%20THE%20z-TRANSFORM.md) — 5.2 SOME [PROPERTIES OF THE](#page-11-0) z-TRANSFORM -- [069_5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/069_5.3%20z-TRANSFORM%20SOLUTION%20OF%20LINEAR%20DIFFERENCE%20EQUATIONS.md) — (d) -- [070_5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/070_5.5%20FREQUENCY%20RESPONSE%20OF%20DISCRETE-TIME%20SYSTEMS.md) — 5.5 FREQUENCY RESPONSE OF [DISCRETE-TIME](#page-11-0) SYSTEMS -- [071_5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/071_5.6%20FREQUENCY%20RESPONSE%20FROM%20POLE-ZERO%20LOCATIONS.md) — 5.6 FREQUENCY [RESPONSE FROM](#page-11-0) POLE-ZERO LOCATIONS -- [072_5.7 DIGITAL PROCESSING OF ANALOG SIGNALS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/072_5.7%20DIGITAL%20PROCESSING%20OF%20ANALOG%20SIGNALS.md) — 5.7 DIGITAL [PROCESSING OF](#page-11-0) ANALOG SIGNALS -- [073_5.8 THE BILATERAL z-TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/073_5.8%20THE%20BILATERAL%20z-TRANSFORM.md) — 5.8 THE BILATERAL z[-TRANSFORM](#page-11-0) -- [074_5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/074_5.9%20CONNECTING%20THE%20LAPLACE%20AND%20z-TRANSFORMS.md) — [5.9 CONNECTING THE](#page-11-0) LAPLACE AND z-TRANSFORMS -- [075_5.10 MATLAB - DISCRETE-TIME IIR FILTERS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/075_5.10%20MATLAB%20-%20DISCRETE-TIME%20IIR%20FILTERS.md) — [5.10 MATLAB: DISCRETE-TIME](#page-11-0) IIR FILTERS -- [076_5.11 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/076_5.11%20SUMMARY.md) — [5.11 SUMMARY](#page-12-0) -- [077_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/077_REFERENCES.md) — $$ -- [078_6 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER SERIES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/078_6%20CONTINUOUS-TIME%20SIGNAL%20ANALYSIS%20-%20THE%20FOURIER%20SERIES.md) — [CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER SERIES -- [079_6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/079_6.1%20PERIODIC%20SIGNAL%20REPRESENTATION%20BY%20TRIGONOMETRIC%20FOURIER%20SERIES.md) — 6.1 PERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY TRIGONOMETRIC FOURIER SERIES -- [080_6.2 EXISTENCE AND CONVERGENCE OF THE FOURIER SERIES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/080_6.2%20EXISTENCE%20AND%20CONVERGENCE%20OF%20THE%20FOURIER%20SERIES.md) — [6.2 EXISTENCE AND](#page-12-0) CONVERGENCE OF THE FOURIER SERIES -- [081_6.3 EXPONENTIAL FOURIER SERIES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/081_6.3%20EXPONENTIAL%20FOURIER%20SERIES.md) — EXAMPLE 6.6 Exponential Fourier Series of Periodic Exponential Wave -- [082_6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/082_6.4%20LTIC%20SYSTEM%20RESPONSE%20TO%20PERIODIC%20INPUTS.md) — [6.4 LTIC SYSTEM](#page-12-0) RESPONSE TO PERIODIC INPUTS -- [083_6.5 GENERALIZED FOURIER SERIES - SIGNALS AS VECTORS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/083_6.5%20GENERALIZED%20FOURIER%20SERIES%20-%20SIGNALS%20AS%20VECTORS.md) — [6.5 GENERALIZED](#page-12-0) FOURIER SERIES: SIGNALS AS VECTORS -- [084_6.6 NUMERICAL COMPUTATION OF D_n.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/084_6.6%20NUMERICAL%20COMPUTATION%20OF%20D_n.md) — 6.6 NUMERICAL [COMPUTATION OF](#page-12-0) Dn -- [085_6.7 MATLAB - FOURIER SERIES APPLICATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/085_6.7%20MATLAB%20-%20FOURIER%20SERIES%20APPLICATIONS.md) — [6.7 MATLAB: FOURIER](#page-12-0) SERIES APPLICATIONS -- [086_6.8 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/086_6.8%20SUMMARY.md) — [6.8 SUMMARY](#page-12-0) -- [087_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/087_REFERENCES.md) — [REFERENCES](#page-12-0) -- [088_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/088_PROBLEMS.md) — [CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER TRANSFORM -- [089_7 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/089_7%20CONTINUOUS-TIME%20SIGNAL%20ANALYSIS%20-%20THE%20FOURIER%20TRANSFORM.md) — 696 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM -- [090_7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/090_7.2%20TRANSFORMS%20OF%20SOME%20USEFUL%20FUNCTIONS.md) — [7.2-1 Connection Between the Fourier and Laplace Transforms](#page-12-0) -- [091_7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/091_7.3%20SOME%20PROPERTIES%20OF%20THE%20FOURIER%20TRANSFORM.md) — 7.3 Some Properties of the Fourier Transform 701 -- [092_7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/092_7.4%20SIGNAL%20TRANSMISSION%20THROUGH%20LTIC%20SYSTEMS.md) — 7.4 SIGNAL [TRANSMISSION](#page-12-0) THROUGH LTIC SYSTEMS -- [093_7.5 IDEAL AND PRACTICAL FILTERS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/093_7.5%20IDEAL%20AND%20PRACTICAL%20FILTERS.md) — [7.5 IDEAL AND](#page-13-0) PRACTICAL FILTERS -- [094_7.6 SIGNAL ENERGY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/094_7.6%20SIGNAL%20ENERGY.md) — [7.6 SIGNAL](#page-13-0) ENERGY -- [095_7.7 APPLICATION TO COMMUNICATIONS - AMPLITUDE MODULATION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/095_7.7%20APPLICATION%20TO%20COMMUNICATIONS%20-%20AMPLITUDE%20MODULATION.md) — [7.7 APPLICATION TO](#page-13-0) COMMUNICATIONS: AMPLITUDE MODULATION -- [096_7.8 DATA TRUNCATION - WINDOW FUNCTIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/096_7.8%20DATA%20TRUNCATION%20-%20WINDOW%20FUNCTIONS.md) — 7.8 DATA [TRUNCATION: WINDOW](#page-13-0) FUNCTIONS -- [097_7.9 MATLAB - FOURIER TRANSFORM TOPICS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/097_7.9%20MATLAB%20-%20FOURIER%20TRANSFORM%20TOPICS.md) — [7.9 MATLAB: FOURIER](#page-13-0) TRANSFORM TOPICS -- [098_7.10 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/098_7.10%20SUMMARY.md) — [7.10 SUMMARY](#page-13-0) -- [099_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/099_REFERENCES.md) — [REFERENCES](#page-13-0) -- [100_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/100_PROBLEMS.md) — [PROBLEMS](#page-13-0) -- [101_8 SAMPLING - THE BRIDGE FROM CONTINUOUS TO DISCRETE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/101_8%20SAMPLING%20-%20THE%20BRIDGE%20FROM%20CONTINUOUS%20TO%20DISCRETE.md) — SAMPLING: THE BRIDGE FROM [CONTINUOUS TO](#page-13-0) DISCRETE -- [102_8.1 THE SAMPLING THEOREM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/102_8.1%20THE%20SAMPLING%20THEOREM.md) — 8.1 THE [SAMPLING](#page-13-0) THEOREM -- [103_8.2 SIGNAL RECONSTRUCTION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/103_8.2%20SIGNAL%20RECONSTRUCTION.md) — 8.2 SIGNAL [RECONSTRUCTION](#page-13-0) -- [104_8.3 ANALOG-TO-DIGITAL (A - D) CONVERSION.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/104_8.3%20ANALOG-TO-DIGITAL%20(A%20-%20D)%20CONVERSION.md) — [8.3 ANALOG-TO-DIGITAL](#page-13-0) (A/D) CONVERSION -- [105_8.4 DUAL OF TIME SAMPLING - SPECTRAL SAMPLING.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/105_8.4%20DUAL%20OF%20TIME%20SAMPLING%20-%20SPECTRAL%20SAMPLING.md) — 8.4 DUAL OF TIME [SAMPLING: SPECTRAL](#page-13-0) SAMPLING -- [106_8.5 NUMERICAL COMPUTATION OF THE FOURIER TRANSFORM - THE DISCRETE FOURIER TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/106_8.5%20NUMERICAL%20COMPUTATION%20OF%20THE%20FOURIER%20TRANSFORM%20-%20THE%20DISCRETE%20FOURIER%20TRANSFORM.md) — 8.5 NUMERICAL [COMPUTATION OF THE](#page-13-0) FOURIER TRANSFORM: THE -- [107_8.6 THE FAST FOURIER TRANSFORM (FFT).md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/107_8.6%20THE%20FAST%20FOURIER%20TRANSFORM%20(FFT).md) — 8.6 THE FAST FOURIER [TRANSFORM](#page-13-0) (FFT) -- [108_8.7 MATLAB - THE DISCRETE FOURIER TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/108_8.7%20MATLAB%20-%20THE%20DISCRETE%20FOURIER%20TRANSFORM.md) — [8.7 MATLAB: THE](#page-13-0) DISCRETE FOURIER TRANSFORM -- [109_8.8 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/109_8.8%20SUMMARY.md) — [8.8 SUMMARY](#page-13-0) -- [110_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/110_REFERENCES.md) — [REFERENCES](#page-13-0) -- [111_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/111_PROBLEMS.md) — FOURIER ANALYSIS OF [DISCRETE-TIME](#page-14-0) SIGNALS -- [112_9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/112_9%20FOURIER%20ANALYSIS%20OF%20DISCRETE-TIME%20SIGNALS.md) — 852 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS -- [113_9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/113_9.2%20APERIODIC%20SIGNAL%20REPRESENTATION%20BY%20FOURIER%20INTEGRAL.md) — 9.2 APERIODIC SIGNAL REPRESENTATION BY FOURIER INTEGRAL -- [114_9.3 PROPERTIES OF THE DTFT.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/114_9.3%20PROPERTIES%20OF%20THE%20DTFT.md) — FREQUENCY-SHIFTING PROPERTY -- [115_9.4 LTI DISCRETE-TIME SYSTEM ANALYSIS BY DTFT.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/115_9.4%20LTI%20DISCRETE-TIME%20SYSTEM%20ANALYSIS%20BY%20DTFT.md) — [9.4 LTI DISCRETE-TIME](#page-14-0) SYSTEM ANALYSIS BY DTFT -- [116_9.5 DTFT CONNECTION WITH THE CTFT.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/116_9.5%20DTFT%20CONNECTION%20WITH%20THE%20CTFT.md) — [9.5 DTFT CONNECTION WITH THE](#page-14-0) CTFT -- [117_9.6 GENERALIZATION OF THE DTFT TO THE z-TRANSFORM.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/117_9.6%20GENERALIZATION%20OF%20THE%20DTFT%20TO%20THE%20z-TRANSFORM.md) — [9.6 GENERALIZATION OF THE](#page-14-0) DTFT TO THE z-TRANSFORM -- [118_9.7 MATLAB - WORKING WITH THE DTFS AND THE DTFT.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/118_9.7%20MATLAB%20-%20WORKING%20WITH%20THE%20DTFS%20AND%20THE%20DTFT.md) — [9.7-1 Computing the Discrete-Time Fourier Series](#page-14-0) -- [119_9.8 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/119_9.8%20SUMMARY.md) — [9.8 SUMMARY](#page-14-0) -- [120_REFERENCE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/120_REFERENCE.md) — [REFERENCE](#page-14-0) -- [121_PROBLEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/121_PROBLEMS.md) — $$ -- [122_10 STATE-SPACE ANALYSIS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/122_10%20STATE-SPACE%20ANALYSIS.md) — [STATE-SPACE](#page-14-0) ANALYSIS -- [123_10.1 MATHEMATICAL PRELIMINARIES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/123_10.1%20MATHEMATICAL%20PRELIMINARIES.md) — 910 CHAPTER 10 STATE-SPACE ANALYSIS -- [124_10.2 INTRODUCTION TO STATE SPACE.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/124_10.2%20INTRODUCTION%20TO%20STATE%20SPACE.md) — [10.2 INTRODUCTION TO](#page-14-0) STATE SPACE -- [125_10.3 A SYSTEMATIC PROCEDURE TO DETERMINE STATE EQUATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/125_10.3%20A%20SYSTEMATIC%20PROCEDURE%20TO%20DETERMINE%20STATE%20EQUATIONS.md) — [10.3 A SYSTEMATIC](#page-14-0) PROCEDURE TO DETERMINE STATE EQUATIONS -- [126_10.4 SOLUTION OF STATE EQUATIONS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/126_10.4%20SOLUTION%20OF%20STATE%20EQUATIONS.md) — [10.4 SOLUTION OF](#page-14-0) STATE EQUATIONS -- [127_10.5 LINEAR TRANSFORMATION OF A STATE VECTOR.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/127_10.5%20LINEAR%20TRANSFORMATION%20OF%20A%20STATE%20VECTOR.md) — 10.5 LINEAR [TRANSFORMATION OF A](#page-14-0) STATE VECTOR -- [128_10.6 CONTROLLABILITY AND OBSERVABILITY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/128_10.6%20CONTROLLABILITY%20AND%20OBSERVABILITY.md) — [10.6 CONTROLLABILITY AND](#page-14-0) OBSERVABILITY -- [129_10.7 STATE-SPACE ANALYSIS OF DISCRETE-TIME SYSTEMS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/129_10.7%20STATE-SPACE%20ANALYSIS%20OF%20DISCRETE-TIME%20SYSTEMS.md) — [10.7 STATE-SPACE](#page-15-0) ANALYSIS OF DISCRETE-TIME SYSTEMS -- [130_10.8 MATLAB - TOOLBOXES AND STATE-SPACE ANALYSIS.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/130_10.8%20MATLAB%20-%20TOOLBOXES%20AND%20STATE-SPACE%20ANALYSIS.md) — [10.8 MATLAB: TOOLBOXES AND](#page-15-0) STATE-SPACE ANALYSIS -- [131_10.9 SUMMARY.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/131_10.9%20SUMMARY.md) — [10.9 SUMMARY](#page-15-0) -- [132_REFERENCES.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/132_REFERENCES.md) — $$ -- [133_INDEX.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/133_INDEX.md) — [INDEX](#page-15-0) \ No newline at end of file diff --git a/engineering/llm.txt b/engineering/llm.txt deleted file mode 100644 index 4a4607913f951a3e509eebe2eee58e65ae92d6ad..0000000000000000000000000000000000000000 --- a/engineering/llm.txt +++ /dev/null @@ -1,24 +0,0 @@ -# Category Index: Engineering -> Total Books in Category: 2 | Hugging Face Storage Backend - -## Category Overview -This index lists all books in **Engineering** stored on Hugging Face Dataset (learner20011/CloverTexts-Data). - -## Books Directory -### 1. Fundamentals of Electric Circuits -- **Doc ID**: fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku -- **Author**: Charles K. Alexander and Matthew N.O. Sadiku -- **Pages**: 990 | **Year**: 2015 -- **Tags**: 6th, alexander, and, charles, circuits, electric, fundamentals, matthew -- **Book LLM Text**: [https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/llm.txt](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/llm.txt) -- **PDF Download**: [https://huggingface.co/datasets/learner20011/CloverTexts-Data/resolve/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku.pdf](https://huggingface.co/datasets/learner20011/CloverTexts-Data/resolve/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku.pdf) -- **Chapters Count**: 234 - -### 2. LINEAR SYSTEMS AND SIGNALS -- **Doc ID**: linear-systems-and-signals-3rd-edition-bp-lathi -- **Author**: B. P. Lathi and R. A. Green -- **Pages**: 1010 | **Year**: 2017 -- **Tags**: 3rd, and, lathi, linear, signals, systems -- **Book LLM Text**: [https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/llm.txt](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/llm.txt) -- **PDF Download**: [https://huggingface.co/datasets/learner20011/CloverTexts-Data/resolve/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.pdf](https://huggingface.co/datasets/learner20011/CloverTexts-Data/resolve/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/linear-systems-and-signals-3rd-edition-bp-lathi.pdf) -- **Chapters Count**: 133 diff --git a/llms.txt b/llms.txt deleted file mode 100644 index 5fecb3f3ecf00f0e965db08c328901a6d717ea8c..0000000000000000000000000000000000000000 --- a/llms.txt +++ /dev/null @@ -1,19 +0,0 @@ -# CloverTexts Multi-Book Academic Library Index -> Position-Aware Multi-Book PDF & AI Academic Platform -> Hosted on GitHub Pages | Storage Backend: Hugging Face Datasets (learner20011/CloverTexts-Data) -> 100% Scrapable (Zero Bot Challenge / Direct HTTP GET) - -## Available Categories & Category LLM Indices -- [Category: Engineering](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/llm.txt): 2 book(s) -- [Category: trading](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/trading/llm.txt): 2 book(s) - -## All Books Directory & Hugging Face LLM Indices -- [Fundamentals of Electric Circuits (Engineering)](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/llm.txt) — Author: Charles K. Alexander and Matthew N.O. Sadiku, 990 pages, 234 chapters -- [LINEAR SYSTEMS AND SIGNALS (engineering)](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/llm.txt) — Author: B. P. Lathi and R. A. Green, 1010 pages, 133 chapters -- [Technical Analysis of the Financial Markets (trading)](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/llm.txt) — Author: John J. Murphy, 494 pages, 32 chapters -- [The New Trading for a Living (trading)](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/trading/The New Trading for a Living/llm.txt) — Author: Alexander Elder, 303 pages, 21 chapters - -## Library API Endpoints -- Aggregate Catalog: [catalog.json](/catalog.json) (or Hugging Face: https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/catalog.json) -- Master LLM Index: [llms.txt](/llms.txt) (or Hugging Face: https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/llms.txt) -- Hugging Face Dataset Hub: https://huggingface.co/datasets/learner20011/CloverTexts-Data \ No newline at end of file diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/001_Title Page.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/001_Title Page.md deleted file mode 100644 index 64093f0ee6ef2fce0318541a26b7eb7d51ad1ba6..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/001_Title Page.md +++ /dev/null @@ -1,159 +0,0 @@ -JOHN J. MURPHY - -#### **NEW YORK INSTITUTE OF FINANCE** - -#### **Published by the Penguin Group** - -**Penguin Group (USA) Inc.** - -#### **375 Hudson Street, New York, New York 10014, USA** - -Penguin Group (Canada), 90 Eglinton Avenue East, Suite 700, Toronto, Ontario M4P 2Y3, Canada (a division of Pearson Penguin Canada Inc.) - -Penguin Books Ltd., 80 Strand, London WC2R 0RL, England - -Penguin Group Ireland, 25 St. Stephen's Green, Dublin 2, Ireland (a division of Penguin Books Ltd.) Penguin Group (Australia), 250 Camberwell Road, Camberwell, Victoria 3124, Australia (a division of Pearson Australia Group Pty. Ltd.) - -Penguin Books India Pvt. Ltd., 11 Community Centre, Panchsheel Park, New Delhi—110 017, India Penguin Group (NZ), 67 Apollo Drive, Rosedale, Auckland 0632, New Zealand (a division of Pearson New Zealand Ltd.) - -Penguin Books (South Africa) (Pty.) Ltd., 24 Sturdee Avenue, Rosebank, Johannesburg 2196, South Africa - -Penguin Books Ltd., Registered Offices: 80 Strand, London WC2R 0RL, England - -While the author has made every effort to provide accurate telephone numbers and Internet addresses at the time of publication, neither the publisher nor the author assumes any responsibility for errors or for changes that occur after publication. Further, the publisher does not have any control over and does not assume any responsibility for author or third-party websites or their content. - -Portions of this book were previously published as *Technical Analysis of the Futures Markets* (New York Institute of Finance, 1985). - -Copyright © 1999 by John J. Murphy - -All rights reserved. - -No part of this book may be reproduced, scanned, or distributed in any printed or electronic form without permission. Please do not participate in or encourage piracy of copyrighted materials in violation of the author's rights. Purchase only authorized editions. - -NEW YORK INSTITUTE OF FINANCE and NYIF are trademarks of Executive Tax Reports, Inc., used under license by Penguin Group (USA) Inc. - -First edition: January 1999 - -Library of Congress Cataloging-in-Publication Data - -Murphy, John J. - -Technical analysis of the financial markets / John J. Murphy. p. cm. Rev. ed. of: Technical analysis of the futures markets. c1986. Includes bibliographical references and index. ISBN: 978-1-101-65919-9 1. Futures market. 2. Commodity exchanges. I. Murphy, John J. Technical analysis of the future markets. II. Title. HG6046.M87 1999 98-38531 332.64'4—dc21 CIP - -PUBLISHER'S NOTE: This publication is designed to provide accurate and authoritative information in regard to the subject matter covered. It is sold with the understanding that the publisher is not engaged in rendering legal, accounting, or other professional services. If you require legal advice or - -other expert assistance, you should seek the services of a competent professional. - -Most New York Institute of Finance books are available at special quantity discounts for bulk purchases for sales promotions, premiums, fund-raising, or educational use. Special books, or book excerpts, can also be created to fit specific needs. For details, write: Special Markets, Penguin Group (USA) Inc., 375 Hudson Street, New York, New York 10014. - -Version\_2 - -*To my parents, Timothy and Margaret and To Patty, Clare, and Brian* - -# **Contents** - -**About the Author About the Contributors [Introduction](#page-18-0) [Acknowledgments](#page-19-0)** - -# **1 [Philosophy](#page-22-0) of Technical Analysis** - -Introduction [Philosophy](#page-23-0) or Rationale [Technical](#page-23-1) versus Fundamental Forecasting [Analysis](#page-23-2) versus Timing Flexibility and [Adaptability](#page-26-0) of Technical Analysis [Technical](#page-27-0) Analysis Applied to Different Trading Mediums [Technical](#page-28-0) Analysis Applied to Different Time Dimensions Economic [Forecasting](#page-29-0) [Technician](#page-29-1) or Chartist? A Brief [Comparison](#page-30-0) of Technical Analysis in Stocks and Futures Less [Reliance](#page-30-1) on Market Averages and Indicators Some [Criticisms](#page-31-0) of the Technical Approach [Random](#page-34-0) Walk Theory Universal [Principles](#page-34-1) - -# **2 Dow Theory** - -Introduction Basic [Tenets](#page-40-0) The [Use](#page-40-1) of Closing Prices and the Presence of Lines Some [Critici](#page-41-0)sms of Dow Theory Stocks as [Economic](#page-46-0) Indicators Dow Theory [Applied](#page-47-0) to Futures Trading [Conclusion](#page-48-0) - -# **[3](#page-49-0) Chart Construction** - -Introduction Types of Charts [Available](#page-50-0) [Candlesticks](#page-50-1) [Arithmetic](#page-50-2) versus Logarithmic Scale [Construction](#page-52-0) of the Daily Bar Chart [Volume](#page-53-0) [Futures](#page-54-0) Open Interest [Weekly](#page-55-0) and Monthly Bar Charts [Conclusion](#page-56-0) - -# **[4](#page-60-0) Basic Concepts of Trend** - -Definition of Trend Trend Has Three [Directions](#page-61-0) [Trend](#page-61-1) Has Three Classifications Support and [Resistance](#page-63-0) [Trendlines](#page-64-0) The Fan [Principle](#page-66-0) - -The Importance of the Number Three The Relative Steepness of the Trendline The [Channel](#page-84-0) Line Percentage [Retracements](#page-84-1) Speed [Resistance](#page-88-0) Lines Gann and [Fibonacci](#page-93-0) Fan Lines Internal [Trendlines](#page-95-0) [Reversal](#page-97-0) Days Price [Gaps](#page-98-0) [Conclusion](#page-98-1) - -# **[5](#page-104-0) Major Reversal Patterns** - -Introduction Price [Patterns](#page-105-0) Two [Types](#page-105-1) of Patterns: Reversal and Continuation The [Head](#page-105-2) and Shoulders Reversal Pattern The [Importance](#page-106-0) of Volume Finding a Price [Objective](#page-108-0) The [Inverse](#page-112-0) Head and Shoulders [Complex](#page-112-1) Head and Shoulders Patterns Triple Tops and [Bottoms](#page-114-0) Double Tops and [Bottoms](#page-116-0) [Variations](#page-118-0) from the Ideal Pattern [Saucers](#page-120-0) and Spikes [Conclusion](#page-123-0) - -# **[6](#page-128-0) Continuation Patterns** - -Introduction - -Triangles The Symmetrical Triangle The [Asce](#page-130-0)nding Triangle The [Descending](#page-131-0) Triangle The [Broadening](#page-134-0) Formation Flags and [Pennants](#page-136-0) The Wedge [Formation](#page-138-0) The [Rectangle](#page-139-0) Formation The [Measured](#page-143-0) Move The [Continuation](#page-144-0) Head and Shoulders Pattern [Confirmation](#page-148-0) and Divergence [Conclusion](#page-149-0) - -# **[7](#page-152-0) Volume and Open Interest** - -Introduction [Volume](#page-153-0) and Open Interest as Secondary Indicators [Interpretatio](#page-153-1)n of Volume for All Markets [Interpretation](#page-153-2) of Open Interest in Futures [Summary](#page-157-0) of Volume and Open Interest Rules Blowoffs and Selling [Climaxes](#page-163-0) [Commitments](#page-168-0) of Traders Report Watch the [Commercials](#page-168-1) Net Trader [Positions](#page-168-2) Open [Interest](#page-169-0) In Options [Put/Call](#page-170-0) Ratios [Combine](#page-170-1) Option Sentiment With Technicals [Conclusion](#page-171-0) - -# **8 Long Term Charts** - -Introduction The [Importance](#page-174-0) of Longer Range Perspective [Construction](#page-174-1) of Continuation Charts for Futures The Perpetual [Contract™](#page-174-2) Long Term Trends Dispute [Randomness](#page-175-0) [Patterns](#page-176-0) on Charts: Weekly and Monthly Reversals Long Term to Short Term [Charts](#page-177-0) Why Should Long Range Charts Be [Adjusted](#page-177-1) for Inflation? Long Term Charts Not [Intended](#page-177-2) for Trading Purposes [Examples](#page-178-0) of Long Term Charts - -# **9 Moving [Averages](#page-180-0)** - -Introduction The Moving [Average:](#page-186-0) A Smoothing Device with a Time Lag [Moving](#page-186-1) Average Envelopes [Bollinger](#page-187-0) Bands Using [Bollinger](#page-196-0) Bands as Targets Band [Width](#page-197-0) Measures Volatility Moving [Averages](#page-199-0) Tied to Cycles [Fibonacci](#page-199-1) Numbers Used as Moving Averages Moving [Averages](#page-200-0) Applied to Long Term Charts The [Weekly](#page-201-0) Rule To [Optimize](#page-201-1) or Not [Summary](#page-203-0) The [Adaptive](#page-208-0) Moving Average [Alternativ](#page-208-1)es to the Moving Average - -# **10 Oscillators and Contrary Opinion** - -Introduction Oscillator Usage in [Conjunction](#page-211-0) with Trend [Measuring](#page-211-1) Momentum [Measuring](#page-212-0) Rate of Change (ROC) [Constructing](#page-213-0) an Oscillator Using Two Moving Averages [Commodity](#page-218-0) Channel Index The Relative [Strength](#page-218-1) Index (RSI) [Using](#page-221-0) the 70 and 30 Lines to Generate Signals [Stochastics](#page-223-0) (K%D) Larry [Williams](#page-229-0) %R The [Importance](#page-229-1) of Trend When [Oscillators](#page-232-0) are Most Useful Moving [Average](#page-233-0) Convergence/Divergence (MACD) MACD [Histogram](#page-234-0) [Combine](#page-235-0) Weeklies and Dailies The [Principle](#page-237-0) of Contrary Opinion in Futures Investor [Sentiment](#page-238-0) Readings Investors [Intelligence](#page-239-0) Numbers - -# **[11](#page-243-0) Point and Figure Charting** - -Introduction The Point and Figure [Versus](#page-245-0) the Bar Chart [Construction](#page-245-1) of the Intraday Point and Figure Chart The [Horizontal](#page-245-2) Count Price [Patterns](#page-249-0) 3 Box [Reversal](#page-252-0) Point and Figure Charting [Construction](#page-253-0) of the 3 Point Reversal Chart - -The Drawing of Trendlines Measuring Techniques [Trading](#page-260-0) Tactics [Advantages](#page-263-0) of Point and Figure Charts P&F [Technical](#page-264-0) Indicators [Computerized](#page-265-0) P&F Charting P&F Moving [Averages](#page-269-0) [Conclusion](#page-269-1) - -# **[12](#page-272-0) Japanese Candlesticks** - -Introduction [Candlestick](#page-273-0) Charting Basic [Candle](#page-273-1)sticks [Candle](#page-273-2) Pattern Analysis [Filtered](#page-275-0) Candle Patterns [Conclusion](#page-277-0) Candle [Patterns](#page-282-0) - -# **[13](#page-284-0) Elliott Wave Theory** - -Historical Background Basic [Tenets](#page-293-0) of the Elliott Wave Principle [Connection](#page-293-1) Between Elliott Wave and Dow Theory [Corrective](#page-293-2) Waves The Rule of [Alternation](#page-297-0) [Channeling](#page-298-0) Wave 4 as a [Support](#page-306-0) Area [Fibonacci](#page-307-0) Numbers as the Basis of the Wave Principle [Fibonacci](#page-308-0) Ratios and Retracements - -Fibonacci Time Targets Combining All Three Aspects of Wave Theory Elliott Wave [Applied](#page-312-0) to Stocks Versus Commodities Summary and [Conclusions](#page-312-1) [Reference](#page-314-0) Material - -# **[14](#page-316-0) Time Cycles** - -Introduction [Cycles](#page-317-0) How [Cyclic](#page-317-1) Concepts Help Explain Charting Techniques [Domin](#page-318-0)ant Cycles [Combining](#page-327-0) Cycle Lengths The [Importance](#page-330-0) of Trend Left and Right [Translation](#page-332-0) How to Isolate [Cycles](#page-333-0) [Seasonal](#page-334-0) Cycles Stock [Market](#page-334-1) Cycles The [January](#page-339-0) Barometer The [Presidential](#page-343-0) Cycle [Combining](#page-343-1) Cycles with Other Technical Tools [Maximum](#page-344-0) Entropy Spectral Analysis Cycle Reading and [Software](#page-344-1) - -# **[15](#page-345-0) Computers and Trading Systems** - -Introduction Some [Computer](#page-346-0) Needs [Grouping](#page-346-1) Tools and Indicators [Using](#page-347-0) the Tools and Indicators - -Welles Wilder's Parabolic and Directional Movement Systems Pros and Cons of System Trading Need [Expert](#page-349-0) Help? Test [Systems](#page-355-0) or Create Your Own [Conclusion](#page-356-0) - -# **[16](#page-357-0) Money Management and Trading Tactics** - -Introduction The Three Elements of [Successful](#page-359-0) Trading [Money](#page-359-1) Management [Reward](#page-359-2) to Risk Ratios Trading [Multiple](#page-360-0) Positions: Trending versus Trading Units [What](#page-362-0) to Do After Periods of Success and Adversity [Trading](#page-363-0) Tactics [Combining](#page-363-1) Technical Factors and Money Management Types of [Tradin](#page-364-0)g Orders From Daily Charts to [Intraday](#page-366-0) Price Charts The Use of [Intraday](#page-367-0) Pivot Points Summary of Money [Management](#page-368-0) and Trading Guidelines [Application](#page-370-0) to Stocks Asset [Allocation](#page-371-0) Managed [Accounts](#page-372-0) and Mutual Funds [Market](#page-372-1) Profile - -## **[17](#page-373-0) The Link Between Stocks and Futures: Intermarket Analysis** - -[Intermarket](#page-374-0) Analysis Program Trading: The Ultimate Link The Link Between Bonds and Stocks The Link Between Bonds and Commodities The Link Between [Commodities](#page-377-0) and the Dollar Stock Sectors and [Industry](#page-378-0) Groups The [Dollar](#page-378-1) and Large Caps [Intermarket](#page-380-0) Analysis and Mutual Funds Relative Strength [Analysis](#page-381-0) Relative [Strength](#page-381-1) and Sectors Relative [Strength](#page-382-0) and Individual Stocks [Top-Down](#page-384-0) Market Approach [Deflation](#page-385-0) Scenario [Intermarket](#page-385-1) Correlation [Intermarket](#page-386-0) Neural Network Software [Conclusion](#page-387-0) - -# **[18](#page-388-0) Stock Market Indicators** - -Measuring Market Breadth [Sample](#page-390-0) Data [Comparing](#page-390-1) Market Averages The [Advance](#page-390-2)-Decline Line AD [Divergence](#page-391-0) Daily Versus [Weekly](#page-392-0) AD Lines [Variations](#page-393-0) in AD Line [McClellan](#page-393-1) Oscillator McClellan [Summation](#page-394-0) Index New Highs [Versus](#page-394-1) New Lows New [High-New](#page-395-0) Low Index Upside Versus [Downside](#page-396-0) Volume The Arms Index TRIN Versus TICK [Smoothing](#page-399-0) the Arms Index Open [Arms](#page-400-0) [Equivolume](#page-400-1) Charting [Candlepowe](#page-401-0)r [Comparing](#page-401-1) Market Averages [Conclusion](#page-403-0) - -# **[19](#page-406-0) Pulling It All Together—A Checklist** - -Technical Checklist How to Coordinate Technical and [Fundamental](#page-407-0) Analysis [Chartered](#page-407-1) Market Technician (CMT) Market [Technicians](#page-409-0) Association (MTA) The Global Reach of [Technical](#page-409-1) Analysis [Technical](#page-410-0) Analysis by Any Name Federal Reserve Finally [Approves](#page-410-1) [Conclusion](#page-411-0) - -# **[A](#page-412-0)Advanced Technical Indicators** - -Demand Index (DI) [Herrick](#page-415-0) Payoff Index (HPI) Starc [Bands](#page-415-1) and Keltner Channels [Formula](#page-417-0) for Demand Index - -# **[B](#page-422-0) Market Profile** - -Introduction - -Market Profile Graphic Market Structure Market Profile [Organizi](#page-426-0)ng Principles Range [Developm](#page-427-0)ent and Profile Patterns [Tracking](#page-428-0) Longer Term Market Activity [Conclusion](#page-432-0) - -# **[C](#page-437-0)The Essentials of Building a Trading System** - -5-Step Plan Step 1: Start with a [Concept](#page-439-0) (an Idea) [Step](#page-440-0) 2: Turn Your Idea into a Set of Objective Rules Step 3: [Visually](#page-440-1) Check It Out on the Charts Step 4: Formally Test It with a [Computer](#page-442-0) Step 5: [Evaluate](#page-442-1) Results Money [Management](#page-442-2) [Conclusion](#page-445-0) - -# **[D](#page-446-0)Continuous Futures Contracts** - -Nearest Contract Next [Contract](#page-449-0) Gann [Contract](#page-449-1) [Continuous](#page-450-0) Contracts [Constant](#page-450-1) Forward Continuous Contracts **[Glossary](#page-451-0) Selected [Bibliography](#page-451-1) [Selected](#page-453-0) Resources [Index](#page-461-0)** diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/002_Copyright.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/002_Copyright.md deleted file mode 100644 index b1e3218395aa9ae3e1e2df20ff4a15121f790979..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/002_Copyright.md +++ /dev/null @@ -1,9 +0,0 @@ -# **About the Author** - -**John J. Murphy** has been applying technical analysis for three decades. He was formerly Director of Futures Technical Research and the senior managed account trading advisor with Merrill Lynch. Mr. Murphy was the technical analyst for CNBC-TV for seven years. He is the author of three books, including *Technical Analysis of the Futures Markets*, the predecessor to this book. His second book, *Intermarket Technical Analysis*, opened up a new branch of analysis. His third book, *The Visual Investor*, applies technical work to mutual funds. - -In 1996, Mr. Murphy founded MURPHYMORRIS, Inc., along with software developer Greg Morris, to produce interactive educational products and online analysis for investors. Their Web site address is: - -#### www.murphymorris.com. - -He is also head of his own consulting firm, JJM Technical Advisors, located in Oradell, New Jersey. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/003_About the Contributors.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/003_About the Contributors.md deleted file mode 100644 index 0781b99b8233c737ec53df3dd9387e902a71b79d..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/003_About the Contributors.md +++ /dev/null @@ -1,9 +0,0 @@ -# **About the Contributors** - -**Thomas E. Aspray** (Appendix A) is a Capital Market Analyst with Princeton Economic Institute Ltd., located in Princeton, New Jersey. Mr. Aspray has been trading markets since the 1970s. Many of the techniques he pioneered in the early 1980s are now [used](#page-415-0) by other professional traders. - -**Dennis C. Hynes** (Appendix B) is Managing Director and cofounder of R.W. Pressprich & Co., Inc., a fixed income broker/dealer located in New York City. He also serves as the firm's Chief Market Strategist. Mr. Hynes is a futures and options trader and a [CTA](#page-424-0) (Commodity Trading Advisor). He has an MBA in Finance from the University of Houston. - -**Greg Morris** (Chapter 12 and Appendix D) has been developing trading systems and indicators for 20 years for investors and traders to be used with major technical analysis software programs. He is the author of two books on candlestick charting (see [Chapt](#page-273-0)er 12). In [August](#page-449-0) 1996, Mr. Morris teamed with John Murphy to found MURPHYMORRIS Inc., a Dallas-based firm dedicated to educating investors. - -**Fred G. Schutzman, [CMT](#page-273-0)** (Appendix C) is the President and Chief Executive Officer of Briarwood Capital Management, Inc., a New York-based Commodity Trading Advisor. He is also responsible for technical research and trading system development at [Emcor](#page-439-0) Eurocurrency Management Corporation, a risk management consulting firm. Mr. Schutzman is a member of the Market Technicians Association and is currently serving on their Board of Directors. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/004_Introduction.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/004_Introduction.md deleted file mode 100644 index 3d96d9a0047146513097ff071e47b51f9bde8504..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/004_Introduction.md +++ /dev/null @@ -1,19 +0,0 @@ -# **Introduction** - -I had no idea when *Technical Analysis of the Futures Markets* was published in 1986 that it would create such an impact on the industry. It has been referred to by many in the field as the "Bible" of technical analysis. The Market Technicians Association uses it as a primary source in their testing process for the Chartered Market Technician program. The Federal Reserve has cited it in research studies that examine the value of the technical approach. In addition, it has been translated into eight foreign languages. I was also unprepared for the long shelf life of the book. It continues to sell as many copies ten years after it was published as it did in the first couple of years. - -It became clear, however, that a lot of new material had been added to the field of technical analysis in the past decade. I added some of it myself. My second book, *Intermarket Technical Analysis* (Wiley, 1991), helped create that new branch of technical analysis, which is widely used today. Old techniques like Japanese candlestick charting and newer ones like Market Profile have become part of the technical landscape. Clearly, this new work needed to be included in any book that attempted to present a comprehensive picture of technical analysis. The focus of my work changed as well. - -While my main interest ten years ago was in the futures markets, my recent work has dealt more with the stock market. That also brought me full circle, since I began my career as a stock analyst thirty years ago. That was also one of the side effects of my being the technical analyst for CNBC-TV for seven years. That focus on what the general public was doing also led to my third book, *The Visual Investor* (Wiley, 1996). That book focused on the use of technical tools for market sectors, primarily through mutual funds, which have become extremely popular in the 1990s. - -Many of the technical indicators that I wrote about ten years ago, which had been used primarily in the futures markets, have been incorporated into stock market work. It was time to show how that was being done. Finally, like any field or discipline, writers also evolve. Some things that seemed very important to me ten years ago aren't as important today. As my work has evolved into a broader application of technical principles to all financial - -markets, it seemed only right that any revision of that earlier work should reflect that evolution. - -I've tried to retain the structure of the original book. Therefore, many of the original chapters remain. However, they have been revised with new material and updated with new graphics. Since the principles of technical analysis are universal, it wasn't that difficult to broaden the focus to include all financial markets. Since the original focus was on futures, however, a lot of stock market material has been added. - -Three new chapters have been added. The two previous chapters on point and figure charting (Chapters 11 and 12) have been merged into one. A new Chapter 12 on candlestick charting has been inserted. Two additional chapters have also been added at the end of the book. Chapter 17 is an introduction to my work on [intermarke](#page-245-0)t an[aly](#page-273-0)sis. Chapter 18 deals with stock market [indicator](#page-273-0)s. We've replaced the previous appendices with new ones. Market Profile is introduced in Appendix B. The other [appendice](#page-374-0)s show some of the more advanced technical indicators and ex[plain](#page-390-0) how to build a technical trading system. There's also a glossary. - -I approached this revision with [some](#page-424-0) trepidation. I wasn't sure redoing a book considered a "classic" was such a good idea. I hope I've succeeded in making it even better. I approached this work from the perspective of a more seasoned and mature writer and analyst. And, throughout the book, I tried to show the respect I have always had for the discipline of technical analysis and for the many talented analysts who practice it. The success of their work, as well as their dedication to this field, has always been a source of comfort and inspiration to me. I only hope I did justice to it and to them. - -*John Murphy* diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/005_Acknowledgments.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/005_Acknowledgments.md deleted file mode 100644 index 8087c231633d18d41bec435c623f305e349589d9..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/005_Acknowledgments.md +++ /dev/null @@ -1,3 +0,0 @@ -# **Acknowledgments** - -The person who deserves the most credit for the second edition of this book is Ellen Schneid Coleman, Executive Editor at Simon & Schuster. She convinced me that it was time to revise *Technical Analysis of the Futures Markets* and broaden its scope. I'm glad she was so persistent. Special thanks go to the folks at Omega Research who provided me with the charting software I needed and, in particular, Gaston Sanchez who spent a lot of time on the phone with me. The contributing authors—Tom Aspray, Dennis Hynes, and Fred Schutzman—added their particular expertise where it was needed. In addition, several analysts contributed charts including Michael Burke, Stan Ehrlich, Jerry Toepke, Ken Tower, and Nick Van Nice. The revision of Chapter 2 on Dow Theory was a collaborative effort with Elyce Picciotti, an independent technical writer and market consultant in New Orleans, Louisiana. Greg Morris deserves special mention. He wrote the chapter on [candlestic](#page-40-0)k charting, contributed the article in Appendix D, and did most of the graphic work. Fred Dahl of Inkwell Publishing Services (Fishkill, NY), who handled production of the first edition of this book, did this one as well. It was great working with him again. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/006_1 Philosophy of Technical Analysis.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/006_1 Philosophy of Technical Analysis.md deleted file mode 100644 index 89472f03384e104297b7e0290c3dba45244fd5f6..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/006_1 Philosophy of Technical Analysis.md +++ /dev/null @@ -1,300 +0,0 @@ -### **INTRODUCTION** - -Before beginning a study of the actual techniques and tools used in technical analysis, it is necessary first to define what technical analysis is, to discuss the philosophical premises on which it is based, to draw some clear distinctions between technical and fundamental analysis and, finally, to address a couple of criticisms frequently raised against the technical approach. - -The author's strong belief is that a full appreciation of the technical approach must begin with a clear understanding of what technical analysis claims to be able to do and, maybe even more importantly, the philosophy or rationale on which it bases those claims. - -First, let's define the subject. *Technical analysis is the study of market action, primarily through the use of charts, for the purpose of forecasting future price trends.* The term "market action" includes the three principal sources of information available to the technician—price, volume, and open interest. (Open interest is used only in futures and options.) The term "price action," which is often used, seems too narrow because most technicians include volume and open interest as an integral part of their market analysis. With this distinction made, the terms "price action" and "market action" are used interchangeably throughout the remainder of this discussion. - -### **PHILOSOPHY OR RATIONALE** - -There are three premises on which the technical approach is based: - -- 1. Market action discounts everything. -- 2. Prices move in trends. -- 3. History repeats itself. - -#### **Market Action Discounts Everything** - -The statement "market action discounts everything" forms what is probably the cornerstone of technical analysis. Unless the full significance of this first premise is fully understood and accepted, nothing else that follows makes much sense. The technician believes that anything that can possibly affect the price—fundamentally, politically, psychologically, or otherwise—is actually reflected in the price of that market. It follows, therefore, that a study of price action is all that is required. While this claim may seem presumptuous, it is hard to disagree with if one takes the time to consider its true meaning. - -All the technician is really claiming is that price action should reflect shifts in supply and demand. If demand exceeds supply, prices should rise. If supply exceeds demand, prices should fall. This action is the basis of all economic and fundamental forecasting. The technician then turns this statement around to arrive at the conclusion that if prices are rising, for whatever the specific reasons, demand must exceed supply and the fundamentals must be bullish. If prices fall, the fundamentals must be bearish. If this last comment about fundamentals seems surprising in the context of a discussion of technical analysis, it shouldn't. After all, the technician is indirectly studying fundamentals. Most technicians would probably agree that it is the underlying forces of supply and demand, the economic fundamentals of a market, that cause bull and bear markets. The charts do not in themselves cause markets to move up or down. They simply reflect the bullish or bearish psychology of the marketplace. - -As a rule, chartists do not concern themselves with the reasons why prices rise or fall. Very often, in the early stages of a price trend or at critical turning points, no one seems to know exactly why a market is performing a certain way. While the technical approach may sometimes seem overly simplistic in its claims, the logic behind this first premise—that markets discount everything—becomes more compelling the more market experience one gains. It follows then that if everything that affects market price is ultimately reflected in market price, then the study of that market price is all that is necessary. By studying price charts and a host of supporting technical indicators, the chartist in effect lets the market tell him or her which way it is most likely to go. The chartist does not necessarily try to outsmart or outguess the market. All of the technical tools discussed later on are simply techniques used to aid the chartist in the process of studying market action. The chartist knows there are reasons why markets go up or down. He or she just doesn't - -believe that knowing what those reasons are is necessary in the forecasting process. - -#### **Prices Move in Trends** - -The concept of trend is absolutely essential to the technical approach. Here again, unless one accepts the premise that markets do in fact trend, there's no point in reading any further. The whole purpose of charting the price action of a market is to identify trends in early stages of their development for the purpose of trading in the direction of those trends. In fact, most of the techniques used in this approach are trend-following in nature, meaning that their intent is to identify and follow existing trends. (See Figure 1.1.) - -**Figure 1.1** *Example of an uptrend. Technical analysis is based on the premise that markets trend and that those trends tend to persist.* - -There is a corollary to the premise that prices move in trends—*a trend in motion is more likely to continue than to reverse.* This corollary is, of course, an adaptation of Newton's first law of motion. Another way to state this corollary is that a trend in motion will continue in the same direction until it reverses. This is another one of those technical claims that seems almost circular. But the entire trend-following approach is predicated on riding an existing trend until it shows signs of reversing. - -### **History Repeats Itself** - -Much of the body of technical analysis and the study of market action has to - -do with the study of human psychology. Chart patterns, for example, which have been identified and categorized over the past one hundred years, reflect certain pictures that appear on price charts. These pictures reveal the bullish or bearish psychology of the market. Since these patterns have worked well in the past, it is assumed that they will continue to work well in the future. They are based on the study of human psychology, which tends not to change. Another way of saying this last premise—that history repeats itself—is that the key to understanding the future lies in a study of the past, or that the future is just a repetition of the past. - -### **TECHNICAL VERSUS FUNDAMENTAL FORECASTING** - -While technical analysis concentrates on the study of market action, fundamental analysis focuses on the economic forces of supply and demand that cause prices to move higher, lower, or stay the same. The fundamental approach examines all of the relevant factors affecting the price of a market in order to determine the intrinsic value of that market. The intrinsic value is what the fundamentals indicate something is actually worth based on the law of supply and demand. If this intrinsic value is under the current market price, then the market is overpriced and should be sold. If market price is below the intrinsic value, then the market is undervalued and should be bought. - -Both of these approaches to market forecasting attempt to solve the same problem, that is, to determine the direction prices are likely to move. They just approach the problem from different directions. *The fundamentalist studies the cause of market movement, while the technician studies the effect.* The technician, of course, believes that the effect is all that he or she wants or needs to know and that the reasons, or the causes, are unnecessary. The fundamentalist always has to know why. - -Most traders classify themselves as either technicians or fundamentalists. In reality, there is a lot of overlap. Many fundamentalists have a working knowledge of the basic tenets of chart analysis. At the same time, many technicians have at least a passing awareness of the fundamentals. The problem is that the charts and fundamentals are often in conflict with each other. Usually at the beginning of important market moves, the fundamentals do not explain or support what the market seems to be doing. It is at these critical times in the trend that these two approaches seem to differ the most. Usually they come back into sync at some point, but often too late for the trader to act. - -One explanation for these seeming discrepancies is that *market price* - -*tends to lead the known fundamentals.* Stated another way, *market price acts as a leading indicator of the fundamentals* or the conventional wisdom of the moment. While the known fundamentals have already been discounted and are already "in the market," prices are now reacting to the unknown fundamentals. Some of the most dramatic bull and bear markets in history have begun with little or no perceived change in the fundamentals. By the time those changes became known, the new trend was well underway. - -After a while, the technician develops increased confidence in his or her ability to read the charts. The technician learns to be comfortable in a situation where market movement disagrees with the so-called conventional wisdom. A technician begins to enjoy being in the minority. He or she knows that eventually the reasons for market action will become common knowledge. It is just that the technician isn't willing to wait for that added confirmation. - -In accepting the premises of technical analysis, one can see why technicians believe their approach is superior to the fundamentalists. If a trader had to choose only one of the two approaches to use, the choice would logically have to be the technical. Because, by definition, the technical approach includes the fundamental. If the fundamentals are reflected in market price, then the study of those fundamentals becomes unnecessary. Chart reading becomes a shortcut form of fundamental analysis. The reverse, however, is not true. Fundamental analysis does not include a study of price action. It is possible to trade financial markets using just the technical approach. It is doubtful that anyone could trade off the fundamentals alone with no consideration of the technical side of the market. - -### **ANALYSIS VERSUS TIMING** - -This last point is made clearer if the decision making process is broken down into two separate stages—analysis and timing. Because of the high leverage factor in the futures markets, timing is especially crucial in that arena. It is quite possible to be correct on the general trend of the market and still lose money. Because margin requirements are so low in futures trading (usually less than 10%), a relatively small price move in the wrong direction can force the trader out of the market with the resulting loss of all or most of that margin. In stock market trading, by contrast, a trader who finds him or herself on the wrong side of the market can simply decide to hold onto the stock, hoping that it will stage a comeback at some point. - -Futures traders don't have that luxury. A "buy and hold" strategy doesn't apply to the futures arena. Both the technical and the fundamental approach can be used in the first phase—the forecasting process. However, the question - -of timing, of determining specific entry and exit points, is almost purely technical. Therefore, considering the steps the trader must go through before making a market commitment, it can be seen that the correct application of technical principles becomes indispensable at some point in the process, even if fundamental analysis was applied in the earlier stages of the decision. Timing is also important in individual stock selection and in the buying and selling of stock market sector and industry groups. - -### **FLEXIBILITY AND ADAPTABILITY OF TECHNICAL ANALYSIS** - -One of the great strengths of technical analysis is its adaptability to virtually any trading medium and time dimension. There is no area of trading in either stocks or futures where these principles do not apply. - -The chartist can easily follow as many markets as desired, which is generally not true of his or her fundamental counterpart. Because of the tremendous amount of data the latter must deal with, most fundamentalists tend to specialize. The advantages here should not be overlooked. - -For one thing, markets go through active and dormant periods, trending and nontrending stages. The technician can concentrate his or her attention and resources in those markets that display strong trending tendencies and choose to ignore the rest. As a result, the chartist can rotate his or her attention and capital to take advantage of the rotational nature of the markets. At different times, certain markets become "hot" and experience important trends. Usually, those trending periods are followed by quiet and relatively trendless market conditions, while another market or group takes over. The technical trader is free to pick and choose. The fundamentalist, however, who tends to specialize in only one group, doesn't have that kind of flexibility. Even if he or she were free to switch groups, the fundamentalist would have a much more difficult time doing so than would the chartist. - -Another advantage the technician has is the "big picture." By following all of the markets, he or she gets an excellent feel for what markets are doing in general, and avoids the "tunnel vision" that can result from following only one group of markets. Also, because so many of the markets have built-in economic relationships and react to similar economic factors, price action in one market or group may give valuable clues to the future direction of another market or group of markets. - -### **TECHNICAL ANALYSIS APPLIED TO DIFFERENT TRADING MEDIUMS** - -The principles of chart analysis apply to both *stocks* and *futures.* Actually, technical analysis was first applied to the stock market and later adapted to futures. With the introduction of *stock index futures*, the dividing line between these two areas is rapidly disappearing. *International stock markets* are also charted and analyzed according to technical principles. (See Figure 1.2.) - -*Financial futures*, including *interest rate markets* and *foreign currencies*, have become enormously popular over the past decade and have proven to be excellent subjects for chart analysis. - -Technical principles play a role in *options trading.* Technical forecasting can also be used to great advantage in the *hedging process.* - -**Figure 1.2** *The Japanese stock market charts very well as do most stock markets around the world.* - -### **TECHNICAL ANALYSIS APPLIED TO DIFFERENT TIME DIMENSIONS** - -Another strength of the charting approach is its ability to handle different time dimensions. Whether the user is trading the intraday tic-by-tic changes for *day trading purposes* or *trend trading* the intermediate trend, the same principles apply. A time dimension often overlooked is *longer range technical* - -*forecasting.* The opinion expressed in some quarters that charting is useful only in the short term is simply not true. It has been suggested by some that fundamental analysis should be used for long term forecasting with technical factors limited to short term timing. The fact is that longer range forecasting, using weekly and monthly charts going back several years, has proven to be an extremely useful application of these techniques. - -Once the technical principles discussed in this book are thoroughly understood, they will provide the user with tremendous flexibility as to how they can be applied, both from the standpoint of the medium to be analyzed and the time dimension to be studied. - -### **ECONOMIC FORECASTING** - -Technical analysis can play a role in economic forecasting. For example, the direction of commodity prices tells us something about the direction of inflation. They also give us clues about the strength or weakness of the economy. Rising commodity prices generally hint at a stronger economy and rising inflationary pressure. Falling commodity prices usually warn that the economy is slowing along with inflation. The direction of interest rates is affected by the trend of commodities. As a result, charts of commodity markets like gold and oil, along with Treasury Bonds, can tell us a lot about the strength or weakness of the economy and inflationary expectations. The direction of the U.S. dollar and foreign currency futures also provide early guidance about the strength or weakness of the respective global economies. Even more impressive is the fact that trends in these futures markets usually show up long before they are reflected in traditional economic indicators that are released on a monthly or quarterly basis, and usually tell us what has already happened. As their name implies, futures markets usually give us insights into the future. The S&P 500 stock market index has long been counted as an official leading economic indicator. A book by one of the country's top experts on the business cycle, *Leading Indicators for the 1990s* (Moore), makes a compelling case for the importance of commodity, bond, and stock trends as economic indicators. All three markets can be studied employing technical analysis. We'll have more to say on this subject in Chapter 17, "The Link Between Stocks and Futures." - -### **[TECHN](#page-374-0)ICIAN OR CHARTIST?** - -There are several different titles applied to practitioners of the technical approach: technical analyst, chartist, market analyst, and visual analyst. Up until recently, they all meant pretty much the same thing. However, with increased specialization in the field, it has become necessary to make some further distinctions and define the terms a bit more carefully. Because virtually all technical analysis was based on the use of charts up until the last decade, the terms "technician" and "chartist" meant the same thing. This is no longer necessarily true. - -The broader area of technical analysis is being increasingly divided into two types of practitioners, the traditional chartist and, for want of a better term, statistical technicians. Admittedly, there is a lot of overlap here and most technicians combine both areas to some extent. As in the case of the technician versus the fundamentalist, most seem to fall into one category or the other. - -Whether or not the traditional chartist uses quantitative work to supplement his or her analysis, charts remain the primary working tool. Everything else is secondary. Charting, of necessity, remains somewhat subjective. The success of the approach depends, for the most part, on the skill of the individual chartist. The term "art charting" has been applied to this approach because chart reading is largely an art. - -By contrast, the statistical, or quantitative, analyst takes these subjective principles, quantifies, tests, and optimizes them for the purpose of developing mechanical trading systems. These systems, or trading models, are then programmed into a computer that generates mechanical "buy" and "sell" signals. These systems range from the simple to the very complex. However, the intent is to reduce or completely eliminate the subjective human element in trading, to make it more scientific. These statisticians may or may not use price charts in their work, but they are considered technicians as long as their work is limited to the study of market action. - -Even computer technicians can be subdivided further into those who favor mechanical systems, or the "black box" approach, and those who use computer technology to develop better technical indicators. The latter group maintains control over the interpretation of those indicators and also the decision making process. - -One way of distinguishing between the chartist and the statistician is to say that all chartists are technicians, but not all technicians are chartists. Although these terms are used interchangeably throughout this book, it should be remembered that charting represents only one area in the broader subject of technical analysis. - -### **A BRIEF COMPARISON OF TECHNICAL** - -## **ANALYSIS IN STOCKS AND FUTURES** - -A question often asked is whether technical analysis as applied to futures is the same as the stock market. The answer is both yes and no. The basic principles are the same, but there are some significant differences. The principles of technical analysis were first applied to stock market forecasting and only later adapted to futures. Most of the basic tools—bar charts, point and figure charts, price patterns, volume, trendlines, moving averages, and oscillators, for example—are used in both areas. Anyone who has learned these concepts in either stocks or futures wouldn't have too much trouble making the adjustment to the other side. However, there are some general areas of difference having more to do with the different nature of stocks and futures than with the actual tools themselves. - -### **Pricing Structure** - -The pricing structure in futures is much more complicated than in stocks. Each commodity is quoted in different units and increments. Grain markets, for example, are quoted in cents per bushel, livestock markets in cents per pound, gold and silver in dollars per ounce, and interest rates in basis points. The trader must learn the contract details of each market: which exchange it is traded on, how each contract is quoted, what the minimum and maximum price increments are, and what these price increments are worth. - -### **Limited Life Span** - -Unlike stocks, futures contracts have expiration dates. A March 1999 Treasury Bond contract, for example, expires in March of 1999. The typical futures contract trades for about a year and a half before expiration. Therefore, at any one time, at least a half dozen different contract months are trading in the same commodity at the same time. The trader must know which contracts to trade and which ones to avoid. (This is explained later in this book.) This limited life feature causes some problems for longer range price forecasting. It necessitates the continuing need for obtaining new charts once old contracts stop trading. The chart of an expired contract isn't of much use. New charts must be obtained for the newer contracts along with their own technical indicators. This constant rotation makes the maintenance of an ongoing chart library a good deal more difficult. For computer users, it also entails greater time and expense by making it necessary to be constantly obtaining new historical data as old contracts expire. - -### **Lower Margin Requirements** - -This is probably the most important difference between stocks and futures. - -All futures are traded on margin, which is usually less than 10% of the value of the contract. The result of these low margin requirements is tremendous leverage. Relatively small price moves in either direction tend to become magnified in their impact on overall trading results. For this reason, it is possible to make or lose large sums of money very quickly in futures. Because a trader puts up only 10% of the value of the contract as margin, then a 10% move in either direction will either double the trader's money or wipe it out. By magnifying the impact of even minor market moves, the high leverage factor sometimes makes the futures markets seem more volatile than they actually are. When someone says, for example, that he or she was "wiped out" in the futures market, remember that he or she only committed 10% in the first place. - -From the standpoint of technical analysis, the high leverage factor makes timing in the futures markets much more critical than it is in stocks. The correct timing of entry and exit points is crucial in futures trading and much more difficult and frustrating than market analysis. Largely for this reason, technical trading skills become indispensable to a successful futures trading program. - -#### **Time Frame Is Much Shorter** - -Because of the high leverage factor and the need for close monitoring of market positions, the time horizon of the commodity trader is much shorter of necessity. Stock market technicians tend to look more at the longer range picture and talk in time frames that are beyond the concern of the average commodity trader. Stock technicians may talk about where the market will be in three or six months. Futures traders want to know where prices will be next week, tomorrow, or maybe even later this afternoon. This has necessitated the refinement of very short term timing tools. One example is the moving average. The most commonly watched averages in stocks are 50 and 200 days. In commodities, most moving averages are under 40 days. A popular moving average combination in futures, for example, is 4, 9, and 18 days. - -#### **Greater Reliance on Timing** - -*Timing is everything in futures trading.* Determining the correct direction of the market only solves a portion of the trading problem. If the timing of the entry point is off by a day, or sometimes even minutes, it can mean the difference between a winner or a loser. It's bad enough to be on the wrong side of the market and lose money. Being on the right side of the market and still losing money is one of the most frustrating and unnerving aspects of futures trading. It goes without saying that timing is almost purely technical in nature, because the fundamentals rarely change on a day-to-day basis. - -### **LESS RELIANCE ON MARKET AVERAGES AND INDICATORS** - -Stock market analysis is based heavily on the movement of broad market averages—such as the Dow Jones Industrial Average or the S&P 500. In addition, technical indicators that measure the strength or weakness of the broader market—like the NYSE advance-decline line or the new highs-new lows list—are heavily employed. While commodity markets can be tracked using measures like the Commodity Research Bureau Futures Price Index, less emphasis is placed on the broader market approach. Commodity market analysis concentrates more on individual market action. That being the case, technical indicators that measure broader commodity trends aren't used much. With only about 20 or so active commodity markets, there isn't much need. - -### **Specific Technical Tools** - -While most of the technical tools originally developed in the stock market have some application in commodity markets, they are not used in the exact same way. For example, chart patterns in futures often tend not to form as fully as they do in stocks. - -Futures traders rely more heavily on shorter term indicators that emphasize more precise trading signals. These points of difference and many others are discussed later in this book. - -Finally, there is another area of major difference between stocks and futures. Technical analysis in stocks relies much more heavily on the use of *sentiment indicators* and *flow of funds* analysis. *Sentiment indicators* monitor the performance of different groups such as odd lotters, mutual funds, and floor specialists. Enormous importance is placed on sentiment indicators that measure the overall market bullishness and bearishness on the theory that the majority opinion is usually wrong. *Flow of funds* analysis refers to the cash position of different groups, such as mutual funds or large institutional accounts. The thinking here is that the larger the cash position, the more funds that are available for stock purchases. - -*Technical analysis in the futures markets is a much purer form of price analysis.* While contrary opinion theory is also used to some extent, much more emphasis is placed on basic trend analysis and the application of traditional technical indicators. - -### **SOME CRITICISMS OF THE TECHNICAL APPROACH** - -A few questions generally crop up in any discussion of the technical approach. One of these concerns is the *self-fulfilling prophecy.*Another is the question of whether or not past price data can really be used to forecast future price direction. The critic usually says something like: "Charts tell us where the market has been, but can't tell us where it is going." For the moment, we'll put aside the obvious answer that a chart won't tell you anything if you don't know how to read it. The Random Walk Theory questions whether prices trend at all and doubts that any forecasting technique can beat a simple *buy and hold* strategy. These questions deserve a response. - -#### **The Self-Fulfilling Prophecy** - -The question of whether there is a self-fulfilling prophecy at work seems to bother most people because it is raised so often. It is certainly a valid concern, but of much less importance than most people realize. Perhaps the best way to address this question is to quote from a text that discusses some of the disadvantages of using chart patterns: - -- a. The use of most chart patterns has been widely publicized in the last several years. Many traders are quite familiar with these patterns and often act on them in concert. This creates a "self-fulfilling prophecy," as waves of buying or selling are created in response to "bullish" or "bearish" patterns… -- b. Chart patterns are almost completely subjective. No study has yet succeeded in mathematically quantifying any of them. They are literally in the mind of the beholder…. (Teweles et al.) - -These two criticisms contradict one another and the second point actually cancels out the first. If chart patterns are "completely subjective" and "in the mind of the beholder," then it is hard to imagine how everyone could see the same thing at the same time, which is the basis of the self-fulfilling prophecy. Critics of charting can't have it both ways. They can't, on the one hand, criticize charting for being so objective and obvious that everyone will act in the same way at the same time (thereby causing the price pattern to be fulfilled), and then also criticize charting for being too subjective. - -The truth of the matter is that charting is very subjective. Chart reading is an art. (Possibly the word "skill" would be more to the point.) Chart patterns are seldom so clear that even experienced chartists always agree on their interpretation. There is always an element of doubt and disagreement. As this book demonstrates, there are many different approaches to technical analysis that often disagree with one another. - -Even if most technicians did agree on a market forecast, they would not all necessarily enter the market at the same time and in the same way. Some - -would try to anticipate the chart signal and enter the market early. Others would buy the "breakout" from a given pattern or indicator. Still others would wait for the pullback after the breakout before taking action. Some traders are aggressive; others are conservative. Some use stops to enter the market, while others like to use market orders or resting limit orders. Some are trading for the long pull, while others are day trading. Therefore, the possibility of all technicians acting at the same time and in the same way is actually quite remote. - -Even if the self-fulfilling prophecy were of major concern, it would probably be "self-correcting" in nature. In other words, traders would rely heavily on charts until their concerted actions started to affect or distort the markets. Once traders realized this was happening, they would either stop using the charts or adjust their trading tactics. For example, they would either try to act before the crowd or wait longer for greater confirmation. So, even if the self-fulfilling prophecy did become a problem over the near term, it would tend to correct itself. - -It must be kept in mind that bull and bear markets only occur and are maintained when they are justified by the law of supply and demand. Technicians could not possibly cause a major market move just by the sheer power of their buying and selling. If this were the case, technicians would all become wealthy very quickly. - -Of much more concern than the chartists is the tremendous growth in the use of computerized technical trading systems in the futures market. These systems are mainly trend-following in nature, which means that they are all programmed to identify and trade major trends. With the growth in professionally managed money in the futures industry, and the proliferation of multimillion-dollar public and private funds, most of which are using these technical systems, tremendous concentrations of money are chasing only a handful of existing trends. Because the universe of futures markets is still quite small, the potential for these systems distorting short term price action is growing. However, even in cases where distortions do occur, they are generally short term in nature and do not cause major moves. - -Here again, even the problem of concentrated sums of money using technical systems is probably self-correcting. If all of the systems started doing the same thing at the same time, traders would make adjustments by making their systems either more or less sensitive. - -The self-fulfilling prophecy is generally listed as a criticism of charting. It might be more appropriate to label it as a compliment. After all, for any forecasting technique to become so popular that it begins to influence events, it would have to be pretty good. We can only speculate as to why this concern is seldom raised regarding the use of fundamental analysis. - -#### **Can the Past Be Used to Predict the Future?** - -Another question often raised concerns the validity of using past price data to predict the future. It is surprising how often critics of the technical approach bring up this point because every known method of forecasting, from weather predicting to fundamental analysis, is based completely on the study of past data. What other kind of data is there to work with? - -The field of statistics makes a distinction between *descriptive statistics* and *inductive statistics. Descriptive statistics* refers to the graphical presentation of data, such as the price data on a standard bar chart. *Inductive statistics* refers to generalizations, predictions, or extrapolations that are inferred from that data. Therefore, the price chart itself comes under the heading of the descriptive, while the analysis technicians perform on that price data falls into the realm of the inductive. - -As one statistical text puts it, "The first step in forecasting the business or economic future consists, thus, of gathering observations from the past." (Freund and Williams) Chart analysis is just another form of *time series analysis*, based on a study of the past, which is exactly what is done in all forms of time series analysis. The only type of data anyone has to go on is past data. We can only estimate the future by projecting past experiences into that future. - -So it seems that the use of past price data to predict the future in technical analysis is grounded in sound statistical concepts. If anyone were to seriously question this aspect of technical forecasting, he or she would have to also question the validity of every other form of forecasting based on historical data, which includes all economic and fundamental analysis. - -### **RANDOM WALK THEORY** - -The *Random Walk Theory*, developed and nurtured in the academic community, claims that price changes are "serially independent" and that price history is not a reliable indicator of future price direction. In a nutshell, price movement is random and unpredictable. The theory is based on the *efficient market hypothesis*, which holds that prices fluctuate randomly about their intrinsic value. It also holds that the best market strategy to follow would be a simple "buy and hold" strategy as opposed to any attempt to "beat the market." - -While there seems little doubt that a certain amount of randomness or "noise" does exist in all markets, it's just unrealistic to believe that *all* price movement is random. This may be one of those areas where empirical observation and practical experience prove more useful than sophisticated statistical techniques, which seem capable of proving anything the user has in mind or incapable of disproving anything. It might be useful to keep in mind that randomness can only be defined in the negative sense of an inability to uncover systematic patterns in price action. The fact that many academics have not been able to discover the presence of these patterns does not prove that they do not exist. - -The academic debate as to whether markets trend is of little interest to the average market analyst or trader who is forced to deal in the real world where market trends are clearly visible. If the reader has any doubts on this point, a casual glance through any chart book (randomly selected) will demonstrate the presence of trends in a very graphic way. How do the "random walkers" explain the persistence of these trends if prices are serially independent, meaning that what happened yesterday, or last week, has no bearing on what may happen today or tomorrow? How do they explain the profitable "real life" track records of many trend-following systems? - -How, for example, would a buy and hold strategy fare in the commodity futures markets where timing is so crucial? Would those long positions be held during bear markets? How would traders even know the difference between bull and bear markets if prices are unpredictable and don't trend? In fact, how could a bear market even exist in the first place because that would imply a trend? (See Figure 1.3.) - -**Figure 1.3** *A "random walker" would have a tough time convincing a holder of gold bullion that there's no real trend on this chart.* - -It seems doubtful that statistical evidence will ever totally prove or disprove the Random Walk Theory. However, the idea that markets are - -random is totally rejected by the technical community. If the markets were truly random, no forecasting technique would work. Far from disproving the validity of the technical approach, the *efficient market hypothesis* is very close to the technical premise that *markets discount everything.* The academics, however, feel that because markets quickly discount all information, there's no way to take advantage of that information. The basis of technical forecasting, already touched upon, is that important market information is discounted in the market price long before it becomes known. Without meaning to, the academics have very eloquently stated the need for closely monitoring price action and the futility of trying to profit from fundamental information, at least over the short term. - -Finally, it seems only fair to observe that any process appears random and unpredictable to those who do not understand the rules under which that process operates. An electrocardiogram printout, for example, might appear like a lot of random noise to a layperson. But to a trained medical person, all those little blips make a lot of sense and are certainly not random. The working of the markets may appear random to those who have not taken the time to study the rules of market behavior. *The illusion of randomness gradually disappears as the skill in chart reading improves.* Hopefully, that is exactly what will happen as the reader progresses through the various sections of this book. - -There may even be hope for the academic world. A number of leading American universities have begun to explore Behavioral Finance which maintains that human psychology and securities pricing are intertwined. That, of course, is the primary basis of technical analysis. - -### **UNIVERSAL PRINCIPLES** - -When an earlier version of this book was published twelve years ago, many of the technical timing tools that were explained were used mainly in the futures markets. Over the past decade, however, these tools have been widely employed in analyzing stock market trends. The technical principles that are discussed in this book can be applied universally to all markets—even mutual funds. One additional feature of stock market trading that has gained wide popularity in the past decade has been sector investing, primarily through index options and mutual funds. Later in the book we'll show how to determine which sectors are hot and which are not by applying technical timing tools. - - - -### **INTRODUCTION** - -Charles Dow and his partner Edward Jones founded Dow Jones & Company in 1882. Most technicians and students of the markets concur that much of what we call *technical analysis* today has its origins in theories first proposed by Dow around the turn of the century. Dow published his ideas in a series of editorials he wrote for the *Wall Street Journal.* Most technicians today recognize and assimilate Dow's basic ideas, whether or not they recognize the source. *Dow Theory* still forms the cornerstone of the study of technical analysis, even in the face of today's sophisticated computer technology, and the proliferation of newer and supposedly better technical indicators. - -On July 3, 1884, Dow published the first stock market average composed of the closing prices of eleven stocks: nine railroad companies and two manufacturing firms. Dow felt that these eleven stocks provided a good indication of the economic health of the country. In 1897, Dow determined that two separate indices would better represent that health, and created a 12 stock industrial index and a 20 stock rail index. By 1928 the industrial index had grown to include 30 stocks, the number at which it stands today. The editors of *The Wall Street Journal* have updated the list numerous times in the ensuing years, adding a utility index in 1929. In 1984, the year that marked the one hundredth anniversary of Dow's first publication, the Market Technicians Association presented a Gorham-silver bowl to Dow Jones & Co. According to the MTA, the award recognized "the lasting contribution that Charles Dow made to the field of investment analysis. His index, the forerunner of what today is regarded as the leading barometer of stock market activity, remains a vital tool for market technicians 80 years after his death." - -Unfortunately for us, Dow never wrote a book on his theory. Instead, he set down his ideas of stock market behavior in a series of editorials that *The* - -*Wall Street Journal* published around the turn of the century. In 1903, the year after Dow's death, S.A. Nelson compiled these essays into a book entitled *The ABC of Stock Speculation.* In that work, Nelson first coined the term "Dow's Theory." Richard Russell, who wrote the introduction to a 1978 reprint, compared Dow's contribution to stock market theory with Freud's contribution to psychiatry. In 1922, William Peter Hamilton (Dow's associate and successor at the Journal) categorized and published Dow's tenets in a book entitled *The Stock Market Barometer.* Robert Rhea developed the theory even further in the *Dow Theory* (New York: Barron's), published in 1932. - -Dow applied his theoretical work to the stock market averages that he created; namely the Industrials and the Rails. However, most of his analytical ideas apply equally well to all market averages. This chapter will describe the six basic tenets of Dow Theory and will discuss how these ideas fit into a modern study of technical analysis. We will discuss the ramifications of these ideas in the chapters that follow. - -### **BASIC TENETS** - -#### **1. The Averages Discount Everything.** - -The sum and tendency of the transactions of the Stock Exchange represent the sum of all Wall Street's knowledge of the past, immediate and remote, applied to the discounting of the future. There is no need to add to the averages, as some statisticians do, elaborate compilations of commodity price index numbers, bank clearings, fluctuations in exchange, volume of domestic and foreign trades or anything else. Wall Street considers all these things (Hamilton, pp. 40–41). - -Sound familiar? The idea that the markets reflect every possible knowable factor that affects overall supply and demand is one of the basic premises of technical theory, as was mentioned in Chapter 1. The theory applies to market averages, as well as it does to individual markets, and even makes allowances for "acts of God." While the markets cannot anticipate events such as earthquakes and various other natural [calamit](#page-23-0)ies, they quickly discount such occurrences, and almost instantaneously assimilate their affects into the price action. - -#### **2. The Market Has Three Trends.** - -Before discussing how trends behave, we must clarify what Dow considered a trend. Dow defined an uptrend as a situation in which each successive rally - -closes higher than the previous rally high, and each successive rally low also closes higher than the previous rally low. In other words, an uptrend has a pattern of rising peaks and troughs. The opposite situation, with successively lower peaks and troughs, defines a downtrend. Dow's definition has withstood the test of time and still forms the cornerstone of trend analysis. - -Dow believed that the laws of action and reaction apply to the markets just as they do to the physical universe. He wrote, "Records of trading show that in many cases when a stock reaches top it will have a moderate decline and then go back again to near the highest figures. If after such a move, the price again recedes, it is liable to decline some distance" (Nelson, page 43). - -Dow considered a trend to have three parts, *primary, secondary*, and *minor*, which he compared to the tide, waves, and ripples of the sea. The primary trend represents the tide, the secondary or intermediate trend represents the waves that make up the tide, and the minor trends behave like ripples on the waves. - -An observer can determine the direction of the tide by noting the highest point on the beach reached by successive waves. If each successive wave reaches further inland than the preceding one, the tide is flowing in. When the high point of each successive wave recedes, the tide has turned out and is ebbing. Unlike actual ocean tides, which last a matter of hours, Dow conceived of market tides as lasting for more than a year, and possibly for several years. - -The secondary, or intermediate, trend represents corrections in the primary trend and usually lasts three weeks to three months. These intermediate corrections generally retrace between one-third and two-thirds of the previous trend movement and most frequently about half, or 50%, of the previous move. - -According to Dow, the minor (or near term) trend usually lasts less than three weeks. This near term trend represents fluctuations in the intermediate trend. We will discuss trend concepts in greater detail in Chapter 4, "Basic Concepts of Trends," where you will see that we continue to use the same basic concepts and terminology today. - -#### **3. Major Trends Have Three Phases.** - -Dow focused his attention on primary or major trends, which he felt usually take place in three distinct phases: an accumulation phase, a public participation phase, and a distribution phase. The accumulation phase represents informed buying by the most astute investors. If the previous trend was down, then at this point these astute investors recognize that the market has assimilated all the so-called "bad" news. The public participation phase, where most technical trend-followers begin to participate, occurs when prices - -begin to advance rapidly and business news improves. The distribution phase takes place when newspapers begin to print increasingly bullish stories; when economic news is better than ever; and when speculative volume and public participation increase. During this last phase the same informed investors who began to "accumulate" near the bear market bottom (when no one else wanted to buy) begin to "distribute" before anyone else starts selling. - -Students of Elliott Wave Theory will recognize this division of a major bull market into three distinct phases. R. N. Elliott elaborated upon Rhea's work in *Dow Theory*, to recognize that a bull market has three major, upward movements. In Chapter 13, "Elliott Wave Theory," we'll show the close similarity between Dow's three phases of a bull market and the five wave Elliott sequence. - -#### **4. The Averages Must Confirm Each Other.** - -Dow, in referring to the Industrial and Rail Averages, meant that no important bull or bear market signal could take place unless both averages gave the same signal, thus confirming each other. He felt that both averages must exceed a previous secondary peak to confirm the inception or continuation of a bull market. He did not believe that the signals had to occur simultaneously, but recognized that a shorter length of time between the two signals provided stronger confirmation. When the two averages diverged from one another, Dow assumed that the prior trend was still maintained. (Elliott Wave Theory only requires that signals be generated in a single average.) Chapter 6, "Continuation Patterns," will cover the key concepts of confirmation and divergence. (See Figures 2.1 and 2.2.) - -#### **5. Volume Must Confirm the Trend.** - -Dow recognized [volume](#page-44-0) as a sec[ond](#page-45-0)ary but important factor in confirming price signals. Simply stated, *volume should expand or increase in the direction of the major trend.* In a major uptrend, volume would then increase as prices move higher, and diminish as prices fall. In a downtrend, volume should increase as prices drop and diminish as they rally. Dow considered volume a secondary indicator. He based his actual buy and sell signals entirely on closing prices. In Chapter 7, "Volume and Open Interest," we'll cover the subject of volume and build on Dow's ideas. Today's sophisticated volume indicators help determine whether volume is increasing or falling off. Savvy traders then compare this [inform](#page-153-0)ation to price action to see if the two are confirming each other. - -**Figure 2.1** *A long term view of the Dow Theory at work. For a major bull trend to continue, both the Dow Industrials and the Dow Transports must advance together.* - -#### **6. A Trend Is Assumed to Be in Effect Until It Gives Definite Signals That It Has Reversed.** - -This tenet, which we touched upon in Chapter 1, forms much of the foundation of modern trend-following approaches. It relates a physical law to market movement, which states that an object in motion (in this case a trend) tends to continue in motion until some [external](#page-23-0) force causes it to change direction. A number of technical tools are available to traders to assist in the difficult task of spotting reversal signals, including the study of support and resistance levels, price patterns, trendlines, and moving averages. Some indicators can provide even earlier warning signals of loss of momentum. All of that not withstanding, the odds usually favor that the existing trend will continue. - -**Figure 2.2** *Examples of two Dow Theory confirmations. At the start of 1997 (point 1), the Dow Transports confirmed the earlier breakout in the Industrials. The following May (point 2), the Dow Industrials confirmed the earlier new high in the Transports.* - -The most difficult task for a Dow theorist, or any trend-follower for that matter, is being able to distinguish between a normal secondary correction in an existing trend and the first leg of a new trend in the opposite direction. Dow theorists often disagree as to when the market gives an actual reversal signal. Figures 2.3a and 2.3b show how this disagreement manifests itself. - -Figures 2.3a and 2.3b illustrate two different market scenarios. In Figure 2.3a, notice that the rally at point C is lower than the previous peak at A. Price then declines [below](#page-45-1) point [B.](#page-46-1) The presence of these two lower peaks and two lower troughs [giv](#page-45-1)es a [clear](#page-46-1)-cut sell signal at the point where the low at B is [brok](#page-45-1)en (point S). This reversal pattern is sometimes referred to as a "failure swing." - -**Figure 2.3a** *Failure Swing. The failure of the peak at C to overcome A, followed by the violation of the low at B, constitutes a "sell" signal at S.* - -**Figure 2.3b** *Nonfailure Swing. Notice that C exceeds A before falling below B. Some Dow theorists would see a "sell" signal at S1, while others would need to see a lower high at E before turning bearish at S2.* - -In Figure 2.3b, the rally top at C is higher than the previous peak at A. Then price declines below point B. Some Dow theorists would not consider the clear violation of support, at S1, to be a bona fide sell signal. They would point out that [only](#page-46-1) lower lows exist in this case, but not lower highs. They would prefer to see a rally to point E which is lower than point C. Then they would look for another new low under point D. To them, S2 would represent the actual sell signal with two lower highs and two lower lows. - -The reversal pattern shown in Figure 2.3b is referred to as a "nonfailure swing." A failure swing (shown in Figures 2.3a) is a much weaker pattern than the nonfailure swing in Figure 2.3b. Figures 2.4a and 2.4b show the same scenarios at a market bottom. - -### **THE USE OF CLOSING PRICES AND THE PRESENCE OF LINES** - -Dow relied exclusively on *closing prices.* He believed that averages had to *close* higher than a previous peak or lower than a previous trough to have significance. Dow did not consider intraday penetrations valid. - -**Figure 2.4a** *Failure Swing Bottom. The "buy" signal takes place when point B is exceeded (at B1).* - -**Figure 2.4b** *Nonfailure Swing Bottom. "Buy" signals occur at points B1 or B2.* - -When traders speak of *lines* in the averages, they are referring to horizontal patterns that sometimes occur on the charts. These sideways trading ranges usually play the role of corrective phases and are usually referred to as consolidations. In more modern terms, we might refer to such lateral patterns as "rectangles." diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/007_2 Dow Theory.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/007_2 Dow Theory.md deleted file mode 100644 index 2f4f49b213b70932d03931c596481d88967af3d3..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/007_2 Dow Theory.md +++ /dev/null @@ -1,63 +0,0 @@ -### **SOME CRITICISMS OF DOW THEORY** - -Dow Theory has done well over the years in identifying major bull and bear markets, but has not escaped criticism. On average, Dow Theory misses 20 to 25% of a move before generating a signal. Many traders consider this to be too late. A Dow Theory buy signal usually occurs in the second phase of an uptrend as price penetrates a previous intermediate peak. This is also, incidentally, about where most trend-following technical systems begin to - -identify and participate in existing trends. - -In response to this criticism, traders must remember that Dow never intended to anticipate trends; rather he sought to recognize the emergence of major bull and bear markets and to capture the large middle portion of important market moves. Available records suggest that Dow's Theory has performed that function reasonably well. From 1920 to 1975, Dow Theory signals captured 68% of the moves in the Industrial and Transportation Averages and 67% of those in the S&P 500 Composite Index (Source: Barron's). Those who criticize Dow Theory for failing to catch actual market tops and bottoms lack a basic understanding of the trend-following philosophy. - -## **STOCKS AS ECONOMIC INDICATORS** - -Dow apparently never intended to use his theory to forecast the direction of the stock market. He felt its real value was to use stock market direction as a barometric reading of general business conditions. We can only marvel at Dow's vision and genius. In addition to formulating a great deal of today's price forecasting methodology, he was among the first to recognize the usefulness of stock market averages as a leading economic indicator. - -## **DOW THEORY APPLIED TO FUTURES TRADING** - -Dow's work considered the behavior of stock averages. While most of that original work has significant application to commodity futures, there are some important distinctions between stock and futures trading. For one thing, Dow assumed that most investors follow only the major trends and would use intermediate corrections for timing purposes only. Dow considered the minor or near term trends to be unimportant. Obviously, this is not the case in futures trading in which most traders who follow trends trade the intermediate instead of the major trend. These traders must pay a great deal of attention to minor swings for timing purposes. If a futures trader expected an intermediate uptrend to last for a couple of months, he or she would look for short term dips to signal purchases. In an intermediate downtrend, the trader would use minor bounces to signal short sales. The minor trend, therefore, becomes extremely important in futures trading. - -### **NEW WAYS TO TRADE THE DOW AVERAGES** - -For the first 100 years of its existence, the Dow Jones Industrial Average could only be used as a market indicator. That all changed on October 6, 1997 when futures and options began trading on Dow's venerable average for the first time. The Chicago Board of Trade launched a futures contract on the Dow Jones Industrial Average, while options on the Dow (symbol: DJX) started trading at the Chicago Board Options Exchange. In addition, options were also launched on the Dow Jones Transportation Average (symbol: DJTA) and the Dow Jones Utility Index (symbol: DJUA). In January 1998, the American Stock Exchange started trading the Diamonds Trust, a unit investment trust that mimics the 30 Dow industrials. In addition, two mutual funds were offered based on the 30 Dow benchmark. Mr. Dow would probably be happy to know that, a century after their creation, it would now be possible to trade his Dow averages, and actually put his Dow Theory into practice. - -### **CONCLUSION** - -This chapter presented a relatively quick review of the more important aspects of the Dow Theory. It will become clear, as you continue through this book, that an understanding and appreciation of Dow Theory provides a solid foundation for any study of technical analysis. Much of what is discussed in the following chapters represents some adaptation of Dow's original theory. The standard definition of a trend, the classification of a trend into three categories and phases, the principles of confirmation and divergence, the interpretation of volume, and the use of percentage retracements (to name a few), all derive, in one way or another, from Dow Theory. - -In addition to the sources already cited in this chapter, an excellent review of the principles of Dow Theory can be found in *Technical Analysis of Stock Trends* (Edwards & Magee). - -### **INTRODUCTION** - -This chapter is primarily intended for those readers who are unfamiliar with bar chart construction. We'll begin by discussing the different types of charts available and then turn our focus to the most commonly used chart—*the daily bar chart.* We'll look at how the price data is read and plotted on the chart. *Volume and open interest* are also included in addition to price. We'll then look at other variations of the bar chart, including *longer range weekly and monthly charts.* Once that has been completed, we'll be ready to start looking at some of the analytical tools applied to that chart in the following chapter. Those readers already familiar with the charts themselves might find this chapter too basic. Feel free to move on to the next chapter. - -### **TYPES OF CHARTS AVAILABLE** - -The daily bar chart has already been acknowledged as the most widely used type of chart in technical analysis. There are, however, other types of charts also used by technicians, such as line charts, point and figure charts, and more recently, candlesticks. Figure 3.1 shows a standard daily bar chart. It's called a bar chart because each day's range is represented by a vertical bar. The bar chart shows the open, high, low, and closing prices. The tic to the right of the vertical bar is the closing [price.](#page-51-0) The opening price is the tic to the left of the bar. - -Figure 3.2 shows what the same market looks like on a line chart. In the line chart, only the closing price is plotted for each successive day. Many chartists believe that because the closing price is the most critical price of the trading [day,](#page-51-1) a line (or close-only) chart is a more valid measure of price - -activity. - -**Figure 3.1** *A daily bar chart of Intel. Each vertical bar represents one day's action.* - -**Figure 3.2** *A line chart of Intel. This type of chart produces a solid line by connecting the successive closing prices.* - -A third type of chart, the point and figure chart, is shown in Figure 3.3. Notice here that the point and figure chart shows the same price action but in a more compressed format. Notice the alternating column of x's and o's. The x columns show rising prices and the o columns, declining prices. [Buy](#page-52-1) and sell signals are more precise and easier to spot on the point and figure chart than on the bar chart. This type of chart also has a lot more flexibility. Chapter 11 covers point and figure charts. - -### **[CANDLESTICKS](#page-245-0)** - -Candlestick charts are the Japanese version of bar charting and have become very popular in recent years among western chartists. The Japanese candlestick records the same four prices as the traditional bar chart—the open, the close, the high, and the low. The visual presentation differs however. On the candlestick chart, a thin line (called the *shadow)* shows the day's price range from the high to the low. A wider portion of the bar (called the *real body)* measures the distance between the open and the close. If the close is higher than the open, the real body is white (positive). If the close is lower than the open, the real body is black (negative). (See Figure 3.4.) - -**Figure 3.3** *A point and figure chart of Intel. Notice the alternating columns of x's and o's. The x column shows rising prices. The o column shows falling prices. Buy and sell signals are more precise on this type of chart.* - -The key to candlestick charts is the relationship between the open and - -the close. Possibly because of the growing popularity of candlesticks, western chartists now pay a lot more attention to the opening tic on their bar charts. You can do everything with a candlestick chart that you can do with a bar chart. In other words, all the technical tools and indicators we'll be showing you for the bar chart can also be used on candlesticks. We'll show you a bit later in the chapter how to construct bar charts for weekly and monthly periods. You can do the same with candlesticks. Chapter 12, "Japanese Candlesticks," provides a more thorough explanation of candlestick charting. - -**Figure 3.4** *A candlestick chart of Intel. The color of the candlestick is determined by the relationship between the open and the close. White candlesticks are positive, while black candlesticks are negative.* - -### **ARITHMETIC VERSUS LOGARITHMIC SCALE** - -Charts can be plotted using arithmetic or logarithmic price scales. For some types of analysis, particularly for very long range trend analysis, there may be some advantage to using logarithmic charts. (See Figures 3.5 and 3.6.) Figure 3.5 shows what the different scales would look like. On the arithmetic scale, the vertical price scale shows an equal distance for each price unit of change. Notice in this example that each point on the arit[hmetic](#page-54-1) scale is [eq](#page-54-1)[ui](#page-55-1)[distant.](#page-54-1) On the log scale, however, note that the percentage increases get smaller as the price scale increases. The distance from points 1 to 2 is the same as the distance from points 5 to 10 because they both represent the same doubling in - -price. For example, a move from 5 to 10 on an arithmetic scale would be the same distance as a move from 50 to 55, even though the former represents a doubling in price, while the latter is a price increase of only 10%. Prices plotted on ratio or log scales show equal distances for similar percentage moves. For example, a move from 10 to 20 (a 100% increase) would be the same distance on a log chart as a move from 20 to 40 or 40 to 80. Many stock market chart services use log charts, whereas futures chart services use arithmetic. Charting software packages allow both types of scaling, as shown in Figure 3.6. - -**Figure 3.5** *A comparison of an arithmetic and logarithmic scale. Notice the equal spacing on the scale to the left. The log scale shows percentage changes (right scale).* diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/008_3 Chart Construction.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/008_3 Chart Construction.md deleted file mode 100644 index f3683d547b8396d6ec54c0a35d266bd2ebcda5d8..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/008_3 Chart Construction.md +++ /dev/null @@ -1,47 +0,0 @@ -### **CONSTRUCTION OF THE DAILY BAR CHART** - -The construction of the daily bar chart is extremely simple. The bar chart is both a price and a time chart. The vertical axis (the *y* axis) shows a scale representing the price of the contract. The horizontal axis (the *x* axis) records the passage of time. Dates are marked along the bottom of the chart. All the user has to do is plot a vertical bar in the appropriate day from the day's high to the day's low (called the range). Place a horizontal tic to the right of the vertical bar identifying the daily closing price. (See Figure 3.7.) - -The reason for placing the tic to the right of the bar is to distinguish it from the opening price, which chartists record to the left of the bar. Once that day's activity has been plotted, the user moves one day to [the](#page-56-1) right to plot the - -next day's action. Most chart services use five day weeks. Weekends are not shown on the chart. Whenever an exchange is closed during the trading week, that day's space is left blank. The bars along the bottom of the chart measure volume. (See Figure 3.7.) - -**Figure 3.6** *Longer view of Intel using two different price scales. The chart to the left shows the traditional arithmetic scale. The chart on the right shows a logarithmic scale. Notice that the three year up trendline worked better on the log chart.* - -### **VOLUME** - -Another piece of important information should be included on the bar chart volume. *Volume* represents the total amount of trading activity in that market for that day. It is the total number of futures contracts traded during the day or the number of common stock shares that change hands on a given day in the stock market. The volume is recorded by a vertical bar at the bottom of the chart under that day's price bar. A higher volume bar means the volume was heavier for that day. A smaller bar represents lighter volume. A vertical scale along the bottom of the chart is provided to help plot the data, as shown in Figure 3.7. - -**Figure 3.7** *A closer look at the Intel daily bar chart. Each bar measures the day's price range. The opening price is marked by the small tic to the left of each bar. The closing tic is to the right. The bars along the bottom measure each day's volume.* - -### **FUTURES OPEN INTEREST** - -*Open interest* is the total number of outstanding futures contracts that are held by market participants at the end of the day. Open interest is the number of outstanding contracts held by the longs or the shorts, not the total of both. Remember, because we're dealing with futures contracts, for every long there must also be a short. Therefore, we only have to know the totals on one side. Open interest is marked on the chart with a solid line along the bottom, usually just above the volume but below the price. (See Figure 3.8.) - -**Figure 3.8** *A daily line chart of a Treasury Bond futures contract The vertical bars along the bottom measure the total daily volume. The solid line along the middle represents the total outstanding open interest for the Treasury Bond futures market.* - -#### **Total Versus Individual Volume and Open Interest Numbers in Futures** - -Futures chart services, along with most futures technicians, use only the *total* volume and open interest figures. Although figures are available for each individual delivery month, the total figures for each commodity market are the ones that are used for forecasting purposes. There is a good reason for this. - -In the early stages of a futures contract's life, volume and open interest are usually quite small. The figures build up as the contract reaches maturity. In the last couple of months before expiration, however, the numbers begin to drop again. Obviously, traders have to liquidate open positions as the contract approaches expiration. Therefore, the increase in the numbers in the first few months of life and the decline near the end of trading have nothing to do with market direction and are just a function of the limited life feature of a commodity futures contract. To provide the necessary continuity in volume and open interest numbers, and to give them forecasting value, the total numbers are generally used. (Stock charts plot total volume figures, but do not include open interest.) - -Futures volume and open interest numbers are reported a day late. Therefore, the chartist must be content with a day's lag in obtaining and interpreting the figures. The numbers are usually reported during the following day's trading hours, but too late for publication in the day's financial newspapers. Estimated volume figures are available, however, after the markets close and are included in the following morning's paper. Estimated volume numbers are just that, but they do at least give the futures technician some idea of whether trading activity was heavier or lighter the previous day. In the morning paper, therefore, what the reader gets is the last day's futures prices along with an estimated volume figure. Official volume and open interest numbers, however, are given for the day before. Stock chartists don't have that problem. Volume totals for stocks are immediately available. - -#### **The Value of Individual Volume and Open Interest Numbers in Futures** - -The individual open interest numbers in futures do provide valuable information. They tell us which contracts are the most liquid for trading purposes. *As a general rule, trading activity should be limited to those delivery months with the highest open interest. Months with low open interest numbers should be avoided.* As the term implies, higher open interest means that there is more interest in certain delivery months. - -### **WEEKLY AND MONTHLY BAR CHARTS** - -We've focused so far on the daily bar chart. However, be aware that a bar chart can be constructed for any time period. The intraday bar chart measures the high, low, and last prices for periods as short as five minutes. The average daily bar chart covers from six to nine months of price action. For longer range trend analysis, however, weekly and monthly bar charts must be used. The value of using these longer range charts is covered in Chapter 8. But the method of constructing and updating the charts is essentially the same. (See Figures 3.9 and 3.10.) - -On the weekly chart, one bar represents the price acti[vity](#page-174-0) for the entire week. On the monthly chart, each bar shows the entire month's price action. [Obviously,](#page-59-0) wee[kly](#page-59-1) and monthly charts compress the price action to allow for much longer range trend analysis. A weekly chart can go back as much as five years and a monthly chart up to 20 years. It's a simple technique that helps the chartist study the markets from a longer range perspective—a valuable perspective that is often lost by relying solely on daily charts. - -**Figure 3.9** *A weekly bar chart of the U.S. Dollar Index. Each bar represents one week's price data. By compressing the price data, the weekly chart allows for chart analysis of longer range price trends, usually in the vicinity of five years.* - -**Figure 3.10** *A monthly bar chart of the U.S. Dollar Index. Each bar represents one month's price data. By compressing the data even further, the monthly chart allows chart analysis for periods as long as twenty years.* - -## **CONCLUSION** - -Now that we know how to plot a bar chart, and having introduced the three basic sources of information—price, volume, and open interest—we're ready to look at how that data is interpreted. Remember that the chart only records the data. In itself, it has little value. It's much like a paint brush and canvas. By themselves, they have no value. In the hands of a talented artist, however, they can help create beautiful images. Perhaps an even better comparison is a scalpel. In the hands of a gifted surgeon, it can help save lives. In the hands of most of us, however, a scalpel is not only useless, but might even be dangerous. A chart can become an extremely useful tool in the art or skill of market forecasting once the rules are understood. Let's begin the process. In the next chapter, we'll look at some of the basic concepts of trend and what I consider to be the building blocks of chart analysis. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/009_4 Basic Concepts of Trend.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/009_4 Basic Concepts of Trend.md deleted file mode 100644 index 1454cdbc548778ad06554f9841cc7805688f6f83..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/009_4 Basic Concepts of Trend.md +++ /dev/null @@ -1,413 +0,0 @@ -### **DEFINITION OF TREND** - -The concept of *trend* is absolutely essential to the technical approach to market analysis. All of the tools used by the chartist—support and resistance levels, price patterns, moving averages, trendlines, etc.—have the sole purpose of helping to measure the trend of the market for the purpose of participating in that trend. We often hear such familiar expressions as "always trade in the direction of the trend," "never buck the trend," or "the trend is your friend." So let's spend a little time to define what a trend is and classify it into a few categories. - -In a general sense, the trend is simply the direction of the market, which way it's moving. But we need a more precise definition with which to work. First of all, markets don't generally move in a straight line in any direction. Market moves are characterized by a series of *zigzags.* These zigzags resemble a series of successive waves with fairly obvious peaks and troughs. *It is the direction of those peaks and troughs that constitutes market trend.* Whether those peaks and troughs are moving up, down, or sideways tells us the trend of the market. An *uptrend* would be defined as a series of successively higher peaks and troughs; a *downtrend* is just the opposite, a series of declining peaks and troughs; horizontal peaks and troughs would identify a sideways price trend. (See Figures 4.1a-d.) - -**Figure 4.1a** *Example of an uptrend with ascending peaks and troughs.* - -**Figure 4.1b** *Example of a downtrend with descending peaks and troughs.* - -**Figure 4.1c** *Example of a sideways trend with horizontal peaks and troughs. This type of market is often referred to as "trendless."* - -**Figure 4.1d** *Example of a downtrend turning into an uptrend. The first portion to the left shows a downtrend. From April 1996 to April 1997, the market traded sideways. During the summer 1997, the trend turned up.* - -### **TREND HAS THREE DIRECTIONS** - -We've mentioned an uptrend, downtrend, and sideways trend for a very good reason. Most people tend to think of markets as being always in either an uptrend or a downtrend. The fact of the matter is that markets actually move in three directions—up, down, and sideways. It is important to be aware of this distinction because for at least a third of the time, by a conservative estimate, prices move in a flat, horizontal pattern that is referred to as a *trading range.* This type of sideways action reflects a period of equilibrium in the price level where the forces of supply and demand are in a state of relative balance. (If you'll recall, Dow Theory refers to this type of pattern as a *line.*) Although we've defined a flat market as having a sideways trend, it is more commonly referred to as being *trendless.* - -Most technical tools and systems are trend-following in nature, which means that they are primarily designed for markets that are moving up or down. They usually work very poorly, or not at all, when markets enter these lateral or "trendless" phases. It is during these periods of sideways market movement that technical traders experience their greatest frustration, and systems traders their greatest equity losses. A trend-following system, by its very definition, needs a trend in order to do its stuff. The failure here lies not - -with the system. Rather, the failure lies with the trader who is attempting to apply a system designed for trending markets into a nontrending market environment. - -There are three decisions confronting the trader—whether to buy a market (go long), sell a market (go short), or do nothing (stand aside). When a market is rising, the buying strategy is preferable. When it is falling, the second approach would be correct. *However, when the market is moving sideways, the third choice—to stay out of the market—is usually the wisest.* - -### **TREND HAS THREE CLASSIFICATIONS** - -In addition to having three directions, trend is usually broken down into the three categories mentioned in the previous chapter. Those three categories are the *major, intermediate*, and *near term trends.* In reality, there are almost an infinite number of trends interacting with one another, from the very short term trends covering minutes and hours to superlong trends lasting 50 or 100 years. Most technicians, however, limit trend classifications to three. There is a certain amount of ambiguity, however, as to how different analysts define each trend. - -Dow Theory, for example, classifies the *major trend* as being in effect for longer than a year. Because futures traders operate in a shorter time dimension than do stock investors, I would be inclined to shorten the major trend to anything over six months in the commodity markets. Dow defined the intermediate, or secondary, trend as three weeks to as many months, which also appears about right for the futures markets. The near term trend is usually defined as anything less than two or three weeks. - -Each trend becomes a portion of its next larger trend. For example, the intermediate trend would be a *correction* in the major trend. In a long term uptrend, the market pauses to correct itself for a couple of months before resuming its upward path. That secondary correction would itself consist of shorter waves that would be identified as near term dips and rallies. This theme recurs many times—that each trend is part of the next larger trend and is itself comprised of smaller trends. (See Figures 4.2a and b.) - -In Figure 4.2a, the major trend is up as reflected by the rising peaks and troughs (points 1, 2, 3, 4). The corrective phase (2-3) represents an intermediate correction within the major [uptrend.](#page-65-0) But notic[e](#page-65-1) that the wave 2-3 also breaks [down](#page-65-0) into three smaller waves (A, B, C). At point C, the analyst would say that the major trend was still up, but the intermediate and near term trends were down. At point 4, all three trends would be up. It is important to understand the distinction between the various degrees of trend. When someone asks what the trend is in a given market, it is difficult, if not - -impossible, to respond until you know which trend the person is inquiring about. You may have to respond in the manner previously discussed by defining the three different trend classifications. - -**Figure 4.2a** *Example of the three degrees of trend: major, secondary, and near term. Points 1, 2, 3, and 4 show the major uptrend. Wave 2-3 represents a secondary correction within the major uptrend. Each secondary wave in turn divides into near term trends. For example, secondary wave 2-3 divides into minor waves A-B-C.* - -**Figure 4.2b** *The major trend (over a year) is up during 1997. A short term correction occurred during March. An intermediate correction lasted from August to November (three months). The intermediate correction broke down* - -#### *into three short term trends.* - -Quite a bit of misunderstanding arises because of different traders' perceptions as to what is meant by a trend. To long term position traders, a few days' to a few weeks' price action might be insignificant. To a day trader, a two or three day advance might constitute a major uptrend. It's especially important, then, to understand the different degrees of trend and to make sure that all involved in a transaction are talking about the same ones. - -As a general statement, most trend-following approaches focus on the intermediate trend, which may last for several months. The near term trend is used primarily for timing purposes. In an intermediate uptrend, short term setbacks would be used to initiate long positions. - -### **SUPPORT AND RESISTANCE** - -In the previous discussion of trend, it was stated that prices move in a series of peaks and troughs, and that the direction of those peaks and troughs determined the trend of the market. Let's now give those peaks and troughs their appropriate names and, at the same time, introduce the concepts of *support* and *resistance.* - -The troughs, or reaction lows, are called *support.* The term is selfexplanatory and indicates that support is a level or area on the chart *under the market* where buying interest is sufficiently strong to overcome selling pressure. As a result, a decline is halted and prices turn back up again. Usually a support level is identified beforehand by a previous reaction low. In Figure 4.3a, points 2 and 4 represent support levels in an uptrend. (See Figures 4.3a and b.) - -*Resistance* is the opposite of support and represents a price level or area *[over](#page-67-0) the market* where selling pressure overcomes buying pressure and a price advance [is](#page-67-0) turned [b](#page-67-1)ack. Usually a resistance level is identified by a previous peak. In Figure 4.3a, points 1 and 3 are resistance levels. Figure 4.3a shows an uptrend. In an uptrend, the support and resistance levels show an ascending pattern. Figure 4.3b shows a downtrend with descending peaks and troughs. In the d[owntrend,](#page-67-0) points 1 and 3 are support levels under the [market](#page-67-0) and points 2 and 4 are resistance levels over the market. - -**Figure 4.3a** *Shows rising support and resistance levels in uptrend. Points 2 and 4 are support levels which are usually previous reaction lows. Points 1 and 3 are resistance levels, usually marked by previous peaks.* - -**Figure 4.3b** *Shows support and resistance in a downtrend.* - -In an uptrend, the resistance levels represent pauses in that uptrend and are usually exceeded at some point. In a downtrend, support levels are not sufficient to stop the decline permanently, but are able to check it at least temporarily. - -A solid grasp of the concepts of support and resistance is necessary for a full understanding of the concept of trend. For an uptrend to continue, each successive low (support level) must be higher than the one preceding it. Each rally high (resistance level) must be higher than the one before it. If the corrective dip in an uptrend comes all the way down to the previous low, it - -may be an early warning that the uptrend is ending or at least moving from an uptrend to a sideways trend. If the support level is violated, then a trend *reversal* from up to down is likely. - -Each time a previous resistance peak is being tested, the uptrend is in an especially critical phase. Failure to exceed a previous peak in an uptrend, or the ability of prices to bounce off the previous support low in a downtrend, is usually the first warning that the existing trend is changing. Chapters 5 and 6 on *price patterns* show how the testing of these support and resistance levels form pictures on the charts that suggest either a trend reversal in progress or merely a pause in the existing trend. But the basic building [blocks](#page-105-0) on whic[h](#page-129-0) those price patterns are based are support and resistance levels. - -Figures 4.4a-c are examples of a classic trend reversal. Notice, in Figure 4.4a, that at point 5 prices failed to exceed the previous peak (point 3) before turning down to violate the previous low at point 4. This trend reversal could have been [identifi](#page-68-0)[ed](#page-69-0) simply by watching the support and [resistance](#page-68-0) levels. In our coverage of price patterns, this type of reversal pattern will be identified as a *double top.* - -#### **How Support and Resistance Levels Reverse Their Roles** - -So far we've defined "support" as a previous low and "resistance" as a previous high. However, this is not always the case. This leads us to one of the more interesting and lesser known aspects of support and resistance their reversal of roles. *Whenever a support or resistance level is penetrated by a significant amount, they reverse their roles and become the opposite.* In other words, a resistance level becomes a support level and support becomes resistance. To understand why this occurs, perhaps it would be helpful to discuss some of the psychology behind the creation of support and resistance levels. - -**Figure 4.4a** *Example of a trend reversal. The failure of prices at point 5 to exceed the previous peak at point 3 followed by a downside violation of the previous low at point 4 constitutes a downside trend reversal. This type of* - -*pattern is called a double top.* - -**Figure 4.4b** *Example of a bottom reversal pattern. Usually the first sign of a bottom is the ability of prices at point 5 to hold above the previous low at point 3. The bottom is confirmed when the peak at 4 is overcome.* - -**Figure 4.4c** *Example of a bottom reversal. During January 1998 prices retested the December support low and bounced off it, forming a second support level. The upside penetration of the middle resistance peak signaled a new uptrend.* - -#### **The Psychology of Support and Resistance** - -To illustrate, let's divide the market participants into three categories—the longs, the shorts, and the uncommitted. The longs are those traders who have already purchased contracts; the shorts are those who have already committed themselves to the sell side; the uncommitted are those who have either gotten out of the market or remain undecided as to which side to enter. - -Let's assume that a market starts to move higher from a support area where prices have been fluctuating for some time. The longs (those who bought near the support area) are delighted, but regret not having bought more. If the market would dip back near that support area again, they could add to their long positions. The shorts now realize (or strongly suspect) that they are on the wrong side of the market. (How far the market has moved away from that support area will greatly influence these decisions, but we'll come back to that point a bit later.) The shorts are hoping (and praying) for a dip back to that area where they went short so they can get out of the market where they got in (their break even point). - -Those sitting on the sidelines can be divided into two groups—those who never had a position and those who, for one reason or another, liquidated previously held long positions in the support area. The latter group are, of course, mad at themselves for liquidating their longs prematurely and are hoping for another chance to reinstate those longs near where they sold them. - -The final group, the undecided, now realize that prices are going higher and resolve to enter the market on the long side on the next good buying opportunity. All four groups are resolved to "buy the next dip." They all have a "vested interest" in that support area under the market. Naturally, if prices do decline near that support, renewed buying by all four groups will materialize to push prices up. - -The more trading that takes place in that support area, the more significant it becomes because more participants have a vested interest in that area. The amount of trading in a given support or resistance area can be determined in three ways: the amount of time spent there, volume, and how recently the trading took place. - -*The longer the period of time that prices trade in a support or resistance area, the more significant that area becomes.* For example, if prices trade sideways for three weeks in a congestion area before moving higher, that support area would be more important than if only three days of trading had occurred. - -*Volume is another way to measure the significance of support and resistance.* If a support level is formed on heavy volume, this would indicate that a large number of units changed hands, and would mark that support level as more important than if very little trading had taken place. Point and figure charts that measure the intraday trading activity are especially useful in identifying these price levels where most of the trading took place and, consequently, where support and resistance will be most likely to function. - -*A third way to determine the significance of a support or resistance area is how recently the trading took place.* Because we are dealing with the - -reaction of traders to market movement and to positions that they have already taken or failed to take, it stands to reason that the more recent the activity, the more potent it becomes. - -Now let's turn the tables and imagine that, instead of moving higher, prices move lower. In the previous example, because prices advanced, the combined reaction of the market participants caused each downside reaction to be met with additional buying (thereby creating new support). However, if prices start to drop and move below the previous support area, the reaction becomes just the opposite. All those who bought in the support area now realize that they made a mistake. For futures traders, their brokers are now calling frantically for more margin money. Because of the highly leveraged nature of futures trading, traders cannot sit with losses very long. They must put up additional margin money or liquidate their losing positions. - -What created the previous support in the first place was the predominance of buy orders under the market. Now, however, all of the previous buy orders under the market have become sell orders over the market. *Support has become resistance.* And the more significant that previous support area was—that is, the more recent and the more trading that took place there—the more potent it now becomes as a resistance area. All of the factors that created support by the three categories of participants—the longs, the shorts, and the uncommitted—will now function to put a ceiling over prices on subsequent rallies or bounces. - -It is useful once in a while to pause and reflect on why the price patterns used by chartists, and concepts like support and resistance, actually do work. It's not because of some magic produced by the charts or some lines drawn on those charts. These patterns work because they provide pictures of what the market participants are actually doing and enable us to determine their reactions to market events. Chart analysis is actually a study of human psychology and the reactions of traders to changing market conditions. Unfortunately, because we live in the fast-paced world of financial markets, we tend to rely heavily on chart terminology and shortcut expressions that overlook the underlying forces that created the pictures on the charts in the first place. There are sound psychological reasons why support and resistance levels can be identified on price charts and why they can be used to help predict market movements. - -#### **Support Becoming Resistance and Vice Versa: Degree of Penetration** - -A support level, penetrated by a significant margin, becomes a resistance level and vice versa. Figures 4.5a-c are similar to Figures 4.3a and b but with one added refinement. Notice that as prices are rising in Figure 4.5a the reaction at point 4 stops at or above the top of the peak at point 1. That previous peak at - -point 1 had been a resistance level. But once it was decisively penetrated by wave 3, that previous resistance peak became a support level. All of the previous selling near the top of wave 1 (creating the resistance level) has now become buying under the market. In Figure 4.5b, showing declining prices, point 1 (which had been a previous support level under the market) has now become a resistance level over the market [acting](#page-72-0) as a ceiling at point 4. - -**Figure 4.5a** *In an uptrend, resistance levels that have been broken by a significant margin become support levels. Notice that once resistance at point 1 is exceeded, it provides support at point 4. Previous peaks function as support on subsequent corrections.* - -**Figure 4.5b** *In a downtrend, violated support levels become resistance levels on subsequent bounces. Notice how previous support at point 1 became resistance at point 4.* - -**Figure 4.5c** *Role reversal at play. Once the early 1997 resistance peak was broken, it reversed roles to become a support level. A year later, the intermediate price decline found support right at that prior resistance peak which had become new support.* - -It was mentioned earlier that the distance prices traveled away from support or resistance increased the significance of that support or resistance. This is particularly true when support and resistance levels are penetrated and reverse roles. For example, it was stated that support and resistance levels reverse roles only after a significant penetration. But what constitutes significant? There is quite a bit of subjectivity involved here in determining whether a penetration is significant or not. As a benchmark, some chartists use a 3% penetration as a criteria, particularly for major support and resistance levels. Shorter term support and resistance areas would probably require a much smaller number, like 1%. In reality, each analyst must decide for himself or herself what constitutes a significant penetration. It's important to remember, however, that support and resistance areas only reverse roles when the market moves far enough away to convince the market participants that they have made a mistake. The farther away the market moves, the more convinced they become. - -#### **The Importance of Round Numbers as Support and Resistance** - -There is a tendency for round numbers to stop advances or declines. Traders tend to think in terms of important round numbers, such as 10, 20, 25, 50, 75, 100 (and multiples of 1000), as price objectives and act accordingly. These - -round numbers, therefore, will often act as "psychological" support or resistance levels. A trader can use this information to begin taking profits as an important round number is approached. - -The gold market is an excellent example of this phenomenon. The 1982 bear market low was right at \$300. The market then rallied to just above \$500 in the first quarter of 1983 before falling to \$400. A gold rally in 1987 stopped at \$500 again. From 1990 to 1997, gold failed each attempt to break through \$400. The Dow Jones Industrial Average has shown a tendency to stall at multiples of 1000. - -One trading application of this principle is to *avoid placing trading orders right at these obvious round numbers.* For example, if the trader is trying to buy into a short term market dip in an uptrend, it would make sense to place limit orders just above an important round number. Because others are trying to buy the market at the round number, the market may never get there. Traders looking to sell on a bounce should place resting sell orders just below round numbers. The opposite would be true when placing protective stops on existing positions. As a general rule, *avoid placing protective stops at obvious round numbers.* - -In other words, protective stops on long positions should be placed below round numbers and on short positions, above such numbers. The tendency for markets to respect round numbers, and especially the more important round numbers previously referred to, is one of those peculiar market characteristics that can prove most helpful in trading and should be kept in mind by the technically oriented trader. - -### **TRENDLINES** - -Now that we understand support and resistance, let's add another building block to our arsenal of technical tools—*the trendline.* (See Figures 4.6a-c.) The basic trendline is one of the simplest of the technical tools employed by the chartist, but is also one of the most valuable. An *up trendline* is a straight line drawn upward to the right along successive reaction lows as [shown](#page-75-0) [b](#page-76-0)y the solid line in Figure 4.6a. A *down trendline* is drawn downward to the right along successive rally peaks as shown in Figure 4.6b. - -**Figure 4.6a** *Example of an up trendline. The up trendline is drawn under the rising reaction lows. A tentative trendline is first drawn under two successively higher lows (points 1 and 3), but needs a third test to confirm the validity of the trendline (point 5).* - -**Figure 4.6b** *A down trendline is drawn over the successively lower rally highs. The tentative down trendline needs two points (1 and 3) to be drawn and a third test (5) to confirm its validity.* - -**Figure 4.6c** *Long term up trendline at work. The up trendline was drawn upward and to the right along the first two reaction lows (see arrows). The third low at the start of 1998 bounced right off the rising trendline, thereby keeping the uptrend intact.* - -#### **Drawing a Trendline** - -The correct drawing of trendlines is a lot like every other aspect of charting and some experimenting with different lines is usually necessary to find the correct one. Sometimes a trendline that looks correct may have to be redrawn. But there are some useful guidelines in the search for that correct line. - -First of all, there must be evidence of a trend. This means that, for an up trendline to be drawn, there must be at least two reaction lows with the second low higher than the first. Of course, it always takes two points to draw any straight line. In Figure 4.6a, for example, only after prices have begun to move higher from point 3 is the chartist reasonably confident that a reaction low has been formed, and only then can a tentative up trendline be drawn under points 1 [and](#page-75-0) 3. - -Some chartists require that the peak at point 2 be penetrated to confirm the uptrend before drawing the trendline. Others only require a 50% retracement of wave 2-3, or that prices approach the top of wave 2. While the criteria may differ, the main point to remember is that the chartist wants to be reasonably sure that a reaction low has been formed before identifying a valid reaction low. Once two ascending lows have been identified, a straight line is drawn connecting the lows and projected up and to the right. - -#### **Tentative Versus the Valid Trendline** - -So far, all we have is a *tentative trendline.* In order to confirm the validity of a trendline, however, that line should be touched a third time with prices bouncing off of it. Therefore, in Figure 4.6a, the successful test of the up trendline at point 5 confirmed the validity of that line. Figure 4.6b shows a downtrend, but the rules are the same. The successful test of the trendline occurs at point 5. To summarize, two [points](#page-75-0) are needed to [draw](#page-75-1) the trendline, and a third point to make it a *valid trendline.* - -#### **How to Use the Trendline** - -Once the third point has been confirmed and the trend proceeds in its original direction, that trendline becomes very useful in a variety of ways. One of the basic concepts of trend is that a trend in motion will tend to remain in motion. As a corollary to that, once a trend assumes a certain slope or rate of speed, as identified by the trendline, it will usually maintain the same slope. The trendline then helps not only to determine the extremities of the corrective phases, but maybe even more importantly, tells us when that trend is changing. - -In an uptrend, for example, the inevitable corrective dip will often touch or come very close to the up trendline. Because the intent of the trader is to buy dips in an uptrend, that trendline provides a support boundary under the market that can be used as a buying area. A down trendline can be used as a resistance area for selling purposes. (See Figures 4.7a and b.) - -As long as the trendline is not violated, it can be used to determine buying and selling areas. However, at point 9 in Figures 4.7a-b, the violation of the trendline signals a trend change, c[alling](#page-77-0) for liquidat[io](#page-78-0)n of all positions in the direction of the previous trend. Very often, *the breaking of the trendline is one of the best early warnings of a change in [trend.](#page-77-0)* - -**Figure 4.7a** *Once the up trendline has been established, subsequent dips near* - -*the line can be used as buying areas. Points 5 and 7 in this example could have been used for new or additional longs. The breaking of the trendline at point 9 called for liquidation of all longs by signaling a downside trend reversal.* - -**Figure 4.7b** *Points* 5 *and 7 could have been used as selling areas. The breaking of the trendline at point 9 signaled an upside trend reversal.* - -#### **How to Determine the Significance of a Trendline** - -Let's discuss some of the refinements of the trendline. First, what determines the significance of a trendline? The answer to that question is twofold—*the longer it has been intact and the number of times it has been tested.* A trendline that has been successfully tested eight times, for example, that has continually demonstrated its validity, is obviously a more significant trendline than one that has only been touched three times. Also, a trendline that has been in effect for nine months is of more importance than one that has been in effect for nine weeks or nine days. The more significant the trendline, the more confidence it inspires and the more important is its penetration. - -#### **Trendlines Should Include All Price Action** - -Trendlines on bar charts should be drawn over or under the entire day's price range. Some chartists prefer to draw the trendline by connecting only the closing prices, but that is not the more standard procedure. The closing price may very well be the most important price of the day, but it still represents only a small sample of that day's activity. The technique of including the day's price range takes into account all of the activity and is the more common usage. (See Figure 4.8.) - -**Figure 4.8** *The correct drawing of a trendline should include the entire day's trading range.* - -#### **How to Handle Small Trendline Penetrations** - -Sometimes prices will violate a trendline on an intraday basis, but then close in the direction of the original trend, leaving the analyst in some doubt as to whether or not the trendline has actually been broken. (See Figure 4.9.) Figure 4.9 shows how such a situation might look. Prices did dip under the trendline during the day, but closed back above the up trendline. Should the [tren](#page-79-0)dline be [redrawn?](#page-79-0) - -**Figure 4.9** *Sometimes an intraday violation of a trendline will leave the chartist in doubt as to whether the original trendline is still valid or if a new line should be drawn. A compromise is to keep the original trendline, but draw a new dotted line until it can be better determined which is the truer line.* - -Unfortunately, there's no hard and fast rule to follow in such a situation. Sometimes it is best to ignore the minor breach, especially if subsequent market action proves that the original line is still valid. - -#### **What Constitutes a Valid Breaking of a Trendline?** - -As a general rule, *a close beyond the trendline is more significant than just an intraday penetration.* To go a step further, sometimes even a closing penetration is not enough. Most technicians employ a variety of time and price filters in an attempt to isolate valid trendline penetrations and eliminate bad signals or "whipsaws." One example of a price filter is the 3% *penetration criteria.* This price filter is used mainly for the breaking of longer term trendlines, but requires that the trendline be broken, on a closing basis, by at least 3%. (The 3% rule doesn't apply to some financial futures, such as the interest rate markets.) - -If, for example, gold prices broke a major up trendline at \$400, prices would have to close below that line by 3% of the price level where the line was broken (in this case, prices would have to close \$12 below the trendline, or at \$388). Obviously, a \$12 penetration criteria would not be appropriate for shorter term trading. Perhaps a 1% criterion would serve better in such cases. The % rule represents just one type of price filter. Stock chartists, for example, might require a full point penetration and ignore fractional moves. There is tradeoff involved in the use of any type of filter. If the filter is too small, it won't be very useful in reducing the impact of whipsaws. If it's too big, then much of the initial move will be missed before a valid signal is given. Here again, the trader must determine what type of filter is best suited to the degree of trend being followed, always making allowances for the differences in the individuals markets. - -An alternative to a price filter (requiring that a trendline be broken by some predetermined price increment or percentage amount) is a *time filter.* A common time filter is the *two day rule.* In other words, to have a valid breaking of a trendline, prices must close beyond the trendline for two successive days. To break an up trendline, therefore, prices must close under the trendline two days in a row. A one day violation would not count. The 1- 3% rule and the two day rule are also applied to the breaking of important support and resistance levels, not just to major trendlines. Another filter would require a Friday close beyond a major breakout point to ensure a weekly signal. - -#### **How Trendlines Reverse Roles** - -It was mentioned earlier that support and resistance levels became the opposite once violated. The same principle holds true of trendlines. (See Figures 4.10a-c.) In other words, an up trendline (a support line) will usually become a resistance line once it's decisively broken. A down trendline (a resistance line) will often become a support line once it's decisively broken. [This](#page-81-0) is why it['s](#page-82-0) usually a good idea to project all trendlines as far out to the - -right on the chart as possible even after they've been broken. It's surprising how often old trendlines act as support and resistance lines again in the future, but in the opposite role. - -### **Measuring Implications of Trendlines** - -Trendlines can be used to help determine price objectives. We'll have a lot more to say about price objectives in the next two chapters on price patterns. In fact, some of the price objectives addressed that are derived from various price patterns are similar to the one we'll cover here with trendlines. Stated briefly, once a trendline is broken, prices will usually move a distance beyond the trendline equal to the vertical distance that prices achieved on the other side of the line, prior to the trend reversal. - -**Figure 4.10a** *Example of a rising support line becoming resistance. Usually a support line will function as a resistance barrier on subsequent rallies, after it has been broken on the downside.* - -**Figure 4.10b** *Very often a down trendline will become a support line once it's been broken on the upside.* - -**Figure 4.10c** *Trendlines also reverse roles. On this chart, the broken up trendline became a resistance barrier on the following rally attempt.* - -In other words, if in the prior uptrend, prices moved \$50 above the up trendline (measured vertically), then prices would be expected to drop that same \$50 below the trendline after it's broken. In the next chapter, for example, we'll see that this measuring rule using the trendline is similar to that used for the well-known *head and shoulders* reversal pattern, where the distance from the "head" to the "neckline" is projected beyond that line once it's broken. - -### **THE FAN PRINCIPLE** - -This brings us to another interesting use of the trendline—the *fan principle.* (See Figures 4.11a-c.) Sometimes after the violation of an up trendline, prices will decline a bit before rallying back to the bottom of the old up trendline (now a resistance line). In Figure 4.11a, notice how prices rallied to but failed to pe[netrate](#page-83-0) line 1. [A](#page-83-1) second trendline (line 2) can now be drawn, which is also broken. After another failed rally attempt, a third line is drawn (line 3). *The breaking of that third [trendline](#page-83-0) is usually an indication that prices are headed lower.* In Figure 4.11b, the breaking of the third down trendline (line 3) constitutes a new uptrend signal. Notice in these examples how previously broken support lines became resistance and resistance lines became support. The term "fan pr[inciple"](#page-83-1) derives from the appearance of the lines that gradually flatten out, resembling a fan. *The important point to remember here* - -*is that the breaking of the third line is the valid trend reversal signal.* - -**Figure 4.11a** *Example of the fan principle. The breaking of the third trendline signals the reversal of a trend. Notice also that the broken trendlines 1 and 2 often become resistance lines.* - -**Figure 4.11b** *The fan principle at a bottom. The breaking of the third trendline signals the upside trend reversal. The previously broken trendlines (1 and 2) often become support levels.* - -**Figure 4.11c** *Fan lines are drawn along successive peaks as shown in this chart. The breaking of the third fan line usually signals the start of the uptrend.* - -### **THE IMPORTANCE OF THE NUMBER THREE** - -In examining the three lines in the fan principle, it's interesting to note how often the number three shows up in the study of technical analysis and the important role it plays in so many technical approaches. For example, the fan principle uses three lines; major bull and bear markets usually have three major phases (Dow Theory and Elliott Wave Theory); there are three kinds of *gaps* (to be covered shortly); some of the more commonly known reversal patterns, such as the *triple top* and the *head and shoulders*, have three prominent peaks; there are three different classifications of trend (major, secondary, and minor) and three trend directions (up, down, and sideways); among the generally accepted continuation patterns, there are three types of *triangles*—the symmetrical, ascending, and descending; there are three principle sources of information—price, volume, and open interest. For whatever the reason, the number three plays a very prominent role throughout the entire field of technical analysis. - -### **THE RELATIVE STEEPNESS OF THE** - -### **TRENDLINE** - -The relative steepness of the trendline is also important. In general, most important up trendlines tend to approximate an average slope of 45 degrees. Some chartists simply draw a 45 degree line on the chart from a prominent high or low and use this as a major trendline. The 45 degree line was one of the techniques favored by W. D. Gann. Such a line reflects a situation where prices are advancing or declining at such a rate that price and time are in perfect balance. - -**Figure 4.12** *Most valid trendlines rise at an angle approximating 45 degrees (see line 2). If the trendline is too steep (line 1), it usually indicates that the rate of ascent is not sustainable. A trendline that is too flat (line 3) suggests that the uptrend is too weak and probably suspect. Many technicians use 45 degree lines from previous tops or bottoms as major trendlines.* - -If a trendline is too steep (see line 1 in Figure 4.12), it usually indicates that prices are advancing too rapidly and that the current steep ascent will not be sustained. The breaking of that steep trendline may be just a reaction back to a more sustainable slope closer to the 45 [degree](#page-85-0) line (line 2). If a trendline is too flat (see line 3), it may indicate that the uptrend is too weak and not to be trusted. - -#### **How to Adjust Trendlines** - -Sometimes trendlines have to be adjusted to fit a slowing or an accelerating trend. (See Figure 4.13 and Figures 4.14a and b.) For example, as shown in the previous case, if a steep trendline is broken, a slower trendline might have to be drawn. If the original trendline is too flat, it may have to be redrawn at a steeper angle. [Figure](#page-86-0) 4.13 shows a [situatio](#page-87-0)n w[he](#page-87-1)re the breaking of the steeper - -trendline (line 1) necessitated the drawing of a slower line (line 2). - -**Figure 4.13** *Example of a trendline that is too steep (line 1). The original up trendline proved too steep. Often the breaking of a steep trendline is only an adjustment to a slower and more sustainable up trendline (line 2).* - -In Figure 4.14a, the original trendline (line 1) is too flat and has to be redrawn at a steeper angle (line 2). The uptrend accelerated, requiring a steeper line. A trendline that is too far away from the price action is obviously of little use in [trackin](#page-87-0)g the trend. - -In the case of an accelerating trend, sometimes several trendlines may have to be drawn at increasingly steeper angles. In my experience, however, where steeper trendlines become necessary, it is best to resort to another tool —the moving average—which is the same as a curvilinear trendline. One of the advantages of having access to several different types of technical indicators is being able to choose the one most appropriate for a given situation. All of the techniques covered in this book work well in certain situations, but not so well in others. By having an arsenal of tools to fall back on, the technician can quickly switch from one tool to another that might work better in a given situation. An accelerated trend is one of those cases where a moving average would be more useful than a series of steeper and steeper trendlines. - -**Figure 4.14a** *Example of an up trendline that is too flat (line 1). Line 1 proved too slow as the uptrend accelerated. In this case, a second and steeper trendline (line 2) should be drawn to more closely track the rising trend.* - -**Figure 4.14b** *An accelerating uptrend requires the drawing of steeper trendlines as shown in this chart. The steepest trendline becomes the most important one.* - -Just as there are several different degrees of trend in effect at any one time, so is there a need for different trendlines to measure those various trends. A major up trendline, for example, would connect the low points of the major uptrend, while a shorter and more sensitive line might be used for secondary swings. An even shorter line can measure the short term - -#### movements. (See Figure 4.15.) - -**Figure 4.15** *Different trendlines are used to define the different degrees of trend. Line 1 in the above example is the major up trendline, defining the major uptrend. Lines 2, 3, and 4 define the intermediate uptrends. Finally, line* 5 *defines a shorter term advance within the last intermediate uptrend. Technicians use many different trendlines on the same chart.* - -### **THE CHANNEL LINE** - -The *channel line*, or the *return line* as it is sometimes called, is another useful variation of the trendline technique. Sometimes prices trend between two parallel lines—the basic trendline and the channel line. Obviously, when this is the case and when the analyst recognizes that a channel exists, this knowledge can be used to profitable advantage. - -The drawing of the channel line is relatively simple. In an uptrend (see Figure 4.16a), first draw the basic up trendline along the lows. Then draw a dotted line from the first prominent peak (point 2), which is parallel to the basic up trendline. Both lines move up to the right, forming a channel. If the next [rally](#page-89-0) reaches and backs off from the channel line (at point 4), then a channel may exist. If prices then drop back to the original trendline (at point - -5), then a channel probably does exist. The same holds true for a downtrend (Figure 4.16b), but of course in the opposite direction. - -**Figure 4.16a** *Example of a trend channel. Once the basic up trendline is drawn (below points 1 and 3) a channel, or return, line (dotted line) can be projected over the first peak at* 2, *which is parallel to the basic up trendline.* - -**Figure 4.16b** *A trend channel in a downtrend. The channel is projected downward from the first low at point 2, parallel to the down trendline along the 1 and 3 peaks. Prices will often remain within such a trend channel.* - -**Figure 4.16c** *Notice how prices fluctuated between the upper and lower parallel channels over a period of 25 years. The 1987, 1989, and 1993 tops occurred right at the upper channel line. The 1994 bottom bounced off the lower trendline.* - -The reader should immediately see the value of such a situation. The basic up trendline can be used for the initiation of new long positions. The channel line can be used for short term profit taking. More aggressive traders might even use the channel line to initiate a countertrend short position, although trading in the opposite direction of the prevailing trend can be a dangerous and usually costly tactic. As in the case of the basic trendline, the longer the channel remains intact and the more often it is successfully tested, the more important and reliable it becomes. - -The breaking of the major trendline indicates an important change in trend. But the breaking of a rising channel line has exactly the opposite meaning, and signals an acceleration of the existing trend. Some traders view the clearing of the upper line in an uptrend as a reason to add to long positions. - -**Figure 4.17** *The failure to reach the upper end of the channel is often an early warning that the lower line will be broken. Notice the failure to reach the upper line at point* 5 *is followed by the breaking of the basic up trendline at point 6.* - -Another way to use the channel technique is to spot failures to reach the channel line, usually a sign of a weakening trend. In Figure 4.17, the failure of prices to reach the top of the channel (at point 5) may be an early warning that the trend is turning, and increases the odds that the other line (the basic up trendline) will be broken. As a general rule of thu[mb,](#page-91-0) the failure of any move within an established price channel to reach one side of the channel usually indicates that the trend is shifting, and increases the likelihood that the other side of the channel will be broken. - -The channel can also be used to adjust the basic trendline. (See Figures 4.18 and 4.19.) If prices move above a projected rising channel line by a significant amount, it usually indicates a strengthening trend. Some chartists then draw a steeper basic up trendline from the last reaction low parallel to the new [chann](#page-92-1)el line (as [demonstrated](#page-92-0) in Figure 4.18). Often, the new steeper support line functions better than the old flatter line. Similarly, the failure of an uptrend to reach the upper end of a channel justifies the drawing of a new support line under the last reaction low p[arallel](#page-92-0) to the new resistance line over the past two peaks (as shown in Figure 4.19). - -**Figure 4.18** *When the upper channel line is broken (as in wave 5), many chartists will redraw the basic up trendline parallel to the new upper channel line. In other words, line 4-6 is drawn parallel to line 3-5. Because the uptrend is accelerating, it stands to reason that the basic up trendline will do likewise.* - -**Figure 4.19** *When prices fail to reach the upper channel line, and a down trendline is drawn over the two declining peaks (line 3-5), a tentative channel line can be drawn from the low at point 4 parallel to line 3-5. The lower* - -*channel line sometimes indicates where initial support will be evident.* - -Channel lines have measuring implications. *Once a breakout occurs from an existing price channel, prices usually travel a distance equal to the width of the channel.* Therefore, the user has to simply measure the width of the channel and then project that amount from the point at which either trendline is broken. - -It should always be kept in mind, however, that of the two lines, the basic trendline is by far the more important and the more reliable. The channel line is a secondary use of the trendline technique. But the use of the channel line works often enough to justify its inclusion in the chartist's toolkit. - -### **PERCENTAGE RETRACEMENTS** - -In all of the previous examples of uptrends and downtrends, the reader has no doubt noticed that after a particular market move, prices retrace a portion of the previous trend before resuming the move in the original direction. These countertrend moves tend to fall into certain predictable percentage parameters. The best known application of the phenomenon is the 50% *retracement.* Let's say, for example, that a market is trending higher and travels from the 100 level to the 200 level. Very often, the subsequent reaction retraces about half of the prior move, to about the 150 level, before upward momentum is regained. This is a very well-known market tendency and happens quite frequently. Also, these percentage retracements apply to any degree of trend—major, secondary, and near term. - -Besides the 50% retracement, there are minimum and maximum percentage parameters that are also widely recognized—*the one-third and the two-thirds retracements.* In other words, the price trend can be divided into thirds. Usually, a minimum retracement is about 33% and a maximum about 66%. What this means is that, in a correction of a strong trend, the market usually retraces at least a third of the previous move. This is very useful information for a number of reasons. If a trader is looking for a buying area under the market, he or she can just compute a 33-50% zone on the chart and use that price zone as a general frame of reference for buying opportunities. (See Figures 4.20a and b.) - -**Figure 4.20a** *Prices often retrace about half of the prior trend before resuming in the original direction. This example shows a 50% retracement. The minimum retracement is one-third and the maximum, two-thirds of the prior trend.* - -The maximum retracement parameter is 66%, which becomes an especially critical area. If the prior trend is to be maintained, the correction must stop at the two-thirds point. This then becomes a relatively low risk buying area in an uptrend or selling area in a downtrend. If prices move beyond the two-thirds point, the odds then favor a trend reversal rather than just a retracement. The move usually then retraces the entire 100% of the prior trend. - -You may have noticed that the three percentage retracement parameters we've mentioned so far—50%, 33%, and 66%—are taken right from the original Dow Theory. When we get to the Elliott Wave Theory and Fibonacci ratios, we will see that followers of that approach use percentage retracements of 38% and 62%. I prefer to combine both approaches for a minimum retracement zone of 33-38% and a maximum zone of 62-66%. Some technicians round off these numbers even further to arrive at a 40-60% retracement zone. - -**Figure 4.20b** *The three horizontal lines mark the 38%, 50%, and 62% retracement levels measured from the April 1997 low to the August high. The first decline fell to the 38% line, the second decline to the 62% line, and the third near the 50% line. Most corrections will find support in the 38% to 50% retracement zones. The 38% and 62% lines are Fibonacci retracements and are popular with chartists.* - -Students of W. D. Gann are aware that he broke down the trend structure into eighths—1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8, 8/8. However, even Gann attached special importance to the 3/8 (38%), 4/8 (50%), and 5/8 (62%) retracement numbers and also felt it was important to divide the trend into thirds—1/3 (33%) and 2/3 (66%). - -### **SPEED RESISTANCE LINES** - -Speaking of thirds, let's touch on another technique that combines the trendline with percentage retracements—*speedlines.* This technique, developed by Edson Gould, is actually an adaptation of the idea of dividing the trend into thirds. The main difference from the percentage retracement concept is that the speed resistance lines (or speedlines) measure the rate of ascent or descent of a trend (in other words, its speed). - -To construct a bullish *speedline*, find the highest point in the current uptrend. (See Figure 4.21a.) From that high point on the chart, a vertical line is drawn toward the bottom of the chart to where the trend began. That - -vertical line is then divided into thirds. A trendline is then drawn from the beginning of the trend through the two points marked off on the vertical line, representing the one-third and two-thirds points. In a downtrend, just reverse the process. Measure the vertical distance from the low point in the downtrend to the beginning of the trend, and draw two lines from the beginning of the trend through the one-third and two-thirds points on the vertical line. (See Figures 4.21a and b.) - -**Figure 4.21a** *Examples of speed resistance lines in an uptrend. The vertical distance from the peak to the beginning of the trend is divided into thirds. Two trendlines are then drawn from point 1 through points 2 and 3. The upper line is the 2/3 speedline and the lower, the 1/3*. *The lines should act as support during market corrections. When they're broken, they revert to resistance lines on bounces. Sometimes these speedlines intersect price action.* - -**Figure 4.21b** *Speedlines in a downtrend.* - -Each time a new high is set in an uptrend or a new low in a downtrend, a new set of lines must be drawn (because there is now a new high or low point). Because the *speedlines* are drawn from the beginning of the trend to the one-third and two-thirds points, those trendlines may sometimes move through some of the price action. This is one case where trendlines are not drawn under lows or over highs, but actually through the price action. - -If an uptrend is in the process of correcting itself, the downside correction will usually stop at the higher speedline (the 2/3 speedline). If not, prices will drop to the lower speedline (the 1/3 speedline). If the lower line is also broken, prices will probably continue all the way to the beginning of the prior trend. In a downtrend, the breaking of the lower line indicates a probable rally to the higher line. If that is broken, a rally to the top of the prior trend would be indicated. - -As with all trendlines, speedlines reverse roles once they are broken. Therefore, during the correction of an uptrend, if the upper line (2/3 line) is broken and prices fall to the 1/3 line and rally from there, that upper line becomes a resistance barrier. Only when that upper line is broken would a signal be given that the old highs will probably be challenged. The same principle holds true in downtrends. - -### **GANN AND FIBONACCI FAN LINES** - -Charting software also allows the drawing of *Gann* and *Fibonacci* fan lines. *Fibonacci* fan lines are drawn in the same fashion as the speedline. Except that Fibonacci lines are drawn at 38% and 62% angles. (We'll explain where those 38% and 62% numbers come from in Chapter 13, "Elliott Wave Theory.") *Gann* lines (named after the legendary commodity trader, W.D. Gann) are trendlines drawn from prominent tops or bottoms at specific geometric angles. The most important Gann [line](#page-293-0) is drawn at a 45 degree angle from a peak or trough. Steeper Gann lines can be drawn during an uptrend at 63 3⁄ 4 degree and 75 degree angles. Flatter Gann lines can be drawn at 26 1⁄ 4 and 15 degree lines. It's possible to draw as many as nine different Gann lines. - -Gann and Fibonacci lines are used in the same way as speedlines. They are supposed to provide support during downward corrections. When one line is broken, prices will usually fall to the next lower line. Gann lines are somewhat controversial. Even if one of them works, you can't be sure in advance which one it will be. Some chartists question the validity of drawing geometric trendlines at all. - -## **INTERNAL TRENDLINES** - -These are variations of the trendline that don't rely on extreme highs or lows. Instead, *internal* trendlines are drawn through the price action and connect as many internal peaks or troughs as possible. Some chartists develop a good eye for this type of trendline and find them useful. The problem with internal trendlines is that their drawing is very subjective; whereas the rules for drawing of more traditional trendlines along the extreme highs and lows are more exact. (See Figure 4.21c.) - -## **REVERSAL [DAYS](#page-98-2)** - -Another important building block is the *reversal day.* This particular chart formation goes by many names—the top reversal day, the bottom reversal day, the *buying* or *selling climax*, and the key reversal day. By itself, this formation is not of major importance. But, taken in the context of other technical information, it can sometimes be significant. Let's first define what a reversal day is. - -**Figure 4.21c** *Internal trendlines are drawn through the price action connecting as many highs and lows as possible. This internal trendline drawn along the early 1996 highs provided support a year later during the spring of 1997.* - -A *reversal day* takes place either at a top or a bottom. The generally - -accepted definition of a *top reversal day* is the setting of a new high in an uptrend, followed by a lower close on the same day. In other words, prices set a new high for a given upmove at some point during the day (usually at or near the opening) then weaken and actually close lower than the previous day's closing. A *bottom reversal day* would be a new low during the day followed by a higher close. - -The wider the range for the day and the heavier the volume, the more significant is the signal for a possible near term trend reversal. Figures 4.22ab show what both would look like on a bar chart. Note the heavier volume on the reversal day. Also notice that both the high and low on the reversal day exceed the range of the previous day, forming an *outside day.* [While](#page-99-0) an [o](#page-100-0)utside day is not a requirement for a reversal day, it does carry more significance. (See Figure 4.22c.) - -The bottom reversal day is sometimes referred to as a *selling climax.* This is usually a dramatic turnaround at the bottom of a down move where all the discouraged longs [have](#page-100-1) finally been forced out of the market on heavy volume. The subsequent absence of selling pressure creates a vacuum over the market, which prices quickly rally to fill. The selling climax is one of the more dramatic examples of the reversal day and, while it may not mark the final bottom of a falling market, it usually signals that a significant low has been seen. - -**Figure 4.22a** *Example of a top reversal day. The heavier the volume on the reversal day and the wider the range, the more important it becomes.* - -**Figure 4.22b** *Example of a bottom reversal day. If volume is especially heavy, bottom reversals are often referred to as "selling climaxes."* - -**Figure 4.22c** *The chart action of October 28, 1997 was a classic example of an upside reversal day or a "selling climax." Prices opened sharply lower and closed sharply higher. The unusually heavy volume bar for that day added to its importance. Two less dramatic upside reversal days (see arrows) also marked price bottoms.* - -#### **Weekly and Monthly Reversals** - -This type of reversal pattern shows up on weekly and monthly bar charts, and with much greater significance. On a weekly chart, each bar represents the entire week's range with the close registered on Friday. An *upside weekly* - -*reversal*, therefore, would occur when the market trades lower during the week, makes a new low for the move, but on Friday closes above the previous Friday's close. - -Weekly reversals are much more significant than daily reversals for obvious reasons and are watched closely by chartists as signaling important turning points. By the same token, monthly reversals are even more important. - -### **PRICE GAPS** - -Price *gaps* are simply areas on the bar chart where no trading has taken place. In an uptrend, for example, prices open above the highest price of the previous day, leaving a gap or open space on the chart that is not filled during the day. In a downtrend, the day's highest price is below the previous day's low. Upside gaps are signs of market strength, while downside gaps are usually signs of weakness. Gaps can appear on long term weekly and monthly charts and, when they do, are usually very significant. But they are more commonly seen on daily bar charts. - -Several myths exist concerning the interpretation of gaps. One of the maxims often heard is that "gaps are always filled." This is simply not true. Some should be filled and others shouldn't. We'll also see that gaps have different forecasting implications depending on which types they are and where they occur. - -### **Three Types of Gaps** - -There are three general types of gaps—the *breakaway, runaway (or measuring)*, and *exhaustion gaps.* - -*The Breakaway Gap.* The *breakaway gap* usually occurs at the completion of an important price pattern, and usually signals the beginning of a significant market move. After a market has completed a major basing pattern, the breaking of resistance often occurs on a breakaway gap. Major breakouts from topping or basing areas are breeding grounds for this type of gap. The breaking of a major trendline, signaling a reversal of trend, might also see a breakaway gap. - -Breakaway gaps usually occur on heavy volume. More often than not, breakaway gaps are not filled. Prices may return to the upper end of the gap (in the case of a bullish breakout), and may even close a portion of the gap, but some portion of the gap is often left unfilled. As a rule, the heavier the volume after such a gap appears, the less likely it is to be filled. Upside gaps usually act as support areas on subsequent market corrections. It's important - -that prices not fall below gaps during an uptrend. In all cases a close below an upward gap is a sign of weakness. (See Figures 4.23a and b.) - -**Figure 4.23a** *The three types of gaps. The breakaway gap signaled the completion of the basing pattern. The runaway gap occurred at about the midway point (which is why it is also called the measuring gap). An exhaustion gap to the upside, followed within a week by a breakaway gap to the downside, left an island reversal top. Notice that the breakaway and runaway gaps were not filled on the way up, which is often the case.* - -*The Runaway or Measuring Gap.* After the move has been underway for awhile, somewhere around the middle of the move, prices will leap forward to form a second type of gap (or a series of gaps) called the *runaway gap.* This type of gap reveals a situation where the market is moving effortlessly on moderate volume. In an uptrend, it's a sign of market strength; in a downtrend, a sign of weakness. Here again, runaway gaps act as support under the market on subsequent corrections and are often not filled. As in the case of the breakaway, a close below the runaway gap is a negative sign in an uptrend. - -**Figure 4.23b** *The first box shows an "exhaustion" gap near the end of the rally. Prices falling below that gap signaled a top. The second box is a "measuring" gap about halfway through the downtrend. The third box is another "exhaustion" gap at the bottom. The move back above that gap signaled higher prices.* - -This variety of gap is also called a *measuring gap* because it usually occurs at about the halfway point in a trend. By measuring the distance the trend has already traveled, from the original trend signal or breakout, an estimate of the probable extent of the remaining move can be determined by doubling the amount already achieved. - -*The Exhaustion Gap.* The final type of gap appears near the end of a market move. After all objectives have been achieved and the other two types of gaps (breakaway and runaway) have been identified, the analyst should begin to expect the *exhaustion gap.* Near the end of an uptrend, prices leap forward in a last gasp, so to speak. However, that upward leap quickly fades and prices turn lower within a couple of days or within a week. When prices close under that last gap, it is usually a dead giveaway that the exhaustion gap has made its appearance. This is a classic example where falling below a gap in an uptrend has very bearish implications. - -#### **The Island Reversal** - -This takes us to *the island reversal pattern.* Sometimes after the upward exhaustion gap has formed, prices will trade in a narrow range for a couple of days or a couple of weeks before gapping to the downside. Such a situation leaves the few days of price action looking like an "island" surrounded by space or water. The exhaustion gap to the upside followed by a breakaway gap to the downside completes the island reversal pattern and usually indicates a trend reversal of some magnitude. Of course, the major significance of the reversal depends on where prices are in the general trend structure. (See Figure 4.23c.) - -**Figure 4.23c** *The two gaps on this daily chart form an "island reversal" top. The first box shows an up gap after a rally. The second box shows a down gap three weeks later. That combination of gaps usually signals an important top.* - -### **CONCLUSION** - -This chapter introduced introductory technical tools that I consider to be the building blocks of chart analysis—support and resistance, trendlines and channels, percentage retracements, speed resistance lines, reversal days, and gaps. Every technical approach covered in later chapters uses these concepts and tools in one form or another. Armed with a better understanding of these concepts, we're now ready to begin a study of price patterns. - -### **INTRODUCTION** - -So far we've touched on Dow Theory, which is the basis of most trend following work being used today. We've examined the basic concepts of trend, such as support, resistance, and trendlines. And we've introduced volume and open interest. We're now ready to take the next step, which is a study of chart patterns. You'll quickly see that these patterns build on the previous concepts. - -In Chapter 4, the definition of a trend was given as a series of ascending or descending peaks and troughs. As long as they were ascending, the trend was up; if they were descending, the trend was down. It was stressed, however, [that](#page-61-0) markets also move sideways for a certain portion of the time. It is these periods of sideways market movement that will concern us most in these next two chapters. - -It would be a mistake to assume that most changes in trend are very abrupt affairs. The fact is that important changes in trend usually require a period of transition. The problem is that these periods of transition do not always signal a trend reversal. Sometimes these sideways periods just indicate a pause or consolidation in the existing trend after which the original trend is resumed. - -### **PRICE PATTERNS** - -The study of these transition periods and their forecasting implications leads us to the question of price patterns. First of all, what are price patterns? Price patterns are pictures or formations, which appear on price charts of stocks or commodities, that can be classified into different categories, and that have diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/010_5 Major Reversal Patterns.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/010_5 Major Reversal Patterns.md deleted file mode 100644 index 376c1e9df421d0c106c9a0c20680e8085e5a2f45..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/010_5 Major Reversal Patterns.md +++ /dev/null @@ -1,241 +0,0 @@ -## **TWO TYPES OF PATTERNS: REVERSAL AND CONTINUATION** - -There are two major categories of price patterns—reversal and continuation. As these names imply, reversal patterns indicate that an important reversal in trend is taking place. The continuation patterns, on the other hand, suggest that the market is only pausing for awhile, possibly to correct a near term overbought or oversold condition, after which the existing trend will be resumed. The trick is to distinguish between the two types of patterns as early as possible during the formation of the pattern. - -In this chapter, we'll be examining the five most commonly used major reversal patterns: the head and shoulders, triple tops and bottoms, double tops and bottoms, spike (or V) tops and bottoms, and the rounding (or saucer) pattern. We will examine the price formation itself, how it is formed on the chart, and how it can be identified. We will then look at the other important considerations—the accompanying *volume pattern* and *measuring implications.* - -*Volume* plays an important confirming role in all of these price patterns. In times of doubt (and there are lots of those), a study of the volume pattern accompanying the price data can be the deciding factor as to whether or not the pattern can be trusted. - -Most price patterns also have certain *measuring techniques* that help the analyst to determine minimum price objectives. While these objectives are only an approximation of the size of the subsequent move, they are helpful in assisting the trader to determine his or her reward to risk ratio. - -In Chapter 5, we'll look at a second category of patterns—the continuation variety. There we will examine triangles, flags, pennants, wedges, and rectangles. These patterns usually reflect pauses in the existing trend rather [than](#page-105-0) trend reversals, and are usually classified as intermediate and minor as opposed to major. - -### **Preliminary Points Common to All Reversal Patterns** - -Before beginning our discussion of the individual major reversal patterns, there are a few preliminary points to be considered that are common to all of these reversal patterns. - -- 1. A prerequisite for any reversal pattern is the existence of a prior trend. -- 2. The first signal of an impending trend reversal is often the breaking of an - -important trendline. - -- 3. The larger the pattern, the greater the subsequent move. -- 4. Topping patterns are usually shorter in duration and more volatile than bottoms. -- 5. Bottoms usually have smaller price ranges and take longer to build. -- 6. Volume is usually more important on the upside. - -*The Need for a Prior Trend.* The existence of a prior major trend is an important prerequisite for any reversal pattern. A market must obviously have something to reverse. A formation occasionally appears on the charts, resembling one of the reversal patterns. If that pattern, however, has not been preceded by a trend, there is nothing to reverse and the pattern is suspect. Knowing where certain patterns are most apt to occur in the trend structure is one of the key elements in pattern recognition. - -A corollary to this point of having a prior trend to reverse is the matter of measuring implications. It was stated earlier that most of the measuring techniques give only *minimum* price objectives. The *maximum* objective would be the total extent of the prior move. If a major bull market has occurred and a major topping pattern is being formed, the maximum implication for the potential move to the downside would be a 100% retracement of the bull market, or the point at which it all began. - -*The Breaking of Important Trendlines.* The first sign of an impending trend reversal is often the breaking of an important trendline. Remember, however, that the violation of a major trendline does not necessarily signal a trend reversal. What is being signaled is a change in trend. The breaking of a major up trendline might signal the beginning of a sideways price pattern, which later would be identified as either the reversal or consolidation type. Sometimes the breaking of the major trendline coincides with the completion of the price pattern. - -*The Larger the Pattern, the Greater the Potential.* When we use the term "larger," we are referring to the height and the width of the price pattern. The height measures the volatility of the pattern. The width is the amount of time required to build and complete the pattern. The greater the size of the pattern —that is, the wider the price swings within the pattern (the volatility) and the longer it takes to build—the more important the pattern becomes and the greater the potential for the ensuing price move. - -Virtually all of the measuring techniques in these two chapters are based on the *height* of the pattern. This is the method applied primarily to bar charts, which use a *vertical* measuring criteria. The practice of measuring the *horizontal* width of a price pattern usually is reserved for point and figure charting. That method of charting uses a device known as the *count*, which - -assumes a close relationship between the width of a top or bottom and the subsequent price target. - -*Differences Between Tops and Bottoms.* Topping patterns are usually shorter in duration and are more volatile than bottoms. Price swings within the tops are wider and more violent. Tops usually take less time to form. Bottoms usually have smaller price ranges, but take longer to build. For this reason it is usually easier and less costly to identify and trade bottoms than to catch market tops. One consoling factor, which makes the more treacherous topping patterns worthwhile, is that *prices tend to decline faster than they go up.* Therefore, the trader can usually make more money a lot faster by catching the short side of a bear market than by trading the long side of a bull market. Everything in life is a tradeoff between reward and risk. The greater risks are compensated for by greater rewards and vice versa. Topping patterns are harder to catch, but are worth the effort. - -*Volume is More Important on the Upside.* Volume should generally increase in the direction of the market trend and is an important confirming factor in the completion of all price patterns. The completion of each pattern should be accompanied by a noticeable increase in volume. However, in the early stages of a trend reversal, *volume is not as important at market tops.* Markets have a way of "falling of their own weight" once a bear move gets underway. Chartists like to see an increase in trading activity as prices drop, but it is not critical. At bottoms, however, the volume pickup is absolutely essential. If the volume pattern does not show a significant increase during the upside price *breakout*, the entire price pattern should be questioned. We will be taking a more in-depth look at volume in Chapter 7. - -### **THE HEAD AND SH[OULDE](#page-153-0)RS REVERSAL PATTERN** - -Let's take a close look now at what is probably the best known and most reliable of all major reversal patterns—*the head and shoulders reversal.* We'll spend more time on this pattern because it is important and also to explain all the nuances involved. Most of the other reversal patterns are just variations of the head and shoulders and will not require as extensive a treatment. - -This major reversal pattern, like all of the others, is just a further refinement of the concepts of trend covered in Chapter 4. Picture a situation in a major uptrend, where a series of ascending peaks and troughs gradually begin to lose momentum. The uptrend then levels off for awhile. During this time the forces of supply and demand are in re[lative](#page-61-0) balance. Once this distribution phase has been completed, support levels along the bottom of the horizontal trading range are broken and a new downtrend has been established. That new downtrend now has descending peaks and troughs. - -Let's see how this scenario would look on a *head and shoulders* top. (See Figures 5.1a and b.) At point A, the uptrend is proceeding as expected with no signs of a top. Volume expands on the price move into new highs, which is normal. The corrective dip to point B is on lighter volume, which is also to be [expecte](#page-109-0)d. A[t](#page-110-0) point C, however, the alert chartist might notice that the volume on the upside breakout through point A is a bit lighter than on the previous rally. This change is not in itself of major importance, but a little yellow caution light goes on in the back of the analyst's head. - -**Figure 5.1a** *Example of a head and shoulders top. The left and right shoulders (A and E) are at about the same height. The head (C) is higher than either shoulder. Notice the lighter volume on each peak. The pattern is completed on a close under the neckline (line 2). The minimum objective is the vertical distance from the head to the neckline projected downward from the breaking of the neckline. A return move will often occur back to the neckline, which should not recross the neckline once it has been broken.* - -**Figure 5.1b** *A head and shoulders top. The three peaks show the head higher than either shoulder. The return move (see arrow) back to the neckline occurred on schedule.* - -Prices then begin to decline to point D and something even more disturbing happens. The decline carries below the top of the previous peak at point A. Remember that, in an uptrend, a penetrated peak should function as support on subsequent corrections. The decline well under point A, almost to the previous reaction low at point B, is another warning that something may be going wrong with the uptrend. - -The market rallies again to point E, this time on even lighter volume, and isn't able to reach the top of the previous peak at point C. (That last rally at point E will often retrace one-half to two-thirds of the decline from points C to D.) To continue an uptrend, each high point must exceed the high point of the rally preceding it. The failure of the rally at point E to reach the previous peak at point C fulfills half of the requirement for a new downtrend—namely, descending peaks. - -By this time, the major up trendline (line 1) has already been broken, usually at point D, constituting another danger signal. But, despite all of these warnings, all that we know at this point is that the trend has shifted from up to sideways. This might be sufficient cause to liquidate long positions, but not necessarily enough to justify new short sales. - -#### **The Breaking of the Neckline Completes the Pattern** - -By this time, a flatter trendline can be drawn under the last two reaction lows - -(points B and D), which is called a *neckline* (see line 2). This line generally has a slight upward slope at tops (although it's sometimes horizontal and, less often, tilts downward). *The deciding factor in the resolution of the head and shoulders top is a decisive closing violation of that neckline.* The market has now violated the trendline along the bottom of points B and D, has broken under support at point D, and has completed the requirement for a new downtrend—descending peaks and troughs. The new downtrend is now identified by the declining highs and lows at points C, D, E, and F. Volume should increase on the breaking of the neckline. A sharp increase in downside volume, however, is not critically important in the initial stages of a market top. - -### **The Return Move** - -Usually a *return move* develops which is a bounce back to the bottom of the neckline or to the previous reaction low at point D (see point G), both of which have now become overhead resistance. The return move does not always occur or is sometimes only a very minor bounce. Volume may help determine the size of the bounce. If the initial breaking of the neckline is on very heavy trading, the odds for a return move are diminished because the increased activity reflects greater downside pressure. Lighter volume on the initial break of the neckline increases the likelihood of a return move. That bounce, however, should be on light volume and the subsequent resumption of the new downtrend should be accompanied by noticeably heavier trading activity. - -### **Summary** - -Let's review the basic ingredients for a head and shoulders top. - -- 1. A prior uptrend. -- 2. A left shoulder on heavier volume (point A) followed by a corrective dip to point B. -- 3. A rally into new highs but on lighter volume (point C). -- 4. A decline that moves below the previous peak (at A) and approaches the previous reaction low (point D). -- 5. A third rally (point E) on noticeably light volume that fails to reach the top of the head (at point C). -- 6. A close below the neckline. -- 7. A return move back to the neckline (point G) followed by new lows. - -What has become evident is three well defined peaks. The middle peak (the head) is slightly higher than either of the two shoulders (points A and E). The pattern, however, is not complete until the neckline is decisively broken on a closing basis. Here again, the 1-3% penetration criterion (or some variation thereof) or the requirement of two successive closes below the neckline (the two day rule) can be used for added confirmation. Until that downside violation takes place, however, there is always the possibility that the pattern is not really a head and shoulders top and that the uptrend may resume at some point. - -### **THE IMPORTANCE OF VOLUME** - -The accompanying volume pattern plays an important role in the development of the head and shoulders top as it does in all price patterns. As a general rule, the second peak (the head) should take place on lighter volume than the left shoulder. This is not a requirement, but a strong tendency and an early warning of diminishing buying pressure. The most important volume signal takes place during the third peak (the right shoulder). Volume should be noticeably lighter than on the previous two peaks. Volume should then expand on the breaking of the neckline, decline during the return move, and then expand again once the return move is over. - -As mentioned earlier, volume is less critical during the completion of market tops. But, at some point, volume should begin to increase if the new downtrend is to be continued. Volume plays a much more decisive role at market bottoms, a subject to be discussed shortly. Before doing so, however, let's discuss the measuring implications of the head and shoulders pattern. - -### **FINDING A PRICE OBJECTIVE** - -The method of arriving at a price objective is based on the *height* of the pattern. Take the vertical distance from the head (point C) to the neckline. Then project that distance from the point where the neckline is broken. Assume, for example, that the top of the head is at 100 and the neckline is at 80. The vertical distance, therefore, would be the difference, which is 20. That 20 points would be measured downward from the level at which the neckline is broken. If the neckline in Figure 5.1a is at 82 when broken, a downside objective would be projected to the 62 level (82 – 20=62). - -Another technique that accomplishes about the same task, but is a bit easier, is to simply measure the [length](#page-109-0) of the first wave of the decline (points C to D) and then double it. In either case, the greater the height or volatility of the pattern, the greater the objective. Chapter 4 stated that the measurement taken from a trendline penetration was similar to that used in the head and - -shoulders pattern. You should be able to see that now. Prices travel roughly the same distance below the broken neckline as they do above it. You'll see throughout our entire study of price patterns that *most price targets on bar charts are based on the height or volatility of the various patterns.* The theme of measuring the height of the pattern and then projecting that distance from a breakout point will be constantly repeated. - -*It's important to remember that the objective arrived at is only a minimum target.* Prices will often move well beyond the objective. Having a minimum target to work with, however, is very helpful in determining beforehand whether there is enough potential in a market move to warrant taking a position. If the market exceeds the price objective, that's just icing on the cake. The *maximum* objective is the size of the prior move. If the previous bull market went from 30 to 100, then the maximum downside objective from a topping pattern would be a complete retracement of the entire upmove all the way down to 30. Reversal patterns can only be expected to reverse or retrace what has gone before them. - -#### **Adjusting Price Objectives** - -A number of other factors should be considered while trying to arrive at a price objective. The measuring techniques from price patterns, such as the one just mentioned for the head and shoulders top, are only the first step. There are other technical factors to take into consideration. For example, where are the prominent support levels left by the reaction lows during the previous bull move? Bear markets often pause at these levels. What about percentage retracements? The *maximum objective* would be a 100% retracement of the previous bull market. But where are the 50% and 66% retracement levels? Those levels often provide significant support under the market. What about any prominent gaps underneath? They often function as support areas. Are there any long term trendlines visible below the market? - -The technician must consider other technical data in trying to pinpoint price targets taken from price patterns. If a downside price measurement, for example, projects a target to 30, and there is a prominent support level at 32, then the chartist would be wise to adjust the downside measurement to 32 instead of 30. As a general rule, when a slight discrepancy exists between a projected price target and a clearcut support or resistance level, it's usually safe to adjust the price target to that support or resistance level. It is often necessary to adjust the measured targets from price patterns to take into account additional technical information. The analyst has many different tools at his or her disposal. The most skillful technical analysts are those who learn to blend all of those tools together properly. - -## **THE INVERSE HEAD AND SHOULDERS** - -The head and shoulders bottom, or the *inverse head and shoulders* as it is sometimes called, is pretty much a mirror image of the topping pattern. As Figure 5.2a shows, there are three distinct bottoms with the head (middle trough) a bit lower than either of the two shoulders. A decisive close through the neckline is also necessary to complete the pattern, and the measuring [technique](#page-114-1) is the same. One slight difference at the bottom is the greater tendency for the return move back to the neckline to occur after the bullish breakout. (See Figure 5.2b.) - -**Figure 5.2a** *Example of an inverse head and shoulders. The bottom version of this pattern is a mirror image of the top. The only significant difference is the volume pattern in the second half of the pattern. The rally from the head should see heavier volume, and the breaking of the neckline should see a burst of trading activity. The return move back to the neckline is more common at bottoms.* - -The most important difference between the top and bottom patterns is the volume sequence. Volume plays a much more critical role in the identification and completion of a head and shoulders bottom. This point is generally true of all bottom patterns. It was stated earlier that markets have a tendency to "fall of their own weight." At bottoms, however, markets require a significant increase in buying pressure, reflected in greater volume, to launch a new bull market. - -**Figure 5.2b** *A head and shoulders bottom. The neckline has a slight downward slant, which is normally the case. The pullback after the breakout (see arrow) nicked the neckline a bit, but then resumed the uptrend.* - -A more technical way of looking at this difference is that a market can fall just from inertia. Lack of demand or buying interest on the part of traders is often enough to push a market lower; but a market does not go up on inertia. Prices only rise when demand exceeds supply and buyers are more aggressive than sellers. - -The volume pattern at the bottom is very similar to that at the top for the first half of the pattern. That is, the volume at the head is a bit lighter than that at the left shoulder. The rally from the head, however, should begin to show not only an increase in trading activity, but the level of volume often exceeds that registered on the rally from the left shoulder. The dip to the right shoulder should be on very light volume. The critical point occurs at the rally through the neckline. This signal must be accompanied by a sharp burst of trading volume if the breakout is for real. - -This point is where the bottom differs the most from the top. At the bottom, heavy volume is an absolutely essential ingredient in the completion of the basing pattern. The return move is more common at bottoms than at tops and should occur on light volume. Following that, the new uptrend should resume on heavier volume. The measuring technique is the same as at the top. - -The neckline at the top usually slopes slightly upward. Sometimes, however, it is horizontal. In either case, it doesn't make too much of a difference. Once in a while, however, a top neckline slopes downward. This slope is a sign of market weakness and is usually accompanied by a weak right shoulder. However, this is a mixed blessing. The analyst waiting for the breaking of the neckline to initiate a short position has to wait a bit longer, because the signal from the down sloping neckline occurs much later and only after much of the move has already taken place. For basing patterns, most necklines have a slight downward tilt. A rising neckline is a sign of greater market strength, but with the same drawback of giving a later signal. - -### **COMPLEX HEAD AND SHOULDERS PATTERNS** - -A variation of the head and shoulders pattern sometimes occurs which is called the *complex head and shoulders pattern.* These are patterns where two heads may appear or a double left and right shoulder. These patterns are not that common, but have the same forecasting implications. A helpful hint in this regard is the strong tendency toward symmetry in the head and shoulders pattern. This means that a single left shoulder usually indicates a single right shoulder. A double left shoulder increases the odds of a double right shoulder. - -### **Tactics** - -Market tactics play an important role in all trading. Not all technical traders like to wait for the breaking of the neckline before initiating a new position. As Figure 5.3 shows, more aggressive traders, believing that they have correctly identified a head and shoulders bottom, will begin to probe the long side during the formation of the right shoulder. Or they will buy the first tec[hnical](#page-117-0) signal that the decline into the right shoulder has ended. - -Some will measure the distance of the rally from the bottom of the head (points C to D) and then buy a 50% or 66% retracement of that rally. Still others would draw a tight down trendline along the decline from points D to E and buy the first upside break of that trendline. Because these patterns are reasonably symmetrical, some will buy into the right shoulder as it approaches the same level as the bottom of the left shoulder. A lot of anticipatory buying takes place during the formation of the right shoulder. If the initial long probe proves to be profitable, additional positions can be added on the actual penetration of the neckline or on the return move back to the neckline after the breakout. - -#### **The Failed Head And Shoulders Pattern** - -Once prices have moved through the neckline and completed a head and shoulders pattern, *prices should not recross the neckline again.* At a top, once the neckline has been broken on the downside, any decisive close back above the neckline is a serious warning that the initial breakdown was probably a bad signal, and creates what is often called, for obvious reasons, *a failed head and shoulders.* This type of pattern starts out looking like a classic head and shoulders reversal, but at some point in its development (either prior to the breaking of the neckline or just after it), prices resume their original trend. - -**Figure 5.3** *Tactics for a head and shoulders bottom. Many technical traders will begin to initiate long positions while the right shoulder (E) is still being formed. One-half to two-thirds pullback of the rally from points C to D, a decline to the same level as the left shoulder at point A*, *or the breaking of a short term down trendline (line 1) all provide early opportunities for market entry. More positions can be added on the breaking of the neckline or the return move back to the neckline.* - -There are two important lessons here. The first is that none of these chart patterns are infallible. They work most of the time, but not always. The second lesson is that technical traders must always be on the alert for chart signs that their analysis is incorrect. One of the keys to survival in the financial markets is to keep trading losses small and to exit a losing trade as quickly as possible. One of the greatest advantages of chart analysis is its ability to quickly alert the trader to the fact that he or she is on the wrong side of the market. The ability and willingness to quickly recognize trading errors and to take defensive action immediately are qualities not to be taken lightly in the financial markets. - -#### **The Head And Shoulders as a Consolidation Pattern** - -Before moving on to the next price pattern, there's one final point to be made on the head and shoulders. We started this discussion by listing it as the best known and most reliable of the major reversal patterns. You should be warned, however, that this formation can, on occasion, act as a consolidation rather than a reversal pattern. When this does happen, it's the exception rather than the rule. We'll talk more about this in Chapter 6, "Continuation Patterns." - -### **TRIPLE TOPS AND BOTTOMS** - -Most of the points covered in the treatment of the head and shoulders pattern are also applicable to other types of reversal patterns. (See Figures 5.4a-c.) The *triple top* or *bottom*, which is much rarer in occurrence, is just a slight variation of that pattern. The main difference is that the three peaks or troughs in the *triple top* or *bottom* are at about the same level. (See [Figure](#page-118-1) 5.4a.[\)](#page-119-0) Chartists often disagree as to whether a reversal pattern is a head and shoulders or a triple top. The argument is academic, because both patterns imply the exact same thing. - -The volume tends to decline with each successive peak at the top and should increase at the breakdown point. The triple top is not complete until support levels along both of the intervening lows have been broken. Conversely, prices must close through the two intervening peaks at the bottom to complete a triple bottom. (As an alternate strategy, the breaking of the nearest peak or trough can also be used as a reversal signal.) Heavy upside volume on the completion of the bottom is also essential. - -**Figure 5.4a** *A triple top. Similar to the head and shoulders except that all peaks are at the same level. Each rally peak should be on lighter volume. The pattern is complete when both troughs have been broken on heavier volume. The measuring technique is the height of the pattern projected downward from the breakdown point. Return moves back to the lower line are not unusual.* - -**Figure 5.4b** *A triple bottom. Similar to a head and shoulders bottom except that each low is at the same level. A mirror image of the triple top except that volume is more important on the upside breakout.* - -**Figure 5.4c** *A triple bottom reversal pattern. Prices found support just* - -*below 12 three times on this chart before launching a major advance. The bottom formation on this weekly chart lasted two full years, thereby giving it major significance.* - -The measuring implication is also similar to the head and shoulders, and is based on the height of the pattern. Prices will usually move a minimum distance from the breakout point at least equal to the height of the pattern. Once the breakout occurs, a return move to the breakout point is not unusual. Because the triple top or bottom represents only a minor variation of the head and shoulders pattern, we won't say much more about it here. - -### **DOUBLE TOPS AND BOTTOMS** - -A much more common reversal pattern is the *double top or bottom.* Next to the *head and shoulders*, it is the most frequently seen and the most easily recognized. (See Figures 5.5a-e.) Figures 5.5a and 5.5b show both the top and bottom variety. For obvious reasons, the top is often referred to as an "M" and the bottom as a "W." The general characteristics of a *double top* are similar to that of the head and [shoulders](#page-120-1) [a](#page-123-1)nd [triple](#page-120-1) top except [that](#page-121-0) only two peaks appear instead of three. The volume pattern is similar as is the measuring rule. - -**Figure 5.5a** *Example of a double top. This pattern has two peaks (A and C) at about the same level. The pattern is complete when the middle trough at point B is broken on a closing basis. Volume is usually lighter on the second peak (C) and picks up on the breakdown (D). A return move back to the lower line is not unusual. The minimum measuring target is the height of the top projected downward from the breakdown point.* - -**Figure 5.5b** *Example of a double bottom. A mirror image of the double top. Volume is more important on the upside breakout. Return moves back to the breakout point are more common at bottoms.* - -**Figure 5.5c** *Example of a double bottom. This stock bounced sharply off the 68 level twice over a span of three months. Note that the second bottom was also an upside reversal day. The breaking of resistance at 80 completed the bottom.* - -In an uptrend (as shown in Figure 5.5a), the market sets a new high at point A, usually on increased volume, and then declines to point B on declining volume. So far, everything is proceeding as expected in a normal uptrend. The next rally to point C, [however,](#page-120-1) is unable to penetrate the previous peak at A on a closing basis and begins to fall back again. A - -potential *double top* has been set up. I use the word "potential" because, as is the case with all reversal patterns, the reversal is not complete until the previous support point at B is violated on a closing basis. Until that happens, prices could be in just a sideways consolidation phase, preparing for a resumption of the original uptrend. - -**Figure 5.5d** *Example of a double top. Sometimes the second peak doesn't quite reach the first peak as in this example. This two month double top signaled a major decline. The actual signal was the breaking of support near 46 (see box).* - -The ideal top has two prominent peaks at about the same price level. Volume tends to be heavier during the first peak and lighter on the second. A decisive close under the middle trough at point B on heavier volume completes the pattern and signals a reversal of trend to the downside. A return move to the breakout point is not unusual prior to resumption of the downtrend. - -#### **Measuring Technique for the Double Top** - -The measuring technique for the double top is the height of the pattern projected from the breakdown point (the point where the middle trough at point B is broken). As an alternative, measure the height of the first downleg (points A to B) and project that length downward from the middle trough at point B. Measurements at the bottom are the same, but in the other direction. - -**Figure 5.5e** *Price patterns show up regularly on the charts of major stock averages. On this chart, the Nasdaq Composite Index formed a double bottom near the 1470 level before turning higher. The break of the down trendline (see box) confirmed the upturn.* - -### **VARIATIONS FROM THE IDEAL PATTERN** - -As in most other areas of market analysis, real-life examples are usually some variation of the ideal. For one thing, sometimes the two peaks are not at exactly the same price level. On occasion, the second peak will not quite reach the level of the first peak, which is not too problematical. What does cause some problems is when the second peak actually exceeds the first peak by a slight margin. What at first may appear to be a valid upside breakout and resumption of the uptrend may turn out to be part of the topping process. To help resolve this dilemma, some of the filtering criteria already mentioned may come in handy. - -#### **Filters** - -Most chartists require a close beyond a previous resistance peak instead of just an intraday penetration. Second, a price filter of some type might be used. One such example is a percentage penetration criterion (such as 1% or 3%). Third, the two day penetration rule could be used as an example of a time filter. In other words, prices would have to close beyond the top of the first peak for two consecutive days to signal a valid penetration. Another time - -filter could be a Friday close beyond the previous peak. The volume on the upside breakout might also provide a clue to its reliability. - -These filters are certainly not infallible, but do serve to reduce the number of false signals (or whipsaws) that often occur. Sometimes these filters are helpful, and sometimes they're not. The analyst must face the realization that he or she is dealing with percentages and probabilities, and that there will be times when bad signals occur. That's simply a fact of trading life. - -It's not that unusual for the final leg or wave of a bull market to set a new high before reversing direction. In such a case, the final upside breakout would become a "bull trap." (See Figures 5.6a and b.) We'll show you some indicators later on that may help you spot these false breakouts. - -#### **The Term "Double Top" Greatly Overuse[d](#page-124-0)** - -The terms "double top and bottom" are greatly overused in the financial markets. Most potential double tops or bottoms wind up being something else. The reason for this is that prices have a strong tendency to back off from a previous peak or bounce off a previous low. These price changes are a natural reaction and do not in themselves constitute a reversal pattern. Remember that, at a top, prices must actually violate the previous reaction low before the double top exists. - -**Figure 5.6a** *Example of a false breakout, usually called a bull trap. Sometimes near the end of a major uptrend, prices will exceed a previous peak before failing. Chartists use various time and price filters to reduce such whipsaws. This topping pattern would probably qualify as a double top.* - -**Figure 5.6b** *Example of a false breakout. Notice that the upside breakout was on light volume and the subsequent decline on heavy volume—a negative chart combination. Watching the volume helps avoid some false breakouts, but not all.* - -Notice in Figure 5.7a that the price at point C backs off from the previous peak at point A. This is perfectly normal action in an uptrend. Many traders, however, will immediately label this pattern as a double top as soon as prices fail to [clear](#page-126-0) the first peak on the first attempt. Figure 5.7b shows the same situation in a downtrend. It is very difficult for the chartist to determine whether the pullback from the previous peak or the bounce from the previous low is just a temporary setback in the existing trend or the [start](#page-126-1) of a double top or bottom reversal pattern. Because the technical odds usually favor continuation of the present trend, it is usually wise to await completion of the pattern before taking action. - -**Figure 5.7a** *Example of a normal pullback from a previous peak before resumption of the uptrend. This is normal market action and not to be confused with a double top. The double top only occurs when support at point B is broken.* - -**Figure 5.7b** *Example of a normal bounce off a previous low. This is normal market action and not to be confused with a double bottom. Prices will normally bounce off a previous low at least once, causing premature calls for a double bottom.* - -#### **Time Between Peaks or Troughs Is Important** - -Finally, the size of the pattern is always important. The longer the time period between the two peaks and the greater the height of the pattern, the greater the potential impending reversal. This is true of all chart patterns. In general, most valid double tops or bottoms should have at least a month between the two peaks or troughs. Some will even be two or three months apart. (On longer range monthly and weekly charts, these patterns can span several years.) Most of the examples used in this discussion have described market tops. The reader should be aware by now that bottoming patterns are mirror images of tops except for some of the general differences already touched - -### **SAUCERS AND SPIKES** - -Although not seen as frequently, reversal patterns sometimes take the shape of saucers or rounding bottoms. The *saucer bottom* shows a very slow and very gradual turn from down to sideways to up. It is difficult to tell exactly when the saucer has been completed or to measure how far prices will travel in the opposite direction. Saucer bottoms are usually spotted on weekly or monthly charts that span several years. The longer they last, the more significant they become. (See Figure 5.8.) - -*Spikes* are the hardest market turns to deal with because the spike (or V pattern) happens very quickly with little or no transition period. They usually take place in a [market](#page-127-0) that has gotten so overextended in one direction, that a sudden piece of adverse news causes the market to reverse direction very abruptly. A daily or weekly reversal, on very heavy volume, is sometimes the only warning they give us. That being the case, there's not much more we can say about them except that we hope you don't run into too many of them. Some technical indicators we discuss in later chapters will help you determine when markets have gotten dangerously over-extended. (See Figure 5.9.) - -**Figure 5.8** *This chart shows what a saucer (or rounding) bottom looks like. They're very slow and gradual, but usually mark major turns. This bottom lasted four years.* - -**Figure 5.9** *Example of a v reversal pattern. These sudden reversals take place with little or no warning. A sudden price drop on heavy volume is usually the only telltale sign. Unfortunately, these sudden turns are hard to spot in advance.* - -### **CONCLUSION** - -We've discussed the five most commonly used major reversal patterns—the head and shoulders, double and triple tops and bottoms, the saucer, and the V, or spike. Of those, the most common are the head and shoulders, and double tops and bottoms. These patterns usually signal important trend reversals in progress and are classified as major reversal patterns. There is another class of patterns, however, which are shorter term in nature and usually suggest trend consolidations rather than reversals. They are aptly called *continuation* patterns. Let's look at this other type of pattern in Chapter 6. - -### **INTRODUCTION** - -The chart patterns covered in this chapter are called *continuation* patterns. These patterns usually indicate that the sideways price action on the chart is nothing more than a pause in the prevailing trend, and that the next move will be in the same direction as the trend that preceded the formation. This distinguishes this group of patterns from those in the previous chapter, which usually indicate that a major trend reversal is in progress. - -Another difference between reversal and continuation patterns is their time duration. Reversal patterns usually take much longer to build and represent major trend changes. Continuation patterns, on the other hand, are usually shorter term in duration and are more accurately classified as near term or intermediate patterns. - -Notice the constant use of the term "usually." The treatment of all chart patterns deals of necessity with general tendencies as opposed to rigid rules. There are always exceptions. Even the grouping of price patterns into different categories sometimes becomes tenuous. Triangles are usually continuation patterns, but sometimes act as reversal patterns. Although triangles are usually considered intermediate patterns, they may occasionally appear on long term charts and take on major trend significance. A variation of the triangle—the inverted variety—usually signals a major market top. Even the head and shoulders pattern, the best known of the major reversal patterns, will on occasion be seen as a consolidation pattern. - -Even with allowances for a certain amount of ambiguity and the occasional exception, chart patterns do generally fall into the above two categories and, if properly interpreted, can help the chartist determine what the market will probably do most of the time diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/011_6 Continuation Patterns.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/011_6 Continuation Patterns.md deleted file mode 100644 index 56295fe5dcec3b1cbcccd2ea281cac91000bc18e..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/011_6 Continuation Patterns.md +++ /dev/null @@ -1,239 +0,0 @@ -### **TRIANGLES** - -Let's begin our treatment of continuation patterns with the *triangle.* There are three types of triangles—*symmetrical, ascending*, and *descending.* (Some chartists include a fourth type of triangle known as an *expanding triangle*, or *broadening formation.* This is treated as a separate pattern later.) Each type of triangle has a slightly different shape and has different forecasting implications. - -Figures 6.1a-c show examples of what each triangle looks like. The symmetrical triangle (see Figure 6.1a) shows two converging trendlines, the upper line descending and the lower line ascending. The vertical line at the left, [measuring](#page-130-1) th[e](#page-131-1) height of the pattern, is called the *base.* The point of intersection at the right, [where](#page-130-1) the two lines meet, is called the *apex.* For obvious reasons, the symmetrical triangle is also called a *coil.* - -The ascending triangle has a rising lower line with a flat or horizontal upper line (see Figure 6.1b). The descending triangle (Figure 6.1c), by contrast, has the upper line declining with a flat or horizontal bottom line. Let's see how each one is interpreted. - -**Figure 6.1a** *Example of a bullish symmetrical triangle. Notice the two converging trendlines. A close outside either trendline completes the pattern. The vertical line at the left is the base. The point at the right where the two lines meet is the apex.* - -**Figure 6.1b** *Example of an ascending triangle. Notice the flat upper line and the rising lower line. This is generally a bullish pattern.* - -**Figure 6.1c** *Example of a descending triangle. Notice the flat bottom line and the declining upper line. This is usually a bearish pattern.* - -### **THE SYMMETRICAL TRIANGLE** - -The *symmetrical triangle* (or the *coil)* is usually a continuation pattern. It represents a pause in the existing trend after which the original trend is resumed. In the example in Figure 6.1a, the prior trend was up, so that the percentages favor resolution of the triangular consolidation on the upside. If the trend had been down, then the symmetrical triangle would have bearish implications. - -The minimum requirement for a triangle is four reversal points. Remember that it always takes two points to draw a trendline. Therefore, in order to draw two converging trendlines, each line must be touched at least twice. In Figure 6.1a, the triangle actually begins at point 1, which is where the consolidation in the uptrend begins. Prices pull back to point 2 and then rally to point 3. Point 3, however, is lower than point 1. The upper trendline can only be [drawn](#page-130-1) once prices have declined from point 3. - -Notice that point 4 is higher than point 2. Only when prices have rallied from point 4 can the lower upslanting line be drawn. It is at this point that the analyst begins to suspect the he or she is dealing with the symmetrical triangle. Now there are four reversal points (1, 2, 3, and 4) and two converging trendlines. - -While the minimum requirement is four reversal points, many triangles have six reversal points as shown in Figure 6.1a. - -This means that there are actually three peaks and three troughs that combine to form five waves within the triangle [before](#page-130-1) the uptrend resumes. (When we get to the Elliott Wave Theory, we'll have more to say about the five wave tendency within triangles.) - -#### **Time Limit for Triangle Resolution** - -There is a time limit for the resolution of the pattern, and that is the point where the two lines meet—at the apex. As a general rule, prices should break out in the direction of the prior trend somewhere between two-thirds to threequarters of the horizontal width of the triangle. That is, the distance from the vertical base on the left of the pattern to the apex at the far right. Because the two lines must meet at some point, that time distance can be measured once the two converging lines are drawn. An upside breakout is signaled by a penetration of the upper trendline. If prices remain within the triangle beyond the three-quarters point, the triangle begins to lose its potency, and usually means that prices will continue to drift out to the apex and beyond - -The triangle, therefore, provides an interesting combination of price and time. The converging trendlines give the price boundaries of the pattern, and indicate at what point the pattern has been completed and the trend resumed by the penetration of the upper trendline (in the case of an uptrend). But these trendlines also provide a time target by measuring the width of the pattern. If the width, for example, were 20 weeks long, then the breakout should take place sometime between the 13th and the 15th week. (See Figure 6.1d.) - -The actual trend signal is given by a closing penetration of one of the trendlines. Sometimes a return move will occur back to the penetrated trendline after the breakout. In an uptrend, that line has be[come](#page-133-0) a support line. In a downtrend, the lower line becomes a resistance line once it's broken. The apex also acts as an important support or resistance level after the breakout occurs. Various penetration criteria can be applied to the breakout, similar to those covered in the previous two chapters. A minimum penetration criterion would be a closing price outside the trendline and not just an intraday penetration. - -**Figure 6.1d** *Dell formed a bullish symmetrical triangle during the fourth quarter of 1997. Measured from left to right, the triangle width is 18 weeks. Prices broke out on the 13th week (see circle), just beyond the two-thirds point.* - -#### **Importance of Volume** - -Volume should diminish as the price swings narrow within the triangle. This tendency for volume to contract is true of all consolidation patterns. But the volume should pick up noticeably at the penetration of the trendline that completes the pattern. The return move should be on light volume with heavier activity again as the trend resumes. - -Two other points should be mentioned about volume. As is the case with reversal patterns, volume is more important on the upside than on the downside. An increase in volume is essential to the resumption of an uptrend in all consolidation patterns. - -The second point about volume is that, even though trading activity diminishes during formation of the pattern, a close inspection of the volume usually gives a clue as to whether the heavier volume is occurring during the upmoves or down-moves. In an uptrend, for example, there should be a slight tendency for volume to be heavier during the bounces and lighter on the price dips. - -#### **Measuring Technique** - -Triangles have measuring techniques. In the case of the symmetrical triangle, - -there are a couple of techniques generally used. The simplest technique is to measure the height of the vertical line at the widest part of the triangle (the base) and measure that distance from the breakout point. Figure 6.2 shows the distance projected from the breakout point, which is the technique I prefer. - -The second method is to draw a trendline from the top of the base (at point A) parallel to the lower trendline. This upper channel [line](#page-134-1) then becomes the upside target in an uptrend. It is possible to arrive at a rough time target for prices to meet the upper channel line. Prices will sometimes hit the channel line at the same time the two converging lines meet at the apex. - -**Figure 6.2** *There are two ways to take a measurement from a symmetrical triangle. One is to measure the height of the base (AB); project that vertical distance from the breakout point at C. Another method is to draw a parallel line upward from the top of the baseline (A) parallel to the lower line in the triangle.* - -### **THE ASCENDING TRIANGLE** - -The ascending and descending triangles are variations of the symmetrical, but have different forecasting implications. Figures 6.3a and b show examples of an *ascending triangle.* Notice that the upper trendline is flat, while the lower line is rising. This pattern indicates that buyers are more aggressive than sellers. It is considered a bullish pattern and is [usuall](#page-135-0)y resolved with a breakout to the upside. - -Both the ascending and descending triangles differ from the symmetrical in a very important sense. No matter where in the trend structure the ascending or descending triangles appear, they have very definite forecasting implications. The ascending triangle is bullish and the descending triangle is - -bearish. The symmetrical triangle, by contrast, is inherently a neutral pattern. This does not mean, however, that the symmetrical triangle does not have forecasting value. On the contrary, because the symmetrical triangle is a continuation pattern, the analyst must simply look to see the direction of the previous trend and then make the assumption that the previous trend will continue. - -**Figure 6.3a** *An ascending triangle. The pattern is completed on a decisive close above the upper line. This breakout should see a sharp increase in volume. That upper resistance line should act as support on subsequent dips after the breakout. The minimum price objective is obtained by measuring the height of the triangle (AB) and projecting that distance upward from the breakout point at C.* - -**Figure 6.3b** *The Dow Transports formed a bullish ascending triangle near the end of 1997. Notice the flat upper line at 3400 and the rising lower line. This is normally a bullish pattern no matter where it appears on the chart.* - -Let's get back to the ascending triangle. As already stated, more often than not, the ascending triangle is bullish. The bullish breakout is signaled by a decisive closing above the flat upper trendline. As in the case of all valid upside breakouts, volume should see a noticeable increase on the breakout. A return move back to the support line (the flat upper line) is not unusual and should take place on light volume. - -#### **Measuring Technique** - -The measuring technique for the ascending triangle is relatively simple. Simply measure the height of the pattern at its widest point and project that vertical distance from the breakout point. This is just another example of using the volatility of a price pattern to determine a minimum price objective. - -#### **The Ascending Triangle as a Bottom** - -While the ascending triangle most often appears in an uptrend and is considered a continuation pattern, it sometimes appears as a bottoming pattern. It is not unusual toward the end of a downtrend to see an ascending triangle develop. However, even in this situation, the interpretation of the pattern is bullish. The breaking of the upper line signals completion of the base and is considered a bullish signal. Both the ascending and descending triangles are sometimes also referred to as *right angle* triangles. - -### **THE DESCENDING TRIANGLE** - -The *descending triangle* is just a mirror image of the ascending, and is generally considered a bearish pattern. Notice in Figures 6.4a and b the descending upper line and the flat lower line. This pattern indicates that sellers are more aggressive than buyers, and is usually resolved on the downside. The downside signal is registered by a [decisive](#page-137-0) close u[nd](#page-137-1)er the lower trendline, usually on increased volume. A return move sometimes occurs which should encounter resistance at the lower trendline. - -The measuring technique is exactly the same as the ascending triangle in the sense that the analyst must measure the height of the pattern at the base to the left and then project that distance down from the breakdown point. - -While the descending triangle is a continuation pattern and usually is found within downtrends, it is not unusual on occasion for the descending triangle to be found at market tops. This type of pattern is not that difficult to recognize when it does appear in the top setting. In that case, a close below the flat lower line would signal a major trend reversal to the downside. - -**Figure 6.4a** *A descending triangle. The bearish pattern is completed with a decisive close under the lower flat line. The measuring technique is the height of the triangle (AB) projected down from the breakout at point C.* - -**Figure 6.4b** *A bearish descending triangle formed in Du Pont during the autumn of 1997. The upper line is descending while the lower line is flat. The break of the lower line in early October resolved the pattern to the downside.* - -#### **The Volume Pattern** - -The volume pattern in both the ascending and descending triangles is very similar in that the volume diminishes as the pattern works itself out and then increases on the breakout. As in the case of the symmetrical triangle, during the formation the chartist can detect subtle shifts in the volume pattern coinciding with the swings in the price action. This means that in the ascending pattern, the volume tends to be slightly heavier on bounces and lighter on dips. In the descending formation, volume should be heavier on the downside and lighter during the bounces. - -#### **The Time Factor in Triangles** - -One final factor to be considered on the subject of triangles is that of the time dimension. The triangle is considered an intermediate pattern, meaning that it usually takes longer than a month to form, but generally less than three months. A triangle that lasts less than a month is probably a different pattern, such as a pennant, which will be covered shortly. As mentioned earlier, triangles sometimes appear on long term price charts, but their basic meaning is always the same. - -### **THE BROADENING FORMATION** - -This next price pattern is an unusual variation of the triangle and is relatively rare. It is actually an inverted triangle or a triangle turned backwards. All of the triangular patterns examined so far show converging trendlines. The *broadening formation*, as the name implies, is just the opposite. As the pattern in Figure 6.5 shows, the trendlines actually diverge in the broadening formation, creating a picture that looks like an expanding triangle. It is also called a megaphone top. - -[The](#page-139-1) volume pattern also differs in this formation. In the other triangular patterns, volume tends to diminish as the price swings grow narrower. Just the opposite happens in the broadening formation. *The volume tends to expand along with the wider price swings.* This situation represents a market that is out of control and unusually emotional. Because this pattern also represents an unusual amount of public participation, it most often occurs at major market tops. *The expanding pattern, therefore, is usually a bearish formation.* It generally appears near the end of a major bull market. - -**Figure 6.5** *A broadening top. This type of expanding triangle usually occurs at major tops. It shows three successively higher peaks and two declining troughs. The violation of the second trough completes the pattern. This is an unusually difficult pattern to trade and fortunately is relatively rare.* - -### **FLAGS AND PENNANTS** - -The *flag* and *pennant* formations are quite common. They are usually treated together because they are very similar in appearance, tend to show up at about the same place in an existing trend, and have the same volume and measuring criteria. - -The *flag* and *pennant* represent brief pauses in a dynamic market move. In fact, one of the requirements for both the flag and the pennant is that they be preceded by a sharp and almost straight line move. They represent situations where a steep advance or decline has gotten ahead of itself, and where the market pauses briefly to "catch its breath" before running off again in the same direction. - -Flags and pennants are among the most reliable of continuation patterns and only rarely produce a trend reversal. Figures 6.6a-b show what these two patterns look like. To begin with, notice the steep price advance preceding the formations on heavy volume. Notice also the dramatic drop off in activity as the consolidation patterns form and then the [sudden](#page-140-0) b[ur](#page-140-1)st of activity on the upside breakout. - -The construction of the two patterns differs slightly. The flag resembles a parallelogram or rectangle marked by two parallel trendlines that tend to slope against the prevailing trend. In a downtrend, the flag would have a slight upward slope. - -**Figure 6.6a** *Example of a bullish flag. The flag usually occurs after a sharp move and represents a brief pause in the trend. The flag should slope against the trend. Volume should dry up during the formation and build again on the breakout. The flag usually occurs near the midpoint of the move.* - -**Figure 6.6b** *A bullish pennant. Resembles a small symmetrical triangle, but* - -### *usually lasts no longer than three weeks. Volume should be light during its formation. The move after the pennant is completed should duplicate the size of the move preceding it.* - -The pennant is identified by two converging trendlines and is more horizontal. It very closely resembles a small symmetrical triangle. An important requirement is that volume should dry up noticeably while each of the patterns is forming. - -Both patterns are relatively short term and should be completed within one to three weeks. Pennants and flags in downtrends tend to take even less time to develop, and often last no longer than one or two weeks. Both patterns are completed on the penetration of the upper trendline in an uptrend. The breaking of the lower trendline would signal resumption of downtrends. The breaking of those trendlines should take place on heavier volume. As usual, upside volume is more critically important than downside volume. (See Figures 6.7a-b.) - -#### **Measuring Implications** - -The [measuri](#page-141-0)[ng](#page-142-0) implications are similar for both patterns. Flags and pennants are said to "fly at half-mast" from a *flagpole.* The flagpole is the prior sharp advance or decline. The term "half-mast" suggests that these minor continuation patterns tend to appear at about the halfway point of the move. In general, the move after the trend has resumed will duplicate the flagpole or the move just prior to the formation of the pattern. - - - -### **Figure 6.7a** *A bullish flag in International Paper. The flag looks like a downsloping parallelogram. Notice that the flag occurred right at the halfway point of the uptrend.* - -To be more precise, measure the distance of the preceding move from the original breakout point. That is to say, the point at which the original trend signal was given, either by the penetration of a support or resistance level or an important trendline. That vertical distance of the preceding move is then measured from the breakout point of the flag or pennant—that is, the point at which the upper line is broken in an uptrend or the lower line in a downtrend. - -#### **Summary** - -Let's summarize the more important points of both patterns. - -**Figure 6.7b** *A couple of pennants are flying on this Caterpillar chart. Pennants are short term continuation patterns that look like small symmetrical triangles. The pennant to the left continued the uptrend, while the one to the right continued the downtrend.* - -- 1. They are both preceded by an almost straight line move (called a flagpole) on heavy volume. -- 2. Prices then pause for about one to three weeks on very light volume. -- 3. The trend resumes on a burst of trading activity. -- 4. Both patterns occur at about the midpoint of the market move. -- 5. The pennant resembles a small horizontal symmetrical triangle. - -- 6. The flag resembles a small parallelogram that slopes against the prevailing trend. -- 7. Both patterns take less time to develop in downtrends. -- 8. Both patterns are very common in the financial markets. - -### **THE WEDGE FORMATION** - -The *wedge* formation is similar to a symmetrical triangle both in terms of its shape and the amount of time it takes to form. Like the symmetrical triangle, it is identified by two converging trendlines that come together at an *apex.* In terms of the amount of time it takes to form, the wedge usually lasts more than one month but not more than three months, putting it into the intermediate category. - -What distinguishes the *wedge* is its noticeable slant. The wedge pattern has a noticeable slant either to the upside or the downside. As a rule, like the flag pattern, the wedge slants against the prevailing trend. Therefore, a *falling wedge is considered bullish and a rising wedge is bearish.* Notice in Figure 6.8a that the bullish wedge slants downward between two converging trendlines. In the downtrend in Figure 6.8b, the converging trendlines have an [unmistakable](#page-143-1) upward slant. - -**Figure 6.8a** *Example of a bullish falling wedge. The wedge pattern has two converging trendlines, but slopes against the prevailing trend. A falling wedge is usually bullish.* - -**Figure 6.8b** *Example of a bearish wedge. A bearish wedge should slope upward against the prevailing downtrend.* - -#### **Wedges as Tops and Bottom Reversal Patterns** - -*Wedges* show up most often within the existing trend and usually constitute continuation patterns. The wedge can appear at tops or bottoms and signal a trend reversal. But that type of situation is much less common. Near the end of an uptrend, the chartist may observe a clearcut rising wedge. Because a continuation wedge in an uptrend should slope downward against the prevailing trend, the rising wedge is a clue to the chartist that this is a bearish and not a bullish pattern. At bottoms, a falling wedge would be a tip-off of a possible end of a bear trend. - -Whether the wedge appears in the middle or the end of a market move, the market analyst should always be guided by the general maxim that *a rising wedge is bearish and a falling wedge is bullish.* (See Figure 6.8c.) - -### **THE RECTANGLE FORMATION** - -The *rectangle formation* often goes by other names, but is usually easy to spot on a price chart. It represents a pause in the trend during which prices move sideways between two parallel horizontal lines. (See Figures 6.9a-c.) - -The rectangle is sometimes referred to as a *trading range* or a *congestion area.* In Dow Theory parlance, it is referred to as a *line.* Whatever it is called, it usually represents just a consolidation period in the [existing](#page-145-0) tre[nd](#page-146-0), and is usually resolved in the direction of the market trend that preceded its occurrence. In terms of forecasting value, it can be viewed as being similar to the symmetrical triangle but with flat instead of converging trendlines. - -**Figure 6.8c** *Example of a bearish rising wedge. The two converging trendlines have a definite upward slant. The wedge slants against the prevailing trend. Therefore, a rising wedge is bearish, and a falling wedge is bullish.* - -**Figure 6.9a** *Example of a bullish rectangle in an uptrend. This pattern is also called a trading range, and shows prices trading between two horizontal trendlines. It is also called a congestion area.* - -**Figure 6.9b** *Example of a bearish rectangle. While rectangles are usually considered continuation patterns, the trader must always be alert for signs that it may turn into a reversal pattern, such as a triple bottom.* - -**Figure 6.9c** *A bullish rectangle. Compaq's uptrend was interrupted for four months while it traded sideways. The break above the upper line in early May completed the pattern and resumed the uptrend. Rectangles are usually continuation patterns.* - -A decisive close outside either the upper or lower boundary signals completion of the rectangle and points the direction of the trend. The market analyst must always be on the alert, however, that the rectangular consolidation does not turn into a reversal pattern. In the uptrend shown in Figure 6.9a, for example, notice that the three peaks might initially be viewed as a possible triple top reversal pattern. - -#### **The Importance of the Volume Pattern** - -One important clue to watch for is the volume pattern. Because the price swings in both directions are fairly broad, the analyst should keep a close eye on which moves have the heavier volume. If the rallies are on heavier and the setbacks on lighter volume, then the formation is probably a continuation in the uptrend. If the heavier volume is on the downside, then it can be considered a warning of a possible trend reversal in the works. - -#### **Swings Within the Range Can Be Traded** - -Some chartists trade the swings within such a pattern by buying dips near the bottom and selling rallies near the top of the range. This technique enables the short term trader to take advantage of the well defined price boundaries, and profit from an otherwise trendless market. Because the positions are being taken at the extremes of the range, the risks are relatively small and well defined. If the trading range remains intact, this countertrend trading approach works quite well. When a breakout does occur, the trader not only exits the last losing trade immediately, but can reverse the previous position by initiating a new trade in the direction of the new trend. Oscillators are especially useful in sideways trading markets, but less useful once the breakout has occurred for reasons discussed in Chapter 10. - -Other traders assume the rectangle is a continuation pattern and take long positions near the lower end of the price band in an uptrend, or initiate short positions near the top of the range in downtrends. [Others](#page-211-0) avoid such trendless markets altogether and await a clearcut breakout before committing their funds. Most trend-following systems perform very poorly during these periods of sideways and trendless market action. - -#### **Other Similarities and Differences** - -In terms of duration, the rectangle usually falls into the one to three month category, similar to triangles and wedges. The volume pattern differs from other continuation patterns in the sense that the broad price swings prevent the usual dropoff in activity seen in other such patterns. - -The most common measuring technique applied to the rectangle is based on the height of the price range. Measure the height of the trading range, from top to bottom, and then project that vertical distance from the breakout point. This method is similar to the other vertical measuring techniques already mentioned, and is based on the volatility of the market. When we cover the count in point and figure charting, we'll say more on the question of horizontal price measurements. - -Everything mentioned so far concerning volume on breakouts and the probability of return moves applies here as well. Because the upper and lower boundaries are horizontal and so well defined in the rectangle, support and resistance levels are more clearly evident. This means that, on upside breakouts, the top of the former price band should now provide solid support on any selloffs. After a downside breakout in downtrends, the bottom of the trading range (the previous support area) should now provide a solid ceiling over the market on any rally attempts. - -### **THE MEASURED MOVE** - -The *measured move*, or the *swing* measurement as it is sometimes called, describes the phenomenon where a major market advance or decline is divided into two equal and parallel moves, as shown in Figure 6.10a. For this approach to work, the market moves should be fairly orderly and well defined. The measured move is really just a variation of some of the techniques we've already touched on. We've seen that [some](#page-148-1) of the consolidation patterns, such as flags and pennants, usually occur at about the halfway point of a market move. We've also mentioned the tendency of markets to retrace about a third to a half of a prior trend before resuming that trend. - -**Figure 6.10a** *Example of a measured move (or the swing measurement) in an uptrend. This theory holds that the second leg in the advance (CD) duplicates the size and slope of the first upleg (AB). The corrective wave (BC) often retraces a third to a half of AB before the uptrend is resumed.* - -**Figure 6.10b** *A measured move takes the prior upleg (AB) and adds that value to the bottom of the correction at C. On this chart, the prior uptrend (AB) was 20 points. Adding that to the lowpoint at C (62) yielded a price target to 82 (D).* - -In the measured move, when the chartist sees a well-defined situation, such as in Figure 6.10a, with a rally from point A to point B followed by a countertrend swing from point B to point C (which retraces a third to a half of wave AB), it is assumed that the next leg in the uptrend (CD) will come close to duplicating the [first](#page-148-1) leg (AB). The height of wave (AB), therefore, is simply measured upward from the bottom of the correction at point C. - -### **THE CONTINUATION HEAD AND SHOULDERS PATTERN** - -In the previous chapter, we treated the head and shoulders pattern at some length and described it as the best known and most trustworthy of all reversal patterns. The head and shoulders pattern can sometimes appear as a continuation instead of a reversal pattern. - -In the continuation head and shoulders variety, prices trace out a pattern that looks very similar to a sideways rectangular pattern except that the middle trough in an uptrend (see Figure 6.11a) tends to be lower than either of the two shoulders. In a downtrend (see Figure 6.11b), the middle peak in the consolidation exceeds the other two peaks. The result in both cases is a head - -and shoulders pattern turned upside down. Because it is turned upside down, there is no chance of confusing it with the reversal pattern. - -**Figure 6.11a** *Example of a bullish continuation head and shoulders pattern.* - -**Figure 6.11b** *Example of a bearish continuation head and shoulders pattern.* - -**Figure 6.11c** *General Motors formed a continuation head and shoulders pattern during the first half of 1997. The pattern is very clear but shows up in an unusual place. The pattern was completed and the uptrend resumed with the close above the neckline at 60.* - -### **CONFIRMATION AND DIVERGENCE** - -The principle of *confirmation* is one of the common themes running throughout the entire subject of market analysis, and is used in conjunction with its counterpart—*divergence.* We'll introduce both concepts here and explain their meaning, but we'll return to them again and again throughout the book because their impact is so important. We're discussing confirmation here in the context of chart patterns, but it applies to virtually every aspect of technical analysis. *Confirmation* refers to the comparison of all technical signals and indicators to ensure that most of those indicators are pointing in the same direction and are confirming one another. - -*Divergence* is the opposite of confirmation and refers to a situation where different technical indicators fail to confirm one another. While it is being used here in a negative sense, divergence is a valuable concept in market analysis, and one of the best early warning signals of impending trend reversals. We'll discuss the principle of divergence at greater length in Chapter 10, "Oscillators and Contrary Opinion." - -## **CONCLUSION** - -This concludes our treatment of price patterns. We stated earlier that the three pieces of raw data used by the technical analyst were *price, volume*, and *open interest.* Most of what we've said so far has focused on price. Let's take a closer look now at volume and open interest and how they are incorporated into the analytical process. - -## **INTRODUCTION** - -Most technicians in the financial markets use a multidimensional approach to market analysis by tracking the movement of three sets of figures—*price, volume*, and *open interest.* Volume analysis applies to all markets. Open interest applies primarily to futures markets. Chapter 3 discussed the construction of the daily bar chart and showed how the three figures were plotted on that type of chart. It was stated then that even though volume and open interest figures are available for each d[elivery](#page-50-0) month in futures markets, the *total* figures are the ones generally used for forecasting purposes. Stock chartists simply plot total volume along with the accompanying price. - -Most of the discussion of charting theory to this point has concentrated mainly on price action with some mention of volume. In this chapter, we'll round out the three dimensional approach by taking a closer look at the role played by volume and open interest in the forecasting process. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/012_7 Volume and Open Interest.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/012_7 Volume and Open Interest.md deleted file mode 100644 index 2c7758bef49e9487459678532a371e8da4aa1151..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/012_7 Volume and Open Interest.md +++ /dev/null @@ -1,262 +0,0 @@ -### **VOLUME AND OPEN INTEREST AS SECONDARY INDICATORS** - -Let's begin by placing volume and open interest in their proper perspective. *Price* is by far the most important. *Volume* and *open interest* are secondary in importance and are used primarily as confirming indicators. Of those two, volume is the more important. - -**Volume** - -*Volume* is the number of entities traded during the time period under study. Because we'll be dealing primarily with daily bar charts, our main concern is with daily volume. That daily volume is plotted by a vertical bar at the bottom of the chart under the day's price action. (See Figure 7.1.) - -**Figure 7.1** *Notice that the volume bars are noticeably larger as prices are rallying (see circles). That means that volume is confirming the price rise and is bullish.* - -Volume can be plotted for *weekly* bar charts as well. In that case, total volume for the week would simply be plotted under the bar representing that week's price action. Volume is usually not used, however, on *monthly* bar charts. - -#### **Open Interest in Futures** - -The total number of outstanding or unliquidated contracts at the end of the day is *open interest.* In Figure 7.2, open interest is the solid line plotted on the chart under its corresponding price data for the day, but above the volume bars. Remember that official volume and open interest figures are reported a day late in the futures [markets](#page-156-0) and are, therefore, plotted with a one day lag. (Only estimated volume figures are available for the last trading day.) That means that each day the chartist plots the high, low, and closing price bar for the last day of trading, but plots the official volume and open interest figures for the previous day. - -Open interest represents the total number of outstanding longs or shorts - -in the market, *not the sum of both.* Open interest is the number of contracts. A contract must have both a *buyer* and a *seller.* Therefore, two market participants—a *buyer* and a *seller—*combine to create only one contract. The open interest figure reported each day is followed by either a positive or negative number showing the increase or decrease in the number of contracts for that day. It is those changes in the open interest levels, either up or down, that give the chartist clues as to the changing character of market participation and give open interest its forecasting value. - -*How Changes in Open Interest Occur.* In order to grasp the significance of how changes in the open interest numbers are interpreted, the reader must first understand how each trade produces a change in those numbers. - -Every time a trade is completed on the floor of the exchange, the open interest is affected in one of three ways—it increases, decreases, or stays unchanged. Let's see how those changes occur. - -| Buyer | Seller | Change in
Open Interest | | -|-------------------|-----------------|----------------------------|--| -| 1. Buys new long | Sells new short | Increases | | -| 2. Buys new long | Sells old long | No change | | -| 3. Buys old short | Sells new short | No change | | -| 4. Buys old short | Sells old long | Decreases | | - -In the first case, both the buyer and seller are initiating a new position and a new contract is established. In case 2, the buyer is initiating a new long position, but the seller is merely liquidating an old long. One is entering and the other exiting a trade. The result is a standoff and no change takes place in the number of contracts. In case 3, the same thing happens except this time it is the seller who is initiating a new short and the buyer who is only covering an old short. Because one of the traders is entering and the other exiting a trade, again no change is produced. In case 4, both traders are liquidating an old position and the open interest decreases accordingly. - -To sum up, if both participants in a trade are initiating a new position, the open interest will increase. If both are liquidating an old position, the open interest will decline. If, however, one is initiating a new trade while the other is liquidating an old trade, open interest will remain unchanged. By looking at the net change in the total open interest at the end of the day, the chartist is able to determine whether money is flowing into or out of the market. This information enables the analyst to draw some conclusions about the strength or weakness of the current price trend. - -#### **General Rules for Interpreting Volume and Open Interest** - -The futures technician incorporates volume and open interest information into market analysis. The rules for the interpretation of volume and open interest are generally combined because they are so similar. There are, however, some distinctions between the two that should be addressed. We'll begin here with a statement of the general rules for both. Having done that, we'll then treat each one separately before combining them again at the end. - -| separately
before | combining | them
again
at
the
end. | | -|----------------------|-----------|------------------------------------|--| -| | | | | -| | | | | -| | | | | - -If volume and open interest are both increasing, then the current price trend will probably continue in its present direction (either up or down). If, however, volume and open interest are declining, the action can be viewed as a warning that the current price trend may be nearing an end. Having said that, let's now take a look at volume and open interest separately. (See Figure 7.2.) - -**Figure 7.2** *A daily chart of crude oil futures shows volume and open interest (solid line). The open interest line is rising as prices are falling, which is bearish.* - -### **INTERPRETATION OF VOLUME FOR ALL MARKETS** - -The level of volume measures the intensity or urgency behind the price move. Heavier volume reflects a higher degree of intensity or pressure. By monitoring the level of volume along with price action, the technician is better able to gauge the buying or selling pressure behind market moves. This information can then be used to confirm price movement or warn that a price move is not to be trusted. (See Figures 7.3 and 7.4.) - -To state the rule more concisely, *volume should increase or expand in the direction of the existing price trend.* In an uptrend, volume should be heavier as the price moves higher, and [should](#page-157-1) decrease [or](#page-158-0) contract on price dips. As long as this pattern continues, volume is said to be confirming the price trend. - -**Figure 7.3** *The upside price breakout by McDonalds through the November 1997 peak was accompanied by a noticeable burst of trading activity. That's bullish.* - -**Figure 7.4** *The volume bars are following Intel's price uptrend. Volume is heavier as prices are rising, and drops off as prices weaken. Notice the burst of trading activity during the last three days' price jump.* - -The chartist is also watching for signs of *divergence* (there's that word again). Divergence occurs if the penetration of a previous high by the price trend takes place on declining volume. This action alerts the chartist to diminishing buying pressure. If the volume also shows a tendency to pick up on price dips, the analyst begins to worry that the uptrend is in trouble. - -#### **Volume as Confirmation in Price Patterns** - -During our treatment of price patterns in Chapters 5 and 6, volume was mentioned several times as an important confirming indicator. One of the first signs of a *head and shoulders* top occurred when prices moved into new highs during the formation of the *head* on light [volume](#page-105-0) with h[ea](#page-129-0)vier activity on the subsequent decline to the *neckline.* The *double* and *triple tops* saw lighter volume on each successive peak followed by heavier downside activity. Continuation patterns, like the *triangle*, should be accompanied by a gradual drop off in volume. As a rule, the resolution of all price patterns (the breakout point) should be accompanied by heavier trading activity if the signal given by that breakout is real. (See Figure 7.5.) - -In a downtrend, the volume should be heavier during down moves and lighter on bounces. As long as that pattern continues, the selling pressure is greater than buying pressure [and](#page-159-0) the downtrend should continue. It's only when that pattern begins to change that the chartist starts looking for signs of - -### a bottom. - -### **Volume Precedes Price** - -By monitoring the price and volume together, we're actually using two different tools to measure the same thing—pressure. By the mere fact that prices are trending higher, we can see that there is more buying than selling pressure. It stands to reason then that the greater volume should take place in the same direction as the prevailing trend. Technicians believe that *volume precedes price*, meaning that the loss of upside pressure in an uptrend or downside pressure in a downtrend actually shows up in the volume figures before it is manifested in a reversal of the price trend. - -**Figure 7.5** *The first half of this chart shows a positive trend with heavier volume on up days. The box at the top shows a sudden downturn on heavy volume—a negative sign. Notice the increase in trading as the continuation triangle is broken on the downside.* - -### **On Balance Volume** - -Technicians have experimented with many volume indicators to help quantify buying or selling pressure. Trying to "eyeball" the vertical volume bars along the bottom of the chart is not always precise enough to detect significant shifts in the volume flow. The simplest and best known of these volume indicators is *on balance volume* or *OBV.* Developed and popularized by Joseph Granville in his 1963 book, *Granville's New Key to Stock Market Profits*, OBV actually produces a curving line on the price chart. This line can - -be used either to confirm the quality of the current price trend or warn of an impending reversal by diverging from the price action. - -Figure 7.6 shows the price chart with the OBV line along the bottom of the chart instead of the volume bars. Notice how much easier it is to follow the volume trend with the OBV line. - -The [constr](#page-160-0)uction of the OBV line is simplicity itself. The total volume for each day is assigned a plus or minus value depending on whether prices close higher or lower for that day. A higher close causes the volume for that day to be given a plus value, while a lower close counts for negative volume. A running cumulative total is then maintained by adding or subtracting each day's volume based on the direction of the market close. - -It is the direction of the OBV line (its trend) that is important and not the actual numbers themselves. The actual OBV values will differ depending on how far back you are charting. Let the computer handle the calculations. Concentrate on the direction of the OBV line. - -The *on balance volume* line should follow in the same direction as the price trend. If prices show a series of higher peaks and troughs (an uptrend), the OBV line should do the same. If prices are trending lower, so should the OBV line. It's when the volume line fails to move in the same direction as prices that a divergence exists and warns of a possible trend reversal. - -**Figure 7.6** *The line along the bottom shows on balance volume (OBV) for the same Compaq chart. Notice how much easier it was to spot the downturn in October 1997.* - -#### **Alternatives to OBV** - -The *on balance volume* line does its job reasonably well, but it has some shortcomings. For one thing, it assigns an entire day's volume a plus or minus value. Suppose a market closes up on the day by some minimal amount such as one or two tics. Is it reasonable to assign all of that day's activity a positive value? Or consider a situation where the market spends most of the day on the upside, but then closes slightly lower. Should all of that day's volume be given a negative value? To resolve these questions, technicians have experimented with many variations of OBV in an attempt to discover the true upside and downside volume. - -One variation is to give greater weight to those days where the trend is the strongest. On an up day, for example, the volume is multiplied by the amount of the price gain. This technique still assigns positive and negative values, but gives greater weight to those days with greater price movement and reduces the impact of those days where the actual price change is minimal. - -There are more sophisticated formulas that blend volume (and open interest) with price action. James Sibbet's Demand Index, for example, combines price and volume into a leading market indicator. The Herrick Payoff Index uses open interest to measure money flow. (See Appendix A for an explanation of both indicators.) - -It should be noted that volume reporting in the stock market is much more useful than in the futures markets. Stock trading volume is [reported](#page-415-0) immediately, while it is reported a day late for futures. Levels of upside and downside volume are also available for stocks, but not in futures. The availability of volume data for stocks on each price change during the day has facilitated an even more advanced indicator called Money Flow, developed by Laszlo Birinyi, Jr. This real-time version of OBV tracks the level of volume on each price change in order to determine if money is flowing into or out of a stock. This sophisticated calculation, however, requires a lot of computer power and isn't readily available to most traders. - -These more sophisticated variations of OBV have basically the same intent—to determine whether the heavier volume is taking place on the upside (bullish) or the downside (bearish). Even with its simplicity, the OBV line still does a pretty good job of tracking the volume flow in a market—either in futures or stocks. And OBV is readily available on most charting software. Most charting packages even allow you to plot the OBV line right over the price data for even easier comparison. (See Figures 7.7 and 7.8.) - -**Figure 7.7** *An excellent example of how a bearish divergence between the on balance volume line (bottom) and the price of Intel correctly warned of a major downturn.* - -#### **Other Volume Limitations in Futures** - -We've already mentioned the problem of the one day lag in reporting futures volume. There is also the relatively awkward practice of using total volume numbers to analyze individual contracts instead of each contract's actual volume. There are good reasons for using total volume. But how does one deal with situations when some contracts close higher and others lower in the same futures market on the same day? *Limit* days produce other problems. Days when markets are locked *limit up* usually produce very light volume. This is a sign of strength as the numbers of buyers so overwhelm the sellers that prices reach the maximum trading limit and cease trading. According to the traditional rules of interpretation, light volume on a rally is bearish. The light volume on *limit* days is a violation of that principle and can distort OBV numbers. - -**Figure 7.8** *Overlaying the OBV (solid line) right over the price bars makes for easier comparison between price and volume. This chart of McDonalds shows the OBV line leading the price higher and warning in advance of the bullish breakout.* - -Even with these limitations, however, volume analysis can still be used in the futures markets, and the technical trader would be well advised to keep a watchful eye on volume indications. - -### **INTERPRETATION OF OPEN INTEREST IN FUTURES** - -The rules for interpreting open interest changes are similar to those for volume, but require additional explanation. - -1. With prices advancing in an uptrend and total open interest increasing, *new money is flowing into the market reflecting aggressive new buying, and is considered bullish.* (See Figure 7.9.) - -**Figure 7.9** *The uptrend in silver prices was confirmed by a similar rise in the open interest line. The boxes to the right show some normal liquidation of outstanding contracts as prices start to correct downward.* - -- 2. If, however, prices are rising and open interest declines, *the rally is being caused primarily by short covering* (holders of losing short positions being forced to cover those positions). *Money is leaving rather than entering the market.* This action is considered bearish because the uptrend will probably run out of steam once the necessary short covering has been completed. (See Figure 7.10.) -- 3. With prices in a downtrend and open interest rising, the technician knows that *new money is flowing into the market, reflecting aggressive new short selling.* This action increases the [odds](#page-165-0) that the downtrend will continue and is considered bearish. (See Figure 7.11.) -- 4. If, however, total open interest is declining along with declining prices, *the price decline is being caused by discouraged or losing longs being forced to liquidate their positions.* This [action](#page-166-0) is believed to indicate a strengthening technical situation because the downtrend will probably end once open interest has declined sufficiently to show that most losing longs have completed their selling. - -**Figure 7.10** *An example of a weak price rebound in gold futures. The price rise is accompanied by falling open interest, while the price decline shows rising open interest. A strong trend would see open interest trending with price, not against it.* - -Let's summarize these four points: - -- 1. *Rising open interest in an uptrend is bullish.* -- 2. *Declining open interest in an uptrend is bearish.* -- 3. *Rising open interest in a downtrend is bearish.* -- 4. *Declining open interest in a downtrend is bullish.* - -**Figure 7.11** *The downturn in copper during the summer of 1997 and the subsequent price decline was accompanied by rising open interest. Rising open interest during a price decline is bearish because it reflects aggressive short selling.* - -#### **Other Situations Where Open Interest Is Important** - -In addition to the preceding tendencies, there are other market situations where a study of open interest can prove useful. - -1. Toward the end of major market moves, where open interest has been increasing throughout the price trend, *a leveling off or decline in open interest is often an early warning of a change in trend.* - -2. *A high open interest figure at market tops can be considered bearish if the price drop is very sudden.* This means that all of the new longs established near the end of the uptrend now have losing positions. Their forced liquidation will keep prices under pressure until the open interest has declined sufficiently. As an example, let's assume that an uptrend has been in effect for some time. Over the past month, open interest has increased noticeably. Remember that every new open interest contract has one new long and one new short. Suddenly, prices begin to drop sharply and fall below the lowest price set over the past month. Every single new long established during that month now has a loss. - -The forced liquidation of those longs keeps prices under pressure until they have all been liquidated. Worse still, their forced selling often begins to feed on itself and, as prices are pushed even lower, causes additional margin selling by other longs and intensifies the new price decline. As a corollary to the preceding point, *an unusually high open interest in a bull market is a danger signal.* - -3. *If open interest builds up noticeably during a sideways consolidation or a horizontal trading range, the ensuing price move intensifies once the breakout occurs.* This only stands to reason. The market is in a period of indecision. No one is sure which direction the trend breakout will take. The increase in open interest, however, tells us that a lot of traders are taking positions in anticipation of the breakout. Once that breakout does occur, a lot of traders are going to be caught on the wrong side of the market. - -Let's assume we've had a three month trading range and that the open interest has jumped by 10,000 contracts. This means that 10,000 new long positions and 10,000 new short positions have been taken. Prices then break out on the upside and new three month highs are established. Because prices are trading at the highest point in three months, every single short position (all 10,000 of them) initiated during the previous three months now shows a loss. The scramble to cover those losing shorts naturally causes additional upside pressure on prices, producing even more panic. Prices remain strong until all or most of those 10,000 short positions have been offset by buying into the market strength. If the breakout had been to the downside, then it would have been the longs doing the scrambling. - -The early stage of any new trend immediately following a breakout is usually fueled by forced liquidation by those caught on the wrong side of the market. The more traders caught on the wrong side (manifested in the high open interest), the more severe the response to a sudden adverse market move. On a more positive note, the new trend is further aided by those on the right side of the market whose judgment has been vindicated, and who are now using accumulated paper profits to finance additional positions. It can be seen why *the greater the increase in open interest during a trading range (or any price formation for that matter), the greater the potential for the subsequent price move.* - -4. *Increasing open interest at the completion of a price pattern is viewed as added confirmation of a reliable trend signal.* The breaking of the *neckline*, for example, of a *head and shoulders* bottom is more convincing if the breakout occurs on increasing open interest along with the heavier volume. The analyst has to be careful here. Because the impetus following the initial trend signal is often caused by those on the wrong side of the market, *sometimes the open interest dips slightly at the beginning of a new trend.* This initial dip in the open interest can mislead the unwary chart reader, and argues against focusing too much attention on the open interest changes over the very short term. - -### **SUMMARY OF VOLUME AND OPEN INTEREST RULES** - -Let's summarize some of the more important elements of price, volume, and open interest. - -- 1. Volume is used in all markets; open interest mainly in futures. -- 2. Only the *total* volume and open interest are used for futures. -- 3. Increasing volume (and open interest) indicate that the current price trend will probably continue. -- 4. Declining volume (and open interest) suggest that the price trend may be changing. -- 5. Volume precedes price. Changes in buying or selling pressure are often detected in volume before price. -- 6. On balance volume (OBV), or some variation thereof, can be used to more accurately measure the direction of volume pressure. -- 7. Within an uptrend, a sudden leveling off or decline in open interest often warns of a change in trend. (This applies only to futures.) -- 8. Very high open interest at market tops is dangerous and can intensify downside pressure. (This applies only to futures.) -- 9. A buildup in open interest during consolidation periods intensifies the ensuing breakout. (This applies only to futures.) -- 10. Increases in volume (and open interest) help confirm the resolution of price patterns or any other significant chart developments that signal the beginning of a new trend. - -## **BLOWOFFS AND SELLING CLIMAXES** - -One final situation not covered so far that deserves mention is the type of dramatic market action that often takes place at tops and bottoms—*blowoffs* and *selling climaxes. Blowoffs* occur at major market tops and *selling climaxes* at bottoms. In futures, blowoffs are often accompanied by a drop in open interest during the final rally. In the case of a blowoff at market tops, prices suddenly begin to rally sharply after a long advance, accompanied by a large jump in trading activity and then peak abruptly. (See Figure 7.12.) In a selling climax bottom, prices suddenly drop sharply on heavy trading activity and rebound as quickly. (Refer back to Figure 4.22c.) - -## **COMMITMENTS OF T[RADERS](#page-100-1) REPORT** - -Our treatment of open interest would not be complete without mentioning the *Commitments of Traders (COT) Report*, and how it is used by futures technicians as a forecasting tool. The report is released by the Commodity Futures Trading Commission (CFTC) twice a month—a mid-month report and one at month's end. The report breaks down the open interest numbers into three categories—large hedgers, large speculators, and small traders. The large hedgers, also called commercials, use the futures markets primarily for hedging purposes. Large speculators include the large commodity funds, who rely primarily on mechanical trend-following systems. The final category of small traders includes the general public, who trade in much smaller amounts. - -**Figure 7.12** *A couple of blowoff tops in coffee futures. In both cases, prices rallied sharply on heavy volume. The negative warnings came from the decline in open interest (solid line) during both rallies (see arrows).* - -### **WATCH THE COMMERCIALS** - -The guiding principle in analyzing the Commitments Report is the belief that the large commercial hedgers are usually right, while the traders are usually wrong. That being the case, the idea is to place yourself in the same positions as the hedgers and in the opposite positions of the two categories of traders. For example, a bullish signal at a market bottom would occur when the commercials are heavily net long while the large and small traders are heavily net short. In a rising market, a warning signal of a possible top would take - -place when the large and small traders become heavily net long at the same time that the commercials are becoming heavily net short. - -### **NET TRADER POSITIONS** - -It is possible to chart the trends of the three market groups, and to use those trends to spot extremes in their positions. One way to do that is to study the net trader positions published in *Futures Charts* (Published by Commodity Trend Service, PO Box 32309, Palm Beach Gardens, FL 33420). That charting service plots three lines that show the net trader positions for all three groups on a weekly price chart for each market going back four years. By providing four years of data, historical comparisons are easily done. Nick Van Nice, the publisher of that chart service, looks for situations where the commercials are at one extreme, and the two categories of traders at the other, to find buying and selling opportunities (as shown in Figures 7.13 and 7.14). Even if you don't use the COT Report as a primary input in your trading decisions, it's not a bad idea to keep an eye on what those three groups are doing. - -### **OPEN INTEREST IN OPTIONS** - -Our coverage of open interest has concentrated on the *futures* markets. Open interest plays an important role in *options* trading as well. Open interest figures are published each day for *put* and *call* options on futures markets, stock averages, industry indexes, and individual stocks. While open interest in options may not be interpreted in exactly the same way as in futures, it tells us essentially the same thing—where the interest is and the liquidity. Some option traders compare *call* open interest (bulls) to *put* open interest (bears) in order to measure market sentiment. Others use option volume. - -**Figure 7.13** *This weekly chart of S&P 500 futures shows three buy signals (see arrows). The lines along the bottom show the commercials (solid line) heavily net long and the large speculators (dashed line) heavily net short at each buy signal.* - -### **PUT/CALL RATIOS** - -Volume figures for the options markets are used essentially the same way as in futures and stocks—that is, they tell us the degree of buying or selling pressure in a given market. Volume figures in options are broken down into *call* volume (bullish) and *put* volume (bearish). By monitoring the volume in calls versus puts, we are able to determine the degree of bullishness or bearishness in a market. One of the primary uses of volume data in options trading is the construction of put/call volume ratios. When options traders are bullish, call volume exceeds put volume and the put/call ratio falls. A bearish attitude is reflected in heavier put volume and a higher put/call ratio. The put/call ratio is usually viewed as a contrary indicator. A very high ratio signals an oversold market. A very low ratio is a negative warning of an overbought market. - -**Figure 7.14** *This weekly chart of copper futures shows three sell signals marked by the arrows. Each sell signal shows net long positions by the two categories of speculators and a net short position by the commercials. The commercials were right.* - -### **COMBINE OPTION SENTIMENT WITH TECHNICALS** - -Options traders use open interest and volume put/call figures to determine extremes in bullish or bearish sentiment. These sentiment readings work best when combined with technical measures such as support, resistance, and the trend of the underlying market. Since timing is so crucial in options, most option traders are technically oriented. - -### **CONCLUSION** - -That concludes our coverage of volume and open interest, at least for now. Volume analysis is used in all financial markets—futures, options, and stocks. Open interest applies only to futures and options. But, since futures and options are traded on so many stock market vehicles, some understanding of how open interest works can be useful in all three financial arenas. In most of our discussions so far, we've concentrated on daily bar charts. The next step is to broaden our time horizon and to learn how to apply the tools we've - -learned to weekly and monthly charts in order to perform long range trend analysis. We'll accomplish that in the next chapter. - -### **INTRODUCTION** - -Of all the charts utilized by the market technician for forecasting and trading the financial markets, the *daily* bar chart is by far the most popular. The daily bar chart usually covers a period of only six to nine months. However, because most traders confine their interest to relatively short term market action, daily bar charts have gained wide acceptance as the primary working tool of the chartist. - -The average trader's dependence on these daily charts, however, and the preoccupation with short term market behavior, cause many to overlook a very useful and rewarding area of price charting—*the use of weekly and monthly charts for longer range trend analysis and forecasting.* - -The daily bar chart covers a relatively short period of time in the life of any market. A thorough trend analysis of a market, however, should include some consideration of how the daily market price is moving in relation to its long range trend structure. To accomplish that task, *longer range charts must be employed.* Whereas on the daily bar chart each bar represents one day's price action, on the weekly and monthly charts each price bar represents one week's and one month's price action, respectively. *The purpose of weekly and monthly charts is to compress price action in such a way that the time horizon can be greatly expanded and much longer time periods can be studied.* - -### **THE IMPORTANCE OF LONGER RANGE PERSPECTIVE** - -*Long range price charts provide a perspective on the market trend that is impossible to achieve with the use of daily charts alone.* During our - -introduction to the technical philosophy in Chapter 1, it was pointed out that one of the greatest advantages of chart analysis is the application of its principles to virtually any time dimension, including long range forecasting. We also addressed the fallacy, espoused by [some,](#page-23-0) that technical analysis should be limited to short term "timing" with longer range forecasting left to the fundamental analyst. - -The accompanying charts will demonstrate that the principles of technical analysis—including trend analysis, support and resistance levels, trendlines, percentage retracements, and price patterns—lend themselves quite well to the analysis of long range price movements. *Anyone who is not consulting these longer range charts is missing an enormous amount of valuable price information.* - -### **CONSTRUCTION OF CONTINUATION CHARTS FOR FUTURES** - -The average futures contract has a trading life of about a year and a half before expiration. This *limited life* feature poses some obvious problems for the technician interested in constructing a long range chart going back several years. Stock market technicians don't have this problem. Charts are readily available for individual common stocks and the market averages from the inception of trading. How then does the futures technician construct longer range charts for contracts that are constantly expiring? - -The answer is the *continuation* chart. Notice the emphasis on the word "continuation." The technique most commonly employed is simply to link a number of contracts together to provide continuity. When one contract expires, another one is used. In order to accomplish this, the simplest method, and the one used by most chart services, is to *always use the price of the nearest expiring contract.* When that nearest expiring contract stops trading, the next in line becomes the nearest contract and is the one plotted. - -#### **Other Ways to Construct Continuation Charts** - -The technique of linking prices of the nearest expiring contracts is relatively simple and does solve the problem of providing price continuity. However, there are some problems with that method. Sometimes the expiring contract may be trading at a significant premium or discount to the next contract, and the changeover to the new contract may cause a sudden price drop or jump on the chart. Another potential distortion is the extreme volatility experienced by some spot contracts just before expiration. - -Futures technicians have devised many ways to deal with these - -occasional distortions. Some will stop plotting the nearest contract a month or two before it expires to avoid the volatility in the spot month. Others will avoid using the nearest contract altogether and will instead chart the second or third contract. Another method is to chart the contract with the highest open interest on the theory that that delivery month is the truest representation of market value. - -Continuation charts can also be constructed by linking specific calendar months. For example, a November soybean continuation chart would combine only the historic data provided by each successive year's November soybean contract. (This technique of linking specific delivery months was favored by W.D. Gann.) Some chartists go even further by averaging the prices of several contracts, or constructing indices that attempt to smooth the changeover by making adjustments in the price premium or discount. - -### **THE PERPETUAL CONTRACT™** - -An innovative solution to the problem of price continuity was developed by Robert Pelletier, president of Commodity Systems, Inc., a commodity and stock data service (CSI. 200 W. Palmetto Park Road, Boca Raton, FL 33422), called the *Perpetual Contract.*™ ("Perpetual Contract™" is a registered trademark of that firm.) - -The purpose of the Perpetual Contract™ is to provide years of futures price history in one continuous time series. That is accomplished by constructing a time series based on a constant forward time period. For example, the series would determine a value three months or six months into the future. The time period varies and can be chosen by the user. The Perpetual Contract™ is constructed by taking a weighted average of two futures contracts that surround the time period desired. - -The value for the Perpetual Contract™ is not an actual price, but a weighted average of two other prices. The main advantage of the Perpetual Contract™ is that it eliminates the need for using only the nearest expiring contract and smoothes out the price series by eliminating the distortions that can take place during the transition between delivery months. For chart analysis purposes, the nearest-month continuation charts published by chart services are more than adequate. A continuous price series, however, is more useful for back-testing trading systems and indicators. A more complete explanation of ways to construct continuous futures contracts is provided by Greg Morris in Appendix D. - -### **LONG TERM TRENDS DISPUTE RANDOMNESS** - -The most striking features of long range charts is that not only are trends very clearly defined, but that long range trends often last for years. Imagine making a forecast based on one of these long range trends, and not having to change that forecast for several years! - -The persistence of long range trends raises another interesting question that should be mentioned—the question of randomness. While technical analysts do not subscribe to the theory that market action is random and unpredictable, it seems safe to observe that whatever randomness does exist in price action is probably a phenomenon of the very short term. *The persistence of existing trends over long periods of time, in many cases for years, is a compelling argument against the claims of Random Walk Theorists that prices are serially independent and that past price action has no effect on future price action.* - -### **PATTERNS ON CHARTS: WEEKLY AND MONTHLY REVERSALS** - -*Price patterns* appear on the long range charts, which are interpreted in the same way as on the daily charts. *Double tops and bottoms* are very prominent on these charts, as are head and shoulder reversals. *Triangles*, which are usually continuation patterns, are frequently seen. - -Another pattern that occurs quite frequently on these charts is the *weekly and monthly reversal.* For example, on the monthly chart, a new monthly high followed by a close below the previous month's close often represents a significant turning point, especially if it occurs near a major support or resistance area. Weekly reversals are quite frequent on the weekly charts. These patterns are the equivalent of the *key reversal day* on the daily charts, except that on the long range charts these reversals carry a great deal more significance. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/013_8 Long Term Charts.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/013_8 Long Term Charts.md deleted file mode 100644 index 1eae5b951f9ccd27275e411debde9effbbf56936..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/013_8 Long Term Charts.md +++ /dev/null @@ -1,133 +0,0 @@ -### **LONG TERM TO SHORT TERM CHARTS** - -It's especially important to appreciate the order in which price charts should be studied in performing a thorough trend analysis. The proper order to follow in chart analysis is to begin with the long range and gradually work to the near term. The reason for this should become apparent as one works with the different time dimensions. If the analyst begins with only the near term - -picture, he or she is forced to constantly revise conclusions as more price data is considered. A thorough analysis of a daily chart may have to be completely redone after looking at the long range charts. By starting with the big picture, going back as far as 20 years, all data to be considered are already included in the chart and a proper perspective is achieved. Once the analyst knows where the market is from a longer range perspective, he or she gradually "zeros in" on the shorter term. - -The first chart to be considered is the 20 year monthly chart. The analyst looks for the more obvious chart patterns, major trendlines, or the proximity of major support or resistance levels. He or she then consults the most recent five years on the weekly chart, repeating the same process. Having done that, the analyst narrows his or her focus to the last six to nine months of market action on the daily bar chart, thus going from the "macro" to the "micro" approach. If the trader wants to proceed further, intraday charts can then be consulted for an even more microscopic study of recent action. - -### **WHY SHOULD LONG RANGE CHARTS BE ADJUSTED FOR INFLATION?** - -A question often raised concerning long term charts is whether or not historic price levels seen on the charts should be adjusted for inflation. After all, the argument goes, do these long range peaks and troughs have any validity if not adjusted to reflect the changes in the value of the U.S. dollar? This is a point of some controversy among analysts. - -I do not believe that any adjustment is necessary on these long range charts for a number of reasons. The main reason is my belief that the markets themselves have already made the necessary adjustments. A currency declining in value causes commodities quoted in that currency to increase in value. The declining value of the dollar, therefore, would contribute to rising commodity prices. A rising dollar would cause the price of most commodities to fall. - -The tremendous price gains in commodity markets during the 1970s and declining prices in the 1980s and 1990s are classic examples of inflation at work. To have suggested during the 1970s that commodity price levels that had doubled and tripled in price should then be adjusted to reflect rising inflation would make no sense at all. The rising commodity markets already were a manifestation of that inflation. Declining commodity markets since the 1980s reflect a long period of disinflation. Should we take the price of gold, which is now worth less than half of its value in 1980, and adjust it to reflect the lower inflation rate? The market has already taken care of that. - -The final point in this debate goes to the heart of the technical theory, which states that price action discounts everything, even inflation. All financial markets adjust to periods of inflation and deflation and to changes in currency values. The real answer to whether long range charts should be adjusted for inflation lies in the charts themselves. Many markets fail at historic resistance levels set several years earlier and then bounce off support levels not seen in several years. It's also clear that falling inflation since the early 1980s has helped support bull markets in bonds and stocks. It would seem that those markets have already made their own inflation adjustment. (See Figure 8.1.) - -**Figure 8.1** *The gold price peak in 1980 ushered in two decades of low inflation. Low inflation normally causes falling gold prices and rising stock prices as this chart shows. Why adjust the charts again for inflation? It's already been done.* - -### **LONG TERM CHARTS NOT INTENDED FOR TRADING PURPOSES** - -Long term charts are not meant for trading purposes. A distinction has to be made between market *analysis* for forecasting purposes and the *timing* of market commitments. Long term charts are useful in the analytical process to help determine the major trend and price objectives. They are not suitable, however, for the timing of entry and exit points and should not be used for - -that purpose. For that more sensitive task, daily and intraday charts should be utilized. - -## **EXAMPLES OF LONG TERM CHARTS** - -The following pages contain examples of long term weekly and monthly charts (Figures 8.2–8.12). The drawings on the charts are limited to long term support and resistance levels, trendlines, percentage retracements, weekly reversals, and an occasional price pattern. Be aware, however, that anything that can be [done](#page-180-1) on a [dai](#page-185-0)ly chart can also be done on a weekly or monthly. We'll show you later in the book how the application of various technical indicators to these long term charts is accomplished, and how signals on weekly charts become valuable filters for shorter term timing decisions. Remember also that *semilog* chart scaling becomes more valuable when studying long range price trends. - -**Figure 8.2** *This chart of semiconductor stocks shows the valuable perspective of a weekly chart. The late 1997 price fall stopped right at the 62% retracement level and bounced off chart support formed the previous spring (see circle).* - -**Figure 8.3** *The early 1998 bottom in General Motors began right at the trendline drawn along the 1995-1996 lows. That's why it's a good idea to track weekly charts.* - -**Figure 8.4** *This monthly chart shows the 1997 rally in Burlington Resources stopping right at the same level that stopped the 1989 and 1993 rallies. The 1995 bottom was at the same level as the 1991 bottom. Who says charts don't have a memory?* - -**Figure 8.5** *An investor in Inco Ltd. during the 1997 rally could have benefited from the knowledge that the 1989, 1991, and 1995 tops occurred right at 38.* - -**Figure 8.6** *Do long term charts matter? The 1993 bottom in IBM was at the same level as the bottom formed 20 years earlier in 1974. The break of an 8 year down trendline (see box) in 1995 confirmed the new major uptrend.* - -**Figure 8.7** *Helmerich & Payne finally broke out above 19 in 1996 after failing in 1987, 1990, and 1993. The late 1996 pullback at 28 occurred near the 1980 peak.* - -**Figure 8.8** *This monthly chart of Dow Jones shows a head and shoulders bottom forming for 10 years from 1988 to 1997. The right shoulder also has the shape of a bullish ascending triangle. The breakout over the neckline at 42 completed the bottom.* - -**Figure 8.9** *A bullish symmetrical triangle was easy to spot on the monthly chart of Southwest Airlines. But you probably wouldn't have spotted it on a daily chart.* - -**Figure 8.10** *The 1994 bottom in the Dow Utilities bounced off a trendline lasting 20 years. There are those who claim that past price action has no bearing on the future. If you still believe that, go back and look at these long term charts again.* - -**Figure 8.11** *On this linear-scaled chart of the Japanese stock market, the long term up trendline (line 1) drawn under the 1982 and 1984 lows was broken in early 1992 (see circle) near 22,000. That was two years after the actual peak.* - -**Figure 8.12** *The same Japanese chart from Figure 8.11 using log scaling. Line 1 is the trendline from the previous figure. The steeper line 2 was broken in mid-1990 (see box) at 30,000. Up trendlines on log charts are broken sooner than linear up t[rendlines.](#page-185-1)* - -### **INTRODUCTION** - -The *moving average* is one of the most versatile and widely used of all technical indicators. Because of the way it is constructed and the fact that it can be so easily quantified and tested, it is the basis for many mechanical trend-following systems in use today. - -Chart analysis is largely subjective and difficult to test. As a result, chart analysis does not lend itself that well to computerization. Moving average rules, by contrast, can easily be programmed into a computer, which then generates specific buy and sell signals. While two technicians may disagree as to whether a given price pattern is a *triangle* or a *wedge*, or whether the volume pattern favors the bull or bear side, moving average trend signals are precise and not open to debate. - -Let's begin by defining what a *moving average* is. As the second word implies, it is an *average* of a certain body of data. For example, if a 10 day average of closing prices is desired, the prices for the last 10 days are added up and the total is divided by 10. The term *moving* is used because only the latest 10 days' prices are used in the calculation. Therefore, the body of data to be averaged (the last 10 closing prices) moves forward with each new trading day. The most common way to calculate the moving average is to work from the total of the last 10 days' closing prices. Each day the new close is added to the total and the close 11 days back is subtracted. The new total is then divided by the number of days (10). (See Figure 9.1a.) - -The above example deals with a simple 10 day moving average of closing prices. There are, however, other types of moving averages that are not simple. There are also many questions as to the [best](#page-187-1) way to employ the moving average. For example, how many days should be averaged? Should a short term or a long term average be used? Is there a *best* moving average for - -all markets or for each individual market? Is the closing price the best price to average? Would it be better to use more than one average? - -**Figure 9.1a** *A 10 day moving average applied to a daily bar chart of the S&P 500. Prices crossed the average line several times (see arrows) before finally turning higher. Prices stayed above the average during the subsequent rally.* - -Which type of average works better—a simple, linearly weighted or exponentially smoothed? Are there times when moving averages work better than others? - -There are many questions to be considered when using moving averages. We'll address many of these questions in this chapter and show examples of some of the more common usages of the moving average. - -### **THE MOVING AVERAGE: A SMOOTHING DEVICE WITH A TIME LAG** - -The *moving average* is essentially a trend following device. Its purpose is to identify or signal that a new trend has begun or that an old trend has ended or reversed. Its purpose is to track the progress of the trend. It might be viewed as a curving trendline. It does not, however, predict market action in the same sense that standard chart analysis attempts to do. The moving average is a follower, not a leader. It never anticipates; it only reacts. The moving average follows a market and tells us that a trend has begun, but only after the fact. - -The moving average is a smoothing device. By averaging the price data, a smoother line is produced, making it much easier to view the underlying trend. By its very nature, however, the moving average line also lags the market action. A shorter moving average, such as a 20 day average, would hug the price action more closely than a 200 day average. The time lag is reduced with the shorter averages, but can never be completely eliminated. Shorter term averages are more sensitive to the price action, whereas longer range averages are less sensitive. In certain types of markets, it is more advantageous to use a shorter average and, at other times, a longer and less sensitive average proves more useful. (See Figure 9.1b.) - -#### **Which Prices to Average** - -We have been using the closing price in all of our [exam](#page-188-0)ples so far. However, while the closing price is considered to be the most important price of the trading day and the price most commonly used in moving average construction, the reader should be aware that some technicians prefer to use other prices. Some prefer to use a *midpoint* value, which is arrived at by dividing the day's range by two. - -**Figure 9.1b** *A comparison of a 20 day and a 200 day moving average. During the sideways period from August to January, prices crossed the shorter average several times. However, they remained above the 200 day average throughout the entire period.* - -Others include the closing price in their calculation by adding the high, - -low, and closing prices together and dividing the sum by three. Still others prefer to construct *price bands* by averaging the high and low prices separately. The result is two separate moving average lines that act as a sort of volatility buffer or neutral zone. Despite these variations, the closing price is still the price most commonly used for moving average analysis and is the price that we'll be focusing most of our attention on in this chapter. - -#### **The Simple Moving Average** - -The *simple moving average*, or the arithmetic mean, is the type used by most technical analysts. But there are some who question its usefulness on two points. The first criticism is that only the period covered by the average (the last 10 days, for example) is taken into account. The second criticism is that the simple moving average gives equal weight to each day's price. In a 10 day average, the last day receives the same weight as the first day in the calculation. Each day's price is assigned a 10% weighting. In a 5 day average, each day would have an equal 20% weighting. Some analysts believe that a heavier weighting should be given to the more recent price action. - -#### **The Linearly Weighted Moving Average** - -In an attempt to correct the weighting problem, some analysts employ a *linearly weighted moving average.* In this calculation, the closing price of the 10th day (in the case of a 10 day average) would be multiplied by 10, the ninth day by nine, the eighth day by eight, and so on. The greater weight is therefore given to the more recent closings. The total is then divided by the sum of the multipliers (55 in the case of the 10 day average: 10 + 9 + 8 +…+ 1). However, the linearly weighted average still does not address the problem of including only the price action covered by the length of the average itself. - -#### **The Exponentially Smoothed Moving Average** - -This type of average addresses both of the problems associated with the simple moving average. First, the exponentially smoothed average assigns a greater weight to the more recent data. Therefore, it is a weighted moving average. But while it assigns lesser importance to past price data, it does include in its calculation all of the data in the life of the instrument. In addition, the user is able to adjust the weighting to give greater or lesser weight to the most recent day's price. This is done by assigning a percentage value to the last day's price, which is added to a percentage of the previous day's value. The sum of both percentage values adds up to 100. For example, the last day's price could be assigned a value of 10% (.10), which is added to the previous day's value of 90% (.90). That gives the last day 10% of the total weighting. That would be the equivalent of a 20 day average. By giving the last day's price a smaller value of 5% (.05), lesser weight is given to the last day's data and the average is less sensitive. That would be the equivalent of a 40 day moving average. (See Figure 9.2.) - -**Figure 9.2** *The 40 day exponential moving average (dotted line) is more sensitive than the simple arithmetic 40 day moving average (solid line).* - -The computer makes this all very easy for you. You just have to choose the number of days you want in the moving average—10, 20, 40, etc. Then select the type of average you want—simple, weighted, or exponentially smoothed. You can also select as many averages as you want—one, two, or three. - -### **The Use of One Moving Average** - -The simple moving average is the one most commonly used by technicians, and is the one that we'll be concentrating on. Some traders use just one moving average to generate trend signals. The moving average is plotted on the bar chart in its appropriate trading day along with that day's price action. When the closing price moves above the moving average, a buy signal is generated. A sell signal is given when prices move below the moving average. For added confirmation, some technicians also like to see the moving average line itself turn in the direction of the price crossing. (See Figure 9.3.) - -If a very short term average is employed (a 5 or 10 day), the average - -tracks prices very closely and several crossings occur. This action can be either good or bad. The use of a very sensitive average produces more trades (with higher commission costs) and results in many false signals (whipsaws). If the average is too sensitive, some of the short term random price movement (or "noise") activates bad trend signals. - -**Figure 9.3** *Prices fell below the 50 day average during October (see left circle). The sell signal is stronger when the moving average also turns down (see left arrow). The buy signal during January was confirmed when the average itself turned higher.* - -While the shorter average generates more false signals, it has the advantage of giving trend signals earlier in the move. It stands to reason that the more sensitive the average, the earlier the signals will be. So there is a tradeoff at work here. The trick is to find the average that is sensitive enough to generate early signals, but insensitive enough to avoid most of the random "noise." (See Figure 9.4.) - -**Figure 9.4** *A shorter average gives earlier signals. The longer average is slower, but more reliable. The 10 day turned up first at the bottom. But it also gave a premature buy signal during November and an untimely sell signal during February (see boxes).* - -Let's carry the above comparison a step further. While the longer average performs better while the trend remains in motion, it "gives back" a lot more when the trend reverses. The very insensitivity of the longer average (the fact that it trailed the trend from a greater distance), which kept it from getting tangled up in short term corrections during the trend, works against the trader when the trend actually reverses. Therefore, we'll add another corollary here: The longer averages work better as long as the trend remains in force, but a shorter average is better when the trend is in the process of reversing. - -It becomes clearer, therefore, that the use of one moving average alone has several disadvantages. It is usually more advantageous to employ two moving averages. - -#### **How to Use Two Averages to Generate Signals** - -This technique is called the *double crossover method.* This means that a buy signal is produced when the shorter average crosses above the longer. For example, two popular combinations are the 5 and 20 day averages and the 10 and 50 day averages. In the former, a buy signal occurs when the 5 day average crosses above the 20, and a sell signal when the 5 day moves below the 20. In the latter example, the 10 day crossing above the 50 signals an - -uptrend, and a downtrend takes place with the 10 slipping under the 50. This technique of using two averages together lags the market a bit more than the use of a single average but produces fewer whipsaws. (See Figures 9.5 and 9.6.) - -**Figure 9.5** *The double crossover method uses two moving averages. The 5 and 20 day combination is popular with futures traders. The 5 day fell below the 20 day during October (see circle) and caught the entire downtrend in crude oil prices.* - -#### **The Use of Three Averages, or the Triple Crossover Method** - -That brings us to the *triple crossover method.* The most widely used triple crossover system is the popular *4-9-18-day moving average combination.* The 4-9-18 method is used mainly in futures trading. This concept was first mentioned by R.C. Allen in his 1972 book, *How to Build a Fortune in Commodities* and again later in a 1974 work by the same author, *How to Use the 4-Day, 9-Day and 18-Day Moving Averages to Earn Larger Profits from Commodities.* The 4-9-18-day system is a variation on the 5, 10, and 20 day moving average numbers, which are widely used in commodity circles. Many commercial chart services publish the 4-9-18-day moving averages. (Many charting software packages use the 4-9-18-day combination as their default values when plotting three averages.) - -**Figure 9.6** *Stock traders use 10 and 50 day moving averages. The 10 day fell below the 50 day in October (left circle), giving a timely sell signal. The bullish crossover in the other direction took place during January (lower circle).* diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/014_9 Moving Averages.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/014_9 Moving Averages.md deleted file mode 100644 index 4de2be53768d6e55a4098b2dad7ede6170f9302c..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/014_9 Moving Averages.md +++ /dev/null @@ -1,172 +0,0 @@ -#### **How to Use the 4-9-18-Day Moving Average System** - -It's already been explained that the shorter the moving average, the closer it follows the price trend. It stands to reason then that the shortest of the three averages—the 4 day—will follow the trend most closely, followed by the 9 day and then the 18. In an uptrend, therefore, the proper alignment would be for the 4 day average to be above the 9 day, which is above the 18 day average. In a downtrend, the order is reversed and the alignment is exactly the opposite. That is, the 4 day would be the lowest, followed by the 9 day and then the 18 day average. (See Figures 9.7a-b.) - -A buying alert takes place in a downtrend when the 4 day crosses above both the 9 and the 18. A confirmed buy signal occurs when the 9 day then crosses above the 18. This pla[ces](#page-195-0) the 4 day [o](#page-196-1)ver the 9 day which is over the 18 day. Some intermingling may occur during corrections or consolidations, but the general uptrend remains intact. Some traders may take profits during the intermingling process and some may use it as a buying opportunity. There is obviously a lot of room for flexibility here in applying the rules, depending on how aggressively one wants to trade. - -**Figure 9.7a** *Futures traders like the 9 and 18 day moving average combination. A sell signal was given in late October (first circle) when the 9 day fell below the 18. A buy signal was given in early 1998 when the 9 day crossed back above the 18 day.* - -When the uptrend reverses to the downside, the first thing that should take place is that the shortest (and most sensitive) average—the 4 day—dips below the 9 day and the 18 day. This is only a selling alert. Some traders, however, might use that initial crossing as reason enough to begin liquidating long positions. Then, if the next longer average—the 9 day—drops below the 18 day, a confirmed sell short signal is given. - -**Figure 9.7b** *The 4-9-18 day moving average combo is also popular with futures traders. At a bottom, the 4 day (solid line) turns up first and crosses the other two lines. Then the 9 day crosses over the 18 day (see circle), signaling a bottom.* - -### **MOVING AVERAGE ENVELOPES** - -The usefulness of a single moving average can be enhanced by surrounding it with envelopes. *Percentage envelopes* can be used to help determine when a market has gotten overextended in either direction. In other words, they tell us when prices have strayed too far from their moving average line. In order to do this, the envelopes are placed at fixed percentages above and below the average. Shorter term traders, for example, often use 3% envelopes around a simple 21 day moving average. When prices reach one of the envelopes (3% from the average), the short term trend is considered to be overextended. For long range analysis, some possible combinations include 5% envelopes around a 10 week average or a 10% envelope around a 40 week average. (See Figures 9.8a-b.) - -**Figure 9.8a** *3% envelopes placed around a 21 day moving average of the Dow. Moves outside the envelopes suggest an overextended stock market.* - -**Figure 9.8b** *For longer range analysis, 5% envelopes can be placed around a 10 week average. Moves outside the envelopes helped identify market extremes.* - -### **BOLLINGER BANDS** - -This technique was developed by John Bollinger. Two trading bands are - -placed around a moving average similar to the envelope technique. Except that Bollinger Bands are placed two standard deviations above and below the moving average, which is usually 20 days. *Standard deviation* is a statistical concept that describes how prices are dispersed around an average value. Using two standard deviations ensures that 95% of the price data will fall between the two trading bands. As a rule, prices are considered to be overextended on the upside (overbought) when they touch the upper band. They are considered overextended on the downside (oversold) when they touch the lower band. (See Figures 9.9a-b.) - -**Figure 9.9a** *Bollinger bands plotted around a 20 day moving average. During the sideways period from August to January, prices kept touching the outer bands. Once the uptrend resumed, prices traded between the upper band and 20 day average.* - -**Figure 9.9b** *Bollinger bands work on weekly charts as well, by using a 20 week average as the middle line. Each touch of the lower band (see circles) signaled an important market bottom and a buying opportunity.* - -### **USING BOLLINGER BANDS AS TARGETS** - -The simplest way to use Bollinger Bands is to use the upper and lower bands as price targets. In other words, if prices bounce off the lower band and cross above the 20 day average, the upper band becomes the upper price target. A crossing below the 20 day average would identify the lower band as the downside target. In a strong uptrend, prices will usually fluctuate between the upper band and the 20 day average. In that case, a crossing below the 20 day average warns of a trend reversal to the downside. - -### **BAND WIDTH MEASURES VOLATILITY** - -Bollinger Bands differ from envelopes in one major way. Whereas the envelopes stay a *constant* percentage width apart, Bollinger Bands *expand* and *contract* based on the last 20 days' volatility. During a period of rising price volatility, the distance between the two bands will widen. Conversely, during a period of low market volatility, the distance between the two bands will contract. There is a tendency for the bands to alternate between expansion and contraction. When the bands are unusually far apart, that is - -often a sign that the current trend may be ending. When the distance between the two bands has narrowed too far, that is often a sign that a market may be about to initiate a new trend. Bollinger Bands can also be applied to weekly and monthly price charts by using 20 *weeks* and 20 *months* instead of 20 *days.* Bollinger Bands work best when combined with overbought/oversold oscillators that are explained in the next chapter. (See Appendix A for additional band techniques.) - -#### **Centering the Average** - -The more statistically correct way to plot a moving average is to *center* it. That means to place it in the middle of the time period it covers. A 10 day average, for example, would be placed five days back. A 20 day average would be plotted 10 days back in time. *Centering* the average, however, has the major flaw of producing much later trend change signals. Therefore, moving averages are usually placed at the end of the time period covered instead of the middle. The centering technique is used almost exclusively by cyclic analysts to isolate underlying market cycles. - -### **MOVING AVERAGES TIED TO CYCLES** - -Many market analysts believe that *time cycles* play an important role in market movement. Because these time cycles are repetitive and can be measured, it is possible to determine the approximate times when market tops or bottoms will occur. Many different time cycles exist simultaneously, from a short term 5 day cycle to Kondratieff's long 54 year cycle. We'll delve more into this fascinating branch of technical analysis in Chapter 14. - -The subject of cycles is introduced here only to make the point that there seems to be a relationship between the underlying cycles that affect a certain market and the correct moving averages to use. In other [words](#page-317-0), the moving averages can be adjusted to fit the dominant cycles in each market. - -There appears to be a definite relationship between moving averages and cycles. For example, the *monthly cycle* is one of the best known cycles operating throughout the commodity markets. A month has 20-21 trading days. Cycles tend to be related to their next longer and shorter cycles *harmonically*, or by a factor of two. That means that the next longer cycle is double the length of a cycle and the next shorter cycle is half its length. - -The monthly cycle, therefore, may explain the popularity of the 5, 10, 20, and 40 day moving averages. The 20 day cycle measures the monthly cycle. The 40 day average is double the 20 day. The 10 day average is half of 20 and the 5 day average is half again of 10. - -Many of the more commonly used moving averages (including the 4, 9, and 18 day averages, which are derivatives of 5, 10, and 20) can be explained by cyclic influences and the harmonic relationships of neighboring cycles. Incidentally, the 4 week cycle may also help explain the success of the *4 week rule*, covered later in the chapter, and its shorter counterpart—the *2 week rule.* - -### **FIBONACCI NUMBERS USED AS MOVING AVERAGES** - -We'll cover the Fibonacci number series in the chapter on Elliott Wave Theory. However, I'd like to mention here that this mysterious series of numbers—such as 13, 21, 34, 55, and so on—seem to lend themselves quite well to moving average analysis. This is true not only of daily charts, but for weekly charts as well. The *21 day moving average* is a Fibonacci number. On the weekly charts, the 13 week average has proven valuable in both stocks and commodities. We'll postpone a more in depth discussion of these numbers until Chapter 13. - -### **MOVING [AVERA](#page-293-0)GES APPLIED TO LONG TERM CHARTS** - -The reader should not overlook using this technique in longer range trend analysis. Longer range moving averages, such as 10 or 13 weeks, in conjunction with the 30 or 40 week average, have long been used in stock market analysis, but haven't been given as much attention in the futures markets. The 10 and 40 week moving averages can be used to help track the primary trend on weekly charts for futures and stocks. (See Figure 9.10.) - -**Figure 9.10** *Moving averages are valuable on weekly charts. The 40 week moving average should provide support during bull market corrections as it did here.* - -#### **Some Pros and Cons of the Moving Average** - -One of the great advantages of using moving averages, and one of the reasons they are so popular as trend-following systems, is that they embody some of the oldest maxims of successful trading. They trade in the direction of the trend. They let profits run and cut losses short. The moving average system forces the user to obey those rules by providing specific buy and sell signals based on those principles. - -Because they are trend-following in nature, however, moving averages work best when markets are in a trending period. They perform very poorly when markets get choppy and trade sideways for a period of time. And that might be a third to a half of the time. - -The fact that they do not work that well for significant periods of time, however, is one very compelling reason why it is dangerous to rely too heavily on the moving average technique. In certain trending markets, the moving average can't be beat. Just switch the program to automatic. At other times, a nontrending method like the overbought–oversold oscillator is more appropriate. (In Chapter 15, we'll show you an indicator called ADX that tells you when a market is trending and when it is not, and whether the market climate favors a trending moving average technique or a nontrending oscillator appro[ach.\)](#page-346-0) - -#### **Moving Averages As Oscillators** - -One way to construct an oscillator is to compare the difference between two moving averages. The use of two moving averages in the double crossover method, therefore, takes on greater significance and becomes an even more useful technique. We'll see how this is done in Chapter 10. One method compares two exponentially smoothed averages. That method is called Moving Average Convergence/Divergence (MACD). It is used partially as an oscillator. Therefore, we'll postpone our expla[nation](#page-211-0) of that technique until we deal with the entire subject of oscillators in Chapter 10. - -#### **The Moving Average Applied to Other Technical Data** - -The moving average can be applied to virtually any [technic](#page-211-0)al data or indicator. It can be used on open interest and volume figures, including on balance volume. The moving average can be used on various indicators and ratios. It can be applied to oscillators as well. - -### **THE WEEKLY RULE** - -There are other alternatives to the moving average as a trend-following device. One of the best known and most successful of these techniques is called the *weekly price channel* or, simply, *the weekly rule.* This technique has many of the benefits of the moving average, but is less time consuming and simpler to use. - -With the improvements in computer technology over the past decade, a considerable amount of research has been done on the development of technical trading systems. These systems are mechanical in nature, meaning that human emotion and judgment are eliminated. These systems have become increasingly sophisticated. At first, simple moving averages were utilized. Then, double and triple crossovers of the averages were added. The averages were then linearly weighted and exponentially smoothed. These systems are primarily trend-following, which means their purpose is to identify and then trade in the direction of an existing trend. - -With the increased fascination with fancier and more complex systems and indicators, however, there has been a tendency to overlook some of the simpler techniques that continue to work quite well and have stood the test of time. We're going to discuss one of the simplest of these techniques—the weekly rule. - -In 1970, a booklet entitled the *Trader's Notebook* was published by Dunn & Hargitt's Financial Services in Lafayette, Indiana. The best known commodity trading systems of the day were computer-tested and compared. - -The final conclusion of all that research was that the most successful of all the systems tested was the *4 week rule*, developed by Richard Donchian. Mr. Donchian has been recognized as a pioneer in the field of commodity trend trading using mechanical systems. (In 1983, *Managed Account Reports* chose Donchian as the first recipient of the Most Valuable Performer Award for outstanding contributions to the field of futures money management, and presents The Donchian Award to other worthy recipients.) - -More recent work done by Louis Lukac, former research director at Dunn & Hargitt and currently president of Wizard Trading (a Massachusetts CTA) supports the earlier conclusions that breakout (or channel) systems similar to the weekly rule continue to show superior results. (Lukac et al.)\* - -Of the 12 systems tested from 1975-84, only 4 generated significant profits. Of those 4, 2 were channel breakout systems and one was a dual moving average crossover system. A later article by Lukac and Brorsen in *[T](#page-210-0)he Financial Review* (November 1990) published the results of a more extensive study done on data from 1976–86 that compared 23 technical trading systems. Once again, channel breakouts and moving average systems came out on top. Lukac finally concluded that a channel breakout system was his personal choice as the best starting point for all technical trading system testing and development. - -#### **The 4 Week Rule** - -The 4 week rule is used primarily for futures trading. The system based on the 4 week rule is simplicity itself: - -- 1. Cover short positions and buy long whenever the price exceeds the highs of the four preceding full calendar weeks. -- 2. Liquidate long positions and sell short whenever the price falls below the lows of the four preceding full calendar weeks. - -The system, as it is presented here, is continuous in nature, which means that the trader always has a position, either long or short. As a general rule, continuous systems have a basic weakness. They stay in the market and get "whipsawed" during trendless market periods. It's already been stressed that trend-following systems do not work well when markets are in these sideways, or trendless phases. - -The 4 week rule can be modified to make it noncontinuous. This can be accomplished by using a shorter time span—such as a one or two week rule for liquidation purposes. In other words, a four week "breakout" would be necessary to initiate a new position, but a one or two week signal in the opposite direction would warrant liquidation of the position. The trader would then remain out of the market until a new four week breakout is registered. - -The logic behind the system is based on sound technical principles. Its signals are mechanical and clearcut. Because it is trend following, it virtually guarantees participation on the right side of every important trend. It is also structured to follow the often quoted maxim of successful trading—"let profits run, while cutting losses short." Another feature, which should not be overlooked, is that this method tends to trade less frequently, so that commissions are lower. Another plus is that the system can be implemented with or without the aid of a computer. - -The main criticism of the weekly rule is the same one leveled against all trend-following approaches, namely, that it does not catch tops or bottoms. But what trend-following system does? The important point to keep in mind is that the four week rule performs at least as well as most other trendfollowing systems and better than many, but has the added benefit of incredible simplicity. - -#### **Adjustments to the 4 Week Rule** - -Although we're treating the four week rule in its original form, there are many adjustments and refinements that can be employed. For one thing, the rule does not have to be used as a trading system. Weekly signals can be employed simply as another technical indicator to identify breakouts and trend reversals. Weekly breakouts can be used as a confirming filter for other techniques, such as moving average crossovers. One or 2 week rules function as excellent filters. A moving average crossover signal could be confirmed by a two week breakout in the same direction in order for a market position to be taken. - -#### **Shorten or Lengthen Time Periods for Sensitivity** - -The time period employed can be expanded or compressed in the interests of risk management and sensitivity. For example, the time period could be shortened if it is desirable to make the system more sensitive. In a relatively high priced market, where prices are trending sharply higher, a shorter time span could be chosen to make the system more sensitive. Suppose, for example, that a long position is taken on a 4 week upside breakout with a protective stop placed just below the low of the past 2 weeks. If the market has rallied sharply and the trader wishes to trail the position with a closer protective stop, a one week stopout point could be used. - -In a trading range situation, where a trend trader would just as soon stay on the sidelines until an important trend signal is given, the time period could be expanded to eight weeks. This would prevent taking positions on shorter term and premature trend signals. - -#### **The 4 Week Rule Tied to Cycles** - -Earlier in the chapter reference was made to the importance of the monthly cycle in commodity markets. The 4 week, or 20 day, trading cycle is a dominant cycle that influences all markets. This may help explain why the 4 week time period has proven so successful. Notice that mention was made of 1, 2, and 8 week rules. The principle of *harmonics* in cyclic analysis holds that each cycle is related to its neighboring cycles (next longer and next shorter cycles) by 2. - -In the previous discussion of moving averages, it was pointed out how the monthly cycle and harmonics explained the popularity of the 5, 10, 20, and 40 day moving averages. The same time periods hold true in the realm of weekly rules. Those daily numbers translated into weekly time periods are 1, 2, 4, and 8 weeks. Therefore, adjustments to the 4 week rule seem to work best when the beginning number (4) is divided or multiplied by 2. To shorten the time span, go from 4 to 2 weeks. If an even shorter time span is desired, go from 2 to 1. To lengthen, go from 4 to 8. Because this method combines price and time, there's no reason why the cyclic principle of harmonics should not play an important role. The tactic of dividing a weekly parameter by 2 to shorten it, or doubling it to lengthen it, does have cycle logic behind it. - -The 4 week rule is a simple breakout system. The original system can be modified by using a shorter time period—a 1 or 2 week rule—for liquidation purposes. If the user desires a more sensitive system, a 2 week period can be employed for entry signals. Because this rule is meant to be simple, it is best addressed on that level. The 4 week rule is simple, but it works. (Charting packages allow you to plot *price channels* above and below current prices to spot channel breakouts. Price channels can be used on daily, weekly, or monthly charts. See Figures 9.11 and 9.12.) - -**Figure 9.11** *A 20 day (4 week) price channel applied to Treasury Bond futures prices. A buy signal was given when prices closed above the upper channel (see circle). Prices have to close beneath the lower channel to reverse the signal.* - -**Figure 9.12** *A 4 month price channel applied to the S&P 500 Index. Prices crossed the upper channel in early 1995 (see circle) to give a buy signal which remains in effect 3 years later. A close beneath the lower line is needed to give a sell signal.* - -## **TO OPTIMIZE OR NOT** - -The first edition of this book included the results of extensive research produced by Merrill Lynch, which published a series of studies on computerized trading techniques applied to the futures markets from 1978-82. Extensive testing of various moving average and channel breakout parameters was performed to find the best possible combinations in each futures market. The Merrill Lynch researchers produced a different set of optimized indicator values for each market. - -Most charting packages allow you to optimize systems and indicators. Instead of using the same moving average in all markets, for example, you could ask the computer to find the moving average, or moving average combinations, that have worked the best in the past for that market. That could also be done for daily and weekly breakout systems and virtually all technical indicators included in this book. Optimization allows technical parameters to adapt to changing market conditions. - -Some argue that optimization helps their trading results and others that it doesn't. The heart of the debate centers on how the data is optimized. Researchers stress that the correct procedure is to use only part of the price data to choose the best parameters, and another portion to actually test the results. Testing the optimized parameters on "out of sample" price data helps ensure that the final results will be closer to what one might experience from actual trading. - -The decision to optimize or not is a personal one. Most evidence, however, suggests that optimization is not the Holy Grail some think it to be. I generally advise traders following only a handful of markets to experiment with optimization. Why should Treasury Bonds or the German mark have the exact same moving averages as corn or cotton? Stock market traders are a different story. Having to follow thousands of stocks argues against optimizing. If you specialize in a handful of markets, try optimizing. If you're a generalist who follows a large number of markets, use the same technical parameters for all of them. - -### **SUMMARY** - -We've presented a lot of variations on the moving average approach. Let's try to simplify things a bit. Most technicians use a combination of two moving averages. Those two averages are usually simple averages. Although exponential averages have become popular, there's no real evidence to prove that they work any better than the simple average. The most commonly used daily moving average combinations in futures markets are 4 and 9, 9 and 18, - -5 and 20, and 10 and 40. Stock traders rely heavily on a 50 day (or 10 week) moving average. For longer range stock market analysis, popular weekly moving averages are 30 and 40 weeks (or 200 days). Bollinger Bands make use of 20 day and 20 week moving averages. The 20 week average can be converted to daily charts by utilizing a 100 day average, which is another useful moving average. Channel breakout systems work extremely well in trending markets and can be used on daily, weekly, and monthly charts. - -### **THE ADAPTIVE MOVING AVERAGE** - -One of the problems encountered with the moving average is choosing between a fast or a slow average. While one may work better in a trading range market, the other may be preferable in a trending market. The answer to the problem of choosing between the two may lie with an innovative approach called the "adaptive moving average." - -Perry Kaufman presents this technique in his book *Smarter Trading.* The speed of Kaufman's "adaptive moving average" automatically adjusts to the level of noise (or volatility) in a market. The AMA moves more slowly when markets are trending sideways, but then moves more swiftly when the market is trending. That avoids the problem of using a faster moving average (and getting whipsawed more frequently) during a trading range, and using a slower average that trails too far behind a market when it is trending. - -Kaufman does that by constructing an Efficiency Ratio that compares price direction with the level of volatility. When the Efficiency Ratio is high, there is more direction than volatility (favoring a faster average). When the ratio is low, there's more volatility than direction (favoring a slower average). By incorporating the Efficiency Ratio, the AMA automatically adjusts to the speed most suitable for the current market. - -### **ALTERNATIVES TO THE MOVING AVERAGE** - -Moving averages don't work all of the time. They do their best work when the market is in a trending phase. They're not very helpful during trendless periods when prices trade sideways. Fortunately, there's another class of indicator that performs much better than the moving average during those frustrating trading ranges. They're called *oscillators* and we'll explain them in the next chapter. - -\*See Bibliography - -### **INTRODUCTION** - -In this chapter, we're going to talk about an alternative to trend-following approaches—the *oscillator.* The oscillator is extremely useful in nontrending markets where prices fluctuate in a horizontal price band, or trading range, creating a market situation where most trend-following systems simply don't work that well. The oscillator provides the technical trader with a tool that can enable him or her to profit from these periodic sideways and trendless market environments. - -The value of the oscillator is not limited to horizontal trading ranges, however. Used in conjunction with price charts during trending phases, the oscillator becomes an extremely valuable ally by alerting the trader to short term market extremes, commonly referred to as *overbought* or *oversold* conditions. The oscillator can also warn that a trend is losing momentum before that situation becomes evident in the price action itself. Oscillators can signal that a trend may be nearing completion by displaying certain divergences. - -We'll begin by explaining first what an oscillator is and the basis for its construction and interpretation. We'll then discuss the meaning of momentum and its implications for market forecasting. Some of the more common oscillator techniques will be presented from the very simple to the more complicated. The important question of divergence will be covered. We'll touch on the value of coordinating oscillator analysis with underlying market cycles. Finally, we'll discuss how oscillators should be used as part of the overall technical analysis of a market. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/015_10 Oscillators and Contrary Opinion.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/015_10 Oscillators and Contrary Opinion.md deleted file mode 100644 index 76f61de90aa7d0d20c90173687af97c153023484..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/015_10 Oscillators and Contrary Opinion.md +++ /dev/null @@ -1,327 +0,0 @@ -## **OSCILLATOR USAGE IN CONJUNCTION WITH TREND** - -The oscillator is only a secondary indicator in the sense that it must be subordinated to basic trend analysis. As we go through the various types of oscillators used by technicians, the importance of trading in the direction of the overriding market trend will be constantly stressed. The reader should also be aware that there are times when oscillators are more useful than at others. For example, near the beginning of important moves, oscillator analysis isn't that helpful and can even be misleading. Toward the end of market moves, however, oscillators become extremely valuable. We'll address these points as we go along. Finally, no study of market extremes would be complete without a discussion of Contrary Opinion. We'll talk about the role of the contrarian philosophy and how it can be incorporated into market analysis and trading. - -### **Interpretation of Oscillators** - -While there are many different ways to construct momentum oscillators, the actual interpretation differs very little from one technique to another. Most oscillators look very much alike. They are plotted along the bottom of the price chart and resemble a flat horizontal band. The oscillator band is basically flat while prices may be trading up, down, or sideways. However, the peaks and troughs in the oscillator coincide with the peaks and troughs on the price chart. Some oscillators have a midpoint value that divides the horizontal range into two halves, an upper and a lower. Depending on the formula used, this midpoint line is usually a *zero line.* Some oscillators also have upper and lower boundaries ranging from 0 to 100. - -#### **General Rules for Interpretation** - -As a general rule, when the oscillator reaches an extreme value in either the upper or lower end of the band, this suggests that the current price move may have gone too far too fast and is due for a correction or consolidation of some type. As another general rule, the trader should be buying when the oscillator line is in the lower end of the band and selling in the upper end. The crossing of the midpoint line is often used to generate buy and sell signals. We'll see how these general rules are applied as we deal with the various types of oscillators. - -#### **The Three Most Important Uses for the Oscillator** - -There are three situations when the oscillator is most useful. You'll see that these three situations are common to most types of oscillators that are used. - -- 1. The oscillator is most useful when its value reaches an extreme reading near the upper or lower end of its boundaries. The market is said to be *overbought* when it is near the upper extreme and *oversold* when it is near the lower extreme. This warns that the price trend is overextended and vulnerable. -- 2. A divergence between the oscillator and the price action when the oscillator is in an extreme position is usually an important warning. -- 3. The crossing of the zero (or midpoint) line can give important trading signals in the direction of the price trend. - -**Figure 10.1a** *The 10 day momentum line fluctuates around a zero line. Readings too far above the zero line are overbought, while values too far below the line are oversold. Momentum should be used in conjunction with the trend of the market.* - -### **MEASURING MOMENTUM** - -The concept of *momentum* is the most basic application of oscillator analysis. Momentum measures the velocity of price changes as opposed to the actual price levels themselves. Market momentum is measured by continually taking price differences for a fixed time interval. To construct a 10 day momentum line, simply subtract the closing price 10 days ago from the last closing price. This positive or negative value is then plotted around a zero line. The formula for momentum is: - -#### M=V – V x - -where V is the latest closing price and V x is the closing price x days ago. - -**Figure 10.1b** *A comparison of 10 and 40 day momentum lines. The longer version is more helpful in catching major market turns (see circles).* - -If the latest closing price is greater than that of 10 days ago (in other words, prices have moved higher), then a positive value would be plotted above the zero line. If the latest close is below the close 10 days earlier (prices have declined), then a negative value is plotted below the zero line. - -While the 10 day momentum is a commonly used time period for reasons discussed later, any time period can be employed. (See Figure 10.1a.) A shorter time period (such as 5 days) produces a more sensitive line with more pronounced oscillations. A longer number of days (such as 40 days) results in a much smoother line in which the oscillator swings are [less](#page-213-1) volatile. (See Figure 10.1b.) - -#### **Momentum Measures Rates of Ascent or Descent** - -Let's talk a bit more [about](#page-214-0) just what this momentum indicator is measuring. By plotting price differences for a set period of time, the chartist is studying rates of ascent or descent. If prices are rising and the momentum line is above the zero line and rising, this means the uptrend is accelerating. If the upslanting momentum line begins to flatten out, this means that the new gains being achieved by the latest closes are the same as the gains 10 days earlier. While prices may still be advancing, the rate of ascent (or the velocity) has - -leveled off. When the momentum line begins to drop toward the zero line, the uptrend in prices is still in force, but at a decelerating rate. The uptrend is losing momentum. - -When the momentum line moves below the zero line, the latest 10 day close is now under the close of 10 days ago and a near term downtrend is in effect. (And, incidentally, the 10 day moving average also has begun to decline.) As momentum continues to drop farther below the zero line, the downtrend gains momentum. Only when the line begins to advance again does the analyst know that the downtrend is decelerating. - -It's important to remember that momentum measures the differences between prices at two time intervals. In order for the line to advance, the price gains for the last day's close must be greater than the gains of 10 days ago. If prices advance by only the same amount as 10 days ago, the momentum line will be flat. If the last price gain is less than that of 10 days ago, the momentum line begins to decline even though prices are still rising. This is how the momentum line measures the acceleration or deceleration in the current advance or decline in the price trend. - -#### **The Momentum Line Leads the Price Action** - -Because of the way it is constructed, the momentum line is always a step ahead of the price movement. It leads the advance or decline in prices, then levels off while the current price trend is still in effect. It then begins to move in the opposite direction as prices begin to level off. - -#### **The Crossing of the Zero Line as a Trading Signal** - -The momentum chart has a *zero line.* Many technicians use the crossing of the zero line to generate buy and sell signals. A crossing above the zero line would be a buy signal, and a crossing below the zero line, a sell signal. It should be stressed here again, however, that basic trend analysis is still the overriding consideration. Oscillator analysis should not be used as an excuse to trade against the prevailing market trend. Buy positions should only be taken on crossings above the zero line if the market trend is up. Short positions should be taken on crossings below the zero line only if the price trend is down. (See Figures 10.2a and b.) - -**Figure 10.2a** *The trendlines on the momentum chart are broken sooner than those on the price chart. The value of the momentum indicator is that it turns sooner than the market itself, making it a leading indicator.* - -**Figure 10.2b** *Some traders regard a crossing above the zero line as a buy signal and a crossing below the line as a sell signal (see circles). A moving average is helpful to confirm trend changes. The momentum line peaked before the price (see arrows).* - -#### **The Need for an Upper and Lower Boundary** - -One problem with the momentum line, as it is described here, is the absence of a fixed upper and lower boundary. It was stated earlier that one of the major values of oscillator analysis is being able to determine when markets are in extreme areas. But, how high is too high and how low is too low on the momentum line? The simplest way to solve this problem is by visual inspection. Check the back history of the momentum line on the chart and draw horizontal lines along its upper and lower boundaries. These lines will have to be adjusted periodically, especially after important trend changes have occurred. But it is the simplest and probably the most effective way of identifying the outer extremities. (See Figures 10.3 and 10.4.) - -**Figure 10.3** *By visual inspection, the analyst can find the upper and lower momentum boundaries that are suitable for each market (see horizontal lines).* - -**Figure 10.4** *A 13 week momentum line on a weekly chart of Treasury Bonds. The arrows mark the turning points from momentum extremes. The momentum line changed direction before the price at each major turn (points 1, 2, and 3).* - -### **MEASURING RATE OF CHANGE (ROC)** - -To measure the *rate of change*, a ratio is constructed of the most recent closing price to a price a certain number of days in the past. To construct a 10 day rate of change oscillator, the latest closing price is divided by the close 10 days ago. The formula is as follows: - -Rate of change=100 (V/V*x*) - -where V is the latest close and V*x* is the closing price *x* days ago. - -In this case, the 100 line becomes the midpoint line. If the latest price is higher than the price 10 days ago (prices are rising), the resulting rate of change value will be above 100. If the last close is below 10 days ago, the ratio would be below 100. (Charting software sometimes uses variations of the preceding formulas for momentum and rate of change. While the construction techniques may vary, the interpretation remains the same.) - -### **CONSTRUCTING AN OSCILLATOR USING TWO MOVING AVERAGES** - -Chapter 9 discussed two moving averages being used to generate buy and sell - -signals. The crossing of the shorter average above or below the longer average registered buy and sell signals, respectively. It was mentioned at that time that these dual moving average combinations could also be used to construct oscillator charts. This can be done by plotting the difference between the two averages as a histogram. These histogram bars appear as a plus or minus value around a centered zero line. This type of oscillator has three uses: - -- 1. To help spot divergences. -- 2. To help identify short term variations from the long term trend, when the shorter average moves too far above or below the longer average. -- 3. To pinpoint the crossings of the two moving averages, which occur when the oscillator crosses the zero line. - -The shorter average is divided by the longer. In both cases, however, the shorter average oscillates around the longer average, which is in effect the zero line. If the shorter average is above the longer, the oscillator would be positive. A negative reading would be present if the shorter average were under the longer. (See Figures 10.5-10.7.) - -When the two moving average lines move too far apart, a market extreme is created calling for a pause in the trend. (See Figure 10.6.) Very often, the trend remains [stalled](#page-220-0) until [the](#page-221-1) shorter average line moves back to the longer. When the shorter line approaches the longer, a critical point is reached. In an uptrend, for example, the shorter line dips [back](#page-220-1) to the longer average, but should bounce off it. This usually represents an ideal buying area. It's much like the testing of an up trendline. If the shorter average crosses below the longer average, however, a trend reversal is signaled. - -**Figure 10.5** *The histogram lines measure the difference between the two moving averages. Crossing above and below the zero line give buy and sell signals (see arrows). Notice that the histogram turns well before the actual signals (see circles).* - -**Figure 10.6** *A histogram measuring the difference between the 10 and 50 day averages. The histogram always turns well before the zero line crossover. In an uptrend, the histogram will find support at the zero line and turn up again (third arrow).* - -In a downtrend, a rise in the shorter average to the longer usually represents an ideal selling area unless the longer line is crossed, in which case a trend reversal signal would be registered. The relationships between the two averages can be used, therefore, not only as an excellent trend-following system, but also to help identify short term overbought and oversold conditions. - -**Figure 10.7** *A histogram plotting the difference between 2 weekly averages. The histogram turned in the direction of the new price trend weeks before the actual zero line crossings on the histogram. Notice how easily the overbought and oversold levels are seen.* - -### **COMMODITY CHANNEL INDEX** - -It is possible to normalize an oscillator by dividing the values by a constant divisor. In the construction of his Commodity Channel Index (CCI), Donald R. Lambert compares the current price with a moving average over a selected time span—usually 20 days. He then normalizes the oscillator values by using a divisor based on mean deviation. As a result, the CCI fluctuates in a constant range from +100 on the upside to -100 on the downside. Lambert recommended long positions in those markets with values over +100. Markets with CCI values below -100 were candidates for short sales. - -It seems, however, that most chartists use CCI simply as an overbought/oversold oscillator. Used in that fashion readings over +100 are considered overbought and under -100 are oversold. While the Commodity Channel Index was originally developed for commodities, it is also used for trading stock index futures and options like the S&P 100 (OEX). Although 20 days is the common default value for CCI, the user can vary the number to adjust its sensitivity. (See Figures 10.8 and 10.9.) - -**Figure 10.8** *A 20 day Commodity Channel Index. The original intent of this indicator was to buy moves above +100 and sell moves below -100 as shown here.* - -**Figure 10.9** *The Commodity Channel Index can be used for stock indexes like this one and can also be used like any other oscillator to measure market extremes. Notice that the CCI turns before prices at each top and bottom. The default length is 20 days.* - -### **THE RELATIVE STRENGTH INDEX (RSI)** - -The RSI was developed by J. Welles Wilder, Jr. and presented in his 1978 book, *New Concepts in Technical Trading Systems.* We're only going to cover the main points here. A reading of the original work by Wilder himself is recommended for a more in-depth treatment. Because this particular oscillator is so popular among traders, we'll use it to demonstrate most of the principles of oscillator analysis. - -As Wilder points out, one of the two major problems in constructing a momentum line (using price differences) is the erratic movement often caused by sharp changes in the values being dropped off. A sharp advance or a decline 10 days ago (in the case of a 10 day momentum line) can cause sudden shifts in the momentum line even if the current prices show little change. Some smoothing is therefore necessary to minimize these distortions. The second problem is that there is the need for a constant range for comparison purposes. The RSI formula not only provides the necessary smoothing, but also solves the latter problem by creating a constant vertical range of 0 to 100. - -The term "relative strength," incidentally, is a misnomer and often - -causes confusion among those more familiar with that term as it is used in stock market analysis. *Relative strength* generally means a ratio line comparing two different entities. A ratio of a stock or industry group to the S&P 500 Index is one way of gauging the *relative strength* of different stocks or industry groups against one objective benchmark. We'll show you later in the book how useful *relative strength* or *ratio* analysis can be. Wilder's *Relative Strength Index* doesn't really measure the relative strength between different entities and, in that sense, the name is somewhat misleading. The RSI, however, does solve the problem of erratic movement and the need for a constant upper and lower boundary. The actual formula is calculated as follows: - -Fourteen days are used in the calculation; 14 weeks are used for weekly charts. To find the average up value, add the total points gained on up days during the 14 days and divide that total by 14. To find the average down value, add the total number of points lost during the down days and divide that total by 14. Relative strength (RS) is then determined by dividing the *up* average by the *down* average. That RS value is then inserted into the formula for RSI. The number of days can be varied by simply changing the value of *x.* - -Wilder originally employed a 14 day period. *The shorter the time period, the more sensitive the oscillator becomes and the wider its amplitude.* RSI works best when its fluctuations reach the upper and lower extremes. Therefore, if the user is trading on a very short term basis and wants the oscillator swings to be more pronounced, the time period can be shortened. The time period is lengthened to make the oscillator smoother and narrower in amplitude. The amplitude in the 9 day oscillator is therefore greater than the original 14 day. While 9 and 14 day spans are the most common values used, technicians experiment with other periods. Some use shorter lengths, such as 5 or 7 days, to increase the volatility of the RSI line. Others use 21 or 28 days to smooth out the RSI signals. (See Figures 10.10 and 10.11.) - -**Figure 10.10** *The 14 day Relative Strength Index becomes overbought over 70 and oversold below 30. This chart shows the S&P 100 being oversold in October and overbought during February.* - -**Figure 10.11** *The amplitude of the RSI line can be widened by shortening the time period. Notice that the 7 day RSI reaches the outer extremes more frequently than the 14 day RSI. That makes the 7 day RSI more useful to short term traders.* - -#### **Interpreting RSI** - -RSI is plotted on a vertical scale of 0 to 100. Movements above 70 are considered overbought, while an oversold condition would be a move under 30. Because of shifting that takes place in bull and bear markets, the 80 level usually becomes the overbought level in bull markets and the 20 level the oversold level in bear markets. - -"Failure swings," as Wilder calls them, occur when the RSI is above 70 or under 30. A *top failure swing* occurs when a peak in the RSI (over 70) fails to exceed a previous peak in an uptrend, followed by a downside break of a previous trough. A *bottom failure swing* occurs when the RSI is in a downtrend (under 30), fails to set a new low, and then proceeds to exceed a previous peak. (See Figures 10.12a-b.) - -**Figure 10.12a** *A bottom failure swing in the RSI line. The second RSI trough (point 2) is higher than the first (point 1) while it is below 30 and prices are still falling. The upside penetration of the RSI peak (point 3) signals a bottom.* - -**Figure 10.12b** *A top failure swing. The second peak (2) is lower than the first (1) while the RSI line is over 70 and prices are still rallying. The break by the RSI line below the middle trough (point 3) signals the top.* - -Divergence between the RSI and the price line, when the RSI is above 70 or below 30, is a serious warning that should be heeded. Wilder himself considers divergence "the single most indicative characteristic of the Relative Strength Index" [Wilder, p. 70]. - -Trendline analysis can be employed to detect changes in the trend of the RSI. Moving averages can also be used for the same purpose. (See Figure 10.13.) - -**Figure 10.13** *Trendlines work very effectively on the RSI line. The breaking of the two RSI trendlines gave timely buy and sell signals on this chart (see arrows).* - -In my own personal experience with the RSI oscillator, its greatest value lies in failure swings or divergences that occur when the RSI is over 70 or under 30. Let's clarify another important point on the use of oscillators. Any strong trend, either up or down, usually produces an extreme oscillator reading before too long. In such cases, claims that a market is overbought or oversold are usually premature and can lead to an early exit from a profitable trend. In strong uptrends, overbought markets can stay overbought for some time. Just because the oscillator has moved into the upper region is not reason enough to liquidate a long position (or, even worse, short into the strong uptrend). - -The first move into the overbought or oversold region is usually just a warning. The signal to pay close attention to is the second move by the oscillator into the *danger zone.* If the second move fails to confirm the price move into new highs or new lows (forming a double top or bottom on the oscillator), a possible divergence exists. At that point, some defensive action can be taken to protect existing positions. If the oscillator moves in the opposite direction, breaking a previous high or low, then a divergence or failure swing is confirmed. - -The 50 level is the RSI midpoint value, and will often act as support during pullbacks and resistance during bounces. Some traders treat RSI crossings above and below the 50 level as buying and selling signals - -respectively. - -## **USING THE 70 AND 30 LINES TO GENERATE SIGNALS** - -Horizontal lines appear on the oscillator chart at the 70 and 30 values. Traders often use those lines to generate buy and sell signals. We already know that a move under 30 warns of an oversold condition. Suppose the trader thinks a market is about to bottom and is looking for a buying opportunity. He or she watches the oscillator dip under 30. Some type of divergence or double bottom may develop in the oscillator in that oversold region. A crossing back above the 30 line at that point is taken by many traders as a confirmation that the trend in the oscillator has turned up. Accordingly, in an overbought market, a crossing back under the 70 line can often be used as a sell signal. (See Figure 10.14.) - -**Figure 10.14** *The RSI oscillator can be used on monthly charts. Notice the two major oversold buy signals in 1974 and 1994. The overbought peaks in the RSI line did a pretty good job of pinpointing important tops in the utilities.* - -## **STOCHASTICS (K%D)** - -The *Stochastic* oscillator was popularized by George Lane (president of - -Investment Educators, Inc., Watseka, IL). It is based on the observation that as prices increase, closing prices tend to be closer to the upper end of the price range. Conversely, in downtrends, the closing price tends to be near the lower end of the range. Two lines are used in the Stochastic Process—the %K line and the %D line. The %D line is the more important and is the one that provides the major signals. - -The intent is to determine where the most recent closing price is in relation to the price range for a chosen time period. Fourteen is the most common period used for this oscillator. To determine the K line, which is the more sensitive of the two, the formula is: - -%K=100 [(C - L14) / (H14 - L14)] - -where C is the latest close, L14 is the lowest low for the last 14 periods, and H14 is the highest high for the same 14 periods (14 periods can refer to days, weeks, or months). - -The formula simply measures, on a percentage basis of 0 to 100, where the closing price is in relation to the total price range for a selected time period. A very high reading (over 80) would put the closing price near the top of the range, while a low reading (under 20) near the bottom of the range. - -The second line (%D) is a 3 period moving average of the %K line. This formula produces a version called *fast* stochastics. By taking another 3 period average of %D, a smoother version called *slow* stochastics is computed. Most traders use the *slow* stochastics because of its more reliable signals.\* - -These formulas produce two lines that oscillate between a vertical scale from 0 to 100. The K line is a faster line, while the D line is a slower line. The major signal to watch for is a divergence between the D line and th[e](#page-244-0) price of the underlying market when the D line is in an overbought or oversold area. The upper and lower extremes are the 80 and 20 values. (See Figure 10.15.) - -A bearish divergence occurs when the D line is over 80 and forms two declining peaks while prices continue to move higher. A bullish divergence is present when the D line is under 20 and forms two rising bottoms [while](#page-231-0) prices continue to move lower. Assuming all of these factors are in place, the actual buy or sell signal is triggered when the faster K line crosses the slower D line. - -There are other refinements in the use of Stochastics, but this explanation covers the more essential points. Despite the higher level of sophistication, the basic oscillator interpretation remains the same. An alert or set-up is present when the %D line is in an extreme area and diverging from the price action. The actual signal takes place when the D line is crossed by the faster K line. - -The Stochastic oscillator can be used on weekly and monthly charts for longer range perspective. It can also be used effectively on intraday charts for shorter term trading. (See Figure 10.16.) - -One way to combine daily and weekly stochastics is to use weekly signals to determine market direction and daily signals for timing. It's also a good idea to combine stochastics with RSI. (See Figure 10.17.) - -**Figure 10.15** *The down arrows show two sell signals which occur when the faster %K line crosses below the slower %D line from above the 80 level. The %K line crossing above the %D line below 20 is a buy signal (up arrow).* - -**Figure 10.16** *Turns in the 14 week stochastics from above 80 and below 20 did a nice job of anticipating major turns in the Treasury Bond market. Stochastics charts can be constructed for 14 days, 14 weeks, or 14 months.* - -**Figure 10.17** *A comparison of the 14 week RSI and stochastics. The RSI line is less volatile and reaches extremes less frequently than stochastics. The best signals occur when both oscillators are in overbought or oversold territory.* - -### **LARRY WILLIAMS %R** - -Larry Williams %R is based on a similar concept of measuring the latest close in relation to its price range over a given number of days. Today's close is subtracted from the price high of the range for a given number of days and that difference is divided by the total range for the same period. The concepts already discussed for oscillator interpretation are applied to %R as well, with the main factors being the presence of divergences in overbought or oversold areas. (See Figure 10.18.) Since %R is subtracted from the high, it looks like an upside down stochastics. To correct that, charting packages plot an inverted version of %R. - -**Figure 10.18** *Larry Williams %R oscillator is used in the same fashion as other oscillators. Readings over 80 or under 20 identify market extremes.* - -#### **Choice of Time Period Tied to Cycles** - -Oscillator lengths can be tied to underlying market cycles. A time period of 1/2 the cycle length is used. Popular time inputs are 5, 10, and 20 days based on calendar day periods of 14, 28, and 56 days. Wilder's RSI uses 14 days, which is half of 28. In the previous chapter, we discussed some reasons why the numbers 5, 10, and 20 keep cropping up in moving average and oscillator formulations, so we won't repeat them here. Suffice it to mention here that 28 calendar days (20 trading days) represent an important dominant monthly trading cycle and that the other numbers are related harmonically to that monthly cycle. The popularity of the 10 day momentum and the 14 day RSI lengths are based largely on the 28 day trading cycle and measure 1/2 of the value of that dominant trading cycle. We'll come back to the importance of cycles in Chapter 14. - -### **THE [IMPORT](#page-317-0)ANCE OF TREND** - -In this chapter, we've discussed the use of the oscillator in market analysis to help determine near term overbought and oversold conditions, and to alert traders to possible divergences. We started with the momentum line. We discussed another way to measure rates of change (ROC) by using price ratios instead of differences. We then showed how two moving averages could be - -compared to spot short term extremes and crossovers. Finally, we looked at RSI and Stochastics and considered how oscillators should be synchronized with cycles. - -Divergence analysis provides us with the oscillator's greatest value. However, the reader is cautioned against placing too much importance on divergence analysis to the point where basic trend analysis is either ignored or overlooked. Most oscillator buy signals work best in uptrends and oscillator sell signals are most profitable in downtrends. The place to start your market analysis is always by determining the general trend of the market. If the trend is up, then a buying strategy is called for. Oscillators can then be used to help time market entry. Buy when the market is oversold in an uptrend. Sell short when the market is overbought in a downtrend. Or, buy when the momentum oscillator crosses back above the zero line when the major trend is bullish and sell a crossing under the zero line in a bear market. - -The importance of trading in the direction of the major trend cannot be overstated. The danger in placing too much importance on oscillators by themselves is the temptation to use divergence as an excuse to initiate trades contrary to the general trend. This action generally proves a costly and painful exercise. The oscillator, as useful as it is, is just one tool among many others and must always be used as an aid, not a substitute, for basic trend analysis. - -### **WHEN OSCILLATORS ARE MOST USEFUL** - -There are times when oscillators are more useful than at others. During choppy market periods, as prices move sideways for several weeks or months, oscillators track the price movement very closely. The peaks and troughs on the price chart coincide almost exactly with the peaks and troughs on the oscillator. Because both price and oscillator are moving sideways, they look very much alike. At some point, however, a price breakout occurs and a new uptrend or downtrend begins. By its very nature, the oscillator is already in an extreme position just as the breakout is taking place. If the breakout is to the upside, the oscillator is already overbought. An oversold reading usually accompanies a downside breakout. The trader is faced with a dilemma. Should he or she buy the bullish breakout in the face of an overbought oscillator reading? Should the downside breakout be sold into an oversold market? - -In such cases, the oscillator is best ignored for the time being and the position taken. The reason for this is that in the early stages of a new trend, following an important breakout, oscillators often reach extremes very quickly and stay there for awhile. Basic trend analysis should be the main consideration at such times, with oscillators given a lesser role. Later on, as the trend begins to mature, the oscillator should be given greater weight. (We'll see in Chapter 13, that the fifth and final wave in Elliott Wave analysis is often confirmed by bearish oscillator divergences.) Many dynamic bull moves have been missed by traders who saw the major trend signal, but decided to wait for [their](#page-293-0) oscillators to move into an oversold condition before buying. To summarize, give less attention to the oscillator in the early stages of an important move, but pay close attention to its signals as the move reaches maturity. - -### **MOVING AVERAGE CONVERGENCE/DIVERGENCE (MACD)** - -We mentioned in the previous chapter an oscillator technique that uses 2 exponential moving averages and here it is. The Moving Average Convergence/Divergence indicator, or simply MACD, was developed by Gerald Appel. What makes this indicator so useful is that it combines some of the oscillator principles we've already explained with a dual moving average crossover approach. You'll see only two lines on your computer screen although three lines are actually used in its calculation. The faster line (called the MACD line) is the difference between two exponentially smoothed moving averages of closing prices (usually the last 12 and 26 days or weeks). The slower line (called the signal line) is usually a 9 period exponentially smoothed average of the MACD line. Appel originally recommended one set of numbers for buy signals and another for sell signals. Most traders, however, utilize the default values of 12, 26, and 9 in all instances. That would include daily and weekly values. (See Figure 10.19a.) - -The actual buy and sell signals are given when the two lines cross. A crossing by the faster MACD line above the slower signal line is a buy signal. A crossing by the faster line below the slower is a sell [signa](#page-236-0)l. In that sense, MACD resembles a dual moving average crossover method. However, the MACD values also fluctuate above and below a zero line. That's where it begins to resemble an oscillator. An overbought condition is present when the lines are too far above the zero line. An oversold condition is present when the lines are too far below the zero line. The best buy signals are given when prices are well below the zero line (oversold). Crossings above and below the zero line are another way to generate buy and sell signals respectively, similar to the momentum technique we discussed previously. - -**Figure 10.19a** *The Moving Average Convergence Divergence system shows two lines. A signal is given when the faster MACD line crosses the slower signal line. The arrows show five trading signals on this chart of the Nasdaq Composite Index.* - -Divergences appear between the trend of the MACD lines and the price line. A negative, or bearish, divergence exists when the MACD lines are well above the zero line (overbought) and start to weaken while prices continue to trend higher. That is often a warning of a market top. A positive, or bullish, divergence exists when the MACD lines are well below the zero line (oversold) and start to move up ahead of the price line. That is often an early sign of a market bottom. Simple trendlines can be drawn on the MACD lines to help identify important trend changes. (See Figure 10.19b.) - -**Figure 10.19b** *The MACD lines fluctuate around a zero line, giving it the quality of an oscillator. The best buy signals occur below the zero line. The best sell signals come from above. Notice the negative divergence given in October (see down arrow).* - -### **MACD HISTOGRAM** - -We showed you earlier in the chapter how a histogram could be constructed that plots the difference between two moving average lines. Using that same technique, the two MACD lines can be turned into an MACD histogram. The histogram consists of vertical bars that show the difference between the two MACD lines. The histogram has a zero line of its own. When the MACD lines are in positive alignment (faster line over the slower), the histogram is above its zero line. Crossings by the histogram above and below its zero line coincide with actual MACD crossover buy and sell signals. - -The real value of the histogram is spotting when the spread between the two lines is widening or narrowing. When the histogram is over its zero line (positive) but starts to fall toward the zero line, the uptrend is weakening. Conversely, when the histogram is below its zero line (negative) and starts to move upward toward the zero line, the downtrend is losing its momentum. Although no actual buy or sell signal is given until the histogram crosses its zero line, the histogram turns provide earlier warnings that the current trend is losing momentum. Turns in the histogram back toward the zero line always precede the actual crossover signals. Histogram turns are best used for - -spotting early exit signals from existing positions. It's much more dangerous to use the histogram turns as an excuse to initiate new positions against the prevailing trend. (See Figure 10.20a.) - -**Figure 10.20a** *The MACD histogram plots the difference between the two MACD lines. Signals are given on the zero line crossings. Notice that the histogram turns earlier than the crossover signals, giving the trader some advanced warning.* - -### **COMBINE WEEKLIES AND DAILIES** - -As with all technical indicators, signals on weekly charts are always more important than those on daily charts. The best way to combine them is to use weekly signals to determine market direction and the daily signals to fine-tune entry and exit points. A daily signal is followed only when it agrees with the weekly signal. Used in that fashion, the weekly signals become trend filters for daily signals. That prevents using daily signals to trade against the prevailing trend. Two crossover systems in which this principle is especially true are MACD and Stochastics. (See Figure 10.20b.) - -**Figure 10.20b** *The MACD histogram works well on weekly charts. At the middle peak, the histogram turned down 10 weeks before the sell signal (down arrow). At the two upturns, the histogram turned up 2 and 4 weeks before the buy signals (up arrows).* - -### **THE PRINCIPLE OF CONTRARY OPINION IN FUTURES** - -Oscillator analysis is the study of market extremes. One of the most widely followed theories in measuring those market extremes is the principle of Contrary Opinion. At the beginning of the book, two principal philosophies of market analysis were identified—fundamental and technical analysis. Contrary Opinion, although it is generally listed under the category of technical analysis, is more aptly described as a form of psychological analysis. Contrary Opinion adds the important third dimension to market analysis—the psychological—by determining the degree of bullishness or bearishness among participants in the various financial markets. - -The principle of *Contrary Opinion* holds that when the vast majority of people agree on anything, they are generally wrong. A true contrarian, therefore, will first try to determine what the majority are doing and then will act in the opposite direction. - -Humphrey B. Neill, considered the dean of contrary thinking, described his theories in a 1954 book entitled, *The Art of Contrary Thinking.* Ten years later, in 1964, James H. Sibbet began to apply Neill's principles to commodity futures trading by creating the Market Vane advisory service, which includes the Bullish Consensus numbers (Market Vane, P.O. Box 90490, Pasadena, CA 91109). Each week a poll of market letters is taken to determine the degree of bullishness or bearishness among commodity professionals. The purpose of the poll is to quantify market sentiment into a set of numbers that can be analyzed and used in the market forecasting process. The rationale behind this approach is that most futures traders are influenced to a great extent by market advisory services. By monitoring the views of the professional market letters, therefore, a reasonably accurate gauge of the attitudes of the trading public can be obtained. - -Another service that provides an indication of market sentiment is the "Consensus Index of Bullish Market Opinion," published by *Consensus National Commodity Futures Weekly* (Consensus, Inc., 1735 McGee Street, Kansas City, MO 64108). These numbers are published each Friday and use 75% as an overbought and 25% as an oversold measurement. - -#### **Interpreting Bullish Consensus Numbers** - -Most traders seem to employ a fairly simple method of analyzing these weekly numbers. If the numbers are above 75%, the market is considered to be overbought and means that a top may be near. A reading below 25% is interpreted to warn of an oversold condition and the increased likelihood that a market bottom is near. - -#### **Contrary Opinion Measures Remaining Buying or Selling Power** - -Consider the case of an individual speculator. Assume that speculator reads his or her favorite newsletter and becomes convinced that a market is about to move substantially higher. The more bullish the forecast, the more aggressively that trader will approach the market. Once that individual speculator's funds are fully committed to that particular market, however, he or she is overbought—meaning there are no more funds to commit to the market. - -Expanding this situation to include all market participants, if 80-90% of market traders are bullish on a market, it is assumed that they have already taken their market positions. Who is left to buy and push the market higher? This then is one of the keys to understanding Contrary Opinion. If the overwhelming sentiment of market traders is on one side of the market, there simply isn't enough buying or selling pressure left to continue the present trend. - -#### **Contrary Opinion Measures Strong Versus Weak Hands** - -A second feature of this philosophy is its ability to compare strong versus weak hands. Futures trading is a zero sum game. For every long there is also a short. If 80% of the traders are on the long side of a market, then the remaining 20% (who are holding short positions) must be well financed enough to absorb the longs held by the other 80%. The shorts, therefore, must be holding much larger positions than the longs (in this case, 4 to 1). - -This means further that the shorts must be well capitalized and are considered to be strong hands. The 80%, who are holding much smaller positions per trader, are considered to be weaker hands who will be forced to liquidate those longs on any sudden turn in prices. - -#### **Some Additional Features of the Bullish Consensus Numbers** - -Let's consider a few additional points that should be kept in mind when using these numbers. The norm or equilibrium point is at 55%. This allows for a built-in bullish bias on the part of the general public. The upper extreme is considered to be 90% and the lower extreme, 20%. Here again, the numbers are shifted upward slightly to allow for the bullish bias. - -A contrarian position can usually be considered when the bullish consensus numbers are above 90% or under 20%. Readings over 75% or under 25% are also considered warning zones and suggest that a turn may be near. However, it is generally advisable to await a change in the trend of the numbers before taking action against the trend. A change in the direction of the Bullish Consensus numbers, especially if it occurs from one of the danger zones, should be watched closely. - -#### **The Importance of Open Interest (Futures)** - -Open interest also plays a role in the use of Bullish Consensus numbers. In general, the higher the open interest figures are, the better the chance that the contrarian positions will prove profitable. A contrarian position should not be taken, however, while open interest is still increasing. A continued rise in open interest numbers increases the odds that the present trend will continue. Wait for the open interest numbers to begin to flatten out or to decline before taking action. - -Study the Commitments of Traders Report to ensure that hedgers hold less than 50% of the open interest. Contrary Opinion works better when most of the open interest is held by speculators, who are considered to be weaker hands. It is not advisable to trade against large hedging interests. - -#### **Watch the Market's Reaction to Fundamental News** - -Watch the market's reaction to fundamental news very closely. The failure of - -prices to react to bullish news in an overbought area is a clear warning that a turn may be near. The first adverse news is usually enough to quickly push prices in the other direction. Correspondingly, the failure of prices in an oversold area (under 25%) to react to bearish news can be taken as a warning that all the bad news has been fully discounted in the current low price. Any bullish news will push prices higher. - -#### **Combine Contrarian Opinion with Other Technical Tools** - -As a general rule, trade in the same direction as the trend of the consensus numbers until an extreme is reached, at which time the numbers should be monitored for a sign of a change in trend. It goes without saying that standard technical analytical tools can and should also be employed to help identify market turns at these critical times. The breaking of support or resistance levels, trendlines, or moving averages can be utilized to help confirm that the trend is in fact turning. Divergences on oscillator charts are especially useful when the Bullish Consensus numbers are overbought or oversold. - -### **INVESTOR SENTIMENT READINGS** - -Each weekend *Barron's* includes in its Market Laboratory section a set of numbers under the heading "Investor Sentiment Readings." In that space, four different investor polls are included to gauge the degree of bullishness and bearishness in the stock market. The figures are given for the latest week and the period two and three weeks back for comparison purposes. Here's a random sample of what the latest week's figures might look like. Remember that these numbers are contrary indicators. Too much bullishness is bad. Too much bearishness is good. - -| investor's intelligence | | -|--------------------------------------------------------------------------------------------------------------|-----| -| Bulls | 48% | -| Bears | 27 | -| Correction | 24 | -| Consensus Index | | -| Bullish Opinion | 77% | -| AAII Index
(American Association of
Individual Investors
625 N. Michigan Ave.
Chicago, IL 60611) | | -| Bullish | 53% | -| Bearish | 13 | -| Neutral | 34 | -| Market Vane | | -| Bullish Consensus | 66% | - -### **INVESTORS INTELLIGENCE NUMBERS** - -Investors Intelligence (30 Church Street, New Rochelle, NY 10801) takes a weekly poll of investment advisors and produces three numbers—the percent of investment advisors that are bullish, those that are bearish, and those that are expecting a market correction. Bullish readings over 55% warn of too much optimism and are potentially negative for the market. Bullish readings below 35% reflect too much pessimism and are considered positive for the market. The correction figure represents advisers who are bullish but expecting short term weakness. - -Investors Intelligence also publishes figures each week that measure the number of stocks that are above their 10 and 30 week moving averages. Those numbers can also be used in a contrary fashion. Readings above 70% suggest an overbought stock market. Readings below 30% suggest an oversold market. The 10 week readings are useful for measuring short to intermediate market turns. The 30 week numbers are more useful for measuring major market turns. The actual signal of a potential change in trend takes place when the numbers rise back above 30 or fall back below 70. - -\*The second smoothing produces 3 lines. Fast stochastics uses the first 2 lines. Slow stochastics uses the last 2 lines. - -### **INTRODUCTION** - -The first charting technique used by stock market traders before the turn of the century was point and figure charting. The actual name "point and figure" has been attributed to Victor deVilliers in his 1933 classic, *The Point and Figure Method of Anticipating Stock Price Movements.* The technique has had various names over the years. In the 1880s and 1890s, it was known as the "book method." This was the name Charles Dow gave it in a July 20, 1901 editorial of *The Wall Street Journal.* - -Dow indicated that the book method had been used for about 15 years, giving it a starting date of 1886. The name "figure charts" was used from the 1920s until 1933 when "point and figure" became the accepted name for this technique of tracking market movement. R.D. Wyckoff also published several works dealing with the point and figure method in the early 1930s. - -*The Wall Street Journal* started publishing daily high, low, and closing stock prices in 1896, which is the first reference to the more commonly known bar chart. Therefore, it appears that the point and figure method predates bar charting by at least 10 years. - -We're going to approach point and figure charting in two steps. We'll look at the original method that relies on intraday price moves. Then we'll show you a simpler version of point and figure charting that can be constructed by using only the high and low prices for any market. - -### **THE POINT AND FIGURE VERSUS THE BAR** diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/016_11 Point and Figure Charting.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/016_11 Point and Figure Charting.md deleted file mode 100644 index 58ef570aab7fefff16436e91c5e8e8fdea1b1f68..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/016_11 Point and Figure Charting.md +++ /dev/null @@ -1,280 +0,0 @@ -### **CHART** - -Let's begin with some of the basic differences between point and figure charting and bar charting and look at a couple of chart examples. - -*The point and figure chart is a study of pure price movement.* That is to say, it does not take time into consideration while plotting the price action. A bar chart, by contrast, combines both price and time. Because of the way the bar chart is constructed, the vertical axis is the price scale and the horizontal axis, a time scale. On a daily chart, for example, each successive day's price action moves one space or bar to the right. This happens even if prices saw little or no change for that day. Something must always be placed in the next space. On the point and figure chart, only the price changes are recorded. If no price change occurs, the chart is left untouched. During active market periods, a considerable amount of plotting may be required. During quiet market conditions, little or no plotting will be needed. - -*An important difference is the treatment of volume.* Bar charts record volume bars under the day's price action. Point and figure charts ignore volume numbers, as a separate entity. This last phrase, "as a separate entity," is an important one. Although the volume numbers are not recorded on the point and figure chart, it does not necessarily follow that volume, or trading activity, is totally lost. On the contrary, since intraday point and figure charts record all price change activity, the heavier or lighter volume is reflected in the amount of price changes recorded on the chart. Because volume is one of the more important ingredients in determining the potency of support and resistance levels, point and figure charts become especially useful in determining at which price levels most of the trading activity took place and, hence, where the important support and resistance numbers are. - -Figure 11.1 compares a bar chart and a point and figure chart covering the same time span. In one sense, the charts look similar, but, in another sense, quite different. The general price and trend picture is captured on both charts, [but](#page-247-0) the method of recording prices is different. Notice in Figure 11.2 the alternating columns of x's and o's. The *x columns* represent rising prices, while the *o columns* show declining prices. Each time a column of x's moves one box above a previous column of x's, an upside breakout occurs. [\(See](#page-247-1) arrows in Figure 11.2.) - -**Figure 11.1** *A comparison of a daily bar chart for the S&P 500 Index (left) and a point and figure chart (right) for the same time period. The point and figure chart uses x columns for rising prices and o columns for declining prices.* - -**Figure 11.2** *A buy signal is given when one x column rises above the top of a previous x column (see up arrows). A sell signal is given when a column of o's falls below a previous o column (see down arrows). Signals are more precise* - -Correspondingly, when a column of o's declines one box under a previous column of o's, a downside breakout occurs. Notice how much more precise these breakouts are than those on the bar chart. These breakouts can, of course, be used as buy and sell signals. We'll have more to say on buy and sell signals a bit later. But the charts demonstrate one of the advantages of the point and figure chart, mainly the greater precision and ease in recognizing trend signals. - -Figures 11.3 and 11.4 reveal another major advantage of the point and figure chart: flexibility. While all three of the p&f charts cover the same price action, we can make them look very different to serve different purposes. One way to [change](#page-248-0) the p&f [cha](#page-249-1)rt is to vary the *reversal criteria* (let's say from a 3 box reversal to a 5 box reversal). The larger the number of boxes required for a reversal, the less sensitive the chart becomes. The second way to vary the chart is to change the *box size*. Figure 2 uses a box size of 5 points. Figure 11.3 changes the box size from 5 points to 10 points. The number of columns has been reduced from 44 in the 5×3 chart in Figure 11.2 to only 16 columns in Figure 11.3. By using the larger box size in Figure 11.3, fewer signals are given. That allows the investor to [concentrate](#page-248-0) on the major trend of a market by avoiding all the short term sell signals that are [elimina](#page-247-1)ted from the less se[nsitive](#page-248-0) chart. - -**Figure 11.3** *Increasing the box size from 5 points to 10 makes the point and figure chart less sensitive and fewer signals are given. This is more suitable for a long term investor.* - -**Figure 11.4** *Reducing the box size to 3 points produces more signals. This is better for shorter term trading. The last rally from 920 to 1060 produced 6 different buy signals. Protective sell stops can be placed under the highest column of o's (see S1-S5).* - -Figure 11.4 reduces the box size from 5 to 3. That increases the sensitivity of the chart. Why would anyone want to do that? Because it's better for shorter term trading. Compare the last rally from 920 to 1060 in all three [charts.](#page-249-1) The 10×3 chart (Figure 11.3) shows the last column as a series of x's with no o columns. The 5×3 chart (Figure 11.2) shows the last upleg in 5 columns—3 x columns and 2 o columns. The 3×3 chart (Figure 11.4) breaks the last upleg into 11 columns—6 x [colum](#page-248-0)ns and 5 o columns. By increasing the number of corrections during the u[ptrend](#page-247-1) (by increasing the number of o columns), more repeat buy signals are given either for later [entry](#page-249-1) or for adding to winning positions. It also allows the trader to raise protective sell stops below the latest columns of o's. The bottom line is that you can alter the look of the point and figure chart to adjust its sensitivity to suit your own needs. - -### **CONSTRUCTION OF THE INTRADAY POINT AND FIGURE CHART** - -We've already stated that the intraday chart was the original type used by point and figure chartists. The technique was originally used to track stock market movement. The intent was to capture and record on paper each one point move of the stocks under consideration. It was felt that accumulation - -(buying) and distribution (selling) could be better detected in this manner. Only whole numbers were employed. Each box was given a value of one point and each one point move in either direction was recorded. Fractions were largely ignored. When the technique was later adopted to commodity markets, the value of the box had to be adjusted to fit each different commodity market. Let's construct an intraday chart using some actual price data. - -The following numbers describe 9 actual days of trading in a Swiss franc futures contract. The box size is 5 points. Therefore, every 5 point swing in either direction is plotted. We'll start with a 1 box reversal chart. - -Figure 11.5a is what the previously listed numbers would look like on the chart. Let's begin on the left side of the chart. First the chart is scaled to reflect a 5 point increment for every box. - -- Column 1: [Put](#page-251-0) a dot at 4875. Because the next number—4880—is higher, fill in the next box up to 4880. -- Column 2: The next number is 4860. Move 1 column to the right, go down 1 box, and fill in all the o's down to 4860. -- Column 3: The next number is 4865. Move 1 column to the right, move up 1 box and put an x at 4865. Stop here. So far you have only 1 x marked in column 3 because prices have only moved up 1 box. On a 1 box reversal chart, there must always be at least 2 boxes filled in each column. Notice that the next number is 4850, calling for o's down to that number. Do you go to the next column to record the column of declining o's? The answer is no because that would leave only 1 mark, the x, in column 3. Therefore, in the column with the lone x (column 3) fill in o's down to 4850. - -**Figure 11.5a** *A 5×1 point and figure chart of a Deutsche mark contract is shown in the upper chart. The blackened boxes show the end of each day's trading. Figure 11.5b shows the same price data with a 3 box reversal. Notice the compression. Figure 11.5c shows a 5 box reversal.* - -- Column 4: The [next](#page-252-1) number is 4860. Move to the next column, move 1 box up, and plot in the x's up to [4860.](#page-252-1) -- Column 5: The next number is 4855. Because this is a move down, go to the next column, move down a box, and fill the o at 4860. Notice on the table that this is the last price of the day. Let's do one more. -- Column 6: The first number on 5/2 is 4870. So far, you only have one o in column 5. You must have at least 2 marks in each column. Therefore, fill in x's (because prices are advancing) up to 4870. But notice that the last price on the previous day is blacked out. This is to help keep track of time. By blacking in the last price each day, it's much easier to keep track of the separate days' trading. - -Feel free to continue through the remainder of the chart to sharpen your understanding of the plotting process. Notice that this chart has several columns where both x's and o's are present. This situation will only develop on the 1 point reversal chart and is caused by the necessity of having at least 2 boxes filled in each column. Some purists might argue with combining the x's and o's. Experience will show, however, that this method of plotting prices makes it much easier to follow the order of the transactions. - -Figure 11.5b takes the same data from Figure 11.5a and transforms it into a 3 box reversal chart. Notice that the chart is condensed and a lot of data is lost. Figure 11.5c shows a 5 box reversal. These are the 3 reversal criteria that have [tradition](#page-252-1)ally been used—the 1, 3, and 5 [box](#page-251-0) reversal. The 1 box - -reversal is generally used for very short term activity and the 3 box for the study of the intermediate trend. The 5 box reversal, because of its severe condensation, is generally used for the study of long term trends. The correct order to use is the one shown here, that is, begin with the 1 point reversal chart. The 3 and 5 box reversals can then be constructed right off the first chart. For obvious reasons, a 1 point reversal chart could not possibly be constructed from a 3 or 5 box reversal. - - - -### **THE HORIZONTAL COUNT** - -One principal advantage of the intraday 1 box reversal chart is the ability to obtain price objectives through use of the *horizontal count.* If you think back to our coverage of bar charts and price patterns, the question of price objectives was discussed. However, virtually all methods of obtaining price objectives off bar charts were based on what we call *vertical measurements.* This meant measuring the height of a pattern (the volatility) and projecting that distance upward or downward. For example, the head and shoulders pattern measured the distance from the head to the neckline and swung that objective from the break of that neckline. - -### **Point and Figure Charts Allow Horizontal Measurement** - -The principle of the horizontal count is based on the premise that there is a direct relationship between the width of a congestion area and the subsequent move once a breakout occurs. If the *congestion area* represents a basing pattern, some estimate can be made of the upside potential once the base is completed. Once the uptrend has begun, subsequent congestion areas can be used to obtain additional counts which can be utilized to confirm the original counts from the base. (See Figure 11.6.) - -The intent is to measure the width of the pattern. Remember we're talking here of intraday 1 box reversal charts. The technique requires some modifications for other types of [charts](#page-253-1) that we'll come back to later. Once a topping or basing area has been identified, simply count the number of columns in that top or base. If there are 20 columns, for example, the upside or downside target would be 20 boxes from the measuring point. The key is to determine which line to measure from. Sometimes this is easy and, at other times, more difficult. - -Usually, the horizontal line to count across is near the middle of the congestion area. A more precise rule is to use the line that has the least number of empty boxes in it. Or put the other way, the line with the most number of filled in x's and o's. Once you find the correct line to count across, it's important that you include every column in your count, even the ones that are empty. Count the number of columns in the congestion area and then project that number up or down from the line that was used for the count. - -**Figure 11.6** *By counting the number of columns across the horizontal congestion area, price objectives can be determined. The wider the congestion area, the greater the objective.* - -### **PRICE PATTERNS** - -Pattern identification is also possible on point and figure charts. Figure 11.7 shows the most common types. - -As you can see, they're not much different from ones already discussed on bar charting. Most of the patterns are variations on the double and triple tops and bottoms, head and shoulders, V's and inverted V's, and saucers. The term "fulcrum" shows up quite a bit in the point and figure literature. Essentially, the *fulcrum* is a well defined congestion area, occurring after a significant advance or decline, that forms an accumulation base or a distribution top. In a base, for example, the bottom of the area is subjected to repeated tests, interrupted by intermittent rally attempts. Very often, the fulcrum takes on the appearance of a double or triple bottom. 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Service | | -| | | -| | | - -**Figure 11.7** *Reversal patterns. (Source: Alexander H. Wheelan*, Study Helps in Point and Figure Technique *[New York, NY: Morgan, Rogers and Roberts,* - -### *Inc., 1954] p. 25.) Reprinted in 1990 by Traders Press, P.O. Box 6206, Greenville, SC 29606.]* - -Those reversal patterns with the most pronounced horizontal ranges obviously lend themselves quite well to the taking of count measurements. The V base, in contrast, because of the absence of a significant horizontal price area, would not be amenable to the taking of a horizontal count. The blackened boxes in the chart examples in Figure 11.7 represent suggested buying and selling points. Notice that those entry points generally coincide with the retesting of support areas in a base or resistance areas in a top, breakout points, and the breaking of tren[dlines.](#page-254-0) - -#### **Trend Analysis and Trendlines** - -The price patterns in Figure 11.7 show trendlines drawn as part of those patterns. Trendline analysis on intraday charts is the same as that applied to bar charts. Up trendlines are drawn under successive lows and down trendlines are drawn over [success](#page-254-0)ive peaks. This is not true of the simplified point and figure chart, which we're going to study next. It utilizes 45 degree lines and plots them differently. - -### **3 BOX REVERSAL POINT AND FIGURE CHARTING** - -In 1947, a book on point and figure was written by A.W. Cohen entitled, *Stock Market Timing.* The following year, when the *Chartcraft Weekly Service* was started, the book's name was changed to *The Chartcraft Method of Point & Figure Trading.* Several revised editions have been published since then to include commodities and options. In 1990, Michael Burke wrote *The All New Guide to the Three-Point Reversal Method of Point & Figure Construction and Formations* (Chartcraft, New Rochelle, NY). - -The original 1 box reversal method of plotting markets required intraday prices. The 3 box reversal was a condensation of the 1 box and was meant for intermediate trend analysis. Cohen reasoned that because so few 3 box reversals occurred in stocks during the day that it was not necessary to use intraday prices to construct the 3 box reversal chart. Hence the decision to use only the high and low prices, which were readily available in most financial newspapers. This modified technique, which is the basis of the Chartcraft service, greatly simplified point and figure charting and made it accessible to the average trader. - -### **CONSTRUCTION OF THE 3 POINT REVERSAL CHART** - -The construction of the chart is relatively simple. First, the chart must be scaled in the same way as the intraday chart. A value must be assigned to each box. These tasks are performed for subscribers to the *Chartcraft* service because the charts are already constructed and the box values assigned. The chart shows a series of alternating columns with x's representing rising prices and the o columns showing falling prices. (See Figure 11.8.) - -The actual plotting of the x's and o's requires only the high and low prices for the day. If the last column is an x column (showing rising prices), then look at the high price for the day. If the daily [high](#page-257-0) permits the filling in of 1 or more x's, then fill in those boxes and stop. That's all you do for that day. Remember that the entire value of the box must be filled. Fractions or partial filling of the box don't count. Repeat the same process the next day, looking only at the high price. As long as prices continue to rise, permitting the plotting of at least one x, continue to fill in the boxes with x's, ignoring the low price. - -The day finally comes when the daily high price is not high enough to fill the next x box. At that point, look at the low price to determine if a 3 box reversal has occurred in the other direction. If so, move one column to the right, move down one box, and fill the next 3 boxes with o's to signify a new down column. Because you are now in a down column, the next day consult the low price to see if that column of o's can be continued. If one or more o's can be filled in, then do so. Only when the daily low does not permit the filling in of any more o's do you look at the daily high to see if a 3 box reversal has occurred to the upside. If so, move 1 column to the right and begin a new x column. - -**Figure 11.8** *Source: Courtesy of Chartcraft, Inc., New Rochelle, NY.* - -#### **Chart Patterns** - -Figure 11.9 shows 16 price patterns most common to this type of point and figure chart—8 buy signals and 8 sell signals. - -Let's take a look at the patterns. Since column 2, showing signals S-1 [through](#page-259-0) S-8, is just a mirror image of column 1, we'll concentrate on the buy side. The first 2 signals, B-1 and B-2, are simple formations. All that is required for the *simple bullish buy signal* is 3 columns, with the second column of x's moving 1 box above the previous column of x's. B-2 is similar to B-1 with one minor difference—there are now 4 columns, with the bottom of the second column of o's higher than the first. B-1 shows a simple breakout through resistance. B-2 shows the same bullish breakout but with the added bullish feature of rising bottoms. B-2 is a slightly stronger pattern than B-1 - -for that reason. - -The third pattern (B-3), *breakout of a triple top*, begins the complex formations. Notice that the simple bullish buy signal is a part of each complex formation. Also, as we move down the page, these formations become increasingly stronger. The triple top breakout is stronger because there are 5 columns involved and 2 columns of x's have been penetrated. Remember that the wider the base, the greater the upside potential. The next pattern (B-4), *ascending triple top*, is stronger than B-3 because the tops and bottoms are both ascending. The *spread triple top* (B-5) is even stronger because there are 7 columns involved, and 3 columns of x's are exceeded. - -The *upside breakout above a bullish triangle* (B-6) combines two signals. First, a simple buy signal must be present. Then the upper trendline must be cleared. (We'll cover the drawing of trendlines on these charts in the next section). Signal B-7, *upside breakout above a bullish resistance line*, is self-explanatory. Again, two things must be present. A buy signal must have already been given; and the upper channel line must be completely cleared. The final pattern, the *upside breakout above a bearish resistance line* (B-8), also requires two elements. A simple buy signal must be combined with a clearing of the down trendline. Of course, everything we've said regarding patterns B-1 through B-8 applies equally to patterns S-1 through S-8 except that, in the latter case, prices are headed down instead of up. - -**Figure 11.9** *Source: K.C. Zieg, Jr., and P.J. Kaufman*, Point and Figure Commodity Trading Techniques *(New Rochelle, NY: Investors Intelligence) p. 73.* - -*There is a difference between how these patterns are applied to commodity markets as opposed to common stocks.* In general, all 16 signals can be used in stock market trading. However, because of the rapid movement so characteristic of the futures markets, the *complex* patterns are not as common in the commodity markets. Much greater emphasis is therefore placed on the *simple* signals. Many futures traders utilize the simple signals alone. If the trader chooses to wait for the more complex and stronger patterns, many profitable trading opportunities will be missed. - -## **THE DRAWING OF TRENDLINES** - -In our discussion of intraday charts, it was pointed out that trendlines were drawn in the conventional way. This is not the case on these 3 point reversal charts. Trendlines on these charts are drawn at 45 degree angles. Also, trendlines do not necessarily have to connect previous tops or bottoms. - -#### **The Basic Bullish Support Line and Bearish Resistance Line** - -These are your basic up and down trendlines. Because of the severe condensation on these charts, it would be impractical to try to connect rally tops or reaction lows. The 45 degree line is, therefore, used. In an uptrend, the *bullish support line* is drawn at a *45 degree angle* upward to the right from under the lowest column of o's. As long as prices remain above that line, the major trend is considered to be bullish. In a downtrend, the bearish resistance line is drawn at a 45 degree angle downward to the right from the top of the highest column of x's. As long as prices remain below that down trendline, the trend is bearish. (See Figures 11.10-11.12.) - -At times, those lines may have to be adjusted. For example, sometimes a correction in an uptrend breaks below the rising support line after which the uptrend resumes. In such [cases,](#page-261-0) a new s[uppor](#page-263-1)t line must be drawn at a 45 degree angle from the bottom of that reaction low. Sometimes a trend is so strong that the original up trendline is simply too far away from the price action. In that case, a tighter trendline should be drawn in an attempt to arrive at a "best fitting" support line. - -**Figure 11.10** *Examples of the Chartcraft three point reversal stock charts. Notice that the trendlines are drawn at 45 degree angles. (Source: Courtesy of Chartcraft, New Rochelle, NY.)* - -**Figure 11.11** *Two more examples of the Chartcraft 3 point reversal method of point and figure charting. Trendlines on these charts are drawn at 45 degree angles. (Source: Courtesy of Chartcraft, New Rochelle, NY.)* - -**Figure 11.12** *The box to the bottom left shows a horizontal target to 92 in British Telecomm PLC arrived at by tripling the base and adding to 50. To the right, a vertical target to 102 is arrived at by tripling the x column and adding to 63. (Source: Courtesy of Chartcraft, New Rochelle, NY.)* - -### **MEASURING TECHNIQUES** - -Three point reversal charts allow the use of two different measuring techniques—the *horizontal* and the *vertical.* For the horizontal, count the number of columns in a bottom or topping pattern. That number of columns must then be multiplied by the value of the reversal or the number of boxes needed for a reversal. For example, let's assign a \$1.00 box value to a chart with a 3 box reversal. We count the number of boxes across a base and come up with 10. Because we're using a 3 box reversal, the value of that reversal is \$3.00 (3x\$1.00). Multiply the 10 columns across the base by \$3 for a total of \$30. That number is then added to the bottom of the basing pattern or subtracted from the top of a topping pattern to arrive at the price objective. - -The *vertical* count is a bit simpler. Measure the number of boxes in the first column of the new trend. In an uptrend, measure the first up column of x's. In a downtrend, measure the first down column of o's. Multiply that value by 3 and add that total to the bottom or subtract it from the top of the column. What you're doing in effect with a 3 box reversal chart is tripling the size of the first leg. If a double top or bottom occurs on the chart, use the second column of o's or x's for the vertical count. (See Figure 11.12.) - -### **TRADING TACTICS** - -Let's look at the various ways that these point and figure charts can be used to determine specific entry and exit points. - -- 1. A simple buy signal can be used for the covering of old shorts and/or the initiation of new longs. -- 2. A simple sell signal can be used for the liquidation of old longs and/or the initiation of new shorts. -- 3. The simple signal can be used only for liquidation purposes with a complex formation needed for a new commitment. -- 4. The trendline can be used as a filter. Long positions are taken above the trendline and short positions below the trendline. -- 5. For stop protection, always risk below the last column of o's in an uptrend and over the last column of x's in a downtrend. -- 6. The actual entry point can be varied as follows: - - a. Buy the actual breakout in an uptrend. - - b. Buy a 3 box reversal after the breakout occurs to obtain a lower entry point. - - c. Buy a 3 box reversal in the direction of the original breakout after a correction occurs. Not only does this require the added confirmation of a positive reversal in the right direction, but a closer stop point can now be used under the latest column of o's. - - d. Buy a second breakout in the same direction as the original breakout signal. - -As you can readily see from the list, there are many different ways that the point and figure chart can be used. Once the basic technique is - -understood, there is almost unlimited flexibility as to how to best enter and exit a market using this approach. - -### **Adjusting Stops** - -The actual buy or sell signal occurs on the first signal. However, as the move continues, several other signals appear on the chart. These repeat buy or sell signals can be used for additional positions. Whether or not this is done, the protective stop point can be raised to just below the latest o column in an uptrend and lowered to just over the latest x column in a downtrend. This use of a *trailing stop* allows the trader to stay with the position and protect accumulated profits at the same time. - -### **What to Do After a Prolonged Move** - -Intermittent corrections against the trend allow the trader to adjust stops once the trend has resumed. How is this accomplished, however, if no 3 box reversals occur during the trend? The trader is then faced with a long column of x's in an uptrend or o's in a downtrend. This type of market situation creates what is called a *pole*, that is, a long column of x's and o's without a correction. The trader wants to stay with the trend but also wants some technique to protect profits. There is at least one way to accomplish this. After an uninterrupted move of 10 or more boxes, place a protective stop at the point where a 3 box reversal would occur. If the position does get stopped out, reentry can be done on another 3 box reversal in the direction of the original trend. In that case, an added advantage is the placement of the new stop under the most recent column of o's in an uptrend or over the latest column of x's in a downtrend. - -## **ADVANTAGES OF POINT AND FIGURE CHARTS** - -Let's briefly recap some of the advantages of point and figure charting. - -- 1. By varying the box and reversal sizes, these charts can be adapted to almost any need. There are also many different ways these charts can be used for entry and exit points. -- 2. Trading signals are more precise on point and figure charts than on bar charts. -- 3. By following these specific point and figure signals, better trading discipline can be achieved. (See Figures 11.13-11.18.) - -**Figure 11.13** *This chart of Treasury Bond futures prices covers more than two years. The arrows mark the buy and sell signals. Most of the signals captured the market trend very well. Even when a bad signal is given, the chart quickly corrects itself.* - -**Figure 11.14** *The early 1994 sell signal (first down arrow) lasted all the way through 1994. The buy signal at the start of 1995 (first up arrow) lasted for two years until 1997. A sell signal in mid-1997 turned into a buy at the start of 1998.* - -**Figure 11.15** *This chart condenses the previous dollar chart by doubling the box size. Only two signals are given on this less sensitive version. The last signal was a buy (see up arrow) in mid-1995 near 85, which has lasted for almost three years.* - -**Figure 11.16** *This point and figure chart of gold gave a sell signal (see down arrow) near \$380 during 1996. Gold prices fell another \$100 over the next two years.* - -**Figure 11.17** *The crude oil point and figure chart gave a sell signal (see down arrow) near \$20 during October 1997 and caught the subsequent \$6 tumble. Crude oil prices would have to rise above the last x column at 16.50 to reverse the downtrend.* - -**Figure 11.18** *This point and figure chart of the Semiconductor Index gave four signals over a period of two and a half years. The down arrows mark two timely sell signals in 1995 and 1997. The buy signal during 1996 (first up arrow) caught most of the ensuing rally.* - -## **P&F TECHNICAL INDICATORS** - -In his 1995 book, *Point & Figure Charting* (John Wiley & Sons), Thomas J. Dorsey espouses the Chartcraft method of 3 point reversal charting of stocks. He also discusses point and figure application to commodity and options trading. In addition to explaining how to construct and read the charts, Dorsey also shows how the P&F technique can be applied to relative strength analysis, sector analysis, and in the construction of an NYSE Bullish Percent Index. He shows how p&f charts can be constructed for the NYSE advance decline line, the NYSE High-Low Index, and the percentage of stocks over their 10 and 30 week averages. Dorsey credits Michael Burke, the publisher of Chartcraft, (Chartcraft, Inc., Investors Intelligence, 30 Church Street, New Rochelle, N.Y. 10801) with the actual development of these innovative p&f indicators which are available in that chart service. - -### **COMPUTERIZED P&F CHARTING** - -Computers have taken the drudgery out of point and figure charting. The days of laboriously constructing columns of x's and o's are gone. Most charting software packages do the charting for you. In addition, you can vary the box and reversal sizes with a keystroke to adjust the chart for shorter or longer term analysis. You can construct p&f charts from real-time (intraday) and end of day data, and you can apply them to any market you want. But you can do a lot more with a computer. - -Kenneth Tower (CMT), technical analyst for UST Securities Corporation, (5 Vaughn Drive, CN5209, Princeton, N.J. 08543) uses a logarithmic method of point and figure charting. A screening process that measures the volatility of a stock over the last 3 years determines the right percentage box size for each stock. Figures 11.19 and 11.20 show examples of Tower's logarithmic p&f charts applied to America Online and Intel. The box size for AOL in Figure 11.19 is 3.6%. A 1 box reversal, therefore, would require a retracement of 3.6%. Since that [happens](#page-270-0) to be a [2](#page-270-1) box reversal chart, prices would have to retrace 7.2% to start a new column. Each box size for the Intel chart shown in [Figur](#page-270-0)e 11.20 is worth 3.2%. - -**Figure 11.19** *A logarithmic point and figure chart of America Online. The reversal criteria is based on percentages. Each box is worth 3.6%. Since this is a two box reversal chart, a reversal is worth 7.2%. Notice the horizontal upside counts to 69.7 and 136.5 (see arcs). (Chart courtesy of UST Securities Corp.)* - -**Figure 11.20** *A one box reversal point and figure chart of Intel using percentages. A reversal of 3.2% is needed to move into the next column. Measuring horizontally from right to left along the base, upside counts can be made to 33 and then to 87.6 (see arcs). (Chart courtesy of UST Securities Corp.)* - -The arcs you see on both charts are examples of using horizontal price counts across a price base to arrive at short and long term price objectives. The Intel chart, for example, shows a short term objective to 33, arrived at by measuring halfway across the price base (lower arc). The larger arc, which measures to 87.6, is arrived at by measuring across the entire price base and - -projecting that distance upward. If you look closely at Figures 11.19 and 11.20, you'll also see price dots trailing the price action. Those dots happen to be moving averages. - -### **P&F MOVING AVERAGES** - -Moving averages are usually applied to bar charts. But here they are on point & figure charts, courtesy of Ken Tower and UST Securities. Tower uses two moving averages on his charts, a 10 column and a 20 column moving average. The dots you see in Figures 11.19 and 11.20 are 10 column averages. These moving averages are constructed by first finding an average price for each column. That is done by simply adding up the prices in each column and dividing the total by the nu[mber](#page-270-0) of x's or [o's](#page-270-1) in that column. The resulting numbers are then averaged over 10 and 20 columns. The moving averages are used in the same way as on bar charts. - -Figure 11.21 shows two point and figure charts of the same stock with 10 column averages (dots) and 20 column averages (dashes). The bottom chart is a 2.7% reversal logarithmic chart of Royal Dutch Petroleum going back to 1992. [Notice](#page-272-1) that the faster moving average stayed above the slower moving average from 1993 to the 1997 during the four year uptrend. You can see the two moving averages coming together during the second half of 1997 in what turned out to be a consolidation year for that stock. To the far right, you can see that Royal Dutch may be on the verge of resuming its major uptrend. A closer look at that potential upside breakout is seen in the upper chart in Figure 11.21. - -The upper chart is a traditional one point reversal linear chart of the same stock. The time frame covered in the linear chart is much shorter than the long [chart.](#page-272-1) But you get a closer look at the late 1997 and early 1998 price action and can see the short term upside breakout at the start of 1998. The stock still needs to close through 60 to confirm a major bullish breakout. The moving averages haven't been much help during the trading range (they never are), but should begin to trend higher once again if the bullish breakout materializes. By adding moving averages to point and figure charts, Ken Tower brings another valuable technical indicator to p&f charting. The use of logarithmic charts also adds a modern wrinkle to this old charting method. - -**Figure 11.21** *Two point and figure versions of Royal Dutch Petroleum. The bottom chart is a log chart spanning several years. The upper chart is a linear chart for one year. The dots and dashes represent 10 and 20 column moving averages, respectively. (Prepared by UST Securities Corp. Updated through March 26, 1998.)* - -### **CONCLUSION** - -Point and figure charting isn't the oldest technique in the world. That credit goes to the Japanese candlestick chart, which has been used in that country for centuries. In the next chapter Greg Morris, author of two books on candlesticks, will introduce that ancient technique that has gained new popularity in recent years among Western technical analysts. - -### **INTRODUCTION** - -While the Japanese have used this charting and analysis technique for centuries, only in recent years has it become popular in the West. The term, candlesticks, actually refers to two different, but related subjects. First, and possibly the more popular, is the method of displaying stock and futures data for chart analysis. Secondly, it is the art of identifying certain combinations of candlesticks in defined and proven combinations. Fortunately, both techniques can be used independently or in combination. - -### **CANDLESTICK CHARTING** - -Charting market data in candlestick form uses the same data available for standard bar charts; open, high, low, and close prices. While using the exact same data, candlestick charts offer a much more visually appealing chart. Information seems to jump off the page (computer screen). The information displayed is more easily interpreted and analyzed. The box below is a depiction of of a single day of prices showing the difference between the bar (left) and the candlestick(s). (See Figure 12.1.) - -#### **Figure 12.1** - -You can see how the name "candlesticks" came about. They look somewhat like a candle with a wick. The rectangle represents the difference between the open and close price for the day, and is called the *body.* Notice that the body can be either black or white. A *white body* means that the close price was greater (higher) than the open price. Actually, the body is not white, but open (not filled), which makes it work better with computers. This is so that it will print correctly when printing charts on a computer. This is one of the adaptations that have occurred in the West; the Japanese use red for the open body. The *black body* means that the close price was lower than the open price. The open and close prices are given much significance in Japanese candlesticks. The small lines above and below the body are referred to as *wicks* or *hairs* or *shadows.* Many different names for these lines appear in Japanese reference literature, which is odd since they represent the high and low prices for the day and are normally not considered vital in the analysis by the Japanese. (See Figure 12.2.) - -Figure 12.2 shows the same data in both the popular bar chart and in a Japanese candlestick format. You can quickly see that information not readily available on the bar [chart](#page-274-0) seems to jump from the page (screen) on the cand[lestick](#page-274-0) chart. Initially, it takes some getting use to, but after a while you may prefer it. - -**Figure 12.2** - -The different shapes for candlesticks have different meanings. The Japanese have defined different primary candlesticks, based upon the relationship of open, high, low, and close prices. Understanding these basic candlesticks is the beginning of candlestick analysis. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/017_12 Japanese Candlesticks.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/017_12 Japanese Candlesticks.md deleted file mode 100644 index fce0d5eef509f4305a1ec1e237c73e365a1cf2da..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/017_12 Japanese Candlesticks.md +++ /dev/null @@ -1,137 +0,0 @@ -### **BASIC CANDLESTICKS** - -Different body/shadow combinations have different meanings. Days in which the difference between the open and close prices is great are called *Long Days.* Likewise, days in which the difference between the open and close price is small, are called *Short Days.* Remember, we are only talking about the size of the body and no reference is made to the high and/or low prices. (See Figure 12.3.) - -*Spinning Tops* are days in which the candlesticks have small bodies with upper and lower shadows that are of greater length than that of the body. The body [color](#page-275-1) is relatively unimportant in spinning top candlesticks. These candlesticks are considered as days of indecision. (See Figure 12.4.) - -**Figure 12.3** - -**Figure 12.4** *Spinning tops.* - -When the open price and the close price are equal, they are called Doji lines. *Doji candlesticks* can have shadows of varying length. When referring to Doji candlesticks, there is some consideration as to whether the open and close price must be exactly equal. This is a time when the prices must be almost equal, especially when dealing with large price movements. - -There are different Doji candlesticks that are important. The Longlegged Doji has long upper and lower shadows and reflects considerable indecision on the part of market participants. The Gravestone Doji has only a long upper shadow and no lower shadow. The longer the upper shadow, the more bearish the interpretation. The Dragonfly Doji is the opposite of the Gravestone Doji, the lower shadow is long and there is no upper shadow. It is usually considered quite bullish. (See Figure 12.5.) - -**Figure 12.5** *Doji candlesticks.* - -The single candlestick lines are essential to Japanese candlestick analysis. You will find that all Japanese candle patterns are made from combinations of these basic candlesticks. - -### **CANDLE PATTERN ANALYSIS** - -A Japanese candle pattern is a psychological depiction of traders' mentality at the time. It vividly shows the actions of the traders as time unfolds in the market. The mere fact that humans react consistently during similar situations makes candle pattern analysis work. - -A Japanese candle pattern can consist of a single candlestick line or be a combination of multiple lines, normally never more than five. While most candle patterns are used to determine reversal points in the market, there are a few that are used to determine trend continuation. They are referred to as reversal and continuation patterns. Whenever a reversal pattern has bullish implications, an inversely related pattern has bearish meaning. Similarly, whenever a continuation pattern has bullish implications, an opposite pattern gives bearish meaning. When there is a pair of patterns that work in both bullish and bearish situations, they usually have the same name. In a few cases, however, the bullish pattern and its bearish counterpart have completely different names. - -#### **Reversal Patterns** - -A *reversal candle pattern* is a combination of Japanese candlesticks that normally indicate a reversal of the trend. One serious consideration that must be used to help identify patterns as being either bullish or bearish is the trend of the market preceding the pattern. You cannot have a bullish reversal pattern in an uptrend. You can have a series of candlesticks that resemble the bullish pattern, but if the trend is up, it is not a bullish Japanese candle pattern. Likewise, you cannot have a bearish reversal candle pattern in a downtrend. - -This presents one of the age-old problems when analyzing markets: What is the trend? You must determine the trend, before you can utilize Japanese candle patterns effectively. While volumes have been written on the subject of trend determination, the use of a moving average will work quite well with Japanese candle patterns. Once the short term (ten periods or so) trend has been determined, Japanese candle patterns will significantly assist in identifying the reversal of that trend. - -Japanese literature consistently refers to approximately forty reversal candle patterns. These vary from single candlestick lines to more complex patterns of up to five candlestick lines. There are many good references on candlesticks, so only a few of the more popular patterns will be discussed here. - -*Dark Cloud Cover.* This is a two day reversal pattern that only has bearish implications. (See Figure 12.6.) This is also one of the times when the pattern's counterpart exists but has a different name (see Piercing Line). The first day of this pattern is a long white candlestick. This reflects the current trend of the market and helps confirm the uptrend to traders. The next day opens above the high price of the previous day, again adding to the bullishness. However, trading for the rest of the day is lower with a close price at least below the midpoint of the body of the first day. This is a significant blow to the bullish mentality and will force many to exit the market. Since the close price is below the open price on the second day, the body is black. This is the dark cloud referred to in the name. - -*Piercing Line.* The opposite of the Dark Cloud Cover, the Piercing Line, has bullish implications. (See Figure 12.7.) The scenario is quite similar, but opposite. A downtrend is in place, the first candlestick is a long black day which solidifies traders' confidence in the downtrend. The next day, prices open at a new low and then [trade](#page-279-0) higher all day and close above the midpoint of the first candlestick's body. This offers a significant change to the downtrend mentality and many will reverse or exit their positions. - -**Figure 12.6** *Dark cloud cover -.* - -*Evening Star and Morning Star.* The Evening Star and its cousin, the Morning Star, are two powerful reversal candle patterns. These are both three day patterns that work exceptionally well. The scenario for understanding the change in trader psychology for the Evening Star will be thoroughly discussed here since the opposite can be said for the Morning Star. (See Figures 12.8 and 12.9.) - -**Figure 12.7** *Piercing line +.* - -**Figure 12.8** *Evening star -.* - -**Figure 12.9** *Morning star +.* - -The Evening Star is a bearish reversal candle pattern, as its name suggests. The first day of this pattern is a long white candlestick which fully enforces the current uptrend. On the open of the second day, prices gap up above the body of the first day. Trading on this second day is somewhat restricted and the close price is near the open price while remaining above the body of the first day. The body for the second day is small. This type of day following a long day is referred to as a Star pattern. A Star is a small body day that gaps away from a long body day. The third and last day of this pattern opens with a gap below the body of the star and closes lower with the close price below the midpoint of the first day. - -The previous explanation was the perfect scenario. Many references will accept as valid, an Evening Star which does not meet each detail exactly. For instance, the third day might not gap down or the close on the third day might not be quite below the midpoint of the first day's body. These details are subjective when viewing a candlestick chart, but not when using a computer program to automatically identify the patterns. That is because computer programs require explicit instructions to read the candle chart, and don't allow for subjective interpretation. - -#### **Continuation Patterns** - -Each trading day, a decision needs to be made, whether it is to exit a trade, enter a trade, or remain in a trade. A candle pattern that helps identify the fact that the current trend is going to continue is more valuable than may first appear. It helps answer the question as to whether or not you should remain in a trade. Japanese literature refers to 16 continuation candle patterns. One continuation pattern and its related opposite cousin are particularly good at - -trend continuation identification. - -*Rising and Falling Three Methods.* The Rising Three Methods continuation candle pattern is the bullish counterpart to this duo and will be the subject of this scenario building. A bullish continuation pattern can only occur in an uptrend and a bearish continuation pattern can only occur in a downtrend. This restates the required relationship to the trend that is so necessary in candle pattern analysis. (See Figures 12.10 and 12.11.) - -**Figure 12.10** *Rising Three Methods +.* - -**Figure 12.11** *Falling Three Methods -.* - -The first day of the Rising Three Methods pattern is a long white day which fully supports the uptrending market. However, over the course of the next three trading periods, small body days occur which, as a group, trend downward. They all remain within the range of the first day's long white body and at least two of these three small-bodied days have black bodies. This period of time when the market appears to have gone nowhere is considered by the Japanese as a "period of rest." On the fifth day of this pattern, another long white day develops which closes at a new high. Prices have finally broken out of the short trading range and the uptrend will continue. - -A five day pattern such as the Rising Three Methods requires a lot of detail in its definition. The above scenario is the perfect example of the Rising Three Methods pattern. Flexibility can be applied with some success and this only comes with experience. For example, the three small reaction days could remain within the first day's high-low range instead of the body's range. The small reaction days do not always have to be predominantly black. And finally, the concept of the "period of rest" could be expanded to include more than three reaction days. Don't ignore the Rising and Falling Three Methods pattern; it can give you a feeling of comfort when worrying about protecting profits in a trade. - -#### **Using Computers for Candle Pattern Identification** - -A personal computer with software designed to recognize candle patterns is a great way to remove emotion, especially during a trade. However, there are a couple of things to keep in mind when viewing candlesticks on a computer screen. A computer screen is made up of small light elements called pixels. There are only so many pixels on your computer screen, with the amount based upon the resolution of your video card/monitor combination. If you are viewing price data that has a large range of prices in a short period of time, you may think that you are seeing many Doji days (open and close price are equal) when in fact, you are not. With a large range of prices on the screen, each pixel element will have a price range of its own. A computer software program that identifies patterns based on a mathematical relationship will overcome this visual anomaly. Hopefully, the above explanation will keep you from thinking that your software isn't working. - -### **FILTERED CANDLE PATTERNS** - -A revolutionary concept developed by Greg Morris in 1991, called candle pattern filtering, provides a simple method to improve the overall reliability of candle patterns. While the short term trend of the market must be identified before a candle pattern can exist, determination of overbought and oversold markets using traditional technical analysis will enhance a candle pattern's predictive ability. Concurrently, this technique helps eliminate bad or premature candle patterns. - -One must first grasp how a traditional technical indicator responds to price data. In this example, Stochastics %D will be used. The stochastic indicator oscillates between 0 and 100, with 20 being oversold and 80 being overbought. The primary interpretation for this indicator is when %D rises above 80 and then falls below 80, a sell signal has been generated. Similarly, when it drops below 20 and then rises above 20, a buy signal is given. (See Chapter 10 for more on Stochastics.) - -Here is what we know about stochastics %D: When it enters the area above 80 or below 20, it will eventually generate a signal. In other words, it is just a [matte](#page-211-0)r of time until a signal is given. The area above 80 and below 20 is called the presignal area and represents the area that %D must get to before it can give a trading signal of its own. (See Figure 12.12.) - -**Figure 12.12** - -The filtered candle pattern concept uses this presignal area. Candle patterns are considered *only* when %D is in its presignal area. If a candle pattern occurs when stochastics %D is at, say 65, the pattern is ignored. Also, only reversal candle patterns are considered using this concept. - -Candle pattern filtering is not limited to using stochastics %D. Any technical oscillator that you might normally use for analysis can be used to filter candle patterns. Wilder's RSI, Lambert's CCI, and Williams' %R are a few that will work equally as well. (These oscillators are explained in Chapter 10.) - -### **[CONCLUSION](#page-211-0)** - -Japanese candlestick charting and candle pattern analysis are essential tools for making market timing decisions. One should use Japanese candle patterns in the same manner as any other technical tool or technique; that is, to study - -the psychology of market participants. Once you become used to seeing your price charts using candlesticks, you may not want to use bar charts again. Japanese candle patterns, used in conjunction with other technical indicators in the filtering concept, will almost always offer a trading signal prior to using other price-based indicators. - -### **CANDLE PATTERNS** - -The candle patterns listed below comprise the library that is used to identify candlestick signals. The number in parentheses at the end of each name represents the number of candles that are used to define that particular pattern. The bullish and bearish patterns are divided into two groups signifying either reversal or continuation patterns. - -Long White Body (1) Long Black Body (1) Hammer (1) Hanging Man (1) Inverted Hammer (1) Shooting Star (1) Belt Hold (1) Belt Hold (1) Engulfing Pattern (2) Engulfing Pattern (2) Harami (2) Harami (2) Harami Cross (2) Harami Cross (2) Piercing Line (2) Dark Cloud Cover (2) Doji Star (2) Doji Star (2) Meeting Lines (2) Meeting Lines (2) Three White Soldiers (3) Three Black Crows (3) Morning Star (3) Evening Star (3) Morning Doji Star (3) Evening Doji Star (3) Abandoned Baby (3) Abandoned Baby (3) Tri-Star (3) Tri-Star (3) Breakaway (5) Breakaway (5) Three Inside Up (3) Three Inside Down (3) Three Outside Up (3) Three Outside Down (3) Kicking (2) Kicking (2) Unique Three Rivers Bottom (3) Latter Top (5) Three Stars in the South (3) Matching High (2) Concealing Swallow (4) Upside Gap Two Crows (3) Stick Sandwich (3) Identical Three Crows (3) - -### **Bullish Reversals Bearish Reversals** - -Homing Pigeon (2) Deliberation (3) Ladder Bottom (5) Advance Block (3) Matching Low (2) Two Crows (3) - -Separating Lines (2) Separating Lines (2) Rising Three Methods (5) Falling Three Methods (5) Upside Tasuki Gap (3) Downside Tasuki Gap (3) Side by Side White Lines (3) Side by Side White Lines (3) Three Line Strike (4) Three Line Strike (4) On Neck Line (2) On Neck Line (2) In Neck Line (2) In Neck Line (2) - -### **Bullish Continuation Bearish Continuation** - -Upside Gap Three Methods (3) Downside Gap Three Methods (3) - -\*This chapter was contributed by Gregory L. Morris. - -### **HISTORICAL BACKGROUND** - -In 1938, a monograph entitled *The Wave Principle* was the first published reference to what has come to be known as the *Elliott Wave Principle.* The monograph was published by Charles J. Collins and was based on the original work presented to him by the founder of the Wave Principle, Ralph Nelson (R.N.) Elliott. - -Elliott was very much influenced by the Dow Theory, which has much in common with the Wave Principle. In a 1934 letter to Collins, Elliott mentioned that he had been a subscriber to Robert Rhea's stock market service and was familiar with Rhea's book on Dow Theory. Elliott goes on to say that the Wave Principle was "a much needed complement to the Dow Theory." - -In 1946, just two years before his death, Elliott wrote his definitive work on the Wave Principle, *Nature's Law—The Secret of the Universe.* - -Elliott's ideas might have faded from memory if A. Hamilton Bolton hadn't decided in 1953 to publish the *Elliott Wave Supplement* to the *Bank Credit Analyst*, which he did annually for 14 years, until his death in 1967. A.J. Frost took over the Elliott Supplements and collaborated with Robert Prechter in 1978 on the *Elliott Wave Principle.* Most of the diagrams in this chapter are taken from Frost and Prechter's book. Prechter went a step further and in 1980 published *The Major Works of R.N. Elliott*, making available the original Elliott writings that had long been out of print. - -### **BASIC TENETS OF THE ELLIOTT WAVE PRINCIPLE** - -There are three important aspects of wave theory—*pattern, ratio*, and *time* in that order of importance. *Pattern* refers to the wave patterns or formations that comprise the most important element of the theory. *Ratio analysis* is useful in determining retracement points and price objectives by measuring the relationships between the different waves. Finally, *time* relationships also exist and can be used to confirm the wave patterns and ratios, but are considered by some Elliotticians to be less reliable in market forecasting. - -Elliott Wave Theory was originally applied to the major stock market averages, particularly the Dow Jones Industrial Average. In its most basic form, the theory says that the stock market follows a repetitive rhythm of a five wave advance followed by a three wave decline. Figure 13.1 shows one complete cycle. If you count the waves, you will find that one complete cycle has eight waves—five up and three down. In the advancing portion of the cycle, notice that each of the five waves are numbered. [Waves](#page-295-0) 1, 3, and 5 called *impulse* waves—are rising waves, while waves 2 and 4 move against the uptrend. Waves 2 and 4 are called *corrective* waves because they correct waves 1 and 3. After the five wave numbered advance has been completed, a three wave correction begins. The three corrective waves are identified by the letters a, b, c. - -Along with the constant form of the various waves, there is the important consideration of degree. There are many different degrees of trend. Elliott, in fact, categorized nine different degrees of trend (or magnitude) ranging from a *Grand Supercycle* spanning two hundred years to a *subminuette* degree covering only a few hours. The point to remember is that the basic eight wave cycle remains constant no matter what degree of trend is being studied. - -**Figure 13.1** *The Basic Pattern. (A.J. Frost and Robert Prechter*, Elliott Wave Principle *[Gainesville, GA: New Classics Library, 1978], p. 20. Copyright © 1978 by Frost and Prechter.)* - -Each wave subdivides into waves of one lesser degree that, in turn, can also be subdivided into waves of even lesser degree. It also follows then that each wave is itself part of the wave of the next higher degree. Figure 13.2 demonstrates these relationships. The largest two waves—1 and 2—can be subdivided into eight lesser waves that, in turn, can be subdivided into 34 even lesser waves. The two largest waves—1 and 2—are only the [first](#page-296-0) two waves in an even larger five wave advance. Wave 3 of that next higher degree is about to begin. The 34 waves in Figure 13.2 are subdivided further to the next smaller degree in Figure 13.3, resulting in 144 waves. - -**Figure 13.2** *(Frost and Prechter, p. 21. Copyright © 1978 by Frost and Prechter.)* - -The numbers shown so far 1,2,3,5,8,13,21,34,55,89,144—are not just random numbers. They are part of the *Fibonacci number sequence*, which forms the mathematical basis for the Elliott Wave Theory. We'll come back to them a little later. For now, look at Figures 13.1-13.3 and notice a very significant characteristic of the waves. Whether a given wave divides into five waves or three waves is determined by the direction of the next larger wave. For example, in Figure 13.2, waves [\(1\),](#page-295-0) (3), and [\(5\)](#page-297-1) subdivide into five waves because the next larger wave of which they are part—wave 1—is an advancing wave. Because waves (2) and (4) are moving against the trend, they subdivide into [only](#page-296-0) three waves. Look more closely at corrective waves (a), (b), and (c), which comprise the larger corrective wave 2. Notice that the two declining waves—(a) and (c)—each break down into five waves. This is because they are moving in the same direction as the next larger wave 2. Wave (b) by contrast only has three waves, because it is moving against the next larger wave 2. - -**Figure 13.3** *(Frost and Prechter, p. 22. Copyright © 1978 by Frost and Prechter.)* - -Being able to determine between threes and fives is obviously of tremendous importance in the application of this approach. That information tells the analyst what to expect next. A completed five wave move, for example, usually means that only part of a larger wave has been completed and that there's more to come (unless it's a fifth of a fifth). *One of the most important rules to remember is that a correction can never take place in five waves.* In a bull market, for example, if a five wave decline is seen, this means that it is probably only the first wave of a three wave (a-b-c) decline and that there's more to come on the downside. In a bear market, a three wave advance should be followed by resumption of the downtrend. A five wave rally would warn of a more substantial move to the upside and might possibly even be the first wave of a new bull trend. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/018_13 Elliott Wave Theory.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/018_13 Elliott Wave Theory.md deleted file mode 100644 index 9d3b61f77d2aa093b2cd5e427e72ecf72b183c70..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/018_13 Elliott Wave Theory.md +++ /dev/null @@ -1,203 +0,0 @@ -### **CONNECTION BETWEEN ELLIOTT WAVE AND DOW THEORY** - -Let's take a moment here to point out the obvious connection between Elliott's idea of five advancing waves and Dow's three advancing phases of a bull market. It seems clear that Elliott's idea of three up waves, with two intervening corrections, fits nicely with the Dow Theory. While Elliott was no doubt influenced by Dow's analysis, it also seems clear that Elliott believed - -he had gone well beyond Dow's theory and had in fact improved on it. It's also interesting to note the influence of the sea on both men in the formulation of their theories. Dow compared the major, intermediate, and minor trends in the market with the tides, waves, and ripples on the ocean. Elliott referred to "ebbs and flows" in his writing and named his theory the "wave" principle. - -### **CORRECTIVE WAVES** - -So far, we've talked mainly about the impulse waves in the direction of the major trend. Let's turn our attention now to the corrective waves. In general, corrective waves are less clearly defined and, as a result, tend to be more difficult to identify and predict. One point that is clearly defined, however, is that corrective waves can never take place in five waves. Corrective waves are threes, never fives (with the exception of triangles). We're going to look at three classifications of corrective waves—zig-zags, flats, and triangles. - -### **Zig-Zags** - -A zig-zag is a three wave corrective pattern, against the major trend, which breaks down into a 5-3-5 sequence. Figures 13.4 and 13.5 show a bull market zig-zag correction, while a bear market rally is shown in Figures 13.6 and 13.7. Notice that the middle wave B falls short of the beginning of wave A and that wave C moves well beyond the [end](#page-298-1) of wave [A.](#page-299-0) - -A less common variation of the zig-zag is the double [zigzag](#page-299-1) shown in [Figu](#page-299-2)re 13.8. This variation sometimes occurs in larger corrective patterns. It is in effect two different 5-3-5 zig-zag patterns connected by an intervening a-bc pattern. - -**Figure 13.4** *Bull Market Zig-Zag (5-3-5). (Frost and Prechter, p. 36. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.5** *Bull Market Zig-Zag (5-3-5). (Frost and Prechter, p. 36. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.6** *Bear Market Zig-Zag (5-3 5). (Frost and Prechter, p. 36. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.7** *Bear Market Zig-Zag (5-3-5). (Frost and Prechter, p. 36* - -*Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.8** *Double Zig-Zag. (Frost and Prechter, p. 37. Copyright © 1978 by Frost and Prechter.)* - -**Flats** - -What distinguishes the flat correction from the zig-zag correction is that the flat follows a 3-3-5 pattern. Notice in Figures 13.10 and 13.12 that the A wave is a 3 instead of a 5. In general, the flat is more of a consolidation than a correction and is considered a sign of strength in a bull market. Figures 13.9- 13.12 show examples of normal flats. In a bull [mark](#page-301-0)et, for [exa](#page-302-0)mple, wave B rallies all the way to the top of wave A, showing greater market strength. The final wave C terminates at or just below the bottom of wave A in contrast [to](#page-301-1) a [zig-za](#page-302-0)g, which moves well under that point. - -There are two "irregular" variations of the normal *flat* correction. Figures 13.13-13.16 show the first type of variation. Notice in the bull market example (Figures 13.13 and 13.14) that the top of wave B exceeds the top of A and that wave C violates the bottom of A. - -[Anothe](#page-303-0)r [variation](#page-302-1) occurs when wave B reaches the top of A, but wave C fails to reach the [bottom](#page-302-1) of A. [Nat](#page-302-2)urally, this last pattern denotes greater market strength in a bull market. This variation is shown in Figures 13.17- 13.20 for bull and bear markets. - -**Figure 13.9** *Bull Market Flat (3-3-5), Normal Correction. (Frost and Prechter, p. 38. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.10** *Bull Market Flat (3-3-5), Normal Correction. (Frost and Prechter, p. 38. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.11** *Bear Market Flat (3-3-5), Normal Correction. (Frost and Prechter, p. 38. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.12** *Bear Market Flat (3-3-5), Normal Correction. (Frost and Prechter, p. 38. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.13** *Bull Market Flat (3-3-5), Irregular Correction. (Frost and Prechter,p. 39. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.14** *Bull Market Flat (3-3-5), Irregular Correction. (Frost and* - -*Prechter, p. 39. Copyright © 1918 by Frost and Prechter.)* - -**Figure 13.15** *Bear Market Flat (3-3-5), Irregular Correction. (Frost and Prechter, p. 39. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.16** *Bear Market Flat (3-3-5), Irregular Correction. (Frost and Prechter, p. 39. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.17** *Bull Market Flat (3-3-5), Inverted Irregular Correction. (Frost and Prechter, p. 40. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.18** *Bull Market flat (3-3-5), Inverted Irregular Correction. (Frost and Prechter, p. 40. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.19** *Bear Market Flat (3-3-5), Inverted Irregular Correction (Frost and Prechter, p. 40. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.20** *Bear Market Flat (3-3-5), Inverted Irregular Correction. (Frost and Prechter, p. 40. Copyright © 1978 by Frost and Prechter.)* - -#### **Triangles** - -Triangles usually occur in the fourth wave and precede the final move in the direction of the major trend. (They can also appear in the b wave of an a-b-c correction.) In an uptrend, therefore, it can be said that triangles are both bullish and bearish. They're bullish in the sense that they indicate resumption of the uptrend. They're bearish because they also indicate that after one more wave up, prices will probably peak. (See Figure 13.21.) - -**Figure 13.21** *Corrective Wave (Horizontal) Triangles. (Frost and Prechter, p. 43. Copyright © 1978 by Frost and Prechter.)* - -Elliott's interpretation of the triangle parallels the classical use of the pattern, but with his usual added precision. Remember from Chapter 6 that the triangle is usually a continuation pattern, which is exactly what Elliott said. Elliott's triangle is a sideways consolidation pattern that breaks down into five waves, each wave in turn having three waves of its [own.](#page-129-0) Elliott also classifies four different kinds of triangles—*ascending, descending, symmetrical*, and *expanding*—all of which were seen in Chapter 6. Figure 13.21 shows the four varieties in both uptrends and downtrends. - -Because chart patterns in commodity futures contracts sometimes don't form as fully as they do in the stock market, it is not unusual for triangles in the futures markets to have only three waves instead of five. (Remember, however, that the minimum requirement for a triangle is still four points—two upper and two lower—to allow the drawing of two converging trendlines.) Elliott Wave Theory also holds that the fifth and last wave within the triangle sometimes breaks its trendline, giving a false signal, before beginning its "thrust" in the original direction. - -Elliott's measurement for the fifth and final wave after completion of the triangle is essentially the same as in classical charting—that is, the market is expected to move the distance that matches the widest part of the triangle (its height). There is another point worth noting here concerning the timing of the final top or bottom. According to Prechter, the apex of the triangle (the point where the two converging trendlines meet) often marks the timing for the completion of the final fifth wave. - -### **THE RULE OF ALTERNATION** - -In its more general application, this rule or principle holds that the market usually doesn't act the same way two times in a row. If a certain type of top or bottom occurred the last time around, it will probably not do so again this time. The rule of alternation doesn't tell us exactly what will happen, but tells us what probably won't. In its more specific application, it is most generally used to tell us what type of corrective pattern to expect. Corrective patterns tend to alternate. In other words, if corrective wave 2 was a simple a-b-c pattern, wave 4 will probably be a complex pattern, such as a triangle. Conversely, if wave 2 is complex, wave 4 will probably be simple. Figure 13.22 gives some examples. - -### **CHANNELING** - -Another important aspect of wave theory is the use of *price channels.* You'll recall that we covered trend channeling in Chapter 4. Elliott used price channels as a method of arriving at price objectives and also to help confirm the completion of wave counts. Once an uptrend has been established, an initial trend channel is constructed by dra[wing](#page-61-0) a basic up trendline along the bottoms of waves 1 and 2. A parallel channel line is then drawn over the top of wave 1 as shown in Figure 13.23. The entire uptrend will often stay within those two boundaries. - -If wave 3 begins to accelerate to the point that it exceeds the upper channel line, the lines [have](#page-307-1) to be redrawn along the top of wave 1 and the bottom of wave 2 as shown in Figure 13.23. The final channel is drawn under the two corrective waves—2 and 4—and usually above the top of wave 3 as shown in Figure 13.24. If wave 3 is unusually strong, or an extended wave, the upper line may have to be [drawn](#page-307-1) over the top of wave 1. The fifth wave should come close to the upper channel line before terminating. For the drawing of [channel](#page-308-1) lines on long term trends, it's recommended that semilog charts be employed along with arithmetic charts. - -**Figure 13.23** *Old and New Channels. (Frost and Prechter, p. 62. Copyright © 1978 by Frost and Prechter.)* - -**Figure 13.24** *Final Channel. (Frost and Prechter, p. 63. Copyright © 1978 by Frost and Prechter.)* - -### **WAVE 4 AS A SUPPORT AREA** - -In concluding our discussion of wave formations and guidelines, one important point remains to be mentioned, and that is the significance of wave 4 as a support area in subsequent bear markets. Once five up waves have been completed and a bear trend has begun, that bear market will usually not move below the previous fourth wave of one lesser degree; that is, the last fourth wave that was formed during the previous bull advance. There are exceptions to that rule, but usually the bottom of the fourth wave contains the bear market. This piece of information can prove very useful in arriving at a maximum downside price objective. - -### **FIBONACCI NUMBERS AS THE BASIS OF THE WAVE PRINCIPLE** - -Elliott stated in *Nature's Law* that the mathematical basis for his Wave Principle was a number sequence discovered by Leonardo Fibonacci in the thirteenth century. That number sequence has become identified with its discoverer and is commonly referred to as the *Fibonacci numbers.* The number sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, and so on to - -infinity. - -The sequence has a number of interesting properties, not the least of which is an almost constant relationship between the numbers. - -- 1. The sum of any two consecutive numbers equals the next higher number. For example, 3 and 5 equals 8, 5 and 8 equals 13, and so on. -- 2. The ratio of any number to its next higher number approaches .618, after the first four numbers. For example, 1/1 is 1.00, 1/2 is .50, 2/3 is .67, 3/5 is .60, 5/8 is .625, 8/13 is .615, 13/21 is .619, and so on. Notice how these early ratio values fluctuate above and below .618 in narrowing amplitude. Also, notice the values of 1.00, .50, .67. We'll comment further on these values when we talk more about ratio analysis and percentage retracements. -- 3. The ratio of any number to its next lower number is approximately 1.618, or the inverse of .618. For example, 13/8 is 1.625, 21/13 is 1.615, 34/21 is 1.619. The higher the numbers become, the closer they come to the values of .618 and 1.618. -- 4. The ratios of alternate numbers approach 2.618 or its inverse, .382. For example, 13/34 is .382, 34/13 is 2.615. - -### **FIBONACCI RATIOS AND RETRACEMENTS** - -It was already stated that wave theory is comprised of three aspects—wave form, ratio, and time. We've already discussed wave form, which is the most important of the three. Let's talk now about the application of the *Fibonacci ratios and retracements.* These relationships can apply to both price and time, although the former is considered to be the more reliable. We'll come back later to the aspect of time. - -First of all, a glance back at Figures 13.1 and 13.3 shows that the basic wave form always breaks down into Fibonacci numbers. One complete cycle comprises eight waves, five up and three down—all Fibonacci numbers. Two further subdivisions will produce 34 and [144](#page-295-0) wav[es—a](#page-297-1)lso Fibonacci numbers. The mathematical basis of the wave theory on the Fibonacci sequence, however, goes beyond just wave counting. There's also the question of proportional relationships between the different waves. The following are among the most commonly used Fibonacci ratios: - -- 1. One of the three impulse waves sometimes extends. The other two are equal in time and magnitude. If wave 5 extends, waves 1 and 3 should be about equal. If wave 3 extends, waves 1 and 5 tend toward equality. -- 2. A minimum target for the top of wave 3 can be obtained by multiplying - -the length of wave 1 by 1.618 and adding that total to the bottom of 2. - -- 3. The top of wave 5 can be approximated by multiplying wave 1 by 3.236 (2×1.618) and adding that value to the top or bottom of wave 1 for maximum and minimum targets. -- 4. Where waves 1 and 3 are about equal, and wave 5 is expected to extend, a price objective can be obtained by measuring the distance from the bottom of wave 1 to the top of wave 3, multiplying by 1.618, and adding the result to the bottom of 4. -- 5. For corrective waves, in a normal 5-3-5 zig-zag correction, wave c is often about equal to the length of wave a. -- 6. Another way to measure the possible length of wave c is to multiply .618 by the length of wave a and subtract that result from the bottom of wave a. -- 7. In the case of a flat 3-3-5 correction, where the b wave reaches or exceeds the top of wave a, wave c will be about 1.618 the length of a. -- 8. In a symmetrical triangle, each successive wave is related to its previous wave by about .618. - -#### **Fibonacci Percentage Retracements** - -The preceding ratios help to determine price objectives in both impulse and corrective waves. Another way to determine price objectives is by the use of *percentage retracements.* The most commonly used numbers in retracement analysis are 61.8% (usually rounded off to 62%), 38%, and 50%. Remember from Chapter 4 that markets usually retrace previous moves by certain predictable percentages—the best known ones being 33%, 50%, and 67%. The Fibonacci sequence refines those numbers a bit further. In a strong trend, a mi[nimum](#page-61-0) retracement is usually around 38%. In a weaker trend, the maximum percentage retracement is usually 62%. (See Figures 13.25 and 13.26.) - -**Figure 13.25** *The three horizontal lines show Fibonnaci retracement levels of 38%, 50%, and 62% measured from the 1981 bottom to the 1993 peak in Treasury Bonds. The 1994 correction in bond prices stopped right at the 38% retracement line.* - -**Figure 13.26** *The three Fibonacci percentage lines are measured from the 1994 bottom in bond prices to the early 1996 top. Bond prices corrected to the 62% line.* - -It was pointed out earlier, that the Fibonacci ratios approach .618 only after the first four numbers. The first three ratios are 1/1 (100%), 1/2 (50%), and 2/3 (66%). Many students of Elliott may be unaware that the famous 50% retracement is actually a Fibonacci ratio, as is the two-thirds retracement. A complete retracement (100%) of a previous bull or bear market also should mark an important support or resistance area. - -### **FIBONACCI TIME TARGETS** - -We haven't said too much about the aspect of time in wave analysis. Fibonacci time relationships exist. It's just that they're harder to predict and are considered by some Elliotticians to be the least important of the three aspects of the theory. Fibonacci time targets are found by counting forward from significant tops and bottoms. On a daily chart, the analyst counts forward the number of trading days from an important turning point with the expectation that future tops or bottoms will occur on Fibonacci days—that is, on the 13th, 21st, 34th, 55th, or 89th trading day in the future. The same technique can be used on weekly, monthly, or even yearly charts. On the weekly chart, the analyst picks a significant top or bottom and looks for weekly time targets that fall on Fibonacci numbers. (See Figures 13.27 and 13.28.) - -### **[COM](#page-313-0)BINING ALL THREE ASPECTS OF WAVE THEORY** - -The ideal situation occurs when wave form, ratio analysis, and time targets come together. Suppose that a study of waves reveals that a fifth wave has been completed, that wave 5 has gone 1.618 times the distance from the bottom of wave 1 to the top of wave 3, and that the time from the beginning of the trend has been 13 weeks from a previous low and 34 weeks from a previous top. Suppose further that the fifth wave has lasted 21 days. Odds would be pretty good that an important top was near. - -**Figure 13.27** *Fibonacci time targets measured in months from the 1981 bottom in Treasury Bonds. It may be coincidence, but the last four Fibonacci time targets (vertical bars) coincided with important turns in bond prices.* - -**Figure 13.28** *Fibonacci time targets in months from the 1982 bottom in the Dow. The last three vertical bars coincide with bear market years in stocks— 1987, 1990, and 1994. The 1987 peak was 13 years from the 1982 bottom—a Fibonacci number.* - -A study of price charts in both stocks and futures markets reveals a number of Fibonacci time relationships. Part of the problem, however, is the variety of possible relationships. Fibonacci time targets can be taken from top to top, top to bottom, bottom to bottom, and bottom to top. These relationships can always be found after the fact. It's not always clear which of the possible relationships are relevant to the current trend. - -### **ELLIOTT WAVE APPLIED TO STOCKS VERSUS COMMODITIES** - -There are some differences in applying wave theory to stocks and commodities. For example, wave 3 tends to extend in stocks and wave 5 in commodities. The unbreakable rule that wave 4 can never overlap wave 1 in stocks is not as rigid in commodities. (Intraday penetrations can occur on futures charts.) Sometimes charts of the cash market in commodities give a clearer Elliott pattern than the futures market. The use of continuation charts in commodity futures markets also produces distortions that may affect long term Elliott patterns. - -Possibly the most significant difference between the two areas is that major bull markets in commodities can be "contained," meaning that bull market highs do not always exceed previous bull market highs. It is possible in commodity markets for a completed five wave bull trend to fall short of a previous bull market high. The major tops formed in many commodity markets in the 1980 to 1981 period failed to exceed major tops formed seven and eight years earlier. As a final comparison between the two areas, it appears that the best Elliott patterns in commodity markets arise from breakouts from long term extended bases. - -It is important to keep in mind that wave theory was originally meant to be applied to the stock market averages. It doesn't work as well in individual common stocks. It's quite possible that it doesn't work that well in some of the more thinly traded futures markets as well because mass psychology is one of the important foundations on which the theory rests. Gold, as an illustration, is an excellent vehicle for wave analysis because of its wide following. - -### **SUMMARY AND CONCLUSIONS** - -Let's briefly summarize the more important elements of wave theory and then try to put it into proper perspective. - -- 1. A complete bull market cycle is made up of eight waves, five up waves followed by three down waves. -- 2. A trend divides into five waves in the direction of the next longer trend. - -- 3. Corrections always take place in three waves. -- 4. The two types of simple corrections are zig-zags (5-3-5) and flats (3-3- 5). -- 5. Triangles are usually fourth waves, and always precede the final wave. Triangles can also be B corrective waves. -- 6. Waves can be expanded into longer waves and subdivided into shorter waves. -- 7. Sometimes one of the impulse waves extends. The other two should then be equal in time and magnitude. -- 8. The Fibonacci sequence is the mathematical basis of the Elliott Wave Theory. -- 9. The number of waves follows the Fibonacci sequence. -- 10. Fibonacci ratios and retracements are used to determine price objectives. The most common retracements are 62%, 50%, and 38%. -- 11. The rule of alternation warns not to expect the same thing twice in succession. -- 12. Bear markets should not fall below the bottom of the previous fourth wave. -- 13. Wave 4 should not overlap wave 1 (not as rigid in futures). -- 14. The Elliott Wave Theory is comprised of wave forms, ratios, and time, in that order of importance. -- 15. The theory was originally applied to stock market averages and does not work as well on individual stocks. -- 16. The theory works best in those commodity markets with the largest public following, such as gold. -- 17. The principal difference in commodities is the existence of contained bull markets. - -The Elliott Wave Principle builds on the more classical approaches, such as Dow Theory and traditional chart patterns. Most of those price patterns can be explained as part of the Elliott Wave structure. It builds on the concept of "swing objectives" by using Fibonacci ratio projections and percentage retracements. The Elliott Wave Principle takes all of these factors into consideration, but goes beyond them by giving them more order and increased predictability. - -### **Wave Theory Should Be Used in Conjunction with Other Technical Tools** - -There are times when Elliott pictures are clear and other times when they are not. Trying to force unclear market action into an Elliott format, and ignoring other technical tools in the process, is a misuse of the theory. The key is to view Elliott Wave Theory as a partial answer to the puzzle of market - -forecasting. Using it in conjunction with all of the other technical theories in this book will increase its value and improve your chances for success. - -### **REFERENCE MATERIAL** - -Two of the best sources of information on Elliott Wave Theory and the Fibonacci numbers are *The Major Works of R.N. Elliott*, (Prechter, Jr.) and the *Elliott Wave Principle* (Frost and Prechter). All of the diagrams used in Figures 13.1-13.24 are from the *Elliott Wave Principle* and are reproduced in this chapter through the courtesy of New Classics Library. - -A primer booklet on the Fibonacci numbers, *Understanding Fibonacci [Numbers](#page-295-0)* by [Edwar](#page-308-1)d D. Dobson, is available from Traders Press (P.O. Box 6206, Greenville, S.C. 29606 (800-927-8222). - - - -### **INTRODUCTION** - -Our main focus up to this point has been on price movement, and not too much has been said about the importance of *time* in solving the forecasting puzzle. The question of time has been present by implication throughout our entire coverage of technical analysis, but has generally been relegated to secondary consideration. In this chapter, we're going to view the problem of forecasting through the eyes of cyclic analysts who believe that *time cycles* hold the ultimate key to understanding why markets move up or down. In the process, we're going to add the important dimension of time to our growing list of analytical tools. Instead of just asking ourselves *which way* and *how far* a market will go, we'll start asking *when* it will arrive there or even *when* the move will begin. - -Consider the standard daily bar chart. The vertical axis gives the price scale. But that's only half of the relevant data. The horizontal scale gives the time horizon. Therefore, the bar chart is really a time and price chart. Yet, many traders concentrate solely on price data to the exclusion of time considerations. When we study chart patterns, we're aware that there is a relationship between the amount of time it takes for those patterns to form and the potential for subsequent market moves. The longer a trendline or a support or resistance level remains in effect, the more valid it becomes. Moving averages require input as to the proper time period to use. Even oscillators require some decision as to how many days to measure. In the previous chapter, we considered the usefulness of Fibonacci time targets. - -It seems clear then that all phases of technical analysis depend to some extent on time considerations. Yet those considerations are not really applied in a consistent and dependable manner. That's where time cycles come into play. Instead of playing a secondary or supporting role in market movement, - -cyclic analysts hold that time cycles are the determining factor in bull and bear markets. Not only is time the dominant factor, but all other technical tools can be improved by incorporating cycles. Moving averages and oscillators, for example, can be optimized by tying them to dominant cycles. Trendline analysis can be made more precise with cyclic analysis by determining which are valid trendlines and which are not. Price pattern analysis can be enhanced if combined with cyclic peaks and troughs. By the use of "time windows," price movement can be filtered in such a way that extraneous action can be ignored and primary emphasis placed only on such times when important cycle tops and bottoms are due to occur. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/019_14 Time Cycles.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/019_14 Time Cycles.md deleted file mode 100644 index dacd7092a4737c4b79d997b7ec86fd75f3ef6ece..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/019_14 Time Cycles.md +++ /dev/null @@ -1,235 +0,0 @@ -### **CYCLES** - -The most intriguing book I've ever read on the subject of cycles was written by Edward R. Dewey, one of the pioneers of cyclic analysis, with Og Mandino entitled *Cycles: The Mysterious Forces That Trigger Events.* Thousands of seemingly unrelated cycles were isolated spanning hundreds and, in some cases, thousands of years. Everything from the 9.6 year cycle in Atlantic salmon abundance to the 22.20 year cycle in international battles from 1415 to 1930 was tracked. An average cycle of sunspot activity since 1527 was found to be 11.11 years. Several economic cycles, including the 18.33 year cycle in real estate activity and a 9.2 year stock market cycle, were presented. (See Figures 14.1 and 14.2.) - -**Figure 14.1** *The 22.2 year cycle of incidence of sunspots. Drought often follows two years after the sunspot minima which last occurred in the early 1970s, and is due again in the mid 1990s. In the chart, the dotted line is the "ideal" cycle, and the solid line is the actual detrended data. (Courtesy of the Foundation for the Study of Cycles, Wayne, PA.)* - -**Figure 14.2** *The 22.2 year cycle in international battles was due to top in 1982. In the chart, the dotted line is the "ideal" cycle, and the solid line is the actual detrended data. (Courtesy of the Foundation for the Study of Cycles, Wayne, PA.* - -Two startling conclusions are discussed by Dewey. First, that many of the cycles of seemingly unrelated phenomena clustered around similar periods. On p. 188 of his book, Dewey listed 37 different examples of the 9.6 year cycle, including caterpillar abundance in New Jersey, coyote abundance in Canada, wheat acreage in the U.S., and cotton prices in the U.S. Why should such unrelated activities show the same cycles? - -The second discovery was that these similar cycles acted in synchrony, that is, they turned at the same time. Figure 14.3 shows 12 different examples of the 18.2 year cycle including marriages, immigration, and stock prices in the U.S. Dewey's startling conclusion was that something "out there" in the universe must be causing these cycles; that [there](#page-320-0) seemed to be a sort of *pulse* to the universe that accounted for the pervasive presence of these cycles throughout so many areas of human existence. - -**Figure 14.3** *The 18.2 year cycles on parade. (Source: Dewey, Edward R.*, Cycles: The Mysterious Forces That Trigger Events *(New York: Manor Books, 1973.)* - -In 1941, Dewey organized the Foundation for the Study of Cycles (900 W. Valley Rd., Suite 502, Wayne, PA 19087). It is the oldest organization engaged in cycles research and the recognized leader in the field. The Foundation publishes *Cycles* magazine, which presents research in many different areas including economics and business. It also publishes a monthly report, *Cycle Projections*, which applies cyclical analysis to stocks, commodities, real estate and the economy. - -#### **Basic Cyclic Concepts** - -In 1970, J.M. Hurst authored *The Profit Magic of Stock Transaction Timing.* Although it deals mainly with stock market cycles, this book represents one of the best explanations of cycle theory available in print, and is highly recommended reading. The following diagrams are derived from Hurst's original work. - -First, let's see what a cycle looks like and discuss its three main - -characteristics. Figure 14.4 shows two repetitions of a price cycle. The cycle bottoms are called *troughs* and the tops referred to as *crests.* Notice that the two waves shown here are measured from trough to trough. *Cyclic analysts prefer to measure [cycle](#page-321-0) lengths from low to low.* Measurements can be taken between crests, but they are not considered to be as stable or reliable as those taken between the troughs. Therefore, common practice is to measure the beginning and end of a cyclic wave at a low point, as shown in this example. - -The three qualities of a cycle are *amplitude, period*, and *phase.* Amplitude measures the height of the wave as shown in Figure 14.5, and is expressed in dollars, cents, or points. The period of a wave, as shown in Figure 14.6, is the time between troughs. In this example, the period is 20 days. The phase is a measure of the time location of a wave [trough.](#page-322-0) In Figure 14.7, the phase difference between two waves is shown. Because there are [several](#page-322-1) different cycles occurring at the same time, *phasing* allows the cyclic analyst to study the [relationships](#page-323-0) between the different cycle lengths. Phasing is also used to identify the date of the last cycle low. If, for example, a 20 day cycle bottomed 10 days earlier, the date of the next cycle low can be determined. Once the amplitude, period, and phase of a cycle are known, the cycle can theoretically be extrapolated into the future. Assuming the cycle remains fairly constant, it can then be used to estimate future peaks and troughs. That is the basis of the cyclic approach in its simplest form. - -**Figure 14.4** *Two cycles of a price wave. A simple, single price wave of the* - -*kind that combines to form stock and commodity price action. Only two cycles of this wave are shown, but the wave itself extends infinitely far to the left and to the right. Such waves repeat themselves cycle after cycle. As a result, once the wave is identified, its value can be determined at any past or future time. It is this characteristic of waves that provides a degree of predictability for equity price action.* - -**Figure 14.5** *The amplitude of a wave. In this figure, the wave has an amplitude of ten dollars (from minus five dollars to plus five dollars). Amplitude is always measured from wave trough to wave crest.* - -**Figure 14.6** *The period of a wave. In this figure, the wave has a period of 20 days, which is shown measured between two consecutive wave troughs. The period could just as well have been measured between wave crests. But in the case of price waves, the wave troughs are usually more clearly defined than the wave crests for reasons that will be discussed later. Consequently, price wave periods are most often measured from trough to trough.* - -**Figure 14.7** *The phase difference between two waves. The phase difference between the two waves shown is 6 days. This phase difference is measured between the troughs of the two waves because, again, wave troughs are the most convenient points to identify in the case of price waves.* - -#### **Cyclic Principles** - -Let's take a look now at some of the principles that underlie the cyclic philosophy. The four most important ones are the Principles of Summation, Harmonicity, Synchronicity, and Proportionality. - -The Principle of Summation holds that all price movement is the simple addition of all active cycles. Figure 14.8 demonstrates how the price pattern on the top is formed by simply adding together the two different cycles at the bottom of the chart. Notice, in particular, the appearance of the double top in composite wave C. Cycle theory [holds](#page-324-0) that all price patterns are formed by the interaction of two or more different cycles. We'll come back to this point again. The Principle of Summation gives us an important insight into the rationale of cyclic forecasting. Let's assume that all price action is just the sum of different cycle lengths. Assume further that each of those individual cycles could be isolated and measured. Assume also that each of those cycles will continue to fluctuate into the future. Then by simply continuing each cycle into the future and summing them back together again, the future price trend should be the result. Or, so the theory goes. - -**Figure 14.8** *The summation of two waves. The dotted lines show how, at each point in time, the value of wave A is added to the value of wave B to produce the value of composite wave C.* - -The Principle of Harmonicity simply means that neighboring waves are usually related by a small, whole number. That number is usually *two.* For example, if a 20 day cycle exists, the next shorter cycle will usually be half its length, or 10 days. The next longer cycle would then be 40 days. If you'll remember back to the discussion on the 4 *week rule* (Chapter 9), the principle of harmonics was invoked to explain the validity of using a shorter 2 week rule and a longer 8 weeks. - -The Principle of Synchronicity refers to the strong [tendenc](#page-186-0)y for waves of differing lengths to bottom at about the same time. Figure 14.9 is meant to show both harmonicity and synchronicity. Wave B at the bottom of the chart is half the length of wave A. Wave A includes two repetitions of the smaller wave B, showing harmonicity between the two waves. [Notice](#page-326-0) also that when wave A bottoms, wave B tends to do the same, demonstrating synchronicity between the two. Synchronicity also means that similar cycle lengths of different markets will tend to turn together. - -The Principle of Proportionality describes the relationship between cycle period and amplitude. Cycles with longer periods (lengths) should have proportionally wider amplitudes. The amplitude, or height, of a 40 day cycle, for example, should be about double that of a 20 day cycle. - -#### **The Principles of Variation and Nominality** - -There are two other cyclic principles that describe cycle behavior in a more general sense—*The Principles of Variation and Nominality.* - -The Principle of Variation, as the name implies, is a recognition of the fact that all of the other cyclic principles already mentioned—summation, harmonicity, synchronicity, and proportionality—are just strong tendencies and not hard and fast rules. Some "variation" can and usually does occur in the real world. - -The Principle of Nominality is based on the premise that, despite the differences that exist in the various markets and allowing for some variation in the implementing of cyclic principles, there seems to be a nominal set of harmonically related cycles that affect all markets. And that nominal model of cycle lengths can be used as a starting point in the analysis of any market. Figure 14.10 shows a simplified version of that nominal model. The model begins with an 18 year cycle and proceeds to each successively lower cycle *half its* length. The only exception is the relationship between 54 and 18 [months](#page-326-1) which is a *third* instead of a *half.* - -When we discuss the various cycle lengths in the individual markets, we'll see that this nominal model does account for most cyclic activity. For now, look at the "Days" column. Notice 40, 20, 10, and 5 days. You'll recognize immediately that these numbers account for most of the popular moving average lengths. Even the well known 4, 9, and 18 day moving average technique is a variation of the 5, 10, and 20 day numbers. Many oscillators use 5, 10, and 20 days. Weekly rule breakouts use the same numbers translated into 2, 4, and 8 weeks. - -**Figure 14.9** *Harmonicity and synchronicity.* - - - -| Days | Weeks | Months | Years | -|------|-----------------|--------|-------| -| | | | 18 | -| | | | 9 | -| | | 54 | | -| | | 18 | | -| | 40 | | | -| | $\overline{20}$ | | | -| 80 | | | | -| 40 | | | | -| 20 | | | | -| 10 | | | | -| | | | | - -### **HOW CYCLIC CONCEPTS HELP EXPLAIN CHARTING TECHNIQUES** - -Chapter 3 in Hurst's book explains in great detail how the standard charting techniques—trendlines and channels, chart patterns, and moving averages can be better understood and used to greater advantage when coordinated with [cyclic](#page-50-0) principles. Figure 14.11 helps explain the existence of trendlines and channels. The flat cycle wave along the bottom becomes a rising price channel when it is summed with a rising line representing the long term uptrend. Notice how [much](#page-328-0) the horizontal cycle along the bottom of the chart resembles an oscillator. - -Figure 14.12 from the same chapter shows how a *head and shoulders* topping pattern is formed by combining two cycle lengths with a rising line representing the sum of all longer duration components. Hurst goes on to explain [double](#page-329-0) tops, triangles, flags, and pennants through the application of cycles. The "V" top or bottom, for example, occurs when an intermediate cycle turns at the exact same time as its next longer and next shorter duration cycles. - -**Figure 14.11** *Channel formation. (Source: Hurst, J.M.*, The Profit Magic of Stock Transaction Timing *[Englewood Cliffs, N.J.: Prentice-Hall, Inc., 1970].)* - -**Figure 14.12a** *Adding another component. (Source: Hurst, J.M.*, The Profit Magic of Stock Transaction Timing *[Englewood Cliffs, N.J.: Prentice-Hall, Inc., 1970].)* - -**Figure 14.12b** *The Summation principle applied. (Source: Hurst, J.M.*, The Profit Magic of Stock Transaction Timing *[Englewood Cliffs, N.J.: Prentice Hall. Inc., 19701].)* - -Hurst also addresses how moving averages can be made more useful if their lengths are synchronized with dominant cycle lengths. Students of traditional charting techniques should gain additional insight into how these popular chart pictures form and maybe even why they work by reading - -Hurst's chapter, entitled "Verify Your Chart Patterns." - -### **DOMINANT CYCLES** - -There are many different cycles affecting the financial markets. The only ones of real value for forecasting purposes are the *dominant cycles.* Dominant cycles are those that consistently affect prices and that can be clearly identified. Most futures markets have at least five dominant cycles. In an earlier chapter on the use of long term charts, it was stressed that all technical analysis should begin with the long term picture, gradually working toward the shorter term. That principle holds true in the study of cycles. The proper procedure is to begin the analysis with a study of long term dominant cycles, which can span several years; then work toward the intermediate, which can be several weeks to several months; finally, the very short term cycles, from several hours to several days, can be used for timing of entry and exit points and to help confirm the turning points of the longer cycles. - -**Figure 14.13** *(Source: The Power of Oscillator/Cycle Combinations by Walt* - -### *Bressert.)* - -#### **Classification of Cycles** - -The general categories are: *long term cycles* (2 or more years in length), the *seasonal cycle* (1 year), the *primary or intermediate cycle* (9 to 26 weeks), and the *trading cycle* (4 weeks). The trading cycle breaks down into two shorter *alpha* and *beta* cycles, which average 2 weeks each. (The labels Primary, Trading, Alpha, and Beta are used by Walt Bressert to describe the various cycle lengths.) (See Figure 14.13.) - -#### **The Kondratieff Wave** - -There are even longer range [cycles](#page-330-1) at work. Perhaps the best known is the approximate 54 year Kondratieff cycle. This controversial long cycle of economic activity, first discovered by a Russian economist in the 1920s by the name of Nikolai D. Kondratieff, appears to exert a major influence on virtually all stock and commodity prices. In particular, a 54 year cycle has been identified in interest rates, copper, cotton, wheat, stocks, and wholesale commodity prices. Kondratieff tracked his "long wave" from 1789 using such factors as commodity prices, pig iron production, and wages of agricultural workers in England. (See Figure 14.14.) The Kondratieff cycle has become a popular subject of discussion in recent years, primarily owing to the fact that its last top occurred in the 1920s, and its next top is long overdue. Kondratieff himself paid a heavy price for his [cycli](#page-332-1)c view of capitalistic economies. He is believed to have died in a Siberian labor camp. For more information, see *The Long Wave Cycle* (Kondratieff), translated by Guy Daniels. (Two other books on the subject are *The K Wave* by David Knox Barker and *The Great Cycle* by Dick Stoken.) - -**Figure 14.14** *Kondratieff's long wave. For more information, see* The Long Wave Cycle *by Nikolai Kondratieff, translated by Guy Daniels (New York: Richardson and Snyder, 1984). That translation is the first ever from the original Russian text. (Copyright ©1984 by The New York Times Company. Reprinted by permission [May 27, 1984, p. F11.])* - -### **COMBINING CYCLE LENGTHS** - -As a general rule, long term and seasonal cycles determine the major trend of a market. Obviously, if a two year cycle has bottomed, it can be expected to advance for at least a year, measured from its trough to its crest. Therefore, the long term cycle exerts major influence on market direction. Markets also have annual seasonal patterns, meaning that they tend to peak or trough at certain times of the year. Grain markets, for example, usually hit their low point around harvest time and rally from there. Seasonal moves usually last for several months. - -For trading purposes, *the weekly primary cycle is the most useful.* The 3 to 6 month primary cycle is the equivalent of the intermediate trend, and generally determines which side of a market to trade. The next shorter cycle, the 4 week trading cycle, is used to establish entry and exit points in the direction of the primary trend. If the primary trend is up, troughs in the trading cycle are used for purchases. If the primary trend is down, crests in the trading cycles should be sold short. The 10 day *alpha* and *beta* cycles can be used for further fine tuning. (See Figure 14.13.) - -### **THE IMPORTANCE OF TREND** - -The concept of trading in the direction of the trend is stressed throughout the body of technical analysis. In an earlier chapter, it was suggested that short term dips should be used for purchases if the intermediate trend was up, and that short term bulges be sold in downtrends. In the chapter on Elliott Wave Theory, it was pointed out that five wave moves only take place in the direction of the next larger trend. Therefore, it is necessary when using any short term trend for timing purposes to first determine the direction of the next longer trend and then trade in the direction of that longer trend. That concept holds true in cycles. *The trend of each cycle is determined by the direction of its next longer cycle.* Or stated the other way, once the trend of a longer cycle is established, the trend of the next shorter cycle is known. - -#### **The 28 Day Trading Cycle in Commodities** - -There is one important short term cycle that tends to influence most commodity markets—*the 28 day trading cycle.* In other words, most markets have a tendency to form a trading cycle low every 4 weeks. One possible explanation for this strong cyclic tendency throughout all commodity markets is the *lunar cycle.* Burton Pugh studied the 28 day cycle in the wheat market in the 1930s *(Science and Secrets of Wheat Trading*, Lambert-Gann, Pomeroy, WA, 1978, orig., 1933) and concluded that the moon had some influence on market turning points. His theory was that wheat should be bought on a full moon and sold on a new moon. Pugh acknowledged, however, that the lunar effects were mild and could be overridden by the effects of longer cycles or important news events. - -Whether or not the moon has anything to do with it, the average 28 day cycle does exist and explains many of the numbers used in the development of shorter term indicators and trading systems. First of all, the 28 day cycle is based on calendar days. Translated into actual trading days, the number becomes 20. We've already commented on how many popular moving averages, oscillators, and weekly rules are based on the number 20 and its harmonically related shorter cycles, 10 and 5. The 5, 10, and 20 day moving averages are widely used along with their derivatives, 4, 9, and 18. Many traders use 10 and 40 day moving averages, with the number 40 being the next harmonically related longer cycle at twice the length of 20. - -In Chapter 9, we discussed the profitability of the 4 week rule developed by Richard Donchian. Buy signals were generated when a market set new 4 week highs and a sell signal when a 4 week low was established. Knowledge of the e[xistence](#page-186-0) of a 4 week trading cycle gives a better insight into the significance of that number and helps us to understand why the 4 week rule has worked so well over the years. When a market exceeds the high of the - -previous 4 weeks, cycle logic tells us that, at the very least, the next longer cycle (the 8 week cycle) has bottomed and turned up. - -### **LEFT AND RIGHT TRANSLATION** - -The concept of translation may very well be the most useful aspect of cycle analysis. Left and right translation refers to the shifting of the cycle peaks either to the left or the right of the ideal cycle midpoint. For example, a 20 day trading cycle is measured from low to low. The ideal peak should occur 10 days into the cycle, or at the halfway point. That would allow for a 10 day advance followed by a 10 day decline. Ideal cycle peaks, however, rarely occur. Most variations in cycles occur at the peaks (or crests) and not at the troughs. That's why cycle troughs are considered more reliable and are used to measure cycle lengths. - -The cycle crests act differently depending on the trend of the next longer cycle. If the trend is up, the cycle crest shifts to the right of the ideal midpoint, causing right translation. If the longer trend is down, the cycle crest shifts to the left of the midpoint, causing left translation. Therefore, right translation is bullish and left translation is bearish. Stop to think about it. All we're saying here is that in a bull trend, prices will spend more time going up than down. In a bear trend, prices spend more time going down than up. Isn't that the basic definition of a trend? Only, in this case, we're talking about time instead of price. (See Figure 14.15.) - -## **HOW TO [ISOLAT](#page-335-0)E CYCLES** - -In order to study the various cycles affecting any given market, it is necessary to first isolate each dominant cycle. There are various ways of accomplishing this task. The simplest is by visual inspection. By studying daily bar charts, for example, it is possible to identify obvious tops and bottoms in a market. By taking the average time periods between those cyclic tops and bottoms, certain average lengths can be found. - -There are tools available to make that task a bit easier. One such tool is the *Ehrlich Cycle Finder*, named after its inventor, Stan Ehrlich (ECF, 112 Vida Court, Novato, CA 94947 [415] 892-1183). The Cycle Finder is an accordion-like device that can be placed on the price chart for visual inspection. The distance between the points is always equidistant and can be expanded or contracted to fit any cycle length. By plotting a distance between any two obvious cycle lows, it can be quickly determined if other cycle lows of the same length exist. An electronic version of that device, called the - -*Ehrlich Cycle Forecaster*, is now available as an analysis technique on Omega Research's Trade Station and Super Charts (#Omega Research, 8700 West Flagler Street, Suite 250, Miami, FL 33174, [305] 551-9991, www.omegaresearch.com). (See Figures 14.16-14.18.) - -**Figure 14.15** *Example of left and right translation. Figure A shows a simple cycle. Figure B shows the trend of the larger cycle. Figure C shows the combined effect. When the longer trend is up, the midpeak shifts to the right. When the longer trend is down, the midpeak shifts to the left. Right translation is bullish, left translation is bearish. (Source: The Power of Oscillator/Cycle Combination by Walt Bressert.)* - -Computers can help you find cycles by visual inspection. The user first puts a price chart on the screen. The next step is to pick a prominent bottom - -on the chart as a starting point. Once that is done, vertical lines (or arcs) appear every 10 days (the default value). The cycle periods can be lengthened, shortened, or moved left or right to find the right cycle fit on the chart. (See Figures 14.19 and 14.20.) - -**Figure 14.16** *The 4 year presidential cycle is clearly identified with the Ehrlich Cycle Forecaster (see vertical lines). If the cycle is still working, the next major low would be expected to occur during 1998.* - -**Figure 14.17** *The Ehrlich Cycle Forecaster has identified a 49 day trading cycle in S&P 500 futures prices (see vertical lines). The ECF estimates that the next cycle low will be formed 49 days from the last cycle low, which would be on March 30, 1998.* - -**Figure 14.18** *The ECF has uncovered a 133 day cycle in Boeing (see vertical lines). Since the last cycle low occurred during November, 1997, the ECF estimates that the next cycle low is due to occur 133 days later on June 3, 1998.* - -**Figure 14.19a** *The bottoms in the cycle arcs coincide with important reaction lows in the Dow when spaced 40 weeks apart. That suggests a 40 week cycle in the Dow. The last two cycle troughs were in the spring of 1997 and the start* - -*of 1998 (see arrows).* - -**Figure 14.19b** *The daily cycle arcs reveal the presence of 50 day cycle bottoms in the Dow during the second half of 1997 and the start of 1998. The idea is to shift the arcs until their lows coincide with a number of reaction lows on the price chart.* - -**Figure 14.20a** *Beginning with the major bottom in 1981, the cycle finder arcs reveal that bonds have shown a tendency to form important bottoms every 75 months (6.25 years). These numbers may shift with time, but still provide useful trading information.* - -**Figure 14.20b** *Applied to this daily chart, the cycle arcs showed a tendency for bond prices to bottom every 55 trading days during this time span (see arrows).* - -### **SEASONAL CYCLES** - -All markets are affected to some extent by an annual seasonal cycle. The seasonal cycle refers to the tendency for markets to move in a given direction at certain times of the year. The most obvious seasonals involve the grain markets where seasonal lows usually occur around harvest time when supply is most plentiful. In soybeans, for example, most seasonal tops occur between April and June with seasonal bottoms taking place between August and October. (See Figure 14.21.) One well known seasonal pattern is the "February Break" where grain and soybean prices usually drop from late December or early January into February. - -**Figure 14.21** *Soybeans usually peak in May and bottom in October.* - -Although the reasons for seasonal tops and bottoms are more obvious in the agricultural markets, virtually all markets experience seasonal patterns. Copper, for example, shows a strong seasonal uptrend from the January/February period with a tendency to top in March or April. (See Figure 14.22.) Silver has a low in January with higher prices into March. Gold shows a tendency to bottom in August. Petroleum products have a tendency to peak during October and usually don't bottom until the end of the [winter.](#page-341-0) (See Figure 14.23.) Financial markets also have seasonal patterns. - -**Figure 14.22** *Copper usually bottoms during October and February, but peaks during the April-May period.* - -**Figure 14.23** *Crude oil prices peak during October and turn up during* - -#### *March.* - -The U.S. Dollar has a tendency to bottom during January. (See Figure 14.24.) Treasury Bond prices usually hit important highs during January. Over the entire year, Treasury Bond prices are usually weaker during the first half of the year and stronger during the second half. (See Figure 14.25.) The [examples](#page-342-0) of seasonal charts are provided by the Moore Research Center (Moore Research Center, 321 West 13th Avenue, Eugene, OR 97401, (800) 927-7259), which specializes in seasonal analysis of [futures](#page-343-2) markets. - -**Figure 14.24** *The peak in the German mark during January coincides with a lowpoint in the U.S. dollar that usually occurs at the start of the new year.* - -**Figure 14.25** *Treasury Bonds prices usually peak around the new year, and then remain weak for most of the first half. The second half of the year is better for bond bulls.* - -### **STOCK MARKET CYCLES** - -Did you know that the strongest three month span for the stock market is November through January? February is then weaker, but is followed by a strong March and April. After a soft June, the market turns strong during July (the start of the traditional summer rally). The weakest month of the year is September. The strongest month is December (ending with the well known Santa Claus rally just after Christmas). That information, and a whole lot more about stock market cycles, can be found in Yale Hirsch's annual *Stock Trader's Almanac* (The Hirsch Organization, 184 Central Avenue, Old Tappen, NJ 07675). - -### **THE JANUARY BAROMETER** - -According to Hirsch: "as January goes, so goes the year." The well known January Barometer holds that what the S&P 500 does during January will determine what kind of year the market as a whole will have. Another variation on that theme is the belief that the direction of the S&P 500 during the first 5 trading days of the year gives some hint of what's ahead for the year. The January Barometer shouldn't be confused with the January Effect, which is the tendency for smaller stocks to outperform larger stocks during - -January. - -### **THE PRESIDENTIAL CYCLE** - -Another well known cycle that affects stock market behavior is the 4 year cycle, also called the Presidential Cycle, because it coincides with the elected term of U.S. presidents. Each of the 4 years has a different historical return. The election year (1) is normally strong. The postelection and midyears (2 and 3) are normally weak. The preelection year (4) is normally strong. According to Hirsch's *Trader's Almanac*, election years since 1904 have seen averages gains of 224%; postelection years, gains of 72%; midterm years, gains of 63%; and preelection years, gains of 217%. (See Figure 14.16.) - -### **COMBINING CYCLES WITH OTH[ER](#page-336-0) TECHNICAL TOOLS** - -Two of the most promising areas of overlap between cycles and traditional technical indicators are in the use of moving averages and oscillators. It is believed that the usefulness of both indicators can be enhanced if the time periods used are tied to each market's dominant cycles. Let's assume that a market has a dominant 20 day trading cycle. Normally, when constructing an oscillator, it's best to use half the length of the cycle. In this case, the oscillator period would be 10 days. To trade a 40 day cycle, use a 20 day oscillator. Walt Bressert discusses in his book, *The Power of Oscillator/Cycle Combinations*, how cycles can be used to adjust time spans for the Commodity Channel Index, the Relative Strength Index, Stochastics, and Moving Average Convergence Divergence (MACD). - -Moving averages can also be tied to cycles. You could use different moving averages to track different cycle lengths. To generate a moving average crossover system for a 40 day cycle, you could use a 40 day moving average in conjunction with a 20 day average (one-half of the 40 day cycle) or a 10 day average (one-quarter of the 40 day cycle). The main problem with this approach is determining what the dominant cycles are at a particular point in time. - -## **MAXIMUM ENTROPY SPECTRAL ANALYSIS** - -The search for the right dominant cycles in any market is complicated by the belief that cycle lengths aren't static; in other words, they keep changing over time. What worked a month ago may not work a month from now. In his book, *MESA and Trading Market Cycles*, John Ehlers uses a statistical approach called Maximum Entropy Spectral Analysis (MESA). Ehlers explains that one of the main advantages of MESA is its high-resolution measurement of cycles with relatively small time periods, which is crucial for shorter term trading. Ehlers also explains how cycles can be used to optimize moving average lengths and many of the oscillator-type indicators we've already mentioned. Uncovering cycles allows for the dynamic adjusting of technical indicators to fit current market conditions. Ehlers also addresses the problem of distinguishing between a market in a *cycle* mode versus one that is in a *trend* mode. When a market is in a trend mode, a trend-following indicator like a moving average is needed to implement trades. A cycle mode would favor the use of oscillator-type indicators. Cycle measurement can help determine which mode the market is currently in, and which type of technical indicator is more appropriate to use for trading strategies. - -### **CYCLE READING AND SOFTWARE** - -Most of the books referred to in this chapter on cycles can be obtained through mail order firms like Traders Press (see reference in previous chapter) or Traders' Library, P.O. Box 2466, Ellicott City, MD 21041, [800] 272- 2855). There's also a lot more software to help you perform cycle analysis with your computer. The *Ehrlich Cycle Forecaster* and Walt Bressert's *CycleTrader* are both available as add-on options to run with charting software provided by Omega Research. Bressert's *CycleTrader* integrates the concepts he describes in his book, *The Power of Oscillator/Cycle Combinations.* (Bressert Marketing Group, 100 East Walton, Suite 200, Chicago, IL 60611 (312) 867-8701). More information on the MESA computer program can be gotten from John Ehlers (Box 1801, Goleta, CA 93116 (805) 969-6478). For ongoing cycle research and analysis, don't forget the Foundation for the Study of Cycles. - -### **INTRODUCTION** - -The computer has played an increasingly important role in the field of technical analysis. In this chapter, we'll see how the computer can make the technical trader's task a good deal easier by providing quick and easy access to an arsenal of technical tools and studies that would have required an enormous amount of work just a few years earlier. This assumes, of course, that the trader knows how to use these tools, which brings us to one of the disadvantages of the computer. - -The trader not properly schooled in the concepts that underlie the various indicators, and who is not comfortable with how each indicator is interpreted, may find him- or herself overwhelmed with the vast array of computer software currently available. Even worse, the amount of impressive technical data at one's fingertips sometimes fosters a false sense of security and competence. Traders mistakenly assume that they are automatically better simply because they have access to so much computer power. - -The theme emphasized in this discussion is that the computer is an extremely valuable tool in the hands of a technically oriented trader who has already done his or her basic homework. When we review many of the routines available in the computer, you'll see that a fair number of the tools and indicators are quite basic and have already been covered in previous chapters. There are, of course, more sophisticated tools that require more advanced charting software. - -Much of the work involved in technical analysis can be performed without the computer. Certain functions can be more easily performed with a simple chart and ruler than with a computer printout. Some types of longer - -range analysis don't require a computer. As useful as it is, the computer is only a tool. It can make a good technical analyst even better. It won't, however, turn a poor technician into a good one. - -### **Charting Software** - -Several of the technical routines available in charting software have been covered in previous chapters. We'll review some of the tools and indicators currently available. We'll then address some additional features such as the ability to automate the various functions chosen by the user. In addition to providing us with the various technical studies, the computer also enables us to test various studies for profitability, which may be the most valuable feature of the program. Some software allows the user, with little or no programming background, to construct indicators and systems. - -#### **Welles Wilder's Directional Movement and Parabolic Systems** - -We'll take a close look at a couple of Welles Wilder's more popular systems, the *Directional Movement System* and the *Parabolic System.* We'll use those two systems in our discussion of the relative merits of relying on mechanical trading systems. It will be demonstrated that mechanical trend following systems only work well in certain types of market environments. It will also be shown how a mechanical system can be incorporated into one's market analysis and used simply as a confirming technical indicator. diff --git a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/020_15 Computers and Trading Systems.md b/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/020_15 Computers and Trading Systems.md deleted file mode 100644 index 5441c9456b8a4bcf90ee5e94c88a711f2d29da8e..0000000000000000000000000000000000000000 --- a/trading/Technical_Analysis_of_the_Financial_Market_-_John_J_Murphy/020_15 Computers and Trading Systems.md +++ /dev/null @@ -1,144 +0,0 @@ -#### **Too Much of a Good Thing** - -It may strike you that there are *too many* indicators from which to choose. Instead of simplifying our lives, has the computer only served to complicate things by giving us so much more to look at? Charting packages offer 80 different studies that are available to the technician. How does one possibly reach any conclusions (and find the time to trade) with so much data to contend with? We'll say a few words about some work being done in that direction. - -### **SOME COMPUTER NEEDS** - -Charting software can be applied to virtually any financial market. Most software is user-friendly, meaning that it can be easily implemented by choosing from successive lists of available routines. The place to start is with a charting software package that works for the computer you already own or are thinking of buying. Bear in mind that most charting software has been written for IBM-compatible computers. - -Charting packages do not provide daily market data. The user must obtain that data elsewhere. Data can be collected automatically from a data service over telephone lines (requiring a phone modem). Charting packages provide the names of various data vendors from which to choose. These data vendors provide all the software and instructions needed to set up and collect the data files. - -When first starting out, the user must collect historical data going back for at least several months to have something to work with. After that, data should be collected daily. It is possible to analyze "on line" data during the trading day by hooking up to a quote service. However, in our use of daily data, we will be referring to end-of-day data, which is available after the markets close. The final piece of equipment you might want is a printer to obtain a copy of whatever appears on the terminal screen. CD-Rom capability is highly recommended since some software vendors provide you with several years of historical data on a CD-Rom disk to get you started. There are some data vendors that also provide charting capability, which simplifies your task even more. One such service is Telescan (5959 Corporate Drive, Suite 2000, Houston, TX 77036, (800) 324-8246, www.telescan.com). - -### **GROUPING TOOLS AND INDICATORS** - -The following list groups some of the chart and indicator options. - -- *Basic Charts:* Bar, line, point and figure, and candlesticks -- *Chart Scales:* Arithmetic and semilogarithmic -- *Bar Chart:* Price, volume, and open interest (for futures) -- *Volume:* Bars, on balance, and Demand Index -- *Basic Tools:* Trendlines and channels, percentage retracements, moving averages, and oscillators -- *Moving Averages:* Reference envelopes, Bollinger Bands -- *Oscillators:* Commodity Channel Index, momentum, rate of change, MACD, Stochastic, Williams %R, RSI -- *Cycles:* Cycle Finder -- *Fibonacci Tools:* Fan lines, arcs, time zones and retracements -- *Wilder:* RSI, Commodity Selection Index, Directional Movement, Parabolic, Swing Index, ADX line - -### **USING THE TOOLS AND INDICATORS** - -How does one cope with so much from which to choose? A suggestion is to - -first use the basic tools such as price, volume, trendlines, percentage retracements, moving averages, and oscillators. Notice the large number of oscillators available. Pick one or two that you are most comfortable with and go with them. Use such things as cycles and Fibonacci tools as secondary inputs unless you have a special interest in those areas. Cycles can help fine tune moving average and oscillator lengths, but require study and practice. For mechanical trading systems, Wilder's Parabolic and DMI are especially noteworthy. - -### **WELLES WILDER'S PARABOLIC AND DIRECTIONAL MOVEMENT SYSTEMS** - -We're going to spend some time on two studies that are especially useful. Both studies were developed by J. Welles Wilder Jr. and discussed in his book, *New Concepts in Technical Trading Systems.* Three of Wilder's other studies included on the computer menu—Commodity Selection Index, Relative Strength Index, and the Swing Index—are also included in the same book. - -#### **Parabolic System (SAR)** - -Wilder's Parabolic system (SAR) is a time/price reversal system that is always in the market. The letters "SAR" stand for "stop and reverse," meaning that the position is reversed when the protective stop is hit. It is a trend-following system. It gets its name from the shape assumed by the trailing stops that tend to curve like a parabola. (See Figures 15.1-15.4. Notice that as prices trend higher, the rising dots below the price action (the stop and reverse points) tend to start out slower and then accelerate with the trend. In a downtrend, the same thing happens but in the opposite [direction](#page-350-0) ([the](#page-351-0) dots are above the price action). The SAR numbers are calculated and available to the user for the following day. - -Wilder built an acceleration factor into the system. Each day the stop moves in the direction of the new trend. At first, the movement of the stop is relatively slow to allow the trend time to become established. As the acceleration factor increases, the SAR begins to move faster, eventually catching up to the price action. If the trend falters, or fails to materialize, the result is usually a stop and reverse signal. As the accompanying charts show, the Parabolic system works extremely well in trending markets. Notice that while the trending portions were captured well, the system whipsawed constantly during the sideways, nontrending periods. - -**Figure 15.1** *The Parabolic SARs look like dots on the chart. A buy signal was given when the upper SAR was hit (first arrow). Notice how the SARs accelerated upward during the rally and caught most of the uptrend. A small whipsaw occurred to the upper right, which was quickly corrected. This system works when a trend is present.* - -**Figure 15.2** *A longer range version of the previous chart shows the good and bad aspects of Parabolics and any trend-following system. They work during trending periods (to the left and right of the chart). But are useless during the type of trading range that occurred from August to January.* - -**Figure 15.3** *Parabolics can be used on a monthly chart to track the primary trend. A sell signal in early 1994 was followed by a buy in late summer. Except for one whipsaw during 1996, this system has stayed positive for almost four years.* - -**Figure 15.4** *Parabolics applied to weekly chart of Dell Computer. After staying positive through most of 1997, a sell signal was given during October. That sell signal was reversed and a buy signal given as 1997 ended.* - -That demonstrates both the strength and weakness of most trendfollowing systems. They work well during strong trending periods, which Wilder himself estimates occur only about 30% of the time. If that estimate is even close to reality, then a trend-following system will not work for about 70% of the time. How then does one deal with this problem? - -#### **DMI and ADX** - -One possible solution is to use some type of filter or a device to determine if the market is in a trending mode. Wilder's ADX line rates the directional movement of the various markets on a scale of 0 to 100. A rising ADX line means the market is trending and a better candidate for a trend-following system. A falling ADX line indicates a nontrending environment, which would not be suitable for a trend-following approach. (See Figure 15.5.) - -**Figure 15.5** *The ADX line measures the degree of directional movement. A downturn from above 40 (left arrow) signaled the onset of a trading range. The upturn from below 20 (right arrow) signaled the resumption of a trending phase.* - -Because the ADX line is on a scale from 0 to 100, the trend trader could simply trade those markets with the highest trend ratings. Nontrending systems (oscillators, for example) could be utilized on markets with low directional movement. - -Directional Movement can be used either as a system on its own or as a filter on the Parabolic or any other trend-following system. Two lines are generated in the DMI study, +DI and -DI. The first line measures positive (upward) movement and the second number, negative (downward) movement). Figure 15.6 shows the two lines. The darker line is + DI and the lighter line -DI. A buy signal is given when the +DI line crosses over the - DI line and a sell signal when it crosses below the - DI line. - -Figure 15.6 [also](#page-353-0) shows both the Parabolic and Directional Movement - -systems. The Parabolic is clearly a more sensitive system, meaning that more frequent and earlier signals are given. However, by using the Directional Movement as a filter, several of the bad signals in the Parabolic could be avoided by following only those signals in the same direction as the Directional Movement lines. It appears then that the Parabolic and Directional Movement systems should be used together, with Directional Movement acting as a screen or filter on the more sensitive Parabolic. - -**Figure 15.6** *The Directional Movement lines along the bottom of the chart can be used as a filter on Parabolics (upper chart). When the +DI line is above the -DI line (far left and far right of chart), all Parabolic sell signals can be ignored. That would have eliminated several whipsaws during the rally phases.* - -The best time to use a trending system is when the ADX line is rising. (See Figures 15.7 and 15.8.) Be forewarned, however, that when the ADX line starts to drop from above the 40 level, that is an early sign that the trend is weakening. A rise back above the 20 level is often a sign of the start of a new [trend.](#page-354-0) (The ADX [line](#page-354-1) is essentially a smoothed difference between the +DI and -DI lines.) - -**Figure 15.7** *The 14 week ADX line peaked in early 1996 from well over 40, and initiated an 18 month trading range in utilities. The ADX upturn during the summer of 1997 from below 20 signaled that utilities were starting to trend.* - -**Figure 15.8** *An ADX line overlaid over a monthly chart of the AMEX Oil Index (XOI). The ADX peaked above 40 in 1990, ending the oil stock rally. The upturn in the ADX line from below 20 at the start of 1995 signaled the end of a 4 year trading range in oil stocks, and correctly spotted the start of a* - -### *new upleg.* - -### **PROS AND CONS OF SYSTEM TRADING** - -#### **Advantages of Mechanical Systems** - -- 1. Human emotion is eliminated. -- 2. Greater discipline is achieved. -- 3. More consistency is possible. -- 4. Trades are taken in the direction of the trend. -- 5. Participation is virtually guaranteed in the direction of every important trend. -- 6. Profits are allowed to run. -- 7. Losses are minimized. - -#### **Disadvantages of Mechanical Systems** - -- 1. Most mechanical systems are trend-following. -- 2. Trend-following systems rely on major trends in order to be profitable. -- 3. Trend-following systems are generally nonprofitable when markets are not trending. -- 4. There are long periods of time when markets are not trending and, therefore, not suitable for a trending approach. - -The major problem is the failure of the system to recognize when the market is not trending and its inability to turn itself off. The measure of a good system is not only its ability to make money in trending markets, but its ability to preserve capital during nontrending periods. It is this inability of the system to monitor itself that is its greatest weakness. This is where some overriding filtering device, such as Welles Wilder's Directional Movement system or the ADX line could prove especially useful by allowing the trader to determine which markets are most suitable for a trending system. - -Another drawback is that no allowance is generally made for anticipating market reversals. Trend-following systems ride with the trend until it turns. They don't recognize when a market has reached a long term support or resistance level, when oscillator divergences are being given, or when an Elliott Wave fifth pattern is clearly visible. Most traders would get more defensive at that point, and begin taking some profits. The system, however, will stay with the position until well after the market has changed direction. Therefore, it's up to the trader to determine how best to employ the system. - -That is to say, whether it should be followed blindly or whether it should be incorporated into a trading plan with other technical factors. That brings us to our next section on how a mechanical system can be used as just another technical input into the forecasting and trading process. - -#### **Using System Signals as a Disciplining Device** - -The system signals can be used simply as a mechanical confirmation along with other technical factors. Even if the system is not being traded mechanically, and other technical factors are being employed, the signals could be used as a disciplined way to keep the trader on the right side of the major trend. No short positions would be taken as long as the computer trend was up. No longs would be taken in a computer downtrend. (This would be a simple way for fundamentally oriented traders to use a technical device as a filter or trigger on their own trading ideas.) Trend direction can be a matter of judgment. The computer signals relieve the trader of some degree of uncertainty. They can prevent him or her from falling into the trap of "top and bottom picking." - -### **Using Signals as Alerts** - -System signals can also be used as an excellent screening device to alert the trader to recent trend changes. The trader can simply glance at the trend signals and instantly has several trading candidates. The same information could be found by studying all of the charts. The computer just makes that task quicker, easier, and more authoritative. The ability of the computer to automate system signals and then alert the trader when signals are triggered is an enormous asset, especially when the universe of financial markets has grown so large. - -### **NEED EXPERT HELP?** - -One of the products offered by Omega Research called TradeStation offers a variety of Expert Features (Omega Research, Miami, FL 33174, (305) 551- 9991). You can call up its Expert Commentary, which interprets indicators for you based on current market conditions. Omega's Expert Analyst will determine which indicators should work best in the current market and interpret them for you. In addition, it has two Expert Tools. The Trendlines Automatic Indicator actually draws trendlines for you. The Candlestick Patterns Indicator reads the more common candlestick chart patterns. - -### **TEST SYSTEMS OR CREATE YOUR OWN** - -Omega Research also includes a library of the most popular trading systems used by traders. You can test them, change them, or create your own if you wish. All of Omega's charting tools, indicators, and trading systems are written in a relatively simple language called EasyLanguage. EasyLanguage takes trading ideas that you have described in plain language and converts them into the machine code needed to run the program. It's hard to overestimate the value of being able to develop, test, optimize if you wish, and then automate your own trading ideas—without being a computer programmer. The computer will even generate the appropriate trading orders for you and alert you via your alphanumeric pager that signals have been triggered. (In Appendix C, we'll use Omega Research's EasyLanguage and TradeStation to show you how to go about creating a trading system of your own.) - -### **CONCLUSION** - -This chapter introduced a couple more of Welles Wilder's systems to you— Parabolics and Directional Movement (DMI). Parabolics can generate useful trading signals, but probably shouldn't be used alone. The two Dl lines can be used as a filter on Parabolics or any other sensitive trend-following trading system. The ADX line, which is part of the DMI system, provides one way to determine which type of market you're dealing with—a trending or a trading market. A rising ADX line suggests a trend and favors moving averages. A falling ADX line suggests a trading range and favors oscillators. We also used the Parabolic examples to show the good and bad sides of most trendfollowing systems. They work well when a trend is present. They're useless during a trading range. You have to be able to tell the difference. We also touched on the merits of mechanical trading systems. These systems remove human emotion and can be very helpful in the right market climate. They can also be used as technical alerts and used in conjunction with fundamental analysis. (See Appendix C for more on system trading.) - -There's no question that the computer has revolutionized financial market analysis and trading. While our interest is primarily in technical analysis, software [program](#page-439-0)s also allow you to blend fundamental analysis with the technical. When the first edition of this book was published in 1986, it cost about \$5,000 to outfit yourself with the necessary computer hardware to perform serious technical analysis. The leading software package of the day cost close to \$2,000. How things have changed. You can now obtain incredibly powerful computers for less than \$2,000. Most software packages - -can be had for less than \$300. The better ones provide you with up to 20 years of historical price data on a CD-Rom disk at little or no additional cost. - -Another big benefit is the amount of educational help that you can obtain with those software packages. The user manuals alone are the size of a book and include technical formulas and all kinds of useful explanations. The screening and alert capabilities of today's computer are especially helpful to those monitoring global bond and stock markets and thousands of individual common stocks, not to mention mutual funds. In Chapter 17, we'll talk about an even more sophisticated use of computer technology for developing *neural networks.* But the message to you is clear. If you are serious about investing or trading financial markets, get a computer and [learn](#page-374-0) how to use it. You'll be glad you did. - -## **INTRODUCTION** - -The previous chapters presented the major technical methods used to forecast and trade financial markets. In this chapter, we'll round out the trading process by adding to the task of *market forecasting* the crucial elements of *trading tactics* (or timing) and the often overlooked aspect of *money management.* No trading program can be complete without all three elements. - -## **THE THREE ELEMENTS OF SUCCESSFUL TRADING** - -Any successful trading program must take into account three important factors: price forecasting, timing, and money management. - -- 1. *Price forecasting* indicates which way a market is expected to trend. It is the crucial first step in the trading decision. The forecasting process determines whether the trader is bullish or bearish. It provides the answer to the basic question of whether to enter the market from the long or short side. If the price forecast is wrong, nothing else that follows will work. -- 2. *Trading tactics*, or timing, determines specific entry and exit points. Timing is especially crucial in futures trading. Because of the low margin requirements and the resulting high leverage, there isn't much room for error. It's quite possible to be correct on the direction of the market, but still lose money on a trade if the timing is off. Timing is - -almost entirely technical in nature. Therefore, even if the trader is fundamentally oriented, technical tools must be employed at this point to determine specific entry and exit points. - -3. *Money management* covers the allocation of funds. It includes such areas as portfolio makeup, diversification, how much money to invest or risk in any one market, the use of stops, reward-to-risk ratios, what to do after periods of success or adversity, and whether to trade conservatively or aggressively. - -The simplest way to summarize the three different elements is that price forecasting tells the trader *what* to do (buy or sell), timing helps decide *when* to do it, and money management determines *how much* to commit to the trade. The subject of price forecasting has been covered in the previous chapters. We'll deal with the other two aspects here. We'll discuss money management first because that subject should be taken into consideration when deciding on the appropriate trading tactics.